Continuum Mechanics and Thermodynamics of Matter

Aimed at advanced undergraduate and graduate students, this book provides a clearunified view of continuum mechanics that will be a welcome addition to the literature.Samuel Paolucci provides a well-grounded mathematical structure and also gives thereader a glimpse of how this material can be extended in a variety of directions, fur-nishing young researchers with the necessary tools to venture into brand new territory.Particular emphasis is given to the roles that thermodynamics and symmetries play inthe development of constitutive equations for different materials.

Continuum Mechanics and Thermodynamics of Matter is ideal for a one-semestercourse in continuum mechanics, with 250 end-of-chapter exercises designed to test anddevelop the reader’s understanding of the concepts covered. Six appendices enhancethe material further, including a comprehensive discussion of the kinematics, dynamics,and balance laws applicable in Riemann spaces.

S. Paolucci is currently Professor of Aerospace and Mechanical Engineering and Direc-tor of C-SWARM at the University of Notre Dame.

Continuum Mechanics andThermodynamics of Matter

S. PAOLUCC IUniversity of Notre Dame

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© Samuel Paolucci 2016

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Library of Congress Cataloguing in Publication dataNames: Paolucci, S., author.Title: Continuum mechanics and thermodynamics of matter / S. Paolucci,University of Notre Dame.Description: New York, NY : Cambridge University Press, 2016. | © 2016 |Includes bibliographical references and index.Identifiers: LCCN 2015034141 | ISBN 9781107089952 (Hardback : alk. paper) |ISBN 1107089956 (Hardback : alk. paper)Subjects: LCSH: Continuum mechanics. | Thermodynamics.Classification: LCC QA808.2 .P36 2016 | DDC 531–dc23LC record available at http://lccn.loc.gov/2015034141

ISBN 978-1-107-08995-2 Hardback

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Contents

Preface Page xi

1 Introduction 11.1 Continuum mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Continuum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3.1 Deformation and strain . . . . . . . . . . . . . . . . . . . . . . . 51.3.2 Stress field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.4 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.5 Constitutive theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.5.1 Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.5.2 Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.6 Pioneers of continuum mechanics . . . . . . . . . . . . . . . . . . . . . 10Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2 Tensor analysis 132.1 Review of linear algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.2 Tensor algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.3 The metric tensor and its properties . . . . . . . . . . . . . . . . . . . 222.4 General polyadic tensor of order m . . . . . . . . . . . . . . . . . . . . 242.5 Scalar product of two vectors . . . . . . . . . . . . . . . . . . . . . . . 272.6 Vector product of two vectors . . . . . . . . . . . . . . . . . . . . . . . 272.7 Tensor product of two vectors . . . . . . . . . . . . . . . . . . . . . . . 292.8 Contraction of tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.9 Transpose of a tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312.10 Symmetric and skew-symmetric tensors . . . . . . . . . . . . . . . . . 322.11 Dual of a tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.12 Exterior product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.13 Tensor fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2.13.1 Cartesian coordinate system . . . . . . . . . . . . . . . . . . . . 412.13.2 Curve in space . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.13.3 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.13.4 Surface in space . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.13.5 Curvilinear coordinate system . . . . . . . . . . . . . . . . . . . 45

2.14 Gradient of a scalar field . . . . . . . . . . . . . . . . . . . . . . . . . . 492.15 Gradient of a vector field . . . . . . . . . . . . . . . . . . . . . . . . . . 50

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vi CONTENTS

2.16 Covariant differentiation of a vector . . . . . . . . . . . . . . . . . . . . 522.17 Divergence of a vector field . . . . . . . . . . . . . . . . . . . . . . . . . 552.18 Curl of a vector field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 562.19 Orthogonal curvilinear coordinate system . . . . . . . . . . . . . . . . 58

2.19.1 Physical components . . . . . . . . . . . . . . . . . . . . . . . . 592.19.2 Gradient of a scalar field . . . . . . . . . . . . . . . . . . . . . . 602.19.3 Gradient and divergence of a vector field . . . . . . . . . . . . 602.19.4 Curl of a vector field . . . . . . . . . . . . . . . . . . . . . . . . 612.19.5 Laplacian of a scalar field . . . . . . . . . . . . . . . . . . . . . 612.19.6 Divergence of a dyadic tensor field . . . . . . . . . . . . . . . . 62

2.20 Integral theorems and generalizations . . . . . . . . . . . . . . . . . . . 622.20.1 Regions with discontinuous surfaces, curves, and

points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

3 Kinematics 733.1 Deformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

3.1.1 Deformation gradient . . . . . . . . . . . . . . . . . . . . . . . . 763.1.2 Transformation of linear elements . . . . . . . . . . . . . . . . 773.1.3 Transformation of a surface element . . . . . . . . . . . . . . . 803.1.4 Transformation of a volume element . . . . . . . . . . . . . . . 823.1.5 Relations between deformation and inverse

deformation gradients . . . . . . . . . . . . . . . . . . . . . . . . 833.1.6 Identities of Euler–Piola–Jacobi . . . . . . . . . . . . . . . . . . 843.1.7 Cayley–Hamilton theorem . . . . . . . . . . . . . . . . . . . . . 853.1.8 Real symmetric matrices . . . . . . . . . . . . . . . . . . . . . . 883.1.9 Polar decomposition theorem . . . . . . . . . . . . . . . . . . . 913.1.10 Strain kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . 953.1.11 Compatibility conditions . . . . . . . . . . . . . . . . . . . . . . 98

3.2 Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1003.2.1 Velocity and acceleration . . . . . . . . . . . . . . . . . . . . . . 1013.2.2 Path lines, stream lines, and streak lines . . . . . . . . . . . . 1043.2.3 Relative deformation . . . . . . . . . . . . . . . . . . . . . . . . 1063.2.4 Stretch and spin . . . . . . . . . . . . . . . . . . . . . . . . . . . 1103.2.5 Kinematical significance of D and W . . . . . . . . . . . . . . 1123.2.6 Kinematics and dynamical systems . . . . . . . . . . . . . . . . 1153.2.7 Internal angular velocity and acceleration . . . . . . . . . . . . 119

3.3 Objective tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1203.3.1 Apparent velocity . . . . . . . . . . . . . . . . . . . . . . . . . . 1223.3.2 Apparent acceleration . . . . . . . . . . . . . . . . . . . . . . . 1243.3.3 Properties of kinematic quantities . . . . . . . . . . . . . . . . 1253.3.4 Corotational and convected derivatives . . . . . . . . . . . . . 1283.3.5 Push-forward and pull-back operations . . . . . . . . . . . . . 129

3.4 Transport theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1313.4.1 Material derivative of a line integral . . . . . . . . . . . . . . . 1313.4.2 Material derivative of a surface integral . . . . . . . . . . . . . 1343.4.3 Material derivative of a volume integral . . . . . . . . . . . . . 136

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

CONTENTS vii

4 Mechanics and thermodynamics 1494.1 Balance law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1494.2 Fundamental axioms of mechanics . . . . . . . . . . . . . . . . . . . . . 1514.3 Fundamental axioms of thermodynamics . . . . . . . . . . . . . . . . . 1544.4 Forces and moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1564.5 Rigid body dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1584.6 Stress and couple stress hypotheses . . . . . . . . . . . . . . . . . . . . 161

4.6.1 Stress and couple stress tensors . . . . . . . . . . . . . . . . . . 1634.7 Local forms of axioms of mechanics . . . . . . . . . . . . . . . . . . . . 1654.8 Properties of stress vector and tensor . . . . . . . . . . . . . . . . . . . 169

4.8.1 Principal stresses and principal stress directions . . . . . . . . 1694.8.2 Mean stress and deviatoric stress tensor . . . . . . . . . . . . . 1724.8.3 Lamé’s stress ellipsoid . . . . . . . . . . . . . . . . . . . . . . . 1724.8.4 Mohr’s circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

4.9 Work and heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1744.10 Heat flux hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1754.11 Entropy flux hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . . 1764.12 Local forms of axioms of thermodynamics . . . . . . . . . . . . . . . . 1784.13 Field equations in Euclidean frames . . . . . . . . . . . . . . . . . . . . 1814.14 Jump conditions in Euclidean frames . . . . . . . . . . . . . . . . . . . 183Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188

5 Principles of constitutive theory 1915.1 General constitutive equation . . . . . . . . . . . . . . . . . . . . . . . 1925.2 Frame indifference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1945.3 Temporal material smoothness . . . . . . . . . . . . . . . . . . . . . . . 1965.4 Spatial material smoothness . . . . . . . . . . . . . . . . . . . . . . . . 1965.5 Spatial and temporal material smoothness . . . . . . . . . . . . . . . . 1985.6 Material symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1995.7 Reduced constitutive equations . . . . . . . . . . . . . . . . . . . . . . 208

5.7.1 Constitutive equation for a simple isotropic solid . . . . . . . 2125.7.2 Constitutive equation for a simple (isotropic) fluid . . . . . . 212

5.8 Isotropic and hemitropic representations . . . . . . . . . . . . . . . . . 2135.9 Expansions of constitutive equations . . . . . . . . . . . . . . . . . . . 2155.10 Thermodynamic considerations . . . . . . . . . . . . . . . . . . . . . . 216

5.10.1 Thermodynamic states . . . . . . . . . . . . . . . . . . . . . . . 2165.10.2 Thermodynamic potentials . . . . . . . . . . . . . . . . . . . . 2265.10.3 Thermodynamic processes . . . . . . . . . . . . . . . . . . . . . 2325.10.4 Thermodynamic equilibrium and stability . . . . . . . . . . . 2355.10.5 Potential energy and strain energy . . . . . . . . . . . . . . . . 239

5.11 Entropy and nonequilibrium thermodynamics . . . . . . . . . . . . . . 2425.11.1 Coleman–Noll procedure . . . . . . . . . . . . . . . . . . . . . . 2425.11.2 Müller–Liu procedure and Lagrange multipliers . . . . . . . . 242

5.12 Jump conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2445.12.1 Characterization of jump conditions . . . . . . . . . . . . . . . 2455.12.2 Material singular surface . . . . . . . . . . . . . . . . . . . . . . 2485.12.3 Equilibrium jump conditions . . . . . . . . . . . . . . . . . . . 251

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264

viii CONTENTS

6 Spatially uniform systems 2716.1 Material with no memory . . . . . . . . . . . . . . . . . . . . . . . . . . 2726.2 Material with short memory of volume . . . . . . . . . . . . . . . . . . 2756.3 Material with longer memory of volume . . . . . . . . . . . . . . . . . 2766.4 Material with short memory . . . . . . . . . . . . . . . . . . . . . . . . 278Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281

7 Thermoelastic solids 2837.1 Clausius–Duhem inequality . . . . . . . . . . . . . . . . . . . . . . . . . 2847.2 Material symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2897.3 Linear deformations of anisotropic materials . . . . . . . . . . . . . . 296

7.3.1 Propagation of elastic waves in crystals . . . . . . . . . . . . . 2997.4 Nonlinear deformations of anisotropic

materials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3047.5 Linear deformations of isotropic materials . . . . . . . . . . . . . . . . 3057.6 Nonlinear deformations of isotropic materials . . . . . . . . . . . . . . 306

7.6.1 Special nonlinear deformations . . . . . . . . . . . . . . . . . . 309Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335

8 Fluids 3398.1 Coleman–Noll procedure . . . . . . . . . . . . . . . . . . . . . . . . . . 3408.2 Müller–Liu procedure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3428.3 Representations of qd and σd . . . . . . . . . . . . . . . . . . . . . . . 3478.4 Propagation of sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3578.5 Classifications of fluid motions . . . . . . . . . . . . . . . . . . . . . . . 360

8.5.1 Restrictions on the type of motion . . . . . . . . . . . . . . . . 3608.5.2 Specializations of the equations of motion . . . . . . . . . . . 3718.5.3 Specializations of the constitutive equations . . . . . . . . . . 372

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380

9 Viscoelasticity 3839.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3839.2 Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386

9.2.1 Motion with constant stretch history . . . . . . . . . . . . . . 3909.3 Constitutive equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396

9.3.1 Constitutive equations for motion with constantstretch history . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399

9.3.2 Fading memory . . . . . . . . . . . . . . . . . . . . . . . . . . . 4069.3.3 Constitutive equations of differential type . . . . . . . . . . . . 4089.3.4 Constitutive equations of integral type . . . . . . . . . . . . . 4109.3.5 Constitutive equations of rate type . . . . . . . . . . . . . . . . 417

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432

Appendices 437A Summary of Cartesian tensor notation . . . . . . . . . . . . . . . . . . 439

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 441B Isotropic tensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 446C Balance laws in material coordinates . . . . . . . . . . . . . . . . . . . 447

CONTENTS ix

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448D Curves and surfaces in space . . . . . . . . . . . . . . . . . . . . . . . . 449

D.1 Space curve . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 451D.2 Balance law for a space curve . . . . . . . . . . . . . . . . . . . 454D.3 Space surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455D.4 Balance law for a flux through a space surface . . . . . . . . . 473Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 482

E Representation of isotropic tensor fields . . . . . . . . . . . . . . . . . 483E.1 Scalar function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483E.2 Vector function . . . . . . . . . . . . . . . . . . . . . . . . . . . 483E.3 Symmetric tensor function . . . . . . . . . . . . . . . . . . . . . 484Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485

F Legendre transformations . . . . . . . . . . . . . . . . . . . . . . . . . . 487Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 489

Index 491

Preface

The goal of this text is to introduce students to the topic of continuum mechan-ics, with analysis of the kinematic and mechanical behavior of materials modeledunder the continuum assumption. This includes the derivation of fundamentalbalance equations, based on the classical laws of physics, and the developmentof constitutive equations characterizing the behavior of idealized materials. Suchbackground provides the starting point for the studies of thermoelasticity, fluidmechanics, and viscoelasticity that are provided in the text. Furthermore, thematerial covered also imparts students with sufficient background for studyingmore advanced topics in continuum mechanics, such as wave propagation, polarmaterials, mixture theory, shell theory, piezoelectricity, and electromagnetic andmagnetohydrodynamic fluid mechanics.

A few years ago, I was involved in a project that required fundamental under-standing of immiscible multiphase mixtures. I was not (and am still not) satisfiedwith the current formulations of continuum mechanics in this area, but this is asubject that will be taken up in future publications. Nevertheless, my extensivestudies necessitated a deeper understanding of many aspects of single-phase con-tinuum mechanics. Such studies provided me valuable insights and have enabledme to write the present book as an outgrowth of my efforts. At the same time,they have enabled me to become a better teacher of the subject. I hope thatthe results might be useful to other teachers and students as well. The book isintended for use by students in engineering, science, and applied mathematics. Aspre-requisites, a student should have knowledge of multivariable calculus, linearalgebra, and differential equations, which are standard in undergraduate programsof engineering and science.

I started writing this book in 2004 to fill a number of gaps that I felt limited myunderstanding and application of the beautiful theory of continuum mechanics –especially on the relation between continuum mechanics and thermodynamics. Ibecame quite dissatisfied with existing textbooks. Some were delightful but super-ficial, others wonderful but ancient. Of course many excellent monographs existed,such as The Classical Field Theories by Truesdell and Toupin, The Non-LinearField Theories of Mechanics by Truesdell and Noll, and Mechanics of Continuaby Eringen. Unfortunately, such books were and have been out of print for quitea while and, in the case of the first two, they are challenging works that are notintended for use in a classroom. In the end, as I started to research the material,I fell in love with the subject. I sensed a unifying approach to teaching it that Iwanted to develop and then to share. Since then, a number of good texts haveappeared, but I feel that the need for the present book still exists. The presenttext is designed for a one-semester course in continuum mechanics. While cover-

xi

xii PREFACE

ing the standard material, the book also provides a well-grounded mathematicalstructure and glimpses of how such material can be extended in a variety of direc-tions. Thus, a major aim of the present text is not only providing a sound basisof continuum mechanics but, just as importantly, providing the tools for someoneto venture into new territory. I hope that this aspect does not detract from thepresentation and does not confuse the student.

Particularly in a subject such as continuum mechanics, many symbols and manyfonts are used to refer to each specific quantity introduced. This is done to makethe presentation clear. However, I have found this to be an absolutist approachthat often burdens the reader to recall the meaning of way too many symbols.While I have retained the rigor, I have not tried to be an absolutist in this respect.Any symbol that is re-used, its meaning is made clear from the context. The textis divided into nine chapters, and each chapter includes exercise problems to testand extend the understanding of concepts presented.

Chapter 1 provides the essential understanding for the need of treating the be-havior of common materials through the mathematical artifice of a continuumdescription. In addition, the different subject areas that make up continuum me-chanics are introduced.

In Chapter 2, the essential mathematics for treating continuum problems isprovided. Here, we define tensors and cover the algebra and multivariate calculusassociated with these objects. In addition, we discuss integral theorems and theirgeneralizations when discontinuous surfaces are present in a continuum region.Here, and throughout the text, we try to make the student comfortable in dealingwith three different forms of representing tensors and the associated equations theyenter in: by their representation of a coordinate-independent geometrical object,A; by the matrix representing its components, A; and by the specific componentelements, aij.

Chapter 3 provides a comprehensive discussion of the kinematics of a continuumbody. The deformation and motion of such a body are treated using Lagrangianand Eulerian descriptions. In addition, generalized balance laws are formulatedand the important concept of frame-invariance is introduced and utilized.

Chapter 4 is devoted to the fundamental laws of mechanics and thermodynamics.The corresponding global and local forms of the governing equations are developed,and the role that discontinuous surfaces embedded within a continuum region playis discussed. In this chapter, we also consider the effects of the microstructure thatunderlies the continuum body and subsequently write the balance equations forpolar materials. This is done to provide the student interested in this topic astarting point from which to pursue further studies (e.g., the modeling of liquidcrystals). In order to focus on major concepts, the text following this chapteronly deals with non-polar materials. In this chapter, the stress and couple stresstensors as well as the heat and entropy fluxes are naturally introduced and theirproperties discussed. In addition, we examine the local equations resulting fromEuclidean and Galilean transformations and their implications.

Chapter 5 covers the principles of constitutive theory, where thermodynamicsplays an essential role. This chapter provides a unifying theory regardless of thetype of matter and forms the centerpiece of the text. The constitutive equationsrepresent macro thermo-mechanical models of real materials. Here, the principlesof frame-indifference, causality, equipresence, material smoothness, memory, sym-

PREFACE xiii

metry, and thermodynamics are systematically utilized to obtain reduced formsof constitutive equations for general materials, and for solids and fluids in partic-ular. Tables are provided for expedient formulations of constitutive equations ofisotropic materials. The development of such tables is clearly described and illus-trated. Thermodynamics plays an integral part of constitutive theory and manythermodynamic tensor quantities are developed – these reduce to well-known scalarquantities encountered in classical thermodynamics. Formulations are given us-ing the different thermodynamic potentials of internal energy, entropy, Helmholtzfree energy, Gibbs free energy, and enthalpy, and the corresponding Maxwell re-lations provide very useful relations among thermodynamic quantities. Here wealso discuss the concepts of thermodynamic equilibrium and stability. In addition,the critical role that the second law of thermodynamics plays in the reductionof constitutive equations is explored using the conventional Coleman–Noll pro-cedure and the more general Müller–Liu procedure that makes use of Lagrangemultipliers. Lastly, in this chapter we provide a comprehensive discussion of jumpconditions across discontinuous surfaces, and their role in describing material andnon-material singular surfaces, including boundary conditions, shocks, and phase-change interfaces.

Chapter 6 is provided to clearly illustrate many of the constitutive theory con-cepts to the case of spatially uniform material bodies. Here, many of the thermo-dynamic concepts can easily be applied within a mathematical setting that is notoverly burdensome to the student.

This is followed by Chapter 7 where constitutive equations of thermoelasticsolids are rigorously developed using the Coleman–Noll procedure. Material sym-metries and crystal microscopic structures are fully discussed and linear and non-linear constitutive equations for non-isotropic and isotropic thermoelastic solidsare considered. In addition, we examine a number of fundamental nonlinear equi-librium deformations.

Fluids are discussed in Chapter 8. Here, we provide a rigorous developmentof constitutive equations using both the Coleman–Noll and the Müller–Liu pro-cedures. General representations of the stress tensor and heat flux are provided.Their simplifications leading to Euler equations, Newtonian equations, the second-order representation, and the Reiner–Rivlin fluid are fully developed. Lastly, com-prehensive classifications of fluid motions are provided. The classifications are inthe general areas of kinematically restricted types of motions, specialized equationsof motion, and specialized constitutive equations.

In Chapter 9, we treat the subject of viscoelasticity. The additional kinemat-ics considerations, aspects of constitutive theory, and general classes of motionsof materials having memory are provided. Lastly, the concept of fading mem-ory and application of finite linear viscoelasticity undergoing simple deformationsare considered. The treatment of phenomenological constitutive equations, whileimportant, is intentionally left out.

The book includes six appendices that enhance the material presented in thechapters. Of particular note is an appendix that provides a comprehensive dis-cussion of the kinematics, dynamics, and balance laws applicable in Riemannianspaces, such as arbitrary surfaces and curves embedded in the three-dimensionalEuclidean space.

Lastly, bibliographies pertinent to material provided in each individual chapter

xiv PREFACE

is given at the end of the specific chapter.I would like to conclude by thanking Jim Jenkins, who first introduced me

to continuum mechanics while I was a graduate student in Theoretical and Ap-plied Mechanics at Cornell University, and a number of authors who provided meguidance and inspiration throughout my journey in understanding continuum me-chanics and thermodynamics of material bodies. Foremost among them, in alpha-betical order, R. Aris, R.M. Bowen, H.B. Callen, D.B. Coleman, D.G.B. Edelen,J.L. Ericksen, A.C. Eringen, I-S. Liu, I. Müeller, W. Noll, R.S. Rivlin, G.F. Smith,A.J.M. Spencer, R. Toupin, and C. Truesdell. In addition, I would like to thankthe many students who have taken my course in Continuum Mechanics at the Uni-versity of Notre Dame; they have provided me useful feedback through multipleversions of the material. In particular, I would like to thank Dr. Gianluca Pulitiwho drew most of the figures in the text.

Notre Dame, Indiana S. PAOLUCCI

1

Introduction

Classical continuum physics deals with media without a visible microstructure.That is, the scale of observation is large compared to the molecular scale, butsmall relative to other heterogeneities within the system. More modern continuumtheories consider more directly the influence of microstructures; among them aremicromorphic, mixture, and nonlocal theories. On the smallest scale, individualmolecules are observed. Statistical mechanical theories and some micromorphicfield theories may be applicable on this scale. On a slightly larger scale, thematerial body appears locally uniform with no distinct microstructure. This isthe scale of observation on which classical continuum theories apply. On yet alarger scale of observation, large heterogeneities in space and/or time are evident.Such heterogeneities are well characterized by the solution of continuum mechanicsproblems at these scales.

From the atomic point of view, a macroscopic sample of matter is an agglom-erate of an enormous number of nuclei and electrons. A complete mathematicaldescription of a sample consists of the specification of suitable coordinates for eachnucleus and electron; the number of such coordinates is enormous considering themagnitude of Avogadro’s number of 6.0221×1023 mol−1 which gives us the numberof molecules in one mole of a substance.

In contrast to the atomistic description, only a few parameters are requiredto describe the system macroscopically. The key to this reduction is the slownessand large scale of macroscopic measurements in comparison to the speed of atomicmotions (typically of the order of 10−15 s) and atomic distance scales (typicallyof the order of 10−10 m). For example, some of our fastest macroscopic measure-ments are of the order of 10−6 s. Consequently, macroscopic measurements senseonly averages of the atomic coordinates. The mathematical process of averagingeliminates coordinates and thus reduces the level of description in going from theatomic to the macroscopic level.

Of the enormous number of atomic coordinates, a very few, with unique symme-try properties, survive the statistical averaging. Certain of these are mechanical innature (e.g., volume, shape, and components of elastic strain), others are thermalin nature (e.g., temperature and internal energy), or electrical/magnetical in na-ture (e.g., electric and magnetic dipole moments). The subject of mechanics (e.g.,elasticity and fluid mechanics) is the study of one set of surviving coordinates,the subject of thermal sciences (e.g., thermodynamics and heat transfer) is the

1

2 INTRODUCTION

study of another set of surviving coordinates, and the subject of electricity andmagnetism is the study of still another set of coordinates. In general, all thesesets of coordinates are coupled, and the study of continuum mechanics providesthe framework to study the coupling between these coordinates.

For many materials the behavior of large samples can be studied without re-course to the details of the atomic level structure. We can describe fluids, solids,glasses, bio-materials, mixtures, etc., by making use of the framework provided bycontinuum mechanics.

1.1 Continuum mechanics

Continuum mechanics is the study of the macroscopic consequences of the largenumber of atomic coordinates, which, by virtue of statistical averaging, do notappear explicitly in the macroscopic description of a system. It is a branch ofphysics that deals with materials. The fact that matter is made of atoms and thatit commonly has some sort of heterogeneous microstructure is mostly ignored inthe simplifying approximation that physical quantities, such as mass, momentum,and energy, can be handled in the infinitesimal limit. For most materials, this ispossible as long as the characteristic length scale is far larger than 10−9 m andthe characteristic speed is much less than the speed of light (3 × 108 m/s). If thelength scale is of the order of 10−9 m or less, then quantum mechanics applies.If the speed is near the speed of light, then relativistic mechanics applies. If thelength scale is of the order of 10−9 or less and the speed is near that of light, thenquantum field theory applies.

What are the consequences of the existence of the “hidden” atomic motion? Re-call that in mechanics, thermal sciences, and electricity and magnetism we aremuch concerned with the concept of energy. Energy transferred to a mechanicalmode of a system is called mechanical work δW . Similarly, energy can be trans-ferred to an electrical mode of the system. Mechanical work is typified by theterm −pdV (p is pressure and V is volume), and electrical work is typified by theterm −E dP (E is the electric field and P is the electric dipole moment). It isequally possible to transfer energy to the hidden atomic modes of motion as wellas to those which happen to be macroscopically observable. Energy transfer tothe hidden atomic modes is called heat. The energy residing in the hidden atomicmotions we call internal energy. Heat transfer and internal energy are typified byterms such as δQ and dU .

Continuum mechanics is very general; it applies to complicated systems with me-chanical, thermal, and electrical/magnetical properties. In this book, we will focuson mechanical and thermal properties of materials, keeping in mind that this is nota limitation of continuum mechanics theory. Differential equations are employedin solving problems in continuum mechanics. Some of these differential equationsare specific to the materials being investigated, while others capture fundamentalphysical laws, such as conservation of mass or conservation of momentum.

The physical laws of a material’s response to forces do not depend on the coor-dinate system in which they are observed. Continuum mechanics is thus describedby tensors, which are mathematical objects that are independent of a coordinatesystem. Such tensors can be expressed in coordinate systems for computationalconvenience.

1.2. CONTINUUM 3

V0V1V2

P

Figure 1.1: Limit at a point P .

1.2 Continuum

A continuum is a classical concept derived from mathematics:

a) the real number system is a continuum;

b) time can be represented by a real number system;

c) three-dimensional space can be represented by three real number systems;

d) time-space together is identified as a four-dimensional continuum.

A material continuum is characterized by quantities such as mass, momentum,energy, and state variables.

Matter, as measured by its mass m, is assumed to have a continuous distributionin space. A certain amount of mass occupies a definite volume V . As illustratedin Fig. 1.1, we define the mass density at an arbitrary point P by

ρ(P ) = limn→∞Vn→0

mn

Vn, (1.1)

where mn is the mass contained in the averaging volume Vn.Since the averaging volume must be sufficiently larger than molecular scales, to

conform to the real world, we take the definition of the density of the material atP with an acceptable variability ǫ > 0 in a defining limit volume δ > 0:

limn→∞

Vn→δ≪1

∣ ρ(P )mn/Vn − 1∣ < ǫ≪ 1. (1.2)

It is our responsibility to make sure that δ is sufficiently large and ǫ sufficientlysmall for the concepts of a continuum to make sense. For example, δ should belarge enough in the four-dimensional time-space continuum to include a sufficientlylarge number of molecules so that the number of molecules entering or leaving δis such as to lead to ǫ sufficiently small. Similarly, we define densities of momen-tum and energy. For vector quantities, the definition applies to each componentindividually. Note that in general the size of the limit volume δ for a fixed accept-able variability ǫ is different for different physical quantities. Thus, again, it is

4 INTRODUCTION

our responsibility to understand that the continuum description only makes sensewhen describing average properties at scales larger than the largest δ among allquantities that we are interested in describing within the acceptable variability ǫ.

Continuum mechanics ignores all the fine detail of atomic and molecular (orparticle) level structure and assumes that

- the highly discontinuous structure of real materials can be replaced by asmoothed hypothetical continuum;

- every portion of the continuum, however small, exhibits the macroscopicphysical properties of the bulk material.

In any branch of continuum mechanics, the field variables (i.e., density, displace-ment, and velocity) are conceptual constructs. They are taken to be defined atall points of the imagined continuum and their values are calculated via axiomaticrules of procedure.

The continuum model breaks down over distances comparable to interatomicspacing (in solids about 10−10 m). Nonetheless, the average of a field variableover a small but finite region is meaningful. Such an average can, in principle, becompared directly to its nominal counterpart found by experiment, which will itselfrepresent an average of a kind taken over a region containing many atoms, becauseof the finite physical size of any measuring probe. For solids, the continuum modelis valid in this sense down to a scale of order 10−8 m which is the side of a cubecontaining a million or so atoms. Further, when field variables change slowly withposition at a microscopic level ∼10−6 m, their averages over such volumes (10−20

m3 say) differ insignificantly from their centroidal values. In this case, pointwisevalues can be compared directly to observations. Such behaviors are illustrated inFig. 1.2 for the mass density at point P as a function of the size of the averagingvolume.

Within the continuum we take the behavior to be determined by balance lawsfor mass, linear momentum, angular momentum, energy, and the second law ofthermodynamics. The continuum hypothesis enables us to apply these laws on alocal as well as a global scale.

1.3 Mechanics

Classical mechanics is the study of the motion and deformation changes in a bodycomposed of matter due to the action of forces. It is often referred to as Newtonianmechanics after Newton and his laws of motion. Classical mechanics is subdividedinto statics (which models objects at rest), kinematics (which models objects inmotion), and dynamics (which models objects subjected to forces). In continuummechanics, we deal with all three aspects that are based on the concepts of time,space, and forces. To understand the concept of forces, knowledge is needed fromall branches of engineering, physics, chemistry, and biology.

Classical mechanics produces very accurate results within the domain of every-day experience. It is superseded by relativistic mechanics for systems moving atlarge velocities (near the speed of light), quantum mechanics for systems at smallspatial scales (atomic or subatomic scales), and relativistic quantum field theoryfor systems with both properties. Nevertheless, classical mechanics is still very

1.3. MECHANICS 5

Inhomogenoeus – cannot do statistics

Inhomogenoeus – materials with structure;

treat as multiphase (micromorphic)

Homogeneous – treat as one phase

ρ(P )

Molecular

and atomic

Micro

continuum

Classical

continuum

V∗ Vmin Vmax V

Size effect

Figure 1.2: Density limit with acceptable variability at a point P as a function ofaveraging volume.

useful, because (i) it is much simpler and easier to apply than these other theories,and (ii) it has a very large range of approximate validity. Classical mechanics canbe used to describe the motion of human-sized objects (i.e., tops and baseballs),many astronomical objects (i.e., planets and galaxies), and certain microscopicobjects (i.e., sand grains and organic molecules.)

1.3.1 Deformation and strain

If we take a solid cube and subject it to some deformation, the most obviouschange in external characteristics will be a modification of the shape.

The specification of the deformation is thus a geometrical problem and may becarried out from two different viewpoints: relate the deformation

1. with respect to the undeformed state (Lagrangian), or

2. with respect to the deformed state (Eulerian).

Locally, the mapping from the deformed to the undeformed state can be assumedto be linear and described by a differential relation, which is a combination of purestretch (a rescaling of each coordinate) and a pure rotation.

The mechanical effects of the deformation are confined to the stretch and it isconvenient to characterize this by a strain measure. For example, for a wire under

6 INTRODUCTION

Material B in V

Closed surface S

P

n∆fn

∆Sn ≪ 1

Figure 1.3: Traction with acceptable variability at a point P .

load the strain would be the relative extension, i.e.,

linear strain = change in length

initial length.

The generalization of this idea requires us to introduce a strain tensor at eachpoint of the continuum.

1.3.2 Stress field

Stress is a measure of force intensity or density. As illustrated in Fig. 1.3, thetraction or stress vector t at an arbitrary point P on a surface with normal vectorn with an acceptable variability ǫ > 0 is defined by

limn→∞

∆Sn→α≪1

∣ t(P,n)∆fn/∆Sn

− 1∣ < ǫ≪ 1, (1.3)

where α > 0 is sufficiently small, or more simply

t(P,n) = dfdS

.

Within a deformed continuum there will be a force system acting. If we wereable to cut the continuum in the neighborhood of a point P as illustrated inFig. 1.4, we would find a force acting on the cut surface, which would depend onthe inclination of the surface and is not necessarily perpendicular to the surface.This force system can be described by introducing a stress tensor σ at each pointwhose components describe the loading characteristics.

1.4 Thermodynamics

Thermodynamics is the physics of energy, heat, work, entropy, and the spontaneityof processes.

1.4. THERMODYNAMICS 7

t

n P

Figure 1.4: Stress vector at a point P .

While dealing with processes in which systems exchange matter or energy, clas-sical thermodynamics is not concerned with the rate at which such processes takeplace, termed kinetics. For this reason, the use of the term thermodynamics usu-ally refers to equilibrium thermodynamics. In this connection, a central concept inthermodynamics is that of quasistatic processes, which are idealized infinitely slowprocesses. Because thermodynamics is not concerned with the concept of time,it has been suggested that a better name for equilibrium thermodynamics wouldhave been thermostatics. Time-dependent thermodynamic processes are studiedby nonequilibrium thermodynamics. In continuum mechanics, we deal with bothequilibrium and non-equilibrium thermodynamics.

Thermodynamic laws are of very general validity, and they do not depend onthe details of the interactions or the systems being studied. This means they canbe applied to systems about which one knows nothing other than the balance ofenergy and matter transfer between them and the environment.

The quantities that set thermostatics apart from classical particle mechanics aretemperature and entropy. The significance of entropy can be illustrated as follows.Consider a flowing fluid. The fluid molecules possess kinetic energy which can bebroken into two components, a part which is ordered and contributes to the bulkvelocity, and another part which is random. The ordered energy is similar to themacroscopic kinetic energy of particle mechanics, and is mechanical in form. It iscapable of being converted to work. Extraction of the ordered kinetic energy wouldleave only the random (thermal) energy in the fluid. The random component of theenergy would contribute nothing to the work, as molecules would impact with suchforces so as to cancel each other. Theoretically, one could extract all the orderedenergy from the fluid leaving only the random energy. Now suppose that ratherthan extracting the organized energy, we somehow bring the convective fluid to astop. The total energy of the fluid would remain unchanged, but there would nolonger be any ordered component. All the kinetic energy of the molecules is nowcoming from random motions, and any attempt to convert this energy to work isfruitless. Entropy is a measure of the randomness, or of the energy’s inability todo work – except through transfer of randomness from one body to another.

The random thermal energy will not freely convert back to mechanical form.That is, the likelihood that the molecules will realign to travel in some preferreddirection is extremely small. Thus, since entropy is a measure of the randomness,

8 INTRODUCTION

it will not decrease without some external interactions. The only way one candecrease molecular randomness is to transfer some of this randomness to anotherbody, and thereby increasing the randomness (entropy) of the other body. Thus,transfer of thermal energy from one body to another effectively transfers entropy.The transfer of thermal energy (randomness) as heat in this fashion is the onlyknown way by which it is possible to reduce a body’s entropy.

1.5 Constitutive theory

The specification of the stress and strain states of a body is insufficient to describeits full behavior; we need in addition to link these two fields.

This is achieved by introducing a constitutive relation, which prescribes theresponse of the continuum to arbitrary loading and thus defines the connectionbetween the stress and strain tensors for the particular material.

At best a mathematical expression provides an approximation to the actualbehavior of the material, but as we shall see we can simulate the behavior of awide class of media.

In general, we think of materials as existing in either a solid or fluid state. Thedistinction between solid and fluid matter is relative; it depends on time scalesover which the material deforms. In turn, we can view a solid as either hard orsoft. Hard solids tend to respond elastically to an applied force, they have largeacoustic speeds, their energy character is enthalpic, they tend to be anisotropic,they usually rupture upon yielding, and they retain perfect memory of only theirinitial state. On the other hand, soft solids tend to be dissipative, have a lowacoustic speed, their energy behavior is entropic, they tend to be isotropic, theyfail through plastic deformation (fluid like), and they behave as viscous on a shorttime scale and elastic on a long time scale. Fluids, in turn, can be classified as eitherisotropic or anisotropic. Isotropic fluids can exhibit time scale effects. In general,they respond elastically at short times and viscous at long times. Anisotropicfluids, such as liquid crystals, behave solid-like and exhibit elastic behavior insome directions.

1.5.1 Solids

To get an idea of the behavior of solids, we consider extension of a wire underloading. The tensile stress σ and tensile strain e are then typically related. Atypical stress-strain curve is illustrated in Fig. 1.5.

a) Elasticity: If the wire returns to its original configuration when the load isremoved, the behavior is said to be elastic.

i) for linear elasticity σ = E e – called Hooke’s law and is usually validfor small strains (E is the elastic modulus);

ii) for nonlinear elasticity σ = f(e) – it is important for rubber-like mate-rials.

b) Plasticity: Once the yield point is exceeded, permanent deformation occursand there is no unique stress-strain curve, but a unique dσ-de relation. Dueto microscopic processes, the yield stress rises with σ (work hardening).

1.5. CONSTITUTIVE THEORY 9

X

σ

e

Elasticregion

Yield point

Plastic region Ultimate stressor

Fracture point

Figure 1.5: A typical stress-strain curve.

c) Viscoelasticity (rate-dependent behavior): Materials may creep and showslow long-term deformation, e.g., plastics and metals at elevated tempera-tures. Simple models of viscoelasticity are

i) Maxwell model:

σ +E

µσ = E e

which allows for instantaneous elasticity and represents a crude descrip-tion of a fluid (µ is the viscosity of the material).

ii) Kelvin–Voigt model:

σ = E e + µ e

which displays long-term elasticity.

More complex models can be written down, but all have the same charac-teristic of depending on the time history of deformation.

1.5.2 Fluids

The simplest constitutive equation encountered in continuum mechanics is that ofan ideal fluid:

σ = −p(ρ,T ) 1,where ρ is the density, T is the absolute temperature, the pressure field p is isotropicand depends on density and temperature, and 1 is the unit tensor. If the fluidis incompressible, ρ is a constant. The next level of complication is to allow the

10 INTRODUCTION

stress to depend on the flow of the fluid. The simplest such form, a Newtonianviscous fluid, includes a linear dependence on strain rate

σ = −p(ρ,T ) 1 + µ(ρ,T ) e.The quantity µ is called the shear viscosity.

1.6 Pioneers of continuum mechanics

The study of continuum mechanics originated from the works of James and JohnBernoulli, Euler, and Cauchy. The field remained stagnant for a very long periodof time after them. It was only after World War II that interest in the field wasrenewed. The modern field of continuum mechanics is the result of pioneeringworks from Truesdell, Noll, Toupin, Rivlin, Coleman, Ericksen, Müller, Eringen,Gurtin, and Liu, among others. Clifford Truesdell is considered the father ofmodern continuum mechanics.

Bibliography

B.D. Coleman. Thermodynamics of materials with memory. Archive for RationalMechanics and Analysis, 17:1–46, 1964.

B.D. Coleman and W. Noll. The thermodynamics of elastic materials withheat conduction and viscosity. Archive for Rational Mechanics and Analysis,13(1):167–178, 1963.

A.C. Eringen. Basic principles: Balance laws. In A.C. Eringen, editor, ContinuumPhysics, volume II, pages 69–88. Academic Press, Inc., New York, NY, 1975.

A.C. Eringen. Basic principles: Deformation and motion. In A.C. Eringen, editor,Continuum Physics, volume II, pages 3–67. Academic Press, Inc., New York, NY,1975.

A.C. Eringen. Basic principles: Thermodynamics of continua. In A.C. Eringen,editor, Continuum Physics, volume II, pages 89–127. Academic Press, Inc., NewYork, NY, 1975.

A.C. Eringen. Constitutive equations for simple materials: General theory. InA.C. Eringen, editor, Continuum Physics, volume II, pages 131–172. AcademicPress, Inc., New York, NY, 1975.

A.C. Eringen. Mechanics of Continua. R.E. Krieger Publishing Company, Inc.,Melbourne, FL, 1980.

Y.C. Fung. A First Course in Continuum Mechanics. Prentice Hall, Inc., Engle-wood Cliffs, NJ, 3rd edition, 1994.

M.E. Gurtin. Modern continuum thermodynamics. In S. Nemat-Nasser, editor,Mechanics Today, volume 1, pages 168–213. Pergamon Press, New York, 1974.

I. Müller. Thermodynamics. Pitman Publishing, Inc., Boston, MA, 1985.

BIBLIOGRAPHY 11

W. Noll. Lectures on the foundations of continuum mechanics and thermody-namics. Archive for Rational Mechanics and Analysis, 52(1):62–92, 1973.

W. Noll. The Foundations of Mechanics and Thermodynamics – Selected Papers.Springer-Verlag, New York, 1974.

C. Truesdell. The mechanical foundations of elasticity and fluid dynamics. Jour-nal of Rational Mechanics and Analysis, 1(1):125–300, 1952.

C. Truesdell. Rational Thermodynamics. Springer-Verlag, New York, NY, 2ndedition, 1984.

C. Truesdell and W. Noll. The non-linear field theories of mechanics. In S. Flügge,editor, Handbuch der Physik, volume III/3. Springer, Berlin-Heidelberg-NewYork, 1965.

C. Truesdell and R.A. Toupin. The classical field theories. In S. Flügge, editor,Handbuch der Physik, volume III/1. Springer, Berlin-Heidelberg-New York, 1960.

2

Tensor analysis

Tensor analysis is the language used to describe continuum mechanics. Physicallaws, if they really describe the real world, should be independent of the positionand orientation of the observer. Two individuals using two coordinate systemsin the same reference frame should observe the same physical event. For thisreason, the equations of physical laws, which are tensor equations, should holdin any coordinate system. Invariance of physical laws to two frames of referencein accelerated motion relative to each other is more difficult and requires generalrelativity theory (tensors in four-dimensional space-time). For simplicity, we limitourselves to tensors in three-dimensional Euclidean space. We will not considertensors on manifolds (curved spaces) in the main text. This topic is discussed inAppendix D.

We will discuss general tensors on an arbitrary curvilinear coordinate system,although for the development of continuum mechanics theory, we will use Cartesiantensors. For the solution of specific problems, orthogonal curvilinear coordinatesand indeed rectangular coordinates will be used.

In three-dimensional space, a scalar quantity has the same magnitude irrespec-tive of the coordinate system. A vector may be visualized as an arrow that haslength equal to the magnitude of the vector and that is pointing in the directionof the vector. It has an existence independent of any coordinate system in whichit is observed. Thus a vector is unchanged if it is moved parallel with itself. Twovectors that have the same direction and length are equal. They do not have tohave the same origin. In vector operations, we shift the vectors so that they havethe same origin. A vector of unit length is called a unit vector. A second-ordertensor is a quantity with two directions associated with it. Although it cannot bevisualized simply as an arrow, it also has an existence independent of the coordi-nate system. This remains true for tensors of arbitrary order. It is precisely thisindependence of the coordinate system that motivates us to study tensors.

2.1 Review of linear algebra

Notation: All scalar quantities will be denoted using the lowercase italic font.Matrices will be denoted by the uppercase (not bold) italic font. Vectors and higherorder tensors are denoted by the lowercase and uppercase bold fonts, respectively.

13

14 TENSOR ANALYSIS

Volumes, surfaces, and curves will be denoted using the script font. Lastly, wewill use the fraktur font to denote functionals. In all cases, exceptions are madeto respect the notation used historically.

Examples

Scalars ∶ a, b, c

Vectors ∶ u,v,w

Tensors ∶ T,D,W,1,ǫ,σ

Matrices ∶ A,B,C;

Geometric objects ∶ V ,S

Functionals ∶ F,G

Definition: The vector u is said to be a linear combination of the vectorsv1,v2, . . . ,vn if

u = c1v1 + c2v2 +⋯ + cnvn,

where the ci’s are real numbers.Definition: A set of vectors v1,v2, . . .vn is said to be linearly independent

if

c1v1 + c2v2 +⋯ + cnvn = 0implies that c1 = c2 = ⋯ = cn = 0; i.e., the only linear combination that is equal tozero is the trivial linear combination.

Definition: A vector space V is a collection of vectors, which is closed underlinear combinations; i.e., if (w1,w2) ∈ V , then (c1w1 + c2w2) ∈ V for all real(c1, c2).

Definition: A basis of V is a set u1,u2, . . . ,un, which

i) is linearly independent, and

ii) spans V (i.e., if w ∈ V , then ∃ ci’s such that w = c1u1 + c2u2 +⋯ + cnun).

Definition: The dimension of a vector space V corresponds to the number ofvectors (unique) in a basis (not unique).

From now on, if the dimension of the vector space is n, we will indicate this bya superscript, i.e., Vn.

To connect some of the later discussions with vectors and matrices, we use thefollowing convention. If A denotes a matrix [aij], then aij denotes the entry in

the ith row and jth column of the matrix. Sometimes the notation aij or aij

is used instead. These symbols also refer to the entry in the ith row and jthcolumn. The transpose of the matrix A is denoted by AT , and the entries denotedby aji, while the trace of A is denoted by tr A = tr AT and the entries by aii.Note that if α is a scalar, then tr (αA) = α trA. The transpose of the product ofmatrices A and B is given by (AB)T = BTAT . Also note that tr (AB) = tr (BA).The determinant of matrix A is denoted as detA = detAT . We also have that

2.1. REVIEW OF LINEAR ALGEBRA 15

det (αA) = αn detA, where the dimension of A is n × n. The determinant of theproduct of two matrices A and B is given by the product of the determinant of eachmatrix, i.e., det (AB) = (detA) (detB). The determinant of the exponential of amatrix is equal to the exponential of the trace of the matrix, i.e., det (eA) = etrA.A matrix is said to be symmetric if aij = aji and skew-symmetric if aij = −aji.Note that a symmetric three-dimensional matrix has six independent componentswhile a skew-symmetric one has only three independent components since it hasa zero trace. Any matrix can be decomposed as a sum of its symmetric andskew-symmetric parts. This is denoted symbolically as

A = sym A + skw A, (2.1)

where

sym A = 1

2(A +AT ) and skw A = 1

2(A −AT ) . (2.2)

If D is a symmetric matrix and W a skew-symmetric matrix, then it is evidentthat

DT =D and WT = −W. (2.3)

The inverse of matrix A is defined by

AA−1 = A−1A = I, (2.4)

where I is the identity matrix with entries of unity along the main diagonal andzero for all other entries. The entry in the ith row and jth column of the identitymatrix I is denoted by

δij = 1, i = j0, i ≠ j , (2.5)

where δij is called the Kronecker delta symbol. We note that (A−1)−1 = A and(αA−1)−1 = α−1A. A matrix is said to be orthogonal if A−1 = AT . The inverse ofthe product of matrices A and B is given by (AB)−1 = B−1A−1. The determinantof the inverse of matrix A is given by the reciprocal of the determinant of thematrix, i.e., detA−1 = (detA)−1. For an orthogonal matrix AAT = ATA = I, sodet(AAT ) = (detA)(detAT ) = det I = 1 and subsequently detA = ±1. Note thatfor a three-dimensional system, tr I = 3. An orthogonal matrix whose determinantequals +1 is called a proper orthogonal matrix, and one whose determinant equals−1 is called an improper orthogonal matrix. A proper orthogonal matrix, whenviewed as a transformation, transforms a right-handed set of axes into a right-handed set of axes, whereas an improper orthogonal matrix transforms a right-handed set of axes into a left-handed set of axes or vice versa. In either case,the transformation preserves vector lengths and angles between vectors. If U is acolumn matrix, the entry in the ith row is denoted by ui, where it is understoodthat the column entry is j = 1. Similarly, a matrix V having only one row is calleda row matrix and the superscript i = 1 is understood, so the jth column is denotedby vj .

It is straightforward to show that the determinant of a 3 × 3 matrix A is givenby

a ≡ detA = det [aij] = 3∑i=1

3∑j=1

3∑k=1

ǫijkai1a

j2ak3 =

3∑i=1

3∑j=1

3∑k=1

ǫijka1i a2

ja3

k. (2.6)

16 TENSOR ANALYSIS

These equations are obtained by expanding by minors the determinant of A alonga column or row, respectively. The entries ǫijk and ǫijk correspond to the Levi–Civita (or permutation) symbol

ǫijk = ǫijk =⎧⎪⎪⎪⎨⎪⎪⎪⎩

1 if (i, j, k) is an even permutation of (1,2,3),−1 if (i, j, k) is an odd permutation of (1,2,3),0 if any two labels are the same,

(2.7)

i.e., ǫijk = ǫjki = ǫkij = −ǫikj = −ǫkji = −ǫjik.

Example

The determinant obtained by expanding by minors along the first columnis given by

detA = det

⎡⎢⎢⎢⎢⎢⎣a11 a12 a13a21 a22 a23a31 a32 a33

⎤⎥⎥⎥⎥⎥⎦= a11 (a22a33 − a32a23) − a21 (a12a33 − a32a13) + a31 (a12a23 − a22a13)= 3∑

j=1

3∑k=1

a11 (ǫ1jkaj2ak3) + a21 (ǫ2jkaj2ak3) + a31 (ǫ3jkaj2ak3)= 3∑

i=1

3∑j=1

3∑k=1

ǫijkai1a

j2ak3 .

Some properties of the Kronecker delta and Levi–Civita symbols are given inAppendix A.

2.2 Tensor algebra

Think of curvilinear coordinates as an application. At a point P ∈ E3, where E3

denotes the three-dimensional Euclidean space, one could define three arbitrarycoordinates. The length (or norm) of a vector v in Euclidean space is given by

∣∣v∣∣ =√v ⋅ v ≥ 0, (2.8)

where u⋅v =∑3

i=1 uivi denotes the usual scalar inner product. As shown in Fig. 2.1,

we could choose a basis, called the natural basis, at a point P as vectors tangentto any three coordinate lines (x1, x2, x3) at P (not unique vectors): e1,e2,e3 =ei. The natural basis is a basis as long as the scalar triple product is nonzero,e1 ⋅ (e2 × e3) ≠ 0. The quantity u × v is a vector that is orthogonal to both u andv, and is obtained from the vector, or cross, product of two vectors. We will saymore about this shortly. From now on, unless stated otherwise, italic subscriptsand superscripts are understood to take on values from 1 to 3.

We could have chosen as a basis the normals to the coordinate surfaces: e1,e2,e3 = ei. Again, this is a basis as long as their scalar triple product is nonzero,e1 ⋅ (e2 × e3) ≠ 0. For orthogonal coordinates, ei = ei; otherwise, they are notequal. The basis ei is called a reciprocal basis relative to ei. Now e1 and e2

2.2. TENSOR ALGEBRA 17

P

x1x2

x3

e1

e2

e3

Figure 2.1: Basis at point P .

are tangent to the x3 surface, and e3 is normal to the x3 surface, i.e.,

e1 ⋅ e3 = 0 and e2 ⋅ e

3 = 0.We require that the length of e3 be chosen such that

e3 ⋅ e3 = 1.

Given a basis e1,e2,e3 ∈ E3, we can always find the reciprocal basis e1,e2,e3with the properties

ei ⋅ ej = δji . (2.9)

In E3, e1(e2,e3); therefore, e1 = const. e2×e3. To find the constant, we require

e1 ⋅ e1 = 1 = const. e1 ⋅ (e2 × e3) ,

or abbreviating the scalar triple product by

[e1,e2,e3] ≡ e1 ⋅ (e2 × e3) , (2.10)

const. = 1[e1,e2,e3] ,thus

e1 = e2 × e3[e1,e2,e3] . (2.11)

Similarly

e2 = e3 × e1[e1,e2,e3] , (2.12)

ande3 = e1 × e2[e1,e2,e3] . (2.13)

Using the above results, it can be easily shown that

[e1,e2,e3] = 1[e1,e2,e3] . (2.14)

It is straightforward to show that the scalar triple product can be rewritten in theequivalent forms

[u,v,w] = [v,w,u] = [w,u,v] . (2.15)

18 TENSOR ANALYSIS

e1

e2

e1e2

(0,5)

(−1,0) (1,0)Figure 2.2: Basis in example.

Example

In E2, given e1 = (1,5) and e2 = (−1,2), find e1,e2.Let e1 = (a, b) and e2 = (c, d), so that

e1 ⋅ e1 = 1 = a + 5b,

e1 ⋅ e2 = 0 = c + 5d,

e2 ⋅ e1 = 0 = −a + 2b,

e2 ⋅ e2 = 1 = −c + 2d.

We easily find that e1 = 1

7(2,1) and e2 = 1

7(−5,1). The results are displayed

in Fig. 2.2.

If we have 2 bases at a point in E3, ei and ei, given a vector v in the samespace, we can write it with respect to either basis:

v = v1e1 + v2e2 + v3e3 = 3∑i=1

viei = viei, (2.16)

or

v = v1e1 + v2e2 + v3e3 = 3∑i=1

viei = viei. (2.17)

The summation convention used (Einstein’s) is that we implicitly sum on repeatedindices from 1 to 3, one variable with a superscript index and the other variablewith the subscript index. The variables vi are called contravariant components,and vi are called covariant components. The summed index, i in this case, isalso called a dummy index since it can be replaced by any other symbol, say j,

2.2. TENSOR ALGEBRA 19

without changing the value of the expression. An index appearing in an expressionthat does not sum is called a free index. The indicial notation and summationconvention permits us to write equations of continuum mechanics in a much shorterform than would otherwise be possible. This brevity makes the equations easier toremember and understand, once the notation is learned. For example, our previousequations for the reciprocal basis can be combined into the single equation

ei = ǫijkej × ek[e1,e2,e3] . (2.18)

Note that in the above equation i is a free index, while j and k are dummy indicesthat can be replaced by other indices. In using Einstein’s notation, it is imperativethat an index appear no more than twice on either side of an equation, and thatthe same free row and column indices appear on both sides of the equation.

We now also note that ǫijk, and similarly ǫijk, are skew-symmetric since theinterchange of any two indices changes the sign of the quantity, i.e.,

ǫijkaima

jna

kp = ǫjikajmainakp = −ǫijkainajmakp.

In the same way, we can show that interchanging any two of the indices m, n, andp alters the sign. Furthermore, in view of (2.6), this suggests that we may writemore generally, for an expansion by columns,

ǫijkaima

jna

kp = a ǫmnp.

The same type of argument may be used to infer that for an expansion by rows,

ǫijkami anj a

pk= a ǫmnp.

The number of superscripts or subscripts that the components of a variable hasis called the order (or rank) of the tensor. A scalar variable is a zeroth ordertensor, a vector variable is a first order tensor. The variable τ such that

τ = τ ijeiej or τ = τlmelem (2.19)

is a second order tensor, also called a dyadic, and τ ij and τlm are its contravari-ant and covariant components, respectively. Note that the variables used in thesummation are arbitrary, but it is emphasized that there should be no more thantwo such index variables in each expression.

The main issue of tensor analysis is the requirement that scalars, vectors, andhigher order tensor quantities remain invariant when a different curvilinear coordi-nate system is used, i.e., when a new basis e1, e2, e3 is used instead of e1,e2,e3.Note that corresponding to the new basis e1, e2, e3, there will be a reciprocalbasis e1, e2, e3, where we also take ei ⋅ e

j = δji . Now at a point P we have 4

bases:

ei ∶ the given basis, (2.20)ei ∶ the reciprocal to the given basis, (2.21)ei ∶ the new basis, (2.22)ei ∶ the reciprocal to the new basis. (2.23)

20 TENSOR ANALYSIS

A change of coordinate systems can be made by a translation of axes withoutrotation followed by a rotation of the axes keeping the origin fixed (and possiblyfollowed by a reflection in a coordinate plane if we allow transformations betweenright-handed and left-handed systems). The components of a vector (and similarlyhigher order tensors) are not changed by translation of axes, since the projectionsof a vector onto parallel axes are the same in magnitude and sense. Hence we needonly to consider transformations with fixed origin.

What conditions do the invariance requirement impose on the tensor compo-nents?

A scalar imposes no condition since it has the same value in any coordinatesystem.

Consider the vector v, which we can write in any of the following forms:

v = viei = viei = viei = viei. (2.24)

Since v = viei = vj ej , how are the components vi and vj related? First we must askhow are ei and ej related at a point? The answer is by a coordinate transformation.As illustrated in Fig. 2.3, in general we have

ej = aijei, (2.25)

or

New basis

Given basis

e1 e2 e3

e1 a11 a12 a13e2 a21 a22 a23e3 a31 a32 a33

The matrix component aij is the cosine of the angle between the new coordinatevector ej and the given coordinate vector ei, i.e., it is the direction cosine. Wenote that in general [aij] ≠ [aij]T . The nine components aij are not all independent

if the coordinates are orthogonal; only 6 are. Furthermore, in such case [aij]−1 =[aij]T = [aji ] so that ei = aji ej . Now, returning to the general case, we have

v = viei = vj ej = vjaijei,so that

aij vj = vi. (2.26)

Components with superscript indices are called contravariant because the trans-formation equation for components is contrary (or opposite) to the transformationequation of the basis (2.25). How are vi and vi related? To see this, we write

v = viei = vj ej .Dotting from the right with ek, we have

viei⋅ ek = vj ej ⋅ ek = vjδjk = vk.

2.2. TENSOR ALGEBRA 21

x3

x2

x1

x2

x1

e1e1

e2

e2

e3

e3

x3

cos−1a32

cos−1a12

cos−1a22

Figure 2.3: Coordinate transformation.

Now using (2.25) on the left-hand side of the above equation, we have

vk = viei ⋅ ajkej = viajkδij ,or

vk = aikvi. (2.27)

Components with subscript indices are called covariant because the transformationequation for components is the same as the transformation equation of the basis(2.25).

For a second-order tensor τ , we have

τ = τijeiej = τlmelem = τrseres = τpq epeq. (2.28)

A product such as eiej is understood to mean a tensor (in this case a dyadic)product. Sometimes the symbol ⊗ is used to make this dyadic product explicitso that one would write ei ⊗ ej , but we will not use such notation. Products oftensors without any explicit operator between the tensors are always understoodto mean tensor products in the present text. Now consider

τijeiej = τlmelem.

Dotting the above equation from the right with ek and ep consecutively, we have

τij(ei ⋅ ep)(ej ⋅ ek) = τlm(el ⋅ ep)(em ⋅ ek) = τlmδlpδmk = τpk.Now, using (2.25) on the left-hand side of the above equation, we can write

τpk = τij(ei ⋅ arper)(ej ⋅ aqkeq) = τijaipajk.

22 TENSOR ANALYSIS

Thus the covariant components of a second-order tensor transform as

τpk = aipajkτij . (2.29)

We also have

τrseres = τpq epeq = τpqarperasqes = τpqarpasqeres,and thus the contravariant components of a second-order tensor transform as

arpasq τ

pq = τrs. (2.30)

A similar procedure can be used to obtain relations between covariant and con-travariant components of a tensor of any order.

2.3 The metric tensor and its properties

How are vi and vj related? To answer this question, we first note that

v = viei = vjej .Dotting the above equation from the right with ek, we have

viei⋅ ek = viδik = vk = vjej ⋅ ek,

orvk = gjkvj , (2.31)

where we have definedgij ≡ ei ⋅ ej = gji. (2.32)

If we think of G = [gij] as a matrix, then we see that this matrix is symmetricsince it is equal to its transpose, i.e., G = GT .

Is the use of gij as covariant components of some tensor g justified? That is, dothe gij transform like covariant components of a second-order tensor? To see this,we first define analogously

glm ≡ el ⋅ em = gml. (2.33)

Subsequently

glm = el ⋅ em = ailei ⋅ ajmej = ailajmei ⋅ ej = ailajmgij .We conclude that the second-order tensor

g = gijeiej (2.34)

indeed transforms like a second-order tensor with covariant components gij . Thetensor g is called the metric tensor.

We could also write it as

g = glmelem = gijeiej. (2.35)

2.3. THE METRIC TENSOR AND ITS PROPERTIES 23

Now dotting the above equation from the right with er and ep consecutively, wehave

glm(el ⋅ ep)(em ⋅ er) = gij(ei ⋅ ep)(ej ⋅ er),or

glmδplgmr = gij(ei ⋅ ep)δjr ,

or

gpmgmr = gir(ei ⋅ ep) = gmr(em ⋅ ep).Thus

gpm = em ⋅ ep = gmp. (2.36)

Now consider the equation

v = viei = vjej,and dot both sides from the right by ek to obtain

viei⋅ ek = vjej ⋅ ek,

or

vigik = vjδkj ,

orvk = gikvi. (2.37)

We see that the covariant and contravariant components gij and gij of the metrictensor provide the means of computing the covariant or contravariant components(vi or vj) of vector v if we know the contravariant or covariant components (vj

or vi). More simply, the components of the metric tensor provide us the means ofraising or lowering indices.

This is also true when operating on the basis vectors:

ei = gijej and ei = gijej , (2.38)

since we see that if we take the dot product of ek from the right with the firstequation, we have the identity

ei ⋅ ek = gijej ⋅ ek = gik,while if we take the dot product of ek from the right with the second equation, wehave the identity

ei ⋅ ek = gijej ⋅ ek = gik.Alternately, if we take the dot product of ek from the right with the first equation,we obtain

ei ⋅ ek = gijej ⋅ ek,

24 TENSOR ANALYSIS

or

δki = gijgjk. (2.39)

In matrix form, this last equation is

I = G [gjk],thus we see that

[gjk] = G−1,i.e., the covariant and contravariant components of the metric tensor are inversesof each other. In matrix form, the equations would be written as

I = GG−1 = G−1G.

2.4 General polyadic tensor of order m

In general, a tensor τ i1i2⋯irj1j2⋯jnei1ei2⋯eire

j1ej2⋯ejn with (rn) components such that

r + n =m is called a polyadic tensor of order m. For example, a tensor τ of orderm = 5 with (5

0) components is written as τ ijklmeiejekelem. When r and n are not

zero, then the components are said to be mixed.

Consider the dyadic tensor

τ = τijeiej = τklekel = τrs eres. (2.40)

We can rewrite the above as

τiseies = τrlerel = τrs eres.

Transforming coordinates by using the metric tensor

τisgirere

s = τrlerglses = τrs eres,we obtain

τisgir = τrlgls = τrs . (2.41)

Again we see that the metric tensor provides us with the means of “raising andlowering” component indices for a dyadic tensor. This remains true for tensors ofany order.

2.4. GENERAL POLYADIC TENSOR OF ORDER M 25

Example

Suppose we have the equation

ai = τijbj ,and suppose that we wanted to obtain ar (contravariant components).Then to raise the subscript, we would multiply the equation by gir:

aigir = τijgirbj ,

or

ar = τrj bj.We recall from before that the covariant and contravariant components of a

second-order tensor under a change of basis transform as

τij = τklaki aljand

τ ijaki alj = τkl.

How does τ ij transform when the basis is changed? To see this, we first write

ei = ajiej ,and

τ = τ ijeiej = τ lkelek.Now if we take the dot product of the above equation with er from the right, wehave

τ ijei(ej ⋅ er) = τ lk el(ek ⋅ er) = τ lkelδkr = τ lrel.Now, changing the basis, we have

τ ijeiaprδ

jp = τ lrailei,

or

τ ipaprei = τ lrailei.

Thus, the mixed components transform as

τ lrail = τ ipapr . (2.42)

That is, the respective contravariant and covariant components transform as onewould expect contravariant and covariant components would. This remains truefor a tensor of any order, i.e.,

τ i1i2⋯irj1j2⋯jnak1

i1ak2

i2⋯akr

ir= τk1k2⋯kr

l1l2⋯lnal1j1a

l2j2⋯alnjn . (2.43)

26 TENSOR ANALYSIS

All tensors that we have been discussing up to this point are called absolute ten-sors. We now extend the definition and call a tensor whose components transformas

τ i1i2⋯irj1j2⋯jnak1

i1ak2

i2⋯akr

ir= sgnas ∣a∣w τk1k2⋯kr

l1l2⋯lnal1j1a

l2j2⋯alnjn (2.44)

a relative or weighted tensor of order m = r+n, where a = det[aij] is the determinantof the transformation, which is assumed to be non-singular. If s = w = 0, τ is calledan absolute tensor. If s = 0 and w ≠ 0, τ is called a relative tensor of weight w.If s = 1 and w = 0, τ is called an axial tensor. Lastly, if s = 1 and w ≠ 0, τ iscalled an axial relative tensor of weight w. When the word tensor is used from nowon, it is understood to mean an absolute tensor unless explicitly stated otherwise.Relative or weighted tensors are also referred to as pseudo-tensors.

Does the Kronecker delta δij transform like mixed components of a tensor oforder two? To see if it does, we take this to be the components of the unit, oridentity, tensor

1 = δijeiej = δkl ekel. (2.45)

It is immediately obvious that if we let τ ij → δij, the previous derivation appliesand thus the Kronecker delta transforms like mixed components of an absolutesecond-order tensor. Since it has the same value in all coordinate systems, i.e.,δij = δij, such a tensor is called isotropic. More specifically, a tensor is isotropic ifits components are the same under arbitrary rotations of the basis vectors. Thus,e.g., if we have

Bij = aki amj Bkm,

then the tensor B is isotropic if Bij = Bij .A simple example of a third-order isotropic tensor is the Levi–Civita tensor

ǫ = ǫijkeiejek. (2.46)

Since the tensor is isotropic, we have that

ǫijk = ǫijk. (2.47)

In addition, as noted earlier (see (2.6)), since

a = ǫrstar1as2at3, (2.48)

it then follows that

ǫijka = ǫrstari asjatk, (2.49)

or

ǫijk = a−1ǫrstari asjatk. (2.50)

We see that ǫijk transform like covariant components of a third-order relativetensor with weight w = −1. Note that a is nothing more than the determinant ofthe transformation. In a similar fashion, it can be shown that

ǫijkariasja

tk = aǫrst, (2.51)

so that ǫrst transforms like contravariant components of a third-order relativetensor with weight w = 1.

A general discussion of isotropic tensors is given in Appendix B.

2.5. SCALAR PRODUCT OF TWO VECTORS 27

2.5 Scalar product of two vectors

Let

u = uiei = ujej ,and

v = vlel = vmem.

Now the scalar (or dot or inner) product of the two vectors is given by

u ⋅ v = (uiei) ⋅ (vlel) = uivl(ei ⋅ el),or

u ⋅ v = giluivl. (2.52)

Similarly

u ⋅ v = gjmujvm. (2.53)

Also,

u ⋅ v = (uiei) ⋅ (vmem) = uivm(ei ⋅ em) = uivmδmi ,so we can see that

u ⋅ v = uivi = uivi. (2.54)

2.6 Vector product of two vectors

Consider first the scalar triple product

[e1,e2,e3] = (g1iei) ⋅ [(g2jej) × (g3kek)] ,= g1ig2jg3k [ei ⋅ (ej × ek)] ,= ǫijkg1ig2jg3k (e1 ⋅ e2 × e3) ,= ǫijkg1ig2jg3k

e1 ⋅ e2 × e3,

= detG[e1,e2,e3] ,where we have used (2.18) and (2.6). We now see that

[e1,e2,e3] =√g, (2.55)

and subsequently

[e1,e2,e3] = 1√g, (2.56)

28 TENSOR ANALYSIS

where

g ≡ detG. (2.57)

The vector (or cross) product of two vectors, using the definition of the reciprocalbasis, is now given by

u × v = uivjei × ej =√gǫijkuivjek = wkek =w, (2.58)

thus

wk =√gǫijkuivj =√gǫkijuivj . (2.59)

Similarly, we find that

w = wkek, (2.60)

where

wk = 1√gǫijkuivj = 1√

gǫkijuivj . (2.61)

Note that v × u = −u × v since this corresponds to an interchange of two adjacentindices in the Levi–Civita symbol.

We now define

εijk ≡√g ǫijk and εijk ≡√g ǫijk. (2.62)

Then, using our earlier result, we have

εijk =√g ǫijk =√g ǫrstari asjatk, (2.63)

or

εijk = εrstariasjatk. (2.64)

Thus εijk transform like the covariant components of a third-order absolute tensor.Similarly, we can show that

εijkari asja

tk = εrst, (2.65)

transform like contravariant components of a third-order absolute tensor, where

εrst ≡ ǫrst√g. (2.66)

Subsequently, we define the third-order absolute Levi–Civita tensor

ε = εijkeiejek = εijkeiejek. (2.67)

Note that with the above definitions, the vector, or cross, product of two vectorscan now be rewritten as

w = u × v = wkek = wkek, (2.68)

2.7. TENSOR PRODUCT OF TWO VECTORS 29

where

wk = εkijuivj and wk = εkijuivj . (2.69)

From the above, formally we can also define the following vector operator that isoccasionally found to be useful:

×v = −v × . (2.70)

The proof is obtained by simply operating on an arbitrary vector from the rightand left.

2.7 Tensor product of two vectors

The tensor product of vectors u and v is simply uv. If w is a vector and A is asecond rank tensor, then it is clear that

tr (uv) = u ⋅ v, (2.71)(uv) ⋅w = u (v ⋅w) = (v ⋅w)u, (2.72)

A ⋅ (uv) = (A ⋅u)v, (2.73)(uv) ⋅A = u (v ⋅A) = u (AT⋅ v) . (2.74)

Note that while the product of u and v yields a second-order tensor, it is impor-tant to realize that a general second-order tensor, say A, cannot be representedby the tensor product of two vectors, say uv. This should be immediately ob-vious by noting that the components of a general second-order tensor has nineindependent elements, while the components of the second-order tensor resultingfrom the tensor product of two vectors has only six independent entries (the threecomponents of the two vectors). Furthermore, while in general detA ≠ 0, we havethat detuv = 0. However, if A is symmetric, then it can be shown that it can berepresented as a tensor product of the vectors u and v, the components of whichrepresents a quadric surface (see Section 2.10). Lastly, we note that the scalar andvector products of u and v are composed of particular linear combinations of thecomponents of their tensor product.

2.8 Contraction of tensors

Contraction enables us to obtain an (n−2) rank tensor from a given tensor of rankn. This is accomplished by placing a dot product between any polyadic expression.For example, with the tetrad abcd (where a, b, c and d are vectors), contractioncan be done in three ways with adjacent vectors to yield

a ⋅ bcd = (a ⋅ b)cd, (2.75)

ab ⋅ cd = (b ⋅ c)ad, (2.76)

abc ⋅ d = (c ⋅ d)ab, (2.77)

where the dot product in parentheses is a scalar function multiplying the resultingdyad. Note that

ab ⋅ cd ≠ cd ⋅ ab. (2.78)

30 TENSOR ANALYSIS

There are two possible scalar (double-dot) products of two dyads and they aredefined by

ab ∶ cd ≡ (a ⋅ c)(b ⋅ d), (2.79)

ab ⋅ ⋅ cd ≡ (b ⋅ c)(a ⋅ d). (2.80)

It is easily shown that the scalar product is commutative as it should be sincea scalar is invariant in any coordinate system, although these products would bedifferent if higher order tensors were involved. For example, if u and v are vectorsand A is a second-order tensor, then we note that

u ⋅A ⋅ v ≠ v ⋅A ⋅ u, (2.81)

even though both sides are scalar quantities. The two sides are equal only if A issymmetric, i.e., A =AT . If A is skew-symmetric, i.e., A = −AT , then

u ⋅A ⋅ v = −v ⋅A ⋅ u. (2.82)

Note that the contraction of a second rank tensor A with the unit tensor yieldsthe scalar corresponding to the trace of the tensor:

trA = 1 ∶A =A ∶ 1 or Aii = δij Aij . (2.83)

Example

Let τ = τ ijkleieje

kel be a tensor of rank 4. Placing a dot product betweenany two basis vectors gives a new tensor of rank 2. For example,

β = τ ijkl(ei ⋅ ej)ekel = τ ijkl gijekel = τ iiklekel.

Alternatively, we could have written τ = τ ijkleiejekel so that by contractingwe have

β = τ ijkl(ei ⋅ ej)ekel = τ ijklδji ekel = τ iiklekel.Since the new tensor β should be independent of any coordinate system,we also have

β = τ iiklekel.

2.9. TRANSPOSE OF A TENSOR 31

Example

We show that the cross product of two vectors can also be written as

w = u × v = ǫ ∶ (uv) = (ǫijkeiejek) ∶ (ulel) (vmem)= ǫijkulvm (ej ⋅ el) (ek ⋅ em)ei= ǫijkgjlgkmu

lvmei

= ǫijkgingjlgkmulvmen

= √gǫnlmu

lvmen

= wnen,

or alternatively as

w = u × v = ǫ ⋅ ⋅(vu) = (ǫ ⋅ v) ⋅ u= [(ǫijkeiejek) ⋅ (vlel)] ⋅ (umem)= ǫijkgklgjmv

lumei

= ǫijkgingjmgklumvlen

= √gǫnmlu

mvlen

= wnen,

recovering our previous result.

2.9 Transpose of a tensor

The transpose of a tensor of any rank is given by the permutation of any twocomponents. For example, if A = aijkeiejek, then B = bijkeiejek is a transposeof A if bijk = aikj . The superscript of T on a matrix A = [aij], AT is understoodto denote the permutation of the two indices, i.e., AT = [aji]. The transposeoperation for a second rank tensor can also be viewed as the unique transpositionsatisfying

(v ⋅AT ) ⋅ u = (A ⋅ v) ⋅ u = u ⋅ (A ⋅ v) (2.84)

for all vector u and v. Note that (AT )T = A, and if A−1 = AT , then A ⋅AT =AT⋅A = 1 and A is an orthogonal tensor. The above rule applies to vectors as

well, since a column or row vector is understood to correspondingly have unity inthe row or column entry, so that upon transposition it becomes a row or columnvector correspondingly. If A is a tensor of rank 4, generalizing the above definitionfor tensors of rank 2, the unique transpose of A, which we denote simply as A

T ,satisfies

(B ∶AT ) ∶C = (A ∶ B) ∶C =C ∶ (A ∶B) (2.85)

for all second rank tensors B and C. Note that (AT )klij = [aijkl], (AT )T = A,and (BC)T =CTBT .

32 TENSOR ANALYSIS

For more general transposition, the superscript Tm,n will denote that the m

and n entries are to be transposed. For example, if A = [aijkl], then AT2,3 =[aikjl]. If multiple transpositions are required, the operator is applied repeatedly,e.g., AT2,3T1,4 = [alkji].2.10 Symmetric and skew-symmetric tensors

A tensor A is said to be symmetric in two indices of the same type (both covariantor both contravariant) if the value of any component is not changed by permutingthem. The corresponding indices are enclosed in parentheses, the unaffected in-dices being separated by vertical bars on either side. Thus, if aijkm = amjki, thenwe write a(i∣jk∣m) to denote the symmetry. A tensor A is said to be completelysymmetric in any set of upper or lower indices if its components are not altered invalue by any permutation of the set, e.g.,

aijkm = ajikm = ajimk =⋯. (2.86)

This is often denoted by sym A or its indices enclosed in parentheses, e.g., a(ijkm).The validity of this property in one coordinate system ensures it in all coordinatesystems.

A tensor A is skew-symmetric (or anti-symmetric) if its components are notaltered in value by any even permutation of the indices and are merely changedin sign by an odd permutation of these indices. The corresponding indices areenclosed in brackets, the unaffected indices being separated by vertical bars oneither side. Thus, if aijkm = −amjki, then we write a[i∣jk∣m]. A tensor A of rank kthat is skew-symmetric in all indices is called completely skew-symmetric, e.g.,

aijkm = −ajikm = ajimk = ⋯. (2.87)

This is often denoted by skw A or its indices enclosed in brackets, e.g., a[ijkm].In such tensors, all components having two equal indices are zero. A covariantor contravariant completely skew-symmetric tensor of rank k is called a k-vector.Note that 0-vectors are scalars and 1-vectors are vectors.

We point out that the generalized Levi–Civita (or permutation) symbol

ǫijk... = ǫijk... =⎧⎪⎪⎪⎨⎪⎪⎪⎩+1 if (i, j, k, . . .) is an even permutation of (1,2,3, . . .),−1 if (i, j, k, . . .) is an odd permutation of (1,2,3, . . .),0 if any two labels are the same,

(2.88)

is completely skew-symmetric (see Appendix A). Note that ǫijk... and ǫijk... arerelative isotropic tensors of weight +1 and −1, respectively. The correspondinggeneralized absolute contravariant and covariant components of the Levi–Civitatensors are defined by

εijk... ≡ ǫijk...√g

and εijk... ≡√g ǫijk..., (2.89)

where

g ≡ 1

n!ǫi1...inǫj1...jngi1j1⋯ginjn (2.90)

2.10. SYMMETRIC AND SKEW-SYMMETRIC TENSORS 33

is the contracted product of two epsilons of weight +1 and n absolute dyadics.Hence the discriminant g of the fundamental quadratic form is a relative scalar ofweight 2.

Associated with the components of the generalized Levi–Civita tensor is thegeneralized Kronecker delta

δi1...ikj1...jk= det

⎡⎢⎢⎢⎢⎢⎣δi1j1 ⋯ δi1jk⋮ ⋮ ⋮

δikj1 ⋯ δikjk

⎤⎥⎥⎥⎥⎥⎦, k ≤ n. (2.91)

We note that the k superscripts and subscripts in the component of the generalizedKronecker delta can range from 1 to n. This tensor has the properties that

δi1...ikj1...jk= ǫi1...ikǫj1...jk , (2.92)

(n − k)! δi1...ikj1...jk= ǫi1...ikik+1...inǫj1...jkik+1...in , (2.93)

δi1...ikik+1...ini1...ikjk+1...jn

= k! δik+1...injk+1...jn, (2.94)

and

δi1...ini1...in= n! (2.95)

It can be shown that δi1...ikj1...jkis the component of an absolute tensor of rank 2k,

while each Levi–Civita tensor is of rank n. If both the upper and lower indicesconsist of the same set of distinct numbers, chosen from 1, . . . , n, the Kroneckerdelta is +1 or −1 according to whether the upper indices form an even or oddpermutation of the lower indices; in all other cases, it is zero. This rule is a directconsequence of the epsilons. For example, in the case of n = 3 and k = 2, we have

δ1212 = +1, δ1221 = −1, δ3223 = −1, δ2311 = δ1321 = 0, (2.96)

while with k = 3 we have

δ123123 = δ231123 = +1, δ213123 = δ321123 = −1, δ322123 = 0. (2.97)

In the case n = 4, we can write

δip = 1

3!ǫiqrsǫpqrs, δijpq = 1

2!ǫijrsǫpqrs, δijkpqr = 1

1!ǫijksǫpqrs, δijklpqrs = ǫijklǫpqrs. (2.98)

Symmetric and skew-symmetric parts of any tensor can be constructed by anappropriate linear combination of the tensor and its transposes. For example,to construct symmetric parts of the rank-2 tensor A and rank-3 tensor B, oneaverages the components of the tensor with all of its transposes:

a(ij) = 1

2!(aij + aji) , (2.99)

b(ijk) = 1

3!(bijk + bjki + bkij + bjik + bikj + bkji) . (2.100)

34 TENSOR ANALYSIS

The skew-symmetric parts are constructed similarly:

a[ij] = 1

2!(aij − aji) , (2.101)

b[ijk] = 1

3!(bijk + bjki + bkij − bjik − bikj − bkji) . (2.102)

Note that

b(ij)k = 1

2!(bijk + bjik) , (2.103)

and

b[ij]k = 1

2!(bijk − bjik) . (2.104)

It can be shown that in an n-dimensional space, a completely symmetric tensorof rank r can have at most [(n − 1) + r]!/(n − 1)!r! distinct components. Simi-larly, a completely skew-symmetric tensor can have at most n!/r!(n − r)! distinctcomponents, and if all particular indices are different, the corresponding compo-nents differ from each other only in sign, while if even two such indices are equal,the components vanish. Clearly for n = 3 and r = 2, the symmetric tensor has 6components while the skew-symmetric one has 3 components. Also note that (i)if n = 3 and r = 3, then the symmetric tensor has 10 components while the skew-symmetric one has 1 component; (ii) if n = 3 all components of skew-symmetrictensors of rank r ≥ 4 vanish; and (iii) if n = r, then the skew-symmetric tensor hasone component. Subsequently, we can write the above completely skew-symmetricthird rank tensor in three-dimensional space as

b[ijk] = c ǫijk, (2.105)

where c is a scalar.As remarked earlier, and as easily verified, any rank-2 tensor A can be decom-

posed into symmetric and skew-symmetric parts:

A = sym A + skw A or aij = a(ij) + a[ij]. (2.106)

Note that if D = sym A, W = skw A, and C is a general rank-2 tensor, thentr W = 0, D =DT , W = −WT , D ∶W = 0, and

D ∶C = DT∶C =D ∶ 1

2(C +CT ) , (2.107)

W ∶C = −WT∶C =W ∶

1

2(C −CT ) . (2.108)

The symmetric part of A can also be made traceless by subtracting 1

3(tr A)1 from

it, i.e.,

A = sym A −1

3(tr A)1 or a(ij) = a(ij) − 1

3akkδij , (2.109)

so that A can be uniquely decomposed as a linear superposition of three contri-butions

A = 1

3(tr A)1 + A + skw A or aij = 1

3akkδij + a(ij) + a[ij] (2.110)

2.10. SYMMETRIC AND SKEW-SYMMETRIC TENSORS 35

each one of which has unique symmetry properties: the first term is isotropic, thesecond is a traceless symmetric rank-2 tensor, and the third is a skew-symmetricrank-2 tensor. The first term is also called the spherical part of A, while thesum of the second and third terms is also called the deviatoric part of A and isdenoted as A′. It is clear that in such case A is decomposed into three orthogonalcomponents since

1 ∶ (skw A) = 0, 1 ∶ A = 0, A ∶ (skw A) = 0. (2.111)

This leads to the following decomposition of the inner product between two arbi-trary rank-2 tensors:

A ∶C = 1

3(tr A)(tr C) + A ∶ C + (skw A) ∶ (skw C). (2.112)

We note that any symmetric second rank tensor [Sij] = [Sij]T = [Sji] can berepresented by a quadric surface. Consider the equation

x ⋅S ⋅ x = 1 or xiSijxj = 1. (2.113)

Performing the summations and using the symmetry of the tensor, we obtain

S11x2

1 + S22x2

2 + S33x2

3 + 2 (S23x2x3 + S31x3x1 + S12x1x2) = 1. (2.114)

This is a general equation of a second degree surface, a quadric, referred to itscenter as origin. An important property of a symmetric tensor that we shallexplore in Sections 3.1.7 and 3.1.8 is the possession of principal axes. These arethree directions at right angles to each other such that, when the general quadric(2.113) is referred to them as axes, the equation takes the simpler form

S1x2

1 + S2x2

2 + S3x2

3 = 1. (2.115)

It is clear that the semi-axes of the representation quadric are of lengths S−1/21

,

S−1/22

, and S−1/23

. If S1, S2, and S3 are all positive, the surface is an ellipsoid. Iftwo coefficients are positive and one negative, it is a hyperboloid of one sheet. Ifone coefficient is positive and two are negative, it is a hyperboloid of two sheets.If all three coefficients are negative, the surface is an imaginary ellipsoid.

Now, it can also be easily shown that a general rank-3 tensor B with componentsbijk cannot be decomposed into its symmetric and skew-symmetric parts b(ijk) andb[ijk], respectively, since they together contain less information than bijk. However,it can be shown that B can be written as

B = S +A, (2.116)

where S is a completely symmetric tensor and A is a skew-symmetric tensor definedby the properties

aiii = 0, aiij + aiji + ajii = 0, ∣ǫijk ∣aijk = 0, (2.117)

where there is no summation over the underlined indices. The component of S

and A are given by

sijk = 1

6(bijk + bjki + bkij + bjik + bkji + bikj) , (2.118)

aijk = 1

6(5 bijk − bjki − bkij − bjik − bkji − bikj) . (2.119)

36 TENSOR ANALYSIS

It should be noted that sijkaijk = 0. Now it can also be shown that the skew-symmetrix tensor A can be represented as

aijk = cimǫmjk − ckmǫmij + djmǫmik − dkmǫmij + c ǫijk, (2.120)

where cij and dij are traceless rank-2 tensors, and c is a constant. The tensors cijand dij , in turn, may be expressed by the components of aijk as

cjn = 1

3(aijk + aikj) ǫink, (2.121)

djn = aijkǫikn +1

3aimkǫimkδjn, (2.122)

c = 1

6aijkǫijk. (2.123)

Alternatively, the third rank tensor B can be decomposed as

bijk = b(ij)k + b[ij]k = δijnk + b′(ij)k + b[ij]k, (2.124)

where

nk = 1

3bllk, and b′(ij)k = b(ij)k − δijnk. (2.125)

As is clear, the decomposition of a third rank tensor is not unique even in thiscase since one can choose any of the pairs (i, j), (j, k), or (i, k) in the indices ofbijk to perform the decomposition. However, if for physical reasons bijk happensto be symmetric or skew symmetric with respect to a specific pair of indices, thenthe above decomposition becomes very useful.

While a decomposition may not be unique, in general it is very advantageousto decompose a tensor into irreducible invariant subspaces. To make this clear, wenote that we could have written (2.106) as

A = 2∑s=1

Is ⋅A = I1 ⋅A + I2 ⋅A =A1 +A2, (2.126)

where the components of the symmetrizing and anti-symmetrizing linear operatorsI1 and I2 are given by

(I lmij )1 = 1

2!(δliδmj + δmi δlj) = δ(li δm)j and (I lmij )2 = 1

2!(δliδmj − δmi δlj) = δ[li δm]j = 1

2!δlmij ,

(2.127)and they lead to the symmetric and skew-symmetric decompositions A1 and A2

with components

(aij)1 = 1

2!(aij + aji) = a(ij) and (aij)2 = 1

2!(aij − aji) = a[ij]. (2.128)

Furthermore, it should be noted that I1 and I2 have the following properties:

Ip ⋅ Iq = Ip if p = q0 if p ≠ q and

n∑s=1

Is = I, (2.129)

2.11. DUAL OF A TENSOR 37

where I is the identity operator and n = 2 in this case. It turns out that adecomposition satisfying (2.129) can always be accomplished for tensors of anyrank. For tensor B of rank 3, there are four symmetry operators, n = 4, and theircomponents are given by

(I lmnijk )1 = δ(li δmj δn)k , (2.130)

(I lmnijk )2 = δ[li δmj δn]k = 1

3!δlmnijk , (2.131)

(I lmnijk )3 = 2

3![(δliδmj + δmi δlj) δnk − (δmi δnj + δni δmj ) δlk]

= 1

3![δ(li δm)j δnk − δ

(mi δ

n)j δlk] , (2.132)

(I lmnijk )4 = 2

3![(δliδnk + δni δlk) δmj − (δmi δnk + δni δmk ) δlj]

= 1

3![δ(li δn)k δmj − δ

(mi δ

n)kδlj] . (2.133)

Clearly, I1 and I2 are the symmetrizing and anti-symmetrizing operators, while I3and I4 are operators with mixed symmetry. These operators lead to the followingdecomposition of B:

B = 4∑s=1

Is ⋅B = 4∑s=1

Bs, (2.134)

where the components of Bs are given by

(bijk)1 = 1

3!(bijk + bjki + bkij + bjik + bkji + bikj) = b(ijk), (2.135)

(bijk)2 = 1

3!(bijk + bjki + bkij − bjik − bkji − bikj) = b[ijk], (2.136)

(bijk)3 = 2

3![(bijk + bjik) − (bjki + bkji)] = 1

3![b(ij)k − b(jk)i] , (2.137)

(bijk)4 = 2

3![(bijk + bkji) − (bjik + bkij)] = 1

3![b(i∣j∣k) − b(j∣i∣k)] . (2.138)

Note that in three-dimensional space B has 33 = 27 components, and in con-formance with this fact, B1, the symmetric part, has 10 components, B2, theanti-symmetric part, has 1 component, and B3 and B4, the parts with mixedsymmetry, each have 8 components.

2.11 Dual of a tensor

As noted previously, an n-vector is a completely skew-symmetric tensor of rankn. Within the space n, it is possible to associate with any k-vector, 0 ≤ k ≤ n, an(n − k)-vector, its dual, as follows. If Vj1⋯jk and W i1⋯ik are k-vectors of weightN , their duals, also called Hodge duals, are the (n − k)-vectors

vi1⋯in−k ≡ (dualV)i1⋯in−k = 1

k!εi1⋯in−kj1⋯jkVj1⋯jk (2.139)

38 TENSOR ANALYSIS

or

wj1⋯jn−k ≡ (dualW)j1⋯jn−k = 1

k!W i1⋯ikεi1⋯ikj1⋯jn−k (2.140)

of weightsN+1 andN−1, respectively. Note that the epsilons have n indices, in thecontracted products the contravariant tensor is written first, and the summationindices are adjacent. It can be easily shown that these definitions imply that

dual (dual W) = s (−1)k(n−k)W, (2.141)

so that w and W contain exactly the same information (aside from the sign). Thesignature s is the sign of the determinant of the inner product tensor. For ordinaryEuclidean spaces, the signature is always positive, and so s = +1.

Note that in E3 if w is a 1-vector, W is a 2-vector, and ε is the rank-3 absoluteLevi–Civita tensor, then

wi = 1

2εijkWjk or wk = 1

2W ijεijk, (2.142)

and

Wjk = wiεijk or W ij = εijkwk. (2.143)

As shorthand notation, from now on we will also use the angle bracket notationto denote the dual of a tensor field, i.e.,

w = ⟨W⟩ . (2.144)

The components wi and Wjk are related as follows:

W = [Wjk] =⎡⎢⎢⎢⎢⎢⎣

0 w3−w2

−w3 0 w1

w2−w1 0

⎤⎥⎥⎥⎥⎥⎦. (2.145)

It is easy to show that if W1 and W2 are two skew-symmetric tensors with corre-sponding axial vectors w1 and w2, then

W1 ∶W2 = 2w1 ⋅w2, (2.146)

and subsequently ∣W∣ =√2 ∣w∣. In addition, if a is an arbitrary vector, we have

W ⋅w = 0 and W ⋅ a = −w × a. (2.147)

Remark: To avoid the minus sign in the last expression, some authors use thedirect relation often used in rigid body dynamics,

W ⋅ a =w × a, (2.148)

and subsequently the axial vector w is chosen to be the negative of that given in(2.142)–(2.145). In this case, one obtains

wi = −12εijkWjk or wk = −1

2W ijεijk, (2.149)

2.12. EXTERIOR PRODUCT 39

a

aa

bbb

Figure 2.4: Vectors and bivectors.

and

Wjk = −wiεijk or W ij = −εijkwk, (2.150)

where now the components wi and Wjk are related as follows:

W = [Wjk] =⎡⎢⎢⎢⎢⎢⎣

0 −w3 w2

w3 0 −w1

−w2 w1 0

⎤⎥⎥⎥⎥⎥⎦. (2.151)

One should be aware of such sign differences when comparing references.

2.12 Exterior product

The exterior or wedge product provides a generalization of the standard vectorproduct, which is restricted to three-dimensional vector spaces. For any two scalarsa and b and any three multivectors A, B, and C, all familiar rules of addition andmultiplication hold, such as

(aA + bB) ∧C = aA ∧C + bB ∧C, (2.152)(A ∧B) ∧C =A ∧ (B ∧C) =A ∧B ∧C, (2.153)

except for a modified commutation law between a p-vector A and a q-vector B:

A ∧B = (−1)pq B ∧A. (2.154)

Application to 1-vectors a and b results in a 2-vector, or bivector,

a ∧ b = −b ∧ a and a ∧ a = 0, (2.155)

which is clearly bilinear and anti-symmetric. A bivector represents an orientedtwo-dimensional area as indicated in Fig. 2.4. The order of the product matters,as indicated in the figure. In addition,

a ∧ b = (ajej) ∧ (bkek) = aj bk ej ∧ ek = 1

2(aj bk − bj ak) ej ∧ ek. (2.156)

If a and b are vectors in E3, then we have

a∧b = (a2 b3 − b2 a3) e2∧e3+(a3 b1 − b3 a1) e3∧e1+(a1 b2 − b1 a2) e1∧e2 (2.157)

40 TENSOR ANALYSIS

a

a

b

b

cc

Figure 2.5: Trivectors.

and we note that the components of the bivector are the same as those of thecross product a × b. Similarly, the product of three 1-vectors a, b, and c in E3,illustrated in Fig. 2.5, yields the 3-vector, or trivector,

a ∧ b ∧ c = (a1b2c3 + a2b3c1 + a3b1c2 − a1b3c2 − a2b1c3 − a3b2c1)e1 ∧ e2 ∧ e3,(2.158)

noting that the magnitude of the trivector is just det (abc) and is equal to thescalar triple product a ⋅ (b × c).

In En, the number of basis k-vector elements is

( nk) (2.159)

and the total number of elements is 2n. Thus, in E3 the basis elements are

1,e1,e2,e3,e12,e23,e31,e123 , (2.160)

where we have used the shorthand notation

eij = ei ∧ ej and eijk = ei ∧ ej ∧ ek. (2.161)

Note that in E3 no higher k-vector exists than e123. The highest vector in thespace is usually denoted as I and is a pseudoscalar. Thus in E3 we have thatI = e123 and its inverse is given by I−1 = e321.

The wedge product is defined as

ei ∧ ej = eij = 0 for i = j,ei ej for i ≠ j, (2.162)

so that ei ej = −ej ei if i ≠ j, with the defining equation of the algebra given by

ei ei = 1. (2.163)

This implies, e.g., that

(ei ej ek)ek = (ei ej) (ek ek) = (ei ej)1 = ei ej, (2.164)

2.13. TENSOR FIELDS 41

and (ei ej ek)ej = (ei ej) (ek ej) = −ei (ej ej)ek = −ei ek. (2.165)

From (2.156), this also implies that

a ∧ b = 1

2(ab − ba) . (2.166)

Now the dual of a multivector A is given by

(dualA) =A I−1. (2.167)

In E3, it now follows from (2.157) and (2.158) that

[dual (a ∧ b)] = (a ∧ b) I−1 = a × b (2.168)

and [dual (a ∧ b ∧ c)] = (a ∧ b ∧ c) I−1 = a ⋅ (b × c) , (2.169)

since

(e2 e3) (e3 e2 e1) = e1,(e3 e1) (e3 e2 e1) = e2,(e1 e2) (e3 e2 e1) = e3,(e1 e2 e3) (e3 e2 e1) = 1.In summary, and in full accordance with (2.142) and (2.143), the exterior prod-

uct of two 1-vectors u and v in E3 is a 2-vector or skew-symmetric tensor W withassociated dual (also called axial vector) w corresponding to the vector (or crossproduct) of the two vectors u × v, i.e., if we take

W = u ∧ v, (2.170)

then it easily follows from (2.168) that

w = u × v. (2.171)

2.13 Tensor fields

2.13.1 Cartesian coordinate system

Let (ξ1, ξ2, ξ3) be coordinates in the Euclidean space E3, and r the position vector

r = ξ1i1 + ξ2i2 + ξ3i3 = ξkik, (2.172)

where (i1, i2, i3) = (i1, i2, i3) is the Cartesian (or rectangular) coordinate unit basis,as shown in Fig. 2.6. Quite often, the Cartesian coordinate system is used as areference coordinate system for general curvilinear coordinate systems. In thissystem, the basis is taken to be constant, i.e., it is independent of coordinates.The Cartesian coordinates for a vector u are given by

uj = u ⋅ ij = u ⋅ ij = u lj, (2.173)

where u = ∣∣u∣∣ and lj = cosαj ’s are the direction cosines. Note that if we have aunit vector (u = 1), then lj is just the component of the unit vector along the ξj

direction and lj lj = 1, which is just a standard trigonometric identity.

42 TENSOR ANALYSIS

i1

i2

i3 ξ1

ξ2

ξ3

u

i1

i2

i3

u1

u2

u3

α1

α2

α3

Figure 2.6: Cartesian coordinates and basis.

2.13.2 Curve in space

Let t be some arbitrary parameter and take r = r(t) or ξi = ξi(t). The vectorfunction r = r(t) represents a curve in Cartesian coordinates as illustrated inFig. 2.7. Define the derivative

dr

dt≡ lim

∆t→0

r(t +∆t) − r(t)∆t

= lim∆t→0

∆r

∆t. (2.174)

Now we see that dr/dt is tangent to the curve r(t) at t. We could also define thecurve in terms of the arc length s of the curve r = r(s). Then dr/ds is the unitvector tangent to the curve since

ds2 = dr ⋅ dr = drds⋅dr

dsds2 (= dr

dt⋅dr

dtdt2) , (2.175)

so

1 = drds⋅dr

ds(and alternatively (ds

dt)2 = dr

dt⋅dr

dt) . (2.176)

It also follows that dr = dξkik since the Cartesian basis ik is constant.

2.13.3 Derivatives

Below we briefly mention generalizations of the standard derivative of a real-valuedfunction of a single real variable and the directional derivative of such a functionto corresponding derivatives of a tensor-valued function of a real tensor variable.To illustrate this, let f(v) be a scalar function of a vector v. The derivative of f

2.13. TENSOR FIELDS 43

i1

i2

i3ξ1

ξ2

ξ3

∆rr(t)

r(t +∆t)

Figure 2.7: Curve in Cartesian coordinate space.

with respect to v is a vector denoted by ∂f/∂v and defined by its scalar productwith an arbitrary vector a:

∂f

∂v⋅ a ≡ lim

s→0

f(v + sa) − f(v)s

. (2.177)

In a Cartesian coordinate system, this expression takes the form

∂f

∂v⋅ a = ∂f

∂viai.

For the particular case of a = ej , we have

∂f

∂v⋅ ej = ∂f

∂vj

and we conclude that∂f

∂v= ∂f∂vi

ei. (2.178)

In a similar manner, if f(D) is a scalar function of a tensor D of rank 2, thederivative of f with respect to D is a tensor of rank 2 denoted by ∂f/∂D anddefined by

tr ( ∂f∂D⋅AT) ≡ lim

s→0

f(D + sA) − f(D)s

, (2.179)

where A is an arbitrary tensor of rank 2. We conclude that

∂f

∂D= ∂f

∂Dij

eiej. (2.180)

Such generalizations are essential in defining different quantities that we willencounter later on (e.g., the deformation gradient). Nevertheless, we will not delve

44 TENSOR ANALYSIS

into mathematical details associated with limits in different functional spaces norprove the theorem below.

Specifically, the Fréchet derivative is commonly used to generalize the standardderivative to the case of differentiating tensor-valued functions of tensor variables.

Definition: A tensor function A(X) is Fréchet differentiable at X0 if there isa linear operator L such that in a neighborhood V of X0

∣∣A(X) −A(X0) −L ⋅ (X −X0)∣∣ = o (∣∣X −X0∣∣) . (2.181)

In this case, we write L =DXA(X0), and DXA(X0) is called the Fréchet deriva-tive of A(X) at X0.

That is, if A is a tensor function of rank n that depends on the scalar x, vectorx, and rank-2 tensor X, then the corresponding Fréchet derivatives or ranks n,n + 1, and n + 2 are given by

DxA(x,x,X) = ∂A(x,x,X)∂x

, (2.182)

DxA(x,x,X) = ∂A(x,x,X)∂x

, (2.183)

DXA(x,x,X) = ∂A(x,x,X)∂X

. (2.184)

The derivative of a quantity represented by a tensor of rank p with respect to atensor of rank q is defined as a tensor of rank n = p+q whose components are equalto the derivatives of the quantity with respect to the corresponding components ofthe tensor of rank q. For example, the derivative of a scalar quantity with respectto a vector is given by the vector whose components are equal to the derivativesof the scalar with respect to the corresponding components of the vector.

Analogously, the Gateaux derivative provides a generalization of the classicaldirectional derivative.

Definition: A tensor function A(X) is Gateaux differentiable at X0 if there isan operator DXA(X0,H) such that

lims→0∣∣A(X0 + sH) −A(X0) − sDXA(X0,H)∣∣ = 0 (2.185)

for (X0 + sH) ∈ V, a neighborhood of X0. Furthermore, DXA(X0,H) is calledthe Gateaux derivative of A(X) at X0, and we write

DXA(X0,H) = d

dsA(X0 + sH)∣

s=0. (2.186)

Theorem: If A(X) is Fréchet differentiable at X0, it is Gateaux differentiableat X0. Conversely, if the Gateaux derivative of A(X) at X0, DXA(X0,H), islinear in H, i.e., DXA(X0, ⋅) ∈ L and is continuous in X, then A(X) is Fréchetdifferentiable at X0. In either case, we have the formula

DXA(X0) ⋅H =DXA(X0,H). (2.187)

2.13. TENSOR FIELDS 45

That is,

DxA(x,x,X) u = dA(x + su,x,X)ds

∣s=0

, (2.188)

DxA(x,x,X) ⋅ u = dA(x,x + su,X)ds

∣s=0

, (2.189)

DXA(x,x,X) ∶U = dA(x,x,X + sU)ds

∣s=0

, (2.190)

where u, u, and U are arbitrary scalar, vector, and rank-2 tensors.

2.13.4 Surface in space

A surface in space is defined by

φ(ξ1, ξ2, ξ3) = φ(ξk) = const., (2.191)

or alternatively

f(ξ1, ξ2, ξ3) = f(ξk) = 0. (2.192)

On this surface

df = 0 = ∂f

∂ξ1dξ1 +

∂f

∂ξ2dξ2 +

∂f

∂ξ3dξ3 = ∂f

∂ξidξi = ∂f

∂ξiii ⋅ dξkik = ∇f ⋅ dr, (2.193)

where we define the gradient operator in Cartesian coordinates as

∇ ≡ ii ∂∂ξi

. (2.194)

We note that a superscript index in a denominator should be understood as asubscript for the term; i.e., Fi = ∂f/∂ξi. Now we readily see that dr is a vectorthat is tangent to the surface while ∇f is a vector normal to the surface.

2.13.5 Curvilinear coordinate system

A general curvilinear coordinate system is given by

xi = xi(ξ1, ξ2, ξ3) = xi(ξk) (2.195)

and

ξk = ξk(x1, x2, x3) = ξk(xi) (2.196)

with the transformation being non-singular, i.e.,

det [∂xi∂ξk] ≠ 0,±∞, (2.197)

where i and k are the row and column indices, respectively. Now xi = const.

is the xi coordinate surface (whose normal points in the xi direction), and theintersection of x1 = const. and x2 = const. is the x3 coordinate curve.

46 TENSOR ANALYSIS

Example

For cylindrical polar coordinates:

x1 = [(ξ1)2 + (ξ2)2]1/2, ξ1 = x1 cosx2,

x2 = tan−1 (ξ2ξ1), ξ2 = x1 sinx2,

x3 = ξ3, ξ3 = x3.At each point in space, we can define two sets of “natural” base vectors:

(i) Let

ei = ∂r

∂xi= ∂

∂xi(ξkik) = ∂ξk

∂xiik (2.198)

which is tangent to the xi coordinate curve and not necessarily a unit vector.Note that for the Cartesian basis, ik = ik and ik is not a function of thecoordinates.

Example

For cylindrical polar coordinates:

e1 = cosx2i1 + sinx2i2,

e2 = −x1 sinx2i1 + x1 cosx2i2,

e3 = i3.

Note that ei is not necessarily a unit vector; e.g., by inspection e2 isnot.

(ii) Let

ei = ∇xi = ∂xi∂ξk

ik (2.199)

which is normal to the xi coordinate surface, i.e., it is normal to xi(ξk) =const.

Example

For cylindrical polar coordinates:

e1 = ξ1[(ξ1)2 + (ξ2)2]1/2 i1 + ξ2[(ξ1)2 + (ξ2)2]1/2 i2 = cosx2i1 + sinx2i2.Now since tanx2 = ξ2/ξ1, differentiating with respect to ξ1 and ξ2 inturn we have

sec2 x2dx2 = − ξ2(ξ1)2 dξ1 = 1

ξ1dξ2,

2.13. TENSOR FIELDS 47

so

dx2

dξ1= −ξ2 cos2 x2(ξ1)2 = −x1 sinx2 cos2 x2(x1)2 cos2 x2 = − sinx2

x1

and

dx2

dξ2= cos2 x2

ξ1= cos2 x2

x1 cosx2= cosx2

x1.

Thus

e2 = −sinx2

x1i1 +

cosx2

x1i2,

e3 = i3.

Are such bases reciprocal? The answer is yes since

ei ⋅ ej = (∂ξk

∂xiik) ⋅ (∂xj

∂ξlil) = ∂ξk

∂xi∂xj

∂ξk= ∂xj∂xi= δji

and the components of the metric tensor are given by

gij = ei ⋅ ej = ∂ξk∂xi

∂ξk

∂xj= gij(xl) (2.200)

and

gij = ei ⋅ ej = ∂xi∂ξk

∂xj

∂ξk= gij(xl). (2.201)

The reason why gij is called a metric tensor is now evident since

ds2 = dξkik ⋅ dξlil = ∂ξk∂xi

dxiik ⋅∂ξl

∂xjdxj il = ei ⋅ ejdxidxj = gijdxidxj .

What are the components aji in the basis coordinate transformation

ei = ajiej?Suppose we have another curvilinear coordinate system such that

xi = xi(ξk) and ξk = ξk(xi).Then

ei = ∂r

∂xi= ∂ξk∂xi

ik = ∂xj∂xi

∂ξk

∂xjik = ∂xj

∂xiej,

so that at a point in space we have

aji = ∂x

j

∂xi. (2.202)

48 TENSOR ANALYSIS

This result also allows us to bypass the rectangular coordinate system when trans-forming between two curvilinear systems, i.e., instead of having

xi = xi(ξk) and ξk = ξk(xj),we can go directly to

xi = xi(xj).Example

The mapping from cylindrical polar coordinates

x1 = [(ξ1)2 + (ξ2)2]1/2 ,x2 = tan−1 (ξ2

ξ1) ,

x3 = ξ3,

to spherical coordinates is given by

x1 = [(x1)2 + (x3)2]1/2 ,x2 = x2,

x3 = tan−1x1

x3.

The inverse mapping is

x1 = x1 sin x3,

x2 = x2,

x3 = x1 cos x3.

To transform basis vectors in the two coordinate systems, the covariant(contravariant) coefficients can be calculated using the above transforma-tions.

Now

g = detG = det [gij] = det [∂ξk∂xi]det [∂ξk

∂xj] = (det [ ∂ξi

∂xj])2 , (2.203)

or

√g = det [ ∂ξi

∂xj] . (2.204)

Similarly

√g = det [ ∂ξi

∂xj] . (2.205)

2.14. GRADIENT OF A SCALAR FIELD 49

Subsequently, we have that

det [ ∂xi∂xj] = det [ ∂xi

∂ξk]det [∂ξk

∂xj] =√ g

g= det [aij] = detA = a. (2.206)

Note that subscripts inside determinant quantities are understood not to be freeindices. From before, we recall that gij transform like covariant components of asecond-order tensor; thus

gij = ∂xr∂xi

∂xs

∂xjgrs. (2.207)

2.14 Gradient of a scalar field

We now take

r = xiei and dr = dxiei, (2.208)

and define the scalar field

φ = φ(xk) = φ(xk) (2.209)

which is a tensor of order zero (a scalar), or a relative tensor of order zero andweight zero.

The gradient of a scalar field φ is defined by

∇φ ≡ ∂φ

∂xjej , (2.210)

where now

∇ ≡ ej ∂

∂xj. (2.211)

Note that the Cartesian definition is recovered when we have an identity coordinatetransformation, in which case (x1, x2, x3) = (ξ1, ξ2, ξ3) and (e1,e2,e3) = (i1, i2, i3).Does this definition give a tensor (in this case a vector)? In other words, does∂φ/∂xj transform like a covariant component of a tensor of order one? Recall thatvector covariant components transform as

vi = ajivj .Now if we take

vj → ∂φ

∂xj, vi → ∂φ

∂xi, a

ji = ∂x

j

∂xi,

do we get an equality upon substitution:

∂φ

∂xi= ∂xj∂xi

∂φ

∂xj?

The answer is yes by the chain rule, and so the gradient of a scalar defines a propertensor (a vector).

50 TENSOR ANALYSIS

2.15 Gradient of a vector field

Assume that we have the following vector field

v = vi(xk)ei. (2.212)

Consider the following definition for the gradient of this vector field (a dyad)

∇v = ∂vi∂xj

eiej . (2.213)

Is this a tensor, i.e., do the components transform like (11) tensor components?

Now given the dyadic tensor

τ = τ ijeiej = τ lmelem

recall that (11) components transform as

τ lmail = τ ijajm.

Now take

τ ij → ∂vi

∂xj, τ lm → ∂vl

∂xm, aij = ∂x

i

∂xj.

Is the following equality true:

∂vl

∂xm∂xi

∂xl= ∂vi∂xj

∂xj

∂xm?

Since v is a vector, from before we have that

vi = vlail = vl ∂xi

∂xl,

so that

∂vi

∂xj= ∂

∂xj(vl ∂xi

∂xl) = ∂

∂xm(vl ∂xi

∂xl) ∂xm∂xj

,

or

∂vi

∂xj∂xj

∂xm= ∂

∂xm(vl ∂xi

∂xl) = ∂vl

∂xm∂xi

∂xl+ vl

∂2xi

∂xm∂xl.

So we see that the equality is not true due to the presence of the second term inthe above equation. Thus (∂vi/∂xj)eiej is not a tensor since it does not remaininvariant under coordinate transformations.

Now let us try to define the gradient of the vector field by

∇v ≡ ∂v

∂xjej = ∂

∂xj(viei)ej = ∂vi

∂xjeie

j+ vi

∂ei

∂xjej . (2.214)

Now we have just shown that the first term on the right-hand side is not a tensor.It can also be shown that the second term on the right-hand side is not a tensor

2.15. GRADIENT OF A VECTOR FIELD 51

either. However, the sum of the two terms is indeed a tensor! We note that∂ei/∂xj is a vector for fixed j, and thus can be written as a linear combination ofek, i.e.,

∂ei

∂xj= Γk

ijek = kij

ek, (2.215)

where Γkij are not components of a tensor. It is called the Christoffel symbol of the

second kind. If we take the dot product of the above equation with er, we obtain

∂ei

∂xj⋅ er = Γk

ijek ⋅ er = Γk

ijδrk = Γr

ij . (2.216)

However, we recall that ei = ∂r/∂xi so that ∂ei/∂xj = ∂2r/∂xj∂xi, and

∂ei

∂xj⋅ er = ∂2r

∂xj∂xi⋅∇xr = ∂2ξk

∂xj∂xiik ⋅

∂xr

∂ξlil = ∂2ξl

∂xj∂xi∂xr

∂ξl.

Thus

Γkij = ∂2ξl

∂xj∂xi∂xk

∂ξl. (2.217)

Note that the Christoffel symbol of the second kind is symmetric with respect tothe exchange of lower indices i and j. It can be shown that it can be expressedfully in terms of components of the metric tensor g, i.e.,

Γkij = 1

2gkl (∂gil

∂xj+∂gjl

∂xi−∂gij

∂xl) . (2.218)

Furthermore, it can also be rewritten as

Γkij = gklΓijl = gkl [ij, l] , (2.219)

where Γijl is the Christoffel symbol of the first kind and are not components of atensor. Because of the symmetry of the Christoffel symbol of the second kind, wenote that the Christoffel symbol of the first kind is also symmetric with respect tothe exchange of the indices i and j, and it is given by

Γijk = 1

2(∂gik∂xj

+∂gjk

∂xi−∂gij

∂xk) . (2.220)

Returning to our definition of the gradient of a vector field, we now have

∇v = ∂vi∂xj

eiej+ viΓk

ijekej = ( ∂vi

∂xj+ Γi

kjvk)eiej

or

∇v = vi,jeiej , (2.221)

where

vi,j ≡ ∂vi

∂xj+ Γi

kjvk (2.222)

52 TENSOR ANALYSIS

is called the covariant derivative. In general, all subscripts following a commadenote covariant derivatives with respect to the corresponding components.

We note that the definition of the gradient of a vector as

∇v = ∂v

∂xjej

is consistent with that of the gradient of a scalar when we take v → φ. Indeed, thisdefinition remains true for a tensor T of arbitrary order, i.e., we can write moregenerally

∇T = ∂T∂xj

ej.

2.16 Covariant differentiation of a vector

If instead we write our vector field in terms of covariant components

v = viei,we then have

∇v ≡ ∂

∂xj(viei)ej = ∂vi

∂xjeiej + vi

∂ei

∂xjej . (2.223)

Now since

ei ⋅ ej = δji ,

upon differentiating we have

∂ei

∂xk⋅ ej + ei ⋅

∂ej

∂xk= 0,

or

ei ⋅∂ej

∂xk= −Γr

iker ⋅ ej = −Γr

ikδjr = −Γj

ik,

and

∂ej

∂xk= −Γj

lkel.

Now

∇v = ∂vi∂xj

eiej − viΓikje

kej = ( ∂vi∂xj− vkΓ

kij)eiej ,

or

∇v = vi,jeiej, (2.224)

where

vi,j ≡ ∂vi∂xj− Γk

ijvk. (2.225)

2.16. COVARIANT DIFFERENTIATION OF A VECTOR 53

Note that

vi,j − vj,i = ∂vi∂xj−∂vj

∂xi. (2.226)

Furthermore,

∇v ⋅ dr = ( ∂v∂xk

ek) ⋅ (dxiei) = ∂v∂xi

dxi = dv. (2.227)

These results generalize to any m = r + n order tensor with (rn) components.

Example

Assume we have the third-order (triadic) tensor

T = T ijkeie

jek. (2.228)

Then it follows that

∇T = T ijk,leie

jekel, (2.229)

where

T ijk,l = ∂T

ijk

∂xl+ Γi

rlTrjk − Γ

rjlT

irk − Γ

rklT

ijr. (2.230)

The rules for covariant differentiation are mostly the same as for ordinary differ-entiation. That is, (i) the covariant derivative of a sum of two tensors is the sumof the covariant derivatives of the tensors; (ii) the covariant derivative of a productof two tensors is the covariant derivative of the first tensor times the second plusthe first tensor times the covariant derivative of the second; (iii) higher covariantderivatives are defined as covariant derivatives of covariant derivatives – however,one should be careful in calculating these higher order derivatives since in general

vi,jk ≠ vi,kj . (2.231)

To see this, we calculate the components of the second covariant derivative of v,i.e., vi,jk. As we have shown, the components of the covariant derivative of v aregiven by

vi,j = ∂vi∂xj− Γl

ijvl. (2.232)

By definition, the components of the second covariant derivative are those of thecovariant derivative of the covariant derivative:

vi,jk = (vi,j),k = ∂

∂xk[ ∂

∂xj− Γl

ijvl] − Γmikvm,j − Γ

mjkvi,m. (2.233)

Expanding the expression, we have

vi,jk = ∂2vi

∂xj∂xk−Γl

ij

∂vl

∂xk−∂Γl

ij

∂xkvl−[∂vm

∂xj− Γl

mjvl]Γmik−[ ∂vi

∂xm− Γl

imvl]Γmjk. (2.234)

54 TENSOR ANALYSIS

Rearranging terms, the components of the second covariant derivative of the vectorv can be expressed in the form

vi,jk = ∂2vi

∂xj∂xk− Γl

ij

∂vl

∂xk− Γm

ik

∂vm

∂xj− Γm

jk

∂vi

∂xm−

⎡⎢⎢⎢⎣∂Γl

ij

∂xk− Γl

mjΓmik − Γ

limΓm

jk

⎤⎥⎥⎥⎦ vl.(2.235)

Now it is easy to see that

vi,jk − vi,kj = Rlijkvl, (2.236)

where

Rlijk ≡ ∂Γ

lik

∂xj−∂Γl

ij

∂xk+ Γl

mjΓmik − Γ

lmkΓ

mij (2.237)

is called the Riemann–Christoffel tensor. The covariant form of this tensor is

Rmijk = glmRlijk. (2.238)

It is an easy exercise to show that the covariant form can be expressed as

Rmijk = ∂Γikm

∂xj−∂Γijm

∂xk+ ΓmknΓ

nij − ΓmjnΓ

nik (2.239)

or

Rmijk = 1

2( ∂gmk

∂xi∂xj−

∂gik

∂xm∂xj−

∂gmj

∂xi∂xk+

∂gij

∂xm∂xk) + gpq (ΓmkpΓijq − ΓmjpΓikq) ,

(2.240)from which we see that the Riemann–Christoffel tensor is skew-symmetric in thefirst two indices and the last two indices, and symmetric in the interchange of thefirst two and last two indices. Consequently,

Rimjk = −Rmijk, Rmikj = −Rmijk, Rjkmi = Rmijk. (2.241)

Now, using these symmetry conditions, it is easy to show that we can write

Rmijk = εpmiεqjkSpq and Spq = 1

4εpmiεqjkRmijk, (2.242)

where S is a symmetric rank 2 tensor which clearly has only six components.However, these six components are not all independent as the Riemann–Christoffeltensor satisfies Bianchi’s identities:

Rmijk,l +R

mikl,j +R

milj,k = 0 or Rmijk,l +Rmikl,j +Rmilj,k = 0. (2.243)

Bianchi’s identities provide three additional equations that restrict the six compo-nents of S.

Note that if the metric tensor is constant throughout the space, then Rmijk = 0,and we have a flat or Euclidean space so that vi,jk = vi,kj . It can also be shownthat if Rmijk = 0, then we can find a coordinate system in which the componentsof the metric tensor are constant throughout space. Thus, Rmijk = 0 is a necessaryand sufficient condition for a (flat) Euclidean space. If Rmijk ≠ 0, then the spaceis curved and is called a Riemannian space.

2.17. DIVERGENCE OF A VECTOR FIELD 55

2.17 Divergence of a vector field

Consider the vector field

v = vieiso that

∇v = vi,jeiej .Now contraction of ∇v gives ∇ ⋅ v, i.e., the divergence of v. Thus we define

∇ ⋅ v ≡ vi,jei ⋅ ej = vi,jδji = vi,i, (2.244)

where now

vi,i = ∂vi

∂xi+ Γi

kivk. (2.245)

We recall that

εijk =√gǫijk.Now noting that

∂ǫijk

∂xp= 0,

we can easily show that εijk is also an isotropic tensor, so that we have

εijk,p = 0, (2.246)

or

ǫijk∂√g

∂xp−√g Γl

ipǫljk −√g Γl

jpǫilk −√g Γl

kpǫijl = 0.If we take i = 1, j = 2, and k = 3, then

1√g

∂√g

∂xp= Γ1

1p + Γ2

2p + Γ3

3p = Γkkp = Γk

pk.

Subsequently we can rewrite

∇ ⋅ v = vi,i = ∂vi

∂xi+

1√g

∂√g

∂xivi = 1√

g

∂

∂xi(√gvi) . (2.247)

From before we obtained the vector field

∇φ = ∂φ∂xi

ei = ∂φ

∂xigijej .

Now the divergence of this vector field is given by

∇ ⋅∇φ = ∇2φ = ( ∂φ∂xi

gij),j

,

56 TENSOR ANALYSIS

where the operator ∇2 is called the Laplacian. It can be shown that

gij,l = 0 (2.248)

(called Ricci’s theorem), i.e., the metric tensor behaves like a constant under co-variant differentiation. Subsequently, we can write

∇2φ = gij ( ∂φ

∂xi),j

= gij ( ∂2φ

∂xj∂xi− Γk

ij

∂φ

∂xk) . (2.249)

Using the previous result of the divergence of a vector field, with

vi = gij ∂φ∂xj

, (2.250)

we can rewrite this last result as

vi,i = ∇2φ = 1√g

∂

∂xi(√ggij ∂φ

∂xj) . (2.251)

2.18 Curl of a vector field

The curl of a vector field is given by

w = curl v = ∇ × v = (ei ∂

∂xi) × (vjej)

= ei ×∂

∂xi(vjej)

= ei × vj,iej

= vj,iei× ej

= vj,iǫijk√gek

= wkek,

where we can write

wk = εkijvj,i. (2.252)

Similarly, in terms of covariant components, we find that

w = curl v = wlel,

where

wl = εkjlgkivj,i. (2.253)

2.18. CURL OF A VECTOR FIELD 57

Example

The curlv can be computed by an alternate method that is sometimesconvenient. This is done through a double contraction of the absoluteLevi–Civita tensor:

w = curlv = ε ∶ (∇v) = (εijkeiejek) ∶ (vm,lelem)

= εijkvm,lδljδ

mk ei

= εijkvk,jei,

recovering the result above. The analogous result in terms of covariantcomponents can be obtained in a similar fashion.

Vectors (like moment of a force, angular velocity, vorticity, etc.) whose directionis established by convention, and which therefore change direction when the “hand-edness” of the coordinate system is changed (from right-handed to left-handed, say)are called axial vectors (or pseudo-vectors). Note that to discuss such vectors weresort to “right-hand-screw direction” or “left-hand-screw direction.” Vectors (likeforce, velocity, etc.) whose direction depends only on their physical meaning, andwhich therefore do not change direction when the “handedness” of the coordinatesystem is changed, are called polar vectors. Such vectors can be represented with-out ambiguity by an arrow pointing in a certain direction. Thus, a polar vector issymbolized by a line with an arrow indicating the direction, while an axial vectorby a line with a sense of rotation around the line with the line forming its axis (orcenter of rotation). To determine the nature of a vector, imagine it reflected in amirror perpendicular to itself as illustrated in Fig. 2.8. If the reflection preservesthe direction of the quantity describing a physical phenomenon, then the vector isaxial. The vector product of two polar vectors is not a true polar vector but ananti-symmetrical tensor. It happens, as a coincidence, that in three dimensions ananti-symmetrical tensor of the second rank has the same number of independentcomponents as a vector, but this is not true in any other number of dimensions.For instance, in four dimensions an anti-symmetrical tensor of the second rankhas 4× 4 components with only 6 being independent; a vector, on the other hand,has only four components in this case. Lastly, we note that the components ofa skew-symmetric second rank tensor transform like the components of an axialvector, whose components change sign when the “handedness” of the coordinatesystem is changed. In a similar fashion, any triple scalar product [u,v,w] is anaxial- or pseudo-scalar since basis reversal in any component formulation resultsin a change of sign. A polar or genuine scalar has a value completely independentof any basis. For more on this, see Section 2.11.

Example

If ε is the absolute Levi–Civita tensor and W is a skew-symmetric second-order tensor, then its dual is given by the axial vector

= 1

2ε ∶W. (2.254)

58 TENSOR ANALYSIS

e3e3e3

e2e2e2e1e1e1

d c = a × bc = a × baa bb

InversionInversion

Polar vector Axial vectorAxial vector

d = −d c = cc = c(a = −a,b = −b)(a = −a,b = −b)

e3e3e3

e2e2e2

e1e1e1d

c = a × bc = a × b aa

b b

Figure 2.8: Polar vectors a, b, and d, and axial vector c.

The vorticity vector ω will be shown later to be related to the angular veloc-ity , the axial vector corresponding to the second-order skew-symmetricspin tensor W, i.e., ω = curl v = −2, where v is the linear velocity.

2.19 Orthogonal curvilinear coordinate system

In the special case of an orthogonal curvilinear coordinate system, the componentsof the metric tensor are such that

gij = ei ⋅ ej = 0 if i ≠ j,≠ 0 if i = j, (2.255)

or in matrix notation

G = [gij] =⎡⎢⎢⎢⎢⎢⎣g11 0 0

0 g22 0

0 0 g33

⎤⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎣h21 0 0

0 h22 0

0 0 h23

⎤⎥⎥⎥⎥⎥⎦, (2.256)

where

hi ≡√gii, (2.257)

and the underline under an index means that that index does not sum. Similarly

gij = ei ⋅ ej = 0 if i ≠ j,≠ 0 if i = j. (2.258)

2.19. ORTHOGONAL CURVILINEAR COORDINATE SYSTEM 59

From earlier results it follows that

gii = 1

gii, (2.259)

and

ds2 = h21 (dx1)2 + h22 (dx2)2 + h23 (dx3)2 = h2i (dxi)2 . (2.260)

Example

In cylindrical polar coordinates

(h1, h2, h3) = (1, x1,1) , (2.261)

and in spherical coordinates

(h1, h2, h3) = (1, x1, x1 sinx2) . (2.262)

For orthogonal coordinates, it can be readily seen that Γijk = Γi

kj = 0 if i ≠ j ≠ k,e.g., for i = 1, j = 2, and k = 3, we have

Γ1

23 = 1

2g11 (∂g31

∂x2+∂g21

∂x3−∂g23

∂x1) + 0 + 0 = 0. (2.263)

We also have

Γijj = −hj

h2i

∂hj

∂xi(i ≠ j), (2.264)

and

Γi

ji = Γi

ij = 1

hi

∂hi

∂xj. (2.265)

Example

In cylindrical polar coordinates the only nonzero components are

Γ1

22 = −h2

h21

∂h2

∂x1= −x1, (2.266)

Γ2

12 = Γ2

21 = 1

h2

∂h2

∂x1= 1

x1. (2.267)

For rectangular or Cartesian coordinates, Γijk = 0 for any i, j, and k so that

covariant differentiation reduces to standard partial differentiation.

2.19.1 Physical components

We now normalize the basis,

ei ≡ ei∣∣ei∣∣ = ei√ei ⋅ ei

= ei

hi(2.268)

60 TENSOR ANALYSIS

and similarly

ei = gijej = giiei = 1

h2iei = 1

hiei. (2.269)

We then see that

v = viei = viei = v<i>ei, (2.270)

where we have defined the physical components by

v<i> = hivi = 1

hivi. (2.271)

2.19.2 Gradient of a scalar field

We now note that

∇φ = ∂φ∂xi

ei = ∂φ

∂xigijej,

so

∇φ = ∂φ

∂xi1

h2iei,

or using physical components

∇φ = 1

hi

∂φ

∂xiei. (2.272)

Example

In the cylindrical polar coordinate system

∇φ = ∂φ

∂x1e1 +

1

x1∂φ

∂x2e2 +

∂φ

∂x3e3. (2.273)

2.19.3 Gradient and divergence of a vector field

In terms of physical components, the covariant derivative of a vector field is nowgiven by

∇v = vi,jeiej = ( ∂vi∂xj+ Γi

kjvk)eiej = v<i,j>eiej , (2.274)

where we have defined

v<i,j> ≡ hihj[ ∂

∂xj(v<i>hi) + Γi

kj

v<k>

hk] . (2.275)

The divergence of a vector field is now easily obtained by contracting the aboveresult

∇ ⋅ v = vi,i = 1√g

∂

∂xi(√gvi) = 1

h1h2h3

∂

∂xi(h1h2h3

hiv<i>) . (2.276)

2.19. ORTHOGONAL CURVILINEAR COORDINATE SYSTEM 61

Example

In the cylindrical polar coordinate system

∇ ⋅ v = 1

x1∂

∂xi(x1hiv<i>) = 1

x1∂

∂x1(x1v<1>) + 1

x1∂v<2>

∂x2+∂v<3>

∂x3. (2.277)

2.19.4 Curl of a vector field

Using our earlier results and following a similar procedure, we find that

∇ × v = ǫijk 1

hihj

∂

∂xi(hjv<j>) ek. (2.278)

Example

In the cylindrical polar coordinate system

∇ × v = ( 1

x1∂v<3>

∂x2−∂v<2>

∂x3) e1 + (∂v<1>

∂x3−∂v<3>

∂x1) e2 +

1

x1[ ∂

∂x1(x1v<2>) − ∂v<1>

∂x2] e3. (2.279)

2.19.5 Laplacian of a scalar field

Using our earlier result

∇2φ = 1√

g

∂

∂xi(√ggij ∂φ

∂xj) (2.280)

in an orthogonal curvilinear system, we have

∇2φ = 1√

g

∂

∂xi(√ggii ∂φ

∂xi) = 1

h1h2h3

∂

∂xi

⎛⎝h1h2h3h2i

∂φ

∂xi

⎞⎠ . (2.281)

Example

In the cylindrical polar coordinate system

∇2φ = 1

x1∂

∂xi

⎛⎝x1

h2i

∂φ

∂xi

⎞⎠ = 1

x1∂

∂x1(x1 ∂φ

∂x1) + 1(x1)2 ∂2φ

∂(x2)2 + ∂2φ

∂(x3)2 .(2.282)

62 TENSOR ANALYSIS

2.19.6 Divergence of a dyadic tensor field

It will also be useful to write the divergence of a second-order tensor in terms ofphysical components:

∇ ⋅ τ = τij,j ei,

= (∂τ ij∂xj

+ Γijkτ

kj+ Γ

jjkτ ik)ei,

= [ 1√g

∂

∂xj(√gτ ij) + Γi

jkτkj]ei (since Γ

jjk= 1√

g

∂√g

∂xk) ,

= hi

⎡⎢⎢⎢⎢⎣1

h1h2h3

∂

∂xj

⎛⎝h1h2h3 τ<ij>hihj

⎞⎠ + Γijk

τ<kj>

hkhj

⎤⎥⎥⎥⎥⎦ ei,or

∇ ⋅ τ = τ<ij,j>ei, (2.283)

where

τ<ij,j> ≡ hi

h1h2h3

∂

∂xj

⎛⎝h1h2h3hihjτ<ij>⎞⎠ + hi

hkhjΓijkτ<kj>, (2.284)

and

τ<ij> ≡ hihjτ ij = hihjτ ij = 1

hihjτij . (2.285)

Example

In the cylindrical polar coordinate system

τ<ij,j> = hix1

∂

∂xj

⎛⎝ x1

hihjτ<ij>⎞⎠ + hi

hkhjΓijkτ<kj>. (2.286)

2.20 Integral theorems and generalizations

If V is any composite volume with piecewise smooth bounding surface S andoutward normal n = nie

i, and F is any continuous function in V whose gradient∇F is also continuous in V , then

∫V∇F dV = ∫

SF dS = ∫

SF ndS or ∫

VF,i dV = ∫

SF dSi = ∫

SF ni dS. (2.287)

We will not prove this general form of the Gauss–Green theorem here.The utility of the theorem stems largely from the observation that the function

F may be either a scalar or a tensor of any order. For instance, replacing thefunction F with an arbitrary vector v = vjej gives us

∫V∇v dV = ∫

SvndS or ∫

Vvj,i dV = ∫

Svj ni dS. (2.288)

2.20. INTEGRAL THEOREMS AND GENERALIZATIONS 63

If we contract the above result, we obtain the usual divergence theorem

∫V∇ ⋅ v dV = ∫

Sv ⋅ ndS or ∫

Vvi,i dV = ∫

Svi ni dS. (2.289)

Replacing the function F with ×v = ×(vjej), and recalling that ×v = −v×, givesus the result

∫V∇ × v dV = ∫

Sn × v dS or ∫

Vεkijvj,i dV = ∫

Sεkijni v

j dS. (2.290)

A variety of useful results relating integrals over a closed curve C to integralsover an open surface A bounded by C can be obtained using the two-dimensionalform of the theorem, i.e.,

∫A∇GdA = ∫

CGn′ dl or ∫

AG,i dA = ∫

CGn′i dl, (2.291)

where G is a continuous function defined on an open surface with bounding curveC and n′ = n′iei is the outward normal to the curve C lying on a plane tangent toA. A result similar to the above can be written when A is a curved surface, butthis would require a discussion of tensors on curved, or Riemannian, spaces whichis beyond the scope of the present discussion, but see Appendix D.

We now replace G by ×v ⋅ n = ×(viei) ⋅ (njej), where n is the normal to the

surface A and v is a continuous vector function. Then we have

∫A∇ × v ⋅ndA = ∫

C×v ⋅nn′ dl,

= −∫Cn ⋅ v × n′ dl,

= ∫Cn ⋅ n′ × v dl,

= ∫Cv ⋅n × n′ dl.

Now since n and n′ are normal to both C and to each other, the vector productn × n′ in the integral is equal to the vector t = n × n′ = tkek tangent to C, wheretk = εkijnin

′j . Subsequently, using our previous result of the curl of a vector field,

and noting that tdl = dr, where dr = dtkek is the oriented differential length of theline tangent to C, we arrive at the result

∫A∇ × v ⋅ ndA = ∫

Cv ⋅ dr or ∫

Aεkijnkvj,i dA = ∫

Cvk dt

k, (2.292)

which is known as Stokes’ theorem.In addition, setting v = φa where a is an arbitrary constant vector, and using

the property of invariance to cyclic permutations of the triple scalar product, weobtain the following useful identity:

∫An ×∇φdA = ∫

Cφdr or ∫

Aεkijniφ,j dA = ∫

Cφdtk. (2.293)

2.20.1 Regions with discontinuous surfaces, curves, andpoints

In the above theorems, we have assumed that the tensor field F is continuous in V .When V contains a discontinuous surface across which F undergoes a jump, we can

64 TENSOR ANALYSIS

decompose the volume into two subvolumes separated by the discontinuous surfaceζ with unit normal ν as illustrated in Fig. 2.9. Then within each subvolume, the

n

V +

V −

S+

S−

ζ

ζ+

ζ−

ν

n

Figure 2.9: Arbitrary volume V containing a discontinuous surface.

field F is continuous and our previous results apply. Subsequently, we can write

∫V +∇F dV = ∫

S+F dS +∫

ζ+F +dζ

+ (2.294)

and

∫V −∇F dV = ∫

S−F dS +∫

ζ−F −dζ−. (2.295)

Now adding the above equations, letting ζ+ and ζ− approach ζ, and recognizingthat in this limit dζ+ = −dζ− = −dζ = −ν dζ, we obtain

∫V −ζ∇F dV = ∫

S−ζF dS −∫

ζJF K dζ, (2.296)

where we have defined the jump operator

JF K ≡ F + −F −, (2.297)

and

F +(x) ≡ limx↓ζ

F (x) and F −(x) ≡ limx↑ζ

F (x). (2.298)

For example, if we have a vector field v, the generalized divergence theorem isobtained by taking F → v and contracting the result to obtain

∫V −ζ∇ ⋅ v dV = ∫

S−ζv ⋅ dS −∫

ζJvK ⋅ dζ. (2.299)

Analogously, if area A with unit normal n contains a discontinuous line γ withtangential unit vector t, as illustrated in Fig. 2.10, across which v changes sud-denly, then using the same procedure as before that led to Stokes’ theorem, and

2.20. INTEGRAL THEOREMS AND GENERALIZATIONS 65

dr

A+

A−

C+

C−

γ

γ+

γ−

tdr+

dr−

dr

Figure 2.10: Arbitrary surface A containing a discontinuous curve.

noting that dr+ = −dr− = −dγ = −tdγ, we now obtain the generalized Stokes theo-rem:

∫A−γ∇ × v ⋅ dA = ∫

C−γv ⋅ dr −∫

γJvK ⋅ dγ. (2.300)

For reference, we also note that in one dimension we have an analogous versionof the generalized divergence theorem. In this case, as illustrated in Fig. 2.11, ifa function v(ξ) defined on the curve C ∶ ξ1 ≤ ξ ≤ ξ2 is discontinuous at the point σin the interval of C, but continuous in the subintervals ξ1 < ξ < σ and σ < ξ < ξ2,then

∫C(t)−σ(t)

dv

dξdξ = [v(ξ)]ξ2ξ1 − Jv(σ)K, (2.301)

whereJv(σ)K = v+(σ) − v−(σ), (2.302)

and

v+(σ) ≡ limξ↓σ

v(ξ) and v−(σ) ≡ limξ↑σ

v(ξ). (2.303)

Problems

1. The scalar triple product of three vectors [u,v,w] is given by u ⋅ (v ×w).Establish the following property of the scalar triple product:

u ⋅ (v ×w) = v ⋅ (w × u) =w ⋅ (u × v) = −u ⋅ (w × v) = −v ⋅ (u ×w) =−w ⋅ (v × u)

for all (u,v,w) ∈ E3.

66 TENSOR ANALYSIS

C

σξ1

ξ

ξ2

Figure 2.11: Arbitrary curve C containing a discontinuous point.

2. Show that

ei × ej = ±ǫijkek,and, subsequently,

u ⋅ (v ×w) = ±ǫijkuivjwk.

3. Let u, v, w, and x be arbitrary vectors, A, B, and C second rank tensors, 1the second-rank unit tensor, trA the trace of A, and detA the determinantof A. Show that

a) (uv)T = vu,

b) tr(uv) = 1 ∶ uv = u ⋅ v,

c) det (uv) = 0,d) (uv) ∶ (wx) = (u ⋅w)(v ⋅ x),e) v ⋅AT

⋅ u = (A ⋅ v) ⋅ u = u ⋅A ⋅ v,

f) A ∶ uv = uv ∶A = u ⋅A ⋅ v,

g) ∣∣A∣∣ = (A ∶A)1/2 ≥ 0,h) trA = 1 ∶A =A ∶ 1,

i) det (A−1) = (detA)−1,j) (A−1)T = (AT )−1,k) det (A ⋅B) = detAdetB,

l) (A ⋅B)T = BT⋅AT ,

m) A ∶ B = B ∶A = tr(AT⋅B) = tr(A ⋅BT ) = tr(BT

⋅A) = tr(B ⋅AT ),n) A ∶ (B ⋅C) = (BT

⋅A) ∶C = (A ⋅CT ) ∶ B.

4. Show that B = F−1 ⋅A ⋅ (F−1)T is symmetric if A is symmetric.

5. Under the assumption that A is symmetric, construct the partial derivativeof ∣∣A∣∣ with respect to A.

2.20. INTEGRAL THEOREMS AND GENERALIZATIONS 67

6. Let A be a tensor of rank 4, and B and C tensors of rank 2. Show that(AT )T =A, and

B ∶AT∶C =C ∶A ∶ B = (A ∶ B) ∶C.

7. Let A be a tensor of rank 2, the deviatoric part of A be defined by A′ ≡A − 1

3(1 ∶A)1, 1 be the unit tensor of rank 4 such that (1)ijkl = δikδjl, and

1 be the unit tensor of rank 4 such that (1)ijkl = δilδjk. Show that

a) trA′ = 0,b) 1 ≠ 1T ,

c) A = 1 ∶A and AT = 1 ∶A,

d) 1

2(A +AT ) = D ∶ A and 1

2(A −AT ) =W ∶ A, where D ≡ 1

2(1 + 1) and

W ≡ 1

2(1 − 1) are symmetric and skew-symmetric tensors of rank 4.

e) A′ = P ∶A, where the projector P is given by P = 1− 1

311 with compo-

nents Pijkl = δikδjl − 1

3δijδkl,

f) 1, 1, and 11 are isotropic tensors, and the most general isotropic tensorof rank 4 is of the form α11 + β1 + γ1 with components

(α11 + β1 + γ1)ijkl = αδijδkl + βδikδjl + γδilδjk.8. How many distinct components are there in the completely symmetric tensor

of rank 3 in E3?

9. Show that a completely skew-symmetric tensor of rank 3 in E3 has only onenonzero distinct component.

10. Assuming that u, v, and a are arbitrary vectors, and w is the axial vectorcorresponding to W = u ∧ v. By using (2.147), show that

w = u × v. (2.304)

11. Let W be an arbitrary rank-2 three-dimensional skew-symmetric tensor, wthe corresponding axial vector, and v an arbitrary three-dimensional vectorsatisfying (2.147). Using the identity (a × b) × c = (a ⋅ c)b − (b ⋅ c)a, showthat

W2 =ww − (w ⋅w)1 (2.305)

and hence that ∣w∣2 = −12trW2. (2.306)

12. Let wij be the components of a skew-symmetric tensor and sij the compo-nents of a symmetric tensor.

a) Provide justifications for each of the following equal signs:

wijsij = −wjis

ij = −wjisji = −wkls

kl = −wijsij = 0.

68 TENSOR ANALYSIS

b) Establish the following two identities for any arbitrary tensor of rank 2with components vij :

vijwij = 1

2(vij − vji)wij , vijsij = 1

2(vij + vji)sij .

13. Let Qijr be the components of a tensor of rank 3 that is skew-symmetric inits first two indices, i.e., Q[ij]r . Show that the tensor can be decomposedinto three orthogonal parts

Q[ij]r =Q(1)ijr +Q(2)ijr +Q

(3)ijr ,

where

Q(1)ijr = 1

3(Qijr +Qrij +Qjri) ,

Q(2)ijr = 1

6(4 Qijr − 2 Qrij − 2 Qjri − 3 Qikk δjr − 3 Qjkk δir) ,

Q(3)ijr = 1

2(Qikk δjr +Qkjk δir) .

14. Let e1 = (1,0) and e2 = (1,1) be a basis for E2.

a) Find the reciprocal basis (e1,e2).b) Compute [gij] and [gij] directly from inner products involving basis

vectors, and show that [gij] = [gij]−1.c) Let v be a vector in E2 with contravariant components (relative to the

above basis) v1 = 1 and v2 = 2. Find the covariant components of v

relative to this basis.

d) Compute ∣∣v∣∣.15. In the treatment of the metric tensor, both covariant gij and contravariant

gij components are discussed. The mixed components gij are, however, not

mentioned. Explain why. (Hint: Compute gij .)

16. a) Show that the triadic tensor tijkeiejek can be contracted to a vectorby the formula tijkei ⋅ ejek and that the result is independent of thebasis used (i.e., show that tijkgijek = tijk gij ek).

b) Generalize the result for any order tensor. Notice that the result de-pends on where the dot is placed.

17. Show that in cylindrical polar coordinates

e1 = cosx2i1 + sinx2i2,

e2 = −sinx2

x1i1 +

cosx2

x1i2,

e3 = i3.

18. Calculate gik for the cylindrical polar coordinate system.

2.20. INTEGRAL THEOREMS AND GENERALIZATIONS 69

19. Show that

vi = ∂xi∂xj

vj .

20. Show that

v ⋅ v = vivjgijand

vi = vkgik.21. Show that

ǫijkǫrst = δirδjsδkt + δisδjtδkr + δitδjrδks − δisδjrδkt − δirδjtδks − δitδjsδkr,and that then

a) ǫijkǫist = δjsδkt − δjtδks,b) ǫijkǫijt = 2δkt,c) ǫijkǫijk = 6,d) and compute ∇×∇ × v.

22. From the definition

gij = ∂r

∂xi⋅∂r

∂xj,

verify the formula

Γijk = 1

2gir (∂grj

∂xk+∂grk

∂xj−∂gjk

∂xr) .

23. Show that the Christoffel symbol Γijk follows the transformation law

Γijk = ∂x

i

∂xr∂xs

∂xk∂xt

∂xjΓrst +

∂xi

∂xr∂2xr

∂xk∂xj,

which, in view of the second term, shows that Γijk is not of the form of

components of a tensor with respect to an arbitrary coordinate system.

24. Prove that

ǫijk ∣∂x∂x∣ = ǫrst ∂xi

∂xr∂xj

∂xs∂xk

∂xt.

25. Show that T ik,l transforms like a third-order tensor.

26. Show that

vi,j − vj,i = ∂vi∂xj−∂vj

∂xi.

70 TENSOR ANALYSIS

27. a) Show that gij,l = 0 by taking the covariant derivative of the (02) compo-

nent form of a second-order tensor and using the definition of Γijk.

b) Take the covariant derivative of gijgjk = δki and use part a) to show

that gij,l= 0.

28. Show that

εijk,p = 0.29. Given a set of curvilinear coordinates xi defined by

ξ1 = x1x2, ξ2 = x1 + x3, ξ3 = x3.a) Find the inverse transformation xi = xi(ξj).b) Find the ei’s.

c) Find the ei’s.

d) Find the gij ’s.

e) Find the gij ’s.

f) Find the Γijk’s (there are 27 of them).

g) Show that gij,l = 0 by direct computation.

h) Write out the equation ∇2φ = 0 in xi coordinates, where φ = φ(xi) is ascalar field.

30. Calculate Γijk for cylindrical polar coordinates.

31. Assume that in cylindrical polar coordinates a rank-2 tensor T has mixedcomponents

[T ij ] =⎡⎢⎢⎢⎢⎢⎣2 −1 1

0 1 2

3 0 −2

⎤⎥⎥⎥⎥⎥⎦at the point (x1, x2, x3) = (1, π/4,−√3). Find the component T

1

2 of T inspherical coordinates.

Bibliography

R. Aris. Vectors, Tensors and the Basic Equations of Fluid Mechanics. DoverPublications, Inc., Mineola, NY, 1962.

H.D. Block. Introduction to Tensor Analysis. Charles E. Merrill Books, Inc.,Columbus, Ohio, 1978.

A.I. Borisenko and I.E. Tarapov. Vector and Tensor Analysis with Applications.Dover Publications, Inc., New York, NY, 1968.

R.M. Bowen and C.-C. Wang. Introduction to Vectors and Tensors – Linear andMultilinear Algebra, volume 1. Plenum Press, New York, NY, 1976.

BIBLIOGRAPHY 71

R.M. Bowen and C.-C. Wang. Introduction to Vectors and Tensors – Vector andTensor Analysis, volume 2. Plenum Press, New York, NY, 1976.

L. Brand. Vector and Tensor Analysis. John Wiley & Sons, Inc., New York, NY,1955.

L. Brand. Vector Analysis. John Wiley & Sons, Inc., New York, NY, 1957.

J.H. Heinbockel. Introduction to Tensor Calculus and Continuum Mechanics.Trafford Publishing, Victoria, B.C., Canada, 2001.

M. Itskov. Tensor Algebra and Tensor Analysis for Engineers. Springer-Verlag,Berlin, 2nd edition, 2009.

H. Jeffreys. Cartesian Tensors. Cambridge University Press, London, 1969.

J.K. Knowles. Linear Vector Spaces and Cartesian Tensors. Oxford UniversityPress, New York, 1998.

A.J. McConnell. Applications of Tensor Analysis. Dover Publications, Inc., NewYork, NY, 1957.

C. Perwass. Geometric Algebra with Applications in Engineering. Springer-Verlag, Berlin, 2009.

J.G. Simmonds. A Brief on Tensor Analysis. Springer-Verlag, New York, NY,1994.

G. Temple. Cartesian Tensors. Methuen & Co. Ltd., London, 1960.

J. Vince. Rotation Transforms for Computer Graphics. Springer-Verlag, London,2011.

T.L. Wade. Tensor algebra and Young’s symmetry operators. American Journalof Mathematics, 63(3):645–657, 1941.

T.L. Wade and R.H. Bruck. Types of symmetries. The American MathematicalMonthly, 51(3):123–129, 1944.

S. Winitzki. Linear Algebra via Exterior Products. lulu.com, 2010.

3

Kinematics

Kinematics is the study of deformation and motion of material bodies. The rela-tionship between the initial position of material points or material particles of abody and their subsequent places is essential in the description of the local lengthand angle changes and translations and rotations of elements of the body. Weare concerned with such changes and their measures irrespective of the type ofmaterial and the external effects.

To describe the positions of material points, we introduce two sets of coordinatesystems, one for the undeformed body and one for the deformed body. The defor-mation of a point is then described by the relation of the coordinates of the samematerial point in the undeformed and deformed states.

The material points of a continuum medium, at a certain time, occupy a regionB in space. In order to describe the body in space, we will identify it with a regionin a three-dimensional Euclidean space E3 relative to a frame of reference. We calla one-to-one mapping from B into E3, or a complete specification of the positionsof particles of a body, a configuration of B. It is usually convenient to choose aparticular configuration of B, say κ, as a reference:

κ ∶ B → E3 or κ(X) =X, (3.1)

where X ∈ B labels a material point. We call κ a reference configuration of B. Ingeneral, the initial or undeformed configuration need not be chosen to be the sameas the reference configuration. The coordinate X, with components XK (K =1,2,3), is called the reference coordinate or material coordinate since the point X

in the reference configuration is identified with the material point labeled X ofthe body B. We note that the set of all points in the material body defines thevolume in the reference configuration, i.e.,

V ≡ κ(B), (3.2)

so that X ∈ V .Let κ be a reference configuration and χ an arbitrary deformed configuration

of B. Then the mapping of the material point labeled X

x = χκ(X) = χ(κ−1(X)) = χκ(X) (3.3)

73

74 KINEMATICSreplacemen

B

V ≡ κ(B)V ≡ χκ(B)

χκ(X)κ(X)X X x

Figure 3.1: Material body, reference configuration, and deformed configuration.

is called the deformation of B from κ to χ. In terms of coordinate systems in thedeformed, xi (i = 1,2,3), and the reference, XK (K = 1,2,3), configurations, thedeformation χκ can be expressed as

xi = χiκ(XK), (3.4)

where χκ is called the deformation function. After deformation takes place, thevolume V , with surface S, occupies a region consisting of the deformed volumeV ≡ χκ(B) with surface S . In this deformed state, the material point X occupiesthe spatial location x ∈ V in the deformed configuration, with components xi (i =1,2,3). The configurations are illustrated in Fig. 3.1. We call XK the material orLagrangian coordinates of a particle or material point and xi the spatial or Euleriancoordinates. They both have dimensions of length, [L]. The brackets denote thedimensions of the quantity enclosed inside of them. The deformation of the bodycarries various materials points through various spatial points expressed by thedeformation function. The aim of continuum mechanics is the determination ofthe explicit form of the deformation function when the external effects and theinitial and boundary conditions of a prescribed body are known.

From now on, for notational simplicity, we shall write the deformation functionas χ where it is understood that this is a deformation relative to the configurationκ, unless otherwise noted. We also note that we will use quite interchangeablythe following nomenclature to denote a tensor. Say that T is a second-ordertensor. Then, using the contravariant component form, we write that T = T ijeiej ,where T ij is understood to be the component in the directions eiej . We shallalso write the components as the matrix T = [T ij]. Lastly, we shall take thereference as well as deformed coordinate systems to be Cartesian so that ek → ikand T ij → Tij . However, to distinguish between quantities in the reference anddeformed states, we will use uppercase letters to denote quantities associated withthe reference configuration and lowercase letters with those associated with thedeformed configuration, so that, e.g., IK and ik are the respective basis vectors inthe reference and deformed coordinate systems. Sometimes it is advantageous toselect two different reference frames for the reference and deformed configurations,particularly when curvilinear coordinates are used. In this case, the general pictureis described as in Fig. 3.2. Note that

IK ⋅ IL = δKL, ik ⋅ il = δkl, IK ⋅ ik = gKk = gkK . (3.5)

Subsequently, we write

x = χ(X) or xk = χk(XK) (3.6)

75

S

S

b

x3

x2

x1

X3

X1

X2

I1

I2

I3

i1

i2

i3

X

XV

V

X

x

O

o

Figure 3.2: Two reference frames.

for every point in the reference configuration. Conversely, we write

X = χ−1(x) or XK = χ−1K (xk). (3.7)

We assume that the mappings are single valued and possess continuous derivativeswith respect to their arguments of whatever order is desired, except possibly atsome singular points, curves, and surfaces. Furthermore, we assume that the aboveare unique inverses of each other in a neighborhood of any material point. Thisassumption is known as the axiom of continuity. It expresses the fact that matteris indestructible, i.e., no region of a finite volume of matter can be deformed intoone of zero volume. Furthermore, it implies that matter is impenetrable, i.e., themotion carries every region into a region, every surface into a surface, and everycurve into a curve. This assumption is embodied into the requirement that

J ≡ det [ ∂xk∂XK

] ≠ (0,±∞) (3.8)

is satisfied for all material points X ∈ B, except possibly at some singular points,curves, and surfaces. Note that as long as the handedness of the coordinate systemin the reference and deformed configurations are the same, then J > 0. Unlessspecifically noted otherwise, we will assume this to be the case. If J = 1, thedeformation is said to be isochoric.

Components of a tensor field quantity of any order can be written using eitherthe material or the spatial descriptions

ψ(X) = ψ[χ−1(x)] = ψ(x), (3.9)

76 KINEMATICS

χκ

κF(X)

x

dx

dX

X

O

Figure 3.3: Mapping of neighborhood by the deformation gradient.

or

ψ...(XK) = ψ...[χ−1K (xk)] = ψ...(xk). (3.10)

For notational simplicity we drop the hat on the function since the functionaldependence should be obvious in applications or will be explicitly displayed whennecessary.

3.1 Deformation

3.1.1 Deformation gradient

The material deformation gradient at X, as illustrated in Fig. 3.3, is a lineartransformation defined by

F = F(X) ≡ (Grad x)T = (∇Xx)T or FkK ≡ xk,K = ∂xk

∂XK

, (3.11)

where it is obvious that J = det [FkK] ≠ (0,±∞). Above we have denoted thegradient with respect to the material coordinates by “Grad.” Similarly, we willdenote the divergence with respect to the material coordinates by “Div” and thecurl by “Curl.” The analogous gradient, divergence, and curl operators with respectto spatial coordinates will be denoted by “grad”, “div”, and “curl” respectively. Notethat F depends on the reference configuration, but the κ subscript is dropped sincethis leads to no confusion at the moment. The deformation gradient is not a truetensor since it relates two points in different coordinate systems, the current tothe reference. Subsequently, it is often referred to as a two-point tensor or a doublevector.

3.1. DEFORMATION 77

VV

Θ

θdX(j) dX(i)

X

x

dx(j) dx(i)F(X)

Figure 3.4: Transformation of a vector element.

3.1.2 Transformation of linear elements

It follows from the definition of the deformation gradient that

dx = F ⋅ dX or dxk = FkK(X) dXK . (3.12)

This equation represents the transformation for infinitesimal linear elements ofmaterial under the deformation x = χ(X) illustrated in Fig. 3.4. Clearly theinverse transformation is given by

dX = F−1 ⋅ dx or dXK = F −1Kk(x) dxk. (3.13)

In the deformed configuration, the length of a differential vector element is

(dxkdxk)1/2 = (FkKFkLdXKdXL)1/2 . (3.14)

Now introducing the direction unit vectors

dx = tdx or dxk = tkdx (3.15)

anddX = TdX or dXK = TKdX, (3.16)

we have

(tkdx tkdx)1/2 = (FkKFkLTKdX TLdX)1/2 , (3.17)

or, noting that tktk = 1,dx = (CKLTKTL)1/2 dX, (3.18)

which relates the length of a differential element before and after deformation.Above we have defined the symmetric second-order tensor

C ≡ FT⋅F or CKL ≡ FkKFkL, (3.19)

which is called the right Cauchy–Green strain tensor. The quantity

λ ≡ dx

dX= (CKLTKTL)1/2 = (C ∶ TT)1/2 (3.20)

78 KINEMATICS

is called the length stretch ratio.To examine the change in orientation of the differential line element, we note

that

tk = dxkdx= FkKTKdX(CLMTLTM)1/2 dX =

FkKTK

λor t = 1

λF ⋅T. (3.21)

Example

We recall that dxi = FiIdXI where FiI = ∂xi/∂XI , and the deformationfunction is given by xi = χi(XI). Let us assume that we have the deforma-tion

x1 = X1 −AX1,

x2 = X2 −AX2,

x3 = X3 +BX3,

where A and B are positive constants. The deformation gradient is thengiven by

FiI =⎡⎢⎢⎢⎢⎢⎣1 −A 0 0

0 1 −A 0

0 0 1 +B

⎤⎥⎥⎥⎥⎥⎦.

In this example FiI is independent of XK since the deformation function islinear. When this is so, it is called a homogeneous deformation. The rightCauchy–Green strain tensor is then given by

CKL = FiKFiL,

so

CKL =⎡⎢⎢⎢⎢⎢⎣(1 −A)2 0 0

0 (1 −A)2 0

0 0 (1 +B)2⎤⎥⎥⎥⎥⎥⎦.

Now let us examine how a differential material element with direction co-sine T (1)K = (0,0,1) in the reference configuration stretches when deformed.Since

dx(1) = (CKLTKTL)1/2dX(1),then λ(1) = (1 + B). If we look in the direction T

(2)K = (0,1,0), then

λ(2) = (1 −A). Let us examine the stretch and change in orientation for a

material differential element with direction cosine T (3)K = (1/√2,1/√2,0).The material is stretched by

λ(3) = (CKLT(3)K T

(3)L )1/2 = (1 −A),

and the new orientation is given by

t(3)k= FkKT

(3)K

λ(3)= [(1 −A)/

√2, (1 −A)/√2,0](1 −A) = ( 1√

2,1√2,0) ;

3.1. DEFORMATION 79

V

VΘ

θdX(2) dX(1)

X

x

dx(2) dx(1)F(X)

Figure 3.5: Reference and current configurations of elementary arcs on intersectingmaterial curves.

thus we see that there is no change in orientation for a vector in that

direction. Lastly, we look at a material line element in the direction T (4)K =(1/√3,1/√3,1/√3), then

λ(4) = (CKLT(4)K T

(4)L )1/2 = (C11 +C22 +C33

3)1/2 = [2

3(1 −A)2 + 1

3(1 +B)2]1/2

and

t(4)k = FkKT

(4)K

λ(4)= 1

λ(4)(1 −A√

3,1 −A√

3,1 +B√

3) .

Now we take two oriented differential line elements at two different orientations,as illustrated in Fig. 3.5, and examine the change in angle between the two lineelements before and after deformation. First we take

dx(1)i = t

(1)i dx(1), (3.22)

dx(2)j = t

(2)j dx(2), (3.23)

dX(1)I

= T(1)IdX(1), (3.24)

dX(2)J

= T(2)JdX(2), (3.25)

and since

t(α)i = FiIT

(α)I

λ(α), (3.26)

we then have

cos θ(ii) = t(1)i t(2)i = CIJ

λ(1)λ(2)cosΘ(IJ). (3.27)

Note that cosΘ(IJ) = T (1)I T(2)J .

At this point, we emphasize that FkK contains all the information about anydeformation. Additionally, we note that the integral of any arbitrary tensor fieldG(x) along an arbitrary curve is easily obtained from

∫CG(xi)dxk = ∫

CG(XI)FkK(XJ)dXK , (3.28)

80 KINEMATICS

since in the second integral C is independent of the deformation.

3.1.3 Transformation of a surface element

Oriented differential area elements in the reference and deformed coordinate frames,as illustrated in Fig. 3.6, are given by

dS = dX(1) × dX(2) and ds = dx(1) × dx(2), (3.29)

or

dSI = ǫIJKdX(1)J dX(2)K and dsi = ǫijkdx(1)j dx

(2)k. (3.30)

Since dxj = FjJdXJ = xj,JdXJ , we have

dsi = ǫijkFjJFkKdX(1)J dX

(2)K = ǫijkxj,Jxk,KdX(1)J dX

(2)K . (3.31)

Now recalling that J = det [FjJ ], using (2.49), we can write

ǫLJKJ = ǫljkxl,Lxj,Jxk,K (3.32)

or, since ǫLJKǫLJK = 6,J = 1

6ǫLJKǫljkxl,Lxj,Jxk,K , (3.33)

and

ǫLJKXL,iJ = ǫljk (XL,ixl,L)xj,Jxk,K = ǫljkδlixj,Jxk,K = ǫijkxj,Jxk,K . (3.34)

Thus

dsi = JXL,iǫLJKdX(1)JdX

(2)K, (3.35)

or finally

ds = J (F−1)T ⋅ dS or dsi = JXL,idSL, (3.36)

and since we can also write

xi,Ldsi = JdSL, (3.37)

we additionally have

dS = J−1FT⋅ ds or dSL = J−1xi,Ldsi. (3.38)

From above we also see that (F−1)T = (FT )−1 (also written quite often as F−T ),and we have used the fact that J−1 = 1/J since

δik = ∂xi

∂XJ

∂XJ

∂xk, (3.39)

and taking the determinant of both sides,

1 = det [ ∂xi∂XJ

∂XJ

∂xk] = det [ ∂xi

∂XJ

]det [∂XJ

∂xk] = JJ−1. (3.40)

3.1. DEFORMATION 81

V

V

dX(2)

xX

N

dX(1)

dx(1)

n

dx(2)

F(X)Figure 3.6: Reference and current configurations of an element of a material sur-face.

Lastly, sinceds = nds or dsi = ni ds (3.41)

anddS =NdS or dSI = NI dS, (3.42)

and using the fact that nini = 1, we can also write

ds = J (XJ,lXK,lNJNK)1/2 dS. (3.43)

The quantity

η = dsdS= J (XJ,lXK,lNJNK)1/2 = J (C−1 ∶NN)1/2 (3.44)

is called the area stretch ratio, and using (3.36), (3.41), (3.42), and (3.44), it iseasy to see that the area normals are related by

n = Jη(F−1)T ⋅N. (3.45)

As we can see from (3.21) and (3.45), the linear and area stretches are related fort = n and T =N by

λη = J, (3.46)

or, equivalently,dxds = dv, (3.47)

since dX dS = dV . For isochoric deformations, we see that λη = 1.We also see that the surface integral of an arbitrary tensor field G(x) can be

written as

∫S

G(xi) ds = ∫SG(XI)J(XL) (XJ,lXK,lNJNK)1/2 dS, (3.48)

where the second integral is easier to compute since S is independent of the defor-mation.

82 KINEMATICS

V

V

X

xdX(2) dX(3)

dX(1)

dx(3)

dx(2)

dx(1)F(X)

Figure 3.7: Reference and current configurations of an element of a material vol-ume.

3.1.4 Transformation of a volume element

By definition, the differential volume elements in the reference and deformed co-ordinate frames illustrated in Fig. 3.7 are respectively given by

dV = ∣[dX(1), dX(2), dX(3)]∣ (3.49)

and

dv = ∣[dx(1), dx(2), dx(3)]∣ . (3.50)

Now using our previous results, we have

dv = ∣ǫijkdx(1)i dx(2)j dx

(3)k∣ ,

= ∣ǫijkx(1)i,Ix(2)j,Jx(3)k,K

dX(1)IdX

(2)JdX

(3)K∣ ,

= ∣ǫIJKJdX(1)I dX(2)J dX

(3)K ∣ ,

where we used (3.32) in the last step, or since J > 0, we finally have

dv = J dV. (3.51)

We see that the quantity

J = dvdV

(3.52)

also represents the volume stretch ratio. For this reason, it is sometimes convenientto perform a multiplicative decomposition of the deformation gradient tensor,

F = J1/3F or Fij = J1/3F ij , (3.53)

into the dilatational part, J1/3 1, and the isochoric part, F, since detF = 1.Lastly, we note that a volume integral of a tensor field G(x) can now be written

as

∫V

G(x)dv = ∫VG(X)J(X)dV, (3.54)

where V is independent of the deformation.

3.1. DEFORMATION 83

3.1.5 Relations between deformation and inversedeformation gradients

If we have the inverse deformation function XI,i = XI,i(xj,J), we would like theability to express it and its gradient as explicit functions of xj,J or of XK,k. Simi-larly, if we are given the deformation function xi,I = xi,I(XJ,j), we would like theability to express it and its gradient as explicit functions of XJ,j or of xk,K . Todo this, we start with our previous result (3.34):

ǫLJKJXL,i = ǫijkxj,Jxk,K . (3.55)

Multiplying both sides by ǫIJK and noting that ǫIJKǫLJK = 2δIL, we have

2δILJXL,i = ǫijkǫIJKxj,Jxk,K , (3.56)

or

XI,i = 1

2J−1ǫijkǫIJKxj,Jxk,K . (3.57)

In a similar fashion we can show that

xi,I = 1

2JǫijkǫIJKXJ,jXK,k. (3.58)

Also from (3.33), we have

J = 1

6ǫijkǫIJKxi,Ixj,Jxk,K . (3.59)

Now differentiating with respect to xl,L, we have

6∂J

∂xl,L= ǫljkǫLJKxj,Jxk,K + ǫilkǫILKxi,Ixk,K + ǫijlǫIJLxi,Ixj,J ,

= 3ǫljkǫLJKxj,Jxk,K ,

or, using (3.56),

∂J

∂xl,L= 1

2ǫljkǫLJKxj,Jxk,K = JXL,l. (3.60)

Similarly, it can be shown that

∂J−1

∂XL,l

= 1

2ǫLJKǫljkXJ,jXK,k = J−1xl,L. (3.61)

Lastly, since

xi,JXJ,k = δik, (3.62)

differentiating both sides with respect to xl,L,

δilδJLXJ,k + xi,J∂XJ,k

∂xl,L= 0. (3.63)

Now multiplying both sides by XK,i,

∂XK,k

∂xl,L= −XK,lXL,k. (3.64)

Similarly, it can be shown that

∂xl,L

∂XK,k

= −xk,Lxl,K . (3.65)

84 KINEMATICS

3.1.6 Identities of Euler–Piola–Jacobi

We want to show that

(JXI,i),I = 0 (3.66)

and

(J−1xi,I),i = 0, (3.67)

which are known as Euler–Piola–Jacobi identities.Now

(JXI,i),I = ∂J

∂XI

XI,i + J∂XI,i

∂XI

,

= ∂J

∂xj,J

∂xj,J

∂XI

XI,i + J∂XI,i

∂xj,J

∂xj,J

∂XI

,

= ∂J

∂xj,J

∂2xj

∂XI∂XJ

XI,i + J∂XI,i

∂xj,J

∂2xj

∂XI∂XJ

, (3.68)

but from (3.60)

∂J

∂xj,J= JXJ,j (3.69)

and from (3.64)

∂XI,i

∂xj,J= −XI,jXJ,i, (3.70)

so

(JXI,i),I = J (XJ,jXI,i −XI,jXJ,i)xj,IJ , (3.71)

or

(JXI,i),I = JSIJ,ijDj,IJ , (3.72)

where we have defined

SIJ,ij =XJ,jXI,i −XI,jXJ,i = −SJI,ij (3.73)

and

Dj,IJ = xj,IJ =Dj,JI . (3.74)

Now since

SIJ,ijDj,IJ = −SJI,ijDj,IJ = −SJI,ijDj,JI = −SIJ,ijDj,IJ = 0, (3.75)

then

(JXI,i),I = 0. (3.76)

In a similar fashion, it can be shown that

(J−1xi,I),i = 0. (3.77)

3.1. DEFORMATION 85

3.1.7 Cayley–Hamilton theorem

A scalar λ is called an eigenvalue of the matrix A of size n × n if there exists anonzero vector v such that it satisfies the following eigenvalue problem:

(A − λ1) ⋅ v = 0 or (A − λI)v = 0 or (aik − λδik)vk = 0. (3.78)

It follows that λ is an eigenvalue if and only if

det (A − λI) = 0 or det (aik − λδik) = 0, (3.79)

or more explicitly, if λ is a root of the following characteristic polynomial equation:

f(λ) = (−λ)n +A(1) (−λ)n−1 +⋯ +A(n−1) (−λ) +A(n) = 0. (3.80)

The coefficients A(1), . . . ,A(n) are scalar functions of A, called the principal in-variants of A, and are given by (see (2.91)–(2.95))

A(1) = 1

1!δi1j1ai1j1 = tr A, (3.81)

A(2) = 1

2!δi1j1i2j2ai1j1ai2j2 = 1

2(A(1)tr A − tr A2) , (3.82)

A(3) = 1

3!δi1j1i2j2i3j3ai1j1ai2j2ai3j3 = 1

3(A(2)tr A −A(1)tr A2

+ tr A3) ,(3.83)

⋮

A(k) = 1

k!δi1j1⋯ikjkai1j1⋯aikjk ,

= 1

k(A(k−1)tr A −⋯ + (−1)k−1 tr Ak) , 1 < k ≤ n, (3.84)

⋮

A(n) = 1

n!δi1j1⋯injnai1j1⋯ainjn ,

= 1

n(A(n−1)tr A −A(n−2)tr A2

+A(n−3)tr A3−⋯+ (−1)n−1 tr An) ,

= detA. (3.85)

To determine the gradients of the principal invariants with respect to A, it isuseful to rewrite the above relation

det (A − λI) = n

∑k=0

(−λ)n−kA(k), (3.86)

where A(0) = 1, and it can be shown that

∂A(n)∂A

= ∂ detA∂A

= detA (A−1)T . (3.87)

Using this result in conjunction with (3.86), the following recursion can be ob-tained:

∂A(k+1)∂A

= A(k)I −AT∂A(k)∂A

, k = 0,1, . . . , n, (3.88)

86 KINEMATICS

where A(n+1) = 0. By induction, the above recursion can also be written in theform

∂A(k)∂A

= ⎡⎢⎢⎢⎣k−1

∑j=0

(−1)j A(k−j−1)Aj⎤⎥⎥⎥⎦T

. (3.89)

The Cayley–Hamilton theorem states that the matrix A satisfies its own char-acteristic polynomial equation, i.e.,

f(A) = (−A)n +A(1) (−A)n−1 +⋯ +A(n−1) (−A) +A(n)I = 0. (3.90)

The proof is straightforward. First rewrite the eigenvalue problem as

Av = λv.We note that for any r = 1, . . . , n

Arv = Ar−1(Av) = Ar−1λv = λAr−1v = ⋯ = λrv.Subsequently, since the actions of λr and Ar on the nonzero vector v are the same,replacing λr by Ar in the characteristic polynomial equation (3.80) establishes thetheorem.

The case with n = 3 is most relevant to our discussions. In this case, we write

−λ3 +A(1)λ2−A(2)λ +A(3) = 0, (3.91)

where

A(1) = aii = trA,= λ(1) + λ(2) + λ(3), (3.92)

A(2) = 1

2(aiiakk − aikaki) = 1

2[(trA)2 − trA2] ,

= λ(1)λ(2) + λ(2)λ(3) + λ(3)λ(1), (3.93)

A(3) = 1

6ǫijkǫrstairajsakt = 1

6[(trA)3 − 3 trA trA2

+ 2 trA3] = detA,= λ(1)λ(2)λ(3). (3.94)

The eigenvalues λ(k) (k = 1,2,3) are the zeros of the characteristic polynomialequation (3.91), and for each eigenvalue, there corresponds an associated eigen-

vector v(k) = v(k)i ii. Use of the Cayley–Hamilton theorem provides

−A3+A(1)A2

−A(2)A +A(3)I = 0 (3.95)

and

−A2+A(1)A −A(2)I +A(3)A−1 = 0. (3.96)

It should be noted that by taking the trace of (3.95), we obtain

−trA3+A(1)trA2

−A(2)trA + 3A(3) = 0, (3.97)

3.1. DEFORMATION 87

so that subsequently, using (3.92)–(3.94), we have

trA = A(1), (3.98)

trA2 = A2

(1) − 2A(2), (3.99)

trA3 = A3

(1) − 3A(1)A(2) + 3A(3). (3.100)

In addition, by taking the trace of (3.96) and using (3.93), it is easy to show thatwe can alternatively write

A(2) = detA trA−1. (3.101)

Lastly, from (3.89), we also have that

∂A(1)∂A

= I, ∂A(2)∂A

= A(1)I−AT ,∂A(3)∂A

= (A2−A(1)A +A(2)I)T = A(3) (A−1)T .

(3.102)In addition, any matrix A can be decomposed into a spherical or mean part,

1

3A(1)I, and a deviatoric part, A′:

A = 1

3A(1)I +A

′. (3.103)

Note that any tensor of the form α1, where α is a scalar, is known as a sphericaltensor. Now it is easy to show that

trA′ = 0, (3.104)

trA′2 = 2

3A2

(1) − 2A(2), (3.105)

trA′3 = 2

9A3

(1) −A(1)A(2) + 3A(3), (3.106)

and the invariants of the deviatoric part A′ are related1 to those of A as follows:

A′(1) = trA′ = 0, (3.107)

A′(2) = −1

2trA′

2 = −13A2

(1) +A(2), (3.108)

A′(3) = 1

3trA′

3 = 2

27A3

(1) −1

3A(1)A(2) +A(3). (3.109)

We note that the eigenvectors or principal directions of A′ are the same as thoseof A, while, if we denote the eigenvalues or principal values of A′ as s(i), then itis easy to show that

s(i) = λ(i) − 1

3(λ(1) + λ(2) + λ(3)) . (3.110)

Subsequently, we can also write the principal scalar invariants of A′ as

A′(1) = s(1) + s(2) + s(3) = 0, (3.111)

A′(2) = (s(1)s(2) + s(2)s(3) + s(3)s(1)) = −12(s(1)2 + s(2)2 + s(3)2) , (3.112)

A′(3) = s(1)s(2)s(3). (3.113)

1Some authors define A′(2)

as the negative of our definition. Our convention is consistent with

the definitions of the invariants of A given in (3.91)–(3.94).

88 KINEMATICS

We note that we can also write A′(2) in terms of the eigenvalues of A:

A′(2) = −16 [(λ(1) − λ(2))2 + (λ(2) − λ(3))2 + (λ(3) − λ(1))2] . (3.114)

3.1.8 Real symmetric matrices

Let A = [aik] = [aki] be a real symmetric matrix, and

A = aikiiik. (3.115)

We would like to transform it to

A = apqipiq, (3.116)

such that

[apq] = [aqp] =⎡⎢⎢⎢⎢⎢⎣λ(1) 0 0

0 λ(2) 0

0 0 λ(3)

⎤⎥⎥⎥⎥⎥⎦. (3.117)

Noting that dξi = (∂ξi/∂ξk)dξk and ds2 = dξidξi = dξkdξk, and since

∂ξi

∂ξk

∂ξi

∂ξldξkdξl = dξidξi, (3.118)

we have that

∂ξi

∂ξk

∂ξi

∂ξl= δkl. (3.119)

If we let

Rik ≡ ∂ξi∂ξk

, (3.120)

and R = [Rik], then we can write

RTR = I or RikRil = δkl. (3.121)

This shows that the matrix R is orthogonal since R−1 = RT . The matrix R is saidto be a proper orthogonal matrix if detR = 1 and an improper orthogonal matrix ifdetR = −1. In the following we will always assume that R is a proper orthogonalmatrix.

According to the transformation rule for components of a second-order tensor,we can now write that

A = RART or apq = RpiRqkaik. (3.122)

We now establish the following results for our matrix A:

3.1. DEFORMATION 89

1) If A is real and symmetric, then the eigenvalues λ(k) are real. To see this,we write

aikvk = λvi, (3.123)

a∗ikv∗k = λ∗v∗i , (3.124)

where the star superscripts denote complex conjugates. Now multiplying thefirst equation by v∗i and the second equation by vi, subtracting, and notingthat aik is symmetric and real and viv∗i is real, we obtain

(λ − λ∗)viv∗i = 0. (3.125)

Since vi ≠ 0 (and hence v∗i ≠ 0), then λ∗ = λ and thus λ is real.

2) Eigenvectors corresponding to distinct eigenvalues are orthogonal. To seethis, we write

aikv(α)k

= λ(α)v(α)i , (3.126)

aikv(β)k

= λ(β)v(β)i . (3.127)

Now multiplying the first equation by v(β)i and the second by v(α)i and pro-ceeding as before, we obtain

(λ(α) − λ(β)) v(α)i v(β)i = 0. (3.128)

If λ(α) ≠ λ(β), then v(α)i and v(β)i are orthogonal, i.e., v(α) ⋅ v(β) = 0.Note that if the eigenvalues are not distinct, say λ(1) = λ(2) = λ, then

aikv(1)k

= λv(1)i , (3.129)

aikv(2)k

= λv(2)i , (3.130)

and in this case the linear combination αv(1) + βv(2) is also an eigenvectorof λ for arbitrary α and β.

The matrix A is said to be positive definite if

aikv(α)i v

(α)k> 0 or v(α) ⋅A ⋅ v(α) > 0. (3.131)

Subsequently, since A ⋅ v(α) = Av(α) = λ(α)v(α), and since for any v(α) ≠ 0 wehave that v(α) ⋅v(α) > 0, it then follows that A is positive definite if λ(α) > 0. It issemi-definite if and only if λ(α) ≥ 0, in which case

aikv(α)i v

(α)k ≥ 0 or v(α) ⋅A ⋅ v(α) ≥ 0. (3.132)

The terms negative definite and negative semi-definite apply to tensors A whoseeigenvalues are negative definite and negative semi-definite, respectively. It canbe shown that if A is an arbitrary square matrix, then AAT = ATA are positivesemi-definite, and if A is invertible, then they are positive definite.

90 KINEMATICS

v1

v2

v3

v1

v2

v3

x

Figure 3.8: Rotation of principal axes.

If we now return to our transformation of matrix A, take v(i)k

to be the normal-ized eigenvectors of A, and define the orthogonal matrix as

Rik ≡ v(i)k, (3.133)

we then have

apq = RpiRqkaik

= Rpiv(q)kaik

= Rpiλ(q)v(q)i

= v(p)i λ(q)v(q)i

= λ(q)v(p)i v(q)i

=⎡⎢⎢⎢⎢⎢⎣λ(1) 0 0

0 λ(2) 0

0 0 λ(3)

⎤⎥⎥⎥⎥⎥⎦. (3.134)

The normalized eigenvectors correspond to the principal directions in the newcoordinate system as illustrated in Fig. 3.8, while the eigenvalues are denoted asprincipal values. Furthermore, in the new coordinate system, the invariants of A,given by (3.92)–(3.94), can now be written as

A(1) = λ(1) + λ(2) + λ(3), (3.135)

A(2) = λ(1)λ(2) + λ(2)λ(3) + λ(3)λ(1), (3.136)

A(3) = λ(1)λ(2)λ(3). (3.137)

We note that a symmetric tensor obviously has six independent componentsor degrees of freedom. When referred to principal axes, it still has 6 degreesof freedom: 3 are the directions of the principal axes and the other 3 are themagnitudes of the principal components.

3.1. DEFORMATION 91

It then follows that any symmetric tensor S of rank 2 can be written in termsof its principal values and directions:

S = 3

∑i=1

λ(i) vivi, (3.138)

and subsequently we can write

√S = 3

∑i=1

√λ(i) vivi, (3.139)

S−1 = 3

∑i=1

λ(i)−1

vivi, (3.140)

eS = 3

∑i=1

eλ(i)

vivi, (3.141)

logS = 3

∑i=1

logλ(i) vivi. (3.142)

3.1.9 Polar decomposition theorem

It is clear that any second-order tensor F can be decomposed into the sum ofsymmetric and skew-symmetric tensors:

F = sym F + skw F = 1

2(F +FT ) + 1

2(F −FT ) . (3.143)

This is sometimes called the Cartesian decomposition of a tensor.Another useful decomposition is given by the polar decomposition theorem which

states that for any real non-singular second-order tensor F, which we take here tobe the material deformation gradient, there exist real symmetric positive-definitetransformations U and V and a real orthogonal transformation R such that

F =R ⋅U =V ⋅R or F = R U = V R. (3.144)

Note that a positive-definite symmetric tensor represents a pure stretch deforma-tion along three mutually orthogonal axes, i.e., in the directions of the eigenvectors,while an orthogonal tensor represents a rotation. Therefore, the above states thatany local deformation is a combination of a pure stretch and a rotation: first stretchU and then rotate R, or first rotate R and then stretch V . The two decomposi-tions of the deformation gradient are illustrated in Fig. 3.9. We call R the rotationtensor, while U and V are called the right and the left stretch tensors. From thisdecomposition, we see that no deformation corresponds to U = V = 1 in whichcase the most general deformation gradient that does not lead to a deformation isgiven by F =R, i.e., a pure rotation.

The above theorem is proved as follows. Clearly we have, assuming that R is aproper orthogonal transformation,

U2 = FTF, V 2 = FFT , detU = detV = detF. (3.145)

Let the eigenvalues and eigenvectors of U be λ(k) and ek, respectively, so that

Uek = λ(k)ek. (3.146)

92 KINEMATICS

ds

ds

λ2ds

λ1ds

λ2ds

λ1ds

ds

ds

ds

ds

λ2ds

λ1ds

UR

R V

Figure 3.9: Polar decomposition of deformation gradient.

Then, since V = RURT , we have that

V (Rek) = RURT (Rek) = RUek = λ(k) (Rek) . (3.147)

In other words, V and U have the same eigenvalues and their eigenvectors differonly by the rotation R. The eigenvalues λ(k) are called the principal stretches,and the corresponding mutually orthogonal eigenvectors ek are called the principaldirections of stretch.

Given a non-singular deformation gradient F, it is convenient to introduce theright and left Cauchy–Green strain tensors respectively defined by

C = U2 = FTF and B = V 2 = FFT . (3.148)

Now it is clear that C (analogously B) is symmetric since

CT = (FTF )T = FTF = C, (3.149)

and positive definite since

CIK = FlIFlK = ∂xl

∂XI

∂xl

∂XK

, (3.150)

and for any vI ≠ 0CIKvIvK = FlIvIFlKvK =WlWl > 0, (3.151)

where Wl = FlIvI . Since C is a symmetric positive-definite matrix, then it haspositive eigenvalues, and the corresponding eigenvectors form an orthogonal basissuch that C can be written in the form

C =QTΛ2Q, (3.152)

3.1. DEFORMATION 93

where

Λ2 = [(λ(k))2] , (3.153)

Q = [ek] = [ekj] . (3.154)

The eigenvalues of U are the positive square roots of those of C associated withthe same eigenvectors. Subsequently we write

U ≡ C1/2 (3.155)

and call U the “square root” of C. In other words,

U = C1/2 = QTΛQ. (3.156)

Lastly, we define

R ≡ FU−1. (3.157)

Clearly R is orthogonal, since

RT = (FU−1)T = (U−1)T FT = (UT )−1 FT = (FU−1)−1 = R−1, (3.158)

where the equality in the next to the last step follows since

[(UT )−1 FT ] (FU−1) = U−1CU−1 = U−1UUU−1 = I, (3.159)

and we have also used the fact that

(U−1)T = (UT )−1 = U−1, (3.160)

which can be easily proved. Using a similar procedure, we can also demonstratethe decomposition

F = V R. (3.161)

Thus, if detF ≠ 0, then F = RU = V R, where R−1 = RT , and U = UT , and V = V T

are positive definite. We can also prove that the decompositions are unique, i.e.,if F = V R, then V = V , and R = R.

The above decomposition is a consequence of a more general theorem whichstates that every non-singular complex matrix can be uniquely written as a productof a positive-definite Hermitian matrix A and a unitary matrix B. To clarify theabove statement, we recall that a matrix A is Hermitian (or self-adjoint) if it isequal to its adjoint matrix A†, i.e., A = A†, where the adjoint matrix A† is definedas the transpose of the complex conjugate of A, i.e., A† = (A⋆)T , and A⋆ denotesthe complex conjugate of A. Furthermore, a matrix B is normal if BB† = B†B

and is unitary if BB† = B†B = I. Clearly, if such matrices are real, then weobtain the above enunciated polar decomposition theorem. We also note that ifthe complex matrix is of unit dimension, we then obtain the polar representationof a complex number: z = reiθ. The number r is a positive real number, which isthe magnitude of z, and the quantity eiθ is a complex number of magnitude unityrepresenting an angular rotation.

94 KINEMATICS

Example

Let C be given by

C = [ 3√2√

2 2] ,

which has eigenvalues (λ(1))2 = 4 and (λ(2))2 = 1 and corresponding eigen-

vectors e1 = (√2/3,√1/3) and e2 = (−√1/3,√2/3). Therefore, we have

Λ = [ 2 0

0 1] ,

Q = 1√3[ √2 1

−1√2] ,

and subsequently

U = C1/2 = QTΛQ = 1

3[ 5

√2√

2 4] .

One can easily verify that U2 = C.

Example

Consider the deformation x = χ(X) given in Cartesian coordinates, in boththe reference and the deformed configurations, by

x1 = X1 + κX2, (3.162)

x2 = X2, (3.163)

x3 = X3. (3.164)

This deformation, illustrated in Fig. 3.10, is called a simple shear and κ > 0is called the amount of shear. Now we have the deformation gradient

F =⎡⎢⎢⎢⎢⎢⎣1 κ 0

0 1 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦, (3.165)

and since J = detF = 1, simple shear is a volume-preserving, or isochoric,deformation. The right Cauchy–Green tensor is given by

C =⎡⎢⎢⎢⎢⎢⎣

1 κ 0

κ 1 + κ2 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦. (3.166)

From the eigenvalues and eigenvectors of C, we find the principal stretches

(λ(1,2))2 = 1 + 1

2κ2 ± κ

√1 +

1

4κ2, (λ(3))2 = 1, (3.167)

3.1. DEFORMATION 95

with corresponding principal directions of stretches

e1,2 = (12κ ±

1

2

√4 + κ2) i1 + i2, e3 = i3. (3.168)

We note that (λ(2))2 = 1/ (λ(1))2. From the square root of C, we obtainthe right stretch tensor

U =⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

2√4 + κ2

κ√4 + κ2

0

κ√4 + κ2

2 + κ2√4 + κ2

0

0 0 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦. (3.169)

Note that in the principal direction e1, the principal stretch, λ(1) > 1, is anextension, while in the direction e2, the stretch, λ(2) < 1, is a contraction.Similarly, we have

B =⎡⎢⎢⎢⎢⎢⎣1 + κ2 κ 0

κ 1 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦and V =

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

2 + κ2√4 + κ2

κ√4 + κ2

0

κ√4 + κ2

2√4 + κ2

0

0 0 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦. (3.170)

The rotation tensor can be calculated from R = FU−1:

R =⎡⎢⎢⎢⎢⎢⎢⎢⎢⎣

2√4 + κ2

κ√4 + κ2

0

−κ√4 + κ2

2√4 + κ2

0

0 0 1

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎦. (3.171)

If we denote θ = tan−1(κ/2), then R becomes

R =⎡⎢⎢⎢⎢⎢⎣

cosθ sin θ 0

− sin θ cosθ 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦, (3.172)

which is a clockwise rotation about the x3-axis by the angle θ.

3.1.10 Strain kinematics

From before we recall that from the deformation of a material line segment, wehave

dx(1) ⋅ dx(2) = (F ⋅ dX(1)) ⋅ (F ⋅ dX(2)) = (FT⋅F) ⋅ dX(1) ⋅ dX(2) =

C ⋅ dX(1) ⋅ dX(2). (3.173)

96 KINEMATICS

x2,X2

x1,X1

κ κ

2

θ

O 1

1

Figure 3.10: Simple shear.

Now the change in length and orientation between the current and reference con-figurations is

dx(1) ⋅ dx(2) − dX(1) ⋅ dX(2) = 2E ⋅ dX(1) ⋅ dX(2), (3.174)

where

E = 1

2(C − 1) or E = 1

2(C − I) (3.175)

is called the Green–St. Venant strain tensor, or the finite strain tensor in thereference configuration. Similarly, since dX = F−1 ⋅ dx, we also have

dX(1) ⋅ dX(2) = (F−1 ⋅ dx(1)) ⋅ (F−1 ⋅ dx(2)) = (F ⋅FT )−1 ⋅ dx(1) ⋅ dx(2) =B−1 ⋅ dx(1) ⋅ dx(2), (3.176)

and then

dx(1) ⋅ dx(2) − dX(1) ⋅ dX(2) = 2e ⋅ dx(1) ⋅ dx(2), (3.177)

where

e = 1

2(1 −B−1) or e = 1

2(I −B−1) (3.178)

is called the Almansi–Hamel strain tensor, or the finite strain tensor in the currentconfiguration.

We note that the mapping from the reference configuration frame to the de-formed configuration frame is given by the rigid frame transformation

x = b +Q ⋅X, (3.179)

where b and Q are the constant frame translation vector and orthogonal rotationtensor, respectively. Note that Q−1 = QT . Subsequently, the deformation in thecurrent configuration by displacement of a material point from X to X + u(X) isgiven by

x = b +Q ⋅ [X + u(X)], (3.180)

3.1. DEFORMATION 97

where u(X) is the displacement in the reference configuration:

u(X) =QT⋅ (x − b) −X. (3.181)

The material displacement gradient at X is a linear transformation defined by

H =H(X) ≡ (Grad u)T or HIJ = uI,J . (3.182)

It easily follows that

H =QT⋅F − 1. (3.183)

Analogously, the deformation in the reference configuration by displacement ofa material point in the current configuration from x to x − u(x) is given by

[x − u(x)] = b +Q ⋅X, (3.184)

where u(x) is the displacement in the current configuration:

u(x) = x − (b +Q ⋅X). (3.185)

Now, the spatial displacement gradient at x is a linear transformation defined by

h = h(x) ≡ (grad u)T or hij = ui,j . (3.186)

It easily follows that

h = 1 −Q ⋅F−1. (3.187)

Both strain tensors E and e vanish when there is no deformation, i.e., whenU = V = I, in which case we also see that F = R = Q and H = h = 0. For smalldeformations, these strains are, therefore, expected to be small. Since from abovewe have that the deformation gradient and its inverse, in terms of the materialand spatial displacement gradients, are given by

F =Q (I +H) and F −1 = QT (I − h), (3.188)

the finite strain tensors (3.175) and (3.178) can be rewritten in terms of the dis-placement gradients as

E = 1

2(H +HT

+HTH) and e = 1

2(h + hT − hTh) . (3.189)

Note that, using (3.175), (3.178), and (3.189), we can also write

C = I +H +HT+HTH and B−1 = I − h − hT + hTh. (3.190)

For small deformations, HTH and hTh are second-order quantities; thus ne-glecting these terms, we obtain the infinitesimal strain tensors

E ≡ 1

2(H +HT ) = 1

2[(Grad u) + (Grad u)T ] (3.191)

and

e ≡ 1

2(h + hT ) = 1

2[(grad u) + (grad u)T ] . (3.192)

98 KINEMATICS

The linear strain tensor E was introduced by Cauchy in the classical theory ofelasticity.

For small displacement gradients, the right stretch tensor and the rotation tensorcan be approximated by

U = (FTF )1/2 ≈ I + 1

2(H +HT ) = I + E, (3.193)

R = FU−1 ≈ Q [I + 1

2(H −HT )] = Q (I + R), (3.194)

where

R ≡ 1

2(H −HT ) = 1

2[(Grad u) − (Grad u)T ] (3.195)

is called the infinitesimal rotation tensor. Note that the infinitesimal strain androtation tensors correspond to the symmetric and skew-symmetric parts of thefinite displacement gradient! Also note that when we have no deformation, wehave that E = e = R = 0 and R = Q.

3.1.11 Compatibility conditions

In three dimensions the right and left Cauchy–Green tensors C and B and thefinite strain tensors E and e are all symmetric and hence have six independentcomponents, which can be written in terms of the gradient of the displacement u,which has three components:

2EIJ = CIJ − δIJ =HIJ +HJI +HKIHKJ = uI,J + uJ,I + uK,IuK,J , (3.196)

2eij = δij −B−1ij = hij + hji − hkihkj = ui,j + uj,i − uk,iuk,j . (3.197)

If we are given a displacement vector u which is differentiable, by differentiationand substitution in the previous equations we can obtain C, B, E, and e. If, onthe other hand, C, B, E, or e is given, then it’s not clear that we can obtainthe corresponding single-valued continuous displacement field u. From this stand-point, (3.196) or (3.197) correspond to an over determined system of six partialdifferential equations that may not possess a unique solution for u unless specificintegrability conditions are satisfied. We emphasize that if the displacement fieldis given, then the compatibility conditions are not needed. On the other hand,if the problem is formulated in terms of strain tensors, then such conditions arerequired to ensure compatibility with a single-valued differentiable displacementfield. We note that if the compatibility conditions are violated, the correspond-ing displacement field in the body is not unique and then the body may possessdislocations.

To find such conditions, we first note that both the undeformed and deformedbodies are embedded in a three-dimensional Euclidean space. If we consider thatthe deformation and inverse deformation

xi = χi(XI) and XI = χ−1I (xi) (3.198)

are nothing more than coordinate transformations from X to x and from x to X,then it is clear that the right Cauchy–Green tensor C plays the role of the metric

3.1. DEFORMATION 99

tensor in the curvilinear coordinate X while the inverse of the left Cauchy–Greentensor B−1 plays the role of the metric tensor in the curvilinear coordinate x. InSection 2.16 we have shown that in a Euclidean space the Riemann–Christoffeltensor vanishes. Thus, if we now correspondingly replace g and x with C and X

or g with B−1 in the Riemann–Christoffel tensor (2.240) and subsequently in theChristoffel symbol (2.220), we have

R(C)MIJK

= 0 and R(B)mijk

= 0. (3.199)

These correspond to 34 = 81 equations, respectively. If we now note the symmetryconditions (2.241), we can use (2.242) to write

S(C)PQ = 1

4ǫPMIǫQJKR

(C)MIJK = 0 and S(B)pq = 1

4ǫpmiǫqjkR

(B)mijk

= 0, (3.200)

where S(C) and S(B) are easily shown to be symmetric tensors; subsequently, theseyield six equations. However, these six equations are not all independent since theRiemann–Christoffel tensor satisfies Bianchi’s identities (2.243):

R(C)MIJK,L

+R(C)MIKL,J

+R(C)MILJ,K

= 0, (3.201)

with analogous identities for R(B)mijk

. Bianchi’s identities provide three additional

equations which restrict the six components of S(C) and S(B) to three degrees offreedom. If we subsequently use (3.175) or (3.178), the six equations are givenexplicitly by

EMI,JK +EJK,MI −EMJ,IK −EIK,MJ +

C−1PQ [(EIP,M +EMP,I −EMI,P ) (EJQ,K +EKQ,J −EJK,Q)−(EIP,K +EKP,I −EIK,P ) (EJQ,M +EMQ,J −EJM,Q)] = 0 (3.202)

and

emi,jk + ejk,mi − emj,ik − eik,mj −

Bpq [(eip,m + emp,i − emi,p) (ejq,k + ekq,j − ejk,q)−(eip,k + ekp,i − eik,p) (ejq,m + emq,j − ejm,q)] = 0. (3.203)

For the case of infinitesimal strains, all quadratic terms are small and thus weobtain, say,

emi,jk + ekj,mi − emj,ik − eki,mj = 0. (3.204)

Note that if we set

j(e)mkji

≡ emi,jk + ekj,mi − emj,ik − eki,mj = 0, (3.205)

then we have identically

j(e)mkji,l

+ j(e)mkil,j

+ j(e)mklj,i

= 0, (3.206)

100 KINEMATICS

a relation formally analogous to Bianchi’s identities (2.243) in flat Euclidean space.

The compatibility conditions (3.204) may be rewritten in the form j(e)mkji

= 0. Sub-

sequently, the conditions (3.204) may be divided into two sets of three conditions,

j(e)1212= j(e)

2323= j(e)

3131= 0 and j(e)

1213= j(e)

2321= j(e)

3132= 0, or, more explicitly,

e11,22 + e22,11 − 2e12,12 = 0, (3.207)

e22,33 + e33,22 − 2e23,23 = 0, (3.208)

e33,11 + e11,33 − 2e31,31 = 0, (3.209)

and

e12,23 + e23,12 − e22,31 − e31,22 = 0, (3.210)

e23,31 + e31,23 − e33,12 − e12,33 = 0, (3.211)

e31,12 + e12,31 − e11,23 − e23,11 = 0, (3.212)

such that if both sets are satisfied upon the boundary of a region, then the van-ishing of either set in the interior implies the vanishing of both sets.

When these conditions are satisfied, then the single-valued integral of the linearform of (3.197),

eij = 1

2(ui,j + uj,i) , (3.213)

exists and is given by

ui = u0i + Qjixj + bi, (3.214)

where u0i is any solution of (3.213), Qij is a skew-symmetric rotation tensor inde-pendent of xi, and bi is an arbitrary vector independent of xi. This means thatthe displacement field u is single valued and uniquely determined to within a rigidmotion.

3.2 Motion

The motion of a body B is regarded as a continuous sequence of configurations intime. Thus the motion χκ can be expressed as the map

χκ ∶ Bκ ×R → E3 or x = χκ(X, t). (3.215)

It represents a one-parameter family of deformations. The quantity of time hasphysical dimension of [T ], which is independent of [L]. More simply, by droppingthe subscript κ referring to the reference configuration for the time being, we write

x = χ(X, t) or xk = χk(XK , t) (3.216)

for every point in the reference configuration. Conversely, we write

X = χ−1(x, t) or XK = χ−1K (xk, t). (3.217)

For a fixed material point X with coordinate X, χ ∶ R → E3 is a curve called apath or trajectory of the material point.

3.2. MOTION 101

The coordinates X are assumed to be assigned once and for all to given parti-cles in the material. Since they are the coordinates of the particles at an arbitraryinitial time t0, they serve for all time as names for the particles of the material.The coordinates x, on the other hand, are thought of as assigned once and forall to a point in the Euclidean space where material body resides. They are thenames of places. The motion x = χ(X, t) chronicles the places that x is occupiedby the particles X in the course of time. Problems for which X and t are takenas independent variables are said to be set in the material description; those in x

and t, the spatial description. For purposes of interpretation, x = χ(X, t) shouldbe thought of purely as a continuous coordinate transformation. The materialdescription is an immediate extension of the scheme used in the mechanics ofmass-points, where the paths of the several distinct masses are traced, while thespatial description has no counterpart in elementary mechanics. While the mate-rial description is more fundamental, it leads to mathematical difficulties; this isthe reason why the spatial description is usually preferred. A fully general spatialdescription was first given by Euler; that is why it’s referred to as the Euleriandescription. The material description is called the Lagrangian description, eventhough it was also Euler who first formulated such description.

Analogously, components of a tensor of any order can now be written usingeither the material or the spatial descriptions

ψ(X, t) = ψ[χ−1(x, t), t] = ψ(x, t), (3.218)

or

ψ...(XK , t) = ψ...[χ−1K (xk, t), t] = ψ...(xk, t). (3.219)

Again, for notational simplicity, we drop the hat on the function and thus do notdifferentiate between a function and its values since the functional dependenceshould be obvious in applications or will be explicitly displayed when necessary.

Since J = detF ≠ (0,±∞), when we have motion, J has to remain of the samesign. Thus, as before, and without loss of generality, we assume that 0 < J <∞.

3.2.1 Velocity and acceleration

Given the motion χ, we can calculate the velocity of the particle at X in thereference configuration by

v ≡ x ≡ ∂x∂t∣X

= ∂χ(X, t)∂t

∣X

or vk ≡ xk ≡ ∂xk∂t∣XK

= ∂χk(XK , t)∂t

∣XK

, (3.220)

whose components can be written as

vi = vi(XK , t), in the material (Lagrangian) description,vi(xk, t), in the spatial (Eulerian) description, (3.221)

and we have introduced the material derivative of a tensor quantity z as

z ≡ dzdt. (3.222)

102 KINEMATICS

To make these descriptions clear, if ψ is a tensor field such that ψ = ψ(X, t), then

ψ(X, t) = ∂ψ(X, t)∂t

∣X

+ X ⋅Grad ψ(X, t) = ∂ψ(X, t)∂t

∣X

(3.223)

since X = 0 because X moves with the particle associated with X. On the otherhand, if we write ψ = ψ(x, t), then

ψ(x, t) = ∂ψ(x, t)∂t

∣x

+ x ⋅ grad ψ(x, t) = ∂ψ(x, t)∂t

∣x

+ v ⋅ grad ψ(x, t) (3.224)

since x = χ(X, t) and x = ∂χ/∂t∣X = v.Now, more generally, we define the material derivative for the components of a

tensor field of any order as

ψ... ≡ ∂ψ...(XK , t)∂t

∣XK

= ∂ψ...(xk, t)∂t

∣xk

+∂ψ...(xk, t)

∂xl∣t

∂xl

∂t∣XK

,

or, dropping the explicit indication of what is kept fixed in the differentiations,

ψ ≡ ∂ψ(X, t)∂t

= ∂ψ(x, t)∂t

+ v(x, t) ⋅ grad ψ(x, t) (3.225)

or

ψ... ≡ ∂ψ...(XK , t)∂t

= ∂ψ...(xl, t)∂t

+ vk(xl, t) ∂ψ...(xl, t)∂xk

, (3.226)

or using general tensor notation

ψ...... ≡ ∂ψ...

...

∂t+ vk ψ...

...,k = ∂ψ......

∂t+ vk [∂ψ...

...

∂xk+ Γ⋅k⋅ψ

...

... +⋯− Γ⋅

⋅kψ...... −⋯] . (3.227)

We note that sometimes the operator

D

Dt≡ ∂

∂t+ v(x, t) ⋅ grad (3.228)

is defined to indicate the material derivative when written in spatial coordinates.We will not use such convention here. We will just use the overdot accent todenote the material derivative; the form that it will take in the material or spatialdescription will be understood as indicated in (3.225). A motion is homogeneousif it is of the form

v = x = b(t) +Q(t) ⋅ x. (3.229)

We note that when v is given, we can also determine the identity of the materialpoint at x, but this requires solving the ordinary differential equations

dx

dt= v(x, t), (3.230)

subject to the initial conditions

x(t0) = χ(X, t0) =X. (3.231)

3.2. MOTION 103

A surface f(x, t) = 0 consisting of a set of particles is said to be a materialsurface if

f = ∂f∂t+ v ⋅ grad f = 0. (3.232)

A material boundary is a surface which the material does not cross. The materialinside a material surface is called a body. At a boundary, the normal componentof velocity xn is equal to the normal velocity of the boundary cn,

xn = x ⋅ n = c ⋅ n = cn on S . (3.233)

We will denote by N the normal component to the boundary S in the materialconfiguration. If cn = 0, the boundary is said to be stationary. If the velocityvector x is equal to the velocity of the surface c, i.e.,

x = c on S , (3.234)

then the material is said to adhere to the boundary, or is referred to as a no-slip/no-penetration condition. If in addition the boundary is stationary, then thiscondition becomes

x = 0 on S . (3.235)

A more extensive discussion of boundary conditions is given later in Section 5.12.If we take ψ → v in (3.225), we obtain the acceleration

a ≡ x ≡ ∂2x∂t2∣X

≡ v = ∂v∂t+ (v ⋅ grad)v (3.236)

or

ak ≡ xk ≡ ∂2xk∂t2∣XK

≡ vk = ∂vk∂t+ vj

∂vk

∂xj. (3.237)

The spatial velocity gradient at (x, t) is a linear transformation given by

L = L(x, t) ≡ (grad x)T = (grad v)T or Lij = vi,j . (3.238)

We now note that

F = ˙(Grad x)T ,= ˙(Grad χ(X, t))T ,= (Grad

∂χ(X, t)∂t

)T ,= (Grad x ⋅ grad

∂χ(X, t)∂t

)T ,= (grad x)T ⋅ (Grad x)T ,

104 KINEMATICS

or

F = L ⋅F. (3.239)

Above, we have used the overbar to span the terms affected by the dot operation.We now see that the spatial gradient of velocity is related to the rate of deformationby

(grad v)T = L = F ⋅F−1. (3.240)

Using (3.239), it is easy to obtain the following interesting result

J = J divv. (3.241)

A motion such that the volume occupied by any material region is unaltered, i.e.,J = 1, is called isochoric. It immediately follows that a motion is isochoric if andonly if its velocity is solenoidal, i.e., divv = 0.3.2.2 Path lines, stream lines, and streak lines

A point where x = 0 is called a stagnation point. A motion such that the velocityfield does not change with time, i.e., x = χ(x) is said to be steady. More generally,any quantity which is independent of time is said to be steady. For example, itis sometimes convenient to have the observer move at a constant speed V in thex-direction, in which case a quantity f can be rewritten as

f(x, y, z, t) = g(ξ, y, z), ξ = x − V t. (3.242)

Subsequently, we have that∂f

∂t= −V ∂g

∂ξ. (3.243)

The curve in space traversed by X as t varies is the path line of X:

x = χ(X, t) for X fixed and −∞ < t <∞. (3.244)

It also corresponds to the integral curve of the system

dx = v dt (3.245)

which passes through X at t = t0.Vector lines of the field v at time t are the stream lines. They correspond to

the integral curves

f1(x, t) = 0 and f2(x, t) = 0 (3.246)

of the system

dx1 ∶ dx2 ∶ dx3 = v1 ∶ v2 ∶ v3 for t = const. (3.247)

To see this, we note that we can rewrite (3.245) or (3.247) as

dx1

v1= dx2v2= dx3v3

, (3.248)

3.2. MOTION 105

or

v2dx1 − v1dx2 = 0, (3.249)

v3dx2 − v2dx3 = 0, (3.250)

v1dx3 − v3dx1 = 0, (3.251)

or more succinctly

ǫijkvjdxk = 0. (3.252)

Because only two of the equations (3.252) can be independent, we can write

gij(xk)dxj = 0, i = 1,2, j = 1,2,3. (3.253)

The solution of these equations are given by the integral curves (3.246).The streak line through x at time t is the locus at time t of all particles that at

any time, past or future, will occupy or have occupied the place x. If we write themotion in the forms x = χ(X, t) and X = χ−1(x, t), then the streak line throughx at time t is given parametrically by the locus of x, where

x = χ(χ−1(x, t′), t) for −∞ < t′ <∞. (3.254)

At a given place x and time t, the stream line through x, the path line of theparticle occupying x, and the streak line through x all have a common tangent.When the motion is steady, all three curves coincide, but in general for unsteadymotion, they are distinct. A stream line never crosses itself nor ends, exceptpossibly at a stagnation point. Since the stream lines are determined at a fixedinstant, singularities such as stagnation points may be generated or destroyed inthe course of time. The path lines and streak lines of an unsteady motion maycross themselves or double back upon themselves.

Example

Consider the plane motion whose spatial description is given by

v = x = (x1, x2) = ( x11 + t

,1) . (3.255)

By integrating the equivalent of (3.247) for this planar case, we get theequation for the stream lines, which at t0 = 0 go through x0 = (x10, x20):

(x2 − x20) − (1 + t) ln ∣ x1x10∣ = 0, (3.256)

which corresponds to (3.246) for the planar case. Equivalently, a parametricequation for the stream line at time t is obtained by integrating

dx

dτ= v(x, t), (3.257)

the solution of which is given by

x = (x1, x2) = (x10eτ/(1+t), x20 + τ) . (3.258)

106 KINEMATICS

Also note that if we eliminate the parameter τ in (3.258), we obtain (3.256).By integrating (3.245) with t0 = 0, we get the material description by takingx(0) =X:

x = (x1, x2) = (X1(1 + t),X2 + t) . (3.259)

Note that x = (x1, x2) = (X1,1), which upon substituting for X1 from(3.259), we recover (3.255). By eliminating t from the above, we get thepath line of the particle X:

x1 −X1x2 =X1 (1 −X2) . (3.260)

Thus, each particle moves in a straight line at constant speed, but thestream lines change in time according to (3.256). To get the streak linethrough x when t = 0, we need only hold x fixed in (3.260):

X1X2 −X1 (1 + x2) + x1 = 0. (3.261)

This is a hyperbola. To get the streak line through x at time t, we firstinvert (3.259) at time t′:

X = (X1,X2) = ( x1

1 + t′, x2 − t

′) . (3.262)

This gives the particle that occupies the place x at time t′. The place x

occupied by this particle at time t follows from (3.259):

x = (x1,x2) = (x1 1 + t

1 + t′, x2 + t − t

′) , (3.263)

which, by eliminating t′, corresponds to the hyperbolic curve

x1x2 − x1 (1 + x2 + t) + x1 (1 + t) = 0 (3.264)

that includes (3.261) as the special case with t = 0. The stream, path, andstreak lines are illustrated in Fig. 3.11.

3.2.3 Relative deformation

In practice, for a given motion, the reference configuration is often chosen as theconfiguration at some instant t = t0. For some media (e.g., fluids) this choice is notonly unnecessary but also inconvenient. The configuration can be chosen indepen-dently of any motion. It is more convenient to choose the current configurationat time t as the reference configuration and measure changes from an earlier timeτ ≤ t to this configuration.

Thus we denote the position of the material point X at time τ by ξ:

ξ = χ(X, τ). (3.265)

Subsequently we can write, with a little abuse of functional notation,

ξ = χ(χ−1(x, t), τ) ≡ (t)χ(x, τ) or ξα = χα(χ−1K (xk, t), τ) ≡ (t)χα (xk, τ) , (3.266)

3.2. MOTION 107

0 0.5 1 1.5 2 2.5 3−5

−4

−3

−2

−1

0

1

0 0.5 1 1.5 2 2.5 3−5

−4

−3

−2

−1

0

1

−1 0 1 2 3 4 5 6

0

1

2

3

4

5

6

0 1 2 3 4 5 6−1

−0.5

0

0.5

1

1.5

2

2.5

3

0 1 2 3 4 5 6−0.5

0

0.5

1

1.5

2

2.5

3

3.5

4

x2x2

x1x1

x2,X2x2,X2

x2,X2

x1,X1x1,X1

x1,X1

x10= 1

x10 = 1x10= 2x10 = 2

x10 = 4x10 = 4

x10= 6

x10 = 6

x10= 10

x10 = 10

x10= 14

x10 = 14

(0.5,2)

(1,2)

(1,2)

(2,2)

(2,2)

(3,2)

(3,2)(0.5,1)

(1,1)

(1,1)

(1,1)

(2,1)

(2,1)

(0.5,0)

(1,0)

(1,0)

(3,1)

(3,1)

(2,0)

(2,0)

(3,0)

(3,0)

path

line

stream line

streak line

(a) (b)

(c)

(d) (e)

Figure 3.11: Stream, path, and streak lines corresponding to the velocity field(3.255): (a) stream lines at t = 0, with x10 as indicated and x20 = 0; (b) streamlines at t = 1, with x10 as indicated and x20 = 0; (c) path lines of the particles X

given; (d) streak lines at t = 0 of all particles ever occupying x; (e) path line of theparticle (1,1), stream line at t = 0, and initial streak line through (1,1).

108 KINEMATICS

(a)

(b)

(c)

X, t0

ξ, τ

x, t

Figure 3.12: (a) Initial configuration at t = t0, (b) current configuration at timeτ ≤ t, and (c) reference configuration at time t.

where (t)χ is called the relative motion. The different configurations are illustratedin Fig. 3.12. Note that we use lowercase Greek subscripts for indices associatedwith the current configuration at time τ , and these indices become Latin lowercaseletters when we take the limit τ → t. Thus ξ is the place at time τ of the materialpoint, which at time t is located at x and at time t0 is located at X. In the fol-lowing, unless explicitly given, it will be understood that the relative deformationgradient is to be evaluated as x, but for notational simplicity, we will not displaythis functional dependence explicitly.

As before, we assume that the mapping from τ to t is one-to-one, i.e.,

(t)J(τ) = det [∂ξα∂xl] > 0. (3.267)

Note that (t)J(t) = 1.The relative deformation gradient at τ relative to time t, (t)F(τ), is defined by

(t)F(τ) ≡ (grad ξ)T or (t)Fαk(τ) ≡ ∂ξα∂xk

. (3.268)

We note that

(t)F(t) = 1 or (t)Fik(t) = δik, (3.269)

and if (t)F(τ) and (t′)F(τ) are two deformation gradients, where τ ≤ t′ ≤ t, thenthey are related by

(t)F(τ) = (t′)F(τ) ⋅ (t)F(t′). (3.270)

This follows since, by definition, we have

ξ = (t)χ(x, τ) = (t′)χ(x′, τ), (3.271)

and if x at time t is a reference configuration relative to x′ at time t′, then we alsohave

x = (t′)χ(x′, t). (3.272)

3.2. MOTION 109

The chain rule now gives

∂ξα

∂xl= ∂ξα∂x′

k

∂x′k∂xl

, (3.273)

which corresponds to (3.270). Note that if t → t0 corresponding to the originalconfiguration, and t′ → t, then we have

(t0)F(τ) = (t)F(τ) ⋅ (t0)F(t), (3.274)

or more explicitly

F (X, τ) = (t)F (x, τ) F (X, t). (3.275)

Example

Consider the following motion in the x1-x2 plane relative to the referenceconfiguration κ in the Cartesian coordinate system:

x = χ(X, t) = (X1et,X2(t + 1)) , (3.276)

with inverse

X = χ−1(x, t) = (x1e−t, x2t + 1) . (3.277)

Note that for this motion the reference configuration is the configurationof the body at the instant t = t0 = 0.Then, the relative deformation is given by

ξ = (t)χ (x, τ) = χ(χ−1(x, t), τ) = χ((x1e−t, x2t + 1) , τ) ,

= (x1eτ−t, x2 (τ + 1t + 1

)) . (3.278)

From the above, we now calculate the deformation gradients

F(t) = et i1i1 + (t + 1) i2i2, (3.279)

(t)F(τ) = eτ−t i1i1 + (τ + 1t + 1

) i2i2. (3.280)

One can also obtain the path of an arbitrary material point X0 in thismotion, by eliminating time t from the deformation function,

x2 = X20 (1 + ln x1

X10

) for X10 ≠ 0,x1 = 0 for X10 = 0. (3.281)

Note that by choosing four points X0 which would correspond to cornersof a square in the reference configuration, one can examine the image ofthis square at any instant in the motion.

110 KINEMATICS

Since the mapping is invertible, det[(t)Fik(τ)] ≠ (0,±∞), and using the polardecomposition theorem,

(t)F (τ) = (t)R(τ)(t)U(τ) = (t)V (τ)(t)R(τ). (3.282)

We note that (t)R(t) = (t)U(t) = (t)V (t) = I.We can also introduce the relative right and left Cauchy–Green strain tensors

by

(t)C(τ) = (t)FT (τ) ⋅ (t)F(τ) or (t)Ckl = ξα,kξβ,lδαβ , (3.283)

(t)B(τ) = (t)F(τ) ⋅ (t)FT (τ) or (t)Bαβ = ξα,kξβ,lδkl, (3.284)

and recognize that

(t)C(t) = (t)B(t) = 1. (3.285)

We also note that since

CKL(τ) = ξα,Kξβ,Lδαβ = ξα,kξβ,lxk,Kxl,Lδαβ = xk,K (t)Ckl xl,L, (3.286)

Bαβ(τ) = ξα,Kξβ,LδKL = ξα,kξβ,lxk,Kxl,LδKL = (t)Fα,kBkl (t)Fβ,l, (3.287)

the relationships between the absolute and relative Cauchy–Green tensors are givenby

C(τ) = FT (t) ⋅ (t)C(τ) ⋅F(t) and B(τ) = (t)F(τ) ⋅B(t) ⋅ (t)FT (τ), (3.288)

which reduce to the standard definitions when τ = t.3.2.4 Stretch and spin

While the deformation gradient measures the local deformation, the material timederivative of the deformation gradient measures the rate at which such changesoccur. Another measure for the rate of deformation is the spatial gradient ofvelocity. From before, they are related by L = (grad v)T = F ⋅F−1, where F is therate of change of deformation relative to the reference configuration. Similarly, wecan define the rate of change of deformation relative to the current configurationby

Lik(t) ≡ ∂vi∂xk

= ∂ξα∂xk∣τ=t

= ∂

∂xk(∂ξα∂τ)∣

τ=t= ∂

∂τ(∂ξα∂xk)∣

τ=t

=(t)Fαk(τ)∣τ=t = (t)Fik(t), (3.289)

or

L(x, t) = (t)F(t). (3.290)

In other words, the velocity gradient can also be interpreted as the rate of changeof deformation relative to the current configuration. This can also be seen by usingthe derivative of (3.275) with respect to τ .

3.2. MOTION 111

If we hold x and t fixed and take the derivative of the relative deformationgradient with respect to τ , using the polar decomposition, we obtain

(t)F(τ) = (t)R(τ) ⋅ (t)U(τ) + (t)R(τ) ⋅ (t)U(τ), (3.291)

and setting τ = t, we have

L(t) = (t)U(t) + (t)R(t) =D(t) +W(t). (3.292)

where we define

Dik(t) ≡ (t)Uik(t) = ∂

∂τ(t)Uαk(τ)∣τ=t (3.293)

and

Wik(t) ≡ (t)Rik(t) = ∂

∂τ(t)Rαk(τ)∣τ=t . (3.294)

The quantity D is called the stretch or rate of strain tensor, while the quantity W

is called the spin or rate of rotation tensor.Now we can easily see that since (t)Uαk(τ)∣τ=t is symmetric, so is (t)Uik(t), thus

the stretching tensor is symmetric, i.e.,

DT (t) =D(t) or Dki(t) =Dik(t). (3.295)

Furthermore, to see that the spin tensor is skew-symmetric, we first note that sinceR−1 = RT , we can write

(t)Rαk(τ) (t)Rαl(τ) = δkl, (3.296)

and differentiating with respect to τ , we have

(t)Rαk(τ) (t)Rαl(τ) + (t)Rαk(τ) (t)Rαl(τ) = 0. (3.297)

Now evaluating at τ = t, we obtain

Wik(t)δil + δikWil(t) = 0. (3.298)

Thus

WT (t) = −W (t) or Wki(t) = −Wik(t). (3.299)

Therefore, the decomposition of L(t),L(t) =D(t) +W(t) or Lik =Dik +Wik, (3.300)

corresponds to the Cartesian decomposition into symmetric and skew-symmetricparts of (gradv)T where we see that

D(t) = 1

2(L +LT ) = 1

2((grad v)T + grad v) or Dik = 1

2( ∂vi∂xk+∂vk

∂xi) , (3.301)

W(t) = 1

2(L −LT ) = 1

2((grad v)T − grad v) or Wik = 1

2( ∂vi∂xk−∂vk

∂xi) . (3.302)

112 KINEMATICS

3.2.5 Kinematical significance of D and W

We can write

dξα = (t)Fαk(τ)dxk (3.303)

or differentiating with respect to τ ,

˙dξα = (t)Fαk(τ)dxk. (3.304)

Evaluating the above at τ = t, we have

˙dxi = Lik(t)dxk. (3.305)

If we let dξα = lαdξ and dxk = lkdx, then

dξ = [(t)Cαk(τ)lαlk]1/2 dx, (3.306)

where (t)Cαk(τ) is the relative right Cauchy–Green tensor. Now

˙dξ = 1

2

(t)Cαk(τ)lαlk[(t)Cβn(τ)lβln]1/2 dx, (3.307)

and evaluating at τ = t, we obtain

˙dx = 1

2(t)Cik(t)lilkdx. (3.308)

Since

(t)C(τ) = (t)FT (τ) (t)F (τ), (3.309)

differentiating with respect to τ , we have

(t)C(τ) = (t)FT (τ) (t)F (τ) + (t)FT (τ) (t)F (τ), (3.310)

which, when evaluated at τ = t, gives

(t)C(t) = LT (t) +L(t) = 2D(t) or (t)Cik(t) = ∂vk∂xi+∂vi

∂xk= 2Dik(t). (3.311)

Thus

˙dx =Dik(t)lilkdx. (3.312)

Example

Suppose that we have li = (1,0,0). Then

D11 =˙dx

dx= ˙ln ∣dx∣. (3.313)

The quantity D11 is the rate of change in length per unit length of an

3.2. MOTION 113

element parallel to the x1 axis.

The relation between differential volumes at t and at τ is given by

dv(τ) = (t)J(τ) dv(t), (3.314)

and

˙dv(τ) = [ ∂

∂τ(t)J(τ)]dv(t). (3.315)

Now using the chain rule and (3.60), we can write

∂(t)J(τ)∂τ

= ∂(t)J(τ)∂ξα,k

∂2ξα

∂τ∂xk= (t)J(τ)∂xk

∂ξα

∂2ξα

∂xk∂τ, (3.316)

and at τ = t,˙

dv(t) = δki ∂vi∂xk

dv(t), (3.317)

so

˙dv(t) = ∂vi

∂xidv(t) =Dii(t) dv(t), (3.318)

or

˙dv(t)dv(t) =Dii(t). (3.319)

Thus, the trace of the stretch tensor corresponds to the local time rate of changeof the volume per unit volume.

In addition,

dξ(1)α dξ(2)α = (t)Fαk(τ)(t)Fαl(τ)dx(1)kdx(2)l, (3.320)

so

˙dξ(1)α dξ

(2)α = ((t)Fαk(τ)(t)Fαl(τ) + (t)Fαk(τ)(t)Fαl(τ))dx(1)k

dx(2)l, (3.321)

and at τ = t,˙

dx(1)i dx

(2)i = ( ∂vl

∂xk+∂vk

∂xl)dx(1)

kdx(2)l= 2Dkl dx

(1)kdx(2)l. (3.322)

Also at τ = t, and since l(1)α l(2)α = cos θ,

˙dx(1)i dx

(2)i = ( ˙

dx(1)dx(2) + dx(1) ˙dx(2)) cos θ − dx(1)dx(2)θ sin θ, (3.323)

and if we let θ = π/2, then we get

∣Dij ∣ = − θ2, i ≠ j, (3.324)

114 KINEMATICS

e3

ω

d33l3

d22l2

e2e1

d11l1l2 l1

l3

Figure 3.13: Decomposition of relative rate of deformation into rates of stretchand spin.

indicating that the off-diagonal terms of the stretch tensor provide information onthe local rate of relative rotation between material line elements.

Lastly, we note that

dξα = (t)Fαk(τ)dxk (3.325)

and

˙dξα = (t)Fαk(τ)dxk, (3.326)

so that at τ = t,˙dxi = ∂vi

∂xkdxk = (Dik +Wik)dxk, (3.327)

or

˙dxi

dxk=Dik +Wik. (3.328)

The decomposition is illustrated in Fig. 3.13.If the body is rigid (in this case, it is also isochoric), Dik = 0, and using (2.143)

we have

˙dxi =Wikdxk = ǫiklldxk = −ǫilkldxk = − ( × dx)i , (3.329)

so

˙dx = − × dx, (3.330)

3.2. MOTION 115

where is the rigid body angular velocity (see (2.147)). Above, we have used thefact that it is always possible to associate an axial vector, in our case the angularvelocity , with any second-order skew-symmetric tensor, in our case the spintensor W, by the relation

W = ⋅ ǫ or Wik =lǫlik. (3.331)

Subsequently, since

ǫpikWik = ǫpikǫlikl = 2δpll = 2p, (3.332)

we also have, as in (2.142),

= 1

2ǫ ∶W or p = 1

2ǫpikWik. (3.333)

Note that in this case stretching occurs in a direction perpendicular to the planecontaining and dx and forms a right-handed system.

From above, we have that

2p = ǫpikWik,

= −ǫpki [12( ∂vi∂xk−∂vk

∂xi)] ,

= −1

2(ǫpkivi,k − ǫpkivk,i) ,

= −ǫpkivi,k,

= − (∇ × v)p ,or, taking ω = −2,

ω = ∇ × v. (3.334)

The quantity ω is called the vorticity, and we see that its magnitude is twice theangular velocity . A motion for which the spin vanishes, i.e., W = 0, is calledirrotational.

Lastly, it is easy to see that

W(1) = trW = 0, W(2) = ⋅ = 1

4ω ⋅ω, and W(3) = detW = 0. (3.335)

The vorticity squared is called enstrophy and is a quantity directly related to thecontribution of spin in the local kinetic energy of the associated motion.

3.2.6 Kinematics and dynamical systems

We will presently extend the kinematical tools that we have developed and applythem to the study of dynamical systems. As we will see, this leads to an interestingapproach to understanding the dynamics of such systems.

116 KINEMATICS

The dynamical system

Consider the autonomous dynamical system

dx

dt= v(x(t)), x(t0) =X, (3.336)

where x is considered a spatial curvilinear coordinate in an n-dimensional Eu-clidean space En, and v(x) is the velocity at x(t). Under very general conditions,the solution of (3.336) is given by

x = χ(X, t), (3.337)

where we now think of X as coordinates from a reference frame in En, which wetake to be the same as that for x(t). We consider x(t) to be a field generatedby the reference field X of all possible initial conditions. Thus, our phase spacerepresents a continuum of all possible fields x ∈ En emanating from the continuumof all possible initial conditions X ∈ En.

Now note from (3.336) that the velocity is tangential to the trajectory. Differ-entiating the velocity (taking the material derivative), we have

dv

dt= (gradv)T ⋅ v = L ⋅ v, (3.338)

where L denotes the velocity gradient. Using the Cartesian decomposition, thevelocity gradient can be rewritten as

L =D +W. (3.339)

Locally, D and W represent the stretching and twisting of the tangent bundle oftrajectories in the neighborhood of the specific trajectory x(t) evolving from anypoint X and points near it.

Local tangent and normal spaces

As noted, since v(x) is tangential to the trajectory x(t), we can define the unittangential vector t as

t ≡ v∣v∣ . (3.340)

Subsequently, dividing (3.338) by ∣v∣ and manipulating the equation, we arrive atthe evolution equation for the tangent vector:

dt

dt= L ⋅ t − (t ⋅L ⋅ t) t = L ⋅ t − (t ⋅D ⋅ t) t, (3.341)

since t ⋅W ⋅ t = 0. Note that since t ⋅ t = 1, and thus t ⋅dt/dt = 0, we see that takingthe inner product of t with (3.341), the above equation is identically satisfied, asit should be.

Now, any unit vector normal to t, say ni for i = 1, . . . , n− 1, can be obtained byrequiring that

ni ⋅ t = 0 and ni ⋅ nj = 1 if i = j,0 if i ≠ j. (3.342)

3.2. MOTION 117

Note that dni/dt ⋅ nj = 0 for any i, j = 1, . . . , n − 1. Differentiating (taking thematerial derivative of) (3.342)1, and using (3.341) and (3.342)1, we obtain

dni

dt⋅ t = −ni ⋅

dt

dt= −ni ⋅ [L ⋅ t − (t ⋅D ⋅ t) t] = −ni ⋅L ⋅ t. (3.343)

Subsequently, it follows that

dni

dt= −ni ⋅L +Gni, (3.344)

where G is an arbitrary scalar function. To determine it, we take the inner productof this last equation with nj , from which we obtain

dni

dt⋅nj = −ni ⋅L ⋅ nj +Gni ⋅ nj . (3.345)

Now, using (3.342)2 and its derivative (see above), we see that G = nj ⋅ L ⋅ nj =nj ⋅D ⋅nj , nj ⋅W ⋅nj = 0, and ni ⋅L ⋅nj = 0 when i ≠ j. Subsequently, the evolutionequation for the normal vector (3.344) becomes

dni

dt= −ni ⋅L + (nj ⋅D ⋅ nj) ni. (3.346)

We note from (3.341) and (3.346) that the unit tangent and normal vectors canbe taken as their negatives (t → −t and ni → −ni) without changing the respectiveevolution equations.

Local deformations along the trajectory

We now examine the local deformations in the neighborhood of a specified trajec-tory x(t) by examining neighboring trajectories.

Three useful measures of deformation are the stretch in the tangential direction,the deformation of an area whose normal is in the tangential direction (say, howan initial circular area deforms into an ellipsoidal area as we move along thetrajectory), and how a local volume deforms (say, from an initial spherical volumeinto an ellipsoidal volume as we move along the trajectory).

To examine the stretch in the tangential direction, we look at the square of thearc length dx(t) (local differential element in En):

(dx)2 = dx ⋅ dx. (3.347)

Taking the material derivative, we find that

d

dt(dx)2 = 2dx ⋅ d

dt(dx) = 2dx ⋅ dv = 2dx ⋅ (∂v

∂x)T ⋅ dx = 2dx ⋅L ⋅ dx = 2dx ⋅D ⋅ dx.

(3.348)

Now, dividing through by (dx)2, and noting that

t = dxdx, (3.349)

we obtain the equation for the local relative tangent stretch rate:

ωt ≡ d

dt(ln dx) = d

dt(ln λ) = t ⋅D ⋅ t, (3.350)

118 KINEMATICS

where we have noted from (3.20) that the arc length is related to the length stretchratio by λ = dx/dX . In addition, we should recognize that ∣t ⋅D ⋅ t∣ = ∣D ∶ tt∣ ≤ ∣D∣,since ∣tt∣ = 1, so that

ωt ≤ ∣D∣ . (3.351)

It is noted that this equation is also readily obtained from (3.305) or (3.312). Also,it is obvious that ωt = λD is an extremum (or principal stretch) if t = tD, wheretD is a principal direction of D:

(D − λD 1) ⋅ tD = 0, (3.352)

since (3.350) corresponds to the necessary and sufficient conditions for the solutionof (3.352) when λD = ωt. In such case, the corresponding instantaneous rate ofrotation, from (3.352) and (3.341), is given by

dtD

dt=W ⋅ tD =w × tD, (3.353)

where w = ⟨W⟩ is the axial vector associated with the spin tensor W (see (2.148)).To examine the local relative change in the differential volume dv(t), one follows

a similar procedure as above, or alternatively, using (3.319), to obtain the localrelative volume stretch rate

ωV ≡ d

dt(ln dv) = d

dt(ln J) = tr D, (3.354)

where we have recognized from (3.52) that the volume stretch ratio is given byJ = dv/dV . Note that ∣trD∣ = ∣D ∶ 1∣ ≤ ∣D∣, since ∣1∣ = 1, so that

ωV ≤ ∣D∣ . (3.355)

Now, the easiest way to obtain the local relative change in the differential areads(t) whose normal is in the tangential direction is to first use (3.47):

dv = dxds so that ln dv = ln dx + ln ds. (3.356)

Then,d

dt(ln ds) = d

dt(ln dv) − d

dt(ln dx) = tr D − t ⋅D ⋅ t. (3.357)

But

tr D−t⋅D⋅t =D ∶ 1−D ∶ t t =D ∶ (1 − t t) =D ∶ nn = n⋅D⋅n = n−1

∑i=1

ni ⋅D⋅ni. (3.358)

Subsequently, we have that the local relative area stretch rate

ωn ≡ d

dt(ln ds) = d

dt(ln η) = n ⋅D ⋅ n = n−1

∑i=1

ni ⋅D ⋅ ni, (3.359)

where we have recognized from (3.44) that the area stretch ratio is given by η =ds/dS. It is noted that ∣n ⋅D ⋅n∣ = ∣D ∶ nn∣ ≤ ∣D∣, since ∣nn∣ = 1, so that

ωn ≤ ∣D∣ . (3.360)

3.2. MOTION 119

3.2.7 Internal angular velocity and acceleration

In the next chapter, we will formulate balance laws for a polar material. Suchmaterial is characterized kinematically by an internal angular velocity (an axialvector field), ν, that is independent of the translational velocity field. As notedearlier, for an ordinary continuum, the angular velocity field is equal to one-halfof the vorticity (or the curl of the velocity field):

= −12ω = −1

2∇ × v or i = −1

2ωi = 1

2ǫijk vj,k. (3.361)

We interpret the usual angular velocity as an average angular velocity at lo-cation x and time t, while the internal angular velocity ν represents the angularvelocity of the polar-material particle at the same location and time. The inter-nal angular velocity can also be represented by the second rank skew-symmetricrotation tensor

Υ = ν ⋅ ǫ or Υjk = Υ[jk] = νi ǫijk. (3.362)

The internal or particle angular velocity relative to the average local angular ve-locity is given by

Θ ≡ ν − or Θi = νi −i. (3.363)

The quantity Θ is called the relative angular velocity. In the case of irrotationalinternal motion, the particle angular velocity, ν, vanishes, and we note that in thiscase Θ = −. This type of motion is more restrictive than ordinary irrotationalmotion, which requires that vanish. Stationary motion requires that the velocityfield, and subsequently the average angular velocity, , vanish. In this case,Θ = ν, so that the material particle is stationary but rotating. The necessary andsufficient conditions for rigid-body motion are that the stretch tensor D and therelative angular velocity Θ both vanish.

Analogous to the translational acceleration, v, and the velocity gradient, L =(gradv)T , we define the internal angular acceleration by ν, and the internal an-gular velocity gradient by

Ξ ≡ (grad ν)T or Ξkl ≡ νk,l. (3.364)

Note that Ξkl are components of an axial second rank tensor and associated withit are the components of a third-rank tensor that is anti-symmetric with respectto its first pair of indices,

Ξijl = Ξkl ǫkij and Ξijl = −Ξjil, (3.365)

and thus does not correspond to a general tensor of rank 3. Its irreducible symme-try parts can be found by substituting Ξ in the relations (2.135)–(2.138) to findthat

(Ξijl)1= (Ξ(ijl)) = 0, (3.366)

(Ξijl)2= (Ξ[ijl]) = ξ ǫijl, (3.367)

(Ξijl)3= 1

3![Ξ(ij)l − Ξ(jl)i] = 0, (3.368)

(Ξijl)4= 1

3![Ξ(i∣j∣l) − Ξ(j∣i∣l)] = Ξ′kl ǫkij , (3.369)

120 KINEMATICS

where we see that (Ξijl)2

has one component and (Ξijl)4

has eight components.Furthermore, we note that

ν(0) = 1

3trΞ = 1

3divν and Ξ′ = Ξ − 1

3(trΞ)1, (3.370)

so that ν(0) 1 and Ξ′ are the spherical and deviatoric parts of Ξ.

3.3 Objective tensors

We use concepts associated with space and time to describe the motion of materialobjects based on our experience. Based on this experience, we ascribe certainfundamental properties to space and time. Specifically, we consider the propertiesof homogeneity and isotropy of space and time.

Space homogeneity means that a location in space is identical to any other lo-cation of space. That is, any physical process will occur the same way no matterwhere it occurs; i.e., under identical initial conditions, an experiment will yieldthe same result no matter where it is conducted. Shifting the origin means dis-placing the system. The implication of this is that we can choose the origin of ourcoordinate system anywhere we wish without affecting processes.

Space isotropy means that one direction in space is equivalent to any other di-rection. A particular experiment will yield the same result whether the laboratoryis pointing north or east. That is, an arbitrary rotation of our coordinate systemshould not change the internal state of an isolated system, and thus, we should beable to orient our coordinate system any way we wish.

Time homogeneity means that one instant (or duration) of time is identical toany other instant (or duration) of time. An experiment should yield the sameresult when performed under the same conditions independent of the time of theday or the day of the year it is performed. Homogeneity of time implies thatwe can choose the initial time for an observation of a physical process to be anyinstant of time we desire.

Time isotropy means the equivalence of time directions. That is, the futuredirection is equivalent to the past direction. While such isotropy applies to thedynamics of Newtonian particles, processes consisting of a large number of particlesdo not show time reversibility and thus do not happen in nature. For complexmacroscopic systems, this concept is replaced by the concept of entropy and thesecond law of thermodynamics, which are discussed in Chapters 4 and 5.

A frame of reference can be interpreted as an observer who observes an event interms of positions and time with a ruler, a protractor, and a clock. Different ob-servers may use different rulers, protractors, and clocks and come up with differentresults for the same event. However, if the same units of measure for their rulers,protractors, and clocks are used, they should obtain the same distance, angle, andtime lapse between any two events under observation, even though the values oftheir observations may still be different. We shall impose these requirements on achange of frame from one to another.

In the formulation of physical laws, it is desirable to use quantities that areindependent of the motion of the observer. Such quantities are called objectiveor frame indifferent or frame invariant since they reflect the objective propertiesof the object they embody. Such quantities are represented by tensors of various

3.3. OBJECTIVE TENSORS 121

orders and they should be independent of the choice of the coordinate systemin which they are expressed. Within the realm of classical mechanics, the mostgeneral transformation which represents the homogeneity and isotropy of spaceand homogeneity of time is a time-dependent rigid transformation, referred to asthe Euclidean transformation. Under such transformation, lengths, angles, andtime lapses are preserved. Since we are staying within the realm of classical me-chanics, we require that physical quantities must be objective with respect totime-dependent rigid motions of the spatial frame of reference. This is known asthe principle of objectivity. Let a Cartesian frame F be in relative rigid motionwith respect to another frame, F ′. A point with Cartesian coordinates x at time tin F will have the Cartesian coordinates x′ at time t′ in F ′. Since the frames arein rigid motion with respect to one another, the two motions x(X, t) and x′(X, t′)are equivalent if

x′(X, t′) = b(t) +Q(t) ⋅ x(X, t) or x′i(Xj, t′) = bi(t) +Qik(t)xk(Xj , t) (3.371)

with

t′ = a + t, (3.372)

where

b(t0) = x′0 −Q(t0) ⋅ x0 and a = t′0 − t0 (3.373)

represent the position vector between the origins of the two frames and the initialtime shift, respectively, x0 ∈ E3 and x′0 ∈ E3 correspond to the absolute coordinatesof the origins of F and F ′, t0 ∈R and t′0 ∈R represent the absolute time coordinatesin the two frames, and Q(t) = grad x′ is the orthogonal rotation tensor, so that

Q ⋅QT =QT⋅Q = 1 or QijQkj =QjiQjk = δik. (3.374)

The transformation between the two frames is illustrated in Fig. 3.14. Note that band Q are functions of time only, and detQ(t) = ±1. We recognize that the matrixQ(t) has components Qij corresponding to direction cosines between the axes x′iand xj . We note that under this transformation, lengths, angles, and time lapsesremain invariant; e.g.,

ds′2 = dx′kdx′k = QkidxiQkjdxj = δijdxidxj = dxidxi = ds2. (3.375)

Now any tensorial quantity is said to be objective, or frame indifferent, if in anytwo objectively equivalent rigid motions, it obeys the appropriate tensor trans-formation law for all times. In such case, we say that objective quantities areinvariant under a change of observers. Thus, we recall that if u is a vector andT is a second-order tensor, then for them to be objective, they must satisfy therelations

u′(X, t′) =Q(t) ⋅ u(X, t) or u′k(X, t′) = Qkl(t)ul(X, t) (3.376)

and

T′(X, t′) =Q(t) ⋅T(X, t) ⋅QT (t) or T ′kl(X, t′) = Qkm(t)Qln(t)Tmn(X, t).(3.377)

122 KINEMATICS

i3

i2

i1

i′3

i′1

i′2

b

x′

x

Figure 3.14: Rigid translation and rotation of Cartesian frames.

More generally, a tensor ψ of order n must satisfy the relation

ψ′i1i2⋯in(X, t′) =Qi1j1(t)Qi2j2(t)⋯Qinjn(t)ψj1j2⋯jn(X, t). (3.378)

Example

For any pair of coordinates x1 and x2 in F , we can associate the distancevector u such that

u = x2 − x1. (3.379)

Alternately, in F ′ we have

u′ = x′2 − x′1. (3.380)

Subsequently, using the Euclidean transformation (3.371), we have

u′ =Q(t) ⋅ (x2 − x1) =Q(t) ⋅ u, (3.381)

verifying that indeed distance vectors are objective quantities.

3.3.1 Apparent velocity

The velocity in F ′ is related to that in F by

x′i = bi(t) + Qik(t)xk +Qik(t)xk. (3.382)

We immediately recognize that the velocity does not transform like a tensor, so itis not objective. Now we can rewrite

x′i = bi + QimQlmQlkxk +Qikxk = bi +Ωil (x′l − bl) +Qikxk, (3.383)

3.3. OBJECTIVE TENSORS 123

where we have used (3.371) and (3.374), and have defined the relative frame rota-tion tensor, or frame spin tensor, by

Ω(t) ≡ Q(t) ⋅QT (t) or Ωil ≡ QikQlk. (3.384)

We want to show that ΩT = −Ω or Ωil = −Ωli. Now

QikQlk = δil, (3.385)

and differentiating,

QikQlk +QikQlk = 0, (3.386)

or

Ωil = −Ωli. (3.387)

Subsequently, using (2.142) and (2.143), we can write

Ωil = wpǫpil and wp = 1

2ǫplmΩlm, (3.388)

where w is the frame angular velocity of F ′ with respect to frame F . Thus, wewrite

x′ = ˙x + v′ or x′i = ˙xi + v′

i, (3.389)

where

(x′ − v′) ≡ ˙x =Q ⋅ x or (x′i − v′i) ≡ ˙xi = Qikxk, (3.390)

and

v′ ≡ b −w × (x′ − b) or v′i ≡ bi − ǫiplwp (x′l − bl) . (3.391)

We note that x′ is the true velocity while ˙x is called the apparent velocity. Thedifference between these two velocities is given by the inertial velocity v′ of frameF ′ relative to frame F . Note that while the true velocity is not an objectivequantity, the apparent velocity is.

Furthermore, the internal angular velocity ν (an axial vector) at a material pointin frame F is related to that in frame F ′ by

(ν ′ −w) ≡ ν = (detQ)Q ⋅ ν. (3.392)

It is pointed out that ν′ is the true internal angular velocity while ν is called theapparent internal angular velocity. The difference between these two velocities isgiven by the frame angular velocity w of frame F ′ relative to frame F . Note thatwhile the true internal angular velocity is not an objective quantity, the apparentinternal angular velocity is.

In general, the true velocity and angular velocity are objective if and only ifv′ = 0 and w = 0, which can be easily shown to lead to the requirements thatb = b0 = const. and Q =Q0 = const. Subsequently,

x′(X, t′) = b0 +Q0 ⋅ x(X, t′), (3.393)

t′ = a + t, (3.394)

which is called a time-independent rigid transformation.

124 KINEMATICS

3.3.2 Apparent acceleration

In transforming the acceleration from frame F to frame F ′, we obtain

x′i = bi + Qikxk + 2Qikxk +Qikxk, (3.395)

which again does not transform like a tensor. Now, using (3.374), we have

x′i = bi + QimQlmQlkxk + 2QimQlmQlkxk +Qikxk. (3.396)

Differentiating the frame spin tensor, we also have

Ωil = QikQlk + QikQlk,

= QikQlk + QipQspQskQlk,

= QikQlk +ΩisΩls,

so that, also using (3.371) and (3.384), (3.396) becomes

x′i = bi + (Ωil −ΩisΩls) (x′l − bl) + 2Ωil(x′l − bl) + 2ΩisΩls (x′l − bl) +Qikxk,

= bi + Ωil (x′l − bl) + 2Ωil(x′l − bl) −ΩisΩsl (x′l − bl) +Qikxk. (3.397)

Thus, in terms of the frame’s angular velocity, we have

x′ = x + i′, (3.398)

where

(x′ − i′) ≡ x =Q(t) ⋅ x, (3.399)

and

i′ ≡ b − w × (x′ − b) − 2w × (x′ − b) −w × [w × (x′ − b)] . (3.400)

We note that x′ is the true acceleration while x is called the apparent acceleration.The difference between these two accelerations is given by the inertial accelerationi′, which consists respectively of the inertial acceleration of relative translation ofthe frames, the Euler acceleration, the Coriolis acceleration, and the centripetalacceleration (its negative is also called the centrifugal acceleration).

We also note that the internal angular acceleration ν at a material point inframe F is related to that in frame F ′ by

(ν ′ − υ′) ≡ ν = (detQ)Q ⋅ ν, (3.401)

whereυ′ ≡ w +Ω ⋅ (ν′ −w) = w −w × (ν ′ −w) . (3.402)

It is pointed out that ν ′ is the true internal angular acceleration while ν is calledthe apparent internal angular acceleration. The difference between these two an-gular accelerations is given by the inertial internal angular acceleration υ′, whichconsists respectively of the angular acceleration of relative rotation of the framesand the Coriolis angular acceleration.

3.3. OBJECTIVE TENSORS 125

Now we know that the location of a point will appear different to observerslocated at different places. Similarly, as we have seen, the velocity of a pointis dependent upon the velocity of the observer. Therefore, these quantities arenot objective (however, we note that the apparent quantities are). On the otherhand, the distance between two points and the angles between two directions areindependent of the rigid motion of the frame of reference (observer).

Note that in general the translational and internal angular accelerations areobjective if and only if i′ = 0 and υ′ = 0, which can be shown to lead to Q =Q0 =const. and b(t) = b0+Vt, where b0 = const. and V = const. Subsequently, we have

x′(X, t′) = b0 +Vt +Q0 ⋅ x(X, t), (3.403)

t′ = a + t, (3.404)

which is called a Galilean transformation. A Galilean frame differs from a time-independent rigid frame by the constant translation velocity V. Newton’s secondlaw of motion is known to be valid only in this special frame of reference, alsoknown as the inertial frame. Note that the velocity is not frame indifferent withrespect to the Galilean transformation, but is so only under the more restrictivetime-independent rigid transformation.

Einstein removed these restrictions by examining frame-invariance propertiesof arbitrary four-dimensional space-time transformations in his work on generalrelativity.

3.3.3 Properties of kinematic quantities

In classical mechanics, physical properties of materials should not depend on thecoordinate frame selected. Properties should be the same whether or not theobserver is in motion. Thus, the evolution equations as well as the constitutiveequations that we address later must be objective, or frame invariant, with respectto rigid motions of the spatial, or Euclidean, frame of reference. We refer to thisas objectivity or frame indifference.

In order to see how a reference configuration may be affected by a change offrame, we choose the reference configuration as the configuration occupied by somebody at time t0, so that for some arbitrary point XK

xi = χi(XK , t).By noting that at t = t0 we have that x =X, it now follows that in the new frame

X ′L = χ′L(XK , t′

0) = bL(t0) +QLK(t0)XK .

On the other hand, the motion relative to the change of frame is given by

x′i(X′, t′) = bi(t) +Qik(t)xk(X, t). (3.405)

The deformation gradient in the new frame is then given by

F ′iK(X′, t′) = ∂x′i∂X ′K

= ∂x′i∂xj

∂xj

∂XL

∂XL

∂X ′K.

126 KINEMATICS

Subsequently, and more simply, we see that the deformation gradient in the newframe is given by

F′(t) =Q(t) ⋅F ⋅QT (t0) or F ′iK = Qij(t)FjLQKL(t0), (3.406)

and is thus not objective. The deformation gradient is not an absolute tensor; itis referred to as a two-point tensor or a double vector, since it is a quantity thattransforms as a vector with respect to each of the indices.

With polar decompositions of F and F′, we also have

R′ ⋅U′ =Q(t) ⋅R ⋅U ⋅QT (t0) and V′ ⋅R′ =Q(t) ⋅V ⋅R ⋅QT (t0).By the uniqueness of such decompositions, we find that

U′ =Q(t0) ⋅U ⋅QT (t0), V′ =Q(t) ⋅V ⋅QT (t), R′ =Q(t) ⋅R ⋅QT (t0), (3.407)

and subsequently

C′ =Q(t0) ⋅C ⋅QT (t0), B′ =Q(t) ⋅B ⋅QT (t). (3.408)

Therefore, we conclude that V and B are objective tensors, while R, U, and C

are not objective tensors.Moreover, if we take the material derivative of the deformation gradient in the

moving frame, we have

F′ =Q(t) ⋅ F ⋅QT (t0) + Q(t) ⋅F ⋅QT (t0),and since F = L ⋅F, we have

L′ ⋅F′ = Q(t) ⋅L ⋅F ⋅QT (t0) + Q(t) ⋅F ⋅QT (t0),= Q(t) ⋅L ⋅QT (t) ⋅F′ + Q(t) ⋅QT (t) ⋅F′,

or, using (3.384) and since F′ is non-singular,

L′ =Q(t) ⋅L ⋅QT (t) +Ω(t). (3.409)

Lastly, since L =D +W, the above becomes

D′ +W′ =Q(t) ⋅ (D +W) ⋅QT (t) +Ω(t).By separating symmetric and skew-symmetric parts, we obtain

D′ =Q(t) ⋅D ⋅QT (t) or D′ik = QipQkqDpq, (3.410)

and

W′ =Q(t) ⋅W ⋅QT (t) +Ω(t) or W ′

ik = QipQkqWpq +Ωik. (3.411)

Therefore, the rate of strain tensor D is an objective quantity, while the velocitygradient L and the rate of rotation tensor W are not objective quantities.

Note that since the average angular velocity and the internal angular velocitytransform as

′ = (detQ)Q ⋅ +w and ν ′ = (detQ)Q ⋅ ν +w, (3.412)

3.3. OBJECTIVE TENSORS 127

the relative angular velocity and internal angular velocity gradient transform asobjective axial quantities (see (3.363) and (3.364)):

Θ′ = (detQ)Q ⋅Θ and Ξ′ = (detQ)Q ⋅Ξ ⋅QT . (3.413)

Now assume that we have an objective vector field u(x, t), so that

u′(x′, t′) =Q(t) ⋅u(x, t).Taking the gradient with respect to x, we have

grad u′(x′, t′) = (grad′ u′) ⋅ (grad x′) =Q(t) ⋅ grad u(x, t).But since Q(t) = grad x′, we have

(grad u)′ =Q(t) ⋅ (grad u(x, t)) ⋅QT (t). (3.414)

On the other hand, if we express this vector field in the material coordinate

u′(X′, t′) =Q(t) ⋅u(X, t),then by taking the gradient with respect to X, we easily find that

(Grad u)′ =Q(t) ⋅ (Grad u) ⋅QT (t0). (3.415)

Furthermore, if we take the material derivative of the vector field, we have

u′ = Q(t) ⋅ u + Q(t) ⋅ u,= Q(t) ⋅ u + Q(t) ⋅QT (t) ⋅ u′,= Q(t) ⋅ u +Ω(t) ⋅ u′. (3.416)

Therefore, if u is an objective vector field, then its spatial gradient, grad u, isan objective quantity, while its material gradient, Grad u, and its material timederivative, u, are not objective quantities.

If φ is an objective scalar field, then we easily find that

φ′ = φ, (grad φ)′ =Q(t) ⋅ (grad φ) , (Grad φ)′ =Q(t0) ⋅ (Grad φ) , (3.417)

so that the material derivative and the spatial gradient are objective, while thematerial gradient is not.

Similarly, we can show that if ψ is an objective tensor field of order n, then thematerial derivative ψ is not objective for n > 0, the spatial gradient grad ψ is anobjective tensor field of order n + 1, while the material gradient Grad ψ is not anobjective tensor quantity.

It should be noted that our analysis of frame invariance is based on the Eu-clidean transformation (3.371)–(3.372) since it is expected that physical quantitiesshould be invariant under such transformation. We observe that if the referenceconfiguration is unaffected by the change of frame, then Q(t0) = 1. Lastly, it iseasy to see that a number of quantities (e.g., F and C) that are not invariant un-der the Euclidean transformation are invariant under the more restrictive Galileantransformation (3.403)–(3.404), since then Q(t) =Q(t0) =Q0.

128 KINEMATICS

3.3.4 Corotational and convected derivatives

Suppose we have the objective vector field u so that

u′i = Qik(t)uk. (3.418)

As we have seen, the material derivative of an objective vector field u is notobjective. However, let’s look at the vector

u ≡ u −W ⋅u or

ui ≡ ui −Wikuk, (3.419)

and see how it transforms. Using (3.411), we have

u′

i = u′i −W ′

iku′

k = Qikuk +Qikuk −ΩikQklul −QipQklQksWplus,

= Qikuk +Qikuk − QipQkpQklul −QipδlsWplus,

= Qikuk +Qikuk − Qipup −QipWplul,

= Qik (uk −Wklul) ,= Qik

uk. (3.420)

Thus we find that the quantityu, called the corotational time derivative, does

transform like a tensor. Such derivative of an objective vector field is not unique.For example, one can easily verify that the convected time derivative

⊙u ≡ u −L ⋅u or

⊙ui ≡ ui −Likuk (3.421)

is also objective.

Analogously, for a second-order objective tensor field T, using the same proce-dure as above, one can show that the quantity

T ≡ T −W ⋅T +T ⋅W or

T ij ≡ Tij −WikTkj + TikWkj , (3.422)

called the corotational or Jaumann derivative, transforms like an objective second-order tensor. Again we note that the definitions of such derivatives are not unique.For example, it can also be shown that the convected rates given by the Oldroydtensor

⋆

T ≡ T −T ⋅LT−L ⋅T or

⋆

T ij ≡ Tij − TikLjk −LikTkj , (3.423)

the Truesdell tensor

T ≡ T −LT⋅T −T ⋅L + (tr L)T or

T ij ≡ Tij −LkiTkj − TikLkj +LkkTij , (3.424)

as well as the Cotter–Rivlin tensor

T ≡ T +LT⋅T +T ⋅L or

T ij ≡ Tij +LkiTkj + TikLkj (3.425)

are also objective.

3.3. OBJECTIVE TENSORS 129

3.3.5 Push-forward and pull-back operations

Transformations between material and spatial descriptions are sometimes calledpush-forward and pull-back operations. A push-forward operation transforms atensor-valued quantity based on the reference configuration to the current config-uration. A pull-back operation transforms a tensor-valued quantity based in thecurrent configuration to the reference configuration. A pull-back operation is aninverse of the push-forward operation.

Consider the Green–St. Venant strain tensor E, which is defined in the referenceconfiguration. From it, it is possible to compute the corresponding Almansi–Hamelstrain tensor e in the current configuration by a push-forward operation. To affectthis, we rewrite (3.178) as follows:

e = 1

2(1 −B−1)

= 1

2[1 − (F ⋅FT )−1]

= 1

2[1 −F−T ⋅F−1]

= 1

2F−T ⋅ [FT

⋅ (1 −F−T ⋅F−1) ⋅F] ⋅F−1= 1

2F−T ⋅ (FT

⋅F − 1) ⋅F−1= 1

2F−T ⋅ (C − 1) ⋅F−1

= F−T ⋅E ⋅F−1

≡ χ⋆(E) , (3.426)

where we have used (3.175). Note that F−1 maps the current configuration intothe reference configuration, E maps the reference configuration into the referenceconfiguration, and F−T maps the reference configuration into the current configu-ration. The operator χ

⋆() is the push-forward operator.

The inverse, or pull-back operation of e, from (3.175), is given from

E = 1

2(C − 1)

= 1

2(FT

⋅F − 1)= 1

2FT⋅ [F−T ⋅ (FT

⋅F − 1) ⋅F−1] ⋅F= 1

2FT⋅ (1 −F−T ⋅F−1) ⋅F

= 1

2FT⋅ [1 − (F ⋅FT )−1] ⋅F

= 1

2FT⋅ (1 −B−1) ⋅F

= FT⋅ e ⋅F

≡ χ−1⋆(e) , (3.427)

130 KINEMATICS

where we have used (3.178). Note that F maps the reference configuration into thecurrent configuration, e maps the current configuration into the current configura-tion, and FT maps the current configuration into the reference configuration. Theoperator χ−1

⋆() is the pull-back operator.

The above push-forward and pull-back operators can be applied to push-forwardor pull-back other corresponding rank 2 quantities. Analogous push-forward andpull-back operators can be obtained for tensors of other ranks. The push-forwardand pull-back operators provide relationships between the same type of compo-nents (contravariant, covariant, or mixed) between the current and reference con-figurations. Thus, for a vector v in the reference configuration, there are twopossible push-forward operators (F−T ⋅ v for covariant components and F ⋅ v forcontravariant components) and two corresponding push-back operators (FT

⋅v forcovariant components and F−1 ⋅ v for contravariant components). For relationsinvolving a second rank tensor A in the reference configuration, there are fourpossible push-forward operators (F−T ⋅A ⋅F−1 for covariant components, F ⋅A ⋅FT

for contravariant components, and F ⋅A ⋅ F−1 and F−T ⋅A ⋅ FT between the twomixed components) and four corresponding push-back operators (FT

⋅A ⋅ F forcovariant components, F−1 ⋅A ⋅F−T for contravariant components, and F−1 ⋅A ⋅F

and FT⋅A ⋅ F−T between the two mixed components). Note that the definitions

(3.426) and (3.427) provide the corresponding relations for E and e written usingcovariant components.

Lastly, we point out that the Lie time derivative can be defined using pull-backand push-forward operations. For example, if v = vi ei is a spatial vector writtenusing contravariant components, the material time derivative is given by

v = vi ei + vi ei. (3.428)

The Lie time derivative is a material derivative holding the deformed basis con-stant, i.e., it corresponds to the first term on the right hand side of the aboveequation:

Lv = vi ei. (3.429)

In terms of pull-back and push-forward operations, it is given by

Lv = χ⋆( ddt[χ−1⋆(v)]) . (3.430)

The spatial vector is first pulled back to the reference configuration, there thedifferentiation is carried out, where the base vectors are constant, and then thevector is pushed forward again to the current configuration.

One of the most important uses of the Lie time derivative is that Lie timederivatives of objective spatial tensors are objective spatial tensors. For example,it can be easily shown that the Lie time derivative of a rank 2 spatial tensor A

written using covariant components is given by

LA = A +LT⋅A +A ⋅L, (3.431)

which we recognize as the Cotter–Rivlin tensor (3.425).

3.4. TRANSPORT THEOREMS 131

C(t0)A = a(t0)

X = x(t0)B = b(t0)

C (t)a(t)

x(t)b(t)

Figure 3.15: Material curve segment.

3.4 Transport theorems

3.4.1 Material derivative of a line integral

Let ψ(x, t) be a tensor field of arbitrary order that is continuous on the arbitraryoriented curve C (t) illustrated in Fig. 3.15. Then, using (3.12), the materialderivative of the line integral of field ψ over the material curve is given by

d

dt∫

C (t)ψ(xk, t) dxi = d

dt∫Cψ(XL, t) ∂xi

∂XK

dXK ,

= ∫C

∂

∂t[ψ(XL, t) ∂xi

∂XK

]dXK ,

= ∫C(ψ(XL, t) ∂xi

∂XK

+ψ(XL, t) ∂2xi

∂t∂XK

)dXK ,

= ∫C(ψ(XL, t) ∂xi

∂XK

+ ψ(XL, t)vi,l ∂xl∂XK

)dXK .

Now we can rewrite

d

dt∫

C (t)ψ(xk, t) dxi = ∫

C (t)ψ dxi + ∫

C (t)ψ vi,l dxl, (3.432)

or

d

dt∫

C (t)ψ(x, t)dx = ∫

C (t)ψ dx +∫

C (t)ψ (gradv)T ⋅ dx, (3.433)

and note that

ψ = ∂ψ(X, t)∂t

= ∂ψ(x, t)∂t

+ v(x, t) ⋅ gradψ(x, t), (3.434)

where v = x. If ψ is a vector field, i.e., ψ → ui, then if we project it on the curve,we have

d

dt∫

C (t)ui(xk, t) dxi = ∫

C (t)(ui + ul vl,i) dxi, (3.435)

or

d

dt∫

C (t)u(x, t) ⋅ dx = ∫

C (t)[u + u ⋅ (gradv)T ] ⋅ dx. (3.436)

132 KINEMATICS

Example

As an illustration on the use of (3.436), we examine the evolution of thelength of the curve between the points x1 and x2 by taking the vector u

to be the unit tangent vector t along the curve. In this case, if we call thelength of the material curve L, using (3.436) and the fact that t ⋅ t = 0, wehave

dL

dt= d

dt∫

C (t)dx = d

dt∫

C (t)t ⋅ dx = ∫

C (t)(t + t ⋅L) ⋅ dx =∫

C (t)t ⋅L ⋅ tdx. (3.437)

But note from (3.15) that

L ⋅ t = (grad v)T ⋅ dxdx= dvdx

; (3.438)

thus we obtain

dL

dt= ∫

C (t)t ⋅ dv = ∫

C (t)d (v ⋅ t) −∫

C (t)v ⋅ dt =

v ⋅ t∣x2

x1

−∫C (t)

κL v ⋅ ndx, (3.439)

where dt/dx = κL n, n is the principal normal to t, and κL is the principalcurvature. Here we note that since dt/dx ⋅ t = 0, then dt/dx is orthogonalto t, and thus one chooses dt/dx = κL n with dt/dx ⋅ dt/dx = κ2L so that n

is a unit vector.We note that if C (t) is a closed curve (x1 = x2) or v ⋅ t = 0 at x1 and x2,we then have that

dL

dt= −∫

C (t)κL v ⋅ndx. (3.440)

Note that if the tensor field ψ depends on one spatial dimension, i.e., ψ = ψ(ξ, t),where ξ is measured along the curve, and C (t) ∶ ξ1(t) ≤ ξ ≤ ξ2(t), then takingv(ξ, t) = ξ(t), the above result leads to the Leibniz rule since

d

dt∫

ξ2(t)

ξ1(t)ψ(ξ, t)dξ = ∫

ξ2(t)

ξ1(t)(∂ψ∂t+ v

∂ψ

∂ξ)dξ + ∫ ξ2(t)

ξ1(t)ψ∂v

∂ξdξ, (3.441)

= ∫ξ2(t)

ξ1(t)∂ψ

∂tdξ +∫

ξ2(t)

ξ1(t)∂(ψv)∂ξ

dξ, (3.442)

= ∫ξ2(t)

ξ1(t)∂ψ

∂tdξ +ψ(ξ2, t)v(ξ2, t) −ψ(ξ1, t)v(ξ1, t), (3.443)

= ∫ξ2(t)

ξ1(t)∂ψ

∂tdξ +ψ(ξ2, t)ξ2(t) − ψ(ξ1, t)ξ1(t). (3.444)

In the above steps, we have assumed that ψ is differentiable on C (t). To removethis assumption, and thus allow for ψ to be discontinuous, as indicated in Fig. 3.16,

3.4. TRANSPORT THEOREMS 133

ψ ψ

Ξ1 Γ Ξ2 ξ1 γ ξ2

t0 ∶ t ∶

Ξ ξ

Figure 3.16: Material line segment with discontinuity of ψ at γ(t).assume that it is discontinuous at the point γ(t) in the interval C (t). Now ψ iscontinuous in the subintervals ξ1(t) ≤ ξ < γ(t) and γ(t) < ξ ≤ ξ2(t), so we canapply the above result in these subintervals to obtain

d

dt∫

γ(t)

ξ1(t)ψ(ξ, t)dξ = ∫ γ(t)

ξ1(t)∂ψ

∂tdξ + ψ−(γ, t)γ(t)− ψ(ξ1, t)ξ1(t) (3.445)

and

d

dt∫

ξ2(t)

γ(t)ψ(ξ, t)dξ = ∫ ξ2(t)

γ(t)∂ψ

∂tdξ +ψ(ξ2, t)ξ2(t) −ψ+(γ, t)γ(t), (3.446)

where we have defined

ψ+(γ, t) ≡ limξ↓γ(t)

ψ(ξ, t) and ψ−(γ, t) ≡ limξ↑γ(t)

ψ(ξ, t). (3.447)

Now adding the above results, we obtain the generalized Leibniz rule

d

dt∫

C (t)−γ(t)ψ(ξ, t)dξ = ∫

C (t)−γ(t)∂ψ

∂tdξ + [ψ(ξ, t)v(ξ, t)]ξ2ξ1 − Jψ(γ, t)Kγ(t),(3.448)

where we have denoted the jump in ψ by

Jψ(γ, t)K ≡ ψ+(γ, t) −ψ−(γ, t). (3.449)

Lastly, using the generalized divergence theorem (2.301) on a line, we can rewritethe generalized Leibniz rule in the form

d

dt∫

C (t)−γ(t)ψ(ξ, t)dξ = ∫

C (t)−γ(t)[∂ψ∂t+∂(ψ v)∂ξ

] dξ +Jψ(γ, t) [v(γ, t) − γ(t)]K. (3.450)

134 KINEMATICS

3.4.2 Material derivative of a surface integral

Let ψ(x, t) be a tensor field of arbitrary order that is continuous on an arbitrarymaterial surface S (t), which is moving with velocity v(t) and is bounded by aclosed curve C (t). Then, using (3.36), the material derivative of the differentialelement of surface area is given by

d

dt∫

S (t)ψ(xk, t) dsi = d

dt∫Sψ(XK , t)J(XK , t)XL,i dSL,

= ∫S

∂

∂t[ψ(XK , t)J(XK , t)XL,i] dSL,

= ∫S[∂ψ(XK , t)

∂tJ(XK , t)XL,i+

ψ(XK , t)∂J(XK , t)∂t

XL,i+

ψ(XK , t)J(XK , t)∂XL,i

∂t]dSL.

However, using (3.60), we have

∂J

∂t∣XK

= ∂J

∂xi,K

∂2xi

∂t∂XK

= JXK,i

∂vi

∂XK

= J ∂vi∂xi

, (3.451)

and it can be shown that

˙F−1 = −F−1 ⋅ F ⋅F−1 = −F−1 ⋅L. (3.452)

Subsequently, we have

d

dt∫

S (t)ψ(xk, t) dsi = ∫

S[∂ψ(XK , t)

∂tJ(XK , t)XL,i+

ψ(XK , t)J(XK , t) ∂vj∂xj

XL,i − ψ(XK , t)J(XK , t)XL,j

∂vj

∂xi]dSL,

and, using (3.36), we arrive at

d

dt∫

S (t)ψ(xk, t) dsi = ∫

S (t)(ψ +ψvj,j)dsi −∫

S (t)ψvj,i dsj , (3.453)

or

d

dt∫

S (t)ψ(x, t) ds = ∫

S (t)(ψ +ψ divv)ds −∫

S (t)ψ (gradv)T ⋅ ds. (3.454)

Note that if ψ is a vector field projected on the surface, i.e., ψ → ui, then wehave

d

dt∫

S (t)ui(xk, t) dsi = ∫

S (t)(ui + uivk,k − ukvi,k)dsi, (3.455)

3.4. TRANSPORT THEOREMS 135

γ(t)C+

C−

γ+

γ−

t

nc n−

S+

S−

n+

Figure 3.17: Material surface S (t) with discontinuity along curve γ(t).or

d

dt∫

S (t)u(x, t) ⋅ ds = ∫

S (t)[∂u∂t+ (v ⋅ grad)u + u (divu) − (u ⋅ grad)v] ⋅ ds,

= ∫S (t)[∂u∂t+ v divu + curl (u × v)] ⋅ ds, (3.456)

= ∫S (t)(∂u∂t+ v divu) ⋅ ds +∫

C (t)u × v ⋅ dx, (3.457)

where we have used Stokes’ theorem (2.292) in the last step. We note from (3.456)that in order that the flux of the vector field u(x, t) across every material surfaceremain constant in time, it is necessary and sufficient that Zorawski’s criterion besatisfied:

∂u

∂t+ v divu + curl (u × v) = 0. (3.458)

A similar argument can be extended to a surface S (t) intersected by a dis-continuity line γ(t) moving with a velocity c(t) on the surface, as illustrated inFig. 3.17. Applying our result (3.457) of the transport of vector quantity u tothe two subsurfaces separated by γ(t), adding the results, letting γ+(t) and γ−(t)approach γ(t), noting that dγ+ = −dγ− = −dγ and c+ = c− = c, and using thegeneralized Stokes theorem (2.300), we obtain

d

dt∫

S (t)−γ(t)u(x, t) ⋅ ds = ∫

S (t)−γ(t)[∂u∂t+ v divu + curl (u × v)] ⋅ ds +

∫γ(t)

Ju × (v − c)K ⋅ dγ. (3.459)

Example

As an illustration on the use of (3.457), we examine the rate of increase ofa surface area by taking the vector u to be the unit vector n normal to thesurface. In this case, if we call the area of the material surface A, using

136 KINEMATICS

(3.457) and the fact that ∂n/∂t ⋅n = 0, we have

dA

dt= d

dt∫

S (t)ds (3.460)

= d

dt∫

S (t)n ⋅ ds (3.461)

= ∫S (t)

[∂n∂t+ v (divn)] ⋅ ds + ∫

C (t)n × v ⋅ dx (3.462)

= ∫S (t)

(divn)v ⋅ ds + ∫C (t)

n × v ⋅ dx. (3.463)

We now note from (D.217) that divn = −2KM , where KM is the meansurface curvature. Subsequently, we obtain

dA

dt= −∫

S (t)KM v ⋅nds + ∫

C (t)n × v ⋅ tdx. (3.464)

3.4.3 Material derivative of a volume integral

Let ψ(x, t) be a tensor field of arbitrary order that is continuous in V (t), anarbitrary volume bounded by the closed surface S (t), and let a material point inV (t) move with velocity v(t). Then, using (3.51) the material derivative of thefield ψ over the volume is given by

d

dt∫

V (t)ψ(xk, t) dv = d

dt∫Vψ(XK , t)J(XK , t) dV,

= ∫V

∂

∂t[ψ(XK , t)J(XK , t)] dV,

= ∫V[∂ψ(XK , t)

∂tJ(XK , t) +ψ(XK , t)∂J(XK , t)

∂t]dV.

However, using (3.451), we have

d

dt∫

V (t)ψ(xk, t) dv = ∫

V[ψ(XK , t) +ψ(XK , t)∂XK

∂xi

∂vi(XK , t)∂XK

]J(XK , t) dV,= ∫

V (t)(ψ + ψ ∂vi

∂xi)dv,

= ∫V (t)(∂ψ∂t+ vi

∂ψ

∂xi+ ψ

∂vi

∂xi)dv.

Thus we arrive at

d

dt∫

V (t)ψ(xk, t) dv = ∫

V (t)[∂ψ∂t+

∂

∂xi(viψ)]dv =

∫V (t)

∂ψ

∂tdv + ∫

S (t)ψvi dsi, (3.465)

3.4. TRANSPORT THEOREMS 137

n

V+(t)

V−(t)

S+(t)

S−(t)

ζ(t)

ζ+(t)ζ−(t)

ν

n

c

Figure 3.18: Material volume V (t) with discontinuity along surface ζ(t).or

d

dt∫

V (t)ψ(x, t) dv = ∫

V (t)[∂ψ∂t+ div (vψ)]dv =

∫V (t)

∂ψ

∂tdv + ∫

S (t)ψv ⋅ ds, (3.466)

where the last step is obtained by the use of the divergence theorem (2.289). Thisresult is known as Reynolds’ transport theorem.

Example

Let ψ = 1 and V (t) be an arbitrary subvolume of a continuous materialbody which is in motion. It then follows from (3.466) that

dV

dt= d

dt∫

V (t)dv = ∫

V (t)divv dv = ∫

S (t)v ⋅ ds = ∫

S (t)v ⋅ nds. (3.467)

We see that the volume remains constant, i.e., the motion is incompressible,if and only if divv = 0.

Note that in the last step, the conventional divergence theorem is used, which as-sumes that the quantity ψ v is continuous in V (t). However, time rates of integralsover regions containing a discontinuity surface are common occurrences. Thus, wegeneralize the above result by assuming that the volume V (t) is intersected bya surface of discontinuity ζ(t) with unit normal ν moving with velocity c(t), asillustrated in Fig. 3.18. Now applying the above result to the two subvolumes inwhich the quantity ψv is continuous, we have

d

dt∫

V +(t)ψ(x, t) dv = ∫

V +(t)∂ψ

∂tdv +∫

S +(t)ψv ⋅n ds −∫

ζ+(t)ψ+c+ ⋅ ν+dζ (3.468)

138 KINEMATICS

and

d

dt∫

V −(t)ψ(x, t) dv = ∫

V −(t)∂ψ

∂tdv + ∫

S −(t)ψv ⋅ n ds +∫

ζ−(t)ψ−c− ⋅ ν−dζ,(3.469)

where we have defined

ψ+(x, t) ≡ limx↓ζ(t)

ψ(x, t) and ψ−(x, t) ≡ limx↑ζ(t)

ψ(x, t), (3.470)

and analogously for c and ν. Now upon adding the two equations, letting ζ+ andζ− approach ζ, and noting that dζ+ = −dζ− = −dζ = −νdζ and c+ = c− = c, weobtain

d

dt∫

V (t)−ζ(t)ψ(x, t) dv = ∫

V (t)−ζ(t)∂ψ

∂tdv +∫

S (t)−ζ(t)ψv ⋅ ds −

∫ζ(t)

JψcK ⋅ dζ, (3.471)

where we define

JAK ≡A+ −A−. (3.472)

If we use the generalized divergence theorem (2.299) to replace the second termon the right hand side, we obtain the generalized Reynolds’ transport theorem

d

dt∫

V (t)−ζ(t)ψ(x, t) dv = ∫

V (t)−ζ(t)[∂ψ∂t+ div (ψv)]dv +∫ζ(t)

Jψ(v − c)K ⋅ dζ. (3.473)

Problems

1. Given the deformation function

x1 = X1 + κX2,

x2 = X2,

x3 = X3,

where κ is a positive constant, obtain the stretch ratio in the directions

L(1)K = (1,0,0), L(2)K = (0,1,0), L(3)K = (0,0,1), L(4)K = (1/√2,1/√2,0), L(5)K =

1/√3(1,1,1), and the changes in orientation of line elements in the last twodirections.

2. Show that

xi,I = 1

2JǫijkǫIJKXJ,jXK,k.

3. Show that

∂J−1

∂XI,i

= 1

2ǫIJKǫijkXJ,jXK,k.

3.4. TRANSPORT THEOREMS 139

4. Show that

∂xi,K

∂XI,k

= −xk,Kxi,I .5. Show that

(J−1xi,I),i = 0.6. If A is a second rank tensor, show that

˙A−1 = −A−1AA−1, (3.474)

and

∂ detA

∂A= detA (A−1)T . (3.475)

7. If A is a second rank tensor, show that

˙detA = detA tr (AA−1) . (3.476)

8. If A is a second rank tensor, show that

∂ trA

∂A= I, ∂ trA2

∂A= 2AT , and

∂ trA3

∂A= 3 (A2)T , (3.477)

or in general

∂ trAk

∂A= k (Ak−1)T . (3.478)

9. If A is a second rank tensor, use (3.86) and (3.87) to obtain the followingrecursion relation:

∂A(k+1)∂A

= A(k)I −AT∂A(k)∂A

, k = 0,1, . . . , n, (3.479)

where A(0) = I and A(n+1) = 0. By induction, show that the above recursioncan also be written in the following form:

∂A(k)∂A

= ⎡⎢⎢⎢⎣k−1

∑j=0

(−1)j A(k−j−1)Aj⎤⎥⎥⎥⎦T

. (3.480)

10. Evaluate the invariants of a three-dimensional skew-symmetric second ranktensor A.

11. Show that the characteristic polynomial for A = [aik] is given by

λ3 −A(1)λ2 +A(2)λ −A(3) = 0, (3.481)

where

A(1) = aii, (3.482)

A(2) = 1

2[(aii)2 − aikaki] , (3.483)

A(3) = det[aik]. (3.484)

140 KINEMATICS

a) Show that the eigenvalues of A are given by

λ(k) = 1

3A(1) + 2 (A2

(1) − 3A(2))1/2 cos [13 (θ + 2π k)] , k = 1,2,3,(3.485)

where

cosθ = 2A3

(1) − 9A(1)A(2) + 27A(3)

2 (A2

(1) − 3A(2))3/2(3.486)

b) As noted in (3.138), the spectral representation of a symmetric matrixA is given by

A = 3

∑1

λ(i)Ai, Ai = vivi, (3.487)

where vi are the normalized eigenvectors of A. If the eigenvalues areall distinct, show that

Ak = (A − λ(l)I) (A − λ(m)I)(λ(k) − λ(l)) (λ(l) − λ(m)) , (3.488)

where (k, l,m) represents a cyclic permutation of (1,2,3).c) In the case of coalescence of two eigenvalues (λ(1) ≠ λ(2) = λ(3) = λ),

show that

A1 = (A − λI)(λ(1) − λ) . (3.489)

d) For the case of coalescence of all eigenvalues (λ(1) = λ(2) = λ(3) = λ),show that

A = λI. (3.490)

12. Show that if W is a 3 × 3 skew-symmetric matrix, then trA4 = 1

2(trA2)2.

[Hint: Use Cayley–Hamilton theorem.]

13. Given the right Cauchy–Green tensor C, show that the necessary and suffi-cient conditions for a rigid deformation are C(1) = C(2) = 3 and C(3) = 1.

14. Show that (U−1)T = (UT )−1.15. Show that F =V ⋅R.

16. Prove that the decomposition F =V ⋅R is unique.

17. Determine the eigenvalues and normalized eigenvectors of A given by

A =⎡⎢⎢⎢⎢⎢⎣

11

6−

2

3−

1

6

−2

3

7

3−

2

3

−1

6−

2

3

11

6

⎤⎥⎥⎥⎥⎥⎦.

18. For the two-dimensional small strain theory, the strains in a beam are givenby

e11 = aX1X2, e22 = −a bX1X2, e12 = 1

2a (1 + b)(c2 −X2

2) ,

3.4. TRANSPORT THEOREMS 141

where a, b, and c are positive constants, a ≪ 1, and b ≤ 1/2. Assume thatdisplacements u1 and u2 along the X1 and X2 axes are functions of X1 andX2.

a) Using compatibility conditions, show that continuous single-valued dis-placements u1 and u2 are possible.

b) Subsequently, derive expressions for u1 and u2 in terms of X1 and X2

with the conditions u1 = u2 = u1,2 = 0 at the point X1 = L, X2 = 0.19. Show that the stretching tensor D(t) and the spin tensor W(t) may be

expressed in terms of the polar decomposition F(t) =R(t) ⋅U(t) by

D(t) = 1

2R(t) ⋅ [U(t) ⋅U−1(t) +U−1(t) ⋅ U(t)] ⋅RT (t), (3.491)

W(t) = R(t) ⋅RT (t) + 1

2R(t) ⋅ [U(t) ⋅U−1(t)−

U−1(t) ⋅ U(t)] ⋅RT (t). (3.492)

20. Derive (3.202) from (3.200).

21. Derive (3.203) from (3.200).

22. Derive (3.207)–(3.212) from (3.204).

23. Show that the material derivative of the differential area (3.36) is given by

˙ds = trDds −LT

⋅ ds. (3.493)

24. Using (3.239), show that the evolution equation for f ≡ F−1 is given by

∂f

∂t+∇ (f ⋅ v) = 0. (3.494)

Is this equation objective?

25. Using (3.239), show that the evolution equation for right Cauchy–Greentensor C is given by

C = 2FT⋅D ⋅F. (3.495)

Is this equation objective?

26. Using (3.239), show that the evolution equation for left Cauchy–Green tensorB is given by

⋆

B = 0, (3.496)

where⋆

B is the Oldroyd tensor defined in (3.423). Is this equation objective?

27. A deformation of the form

x1 = f1(X1,X2),x2 = f2(X1,X2),x3 = X3,

142 KINEMATICS

where f1 and f2 are smooth functions, is called plane strain. Show thatfor such a deformation, the principal stretch λ3 (in the X3 direction) isunity. Show further that the deformation is isochoric if and only if the otherprincipal stretches, λ1 and λ2, satisfy

λ1 = 1

λ2.

28. A motion is plane if the velocity field has the form

v(x, t) = v1(x1, x2, t)i1 + v2(x1, x2, t)i2,in some Cartesian frame. Show that in a plane motion

W ⋅D +D ⋅W = (divv)W,

where D and W are the stretch and spin tensors, respectively.

29. Later we will be making use of the convected tensor T defined by

T = FT⋅ T ⋅F. (3.497)

a) Show that the spin tensor satisfies the differential equation

W +D ⋅W +W ⋅D = J, (3.498)

where

J = 1

2[(grad v) − (grad v)T ] (3.499)

is the skew-symmetric part of the acceleration gradient.

b) Show that convected spin tensor satisfies the differential equation

˙W = J. (3.500)

30. Prove thatD(t) = (t)U(t) = (t)V(t). (3.501)

31. Consider the motion

x1 = 0, x2 = κx1, x3 = 0.Calculate ξ(x, τ ; t), (t)R(τ), and W(t).

32. A motion of a body is given by

x1 = e−at [X1 cos (bX3t) −X2 sin (bX3t)] ,x2 = e−at [X1 sin (bX3t) +X2 cos (bX3t)] ,x3 = φ(t)X3,

where a, b > 0 are constants. Find the form of φ(t) for which the motion isisochoric. Determine, for the isochoric motion, the components of velocityand acceleration in the spatial description, and show that both the particlepaths and the stream lines lie on the surfaces

(x21 + x22)x3 = constant.

3.4. TRANSPORT THEOREMS 143

33. A plane circular shearing motion of a body is given by

x1 = X1 + φ(X3) cos (ωt) +ψ(X3) sin (ωt) ,x2 = X2 + φ(X3) sin (ωt) −ψ(X3) cos (ωt) ,x3 = X3,

where ω > 0 is a constant, the functions φ and ψ are differentiable, and thereferential and spatial coordinates refer to a common Cartesian system.

a) Show that the motion is isochoric and the particle paths and streamlines are circles.

b) Discuss the stretch and rotation undergone by material line elements,which, in the reference configuration, lie parallel and orthogonal to theX3 direction.

34. In a plane motion of a continuum, the velocity field is given by

x1 = −V [1 − a2(x21 − x22)(x21+ x2

2)2 ] , x2 = 2V a2x1x2

r4, x3 = 0,

where a and V are constants, (x1, x2, x3) are Cartesian coordinates, andr2 = x21 + x22 + x23. Determine the stream lines.

35. The velocity field of a continuum is given by

x1 = V a2 (x21 − x22)(x21+ x2

2)2 , x2 = 2V a2 x1x2(x2

1+ x2

2)2 , x3 = 0,

where a and V are constants. Determine the path lines.

36. In a plane motion of a continuum, the velocity field is given by

x1 = − k4π

x21x2(x21 + x22) , x2 = k

4π

x22x1(x21 + x22) , x3 = 0,

where k is a constant. Determine the stream lines and vortex lines.

37. Give a geometrical description of the deformation

x1 = X1 − τX2X3,

x2 = X2 + τX1X3,

x3 = X3,

where τ is a constant, and calculate the components of F and C. Is thedeformation isochoric? Find the surface into which the cylinder X2

1 +X22 = a2

(a = const.) deforms.

38. A steady two-dimensional flow (pure straining) is given by

v1 = αx1 and v2 = −αx2,with α = const.

144 KINEMATICS

a) Find the equation for a general stream line of the flow, and sketch someof them.

b) At t = 0 the fluid on the curve x21+x22 = a2 is marked by some technique.

Find the equation for this material fluid curve for t > 0.c) Does the area within the curve change in time, and why?

39. Do Problem 38, but for the two-dimensional flow (simple shear) given by

v1 = γ x2 and v2 = 0,with γ = const. Which of the two flows stretches the curve faster at longtimes?

40. Verify that the two-dimensional flow given by

v1 = x2 − c2(x1 − c1)2 + (x2 − c2)2 and v2 = c1 − x1(x1 − c1)2 + (x2 − c2)2 ,where c1 and c2 are constants, satisfies div v = 0, and then find the streamfunction ψ(x1, x2) such that

v1 = ∂ψ

∂x2and v2 = − ∂ψ

∂x1.

Sketch the stream lines.

41. Verify that the two-dimensional flow given in cylindrical polar coordinatesby

vr = U (1 − a2r2) cosθ and vθ = −U (1 + a2

r2) sin θ

satisfies div v = 0, and find the stream function ψ(r, θ) such that

vr = 1

r

∂ψ

∂θand vθ = −∂ψ

∂r.

Sketch the stream lines.

42. Verify that the axisymmetrical flow (uniaxial straining) given in cylindricalpolar coordinates by

vr = −12αr and vz = αz

satisfies div v = 0, and find the Stokes stream function ψ(r, z) such that

vr = −1r

∂ψ

∂zand vz = 1

r

∂ψ

∂r.

Sketch the stream lines.

43. Show that if the first invariant of the stretch tensor is zero, i.e., D(1) = 0,then the motion is isochoric.

3.4. TRANSPORT THEOREMS 145

44. When the stretching tensor is spherical, i.e., all three principal stretches areequal, the motion is purely dilatational. Show that in this case

(13D(1))3 = (1

3D(2))3/2 =D(3).

45. Show that for the simple shearing motion x1 = κx2, x2 = x3 = 0,a) D(1) =D(3) = 0;b) the principal stretches are such that d1 = −d3 and d2 = 0;c) vorticity is parallel to the principal axis of stretch along which the

stretch is zero;

d) the amount of shear is given by κ = ω =√−4D(2), where ω = ∣ω∣.46. The kinematical vorticity number A is the dimensionless ratio of the magni-

tudes of the spin and stretch tensors. Show that

A =√

W ∶W

D ∶D=¿ÁÁÀ−W(2)

D(2)= ω√

2D(2)= ω√

2(d21+ d2

2+ d2

3) , (3.502)

where di’s are the principal stretches and ω = ∣ω∣. Note that if A ≪ 1, thenthe motion is nearly irrotational.

47. Let the exponential of a matrix A be defined by the series

exp(A) = 1 +A + A2

2!+A3

3!+⋯.

Show that

i) if AB = BA, then exp(A) exp(B) = exp(A +B);ii) if A is skew-symmetric, then exp(A) is orthogonal;

iii) for any skew-symmetric matrixW , the matrix function Q(t) = exp[−(t−t0)W ] satisfies the following differential equation

Q +QW = 0 and Q(t0) = 1.48. Suppose that ψ(x, t) transforms like a scalar field under a change of frame.

Is ∂ψ/∂t objective? Is ψ objective?

49. Suppose that q is a frame-indifferent vector, i.e., q′ = Q ⋅ q. Is the quantityq −L ⋅ q frame indifferent?

50. Suppose that σ is a symmetric frame indifferent second rank tensor, i.e.,σ = σT and σ′ =Q ⋅σ ⋅QT . Is the quantity σ − 2D ⋅σ frame indifferent?

51. Show that (3.422) is an objective tensor.

52. Show that (3.423) is an objective tensor.

53. Show that (3.424) is an objective tensor.

54. Show that (3.425) is an objective tensor.

146 KINEMATICS

Bibliography

R.C. Batra. Elements of Continuum Mechanics. AIAA, Reston, VA, 2006.

B.S. Berger and M. Rokni. Lyapunov exponents and continuum kinematics.International Journal of Engineering Science, 25(10):1251–1257, 1987.

R.M. Bowen. Introduction to Continuum Mechanics for Engineers. Plenum Press,New York, NY, 1989.

A.C. Eringen. Nonlinear Theory of Continuous Media. McGraw-Hill Book Com-pany, Inc., New York, NY, 1962.

A.C. Eringen. Basic principles: Deformation and motion. In A.C. Eringen, editor,Continuum Physics, volume II, pages 3–67. Academic Press, Inc., New York, NY,1975.

A.C. Eringen. Mechanics of Continua. R.E. Krieger Publishing Company, Inc.,Melbourne, FL, 1980.

M.E. Gurtin. An Introduction to Continuum Mechanics. Academic Press, SanDiego, CA, 2003.

M.E. Gurtin, E. Fried, and L. Anand. The Mechanics and Thermodynamics ofContinua. Cambridge University Press, Cambridge, UK, 2010.

P. Haupt. Continuum Mechanics and Theory of Materials. Springer-Verlag,Berlin, 2000.

G.A. Holzapfel. Nonlinear Solid Mechanics. John Wiley & Sons, Ltd., Chichester,England, 2005.

K. Hutter and K. Jöhnk. Continuum Methods of Physical Modeling. Springer-Verlag, Berlin, 1981.

W.M. Lai, D. Rubin, and E. Krempl. Introduction to Continuum Mechanics.Butterworth-Heinemann, Burlington, MA, 2010.

I.-S. Liu. Continuum Mechanics. Springer-Verlag, Berlin, 2002.

L.E. Malvern. Introduction to the Mechanics of a Continuous Medium. Prentice-Hall, Inc., Upper Saddle River, NJ, 1969.

A.G. McLellan. The Classical Thermodynamics of Deformable Materials. Cam-bridge University Press, 1980.

W. Noll. On the continuity of the solid and fluid states. Journal of RationalMechanics and Analysis, 4(1):3–81, 1955.

W. Noll. A mathematical theory of the mechanical behavior of continuous media.Archive for Rational Mechanics and Analysis, 2(1):197–226, 1958.

J.M. Ottino. The Kinematics of Mixing: Stretching, Chaos, and Transport. Cam-bridge University Press, 1989.

BIBLIOGRAPHY 147

C. Truesdell and W. Noll. The non-linear field theories of mechanics. In S. Flügge,editor, Handbuch der Physik, volume III/3. Springer, Berlin-Heidelberg-NewYork, 1965.

C. Truesdell and R.A. Toupin. The classical field theories. In S. Flügge, editor,Handbuch der Physik, volume III/1. Springer, Berlin-Heidelberg-New York, 1960.

K. Washizu. A note on the conditions of compatibility. Journal of Mathematicsand Physics, 36:306–312, 1958.

A. Wintner and F.D. Murnaghan. On a polar representation of non singularsquare matrices. Proceedings of the National Academy of Sciences of the UnitedStates of America, 17:676–678, 1931.

4

Mechanics and thermodynamics

4.1 Balance law

Consider an arbitrary material body having volume V (t) and surface S (t) whichis separated into two parts V

+(t) and V−(t), or V (t)− ζ(t), by a singular surface

whose intersection with V is denoted by the surface ζ(t), as shown in Fig. 4.1.The surface ζ(t) has a unit normal vector n pointing into V +(t) and velocity c.The line of intersection of S (t) and ζ(t) will be denoted by C (t) and the unitvector normal to C (t) which is tangential to ζ(t) and pointing externally to V (t)is called µ. The parts of S (t) not on C (t) will be denoted as S +(t) and S −(t),or S (t) − C (t). Note that S

+(t) and S−(t) also include the surface areas of

V+(t) and V

−(t) adjacent to ζ(t), and there they have exterior normals n+ andn−, respectively. The symbol n also denotes the unit normal on S (t)−C (t), whichpoints out of V (t). The material in V (t) − ζ(t) moves with particle velocity v.

Generally, from the physical point of view, tensor fields or one of their deriva-tives would change very fast within a small layer near ζ(t). We idealize suchsituation by considering this layer to be of zero thickness and subsequently allowthe tensor fields to be singular there. The singularity will appear as a discontinuityin functions or their derivatives. A surface that is singular with respect to somequantity and that has a nonzero speed of propagation is said to be a propagatingsingular surface or wave. Singularities of the first type are usually associated withmaterial singular surfaces which are formed by the same material particles at alltimes. Examples of the second type are nonmaterial singular surfaces, or shocks,or acceleration waves. We note that in addition to singular surfaces, singular linesand singular points are also idealizations of common occurrences.

The general balance statement of a tensor quantity Ψ associated with the bodyis given in the form

dΨ

dt= T (Ψ) + G(Ψ), (4.1)

where T (Ψ) denotes the flux of Ψ through the surface of the body, and G(Ψ) thecombined external supply of Ψ to the body and internal production of Ψ withinthe body. One should note that while we represent symbolically the sum of supplyand production by a single term in the present derivation, physically supply isdifferent from production because it may be controlled from the exterior of the

149

150 MECHANICS AND THERMODYNAMICS

S+

S−

n

V−

V+

C (t)µ

n− n v

n+

c

ζ(t)

Figure 4.1: Arbitrary volume V intersected by a discontinuous surface.

body. Subsequently, in later applications, we will recognize this difference by writ-ing their contributions separately. By the Radón–Nikodym theorem, we deducethat additive densities of Ψ, T (Ψ), and G(Ψ) exist, and denote the correspondingquantities that are defined within V (t)− ζ(t) by ψ, t, and g, respectively. Analo-gously, we also infer the existence of those quantities that are defined only on thesingular surface ζ(t) by a tilde superscript, i.e., ψ, t, and g. Subsequently, (4.1)can be rewritten more explicitly as

d

dt(∫

V (t)−ζ(t)ψ dv + ∫

ζ(t)ψ ds) = (∫

S (t)−C (t)t ⋅nds + ∫

C (t)t ⋅ µdl) +

(∫V (t)−ζ(t)

g dv + ∫ζ(t)

g ds) . (4.2)

Now using the generalized Reynolds and surface transport theorems (3.473) and(D.224), and the generalized divergence and surface divergence theorems (2.299)and (D.216), we have

∫V (t)−ζ(t)

[∂ψ∂t+ div (ψv) − div t − g]dv +

∫ζ(t)[∂ψ∂t+ ∇ ⋅ (ψv) − 2KM ψ v(n) − ∇ ⋅ t − g + Jψ(v − c) − tK ⋅ n]ds = 0, (4.3)

where ∇ is the surface gradient operator, v(n) is the velocity component normalto ζ(t), and KM is the mean curvature of the surface ζ(t) (see Appendix D and(D.183)).

Now evaluating the balance law over an arbitrary volume in the continuousregion V (t)− ζ(t), we obtain the local form of the balance law for the volumetric

4.2. FUNDAMENTAL AXIOMS OF MECHANICS 151

tensor quantity ψ over the region

∂ψ

∂t+ div (ψv) − div t − g = 0 (4.4)

since this holds for all sufficiently regular volumes V (t) − ζ(t), however small.Alternately, taking the limit by shrinking a volume down to ζ(t) in such a waythat the volume tends to zero, while the area of ζ(t) remains unchanged, andassuming that ∂ψ/∂t and g remain bounded, then the volume integral vanishesin the limit. In addition, since the integrand of the remaining surface integral issmooth on ζ(t), and thus holds for any surface area no matter how small, then theintegrand of the surface integral must vanish. Subsequently, we obtain the balancelaw for the areal tensor quantity ψ defined on the singular surface:

∂ψ

∂t+ ∇ ⋅ (ψv) − 2KM ψ v(n) − ∇ ⋅ t − g + Jψ (v − c) − tK ⋅ n = 0. (4.5)

This surface balance law is very important in the continuum mechanics of mem-branes or thin shells. However, from here on, we shall assume that the surfaceof discontinuity does not possess any properties of its own, i.e., ψ = 0, t = 0, andg = 0. Subsequently, the integral balance law (4.3) becomes

∫V (t)−ζ(t)

[∂ψ∂t+ div (ψv) − div t − g]dv +∫

ζ(t)Jψ (v − c) − tK ⋅nds = 0, (4.6)

and the local balance equation for the surface of discontinuity (4.5) reduces to ajump condition on volumetric quantities that has to be satisfied across the surfaceζ(t):

Jψ (v − c) − tK ⋅ n = 0. (4.7)

In the rest of this chapter, we shall only make use of the integral balance laws(4.1) and (4.6), and the local balance laws (4.4) and (4.7), applied to an arbitrarymaterial volume to obtain field equations and jump conditions in the spatial de-scription. The corresponding balance laws in material coordinates are given inAppendix C, while the balance laws pertaining to material surfaces and materiallines are discussed in Appendix D.

4.2 Fundamental axioms of mechanics

Associated with each material body, there is a quantity called mass, which has adimension [M] independent of the dimensions of length, [L], and time, [T ]. Thisquantity is positive definite and additive, i.e., it is an extensive property, and isabsolutely continuous in the space variables. Subsequently, there exists a densityρ called the mass density with dimensions [ρ] = [M]/[L]3. The quantity of mass isassigned to a set of particles having positive volume. Every finite body has finitemass, and zero volume implies zero mass. However, to deal with singular surfaces,we allow 0 ≤ ρ < ∞. The specific volume is given by the reciprocal of the massdensity, v = 1/ρ, and the volume of the body is given by

V = ∫V

dv. (4.8)

152 MECHANICS AND THERMODYNAMICS

Definition: The total mass M of a material body having volume V is deter-mined by

M ≡ ∫V

ρdv. (4.9)

Definition: The center of massM of a material body having volume V is givenby

xc ≡ xP +1

M∫

V

ρ (x − xP ) dv, (4.10)

where xP is an arbitrary point fixed in the reference frame. It is easily verifiedthat xc is independent of the choice of xP . In rigid bodies, the center of massis fixed relative to the body’s geometry. This is not the case when deformationsoccur. In a deformable body, the center of mass moves about within (or possiblyoutside) the body in the course of time.

Definition: The linear momentum P of a continuous mass medium containedin V is given by

P ≡ ∫V

ρv dv, (4.11)

where v = x is the velocity of the material particle.Definition: The barycentric velocity vc = xc of a continuous mass medium

contained in V is given by

vc ≡ 1

M∫

V

ρv dv. (4.12)

In the following, we will formulate the balance laws for a polar material. Polarmaterials arise from a statistical mechanics model that assumes noncentral forcesof interaction between particles. When such forces are noncentral, an interparticlecouple, in addition to an interparticle force, manifests itself. Under the actionof the couple, the material particle will have a tendency to rotate relative toits neighbors. The essential idea of a polar material is obtained by introducing akinematic variable to model the rotation of the particle relative to its neighbors anda skew-symmetric tensor to model the forces that balance the action of the couple.Subsequently, an internal (particle) angular velocity and an associated internalspin are defined independently of the velocity field, along with other pertinentquantities.

Definition: The internal spin s of a continuous medium contained in V is givenby

s ≡ ∫V

ρϕdv, (4.13)

where the internal spin per unit mass ϕ is assumed to exist and to be linearlyrelated to the internal angular velocity ν by

ϕ = i ⋅ ν, (4.14)

where i is the symmetric positive-definite rank-2 internal inertia tensor per unitmass. The internal inertia tensor describes the average inertia of material particles

4.2. FUNDAMENTAL AXIOMS OF MECHANICS 153

relative to the position vector x, where x is viewed as the center of mass of particlesnear x. The internal spin and the internal spin per unit mass can be taken tocorrespond to axial vectors of the respective skew-symmetric internal spin tensorsS = s ⋅ ǫ and Φ = ϕ ⋅ ǫ so that the internal spin tensors are related by

S ≡ ∫V

ρΦ dv. (4.15)

Definition: The moment of momentum h, about point xP , of a continuousmass medium contained in V is defined by

h ≡ ∫V

r × ρv dv, (4.16)

where r = x − xP , x is the position vector, and r is the vector from the moment-center xP to the point x on the line of action of the force. Note that if we takethe moment-center to be the origin, then r = x; most often the moment-center istaken to be the center of mass, in which case r = x − xc. Since h can be taken tocorrespond to the axial vector of the skew-symmetric tensor H, i.e., H = h ⋅ ǫ, wecan also write (see (2.170) and (2.171))

H ≡ ∫V

(r ∧ ρv) dv. (4.17)

Definition: The kinetic energy K of the continuous mass medium in V is givenby the sum of the internal spin and translational kinetic energies:

K ≡ 1

2∫

V

ρ (ν ⋅ϕ + v ⋅ v) dv. (4.18)

We now state the three fundamental laws of mechanics.Axiom 1 – Conservation of mass: The total mass of a body is unchanged

during motion. When this is valid for an arbitrarily small neighborhood of eachmaterial point, we say that the mass is conserved locally.

The total or global mass conservation may be expressed by

M = ∫V

ρ dv = ∫VρR dV ≡MR, (4.19)

where ρR(X, t0) is the mass density in the reference configuration. Note thatthe above volume integrals can both be expressed in either the reference materialcoordinate system or the spatial coordinate system:

∫V(ρJ − ρR)dV or ∫

V

(ρ − ρRJ−1)dv.As a result, global mass conservation can be rewritten in the reference configurationas

M −MR = ∫V(ρJ − ρR)dV = 0. (4.20)

If the volume is taken to correspond to an arbitrary volume element within thebody, then the integrand must be identically zero to satisfy the equation. Settingthe integrand to zero, we obtain the local mass balance

ρ(X, t)J(X, t) = ρR(X, t0) or ρ(x, t) = ρR(X, t0)J−1(x, t). (4.21)

154 MECHANICS AND THERMODYNAMICS

If J = 1, then ρ(X, t) = ρR(X, t0) and the motion is isochoric. If in additionρR = const., then the motion is homochoric. The equation above is the generalsolution of the conservation of mass equation. However, it is expressed within thematerial description, and in order to render it explicit, we must know the motion.A spatial form, for application of which the motion itself need not be given, ispreferable. This will be given later. The material derivative of the global massconservation equation (4.19) in the current configuration is

dM

dt= d

dt∫

V

ρdv = 0. (4.22)

Axiom 2 – Balance of linear momentum: The time rate of change of linearmomentum P is equal to the resultant force F acting on the body. We postulatethat this is valid for an arbitrarily small neighborhood of each material point, thusgiving rise to the local form of the linear momentum balance.

This statement is expressed by the following equation:

dP

dt= F or

d

dt∫

V

ρv dv = F . (4.23)

Axiom 3 – Balance of angular momentum: The time rate of angularmomentum is given by the change of the sum of the internal spin and the momentof momentum of a body about a fixed point, and is equal to the resultant momentm about the point. Analogously, we postulate that this is valid for an arbitrarilysmall neighborhood of each material point in the body, thus giving rise to the localform of the angular momentum balance.

This statement is expressed by the following equation:

d

dt(s + h) =m or

d

dt∫

V

ρ (ϕ + r × v) dv =m. (4.24)

In terms of an equivalent resultant-moment skew-symmetric tensor M = m ⋅ ǫ,we have

d

dt(S +H) =M or

d

dt∫

V

ρ (Φ + r ∧ v) dv =M. (4.25)

The equations for linear and angular momenta are called Euler’s equations ofmotion and are considered extensions of Newton’s second and third laws of motionof a particle.

4.3 Fundamental axioms of thermodynamics

Internal energy is defined to be the energy of material particles due to internalmechanisms not explicitly modeled. From correspondence between kinetic theoryand continuum mechanics, this energy includes classically the kinetic energy dueto the motion of molecules (translational, rotational, vibrational) and the poten-tial energy associated with the vibrational and electric energy of atoms withinmolecules or crystals (it includes the energy in all the chemical bonds, and the en-ergy of the free conduction electrons in metals). More generally, it is understood toinclude all forms of energies of material particles due to mechanisms whose scales

4.3. FUNDAMENTAL AXIOMS OF THERMODYNAMICS 155

are not modeled. It is an extensive property that, in classical thermodynamics,is a state function, i.e., it is independent of the process followed in changing thestate of the body.

Definition: The internal energy, E , of a continuous mass medium contained inV is given by

E ≡ ∫V

ρedv, (4.26)

where a local internal energy density per unit mass, e, has been assumed to exist.Axiom 4 – Balance of energy: The time rate of change of the internal energy

plus kinetic energy is equal to the heat energy that enters or leaves the body perunit time plus the sum of the rates of work of the external forces per unit time. Wepostulate that this statement remains true for an arbitrarily small neighborhood ofa material point.

The above statement is expressed by the equation

d

dt(E +K) = Q +∑

α

Wα ord

dt∫

V

ρ [e + 1

2(ν ⋅ϕ + v ⋅ v)]dv = Q +∑

α

Wα, (4.27)

where Q is the heat energy per unit time, and Wα is the αth kind of work ofexternal forces per unit time (mechanical, chemical, electrical, magnetical, etc.).

The above axiom implies that the energies are additive, and that if properaccounting is made of all the energies due to external effects, what is left over tobalance is the rate of the internal energy.

In particle mechanics, the conservation of energy is obtained as the first integralof Newton’s second law of motion when the forces are not explicit functions ofvelocity and time. In thermodynamics, it is stated as the first principle of ther-mostatics, applicable to systems in equilibrium. The axiom stated above is anextension of both that of classical mechanics and that of thermostatics. It is validfor every system, including dissipative ones, in which the energy principle of bothclassical mechanics and thermostatics fails to apply.

The first law of thermodynamics states that energy is conserved. However,there are many thermodynamic processes that conserve energy but that actuallynever occur. Furthermore, the first law of thermodynamics does not restrict ourability to convert work into heat or heat into work, except that energy must beconserved in the process. And yet in practice, although we can convert a givenquantity of work completely into heat, we have never been able to find a schemethat converts a given amount of heat completely into work. A quantity that wecall entropy provides a measure of a system’s thermal energy that is unavailablefor doing useful work. In terms of statistical mechanics, the entropy of a systemdescribes the number of possible microscopic configurations the particles in thesystem can have. The statistical definition of entropy is generally thought to bethe more fundamental definition, from which all other properties of entropy follow.Entropy is an extensive quantity that represents a state function of the system.

Definition: The entropy, S, of a continuous mass medium contained in V isgiven by

S ≡ ∫V

ρη dv, (4.28)

156 MECHANICS AND THERMODYNAMICS

where a local entropy density per unit mass, η, has been assumed to exist.Axiom 5 – Second law of thermodynamics: The time rate of change of the

total entropy S in an arbitrary material body having volume V enclosed by surfaceS with exterior normal n is never less than the sum of the contact entropy supplyΠ through the surface of the body and the entropy B produced by external sources.It is postulated that this is true for all parts of the body and for all independentprocesses.

According to the first part of the above axiom, also referred to as the entropyinequality, we have

Γ ≡ dSdt− B −Π ≥ 0 or Γ ≡ d

dt∫

V

ρη dv −B −Π ≥ 0, (4.29)

where Γ is the total entropy production.

4.4 Forces and moments

Generically, we will refer to forces and moments (or couples) as loads. In contin-uum mechanics, as opposed to particle mechanics, loads may depend on spatialgradients of various orders, their various time rates and integrals, as well as othervariables pertaining to various loads of mechanical, electrical, or some other origin.(While they may be included, we assume that there are no concentrated loads act-ing at points.) A body will undergo a deformation when subjected to loads thatmay be either external (acting on the body) or internal (acting between two partsof the same body) in character. By a suitable choice of a free body imagined to becut out of the complete body, any internal load in the original body may becomean external load on the isolated body. The term free body denotes a portion of thecomplete body instantaneously bounded by an arbitrary closed surface.

External loads arise from external effects and are classified as being either volu-metric (or body) loads or surface (or contact) loads:

i) Body loads act on elements of the volume or mass inside the body (e.g., grav-ity). These are “action-at-a-distance” forces. A body load density per unitmass is assumed to exist. External body loads are assumed to be objective.

ii) Contact loads arise from the action of one body upon another through thebounding surface contact. A surface force density per unit area, called thesurface traction, and a surface couple density per unit area called the surfacecouple, are assumed to exist. Surface tractions and couples depend on theorientation of the surface on which they act.

Internal (or mutual) loads are the result of the mutual interaction of pairs ofmaterial particles that are located in the interior of the body. According to New-ton’s third law, the mutual action of a pair of particles consists of two forces actingalong the line connecting the particles (central forces), equal in magnitude, andopposite in direction to one another. Therefore, the resultant internal force andcouple are zero. Subsequently, such mutual loads are objective. However, we shallnot make this assumption for the moment and thus allow the existence of a bodycouple and a couple stress vector. The effect of interparticle forces in a continuumappears in the form of a resultant effect of one part of the body on another part

4.4. FORCES AND MOMENTS 157

of the body through the latter’s bounding surface. This concept gives rise to thestress hypothesis, and the existence of quantities called stress and couple stress,which will be discussed later.

In mechanics, real forces are always exerted by one body on another body (pos-sibly by one part of a body acting on another part), regardless of whether they arebody forces or surface forces. Now let f be the body force per unit mass acting onan infinitesimal volume dv of the body, so that body force on the volume is ρf dv.In general, the vector f varies from point to point in the body at any given timeand may also vary with time at any given point, thus f(x, t). The vector sum ofthe body forces acting on an arbitrary finite volume V is then given by the spacialintegral over the volume

∫V

ρf dv. (4.30)

Let t(n) be the surface traction per unit area acting on the differential surfaceds of the body with exterior normal n. In general, in addition to being dependenton the surface normal n, the vector t varies from point to point on the body surfaceat any given time and may also vary with time at any given point, thus t(n,x, t).However, here we only display the dependence on n, while the other dependencesare suppressed for the moment. The force exerted across the differential areaelement is then t(n)ds, and the vector sum of the forces across the boundingsurface S of the arbitrary volume V is given by the vector surface integral

∫S

t(n)ds. (4.31)

Subsequently, the resultant force acting on a body is given by

F = ∫S

t(n)ds +∫V

ρf dv. (4.32)

The moment of a force g about a point xP is given by the vector product r×g.The total moment on a given body of volume V bounded by a closed surface S

is the sum of the total moments due to external and internal forces and couples.If the force is the body force ρf , then the moments about the three axes are givenby r × ρf dv, and the total moment of all forces acting on a finite volume is givenby

∫V

r × ρf dv. (4.33)

The moment of surface traction t(n) on an element of area ds can be expressedin a similar fashion, and the total of the distributed force on a finite surface S isobtained by means of the surface integral

∫S

r × t(n)ds. (4.34)

To account for the internal spin, we let l and m(n), respectively, denote theexternal supply of spin per unit mass and the contact couple stress vector perunit surface area. In general, the axial vector l (or its associated skew-symmetricsecond-rank external supply tensor L = l ⋅ǫ) varies from point to point in the body

158 MECHANICS AND THERMODYNAMICS

at any given time and may also vary with time at any given point, so that l(x, t),and in general, the axial vector m (or its associated skew-symmetric second-rankcouple stress tensor M = m ⋅ ǫ), in addition to being dependent on the surfacenormal n, also varies from point to point on the body surface at any given timeand may also vary with time at any given point, so that m(n,x, t); however, wewill suppress the other dependences for the moment. The vector sum of all thebody couples l acting on an arbitrary finite volume V is given by the space integralover the volume

∫V

ρldv, (4.35)

and the vector sum of all surface couples m(n) acting on an arbitrary finite surfaceS is given by the space integral over the surface

∫S

m(n)ds. (4.36)

Subsequently, the resultant moment about xc is given by

m = ∫S

[m(n) + r × t(n)]ds + ∫V

ρ (l + r × f) dv (4.37)

or

M = ∫S

[M(n) + r ∧ t(n)]ds +∫V

ρ (L + r ∧ f)dv. (4.38)

Now, using (4.32), (4.37), and (4.38), the basic Axioms 2 and 3 (see (4.23)–(4.25))become

d

dt∫

V

ρv dv = ∫S

t(n)ds +∫V

ρf dv (4.39)

and

d

dt∫

V

ρ (ϕ + r × v) dv = ∫S

[m(n) + r × t(n)]ds +∫V

ρ (l + r × f)dv (4.40)

or

d

dt∫

V

ρ (Φ + r ∧ v) dv = ∫S

[M(n) + r ∧ t(n)]ds +∫V

ρ (L + r ∧ f) dv. (4.41)

These are Euler’s equations that govern the global motion of the body. Note thatthe volume and surface integrals can be taken over the space occupied instan-taneously by the deformed configuration of the body, as is usually done in solidmechanics, or applied to a given fixed volume of space (with the closed surfacecalled a control surface), which at different times does not usually contain thesame material, as is usually done in fluid mechanics.

4.5 Rigid body dynamics

A deformation is rigid if the distance between any two material points remainsconstant during the motion, i.e.,

∣x(X, t) − x(Y, t)∣2 = ∣X −Y∣2 , (4.42)

4.5. RIGID BODY DYNAMICS 159

where X and Y denote two points in the reference configuration. By differentiatingwith respect to X, we obtain

FT (X, t) ⋅ [x(X, t) − x(Y, t)] =X −Y, (4.43)

where F(X, t) is the usual deformation gradient. Repeating the procedure bydifferentiating with respect to Y, we obtain

FT (Y, t) ⋅ [x(X, t) − x(Y, t)] =X −Y. (4.44)

Subsequently, it follows that

F(X, t) ⋅FT (Y, t) = 1. (4.45)

Since this result is valid for any arbitrary point Y, then we must have that thedeformation gradient is independent of the spacial location, i.e., F(X, t) = F(t),so

F(t) ⋅FT (t) = 1, (4.46)

and thus the deformation gradient is orthogonal. From the polar decomposition,we subsequently have that F(t) = R(t), U = V = 1, and we can rewrite (4.43) asthe rigid motion

x(X, t) = b(t) +R(t) ⋅X, (4.47)

where b(t) = x(Y, t)−R(t) ⋅Y. It is easily verified that the motion (4.47) satisfies(4.42). In particular, if Xc is the center of mass in the reference configuration, i.e.,

Xc =XP +∫Vρ(X) (X −XP ) dV, (4.48)

and XP is an arbitrary point, then

xc(t) = b(t) +R(t) ⋅Xc, (4.49)

where xc(t) ≡ x(Xc, t). Now subtracting (4.49) from (4.47), we obtain

[x(X, t) − xc(t)] =R(t) ⋅ (X −Xc) . (4.50)

Differentiating with respect to time and substituting (4.50), we obtain

υ(x, t) =W(t) ⋅ r(t) = −(t) × r(t), (4.51)

where υ(x, t) = v(x, t)−vc(t) is the particle velocity relative to the velocity of thecenter of mass (barycentric velocity), W(t) = R(t) ⋅RT (t) is the skew-symmetricrigid spin tensor, and (t) = ⟨W(t)⟩ is the angular velocity, which is the axialvector corresponding to the spin tensor (see (2.147)). From (3.491) and (3.492), wealso see that D(x, t) = 0, W(x, t) =W(t), and then it follows that L(x, t) =W(t),or from (4.51) that

W(t) = 1

2((gradv)T − gradv) and (t) = −1

2curlv. (4.52)

We also note from (4.12) that

∫V

ρυ dv = 0. (4.53)

160 MECHANICS AND THERMODYNAMICS

Now, using (4.12) and (4.22), we can write the linear momentum equation (4.23)in the form

Mac = F , (4.54)

where ac = vc = xc is the acceleration of the center of mass. Equation (4.54) isjust Newton’s equation of motion of the center of mass of the material body.

Below we will take the material to be nonpolar. The moment of momentumequation (4.16) can then be rewritten by separating the contributions of the centerof mass and that relative to the center of mass:

h = rc ×Mvc +∫V

(x − xc) × ρ (v − vc)dv, (4.55)

where rc = xc−xP is the moment arm about the fixed position xP . In the followingdevelopment, we select the point xP to correspond to the center of mass, i.e.,r = x − xc and rc = 0. Subsequently, we then have

h = ∫V

r × ρυ dv and H = ∫V

r ∧ ρυ dv. (4.56)

Substituting (4.51) into (4.56), we obtain

h = −Ic ⋅ and H = − (I ⋅W +W ⋅ I) , (4.57)

where we have defined the inertia tensors

I(t) ≡ ∫V

ρ rrdv and Ic(t) ≡ (tr I(t))1 − I(t). (4.58)

Principal moments of inertia, I(i)c for i = 1,2,3, can be readily obtained from

det (Ic − Ic1) = 0. (4.59)

Similarly, we can rewrite the kinetic energy by separating the contributions ofthe center of mass and that relative to it:

K = 1

2Mvc ⋅ vc +

1

2∫

V

ρυ ⋅ υ dv (4.60)

and subsequently it can be rewritten in the following forms:

K = 1

2Mvc ⋅ vc +

1

2 ⋅ Ic ⋅ = 1

2Mvc ⋅ vc +

1

2W ⋅ I ⋅W. (4.61)

We now note that

I ≡ I −W ⋅ I + I ⋅W = 0 and Ic = −I, (4.62)

where

I is the corotational or Jaumann derivative (see (3.422)), so that subse-quently, by taking the time derivative of equations (4.56), we can write the equa-tions for the moment of momentum of a rigid body as

h = −Ic⋅ ˙ +×(Ic ⋅) =m and H = − (I ⋅ W + W ⋅ I)+(I ⋅W2−W2

⋅ I) =M.

(4.63)

4.6. STRESS AND COUPLE STRESS HYPOTHESES 161

In terms of the principal moments of inertia, we can rewrite the above equation as

I(1)c ˙ 1 +23 (I(2)c − I(3)c ) = −m1, (4.64)

I(2)c ˙ 2 +31 (I(3)c − I(1)c ) = −m2, (4.65)

I(3)c ˙ 3 +12 (I(1)c − I(2)c ) = −m3. (4.66)

These correspond to the Euler equations of motion for a rigid body about thecenter of mass.

Lastly, taking the time derivative of the kinetic energy (4.61), and using (4.54),(4.58), and (4.62), we readily see that

K = vc ⋅F + ⋅m = vc ⋅F +W ∶M. (4.67)

That is, the time rate of change of the kinetic energy is equal to the total powerexpended by the external forces and couples.

4.6 Stress and couple stress hypotheses

Internal loads and their connection to surface loads may be understood by theapplication of the balance of momenta on a small continuous region of volume ∆v

bounded by the closed surface ∆s fully contained in a body. At a point on thesurface, the effect of the other part of the body is equivalent to a system of forcest(n) called stress vectors and surface couples m(n) called couple stress vectors.Both stress and couple stress vectors are objective. On surfaces passing throughthe same point, but oriented differently, the stress and couple stress vectors arein general different. Thus these loads depend not only on their location on thesurface but also on the exterior normal vector of the surface. This is why we areindicating this dependence explicitly. To determine the dependence of the stressand couple stress vectors on n, we apply the balance of momenta to a volume in theshape of a small tetrahedron with its vertex fixed at r and having three of its faceson the Cartesian coordinate surfaces and the fourth face being a curved surface,as illustrated in Fig. 4.2. The stress vector on the coordinate surface xk = const.(whose exterior normal is in the −ik direction) is denoted by t(−ik).

Now from the mean value theorem we can define the mean density

ρ⋆ ≡ 1

∆v∫∆vρ dv, (4.68)

the mean linear momentum

ρ⋆v⋆ ≡ 1

∆v∫∆vρv dv, (4.69)

the mean body force

ρ⋆f⋆ ≡ 1

∆v∫∆vρf dv, (4.70)

and the mean stress vector

t⋆(n) ≡ 1

∆s∫∆s

t(n) ds. (4.71)

162 MECHANICS AND THERMODYNAMICS

x1

x2

x3

t⋆(n)n

t⋆(−i1)−i1

t⋆(−i2)−i2

t⋆(−i3)−i3

i3

i2

i1

∆l 1

∆l2

∆l3

r

Figure 4.2: Infinitesimal tetrahedral volume ∆v with surface ∆s.

It is stressed that from the mean value theorem, there exist points within ∆v wherevolumetric mean quantities are defined, and points within the curved surface ∆s

or the coordinate surface ∆sk where corresponding mean surface quantities aredefined. Subsequently, the balances of mass (4.22) and linear momentum (4.39)applied to the tetrahedron, by making use of the mean quantities, become

d

dt(ρ⋆∆v) = 0, (4.72)

and

d

dt(ρ⋆v⋆∆v) = t⋆(n)∆s + t⋆(−ik)∆sk + ρ⋆f⋆∆v. (4.73)

We note that by using the mass balance, the balance of linear momentum can berewritten in the form

ρ⋆ (v⋆ − f⋆)∆v = t⋆(n)∆s + t⋆(−ik)∆sk. (4.74)

Now dividing both sides of the equation by ∆s, letting ∆v → 0 and ∆s → 0 sothat the curved surface approaches r, noting that ∆v/∆s → 0, and assuming thatthe quantities ρ⋆, v⋆, f⋆, and ∆sk/∆s remain bounded, we obtain

t(n)ds = −t(−ik)dsk. (4.75)

If we denote by ∆lk a typical dimension of our tetrahedron in the xk direction (seeFig. 4.2), and if in the above limiting process we take ∆v → 0 and ∆lk → 0, weobtain

t(ik) = −t(−ik), (4.76)

4.6. STRESS AND COUPLE STRESS HYPOTHESES 163

which demonstrates that the stress vector acting on opposite sides of the samesurface at a given point is equal in magnitude and opposite in sign. This resultis known as Cauchy’s lemma. The four faces of the tetrahedron form a closedsurface. Therefore, in the limit, the vector sum of the coordinate surfaces mustadd up to the area vector ds, i.e.,

ds = nds or dsk = nk ds. (4.77)

Substituting the last two expressions in our limiting result, we obtain

t(n) = t(ik)nk. (4.78)

This proves that the stress vector at a point on a surface with an exterior normal nis a linear function of the stress vectors acting on the coordinate surfaces throughthe same point, the coefficients being the direction cosines of n.

Analogously, from the mean value theorem, we can define the mean internalspin

ρ⋆ϕ⋆ ≡ 1

∆v∫∆vρϕ dv, (4.79)

the mean body couple

ρ⋆l⋆ ≡ 1

∆v∫∆vρl dv, (4.80)

and the mean surface couple

m⋆(n) ≡ 1

∆s∫∆s

m(n) ds. (4.81)

Subsequently, using the mass balance (4.72) and the linear momentum balance(4.74) on the tetrahedron, the angular momentum balance (4.40) becomes a bal-ance for the internal spin:

ρ⋆ (ϕ⋆ − l⋆)∆v =m⋆(n)∆s +m⋆(−ik)∆sk. (4.82)

Now, assuming that the quantities ρ⋆, ϕ⋆, and l⋆ remain bounded in the limit∆v → 0, it is immediately clear that application of the same procedure to thebalance of internal spin leads to the equivalent statements for the couple stressaxial vector:

m(ik) = −m(−ik), (4.83)

and

m(n) =m(ik)nk. (4.84)

4.6.1 Stress and couple stress tensors

We now define the lth component of the stress vector t(ik) acting on the positiveside of the kth coordinate surface as

t(ik) ≡ tk ≡ σlkil. (4.85)

164 MECHANICS AND THERMODYNAMICS

O

x1

x2

x3

x

t3

t2

−t3

−t2

−σ22

−σ32

−σ12

−σ33

−σ13

−σ23

σ22

σ32

σ12

σ13σ23

σ33

Figure 4.3: Components of stress tensor.

The components σlk correspond to the components of a second-order tensor σ thatwe call the stress tensor:

σ = σlkilik. (4.86)

It is a simple exercise to show that the stress tensor is objective. As we see, nowwriting complete explicit dependencies, the stress vector t(n,x, t) is related to thestress tensor σ(x, t) by the linear transformation

t(n,x, t) = σ(x, t) ⋅ n or tl(ik,x, t) = σlk(x, t). (4.87)

This result is known as Cauchy’s theorem, and it asserts that t(n,x, t) dependsupon the surface orientation only in a linear fashion. Note that the second sub-script in σlk indicates the coordinate surface xk = const. on which the stress vectortk acts, and the first subscript indicates the direction of the component of tk, as il-lustrated in Fig. 4.3. Some authors use the opposite convention for the subscripts.

The components σkk are called normal stresses and the mixed components σlkwith l ≠ k are called shearing stresses. In matrix form they are given by

[σlk] =⎡⎢⎢⎢⎢⎢⎣σ11 σ12 σ13σ21 σ22 σ23σ31 σ32 σ33

⎤⎥⎥⎥⎥⎥⎦. (4.88)

4.7. LOCAL FORMS OF AXIOMS OF MECHANICS 165

A stress is called hydrostatic if and only if

t(n,x, t) = σ(x, t) ⋅ n = −p(x, t)n (4.89)

for all n, where p is a scalar independent of n (the negative sign is used by con-vention). Alternately, the stress is hydrostatic if and only if every component ofσ transforms like a scalar.

It can be shown in a similar manner that the axial couple stress vector m(n,x, t)is related to an axial tensor Σ(x, t) by the linear transformation

m(n,x, t) =Σ(x, t) ⋅ n or ml(ik,x, t) = Σlk(x, t), (4.90)

where

Σ = Σlkilik (4.91)

is an objective second-order axial tensor called the couple stress tensor. This tensoris dual to the third-order tensor

Σ =Σ ⋅ ǫ (4.92)

that is skew-symmetric with respect to the first pair of indices, i.e., Σjkl = −Σkjl.

Its irreducible symmetry is analogous to that of Ξ given by (3.366)–(3.370).With the introductions of the stress and the couple stress tensors through (4.87)

and (4.90), the balances of linear and angular momenta (4.39) and (4.40) give riseto Cauchy’s first and second laws of motion:

d

dt∫

V

ρv dv = ∫S

σ ⋅ ds +∫V

ρf dv (4.93)

and

d

dt∫

V

ρ (ϕ + r × v) dv = ∫S

(Σ + r ×σ) ⋅ ds + ∫V

ρ (l + r × f) dv (4.94)

or

d

dt∫

V

ρ (Φ + r ∧ v) dv = ∫S

(Σ + r ∧σ) ⋅ ds +∫V

ρ (L + r ∧ f) dv. (4.95)

4.7 Local forms of axioms of mechanics

The local form of the mass conservation equation is obtained by comparing theintegral form of the equation (4.22), which reflects the statement of Axiom 1, withthe general balance law (4.1). In this comparison, we note that ψ → ρ, t → 0, andg → 0. Subsequently, from (4.6) we have

∫V (t)−ζ(t)

[∂ρ∂t+ div (ρv)]dv +∫

ζ(t)Jρ(v − c)K ⋅ndζ = 0. (4.96)

Setting the integrand of the first integral to zero, we obtain the local mass conser-vation which applies at a regular point in a continuous region

∂ρ

∂t+ div (ρv) = 0, (4.97)

166 MECHANICS AND THERMODYNAMICS

which can also be rewritten in the form

ρ + ρ divv = 0 or ρ + ρvk,k = 0, (4.98)

where we use the definition of the material derivative:

ρ = ∂ρ∂t+ (v ⋅ grad)ρ. (4.99)

It should be noted that, in contrast to (4.21)2, in general ρ can be determinedfrom v without knowing the motion χ explicitly. If we set the integrand of thesecond integral to zero, we obtain the jump condition across the singular movingsurface ζ:

Jρ(v − c)K ⋅ n = 0 or Jρ(vk − ck)Knk = 0. (4.100)

The jump condition allows us to define the mass flux (per unit area)

m = ρ (v − c) ⋅n, (4.101)

so we see that the mass flux is continuous across the singular surface:

JmK = 0. (4.102)

Note that m is continuous at the jump, so m+ =m− =m, where m is the value atthe singular surface. Furthermore, since

JvK ⋅ n = Jv − cK ⋅ n =s1

ρρ (v − c) ⋅ n =m JvK , (4.103)

where v = 1/ρ is the specific volume, we can decompose the jump in velocity intotangential and normal components,

JvK = Jv(n)K + Jv(s)K =m JvKn + Jv(s)K s, (4.104)

where

v(n) = v(n) n = (v ⋅n)n and v(s) = v(s) s = (v ⋅ s)s = v ⋅ (1 − nn) (4.105)

are the normal and tangential velocity components, Jv(s)K is the slip, and s is avector tangent to the singular surface.

It will be particularly useful to take advantage of the local mass balances (4.97)and (4.102) in conjunction with taking ψ → ρψ and g → ρg in the general balancelaws (4.1) and (4.6) for an arbitrary volume. In such case, the balance laws reduceto the more suggestive forms

d

dt∫

V

ρψ dv = ∫S

t ⋅ ds +∫V

ρg dv (4.106)

and

∫V (t)−ζ(t)

[ρ(ψ − g) − div t]dv + ∫ζ(t)(mJψK − JtK ⋅ n)dζ = 0. (4.107)

4.7. LOCAL FORMS OF AXIOMS OF MECHANICS 167

The local form of the balance of linear momentum is obtained by comparing theintegral form of equation (4.93), which reflects the statement of Axiom 2, with thereformulated general balance laws given above. In this comparison, we note thatψ → v, t → σ, and g → f . Subsequently, we write

∫V (t)−ζ(t)

[ρ (a − f) − divσ]dv +∫ζ(t)(mJvK − JσK ⋅ n)dζ = 0, (4.108)

where a = v is the material particle acceleration.Setting the integrand of the first integral to zero, we obtain the local balance of

linear momentum, which applies at a regular point in a continuous region

ρ (a − f) = divσ or ρ (al − fl) = σlk,k, (4.109)

and setting the integrand of the second integral to zero, we obtain the jump con-dition across the singular moving surface ζ

mJvK − JσK ⋅ n = 0 or mJvlK − JσlkKnk = 0. (4.110)

The above equation valid at a regular point is sometimes called Cauchy’s first lawof motion. Taking the inner product of (4.110) with n and s, and using (4.104)and (4.105), we have

m2JvK − n ⋅ JσK ⋅n = 0 and mJv(s)K − s ⋅ JσK ⋅ n = 0. (4.111)

Lastly, it should be emphasized that in general a can be determined from v withoutknowing the motion χ explicitly. In arriving at (4.109) from (4.108), we haveassumed that the body force f is continuous.

Lastly, the local form of the equation stating the balance of angular momentumis obtained by comparing the integral form of equation (4.94), that reflects thestatement of Axiom 3, with the reformulated balance laws (4.106) and (4.107). Inthis comparison, we note that ψ → (ϕ + r × v), t→ (Σ+ r×σ), and g → (l+ r× f).Subsequently, we have

∫V (t)−ζ(t)

ρ [(ϕ + r × a) − (l + r × f)] − div (Σ + r ×σ)dv +∫ζ(t)(mJϕ + r × vK − JΣ + r ×σK ⋅n) dζ = 0. (4.112)

We now recognize that

div (r ×σ) = r × (divσ) − ε ∶ σ, (4.113)

so that, by rearranging the above integrals, we have

∫V (t)−ζ(t)

[ρ (ϕ − l)− divΣ + ε ∶ σ] + r × [ρ (a − f) − divσ]dv +∫ζ(t)[(mJϕK − JΣK ⋅ n) + r × (mJvK − JσK ⋅ n)]dζ = 0. (4.114)

Using the balances of linear momentum at regular and singular points (4.109) and(4.110), we finally obtain

∫V (t)−ζ(t)

[ρ (ϕ − l)− divΣ + ε ∶ σ]dv + ∫ζ(t)(mJϕK − JΣK ⋅ n)dζ = 0. (4.115)

168 MECHANICS AND THERMODYNAMICS

Now setting the integrand of the first integral to zero, we obtain the local balanceof angular momentum, which applies at a regular point in a continuous region

ρ (ϕ − l) = divΣ − ε ∶ σ or ρ (ϕl − ll) = Σlk,k − εlkmσkm, (4.116)

and setting the integrand of the second integral to zero, we obtain the jump con-dition across the singular moving surface ζ

mJϕK − JΣK ⋅n = 0 or mJϕlK − JΣlkKnk = 0. (4.117)

The above equation valid at a regular point is sometimes called Cauchy’s secondlaw of motion. In arriving at (4.116) from (4.115), we have assumed that theinternal body couple l is continuous. Note that we can also write (4.116) in theform

ρ (Φ − L) = div Σ − skwσ or ρ (Φ[ij] − L[ij]) = Σ[ij]k,k − σ[ij]. (4.118)

We see that if ϕ = 0 (from (4.14) when i = 0), l = 0, and Σ = 0, then (4.116)becomes

ε ∶ σ = 0, (4.119)

or

σ = σT or σlk = σkl, (4.120)

i.e., the stress tensor is symmetric. Materials in which internal spin, body couples,and couple stresses occur are called polar materials, and those in which they do notoccur are called nonpolar materials. A couple, in classical mechanics, is viewed asa pair of parallel (bounded) forces having equal magnitude and opposite direction,separated by a moment arm. If we let the moment arm approach zero, whichoccurs when we previously took ∆v → 0 and ∆s → 0, the moment of the coupleapproaches zero. Thus, in the classical continuum mechanics picture of centralforces, one cannot have a body couple or a couple stress vector without havingsome other forces acting on the body (e.g., due to a magnetic field). Nevertheless,if the infinitesimals dv and ds are considered to represent some physical volumeand surface small enough, but with an acceptable variability, as noted in theintroduction, then the existence of an internal spin density, a body couple, and acouple stress vector may be admitted even without the need of additional forces.

By taking the inner product of v with (4.109) and ν with (4.116), we find thatthe local balance of kinetic energy is given by

1

2ρ ( ˙ν ⋅ϕ + ˙v ⋅ v) = div (ν ⋅Σ + v ⋅σ) + ρ (ν ⋅ l + v ⋅ f) −Φ (4.121)

or

1

2ρ ( ˙νlϕl +

˙vlvl) = (νlΣlk + vlσlk),k + ρ (νlll + vlfl) −Φ, (4.122)

where, using (3.362), we have

Φ ≡ 1

2ρν ⋅ i ⋅ ν +Ξ ∶Σ + (Υ +L) ∶ σ (4.123)

4.8. PROPERTIES OF STRESS VECTOR AND TENSOR 169

or

Φ ≡ 1

2ρνk iklνl +ΞklΣkl + (Υkl +Lkl)σkl (4.124)

which is called the mechanical energy, also called the stress power, that may beregarded as an internal production of energy, and we recall from (4.14), (3.238),

and (3.364) that ϕ = i ⋅ν, Ξ ≡ (grad ν)T , and L = (grad v)T . The term 1

2ρν ⋅ i ⋅ν

represents the internal spin production. If the internal specific inertia tensor is suchthat i = 0, the material is called micropolar, and if i ≠ 0, it is called micromorphic;it is called a microstretch material if i = i1.

The kinetic energy equation (4.121) can also be rewritten in integral form ifit is integrated over an arbitrary continuous volume and use is made of (4.106),(4.107), (2.289), (4.87), and (4.90):

d

dt∫

V

1

2ρ (ν ⋅ϕ + v ⋅ v) dv = ∫

S

[ν ⋅m(n) + v ⋅ t(n)]ds +∫

V

ρ (ν ⋅ l + v ⋅ f) dv −∫V

Φdv. (4.125)

4.8 Properties of stress vector and tensor

The discussions in this section apply to nonpolar materials for which the Cauchystress tensor σ is symmetric.

4.8.1 Principal stresses and principal stress directions

The stress vector t(n) acting at a point on a surface with exterior normal n canbe decomposed into a component normal to the surface and a component that istangent to it:

t(n) = t(n)(n) + t(s)(n), (4.126)

wheret(n)(n) = σn n and t(s)(n) = τ s, (4.127)

and the normal and tangential vectors are such that

n ⋅ n = 1, s ⋅ s = 1, and n ⋅ s = 0. (4.128)

Note that, using (4.87), we have that

σn = n⋅t(n) = n⋅σ ⋅n, τ = s⋅t(n) = s⋅σ ⋅n, and τ2 = t(n)⋅t(n)−σ2

n. (4.129)

Furthermore, the stress tensor can be written in terms of the normal and shearcomponents as follows:

σ = σn nn + τ (ns + sn) . (4.130)

In analyzing internal (mutual) tractions at a material point, it is useful to findthe direction for which the traction vector is purely normal. The traction vector ispurely normal in the direction n when t(n) = σ ⋅ n = σn for some σ. Subsequently,we can write

(σ − σ1) ⋅ n = 0 (4.131)

170 MECHANICS AND THERMODYNAMICS

to find σ and n. This is just an eigenvalue problem. The normalized eigenvectors±n(l) are called the principal axes of stress and the eigenvalues σ(l), given by theroots of det[σij − σδij] = 0, are called the principal stresses. Since for nonpolarmaterials σij is symmetric, the principal stresses σ(l) are real and the principalaxes are orthogonal (see Section 3.1.7). The characteristic equation is given bythe cubic

σ3− σ(1)σ2

+ σ(2)σ − σ(3) = 0, (4.132)

where the stress invariants are given by (see (3.92)–(3.94) and (3.135)–(3.137))

σ(1) = σ(1) + σ(2) + σ(3) = σii, (4.133)

σ(2) = σ(1)σ(2) + σ(2)σ(3) + σ(3)σ(1) = 1

2(σiiσjj − σijσji) , (4.134)

σ(3) = σ(1)σ(2)σ(3) = det [σij] . (4.135)

The above invariants provide formal expressions for the principal stresses in termsof stress tensor components. Here we follow the convention of ordering the princi-pal stresses so that σ(1) ≥ σ(2) ≥ σ(3). If σ(k) is positive, it is said to be a tension,and if it is negative, it is said to be a compression. If σ(1) ≠ 0 and σ(2) = σ(3) = 0,then we have a simple tension or compression. If one and only one principal stressvanishes, the state of stress is called biaxial. We have a triaxial stress otherwise.When the principal axes and stresses are known, stress components are then givenby the following spectral decomposition:

σ =NT⋅Λσ ⋅N, (4.136)

where Njl = n(l)j (note that N−1 = NT ) and Λσ is the diagonal matrix composedof principal stresses.

Suppose that σn is kept fixed and the orientation n of the surface at a point isvaried. This is accomplished by defining different volumes with different surfacesall going through the same point. Then

σn = t(n) ⋅ n = σlknlnk = const. (4.137)

represents a quadric surface (see Section 2.10) called the stress quadric of Cauchy.We would like to find the extremal values of σn, which are in the directions n.Accordingly, we can write

f(nk) = σij ninj − σ(nini − 1), (4.138)

where σ serves the role of a Lagrange multiplier to impose the normalizationconstraint. Now differentiating f(nk) with respect to nk, noting that ∂ni/∂nk =δik, and taking advantage of the symmetry of σij , the extremal problem reducesto the solution of the following problem:

(σij − σδij) nj = 0, (4.139)

which is identical to the eigenvalue problem for the principal stresses. Thus, theLagrange multiplier σ is the same as a principal stress. Furthermore, the principal

4.8. PROPERTIES OF STRESS VECTOR AND TENSOR 171

stresses σ(l) include both the maximum and minimum values of normal stress, andthe vectors n(l) can also be thought of geometrically as the principal axes of thestress quadric surface. Note that referring to the principal axes with n = nkik,where nk are direction cosines, the stress vector can be rewritten as

t(n) = σ ⋅ n = σ(1)n1i1 + σ(2)n2i2 + σ

(3)n3i3, (4.140)

and, subsequently, the normal component takes the form

σn = t(n) ⋅ n = σ(1)n2

1 + σ(2)n2

2 + σ(3)n2

3. (4.141)

With regard to the extreme values of the shear component τ , it is easiest toexpress it in terms of the principal stresses:

τ2 = t(n) ⋅ t(n) − σ2

n,

= σ(1)2

n2

1 + σ(2)2n2

2 + σ(3)2n2

3 − (σ(1)n2

1 + σ(2)n2

2 + σ(3)n2

3)2 . (4.142)

Now since nknk = 1, we can eliminate one of the direction cosines from the aboveexpression, say n3, to obtain an expression in terms of n1 and n2:

τ2 = (σ(1)2 − σ(3)2) n2

1 + (σ(2)2 − σ(3)2) n2

2 + σ(3)2

− [(σ(1) − σ(3)) n2

1 + (σ(2) − σ(3)) n2

2 + σ(3)]2 . (4.143)

To find extreme values of τ(n1, n2), we differentiate the above expression withrespect to n1 and n2 and set the results equal to zero. After some algebra, wehave

∂τ2

∂n1

= n1 (σ(1) − σ(3))σ(1) − σ(3) − 2 [(σ(1) − σ(3)) n2

1+

(σ(2) − σ(3)) n2

2] = 0, (4.144)

∂τ2

∂n2

= n2 (σ(2) − σ(3))σ(2) − σ(3) − 2 [(σ(1) − σ(3)) n2

1+

(σ(2) − σ(3)) n2

2] = 0. (4.145)

An obvious solution is n1 = n2 = 0, for which n3 = ±1 and then τ = 0. This is theexpected result since n3 = ±1 is the principal plane where σn is an extreme valuefor which the shear component is zero. If we had eliminated n1 or n2 instead onn3, similar calculations would lead to the other two principal planes for which theshear component is also zero.

A second solution is obtained by taking n1 = 0 and solving the resulting secondquadratic equation for n2. The result is n2 = ±1/√2 and, from the normalization,we find n3 = ±1/√2. For this solution, the shear component is given by

τ = ±12(σ(2) − σ(3)) . (4.146)

As before, if we had taken n2 = 0 or n3 = 0, we would obtain equivalent solutions.The complete solutions are

τ =⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩±

1

2(σ(2) − σ(3)) when n1 = 0, n2 = ± 1√

2, n3 = ± 1√

2,

±1

2(σ(3) − σ(1)) when n1 = ± 1√

2, n2 = 0, n3 = ± 1√

2,

±1

2(σ(1) − σ(2)) when n1 = ± 1√

2, n2 = ± 1√

2, n3 = 0.

(4.147)

172 MECHANICS AND THERMODYNAMICS

Because of our ordering of the principal stresses, it is clear that the maximumshear stress is subsequently given by

τmax = 1

2∣σ(3) − σ(1)∣ . (4.148)

It may be shown that for distinctive principal stresses, the above are the only twopossible maximal solutions.

4.8.2 Mean stress and deviatoric stress tensor

It is often convenient to define the spherical (or mean) stress as

σ(0) = 1

3(σ(1) + σ(2) + σ(3)) = 1

3σkk = 1

3σ(1). (4.149)

Subsequently, using the relation (3.103), the deviatoric stress is defined as

σ′ij = σij − 1

3σ(1)δij . (4.150)

Now, recalling (3.107)–(3.109), we see that the invariants of the deviatoric stresstensor are related to those of the stress tensor by

σ′(1) = 0, (4.151)

σ′(2) = −1

3σ2

(1) + σ(2), (4.152)

σ′(3) = 2

27σ(1) −

1

3σ(1)σ(2) + σ(3). (4.153)

4.8.3 Lamé’s stress ellipsoid

Let the principal axes of the stress tensor be chosen as the coordinate axes (n1, n2,

n3) in the directions (i1, i2, i3), respectively. Then the components of stress vectort(n) are given by

t1(n) = σ(1)n1, t2(n) = σ(2)n2, t3(n) = σ(3)n3. (4.154)

Now solving the above equations for nk and substituting them into the unit vectornormalization condition

n2

1 + n2

2 + n2

3 = 1, (4.155)

we obtain the following equation:

( t1(n)σ(1)

)2 + ( t2(n)σ(2)

)2 + ( t3(n)σ(3)

)2 = 1. (4.156)

This is just the equation for an ellipsoid, called the Lamé’s stress ellipsoid, withreference to a system of rectangular coordinates with axes (t1(n), t2(n), t3(n))having semi-axes (σ(1), σ(2), σ(3)) with the stress vector t(n) issuing from theorigin.

4.8. PROPERTIES OF STRESS VECTOR AND TENSOR 173

4.8.4 Mohr’s circles

We note that the equations for given normal and shear components of the stressvector along with the normalization condition referred to the principal axes makeup a system of equations for the direction cosines nk:

σn = σ(1)n2

1 + σ(2)n2

2 + σ(3)n2

3, (4.157)

σ2

n + τ2 = σ(1)2n2

1 + σ(2)2n2

2 + σ(3)2n2

3, (4.158)

n2

1 + n2

2 + n2

3 = 1. (4.159)

The solution of this system, assuming that the principal stresses are distinct, isgiven by

n2

1 = (σn − σ(2)) (σn − σ(3)) + τ2(σ(1) − σ(2)) (σ(1) − σ(3)) , (4.160)

n2

2 = (σn − σ(3)) (σn − σ(1)) + τ2(σ(2) − σ(3)) (σ(2) − σ(1)) , (4.161)

n2

3 = (σn − σ(1))(σn − σ(2)) + τ2(σ(3) − σ(1)) (σ(3) − σ(2)) . (4.162)

These solutions can be interpreted graphically by using σn as the abscissa andτ as the ordinate as illustrated in Fig. 4.4. First note that, due to the orderingof the principal stresses, the denominators in the first and third solutions arepositive, while the denominator in the second solution is negative. Subsequently,the numerators must be such that

(σn − σ(2))(σn − σ(3)) + τ2 ≥ 0, (4.163)

(σn − σ(3))(σn − σ(1)) + τ2 ≤ 0, (4.164)

(σn − σ(1)) (σn − σ(2)) + τ2 ≥ 0, (4.165)

which can be rearranged in the following geometric quadratic forms:

[σn − 1

2(σ(2) + σ(3))]2 + τ2 ≥ [1

2(σ(2) − σ(3))]2 , (4.166)

[σn − 1

2(σ(1) + σ(3))]2 + τ2 ≤ [1

2(σ(1) − σ(3))]2 , (4.167)

[σn − 1

2(σ(1) + σ(2))]2 + τ2 ≥ [1

2(σ(1) − σ(2))]2 . (4.168)

Clearly, admissible stress vectors in the (σn, τ) plane lie in the region exteriorto the circles defined by the first and third equations and interior to the circledefined by the second equation. Furthermore, since the radius of the circle definedby the second equation is τmax, it must be such that this circle encloses the othertwo circles. These circles are known as Mohr’s circles.

174 MECHANICS AND THERMODYNAMICS

τ

σnσ(3) σ(2) σ(1)

Figure 4.4: Mohr’s circles.

4.9 Work and heat

For the sake of simplicity, we consider thermomechanical systems in which onlythe mechanical work W is present, i.e., ∑α Wα → W . Thus, the conservation ofenergy statement (4.27) takes the form

E + K = Q + W . (4.169)

In thermostatics, we have K = 0, and if we write dE ≡ Edt, dQ ≡ Qdt, and dW ≡Wdt, we arrive at the first law of thermodynamics:

dE = dQ + dW , (4.170)

where we make a distinction between exact differentials between states that arepath independent such as dE , and inexact differentials that are path dependentsuch as dQ and dW . Quantities that are path independent denote properties of thematerial body. Here we recognize that (4.170) should more accurately be calledthe first law of thermostatics.

The mechanical energy consists of the work done by the surface and body forcesper unit time (see (4.125)):

W = ∫S

[ν ⋅m(n) + v ⋅ t(n)]ds + ∫V

ρ (ν ⋅ l + v ⋅ f)dv, (4.171)

where

ν ⋅m(n) = (ν ⋅Σ) ⋅ n and v ⋅ t(n) = (v ⋅σ) ⋅ n. (4.172)

In a continuous medium, heat may enter an arbitrary region through contactheat supply per unit area H(n) through the surface with exterior normal n, or itmay be supplied volumetrically from external sources (e.g., thermal radiation) perunit mass of the body so that r denotes the energy supply density. In general, r

4.10. HEAT FLUX HYPOTHESIS 175

varies from point to point in the body at any given time and may also vary withtime at any given point, so that r(x, t), and in general, H , in addition to beingdependent on the surface normal n, also varies from point to point on the bodysurface at any given time and may also vary with time at any given point, so thatH(n,x, t); however, we will suppress the other dependences for the moment. Thenthe total heat input per unit time is given by

Q = ∫S

H(n) ds + ∫V

ρr dv. (4.173)

Subsequently, the balance of energy statement (4.169) becomes

d

dt∫

V

ρ(e + 1

2ν ⋅ϕ +

1

2v ⋅ v)dv = ∫

S

[ν ⋅m(n) + v ⋅ t(n) +H(n)]ds +∫

V

ρ (ν ⋅ l + v ⋅ f + r) dv. (4.174)

Note that by subtracting (4.125), we can also write the integral balance of internalenergy:

d

dt∫

V

ρedv = ∫S

H(n)ds +∫V

(Φ + ρr) dv. (4.175)

4.10 Heat flux hypothesis

To determine the dependence of the contact heat supply on n, we apply the inter-nal energy balance (4.175) to the same infinitesimal tetrahedral volume shown inFig. 4.2. From the mean value theorem, we define the mean internal energy

ρ⋆e⋆ ≡ 1

∆v∫∆vρedv, (4.176)

the mean external body energy source

ρ⋆r⋆ ≡ 1

∆v∫∆vρr dv, (4.177)

the mean contact heat supply

H⋆(n) ≡ 1

∆s∫∆sH(n)ds, (4.178)

and the mean mechanical energy

Φ⋆ ≡ 1

∆v∫∆v

Φdv. (4.179)

Subsequently, the internal energy balance (4.175) applied to the tetrahedron, bymaking use of mean quantities and the mass balance (4.72), becomes

ρ⋆ (e⋆ − r⋆)∆v = Φ⋆∆v +H⋆(n)∆s +H⋆(−ik)∆sk. (4.180)

Now dividing both sides of the equation by ∆s, letting ∆v → 0 and ∆s → 0 so thata point on the curved surface approaches r, noting that ∆v/∆s → 0, and assumingthat the quantities ρ⋆, e⋆, r⋆, Φ⋆, and ∆sk/∆s remain bounded, we obtain

H(n)ds = −H(−ik)dsk. (4.181)

176 MECHANICS AND THERMODYNAMICS

If we denote by ∆lk a typical dimension of our tetrahedron in the xk direction,and if in the above limiting process we take ∆v → 0 and ∆lk → 0, we obtain

H(ik) = −H(−ik), (4.182)

which demonstrates that the contact heat supply acting on opposite sides of thesame surfaces at a given point are equal in magnitude and opposite in sign. Thefour faces of the tetrahedron form a closed surface. As before, in the limit, thevector sum of the coordinate surfaces must add up to the area vector ds, i.e.,

ds = nds or dsk = nk ds. (4.183)

Substituting the last two expressions in our limiting result, we obtain

H(n) =H(ik)nk. (4.184)

This proves that the contact heat supply at a point on a surface with an exteriornormal n is a linear function of the heat supply acting on the coordinate surfacethrough the same point, the coefficient being the direction cosine of n.

We now define the component −qk as the component of the contact heat supplyH(ik) acting on the positive side of the kth coordinate surface (the negative signis used by convention):

H(ik) ≡ −qk. (4.185)

As we see, now writing the complete explicit dependence, the heat supplyH(n,x, t)is related to the heat flux vector q(x, t) by the linear transformation

H(n,x, t) = −q(x, t) ⋅ n or H(ik,x, t) = −qk(x, t). (4.186)

This result is the counterpart of Cauchy’s postulate and Cauchy’s theorem forcontact heat supply and is called the Fourier–Stokes heat flux theorem. It assertsthat H(n,x, t) depends upon the surface orientation only in a linear fashion.

Using the above result with (4.172), the global balance of energy (4.174) becomes

d

dt∫

V

ρ(e + 1

2v ⋅ v +

1

2ν ⋅ϕ)dv = ∫

S

(v ⋅σ + ν ⋅Σ − q) ⋅ ds +∫

V

ρ (v ⋅ f + ν ⋅ l + r) dv. (4.187)

4.11 Entropy flux hypothesis

The entropy inequality, which is reflected in Axiom 5, applied to an arbitrary bodyis given by

∫V

γ dv ≡ d

dt∫

V

ρη dv −∫V

ρb dv −∫S

π(n)ds ≥ 0, (4.188)

where, for a continuous material mass, local entropy supply per unit mass, contactentropy supply per unit area, and entropy production per unit volume have been

4.11. ENTROPY FLUX HYPOTHESIS 177

assumed to exist and given by

B ≡ ∫V

ρb dv, (4.189)

Π ≡ ∫S

π(n)ds, (4.190)

Γ ≡ ∫V

γ dv. (4.191)

In addition, we have assumed that the contact entropy supply is a function of thesurface exterior normal direction.

As before, from the mean value theorem, we can define the mean entropy density

ρ⋆η⋆ = 1

∆v∫∆vρη dv, (4.192)

the mean external entropy supply

ρ⋆b⋆ = 1

∆v∫∆vρb dv, (4.193)

and the mean contact entropy supply

π⋆(n) = 1

∆s∫∆sπ(n)ds. (4.194)

Subsequently, using the mass balance (4.72) applied to the infinitesimal tetrahe-dron in the positive quadrant shown in Fig. 4.5, we obtain

ρ⋆ (η⋆ − b⋆)∆v − π⋆(n)∆s − π⋆(−ik)∆sk ≥ 0. (4.195)

Now proceeding as before by taking ∆v → 0 and ∆s → 0 with ∆v/∆s → 0, weobtain

π(n)ds + π(−ik)dsk ≤ 0, (4.196)

and upon taking ∆v → 0 and ∆lk → 0, we obtain that

π(ik) = −π(−ik). (4.197)

Using the fact that dsk = nkds leads to the result that

π(n) − π(ik)nk ≤ 0. (4.198)

Repeating the above procedure over the tetrahedron in the negative quadrantshown in Fig. 4.5, we also obtain

π(−n)ds + π(ik)dsk ≤ 0, (4.199)

or

−π(n) + π(ik)nk ≤ 0. (4.200)

Now the only way that the quantity on the left-hand side of (4.198) and its negativeon the left-hand side of (4.200) be both less than or equal to zero is only if thequantity is equal to zero, i.e.,

π(n) = π(ik)nk. (4.201)

178 MECHANICS AND THERMODYNAMICS

x3

x2

n

x1

−n

Figure 4.5: Tetrahedral infinitesimal volumes.

This proves that the contact entropy supply at a point on a surface with an exteriornormal n is a linear function of the entropy supply acting on the coordinate surfacethrough the same point, the coefficient being the direction cosine of n.

We now define the component −hk as the component of the contact entropysupply P (ik) acting on the positive side of the kth coordinate surface (the negativesign is used by convention):

π(ik) ≡ −hk. (4.202)

As we see, now writing the complete explicit dependence, the entropy supplyP (n,x, t) is related to the entropy flux vector h(x, t) by the linear transformation

π(n,x, t) = −h(x, t) ⋅ n or π(ik,x, t) = −hk(x, t). (4.203)

This result is the counterpart of Cauchy’s theorem, but for contact entropy supply.It asserts that π(n,x, t) depends upon the surface orientation only in a linearfashion.

Using the above result in (4.188), we obtain the entropy inequality

∫V

γ dv ≡ d

dt∫

V

ρη dv −∫V

ρb dv + ∫S

h ⋅ ds ≥ 0. (4.204)

4.12 Local forms of axioms of thermodynamics

The local form of conservation of energy is obtained by comparing (4.187) withthe reformulated general balance laws (4.106). In this comparison, we note thatψ → (e+ 1

2v ⋅v+ 1

2ν ⋅ϕ), t → (v ⋅σ+ν ⋅Σ−q), and g → (v ⋅f +ν ⋅l+r). Subsequently,

4.12. LOCAL FORMS OF AXIOMS OF THERMODYNAMICS 179

from (4.107), we have

∫V (t)−ζ(t)

ρ [(e + 1

2˙v ⋅ v +

1

2˙ν ⋅ϕ) − (v ⋅ f + ν ⋅ l + r)]−div (v ⋅σ + ν ⋅Σ − q) dv +

∫ζ(t)(ms

e +1

2v ⋅ v +

1

2ν ⋅ϕ

− Jv ⋅σ + ν ⋅Σ − qK ⋅ n)dζ = 0. (4.205)

The above equation, upon subtracting the local equation of kinetic energy (4.121),can be rewritten in the form

∫V (t)−ζ(t)

ρ (e − r) −Φ + divqdv +∫ζ(t)(ms

e +1

2v ⋅ v +

1

2ν ⋅ϕ

− Jv ⋅σ + ν ⋅Σ − qK ⋅n)dζ = 0. (4.206)

Now setting the integrand of the first integral to zero, we obtain the local internalenergy balance, which applies at a regular point in a continuous region

ρ (e − r) = Φ − divq or ρ (e − r) = Φ − qk,k, (4.207)

and setting the integrand of the second integral to zero, we obtain the energy jumpcondition across the singular moving surface ζ

m

se +

1

2v ⋅ v +

1

2ν ⋅ϕ

− Jv ⋅σ + ν ⋅Σ − qK ⋅ n = 0 (4.208)

or

m

se +

1

2vlvl +

1

2νlϕl

− Jvlσlk + νlΣlk − qkKnk = 0. (4.209)

In arriving at (4.207) from (4.206), we have assumed that the external heat supplyr is continuous. Equation (4.207), which applies at a regular point in a continuousmedium, corresponds to the local first law of thermodynamics stating that theinternal energy change per unit time is due to the stress power, the heat flux,and the external heat supply. By decomposing the tensors appearing in the stresspower term (given in (4.124)) into spherical, deviatoric (indicated by primes),and skew-symmetric parts (see (2.110)), the stress power can be rewritten in thefollowing illuminating form:

Φ = [12ρ i(0) νkνk +m(0)νk,k + σ(0)vk,k]´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

(1)

+

[12ρ i′(kl)νkνl +Ξ

′

(kl)Σ′

(kl) +D′

(kl)σ′

(kl)]´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶(2)

+

[Ξ[kl]Σ[kl] + (Υ[kl] +W[kl])σ[kl]]´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶(3)

, (4.210)

180 MECHANICS AND THERMODYNAMICS

where σ(0) is defined in (4.149), and we have defined the spherical components ofthe internal inertia and couple stress tensors by

i(0) = 1

3tr i and m(0) = 1

3trΣ. (4.211)

We see that the stress power is given by three terms which in order represent(1) the local work of changing the volume (dilatation), (2) changing the shape(distortion), and (3) rotating the material element. Furthermore, for σ(0) = −p,with p a mechanical pressure, the third term in (1) represents the −p dV workterm, and the third term in (2) is just the product of the rate of distortion inshape and the shearing stresses. These are the only terms that apply for nonpolarmaterials since σ[kl] = 0 in this case.

The above jump condition can be rewritten in a more convenient and simplerform if we first note that, for any tensors A and B, we have that

JABK = JAK⟪B⟫ + ⟪A⟫JBK, (4.212)

where we have defined the average operator

⟪A⟫ ≡ 1

2(A+ +A−) . (4.213)

Subsequently, we first note that, using (4.212) and (4.110), we have

1

2m Jv ⋅ vK − Jv ⋅σK ⋅n = −JvK ⋅ ⟪σ⟫ ⋅n (4.214)

and, using (4.212) and (4.117), we have

1

2m Jν ⋅ϕK − Jν ⋅ΣK ⋅ n = −JνK ⋅ ⟪Σ⟫ ⋅ n. (4.215)

In obtaining this last equality, we have used the fact that ϕ = i ⋅ν (see (4.14)), andnoted that the internal inertia tensor i is symmetric and continuous across a jumpdiscontinuity since it is only a function of the local coordinates. Subsequently, wecan rewrite (4.208) as

m JeK − (JνK ⋅ ⟪Σ⟫ + JvK ⋅ ⟪σ⟫ − JqK) ⋅ n = 0. (4.216)

Using (4.104) and (4.129), we can also rewrite the above equation as

m (JeK − JvK⟪σn⟫) − (JνK ⋅ ⟪Σ⟫ ⋅ n + Jv(s)K⟪τ⟫ − JqK ⋅ n) = 0. (4.217)

The local form of the entropy inequality is obtained by comparing (4.204) withthe reformulated balance law (4.106). In this comparison we note that ψ → η,t → −h, and g → b. Subsequently, from (4.107), we have

∫V (t)

γ dv ≡ ∫V (t)−ζ(t)

[ρ(η − b) + divh]dv +∫ζ(t)(mJηK + JhK ⋅ n)dζ ≥ 0. (4.218)

Since the volume is arbitrary, choosing a volume in the regular region leads to thelocal entropy inequality, which applies at a regular point in the continuous region

γv ≡ ρ(η − b) + divh ≥ 0 or γv ≡ ρ(η − b) + hk,k ≥ 0, (4.219)

4.13. FIELD EQUATIONS IN EUCLIDEAN FRAMES 181

and setting the integrand of the last integral to zero, we obtain the jump conditionacross the singular moving surface ζ:

γs ≡mJηK + JhK ⋅n ≥ 0 or γs ≡mJηK + JhkKnk ≥ 0. (4.220)

In arriving at (4.219) from (4.218), we have assumed that the external entropysupply b is continuous. Note that γv and γs represent the entropy productions perunit volume in the continuous region and per unit area on the singular surface,respectively. Equation (4.219) corresponds to the local form of the second law ofthermodynamics.

4.13 Field equations in Euclidean frames

In the previous sections we have obtained the local field equations in an inertialframe. Nevertheless, we require that the density ρ, the forces f and t (and thereforethe stress tensor σ), the internal inertia tensor i, the couples l and m (and thereforethe couple stress tensor Σ), the internal energy e, the energy supplies r and h (andtherefore the heat flux q), the entropy η, and the entropy productions γv,s andsupplies b and P (and therefore the entropy flux h) be all frame indifferent undera Euclidean transformation, i.e.,

ρ′ = ρ, e′ = e, η′ = η, r′ = r, b′ = b, γ′v,s = γv,s,f ′ =Q ⋅ f , l′ = (detQ)Q ⋅ l, q′ =Q ⋅ q, h′ =Q ⋅ h, (4.221)

σ′ =Q ⋅σ ⋅QT , i′ =Q ⋅ i ⋅QT , Σ′ = (detQ)Q ⋅Σ ⋅QT .

We note that the frame indifference requirement of the internal spin, ϕ′, isconnected (see (4.14)) with the frame-indifference requirements of the internalinertia tensor, i′, and the internal angular velocity, ν′. Furthermore, as with theinternal angular velocity, the internal spin rate ϕ is not frame invariant since

ϕ′ = i′ ⋅ ν ′ + i′ ⋅ ν ′ = (detQ)Q ⋅ ϕ + j′, (4.222)

or

(ϕ′ − j′) ≡ ϕ = (detQ)Q ⋅ ϕ, (4.223)

where we have defined the inertial internal spin (see (2.147))

j′ ≡ φ′ +Ω ⋅ (ϕ′ −φ′) = φ′ −w × (ϕ′ −φ′) , (4.224)

whereφ′ = i′ ⋅w (4.225)

is the spin rate of the second frame relative to the first. We note that ϕ′ is the trueinternal spin rate while ϕ is called the apparent internal spin rate. The differencebetween these two internal spin rates is given by the inertial angular spin rate j′.

We recall that the material derivative of an objective scalar field is objective(not true for tensor fields in general); therefore,

(ρ)′ = ρ, (e)′ = e, (η)′ = η. (4.226)

182 MECHANICS AND THERMODYNAMICS

In addition, it is easy to show that the spatial divergence of velocity, heat flux,and entropy flux are frame invariant, i.e.,

(divv)′ = divv, (divq)′ = divq, (divh)′ = divh, (4.227)

and so is the stress power (4.123), i.e.,

Φ′ = Φ. (4.228)

Moreover, we can show that the spatial divergence of an objective second ranktensor field is also objective, and in particular

(divσ)′ =Q ⋅ (divσ), (divΣ)′ = (detQ)Q ⋅ (divΣ). (4.229)

Subsequently, from the above transformation properties, it follows immediatelythat the balances of mass, energy, and entropy are objective scalar equations andare thus valid in arbitrary Euclidean frames. On the other hand, the balancesof linear and angular momenta are only Galilean invariant since they contain theacceleration and the internal spin rate. In an arbitrary Euclidean frame, the linearand angular momenta take the forms

ρ′ [(a − i)′ − f ′] = (divσ)′ (4.230)

and

ρ′ [(ϕ − j)′ − l′] = (divΣ)′ − ε ∶ σ′, (4.231)

where i′ is given by (3.400), j′ is given by (4.224), and we have used the fact thatε is an isotropic axial tensor. However, we note that the apparent acceleration andinternal spin rate are given by (see (3.398) and (4.223))

(a − i)′ =Q ⋅ a and (ϕ − j)′ = (detQ)Q ⋅ ϕ. (4.232)

Subsequently, it can be easily shown that the apparent acceleration and the ap-parent internal spin rate in two non-inertial frames are related as

(a − i)′ =Q ⋅ (a − i) and (ϕ − j)′ = (detQ)Q ⋅ (ϕ − j) , (4.233)

where

i ≡ b − w × (x − b) − 2w × (x − b) −w × [w × (x − b)] , (4.234)

j ≡ φ −w × (ϕ −φ) , (4.235)

and w and φ = i ⋅w are the frame’s angular velocity and spin rate relative to thesecond frame.

We can interpret the vector fields i and j as inertial (apparent) body force andinertial (apparent) body couple. Now the individual terms retain the same names,but are interpreted as forces and couples. If we combine the inertial force andcouple with the real body forces f and couples l, so that we now think of the bodyforces and couples as including these apparent forces and couples, then the balanceof linear and angular momenta are also seen to be invariant under a Euclideantransformation. From now on, we shall assume this to be the case and interpretthe body forces f to mean (f + i) and the body couples l to mean (l + j).

Subsequently, we take b = 0 and w = 0 in an inertial frame and b and w

to be nonzero in a non-inertial frame so that i and j will be zero or nonzerocorrespondingly. We remark that, with this understanding, the correspondingintegral balance laws are also invariant under a general Euclidean transformation.

4.14. JUMP CONDITIONS IN EUCLIDEAN FRAMES 183

4.14 Jump conditions in Euclidean frames

The jump conditions should be invariant with respect to Euclidean transforma-tions. We shall examine whether their invariance places any restrictions on themotion in Cartesian frames.

We first recall that the velocity in a Euclidean frame is given by (3.382),

v′ = b(t) + Q(t) ⋅ x +Q(t) ⋅ v, (4.236)

where b is the position vector between the two Cartesian frames F and F ′, andQ is the orthogonal rotation tensor between the two frames. To examine thejump in velocity in a new frame, Jv′K = (v′)+ − (v′)−, we need to examine thevelocity differences at the same location x on a discontinuous surface. Under suchtransformation, we have

(v′)+ = b(t) + Q(t) ⋅ x +Q(t) ⋅ v+ and (v′)− = b(t) + Q(t) ⋅ x +Q(t) ⋅ v−. (4.237)

We clearly see that velocity differences, and subsequently jumps in velocities, arealways objective:

Jv′K = (v′)+ − (v′)− =Q(t) ⋅ (v+ − v−) =Q(t) ⋅ JvK. (4.238)

Now, since from (4.221) we have that ρ′ = ρ, and since n′ = Q ⋅ n, the trans-formation of the mass balance jump condition (4.102) across the singular movingsurface ζ is given by

Jm′K = Jρ′ (v′ − c′)K ⋅ n′ = JρQ ⋅ (v − c)K ⋅Q ⋅n = Jρ (v − c)K ⋅QT⋅Q ⋅n =

Jρ (v − c)K ⋅ n = JmK = 0. (4.239)

Subsequently, the mass balance jump condition is frame indifferent.Using the transformations (4.221), the linear momentum jump condition (4.110)

takes the form:

m′Jv′K − Jσ′K ⋅ n′ =Q ⋅ (m JvK − JσK ⋅QT⋅Q ⋅ n) =m JvK − JσK ⋅ n = 0. (4.240)

Thus, the linear momentum balance jump condition is frame indifferent.Similarly, using the transformations (4.221), the angular momentum jump con-

dition (4.117) takes the form:

m′Jϕ′K − JΣ′K ⋅ n′ = (detQ)Q ⋅ (m JϕK − JΣK ⋅QT⋅Q ⋅ n) =

m JϕK − JΣK ⋅ n = 0, (4.241)

and thus, the angular momentum balance jump condition is frame indifferent aswell.

Now, using the transformations (4.221), for the transformation of the energyjump condition, we use the form (4.216):

m′ Je′K − (Jv′K ⋅ ⟪σ′⟫ + Jν ′K ⋅ ⟪Σ′⟫ − Jq′K) ⋅ n′ =m JeK − (JvK ⋅QT

⋅Q ⋅ ⟪σ⟫ + (detQ)2JνK ⋅QT⋅Q ⋅ ⟪Σ⟫ − JqK) ⋅QT

⋅Q ⋅n =m JeK − (JvK ⋅ ⟪σ⟫ + JνK ⋅ ⟪Σ⟫ − JqK) ⋅ n = 0. (4.242)

184 MECHANICS AND THERMODYNAMICS

Thus, the energy balance jump condition is frame indifferent.Lastly, using the transformations (4.221), the transformation of the entropy

jump condition (4.220) is given by

γs =m′Jη′K + Jh′K ⋅ n′ =mJηK + JhK ⋅QT⋅Q ⋅n =mJηK + JhK ⋅n ≥ 0. (4.243)

Thus, the entropy inequality jump condition is also frame indifferent.We close this chapter by noting that while we have arrived at the appropriate

global and local balance laws and jump conditions for a polar material, from nowon we shall focus all further discussions on nonpolar materials. Specifically, wetake ϕ = 0 (or i = 0), l = 0, and Σ = 0 in the balance equation and jump conditionof angular momentum so that σ = σT . The energy equation and its jump conditionsubsequently simplify accordingly.

Problems

1. Show thatd(lnJ)dt

= divv. (4.244)

2. Obtain the local mass balance equation (4.98) by using (4.21) and each ofthe following evolution equations for the

ii) deformation gradient (3.239),

ii) inverse deformation gradient (3.494),

iii) right Cauchy–Green tensor (3.495), and

iv) left Cauchy–Green tensor (3.496).

3. Show that the integral of the evolution equation for the deformation gradient(3.239) is given by

F(x, t) = exp [∇∫ t

t0v(x, t′)dt′]T ⋅F(x, t0). (4.245)

4. Show that (4.244) can also be obtained from (4.245).

5. Show that for steady motion (∂v/∂t = 0) of a continuum, the stream linesand path lines coincide.

6. Show that divv, divq, and divh are frame indifferent under a Euclideantransformation.

7. Show that divσ and divΣ are frame indifferent under a Euclidean transfor-mation.

8. Show that the stress power Φ given in (4.123) is frame indifferent under aEuclidean transformation.

9. If A and B are two tensors, show that

i)JABK = JAK⟪B⟫ + ⟪A⟫JBK, (4.246)

4.14. JUMP CONDITIONS IN EUCLIDEAN FRAMES 185

ii)

⟪A⟫ = A∓ ± 1

2JAK, (4.247)

where ⟪A⟫ ≡ 1

2(A+ +A−) . (4.248)

10. Starting from (4.216), derive (4.217).

11. Show that for any tensor quantity ψ, the mass conservation statement (4.98)implies that

ρψ = ∂ρψ∂t+ div (ρψv) . (4.249)

12. Show that the acceleration vector a = v may be expressed in the form

a = ∂v∂t+ω × v +

1

2grad (v ⋅ v) , (4.250)

where ω = curlv is the vorticity vector. Show that the above form is validfor any arbitrary curvilinear coordinates.

13. Consider a flow with the hydrostatic stress tensor σ = −p1 and the conser-vative body force f = −gradφ.

i) Show that if the flow is steady (∂v/∂t = 0), then

ρv ⋅ grad(12v ⋅ v + φ) + v ⋅ gradp = 0. (4.251)

ii) Show that if the flow is steady and irrotational (∂v/∂t = 0 and curlv =0), then

ρgrad(12v ⋅ v + φ) + gradp = 0. (4.252)

14. Consider a cubic block with its sides parallel to the Cartesian coordinateaxes in the reference configuration. Suppose that the stress tensor is givenby σ = −p1 + µB, where p and µ are scalar quantities and B is the leftCauchy–Green strain tensor. Determine the normal and shear componentsof the tractions on the surface of the block in the deformed configuration,under the simple shear deformation given by (3.162)–(3.164).

15. At a specific point in a deformable body, the components of the Cauchystress tensor with respect to the Cartesian coordinate system are given by

[σ] =⎡⎢⎢⎢⎢⎢⎢⎢⎣

1 4 −2

4 0 0

−2 0 3

⎤⎥⎥⎥⎥⎥⎥⎥⎦. (4.253)

i) Find the components of the traction vector tn on the normal to theplane that passes through the point and that is parallel to the plane2x1 + 3x2 + x3 = 5.

186 MECHANICS AND THERMODYNAMICS

ii) Find the length of tn and the angle that tn makes with the normal tothe plane.

iii) Find the stress components of σ along the new normal basis

e1 = e1, e2 = 1√2(e1 − e3) , e3 = 1

3(2e1 − e2 + 2e3) . (4.254)

16. The components of the symmetric Cauchy stress tensor referred to the Carte-sian coordinate system are given by

[σ] =⎡⎢⎢⎢⎢⎢⎣

0 0 αx20 0 −βx3αx2 −βx3 0

⎤⎥⎥⎥⎥⎥⎦, (4.255)

where α and β are constants. For the point x = (0, β2, α),i) find the three principal stress invariants of tensor σ;

ii) compute the principal stress components and their associated principaldirections;

iii) compute the maximum magnitude of shear stress and the plane onwhich it acts.

17. Assume a plane stress state in a parallelipiped bounded by the planes x1 = ±a,x2 = ±b, and x3 = ±c, so that the components of the symmetric Cauchy stresstensor referred to the Cartesian coordinate system are given by

[σ] =⎡⎢⎢⎢⎢⎢⎣α(x1 − x2) βx21x2 0

βx21x2 −α(x1 − x2) 0

0 0 0

⎤⎥⎥⎥⎥⎥⎦, (4.256)

where α and β are constants. For the point x = (a/2,−b/2,0), find

i) the principal normal stresses and the associated principal directions;

ii) the planes, characterized by the unit normal n, that give the maximumand minimum shear stresses and the magnitude of the extremal shearstress;

iii) the total Cauchy traction vector on each face of this parallelipiped.

18. A dynamical process is described by the motion

x1 = etX1 − e−tX2, (4.257)

x2 = etX1 + e−tX2, (4.258)

x3 = X3, (4.259)

for t > 0, and the symmetric Cauchy stress tensor with components

[σij] =⎡⎢⎢⎢⎢⎢⎣

x21 αx2x23 0

αx2x23 x22 0

0 0 βx31

⎤⎥⎥⎥⎥⎥⎦, (4.260)

where α and β are scalar constants. Find the system of forces so that themass conservation and linear momentum balance equations are satisfied. TheCauchy traction vector t is assumed to act at a point x of a plane tangentialto the sphere given by φ = x21 + x22 + x23.

4.14. JUMP CONDITIONS IN EUCLIDEAN FRAMES 187

19. Suppose that σ = −p1, Σ = −m1, and q = 0. Obtain the simplified jumpconditions.

20. If in a motion of a body the material points crossing a surface ζ(t) gain orlose mass, what would the jump condition at ζ(t) be?

21. Across a moving surface ζ(t), mass and linear momentum of the materialpoints of a continuum are seen to undergo jumps because of the creation ordestruction of mass. Find the jump conditions at ζ(t).

22. Across a moving surface ζ(t), radiation of heat energy is causing suddenenergy loss. Express the jump condition across a moving surface ζ(t).

23. For the velocity field given in Problem 3.34, determine

i) components of the deformation rate tensor;

ii) components of the spin tensor and vorticity vector;

iii) invariants of the deformation rate tensor.

24. For the velocity field given in Problem 3.35, determine

i) components of the deformation rate tensor;

ii) components of the spin tensor and vorticity vector;

iii) invariants of the deformation rate tensor.

25. For the velocity field given in Problem 3.36, determine

i) components of the deformation rate tensor;

ii) components of the spin tensor and vorticity vector;

iii) invariants of the deformation rate tensor.

26. For the isochoric and irrotational motion of a body, show that the kineticenergy of a nonpolar material is given by

K = 1

2ρ∫

S

φgradφ ⋅ nds, (4.261)

where φ is the velocity potential, i.e., v = −gradφ, and n is the exteriornormal of the closed surface S .

27. If we have a nonpolar ideal fluid, so that σ = −p1, show that the stress powermay be expressed by the equation

Φ = pρ

dρ

dt. (4.262)

28. Show that the balance equation for the vorticity vector ω = curlv, valid foran ideal fluid subject to a conservative body force (σ = −p1 and f = −gradφ),is given by

d

dt(ωρ) = (gradv) ⋅ (ω

ρ) + gradρ × gradp

ρ3, (4.263)

where L = (gradv)T .

188 MECHANICS AND THERMODYNAMICS

29. Show that the balance equation for the vorticity vector ω = curlv, valid fora constant density ideal fluid (ρ = const. and σ = −p1), is given by

ω = L ⋅ω, (4.264)

where L = (gradv)T .

30. Show that the balance equation for a barotropic ideal fluid subject to aconservative body force (ρ = ρ(p), σ = −p1, and f = −gradφ) is given by

ξ = L ⋅ ξ, (4.265)

where ξ = ω/ρ is the specific vorticity (vorticity per unit mass), ω = curlv,

and L = (gradv)T .

i) Now if ξR is the specific vorticity in the reference configuration, and F

is the deformation gradient, show that

ξ = F ⋅ ξR (4.266)

integrates the equation.

ii) Use the polar decomposition F = R ⋅ U and the decomposition L =D +W to show that a vortex filament is stretched and rotated duringits motion.

iii) What can you say about the dynamics of vortex filaments in planemotion?

Bibliography

R.G. Bartle. The Elements of Integration. John Wiley & Sons, New York, NY,1966.

R.M. Bowen. Introduction to Continuum Mechanics for Engineers. Plenum Press,New York, NY, 1989.

H.B. Callen. Thermodynamics. John Wiley & Sons, Inc., New York, NY, 1962.

P. Chadwick. Continuum Mechanics – Concise Theory and Problems. DoverPublications, Inc., Mineola, NY, 2nd edition, 1999.

K. Denbigh. The Principles of Chemical Equilibrium. Cambridge UniversityPress, Cambridge, England, 1981.

A.C. Eringen. Nonlinear Theory of Continuous Media. McGraw-Hill Book Com-pany, Inc., New York, NY, 1962.

A.C. Eringen. Basic principles: Balance laws. In A.C. Eringen, editor, ContinuumPhysics, volume II, pages 69–88. Academic Press, Inc., New York, NY, 1975.

A.C. Eringen. Basic principles: Thermodynamics of continua. In A.C. Eringen,editor, Continuum Physics, volume II, pages 89–127. Academic Press, Inc., NewYork, NY, 1975.

BIBLIOGRAPHY 189

A.C. Eringen. Mechanics of Continua. R.E. Krieger Publishing Company, Inc.,Melbourne, FL, 1980.

M.E. Gurtin. An Introduction to Continuum Mechanics. Academic Press, SanDiego, CA, 2003.

M.E. Gurtin, E. Fried, and L. Anand. The Mechanics and Thermodynamics ofContinua. Cambridge University Press, Cambridge, UK, 2010.

P. Haupt. Continuum Mechanics and Theory of Materials. Springer-Verlag,Berlin, 2000.

G.A. Holzapfel. Nonlinear Solid Mechanics. John Wiley & Sons, Ltd., Chichester,England, 2005.

K. Hutter and K. Jöhnk. Continuum Methods of Physical Modeling. Springer-Verlag, Berlin, 1981.

W. Jaunzemis. Continuum Mechanics. The Macmillan Company, New York, NY,1967.

J. Kestin. A Course in Thermodynamics, volume 1. McGraw-Hill Book Company,New York, NY, 1979.

J. Kestin. A Course in Thermodynamics, volume 2. McGraw-Hill Book Company,New York, NY, 1979.

I.-S. Liu. Continuum Mechanics. Springer-Verlag, Berlin, 2002.

I. Müller. Thermodynamics. Pitman Publishing, Inc., Boston, MA, 1985.

W. Noll. On the continuity of the solid and fluid states. Journal of RationalMechanics and Analysis, 4(1):3–81, 1955.

R.S. Rivlin. The fundamental equations of nonlinear continuum mechanics. In S.I.Pai, A.J. Faller, T.L. Lincoln, D.A. Tidman, G.N. Trytten, and T.D. Wilkerson,editors, Dynamics of Fluids in Porous Media, pages 83–126, Academic Press,New York, 1966.

M. Silhavy. The Mechanics and Thermodynamics of Continuous Media. Springer-Verlag, Berlin, 1997.

C. Truesdell. Thermodynamics for beginners. In M. Parkus and L.I. Sedov,editors, Irreversible Aspects of Continuum Mechanics and Transfer of PhysicalCharacteristics of Moving Fluids, pages 373–389. Springer, Wien, 1968.

C. Truesdell. A First Course in Rational Continuum Mechanics, volume 1. Aca-demic Press, New York, NY, 1977.

C. Truesdell and W. Noll. The non-linear field theories of mechanics. In S. Flügge,editor, Handbuch der Physik, volume III/3. Springer, Berlin-Heidelberg-NewYork, 1965.

C. Truesdell and R.A. Toupin. The classical field theories. In S. Flügge, editor,Handbuch der Physik, volume III/1. Springer, Berlin-Heidelberg-New York, 1960.

5

Principles of constitutive theory

Conservation of mass, the balance of linear momentum, the balance of angularmomentum, the balance of energy, and the entropy inequality are valid for allcontinuous media. However, different material bodies having the same mass andgeometry, when subjected to identical external effects, respond differently. Theinternal constitution of matter is responsible for the different responses. To un-derstand these differences, we need equations that reflect structural differencesbetween materials. This is the subject of constitutive theory.

Constitutive relations can be regarded as mathematical models for materialbodies, and as such, they define ideal materials. Since real materials always con-tain irregularities and defects, the validity of a model should be verified throughexperiments on the results it predicts. On the contrary, some experiments maysuggest certain functional dependence of the constitutive relations on its variablesto within a reasonable satisfaction for certain materials. However, experimentsalone are rarely, if ever, sufficient to determine constitutive relations of a materialbody.

There are some universal requirements that a model should obey lest its conse-quences be contradictory to some well-known physical experience. Therefore, insearch of a correct formulation of a mathematical model, in general, we shall firstimpose these requirements on the proposed model. The most important universalrequirements of this kind are:

• Principle of causality: causality describes the relationship between causesand effects as governed by the laws of nature. In classical physics, we assumethat a cause should always precede its effect; i.e., the cause and its effect areseparated by a time interval, and the effect belongs to the future of its cause.

• Principle of equipresence: all constitutive functions should be expressed interms of the same set of independent constitutive variables until the contraryis deduced.

• Principle of frame indifference: the constitutive equations must be objectivewith respect to Euclidean motions of the spatial frame of reference – theymust be the same as seen from inertial and non-inertial frames of reference.

• Principles of material smoothness and memory: the constitutive functions

191

192 PRINCIPLES OF CONSTITUTIVE THEORY

at regular points should be spatially and temporally smooth so that materialgradients and time derivatives up to some orders exist.

• Principle of material symmetry: a body subjected to the same thermo-mechanical history at two different configurations in general have differentresults. However, it may happen that the results are exactly the same if thematerial possesses a certain symmetry that makes it indistinguishable in thetwo configurations.

• Thermodynamic principles: the constitutive equations must be consistentwith thermodynamic concepts such as thermodynamic states and processes,and must obey the second law of thermodynamics.

These requirements impose severe restrictions on the model and hence lead to agreat simplification for general constitutive relations. The reduction of constitutiverelations from very general to more specific and mathematically simpler ones for agiven class of materials is the main objective of constitutive theories in continuummechanics.

5.1 General constitutive equation

The principle of causality amounts to the selection of physical independent vari-ables on which the response of a material depends on. For nonpolar thermo-mechanical materials, in addition to motion, we include temperature in this setof fields. Its existence is postulated through the zeroth law of thermodynamics,which speaks to thermal equilibrium between bodies in contact, and also impliesthe existence of an empirical temperature or coldness function, which, without lossof generality, can always be related to the absolute temperature θ with dimension[Θ] that is different than the dimensions of length [L], mass [M], and time [T ].We shall suppose that it is always possible to assign a positive-definite tempera-ture θ > 0 to each material point X of a body. As will be noted in Section 5.10,the absolute positive-definite thermodynamic temperature and entropy are inex-tricably connected with each other. If the existence of a quantity we call entropyis postulated, then temperature appears naturally as a dual quantity of entropythrough an equation of state. In such case, it is not necessary to invoke the zerothlaw of thermodynamics. The choice of fields limits the class of physical and chem-ical phenomena observable in the material. Mechanics, electricity and magnetism,and thermodynamics are three parallel divisions of classical macroscopic physicswhich continuum mechanics speaks to. In this book we have not dealt with thebalance laws of electricity and magnetism, and thus we will not consider how ma-terials respond to such fields. Here we focus exclusively on the thermomechanicalresponse of materials. Subsequently, the behavior of the material particle locatedat X at time t is characterized by a description of the set of independent fields

I(X, t) ≡ x(X, t), θ(X, t) (5.1)

called the basic fields. It would appear that density should be included in the setof basic fields. However, from the mass balance, ρ(X, t) = ρR(X, t0)J−1(X, t), wesee that ρ can be obtained if we know the density in the reference configuration,

5.1. GENERAL CONSTITUTIVE EQUATION 193

ρR(X, t0), and the motion, x(X, t). We note that ρR(X, t0) is initially given asan explicit function of X.

Given a reference configuration, the conservation of mass and the balances ofmomenta and energy alone are not sufficient to determine the basic fields sincethese contain other, unknown, field quantities: the stress tensor σ, the heat fluxq, the entropy flux h, the body force f , the energy supply r, and the entropysupply b. The external supplies f and r are regarded as known functions whichare provided by the environment that the body encounters. It is assumed thatmaterial properties are independent of external supplies. Furthermore, while notnecessary, it is convenient to reformulate the entropy inequality (4.219) in terms ofthe Helmholtz free energy density ψ instead of the internal energy density e (i.e.,ψ = e − θ η). Subsequently, the unknown quantities consist of the stress tensor,the heat flux, the entropy flux, the Helmholtz free energy density, and the entropydensity, and they will depend not only on the behavior of the body but also onthe kind of material that constitutes the body. Equations for this set of dependentfields,

C(X, t) ≡ σ(X, t),q(X, t),h(X, t), ψ(X, t), η(X, t), (5.2)

called constitutive quantities, must depend on the basic fields, in addition to thelocation of a material point and the current time X, t, to characterize thermo-mechanical responses of a particular material body.

The principle of causality also requires that the response of a material is notinfluenced by future values of the basic fields or by material particles outside of(not in contact with) the body except for their effects as included in external bodysupplies. Subsequently, we postulate that, in general, the history of the behaviorup to the present time determines the present response of the body at any materialpoint located at X ∈ V = κ(B) and time t, i.e.,

FY∈V−∞<τ≤t

C(Y, τ);I(Y, τ),X, t = 0, (5.3)

where F is a functional. A functional is simply a function whose arguments arefunctions and whose values are tensors. It should be noted that the responsefunctional F depends on the reference configuration κ since both the motion χ andthe material particle coordinate X depend on the reference configuration. Thus,we should write Fκ for the response functional relative to reference configurationκ. For simplicity, we continue to drop the subscript denoting this dependence.It will be necessary to make this dependence explicit later in the discussion ofmaterial symmetries.

Let a tensor field ϕ be a function of time. If we take τ = t − s, then the historyof ϕ up to time t is defined by

ϕ(t)(s) ≡ ϕ(t − s) = ϕ(τ), (5.4)

where 0 ≤ s < ∞ denotes the time coordinate pointing into the past from thepresent time t. Clearly, when a material has no memory, ϕ(t)(0) = ϕ(t). Now wecan rewrite

FY∈V

0≤s<∞

C(Y, t − s);I(Y, t − s),X, t = FY∈V

0≤s<∞

C(t)(Y, s);I(t)(Y, s),X, t = 0. (5.5)

194 PRINCIPLES OF CONSTITUTIVE THEORY

Such response functionals are sufficiently general so that, e.g., the stress coulddepend on the histories of stress, heat flux, entropy flux, free energy, entropy, andtheir gradients and rates of varying orders as well as on the motion and temperatureat all other points in the body. Indeed, to produce a theory of heat conductionfor which thermal disturbances propagate with finite, rather than infinite, speed,it is necessary to consider such formulation. The functional can be solved for theconstitutive quantities if it is non-singular and single valued. We assume that thisis the case for the class of materials that we wish to consider. So, if we let thetensor T (X, t) ∈ C, of arbitrary rank, represent a constitutive function of any ofthe unknown fields, then, with some abuse of functional notation, T is given by

T (X, t) = FY∈V

0≤s<∞

x(t)(Y, s), θ(t)(Y, s),X, t. (5.6)

While we call F the constitutive function of T , we recognize that in reality F is afunctional. Such a functional allows the description of arbitrary nonlocal effectsof any inhomogeneous material body with a perfect memory of the past. Notethat (obviously) the functionals in (5.5) and (5.6) are different. Here, and below,we re-use some of the symbols, such as F, so as not to unnecessarily cloud thepresentation with too many symbols.

We note that for given x and θ at all material points X ∈ V and for all time t,the above response functionals provide all T ∈ C. Subsequently, for given ρR(X, t0)and boundary conditions consistent with the given motion and temperature, themass balance provides the density ρ, the linear momentum balance provides thebody force density f , the angular momentum balance is identically satisfied fornonpolar materials with σ = σT , and the energy balance provides the energysupply density r. In some sense, the above procedure assures us of the existence ofa solution satisfying all the balance laws. The use of the resulting body force andenergy supply densities obtained from such procedure can be effectively utilizedto verify the correct implementation of a numerical algorithm intended to producean approximation to the given motion and temperature fields. This procedureis called the method of manufactured solutions. In reality, the program for us ismore challenging since, in general, we are given the body force f and energy supplyr along with ρR(X, t0) and boundary conditions, and are asked to obtain themotion x and temperature θ. Nevertheless, in either case not all such solutions arephysically realized. It is postulated that all solutions that are physically realizableare such that the response functionals satisfy the entropy inequality as well as theother previously noted accepted principles of continuum mechanics.

5.2 Frame indifference

Assuming that the constitutive function T is a scalar field, and recognizing that θ isa scalar as well, then under an arbitrary Euclidean transformation (see Section 3.3)with arbitrary scalar a, vector b(t), and orthogonal second-order rotation tensorQ(t), in order for the field to be frame indifferent or objective we must have

FY∈V

0≤s<∞

b(t − s) +Q(t − s) ⋅ x(Y, t − s), θ(Y, t − s),X, t + a =F

Y∈V0≤s<∞

x(Y, t − s), θ(Y, t − s),X, t. (5.7)

5.2. FRAME INDIFFERENCE 195

Now if we choose the special frame with b(t − s) = 0 and Q(t − s) = 1, corre-sponding to a time shift, the restriction becomes

FY∈V

0≤s<∞

x(Y, t − s), θ(Y, t − s),X, t + a = FY∈V

0≤s<∞

x(Y, t − s), θ(Y, t − s),X, t. (5.8)

We now see that if we take a = −t, then the scalar field is seen to be objective onlyif it contains no explicit dependence on time, i.e.,

T (X, t) = FY∈V

0≤s<∞

x(t)(Y, s), θ(t)(Y, s),X. (5.9)

Accounting for the independence on t, if we now choose the special frame withb(t− s) = −x(X, t− s) and Q(t− s) = 1, corresponding to a rigid translation of theframe, the restriction becomes

FY∈V

0≤s<∞

[x(Y, t − s) − x(X, t − s)], θ(Y, t − s),X =F

Y∈V0≤s<∞

x(Y, t − s), θ(Y, t − s),X. (5.10)

Subsequently, in order for the constitutive function to be objective, it must be ofthe form

T (X, t) = FY∈V

0≤s<∞

[x(Y, t − s) − x(X, t − s)], θ(Y, t − s),X. (5.11)

Lastly, again accounting for the independence on t, if we now choose the specialframe with b(t − s) = 0 and Q(t − s) arbitrary, corresponding to a rigid rotationof the frame, the restriction for a second-order tensor, e.g., becomes

FY∈V

0≤s<∞

Q(t − s) ⋅ [x(Y, t − s) − x(X, t − s)], θ(Y, t − s),X =Q(t) ⋅ F

Y∈V0≤s<∞

[x(Y, t − s) − x(X, t − s)], θ(Y, t − s),X ⋅QT (t), (5.12)

or

FY∈V

0≤s<∞

Q(t)(s) ⋅ [x(t)(Y, s) − x(t)(X, s)], θ(t)(Y, s),X =Q(t) ⋅ F

Y∈V0≤s<∞

[x(t)(Y, s) − x(t)(X, s)], θ(t)(Y, s),X ⋅QT (t). (5.13)

Since any general rigid motion of a Euclidean frame and time shift can beobtained by a sequence of the above three transformations, then the general con-stitutive equation is of the form

T (X, t) = FY∈V

0≤s<∞

[x(t)(Y, s) − x(t)(X, s)], θ(t)(Y, s),X, (5.14)

subject to the restriction arising from the rigid rotation of the frame, such as (5.13)for a second-order tensor.

196 PRINCIPLES OF CONSTITUTIVE THEORY

5.3 Temporal material smoothness

Suppose that there exists a time τ < t such that the histories x(t)(Y, s), x(t)(X, s),and θ(t)(Y, s) for s > 0 possess Taylor series expansions about s = 0 for all Y ∈ B:

x(t)(Y, s) = x(Y, t) − x(Y, t)s + 1

2!x(Y, t)s2 +⋯, (5.15)

x(t)(X, s) = x(X, t) − x(X, t)s + 1

2!x(X, t)s2 +⋯, (5.16)

θ(t)(Y, s) = θ(Y, t) − θ(Y, t)s + 1

2!θ(Y, t)s2 +⋯. (5.17)

Now, assume that derivatives up to orders p and q exist for x(Y, t) and θ(Y, t).In such case, materials satisfy the constitutive equations of the form

T (X, t) = GY∈V[(k)x (Y, t) − (k)x (X, t)], (l)θ (Y, t),X, k = 0,1, . . . ,p, l = 0,1, . . . ,q

(5.18)and are called materials of rate type.

Definition: A material is said to be of mechanical rate p and thermal rate q

if and only if the constitutive functional depends on the mechanical and thermalrates up to order p and q, respectively.

The condition of objectivity restricts the constitutive functional (5.18). Forexample, if T is a second-order tensor, we must require that

GY∈VQ(t) ⋅ [(k)x (Y, t) − (k)x (X, t)], (l)θ (Y, t),X =

Q(t) ⋅ GY∈V[(k)x (Y, t) − (k)x (X, t)], (l)θ (Y, t),X ⋅QT (t), (5.19)

where k = 0,1, . . . ,p and l = 0,1, . . . ,q.

5.4 Spatial material smoothness

Histories of any part of a material body can affect the response at any other pointof the body. In most applications, such nonlocal effect is rarely important. Itis usually assumed that only gradients up to some order in an arbitrarily smallneighborhood of X affect the material response at the point X. Hence, the motionand temperature can be approximated to some order by a Taylor series expansionabout point X:

x(t)k(Y, s) = x

(t)k(X, s) + x(t)

k,K1(X, s)(YK1

−XK1) +

1

2!x(t)k,K1K2

(X, s)(YK1−XK1

)(YK2−XK2

) +⋯, (5.20)

θ(t)(Y, s) = θ(t)(X, s) + θ(t),K1(X, s)(YK1

−XK1) +

1

2!θ(t),K1K2

(X, s)(YK1−XK1

)(YK2−XK2

) +⋯, (5.21)

5.4. SPATIAL MATERIAL SMOOTHNESS 197

so that the constitutive functional (5.14), up to gradient orders P and Q, for x(t)

and θ(t) respectively, can be rewritten as

T (X, t) = F0≤s<∞

iF(t)(X, s), θ(t)(X, s), jG(t)(X, s),DK ,X, (5.22)

where i = 1, . . . ,P, j = 1, . . . ,Q and we define deformation gradients of grade i andtemperature gradients of grade j by

iF(t)kK1K2⋯Ki

≡ x(t)k,K1K2⋯Ki

and jG(t)K1K2⋯Kj

≡ θ(t),K1K2⋯Kj, (5.23)

K = Kn = 1,2,3, and the three vectors DK represent the decomposition of alldirectional vectors (Y −X) originating from X. Note that

jG(t)K1K2⋯Kj

= F (t)k1K1

F(t)k2K2⋯F

(t)kjKj

jg(t)k1k2⋯kj

, (5.24)

wherejg(t)k1k2⋯kj

≡ θ(t),k1k2⋯kj

. (5.25)

Furthermore, we observe that

1F(t)(X, s) = F(t)(X, s) = Gradx(t)(X, s) (5.26)

is the conventional deformation gradient,

1G(t)(X, s) =G(t)(X, s) = Gradθ(t)(X, s) (5.27)

is the conventional temperature gradient, and

G(t)(X, s) = FT (X, t) ⋅ g(t)(x, s), (5.28)

where1g(t)(x, s) = g(t)(x, s) = gradθ(t)(x, s) (5.29)

is the conventional temperature gradient in the spatial coordinates.The presence of DK and X in the response functional indicate material anisotro-

py and material inhomogeneity, respectively. The vectors DK are called materialdescriptors and express the directional dependence of the material properties atmaterial point X. Without loss of generality, one may replace DK by 1K , whichare the unit vectors of coordinates Xk. The presence of DK is an indicationthat the form of the response functional depends on the choice of the materialreference configuration. Since, as already noted earlier, we recognize that theresponse functional depends on the reference configuration, then there is no loss ofgenerality in dropping the dependence on DK from the arguments of the responsefunctional. Changes in the material reference configuration are discussed later.Subsequently, we write

T (X, t) = F0≤s<∞

iF(t)(X, s), θ(t)(X, s), jG(t)(X, s),X, (5.30)

where i = 1, . . . ,P and j = 1, . . . ,Q.Definition: A material is said to be of mechanical grade P and thermal grade

Q if and only if the constitutive functional depends on the deformation gradientsup to order P and temperature gradients up to order Q.

198 PRINCIPLES OF CONSTITUTIVE THEORY

Definition: Thermomechanical materials of grade one (P = Q = 1) are calledsimple materials and their constitutive function takes the form

T (X, t) = F0≤s<∞

F(t)(X, s), θ(t)(X, s),G(t)(X, s),X. (5.31)

This class of materials has perfect temporal memory but only very limited non-local sensitivity since we only retain first-order gradients. Nevertheless, this classis general enough to include most material bodies of practical interest. Non-simple material bodies, corresponding to P > 1 and/or Q > 1, are by no meansunimportant. As an example, in theories of mixtures and porous media, the densitygradient, which is related to the second gradient of deformation, must be takeninto account to obtain a consistent theory. Note from (5.20) and (5.21) that P = 0corresponds to simple (Q = 1) or non-simple (Q > 1) rigid materials and Q = 0 tosimple (P = 1) or non-simple (P > 1) non-heat-conducting materials.

The condition of objectivity restricts the constitutive functional (5.30). Forexample, if T is a second-order tensor, we must require that

F0≤s<∞

Q(t)(s) ⋅ iF(t)(X, s), θ(t)(X, s), jG(t)(X, s),X =Q(t) ⋅ F

0≤s<∞iF(t)(X, s), θ(t)(X, s), jG(t)(X, s),X ⋅QT (t), (5.32)

where i = 1, . . . ,P and j = 1, . . . ,Q.As noted earlier, a body is called homogeneous if the constitutive function does

not depend on X explicitly. Therefore, for a homogeneous simple material body,the above constitutive function reduces to

T (X, t) = F0≤s<∞

F(t)(X, s), θ(t)(X, s),G(t)(X, s). (5.33)

5.5 Spatial and temporal material smoothness

As in Section 5.3, if the constitutive functional in addition possesses continuousmaterial derivatives with respect to s at s = 0, then the gradients of motion andtemperature can be approximated up to some order by a Taylor series expansion:

x(t)k,K1(X, s) = xk,K1

(X, t) − xk,K1(X, t)s + 1

2!xk,K1

(X, t)s2 +⋯, (5.34)

⋮

θ(t)(X, s) = θ(X, t) − θ(X, t)s + 1

2!θ(X, t)s2 +⋯, (5.35)

θ(t),K1(X, s) = θ,K1

(X, t) − θ,K1(X, t)s + 1

2!θ,K1(X, t)s2 +⋯, (5.36)

⋮

Subsequently, the constitutive functional (5.30) becomes

T (X, t) = T i(k)F (X, t), (l)θ (X, t), j(m)G (X, t),X, (5.37)

where i = 1,2, . . . ,P, j = 1,2, . . . ,Q, k = 0,1, . . . ,p, l = 0,1, . . . ,q, and m = 0,1, . . . , r.Note that now we no longer have a functional, but a tensor-valued function. A

5.6. MATERIAL SYMMETRY 199

simple material (P = Q = 1) involving time rates of the deformation gradients upto order p and temperature and its gradients up to order q and r, respectively, isdescribed by

T (X, t) = T (k)F (X, t), (l)θ (X, t), (m)G (X, t),X (5.38)

with k = 0,1, . . . ,p, l = 0,1, . . . ,q, and m = 0,1, . . . , r. Clearly, for a simple materialwith no memory (p = q = r = 0), we have

T (X, t) = T F(X, t), θ(X, t),G(X, t),X. (5.39)

For a simple material that includes heat conduction and viscous dissipation, oneshould take p = 1 and q = r = 0, in which case the constitutive function is

T (X, t) = T F(X, t), F(X, t), θ(X, t),G(X, t),X. (5.40)

In general, the set of variables defines the class of viscous heat-conducting materialfor the purpose of the theory, and the form of the constitutive functions defines aparticular material within that class.

5.6 Material symmetry

At this stage, we wish to make a distinction between property tensors and fieldtensors. A property tensor depends on the structure of the material. A field tensordepends on the field applied to the material; it can have any arbitrary form anddoes not depend on the structure of the material. In constitutive representations,we usually find that two specific field tensors are related (linearly, quadratically,etc.) through property tensors. In this relation, the property tensor expressesthe response to a generalized force by yielding a generalized displacement. Forexample, mass and volume of an object are field tensors of rank 0; they are linearlyrelated through density, which is a rank 0 property tensor. The heat flux andthe temperature gradient are field tensors of rank 1. As we will see, for a solidthe temperature gradient is linearly related to the heat flux through the rank 2property tensor of thermal conductivity. The stress and strain tensors are rank 2field tensors. Again, for a solid we will see that they are linearly related throughthe rank 4 stiffness or compliance tensors. Clearly, all property and field tensorsmust transform appropriately (as discussed in Chapter 2) to be frame indifferent.

The rank of a tensor determines the number of components. However, sym-metries reduce the number of independent components considerably. Symmetriesinherent in field tensors are usually obtained from application of physical princi-ples, such as application of angular momentum (for nonpolar materials) yieldingthe symmetry of the stress tensor, or from definitions involving second or higherderivatives, where the tensor is not affected from interchange of the order that suchderivatives are taken, or from application of thermodynamical reasoning, such asthe requirement of reversible changes at equilibrium or application of Onsager’sprinciple near equilibrium. On the other hand, symmetries of property tensors arediscovered from examining the intrinsic symmetries of the material at equilibriumconditions. The study of symmetries of materials is the subject of this section.

200 PRINCIPLES OF CONSTITUTIVE THEORY

The constitutive functional of a simple material relative to a reference configu-ration κ can be written in the form

T (X, t) = Fκ0≤s<∞

F(t)(X, s), θ(t)(X, s),G(t)(X, s),X. (5.41)

As the notation indicates, the response functional depends on the reference con-figuration κ because X depends on κ and F depends on the motion χκ, which inturn depends on the reference configuration κ.

The concept of material symmetry arises when one attempts to determine inwhat fashion the response functional depends on the choice of reference configu-ration. Note that

X = κ(X) (5.42)

indicates the position occupied by the material particle labeled X in the referenceconfiguration κ. Suppose that κ is another reference configuration of the samematerial point. Then

X = κ(X) (5.43)

is the position occupied by X in the reference configuration κ. Note that

X = κ(κ−1(X)) = κκ(X) (5.44)

represents a change of reference configuration. Now define

P ≡ Grad κκ(X) = ∂X∂X

, (5.45)

and recall that the motion is defined by

x = χκ(X, t). (5.46)

Likewise, we can define the motion χκ by

x = χκ(X, t) = χ(κ−1(X), t) = χκ(X, t) (5.47)

= χ(κ−1(X), t) = χκ(X, t), (5.48)

where χ is the motion of B. The general concept is illustrated in Fig. 5.1. Ofcourse if the motion χ is the same as χ, the spatial regions χ(B, t) and χ(B, t)for all X ∈ B are the same. Then, it easily follows that

F = Gradχκ(X, t) = Grad χκ(X, t) = F ⋅P or FkK = FkLPLK . (5.49)

This equation relates the deformation gradients constructed from viewing the mo-tion from two different reference configurations. Analogously, if the tempera-ture at a material point is the same in two different reference configurations, i.e.,θκ(X, t) = θκ(X, t), then it is easy to see that the relation between temperaturegradients between the two different reference configurations is given by

G = PT⋅ G or θ,K = PLK θ,L. (5.50)

5.6. MATERIAL SYMMETRY 201

B κ

χκ

κ χκ

P

x

x

X

X

X

V = κ(B)

V = κ(B) V = χκ(B, t)

V = χκ(B, t)Figure 5.1: Material body, and reference and deformed configurations.

The response functional of particle X in a simple material in reference configura-tion κ is given by

T (X, t) = Fκ0≤s<∞

F(t), θ(t),G(t),X. (5.51)

The response functional of the same particle in the reference configuration κ isgiven by

T (X, t) = Fκ0≤s<∞

F(t), θ(t), G(t),X, (5.52)

where we have used the fact that, for a fixed material point, θ(t)(X, t) = θ(t)(X, t) =θ(t)(X, t).

From the above discussion, we can see that the response functional only dependson κ in a neighborhood of X . This neighborhood is called a local reference con-figuration. Our response functional applies to the local reference configuration atmaterial point X and we look for the set of all reference configurations that areequivalent at X . Subsequently, from now on, our reference configuration will beunderstood to be the local reference configuration and so from (5.44) we take thespecific linear transformation

X =A +H ⋅X, (5.53)

where A now represents a constant translation vector and H a constant rotationor inversion tensor. Note that A and H are time independent since the change oflocal reference configuration is time independent, and in this case, P =H.

The concept of material symmetry arises when one tries to characterize thosechanges of reference configuration which do not affect the response of the material.We now want to characterize those linear transformations which produce the samevalue of the response functional at material point X independent of the referenceconfiguration, i.e., T (X, t) = T (X, t) or

Fκ0≤s<∞

F(t), θ(t), G(t), X = Fκ0≤s<∞

F(t) ⋅H, θ(t),HT⋅ G(t),H−1 ⋅ (X −A), (5.54)

where we have used (5.49) to relate F to F, (5.50) to relate G to G, and (5.53)to relate X to X. From above, it follows that a response functional relative to

202 PRINCIPLES OF CONSTITUTIVE THEORY

a reference configuration determines the response functional relative to any otherreference configuration.

A material body subjected to the same history using two different configurations,in general, yields different results. However, it may happen that the results areexactly the same if the material possesses a certain symmetry that makes it unableto distinguish between the two configurations. The reference configurations arethen said to be materially indistinguishable and κ = κ. The consequences ofthese conditions are obtained by requiring objectivity with respect to materialcoordinates in any reference configuration. Thus, for a tensor of any order, wehave from (5.54) that

Fκ0≤s<∞

F(t), θ(t),G(t),X = Fκ0≤s<∞

F(t) ⋅H, θ(t),HT⋅G(t),H−1 ⋅ (X −A)

for all (F(t), θ(t),G(t)), (5.55)

where H now indicates a material symmetry transformation. If a material is suchthat if we take A =X for all material points in the body, the above remains valid,then we see that the response functional (5.55) becomes independent of X. Suchmaterials are called homogeneous. From now on, we will only consider homoge-neous materials. Furthermore, we will consider only the class of transformationsH such that the mass density of the material remains the same. Thus, since from(4.21) we have that

ρ = ρRJ−1 = ρR[detF ]−1 = ρR[det(FH)]−1, (5.56)

and since det(FH) = (detF )(detH), we require that changes in the referenceconfiguration satisfy detH = ±1 (recall that following (3.8) we assumed that J =detF > 0; more generally J could be of either sign). Note that F and H areboth non-singular so that their inverses always exist. A material transformationH satisfying these properties for the constitutive quantity T is called a unimodulartransformation. The set of all unimodular transformations forms a group calledthe unimodular group U (V ). Subsequently, we have that H ∈ U (V ).

Definition: A collection of all symmetry transformations with respect to ref-erence configuration κ forms the symmetry group of a specific material, denotedGκ, if

i) for every (H1,H2) ∈ Gκ, H1 ⋅H2 ∈ Gκ (closure under multiplication);

ii) for every (H1,H2,H3) ∈ Gκ, (H1 ⋅H2) ⋅H3 =H1 ⋅ (H2 ⋅H3) (associative lawfor products);

iii) the set Gκ contains a unit element 1 such that for all H ∈ Gκ, H⋅1 = 1⋅H =H;

iv) for every H ∈ Gκ, there exists an element H−1 ∈ Gκ called the inverse of Hsuch that H ⋅H−1 =H−1 ⋅H = 1.

If H1 and H2 are symmetry transformations, then H1 ⋅H2 is also a symmetrytransformation, since by hypothesis

Fκ0≤s<∞

F(t), θ(t),G(t) = Fκ0≤s<∞

F(t) ⋅H2, θ(t),HT

2 ⋅G(t)

for all (F(t), θ(t),G(t)), (5.57)

5.6. MATERIAL SYMMETRY 203

and in particular for F(t) = F(t) ⋅H1 and G

(t) =HT1 ⋅G

(t), so

Fκ0≤s<∞

F(t) ⋅H1, θ(t),HT

1 ⋅G(t) = Fκ

0≤s<∞F(t) ⋅H1 ⋅H2, θ

(t),HT2 ⋅H

T1 ⋅G

(t),and since by assumption H1 is a symmetry transformation,

Fκ0≤s<∞

F(t), θ(t),G(t) = Fκ0≤s<∞

F(t) ⋅ (H1 ⋅H2), θ(t), (H1 ⋅H2)T ⋅G(t)for all (F(t), θ(t),G(t)). (5.58)

If H is a symmetry transformation, then H−1, where H ⋅H−1 = H−1 ⋅H = 1, isalso a symmetry transformation, since if

Fκ0≤s<∞

F(t), θ(t),G(t) = Fκ0≤s<∞

F(t) ⋅H, θ(t),HT⋅G(t)

for all (F(t), θ(t),G(t)), (5.59)

then in particular for F(t) = F(t) ⋅H−1 and G

(t) = (H−1)T ⋅G(t), we have

Fκ0≤s<∞

F(t) ⋅H−1, θ(t), (H−1)T ⋅G(t) = Fκ0≤s<∞

F(t), θ(t),G(t)for all (F(t), θ(t),G(t)). (5.60)

A group is called Abelian (or commutative) if for every pair of elements (H1,H2) ∈Gκ, H1 ⋅H2 =H2 ⋅H1.

Any finite set of elements satisfying the four group axioms is said to form afinite group, the order of the group being equal to the number of elements in theset. If the group does not have a finite number of elements, it is called an infinitegroup.

A subgroup of a group is a subset of a group such that the subset itself is a group.The term proper subgroup is defined to be consistent with the terms of subgroupand proper subset.

Example

The set of four numbers 1, i,−1,−i, where i is the imaginary number,forms a group of order 4 under multiplication. Group property (i) is clearlysatisfied since the product of any two scalar elements (and square of eachelement) are elements of the set (e.g., 1i = i, i(−i) = 1, i2 = −1, (−i)2 = −1,etc.). The associative law (ii) also holds for the multiplication of numbers.The unit element is taken as the number 1. Finally, if the inverse of everyelement is taken as its reciprocal (e.g., 1/i = −i, 1/(−1) = −1, etc.), thengroup property iv) is satisfied.

Example

An example of a finite group of matrices which is Abelian is given by

( 1 0

0 1) , ( 0 1

−1 0) , ( −1 0

0 −1) , ( 0 −1

1 0) , (5.61)

204 PRINCIPLES OF CONSTITUTIVE THEORY

which form a group of order 4 under matrix multiplication.

Sets of matrices which form groups with respect to matrix multiplication areusually called matrix groups, and are of extreme importance for us.

Example

Another example of a matrix group of order 6 is given by

H1 = ( 1 0

0 1) , H2 = ⎛⎝ −

1

2

√3

2

−

√3

2−

1

2

⎞⎠ , H3 = ⎛⎝ −1

2−

√3

2√3

2−

1

2

⎞⎠ ,H4 = ( 1 0

0 −1) , H5 = ⎛⎝ −

1

2

√3

2√3

2

1

2

⎞⎠ , H6 = ⎛⎝ −1

2−

√3

2

−

√3

2

1

2

⎞⎠ . (5.62)

It may be easily verified by examining their products, etc. For example,

H4H2 = ( 1 0

0 −1)⎛⎝ −

1

2

√3

2

−

√3

2−

1

2

⎞⎠ = ⎛⎝ −1

2−

√3

2

−

√3

2

1

2

⎞⎠ =H6, (5.63)

and so on.

For a given material, the set of all material symmetry transformations for theconstitutive quantity T (X, t) with respect to κ, denoted by Gκ(T ), is a subgroupof the unimodular group,

Gκ(T ) ⊆U (V ). (5.64)

We call Gκ the material symmetry group of T with respect to the reference con-figuration κ. It is important to note that the symmetry group depends on thereference configuration κ. It is clear that Gκ depends as well on the constitutivequantity T . In other words, we may have different symmetry groups for differentconstitutive quantities of the same material body. The largest group contained inthe symmetry groups of all constitutive quantities of the material body B is calledthe material symmetry group of B and is denoted by Gκ(V ).

For any κ and κ such that

P = ∂X∂X

, (5.65)

the following relation, known as Noll’s rule, between the symmetry groups withrespect to the two different reference configurations, holds:

Gκ = P ⋅ Gκ ⋅P−1. (5.66)

To prove the above result, we note that since κ is a different configuration than

5.6. MATERIAL SYMMETRY 205

κ, from (5.49) and (5.50), for any H ∈ Gκ, we have

Fκ0≤s<∞

F(t), θ(t), G(t) = Fκ0≤s<∞

F(t) ⋅P, θ(t),PT⋅ G(t) =

Fκ0≤s<∞

(F(t) ⋅P) ⋅H, θ(t),HT⋅ (PT

⋅ G(t)) =Fκ

0≤s<∞F(t) ⋅ (P ⋅H ⋅P−1) ⋅P, θ(t),PT

⋅ (P ⋅H ⋅P−1)T ⋅ G(t) =Fκ

0≤s<∞F(t) ⋅ (P ⋅H ⋅P−1), θ(t), (P ⋅H ⋅P−1)T ⋅ G(t),

which implies that H = (P ⋅H ⋅P−1) ∈ Gκ.Suppose that P = α1, where α ≠ 0 is a scalar constant. If α > 1, this transfor-

mation represents a dilatation; if 0 < α < 1, it is a contraction; and if α = −1, it isa central inversion. Then P−1 = α−11 and so

Gκ = (α1) ⋅ Gκ ⋅ (α1)−1 = Gκ, (5.67)

so that the material symmetry group is invariant under a uniform transformation.

Example

We want to find the generators of the symmetry group for the geometryshown in Fig. 5.2. By inspection, we have

H(1) =⎡⎢⎢⎢⎢⎢⎣1 0 0

0 1 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦,H(2) =

⎡⎢⎢⎢⎢⎢⎣0 1 0

−1 0 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦,H(3) =

⎡⎢⎢⎢⎢⎢⎣1 0 0

0 −1 0

0 0 −1

⎤⎥⎥⎥⎥⎥⎦,

H(4) =⎡⎢⎢⎢⎢⎢⎣−1 0 0

0 1 0

0 0 −1

⎤⎥⎥⎥⎥⎥⎦,H(5) =

⎡⎢⎢⎢⎢⎢⎣0 1 0

1 0 0

0 0 −1

⎤⎥⎥⎥⎥⎥⎦,(5.68)

where

XI =HIKXK . (5.69)

We note that the above generators are not all the generators that formthe complete discrete symmetry group for this cubic geometry. A morecomplete discussion of specific symmetries is given in Chapter 7.If we call this group G0 = H(1),H(2),H(3),H(4),H(5), we want to findthe group G

0for the transformation shown in Fig. 5.3 and given by

X1 = X1 +KX2, (5.70)

X2 = X2, (5.71)

X3 = X3. (5.72)

Now

P = [ ∂XI

∂XJ

] =⎡⎢⎢⎢⎢⎢⎣1 K 0

0 1 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦and P −1 =

⎡⎢⎢⎢⎢⎢⎣1 −K 0

0 1 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦. (5.73)

206 PRINCIPLES OF CONSTITUTIVE THEORY

Note that P is not orthogonal since P −1 ≠ PT . We can obtain the newgroup by the transformation given by Noll’s rule:

G0= PG0P

−1 = H(1), H(2), H(3), H(4), H(5), (5.74)

where the H(i), i = 1, . . . ,5, satisfy the symmetry transformation

XI = HIKXK . (5.75)

It is easy to verify that H(2) is a symmetry transformation of G0. For

example, for H(2) ∈ G0, we have

H(2) =⎡⎢⎢⎢⎢⎢⎣1 K 0

0 1 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣

0 1 0

−1 0 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣1 −K 0

0 1 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎣−K 1 +K2 0

−1 K 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦. (5.76)

Now, for H(2) to be a symmetry transformation of (5.70)–(5.72), it mustreproduce the same geometry shown in Fig. 5.3 after transforming the indi-cated material vector elements. Transforming the vectors a = (1,0,0) andb = (K,1,0), we obtain

H(2) ⋅ a =⎡⎢⎢⎢⎢⎢⎣−K 1 +K2 0

−1 K 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣1

0

0

⎤⎥⎥⎥⎥⎥⎦= −⎡⎢⎢⎢⎢⎢⎣K

1

0

⎤⎥⎥⎥⎥⎥⎦= −b, (5.77)

H(2) ⋅ b =⎡⎢⎢⎢⎢⎢⎣−K 1 +K2 0

−1 K 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣K

1

0

⎤⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎣1

0

0

⎤⎥⎥⎥⎥⎥⎦= a, (5.78)

thus obtaining the same geometry. Subsequently, H(2) is a symmetry trans-formation for (5.70)–(5.72).

The constitutive equations can be specialized to specific classes of homogeneousmaterials, each class different in the invariance properties assigned to the consti-tutive functional. According to the admitted symmetry group, i.e., according tothe set of all second-order invertible tensors H for which

Fκ0≤s<∞

F(t), θ(t),G(t) = Fκ0≤s<∞

F(t) ⋅H, θ(t),HT⋅G(t)

for all (F(t), θ(t),G(t)), (5.79)

we distinguish the different classes of materials. In the sequel, Gκ will denotethe corresponding class of materials. This classification is important since it leadsto mathematical definitions for various types of real materials, and once Gκ isspecified, i.e., the material symmetry is selected, particular restrictive conditionson the form of the constitutive functional result.

The choice of the type of simple material, along with the imposed material sym-

5.6. MATERIAL SYMMETRY 207

X1

X2

0 1 2 3

1

2

3

Figure 5.2: Simple material geometry with the X3 axis pointing out of the page.

X1

X2

01 2 3

1

2

3

b

aK

Figure 5.3: Transformed material geometry.

metry, leads to very useful representation theorems for the tensor T . It is thenpossible to specify the constitutive functional of the material by means of a certainnumber of independent scalar functions of the history of deformation, called the re-sponse coefficients, which characterize the different materials. These quantities aremainly to be observed in suitable experiments. The mathematical results soughtby experimentalists to guide their design of tests for a practical evaluation of theresponse coefficients are mainly those concerning problems investigated withoutfurther specifications on the constitutive functional. These analytical results areknown as the universal solutions and the universal relations. A controllable so-lution (one satisfying the balance laws) which is the same for all materials in agiven class Gκ is a universal solution. In correspondence with a given deformationor motion, a universal relation is an equation between, say, the stress componentsand the position vector components which holds for all X and t and which is thesame for any material in an assigned class.

Definition: A material is called solid if there exists a reference configuration κ

208 PRINCIPLES OF CONSTITUTIVE THEORY

such that Gκ is the full orthogonal group

O = H ∣ H−1 =HT ,detH = ±1, (5.80)

or a subgroup of it, i.e.,

Gκ ⊆ O(V ). (5.81)

Such a configuration is called an undistorted configuration for the solid.In general, only certain particular reference configurations are undistorted for a

solid body. For a distorted configuration of the solid, the symmetry group neithercontains the orthogonal group nor is contained within it.

The orthogonal group O is a proper subgroup of the unimodular group U , i.e.,O ⊂ U . One can easily construct examples of unimodular linear transformationswhich are not orthogonal.

Definition: A material is called a fluid if for a reference configuration κ, thesymmetry group is the full unimodular group, i.e.,

Gκ =U (V ). (5.82)

For a fluid, Noll’s rule implies that Gκ = P ⋅U ⋅P−1 = U = Gκ for all κ and κ.Therefore, a fluid has the same symmetry group with respect to any configuration.In other words, a fluid does not have a preferred configuration.

Definition: A material that is neither a fluid nor a solid is called a fluid crystal.In other words, for a fluid crystal, there does not exist a reference configuration κfor which either Gκ ⊆ O(V ) or Gκ = U (V ).

Definition: A material is called hemitropic if there exists a reference configu-ration κ such that Gκ = O

+(V ), where O+ is the proper orthogonal group, i.e.,

O+ = H ∣ H−1 =HT ,detH = 1. (5.83)

Definition: A material is called isotropic if there exists a configuration κ suchthat

Gκ = O(V ) or Gκ = U (V ). (5.84)

Such a configuration is called undistorted for the isotropic material body, or simplyan isotropic configuration.

The unimodular, orthogonal, and proper orthogonal groups are clearly all infi-nite groups.

It can be readily seen that the only isotropic materials are isotropic solids, Gκ =O(V ), and fluids, Gκ = U (V ). Any other materials are anisotropic. Anisotropicmaterials include anisotropic solids, Gκ ⊂ O(V ), and fluid crystals, O(V ) ⊂ Gκ ⊂U (V ). The relationship between the different materials is illustrated in Fig. 5.4.

5.7 Reduced constitutive equations

The frame-indifference requirement allows a reduction of the form of the consti-tutive equations to a simpler form. In the discussion of this section, we consider

5.7. REDUCED CONSTITUTIVE EQUATIONS 209

(Isotropic) fluids

Gκ = U (V )(Anisotropic) fluid crystals

O(V ) ⊂ Gκ ⊂ U (V )Isotropic solids

Gκ = O(V )Anisotropic solids

Gκ ⊂ O(V )Figure 5.4: Illustration of the classification of material symmetry groups.

simple homogeneous materials. In addition, dependencies of the constitutive func-tional on X and s are understood; therefore, we shall omit them from the argumentlist and, for simplicity, rewrite (5.33) as

T = F0≤s<∞

F(t), θ(t),G(t). (5.85)

We note that θ(t) and G(t) impose no restrictions on the constitutive functionalsince they are not affected by a Euclidean transformation. We now examine therestrictions imposed by the functional dependence on F(t). In particular, we takeT to represent a second-order tensor, so that objectivity requires that

F0≤s<∞

Q(t) ⋅F(t), θ(t),G(t) =Q ⋅ F0≤s<∞

F(t), θ(t),G(t) ⋅QT

for all Q ∈ O(V ), (5.86)

and material symmetry requires that

F0≤s<∞

F(t), θ(t),G(t) = F0≤s<∞

F(t) ⋅H, θ(t),HT⋅G(t) for all H ∈ Gκ(T ). (5.87)

For ease of discussion, we shall temporarily omit the arguments that are not af-fected by the frame-indifference transformation and rewrite (5.86) as

F0≤s<∞

Q(t) ⋅F(t) =Q ⋅ F0≤s<∞

F(t) ⋅QT . (5.88)

Now using the polar decomposition F(t) =R(t) ⋅U(t) on the left-hand side, we have

F0≤s<∞

Q(t) ⋅R(t) ⋅U(t) =Q ⋅ F0≤s<∞

F(t) ⋅QT , (5.89)

where R ∈ O(V ). If we choose Q(t) =R(t)T , then we obtain

F0≤s<∞

U(t) =RT⋅ F0≤s<∞

F(t) ⋅R. (5.90)

Conversely, if we originally take

F0≤s<∞

F(t) =R ⋅ F0≤s<∞

U(t) ⋅RT , (5.91)

210 PRINCIPLES OF CONSTITUTIVE THEORY

then the transformed quantity is seen to be objective since, if we take R(t) =Q(t)T ,we have

Q ⋅ F0≤s<∞

F(t) ⋅QT = Q ⋅ (R ⋅ F0≤s<∞

U(t) ⋅RT) ⋅QT

= F0≤s<∞

Q(t) ⋅R(t) ⋅U(t)= F

0≤s<∞Q(t) ⋅F(t).

As a result of the above, we can rewrite our constitutive functional in the form

T (X, t) =R(t) ⋅ F0≤s<∞

U(t)(X, s), θ(t)(X, s),G(t)(X, s) ⋅RT (t). (5.92)

This equation represents the general solution to the frame-indifference restrictionrequirement. Note that the stress at time t is only affected by the rotation at timet, R(t), and not by the history of rotation, R(t)(s) for s > 0.

The representation (5.92) is not convenient since for practical applications thepolar decomposition of F would have to be worked out. To derive a more practicalrepresentation, we now note that, using once again the polar decomposition F =R ⋅U, we can rewrite the constitutive function in the form

FT⋅ F0≤s<∞

F(t) ⋅F = UT⋅RT

⋅ (R ⋅ F0≤s<∞

U(t) ⋅RT) ⋅R ⋅U= U ⋅ F

0≤s<∞U(t) ⋅U.

Subsequently, since C =U2, we can write

S0≤s<∞

C(t) ≡U ⋅ F0≤s<∞

U(t) ⋅U = FT⋅ F0≤s<∞

F(t) ⋅F. (5.93)

Thus we finally arrive at the more convenient result that the constitutive functional

for a second-order tensor quantity is objective if and only if Q(t) =R(t)T and

T (X, t) = S0≤s<∞

C(t)(X, s), θ(t)(X, s),G(t)(X, s), (5.94)

where

T ≡ FT⋅ T ⋅F (5.95)

is called the convected tensor. Constitutive functionals of this form, which are notsubject to any further restrictions from the objectivity condition, are said to be inreduced form.

We shall next develop a form of the constitutive equation in which the referenceconfiguration is taken to be the current configuration. This form is most usefulfor the discussion of fluids. First, we recall, upon using (3.288), (5.4), and (5.28),that

C(t)(X, s) = C(X, τ) = FT (X, t) ⋅ (t)C(x, τ) ⋅F(X, t)= FT (X, t) ⋅ (t)C(t)(x, s) ⋅F(X, t), (5.96)

G(t)(X, s) = G(X, τ) = FT (X, t) ⋅ (t)g(x, τ) = FT (X, t) ⋅ (t)g(t)(x, s),(5.97)

5.7. REDUCED CONSTITUTIVE EQUATIONS 211

where we have taken τ = t − s, and note that

(t)C(x, t) = 1 and (t)g(x, t) = g(x, t) = grad θ(x, t). (5.98)

Subsequently, from (5.94) and (5.95), the constitutive functional becomes

T (X, t) = (FT (X, t))−1 ⋅ S0≤s<∞

FT (X, t) ⋅ (t)C(t)(x, s) ⋅F(X, t), θ(t)(X, s),FT (X, t) ⋅ (t)g(t)(x, s) ⋅F−1(X, t),

or, with a little abuse of notation, we have

T (x, t) = F0≤s<∞

F(x, t), (t)C(t)(x, s), θ(t)(x, s), (t)g(t)(x, s). (5.99)

Now, assuming that T is a second-order tensor, for frame indifference we requirethat

F0≤s<∞

F, (t)C(t), θ(t), (t)g(t) =Q ⋅ F0≤s<∞

F, (t)C(t), θ(t), (t)g(t) ⋅QT . (5.100)

Since

(t)C(t) = (t)F(t)T ⋅ (t)F(t) and (t)g

(t) = (FT )−1 ⋅G(t), (5.101)

and since θ(t) = θ(t) and G

(t) = G(t), if we take F(t) = Q(t) ⋅F(t) and use (3.275),

we have

(t)F(t) = Q(t) ⋅F(t) ⋅F−1 ⋅QT =Q(t) ⋅ (t)F(t) ⋅QT , (5.102)

(t)C(t) = (Q(t) ⋅ (t)F(t) ⋅QT )T ⋅ (Q(t) ⋅ (t)F(t) ⋅QT )

= Q ⋅ (t)F(t)T⋅Q(t)

T⋅Q(t) ⋅ (t)F(t) ⋅QT

= Q ⋅ (t)C(t)⋅QT , (5.103)

(t)g(t) = Q ⋅ (FT )−1 ⋅G(t) =Q ⋅ (t)g(t). (5.104)

Subsequently, the frame indifference condition becomes

F0≤s<∞

Q ⋅F,Q ⋅ (t)C(t) ⋅QT , θ(t),Q ⋅ (t)g(t) =Q ⋅ F

0≤s<∞F, (t)C(t), θ(t), (t)g(t) ⋅QT for all Q =RT ∈ O(V ). (5.105)

For material symmetry, with T being an arbitrary tensor, we have

F0≤s<∞

F, (t)C(t), θ(t), (t)g(t) = F0≤s<∞

F, (t)C(t), θ(t), (t)g(t), (5.106)

for H ∈ Gκ(T ). We now note that if we take F(t) = F(t) ⋅H and G

(t) =HT⋅G(t),

and use (3.275) and (5.101), we have

(t)F(t) = F(t) ⋅H ⋅H−1 ⋅F−1 = F(t) ⋅F−1 = (t)F(t),

(t)C(t) = (t)C

(t),

(t)g(t) = (FT )−1 ⋅ (HT )−1 ⋅HT⋅G(t) = (FT )−1 ⋅G(t) = (t)g(t).

212 PRINCIPLES OF CONSTITUTIVE THEORY

Subsequently, the symmetry condition becomes

F0≤s<∞

F, (t)C(t), θ(t), (t)g(t) = F0≤s<∞

F ⋅H, (t)C(t), θ(t), (t)g(t)for all H ∈ Gκ(T ). (5.107)

5.7.1 Constitutive equation for a simple isotropic solid

Different symmetry groups correspond to solids having different material symme-tries. For a simple isotropic solid, Gκ(T ) = O(V ), so that by using the polardecomposition F = R ⋅U and taking H = RT ∈ O(V ), the symmetry condition(5.107), using the current reference configuration, becomes

F0≤s<∞

F, (t)C(t), θ(t), (t)g(t) = F0≤s<∞

R ⋅U ⋅RT , (t)C(t), θ(t), (t)g(t). (5.108)

Now since V = R ⋅U ⋅RT and B = V2, with a little abuse of notation, we canrewrite (5.99) as

T (x, t) = F0≤s<∞

B, (t)C(t), θ(t), (t)g(t), (5.109)

which is required to satisfy the appropriate invariance condition depending onthe order of tensor T . For a second-order tensor, using (5.103) and (5.104), suchcondition is

F0≤s<∞

Q ⋅B ⋅QT ,Q ⋅ (t)C(t) ⋅QT , θ(t),Q ⋅ (t)g(t) =Q ⋅ F

0≤s<∞B, (t)C(t), θ(t), (t)g(t) ⋅QT , (5.110)

for all Q =RT ∈ O(V ). The above equation corresponds to the reduced form of theconstitutive function for a simple isotropic solid. From this representation, we seethat in the current configuration the constitutive dependence on the deformationgradient F reduces to the dependence on the left Cauchy–Green tensor B. It isnoted that (5.109) remains valid for hemitropic solids as long as now H = RT ∈O+(V ).

5.7.2 Constitutive equation for a simple (isotropic) fluid

Now, from the requirement that the mass density remain the same, we must havethat detF = det(FH). Subsequently, assume that H = aA, where a > 0 is a scalarand A is a matrix. Now since det(aA) = a3 detA, the above requirement becomesdetF = detF a3 detA. Thus we see that if we take A = F −1 and a = (detF )1/3,the requirement is satisfied identically, and the transformation

H = (detF )1/3F−1 (5.111)

is such that Gκ = U (V ). Subsequently, using (5.107), for a simple fluid the abovesymmetry condition becomes

F0≤s<∞

F, (t)C(t), θ(t), (t)g(t) = F0≤s<∞

(detF )1/31, (t)C(t), θ(t), (t)g(t) (5.112)

5.8. ISOTROPIC AND HEMITROPIC REPRESENTATIONS 213

or, since detF = J = ρR/ρ, with abuse of notation, we have

F0≤s<∞

F, (t)C(t), θ(t), (t)g(t) = F0≤s<∞

ρ, (t)C(t), θ(t), (t)g(t). (5.113)

Thus, the constitutive functional (5.99) reduces to

T (x, t) = F0≤s<∞

ρ, (t)C(t), θ(t), (t)g(t), (5.114)

which is required to satisfy the appropriate invariance condition depending onthe order of tensor T . For a second-order tensor, using (5.103) and (5.104), suchcondition is

F0≤s<∞

ρ,Q ⋅ (t)C(t) ⋅QT , θ(t),Q ⋅ (t)g(t) =Q ⋅ F

0≤s<∞ρ, (t)C(t), θ(t), (t)g(t) ⋅QT , (5.115)

for all Q = RT ∈ O(V ). Equation (5.114) corresponds to the reduced form of theconstitutive function for a simple fluid. From this representation, we see that theconstitutive dependence on the deformation gradient F reduces to the dependenceon its determinant only, or equivalently ρ.

5.8 Isotropic and hemitropic representations

Below we consider constitutive equations of different rates and grades as given by(5.37). Furthermore, we consider the representation of isotropic and hemitropicfunctions. Since representation of more general non-isotropic functions applyspecifically to solids, we shall discuss this topic in Chapter 7, which deals withthermoelastic solids.

The main problem of invariant theory is the representation of tensor-valuedfunctions. Let φα, vβ , Aγ , and Wδ denote scalars, vectors, symmetric tensors,and skew-symmetric tensors, respectively. Let ψ, h, and T be scalar-, vector-,and tensor-valued functions of φα, vβ , Aγ , and Wδ, respectively. An orthogonaltensor Q is said to be a symmetry transformation of the functions ψ, h, and T,respectively, if

ψ(φ′α,v′β ,A′γ ,W′

δ) = ψ(φα,vβ ,Aγ ,Wδ), (5.116)

h(φ′α,v′β ,A′γ ,W′

δ) = Q ⋅ h(φα,vβ ,Aγ ,Wδ), (5.117)

T(φ′α,v′β ,A′γ ,W′

δ) = Q ⋅T(φα,vβ ,Aγ ,Wδ) ⋅QT , (5.118)

where

φ′α = φα, (5.119)

v′β = Q ⋅ vβ , (5.120)

A′γ = Q ⋅Aγ ⋅QT , (5.121)

W′

δ = (detQ)Q ⋅Wδ ⋅QT . (5.122)

214 PRINCIPLES OF CONSTITUTIVE THEORY

The skew-symmetric tensors Wδ = wδ ⋅ ε may be replaced by the axial vectorswδ = 1

2ε ∶Wδ ≡ ⟨Wδ⟩, where the axial vectors transform according to

w′δ = (detQ)Q ⋅wδ, (5.123)

and note that in general detQ = ±1. Here we also note that the axial scalars(pseudoscalars) wδ are frame indifferent if the following holds:

w′δ = (detQ)wδ. (5.124)

The symmetry groups of the functions ψ, h, and T consist of the sets of all sym-metry transformations of ψ, h, and T, respectively.

Definition: We say that ψ, h, and T are scalar-, vector-, and tensor-valuedisotropic functions, respectively, if for any number of scalars φα, vectors vβ , sym-metric tensors Aγ , and skew-symmetric tensors Wδ, they are invariant with re-spect to all Q ∈ O(V ).

Definition: We say that ψ, h, and T are scalar-, vector-, and tensor-valuedhemitropic functions, respectively, if for any number of scalars φα, vectors vβ ,symmetric tensors Aγ , and skew-symmetric tensors Wδ, they are invariant withrespect to all Q ∈ O

+(V ).A material property is isotropic if that property at a point is the same in all

directions, and transversely isotropic if that property is the same in all direc-tions in a plane. It is called anisotropic (or non-isotropic) otherwise. Isotropicfunctions are also called isotropic invariants. From the definition, the conditionsimpose no restrictions on the scalar variables which a function depends on. Hence,scalar variables are irrelevant as far as the representation of isotropic invariantsare concerned. A scalar-valued function basis for isotropic or hemitropic func-tions of vβ , Aγ , and Wδ consists of a set of isotropic or hemitropic scalar-valuedfunctions I1, . . . , In of vβ , Aγ , and Wδ such that any isotropic or hemitropicscalar-valued function of vβ , Aγ , and Wδ can be expressed as a single-valuedfunction of I1, . . . , In, i.e.,

ψ = ψ(φα,vβ ,Aγ ,Wδ) = ψ(φα, I1, . . . , In). (5.125)

Let h0, . . . ,hR and T0, . . . ,TS denote respectively vector- and tensor-valuedisotropic or hemitropic functions of vβ , Aγ , and Wδ. Then, if any vector-valued and tensor-valued isotropic or hemitropic functions h(φα,vβ ,Aγ ,Wδ) andT(φα,vβ ,Aγ ,Wδ) may be expressed respectively in the forms

h = R

∑r=0

arhr and T = S

∑s=0

bsTs, (5.126)

where ar and bs are functions of φα and the isotropic or hemitropic scalar invariantsof vβ , Aγ , and Wδ (i.e., I1, . . . , In), we say that h0, . . . ,hR and T0, . . . ,TSare vector and tensor generators of vβ , Aγ , and Wδ.

The general representation for isotropic and hemitropic scalar-, vector-, andtensor-valued functions are given by Zheng (1994) and are reproduced in Ta-bles 5.1–5.8. To clarify the nomenclature in the tables, we note that A and W

are component matrices of the second-order tensors A and W, respectively, AWis the component matrix of A ⋅W = AijWjk, u ⋅Av = uiAijvj , etc.

5.9. EXPANSIONS OF CONSTITUTIVE EQUATIONS 215

As an illustration on the use of the tables, we examine the isotropic represen-tation of T = T(A), where T and A are symmetric second-order tensors, and itobeys the invariance condition

T(Q ⋅A ⋅QT ) =Q ⋅T(A) ⋅QT (5.127)

for Q ∈ O(V ). Then its irreducible representation from Table 5.3 is given by

T = b01 + b1A + b2A2, (5.128)

where b0, b1, and b2 are functions of the scalar invariants I1 = trA, I2 = trA2, andI3 = trA3 obtained from Table 5.1. The tensors 1, A, and A2 are the generatorsof the tensor T.

As a second illustration, the hemitropic representation of h = h(v,A), wherev is a vector and A is a symmetric second-order tensor, that satisfies the frame-invariance condition

h(Q ⋅ v,Q ⋅A ⋅QT ) =Q ⋅ h(v,A), (5.129)

for Q ∈ O+(V ), is obtained from Table 5.6:

h = a0v + a1Av + a2v ×Av, (5.130)

where a0, a1, and a2 are functions of the scalar invariants I1, . . . , I7 given by v ⋅v,tr A, tr A2, tr A3, v ⋅ Av, v ⋅ A2v, and the scalar triple product [v,Av,A2v]obtained from Table 5.5.

Appendix E provides details illustrating a procedure for obtaining representa-tions of isotropic scalar, vector, and symmetric second-order tensor functions ofa vector and a symmetric second-order tensor. While such a procedure can beexpanded to obtain more general results, the results presented in the tables havebeen obtained by a more systematic procedure.

5.9 Expansions of constitutive equations

Let us assume that we have a constitutive function for an arbitrary rank tensorquantity T that is a function of scalar fields φα, vector fields vβ , rank-2 tensorfields Aγ , etc.:

T = T (φα,vβ ,Aγ , . . .) . (5.131)

Let us also assume that at (φ0α,v0

β ,A0γ , . . .) the value of the function T is defined

and derivatives of T to all orders exist at this point. Then, in the neighborhoodof such point, we can write the Taylor series expansion

T (φα,vβ ,Aγ , . . .) = ∞∑n=0

1

n!(φα − φ0α) ∂

∂φα+ (vβ − v

0

β) ⋅ ∂

∂vβ

+

(Aγ −A0

γ) ∶ ∂

∂Aγ

+⋯n T 0, (5.132)

where T 0 ≡ T (φ0α,v0

β ,A0γ , . . .) signifies that after the function is operated on, the

resulting terms are evaluated at the reference values of φα = φ0α, vβ = v0

β , Aγ =A0γ ,

216 PRINCIPLES OF CONSTITUTIVE THEORY

etc., and the resulting terms are understood as

∂T 0

∂φα∣φ′α,vβ,Aγ

,∂T 0

∂vβ

∣φα,v′

β,Aγ

,∂T 0

∂Aγ

∣φα,vβ,A′γ

, ⋯,

which correspond to tensors of appropriate orders, and φ′α indicates that all φ1, φ2,... are kept constant with the exception of φα; the meaning for the other variablesis analogous. It is noted that typically the reference state corresponds to one ofthermodynamic equilibrium (see below), in which case the above terms are relatedto equilibrium or near-equilibrium tensor properties.

5.10 Thermodynamic considerations

A system is a region containing matter and energy that is separated from itssurroundings by arbitrarily imposed walls or boundaries. In a thermodynamicanalysis, the system is the subject of the investigation. A boundary is a closedsurface surrounding the system through which mass and/or energy may enter orleave the system. Everything external to the system is the surroundings.

If no mass can cross the complete boundary of a system (but work and heat can),then we have a closed system. If in addition energy does not cross the completeboundary, then we have a mechanically and thermally isolated system. An opensystem is one in which both mass and energy can cross the system’s boundary, andsuch boundary is a nonmaterial surface.

The condition of the system at any instant of time is called its state. The stateat a given instant of time is described by the properties of the system.

5.10.1 Thermodynamic states

Definition: A thermodynamic state corresponds to a set of values of propertyvariables of a system that must be specified to reproduce the system uniquely.The number of values required to specify the state depends on the system. Once asufficient number has been specified, the values of all other variables are uniquelydetermined.

Property variables are classified as being extensive or intensive. Extensive prop-erties are additive and thus depend on the amount of matter in the system (e.g.,internal energy, volume). Intensive properties are independent of the amount ofmatter in the system (e.g., temperature, pressure).

We make a special note that, quite often, in a homogeneous system we write anextensive property, say Φ, as Φ = mφ, where m is the mass of the system, thusreferring to φ as a property density or specific property. For example, we write theinternal energy as me, where e is the internal energy density or specific internalenergy. Similarly, we write the extensive properties of volume V as m v, where v

is the volume density or specific volume, and number of moles Ni of a chemicalcomponent i as mni, where ni is the mole number density of component i. Notethat yi = Mi ni is the mass fraction of component i, where Mi is the molecularmass of the component. Alternately, specific quantities are referred to one molerather than to a unit of mass of the system. In such case, Φ = Nφ, where N =∑i Ni

denotes the total number of moles of matter in the system and φ is referred as

5.10. THERMODYNAMIC CONSIDERATIONS 217

a molar density or specific molar quantity. For example, corresponding to thenumber of moles Ni of chemical component i, one writes Nni, where ni is the molefraction of component i. Property densities also do not depend on the system size.But such quantities should not be confused with intensive quantities. In applyinga physical statement, it is always best to recall the extensive quantity Φ beforeapplying the statement.

It is postulated that there exists an extensive quantity, called entropy, that is afunction of a number of extensive quantities that describe any composite system,which is defined for all states and has the following properties. The values assumedby the extensive variables in the absence of an internal constraint are those thatmaximize the entropy at an equilibrium state. Furthermore, it is assumed thatentropy is continuous, is differentiable, and is a monotonically increasing functionof internal energy.

Writing the above quantities per unit mass, we assume that the specific entropyη is a function of the set of specific state variables (e,να), with α = 1, . . . , n, andthese quantities specify a thermodynamic state. In addition, we postulate thatsuch state characterizes completely the entropy density η of a material point Xand occupying coordinate x at time t. The choice of να, which are in generaltensors of different ranks, and which we will generically call specific thermostaticvolumes, depend on the system under consideration and define the thermodynamiccharacter of the system. For example, for a single-component simple system, n = 2,ν1 → v, and ν2 → n. Here we recognize ν1 as the specific volume, v = 1/ρ, andν2 as the mole number density, n (note that y = Mn = 1 and n = 1 in this case).More generally, and as will become clear later, να are related to F, C, and g for asimple solid (see (5.99)), and ρ, C, and g for a simple fluid (see (5.114)). Once eand να are selected, we will have the thermodynamic state of the system defined.Equilibrium states are, macroscopically, characterized completely by η, e, and να.Note that microscopically there will be fluctuations about an equilibrium state,but macroscopic measurements do not see them.

A basic problem is the determination of an equilibrium state. The problem ofthermodynamic equilibrium can be completely solved with the aid of the aboveextremum principle if the entropy of the system is known as a function of the statevariables, i.e.,

η = η(e,να,X), α = 1, . . . , n. (5.133)

Such constitutive relation is called a fundamental relation and η is considereda thermodynamic potential. Thus, if the fundamental relation is known for aparticular system, then all thermodynamic information about the system can beobtained. In addition, the continuity, differentiability, and monotonic property of∂η/∂e∣να,X > 0 imply that the entropy function can be inverted with respect tointernal energy and that also the internal energy is a single-valued, continuous, anddifferentiable function of (η,να) for fixed X . Thus (5.133) can be solved uniquelyfor e in the form

e = e(η,να,X), α = 1, . . . , n, (5.134)

where now e is considered the thermodynamic potential. The set (η,να) describesthe thermodynamic state at the material point X . Constitutive equations (5.133)and (5.134) are alternative forms of the fundamental relation, and each contains

218 PRINCIPLES OF CONSTITUTIVE THEORY

all thermodynamic information about the system. Specific choices of functionsdefine different thermodynamic substances.

For a thermodynamic state to represent a physical one, it must not contradictthe basic axioms of mechanics, thermodynamics, and constitutive representations.Such restrictions provide conditions of admissibility on the thermodynamic stateof the system. The specific functional form defines different thermodynamic sub-stances.

Definition: The changes that occur in e due to changes of η and να are calleda thermodynamic process.

Definition: A thermodynamic process is called thermodynamically homoge-neous if e is independent of X .

For a thermodynamically homogeneous state, the functional form of the internalenergy density is the same at all points of the material body. Such equationis a thermodynamical constitutive equation for the internal energy density. It issubject to the restrictions of the the second law of thermodynamics as well asother constitutive principles to be discussed. The selection of να depends on thethermodynamic state of the body. A certain class of variables cannot be admittedinto the class of να since e is a thermodynamic property of a material. For example,it cannot be an explicit function of t, x, v, and a. On the other hand, η and να

do depend on time and coordinate, i.e.,

η = η(x, t) and να = να(x, t). (5.135)

Since for a given motion we have x = χ(X, t), we see that η = η(X, t) and να =να(X, t), and therefore, e = e(X, t), where X is the coordinate in the referenceconfiguration of the material point X . Nevertheless, we again note that e is notan explicit function of x and t, but depends on the values of η and να at x and t.

Definition: The thermostatic temperature θ and thermostatic tensions τα aredefined by

θ ≡ ∂e

∂η∣νγ ,X

and τα ≡ ∂e

∂να

∣η,ν′α,X

. (5.136)

It is noted that the postulation of the existence of entropy satisfying the pre-viously mentioned properties leads to the above definition of temperature that isindependent of that introduced by the zeroth law of thermodynamics. Further-more, the requirements that the entropy function can be inverted with respect tointernal energy and that also the internal energy can be inverted with respect tothe entropy requires that 0 < θ <∞.

From now on, since all partial derivatives are for fixed material point X , wewill suppress the subscript designations of fixed X in all such derivatives. We alsodefine the additional thermodynamic quantities

ϕα ≡ ∂θ

∂να

∣η,ν′α

and φαβ ≡ ∂τα

∂νβ

∣η,ν′

β

, (5.137)

where ϕα is the isentropic thermal stiffness tensor and φαβ is the isentropic elasticstiffness tensor. When ν1 is the specific volume, v, then −τ 1 is the thermodynamicpressure, p, and v2φ1 1 corresponds to the square of the isentropic speed of sound.

5.10. THERMODYNAMIC CONSIDERATIONS 219

For a mixture, if we take ν2, . . . ,νn as the mole number densities of the con-stituents, then τ 2, . . . ,τn are known as the chemical potentials. If ν1 is related tothe strain tensor, then τ 1 will be seen to be the stress tensor.

Taking the first differential of the first and second forms of the fundamentalrelations (5.133) and (5.134), we see that for an arbitrary change of the thermo-dynamic state at a given material point X , we have

dη = 1

θde −

τα

θ⋅ dνα (5.138)

and

de = θ dη + τα ⋅ dνα, (5.139)

where the inner products denote full contractions between tensors τα and να.Equation (5.139) is the local form of the Gibbs equation. The variables (1/θ, e)and (−τα/θ,να), and (θ, η) and (τα,να) are considered to be conjugate variablepairs of intensive variables and specific extensive variables in the correspondingfundamental relations.

From (5.136), it is clear that

θ = θ(η,νβ ,X) and τα = τα(η,νβ ,X). (5.140)

Such relations expressing intensive properties in terms of the state variables (thethermodynamic state) are called equations of state. Knowledge of a single equa-tion of state does not give complete knowledge of thermodynamic properties ofa system; knowledge of all the equations of state is equivalent to knowledge ofthe fundamental equation and thus, is thermodynamically complete. Equations ofstate can be derived from the fundamental relation.

If the equation for thermostatic tensions in (5.140)2 is solvable for νβ , then wecan select τα as a new state variable to replace νβ . A sufficient condition for thisto be possible is that φαβ is continuous and does not vanish in some neighborhoodof νβ . In this case, we have

να = να(η,τ β ,X) (5.141)

and subsequently, we can write

η = η(e, να(η,τβ ,X),X) = η(e,τ β ,X) (5.142)

ore = e(η, να(η,τ β ,X),X) = e(η,τ β ,X). (5.143)

Effectively, we have eliminated να using equations of state (5.140)2. We also definethe additional thermodynamic quantities

ζα ≡ ∂να

∂η∣τγ

and χαβ ≡ ∂να

∂τβ

∣η,τ ′

β

, (5.144)

where ζα is the isopiestic thermal expansion tensor and is related to the thermalexpansion tensor and specific heat at constant tension (see below) and χαβ is theisentropic elastic compliance tensor.

220 PRINCIPLES OF CONSTITUTIVE THEORY

Alternately, if the equation for temperature in (5.140)1 is solvable for η, thenwe can select θ as a new state variable to replace η. A sufficient condition forthis to be possible is that ∂θ/∂η∣νβ

is continuous and does not vanish in someneighborhood of η. In this case, we have

η = η(θ,να,X) and e = e(η(θ,νβ ,X),να,X) = e(θ,να,X), (5.145)

and subsequently,

τα = τα(θ,νβ ,X) (5.146)

or

να = να(θ,τ β ,X). (5.147)

Effectively, we have eliminated η between the equations of state (5.140). Equations(5.145) are referred to as caloric equations of state, while equations (5.146) and(5.147) are referred to as thermal equations of state. Note that if ν1 is the specificvolume and −τ 1 the thermodynamic pressure, (5.146) is nothing more than p =p(θ, v) at material point X , which reduces to the ideal gas equation, or van derWaal’s equation of state, etc., depending on the specific fundamental relation.Furthermore, if equation (5.147) is substituted in the relations (5.145), we have

η = η(θ,τα,X) and e = e(θ,τα,X). (5.148)

Differentiating (5.146) and (5.147) at an arbitrary material point X , we have

dτα = βαdθ + ξαβ ⋅ dνβ and dνα = ααdθ + υαβ ⋅ dτ β, (5.149)

where

βα ≡ ∂τα

∂θ∣νγ

, αα ≡ ∂να

∂θ∣τγ

, ξαβ ≡ ∂τα

∂νβ

∣θ,ν′

β

, υαβ ≡ ∂να

∂τβ

∣θ,τ ′

β

, (5.150)

and βα is the isochoric thermal tension tensor, αα is the thermal strain tensor(also called the piezocaloric tensor), ξαβ is the isothermal elastic stiffness tensor,and υαβ is the isothermal elastic compliance tensor. The quantities (5.150) arerelated by the identities

βα+ξαβ ⋅αβ = 0, αα+υαβ ⋅ββ = 0, υαγ ⋅ξγ β = 1αβ, ξαγ ⋅υγ β = 1αβ, (5.151)

where 1αβ is the unit tensor. Since the tensors (5.150) provide changes of mea-surable quantities, they are useful for inferring the forms of the thermal equationsof state from experiments. For example, when ν1 is the specific volume, v, and−τ 1 the thermodynamic pressure, p, then β1/p is the isochoric pressure coefficient,α1/v is the coefficient of thermal or volume expansion, and υ1 1/v is the isothermalcompressibility.

When compared with the fundamental relations, equations of state offer theadvantage of connecting easily measurable quantities, but the disadvantage ofbeing insufficient in providing all thermodynamic properties of a material. Weelaborate on this aspect below.

5.10. THERMODYNAMIC CONSIDERATIONS 221

Definition: We say that f(x1, . . . , xn) is a homogeneous function of degree nfor all xi > 0 and λ > 0 if

f(λx1, . . . , λxn) = λn f(x1, . . . , xn). (5.152)

In thermodynamics, intensive functions are homogeneous of degree 0 and exten-sive functions are homogeneous of degree 1.

Euler’s theorem: Let f(x1, . . . , xn), with xi > 0, i = 1, . . . , n, be a continuousand differentiable function. Then f is a homogeneous function of degree n if andonly if

nf(x1, . . . , xn) = ∂f(x1, . . . , xn)∂xi

∣x′i

xi. (5.153)

Proof: Simply differentiate the homogeneity condition (5.152) with respect toλ:

d

dλf(λx1, . . . , λxn) = d

dλ[λn f(x1, . . . , xn)]

to obtain∂f(λx1, . . . , λxn)

∂(λxi) ∣x′i

xi = nλn−1f(x1, . . . , xn).Now setting λ = 1, we have our proof.

Since the internal energy is an extensive quantity, which, through the fundamen-tal relation (5.134), is a function of extensive quantities of entropy and thermo-static volumes, and since extensive quantities are homogeneous functions of degree1, from (5.134), Euler’s theorem (5.153), and the definitions (5.136), we obtain

e = θ η + τα ⋅ να. (5.154)

This result is known as Euler’s equation. Now, taking the differential of Euler’sequation (5.154) and subtracting the Gibbs equation (5.139), we obtain the Gibbs–Duhem equation:

η dθ + να ⋅ dτα = 0. (5.155)

Again we note that the inner products in the above equations denote full contrac-tions.

Before continuing, it is worthwhile to summarize the formal structure of ther-modynamics. The fundamental equation (5.133) or (5.134) contains all thermo-dynamic information about a system. To be specific, we use the internal energyrepresentation (5.134) as the fundamental relation. With the definitions of thethermostatic temperature and thermostatic tensions (5.136), the fundamental re-lation implies the equations of state (5.140). If all equations of state are known,they may be substituted into Euler’s equation (5.154) to recover the fundamentalequation (5.134). Thus, the totality of the equations of state is equivalent to thefundamental equation. Any single equation of state contains less thermodynamicinformation. If all equations of state minus one are known, the Gibbs–Duhemequation (5.155) may be integrated to obtain the missing one, but such equa-tion will contain an undetermined integration constant. Thus, just one missingequation of state suffices to determine the fundamental equation except for anundetermined constant. Note that, as shown, it is always possible to express e as

222 PRINCIPLES OF CONSTITUTIVE THEORY

a function of variables other than η and να. Thus, we could eliminate η betweenthe fundamental relation and the equation of state for θ in (5.140) to obtain anequation of the form (5.145)2. However, this is not a fundamental relation sinceit does not contain all possible thermodynamic information about a system. Infact, recalling the definition of θ in (5.136)1, we see that e = e(θ,να,X) is a par-tial differential equation. Even if this equation was integrable, it would yield afundamental relation with undetermined functions.

Definition: The heat and work increments are defined by

dq ≡ θ dη and dw ≡ τα ⋅ dνα. (5.156)

We note that the Gibbs equation (5.139) can now be rewritten in the familiarform

de = dq + dw, (5.157)

which is the local form of the first law of thermodynamics (4.170).Using the Gibbs equation (5.139), the specific heat c, and the caloric stiffness

λναand caloric compliance λτα

for fixed X are subsequently defined by

c ≡ dq

dθ= θdη

dθ= 1

dθ(de − τα ⋅ dνα) , (5.158)

λνα≡ dq

dνα

= θ dηdνα

= 1

dνα

(de − τ β ⋅ dνβ) , (5.159)

λτα≡ dq

dτα

= θ dηdτα

= 1

dτα

(de − τβ ⋅ dνβ) . (5.160)

In practical applications, experimental measurements frequently dictate thata partial derivative be evaluated. For example, we may be concerned with theanalysis of the temperature change, which is required to maintain the volumeof a single component system constant if the pressure is increased slightly. Insuch case, we require ∂θ/∂p∣v. A general feature of the derivatives that arise isthat they generally involve both intensive and extensive properties. Of all suchderivatives, only N = (n + 1)(n + 2)/2 can be independent, and this number canbe shown to correspond to the number of unique second derivatives. As secondderivative quantities are associated with special material properties, then one canchoose N such properties as conventional and then any other property can bewritten in terms of these conventional properties. For example, in the case of asingle-component simple system where the mole number density is constant, wehave n = 1 and ν1 is the specific volume, the number of independent propertiesare N = 3. The three conventional properties, which are defined in terms of thethree unique second derivatives, are the specific heat at constant pressure cp, thecoefficient of thermal expansion α, and the isothermal compressibility κθ:

cp = θ ∂η

∂θ∣p

, α = 1

v

∂v

∂θ∣p

, κθ = −1v

∂v

∂p∣θ

. (5.161)

Any other property can then be subsequently written in terms of these three. Forexample, the specific heat at constant volume and the adiabatic compressibility,

cv = θ ∂η

∂θ∣v

and κη = − 1

v

∂v

∂p∣η

, (5.162)

5.10. THERMODYNAMIC CONSIDERATIONS 223

can then be obtained from the difference and ratio of specific heats relations:

cp − cv = θ vα2

κθ(5.163)

and

γ = cpcv= κθκη. (5.164)

More generally, for experimental measurements, it is convenient to keep να =const. or τα = const. Thus, the specific heats at constant thermostatic volumes andthe specific heats at constant thermostatic tensions (using (5.150)2) follow from

cνα= θ ∂η

∂θ∣να

= ∂e∂θ∣να

and cτα= θ ∂η

∂θ∣τα

= ∂e∂θ∣τα

− τα ⋅αα. (5.165)

To determine the specific heats at constant tensions in a form more amenable toexperimental measurements, we regard e as a function of θ and νβ at fixed X in(5.158) (see (5.145)2). Then, from (5.158), we have

cτα= 1

dθ[ ∂e∂θ∣να

dθ + ( ∂e

∂να

∣θ,ν′α

− τα) ⋅ dνα] . (5.166)

If we now regard να as a function of θ and τβ at fixed X , as in (5.147), and holdthe tensions τ β = const., using (5.149)2, (5.150), and (5.165)1, we get

cτα− cνα

= ( ∂e

∂να

∣θ,ν′α

− τα) ⋅αα, (5.167)

which, upon using (5.159), becomes

cτα− cνα

= λνα⋅αα. (5.168)

Analogously, by regarding e as a function of θ and τα at fixed X , it is easy toshow that also

cτα− cνα

= −λτα⋅ βα. (5.169)

Subsequently, we see that

λνα⋅αα = −λτα

⋅ βα, (5.170)

or, using (5.151)1,2,

λνβ= λτα

⋅ ξαβ or λτβ= λνα

⋅υαβ. (5.171)

We also note from (5.159) and (5.160) (using (5.149)2) that when η = η(θ,να) andη = η(θ,τα), we respectively have

λνα= θ ∂η

∂να

∣θ,ν′α

= ∂e

∂να

∣θ,ν′α

− τα, (5.172)

λτα= θ ∂η

∂τα

∣θ,τ ′α

= ∂e

∂τα

∣θ,τ ′α

− τβ ⋅υβ α, (5.173)

224 PRINCIPLES OF CONSTITUTIVE THEORY

and

dq ≡ θ dη = cναdθ +λνα

⋅ dνα and dq ≡ θ dη = cταdθ +λτα

⋅ dτα. (5.174)

The ratio of specific heats

γα ≡ cτα

cνα

, (5.175)

which represents a more general quantity than (5.164), is important and it arisesin many branches of physics. It, as well as the specific heats, is connected withother quantities using the following identities (which can be easily proved):

φαγ ⋅υγ β +ϕααβ = 1αβ, ϕβ ⋅ υβ α + γαζα = 0, ζβ ⋅φβ α +ϕα = 0,γα +ϕα ⋅αα = 1, (5.176)

where we have used (5.175) and

ζα = (θ/cτα)αα. (5.177)

Results (5.168) and (5.169) and similar ones may be obtained easily by consid-ering the following. Assume that we have two specific extensive tensor quantitiesa and b given in terms of two intensive tensor quantities a and b:

a = a(a, b) and b = b(a, b). (5.178)

Then

da = ∂a∂a∣b

⋅ da +∂a

∂b∣a

⋅ db and db = ∂b∂a∣b

⋅ da +∂b

∂b∣a

⋅ db. (5.179)

Now, holding b constant and taking the inner product of the second equation with∂b/∂b∣a, we obtain

0 = [ ∂b∂b∣a

⋅∂b

∂a∣b

⋅ da +∂b

∂b∣a

⋅∂b

∂b∣a

⋅ db] (5.180)

or

db = − ∂b∂b∣a

⋅∂b

∂a∣b

⋅ da. (5.181)

Introducing this result into the first equation in (5.179) gives

da = ∂a∂a∣b

⋅ da −∂a

∂b∣a

⋅∂b

∂b∣a

⋅∂b

∂a∣b

⋅ da. (5.182)

Now dividing this equation by da at constant b, leads to the final result

∂a

∂a∣b

−∂a

∂a∣b

= − ∂a∂b∣a

⋅∂b

∂a∣b

. (5.183)

To illustrate the use of (5.183), if we take a → e, b → να, a → θ, and b → τα,we easily obtain (5.167) and subsequently (5.168). If we take a → να, b → η,

5.10. THERMODYNAMIC CONSIDERATIONS 225

a → τα, and b → θ, we obtain an equation that provides the difference betweenthe isentropic and isothermal elastic compliance tensors:

∂να

∂τβ

∣η,τ ′

β

−∂να

∂τβ

∣θ,τ ′

β

= − ∂να

∂η∣τγ

∂η

∂τβ

∣θ,τ ′

β

, (5.184)

or, using (5.144), (5.150)4, and (5.173),

χαβ − υαβ = −1θζαλτβ

. (5.185)

We note that in the case where ν1 is the specific volume, this relationship leadsto (5.163).

The Gibbs–Duhem relation (5.155) expresses the existence of a relationshipamong the first derivatives of the fundamental relation (5.134). Similarly, thereare relations among the second derivatives. They are associated with the equalityof the various mixed second derivatives. Such identities are known as Maxwellrelations. Specifically, using the fundamental relation (5.134) and the definitions(5.136), we observe that

∂

∂να

( ∂e∂η∣νγ

)∣η,ν′α

= ∂

∂η( ∂e

∂να

∣η,ν′α

)∣νγ

and

∂

∂νβ

( ∂e

∂να

∣η,ν′α

)∣η,ν′

β

= ∂

∂να

⎛⎜⎝∂e

∂νβ

∣η,ν′

β

⎞⎟⎠RRRRRRRRRRRRRη,ν′α

, (5.186)

or∂θ

∂να

∣η,ν′α

= ∂τα

∂η∣νγ

and∂τα

∂νβ

∣η,ν′

β

= ∂τβ

∂να

∣η,ν′α

. (5.187)

Given a fundamental relation expressed in terms of its (n + 1) natural variables,there are n(n+1)/2 separate pairs of mixed second derivatives. For example, in thecase of a single-component simple system where n = 2 and our specific thermostaticvolumes correspond to the specific volume, v, and mole number density, n, andthe thermostatic tensions correspond to the negative pressure, −p, and chemicalpotential, µ, then we have three Maxwell relations and they are given by

∂θ

∂v∣η,n

= − ∂p∂η∣v,n

,∂θ

∂n∣η,v

= ∂µ∂η∣v,n

, and −∂p

∂n∣η,v

= ∂µ∂v∣η,n

.

Now it is easy to show that the differentials of the equations of state (5.140),using (5.187) with (5.137) and (5.165)1, can be written as follows:

dθ = θ

cνα

dη +ϕα ⋅ dνα and dτα = ϕα dη +φαβ ⋅ dνβ . (5.188)

All thermodynamic properties that we have introduced are summarized in Ta-ble 5.9. In the table, we also denote their physical description, provide their cor-responding values for the special case of a single-component simple system withconstant mole number density (in which case n = 1, where ν1 → v and τ 1 → −p),and indicate their units using the International System of units (SI).

226 PRINCIPLES OF CONSTITUTIVE THEORY

5.10.2 Thermodynamic potentials

In both the internal energy and entropy representations, extensive properties playthe roles of independent variables, whereas intensive quantities arise as derivedconcepts. This situation is in contrast to the situation encountered in an experi-ment. The experimenter usually finds that the intensive quantities are more easilymeasured and controlled, and therefore, it is easier to think of the intensive quan-tities as independent variables and the extensive quantities as derived properties.

The question arises as to the possibility of recasting the fundamental relations insuch a way that intensive quantities will replace some or all extensive properties asindependent variables. This is possible by the use of Legendre transformations (seeAppendix F). The Legendre transformed functions of the fundamental relation arealso called thermodynamic potentials and particular transformations subsequentlylead to alternate fundamental representations.

To affect such representations, we use the energy form of the fundamental re-lation, and to simplify the presentation, we first take ν0 to signify the entropy ηand τ 0 to signify the temperature θ, so that the fundamental relation (5.134) isrewritten in the form

e = e(ν0, . . . ,νn). (5.189)

Subsequently, returning to explicit summation convention, the first differential ofthe fundamental relation (5.139) is now given by

de = n

∑α=0

τα ⋅ dνα, (5.190)

where the equations of state (5.136) are given by

τα = ∂e

∂να

∣ν′α

, α = 0, . . . , n. (5.191)

Note that now the heat increment is given by dq = τ 0 ⋅dν0 and the work incrementby dw = ∑n

α=1 τα ⋅ dνα. Furthermore, the Euler relation (5.154) is now written as

e = n

∑α=0

τα ⋅ να (5.192)

and the Gibbs–Duhem relation (5.155) as

n

∑α=0

να ⋅ dτα = 0. (5.193)

The Maxwell relations (5.187) are subsequently given by

∂τα

∂νβ

= ∂τβ

∂να

for 0 ≤ α,β ≤ n, (5.194)

where in each of these partial derivatives, the variables to be held constant areall those of the set ν0, . . . ,νn except the variables with respect to which thederivative is taken.

5.10. THERMODYNAMIC CONSIDERATIONS 227

A partial Legendre transformation can subsequently be made by replacing thethermostatic volumes ν0, . . . ,νm by the thermostatic tensions τ 0, . . . ,τm withm ≤ n. The Legendre transformed function is

F ≡ e[τ 0, . . . ,τm] = F (τ 0, . . . ,τm,νm+1, . . . ,νn) = e − m

∑α=0

τα ⋅ να, (5.195)

where the quantities inside the brackets denote the new independent intensivequantities in the transformed energy fundamental relation that replace correspond-ing conjugate variables. The corresponding equations of state are

− να = ∂F

∂τα

∣τ ′α

, α = 0, . . . ,m, and τα = ∂F

∂να

∣ν′α

, α =m + 1, . . . , n,(5.196)

with the first differential of F given by

dF = m

∑α=0

(−να) ⋅ dτα +

n

∑α=m+1

τα ⋅ dνα. (5.197)

Furthermore, the equilibrium values of any unconstrained extensive parameters ina system in contact with reservoirs of constant τ 0, . . . ,τm minimize F at constantτ 0, . . . ,τm.

Now, as before, given a thermodynamic potential expressed in terms on its n+1natural variables, there are n(n + 1)/2 separate pairs of mixed partial derivativesthat yield n(n + 1)/2 Maxwell relations. The corresponding equality provided bythe Maxwell relations of mixed second derivatives of the potential F become

∂να

∂τβ

= ∂νβ

∂τα

for 0 ≤ α,β ≤m, (5.198)

∂να

∂νβ

= −∂τ β

∂τα

for 0 ≤ α ≤m and m < β ≤ n, and (5.199)

∂τα

∂νβ

= ∂τβ

∂να

for m < α,β ≤ n. (5.200)

In each of these partial derivatives, the variables to be held constant are all thoseof the set τ 0, . . . ,τm,νm+1, . . . ,νn except the variables with respect to whichthe derivative is taken.

Now, using the fundamental relation (5.189) and reverting from ν0 to η and τ 0

to θ, the following Legendre transformations have been found to be useful:

ψ = e[θ] = ψ(θ,να,X) = e − θ η for α = 1, . . . , n, (5.201)

h = e[τ 1] = h(η,τ 1,να,X) = e − τ 1 ⋅ ν1 for α = 2, . . . , n, (5.202)

g = e[θ,τ 1] = g(θ,τ 1,να,X) = e − θ η − τ 1 ⋅ ν1 for α = 2, . . . , n, (5.203)

where ψ is the specific Helmholtz potential or free energy, h is the specific enthalpypotential, and g the specific Gibbs potential, respectively. Note that the potentialsare related through the identity

e − ψ + g − h = 0. (5.204)

Other potentials are used infrequently and are mostly unnamed.

228 PRINCIPLES OF CONSTITUTIVE THEORY

Helmholtz potential

In experiments, entropy is generally not a controllable parameter, but temperatureoften is controllable. As such, it is a more appropriate choice for an independentvariable. To change to a temperature representation, we replace the internal energywith its Legendre transform, the Helmholtz potential. Thus, given the fundamentalrelation (5.134), define the specific Helmholtz potential or free energy through theLegendre transformation (5.201), whose first differential for α = 1, . . . , n is givenby

dψ = ∂ψ∂θ∣νγ

dθ +∂ψ

∂να

∣θ,ν′α

⋅ dνα, (5.205)

where we have reverted to the implied summation convention and full contractionis also implied. Then, differentiating (5.201) and using (5.139), we obtain

dψ = de − θ dη − η dθ = −η dθ + τα ⋅ dνα. (5.206)

Subsequently, comparing the above expressions, we see that

−η = ∂ψ∂θ∣νγ

and τα = ∂ψ

∂να

∣θ,ν′α

, (5.207)

so that, from (5.201) and (5.207)1, we have

e = ψ − θ ∂ψ∂θ∣νγ

, (5.208)

and then from (5.165) and (5.207)1, we see that

cνγ= −θ ∂2ψ

∂θ2∣νγ

. (5.209)

The corresponding Maxwell relations are given by

∂

∂να

( ∂ψ∂θ∣νγ

)∣θ,ν′α

= ∂

∂θ( ∂ψ∂να

∣θ,ν′α

)∣νγ

and

∂

∂νβ

( ∂ψ∂να

∣θ,ν′α

)∣θ,ν′

β

= ∂

∂να

⎛⎜⎝∂ψ

∂νβ

∣θ,ν′

β

⎞⎟⎠RRRRRRRRRRRRRθ,ν′α

(5.210)

which, upon using (5.207), become

−∂η

∂να

∣θ,ν′α

= ∂τα

∂θ∣νγ

and∂τα

∂νβ

∣θ,ν′

β

= ∂τβ

∂να

∣θ,ν′α

. (5.211)

Now it is easy to show that the differentials of the equations of state (5.207), using(5.150), (5.165) and (5.211), can be written as follows:

dη = cνα

θdθ − βα ⋅ dνα and dτα = βα dθ + ξαβ ⋅ dνβ . (5.212)

5.10. THERMODYNAMIC CONSIDERATIONS 229

In addition, substituting (5.212)1 into (5.139), we obtain

de = cναdθ + (τα − θβα) ⋅ dνα. (5.213)

The name of free energy is appropriate for ψ because, as follows from (5.206), itis the portion of the energy available for doing work at constant temperature.

Using (5.211)1 with (5.151), (5.171), and (5.172), we see that we can write

λνα= −θβα and λτα

= θαα, (5.214)

and subsequently, using (5.151), we can rewrite (5.168) and (5.169) as

cτα− cνα

= −θαα ⋅βα = θαα ⋅ ξαβ ⋅αβ = θβα ⋅ υαβ ⋅ ββ . (5.215)

This result represents the generalization of relation (5.163).An additional thermodynamic property that is often found useful is the Grüneisen

parameter, a generalization of which is provided by the following definition of theGrüneisen tensor:

Γαβ = να

cνβ

∂η

∂νβ

∣θ,νγ≠β

= − να

cνβ

∂τβ

∂θ∣νγ

= −να ββ

cνβ

, (5.216)

where we have used (5.211)1 and (5.212)2. Now using (5.214)1 and (5.151)1, it iseasy to also show that

Γαβ = να λνβ

θ cνβ

= να ξβ γ ⋅αγ

cνβ

. (5.217)

Enthalpy potential

In experiments, the thermostatic volumes may not be controllable parameters, butthe thermostatic tensions may be controllable. As such, they would be more ap-propriate choices for independent variables. To change to a thermostatic tensionsrepresentation, we replace the internal energy with its Legendre transform, the en-thalpy potential. Thus, given the fundamental relation (5.134), define the specificenthalpy potential as the Legendre transformation (5.202), whose first differentialfor α = 2, . . . , n is given by

dh = ∂h∂η∣τ1,νγ

dη +∂h

∂τ 1

∣η,νγ

⋅ dτ 1 +∂h

∂να

∣η,τ1,ν′α

⋅ dνα, (5.218)

where we have reverted to the implied summation convention. Then, differentiat-ing (5.202) and using (5.139), we obtain

dh = de − ν1 ⋅ dτ 1 − τ 1 ⋅ dν1 = θ dη − ν1 ⋅ dτ 1 + τα ⋅ dνα. (5.219)

Subsequently, comparing the above expressions, we see that

θ = ∂h∂η∣τ1,νγ

, −ν1 = ∂h

∂τ 1

∣η,νγ

, and τα = ∂h

∂να

∣η,τ 1,ν′α

, (5.220)

so that

e = h − τ 1 ⋅∂h

∂τ 1

∣η,νγ

, (5.221)

230 PRINCIPLES OF CONSTITUTIVE THEORY

and then from (5.165)2, using (5.220)2 and (5.150)2, we see that

cτ1= ∂h∂θ∣τ1

. (5.222)

As can be seen from (5.219), the enthalpy h provides the portion of the energythat can be released as heat when the thermostatic tension τ 1 and the specificthermostatic volumes να for α ≥ 2 are kept constant.

The corresponding Maxwell relations are given by

∂

∂τ 1

( ∂h∂η∣τ1,νγ

)∣η,νγ

= ∂

∂η( ∂h∂τ 1

∣η,νγ

)∣τ1,νγ

,

∂

∂να

( ∂h∂η∣τ1,νγ

)∣η,τ 1,ν′α

= ∂

∂η( ∂h

∂να

∣η,τ1,ν′α

)∣τ 1,νγ

,

∂

∂τ 1

( ∂h

∂να

∣η,τ1,ν′α

)∣η,νγ

= ∂

∂να

( ∂h

∂τ 1

∣η,νγ

)∣η,τ1,ν′α

,

and∂

∂νβ

( ∂h

∂να

∣η,τ1,ν′α

)∣η,τ 1,ν

′β

= ∂

∂να

⎛⎜⎝∂h

∂νβ

∣η,τ1,ν

′β

⎞⎟⎠RRRRRRRRRRRRRη,τ1,ν′α

, (5.223)

which become

∂θ

∂τ 1

∣η,νγ

= − ∂ν1

∂η∣τ1,νγ

,∂θ

∂να

∣η,τ1,ν′α

= ∂τα

∂η∣τ1,νγ

,

∂τα

∂τ 1

∣η,νγ

= − ∂ν1

∂να

∣η,τ1,ν′α

, and∂τα

∂νβ

∣η,τ 1,ν

′β

= ∂τβ

∂να

∣η,τ 1,ν′α

. (5.224)

Now it is easy to show that the differentials of the equations of state (5.220),using the Maxwell relations (5.224), can be written as follows:

dθ = ∂θ∂η∣τ1,νγ

dη −∂ν1

∂η∣τ1,νγ

⋅ dτ 1 +∂τα

∂η∣τ1,νγ

⋅ dνα,

dν1 = − ∂θ

∂τ 1

∣η,νγ

dη +∂ν1

∂τ 1

∣η,νγ

⋅ dτ 1 −∂τα

∂τ 1

∣η,νγ

⋅ dνα,

dτα = ∂θ

∂να

∣η,τ1,ν′α

dη −∂ν1

∂να

∣η,τ1,ν′α

⋅ dτ 1 +∂τα

∂νβ

∣η,τ 1,ν

′β

⋅ dνα. (5.225)

In addition, using (5.202), (5.158)–(5.160), (5.137)2, and (5.144)1, it can be shownthat the differential of the internal energy is given by

de = (θ + τ 1 ⋅ ζ1)dη + (τ 1 ⋅φ−1

11) ⋅ dτ 1 + (τα − τ 1 ⋅

∂τα

∂τ 1

∣η,νγ

) ⋅ dνα. (5.226)

Gibbs potential

In some experiments, neither the entropy nor the thermostatic volumes are control-lable parameters, but temperature and thermostatic stresses may be controllable.

5.10. THERMODYNAMIC CONSIDERATIONS 231

As such, they would be more appropriate choices for independent variables. Tochange to a temperature and thermostatic stresses representation, we replace theinternal energy with its Legendre transform, the Gibbs potential. Thus, given thefundamental relation (5.134), define the specific Gibbs potential as the Legendretransformation (5.203), whose first differential for α = 2, . . . , n is given by

dg = ∂g∂θ∣τ1,νγ

dθ +∂g

∂τ 1

∣θ,νγ

⋅ dτ 1 +∂g

∂να

∣θ,τ1,ν′α

⋅ dνα, (5.227)

where we have reverted to the implied summation convention. Then, differentiat-ing (5.203) and using (5.139), we obtain

dg = de − η dθ − θ dη − τ 1 ⋅ dν1 − ν1 ⋅ dτ 1 = −η dθ − ν1 ⋅ dτ 1 + τα ⋅ dνα. (5.228)

Subsequently, comparing the above expressions, we see that

−η = ∂g∂θ∣τ 1,νγ

, −ν1 = ∂g

∂τ 1

∣θ,νγ

and τα = ∂g

∂να

∣θ,τ1,ν′α

, (5.229)

so that

e = g − θ ∂g∂θ∣τ 1,νγ

− τ 1 ⋅∂g

∂τ 1

∣θ,νγ

, (5.230)

and then from (5.165), using (5.229)2 and (5.150)2, we see that

cτ 1= −θ ∂2g

∂θ2∣τ1

. (5.231)

The corresponding Maxwell relations are given by

∂

∂τ 1

( ∂g∂θ∣τ 1,νγ

)∣θ,νγ

= ∂

∂θ( ∂g

∂τ 1

∣θ,νγ

)∣τ1,νγ

,

∂

∂να

( ∂g∂θ∣τ1,νγ

)∣θ,τ1,ν′α

= ∂

∂θ( ∂g

∂να

∣θ,τ1,ν′α

)∣τ1,νγ

,

∂

∂να

( ∂g

∂τ 1

∣θ,νγ

)∣θ,τ1,ν′α

= ∂

∂τ 1

( ∂g

∂να

∣θ,τ1,ν′α

)∣θ,νγ

,

and∂

∂νβ

( ∂g

∂να

∣θ,τ1,ν′α

)∣θ,τ1,ν

′β

= ∂

∂να

⎛⎜⎝∂g

∂νβ

∣θ,τ1,ν

′β

⎞⎟⎠RRRRRRRRRRRRRθ,τ1,ν′α

, (5.232)

which become

∂η

∂τ 1

∣θ,νγ

= ∂ν1

∂θ∣τ1,νγ

, −∂η

∂να

∣θ,τ1,ν′α

= ∂τα

∂θ∣τ1,νγ

,

−∂ν1

∂να

∣θ,τ1,ν′α

= ∂τα

∂τ 1

∣θ,νγ

, and∂τα

∂νβ

∣θ,τ1,ν

′β

= ∂τβ

∂να

∣θ,τ1,ν′α

. (5.233)

232 PRINCIPLES OF CONSTITUTIVE THEORY

Now it is easy to show that the differentials of the equations of state (5.229),using the Maxwell relations (5.233), can be written as follows:

dη = ∂η∂θ∣τ1,νγ

dθ +∂ν1

∂θ∣τ1,νγ

⋅ dτ 1 −∂τα

∂θ∣τ1,νγ

⋅ dνα,

dν1 = ∂η

∂τ 1

∣θ,νγ

dθ +∂ν1

∂τ 1

∣θ,νγ

⋅ dτ 1 −∂τα

∂τ 1

∣θ,νγ

⋅ dνα,

dτα = − ∂η

∂να

∣θ,τ1,ν′α

dθ −∂ν1

∂να

∣θ,τ1,ν′α

⋅ dτ 1 +∂τα

∂νβ

∣θ,τ1,ν

′β

⋅ dνβ . (5.234)

In addition, using (5.203), (5.158)–(5.160), and (5.150), it can be shown that thedifferential of the internal energy is given by

de = (cτ1+ τ 1 ⋅α1) dθ + (λτ1

+ τ 1 ⋅υ11) ⋅ dτ 1 + (λνα+ τα − τ 1 ⋅

∂τα

∂τ 1

∣θ,νγ

) ⋅ dνα.

(5.235)

5.10.3 Thermodynamic processes

When any property of a system changes in value, there is a change in state, andthe system is said to undergo a process.

Definition: For a given material pointX , the thermodynamic state (η(t),να(t)),for variable t, defines a thermodynamic path. The path on which η = const. is calledisentropic and that with θ = const. is called isothermal.

Now, for fixed X , from the Gibbs equation (5.139), we may also write

e = θ η + τα ⋅ να. (5.236)

For thermal changes, diffusion, and chemical phenomena, different types of ef-fects make up the entropy flux h and the entropy source b in (4.219) and (4.220).It is always possible to express the entropy flux and entropy source as

h = q

θ+ h1 and b = r

θ+ b1, (5.237)

where q/θ is the entropy flux due to heat input, r/θ is the entropy source suppliedby the energy source, and the remaining terms h1 and b1 are respectively theentropy flux and source due to all other effects. Above, we have assumed theentropy flux and source due to heat to be of a specific form. Later we shall provethat these forms are indeed correct for a simple material.

Using (5.237) to substitute for r in the energy equation (4.207) and solving theresulting equation for ρb, we obtain

ρb = ρb1 + ρ eθ−Φ

θ+1

θdivq. (5.238)

Now substituting this result as well as the expression for h into the local entropyinequalities (4.219) and (4.220), we have

γv ≡ ρ(η − eθ) + Φ

θ−

1

θ2g ⋅ q + divh1 − ρb1 ≥ 0, (5.239)

5.10. THERMODYNAMIC CONSIDERATIONS 233

where we recall thatg ≡ gradθ, (5.240)

and

γs ≡ rρη(v − c) + q

θ+ h1

z⋅ n ≥ 0. (5.241)

Definition: A thermodynamic process in which h1 = 0 and b1 = 0 is called asimple thermomechanical process.

A consequence of the above definition is an expression for temperature as thecommon ratio of

θ = ∣q∣∣h∣ = rb . (5.242)

For a simple thermomechanical process, the global form of the entropy inequal-ity, using (4.28), (4.29), (4.189)–(4.191), (4.204), and (5.242), is given by

Γ ≡ dSdt+∫

S

q

θ⋅ ds −∫

V

ρr

θdv ≥ 0. (5.243)

For an adiabatic process with no external energy sources q = 0 and r = 0, we havethat S ≥ 0, i.e., entropy cannot decrease. The local forms of the entropy inequalityfor a simple thermomechanical process become

γv ≡ ρ(η − eθ) + 1

θL ∶ σ −

1

θ2g ⋅ q ≥ 0

or γv ≡ ρ(η − eθ) + 1

θLlkσlk −

1

θ2θ,kqk ≥ 0, (5.244)

and

γs ≡ ρ JηK (v − c) + rqθ

z⋅n ≥ 0 or γs ≡ ρ JηK (vk − ck) + rqk

θ

znk ≥ 0. (5.245)

Definition: A thermodynamic process is said to be mechanically admissible ifit obeys the conservation of mass, the balance of momenta, and the balance ofenergy. It is called constitutively admissible if it satisfies constitutive restrictions,such as material frame indifference, symmetries, etc.

Definition: A process will be called thermodynamically admissible if and onlyif it obeys the local entropy inequalities and possesses a positive-definite temper-ature, i.e.,

γv ≥ 0, γs ≥ 0 and 0 < θ <∞. (5.246)

Definition: A process will be called a reversible process if and only if γv = γs = 0.Note that, for an isolated system, a reversible adiabatic process is an isentropicprocess.

In a simple mechanically admissible process, ρ, v, σ, e, and q must satisfy theequations of mass, balance of momenta, and balance of energy. Note that ourequations do not account for interactions between matter in our system and theexternal body and energy sources. For a constitutively admissible process, variousrestrictions also have to be satisfied. One of these is the principle of equipresence.

234 PRINCIPLES OF CONSTITUTIVE THEORY

Thus, e.g., when e is given by (5.134), the constitutive equations for the stresstensor and the heat flux must also use the same independent variables, i.e.,

σ = σ(η,να,X) and q = q(η,να,X). (5.247)

The above functions, along with that for e, are subject to additional constitutiverequirements, so that some of the variables from the chosen list may be shown notto be present in the arguments of some of the constitutive functions in view ofother restrictions.

To illustrate this, using the constitutive equation provided by the fundamentalrelation e = e(η,να,X), we can rewrite the entropy inequality (5.244) in the form

γvθ ≡ ρη (θ − ∂e

∂η∣να

) − ρνα ⋅∂e

∂να

∣η,ν′α

+L ∶ σ − grad(log θ) ⋅ q ≥ 0. (5.248)

The quantity γvθ is called the dissipation. Now at material point X , e, θ, σ, andq are functions of η and να. This inequality, which is linear in η, must be validfor all values of η. The independent variable η appears only in the first term. Forthe equation to be valid for all values of η, we must set its coefficient to zero:

θ = ∂e∂η∣να

. (5.249)

The above is the same expression as that for the thermostatic temperature (5.136).Thus, we have shown that for a thermodynamically admissible simple thermo-mechanical process characterized by the set of state variables (η,να) with να

being functionally independent of η and not containing time rates or integrals of η,the temperature and the thermostatic temperature are the same. It subsequentlyfollows that the tensions

τα = ∂e

∂να

∣η,ν′α

(5.250)

and the thermostatic tensions (5.136) are also the same, and the entropy inequalitybecomes

γvθ ≡ −ρνα ⋅ τα +L ∶ σ − grad(log θ) ⋅ q ≥ 0. (5.251)

In the expression of the production of entropy, inner products of vectorial andtensorial quantities occur in pairs. In the thermodynamics of irreversible processes,it has become customary to refer to one set of the pairs as thermodynamic forcesor affinities and the conjugate set multiplying the forces as the thermodynamicfluxes. For example, from the above entropy inequality, one may select the pairs

Force F lux

−ρνα τα

L σ

−grad(log θ) q

Linear constitutive equations are then set up between any one of the fluxes and allof the forces with symmetric constitutive coefficients. This is known as Onsager’s

5.10. THERMODYNAMIC CONSIDERATIONS 235

principle. We will not follow this thermodynamic approach to constitutive theorysince it is fraught with potential problems, among them the fact that the choiceof forces and fluxes is not unique.

We note that for an isolated system, it is more convenient to take the constitutiveequation corresponding to the entropy form of the fundamental relation (5.133):η = η(e,να,X), where now e, να, and X are the independent variables. For asystem in contact with a heat bath at a given temperature, θ replaces e to becomean independent variable, or control parameter. The energy e and entropy η willthen vary, so that e and η become dependent variables given by equations of state.In this case, it is more appropriate to consider the Helmholtz free energy formof the fundamental relation. A system with fixed external parameters in thermalcontact with a heat reservoir at equilibrium has a minimum Helmholtz free energyform of the fundamental relation. Subsequently, a different form of the entropyinequality valid at a regular point is obtained by introducing the specific Helmholtzfree energy. Using (5.201) and noting that ψ = e − θη − θη, the entropy inequality(5.239) for a general process, called the Clausius–Duhem inequality, becomes

−γvθ ≡ ρ (ψ + ηθ) −L ∶ σ + g ⋅ qθ− θ div (h − q

θ) + ρθ (b − r

θ) ≤ 0. (5.252)

Note that the above equation is more general than the classical Clausius–Duheminequality. Classically, the last two terms on the left-hand side of the inequalityare set to zero. This implies the fundamental assumptions that for a simple ther-momechanical process, h = q/θ and b = r/θ. One can show that generally theseassumptions are not valid and a more careful analysis is necessary. However, per-forming the analysis more carefully, we will prove that they are valid for simplethermoelastic solids and fluids.

5.10.4 Thermodynamic equilibrium and stability

A nonequilibrium state is a state where spatial and/or temporal variations of ve-locity or temperature exist. Alternately, an equilibrium state is a persistent statein which no spatial or temporal variations of velocity or temperature exist. Asystem at equilibrium has no tendency to change when it is isolated from thesurroundings. If the complete body is at an equilibrium state, then the body isin thermodynamic equilibrium. More specifically, in considering the equilibriumstate, we consider the body as being isolated, i.e., free of external supplies, f = 0,r = b = 0, and with no mass, momentum, or heat exchange with the surroundings,i.e.,

v ⋅ n = 0, σ ⋅n = 0, q ⋅n = 0, h ⋅ n = 0 on S . (5.253)

In such case, using (4.187) for a nonpolar material and (4.204) (and using thedefinitions (4.18), (4.26), and (4.28)), we have that

E + K = 0 and S ≥ 0. (5.254)

In other words, for an isolated body with constant total energy, the total entropymust not decrease in time.

Definition: A body is said to be in thermodynamic equilibrium if, when freeof external supplies, it is in mechanical and thermal equilibrium and the totalentropy remains constant.

236 PRINCIPLES OF CONSTITUTIVE THEORY

Clearly, the above implies that at thermodynamic equilibrium we have zeroentropy production, i.e., γv = γs = 0. In contradistinction, it should be notedthat for a reversible process, γv = γs = 0 but η ≠ 0, and in such case, we have atime-dependent nonequilibrium process.

Frequently, it is convenient to study the conditions for equilibrium using a some-what different perspective. Instead of considering the approach to equilibrium, wefocus on the equilibrium state itself. To establish the conditions that are satis-fied subject to internal or external constraints, we ask about the characteristicsof neighboring imaginary states that can conceivably be reached by displacingthe system from equilibrium. Such displacements from equilibrium are called vir-tual. A virtual displacement is a reversible process whereby conditions are createdwhich permit the imaginary insertion of a supplementary internal constraint intothe system. The subsequent removal of this internal constraint would induce thesystem to reach the equilibrium state, assuming that the external constraints haveremained unchanged. We employ the symbol δ to denote a small, virtual changeof a given quantity associated with the internal constraints, and make use of theequations of thermodynamics in a form appropriate to the reversible process.

We mention five criteria of equilibrium in terms of virtual displacements. Eachcriterion applies to a different set of constraints. In each case, we use the ap-propriate thermodynamic potential. To begin with, a fundamental postulate ofthermodynamics is that the values assumed by the independent quantities, in theabsence of internal constraints, are such as to maximize the entropy. This basicextremum principle implies that δη = 0 and δ2η < 0, i.e., entropy is a maximum atthe thermodynamic equilibrium.

Entropy maximum principle: For a given value of the total internal energy,the equilibrium value of any unconstrained internal quantity is such as to maximizethe entropy.

An equivalent statement of thermodynamic stability is that the internal energyis a minimum with respect to all virtual displacements at equilibrium: δe = 0 andδ2e > 0.

Internal energy minimum principle: For a given value of the total entropy,the equilibrium value of any unconstrained internal quantity is such as to minimizethe internal energy.

The first condition,

δe = θ δη + τα ⋅ δνα = 0, (5.255)

for any δη and δνα, except the trivial case δη = 0 and δνα = 0, leads to the equalityof temperatures and tensions of unconstrained subsystems and the environmentand confirms our original assumption of equilibrium corresponding to a state whereno temperature and velocity variations exist. To examine the consequences of thesecond condition, using the fundamental relation (5.134), for any infinitesimal

5.10. THERMODYNAMIC CONSIDERATIONS 237

variation, we have

δ2e = ∂

∂η( ∂e∂η∣νγ

)∣νγ

(δη)2 + 2 ∂

∂να

( ∂e∂η∣νγ

)∣η,ν′α

⋅ δηδνα +

∂

∂νβ

( ∂e

∂να

∣η,ν′α

)∣η,ν′

β

∶ δναδνβ

= ∂θ

∂η∣νγ

(δη)2 + 2 ∂θ

∂να

∣η,ν′α

⋅ δηδνα +∂τα

∂νβ

∣η,ν′

β

∶ δναδνβ

= θ

cνα

(δη)2 + 2ϕα ⋅ δηδνα +φαβ ∶ δναδνβ > 0, (5.256)

where we have used (5.136), (5.137), and (5.165). The stability condition is thatthe right-hand side must be positive definite for any δη and δνα, except the trivialcase δη = 0 and δνα = 0. The right-hand side is a homogeneous quadratic form inthe variables δη and δνα. The coefficients of the quadratic form and the stabilityrequirement can be rewritten in the symmetric matrix form

He = ⎛⎜⎝θ/cνα

ϕα

ϕα φαβ

⎞⎟⎠ > 0. (5.257)

From previous results, we know that a symmetric matrix has real eigenvaluesand it is positive definite if all its eigenvalues are positive. Alternately, we useSylvester’s criterion, which states that a matrix He = [hij] is positive definite ifand only if the determinants of all of its principal minors are positive, i.e.,

h11 > 0, ∣ h11 h12h21 h22

∣ > 0, RRRRRRRRRRRRRh11 h12 h13h21 h22 h23h31 h32 h33

RRRRRRRRRRRRR > 0, etc. (5.258)

To illustrate the application of the stability condition, we take the case wheren = 1 with ν1 being the specific volume v and thus write

He = ( θ/cv ϕ

ϕ φ) > 0. (5.259)

The trace and determinant of the matrix He correspond to the sum and productsof the eigenvalues; thus we can write

θ

cv+ φ = λ1 + λ2 > 0 and

θ

cvφ −ϕ2 = λ1 λ2 > 0. (5.260)

Now, solving for λ1 and λ2, and requiring that λ1 > 0 and λ2 > 0, we obtain thethermodynamic restrictions. Alternately, using Sylvester’s criterion, we easily seethat these conditions are satisfied if and only if

θ

cv> 0, φ > 0, and

θ

cvφ > ϕ2. (5.261)

238 PRINCIPLES OF CONSTITUTIVE THEORY

Since θ > 0, the first condition results in

cv > 0. (5.262)

The second condition, using (5.136), (5.137), and (5.162), results in

φ = ∂2e∂v2∣η

= − ∂p∂v∣η

= 1

vκη> 0, (5.263)

or, since v > 0,κη > 0. (5.264)

To examine the third and last condition, using (5.150), (5.164), (5.176)3, and(5.177), we easily see that

ϕ = θ α

κθ cv. (5.265)

Subsequently, the condition becomes

θ

vκηcv> ( θ α

κθcv)2 . (5.266)

Now, using relations (5.163) and (5.164) along with the previous result that κη > 0,it is easy to show that the third condition yields

κθ > 0. (5.267)

Finally, using the relations (5.163) and (5.164) once more, it is easy to see that

cp > cv > 0 and κθ > κη > 0. (5.268)

These results, with the use of (5.163), also imply that the magnitude of the coef-ficient of volume expansion is bounded by

α2 < cp κθθ v

, (5.269)

and, with the use of (5.164), that the ratio of specific heats or compressibilitymoduli is bounded by

γ > 1. (5.270)

We now formally state the additional equivalent principles pertaining to thestability of thermodynamic equilibrium.

Helmholtz free energy minimum principle: In a system in diathermalcontact with a heat reservoir, the equilibrium value of any unconstrained internalquantity is such as to minimize the Helmholtz free energy at constant temperature(equal to that of the heat reservoir).

Enthalpy minimum principle: In a system in contact with reservoirs ofthermostatic tensions, the equilibrium value of any unconstrained internal quantityis such as to minimize the enthalpy at constant thermostatic tensions (equal tothose of the thermostatic tensions reservoirs).

Gibbs free energy minimum principle: In a system in contact with atemperature and thermostatic tensions reservoirs, the equilibrium value of any

5.10. THERMODYNAMIC CONSIDERATIONS 239

unconstrained internal quantity is such as to minimize the Gibbs free energy atconstant temperature and thermostatic tensions (equal to those of the respectivereservoirs).

A process in which a system that is in a thermodynamic equilibrium state pro-ceeds by an infinitely slow (in imagination) evolution to another neighboring stateof thermodynamic equilibrium is called a quasi-static or quasi-equilibrium process.

Definition: A process will be called an equilibrium process if and only if thesystem is at thermodynamic equilibrium.

Definition: A process will be called a thermostatic process if the process consistsof an infinite sequence of quasi-static processes.

5.10.5 Potential energy and strain energy

When the body forces are steady and derivable from a potential U(x), i.e.,

f = −grad U or fk = −U,k, (5.271)

then the work done by body forces, in using Reynolds’ transport theorem (3.466),can be rewritten as

∫V

ρvkfkdv = −∫V

ρvkU,kdv = −U , (5.272)

where

U ≡ ∫V

ρUdv (5.273)

is called the potential energy. Upon substituting this into the energy balance(4.187) for a nonpolar material, we have

E + K + U = ∫S

(v ⋅σ − q) ⋅ ds +∫V

ρr dv. (5.274)

When the work of the surface tractions is zero, the body is insulated, and there isno external energy supply, we have the balance of mechanical energy, which statesthat the sum of internal, kinetic, and potential energies is constant:

E +K + U = const. (5.275)

If E = 0, we obtain the local principle of conservation of energy of classical me-chanics, which is obtained by integrating Newton’s second law for a particle whenthe force is derived from a potential.

It is natural to inquire whether a similar situation regarding the existence of apotential which leads to the appearance of a recoverable work term in the formof potential energy can exist in the presence of surface tractions. To this end, weassume that the stress tensor can be rewritten as the sum of two symmetric stresstensors

σ = σe+σd, (5.276)

where σe is the elastic, recoverable or reversible, part, and σd is the dissipative,or irreversible, part. We assume that the recoverable stress is derivable from a

240 PRINCIPLES OF CONSTITUTIVE THEORY

potential τ = τ(F) called the strain energy function or elastic potential function,i.e.,

σe = ρ∂τ(F)∂F

⋅FT or σekl = ρ∂τ(F)

∂FkL

FlL. (5.277)

We now note that

ρτ = ρ∂τ(F)∂FkL

FkL = ρ∂τ(F)∂FkL

LklFlL = Lklσekl = (Dkl +Wkl)σe

kl (5.278)

or, since σe is symmetric,

ρτ =D ∶ σe. (5.279)

Subsequently, again using the fact that the stress tensor is symmetric, we have

∫S

(v ⋅σ) ⋅ ds = ∫V

div(v ⋅σ)dv= ∫

V

[(gradv) ∶ σ + v ⋅ (divσ)]dv= ∫

V

[LT∶ σ + v ⋅ (divσ)]dv

= ∫V

[D ∶ (σe+σd) + v ⋅ (divσ)]dv

or

∫S

(v ⋅σ) ⋅ ds = ∫V

[ρτ +D ∶ σd+ v ⋅ (divσ)]dv. (5.280)

Upon substituting the above into the energy balance (4.187) for a nonpolarmaterial (and using (4.173) and (4.186)), we have

E + K = ∫V

[ρτ +D ∶ σd+ v ⋅ (divσ)]dv + ∫

V

ρv ⋅ f dv + Q. (5.281)

Now assuming that mass is conserved, we have

K = 1

2

d

dt∫

V

ρv ⋅ v dv = ∫V

ρv ⋅ adv, (5.282)

and define the total strain energy, the total dissipative power, and the total thermalenergy as

T ≡ ∫V

ρτ dv, D ≡ ∫V

D ∶ σd dv, and Q ≡ ∫V

(−divq + ρr) dv. (5.283)

Subsequently, rearranging the energy balance, we have

E = T +D + Q − ∫V

v ⋅ [ρ(a − f) − divσ]dv. (5.284)

or, making use of the balance of linear momentum (4.109), we obtain

E = T +D + Q. (5.285)

5.10. THERMODYNAMIC CONSIDERATIONS 241

In the special case where τ = e, the above equation reduces to D+ Q = 0, statingthat energy dissipation is totally converted to heat.

Another special case is obtained when the strain energy function depends onJ = detF only. Then, using (3.60), we have

σekl = ρ∂τ(J)∂J

∂J

∂FlL

FkL = ρ∂τ(J)∂J

JF −1Ll FkL, (5.286)

or since ρ = ρRJ−1, we can write more simply

σe = −p1 or σekl = −p δkl, (5.287)

where we have defined the elastic hydrostatic pressure by

p ≡ −ρR ∂τ(J)∂J

. (5.288)

Subsequently, from (5.279), we see that

ρτ = −pDkk = −pvk,k. (5.289)

We now show how the first principle of thermostatics is deduced as a special case ofthe above result. Classical thermostatics deals with homogeneous systems and thestress consists of a purely hydrostatic pressure that is constant (σd = 0). Under

these conditions, since from (3.318) we have ˙dv =Dkk dv, we have

T = ∫V

ρτ dv = −∫V

pDkk dv = −p∫V

˙dv = −p d

dt∫

V

dv = −p ˙V , (5.290)

where V is the volume of the body. Since D = 0, upon integration of (5.285), wehave

dE = dQ + dW (5.291)

with the interpretation that the above are considered to be changes between neigh-boring states, and we wrote T =W ,

dQ = Qdt and dW = W dt = −pdV . (5.292)

Equation (5.291) is the expression for the first law of thermostatics, which is validwhen the system is uniform and explicitly independent of time, the dissipative partof the stress vanishes, and the elastic part of the stress consists of the hydrostaticpressure only. It may be rewritten into a differential form if an integrating factor1/θ can be found so that we may express dQ as a total differential

dQ

θ= dS, (5.293)

where S is the total entropy. The variable θ is the absolute temperature, whichis defined to be positive definite, i.e., θ > 0. With this substitution, the equationtakes the form

dE = θ dS − pdV . (5.294)

242 PRINCIPLES OF CONSTITUTIVE THEORY

5.11 Entropy and nonequilibrium thermodynamics

5.11.1 Coleman–Noll procedure

Most of the classical work involving the derivation of reduced constitutive functionsin nonequilibrium thermodynamics is based on the use of the more restrictedClausius–Duhem inequality (5.252), which makes use of the assumptions that h =q/θ and b = r/θ. In our description and application of the Coleman–Noll procedure,we will not make these assumptions.

The analysis proceeds as follows. First, one writes the constitutive equationsfor C = σ,q,h, ψ, η (see (5.2)) as functions of the appropriate reduced indepen-dent variables I (see (5.1)). For example, for a simple homogeneous thermoelasticsolid, these would be I = F, (t)C(t), θ(t), (t)g(t) (see (5.99)), while for a simple

fluid, they would be I = ρ, (t)C(t), θ(t), (t)g(t) (see (5.114)). These constitutiveequations are then introduced into the Clausius–Duhem inequality (5.252) andall differentiations performed so that, with the exception of two sets of terms,the resulting inequality involves a number of terms each multiplied linearly by anindependent variable or a derivative of an independent variable. Now since theinequality must be satisfied for arbitrary external sources and arbitrary variationsof the independent variables, the terms that involve the sources or that are mul-tiplied linearly by an independent variable or one of its derivatives must be set tozero. Subsequently, one arrives at what is called the residual inequality. Analysisof the terms that are set to zero, along with the use of integrability conditions,the residual inequality, and its analysis in the equilibrium limit, provides specificconstitutive forms for C = σ,q,h, ψ, η. Details of the procedure are given inChapter 7 for thermoelastic solids and Chapter 8 for fluids.

5.11.2 Müller–Liu procedure and Lagrange multipliers

The classical Coleman–Noll procedure based on the standard Clausius–Duhem in-equality (5.244) makes use of fundamental assumptions. The first is the necessarypresence of energy and entropy external sources. Their presence is essential inbeing able to manipulate the energy and entropy balance equations (4.207) and(4.219) to arrive at the Clausius–Duhem inequality given in (5.244) (see the dis-cussion surrounding (5.239)). This is perplexing from the standpoint that sincethe constitutive equations describe material properties, these properties should beindependent of external sources, let alone their presence. Second, as noted previ-ously, in the standard Clausius–Duhem inequality, the entropy flux and externalsource are assumed to be related to the energy flux and external source by h = q/θand b = r/θ. The above assumptions are not necessary in the Müller–Liu proce-dure, which we outline next. The procedure is based on the application of fourbasic principles and applied through the use of Lagrange multipliers.

Müller’s entropy principles:

1. The specific entropy η and the entropy flux h are constitutive quantities.

2. The entropy production must be nonnegative, i.e., γv ≥ 0, for all thermody-namic processes corresponding to solutions of the field equations. Solutions

5.11. ENTROPY AND NONEQUILIBRIUM THERMODYNAMICS 243

of the field equations are assumed to exist and to satisfy all the balance equa-tions, the constitutive equations, and initial and boundary conditions.

3. The external supply terms appearing in the balance equations cannot influ-ence the material behavior.

4. An empirical temperature, or coldness function, exists and such temperature,along with the tangential velocity components, is continuous across materialsingular surfaces called ideal boundaries, i.e., JvK = 0 and JθK = 0. At idealboundaries no entropy is produced: γs = 0.

Note that continuity of the normal velocity component at a material singularsurface is required for mass conservation (see below).

Implementation of the above principles is facilitated through the use of a lemmaproved by Liu and which we now state without proof.

Liu’s Lemma: Let β be a scalar, α, a, and b vectors, and A a second-order tensor,all of which are given and whose dimensions are consistent with each other. Thenthe following three statements are equivalent:

(a) The inequalityα ⋅ a + β ≥ 0 (5.295)

holds for all vectors a that satisfy the equation

A ⋅ a + b = 0. (5.296)

(b) There exists a vector quantity λ such that for all a, the inequality

α ⋅ a + β −λ ⋅ (A ⋅ a + b) ≥ 0 (5.297)

holds.

(c) There exists a vector quantity λ such that

α = λ ⋅A (5.298)

andβ −λ ⋅ b ≥ 0 (5.299)

hold.

In the above lemma, λ is just a Lagrange multiplier and the contractions areunderstood to be full contractions.

The application of the principles begins with the statement that the local entropyinequality (4.219) must be satisfied for any fields satisfying the conservation ofmass, and balances of momentum and energy, (4.98), (4.109), and (4.207), usingthe appropriate reduced constitutive equations, and for given initial and boundaryconditions. This is nothing but a restatement of the second principle of Müller. Itshould be noted that such statement also corresponds to the first statement of Liu’slemma where, after introduction of the appropriate reduced constitutive equationsinto (4.219), one associates the solution fields and their derivatives with a, thelocal entropy inequality (4.219) with (5.295), and the local conservation of mass,

244 PRINCIPLES OF CONSTITUTIVE THEORY

and momentum and energy balance equations, (4.98), (4.109), and (4.207) with(5.296). Equivalently, using the lemma’s second statement, (5.297), the entropyinequality can be rewritten as a new inequality that explicitly accounts for theconstraints which the solution must satisfy; i.e.,

γv ≡ [ρ(η − b) + divh] − λρ [ρ + ρ divv] − λv⋅ [ρ (v − f) − divσ] −

λe [ρ (e − r) −Φ + divq] ≥ 0 (5.300)

must hold for all fields. The λ’s are Lagrange multipliers, which also dependon the appropriate reduced independent variables. Now, as in the Coleman–Nollprocedure, and consistent with Müller’s first and fourth principles, one writes theconstitutive equations for C = e, η,q,h,σ as functions of the appropriate reducedindependent variables I in which θ is taken to be the empirical temperature.Subsequently, these constitutive equations are introduced into the above modifiedentropy inequality and all differentiations performed so that, with the exceptionof two sets of terms, the resulting inequality involves a number of terms eachmultiplied linearly by an independent variable or a derivative of an independentvariable. The definition of the independent vector a in the lemma will depend onthe specific reduced independent variables. Now, application of the lemma’s thirdstatement (5.298) leads to a number of simplifying relations, while (5.299) leadsto the residual inequality. It is noted that in the simplification process, Müller’sthird and fourth principles are used to help solve for the Lagrange multipliersand to define the empirical temperature. Further analysis in the equilibrium limitprovides specific constitutive forms for C = e, η,q,h,σ. Details of the procedureare given in Chapter 8 for fluids.

5.12 Jump conditions

A surface within a material body across which we have discontinuous fields is oneof two types depending on the mass flux m (see (4.101)):

Material singular surface – a surface which is formed by the same materialparticles at all times; on such a surface, we have that m = 0, so that thenormal velocity of the medium moves with the normal velocity of the surface.Note that on a material singular surface, the tangential velocity componentis generally allowed to slip.

Nonmaterial singular surface – a surface which is not formed by the samematerial particles, but for which a field variable experiences a jump acrossit; on such a surface, m ≠ 0, so that the medium and the surface generallymove with different velocities.

The discontinuities of some field variables can have different degrees. For exam-ple, a variable can experience a finite jump across such surface. When the motionx = χ(X, t) is discontinuous across the surface, dislocations are formed. Also,higher derivatives of the motion (e.g., v, F) can be discontinuous. Examples ofsuch singular surfaces are:

1. A surface separating two immiscible fluids, or two pure solids, or a fluid anda solid without phase change, is a material surface. On such surface, the

5.12. JUMP CONDITIONS 245

normal velocity is continuous. Within this context, we note that a physi-cal boundary on which conditions have to be prescribed can be viewed as asingular surface separating two immiscible materials. Below, we shall takeadvantage of this fact to elaborate on boundary conditions for general prob-lems.

2. A perfectly sliding surface between two materials is a material singular sur-face. Such a surface is referred to as a vortex sheet and the following condi-tions are satisfied: JvK ⋅n = Jv(n)K = 0 and JvK ⋅ s = Jv(s)K ≠ 0 (see (4.104) and(4.105)), where Jv(s)K is the amount of slip.

3. A singular surface in a pure material across which the phase of the materialchanges between gas, liquid, and solid is generally a nonmaterial surfacesince the surface does not consist of the same material particles for all times.

4. A shock is a surface across which the normal component of velocity experi-ences a jump and is a nonmaterial singular surface. Such a surface is alsoreferred to as a shock wave across which we have Jv(n)K ≠ 0 and JvK ≠ 0 (notethat v is the specific volume).

5. An acceleration wave is a singular surface across which v and F are contin-uous, but v, grad F, and F are not.

6. It is possible to have a nonmaterial vortex surface that is also an accelerationwave. In such case, from the mass balance jump condition, it is easy to seethat the specific volume is continuous across such a surface (see (4.103)). Onsuch a surface, m ≠ 0, JvK = 0, Jv(n)K = 0, and Jv(s)K ≠ 0.

5.12.1 Characterization of jump conditions

As noted earlier, when the field variables do not satisfy continuity conditions onarbitrary surfaces within a body, then the global balance laws yield jump condi-tions that must hold across such surfaces. Below, we summarize the general jumpconditions for nonpolar materials (see Chapter 4 for their derivations).

Mass: Across a singular surface moving with velocity c, we have that

JmK = 0, (5.301)

wherem = ρ (v − c) ⋅ n (5.302)

is the mass flux (per unit surface area). Note that m is continuous at thejump, so m+ =m− =m, where we now take m to be the value at the singularsurface. Furthermore, we have shown that

JvK =m JvKn + Jv(s)K s, (5.303)

so thatJvK ⋅ n =m JvK and JvK ⋅ s = q

v(s)y, (5.304)

where v = 1/ρ is the specific volume,qv(s)

yis the slip, and n and s are the

surface unit normal and tangential vectors, respectively. Note that JvK ⋅n = 0if and only if JvK = 0 at nonmaterial surfaces (m ≠ 0).

246 PRINCIPLES OF CONSTITUTIVE THEORY

Linear momentum: The jump condition across the singular moving surface is

m JvK − JσK ⋅n = 0. (5.305)

Taking the inner product of it with n and s respectively, and using (5.304),(4.105), and (4.129), we obtain

m2JvK − JσnK = 0 and m Jv(s)K − JτK = 0. (5.306)

Note that for nonpolar materials, the jump condition for the angular mo-mentum, (4.117), is trivially satisfied.

Energy: We recall that the jump condition for energy, (4.217), can be rewrittenas

m (JeK − JvK⟪σn⟫) − Jv(s)K⟪τ⟫ + JqK ⋅ n = 0, (5.307)

or, using (4.212), (5.304), and (5.306), and the fact that e = ψ + θ η (see(5.201)), we have

m (ε + Jθ ηK) − Jv(s)K⟪τ⟫ + JqK ⋅ n = 0, (5.308)

where

ε = JψK − JvK⟪σn⟫ = JψK − JvσnK + 1

2m2

qv2

y = n ⋅ JµK ⋅n (5.309)

is the specific energy release rate, and ⟪τ⟫ is the mean shear stress or friction,over the jump. Note that

µ =Π + 1

2m2v2 1 (5.310)

is a tensorial dynamic nonequilibrium chemical potential, and

Π = ψ 1 − vσ (5.311)

is the Eulerian Eshelby energy-momentum tensor.

Entropy: The jump condition across the singular moving surface is

γs ≡m JηK +rqθ

z⋅n ≥ 0, (5.312)

where γs represents the entropy production (per unit area) on the singularsurface. In writing the above condition, we have assumed a simple thermo-dynamic process so that h = q/θ (see (5.242)). We will show in Chapters 7and 8 that both simple thermoelastic solids and fluids satisfy such condition.

Shock surface in an elastic medium

In an elastic medium (where momentum diffusion is neglected), we have thatσ = −p1 (so that σn = −p and τ = 0), from which it follows from (5.306) that

m2JvK + JpK = 0 and Jv(s)K = 0, (5.313)

5.12. JUMP CONDITIONS 247

so that there is no slip at the interface. Note that at a shock surface, m ≠ 0. Inaddition, from this fact and (5.303), we now have that

JvK =m JvKn. (5.314)

In this case, using (5.313), (5.307) simplifies to

m (JeK + JvK⟪p⟫)+ JqK ⋅n =m (JhK − ⟪v⟫ JpK)+ JqK ⋅n =msh +

1

2m2v2

+ JqK ⋅n = 0,

(5.315)where the enthalpy is given by h = e+vp (see (5.202)). Lastly, by substituting thisexpression into (5.312), we see that the entropy production across a shock surfaceis given by

γs ≡msη −

1

θ(h + 1

2m2v2) ≥ 0. (5.316)

It should be pointed out that it is often the case that for an elastic medium, onealso neglects thermal diffusion (so that q = 0). In this case, from (5.315) and(5.312), we have the following jump conditions for energy and entropy productionat the surface:

sh +

1

2m2v2

= 0 and γs ≡m JηK ≥ 0. (5.317)

Note that at an ideal surface, the entropy jump is zero.

Phase change surface

Here we discuss the jump conditions corresponding to a phase change surfaceseparating the same material, i.e., the two phases correspond to the same materialin two different forms of molecular state. For example, such conditions wouldapply when liquid water’s temperature locally falls below 0C in which case waterwould change to the solid phase of ice, while if locally the temperature rises above100C, the liquid would change into the gaseous phase of water vapor.

In general, there are many types of phase transitions. In addition to melting-solidification and evaporation-condensation, there are also solid-solid as well asother types of transitions. Phase transitions are generally classified according tothe Ehrenfest classification. The order of a phase transition is defined to be theorder of the lowest derivative that varies discontinuously at the phase boundary.The first three orders are given in Table 5.10. Below we shall discuss first-orderphase transitions in a simple isotropic material, i.e., those transitions that arecharacterized by jumps in entropy or specific volume. Other types of second-ordertransitions are solid-solid (structural) transitions in crystals.

At a phase change surface, the two phases exchange mass. Thus, such a surfaceis a nonmaterial moving surface with m ≠ 0. Subsequently, the mass balanceand the linear momentum jump conditions require that (5.304) and (5.306) besatisfied. Furthermore, across a phase change surface, the temperature is knownto be continuous; thus we take JθK = 0. Lastly, the energy jump condition (5.308)and the entropy inequality (5.312) can be rewritten in the following forms:

m (ε + θ JηK) − Jv(s)K⟪τ⟫ + JqK ⋅ n = 0 (5.318)

248 PRINCIPLES OF CONSTITUTIVE THEORY

andγs θ ≡mθ JηK + JqK ⋅ n ≥ 0. (5.319)

Upon substituting the energy balance jump condition into the entropy inequality,we obtain

−γsθ ≡mε − ⟪τ⟫Jv(s)K ≤ 0, (5.320)

or

−γsθ ≡m (JeK − θ JηK − ⟪σn⟫JvK) − ⟪τ⟫Jv(s)K ≤ 0. (5.321)

At equilibrium, γs = 0 and σ = −p1 (i.e., σn = −p and τ = 0), so that the linearmomentum jump condition (5.306) requires that the phase boundary does not slip,Jv(s)K = 0, and so from (5.321), we have that

⟪p⟫ = θ JηK − JeKJvK . (5.322)

If we differentiate the above with respect to the temperature, and use the funda-mental relation de = θ dη − pdv (see (5.139)), we obtain

d⟪p⟫dθ= JηK

JvK +θ

JvKdJηKdθ−

1

JvKdJeKdθ−(θ JηK − JeK)

JvK2dJvKdθ= JηK

JvK . (5.323)

Now if we define the latent heat by

L ≡ θ JηK, (5.324)

from (5.325), we obtain the following Clausius–Clapeyron equation

d⟪p⟫dθ= L

θ JvK . (5.325)

We note that in the limit where the phase change boundary is a material bound-ary, m = 0, the mass jump condition (5.304) further requires that JvK ⋅n = 0 (notethat generally JvK ≠ 0), the linear momentum jump condition becomes JpK = 0 (andthus ⟪p⟫ = p), and subsequently we obtain the standard relations of thermostaticscorresponding to (5.322) and (5.325):

p = θ JηK − JeKJvK and

dp

dθ= L

θ JvK . (5.326)

5.12.2 Material singular surface

The mass balance jump condition (5.301) is trivially satisfied at a material singularsurface since at such surface, we have that m = 0. The condition just states thatthe normal velocity of a material point at the surface is the same as that imposedby the normal velocity of the surface, c(n) = c ⋅ n, i.e.,

v ⋅ n = c(n), (5.327)

and, subsequently, the normal component of velocity is continuous there (see(5.304)):

JvK ⋅n = 0. (5.328)

5.12. JUMP CONDITIONS 249

Note that no condition is imposed on the tangential component; for this reason, itis also called a free-slip condition and Jv(s)K is the amount of slip. If JvK ≠ 0, thenin addition the surface in this case is also a vortex sheet. At the singular surface,using (5.302), the mass balance jump condition requires that

JρvK ⋅ n = JρKc(n). (5.329)

If c = 0, the singular surface is a stationary surface. If JρK = 0, then the density aswell as the linear momentum normal to the surface are continuous. On the otherhand, at a contact discontinuity, JρK ≠ 0 and the linear momentum normal to thesurface is discontinuous.

Subsequently, the linear momentum balance tells us that at a material singularsurface, we must have (see (5.306))

JσnK = 0 and JτK = 0, (5.330)

(note that generally JvK ≠ 0 and Jv(s)K ≠ 0) so that the surface normal and tangen-tial components of the stress are continuous. Hence, on a material interface, thesurface traction t(n) = σ ⋅ n is continuous.

The energy jump condition at the material singular surface requires that (see(5.307) and note that now ⟪τ⟫ = τ)

JqK ⋅ n = Jv(s)K τ, (5.331)

so that the jump in the normal heat flux equals the jump in power of the shearstress. Note that at a contact discontinuity, JeK ≠ 0 and JvK ≠ 0. Clearly if thevelocity is continuous on such surface, JvK = 0, then the normal component of theheat flux must be continuous as well, i.e.,

JqK ⋅ n = 0. (5.332)

Lastly, from (5.312) and (5.331), we see that at the material singular surface,the entropy production is due to the jumps in temperature and slip velocity,

γs ≡s1

θ

⟪q⟫ ⋅ n +⟪1θ⟫ Jv(s)K τ ≥ 0. (5.333)

Furthermore, if JηK = 0, the entropy is continuous there. On the other hand, at acontact discontinuity, we have that JηK ≠ 0.Contact surface in an inviscid fluid

In this case, we have that σ = −p1 (so that σn = −p and τ = 0), from which itfollows from (5.330) that p is continuous, and from (5.331) that JqK ⋅ n = 0, whichimplies that ⟪q⟫ ⋅ n = q ⋅ n is continuous, so that the surface entropy production(5.333) subsequently becomes γs ≡ J1/θKq ⋅ n ≥ 0. Thus, we see that a nonzeroentropy production is only possible if we have a temperature jump at the surface,and the temperature must jump from a low to a high value when q ⋅ n ≥ 0 andvice versa when q ⋅ n ≤ 0. Note that, in general, it is possible to have JvK ≠ 0 andJv(s)K ≠ 0 on such a surface.

250 PRINCIPLES OF CONSTITUTIVE THEORY

Boundary surface

To fully describe a mathematical problem involving a continuous medium, bound-ary conditions are necessary. In general, physical boundaries are nothing butmaterial singular surfaces. Thus, appropriate boundary conditions are obtainedby examining jump conditions at such interfaces. In this subsection, these ma-terial interfaces will be called boundaries and we take the region on the positiveside of boundaries to correspond to the material being studied, while that on thenegative side to the material associated with the boundary. Subsequently, we willthen drop the plus superscript on relevant quantities and replace those quantitieswith the minus superscript by the appropriate prescribed values.

Suppose that we have a boundary that is moving at a prescribed velocity c.Then, since m = 0, from (5.302), we must have that at the boundary

v ⋅ n = c(n), (5.334)

where c(n) is now the normal boundary speed. Equation (5.334) is referred to asthe no-penetration condition. The tangential velocity condition is given by (see(5.304))

JvK ⋅ s = Jv(s)K. (5.335)

If Jv(s)K = 0, then we have what is referred to as the no-slip condition. If Jv(s)K ≠ 0,then we have a slip condition. The combined no-penetration and no-slip conditionsrequire that the material velocity be continuous at the boundary (see (5.303)),JvK = 0, and it is equal to the boundary velocity, v = c. This condition is agenerally accepted boundary condition derived from experimental observations.In fluid problems with solid boundaries, the no-penetration and no-slip conditionsare typically applied.

Now suppose that a force F per unit area acts upon the boundary. Then thejump condition (5.305) requires the stress of the medium at the boundary to satisfy

σ ⋅ n = F. (5.336)

A boundary is called free if F = 0.Furthermore, the energy jump condition (5.331), in lieu of (5.336) and the defini-

tion of shear (4.129), tells us that the heat flux must satisfy the following condition:

JqK ⋅ n = Jv(s)K s ⋅ F = Jv(s)KF(s). (5.337)

Clearly if the boundary is free, F = 0, or the force is normal to the boundary,F(s) = 0 (note that generally Jv(s)K ≠ 0 in such cases), or the boundary does notslip, Jv(s)K = 0 (in which case F ≠ 0), then the normal component of the heat fluxmust be continuous, i.e.,

q ⋅n = G, (5.338)

where G is the normal component of the heat flux imposed at the boundary. Aboundary is called adiabatic or thermally isolated if G = 0, in which case the normalcomponent of the heat flux at the boundary must vanish.

Lastly, from (5.333), we see that the entropy production is given by

γs ≡s1

θ

⟪q⟫ ⋅n +⟪1θ⟫ Jv(s)KF(s) ≥ 0. (5.339)

5.12. JUMP CONDITIONS 251

If we assume that the temperature is continuous, JθK = 0 (in which case ⟪θ⟫ = θ),then

γs ≡ 1

θJv(s)KF(s) ≥ 0. (5.340)

On the other hand, if we assume that the boundary is free, or the force is normalto the boundary, or the boundary does not slip, upon using (5.338), we see that

γs ≡s1

θ

G ≥ 0, (5.341)

which tells us that the direction heat must flow depends on the relative magnitudesof temperatures at the boundary and just inside the boundary. Furthermore, ifthe temperature is continuous and the boundary is free, or the force is normal tothe boundary, or the boundary does not slip, then the boundary does not produceany entropy, γs = 0, and the boundary is an ideal boundary.

5.12.3 Equilibrium jump conditions

At equilibrium, we have that γs = 0, σe = −p1 (so that σn = −p and τ = 0), andqe = 0. Thus the entropy jump condition (5.312) reduces to

m JηK = 0. (5.342)

At a material surface, m = 0, and in general JηK ≠ 0. In this case, the massjump condition (5.304) reduces to JvK ⋅ n = 0 (note that generally JvK ≠ 0) and,in conjunction with the linear momentum jump conditions (5.306), we have thatJpK = 0 and in general Jv(s)K ≠ 0. Subsequently, from (5.303), we have that JvK =Jv(s)K s. We note that the energy jump condition (5.307) is satisfied identically.

At a nonmaterial surface, m ≠ 0, and so we must have that JηK = 0. In thiscase, the mass jump condition (5.304) remains JvK ⋅ n =m JvK and, in conjunctionwith the linear momentum jump conditions (5.306), we have that m2JvK + JpK = 0and Jv(s)K = 0. Using this result and the relation g = e − θ η + p v (see (5.203)), theenergy jump condition (5.307) then requires that

se + p v +

1

2m2v2

=

sg + θ η +

1

2m2v2

= 0. (5.343)

Now, if at the surface JθK → 0 and m → 0, we then have that JgK = JµK = 0, whereµ is the equilibrium chemical potential or the specific Gibbs free energy.

252 PRINCIPLES OF CONSTITUTIVE THEORY

Table 5.1: Complete and irreducible function basis of isotropic scalar invariants ofvectors vβ , symmetric tensors Aγ , and skew-symmetric tensors Wδ.

A. One VariableVariable Invariantsv v ⋅ v

A tr A, tr A2, tr A3

W tr W 2

B. Two Variables, A AssumedVariables Invariantsv1, v2 v1 ⋅ v2

v, A v ⋅Av, v ⋅A2v

v, W v ⋅W 2v

A1, A2 tr A1A2, tr A1A22, tr A

21A2, tr A2

1A22

W1, W2 tr W1W2

A, W tr AW 2, tr A2W 2, tr A2W 2AW

C. Three Variables, B AssumedVariables Invariantsv1, v2, v3 0

v1, v2, A v1 ⋅Av2, v1 ⋅A2v2

v1, v2, W v1 ⋅Wv2, v1 ⋅W2v2

v, A1, A2 v ⋅A1A2v

v, W1, W2 v ⋅W1W2v, v ⋅W1W22 v, v ⋅W 2

1W2v

v, A, W v ⋅AWv, v ⋅A2Wv, v ⋅WAW 2v

A1, A2, A3 tr A1A2A3

W1, W2, W3 tr W1W2W3

A1, A2, W tr A1A2W , tr A1A22W , tr A2

1A2W , tr A1W2A2W

A, W1, W2 tr AW1W2, tr AW1W22 , tr AW 2

1W2

D. Four Variables, C AssumedVariables Invariantsv1, v2, A1, A2 v1 ⋅ (A1A2 −A2A1)v2

v1, v2, W1, W2 v1 ⋅ (W1W2 −W2W1)v2

v1, v2, A, W v1 ⋅ (AW +WA)v2

5.12. JUMP CONDITIONS 253

Table 5.2: Generators for h, a vector-valued isotropic function of vector vβ , sym-metric tensors Aγ , and skew-symmetric tensors Wδ.

A. One VariableVariable Generatorv v

A or W –

B. Two Variables, A AssumedVariables Generatorsv1, v2 –v, A Av, A2v

v, W Wv, W 2v

A1, A2 –W1, W2 –A, W –

C. Three Variables, B AssumedVariables Generatorsv1, v2, A –v1, v2, W –v, A1, A2 (A1A2 −A2A1)vv, W1, W2 (W1W2 −W2W1)vv, A, W (AW +WA)vA1, A2, A3 –

Table 5.3: Generators for T, a symmetric tensor-valued isotropic function of vec-tors vβ , symmetric tensors Aγ , and skew-symmetric tensors Wδ.

A. No VariableVariable Generator0 1

B. One Variable, A AssumedVariable Generatorsv v v

A A, A2

W W 2

C. Two Variables, B AssumedVariables Generatorsv1, v2 v1v2 + v2v1

v, A vAv +Avv, vA2v +A2vv

v, W vWv +Wvv, WvWv, WvW 2v +W 2vWv

A1, A2 A1A2 +A2A1, A21A2 +A2A

21, A1A

22 +A

22A1

W1, W2 W1W2 +W2W1, W1W22 −W

22W1, W

21W2 −W2W

21

A, W AW −WA, WAW , A2W −WA2, WAW 2−W 2AW

D. Three Variables, C AssumedVariables Generatorsv1, v2, A A(v1v2 − v2v1) − (v1v2 − v2v1)Av1, v2, W W (v1v2 − v2v1) + (v1v2 − v2v1)W

254 PRINCIPLES OF CONSTITUTIVE THEORY

Table 5.4: Generators for T, a skew-symmetric tensor-valued isotropic function ofvectors vβ , symmetric tensors Aγ , and skew-symmetric tensors Wδ.

A. One VariableVariable Generatorv or A –W W

B. Two Variables, A AssumedVariables Generatorsv1, v2 v1v2 − v2v1

v, A vAv −Avv, vA2v −A2vv, AvA2v −A2vAv

v, W vWv −Wvv, vW 2v −W 2vv

A1, A2 A1A2 −A2A1, A1A22 −A

22A1, A

21A2 −A2A

21,

A1A2A21 −A

21A2A1, A2A1A

22 −A

22A1A2

W1, W2 W1W2 −W2W1

A, W AW +WA, AW 2−W 2A

C. Three Variables, B AssumedVariables Generatorsv1, v2, A A(v1v2 − v2v1) + (v1v2 − v2v1)Av1, v2, W W (v1v2 − v2v1) − (v1v2 − v2v2)Wv, A1, A2 A1vA2v −A2vA1v + v(A1A2 −A2A1)v−(A1A2 −A2A1)vvA1, A2, A3 A1A2A3 +A2A3A1 +A3A1A2 −A2A1A3 −

A1A3A2 −A3A2A1

5.12. JUMP CONDITIONS 255

Table 5.5: Complete and irreducible function basis of hemitropic scalar invariantsof vectors vβ , symmetric tensors Aγ , and skew-symmetric tensors Wδ.

A. One VariableVariable Invariantsv v ⋅ v

A tr A, tr A2, tr A3

W tr W 2

B. Two Variables, A AssumedVariables Invariantsv1, v2 v1 ⋅ v2

v, A v ⋅Av, v ⋅A2v, [v,Av,A2v]v, W v ⋅ ⟨W ⟩A1, A2 tr A1A2, tr A1A

22, tr A

21A2, tr A

21A

22

W1, W2 tr W1W2

A, W tr AW 2, tr A2W 2, tr A2W 2AW

C. Three Variables, B AssumedVariables Invariantsv1, v2, v3 [v1,v2,v3]v1, v2, A v1 ⋅Av2, [v1,v2,Av1], [v1,v2,Av2]v1, v2, W v1 ⋅Wv2

v, A1, A2 v ⋅ ⟨A1A2⟩, v ⋅ ⟨A1A22⟩, v ⋅ ⟨A2

1A2⟩, [v,A1v,A2v]v, W1, W2 v ⋅ ⟨W1W2⟩v, A, W v ⋅AWv, v ⋅ ⟨AW ⟩, v ⋅ ⟨AW 2⟩A1, A2, A3 tr A1A2A3

W1, W2, W3 tr W1W2W3

A1, A2, W tr A1A2W , tr A21A2W , tr A1A

22W , tr A1W

2A2W

A, W1, W2 tr AW1W2, tr AW 21W2, tr AW1W

22

D. Four Variables, C AssumedVariables Invariantsv1, v2, A1 A2 –v1, v2, W1 W2 –v1, v2, A W –

256 PRINCIPLES OF CONSTITUTIVE THEORY

Table 5.6: Generators for h, a vector-valued hemitropic function of vector vβ ,symmetric tensors Aγ , and skew-symmetric tensors Wδ.

A. One VariableVariable Generatorv v

A –W ⟨W ⟩B. Two Variables, A AssumedVariables Generatorsv1, v2 v1 × v2

v, A Av, v ×Avv, W Wv

A1, A2 ⟨A1A2⟩, ⟨A21A2⟩, ⟨A1A

22⟩, ⟨A1A2A

21⟩, ⟨A2A1A

22⟩

W1, W2 ⟨W1W2⟩A, W ⟨AW ⟩, ⟨AW 2⟩C. Three Variables, B AssumedVariables Generatorv1, v2, A –v1, v2, W –v, A1, A2 –v, W1, W2 –v, A, W –A1, A2, A3 ⟨A1A2A3 +A2A3A1 +A3A1A2⟩

5.12. JUMP CONDITIONS 257

Table 5.7: Generators for T, a symmetric tensor-valued hemitropic function ofvectors vβ , symmetric tensors Aγ , and skew-symmetric tensors Wδ.

A. No VariableVariable Generator0 1

B. One Variable, A AssumedVariable Generatorsv vv

A A, A2

W W 2

C. Two Variables, B AssumedVariables Generatorsv1, v2 v1v2 + v2v1, v1(v1 × v2) + (v1 × v2)v1,

v2(v1 × v2) + (v1 × v2)v2

v, A (W = v ⋅ ǫ) vAv +Avv, AW +WA, A2W +WA2,v(v ×Av) + (v ×Av)v

v, W1 (W2 = v ⋅ ǫ) vW1v +W1vv, W1W2 +W2W1, W21W2 +W2W

21

A1, A2 A1A2 +A2A1, A21A2 +A2A

21, A1A

22 +A

22A1

W1, W2 W1W2 +W2W1, W21W2 −W2W

21 , W1W

22 −W

22W1

A, W AW −WA, AW 2+W 2A, WAW 2

−W 2AW ,A2W −WA2

D. Three Variables, C AssumedVariables Generatorsv1, v2, A –v1, v2, W –

258 PRINCIPLES OF CONSTITUTIVE THEORY

Table 5.8: Generators for T, a skew-symmetric tensor-valued hemitropic functionof vectors vβ , symmetric tensors Aγ , and skew-symmetric tensors Wδ.

A. One VariableVariable Generatorv (W = v ⋅ ǫ) W

A –W W

B. Two Variables, A AssumedVariables Generatorsv1, v2 v1v2 − v2v1

v, A (W = v ⋅ ǫ) AW −WA, vAv −Avvv, W1 (W2 = v ⋅ ǫ) W1W2 −W2W1

A1, A2 A1A2 −A2A1, A1A22 −A

22A1, A

21A2 −A2A

21,

A1A2A21 −A

21A2A1, A2A1A

22 −A

22A1A2

W1, W2 W1W2 −W2W1

A, W AW +WA, AW 2−W 2A

C. Three Variables, B AssumedVariables Generatorsv1, v2, A –v1, v2, W –v, A1, A2 –A1, A2, A3 A1A2A3 +A2A3A1 +A3A1A2 −A2A1A3−

A1A3A2 −A3A2A1

5.12. JUMP CONDITIONS 259

Table 5.9: Thermodynamic property tensors.

General Description Special Case Units(n = 1) (SI)1

η Entropy η J/kg⋅Ke Internal energy e J/kgψ Helmholtz potential ψ J/kgh Enthalpy potential h J/kgg Gibbs potential g J/kgθ Thermostatic temperature θ Kνα Specific thermostatic volume v = 1/ρ m3/kgτα Thermostatic tension −p N/m2, J/m3

βα Isochoric thermal tension −α/κθ J/m3⋅K

αα Thermal strain α/ρ m3/kg⋅Kλνα

Caloric stiffness θ α/κθ J/m3

λταCaloric compliance θ α/ρ m3/kg

cναSpecific heat at const. thermostatic volume cv J/kg⋅K

cταSpecific heat at const. thermostatic tension cp J/kg⋅K

γα Ratio of specific heats γ ≡ cp/cv —ζα Isopiestic thermal expansion (γ − 1)κθ/γ α m3

⋅K/Jϕα Isentropic thermal stiffness −(γ − 1)ρ/α kg⋅K/m3

ξαβ Isothermal elastic stiffness ρ/κθ J⋅kg/m6

υαβ Isothermal elastic compliance κθ/ρ m6/J⋅kgφαβ Isentropic elastic stiffness γ ρ/κθ J⋅kg/m6

χαβ Isentropic elastic compliance κθ/γ ρ m6/J⋅kgΓαβ Grüneisen parameter Γ ≡ α/ρ cv κθ —

1 1 J = 1 N⋅m, 1 N = 1 kg⋅m/s2.

Table 5.10: Ehrenfest classification of phase transitions in a simple isotropic ma-terial.

Discontinuous QuantitiesOrder Differentials Experimental QuantitiesFirst η, v η, v

Second∂η

∂θ∣p

,∂v

∂θ∣p

,∂η

∂p∣θ

,∂v

∂p∣θ

cp, α, κθ

Third∂2η

∂θ2∣p

,∂2v

∂θ2∣p

,∂2η

∂p∂θ,

∂cp

∂θ∣p

,∂α

∂θ∣p

,∂κθ

∂θ∣p

,

∂2v

∂p∂θ,∂2η

∂p2∣θ

,∂2v

∂p2∣θ

∂cp

∂p∣θ

,∂α

∂p∣θ

,∂κθ

∂p∣θ

260 PRINCIPLES OF CONSTITUTIVE THEORY

Problems

1. If we have F (x, y) = 0, under what conditions, can we write y = f(x)?2. What are the necessary and sufficient conditions for the maximum of f(x, y, z)?3. Find the generators of the hemitropic symmetry group as represented by

laminated wood with the geometry shown in Fig. 5.5.

X3

X1

X2

Figure 5.5: Laminated wood geometry.

4. Suppose that a given material is characterized by the constitutive equationfor the stress tensor in the following form:

σ = σ(ρ,v,L),where v is the velocity and L = (gradv)T . Use the principle of materialframe indifference to write the constitutive equation in reduced form.

5. For σ = σ(ρ, θ,g), use objectivity to show that

σ = α1 + β gg,

where α and β are functions of ρ, θ, and g ⋅ g (provide details).

6. Let T = T(A) be a continuous isotropic symmetric tensor of rank 2 whichis a function of the symmetric tensor A of rank 2. Clearly, from (5.126) andTables 5.1 and 5.3, we obtained the representation for T given in (5.128).Obtain this representation by using the Cayley–Hamilton theorem.

7. Consider a material with the stress tensor

σij = −pδij +KijklDkl.

Show that, because of the symmetries of the stress and rate of deformationtensors, the tensor of rank 4 Kijkl has at most 36 independent components.Display the components in a 6 × 6 array.

5.12. JUMP CONDITIONS 261

8. Given the stress tensor in Problem 7, show that objectivity requires that

Kijkl =QipQjqQrsQlsKpqrs,

where Q is any orthogonal tensor.

9. i) If a material is assumed isotropic, show that the rank-4 tensor K givenin Problem 7 only has two independent components. Subsequently,provide the reduced representation of the stress tensor σ.

ii) Write the resulting representation into isotropic, symmetric, and devi-atoric parts (see Section 2.10).

10. If a material has the constitutive equation for the stress tensor

σij = −p δij + αDij + βDikDkj ,

show that if the material is incompressible, then

σii = −3(p + 2

3βD(2)) ,

where D(2) is the second invariant of D.

11. Using the definitions of ζα in (5.144), αα in (5.150), and cταin (5.165),

derive (5.177).

12. Prove (5.169).

13. Using the relations (5.175) and (5.177), and the definitions (5.137), (5.150),and (5.158), prove the identities (5.176).

14. Use (5.183) and (5.217) to show that

φαβ − ξαβ = θ

cνα

βαββ = θ cναΓαΓβ .

15. Verify (5.212) by using (5.207).

16. Verify (5.225) by using (5.220).

17. Verify (5.234) by using (5.229).

18. Prove (5.226).

19. Prove (5.235).

20. A hyperelastic material is described by the consitutive equation (see (5.277))

σe = σe(F) = ρ ∂τ∂F⋅FT ,

where τ = τ(F) is the strain (or stored) energy function. Show that if τsatisfies objectivity, then so does σe.

262 PRINCIPLES OF CONSTITUTIVE THEORY

21. Consider a general macroscale description of flow through porous media de-scribed by the proposed model

∇p = f (γ,u,v,L) ,where ∇p is the pressure gradient and f is a vector-valued function thatdepends on γ, a scalar property function (more generally it depends on anumber of scalar functions representing geometric properties such as porosityand specific surface density, and thermophysical properties of the mediumand fluid such as density, viscosity, and compressibility), u ≡ v − vs is thefluid velocity v relative to the solid velocity vs, and L ≡ (gradv)T , which isthe fluid velocity gradient.

i) Use the principle of frame indifference to show that

∇p = f (γ,u,D) .ii) Subsequently, show that the reduced form of this model is given by

∇p = −H ⋅ u,where

H = α01 + α1D + α2D2

and

αi = αi(γ, ∣u∣,D(1),D(2),D(3),u ⋅Du,u ⋅D2u), i = 1,2,3.We call this the generalized Darcy’s law.

iii) The second law of thermodynamics requires that the viscous dissipationbe such that f ⋅ u ≤ 0. Show that the constraints on the αi’s imposedby this restriction are

α0 > 0, α2 > 0, and α2

1 − 4α0α2 < 0.iv) If the relative velocity is small but the fluid velocity is not small, show

that the generalized Darcy’s law linear in u is given by

∇p = −H ⋅ u,where

H = α01 + α1D + α2D2

and αi = αi(γ,D(1),D(2),D(3)).v) If the relative velocity and the fluid velocity are both small, show that

the generalized Darcy’s law linear in u and v becomes

∇p = −α0 (γ,D(1))u.Furthermore, show that if the fluid is incompressible, this reduces tothe classical Darcy’s law

∇p = −α0 (γ)u,where α0(γ) is the reciprocal of the permeability “coefficient.”

5.12. JUMP CONDITIONS 263

22. Show that

q = q1 (F, θ,g)implies

q = F ⋅ q2 (U, θ,RT⋅ g)

and vice-versa.

23. Show that for a hemitropic material, it is necessary and sufficient that

q = q (B, θ,g) . (5.344)

24. Show that for an isotropic material

q = (α0 1 + α1B + α2B2) ⋅ g.

25. For an isotropic solid, ψ = ψ(B(α), θ). Show that for a fluid, ψ = ψ(B(3), θ).26. In the process of obtaining the fundamental relation, it is very convenient in

experiments where one can control temperature and thermostatic volumesto be able to decompose the Helmholtz free energy into components thatare independent of temperature changes and those that are dependent ontemperature changes. With this aim in mind

i) use the first equation in (5.165) to show that

∫θ

θ0

cνα(θ′,να) dθ′ = e(θ,να) − e(θ0,να);

ii) integrate the first equation in (5.209) and use (5.201) and (5.208) toshow that

∫θ

θ0

θ

θ′cνα(θ′,να)dθ′ = e(θ,να) −ψ(θ,να) − θ η(θ0,να);

iii) subsequently, show that the Hemholtz free energy can be rewritten inthe following convenient form:

ψ(θ,να) = e(θ0,να) − θ η(θ0,να) −∫ θ

θ0

( θθ′− 1) cνα

(θ′,να)dθ′. (5.345)

27. In the process of obtaining the fundamental relation, it is very convenientin experiments where one can control temperature and thermostatic ten-sions to be able to decompose the Gibbs free energy into components thatare independent of temperature changes and those that are dependent ontemperature changes. With this aim in mind

i) use (5.222) to show that

∫θ

θ0

cτα(θ′,τα)dθ′ = h(θ,τα) − h(θ0,τα);

264 PRINCIPLES OF CONSTITUTIVE THEORY

ii) integrate (5.231) and use (5.229), (5.230), (5.202), and (5.203) to showthat

∫θ

θ0

θ

θ′cτα(θ′,τα)dθ′ = h(θ,τα) − g(θ,τα) − θ η(θ0,τα);

iii) subsequently, show that the Gibbs free energy can be rewritten in thefollowing convenient form:

g(θ,τα) = h(θ0,τα) − θ η(θ0,τα) −∫ θ

θ0

( θθ′− 1) cτα

(θ′,τα)dθ′. (5.346)

28. From a set of experiments on a compressible gas at equilibrium, one sees thatthe specific heats at constant volume and at constant pressure are effectivelyconstant. If we assume that they are truly constant and that the gas is non-dissipative, determine the Helmholtz free energy of the gas, ψ = ψ(ρ, θ), interms of unknown constants.

29. Show that

JvK⟪σn⟫ = JvσnK − 1

2m2

qv2

y.

30. Starting from (4.217), derive (5.308).

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266 PRINCIPLES OF CONSTITUTIVE THEORY

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6

Spatially uniform systems

As a means of exercising the use of our balance equations, initial and boundaryconditions, constitutive principles, and thermodynamics relations, we first examinesystems that are uniform in space. Note that in order for a system to be spatiallyuniform, it is necessary but not sufficient for the material to be homogeneous andisotropic. Let us first summarize the mass, momenta, and energy balance laws fornonpolar materials obtained earlier:

ρ = −ρdivv, (6.1)

ρa = divσ + ρ f , (6.2)

σ = σT , (6.3)

ρ e = −divq +L ∶ σ + ρr. (6.4)

In addition, we have the Clausius–Duhem inequality

−γvθ ≡ ρ (ψ + ηθ) −L ∶ σ + g ⋅ qθ− θ div (h − q

θ) + ρθ (b − r

θ) ≤ 0. (6.5)

Now since a boundary is a material surface, the conditions there must be

v ⋅n = cn, σ ⋅ n = F, q ⋅ n = G, h ⋅ n = R. (6.6)

We consider a system that is closed, and having a boundary that is stationary(cn = 0), free (F = 0), adiabatic (G = 0, thermally isolated), and isentropic (R = 0, noentropy exchange with surroundings). We are thinking of a body having constantmass, a composition that is uniform in space, and state functions that only varywith time. Since the mass is constant, it is convenient to introduce the specificvolume v = 1/ρ; subsequently, since divv = trD, the mass balance (6.1) providesthe relation

trD = v

v. (6.7)

Furthermore, since we are considering a thermally isolated uniform system under-going a simple thermomechanical process, then, in conjunction with the boundaryconditions and the Clausius–Duhem inequality, we must have

σ = −p1, h = q

θ= 0, b = r

θ= 0. (6.8)

271

272 SPATIALLY UNIFORM SYSTEMS

Now our equations and the Clausius–Duhem inequality (6.2)–(6.6) reduce to

a = f , (6.9)

e = −p v, (6.10)

−γv θ v ≡ ψ + η θ + p v ≤ 0. (6.11)

The external momentum body source can be an arbitrarily prescribed function oftime:

f = f(t). (6.12)

Note that (6.9) and (6.12) require that the acceleration of the body be uniform,a = a(t), and from the linear momentum equation (6.9) (this is nothing morethan the conventional Newton’s second law of motion), the acceleration can beintegrated to obtain the velocity, v = v(t), as well as the motion, x = χ(t), forgiven initial conditions. We further note that in this case, where the system isspatially uniform, the mechanics and thermodynamics of the system are completelydecoupled. In addition, we note that equation (6.10) is nothing more than the firstlaw of thermostatics for a thermally isolated system, and (6.11) is the Clausius–Duhem inequality.

Subsequently, enforcement of the Clausius–Duhem inequality, in conjunctionwith the balance laws, should provide relations between the dependent constitutivevariables

ψ = F0≤s<∞

(v, θ), η = G0≤s<∞

(v, θ), p = H0≤s<∞

(v, θ), (6.13)

and the independent variables describing the thermodynamic process

v = v(t), θ = θ(t). (6.14)

In (6.13) our constitutive variables are taken to be functionals of the independentvariables (6.14) (see (5.31)), and we note that v > 0 and θ > 0. Also, the dependenceon the deformation gradient, F, here reduces to the dependence on trD, and thusto the dependence on v through the mass balance equation. Furthermore, we haveno dependence on the heat flux, g, in thermally isolated uniform systems.

Now the same material may have different constitutive equations (6.13), whichdepend on the level of description of the thermodynamic process (6.14). Sincewe are dealing with a uniform system, what we mean here is the choice of dif-ferent rate-type materials, i.e., how much memory does the material have (seeSection 5.3).

6.1 Material with no memory

Here we take p = q = 0 in (5.38) so that

ψ = ψ(v, θ), η = η(v, θ), p = p(v, θ). (6.15)

Subsequently, (6.11) becomes

−γv θ v ≡ [∂ψ∂v(v, θ) + p(v, θ)] v + [∂ψ

∂θ(v, θ) + η(v, θ)] θ ≤ 0. (6.16)

6.1. MATERIAL WITH NO MEMORY 273

Now, since v > 0 and θ > 0, for arbitrary processes in which v and θ can be pre-scribed to be of either sign, and since the terms in square brackets are independentof v and θ, (6.16) requires that

∂ψ

∂v= −p and

∂ψ

∂θ= −η, (6.17)

in which case we have from (5.206) that the Helmholtz free energy is a thermody-namic potential function of specific volume and temperature,

ψ = −p v − η θ, (6.18)

and in conjunction with the definition of the Helmholtz potential (5.201), we obtainthe Gibbs equation

e = −p v + θ η, (6.19)

where clearly

e = e(v, θ) = ψ − θ ∂ψ∂θ. (6.20)

Thus, since (6.16) is identically satisfied, processes without memory have zeroentropy production, i.e., they are reversible. In addition, since the constitutive re-lations are independent of v and θ, such processes are also equilibrium processes. Infact, classical thermodynamics deals exclusively with reversible equilibrium pro-cesses, which, as we have seen, result from spatially uniform systems in whichmaterials have no memory.

Now the internal energy minimum principle tells us that for a given value ofthe total entropy, the equilibrium value of any unconstrained internal quantity issuch as to minimize the internal energy. Thus, it is convenient to substitute thetemperature dependence by the entropy dependence:

e = e(η, v). (6.21)

We now represent the total variation in internal energy as a Taylor series expansionaround the equilibrium state. The first-order terms cancel, and for the quadraticterm, we have the stability condition:

δ2e = (∂2e∂η2)

v

(δη)2 + 2 [ ∂∂v(∂e∂η)

v

]η

δη δv + (∂2e∂v2)η

(δv)2 > 0. (6.22)

The stability condition is that the right-hand side must be positive definite for anyδη and δv, except of course the trivial case δη = δv = 0, in which case third-orderterms must be taken into account. This is a quadratic form which is positivedefinite if and only if the following two conditions are satisfied:

(∂2e∂η2)

v

> 0 and (∂2e∂η2)

v

(∂2e∂v2)η

− [ ∂∂v(∂e∂η)

v

]2η

> 0. (6.23)

Now, using standard thermodynamic relations, the first condition becomes

(∂2e∂η2)

v

= (∂θ∂η)

v

= θ

cv> 0, (6.24)

274 SPATIALLY UNIFORM SYSTEMS

which, since θ > 0, implies that the specific heat at constant volume

cv ≡ θ (∂η∂θ)

v

(6.25)

is positive definite: cv > 0. It is easy to see that the two conditions (6.23) alsoimply that

(∂2e∂v2)η

> 0 (6.26)

and thus

(∂2e∂v2)η

= −(∂p∂v)η

= 1

vκη> 0. (6.27)

Subsequently, since v > 0, the condition implies that the adiabatic compressibility

κη ≡ −1v(∂v

∂p)η

(6.28)

is positive definite: κη > 0. Lastly, the second condition in (6.23) can be rewrittenas

(∂2e∂η2)

v

(∂2e∂v2)η

− [ ∂∂v(∂e∂η)

v

]2η

= −(∂θ∂η)

v

(∂p∂v)η

− (∂θ∂v)2η

(6.29)

and since

(∂θ∂v)η

= −(∂θ∂η)

v

(∂η∂v)θ

, (6.30)

this condition becomesθ

vκη cv− ( θ α

κθ cv)2 > 0, (6.31)

where the isothermal compressibility and the thermal expansion coefficient aredefined by

κθ ≡ −1v(∂v

∂p)θ

, (6.32)

and

α ≡ 1

v(∂v

∂θ)p

. (6.33)

This condition, in conjunction with the previous conditions, can be easily shownto lead to

cp > cv > 0 and κθ > κη > 0, (6.34)

where the specific heat at constant pressure is defined by

cp ≡ θ (∂η∂θ)p

, (6.35)

and note that the sign of α is arbitrary.

6.2. MATERIAL WITH SHORT MEMORY OF VOLUME 275

6.2 Material with short memory of volume

Here we take p = 1 and q = 0 in (5.38) so that

ψ = ψ(v, v, θ), η = η(v, v, θ), p = p(v, v, θ). (6.36)

Subsequently, (6.11) becomes

−γv θ v ≡ [∂ψ∂v(v, v, θ) + p(v, v, θ)] v + ∂ψ

∂v(v, v, θ) v +

[∂ψ∂θ(v, v, θ) + η(v, v, θ)] θ ≤ 0. (6.37)

Now, since v > 0 and θ > 0, for arbitrary processes in which v, v, and θ can be ofeither sign, and since ψ is not a function of v, (6.37) requires that

∂ψ

∂v= 0 or ψ = ψ(v, θ), (6.38)

so that we can rewrite (6.37) as

−γv θ v ≡ [∂ψ∂v(v, θ) + p(v, v, θ)] v + [∂ψ

∂θ(v, θ) + η(v, v, θ)] θ ≤ 0, (6.39)

from which, for arbitrary θ, since the terms in square brackets are independent ofθ, we must require that

∂ψ

∂θ= −η. (6.40)

Thus we have thatη = η(v, θ) and e = e(v, θ), (6.41)

and the reduced Clausius–Duhem inequality becomes

−γv θ v ≡ [∂ψ∂v(v, θ) + p(v, v, θ)] v ≤ 0. (6.42)

Now at thermodynamic equilibrium (v = 0), the entropy production must bezero, and clearly it is trivially so. But, in addition, from (6.42) we also note thatthe entropy production must be a minimum at equilibrium. Thus, at equilibriumwe must have that

γv(v, v, θ)∣v=0 = 0, ∂γv

∂v(v, v, θ)∣

v=0= 0, and

∂2γv

∂v2(v, v, θ)∣

v=0

> 0. (6.43)

Now, from the first derivative condition, we obtain

∂ψ

∂v= −pe, (6.44)

where we have defined the equilibrium pressure

pe = pe(v, θ) ≡ p(v,0, θ). (6.45)

276 SPATIALLY UNIFORM SYSTEMS

Defining the nonequilibrium residual pressure as

pr = pr(v, v, θ) ≡ p(v, v, θ) − pe(v, θ), (6.46)

so that pr(v,0, θ) = 0, the second derivative condition requires that

∂pr

∂v(v,0, θ) < 0. (6.47)

Obviously, outside of equilibrium, pr can have an arbitrary dependence on v. Toa leading order approximation, it depends linearly on v; thus we write it as

pr = −λ v

v, (6.48)

where we have defined the dilatational viscosity

λ = λ(v, θ) > 0. (6.49)

Note that the positive-definite requirement on the dilatational viscosity is dictatedby the second derivative condition.

Summarizing the results of this system, we see that the constitutive quantitiesare given by

ψ = ψ(v, θ), η = −∂ψ∂θ, and p = −∂ψ

∂v− λ(v, θ) v

v. (6.50)

In addition, we find that the Helmholtz free energy is a potential for entropy butonly for the equilibrium pressure,

ψ = −pe v − η θ, (6.51)

and, in conjunction with (5.201), the Gibbs equation is given by

e = −pe v + θ η, (6.52)

where

e = e(v, θ) = ψ − θ ∂ψ∂θ. (6.53)

For this material, none of the Helmholtz free energy, entropy, or internal energyconstitutes fundamental relations since they cannot fully describe the material.On the other hand, the Gibbs free energy, given by

g = g(v, v, θ) = e + p v − θ η = ψ − ∂ψ∂v

v − λ(v, θ) v, (6.54)

does provide the fundamental relation for this material.

6.3 Material with longer memory of volume

Here we take p = 2 and q = 0 in (5.38) so that

ψ = ψ(v, v, v, θ), η = η(v, v, v, θ), and p = p(v, v, v, θ). (6.55)

6.3. MATERIAL WITH LONGER MEMORY OF VOLUME 277

Subsequently, (6.11) becomes

−γv θ v ≡ [∂ψ∂v(v, v, v, θ) + p(v, v, v, θ)] v + ∂ψ

∂v(v, v, v, θ) v + ∂ψ

∂v(v, v, v, θ)...v +

[∂ψ∂θ(v, v, v, θ) + η(v, v, v, θ)] θ ≤ 0. (6.56)

Now, since v > 0 and θ > 0, for arbitrary processes in which v, v,...v, and θ can be

of either sign, (6.56) requires that (because of the linearity in...v and θ)

∂ψ

∂v= 0 and

∂ψ

∂θ= −η. (6.57)

Thus, from the above, as well as (5.201), we have that

ψ = ψ(v, v, θ), η = η(v, v, θ), and e = e(v, v, θ), (6.58)

and the reduced Clausius–Duhem inequality becomes

−γv θ v ≡ [∂ψ∂v(v, v, θ) + p(v, v, v, θ)] v + ∂ψ

∂v(v, v, θ) v ≤ 0. (6.59)

Note that the above inequality is nonlinear in v; thus to make further progress, weexamine the equilibrium behavior of this material. At equilibrium, we must havev = 0 and v = 0, and γv = 0 is a minimum:

γv(v, v, v, θ)∣v=v=0 = 0, (6.60)

∂γv

∂v(v, v, v, θ)∣

v=v=0= 0, ∂γv

∂v(v, v, v, θ)∣

v=v=0= 0, (6.61)

∂2γv

∂v2(v, v, v, θ)∣

v=v=0

> 0, ∂2γv

∂v2∂2γv

∂v2(v, v, v, θ)∣

v=v=0

−

[∂2γv∂v∂v

(v, v, v, θ)]2v=v=0

> 0. (6.62)

Clearly γv = 0 is trivially satisfied at equilibrium. As in the previous section, thefirst and second derivative conditions sequentially give us that

∂ψr

∂v(v,0, θ) = 0 and

∂ψe

∂v= −pe, (6.63)

where we have defined the equilibrium and residual free energy and pressure asfollows:

ψe = ψe(v, θ) ≡ ψ(v,0, θ), (6.64)

pe = pe(v, θ) ≡ p(v,0,0, θ), (6.65)

ψr = ψr(v, v, θ) ≡ ψ(v, v, θ) − ψe(v, θ), (6.66)

pr = pr(v, v, v, θ) ≡ p(v, v, v, θ) − pe(v, θ), (6.67)

and we must have that ψr(v,0, θ) = 0 and pr(v,0,0, θ) = 0. Subsequently, theClausius–Duhem inequality can be rewritten as

−γv θ v ≡ [∂ψr

∂v(v, v, θ) + pr(v, v, v, θ)] v + ∂ψr

∂v(v, v, θ) v ≤ 0. (6.68)

278 SPATIALLY UNIFORM SYSTEMS

The resulting second derivative condition

∂2ψr

∂v∂v(v,0, θ) + ∂pr

∂v(v,0,0, θ) < 0 (6.69)

does not yield additional simplifications or reduction of (6.68) unless we approxi-mate ψr up to some order in v, and pr up to some order in v and v.

In summary, the constitutive equations of this material are given by

ψ = ψe(v, θ) +ψr(v, v, θ), η = −∂ψ∂θ, and p = −∂ψe

∂v+ pr(v, v, v, θ). (6.70)

In addition, we find that the Helmholtz free energy is a potential for entropy butonly the equilibrium part of the free energy,

ψ = −pe v +∂ψr

∂vv +

∂ψr

∂vv − η θ, (6.71)

and, in conjunction with (5.201), the Gibbs equation is now given by

e = −pe v +∂ψr

∂vv +

∂ψr

∂vv + θ η, (6.72)

where

e = e(v, v, θ) = ψ − ∂ψ∂θ. (6.73)

Note that the terms involving derivatives of ψr contribute to entropy productionwhen the system is not in equilibrium.

Again, for this material, none of the Helmholtz free energy, entropy, or inter-nal energy constitutes fundamental relations since they cannot fully describe thismaterial. On the other hand, the Gibbs free energy

g = g(v, v, v, θ) = ψ − [∂ψe

∂v− pr(v, v, v, θ)] v (6.74)

does provide the fundamental relation for the material.

6.4 Material with short memory

Here we take p = q = 1 in (5.38) so that

ψ = ψ(v, v, θ, θ), η = η(v, v, θ, θ), and p = p(v, v, θ, θ). (6.75)

Subsequently, (6.11) becomes

−γv θ v ≡ [∂ψ∂v(v, v, θ, θ) + p(v, v, θ, θ)] v + ∂ψ

∂v(v, v, θ, θ) v +

[∂ψ∂θ(v, v, θ, θ) + η(v, v, θ, θ)] θ + ∂ψ

∂θ(v, v, θ, θ) θ ≤ 0. (6.76)

Now, since v > 0 and θ > 0, for arbitrary processes in which v, v, θ, and θ can beof either sign, (6.76) requires that (because of the linearity in v and θ)

∂ψ

∂v= 0 and

∂ψ

∂θ= 0, (6.77)

6.4. MATERIAL WITH SHORT MEMORY 279

so thatψ = ψ(v, θ) (6.78)

and we can rewrite (6.76) as

−γv θ v ≡ [∂ψ∂v(v, θ) + p(v, v, θ, θ)] v + [∂ψ

∂θ(v, θ) + η(v, v, θ, θ)] θ ≤ 0. (6.79)

Now, at thermodynamic equilibrium, v = 0 and θ = 0, and the entropy productionmust be zero, and clearly it is trivially so. But, in addition, from (6.79) we alsonote that the entropy production must be a minimum at equilibrium. Thus, atequilibrium, we must have that

γv(v, v, θ, θ)∣v=θ=0

= 0, (6.80)

∂γv

∂v(v, v, θ, θ)∣

v=θ=0= 0, ∂γv

∂θ(v, v, θ, θ)∣

v=θ=0

= 0, (6.81)

∂2γv

∂v2(v, v, θ, θ)∣

v=θ=0

> 0, ∂2γv

∂v2∂2γv

∂θ2(v, v, θ, θ)∣

v=θ=0

−

[ ∂2γv∂v∂θ

(v, v, θ, θ)]2v=θ=0

> 0. (6.82)

Satisfaction of the first derivative conditions requires that

∂ψ

∂v= −pe and

∂ψ

∂θ= −ηe, (6.83)

and allows the inequality to be rewritten as

−γv θ v ≡ pr(v, v, θ, θ) v + ηr(v, v, θ, θ) θ ≤ 0, (6.84)

where we have defined the equilibrium and residual pressure and entropy as follows:

pe = pe(v, θ) ≡ ψ(v,0, θ,0), (6.85)

ηe = ηe(v, θ) ≡ ψ(v,0, θ,0), (6.86)

pr = pr(v, v, θ, θ) ≡ p(v, v, θ, θ) − pe(v, θ), (6.87)

ηr = ηr(v, v, θ, θ) ≡ ψ(v, v, θ, θ) − ψe(v, θ), (6.88)

and we have that pr(v,0, θ,0) = 0 and ηr(v,0, θ,0) = 0. The second derivativeconditions now require that

∂pr

∂v(v,0, θ,0) < 0. (6.89)

Additional simplifications and reduction of (6.84) result when pr and ηr are ap-proximated up to some order in v and θ.

To summarize, the constitutive equations of this material are given by

ψ = ψ(v, θ), η = −∂ψ∂θ+ ηr(v, v, θ, θ), and p = −∂ψ

∂v+ pr(v, v, θ, θ). (6.90)

In addition, we find that the Helmholtz free energy is a potential only for theequilibrium parts of the pressure and entropy,

ψ = −pe v − ηe θ, (6.91)

280 SPATIALLY UNIFORM SYSTEMS

and, in conjunction with (5.201), the Gibbs equation is now given by

e = −pe v + ηr θ + θ η, (6.92)

where

e = e(v, v, θ, θ) = ψ − θ (∂ψ∂θ− ηr) . (6.93)

Note that the term involving ηr contributes to entropy production when the systemis not in equilibrium.

While the Helmholtz free energy cannot fully describe this material, the entropy,the internal energy, and the Gibbs free energy

g = g(v, v, θ, θ) = ψ − [∂ψ∂v− pr(v, v, θ, θ), ] v (6.94)

provide fundamental relations since they can fully describe the material.

Problems

1. The Helmholtz free energy for an ideal gas is given by

ψ(v, θ) = ψ0 + (αR − η0) (θ − θ0) −Rθ ln [( θθ0)α v

v0] ,

where all quantities with zero subscripts are reference constant quantities,R = R/M is the specific gas constant with R the universal gas constantand M the molecular weight of the gas, and α is a parameter related to theinternal degrees of freedom of the gas molecule (α = 3/2 for monatomic gases,α = 5/2 for diatomic gases, and α = 3 for polyatomic gases).

Assuming that an ideal gas is a material with no memory, obtain the corre-sponding constitutive expressions for entropy, internal energy, and pressurefrom the above fundamental equation.

2. The following fundamental equation for the Helmholtz free energy has beenused to determine the properties of water:

ψ(v, θ) = ψ0(θ) −Rθ [ln v −1

vQ(v, θ)] , (6.95)

where ψ0(θ) and Q(v, θ) are usually given as expansions in their respectivearguments that contain adjustable constants.

Assuming that water has no memory, obtain expressions for the entropy,internal energy, and pressure from the above relation.

3. Obtain the constitutive equations for the material with longer memory ofvolume assuming the linear variations ψr = a(v, θ) v and pr = b(v, θ) v +c(v, θ) v. What are the constraints on a(v, θ), b(v, θ), and c(v, θ)?

4. Obtain the constitutive equations for the material with short memory of vol-ume and temperature assuming the linear variations ηr = a(v, θ) v + b(v, θ) θand pr = c(v, θ) v + d(v, θ) θ. What are the constraints on a(v, θ), b(v, θ),c(v, θ), and d(v, θ)?

BIBLIOGRAPHY 281

Bibliography

M. Bailyn. A Survey of Thermodynamics. AIP Press, New York, NY, 1994.

H.B. Callen. Thermodynamics. John Wiley & Sons, Inc., New York, NY, 1962.

M.R. El-Saden. A thermodynamic formalism based on the fundamental relationand the Legendre transformation. International Journal of Mechanical Sciences,8(1):13–24, 1966.

J. Kestin. A Course in Thermodynamics, volume 1. McGraw-Hill Book Company,New York, NY, 1979.

I. Samohýl. Thermodynamics of Irreversible Processes in Fluid Mixtures. B.G.Teubner Verlagsgesellschaft, Leipzig, 1987.

7

Thermoelastic solids

We have shown that the constitutive functional of an objective simple homogeneoussolid in the current configuration is given by (5.99):

T (x, t) = F0≤s<∞

F(x, t), (t)C(t)(x, s), θ(t)(x, s), (t)g(t)(x, s). (7.1)

From (5.96) and (5.97), and suppressing spatial dependencies, we note that

(t)C(t)(s) = (FT (t))−1 ⋅C(t)(s) ⋅F−1(t), (7.2)

(t)g(t)(s) = (FT (t))−1 ⋅G(t)(s). (7.3)

Now, assuming continuous material derivatives with respect to s at s = 0, we canwrite

C(t)(s) = C(t) − C(t) s +⋯, (7.4)

θ(t)(s) = θ(t) − θ(t) s +⋯, (7.5)

G(t)(s) = G(t) − G(t) s +⋯, (7.6)

where we recall that C(t) = FT (t) ⋅F(t) and G(t) = FT (t) ⋅ g(t), so that

(t)C(t)(s) = (FT (t))−1 ⋅ [C(t) − C(t) s +⋯] ⋅F−1(t)= 1 − (FT (t))−1 ⋅ [FT (t) ⋅F(t) +FT (t) ⋅ F(t)] ⋅F−1(t) s +⋯= 1 − [(F(t) ⋅F−1(t))T + (F(t) ⋅F−1(t))] s +⋯= 1 − [LT (t) +L(t)] s +⋯= 1 − 2 D(t) s +⋯, (7.7)

and

(t)g(t)(s) = (FT (t))−1 ⋅ [G(t) − G(t) s +⋯]= g(t) − (FT (t))−1 ⋅ [FT (t) ⋅ g(t) +FT (t) ⋅ g(t)] s +⋯= g(t) − [(F(t) ⋅F(t)−1)T ⋅ g(t) + g(t)] s +⋯= g(t) − [LT (t) ⋅ g(t) + g(t)] s +⋯. (7.8)

283

284 THERMOELASTIC SOLIDS

Subsequently, using (7.5), (7.7), and (7.8) in (7.1), and considering a material withno memory (p = q = r = 0), we obtain the following reduced constitutive equationfor a simple homogeneous solid (see (5.39)):

T = T (F, θ,g) . (7.9)

Note that there is no loss of generality in expressing this relation as a function ofg or G since they are related by G = FT

⋅ g and since the constitutive functionalready depends on F.

A homogeneous solid material which only remembers its natural state is calleda thermoelastic solid and is subsequently defined by

ψ = ψ(F, θ,g), (7.10)

η = η(F, θ,g), (7.11)

q = q(F, θ,g), (7.12)

h = h(F, θ,g), (7.13)

σ = σ(F, θ,g). (7.14)

The above equations are required to satisfy the following frame-invariance condi-tions:

ψ (Q ⋅F, θ,Q ⋅ g) = ψ (F, θ,g) , (7.15)

η (Q ⋅F, θ,Q ⋅ g) = η (F, θ,g) , (7.16)

q (Q ⋅F, θ,Q ⋅ g) = Q ⋅ q (F, θ,g) , (7.17)

h (Q ⋅F, θ,Q ⋅ g) = Q ⋅h (F, θ,g) , (7.18)

σ (Q ⋅F, θ,Q ⋅ g) = Q ⋅σ (F, θ,g) ⋅QT , (7.19)

for all Q ∈ O(V ). Material symmetry further requires that ψ, η, q, h, and σ besuch that H ∈ Gκ ⊆ O(V ).

A thermoelastic material is a highly idealized material that has perfect memoryof only its natural or preferred state. This material remembers precisely thatstate and when the forces maintaining a different state are removed, it alwaysreturns to its configuration in its natural state. The deformation history of allintermediate states leaves no trace on its memory. Thus, as evident from theabove, the constitutive equations of thermoelastic materials depend only on thepresent deformation gradients relative to the natural state.

7.1 Clausius–Duhem inequality

We restrict the material response functions to only those that satisfy the moregeneral Clausius–Duhem inequality (5.252). For the moment, we define the vectorquantity

K ≡ h − q

θ=K (F, θ,g) , (7.20)

which satisfies the frame-invariance condition

K (Q ⋅F, θ,Q ⋅ g) =Q ⋅K (F, θ,g) (7.21)

7.1. CLAUSIUS–DUHEM INEQUALITY 285

since such condition is satisfied by both q and h. It should be noted that theconstitutive quantities in this case are equivalently given by

C = ψ, η,q,K,σ (7.22)

and are functions of the independent basic fields

I = F, θ,g . (7.23)

Now we can write

ψ = ∂ψ

∂FkK

˙FkK +

∂ψ

∂θθ +

∂ψ

∂θ,k

˙θ,k, (7.24)

and

Kk,k = ∂Kk

∂FlL

FlL,k +∂Kk

∂θθ,k +

∂Kk

∂θ,lθ,lk, (7.25)

and since ˙FkK = vk,lFlK and FlL,k = FlL,KF

−1

Kk, (5.252) becomes

−γv θ ≡ ρ(η + ∂ψ∂θ) θ − (σkl − ρ ∂ψ

∂FkK

FlK)vk,l + ρ ∂ψ∂θ,k

˙θ,k −

1

2θ (F −1Kk

∂Kk

∂FlL

+F −1Lk

∂Kk

∂FlK

)FlL,K −1

2θ (∂Kk

∂θ,l+∂Kl

∂θ,k) θ,lk +

(qkθ− θ

∂Kk

∂θ) θ,k + ρθ (b − r

θ) ≤ 0. (7.26)

In writing the above, we have accounted for the symmetry of second derivativesFlL,K and θ,lk. The above inequality must hold for every thermodynamic process,which means that special cases may be chosen which might result in further re-strictions on the constitutive functions C. Pursuing this fact, it is clear that I canbe independently chosen (after all, that is why they are called independent fields).(Note that for given ρ0(X), by choosing F we are also choosing ρ > 0, since F isnon-singular.) After a choice, it should be clear that C is fixed. Furthermore, thetime and space derivatives of the independent fields, i.e.,

a ≡ θ,gradv, g,GradF,gradg , (7.27)

may then be arbitrarily chosen to be of any magnitude or sign. This is the casebecause we are considering only simple homogenous materials (P = Q = 1 in (5.30);see also (5.33)) with no memory (p = q = r = 0 in (5.38); see also (5.39)). Notethat, for fixed F, choosing gradv is equivalent to choosing F. First consider theequilibrium state where F = 1, θ = θ0,g = 0, where θ0 > 0 is an arbitrary constanttemperature. In this case, from (7.27), we see that a = 0, and thus from (7.26),we must have

(b − r

θ0) ≤ 0. (7.28)

Since r can be chosen in general to be of any magnitude and sign, in order for theinequality to be always satisfied for any θ0, we must have that

b = rθ. (7.29)

286 THERMOELASTIC SOLIDS

If we now define the vector

α ≡ ρ(η + ∂ψ∂θ) ,−(σkl − ρ ∂ψ

∂FkK

FlK) , ρ ∂ψ∂θ,k

,

−1

2θ (F −1Kk

∂Kk

∂FlL

+F −1Lk

∂Kk

∂FlK

) ,−12θ (∂Kk

∂θ,l+∂Kl

∂θ,k) (7.30)

and the scalar

β ≡ (qkθ− θ

∂Kk

∂θ) θ,k, (7.31)

and use the result (7.29), then the inequality (7.26) can be rewritten as

α ⋅ a + β ≤ 0. (7.32)

Since the inequality is linear in a, and since the variables in a can take on valuesof any magnitude and sign, one would be able to violate (7.32) unless

α = 0 and β ≤ 0. (7.33)

More explicitly, this provides

η = −∂ψ∂θ, (7.34)

σkl = ρ ∂ψ

∂FkK

FlK , (7.35)

∂ψ

∂θ,k= 0, (7.36)

(F −1Kk

∂Kk

∂FlL

+F −1Lk

∂Kk

∂FlK

) = 0, (7.37)

(∂Kk

∂θ,l+∂Kl

∂θ,k) = 0, (7.38)

and

(qkθ− θ

∂Kk

∂θ) θ,k ≤ 0. (7.39)

This last inequality is known as the residual entropy inequality. It is straightfor-ward to show that the constraint (7.37) can be rewritten in the following form ifit is multiplied by J , and (3.60) and (3.64) are used:

[∂(JF −1KkKk)∂FlL

+∂(JF −1LkKk)

∂FlK

] = 0. (7.40)

7.1. CLAUSIUS–DUHEM INEQUALITY 287

Now the constraint differential equations (7.38) and (7.40) are of the form

( ∂fi∂yj+∂fj

∂yi) = 0. (7.41)

It can be readily verified that the general solution of this equation is

fi = Ωijyj + ωi, (7.42)

with Ωij = −Ωji and ωi an axial vector independent of yk. Using this result, thesolution of (7.38) and (7.40) is subsequently given by

K = h − q

θ= J−1F ⋅ [Ω(θ) ⋅FT

⋅ g +ω(θ)] , (7.43)

where Ω(θ) = −ΩT (θ) is an arbitrary skew-symmetric rank 2 tensor and ω(θ) isan arbitrary axial vector. Note that at this point, K is not necessarily zero and,subsequently, h does not necessarily equal q/θ! However, most importantly, wenote that there are no objective vectors or skew-symmetric rank 2 tensors that areonly functions of θ, thus

ω(θ) = 0 and Ω(θ) = 0 (7.44)

in (7.43). Subsequently, we have that

h = q

θ. (7.45)

With the above results, we can now rewrite

ψ = ψ(F, θ), (7.46)

η = −∂ψ

∂θ, (7.47)

q = q(F, θ,g), (7.48)

h = h(F, θ,g) = q

θ, (7.49)

σ = ρ∂ψ

∂F⋅FT = ρF ⋅ ∂ψ

∂F, (7.50)

where (7.50) follows from the symmetry of the stress tensor, and using (7.39),(7.26) results in the reduced entropy inequality

−γv θ2 ≡ q ⋅ g ≤ 0. (7.51)

By comparing (7.50) with (5.277), we note that the strain energy function τ as-sociated with the elastic part of the stress tensor σe is nothing more than therecoverable part of the Helmoltz free energy density function. Here, the differen-tial of the free energy density function (7.46), using (7.47) and (7.50), is now givenby

dψ = 1

ρσ ⋅ (FT )−1 ⋅ dF − η dθ, (7.52)

288 THERMOELASTIC SOLIDS

and subsequently, using the definition of the Helmholtz free energy density (5.201),we have the Gibbs equation for thermoelastic materials

de = 1

ρσ ⋅ (FT )−1 ⋅ dF + θ dη. (7.53)

Equations (7.52) and (7.53) should be compared to (5.206) and (5.139).Now the frame-invariance conditions (7.15)–(7.19) require the reduced constitu-

tive relations to be (see (5.94)–(5.95))

ψ = ψ(C, θ), (7.54)

η = −∂ψ

∂θ, (7.55)

q = q(C, θ,G), (7.56)

h = h(C, θ,G) = q

θ, (7.57)

σ = 2ρC ⋅∂ψ

∂C⋅C, (7.58)

where

q(C, θ,G) = FT⋅ q(F, θ,g), (7.59)

h(C, θ,G) = FT⋅ h(F, θ,g), (7.60)

σ(C, θ,G) = FT⋅σ(F, θ,g) ⋅F, (7.61)

are the convected heat flux, entropy flux, and stress tensor. The reduced entropyinequality (7.51) is now given by

−γv θ2 ≡ q(F, θ,g) ⋅ g = q(C, θ,G) ⋅G ≡ A(C, θ,G) ≤ 0 (7.62)

and is called Fourier’s inequality. It states that the angle between a nonzerotemperature gradient and a nonzero heat flux is greater than or equal to 90, orthat heat flows from a high temperature to a lower temperature.

Recall that a state with no entropy production is a thermodynamic equilibriumstate. We now see that, for fixed independent variables I, A is maximum (or γvis a minimum) when G = 0, i.e., at A(C, θ,0). Subsequently, we must have

∂A

∂θ,J∣G=0

= [qJ + ∂qI∂θ,J

θ,I]G=0

= 0 (7.63)

or

qJ(C, θ,0) = 0, (7.64)

and

∂2A

∂θ,K∂θ,J∣G=0

= [ ∂qJ∂θ,K

+∂qK∂θ,J

+∂2qI

∂θ,K∂θ,Jθ,I]∣

G=0

= [ ∂qJ∂θ,K

+∂qK∂θ,J]G=0

≤ 0.(7.65)

When we have no deformation, F =R, J = 1, and C = 1. Then

ψ = ψ(θ), e = e(θ), q = q(θ,g), h = q

θ(θ,g), σ = 0, (7.66)

7.2. MATERIAL SYMMETRIES 289

and the balance equations of mass, momentum, and energy become

ρ = ρR, a = f , ρR cv θ = −div q + ρR r, (7.67)

where cv = ∂e/∂θ∣v is the specific heat at constant volume (see (5.165)). Notethat the linear momentum is just Newton’s equation of motion, while the energyequation is just the equation of heat conduction with an external heat source.

In passing, we note that some materials are defined as elastic in a special sense.In this regard, a material is said to be a hyperelastic material if the stress tensorcan be represented through an energy function, such as in (7.50), where ψ is theHelmholtz free energy function. In the special case where ψ is only a function ofthe deformation gradient or some strain tensor, the Helmholtz free energy functionis referred to as the stored energy function or strain energy function, such as in(7.46) or (5.277). In contrast, a hypo-elastic material is distinct from a hyper-elastic material in that, except under special circumstances, it cannot be derivedfrom a (scalar-valued) energy function.

7.2 Material symmetries

To further reduce the constitutive equations, it is first necessary to consider sym-metries of materials. Material symmetry requires that the constitutive func-tions (7.54)–(7.61) for a homogeneous solid must be symmetric with respect toH ∈ Gκ ⊆ O(V ) for all (C, θ,G). Such transformations are applied to the undis-torted reference state and makes the resulting configuration indistinguishable fromthe original configuration. Different transformation subgroups distinguish the dif-ferent material classes.

Here we will discuss anisotropic solids that satisfy crystal symmetries (32 crystalclasses forming the crystallographic group) and transverse isotropy in which anyrotation of material coordinate axes about a preferred direction does not alter thematerial properties. Most laminated materials exhibit this type of anisotropy.

All possible crystal classes are described by six parameters: unit cell translationvectors (a,b,c) and angles (α,β, γ), as illustrated in Fig. 7.1. The specific choicesof these parameters are indicated in Table 7.1. The possible lattice variationsthat are obtained are called primitive (P), face-centered (F), body-centered (I), andbase-centered (C). These give rise to 14 space lattices called Bravais lattices intowhich all crystal structures fall. The space lattices are generated by the translationvector

t =ma + nb + pc, (7.68)

where m, n, and p are integers. The magnitudes of the translation vectors (a,b,c)are called lattice parameters, which, together with the lattice angles (α,β, γ), definethe unit cell of a crystal.

The symmetry of a crystal is related to the symmetry of its physical properties.A fundamental postulate of crystal physics is known as Neumann’s principle orpostulate:

Neumann’s Principle: The symmetry elements of any physical property ten-sor of a crystal must include all the symmetry elements of the point group of thecrystal.

290 THERMOELASTIC SOLIDS

a

b

c

X2

X3

αβ

γ

X1

Figure 7.1: General space lattice showing translation vectors and angles.

The Neumann principle does not state that the symmetry elements of a physicalproperty are the same as those of the point group. It just says that the symmetryelements of a physical property must include those of the point group. The physicalproperties may, and often do, possess more symmetry than the point group. Forexample, the elasticity property tensor relates the field tensors of stress and strain.The symmetries of both stress and strain tensors require that the elasticity tensorpossess higher symmetries than that possessed by the crystal.

It is well known from group theory that for various crystal classes, every sym-metry operation may be deduced from a few basic symmetry operations. Theapplication of matrices corresponding to these basic operations (the generatingtransformations) are sufficient to obtain the effect due to the symmetry of a crys-tal class on a given property tensor. Table 7.2 lists the generating matrices andTable 7.3 summarizes the symmetries for the crystal classes. All n symmetrytransformations for each of the 32 crystal classes are given in the table.

In Table 7.2, I is the identity transformation, C is the central inversion, Rn

(n = 1,2,3) is the reflection in a plane normal to Xn, Dn rotates the coordinatesystem through 180 about the Xn-axis, Tn is a reflection in the plane through theXn-axis bisecting the angle between the other two axes, M1 and M2 are rotationsof the axes through 120 and 240, respectively, about a line passing through theorigin and the point (1,1,1), and the transformations S1 and S2 are rotations ofthe axes through 120 and 240, respectively, about the X3-axis.

Now, for every property tensor of any rank, the tensor components must satisfyall symmetries associated with a particular crystal class. The procedure that oneuses is as follows. Let’s assume that the rank-2 symmetric property tensor A

relates the two vector fields q and g:

q =A ⋅ g. (7.69)

Now if the material is symmetric with respect to an orthogonal transformation H,then we must have that the property tensor does not change when both q and g

are subjected to the orthogonal symmetry transformation H:

(HT⋅ q) =A ⋅ (HT

⋅ g) ,or

q = (H ⋅A ⋅HT ) ⋅ g. (7.70)

7.2. MATERIAL SYMMETRIES 291

Subsequently, the property tensor must satisfy

A = (H ⋅A ⋅HT ) . (7.71)

Example

To illustrate the method, a 90 rotation about the X3-axis (H = R2T3) isa symmetry element of a tetragonal-pyramidal (Class # 9) material (seeTable 7.3):

A = [aij] =⎡⎢⎢⎢⎢⎢⎣a11 a12 a13a21 a22 a23a31 a32 a33

⎤⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎣

0 1 0

−1 0 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣a11 a12 a13a21 a22 a23a31 a32 a33

⎤⎥⎥⎥⎥⎥⎦⎡⎢⎢⎢⎢⎢⎣0 −1 0

1 0 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦=⎡⎢⎢⎢⎢⎢⎣a22 −a21 a23−a12 a11 −a13a32 −a31 a33

⎤⎥⎥⎥⎥⎥⎦.

The above equality implies that a11 = a22, a21 = −a12, a13 = a23 = 0,a31 = a32 = 0, and a33 = a33. But since A is assumed symmetric, we alsohave that a12 = a21 = 0. Subsequently, we have that

A = [aij] =⎡⎢⎢⎢⎢⎢⎣a11 0 0

0 a11 0

0 0 a33

⎤⎥⎥⎥⎥⎥⎦.

The other symmetry transformations for the tetragonal-pyramidal class(D3 and R1T3) do not reduce the components of A any further. This resultprovides the most general symmetric property tensor for this material class(see Table 7.4).

To arrive at the symmetries of the components, considerable time can be savedby using the method of direct inspection of Fumi, which is a modified procedurethat one can use to arrive at the same result obtained as described above. In thismethod, one can deduce the value of a transformed property tensor component byinspection. However, we should note that this method is not directly applicableto crystal systems whose symmetries include generating matrices S1 and S2 (seeTables 7.2 and 7.3).

Consider the components of a rank-2 tensor [aij]. It is clear that each subscriptin the tensor component is associated with a coordinate direction. Specifically,using the bases, the tensor is fully represented by A = aijeiej , and associatedwith each basis, we have the corresponding coordinates Xi and Xj. To examinematerial symmetries, we apply all coordinate transformations corresponding tosymmetry transformations of a specific crystal class. Subsequently, let’s examinethe product of the coordinate pair XiXj . If we apply a coordinate transformation,the new coordinate pair would become X ′iX

′

j . Now assuming that Hij corresponds

292 THERMOELASTIC SOLIDS

to one of the crystal class symmetries, then we have

X ′iX′

j = (HikXk) (HjlXl) =HikHjlXkXl. (7.72)

Now the components of the objective property tensor A must transform as

aij =QikQjlakl, (7.73)

which is analogous to the way in which the product of coordinates transforms.Note that they are not identical since interchanging Hik and Hjl does not changethe value of the product of the coordinates, but doing so for aij has implicationsassociated with bases in different directions. Therefore, the elements of a tensortransform, upon a change of coordinate axes, in exactly the same way as theproduct of corresponding coordinates provided we maintain the correct order ofterms.

Example

Here we repeat the previous example, but use the modified procedure toarrive at the same result. That is, to illustrate the method, a 90 rotationabout the X3-axis (H = R2T3) is a symmetry element of a tetragonal-pyramidal (Class # 9) material (see Table 7.3). Now the material symmetryfor this crystal class requires that

X ′1 = X2

X ′2 = −X1

X ′3 = X3

or in a more concise way

X1 → −X2, X2 →X1, X3 →X3.

Now to determine the new value for, say, a12, we evaluate

X ′1X′

2 =X2 (−X1) .As the tensor elements transform like the product of coordinates, we seeby inspection that

a′12 = −a21;that is, upon this change of axes, the number that appears in the subscriptsof the new tensor a′ij is the negative of the number that appeared in thesubscripts of aij . Using the above procedure, it easily follows that

11 → 22 12 → −21 13 → −23

21 → −12 22 → 11 23 → 13

31 → −32 32 → 31 33 → 33

that is

a11 → a22 a12 → −a21 a13 → −a23a21 → −a12 a22 → a11 a23 → a13a31 → −a32 a32 → a31 a33 → a33

7.2. MATERIAL SYMMETRIES 293

Then, as a consequence of Newmann’s principle that every componentshould transform into itself, we must have that a11 = a22, a21 = −a12,and a13 = a31 = a23 = a32 = 0. But, since A is assumed to be symmetric, wemust also have that a12 = a21 = 0 and subsequently, we obtain

A = [aij] =⎡⎢⎢⎢⎢⎢⎣a11 0 0

0 a11 0

0 0 a33

⎤⎥⎥⎥⎥⎥⎦.

As before, the other symmetry transformations for the tetragonal-pyramidal class (D3 and R1T3) do not reduce aij further.The method is directly applicable to tensors of all ranks. For example, ifwe wish to determine the new value of, say, the elastic stiffness elementc1213 for this same material, we would examine the transformation of theproduct X1X2X1X3:

X ′1X′

2X′

1X′

3 =X2 (−X1)X2X3;

then the value of c′1213 is thus −c2123.

As noted, we can use either of the above procedures to write the general formsof property tensors of varying ranks for different crystal classes. For example,consider the property tensor A of rank 2 that relates two field vectors q and g:

qi = aij gj, i, j = 1,2,3. (7.74)

If q is the heat flux and g is the temperature gradient, then if we take aij → −kij ,the symmetric property tensor K corresponds to the thermal conductivity. It isalso possible to have a rank-2 property tensor A, which relates a scalar field θ toa rank-2 tensor field e:

eij = aij ∆θ, i, j = 1,2,3. (7.75)

For example, if e is the linear strain tensor and ∆θ is the temperature differ-ence, then the symmetric property tensor A corresponds to the thermal expansiontensor. Now, requiring that the material tensor component matrix satisfies allsymmetries corresponding to each crystal class, in addition to the symmetry dic-tated by physical restrictions on the field tensors, results in the reduced forms ofthe component matrices shown in Table 7.4.

In providing a comprehensive example for a rank-4 tensor, below we introducea convenient notation due to Voigt that will allow us to write the generally sparseproperty tensor components into convenient smaller matrices. The sparsity arisesfrom the fact that quite often, one or both of the field tensors that the propertytensor relates will satisfy symmetry properties arising from physical restrictions.

Consider the property tensor of rank 4, C, that relates two symmetric rank-2field tensors σ and e:

σij = cijkl ekl i, j, k, l = 1,2,3. (7.76)

Since σij = σji and ekl = elk, this requires that cijkl = cmn, which permits us towrite in condensed form as

σm = cmn en m,n = 1, . . . ,6, (7.77)

294 THERMOELASTIC SOLIDS

thereby taking direct advantage of the reduction in the number of componentsfrom 81 to 21. To convert between the tensor notation (7.76) and the matrixnotation (7.77), we adopt the following conventions:

⎡⎢⎢⎢⎢⎢⎣σ11 σ12 σ13

σ22 σ23Sym σ33

⎤⎥⎥⎥⎥⎥⎦⇐⇒

⎡⎢⎢⎢⎢⎢⎣σ1 σ6 σ5

σ2 σ4Sym σ3

⎤⎥⎥⎥⎥⎥⎦, i.e., σij ⇔ σm, (7.78)

⎡⎢⎢⎢⎢⎢⎣e11 e12 e13

e22 e23Sym e33

⎤⎥⎥⎥⎥⎥⎦⇐⇒

⎡⎢⎢⎢⎢⎢⎣e1

1

2e6

1

2e5

e21

2e4

Sym e3

⎤⎥⎥⎥⎥⎥⎦, i.e., eij ⇔ em if i = j,

1

2em if i ≠ j,

(7.79)and

cijkl⇔ cmn. (7.80)

Example

To write the tensor components σ21 in the matrix representation, we pro-ceed as follows:

σ21 = c2111e11 + c2122e22 +⋯+ c2123e23 + c2132e32 +⋯, (7.81)

σ6 = c2111e1 + c2122e2 +⋯ + c2123( 12e4) + c2132( 12e4) +⋯, (7.82)

σ6 = c61e1 + c62e2 +⋯+1

2c64e4 +

1

2c64e4 +⋯, (7.83)

σ6 = c61e1 + c62e2 +⋯+ c64e4 +⋯. (7.84)

The reverse representation of σ6 from matrix form to tensor componentform σ21 is given by

σ6 = c61e1 + c62e2 +⋯+ c64e4 +⋯, (7.85)

σ6 = c61e1 + c62e2 +⋯+1

2c64e4 +

1

2c64e4 +⋯, (7.86)

σ21 = c61e11 + c62e22 +⋯+ c64e23 + c64e32 +⋯, (7.87)

σ21 = c2111e11 + c2122e22 +⋯ + c2123e23 + c2132e32 +⋯. (7.88)

Now, requiring that the material tensor component matrix C satisfy all sym-metries corresponding to each crystal class results in the reduced forms of thecomponent matrices using Voigt’s notation shown in Table 7.5.

The other class of materials that we would like to discuss are those that havetransverse isotropy. A material with a single preferred direction which is the sameat every point is said to be transversely isotropic. For such material, constitutiverelations are invariant under rotations about the preferred direction. If the X3

direction is chosen as the preferred direction as illustrated in Fig. 7.2, then suchrotations are represented by the matrix

MΘ =⎡⎢⎢⎢⎢⎢⎣

cosΘ sinΘ 0

− sinΘ cosΘ 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦, 0 ≤ Θ ≤ 2π. (7.89)

7.2. MATERIAL SYMMETRIES 295

X3

X1

X2

Figure 7.2: Laminated material geometry.

Now, five cases can arise depending on whether or not certain other transforma-tions are permitted. These cases are characterized by the transformation groupsgenerated by the following matrices: (i) MΘ, (ii) MΘ, R1, (iii) MΘ, R3, (iv) MΘ,D2, and (v) MΘ, R1, R3, D2, where the transformations R1, R3, D2 are givenin Table 7.2. Case (i), in which only rotations about the preferred direction areallowed, is sometimes said to characterize rotational symmetry. Now, in view ofthe requirement to satisfy the symmetries (MΘ,R1,R3,D2), it is easy to showthat the corresponding components of the rank-2 and rank-4 property tensors aregiven as shown in Tables 7.4 and 7.5. Note that we respectively have 2 and 5

independent parameters in the corresponding component matrices of the rank-2and rank-4 property tensors of transversely isotropic materials.

An important point that needs to be made is that the symmetry of a crystaldepends on the state of the crystal. We note that our assessment of the materialsymmetries have been based on the state of the material without considering ex-ternal influences (symmetries of property tensors at equilibrium). If, due to someexternal influence (represented by field tensors), there is a change in the state ofthe crystal, there may also be a change in the crystal symmetry. The symmetryof a given state of the crystal may be determined using the Curie principle fromthe symmetry of the crystal free of any external influence (symmetry of prop-erty tensors) and from the symmetry of the external influences (symmetry of fieldtensors).

Curie’s Principle: A crystal under an external influence will exhibit only thosesymmetry elements that are common to the crystal without the influence and thoseof the influence without the crystal.

In the above example, and in Tables 7.4 and 7.5, the symmetries associatedwith external influences have been used through the symmetry assumptions of thecorresponding field tensors.

296 THERMOELASTIC SOLIDS

7.3 Linear deformations of anisotropic materials

For the subsequent discussion, we consider the constitutive equations (7.46)–(7.48),(7.50), and (7.45). Due to the requirement of frame indifference, the dependenceof the Helmholtz free energy density and heat flux on F must reduce to the depen-dence on the right Cauchy–Green strain tensor C (see (7.54)–(7.45)). However,in the linear limit, since C = (1 − 2E) and the left Cauchy–Green strain tensoris B = (1 − 2e)−1 = 1 + 2e + O(e2) (see (3.178)), for convenience in arriving atcorresponding results for isotropic materials, we write the linear approximation interms of the Almansi–Hamel strain tensor e, i.e.,

ψ = ψ(θ,e) and q = q(θ,e,g), (7.90)

where, in relating (7.90) to (5.201), (5.206), and (5.207), we note that the specificthermostatic volume is related to the linear strain tensor by ν1 → e/ρ0, whereρ0 is a constant reference density, which we take here to be that at the referencestate, and the thermostatic tension is related to the stress tensor, τ 1 → σ. Wenow assume that the strain e, the temperature difference θ = θ − θ0, and thetemperature gradient g are small, where θ0 is the temperature of the body at thereference state. Since we address small deformations from the reference state, wewrite the following expansions of the Helmholtz free energy density and heat fluxabout the reference state (see (3.189), (3.190), and (3.192)):

ψ(θ,e) = ψ0 − η0 θ +1

ρ0σ0 ∶ e −

ce0

2 θ0θ2

+1

ρ0e ∶ β0 θ +

1

2ρ20

e ∶ ξ0 ∶ e +⋯ (7.91)

and

q(θ,e,g) = q0 − i0 θ +1

ρ0j0 ∶ e − k0 ⋅ g +⋯, (7.92)

where we have used the following quantities evaluated at the reference state (θ =θ0,e = 0,g = 0), which is denoted by the zero subscript:

ψ0 = ψ (θ = θ0,e = 0) , (7.93)

η0 = −∂ψ

∂θ∣θ=θ0,e=0

, (see (5.207)) (7.94)

σ0 = ρ0∂ψ

∂e∣θ=θ0,e=0

, (see (5.207)) (7.95)

ce0= − θ0

∂2ψ

∂θ2∣θ=θ0,e=0

= θ0 ∂η∂θ∣θ=θ0,e=0

, (see (5.209)) (7.96)

β0 = ρ0∂2ψ

∂e∂θ∣θ=θ0,e=0

= −ρ0 ∂η∂e∣θ=θ0,e=0

= ∂σ∂θ∣θ=θ0,e=0

, (see (5.211)) (7.97)

7.3. LINEAR DEFORMATIONS OF ANISOTROPIC MATERIALS 297

ξ0 = ρ20∂2ψ

∂e2∣θ=θ0,e=0

= ρ0 ∂σ∂e∣θ=θ0,e=0

, (see (5.150)) (7.98)

q0 = q (θ = θ0,e = 0,g = 0) , (7.99)

i0 = −∂q

∂θ∣θ=θ0,e=0,g=0

, (7.100)

j0 = ρ0∂q

∂e∣θ=θ0,e=0,g=0

, (7.101)

k0 = −∂q

∂g∣θ=θ0,e=0,g=0

. (7.102)

We now note that the reference state is an equilibrium state and thus ψ(θ,e) ≥ψ0 ≥ 0, σ0 = 0, q0 = 0, i0 = 0, and j0 = 0. Furthermore, we recognize that

β0 = −ξ0 ∶ α0, (see (5.151)) (7.103)

α0 = 1

ρ0

∂e

∂θ∣θ=θ0,e=0

. (see (5.150)) (7.104)

Subsequently, if we truncate the expansion for the Helmholtz free energy densityto second order and the heat flux to first order, the constitutive equations become

ψ = ψ0 − η0 θ −ce0

2 θ0θ2

−1

ρ0e ∶ ξ0 ∶ α0 θ +

1

2ρ20

e ∶ ξ0 ∶ e, (7.105)

η = −∂ψ

∂θ= η0 + ce0

θ0θ +

1

ρ0e ∶ ξ0 ∶ α0, (7.106)

q = −k0 ⋅ g, (7.107)

h = −k0

θ0⋅ g, (7.108)

σ = ρ0∂ψ

∂e= 1

ρ0ξ0 ∶ (e − ρ0α0 θ) . (7.109)

Now, defining the linear thermal and mechanical strain tensors

eθ = ρ0α0 θ = a0 θ and eM = e − eθ, (7.110)

we can rewrite the above constitutive equations in their final form:

ψ = ψ0 − η0 θ −ce0

2 θ0θ2 −

1

ρ0e ∶C0 ∶ a0 θ +

1

2ρ0e ∶C0 ∶ e

= ψ0 − η0 θ −cσ0

2 θ0θ2

+1

2ρ0eM ∶C0 ∶ eM , (7.111)

η = η0 +ce0

θ0θ +

1

ρ0e ∶C0 ∶ a0

= η0 +cσ0

θ0θ +

1

θeM ∶C0 ∶ eθ, (7.112)

q = −k0 ⋅ g, (7.113)

h = −k0

θ0⋅ g, (7.114)

σ = C0 ∶ (e − a0 θ) =C0 ∶ eM . (7.115)

298 THERMOELASTIC SOLIDS

In the above equations, ce0is the specific heat capacity at constant strain, cσ0

isthe specific heat capacity at constant stress (see (5.215)) given by

cσ0− ce0

= θ0α0 ∶ ξ0 ∶ α0 = θ0ρ0

a0 ∶C0 ∶ a0, (7.116)

a0 and β0 are the symmetric second rank thermal expansion and thermal stresstensors related by

β0 = −C0 ∶ a0 (7.117)

and satisfying the symmetry conditions

(a0)ij = (a0)ji and (β0)ij = (β0)ji, (7.118)

C0 = 1

ρ0ξ0 (7.119)

is the fully symmetric rank-4 elastic stiffness tensor satisfying the following re-quirements due to the symmetries of the stress and mechanical strain tensors,

(c0)ijkl = (c0)jikl = (c0)jilk = (c0)ijlk = (c0)klij , (7.120)

and k0 is the rank-2 thermal conductivity tensor for which we note that, using(7.113) in the residual inequality (7.62), the symmetric part of k0 must be positivesemi-definite:

symk0 ≥ 0. (7.121)

Now, the thermal conductivity can be decomposed into symmetric and skew-symmetric parts, and it can be shown that the divergence of the heat flux com-ponent associated with the skew-symmetric part does not contribute to heat flowand thus is unmeasurable or not observable. Based on this fact, it is generallyassumed that the thermal conductivity tensor is symmetric:

k0 = kT0 and k0 ≥ 0. (7.122)

It should be pointed out that the symmetries in the material tensors reduce thenumber of independent components from 9 to 6 for a0, β0, and k0, and from 81to 21 for C0.

Lastly, the stability condition that the internal energy must be a minimumat equilibrium requires that the matrix (5.257) be positive. The first conditionresulting from this requires that

ce0> 0. (7.123)

The second condition of φ0 > 0, using (5.137) and (7.115), requires that the isother-mal elastic stiffness tensor be positive definite

C0 > 0. (7.124)

Again, since C0 is a real symmetric matrix, this condition requires that all theeigenvalues of C0 be real and positive definite, or alternately, all the determinantsof the principal minors of C0 be positive definite (see (5.258)).

The specific requirements on the components of C0 depend on the materialcrystal class. In general, such restrictions are fairly complex, but, first translating

7.3. LINEAR DEFORMATIONS OF ANISOTROPIC MATERIALS 299

C0 to Voigt’s notation, we see from Table 7.5 that progress can be made for thematerial classes of orthorhombic, tetragonal (Classes # 12–15), hexagonal, cubic,hemitropic and isotropic since in those cases, the component matrix of C0 hasnonzero elements only along the main diagonal and in the principal 3 × 3 minor.For example, consider a material having cubic crystal symmetry. From Table 7.5,we see that the eigenvalues of [C0] are easily found:

λ1 = λ2 = (c0)11 − (c0)12, λ3 = (c0)11 + 2 (c0)12, and λ4 = λ5 = λ6 = (c0)44.(7.125)

Equilibrium stability then requires that

(c0)11 − (c0)12 > 0, (c0)11 + 2 (c0)12 > 0, and (c0)44 > 0. (7.126)

Alternately, the determinants of the principal minors of [C0] lead to

(c0)11 > 0, (c0)211 − (c0)212 > 0, [(c0)11 − (c0)12]2 [(c0)11 + 2 (c0)12] > 0, (c0)44 > 0.(7.127)

These are equivalent to (7.126) since the first three inequalities in (7.127) can bereplaced by

(c0)11 + (c0)12 > 0, (c0)11 − (c0)12 > 0, (c0)11 + 2 (c0)12 > 0, (7.128)

which in turn can be replaced by the first two inequalities in (7.126). For thesecrystal classes, the positive semi-definite condition on the thermal conductivityalso requires that

(k0)11 ≥ 0, (k0)22 ≥ 0, and (k0)33 ≥ 0. (7.129)

For a cubic crystal, this reduces to the restriction that (k0)11 ≥ 0.7.3.1 Propagation of elastic waves in crystals

We consider only the case of isothermal conditions (θ = 0). The procedure isreadily extended to non-isothermal conditions.

Since the rank-2 stress and strain field tensors σ and e are symmetric, usingVoigt’s notation, we observe from Table 7.5 that the elasticity or stiffness propertytensor C relations are given by

σ =C ⋅ e, (7.130)

where σ = (σ1, σ2, σ3, σ4, σ5, σ6)T , e = (e1, e2, e3, e4, e5, e6)T , and, to simplify nota-tions, we have suppressed the zero subscripts indicating the reference density andthe linearity of the isothermal elastic stiffness tensor (not to be confused with theright Cauchy–Green strain tensor). As we want an equation which involves dis-placements, we substitute the definition of em in terms of displacements gradientshij = ui,j (see (3.192) and (7.79)):

e =T⋅ u, (7.131)

where u = (u1, u2, u3)T is the material displacement vector, and, using the abbre-viated notation ∂i = ∂/∂xi (i = 1,2,3),

≡ ⎛⎜⎝∂1 0 0 0 ∂3 ∂20 ∂2 0 ∂3 0 ∂10 0 ∂3 ∂2 ∂1 0

⎞⎟⎠ . (7.132)

300 THERMOELASTIC SOLIDS

Substituting (7.131) into (7.130), we obtain

σ =C ⋅ T⋅ u. (7.133)

Now using the linear momentum balance (4.109), assuming that we have nobody forces, using Voigt’s notation, and considering only linear terms, upon usingthe abbreviation ∂tt = ∂2/∂t2, we have

ρ∂ttu = ⋅σ, (7.134)

or, using (7.133),

ρ∂ttu = C ⋅ u, (7.135)

where we have defined the symmetric operator

C ≡ ⋅C ⋅ T . (7.136)

The system of partial differential equations (7.135) represents the equationsgoverning the displacement of a differential volume element. While the equationslook formidable, they are nothing more than wave equations for each componentof the displacement vector. These equations involve components of the propertytensor. Now we assume a solution in the form of a wave propagating in threedimensions, i.e.,

u(x, t) = ei(ωt−k⋅x)U, (7.137)

where ω is the wave frequency, k = (k1, k2, k3) is the wave vector corresponding tothe direction of wave propagation, and U = (U1, U2, U3)T is the displacement wavevector. Substituting this assumed solution form into the wave equation (7.135),we obtain

(k2A − ρω21) ⋅ U = 0, (7.138)

where

A = L ⋅C ⋅LT , (7.139)

L ≡ ⎛⎜⎝l1 0 0 0 l3 l20 l2 0 l3 0 l10 0 l3 l2 l1 0

⎞⎟⎠ , (7.140)

l = (l1, l2, l3)T = k/k is the wave direction cosine vector with k = ∣k∣, and U =(U1, U2, U3)T =U/∣U∣ is the displacement direction cosine vector, as illustrated inFig. 7.3. The wave (or phase) velocity is given by

cw = ωk= ωkl. (7.141)

Note that the crystal gives us C in one of the forms given in Table 7.5. For a

7.3. LINEAR DEFORMATIONS OF ANISOTROPIC MATERIALS 301

x1

x3

x2

cos−1 l1

cos−1

l3

cos−1 l2

x′2

x′3

x′1

l

cos−1 U2

cos−1 U1

cos−1

U3

U

Figure 7.3: The relationship between the wave and displacement direction cosinevectors.

triclinic crystal, the symmetric matrix A has components

a11 = c11l21 + c66l22 + c55l23 + 2 (c56l2l3 + c15l3l1 + c16l1l2) , (7.142)

a22 = c66l21 + c22l22 + c44l23 + 2 (c24l2l3 + c46l3l1 + c26l1l2) , (7.143)

a33 = c55l21 + c44l22 + c33l23 + 2 (c34l2l3 + c35l3l1 + c45l1l2) , (7.144)

a12 = c16l21 + c26l22 + c45l23 + (c46 + c25) l2l3 + (c14 + c56) l3l1 +(c12 + c66) l1l2, (7.145)

a13 = c15l21 + c46l22 + c35l23 + (c45 + c36) l2l3 + (c13 + c55) l3l1 +(c14 + c56) l1l2, (7.146)

a23 = c56l21 + c24l22 + c34l23 + (c44 + c23) l2l3 + (c36 + c45) l3l1 +(c25 + c46) l1l2. (7.147)

Next we pick the direction l (or k) in which we wish to propagate the wave.Subsequently, the wave frequency ω, and thus the wave speed cw = ω/k, and thedirection of the displacement vector U are the unknown variables. The aboveproblem corresponds to a standard eigenvalue problem with ρω2 (or ρ c2w) beingthe eigenvalue and U the eigenvector. A nontrivial solution (U ≠ 0) exists only if

Ω(ω,k) ≡ det [k2A(l1, l2, l3) − ρω21] = 0 (7.148)

(a trivial solution is impossible since U21 + U

22 + U

23 = 1). This gives rise to the

following characteristic cubic polynomial (dispersion relation) for ρω2,

Ω(ω,k) ≡ (k2a11 − ρω2) (k2a22 − ρω2) (k2a33 − ρω2) −k4 [k2 (a11a223 + a22a213 + a33a212 − 2a12a13a23) − ρω2 (a212 + a213 + a223)] = 0, (7.149)

302 THERMOELASTIC SOLIDS

which provides three wave speeds, say c(i)w = ω(i)/k, i = 1,2,3. For each wave speed,we can subsequently obtain the corresponding displacement direction cosine vectorU(i) from (7.138) in the form

U2

U1

= a23 (a11 − ρ c2w) − a12a13a13 (a22 − ρ c2w) − a12a23 and

U3

U1

= (a11 − ρ c2w) (a22 − ρ c2w) − a212a13 (a22 − ρ c2w) − a12a23 . (7.150)

Such displacements, which are functions of the phase velocity, propagation direc-tion, and elastic stiffness constants, will, in general, constitute quasi-longitudinaland quasi-shear waves. Numerical computations are required to obtain the wavespeeds and the displacement directions. However, general features of the disper-sion relation (7.149) can be deduced by considering special propagation directionsfor which the dispersion relation factors.

Another important aspect of elastic wave propagation in crystals is that thedirection of energy propagation may not coincide with the propagation directionl of the wave velocity. Indeed, it can be shown that energy in an elastic mediumpropagates with the group velocity

cg = ∂ω∂k= −1

k

∂Ω/∂l∂Ω/∂ω . (7.151)

The above results for a triclinic crystal can be specialized to the other crystalclasses by appropriately setting some of the elastic parameters to zero. Say thatwe are studying wave propagation in aluminum. Aluminum has a cubic crystalsymmetry so that c14 = c15 = c16 = 0, c24 = c25 = c26 = 0, c34 = c35 = c36 = 0,c45 = c46 = 0, c56 = 0, c11 = c22 = c33, c12 = c13 = c23, and c44 = c55 = c66, with specificvalues of the elastic parameters given by (c11, c12, c44) = (10.80,6.13,2.85)× 1010N/m2. In this case, it is easy to see that

A =⎡⎢⎢⎢⎢⎢⎣c11l

21 + c44 (1 − l21) (c12 + c44) l1l2 (c12 + c44) l3l1(c12 + c44) l1l2 c11l

22 + c44 (1 − l22) (c12 + c44) l2l3(c12 + c44) l3l1 (c12 + c44) l2l3 c11l

23 + c44 (1 − l23)

⎤⎥⎥⎥⎥⎥⎦. (7.152)

If the propagation direction in a cubic crystal is along any crystal axis, we obtain

Ω(ω,k) ≡ (k2c11 − ρω2) (k2c44 − ρω2)2 = 0, (7.153)

and the wave displacements are pure longitudinal and shear modes. Also for suchcrystal, it is easy to see that if the propagation direction is in the (x1, x3)-plane(l2 = 0), the dispersion relation (7.149) becomes a product of a linear term (in ρω2

or, equivalently, in ρ c2w) and a quadratic term

Ω(ω,k) ≡ [k2c44 − ρω2] ⋅[(c11k21 + c44k23) − ρω2] [(c11k23 + c44k21) − ρω2] − (c12 + c44)2 k21k23 = 0. (7.154)

Subsequently, we have that

c(1)w = (c44ρ)1/2 , (7.155)

c(2,3)w =⎡⎢⎢⎢⎢⎢⎣c11 + c44 ∓

√(c11 − c44)2 cos2 2φ + (c11 + c44)2 sin2 2φ2ρ

⎤⎥⎥⎥⎥⎥⎦1/2

, (7.156)

7.3. LINEAR DEFORMATIONS OF ANISOTROPIC MATERIALS 303

where cosφ = l1 (note that in this case l21+l23 = 1). Now it is easy to see that U(1) has

a particle velocity that is normal to the (x1, x3)-plane and is therefore normal to k.This is a pure shear wave polarized along x2. On the other hand, U(2,3) are quasi-shear and quasi-longitudinal waves, respectively, but they reduce to pure modesfor special propagation directions: φ = 0 (l3 = 0) or φ = π/2 (l1 = 0). Pure modesolutions also occur at φ = π/4 (l1 = l3 = 1/√2) and φ = 3π/4 (l1 = −l3 = −1/√2),in which case we respectively have

U1

U3

= ∓ (c12 + c44)(c11 + c44) − 2ρ c2w . (7.157)

But, from (7.155) and (7.156) for φ = π/4 and φ = 3π/4, we have that

c(1)w = (c44

ρ)1/2, pure shear

c(2)w = (c11 − c12

2ρ)1/2, pure shear

c(3)w = (c11 + c12 + 2 c44

2ρ)1/2, pure longitudinal

(7.158)

for these propagation directions. Substitution of (7.158) into (7.157) gives

( U1

U3

)(2,3) = ∓1 at φ = π4

and ( U1

U3

)(2,3) = ±1 at φ = 3π

4. (7.159)

Also, from (7.154), we have that

∂Ω

∂k1= 2 c11k1 (c11k23 + c44k21 − ρω2) + 2 c44k1 (c11k21 + c44k23 − ρω2) −

2 (c12 + c44)2 k1k23 , (7.160)

∂Ω

∂k2= 0, (7.161)

∂Ω

∂k3= 2 c44k3 (c11k23 + c44k21 − ρω2) + 2 c11k3 (c11k21 + c44k23 − ρω2) −

2 (c12 + c44)2 k21k3, (7.162)

∂Ω

∂ω= −2ρω [(c11 + c44)k2 − 2ρω2] . (7.163)

Substitution of these derivatives into (7.151) gives

(cg)1 = l1

ρ cw⋅

[c11 (c11l23 + c44l21 − ρ c2w) + c44 (c11l21 + c44l23 − ρ c2w) − (c12 + c44)2 l23][(c11 + c44) − 2ρ c2w] , (7.164)

(cg)2 = 0, (7.165)

(cg)3 = l3

ρ cw⋅

[c44 (c11l23 + c44l21 − ρ c2w) + c11 (c11l21 + c44l23 − ρ c2w) − (c12 + c44)2 l21][(c11 + c44) − 2ρ c2w] . (7.166)

304 THERMOELASTIC SOLIDS

In summary, (i) there are three different types of waves that are propagatedalong any given direction in the crystal; (ii) the speeds of these waves are differentand are functions of the wave direction l, the density ρ, and the elastic stiffnessC; (iii) the direction U of the displacement amplitude ∣U∣ of the wave is notarbitrary, but is fixed by the direction of propagation l and the elastic stiffness C,and moreover, it is different for the three types of waves. Lastly, we should notethat the eigenvalue problem, and thus the dispersion relation, will take on a muchsimpler form and be more amenable to explicit solution for special directions l inthe crystal. That is, for certain propagation directions, usually along directionsof high crystalline symmetry, the three normal modes are found to be one purelylongitudinal (U(1) ∥ l) and two purely transverse (U(2,3) ⊥ l); these are called puremodes. Except for pure mode directions, the eigenvectors U(i) depend explicitlyon the values of the elastic constants. Lastly, the group velocity direction may notcoincide with the propagation direction l of the wave velocity.

7.4 Nonlinear deformations of anisotropic

materials

We recall from (7.54) that ψ = ψ(C, θ) and the right Cauchy–Green tensor C issymmetric. We assume that the Helmholtz free energy is a polynomial in the sixcomponents of C without limitation on its degree, i.e., we assume that ψ is ananalytic function. The material symmetry determines a polynomial basis I1, . . . , Inin the components of C which are invariant under the specific symmetry group.Hence, any polynomial in I1, . . . , In is also invariant under the symmetry group.Conversely, it can be proved that any polynomial in the components of C that isinvariant under the the material symmetry group is expressible as a polynomial inthe polynomial basis. Hence, we can write

ψ = ψ(θ, I1, . . . , In). (7.167)

Therefore, the constitutive equation for the stress tensor, given by (7.58) and(7.61), can be expressed as

σ = 2ρF ⋅ ∂ψ∂C⋅FT = 2ρ n

∑α=1

∂ψ

∂IαΛα, (7.168)

where

Λα = F ⋅ ∂Iα∂C⋅FT . (7.169)

The functions Λα do not depend on the Helmholtz free enery; they depend onlyon the symmetry of the material. Consequently, for different materials exhibitingthe same type of anisotropy, only the coefficients ∂ψ/∂Iα take different values.

Using known theorems of invariant theory, polynomial bases for the Helmholtzfree energy for various crystal classes can be obtained. The free energy will then bean arbitrary polynomial in the appropriate basis. It has been found that there areonly 11 different sets of bases which characterize the elastic properties of crystalsand they are associated with the 11 types of crystals. Table 7.6 summarizes thebasis for each class of crystals along with those of transverse isotropic and isotropicmaterials.

7.5. LINEAR DEFORMATIONS OF ISOTROPIC MATERIALS 305

Example

The Helmholtz free energy for an orthorhombic material is a polynomialin the invariants c11, c22, c33, c223, c213, c212, c12c23c13. Now, it is straightfor-ward to see that the general form of ψ is

ψ = ψ1 + c12 c23 c13ψ2, (7.170)

where ψ1 and ψ2 are functions of θ and the six invariantsc11, c22, c33, c223, c213, c212. We then obtain from (7.168) that

σij = 2ρ [ ∂ψ∂c11

xi,1xj,1 +∂ψ

∂c22xi,2xj,2 +

∂ψ

∂c33xi,3xj,3+

(2 c12 ∂ψ∂c2

12

+ c23c13ψ2)(xi,1xj,2 + xi,2xj,1) +(2 c23 ∂ψ

∂c223

+ c12c13ψ2)(xi,2xj,3 + xi,3xj,2) +(2 c13 ∂ψ

∂c213

+ c12c23ψ2)(xi,1xj,3 + xi,3xj,1)] . (7.171)

7.5 Linear deformations of isotropic materials

For linear deformations of an isotropic solid, using the symmetry conditions (7.120)and using Voigt’s notation with (c0)11 = 2µ0 + λ0 and (c0)12 = λ0 (see Table 7.5),we must have (see Appendix B)

(c0)ijkl = λ0 δijδkl + µ0 (δikδjl + δilδjk) , (7.172)(a0)kl = a0 δkl, (7.173)(β0)ij = −(c0)ijkl(a0)kl (7.174)

= − [λ0δijδkl + µ0 (δikδjl + δilδjk)]a0 δkl (7.175)

= −3(λ0 + 2

3µ0)a0 δij (7.176)

= 3β0 δij , (7.177)(k0)ij = k0 δij . (7.178)

The constants λ0 and µ0 are known as isothermal Lamé’s constants, and they arerelated to the isothermal Young modulus E0 and the isothermal Poisson ratio ν0 aswell as to the other parameters, as indicated in Table 7.7. The parameter µ0 is theisothermal shear modulus and the parameter λ0 is related to the isothermal bulkmodulus b0 = λ0 + 2/3µ0. Note that the isothermal bulk modulus is the reciprocalof the isothermal compressibility, i.e., b0 = 1/κθ. In addition, we have defined theisothermal coefficient of thermal expansion a0 and the thermal stress coefficient

306 THERMOELASTIC SOLIDS

β0 = −b0 a0. Now, our constitutive relations (7.105)–(7.109) become

ψ = ψ0 − η0 θ −ce0

2 θ0θ2

−3 b0 a0

ρ0(tr e) θ + 1

2ρ0[2µ0 e ∶ e + λ0 (tr e)2] , (7.179)

η = η0 + ce0

θ

θ0+3 b0 a0

ρ0(tr e), (7.180)

q = −k0 g, (7.181)

h = −k0

θ0g, (7.182)

σ = (λ0 tr e − 3 b0 a0 θ)1 + 2µ0 e. (7.183)

Equation (7.183) corresponds to the thermoelastic generalization of Hooke’s law.In addition, it is easy to show that

cσ0− ce0

= 9 b0 a20 θ0

ρ0= 3E0 a

20 θ0

ρ0 (1 − 2 ν0) (7.184)

and that the Grüneisen tensor (5.217) in this case simplifies to yield the conven-tional formula

Γ = 3ρ0 a0 b0

ce0

. (7.185)

Thermodynamic stability at equilibrium in this case requires that

ce0> 0, (c0)11 − (c0)12 > 0, and (c0)11 + 2 (c0)12 > 0. (7.186)

or, equivalently,

ce0> 0, µ0 > 0, and b0 > 0. (7.187)

These in turn restrict the Poisson ratio to be within the range

−1 < ν0 < 1

2. (7.188)

Lastly, the reduced inequality provides a restriction for the thermal conductivity:

k0 ≥ 0. (7.189)

7.6 Nonlinear deformations of isotropic materials

More generally, we have shown that the constitutive functional of an objectivesimple homogeneous solid that is isotropic or hemitropic is given by (see (5.109))

T (x, t) = F0≤s<∞

B, (t)C(t), θ(t), (t)g(t). (7.190)

Subsequently, if we consider a material with no memory (p = q = r = 0), and userelations (7.46)–(7.48), (7.50), and (7.45), the constitutive functional reduces to

7.6. NONLINEAR DEFORMATIONS OF ISOTROPIC MATERIALS 307

the following constitutive functions:

ψ = ψ(B, θ), (7.191)

η = −∂ψ

∂θ, (7.192)

q = q(B, θ,g), (7.193)

h = q

θ, (7.194)

σ = 2ρ∂ψ

∂B⋅B, (7.195)

with H (=RT =Q) ∈ Gκ = O(V ) or H (=RT =Q) ∈ Gκ = O+(V ) for all (B, θ,g).

Recall from (7.50) that

σij = ρ ∂ψ

∂FiI

FjI = 2ρ ∂ψ

∂Bik

Bkj , (7.196)

where we have used the fact that Bik = FiIFkI .From Tables 5.1 and 5.5, and using the relations (3.98)–(3.100), we see that for

isotropic and hemitropic solids,

ψ = ψ(B(γ), θ), (7.197)

where B(γ), γ = 1,2,3, correspond to the invariants of B.From Tables 5.1 and 5.2, we see that the isotropic representation for the heat

flux is given by

q = (α0I + α1B + α2B2)g, (7.198)

with

αj = αj (B(γ), θ,g ⋅ g,g ⋅B g,g ⋅B2g) , (7.199)

while for a hemitropic solid, from Tables 5.5 and 5.6, we have

q = (α0I + α1B)g + α2 g ×B g, (7.200)

with

αi = αi (B(γ), θ,g ⋅ g,g ⋅B g,g ⋅B2g, [g,B g,B2g)]) . (7.201)

Note that since

dψ = ∂ψ∂B⋅ dB +

∂ψ

∂θdθ = 1

2ρσ ⋅B−1 ⋅ dB − η dθ, (7.202)

we have the Gibbs equation

de = 1

2ρσ ⋅B−1 ⋅ dB + θ dη. (7.203)

Equations (7.202) and (7.203) should be compared to (5.206) and (5.139).

308 THERMOELASTIC SOLIDS

Now, using (7.197), we can write

∂ψ

∂Bik

= 3

∑γ=1

∂ψ

∂B(γ)

∂B(γ)∂Bik

. (7.204)

It is easy to see from (3.88) or (3.89) that

∂B(1)∂Bik

= δik, ∂B(2)∂Bik

= B(1)δik −Bki, and∂B(3)∂Bik

= B(3)B−1ki . (7.205)

Subsequently, since B = F ⋅FT , we have

σ = β01 + β1B + β2B2, (7.206)

where

β0 = 2ρB(3) ∂ψ

∂B(3), (7.207)

β1 = 2ρ( ∂ψ

∂B(1)+B(1)

∂ψ

∂B(2)) , (7.208)

β2 = −2ρ ∂ψ

∂B(2). (7.209)

Alternatively, by taking the inner product of Cayley–Hamilton’s relation (3.95),

B3−B(1)B

2+B(2)B −B(3)1 = 0, (7.210)

with B−1, solving the resulting equation for B2, and substituting the result intothe constitutive equation (7.206), we can write equivalently

σ = β01 + β1B + β−1B−1, (7.211)

where

β0 = β0 − β2B(2) = 2ρ(B(2) ∂ψ

∂B(2)+B(3)

∂ψ

∂B(3)) , (7.212)

β1 = β1 + β2B(1) = 2ρ ∂ψ

∂B(1), (7.213)

β−1 = β2B(3) = −2ρB(3) ∂ψ

∂B(2). (7.214)

From (7.211) we see that

σ ⋅B = B ⋅σ. (7.215)

Note that for a deformation such that B13 = B23 = 0, we see that σ13 = σ23 = 0 aswell. Then, the only nontrivial relation from (7.215) can be written in the form

σ11 − σ22

σ12= B11 −B22

B12

. (7.216)

Relations (7.215) and (7.216) between stresses and deformations that are inde-pendent of any specific constitutive relation are called universal relations. Suchrelations play important roles in experimental validation of material models sincethey reflect a direct consequence of material symmetry without having to knowthe constitutive function itself, as long as the material is classified as a simpleisotropic homogeneous thermoelastic one.

7.6. NONLINEAR DEFORMATIONS OF ISOTROPIC MATERIALS 309

7.6.1 Special nonlinear deformations

For isothermal conditions, the steady balance equations are identically satisfied inthe absence of external supplies f and r when divσ = 0. Boundary conditions canalso be satisfied by appropriate application of forces on the boundary of the body.We would like to find all possible material deformations in such case. Specifically,we would like to find solutions of σik,k = 0 for all ψ = ψ(B(γ), θ). First we notethat for the deformation function xi = χi(XI), we can write

σ = J−1σ ⋅FT or σik = J−1σiKxk,K , (7.217)

where the quantity σ, called the first Piola–Kirchhoff stress tensor, is given by

σ = Jσ ⋅ (FT )−1 or σiK = JσikXK,k. (7.218)

Unlike the Cauchy stress tensor, the first Piola–Kirchhoff stress tensor is not sym-metric in general. If follows that

t = σ ⋅ n = J−1σ ⋅FT⋅ n = 1

ησ ⋅N or

ti = σiknk = J−1σiKxk,Knk = 1

ησiKNK , (7.219)

where

N = η J−1FT⋅ n or NK = η J−1xk,Knk, (7.220)

and η is the area stretch ratio (see (3.45)).Now, using mass conservation,

σik = ρ ∂ψ

∂xi,Kxk,K = ρ0J−1 ∂ψ

∂xi,Kxk,K ; (7.221)

thus

σiK = ρ0 ∂ψ

∂xi,K= ∂ψ0

∂xi,K, (7.222)

where ψ0 = ρ0ψ and we have assumed that ρ0 = const. Hence σik,k = 0 implies thatσiK,K = 0, since

σik,k = (J−1σiKxk,K),k = (J−1xk,K),k σiK + J−1σiK,Lxk,KXL,k = J−1σiK,K ,(7.223)

where above we have made use of the Euler–Piola–Jacobi identity (3.67).In summary, for the deformation function xi = χi(XK), the original problem

has been reformulated into

σiK,K = 0, where σiK = ∂ψ0

∂xi,K, for all ψ0 = ψ0(B(γ), θ). (7.224)

In particular, if we choose ψ0 = B(1) = trB = tr (FFT ) = xi,Kxi,K , thenσiK = 2xi,K , and subsequently, the requirement that xi,KK = 0 implies that xiis harmonic and infinitely differentiable.

310 THERMOELASTIC SOLIDS

Another particular choice is ψ0 = B(3) = detB = det (FFT ) = J2. Now us-ing (3.60), we have σiK = 2J2XK,i and subsequently we require that σiK,K =(2J2XK,i),K = 0. Now using the Euler–Piola–Jacobi identity (3.66), we have that2JXK,iJ,K = 0 or J = const.

Now for any ψ0 = Jψ, using (3.60), we get

σiK = ∂(Jψ)∂xi,K

= JXK,iψ + J∂ψ

∂xi,K, (7.225)

and so, using the Euler–Piola–Jacobi identity (3.66) again and our previous results,we have

0 = σiK,K = JXK,i

∂ψ

∂XK

+ J ( ∂ψ

∂xi,K),K

= JXK,i

∂ψ

∂XK

+ JΣiK,K =

JXK,i

∂ψ

∂XK

; (7.226)

thus ψ = const., which implies that B(1) = const., B(2) = const., and B(3) = const.Since B(1),KK = 0, then

0 = (xi,IKxi,I + xi,Ixi,IK),K = 2 (xi,IKxi,I),K = 2 (xi,IKKxi,I + xi,IKxi,IK) =2xi,IKxi,IK ;

thus

xi,IK = 0 or xi,I = const. (7.227)

Hence, the deformations possible in materials subject to surface traction aloneare homogeneous.

A. Homogeneous deformations

A homogeneous deformation is one such that

x = x0 +F ⋅ (X −X0) , (7.228)

where the deformation gradient F is a constant two-point tensor, and X0 and x0

correspond to the origins in the reference and deformed configurations. Such defor-mation is a universal solution of thermoelastic materials since it does not dependon any particular form of the constitutive relation. Therefore, a homogeneous de-formation is an important controllable class of deformation which can be exploitedin conjunction with experiments to determine specific material parameters.

a) Uniaxial stretch: The homogeneous deformation of uniaxial stretch is givenby

x1 = α1X1, (7.229)

x2 = α2X2, (7.230)

x3 = α3X3, (7.231)

7.6. NONLINEAR DEFORMATIONS OF ISOTROPIC MATERIALS 311

where the stretches αi, i = 1,2,3, are constant. Then

F = [Grad x] =⎡⎢⎢⎢⎢⎢⎣α1 0 0

0 α2 0

0 0 α3

⎤⎥⎥⎥⎥⎥⎦, (7.232)

B = FFT =⎡⎢⎢⎢⎢⎢⎣α21 0 0

0 α22 0

0 0 α23

⎤⎥⎥⎥⎥⎥⎦, and B−1 =

⎡⎢⎢⎢⎢⎢⎣α−21 0 0

0 α−22 0

0 0 α−23

⎤⎥⎥⎥⎥⎥⎦. (7.233)

Furthermore,

B(1) = α2

1 + α2

2 + α2

3, (7.234)

B(2) = α2

1α2

2 + α2

2α2

3 + α2

3α2

1, (7.235)

B(3) = α2

1α2

2α2

3. (7.236)

Subsequently,

[σij] = β0

⎡⎢⎢⎢⎢⎢⎣1 0 0

0 1 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦+ β1

⎡⎢⎢⎢⎢⎢⎣α21 0 0

0 α22 0

0 0 α23

⎤⎥⎥⎥⎥⎥⎦+

β−1

⎡⎢⎢⎢⎢⎢⎣α−21 0 0

0 α−22 0

0 0 α−23

⎤⎥⎥⎥⎥⎥⎦. (7.237)

where βk = βk(α2γ , θ), with γ = 1,2,3 and k = 0,1,−1.

For uniaxial stretch in the X1-direction, the lateral stresses must vanish:

σ22 = β0 + β1α2

2 + β−1α−2

2 = 0, (7.238)

σ33 = β0 + β1α2

3 + β−1α−2

3 = 0, (7.239)

so that σ22 − σ33 results in

(α2

2 − α2

3)(β1 − β−11

α22α23

) = 0, (7.240)

which has a symmetric solution, α2 = α3, and an asymmetric solution, α2 ≠α3. Since σ22 = 0, we can also write the uniaxial stress in the form

σ11 = (α2

1 − α2

2)(β1 − β−11

α21α22

) . (7.241)

Now since the stress σ11 and stretches α1 and α2 can be measured, then(7.241) can be used to determine β1 and β

−1, while β0 can subsequently bedetermined from (7.238) and (7.239).

b) Simple extension: The deformation

x1 = αvX1, (7.242)

x2 = αvX2, (7.243)

x3 = vX3, (7.244)

312 THERMOELASTIC SOLIDS

where 0 < α ≤ 1 and v ≥ 1 is called a simple extension in the X3-direction.Note that

F = v⎡⎢⎢⎢⎢⎢⎣α 0 0

0 α 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦, (7.245)

B = v2⎡⎢⎢⎢⎢⎢⎣α2 0 0

0 α2 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦, and B−1 = v−2

⎡⎢⎢⎢⎢⎢⎣α−2 0 0

0 α−2 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦. (7.246)

Furthermore,

B(1) = v2(1 + 2α2), B(2) = α2v4(2 + α2), B(3) = α4v6. (7.247)

Subsequently,

σ11 = σ22 = β0 + α2v2β1 + α

−2v−2β−1, (7.248)

σ33 = β0 + v2β1 + v

−2β−1, (7.249)

σij = 0 for i ≠ j, (7.250)

where βk = βk(B(γ), θ), with γ = 1,2,3 and k = 0,1,−1.For a simple extension, we must have

σ11 = σ22 = 0, (7.251)

so that

α2 = f(v2, θ), (7.252)

and thus

σ33 = β0 + v2β1 + v

−2β−1 = σ33(v2, θ). (7.253)

c) Simple dilatation or compression: The deformation function for a simpledilatation or compression is given by

x1 = vX1, (7.254)

x2 = vX2, (7.255)

x3 = vX3, (7.256)

depending on whether v ≥ 1 or v ≤ 1. Then F = v1, B = v21, B−1 = v−21,B(1) = 3 v2, B(2) = 3 v4, and B(3) = v6, and subsequently, we have

σ11 = σ22 = σ33 = β0 + v2β1 + v

−2β−1 ≡ −p and σij = 0 for i ≠ j, (7.257)

where p = p(v2, θ).

7.6. NONLINEAR DEFORMATIONS OF ISOTROPIC MATERIALS 313

d) Simple shear: The deformation function is given by

x1 =X1 + κX2, (7.258)

x2 =X2, (7.259)

x3 =X3, (7.260)

where κ ≥ 0 is the amount of shear. Then

F =⎡⎢⎢⎢⎢⎢⎣1 κ 0

0 1 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦, (7.261)

B =⎡⎢⎢⎢⎢⎢⎣1 + κ2 κ 0

κ 1 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦, and B−1 =

⎡⎢⎢⎢⎢⎢⎣1 −κ 0

−κ 1 + κ2 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦. (7.262)

Furthermore, B(1) = B(2) = 3 + κ2 and B(3) = 1 so that βk = βk(B(γ), θ) =βk(κ2, θ), for γ = 1,2,3 and k = 0,1,−1. Note that simple shear is an isochoricor incompressible deformation since B(3) = 1. Now, as can be easily verified,we have

[σij] = (β0 + β1 + β−1)⎡⎢⎢⎢⎢⎢⎣1 0 0

0 1 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦+ κ (β1 − β−1)

⎡⎢⎢⎢⎢⎢⎣0 1 0

1 0 0

0 0 0

⎤⎥⎥⎥⎥⎥⎦+

κ2β1

⎡⎢⎢⎢⎢⎢⎣1 0 0

0 0 0

0 0 0

⎤⎥⎥⎥⎥⎥⎦+ κ2β

−1

⎡⎢⎢⎢⎢⎢⎣0 0 0

0 1 0

0 0 0

⎤⎥⎥⎥⎥⎥⎦, (7.263)

and thus

σ21

κ= σ12

κ= β1 − β−1 ≡ µ(κ2), (7.264)

where µ(κ2) is called the generalized shear modulus. Note that for small κ,

µ(κ2) = µ(0) +O(κ2), (7.265)

where µ(0) is called the ordinary shear modulus of the material. The prin-cipal stretches obtained earlier (see (3.167)) are

(λ(1))2 = 1 + 1

2κ2 + κ

√1 +

1

4κ2, (7.266)

(λ(2))2 = 1 + 1

2κ2 − κ

√1 +

1

4κ2 = 1

(λ(1))2 , (7.267)

(λ(3))2 = 1, (7.268)

and the principal stresses are easily seen to be given by

σ(i) = β0 + (λ(i))2 β1 + (λ(i))−2 β−1, i = 1,2,3. (7.269)

314 THERMOELASTIC SOLIDS

It can now be shown that

σ(1) − σ(2)

(λ(1))2 − (λ(2))2 = µ(κ2) (7.270)

or

σ(1) − σ(2) = 2κ√

1 +1

4κ2 µ(κ2), (7.271)

so that µ ≥ 0 only if the direction of the greatest principal stress is the sameas that of the greatest principal stretches. We also see that the shear stressesalone do not suffice to maintain simple shear since

σ11 = β0 + (1 + κ2)β1 + β−1, (7.272)

σ22 = β0 + β1 + (1 + κ2)β−1, (7.273)

σ33 = β0 + β1 + β−1, (7.274)

and σ11 = σ22 = σ33 = 0 if and only if β0 = β1 = β−1 = 0, and thus µ ≡ 0.Furthermore, we have

σ11 − σ33

κ2= β1, (7.275)

σ22 − σ33

κ2= β

−1, (7.276)

σ11 − σ22

κ2= β1 − β−1 = σ12

κ. (7.277)

We see that σ11 = σ22 implies that µ = 0. The existence of a normal stressdifference is usually known as the Poynting effect. The mean hydrostaticpressure is given by

p ≡ −13σii = − (β0 + β1 + β−1) − 1

3(β1 + β−1) κ2. (7.278)

When there is no shear, κ = 0, and so from (7.263), we see that

limκ→0

p = − limκ→0[β0(κ2) + β1(κ2) + β−1(κ2)] = 0. (7.279)

Thus, a mean hydrostatic pressure, which is of second order in the amountof shear, is necessary to generate a simple shear. This is the so-called Kelvineffect. Note that the equation

σ11 − σ22

σ12= κ (7.280)

provides a universal relation for this deformation.

B. Incompressible deformations

For an isothermal incompressible (or isochoric) deformation, we have J = B(3) = 1,and so from (7.197), we have ψ = ψ(B(1),B(2), θ). We note from (3.101) thatB(2) = trB−1. The constitutive equation is given by

σij = −p δij + β1Bij + β−1B−1

ij , (7.281)

7.6. NONLINEAR DEFORMATIONS OF ISOTROPIC MATERIALS 315

where

β1 = 2ρ ∂ψ

∂B(1)and β

−1 = −2ρ ∂ψ

∂B(2). (7.282)

In this case p cannot be determined from the deformation. Subsequently, since themass density is constant and the hydrostatic pressure undetermined, β0 becomessuperfluous.

B.1 Homogeneous strain

Assuming that no external body forces are present, it is clear that any homoge-neous isochoric strain is a steady configuration subject to surface tractions only forany isothermal, homogeneous, incompressible, elastic body, and that the pressurep is an arbitrary constant that is to be determined. To illustrate, consider thesimple extension deformation

x1 = αvX1, (7.283)

x2 = αvX2, (7.284)

x3 = vX3, (7.285)

and then

F = v⎡⎢⎢⎢⎢⎢⎣α 0 0

0 α 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦. (7.286)

For the deformation to be isochoric, J = detF = 1; thus α2v3 = 1, or α = v−3/2.Then

B =⎡⎢⎢⎢⎢⎢⎣v−1 0 0

0 v−1 0

0 0 v2

⎤⎥⎥⎥⎥⎥⎦and B−1 =

⎡⎢⎢⎢⎢⎢⎣v 0 0

0 v 0

0 0 v−2

⎤⎥⎥⎥⎥⎥⎦, (7.287)

and

B(1) = 2v−1 + v2, B(2) = 2v + v−2. (7.288)

Subsequently,

σ11 = σ22 = −p + v−1β1 + vβ−1, (7.289)

σ33 = −p + v2β1 + v−2β

−1, (7.290)

with β1 = β1(B(1),B(2), θ) and β−1 = β−1(B(1),B(2), θ). Clearly this deformation

satisfies the field equation

σik,k = 0. (7.291)

If we require that σ11 = σ22 = 0, we then have

p = v−1β1 + vβ−1. (7.292)

316 THERMOELASTIC SOLIDS

Thus p is determined once we have the solution to the boundary value problem.Then

σ33 = σ33(v) = (v2 − v−1)β1 + (v−2 − v)β−1. (7.293)

The simplest isothermal incompressible deformation is given by a linear variationof the free energy with respect to B(1) and B(2):

ψ = α (B(1) − 3) + β (B(2) − 3) , (7.294)

where α = α(θ) and β = β(θ). Subsequently, we have that β1 = 2ρα and β−1 =

−2ρβ. This defines a Mooney–Rivlin material and is considered to be a good modelfor rubber-like materials. The case β = 0 (and thus β

−1 = 0) defines a neo-Hookeanmaterial, which provides a reasonable approximation for the behavior of rubberunder small strains. Experimental data indicate that both α and β are positive,leading to what are called the empirical inequalities

β1 > 0 and β−1 ≤ 0. (7.295)

B.2 Nonhomogeneous strain

The steady linear momentum equation is given by

divσ + ρf = 0 or σkl,k + ρfl = 0, (7.296)

where

σ = −p1 + β1B + β−1B−1 ≡ −p1 + σ (7.297)

or

σkl = −pδkl + β1B

kl + β−1(B−1)kl ≡ −pδkl + σk

l , (7.298)

Bkl = gmlB

km, and Bkm = gIJF kI F

mJ . Note that since we consider deformations in-

volving non-Cartesian coordinates, we have written the linear momentum equationfor covariant components and have used the mixed-component form of the Cauchystress tensor. Furthermore, for convenience, we have defined the extra stress σ. Ifwe assume that the external body force is conservative, and thus derivable from apotential, i.e.,

f = −grad U or fl = −U,l, (7.299)

then for an incompressible material with ρ0(X) = const., we have

−grad (p + ρU) + div σ = 0 or − (p + ρU),l + σkl,k = 0, (7.300)

where (p+ρU) is just a modified pressure. To eliminate it, we take the curl of theabove equation, so we have

curl div σ = 0 or ǫijlσkl,kj = 0 (7.301)

7.6. NONLINEAR DEFORMATIONS OF ISOTROPIC MATERIALS 317

or

σkl,kj = σk

j,kl. (7.302)

We see that the presence of a conservative body force does not make the solutionof problems for incompressible materials more difficult since all it does is modifythe pressure p. Of course, the surface tractions that must be applied are suchthat the deformation will depend on U , but the compatibility of any given de-formation is independent of whether or not a conservative body force is present.Thus, for simplicity, only steady solutions subjected to no external body forces areusually considered, i.e., a = 0 and f = 0, so that divσ = 0, or in terms of physicalcomponents, from (2.283) and (2.284),

σ<kl,l> = hk

h1h2h3

∂

∂xl(h1h2h3hkhl

σ<kl>) + hk

hmhlΓklmσ<ml> = 0, (7.303)

where from (2.285) we have

σ<kl> = hkhlσkl = −pδ<kl> + σ<kl> (7.304)

and

σ<kl> = hkhlσkl . (7.305)

Given a particular tensor field B, whether or not (7.302) is satisfied will depend,in general, upon β1(B(1),B(2), θ) and β

−1(B(1),B(2), θ). However, there are cer-tain forms of B such that (7.302) is satisfied independent of the specific forms ofβ1 and β

−1. The deformations that lead to such forms of B can be effected in everyhomogeneous, incompressible, isotropic, isothermal, elastic body by application ofsuitable surface tractions. In such case, we shall drop the dependences of β1 andβ−1 on θ. All solutions are such that appropriate physical components of the stress

are constant on each member of a family of parallel planes, coaxial cylinders, orconcentric spheres. There are five families of such deformations. They are thefollowing:

a) Bending, stretching, and shearing of a rectangular block: Let the co-ordinate in the initial reference configuration be Cartesian (X,Y,Z), andthose in the deformed configuration cylindrical, (r, θ, z). Consider the defor-mation

r =√2AX, (7.306)

θ = BY, (7.307)

z = Z

AB−BCY, (7.308)

where A, B, and C are constants, and AB ≠ 0. If C = 0, this deformationcarries the block bounded by the planes X = X1, X = X2, Y = ±Y0, Z =±Z0 into the annular wedge bounded by the cylinders r = r1 = √2AX1 andr = r2 = √2AX2, and the planes θ = ±θ0 = ±BY0, z = ±z0 = ±Z0/(A B). If

318 THERMOELASTIC SOLIDS

we think of B as given, then an arbitrary axial stretch 1/(A B) is allowed,and the radial stretch is adjusted to render the deformation isochoric. Inthe general case, the deformation may be effected in two steps, the first ofwhich is the bending and axial stretch just described, while the second isa homogeneous strain, in fact a simple shear, which carries the body intothe solid bounded by the cylindrical surfaces r = r1 and r = r2, the planesθ = ±θ0, and the helicoidal surfaces z + Cθ = ±z0.

b) Straightening, stretching, and shearing of a sector of a hollowcircular cylinder: Using a cylindrical coordinate system (R,Θ, Z), con-sider a body that in the reference configuration is bounded by the cylindersR = R1 and R = R2, the planes Θ = ±Θ0, and the planes Z = ±Z0. We useCartesian coordinates (x, y, z) for the deformed state. Let the body first bestraightened by the deformation x = 1

2AR2, y = Θ/A, z = Z. It then becomes

the block bounded by the planes x = 1

2AR2

1, x = 1

2AR2

2, y = ±Θ0/A, z = ±Z0.Now stretch the block along the x-axis with equal transverse contractions:x′ = B2x, y′ = y/B, z′ = z/B. Finally, effect a shear in the z-y planes: x = x′,y = y′, z = z′ + Cy′. The resulting deformation is

x = 1

2A B2R2, (7.309)

y = Θ

AB, (7.310)

z = ZB+

CΘ

AB, (7.311)

where R1 ≤ R ≤ R2, −Θ0 ≤ Θ ≤ Θ0, −Z0 ≤ Z ≤ Z0, and AB ≠ 0.c) Inflation, bending, torsion, extension, and shearing of an annular

wedge: The deformation carries the particle with cylindrical coordinates(R,Θ, Z) into the deformed cylindrical coordinates (r, θ, z) as follows:

r =√AR2 +B, (7.312)

θ = CΘ +DZ, (7.313)

z = EΘ + FZ, (7.314)

where B, C, D, E, F are arbitrary constants, A and B have values such thatAR2+B > 0 when R is in some interval R1 ≤ R ≤ R2, and A(CF-DE) = 1. This

deformation may be regarded as resulting from a succession of four simplerones. First, the cylinder defined by R1 ≤ R ≤ R2 is inflated uniformly:r′ = √AR2 +B, θ′ = Θ, z′ = Z/A. Second, the inflated cylinder is subjectedto a uniform longitudinal stretch of amount 1/C, so that r = r′, θ = Cθ′, andz = z′/C. Third, the inflated and stretched cylinder is twisted with a torsionof amount ACD; thus r = r, θ = θ + ACDz, and z = z. Fourth, a kind ofshear of the azimuthal planes is effected: r = r, θ = θ, z = z + Kθ. As theresult of this last deformation, the planes Z = const. are deformed into thehelicoidal surfaces z − Eθ/C = const. The constants E and F of the resultingdeformation are given by E = CK, F = DK+1/(AC). Note that if A > 0, r is anincreasing function of R, while if A < 0, r is a decresing function of R; thusthe case when A < 0 represents a hollow cylinder, or a part of one, which is

7.6. NONLINEAR DEFORMATIONS OF ISOTROPIC MATERIALS 319

turned inside out, or everted. If D = E = 0, the deformation is a plane strainsuperposed upon a uniform extension perpendicular to the plane.

d) Inflation or eversion of a sector of a spherical shell: Let the point withspherical coordinates (R,Θ,Φ) be deformed into one with spherical coordi-nates (r, θ, φ), where

r = (±R3+A)1/3 , (7.315)

θ = ±Θ, (7.316)

φ = Φ. (7.317)

This deformation carries the region between two concentric spheres into aregion between two other concentric spheres; any angular boundaries arepreserved. The ± signs are associated; if both are taken as +, the curvatureof the shell is unchanged in sign, while if both are taken as −, the shell iseverted.

e) Inflation, bending, extension, and azimuthal shearing of an annularwedge: The deformation carries the material particle with cylindrical co-ordinates (R,Θ, Z) into the deformed cylindrical coordinates (r, θ, z) as fol-lows:

r = AR, (7.318)

θ = B ln (R) + CΘ, (7.319)

z = DZ, (7.320)

where A, B, C, and D are arbitrary constants with the restriction thatA2C D = 1, and R1 ≤ R ≤ R2, −Θ0 ≤ Θ ≤ Θ0, and −Z0 ≤ Z ≤ Z0. This

deformation may also be viewed as resulting from a succession of four sim-pler deformations.

Below we will describe in detail only the bending, stretching, and shearing ofa rectangular block as an illustration of the solution of a problem involving non-Cartesian coordinates. The deformation (7.306)–(7.308) looks as shown in Fig. 7.4.The deformation gradient, with (r, θ, z) = (x1, x2, x3), is given by

[F kI ] = [ ∂xk∂XI

] =⎡⎢⎢⎢⎢⎢⎣

A/√2AX 0 0

0 B 0

0 −BC 1/(AB)⎤⎥⎥⎥⎥⎥⎦

=⎡⎢⎢⎢⎢⎢⎣

A/r 0 0

0 B 0

0 −BC 1/(AB)⎤⎥⎥⎥⎥⎥⎦, (7.321)

and subsequently

[Bkm] = [F kI ] [Fm

J ] [gIJ] , = [F kI ] [Fm

I ]T=⎡⎢⎢⎢⎢⎢⎣

A2/r2 0 0

0 B2−B2C

0 −B2C B2C2+ 1/(A2B2)

⎤⎥⎥⎥⎥⎥⎦, (7.322)

320 THERMOELASTIC SOLIDS

X

Y

Z

x

y

z

rθ

Figure 7.4: Bending and stretching of a rectangular block.

where we have made use of the fact that in the undeformed configuration, [gIJ] = I.Now

[Bkl ] = [Bkm] [gml] =

⎡⎢⎢⎢⎢⎢⎣A2/r2 0 0

0 B2r2 −B2C

0 −B2Cr2 B

2C2+ 1/(A2

B2)⎤⎥⎥⎥⎥⎥⎦

(7.323)

and

[(B−1)kl] =⎡⎢⎢⎢⎢⎢⎣r2/A2 0 0

0 (1 +A2B4C2)/(B2

r2) (A2B2C)/r2

0 A2B2C A2B2

⎤⎥⎥⎥⎥⎥⎦, (7.324)

where we note that [gml] = diag (1, r2,1) and [gml] = diag (1, r−2,1). In addition,we have the invariants

B(1) = Bkk = A2

r2+ B2r2 +B2C2

+1

A2B2, (7.325)

B(2) = 1

2[B2

(1) −BkmB

mk ] = r2

A2+1 + A

2B4C2

r2B2+A2B2, (7.326)

B(3) = det [Bkl ] = 1, (7.327)

so that β1 = β1(B(1),B(2),B(3)) = β1(r) and β−1 = β−1(B(1),B(2),B(3)) = β−1(r).

We now have

σkl = −pδkl + β1B

kl + β−1 (B−1)kl = −pδkl + σk

l (r), (7.328)

where

[σkl ](r) = β1

⎡⎢⎢⎢⎢⎢⎣A2/r2 0 0

0 B2r2 −B

2C

0 −B2Cr2 B2C2+ 1/(A2B2)

⎤⎥⎥⎥⎥⎥⎦+

β−1

⎡⎢⎢⎢⎢⎢⎣r2/A2 0 0

0 (1 +A2B4C2)/(B2r2) (A2B2C)/r20 A2B2C A2B2

⎤⎥⎥⎥⎥⎥⎦, (7.329)

7.6. NONLINEAR DEFORMATIONS OF ISOTROPIC MATERIALS 321

or in terms of physical components, using (7.304) and (7.305), we have

σ<kl> = −pδ<kl> + σ<kl>(r), (7.330)

where

[σ<kl>](r) = β1

⎡⎢⎢⎢⎢⎢⎣A2/r2 0 0

0 B2r2 −B2Cr

0 −B2Cr B2C2+ 1/(A2B2)

⎤⎥⎥⎥⎥⎥⎦+

β−1

⎡⎢⎢⎢⎢⎢⎣r2/A2 0 0

0 (1 + A2B4C2)/(B2

r2) (A2B2C)/r

0 (A2B2C)/r A2B2

⎤⎥⎥⎥⎥⎥⎦. (7.331)

In this case, from (7.303), the linear momentum balance becomes

1

r

∂

∂r(rσ<rr>) − 1

rσ<θθ> = 0, (7.332)

1

r2∂

∂r(r2

0σ<rθ>) − 1

r

∂p

∂θ= 0, (7.333)

1

r

∂

∂r(r

0σ<rz>) − ∂p

∂z= 0, (7.334)

where we have used (2.261), (2.266), and (2.267). It is now evident from (7.330),(7.331), (7.333), and (7.334) that p = p(r) and subsequently, from (7.330), σ<kl> =σ<kl>(r). Note that, from (7.330), we can rewrite (7.332) in the form

dσ<rr>

dr+1

r(σ<rr> − σ<θθ>) = 0 (7.335)

(7.336)

so that, for r1 ≠ 0,σ<rr>(r) = σ<rr>(r1) −∫ r

r1

1

ξ[σ<rr>(ξ) − σ<θθ>(ξ)]dξ (7.337)

= σ<rr>(r1) −∫ r

r1

[(A2

ξ3−B2ξ)β1(ξ) − (1 +A2B4C2

B2ξ3

−ξ

A2)β−1(ξ)]dξ

(7.338)

= σ<rr>(r1) + 1

2∫

r

r1

[dB(1)(ξ)dξ

β1(ξ) − dB(2)(ξ)dξ

β−1(ξ)]dξ (7.339)

and, from (7.330),

p = σ<rr>(r) +∫ r

r1

[(A2

ξ3−B2ξ)β1(ξ) − (1 + A2B4C2

B2ξ3

−ξ

A2)β−1(ξ)]dξ −

σ<rr>(r1). (7.340)

Subsequently, from (7.332) and (7.338), we have

σ<θθ> = d

dr(rσ<rr>)

= σ<rr> − (A2

r2− B

2r2)β1(r) + (1 +A2B4C2

B2r2−r2

A2)β−1(r), (7.341)

322 THERMOELASTIC SOLIDS

while from (7.330) and (7.331), we have

σ<zz> = −p + σ<zz>

= σ<rr> + (σ<zz> − σ<rr>)= σ<rr> + (B2C2

+1

A2B2−

A2

r2)β1(r) − ( r2

A2−A2B2)β

−1(r), (7.342)

σ<θz> = σ<zθ> = σ<zθ>= −B2Crβ1(r) + A2B2C

rβ−1(r). (7.343)

Now, since t<k> = σ<kl>n<l>, by the choice of σ<rr>(r1) = 0, the surface r = r1may be rendered free of traction. Subsequently, the resultant normal force R perunit height on the part of the axial plane θ = const. cut off by the cylinders r = r1and r = r2 > r1 is given by

R = ∫ r2

r1

σ<θθ> dr = r2 σ<rr>(r2), (7.344)

where we have used (7.332). Thus, the faces θ = const. are free of resultant normaltraction if and only if the surface r = r2 is free of normal traction. In order thatthis surface also be free of traction, so that the block may be bent and sheared byforces applied to its plane and helicoidal faces only, we see from (7.338) that it isnecessary that

∫r2

r1[(A2

r3−B

2r)β1(r) − (1 +A2B4C2

B2r3−r

A2)β−1(r)] dr = 0. (7.345)

For given r1 and r2, this is a relation among the constants A, B, and C. Whetheror not this equation has any real roots depends on the nature of β1 and β

−1.

It is now clear that the helicoidal faces cannot be free of traction. We readilycalculate the normal and tangential tractions N and T that have to be applied onthese faces to maintain the deformation:

N = 1

1 + C2/r2 [σ<zz> + 2C

rσ<θz> +

C2

r2σ<θθ>] , (7.346)

T = 1

1 + C2/r2 [(1 − C2

r2)σ<θz> + C

r(σ<θθ> − σ<zz>)] , (7.347)

where the values of σ<kl> are to be taken from (7.341)–(7.343). In the case ofpure bending, C = 0, and then N = σ<zz> and T = 0, but in general both normaland tangential tractions must be applied. The presence of these tractions is thePoynting effect for bending.

The tractions that must be applied upon the plane ends θ = ±θ0 may be readoff from (7.341) to (7.343), and the resultant normal force on them is given by(7.344). The resultant moment M per unit height, with respect to a point on the

7.6. NONLINEAR DEFORMATIONS OF ISOTROPIC MATERIALS 323

axis r = 0, exerted by the normal stress acting upon these faces, is given by

M = ∫r2

r1

rσ<θθ>(r) dr= ∫

r2

r1rd

dr[rσ<rr>(r)]dr

= r2σ<rr>(r)∣r2r1 −∫ r2

r1

rσ<rr>(r) dr= 1

2r2σ<rr>(r)∣r2r1 + 1

2∫

r2

r1

r2dσ<rr>(r)

drdr

= 1

2r2σ<rr>(r)∣r2r1 − 1

2∫

r2

r1

r [σ<rr>(r) − σ<θθ>(r)]dr= 1

2r2σ<rr>(r)∣r2r1 − 1

2∫

r2

r1

[(A2

r−B2r3)β1(r)−

(1 +A2B4C2

B2r−r3

A2)β−1(r)] dr, (7.348)

where we have used (7.341) in the last step.All the results are referred to the deformed body but can easily be expressed in

terms of the undeformed body by use of (7.306)–(7.308).The resultant normal traction N acting upon the plane annular wedge r1 ≤ r ≤ r2,

z = const., and −θ0 ≤ θ ≤ θ0 may be calculated as follows. We would like tosolve a problem with boundary conditions t<k> = 0 on r = r1, r2. Now, sincet<k> = σ<kl>n<l>, this requires that σ<rr> = 0 on r = r1, r2. These conditionsdetermine σ<rr>(r1) and provide a relationship between A, B, and C. In the z-direction, we have t<k> = σ<kz>n<z>, so that the conditions t<r> = t<θ> = 0 andt<z> = N become σ<rz> = σ<θz> = 0 and σ<zz>(r) = N. Lastly, at θ = θ0 we requirethat t<r> = t<z> = 0 and t<θ> = L; since t<k> = σ<kθ>n<θ>, we have σ<rθ> = σ<zθ> = 0and σ<θθ>(r) = L. But the resultant normal force is then

R = ∫ r2

r1σ<θθ>dr = rσ<rr>∣r2r1 = 0, (7.349)

since

1

r

∂

∂r(rσ<rr>) − σ<θθ>

r= 0, (7.350)

and we have used the boundary conditions. Thus, in the case of pure bending(C = 0), we subsequently have from (7.348) that

M0 = −1

2∫

r2

r1[(A

2

r2− B

2r3)β1(r) + ( r3

A2−

1

rB2)β−1(r)]dr, (7.351)

where we have integrated by parts in the second step.

324 THERMOELASTIC SOLIDS

Table 7.1: Unit cell axes and angles associated with crystal systems comprisingthe 14 Bravais lattices.

Crystal System Unit Cell Axes Unit Cell Angles Lattice Variations

Triclinic a ≠ b ≠ c α ≠ β ≠ γ PMonoclinic a ≠ b ≠ c α = γ = 90 ≠ β P, C

Orthorhombic a ≠ b ≠ c α = β = γ = 90 P, F, I, CTetragonal a = b ≠ c α = β = γ = 90 P, ITrigonal a = b ≠ c α = β = γ ≠ 90 P

Hexagonal a = b ≠ c α = β = 90, γ = 120 PCubic a = b = c α = β = γ = 90 P, F, I

7.6. NONLINEAR DEFORMATIONS OF ISOTROPIC MATERIALS 325

Table 7.2: Generating transformations.

I =⎡⎢⎢⎢⎢⎢⎣1 0 0

0 1 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦, C =

⎡⎢⎢⎢⎢⎢⎣−1 0 0

0 −1 0

0 0 −1

⎤⎥⎥⎥⎥⎥⎦,

R1 =⎡⎢⎢⎢⎢⎢⎣−1 0 0

0 1 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦, R2 =

⎡⎢⎢⎢⎢⎢⎣1 0 0

0 −1 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦, R3 =

⎡⎢⎢⎢⎢⎢⎣1 0 0

0 1 0

0 0 −1

⎤⎥⎥⎥⎥⎥⎦,

D1 =⎡⎢⎢⎢⎢⎢⎣1 0 0

0 −1 0

0 0 −1

⎤⎥⎥⎥⎥⎥⎦, D2 =

⎡⎢⎢⎢⎢⎢⎣−1 0 0

0 1 0

0 0 −1

⎤⎥⎥⎥⎥⎥⎦, D3 =

⎡⎢⎢⎢⎢⎢⎣−1 0 0

0 −1 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦,

T1 =⎡⎢⎢⎢⎢⎢⎣1 0 0

0 0 1

0 1 0

⎤⎥⎥⎥⎥⎥⎦, T2 =

⎡⎢⎢⎢⎢⎢⎣0 0 1

0 1 0

1 0 0

⎤⎥⎥⎥⎥⎥⎦, T3 =

⎡⎢⎢⎢⎢⎢⎣0 1 0

1 0 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦,

M1 =⎡⎢⎢⎢⎢⎢⎣0 1 0

0 0 1

1 0 0

⎤⎥⎥⎥⎥⎥⎦, M2 =

⎡⎢⎢⎢⎢⎢⎣0 0 1

1 0 0

0 1 0

⎤⎥⎥⎥⎥⎥⎦,

S1 =⎡⎢⎢⎢⎢⎢⎣−1/2 √

3/2 0

−√3/2 −1/2 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦, S2 =

⎡⎢⎢⎢⎢⎢⎣−1/2 −

√3/2 0√

3/2 −1/2 0

0 0 1

⎤⎥⎥⎥⎥⎥⎦.

326 THERMOELASTIC SOLIDS

Table 7.3: Transformations that characterize the 32 crystal classes.

Crystal System Type Class Class # Symmetry Transformations n

Triclinic: 1 Pedial 1 I 1

Pinacoidal 2 I,C 2

Monoclinic: 2 Sphenoidal 3 I,D1 2

Domatic 4 I,R1 2

Prismatic 5 I,C,R1,D1 4

Orthorhombic: 3 Rhombic-disphenoidal 6 I,D1,D2,D3 4

Rhombic-pyramidal 7 I,R2,R3,D1 4

Rhombic-dipyramidal 8 I,C,R1,R2,R3,D1,D2,D3 8

Tetragonal: 4 Tetragonal-pyramidal 9 I,D3, (R1,R2)T3 4

Tetragonal-disphenoidal 10 I,D3, (D1,D2)T3 4

Tetragonal-dipyramidal 11 I,C,R3,D3, (R1,R2,D1,D2)T3 8

5 Tetragonal-trapezohedral 12 I,D1,D2,D3, (C,R1,R2,R3)T3 8

Ditetragonal-pyramidal 13 (I,R1,R2,D3) (I,T3) 8

Tetragonal-scalenohedral 14 (I,D1,D2,D3) (I,T3) 8

Ditetragonal-dipyramidal 15 (I,C,R1,R2,R3,D1,D2,D3)⋅(I,T3) 16

Trigonal: 6 Trigonal-pyramidal 16 I,S1, S2 3

Rhombohedral 17 (I,C) (I,S1, S2) 6

7 Trigonal-trapezohedral 18 (I,D1) (I,S1, S2) 6

Ditrigonal-pyramidal 19 (I,R1) (I,S1, S2) 6

Trigonal-dipyramidal 20 (I,C,R1,D1) (I,S1, S2) 12

Hexagonal: 8 Hexagonal-pyramidal 21 (I,D3) (I,S1, S2) 6

Trigonal-scalenohedral 22 (I,R3) (I,S1, S2) 6

Hexagonal-dipyramidal 23 (I,C,R3,D3) (I,S1, S2) 12

9 Hexagonal-trapezohedral 24 (I,D1,D2,D3) (I,S1, S2) 12

Dihexagonal-pyramidal 25 (I,R1,R2,D3) (I,S1, S2) 12

Ditragonal-dipyramidal 26 (I,R1,R3,D3) (I,S1, S2) 12

Dihexagonal-dipyramidal 27 (I,C,R1,R2,R3,D1,D2,D3)⋅(I,S1, S2) 24

Cubic: 10 Tetartoidal 28 (I,D1,D2,D3) (I,M1,M2) 12

Diploidal 29 (I,C,R1,R2,R3,D1,D2,D3)⋅(I,M1,M2) 24

11 Gyroidal 30 (I,D1,D2,D3) (I,M1,M2),(C,R1,R2,R3) (T1, T2, T3) 24

Hextetrahedral 31 (I,D1,D2,D3)⋅(I,T1, T2, T3,M1,M2) 24

Hexoctahedral 32 (I,C,R1,R2,R3,D1,D2,D3)⋅(I,T1, T2, T3,M1,M2) 48

7.6. NONLINEAR DEFORMATIONS OF ISOTROPIC MATERIALS 327

Table 7.4: Crystal symmetry on the component matrix A of a symmetric propertytensor relating two vector fields or a scalar field to a rank-2 symmetric tensor field.

Triclinic (6 parameters)

⎡⎢⎢⎢⎢⎢⎣a11 a12 a13

a22 a23Sym a33

⎤⎥⎥⎥⎥⎥⎦Monoclinic (4 parameters)

⎡⎢⎢⎢⎢⎢⎣a11 0 a13

a22 0

Sym a33

⎤⎥⎥⎥⎥⎥⎦Orthorhombic (3 parameters)

⎡⎢⎢⎢⎢⎢⎣a11 0 0

a22 0

Sym a33

⎤⎥⎥⎥⎥⎥⎦Tetragonal, Trigonal, Hexagonal, and Transverse isotropic (2 parameters)

⎡⎢⎢⎢⎢⎢⎣a11 0 0

a11 0

Sym a33

⎤⎥⎥⎥⎥⎥⎦Cubic and Isotropic (1 parameter)

⎡⎢⎢⎢⎢⎢⎣a11 0 0

a11 0

Sym a11

⎤⎥⎥⎥⎥⎥⎦

328 THERMOELASTIC SOLIDS

Table 7.5: Crystal symmetry on the symmetric component matrix C of a rank-4property tensor relating two symmetric rank-2 tensor fields using Voigt’s repre-sentation.

Triclinic (21 parameters)

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

c11 c12 c13 c14 c15 c16c22 c23 c24 c25 c26

c33 c34 c35 c36c44 c45 c46

Sym c55 c56c66

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦Monoclinic (13 parameters)

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

c11 c12 c13 0 c15 0

c22 c23 0 c25 0

c33 0 c35 0

c44 0 c46Sym c55 0

c66

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦Orthorhombic (9 parameters)

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

c11 c12 c13 0 0 0

c22 c23 0 0 0

c33 0 0 0

c44 0 0

Sym c55 0

c66

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦Tetragonal (7 or 6 parameters)

Classes 9–11 Classes 12–15⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

c11 c12 c13 0 0 c16c11 c13 0 0 −c16

c33 0 0 0

c44 0 0

Sym c44 0

c66

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦,

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

c11 c12 c13 0 0 0

c11 c13 0 0 0

c33 0 0 0

c44 0 0

Sym c44 0

c66

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

7.6. NONLINEAR DEFORMATIONS OF ISOTROPIC MATERIALS 329

Table 7.5: (continued) Crystal symmetry on the component matrix C of a rank-4property tensor relating two symmetric rank-2 tensor fields using Voigt’s repre-sentation. In the matrices below c66 = 1

2(c11 − c12).

Trigonal (7 or 6 parameters)

Classes 16–17 Classes 18–20⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

c11 c12 c13 c14 c15 0

c11 c13 −c14 −c15 0

c33 0 0 0

c44 0 −c15Sym c44 c14

c66

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦,

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

c11 c12 c13 c14 0 0

c11 c13 −c14 0 0

c33 0 0 0

c44 0 0

Sym c44 c14c66

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦Hexagonal and Transverse isotropic (5 parameters)

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

c11 c12 c13 0 0 0

c11 c13 0 0 0

c33 0 0 0

c44 0 0

Sym c44 0

c66

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦Cubic (3 parameters)

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

c11 c12 c12 0 0 0

c11 c12 0 0 0

c11 0 0 0

c44 0 0

Sym c44 0

c44

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦Isotropic (2 parameters)

⎡⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣

c11 c12 c12 0 0 0

c11 c12 0 0 0

c11 0 0 0

c66 0 0

Sym c66 0

c66

⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦

330 THERMOELASTIC SOLIDS

Table 7.6: Complete irreducible polynomial basis of scalar invariants of the elasticstiffness tensor for different crystal systems.

Crystal System Type Invariants

Triclinic 1 c11, c22, c33, c12, c13, c23

Monoclinic 2 c11, c22, c33, c2

12, c2

13, c23, c13c12

Orthorhombic 3 c11, c22, c33, c2

23, c2

13, c2

12, c12c23c13

Tetragonal 4 c11 + c22, c33, c2

13 + c2

23, c2

12, c11c22, c12(c11 − c22),c13c23(c11 − c22), c12c23c13, c12(c

2

13 − c2

23),c11c

2

23 + c22c2

13, c23c13(c2

13 − c2

23), c2

13c2

23

5 c11 + c22, c33, c2

12, c2

13 + c2

23, c11c22, c12c23c13,

c11c2

23 + c22c2

13, c2

13c2

23

Trigonal 6 c33, c11 + c22, c11c22 − c2

12, c11[(c11 + 3c22)2 − 12c212],

c213 + c2

23, c13(c2

13 − 3c2

23), (c11 − c22)c13 − 2c12c23,(c22 − c11)c23 − 2c12c13, 3c12(c11 − c22)

2 − 4c312,c23(c

2

23 − 3c2

13), c22c2

13 + c11c2

23 − 2c23c13c12,c13[(c11 + c22)

2 + 4(c212 − c2

22)] − 8c11c12c23,c23[(c11 + c22)

2 + 4(c212 − c2

22)] + 8c11c12c13,(c11 − c22)c23c13 + c12(c

2

23 − c2

13)7 c33, c11 + c22, c11c22 − c

2

12, c11[(c11 + 3c22)2 − 12c212],

c2

13 + c2

23, c23(c2

23 − 3c2

13), (c11 − c22)c23 + 2c12c13,c11c

2

13 + c22c2

23 + 2c23c13c12,c23[(c11 + c22)

2 + 4(c212 − c2

22)] + 8c11c12c13Hexagonal 8 c33, c11 + c22, c11c22 − c

2

12, c11[(c11 + 3c22)2 − 12c212],

c213 + c2

23, c2

13(c2

13 − 3c2

23)2, c11c

2

23 + c22c2

13 − 2c23c13c12,c12(c

2

13 − c2

23) + (c22 − c11)c13c23,3c12(c11 − c22)

2 − 4c312,c13c23[3(c

2

13 − c2

23)2 − 4c213c

2

23],c11(c

4

13 + 3c4

23) + 2c22c2

13(c2

13 + 3c2

23) − 8c12c23c3

13,

c2

13[(c11 + c22)2 + 4(c212 − c

2

22)]−2c11[(c11 + 3c22)(c

2

13 + c2

23) − 4c23c13c12],c23c13[(c11 + c22)

2 + 4(c212 − c2

22)] + 4c11c12(c2

23 − c2

13),c12[(c

2

13 + c2

23)2 + 4c223(c

2

13 − c23)] − 4c3

13c23(c11 − c22)9 c33, c11 + c22, c11c22 − c

2

12, c11[(c11 + 3c22)2 − 12c212],

c213 + c2

23, c2

13(c2

13 − 3c2

23)2,

c11c2

23 + c22c2

13 − 2c23c13c12,c11(c

4

13 + 3c4

23) + 2c22c2

13(c2

13 + 3c2

23) − 8c12c23c3

13,

c2

13[(c11 + 3c22)2 + 4(c212 − c

2

22)]−2c11[(c11 + 3c22)(c

2

13 + c2

23) − 4c23c13c12]

7.6. NONLINEAR DEFORMATIONS OF ISOTROPIC MATERIALS 331

Table 7.6: (continued) Complete irreducible polynomial basis of scalar invariantsof the elastic stiffness tensor for different crystal systems.

Crystal System Type Invariants

Cubic 10 c11 + c22 + c33, c22c33 + c33c11 + c11c22, c11c22c33,c223 + c

2

13 + c2

12, c2

13c2

12 + c2

12c2

23 + c2

23c2

13, c23c13c12,

c22c2

12 + c33c2

23 + c11c2

13, c2

13c33 + c2

12c11 + c2

23c22,

c33c2

22 + c11c2

33 + c22c2

11, c2

12c4

13 + c2

23c4

12 + c2

13c4

23,

c11c2

13c2

12 + c22c2

12c2

23 + c33c2

23c2

13, c2

23c22c33+c213c33c11 + c

2

12c11c22, c11c22c2

13 + c22c33c2

12+c33c11c

2

23, c2

23c2

13c22 + c2

13c2

12c33 + c2

12c2

23c1111 c11 + c22 + c33, c22c33 + c33c11 + c11c22, c11c22c33,

c223 + c2

13 + c2

12, c2

13c2

12 + c2

12c2

23 + c2

23c2

13, c23c13c12,

c22(c2

12 + c2

23) + c33(c2

13 + c2

23) + c11(c2

12 + c2

13),c11c

2

13c2

12 + c22c2

12c2

23 + c33c2

23c2

13,

c223c22c33 + c2

13c33c11 + c2

12c11c22

Transverse isotropic tr C, tr C2, tr C3, c33, c2

13 + c2

23

Isotropic tr C, tr C2, tr C3

Table 7.7: Relations for the elastic properties in terms of different independentparameters.

λ0, µ0 (c0)11, (c0)12 E0, ν0 E0, µ0

λ0 λ0 (c0)12 E0 ν0(1 + ν0)(1 − 2 ν0) µ0 (E0 − 2µ0)3µ0 −E0

µ0 µ0

1

2[(c0)11 − (c0)12] E0

2 (1 + ν0) µ0

E0

µ0 (3λ0 + 2µ0)λ0 + µ0

[(c0)11 − (c0)12] [(c0)11 + 2 (c0)12][(c0)11 + (c0)12] E0 E0

b0 λ0 +2

3µ0

1

3[(c0)11 + 2 (c0)12] E0

3 (1 − 2 ν0) E0 µ0

3 (3µ0 −E0)ν0

λ0

2 (λ0 + µ0) [(c0)12][(c0)11 + (c0)12] ν0E0

2µ0

− 1

332 THERMOELASTIC SOLIDS

Problems

1. Show that (7.40) and subsequently (7.43) are true.

2. First show that (7.50) is true and subsequently provide the details to obtain(7.205) and (7.206).

3. The Cartesian coordinates of a point, Xi, transforms upon a change of co-ordinate system according to

X ′i = aijXj,

where aij is the direction cosine which relates the axes of the two coordinatesystems. In the method of direct inspection, we have stated that if onecan write down a relation between the axes of the two coordinate systemsXi and Xj (e.g., X ′1 = −X1, X

′

2 = −X2), then this provides a convenientmethod for extracting the direction cosine associated with material tensors(e.g., c11 = −1, c12 = 0, c13 = 0, etc.).

The above statement is rather cavalier, however, and not strictly correct:the first equation involves coordinates of a point, while the latter equations,written by inspection, involve a vector relationship between reference axesin the two coordinate systems.

Demonstrate (by means of a sketch or otherwise) that the procedure is correct– i.e., that the coordinates of a point transform in exactly the same way asthe axes of the references system.

4. Show that for transversely isotropic materials, the requirement of satisfyingthe symmetries (MΘ,R1,R3,D2), the rank-2 and rank-4 property tensorshave components respectively given in Tables 7.4 and 7.5.

5. The trace of a rank-2 tensor is defined as the sum of its diagonal elements,trC = c11+c22+c33. Show that the trace of a tensor remains invariant upon achange of reference axes which is specified by a general transformation withdirection cosines [aij].

6. The engineering moduli of crystals may all be expressed in terms of theelements of the stiffness matrix constants cij (in Voigt’s notation). Show,for example, that for a cubic crystal the isothermal bulk modulus is given by

b0 = 1

κθ= 1

3(c11 + 2 c12) .

7. Repeat the analysis in Section 7.3.1 of the propagation of elastic waves in acubic crystals, but without making the isothermal assumption.

8. Solve for e in (7.183) to obtain the following result

tr e = 3a0 θ − p

b0= 3a0 θ − 3 (1 − 2 ν0)

E0

p,

7.6. NONLINEAR DEFORMATIONS OF ISOTROPIC MATERIALS 333

and subsequently show that

e = (a0 θ + λ0

2µ0 b0p)1 + 1

2µ0

σ = (a0 θ + 3 ν0

E0

p)1 + 1 + ν0

E0

σ,

where we have defined the mechanical pressure

p = −13trσ.

9. Eliminate θ between (7.180) and (7.183) to show that

σ = [(b′0 − 2

3µ0) tr e − 3 θ0 b0 a0

ce0

(η − η0)]1 + 2µ0 e, (7.352)

where

b′0 = b0 + 9 θ0 b20 a

20

ρ0 ce0

(7.353)

is the isentropic bulk modulus.

10. Solve (7.352) for e to show that

e = (θ0 a′0ce0

θ −ν′0E′

0

trσ)1 + 1 + ν′0E′

0

σ,

where, using (7.353),

a′0 = b0 a0

b′0

,

ν′0 = 3 b′0 − 2µ0

2 (3 b′0+ µ0) ,

E′0 = 9 b′0 µ0

3 b′0+ µ0

,

are the isentropic coefficient of thermal expansion, isentropic Poisson ratio,and isentropic Young modulus, respectively.

11. Show that in an elastic isotropic medium, the three acoustic wave speedsconsist of the longitudinal wave

v(1)w = (c11ρ)1/2 = (λ + 2µ

ρ)1/2

and the two shear waves

v(2,3)w = (c11 − c122ρ

)1/2 = (µρ)1/2 .

12. Use (7.215) to show (7.216) with B13 = B23 = 0.13. Prove (7.270).

334 THERMOELASTIC SOLIDS

14. The representation for the stress tensor for a homogeneous isothermal isotropicelastic solid is given by

σ = β0 (B(1),B(2),B(3))1 + β1 (B(1),B(2),B(3))B + β2 (B(1),B(2),B(3))B2.

(7.354)

i) Show that when B = 1, the above equation yields the residual stress

σ = −p1,where p is the residual pressure given by

p = −β0(3,3,1)− β1(3,3,1)− β2(3,3,1).ii) Expand (7.354) about the state B = 1 and show that the linear approx-

imation is given by

σ = −p1 + 1

2(λ + p)(trB′)1 + (µ − p)B′,

where B′ =B − 1,

λ + p = 2 ( ∂

∂B(1)+ 2

∂

∂B(2)+

∂

∂B(3))(β0 + β1 + β2)∣

B(1)=B(2)=3,B(3)=1

,

andµ − p = β1(3,3,1) + 2β2(3,3,1).

15. Using the Mooney–Rivlin expression for the free energy

ρψ = 1

2µ [(1

2+ β)(B(1) − 3) + (1

2− β) (B(2) − 3)] , µ > 0, −1

2≤ β ≤ 1

2,

look at (i) simple extension, (ii) dilatation or compression, and (iii) simpleshear, to see what is predicted. As a special case, what does the neo-Hookeanmaterial (β = 1/2) say.

16. Use the residual inequality to find the restrictions on the coefficients of

q = α00 g + α01 1 ⋅ g + α10B ⋅ g.

17. Determine C and provide an explicit relationship between A and B emergingfrom the conditions σ<rr> = 0 at r = r1, r2, for a Mooney–Rivlin materialunder the deformation of bending, stretching, and shearing of a rectangularblock given by (7.306)–(7.308). What about if the material is neo-Hookean?

18. Analyze fully the deformation

r =√AR2 +B,

θ = Θ +DZ,

z = FZ.

What are the consequences for a Mooney–Rivlin material? What about fora neo-Hookean?

BIBLIOGRAPHY 335

19. Obtain the stress tensor for the deformation of straightening, stretching, andshearing of a sector of a hollow circular cylinder given by (7.309)–(7.311).

20. Obtain the stress tensor for the deformation of inflation, bending, torsion,extension, and shearing of an annular wedge given by (7.312)–(7.314).

21. Obtain the stress tensor for the deformation of inflation or eversion of asector of a spherical shell given by (7.315)–(7.317).

22. Obtain the stress tensor for the deformation of inflation, bending, extension,and azimuthal shearing of an annular wedge given by (7.318)–(7.320).

23. In an effort to produce a theory of heat conduction for which thermal dis-turbances propagate with finite speed, it is reasonable to investigate a ho-mogeneous material for which p = q = 1 and r = 0 in (5.38), so that (5.40)becomes

T (X, t) = T F(X, t), F(X, t), θ(X, t), θ(X, t),G(X, t).i) Derive the reduced constitutive equations in this case.

ii) Derive the thermodynamic restrictions.

24. In order to include additional memory effects, it is reasonable to investigatea homogeneous material for which p = 2 and q = r = 0 in (5.38), so that (5.40)becomes

T (X, t) = T F(X, t), F(X, t), F(X, t), θ(X, t),G(X, t).i) Derive the reduced constitutive equations in this case.

ii) Derive the thermodynamic restrictions.

25. In order to include additional spatial effects, it is reasonable to investigate ahomogeneous material for which P = 2 and Q = 1 in (5.30) and p = 1, q = r = 0in (5.38), so that (5.33) becomes

T (X, t) = T F(X, t),Grad F(X, t), θ(X, t),G(X, t).i) Derive the reduced constitutive equations in this case.

ii) Show that the thermodynamic restrictions are such as to reduce it tothe case of the thermoelastic material with heat conduction considered.

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M. Bailyn. A Survey of Thermodynamics. AIP Press, New York, NY, 1994.

M.A. Biot. Thermoelasticity and irreversible thermodynamics. Journal of AppliedPhysics, 27(3):240–253, 1956.

336 THERMOELASTIC SOLIDS

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T.J. Chung. General Continuum Mechanics. Cambridge University Press, NewYork, NY, 2007.

S.C. Cowin and M.M. Mehrabadi. On the identification of material symmetry foranisotropic elastic materials. Quarterly of Applied Mathematics, 40(4):451–476,1987.

R.S. Dhaliwal and H.H. Sherief. Generalized thermoelasticity for anisotropicmedia. Quarterly of Applied Mathematics, 38(1):1–8, 1980.

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E. Hartmann. An Introduction to Crystal Physics. International Union of Crys-tallography, 2001.

P. Haupt. Continuum Mechanics and Theory of Materials. Springer-Verlag,Berlin, 2000.

G.A. Holzapfel. Nonlinear Solid Mechanics. John Wiley & Sons, Ltd., Chichester,England, 2005.

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V.A. Lubarda. On thermodynamic potentials in linear thermoelasticity. Inter-national Journal of Solids and Structures, 41(26):7377–7398, 2004.

I. Müller. The coldness, a universal function in thermoelastic bodies. Archive forRational Mechanics and Analysis, 41(5):319–332, 1971.

A.S. Nowick. Crystal Properties via Group Theory. Cambridge University Press,Cambridge, England, 1995.

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J.F. Nye. Physical Properties of Crystals. Oxford University Press, Oxford,England, 2008.

R.W. Ogden. Non-Linear Elastic Deformations. John Wiley & Sons, New York,1984.

P. Podio-Guidugli. A primer in elasticity. Journal of Elasticity, 58(1):1–104,2000.

R.S. Rivlin. The fundamental equations of nonlinear continuum mechanics. In S.I.Pai, A.J. Faller, T.L. Lincoln, D.A. Tidman, G.N. Trytten, and T.D. Wilkerson,editors, Dynamics of Fluids in Porous Media, pages 83–126, Academic Press,New York, 1966.

D.E. Sands. Introduction to Crystallography. Dover Publications, Inc., Mineola,NY, 1975.

J.N. Sharma and H. Singh. Generalized thermoelastic waves in anisotropic media.Journal of the Acoustical Society of America, 85(4):1407–1413, 1989.

R.T. Shield. Deformations possible in every compressible, isotropic, perfectlyelastic material. Journal of Elasticity, 1(1):91–92, 1971.

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G.F. Smith and R.S. Rivlin. The anisotropic tensors. Quarterly of Applied Math-ematics, 15(3):308–314, 1957.

338 THERMOELASTIC SOLIDS

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E.S. Suhubi. Constitutive equations for simple materials: Thermoelastic solids.In A.C. Eringen, editor, Continuum Physics, volume II, chapter 2, pages 173–265.Academic Press, Inc., New York, NY, 1975.

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D.C. Wallace. Thermoelastic theory of stressed crystals and higher-order elasticconstants. Solid State Physics, 25(1):301–404, 1970.

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C.-C. Wang. On the symmetry of the heat conduction tensor. In C. Truesdell,editor, Rational Thermodynamics, pages 396–401. Springer-Verlag, New York,NY, 1984.

8

Fluids

We have shown earlier (see (5.114)) that the response functional for a simple fluidis given by

T (x, t) = F0≤s<∞

ρ(x, t), (t)C(t)(x, s), θ(t)(x, s), (t)g(t)(x, s). (8.1)

We now assume that continuous material derivatives exist, and consider the sim-plest constitutive function of a thermoviscous fluid. From (7.5), (7.7), and (7.8),we have shown that, assuming continuous material derivatives with respect to sat s = 0, we have

θ(t)(s) = θ(t) − θ(t) s +⋯, (8.2)

(t)g(t)(s) = g(t) − [LT (t) ⋅ g(t) + g(t)] s +⋯, (8.3)

(t)C(t)(s) = 1 − 2 D(t) s +⋯. (8.4)

Subsequently, if we consider rates of deformation gradients up to first order (p = 1),and temperature and its gradients up to zero order (q = r = 0), we have the followingconstitutive equations for a thermoviscous fluid:

ψ = ψ (ρ(x, t),D(x, t), θ(x, t),g(x, t)) , (8.5)

η = η (ρ(x, t),D(x, t), θ(x, t),g(x, t)) , (8.6)

q = q (ρ(x, t),D(x, t), θ(x, t),g(x, t)) , (8.7)

h = h (ρ(x, t),D(x, t), θ(x, t),g(x, t)) , (8.8)

σ = σ (ρ(x, t),D(x, t), θ(x, t),g(x, t)) , (8.9)

which are required to satisfy the following frame-invariance conditions:

ψ (ρ,Q ⋅D ⋅QT , θ,Q ⋅ g) = ψ (ρ,D, θ,g) , (8.10)

η (ρ,Q ⋅D ⋅QT , θ,Q ⋅ g) = η (ρ,D, θ,g) , (8.11)

q (ρ,Q ⋅D ⋅QT , θ,Q ⋅ g) = Q ⋅ q (ρ,D, θ,g) , (8.12)

h (ρ,Q ⋅D ⋅QT , θ,Q ⋅ g) = Q ⋅ h (ρ,D, θ,g) , (8.13)

σ (ρ,Q ⋅D ⋅QT , θ,Q ⋅ g) = Q ⋅σ (ρ,D, θ,g) ⋅QT , (8.14)

339

340 FLUIDS

for all Q ∈ O(V ). As noted earlier, material symmetry requires that ψ, η, q,h, and σ be isotropic scalar, vector, and tensor functions that satisfy the aboveframe-indifference equations.

8.1 Coleman–Noll procedure

From (5.252), and since when σ = σT , it follows that L ∶ σ =D ∶ σ, we have

−γv θ ≡ ρ (ψ + η θ) −Dik σik +qi

θgi − θ (hi − qi

θ),i

+ ρθ (b − rθ) ≤ 0. (8.15)

For simplicity, we have suppressed the functional dependencies in the Clausius–Duhem inequality so that only the constitutive dependent and independent vari-ables appear explicitly. Using the chain rule, we can write

ψ = ∂ψ∂ρ

ρ +∂ψ

∂Dik

Dik +∂ψ

∂θθ +

∂ψ

∂gigi, (8.16)

so that substituting this into the inequality, and considering that from the localmass balance (4.98) we have ρ = −ρdivv = −ρ trD, we arrive at

−γv θ ≡ ρ(η + ∂ψ∂θ) θ − (σik + ρ2 ∂ψ

∂ρδik)Dik + ρ

∂ψ

∂Dik

Dik + ρ∂ψ

∂gigi +

qi

θgi − θ (hi − qi

θ),i

+ ρθ (b − rθ) ≤ 0. (8.17)

Now the response functions ψ, η, qi, hi, σik are fixed if we fix the independentvariables ρ,Dik, θ, gi. Subsequently, the above equation will then depend linearlyon θ, gi, and Dik, which can still be made to vary.

By requiring that the inequality hold for arbitrary choices of ρ,Dik, θ, gi andassuming that

hi = qiθ

and b = rθ

(8.18)

(these can in fact be proved as in the previous chapter or as in the next section),we get that

ρ(η + ∂ψ∂θ) = 0, ρ

∂ψ

∂gi= 0, ρ

∂ψ

∂Dik

= 0, −(σik + ρ2 ∂ψ∂ρ

δik)Dik +qi

θgi ≤ 0,(8.19)

or

ψ = ψ(ρ, θ), (8.20)

η = −∂ψ

∂θ, (8.21)

−γv θ ≡ −(σik + ρ2 ∂ψ∂ρ

δik)Dik +qi

θgi ≡ A(ρ,Dik, θ, gi) ≤ 0. (8.22)

Equation (8.22) is called the residual entropy inequality. If we have no deformation,then Dik = 0 and obtain the heat conduction inequality

−qi

θgi ≥ 0, (8.23)

8.1. COLEMAN–NOLL PROCEDURE 341

which states that the angle between a nonzero temperature gradient and a nonzeroheat flux is greater than or equal to 90. If the material is isothermal, then gi = 0and the residual entropy inequality becomes the mechanical dissipation inequality

(σik + ρ2 ∂ψ∂ρ

δik)Dik ≥ 0. (8.24)

We recall that a state with no entropy production is a thermodynamic equi-librium state. At the equilibrium state, Dik = gi = 0 and we denote the equilib-rium stress and heat flux by σe

ik = σeik(ρ,0, θ,0) and qei = qei (ρ,0, θ,0), so that

A(ρ,0, θ,0) = 0. Then, assuming that A is continuous with Dik and gi at theequilibrium state, the necessary conditions for A to be a maximum there are givenby

∂A

∂Dik

∣D=0, g=0

= σeik + ρ

2 ∂ψ

∂ρδik = 0 and

∂A

∂gi∣D=0, g=0

= qeiθ= 0. (8.25)

Thus, at equilibrium, we must have

σeik = −ρ2 ∂ψ

∂ρδik and qei = 0. (8.26)

We note that this agrees with equilibrium thermodynamics, since if we take v tobe the specific volume, i.e., v = 1/ρ, then

dv = − 1

ρ2dρ, (8.27)

and since

σeik = −p δik and p = −∂ψ

∂v, (8.28)

where p is the thermodynamic pressure, we get

σeik = −ρ2 ∂ψ

∂ρδik. (8.29)

The requirement for the matrix of second derivatives of A with respect to D andg for A to be a maximum at a thermodynamic equilibrium state leads to additionalrestrictions. These reduce exactly to the requirement (5.259), whose analysis wehave previously discussed and whose resulting restrictions are given by (5.268).

Subsequently, we take

σik = σeik + σ

dik and qi = qei + qdi , (8.30)

where σdik and qdi represent dissipative, or nonequilibrium, parts of the stress tensor

and heat flux, thus allowing us to rewrite the residual entropy inequality as

−γv θ ≡ −σdikDik +

qdiθgi ≤ 0. (8.31)

This inequality clearly states that heat, in the presence of deformation, does not,in general, flow from higher to lower temperatures. From the above definition of athermodynamic equilibrium state, we note that σd

ik and qdi must vanish when Dik

and gi vanish, respectively.

342 FLUIDS

8.2 Müller–Liu procedure

In the Müller–Liu procedure, we require that the entropy inequality (4.219) besatisfied for all tensor fields satisfying the conservation of mass (4.98), and linearmomentum (4.109), angular momentum (for a nonpolar material) (4.120), andenergy (4.207) balances. Satisfaction of these equations is enforced through theuse of Lagrange multipliers (see (5.300), and note that divv = trD and Φ = L ∶

σ =D ∶ σ):

γv ≡ [ρ(η − b) + divh] − λρ [ρ + ρ trD] −λv⋅ [ρ (v − f) − divσ] −

λe [ρ (e − r) −D ∶ σ + divq] ≥ 0. (8.32)

Here, we take the constitutive quantities to be given by

C = e, η,q,h,σ (8.33)

and which are functions of the independent basic fields

I = ρ,D, θ,g . (8.34)

By the principle of equipresence, the Lagrange multipliers λρ,λv, λe are requiredto be functions of the same independent basic fields.

As in the Coleman–Noll procedure, we rewrite the derivatives of the constitutivedependent variables appearing in (8.32), e, η,gradq,gradh,gradσ, in terms ofderivatives of the independent basic fields, ρ, D, θ, g,grad ρ,gradD,grad θ, gradg:

e = ∂e

∂ρρ +

∂e

∂D∶ D +

∂e

∂θθ +

∂e

∂g⋅ g, (8.35)

η = ∂η

∂ρρ +

∂η

∂D∶ D +

∂η

∂θθ +

∂η

∂g⋅ g, (8.36)

gradq = ∂q

∂ρgradρ +

∂q

∂D∶ gradD +

∂q

∂θgradθ +

∂q

∂g⋅ gradg, (8.37)

gradh = ∂h

∂ρgradρ +

∂h

∂D∶ gradD +

∂h

∂θgradθ +

∂h

∂g⋅ gradg, (8.38)

gradσ = ∂σ

∂ρgradρ +

∂σ

∂D∶ gradD +

∂σ

∂θgradθ +

∂σ

∂g⋅ gradg. (8.39)

Subsequently, identifying (8.32) with (5.297), it is easy to verify that

β = ∂h

∂θ⋅ gradθ − ρ b, (8.40)

λ = (λρ,λv, λe) , (8.41)

α = (ρ ∂η∂ρ,0, ρ

∂η

∂D, ρ∂η

∂θ, ρ∂η

∂g,∂h

∂ρ,∂h

∂D,∂h

∂g) , (8.42)

a = (ρ, v, D, θ, g,grad ρ,gradD,gradg)T , (8.43)

b = (ρ trD,−∂σ

∂θ⋅ gradθ − ρ f ,

∂q

∂θ⋅ gradθ −D ∶ σ − ρr)T (8.44)

8.2. MÜLLER–LIU PROCEDURE 343

and

A =⎛⎜⎜⎜⎜⎜⎝

1 0 0 0 0 0 0 0

0 ρ1 0 0 0 −∂σ

∂ρ−∂σ

∂D−∂σ

∂g

ρ∂e

∂ρ0 ρ

∂e

∂Dρ∂e

∂θρ∂e

∂g

∂q

∂ρ

∂q

∂D

∂q

∂g

⎞⎟⎟⎟⎟⎟⎠. (8.45)

It should be noted from the above definitions that the entropy inequality (4.219)corresponds to (5.295), and the conservation of mass (4.98), and linear momentum(4.109), angular momentum (for a nonpolar material) (4.120), and energy (4.207)balances correspond to the system (5.296). Subsequently, from (5.298), and usingthe fact that ρ > 0, we obtain the result that λv = 0, and the following additionalrelations:

∂η

∂ρ− λρ

1

ρ− λe

∂e

∂ρ= 0, (8.46)

∂η

∂Dkl

− λe∂e

∂Dkl

= 0, (8.47)

∂η

∂θ− λe

∂e

∂θ= 0, (8.48)

∂η

∂gk− λe

∂e

∂gk= 0, (8.49)

∂hj

∂ρ− λe

∂qj

∂ρ= 0, (8.50)

K ′′jkl +K′′

lkj = 0, (8.51)

K ′jk +K′

kj = 0, (8.52)

where, in writing (8.51) and (8.52), we have defined

K ′′jkl ≡ ∂hj

∂Dkl

− λe∂qj

∂Dkl

and K ′jk ≡ ∂hj∂gk− λe

∂qj

∂gk, (8.53)

and used the symmetries Dkl =Dlk and

∂Dkl

∂xj= 1

2(∂vk,l∂xj

+∂vl,k

∂xj) = 1

2( ∂2vk

∂xj∂xl+

∂2vl

∂xj∂xk) and

∂θ,k

∂xj= ∂θ,j∂xk

. (8.54)

Furthermore, we note that K ′′jkl =K ′′jlk and from (8.51), we have

K ′′jkl = −K ′′lkj = −K ′′ljk =K ′′kjl =K ′′klj = −K ′′jlk = −K ′′jkl. (8.55)

344 FLUIDS

Thus, (8.51) further reduces toK ′′jkl = 0. (8.56)

In addition, from (5.299), we obtain the following reduced entropy inequality:

γv ≡ (∂hj∂θ− λe

∂qj

∂θ) ∂θ∂xj− ρ (b − λe r) − (λρ ρ δij − λe σij)Dij ≥ 0. (8.57)

Note that for the vector fields q and h to be frame indifferent, they must satisfy(8.12) and (8.13), or, suppressing for the moment dependencies on ρ and θ, wewrite, e.g.,

fi (Qjk) ≡ hi (QkmQlnDmn,Qmngn) −Qijhj (Dkl, gm) = 0. (8.58)

Now, differentiating with respect to Qpq and subsequently taking Qij = δij , weobtain

∂fi

∂Qpq

∣Q=1

= ∂hi

∂Dpl

Dql +∂hi

∂Dkp

Dkq +∂hi

∂gpgq − δiphq = 0. (8.59)

It is evident that∂fi

∂Qpq

∣Q=1

= ∂fi

∂Qqp

∣Q=1

, (8.60)

so that we can write

(δiphq − δiqhp) = ( ∂hi∂Dpl

Dql +∂hi

∂Dkp

Dkq −∂hi

∂Dql

Dpl −∂hi

∂Dkq

Dkp) +(∂hi∂gp

gq −∂hi

∂gqgp) , (8.61)

and similarly

(δipqq − δiqqp) = ( ∂qi∂Dpl

Dql +∂qi

∂Dkp

Dkq −∂qi

∂Dql

Dpl −∂qi

∂Dkq

Dkp) +( ∂qi∂gp

gq −∂qi

∂gqgp) . (8.62)

Now multiplying the latter equation by λe, subtracting it from the first equation,using the definitions (8.53) and equations (8.52) and (8.56), we obtain

(δipKq − δiqKp) = (K ′ipgq −K ′iqgp) , (8.63)

where we have defined

Kj ≡ hj − λe qj . (8.64)

Using the fact that K ′jk is skew-symmetric (see (8.52)), system (8.63) yields thefollowing equations:

K1 =K ′12 g2, K2 =K ′23 g3, K3 =K ′31 g1, (8.65)

0 =K ′12 g2 +K ′31 g3, 0 =K ′12 g1 +K ′23 g3, 0 =K ′31 g1 +K ′23 g2, (8.66)

0 =K ′23 g1 +K ′12 g3, 0 =K ′31 g2 +K ′12 g3, 0 =K ′23 g1 +K ′31 g2. (8.67)

8.2. MÜLLER–LIU PROCEDURE 345

Now, by appropriately linearly combining the three equations (8.67), we obtain

K ′12 g3 = 0, K ′23 g1 = 0, K ′31 g2 = 0, (8.68)

and since gk ≠ 0 in general, then we must have that

K ′jk = 0, (8.69)

and subsequently from (8.65), we have that

Kj = 0, (8.70)

and from (8.64), we have that

hj = λe(ρ,Dkl, θ, gk) qj . (8.71)

Now, using (8.56), (8.69), and (8.70), equations (8.53) become

K ′′jkl = ∂Kj

∂Dkl

+∂λe

∂Dkl

qj = ∂λe

∂Dkl

qj = 0 and K ′jk = ∂Kj

∂gk+∂λe

∂gkqj = ∂λe

∂gkqj = 0, (8.72)

and since qj ≠ 0 in general, we must have that λe is independent of Dkl and gk,i.e., λe = λe(ρ, θ), and thus

hj = λe(ρ, θ) qj . (8.73)

Using this result in (8.50), we obtain

∂λe

∂ρqj = 0. (8.74)

Now, since qj ≠ 0 in general, we must come to the conclusion that

λe = λe(θ), (8.75)

and subsequently, we can rewrite (8.73) as

hj = λe(θ) qj . (8.76)

Now since λe = λe(θ), from (8.47) and (8.49), we see that the quantity (η−λe e)is independent of D and g. Furthermore, from (8.48), we have that

∂

∂θ(η − λe e) = −e λe

θ, (8.77)

so that it follows that e is independent of D and g (if we exclude the possibilityof λe being a constant, in general). Subsequently, we have that

e = e(ρ, θ) and η = η(ρ, θ), (8.78)

and the relation (8.46) can be rewritten as

de = 1

λe(dη − λρ

ρdρ) , (8.79)

346 FLUIDS

which also shows thatλρ = λρ(ρ, θ). (8.80)

Note that since ψ = e − θ η, we also have that

ψ = ψ(ρ, θ). (8.81)

Using (8.76), the reduced inequality (8.57) becomes

γv ≡ dλedθ

qj gj − ρ (b − λe r) − (λρ ρ δij − λe σij)Dij ≥ 0. (8.82)

We recall that a state with no entropy production is a thermodynamic equilib-rium state. At the equilibrium state, Dij = gj = 0, and we denote the equilibriumstress and heat flux by σe

ij = σeij(ρ,0, θ,0) = −p(ρ, θ) δij and qei = qei (ρ,0, θ,0),

so that γv(ρ,0, θ,0) = 0. Noting that the independent fields are assumed to notdepend on the external energy supply r, and since r can be of arbitrary sign, wesee from (8.82) that we must have that

b − λe r = 0. (8.83)

Subsequently, the reduced entropy inequality becomes

γv ≡ (∂hj∂θ− λe

∂qj

∂θ) ∂θ∂xj− (λρ ρ δij − λe σij)Dij ≥ 0. (8.84)

Then, assuming that γv is continuous with Dij and gj at the equilibrium state,the necessary conditions for γv to be a minimum there are given by

∂γv

∂Dij

∣D=0, g=0

= (λe p + λρ ρ) δij = 0 and∂γv

∂gj∣D=0, g=0

= dλedθ

qej = 0. (8.85)

Subsequently, we have that

λρ = −λe p(ρ, θ)ρ

and qej = 0. (8.86)

Substituting the first equation into (8.79), we have

de = 1

λedη +

p

ρ2dρ. (8.87)

By comparison with the Gibbs equation (5.138) (and noting that in our case n = 1,ν1 → 1/ρ and τ 1 → −p), we see that

λe = λe(θ) = 1

θ, (8.88)

and subsequently,

hj = 1

θqj (8.89)

and from (8.83) we also have that

b = 1

θr. (8.90)

8.3. REPRESENTATIONS OF qd AND σd 347

In addition, we now have from (8.86)1 that

λρ = −p(ρ, θ)ρθ

, (8.91)

which for an ideal gas, where the thermal equation of state is p(ρ, θ) = ρRθ,reduces to λρ = −R, where R is the ideal gas constant. More generally, λρ will bea function of ρ and θ as given by (8.91). We can also rewrite (8.87) in terms ofthe Helmholtz free energy,

dψ = −η dθ + p

ρ2dρ, (8.92)

from which we easily see that

η = − ∂ψ∂θ∣ρ

and p = ρ2 ∂ψ∂ρ∣θ

. (8.93)

The requirement for the matrix of second derivatives of γv (or A in the Coleman–Noll procedure) with respect to D and g be positive definite at equilibrium for theentropy production to be a minimum provides additional restrictions. These reduceexactly to the requirement (5.259), whose analysis we have previously discussedand whose resulting restrictions are given by (5.268).

Finally, taking

σij = σeij + σ

dij and qj = qej + qdj , (8.94)

where σdij and qdj represent dissipative, or nonequilibrium, parts of the stress tensor

and heat flux, and using (8.89), the residual entropy inequality (8.84) becomes

γv θ ≡ σdijDij −

1

θqdj gj ≥ 0. (8.95)

This inequality clearly states that heat, in the presence of deformation, does not,in general, flow from higher to lower temperatures. From the above definition of athermodynamic equilibrium state, we note that σd

ij and qdj must vanish when Dij

and gj vanish, respectively.

8.3 Representations of qd and σd

From Tables 5.1–5.3, we can immediately write down the representations for qdi andσdik corresponding to isotropic vector-valued and symmetric tensor-valued func-

tions of vector θ,i and symmetric second-order tensor Dik. These are

qdi = − (κ0 δil + κ1Dil + κ2DikDkl) θ,l, (8.96)

σdik = α0 δik + α1Dik + α2DilDlk + α3 θ,iθ,k +

1

2α4 (θ,iDkl + θ,kDil) θ,l +

1

2α5 (θ,iDkl + θ,kDil)Dlmθ,m, (8.97)

348 FLUIDS

where κp (p = 0,1,2) and αq (q = 0, . . . ,5) are functions of ρ and θ, and the isotropicscalar invariants I1, . . . , I6, where

I1 =Dii =D(1) = d(1) + d(2) + d(3), (8.98)

I2 =DikDki =D2

(1) − 2D(2) = d(1)2 + d(2)2 + d(3)2, (8.99)

I3 =DikDklDli =D3

(1) − 3D(1)D(2) + 3D(3) = d(1)3 + d(2)3 + d(3)3, (8.100)

I4 = θ,iθ,i, I5 = θ,iDikθ,k, I6 = θ,iDikDklθ,l, (8.101)

and, using (3.135)–(3.137),

D(1) = d(1) +d(2) +d(3), D(2) = d(1)d(2) +d(2)d(3) +d(3)d(1), D(3) = d(1)d(2)d(3),(8.102)

where d(i), i = 1,2,3, are the principal deformation rates. We note that we haveused (3.98)–(3.100) and (8.102) in relating the isotropic scalar invariants I1, I2,and I3 to D(1), D(2), and D(3), and d(1), d(2), and d(3). In the representation(8.97), we require that α0 → 0 as θ,i → 0 and Dik → 0, since as we approachequilibrium, we must have σd

ik → 0.Now, substituting the representations (8.96) and (8.97) into the reduced en-

tropy inequality (8.95), and using the Cayley–Hamilton theorem (3.95) and thedefinitions of the invariants (8.98)–(8.101), we obtain

0 ≤ γv θ ≡ α0I1 + α1I2 + α2I3 + [16α5 (I31 − 3I1I2 + 2I3) + κ0θ ] I4 +

[α3 −1

2α5 (I21 − I2) + κ1

θ] I5 + (α4 + α5I1 +

κ2

θ) I6, (8.103)

= α0D(1) + α1 (D2

(1) − 2D(2)) + α2 (D3

(1) − 3D(1)D(2) + 3D(3)) +(α5D(3) +

κ0

θ) I4 + (α3 − α5D(2) +

κ1

θ) I5 +

(α4 + α5D(1) +κ2

θ) I6. (8.104)

The constitutive equations (8.96) and (8.97) are too complicated for solutionsof fluid problems. Generally, polynomial approximations of various degrees inD and g are used. We note that deg I1 = degD(1) = 1, deg I2 = degD(2) = 2,

deg I3 = degD(3) = 3, deg I4 = 2, deg I5 = 3, and deg I6 = 4. Subsequently, for qdi tobe of order N , κ0, κ1, and κ2 can at most be of degrees N − 1, N − 2, and N − 3,respectively, while for σd

ik to be of order N , α0, α1, α2, α3, α4, and α5 can at mostbe of degrees N , N − 1, N − 2, N − 2, N − 3, and N − 4, respectively.

Below, as alternative to using directly the above general representations, we willconstruct the following specific representations in detail to illustrate the use of theframe-indifference restrictions:

a) Rigid motion, thus Dik = 0b) Isothermal conditions, thus θ,i = 0c) Zeroth-order representations in Dik and θ,i

d) First-order representations in Dik and θ,i

8.3. REPRESENTATIONS OF qd AND σd 349

e) Second-order representations in Dik and θ,i

The results that we obtain will be seen to correspond to specific simplifications of(8.96), (8.97), and (8.103) or (8.104).

a) Rigid motion

In this case, since Dik = 0, we have qdi = qdi (ρ, θ, θ,k) and σdik = σd

ik(ρ, θ, θ,k). FromTables 5.1–5.3, and as can be readily verified from (8.96) and (8.97) when Dik = 0,the most general representations consistent with frame indifference are

qdi = −κ0 θ,i, (8.105)

σdik = α0 δik + α3 θ,iθ,k, (8.106)

where

κ0 = κ0(ρ, θ, I4), α0 = α0(ρ, θ, I4), α3 = α3(ρ, θ, I4). (8.107)

Now, substituting the representations (8.105) and (8.106) into the reduced en-tropy inequality (8.95), or from (8.103), we see that the reduced entropy inequalityin this case is

γv θ ≡ κ0θI4 ≥ 0, (8.108)

or, since θ > 0 and I4 > 0,0 ≤ κ0 <∞. (8.109)

Note that the reduced entropy inequality imposes no restrictions on α0 and α3

since Dik = 0. Subsequently, we have

qi = −κ0 θ,i, (8.110)

σik = (−p + α0) δik + α3 θ,iθ,k, (8.111)

which represent the constitutive equations that account for heat conduction andthermoelastic stress.

If we linearize qi in the limit θ,i → 0, we obtain Fourier’s law of heat conduction

qi = −k θ,i, (8.112)

where k = k(ρ, θ) ≡ κ0(ρ, θ,0). In addition, if we linearize σik in the limit θ,i → 0

and require that σdik → 0 as θ,i → 0, we have that α0(ρ, θ,0) = 0, and thus we

obtain the stress tensor for a perfect fluid

σik = −p δik, (8.113)

where p = p(ρ, θ). Balance equations using (8.112) and (8.113) are referred to asEuler–Fourier equations.

350 FLUIDS

b) Isothermal conditions

In this case, since θ,i = 0, it is easy to see that

qi = qdi = qdi (ρ,Dik, θ) = 0 and σdik = σd

ik(ρ,Dik, θ). (8.114)

From Tables 5.1–5.3, and as can be readily verified from (8.97) when θ,i = 0, themost general representation consistent with frame indifference is

σdik = α0δik + α1Dik + α2DilDlk, (8.115)

where αi = αi(ρ, θ,D(1),D(2),D(3)), i = 0,1,2, and α0(ρ, θ,0,0,0) = 0 when Dik =0. In order that α0, α1, and α2 be continuous at the coalescence of the principalvalues of Dik, it is required that σd

ik be three times continuously differentiable inthe components of Dik.

Subsequently, suppressing all dependencies of αj ’s on ρ, θ, and D(j)’s, from(8.26) and (8.30), we can write

σik = (−ρ2 ∂ψ∂ρ+ α0) δik + α1Dik + α2DilDlk. (8.116)

This is the constitutive relation for what is referred to as a Reiner–Rivlin fluid. Ifwe let p be the mechanical pressure defined by

p = −13σkk, (8.117)

and since from (8.26) and (8.28), the thermodynamic pressure is

p = ρ2 ∂ψ∂ρ

, (8.118)

we have, using (3.98) and (3.99), that

−p = −p + α0 +1

3α1D(1) +

1

3α2 (D2

(1) − 2D(2)) . (8.119)

Note that the mechanical and thermodynamic pressures are generally different!From (8.104), we see that the reduced entropy inequality in this case is

γv θ ≡ α0D(1) + α1(D2

(1) − 2D(2)) + α2(D3

(1) − 3D(1)D(2) + 3D(3)) ≥ 0. (8.120)

If we expand the coefficients in power series of the invariants

αi = ∞∑p=0

∞

∑q=0

∞

∑r=0

aipqrDp

(1)Dq

(2)Dr(3), (8.121)

where aipqr = aipqr(ρ, θ), then the above inequality will impose certain restrictionson the coefficients. Clearly, we must have that a0000 = 0. However, it is not obvioushow one finds a general inequality that aipqr must satisfy for arbitrary orders.

8.3. REPRESENTATIONS OF qd AND σd 351

c) Zeroth-order representations

From (8.96) and (8.97), with Dik = 0 and θ,i = 0, we see that the zeroth-orderrepresentations are given by

qdi = 0 and σdik = 0; (8.122)

thusqi = 0 and σik = −p δik, (8.123)

where p = p(ρ, θ). This corresponds to inviscid fluids and p is a thermodynamicproperty. The corresponding linear momentum equation is called Euler’s equationof motion. Fluids in which p = p(ρ) are called barotropic fluids.

d) First-order representations

If we linearize (8.96) and (8.97) in the limits θ,i → 0 and Dik → 0, and noting thatfor first-order representations κi = κi(ρ, θ,D(1)) and αi = αi(ρ, θ,D(1)), we thenhave that

qdi = −κ00 θ,i (8.124)

and

σdik = (α00 + α01D(1)) δik + α10Dik, (8.125)

where now κ00, α00, α01, and α10 are only functions of ρ and θ. Since the qdi andσdik must vanish when θ,i and Dik vanish, we have that α00 = 0. From the reduced

entropy inequality (8.104), we now have

γv θ ≡ (α01 + α10)D2

(1) − 2α10D(2) +κ00

θI4 ≥ 0. (8.126)

Rewriting D(2) in terms of the invariant of the deviatoric component using (3.108),we have

γv θ ≡ (α01 +1

3α10)D2

(1) − 2α10D′

(2) +κ00

θI4 ≥ 0, (8.127)

or, rewriting the above in terms of principal invariants using (3.135) and (3.114),we have

γv θ ≡ (α01 +1

3α10)(λ(1) + λ(2) + λ(3))2 +

1

3α10 [(λ(1) − λ(2))2 + (λ(2) − λ(3))2 + (λ(3) − λ(1))2] + κ00

θI4 ≥ 0. (8.128)

Since the inequality must hold for any deformation, then we require that

α01 +1

3α10 ≥ 0, α10 ≥ 0, and κ00 ≥ 0. (8.129)

Now calling α01 = λ, α10 = 2µ, and κ00 = k, where λ = λ(ρ, θ) and µ = µ(ρ, θ) arethe dilatational and shear viscosities, and k = k(ρ, θ) is the thermal conductivity,the restrictions (8.129) become

0 ≤ ζ(ρ, θ) <∞, 0 ≤ µ(ρ, θ) <∞, and 0 ≤ k <∞, (8.130)

352 FLUIDS

where ζ ≡ (λ + 2

3µ) is called the bulk viscosity.

Subsequently, the constitutive equations (8.94) and the reduced entropy inequal-ity (8.127) become

qi = −k θ,i, (8.131)

σik = (−p + λD(1)) δik + 2µDik, (8.132)

and

γv θ ≡ ζ D2

(1) − 4µD′

(2) +k

θI4 ≥ 0. (8.133)

These are the constitutive equations for what is referred to as a thermoviscousNewtonian fluid. The balance equations for linear momentum and energy usingthese constitutive equations are referred as the Newton–Fourier equations. Notefrom (8.119) that for a thermoviscous Newtonian fluid, the mechanical and thermo-dynamic pressures are related by

−p = −p + ζ D(1). (8.134)

In addition, note that the bulk viscosity measures the excess between the mechan-ical and thermodynamic pressures. These pressures are equal under any of thefollowing conditions:

i) D(1) ≠ 0 but the bulk viscosity vanishes, i.e.,

ζ = 0. (8.135)

This represents Stokes’ hypothesis. It is not valid in general (it vanishesfor a monatomic gas), and indeed, there is indication that Stokes himselfquestioned it. Then, the constitutive equation is given by (8.132) with λ =−

2

3µ and the pressure is, as usual, given by p = p(ρ, θ). The linear momentum

equations with this stress tensor is referred to as the Navier–Stokes equations.

ii) D(1) ≠ 0 but the dilatational and shear viscosities vanish, i.e., λ = µ = 0. Inthis case, the constitutive equations reduces to

σik = −p δik, (8.136)

and the pressure is again given by p = p(ρ, θ). The linear momentum equa-tions with this stress tensor is referred to as Euler’s equations of motion.

iii) D(1) = 0, in which case the fluid is isochoric (or incompressible), the consti-tutive equation (8.132) becomes

σik = −p δik + 2µDik, (8.137)

the dilatational viscosity λ drops out from the equations, and p correspondsto the mean or hydrostatic pressure.

iv) Dik = 0, in which case we have rigid motion with

σik = −p δik, (8.138)

both terms involving viscosities drop out, and p also corresponds to the meanor “hydrostatic” pressure.

8.3. REPRESENTATIONS OF qd AND σd 353

e) Second-order representations

If we write second-order expansions of (8.96) and (8.97) in the limits θ,i → 0 andDik → 0, we require that κ0 = κ00 + κ01D(1), κ1 = κ10, κ2 = 0, α0 = α0000 +

α0100D(1)+α0200D2

(1)+α0010D(2)+α0001I4, α1 = α10+α11D(1), α2 = α20, α3 = α30,and α4 = α5 = 0. Note that now κ00, κ01, κ10, α0000, α0100, α0200, α0010, α10, α11,α20, and α30 are just functions of ρ and θ. Subsequently, since qdi and Dd

ik mustvanish as θ,i → 0 and Dik → 0, we have that α0000 = 0 and thus

qdi = − [(κ00 + κ01D(1)) δil + κ10Dil]θ,l, (8.139)

σdik = (α0100D(1) + α0200D

2

(1) + α0010D(2) + α0001I4) δik +(α10 + α11D(1))Dik + α20DilDlk + α30θ,iθ,k. (8.140)

Remark

Before addressing constraints on the coefficients from the above inequality,it is useful to derive the representations (8.139) and (8.140) using simplearguments. We start by noting that since Dik ≠ 0 and θ,l ≠ 0, we have thegeneral representations

qdi = qdi (ρ,Dlm, θ, θ,n), (8.141)

σdik = σd

ik(ρ,Dlm, θ, θ,n). (8.142)

If we limit ourselves to second-order expansions in Dlm and θ,n, we have

qdi = − (Gikθ,k +HiklDkl + Jiklmθ,kDlm +KiklmnDklDmn+

Miklθ,kθ,l) , (8.143)

σdik = Aiklθ,l +BiklmDlm +Ciklmnθ,lDmn +EiklmstDlmDst +

Fiklmθ,lθ,m, (8.144)

where all the coefficients are functions of ρ and θ, and we have used theequilibrium conditions that σd

ik → 0 and qdi → 0 when Dik → 0 and θ,i → 0.Since σd

ik and Dlm are symmetric, we must have

Aikl = Akil, (8.145)

Biklm = Bkilm = Bkiml = Bikml, (8.146)

Ciklmn = Ckilmn = Ckilnm = Ciklnm, (8.147)

Eiklmst = Ekilmst = Ekimlst = Ekimlts = Eikmlts = Eiklmts = Eikstml =Ekistml = Ekitsml = Ekitslm = Eiktslm = Eikstlm, (8.148)

Fiklm = Fkilm = Fkiml = Fikml, (8.149)

Hikl =Hilk, (8.150)

Jiklm = Jikml, (8.151)

Kiklmn =Kilkmn =Kiklnm =Kilknm, (8.152)

Mikl =Milk. (8.153)

Frame indifference requires that qd and σd satisfy

qd(ρ,Q ⋅D ⋅QT , θ,Q ⋅ grad θ) =Q ⋅ qd(ρ,D, θ,grad θ), (8.154)

σd(ρ,Q ⋅D ⋅QT , θ,Q ⋅ grad θ) =Q ⋅σd(ρ,D, θ,grad θ) ⋅QT , (8.155)

354 FLUIDS

for all Q ∈ O(V ). From objectivity, we have seen that in general

Aik⋯st = QipQkq⋯QslQtmApq⋯lm. (8.156)

In addition, material symmetry requires that σd and qd be isotropic ten-sors. Below, we shall make use of the representations of isotropic tensorsgiven in Appendix B.We examine first the constitutive equation for the heat flux. Now

Gik = QipQkqGpq, (8.157)

and since δik is the only isotropic second-order tensor as indicated in (B.15),we must have

Gik = gδik. (8.158)

Similarly, from (B.18), we have

Hikl = hǫikl (8.159)

since ǫikl is the only isotropic third-order tensor. Now, from the symmetries(8.150) and (8.153),

Hikl = hǫikl =Hilk = hǫilk = 0, (8.160)

and

Mikl =mǫikl =Milk =mǫilk = 0. (8.161)

In addition, from (B.20) to (B.22), we have

Jiklm = j1δikδlm + j2δilδkm + j3δimδkl, (8.162)

so that from the symmetry (8.151), it follows that j2 = j3 and thus

Jiklm = j1δikδlm + j2 (δilδkm + δimδkl) . (8.163)

Furthermore, using the symmetries (8.152), after some algebra we find that

Kiklmn = k (ǫilnδkm + ǫilmδkn) =Kilknm = k (ǫikmδln + ǫiknδlm) = 0. (8.164)

Thus, substituting the above results into (8.143), we obtain

qdi = −gθ,kδik − [j1δikδlm + j2 (δilδkm + δimδkl)]θ,kDlm, (8.165)

or finally

qdi = − (g + j1D(1)δik + 2j2Dik)θ,k, (8.166)

where g = g(ρ, θ), j1 = j1(ρ, θ), and j2 = j2(ρ, θ).

8.3. REPRESENTATIONS OF qd AND σd 355

We now proceed in a similar fashion to obtain the coefficients of σdik. Apply-

ing the symmetries (8.145)–(8.149), after some lengthy algebra, we obtain

Aikl = aǫikl = 0, (8.167)

Biklm = b1δikδlm + b2 (δilδkm + δimδkl) , (8.168)

Ciklmn = c (ǫklnδim + ǫilnδkm + ǫklmδin + ǫilmδkn) = 0, (8.169)

Eiklmst = e1δikδlmδst + e2 (δlsδmt + δmsδlt) δik +e3 [(δilδkm + δimδkl) δst + (δisδkt + δitδks) δlm] +e4 [(δisδmt + δitδms) δkl + (δksδmt + δktδms) δil+(δitδls + δisδlt) δkm + (δksδlt + δktδls) δim] , (8.170)

Fiklm = f1δikδlm + f2 (δilδkm + δimδkl) . (8.171)

Substituting the above coefficients in the constitutive equation (8.144), weobtain

σdik = [b1D(1) + (e1 + 2e2)D2

(1) − 4e2D(2) + f1θ,jθ,j] δik +2 (b2 + 2e3D(1))Dik + 8e4DilDlk + f2θ,iθ,k, (8.172)

where the coefficients are functions of ρ and θ. It is clear that equations(8.166) and (8.172) that we have just derived are the same as (8.139) and(8.140), respectively.

Now, using (3.108), the residual entropy inequality (8.104) for this order ofapproximation becomes

γv θ ≡ (α0100 +1

3α10)D2

(1) − 2α10D′

(2) + (α0200 +1

3α11 +

1

3α0010)D3

(1) +

(α0010 − 2α11 − 3α20)D(1)D′(2) + 3α20D(3) +

[κ00θ+ (α0001 +

κ01

θ)D(1)] I4 + (α30 +

κ10

θ) I5 ≥ 0. (8.173)

We now examine the restrictions imposed by the reduced entropy inequality (8.173)on the coefficients in (8.139) and (8.140). The inequality must be satisfied for anyDik and any θ,i. Specifically, it must be satisfied for incompressible constantdensity isothermal deformations for which D(1) = 0 and θ,i = 0. In this case,(8.173) becomes

γv θ ≡ −2α10D′

(2) + 3α20D(3) ≥ 0. (8.174)

Now, since D′(2) ≤ 0 (see (3.108)) and D(3) can be of any magnitude and signdepending on the deformation, in order for the above inequality to be alwayssatisfied, we must have that

α10 ≥ 0 and α20 = 0. (8.175)

If we now consider an incompressible constant density non-isothermal motion anduse the above results, then we must satisfy

γv θ ≡ −2α10D′

(2) +κ00

θI4 + (α30 +

κ10

θ) I5 ≥ 0. (8.176)

356 FLUIDS

Recalling that −2α10D′

(2) ≥ 0, noting that I4 ≥ 0, and since I5 can be of anymagnitude and sign depending on Dik and θ,i, then we must have that

κ00 ≥ 0 and α30 = −κ10θ. (8.177)

If we next consider a compressible isothermal motion and again use previous re-sults, the reduced entropy inequality (8.173) reduces to

γv θ ≡ (α0100 +1

3α10)D2

(1) − 2α10D′

(2) + [(α0200 +1

3α11 +

1

3α0010)D2

(1)+

(α0010 − 2α11)D′(2)]D(1) ≥ 0. (8.178)

Now, since −2α10D′

(2) ≥ 0 and since D(1) can be chosen to be of any magnitudeand sign, for arbitrary motions, we must have that

α0100+1

3α10 ≥ 0, α0200+

1

3α11+

1

3α0010 = 0, and α0010−2α11 = 0. (8.179)

Lastly, considering compressible non-isothermal motion and upon using the previ-ous results, the reduced entropy inequality becomes

γv θ ≡ (α0100 +1

3α10)D2

(1) − 2α10D′

(2) + [κ00θ + (α0001 +κ01

θ)D(1)] I4 ≥ 0. (8.180)

It is now easy to see that for the inequality to be satisfied for arbitrary motions,we must have that

α0001 = −κ01θ. (8.181)

In summary, we have that

κ00 ≥ 0, α0100 +1

3α10 ≥ 0, α10 ≥ 0, α0010 = 2α11, α20 = 0,

α0200 = −α11, α0001 = −κ01θ, α30 = −κ10

θ, (8.182)

and κ01, κ10, and α11 can be of any sign. If we rewrite the above by takingα0100 = λ, α10 = 2µ, α11 = µ′, κ00 = k, κ01 = k′, and κ10 = k′′, then we write thequadratic constitutive representations (8.94), (8.99), (8.139), and (8.140) as

σik = (−p + λD(1) − µ′DjlDlj −k′

θθ,jθ,j) δik + (2µ + µ′D(1))Dik −

k′′

θθ,iθ,k,(8.183)

andqi = − [(k + k′D(1)) δil + k′′Dil] θ,l, (8.184)

and the reduced entropy inequality remains (8.133), where

ζ = λ + 2

3µ ≥ 0, µ ≥ 0, k ≥ 0, (8.185)

are the same restrictions obtained from the linear theory, and we have one ad-ditional viscous property, µ′, and two additional thermal properties, k′ and k′′;

8.4. PROPAGATION OF SOUND 357

these last three properties can be of any sign and do not contribute to entropyproduction. However, note that they do contribute to the mechanical pressuresince

−p = −p + ζ D(1) − 2

3µ′D2

(1) + 2µ′D(2) −

(k′ + 1

3k′′)

θθ,iθ,i, (8.186)

and in this case, the mechanical pressure is not equal to the thermodynamic pres-sure even when the fluid is incompressible (i.e., when D(1) = 0). Furthermore,while in the classical linear theory, we have that p ≥ p in an expansion and p ≤ p ina contraction motion (see (8.134)), the above equation shows that the sign of (p−p)does not necessarily follow the intuitive notion of being positive on expansion andnegative on contraction motion.

8.4 Propagation of sound

The fundamental local laws of conservation of mass and balances of linear mo-mentum and energy with linear constitutive equations that satisfy the entropyinequality for an isotropic fluid, ignoring external sources of momentum and en-ergy, are given by

ρ + ρdiv v = 0, (8.187)

ρ v − divσ = 0, (8.188)

ρ e −Φ + divq = 0, (8.189)

where

σ = −p1 +σd = −p1 + [2µD + (ζ − 2

3µ)D(1)1] , (8.190)

q = −k grad θ, (8.191)

Φ = D ∶ σ = −pD(1) +D ∶ σd = −pD(1) + [2µD ∶D + (ζ − 2

3µ)D2

(1)] ,(8.192)

andµ(ρ, θ) ≥ 0, ζ(ρ, θ) ≥ 0, and k(ρ, θ) ≥ 0 (8.193)

are the dynamic shear viscosity, dynamic bulk viscosity, and thermal conductivity,respectively.

There are too many unknowns in the above equations. To eliminate these addi-tional unknowns, we need thermodynamic constitutive relations. To this end, froma given fundamental relation ψ = ψ(ρ, θ), it would be useful to introduce thermaland caloric equations of state for the pressure and internal energy by writing (see(5.207)2, (5.208), and Table 8.1 for n = 1)

p = p (ρ, θ) = ρ2 ∂ψ

∂ρ∣θ

and e = e (ρ, θ) = ψ − θ ∂ψ

∂θ∣ρ

. (8.194)

In general, we can easily show that (see (5.212)2, (5.213), and Table 8.1 for n = 1)dp = 1

κθ(1ρdρ + αdθ) and ρde = 1

ρ(p − α

κθθ)dρ + ρ cv dθ, (8.195)

358 FLUIDS

where (see (5.161))

cv ≡ ∂e∂θ∣ρ

, κθ ≡ 1

ρ

∂ρ

∂p∣θ

, and α ≡ −1ρ

∂ρ

∂θ∣p

(8.196)

are the coefficients of specific heat at constant volume, isothermal compressibility,and volume expansion, respectively. From equilibrium thermodynamics consider-ations, we also have the restrictions (see (5.268)–(5.270))

cv > 0, κθ > 0, γ ≡ cpcv> 1, and α2 < ρ cp κθ

θ, (8.197)

where

cp ≡ ∂h∂θ∣p

(8.198)

is the specific heat at constant pressure.Upon substituting the thermodynamic relations (8.195) into (8.187)–(8.189), it

is straightforward to see that we obtain the following closed set of equations forthe unknowns ρ, v, and θ once all properties, which are functions of ρ and θ, arespecified:

ρ + ρdivv = 0, (8.199)

ρ v +1

κθ(1ρgradρ + α gradθ) − divσd = 0, (8.200)

ρ cv θ +α

κθθD(1) −D ∶ σd

+ divq = 0. (8.201)

These equations are fairly complicated. To investigate the propagation of acous-tics in a fluid, we make the following simplifying assumption. The density andtemperature are assumed to deviate only slightly from constant equilibrium valuesof ρ0 and θ0, while the equilibrium value of velocity is taken to be zero:

ρ = ρ0 + ρ′, v = 0 + v′, and θ = θ0 + θ′, (8.202)

where ρ′/ρ0 ≪ 1, v′ ≪ 1, and θ′/θ0 ≪ 1. Note that due to Galilean frame indiffer-ence, the results will also apply to the case where the equilibrium value of velocityis a constant. Substituting (8.202) into (8.190)–(8.191) and (8.199)–(8.201), andlinearizing the resulting equations, we have

∂ρ′

∂t+ ρ0 divv

′ = 0, (8.203)

ρ0∂v′

∂t+

1

κθ0( 1

ρ0gradρ′ + α0 gradθ

′)−µ0 div (gradv′)−(ζ0 + 1

3µ0)grad(divv′) = 0,

(8.204)

ρ0 cv0∂θ′

∂t+α0 θ0

κθ0divv′ − k0 div (gradθ′) = 0, (8.205)

where all properties with zero subscripts are evaluated at ρ0 and θ0.These linearized Newton–Fourier equations are the basic tool to understanding

acoustic propagation in fluids. Their structure is rather remarkable. We canclearly recognize two types of terms. In a first class, the coefficients of ρ′, v′, and

8.4. PROPAGATION OF SOUND 359

θ′ depend only on the equilibrium thermodynamic properties of ρ0, θ0, α0, κθ0,and cv0. In the second class, the terms depend on the nonequilibrium transportproperties of µ0, ζ0, and k0. If the latter are set equal to zero, we obtain theideal or non-dissipative equations. The additional terms are responsible for thedissipation through the viscosities and heat conduction. They clearly are also theonly terms giving rise to nonzero entropy production.

Of special interest to us are those motions with very long characteristic spatialscales. This is certainly consistent with the use of constitutive equations that arelinear in the deformation and temperature gradients. The study of such motionswill lead to an eigenvalue problem whose solution is given by a superpositionof elementary long-range motions. With such study in mind, we assume thatour linear equations (8.203)–(8.205) describe the motion of a fluid in an infinitedomain. We investigate their solution in the forms

ρ′(x, t) = eik⋅x+λt ρk, (8.206)

v′(x, t) = eik⋅x+λt vk, (8.207)

θ′(x, t) = eik⋅x+λt θk, (8.208)

where k is the (real) wavenumber vector and λ is the (complex) frequency. Sub-stituting these forms in (8.203)–(8.205), and using (5.163) and (5.164), we obtainthe following set of linear algebraic equations:

λρk + ρ0 ik ⋅ vk = 0, (8.209)

λvk +1

ρ20κθ0

ikρk +α0

ρ0 κθ0ik θk + ν0 k

2 vk + (ν′0 + 1

3ν0)k (k ⋅ vk) = 0, (8.210)

λθk +γ0 − 1

α0

ik ⋅ vk + γ0 χ0 k2 θk = 0, (8.211)

where we have used the following abbreviations:

γ0 = cp0cv0

, ν0 ≡ µ0

ρ0, ν′0 ≡ ζ0

ρ0, χ0 ≡ k0

ρ0 cp0, (8.212)

with γ0 the ratio of specific heats, ν0 the kinematic shear viscosity, ν′0 the kinematicbulk viscosity, and χ0 the thermal diffusivity.

Now, without loss of generality, we can choose the x-axis to be oriented alongthe wavenumber vector k, thus taking

k = (k,0,0) and vk = (uk, vk,wk) , (8.213)

we can rewrite the system (8.209)–(8.211) in the following form:

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

λ ρ0 i k 0 0 01

ρ20κθ0

i k λ + ( 43ν0 + ν

′

0)k2 0 0α0

ρ0 κθ0i k

0 0 λ + ν0 k2 0 0

0 0 0 λ + ν0 k2 0

0γ0 − 1

α0

i k 0 0 λ + γ0 χ0 k2

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎝

ρkukvkwk

θk

⎞⎟⎟⎟⎟⎟⎠= 0.

(8.214)

360 FLUIDS

For a nontrivial solution to exist, the determinant of the square matrix mustbe zero, giving rise to a characteristic polynomial for λ, which represents thedispersion relation,

(λ + ν0 k2)2 λ3 + (43ν0 + ν

′

0 + γ0 χ0)k2 λ2+[c20 + γ0 χ0 (4

3ν0 + ν

′

0)k2]k2 λ + c20 χ0 k4 = 0, (8.215)

where

c0 ≡ ( γ0

ρ0 κθ0)1/2 (8.216)

is the fluid’s sound speed.Two of the roots of (8.215) are the same and immediately obvious: λ3,4 = −ν0 k2.

These two roots are real and correspond to purely dissipative shear modes sincethey depend only on the shear viscosity. The other three roots satisfy the cubicequation within the braces. Two of them are a complex conjugate pair and corre-spond to dissipative acoustic modes since they involve the sound speed to leadingorder, but also involve dissipative terms at higher orders due to shear viscosity,bulk viscosity, and thermal diffusivity. These two acoustic modes travel in op-posite directions. The third root is real and corresponds to a purely dissipativethermal mode since it only involves the thermal diffusivity. These roots can bewritten explicitly, but they are fairly complicated algebraically. Since the consti-tutive equations are linear in the deformation and temperature gradients, it makessense to look at these three roots in the small wavenumber limit. In such limit,these roots are given in Table 8.1 up to second order in k. In the table, we alsosee that when the fluid is ideal, the eigenvalues of the shear and thermal modesare zero, and the acoustic modes travel at the sound speed without any damping.

8.5 Classifications of fluid motions

8.5.1 Restrictions on the type of motion

Restrictions on geometry

Various special motions are defined by geometric restrictions on the velocity field.

1. Homogeneous motion:

v = x =A(t) ⋅ x + b(t). (8.217)

Since any continuous motion may be approximated in the neighborhoodof any point by an appropriate homogeneous motion, this motion is veryimportant. In this case, using (8.217), the stretch and spin tensors are givenby

D = symA and W = skwA. (8.218)

It now follows that any homogenous motion may be regarded at any instantas a rigid translation, a rigid rotation, and an irrotational homogeneousstretchings along the three principal axes of D.

8.5. CLASSIFICATIONS OF FLUID MOTIONS 361

2. Lineal motion:

vx = vx(x, t), vy = 0, vz = 0, ρ = ρ(x, t). (8.219)

In this case, the velocity is constant over each member of a family of parallelplanes, and normal to those planes, the motion is lineal, where in the abovecharacterization, these planes are the surfaces x = const.

3. Pseudo-lineal motion of the first kind:

vx = vx(x, y, z, t), vy = 0, vz = 0, ρ = ρ(x, y, z, t). (8.220)

4. Pseudo-lineal motion of the second kind:

vx = vx(x, t), vy = vy(x, t), vz = vz(x, t), ρ = ρ(x, t). (8.221)

We note that for lineal and pseudo-lineal motions of the first and secondkind, the local mass balance equation becomes

∂ρ

∂t+∂

∂x(ρvx) = 0. (8.222)

Thus, there always exists a function Q(x, t) (or Q(x, y, z, t)) such that

ρ = −∂Q∂x

, ρ vx = ∂Q∂t. (8.223)

5. Plane motion:

vx = vx(x, y, t), vy = vy(x, y, t), vz = 0, ρ = ρ(x, y, t). (8.224)

A motion is said to be plane if its velocity field is a plane field and thus thevorticity ω satisfies

v ⋅ω = 0, (8.225)

with the surfaces z = const. being a family of parallel planes. The thirddimension is obtained by rotating the plane of motions about the z-axis.

The idea of plane motion is easily generalized by replacing z = const. withany curved surface x3 = 0, upon which x1 and x2 may be any curvilinearcoordinates. The equations

xk = xk(x1, x2, t), k = 1,2, (8.226)

now describe a strictly two-dimensional motion. To generate a three-dimen-sional motion in a region of space near x3 = 0, choose a coordinate systemin which the surface x3 = const. are surfaces parallel to x3 = 0, while thesurfaces x1 = const. and x2 = const. are those swept out by the normals tox3 = 0 along the x1 and x2 coordinate curves upon it.

362 FLUIDS

6. Pseudo-plane motion of the first kind:

vx = vx(x, y, z, t), vy = vy(x, y, z, t), vz = 0, ρ = ρ(x, y, z, t). (8.227)

Such motions result if the first and second conditions (8.224)1,2 are removedand the third condition (8.224)3 is retained.

7. Pseudo-plane motion of the second kind:

vx = vx(x, y, t), vy = vy(x, y, t), vz = vz(x, y, t), ρ = ρ(x, y, t). (8.228)

Such motions result if the first and second conditions (8.224)1,2 are retainedwhile the third condition (8.224)3 is removed.

8. Rotationally symmetric motion:

vr = vr(r, z, t), vθ = 0, vz = vz(r, z, t), ρ = ρ(r, z, t). (8.229)

A motion is said to be rotationally symmetric if its velocity field is a symmet-rically rotational field with the surfaces z = const. being a family of coaxialplanes.

Restrictions on steadiness

1. Steady motion:

v = v(x). (8.230)

The ultimate steady flow is when v = 0, in which case the problem reducesto one of hydrostatics.

Here, we just note that the definition of steady motion refers to a particularframe and thus is not frame invariant.

2. Steady motion with steady density:

The requirement of steady motion does not require that ρ be steady, but ˙log ρ

must be steady. Furthermore, if g is any steady quantity that is constant oneach stream line, then g = 0 and

div (ρg v) = ρv ⋅ gradg + g div (ρv) = 0. (8.231)

Thus, by assigning a steady quantity g a constant value on each stream line,we obtain a steady density ρg, and thus

v = v(x), ρ g = f(x), g = 0, (8.232)

where g is an arbitrary function.

8.5. CLASSIFICATIONS OF FLUID MOTIONS 363

3. Motion with steady stream lines:

We first note that in order that the flux of a vector field u(x, t) acrossevery material surface remains constant in time, it is necessary and sufficientthat Zorawski’s criterion (3.458) be satisfied. It follows from the Helmholtz–Zorawski criterion that the necessary and sufficient condition that the vectorlines of u be material lines is that they satisfy

u × [∂u∂t+ vdivu + curl (u × v)] = 0. (8.233)

By taking u = v, we have our result

v ×∂v

∂t= 0, (8.234)

or, equivalently,∂v

∂t= C(x, t)v. (8.235)

4. D’Alembert motion:

v(x, t) = f(t)u(x). (8.236)

These motions have steady stream lines, they are isochoric if and only if uis solenoidal, and their kinematic vorticity number A (see (3.502)) is steady.Now, since

v = f u + f2 u ⋅ gradu, (8.237)

we haveω⋆ ≡ curl v = f curlu + f2 curl [(curlu) × u] . (8.238)

Now, if f/f2 ≠ const., then ω⋆ = 0 cannot be satisfied unless ω = 0. Thus,besides the case when

f = 1

a t + b, a, b = const., a b ≠ 0, (8.239)

a D’Alembert motion is circulation-preserving if and only if it is irrotational.In the case where (8.239) holds, the condition for ω⋆ = 0 is that a curlu =curl [(curlu) × u]. If a = 0, this is just a condition that u be the velocityfield of a circulation-preserving motion, and this is the case if and only if f =const.

5. Motion without acceleration:

v = 0. (8.240)

In this case, every particle initially at X travels in a straight line at a uniformvelocity v(X). Steady motion without acceleration is typified by rectilinearshear motions. But when the stream lines are not steady, they are notstraight, and the straight path lines may cross each other in a very largenumber of ways. A functional form of (8.240) is given by

x = f(X) t +X, (8.241)

364 FLUIDS

and since v = f(X), this functional equation may be rewritten in the form

v = f(x − v t). (8.242)

Homogeneous motion without acceleration is a special case. In this case,substituting (8.217) into

v = ∂v∂t+ v ⋅ (gradv) , (8.243)

we obtainA +A2 = 0 and b =A ⋅b. (8.244)

In the steady case, we have that

A2 = 0 and A ⋅b = 0. (8.245)

In the unsteady case, the general solution is given by the algebraic system

A = (1 +A0t)−1 ⋅A0 and b = (1 +A0t)−1 ⋅b0, (8.246)

where A0 and b0 are the initial values of A and b. In general, these motionsdevelop singularities in a finite time, determined by the initial velocity andvelocity gradient.

Using (8.217) and (8.241), we also have that

x = (A ⋅ x + b) t +X. (8.247)

Now, multiplying both sides by (1 +A0t) and using (8.246),

x = (1 +A0t) ⋅X + b0t, (8.248)

which results inv =A0 ⋅X + b0, (8.249)

so that v is linear in X as well as in x.

Example

Take

A(t) = k(t)⎛⎜⎝−σ(t) 0 0

0 −σ(t) 0

0 −1

⎞⎟⎠ , b(t) = 0, (8.250)

where, we see that to satisfy (8.244), we must have

k(t) = k0

1 + k0 tand σ(t) = σ0 1 + ko t

1 − k0 σ0 t. (8.251)

In this motion, a rectangular region of fluid with edges parallel tothe coordinate planes is extended along the x3-direction at the ratek(t) and is contracted transversely in the ratio σ(t). The stream

lines in the x2 = 0 plane are the curves x3 x1/σ1= const., and the

8.5. CLASSIFICATIONS OF FLUID MOTIONS 365

ratio of the current volume v to the initial volume V is given byv/V = (1 + k0 t)(1 − k0 σ0 t)2. We see that the motion develops asingularity when the volume is reduced to zero.

Restrictions on stretch

1. Rigid motion:

A motion is rigid if all material lengths remain unchanged by the motion.The necessary and sufficient condition for this to be the case is given by

D = 0. (8.252)

Its general solution is

v = c +W(t) ⋅ (x − c(t)) , (8.253)

where W is skew-symmetric and the corresponding axial vector is the angularvelocity of the motion, and c is the linear velocity of the origin of the framerelative to an arbitrary frame. If W = 0, the rigid motion is just a translation.

2. Isochoric motion:

A motion such that the volume occupied by any material region is unaltered,however, that region may change its shape in the course of time, is isochoricand subsequently

D(1) = divv = 0. (8.254)

3. Dilatational motion:

Since D is a real symmetric tensor, its real and orthogonal principal axesare the axes of stretching, and its real eigenvalues d(i) are the principalstretchings. When D is spherical, then the three principle stretchings areequal and the motion is a dilatation:

d(1) = d(2) = d(3) = d. (8.255)

Equivalently, for a dilatation, the conditions on the invariants of D are (see(3.135)–(3.137))

(13D(1))3 = (1

3D(2))3/2 =D(3). (8.256)

For a general motion, the six components of D cannot be assigned arbitrarily.In order that there exists a field v such that D is symmetric, the integrabilitycondition for the components is given by

δkmpq δlnrsDkl,mn = 0. (8.257)

Now, substituting (8.255) into (8.257), it becomes evident that D is a linearfunction, and therefore, the velocity is given, to within a rigid motion, by

v = [a(t) ⋅ x + b(t)]x − 1

2(x ⋅ x)a(t). (8.258)

366 FLUIDS

4. Shearing motion:

A rectilinear shearing is a steady motion in which the paths of the materialparticles are parallel straight lines and the speed of each particle is constantin time. Thus, in the rectangular Cartesian system, we have

vx = vx(y, z), vy = 0, vz = 0. (8.259)

We see that a pseudo-lineal motion of the first kind is isochoric if and only if itis a rectilinear shearing motion. Elements parallel or normal to the directionof motion are not stretched. However, elements normal to the direction ofmotion are rotated, and thus at a later instant are no longer normal. Verytypical of this motion is simple shearing (see (3.162)–(3.164)):

vx = κy, vy = 0, vz = 0. (8.260)

For this special case, we recall that the principal axes of stretching are thez-axis and the bisectors of the x- and y-axes, the principal stretchings are±

1

2κ, the maximum orthogonal shears are experienced by elements in the

x and y coordinate directions, the axis of spin is the z-axis, and both thevorticity and the maximum shearing have the value κ. Note that simpleshearing is a special case of homogeneous motion.

For shearing, we must have that

d(1) = −d(3) and d(2) = 0. (8.261)

But this is not enough, since in simple shearing motion the spin is parallel tothe second principal axis of stretching, and its magnitude satisfies ∣ω∣ = 2d(1).Therefore, the necessary and sufficient conditions for a shearing motion arethat:

i) D(1) =D(3) = 0,ii) ω is parallel to the principal axis of the stretch along which the stretch-

ing is zero, and

iii) ∣ω∣ =√−4D(2) = 2d(1) = amount of shearing,

or, equivalently, note that any shearing motion may be regarded locally as asimple shearing superposed upon a rigid rotation.

Restrictions on spin

1. Irrotational motion:

A motion is irrotational if and only if its velocity field is lamellar, i.e.,

v = −gradφ, (8.262)

where φ is called the velocity potential. Thus, it follows that a motion forwhich the spin vanishes is irrotational:

ω = curlv = −curl (gradφ) = 0 (8.263)

8.5. CLASSIFICATIONS OF FLUID MOTIONS 367

or, equivalently,W = 0. (8.264)

Motions for which W ≠ 0 are called rotational.

Since L ≠ 0 is symmetric if and only if W = 0, we conclude that a necessaryand sufficient condition for a motion to be instantaneously irrotational ata point is that there exist three mutually orthogonal directions undergoingno instantaneous rotation. This follows from the fact that the symmetrictensor has three distinct eigenvalues. From this it also follows that a motionis locally a translation if and only if it is irrotational and rigid.

We note that the derivative of the velocity potential in the direction of thevelocity is given by

dφ

ds= v∣v∣ ⋅ gradφ = −∣v∣ ≤ 0. (8.265)

Hence, the velocity potential can never increase in the direction of motionalong a stream line, and in steady irrotational motion, a particle alwaysmoves toward a region of lower velocity potential. A stagnation point is astationary point for the velocity potential along the stream line on whichit occurs. In addition, using the definition of the material derivative in theEuler description, we also have that

∂φ

∂t− φ = ∣v∣2 ≥ 0. (8.266)

Hence, the squared speed is the excess of the local over the material timederivative of the velocity potential.

In an irrotational motion, since

D = −grad(gradφ) , (8.267)

we easily see upon contraction that

D(1) = −∇2φ. (8.268)

Now, φ is sub-harmonic, harmonic, or super-harmonic according to D(1) < 0,D(1) = 0, or D(1) > 0. Therefore, in the interior of a region where the materialvolumes are not decreasing, the velocity potential cannot have a minimum,and in a region where the material volumes are not increasing, the velocitypotential cannot have a maximum. Lastly, we see that an irrotational mo-tion is isochoric if and only if the velocity potential is harmonic. Thus, allproperties of isochoric irrotational motions follow from potential theory.

2. Complex-lamellar motion:

A motion is complex-lamellar if the velocity field is complex-lamellar:

v = −f gradφ (8.269)

or, analogously,ω ⋅ v = 0. (8.270)

368 FLUIDS

Hence, a rotational motion is complex-lamellar if and only if its vortex linesare orthogonal to its stream lines. Both plane motions and rotationally sym-metric motions are complex-lamellar. Complex-lamellar motions have someof the same properties as irrotational motions. The function φ is somewhatanalogous to the velocity potential. Equations

f∂φ

∂s= −∣v∣ ≤ 0 and f (∂φ

∂t− φ) = ∣v∣2 ≥ 0 (8.271)

are generalizations of (8.265) and (8.266). Thus, the conclusions regardingthe stream lines of an irrotational motion derived from these equations carryover to a region of complex-lamellar motion where f is of one sign. Now,by continuity of the velocity field, f can be of opposite sign at two pointson a stream line only if there is an intermediate point where f = 0, andsubsequently, such point is a stagnation point. Consequently, in a region ofcomplex-lamellar motion without stagnation points, the conclusions regard-ing the speed and the stream line pattern of an irrotational motion continueto hold, provided that −φ is substituted for φ if f < 0.

3. Screw motion:

To understand screw motions, we first note that we can write the accelerationfield in the following form known as Lagrange’s formula:

v = ∂v∂t+ v ⋅ gradv = ∂v

∂t+ 2W ⋅ v + grad(1

2v ⋅ v)

= ∂v

∂t+ω × v + grad(1

2v ⋅ v) . (8.272)

This equation shows that in three-dimensional space, the acceleration is ex-pressed in terms of four vectors: the local acceleration ∂v/∂t, the vorticityvector ω, the velocity vector v, and the gradient of the specific kinetic energygrad( 1

2v ⋅ v). The vector ω × v is called the Lamb vector. In a rotational

motion with vanishing Lamb vector,

ω × v = 0 and ω ≠ 0, (8.273)

the velocity field is a screw field and the motion is called a screw motion. Asreadily apparent, screw motions and complex-lamellar motions are mutuallyexclusive motions, but each shares some of the properties of irrotationalmotions. In a screw motion, just as in an irrotational motion, Lagrange’sequation reduces to

v = ∂v∂t+ grad(1

2v ⋅ v) . (8.274)

Subsequently, we see that the vortex lines in this case coincide with thestream lines.

4. Complex-screw motion:

To understand complex-screw motions, it is useful to introduce a generalizedconvection vector field that is proportional to the velocity field

v = v0 υ, (8.275)

8.5. CLASSIFICATIONS OF FLUID MOTIONS 369

where v0 is any non-vanishing scalar field whose substantial derivative iszero:

v0 = 0, v0 ≠ 0. (8.276)

It is now easy to see that

ω = curl(v0 υ) = v0 + gradv0 × υ, (8.277)

where we have defined ≡ curlυ. (8.278)

We now see thatv ⋅ω = v20 υ ⋅. (8.279)

It follows that the generalized convection vector is complex-lamellar if andonly if the motion is complex-lamellar.

Now motions in whichυ ⋅ = 0 (8.280)

are called complex-screw motions. To clarify such motions, first note that

v ×ω = v20 υ × + v ⋅ v grad(log v0) + (∂ log v0∂t

)v. (8.281)

From this identity, we note that in a complex-screw motion, for which

v ×ω = v ⋅ v grad(log v0) + (∂ log v0∂t

)v, (8.282)

if v0 is either uniform or steady, we have an irrotational or screw motionif and only if v0 is both uniform and steady. This result implies that theclass of complex-screw motions is more extensive than that of irrotationaland screw motions. If we substitute this equation into Lagrange’s formula(8.272), we obtain

v = ∂v

∂t− v ⋅ v grad(log v0) − (∂ log v0

∂t)v + grad(1

2v ⋅ v)

= v0∂υ

∂t+ v20 grad(12υ ⋅υ) . (8.283)

Hence, in a complex-screw motion whose convection vector is steady, theacceleration is complex-lamellar.

Taking the curl of the above equation, we have

curl v = gradv0 × ∂υ∂t+ v0

∂

∂t+1

2gradv20 × grad(υ ⋅ υ) . (8.284)

If we assume that the local time derivatives are zero: (i) if v0 is uniform,then from × υ = 0 it follows that ω × v = 0; (ii) we may have υ ⋅ υ =const.; (iii) if the surfaces v0 = const. coincide with the surfaces ∣υ∣ = const.,these in turn are surfaces of constant speed. In summary, a complex-screwmotion whose convection vector is steady is a circulation-preserving motionif and only if (a) it is an irrotational motion, or (b) its convection vector isof uniform magnitude, or (c) at each fixed time, v0 is a function of the speedalone. When the motion itself is steady, then using a previous result, (c) isreplaced by (c′) the surfaces of constant speed are Lamb surfaces, and theacceleration is normal to them.

370 FLUIDS

5. Motion with steady vorticity:

The vortex lines are steady if and only if

∂ω

∂t×ω = 0. (8.285)

For the vortex lines to be steady, it is sufficient, but not necessary, that thevorticity itself be steady:

∂ω

∂t= curl(∂v

∂t) = 0. (8.286)

From (8.285), it follows that

∂ω

∂t= C ω, (8.287)

where C is a scalar quantity. Thus, we have

curl(∂v∂t) = curl (C v) − gradC × v. (8.288)

Taking the divergence of this equation, we obtain

0 = div (gradC × v) = −gradC ⋅ω. (8.289)

Hence, it follows that in a motion where the vortex lines are steady but thevorticity is not steady, the surfaces upon which ∂ω/∂t has a constant ratioto ω are vortex surfaces.

Now it is clear that (8.285) is satisfied if and only if there exists a scalar fieldU(x, t) such that

∂v

∂t= gradU. (8.290)

This relation may be substituted into Lagrange’s equation (8.272) to obtain

v = ω × v + grad(12v ⋅ v +U) . (8.291)

6. Circulation-preserving motion:

In order to investigate the possibility that the vortex lines be material lines,we substitute u = ω into the Helmholtz–Zorawski criterion (8.233) to obtain

ω × [∂ω∂t+ curl(ω × v)] = 0. (8.292)

Also, by taking the curl of Lagrange’s formula (8.272), we obtain

ω⋆ ≡ curl v = ∂ω∂t+ curl (ω × v) (8.293)

for any motion. Hence, (8.292) becomes

ω ×ω⋆ = 0. (8.294)

8.5. CLASSIFICATIONS OF FLUID MOTIONS 371

Thus, a necessary and sufficient condition that the vortex lines be materiallines is that the acceleration be lamellar or that its curl be parallel to thevorticity. By putting u = ω into the Zorawski criterion (3.458), we similarlyobtain

ω⋆ = 0. (8.295)

Subsequently, a necessary and sufficient condition that the strengths of allvortex tubes remain constant in time is that the acceleration be lamellar;this condition is also sufficient that the vortex tubes be material tubes.

8.5.2 Specializations of the equations of motion

Creeping flow

It is sometimes justifiable to assume that the velocity is small so that terms in-volving velocity of second and higher order are negligible in comparison to linearterms of velocity. Invoking this approximation leads to a linear equation for thevelocity field, which can be solved more readily, particularly in light of the factthat such flows are often isochoric. For the Navier–Stokes equations, this leads toStokes’ flow about various bodies.

Perturbed flow

Another similar specialization arises in stability theory when a basic flow is knownand is perturbed by a small amount. Here, the squares and higher products of thesmall perturbations are regarded as negligible. Again, a linear problem results,which for the Navier–Stokes equations leads to problems such as in acoustics, aswe have seen earlier, and Orr–Sommerfeld equations in hydrodynamic stability.

Boundary layer flow

When large gradients of dependent variables are confined to the neighborhood ofa boundary, we refer to the region close to the boundary as a boundary layer. Theboundary layer is characterized by a sharp change in the velocity in the directionnormal to the boundary (shear layer) or a sharp change in the temperature (ther-mal boundary layer). It usually happens that the thickness of this layer is small incomparison with the longitudinal dimensions. Advantage is taken of this fact toreduce the equations to simpler forms by neglecting certain terms in the balanceequations by comparison with others. The simplifications lead to approximateequations called the boundary layer equations. More generally, the boundary layerphenomenon arises in narrow zones near a boundary as well as in the interior of afluid where such surfaces of “discontinuity” are referred to as shocks.

The concept of a boundary layer was introduced by L. Prandtl in connectionwith the hydrodynamics of viscous liquids. Although the concept has its origin influid dynamics, the mathematical developments of this concept has led to the fieldof perturbation theory.

372 FLUIDS

8.5.3 Specializations of the constitutive equations

Newtonian fluid

Here, the assumption of a linear relation between stress and strain rate, as we haveseen, leads to the constitutive equation

σ = (−p + λdivv)1 + 2µD, (8.296)

where µ = µ(ρ, θ) and λ = λ(ρ, θ). The equations of motion then become theNavier equations.

Stokesian fluid

A Newtonian fluid that incorporates the Stokes hypothesis of a vanishing bulkviscosity (it vanishes for a monatomic gas), i.e.,

ζ = λ + 2

3µ = 0, (8.297)

is referred to as a Stokesian fluid and the linear momentum equations with thissimplification of the stress tensor is referred to as the Navier–Stokes equations.

Ideal fluid

An ideal fluid has no viscosity, i.e., λ = µ = 0, so that

σ = −p1 (8.298)

and

ρ (a − f) = −gradp. (8.299)

This is known as Euler’s equations of motion. In addition, the ideal fluid haszero conductivity, k, and if there is no external heat supply, the energy equationbecomes

η = 0 (8.300)

so that the flow is isentropic.

Ideal gas

The free energy of an ideal gas is given by

ψ(ρ, θ) = e0 − (θ − θ0)η0 + θ [∫ θ

θ0(1θ−

1

θ′) cv(θ′)dθ′ +R ln ρ] , (8.301)

where θ0, e0, and η0 are constants, and R = cp − cv. In this case, we have that

η(ρ, θ) = η0 +∫ θ

θ0

cv(θ′)θ′

dθ′ −R ln ρ, (8.302)

and it is easy to see that the caloric and thermal equations of state are

e(ρ, θ) = e(θ) = e0 +∫ θ

θ0

cv(θ′)dθ′ and p(ρ, θ) = Rρθ. (8.303)

8.5. CLASSIFICATIONS OF FLUID MOTIONS 373

Ideal gases do not exist in nature, but the properties of all known gases approachthose of the respective ideal gas as the pressure is extrapolated to zero. We say thatthese equations of state constitute the asymptotic forms for all real gases. In spiteof this, the ideal gas equations of state can be used as approximations to representthe properties of real gases even at comparatively high pressures. When usedjudiciously, the equations provide a high degree of accuracy, and many practicalcalculations are based on them. Values of the gas constant and several otherproperties of ideal gases are usually collected in tables to facilitate calculations.

Piezotropic fluid and barotropic flow

A fluid is said to be piezotropic if

e = e(η, ρ) = e1(η) + e2(ρ), (8.304)

in which case we have that

θ = e′1(η) and p = −e′2(ρ), (8.305)

where the primes indicate derivatives with respect to the corresponding indepen-dent quantities. Many liquids can be represented in a first approximation aspiezotropic. A fluid in which p ≡ 0, i.e., e2 = const., is incompressible; what iscalled “pressure” in an incompressible fluid is not given by the thermodynamicdefinition and is not a thermodynamic variable at all. It is easy to see also that afluid is piezotropic if and only if γ = 1, or, equivalently, α = 0 (see (5.161)–(5.164)).

If the motion is such that the density and pressure are directly related (e.g., asin (8.305)2) the motion is called barotropic. Thus, all piezotropic fluids (whichincludes incompressible fluids as a special case) flow barotropically, but other fluidsmay also do so. The terms piezotropic and barotropic thus stand in the samerelationship as incompressible and isochoric. The relation between p and ρ allowsus to write

1

ρgradp = gradP (p) = grad ∫ 1

ρdp, (8.306)

and such relation becomes useful in irrotational motion. The quantity P is calledthe pressure function.

Incompressible fluid

All real fluids subjected to a sufficiently large uniform pressure change their specificvolume, and hence their density. Depending on the properties of the fluid and therange of variation of pressure in a particular process, the specific volume mayundergo small changes. Sometimes this change is so small that disregarding itdoes not lead to serious errors in the solution of a particular problem. In thecase of liquids, it is necessary to apply very large changes in pressure to effect asignificant change in specific volume. In the case of gases, negligible changes inspecific volume result only from sufficiently small changes in pressure. Similarly,in many cases, the temperature changes during a process are so small that theireffect upon the specific volume can also be disregarded. In all such cases, we thensay that the fluid is incompressible. In terms of thermodynamics, this is equivalentto replacing the thermal equation of state of the system by

v = const. or ρ = const. (8.307)

374 FLUIDS

The constant is selected as a suitable average in a particular problem with referenceto the true thermal equation of state

F (ρ, θ, p) = 0. (8.308)

Geometrically, the assumption (8.307) is equivalent to saying that the surfaceof states is a plane perpendicular to the ρ-axis. Thus, the real surface F (ρ, θ, p) =0 is replaced by the plane ρ = const.; in general, the latter is not tangent tothe former and does not constitute a mathematical approximation in the samesense that a tangent surface would, and it is necessary to realize that we are justdealing with a physical approximation. Consequently, the relations between thethermodynamic properties of the approximation need not be the same as thosebetween the thermodynamic properties of the real fluid.

Writing the total differential

dρ = ∂ρ∂θ∣p

dθ +∂ρ

∂p∣θ

dp, (8.309)

we notice that (8.307) implies dρ = 0 for any values of the increments of dθ anddp, so that

∂ρ

∂θ∣p

= ∂ρ∂p∣θ

= 0, (8.310)

which says that ρ remains constant regardless of the variation in pressure andtemperature. It follows that for an incompressible fluid, the coefficient of thermalexpansion, the isothermal compressibility, and the isochoric pressure coefficient(or coefficient of tension) acquire the singular values

α = −1ρ

∂ρ

∂θ∣p

= 0, κθ = −1ρ

∂ρ

∂p∣θ

= 0, π = 1

p

∂p

∂θ∣ρ

= indeterminate. (8.311)

In solving a problem, the above values must be used for consistency.The internal energy of an incompressible fluid is a function of temperature alone,

e = e(θ), (8.312)

and the specific heat at constant volume also becomes a function of temperaturealone,

cv = ∂e∂θ∣ρ

= cv(θ). (8.313)

The form of the function must be obtained from experiments. Since the enthalpyis related to the internal energy by the relation h = e+ p/ρ, and since the quantityp/ρ is independent of temperature, the specific heat at constant pressure can becalculated by direct differentiation to obtain

cp = ∂h∂θ∣p

= ∂e∂θ∣p

. (8.314)

Since, however, the internal energy is a function of temperature alone, we have

∂e

∂θ∣p

= ∂e∂θ∣ρ

= cv, (8.315)

8.5. CLASSIFICATIONS OF FLUID MOTIONS 375

and thus

cp = cv = c. (8.316)

Hence, there is no distinction between the two specific heats and the symbol c isusually used to replace both cp and cv.

In some problems, it is found necessary to take into account the effect of a changein temperature on the specific volume, while the effect of change in pressure maystill be disregarded:

∂ρ

∂θ∣p

≠ 0 and∂ρ

∂p∣θ

= 0. (8.317)

Consequently,

α = −1ρ

∂ρ

∂θ∣p

≠ 0, (8.318)

and since in this case the density is a function of temperature only, we must alsoassume that

α = α(θ). (8.319)

In other words, the coefficient of thermal expansion, as well as the density, mustbe averaged with respect to the variation of pressure in the problem, but theirdependence on temperature is accounted for. This fluid model is also described asincompressible. The difference between this and the previous case is that now theeffect of thermal expansion is allowed for.

Geometrically, the equation F (ρ, θ, p) = 0 of an incompressible but thermallyexpanding fluid is described by a cylindrical surface, since ρ = const. at θ = const.This surface averages, but does not approximate, the real ρ, θ, p surface.

In addition, we note that in the first model of incompressible flow, the work perunit mass in any reversible process vanishes since from (5.156) with τ 1 = −p andν1 = v, we have

dw = −pdv = p

ρ2dρ = 0. (8.320)

In the second model, where we allow thermal expansion, the work per unit massis nonzero:

dw = −pdv = −αpρdθ. (8.321)

The motion of an incompressible fluid for which ρ = const. is always isochoricand previous considerations of isochoric motion apply. Subsequently, the linearmomentum equations with this simplification of the stress tensor is referred to asthe Navier–Stokes equations. It should be remembered that for an incompressiblemedium, the pressure is not defined thermodynamically, but is an independentvariable of the motion.

Linear liquid

For some substances, including many liquids (and solids), since their volumechange is small, it is useful to use a thermal equation of state which is linearizedabout the fixed state (θ0, p0):

ρ = ρ(θ, p) = ρ0 [1 − α0 (θ − θ0) + κθ0 (p − p0)] , (8.322)

376 FLUIDS

where the reference density, coefficient of volume expansion, and isothermal com-pressibility are given by

ρ0 = ρ0(θ0, p0), α0 = −1ρ

∂ρ

∂θ∣θ0,p0

, and κθ0 = 1

ρ

∂ρ

∂p∣θ0,p0

. (8.323)

The corresponding caloric equation of state is given by

e = e(θ, p) = e0 +∫ θ

θ0

cv(θ′)dθ′ − 1

ρ0 κθ0α0 θ − κθ0 p+

(1 + α0 θ0 − κθ0 p0) ln [1 − α0 (θ − θ0) + κθ0 (p − p0)] . (8.324)

To be able to obtain all properties, it is now just necessary to specify cv(θ) orcp(θ). For liquids, values are usually obtained experimentally and subsequentlytabled (for solids, approximate formulas from statistical mechanics exist). Herewe only note that for many liquids, cp(θ) is almost independent of temperature(for liquid water cp ≈ 4.1855 J/g⋅K). Then, the temperature dependence of cv(θ)is obtained from

cp − cv = α20 θ

ρ0 κθ0 [1 − α0 (θ − θ0) + κθ0 (p − p0)] . (8.325)

Note that even for those liquids with constant cp, there consequently is a temper-ature dependence of cv(θ).

8.5. CLASSIFICATIONS OF FLUID MOTIONS 377

Table 8.1: Normal modes of acoustic propagation.

Ideal Fluid Dissipative Fluid Normal Mode

λ1 = i c0 k λ1 = i c0 k − 1

2[43ν0 + ν

′

0 + (γ0 − 1)χ0]k2 +O(k3) Acousticλ2 = −i c0 k λ2 = −i c0 k − 1

2[43ν0 + ν

′

0 + (γ0 − 1)χ0]k2 +O(k3) Acousticλ3 = 0 λ3 = −ν0 k2 Shearλ4 = 0 λ4 = −ν0 k2 Shearλ5 = 0 λ5 = −χ0 k

2+O(k4) Thermal

378 FLUIDS

Problems

1. For a scalar quantity such as entropy, frame invariance requires that

η (ρ,Q ⋅D ⋅QT , θ,Q ⋅ g) = η (ρ,D, θ,g) .Use the same procedure used for the frame invariance of a vector quantity(see (8.58)) to obtain a result analogous to (8.61).

2. Derive (8.167)–(8.171).

3. Determine σ11 − σ22 and σ11 − σ33 for a simple shear flow, where

σ = −p1 + 2µD + νD2.

4. i) Show that for a linear compressible viscous fluid described by the stresstensor

σ = −p(ρ, θ)1 + λ(ρ, θ) (trD)1 + 2µ (ρ, θ)D,the viscous dissipation is given by

Φ = λ (trD)2 + 2µ tr (D2). (8.326)

ii) Show that (8.326) can be rewritten as

Φ = (λ + 2

3µ) (trD)2 + 2µ tr [D − 1

3(trD)1]2 . (8.327)

iii) Show that (8.326) can be rewritten as

Φ = (λ + 2µ)D2

(1) − 4µD(2), (8.328)

where D(1) and D(2) are the first two invariants of D.

5. For a Newtonian fluid, show that the mechanical energy can be written as

Φ = ζ D2

(1) + 2µD′

ikD′

ik.

Furthermore, writing the Cauchy stress tensor as σij = −p δij +σdij , show that

∂Φ/∂Dij = σdij/2.

6. Show that for an incompressible, irrotational flow with velocity potential φgiven by v = −∇φ, the mechanical energy for a Newtonian fluid is

Φ = ∇2 (∇φ)2 = 1

2∇

4φ2.

7. i) Verify (8.199)–(8.201) starting from (8.187)–(8.189).

ii) Verify (8.203)–(8.205) starting from (8.199)–(8.201).

iii) Verify (8.209)–(8.211) starting from (8.203)–(8.205).

iv) Rewrite (8.209)–(8.211) in the from (8.214) and subsequently verify thedispersion relation (8.215).

8.5. CLASSIFICATIONS OF FLUID MOTIONS 379

v) Plot λi (i = 1, . . . ,5) vs. k for air at a temperature of 300 K and apressure of 1 atm, treating air as an ideal gas, and compare with theapproximate expressions for λi.

8. If a fluid motion is very slow so that higher order terms in the velocity arenegligible, the limiting case of creeping flow results. For this case, showthat in a steady incompressible flow with zero body forces, the pressure is aharmonic function, i.e., ∇2p = 0.

9. Assuming constant viscosity, show that the Navier–Stokes equations for iso-choric irrotational flow reduce to the Euler equations of motion.

10. Express the mass conservation and Navier–Stokes equations in terms of thevelocity potential φ for an irrotational motion assuming constant viscosities.

11. Show that if the body force is conservative so that f = −grad Ω, the com-pressible Navier–Stokes equations for the irrotational motion of an inviscidbarotropic fluid may be integrated to yield

−∂φ

∂t+1

2(∇φ)2 +P +Ω = f(t),

where φ is the velocity potential given by v = −∇φ, P is the pressure function,and f is an arbitrary function of time. This equation is known as Bernoulli’sequation. Subsequently, show that for steady flow of an ideal gas, it takesthe form

a) for isothermal flow (p/ρ = const.)

1

2v2 +Ω +

p

ρln p = const.

and

b) for isentropic flow (p/ργ = const.)

1

2v2 +Ω + ( γ

γ − 1)(p

ρ) = const.

12. Starting from the Euler equations of motion for a fluid of constant densitywith a potential force per unit mass f = −grad Ω, show that for a fixedvolume V enclosed by a surface S , we have

d

dt∫

V

1

2ρv2dv +∫

S

ρH v ⋅ ds,

where

H = 1

2v2 +Ω +

p

ρ

is the Bernoulli quantity.

13. A two-dimensional irrotational flow given by the velocity potential φ =ekx2 sink x1, with k > 0, occupies the half-space x2 < 0. Show that the flowis incompressible. Calculate the velocity field v and the stream functionψ(x1, x2). Sketch the stream lines.

380 FLUIDS

14. Calculate the vorticity ω = curlv of the velocity field

v1 = −αx1 − r x2 f(t), v2 = −αx2 + r x1 f(t), v3 = 2αx3,where x21 + x

22 = r2. Show that divv = 0 for any function f(t), and that the

vorticity equation for a Newtonian fluid is satisfied if and only if f(t)∝ e2αt.

15. Show that for an inviscid flow with a conservative body force and constantdensity, the following relation between vorticity and velocity is satisfied:

ωi − ωjvi,j = 0.For steady flow, show that

vjωi,j = ωivi,j .

16. Determine the pressure function P (p) for a piezotropic fluid having the equa-tion of state p = k ργ , where k and γ are constants.

17. A non-dissipative van der Waal compressible gas is characterized by theHelmholtz free energy

ψ(v, θ) = −Rθ log(v − b) − av− cv θ log ( θ

θ0) ,

where v is the specific volume, and a and b are constants. Determine explicitexpressions for e(v, θ), p(v, θ), and η(v, θ).

18. Show that for an ideal gas under isothermal conditions and with body forcef = −g i3, we have

ρ

ρ0= p

p0= e−(g/Rθx3),

where ρ0 and p0 are the density and pressure at x3 = 0.

Bibliography

R. Aris. Vectors, Tensors and the Basic Equations of Fluid Mechanics. DoverPublications, Inc., Mineola, NY, 1962.

T.S. Chang. Constitutive equations for simple materials: Thermoelastic fluids. InA.C. Eringen, editor, Continuum Physics, volume II, chapter 3, pages 267–281.Academic Press, Inc., New York, NY, 1975.

B.D. Coleman and W. Noll. The thermodynamics of elastic materials withheat conduction and viscosity. Archive for Rational Mechanics and Analysis,13(1):167–178, 1963.

A.C. Eringen. Mechanics of Continua. R.E. Krieger Publishing Company, Inc.,Melbourne, FL, 1980.

C.B. Kadafar. Methods of solution: Exact solutions in fluids and solids. In E.H.Dill, editor, Continuum Physics, volume II, pages 407–448. Academic Press, Inc.,New York, NY, 1975.

BIBLIOGRAPHY 381

I.-S. Liu. On entropy flux-heat flux relation in thermodynamics with Lagrangemultipliers. Continuum Mechanics and Thermodynamics, 8(4):247–256, 1996.

I.-S. Liu. Continuum Mechanics. Springer-Verlag, Berlin, 2002.

C. Truesdell and W. Noll. The non-linear field theories of mechanics. In S. Flügge,editor, Handbuch der Physik, volume III/3. Springer, Berlin-Heidelberg-NewYork, 1965.

C. Truesdell and K.R. Rajagopal. An Introduction to the Mechanics of Fluids.Birkhauser, Boston, MA, 2000.

9

Viscoelasticity

9.1 Introduction

The stress tensor for an incompressible Newtonian fluid is given by

σik = −p δik + 2µDik, (9.1)

where

Dik = 1

2( ∂vi∂xk+∂vk

∂xi) (9.2)

and vi = vi(xj , t). For the simple shear (Couette) flow illustrated in Fig. 9.1, wehave

v1 = 0, v2 = κx1, v3 = 0, (9.3)

where κ = V /d is the amount of shear, so that σ12 = σ21 = µκ with all otherdissipative stress components being zero.

In fact, when we do an experiment, we find that

σ12 = σ21 = µ (κ) κ, (9.4)

where µ = µ (κ) is called the apparent shear viscosity. Different variations of viscos-ity have been observed as shown in Fig. 9.2. Shear thinning and shear thickening

x1

x2

0d

v2

V

Figure 9.1: Simple shear flow.

383

384 VISCOELASTICITY

κ

µ(κ)

0

(a)

(b)

(c)

Figure 9.2: Variations of shear viscosity: (a) Newtonian fluid; (b) shear thinningfluid; and (c) shear thickening fluid.

fluids are also referred to as pseudoplastic and dilatant fluids, respectively, andsuch fluids are called non-Newtonian fluids.

We note that Newtonian fluids, such as air and water, tend to be low-molecular-weight substances with little internal structure. On the other hand, non-Newtonianfluids, including plastics, food materials, paints, inks, biological fluids, and lubri-cants just to name a few, tend to have some internal structure that causes non-Newtonian behavior. Other anomalous effects of non-Newtonian fluids are theclimbing on a rotating rod, called the Weissenberg effect, and the die swell effectof a fluid exiting from a tube. In contrast to simple shear, both of these phenomenaare normal stress effects.

To reconcile with experiments, one might consider the Reiner–Rivlin fluid whosestress tensor for an incompressible fluid is given by (see (8.116))

σik = σik (Dlm) = −p δik + α1Dik + α2DilDlk, (9.5)

where σik = σki, and α1 and α2 are functions of the invariants D(2) and D(3) (notethat D(1) = 0 for an incompressible fluid). Now, for the simple shear flow given by(9.3), we find that

[σik] = −p ⎛⎜⎝1 0 0

0 1 0

0 0 1

⎞⎟⎠ + α1

⎛⎜⎝0 1

2κ 0

1

2κ 0 0

0 0 0

⎞⎟⎠ + α2

⎛⎜⎝1

4κ2 0 0

0 1

4κ2 0

0 0 0

⎞⎟⎠ , (9.6)

the invariants are D(1) = D(3) = 0, D(2) = − 1

4κ2, and thus α1 = α1(κ2) and

α2 = α2(κ2). For this flow, assuming that no body forces are present, the mass andlinear momentum balances are automatically satisfied. To maintain this motion,we need to apply the following stresses:

σ12 = σ21 = 1

2α1(κ2)κ, σ11 = σ22 = −p + 1

4α2(κ2)κ2, σ33 = −p. (9.7)

However, experimentally it is found that σ11 and σ22 are not equal. Consequently,this extension of Newtonian flow fails.

In passing, we note that special flows that enable the measurement of viscositiesare referred to as viscometric flows. In addition to the simple shearing flows,

9.1. INTRODUCTION 385

there are other incompressible steady flows that can be solved exactly, and theresulting stress and velocity profiles are related to the same material functions.Such problems include the flow through a circular pipe (Poiseuille flow), the flowbetween two concentric pipes, and the flow between two such pipes but in whichthe pipes rotate with different constant velocities.

To gain some insight on why and when the Reiner–Rivlin fluid fails, let usexamine the incompressible elongational or stagnation flow

v1 = −ax1, v2 = ax2, v3 = 0. (9.8)

Now we find that

[σik] = −p ⎛⎜⎝1 0 0

0 1 0

0 0 1

⎞⎟⎠ + α1

⎛⎜⎝−a 0 0

0 a 0

0 0 0

⎞⎟⎠ + α2

⎛⎜⎝a2 0 0

0 a2 0

0 0 0

⎞⎟⎠ , (9.9)

where D(1) = D(3) = 0, D(2) = −a2, and thus α1 = α1(a2) and α2 = α2(a2).Subsequently, we have the following normal stress differences:

σ22 − σ11 = 2α1a, σ33 − σ11 = α1a −α2 a2, σ33 − σ22 = −α1a −α2 a

2. (9.10)

In this case, we see normal stress differences that are observed in experiments. Wenote that the above flow is equivalent to the shearing flow

v1 = κx2, v2 = κx1, v3 = 0 (9.11)

with κ = −a. This is seen by rotating the x1-x2 axes by π/2.At this stage, to gain additional insight, we could superpose the shearing and

elongational flows as in

v1 = −ax1, v2 = κx1 + ax2, v3 = 0. (9.12)

However, although the above generalization is interesting, in practice it is foundthat markedly non-Newtonian fluids have a more complex behavior than is permit-ted by the Reiner–Rivlin constitutive equation. Subsequently, we wish to introducethe effect of history on such flow. Thus, we examine the flow whose material pointswere at ξi(τ) at time τ , which at time t are at xi(t), i.e., we examine the motionξi = ξi(τ) satisfying the following flow:

dξ1

dτ= −a ξ1, dξ2

dτ= κξ1 + a ξ2, dξ3

dτ= 0, (9.13)

with end conditionsξi(t) = xi. (9.14)

The solution is easily obtained by solving the system and applying the end condi-tions:

ξ1 = x1 ea(t−τ), (9.15)

ξ2 = (x2 + κx12a) e−a(t−τ) − κx1

2aea(t−τ)

= x2 e−a(t−τ)

−κx1

asinh [a(t − τ)] , (9.16)

ξ3 = x3. (9.17)

Before examining the effects of history on the constitutive equations, we pauseto examine such effects on kinematic quantities.

386 VISCOELASTICITY

9.2 Kinematics

Recall that a material point X located at x at time t, which was at ξ at time τ ,is described by the relative motion (see (3.266))

ξ = (t)χ(x, τ). (9.18)

Then, the velocity field given as a function of the position ξ and time τ ,

v = v(ξ(τ), τ), (9.19)

can be found if its position x at time t is known. In such case, it is given by

dξ(τ)dτ

= v(ξ(τ), τ) (9.20)

satisfying the end condition

ξ(τ)∣τ=t = (t)χ(x, t) = x. (9.21)

Thus, knowledge of the position x of each material point at time t and the velocityfield (9.19) for all τ is equivalent to knowledge of the relative motion (9.18).

The gradient of the relative motion (9.18) yields the relations (3.268) and (3.303)for the history of the relative deformation gradient:

(t)F(x, τ) = (gradξ)T and dξ = (t)F(x, τ) ⋅ dx. (9.22)

We also recall that the history of the relative deformation gradient (t)F(τ) isrelated to the deformation gradient in the reference configuration by (see (3.274);here we take t0 = 0)

(t)F(x, τ) = (0)F(X, τ) ⋅ (0)F−1(X, t) = F(X, τ) ⋅F−1(X, t), (9.23)

where (t)F(x, t) = 1 (see (3.269)).In what follows, to simplify notations, we will not display the explicit dependence

of quantities on the motion x; this dependence will be made explicit later when itbecomes relevant. Furthermore, as in Section 5.1, it is often convenient to writethe history of quantities in terms of the time shift or history variable s = t − τinstead of the current variable τ , in which case 0 ≤ s <∞. That is, we can write

(t)F(τ) = (t)F(t − s) = (t)F(t)(s), (9.24)

where we recall from Sections 3.2.3 and 5.1 that the preceding subscript “(t)”denotes the relative description of the motion (relative to t) while the superscript“(t)” denotes the history up to time t. Note that we are and will be using thesame symbols for the functions of τ and s. This should lead to no confusion sincewe make the function dependencies explicit.

Now, from (9.22)2, we have that the length of a material element is provided by

dξT⋅dξ = dxT

⋅(t)FT (τ)⋅(t)F(τ)⋅dx = dxT⋅(t)C(τ)⋅dx = dxT

⋅(t)C(t)(s)⋅dx, (9.25)

9.2. KINEMATICS 387

where we have used (3.283). If we assume that the deformation gradient is infinitelydifferentiable in time, we can write the history of the relative right Cauchy–Greentensor as

(t)C(τ) = ∞∑n=0

1

n!

∂n(t)C(τ)∂τn

∣τ=t

(τ − t)n = ∞∑n=0

1

n!A(n)(t) (τ − t)n =

∞

∑n=0

(−s)nn!

A(n)(t), (9.26)

where we have defined the nth Rivlin–Ericksen tensor

A(n)(t) ≡ ∂n(t)C(τ)∂τn

∣τ=t

= (−1)n ∂n(t)C(t)(s)∂sn

RRRRRRRRRRRs=0 , n = 0,1,2, . . . (9.27)

Subsequently, we can write that

dξT ⋅ dξ = dxT⋅

∞

∑n=0

(−s)nn!

A(n)(t) ⋅ dx. (9.28)

Note that if A(n)(t) = 0 for all t, then (t)C(t)(s) is a polynomial in s of degreeless than n.

From the definition of the relative right Cauchy–Green tensor (3.283), it is easyto show that

∂n(t)C(τ)∂τn

= n

∑k=0

(nk) ∂k(t)FT (τ)

∂τk⋅∂n−k(t)F(τ)

∂τn−k, n = 0,1,2, . . . , (9.29)

where the binomial coefficients are given by

(nk) = n!(n − k)!k! . (9.30)

Evaluating (9.29) at τ = t, and using (9.27), we subsequently have that

A(n)(t) = n

∑k=0

(nk) (t)(k)F T (t) ⋅ (t)(n−k)F (t), n = 0,1,2, . . . (9.31)

We also define the nth acceleration gradient by

L(n)(t) ≡ (grad (n)x (t))T , n = 0,1,2, . . . (9.32)

Note that(0)x = x,

(1)x = x = v and

(2)x = x = a, and that L(0) = 1, L(1) = L, and the

spatial gradient of the acceleration is given by

L(2)(t) ≡ (grad x(t))T . (9.33)

Also note that since (see (3.290))

L(t) = (t)F(t) = ∂

∂τ(t)F(τ)∣

τ=t= − ∂

∂s(t)F

(t)(s)∣s=0

, (9.34)

388 VISCOELASTICITY

the spatial gradient of the acceleration is related to (t)F(t) by

L(2)(t) = (t)F(t). (9.35)

Subsequently, we see that the nth acceleration gradient is related to (t)F(t) by

L(n)(t) = (t)(n)F (t) = ∂n

∂τn(t)F(τ)∣

τ=t= (−1)n ∂n

∂sn(t)F(t)(s)∣

s=0, n = 0,1, . . . .

(9.36)Now using (9.27) and (9.29), we easily see that at τ = t the nth Rivlin–Ericksentensor is also given by

A(n)(t) = n

∑k=0

(nk) [L(k)(t)]T ⋅L(n−k)(t), n = 0,1,2, . . . . (9.37)

Note that A(0) = 1 and A(1) = L +LT = 2D.It follows from (9.25) that

dn

dτn(dξT ⋅ dξ) = dxT

⋅∂n(t)C(τ)

∂τn⋅ dx = (−1)n dxT

⋅∂n(t)C(t)(s)

∂sn⋅ dx. (9.38)

Subsequently, evaluating this result at τ = t or s = 0 , we have that

(n)dxT ⋅ dx = dn

dtn(dxT

⋅ dx) = dxT⋅A(n)(t) ⋅ dx. (9.39)

Differentiating (9.39) with respect to t, we obtain

(n+1)dxT ⋅ dx = ˙

dxT ⋅A(n)(t) ⋅ dx + dxT⋅ A(n)(t) ⋅ dx + dxT

⋅A(n)(t) ⋅ ˙dx. (9.40)

Now, using (9.39) and since ˙dx = L ⋅ dx (see (3.305)), we have that

A(n+1)(t) = A(n)(t) +LT (t) ⋅A(n)(t) +A(n)(t) ⋅L(t), n = 0,1,2, . . . , (9.41)

where

A(n)(t) = ∂A(n)(t)∂t

+ (v(t) ⋅ grad)A(n)(t). (9.42)

Recalling the Cotter–Rivlin tensor (3.425), we see that

A(n+1)(t) =

A(n)(t). (9.43)

Furthermore, using the polar decomposition expressed through (3.282)1, andgeneralizing (3.293) and (3.294) by defining the nth stretch (or rate of strain) andspin tensors by

D(n)(t) ≡ (t)(n)U (t) = ∂n

∂τn(t)U(τ)∣

τ=t= (−1)n ∂n

∂sn(t)U

(t)(s)∣s=0

, n = 0,1, . . . ,(9.44)

9.2. KINEMATICS 389

and

W(n)(t) ≡ (t)(n)R (t) = ∂n

∂τn(t)R(τ)∣

τ=t= (−1)n ∂n

∂sn(t)R

(t)(s)∣s=0

, n = 0,1, . . . ,(9.45)

it is easy to show that the nth acceleration gradient and stretch and spin tensorsare respectively given by

L(n)(t) = n

∑k=0

(nk)W(k)(t) ⋅D(n−k)(t), n = 0,1,2, . . . , (9.46)

D(n)(t) = 1

2[A(n)(t) − n−1

∑k=1

(nk)D(k)(t) ⋅D(n−k)(t)] , n = 1,2, . . . , (9.47)

and

W(n)(t) = L(n)(t)−D(n)(t)−n−1

∑k=1

(nk)W(k)(t) ⋅D(n−k)(t), n = 1,2, . . . . (9.48)

Note that D(0) = 1, W(0) = 1, D(1) = D, and W(1) = W. Equation (9.47)is a recursion formula which can be used to find explicit expressions for D(n)

as polynomials in A(k), k = 1,2, . . . , n. Hence, after substitution of (9.37), one

would also obtain explicit expressions for D(n) as polynomials in L(k) and L(k)T,

k = 1,2, . . . , n. Equation (9.48) is also a recursion formula which permits us toexpress W(n) as a polynomial in L(k) and D(k). Since D(k) are polynomials in L(k)

and L(k)T, we can also find expressions for W(n) as polynomials in L(k) and L(k)

T,

k = 1,2, . . . , n. In the special case n = 1, we find that D(1) = 1

2A(1) = 1

2(L +LT )

and W(1) = L(1) −D(1) = L − 1

2(L +LT ) = 1

2(L −LT ).

Note that while the spin W is skew-symmetric, the higher spins W(n), n > 1,are not necessarily skew-symmetric. Also note that the symmetric Rivlin–Ericksentensors A(n) and stretching tensors D(n) are frame indifferent, while the velocitygradients L(n) and spin tensors W(n) are not.

In the above definitions of A(n), D(n), and W(n), the present configuration hasbeen chosen as the reference configuration. If a fixed reference configuration isused, more complicated formulae result (e.g., see (3.491) and (3.492)).

Example

Here we obtain A(2), D(2), and W(2) in terms of acceleration gradients.We readily see from (9.37), (9.47), and (9.48) that

390 VISCOELASTICITY

A(2) = L(2) +L(2)T+ 2LT

⋅L, (9.49)

D(2) = 1

2(A(2) − 2D(1)2)

= 1

2(A(2) − 1

2A(1)

2)= 1

2(L(2) +L(2)T) − 1

4(L +LT )2 +LT

⋅L, (9.50)

W(2) = L(2) −D(2) − 2W(1)D(1)

= 1

2(L(2) −L(2)T) + 1

4(L +LT )2 −LT

⋅L. (9.51)

Lastly, we define the right Cauchy–Green tensor difference history, or historytensor for short, as follows:

G(t)(s) ≡ (t)C(t)(s)−1 or G

(t)(s) = (t)C(t)(s)− I or G(t)ik(s) ≡ (t)C(t)ik

(s)− δik,(9.52)

where we see that G(t)(0) = 0. Then, from (9.26) and (9.52), we have

G(t)(s) = ∞∑

n=1

(−s)nn!

A(n)(t), (9.53)

where

A(n)(t) ≡ (−1)n ∂nG (t)(s)∂sn

∣s=0

, n = 1,2, . . . . (9.54)

9.2.1 Motion with constant stretch history

In many motions, that are important theoretically and experimentally, the historiesfrequently take particular forms enabling considerable simplifications. A motionwith constant stretch history (MWCSH) results when the present deformation isobtained through a sequence of past deformations that remain constant. Thatis, a motion, along the path of a material point, has constant stretch history ifthere exists an orthogonal tensor function Q(t) such that the histories of the rightCauchy–Green tensors (t)C(t)(s) and (0)C(0)(s) are related as

(t)C(t)(s) =Q(t) ⋅ (0)C(0)(s) ⋅QT (t), 0 ≤ s <∞, (9.55)

with Q(0) = 1.It can be proved that a motion is an MWCSH if and only if there exist an

orthogonal tensor Q(t), a scalar κ (called the shearing), and a constant secondrank tensor N0 such that

(0)F(τ) = F(τ) =Q(τ) ⋅ eτ κN0 , Q(0) = 1, ∣N0∣ = 1, (9.56)

where ∣N0∣2 = tr (NT0 ⋅ N0). From the polar decomposition of the deformation

gradient, it is seen that Q(τ) corresponds to the rotation tensor R(τ) and eτ κN0

corresponds to the right stretch tensor U(τ). It is easy to note that in the MWCSH

9.2. KINEMATICS 391

deformation, the principal stretch histories remain constant but the principal axesrotate.

We remark that since (see (3.239))

F(t) = L(t) ⋅F(t), (9.57)

if we have no deformation initially, i.e., F(0) = 1, and if the velocity gradient L isconstant, then we have

F(t) = etL (9.58)

and

(t)F(τ) = e(τ−t)L. (9.59)

Now it is straightforward to show that

(t)C(t)(s) = (0)C(0)(s) (9.60)

so, when compared to the more general definition, we see that in this case Q(t) = 1and κN0 = L. When Q(t) = 1, the rotation associated with the relative deforma-tion gradient reduces to the rotation of the eigenvectors associated with the rateof strain.

Before proceeding with our discussion, it is important to recall the followingrelations involving the exponential of a tensor:

eM = ∞∑n=0

1

n!Mn, (9.61)

(eM)T = eMT

, (eM)−1 = e−M, e0 = 1, (9.62)

eM+N = eMeN if M ⋅N =N ⋅M, (9.63)

(Q ⋅M ⋅QT )n =Q ⋅Mn⋅QT if Q ∈ O, (9.64)

eQ⋅M⋅QT =Q ⋅ eM ⋅QT if Q ∈ O, (9.65)

d

dτeτ M =M ⋅ eτ M = eτ M

⋅M. (9.66)

Furthermore, a tensor M is nilpotent if Mn = 0 for some integer n ≥ 0. Theexponential eτ M is a finite polynomial in τ if and only if M is nilpotent. It canbe shown that in three dimensions, if M is nilpotent, then n ≤ 3, and in this case,

eτ M = 1 + τM + 1

2τ2M2. (9.67)

Moreover, it can be shown that the components of M relative to an appropriateorthonormal basis have the form

M = ⎛⎜⎝0 0 0

χ 0 0

λ ν 0

⎞⎟⎠ . (9.68)

If M2 = 0, then the basis may be chosen such that

M = χ⎛⎜⎝0 0 0

1 0 0

0 0 0

⎞⎟⎠ . (9.69)

392 VISCOELASTICITY

Subsequently, from (9.23), (9.56), and (9.65), we have

(t)F(τ) =Q(τ) ⋅e(τ−t)κN0 ⋅QT (t) =Q(τ) ⋅QT (t) ⋅e(τ−t)κN =Q(τ) ⋅QT (t) ⋅e(τ−t)M,(9.70)

and recalling that τ = t − s, we can also write

(t)F(t)(s) =Q(t − s) ⋅QT (t) ⋅ e−sM, (9.71)

where the tensors N and M are defined by

M ≡ κN ≡ κQ(t) ⋅N0 ⋅QT (t), ∣N∣ = 1. (9.72)

We now observe that

(0)C(τ) = (0)FT (τ) ⋅ (0)F(τ) = eτ κN0T

⋅ eτ κN0 , (9.73)

(0)C(0)(s) = ((0)F(0)(s))T ⋅ (0)F(0)(s) = e−sκN0T

⋅ e−sκN0 , (9.74)

and, using (9.55),

(t)C(τ) = (t)C(t)(s) =Q(t) ⋅ e−sκNT0 ⋅ e−sκN0 ⋅QT (t) = e−sMT

⋅ e−sM. (9.75)

Furthermore, associated with the relative deformation gradient, we have that therelative right stretch tensor is given by

(t)U(τ) = (t)U(t)(s) = ((t)C(t)(s))1/2 = e− 1

2sMT

⋅ e−1

2sM =

Q(t) ⋅ (0)U(0)(s) ⋅QT (t), (9.76)

the relative rate of rotation tensor by

(t)R(τ) = (t)R(t)(s) = (t)F(t)(s) ⋅ ((t)U(t)(s))−1 =Q(τ) ⋅QT (t) ⋅ e− 1

2sM⋅ e

1

2sMT =Q(τ) ⋅ (0)R(0)(s) ⋅QT (t), (9.77)

the velocity gradient by (see (3.289))

L(t) = (t)F(τ)∣τ=t =M +Ω, with Ω = Q(t) ⋅QT (t), (9.78)

the stretch and spin tensors by (see (3.293) and (3.294))

D(t) = (t)U(τ)∣τ=t = 1

2(M +MT ) = κ

2Q(t) ⋅ (N0 +N

T0) ⋅QT (t) (9.79)

and

W(t) = (t)R(τ)∣τ=t = 1

2(M −MT ) +Ω = κ

2Q(t) ⋅ (N0 −N

T0) ⋅QT (t) +Ω, (9.80)

and, since for an MWCSH A(n)(t) = 0, the Rivlin–Ericksen tensors by

A(n+1)(t) = LT (t) ⋅A(n)(t) +A(n)(t) ⋅L(t). (9.81)

9.2. KINEMATICS 393

Note that by using (9.37), we also have that

A(1)(t) = LT+L =MT

+M, (9.82)

A(2)(t) = MT⋅A(1) +A(1) ⋅M = (MT )2 + 2MT

⋅M +M2, (9.83)

⋮

A(n)(t) = MT⋅A(n−1) +A(n−1) ⋅M = n

∑k=0

(nk) (Mn−k)T ⋅Mk,

= Q(t) ⋅A(n)(0) ⋅QT (t). (9.84)

Since, as remarked earlier, if M is nilpotent, then necessarily M3 = 0 in a three-dimensional Euclidean space, then it can be shown that all MWCSH flows can bedivided in the following three classes depending on the properties of M (or N orN0):

I. M2 = 0;

II. M2 ≠ 0 but M3 = 0;

III. Mn ≠ 0 for all n = 1,2, . . ..The consequences of this result are as follows. If M3 = 0, then A(n) = 0 for

n ≥ 5, and e−sM is at most a quadratic polynomial in M. If M2 = 0, then A(n) = 0for n ≥ 3, and e−sM is at most a linear polynomial in M. If M is not nilpotent,then e−sM is not a finite polynomial in M.

Class I includes all steady viscometric flows (e.g., simple shear) and in particularflows defined by a steady velocity field. From (9.75), in this case we have

(t)C(t)(s) = [1 − sMT ] ⋅ [1 − sM] . (9.85)

For Class II, we have

(t)C(t)(s) = [1 − sMT+1

2s2 (MT )2] ⋅ [1 − sM + 1

2s2M2] . (9.86)

This class contains the subclass of doubly superposed viscometric flows. Theseflows are generated by a superposition of two steady viscometric flows; however,we note that not all such superpositions belong to Class II.

Class III, for which

(t)C(t)(s) = [ ∞∑n=0

(−1)nn!

sn (MT )n] ⋅ [ ∞∑n=0

(−1)nn!

snMn] , (9.87)

contains the subclass of triply superposed viscometric flows (e.g., steady pure shear,steady simple extension).

We note that a motion is irrotational if the velocity gradient is symmetric.All MWCSH described by a constant symmetric velocity gradient are customarilycalled extensional or elongational flows. Since every symmetric tensor L can alwaysbe represented in a diagonal form, then one finds that all irrotational MWCSHare extensional or elongational flows and belong to Class III. For such flows, in theappropriate reference frame for which L is diagonal, we have

L =M = κN = κN0, Q(t) = 1, (9.88)

394 VISCOELASTICITY

with N0 = NT0 (thus N = NT , M = MT , and L = LT ), and subsequently, from

(9.75), we have that

(t)C(t)(s) = (0)C(0)(s) = e−2 sM, (9.89)

and, from (9.54) and (9.82), we see that

A(n) = (A(1))n = 2nM, n = 1,2, . . . . (9.90)

Furthermore, an MWCSH is isochoric if trL = 0, and thus if and only if trM =trN = trN0 = 0. Now, if M3 = 0, from the Cayley–Hamilton theorem (see (3.95)),we have that (trM)M2 = 1

2[(trM)2 − trM2]M. (9.91)

If M ≠ 0 and M2 = 0, it implies that trM2 = 0 and trM = 0, while if M2 ≠ 0, then

(trM)M3 = 0 = 1

2[(trM)2 − trM2]M2, (9.92)

which implies that (trM)2 = trM2, which in turn implies that trM = 0 and thustrM2 = 0. Subsequently, we deduce that all flows corresponding to Classes I andII are isochoric.

Before presenting an import theorem, the following lemma is necessary.

Lemma: Suppose that S is a 3 × 3 diagonal matrix, and W is a 3 × 3 skew-symmetric matrix, i.e.,

S = ⎛⎜⎝a 0 0

0 b 0

0 0 c

⎞⎟⎠ and W = ⎛⎜⎝0 x y

−x 0 z

−y −z 0

⎞⎟⎠ . (9.93)

Now if

i) a ≠ b ≠ c, then SW =W S if and only if x = y = z = 0 (i.e., W = 0);ii) a = b ≠ c, then SW =W S if and only if y = z = 0 (where x is arbitrary);

iii) a = b = c, then SW =W S for all x, y, and z (i.e., any W ).

Proof: By multiplication

SW = ⎛⎜⎝0 ax ay

−bx 0 b z

−c y −c z 0

⎞⎟⎠ and W S = ⎛⎜⎝0 bx c y

−ax 0 c z

−ay −b z 0

⎞⎟⎠ . (9.94)

Therefore, SW =W S if and only if

(a − b)x = 0, (a − c)y = 0, (b − c) z = 0. (9.95)

Hence, the lemma follows.

Theorem: For an MWCSH, the relative history of (t)C(t)(s) is determined

uniquely by A(1), A(2), and A(3).

9.2. KINEMATICS 395

Proof:

i) If A(1) has three distinct eigenvalues, then it is claimed that M is determineduniquely by A(1) and A(2). To see this, suppose it is false. Then, from (9.82)and (9.83), we have that

A(1) =MT+M =MT

+M and

A(2) =MT⋅A(1) +A(1) ⋅M =MT

⋅A(1) +A(1) ⋅M. (9.96)

But, from the first relation, we see that

M −M = − (M −M)T (9.97)

is skew-symmetric, and from the second relation, we have

(M −M) ⋅A(1) =A(1) ⋅ (M −M) , (9.98)

i.e., the symmetric tensor A(1) commutes with M −M; hence M −M = 0

since the eigenvalues are distinct. That is, M is uniquely determined by A(1)

and A(2).

ii) If A(1) has only two distinct eigenvalues, then, relative to a principal or-thonormal basis, the component matrix of A(1) can be written in the follow-ing form

A(1) = ⎛⎜⎝a 0 0

0 a 0

0 0 b

⎞⎟⎠ , a ≠ b. (9.99)

We consider two subcases.

a) Relative to the principal orthonormal basis of A(1), the componentmatrix of A(2) is of the form

A(2) = ⎛⎜⎝u 0 0

0 u 0

0 0 v

⎞⎟⎠ , (9.100)

where u may or may not be equal to v. In this case, we deduce thatu = a2, v = b2, and thus (t)C(t)(s) is determined by A(1) (M is not

unique, but (t)C(t)(s) is unique), and then it can be easily shown that

(t)C(t)(s) = e−sA(1) . (9.101)

b) If A(2) is not of the above form, then (t)C(t)(s) is a function of A(1),A(2), and A(3), and M is unique, i.e., the solution of the system

A(1) =MT+M, (9.102)

A(2) =MT⋅A(1) +A(1) ⋅M, (9.103)

A(3) =MT⋅A(2) +A(2) ⋅M, (9.104)

396 VISCOELASTICITY

is unique. To see this, suppose that M is another solution of the system.Then, from (9.102), the difference between M and M is a skew-symmetrictensor. From (9.103), (M−M) commutes with A(1). Thus, from the previouslemma, we have

M −M = ⎛⎜⎝0 x 0

−x 0 0

0 0 0

⎞⎟⎠ . (9.105)

From (9.104), (M−M) also commutes with A(2). Since by assumption A(2)

is not of the form (9.100), by multiplication it is easily seen that x = 0. ThusM =M.

iii) Here we claim that if all eigenvalues of A(1) are equal, then A(1) = a1. It iseasily seen that in this case

(t)C(t)(s) = e−as1. (9.106)

Again, the tensor M is not unique, but any solution M of (9.102) yields thesame relative right Cauchy–Green tensor (9.106).

Thus, the theorem is proved.

9.3 Constitutive equations

The most general constitutive equation that we have considered is of the form (see(5.14))

T (X, t) = FY∈V

0≤s<∞

[x(t)(Y, s) − x(t)(X, s)], θ(t)(Y, s),X, (9.107)

subject to restrictions arising from the rigid rotation of the frame, such as (5.13)for a second order tensor. Reasonable assumptions can be made that values ofconstitutive variables at distant materials points from X do not affect appreciablythe values of the constitutive variables at X. Under such assumption, two possibleavenues for developing more constructive forms of the constitutive equation (9.107)have been pursued. The first is based on smoothness of neighborhood, wherederivatives up to some order in space are assumed to exist (see Section 5.4). Insuch case, considering homogeneous materials and applying the frame-invariancerequirement as done in Section 5.7, one obtains the following reduced constitutiveequation

T (X, t) = G0≤s<∞

iC(t)(X, s), θ(t)(X, s),jG(t)(X, s), (9.108)

where

T ≡ FT⋅ T ⋅F (9.109)

is the convected tensor, and i = 1,2, . . . ,P, j = 1,2, . . . ,Q. Using (5.23), we notethat

iC(t) ≡ FT⋅iF(t) or iC

(t)KK1K2⋯Ki

≡ FkKiF(t)kK1K2⋯Ki

. (9.110)

9.3. CONSTITUTIVE EQUATIONS 397

The above constitutive functional generalizes the expression given in (5.94), whichapplies to a simple material. Subsequently, using the same procedure as in Sec-tion 5.7, one finds that the corresponding constitutive functional for a homoge-neous non-simple isotropic solid in the current configuration is given by

T (x, t) = F0≤s<∞

B, (t)iC(t)(x, s), θ(t)(x, s), (t)jg(t)(x, s) (9.111)

and that for a non-simple (isotropic) fluid by

T (x, t) = F0≤s<∞

ρ, (t)iC(t)(x, s), θ(t)(x, s), (t)jg(t)(x, s), (9.112)

where

(t)iC(t)(x, s) = (t)iC(x, τ) and (t)jg(t)(x, s) = (t)jg(x, τ). (9.113)

The constitutive functionals (9.111) and (9.112) are required to satisfy appropri-ate frame invariance conditions depending on the order of tensor T analogous to(5.110) and (5.115) for a simple solid and fluid, respectively.

For the rest of the chapter, we will limit our discussion to constitutive equationsof isotropic simple materials undergoing isothermal deformations, and more specif-ically, we focus on the constitutive equation of the stress tensor and suppress thedependency on θ. Furthermore, continuing in suppressing the dependencies of in-dependent quantities on x, we recall that the constitutive equation of an isotropicsimple solid (see (5.109) or (9.111) for P = 1) is given by

σ(t) = F0≤s<∞

B(t), (t)C(t)(s), (9.114)

while for a simple fluid the dependence on B(t) reduces to a dependence on thedensity, ρ(t)1 (see (5.114)). Here, F is a tensor-valued functional (i.e., an operator,not necessarily linear, mapping tensor-valued functions into tensors). The aboveconstitutive functional is required to satisfy the invariance condition

Q(t) ⋅ F0≤s<∞

B(t), (t)C(t)(s) ⋅QT (t) =F

0≤s<∞Q(t) ⋅B(t) ⋅QT (t),Q(t) ⋅ (t)C(t)(s) ⋅QT (t). (9.115)

It is useful to rewrite (9.114) in a slightly different form by decomposing theright-hand side into an “equilibrium” function and a functional that vanishes whenthe material has always been at rest:

σ(t) = hB(t) + J0≤s<∞

B(t),G (t)(s), (9.116)

where, having used the definition (9.52), we require that

J0≤s<∞

B(t),0(s) = 0, (9.117)

and it is understood that if the material has always been in equilibrium, then

G(t)(s) = 0 for 0 ≤ s <∞. (9.118)

398 VISCOELASTICITY

The function0(s) = 0, 0 ≤ s <∞, (9.119)

is called the zero function.Now the frame invariance condition for the equilibrium state requires that

Q(t) ⋅ hB(t) ⋅QT (t) = hQ(t) ⋅B(t) ⋅QT (t). (9.120)

Subsequently, satisfaction of this condition yields the representation

hB(t) = h0 1 + h1B + h2B2, (9.121)

where hi = hi(trB, trB2, trB3), i = 0,1,2, (see Tables 5.1 and 5.3).Correspondingly, for a fluid the frame invariance condition becomes

Q(t) ⋅ hρ(t) ⋅QT (t) = hρ(t), (9.122)

whose representation is given by

hρ(t) = −p (ρ(t))1, (9.123)

and (9.116) becomes

σ(t) = −p (ρ(t))1 + J0≤s<∞

ρ(t),G (t)(s) (9.124)

or

σd(t) = σ(t) + p (ρ(t))1 = J0≤s<∞

ρ(t),G (t)(s), (9.125)

where σd(t) is the dissipative or extra stress.If the fluid is incompressible, then ρ(t) = ρR, where we assume that ρR = const.,

and the stress tensor is determined by the history of the motion only up to a scalarpressure p. Thus, for incompressible fluids, we can replace (9.125) by

σd(t) = σ(t) + p1 = J0≤s<∞

G (t)(s), (9.126)

and the tensor-valued functional is now determined only up to an arbitrary scalar-valued functional of G

(t)(s). The indeterminacy is removed by the normalization

trσd(t) = tr J0≤s<∞

G (t)(s) = 0. (9.127)

If follows from (9.126) and (9.127) that p is given by the mean pressure

p = −13trσ, (9.128)

and the extra stress is equal to the deviatoric part of σ, i.e., σd = σ′.Note the frame invariance condition (9.115) and the equilibrium condition (9.117)

reduce to

Q(t) ⋅ F0≤s<∞

ρ(t),G (t)(s) ⋅QT (t) = F0≤s<∞

ρ(t),Q(t) ⋅G (t)(s) ⋅QT (t) (9.129)

and

J0≤s<∞

ρ(t),0(s) = 0. (9.130)

9.3. CONSTITUTIVE EQUATIONS 399

9.3.1 Constitutive equations for motion with constantstretch history

For an MWCSH, the constitutive equation (9.116), upon using (9.52) and (9.75),becomes

σ(t) = hB + f (B,M) , (9.131)

where

f (B,M) = J0≤s<∞

B(t), e−sMT

⋅ e−sM − 1, (9.132)

and, from (9.117),

f (B,0) = 0. (9.133)

The implication of the theorem in Section 9.2.1 is that we can write (9.131) inthe most general form as

σ(t) = hB + f (B,A(1),A(2),A(3)) . (9.134)

For a fluid, this becomes

σ(t) = −p (ρ)1 + f (ρ,A(1),A(2),A(3)) , (9.135)

and if the fluid is incompressible (see (9.126)),

σd(t) = f (A(1),A(2),A(3)) . (9.136)

Note that for simplicity we have used the same tensor function symbol f in all ofthe above cases; clearly they would correspond to a different function in each case.Now objectivity requires that

Q ⋅ f (B,A(1),A(2),A(3)) ⋅QT =f (Q ⋅B ⋅QT ,Q ⋅A(1) ⋅QT ,Q ⋅A(2) ⋅QT ,Q ⋅A(3) ⋅QT ) (9.137)

or for a fluid

Q ⋅ f (ρ,A(1),A(2),A(3)) ⋅QT = f (ρ,Q ⋅A(1) ⋅QT ,Q ⋅A(2) ⋅QT ,Q ⋅A(3) ⋅QT ) .(9.138)

The representation of (9.135) satisfying (9.138) is fairly lengthy. However, fromthe theorem, we see that for a large majority of MWCSH flows, we have thatthe relative history is uniquely determined by A(1) and A(2). In such cases, therepresentation for a fluid is given by (see Table 5.3)

f = f (ρ,A(1),A(2))= α0 1 + α1A

(1)+ α2A

(2)+ α3 (A(1))2 + α4 (A(2))2 +

α5 (A(1) ⋅A(2) +A(2) ⋅A(1)) + α6 [(A(1))2 ⋅A(2) +A(2) ⋅ (A(1))2] +α7 [A(1) ⋅ (A(2))2 + (A(2))2 ⋅A(1)] , (9.139)

400 VISCOELASTICITY

where α0, . . . , α7 are functions of ρ and the invariants (see Table 5.1)

trA(1), tr (A(1))2, tr (A(1))3, trA(2), tr (A(2))2, tr (A(2))3,tr (A(1) ⋅A(2)), tr [A(1) ⋅ (A(2))2], tr [(A(1))2 ⋅A(2)], tr [(A(1))2 ⋅ (A(2))2].(9.140)

From (9.127) and (9.130), we must require that

tr f (ρ,A(1),A(2)) = 0 and f (ρ,0,0) = 0. (9.141)

Example

To illustrate the concepts introduced in previous sections, we reconsiderthe isochoric simple shear flow discussed in Section 9.1:

v1 = 0, v2 = κx1, v3 = 0. (9.142)

Subsequently, we write

dξ1

dτ= 0, dξ2

dτ= κξ1, dξ3

dτ= 0, (9.143)

with final conditionsξi(t) = xi. (9.144)

The solutions are

ξ1 = x1, ξ2 = x2 + (τ − t)κx1 = x2 − sκx1, ξ3 = x3. (9.145)

Note from (9.142) that since

L = κ ⎛⎜⎝0 0 0

1 0 0

0 0 0

⎞⎟⎠ (9.146)

is constant, then from (9.59), (9.71), (9.72), and (9.78), we have anMWCSH with Q(t) = I,

L =M = κN = κN0 = κ⎛⎜⎝0 0 0

1 0 0

0 0 0

⎞⎟⎠ , M2 = N2 =N2

0 = 0, (9.147)

NT0 N0 = ⎛⎜⎝

1 0 0

0 0 0

0 0 0

⎞⎟⎠ , and ∣N0∣2 = tr (NT0 N0) = 1. (9.148)

From (9.82)–(9.84), we have

A(1) = κ ⎛⎜⎝0 1 0

1 0 0

0 0 0

⎞⎟⎠ , A(2) = 2κ2 ⎛⎜⎝1 0 0

0 0 0

0 0 0

⎞⎟⎠ , and A(n) = 0 for n ≥ 3.(9.149)

9.3. CONSTITUTIVE EQUATIONS 401

Note that since M2 = 0 , then e−sM is at most a linear polynomial in M .Furthermore, since A(1) has three distinct eigenvalues, (0, ±κ), then M isuniquely determined by A(1) and A(2). Subsequently, from (9.71), we havethat

(t)F (t)(s) = [ ∂ξi∂xk] = e−sM = I − sM = ⎛⎜⎝

1 0 0

−κs 1 0

0 0 1

⎞⎟⎠ , (9.150)

from (9.85) we have that

(t)C(t)(s) = (I − sMT ) (I + sM) = ⎛⎜⎝1 −κs 0

−κs 1 + (κs)2 0

0 0 1

⎞⎟⎠ , (9.151)

and from (9.52) we have that

G(t)(s) = ⎛⎜⎝

0 −κs 0

−κs (κs)2 0

0 0 0

⎞⎟⎠ . (9.152)

Hence, we see that simple shearing flow is completely determined by

G(t)(x, s) = −sA(1)(x1) + 1

2s2A(2)(x1). (9.153)

Since simple shear flow is isochoric, from (9.82) and (9.83), we have that

trA(1) = trA(2) − tr (A(1))2 = 0. (9.154)

Hence, from (9.139) and (9.140), absorbing α0 in the arbitrary pressure p,the constitutive equation is

σ = −p1 + f (A(1),A(2)) = −p1 + β1A(1) + β2A(2) + β3 (A(1))2 , (9.155)

where β1, β2, and β3 are functions of κ, given by

β1 = α1 + 2α5 κ2+ 4α7 κ

4, β2 = α2 + 2 (α4 + α6) κ2, β3 = α3 (9.156)

with (using (9.140))

αi = αi (0,2κ2,0,2κ2,4κ4,8κ6,0,0,2κ4,4κ6) , i = 1, . . . ,7. (9.157)

It follows from (9.155) that since A(1) and A(2) have the form (9.149), thenthe components of σd(t) must have the form

[σdjk(t)] = ⎛⎜⎝

σd11 σd

12 0

σd21 σd

22 0

0 0 σd33

⎞⎟⎠ , (9.158)

where σd21 = σd

12 and σdjk are only functions of κ. Furthermore, from (9.141)1

we must have thatσd11 + σ

d22 + σ

d33 = 0. (9.159)

402 VISCOELASTICITY

We have taken the liberty to write (9.158) and (9.159) in a slightly moregeneral form that applies to isochoric flows (more general than for (9.142),for which σd

33 = 0). Now (9.155), (9.158), and (9.159) allow us to write thecomponents of the extra stress in terms of three independent functions ofκ:

σd21 = σd

12 = κβ1(κ) ≡ µ1(κ), (9.160)

σd11 − σ

d33 = κ2 (2β2(κ) + β3(κ)) ≡ µ2(κ), (9.161)

σd22 − σ

d33 = κ2 β3(κ) ≡ µ3(κ). (9.162)

Furthermore, we see that the components of the stress tensor itself aregiven by

σ21 = σ12 = µ1(κ), σ11 − σ33 = µ2(κ), σ22 − σ33 = µ3(κ). (9.163)

We note that because of the isotropy condition, µ1, µ2, and µ3 cannotdepend on any Cartesian coordinate or which direction the material issheared. Thus, these three functions are material functions. These func-tions are not completely arbitrary since it can be seen from (9.156), (9.157),and (9.161)–(9.162) that the representation restricts the shear and normalcomponent functions to be odd and even functions, respectively:

µ1(−κ) = −µ1(κ) and µ2,3(−κ) = µ2,3(κ). (9.164)

Furthermore, (9.141)2 implies that these functions must vanish for κ = 0:µ1,2,3(0) = 0. (9.165)

Lastly, it can be easily shown that the reduced entropy inequality (8.31)requires that the mechanical work must be positive semi-definite:

κµ1(κ) ≥ 0, (9.166)

i.e., µ1(κ) must have the same sign as κ. If also µ′1(0) > 0, it is oftenconvenient to introduce the inverse function ζ:

κ = ζ(µ1) (9.167)

called the shear-rate function.A shear-dependent viscosity can be defined as

µ(κ) = µ1(κ)κ≥ 0. (9.168)

It follows that µ must be an even function of κ. If we assume that µ istwice differentiable at κ = 0, and since µ1(0) = 0, then µ is differentiable atκ = 0 and

µ′(0) = 0. (9.169)

If we also assume that µ2,3 are differentiable, then we also have that

µ′2,3(0) = 0. (9.170)

9.3. CONSTITUTIVE EQUATIONS 403

If approximations based on Taylor series expansions of µ and µ2,3 aboutκ = 0 are used to fit experimental data, it can be noted that only evenpowers of κ can occur. This is consistent with our results (9.156) and(9.157).Let us compare the forms taken by the general material functions µ1,2,3

with those obtained from other approximations of an incompressible fluid.Specifically, for a perfect fluid, we have

µ1,2,3(κ) = 0for all κ. For a Newtonian fluid, we have

µ1(κ) = µκ and µ2,3 = 0,where µ > 0 is constant. Note than in both of the above cases, the normalstress differences in (9.163) are zero. Lastly, the incompressible Reiner–Rivlin theory (see (8.116)) places no restriction on µ(κ), but yields theresult that σ11 = σ22, which in turn gives

µ2(κ) = µ3(κ).Example

Consider the following elongational flow:

v1 = ax1, v2 = ax2, v3 = −2ax3, (9.171)

so thatdξ1

dτ= a ξ1, dξ2

dτ= a ξ2, dξ3

dτ= −2a ξ3, (9.172)

with final conditions ξi(t) = xi. The solutions are

ξ1 = ea(τ−t)x1 = e−asx1, ξ2 = e−asx2, ξ3 = e2asx3. (9.173)

In this case,

L = a ⎛⎜⎝1 0 0

0 1 0

0 0 −2

⎞⎟⎠ = LT , (9.174)

the flow is isochoric since trL = 0, L is constant and thus we have anMWCSH with Q = I, and

L =M = aN = κN0 = a ⎛⎜⎝1 0 0

0 1 0

0 0 −2

⎞⎟⎠ , (9.175)

where

κ =√6a and N0 = NT0 = 1√

6

⎛⎜⎝1 0 0

0 1 0

0 0 −2

⎞⎟⎠ . (9.176)

404 VISCOELASTICITY

We see that in this case M is not nilpotent. It then follows from (9.71)that

(t)F (t)(s) = [ ∂ξi∂xk] = e−sM = ⎛⎜⎝

e−as 0 0

0 e−as 0

0 0 e2as

⎞⎟⎠ , (9.177)

from (9.87) that

(t)C(t)(s) = ⎛⎜⎝e−2as 0 0

0 e−2as 0

0 0 e4as

⎞⎟⎠ , (9.178)

and from (9.82) to (9.84) that

A(1) = a ⎛⎜⎝2 0 0

0 2 0

0 0 −4

⎞⎟⎠ and A(n) = (A(1))n for n ≥ 2, (9.179)

since A(1) has two distinct eigenvalues, (2a and −4a). Subsequently, sincethe flow is isochoric, from (9.82) we have that

trA(1) = 0 (9.180)

and the constitutive equation is uniquely determined by A(1):

σ = −p1 + α1A(1)+ α2 (A(1))2 , (9.181)

where we have absorbed α0 in the arbitrary pressure p, and where α1 andα2 are functions of of the invariants trA(1), tr (A(1))2, and tr (A(1))3:

αi = αi (0,24a2,−48a3) . (9.182)

Furthermore, we see that the components of the stress tensor itself aregiven by

σ21 = σ12 = 0, σ11 − σ33 = σ22 − σ33 = 4a (α1 − 3aα2) ≡ µ(a). (9.183)

In this case, we have only one material function.Note that, since A(1) = 2D, we can represent the constitutive equation inthe form

σ = −p1 + f (D) = −p1 + β1D + β2D2, (9.184)

where β1 and β2 are functions of trD2 and detD, i.e., functions of a.

Example

Here we combine the elongation and shear flows:

v1 = ax1, v2 = κx1 + ax2, v3 = −2ax3 (9.185)

9.3. CONSTITUTIVE EQUATIONS 405

so thatdξ1

dτ= a ξ1, dξ2

dτ= κξ1 + a ξ2, dξ3

dτ= −2a ξ3, (9.186)

with final conditions ξi(t) = xi. Then, if we let s = t − τ , the solutions canbe written as

ξ1 = e−asx1, ξ2 = −sκe−asx1 + e−asx2, ξ3 = e2asx3. (9.187)

Note from (9.185) that since

L = ⎛⎜⎝a 0 0

κ a 0

0 0 −2a

⎞⎟⎠ (9.188)

is constant, then from (9.59), (9.71), (9.72), and (9.78), we have that Q(t) =I,

L =M = ⎛⎜⎝a 0 0

κ a 0

0 0 −2a

⎞⎟⎠ , (9.189)

and M is not nilpotent. From (9.82) and (9.83), we have

A(1) = ⎛⎜⎝2a κ 0

κ 2a 0

0 0 −4a

⎞⎟⎠ and A(2) = ⎛⎜⎝2κ2 + 4a2 4κa 0

4κa 4a2 0

0 0 16a2

⎞⎟⎠ .(9.190)

Now since A(1) has three distinct eigenvalues, (−4a, 2a ± κ), then M isuniquely determined by A(1) and A(2). Subsequently, from (9.71) we havethat

(t)F (t)(s) = [ ∂ξi∂xk] = e−sL = ⎛⎜⎝

e−as 0 0

−sκe−as e−as 0

0 0 e2as

⎞⎟⎠ , (9.191)

so that

(t)C(t)(s) = ⎛⎜⎜⎝

e−2as−sκe−2as 0

−sκe−2as [1 + (sκ)2] e−2as 0

0 0 e4as

⎞⎟⎟⎠ . (9.192)

For the simple shear-elongational flow, we have (see (9.139))

σd(t) = σ(t) + p1 = α1A(1)+ α2A

(2)+ α3 (A(1))2 + α4 (A(2))2 +

α5 (A(1) ⋅A(2) +A(2) ⋅A(1)) + α6 [(A(1))2 ⋅A(2) +A(2) ⋅ (A(1))2] +α7 [A(1) ⋅ (A(2))2 + (A(2))2 ⋅A(1)] ,

(9.193)

406 VISCOELASTICITY

where we have absorbed α0 in the arbitrary pressure p, and αi are functionsof the invariants (see (9.140))

trA(1), tr (A(1))2, tr (A(1))3, trA(2), tr (A(2))2, tr (A(2))3,tr (A(1) ⋅A(2)), tr [A(1) ⋅ (A(2))2], tr [(A(1))2 ⋅A(2)],

tr [(A(1))2 ⋅ (A(2))2], (9.194)

which in turn are functions of κ and a. In this case, we see that

σ12 = σ21 = µ1(κ, a), σ11 − σ33 = µ2(κ, a), σ22 − σ33 = µ3(κ, a).(9.195)

From (9.141) and (9.190), we also have the requirements that

trσd(κ, a) = 0 and σd(0,0) = 0. (9.196)

9.3.2 Fading memory

We assume that the memory of a material fades in time. This assumption isappropriate for most simple materials (a notable exception are materials that arehypo-elastic). To characterize how the memory fades, we introduce the memoryinfluence function h(s) of order r > 0 satisfying the following conditions:

a) h(s) is defined for 0 ≤ s <∞ and is real and positive definite, h(s) > 0;b) h(s) is normalized by the condition h(0) = 1;c) h(s) decays to zero monotonically for large s according to

lims→∞

srh(s) = 0.For example,

h(s) = (1 + s)−p , p > 1,is a memory influence function of order r if r < p. The exponential

h(s) = e−αs, α > 0,is a memory influence function of any order.

Now, let a memory influence function be given. We define the magnitude of thehistory tensor G

(t)(s) by

∣G (t)(s)∣2 ≡ tr [(G (t)(s))T ⋅G (t)(s)] (9.197)

and its norm by

∥G (t)(s)∥2h= ∫ ∞

0

∣G (t)(s)∣2 h2(s)ds. (9.198)

The influence function h(s) determines the influence of the values of G(t)(s) in

computing the norm ∥G (t)(s)∥h. Since h(s) → 0 as s →∞, the values of G

(t)(s)for small s (recent past) have a greater weight than the values for large s (distant

9.3. CONSTITUTIVE EQUATIONS 407

past). The collection of all histories in the constitutive functional with finite norm(9.198) forms a Hilbert space Hh with weight h(s) and inner product defined by

⟨G (t)1(s) ⋅G (t)

2(s)⟩

h= ∫ ∞

0

tr [(G (t)1(s))T ⋅G (t)

2(s)]h2(s)ds. (9.199)

The history tensor G(t)(s) belongs to the space Hh if it does not grow too fast as

s →∞.Now consider an influence function h(s) and the functional in (9.116), i.e.,

J0≤s<∞

B(t),G (t)(s),which is defined in a neighborhood of the zero history in the Hilbert space Hh andwhose value is a rank 2 tensor quantity that at zero history is zero (see (9.117)),i.e.,

J0≤s<∞

B(t),0(s) = 0.If the history tensor G

(t)(s) is sufficiently smooth, then we say that J is Fréchetdifferentiable at the zero history (see (2.190)) so that

J0≤s<∞

B(t),G (t)(s) = m

∑n=1

1

n!δn J

0≤s<∞B(t),G (t)(s) + o(∥G (t)(s)∥m

h) (9.200)

= m

∑n=1

F(n) B(t),0 ∣ J

0≤s1<∞G(t)(s1)∣⋯ ∣ J

0≤sn<∞G(t)(sn) +

o(∥G (t)(s)∥mh) , (9.201)

where

lim∥G(t)(s)∥

h→0

o(∥G (t)(s)∥mh)

∥G (t)(s)∥h

= 0. (9.202)

The linear functional δn J is called the nth Fréchet derivative of J at the zerohistory (it is essentially a generalization of the derivative of a real-valued functionto the derivative of a functional), and it can be shown that the functional is Fréchetdifferentiable at the zero history for any value of the tensor B (or ρ1 for a simplefluid) for an influence function h(s) of order r > m + 1/2. The Fréchet functional

derivative of order n evaluated on G(t)(s) = 0 (the second variable) enables us

to define F(n), a tensor-valued function of B(t) (or ρ1 for a simple fluid) and a

multilinear continuous tensor-valued form in G(t)(), whose argument functions

are histories. A material obeying a constitutive equation of the form (9.201) iscalled a simple material of order m.

To clarify the notation, suppose G (t)(s),J(t)(s),H(t)(s) ∈Hh. Then

δ J0≤s<∞

B(t),G (t)(s) = ∂

∂λ1J

0≤s<∞B(t),G (t)(s) + λ1 J

(t)(s)∣λ1=0≡

F(1) B(t),G (t)(s) ∣ J

0≤s1<∞J(t)(s1) (9.203)

408 VISCOELASTICITY

is linear and continuous in J(t)(s). The second derivative is defined by

δ2 J0≤s<∞

B(t),G (t)(s) =∂2

∂λ1 ∂λ2J

0≤s<∞B(t),G (t)(s) + λ1 J

(t)(s) + λ2 K(t)(s)∣

λ1=λ2=0≡

F(2) B(t),G (t)(s) ∣ J

0≤s1<∞J(t)(s1)∣ J

0≤s1<∞K(t)(s2) , (9.204)

where F(2) is linear and continuous in J(t)(s) and K

(t)(s). Higher derivatives arecomputed in the same way. We obtain derivatives on the zero history by lettingG(t)(s) → 0, J

(t)(s) → G(t)(s), and K

(t)(s) → G(t)(s) in the above equations

and subsequently obtain (9.201). It is obvious that the functional derivatives aresymmetric to all transpositions of their linear arguments in (9.201).

9.3.3 Constitutive equations of differential type

For slow motions which have derivatives of G(t)(s) at s = 0, we can use the Taylor

series (9.53) to expand (9.201):

F(n) B(t),0(s) ∣ J

0≤s1<∞G(t)(s1)∣⋯ ∣ J

0≤sn<∞G(t)(sn) =

F(n)⎧⎪⎪⎨⎪⎪⎩B(t),0(s)

RRRRRRRRRRR J0≤s1<∞

∞

∑j1=1

(−1)j1j1!

A(j1)(t) sj11

RRRRRRRRRRR⋯RRRRRRRRRRR J0≤sn<∞

∞

∑jn=1

(−1)jnjn!

A(jn)(t) sjnn ⎫⎪⎪⎬⎪⎪⎭ =∑

(j1,...,jn)f(j1,...,jn) B(t), [A(j1)(t), . . . ,A(jn)(t)] , (9.205)

where each f(j1,...,jn) B(t), [A(j1)(t), . . . ,A(jn)(t)], for each choice of B (or ρ(t)1for a fluid), is a multilinear isotropic tensor-valued function of n tensor variables.The summation is to be extended over all sets of n indices (j1, . . . , jn) satisfyingthe inequalities

1 ≤ j1 ≤ ⋯ ≤ jn ≤m, j1 +⋯ + jn ≤m. (9.206)

Subsequently, for an isotropic and homogeneous simple solid, we have

σ = hB(t) + m

∑n=1

∑(j1,...,jn)

f(j1,...,jn) B(t), [A(j1)(t), . . . ,A(jn)(t)] , (9.207)

while for a fluid we have

σ = −p(ρ(t))1 + m

∑n=1

∑(j1,...,jn)

f(j1,...,jn) ρ(t), [A(j1)(t), . . . ,A(jn)(t)] . (9.208)

Materials satisfying (9.207) or (9.208) are called materials of differential type oforder m.

9.3. CONSTITUTIVE EQUATIONS 409

Example

We would like to write the explicit form of the constitutive equation (9.207)for an isotropic and homogeneous simple viscoelastic solid of order m = 1.In this case, we have

σ(t) = hB(t) + f(1)B(t), [A(1)] = hB(t) + lB(t),D, (9.209)

where we have exploited the fact that A(1) = 2D(1) = 2D. Now, using(9.121) and Tables 5.1 and 5.3, we can write

σ(t) = (h0 1 + h1B + h2B2) + [l0 1 + l1B + l2B2+ l3D + l4D

2] +l5 (B ⋅D +D ⋅B) + l6 (B2

⋅D +D ⋅B2) + l7 (B ⋅D2+D2

⋅B) , (9.210)

where the hi’s are functions of the invariants of B, i.e.,

trB, trB2, trB3, (9.211)

and the lj’s are functions of the invariants of B and D, i.e.,

trB, trB2, trB3, trD, trD2, trD3, tr (B ⋅D) , tr (B ⋅D2) ,tr (B2

⋅D) , tr (B2⋅D2) . (9.212)

Example

Correspondingly, the explicit form of the constitutive equation (9.208) fora simple fluid of order m = 1 is given by

σ(t) = −p(ρ(t))1 + f(1)ρ(t), [A(1)] = −p(ρ(t))1 + lρ(t),D, (9.213)

where we have exploited the fact that A(1) = 2D(1) = 2D. Now, usingTables 5.1 and 5.3, we can write

σ(t) = [−p(ρ(t))+ l0]1 + l1D + l2D2, (9.214)

where the li’s are functions of ρ and the invariants of D, i.e.,trD, trD2, trD3. As we see from (8.116) and (8.118), this is just therepresentation of a Reiner–Rivlin fluid.

In general, (9.207) and (9.208) are not good representations, because (9.53) isnot a convenient way to represent the history when higher order terms are notnegligible. The series representation (9.53) is made useful for retarded motion in

which the first few terms dominate. The retarded history of G(t)(s) is obtained

by replacing s with ǫs:

G(t)ǫ (s) = G

(t)(ǫs) = ∞∑n=1

(−ǫs)nn!

A(n)(t) (9.215)

and thus diminish the importance of the higher A(n) when ǫ is small. Then,

410 VISCOELASTICITY

following the above procedure, constitutive equations for retarded motion (nearlysteady slow motion) of order m are obtained by identifying increasing powers of ǫ.

Example

The retarded motion of order m = 1 of a simple isotropic solid is obtainedby retaining terms in (9.210)–(9.212) that are of O(ǫ) in A(1) or D, i.e.,

σ(t) = (h0 1 + h1B + h2B2) + [k1 trD + k2 tr (B ⋅D)+k3 tr (B2

⋅D)]1 + [k4 trD + k5 tr (B ⋅D) + k6 tr (B2⋅D)]B +

[k7 trD + k8 tr (B ⋅D) + k9 tr (B2⋅D)]B2

+

k10D + k11 (B ⋅D +D ⋅B) + k12 (B2⋅D +D ⋅B2) , (9.216)

where the hi’s and kj ’s are functions of the invariants of B, i.e.,trB, trB2, trB3.Example

The corresponding explicit form of the constitutive equation for retardedmotion of a simple fluid of order m = 1 is obtained by retaining the termsthat are liner in D in (9.214). In this case, we obtain the Newtonian stresstensor

σ(t) = [−p(ρ(t)) + λ trD]1 + 2µD, (9.217)

where λ and µ are functions of ρ.

9.3.4 Constitutive equations of integral type

More generally, if we assume that the motion is not slow or differentiable, wecan make use of Riesz’ representation theorem, which states that every continuousfunctional may be written as an inner product. This allows us to write (9.200)and (9.201) in the form

1

n!δn J

0≤s<∞B(t),G (t)(s) = F(n) B(t),0 ∣ J

0≤s1<∞G(t)(s1)∣⋯ ∣ J

0≤sn<∞G(t)(sn) =

∫∞

0

⋯∫∞

0

K(n) (s1, . . . , sn;B(t)) ⋅ [G (t)(s1)⋯G(t)(sn)]ds1⋯dsn. (9.218)

Substitution of (9.201) and (9.218) into (9.116) provides an approximation to theconstitutive equation of a simple material with fading memory in the form

σ(t) = hB(t) +m

∑n=1∫∞

0

⋯∫∞

0

K(n) (s1, . . . , sn;B(t)) ⋅ [G (t)(s1)⋯G(t)(sn)]ds1⋯dsn. (9.219)

The tensor functions K(n) (s1, . . . , sn;B(t)) ⋅ [G (t)(s1)⋯G(t)(sn)] are bounded

isotropic second-order tensor polynomials that are multilinear in the n tensor vari-ables G

(t)(s1)⋯G(t)(sn). Furthermore, they are completely symmetric under any

9.3. CONSTITUTIVE EQUATIONS 411

permutations of 1, . . . , n. The values of K(n) (s1, . . . , sn;B(t)) are isotropic tensorsof order 2(n+1) that are continuous with the tensor parameter B(t). The error ofthis approximation approaches zero faster than the mth power of the history ten-sor norm (9.198). A material obeying a constitutive function of the form (9.219)is called a simple material of integral type of order m.

For example, when m = 1 we have that

σ(t) = hB(t) + ∫ ∞

0

K(1) (s1;B(t)) ⋅ [G (t)(s1)]ds1, (9.220)

where K(1) (s1;B(t)) ⋅ [G (t)(s1)], for each choice of s1 ≥ 0 and B(t), is a linear

function of the tensor variable G(t)(s1). The quantity K(1) (s1;B(t)) is a fourth-

order tensor whose magnitude satisfies

∫∞

0

∣K(1) (s1;B(t))∣2 h−2(s1)ds1 <∞. (9.221)

Simple materials satisfying the constitutive equation (9.220) are called finite linearviscoelastic materials.

For m = 2 we have

σ(t) = hB(t) +∫ ∞

0

K(1) (s1;B(t)) ⋅ [G (t)(s1)]ds1 +∫∞

0∫∞

0

K(2) (s1, s2;B(t)) ⋅ [G (t)(s1)G (t)(s2)]ds1ds2, (9.222)

where, in addition to the conditions on K(1) noted above, we have that K(2) (s1, s2;B(t)) ⋅[G (t)(s1)G (t)(s2)], for each choice of s1 ≥ 0, s2 ≥ 0, and B(t), is a function

of the tensor variables G(t)(s1) and G

(t)(s2). The quantity K(2) (s1, s2;B(t)) isa sixth-order tensor whose magnitude satisfies

∫∞

0∫∞

0

∣K(2) (s1, s2;B(t))∣2 h−2(s)ds1 ds2 <∞. (9.223)

Correspondingly, the general constitutive equation of integral type for a com-pressible fluid is given by

σ(t) = −p(ρ(t))1 +m

∑n=1∫∞

0

⋯∫∞

0

K(n) (s1, . . . , sn;ρ(t)) ⋅ [G (t)(s1)⋯G(t)(sn)]ds1⋯dsn (9.224)

and for an incompressible fluid by

σ(t) = −p1 +m

∑n=1∫∞

0

⋯∫∞

0

K(n) (s1, . . . , sn) ⋅ [G (t)(s1)⋯G(t)(sn)]ds1⋯dsn. (9.225)

412 VISCOELASTICITY

Example

For a linear viscoelastic incompressible fluid, we have

σik(t) = −p δik +∫ ∞

0

K(1)iklm(s)G(t)

lm(s)ds. (9.226)

From objectivity, K(1)iklm

is isotropic, and since the stress tensor and G(t)lm

are symmetric, we must have

K(1)iklm=K(1)

kilm=K(1)

ikml=K(1)

kiml. (9.227)

Furthermore, K(1)iklm

is a symmetric linear transformation if and only if thecomponents obey the relations

K(1)iklm=K(1)

lmik. (9.228)

The most general fourth rank tensor satisfying the above symmetries canbe written as (see (B.20)–(B.22))

K(1)iklm(s) =K(1)1

(s) δikδlm +K(1)2(s) (δilδkm + δklδim) . (9.229)

For an incompressible fluid, K(1)1

can be combined with the term involving

p, so setting K(1)2= 1

2β(1), we can write

σik(t) = −p δik +∫ ∞

0

β(1) (s) G(t)ik(s)ds, (9.230)

for β(1)(s) decaying sufficiently fast with s. Now substituting (9.53), wehave

σik(t) = −p δik + ∞∑n=1

(−1)nn!

A(n)ik(t)∫ ∞

0

β(1)(s) snds. (9.231)

Possible choices of β(1)(s) are

β(1)(s) =⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

Ke−λs,r

∑l=1

Kl e−λls,

∫∞

0

K(r) e−λ(r)sdr,(9.232)

where the λ’s represent relaxation times of the material. If we assume thatβ(1)(s) =Ke−λs, then

σik(t) = −p δik + ∞∑n=1

(−1)nλn+1

KA(n)ik(t). (9.233)

Using the isotropy and symmetry conditions, there exist certain relations among

9.3. CONSTITUTIVE EQUATIONS 413

the kernels K(n) (s1, . . . , sn;ρ(t)) ⋅ [G (t)(s1)⋯G(t)(sn)] that reduce substantially

the number of independent stress-relaxation moduli that can appear in a givenorder of approximation. It can be shown that such product can be expressed as asum of multilinear products of tensors, trace of tensors, and trace of products oftensors:

α(n)i (s1, . . . , sn;ρ(t)) tr [G (t)(s1)⋯G

(t)(sn1)] tr [G (t)(sn1+1)⋯G

(t)(sn2)]⋯

tr [G (t)(snl−1+1)⋯G(t)(snl

)] G(t)(snl+1)⋯G

(t)(sn), (9.234)

where 1 ≤ n1 < n2 < ⋯ < nl ≤ n, and where n1 ≤ 6, nk − nk−1 ≤ 6, n − nl ≤ 5. Ifnl = n, then the product at the end must be replaced by the unit tensor. Thus,using (9.234) in (9.224), we have

σ(t) = −p(ρ(t)) + m

∑n=1∫∞

0

⋯∫∞

0

∑i

α(n)i (s1, . . . , sn;ρ(t)) tr [G (t)(s1)⋯

G(t)(sn1

)] tr [G (t)(sn1+1)⋯G(t)(sn2

)]⋯ tr [G (t)(snl−1+1)⋯G(t)(snl

)]G(t)(snl+1)⋯G

(t)(sn)ds1⋯dsn. (9.235)

For fluids of first order (m = 1), the constitutive equation is explicitly given by

σ(t) = −p(ρ(t))1 +∫ ∞

0

[α(1)1(s1;ρ(t)) trG (t)(s1)1 + α(1)2

(s1;ρ(t))G (t)(s1)]ds1.(9.236)

For a second-order fluid (m = 2), we have

σ(t) = −p(ρ(t))1 +∫ ∞

0

[α(1)1(s1;ρ(t)) trG (t)(s1)1 + α(1)2

(s1;ρ(t))G (t)(s1)]ds1 +∫∞

0∫∞

0

[α(2)1(s1, s2;ρ(t)) trG (t)(s1) trG (t)(s2)+

α(2)2(s1, s2;ρ(t)) tr (G (t)(s1) ⋅G (t)(s2))]1 +

α(2)3(s1, s2;ρ(t)) trG (t)(s1)G (t)(s2) +

α(2)4(s1, s2;ρ(t))G (t)(s1) ⋅G (t)(s2)ds1ds2. (9.237)

For an incompressible fluid, all terms in (9.236) and (9.237) that are scalarmultiples of the unit tensor can be absorbed in the pressure term resulting inthe following corresponding constitutive equations for orders m = 1 and m = 2,respectively:

σ(t) = −p1 +∫ ∞

0

β(1)1(s1)G (t)(s1)ds1 (9.238)

and

σ(t) = −p1 +∫ ∞

0

β(1)1(s1)G (t)(s1)ds1 +

∫∞

0∫∞

0

β(2)1(s1, s2) trG (t)(s1)G (t)(s2)+

β(2)2(s1, s2)G (t)(s1) ⋅G (t)(s2)ds1ds2. (9.239)

414 VISCOELASTICITY

ε

ε1 ε2

k µ

σ

Figure 9.3: Schematic representation of the Maxwell model.

Note that (9.238) is the same equation as (9.230), which was obtained by followingan alternate procedure.

Explicit representations may also be obtained for general isotropic materialsof the integral type provided that the tensor functions K(n) (s1, . . . , sn;B(t)) ⋅[G (t)(s1)⋯G

(t)(sn)] appearing in (9.219) are polynomial functions of B. For

example, the constitutive function for an isotropic material of first order (m = 1)is given by

σ(t) = hB(t) + ∫ ∞

0

α(1)1(s1;B(t)) ⋅G (t)(s1) +G

(t)(s1) ⋅α(1)1(s1;B(t))+

tr [G (t)(s1) ⋅α(1)2(s1;B(t))] 1 + tr [G (t)(s1) ⋅α(1)3

(s1;B(t))] B +tr [G (t)(s1) ⋅α(1)4

(s1;B(t))] B2ds1, (9.240)

where hB(t) is given by (9.121), and the isotropic tensor functions α(1)k

havethe representations

α(1)ks1;B(t) = α(1)1k

1 + α(1)2k

B + α(1)3k

B2, k = 1,2,3,4, (9.241)

with α(1)ik= α(1)

ik(s1;B(1),B(2),B(3)), i = 1,2,3. This is readily verified by using

Tables 5.1 and 5.3.

Example

A stress tensor approximation for an incompressible fluid, referred as theMaxwell model, is given by

σ = −p1 −∫ ∞

0

k e−s/λd

dsG(t)(s)ds. (9.242)

It is noted, by integrating by parts, that

σ = −p1 −

0

k e−s/λ G(t)(s)RRRRRRRRRRRR

∞

0

−∫∞

0

k

λe−s/λ G

(t)(s)ds= −p1 −∫

∞

0

k

λe−s/λ G

(t)(s)ds, (9.243)

which is just (9.230) or (9.238) with

β(1)(s) = −kλe−s/λ. (9.244)

9.3. CONSTITUTIVE EQUATIONS 415

The model in one dimension is illustrated in Fig. 9.3 where σ is the tensionforce, k the spring constant, µ the dashpot damping coefficient, and ε =ε1 + ε2 the total displacement, with ε1 and ε2 being the displacements dueto the spring and dashpot, respectively. Now we can write

σ = k ε1 = µ ε2 (9.245)

soσ = µ (ε − ε1) = µ ε − µ

kσ. (9.246)

If we let λ = µ/k, then we can write

σ +1

λσ = k ε(t) (9.247)

and we take σ(−∞) = 0. So assuming we know ε(t), we have

σ(t) = ∫ t

−∞

k e−(t−τ)/λε(τ)dτ, (9.248)

or, upon taking s = t − τ ,σ(t) = −∫ ∞

0

k e−s/λd

dsε(t − s)ds, (9.249)

which is a one-dimensional model of (9.242) with G(t)(s)→ ε(t − s).

Example

We would like to examine the application of the Maxwell model to simpleshear. For simple shear, we have

v1 = 0, v2 = κx1, v3 = 0. (9.250)

Thus,dξ1

dτ= 0, dξ2

dτ= κξ1, dξ3

dτ= 0, (9.251)

and so

ξ1 = x1, ξ2 = x2 + κ (τ − t) x1 = x2 − κsx1, ξ3 = x3. (9.252)

Now

(t)F(t)ik(s) = ∂ξi

∂xk= ⎛⎜⎝

1 0 0

−κs 1 0

0 0 1

⎞⎟⎠ , (9.253)

(t)C(t)(s) = ((t)F (t)(s))T (t)F (t)(s) = ⎛⎜⎝1 −κs 0

−κs 1 + (κs)2 0

0 0 1

⎞⎟⎠ , (9.254)

416 VISCOELASTICITY

and from (9.52) we have that

G(t)(s) = ⎛⎜⎝

0 −κs 0

−κs (κs)2 0

0 0 0

⎞⎟⎠ . (9.255)

Now, using Maxwell’s model (9.243), for this incompressible fluid, we have

σ = −p1 −∫ ∞

0

k

λe−s/λ G

(t)(s)ds, (9.256)

where λ is the material relaxation time. Thus,

σ12 = σ21 = k κλ∫∞

0

e−s/λsds = k κλ = µκ (9.257)

and

σ22 + p = −k κ2λ∫∞

0

e−s/λs2 ds = −2k κ2 λ2 = −2µλκ2. (9.258)

Example

Here we would like to apply the Maxwell model to the elongational flow

v1 = 2ax1, v2 = −ax2, v3 = −ax3. (9.259)

In this case, we have

dξ1

dτ= 2a ξ1, dξ2

dτ= −a ξ2, dξ3

dτ= −a ξ3, (9.260)

and

ξ1 = e2a(τ−t)x1 = e−2asx1, ξ2 = easx2, ξ3 = easx3. (9.261)

Thus

(t)C(t)(s) = ⎛⎜⎝

e−4as 0 0

0 e2as 0

0 0 e2as

⎞⎟⎠ and

G(t)(s) = ⎛⎜⎝

e−4as− 1 0 0

0 e2as− 1 0

0 0 e2as− 1

⎞⎟⎠ , (9.262)

and so on. We finally get

σ11(t) = −p + 4ak

λ−1 + 4aand σ22(t) = σ33(t) = −p + 2ak

λ−1 − 2a. (9.263)

If a = 1/(2λ), then σ22, σ33 →∞, so we expect this model to fail. However,the model can be fixed up.

9.3. CONSTITUTIVE EQUATIONS 417

Many other first-order models, more complex than the Maxwell model, can alsobe shown to correspond to specific cases of (9.236) or (9.240) for different choices

of α(1)i (s1;ρ) or α(1)i (s1;B). Their functional dependence on ρ for a compressiblefluid and on B and its invariants for an isotropic solid (see (9.241)) is exploitedin many of the more complex models where specific choices are made (e.g., theBird–Carreau model).

9.3.5 Constitutive equations of rate type

We recall that the response functional (5.3) was sufficiently general so that, e.g.,the stress could depend on the histories of stress, heat flux, entropy flux, freeenergy, entropy, as well as the motion and temperature at all other points in thebody and their rates. Here, we continue to limit the discussion to constitutiveequations of homogeneous simple materials that are independent of temperaturegradient, and suppress the functional dependence on temperature. Subsequently,we focus on the constitutive equation for the stress tensor. If we now considermaterials of rate type, and more specifically, the constitutive equation of rate n

and mechanical rate p, the response functional (5.3) takes the form

F0≤s<∞

σ(X, s), σ(X, s), . . . , (n)σ (X, s);F(X, s), F(X, s), . . . , (p)F (X, s) = 0.(9.264)

Assuming that the functional for(n)σ is non-singular and single valued, we have

the following equivalent formulation for the stress tensor of a homogeneous simplematerial at material point X (see (5.33)):

σ(t) = F0≤s<∞

F(t)(s), (9.265)

where it is now assumed that σ = σ(t) satisfies a differential equation of the form

(n)σ (t) = g(σ(t), σ(t), . . . , (n−1)σ (t);F(t), F(t), . . . , (p)F (t)) (9.266)

with initial conditions

σ(t0) = σ0, σ(t0) = σ0, . . . ,(n−1)σ (t0) = (n−1)σ0 . (9.267)

It should be noted that (9.266) is not really a complete constitutive equation. Inaddition to the initial conditions (9.267) at some initial time t0, we also requirethe history of the deformation from (9.265) up to time t0. Nevertheless, theconstitutive differential equation (9.266) is called a constitutive equations of ratetype. Above, it is assumed that g is a tensor-valued function that is sufficientlysmooth, and that the differential equation has a unique solution. It should benoted that the stress tensor σ(t) depends not only on the initial data but also onvalues of F(τ) for t0 ≤ τ ≤ t. It is also noted that the constitutive equation whichis not of the rate type corresponds to the case where n = 0.

Frame-invariant forms of (9.265) under a Euclidean transformation for a homo-geneous isotropic simple viscoelastic material have been discussed in Sections 9.3

418 VISCOELASTICITY

and 9.3.3. Subsequently, (9.265) takes the form given in (9.207). We also re-call from Section 3.3 that rates of second rank tensors are not objective undersuch transformation. Nevertheless, the corresponding frame-indifferent constitu-tive equation of rate type (9.266) is obtained by following the same procedure asin Sections 5.7 and 9.3. In doing this, by using the polar decomposition of F andchoosing Q =RT , we arrive at the following objective form:

σn(t) = f (σ(t), σ1(t), . . . , σn−1(t);B(t);A(1)(t),A(2)(t), . . . ,A(p)(t)) , (9.268)

where

σi is the ith convected stress rate defined by

σi(t) ≡ ∂i

∂τ i[(t)FT (τ) ⋅σ(τ) ⋅ (t)F(τ)]∣

τ=t

=∑

a+b+c=ia,b,c=0,1,...,i

i!

a! b! c!(L(a)(t))T ⋅ (b)σ(t) ⋅L(c)(t). (9.269)

Note that for i = 0 we have

σ0(t) = σ(t) and for i = 1 we obtain the Cotter–Rivlinconvected stress rate (see (3.425)):

σ1 =

σ = σ +LT⋅σ +σ ⋅L. (9.270)

The corresponding constitutive equation for a simple fluid of rate type is of thefollowing form:

σn(t) = f(σ(t), σ1(t), . . . , σn−1(t);ρ(t);A(1)(t),A(2)(t), . . . ,A(p)(t)) . (9.271)

The simplest form for a fluid of rate type (n = 1) is given by

σ1(t) = f (σ(t);ρ(t);A(1)(t)) , (9.272)

which can be recast as

σd1(t) = g (σd(t);ρ(t);A(1)(t)) (9.273)

in terms of the extra, or dissipative part, of the stress by writing

σd(t) = σ(t) + p(ρ(t))1. (9.274)

Note that under hydrostatic conditions, we must have that

g (0;ρ(t);0) = 0. (9.275)

This class of materials include materials which have properties of both fluids andsolids and are called hygrosteric materials.

Now, using Tables 5.1 and 5.3, and the fact that A(1) = 2D, (9.273) has thefollowing representation:

σd1(t) = α0 1 + α1σ

d+ α2 (σd)2 + α3D + α4D

2+ α5 (σd

⋅D +D ⋅σd) +α6 ((σd)2 ⋅D +D ⋅ (σd)2) + α7 (σd

⋅D2+D2

⋅σd) , (9.276)

where αi, i = 0, . . . ,7, are functions of ρ and the invariants trD, trD2, trD3, trσd,tr (σd)2, tr (σd)3, tr (σd

⋅D), tr (σd⋅D2), tr ((σd)2 ⋅D), and tr ((σd)2 ⋅D2). From

(9.275), it follows that α0 = 0 for σd =D = 0 and ρ arbitrary.

9.3. CONSTITUTIVE EQUATIONS 419

Example

We develop the linear constitutive equation of rate n = 1. In this case,we assume that g (σd;ρ(t);D) is linear in σd and D. Subsequently, from

(9.276), we see that

σd1 must have the form

σd1(t) =

σd = σd+LT

⋅σd+σd

⋅L = α0 1 + α1σd+ α3D +

α5 (σd⋅D +D ⋅σd) , (9.277)

where

α0 = λ2 trσd+ µ2 trD + β1 tr (D ⋅σd) + β4 (trσd) (trD) ,α1 = λ1 + β5 trD, α3 = µ1 + β2 trσ

d, α5 = β3, (9.278)

and µi, λi, and βi are functions of ρ.It should be noted that, since L =D+W, then it follows that

σd = σd+σd⋅

D+D ⋅σd, whereσd is the Jaumann corotational stress rate (see (3.422)),

and thus we can also write, without loss of generality (by subsequentlytaking (α5 − 1)→ α5),

σd = σd

−W ⋅σd+σd

⋅W = α0 1 + α1σd+ α3D +

α5 (σd⋅D +D ⋅σd) . (9.279)

Such constitutive equation models linear fluent materials, where by requir-ing that λ2(trσd) + λ1σd ≠ 0 for σd ≠ 0, we must have that

λ1 ≠ 0, λ1 + 3λ2 ≠ 0, and p = p(ρ). (9.280)

It also models linear hypo-elastic materials, where the material is not afunction of ρ, in which case we have

λ1 = λ2 = 0 and p = 0. (9.281)

Note that the requirement of p = 0 for linear hypo-elastic materials is equiv-alent to having σd = σ.

Example

Consider the steady linear flow in the Cartesian coordinates (x1, x2, x3) ofthe form

v = (0, v2(x1),0) . (9.282)

Note that for this flow the mass conservation equation becomes

divv = 0, (9.283)

i.e., the fluid is incompressible where we take ρ = ρR = const., and thus thepressure is an unknown function unrelated to thermodynamic equations of

420 VISCOELASTICITY

state. Furthermore, since v = 0, the linear momentum equation becomes

divσd− gradϕ = 0, (9.284)

whereϕ = p + ρR U, (9.285)

and we have assumed that the body force derives from a potential, i.e.,

g = −gradU. (9.286)

Now, it is clear that the extra stress in this case is of the form

σd = ⎛⎜⎝t1(x1) s(x1) 0

s(x1) t2(x1) 0

0 0 t3(x1)⎞⎟⎠ , (9.287)

where ti’s are normal stresses and s is the shear stress. Subsequently,

divσd = (t′1, s′,0) , (9.288)

where the primes denote derivatives with respect to x1, and the linearmomentum equation (9.284) reduces to

t′1(x1) − ∂ϕ

∂x1= 0, s′(x1) − ∂ϕ

∂x2= 0, ∂ϕ

∂x3= 0. (9.289)

From these equations, we infer that

s = −ax1 + b and ϕ = −ax2 + t1(x1) + c, (9.290)

where a, b, and c are constants that depend on constitutive parameters.Now, the velocity gradient is given by

L = v′2 ⎛⎜⎝0 0 0

1 0 0

0 0 0

⎞⎟⎠ , (9.291)

and it then follows from (9.277) and (9.278) that

v′2

⎛⎜⎝2 s t2 0

t2 0 0

0 0 0

⎞⎟⎠ = (λ2 t0 + β1 s v′2)⎛⎜⎝

1 0 0

0 1 0

0 0 1

⎞⎟⎠ + λ1⎛⎜⎝t1 s 0

s t2 0

0 0 t3

⎞⎟⎠ +1

2v′2 (µ1 + β2 t0) ⎛⎜⎝

0 1 0

1 0 0

0 0 0

⎞⎟⎠ +1

2β3 v

′

2

⎛⎜⎝2 s t1 + t2 0

t1 + t2 2 s 0

0 0 0

⎞⎟⎠ , (9.292)

wheret0 = t1 + t2 + t3 = trσd. (9.293)

Note that since the material is incompressible, λi, βj , and µk are purelyconstant. The system (9.292) provides four nontrivial coupled equations.

9.3. CONSTITUTIVE EQUATIONS 421

To simplify the solution of such equations, and without loss of generality,we take β3 → 1 + β3. Subsequently, the four independent equations are

λ1 t1 + λ2 t0 + (β1 + β3 − 1) s v′2 = 0, (9.294)

λ1 t2 + λ2 t0 + (β1 + β3 + 1) s v′2 = 0, (9.295)

λ1 t3 + λ2 t0 + β1 s v′

2 = 0, (9.296)

λ1 s +1

2[µ1 + β2 t0 + β3 (t1 + t2) + (t1 − t2)] v′2 = 0. (9.297)

Adding (9.294)–(9.296) results in

(λ1 + 3λ2) t0 + (3β1 + 2β3) s v′2 = 0. (9.298)

Now, since from (9.280) λ1 ≠ 0, we can solve for t1 + t2 and t1 − t2 from(9.294) and (9.295) and substitute these into (9.297), to obtain

λ1 s +1

2µ1 +

1

λ1(λ1 β2 − 2λ2 β3) t0 − 2

λ1[β3 (β1 + β3) − 1] s v′2 v′2 = 0.

(9.299)Again, since from (9.280) λ1 + 3λ2 ≠ 0, we can solve (9.298) for t0 andsubstitute this into (9.299), to obtain the following expressions for v′2 andthe shear stress

v′2 = 4 s

η

1

1 ±√1 − 16ατ2

dη−2s2

(9.300)

and

s = 1

2η v′2

1

1 + β τ2d(v′

2)2 , (9.301)

where

η = −µ1

λ1, β = 1 − λ1 β2 − 2λ2 β3

2 (λ1 + 3λ2) (3β1 + 2β3) − β3 (β1 + β3) , τd = − 1

λ1.

(9.302)In addition, from (9.294) to (9.296) and (9.298), we obtain the normalstresses

t1 = τd (−β0 + β1 + β3 − 1) s v′2, (9.303)

t2 = τd (−β0 + β1 + β3 + 1) s v′2, (9.304)

t3 = τd (−β0 + β1) s v′2, (9.305)

where

β0 = λ2

λ1 + 3λ2(3β1 + 2β3) . (9.306)

For simple shearing flow, where v2(0) = 0 and v2(d) = V and with ϕ havingno gradient in the flow direction, we have that v2(x1) = κx1, with shearrate κ = V /d = const. and v′2 = κ, so that from (9.301) and (9.290) we musthave that a = 0 and from (9.303) to (9.305) that the normal stresses areconstant. Subsequently,

s = b = const. and ϕ = p + ρR U = t1 + c = const. (9.307)

422 VISCOELASTICITY

Let us examine the shear stress given by (9.301). We first note that theclassical result of s = b = 1

2η κ is obtained when β = 0. Now, β can be

either positive or negative. If β > 0, and defining a shearing yield rate byτs =√β τd, we can rewrite (9.301) in the form

s = b = 1

2η κ

1

1 + τ2s κ2, (9.308)

as illustrated in Fig. 9.4(a). We note that for κ ≤ 1

4τ−1s the error in the

classical formula is less than 6%. For 1

4τ−1s ≤ κ ≤ τ−1s , the rate of shearing

is greater than for that provided by the classical result, and indeed atκc = τ−1s , a yield-like phenomenon is observed and there we have that theshear stress is given by sc = 1

4η τ−1s .

On the other hand, if β < 0, and defining a shearing limit time by τ s =√∣β∣ τd, we can rewrite (9.301) in the form

s = b = 1

2η κ

1

1 − τ2s κ2, (9.309)

as illustrated in Fig. 9.4(b). In this case, for κ ≤ 1

4τ−1s the error in the

classical formula is less than 7% and for 1

4τ−1s ≤ κ ≤ τ−1s the shearing stress

is substantially higher than the classical result. In this case, the materialbecomes more and more resistant to shearing stress. We note that theregion where κ > κl ≡ τ−1s is presumed to be unphysical since the rate ofshear in this branch cannot be reached through a continuous increase ofthe rate.Since s and ti are constant, it follows from (9.303) to (9.305) that thenormal extra stresses, ti are also constant. In the case where we have noexternal body forces (U = 0), from (9.285) and (9.290), the pressure p is anarbitrary constant, and thus the normal stresses are also constant. We canchoose

p = t2 (9.310)

so that there is no normal stress in the flow direction. Thus, from (9.303)to (9.305), we can write that

σ11 = −2 τd sκ, σ22 = 0, σ33 = −τd (1 + β3) sκ. (9.311)

Now in the case of a classical viscous fluid, no normal stresses are necessaryfor steady flow. Here, if τd > 0, the stresses σ11 normal to the plates mustbe negative. This indicates that the plates would tend to spread apart inthe absence of such stresses and leads to flow swelling at the exit of suchplates. The stress σ33 normal to the direction of flow and parallel to theplates is also nonzero – its sign depends on the sign of 1+β3. For slow flow,where κ≪ τ−1d , the normal stresses will be small compared to the shearingstresses (assuming for σ33 that β3 = O(1)).

9.3. CONSTITUTIVE EQUATIONS 423

s

κ1

4τ−1s

1

2τ−1s κc = τ−1s

sc = 1

4η τ−1s

(a)

s

κ1

4τ−1s κl = τ−1s

sl = 1

2η τ−1s

(b)

Figure 9.4: Shear stress, s, as a function of shear rate, κ, for the shear flow resultingfrom a linear constitutive equation of rate type of order one: (a) β > 0, (b) β < 0.

424 VISCOELASTICITY

Problems

1. From the definition of the relative right Cauchy–Green tensor (3.283), showthat

∂n(t)C(τ)∂τn

= n

∑k=0

(nk) ∂k(t)FT (τ)

∂τk⋅∂n−k(t)F(τ)

∂τn−k, n = 0,1,2, . . . ,

where the binomial coefficients are given by

(nk) = n!(n − k)!k! .

2. Show that the nth acceleration gradient and stretch and spin tensors can berespectively written as

L(n)(t) = n

∑k=0

(nk)W(k)(t) ⋅D(n−k)(t), n = 0,1,2, . . . ,

D(n)(t) = 1

2[A(n)(t) − n−1

∑k=1

(nk)D(k)(t) ⋅D(n−k)(t)] , n = 1,2, . . . ,

and

W(n)(t) = L(n)(t) −D(n)(t) − n−1

∑k=1

(nk)W(k)(t) ⋅D(n−k)(t), n = 1,2, . . . .

3. Show that the ith corotational rate of vector u, defined by

ui(t) ≡ ∂i

∂τ i[(t)RT (τ) ⋅ u(τ)]∣

τ=t

= ∑a+b=i

a,b=0,1,...,i

i!

a! b!(W(a)(t))T ⋅(b)u(t), (9.312)

is objective. Note that for i = 1 it corresponds to the corotational rate of thevector (see (3.419)):

u1 =

u = u −W ⋅ u. (9.313)

4. Show that the ith convected rate of vector u, defined by

⊗ui(t) ≡ ∂i

∂τ i[(t)FT (τ) ⋅u(τ)]∣

τ=t

= ∑a+b=i

a,b=0,1,...,i

i!

a! b!(L(a)(t))T ⋅(b)u(t), (9.314)

is objective. Note that for i = 1 it corresponds to the convected vector rate

⊗u1 = ⊗

u = u +LT⋅u. (9.315)

5. Show that the ith corotational stress rate, defined by

σi(t) ≡ ∂i

∂τ i[(t)RT (τ) ⋅σ(τ) ⋅ (t)R(τ)]∣

τ=t

=∑

a+b+c=ia,b,c=0,1,...,i

i!

a! b! c!(W(a)(t))T ⋅ (b)σ(t) ⋅W(c)(t), (9.316)

9.3. CONSTITUTIVE EQUATIONS 425

is objective. Note that for i = 1 it corresponds to the Jaumann corotationalstress rate (see (3.422)):

σ1 =

σ = σ −W ⋅σ +σ ⋅W. (9.317)

6. Show that the ith convected stress rate, defined by

σi(t) ≡ ∂i

∂τ i[(t)FT (τ) ⋅σ(τ) ⋅ (t)F(τ)]∣

τ=t

=∑

a+b+c=ia,b,c=0,1,...,i

i!

a! b! c!(L(a)(t))T ⋅ (b)σ(t) ⋅L(c)(t), (9.318)

is objective. Note that for i = 1 it corresponds to the Cotter–Rivlin convectedstress rate (see (3.425)):

σ1 =

σ = σ +LT⋅σ +σ ⋅L. (9.319)

7. Prove that a motion is an MWCSH if and only if there exist an orthogonaltensor Q(t), a scalar κ (called the shearing), and a constant second ranktensor N0 such that

(0)F(τ) = F(τ) =Q(τ) ⋅ eτ κN0 , Q(0) = 1, ∣N0∣ = 1,where ∣N0∣2 = tr (NT

0 ⋅N0).8. Show that in any MWCSH,

A2 −A2

1 = κ2 (NT⋅N −N ⋅NT ) , (9.320)

and hencetrA2

1 = trA2 = 2κ2 (1 + trN2) . (9.321)

Subsequently, show that in a viscometric flow and in a motion of Class II,

κ2 = 1

2trA2

1 = 1

2trA2. (9.322)

9. A tensor M is nilpotent if Mn = 0 for some integer n ≥ 0.i) Show that the exponential eτ M is a finite polynomial in τ if and only

if M is nilpotent.

ii) Show that in three dimensions, if M is nilpotent, then n ≤ 3, and inthis case,

eτ M = 1 + τM + 1

2τ2M2.

iii) Show that the components of M relative to an appropriate orthonormalbasis have the form

M = ⎛⎜⎝0 0 0

χ 0 0

λ ν 0

⎞⎟⎠ ,and if M2 = 0, then the basis may be chosen such that

M = χ⎛⎜⎝0 0 0

1 0 0

0 0 0

⎞⎟⎠ .

426 VISCOELASTICITY

10. As shown in Problem 9, if M is nilpotent, then necessarily M3 = 0 in a three-dimensional Euclidean space. Subsequently, show that all MWCSH flows canbe divided in the following three classes depending on the properties of M:

I. M2 = 0;

II. M2 ≠ 0 but M3 = 0;

III. Mn ≠ 0 for all n = 1,2, . . ..11. The representation for the stress for a third-order linear viscoelastic fluid is

of the formσ = −p1 + S(1) + S(2) + S(3),

where it was shown that

S(1) = µA(1),S(2) = b1trA(2)1 + b2A(2) + c1A(1) ∶A(1) 1 + c2A(1) ⋅A(1).

Determine S(3).

12. For the flow

v1 = ax1, v2 = −ax2, v3 = 0,show that the stress tensor is of the form

σ = −p1 + β1A(1) + β2 (A(1))2and calculate A(1).

13. Show that in an isochoric motion, the first three invariants of A(n) are givenby

trA(1) = 0, (9.323)

trA(2) = tr (A(1))2 , (9.324)

trA(3) = −2 tr (A(1))3 + 3 tr (A(1) ⋅A(2)) , (9.325)

and in general trA(n) is a linear combination of traces of products formedfrom A(1), . . . ,A(n−1).

14. Show that the most general constitutive equation of the stress tensor of asimple fluid for the viscometric flow v = v(x1) i2 is given by

σ = −p(κ)1 + µ1(κ) (N +NT ) + µ2(κ)N ⋅NT+ µ3(κ)NT

⋅N, (9.326)

where

κ ≡ dv(x1)dx1

. (9.327)

Find N and show that the three viscometric functions µ1, µ2, and µ3 satisfythe conditions

µ1(−κ) = −µ1(κ) and µ2,3(−κ) = µ2,3(κ). (9.328)

9.3. CONSTITUTIVE EQUATIONS 427

15. Consider the steady Poiseuille flow of a simple fluid in an infinite circulartube of radius R, and use cylindrical coordinates with the axis of the tubecoincident with the z-axis. We take the velocity field v = (vr, vz , vθ) to havethe form

vr = 0, vz = v(r), vθ = 0,and the fluid to satisfy the no-penetration and no-slip conditions along thetube wall, v(R) = 0. Take the body force to act in the z-direction.

i) Use the linear momentum balance to show that the driving force densityalong the tube (pressure gradient) is given by

f = F

πR2 (z2 − z1) ,where the total force is given by

F = ∫A[(σzz − ρR U)∣z=z2 − (σzz − ρR U)∣z=z1]dA,

where U is the gravitational potential. Prove that f is constant; i.e., itis independent of the choice of z1 and z2.

ii) Show that the velocity profile is given by

v(r) = ∫ R

rζ (1

2f r′) dr′,

where ζ is the shear-rate function (see (9.167)), and from this it followsthat the volume discharge per unit time through a cross-section of thetube,

Q = 2π∫ R

0

v(r) r dr,is given by

Q = 8π

f3 ∫fR/2

0

ζ(r) r2dr.iii) Show that the stress tensor has the form

[σjk] = ⎛⎜⎝σrr σrz 0

σzr σzz 0

0 0 σθθ

⎞⎟⎠ ,where

σzr = σrz = −12f r,

σrr = f z + ρR U + ∫ R

r

1

r′µ2 [ζ (1

2f r′)]dr′ + c,

σrr − σθθ = µ2 [ζ (12f r)] ,

σzz − σθθ = µ3 [ζ (12f r)] ,

428 VISCOELASTICITY

and c is a constant. Note that because of the assumed incompressibilityof the fluid, when only f , R, and U are given, the normal pressures aredetermined only up to a constant hydrostatic pressure of magnitude c.Also note that the normal stress in the axial direction,

σzz = σrr + µ3 [ζ (12f r)] − µ2 [ζ (1

2f r)] ,

generally depends on r.

iv) What is the implication that σzz ≠ σrr upon a stream of fluid exitingthe tube?

16. Consider the steady flow of an incompressible simple fluid between two fixedcoaxial circular cylinders of radii R1 and R2 (R1 < R2) and use cylindricalcoordinates with the axis of the tubes coincident with the z-axis. We takethe velocity field v = (vr, vz , vθ) to have the form

vr = 0, vz = v(r), vθ = 0,and the fluid to satisfy the no-penetration and no-slip conditions along thetube walls, v(R1) = v(R2) = 0. Take the body force to act in the z-direction.

i) Use the linear momentum balance to show that the driving force densityalong and between the tubes (pressure gradient) is given by

f = F

π (R22−R2

1) (z2 − z1) ,

where the total force is given by

F = ∫A[(σzz − ρR U)∣z=z2 − (σzz − ρRU)∣z=z1]dA,

where U is the gravitational potential. Prove that f is constant; i.e., itis independent of the choice of z1 and z2.

ii) Show that the velocity profile is given by

v(r) = ∫ r

R1

ζ [α (r′)]dr′,where ζ is the shear-rate function (see (9.167)),

α(r) = ar−1

2f r2,

and the constant a is chosen so that

∫R2

R1

ζ [α (r)]dr = 0.From this, show that it follows that the volume discharge per unit timethrough a cross-section of the tube,

Q = 2π∫ R2

R1

v(r) r dr,is given by

Q = −π∫ R2

R1

ζ [α(r)] r2dr.

9.3. CONSTITUTIVE EQUATIONS 429

iii) Show that the stress tensor has the form

[σjk] = ⎛⎜⎝σrr σrz 0

σzr σzz 0

0 0 σθθ

⎞⎟⎠ ,where

σzr = σrz = α(r),σrr = f z + ρR U +∫ r

R1

1

r′µ2 ζ [α (r′)]dr′ + c,

σrr − σθθ = µ2 ζ [α (r′)] ,σzz − σθθ = µ3 ζ [α (r′)] ,

and c is a constant. Note that because of the assumed incompressibilityof the fluid, when only f , R, and U are given, the normal pressures aredetermined only up to a constant hydrostatic pressure of magnitude c.Also note that the stress difference in the radial direction between thetwo radii is given by

σrr(R2) − σrr(R1) = −∫ R2

R1

1

rµ2 ζ [α (r)]dr.

iv) What is the implication that σrr(R2) ≠ σrr(R1) upon a stream of fluidexiting the tubes?

17. Consider the steady Couette flow of an incompressible simple fluid betweentwo coaxial circular cylinders of radii R1 andR2 (R1 < R2) and use cylindricalcoordinates with the axis of the tubes coincident with the z-axis. The twotubes rotate with constant angular velocities Ω1 and Ω2, respectively. Wetake the velocity field v = (vr, vz , vθ) to have the form

vr = 0, vz = 0, vθ = ω(r),and the fluid to satisfy the no-penetration and no-slip conditions along thetube walls,

ω(R1) = Ω1 and ω(R2) = Ω2.

i) Show that the velocity profile obeys the equation

dω(r)dr

= 1

rζ ( M

2π r) ,

where ζ is the shear-rate function (see (9.167)) and M is the torqueper unit length required to maintain the relative motion between thecylinders.

ii) By integrating the above equation, show that the angular velocity dif-ference and the torque are related by

Ω2 −Ω1 = 1

2∫

M/(2πR2

1)

M/(2πR2

2)

1

sζ(s)ds.

430 VISCOELASTICITY

iii) Show that the stress tensor has the form

[σjk] = ⎛⎜⎝σrr 0 σrθ0 σzz 0

σθr 0 σθθ

⎞⎟⎠ ,where

σθr = σrθ = M

2π r2,

σrr = ρR U −∫ r

R1

ρRr′ω2(r′) + 1

r′µ2 [ζ ( M

2π r′2)]−

1

r′µ3 [ζ ( M

2π r′2)]dr′ + c,

σrr − σzz = µ2 [ζ ( M

2π r2)] ,

σθθ − σzz = µ3 [ζ ( M

2π r2)] ,

where U is the gravitational potential and c is a constant. Note that be-cause of the assumed incompressibility of the fluid, the normal pressuresare determined only up to a constant hydrostatic pressure of magnitudec. Also note that the stress difference in the radial direction betweenthe two radii given by

σrr(R2) − σrr(R1) = −∫ R2

R1

ρRr ω2(r) − 1

rµ2 [ζ ( M

2π r2)]−

1

rµ3 [ζ ( M

2π r2)]dr

generally depends on r.

iv) What is the implication that σrr(R2) ≠ σrr(R1)?18. Consider the steady Poiseuille flow of a simple fluid in an infinite circular

tube of radius R, and use cylindrical coordinates with the axis of the tubecoincident with the z-axis. Take the velocity field v = (vr, vθ, vz) to have theform

vr = 0, vθ = 0, vz = v(r),with the fluid moving in the positive z-direction and satisfying the no-penetration and no-slip conditions along the tube wall, v(R) = 0. Takethe body force to act in the z-direction. In addition, take the stress tensorto be linear of rate type and of rate 1 and the extra stress given in the form

[σdjk] = ⎛⎜⎝

tr 0 s

0 r−2tθ 0

0 0 tz

⎞⎟⎠ ,where ti = ti(r) (i = r, θ, z) and s = s(r).

i) Show that s = − 1

2ar, where a is a positive constant.

9.3. CONSTITUTIVE EQUATIONS 431

ii) Show that p = −(a − ρg)z + f(r).iii) Show that the velocity is given by

v = 2

ν η a[(u − uR) − log( u

uR)] ,

where

u = 1+√1 − νa2r2, uR = 1+√1 − νa2R2, ν = 4βτ2dη−2 = 1

4κ, a = ρg+∆pz

l,

η, β, and τd are given in (9.302), ∆pz = pz1 − pz2, and pz1 and pz2 arethe average pressures at two cross-sections of the tube separated by adistance l. Note that for steady flow to be possible, it is necessary thata2 ≤ ν−1R−2.

iv) Show that the volume discharge per unit time and the average velocityare given by

Q = πR4 a (2uR − 1)3 η u2R

and v = Q

πR2.

Note that when β = 0, uR = 2, and then

v = vc ≡ R2 a

4 η,

which is the same equation as for classical viscous fluids. For the sakeof simplicity, from now on, assume that we have no body forces so thata =∆pz/l is the pressure gradient in the flow direction.

v) Show that for β > 0v

vc= 1 + ξ2 (2 − 9

4ξ2)

6 [(1 − ξ2)3/2 + (1 − 3

2ξ2 + 3

8ξ4)] ,

where

ξ = 2τs

ηaR and τs =√β τd,

with τs being the shearing yield time. Note that real values of v areobtained only if ∣ξ∣ ≤ 1, so that the largest possible values of the pressuregradient and mean velocity are given by

acr = η

2τsRand vcr = R

6τs.

When the pressure gradient exceeds acr, the flow necessarily becomesunsteady. Also, when a = acr, we have that v/vcr = 4/3. Plot v/vc as afunction of ξ and comment on the accuracy of the classical solution.

vi) Show that for β < 0v

vc= 1 − ζ2 (2 + 9

4ζ2)

6 [(1 + ζ2)3/2 + (1 + 3

2ζ2 + 3

8ζ4)] ,

432 VISCOELASTICITY

where

ζ = 2 τ ′sηaR and τ ′s =√∣β∣ τd,

with τ ′s being the shearing limit time. Note that no matter how largethe pressure gradient is, the mean velocity never exceeds the limitingvalue

vlim = R

3τ ′s.

Plot v/vc as a function of ζ and comment on the accuracy of the classicalsolution.

vii) Show that the normal stresses are given by

σrr = −p + tr = az + 1

2a τd (β3 − 1)v,

σθθ = −p + tθ = az + 1

2a τd (β3 − 1)(v − 2a r2

η u) ,

σzz = −p + tz = az + 1

2a τd [(β3 − 1)v + 4a r2

η u] ,

and note that these stresses deviate from those of the classical normalstresses, which are given by

σrr = σθθ = σzz = −p = az.

Bibliography

R.B. Bird, R.C. Armstrong, and O. Hassager. Dynamics of Polymeric Liquids,volume 1. Wiley, New York, 1977.

T.S. Chang. Constitutive equations for simple materials: Simple materials withfading memory. In E.H. Dill, editor, Continuum Physics, volume II, pages 283–403. Academic Press, Inc., New York, NY, 1975.

B.D. Coleman. Kinematical concepts with applications in the mechanics andthermodynamics of incompressible viscoelastic fluids. Archive for Rational Me-chanics and Analysis, 9(1):273–300, 1962.

B.D. Coleman. Substantially stagnant motions. Transactions of the Society ofRheology, 6(1):293–300, 1962.

B.D. Coleman. On thermodynamics, strain impulses, and viscoelasticity. Archivefor Rational Mechanics and Analysis, 17(3):230–254, 1964.

B.D. Coleman. Thermodynamics of materials with memory. Archive for RationalMechanics and Analysis, 17:1–46, 1964.

B.D. Coleman, H. Markovitz, and W. Noll. Viscometric Flows of Non-NewtonianFluids. Springer-Verlag, Berlin, 1966.

BIBLIOGRAPHY 433

B.D. Coleman and W. Noll. On certain steady flows of general fluids. Archivefor Rational Mechanics and Analysis, 3(1):289–303, 1959.

B.D. Coleman and W. Noll. An approximation theorem for functionals, with ap-plications in continuum mechanics. Archive for Rational Mechanics and Analysis,6(1):355–370, 1960.

B.D. Coleman and W. Noll. Foundations of linear viscoelasticity. Reviews ofModern Physics, 33(2):239–249, 1961.

B.D. Coleman and W. Noll. Recent results in the continuum theory of viscoelasticfluids. Annals of the New York Academy of Sciences, 89(4):672–714, 1961.

B.D. Coleman and W. Noll. Steady extension of incompressible simple fluids.Physics of Fluids, 5(7):840–843, 1962.

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W.O. Criminale, J.L. Ericksen, and G.L. Filbey. Steady shear flow of non-Newtonian fluids. Archive for Rational Mechanics and Analysis, 1(1):410–417,1957.

W.A. Day. The Thermodynamics of Simple Materials with Fading Memory.Springer-Verlag, Berlin, 1972.

M.O. Deville and T.B. Gatski. Mathematical Modeling for Complex Fluids andFlows. Springer-Verlag, Berlin, 2012.

A.C. Eringen. Constitutive equations for simple materials: General theory. InA.C. Eringen, editor, Continuum Physics, volume II, pages 131–172. AcademicPress, Inc., New York, NY, 1975.

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R.L. Fosdick and K.R. Rajagopal. Thermodynamics and stability of fluids of thirdgrade. Proceedings of the Royal Society of London A: Mathematical, Physical andEngineering Sciences, 369(1738):351–377, 1980.

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R.R. Huilgol. Recent advances in the continuum mechanics of viscoelastic liquids.International Journal of Engineering Science, 24(2):161–251, 1986.

D.D. Joseph. Instability of the rest state of fluids of arbitrary grade greater thanone. Archive for Rational Mechanics and Analysis, 75(3):251–256, 1981.

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434 VISCOELASTICITY

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W. Noll. Motions with constant stretch history. Archive for Rational Mechanicsand Analysis, 11(1):97–105, 1962.

A. Pipkin. Small finite deformation of viscoelastic solids. Reviews of ModernPhysics, 36(4):1034–1041, 1964.

M. Reiner. Second-order effects. In R.J. Seeger and G. Temple, editors, ResearchFrontiers in Fluids Dynamics, pages 193–211. Interscience Publishers, New York,1965.

R.S. Rivlin. Viscoelastic fluids. In R.J. Seeger and G. Temple, editors, ResearchFrontiers in Fluids Dynamics, pages 144–192. Interscience Publishers, New York,1965.

R.S. Rivlin. The fundamental equations of nonlinear continuum mechanics. In S.I.Pai, A.J. Faller, T.L. Lincoln, D.A. Tidman, G.N. Trytten, and T.D. Wilkerson,editors, Dynamics of Fluids in Porous Media, pages 83–126, Academic Press,New York, 1966.

R.S. Rivlin. An introduction to non-linear continuum mechanics. In R.S. Rivlin,editor, Non-linear Continuum Theories in Mechanics and Physics and Their Ap-plications, pages 151–310. Springer-Verlag, Berlin, 1969.

R.S. Rivlin and K.N. Sawyers. Nonlinear continuum mechanics of viscoelasticfluids. Annual Review of Fluid Mechanics, pages 117–146, 1971.

J.C. Saut and D.D. Joseph. Fading memory. Archive for Rational Mechanics andAnalysis, 81(1):53–95, 1983.

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C. Truesdell and W. Noll. The non-linear field theories of mechanics. In S. Flügge,editor, Handbuch der Physik, volume III/3. Springer, Berlin-Heidelberg-NewYork, 1965.

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C.-C. Wang. A representation theorem for the constitutive equation of a simplematerial in motions with constant stretch history. Archive for Rational Mechanicsand Analysis, 20(5):329–340, 1965.

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BIBLIOGRAPHY 435

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Appendices

A. SUMMARY OF CARTESIAN TENSOR NOTATION 439

A Summary of Cartesian tensor notation

The standard Kronecker delta symbol is defined as

δij = 1 if i = j,0 if i ≠ j. (A.1)

Note that

δii = 3 (A.2)

and

δijδij = δii = 3. (A.3)

The standard Levi–Civita symbol is given by

ǫijk =⎧⎪⎪⎪⎨⎪⎪⎪⎩

1 if (i, j, k) is an even permutation of (1,2,3),−1 if (i, j, k) is an odd permutation of (1,2,3),0 if any two labels are the same.

(A.4)

Useful relations between the Levi–Civita symbol and the generalized Kroneckerdelta symbol (see (2.91)) are

ǫijkǫlmn = δiljmkn = det⎡⎢⎢⎢⎢⎢⎣δil δim δinδjl δjm δjnδkl δkm δkn

⎤⎥⎥⎥⎥⎥⎦,

= δilδjmδkn + δimδjnδkl + δinδjlδkm − δimδjlδkn − δilδjnδkm −

δinδjmδkl, (A.5)

ǫijkǫlmk = 1! δiljm = det[ δil δimδjl δjm

] = δilδjm − δimδjl, (A.6)

ǫijkǫijl = 2! δkl = 2 δkl, (A.7)

ǫijkǫijk = 3! = 6. (A.8)

The generalized Levi–Civita (or permutation) symbol (see (2.88)) is given by

ǫijkl... =⎧⎪⎪⎪⎨⎪⎪⎪⎩+1 if (i, j, k, l, . . .) is an even permutation of (1,2,3,4, . . .),−1 if (i, j, k, l, . . .) is an odd permutation of (1,2,3,4, . . .),0 if any two labels are the same.

(A.9)

Furthermore, it can be shown that

ǫi1i2i3...ǫj1j2j3... = δi1j1i2j2i3j3... = det⎡⎢⎢⎢⎢⎢⎢⎢⎣

δi1j1 δi1j2 δi1j3 ⋯

δi2j1 δi2j2 δi2j3 ⋯

δi3j1 δi3j2 δi3j3 ⋯

⋮ ⋮ ⋮ ⋱

⎤⎥⎥⎥⎥⎥⎥⎥⎦, (A.10)

ǫi1,...,inǫi1,...,in = n!, (A.11)

440 APPENDIX A

and

a = detA = ǫj1...jna1j1⋯anjn = ǫi1...inai11⋯ainn = 1

n!ǫi1...inǫj1...jnai1j1⋯ainjn . (A.12)

From the aforementioned, it is easy to show, e.g., that

ǫijǫlm = δiljm = det [ δil δimδjl δjm

] = δilδjm − δimδjl, (A.13)

ǫijǫil = 1! δjl = δjl, (A.14)

ǫijǫij = 2! = 2. (A.15)

A vector is represented by its typical components relative to three mutuallyorthogonal unit vectors of a right-handed Cartesian coordinate system with com-ponents (x1, x2, x3). Let

u = uiii and v = vjij.Then

u ⋅ v = uivi and u × v = ǫijkujvk,where repeated subscripts are summed from 1 to 3, i.e., uivi = ∑3

i=1 uivi.Repeated subscripts are dummy subscripts, in the sense that they may be re-

placed by another letter without affecting the value of the sum:

uivi = ujvj = ukvk.If δij appears in an expression summing both i and j, j may be set equal to i

and the δij removed:

uivjδij = uivi.Example

Prove that u × (v ×w) = v (u ⋅w) −w (u ⋅ v):u × (v ×w) = ǫijkuj(ǫklmvlwm)

= ǫijkǫklmujvlwm

= ǫijkǫlmkujvlwm

= (δilδjm − δimδjl)ujvlwm

= viujwj −wiujvj

= v (u ⋅w) −w (u ⋅ v) .Let

∇ = ii ∂

∂xi= ii∂i.

Then

gradφ = ∇φ = ii∂iφ,

BIBLIOGRAPHY 441

divu = ∇ ⋅ u = ∂iui,

curlu = ∇ × u = ǫijk∂juk.

Example

Prove that ∇ ×∇φ = 0:

∇ ×∇φ = ǫijk∂j∂kφ

= ǫijk∂k∂jφ (order of differentiation may be changed)= ǫikj∂j∂kφ (j and k are dummy indices)= −ǫijk∂j∂kφ (ǫijk = −ǫikj)= −∇ ×∇φ.

Subsequently, we must have that ∇ ×∇φ = 0.

Problems

1. Show that (a × b) ⋅ (c × d) = (a ⋅ c)(b ⋅ d) − (a ⋅ d)(b ⋅ c).2. Show that ∇ ⋅ (∇ × u) = 0.3. Show that ∇ × (u × v) = v ⋅∇u − v(∇ ⋅ u) + u(∇ ⋅ v) − u ⋅∇v.

4. Show that ∇ × (∇ × u) = ∇(∇ ⋅ u) −∇2u.

Bibliography

H. Jeffreys. Cartesian Tensors. Cambridge University Press, London, 1969.

G. Temple. Cartesian Tensors. Methuuen & Co. Ltd., London, 1960.

442 APPENDIX B

B Isotropic tensors

Let’s assume that we have a linear relationship between the second rank tensorsfields σ and D:

σik = AiklmDlm. (B.1)

Now if the material is homogeneous and isotropic, its properties will be the sameat all points and in all frames of reference. Hence, the above equation mustremain invariant under rigid rotations of the frame of reference. Now under sucha transformation, the tensors σ and D become σ′ and D′ where

σ′pq = QpiQqkσik, (B.2)

D′rs = QrlQsmDlm, (B.3)

or

Dlm =QmsQlrD′

rs, (B.4)

since Q−1 =QT . Hence,

σ′pq = QpiQqkσik = QpiQqkAiklmDlm = QpiQqkAiklmQmsQlrD′

rs = A′pqrsD′rs, (B.5)

where

A′pqrs = QpiQqkQrlQsmAiklm. (B.6)

The coefficients A′pqrs are therefore the components of a tensor of rank 4. But ifthe linear relations remain invariant, the coefficients must remain unaltered, i.e.,

A′pqrs = Apqrs. (B.7)

Thus the tensor A′pqrs must have the same set of coefficients in all bases. Suchtensors are called isotropic tensors.

In general, we define an isotropic tensor of any rank by the criterion that itscomponents form the same set of numbers in all bases, or that it is invariant underany rotation.

It is evident that all rank-0 tensors (scalars) are isotropic. There are no isotropicrank-1 tensors (vectors). To see that this is the case, consider a small rotation,expressed by the skew-symmetric tensor Qik, of the vector v. Then in the newcoordinate system,

v′i = (δij −Qij)vj = vi −Qijvj , (B.8)

and this can be equal to vi only if

Qijvj = 0. (B.9)

Now the above corresponds to a system of three linear equations. But Qii = 0 andQij = −Qji, j ≠ i, are linearly independent. Hence, the above can be satisfied onlyif vi = 0, and therefore, there is no isotropic tensor of first order other than zero.

B. ISOTROPIC TENSORS 443

The number of linearly independent isotropic tensors of rank n = 0,1,2,3,4,5,6, . . . are an = 1,0,1,1,3,6,15, . . .. These numbers are called the Motzkin sumnumbers and are given by the recurrence relation

an = n − 1n + 1

(2an−1 + 3an−2) with a0 = 1, a1 = 0. (B.10)

Starting at rank 5, syzygies play a role in restricting the number of independentisotropic tensors. In particular, syzygies occur at rank 5, 7, 8, and all higherranks. A syzygy is a mathematical object defined in terms of a polynomial ring ofn variables. An example of a rank-5 syzygy is

ǫijkδlm − ǫjklδim + ǫiklδjm − ǫijlδkm = 0, (B.11)

which can be easily verified.One effective approach of generating isotropic tensors of arbitrary rank in three

dimensions is based on Weyl’s theory of invariant polynomials. This approachreduces to noting that any even rank isotropic tensor must be expressed as alinear combination of products of the unit tensor 1, and odd tensors as a linearcombination of products of the unit tensor and the alternating tensor ǫ. In practice,this means finding every possible way of writing inner products between pairs ofvectors. For example, for a rank-2 isotropic tensor A, we have only one possibleinner product between two vectors, so

A ∶ uv = α(u ⋅ v) (B.12)

or

Aijuivj = αukvk = αδikδjkuivj = αδijuivj . (B.13)

Subsequently, Aij is given by a linear relation with the one isotropic tensor ofsecond rank I(2,1) = 1:

Aij = αI(2,1)ij , (B.14)

where

I(2,1)ij = δij . (B.15)

The unit tensor is the only isotropic tensor of second rank.For a rank-3 isotropic tensor A, again we only have one linear combination of

products of three vectors, so

A⋮uvw = α [u ⋅ (v ×w)] (B.16)

or

Aijkuivjwk = αukǫklmvlwm = αǫklmδikδjlδkmuivjwk = αǫijkuivjwk. (B.17)

Subsequently, we have that the Levi–Civita tensor is the only isotropic tensor ofrank-3, i.e., I(3,1) = ǫ or

I(3,1)ijk

= ǫijk. (B.18)

444 APPENDIX B

For a rank-4 isotropic tensor A, we have three linear combinations of innerproducts of four vectors:

A ∶∶ uvwx = α1(u ⋅ v) (w ⋅ x) + α2(u ⋅w) (v ⋅ x) + α3(u ⋅ x) (v ⋅w). (B.19)

Following the same procedure as above, we obtain the three isotropic tensors

I(4,1)ijkl

= δijδkl, (B.20)

I(4,2)ijkl

= δikδjl, (B.21)

I(4,3)ijkl

= δilδjk. (B.22)

For a rank-5 isotropic tensor A, we have ten linear combinations of inner prod-ucts of five vectors:

A⋮ ∶ uvwxy = α1(u ⋅ v) (w ⋅ x × y) + α2(u ⋅w) (v ⋅ x × y) +α3(u ⋅ x) (v ⋅w × y) + α4(u ⋅ y) (v ⋅w × x) +α5(v ⋅w) (u ⋅ x × y) + α6(v ⋅ x) (u ⋅w × y) +α7(v ⋅ y) (u ⋅w × x) + α8(w ⋅ x) (u ⋅ v × y) +α9(w ⋅ y) (u ⋅ v × x) + α10(x ⋅ y) (u ⋅ v ×w). (B.23)

Following the same procedure as above, we can basically read off the ten isotropictensors ǫklmδij , ǫjlmδik, ǫjkmδil, ǫjklδim, ǫilmδjk, ǫikmδjl, ǫiklδjm, ǫijmδkl, ǫijlδkm,ǫijkδlm corresponding to the above ten products. However, not all these isotropictensors are independent. Specifically, using the syzygy that we noted earlier,we see that ǫijlδkm is related to ǫijkδlm, ǫjklδim, and ǫiklδjm. Subsequently, wecan reduce the isotropic tensors to the following nine: ǫklmδij , ǫjlmδik, ǫjkmδil,ǫjklδim, ǫilmδjk, ǫikmδjl, ǫiklδjm, ǫijmδkl, ǫijkδlm. We note that by appropriatelyinterchanging subscripts m and l in the previous syzygy, we have the syzygy

ǫijkδml − ǫjkmδil + ǫikmδjl − ǫijmδkl = 0. (B.24)

Now we see that the isotropic tensor ǫijmδkl can be written in terms of the isotropictensors ǫijkδml, ǫjkmδil, and ǫikmδjl. Subsequently, since δml = δlm, we can reducethe number of isotropic tensors to the following eight: ǫklmδij , ǫjlmδik, ǫjkmδil,ǫjklδim, ǫilmδjk, ǫikmδjl, ǫiklδjm, ǫijkδlm. Now interchanging l and j in the pre-vious syzygy, we have the syzygy

ǫilkδmj − ǫlkmδij + ǫikmδlj − ǫilmδkj = 0, (B.25)

or

−ǫiklδmj + ǫklmδij + ǫikmδlj − ǫilmδkj = 0. (B.26)

Noting that ǫilmδkj can be written in terms of ǫiklδmj , ǫklmδij , and ǫikmδlj , andusing the symmetry of the Kronecker delta, we can reduce the number of isotropictensors to the following seven: ǫklmδij , ǫjlmδik, ǫjkmδil, ǫjklδim, ǫikmδjl, ǫiklδjm,ǫijkδlm. We can reduce the number of isotropic tensor again by using the followingsyzygy, which is obtained by interchanging the subscripts j and i in the last syzygy:

−ǫjklδmi + ǫklmδji + ǫjkmδli − ǫjlmδki = 0. (B.27)

B. ISOTROPIC TENSORS 445

Now noting that ǫjlmδki can be written in terms of ǫjklδmi, ǫklmδji, and ǫjkmδli,and using the symmetry of the Kronecker delta, allows us to reduce the numberof independent isotropic tensors to the following six: ǫklmδij , ǫjkmδil, ǫjklδim,ǫikmδjl, ǫiklδjm, ǫijkδlm. No further reductions are possible. Thus, for a rank-5tensor, we obtain the following six linearly independent isotropic tensors:

I(5,1)ijklm

= ǫklmδij , (B.28)

I(5,2)ijklm

= ǫjkmδil, (B.29)

I(5,3)ijklm

= ǫjklδim, (B.30)

I(5,4)ijklm = ǫikmδjl, (B.31)

I(5,5)ijklm

= ǫiklδjm, (B.32)

I(5,6)ijklm

= ǫijkδlm. (B.33)

For a rank-6 isotropic tensor A, we have fifteen linear combinations of innerproducts of six vectors:

A⋮⋮uvwxyz = α1(u ⋅ v) (w ⋅ x) (y ⋅ z) + α2(u ⋅w) (v ⋅ x) (y ⋅ z) +α3(u ⋅ x) (v ⋅w) (y ⋅ z) + α4(u ⋅ y) (v ⋅w) (x ⋅ z) +α5(u ⋅ z) (v ⋅w) (x ⋅ y) + α6(u ⋅w) (v ⋅ y) (x ⋅ z) +α7(u ⋅w) (v ⋅ z) (x ⋅ y) + α8(u ⋅ v) (w ⋅ y) (x ⋅ z) +α9(u ⋅ v) (x ⋅ y) (w ⋅ z) + α10(u ⋅ y) (w ⋅ x) (v ⋅ z) +α11(u ⋅ z) (w ⋅ x) (v ⋅ y) + α12(u ⋅ z) (v ⋅ x) (w ⋅ y) +α13(u ⋅ y) (v ⋅ x) (w ⋅ z) + α14(u ⋅ x) (v ⋅ y) (w ⋅ z) +α15(u ⋅ x) (v ⋅ z) (w ⋅ y). (B.34)

Following the same procedure as above, since no syzygies exist for rank-6 tensors,we can read off the fifteen linearly independent isotropic tensors

I(6,1)ijklmn

= δijδklδmn, (B.35)

I(6,2)ijklmn

= δikδjlδmn, (B.36)

I(6,3)ijklmn = δilδjkδmn, (B.37)

I(6,4)ijklmn

= δimδjkδln, (B.38)

I(6,5)ijklmn

= δinδjkδlm, (B.39)

I(6,6)ijklmn

= δikδjmδln, (B.40)

I(6,7)ijklmn

= δikδjnδlm, (B.41)

I(6,8)ijklmn

= δijδkmδln, (B.42)

I(6,9)ijklmn

= δijδlmδkn, (B.43)

I(6,10)ijklmn

= δimδklδjn, (B.44)

I(6,11)ijklmn

= δinδklδjm, (B.45)

446 APPENDIX B

I(6,12)ijklmn

= δinδjlδkm, (B.46)

I(6,13)ijklmn

= δimδjlδkn, (B.47)

I(6,14)ijklmn = δilδjmδkn, (B.48)

I(6,15)ijklmn

= δilδjnδkm. (B.49)

We conclude by noting that while the above representations of isotropic tensorsare linearly independent, these representations are not unique. For example, forthe isotropic tensor of rank-4, it is typical to also see the following representation:

I(4,1)ijkl

= δijδkl, (B.50)

I(4,2)ijkl

= δikδjl + δilδjk, (B.51)

I(4,3)ijkl

= δikδjl − δilδjk. (B.52)

Clearly, while one may find the above representation more convenient, neverthelessit is equivalent to our previous representation.

Problems

1. Verify (B.11).

2. Verify (B.19) and (B.20)–(B.22).

3. Verify (B.23) and (B.28)–(B.33).

4. Verify (B.34) and (B.35)–(B.49).

Bibliography

B.C. Eu. A complete set of irreducible isotropic tensors of rank six. CanadianJournal of Physics, 58(7):931–932, 1980.

A.J.M. Spencer. Theory of invariants. In A.C. Eringen, editor, ContinuumPhysics, volume I. Academic Press, New York, 1971.

C. BALANCE LAWS IN MATERIAL COORDINATES 447

C Balance laws in material coordinates

Sometimes, for solid bodies, it is more convenient to use the material descriptionof balance laws. The corresponding relations for the general balance equationand jump condition obtained in Section 4.1 can be derived in a similar manner.Specifically, using the relationships between differential surface and volume ele-ments (3.36) and (3.51) between the current and reference configurations, we have

d

dt∫VψJ dV = ∫

St ⋅ J (F−1)T ⋅ dS +∫

VgJ dV. (C.1)

Now writing

Ψ = Jψ, T = J t ⋅ (F−1)T , and G = Jg, (C.2)

we can rewrited

dt∫VΨdV = ∫

ST ⋅ dS +∫

VGdV. (C.3)

The transport theorem (3.466) remains valid for Ψ(X, t) in a moving region V (t):d

dt∫VΨdV = ∫

VΨdV +∫

SΨv ⋅ dS, (C.4)

where v(X, t) ⋅N is the outward speed of a surface point X on the surface S withunit normal N. Now, since the material region illustrated in Fig. 4.1 is fixed in thereference configuration, the corresponding generalized transport theorem becomes

d

dt∫V −ζ

ΨdV = ∫V −ζ

ΨdV + ∫ζJΨcK ⋅ dζ, (C.5)

where c(X, t) is the speed of the singular surface ζ(t) with unit normal N, andsince the divergence theorem (2.299) is independent of the current or referenceconfiguration, we have

∫V −ζ

DivT dV = ∫S−ζ

T ⋅ dS −∫ζJTK ⋅ dζ. (C.6)

Subsequently, using (C.5) and (C.6), the integral balance law in material coordi-nates (C.3) for a singular surface which does not possess any properties of its ownbecomes

∫V −ζ[Ψ −DivT −G]dV + ∫

ζJΨc −TK ⋅ dζ = 0. (C.7)

Using the same arguments used in Section 4.1, we obtain the local balance equationand jump condition in material coordinates:

Ψ −DivT −G = 0, (C.8)

JΨc −TK ⋅N = 0. (C.9)

Now the corresponding local conservation of mass, and balances of linear mo-mentum, angular momentum, energy, and entropy for a nonpolar material can be

448 APPENDIX C

easily written down:

ρR = 0, (C.10)

ρR (x − f) = Divσ, (C.11)

σ ⋅FT = F ⋅σT , (C.12)

ρR (e − r) = ΦR −DivqR, (C.13)

γv ≡ ρR (η − b) +DivhR ≥ 0, (C.14)

where we recall that ρR = Jρ, and the following definitions have been introduced:

σ ≡ F ⋅σ ≡ Jσ ⋅ (F−1)T , ΦR ≡ F ∶ σ ≡ E ∶ σ, qR ≡ JF−1 ⋅ q, hR ≡ JF−1 ⋅ h,(C.15)

which correspond to the first and second Piola–Kirchhoff stress tensors, the ma-terial stress power, the Piola–Kirchhoff heat flux, and the Piola–Kirchhoff entropyflux. Note that, in a nonpolar medium, unlike the Cauchy stress tensor σ, the firstPiola–Kirchhoff stress tensor σ is not symmetric. We also note that the secondPiola–Kirchhoff stress tensor σ is symmetric.

Similarly, we obtain the following jump conditions for a nonpolar material at asingular surface for the conservation of mass, and balances of linear momentum,energy, and entropy in material coordinates :

JρRcK ⋅N = 0, (C.16)

JρRx c +σK ⋅N = 0, (C.17)sρR (e + 1

2x ⋅ x) c +σT

⋅ x − qR

⋅N = 0, (C.18)

γs ≡ JρRη c − hRK ⋅N ≥ 0. (C.19)

Bibliography

R.M. Bowen. Introduction to Continuum Mechanics for Engineers. Plenum Press,New York, NY, 1989.

P. Haupt. Continuum Mechanics and Theory of Materials. Springer-Verlag,Berlin, 2000.

G.A. Holzapfel. Nonlinear Solid Mechanics. John Wiley & Sons, Ltd., Chichester,England, 2005.

K. Hutter and K. Jöhnk. Continuum Methods of Physical Modeling. Springer-Verlag, Berlin, 1981.

D. CURVES AND SURFACES IN SPACE 449

D Curves and surfaces in space

Quite often we are required to deal with the continuum mechanics of materialcurves and surfaces that are embedded within the three-dimensional Euclideanspace E3. In such case, we have to consider fields that are defined on such curvesand surfaces as functions of time. With this goal in mind, we first require adescription of their geometry.

One’s first thought might be that a curve or surface can be thought of as a E1 orE2 Euclidean space, which is a subset of E3. This certainly works well for straightlines or plane surfaces since they are Euclidean spaces. But arbitrary curves andsurfaces are non-Euclidean or Riemannian, because, e.g., a vector consisting of thesum of two surface vectors generally does not lie on the surface. Furthermore, thedistance between two points measured along a curved line or surface is in generalnot equal to the distance between these same two points measured in the Euclideanspace E3. Subsequently, we have to consider the Riemannian geometry of spacesV1 and V2. In passing, we note that consideration of the Riemannian geometry ofspace-time V4 is essential in general relativity.

In the following sections, we will examine some fundamental properties of curvesand surfaces. For example, at each point of a space curve, we can construct amoving coordinate system consisting of a tangent vector, a normal vector, and abinormal vector which is perpendicular to both the tangent and normal vectors.How these vectors change as we move along the curve in space brings up thesubjects of curvature and torsion associated with the space curve. The curvatureis a measure of how the tangent vector to the curve is changing and the torsion isa measure of the twisting of the curve out of a plane. We will find that straightlines have zero curvature and plane curves have zero torsion.

In a similar fashion, associated with every smooth surface there are two coordi-nate surface curves and a normal surface vector through each point on the surface.The coordinate surface curves have tangent vectors which together with the nor-mal surface vector, create a set of basis vectors and form a coordinate system ateach point of the surface. These vectors can be used to define such things as atwo-dimensional surface metric and a second-order curvature tensor. How thesesurface vectors change brings into consideration two different curvatures: a normalcurvature and a tangential curvature. How these curvatures are related to the cur-vature tensor and to the Riemann–Christoffel tensor, as well as other interestingrelationships between the various surface vectors and curvatures, is the subject ofthe differential geometry of curves and surfaces, which we discuss below.

Before embarking on this discussion, we find it convenient to define the intrinsicor absolute derivative of a vector u = ui(xj(yα))ei = ui(xj(yα))ei taken along thedirection yα:

δui

δyα= ui,jajα = [ ∂ui

∂xj− Γk

ijuk]ajα = ∂ui∂yα− Γk

ijajαuk, (D.1)

and similarly

δui

δyα= ui,jajα = [ ∂ui

∂xj+ Γi

jkuk]ajα = ∂ui

∂yα+ Γi

jkajαu

k, (D.2)

450 APPENDIX D

where

ajα ≡ ∂xj

∂yα(D.3)

are the contravariant components of the coordinate transformation. The absolutederivative of higher order tensors is similarly defined. For example, to differentiatethe mixed components T ij

klm= T ij

klm(xp(yα)) in the direction tangent to the curve

xp = xp(yα), we have

δTijklm

δyα= T ij

klm,papα. (D.4)

In addition, the rule for taking the absolute derivatives of sums and products oftensors is the same as for ordinary derivatives. For example, the second abso-lute derivative is given by the absolute derivative of the absolute derivative. Toillustrate, if we have the scalar field f = f(xj(yα)), then

δf

δyα= f,jajα, (D.5)

and since

ajγaγl= δj

l, and thus

∂ajα∂xk

= −∂aγl∂xk

alαajγ , (D.6)

it is easy to show that

δ

δyβ( δfδyα) = f,jkajαakβ − ∂aγl

∂xkalαa

kβ

δf

δyγ. (D.7)

Note that, since j and k are dummy indices, we have

δ

δyβ( δfδyα) − δ

δyα( δfδyβ) = (∂aγk

∂xj−∂a

γj

∂xk)ajαakβ δf

δyγ≡F

γαβ

δf

δyγ, (D.8)

where the geometrical tensor of rank 3 whose components are Fγαβ

is called theobject of anholonominity of yγ . The components are skew-symmetric with respectto the indices α and β. Subsequently, the components of the associated axial

tensor of rank 2 are given by Fδγ = 1

2εδαβF

γαβ

. This object arises since, in

general, it is not true that there exists a coordinate system yγ in E3 such that theset e1,e2,e3 is a set of covariant measuring vectors for this coordinate system.

In fact, this (holonomic) coordinate system will exist if and only if Fδγ = 0 for

an arbitrary function f . Measuring vectors that do not satisfy this conditionare called anholonomic measuring vectors. Arc lengths along the vector lines ofthese vectors are called anholonomic coordinates. In general, there does not exista one-to-one mapping between a (holonomic) coordinate system in E3 and ananholonomic coordinate system in E3. Thus, given a function f with values f(xj)for all xj in E3, there does not exist a function g such that f(xj) = g(yγ(xj)) in

E3. Subsequently, Fδγ

often appears as part of a “correction term” when identitieswhich are familiar in standard (holonomic) coordinates are derived for the case ofanholonomic coordinates.

D. CURVES AND SURFACES IN SPACE 451

D.1 Space curve

We recall the position vector in the Euclidean space E3

r = xiei = xjej . (D.9)

A curve C embedded in E3 can be represented by the pair of scalar equations

f1(xk, t) = 0 and f2(xk, t) = 0. (D.10)

Then two unit vectors normal to the curve and to each other are given by

a1 = gradf1∣gradf1∣ , a1 ⋅ a1 = 1, (D.11)

a2 = gradf2∣gradf2∣ , a2 ⋅ a2 = 1, (D.12)

gradf1 ⋅ gradf2 = 0, a1 ⋅ a2 = 0. (D.13)

Note that the last requirement restricts f2(xk, t) for a given f1(xk, t).Alternatively, r = xi(s, t)ei represents the three-dimensional space curve C as a

function of the arc length parameter s and time t. From now until near the end ofthis section, we suppress the dependence of the curve on t, since when consideringthe geometry of a curve, the value of t is fixed. Subsequently, all the followingderivatives with respect to s are to be interpreted later as partial derivatives withrespect to s while keeping t fixed.

The tangent vector to the curve C at point s is given by

a3 = drds= dxids

ei = tiei, (D.14)

where

ti ≡ dxids= ei ⋅ a3 (D.15)

is the contravariant component. It should be noted that the magnitude of ti isunity since

(ds)2 = dx ⋅ dx = dxidxjei ⋅ ej = gij dxids

dxj

ds(ds)2, (D.16)

so thatgijt

itj = 1, (D.17)

and subsequently,a3 ⋅ a3 = 1. (D.18)

If we now take the absolute derivative of (D.17) with respect to the arc lengths and use Ricci’s theorem (2.248), we have

gijδti

δstj + gijt

i δtj

δs= 0, (D.19)

which implies that

gijti δt

j

δs= 0. (D.20)

452 APPENDIX D

Thus, the vector with components δtj/δs is orthogonal to the tangent vector a3with components ti. Define the unit normal vector

a1 = nj ej = gradf1∣gradf1∣ (D.21)

to the space curve to be in the same direction as the vector δtj/δs and write

nj = 1

κ

δtj

δs, (D.22)

where κ is a scale factor, called the curvature, and is selected such that

a1 ⋅ a1 = gijninj = 1, (D.23)

which implies that

gijδti

δs

δtj

δs= κ2. (D.24)

The reciprocal of the curvature, κ−1, is called the radius of curvature. The cur-vature measures the rate of change of the tangent vector to the curve as the arclength varies. By taking the absolute derivative of (D.20), or equivalently of

a3 ⋅ a1 = gijtinj = 0, (D.25)

with respect to the arc length (and use Ricci’s theorem (2.248)), we have that

gijδti

δsnj+ gijt

i δnj

δs= 0. (D.26)

Consequently, the curvature can be determined from

−gijti δn

j

δs= gij δti

δsnj = κgijninj = κ, (D.27)

which also defines the sign of the curvature.In a similar fashion, if we take the absolute derivative of (D.23) with respect to

the arc length parameter s (and use Ricci’s theorem (2.248)), we find that

gijni δn

j

δs= 0. (D.28)

This indicates that the vector (δnj/δs)ej is orthogonal to the unit normal vectora1. Equation (D.25) indicates that a3 is also orthogonal to a1, and hence, anylinear combination of these vectors will also be orthogonal to a1. The unit binormalvector

a2 = bj ej = gradf2∣gradf2∣ (D.29)

is defined to be the unit vector chosen to be in the direction of the linear combi-nation of (δnj/δs + κtj)ej , i.e.,

bj = 1

τ(δnj

δs+ κtj) , (D.30)

D. CURVES AND SURFACES IN SPACE 453

where τ is a scalar called the torsion. The reciprocal of the torsion is calledthe radius of torsion. The sign of the torsion is selected such that the vectorsa1,a2,a3 form a right-handed coordinate system:

a1 ⋅ (a2 × a3) = 1 or εijknibjtk = 1, (D.31)

and the magnitude is selected such that bi is a unit vector satisfying

a2 ⋅ a2 = gijbibj = 1. (D.32)

By using (D.30) it is easily shown that a2 is orthogonal to both a1 and a3 since

a2 ⋅ a1 = gijbinj = 0 and a2 ⋅ a3 = gijbitj = 0. (D.33)

The vectors a1,a2,a3 form a right-handed orthogonal system at a point on thespace curve and satisfy the relation

a2 = a3 × a1, or bi = εijktjnk. (D.34)

Additionally, it is easy to show that the binormal vector satisfies the relationδa2/δs = −τa1.

The triad of vectors a1,a2,a3 form three planes at a point on the curve C .The plane containing a2 and a3 is called the rectifying plane. The plane containinga1 and a2 is called the normal plane. The plane containing a1 and a3 is called theosculating plane. The torsion measures the rate of change of the osculating plane.The three relations

δa1

δs= τa2 − κa3, δa2

δs= −τa1, δa3

δs= κa1, (D.35)

or

δni

δs= τbi − κti, δbi

δs= −τni,

δti

δs= κni, (D.36)

are known as the Frenet–Serret formulas.In general, instead of considering the arc distance as a coordinate, we can take

r = xi(ξ)ei to represent the three-dimensional space curve as a function of theconvected coordinate ξ along the curve C . In such case, we have the correspondingvectors d1,d2,d3 where

d1 = a1, d2 = a2, andd3(d33)1/2 = a3, (D.37)

whered33 = d3 ⋅ d3 (D.38)

is not unity in this case, and

δ

δs= δξδs

δ

δξ= 1(d33)1/2

δ

δξ. (D.39)

Subsequently, the Frenet–Serret formulas become

δd1

δξ= τ(d33)1/2d2 − κd3,

δd2

δξ= −τ (d33)1/2d1,

δd3

δξ= κd33 d1 +

1

2d33

d(d33)dξ

d3, (D.40)

454 APPENDIX D

or

δni

δξ= τ(d33)1/2bi − κti, δbi

δξ= −τ (d33)1/2ni,

δti

δξ= κ (d33)1/2ni

+1

2d33

d(d33)dξ

ti. (D.41)

Now taker = r(ξ, t)d3 + θ

α(ξ, t)dα, (D.42)

where r denotes the component of r in the d3 direction, and θα(ξ, t) denotes thecomponent of r in the dα direction. We use the convention that Greek subscriptsor superscripts only run from 1 to 2, and take dα to have units of length so thatθα is dimensionless. Now

v = r = u +w = u(ξ, t)d3 +wα(ξ, t)dα, (D.43)

whereu(ξ, t) = r(ξ, t) and wα(ξ, t) = θα(ξ, t), (D.44)

and the superposed dot denotes the material derivative.

D.2 Balance law for a space curve

Consider an arbitrary material curve C (t) ∶ ξ1(t) ≤ ξ ≤ ξ2(t), which is separatedinto two parts C +(t) and C −(t), or C (t) − γ(t), by a discontinuity located atξ1(t) < γ(t) < ξ2(t). The singularity at γ(t) moves with velocity γ = γ d3. Thegeneral balance law for the material curve is given by

d

dt∫

C (t)ψ(ξ, t)dξ = [φ(ξ, t)]ξ2(t)

ξ1(t) + ∫C (t)

g(ξ, t)dξ, (D.45)

or more specifically,

d

dt∫

C (t)−γ(t)ψl(ξ, t)dξ + ψp(γ, t) = [φ(ξ, t)]ξ2(t)ξ1(t) + ∫

C (t)−γ(t)gl(ξ, t)dξ + gp(γ, t),

(D.46)where ψ(ξ, t) is an additive tensor quantity per unit length, φ(ξ, t) is the flux of ψthrough the curve’s endpoints, and g(ξ, t) denotes the combined external supplyand internal production of ψ. Their specific contributions should be distinguishedwhen the general balance law is applied to different physical balances since theexternal supply and internal production represent different physical contributions.The tensors are decomposed appropriately between the regions C (t)−γ(t), denotedby the subscript l, and γ(t), denoted by the subscript p.

Now using the generalized Leibnitz rule (3.450) and the generalized divergencetheorem on a curve (2.301), we can rewrite the above in the following form:

∫C (t)−γ(t)

[∂ψl

∂t+∂(ψlv)∂ξ

−∂φl

∂ξ− gl]dξ +

ψp(γ, t) + Jψl(γ, t) [v(γ, t) − γ(t)] − φl(γ, t)K − gp(γ, t) = 0. (D.47)

D. CURVES AND SURFACES IN SPACE 455

Evaluating the balance law over an arbitrary point in the continuous region C (t)−γ(t), we deduce the local form of the balance law for the tensor quantity ψl overthe region:

∂ψl

∂t+∂(ψlv)∂ξ

−∂φl

∂ξ− gl = 0. (D.48)

Alternatively, evaluating the balance law over the singular region γ(t), we havethe evolution equation for the quantity ψp defined at the singular point:

ψp(γ, t) + Jψl(γ, t) [v(γ, t) − γ(t)] − φl(γ, t)K − gp(γ, t) = 0. (D.49)

D.3 Space surface

A surface embedded in E3 can be represented by the scalar equation

f(xk, t) = 0. (D.50)

From now until near the end of this section, we suppress the dependence of thesurface on the parameter t (which indicates time), since when considering thegeometry of a surface, the value of t is fixed. Then the unit vector normal to thesurface is given by

n = gradf∣gradf ∣ , n ⋅n = 1, (D.51)

with components

ni = n ⋅ ei = gijf,j∣gradf ∣ , ni = n ⋅ ei = f,i∣gradf ∣ . (D.52)

Alternatively, if we define a surface coordinate system, with coordinates (y1, y2),say, then a point on the surface can be represented as a function of these surfacecoordinates:

r = xi(y1, y2)ei = xi(y1, y2)ei. (D.53)

Note that for (y1, y2) to uniquely determine a point on the surface, y1 should bea coordinate curve on the surface along which y1 varies while y2 is fixed, and y2

a coordinate curve on the surface along which y2 varies while y1 is fixed. For anysurface, there are an infinite number of coordinate systems that might be used.Any two families of lines may be chosen as coordinate curves, as long as eachmember of one family intersects each member of the other at one and only onepoint.

We now introduce a modification to our index notation. An italic index willcontinue to indicate a value from 1 to 3, while a Greek index will range from 1 to2 only. Then, the equations of a space surface can be written in the parametricform xi = xi(yα), were yα is called a curvilinear coordinate of the surface. Nowthe basis a1,a2, known as the surface natural basis, corresponds to vectors thatare tangent to the coordinate system (y1, y2) lying on the surface, and is relatedto the spatial natural basis e1,e2,e2 in E3 with spatial coordinates (x1, x2, x3)by the transformation

aα = ∂r

∂yα= ∂xi∂yα

ei = aiαei, (D.54)

456 APPENDIX D

where it is noted that [aiα] is a 3×2 transformation matrix having a rank of 2. Thecomponents of the tangent vector to the coordinate curves defining the surface aregiven by

aiα = ei ⋅ aα, (D.55)

and can be viewed as either the components of a covariant surface vector or thecomponents of a contravariant spatial vector. Alternately, we could have chosenvectors on the surface that are normal to the coordinate lines as a basis: a1,a2.This surface reciprocal, or dual, basis, when normalized to unit length has theproperty that

aα ⋅ aβ = δαβ , (D.56)

where δαβ is the Kronecker delta symbol, corresponding to the components of the

two-dimensional rank-2 isotropic unit tensor 1. In this section, we will representtangential tensor fields by a superposed tilde symbol when written in bold notation.The obvious exceptions are surface coordinate bases and the associated surfacemetric tensor. All other tensor fields are understood to represent spatial fields.

Now since n is a unit vector normal to the surface, we have aα ⋅ n = aβ ⋅ n = 0.Because the vector fields a1,a2,n or a1,a2,n are linearly independent, theyform a basis for spatial vector fields on the surface. Note that the normal vectorand its negative are both orthogonal to a1 and a2. We choose n so that a1,a2,nforms a right-handed system. Subsequently, we can decompose ei with respect tothe bases a1,a2,n:

ei = aiαaα + nin. (D.57)

We now define the surface metric tensor as

h = hαβaαaβ = hαβaαaβ , (D.58)

where the components are given by

hαβ = aα ⋅ aβ and hαβ = aα ⋅ aβ . (D.59)

Furthermore, we note that the matrix of the components is symmetric: [hαβ] =[hβα]. We also define the determinant of the surface metric components matrix:

h = det[hαβ]. (D.60)

It is easy to show that the determinant can be obtained, respectively, by expandingby columns or rows:

ǫαβhαγhβδ = hǫγδ or ǫαβhγαhδβ = hǫγδ, (D.61)

where the two-dimensional Levi–Civita, or permutation, symbol is defined as

ǫαβ = ǫαβ =⎧⎪⎪⎪⎨⎪⎪⎪⎩

1 if αβ = 12,−1 if αβ = 21,0 if α = β, (D.62)

or in matrix form

[ǫαβ] = [ǫαβ] = [ 0 1

−1 0] . (D.63)

D. CURVES AND SURFACES IN SPACE 457

The absolute tangential Levi–Civita tensor is of rank 2 and is given by

ε = εαβaαaβ = εαβaαaβ , (D.64)

where the contravariant and covariant components are related to the permutationsymbols as follows:

εαβ ≡ 1√hǫαβ and εαβ ≡√hǫαβ . (D.65)

Now, sincedxi = aiαdyα, (D.66)

then a small change in dyα on the surface coordinates results in change dxi inthe space coordinates. Hence, an element of arc length on the surface can berepresented in terms of the curvilinear coordinates of the space or curvilinearcoordinates of the surface:

ds2 = gij dxidxj = gij aiαajβdyαdyβ = hαβ dyαdyβ, (D.67)

from which we can relate the spatial and surface metrics:

gij aiαa

jβ= hαβ. (D.68)

The quadratic scalar A = hαβ dyαdyβ written in surface coordinates is also calledthe first fundamental form of the surface, while the metric tensor h is also called thefirst fundamental tensor of the surface. The first fundamental form is connectedwith distance on the surface.

We are particularly interested in describing vector fields defined on a surface. Atangential vector field is a two-dimensional subspace of the spatial vector field onthe surface in the sense that every tangential vector field v can be expressed as alinear combination of a1 and a2 or their duals:

v = vαaα = vβaβ . (D.69)

Now dotting the above equation from the right with aγ , we see that

vγ = hαγvα. (D.70)

On the other hand, dotting both sides from the right by aγ , we have that

vγ = hαγvα. (D.71)

The above results can be easily shown to remain valid when operating on thesurface bases vectors:

aγ = hαγaα and aγ = hαγaα. (D.72)

Subsequently, it is easy to verify that

hαβhβγ = δγα, (D.73)

[hγβ]−1 = [hβγ], (D.74)

458 APPENDIX D

and

h ⋅h−1 = h−1 ⋅ h = 1. (D.75)

If we define the surface gradient by

∇ ≡ aα ∂

∂yα= hαβaβ ∂

∂yα, (D.76)

then it is clear that the reciprocal basis is given by the surface gradient of thesurface coordinates:

aα = ∇yα = hβγaγ ∂yα∂yβ

= hαγaγ . (D.77)

We will now require that physical quantities described by scalars, vectors, andhigher order tensors remain invariant when a different surface curvilinear coordi-nate system is used, i.e., when a new basis aα is used instead of aα. Notethat corresponding to the new basis, we also have its reciprocal basis aβ, whereas before we require that aα ⋅ a

β = δβα. Now since a surface vector can be writtenin the alternate forms

v = vαaα = vβaβ = vγaγ = vδaδ, (D.78)

it is easy to show that the covariant and contravariant surface bases transform as

aγ = ∂yα∂yγ

aα and aδ∂yβ

∂yδ= aβ (D.79)

as long as the transformation is non-singular, i.e.,

det [∂yα∂yγ] ≠ 0,±∞. (D.80)

It should be recognized that the natural basis is orthogonal if aα ⋅aβ = 0 when α ≠β. In such case, it is usually more convenient to work in terms of an orthonormalbasis:

a<α> = aα√hαα

= aα√hαα

=√hααaα, (D.81)

where we recall that underlined subscripts are not summed. This basis is referredto as the physical basis for the surface coordinate system. Subsequently, any vectorfield can be expressed in the orthogonal coordinate system in terms of the physicalsurface components

v = v<α>a<α>, (D.82)

where

v<α> =√hααvα = vα√hαα

. (D.83)

A surface tensor S is a particular type of second-order spatial tensor field that isdefined only on the surface, and assigns to each given tangential vector field v on

D. CURVES AND SURFACES IN SPACE 459

a surface another spatial vector field S ⋅ v defined on the surface, and transformsevery spatial vector field normal to the surface into the zero vector. A surfacetensor can be defined by the way it transforms the surface natural basis fielda1,a2:

S ⋅ aα = Siαei. (D.84)

It is easy to show that

S = Siαeia

α = Siαeiaα = Siαeiaα, (D.85)

where

Siα = hαβSiβ and Siα = gijSj

α. (D.86)

Now if we take the surface tensor S = aiαeiaα, the surface vector v can also beviewed as a spatial vector v:

S ⋅ v = (aiαeiaα) ⋅ (vβaβ) = aiαvαei = viei = v.Thus, the relation between the spatial and surface representations is

vi = aiαvα. (D.87)

The surface and spatial representations define the same magnitude and directionsince

v ⋅ v = gij vivj = gij aiαvαajβvβ = gij aiαajβvαvβ = hαβ vαvβ = v ⋅ v. (D.88)

A tangential tensor T is a second-order tangential transformation that projectsevery tangential vector field v on a surface to another tangential vector field T ⋅ v

on the surface and every spatial vector field normal to the surface into the zerovector. The tangential tensor has the standard form

T = Tαβaαaβ = Tαβaαaβ . (D.89)

As in the spatial case, it is easy to show that the covariant and contravariantcomponents of this field transform as

Tµν = ∂yα∂y

µ

∂yβ

∂yν Tαβ and

∂yα

∂yµ

∂yβ

∂yν T

µν = Tαβ. (D.90)

It is usually convenient to introduce an orthogonal surface coordinate system. Ifa<α> is the associated physical basis field, then in terms of physical surfacecomponents, we can write

T = T<γµ>a<γ>a<µ>, (D.91)

where

T<γµ> =√hγγ√hµµT γµ = Tγµ√hγγ√hµµ

. (D.92)

460 APPENDIX D

Now if v is any tangential vector field

v = vαaα, (D.93)

then the tangential vector field that has the same length as v and is orthogonalto it is given by

u = −ε ⋅ v or uα = −εαβvβ . (D.94)

This is easily seen, since

u ⋅ u = (−εαβvβaα) ⋅ (−εγζvζaγ)= εαβεαζvβv

ζ

= δβζvβv

ζ

= v ⋅ v,

and

u ⋅ v = (−εαβvβaα) ⋅ (vγaγ)= −εαβvαvβ

= −1

2(εαβ + εβα)vαvβ

= 0.

Because of this last property, ε is referred to as the tangential cross tensor. Noticethat

(−ε ⋅ a1) ⋅ a2 = εαβhα1hβ2 =√h. (D.95)

It is also easy to show that

−ε ⋅ a1 =√h a2 and ε ⋅ a2 =√h a1, (D.96)

which imply that

∣a2∣ = 1√h∣a1∣ , (D.97)

that a2 is orthogonal to a1, and that the rotation from a1 to a2 is positive. Simi-larly, we have the implications that

∣a1∣ = 1√h∣a2∣ , (D.98)

that a1 is orthogonal to a2, and that the rotation from a2 to a1 is negative.Consider any two surface vectors with components uα and vβ and their spatial

representations ui and vj , where

ui = aiαuα and vj = ajβvβ . (D.99)

These vectors are tangent to the surface and so a unit normal vector to the surfacecan be defined. Actually, the are two normals: they are the negative of each other.

D. CURVES AND SURFACES IN SPACE 461

The normal n is chosen to correspond to the surface normal, which, togetherwith a1 and a2, forms a right-handed coordinate system: (a1 × a2) ⋅ n > 0. Moreexplicitly, from the cross product relations

εαβuαvβni = εijkujvk, (D.100)

which, for arbitrary surface vectors, implies

ni = 1

2εαβεijka

jαa

kβ , (D.101)

we have the definition of a surface unit normal vector in terms of the tangentvectors to the coordinate curves. It is readily seen that

gijninj = 1. (D.102)

The surface projection tensor P is a second-order tangential tensor field thattransforms every tangential vector field into itself:

P ⋅ aβ = aβ = δαβaα. (D.103)

It is easy to see that

P = aαaα = aαaα = hαβaαaβ = hαβaαaβ . (D.104)

Note that the covariant and contravariant components of the surface metric tensorh can also be viewed as covariant and contravariant components of the projectiontensor P.

Spatial vector fields defined on a surface play an important role in continuummechanics. Thus it is often convenient to think in terms of their tangential andnormal components. Subsequently, since the vectors a1, a2 and n are linearlyindependent, they form a basis for a spatial vector field on the surface:

v = v + v(n) = vαaα + v(n)n = vαaα + v(n)n, (D.105)

where v(n) is the normal component of the spatial vector field v. Now, this factcan also be represented in the form

v = (v ⋅ aα)aα + (v ⋅ n)n, (D.106)

or

1 ⋅ v = (P +P(n)) ⋅ v, (D.107)

where 1 is the spatial unit second-order tensor and

P(n) = nn (D.108)

is the normal projection tensor, i.e., it projects a spatial vector field on a surfaceto one in the direction normal to the surface. Since v is an arbitrary spatial vectorfield defined on the surface, this implies that

1 = P +P(n), (D.109)

462 APPENDIX D

and thus

v = v + v(n), where v = P ⋅ v and v(n) = P(n) ⋅ v. (D.110)

Subsequently, we can also write the surface projection tensor in the alternate form

P = 1 −P(n). (D.111)

The contravariant component of this relationship is given by

hαβaiαajβ= gij − ninj . (D.112)

It is important to note that the projection tensor plays the role of the identity orunit tensor for the set of all tangential vector fields. A tensor P is a projectiontensor if it is symmetric and Pm = P for m a positive integer. From above, it canbe readily verified that our projection tensors have the properties

P +P(n) = 1, (D.113)

P ⋅ P = P, (D.114)

P(n) ⋅P(n) = P(n), (D.115)

P ⋅P(n) = 0. (D.116)

The normal unit vector is related to the covariant derivative of the surfacetangents as will be shown next. If we take the covariant derivative of (D.68) withrespect to the surface coordinate yγ and use Ricci’s theorem (2.248), we have

gij aiα,γa

jβ+ gij a

iαa

jβ,γ= hαβ,γ = 0. (D.117)

Interchanging the indices α and β, it is easy to see that

gijaiα,βa

jγ = 0. (D.118)

From the definition of the covariant derivative, we note that

aiα,β = ∂aiα

∂yβ+ Γi

jkajαa

kβ − Γ

γαβaiγ , (D.119)

and that we have two Christoffel symbols, one related to the spatial metric and theother related to the surface metric. Analogous to (2.218), the surface Christoffelsymbol can be written as

Γδαβ = aδ ⋅ ∂aα∂yβ

= 1

2hδγ (∂hαγ

∂yβ+∂hβγ

∂xα−∂hαβ

∂xγ) . (D.120)

The result (D.118) indicates that in terms of space coordinates, the vector aiα,β is

orthogonal to the surface tangent vector ajγ and so must have the same direction

as the unit surface normal ni. Therefore, there must exist a tensor

B = Bαβaαaβ = Bαβaαaβ = Bα

βaαaβ (D.121)

whose components Bαβ are such that

aα,β = Bαβ n or aiα,β = Bαβ ni. (D.122)

D. CURVES AND SURFACES IN SPACE 463

This second-order symmetric tensor is called the curvature tensor, or the secondfundamental form of the surface. It is connected with the rate of change of thetangent vectors. By using (D.101) and (D.102), we can rewrite the above equationin the form

Bαβ = gijaiα,βnj = 1

2εγδεijka

iα,βa

jγa

kδ . (D.123)

Now take the covariant derivative of (D.102) with respect to the surface coordi-nates and use Ricci’s theorem (2.248). From it, it readily follows that

gijninj

,α = 0, (D.124)

where

ni,α = ∂n

i

∂yα+ Γi

jknjakα. (D.125)

The above result shows that the vector ni,α is orthogonal to ni and must lie in the

tangent plane to the surface. It can therefore be expressed as a linear combinationof the surface tangent vector components aiα and written in the form

ni,α = ηβαaiβ , (D.126)

where the coefficients ηβα can themselves be written in terms of the surface metriccomponents hαβ and the curvature components Bαβ . To see this, first recall thatthe unit vector ni is normal to the surface so that

gijniajα = 0. (D.127)

The covariant derivative of this equation with respect to the surface coordinates,upon using Ricci’s theorem (2.248), gives

gijni,βa

jα + gijn

iajα,β= 0. (D.128)

Substituting (D.68), (D.123), and (D.126) in the above equation, we have

Bαβ = −hαγηγβ , (D.129)

or upon solving for the coefficients ηγβ, we find

ηγβ= −hγαBαβ . (D.130)

Substituting this result into (D.126) produces what is known as the Weingartenformula:

ni,α = −hβγBγαa

iβ . (D.131)

This is a relation for the covariant derivative along the surface of the unit normalto the surface in terms of the surface metric, the curvature tensor, and surfacetangents.

A third fundamental form of the surface, connected with the rate of change ofthe normal vector, is given by the symmetric surface tensor

C = Cαβaαaβ , (D.132)

464 APPENDIX D

whereCαβ = gijni

,αnj,β. (D.133)

By using the Weingarten formula and (D.68), the components can be rewritten as

Cαβ = hγδbαγbβδ. (D.134)

We would like to investigate the properties of the Riemann–Christoffel tensorgiven in (2.237) or (2.238) in a two-dimensional space with metric hαβ and coor-dinates yα. In this case, Rαβγδ has only four nonzero components. Furthermore,these four components are either +R1212 or −R1212, since using the symmetry con-ditions (2.241) in the two-dimensional space, it is easy to see that they are allrelated:

R1212 = −R2112 = R2121 = −R1221. (D.135)

From this, it follows that we can write it in terms of the components of the two-dimensional absolute Levi–Civita tensor:

Rαβγδ =KGεαβεγδ. (D.136)

The surface scalar invariantKG is called the Gaussian curvature or total curvature.Now consider the two-dimensional form of the curvature equation (2.236) applied

to xi = xi(yα):aα,βγ − aα,γβ = aδRδ

αβγ or aiα,βγ − aiα,γβ = aiδRδ

αβγ . (D.137)

Using this relation, we now derive interesting relations connected with surfaceproperties. The covariant derivative of (D.122) with respect to surface coordinatesis given by

aiα,βγ = Bαβ,γni+Bαβn

i,γ , (D.138)

where

Bαβ,γ = ∂Bαβ

∂yγ− Γδ

αγBδβ − ΓδβγBαδ. (D.139)

By using the Weingarten formula (D.131), we can rewrite the above equation inthe form

aiα,βγ = Bαβ,γni−Bαβh

σδBσγaiδ, (D.140)

or, upon using (D.137) and (D.139), we see that

aiα,βγ−aiα,γβ = (Bαβ,γ −Bαγ,β)ni

−hσδ (BαβBσγ −BαγBσβ)aiδ = aiδRδαβγ , (D.141)

or

aα,βγ −aα,γβ = (Bαβ,γ −Bαγ,β)n−hσδ (BαβBσγ −BαγBσβ)aδ = aδRδαβγ . (D.142)

Multiplying by gijnj and using (D.102) and (D.127), we obtain what is known as

the Mainardi–Codazzi equation:

Bαβ,γ −Bαγ,β = 0. (D.143)

On the other hand, multiplying (D.141) by gijajσ, using (D.127), and simplifying,

we obtain the Gauss equations of the surface:

Rσαβγ = BαγBσβ −BαβBσγ . (D.144)

D. CURVES AND SURFACES IN SPACE 465

By using the Gauss equations, the equation for the Gaussian curvature (D.136)can be rewritten as

KGεσαεβγ = BαγBσβ −BαβBσγ , (D.145)

orKG = det [Bα

β ]. (D.146)

Still another form for the Gaussian curvature can be obtained by using (D.134)together with the relation hαβ = −εσαεβγhσγ , which can be easily verified,

−KGhαβ = Cαβ − hσγBσγBαβ . (D.147)

If we define the scalar invariant called the mean surface curvature by

KM = 1

2hσγBσγ = 1

2Bσ

σ , (D.148)

then the above equation becomes a relationship between the three surface funda-mental tensor forms:

C − 2KMB +KGh = 0 or Cαβ − 2KMBαβ +KGhαβ = 0. (D.149)

The surface gradient of a spatial vector field v is given by

∇v = ∂v

∂yαaα. (D.150)

Note that if r = r(yβ) = yβaβ is the surface position vector field, then

∇r = ∂r

∂yαaα = aαaα = P. (D.151)

This provides an additional useful expression for the surface projection tensor. Thesurface divergence of the spatial vector field is naturally given by the contraction

∇ ⋅ v = tr (∇v) = ∂v

∂yα⋅ aα. (D.152)

In the above expressions, the spatial vector field v may be an explicit function ofposition in space or it may be an explicit function of position on the surface. Aswill be seen, these lead to different expressions.

If v = vi(xj)ei = vi(xj)ei, then

∇v = ∂v∂xi

aiαaα = ∂v

∂xi(ei ⋅ aα)aα = P ⋅∇v. (D.153)

For v written using contravariant and covariant components, the correspondingexpressions are

∇v = vj,iaiαejaα = vj,iaiαejaα. (D.154)

Subsequently, we can also write

∇ ⋅ v = tr (∇v) = tr (P ⋅∇v) , (D.155)

466 APPENDIX D

with corresponding expressions when using contravariant and covariant compo-nents for v:

∇ ⋅ v = vj,iaiαakβgjkhαβ = vj,iaiαajβhαβ . (D.156)

On the other hand, if w = wi(xj(yα))ei = wi(xj(yα))ei, then from (D.2) wehave

∇w = wi,αeia

α, (D.157)

where

wi,α = ∂w

i

∂yα+ Γi

jkajαw

k, (D.158)

which is the surface covariant derivative of wi, or from (D.1) we have

∇w = wi,αeiaα, (D.159)

where

wi,α = ∂wi

∂yα− Γk

jiajαwk, (D.160)

which is the surface covariant derivative of wi.The corresponding expressions for the surface divergence of w are

∇ ⋅w = wi,αa

jβgijh

αβ = wi,αaiβh

αβ . (D.161)

It is also useful to write w in terms of tangential and normal components:

w =w ⋅ 1 =w ⋅ P +w ⋅P(n) =w ⋅ aαaα +w ⋅ nn = wαaα +w(n)n, (D.162)

or equivalently

w = wαaα+w(n)n. (D.163)

In this case, the surface gradient becomes either

∇w = ∂wα

∂yβaαa

β+wα ∂aα

∂yβaβ +

∂w(n)∂yβ

naβ +w(n)∂n

∂yβaβ (D.164)

or

∇w = ∂wα

∂yβaαaβ +wα

∂aα

∂yβaβ +

∂w(n)∂yβ

naβ +w(n)∂n

∂yβaβ . (D.165)

Now

∂aα

∂yβ= ∂

∂yβ(aiαei)

= ∂aiα∂yβ

ei + aiαa

jβ

∂ei

∂xj

= ∂aiα∂yβ

ei + aiαa

jβΓkjiek

= (∂aiα∂yβ+ a

jβakαΓ

ijk)ei, (D.166)

D. CURVES AND SURFACES IN SPACE 467

and in addition

ei = 1 ⋅ ei

= (P +P(n)) ⋅ ei= (aγaγ + nn) ⋅ ei= (aγhγδalδel + nn) ⋅ ei= hγδgila

lδaγ + nin. (D.167)

Combining the last two expressions, we have

∂aα

∂yβ= Γγ

βαaγ +Bβαn, (D.168)

where

Γγβα= (∂aiα

∂yβ+ a

jβakαΓ

ijk)hγδgilalδ (D.169)

is the surface Christoffel symbol of the second kind, and

Bβα = (∂aiα∂yβ+ a

jβa

kαΓ

ijk)ni = Bαβ (D.170)

are the components of the symmetric second fundamental form tangential tensorfield.

Now note that since aγ ⋅ n = 0 and n ⋅ n = 1, upon differentiating we have

aγ ⋅∂n

∂yβ= −Bγβ and n ⋅

∂n

∂yβ= 0. (D.171)

Thus we see that ∂n/∂yβ is the tangential vector field

∂n

∂yβ= −Bγβa

γ . (D.172)

Similarly, since aγ ⋅ aα = δαγ and n ⋅ aα = 0, upon differentiating we have

aγ ⋅∂aα

∂yβ= −∂aγ

∂yβ⋅ aα = −Γδ

βγaδ ⋅ aα = −Γα

βγ (D.173)

and

n ⋅∂aα

∂yβ= − ∂n

∂yβ⋅ aα = Bγβh

γα, (D.174)

thus concluding that∂aα

∂yβ= −Γα

βγaγ+ hαγBβγn. (D.175)

With (D.168) and (D.172), we can rewrite (D.164) in a simpler form:

∇w = ∂wα

∂yβaαa

β+ Γ

γβαwαaγa

β+Bβαw

αnaβ +∂w(n)∂yβ

naβ −Bγβw(n)aγaβ

= [(wα,β − h

γαBγβw(n))aα + (∂w(n)∂yβ

+Bβαwα)n]aβ

= wα,βaαa

β−w(n)B + n (∇w(n) + B ⋅w)

= P ⋅ ∇ (P ⋅w) −w(n)B + n (∇w(n) + B ⋅w) , (D.176)

468 APPENDIX D

where we have introduced the surface covariant derivative of wα

wα,β = ∂w

α

∂yβ+ Γα

βγwγ (D.177)

and have noted the fact that

P ⋅ ∇(P ⋅w) = wα,βaαa

β . (D.178)

Analogously, using (D.172) and (D.175), we can rewrite (D.165) in the form:

∇w = [(wα,β −Bαβw(n))aα + (∂w(n)∂yβ

+Bβαwα)n]aβ

= wα,βaαaβ −w(n)B + n (∇w(n) + B ⋅w) , (D.179)

where we have introduced the surface covariant derivative of wα,

wα,β = ∂wα

∂yβ− Γ

γβαwγ . (D.180)

Now the corresponding expressions for the surface divergence of w are

∇ ⋅w = wα,α −w(n)h

αβBαβ = wα,α − 2KMw(n), (D.181)

where KM is the mean surface curvature. From the above result, it is now easilyseen that

B = −∇n (D.182)

and

KM = 1

2∇ ⋅ n. (D.183)

There is another measure of surface curvature other than the total and meancurvatures. It is the normal curvature in the λ direction,

κn = λ ⋅ B ⋅λ, (D.184)

where λ is a unit vector tangent to the surface at the point where the normalcurvature is measured. Principal curvatures κ1 and κ2 are the maximum andminimum values of the normal curvature κn. It can be shown that the directionλ, which corresponds to one of the principal curvatures, satisfies

(B − κnP) ⋅λ = 0, (D.185)

or if λ is nonzero, then the curvatures must satisfy the characteristic equation

κ2n − tr[B]κn + det[B] = 0. (D.186)

From this, it can be shown that

κ1 + κ2 = 2KM , κ1κ2 =KG, and ∣B∣2 = tr[B ⋅ B] = 4K2

M − 2KG. (D.187)

As with vector fields, the surface gradient of a rank-2 spatial tensor field T isgiven analogously by

∇T = ∂T∂yα

aα. (D.188)

D. CURVES AND SURFACES IN SPACE 469

Actually, this definition is the same if T is a spatial tensor field of any rank. Thesurface divergence is obtained by contraction:

∇ ⋅T = ∂T∂yα⋅ aα. (D.189)

Also as before, T can be expressed as an explicit function of space coordinates oran explicit function of surface coordinates. If T = T ij(xi)eiej = Tij(xi)eiej , thenthe surface gradient is given by

∇T = P ⋅∇T = T jk,i a

iαejeka

α = Tjk,iaiαejekaα, (D.190)

and the corresponding surface divergence by

∇ ⋅T = T jk,i a

mα a

iβgkmh

αβej = Tjk,iakαaiβhαβej . (D.191)

On the other hand, if T = Tαβ(yγ)aαaβ = Tαβ(yγ)aαaβ , then the surface gradientis given by

∇T = Tαβ,γ aαaβa

γ+ TαβBαγnaβa

γ+ TαβBβγaαna

γ (D.192)

= Tαβ,γaαaβaγ + TδβBαγh

δαnaβaγ + TαδBβγhδβaαnaγ , (D.193)

where the surface covariant derivatives of Tαβ and Tαβ are given by

Tαβ,γ = ∂Tαβ

∂yγ+ Γα

γδTδβ+ Γ

βγδTαδ, (D.194)

Tαβ,γ = ∂Tαβ

∂yγ− Γδ

γαTδβ − ΓδγβTαδ. (D.195)

The corresponding surface divergence of T takes the forms

∇ ⋅ T = Tαβ,β

aα + TαβBαβn (D.196)

= Tαβ,γhβγaα + TδβBαγh

δαhβγn. (D.197)

We will have occasion to perform integration over the surface coordinates y1

and y2. At any point on the surface S , the unit tangent vectors to the twosurface coordinate curves are a1 = ∂r/∂y1 and a2 = ∂r/∂y2. We take da to be thedifferential area of the parallelogram formed from a1 and a2 with sides of lengthdy1 and dy2, so that using (D.96) we have

da = a1 ⋅ (ε ⋅ a2)dy1dy2 =√h dy1dy2, (D.198)

and subsequently,

∫S

F (y1, y2)da =∬S

F (y1, y2)√hdy1dy2. (D.199)

We also need to know the rate of change with respect to time of some geometricalobjects and physical quantities defined on a surface. If we differentiate the equationf(x, t) = 0 with respect to time, then

d

dtf(x, t) = f(x, t) = ∂f

∂t+∂x

∂t⋅∇f = ∂f

∂t+ v ⋅∇f = 0, (D.200)

470 APPENDIX D

where v is the velocity of a particle located at x (or equivalently y) on the surface:

v ≡ x = ∂x(yα, t)∂t

∣yα

, (D.201)

so that x(yα, t) is the material derivative on the surface, or the derivative ofx(yα, t) with respect to t holding yα fixed. It is useful to decompose the velocityfield into tangential and normal components (see Problem 12):

v = vαaα + v(n)n or vi = vαaiα + v(n)ni, (D.202)

where

v(n) = v ⋅ n = − ∂f/∂t∣gradf ∣ . (D.203)

Now the velocity of the surface determines how the surface metric tensor h changeswith time, because

aβ = ∂

∂t( ∂x∂yβ) = ∂v

∂yβ. (D.204)

Subsequently, using (D.176), we have

aβ = ∂vα

∂yβaα + v

α ∂aα

∂yβ+∂v(n)∂yβ

n + v(n)∂n

∂yβ

= (vα,β − hγαBγβv(n))aα + (∂v(n)∂yβ

+Bβαvα)n, (D.205)

or equivalently, using (D.179),

aβ = (vα,β −Bαβv(n))aα + (∂v(n)∂yβ

+Bβαvα)n. (D.206)

Now, since the surface metric hβδ = aβ ⋅ aδ is symmetric, the rate of change of thesurface metric tensor is given by

hβδ = 2aβ ⋅ aδ = 2 (vα,βhαδ −Bδβv(n)) = 2 (vδ,β −Bδβv(n)) . (D.207)

In particular, if the surface particles all move in the normal direction, the rate ofchange of the surface metric tensor is determined by the curvature tensor and thenormal speed. In addition, we have that

h = ∂h

∂hβδhβδ = hhβδhβδ = 2h (vα,α − 2KMv(n)) . (D.208)

Subsequently,

˙da = h

2√hdy1dy2 = (vα,α − 2KMv(n))da. (D.209)

This result shows that if the particles move normal to the surface, then da isunchanged if the surface is flat or at rest.

D. CURVES AND SURFACES IN SPACE 471

Next, we would like to obtain the equivalent relation as the spatial divergencetheorem, but now applied on the surface S . First, it is easy to show that if wehave the tangential vector field w = wαaα, then its surface divergence is given by

∇ ⋅ w = wα,α = 1√

h

∂

∂yα(√hwα) . (D.210)

Subsequently, using (D.199), we have

∫S

∇ ⋅ w da = ∬S

wα,α

√hdy1dy2

= ∬S

∂(√h wα)∂yα

dy1dy2

= ∬S

[∂(√h w1)∂y1

+∂(√h w2)

∂y2]dy1dy2. (D.211)

Now the Gauss–Green theorem for a surface tells us that if P (yα) and Q(yα)are continuous functions having continuous first partial derivatives on the surface,then

∬S

[ ∂P∂y1+∂Q

∂y2]dy1dy2 = ∫

C

(P dy2ds−Q

dy1

ds) ds, (D.212)

where C is a piecewise-smooth simple closed curve bounding S , and s indicatesthe arc length measured along this curve in the positive direction. Subsequently,using the Gauss–Green theorem, we have

∫S

∇ ⋅ w da = ∫C

(w1 dy2

ds−w2 dy

1

ds)√hds = ∫

C

εαβwα dy

β

ds

√hds. (D.213)

Now the unit tangent to the curve C is given by

λ = drds= ∂r

∂yβdyβ

ds= dyβds

aβ . (D.214)

It follows that the unit tangent vector field that is normal to the curve C is givenby

µ = ε ⋅ λ = εαβ dyβds

aα. (D.215)

We note that the rotation from λ to µ is negative. Because of the requirementthat s is measured along C in the positive sense, it is clear that µ is directedoutward with respect to the closed curve. Thus we can rewrite

∫S

∇ ⋅ w da = ∫C

w ⋅ µds = ∫C

w ⋅ ds. (D.216)

This result is known as the surface divergence theorem.If, on the other hand, v is a spatial vector field defined on the surface, from

(D.181) we have∇ ⋅ v = ∇ ⋅ v − 2KMv ⋅n. (D.217)

Subsequently, we arrive at the generalized surface divergence theorem:

∫S

∇ ⋅ v da = ∫C

v ⋅ µds −∫S

2KMv ⋅ nda = ∫C

v ⋅ ds −∫S

2KMv ⋅ da. (D.218)

472 APPENDIX D

In addition, using (D.194) and (D.195), it can be shown that in general

εαβ,γ = εαβ,γ = 0, (D.219)

so that with (D.101) it is not difficult to show that

niεijkvk,j = (εαβvβ),α . (D.220)

Subsequently, if (a1 × a2) ⋅ n > 0, λ and µ are unit tangent vectors such that onthe surface the rotation from λ to µ is positive, and λ is the unit tangent vectorto the curve C bounding S , then

∫S

(∇ × v)⋅nda = ∫S

∇⋅(ε ⋅ v) da = ∫C

(ε ⋅ v)⋅µds = ∫C

v⋅(−ε ⋅ µ)ds = ∫C

v⋅λ ds,

and thus we obtain Stokes’ theorem:

∫S

(∇ × v) ⋅ nda = ∫C

v ⋅ λds, or ∫S

(∇ × v) ⋅ da = ∫C

v ⋅ ds. (D.221)

We will also have occasion to integrate quantities along a curve C (t) on a surfaceso that r = r(yα). In such case, we consider the integrand to be given as an explicitfunction of y2. Subsequently, we have that the arc length s along the curve lyingon the surface is given by

ds = ( ∂r∂y2⋅∂r

∂y2)1/2 dy2 =√a2 ⋅ a2 dy2 =√h22 dy2. (D.222)

Then the line integral takes the form

∫C (t)

F (y2) ds = ∫C (t)

F (y2)√h22 dy2. (D.223)

Next, we would like to obtain the transport theorem of a tensor quantity onthe material surface S (t), which in the reference configuration is given by S.We take the surface density of the quantity to be ψ and note from (D.198) that

da = √h/H dA, where dA and H respectively denote the differential materialsurface and determinant of the metric tensor in the reference configuration. Then,using either (D.208) or (D.209), we have

d

dt∫

S (t)ψ da = d

dt∫Sψ

√h

HdA

= ∫S[ψ +ψ h

2√Hh]dA

= ∫S[∂ψ∂t+ vαψ,α +ψ (vα,α − 2KMv(n))]

√h

HdA,

ord

dt∫

S (t)ψ da = ∫

S (t)[∂ψ∂t+ (ψ vα),α − 2KMψ v(n)] da. (D.224)

This result represents the transport theorem for material surfaces.

D. CURVES AND SURFACES IN SPACE 473

C −

γ1(t)

γ2(t)

γ(t)S− S +

a+2a−2

v

a1

a2a3

µ

n

C+

c(2)a2

Figure D.1: Arbitrary material surface S intersected by a discontinuous surface.

D.4 Balance law for a flux through a space surface

Consider an infinitesimally thin material membrane having surface S (t), which isseparated into two parts S

+(t) and S−(t), or S (t) − γ(t), by a singular surface

whose intersection with S (t) is denoted by the line γ(t) as shown in Fig. D.1. Theunit normal vector to surfaces S +(t) and S −(t) is denoted by n. The parts ofthe boundary of S (t) that are not on the singular line γ(t) are denoted as C

+(t)and C

−(t), and their unit tangent vector by µ. The points on the boundary ofS (t) that are on the singular line γ(t) will be denoted by P(t) ∶ (γ1(t), γ2(t)).We denote by a1 the unit vector normal to the line γ(t), which is tangent to thesingular surface. The singular line moves on the surface S (t) with normal velocityc(2)a2. The unit vector tangent to the singular line γ(t) is denoted by a3 = a1×a2.The material through S (t) moves with particle velocity v.

The general balance statement of the flux of a tensor quantity Φ flowing througha surface S (t) is given in the form

dΦ

dt=H(Φ) + G(Φ), (D.225)

where H(Φ) denotes the external supply flux of Φ through the surface enclosedby the curve C (t), and G(Φ) denotes the combined external supply of Φ to thesurface and internal production of Φ within the surface. One should note that whilewe represent symbolically the sum of supply and production by a single term inthe present derivation, physically supply is different from production because itmay be controlled from the exterior of the surface. Subsequently, in applications,one should recognize this difference by writing their contributions separately. Weassume that additive surface densities of these tensor quantities exist, and denotethe corresponding quantities that are defined on the surface S (t)−γ(t) by a tilde,

474 APPENDIX D

i.e., φ, h, and g, while those that are defined only on the singular line γ(t) by anoverbar, i.e., φ, h, and g. Subsequently, (D.225) can be rewritten more explicitlyas

d

dt(∫

S (t)−γ(t)φ ⋅ nds + ∫

γ(t)φ ⋅ a3 dl) = (∫

C (t)−P(t)h ⋅ µdl + [h(γ(t))]γ2(t)

γ1(t)) +(∫

S (t)−γ(t)g ⋅nds + ∫

γ(t)g ⋅ a3 dl) . (D.226)

Now using (3.459), (3.436), and (2.300), after rearranging the terms, we obtain

∫S (t)−γ(t)

[∂φ∂t+ v div φ + curl (φ × v) − curl h − g] ⋅nds +

∫γ(t)[ ˙φ +φ ⋅ (gradv)T −∇h − g] ⋅ a3 + Jφ × (v − c) − hK ⋅ a2dl = 0. (D.227)

If we now apply the integral balance law over an arbitrary regular materialsurface, then the integrand of the first integral must be zero and thus we obtainthe local balance law:

∂φ

∂t+ v div φ + curl (φ × v) − curl h − g = 0. (D.228)

Subsequently, applying the integral balance law over the singular region, and sincea2 and a3 are orthogonal to each other, we obtain the jump condition across thesingular curve

Jφ × (v − c) − hK ⋅ a2 = 0, (D.229)

and the local balance law along the singular curve

˙φ +φ ⋅ (gradv)T −∇h − g = 0. (D.230)

Note that this last balance law along the singular curve is identical to the balancelaw (D.48) obtained for a curve in space if φ depends only on one spatial dimension,i.e., φ = φ(ξ, t).Problems

1. Given a Cartesian coordinate system and a plane surface ξ3 = const., choose

y1 = ξ1 and y2 = ξ2.a) Show that

h11 = 1

h11= h22 = 1

h22= 1, h12 = h21 = 0,

and

h = 1.b) Prove that all of the surface Christoffel symbols of the second kind are

zero.

D. CURVES AND SURFACES IN SPACE 475

c) Show that B = 0.

d) Conclude that KM =KG = 0.2. Given the cylindrical coordinate system

ξ1 = x1 cosx2 = r cos θ,ξ2 = x1 sinx2 = r sin θ,ξ3 = x3 = z,

and a plane surface ξ3 = const., choose

y1 = x1 = r and y2 = x2 = θ.a) Show that

h11 = 1

h11= 1, h22 = 1

h22= r2, h12 = h21 = 0,

and

h = r2.b) Prove that all of the surface Christoffel symbols of the second kind are

zero.

c) Show that B = 0.

d) Conclude that KM =KG = 0.3. With the cylindrical coordinates given in Problem 2 and a plane surface θ =

const., choose

y1 = x1 = r and y2 = x3 = z.a) Show that

h11 = 1

h11= h22 = 1

h22= 1, h12 = h21 = 0,

and

h = 1.b) Show that all surface Christoffel symbols of the second kind are zero.

c) Show that B = 0.

d) Conclude that KM =KG = 0.4. Given the cylindrical coordinate system defined in Problem 2 and a cylin-

drical surface of radius R, choose

y1 = x2 = θ and y2 = x3 = z.

476 APPENDIX D

a) Show that

h11 = 1

h11= R2, h22 = 1

h22= 1, h12 = h21 = 0,

and

h = R2.

c) Show that all surface Christoffel symbols of the second kind are zero.

b) Show that B11 = −R and B22 = B12 = B21 = 0.d) Conclude that

KM = − 1

2Rand KG = 0.

5. Given the spherical coordinate system

ξ1 = x1 sinx2 cosx3 = r sin θ cosφ,ξ2 = x1 sinx2 sinx3 = r sin θ sinφ,ξ3 = x1 cosx2 = r cos θ,

and a spherical surface of radius R, choose

y1 = x2 = θ and y2 = x3 = φ.a) Show that

h11 = 1

h11= R2, h22 = 1

h22= R2 sin2 θ, h12 = h21 = 0,

and

h = R4 sin2 θ.

b) Show that the only nonzero surface Christoffel symbols of the secondkind are

Γ1

22 = − sin θ cosθ, and Γ2

12 = Γ2

21 = cot θ.c) Show that B11 = −R, B22 = −R sin2 θ, and B12 = B21 = 0.d) Conclude that

KM = − 1R

and KG = 1

R2.

6. Given a Cartesian coordinate system and the surface

ξ3 = g(ξ1, t),choose

y1 = ξ1 and y2 = ξ2.

D. CURVES AND SURFACES IN SPACE 477

a) Show that

h11 = 1 + ( ∂g∂ξ1)2 , h22 = 1, h12 = h21 = 0,

and

h = 1 + ( ∂g∂ξ1)2 .

b) Show that the only nonzero surface Christoffel symbol of the secondkind is

Γ1

11 = ∂g

∂ξ1

∂2g

∂ξ21

[1 + ( ∂g∂ξ1)2]−1 .

c) Show that

B11 = ∂2g∂ξ1[1 + ( ∂g

∂ξ1)2]−1/2 ,

and B22 = B12 = B21 = 0.d) Conclude that

KM = 1

2

∂2g

∂ξ21

[1 + ( ∂g∂ξ1)2]−3/2 and KG = 0.

7. Given the cylindrical coordinates defined in Problem 2 and the axially sym-metric surface ξ = g(r), choose

y1 = r and y2 = θ.a) Show that

h11 = 1 + (dgdr)2 , h22 = r2, h12 = h21 = 0,

and

h = r2 [1 + (dgdr)2] .

b) Show that the only nonzero surface Christoffel symbols of the secondkind are

Γ1

11 = g′g′′[1 + (g′)2]−1, Γ1

22 = −r[1 + (g′)2]−1, and Γ2

12 = Γ2

21 = 1

r.

c) Show that

B11 = g′′[1+(g′)2]−1/2, B22 = rg′[1+(g′)2]−1/2, and B12 = B21 = 0.d) Conclude that

KM = 1

2r[rg′′+g′+(g′)3][1+(g′)2]−3/2 and KG = 1

rg′g′′[1+(g′)2]−2.

478 APPENDIX D

8. Given the cylindrical coordinates defined in Problem 2 and the axially sym-metric surface r = g(z), choose

y1 = x3 = z and y2 = x2 = θ.a) Show that

h11 = 1 + (dgdz)2 , h22 = g2, h12 = h21 = 0,

and

h = g2 [1 + (dgdz)2] .

b) Show that the only nonzero surface Christoffel symbols of the secondkind are

Γ1

11 = g′g′′[1+(g′)2]−1, Γ1

22 = −g′g′′[1+(g′)2]−1, and Γ2

12 = Γ2

21 = g′

g.

c) Show that

B11 = g′′[1+(g′)2]−1/2, B22 = −g′[1+(g′)2]−1/2, and B12 = B21 = 0.d) Conclude that

KM = 1

2g[gg′′−(g′)2−1][1+(g′)2]−3/2 and KG = −g′′

g[1+(g′)2]−2.

9. Given a Cartesian coordinate system and the surface ξ3 = g(ξ1, t) show thatthe Cartesian components of n are

n1 = − ∂g∂ξ1[1 + ( ∂g

∂ξ1)2]−1/2 ,

n2 = 0,n3 = [1 + ( ∂g

∂ξ1)2]−1/2 .

10. Given the cylindrical coordinate system described in Problem 2 and theaxially symmetric surface z = g(r), show that the cylindrical components ofn are

nr = −dgdr[1 + (dg

dr)2]−1/2 ,

nθ = 0,nz = [1 + (dg

dr)2]−1/2 .

D. CURVES AND SURFACES IN SPACE 479

11. Given the cylindrical coordinate system described in Problem 2 and theaxially symmetric surface r = g(z), show that the cylindrical components ofn are

nr = [1 + (dgdr)2]−1/2 ,

nθ = 0,nz = −dg

dr[1 + (dg

dr)2]−1/2 .

12. Prove that

a)

vα = hαβaiβvi;b)

vi = vαaiα + v(n)ni.

13. Let φ be an explicit function of position in space. Show that the surfacegradient of φ is the projection of the spatial gradient of φ:

∇φ = P ⋅∇φ.14. At any point (y1, y2) on a surface, two tangential vector fields λ and µ may

be viewed as forming two edges of a parallelogram. If the rotation from λ toµ is positive, show that λ ⋅ (ε ⋅ µ) determines the area of the correspondingparallelogram.

15. Let λ and µ be unit tangent vector fields.

a) If at any point (y1, y2) on a surface the rotation from λ to µ is positive,show that

λ ⋅ (ε ⋅ µ) = sin θ,where θ is the angle measured between the two directions.

b) Conclude that at this point

n [λ ⋅ (ε ⋅ µ)] = λ × µor

niεαβ = εijkajαakβ,where n is the unit vector normal to the surface.

c) Show that

εαβajαakβ = εijkni.

480 APPENDIX D

16. Show that, upon a change of surface coordinate systems, εαβ and εαβ trans-form according to the rules appropriate to the contravariant and covariantcomponents of a tangential second-order tensor field. Hint: Start with

ǫαβ det[∂yφ∂yγ] = ǫµν ∂yα

∂yµ∂yβ

∂yνand ǫαβ det[∂yγ

∂yφ] = ǫµν ∂yµ

∂yα∂yν

∂yβ.

17. Show that

−ε ⋅ ε = P.18. Show that

−ε ⋅ a1 =√ha2and

−ε ⋅ a2 = −√ha1.This implies that

∣a2∣ = 1√h∣a1∣,

that a2 is orthogonal to a1, and that the rotation from a1 to a2 is positive.In a similar manner, we conclude that

∣a1 ∣ = 1√h∣a2 ∣,

that a1 is orthogonal to a2, and that the rotation from a2 to a1 is negative.It is also interesting to observe that the above two relations imply

h22 = h11h

and h11 = h22h.

19. Prove that det [P] = 1.20. Starting with (D.170), prove that

Bαβ = aiα,βni.

21. Show that (hαβwβ),γ= hαβwβ

,γ ,

implying that hαβ can be treated as a constant with respect to surface co-variant differentiation (Ricci’s theorem).

22. Starting with

(hαβwβ),γ= ∂(hαβwβ)

∂yγ− Γδ

γαhδβwβ ,

rework Problem 21.

D. CURVES AND SURFACES IN SPACE 481

23. a) Prove that∂hαβ

∂yγ= Γδ

γβhαδ + Γδγαhβδ.

b) Deduce that hαβ,γ = 0 (Ricci’s theorem),

∇P = Bαβ [naαaβ + aαnaβ]and

∇ ⋅ P = tr [B]n = 2KMn,

where KM is the mean curvature.

c) Prove that∇P = aiα,βgia

αaβ +Bαβaαnaβ .

d) Conclude thataiα,β = Bαβn

i

orBαβ = aiα,βni.

24. Show that∇ε − εδαBβδ [naαaβ − aαnaβ] = εαβ,γaαaβaγ .

Since the right-hand side indicates that the quantity on the left-hand side isa tangential third-order tensor field, the properties of which are independentof the surface coordinate system chosen, subsequently show that

εαβ,γ = εαβ,γ = 0and

∇ε = εδαBβδ [naαaβ − aαnaβ] .25. Show that

∂ ln√h

∂yγ= Γβ

γβ.

[Hint: Write out in full εαβ,γ = 0 and set α,β = 1,2.]26. Let w be a tangential vector field. Use the result of Problem 25 to show that

∇ ⋅ w = wα,α = 1√

h

∂

∂yα(√hwα) .

27. Let T be a tangential second-order tensor field. Show that

∇ ⋅ T = (aiαTαβ),βei.

28. a) Noting thatvi,αβ = (vi,α),β ,

show thatvi,αβ = vi,βα.

482 APPENDIX D

b) Show that for a tangential vector field v

0 = vi,αβ − vi,βα,

= ni,βBγαv

γ+ niBγα,βv

γ+ aiγv

γ,αβ− ni

,αBγβvγ− niBγβ,αv

γ+ aiγv

γ,βα.

c) Noting thatni,β = −Bδ

βaiδ,

and using the above result and

gijajν (vi,αβ − vi,βα) = 0,

conclude thatvα,βγ − vα,γβ = Rδαβγv

δ,

which means that vα,βγ is not symmetric in β and γ. Here

Rδαβγ =KGεδαεβγ

are known as the components of the surface Riemann–Christoffel tensor,and KG is the total curvature.

29. Starting with the result of Problem 28, show that

Bαβ,γ = Bαγ,β.

This is known as the Mainardi–Codazzi equation for the surface.

Bibliography

R. Aris. Vectors, Tensors and the Basic Equations of Fluid Mechanics. DoverPublications, Inc., Mineola, NY, 1962.

R.M. Bowen and C.-C. Wang. Introduction to Vectors and Tensors – Linear andMultilinear Algebra, volume 1. Plenum Press, New York, NY, 1976.

R.M. Bowen and C.-C. Wang. Introduction to Vectors and Tensors – Vector andTensor Analysis, volume 2. Plenum Press, New York, NY, 1976.

L. Brand. Vector and Tensor Analysis. John Wiley & Sons, Inc., New York, NY,1955.

A.J. McConnell. Applications of Tensor Analysis. Dover Publications, Inc., NewYork, NY, 1957.

I. Müller. Thermodynamics. Pitman Publishing, Inc., Boston, MA, 1985.

J.C. Slattery. Interfacial Transport Phenomena. Springer-Verlag, New York, NY,1990.

E. REPRESENTATION OF ISOTROPIC TENSOR FIELDS 483

E Representation of isotropic tensor fields

In this appendix, we wish to illustrate the procedure for obtaining the representa-tion of isotropic tensor fields. We shall assume that the tensor fields are functionsof a vector v and a symmetric tensor A. If the tensor field is also a functionof a scalar, the representations are the same. If they are functions of more thanone vector or symmetric tensor, or functions of skew-symmetric tensors, then thederivations will be modified accordingly, but the procedure is essentially the same.Comprehensive results are given in Tables 5.1–5.8.

E.1 Scalar function

Let ψ be an isotropic scalar function of the vector v and symmetric tensor A, i.e.,ψ = ψ(v,A). Then, from (5.116), (5.120), and (5.121), it must satisfy

ψ(Q ⋅ v,Q ⋅A ⋅QT ) = ψ(v,A), (E.1)

for all orthogonal tensors Q. If ψ is an isotropic function, then it can only dependon scalar combinations of the components of v and A. The only invariant scalarcombination that can be formed from the components of v alone is v ⋅v. The onlyinvariant scalar combinations that can be formed from the components of A aloneare the three invariants A(1), A(2), and A(3), or what are the same trA, trA2,and trA3 (see (3.98)–(3.100)). Note that all other possible scalars are related tothese through the Cayley–Hamilton theorem. Lastly, we have to consider invari-ant scalars that can be formed from products of v and A. There are only twoindependent such scalars, and they are v ⋅A ⋅v and v ⋅A2

⋅v; all other scalars arerelated to these by using Cayley–Hamilton’s theorem. In summary, the isotropicscalar function ψ can only depend on the following six scalar combinations, whichare also listed in Table 5.1:

v ⋅ v, trA, trA2, trA3, v ⋅A ⋅ v, v ⋅A2⋅ v. (E.2)

E.2 Vector function

Let h be an isotropic vector function of the vector v and symmetric tensor A, i.e.,h = h(v,A). Then, from (5.117), (5.120), and (5.121), it must satisfy

h(Q ⋅ v,Q ⋅A ⋅QT ) =Q ⋅ h(v,A), (E.3)

for all orthogonal tensors Q. The problem of finding a representation can bereduced to the problem of finding a scalar function χ of v and A, and an arbitraryvector u by writing

χ = u ⋅h(v,A). (E.4)

Now χ is an isotropic scalar function of v, A, and u, so it can depend only onindependent scalar combinations of their components. Using the procedure of theprevious subsection, we have

v ⋅ v, u ⋅ v, u ⋅u, (E.5)

trA, trA2, trA3, (E.6)

v ⋅A ⋅ v, v ⋅A2⋅ v, u ⋅A ⋅ v, u ⋅A2

⋅ v, u ⋅A ⋅u, u ⋅A2⋅u. (E.7)

484 APPENDIX E

Now if χ was an arbitrary function of u, then it would depend on all the above com-binations. However, from the definition (E.4), χ can only be a linear homogeneousfunction of u, and thus must be such that

χ = u ⋅ (αv + βA ⋅ v + γA2⋅ v) , (E.8)

where α, β, and γ depend on the scalar combinations given above that do notdepend on u, or what is the same, on (E.2). Now, since u is an arbitrary vector,we must have that the invariant representation of h is given by

h = αv + βA ⋅ v + γA2⋅ v. (E.9)

This result is also noted in Table 5.2.

E.3 Symmetric tensor function

Let T be an isotropic symmetric tensor function of the vector v and symmetrictensor A, i.e., T = T(v,A). Then, from (5.118), (5.120), and (5.121), it mustsatisfy

T(Q ⋅ v,Q ⋅A ⋅QT ) =Q ⋅T(v,A) ⋅QT , (E.10)

for all orthogonal tensors Q. Following a procedure analogous to that in theprevious subsection, we introduce an arbitrary symmetric tensor S and define

ζ = S ∶ T(v,A) (E.11)

so that ζ is an isotropic scalar function of v, A, and S which is homogeneous andlinear in S. We find that the independent and complete set of such combinationsis given by

ζ = S ∶ [α1 + βvv + γA + δA2+ η (vA ⋅ v +A ⋅ vv) + λ (vA2

⋅ v +A2⋅ vv)] ,

(E.12)where α, β, γ, δ, η, and λ depend on the scalar invariants (E.2). Now, since S isan arbitrary symmetric tensor, we must have that the invariant representation ofT is given by

T = α1 + βvv + γA + δA2+ η (vA ⋅ v +A ⋅ vv) + λ (vA2

⋅ v +A2⋅ vv) . (E.13)

This result is also noted in Table 5.3.In arriving at the linear independent list of scalar combinations that are ho-

mogeneous and linear in S, besides making use of the Cayley–Hamilton theorem,use is also made of the following identity, which is called the generalized Cayley–Hamilton theorem:

(X ⋅Y ⋅Z +X ⋅Z ⋅Y +Y ⋅Z ⋅X +Y ⋅X ⋅Z +Z ⋅X ⋅Y +Z ⋅Y ⋅X) −trX (Y ⋅Z +Z ⋅Y) − trY (Z ⋅X +X ⋅Z) − trZ (X ⋅Y +Y ⋅X) −[tr (Y ⋅Z) − trY trZ]X − [tr (Z ⋅X) − trZ trX]Y − [tr (X ⋅Y)−trX trY]Z − [trX trY trZ − trX tr (Y ⋅Z) − trY tr (Z ⋅X)−trZ tr (X ⋅Y) + tr (X ⋅Y ⋅Z) + tr (Z ⋅Y ⋅X)] 1 = 0, (E.14)

BIBLIOGRAPHY 485

where X, Y, and Z are arbitrary second-order tensors. Note that the standardCayley–Hamilton theorem is recovered if one sets X = Y = Z. If in (E.14) we setZ =X, we obtain

X ⋅Y ⋅X +X2⋅Y +Y ⋅X2

− trX (X ⋅Y +Y ⋅X) − trYX2−

[tr (X ⋅Y) − trX trY]X − 1

2[tr (X2) − (trX)2]Y −

tr (X2⋅Y) − trX tr (X ⋅Y) + 1

2trY [trX2

− (trX)2] 1 = 0. (E.15)

In our application to obtain (E.13), we subsequently take X = A and Y = vv in(E.15).

In addition, the following independent general identity can be proved:

V ⋅Y ⋅W +YT⋅V ⋅W +V ⋅W ⋅YT

− trY (V ⋅W) − 1

2tr (W ⋅V)YT

−

[tr (W ⋅V ⋅Y) − 1

2tr (W ⋅V) trY]1 = 0, (E.16)

where Y is an arbitrary second-order tensor, and V and W are second-order skew-symmetric tensors. This identity is useful in obtaining invariant representationsof quantities that depend on skew-symmetric tensors.

Problems

1. Derive (E.13).

Bibliography

J.P. Boehler. On irreducible representations for isotropic scalar functions.Zeitschrift fur Angewandte Mathematik und Mechanik, 57(6):323–327, 1977.

P. Pennisi and M. Trovato. On the irreducibility of Professor G.F. Smith’s rep-resentation for isotropic functions. International Journal of Engineering Science,25(8):1059–1065, 1987.

R.S. Rivlin. Further remarks on the stress-deformation relations for isotropicmaterials. Journal of Rational Mechanics and Analysis, 4(5):681–702, 1955.

R.S. Rivlin and G.F. Smith. On identities for 3 × 3 matrices. Rendiconti diMatematica, Universitá degli studi di Roma, 8:348–353, 1975.

G.F. Smith. On isotropic functions of symmetric tensors, skew-symmetric tensorsand vectors. International Journal of Engineering Science, 9(10):899–916, 1971.

G.F. Smith. Constitutive equations for anisotropic and isotropic materials. InG.C. Sih, editor, Mechanics and Physics of Discrete Systems, volume 3. ElsevierScience B.V., Amsterdam, The Netherlands, 1994.

A.J.M. Spencer. Theory of invariants. In A.C. Eringen, editor, ContinuumPhysics, volume I. Academic Press, New York, 1971.

486 APPENDIX E

C.-C. Wang. A new representation theorem for isotropic functions: An answerto professor G.F. Smith’s criticism of my papers on representations for isotropicfunctions. Part 1. Scalar-valued isotropic functions. Archive for Rational Me-chanics and Analysis, 36(3):166–197, 1970.

C.-C. Wang. A new representation theorem for isotropic functions: An an-swer to professor G.F. Smith’s criticism of my papers on representations forisotropic functions. Part 2. Vector-valued isotropic functions, symmetric tensor-valued isotropic functions, and skew-symmetric tensor-valued isotropic functions.Archive for Rational Mechanics and Analysis, 36(3):198–223, 1970.

C.-C. Wang. Corrigendum to my recent papers on “Representations for isotropicfunctions”. Archive for Rational Mechanics and Analysis, 43(3):392–395, 1971.

Q.-S. Zheng. On the representations for isotropic vector-valued, symmetrictensor-valued and skew-symmetric tensor-valued functions. International Journalof Engineering Science, 31(7):1013–1024, 1993.

Q.-S. Zheng. Theory of representations for tensor functions – A unified invariantapproach to constitutive equations. Applied Mechanics Reviews, 47(11):545–587,1994.

F. LEGENDRE TRANSFORMATIONS 487

F Legendre transformations

The fundamental aspect of Legendre transformations is that we are given an equa-tion (e.g., the fundamental relation) of the form

y = y(x0, x1, . . . , xn) (F.1)

and we would like to find a way whereby the derivatives

pi ≡ ∂y

∂xi(F.2)

could be considered as independent variables without loss of any information (ascontained in the fundamental relation). The technique for doing this is providedby Legendre transformations.

First, consider the case where we have an equation of only a single independentvariable:

y = y(x). (F.3)

This equation just represents a curve with Cartesian coordinates (x, y) and

p = p(x) ≡ dydx

(F.4)

is the slope of the curve, which is a function of x. An intuitive way to consider p asan independent variable would be to eliminate x between the equations for y andp and then we would end up with the equation where p is now the independentvariable:

y = y(p). (F.5)

However, just the knowledge of p and y is not sufficient to reconstruct the curve(F.3). It should be noted that the curve (F.5) is a first-order ordinary differentialequation and the solution can only be determined up to an arbitrary constant.Clearly, one such constant, the y intercept, would provide us the unique solution.If we take ψ to be the y intercept, then a relation of the form

ψ = ψ(p) (F.6)

would enable us to construct a curve such that at x, where the ordinate is y andthe slope is p, the intercept of the line with this slope is given by ψ. In such case,(F.6) would provide the same complete information as (F.3) as the two relationswould be equivalent. To see that these two representations are equivalent, we notethat if ψ is the y intercept and p the slope at x, then we have

p = y −ψx − 0

(F.7)

orψ = y − px. (F.8)

Now suppose that we are given (F.3), and by differentiation, we find (F.4). Byelimination of x and y among equations (F.3), (F.4), and (F.8), we obtain thedesired new relation between ψ and p. In order for us to be able to solve for

488 APPENDIX F

Table F.1: Legendre transformation of one variable.

y = y(x) ψ = ψ(p)p = dy

dx−x = dψ

dpψ = y − px y = ψ + xp

Elimination of y and x from Elimination of ψ and p fromthe above equations yields from the above equations yields

ψ = ψ(p) y = y(x)

x, we require that d2y/dx2 ≠ 0. Thus, we now have that (F.3) is a fundamentalrelation in the y representation, whereas (F.6) is a fundamental relation in the ψrepresentation. Equation (F.8) is taken as the definition of the function ψ, referredto as the Legendre transform of y.

The inverse problem of recovering (F.3) from (F.6) proceeds as follows. Takingthe differential of (F.8) and recalling from (F.4) that dy = pdx, we have

dψ = dy − pdx − xdp = −xdp, (F.9)

or

−x = dψdp. (F.10)

If the two variables ψ and p are eliminated from (F.6), (F.8), and (F.10), werecover equation (F.3). In order for us to be able to solve for p, we require thatd2ψ/dp2 ≠ 0. The symmetry between the Legendre transform and its inverse isevident from Table F.1.

The generalization of the Legendre transformation to functions of more than asingle independent variable is straightforward. In general, the fundamental relationof the form (F.1) represents a hyper-surface in a (n + 2)-dimensional space withCartesian coordinates y, x0, x1, . . . , xn. The derivatives (F.2) are the partial slopesof the hyper-surface. The family of tangent hyper-planes may be characterized bygiving the intercept of a hyper-plane, ψ, as a function of the slopes:

ψ = y − n

∑i=0

pixi. (F.11)

Taking the differential of this equation, we find

dψ = − n

∑i=0

xidpi; (F.12)

thus

−xi = ∂ψ∂pi

. (F.13)

A Legendre transformation

ψ = ψ(p0, p1, . . . , pn) (F.14)

BIBLIOGRAPHY 489

Table F.2: General Legendre transformation.

y = y(x0, x1, . . . , xm, xm+1, . . . , xn) ψ = ψ(p0, p1, . . . , pm, xm+1, . . . , xn)pi = dy

dxi, i ≤m −xi = ∂ψ

∂pi, i ≤m

ψ = y − m

∑i=0

pixi y = ψ + m

∑i=0

xipi

Elimination of y and xi, i ≤m, yields Elimination of ψ and pi, i ≤m, yieldsψ = ψ(p0, p1, . . . , pm, xm+1, . . . , xn) y = y(x0, x1, . . . , xm, xm+1, . . . , xn)

is obtained by eliminating y and the xi from (F.1), the set (F.2), and (F.11). Inorder for us to be able to solve for xi, we require that d2y/dxidxj ≠ 0, where i, j ≤ n.The inverse transformation is obtained by eliminating ψ and pi from (F.14), theset (F.13), and (F.11). In order for us to be able to solve for pi, we require thatd2ψ/dpidpj ≠ 0, where i, j ≤ n.

Finally, a Legendre transformation may be made in only an (m+2)-dimensionalsubspace, m ≤ n, of the full (n + 2)-dimensional space of the relation (F.1). Ofcourse, the subspace must contain the y-coordinate but may involve any choiceof m + 1 coordinates from the set x0, x1, . . . , xn. For convenience of notation,the coordinates are ordered so that the Legendre transformation is made in thesubspace of the first m + 1 coordinates (and of y); the coordinates xm+1, . . . , xnare just treated as constants and thus left untransformed. Subsequently, we canrepresent such transformation and inverse as indicated in Table F.2. We note thatin the Legendre transformation, the variables xi and pi, for i ≤m, are consideredto be conjugate variables.

Bibliography

H.B. Callen. Thermodynamics. John Wiley & Sons, Inc., New York, NY, 1962.

M.R. El-Saden. A thermodynamic formalism based on the fundamental relationand the Legendre transformation. International Journal of Mechanical Sciences,8(1):13–24, 1966.

Index

k-vector, 32

Abelian group, 203Acceleration, 101, 103, 160, 182, 185,

363angular

Coriolis, 124inertial, 124internal, 119, 124

apparent, 124, 182internal, 124

centrifugal, 124centripetal, 124Coriolis, 124Euler, 124gradient, 387

tensor, 388inertial, 124true, 124wave, 245

Acceptable variability, 3, 5, 6, 168Adiabatic

boundary, 250, 271compressibility, 222, 274process, 233

Affinities, 234Almansi–Hamel strain tensor, 96, 296Alternating

symbol, see Levi–Civita symboltensor, see Levi–Civita tensor

Amount of shear, 94, 366Angle, 120Anisotropic

material, 208, 214, 296–305solid, 289

Apparent shear viscosity, 383Area stretch ratio, 81, 118Axes of stretching, 365, 366Axiom of continuity, 75

Azimuthal shearing of annular wedge,319, 335

Balanceangular momentum, 154, 163, 167,

168, 183, 447energy, 155, 175, 176, 179, 184,

239, 248, 447entropy, 447law, 149, 151, 165–167, 447, 454,

473, 474linear momentum, 154, 162, 167,

447mass, 153, 162, 361, 447mechanical energy, 169, 174, 239,

378Barotropic

flow, 373fluid, 188, 351, 379

Barycentric velocity, 152Basic field, 192, 285, 342Basis, 14, 18, 25, 40, 59, 68, 186, 456,

461Cartesian, 41, 42function, 214, 252, 255natural, 16, 17, 19, 20, 46, 455

reciprocal, 16, 19new, 19, 20, 47, 458

reciprocal, 19polynomial, 304, 330, 331reciprocal, 456

Bendingannular wedge, 318, 319rectangular block, 317, 320

Bernoulli’s equation, 379Bianchi’s identities, 54, 99Biaxial stress, 170Bivector, 39Body

491

492 INDEX

coupleapparent, 182inertial, 182

forceapparent, 182inertial, 182

load, 156material, 73, 74, 103, 149, 151

Boundary, 103adiabatic, 250condition, 216, 245, 250–251, 271free, 250ideal, 251no-penetration, 103no-slip, 103stationary, 103surface, 250thermally isolated, 250

Boundary layerequations, 371flow, 371thermal, 371

Bulk modulusisentropic, 333isothermal, 305, 332

Bulk viscosity, 352, 357, 359, 372

Caloriccompliance, 222, 259stiffness, 222, 259

Cartesiancoordinate system, 41–43, 45, 46,

59, 74, 122, 440, 474, 476,478

decomposition, 91, 111, 116Cauchy’s

first law, 165, 167lemma, 163second law, 165, 168stress

quadric, 170tensor, 169, 309, 448

theorem, 164Cauchy–Green

difference history tensorright, 390

strain tensorleft, 92relative, 110, 387

right, 77, 92Causality principle, 191–193Cayley–Hamilton theorem, 85–88, 140,

260, 308, 348, 485generalized, 484

Center of mass, 152Characteristic polynomial, 85, 86, 139,

170, 301, 360Chemical potential, 219, 246, 251Christoffel symbol, 462, 467

first kind, 51second kind, 51, 69

Circulation-preserving motion, 363, 369,370

Clausius–Clapeyron equation, 248Clausius–Duhem inequality, 235, 242,

271, 272, 275, 277, 284–289,340

Closed system, 216Coefficient

binomial, 387, 424damping, 415permeability, 262pressure

isochoric, 220response, 207tension, 374thermal expansion, 220, 222, 274,

374, 375isentropic, 333isothermal, 305

thermal stress, 305thermal tension

isochoric, 220isometric, 220

volume expansion, 220, 238Coldness function, 192, 243Coleman–Noll procedure, 242, 340–341Compression, 170, 312Configuration, 73

current, 79, 81, 82, 106, 108deformed, 74, 94, 96, 201, 310initial, 73, 108reference, 73, 74, 79, 81, 82, 94,

96, 106, 108, 188, 192, 200–202, 204, 207, 310

local, 201undeformed, 73

INDEX 493

undistorted, 208Conservation of mass, 153, 185Constitutive

equation, 192general, 195, 396rate type, 417reduced, 208simple fluid, 212simple isotropic solid, 212thermodynamical, 218

function, 194simple material, 200

principles, 191, 192quantity, 193relation, 8theory, 8, 191, 251

Constitutively admissible process, 233Contact

couple, 156discontinuity, 249entropy supply, 176force, 156heat supply, 174load, 156

Continuum, 3mechanics, 2, 74

Contravariant components, 18Controllable solution, 207Convected

entropy flux, 288heat flux, 288rate, 128stress tensor, 288tensor, 142, 210time derivative, 128

Coordinates, 16anholonomic, 450holonomic, 450

Cotter–Rivlin tensor, 128, 130, 388,418, 425

Couette flow, 383, 429Couple stress

tensor, 158, 165, 180vector, 157, 161

Covariant components, 18Creeping flow, 371Cross product, 16, 28, 31, 40, 41, 440Crystal symmetry, 289, 295, 327–329

Crystallographic group, 289Curie’s principle, 295Curl, 440

vector field, 56, 61Current configuration, 79, 81, 82, 106,

108Curvature, 451, 452

Gaussian, 464, 465mean, 481

surface, 136, 150, 465, 468normal, 451, 468principal, 132, 468radius, 452space, 42, 54tensor, 463total, 464, 482

Curvilinear coordinate system, 16, 19,45

Cylindrical polar coordinate, 46, 48,59–62, 68, 70, 144, 317–319,475, 477–479

D’Alembert motion, 363Darcy’s law, 262

generalized, 262Deformation, 5, 74, 76

function, 74gradient tensor, 76, 82, 83

dilatational part, 82isochoric part, 82relative, 108

homogeneous, 78, 310line element, 77relative, 106rigid, 140, 158surface element, 80volume element, 82

Derivative, 42absolute, 449convected, 128corotational, 128covariant, 52, 53

surface, 466Fréchet, 44, 407Gateaux, 44intrinsic, 449Jaumann, 128, 160, 419, 425material, 102, 131, 134, 136

Determinant, 14–16, 440

494 INDEX

Deviatoriccomponent, 35, 67, 87, 120stress, 172

Die swell effect, 384Differential geometry, 449Dilatant fluid, 384Dilatation, 82, 312, 365Dilatational

motion, 365viscosity, 276, 351

Directioncosine, 20, 41, 78, 171, 173, 301principal, 90, 92, 169vector, 57, 77

Discontinuouscurve, 63point, 63, 132surface, 63, 150, 244, 473

Dislocation, 98, 244Dispersion relation, 301, 302, 360Displacement

gradient tensormaterial configuration, 97spatial configuration, 97

vector, 96, 97, 100, 299, 300Dissipation, 234Dissipation inequality, see Clausius–

Duhem inequality, Residualentropy inequality

Dissipativepower

total, 240stress, 341, 347, 398

Distance, 120, 125, 158Divergence, 440

dyadic tensor, 62surface, 465theorem, 63

generalized, 64, 150surface, 471

vector field, 55, 60Double vector, 76, 126Doubly superposed viscometric flow,

393Dual

multivector, 41tensor, 37

Dummy index, 18

Dyadic product, see Tensor productDynamic nonequilibrium chemical po-

tential, 246Dynamical system, 116Dynamics, 4

Ehrenfest classification, 247, 259Eigenvalue, 85, 86

problem, 85, 86, 170Eigenvector, 86Einstein convention, 18Elastic

compliance tensorisentropic, 219, 225, 259isothermal, 220, 225, 259

hydrostatic pressure, 241potential function, 240properties, 331stiffness tensor, 298, 330, 331

isentropic, 218, 259isothermal, 220, 259

Elasticity, 8linear, 8nonlinear, 8tensor, 299

Electrical work, 2Elongational flow, 385, 393, 416Empirical

inequality, 316temperature, 192, 243

Enstrophy, 115Enthalpy, 247

minimum principle, 238potential, 227, 229, 259

Entropy, 155, 259flux, 178

hypothesis, 176inequality, 156, 178, 180, 194maximum principle, 236production, 156, 176, 181, 236,

242, 246, 247, 249, 250Equation of state, 219, 380

caloric, 220, 357thermal, 220, 357, 373, 374

Equilibriumchemical potential, 251Gibbs free energy, 251jump conditions, 251process, 239

INDEX 495

state, 217, 235, 288thermodynamics, 7, 235

Eshelby energy-momentum tensor, 246Euclidean

frame, 181, 183space, 16, 41, 54, 99transformation, 121, 127, 183, 194

Euler’sequation, 221equation of motion, 154, 158, 351,

352, 372theorem, 221

Euler–Fourier equations, 349Euler–Piola–Jacobi identities, 84Eulerian description, 101Event, 120Eversion of sector of spherical shell,

319Extension of annular wedge, 318, 319Extensional flow, 393Extensive property, 216Exterior product, 39External load, 156Extra stress, 316, 398

Field tensor, 41, 199Finite linear viscoelastic material, 411Finite strain tensor

current configuration, 96, 97material configuration, 96, 97

First fundamental tensor of surface,457

Flat space, 54Flow

barotropic, 373boundary layer, 371creeping, 371perturbed, 371

Fluid, 9, 208, 339crystal, 208ideal, 9, 187, 188, 360, 372, 377incompressible, 373Newtonian, 10, 352, 372piezotropic, 373rate type, 418Stokesian, 372

Force, 156Fourier’s

inequality, 288

law of heat conduction, 349Fourier–Stokes heat flux theorem, 176Frame

angular velocity, 123, 182indifference, 120, 125

principle, 191inertial, 125, 182invariance, 20

scalar, 20tensor, 22, 120vector, 20, 21

reference, 120spin rate, 182

Freebody, 156boundary, 250energy, 227index, 19

Free-slip, 249Frenet–Serret formulas, 453Friction, 246Fumi method of direct inspection, 291Fundamental form of surface

first, 457second, 463, 467third, 463

Fundamental relation, 217, 221, 226,227, 234, 235, 263, 276, 278,280, 357, 487, 488

Galilean transformation, 125, 127Gauss equation, 464Gauss–Green theorem, 62, 471Gaussian curvature, 464, 465General relativity, 13, 125, 449Generalized

Cayley–Hamilton theorem, 484convection vector, 368Darcy’s law, 262divergence theorem, 64, 150Gauss–Green theorem, 62Kronecker delta symbol, 33, 439Leibniz rule, 133Reynolds transport theorem, 138,

150shear modulus, 313Stokes theorem, 65, 135surface

divergence theorem, 150, 471

496 INDEX

transport theorem, 150Generating transformation, 290Genuine scalar, 57Gibbs

equation, 219, 288free energy minimum principle, 238potential, 227, 259

Gibbs–Duhem equation, 221, 226Grüneisen

parameter, 306tensor, 229, 259

Gradient, 440operator, 45scalar field, 49, 60surface, 458, 465vector field, 50, 51, 60

Green–St. Venant strain tensor, 96

Heat, 2conduction, 289

inequality, 340energy, 155flux, 176

hypothesis, 175theorem, 176

increment, 222Helmholtz

free energy, 235minimum principle, 238

potential, 227, 259Helmholtz–Zorawski criterion, 363, 370Hemitropic

function, 214material, 208

History tensor, 390Hodge dual, 37Homochoric motion, 154Homogeneous

deformation, 78, 310function, 221material, 197, 202, 284motion, 102, 360simple material, 198strain, 315

Hooke’s law, 8Hydrostatic stress, 165, 241, 314, 352Hydrostatics, 362Hygrosteric materials, 418Hyperelastic material, 261, 289

Hypo-elastic material, 289, 406, 419

Idealboundary, 243, 247, 251fluid, 9, 187, 188, 360, 372, 377gas, 280, 347, 372, 379, 380

Improperorthogonal

matrix, 15, 88transformation, 88

Incompressibledeformation, 313, 314, 316, 355fluid, 9, 262, 352, 357, 373–375,

383, 384, 398, 399, 411, 413material, 261, 316, 317motion, 137, 355

Inertial frame, 125, 182Infinitesimal

right stretch tensor, 98rotation tensor, 98strain tensor

material configuration, 97spatial configuration, 97

Inflationannular wedge, 318, 319sector of spherical shell, 319

Initial configuration, 73Inner product, see Scalar productIntensive property, 216Internal

angular acceleration, 119apparent, 124inertial, 124

angular velocity, 119, 152apparent, 123gradient, 119

energy, 2, 154, 259balance, 179minimum principle, 236

inertia tensor, 152, 180load, 156spin, 152

apparent rate, 181, 182inertial, 181production, 169tensor, 153

Invariant, see Scalar invariant, Prin-cipal invariant

Inviscid fluid, 351

INDEX 497

contact surface, 249Irreducible invariant subspace, 36Irrotational motion, 115, 145, 366Isentropic path, 232Isochoric

deformation, 75irrotational motion, 367motion, 104, 114, 137, 154, 365pressure coefficient, 374thermal tension, 259

Isolated system, 216Isometric thermal tension coefficient,

220Isopiestic thermal expansion tensor, 219Isothermal

compressibility, 220, 274, 374path, 232

Isotropicfunction, 214invariant, 214material, 208solid, 212tensor, 26, 442, 446

Jump, 133, 138condition, 151, 244

angular momentum, 168characterization, 245energy, 179, 180entropy, 181equilibrium, 251linear momentum, 167mass, 166

operator, 64

Kelvin effect, 314Kelvin–Voigt model, 9Kinematic vorticity number, 145, 363Kinematics, 4, 73Kinetic energy, 153

internal rotational, 153translational, 153

Kronecker delta symbol, 15, 16, 26,439

generalized, 33, 439

Lagrange multiplier, 242Lagrange’s equation, 368Lagrangian

coordinate, 74description, 101

Lamé’sconstants

isothermal, 305stress ellipsoid, 172

Lambsurfaces, 369vector, 368

Lamellaracceleration, 371field, 366

Laplaciangradient field, 56scalar field, 61

Latent heat, 248Lattice

angle, 289base-centered, 289body-centered, 289Bravais, 289face-centered, 289parameter, 289primitive, 289variation, 289

Legendre transformation, 226, 487Leibniz rule, 132

generalized, 133Length stretch ratio, 78, 118Levi–Civita

symbol, 16, 439generalized, 32, 439two-dimensional, 456

tensor, 26, 57absolute, 28generalized, 32surface, 457

Lie time derivative, 130Lineal motion, 361Linear

deformation, 296, 305momentum, 152

Linear liquid, 375Local

reference configuration, 201relative stretch rate

area, 118tangent, 117

498 INDEX

volume, 118

Müller–Liu procedure, 242, 342–347Mainardi–Codazzi equation, 464, 482Mass, 3, 151

conservation equation, 165density, 3, 151total, 152

Materialanisotropy, 197class, 206coordinate, 73deformation gradient, 76derivative

line integral, 131surface integral, 134volume integral, 136

description, 101descriptor, 197displacement gradient, 97frame indifference, 194inhomogeneity, 197micromorphic, 169micropolar, 169microstretch, 169particle, 73point, 73, 74singular surface, 248smoothness principle, 191stress power, 448surface, 103symmetry, 199

group, 204principle, 192transformation, 202

tube, 371Matrix, 14

column, 15determinant, 15, 19groups, 204identity, 15improper orthogonal, 15inverse, 15proper orthogonal, 15row, 15skew-symmetric, 15symmetric, 15transpose, 14

Maximum shear stress, 172

Maxwellmodel, 9relation, 225, 228, 230, 231

Meanshear stress, 246stress, 172surface curvature, 465, 468value theorem, 161, 175, 177

Mechanicaldissipation inequality, 341energy, 169pressure, 180, 350strain tensor, 297work, 2

Mechanically admissible process, 233Mechanics, 4

Newtonian, 4quantum, 4relativistic, 4

quantum, 4Memory

influence function, 406principle, 191

Method of direct inspection, 291, 332Metric tensor, 22, 47, 48Mohr’s circles, 173Mole number density, 217Moment, 156

center, 153momentum, 153

Mooney–Rivlin, 334material, 316, 334

Motion, 100circulation-preserving, 370complex-lamellar, 367, 369complex-screw, 368constant stretch history, 390, 399D’Alembert, 363dilatational, 365homogeneous, 360irrotational, 366isochoric, 365lineal, 361plane, 361pseudo-lineal

first kind, 361second kind, 361

pseudo-plane

INDEX 499

first kind, 362second kind, 362

rigid, 365rotationally symmetric, 362screw, 368shearing, 366steady, 362

with steady density, 362with steady stream lines, 363with steady vorticity, 370without acceleration, 363

Motzkin sum numbers, 443Mutual load, 156

Naturalbase vector, 46state, 284

Navier equations, 372Navier–Stokes equations, 352, 372, 375Negative

definite matrix, 89semi-definite matrix, 89

Neo-Hookean material, 316, 334Neumann’s principle, 289, 293Newton’s law of motion

second, 154third, 154, 156

Newton–Fourier equations, 352Newtonian fluid, 352, 372No-penetration condition, 250No-slip condition, 250Noll’s rule, 204Non-inertial frame, 182Non-isotropic function, 213Non-Newtonian fluid, 384Non-simple material, 198Nonequilibrium

state, 235thermodynamics, 7

Nonhomogeneous strain, 316Nonlinear deformation, 304, 306Nonpolar material, 168, 184Normal

curvature, 468plane, 453stress, 164

Object of anholonominity, 450Objectivity, 125

Observer, 120Oldroyd tensor, 128, 141Onsager’s principle, 235Open system, 216Orr–Sommerfeld equations, 371Orthogonal

coordinates, 16curvilinear coordinate system, 58matrix, 88, 90transformation, 88

Osculating plane, 453

Particle, 74Path, 100

line, 104Perfect fluid, 349Permutation

symbol, see Levi–Civita symboltensor, see Levi–Civita tensor

Perturbation theory, 371Perturbed flow, 371Phase

change surface, 247transition, 259

Physical component, 60Piezocaloric tensor, 220Piezotropic fluid, 373Piola–Kirchhoff

entropy flux, 448heat flux, 448stress tensor, 309, 448

Planemotion, 142, 361strain, 142

Plasticity, 8Poiseuille flow, 427, 430Poisson ratio

isentropic, 333isothermal, 305

Polardecomposition theorem, 91, 93material, 119, 152, 168, 184scalar, 57vector, 57

Positivedefinite matrix, 89semi-definite matrix, 89

Potentialenergy, 239

500 INDEX

theory, 367Poynting effect, 314, 322Prandtl, L., 371Preferred state, 284Pressure coefficient, 220Principal

axes of stress, 170direction, 90invariant, 85, 90

scalar, 87stress, 170stretch, 92, 365

direction, 92value, 90

Principle of objectivity, 121Process, 155Projector, 67Propagation of sound, 357Proper

orthogonalmatrix, 15, 88transformation, 88

subgroup, 203Property tensor, 199, 259Pseudo-

lineal motionfirst kind, 361, 366second kind, 361

plane motionfirst kind, 362second kind, 362

scalar, 57tensor, 26vector, 57

Pseudoplastic fluid, 384Pull-back operation, 129Push-forward operation, 129

Quadric surface, 29, 35, 170, 171Quasi-equilibrium process, 239Quasi-static process, 239

Radón–Nikodym theorem, 150Radius

curvature, 452torsion, 453

Ratematerial type, 196, 418rotation tensor, 111

strain tensor, 111Ratio of specific heats, 224, 259Real symmetric matrix, 88Reciprocal base vector, 46Rectifying plane, 453Rectilinear shearing, 366Reduced

entropy inequality, 306, 344, 346,348–350

form, 210Reference

configuration, 73coordinate, 73

Reiner–Rivlin fluid, 350, 384Relative

angular velocity, 119Cauchy–Green strain tensor

left, 110right, 110

deformation, 106gradient, 108

framerotation tensor, 123spin tensor, 123

motion, 108, 386Residual

entropy inequality, 242, 244, 340,341, 347

stress, 334Response

coefficient, 207functional, 193

Retarded motion, 409Reversible process, 233Reynolds transport theorem, 137, 239

generalized, 138, 150Ricci’s theorem, 56, 462, 463, 480, 481Riemann–Christoffel tensor, 54, 99, 449,

464, 482Riemannian space, 54Riesz representation theorem, 410Rigid

deformation, 140motion, 114, 365transformation, 121

Rigid bodyangular velocity, 115dynamics, 38, 119, 158–161

INDEX 501

Rivlin–Ericksen tensor, 387, 388Rotation tensor, 91Rotationally symmetric motion, 362Rubber, 8Rubber-like material, 316

Scalaraxial, 57field, 49, 70function, 14, 483invariant, 85product, 440

tensor, 30triple, 16, 17, 65vector, 16, 27

Screw motion, 368Second fundamental tensor of surface,

463, 467Shear

layer, 371thickening fluid, 383thinning fluid, 383viscosity, 10, 351

Shear modulusgeneralized, 313isothermal, 305ordinary, 313

Shear-rate function, 402, 427–429Shearing

annular wedge, 318motion, 366rectangular block, 317sector of hollow circular cylinder,

318stress, 164

Shock, 371surface, 245

elastic medium, 246inviscid fluid, 246

wave, 245SI units, 225Simple

compression, 170, 312dilatation, 312extension, 311, 312material, 198shear, 94, 313shear flow, 383shearing, 145, 366

tension, 170thermomechanical process, 233

Skew-symmetric tensor, 30Slip, 166

condition, 250Solid, 8, 207Sound speed, 360Space, 3Spatial

description, 101displacement gradient, 97material smoothness, 196, 198natural basis, 455velocity gradient, 103, 110

Specificenergy release rate, 246heat, 222

constant pressure, 274, 374constant strain, 298constant stress, 298constant thermostatic tension,

223, 259constant thermostatic volume,

223, 259constant volume, 274, 374

volume, 151, 166, 216–218, 220,271, 341

Sphericalcomponent, 87coordinate system, 476coordinates, 48, 59stress, 172stretching tensor, 145tensor, 87

Spin tensor, 111, 388Square root of matrix, 93Stagnation

flow, 385point, 104, 367, 368

State function, 155Statics, 4Stationary surface, 249Steady

motion, 104, 362with steady density, 362

vorticity, 370Stiffness tensor, 299Stokes

502 INDEX

flow, 371hypothesis, 352, 372theorem, 63, 135, 472

generalized, 65, 135Stokesian fluid, 372Stored energy function, 261Straightening sector of hollow circular

cylinder, 318Strain, 5

energy, 239function, 240, 261total, 240

kinematics, 95tensor, 96

Streak line, 105Stream line, 104Stress, 6

extra, 316, 398invariant, 170power, 169quadric of Cauchy, 170tensor, 6, 164vector, 6, 161

Stretchrectangular block, 317, 320sector of hollow circular cylinder,

318tensor, 111, 388

left, 91right, 91

Subgroup, 203Summation convention, 18Surface, 455

Christoffel symbol of second kind,467

coordinate system, 455couple, 156covariant derivative, 466, 468divergence, 465, 468, 469

theorem, 150, 471dual basis, 456embedded in space, 45force, 156gradient, 468integral, 81load, 156metric tensor, 456natural basis, 455

normal, 455physical basis, 458, 459projection tensor, 461, 465reciprocal basis, 456tangential Levi–Civita tensor, 457tensor, 458traction, 156transport theorem, 150unit normal vector, 461

Symmetrygroup, 202, 214

finite, 203infinite, 203

transformation, 213Syzygies, 443

Tangent bundle, 116Tangential tensor, 459Temperature, 233, 234

absolute, 192, 241Temporal material smoothness, 196,

198Tension, 170Tensor

absolute, 26analysis, 13anti-symmetric, 32axial, 26Cartesian, 13, 439contraction, 29contravariant components, 19, 20,

25convected, 142, 210covariant components, 19, 21, 25deviatoric, 35, 67dual, 37dyadic, 19field, 41, 199function, 484generator, 214identity, 26isotropic, 35, 67, 483material description, 75objective, 120order, 19orthogonal, 31polyadic, 24product, 21, 29rank, 19

INDEX 503

relative, 26scalar, 19skew-symmetric, 32

completely, 32spatial description, 75spherical, 35symmetric, 30, 32

completely, 32, 67trace, 30transpose, 31two-point, 76, 126unit, 26, 67vector, 19weighted, 26

Tetrahedron, 161Thermal

conductivity, 351tensor, 298

energy, 240expansion

coefficient, 274isopiestic tensor, 219, 259tensor, 298

stiffness tensorisentropic, 218, 259

strain tensor, 220, 259, 297stress

coefficient, 305tensor, 298

tensionisochoric, 259

Thermally isolated boundary, 250Thermodynamic

constitutive equation, 218equilibrium, 235

stability, 235state, 288, 341, 346

flux, 234force, 234path, 232potential, 217, 226pressure, 218, 350principle, 192process, 218, 232

admissible, 233homogeneous, 218

property, 259state, 216, 219

Thermodynamics, 6, 154, 216first law, 174, 179second law, 156, 181, 192zeroth law, 192

Thermoelastic potential, 288, 296, 297,304, 306, 307

Thermostaticprocess, 239temperature, 218, 259tension, 218, 259volume, 217, 259

Thermostatics, 7, 155, 174first law, 241

Third fundamental tensor of surface,463

Time, 3lapse, 120

Time-independent rigid transformation,123

Torsion, 453annular wedge, 318

Traction, 6Trajectory, 100Transport theorem, 131Transverse isotropy, 289, 294Triaxial stress, 170Triclinic crystal, 301Triply superposed viscometric flow, 393Trivector, 40True

acceleration, 124internal

angular acceleration, 124angular velocity, 123spin rate, 181

velocity, 123Truesdell tensor, 128

Uniaxial stretch, 310Unimodular

group, 202transformation, 202

Universalrelation, 207, 308, 314solution, 207

Vector, 14, 39, 440axial, 41, 57, 115field, 131, 134, 135, 215

504 INDEX

function, 42, 63, 483generator, 214position, 153product, 16, 28space, 14

Velocity, 101angular, 58, 115

internal, 119apparent, 123gradient tensor, 103, 110, 116, 391–

393inertial, 123potential, 366

Virtual displacement, 236Viscoelasticity, 9, 383Viscometric flow, 384, 393Voigt matrix, 293, 299, 305, 332Volume, 3

integral, 82stretch ratio, 82, 118

Vortexsheet, 245tube, 371

Vorticity vector, 58, 115, 119, 185, 187,188, 361, 368, 370, 380

Wedge product, 39Weingarten formula, 463, 464Weissenberg effect, 384Weyl’s theory of invariant polynomi-

als, 443Work, 155

increment, 222

Young modulusisentropic, 333isothermal, 305

Zorawski criterion, 135, 363