Corina Ldes i

CLASSICALMECHANlCS

FOR PHYSICS GRADUATESTUDENTS

Ernesto Corinaldesi

World Scientific Publishing

C l A S S l ~ A l MECHANICS FOR PHYSI~S ~RADUATE STUDENTS

C l A S S l ~ A l MECHANICS FOR PHYSI~S ~RADUATE STUDENTS

CLASSICAL MECHANlCS FOR PHYSICS CRADUATE ~TUDENTS

Ernest0 Carinaldesi

Published by

World Scientific Publishing Co. Pte. Ltd. P 0 Box 128, F m r Road, Singapore 912805 USA office: Suite IB, 1060 Main Street, River Edge, NJ 07661 UK office: 57 Shelton Street, Covent Garden, London WC2H 9HE

Library of Congress ~a~Ioging-~-Publica~ion Data Corinaldesi, E.

Corinaldesi. Classical mechanics for physics graduate students I Emesto

p. cm. lnciudes index. ISBN 9810236255 I . Mechanics. 1. Title.

Q125.C655 1998 531-dc21 98-48670

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Preface

This book evolved from my occasional teaching of graduate Classical Me- chanics at Boston University.

In the late 1960’s I recommended the well-known texts by H. Goldstein, L.D. Landau and E.M. Lifshitz (?), and A. Sommerfeld. In the late 1980’s I learned some modern differential geometry while catching up with gauge field theory. Thus I became able to appreciate the elegance of V.I. Arnold’s Mathematical Methods of Classical Mechanics.

I took the educational risk of presenting Hamiltonian mechanics expressed in Cartan’s notation as a kind of appendix following the traditional treatment. Instructors in other universities also seem to have recognized that the graduate teaching of classical mechanics in physics departments should be updated. In this book I have tried to satisfy both tradition- alists and modernists by approaching each subject at successive levels of abstraction.

My work is designed for a first-year physics graduate student at Boston University, who has taken “Intermediate Mechanics”. Knowledge of curvilinear coordinates, vector analysis, advanced calculus etc. is assumed. The wealth of detail offered should not lull the reader into thinking that the material can be learned by leafing through the book. As much time as possible should be devoted to going through the calculations and solving problems.

Chapter 1 is an introduction meant to establish a rapport between the reader and my way of presenting the material. Chapter 2 builds up a stock of formulae to be drawn from in later chapters. Chapter 3 is mostly devoted to oscillations. It ends with a sketchy account of chaos for discrete maps to introduce the reader to a field founded a century ago by Poincare and cultivated intensively in recent years.

Chapters 4 and 5 cover coordinate systems, inertial forces, and rigid bodies.

Analytical mechanics begins with chapter 6, Lagrangians. Problems already worked out in previous chapters are again solved by using Lagrangian methods. The reader is invited to compare the amount of labor involved in the two versions.

Chapter 7 presents Hamiltonian mechanics. Sections 7.1 to 7.4 give a simple account of the traditional treatment. Section 7.5 introduces Cartan’s notation, which is used throughout the following sections as well as in chapter 8, action-angle variables and adiabatic invariants.

Chapter 9 is on classical perturbation theory with emphasis on the similarity with the perturbation theory of quantum mechanics and field theory.

viii PREFACE

Relativistic dynamics is discussed in chapter 10, with attention to the (‘spinor connection”, the spin, and the Thomas precession.

Chapter 11 illustrates Lagrangian and Hamiltonian methods for continuous systems by discussing two case studies, the vibrating string and the ideal incompressible non-viscous fluid. The Lagrangian description of the latter is not widely known. I worked unsuccessfully trying to formulate it, until I was lucky enough to find D.E. Soper’s Classical Field Theory and reference to the original 1911 paper by G. Herglotz.

Acknowledgements I wish to express my gratitude to my students, especially to Nathaniel R. Greene and Dinesh Loomba, for asking challenging questions and bringing to my attention interesting material. After taking my course Gregg Jaeger and John B. Ross became my graders and helped me in the preparation of the monthly exams from which some of the problems in this text originated.

Adriana Ruth Corinaldesi painstakingly checked parts of my work pointing out errors and lack of clarity. Abner Shimony kindly read the final version of the manuscript and suggested improvements. Neither Abner nor Adriana are responsible for any residual errors and imperfections, which are indeed possible because of the unusual number of formulae presented.

Anthony P. French, John Stachel, and Edwin F. Taylor kindly helped me to dispel doubts about one of the relativity problems.

In addition to the books mentioned in the Preface I learned from many others, but I wish to mention J.D. Jackson Classical Electrodynamics, C.W. Kilmister and J.E. Reeves Rational Mechanics, R.A. Mann The Classical Dynamics of Particles - Galilean and Lorentz relativity, B.F. Schutz Geo- metrical methods of mathematical physics, W.Thirring Classical Dynamical Systems.

I was able to prepare the printed manuscript only because of the gener- ous coaching in Latex given me by J o k Leao. I am also indebted to Jinara Reyes and Guoan Hu for frequent help.

Wei Chen and Lakshmi Narayanan of World Scientific have been an example of amicable and helpful editorship.

Contents

1 INTRODUCTION 1 1.1 Motion in phase space . . . . . . . . . . . . . . . . . . . . . 1 1.2 Motion of a particle in one dimension . . . . . . . . . . . . 2 1.3 Flow in phase space . . . . . . . . . . . . . . . . . . . . . . 5 1.4 The action integral . . . . . . . . . . . . . . . . . . . . . . . 8 1.5 The Maupertuis principle . . . . . . . . . . . . . . . . . . . 12 1.6 Thetime . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.7 Fermat’s principle . . . . . . . . . . . . . . . . . . . . . . . 16 1.8 Chapter 1 problems . . . . . . . . . . . . . . . . . . . . . . 18

2 EXAMPLES OF PARTICLE MOTION 23 2.1 Central forces . . . . . . . . . . . . . . . . . . . . . . . . . . 23 2.2 Circular and quasi-circular orbits . . . . . . . . . . . . . . . 25 2.3 Isotropic harmonic oscillator . . . . . . . . . . . . . . . . . . 26 2.4 The Kepler problem . . . . . . . . . . . . . . . . . . . . . . 27 2.5 The L-R-L vector . . . . . . . . . . . . . . . . . . . . . . . . 31 2.6 Open Kepler-Rutherford orbits . . . . . . . . . . . . . . . . 32 2.7 Integrability . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 2.8 Chapter 2 problems . . . . . . . . . . . . . . . . . . . . . . 36

3 FIXED POINTS. OSCILLATIONS. CHAOS 41 3.1 Fixed points . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 3.2 Small oscillations . . . . . . . . . . . . . . . . . . . . . . . . 44 3.3 Parametric resonance . . . . . . . . . . . . . . . . . . . . . . 47 3.4 Periodically jerked oscillator . . . . . . . . . . . . . . . . . . 50 3.5 Discrete maps. bifurcation. chaos . . . . . . . . . . . . . . . 52 3.6 Chapter 3 problems . . . . . . . . . . . . . . . . . . . . . . 56

ix

X CONTENTS

4 COORDINATE SYSTEMS 67 4.1 Translations and rotations . . . . . . . . . . . . . . . . . . . 67 4.2 Some kinematics . . . . . . . . . . . . . . . . . . . . . . . . 71 4.3 Fictitious forces . . . . . . . . . . . . . . . . . . . . . . . . . 71 4.4 Chapter 4 problems . . . . . . . . . . . . . . . . . . . . . . 73

5 RIGID BODIES 77 5.1 Angular momentum . . . . . . . . . . . . . . . . . . . . . . 77 5.2 Euler’s equations . . . . . . . . . . . . . . . . . . . . . . . . 78 5.3 Euler angles . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 5.4 Spinning top . . . . . . . . . . . . . . . . . . . . . . . . . . 84

5.4.1 Regular precession of top . . . . . . . . . . . . . . . 85 5.4.2 Sleeping top . . . . . . . . . . . . . . . . . . . . . . . 86 5.4.3 Irregular precessions of top . . . . . . . . . . . . . . 87

5.5 Gyrocompass . . . . . . . . . . . . . . . . . . . . . . . . . . 89 5.6 Tilted disk rolling in a circle . . . . . . . . . . . . . . . . . . 91 5.7 Chapter 5 problems . . . . . . . . . . . . . . . . . . . . . . 93

6 LAGRANGIANS 101 6.1 Heuristic introduction . . . . . . . . . . . . . . . . . . . . . 101 6.2 Velocity-dependent forces . . . . . . . . . . . . . . . . . . . 103 6.3 Equivalent Lagrangians . . . . . . . . . . . . . . . . . . . . 105 6.4 Invariance of Lagrange equations . . . . . . . . . . . . . . . 106 6.5 “Proofs” of the Lagrange equations . . . . . . . . . . . . . . 108 6.6 Constraints . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 6.7 Invariance of L and constants of motion . . . . . . . . . . . 115

Invariance of L up to total time derivative . . . . . . 6.7.2 Noether’s theorem . . . . . . . . . . . . . . . . . . . 119

6.8 Chapter 6 problems . . . . . . . . . . . . . . . . . . . . . . 122

7.1 First look . 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 7.2 First look . 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 7.3 H as Legendre transform of L . . . . . . . . . . . . . . . . . 152 7.4 Liouville’s theorem . . . . . . . . . . . . . . . . . . . . . . . 154 7.5 Cartan’s vectors and forms . . . . . . . . . . . . . . . . . . 156 7.6 Lie derivatives . . . . . . . . . . . . . . . . . . . . . . . . . 161 7.7 Time-independent canonical transformations . . . . . . . . 162 7.8 Generating functions . . . . . . . . . . . . . . . . . . . . . . 163 7.9 Lagrange and Poisson brackets . . . . . . . . . . . . . . . . 166 7.10 Hamilton-Jacobi equation revisited . . . . . . . . . . . . . . 168 7.11 Time-deDendent canonical transformations . . . . . . . . . . 170

6.7.1 118

7 HAMILTONIANS 145

CONTENTS xi

7.12 Time-dependent Hamilton-Jacobi equation . . . . . . . . . . 7.13 Stokes’ theorem and some proofs 7.14 Hamiltonian flow as a Lie-Cartan group 7.15 PoincarB-Cartan integral invariant . . . . . . . . . . . . . . 7.16 PoincarB’s invariants . . . . . . . . . . . . . . . . . . . . . . 7.17 Chapter 7 problems . . . . . . . . . . . . . . . . . . . . . .

8.1 One dimension . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Multiply periodic systems . . . . . . . . . . . . . . . . . . . 8.3 Integrability, non-integrability, chaos . . . . . . . . . . . . . 8.4 Adiabatic invariants . . . . . . . . . . . . . . . . . . . . . . 8.5 Outline of rigorous theory . . . . . . . . . . . . . . . . . . . 8.6 Chapter 8 problems . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . .

8 ACTION-ANGLE VARIABLES

9 PERTURBATION THEORY 9.1 The operator ll . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 Perturbation expansions . . . . . . . . . . . . . . . . . . . . 9.3 Perturbed periodic systems . . . . . . . . . . . . . . . . . . 9.4 Chapter 9 problems . . . . . . . . . . . . . . . . . . . . . .

10 RELATIVISTIC DYNAMICS 10.1 Lorentz transformations . . . . . . . . . . . . . . . . . . . . 10.2 Dynamics of a particle . . . . . . . . . . . . . . . . . . . . . 10.3 Formulary of Lorentz transformations 10.4 The spinor connection . . . . . . . . . . . . . . . . . . . . . 10.5 The spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.6 Thomas precession . . . . . . . . . . . . . . . . . . . . . . . 10.7 Charged particle in static em field 10.8 Magnetic moment in static em field . . . . . . . . . . . . . . 10.9 Lagrangian and Hamiltonian . . . . . . . . . . . . . . . . . 10.10Chapter 10 problems . . . . . . . . . . . . . . . . . . . . . .

11.1 Uniform string . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Ideal fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.3 Chapter 11 problems . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . .

. . . . . . . . . . . . . .

11 CONTINUOUS SYSTEMS

171 173 175 178 181 182

193 193 196 200 202 204 206

213 213 215 217 221

227 227 229 232 238 241 244 246 247 248 253

281 261 268 275

INDEX 283

List of Figures

1.1 Arbitrariness of p . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Motion under the force -kq . . . . . . . . . . . . . . . . . . 1.3 Motion under the force +kq . . . . . . . . . . . . . . . . . . 1.4 Motion near potential energy extrema . . . . . . . . . . . . 1.5 Vertical motion under gravity . . . . . . . . . . . . . . . . . 1.6 Elastic collision against a wall . . . . . . . . . . . . . . . . . 1.7 Example of equation (1.14) . . . . . . . . . . . . . . . . . . 1.8 All trajectories in this figure correspond to the same values

1.9 Equation (1.34) . . . . . . . . . . . . . . . . . . . . . . . . .

1.11 Illustration of Maupertuis principle . . . . . . . . . . . . . .

of p , and py . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.10 Real and varied motion under gravity . . . . . . . . . . . .

11 12 14 19

2.1 Isotropic harmonic oscillator . . . . . . . . . . . . . . . . . . 26 2.2 Eccentric anomaly . . . . . . . . . . . . . . . . . . . . . . . 30 2.3 Rutherford orbits . . . . . . . . . . . . . . . . . . . . . . . . 32 2.4 L-R-L vector for Rutherford scattering . . . . . . . . . . . . 33 2.5 Quasi-circular orbits . . . . . . . . . . . . . . . . . . . . . . 37 2.6 Hhon-Heiles potential energy . . . . . . . . . . . . . . . . . 39

3.1 Spiral attractor . . . . . . . . . . . . . . . . . . . . . . . . . 43 3.2 Double pendulum . . . . . . . . . . . . . . . . . . . . . . . . 44 3.3 Instability zones . . . . . . . . . . . . . . . . . . . . . . . . 49 3.4 Stability zones for (3.40) . . . . . . . . . . . . . . . . . . . . 51 3.5 2 = for logistic map . . . . . . . . . . . . . . . . 54 3.6 Lyapunov exponent for logistic map . . . . . . . . . . . . . 55 3.7 Problem3.8 . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 3.8 Triatomic molecule . . . . . . . . . . . . . . . . . . . . . . . 58 3.9 Problem 3.10 . . . . . . . . . . . . . . . . . . . . . . . . . . 59 3.10 Escape from or approach to top position . . . . . . . . . . . 59 3.11 Librational and rotational motion . . . . . . . . . . . . . . . 60

xiii

xiv LIST OF F I ~ ~ ~ S

3.12 Bead on rotating hoop . . . . . . . . . . . . . . . . . . . . . 3.13 Triatomic molecule (A = 6 and X = 12) . . . . . . . . . . . .

61 64

4.1 Example of translation and rotation . . . . . . . . . . . . . 68 4.2 Bug in tilted rotating pipe . . . . . . . . . . . . . . . . . . . 75

5.1 Body-cone and space-cone . . . . . . . . . . . . . . . . . . . 80 5.2 Euler angles . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 5.3 Spinning top . . . . . . . . . . . . . . . . . . . . . . . . . . 84 5.4 Graphof P(u) . . . . . . . . . . . . . . . . . . . . . . . . . 85 5.5 Irregular precessions of top . . . . . . . . . . . . . . . . . . 87 5.6 Gyrocomp~s . . . . . . . . . . . . . . . . . . . . . . . . . . 89 5.7 Rolling disk . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 5.8 6; into page, 6; = &. 4b = 63 . . . . . . . . . . . . . . . . 91 5.9 Thumbtack . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 5.10 Double root of P(u) . . . . . . . . . . . . . . . . . . . . . . 96

6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9

Spherical pendulum . . . . . . . . . . . . . . . . . . . . . . 111 Bead on rotating wire . . . . . . . . . . . . . . . . . . . . . 111 dqi and dqi . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 Block on accelerated cart . . . . . . . . . . . . . . . . . . . 123 Problem 6.13 . . . . . . . . . . . . . . . . . . . . . . . . . . 125 Cylinder rolling inside cylinder . . . . . . . . . . . . . . . . 126 W h ~ l s a n d a x l e . . . . . . . . . . . . . . . . . . . . . . . . 127 Plank. problem 6.2 . . . . . . . . . . . . . . . . . . . . . . . 130 Electron and proton trajectories . . . . . . . . . . . . . . . 133

7.1 Area of parallelogram . . . . . . . . . . . . . . . . . . . . . 159 7.2 Volume of parallelepiped . . . . . . . . . . . . . . . . . . . . 159 7.3 Magnetostatic field example . . . . . . . . . . . . . . . . . . 178 7.4 Equation (7.63) . . . . . . . . . . . . . . . . . . . . . . . . . 180 7.5 Problem 7.17 . . . . . . . . . . . . . . . . . . . . . . . . . . 191

8.1 Ball bouncing up and down . . . . . . . . . . . . . . . . . . 194 8.2 Torus for Kepler motion . . . . . . . . . . . . . . . . . . . . 196 8.3 Particle in box . . . . . . . . . . . . . . . . . . . . . . . . . 206 8.4 Kepler problem in three dimensions . . . . . . . . . . . . . . 206 8.5 hnction q(#), problem 8.2 . . . . . . . . . . . . . . . . . . 209

10.1 Spin precession along circular orbit . . . . . . . . . . . . . . 245 10.2 Spin under sudden acceleration . . . . . . . . . . . . . . . . 254

11.1 Interpreting T, t . . . . . . . . . . . . . . . . . . . . . . . . . 263

Chapter 1

INTRODUCTION

This chapter covers informally material that will be treated extensively in the book. Some of it may have already been learned in undergraduate mechanics. Introduction or review as they may be, the following pages are meant to lead into the substance of analytical mechanics.

1.1 Motion in phase space It is known from General Physics that the dynamics of particles and rigid bodies is governed by second order differential equations. For computational purposes, it is convenient to replace these by systems of first order equations. Thus Newton’s second law in the form

mx = f ( z , k , t ) (1.1)

is replaced by the system

P = f ( q , P , t ) , Q = P 1 (1.2)

where we have renamed the coordinate (q = x), denoted by p the momentum, and taken the mass equal to unity. This technical artifice is epitomized by Hamilton’s equations for the

generalized coordinates q1, . . . qn and their conjugate momenta pl , . . . p, ,

P k = -aHb, Q, t ) / & k I (1.3)

i k = aHb1 q, t ) / a p k >

where H is the “Hamiltonian” (see chapter 7).

1

2 CHAPTER 1. INTRODUCTION

I 4

Figure 1.1: Arbitrariness of p

The above process is not unique. For example, for a free particle ( q = 0) rather than

p = O , q = p , (1.5)

according to which p is constant and q = pt + qo (figure l.l(a)), we might equivalently have written (figure l.l(b))

p = l , q = p - t , p = t + a , q=a t+P . (1.6)

As we shall see in Chapter 6, this arbitrariness is connected with one for the Lagrangian.

Once the second order equations have been replaced by systems of first order ones, it is natural to describe the evolution of a system as the motion of a point in a 2n-dimensional “phase space” for the coordinates q l , . . . q, and the momenta PI, . . . p,.

1.2 Motion of a particle in one dimension It is instructive to study the motion of a particle in one dimension subject to the elastic force - k q and the repulsive force +kq (k > 0), respectively.

In the former case, the equations p = -kq , q = p/m have the general solution (w =

q = acos(wt + a ) , p = -mwasin(wt + a ) , (1.7)

so that the representative point in the phase plane ( p , q ) moves on an ellipse of semi-axes a = a and mwa = (figure 1.2), where E is the energy. The point q = 0 , p = 0 describes a state of stable equilibrium ( a = 0 , E = 0 ) .

1.2. MOTION OF A PARTICLE IN ONE DIMENSZON 3

4

Figure 1.2: Motion under the force -kq

In the case of the repulsive force +kq, the trajectories in the phase plane are hyperbolae. The solutions

q = f a cosh(wt) , p = fmwa sinh(wt) , (1.8)

correspond to a particle with initial velocity towards the center of repulsion coming from q = foo respectively, and stopping at f a before rebounding (figure 1.3). In these two cases the energy E = (p2/2m)-(kq2/2) is negative ( E = -ka2/2).

The solutions

(1.9) q = f a sinh(wt) , p = f m w a cosh(wt) ,

correspond to the positive energy E = +ka2/2. For t = 0 one has q = 0 , p = fmwa, i.e. a particle right on the center of repulsion with a finite velocity (figure 1.3).

The asymptotes p = fmwq separate the regions of positive and negative energies. They intersect at the point of unstable equilibrium p = 0 , q = 0.

Figure 1.3: Motion under the force +kq

4 CHAPTER 1. ~ N T R O ~ ~ C T ~ O N

Figure 1.4: Motion near potential energy extrema

Consider now the one-dimensional motion of a particle subject to a conservative force f ( q ) = -dU(q)/dq. In the vicinity of a potential energy minimum q = qo, one has f ( q ) = - k ( q - qo) with k = (d2V/dq2)q=,o > 0.

If the energy E differs little from U(q0)) the particle moves on an elli se of center q = qo p = 0 in the (p, q) plane with the period T = 2n s" m/k.

In the vicinity of a potential energy maximum the motion resembles that of a particle subject to the repulsive force f ( q ) = +k(q - qo) with k = -(d2U/dq~)q=q~ > 0. Passing through the point (q = qo ) p = 0) there will be a critical curve with tangents p = It;& (q - go), corresponding to the asymptotes of figure 1.3. Such a curve separates regions with E > Ufqo) and E < U(q0).

Returning to the elastic force f ( q ) = -kq, we notice that the area enclosed by an elliptical trajectory in the (p, q plane is

while the period is T = 2 7 ~ m . Clearly 2- = d A / d E .

plane: T = $ dt = $(dt/dq)dq = $(m/p)dq = $m dq/J2m(E - U(q)) ,

A = n a ( m w a ) = 2 n E JL m/k ,

This relation is easily generalized to any closed trajectory in the phase

where A is the area enclosed.

1.3. FLOW IN PHASE SPACE 5

1.3 Flow in phase space Consider a region in the (p,q) plane and assume that each point in the region represents a possible position and momentum of the particle at the time t = 0. If each point moves according to the dynamics of the particle, those points will in general cover a different region at a later time t > 0. We can visualize this as the flow of a “fluid” in two dimensions.

If the forces are conservative, the fluid is incompressible. (See problem 1.5 for a dissipative case.) For example, for the vertical motion of particles under gravity one has q = (gta /2) + b t / m ) + go , p = mgt +po. The points of the rectangular region 0 5 q i Q , 0 5 p 5 P at time t = 0, at a later time t > 0 will cover the parallelogram with vertices a = (g t2 /2 ,mgt ) , b = (gt2 /2 + &,mgt ) , c = ( # / 2 + Pt/m + Q, P + mgt), d = (g t2 /2 + Pt/rn, P + mgt) (see figure 1.5). It can be seen by inspection that the area of this parallelogram is equal to the area PQ of the rectangle at time t = 0.

Figure 1.5: Vertical motion under gravity

As a prelude to the general case, we can also express the area at time t > 0 in the form

But the Jacobian determinant

(1.11)

(1.12)

6

Figure 1.6: Elastic collision against a wall

The same property applies to particles undergoing elastic collision against

For the harmonic oscillator qt = qo cos(wt) + po sin(~t)/mw, a “wall” (see figure 1.6).

pt = -mwqo sin(&) f po cos(wt) one has similarly

and so the area conservation applies also in this case. Can this property be generalized to any conservative force?

Since Pt+at = P$ - ~ d ~ / d q ) ~ dt, qt+dt = qt + (Pt/m) dt, one has

Hence dAt - = o I

dt (1.13)

This is a special case of Liouville’s theorem, which will be proved in Chapter 7.

For a one-dimension~l system with ~ ~ s e r ~ t i v e forces, by Stokes’ theorem area conservation in the ( p , q ) plane is equivalent to conservation of a line integral along the curve enclosing the area,

h$ P d q = j6@ Pdq > (1.14)

where CO and Ct are the boundaries of DO and Dt, respectively.

Ct, where the domain of X does not depend on t , one has Direct proof: If p = p(X, t ) , q = q(A, t ) are the parametric equations of

1.3. FLOW rN PHASE SPACE 7

Figure 1.7: Example of equation (1.14)

As an example, let CO be the circle of equations q(X)=r cosX,p(X)=po-r sinX ,

and Ct the curve p(X, t) = mwq(X) sinh(wt) + p(X) cosh(wt) ,

q(X, t) = q(X) cosh(wt) + ( p ( X ) / w ) sinh(wt) , which is the mapping of CO under the flow induced by the repulsive force f ( q ) = Icq (k = mu’). One has (see figure 1.7)

htp dq = i 2 T [ w r cos X sinh(wt) + (Po - r sin A) cosh (wt)]

[-r sin X cosh(wt) - r cos X sinh(wt)/mw] dX

= -r2n[sinh2(wt) - cosh2(wt)] = +m2 - + d l ’

If CO is a (p,q)-orbit of a periodic motion, then Ct = CO at all times. In the phase flow, the points of CO pursue one another along Co itself. In this case,

is called a “cyclic action variable”. For the harmonic oscillator

q(A, t) = a cos(wt + A), p(X,t) = -mwa sin(wt + A),

(1.15)

E J = fmw2a2 sin2(& + A) dt = w a ’ sin2@ d8 = 2n- . (1.16)

W

8

The energy can be expressed in terms of J ,

WJ 2n

E = - , (1.17)

Since J = A, this agrees with the relation A = 2 r E @ = 2nE/w on page 4.

In the “old quantum mechanics” , Sommerfeld’s quantization condition

J = n h (n=o , l ...) (1.18)

yielded the energy levels for the harmonic oscillator

E = n t w (1.19)

instead of the correct E = (n + l / Z ) h . For a multiply periodic system with n degrees of freedom, the energy can

be expressed as a function of n cyclic action variables, E = E(J1 J2 . . . .In). An example will be presented in the next chapter (Keplerian motion) and the general case will be discussed in chapter 8.

1.4 The action integral The cyclic action variable J originates from ~amilton’s ‘‘characte~stic func-- tion”

S(qz1911E) = /=; Pdq . (1.20)

This is a very useful function. We notice first that the momenta at q1

The time between two positions q1 and q2 is given by and q 2 are given by p2 = dS/dqz and pl = - 8 S / d q l .

In fact

The formula T = dJ/dE for the period of a closed orbit is a special case for q2 = q1.

1.4. THE ACTION INTEGRAL 9

For the harmonic oscillator one has

S(q2, qi , E = ka2 /2 ) = 1; J 2 m ( E - kq2 /2 ) dq = m w I q 1 dq 91

= ( m w / 2 ) q2 a2 - qi + a2sin-'(q2/a) - same for q1 , (1.22) [J- 1 while (1.21) gives

t 2 - tl = [sin-'(qz/a) - sin-'(qI / a ) ] / w . (1.23)

Taking q1 = a, tl = to, q2 = q, t2 = t, we have

t - to = [sin-' (q/a) - 7r/2]/w , q = a cos(w(t - to))

The action S(q, qo, E ) obeys the Hamilton-Jacobi equation

. (1.24)

' ( " ) ' + U ( q ) 2m aq = E , (1.25)

as can be seen by substituting p = aS/aq in the expression for the energy, E = p 2 / 2 m + U(q) .

As an example, verify that the action for a particle of positive energy E subject to the force f ( q ) = +kq, I/ = -kq2/2, is

9

s(e, qo, E ) = Lo &m(E - u(qf)) dq'

and that it satisfies the Hamilton-Jacobi equation for U ( q ) = - k q 2 / 2 . All this may seem somewhat futile. However, the usefulness of action

integrals becomes evident as soon as we extend their definition to more than one dimension, say three dimensions.

Let

S(r,rO, E,a ) = 1: p * dr = 1: J 2 m ( E - U(r)) ds , (1.27)

where the line integral is along a trajectory characterized by the energy E and some integrals of motion summarily referred to by the symbol a. (In one dimension, the only integral of motion was E , hence no a) Note that along a trajectory p dr = lpl ds, ds = )drl.


For the two-dimensional case of projectile motion, with the x-axis horizontal and the y-axis vertically upwards, and ro = 0, in the ascending branch of the trajectory one finds

S(r,O, E , a = p , ) = J2m(E - Ex - mgy') dy'

= p,x 4- [(2m(E - Ex))# - (2m(E - E, - mgy))*]/3m2g , (1.28)

where Ex = pg/2rn.

a s f a x = p , = mvOz and (aS/ay),=, = Jm = rnvo,, it is clear that the trajectory is normal to the curve S = 0 at the r = 0 intersection.

This is not surprising. In the three-dimensional case, the normal to the surface S(r,ro, E,a) = 0 at ro is parallel to the gradient (VS),,, and this in turn is equal to the momentum p at ro.

A method for finding trajectories suggests itself. The action integral satisfies the Hamilton-Jacobi equation

Consider the normal to the curve S = 0 at r = 0. Since

(t/S)2/(2m) + u = E . (1.29)

If S(r) is a solution of this equation, consider the family of surfaces S(r) = const. The trajectories form a family of curves orthogonal to these surfaces. The problem of finding trajectories is similar to that of finding lines of force in electrostatics once the equipotential surfaces are known.

Let us reexamine the problem of projectile motion from this viewpoint. The H ~ i l t o n - ~ a ~ o ~ i equation

(1.30)

can be solved by separation of variables, S(x, y) = Sx(x) + S*(y~, (dS, /dxf2 /(am) = E,, (dS, /dy~2/(2m) + mgy = E,, with Ex + E3 = E .

The function

S(x,y) = d E x - [(2m(E - Ex - mgy))* - (2m(E - Ex)) 9 ]/3m2g (1.31)

is a solution of equation (1.301, ~ r r e s p o n d i n ~ to a curve passing through the origin.

To find the trajectory through the origin corresponding to the energies Ex and EV, we write

1.4. THE ACTION INTEGRAL 11

Figure 1.8: All trajectories in this figure correspond to the same values of P, and P,.

where we have chosen the positive sign for dyldx (ascending branch of the trajectory). This expresses the fact that a displacement (dx, dy) along the trajectory is parallel to the normal to the curve represented by S(x, y).

dy/JE - E, - mgy = dz/& , -(2/mg)dE - E, - mgy = x / a - ( 2 / m g ) J x .

Separating variables, we have

Putting E, = m&/2 and E - E, = mu&/2, and assuming that uox and voY are both positive, this reduces to the familiar formula

(1.32)

This, of course, can be derived by equating the two expressions for t , x/wo,

The Hamilton-Jacobi equation for a free particle of energy E in two

S = p , x k & q z X j y . (1.33)

The curves S = const (solid) and the trajectories (broken) are shown in figure 1.8 for p , > 0 and the plus sign. Of course, the roles of x and y can be interchanged.

However, the Hamilton-Jacobi equation for a free particle of energy E in two dimensions has also the solution

~ = N G E l r - r o l . (1.34)

The curves S=const are circles, the trajectories are straight lines through ro (see figure 1.9). The angular momentum about 10, I = 0, plays the role of the constant of motion a other than the energy (see equation (1.27)).

and (yo9 - d- )/9.

dimensions with momentum component p , has the solution

12 CHAPTER 1 . INTRODUCTION

Figure 1.9: Equation (1.34)

1.5 The Maupertuis principle We have seen that

S(B) - S(A) = ( [P dr) ( l b b ) , (1.35) t

where p = VS, JpI = [VSl = J2m(E - U(r) and “t” is the actual trajectory from A to B.

that Now if “c” is a curve from A to B adjacent to the trajectory, we see

since p and dr are not parallel on “c”. Hence the Maupertuis principle or principle of “least action”: The ac-

tion integral from a point A to a point B, JA J2m(E - U(r) ds, is minimal along a trajectory. Less restrictively, Euler’s formulation of the principle states that

6 h B d m d s = D , (1.37)

B

where 6 indicates variation of the integral when the coordinates r = (2, y, z ) of each point of a trajectory from A to B are changed to r + 6r = (x + 6x,y + 6y,z + Sz), with 6r vanishing at A and B.

It must be possible to derive the equations of trajectories from this variational principle. Let us do so in two dimensions and assuming that U = U(y). With y’ dy/dx, we have

6 1 B d m d s = . A m d z

1.6. THE ~ 1 ~ E 13

Partial integration using &A = &B = 0 yields the trajectory equation

~ u l t i p l y i n ~ by g ' d ( E - U)/( I + gt2) we obtain

so that the quantity in parentheses is a constant, which we denote by C. Hence

The value of the constant can be determined from that of g' at a point of the trajectory. For a projectile (U = m g g ) , we have E = mv$/2,

lution by s e p ~ a t i o n of variables yields equation (1.32) if the value of the integration constant is chosen so that the trajectory may pass through the origin.

(gt)z=w& = V O ~ / U ~ ~ , and 50 c = mv$,/Z, gf = ( ~ - ) / v ~ ~ * so-

1.6 The time What about the time? Equation (1.21) expressed the time interval during which a particle moves from a position to another as the derivative of the action integral with respect to the energy. That formula is easily generalized to three dimensions. In fact

a -S(r2,rl,E,tu) = - BE p - dr = ~~, B f' JZm(E - U(r)) ds

ra = 1; m ds/J2m(E - U(r)) = 1 dslv = t 2 - tl , (1.38)

r l

where v denotes the velocity. It is interesting to compare this time interval with the time it would take

a particle of the same mass, total energy E, and potential energy U(r), to


Figure 1.10: Real and varied motion under gravity

go from rl to r2 along a path other than the trajectory. This latter motion requires a frictionless constraint.

Consider as an example the motion of a projectile. Since

S ( x , u, 0, E, P , = mvoz) = P,X

2 4 4 +[(2E - muOs) - (2E - mu:, - 2 m g y ) ] / 3 f i g , the time t from (0,O) to ( x , y ) is

= (uo, - J-) / g 1

a formula known to high-school students. The time taken to reach the highest point of the trajectory, y = h = u&/2g, x = X h = u02u0,/g, is, of course, t = uo,/g.

Note that instead of choosing p , as a constant of motion a other than the total energy E , we might have chosen the energy of the y-motion, E, = mui /2 + m g y . Then we would have had

t = BS,/BE = B(J2m(E - E,) x ) / B E = xV/m/2(E - E9) = x/uoz. Suppose now that the body, with the same total energy

E = m(uix + vi,)/2, travels from (0,O) to (zh, h) along the straight line of equation y = hx/xh (see figure 1.10) under the influence of gravity and a frictionless constraint.

The time t' taken will be shorter than t = uoy/g,

1.6. THE TIME 15

Does a shortest-time curve from (0,O) to (q,, h) exist for that energy

Its equation is obtained by requiring that E? Yes, it is Bernoulli’s “brachistochrone” .

(1.39)

for variations dr restricted to vanish at A and B.

ential equation A calculation similar to that presented in section 1.5 yields the differ-

Y f 2 1 au ( E - U)(1 + y f 2 )

For U = mgy one obtains by a first integration

= J C - ( H H-Y ’ where C is an integration constant and H = E/mg. Note that 9‘ is infinite (tangent is vertical) for y = H, indicating that v, = 0 for y = H. Thus H is indeed the maximum height reached by a body of energy E.

A second integration putting H - y = C ( l + cosO)/2, and a suitable choice of integration constants, yield

z = H ( 8 + sin8)/2 , y = H(1- cos8)/2 . (1.40)

These are the parametric equations of a cycloid. Thus the brachistochrone is also “tautochrone” (see problem 1.8).

Note that z = y = 0 for 6 = 0, and (z = r H / 2 , y = H ) for 8 = K. The time taken by the projectile of energy E = mgH to go from (0,O) to the point characterized by the value 8 of the parameter is

sin8 d0

where we have used ds = dl + (dz/dy)2dy = a dy, v = d-, and the parametric equations.


We now want to find a brachistochrone through the points (0,O) and (zh, h) of the projectile trajectory. It is easy to show that a brachistochrone for E = mv;4/2 = m(v& + v&,)/2 and H = 4 / 2 9 > v&/2g = h will intersect the projectile trajectory at (q, h) for the value 6 = ~ V O ~ V O , / V ~ of the parameter. The time from (0,O) to (xh, h) will be

1 ‘ U O y < ’ U 0 y = t . t f f = d m 9

It is also easy to show that t” < t’, see problem 1.9.

1.7 Fermat’s principle We must briefly mention the remarkable similarity between geometrical optics and mechanics of a point-mass. In the former, Fermat’s principle

(1.41)

expresses the fact that the time taken by light to travel from A to B along a light ray (“optical path length”) is less than the time it would take along any adjacent path from A to B.

In mechanics, Maupertuis’ principle is the analogue of Fermat’s principle. The action (not the time) is the analogue of the optical path length.

Fermat’s principle is the consequence of the existence of families of “iconal surfaces” to which the families of light rays are orthogonal, in the same way as trajectories are orthogonal to S =const surfaces.

An “iconal equation” is obtained as first approximation of the wave equation for a given frequency u,

4n2 X2

V”+-$=O , (1.42)

where X = Xo/n = c/nu, XO is the vacuum wavelength, and n( r ) is the refractive index.

One begins by expressing $J in the form

$ = exp(2niS(r)/Xo) , ( 1.43)

where S = SO + ( X 0 / 2 7 r ) S 1 + ( X 0 / 2 7 r ) ~ S 2 + . . . iconal equation

Substituting in the wave equation, one finds in first approximation the

( v s O ) ~ = n2 . ( 1.44)

1.7. FERMAT’S PRINCIPLE 17

This is the analogue of the Hamilton-Jacobi equation.

is to geometrical optics. In fact, comparing the Schrodinger equation Wave mechanics (de Broglie) is to classical mechanics what wave optics

2m V2$+ -(E - V)$ = 0 h2

(1.45)

with equation (1.42) one sees that d m plays the role of a refractive index, while the wave-mechanical wavelength is

A = h/J2- = h/p(r).

18 CHAPTER 1 . 1NTRODUCTION

1.8 Chapter 1 problems 1.1 The damped oscillator equation q = -q - q is equivalent to the system q = p , p = -q - yp. Show that these equations cannot be expressed in the Hamiltonian form (1.3-4).

1.2 The damped oscillator equation of the preceding problem is also equivalent to 4 = pexp(-rt), 6 = -q exp(yt). Verify that these equations are Hamilton equations, with H = [p2exp(-rt) + q2exp(yt)]/2.

1.3 The equation q = - (q - qo) for a harmonic oscillator with equilibrium point q = qo, can be converted into the system of equations q = p -pol p = -q + qo, where po is an arbitrary constant. These are Hamilton equations, since a ( p - Po)/& = 0 and a(-q + q o ) / d p = 0. (i) Find the Hamiltonian. (ii) What is the ( p , q ) trajectory for energy E?

1.4 Show that the 4-dimensional phase space volume is conserved in the elastic collision of two particles in one dimension.

1.5 Non-conservative forces invalidate Liouville's theorem. Show that, if = -yq (y > 0), q = p/m, areas in the (p, q)-plane decrease exponentially

with time.

1.6 According to Liouville's theorem, conservative forces cannot change the small-scale phase space density of a system of particles. Yet they can change the large-scale density producing accumulation.

Give a simple one-dimensional example (two-dimesnional phase space). Look up also S. van der Meer, "Stochastic cooling and the accumulation of antiprotons", Revs. Mod. Phys. ,57( 1985)689.

1.7 For the two-dimensional oscillator with T = m(i? + 3i2)/2, U = mw2(x2 + y2)/2, and energy E = E, + E, with E, = mw2a2/2, Ell = mw2b2/2, the action integral is given by S = S, + S,, with S, obtained from equation (1.22) with the replacement q2 + 2, q1 + xo and S, with the replacements a -+ b, q2 + y, q1 + yo.

Show that

t - t o = [sin-' (./a) - sin-'(x~/a)]/w = [sin-' (y/b) - sin-'(yo/b)]/w . Equating these two expressions for t - t o , show that the trajectory equation is

b 2 z 2 + a2y2 - 2abcos(a - P)zy = a2b2sin2(a - /3) (an ellipse) ,

1.8. CHAPTER 1 PROBLEMS 19

Figure 1.11: Illustration of Maupertuis principle

where a = cos - ' (~~ /a ) and p = c0s-l (yo/b).

1.8 A particle, initially at rest, slides down without friction along the brachistochrone (1.40). Show that the time required to reach the bottom is n m , independent of the initial position. Thus the brachistochrone is also tautochrone.

1.9 With the notation of section 1.6, show that tff < t'.

1.10 In figure 1.11, the circle represents an earth meridian, A and B are the initial and final positions of an object released from rest at A and travelling to B along a smooth straight tunnel.

Assuming c infinitesimal, show that

D = (/ &rn( E - V ( P')) ds' ) ACB - ( J JWE - WJ)) ds > O

as expected from the Maupertuis principle.

20 CHAPTER 1 . INTRODUCTION

Solutions to ch. 1 problems S1.l p = -BH/aq, q = a H / a p would give -q - 7p = -aH/aq, p = a H / + . Differentiating the first with respect to p we have a ~ H / ~ ~ q = 7, while differentiating the second with respect to q we have a2 Hlaqap = 0.

S1.3 (i) H = [ ( p -PO)'+ (q - q0)']/2 (ii) q = q o + m cost, p = po- circle of radius

sin t , and center (pa, qo).

and so dAi/dt = -m-'7At, At = AO exp(--m-'yt).

Sl.6 In figure 1.2, think of two small phase space regions around (q = a,p = 0) and (q = -a,p = 0) filled with particles. If the p-axis of the ellipse is much smaller than the q-axis (k very small), after a quarter of a period the two regions will have moved to (q = 0,p = f m w a ) with resulting accumulation around (q = 0,p = 0).

In an antiproton accumulator the empty (phase) spaces between the particles are squeezed outwards, while each tipr rot on is pushed towards the center of the distribution.

S1.7 Regarding Ev (and, therefore, b) as constant of motion other than E , since a = w-'d- , the first of the two expressions for t - to is given by t - t o = OS&fE = ( a S ~ / a a ) ( a a / a ~ ) . Also t-to = as,/aE = ( a S , / ~ b ) ( a b / ~ ~ ) .

The trajectory equation is obtained by equating the two expressions for t - to . This gives sin-'(z/a) - sin-'(y/b) = sin-'(zo/a) - sin-'(yo/b), which yields the desired result by using the identity

sin-'€ - sin-lq= sin-' (q + J ( 1 - e2)(1- , a ) ) .

S1.8 I t is convenient to use the parametric equations of the cycloid in terms of the arc length s from the lowest point, x = . . ., y = s2/4H. Then

ms = -mg dy/ds = -mgs/2H, 5 + wza = 0 with w = m, harmonic motion of the cycloidal pendulum. The time sought is T/4 = ~ / 2 w = ~m.


S1.9 With x = P I O ~ / V O ~ > 0, t"?t' is equivalent to

By simple algebra this yields O?ze + 8x6 + 20x4 + 12%'. Thus "?" is "<".

s1.10 P ( z , 0), P ' ( ~ , b y ) with by = C ~ ( Z + qR)/R, 9 = -Z /~Z[ ,

U(P) = mg(za - 3R2)/2R, U(P') = mg(z2 + (by)' - E = -mgR, d m - d m = -mg(dy)'//(4R

dd - ds N €'&/(2R'),

2m(E - U(P)) ] dz D 1: / l [ d 2 m ( E - U ( P f ) ) - d 2 m ( E - U(P)) + -d Ez 2R"

-_- m3ge' J A z(z + qR)& > - f i R 3 - R J ~

>

since Z(Z + qR) < 0 in the integration interval.

Chapter 2

EXAMPLES OF PARTICLE MOTION

We collect a body of notions to be used in the more formal parts of the book.

2.1 Central forces If a particle moves under the action of a central force f = f ( r ) r / r , its angular momentum 1 with respect to the center of force r = 0 is conserved. This follows from the vanishing of the torque T due to f and r being along the 8ame line. The motion is confined to a plane normal to 1. Expressing Newton’s second law in polar coordinates, we have

mar = f ( r ) and mag = O , (2.1)

where

(2.2) * I d

a, = i: - re2 and a0 = rf? + 2iO = --(r28) . r dt

These second-order equations can be replaced by the following first- order system:

where pe = mr2e has the physical dimensions of an angular momentum. In fact, ps = I = 111 & the magnitude of the conserved angular momentum.

23

24 CHAPTER 2. EXAMPLES OF PARTZCLE MOTION

If the force is conservative, the energy E is a constant of motion. De- noting by U ( r ) the potential energy (f(r) = --U'(r)) , E can be expressed in the form:

where 12

is an "effective potential energy" €or the radial motion. A second-order equation for r can be written in the form

Note the important formulae

and

where the square roots must be taken with the signs of dr( t ) /d t and dr(3)/d@, respectively.

The study of the r-motion is similar to that of the one-dimensional motion of chapter 1 but for the presence of the "centrifugal term" in the effective potential energy.

Thus if f ( r ) is attractive but l i r n r ~ * r 2 1 ~ ( r ) ~ = 0, the centrifugal term prevails at short distances, and there is a mimimum distance of approach to the force center.

On the other hand, if l i m r + ~ r 2 ~ U ( r ) ~ = o (e.g. U = -a / r3 , a > 0) the particle falls into the center of force.

2.2. CIRCULAR AND QUASI-CIRCULAR ORBITS 25

2.2 Circular and quasi-circular orbits Equilibrium points of one-dimensional motion correspond to circular orbits. There will be a circular orbit of radius R if (dUe(r)/dr),=R = 0, namely f (R)+Z2/(mR3) = 0, which is nothing but the elementary m v 2 / R = - f ( R ) ( f ( R ) < 0). The period will be T = 2nd-.

We now study small oscillations about a stable circular orbit. Denoting by R the radius of a circular orbit, for r differing little from R we have

m d2(r - R)/dt2 =mi: = -Ui(r)

N -[Ui(R) + U t ( R ) ( r - R ) ] = - U ~ ( R ) ( T - R ) . (2.9) The circular orbit is stable if Uf(R) > 0. The angular frequency of

small oscillations about a stable circular orbit is

w,, = d m = J(Ut f (R) + 312/mR4) /m (2.10)

wosc = .\/(U"(R) + 3 U ' ( R ) / R ) / m , (2.11)

TmC = SnJm/(U"(R) + 3 U ' ( R ) / R ) , (2.12)

or, using 12/mR3 = - f ( R ) = U'(R) ,

and the period is

while that of the circular orbit is

T = 2 n d m . (2.13)

For an almost circular orbit, the angle Ad between two successive peri- centers is approximately

2nl U ' ( R ) A8 = m,, = RJm(R2U"(R) + 3 R U f ( R ) ) = 2"\/RU''(R) + 3U'(R) '

(2.14) For the Newton force ( f ( r ) = - k / r ) this gives A0 = 2n, and for the

isotropic oscillator ( f ( r ) = -kr) A8 = n. In both cases the orbits are closed.

Note that for U ( r ) = Cra and U( r ) L- Cln(r/ro), Ad does not depend on R.

For V ( r ) = - ( k / r ) - h/r2 the orbits are not closed (rosettes). If h is small

A0 = 2 n d G N 2n (1 + &) (2.15)

26 CHAPTER 2. EXAMPLES OF PARTICLE MOTION

Figure 2.1: Isotropic harmonic oscillator

The average angular velocity of precession is

(2.16)

For h > 0 (attractive cubic force), the precession is in the direction of motion.

2.3 Isotropic harmonic oscillator The force is central and elastic (f(r) = -h), U,(r) = (kr2 + Z2/mrz)/2 is shown in figures 2.1 (left 1 = 0, right 1 > 0).

It is convenient to introduce the variabie s = l/r2. The turning points rl and r2 > r1 are then given by 7-1 = 1 1 6 and r2 = I/&, (31 > 9 2 )

with

(2.17)

The minimum of U, is for r = R, where R is the radius of the stable circular orbit R = G. Since Ue(R) = wd, for given E must be E > wl.

Equation (2.8) gives

l 8 ds (2.18) e = o o - - l o J(s1 - S)(S - 92) ’

Taking SO = (81 + 92)/2 and 80 = ~ 1 4 , so that

(2.20)

2.4. THE KEPLER PROBLEM 27

we find the ellipse.

3 = [31 32 (31 - 9’2) ~ 0 ~ ( 2 8 ) ] / 2 , (2 .21)

T2 = ( 1 + d%cos(%’)) . (2 .22)

Note that T = T I for 8 = 0 and T = ~2 for 8 = n / 2 , in accordance with

We end this section with the calculation of the cyclic action integral what was noted after (2 .14) . See also problems 2.3 and 2.5.

Jr = Jprdr, where pr = m+ = (l/r‘)d~/d8:

277 sin2 (28)d8 Jr = l ( 1 - g) [ l + ~ ~ c o s ( 2 8 ) 1 2 ’

But 2nl = S,2”ld8 = li*ped8, with pe as defined in (2 .3 ) , and SO

(2.23)

(2 .24)

Since $ p a dr = m l ( 5 dz + 6 dy) = m$(+ dr + r28 do), we have Jr + Je = J, + Jv. Therefore by Sommerfeld’s quantization the quantized energy would be E = Aw(n, + nv) or E = hw(nr + no).

2.4 The Kepler problem We discuss negative-energy orbits for

(2 .25)

with h << 12 /2m. Note that the -h/r2 potential cannot be compensated by changing the

value of 1. In fact, 1 appears not only under the square root, but also as a factor multiplying the integral in equation (2 .8) . This gives

(2.26) dr

1 ; v 2 , / 2 m [ E + ( k / r ) + ( 2 m h - 1 2 ) / 2 m ~ 2 ] * 9 = 8 o + l

Introducing the variable s = 1 / ~ , 8 can be expressed in the form

(2 .27) d3

d(31 - 3)(3 - 32) ’

28

where a = t/l - 2mh/l" , (2.28)

(2.29)

The integral is elementary. We find

230 - 81 - 32

s1 - s2 ) -sin-' ( 23 - 81 - 92

6 = $0 - [sin-' ( a 51 - 8 2

Taking 80 = 31 and 60 = 0, we have

a 81 - 3 2 (2.31)

3 = [sl + 32 + - 8 2 ) c o s ( a e ) p . (2.32)

We see at once that the orbit is not closed if a # 1 ( h # 0). The angle of precession per revolution is then 2n(a-' - 1) N 2nmh/Z2 as anticipated in equation (2.16). For a = 1 ( h = 0) the orbit is an ellipse of equation

Here

(2.33)

(2.34)

E = -k/2a c = J1 - 2)ElP/mk2 , a( l - i') = I"/mk , (2.35)

We now calculate the action integral J, = $p,dr, where pp = (Z/r2)dr/d6 = le sin 6/a( 1 - c2), dr = m( 1 - e2) sin O / ( 1 + E cos B)-2dB. A partial integration yields

(2.36)

The evaluation of the integral is a well-known exercise in complex integration: ~~sdB/( l+ccosB) = (2/ie)$dz/(z2f2€-'z+1), where the path integral is along a circle of unit radius and center at the origin. The pole z = (-1 f d m ) / c falls inside this circle.

(2.37)

2.4. THE KEPLER PROBLEM

Hence 2n2mk2

I E ‘ = (J, + J g ) 2 ’ and, by Sommerfeld’s quantization and k = e l ,

29

(2.38)

(2.39)

for the hydrogen energy levels. Note that both here and in equation (2.24) for the isotropic harmonic

oscillator, we would expect to see J, + J e + J4 and n, + ng + ng instead of JT f J.g and n, + ng. The absence of J g is due to our having confined the orbit to a meridian plane so that 4 = 0.

Schwinger (see G. Baym,Lectures on Quantum Mechanics (Addison- Wesley,1969) problem 3, p. 179) noticed that the radial Schrodinger equation for the hydrogen atom can be reduced to the equation for the two- dimensional isotropic harmonic oscillator. In the present context, let us compare the equation for the electron orbit

(2.40)

with that for the isotropic harmonic oscillator

(2.41) P2

where we have renamed the polar angle and the distance from the origin. With the correspondence r + p2 and 8 + 2p, the hydrogen orbit becomes that of an oscillator of energy Eosc = e2 and frequency w = d w .

Therefore, if in Eoac = b ( n , Eos, + e2 and w -+ J-, we find e2 =

The period is given by

(2.42)

yielding Kepler’s third law (k = GmM)

(2.43)

4n2a3 GM

T2 = - (2.44)


X

In contrast to 8, the angle u (“eccentric anomaly”) shown in figure (2.2)

We see from the figure that a cos u = at: + r cos 8. Substituting is simply related to the time t ,

rcos6 = [a(l- e 2 ) - r ] / e (from equation (2.33)), we find r = a(1 -ecogu), 8 = l / a ( I - (SCOSU). NOW

1 1 - €(l - cosu)

€(l + cosu)

- 3 1 - 3 = - - U(1 - €) Q(1 - €COSU) a(1 - €)(I - €COSU) ’

_I

1 --- 1 3 - 8 2 =

a(1 - ecosu) a ( l + c ) a(1 + €)(l - ccosu) ’

JW - ~ ~ ( r ) ) / m

and so dr

where we have taken to = 0 for ug = 0. Hence

u - esinu = nt (2.46)

with

(2.47) 1 n=

m a v m

Comparing (2.47) with (2.43) we see that n = 27r/T is the average angular velocity. Of course, we could have seen this by taking t = T and u = 2n in (2.46).

This equation is useful for the calculation of time averages, since dt/T = (1 - rcos%)dU/Za .

2.5. THE L R - L VECTOR 31

For example, with x = r cos8 equation (2.33) gives r + ex = a(1 - e’). Denoting by ( T ) and (z) the time averages of r and x, we have

c(z) - a(1- c a ) = -(r) = -- $ l T a ( l - ccosu)dt

= +1’= a(1- ccosu)’du = -a(l + c’/2) .

Therefore e(z) = -3a1?/2. Since (g) = 0, we find

(r) = -(3ac/2)6, = - ( 3 a / 2 ) A , where A is a vector of magnitude c pointing from the center of force to the pericenter, the L-R-L vector discussed in the next section.

2.5 The L-R-L vector For U = -k/r , the Laplace-Runge-Lenz (L-R-L) vector

r v x l r k

A = - - + - (2.48)

is a constant of motion (dA/dt = 0) as is easy to prove. At the point of neareat approach, A = (-1 + Z2/kmr)r/r for both closed and open orbits. Using (2.33), the last of (2.35), and anticipating (2.54) and the last of (2.55), it is eaay to show that at the point of nearest approach A = ekr/lklr for both closed and open orbits. Thus IAl = e, and A points from the center of force to the point of nearest approach for k > 0 and E negative or positive, and in the opposite direction for k < 0 and E, of course, positive.

Multiplying vectorially both sides of the equation kr/r = v x 1 - kA by 1 and squaring, one finds

(v - kl x A/Z2)’ = k2/Z2 . (2.49)

Thus the tip of v is on a circle of center kl x All2 and radius kll. Note that 11 x A( = le. For closed orbits v describes the whole circle (2.49) and one has

(2.50) vmax rmax

vmin 1 - e rmin ‘

---=- -

‘The “accidental degeneracy” of the energy levels of the hydrogen atom, i.e. their independence of the angular momentum quantum number, is due to the inverse-square form of the Coulomb force. This endows the Lagrangian with a special symmetry property (eee problem 6.241, which is reaponsible for the constancy of the L-R-L vector.


Figure 2.3: Rutherford orbits

Therefore ZJjnaxTjajii = Z r j n j j l T ~ ~ ~ as one knows from General Physics. For open orbits (Kepler-Rutherford trajectories) see section 2.6.

If the force acting on the particle is f = - k r ~ - ~ + 6f, one finds easily

1 dA dt km k

-- I 6 f x l + - v x ( r x 6 f ) , --

If 6f is central (6f = 6 f ( ~ ) r/r), this reduces to

dA 1 - = - S ~ ( T ) r x 1 . dt kmr

(2.51)

(2.52)

A derivation of equation (2.15) based on this formula is presented in problem 2.6.

The L-R-L vector is "Laplace's second vector", the first being the angular velocity. It was used by W. Lenz who took it from a text boo^ by C. Runge. Neither of these two authors cfaimed it as an original finding. This historical question was discussed by H. Goldstein, Am. J . Phys.,43(1975)737 and 44(1976)1123.

2.6 Open Kepler-Rutherford orbits For U = -k / r we consider the case E > 0, but allow k to be positive (attractive force) or negative (repulsive force). We find

1 (k/JkJ) + ecos8 - = r U ( € 2 - 1)

(2.53)

(2.54)

with t = t/l + 2 E P / m k 2 > 1 , a(e" - 1) = 12//mlkl (2.55)

2.6. OPEN KEPLER-RUTHERFORD ORBITS 33

Figure 2.4: L-R-L vector for Rutherford scattering

The orbits are figure 2.3) (z2/a2) - (y2/b2) = 1, a = Z2/m)k)(e2 -

I 1 A I B I F I I ‘ I

The positron travels from r = 00, 62 = -cos-’(l/e) < 0 to r = 00,

e,’,t = ~os-l(i/e) > 0, IeLI = le,+,,I < ~ / 2 .

C”, = e,+,, + 7 > 0, le i1 = l@;utl > n/2 . The electron travels from r = 00, 6, = 6: - 7 < 0 to r = 00,

As one knows from analytical geometry, b is the “impact parameter”, shortest distance of F from the asymptotes u/x = fb/a. For k > 0 (attractive force), E = mui/2, where WO is the velocity of the incident particle at infinite distance from F. A simple calculation gives 1 = bmuo. In this case the tip of v does not cover the whole of the circle (2.49). One finds umM = k( l+e) / I as for closed orbits. In order to find Vmin = 210, one must substitute v, = - v ~ c o s ~ ~ , , = - W O / E , wY = ~osin@,$,, in (2.49). This gives uo = ( k / Z ) J ? T = b/rmin, Umaxrmin = Uminb.

Zero-energy orbits: These are possible only for k > 0. They are parabolas (e = 1)

(2.56)

The L-R-L vector is a useful tool also in treating open orbits. In order to clarify its meaning in this case, we define

I=)pla= kmA (l-$)r-zp .

If the center of force is inactive, then 2 lim I = r - (r + p)p/)p] = r l

k+O

Using the L-R-L vector, it is easy to show that the scattering angle 0

tan(@/2) = qQ/2Eb , (2.57)

where q is the charge of the scattered particle, Q that of the fixed scattering center (q& > O ) , E is the energy and b the impact parameter.

In fact, A * p = -r*p/r, lim,.+coA.p = -IpjeJ, where pilt is the momentum of the incid~nt particles, COSN = - l i m , . + ~ A . p / ~ A ~ ~ p ~ = l/lAl = l/e, o = T - 2a, sin(0/2) = I/€, tan(@/2) = ~ / v ‘ Z i = /C/ZE~, IC = qQ.

in Rutherford’s scattering is given by

2.7 Integrability The isotropic harmonic oscillator and the Kepler motion share two distinct properties. They are integrable and 4 M.

The former property, integrability, is apparent from equations (2.7) and (2.8) giving t = t ( ~ ) and 0 = 8(r) in the form of definite integrals. Once these formulae have been written, the remaining task is “to do” the integrals. This could be done exactly and equations (2.22) and (2.33) were estab~ished.

The equations of motion of a particle in a conservative central field with potential energy V( r ) are integrable because equations (2.7) and (2.8) are available. If, however, a search in a table of integrals proves unsuccessful, and one must resort to numerical integration, the equations are obviously not solvable exactly.

This definition of exact solvability is a little loose. For instance, an integral may be expressed analytically as a slowly converging series whose use is less convenient than calculating it numerically.

Integrability results from the existence of a sufficient number of integrals of motion, and the existence of an integral of motion is connected with the i n ~ r i a n c e of the equations under a transformation. This will be the leitmotif of Chapters 6 and 7.

Let US consider a particle in three dimensions acted upon by a force f = -VU(r) which does not depend explicitly on the time. The energy E is an integral of motion connected with the invariance of f = ma under time translations t’ = t + 6t (6t constant).

If U = V(p) ( p = d w ) , the n-components of the momenturn and of the angular momentum, pa and l , , are integrals of motion. They

2.7. INTEGRABILITY 35

result from the invariance of the equations under translations z' = z + dz (6% constant) and rotations 4' = 4 + 64 (64 constant). The three integrals of motion, E, p,, and 1, make the equations integrable.

If U = U(T) (T = d-), Z,, Z,, and 1, are integrals of motion resulting from the invariance of the equations under rotations about each of the coordinate axes. They amount to the magnitude of the angular momentum, and a direction which we took as the z axis. Together with E, I , = 111 = I made the equations integrable.

A non-trivial example of integrable two-dimensional problem is the motion with the Toda potential energy

(2.58) uT = -[e2(2-#fi) 1 + e2(++Yfi) + e-421 - - 1 24 8 .

Together with the energy E, the integral of motion

I = 8ai(G2 -3S2)+ (G+&k)e2(z-ufi) + (2j-d&)e2(2+uA) - 22je-*" (2.59)

found by HBnon makes the problem integrable. Expanding UT to the third order in x and 31 one has

the HBnon-Heiles potential energy. It seems that the equations of motion with the potential energy UHH have only one integral of motion, the energy. Therefore, they are not integrable.

2,8 Chapter 2 problems 2.1 For a constant attractive central force (U = Cr, C > 0), equation (2.12) gives To,, = 2 7 ~ d m = T / d . Compare this result with the following.

A puck of mass m on a frictionless horizontal table is pulled by an inextensible string passing through a small hole in the table and attached to a hanging mass M .

2.2 Study the stability of circular orbits for (i) U = C P , and (ii) U = Cln(r/w), where C, CY # 0, and ro are constants.

2.3 Substit~ting U = Cra in equation (2.12), we find

To,, = 27rJmR2-"/a(a + 2)C

For the isotropic oscillator (C = k/2, a = 2) and for the Newton-Coulomb force (C = -k < 0, a = - l) , compare To,, with the period T of the circular orbit, and draw illustrative graphs.

2.4 Study the negative- and zero-energy orbits of a particle acted upon by the attractive force f,. = -2h/r3 (h > 0). Consider only the case 2mh > 1'. For other cases, as well as for a combination of an inverse-square and an ~ n v ~ r s e - ~ u b e force, see G.L. Kotkin and V.G. S ~ b ~ , ~ o ~ ~ e e ~ g o ~ of P r o b ~ e ~ s in Classical ~ e c ~ ~ n z c s (Pergamon Press, 1971).

2.5 With x = r cos8, y = r sin8, equation (2.21) gives

This is the equation of an ellipse

How is this related to the equation

found in problem 1.7?

8222 + s ly2 = 1 .

+ ( Y / B ) ~ = 1 with A = 1/&, B = 1/f i

(x"/u2) + (yt2/@) - (2cosG/ub)x'y' = sin2&

2.6 Use equation (2.52) to evaluate the angle of precession per revolution when &f(r ) is small,

2.7 (i) Show that a particle with energy E < 1/6 and potential energy UHH(X,Y) given by (2.60) is trapped in the triangle of vertices (z = 1,y = O ) , (z = -1/2,y = 4 / 2 ) , (z = -1/2,y = 4 / 2 ) .

(ii) These vertices are saddle points. Show this for the (1,O) vertex.


Figure 2.5: Quasi-circular orbits

Solutions to ch. 2 problems s2.1 The force (7, tension of the string) is constant if the motion of the puck is circular with center at the hole (r = M g ) , the period being T = 2 7 r J w . However, if m executes small oscillations about the circular orbit, one has

mi: = -r + I2/m~', Mi: = r - M g , (m + M)i: = - M g + l'/rn~',

(the tension T ) is given by T = M(i: + g ) , and is clearly not constant.

s2.2 (i) The force must be attractive, and so - ~ T ~ - ' C < 0, aC > 0. Stability requires Ut = U" + 3U'/R > 0, a(a - l)C + 3aC > 0, C a ( a + 2) > 0. Since aC > 0, the condition for stability is a > -2 (a # 0).

Ut = - (C/R2)+(3C/R2) = 2C/R' > 0, indicating stability of all circular orbits. (ii) For u = Cln(T/To) (c > o), f(T) = -c/T,

S2.3 For the isotropic oscillator one finds To,, = 7rm = T/2, where T = 2 1 r G is the period of the circular orbit.

This can be understood analytically by considering the equations of motion x = acos(wt), y = bsin(wt) with a = R + t and 6 = R - E . Since T = Jm N JRa + 2Recos(2wt), we see that wosc = 2w. For a graphic illustration, see figure 2.5(a).

For the Newton-Coulomb force we have Tosc = 2 7 r J m = T, which is evident from figure 2.5(b).

52.4 Inserting U = -h/T2 in (2.8) with the square root taken with a minus sign, we find

38 CHAPTER 2. E ~ A M P ~ ~ S OF PARTICLE MOTION

For simplicity's sake we take 00 = 0, ro = rmW. Then

It is ctear that r + 0 as 9 -+ cot in which case

The particle falh into the center of force after an infinite number of revolutions. On the other hand, from equation (2.7) we have

= t o + ,/mi(,/= - JZZ7 ) . W i n g to = 0 and r a = rmW, we see that the fall into the center of force takes

only the finite time tiail = r m m d m = v ! ! / 2 1 E [ , The vebcity tends to infinity as T tends to zero because the potential energy tends to --03 SO that the kinetic energy tends to +cw.

Circular orbits are E = 0 orbits with the additional requirement h = 12/2m. In fact mv'/R = 2h/R3 gives I' = 2hm and E = snu'/2 - hfR' = 0.

S2.5 The equation for (z ,~ ) reduces to that for (E', y') by a rotation x = x'cos(7/2) + g'sin(7/2) , y = -z'sin(?/Z) + y'cos(712)

if [sl + [SI +

+ (81 - 8 2 ) cos*yJ/~ = l/(a'sin'&), - (31 - sz) cosy]/2 = l/(b'sin26),

and (31 - 9 2 ) sin 7 = 2 cos~/ub sin'&

By working on the previous three equations we find a' + b2 = (31 + SS)/SIS~ =: 2 E / w a and a2b2sin26 = l / s ~ a = 1'/m2w'.

This makes sense. FYom x = a cos(wt - a> and y = b cos(wt - @), we have 5 = [m(& + 9') + fk(z' + y'))l/2 = 7nw'(a' + b'))f2, I = m(z# - 92) = mwab sin&.

S2.6 Define d$,,/dt = (i x A) ~ d A / d t ~ , where i and A are unit vectors in the directions of 1 and A, resqectivtly. Since dA/dt = (dA/dt)/A+term parallel to A, A = IAI, we have (i x A) 4 dA/dt = I-'A-' (I x A) . dA/dt. Therefore

dt kmlrAz - kmrA2 ( k!:) *

.

do,, ( I x A) (r x I) bf ( r ) 1 6 f ( r ) -- y - - -- -


Figure 2.6: HQnon-Heiles potential energy

Aepr N - ' 12= (1 - '> 6f(r)$dO = I[ ke2 (1 - ") kmr 6f(r ) r2 dO , kmcz k m r

where we have replaced A' by its value c2 for 6f = 0. For Sf(r) = -2h/r3, using (2.33) we have

in agreement with (2.16).

S2.7 (i) Expressing UHH in the form

one seea at once that UHH has the value 116 on the sides of the triangle shown in the figure,

UHH(Z = -1 /2 , y) = UHH(X = 1 - &, y) = UHH(X = 1 + *&, y) = 1/6 .

On the other hand, LTHH(O,O) = 0, and UHH < 1/6 inside the triangle (see figure 2.6).

(ii) Near the vertex ( l , O ) , (z = 1 + 6x,y = 6y), the force components F, = -(s - X' + i'), F, = -y( l+ 22)

are given approximately by F, N 62, F, u -36y. For by = 0, we have d26z/dtZ c +6z, so that (1,O) is a point of unstable equilibrium for motion along the z-axis. For 6 x = 0, we have d26y/dt2 = - 3 6 ~ ~ so that (1,O) is a point of stable equilibrium for motion perpendicular to the z-axis.

Chapter 3

FIXED POINTS, OSCILLATIONS, CHAOS

The fixed points of a system of first-order differential equations of motion are defined, and the behavior of solutions in their neighborhood is exam- ined. Oscillations about equilibrium configurations are considered, as well as forced oscillations and parametric resonance. The chapter ends with an outline of chaotic motion.

3.1 Fixed points We shall be concerned with systems of differential equations of first order in the time

j.i = fl(x1,. . . XN) (i = 1 , . . . N = 2n) , (3.1)

where n of the xi’s are coordinates and the other n generalized momenta. Such equations replace n second-order differential equations, as outlined in Chapter 1.

Suppose all the fi’s vanish for xi = zoi (i = 1,. . . N), where the xoi’s are constants. Then q = X O ~ is a solution of the system. The point xo is a “fixed point”.

htample: Defining 2 1 = -9, x2 = 8 , the damped pendulum equation

j.2 = -(g/l)sinxl - 7x2. There are two fixed points, (z1 = 0,zz = 0) and (ZI = X , zp = 0), or, more precisely,

+ -$ + ( g / l ) sin -9 = 0 can be replaced by the system of equations &I = z2,

(21 = 2 n ~ , $2 = 0) and ($1 = (2n + l ) ~ , ~p = 0).

41

42 CHAPTER 3. FIXED POINTS, OSCILLATIONS, CHAOS

The fixed-point solution is stable if any solution x i ( t ) such that ( ~ i ( 0 ) - xoil < e (e infinitesimal) differs infinitesimally from SO$ at all times.

The following is a criterion for stability. Linearize equation (3.1) about (201 , . . . xON). With xi = xi - x g i one has

X i = aijxi where aij = ( a f i / a x j ) , , (3.2)

or, more concisely, x'(t) = Ax'(t) , (3.3)

where

x = ' ( x i ) (3.4) x/N

and A is the N x N matrix with elements ai j . Example: For the damped pendulum, about the fixed point (0,O) one has

011 = 0, 012 = 1, azl = - g / l , a22 = -7, and about the fixed point ( 7 r , O ) one has 011 = 0, u12 = 1, a21 = + g / l , a22 = -7.

In order to study the stability of the fixed point solution x' = 0, we consider solutions of the form

x ' ( t ) = x ext , (3.5)

from which the general solution of equation (3.3) can be constructed. Substituting in (3.3), we find

A X = X X .

The problem is reduced to finding the eigenvalues X i and the eigenvectors Xi of A. If the eigenvalues X i are all different, then the general solution can be expressed in the form

N

i=l

Let us consider the simple case N = 2,

(3.6)

(3.7)

The eigenvalues XI and XZ are the solutions of the quadratic equation

X 2 - S X + A = 0 , (3.8)

3.1. FIXED POINTS 43

Figure 3.1: Spird attractor

where S E t r ~ e ( A ~ = a11 f a22 and A = a1ta22 - 11121121. They are

correeponding to the eigenvectors

(3.10)

We confine ourselves to the cage S2 < 4A (XI and Xz complex numbers,

If S < 0 the fixed point is a “spiral attractor”, if S > 0 a

Example of spiral attractor (figure 3.1): Damped pendulum about lower equi-

t exP(M)l = exPCw2)).

“spird repellor”.

librium pasition, S = -7, A = g/1, while 5’’ < 4A means 7 < 2 Q . The eigenvalues are XI = (-7 4- id-)/2> X2 = XI. Note the solution

One finds e(t) = e0(x2 exp(Xlt) - h1 exp(~2t)) / (~2 - for e(o) = eo, 4(0) = O.

eft) = Q cos(wt i- a) exp(-@/%)

Ocher cases will be studied as problems.


Figure 3.2: Double pendulum

3.2 Small oscillations Let the matrix A in (3.3) be of the form

A = ( -d() I (3.11)

where K and M are n x n real symmetric matrices with constant coefficients, and x' in (3.4) be

Equation (3.3) then gives

q = M - ' p , p = - K q , (3.13)

corresponding to the second order equation

M q = - K q . (3.14)

Restrictions on K and M will be imposed as we go along.

dulum (figure 3.2) about the stable fixed point 41 = 42 = 0, 41 = leave it to the reader to establish the equations

As an example, we consider the small amplitude oscillations, of a double pen- = 0. We

dl = (p1 - pz)/m12 , 4 2 = (-PI + 2p2)/m12 , p l = -2mgl41 , p z = -mg& .

These can be written more concisely in the form (3.13) with

3.2. SMALL OSCILLATIONS 45

For a solution of the form

q(t) = q(0) exp(At) 1

substituting in equation (3.14) we find a system of homogeneous equations for the components (ql(O),qz(O)). This has a non-trivial solution if A is a solution of the equation X4 + (4g/J)X2 + 2g2/12 = 0. The solutions are pure imaginary, X i = kiwi (i = 1 ) 2 ) ,

These are the eigenfrequencies of the system. The eigenvalues of M are

while K is already in diagonal form with elements ~1 = 2mgl and I E ~ = mgl, both poeitive. Notice that the quadratic forms m<j<<<j = mZ2[<: + (el + <2)’] and k i j t i t j = mgl(2t: +ti) are positive definite.

Returning to the general case, one finds at once that the “energy”

E = (qTMq + qTKq)/2 (3.15)

is an integral of motion. In fact

dE/dt = (qTMq + qTMq + qTKq + qTKq)/2

= {[qT(Mq + Kq)] + [. . .IT}/2 = 0 , where we have used the symmetry of the matrices and equation (3.14). Since

the energy can also be expressed in the form

qTMq = pTM-lTp = pTM-’p ,

E = (pTM-’p + qTKq)/2 . (3.16)

We assume that the quadratic form mijti<j ((ti, tj) real) is positive definite. This is natural since the first term in equation (3.15) is the kinetic energy. We assume also that kijti(j is positive definite. This is also natural if equations (3.14) are linearizations about a stable fixed point. Then if the potential energy U has a minimum at the fixed point, (aU/aq i ) , = 0, kfj = (S2U(q)/8qjaqj)o (symmetric), and kijptj > 0 (condition for minimum).


An immediate consequence of these assumptions is that the frequencies are real. In fact, for a solution of the form

q(t) = QeiWt (3.17)

one finds (K-w‘M)Q=O ,

QTKQ ,= = - QTMQ ’

(3.18)

(3.19)

since the numerator and the denominator are both positive. Equations (3.18) have a nontrivial solution if

det(K - w‘M) = 0

(secular equation). For the double pendulum

2 9 - 21w2 -1w’ - lw2 g - 1 w =

= O , 1 2 w 4 - 4 1 g W 2 + 2 g 2 = O ,

w1= 4- , w z = J g ( z - J z , / l , as we have already found.

Our equations can be cast into a canonical form. Let uj be the eigenvectors of M,

Muj = pjuj (no sum over i) . (3.20)

The eigenvalues pi are all positive because of the assumption rnjj<”j > 0. We normalize the eigenvectors so that u7.i = ( l /p j )d j j . Representing q as a linear combination of these eigenvectors, q = QjUi , we have the kinetic energy

K = qTMq/2 = QjQjuTMuj/2 , K = Q i Q j p j ~ T ~ j / 2 = QjQj/2 = QTQ/2 , (3.21)

invariant under rotations in the n-dimensional space of the Qi’s. By a suitable rotation the potential energy can be brought to the form

Equations (3.14) are replaced by

Qi + w:QI (no sum over i)

or by the first-order equations

(3.22)

(3.23)

(3.24)

3.3. PARAMETRIC RESONANCE 47

In some cases it may be convenient to interchange the roles of M and K in the above prescription. For example, U = mgl(2q; + & / 2 , the potential energy of the double pendulum, taka the form U = mgZ(QT + Q $ / 2 by. puttin .q1 = Q l / f i , qz = Q z . The kinetic energy is then K = (QY + Qi + A31Q.)/2.

Performing the rotation Qi = (Q{ + QL)/ 2, Qa = (-Q{ + Q : ) / d , we find

U = rngl(Q<’ + QLa)/2 , K = ml”[(l - l/fi)Q<2 + (1 + 1/ f i )Qi2] /2

and the corresponding equations of motion are

Q: = -96-’(2 + &)Q: , Q: = -g1-’(2 - f i )Q i .

The frequency w1 = d m corresponds to the mode Q: = 0, Qi + = 0, while w 2 = d m corresponds to Q: = 0, Qa = 0 , d q i +

91- 9 2 = 0, &$I- qa = 0.

3.3 Parametric resonance We study the solutions of the equation

q + 2 ( t ) q = 0 (3.25)

in which the time-dependent frequency w(t) is periodic. Example: By retracting and extending the legs, a child sitting on a

swing changes the parameters of the system, and so its w(t), setting it in motion.

Equation (3.25) is equivalent to

i = ( -w2 0 ‘ ) x where x = ( 8 ) . (3.26)

Two solutions q1 and q2 of equation (3.25) are linearly independent if their Wronskian W = q l q 2 - q ~ q 1 ,

0 1 W =xTJx2 with J = ( -1 o ) , (3.27)

is different from zero. It is easy to show that dW/dt = 0, W = constant. Therefore two

solutions ql(t) and q2(t) are independent if their W is different from zero at some arbitrary time.

We choose as a fundamental system set of independent solutions xl(t) and x2(t) such that

(3.28)


namely (ql(0) = 1,q2(0) = 0 ) and (q2(0) = O,q2(0) = 1). A t t = T , where T is the period of w ( t ) ,

x l ( T ) = ( ::: ) and x 2 ( T ) = (3.29)

where a l l = q l (T ) , a12 = q2(T), a21 = q l (T ) , and a22 = 42(T). These can be written in the form

with (3.31)

There will be no occasion to confuse this A with that of equation (3.3). Since any solution can be expressed in the form x ( t ) = clxl ( t ) + czxz(t), for any solution we have

x(T) = Ax(0) . (3.32)

We want to find solutions such that

x ( t + T) = X x ( t ) . (3.33)

(3.34)

for t = 0 we have x(T) = Ax(0) = aAxl (0) + PAx2(O), and x ( T ) = Xx(0) = X [ e x l ( O ) + Pxz(o)]. Hence

a11a + a12P = Xe a21a + a22P = XP

, ,

or ( A - X ) ( F ) = O (3.35)

Note that det A = W = 1 ,

The secular equation det(A - XI) = 0 gives the equation for X

X 2 - S ~ + 1 = 0 , (3.36)

3.3. PARAMETRIC RESONANCE 49

Figure 3.3: Instability zones

where S = a l l + u22. The eigendues are

(3.37)

X I + A2 = s, X l X 2 = 1. If IS1 > 2, the eigenvalues are real and of the same sign (XIX2 = 1).

If they are both positive, we might take A1 > 1, A2 < 1, and write XI = exp(a), A2 = exp(-a) (a real and positive). The solutions are clearly unbounded, x(nT) = exp(na)x(O).

If IS1 < 2, the eigenvalues are complex (A2 = X I ) . Since their product equals unity, they lie on the unit circle, A1 = exp(ir), A2 = exp(-ir) (7 real). Then q(X1; t + T) = exp(i7) q(X1; t). The function

(t) = q(A1; t ) exp(-irt/T) (3.38)

is periodic with period T,

Hence

q(X1; t ) = ei7t/T'pl (t) and q ( X 2 ; t) = e-i7t/T'p2(t) , (3.39)

where cp1 and cp2 are periodic with period T. This result is known to physicists as Bloch's theorem.

S is a function of T. The roots in T of the equations S = f 2 separate the zones of stability from those of instability. For q = -u,2(1 + e cost)q (0 < c (< 1) the instability zones are shown in figure 3.3 .

It may be useful to go step by step through the above mathematics for the harmonic oscillator equation q + w i q = 0 (wo= constant). Here QI (t) = cos(w0t) (qI(0) = 1, QI(O) = o), q z ( t ) = w;'sin(wot) (qz(0) = 0, Q ~ ( O ) = I) ,


, cos(w0T) = ( -WO sin(w0T)

WO' sin(w0T) COS(WOT)

azl x1

x z ( T ) =

while the matrix with elements aij, which we now denote by Ao, is

COS(W~T) WO' sin(woT) -WO sin(w0T) cos(woT) A o = (

Note that here T can be anything, it must not be confused with TO = 2x/wo. The eigenvalues of AD are A1 = exp(iw0T) and Az = exp(-iwoT). The equa-

tion x(t + T) = Ax(t), with x(t) = axi(t) + /3xz(t), gives P = (A - aii)a/oiz. Hence q(A1) = aexp(iw0t) and q(A2) = aexp(-iwot), because in this case 7=i-'InX1 =woT.

SO = (trAo( = (2cos(2xwo)(, and so SO = 2 for wo = k / 2 (k integer), intersection of instability zones with wo axis.

Referring to figure 3.3, for 6 = 0 and T = 2x we have

3.4 Periodically jerked oscillator The equation

00

q + [u; + f c d ( t - nT)]q = 0 (3.40) n=-00

provides an exactly solvable model for equation (3.25). Take q1(0-) = 1, q l ( O - ) = 0, qz (O- ) = 0, q z ( O - ) = 1. A t O+ we must

have q l ( O + ) = ql(O-) , q~(0+) = QZ(O-), and, by integrating equation (3.40) from -e to t: (e + O),

In the interval 0 < t < T we take d.l(O+) - ql(o-) = -fq1(0) = -f, qz(o+) - &(O-) = -fq2(0) = 0.

q l ( t ) = cos(w0t) - fwo ' sin(w0t) , qz( t ) = w0' sin(w0t) . (3.41)

The matrix A, defined by xl(T-) = Ax1 (0-) and xz(T-) = Axz(O-), is

(3.42) > .

c o s ( w o ~ ) - fw;' sin(woT) -wo sin(woT) - f cos(w0T)

w ~ ' sin(woT) cos(w0T) A = (

We have det(A) = 1 and S = trace(A) = 2cos(w0T) - f w i ' sin(w0T). If 15'1 < 2, we find A1 = exp(iy), A2 = X I ,

COST = S/2 = cos(w0T) - fsin(woT)/2wo.

3.4. PERIODICALLY JERKED OSCILLATOR 51

Figure 3.4: Stability zones for (3.40)

In figure 3.4 (S versus T for given W O ) , the thick lines show the stability intervals. In the interval 0 < t < T , the functions q(A; t ) of equation (3.39) axe given by

= a12 cos(wot) + (-a12f + A - a11)w;' sinwot) ,

qo(A; t ) = a12 cos(wot) + (A - ull)wilsin(wot)

Pl(Xi1) = a12q1(t) + (A - a11)qz(t)

and in the interval -T < t < 0 by

As an exercise, verify that gl (A;2' /2) = Xqo(A; -T/2). The graph in figure 3.4 is familiar to physicists. In fact, the present

example is presented for comparison with the Kronig-Penney model of wave mechanics. Our equation corresponds to the Schrodinger equation

.

00

-(R2/2m)d' + vo C 6 ( ~ - ~ C E ) U = EU (3.43) n=--oo

in G. Baym,Lectuws on Quantum Mechantcs(Addison Wesley,1969), eq. 4- 121 (cos(lCa) = cos(qa) + (mvoa/tL2)sin(qa)/(qa)): T/T +) IC, T H a, w: e, 2mE/R2, f e, -2mvoli'.

Let US return to equation (3.30). Putting qn = q(nT-), i n = q(nT-) (value of q and q just before the impulse at time nT), we have

xn+1 = A x n 3 (3.44)

where X n h a components (qn, in) and similarly for xn+1. Since f in the expression (3.42) for A is a constant, the map (qn,qn) -+ (qn+l,qn+l) is manifestly linear.

However, this is equivalent to

(3.45)


as is easy to verify. Consider now the equation

m

G+w;4q+f(q) c & ( t - Q ) q = O I (3.46) n=-m

in general non-linear in q.

non- near map By a procedure identical to that used for f = constant, one finds the

Consider finally the periodically jerked damped rotator do

G+ri+f(q) c J(t-nT)q=O ,

(3.47)

(3.48) n=-do

where q(mod2n) is the angle of the rotator. from (3.46) for the oscillator with w$q replaced by yq, one finds easily

qn+l = qn + y-’[dn - f (qnknl l1 - exp(-F)I in+l = [qn - f (qn)~n] ex~(-yT)

3 (3.49)

For f ( q ) = S[rq + (1 - r ) ] , y -+ 00, J/y + 1, one obtains the “logistic map”

qn+l = rqn(1- qn) * (3.50)

This derives its name from the equation dz/dt = rx(1 - z) describing population growth.

3.5 Discrete maps, bifurcation, chaos Consider a discrete one-dimensional map x , + ~ = f (xn) , such as the map qn+l = rqn(1- qn) derived from the overdamped-overjerked rotator, equation (3.50).

A “fixed point” of the map is a solution of z* = f(z*). The fixed point is stable if, starting from s* + E,

(3.51) If(.* c 6 ) - f(x*)l < I(%* + €) - .*I = /€I *

(3.52)

3.5. DISCRETE MAPS, BIFURCATION, CHAOS 53

For the logistic map zn+l = Tzn(l - xn) (0 < T < 4) in the interval 0 5 x 5 1, there is always the fixed point x* = 0. Since the derivative of rx(1 - x) with respect to x for z = 0 equals r , x* = 0 is stable if T < 1. In this case (T < l), starting from any zo in the interval 0 < 2 < 1, one has limn+ooxn = 0. Of course if zo = 1, x,, = 0 without having to go to the limit .

The other fixed point x* = ( r - l ) / ~ exists only if T > 1, for the simple reason that x* is assumed to be positive. Since the derivative of rz(1 - z) with respect to x for z = ( r - l)/r equals 2 - T , z* = (r - l)/r is stable for 1 < r < 3. Note that the fixed point x* = (T - l)/r is born (at T = 1) when the other, z* = 0, becomes unstable. For 1 < r < 3, starting from an arbitrary 0 < xo < 1 one has lim,=,s,, = (T - l)/r.

What happens at T = 3, when x* = (T - l ) / ~ becomes unstable? A limit cycle is born consisting of only two points since we are in one dimension. The disappearance of a stable fixed point accompanied by the appearance of a “limit cycle” is known as a “Hopf bifurcation”.

For 3 < T < 1 + f i E 3.45, starting from any value 0 < 20 < 1 (xo # (r - l)/r) we have

(3.53)

This means that for large values of n, zn keeps flipping to and fro between two limit values. These are fixed points of the square of the map, defined as f2(z) = f(f(z)) (f2(z) # ( f ( ~ ) ) ~ ! ) . If we start from .x, one application of the map gives 2‘ = rx(1- z), and a second application gives Z“ = rx‘(1 - 5‘) = r2x(1 - z)(l - T Z + m2). To search for a fixed point of f2, put XI’ = x + x*. We get the equation

x*- - T 2 2 * [1 - (1 + T)Z* + 2TX*2 - TZ*3] . (3.54)

Now x* = 0 is a solution. It is a fixed point o f f , and so also of f2, but it is unstable for r > 1 and, o fortiori, for T > 3. Knowing this we first reduce (3.64) to (x* - (r - l ) / r ) [ r ~ * ~ - (r + 1)x* + (1 + r ) / r ] = 0, and, since also z* = (r - l)/r is not acceptable, to the quadratic equation

rx*2 - (T + l)z* + (1 + T ) / T = 0 . (3.55)

This has the roots

(3.56)

(r2 - 2r - 3 > 0 for T > 3). Note that f(z;’) = za and f ( z f ) = z;’ or, more explicitly, xi = rz: (1 - 2:) , 2: = r x i (1 - xi).


Figure 3.5: x = limn+mxn for logistic map

Is each of the above two values a stable fixed point of the f 2 map? In 1 > Idz”/dzl = Idz”/dz‘lldz’/dxl = r21(1 - 2x‘)(1 - 2x)I put z = z;, d = z;, obtaining 14 + 27- - r 2 ( < 1. This is satisfied for 3 < r < 1 + &.

After the first bifurcation for the value r1 = 3 of the control parameter, another occurs at r2 = 1 + fi N 3.45, when each of the two fixed points splits into two, another at 7-3 N 3.54, another at 1-4 N 3.56 etc.

The values of r at the onsets of successive bifurcations are given by Feigenbaum’s empirical formula

T k = T, - c/Sk (k = 1,2,3 ...) (3.57)

where 6 N 4.67, c N 2.64, r , N 3.57. This seems to apply also to other quadratic maps (“universality”). Some authors are so enthusiastic about this formula as if they believed that one day it will be equal in importance to the Balmer formula for the hydrogen spectrum.

From Feigenbaum’s formula one sees that

r k - rk-1 r k + l - rk

= 6 > 0 , (3.58)

showing that the interval between the onsets of bifurcations decreases. We have reached the value r, N 3.57. What happens after that? For r

between rM and 4 there are chaotic intervals interrupted by odd periodic cycles, until chaos prevails at r = 4. Figure 3.5 is the result of a rough calculation performed with the use of a pocket calculator.

When we speak of chaos, we do not only mean that the iterates are dis- tributed at random. Sensitivity to initial conditions is an essential element of chaotic behavior.

3.5. DISCRETE MAPS, BIFURCATION, CHAOS 55

Figure 3.6: Lyapunov exponent for logistic map

Suppose we start from two different values zo and zb = z o + e, with c arbitrarily small. In a chaotic regime, the iterates fn(zo) and fn(zo + e) (fO(zo) = 20 and f0(zo + c) = 20 + c) will diverge from each other as n increases.

If we assume the divergence to be exponential and write

We define the Lyapunov exponent X(z0) as

Note that fn(zo) = f(f(. . . f(zo))),

and so . n-I

(3.60)

(3.61)

(3.62)

Values of X for the logistic map are shown in figure 3.6 (rough rendi- tion from R. Shaw,Z. Nat~rforsch.~36a(l981)80). Positive values of X are indicative of chaotic behavior.

Using a pocket calculator, it is easy to verify that equation (3.59) works perfectly for the tent map of problem 3.11. Why are we not surprised?

In chapter 8 we shall briefly consider chaos for systems with time- independent Hamiltonians.

56 CHAPTER 3. FZXED POINTS, O S C ~ ~ ~ A T Z O ~ S , CHAOS

3.6 Chapter 3 problems 3.1 Show that if y > 2m the fixed point (8 = 0, d = 0) of the damped pendulum is an attractor ( S < 0, A > 0) (overdamped pendulum).

3.2 For the fixed point (8 = n,d = 0) of the damped pendulum one has S = -7, A = -g/Z. The eigenvalues are (XI = (-7 + d m )/’2), positive, and

The fixed point is a “saddle point” or “hyperbolic point”.

drawing graphs in the (8,d) plane. Consider also the case 7 = 0.

= (-7 - J7- )/2), negative.

Study the behavior of the solutions XI exp(X1 t ) and XZ exp(X2t) by

3.3 Consider the case S2 = 4A in which X = XI = A2 = S/2. Show that the general solution can be expressed in the for

x’(t) = [ q X 1 + cz(X1t + XZ)] exp(Xt) , with XI and XZ satisfying AX, = AX1 and (A - X)XZ = XI. Find XI and Xz .

3.4 Appl the result of Problem 3.3 to the damped pendulum with 7 = 2~ at the fixed point (0,O).

3.5 A wire hoop of radius R rotates counterc~ockwise about the vertical z axis with angular velocity w. Let (x2 +z2 = R2, y = 0) be the configuration of the hoop at t = 0. A bead slides on the hoop with coefficient of kinetic friction y.

(i) Show that the angle 8 of the position vector R of the bead with the negative z axis obeys the equation

8 + 74 + ( g / R ) sin 8 - w Z sin B cos 8 = 0 . (ii) Find the fixed points, linearize the equations about them etc.

3.6 The equations 6 = -(U~/u~cos(q/u), q = ~ / r n for a particle of mass rn with potential energy U = U~sin(q/a) (UO > 0) have the fixed points (p = 0,q = -na/2) and 0, = 0,q = na/2).

(i) Linearizing about Cp = 0, q = -na/2), show that the solution can be expressed in the form


Figure 3.7: Problem 3.8

where

and (nz,ay,na), not necessarily real, are components of a unit vector. Use the formula exp(iii - aa) = cos a + i ii + 0 sin a (ii2 = 1) to express p’(t) and q‘(t) as linear combinations of p’(0) and q’ (0).

(ii) Follow the same procedure for the fixed point (p = 0, q = na/2).

3.7 The system of equations

is of the type (3.13). Find the normal modes. PI = mQ1, PZ = m92, lii = -klql - fP;L,lj2 = -k2m - fa1

3.8 Two blocks, each of unit mass, are connected to each other by a spring of strength k and natural length 1, and by springs of unit strength and equal length t l to two walls. They can move on a ~ i ~ i o n I e s s horizontal floor as shown in figure 3.7. Denote by L = 1 + 211 the distance of the walls. Find the normal modes.

3.9 Three particles (a, b, c) of equal mass rn, connected by springs of equal elastic constant k and natural length I , are in equilibrium at the vertexes of an equilateral triangle, a0 = (a*%, ao,) = (-d&/2, - d / 2 ) , bo = ( b o z , b o y ) = ( d f i / 2 , - d / 2 ) , co = ( ~ 0 5 ~ ~ 0 9 ) = (O,d), d = l /&, as shown in figure 3.8. Study the small amplitude oscillations.

3.10 Consider the two ~ystems of axes shown in figure 3.9. The unprim~d system of axes is at rest in the laboratory, with the y axis vertical and pointing downwards. The origin 0‘ of the primed system oscillates with angular frequency i-2 and amplitude a along the 21-axis. 0’ is the point of

58 CHAPTER 3, FIXED POINTS, O S C ~ ~ ~ A ~ ~ O N S , CHAOS

c

Figure 3.8: ~ i a t o m i c molecule 4 4

Figure 3.8: ~ i a t o m i c molecule

suspension of a pendulum of length 1. We have

mi? = --7 sin b, , mg = mg - 7 cos b, + mQ2 c o s ~ ~ t )

Putting XI = 1 sin4 and yl = 1 cost$, we have

Hence X I = 14 cos 4 - ii2 sin 4, d l = -14 sin 4 - 162 cos 4.

T = -mi?,/ sin 4 = --m~($cos t$ - $2 sin #)/ sin 4,

4 + ~ - ‘ [ g + a5b2 c o s ( ~ t ) ] sin 4 = o .

D y putting # = Ip + 5, where Qt is 21 constant and 5 is small. We obtain

< + I-’ (g + afi2 cos(~t~)(s in b, + cos ~r 5) 21 o +

q + I-’ cos b,(g 4- an2 cos(nt))q N 0 , where q = sin Ip + cos Qi, t.

Simulate the above situation by solving the e~uation

1 OD

+ f 23 f-)”d(t - nTf2) q = 0 , n=--oo

where T = 2nfQ.

3.11 For the “tent map” (also “delta map”)

2’ = r ( l - 212 - 1/21) = 2r(z for 0 2 x 5 1/2 , 1 - x for 1/2 5 35 <, 1)

show that the chaotic regime can be expected to begin with T > r, = 1/2.



I

Figure 3.10: Escape from or approach to top position

Solutions to ch. 3 problems S3.1 The eigenvalues XI = (-7+ d m ) / 2 and are real and negative.

= (-7- d m ) / 2

S3.3 Xi exp(Ait) (i = 1,2) correspond to

For 7 = 0, A1 = the straight lines 8 = *m(e - A) are tangent at (0 = r,d = 0) to the curve E = g / l . In fact

o = E - g/i = (i"/2) - g(i + cose)/i = (e2/2) - g[i - cos(e - A ) ] / I

z (Pp) - g(e - n)'/21= [8 + m(e - A ) I [ ~ - &i(e - A)1/2 for e near A.

The curves E = g/1 separate regions of librational motion (e(t + T ) = e( t ) ) from regions of rotational motion (0( t + T ) = O ( t ) + 2 ~ ) (see figure 3.11).

60

Figure 3.11: Librational and rotational motion

s3.3 d[{Xxt + XP) exp(A~~]/dt = xXx~exp(xt) + (XI + AXz) exp(Atf

= A(Xlt + X a ) exp(At>. While equation (3.10) gives

c1 = (1 + y2/4)e0 + ez = ~ ~ e o / z ) + eo. Eliminating c1 we find d + 74/2 = c2 exp(-yt/2). For brevity's sake we do not discuss some pathological cases which occur far A2 = x1.

S3.5 (i) d ( ~ ~ ' d ~ / d ~ = m(w2rz" sin @) cose - ( 7 ~ ~ ) R - mgR sin 8 (iif Putting %I = ff and 2 2 = 6 we have

X I = 22, iz = -7xz - ( g / ~ 9 s i n ~ I +wzs inx~cosx~ . There are two fixed points, (XI = 0, x 2 = 0) and (XI = c0s-l ( g / ~ 2 ) , za = 0)' the latter only for w2 > g/R. Linearize about (0,O):

at1 = 0, a i o = I , azi = w2 - g / R , = -T, A = (g /R) - w2, X = (-7 f Jr2 + 4w" - 49/R ) / Z .

If w2 < (g /R) - 7 2 / 4 , A1 and A 2 are complex conjugate with negative red part, (0,O) is a spiral attractor.

If w2 = g/R, XI = 0, XZ < 0. In this cwe the equation linearized about; (0,O) is If w2 = (g /R) - 7' /4 , XI = A2 = -7/2*

3.6. CHAPTER 3 PROBLEMS f l c30

61

Figure 3.12: Bead on rotating hoop

8 = -74) with solution 8 = 80 + &[l - exp(-$)]/rl 4 = 40 exp(-rt). If w2 > g / R , AX > 0, A2 < 0, (0,O) is unstable. Linearise about ~ ~ - ' ~ g / ~ ' ) , 0):

a11 = 0, a12 = 1, a21 = (ga/~'w2) -w2, a22 = -7, A = w2 - g 2 / R 2 2 w , A = (-7 f Jr2 + (492/&2wZ> - 4#2 )/2.

The Square root is

Note that for 7 = 0, (0,O) becomes unstable and (coK1(g/Rw2),0) comes into existence as a stable fixed point for w = m. 53.6 (i) With p' = p and q' = q + ~ a / 2 , the linearized equations

with

have the formal solution

The matrix A can be expressed in the form A = i(n202 +. nuou)w, n, = (2 - mUo)/2 iaGG, nu = -(aS + m~o>/2aJm';lTic; (n: + nf = 11, w = a-". Then

0) - u - ' ~ s i n ( w t ) ~ ' ( o ) )

) sin(wt)p'(o) + cosfwt)q'(o) .


(ii) Linearizing about (p = 0, q = na/2), we have

p’ ( t ) = cosh(wt)p’(O) + a - ’ a s i n h ( w t ) q ’ ( O ) , q‘(t) = (a/- )sinh(wt)p’(O) + cosh(wt)q’(O)

s3.7 M = ( m O m O ) , K = ( Y L2)

The secular equation det(K - w 2 M ) = 0 gives

with solutions

and w i with a minus before the square root. The kinetic and potential energies are

With QI = f i q 1 v d Qz.= f i q z we have

K = (0’;” + QL2)/2 and an expression for U which, by choosing tan(2a) = - 2 f / ( k l - ha), reduces to U = ( W ~ Q ; ~ + w;Qi2)/2.

The normal modes correspond to Qz’ = 0 and Q1’ = 0.

m2(w2)2 - m(kl+ kcz)wa + (kxk2 - f2) = o

w: = [kl+ kz + J ( k l - 62)’ + 4fz1/2m

K = m(& + &)/2, u = (klqf + kzqj + 2fq1gi)/2.

K = (Q: + Q3/2, U = (klQ’4 + kaQq + 2fQl~2)/2m. The rotation Q1 = &I’ cos a + Qz’ sin a, Qa = -Q: sin a f Q!, cos a gives

53.8 With g1 = xi - II and 4a = xz - t - 81, the equations

take the form q 1 = - ( k -t 1)ql + kqz and qa = kqi - (k + 1 ) q a .

With M = I, the unit matrix, and K having diagonal elements k + 1 and off diagonal elements 4, the equation det(K - w ’ M ) = 0 yields the eigetifrequencies w1 = 1 and w~ = d m i . With Q1 = g1 and Qp = q2 we have K = (6: +0:)/2 and tr = ffk + 1)(Q: + Q i ) - 2kQ~Qz]/2. The n/4 rotation &I = (Q: - QL)/fi

21 = k ( z z - 2 1 - t ) - ( ~ 1 - $ 1 ) and ZZ = - k ( ~ 2 - XI - t ) - (z, - E - 11)

and Qz = (Q: + &!,)/A yields K = (0;’ + &)/2 and U = (wf&’;L + wzQz 2 12 )/ 2

with ~1 = 1 and wz = J%TT. w1 = 1 corresponds to the mode Q2’ = 0, Qa - Q I = 0, qi - qi = 0, xz = 1 + $ I .

The two blocks, at a fixed distance from each other, oscillate together with period


2 r , 01 = I 1 + A cos(t +a), xz = 1 + I 1 + A cos(t +a). w2 = 42WFT corresponds to

Qi’ = 0, Q a + Qi = 0, ~a = -qil 0 1 = 1 1 + A m(J%TTt+a),xz=Z+l1-A cos(dMt++a).

The center of mass of the blocks remains fixed halfway between the walls, (ZI + x 2 ) / 2 = L/2, the blocks oscillate towards and away from each other.

53.9 Denoting by a = a0 +a’, b = bo + b’, and c = co + c’ the positions of the particles, Newton’s second law yields for particle u the (exact) equations

where Tab = la - bl, Xab = a+ - 6+, gab = uy - 6, etc. Similar equations hold for b and c.

1 - I /rab cz (XOabXbb + gOabgtb)/lz etc. One finds If a‘, b‘, and c’ are small, then, with &, = a: - 6: etc. one has

and similar equations for b’ and c’. Denoting by q‘ a column matrix of which, to save space, we write the transpose

we find mq’ = -Kq’, where

k K = - 4

5 -4 -1 6 0 -& -4 5 -1 0 -4 4 -1 -1 2 -4 4 0 fi 0 - f i 3 0 -3 0 - 6 4 0 3 -3

(-6 6 0 -3 -3 6

The determinant of this matrix is zero. This can be seen by summing the elements of the second and third rows to those of the first, thereby getting a row of zeros. Therefore one of the eigenvaluea of M is zero. We shall see that zero is a triple root of the secular equation, and that there are three independent eigenvectors belonging to the eigenvalue zero.

Note at once that d2(a’ + b’ + c’)/dta = 0. Therefore our equations allow for a uniform motion of the center of mass, initially at a + b + c = 0. The two


Figure 3.13: Triatomic molecule (A = 6 and X = 12)

mutually orthogonal eigenvectors of K,

both belonging to the eigenvalue zero, provide for this: g' = vtuf) and q' = vt@) (u =constant) represent uniform motion in the x and y direction, respectively.

K is rather formidable. We can reduce the eigenvalue problem to a simpler one in four dimensions by assuming that the center of mass is at rest, so that c' = -a' - b'. Using this condition we reduce the problem to mq' = -Kq', with q'T = (ua &a ui b i ) and

6 -3 Z& & -3 6 -& -Z&

K = L ( 4 2 f l f l 6 3 -& -z& 3 6

horn mw2u = Ku we obtain the equation (6 - - 36(6 - = 0 for X = 4mw2/k. We see at once that X = 0 and X = 12 are simple roots, while X = fi in R dnrihb rnnt


The eigenvector corresponding to A = 0 is

Sjnceq = qo-(de/2)uy) gives a, = a o , + e a ~ ~ , ay = -ea0++aoy, andsimilar equations for the components of b and c, it clearly describes an infinitesimal rotation about the center of mas8. Note that the angular momentum about the center of mass is a conserved quantity.

The eigenvector corresponding to the eigenvalue X = 12 is

d3

u12= [ 7) It describes oscillations as in figure 3.13a with w = m.

There are two orthogonal eigenvectors for X = 6, w = d q ,

They describe oscillations as in figures 3.13b and 3.13~ , respectively.

S3.10 Starting with ql(O-) = 1, q l ( O - ) = 0, q z ( O - ) = 0, &(O-) = 1, with the notation of section 3.4 we have

x(T'/2) = A+x(O-) , where

cos(w;T/2) - f'wb-' sin(w;T/2) A + = ( -wb sin(wbT/2) - f' cos(wb~/2)

wb-' sin(w;T/2)

w:, = JM, f' = (fcos8)/1. Then

x(T-) = Ax(O-) ,

66 CHAPTER 3. FIXED POINTS, OSCILLATZONS, CHAOS

For our purpose we need

trace(A) = (2 + 5) cos(whT) - - ft2 2wb2

We consider the case cos # < 0 (pendulum rod above horizontal position) for which w& = idg-. For small T we then have

trace(A) = 12 + (f2 cos @)/2gI] c o s h ( d m T) - (f2 cos 9)/291

2: 2 + (g/6)lcos#.(T2 + (f2cos2# T2)/4I2 . A t equilibrium must be y = 0, S = trace(A) = 2, cos# = -41g/f2.

For the frequency of small oscillations about QI = E , we must have

exp(fi7) = x = (S * iJ4-s" 112

with S = 2 + (g/l)T2 - ( f2T2) / (4 l2) ) . Then sin'((rl2) = (1 - cosy)/2 = (2 - S)/4 = (f2T2/16l2) - (gT2/41),

Iy2 N (fZT"1412) - ( g ~ 2 / 1 ) , w2 = +A/T2 N ff"4P) - ($/Z).

53.11 )dz'/dz( = 2 ~ , X = lirnn-)ocn-l In (2~)" = ln(2r).

Chapter 4

COORDINATE SYSTEMS

A detailed treatment of rotations in three dimensions, which will be widely used in the book, is followed by a discussion of the fictitious forces which must be included if the motion of a body is referred to a non-inertial frame of reference.

4.1 Translations and rotations Let P be a point with coordinates xi (i = 1,2,3) with respect to a system of orthogonal axes S with origin at 0. The coordinates xi of P with respect to another system of orthogonal axes S’ with origin at 0’ are related to the xi’s by equations of the form

xi = Rij~j + X O ~ , (4.1)

where xoi are the coordinates of 0’ with respect to S (dummy indices convention). If the axes of S’ are parallel to, and equally oriented as those of S, then Rij = bij (= 1 if i = j, = 0 if i # j ) .

Introducing the notation

r = ( zt ) , r’ = ( ) , ro = ( z:: ) , (4.2) x3 4 503

equation (4.1) can be written in the form

r = Rr’+ ro , (4.3)

67

68 CHAPTER 4. COORDINATE SYSTEMS

Figure 4.1: Example of translation and rotation

where R is the matrix with elements &j. Let Bi denote the unit vector along the i-th axis of S. The distinction

between r = O P = xi&, and the one-column matrix r, though formal, must be remembered.

4

Example (figure 4.1):

XI =xicoscp-x;sin++xocol , x2 =x' ,sin++sacos++xoz , 2 3 = x$ ,

cos$ -sin+ 0

0 0 1 R = ( sin4 cos4 (4.4)

The S' system is obtained by first translating S so that 0 coincides with O', thus obtaining a system S" with origin 0" = 0' and axes parallel to those of S, and then by rotating S" counterclockwise through 4 around zg. The matrix for this rotation is R3($).

Similarly we would have

(4.5) 0 sin4 C O S ~

and

Rz(4) = (4.6) -sin4 0 C O S ~

If 4 + 64 infinitesimal, we have

4.1. TRANSLATIONS AND ROTATIONS 69

where I is the unit matrix and

J I = ( : 0 0 0 0 l ) , J 2 = ( 0 0 0 0 -1 O ) , J 3 = ( - 1 0 0 1 0 0 ) .

-1 0 1 0 0 0 0 0 (4.8)

These skew-symmetric matrices are the “generators of infinitesimal rotations”. Note that their matrix elements are given by (Ji)jk = e i j k ,

where q j k is the Ricci-Levi Civita symbol, skew-symmetric in all indices ((3.g. ckfi = - € i j & ) and €123 = 1.

The generators satisfy the relations

[Ji, Jj] = -eijkJk , (4.9)

where the “commutator” of two matrices A and B is denoted by [A, BJ 3 A 0 - 0A.

Equation (4.9) can be verified by matrix multiplication obtaining [JI, Ja] = -J3 and cyclic permutations, or by comparing the matrix elements of the two sides with the help Of eabcecde = daddbe - Jaedbd.

For an infinitesimal rotation about an arbitrary direction specified by a unit vector ii the matrix R is given by (see problem 4.1)

R = I -J( i i ) 64 , (4.10)

where

(4.11) n2 -n1 0

To obtain a formula for finite rotations, we write

r - ro = lim (I - J(ii)q5/N)Nr‘ = e-’(’Id r‘ N-+W

(4.12)

where, we repeat, J(ii) = niJi, ni being the components of ii with respect to s.

This formal expression can be reduced to a practical formula by using the following properties of the Ji’s.

First of all, one has

JiJjJk + JkJjJi = -Jidjk - &jJk . (4.13)

Multiplying by ninjnb this gives

J(fi)’ = -J(fi) , (4.14)

70

so that (n = 1,Z. . .)

J(ii)"+' = (-)nJ(ii) and J(ii)2n = (-)"+'J(;1)2 . F'urthermore, it is easy to show that

2 (J(ii) )ij = n,nb(J,Jb),j = -6ij + ninj .

Now for R = exp(-J(ii)+) (equation (4.12)) we have

R = l +

and so

do a,

~ ~ - J ( i i ) # ~ 2 n / ( 2 ~ ) ! + ~ [ - J ( ~ ) # ~ 2 ~ + 1 / ( 2 n + l)! , n=O n=O

R = I + J(ii)'(l- cos(b) - J(ii)sin# , (4.15)

&. - - 6 . . 13 cos #J + (1 - cos 4)ninj - Cijknk sin #J . (4.16)

Note that changing the sign of 4 in (4.15) is equivalent to transposing the skew-symmetric matrices Jj. Therefore

R-' = R T . (4.17)

The important formula

R-'J$R = RijJj (4.18)

will be proved in problem 4.2 . The informed reader may expect some grouptheoretical references. The

above rotations are the natural representation of the rotation group R(3) . This is a subgroup of 0(3) , the orthogonal group in three dimensions, defined by r' = Or, rITr' = rTr, OTO = ll OT = 0-'. Besides rotations, O(3) contains the inversion 0 = - I . Since detOT = de t0 , taking the determinant of OTO = I one finds det O2 = 1, det 0 = fl.

The rotation gfoup R(3) is restricted to have det 0 = 1. Thus R(3) = SO(3) c O(3). In SO(3), the S stands for "special".

Expressing a rotation matrix in the form R = exp A, since det(expA) = exp(traceA), detR = 1 implies trace(A) = 0. This is less restrictive than the condition AT = -A, following from (4.17). The trace- lessness of our Ji's is due to their skewsymmetry.

The connection between R(3) and the special unitary group in two dimensions SU(2) will be discussed in chapter 10 as a special case of that of the Lorentz group with S L ( 2 ) , the special linear group in two dimensions.

4.2. SOME KINEMATICS 71

4.2 Some kinematics If S' is in motion with respect to S, R and ro are functions of t . Let P be a moving point. Differentiating r = Rr' + ro with respect to t, we have

i = Rr' + Ri' + io . (4.19)

S' is in translational motion with respect to S (R = 0). The components of the velocity with respect to S, vi = xi, are related to those with respect to S', v: = ki, by the equation vi = R,jv$+voi, where voi = 50i. More concisely v = Rv' + VO.

If the axes of S' are parallel to those of S (R = I), then v = v' + VO. Carefully distinpish v and-v. Even if ,R # I, one has v = v' + VO, which is nothing but dOP/dt = dO'P/dt + dOO'/dt.

$ b e 2: ro = 0, i' = 0, P at rest with respect to S', whose origin coincides with that of S, S' obtained from S by rotation with constant angular velocity w around ii. Then, with J(w) = wiJi,

* -J (w) t f

r = - dt (e r ) = -J(w)r , (4.20)

X i = -Wk(Jk)i.2j = Cikjwkxj (V = w X r in the usual notation).

function oft .

that

Case 8 The point P is at rest with respect to S', ro = 0, but R is a

At each time t one can find an instantaneous angular velocity w(t) such

i = -J(w(t))r . (4.21)

Proof: i = Rr' = Ar with A = RR-' = RRT, d(RRT)/dt = 0, RRT + R k T = 0,

The Ji's form a complete basis for all 3 x 3 skewsymmetric matrices. Hence A + AT = 0, A is skewsymmetric. '

A = -J(w(t ) ) , where -wi( t ) are the components of A in that basis.

4.3 Fictitious forces w: S' in translational motion with respect to the inertial frame S, S and S' axes parallel, i' = 'J' + i'o. The applied force f in Newton's second law in S, mi' = f, is supplemented by an "inertial force" when the motion is referred to S', mi" = f - m'io.

-2: ro = 0, S' in uniform rotational motion with respect to the inertial frame S.

Differentiating r = Rr' twice with respect to t we have i' = Rr' + 2Ri' + Ri'. Multiplying by mR-' and rearranging this gives

72 CHAPTER 4. COORDZNATE SYSTEMS

my‘ = R-I(my) - mR-lRr’ - 2mR-lRi’. Using R = exp(-J(w)t) (w constant) we find

my’ = f‘ - ~ J ( w ) ~ I ‘ + 2mJ(w)i ’ , (4.22)

where

f ‘ = R - ’ f = ( 3) , (4.23)

and fi are the components of the applied force with respect to S’. In problem 4.4 it will be shown that

The second term on the right side of equation (4.22) is the “centrifugal force”

(4.24) fAtf = -mJ(w) r

with components mw2[[z: - (n94)n:l with respect to S‘. These are the S’ components of mw2(r - (ii . r)n) = -mw x (w x r), where r - (n . r)ii is the component or r normal to the rotation axis.

J(w) = wiJi = wnjJ j = wn:Ji = w:Ji.

2 ’

The last term in equation (4.22) is the “Coriolis force”

f&,. 2mJ(w)i ‘ = 2mwjJ j i ’ = 2mw’jJji’ . (4.25)

The i-th component of the Coriolis force with respect to S’ is then

- 2 m e i j 4 i l , = 6: (-2mw x [v - w x r]) . Note that v - w x r is precisely the velocity of P as seen by S’.


4.4 Chapter 4 problems 4.1 Let the system of axes S' be obtained from S by a counterclockwise infinitesimal rotation through the angle d& about the unit vector n . Show that the coordinates xi and xi' of a fixed point with respect to S and S', respectively, are related by the formula

4.2 Show that R-'JiR = Ri jJ j .

4.3 Find eigendues and eigenvectors of J ( n (equation (4.11)).

4.4 In equation (4.10) we have written J(ii) rather than J(fi). Why?

4.5 A thin massless pipe is tilted from the vertical direction by an angle B and forced to rotate about the vertical axis with angular velocity w. A bug crawls inside the pipe at constant speed v heading towards the upper end of the pipe.

(a) Find the magnitude and direction of each force acting on the bug in the frame of the rotating pipe.

(b) Find the power input of friction (Pf), of gravity (Pg), and of the centrifugal force ( Pc) .

(c) Find the power P,,, that must be developed by the motor in order to keep the system rotating.

(d) Verify that P,,, + Pf + Pg = dK/dt ,

where K is the kinetic energy of the bug in the lab frame.

the bug. (e) Verify that P,, = TW, where T is the torque exerted by the motor on

74 CHAPTER 4 . COORDINATE SYSTEMS

Solutions to ch. 4 problems 54.1 The vector b obtained by rotating the vector a counterclockwise about ii through 64 is b N a + ri x a &#. Then the unit vectors &, and 6: along the axes of S and S' itre related by &: N ii, + ii x ztj &+. Therefore

xi = OP , &j N OP + (&: - ii x &: &I#) etc. + *

S4.2 With R = exp(-J(ii)+), we have

d(R-'JtR)/d+ = R-'[J(ii), Jj]R

= njR-'[Jj, Ji]R =: €i jknjR- ' JkR . It is easy to verify that the solution of this equation is R-'liR = &jJj.

s4.3 det(J(fi) - XI) = 0 gives X3 4- X = 0, X = (0,ki). The eigenvalue X = 0 corresponds to the eigenvector VQ = 8. The eigenvaluea f i correspond to the ei~envectors

nl n3 $: in2 v+i = ( 712713 f i n 1 ) .

ng - I

The trace of J(ii} equals the sum of the eigenvalues, 0 + i - i = 0, as we expect. Since the R(3) transformations are defined as real transformations in three-

dimensional real space, the two complex eigenvalues are not acceptable.

s4.4 The components of ii with respect to S and S' are equd: ni = RG'nj = [ ~ i j cos 4 + (1 - cos+)niy + e i j k n k sin $]nj

= nicos++ (1 -cos+)ni -I- zero = nj.

S4.6 As shown in figure 4 . 2 , 4 = 42 cos 6 - 6s sin 8,s; = 82 sin 8 + & cos 8. {a) F; = mg(& sin t? - &; cos 6), FLtf = 71w2r sin 6(&: cos 8 + 6; sin 6) ,

and the normal reactions of the constraints

Verify that the sum of these six forces is zero. (b) Pf = F: I v' = mv(gcos6 - w2rsin26), Pg = F; . V' = -mgvcosO,

One has Pt + Pg + Pctr = 0. (c) P m = N\ . (-wr sin e &I) = 2mw2vr sin2@ (d) From (b) we have P, + Pr + Prr = P, - Pet{.

The kinetic energy of the system is K = (mv2+Zw2)/2, where I is the moment of inertia with respect to the rotation axis. Since v =constant, w =constant, and the change of the moment of inertia is due only to the motion of the bug, we have dK/dt = (I/2)wgd(nr'sin26)/dt = rnw'rv sin2@ = Pm - Pctf. Hence P m = 2mw2rv sin2@.

FOor = 2mwvsin B &it F', = m(gcos6 - w2rsin%)i3g,

Ni = -2mwv sin 6 &:, N; = -m sin 6(g + wzrcos e)&;

P,,r = FLtf . v' = m v d r sin%.


Figure 4.2: Bug in tilted rotating pipe

It is sometimes stated that the power delivered by the motor is half as much,

That we have found the correct expression for Pm is shown by the following: an obvious confusion between Pm and P, - Pctr.

(e) r = w dI/dt = 2mwvr sin2@, r w = 2w'vr sin'@ = Pm

Chapter 5

RIGID BODIES

In this chapter we study the dynamics of rigid bodies intentionally abstain- ing from the use of Lagrangian methods. Some problems requiring a fair amount of work will be proposed again in the following chapter on La- grangians, where it will be easier to establish the equations of motion and to fmd conserved quantities.

5.1 Angular momentum Let us consider two systems of axes, S fixed in the laboratory and S' attached to the body. We assume that the origins 0 and 0' of the systems coincide. Therefore the point 0' of the body is fixed.

Let L denote the angular momentum with respect to 0' = 0. When translated into the notation of Chapter 4, the familiar expression

L = r x v d m J J

reads L = - J(r)v dm

Here J(r) = xiJi (see (4.8)),

L = ( E ! ) , r = ( zi) 2 3 , v = ( ii) . (5.3)

The distinction between bold-faced roman and bold-faced sans-serif should be remembered. Although r and r denote the same vector, r indicates

77

78 CHAPTER 5. RIGID BODIES

its intrinsic geometrical meaning, while r stands for the coordinates with respect to S. Therefore we shall have r', the coordinates with respect to another system S', but not r'.

Below w stands for a column matrix with elements wi. There is no sans-serif omega! Since

where w is the instantaneous angular velocity (see 4.21)), we find

v = -J(w)r , (5.4)

L = 1 J(r)J(w)r dm = - [S 1 J(r)'dm w = 1) w , (5.5)

where 11 = - / J(r)'dm ,

is the inertia matrix with elements

We have used the formula

J(w)r = -J(r)w , ( 5 4

which is easy to prove: (J (w)r ) i = ( J k W k ) i j Z j = e k i j x j W k

- - - Z j € j i k W k = - ( J j Z j ) i k W k = - (J ( r )w) i . Note that the inertia matrix (5.6) depends on the spatial relation of the rigid body with S, and therefore changes during the motion. For this reason the argument of J(o) in (5.6) is r, rather than r.

5.2 Euler's equations In the laboratory system

L = r , (5.9)

the rate of change of the angular momentum equals the applied torque. Let R be the rotation matrix that transforms a', the components of a

vector with respect to the body system S', into a. Differentiating RL' = L, we have RL' + R i ' = r = Rr'. Proceeding as in chapter 4, we find

ii = 7' + R - ' J ( ~ ) L = + J ( ~ ) R - ' L ,

L' = r' + J(w')L' , (5.10)

5.2. EULER’S EQUATIONS 79

in the usual language L = L’ x w‘ + 7’. Here J(w) = wiJi, J(w’) = wfJi, and

L‘ = R-’ L = R-’ 11 w = 11‘ w’

with 11‘ = - 1 J(r’)2dm ,

(5.11)

(5.12)

(5.13)

Note that 11 and its elements do a change during the motion.

and

where Ii are the “principal moments of inertia”.

If the S’ axes are principal inertia axes of the body, then 11’ is diagonal,

L: = 1iw; (no sum over i) , (5.14)

The kinetic energy is

K = wTllw/2 = wTLw/2 . (5.15)

Expressing this in terms of primed quantities, we have

K = wmllfw’/2 , (5.16)

and, if the inertia matrix is diagonal,

K = I iwi2 /2 . (5.17)

We now assume that the origin 0’ = 0 coincides with the center of

L; = Qjk Liwj, . (5.18)

If the S’ axes are principal inertia axes, this equation reduces to Euler’s

mass of a body acted upon by neither forces nor torques. Then

equations 11l.2; = ( 1 2 - 1 3 ) W i W $ , 1 2 4 = ( 1 3 - 1l)w;w; , (5.19)

Fkom these equations one easily verifies that K = I iwi t2 /2 is a constant of motion.

If 11 = 12 (rotational symmetry about the zb-axis) the third of (5.19) yields w; =constant. Then, putting a = [ ( I3 - 11)/11]wS. the first two can be written in the form

w; = -awi and w; = aw; . (5.20)

{ 1 3 4 = (11 - 12)w;w; .


35’

Figure 5.: 1: Body-cone and space-cone

Multiplying the first by w{ and the second by wh, and summing, we find that w : ~ + w p is a constant of motion, which might have been inferred from K and w3’ being constants of motion.

For 11 = 1, the general solution of equations (5.19) is

showing that, as seen by S’, the angular velocity rotates around 6; with period 2?r/lal. For the earth ( 1 3 - 11)/11 N 1/300, w i N w = 2?r/day, so that the period 2?r/a is 300 days = 10 months. This is shorter than the observed value of 14 months (Chandler’s period). The discrepancy is due to earth deformation caused by the polar fluctuations themselves.

The angle formed by the

We wish to discuss more carefully the matter of the rotation of the angular velocity of a rigid body free from torques with 11 = 12 # 1 3 . For such a body the angular momentum L is a fixed vector. On the other hand, the angular velocity w is not.

If 8 is the angle formed by L and w , then cosd = The kinetic energy K is a constant of motion, and w - - also a constant of motion because the sum of the first two terms and”the last term under square root are both constants of motion.

Therefore w is a vector of constant magnitude rotating arounf L on a cone of semiaperture d (“space cone” or, better, “lab cone”).

We have also seen that the angular velocity rotates around 6; on a cone of semiaperture 8’ (“body cone”). Furthermore, at each instant L = 11(wi6: + wh6h) + 13w;&g, w = wi6; + w;&h +ui&$, and 6; lie on the same plane ((L x w ) .Sg = 0) .

5.2. EULER'S EQUATIONS 81

Finally the an le 0" of L with 6; is iven by

For 11 > 13 (rod-like body) one has tan 0" > tan O f , whereas for 11 < 13 (earth, flat disk) tan0" < tan@'. In the former case, w is to be found between L and &:, in the latter L is between w and 6;.

The body-cone rolls on the space-cone. At each instant the angular velocity is along their line of contact (see figure 5.1).

One must not confuse the angular velocity with which w is seen to rotate around 6; in the primed system attached to the 11 = 12 body, with the precessional angular velocity of 6; (the symmetry axis of the body) around the fixed angular momentum L.

We can write L = 11(w{6{ + w!$&) + 13~46; = Zlw + (13 - I I ) w @ $ ,

tan e r r = J + i ~ s / L i = 11 ,/*/13w; = ( I l / I 3 ) tan 01.

w = ( L / I l ) - a&: with a = [(Is - 11)/11]wh . (5.22)

Then dSS/dt = w x 64 = (L x &g)/11, showing that 6; rotates around L with angular velocity

W p r = ILl/11 (5.23)

On the other hand we have seen that with respect to the primed system attached to the body, w appears to rotate around 6; with angular velocity a = [ ( I 3 - 11) /11]4 .

82 CHAPTER 5. RIGID BODZES

i i 3

Figure 5.2: Euler angles

5.3 Euler angles Think of the S' system as being obtained as follows (see figure 5.2).

system S, with basic vectors i l , Pz, and & = 23.

system sb with basic vectors bl = &I, b2, b 3 .

system S' with basic vectors 6; , 6; , 6; = bs.

6; = R($, 6, cp)&

By a counterclockwise rotation through cp around 23, S' becomes the

By a counterclockwise rot3tion througk d about C1, S, becomes the

By a counterclockwise rotation throu4h $ about b 3 , sb becomes the

Thus (5.24)

R($, 8, cp) = e - J ( ~ a ) ~ e - J ( a i ) ~ e - J ( ~ a ) ~ . (5.25) with

Using the formula R-'J(a)R = J(R-'a) we now find

R($,d,cp) = R3(p)R1(d)R3($) > (5.26)

where R3(o) and R1(o) are the matrices for rotations around the S axes given by equations (4.4) and (4.5).

Proof: Putting R(ji i , a ) = exp(-J(fi)a), we have

R(b3, $)R(&i, 6)R(63, CP)

= [R(&,.29)R(&i,$)R(&, -.29)] R(ii , .29)R(b, CP)

= R(i116)R(h, 9) [R(63, - c P ) R ( ~ ~ , $)R(83, CP)]

= R(&, .29)R(b, CP)R(&,$J)

5.3. EULER ANGLES 83

= R ( b , 9) [R(63, - ‘p)R(i i76)R(&, v)] R(63, $J) = R(&, ‘p) R(&, 9) R(&, $J)

= R3((p)R1(6) R3($J)

By equation (5.26)

sincp cos++cos6 coscp sin+ . cos cp cos + - cos 6 sin cp sin +

sin6 sin@ 1 R(h6,cp) ( d ) = ( horn this we read out that

6; = (coscp cos+ - cos6 sincp sin+)&

+(sincp cos+ + cosd coscp sin+)& + sin6 sin+ 63 . Similarly we find

6; = -(coscp sin+ + cosd sincp cos7/I)61

+(- sin cp sin+ + cosd cos cp cos+)62 + sin6 cos+ 63

and 6; = sincp sin6 81 - coscp sin6 &2 + cosd 63 .

We shall frequently refer to the following formulae

81 =61coscp+62sincp, 82=-61sincp+62coscp, ii3=63

61 = P I , 62=ii2cos6++3sin6, 63 =-ii2sint~+ii3cos6

(5.27)

(5.28)

(5.29)

(5.30)

(5.31)

(5.32)

6; = 6, cos+ + 62 sin+ , 6; = -61 sin+ + 6, cos+ , 6; = 63 . (5.33)

Since infinitesimal rotations around different axes are additive in the These can be used to check equations (5.28,29,30).

first order, the angular velocity can be expressed in the form

w = @63 +d&1 +‘$63 . (5.34)

Using repeatedly equations (5.31,32,33), we have

w = (sin6 sin+ ci, + cos+ d)ei + (sin6 cos+ + - sin+ d)a; + (cos6 ci, + 4)s; = (dcoscp + dsind sincp)bl + (dsincp - 1l,sin6coscp)62 + (ci, +~COs6)63 .

The kinetic energy of a rigid body with I1 = I 2 can be written as (5.35)

(5.36)

a4 CHAPTER 5. RZGZD BODIES

0 Figure 5.3: Spinning top

5.4 Spinning top The spinning top (I1 = Iz) shown in figure 5.3 has a fixed point 0 other than the center of mass. Denoting by l the distance of the center of mass and 0 = 0', the torque due to gravity is T = t6; x (-mg63). Since

6 3 = (6; sin$ + 6; cos1(1) sin6 + 6; cosd , we have

T = ems(&', cos $ - 6; sin $J) sin d . (5.37)

By a simple extension of Euler's equations (5.19) to include the torque, we have

11h: = (I1 - Z3)wiw; + lmg sin8 cos$ , Ilw; = (13 - Il)wiwf - ems sin 19 sin 11, , (5.38)

( 13w; = 0 . Multiplying the first of (5.38) by w{ and the second by w i , summing,

Z l (w~2+~~2) /2+ lmgcosd = K - 1 3 ~ ~ ~ / 2 + C m g c o s d = constant . (5.39)

Since w$ is also a constant of motion, so is the total energy E = K + U with U = lmgcosd.

L3 = L a63 = T 153 = 0. A simple calculation yields

and using (5.35), we find

There is another constant of motion, L3 = L * I%. In fact,

~3 = (11sin219 + 13coS219)ci, + 13 C O S ~ 4 .

It is convenient to express the constant of motion w i = cosd @ + 4 in the form w; = a11 /13, and the constant of motion La in the form L3 = 11b. Then from the expression for L3 we find

sin0 ci, = [11b - 1 3 cosd(4 + cosd ci,)]/~l sind = ( b - a c o s ~ ~ ) / s i n d .

5.4. SPINNING TOP 85

Figure 5.4: Graph of P(u)

The energy can be expressed in the form

E = (11/2)d2 + Ve(6) , (5.40)

2 where

+ f?mg cosd (5.41)

is an effective potential energy for the 6 motion. The angles cp and $ have disappeared from this expression for the energy,

in the same way as the angle 8 did from the expression for the energy of a particle moving under the action of a central force.

Introducing the variable u = cos6, one finds u2 = P(u) , where

(1 - u2) - ( b - au)2 (5.42) 2E 11a2 2 t m g ~ P(u) = (11 - - - - 13 I1

is a polynomial of third degree in u.

P(u) to be as in figure 5.4 . Since P(f1) < 0 and lim,,-,*mP(u) = f o o , we expect the graph of

5.4.1 Regular precession of top The “regular precession” is the analogue of circular orbits of Chapter 2. We seek the conditions under which 6 = do(constant) during the motion. For 19 = 60, 8 = 0, d = 0, the first two of Euler’s equations (5.38) read

I1 sineod(sin11 +)/dt = (11-13)sin60cos~ ci,w~+lmgsind~cosII, (5.43)

and

11 sin bod(cos $ +)/dt = (I3 - 11 ) sin 60 sin 11 @ w; - tmg sin 60 sin $ . (5.44)


Assuming 60 # 0, multiplying the first of these equations by C O S ~ , the second by sin $, and subtracting, we obtain 11ci)odo = (11 - 1 3 ) @ 0 W i +lmg,

where we have denoted by @O and do the values of @ and 4 pertaining to the steady precession.

Neglecting the square of do, we find

(5.46)

in agreement with the elementary treatment: dL = T dt with 171 = 4mg sin6 and ldLl N (ILI sin6)+odt yields $0 = lmg/lLI. If the top precesses slowly and spins fast, then ILI 21 1340.

5.4.2 Sleeping top Consider the special case in which the the axis of the top is vertical, 19 = 0, 8 = 0, w3 = w i =$ wo. One has 6 = a,

E = (I3w:)/2 +ems = I?a2/2Z3 +ems. We wish to study the stability of this vertical top. We start from the

equation Ild(wi + iwi)/dt = i(13 - Il)wi(wi + iw;) + lmg sin6 exp(-i@). Putting w{ +iwi = 6w, w& = (Il/I3)a+6w&, sin6 exp(-i$) = iSx, we have linearly in the 6 quantities

IIddw/dt = i(13 - 11)(11/h)a dw + ilmg 6x (5.47)

and i d6x/dt = (Il/I3)a 6x + bw (5.48)

Putting 6w = 0 exp(iXt) and 6x = X exp(iXt) we obtain the equations

(5.49)

This system of equation has a non-trivial solution if the determinant of the coefficients is zero,

1lX2 + a11[2(11/13) - 1]X + (Il - 13)(11a/13)~ + h g = 0 . (5.50)

The vertical spinning is stable if X is real,

[a11(2(11/13) - I)]' -411[(11 - 13)(11a/13)~ + h g ] > 0 , (5.51)

namely a2 > 4emg/1~ . (5.52)

5.4. SPINNING TOP 87

Figure 5.5: Irregular precessions of top

5.4.3 Irregular precessions of top Returning to section 5.4 and figure 5.4 , 1 > u2 > u1 > -1 (02 < el), remember that

(5.53) 6 - uu 1 - u2 1 - 2 2

a(a-'b - U ) $=-=

If a-'6 > u2, then @ never changes sign, and the motion of the axis of the top is as in figure 5.5(a) .

If 212 > a-'6 > u l , then (I? changes sign at u = a-'b, and the axis describes loops as in figure 5.5(b) .

If u2 = a-'6, then ($]u=ua = 0. The axis describes cusps as in figure 5.5(c) . Note that u2 = a-'6 means that a-'b is a root of P(u). This gives

K = E - tmgu = (I:a2/213) + lrng(a-'b - u )

E = (1?a2/213) + h g b / a ,

= (13/2)w3 + h g ( a - 1 6 - u) . (5.54)

At u = u2 (6 = domi,,), the kinetic energy of the top reduces to the spin kinetic energy. If I hold the axis of a spinning top at a certain angle, and then I release it, the axis falls and rises again as in figure 5.5(c).

Suppose we do precisely this, releasing a top spinning with L J ~ = ( 1 1 /13)a from the angle 60 of the axis with the vertical. The total energy will be E = (1?a2/213) + h g cos60. If we put a-'6 = cosdg = ug in P(u) , this becomes

P(U) = (ug - u)[(2emgp1)(i - u2) - a2(uo - u)] . (5.55)

One root of P(u) is ug, the other two are the roots of the quadratic

88

polynomial in the square brackets,

u = u1 = [ha' - JI;a4 - 8tmg11a2uo + 16LZm2g2] / 4 h g

being the smaller of the two. If u2 is very large (top spinning very fast) one finds

u1 uo - 2emg(i - u;)/i,a2 ,

cos81 N cosdo - ( 2 e ~ g / ~ l ~ 2 ) s i n 2 8 ~ , 1 9 ~ - 80 ci ( 2 ~ ~ g ~ ~ /r;w;2) sin 80 .

Thus the axis to the top falls from the release inclination 190 to 1 9 1 , to rise again to 60. The larger the spinning angular velocity, the smaller the difference between dl and 190.

5.5. GYROCOMPASS 89

Figure 5.6: Gyrocompass

5.5 Gyrocompass In its simplest form this consists of a gyroscope rotating about its axis which coincides with the diameter of a horizontal ring. The ring can rotate freely around its own axis (& in figure 5.6). Clearly w{ = 0, w i = &. We anticipate that u$ is a constant of motion, w$ = wo (constant). In the real thing, w$ is kept constant by a motor in order to overcome slowing down by friction.

The gyroscope is in the non-inertial frame of the earth rotating with angular velocity

62 = 0 63 = n[6: sinX -t cosX (-&; sina i- 4; cosa)] . (5.56)

Therefore it will be subject to a centrifugal torque and a Coriolis torque.

be

and those of the torque will be

The components of the centrifugal force acting on an element dm will

dff = dm [n2x: - 0:(0$5$)] drf = €+jkx$dfL = - d ~ ( 0 ~ x ~ ) € j ~ k z $ ~ ~ .

Integrating over the whole gyroscope, and using Jz:z$dm = 0 if i # j , Jxiadm = Jxadm = &/2, Jxh2dm = I , - 13/2, we find

7Ltf = ft2cosa cosx ( I ~ - 11) (5.57)

The components of the Coriolis force acting on drn are

dfi = -2dmcs&i&


and those of the torque are d.: = Z d ~ ~ x ~ ( 2 ~ Q ~ ) - Q ~ ( z ~ ~ ~ ) ~ ,

Since‘ xi = -wax$, xi = W O X ~ , and x i = 0, we find xixi = 0 and z:Q; = Q( - sin a cos X x: c sin X 2; c cos a cos X z;). Integrating over the whole gyroscope, we find

(5.58) ( sia” ) = - f3w& cos X sin a .

Euler’s equations will read

fiw: = (11 - f3)wiw$ +Q2((13 - &)sinX cosX wsa- I ~ w ~ Q s i ~ X ~ ...

= (1, - ~ , )w ;w ; - @(is - i ~ ) c o s 2 ~ s ina cosa- I 3 W o f l C O S X sina , 1 3 w ; = 0 ,

where the dots denote a mechanical torque exerted by the ring. With wi = 0, w; = iu, and w& = WO, we find that the first equation

is satisfied only by virtue of the constraint reaction. The second equation yields

11 i i = -@(13 - 11)sincx cosacoS”~ - 1 ~ ~ ~ f 2 c o s ~ sina . (5.59)

Neglecting the square of the earth angular velocityl the frequ~ncy of small oscillations about the northern direction (stable equilibrium position) is

w = J(I3/J1)WoQcosx (5.60)

]The axes 6: ( i = 1,2,3) are & Euler axes, the coordinates s: of drn are & fixed.

5.6. TZLTED DISK ROLLING IN A CIRCLE

% I Figure 5.7: Rolling disk

91

Figure

3c4

5.8: 6: into page, 6; = 62, 6; =

5.6 Tilted disk rolling in a circle

A uniform disk of radius T rolls on a perfectly rough horizontal plane, the point of contact describing a circle of radius R (figure 5.7). The normal to the disk forms a fixed angle 19 with the vertical direction (see figure 5.8).

The Euler axes are attached to the disk, with 15; normal to it. At each instant the unit vector 62 points from the point of contact to the center of the disk. Figure 5.8 shows the situation at a certain time, say t = 0, when cp = ~ / 2 and 1c, = 0.

Note the rolling condition R@ = - ~ d , with + and 4 both constant.


Since 9 = 0, we have

wi = sind sin$ + , w i =s in8 cos$+ , w$ = cosd $3 4-4 .

Euler's equations give

11 sin 19 cos + $+ = (11 - 13) sin 19 cos + ~ ( c o s 6 + + 4) + T; , { -11 sin4 sin11 4+ = (13 - 11) sin6 sin$ +(cosd $3 + 11) + Ti , (5.6

while the third equation will be satisfied identically since T; will be found to vanish.

At t = 0 (figure 5.8) the forces acting on the disk at the point of contact with the plane are a vertical force F, = mg& and a horizontal force (friction) FI, = -(R - r cosd)m$32&1. The corresponding total torque is

T = -r& x [mg~y, - ( R - r cosb)mG2&l]

= -rm[gcosd - (R -rcos19)+~sin19]6~ . (5.62)

At t = 0, when 'p = n/2 and $ = 0, the second Euler equation is clearly satisfied, while the first, using the rolling condition, gives

. (5.63) rmg cos 19 - 2 - ' - sind~r-1~13 + ( 1 1 -13)cos'lj)+rnr~-mr2cosd]

For a uniform disk (11 = m r 2 / 4 , 13 = mr2/2) we have

4g cos 6 - 2 - ' - sind(6R-5rcosd) ' (5.64)


5.7 Chapter 5 problems 5.1 A uniform sphere of mass m and radius T is uniformly charged with total charge q. It rolls on the horizontal (zl, 2 2 ) plane under the influence of the uniform electrostatic field E = (E, 0,O).

(i) Show that its center of mass moves like a point particle of mass 7m/5.

(ii) Show that the components w1 and w3 of the spin are constants of the motion, while & = 5qE/7mr. The wi’s are the components of w with respect to the fixed axes.

5.2 A sphere of radius a and mass m rolls inside a cylinder of radius a + 6, whose axis is horizontal and coincides with the z-axis of a system of cylindrical coordinates.

Denoting by p = 6 , 4, and z the coordinates of the center of the sphere, show that

mb4 = -(5/7)mg sin 4 , mf = -(2/7)mawp$

and a b , , - $ i = o ,

where wp = w 8,.

5.3 Show that a body whose principal moments of inertia I1 , I2, and I3 are all different, can rotate uniformly around one of them, say 6;. Discuss the stability of such a motion.

5.4 Discuss the regular precession of the spinning top by requiring that E = V,(190) and Vl(d0) = 0.

5.5 Study small oscillations about the regular precession of a spinning top. To simplify the calculations, assume I1 = I2 = I3 + I. (How would you realize this situation? l&& a sphere!)

5.6 Derive the results for the sleeping top (section 5.4.2) by using the equation ti2 = P(u) with P(u) given by (5.42).

5.7 Study the motion of a “skating top”, a spinning symmetrical top constrained to remain in contact with a smooth horizontal surface.

94

Figure 5.9: Thumbtack

5.8 (i) Find the components Q,. = web6 of the angular velocity with respect to the system 6, (i = 1,2,3). Only the third, 6, = 13;, is fixed with respect to the body.

(ii) Find the components hi = L,. bi of the angular momentum with respect to the center of mass for the case I , = 11.

(iii) Express dL/dt = T in terms of the fti's and of the com~onents T = T 64 of the torque about the center of mass.

5.9 Consider the rigid body shaped like a thumbtack with a long pin shown in figure 5.9. The end of the pin is fixed at 0, and the disk can roll on the inclined plane 23 = 0. (i) Find an equation for ip. (ii) Find the frequency of small ~scillations about the equil~brium position.

5.10 ~ s t a ~ l i s ~ the equations of motion for a uniform disk rolling on a h o r i ~ n t a l plane. This is an extension of the special case treated in section 5.6, where the point of contact with the plane was assumed to be on a circle.


Solutions to ch. 5 problems S5.1 Denoting by F1 and FJ the horizontal components of the force exerted by the plane on the sphere at the point of contact, we have

m x l = F1 + q E , mx, = Fj , Iw1 = rF2 , I& = -rF1

with I = 2mr2/5 , and the rolling conditions

rw1 = - x 2 and rwz = X I .

Then F1 = - ( I / r ) & = - ( I / r 2 ) & , F’z = (Z/r)wl = -(I/?)&, and SO

[m + ( I / r 2 ) ] & = qE or (7 /5)m& = q E , and i% = 0 .

Since there is no component of the torque in the vertical direction, ws=constant. We have dso w1 = -x2/r = 0 and w 2 = x l / r = 5qE/7mr.

S5.2 Denoting by F p , F4, and Fz the components of the force exerted by the cylinder on the sphere at the point of contact, ma = m g + F gives

-bmp = F p , b m 4 = -mg sin4 + F# , mi = Fz

Idwldt = r with r = a e p x F = a(F& - FZ&) gives

I ( h p - &+) = 0 , I(&+ + bP) = -aFZ , Z b z = aF4

The rolling condition aw x ( - G p ) = r yields

-awz = b $ , awo = i .

The final results are easily obtained by manipulating these equations.

S5.3 Euler’s equations, Zl&{ = (I2 - I ~ ) w ; w $ and cyclic permutations, are satisfied by w; = wt = 0 and w$ = constant.

and For w; and w; infinitesimal, w$ is constant with an error of the second order,

113: N [ ( Z Z - I ~ ) ( z s - Il)/Iz]WiZW: . The motion is stable if 13 is either greater or smaller than both I1 and Z2 ,

unstable if I S is comprised between I1 and Z Z .

S5.4 It is easy to show that these conditions are equivalent to P(u0) = 0 and P’(u0) = 0, where u = cos19 and P(u) is the polynomial in equation (5.42), which in turn imply that uo is a double root of P(u) (see figure 5.10). They give


Figure 5.10: Double root of P(u)

Using the second of these equations in the first, one has

[Z~a(b - auo) - tmg(1 - ug)](l - ug) - IIuo(b - a u ~ ) ~ = 0 .

But now b-auo = &(l -ug), a = (I3/Zl)(&+@0cos60), and so one finds again (5.45). This is a quadratic equation for 90,

11 C O S ~ O Q O - I~w;@o + tmg = O or 11 cos60& - IIa@o + tmg = 0, with the two roots

* 2

$0 = [a z t daz - (4tmg/11) cos 901/2 cos60 . These are real if a2 > 4tmgcos 60/11. Assuming a very large (w; very large), the roots are

$0 N a/cos60(large) and $0 2: lmg/Zla(sma.ll).

Only the second is obtained by the elementary treatment.

s5.5 Equation (5.45) for the regular precession reduces to I@o& = tmg, while equations (5.38) give

~ ( 6 + sin 8 $4) = tmg sin 6 , d(sin6 +)/dt = d i , { d(@ C O S ~ + $)/dt = 0 .

Put 6 = 80 + 66, Q = QO + 6p, and II, = 40 + 64, where 190, QO, and $JO and their time derivatives have the steady-motion values. Then linearly in the small corrections, we have

64 + sin # O ( + o ~ l ) + &s+) = o , sin 190 J+ + (cos 8 0 +o - &>ad = o , cos6o 6+ - sin60 9069 + s.;G = 0 .


From the second and third equations we find that

sin 80 b+ = ($0 - cos 190 + O ) M + constant

sin 60 63, = (+o - cos 6 0 &)a19 + constant . and

Substituting in the first equation, we have

ad + (9; + 4; - 2+040 cos190)~ = constant , 2 . a .* z pod19 + (z’+: - 2ztmg& cos 190 + t2mm2g2)M = constant

S5.0 Differentiation of this equation with respect to the time gives ii = P’(u)/2. Putting u = 1 - bu we find

d26u/dt2 = [-P’(1) + P”(l)bu]/2 . For b = a and E = (Z:a2/2Zs) + lmg

~ ( u ) = (1 - ~)~[(2mgt/z1)(1 +u) -a2] . The roots of this equation are u = 1 (double) and u = (Z1a’/2lmg) - 1. The former tells us that the curve describing P(u) is tangent to the u-axis at u = 1. Hence we expect P’(1) = 0. Infact

P’(u) = 2(Z1a2 - lmg - 3lmgu)(l - u)/Zl

does vanish for u = 1. We find also that P”(1) = 2(-Z1a2 + 4lmg)/Z1 and so

d2bu/dt2 = (P”(1)/2)6u = [(4lmg/Z1) - a2]JGu

Stability requires that P”(1) < 0, a’ > 4.!!mg/Z1, wo > ( 2 / z s ) f l x . Note that P’(1) = 0 and P”(1) < 0 tell us that P(u) has a maximum for

u = 1. On the other hand the inequality 4lmg - aaZl < 0 tells us that the other root of P(u) = 0 is greater than unity. Figure 5.9a shows this situation, while figure 5.9b shows the reason for instability when 4lmg - a2Z1 > 0.

.

55.7 The only forces acting on the top are gravity and the normal reaction from the plane, N = NQs. Denoting by x i (i = 1,2,3) the coordinates of the center of mass, we have XI = X 2 = 0 and mXs = N - mg.

The torque is T = l N sin 6(&: cos rl, - 6; sin r l , ) , and Euler’s equations axe

Ilk: = (ZI - Z S ) W & W ~ + lNsin6 cosrl, , Zlbi = ( I S - Z~)W&W: - lNsin6 sin+ , { zsa; = o .

These equations are like equations (5.38) with mg replaced by N. However, while mg ia constant, N is not.

98 C ~ A P T ~ R 5. RIGID BODIES

The horizontal component of the velocity of the center of mass and w$ are

From the first two Euler's equations it is easy to obtain that

ZI (wY2 + wI;L) + 2emg cos 8 + me'ssin'8 9' = constant .

constants of motion.

If the top is released from 8 = 80 with d = 0 and + = 0, the "constant" has the value 2 h g cos 80. Hence

(11 + ~ ~ 2 ~ i n 2 8 ) ~ ' -I- II+'ssin28 = 2emg~cos290 - cos 6 ) .

Both to$ and t 3 = (Ilsin'8 + Z~cos'~B)+ + 13 cosfi 4

= Ilsin'$ + + 13 cos8 w: are constants of motion. Hence L3 = Z3 cos @OW;,

sin"$ + = a(cos 80 - COSS) .

If $1 = x2 = 0 (not skating after all, just twirling on the spot with the center of mass going up and down in the vertical direction), the energy is

E = [m& + 119' + ZlsinV +' + z3(4 + + cos 6)*1/2 + emg cos 29. Using

sin'@ + = a(cos290 - cosg), x3 = ecos8, and 4 + +coat9 = w& = (Zl/I3)a, and putting as usual u = cost9, E = (I:a2 1213) + ~ ~ g c o s 8 0 yields 6' = &(u) with

&(uf has two roots of interest to us, u = uo and

u = u1 = [Zla' - JZTa4 - 8emgZla2~o + 16Pmmzgz ]/4emg . This latter is the same value found in section 5.4.3. In both cases the center

f+]slt15.-8, = a(cos80 - cos$l)/(l - u:) has the same value in both cases. It is of mass G falls through e(cos80 - C O S ~ ~ ) .

determined solely by energy conservation.

S5.8 (i) w = 9t;l ++sin@ 6: + t i , + +cos8)t;s, = 9, ~z = +sin@, ft3 = ?j~ + ~ c o s 8 .

(ii) A1 IlCti, A2 = IlQ2, A3 = f3S23 (iii) Deriving d&i/dt (i = 1, ?, 3) either from e q u a t i o ~ f5.31,32) or from the formula dhc/dt = (w - &g) x bi, one finds

Zl& - (I1 - I S ) 0 2 s 2 3 + zln'?) = TI , Z I ~ Z + (Zi - Zs)RsCti - ZiQir) = Tz 13n3 =T3 .

,


56.9 (i) Let b z point from the point of contact P to the center of the disk C, C'o = -AS. With the conditions d = 0 and R+ = -r& Euler's equations for the components of the angular velocity give

sin29 +2[&(c0829 - R/r) - Z I cos29) = TI , Ilsin.29 8 = T2 , { ZS(COS~ - R/r)+ = T3 .

Denoting by f = fibi the force acting on the thumbtack at 0 and by F = F i h that acting at P, the components of the torque are given by

TI = ef2 - rF3, TZ = - t f l l and T3 = r 4 . For the motion of the center of mass r = e b s , the equation mi: = f + F + mg

yields

me sin6(a=mgsinc coscp+f1+4 , mt sin 8 cos 29 G' = -mg(cos e sin 6 + sin c cos 6 sin cp) + { mtsin229+'=mg(cosc cos6-sinesin29 sincp)- f a - F s .

Eliminating the forces f and F, we finally find

+ FA ,

Thecondition (F+f+mg) . t s = Ogives (F'z+f~)sin6+(F3+f3)~0~6 = mgcosc.

[Ilr2 + ( IS + rnrz))e']lc;; = mgRrA sin c cos 'p . (ii) In order to find the frequency of small oscillations about 'p = 7r/2, we put 6 = n/2 - cp, obtaining

[Zlr2 + ( I S + mr2)>e2]Ei' N -mgRr2 sin E 6 . Hence

2 mgRr2 sin e wosc = I 1r + (I3 + mr2)P

'

55.10 Let rc and P denote the position vector of the center of the disk C and the point of contact with the plane. With 6s vertically up, we have

mF, = -mgt3 + F , where F is the reaction of the plane $t P.

is

Expressing this in terms of the bj components of w , one has

If r is the radius of the disk, and b2 is always along P 2 , the rolling condition

+, = -r6, x w .

k, = r(a163 - as&) . Differentiating this, after a moderate calculation F = mi, + mg6s gives

F = n r ( - h s + S 2 1 ~ i ) ~ l + m ( g s i n 2 9 - r ~ ' f - r R ~ + r ~ ~ ~ ) b z + m ( g c o s 6 + r ~ 1 + r ~ z ~ ~ ) 6 ~ .


The components of the torque r = -rb, x F are

Euler's equation give TI = -mr(gcos8 + + rSlzR3), l"2 = 0 , T3 = mr2(--h3 + nln,),

(11 + mr'))ni + (13 + mr' - I I ) ~ z ~ ? c J + 11nz1) = -rmgcosd , I i h + (I1 - 13)%ni - Iinir) = 0 , { (13 mr2)h3 = mr2fli& .

Since f l 3 = w:, the last of these equations gives

(13 + mr')bk = mr2 sin 8 99 .

The second equation gives

13w$h - 211 cosd 89 - 11 sin8 = o .

The first gives

(11 + mr'))d + (13 + mr'))wk+sin 8 - 11 sin 8 cos t@ + rmg cos 8 = o .

Chapter 6

LAGRANGIANS

Praise be to the Lagrangian, labor-saving device and crystal ball. A mechanical problem is encapsuled in the Lagrangian and the constraints (if any). Fkom these the equations of motion are derived by differentiations. In many cases the existence of integrals of motion can be predicted by simple inspection of the Lagrangian.

6.1 Heuristic introduction Consider a system of two particles, a and 6, interacting by a conservative force with potential energy U(lra - r b l ) . Denoting by xai and X b i the coordinates (i = 1,2,3), we have the equations of motion

Fkom these we derive the equations of motion in terms of the coordinates of the center of mass, Xi = (mazai+mbxbi)/M, and of the relative position, %i = z a i - Z b i ,

where M = m, + mb is the total mass and m = mamb/M the reduced mass, and the potential energy is now written as U(l.1).

T = Irl, 8, and 4, and we find We may further introduce relative spherical polar coordinates,

MXi = 0 , ma,. = -dU/dr , mag =mad = O , (6.3)

101

102 CHAPTER 6. LAGRANGIANS

where a, = i: - TB” - r sin"^ J2 , a ~ = ~ ‘ 8 + 2 + d - - ~ s s i n 8 cos@(ji2 , (6.4) a# = r sin B (i; + 2 s i n ~ $13, + 2r costl813, =

F’rom equations (6.2) we see that the total momentum P = MV is a constant of motion, while equations (6.3) tell us that the 3-component of the angular momentum about the center of mass is also a constant of motion if we remember that dl,/dt = WT sin 8 ad. Here V = R is the velocity of the center of mass. The angular momentum about the center of mass is

I = na,(r, - R) X (V, -- V) f m&(rb - R) X (Vb - V) = IIE I” X V.

Let us now introduce the Lagrangian

L z K - U . (6.5)

This can be expressed in terms of the various sets of coordinates used above, reading respectively

L = (mar: t n a ~ r ~ ) / 2 - U(/r, - rat) , L = (MR2 + rnr2)/2 - U(J.1) , (6.6) i L = [ M R ~ + m(i2 + r2tP + +%in2@ @)]/2 - U(T) .

Each of equations (6.1,2,3) can be expressed in the cumpact form of the Lagrange equations

(i = 1, ... 6)

with L as given by the first of (6.6) and

by the second of (6.6) and

by the third and

qi = (2 1,2,3), qi+3 = 2 b z (i = 1,2,3),

pi = Xi (i == 1,2,3), qi+3 =L: (i = 1,2,3),

qi = xi (i = 1,2,3), q4 = T , q5 = 8, qfJ = #. Furthermore the Lagrange equations tell us that from the absence of a

coordinate qi &om L, so that 8 ~ / 8 q ~ = 0, follows at once the conser~tion of the “generalized momentum”

conjugate to that coordinate. Since L in the second and third of (6.6) does not depend explicitly on Xi,

the momentum px, conjugate to X i is seen to be conserved, px* = Pi = 0.

6.2. VELOCZTY-DEPENDENT FORCES 103

Similarly, pg = l3 is conserved because L as given by the third of (6.6) does not depend explicitly on 4.

If a Lagrangian depends explicitly on the time derivative qi of a certain coordinate, but not on the coordinate itself, it is invariant under a transformation qi + qi + dqi (dqi constant) of that coordinate, a translation in the case of Xi, a rotation in the case of 4. The invariance-connection of constants of motion will be presented in section 6.7, culminating in Noether's theorem.

Let us see whether the Lagrange technique works for a rigid body, for instance the spinning top with a fixed point of section 5.4.

Start with the Lagrangian

2 L = K-U = [I19"+Ilsin219d2+r3(d+$cos19) ]/2-lmgcos6 , (6.9)

and put pe = aL/a9, p, = &/a$, pG = aL/ad. Then dpsldt = aL/ad gives

2 118 - sind[II cosd $2 - 135244 + @cos19) + lmg] = 0 , (6.10)

while dp,/dt = 0 and dpQ/dt = 0 yield respectively

p , = (Ilsin'd + I ~ c o ~ % ) + + r3 C O S ~ 4 = constant , (6.11)

and

(We had p, = I lb , p+ = I1a = I3wh :) Multiplying equation (6.10) by 19, and using the other two equations,

one finds easily that dE/dt = 0, where E = 1119~12 + U,(d) (see equation (5.40)).

ptl, = 13(4 + + C O S ~ ) = constant . (6.12)

6.2 Velocity-dep endent forces Consider a particle subject to both a conservative force F = -VU and a force of a different type F'. Newton's second law can be written in the form

a L dt aj., axi 5 (") - - = F,! (i = 1,2,3) ,

In some cases F,! can be expressed in the form

(6.13)

(6.14)

104 CHAPTER 6. LAGRANGZANS

and (6.13) can be replaced by

(6.15)

One such case is that of an electron (charge -e) in a static electromagnetic field (E = -VV, B = V x A), for which

Fl = -(e/c)eijkkjBk . (6.16)

It is easy to verify that in this case 6 L = -(e/c)Ajkj. In fact

dAi a A j dt axi

e C

Therefore the Lagrangian for an electron in an electromagnetic field is

L = mr2/2 -t- eV - ( e / c ) A - r (6.17)

The Lagrangian

L = eV - (e/c)Aj&j - ~ O C 2 4 7 1 - (+/c)

yields the relativistic equations

(7 = 1/41 - (+/c)’ ) for an electron (rest mass mo, charge -e ) in an electromagnetic field.

The quantity 6U = -6L = (e/c)Ajkj is sometimescalled a velocity dependent potential. This terminology is imprecise because 6U is an energy, not a potential. It is also misleading. We are used to think of a potential energy as a function whose space derivatives, taken with a minus sign, are the components of the force. Now

is a the i-th component of the Lorentz force. Conversely, the damping force -yon& can be trivially expressed as

6.3. EQUIVALENT LAGRANGIANS 105

with SU = ymxx. However, the Lagrangian L = mi.'/2 + mwax2/2 - Tmxi.

yields the equation a md(x- -ys)/dt = -m x - ymi , x +w2x = 0

instead of the correct

This might have been expected, since -7mxi. = dA/dt with A = -ymz2/2. We shall see in the following section that the addition of a total time derivative to the Lagrangian doeg not change the equations of motion.

A correct Lagrangian for the damped oscillator is

.

x+yx+w2x=o .

L = e7tm(x2 - w2z2)/2

6.3 Equivalent Lagrangians This may be a convenient point to mention a certain degree of arbitrariness in the choice of the Lagrangian. The reader will already have noticed that the Lagrange equations for a particle subject to conservative forces are ~ o ~ o ~ e n ~ u s in C. Therehe the equ~tjons of motion yielded by the Lagrangians C and kL (k=constant) are identical.

Also identical equations of motion are yielded by the Lagrangians L and L+dh/dt, where A(q , t ) is an arbitrary function of the coordinates and the time. In fact, dA/dt satisfies Lagrange's equations identic~ly:

In section 1.1 we mentioned that the equations @ = 0, q = p ) and (d = 1, q = p - t) are equivalent. The former are yielded by the Lagrangian L = (i2/2, the latter by the Lagrmgian

L' = Q2f2 + qt + q = L + dA/dt with A = qt. This property is particularly interesting when applied to equations (6.15)

for an electron in a static electromagnetic field. In fact, with A depending only on r, we have

6L + dhfdt = -(e/c)&(Ag - ( c / e ) ~ A / ~ ~ ~ ) = -(e/c)&Ai, where A' = A - (c/e)Vh is the gauge-transformed vector potential corresponding to the same magnetic field as A.

106 CHAPTER 6. LAGRANGlANS

6.4 Invariance of Lagrange equations

The Lagrange equations (6.7) with the Lagrangians (6.6) were seen to yield correct results. This is an example of invariance of the Lagrange equations under time-~ndependent coordinate transformations

qi = qi(q;, * . . qk) (i = 1, * . .a). &om

follows that = o .

(6.18)

(6.19)

The Lagrangian is transformed as a scalar, L’(g’, q’) = L(q(q’), q(q’, 4’)). Before presenting a general proof of invariance of the Lagrange equations under time- dependent coordinate transformations g = q(q’,t), we wish to verify this property in the simple case of a tra~formation from an inertial to a uniformly rotating frame of reference, for an electron in a uniform magnetic Geld in the 3-direction. We start from

L = m r 2 / 2 - ( e B / 2 c ) ( z 1 & - m E 1 ) , (6.20) the Lagrangian (6.17) with A1 = - x 2 5 / 2 , A2 = zl B / 2 and A3 = 0, yielding the Lagrange equations

mxl = - (eB/c)xz , mx2 = (eB/c)x l , mx3 = 0 . (6.21)

Transforming to the rotating frame,

z1 = xi cos(wt) - z: sin(wt) , 2 2 = z’, sin(wt) + xi cosfwt) , I 5 3 = 4 ,

with some dgebra we find

mE‘, = w(mw - eB/c)x’, 4- 2m(w - eB/2nac)ii , m x i = w(mw - eB/c)x: - 2m(w - eBl2rnc)x: , (6.22) { rnx; = 0 .

But these are indeed the Lagrange equations derived from the transformed

L’ = mr’2/2+(w/2)(naw-eeB/c)(2:~ +x;’) +(mu -e5 /2c) (z :Ei -z;k:) . (6.23) Lagrangian

Which is the more economical way of obtaining equations (6.22)? I n ~ ~ e n t d i y , Larmor’s theorem states that the action of a uniform magnetic

field on a charged particle can be counter~ted to first order by a rotation. In fact, for w = eB/2mc the above equations reduce to

2 2 mij{ = - (e2B2/4mc2)x{ , m& = - ( e B /4mc2)z: , mi$ = 0 ,

6.4. INVARIANCE OF LAGRANGE EQUATIONS 107

namely equations for a particle acted upon by an elastic force normal to the 3-axis.

We start from a more general form of the equations,

which will be justified later (see (6.27)). The Qj's are generalized non- conservative forces, while L = T(Q, q) - U ( q ) accounts for the conservative force#. Of course, nothing would prevent us from taking L = T and adding to the right side QY) =

Qi = (Bq~/aq(i)qi+aqi/at, and so qi is a function of the Q', the q', and t. Con- versely qi = qi(q, t) , 4; = (aq;/dqj)qj +8q{/8t etc. Note that even if the La- grangian L is a function only of the q's and the gs, regarded as independent variables, the t r~sformed Lagrangian L'(q', Q', t ) = L(q(ql, t ) , Q(Q', q', t ) ) may depend explicitly on t .

The formula a q ~ / a Q j = a q ~ / ~ ~ j will shortly be used together with

We consider transformations of the type qj = qj(q', t). Then

Now

With a little work this gives

Multiplying by aqi/aq;, we find

with Q; = Q$ aqi/aqi. We end this section with a few relations, which will be used in chapter

7. Note first that from a q ~ / ~ q j = aq~ /aq j one has


Hence

(6.24)

6.5 “Proofs” of the Lagrange equations We have deliberately presented the Lagrange equations as just another way of expressing equations of motion which could be derived directly from Newton’s second law.

In light of the previous section, and on the assumption that any physical system, however complex, can be described as a set of point particles, we may “prove” the Lagrange equations in the following simple way.

Each particle obeys Newton’s second law

mara = Fa (a = 1 , . . . N ) ,

which can be expressed in the form

A ( ” ) = & , d t ax,,

where

a

But this is already a Lagrange equation. If instead of the coordinates xaj (a = 1 , . , . N ; i = 1,2,3) we introduce generalized coordinates qi (i = 1,. . .3N)) functions of the ra’s, then the Lagrange equations are transformed into

where

Another popular “proof” starts from Hamilton’s principle derived from the d’Alembert principle, which in turn stems from the “virtual work” principle of statics.

Let q i ( t ) describe the real “trajectory” of a system during the time interval tl to ta, and q:( t ) = qi ( t ) + Gqi(t) a varied trajectory with Gqr(t) infinitesimal, subject to the restriction Gqi((t1) = G q j ( t 2 ) = 0. Thus the

6.5. ‘(PROOFS” OF THE LAGRANGE EQUATIONS 109

real and the varied motion start from, and arrive at the same points at the same times. (It is not so in the case of the ~auper tu is principle, equation (1.37) .)

Let

where

Hamilton’s principle states that

6’ (dK+dW)dt=6LrKdt+1:6Wdt=O , (6.25)

where 6W is the work done by the forces for the disp~acements dqi. Notice that in general we cannot write

i , ta6W dt = d r a w dt . ti

For conservative forces we have

while generalized non-#nser~tive forces can be defined by writing

6W=Qj6qi *

Therefore (6.25) can be expressed in the form

(6.26)


A simple partial integration, using Sqi = dSqi/dt and 6qi(tl) = 6qi(t2) = 0, yields

Since the dqi’s are arbitrary for tl < t < t2, we infer that

d L Qi . (6.27)

A final remark. The first term in equation (6.26) is unchanged if we add dA(q, t ) /dt to the Lagrangian, since

This confirms what we wrote in section 6.3.

the action A = s:12L dt. Extremize, not minimize! For L = (q’ + aq2)/2 (a < o attractor, a > o repellor)

and dq = e ( t ) ( e ( t 1 ) = e ( t 2 ) = 0), using the equation of motion ij = aq we have (exactly)

&S,t:(dA/dt)dt = (aA/%i),,sqi(tz) - (bA/&i),,bqi(tl) = 0.

For Qi = 0, Hamilton’s principle (6.26) states that the real motion extremizes

This is positive for a > 0 (repellor), in which case the action is a minimum. For a < 0 it may be positive or negative.

6.6 Constraints In the preceding chapter we have tacitly assumed that the number of the generalized coordinates equaled the number of degrees of freedom. In deriving Lagrange’s equations from Hamilton’s principle, the final step required the independence of the Sqi’s.

We first consider “holonomic constraints”. A holonomic constraint is a geometrical restriction on the coordinates expressed by an equation

F(q1, . . . qn; t ) = 0 . (6.28)

Examples: Spherical pendulum: A point mass moves frictionlessly on the inner surface

of a hemispherical bowl (figure 6.1). This time-independent constraint can be expressed by the equation

F ( r ) = R - T = 0 in spherical coordinates, or F ( x ~ , x ~ , x ~ ) = R - &- = 0 in Cartesian coordinates.

6.6. CONSTRAINTS 111

Figure 6.1: Spherical pendulum

Figure 6.2: Bead on rotating wire

A time-dependent constraint: A bead can slide frictionlessly on a straight wire which is forced to rotate in the (zlzz)-plane around the origin with constant angular velocity w (figure 6.2). This constraint can be expressed by

F(8 ; t ) = 8 - w t = 0 in plane polar coordinates, or by F(zI, ZZ; t ) = -z1 sin(wt) + xz cos(wt) = 0 in Cartesian coordinates.

A holonomic constraint can be used to eliminate one of the coordinates from the Lagrangian or to express q1,. . . qn in terms of new coordinates

I q; 7 . . . Qn-1. For instance, the Lagrangian for the spherical pendulum

L = (pn/2)(+' + r 2 P + rZsinz8 $1 + pngrcos8

reduces to L = ( r n ~ ' / 2 ) ( P + sin'@ $1 + rngRcose .

With XI = T cos(wt) and x2 = T sin(wt), the Lagrangian L = (m/2)(iT + i:) for a free particle in the (zl,zz)-plane reduces to

L = (m/2)(+2 + TZ"2)

112 CHAPTER 6. LAGRA NGIA NS

for the bead constrained to slide on the frictionless wire. The use of a holonomic constraint to reduce by one the number of

coordinates in the Lagrangian does not provide information on the force implementing the constraint. As we know from General Physics, this is a reaction of varying magnitude normal to the surface F(q1, . . . qn; t) = 0, for instance normal to the hemisphere for the spherical pendulum and to the wire for the bead.

For the spherical pendulum we may try to use the Lagrange equations in the form (6.27), writing

d a~ a L z(z)-EK= Q: , where

L = (m/z)(i: + ki + iz) + mgx3

and

We find

mxl = -Axl/r , mxz = - A x z / r , mx3 = mg - Ax3/r . (6.29)

These equations, together with the constraint equation r = R, are sufficient to determine the four unknowns 2 1 , 2 2 , 2 3 , and A. In fact, differentiating the constraint equation r2 = R2 twice with respect to t we obtain

x l i l + x2x2 + x3x3 = -(&: + i; + ig) = -R'($ + sinz@ 4') = ~ a , ,

Multiplying the first of equations (6.29) by 21, the second by 2 2 , the third by 2 3 ,

and summing we find

m(xl%l + xzxz + 2 3 x 3 ) = -AR + mgxs

mar = -A + mg cos 0

,

, showing that A is the magnitude of the reaction of the constraint.

this case we must take From the constraint - 2 1 sin(&) + x2 cos(wt) = 0 for the bead, we see that in

Q: = -Asin(&) , Qi = Acos(wt) .

Hence the Lagrange equations are

mx1 = -Asin(&) , mxi.2 = Acos(wt) ,

to be used in conjunction with the constraint.

constraint itself, yields Double differentiation of the constraint with respect to t , and use of the

-Pi sin(wt) + xz cos(wt) = 2w[i, cos(wt) + Xz sin(wt)] ,

6.6. CONSTRAINTS 113

and so

Supposing i > 0, this is the magnitude of the Coriolis force, which the con-

In general, taking the variation of a holonomic constraint straint must counteract (see bug with B = ?r/2 problem 4.5).

F(q1 , . . . qn; t) = 0 at a c e r w , we have . .

a16ql + . . . + anc5qn = 0

with ~ i=aF/aqi .

The components of the reaction force are proportional to OF/aqi and so we can write

Qi = X U ~ . Note that the corresponding work 6W in Hamilton’s principle is

6W(reBCt) = Qidqi = Aai6qj = 0, as we expect since the reaction and the “virtual displacement” are at right angles. On the other hand, when we differentiate the constraint with respect to t to supplement Lagrange’s equations, we have (aF/aqj)qi + (aF/at) = 0, aj4.i + at = 0,

ajdqi + atdt = 0 ,

with ai =aF/&i , at = dF/dt .

Figure 1.3 emphasizes the distinction between the 6qi and the dqj = qidt.

In general we may have more than one, say T holonomic constraints

F,(ql , . . . qn; t) = 0 (QI = 1, . . . T )

or, in differential form, a,idqi + a,tdt = 0

with a,i = aF,/aqi and aat = aF,/at .

]In the examples, the constraint equation was written in such a way that the ai’s turned out to be the components of unit vectors normal to the sphere and to the wire, reupectively. Thus X was the magnitude of the reaction. In general we can expect X to be only proportional to the magnitude of the reaction.

114

Figure 6.3: dqi and dqs

Then the L ~ a n g e equations have the general form

where Qi = Xcvaai (sum over a) .

These equations, together with the 9- constraints, are sufficient to determine the qi and the r “Lagrange multipI~ers” A,,

A cons t r~nt in differential form

is ~‘non-~olonomic5’ if it is not integrable, i.e. cannot be expressed in the form dF(q1,. . . qn; t ) = 0.

Aidxi = 0 it is not sufficient to calculate the components of the generalized “curl”, ( a A ~ / a ~ ~ ) - (aA~/axi). Even if one or more of these are different from zero, the ~ ~ e r e n t i a l relation may be integrable after mult~pIication by an “integrating factor”.

For instance, the adiabatic condition C,dT + ( ~ T / V ) d V = 0 for an ideal gas, after multip~ication by l/T becomes dS = 0, where S is the entropy.

A rolling disk provides a good example of non-holonomjc constr~n~. The rolling condition i, = -r&z x w can be expressed in terms of ~ m p o n e n ~ with respect to the fixed axes

In order to determine the integrability of a d~fferential relation

X I = - r ~ c o s d c o s ~ ~ T d s i n ~ s i n ~ - r ~ c o s y , , iz = -rGcosdsiny,- rdsinbcosp- rpjrsintp x3 = r8 co9 6 = dfr sin d)/dt ,

6.7. INVARIANCE OF L AND CONSTANTS OF MOTION 115

where X I , 2 2 , and x3 are the coordinates of the center of the disk. The third constraint is, of course, holonomic, d(x3 - r sin 6) = 0.

We have alcdxl + a1,dcp + alsd6 + al+dqb = 0 )

azcdxz + a2,d'p + a2sdI9 + az*dqb = 0 ) { ascdx3 + a30d6 = 0 )

withai, = I (i = 1,2,3), al, = rcos6cos'p, a2, = rcosdsincp, a ls = -rsindsincp, a26 = r sin 6 cos 'p, a3g = -r cos 6 , a1+ = r cos cp, a2+ = r sin cp.

With the Lagrangian

L = (m/Z)(i: + xi + 2:) + (mr2/8)(d2 + sin26 G2) +(mr2/4) (4 + (i, cos 612 - rmg sin 19

we find mxl = XI.., mx2 = A2 , mx3 = A3 , (mr2/4)(6 - sin6cos6 +2) + (rnr2/2)w;+sin6 +rmgcos6 = k a l e + h a z e + A3036 , (mr2/4)d[sin26 @ + 2 cos I9 w;]/dt = Xlal, + A2azlp , (mr2/2)dw;/dt = Alal* + X2a2+ .

The last equation then gives

(mr2/2)dwb/dt =r(X~cos'p+X2sin'p) =mr2(-ojcos19-~+2d@sin6) . In a similar way one obtains two more equations. Compare with problem 5.10.

The first and the second of the constraints are non-holonomic. For the rolling disk with horizontal axis (19 = ~ / 2 ) problem 6.17 gives a direct proof of the non integrability of the conditions

dxl + r cos 'p dqb = 0 and dx2 + r sin 'p dll, = 0. Non-holonomic constraints are restrictions on the infinitesimal changes

of the coordinates, whose range of allowed values is unrestricted. A rolling disk can reach any point of the (xlxa)-plane, and '19, cp, and + can attain any of their allowed values, in spite of the non-holonomic constraints.

6.7 Invariance of L and constants of motion Consider an infinitesimal transformation

= qi((t) + Sqi( t ) , t' = t + St .

We say that the Lagrangian L is invariant under this transformation if

SL = L(qj(t'), dqi(t')/dt', t') - L(qi(t),dqi(t)/dt, t ) = 0

aSee also A. Sommerfeld, Mechanics (Academic Press, 1964), Problem 11.1, pp. 244, 261.


up to the second order in 6qi and 6t. The first term is obtained from the second by simply replacing qi(t) and dqi(t)/dt by q:(t ’) and dq:(t’)/dt‘, and t by t‘ if the Lagrangian depends explicitly on the time.

Consider transformations 6t = 0, q i ( t ) = q i ( t ) + c f i ( q , q , t ) with c an infinitesimal parameter.

Coordinate transformations

is a constant of motion.

Example: The Lagrangian

L = (mar: + mbii)/2 - U(lra - rbl)

is (exactly) invariant under the transformation

ra 4 ra +6r , rb + rb + 6 r , where 6r = cii (ii constant unit vector). In this case

= T&i(makai + mbkbi) = kffi ‘ v = fi * P . The component of the total momentum in an arbitrary direction is conserved.


L = (mi2 + kr”)/2

is invariant under the transformation

r -+ r + ~ i i x r , xi --t xi + e Eijknjxk

((r + cii x r)2 = r2 + O(c2)). The conserved quantity is

= mki(cijknjxk) = nj[€ijkxk(mki)] = d ‘ 1 , the component of the angular momentum in an arbitrary direction.


L = mi2//2 - c tan-’(m/xI) + x3

( c = constant) is (exactly) invariant under the helical transformation

xi = X I coss - x2 sins , x; = X I s ins+ x2 coss , xi = 23 + cs ,


where s is a constant parameter.

(21 = pcos4,22 = psin4, 23) one has z: = pcos(4 + s), zi = psin(4 + s), c tan-'(z;/z)1) - 25 = c tan-'(tan($ + 8 ) ) - 2 3 - cs = c$ - 23

In fact, xi2 + xk2 = x; + xi and x& = x3, while in cylindrical coordinates

= c tan-'(zz/z1) - 23.

For s -+ E infinitesimal, we have f~ = -XZ, fz = 21, f3 = c, and the conserved quantity is

I = (aL/a+)fi = rn(-zzj., + zliz + ck3) = l3 + cp3 . Time translations

For t' = t - e we have

6L = L(q(t'),dq(t')/dt', t') - L(q(t), dq(t)/dt, t) = -EdL/dt ,

6L = -cdL/dt gives

If a Lagrangian does not depend explicitly on the time, then

a L I = - q i - L &i

is a constant of motion, which will be seen to be the Hamiltonian.

a variational principle, d(a f /By')/dz = af par. One has

If Sf/Sx = 0, one has the Jacobi integral of motion

This is a special case of a general property of the Euler-Lagrange equation of

d(y'af/agt - f)/dz = -af/ax.

&3f/av' - f = constant . As a preparation for Noether's theorem, we want to derive this result

Assume that the Lagrangian does not depend explicitly on the time in a slightly more complicated way.

and that

L(q'(t'),dq'(t')/dt') = L(q(t),dq(t)/dt) + 0 ( e 2 ) , 3For instance L = (1/2)aijqiqj - U(q), with aij and U functions of the pi's, but not

explicitly dependent on t .


where t’ = t - € (c infinitesimal) and ql(t’) = qi(t). We write

Thus we have d L - dL d - dL 0 = -6qi + ---(5qi - 6- dqi dqi dt dt

and we find again the integral of motion I = (aL/dq,)qi - L 5 .

6.7.1 If under an infinitesimal transformation q, + qi + 6qi with 6qi = €fi(ql q , t) the Lagrangian changes by

Invariance of L up to total time derivative

then

is a constant of motion. Proof:

dA = E - d ( - f i ) aL

dt aqi

(6.30)

~

4For instance, if q ( t ) = t2 then q‘(t’) = q( t ) = (t’ + e)21 aq( t ) = q’( t ) - q(t) =

5For the example in footnote 3, Z = (1/2)ai jqi4j + U is clearly the energy. ( t + c ) ~ - t2 = 2et + e 2 , while q’( t ’ ) - q( t ’ ) = t2 - (t - c ) ~ = 2et - c2.


Example: The variation of the Lagrangian

L = (m/2) (2 - w'x')

under the transformation

is

x' = x + e sin(wt)

6~ = (m/2){[kf2 - 21 - w 2 [ z f 2 - x']} = em{wk cos(wt) - u'x sin(wt)} + O(e2) = e dA/dt + O(e')

A = mwxcos(wt) . with

The integral of motion I = mxsin(wt) - w x cos(wt),

sin(wt) I = I dsin(wt)/dt dGdt I '

is proportional to the Wronskian determinant of the solutions sin(wt) and x(t) of the equations of motion.

Consider an infinitesimal Galilean transformation Example:

ra ra + tefi r b l'b + tefi

where 15 is a unit vector and c is an infinitesimal parameter with the dimensions of a velocity.

The variation of the Lagrangian given by the first of (6.6) is 6L = E f i . (mara -k 77Zbi.b) -k O(e') = E dA/dt + O(e2)

A = ii. (mara +mbrb) = Mii. R , where R is the position of the center of mass. The corresponding integral of motion is

I = [t(mara + mbrb) - (mars + mbrb)] ' fi = M(tV - R) . ii = -MR(O). ii

where R(0) is the position of the center of mass at t = 0.

6.7.2 Noether's theorem The following is a version of the theorem for Classical Mechanics.

Consider an infinitesimal transformation

ql(t') = qa(t) + dqz(t) , t' = t + 6t , where


Assume that the action integral is invariant under this transformation with an error of second order in E

Then with a second order error in E we have

giving

d L dqi(t') dqi(t) d6t dt - q&)) + - r3qf ( dt' --) i- Ldt .

We already know that

On the other hand, using

d dt' d dbt d -=-c= I + - dt dt dt' ( dt ) dt'

and

we have

sDo not confuse this with the exact, but trivial identity


The quantity

(6.31)

is a constant of motion. For time translations one has fi = 0 and g = -1.

aL I = -(fj - gqj) + Lg adi

Example: The action

is invariant under

For X = 1 + E we have f = q / 2 and g = t . d(t’) = JT;q(t) , t’ = At ,

The integral of motion is

89 I = E(f - qg) + Lg = -Et + 4912 ,

where E = q 2 / 2 - k / q 2 is the energy integral of motion.

6.8 Chapter 6 problems 0.1 Derive the equation for the thumbtack (problem 5.9) by using a La- g r a n ~ a n ~nvoIvin~ only ‘p, & and the fixed angle @.

0.2 A uniform plank initially leans against a smooth wall and rests on a frictionless floor.

Discuss its motion (i) by an elementary method, and (ii) by using a Lagrangian and constraints.

z = (t?/2) cos8, y = (~ ‘2 ) sin@, I , , = mP/12, IA = IB = mP/3 -+ I

6.3 A particle of charge q moves in a static electromagnetic field with potentials V = -a Inp, A = (-B2//2, Bs /2,0), where p = d m , and a and B are constants.

(i) Express the fields and the force acting on the particle in terms of the unit vectors iip, G#! and 8,.

(ii) Write F = ma in cylindrical coordinates. (iii) Write the Lagrangian and derive the equations of motion, Compare

(iv) Do you expect constants of motion other than the e n e r ~ ? Why? (v) What are they?

with (ii).

6.4 Consider a system of N particles, interacting with one another with velocity-independent forces, Lagrangian

N

a=l

Let (p be an angle about the unit vector fi. We know that

(sum convent~on also for a).

a function of the ra’s and their time derivatives is's ? (i) What is p$ = pd + iip# for the Lagrangian L‘ = C + 6L, where dL is

(ii) What is bp& if SL is the Lagrangian 6L = c-‘qaA&i ?

6.5 Obtain the equation of problem 3.10 by a Lagrangian method.

6.6 Consider the block of mms rn on the frictionless floor of a cart which moves according to X ( t ) a9 shown in figure 6.4. Denote by k the elastic constant of the spring, and by l its natural length.


Figure 6.4: Block on accelerated cart

(i) Write the Lagrangian for the coordinate x of m with respect to a fixed x-axis.

(ii) Change coordinate from x (with respect to the fixed axis) to x’ (with respect to the cart), find the new Lagrangian and the equation of motion with respect to the cart.

(iii) Incorporate the inertial force found in (ii) in the Lagrangian (i) as a potential energy, and find again the equation of motion for 2‘.

6.7 Let L(q, q; t ) be a Lagrangian. A “complementary” Lagrangian L , ( p , ~ , t ) , can be defined by the Legendre transformation

L,(Q, Q; t ) + c.uL(q, 4; t ) = PQ + Q P ~ , where Q = p , P = aq, and Q = fl . Only the case Q = 1 seems to have been considered in the literature.

(i) Show that P = aL,/aQ, P = aL,/aQ, bL,/at = -aaL/at. (ii) (a) Find Lc for the harmonic oscillator, L = (mq2 - lcq2)/2.

(iii) (a) Find L, for the Kepler problem, L = mq2/2 + k/lq( (k > 0). (b) Find the corresponding Lagrange equations.

(b) Find the corresponding Lagrange equations.

6.8 At t = 0 a particle of charge q and mass m starts from rest at the origin of a Cartesian coordinate system, and moves under the influence of the electromagnetic field E = EB, and B = Be, ( E and B constant). While the electrostatic potential is unique up to a constant, q4~ = -Ey, the vector potential is determined up to a gauge transformation.

A’ = -By& and A“ = Bxb,, both giving the same B. (i) Find the gauge transformation connecting the two vector potentials

(ii) Write the Lagrangians L‘ and L“ for the two cases.

124 CHAPTER 6 . LAGRANGIANS

(iii) Check that these Lagrangians differ by a term dF/dt, and so can

(iv) Use L' to write the equations of motion, and solve them for the

(v) Use L" for the same purpose. (vi) Qualitatively compare the paths of an electron and a proton.

be expected to yield equivalent equations of motion.

given initial conditions.

0.9 In Hamilton's principle (equation (6.25)) the variation q( t ) -+ q'(t) = q( t ) + 6q( t ) is synchronous, i.e. the system is at P in the real motion and at P' in the varied motion at the same time t.

f(t) = t + S t , and the varied motion is described by a function of f, q"(0 = q'(t(Q). In this case, one speaks of asynchronous variations.

one bas S*q = 64 - qdSt/dt. Here

On the other hand, if the time is also varied, it is at P' at a time

(i) Show that, while 6q = q'(t) - q(t) and 6*q = q"(@ - q(t) are equal,

(ii) Show that Hamilton's principle for motion under conservative forces takes the form

l l t ' (6*K + 2K(dSt/dt) - 6*U)dt = 0

in terms of asynchronous variations. (iii) The varied motion is necessarily asynchronous if it is required to

take place with the same total energy as the original motion. &om (ii), derive Maupertuis' principle for isoenergetic variations.

6.10 We have a Lagrangian L(q, q, t ) , but want to restrict the coordinates and their derivatives by a holonomic constraint F(q, q, t ) = 0, more general than (6.28), since F depends also on q. Show that the equations of motion can be derived from the modified Lagrangian

L' = L + X ( t ) F ,

where X(t) is regarded as a new coordinate.

6.11 Assume that the holonomic constrai~t F in the preceding problem does not depend explicitly on t and is of the form F = (aij&qi - b ) / 2 = 0, where the symmetric coefficients aij and b are constants.



Show that one finds

8.12 A bead of mass m slides ~ t h o u t f~ct ion on the curve 3 = ffs). The y-axis is vertically up. Find the equation of motion for 2

(i) by using the holonomic constraint, (ii) by using a Lagrange multiplier A,

(iii) find the reaction of the constrit~nt and compare it with A. and

6.13 Consider the system shown in figure 6.5. The pulleys are massless asld frictionless. Assuming that the acceleration A of A4 is known, find the friction force F acting on M and the tension T of the cord

(i) by an element^^ ~ e t h o d , (ii) by a L a ~ ~ ~ a n method with 8 ho~ono~ ic constr~nt relating the

(iii) by expressing the constraint in non-holonomic form and using a po~t ion coordinate z of M with the ~osition coordinate y of m,

Lagrange multiplier.

6.14 A solid cylinder of radius R2 and mass m ( ~ o ~ e n t of inertia 1 = ~ ~ ~ / Z ) rolls without sl~pping inside it c y l i ~ d r i ~ surface of radius R1 as Shawn in figure 6.6. (i) Write the Lagrangian in terms of the angles 81 and $2 shown in figure

6,8. (ii) Derive the equation for 91 using the h o l o n ~ ~ i c con~traint

~~e~ = ~ 2 ( e ~ +e2).

126

Figure 6.6: Cylinder rolling inside cylinder

(iii) Use Lagrange multiplie~s to find the forces exerted by the cylindrical surface on the rolling cylinder.

6.16 A particle of mass m moves under the action of gravity and of the constraint

cosa dp - (sin'@/ coscy)(z da/p) = 0 , where a is a fixed angle, and p and z are cylindricat coordinates. The z-axis is vertically upward.

(i) Write the Lagrange equation in Cartesian coordinates with a La- grange multiplier.

(ii) Show that the constraint, though couched in non-holonomic form, is, in fact, holonomic, and represents the €riction~ess motion on any of a family of hyperboloids.

(iii) For the a > 0 branch of the degenerate hyperboloid that passes through the origin, show the relation of the Lagrange multiplier with the reaction of the frictionless surface on the particle.

6.16 Consider the system of figure 6.7, two wheels of equal radius rigidly attached to an axle, suggested by Goldstein. Find the constraints.

6.17 Show that the constraints for a rolling disk with horizontal axis,

(section 6.6) are non-integrable. dxl +rcosy, d$= 0 and dx2 +rsintpdplt = 0.

6.18 Derive the results of problem 5.1 by the method of Lagrange multipliers, using as initial variables the Euler angles and (zl, 52).


E Figure 6.7: Wheels and axle

6.19 The use of Euler angles is not convenient in treating the motion of a uniform sphere. Derive the results of problem 6.18 by the Lagrange multiplier technique without using Euler angles.

6.20 Derive the results of problem 5.2 by Lagrange equations with multipliers.

6.21 Give a Lagrangian formulation of the problem of a rolling disk (problem 6.10).

6.22 Find the variation of the harmonic oscillator Lagrangian L = m(q2 - w2q2)/2 under the infinitesimal transformation q + q + 6q with 6q = ic/2rn(q - iwq), and determine the corresponding integral of motion according to section 6.7.1.

6.23 Find the variation of the Toda Lagrangian

L = (2 +p2)/2-[exp(2(2-y6)) +exp(2 (2+y~) )+e~p( -42) ] /24+ 118

under the infinitesimal transformation

dz = e[-48i$ + fi(exp(2(a: - y&)) - exp(e(a: + yJ[i))] , I 6~ = ~ [ 1 6 ~ ~ - 24x2 + exp(2(s - yd)) + exp(a(2 + 96)) - 2 exp( - 4 4 , and derive an integral of motion according to section 6.7.1.

6.24 Apply the transformation a:k + Z k 4- 6Xk with 6zk = 7[-22j*k + (2.jkj)dik + 2kkj]/k (q infinitesimal parameter) to the Lagrangian L = mr2/2 + k/r. Show that 6L = qdA/dt with A = - 2 x f / r .

(ii) Show that the ensuing constant of motion is the i-th component of the L-R-L vector.


6.25 Consider a particle of mass m in one dimension with Lagrangian L = mx2/2 - U ( z - ‘ut), where U ( x - ‘ut) represents the interaction with a field travelling with constant velocity ‘u.

(i) Show that L is invariant under the transformation 2‘ = x + EU, t’ = t + c, where E is a constant parameter.

(ii) F’rom this invariance property, derive a constant of motion.

6.26 The Lagrangian

L = (m& + ~ ~ ~ ~ ~ / 2 - U((r, - ql)

is invariant under the seven transformations of the Galilean group (space and time translations: 4 parameters; space rotations: 3 parameters). Use (6.311,

to find the corresponding seven integrals of motion.

6.27 Find the integral of motion at the end of section 6.7.1 by t r a n s f o ~ i n g both coordinates and time so that the action is invariant.

6.28 In problem 6.8 the Lagrangian

L’ = m(x2 + G2 + i 2 ) / 2 + qEy - (qB/c)yx

yields the consta~ts of m o ~ ~ o n

mx-qBylc = constant , rnc+qBx/c -qEt = constant , m i = constant.

While the first and the third originate from the absence of x and z, respectively, from L’, the second has a different origin. Derive it by considering the result of the transformation y + y 4- c ( E infinitesimal constant).


Solutions to ch. 6 problems

U = -mg(xl sine - 23 cos e) , where ( z I , x ~ , x ~ ) are the coordinates of the center of mass with respect to the fixed axes. Since 21 = (R - r cos.9)sincpo, x2 = -(R - r cos.9) coscp, 23 = r sin.9, R - r m.9=t! sin.9,sinfi=e/R,wehave

L = (1/2)(me2 + 11 + e'13/r~)+~sin~.9 + mg(t sin 19 sin t sin cp - r sin 29 cos e)

The Lagrange equation for cp,

.

S6.2 (i) Elementary: mx = FA, m5 = FB - mg, zcmB = ( t / 2 ) ( ~ A sine - FB

Eliminating FA and FB, and expressing x and y in terms of 8, we obtain

~~~e = -(tm9/2) cos e - (me2//4)e , ze = (-emg/2) cos e .

Puzzling? NO, A is not a fixed point. Energy conservation gives

18' tmg sine - emg sineo 2 - 2

1 8 ~ + tmg(sin e - sine,) = o .

-+ 2 1

FA = 0 when x = 0, sin 0 8 + cos 8 8" = 0,

c o d (-3sine+2sinBo) = 0 . Either 0 = 7r/2 (trivial unstable equilibrium) or 8 = sin-' ( ( 2 / 3 ) sin 00).

(ii)

Constraints: L = [m(k2 + 6') + I c m P ] / 2 - mgy

The Lagrange equation

130

Figure 6.8: Plank, problem 6.2

another with z -+ 21, and another with x -+ 6, give

m x - - X A = O , r n ~ t r n g - A X e = O , = e(XASjIl8 A S C O S B ) / 2 -

In the last equation we substitute for AA and Aw their expressions in terms of 8,

A A = m~ = -me(sin B ii + cos B P ) / z , XB = mg + m~ = my + mt(cos B B - sin B $ 7 , ~ ,

obtaining the same equation as in the elementary treatment. A.4 and Aw coincide with FA and FB of the elementary treatment.

(ii)

m l = O .

(iii)

(iv) Yes. The Lagrangian does not depend on 4 and z L = rn@ + p2ij2 -+ i 2 ) / 2 + q[u in p + ~ p ’ # / 2 c ]

( 4 mp’4 + qBp”f2c = $3 + qBp2/2c = constant and i = constant


S6.5 With x = L sin 4, y = a cos(nt) + e cos 4 the Lagrangian

becomes L = (2 + 3i2)/2 + mge cos 4

L = me2d2/2 + maen& sin(S2t) sin 4 + mge cos 9 + aWsin’((~t) /2 . This can be replaced by

L’ = me2d2/2 + ma& cos(at) cos 4 + mge cos 4

since L - L’ = dh/dt with A = -maen sin(i2t) C O S ~ + a2(R[2Rt - sin(2nt)]/8. The Lagrange equation for L’ is 4 + t-’[g + an2 cos(nt)] sin 4 = 0.

S6.6 (i) L = mx2/2 - k(z - x - t)‘/2 (ii)

2 = 2’ + x, L = m ( P + x 2 + 2X5’)/2 - k(z’ - q”2, p’ = ar//ax’ = m(&‘ + x), p’ = aL/axt, mi? = - k ( d - e ) - mX.

L + L = m ( P + X2)/2 - mXx’ - k(z’ - !)‘/2, (iii) Add to the Lagrangian the term -d(mXz’)/dt = -m(Xx’ + kk’). Then

p’ = aE/ax’ = mx‘, p’ = aL/ax‘, giving the same equation as in (ii).

56.7 (aLC/aQ)dQ + (aLc/a4?)dQ + (aLc/+)dt + a[(aL/aq)dq + (aL/ap)dq + ( a ~ / a t ) d t ]

= P dQ + Q d P + a ( p dq+q dp) We find at once a t , / % = -craL/at, P = aLc/aQ, and, using p = aL/aq, we are left with

A t this point, using Q = p, P = crq, and p = aL/aq, we have P = aLc/aQ. (4 (4 Since p = mq, p = mq = -kq, q = -p/k , this gives

(aLc/aQ)dQ + a(aL/aq)dq = Q d P + aq dp.

L, = -(r(mq2 - kq2)/2 + aqp + apq

mQ = -kQ, mp = -kp, d(mq + kq)/dt = 0. (iii) (a)

132 CHAPTER 6. L A G ~ A ~ ~ r A ~ ~

L, == --cr(mq'/2 i- k/lql) i- uq - p + u p * q Using 6 = -Icq/lq13 and p = mq, we find

A" = A' f VA with A = Bzy.

d(mx - qBy/c)/dt = 0 , d(m$)/dt = qE - qBx/c , d(rni)/dt = 0 .

&om the last, and the initial conditions, we have z = 0. From the first, and the initial conditions, i = (qB/mc)y. Substituting in the second equation, we have

@ + w'y = qE/m with w = (qBl/rnc

y =I (qE/rrUt12)[l - C O S ( W ~ ) ] , 5 = (E~/5)[1 - COS(W~)], x = (cE/B)[t - s ~ ( w ~ ) ] / w ] . (v)

r& = ~ q / c ) B ~ , d(m$-t qBx/c)/dt = qE m 2 s O .

From the first and initial conditions, m i = qBy/c. Using the in second, we have again j i + w2g = ~ E / m etc. as in (iv). (vi) Assuming E > 0 and B > 0, a proton (charge +e> is at time t on a circle of center (Ect/B, Ernpc2/eB') and radius Emp&'/eB2, while an electron will be on a circle of center (EctJB, -Em0c2/eB') and radius Em0c2 feB'. No wonder? The proton's trajectory is the cycloid (see figure 6.9)

x = (Ernpc2/eB'>(T -sinr) , g = (Em,cz/eEa)(l -COST) ,


Figure 6.9: Electron and proton trajectories

while the electron's is the cycloid 2

5 = ( ~ m , c /eB2>(r - sin7) , e, = -(Emec2/eB2)(1 - COST) . Rom the equations and the initid conditions, we have

on(&' + &/z - qEe, = constant + o , 2

vmax = (zE/m)Icre/l,,

S6.9 (i) While S q = dSq/dt, we have

Since dYdt = 1 + d&t/dt, for 6t infinitesimal this gives

=sg-- d6t - dqt(t) Z d c j - q d t . d6t . dt dt

(ii) For K = c(q)qa one has

On the other hand, StJ = d*U. Therefore


(iii) For isoen~getic variations, 6*(K + U ) = 0, 6"U = -6*fc,

0 = lsta (2 6°K + 2K dSt/dt~dt , ~ ~ t ~ 6 * ~ ~ K dt) = 0 ?

since 6"dt = r(t + dt) - f(t) - dt = (df/dt)dt - dt = (ddt/dt)dt. Then

In chapter 1, equation (1.37), wy? wrote 6 instead of 6" in order to keep the notation simple at that early stage.

The second equation is nothing but F(q, Q? t) = 0. The first gives

For the simple case F(q,t) = 0, we have

as expected. The gener~jzatjon to the ewe of T constr~nts ~ * ~ g , Q, t) = 0 (a = 1,. . . r ) is trivial.

SB.11 One has

aijqjq2 = b t aijqjtj j = 0 . Multiply the first equation by 4; and integrate with respect to t finding


Since in the present C~IX dL/dt = (aL/a~i )q~ + ( ~ L / a q ~ ) q ~ , we have

56.12 (i) and the Lagrange equation gives

L = m(x2 +$‘)/a - mgy = mil + f’(x)’]a’ - mgf(x)

(ii) $ = f‘(x)k, f’fz) dz - dy = 0 m% - Xf’fx) = 0, m$ f X = -mg and $ - f‘(x)5 - f”(x)&’ = 0

,! = -m($ + g ) = -m[f’(z)i + f”(x)i’ + g ] Use this expression for A in mP = Aft(.) etc. (iii) mx = R,, mfi = €i$ - mg, and so X = -R,,. One has also

IR/ = + gi/J- . Special case:

f ( x ) = --, f yx ) = x / d l K T = tan6 , f”(z) = .!’/(!’ - x2)* = l/lccx~~t9, x = eb cost9 ,

56.15 (i) A: acceleration of M, a: acceleration of m, MA = T - F , ma = mg - 2T, where F is the friction force and T is the tension. Since a = A/2, we write the seccmd equation in the form mA/4 = mg/2 - T. Summing to the first equation we find

F = mg/2 - ( M + m/4)A, T = m(g - A/2) /2 . (ii) If 4’ is the length of the cord, 8’ = d - x + 2y + I , y = (z + t’ - l - d)/2, $ = xf2, U = -mgy,

L = Mx’/2 + m$’f2 + mgy = ( M + m/4)x2/2 + mg(z + constant)/2 6W = -F 62, d(aL/~k)/dt - OL/OX = Q = -F (M + m/4)$ - mg/2 = -F, i! = A, see (i).

(iii) From the constraint we have x - 2$ = 0 and x - 2g = 0, this last to be used together with the Lagrange equations

M X + X + F = O , m$-mg-2X=O . E l i ~ i n a t ~ g X we find (M + m/4)x - mg/2 + F = 0. Clearly X = -T.


S6.14 (9 (ii)

Frequency of small oscillations about lowest position:

(iii)

L = m(R1 - R2)'8:/2 + (mR;/2)@/2 - mg(R1 - Rz)(l - cosf31) L = (3m/4)(& - Rz)28: - mg(R1 - Rz)(l - cos&),

(3m/2)(R1 - Rz)'& = -mg(R1 - Rz) sin 01

w 2 = 2g/3(R1 - Rz), T = PT/W == rJ6(R1 - Rz)/g L = m(P2 + rz4;)/2 + (mR;/2)@/2 - mg(R1 - TCOS &)+constant

Constraints: T = RI - Rz, (RI - Rz)& = Rzez

mi:=mr4:+mgcosel +XI , d(mr'&)/dt = -mgrsin 191 + ~ ( R I - Rz) , { mRi92/2 = -RgXz .

-Al = +N is the normal reaction. Since the first constraint implies i: = 0, we find N = mgcos81 + ~ ( R I - Rz)b:. Aa, from the third equation, is the frictional force.

S6.15 (i) Since dp = (x dx + y dy)/p, the constraint is a, dx + a , dy + ax dz = 0 with a, = cosa x/p, a, = cosa y/p, a, = -(sin'a/cosa)z/p.

If the equations of motions are written as m i = -Xa,, my = -Xay, and mZ = -Xa, - mg, then X turns out to be positive (see (iii)). (ii) By integrating p dp - tan2a z dz = 0, one finds p' - z'tan'a =constant. (iii) For constant = 0 we have a cone x2 + y2 - tanza 2' = 0. The unit vector cormal to the cone has components n, = COSQ x/p, ny = cosa y/p, n, = -(sin's/ cos a)z/p. Then

mi: - ii = m[cos ( ~ ( 2 % + yjj)/p - (sinza/ cos a ) z ~ / p ]

= mg(sinza/cosa)z/p - x c 0 2 a - ~ ( s i n ~ a / c o s ' a ) z ' / p ~ = mgsina - A , having used p = tan a z.

S6.16 x, y: coordinates of midpoint P, xi, yi: coordinates of point i ( i = 1,2)

+i: angle of rotation of i-th wheel, positive direction clockwise ki = a& sine, 5i = -a& cose

Check: If +i increases, the point of contact moves left to right, xi increases, y+ decreases.

. kl +kz a 2

. 51 +$z a 2 2

x=--- - sine (41 + & I ,

y=-=-- c o s e ( & + & ) . Hence dz = (a/2) sin 0 (dh + d&), dg = -(a/2) cos 6 (d+l + d+z), from which

c o d dz +sine dy = 0 ,


sin 8 dx - co9 8 dy = a(dqh + d42)/2 . On the other hand, from 19 = tan-'[(y2 - y1)/(x2 - x1)J one finds

(22 - Zl)(dga - dyl) - ( ~ 2 - y1)(d~2 - dxl) dI9 = (22 - 2d2 + (yz - Y#

0 a = -- (d(6a - d+l) = -d(+i - (62) integrable, hoionomic . b b All the above can be condensed as follows:

z = x + iy, zi = xi + iyi, u = exp(iO), tl = z - bu/2,22 = z + bu/2 ittolling without slipping of each wheel: Im(u d&) = a ddi Displacement of midpoint is normal to axle: Re(% dB) = 0

Rotation of axle: d8 = -Im(u dfi) = -Im(u[df2 - d&J/b) = (a/b)d(+l - (62)

50.17 (i) Assume the existence of two integrating faxtors f (XI , 22, cp, .I) and g(xi, 2 2 , p, +) for the two ~ n ~ t i o n a . Write

f ~ i + O x d x a + O x d p + f r c o s t p d + = O , 0 x &I + g dxs -k 0 x dp+grsincp d+ = 0 .

write "curl= oii conditions: af /ax2 = 0, a f lap = 0, a f /a+ = r c08 p Of /axl, af/& rcoscp - frsincp = 0. The second and the fourth imply f = 0, except for the trivial case cp = 0, rolling along the 21-axis. Similarly 8g/8x1 = 0, ag/alp = 0, ag/a$ = r sin 9 ag f ax2, &/alp r sin cp + gr cos cp = 0. The second and the fourth imply g = 0, except for the trivial case p = n/2, rolling along the XZ-axis.

intagrabls seco sere f

c \ A & Rolling dong the xl-axis: dxl + T d$ = 0, dxz + r sin cp d.I = 0 Rolling along the x2-axis: dxl + r cos cp d+ = 0, dx2 + r d+ = 0

CI'V - 0010 sero intsgrabIe

56.18 Using the expressions for the components of w with respect to fixed Carte- sian .iwre~ (see (5.35)), we have

L = z[S2 + 62sin24 + (4 + + cost~)~1/2 + n(iit: + &:)/2 + qEx1

and the constraints

i .1-r(Qsinp-4sint9cosp)=0 (CI) , i a + r ( S coscp+4 sin4 sinp)=O ( ~ 2 ) .

Hence the equations

mi5 + X i = qE ( 1 ) , t 7 ~ 2 2 + A2 = 0 (2) ,


i ( d + + & sinG)-XIr s i n y + ~ 2 r cosy=o (3) 1 ( 4 + $ C O S I J - @ sing)+Xlr sin8 coscp-i-~zr sin@ sincp=O (4)

d(@ + 4 cosIJ)/dt = 0 (5)

$ + 4 COSIJ - 44 sin$ (5’)

, )

)

namely

Equation 15) tells us that w3 =constant. ~ultiplying (4) by cos cp, and s u b t r ~ t i n g (3) ~ujtjplied by sin 29 sin 9, we have

AIr sin ~ + i ( c o s ‘p 4+cosg COB y $-sin IJ cosy @-sin IJ sin 9 J-sin’IJ sin p +$) = o Subtracting from this the time derivative of (Cl) multiplied by I sin IJ/r, we have

.

sin8 (Xlr - I z 1 / r ) + I cosy COSIJ (8 + li; COSIJ -44 sin@) = 0 , which by (5’) gives

In a similar way we find A2 = (I/r2)x2. Thus we have (7m/5)x1 = qE, xi.2 = 0, WI = 0, and Wi.2 = 5qE/7mr.

56.19 With w1 = 15, w2 = 6, w3 = 7, we haqe the ~agrangian

A1 = (i/.”X1 1

L = m(X: -i- X i ) / 2 + I(&’ + 8’ + 7’)/2 + pExi

with the cons t r~nts

&2+r*=O , k l - r j = ~ .

The Lagrange equations with multipliers A and p read

mxI+p=qE , m&+A=O )

i t i + X r = O , i j i - p r = ~ , I ~ = O . The first and the fourth give rnfl + ( ~ / r ) ~ = qE, and this, using the second cons t r~nt ) gives (m + I/r2& = qE. The second and the third give mxz - (I/r)G = 0 , and, using the first constraint, 32 = 0. The last equation gives w3 = =constant.


S6.20 With ci = w1, B = ~ 2 , + = us, we have

L = m(P$ + i " / 2 + I(ci2 + b2 + +"/2 - h g ( 1 - cos4) ,

w p = h c o s + + j sin4 , wo = -ci s i n 4 + j C O S ~ , { w r = + .

Constraints:

b d + a j = O , & + a & s i n 4 - a b cos+=O . Lagrange equations:

m b 2 J + b X = - b m g sin4 , m S + p = O , ZiL+pa s i n + = O , ZP-pa cosq5=0, Z ; i . + X a = O .

Time derivative of constraints:

b J + a ; i . = O , i + a & s i n 4 - a s cos4+ar$ cos4+a& sin$=O . We have

Aa = -IT = (bI/a)$ , b[mJ + (z/aZ)J] = - m g sin 4 ,

pa = - I ( & sin4 - f l C O S ~ ) = z(i/a + cid, C O S ~ + Bd, sin+) ,

(m + I / a 2 ) i = -(Z/a)&, . From the third and fourth Lagrange equations we have & C O S ~ + fl sin4 = 0. Hence

m i + (I/a>(z/a + cid cos 4 + sin 4) = o ,

G~ = (-ci sin 4 + j cos 4)+ = $/a .

S6.21

L = m(j.: + j.: + j .3 /2 - m g x s + [ 1 1 ( + ~ sin2d + P ) + Z S ( ~ + + co~29)~1/2

Constraints: X Q = r sin29 (holonomic), and the nonholonomic

i .1 = -r$z x w ) = -r(01 x 62) - w

= cos cp + 1;s sin6 sin 91 x 1;2) - w

= -r& co8 cp - 61 sin6 sin cp) w

= -r(% coscp - sin29 sincp) and similarly

xz = -r(% sincp+ sin6 coscp) ,


where Ri = . (see problem 5.8). Since 511 = 6 and 513 = + + + cos19, we have the constraints

d x l + r coscpd++r coscp cos6dcp-r sin8 sincpd6=O d x z + r sincpdpC,+r sincp cos6dcp+r sin6 coscpd6=O

, )

for the Lagrangian

L = m(& + 2; + r2d2 cos28)/2 - mgr sin6

+[zI(+~ sin'e + 13(4 + + cost9)21/2 . The Lagrange equations

give m21 +XI = O ( I ) , m2z + A2 = 0 (2),

(11 + mr2 cos"19)d - 2mr"d2 sin6 cos19 + mgr C O S ~

- 1 1 + ~ sin6 COSQ + Z ~ W & sin6 + r sin6(-AI sincp + XZ coscp) = o (3), 11 sin6(+ sin6 + 2@ C O S ~ ) + 13wL C O S ~

-13w$d sin8 + r cosg(X1 cosy + XZ sincp) = o (4))

+ X1r cosp + Xzr sincp = O (5). Differentiating

i l = -r coscp($++ cosd)+r$ sin8 sincp

$2 = -r sincp(d++cos6) - r 9 sing coscp

we have

A1 =-mx1 =mr(+ coscp++i cosy, cos19-19 sin6 sincp

-+$ sin cp - 6' sin cp cos 0 - d+ cos cp sin 6 - 9' cos 6 sin cp - +d sin 6 cos cp) )

A 2 = -mi2 = mr($ sin cp + (2; sin cp cos 6 + 6 sin .9 cos cp

++$ cos cp + @ cos cp cos 8 - 9+ sin cp sin 6 + d2 cos 0 cos cp - ~9 sin 6 sin cp) )


r sind(-Xl sincp+Xz coscp) =mr" sind(J sin8++1)+$ cos6+9' C O S ~ ) Therefore (3) becomes

(11 + mr2)8 + ( ~ 3 + mr2)w:+ sin 8 - 11 +2 sin 8 cos 8 + mgr cos 8 = o .

Similarly we have

r(X1 COB cp + XZ sin cp) = mr2(g + cos 8 - 29+ sin81 I

and so ( 5 ) becomes

Finally, using ( 5 ) equation (4) becomes (13 + mr2))Lj$ = nar"@d sin 8 .

13w$i - 2119 cost9 - 118 sin8 = 0 .

S6.22 6L = 6 dA/dt + 0 ( c 2 ) with

Hence

.

is an integral of motion. For q(t) = A cos(wt +a), choosing C = -q (O) - iwq(O), one has Z = a.

56.23

Hence

6L = 48 C- d (- c3 - X'L) + o(i2) dt 3

ax = c[8Si(7j2 - 3k2) + (+ + 6 k) exp(2(z - yd)) +(c - 6 X) exp(2(x + y 6 ) ) - 2cexp(-4x)]

is an integral of motion. This integral of motion was found by H4non.

conservation), as one might expect since in the same limit

is invariant under rotations.

In the limit of small displacements I N 12c(z$ - yk)(angular momentum

L N (2 + $"/2 - (2 + y2)/2


S6.24 (i)

Using the equations of motion, we have

With 6kk = dhxk/dt, we have

(ii) Integral of motion:

dbxk/dt = (Q/k)[-xiXk + i 2 & k - 2xixk + r * f 6ik + xkxi]

d6xkldt = (q/k)[-j.ij.k + (i2 - k/mr)6ik + kxixk/mr3].

6L = mxkdxk + k dr-' = (2q/r3)[-r2xi + r . i xi] = q d(-2xi/r)/dt.

aL/axk dXk - A = -2q[-xi/r + rnx;(il'/k - m r . ixi/k] = -2qAi

S6.25 (i) m(dx'/dt')'/2 - U(x' - ut') = m(dx/dt)'/2 - U ( x - ut)

(ii) Since dt' = dt, the action is also invariant. Using equation (6.31) with f = 1 and g = 1, we have

I = mS(u - 5 ) + mx2/2 - U ( X - ut)

= m(u5 - 2 ) / 2 - U ( X - ut) = constant The reader who wishes to work from first principles may proceed as follows:

.

o = 6~ = aL/ax dX + aL/ax 6i + aL/at dt = aqaX eu + o - a u ( x - ut)/at c

But au(x - ut)/ax = a(-L + mx2/2)/8x = -aL/ax = -d(aL/ax)/dt, and SO

= aL/ax eu - du(x - ut)/dt + €S aiqX - ut)/ax

0 = 6L = ed[umx - U(x - ut) - mx2/2]/dt .

= mo*ai(fai - gkai) + mbibi(fbi - gibi) +g[(m& + m&)/2 - U(lra - rb()]

= maXaifai + mbxbifbi - g E , where E is the energy E = K + U. Space translations: fa; = fb; = 1, g = 0

maxai + mbxb; = constant

(i-th component of total momentum conserved) Time translations: fa; = f b i = 0, g = -1

E = (mar: + rnbr;)/2 + U((ra - r b l ) = constant

(energy conservation) Space rotations: fa; = Eijkxak, f b i = Cijkxbk, g = 0

ejk;(maxakxai + mbxbkxbi) = constant

(j-th component of total angular momentum conserved)


58.27

Consider the transformation r: = r, + eBt, ri = rb -+ eiit, t’ = t + {e/E)(m,r, + m#b) t A, where E = KO + & + U is the totd energy, treated as a constant, we

L = (moiz + m&)/2 - U(1.a - rbi)

have dr, + eBdt - --

dt’ dt 11 + (~/E)(ma+o + m4i.b) * fi]

11 . 2

K: = %($$y 2 N y [r: +2E (&.re - I’O[m,r, E +m*rb] t n ,

(and similarly for b)

Kk 4- ‘v & -J- K b 4- €[I - 2(K, + Kb)/E](mo+, -!- mbib) * fi , 6& = (t?/E)(E - 2K0 - 2Kb)(m,i’e 4- mai.6) * .

at once that the action is invariant,

6 L d t + L 6 d t = O . Hence the constant of motion

58.28

with h = gEt - (qB/c)x. Then, according to equation (6.30), dL’ = qEc - (@/c)& =z c dh/dt

aLf - -h=mjr+qBX/e -9Et ali

is a constant of motion.

Chapter 7

HAMILTONIANS

The evolution of a system with n degrees of freedom is described as the motion of a point in a 2n-dimensional space, governed by Hamilton’s equations. These are of the first order in the time. Canonical transformations are discussed at length. The use of Cartan’s differential forms in providing a concise description of mechanics is extolled with missionary zeal.

7.1 First look. 1 Using the notation of chapter 3, the Lagrangian

can be written more concisely in the form

L = 4TAq/2 - U(q) , (7.2)

where the coefficients aij may be functions of the coordinates.

the Lagrange equations According to the program outlined in section 1.1, we wish to replace

with first-order equations. We note first that by inverting the equations

8 L - aji(q14i pj=G- (7.4)

145

146 CHAPTER 7. HAMILTONIANS

we have

where a;’ are the elements of the matrix A-’ l ,

q. - -1 I - a , . P j 1 v

Then we rewrite the Lagrange equations in the form

(7.5)

Defining the Hamiltonian

H ( p , q ) = pTA-’p /2 + U (7.7) we have

Going back to equation (7.5), we see that it can be written in the form

Equations (7.8) and (7.9) are Hamilton’s equations. Example: L = mr”2 + k/r, A = ml, A-’ = m-ll,

(7.10) H = p , p l z k -; with p = m r .

Hamilton’s equations yield

1 m

pj = -k- r 3 , x. * - - p i - . (7.11) Xi

In spherical coordinates L = m(i’ + rZ@ + rzsinzO p)/2 + k / r ,

m O 0 l/m 0 0

O O mr2sin213 0 o l/mr2sina13 0

1 1 1 k H = - (P’ + -pi + - -

2m T’Sh 13 I”

( ; : ) = A ( ! ) P6 ,

(7.12)

‘Of course, in general a;’ # (aij)-’, although the equality may be true in mme cues for some matrix elements.

7.1. FIRST LOOK. 1 147

pr = m i , pe = d f j po = mr'sin2e 4 . (7.13) Hamiiton's equations

ljr =-aH/ar , PO =-a€i/ae , p$ = -aH/a+ (7.14)

give

(7.16)

Since d, does not appear in the Hamiltonian - it is "ignorable" - the momentum

Brmple: conjugate to d,, p#l is a constant of motion.

The Lagrangian for the spinning top c m be written in the form

(7.16)

where I1 0

0 13 coszl, A = ( 0 zlsin26 + z s c o s ~

The Hamiltonian will be

with 0 0

I/ I1 sin' rlt -cosdfZ1sin2d o - cos6/z1sinZrlt cos'6/Z1sin26 + I / Z ~

Therefore 1 ' 1 1

211 ZZxsin'd 21s H = -p* + -@, - p+ cos S ) l + -p$ + i?mg cosd

Since '9 and t,b are ignorable, both p , and p+ are constants of motion. The H ~ i ~ t o n equation pe = -aH/M for pe = 118 gives

(7.18)

(7.19)

'F'rorn the identity E2 = S' + (r x +)'/r2 one can write the kinetic energy in the form K = m+3/2 + 12/2mr2 = (p: + la/rz) /2m. Comparing with the expression for K aa given by equation (7.12), one finds

1' = pi +p;/sin% .


or, with the notation of section 5.4 (pv = I lb , p.,, = I la ) ,

(7.20) .. ( b - acos6)(a - bcos6) tmgsin6 $ 3 = - +

sin38 I1

7.2 First look. 2

Canonical transformations Suppose we are given a system of Hamilton’s equations

Pi = -aH/aqi , di = a H / a p i . (7.21)

We express the “canonical” variables p and q in terms of new variables P and Q , p = p( P, Q ) and q = q(P, Q), and define a new Hamiltonian K(P,Q) = H(p(P,Q),q(P,&)). It is unlikely that the reader will confuse this K with a kinetic energy.

Will Hamilton’s equations

Pi = -aK/aQ, , Qi = aK/api (7.22)

hold? In general, they will not, unless the transformation ( p , q ) + (P,Q) is suitably restricted. Then i t is a “canonical transformation”.

In one dimension, let us express P and Q in the above equations in terms of p and q , and viceversa. Using the scalar transformation property of the Hamiltonian and Hamilton’s equations for H , we obtain the system

Regard this as a system of equations for p and q. Requiring that the determinant of the coefficients be zero, we find

where

and


As we shall see later, these are “Poisson brackets”. In the present one- dimensional case, they are equal to Jacobian determinants,

Using this in equation (7.23)) we have

with the double root

With the Harniltonian transforming as a scalar, equations (7.22) follow from equations (7.21) if the Poisson bracket [Q, PI,,, is unity like that for the identity transformation [q,p],,, = 1. The transformation Q = p, P = -q is canonical, and so is also Q = -p, P = q.

Generalizing to a system with n degrees of freedom, define the Poisson bracket

[QiAjl,,, = 0 ! bi,Pjl,,, = 0 >

[Qi,Pj],,, = dij

A condition for the transformation (p, q) + (P, Q) to be canonical is

[Qi, Qjlp,q = 0 9 [Pi, Pjlp,, = 0 9 [Qi, Pj],,, = 6ij

Let us consider a region “S” of the (p,q) plane bounded by a closed curve “c” . If the transformation (p, q) + (P, Q) is canonical, we have

i (pdq - PdQ) = 0 . This last tells us that the integrand must be a perfect differential,

p d q - P d Q = d G i ,

150 CHAPTER 7. HAMZLTONZANS

where the “generating function” is a function of q and Q. Thus

(7.24)

In section 7.8 we shall e~counter three more types of generating functions.

We check all the above on the harmonic oscillator, H = ($/m + mw’q2)/2, Hamilton’s equations $ = - a H / a q = -mw2q, cj = a H / a p = p/m.

P = (p2/m + mw2qq”)/2w = H / w . This transformation is canonical since [Q, PI,,,, = 1 as is easy to verify, Since the transformed Hamiltonian is K = Pw, Hamilton’s equations read

Example:

Let p = J G Z COSQ, q = J-sin Q , Q = tan-‘(mwq/p),

P=-aK/aQ=O and Q = a K / a P = w .

These tell us that P = constant, namelx H = E = constant, and Q = wt + a, q = A sin(wt +a), where A = d 2 P / m w = ~~.

The generating function is

GI = (mw/2)q2 cotanQ . In fact, aGl/aq = mwq cotan& = p and aGl/BQ = - P .

from the Lagrangian. If Qi = Qi(q, t ) , the Lagrangians L(q, q, t ) and In most textbooks, canonical transformations are introduced starting

M ( Q , O , t ) = L ( q , q , t ) +dA(q,Q,t)ldt

are e q u i ~ ~ e n t . Di~erent ia t in~ we have

Equating coefficients of SQ, and Sgi , we find

Pi=aAi /aQi and pi = -aA/6qi .

Clearly A = -GI, but owing to the time dependence of the coordinate transformation, the H~mi~tonian does not transform as a scalar? but instead

K(P, Q ) = P$Q, - A4 = H ( p , 9) - aA~at

as is easy to verify.


In section 6.4, equation (6.24), we saw that if q = q(Q) then p = (aQ/dq)P and p dq - P dQ = 0. The transformation (p, q) + (P, &) is canonical since [Q, PI,,, = (dQ/dq)(dq/dQ) = 1. In this caae, however, GI would seem to be zero, and so equations 7.24 would give p = 0 and P = 0. However, write GI = G2 - P Q with G2 = PQ(q), dG1 = dGz - PdQ - QdP. Then p dq - P dQ = dG1 = 0 gives

p d q + Q d P = d G z . This tells us that p = 8G2/8q = P aQ/dq and Q = 8G2/8P = Q(q) as we want.

For a function f ( p ( t ) , q(t)) one has Constants of motion

a formula valid also in more than one dimension. Hence f (p(t), q(t)) is a constant of motion if its Poisson bracket with the Hamiltonian is zero.

Before proceeding, the reader should prove that

[A, B + C] = [A, B] + [A, C] , [ A B , C] = [A ,C]B + A[B,C]

and

For brevity’s sake, we omit the subscripts ‘)I, q” in the Poisson brackets. Using these formulae one finds

[si, l j ] = e i j k x k 1 bi, l j ] = EijkPk , [ l i , l j ] = Eijklk , where li = E i j k x j p k is the i-th component of the angular momentum. We prove the last one:

[ti, lj] = CfabCjcd [Xapb, Xcpd] = CiabCjcd(6adXcpb-dbcXaPb) = -EibaCajcXcPb+CiabCbjdXaPd

= -(&jdbc - dic8bj)Xcpb + (dijdod - 6iddaj)Xapd = X i p j - X j p i = Eijklk . The important “Jacobi identity” for three functions ( A , B , C)

[ [A , B1, C] + [[B, CI, A] + [[C, A ] , B] = 0

can be proved by work. In section 7.14 we shall present a neat proof learned from Arnold.


The following is an important application. Let A and B be constants of motion, so that

[ A , H ] = O , [B,H] = O . Then the Jacobi identity tells us that [ [ A , B ] , H] = 0, and so [ A , B] is also a constant of motion.

Example: If the components l i and l j of the angular momentum of a particle are constants of motion (Hamiltonian invariant under rotations around zi and xj axes), so is dso their Poisson bracket [ l d , l j ] = C i j k l k .

7.3 H as Legendre transform of L We pulled out of a hat the expression (7.7) for the Hamiltonian corresponding to a Lagrangian which depends bilinearly on the qi’s. Some readers may ask for a prescription by which the Hamiltonian can be constructed from the Lagrangian.

The prescription is a simple Legendre transformation from the qi’s to the pi’s.

Let A ( x , z , t ) and B(y, z , t ) two functions satisfying the relation

A ( x , z, t ) + B(y, z, t ) = xy .

Partial differentiation with respect to x and y gives

(7.25)

= x , dB ay - a A

- = y , dX

and with respect to z and t

aB bA 8E 8 A az az 7 at at a

-= - - -=- -

Replace (2 , y, z , t ) by (q , p , q, t ) , A by L and B by H . We have at once

a H aL - -- a L . 8 H 8 H 8L 89 at ’

p = - , q = - -- - --

where the arrow denotes equality by Lagrange’s equation. If there are several coordinates, then

H = p i q i - L , (7.26)

30mitting t , replacing (z,v,z) by (u=volume of environment, p=prwure,s=entropy), A by (u=internal energy) and B by (-h=minus enthalpy), we have u - h = pu, 6ulav = p, ahlap = -v , ah/& = au/as(= T ) , as is known from Thermodynamics.

7.3, H AS LEGENDRE TRANSFORM OF L 153

(7.27)

The reader should check that these formulae yield those in section 7.1.

Electron in time-dependent electromagnetic field. &ample:

L = mra/2 - (e/c)r - A(r,t) + eV(r,t), pi = aL/axj = mxj - (e/;)Aj, H =pixi - L=pi(pj + (e/c)Ai)/m - (m/2)[(p+ (e/c)A)/m]

+(e/mc)A. ( p + eA/c) - eV ,

H = -[p + (e/c)AI2 - eV , (7.28) 1 2m

= -eEi - (e/c)ejjkxjBk , m i = -eE - (e/c)r x B .

Fkom the last equation we obtain

where the right side is the rate of work done by the electric field. We now have

(7.29)

(7.30)

where we have used Hamilton’s equations. Thus

av - dH = -+. e - a A -e- a V - -e+.(-VV-E)-e- = - e f . E - e ( g + z i i ) , d t c a t at a t axi

, (7.31) dH d(K + V ) dt dt -- -

where U = -eV is the potential energy. The Legendre transform of the relativistic Lagrangian

L = -moc2d1 - ( i / ~ ) ~ - (e/c)Ajxj + eV

is the Hamiltonian

H = c d m i c 2 + ( p + eA/c)’ - eV .


Example: According to equation (6.23) (but omitting the primes), the Lagrangian for the motion of a particle in a rotating frame is

L = ImiZ + mw"(5: + x$)]/2 + mw(z1X2 - XZX,) 1

Since pl = m(X1 - W X ~ ) , pz = m(x2 -t wxl ) , p3 = m i 3 , a s t r~gh t fo rw~d calculation gives

or, expressing H in terms of the coordinates and their time derivatives,

H =: g2/2m + W ( ~ I X Z - p z ~ i )

H = m(i' - w"xc: f &]/2 .

It is easy to verify that Hamilton's equations give m& = 2mwj.z + P~LJ'ZI

etc.

7.4 Liouville's theorem The conser~tion of phase space volumes under H ~ i l t o n i a n fiow will be expressed in the next section in the language of forms as the invariance of one of Poincard's integrals. However, we now wish to present the standard proof, which is a generalization of that for n = 1 given in the chapter 1.

In an N-dimensional space, let x(t) = (21 (t), . . . a ~ ( t ) ) be the position vector of a point P at time t, x(t + dt) the position vector at the time t i- dt of P which has moved under a flow

x( t + dt) = x ( t ) + A(x(t))dt + . . . Let D(t) be a region in the ~ - d i m e n s i o ~ ~ space at time t , and D(t+dt)

the region,occup~ed at the time t + dt by the points of D(t) if they have moved under the flow.

The volumes v( t ) and v(t + dt) of D(t) and D(t + dt ) are related by the formula

v(t + dt) = u ( t ) + divA dal (t) . . . dzN(t)

Proof: v( t + dt) = / dzl(t + d t ) . . . & N ( t + d t )

Dfti-dt)

7.4. LZO UVILLE 'S THEOREM 155

But now to the first order in dt the Jacobian determinant is

(aAz/azi)dt 1 + (aAz/azz)dt . . . . ..

axr(t +at) 1 + (aAi/azi)dt (aAl/ax2)dt

azj (t) =I ... ...

= 1 + xz aAi) d t+ ... ( If divA = 0, then v ( t + dt) = v(t) + O(dt2).

In phase space (N = 2n), (XI,. . . XN) correspond to (ql, . . . qn,pl . . . pn), qi(t + dt) = qi(t) + (BH/&)dt, pi(t + dt) = p i ( t ) - (BH/&)dt, A has components (BH/Opl,. . . BH/ap,, -BH/Bq1,. . . - BH/8q,). Thus

d i v A = c [ - ( - ) + ~ ( - g ) ] = O a a H a I

i 8% Bpi

Remark: The Lagrangian for a damped oscillator:

L = e%n(i2 - w2x2)/2 , p = BL/6j: = rnxexp(.yt), yields the Lagrange equation 2 + 7x + w2x = 0. This can be replaced by the system of equations x = A,, $ = A, with A, = y and A, = - 7 ~ - wax.

The area change in the (2, y) plane is determined by

BA, BA - + 2 = - 7 . ax ay

On the other hand, working with the canonical variables p and x, since i = A:, p = A6 with A: = (p/rn)exp(-rt), A; = -mw2zexp(rt), the area change in the (p,z) plane is determined by

i3AL aA' - + - - 9 = o + o = o . ax ap

There is no area shrinking in the (p,z) plane.


7.5 Cartan’s vectors and forms The purpose of this section is not to give a smattering of modern differential geometry4, nor to pander to the mathematicians, but rather to introduce the reader to a notation they have invented, which is as useful in Hamilto- nian mechanics as vectors are in General Physics.

We define a vector U as a set of n numbers, its components U’, and write

U = UiGei ,

the Bi’s being basis vectors. A 1-form 3(0) is a linear function on vectors,

3 ( D ) = number .

A 1-form can be expressed as a linear combination of basis 1-forms,

&(.) = WiE”.) .

Assuming the orthonormality condition

B”(ej) = 6; ,

and using the linearity of (zl( o), we have

For example, a force F is a 1-form, a displacement D is a vector,

p(D) = Work (number) . If # = F$ and D = 6xi ~?i , then

Work= Fi 6x’ .

So far so good. The reader may still doubt the usefulness of this notation (why a bar rather than an arrow for vectors?). A mild shock may be experienced when we say that a vector is defined as an operator acting on functions on a manifold, roughly speaking a space, whose points are labeled by coordinates xl, . . . 2”.

4An excellent introduction for physicists is B.F. Schutz, Geometrical methods of mathematical physics (Cambridge University Press, 1980). See also W.L. Burke, Applied Diflerential Geometry (Cambridge University Press,1985), and M. Crampin and F.A.E. Pirani, Applicable Diflerential Geometry (Cambridge University Prw,1986).

7.5. CARTAN'S VECTORS AND FORMS 157

If j (z ' , . . , xn) is one such function, then

The following application of this definition of vector is specially useful. If (xi (t), . . z*(t)f is a curve in our space, and U' = dx'/dt, then

This is the rate of change of f (a? . . . xN) along the curve, taken at the point of coordinates xi(t).

The basis 1-form 2 ( 0 ) will now be written as

2(.> = d&.) ,

so that the c o n d ~ t ~ o ~ E i ( ~ j ) = 6; will now read (greater shock?)

A useful 1-form is the gradient

where h is a function on the manifold. The standard example is h=height above sea level, (xl, s2) coordinates on a map etc.

If Sxj axe the coordinate changes in an i n f i n i t ~ i m ~ displ~ement, define the vector

Then

a S X j

8x = 6x9- .

giving the change of h corresponding to the displacement.

ifold be the phase space with coordinates pi , gg. What does this have to do with Hamiltonian mechanics? Let our man-

Define the ' ' H ~ j l t o ~ ~ a n vector field"

(7.32)

158 CHAPTER 7. HAMZLTONIANS

where a H

and Uqi = - . O H UPi = -- %i aPi

Let f (p, q) be a function on phase space. Then

If A@, q ) and B(p , q ) are functions in phase space, the expression

is, as we know, a Poisson bracket. Clearly df/dt = Uf = [f,H]. We already know that a function f (p, q) is a constant of motion if its Poisson bracket with H is zero, [ j , H ] = 0.

We now proceed to define a 2-form as a function on pairs of vectors. For example, the product (Ai @dxj)(o, 0 ) is a 2-form which, acting on two vectors U and t yields

(dxi @ d d ) ( U , V ) = (JX~(U))(dX~(V)) = u”i . We shall make use of a special kind of 2-forms, the “exterior” 2-forms

G2, which are antisymmetric,

LP(V,U) = -&2(0, P) . A typical example is

3 2 = dxi A &.i &’ 8 Jxj - Jd’ g, &’ 7

so that

If 0 and V are vectors in the x1x2 plane (figure 7.1), then

~ 2 ( 0 , V ) = uivj - ujvi ,

(Jz’ A dx2)(U, V ) = U ’ V ~ - U ~ V ’ = area of parallelogram .

The following in small print will not be needed until later. Generalizing we define an “exterior” m-form as an antisymmetric function on m vectors,

L T ( U 1 . ..uj . . . ui ... Urn) = -ayu1. . . U i . . . U j . . .Urn) .

7.5. CARTAN'S VECTORS AND FORMS 159

Figure 7.1: Area of parallelogram

/ac" Figure 7.2: Volume of parallelepiped

Basis m-forms are

axil . . . A (ixim = X ( - ) ' J x * ' @ Jxia . . . @ (ixi"' , P

where P denotes permutation of the indices. Example: In 3-dimensional space

This is the volume of the parallelepiped shown in figure 7.2. In an n-dimensional space, cjm = 0 (m > n). In fact, any basis form will have

two or more indices with the same numerical value, and will vanish because of the antisymmetry.

The wedge product of two exterior forms, Gr A Gs, is an exterior form with r + 8 slots. For our purpose i t is sufficient to say that Gr A (&" +- bGt) = aGr A G' + &.J~ A cjt, and that the product of two basis forms is

(dd' A , . . Adz'.) A (h"' A .. . A dzk') = dz" A .. . A dx'* A dxk' A .. . A dxk* .

160 CHAPTER 7. HA MILTONIANS

A few words about the operator 8. This is defined to act on an m-form

Gm = fi1...i,,,&” A . . . A dxim (7.33) il ... 1,

by generating the (m + 1)-form

Note that

In fact 28=0 .

(7.34)

(7.35)

a2 a&“ = (mfj, ... i,,, hi A 8x3 A 8xi1 . . . A dxim = 0 ijil ... i ,

because the double derivative indices (symmetric) are contracted with those of b’ A 2xj A . . . (antisymmetric).

We now define the 2-form

(7.36)

operating on pairs of vectors in phase space. If V and W are two vectors,

i

is the total area of the projections of the “parallelogram” formed by V and w on the planes ( P I , q l ) , . . . (pn, qn) . Let us fill the first of the two slots of G2(e, 0 ) with the Hamiltonian vector

field U , equation (7.32). Thus we obtain the very important equation

G2((U,e) = - d H ( e ) . (7.37)

Proof for n = 1: Remembering that Jp(a /aq) = dq(a/ap) = 0, we have

a H - a H - aq DP

- - - -dq(o) - -&(a) = - J H ( e )

The following section 7.6 is included for completeness. It is not necessary for understanding section 7.7.

7.6. LIE DERIVATIVES 161

7.6 Lie derivatives The following is a pedestrian definition of Lie derivative. The Lie derivative L A of a geometrical object such as a vector, tensor, or form, is the change undergone by that object when “transported” through As dong the vector field A, divided by the infinitesimal parameter 3.

If f(s) is a scalar function,

LA = [f(~’ + A’(x ) 3 , . . .) - f(~’, . . .)]/s

(7.38)

~f B is a vector,

LAB = [B(x‘ ) - B ( z ) ] / s ,

a where B(x‘ ) = Bi(x’)- ax‘( 1

If 0 is a l-form,

(7.39)

and so (7.40)

It may be interestin4 to verify equation (7.37). Show first that the Lie

L o f P = O . (7.41)

Then, since &j2 = dJdpA&) = 0 and = 0, using the general formula LAG = d[G(A)] + (&)(A) (B.F. Schutz, Zoc. cit. eq. (4.67), p. 142), we

derivative Lo of 13’ = dp A I& is zero,

have

Since cid = 0, this agrees with equation (7.37).

162 CHAPTER 7. HAMlLTONIANS

7.7 Time-independent canonical transformations

The “canonical variables” (pj,qi) are not unique. A time-independent transformation to new variables (Pi@, q ) , Q i ( p , q) ) is a “canonical transformation”, and the new variables are also “canonical”, if

{ 7.42)

Filling the slots of equation (7.42) with two phase space displacements 6(’) and iiC2), the condition for a tr~sformation to be canonical reads

c ( 6 ( ’ ) p i 6 ( 2 ) q , - 6(1)qi6(2)pi) = X(6(1)$‘$6C2)Qi __ 6(’)QiS(”)Pi) ,

i i

It is given in this form in several textbooks.

transformations, Hamilton’s equations are invariant under time-independent canonical

and using equation (7.37), we have

Hence dK , W,& =- . 8K upi = -- aQi dPi

7.8. GENERATING FUNCTIONS 163

Example: The t r ~ ~ o ~ a t i o n 4 = -qi, Qi = pi is obviously canonical. Note

Example: Coordinates in a plane, (p1,21)(p2,xa) + ( p r , T ) o , @ ) .

the minus sign!

pr = (pixi +pzxa)/v/.: + xX pe = xipz - x2p1, pl = pr case - (pe/r> sin@, pa = p,sine + be/.) cos0.

Using these formulae, and p~rforming a few ~fferentjations, it is easy to verify that

d p ~ Adxi + dpa A 2x2 = 2pr A dv +dpe A 20

7.8 Generating functions In the above two examples we expressed the new canonical variables in terms of the old, and proceeded to verify that c ,JPi A dQ, was indeed equal to Cillrpt A dqi.

Suppose now that we have formulae expressing p and P in terms of q and Q, p = p(q, Q), P = P(q, Q). (For simplicity’s sake we confine ourselves to one ~ ~ ~ s i o n . ) Then, substitutin~

If the functions p = p(q, Q) and P = P(q, Q) can be expressed in the form

(7.43)

where the “generating function” GI is a function of q and Q, the relation Bp/BQ = -BP/Bq is satisfied, and the transformation is canonical.

For the canonical transformation Q = p , P = -9, the generating function is GI = qQ.

Exsmple: The generating function

GI = (qz cos a - 2gQ + Qa cos a)/2sin a


namely the rotation

Q = qcosa -psina , P = qsina+pcosa

in the (p,q)-plane. For a = -7r/2, GI = qQ, Q = p , P = -9. There are altogether four types of generating functions as listed below:

Note the following relations, which can be easily verified:

Jq - P JQ = d~~ = ~ Z ( G ~ - PQ)

= J(G3 + p q ) = J(G4 - PQ + p q ) . This string of Legendre transformations is analogous to the well known

Thermodynamics relations

T ds - p dv = du(s, V ) = d(h(s,p) - pv)

= d(f(T,v) + T3) = d(!J(T,P) + T3 - p v ) I

where v is the volume of the system.

What is special about GI? Just that The generating functions G I and Gz are rather special.

(7.44)

A transformation is canonical if the difference between the action elements &i&i(*) and CiPiJQi(.) is a gradient cZG1(.).

The special role played by Gz(q, P ) is due to the fact that the identity transformation Pi = p i , Q j = qi (i = 1 , . . . n) is induced by G2 = qjPjl where we start using the sum convention of equal indices. Therefore a G2 generating function is convenient to describe infinitesimal transformations.

For the moment, consider

G2 = (qi + 6qi)(Pi - 6pi)

with 6qj and 6pi not necessarily infinitesimal. This yields

7.8. GENERATING FUNCTIONS 165

For a particle, G2 = XjPj + dG2 with dG2 = Pj daj and dG2 = -xi dbi generate the translation Xi = xi + baj and the boost Pi = pi + dbi, respectively.

For both coordinates and momentum components, an orthogonal transformation with coefficients ajj satisfying a k r a k j = di j is induced by

G2 = a k j q j p k *

In fact, Qj = aGz/aPi = aijqj and p , = aGz/dgi = CbkiPk, or, inverting by use of the relations for the adj’s, Pj = ajipi. The inversion Qi = - q j , Pi = -pi is given by the above with

(inv) (id) aij = -6i j , 90 that G2 = -G, .

generates the infinitesimal rotation It is easy to show that G2 = X j P j + 6G2 with dG2 = d a e i jkx ip jnk

xj = xj - 6a e j j k x j n k , Pi = pi - d a E j j k P j n k . In general, infinitesimal transformations about the identity are induced

by a generating function of the form

G2 = qjPj + dG(q, P ) -

and so

with apt = -abG/Bqi and 6qi = adG/aPj. Pi = pi + dpi 1 Qi = qi +ki

The Hamiltonian flow (p(t), q ( t ) ) + (p(t + 6 t ) , q(t + 6 t ) ) can be regarded a8 an infinitesimal canonical transformation with generating function

G2 = q(t)p( t + 6t) + H ( p ( t + fit), q(t))at , where H(p(t+Gt), q ( t ) ) is obtained from H ( p ( t ) , q(t ) ) by replacing p ( t ) by p( t+6 t ) .

and similarly for q(t + 6t ) . Consider now a function f(p, q). Its change if p and q are replaced by

P = p + dp and Q = q + dq with dp = -l?dG/dq and 6Q = bdG/aP is

166 CHAPTER 7. NAMILTONIANS

Here If, SG(q,p)] is the Poisson bracket of f(p, q) and SG(q,p) = SG(q, P)Ip-tp. In particular, Sp = fp, SG] and Sq = [q, SG].

It is interesting that 6Gz = SCK 1, with 1 = (11 = d-, generates infinitesimal rotations about 1. In fact,

bi,l] = 4 ‘ i j k p j , [lj,l] = 0 . Hence Xi = X i + 6CK [Xir l ] = X i - 6Cr t i j k x j l k / l , pi = Pi -k 6Q‘ [pi, 11 =pi - 6a E i j k p j l k 11, Li = 1,.

If the Hamiltonian is invariant under an infinitesimal transformation, then the corresponding SG(q,p) is a constant of motion. In fact,

ddG/dt = [SG, H ] = -6H = 0 .

Therefore, if the Hamiltonian of a particle is invariant under a space translation in the direction of the unit vector ii, the component of the momentum in that direction is a constant of motion; if it is invariant under rotations about ii, then the component of the angular momentum fi 1 is a constant of motion.

7.9 Lagrange and Poisson brackets

In the condition C&pi A dqi = C,& AJQi for time-independent canonical transformations, let us express pi and qj in terms of the new variables (P, Q). Defining a Lagrange bracket as

{ A , B } p , q = - {B ,A}p ,q = 7 (3% - -- (7.45)

7.9. LAGRANGE AND PorssoN BRACKETS 167

The identity t~nsforma~~on &i = qj, Pi = p5 (i = I,. . .n) satisfi~s these relations.

Wa shall show that the above conditions for Lagrange brackets are equivalent to the conditions

for Poisson brackets, also satisfied by the identity transformation. Proof: Put zz+-l = qi, z2i = pi (i = 1,. . . nf

zzn = pn) and define the symplectic metric

all other gij being zero),

= gl, ~2 = plr . . . zZnll = qR, 92i - f ,2 i = +1, gzi,Z;-1 = -1

(912 = 1, Q21 = -1, gs4 = 1, g 4 ~ = -1,. . .,

The Lagrange bracket {A, B),,,q can be written as

and the Poisson bracket fAlZ3],,,q as

1

(7.48)

(7.49)

(7.50)

(7.51)

In one dimension

and so

The proof is straightforward, we only have to use GTG = I (E,$ab$ac = &be) and

Using (7.51), we find

{QIP)P,*[&t PI,,, = 1 *

C, ~ ~ ~ , / ~ X ~ ) ( a X j / ~ ~ a 1 = 6jj

~ ( { Q I , Qif(Qt, tail f { f i . & i I [ 4 , Qjl) = 6ij >

zero - xSit[fi,Qj] = &j , [Qj,Pi] = &j . I

I SimilarIy we find the other Poisson brackets.

The reader can easily show that the Lagrange bracket of two functions (A, B) on phase space is invariant under canonical transformations, namely

{ A , B)p,q = {A , B ) P , Q , and so is also the Poisson bracket

[A) BI,,, = IA*BJP,Q - Therefore from now on we omit the subscripts e),q), and write simply

[A, Bj for the Poisson bracket of A and B.

7.10 Hamilton-Jacobi equation revisited Let H ( p , q) be the (not explicitly time- dependent^ Hamiltonian of a system. Perform a canonical transformation from (pi, qi) to (Pi, Qi) with generating function Gz(q, P) + S(q, P) such that the new momenta are c o n s t ~ t s of motion, Pi = o (i = 1, . . . n) .

Since 4 = - 0 ~ / ~ Q ~ , where K(P, &) is the transformed H a m i ~ t ~ n i ~ , PJ = 0 requires that K be a function of the Pi's, but not of the &its, K = K ( P ) . We have also K = N(p,p) = E.

Let PI = E, and Pi (i > 1) (n - 1) more constants of motion. Then

Gz(q, P ) S(q, Pi = E , P z . . . P,) . { 7.52)

Since pi = aG2/0qi = aS/aqi, substituting in H ( p , q ) = E we obtain the Hami~to~-Jacobi equation

(7.53)

7.10. HAMILTON-JACOB1 EQUATION REVISITED 169

What about the ordinates Qi conjugate to the Pi? From the equations Qi = BK/BPi, since K depends only on the Pi’s, and these are constants of motion, we have that all the Qi ’s are constants of motion. Note that Q1 = 8 K f a P 1 = aPl/aPl = 1. For i > 1 we have Qi = 8PIfaPi = 0. Therefore QI = t - to , while Q2,. . . Qn are (n - 1) constants. We have collected the right number (2n) of constants of motion, (n) Pi’s (including the energy), f& = 1, and (n - 1) Qi’s (i > 1).

We have the equations a S f 8 E = aS/aPl = t - to and, for i > 1,

Going back to chapter 1, equation (1.21) (tz - tl = ifS(qz,ql; E) /BE) agrees

fn the two-dimensional example in section 1.6, PI = E and Pz waa denoted

B S / ~ P * =: Qi.

with t - t o = aS/SE, and there is only one Pi, PI = E.

by a. (In most books the pi’s are denoted by mj-f We have also Qa = as/apz = as/apx,

~a = z - px [ d m - ~ 2 m ( ~ - .E* - m g g ) ] /m’g

= 2 - WO+[VO~ + gt - vov]/g = vo+t - uod = D .

Q1 is indeed a constant, which in this case happens to be zero. ~ t u r ~ i n g to the general case, we now state that

(7.54)

where the integral of the 1-form Cipi dgi is along il trajectory “c” in phase

We must first define the notation, which is also necessary for later use. Divide space from (Po, 90) to (P, 4. “c” into intervals, each of which is represented by a vector

DO not cor?fuse the numbers (dqj, dpj) with the I-forms (&j, dpj). In (7.54) dqi stands for

= C [dqj 6ij + zero] = dqi . j

Thus


Then for an isochronous variation

This tells us that as at as -= as - p o i , - = E . --

aqoi - P i i aqi

7.11 Time-dependent canonical transformations

They leave invariant the 2-form Li2 = A &j - J H A Jt,

G‘ (P ,Q ,K ,T) =G2b iq ,H, t ) ? (7.55)

where H (pi q, t> and K ( P, Q , T ) are the Hamiltonians. This is equivalent to the condition

~ ( p ~ ~ q ~ - P,JQi) - H a t + KdT = JG, , (7.56)

which is an extension of equation (7.44) to the (2% + l)-dimension~ phase space (pi, qj, t ) . If T = t , then

i

(7.57) aG1 K = € € + - at *

&om Hamilton’s equations in terms of the old variables

7.12. TIMEDEPENDENT HAMILTON-JACOB1 EQUATlON 171

follow those in the new variables dPi aK dQi OK dT 8Qi ’ dT 8Pi -=-- -- --

Formal proofs will be given shortly (see section 7.13).

Galilean transformation for a free particle, qi = zi, Qi = z{ (i = 1,2,3) etc. , &ample:

r f = r - u t , p f = p - m u r t ’ = t , H = p 2 / 2 m , K = p ’2 /2m.

Note first that

H = p2/2m = (p’ + mu)’/2rn + K = pf’/2m . We verify that equation (7.55) is satisfied. Since H and K depend only on the momenta, we have c(& Adz: - (aK/ap:)(ip{ A (it’)

= C(dp6 A (bi - uidt) - m-l (Pi - mui)dpi A dt)

= C ( d p i A &i - rn-lpidpi A (it) = x ( d p i A dzi - (aH/api)(ipi A lit) .

Equation (7.56) is satisfied if we take

i

i

4 i

Gl(r,r’,t) = (r - r’) * pl + m u . r - m u’1!/2 - u p’t .

In fact

-- aGi - -mu2/2 - u . p’ = K - H at

7.12 Time-dependent Hamilton-Jacobi equation

If the Hamiltonian H depends explicitly on the time, we perform a canonical transformation with generating function G2(q, P, t ) + S*(q, P, t ) chosen in such a way that the new Hamiltonian K vanishes.

This means H + aGz/8t = K = 0 . We still have pi = 8Gz/aqi. Hence the Hamilton-Jacobi equation

H (E,q*,t) +!g = o . (7.58)

Since K = 0, i)i = -@K/aQi = 0 and t$i = a ~ / ~ P ~ = 0, and so the new coordinates and momenta are constants of motion.

quat ti on (7.58) is a first order differ en ti^^ eq~ation for a functi~n of n + 1 variables, the 41’s and t. Therefore we w o u ~ ~ expect S* to depend on n + 1 constants of motion. Of these, n are provided by the Ps’B, while another can be the value of S* at some particular time, S*[qoi, Pi,to>.

If N does not depend explicitly on t, the relation between Hamilton’s “principal function” S* and Hamilton’s “characteristic function” § is

S * = S - E t . (7.59)

Like S, S* is refated to the action integral,

S*[q,~o,t,to~ = L.dt , Jf where “c” is the trajectory. In fact

dS* as* as* dt aqj at

as* pi@$ f -g- = x p a q i - N = L . - = c -qi + - = a a

Example: Suppose for a particle of mass m we are given

syP, ~ , t ) = fpo4 - ~ ~ t / z ~ ) ~ ( - t ~ + [(po + P ) ~ - fp0 + ~ j 2 t / 2 ~ ] @ ( ~ ) , where PO is a constant, and S([) = 1 for (5 > 0, = 0 for (5 < 0. We have

p = asx/& = po + PB(t) and Q = as*/aP = [q - (po i- P ) t / m ~ e ( t ~ . Thusp =po for t < 0 and p = po -+ P for t > 0. On the other hand, Q = 0 for t < 0 because @(t) = 0, and so, being a constant of motion, Q must be zero also for t > 0. Hence q - (po + P) t /m = 0 for t Here is clearly a particle of mass m with momentum po for t < 0, receiving an impulse P at t = 0, after which it moves with uniform velocity lpo + P)/m,

0.

Now as*

L(”> 2m aq +at = (~p&- t ) + (po + ~ f @ ( t ) j ~ -p&-t) - b0 + ~ j 2 ~ ( ~ ) ) / 2 ~

-(P4 - ~ ~ t / 2 ) ~ ( t ) + [(PO + P)q - (PO + P ~ 2 t / 2 m ) ~ ~ t ) *

Since e ( - t ~ ~ ~ t ) = 0 and t af t ) = 0, we find

where U(q, t ) = - Pq 6(t) , the potential energy for an impulsive force Pd(t).

7.13. STOKES’ THEOREM AND SOME PROOFS 173

We want also to check the invariance of G2:

dp A (iq - dH A dt = (iP A dQ - (iK Adt

The right member is zero, because P and Q are constants and K = 0. Substituting p = p , ~ + PO(t) and H = p2/2m - Pqd(t) in the left member, since d p = Pd(t)dt and dt A dt = 0, we find Pd(t)& A (iq + zero + Pd(t)& A (it = 0.

7.13 Stokes’ theorem and some proofs First a bit of mathematics. In section 7.5 we saw that the operator a acting on a 0-form (a function) h yields the 1-form Jh(.) = Ci(ah/azi)&i(o). The action of d on a 1-form is defined it9 follows (see equation (7.34)). If

G l ( 4 = C X i ( S 1 , . . .zn)dz‘(.) i

is a 1-form, then

For instance, in three dimensions

d ( X & + Y d y + ZJZ)

which corresponds to the elementary vector calculus operation

It is apparent that if GI is a gradient, Xi = ah/axi, then, combining in &’ a pair of terms (aX, /azj)&j A Gi and ( a X j / d z i ) & A ij we have [(82h/8zj8zi) - (82h/Ozi13xj)]Jzj A Jzi = 0. Thus,

if & ’ = a h , then & ‘ = Z h = O ,

which corresponds to curlgradh = 0. We already know from (7.35), in general a2 = 0, the double application of the operator gives zero. We shall encounter later n-forms Gn. These will be defined so that J2Gn = 0.


A special case of Stokes’ integral theorem can now be expressed in the following concise form:

(7.60)

where “c” is a closed curve, and “S1 a surface having “c” as boundary. The left side simply means that we must fill the single slot of ijl ( 0 ) with

the vectors ds defined in section 7.10. The right side means that we must fill the two slots of ~G’(0,o) with pairs of vectors (as, a), each pair forming two sides of an infinitesimal par?llelogfam of-a set covering all “ S ” . In three dimensions, with 6’ = Xdx + Ydy + Zdz,

/G1 = f ( X & + Ydy + Zdz)(ds) = /(Xdz + Ydy + Zdz) e fi a dr , C C C C

while I JS’&’ = [ (g - z) & A&(ds,&) + . . . + . . .

=I [(g-g) (dydz-dzdy)+ . . .+.. . ] LcurlV. ii dA

In our extended (2n + 1)-dimensional phase space, if

3’ = &Jqi - H J t , a

where we have used the definition (7.55) of canonical transformations. Hence

f[G1 (P, Q, H , t> - 3’ (Pl Q, K, T ) ] = 0

7.14. HAMILTONIAN FLOW AS A LIECARTAN GROUP 175

for an arbitrary "c", implying that

6' (p, q, H, t ) - Li' (P, Q , K , T ) = JG1 ,

namely (7.56).

ton's principle for isochronous variations vanishing at tl and t 2 , namely In order to prove the invariance of Hamilton's equations we use Hamil-

2 d l r (c ptqi - H)dt = 0 or d i ij' ( p , q, H , t ) = 0 . i

Since by assumption 6[G1 (2) - G1 (l)] = 0, Hamilton's equations for ( K , P, &, T) follow from the following steps:

G 1 ( P , Q , K , T ) = 6 [Li'(P,Q,K,T) -aG1]

7.14 Hamiltonian flow as a Lie-Cartan group Let

dx. 2 = Ai(x) , X = ( 2 1 , . . . XN) dt

be a system of first order differential equations. In an infinitesimal interval t, a point x = (XI,, . . XN) is shifted to the point

Atx = (21 + A1 (x)t,. . .EN + AN(X)t). The corresponding change of a function f (x) of the coordinates is

d f ( 4 = f ( A W - f(x) = tA(x)f(x) 7

where A(x) is the vector field

Consider now another system

and rewrite the above formulae with A + B and t + 8 .


Let the point x be shifted first by the vector field A acting for the infinitesimal parameter interval t , and then by B for the infinitesimal interval 5 ,

~j + ~j + tAj (x) + xj + tAj (x) + SBj(xl + t A l ( x ) , . . .)

= ~j + ( tAi (x) + s B ~ ( x ) ) + 8tC- a B i ( X ) A j ( x ) +. . . d X j

j

If x is shifted first by B through s and then by A through t , we have

The difference of the values of f ( x ) at BBAt(x ) and AtBs(x ) is

the error being of the third order in the infinitesimals.

field, divided by s t , is denoted by [A, B]. Its components are This difference has the form of a vector field acting on f ( x ) . This vector

[A,BIi = ( -Aj dB, - - B j ) dAi . dXj dXj

j

It is called the “commutator” of A and B. It is skewsymmetric in A and

If and B commute ([A, B] = 0), it does not make any difference in what order they act on a function.

B, [ B , A ] = -[A, B].

Example: A1 = 0, A2 = -x3, A3 = x2 (At induces a rotation through t about XI), B1 = x3, B2 = 0, B3 = -XI ( B s induces a rotation through s about 2 2 )

a a a a - A = -x3- + 22- = iE, , B = xs- - 21- = iLz , ax2 ax3 ax, ax3

- = -iL3 a a [A, B] = 22- - x1- ax1 ax2

Hence [ L I , &2] = i&3 and cyclic permutations. The imaginary unit has been introduced so that (LI&&) are the components of the quantum mechanics angular momentum operator in the coordinate representation (h = 1).

Consider now two Hamiltonian fields U l and UZ,

7.14. ~ A ~ ~ ~ T ~ N ~ A ~ FLOW AS A LIECARTAN GROUP 177

and e d u a t e their commutator. A not-too-long calculation gives

where [HI, Hz] is the Poisson bracket of HI and Nz. Hence

where H = [ H ; , H ; ] f

Summarizing, the commutator of two Hamiltonian fields is also a Hamil- tonian field with a Hamiltonian which is the commutator of the Hamilto- nians in reverse order.

Zn Iight of the above, the condition dfCp,q)/dt = [fl H ] = 0 for a quantity f ( p , q ) to be a constant of motion, can be expressed by stating that f ( p , q ) is a constant of motion if the H ~ i l t o n i a n field with f as a Hm~ltonian commutes with that with H ~ i l t o n i ~ H .

Finally here is the neat proof of the Jacobi identity for Poisson brackets learned from Arnold and promised in section 7.2. The expression

“A, 4, C] + “B, Cl, A] + “C, A), BI

is clearly a sum of terms each containing a second derivative. The terms containing second derivatives of A are the first and the last. Denoting by (dA/dt), and (dA/dt)c the rate of change of A under a flow with Hmiltonians B and C, respectively, we have

= [BA, C] - [CA, B] = (CB - BC)A = [C, B]A , where B and 6 are the Hamiltonian fields with Hamiltonians B and C, respectively, But [C, B] is a & order differential operator. Therefore our original expression cannot contain second derivatives of A. Similarly we show that it cannot contain second derivatives of B and C. Therefore it must be zero.


Figure 7.3: Magnetostatic field example

7.15 Poincar6-Cartan integral invariant In phase space consider a tube formed by flow lines, and two arbitrary closed curves c1 and c2 encircling it.

We shall show that f,G1 = p1 7 (7.61)

where G~ = CipiJqj. Proof: Following Arnold we begin with a similar elementary problem. Con-

sider the curves c1 and c2 encircling a tube of lines of force of a magnetostatic field B = V x A (see figure 7.3).

Let S1 and S.L be two surfaces having c1 and c2 as rims, respectively. Since V . B = 0, Gauss' theorem applied to the closed surface consisting of S1, Sz, and the portion of the tube comprised between c1 and c . ~ , gives

(7.62)

The integral over the tube is zero and so we find (see figure 7.5)

B . i i l dA1 = (-&) dAz . (7.63)

Then Stokes' theorem yields

A . d r = A.dr . 6, I We want to cast the above into a new form. Define

k1 = A& + Avdy + Azdz, kz = &' = Bz& A dz + B,& A dx + BXdx A dy,

where B, = aAx/ay - aAv/az etc. Filling the slots of kz with two vectors and V, we have

G ~ ( ( V , V ) = B . (U x V) = (B x U) .V = (AU,V) ,

7.1 5. POINCAR$-CARTAN INTEGRAL INVARIANT 179

where 0 -B, By

A = ( BL 0 -B.) , -By B, 0

and (AO,P) = ( A j j ~ j ~ ~ . The matrix A has zero determinant. Therefore the secular equation

det(A - XI) = 0 has a solution A = 0, and so A has an eigenvalue zero. This zero eigenvalue corresponds to an eigenvector with components (&, B,, Bz),

A ( $ ) = O .

In our ~ e m e n t ~ t r ea t~en t we applied Gauss’ theorem to a closed surface consisting of SI, SZ, and a surface spanned by vectors 8 which axe eigenvectors of A belonging to the eigenvalue zero (the “flow lines” of B), so that

2 = o because ;t2((v,*) = o .

The above method can be extended to any gaa-dimenrrional space, in particular to the (2n + 1)-dimensional space (pi , . . . pn, qi , . . . qn, t) . We show this for 7~ = 1, the extension to n > 1 is trivial.

lube We have 8‘ = p& - Hdt,

ij2 = &l - - dp h dq - (aH/ap)dp A dt - (BH/8q)& A dt

~ 2 ( ~ , ? ) = u,vq - uqvp - ( a ~ / O p ) ( ~ p ~ - UtVP) - ( a ~ / 8 q ) ( u q ~ - ~ ~ V q )

,

= (AU,V)

Now detA = 0. In fact, the de te rmin~t of any s ~ e ~ y m m e t r i c (2% + 1) x (2% + 1) matrix is zero:

Therefore A has a zero eigenvalue, corresponding to the eigenvector 2n+t detA = detAT = det(-A) = (-) detA.

180

4

CHAPTER 7. HAMILTONIANS

t

Figure 7.4: Equation (7.63)

Hence (see figure 7.4)

If c1 and c2 are the intersections of the planes t = tl and t = t 2 with the tube of flow, then we have, generalizing to n > 1,

and I,, G2 = I,, G2 ,

where G2 = Cid;3i A dQj (no &!), and St, and S t , are the portions of the planes t = t 2 and t = tl having as boundaries cta and q,. Since the tube can be as thin as we want, we have also that G2 = C i J p i A & is invariant under the Hamiltonian flow, as it is under canonical transformations. We already showed in section 7.5 that the Hamiltonian flow is a canonical transformation. This is a mere confirmation of something we already know.

6For n = 1, this is Liouville’s theorem.

7.16. POINCARfi'S INVARIANTS 181

7.16 Poixicad's invariants The invariance of G2 = C,dp&&i under canonical transformations (p, q) 3 (P, Q) implies that

where S is a tw~dimens ion~ surface. This is PoincarB's first integral in-

(7.65)

n m e h

Here

The powers ((;j2)i are also invariant. Note that the highest non-vanishing power is

up to a numerical factor depending on the definition of a product of exterior forms.

(G2)"=&l A. . .dpnA& A . . . d q n

Integrating over a region V of phase space, we have

(7.66)

expressing the invariance of phase space volumes under canonical transformations.

For p -+ p(t), g + q(t) , P + p(t'), Q -+ q(t') we have Liouville's theorem for finite time ~ n ~ r ~ s ?, while the proof in section 7.4 was for the interval (t , t + dt).

71n his Classical Dynarnicol Systems vol.1 of A Course in Mathematical Physics, (Springer,l978) p. 84, W. Thirring writes: Yn the framework of [old] classical mechanics the proof of [Liouville's} theorem requires some effort. But modem concepts are so formulated that there is really nothing to prOV0."


7.1’7 Chapter 7 problems 7.1 (i) Write the non-relativistic Hamiltonian and Hamilton’s equations for a particle of mass m and charge q in the uniform static electromagnetic field E = (0, E,O), B = (O,B,O). (ii) Verify that C, = p,, C, = p, + ( q B / c ) x - qEt, and C, = p, are constants of motion. (iii) Assume that at t = 0 the particle is at rest at z = y = a = 0. Expressing the Hamiltonian in terms of C,, C,, and C,, find the trajectory of the particle.

7.2 (i) What is the Hamiltonian H , derived from the complementary La- grangian L, in problem 6.7?

Comment on the invariance of Hamilton’s equations under the transformation (P, q, H ) + (P, Q, HC).

7.3 Using the canonical transformation Q = p, P = -q, write the Hamilto- nian H = p2/2m - k / ( q ( and Hamilton’s equations in “momentum space”, and show that the Laplace-RungeLenz vector is constant.

7.4 Study the canonical transformation generated by

2qP - sin a (q2 + P 2 ) 2 cos a G2 = 1

containing a parameter a, and the case of a infinitesimal. What happens for a = -7r/2? Use the corresponding GI for transformations differing infinitesimally from p = Q, P = -q.

7.5 (i) Show that the generating function

induces the canonical transformation

(PI, qlr ~2142) + (PI, Q1, S , Q2)

with P I = p l c o s a - p 2 s i n a ’4 = p l s i n a + p ~ c o s a , Q1 =qlcosa-q~s ina ,Q2=qls ina+qzcosa .

(ii) Show that this transformation with

2f w: - w;

tan(2a) = --

7.1 7. CHAPTER 7 PROBLEMS 183

uncouples the two harmonic oscillators with Hamiltonian (w1 > w2)

H = (p; + w;q:)/2 + (pi + wiq;)/2 + fqi@ . Expressing H in terms of the new variables, one has

H = (P," + 02:Q:)/2 + (P," + Q ; Q ; ) / 2

with ~ : = ( w : + w i + J w ) / 2 ,

(iii) What happens is w1 = w2 + w ? Nothing bad, (Y = -7r/4, PI = (PI - p2)/&, 91 = (QI - Q2)/&,

p~ = (Pi + P2)/Jz, ~2 = (Qi + Q2)/Jz, 0:: = w2 + f , 0; = w2 - f .

7.6 Find the generating function for the canonical transformation (T , 8 , 4, P,, pe , P&) + (z, Y, x, P,, P, , PZ) I

7.7 (i) Consider the canonical transformation induced by the generating function

Gz(r,P) = x ~ P ~ + P I P z + ~ ( x ~ P z +x2P1)+Xz1x2

and use it to transform the Hamiltonian

H = lp: + (p2 - Xz1)2 +p3 /2m . (ii) Write and solve Hamilton's equations in the new variables. (iii) Use the result to study the motion of a positron (charge +e) in a

uniform magnetic field of intensity B in the z3 direction (A1 = A3 = 0, A2 = Bsl) .

7.8 (i) Verify that the Poisson bracket of the Toda Hamiltonian

with the Hhon integral of motion

is zero.


(ii) Since I does not explicitly depend on t , its conservation must correspond to the invariance of the Hamiltonian under a transformation p + p + 6p, x + x + bx, with 6p and 6x of the order of an infinitesimal parameter 6, bH = O(Z). Find 6p and 6x.

7.9 (i) What are the infinitesimal transformations for Xk and pk generated by Q A ~ , where 9 is an infinitesimal parameter and

is the i-th component of the Laplace-Runge-Lenz vector.

the first order in 9. (ii) Show that the Hamiltonian H = (Ip12/2m - k / r ) is invariant in

7.10 Let p and q be canonical variables for a particle with Hamiltonian H = p2/2m.

(i) Perform the canonical transformation (p, q) + (P, Q ) with

G ( q , P ) = (9 - at)(P + 6 ) ,

where a and 6 are constants. Find P and Q . (ii) We calculate Q = BH(p(P,Q) ,q(P1Q)) /8P, then we express the

resuly in terms of q and p finding q = p/m + a (strange!). What have we done wrong? What should we have done?

7.11 In a frame X Y Z an electron is subject to the uniform magnetic field B = (O,O,B), A = (-By,O,O). In a frame X ‘ Y ‘ Z ’ moving in the 2 direction withe velocity u (x‘ = x - ut,y’ = y,z‘ = z) , it is subject to the magnetic field B‘ = B and the electric field E’ = (0, - Bu/c , 0) .

(i) Write the Hamiltonians H and K for the electron in X Y Z and X’Y‘Z‘, respectively.

(ii) Find p and p’, and Hamilton’s equations in the two frames. (iii) Verify that

G2(rP,p’,t) = -upLt-mu2t/2 + r . p ’ + m u x

induces the Galilean transformation from (r,p) to (r’,p‘). Note that this corresponds to

GI(r,r‘,t) = G2(r,p‘,t) - r’ ’p’

as we had in section 7.11.


7.12 Suppose we have the integral of motion u(p, q, t) explicitly dependent on t

u(p, q, t) = ln(p + imwq) - iwt

for the harmonic oscillator with Hamiltonian H = p2/2m +mw2q2/2. (For q = Acos(wt + a), p = -mwAsin(wt + a), u = ln(mwA) + i(a + n/2).) Then

which can also be written in the form

Find the corresponding infinitesimal transformation under which

au 6 H = - € - at - 7.13 (i) Find the infinitesimal transformations induced in p and q by the integral of motion I = up - H for a particle in one dimension with Hamil- tonian H = p2/2m + V(Z - ut) (u =constant).

(ii) Show that under this transformation 6 H = -e Or/&.

7.16 Verify the formula

LAG = 2[5(A)] + (&)(A) in the case where G is a 1-form.

7.16 Show that the operators 2 and LA acting on a form commute, JLAG = LA&.

7.17 Consider a flow generated by the vector field V and a closed curve Ct whose points are transported by the flow, so that at the time t + dt they cover the curve Ct+nt. If G1 = Aj&i is a 1-form, one has

f ijl = A t G I + d t A t L p O ' . Ct+dt


Interpret this formula in three dimensions.

7.18 Use the result of problem 7.17 to show that


Solutions to ch. 7 problems

S7.1 (i) H = (p - qA/c)'/2m + qV, V = -Ey, A, = -By, A, = A, = 0 ,

H = [(Po + qBy/c)' + p t + p:]/2m - qEy

pz = mx +qA,/c = rnx - q B y / c , p , = m$ , p , = m i

(ii) C, = par and Cz = pr are constants of motion because x and z do not occur in the Hamiltonian. It is easily verified that C, = p, + (qB/c)x is a constant of motion. In fact C, = -(OH/&) + ( q B / c ) i - qE , (iii) Since C, = C, = C, = 0 at t = 0, we have

At t = 0, p z = p , = p , = 0, x = y = z = 0.

Cu = -(qB/mc)(p= + pBy/c) + qE + qBk/c - qE = 0 .

H = [(qB/c)2y" + (qBx/c - qEt)']/2m - q E y .

F'rom this we see that H = 0. The trajectory is the cycloid

with r = Emca/qBa.

S7.2 (i)

(ii) The equations Hc = PQ - (PQ + a ~ - aL) = -aH(p,q,t)

aHc , p=-- * a H c Q = -

, q = - . i9(-aH) , cud=- - 6 ( - a H ) a H p = -- - --

a p give

a H a(%?) & a P a P

Hamilton's equations are invariant under the canonical transformation Q = p, P = -q without transforming the Hamiltonian, while their invariance under the (Ilpn-canonical) transformation Q = p, P = q requires a change of sign for the Hamiltonian.


s7.4 p = aGz/aq = ( P - q sina)/cosa, Q = aGz /aP = (q - P sina)/cosa,

namely P = p cos a + q sin a, q = Q cos a + P sin a. Note that Gz = G1 -k PQ, where

GI = - (q2 cosa - 2pQ + Q2 cosa) . 2 s ina While GI is singular for a = 0, Gz is singular for a = -n/2. Therefore Gz

with cr + da (infinitesimal) can be used for a transformation differing infinitesimally from the identity P = p, Q = q, in which case P 111 p + q ha, Q N p - p d a . On the other hand, G1 with a = -7~12 + 6a could be used for a transformation differing infinitesimally from p = Q, P = -q.

S7.6 A simple extension of the two-dimensional case, gives

Gz(r,B,+,p,,p,,p,)=r sin0 cos$p,+r sine s i n 4 p , + r C O s e p , . One finds pr = aGz/ar = sin 0 cos t#J p , + sin 8 sin 4 p , + cos 0 p,,

pe = a G z / a e = r cose cost#Jpr+r cose s i n 4 p , - r sinep,, p4 =aGZ/a$= -r sin@ sint#Jp,+r sine C O S ~ ~ , ,

x = aGz/ap+ = r sin 8 cos 4, etc.

p , = C O S ~ (sin 8 pr + cos 8 p8/T) - (sin C#J /r sin 8)p# , p , = sin 4 (sin 0 pr + cos 6 pel r ) + (cos 4 /r sin O)p@ , pr = cos 8 pr - sin 8 p8/T .

The following may be useful:

s7.7 (i) One finds

pl = aGz/axl = LA p l2+~x2 , pa = aGz/axz = Jj; pl+Azl , p3 = aGz/axs = p3,

Qi = aGz/api = & + A x 2 , Qa = aGz/aPZ = P i + f i x l , 93 = aGz/aPs = x3 . Note that p l = 4 Q1. (ii) The transformed Hamiltonian is

x 1 2m 2m K = -(P;” + 9:) + -pi

(harmonic oscillator and free particle in one dimesion). PZ and &a do not appear in K.

Hamilton’s equations yield

x * A m m Pi = -- Q1 , QI = - PI , PA = constant, Qz =constant ,

with the solution Pi + iQ1 = A exp(i(At/m +a))


(A and a real). Then, since

PI + iQ1 = Q 2 + iPz - 4 ( $ 1 - izz) , we have

If X > 0, this gives

z1 - iz2 = [ ~ z + ipz - A exp(i(Xt/m + a ) ) ] / d i .

2 1 = [ Q 2 - A cos(Xt/m +..)]/a , xz = [ -PZ + A sin(Xt/m + *)]/a .

(iii) Comparing the Hamiltonian in (i) with

H = [PT + ( p z - eBz1/cI2 + p i ] / ~ m

we see that the equations of motion for the positron are

21 = [ Q z - A C O S ( U ~ + a ) ] / & Z , z2 = [-PZ + A sin(wt + a)]/& , xg = zs(0) + vt ,

with w = eB/mc.

S7.10 (i) p = aGa/aq = P + b, P = p - b, Q = aG2/aP = q T a t

(ii) H(p(P,Q),q(P,Q)) = (P+b)2/2m. If w-e write (wrongly) Q = aH/aP, we obtain 4 = ( P + b)/m. If in this we substitute P = p - b, Q = q - at, we get q - a = p / m .

We should have remembered that the new Hamiltonian is K = H + BGz/Lh! = ( P + b)’/2m - a(P + b)

Q = a K / a P = ( P + b ) / m - a , q - a = p / m - a . and 80

190 CHAPTER 7. H A ~ ~ L T O ~ I A ~ S

S7.11 (i) H = [(px - eBy/c)' + i t + pi]/2m ,

K = [(p: - eByf/c)2 + p i z + p 3 / 2 m - (eBu/c)y'

p = (mk + ~ B y / c , m ~ l m ~ ) , p' = (mx' t eBy'/c ,m~',mi'~ = p - mu & = 0 , rjy = eB(p, - e ~ y / c } / m c ,gj, = 0

mx = -eB@/c , mQ = eB$/c , m i = 0

x' = (p: - eBy'/c)/m , $' = p i / m , 5' = pL/m

(ii)

x = (pa - e ~ ~ / c ) / m $ = pv/m , i = pL/m

pi = 0 pk = eB(p: - eBy'/c)/mc + eBu/c, p: = 0

Since &' = 2 - a, 2 = 2, @' = $, 2' = i, the equatio~s

equations. We have also

mxf - - -eB$'/c and mz' = 0 are clearly equivalent to the corresponding unprimed

mji = mQ' = eBx'/e - eEk = eB(x - u ) / c + eBzL/c = eBxfc .

im&J 1 , &q = -€ a' = ' p + i w q p+imwq *

S7.13 ( i )

(ii) sp = e[I,pf = -€ E)U(x - ut)fax , &x = Cfl,Xf = .(p/m - u)

p atr(x-ut) 6 U f z - ut) p ax [ m ax + ax ( 7 0 -,,I, dH dH=-dp+-dx= --

8 P @U(x - ut) dU(x - ut) az = -E--

dt at = + E - - -€%

ds ~Iementary remark: In a system moving with velocity u, we would have

rnx"f2 + U[x') = E' =constant. But now xi = E - ut, x' = x - 21, and so E' = ( m / 2 ) ( 2 + u2 - 2uk) + V ( X - ut) = [mx2/2 + U(X - ut)] - um+ + rnlta/2 = H - up + mu2 f 2. Thus 1 = up - H = mu2/2 - E' =constant.

Also, since dG2 = 0 for 3' = dp A dq, the formula in the following problem 7.15 nives



The Bum of the last two expressions is

(2 Aj + w j $$) b' = ( L ~ i 3 ) ~ d x ' .

57.16 Acting with d' on the equation in the previous problem, and remembering that dr = 0, we have

JLAG = d'[(&)(A)] = LA& - (d&)(A) = LA&.

57.17

The above equation simply expresses the vanishing of the solenoidal V x A through the closed surface formed by the surfaces St+dt, St, and the flow tube connecting them (see figure 7.5).


S7.18 For 6' = p &, it is easy to show that

L&' = d"G'(U) - H] .

Since f d( .. .) = 0, we have

Chapter 8

ACTION-ANGLE VARIABLES

Periodic motions are mapped onto uniform circular motions by introducing the action-angle variables. For a system executing periodic motion, a quantity can be defined which is invariant under slow changes of the system parameters. This is an “adiabatic invariant”.

8.1 One dimension In this section we study the mapping of a periodic one-dimensional motion onto a uniform circular motion.

This is realized by a canonical transformation ( p , q ) -) ( J , +/27r), where

J = k p d q = ( I& A J q ) = A ( E ) D

is the action integral along the closed orbit “C” enclosing the region “D” of the (p, $-plane.

Since d p A dq = JJ A d(4/27r) and J is a constant of motion, we have

and so it is clear that 4 is incremented by 27r while C is covered once. Fkom the Hamilton equation

_- - - 27r 8 J 8 J (8.3)

193

194 CHAPTER 8. ACTION-ANGLE VARIABLES

P m

0

-G

Figure 8.1: Ball bouncing up and down

we see that 4 is constant, and the period T is given by

as we already know (see equation (1.10)). Example 1: The mapping of a harmonic motion onto a uniform circular motion

is the inverse of the elementary construction of the harmonic motion by projecting a uniform circular ~ o t i o n onto a diameter.

cos # and q = J;s7;.’;... sin # in the Hamiltonian H = (p2 + m2dq’)/2m, we find E = w J/27r. The transformation is canonical

Substituting p =

because

Example 2: Ball bouncing elastically up and down on floor (see figure 8.1). The canonical trmsformation is

Substitutin~ in the H ~ i l t o n i m H = p’f2m 4- mgq, one finds

3m2gf a E=&(-?-> )

A(E) = J = 2 ( 2 m E ~ ~ / 3 ~ ~ ~ ~ T = d A ( E ) / d ~ = 2v0/g, where E = ~ v ~ / 2 . It is natural to ask: “HOW did you get those canonical transformations?”

The answer is simple. One must find the generating function Gz(q, J). Then p = K’,,/&q and $11271. = &Gz/8J.

The generating ~ n c ~ i o ~ can be obtained by solving the ~ami~ton-Jacob~ equation

8.1. ONE DIMENSION 195

where we have stressed that the energy must be expressed in terms of J . For the harmonic oscillator (Example l ) , Gz(q, J ) is easily obtained

from S(qz,q1, E = E + wJ/2lr, a +

equation (1.22) with the following changes: + q, q1 + 0. One obtains

Gz(q, J ) = [ q { z + lrmw &in-' lrmw (q/F)] . (8.6)

Then

-= - - 8G2 - &n-' (q/F) , q = { x s i n ) , (8.7) 2x d J 2lr lrmw

while

p = aGz = xmw = /Fcos+ . (8.8) 89

For Example 2, we show the convenience of using G4(p, J ) , for which q = -8G4/8p and 4J2lr = 8G4 JOJ. Substituting the former in p2/2m + mgq = E ( J ) , we have (p2/2m) - mgdG4/dp = E ( J ) , and, by an elementary integration,

1 1 G4 = - mg [-b3 6m - P:) - E(J)(p - P O ) ] .

Then 4J2n = dG4/8J = -[(p - p o ) / m g ] d E ( J ) / d J yields

p = m - ( T ) 4 ; 3m2gJ .

If we want p = po = = (3m2gJ/2)* for + = 0, we have

p = (&1-$) 3m2g J .

(8.9)

(8.10)

(8.11)

Substituting this in q = -8G4/8p = -[p'l/2m - E ( J ) ] / m g , we find the equation for q.


_I

Figure 8.2: Torus for Kepler motion

8.2 Multiply periodic systems A ~ u l t i p l y periodic system with n degrees of ffeedom is one in which each of a set of n suitably chosen qj (i = 1,. . . n) is simply periodic with period Ti *

The Ti ’s are in general difierent. If their ratios are irrational, the trajectory in 2~-dimensional phase space is not closed. Remember the Lissajou figures from General Physics!

The motion in the 2n-dimensional phase space can be mapped onto an n-torus, on which n angular coordinates ($1,. , . &)mod 2n can be defined. (For n = 1 the 1-torus is a circle.) Each coordinate changes uniformly with time,

(8.12) d$i - = wi ~ c o ~ s t a n t ) . dt

For each $5 one can define a conjugate momentum Ji, so that n n

i= 1 i= 1

Each Ji will be a constant of motion.

For E < 0 the flow generated by the Hamiltonian Example: Kepler problem in a plane with U = - ( k / r ) - (hlr’)).

(8.14)

is confined to the %-torus shown in figure 8.2. Generalizing what we did in section 2.4, we first evaluate J, = $p,dr and Jo = $pod@, finding (see problem 8.4)

Jt = 27r (A - fa) with cy = ~~ (8.16) %/%-a

8.2. MULTIPLY PERIUDIC SYSTEMS 197

and

Using these, we obtain Je = 2Xpe = 2-d . (8.16)

(8.17) 27r2mk2 PI =

(Jr + J-f '

We solve the H ~ j l t o n - J ~ o b i equation

finding

(8.19) I

+--Je(e-eo) I

2X

We note at once that neither 460 nor c $ ~ is equal to 8. We also note that the r-period is

> (8.20) (Jr + J-)'

4&nka while the 8-period is

(8.21) Te = (%)-I = Tr Je

The l/r2 force has removed the degeneracy. The periods T, and Te are now difFerent, and there is a precession with angular velocity

Wpr = 2s - 5 2n = - 2n ( 1 - /-) Z W ~ F m h , (8.22) Tr

Therefbre A@, = 2 ~ w ~ ~ / w ~ 2 2nmh/12, in agreement with (2.16).


But let us return to the pure Newtonian attraction. We are dying to know what & = $0 is in this case. &om #,/2n = ( ~ ~ / a J r ~ ~ r ~ ~ d ~ ' / p ~ ~ we see that

with la as defined by equation (2'47). Hence, for to = 0,

a), = U - E sin% , (8.24)

where u is the eccentric anomaly. Remark: For the pure Kepler problem ("pure"= only ~ e ~ o n ~ ~ a t t r ~ t j o n ~

some authors evaluate J4 = 2 d 3 , Je = f p e d 6 = fd- dB = 2 4 1 - h),

Jr = $JZrnfE + k / r ) - P/ r2 dr = 2 = [ ( ~ k / ~ ~ ) - 21. Hence

J , . = = 2 7 r ( @ k - V ) ,

from which

The calculation of Jr is presented as an interesting exercise in complex integration. Goidstein attributes to Van Vleck the following method of evaluating Je.

Notice Erst that comparing H = c j p i q i - L = s i p j q j - K + U with H = K + U one has 2 K = Zipiqi. Expressing the kinetic energy in terms of the original (r, 8,Cp) variables and in terms of plane-motion variables (r, $1, one finds

~ r ~ + ~ e ~ + p ~ ~ = p ~ ~ + p ~ ~ . Note that $ can be identified with the Euler angle defined in problem 8.3. Using the above equality one has

Of course, if one likes to catculate integrals, Je can be obtained directly. Using problem 8.3, we have

Je = {J- dB = if Jl - c ~ 2 ~ / s i n ~ B dd .

8.2. MULTIPLY PERIODIC SYSTEMS 199

One can perform two changes of variable,

One gets first cos 0 = sin 6 sin (I (cos 0 = 6; .& = sin 6 sin t/) , and then v = tan t/,

dv Je = - I sin229 1 - sin26 sin2$

= -6 [f- dv -coszt9

) = -1 (f d(I - f d[cos 29 tan-’(cos 6 tan $11

= 6 ( 2 ~ - 2~ ~ 0 ~ 2 9 ) = 2 ~ ( l - 1 s ) . Why fd(I = - 2 i ~ rather than +2n?

We limit ourselves to the statement of the important Liouville’s theorem on integrability ’ :

Let H be the time-independent Hamiltonian of a system with n degrees of freedom (2n-dimensional phase space). If we know n functions Pi@, q) (i = 1,. . .n) such that (i) [H, Pi] = 0 (the Pi’s are integrals of motion), (ii) [Pi, P’] = 0 (the 4 ’ s are “in involution”), (iii) the dPi are linearly independent, then the system is “integrable”, i.e. by algebraic manipulations the mathematical problem can be reduced to integrations.

For a system that satisfies the integrability conditions of Liouville’s theorem is “connected and compact”, Arnold has shown that each conserved Pi corresponds to an angular variable 4i, so that the subspace {Pr = q, i = I , . . .n} can be mapped onto (is diffeomorphic to) an n-torus

Although (Pi, 46) are not symplectic variables, action variables Ji can

Summary of action-angle variables

{ (41 9 * - - &)mod 2 ~ ) .

be constructed such that

i i

A Ji is defined as

(8.25)

(8.26)

‘See section 3.3 in W. Thirring, Classicnl Dynamicol Systems wl.1 of A Course in Mathematical Phyeics (Springer).

aSea Thirring, loc. cit. p. 102.

along a curve in our subspace such that $i changes from zero to 2n, while each of the other $j’s ( j # i ) returns to its initial value. Subject to this condition, ci is arbitrary as can be proved by Stokes’ theorem and the vanishing of G2 in the subspace.

If the Ji’s are known first, then the qii ’s are obtained by a canonical transformation with a generating function Gz(q1,. . . qn, J1 , . . . J n ) solution of a Hamilton-Jacobi equation.

Since $i/2n = dGz /dJ i , the change of qii over a cycle in which q5j

changes by 2n is

(8.27)

8.3 Integrability, non-int egrability, chaos For a conservative system with n = 2, the “energy surface” E = constant is a three-dimensional region embedded in the four-dimensional phase space. A Poincar6 section is the intersection of this region with a plane, for instance the (plrql) plane. The pattern of the points where a given phase space trajectory croases the plane, obtained in most cases by numerical integration, provides useful information about the system. For bounded trajectories, the crossing can be expected to occur at an infinite number of points, unless the motion is periodic. For n = 2, a bounded trajectory for which an integral of motion other than the energy exists, lies on a two-dimensional torus. This intersects the Poincar6 section on a closed curve. For instance, the left side of figure 8.2 shows a torus E = constant, I = constant in the four dimensional space ~ r , ~ ) ~ @ , ~ ) . The right side shows the intersection of the torus with the brt p) plane- There are two frequencies, w1 and w2 (w, and we for figure 8.2). If the motion is simply periodic {u, = w2) there will be a single point on the intersection curve, through which the trajectory passes every time it crosses the PoincarC! plane. If w1 # 02, but W I / W Z rational, there will be a finite number of points. If W I / W ~ is irrational, the curve will be completely covered, telling us that the trajectory on the torus is “ergodic”. In this case, given an arbitrary point on the torus and a number e > 0, the trajectory will pass at a distance less than E from that point. The pure Kepler motion

8.3. ZNTEGRABILZTY, NON-INTEGRABILITY, CHAOS 201

(h == 0) is simply periodic, while the last two cases are possible for T1. # T@ (h # 0, equation (8.21}}.

Another example: For the two-dimensional harmonic oscillator with Hamil- tonian

consider a trajectory with energy E passing through a point PO, for simplicity 0310, q ~ o , QZO = 0). Then

H = [(p: + W:q?) f (g 4- w.2.g:>]/2

p20 = f d 2 E - p;o - W:q&

Thus the trajectory is completely defined by PO up to a sign. The trajectory will crow the (p1 q l ) plane again and again. At which points? If a8 B second integral of motion beaides the energy we take the energy of the “2” oscillator, dl the i n t e r s ~ t i o ~ will lie on the curve

p: + w?g? = 2(E - E2) . Using the solution

ql(t) = g1(0)cos(wlt) + ~l(O)/wI~sin(w,t) I

p ~ ( t ) = p1(0) cos(w1t) -w~q~(O)sin(w~t) , we find for successive intersections the two-dimensional map

qli = q1,i-l cos(2awl/wz) i- (pl,i-~/wl) sin(2lrwl/wa) , { pli = p3,i-i COS(~SW~/W~) - ql,i-1wI sin(2lrul/wz)

In the ( p ~ , w ~ q ~ ) plane, this is a rotation through 2~rwl,fw2. If wl/wz = m/n (m and o) integers), the rotation, repeated n times, is the identity, and the system is multiply periodic. If the ratio of the frequencies is irrational, the intersection curve is covered ergodically.

Keeping E fixed and varying E2 one has a family of nested tori, intersecting the (p1,ql) plane on a family of nested curves,

The Toda Hamiltonian HT = (‘$; -k p 3 f 2 -I- UT (see equation (2.58)) is integrable because of the existence of the Henon integral of motion 1 (2.59). For given E and varying 1 one has a family of nested tori whose intersections with the ( p z , 8 ) plane are nested curves.

In contrast to this, the Hdnon-Heiles Hamiltonian HHH = (pz + p i ) / 2 3- UHR (see equation (2’60), problem 2.7, and figure 2.6) is non-inte~rable, the only integral of motion being the energy E. The nested tori structure of the Toda motion is progressively destroyed with increasing energy, as shown on the Poincartl section by the appearance of islands. A single trajectory jumps from island to island, in much the same way as in the logistic map for r > 3, where sn, starting from a given so, jumps from one branch to another of figure 3.5 as n increases.

It will be enough for us to state that for E = 1/6 all but a few minute islands have disappeared. Apart from these, a single trajectory wanders chaotically in the three-dimensional region E: = 1/6, and crosses the Poinear6 section at random (see,for instance, M. Tabor, Chaos and integrability in nonlinear dynamics (Wiley,1989) p. 122). We note that as E increases with consequent extension of the available (2, g) region, the approximation UHH N UT becomes increasingly bad. We might write HHH = HT + 6 H , and attribute the destruction of the tori to 6N.

We must mention the KAM (Kolmogorov, Arnold, Moser) theorem. For a = 2, part of the theorem states that if, among other technical condition^, the Jacobian / L h i / a J j / is not zero, then those tori whose frequency ratio wz/wl is ~‘sufficient~y irrational’’ are stable under the perturbation 6H for [dH( very small (see, for instance, H.G. Schuster, Deterministic chaos. An introduction (VCH, New YorkJ988).

At this point we rest our brief account of chaos and suggest a viait to any moderately stocked library. The reader will find an abundant supply of books on the subject.

8.4 Adiabatic invariants It is widely known that the s h ~ r t ~ n i n ~ (lengthening) of the string of a pendulum causes the amp~itude t?,,,, to increase (decrease). It may also be known that if the length of a Galilei pendulum (smdl amplitude oscillations) is changed slowly, the amplitude varies according to the law

(8.28)

“Slowly” means take t = T(0) = have ldt/dtl*T(O) << t(0). &om equation (8.28) we infer that

is small, and that t << l / l d t / d t ( . If for t we the initial period of the pendulum, we must

(8.29)

(8.31)

8.4. A ~ ~ A R A T ~ ~ ~ ~ V A ~ A ~ T S 203

where J ( t ) = 2n~(t) /w(t) is the action variable. Thus J ( t ) is an “adiabatic invariant”.

Let us attempt a little theory. Differentiating J = 27rH/w with respect to t , we have

j = -- 2n (- OL + -.> w w at w

(8.32)

since dHldt = -ah/&. For the oscillator with L = m[q2 - ~ ( t ) ~ q ~ ] / 2 this gives

(8.33)

Taking the average of j over a period of the motion with w( t ) and d( t )

( j ) = 2 7 r W - 2 w ( { m w 2 9 2 / 2 ) - ($/2m)) = 0 , (8.34) since the average values of the kinetic and potential energies are equal. This indicates that ( j ) is of the order of L j 2 .

Some authors derive equation (8.29) in the case of the pendulum by considering the work done by the tension of the string when the length C is changed by At , This work is AW N -(mg -t m d 2 - mg$‘/2)At, and its average value is (AW) 2t -mgAt - (E5/2t)At) where Ee is the energy of the oscillations. The second tmm of (AW) changes Ee by AEe = - ( ~ e / Z t ~ A ~ ) and so A ( E e ~ ~ = 0, E9& =constant, E9/w =constarit.

A more sophisticated approach makes use of the generating function GI (q, #42n) = (mwq2(2)cotan4 to transform from the canonical variables (p, q) to (J,4/2n). This is still possible although w is now a function of t, but the transformed Hamiltonian is K ( J , 4, t ) = H + OGl/at,

?rw m W 2

j = -(,2&&2 - p 2 ) .

which already shows that j is small if L j is so.

treated as constants and replaced by their values at t = 0, we have

WJ 3 J 2n 2nw K = - +--sin4 cos4 . (8.35)

Hamilton’s equations for K yield

(8.36) w j = -J- C O S ( ~ ) w

and (8.37)

In chapter 8 on pertubation theory it will be shown that if c j ~ / w ~ is

w 2w

4 = w -+ - sin(24) .

small

(8.38)

Thus the value of J ( t ) oscillates about J ( 0 ) with a small amplitude.


8.5 Outline of rigorous theory We follow sections 51 and 52 of V.I. Arnold, Mathematical Methods of Clas- sical Mechanics (Springer, 1978), but confine ourselves to one-dimensional systems.

Definition: The quantity J ( p , q; A) is an adiabatic invariant of a system with Hamiltonian H ( p , q ; A ) (A = d , E constant) if for every K. > 0 there is an €0 > 0 such that, if E < €0 and 0 < t < l / ~ , then

IJ (P( t ) , d t ) ; 4 - J ( P ( O ) , q ( O ) ; 011 < K.

i = 44 + d(J, 4) , j = dJ, 4)

’

Averaging theorem: Consider the system of equations

, wheref(J,+) =f(J,++%) andg(J ,+) = g ( J , + + 2 ? r ) .

the. equation

where

The gist of the theorem is the comparison of J ( t ) with J ’ ( t ) 3satisfying

j’ = c g ( J ’ ) ,

2%

B ( J ‘ ) = ( 2 d - 1 1 g(J’,4)d+ .

Assuming that in a certain region of the ( J ,+) plane 0 < c < w < CI and I f 1 < c1, 191 < c1, then for J ( 0 ) = J‘(0) and 0 < t < 1 / ~ one has

IJ(t> - J‘(t)l < cof 9

where co depends on c and c1, but not on E .

Quick proof: Define P = J + c k ( J , 4), where * * * * * * * * * *

and assume that this relation can be inverted, J = P + &(P, 4, E ) . Then

3 0 ~ r J and J’ are Arnold’s I and J .

8.5. OUTLINE OF RIGOROUS THEORY 205

= €$(&I) + O(E2) = E$(P) + O ( 2 ) . First compare .? = c$(J’) with P = cg(P) + O(c2). Starting from J’(0) = P(O), it is clear that ( P ( t ) - J’(t)l < ecs for 0 < t < l/e. On the other hand, IJ(t ) - P ( t ) J = elk( < ECZ, since JgJ < c1 and so lkl < (27rcl/c) + (27rcl/c) = 47rcl/c = C Z . Then

IJ(t) - J’(t>l 5 IJ(t) - P(t)I + IP(t) - J’(t)l < cc2 + ccs = eco

for 0 < t < t /e . * * * * * * * * * * Let u9 apply this to the oscillator with w ( t ) = wo + A, X = et , for which

f(J,$,X) = sin(24)/2(wo + A ) and g(J,+,X) = - J COS(~+)/(WO + A ) . Since in this case g(J ‘ ) = 0, we have .? = 0, J‘ ( t ) = J’(0). Therefore .? = 0, J ’ ( t ) = J’(0) . The theorem yields

for 0 < t < l /e.

206

a.

0 ’

4zz

CHAPTER 8. ACTION-ANGLE VARlABLES

L

c 3

?I db a,

4

1 P

Figure 8.4: Kepler problem in three dimensions

8.6 Chapter 8 problems 8.1 Apply the material of section 8.1 to the simple pendulum,

8.2 A particle of energy E bounces elastically to and fro in a one-dimensional box of size L (see figure 8.3). Find the expressions for the (p, q ) + (J, 4/27r) canonical transformation, and a generating function. Which one?

8.3 Perhaps this should have been one of the chapter 7 problems. We place it here in preparation for problem 8.5.

To treat the Kepler problem in three dimensions choose a system of Euler angles (cp, 19,$) such that 1 = 1 6; and r = T &{, see figure 8.4. Denoting by li (i = 1,2,3) the components of 1 with respect to the unprimed


axes, show that (i) [(o,ls] = 1 (not to be confused with [$,/a] = l), and (ii) [WI = 1.

8.4 Perform the calculation leading to equation (8.15).

8.5 For the Kepler problem in three dimensions with U = -(A++) - (h/r2) , find the periods Tr, T ! , and Tv.

8.6 For an electron in the field of a nucleus of charge +Ze, the relativistic Hamiltonian is H = c d m - Z e 2 / r . (i) Find the energy Erel of closed orbits (0 < Erel < moc2) as a function of Jp, Jg, and J4. (ii) Check your result against the non-relativistic expression.

208 CHAPTER 8. ACTION-ANGLE VARlABLES

Solutions to ch. 8 problems 58.1 For 0 < E < 2mge we find

where

The period is

The change of variables t9 = 2 sin-' (sin(eo/2) sin $J) casts this expression into the standard form

. = 4 u J - d$J

with

Using the generating function IC' = sin'((~o/2) = ~ / 2 m g t .

8

G2(B, J e ) = e & 1 JE(J8) - 2mg!sina(B'/2) do'

yielding 9 -is0 ' d$'/ d- - s:d$J'/d- '

Note that the expression for 9 might have been found more simply by noticing that 4 must be constant and pe = me't9. This requires that

(C = constant) ,

Then $ 1 2 ~ = C/S?rml' is equal to 1/T = dE/dJe for C = 2?rme2 dEfdJe. Thus

2 d E


Figure 8.5: Function q(d), problem 8.2

in agreement with the expression previously found. The cases E > 2mge and E = 2mge are left to the reader.

S8.2 Clearly J = A(E) = 2 e m , T = dA(E)/dE = 2 e / v , where v is the velocity. The canonical transformation is

p = J / 2 e , = for 0 5 4 5 T , p = - J / 2 e q = e(2n - d ) / n for 7~ < 4 5 2n . I

It is convenient to use the generating function G3 = - p q(4), where

e 4e Q, C O S ( ( ~ ~ + ~ ) + )

( 2 n + 1y Q(4) = 5 - FC

nmO

is the function shown in figure 8.5. Then 4 = -aGs /aP = 4(4,

J = - a G s / a ( + / 2 ~ ) = 2 ~ p dq(4)/d+ , p = =tJ/Zt , H = p 2 / 2 m = J 2 / 8 m e 2 , .i = - a ~ / a ( 4 / 2 ~ ) = o ,

T = 2 n / 4 = 2eiv . = a H / a J = J /4me2 = e-ld- ,

S8.3

In figure 8.4,11> 0 , Z z < 0, sin cp = 14- > 0, cos cp = - 1 2 / ~ - > 0,

(i) Since cp = - t a n - ' ( 1 1 / 1 2 ) , we have

11 = 1 sincp sin29 , 62 = -1 cos sin29 , 13 = 1 cosd . tan y, = -11/12 > 0.


(ii) It is not surprising that [ $ , 1 ] = 1 , since 1 is the generator of infinitesimal rotations around 1, and 3, is an angle around 1 in a plane normal to 1. If we want to prove it the hard way, we can use the expression

where 2j and lj are the components of r and I with respect to the unprimed system. Thus

ll[XZ, I] - l2[Xl, 11 - -x& + I:) + ls(l1x1 + 12x2) r sin+ d- - - rl sing J- [+,4 = - 9

which leads to the desired result by expressing the coordinates and the angular momentum components in terms of the Euler angles.

S8.4 With the notation of section 2.4 we have p, = -1 ds/dO = la(sl - 8 2 ) sin(af3)/2 = laJ(s1 - s)(s - 92) )

where R = a + bs + cs', a = - 8 1 s ~ < 0, b = 8 1 + 8 2 > 0,

Using eq. 2.267.2 in I.S. Gradshtein and I.M. Ryzhik, Table of integrals, series, and products(Academic Press,l980) pp. 81 to 84, one finds

2 c = -1, A = 4ac - b2 = -(s1 - 8 2 ) < 0 .

etc.

58.5 J , is given by equation (8.15), JIp = 27r 1 3 , and JJ, = 27r 1. The Hamiltunian is

p a l 2 k h 27r2mk2 2m 2mr r2 ( J , + J - ) 2 '

+2----=- H='

The periods are

We expect the last since p is a constant determining the position of the orbital plane.


S8.6 (i) cap2 = (Erei + Z e 2 / r ) 2 - (rnoc2)>”, and, with p a = p : + 12/r2,

(ii) Putting Erel = moc2 + El we have E = p 2 / 2 m - k / r - h / r 2 + . . . with k = Ze’ and h = Z2e4/2moc2. Comparing with equation (8.17), we have

which agrees with the expansion of E,,I found in (i).

Chapter 9

PERTURBATION THEORY

Perturbation exp~sions are presented e m p h ~ i ~ i n ~ their ~ i m i I ~ i t y to those of Quantum Mechanics.

9.1 The operator $2

Let HCp,q) be a Hamiltonian not explicitly dependent on t . Therefore H(p( t ) , q ( t ) ) = H(po,qo) , where po = p ( 0 ) and qo = q(0). As we know, the evolution of a function fb, g) is governed by the equation

dt (94 -==Uf df = [ f , I i ) I )

Unfortunately, the operator

requires the knowledge of p ( t ) and q(t) , i.e. the solution of the dynarnical problem.

In this section we shall express f (p, q) in the form

~ ~ ~ t ) , q( t ) ) = e ~ ~ ~ ~ , Qo) 2 (9.3)

where the operator R acts on the initial values (i.0, a). This is remi~iscent of the Schroedinger-picture formu~ation of Q u ~ t u m

Mechanics, in which the wavefunction at time t is expressed in the form

213

2 14 CHAPTER 9. PERTURBATION THEORY

+(q, t) = exp(-itHCp, q ) ) ~ ( q , ~ ~ , with p and q time-independent operators. (Of course, f ~ ~ t ) , q(t)) has nothing to do with the wavefunction. If we feel the need of a q u ~ t u m mechanical analogue, ~ ( ~ ~ ~ ) , g ( t ) ) can be aaid to correspond to a Heisenberg-picture operator.) Similarly, R is an operator acting on the initial values of the canonical variables and, therefore, time- independent. To save space we shall write z €or (p(t) ,q(t)) and 50 for (PO, qo). Then equation (9.1) can be written more precisely in the form

where the Poisson bracket I., 4, is understood to be in terms of p and q,

(9.5) We now use the property of the Poisson brackets [f, qlp,* = [f, glP,*,

where ( P , Q) are canonical variables like (p, q). Since 030, qo) and @(t), q(t)) are related by a canonical transformation (the Hamiltonian flow), we can write

Then, since the Hamiltonian is a constant of motion, we can also write

(9.7)

where s(zo,t) stands for @Cp~,go , t ) , g~o ,qo , t ) ) , namely p and g as functions of the time and of their initial values. Taking t = 0 in both sides of (9.7), we have (T) = [f(”O,H(50)1,, * (9.8)

t=O

We define the operator R by the equation

A trivial extension of this formula reads

9.2. PERTURBATION EXPANSIONS 215

(9.11)

€%ample: For H = p (a dimensional constant factor omitted) we have

"fbo,9o),pol,Pol = O2f )... (9.13)

a=- 0 , et* 1: et& . (9.14) Hence

&o Example: For H = q we have 52 = -O/dpo. These two examplea are reminiscent of the quantum mechanical momentum

and coordinate operators in the coordinate and in the momentum representation, respectively.

Example: For H = (p2 + ~ ' w 2 q ~ ) / 2 ~ one finds QPO = lpo, Hbo,qo)l= -oIuJzqO, Qqo = ~ / ~ ,

The powers of the matrix (a = w )

HenCe

(9.15)

(9.16)

(9.17)

In terms of J and 4 one has H = wJ/2n, QJo = 0, 5240 = [4o,wJ0/2n] = w, a"& = 0 (71 > I), J ( t ) = exp(tQ)Jo = Jo, $( t ) = exp(tW0 = 40 + wt .

9.2 Perturbation expansions Let H = H(*) + problem for H(Of alone, while H(') is a perturbatio~ (IH(')l < H(O)I),

initid values (PO = ~ ~ O ) ( O ) , q o = q ( O ) ( O ) ) . Thus Q.(o) such that

We assume that we can solve exactly the dynamical

The ~ ~ ~ ~ o n i ~ flow for H(O) yields the functions ~ ( O ) (t) , do) (t)) from

eto"fO)f(Po,qo) = f ~ ( ' ) ( t ) , q ( O ) tt)) (9.18)

216 CHAPTER 9. PERTURBATION TffEORY

is a known operator.

H = H(O) + H(l) . S ~ ~ t i n g from the same initial values as in the unperturbed problem, we write

Let 0 be the corresponding operator for the totd Hamiltonian

e w m , 90) = f(P(t)l d t ) ) * (9.19)

where tn --tR(O) S(t) = e e (9.21)

Differ~ntiating with respect to t we have

where (9.23)

The subscript “I” stands for “interaction” to remind the reader of the kin- ship of this method with the “interaction picture” of Quantum Mechanics.

Note that, in contrast to Q(O), RI is a function of time. The differential equation for S(t) and the initial condition S(0) = 1 can be embodi~d in the integra~ equation

ct

S( t ) = 1 +lo S( t ’ )~ i~ t ’ )d t ’ . (9.24)

9.3. PERTURBED PERIODIC SYSTEMS 217

t etfnfo) cos(w(t - t'))dt' = -7 cos(w(t - t'))dt' I"

= - (7 /~)s in(w~) , (9.29)

et'"''(S1 - do))e-"""'dt') q ( O ) ( t ) = --(1 7 - cos(wt)) . (9.30) (I" mw2

Higher order terms of the perturbation expansion give zero. Therefore

p(O)(t) - (7/w) sin(wt)

Note that the ~ a m j l t o n j ~ H = fp2 + ~ ' w 2 ~ ~ ~ / 2 r n + 7q differs only by an additional constant fiom the Hamiltonian

(9.32)

for a displaced harmonic oscillator, HD = H + 72/2rnw2. Hamilton's equations

are sati&ed by our perturbation solution.

9.3 Perturbed periodic systems In section 7.10 and elsewhere, we glibly stated that all one had to do wi19 to find a canonical transformation to new variables (Pi, Qi) with the Pi's ~ n ~ t ~ t s of motion. This is possible only for a small number of exactly integrable systems.

This section deals with cases where the Hamiltonian H ( p , q) differs by a small perturbation from a Namiltonian HO for which exact periodic solutions of ~ ~ ~ j l t o n ' s equations are known.

We consider a periodic system in one dimension with

H(Jo,$o) = HO(J0) + XHl(J0,d)OPo) (1x1 < 1) * (9.34)

218 C ~ A P T ~ ~ 9. P ~ ~ T U R 3 A T ~ O ~ THEORY

The ~ami~tonian HO depends only on Jo. If the second term were missing (A = 0), the frequency wouid be

and the energy &(Jo) = Ho(J0) . For the harmonic osci~~ator

P2 mw&? WOJO Ho = - f - = - 2m 2 2% ' (9.36)

wo does not depend on Jo, and p and q are the well-known functions of

Our pro~ram is to find new variables (J, #) such that the ~~perturbed" Hamiltonian

is a function of J only with an error O(X"), W O ( J, #I, @o(J, #)) = K ( J , #I (9.37)

H(Jot#o) = K ( J ) + wn> (9.38)

The per turb~t i~n ~ ~ i ~ t o n i a n HI (Jo, &) = H i ~ ~ J o , 40), q(Jo,&)) is The greater n the better is the approximation.

by sumpt ti on periodic in & and can be expanded in Fourier series'

H i ( J o , ~ ) = zh(k Jo)eikbo (9.39) k

(9.40)

Note that the average of f i x over the unperturbed motion is

(4) = hl(0, Jo) ' (9.41)

We now perform a canonical tr~nsformatio~ with a generating function of the G2-type

7 (9-42) ~ S i ( ~ o , J i )

f% Jo = J1 i- 2 r X

'For typographical convenience we do not write k aa a subscript of hi . Our notation

a f ~ = 2 ~ ~ s ( 4 0 , ~ 1 ) / ~ + 0 , S(40tJi) = (40/2n)Jr f X S i ( + o , f ~ ) , where ( ~ o / Z ~ ~ ~ ~ is hi(k, Jo) does not imply that k is a continuous variable.

the generating function of the identity tra~formation.

9.3. PERTURBED PERIODIC SYSTEMS 219

(9.43)

Choose S1 so that

and (9.46)

Then

= Ho(J1) + Ah1 (0,Jl) + 0 ( X 2 ) 1

H ( J o , b) cv Ho(J1) + X(H1 (Joy $o))Jo+ J1 + o(X2) * (9.47)

If we want the approximation quadratic in X we must go one step further with the expansion in powers of A. Neglecting terms O(X3), we have

220 CHAPTER 9. PERTURBATION THEORY

where

We now perform a second transformation (Jl,41) + (J2 ,42 ) etc. If we are content with the approximation quadratic in A, we just rename (J1,41) + ( J , 4) and go no further. We then have

H ( J o , 4 0 ) = H o ( J ) + X(HI(J, $1)

where

Suppose the Hamiltonian depends on several action-angle variables

H(Jo, 40) = Ho(Jo) + ANl(Jo, $0) , (9.48) (Joi , $0,) (i = 1, .

where Jo = (Joi,. . . Jon), 40 = ($01,. . . & o n ) , and

n) ,

Hi(Jo,#o) = Chi(k ,Jo) exp(ik - 40) . (9.49) k

Requiring

one finds

and that the generating function is

Ho(Jo) + XH1(Jo, b) = K(J) + @A2)

K(J) = Ho(J) + A w l (Jo, 40) )

(9.50)

(9.51)

where wo( J) = (v, . . . -)

a Jn (9.53)

The generating function diverges if the zero-order frequencies woi = a H O / a J i are commensurable, i.e. there exists a relation miwoi = 0 with mi integers. In this case, perturbation theory is unsuitable for the integration.


9.4 Chapter 9 problems 9.1 Do the c~culation in section 9.2 ( ~ x ~ p ~ e ) using action-angle variables.

9.2 Oscillator with time-dependent frequency using action-angje variables.

9.3 In section 9.3 we only needed to assume the existence of SI($O,&). Find an explicit expression.

9.4 Let us apply the method of section 9.3 to the harmonic oscillator with the perturbation

9.5 Express the Hamiltonian W = $/2 -t- w2(1 - cos8) for the plane pendulum (8+ w2 sin8 = 0, w = m) in terms of the harmonic oscillator variables

and find the 0; correction to the period.

9.6 Ekpressing the coupled-oscillators Hamil tonian

H = (b; + w:qT)/2 + (pi + ~ ~ q ~ ) / 2 + A9192

in terms of harmonic-oscillator variables, we have

F'ind K(J1, J2) to the second order in A.


Solutions to ch. 9 problems s9.1

L t = - 2 7 d - l et'n(o) cos40 dt' = - 2 7 d - l cos(40 + wt ' ) dt'

+(t) = $0 + 7 J 7 Z z [cos 40 - cos(40 + w t ) ) + Check that Hamilton's equations for J and 4,

J = - a ~ / a ( 4 / 2 . r r ) = - 2 7 J G cos 4 and

are satisfied in the present approximation. = aH/aJ = w / 2 x + \7/2JZi' iZ ] sin 4,


in agreement with eq. (3), solution of problem 13.9 in G.L, Kotkin and V.Ga Serbo, Collection of Problems in Clacsical Mechanics (Pergamon Preaa,l971).

224 CHAPTER 9. PERTURBATlON THEORY

S9.4 We find at once that (HI) = 0, and so there is no correction to the Hamil- tonian linear in A. Furthermore, since wo is a constant, d w o ( J ) / d J = 0, and so we have only

Since

and

we have W O J X2 1 5 J 2 2w 16w2m3w04

K(J ) = - -

,

+ 0 ( x 3 ) . Let us go back to p and q. From the unperturbed solution we have

p = p ( J 0 , 4 0 ) and q = q( Jo, $0). In these formulae we must express JO and 90 in terms of J and 4.

For the harmonic oscillator with the perturbation AH1 = Xq3, in the approximation linear in X we have

4 = /=sin40 wmwo = J-sin (4 - 2nX-

XJ sin 4 - y ( 3 + cos(24)) + 0(X2) . wmwo 2wm wo

Similarly one finds


Check that 07' + m2w2q2))/2m + Xq3 = woJ/2n + O(Xz), and, with [+, J ] = 2n, that [q,pJ = 1 + O(X2).

S9.6 We have

H = - W JOW z 40+wa [ I-cos (c - sin40 )] =--- JOW J' sin440 +. . , 2.rr 2 n 4!n2

From action-angle perturbation theory

J w J 2 2 n 64x2

K ( J ) = - - - -/- .

Substituting J = nwS& where 00 is the amplitude, we have

a H lj= 2 n - a J = fi(1- 16

4 S9.6 We have

w i J i + wz Jz 2n

Ko(J) = Ho(J) = 1

Finally aK x2

41 = 2 n a =w1+ 2Wl (w; - W X ) '


in agreement with problem 7.5,

w: + w.; + J(w: - w;)2 + 4x2

2 sz1 =

u J z - w l + A2 w1 - w; - 2w, (w; - w;,

Chapter 10

RELATIVISTIC DYNAMICS

Rela t i~ t i c ~ i n e ~ a t i ~ and dynamics of a partide, Lorentz tr~sformations and their connection with the SL(2) group, are presented first. The Thomas effect is discussed The equations of motion of a charged particle with magnetic moment in an electromagnetic field are estabrished. The chapter ends with the Lagrangian and Harniltonian equations of a charged particle.

10.1 Lorentz t r a ~ f o r m ~ t i ~ n s Consider an inertial system S characterized by a system of axes XYZ and by a clock C. Let d = 5, z2 = 21, x3 = z be the coordinates of a point particle with respect to XYZ at the time t shown by C.

We regard a? = ct, d, z2, and z3 as the components sp (1.1 = 0,1,2,3)1 of a 4vector

X =

x3

(10.1)

Let x” ( p = 0,1,2,3) be the space-time coordinates of the particle with reapect to another inertial system S’. ‘Here c is the speed of light. In the following, Greek indices take the values from 0

to 3, b m a n indices from 1 to 3. The sum convention will c o n ~ ~ n u ~ I y be used for both kinds of indices.

227

It is well known that x” and x p are related by a Lorentz transformation

= A H vx’’ (10.2) 2 such that the expression (so)2 - - ( x 2 f 2 - ( x 3 ) is invariant,

( p ) 2 - ( x t y - (@) 2 - ( x ‘ 3 f == (.0)2 - (.‘)2 - (x”2 - ( 2 9 3 . A transformation in which S’ = S, ti = t , while (zi,z2, s’) and (d1,x’2,x‘s)

are spatial coordinates with respect to X Y Z and X’Y’Z’, the latter obtained from X Y Z by a rotation, will be regarded as a Lorentz tr~formation. The group of space rotations is a subgroup of the Lorentz group.

The invariant expression can be written more concisely as

(xo)2 - - (sy - fx3)2 = g,@xfiXv = x%, . (10.3)

Here gyp are the components of the metric tensor

goo = 1 , 911 = ~ 2 % = g33 = - I ,grtv = 0 if p # Y (10.4)

and

x p = gbVxv , i.e. xo = x* , zi = -xi (i = I, 2,3) . (10.5)

x, and xfi are called the “covariant” and ‘‘contravariant” components of x. The contravariant components 9,” of the metric tensor can be defined

M solutions of the equations

gfiPgPy = gf’ = (1 if 1.1 = v , 0 if 1.1 # u) (10.~)

These conditions make it possible to invert equation (10.5) to give

x/” =gf ivxv . (10.7)

is defined in equ~tion Of course, one sees at once that ghY = g P y , while g@ (10.6).

(10.8)

is an object s a t i s ~ ~ n ~ all the definitions and relations written above for the space-time coordinates,

at@ = h@ , a, = gcIvuV @La’@ = u,cP.. . (10.9)

A tensor T has c~mponent~ T”” transform~ng ~ c o r d i ~ g to Tfi’ = APPAu,T’@u. Other components are TL”, = TPPgpu, T,” = g,pTpv,

10.2. DYNAMICS OF A PARTICLE 229

10.2 Dynamics of a particle The Pvelocity of a point particle is a 4-vector v with components

Here

(10.10)

ds = c - ' ~ ( d ~ O ) ' - (dx')' - (dz2)2 - ( d ~ ~ ) ~ = e-'J- (10.11)

is an invariant representing an infinitesimal time change as measured by an observer moving with the particle. In fact, if dz' = 0 (i = 1,2,3) and dxo > 0, ds = c-'dxo = dt.

If v = (dx/dt, dyldt, dnldt) is the familiar velocity vector, we have

. hi dzi - yc , v' = - = "I- (v' = "Iv,, . - .) , (10.12) @=-- dxo ds ds dt

(10.13) V

v = 1vI . 1 1 P

where Y = d x ' P = - c '

In a fiame in which the particle is at rest, one has

v o = c , v ' = O ( i = 1 , 2 , 3 ) . (10.14)

The norm of v is vpv@ = c2 in any inertial frame. The acceleration 4-vector is defined as

dv a=--- , ds

Differentiating v p v ~ = c2, we find

(10.15)

aavP = o . (10.16)

We now want to establish the relativistic version of Newton's second law. We start from the experimentally established formula

dP d - = -((nz"Iv) = f dt dt , (10.17)

which tells us that the relativistic momentum is given by the product ma~ls x v e ~ ~ t ~ , but with a velo~ty-dependent mass

m(v) = my = (m = rest mass, constant) . (10.18)

230 CHAPTER 10. RELATZVlSTZC' DYNAMICS

The rest mass was previously denoted by rno. Using dt/ds = 7, we can write

(10.19) d - (my) = yf or mui = f (f' = rfz,. .) .) , ds

From a%yc = 0 we get

ma 0 crj = -ma%$ = -fit14 = 7'(fSvz + . . .) = y2f 9 v . (10.20)

Hence

and so

where

ma=f ,

k( c-$*v j (10.21)

(10.22)

(10.23)

is the 4-force. In terms of a 4-momentum with components po = mcy and pi = mui (pi = my^, , . . .), we have

The relativistic energy is E = cpo = mc2y (E = rn2 +mv2/2+ ... for PI << c).

As we would expect, its rate of change equals the power,

(10.24)

Longitudinal and transversal masses: Suppose f is parallel to the velocity v, f = f r l . Then d(~yv}/d t = fir gives

d. dv dt dt rn-v+my-=q , (10.25)

which tells us that dv/dt is parallel to v, and so dv/dt = (v/v)dv/dt. Since 1 - j3' = fly2, -2pd@/dt = -(Z/y3)d7/dt, dy/dt = (vq/c2)d@/dt, we have

vdv v2 v dv m-- v dt ( 7 3 c2 +.) = = f i t , ( rn73)yz = fil '

10.2. DYNAMICS OF A PARTICLE

v dv mil-- v dt = fll 1

where rill is the “longitudinal mass” mll = myS. If f is normal to v, then

d7 dV

dt dt rn-v + my- = f i

horn dE/dt = f . v = f i . v = 0, we have dy/dt = 0. Then

dv ml- = f i dt

with m l = my (“transversal mass”).

Since fl = mdx” f ds, we see that the tensor Orbital angular momentum

LP” = x@p” - X”Y

satides the equation

-- -xpfu-xYfp . dLPU ds

With 7 = dt/ds and f = y fi etc. , we have

23 1

(10.26)

(10.27)

(10.28)

(10.29)

(10.30)

(10.31)

where

On the other hand L ’ ~ = x1p2 - x’p’ = my(xvv - yv,) . (10.32)

(10.33) -- -xOfl-xlfO . dLo‘ ds

For a free particle this gives

ctp’ - xE/c = const , x = (c2p’/E)t + const = v,t + const . (10.34)

232 CHAPTER 10. RELATIVISTIC DYNAMICS

10.3 Formulary of Lorentz transformations Let a and b be two 4-vectors, and L a Lorentz transformation matrix,

Aoo Ao, Ao2 Ao3 A'o A', A', A', L =

The product a,b@ = g,,,apb" = aTgb

with 1 0 0

g = ( ! ; ;1 jl) must be invariant under Lorentz transformations,

aITgb' = aTgb .

Substituting b = Lb' and aT = alTLT, we find

LTgL=g .

Taking the (0,O) element of both sides of LTgL = g, one finds

2 (Aoo)2 - (A'o)2 - (A20)2 - (A30) = 1

horn equation (10.38) and g2 = I , we have gLTgL = I,

L-' =gLTg ,

Aoo -Ale -A20

-Ao2 A12

-Ao3 A', A23

If the elements of L-' are denoted by we have

(10.35)

(10.36)

(10.37)

(10.38)

(10.39)

(10.40)

(10.41)

(A-') 0 = Aoo , (A-') 0 = - A i , , (A-') i = -Aoi , (LI - ' )~~ = Aji .

Equation (10.39) for L-' gives

(Aoo)2 - (Ao')' - (Ao2)' - (Ao3)2 = 1 . (10.42)

10.3. FORMULARY OF LORENTZ TRANSFORMATIONS 233

How many conditions does LTgL = g impose? Since both sides are symmetric, just 4 + 6 = 10 conditions imposed on the 16 elements of L. Hence a Lorentz transformation depends on 6 parameters.

For these Aoo = 1, Aoi = Aio = 0. The remaining elements Ai j can be expressed in terms of three parameters. We can use the formulae of section 4.1, replacing r by x, ni by ni ', and extending the Ji's to the 4 x 4 matrices

Space rotations

These matrices obey the same commutation relations as their 3 x 3 counterparts, [J1, Jz] = -J3 and cyclic permutations. For finite rotations we have also x = exp(-J(ii)+)x' with J(ii) = n'Jc.

Pure Lorentz transformations are those for which L is symmetric (A"" = A" C ' ) Aoi = A', (i = 1,2,3) are not all zero, and

It follows that L depends only on the three parameters A',, A',, and A3,, and can be expressed in the form

Pure Lorentz transformations

A,, = J1+ (A',)' + ( ~ 2 , ) ~ + ( ~ 3 , ) " .

(10.43)

From the elements Aio one can construct the unit vector

hi, (10.45) n ' = - n -

f - J- ~~~

chapter 4 we did not need to distinguish between covariant and contravariant indices. Here n1 = n,, n2 = nu , n3 = nz .

34j ia the Kronecker symbol (611 = 622 = 633 = 1, 6ij = 0 if i # j).


and the relative velocity

With P = u/c and 7 = l / d m , we have

(10.46)

A % = 7 , Aio=P7n' , Aij = bij + (7 - 1)n'nj . (10.47)

If you are not satisfied by i and j not being in the same position in the

(10.48)

More explicitly the matrix for a pure Lorentz transformation with ve-

two members, write

with g', defined by equation (10.6).

locity u =(un2,uny,un,) , n2 = n' etc. , is

i - i A - g j - (7 - l )n in , 3

L(u) =

(10.49) Special case: The X'Y'Z' axes are parallel to the X Y Z axes. The origin

0' of the X'Y'Z' axes moves with velocity u with respect to X Y Z in the positive z' direction (u > 0, n2 = 1, nu = n, = 0):

(10.50) P7 7 0 0

0 0 1

x = L(u, 0,O)x' gives

xo = r(z'0 + Pz") , 2' = T(2" + Pz'O) , x2 = d 2 , 2 3 = 2'3 ,

2'0 = r(z0 - pz') , 2'' = 7(z' - PZO) , 2 ' 2 = 2 2 , 2'3 = 23 . For u << c, << 1, do = xo, z'l = x' - ut, x f 2 = z2, zf3 = x3 (Galilean

transformation). It is useful to express P and 7 as p = tanh a, 7 = cosh a, /3r = sinh a.

Then

Aoo = cosh a , A', = n'sinha , Aij = 6j.j + (cosha - 1)n'nj . (10.51)


The components a@ of a 4-vector with respect to S are related to the components alp with respect to S‘ by the formulae

(10.52) a” = cosh a a’ + sinh a (njaj) , .Ii = ai - n’[sinh a: ao + (cosh a - l ) (njaj)] .

Remember that njaj = -(n,a’ + . . .). Perform two Lorentz transformations in the same direction, say

coshal sinhal 0 0 0 0 sinhal coshal 0 0

0 1 0 0 0 1

and L(ua, 0, 0). We find

cosh(a1 + a2) sinh(a1 + a2) 0 0 sinh(a1 + az) cosh(a1 + a2) 0 0

0 0 1 0 0 0 0 1

L(Ul,O, 0) L(u21010) =

(10.53)

with

P1 +P2

O = l+PlP2 a

It is easy to remember that when performing two pure Lorentz trans-

A pure infinitesimal Lorentz transformation with u = ii 6a can be formations in the same direction one must simply sum the a’s.

expressed in the form

L(ii6a) = I + n’Ki 6a = I + (nzK1 + . . .)6a , (10.54)

where the Ki’s are the symmetric matrices

0 1 0 0 0 0 1 0 0 0 0 1

K l = ( 1 0 0 0 ) ( o 0 0 0 ) ( o 0 0 0 ) 0 0 0 0 l K 2 = 1 0 0 0 l K 3 = 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0

CHAPTER 10. RELATIVISTIC DYNAMICS 236

Hence for a finite transformation we have

(10.55)

with K(B) = niK,

same result irrespective of the order in which they are performed, While two pure Lorentz transformations in the same direction give the

L(Ul)L(UZ) # L(UZ)L(Ul)

if u1 and u 2 are not parallel. This is similar to the fact that two successive rotations around different

axes yield a result that depends on the order in which they are performed. However, while the product of two space rotations is a space rotation, the product of two pure Lorentz transformations in different directions is not a pure Lorentz transformation.

This is connected with the fact that the Ki’s do not form a closed algebra. One has

(10.56)

(10.57) [ K I , Ka] = J3 , [Ki, K j ] = Cij&Kk 7

[Ji, K2] = -K3 , [Jz, K i ] = K3 9 [Ji, Kj] = - c i j k K k 7

and the already known

[JI, J2] = -J3 and cyclic permutations. (10.58)

It can be shown that

where L(u) is a pure Lorentz transformation and R is a space rotation.

shall prove the approximate formula with an error of the order of lSuI2 In section 10.4, using the SL(2) representation of the Lorentz group, we

L(u -I- SU) L(-U) N L(SW) R(BS4) (10.60)

with 6w = y2Sull + ySul, where dull and S u l are the components of 6u parallel and normal to u, and

(10.61)

Here is a quick check for u << c (7 N- 1) by using the Hausdorff identity

A B - A+B+$[A,B]+ ... (10.62) e e - e


We have

exp (s(ui + 6u’)Ki) exp ( - $ ‘ K i ) N exp ( $u’Ki - -[(d 1 + 6d)Ki, ujKj] 2c2

Other useful relations are

L(-U) L(u + SU) = L(6w) R(-iid$) (10.64)

and

L(u+ SU) R(ii6$) L(-U) = I + y2c-’K * SU + T ~ c - ~ J (U x SU) (10.65)

with K = (KI, K2, K3).

The reader may have wondered why we wrote x = Lx‘ rather than having the “prime” appear on the left side. With our way of writing a pure Lorentz transformation L(u) is a “boost”. If pf0 = mc, pIi = 0 (i = 1,2,3) (particle at rest in S‘), then, taking u in the x1 direction for simplicity’s sake,

Boosts and active space rotations

p 0 =‘ymc , p 1 =‘ymu , p 2 - - -p 3 - 0 - (10.66)

is the momentum with respect to S. The formula a = exp(-J(ii)$)a‘ relating the components of a fixed

vector with respect to X Y Z and X‘Y‘Z‘ (obtained from X Y Z by a counterclockwise rotation through d, around ii) can also be given an “active” meaning. If (a’, u2, u3) are the components of a vector a with respect to XYZ, then ( u ~ ’ , u ~ ~ , u ’ ~ ) are the components, also with respect to X Y Z , of a vector obtained by rotating a around ii through d, dockwig.


10.4 The spinor connection Consider the Hermitian matrix

(10.67) xfO + x13 x l l - ix12 X ' = ( xll + i x 1 2 xlo - x f 3

where a0 is the 2 x 2 unit matrix I, and

(10.68) 0 1 1 0

are the Pauli matrices. These obey the relations

oioj = I6ij + ieijk g k . (10.70)

The Pauli matrices are all traceless. Using this property, we find

( 10.71)

and

Note that

x'fi = (1/2)trace(a,~') . (10.72)

(10.73) 2 det(X') = ( x ' O ) ~ - (z?)~ - ( x ~ ~ ) ~ - ( d 3 ) = x ; x t f i

Let the matrix

M = (1;: :;:) (10.74)

with complex elements have an inverse (det(M) # 0). Then M is an element of L(2), the group of linear transformations in two dimensions.

If we restrict M by requiring that det M = 1, rn l l rn22-rn12rn21 = 1, then M is an element of SL(2) , the group of the special linear transformations in two dimensions.

Consider the matrix X = M X / M + , (10.75)

where M is an element of SL(2) and Mt is the Hermitian conjugate of M,

(10.76)

10.4. THE SPINOR CONNECTION 239

The bar denotes complex conjugation.

quantities

are real. In fact,

X is Hermitian like X’. In fact Xt = MttX’tMt = MX’Mt = X. The

2’’ = (1/2)trace(a,X) (10.77)

Z” = (1/2)trace(~to;) = (1/2)trace(~a,) = ( l / ~ ) t r a c e ( a , ~ ) = x, . Therefore X can be expressed in the form

( 10.78)

But now

det(X) = det(M) . det(X’) . det Mt = I det(M)I2 - det(X’) = det(X‘)

and so

(z‘O)2 - (X’y - (& - ( 2 1 7 2 = ( s o y - ( 2 ’ ) 2 - (22 )2 - (.3)2 . This shows that x’f’ and xfl are related by a Lorentz transformation

xCc = A(M)’,d” . (10.79)

From

s9 = (1/2)trace(a,X) = (1/2)trace(o,MX‘Mt) = (1/2)trace(a,Mt~,M~)s’~

we see that L ~ ( M ) ” ~ = (1/2)trace(a,Ma,Mt) . (10.80)

The following formulae for the traces of products of Pauli matrices will be used

trace(u,) = 0 trace(u,ob) = 26ab trace(uaubuC) = 2ieabc

trze(UaUb0cUd) = 2(&b&d - & d b d + b a d s b c ) . It is easy to verify the correspondence

L(u) = eK(’Ia + M(u) = cosh(a/2) + nisi sinh(a/2) (10.81)

between the pure Lorentz transformation matrix L(u) and the Hermitian matrix M(u). In problem 10.2 it will be shown that

A(M)O0 = cosh a , A( M)Oi = A(M)’, = ni sinh a , (10.82)


A(M)ij = 6, + ninj(cosha - 1) . (10.83)

For a space rotation must be

Since = d o , A(M)O, = 1 , A(M)O~ = A(M)~, = 0.

A(M)O0 = (l/2)trace(MMt) , we see that space rotations correspond to SL(2) matrices satisfying the additional condition Mt = M-l, namely to SU(2) matrices. The special unitary group SU(2) is a subgroup of SL(2), a the group of space rotations is a subgroup of the Lorentz group.

We have

R(A, 4) + M(ii, 4) = cos(4/2) - i ii . c7 sin(4/2) . (10.84)

In section 10.3 the product L(u + Su)L(-u) with lSul << IuI was expressed in the form of equation (10.60). Here we prove that result by using the SL(2) correspondence:

M(u) = cosh(a/2)1+ niui sinh(a/2) = 7+1

7 + 2 67u'a; , CM7 + 1114

M(u+6u) = M(u) +

Since ("T" for Thomas)

L(u + Gu)L(-u) + M(u + Gu)M(-u) = MT , a simple calculation yields

MT rz (I + i 2c2 (7 72 + 1) cr;e,jkuj6uk) (I + Comparing the first factor with

I - i n'u;64/2 + I - n'Ji64

we recognize a rotation around the unit vector in the direction of u x 6u with angle 64 = -[72/(7 + l)c2]lJu x 6111. On the other hand, the second factor can be written as

10.5. THE SPlN 241

where and 6ul are the components of 6u parallel and normal to U. Hence

with 6w = qsu,, + 7sUL *

10.5 The spin A point particle with an intrinsic angular momentu~ (spin) is in non- uniform motion with rkspect to the (inertial) laboratory system K 4, At the time t (lab clock) the particle has velocity U with respect to K . At that time there will be an inertial system K' in which the particle is instantan~usly at rest.

If C ( X Y 2 ) is a system of Cartesian axes in K , let C'(X'Y'2') be the system in K' obtained kom C by a boost with velocity U. The axes of C' are parallel to those of C.

At the time tf&, the particle has veiocity u+du with respect to K , and will be instantaneously at rest in a system K". What system of Cartesian axes are we going to use in K"?

We can either use the system C"(PY"2") obtained from C of K by a boost with velocity U f &U, or a system C'"(X"tY"'2f'f) obtajned from C' of K' by a boost with velocity 6w = r26uil + y h r , where 6ull and 6ul are the com~onents of 6u parallel and normal to U, respectively, and 7 = 1 , f d W . We propose as an exercise to use the formula for addition of velocities to verify that K" (moving with velocity 6w with respect to K') , moves with velocity U + Su with respect to K.

Let a be a $-vector. Its components with respect to K(C), K'fC'), K'(C"), and K"(C"') are related as follows:

a = L(u)a' , a = L(u + 6u)a" , a' = L(6w)a"' . (10.85)

By equation (10.6~), L(-u)L(u f &U) = L ( 6 w ) ~ ~ - ~ 6 # ) , we have also

a' = L(-u)L(u + 6u)a" = L(Sw)R(--iiS#)a" (10.86)

and a = L(u + Su)R(ii6#)a"' . (1O.87)

4We are obliged to denote inertial systems by K rather than the more usual S. In this section there are too many 5"s meaning "spin", and there might be confusion.


Using this last equation, we find

a" = L(-u-Gu)a = L(-u-Gu)L(u+du)R(fidg5)a"' = R(h6g5)a"' . (10.88)

Postulated properties of the spin: (i) The spin S is a 4-vector such that its time component in the rest

frame of the particle is zero (see problem 10.8). (ii) Consider a particle subject to forces and, therefore, in non-uniform

motion, carrying a free spin, namely a spin subject to no torques. Construct a sequence of inertial systems, such that at any time the particle is at rest in one of them. Assume that each of these inertial systems, say that at time t + at, is obtained from that at time t by a pure Lorentz transformation (no space rotation).

Then the components of the free spin with respect to the Cartesian system at the time t + 6t are equal to those with respect to the Cartesian system at the time t.

Let SP (S, = S' = -S1 etc.) be the spin components in a generic system. Property (i) can be cast into the invariant form

UQS, = o , (10.89)

which reduces to SO = 0 if the system is the rest system. Note that with respect to K'(C'), K " ( C f f ) , and K"(C'")

(10.90)

(10.91) s; ( t + at)

S"(t + st> = S p + &) 3

( O ) s:(t + at)

(10.91)

(10.92)

The last two are consistent, since K"(C") and K''(C"') differ by a space rotation.

In the lab system, UPS,, = 0 reduces to uoso - u's, - u2Ss, - u3s, = 0, y(cS0 - u S) = 0,

so = u * s / c (10.93)

10.5. THE spriv 243

Property (ii) implies that5

S"'(t + bt) = S'(t) . Thus S"'fi(t + bt) = S'fi(t).

Rate of change of free spin components in lab system:

S(t + 6t) = L(u + bu)R(fiG+)S"'(t + 6t) = L(u + bu)R(hb4)S'(t)

= L(u + bu)R(hb$)L(-u)S(t) =

having used equation (10.65). This gives

So(t + St ) = So(t) + (7°C) S ( t ) ' du ,

and C

i2 2

S(t + St) = S(t) + +hl+ -@ x (u x Su)

= S( t ) + $ll* S(t))Su + ,,[(S(t) * 6u)u - (u * S(t))6u] , r2 2

- dS(t) = i2 (S(t) * $) u dt

(10.94)

(10.95)

Equations (10.94,95) can be combined in the covariant equation for the free spin (see problem 10.9)

Multiplying this equation by u p , since upup = c2 we have

(10.96)

(10.97)

the 4spin remains normal to the 4-velocity during the motion.

d Note also that

;i;;(S.SC) = o >

as is easily shown multiplying equation (10.96) by S, and remembering that S,u. = 0. The magnitude of S is not constant during the motion.

5Not in contradiction with a"' = L(-Gw)a', since S"' and S' are at different times in the two members of the equation.

10.6 Thomas precession If each of the sequence of inertial systems in which the particle is instantaneously at rest uses Cartesian axes resulting from those of the lab by a pure Lorentz transformation, then the spin appears to precess,

- r2 S’ x (u x a ) , dS’ dt c2((r+ 1) -- (10.98)

where a = du/dt is the particle acceleration with respect to the lab system. In contrast to S, S’ has constant ~ a ~ ~ t u d e during the accelerated motion.

Define 6s’ = S” (t + S t ) - S’(t). Then from

S ” ( t + st) = R(liBq5)Sf”(t + S t ) = R(iiG4)S‘ft)

and (10.61) we have 65’ = -J . fi 64 S’,

ss’ = -S‘ x li 64 = ’y2 c”7 + 1)

s’ x (u x Su) ,

Let us consider the spin precession for a particle moving on a circular orbit, u = w x r, a = -wzr, u x (duldt ) = u2w = [(y2 - I)c2/yz]w,

dS‘ dt

- = - ( 7 - - l ) w x S ’ . (10.99~

By definition, the motion of the particle around w is counterclockwise, That of the spin is also counterclockwise. In one period of the orbital motion, the spin precesses by an angle

1 A’p’ = -27~(y - 1) = -2n (d-- - 1)

(10.100)

(10,tOl)

For simplic~ty’s sake we assume that S‘ lies in the plane of the orbit (zg plane). Then equation (10.99) gives

dS; - dt = (7 - 1>ws; , dS’ -$ = -(y - l)wSf, .

It is easy to verify that 5” = d m is a constant of motion. The solution

S; = S‘ cos((y - 1)wt) , Sh = -S’ sin((y - 1)wt)

(10.102)

(10,103)

10.6. THOMAS PRECESSION 245

Figure 10.1: Spin precession along circular orbit

in conjunction with x = rcos(wt), y = rsin(wt), corresponds to the initial condition at t = 0 with the particle on the x axis and the spin along the x axis pointing in the positive direction.

For t = T, SL = Sfcos(2x(y - l ) ) , SL = -Sf sin(2x(y - 1)). In agreement with (10.100)

tan(Acp‘) = S&/Sk = - tan(2n(y - 1)). What is S doing while S‘ precesses? Notice first that using (10.52) and S’’ = 0

one has

( 10.104)

Then from the above solution for S’ one finds

S, = S’[cos(wt) cos(ywt) + ysin(wt) sin(ywt)] , (10.105) S, = S’[sin(wt) cos(ywt) - y cos(wt) sin(ywt)]

1 ~ 1 ’ = Sf2[cos2(ywt) + y2sin2(ywt)] .

For t = T, S, = S’cos(2n(y - l)) , (S, = -ySfsin(2s(y - 1)). With respect to the lab system, the spin has rotated through an angle &, tan(Acp) = S,/S, = -7 tan(2n(y - l ) ) , and its magnitude has also changed.

Figure 10.la shows what happens for y = 2, u = &c/2. Figure 10.lb is for y = 1.1, ‘u. = 0.42~.

Note that K’(C’) is in motion with velocity u = (0, u) with respect to K ( G ) (lab) both at t = 0 and t = T. One has S, = Sk because u is normal to the x axis. Since Sb = 0, S, = 7s;.

246 CHAPTER 10. RELATIVISTIC DYNAMiCS

10.7 Charged particle in static ern field We study the motion of a point particle with a charge q in a static electromagnetic field of 4-potential

d2xp q dx” q dXU m~ = - FPW - ~ - - 9” FA, - ds c ds c ds

are easily seen to be equivalent to

and

The former gives

(10.106)

(10.107)

(10.lOS)

(10,109)

(10.110)

(10.11 1)

(1 0.112)

the latter dE/dt = qE a u.

10.8. MAGNETIC MOMENT IN STATIC EM FIELD 247

10.8 Magnetic moment in static em field The magnetic moment is expressed as

p = - S ’ SQ . 2mc

(10.113)

A first form of its covariant equations of motion is

Multiplying by up and using uPu@ = C2, we have

= -U SQ S (F’” + Fup) = 0 2mc ’ ”

Using dd’/ds = (q/mc)Ftu”, equation (10.114) can be cast in a second form

- dS’ = -F’LUSu SQ + - ( S p F P “ ~ , ) ~ ’ (9 - 2)Q . ds 2mc 2mc3

(10.115)

For g = 2 the last equation gives in the lab system

dS/dt = (q/mq)[S x B + (l /c)(S * u)E] , dSo/dt = (q/mq) (S E) . ( 10.1 16)

Here are some calculations described by one author as “somewhat lengthy”

By the equation S’ = S - [y/c(y + l)]Sou we find for g = 2 and by another as “some tedious algebra”.

-=- dS’ ‘ [s x B + i(S. u)E] dt mcy C

‘ - mc2(y + 1)

Using S = S’ + [y’/c2(y + l)](S‘ a u)u, we find

-- dS’ - L S ’ x B+ qy (S’ . U)(U x B) dt mcy mc3(y + 1)

+-(S’ 4 * u)E - (S’.E)u mc2 mcyy + 1)

- ’” (S‘ . u)(E - u)u - - ( S ’ s 7 mc4(7 + 1 y C2

248 CHAPTER 10. RELATlVlSTlC DYNAMICS

Since d(yu)/dt = (q/m)(E + (u x B)/c) and

we find - dS' = -S' q x B - ( E . S')u dt mcy mc"(y + 1) - " (S' . u)E + -(S' 4 . u)E ,

mc"y + 1 ) m cz

(10.117)

a formula valid for g = 2. For g = 0 (only for computational convenience, since we do not know of

a charged particle with spin and no magnetic moment) equation (10.115) gives

( 10.11 8)

Now for any g

10.9 Lagrangian and Hamiltonian The equation for a charged particle in an em field

can be derived from the Lagrangian

mdx, dx, qdxv 2 d r d r c dr +--&(x) 7

L = ---

where the parameter r is not necessarily the proper time. In fact

BL BL (B(dx,/dr)) - &? =

(10.120)

( 10.12 1)

10.9. LAGRANGIAN AND HAMZLTONIAN 249

aives

Multiplying this equation by dx@/dr, we find

-- d x f i dxp - - constant. dr dr

0.122

We normalize r so that constant = c2. Thus r = s, the proper time. In the following, differentiation with respect to s will be denoted by a dot. Therefore X,i, = 2.

Some people are disturbed by the fact that the first term in the Lagrangian turns out to be a constant. A cure for this is to use the method of the Lagrangian with constraints presented in problems 6.10 and 6.11.

For the Lagrangian

L = mc’ + (q /c )xpA, with xPxM = c2 , (10.123)

we obtain

mx,, = (q/c)F,,,,xU . Returning to L = (m/2) i# ’ + ( q / c ) i , A , we have

(10.124)

- -A,) (p@ - !A@) , (10.125)

8 L 9 8 X @ C

p,, = - = mi, + -A, ,

1 2m C

H =ppX@ - L = - n

(10.126) m mcL 2 2 .

H = - X p j @ = - Hamilton’s equations read

(10.127)

( 10.128)

I


Hence

Poisson brackets:

[XP,X”I = o , [pfl,Pu] = o , [X” ,P“] =g”, . (10.129)

Hence 1 m

[x,, j.4 = - 9”” , (10.130)

but

The above formulation is based on the canonical form

6‘ = Jp, Adx, ,

in which p , is not a gauge invariant quantity.

(10.131)

The Hamiltonian is invariant under the combined transformations

(10.132)

In fact

p , + -- ” - - ( A , + g) =p, - ;A, Q . caxfi The transformation p , 4 p , + (q/c) dA/dx” is canonical. In fact

while the invariance of the other Poisson brackets is trivial.

10.9. LAGRANGIAN AND HAMILTONIAN 251

Intermezzo: F.J. Dyson, (Am. J. Phys., 58(1990)209) reminisced about the discovery by Feynman that the homogeneous Maxwell equations can be “derived” from Me- chanics.

It amounts to the following:

giving

This requires the Hamiltonian to be of the form

H = (2m)-’(pP + a”)@, + u,) + b(z)

where the scalar field b(z) has been added to Feynman’s formula. Therefore

For a function f (p, q) we have

with 4 q 8A” a + - 1 (p” - ;Ah) Q a . (10.133) UH = - ( p , - -A,> --

mc c ax, &I, m

One may prefer a formulation using the gauge-invariant kinetic momen-

(10.134) Q turn

p; = mx, = p p - - A , . C

For this purpose we start with the Hamiltonian

and the canonical form

(10.135)

(10.136) 4 2c

GI2 = A Jx’ - - F,, Jx’ A axu .


This latter changes the definition of Poisson brackets. Note that the condition d d 2 = 0 gives

FpU,xdxx A dxp A Bx” = 0

(1 /3 ) (Fpu ,x + FUx,,, + FA&,”) dxx A dxp A dxU = 0

F p u V x + F u ~ , p + F ~ p , u = 0

,

, .

Compare this with Feynman’s derivation of the homogeneous Maxwell equations.

If

and

we have

(1 0.137)

(10.138)


10.10 Chapter 10 problems 10.1 A photon of 4-momentum p = ( v , ~ i i ) (c = 1,h = 1) is scattered by a particle at rest, which acquires a kinetic energy K in the collision. Calculate the angle 9 formed by the momentum of the struck particle with the momentum of the incident photon.

10.2 (i) Show that equations (10.82,83) follow from (10.81) and (10.80). (ii) F’rom (10.84) derive the matrix elements for a space rotation.

10.3 (i) Using the SU(2) representation of space rotations, find the rotation which is the product of two space rotations. (ii) Check your result against the product R1 (?r/2)Rz(n/2) calculated by using equations (4.5,6).

10.4 Using the SL(2) representation of the Lorentz group, verify that = L(u)R3(4), where UI = UO&, u2 = 21062, u = UI + U ~ / T O

c = l), and tan+ = -u:70/2.

10.5 (i) Find the generators 6G of infinitesimal space-time translations, space rotations, and Lorentz transformations (PoincarB group). (ii) Find the Poisson brackets of the above (algebra of the Poincar6 group).

10.6 (i) The 4-potential for a constant electromagnetic field FPu is A , = (1/2)z‘Fu, up to a gauge transformation. For a particle of rest-mass m and charge q in this constant em field, verify that the quantities

c, = P, 4- (q/c)A, = P, - (9/2C)F,”ZY

are constants of motion, dC,/ds = 0. Caution: The expressions p , - (q/c)A,, occurring in the Hamiltonian (10.125), differ by a sign from the C,’s. (ii) What is the physical meaning of the (7,’s if the particle is in the electrostatic field E = (O,O, E ) ? (iii) Using the C,’s in the expression (10.126) for the Hamiltonian, by anal- ogy with problem 7.1 show that

C,C. + (2q/c)CpFfi,xu(s) + (q2/c2)xp(s)xu(s)Fp,F/ = rn”’

10.7 (i) In the inertial frame (z‘,g’,z‘,t’) a particle of rest-mass m and charge q > 0 moves in a uniform magnetic field B’ = (0, 0, B‘). Show that if x’ = y’ = z‘ = 0, dx’ldt’ = -d (d > 0), dy‘ldt’ = dz‘/dt‘ = 0 at t‘ = 0, the particle is in uniform circular motion, clockwise, with


Figure 10.2: Spin under sudden acceleration

angular frequency w' = gB'/mcy' (7' = l/d-) and that the orbit equation is

x"+ (y' - T ) = r 2 , z' = o , with T = mcy'v'/qB'. (ii) Perform the Lorentz transformation x' = (x - v't)y', y' = y, z' = z ,

Show that the orbit equation in the (x,y,z , t ) frame is

2

t' = ( t - V'Z/C')~ ' .

(x - v't)2 + (y - rI2/yl2 = Tz/y12 , z = o ,

where mc" E

T = qB2(1 - E2/B2)

and E = y'v'B'/c, B = B'y' are the magnitudes of the electric field E = (0, E , 0) and the magnetic field B = (O,O, B) in the (2, y, z , t) frame.

10.8 Show that s, = C - ' s ' € ~ u p g U " T ~ T ~

has the properties of a spin. Here E , ~ ~ ~ is skewsymmetric with €0123 = 1, u" are the components of the 4-velocity, r1 and r2 are 4-vectors with components (0, l , O , 0) and (0, 0,1,0) in the rest system. In particular show that this S, satisfies equations (10.89,90), (10.93), and (10.104).

10.9 Show that equation (10.96) reduces to equations (10.94,95) in the lab frame.

10.10 Assume that the spin and the velocity are as shown in figure 10.2 (lab system). Show that due to the sudden change of direction of the velocity,


the angIe 'p formed by the spin with the z-axis changes by 6 9 = -y2P2 sin CY cos2y with p = \ul/c.

10.11 Derive equation (10.98) from equations (10.95) and (10.104).

10.12 What is dS'/dt in the following cases? (i) B = 0, E 11 u, (ii) E = 0, B [ I u, (iii) E = 0, B I u (B uniform)

10.13 For an electron (q = -e) in an atom (E = --cIV(r) = -(dV/dr)(r/r)), equation (10,119) gives

S' XI (1=myrxu) . dS' e d V g 2 dt 2m2c2r dr (y l + y ) -=-- ---

Show that for y h* 1 this corresponds to the spin-orbit coupling interaction energy

(g - 1)l' S' . e dV 2m2c2r dr

---

Solutions to ch. 10 problems Sl0 . l Let p' = (u',u'fi'), q = (m,O), q' = (E' = K + m,q') denote the 4- momenta of the photon after the collision and of the particle before and after the collision.

We have p' + q' = p + q, p' - q = p - q'. Squaring the last equation, and using p'a = p 2 = 0, q j a = qz = m2, we have mv' = E'v - vlq'l cos 0,

cos @ = (E'v - mv')/u1q'1 = (E'v - m(m + v - E ' ) ) / v d w = (V + m)(E' - ~ ) / Y J E V ,

c&q@=- v+mJ"TI";; -=- ~~m~~ - v E'+m

510.2 (if With M = M(u), we have

2 A(M)',, = tr(MMt) = tr(M2) = tr([cosha(a/2)+sinh2(cu/2)]1) = tr(cosha1) = 2cosha , 2 A(M)'* = tr(MoiMt) = tr(uiM2) = t r ( ~ i [ 2 n ' ~ j sinh(a/2) cosh(a/2)])

= sinh a dtr(oimj) = 2n'sinh a , 2 A(M)', = tr(uiMMt) = tr(criM') = 2 A(M)Oi ,

2 A(M)'j = tr(aiMcrjMt)

= t r ( c ~ h 2 ( a / 2 ~ u i ~ j fsinh a[U$ (n*m,)Oj +aiaj (ndffd~)j/2+n6ndSinha(a/2fUiasojud) 88f0

= 26ijc0sh2(a/2)+i sinh *(nbcibj d 4- n Eijd)S2 Sinh2(u/2fnbnd(6ie6jd-dij6bd+6*d6jb)

= 2 6ij + 4 sinh2(a/2)n'nj = 2 6ij + 2n'nj(coshcu - 1) .

510.3 (i) We have fcos(#1/2) - i f i 1 - O S ~ ~ ( # ~ / ~ ) ] ~ C O S ( # ~ / ~ ) - i f i ~ . gsin(#2/2)]

= cos(#/2) - i ii usin(#/a) where

COS(#/~) = cos(9+1/2) cos(#2/2) - nl . n2 sin(#q/2)sin(#z/2)

+(81 x ha) sin(#l/2) sin(#2/2)

, hsin(412) = f11 sin(qh/2) cos(#2/2) + i i ~ sin(#2/2) cos(#1/2)

(ii) For ril = (I,O,O), r i a = (O,I,O), #I = $2 = n/2, (i) gives cos(#/2) = 1/2, # = 2n/3, ii = (61 + 6 2 + a,)/&. Hence €&j = (-6,j + 1 - e i j k ) / 2 (k # i and

.


k # j ) , R11 = Ria = R22 = R23 = R3l = RS3 = 0, R13 = R21 = R32 = 1. This agrees with

1 0 0 0 0 1 0 0 1 Rl(7r/2)Rl(n/2)=(0 0 1 0 - I ) ( 0

-1 0 0 1 0 ) = ( 1 0 0 1 0 0 0 )

(x = Rx' : X' = g , ~ ' = Z, Z' = X) .

510.4

S1O.S The generator for xc 4 xlr + Sxfi is 6G = pu6xy. In fact, [xp, dG] = [d',py]dxy = 9': 62" = 6xP.

If L,, = xPpu - xypp, the generators of space rotations are Ji = (1/2)Eijkw", J1 = LZS etc.

In fact, d ~ ' = [x', J3]64 = - X Z & ~ = ~ ~ 6 4 , 6x2 = [x', = xi64 = -~'6$,

6pi = [pi, Js]64 = pd4, 6pz = [pz, Js]64 = -p164. The generators of Lorentz transformations are Ki = Loi = xopi - XiPo. For instance, for an infinitesimal Lorentz transformation dong the l-axis,

X'O = xo + dX0 = xo + [XO, Kl]€ = xo + €Xi, X'l = xi + 62' = xi + [x', Kl]€ = xi + txo. [Pfi,pu] = 0, [ P O , Ji] = 0, [pi, Jj] = -eijkpk,

[PO, Ki] = -Pi, Ipi, Kj] = -PO&, [ ~ 1 , ~ 2 1 = - ~ 3 etc.,

(ii) The Poisson brackets are

[Ji,Kj] = -eijkKk, [Ki,Kj] = EijkJk. Verify that this algebra is realized by putting p , = -8/8x. in all the above generators and replacing Poisson brackets by commutators.


S10.6 (i) dC,,/ds = p p -(q/2c)F,,,x’ = d[mx,,+(q/2c)x”Fu,]/ds -(q/2c)F,,,Xu = mx,, - (q/c)F,,uj.u = 0 (ii) CO = po - (q/2c)FouxY = mci - (q /c )Ez = (Tmc’ - q E z ) / c (total energy divided by c) , C3 = p3 - ( q / 2 ~ ) F 3 ~ x ’ = -mz + qEt = C1 = p l ,

(iii) Since H = mc2/2 , we have c2 = p z

m2c2 = (p,, - qA,,/c)(p’ - qA’/c) = (C,, - 2qA,/c)(CP - 2qA”/c) etc.

S10.7 (i) The equations

imply y‘ =constant. A first integration gives

I dz’ qB’ qB‘ m7 = - y’ - my’v’ , 9

C C

satisfying the initial condition for the velocity components. These can be written in the form dx‘ldt‘ = w’y’ - v‘, dy’/dt’ = -w’x’, with w‘ = qB’/mq‘, from which z’ = T cos(w’t’), y’ = ~ [ l - sin(w’t’)] with T = v’/w’ = mcy’v’/qB’. (ii) xt2 + (y’ - T ) ~ = T’ becomes (z - ~ ’ t ) ~ + (y - ~ ) ~ / - y ” = r2/-y”. Since

we have y‘ = c E / B , 1/-yt2 = 1 - y’2/&’ = 1 - E2/B2,

T = mcy’v’fqB’ = mc2E/qB2( l - E 2 / B 2 ) .

S10.8 In the rest system u’ = (c, 0, O,O), and so Sb = 0 because ~ 0 0 ~ ~ = 0. Only the spatial component S; is different from zero: S; = S’esolz = -S‘, St3 = S’.

In the lab system, uo = rc , u1 = 7uZ, etc., T: = yu3/c , T! = yua,/c, r: = 1 + (7 - l)u:/u2, r: = (7 - 1 ) u 9 u + 2 , T; = (7 - 1 ) U Z U Z / U 2 , T2 1 = (y - l )uzuy /u ’ , T-.: = 1 + (y - l )uJ/uZ, T i = (y - 1)uzu9/u 2 .

s2 = y2(s’u,)uy/c~(7 + I ) , s3 = S’ + 72(S’uz)u,/c2(7 f 1 ) . Using these, one finds SO = yS’uz/c, S Iy = 72(S’uz)7;“Z/cz(y + l) ,

Some authors, for instance J.L. Anderson, Principles of Relativity Physics (Academic Press,l967) p. 250, define the “polarization vector”

where p“ is a component of the momentum, and sup is a skewsymnietric spin tensor with the property s,,,~” = 0. Therefore in the rest system only the space components si j are different from zero. But for trivial factors, w,, is equal to our S,, if

w+. = ( 1 / 2 ) ~ , , ~ ~ ~ s ~ p p ~ ,

sp’ = ( T p ; - T Y T t ) / 2 .


510.9 Expressing S,du”/ds in terms of lab quantities, we have

The rest is trivial.

s10.10 S and u in plane of motion,

From section 10.5, regarding a as small, we have

*om btancp = 6 ( S , / S , ) we find 69 N - ( ~ / C ) ~ I U ~ ~ sina cos2cp. This gives 6cp = -+y2p2 sin0 for cp = 0, and 6p = 0 for cp = h / 2 .

S = (ISlcoscp,(S(sincp), u = (O,lul), b u = (-lulsincu,-IuI(1 -cosa)).

6s = ( ~ / c ) ~ ( s - 6u)u) 65, = 0, &s, N -(y/c)2\S11u12sincu coscp.

Repeating the work for cp‘ one finds bcp‘ = -(y - 1) sin a.

s10.11 Multiply (10.104) by u, finding S.u = y(S’.u). Express S‘ in terms of S and u, obtain dS’/dt in terms of dS/dt, u, and dujdt, use (10.95) and (10.104).

S10.12 (i) dS’/dt = 0

(ii) d(myu)/dt = 0, dS’/dt = (gq/2mcy)S’ x B

(iii) d(myu)/dt = ( q / c ) u x B gives duldt = u x W L , the particle moves on a circle in a plane normal to B with the Larmor angular velocity WL = qB/mcy. Since S’ x (u x ( u x B)) = -[c2(y2 - l)/y2]S’ x B, we have

- = - ( l + @ y ) dS’ q S ’ x B dt mcy 2

This equation is important for the measument of the g - 2 anomaly. For g = 2 u and 9’ are in phase, du/dt = u x WL and dS’/dt = S’ x W L .

S10.13 The correct result is obtained from dS‘/dt = [ S ’ , H ] and the Poisson brackets [SL, Si] = S: etc. Can you justify these latter?

Chapter 11

~ONTINUOUS SYSTEMS

To lust rate the ~ p p ~ ~ c a t ~ ~ n of a n a l y ~ i c ~ mechanics to con~inuous systems, we present two case studies, the uniform string and the ideal non-viscous fluid.

11.1 ~ n i f o r ~ string We msume the reader is familiar with the equation

(11.1)

€or small transversal displacements y(z, t) from the equilibrium configuration (the z-c-axis). The phase velocity is 'v = tl7';t;t, where 7 and p are the tension and the linear density, both constant.

The general solution is

ZI = f(€) + 9(d ? (11.2)

where ( = 5 - vt and 77 = x -i- vt, while f and g are twice differentiable, but otherwise arbitrary.

If f and g are simple-harmonic, the wave number k and the angular frequency w are related by w = vk, and there is no dispersion.

The vibrating string provides a on~dimension~ model for the ~ropa~ation of light in vacuum. Later on we will write equation (11.1) in the form

(8; -d: )g=O , (11.3)

26 1

262 CHAPTER 1 1 . CONTlNUOUS SYSTEMS

where a. = d/8z0, a, = a/ax', zo = vt , and 2' = X. More concisely

a,,oMy=o , (11.4)

Note at once the invariance under zo and x' translations and the Lorentz-like where a. = g@"& (goo = 1, 9" = -1, g" - - g O * = 0).

transformation

z" = cosh a! xo - sinha! 2.' , 2" = cosha! z1 - sinha! xo , a , = c o s h a a ~ - s i n h a 8 : , a, = c o s h a & - s i n h a a h .

In all these cases, a, transforms as a scalar, y'(s'O,~'') = y(so(z'),z'(z')). With at = 8/dt and 3, = a/l?x, define the quantities

Ttt = b ( a t y ) 2 + ~ ( a x y ) ' ] 1 / 2 , Txx = Ttt , (11.5) Txt = -T ax$ 8 , ~ , Ttx = - p at3 &y = V - ~ T X ~ .

Using the wave equation, we find

&Ttt = P atY &tY + 7- aXY a t x v

= T ( 4 Y azxy 4- a,y at,$> = &(78XY 8 t Y ) = -&T,t 7

&Ttt + aXTxt = 0 . (11.6)

This is a continuity equation describing energy conservation. Ttt is the energy density of transversal motion.

Integrating equation (11.6) with respect to x , we have

(11.7)

Thus Txt(a) and Txt(b) can be interpreted as the rates of energy flow into and out of the interval (a, b) . Of course, they may be negative.

For a wave propa~ating in the positive 2 direction, y = f(<), we have

a Txt = -r (s) ( - u s ) = TV(%) > o , (11.8)

while for one propagating in the negative x direction, y = g(q), we have

(11.9)

Tzt is the work per unit time done by the portion of the string left of x on the portion right of 2. In fact (see figure 1.1 ) Tzt = (-7 azg)aty = F,v,, where F,, = -T sin@ 2: -7 tan0 = -T %g, vy =at%.

11 .l. UNIFORM STRING 263

3c

Figure 11,l: Interpreting Tzt

Consider now Ttz. We have

8tTtz = -1.1 at ty ~ Z Y - 1.1 atV a t zy = -I.~v'&~Y - 1.1 atl/ atzV

- ~ z [ T ( ~ z V ) ~ 4- I.1(&%"]/2 = -azTzz

&Ttz + &Tzz = 0 (1 1.10)

The quantity Tt, = -p aty &y is interpreted as a longitudinal momentum density. It is positive/negative for propagation in the positive/negative x direction.

Integration of equation (11.10) over an interval (a, 6) gives

Lagrangian Consider the action

A = L r d t l ' d x C ( y , & y , a t y ) = v - ' ~ d z o d x ' C , ( 1 1.12)

where c = b(ad2 - 7(azY)21/2 = .(a,Y>(a"V)/2 (1 1.13)

is a Lagrangian density. y + 6y with 6y = 0 for t = t l , t = t2

(any 2) and x = a, x = b (any t), or, more briefly, on the boundary of D. From this assumption follows the Lagrange equation

Assume that 6A = 0 under y

(1 1.14)

264 CHAPTER 11. CONTXNUOUS SYSTEMS

or, more concisely,

( 1 1.15)

It is easy to see that this equation reduces to (11.4)

fnvariance properties of L: and conservation laws Consider a Lagrangian density of the form C(y,a,y), where y = y(zo,zl). Assume that C is invariant under dfi = d‘ + dxfi (6s” infinitesimal) up to terms of the second order in 62”. This means

6c = q y ’ , a g ) - qY,aBy) = o ( 1 1.16)

where y‘(z‘O, x‘l) = y ( ~ * ~ x f ) , z1 (2‘)).

For the string, c = (7/2)6,y @@y is invariant under finite xo and 2’

translations, and under finite Lorentz-like transformations. We are inter- ested only in the respective infinitesimal cases, (do = go - E,X’! = s’), (do = z0,xf1 = zt - e), and (do = xo - ez1,d1 = x1 - ezo).

Now dy = y‘(z’) - y(z) 151 by f ( a , y ) 6 ~ ~ , where Jy = y‘(x’) - y(z’) 21 y‘(5) - y(z)

is the change in form of y. Of course, dy = 0 in the present case in which the “field” is a scalar field. Equation (11.16) (dc = 0) gives

Since in the first order &,y = a$y, we have

ac ax” 8c + - dXU = 0 ,

where a,$?/ = 0 *

ac - ac bc=- dy+- aY W , Y > By L ~ ~ a n g e ’ s equation (1 1.15) this can be expressed in the form

11.1. UNIFORM STRING 265

A ~ u m i n g 8,6d = 0, which is trivially true for zo and z1 translations and for Lorentz-like transformations, we have

( 1 1.17)

For so translations do = zo - c (6z0 = -e), y'(zm, d*) = g(zo, d),

Similarly for s1 t r ~ s ~ a t i o n s 8y = e &g. g'(z0 - €, 51) = y(z0, d) , %('(SO, 51) = y(50 + c, d ) , 6y = € 80%

Equation ( 1 1.17) gives

3,Tfiu=0 , ( 1 1.18)

where (1 1.19)

For the string Lagrangian (11.13) this formula yields

Too = Ttt , T" = Tzz =Ti t ,

TO' = VTt, , Po = v-'T,~ , where the quantities Tttl Tax, Tt+, and Tzt are given by equations (11.5). Note that To' = TlO, while Tt, # T,t.

(BtTtt +SZTzt = 0), and &P1+&T1l = 0 is equivalent to equation (11.10) The equation &Po + &T'O = 0 is e q u i ~ e n t to equation (11.6)

(OtTt, + &Tzz = 0). For the Lorentz-like transformation axo = - e d , 6z1 = --ezo,

--e(z'&g + 50B1y) + (-a ) = 0 ,

(11.20) O 4 BC

+& je(als) &[-zO!P' +z1TO0] + &[z~T'* - z ~ T " ] = 0 .

Since BoTaO = &Tal = 0, this gives TIo - To1 = 0. Thus invariance of C under so and x 1 translations, combined with invariance under Lorentz-like transformations, implies Tfi' = T ' p .

~ ~ ~ a t i n g (11.20) and ~ u m i n g that Tfi" vanishes for 2 = foo, we have

i / ( - v 2 t T t , dt + scTtt)dz = 0 ,

266 CHAPTER 11. CONTINUOUS SYSTEMS

Tttx dx = v2t Tt, dx + constant = t TZt dx + constant ) s where the integrals are over the whole x axis.

Since JTttdx = constant, we find for the “energy center”

where P is the total longitudinal momentum.

Define Hamiltonian

BE n(x,t) =

a(atar(x, $1) and the Hamiltonian density

(11.21)

Hamilton’s equations can be derived from the invariance of the action

A = 1.C dtdx = ~ ( ? r 6 t y - 31) dtdz (11.23)

under variations of y and lr vanishing on the boundary of the integration region. In fact, 6A = 0 gives

Using S&y = a&, S6,y = and inte~rating by parts, we have

ax ) 6y] dtdx= 0 .

A ~ u m i n g the in~ependenc~ of 1571. and by, we find H ~ i ~ t o n ’ s equations

(1 1.24)

11 .l. UiVlFORM STRING 267

Poisson brackets etc. Let A and B be two ~nctionals of ~ ( 5 , t ) and gf., t ) . Define the Poisson bracket of A and B as

where the integral is over the whole x axis and & A / 6 ~ ~ ~ , t) etc, denote a “functional derivative” as defined below.

Let #‘[fit] be a functjon~ of f(x, t) , namely a correspondence between the function f(x, t ) of x and t and the function F [ f ( t ] of t alone. The Lagrangian L = J.C(p, atp, as%) dx is a functional of y(x, t ) .

Assume that €or y(z, t) -+ y(x,t) f Sy(x) (not 6y(z,t))

4l.EfItl = ~ A ~ z , t ) M X ) .

We then Say that the functional derivative of F is

For the above Lagrangian

and the Lagrange equation (11.15) can be written in the form

Similarly

where

Writ~n~ p(x‘, t ) =c ly(x, t )S(x - xl)dx, we find at once that


Thus we obtain the Poisson brackets

(11.28)

Similarly Otfl(z, t) = [+, 0, HI - (11.29)

In general, if F is a ~ n c t ~ o n a l of y(z, t) and r(x, t), possibly explicitly dependent on t , we have

For the Hamiltonian dH d H dt at *

-=-

(1 1.30)

11.2 Ideal fluids One-dimensional incompressible ideal fluid

position x and time t obeys the equation In the absence of external forces, the velocity v(a,t) of the fluid at

av av 1 ap - o , - + v -+ - - - at ax p 8x (1 1.31)

where p = constant is the tinear density and p is the pressure" (with the dimensions of a force).

We refer the reader to intermediate mechanics textbooks for the three-dimensional Euler equation

ihr 1 1 - 3- (v. V)V$ - v p = - f bt P P

, (1 1.32)

11.2. IDEAL FLUIDS 269

where f is the force per unit volume and f f p the force per unit mass. Using this together with the continuity equation

- + v * ( p v ) = 0 (11.33) at

one finds f j = - a!:) + aj (pvivj + psij) (1 1.34)

Within the limited scope of this book, we wish to discuss the Lagrangian formulation of this problem. We follow in part D.E. Soper, Classical Field Theory (Wiley, 1976) ch.4, where we learned the existence of the original work by G. Herglotz, Ann. Phys.,36(1911)493.

We must switch from Euler’s description (observation of fluid at a given position) to Lagrange’s description (follow fluid particles in their motion).

If Euler observes a particle at z at time t, let X(x, t) be the position the particle had at t = 0. Therefore X ( z , O ) = x. Let M ( X ) d X be the total mass of particles with coordinates between X and X +dX at t = 0. At time t, these particles will be in the interval (x,x+dx). Since dX = (dX/dx)dz, the linear density at x at time t is

(11.35)

Note now that a particle at x at time t will be at x + w(z, t) dt at time t + dt. When it was at z its X was X ( z , t), when it is at x + w dt its X is X ( z + w dt). The corresponding dX is

ax PL(2,t) = M ( X ) .

ax ax dX = - dt + - dt . ax a t

Since X is just a label for the particle, must be dX = 0, and so

a x p t ax/az ’ y = -- (1 1.36)

and the current is

(1 1.37) J = pw = [ M ( X ) g ] [--I = -M(X)- ax .

a t

The continuity equation is automatically satisfied:

‘For uniform motion we would havex = X+ut, X = x-ut, 8x182 = 1,8X/8t = -v, u = -(sX/8t)/(8X/B~).

Equation (11.31) can be derived from the Lagrangian density

c = c(x,a~~,a~x) = pv2/2 - p U(X) , (11.38)

where X = l/p (“voiume” of the unit mass) and we assume that the “equation of state’’ is p = -dU(X)/dX. Remembering that

(11.39) I

a fairly long calculation using the continuity equation yields

1 = - M ( x ) a,V + axv + - asp) . ( c1

The following is another proof of the one-dimensional version of Euler’s

1 Lb

equation (1 1.31),

a,V + a,v + - axp = o .

Using (11.19) with y replaced by X , and the Lagrangian (11.38,39), we find ($0 = $0 I- t , d = 2 = -21)

Too = pG2/2 + pIJfX) TO1 = pv , TIo = v(Too + p ) , T” = pv2 + p a

Then (11.18) for v = 0 ( ~ ~ * + a,T’* = 0) using the continuity equation flip + ~ x ~ ~ v ) = 0, gives

&atv + V ~ , V + p-laXp) = o

and for v = 1 (a0To1 + alT1l = 0)

Two-dimensional incompressible fluid With x = (d ,z2) and X = (X1,X2) and a natural extension from the one-dimen~iona~ case, we express the density in the form

(I 1.40)


where (11.41)

Then p ( x , t ) d2x = M ( X ) d2X . (1 1.42)

The analogue of the one-dimensional equation (1 1.36) (v&X + & X = 0) is

(11.43) (a,x1)v1 + (&X'>v2 = -a&' , { (81X2)v' + (&X2)v2 = -a,x2 ,

where St = B/ax' (i = 1,2) and at = a/&. Solving for v1 and v2 we find vi = J ' / p (i = 1,2), where

(1 1.44) J' = -M(x)[(atx')(aZx2) - (atx2)(azX')] , J 2 = -M(X)[(&X2)(&X1) - (&X1)(&X2)] .

Continuity equation Let zaflr be skewsymmetric in all indices which take the values (0 ,1 ,2) , and e0l2 = 1. The density and the current components can be written

(1 1.45) p = M€o*qa*x')(aflx2) , J' = Me"*/3(aaX1)(aflX2) .

These satisfy the continuity equation. In fact

a = c ( B M / a X i ) E ~ " p (a,X')(aax') (a ,x2)+Ma, ( P O (8, XI) (80x2)) = o+o = 0

i=1

Lagrangian F'rom the Lagrangian density

c = (p/2)[(v')" + (v2)2] - pV(X) (11.46)

(A = l/p) one might try to derive the two-dimensional version of (11.32) with f = 0. However one can proceed more simply as follows.

Rom

(1 1.47)

p(v ' atv + viv ' a,v + p - l ~ i ~ ~ ~ ~ = 0 *

Circulation theorem We derive the theorem in two dimensions from the assumption that the Lagrangian density L ( X i , a,Xi) is invariant under an infinitesimal transformation X i + X i + @(X) defined below. Then

and, using Lagrange's equations,

Therefore ac , = constant .

(11.48)

(1 1.49)

The Lagrangian density (11.46) is invariant under a transformation X -i X' such that

d(X1, X 2 )

d(X'1, X'2) ' M ( X ' ) = M ( X ) (11.50)

where M(X') is obtained from M ( X ) by replacing Xi by XIi (i = 1,Z). In fact p, J' , and J 2 are invariant:

3(X'l,X'2) a(x"x2) = - M ( X ' ) a(x1, X2) J ' ( x , t ) = - M ( X ) a(t, x2) a(x1, X 2 ) a(t, x2)


and similarly for J 2 .

(11.60) gives M(X)(l - e 8ti/8Xi) = M(X) + e(8M(X)/aXi)Si, For infinitesimal transformations X" = Xi + e S'(X) the condition

w f ( X ) t i ) = o . 8x4 This is satisfied by

(see D.E. Soper, Zoc. cit., p.51, but this is a little more general, M(X) is not regarded as constant). The conserved quantity is

where X = n(7) and x = ~ ( 0 ) . By this rearrangement we have (8d/BOk)(dflk/dr) = dd/dT, while (aL:/8(8tXi))8jX' = - p j , as can be shown by a little calculation. Therefore

&(t) = -fdxj - 1 ( - p j ) = {v-dx . P

Three-dimensional ideal fluids Rather than extending the two-dimensional Lagrangian calculations, we summarize the standard treatment of the three-dimensional fluid.

Define the "vorticity" 62 = V x v. The Euler equation for a non-viscous fluid acted upon by conservative forces (f = -pV@) can be expressed in

a v 1 1 - + - vv2 - v x n+ - vp+ va) = 0 8t 2 P

the form .

Note that for steady irrotational flow of an incompressible non-viscous fluid (n = 0, p = constant) this equation reduces to

8 - ( fv2 + p + pa)) = 0 (Bernoulli equation) . 8x4 2

For a non-viscous fluid the substantial derivative of the vorticity,

+(v*V)M , DQ an -= - Dt a t

obeys the equation

D O 1 - = ( n . V ) v - I l t ( v ~ v ) + - v p x v p . Dt P2

In fact, taking the rotation of the Euler equation we have

1 v x (v x sz) - - vp x v p = 0 . a52 at PS --

Since V ’ 0 = 0, we have

v x (v x 0) = -n(v 4 v) f (52 . V)v - (v * v)n . If the fluid is incompressible ( p = constant, V - v = 0) this equation

D n Dt

reduces to - = (n ’ V)v (11.51)

Let dA be the area.of an infinitesimal surface element, ii a unit vector normal to it, and dS = ii dA. In problem 11.6 we shall prove that

DdSi ad -- - (V * v)dSi - - dSj . Dt ad Using equations (11.51,52) one finds that

= o D ( n * dS) Dt

(11.52)

(1 1.53)

F’rom equation (1 1.53) one infers the Helmoltz theorem for a non-viscous

& L n . d S = O . (1 1.54)

If S is a surface having the curve C as a boundary, since SZ = V x v

Auid

one obtains the Thomson (Kelvin) theorem

(11.55)


11.3 Chapter 11 problems

11.1 (i) Show that the continuity equation (11.33) can be expressed in the form

a at -(fi3) +&~3(v)] = o ,

where a’(., .,a) = & A

(ii) Show also that the continuity equation can be expressed in terms of Lie derivatives as

A Jz, v = v,a/ax + v,a/ay + v,a/az, and 03(P) = a3(V,0, 0 ) = p(v,& A & - V,~X A L + v,& A &.

(a/& + Lv)(@3) = 0 .

11.2 Using problem 11.1 (i) derive the continuity equation in curvilinear coordinates,

ap 1 B -+--( J V ” = O . at f i a x i

11.3 (i) Show that the Euler equation

aV

at - +(v*V)v+ 1

P - vp = -v9

can be expressed in the form

aV 1 1 - + - J(v2) +(JV) (V ) + - Jp++* = 0 at 2 P ,

where V = v,a/Ox + v,,a/ay + v,d/az and 3 = vz& + v,dy + v,&. We are using Cartesian coordinates (d = v1 = v,, . . .), (ii) Show that the Euler equation can also be expressed as

11.4 &om the equation Bv/at + (1/2)2(v2) + JV(P) + ( l / p ) & + d@ = 0 derive Euler’s equation in curvilinear coordinates

276 CHAPTER 1 1 . CONTINUOUS SYSTEMS

11.5 Show that under a general coordinate transformation KIWIi = KWi and

11.6 Show that the continuity equation

i a a t axi 9 + - -(p&Vj) = 0

is invariant under general coordinate transformations in 3-space.

11.7 (i) Consider an infinitesimal dr = (dx', dx2, dz3) and the particles of fluid that are on it at the time t. Show that at the time t + dt the same particles will be found on dr' = (dx" , dx", dxI3),

(ii) Show that the substantial derivative of the oriented-area vector dS = d(')r x d(2)r, dSi = e'jkd(1)xjd(2)xkl is

avj Dt Oxj ax' dSi--dSj . DdSi a d -=-

11.8 Derive the equation aSt/Ot - V x (v x S t ) - (l/p")Vp x V p = 0 from the form version of Euler's equation,

( k + L v ) v + ; d p + d l - ( ; ) 9 - - v 2 = O .

11.9 In three dimensions define

P ( x , t ) = M(X)a~~'(a,X')(apX2)(a,X3) , where X = ( X I , X 2 , X 3 ) and the Greek indices run from zero to three. Verify that

J k p = J O = M ( X ) a(x) v k = - , P

and that the continuity equation is satisfied,

d,,Jp=O .


11.10 (i) A massless scalar field (scalar photons) is described by the La- grangian

I; = (112) J 8#y&p d3x, yielding the wave equation 8,8@(p = 0.

Show that the ener~*momentum tensor

T P V = 8c”@?”cg - g @ V ( 8 ~ y 8 ~ y ~ / 2

satisfies 8,TM” = 0, and derive energy and momentum conservation. (ii) Let

A = I d 3 . d ~ ~ ( 1 / 2 ) ~ ~ y ~ @ ~ + (M/2)X2 S(x - X(t)) + fw(x - X(t))]

be the action for the above field interacting with a particle of mass M , position X(t). Write the Lagrange equations.

(iii) Show that the energy

E = (1/2) / [ (&(P)~/c? + ( V ~ ) ~ ] d ~ x + ~ ~ / 2 ) X ~ - f 1 cp U(X - X) d3x

and the momentum

are conserved.

278 CHAPTER 11. CONTlNUOUS SYSTEMS

Solutions to ch. 11 problems S1l.l (i)

d [ @ 3 ( v ) ] = a(pv,)/ax dx A dy A dz - a(pvY)/ay & A d~ A d~ +a(pv,)/aZ JZ A 2~ A 2y

= v . (pv) G 3 (ii) Using the general formula LAG = d [ G ( A ) ] + (&)(A) we have Lp(pG3) = d[pcj3(V)] + (d(pcj3))(V). The second term is zero since there cannot be forms of order greater than 3 in 3-dimensional space. Using (i) we have V . (pv) = ( i (pcj3((v) ] = Lv(Pcj3).

= J dx' A d x 2 A a x 3 )

where ( X 1 = x , X 2 = y , X 3 = z ) . ax' ax' axk

axl axz a x 3 J = eijk---

Let g = det(g), where g is the matrix with elements

If J is the matrix with elements OXi/axj, we have

Hence G 3 = fi dx' A dx' A dx'. Now with v = Via/axi, we have

The final result follows at once.

T g = JJ , g = det(J) det JT = J'.

&fi &I A dx2 A d x 3 ( v ) ] = vi)/ax') d d A ax2 A dx3

S11.3 (i) d(?) = (av2/axi)dxa)

( B v ) ( V ) = [(v x v) x v]icid = -(i/2)(av2/axi)2xi +vj(av,/axj)dzi

L v P = d[i . (V)] + (dri)(V) = d(v') + (di . ) (V) . (ii) The general formula LAG = J[G(A)] + (&)(A) gives

Thus (i) reduces to (ii).

S11.4 With = vjdxj, we have

Note that one can also start from &/at + (v . V)v + (l/p)Vp = -V+ using the formula (v . V)v = (1/2)Vv2 - v x (V x v). Then in Cartesian coordinates


for which vi = v’, one finds the proposed equation with Vi 4 Vi and V’ + vi. The transition to curvilinear coordinates is then made by using the results of the following problem.

Since the second term in

is symmetric in i and j, it cancels out when we interchange i and j and subtract. The required result follows at once.

S11.6 We must show that

From eq. Applications of the General Theory of Relatavity (Wiley, 1972), we have

(4.4.2) of S. Weinberg, Gravitation and Cosmology: Principles and

I axm axn gij = - axli Qmnazl j ’

from which one finds at once that g‘ = 5’9, where J is the Jacobian determinant. A brief calculation gives

We must show that the second term vanishes. We have

where J is the matrix with elements axk/laxii. This is zero by virtue of Weinberg’s equation (4.7.5),

a In det M (x)


S11.7 (i) During the time dt, the end points a and b of a vector b - a will be transported by the fluid motion to a’ = a + v(a)dt and b’ = b f v(b)dt, and the vector b - a will be transported to b’ - a’ = b - a + (v(b) - v(a))dt. For a = r, b = r + dr, a’ = r’, b’ = r’ + dr’, we have dr’ = dr + (v(r + dr) - v(r))dt, yielding at once the required formula. Thus the substantial derivative of dz’ is

Ddx’ av’ dxj --- - ~t axj

(ii)

avj J a x m

= ( ~ 5 ~ ~ 6 ~ ~ - him6r)- dS, etc.

Sll.8 Acting with the operator 2, we find

S11.10 (i) Proving that apTp” = 0 is straightforward using the wave equation. Space integration gives

c dt / T O ” ~ ~ X + /aiTi”d3x = 0 .

Assuming that cp vanishes at infinity, we find

d / T o u d 3 x = 0 dt ,

E = To0d3x = (1/2) [(atcp)’/c2 + (Vcp)’]d3x = constant , I I P’ = (l/c) JT”~’x = - c - ~ /atcpaicp d3x = constant’ .

(ii) ac ac +ai- = - ac a. -

~ ( ~ o c P ) a(aip) a ‘ ~

11.3. CHAPTER 11 PROBLEMS 28 1

give8 (a: - V2)p = f u(x - X) ,

whereaa

gives

a ac ac atax.( ax.( --=-

M X = vx J f p u(x - X)d3x

= -f J p~,u(x - X)d’x = f U(X - X)V,(P d’x . J (iii) The field energy-momentum tensor T@” (see (i)) now obeys the equation

a,TMy = (a,,a”p)ayp = f u(x - x)a’p . Therefore

J ~~~d~~ = f / u(x - x)atp d S X dt

= f$ /u(x-X)p dSx+ f (X.Vxu)p d3x = . . .-X.( f uVXp d x) = . . .-MX.X ! s J S

INDEX

Index 283

acceleration, relativity 229 action-angle variables 193 action - integral 8 - integral, invariance 120 - least a. principle 12 - variable, cyclic 7 adiabatic invariants 202 angles, Euler 82 angular momentum, relativity 231 anomaly, eccentric a. 30, 198 attractor, spiral 43 axes, principal a. of inertia 79

Bernoulli equation 273 bifurcation 53 Bloch’s theorem 49 boost 237 brachistochrone 15, 19 bracket - Lagrange 166 - Poisson 149, 158, 267

canonical - transformations 148 - variables 148, 162, 170 Cartan - Lie-C. 175 - Poincard - C. 178 - vectors and forms 156 central force 23 centrifugal force 72 chaos 54, 200 characteristic function, Hamilton’s 8, 168, 172 charged particle in em field 104, 153,246 circulation theorem 272, 274

commutator 69, 177 complementary Lagrangian 123, 182 components, covariant , contravariant 228 cone, lab and space 80 constants of motion 115,151,166 constraints 110, 113, 114 continuous systems 261 contravariant (see components) coordinate systems 67 Coriolis force 72 coupled oscillators 183, 221 covariant (see componets) cycle, limit c. 53 cyclic action variable 7

damped oscillator 18 delta map 58 derivative, functional 267 discrete maps 52 disk, rolling 91, 94, 127 double pendulum 44

eccentric anomaly 30, 198 effective potential energy 24 equation - Bernoulli 273 - Hamilton-Jacobi 9, 168, 171 - iconal 16 - Lagrange, string 263 - Schrodinger 17 equations - Euler’s, rigid body 79 - Euler’s, fluids 268 - Hamilton 1, 146 - Lagrange 102, 108 equivalent Lagrangians 105

284 INDEX

Euler angles 82 Euler’s equations - rigid body 79 - fluids 268 expansions, perturbation 215 exterior forms 158

Fermat’s principle 16 flow - Hamiltonian, as Lie-Cartan group 175 - in phase space 5 fluid - Euler’s equation 268 - one-dimensional 268 force - centrifugal, Coriolis 72 - relativity 230 forces, fictitious 71 forms, Cartan’s 156 function - generating 150, 163, 170 - Hamilton’s, characteristic 8, 168, 172 - Hamilton’s, principal 172 functional derivative 267

Galilean group 128 gauge transformation 105 generalized momenta 102 generating function 150,163,170 - infinitesimal rotations 165 - infinitesimal transformations 165 generators - of infinitesinal rotations 69 - of Poincar6 group 253 group - Galilean 128 - Lie-Cartan 175 - Poincard 253 gyrocompass 89

Hamilton-Jacobi equation 9, 168, 171 Hamiltonian 1 - bilinear 146 - damped oscillator 18 - flow as Lie-Cartan group 175 - relativity 249 - string 266 - vector field 157 Hamilton’s - equations 1, 146 - characteristic function 8, 168, 172 - principal function 172 Hamilton- Jacobi equation 9, 168, 171 harmonic oscillator, isotropic 26 holonomic constraint 110, 113 Hopf bifurcation 53 hyperbolic point 56

iconal, surfaces, equation 16 identity, Jacobi 151, 177 inertia - matrix 78 - principal axes 79 infinitesimal rotations - generating function 165 - generator 69 integrability 34, 200 integral invariant, PoincarbCartan 178 integral, Jacobi 117 invari an ce - of Lagrange equations 106 - of Lagrangian up to time derivative 118 - of action integral 120 invariants - adiabatic 202 - Poincard 181

lNDEX 285

irregular precession of top 87 isotropic harmonic oscillator 26

Jacobi - Hamilton-J. equation 9, 168, 171 - identity 151, 177 - integral 117

Kronig-Penney model 51

lab cone 80 Lagrange - brackets 166 - equation, string 263 - equations 102, 108 - equations, invariance 106 - multipliers 114 Lagrangian -complementary 123, 182 - fluid 270, 271 - invariance and constants of motion 115 - invariance up to time derivative 118

- relativity 248 - Toda 127 Lagrangians 101 - equivalent 105 Laplace-Runge- Lenz vector 31, 127 Larmor’s theorem 106 least action principle 12 Legendre transformation 152 Liecartan group, Hamiltonian flow as 175 limit cycle 53 Liouville’a theorem 6, 154, 181 Lorentz transformations 227

magnetic moment in static em field 247

map, delta, tent 58 maps 52 mass, longitudinal, transversal 230 Maupertuis principle 12 metric tensor 228 momenta, generalized 102 multiply periodic system 196

nested tori 201 Noether’s theorem 119 non-holonomic constraints 114 non-integrability 35, 200

operator d 160 operator R 214 orbital angular momentum, relativity 231 oscillations, small 44 oscillator - damped 18 - harmonic isotropic 26 - periodically jerked 50 oscillators, coupled 183;221

parametric resonance 47 particle motion, 1-dim 2-4 pendulum - double 44 - spherical 111 period of closed orbit 4 perturbation theory 213 perturbed periodic systems 217 phase space 2-5 photons, scalar 277 PoincarB-Cartan integral invariant 178 PoincartS - group, generators 253 - invariants 181 - section 200

286 INDEX

point - fixed 41 - hyperbolic, saddle 56 Poisson - brackets 149, 158 - brackets, continuous systems 267 potential energy, effective 24 precession - of pericenter 25 - of top, irregular 87 - of top, regular 85 - Thomas 244 principal inertia axes 79 principle - Fermat's 16 - least action, Maupertuis 12

regular precession, top 85 relativistic - Hamiltonian 249 - Lagrangian 248 repellor, spiral 43 resonance, parametric 47 rolling disk 91, 94, 127 rotations 67, 233 - product of 253

saddle point 56 scalar photons 277 Schrodinger equation 17 section, Poincark 200 skating top 93 sleeping top 86 space - cone 80 - phase s. 5 - rotations 67, 233, 253 spin 241 spinning top 84

spiral attractor, repellor 43 Stokes' theorem 173 string 261 - Lagrangian, Hamiltonian 263,266

tautochrone 15, 19 tensor, metric 228 tent map 58 theorem, Liouville's 6, 154, 181 Thomas precession 244 thumbtack 94, 122 Toda Lagrangian 127 top - spinning 84 - irregular precession 87 - regular precession 85 - skating 93 - sleeping 86 tori, nested 201 transformation - gauge 105 - canonical 148 - infinitesimal canonical 165 - Legendre 152 - Lorentz 227 translations 67

variables, canonical 148, 162, 170 vector field, Hamiltonian 157 vectors - Cartan 156 - relativity 228

spinor connection (SL(2), SU(2)) 238

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