Vector and Tensor Analysis

International Series in Pure and Applied Mathematics

William Ted Martin, CONSULTING EDITOR

VECTOR AND TENSOR ANALYSIS

a

International Series in Pure and Applied 'MathematicsWILI.LA11 TED MARTIN, Consulting Editor

AIILFORS Complex AnalysisBELLMAN Stability Theory of Differential EquationsBucx Advanced CalculusCODDnNUTON AND LEVINSON Theory of Ordinary Differential Equations

G01.011 13 AND SHANKS Elements of Ordinary Differential EquationsGRAVES The Theory of of Real VariablesGRIFFIN Elementary Theory of NumbersIIILDEBRAND - rodurl ion to Numerical Analysis

Principles of Nunurriral AnalysisLAS b;leuu'nts of Pure aunt Applied MathematicsLASS Vector and Tensor AnalysisLEIGIITON An Introduction to the Theory of Differential EquationsNEHAIU Conformal MappingNEWELL Vector AnalysisROSSER Logic for 'MathematiciansRUDIN Principles of Mathematical AnalysisSNEDDON Elciuents of Partial I)iiTerential Equations

SNEDDON Pourier TransformsSTOLL Linear Algebra and 'Matrix TheoryWEINSTOCK Calculus of Variations

VECTOR AND TENSORANALYSIS

BY

HARRY LASSJET PROPULSION LABORATORY

Calif. Institute of Tech.4800 Oak Grove DrivePasadena, California

New York Toronto London

McGRAW-HILL BOOK COMPANY, INC.

1950

VECTOR AND TENSOR. ANALYSIS

Copyright, 1950, by the '.11e(_;ratt-Bill Book Company, Inc. Printed in the

1 nited States of America. All rights reserved. 't'his hook, or parts thereof,may not be reproduced in any form without, perrnis.sion of the publishers.

To MY

MOTHER AND FATHER

PREFACE

This text can be used in a variety of ways. The studenttotally unfamiliar with vector analysis can peruse Chapters 1, 2,and 4 to gain familiarity with the algebra and calculus of vectors.These chapters cover the ordinary one-semester course in vectoranalysis. Numerous examples in the fields of differentialgeometry, electricity, mechanics, hydrodynamics, and elasticitycan be found in Chapters 3, 5, 6, and 7, respectively. Thosealready acquainted with vector analysis who feel that they wouldlike to become better acquainted with the applications of vectorscan read the above-mentioned chapters with little difficulty:only a most rudimentary knowledge of these fields is necessaryin order that the reader be capable of following their contents,which are fairly complete from an elementary viewpoint. Aknowledge of these chapters should enable the reader to furtherdigest the more comprehensive treatises dealing with these sub-jects, some of which are listed in the reference section. It ishoped that these chapters will give the mathematician a briefintroduction to elementary theoretical physics. Finally, theauthor feels that Chapters 8 and 9 deal sufficiently with tensoranalysis and Riemannian geometry to enable the reader to studythe theory of relativity with a minimum of effort as far as themathematics involved is concerned.

In order to cover such a wide range of topics the treatment hasnecessarily been brief. It is hoped, however, that nothing hasbeen sacrificed in the way of clearness of ideas. The author hasattempted to be as rigorous as is possible in a work of this nature.Numerous examples have been worked out fully in the text.The teacher who plans on using this book as a text can surelyarrange the topics to suit his needs for a one-, two-, or even three-semester course.

If the book is successful, it is due in no small measure to thecomposite efforts of those men who have invented and who have

vii

viii PREFACE

applied the vector and tensor analysis. The excellent workslisted in the reference section have been of great aid. Finally, Iwish to thank Professor Charles de Prima of the CaliforniaInstitute of Technology for his kind interest in the developmentof this text.

HARRY LASSURBANA, ILL.

February, 1950

CONTENTS

PREFACE . . . . . . . . . . . . . . . . . . . . . . . . . . vii

CHAPTER 1

THE ALGEBRA OF VECTORS . . . . . . . . . . . . . . . . . 1

1. Definition of a vector 2. Equality of vectors 3. Multipli-cation by a scalar 4. Addition of vectors 5. Subtraction ofvectors 6. Linear functions 7. Coordinate systems 8. Scalar,or dot, product 9. Applications of the scalar product to spacegeometry 10. Vector, or cross, product 11. The distributivelaw for the vector product 12. Examples of the vector product13. The triple scalar product 14. The triple vector product 15.Applications to spherical trigonometry

CHAPTER 2

DIFFERENTIAL VECTOR CALCULUS . . . . . . . . . . . . . 29

16. Differentiation of vectors 17. Differentiation rules 18. Thegradient 19. The vector operator del, V 20. The divergence of avector 21. The curl of a vector 22. Recapitulation 23. Curvi-linear coordinates

CHAPTER 3

DIFFERENTIAL GEOMETRY . . . . . . . . . . . . . . . . . 58

24. Frenet-Serret formulas 25. Fundamental planes 26. In-trinsic equations of a curve 27. Involutes 28. Evolutes 29.Spherical indicatrices 30. Envelopes 31. Surfaces and curvi-linear coordinates 32. Length of arc on a surface 33. Surfacecurves 34. Normal to a surface 35. The second fundamentalform 36. Geometrical significance of the second fundamentalform 37. Principal directions 38. Conjugate directions 39.Asymptotic lines 40. Geodesics

CHAPTER 4

INTEGRATION . . . . . . . . . . . . . . . . . . . . . . . . 89

41. Point-set theory 42. Uniform continuity 43. Some proper-ties of continuous functions 44. Cauchy criterion for sequences45. Regular area in the plane 46. Jordan curves 47. Functionsof bounded variation 48. Arc length 49. The Riemann integral

ix

x CONTENTS

50. Connected and simply connected regions 51. The line inte-gral 52. Line integral (continued) 53. Stokes's theorem 54.Examples of Stokes's theorem 55. The divergence theorem(Gauss) 56. Conjugate functions

CHAPTER 5

STATIC AND DYNAMIC ELECTRICITY . . . . . . . . . . . 127

57. Electrostatic forces 58. Gauss's law 59. Poisson's formula60. Dielectrics 61. Energy of the electrostatic field 62. Dis-continuities of D and E 63. Green's reciprocity theorem 64.Method of images 65. Conjugate harmonic functions 66. Inte-gration of Laplace's equation 67. Solution of Laplace's equationin spherical coordinates 68. Applications 69. Integration ofPoisson's equation 70. Decomposition of a vector into a sum ofsolenoidal and irrotational vectors 71. Dipoles 72. Electricpolarization 73. Magnetostatics 74. Solid angle 75. Movingcharges, or currents 76. Magnetic effect of currents (Oersted)77. Mutual induction and action of two circuits 78. Law of in-duction (Faraday) 79. Maxwell's equations 80. Solution ofMaxwell's equations for electrically free space 81. Poynting'stheorem 82. Lorentz's electron theory 83. Retarded potentials

CHAPTER 6

MECHANICS . . . . . . . . . . . . . . . . . . . . . . . . . 184

84. Kinematics of a particle 85. Motion about a fixed axis 86.Relative motion 87. Dynamics of a particle 88. Equations ofmotion for a particle 89. System of particles 90. Momentumand angular momentum 91. Torque, or force, moment 92. Atheorem relating angular momentum with torque 93. Moment ofmomentum (continued) 94. Moment of relative momentum 95.Kinetic energy 96. Work 97. Rigid bodies 98. Kinematics ofa rigid body 99. Relative time rate of change of vectors 100.Velocity 101. Acceleration 102. Motion of a rigid body withone point fixed 103. Applications 104. Euler's angular coordi-nates 105. Motion of a free top about a fixed point 106. Thetop (continued) 107. Inertia tensor

CHAPTER 7

HYDRODYNAMICS AND ELASTICITY . . . . . . . . . . . . 230

108. Pressure 109. The equation of continuity 110. Equationsof motion for a perfect fluid 111. Equations of motion for anincompressible fluid under the action of a conservative field 112.The general motion of a fluid 113. Vortex motion 114. Appli-cations 115. Small displacements. Strain tensor 116. Thestress tensor 117. Relationship between the strain and stresstensors 118. Navier-Stokes equation

CONTENTS xi

CHAPTER 8

TENSOR ANALYSIS AND RIEMANNIAN GEOMETRY. . . . . 259

119. Summation notation 120. The Kronecker deltas 121.Determinants 122. Arithmetic, or vector, n-space 123. Contra-variant vectors 124. Covariant vectors 125. Scalar product oftwo vectors 126. Tensors 127. The line element 128. Geode-sics in a Riemannian space 129. Law of transformation for theChristoffel symbols 130. Covariant differentiation 131. Geode-sic coordinates 132. The curvature tensor 133. Riemann-Christoffel tensor 134. Euclidean space

CHAPTER 9

FURTHER APPLICATIONS OF TENSOR ANALYSIS . . . . . . 311

135. Frenet-Serret formulas 136. Parallel displacement of vectors137. Parallelism in a subspace 138. Generalized covariant differ-entiation 139. Riemannian curvature. Schur's theorem 140.Lagrange's equations 141. Einstein's law of gravitationTwo-point tensors

142.

REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . 339

INDEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341

CHAPTER 1

THE ALGEBRA OF VECTORS

1. Definition of a Vector. Our starting point for the definitionof a vector will be the intuitive one encountered in elementaryphysics. Any directed line segment will be called a vector. Thelength of the vector will be denoted by the word magnitude. Anyphysical element that has magnitude and direction, and hencecan be represented by a vector, will also be designated as a vector.In Chap. 8 we will give a more mathematically rigorous definitionof a vector.

Elementary examples of vectors are displacements, velocities,forces, accelerations, etc. Physical concepts, such as speed,temperature, distance, and specific gravity, and arithmetic num-bers, such as 2, i, etc., are called scalars to distinguish them fromvectors. We note that no direction is associated with a scalar.

We shall represent vectors byarrows and use boldface type to Ca-, a, a, a, a]indicate that we are speaking ofa vector. In order to distin-guish between scalars and vec- ators, the student will have to Fia. 1.adopt some notation for describ-ing a vector in writing. The student may choose his mode ofrepresenting a vector from Fig. 1 or may adopt his own notation.

To every vector will be associated a real nonnegative numberequal to the length of the vector. This number will depend, ofcourse, on the unit chosen to represent a given class of vectors.A vector of length one will be called a unit vector. If a repre-sents the length of the vector a, we shall write a = jal. Ifjal = 0, we define a as the zero vector.

2. Equality of Vectors. Two vectors will be defined to beequal if, and only if, they are parallel, have the same sense ofdirection, and the same magnitude. The starting points of thevectors are immaterial. It is the direction and magnitude which

1

2 VECTOR AND TENSOR ANALYSIS [SEC. 3

are important. Equal vectors, however, may produce differentphysical effects, as will be seen later. We write a = b if thevectors are equal (see Fig. 2).

Fia. 2.

3. Multiplication by a Scalar. If we multiply a vector a by areal number x, we define the product xa to be a new vectorparallel to a whose magnitude has been multiplied by the factor x.Thus 2a will be a vector which is twice as long as the vector a

Z-aFIG. 3. Fm. 4.

and which has the same direction as a (see Fig. 3). We define-a as the vector obtained from a by reversing its direction (seeFig. 4).

We note thatx(ya) = (xy)a = xya

(x + y)a = xa + yaOa = 0 (zero vector)

It is immediately seen that two vectors are parallel if, and onlyif, one of them can be written as a scalar multiple of the other.

4. Addition of Vectors. Let us suppose we have two vectorsgiven, say a and b. We form a third vector by constructing atriangle with a and b forming two sides of the triangle, b adjoinedto a (see Fig. 5). The vector starting from the origin of a andending at the arrow of b is defined as the vector sum a + b.

We see that a + 0 = a, and if a = b, c = d, then

a+c=b+d

SEc. 6] THE ALGEBRA OF VECTORS 3

From Euclidean geometry we note that

a+b=b+a (1)(a+b)+c=a+(b+c) (2)

x(a+b) =xa+xb (3)

(1) is called the commutative law of vector addition; (2) is calledthe associative law of vector addition; (3) is the distributive lawfor multiplication by a scalar. The reader should have notrouble proving these three results geometrically.

a+b+cFio. 5.

5. Subtraction of Vectors. Given the two vectors a and b,we can ask ourselves the following question: What vector c mustbe added to b to give a? The vector c is defined to be the vectora - b. We can obtain the desired result by two methods.First, construct -b and then add this vector to a, or second, let band a have a common origin and construct the third side of thetriangle. The two possible directions will give a - b and b - a(see Fig. 6). Thus a - b = a + (-b).

a-b

Flo. 6.

a-b

6. Linear Functions. Let us consider all vectors in the two-dimensional Euclidean plane. We choose a basis for this systemof vectors by considering any two nonparallel, nonzero vectors.Call them a and b. Any third vector c can be written as a linear


combination or function of a and b,

c = xa + yb (4)

The proof of (4) is by construction (see Fig. 7).Let us now consider the following problem: Let a and b have

a common origin, 0, and let c be any vector starting from 0 whoseend point lies on the line joining the ends of a and b (see Fig. 8).

B

aFIG. 8.

Let C divide BA in the ratio x: y where x + y = 1. In particu-lar, if C is the mid-point of BA, then x = y = fr. Now

c=OB+BC= b + x(a - b)=xa+(1 - x)b

so that

c = xa + yb (5)

Now conversely, assume c = xa + yb, x + y = 1. Then

c=xa+(1 -x)b =x(a-b)+bWe now note that c is a vector that is obtained by adding to b thevector x(a - b), this latter vector being parallel to the vectora - b. This immediately implies that the end point of c lieson the line joining A to B. We can rewrite (5) as

c-xa-yb=0 (6)1-x-y=0

SEC. 61 THE ALGEBRA OF VECTORS 5

We have proved our first important theorem. A necessaryand sufficient condition that the end points of any three vectorswith common origin be on a straight line is that real constantsit m, n exist such that

la+mb+nc=0l+m+n=O (7)

with l2+m2+n2p,, 0.We shall, however, find (5) more useful for solving problems.Example 1. Let us prove that the medians of a triangle meet

at a point P which divides each median in the ratio 1:2.

C

0FIG. 9.

Let ABC be the given triangle and let A', B', C' be the mid-points. Choose 0 anywhere in space and construct the vectorsfrom 0 to A, B, C, A', B', C', calling them a, b, c, a', b', c' (seeFig. 9). From (5) we have

a'=4b+ic(8)b' = Ja + 4c

Now P (the intersection of two of the medians) lies on the linejoining A and A' and on the line joining B and B'. We shall thusfind it expedient to find a relationship between the four vectorsa, b, a', b' associated with A, B, A', B'. From (8) we eliminatethe vector c and obtain

2a' + a = 2b' + b

or

*a' + '}a = *b' + b (9)

6 VECTOR AND TENSOR ANALYSIS [SFC. 6

But from (5), -sa' + -ia represents a vector whose origin is at 0and whose end point lies on the line joining A to A. Similarly,,'jb' + *b represents a vector whose origin is at 0 and whose endpoint lies on the line joining B to B'. There can only be oneyvector having both these properties, and this is the vector p = OP.Hence p = tea' + is = -b' -fib. Note that P divides AA' andBB' in the ratios 2: 1. Had we considered the median CC' inconnection with AA', we would have obtained that p + J e,and this completes the proof of the theorem.

Example 2. To prove that the diagonals of a parallelogrambisect each other. Let ABCD be the parallelogram and 0 any

C

FIG. 10.

point in space (see Fig. 10). The equation d - a = c - bimplies that ABCD is a parallelogram. Hence

Ja+4c=lb+Id =pso that P bisects AC and BD.

Problems

1. Interpret I aa

2. Give a geometric proof of (3).3. a, b, c are consecutive vectors forming a triangle. What

is the vector sum a + b + c? Generalize this result for anyclosed polygon.

4. Vectors are drawn from the center of a regular polygon toits vertices. From symmetry considerations show that the vec-tor sum is zero.

SEc. 61 THE ALGEBRA OF VECTORS 7

5. a and bare consecutive vectors of a parallelogram. Expressthe diagonal vectors in terms of a and b.

6. a, b, c, d are consecutive vector sides of a quadrilateral.Show that a necessary and sufficient condition that the figure bea parallelogram is that a + c = 0 and show that this impliesb + d = 0.

7. Show graphically that lal + lbl >_ la + bl. From thisshow that la - bl >_ lal - lbi.

8. a, b, c, d are vectors from 0 to A, B, C, D. If

b-a=2(d-c)show that the intersection point of the two lines joining A and Dand B and C trisects these lines.

9. a, b, c, d are four vectors with a common origin. Find anecessary and sufficient condition that their end points lie in aplane.

10. What is the vector condition that the end points of thevectors of Prob. 9 form the vertices of a parallelogram?

11. Show that the mid-points of the lines which join the mid-points of the opposite sides of a quadrilateral coincide. Thefour sides of the quadrilateral are not necessarily coplanar.

12. Show that the line which joins one vertex of a parallelogramto the mid-point of an opposite side trisects the diagonal.

13. A line from a vertex of a triangle trisects the opposite side.It intersects a similar line issuing from another vertex. In whatratio do these lines intersect one another?

14. A line from a vertex of a triangle bisects the opposite side.It is trisected by a similar line issuing from another vertex. Howdoes this latter line intersect the opposite side?

15. Show that the bisectors of a triangle meet in a point.16. Show that if two triangles in space are so situated that the

three points of intersection of corresponding sides lie on a line,then the lines joining the corresponding vertices pass through acommon point, and conversely. This is Desargues's theorem.

17. b = (sin t)a is a variable vector which always remains

parallel to the fixed vector a. What isfib?

Explain geomet-

rically the meaning ofA.


18. Let v, be the velocity of A relative to B and let v2 be thevelocity of B relative to C. What is the velocity of A relativeto C? Of C relative to A? Are these results obvious?

19. Let a, b be constant vectors and let c be defined by theequation

c = (cos t)a + (sin t)b

When is c parallel to a? Parallel to b? Can c ever be paralleldo d2c

to a + b? Perpendicular to a + b? Finddt

9

dt2If a and b

are unit orthogonal vectors with common origin, describe the

positions of c and show that c is perpendicular todc

20. If a and b are not parallel, show that ma + nb = ka + 3bimplies m = k, n = j.

21. Theorem of Ceva. A necessary and sufficient condition thatthe lines which join three points, one on each side of a triangle,to the opposite vertices be concurrent is that the product of thealgebraic ratios in which the three points divide the sides be -1.

22. Theorem of Menelaus. Three points, one on each side of atriangle ABC, are collinear if and only if the product of thealgebraic ratios in which they divide the sides BC, CA, AB isunity.

7. Coordinate Systems. For a considerable portion of thetext we shall deal with the Euclidean space of three dimensions.This is the ordinary space encountered by students of analyticgeometry and the calculus. We choose a right-handed coordi-nate system. If we rotate the x axis into the y axis, a right-handscrew will advance along the positive z axis.

We let i, j, k be the three unit vectors along the positive x, y,and z axes, respectively. The vectors i, j, k form a very simpleand elegant basis for our three-dimensional Euclidean space.From Fig. 11 we observe that

r=xi+yj+zk (10)

The numbers x, y, z are called the components of the vector r.Note that they represent the projections of the vector r on thex, y, and z axes. r is called the position vector of the point P


and will be used quite frequently in what follows. The most gen-eral space-time vector that we shall encounter will be of the form

u =u(x,y,z,t) =a(x,y,z,t)1+#(x,y,z,t)j+ y(x, y, z, t)k (11)

It is of the utmost importance that the student understand themeaning of (11). To be more specific, let us consider a fluid inmotion. At any time t the particle which happens to be at the

z

Fm. 11.

point P(x, y, z) will have a velocity which depends on the coordi-nates x, y, z and on the time t. As time goes on, various particlesarrive at P(x, y, z) and have the velocities u(x, y, z, t) with com-ponents along the x, y, z axes given by a(x, y, z, t), Xx, y) z, t),y(x) y, z, t).

Whenever we have a vector of the type (11), we say that wehave a vector field. An elementary example would be the vectoru = yi - xj. This vector field is time-independent and so iscalled a steady field. At the point P(1, -2, 3) it has the value- 2i - j. Another example would be u = 3xzeq - xyztj + 5xk.We shall have more to say about this type of vector in later


chapters and will, for the present, be interested only in constantvectors (uniform fields).

A moment's reflection shows that if a = ali + a2j + ask,b = b1i + ba + b3k, then

a + b = (ai + b1)i + (a2 + b2)j + (as + b3)k (12)xa + yb = (xai + yb1)i + (xaz + ybz)j + (xaa + yb3)k

8. Scalar, or Dot, Product. We define the scalar or dot prod-uct of two vectors by the identity

a b = JaJJbJ cos 0 (13)

where 0 is the angle between the two vectors when drawn from acommon origin. It makes no difference whether we choose 0 or- 0 since cos 0 = cos (- 0). This definition of the scalar productarose in physics and will play a dominant role in the develop-ment of the text.

Fm. 12.

From (13) we at once verify that

a

a

a b then a b = a

b the a bb

a b = (proj a)b (bI = (proj b)a (al (17)

SEC. 8J THE ALGEBRA OF VECTORS 11

With this in mind we proceed to prove the distributive law, whichstates that

(18)

From Fig. 13 it is apparent that

[proj (b+c)]aIaFf I= (proj b)a Iai + (proj C)a Ial

Fm. 13.

since the projection of the sum is the sum of the projections.Let the reader now prove that

Example 3. To prove that the median to the base of an isos-celes triangle is perpendicular to the base (see Fig. 14). From(5) we see that

m=1 1

so that

0FIG. 14.

which proves that OM is perpendicular to AB.Example 4. To prove that an angle inscribed in a semicircle

is a right angle (see Fig. 15).

12

c,azc

0

E

j'i

a

xam

ple

& Con

e

I° 18.

Lau

,

of T'ì

q°n°

met

r

cb

.

_

-a

C.

C a

b2

(b _a)

Ab

$o that

Exa

mpl

e

6

YE

Cr?

,oR

AN

DB

C-,

a + c

yAC ' c

BC

AC _

$

so that 4$

CA

is

arig

ht

angl

e

0

PIG

.

]a .

C2

-b2+

a2_.

2ab

co$B

_

_j.k

:g=

o

1'1

[SE

c. s

SEC. 8] THE ALGEBRA OF VECTORS

Hence if a = a1i + a2j + ask, b = b1i + b2j + b3k, then

a b = albs + a2b2 + a,b3

13

(19)

Formula (19) is of the utmost importance. Notice that 0 a = 0.Example 7. Cauchy's Inequality

(a - b)(a - b) _ 1a121b12 cost 0 5 1a12Ib12

so that from (19)

(albs + aab2 + aab3)2 S (a12 + a22 + as2)(b12 + b22 + b 32)

In generaln n n

I aaba s (Z aa2)1(I ba2)1a-1 a-1 a-1

(20)

Example 8. Let i' be a unit vector making angles a, fl, 7 withthe x, y, z axes. The projections of i' on the x, y, z axes arecos a, cos ft, cos -y, so that

i' = cosai+cos0 j+cos7k=pli+qlj+rlk (21)

Notice that p12 + q12 + r12 = 1. pi, q1, ri are called the direc-tion cosines of the vector Y. Similarly, let j' and k' be unitvectors with direction cosines p2, qa, r2 and pa, qa, r,. Thus

j' = p2i + q2j + r2k ()k' = pai+qsj+rak

We also impose the condition that i', j', k' be mutually orthogonal,so that the x', y', z' axes form a coordinate system similar to thex-y-z coordinate system with common origin 0 (see Fig. 17).

We have r = r' so that xi + yj + zk = x'i' + y'j' + z'k',where x, y, z are the coordinates of a point P as measured in thex-y-z coordinate system and x', y', z' are the coordinates of thesame point P as measured in the x'-y'-z' coordinate system.Making use of (21) and (22) and equating components, we findthat

x = p1x' + ply' + pgz'

y = q1x' + qty' + qaz' (23)

z = r1x' + r2y' + raz'

14 VECTOR AND TENSOR ANALYSIS 1Src. 8

We now find it more convenient to rename the x-y-z coordinatesystem. Let x = x1, y = x2, z = x3, where the superscripts donot designate powers but are just labels which enable us to differ-entiate between the various axes. Similarly, let x' = x1, y' = x2'

y

FIG. 17.

Z' = 28. Now let a,a represent the cosine of the angle betweenthe xa and V axes. We can write (23) as

3

xa = I a0a.V, a = 1, 2, 3 (24)a-1

By making use of the fact that i' j' = j' k' = k' i' = 0, wecan prove that

3

xa = I Asax#, a = 1, 2, 3 (25)0-1

where Asa = a). We leave this as an exercise for the reader.Let us notice that differentiating (24) yields

axa- = a,a, of, or = 1, 2, 3 (26)


Example 9. The vector a = a'i + a2j + a'k may be repre-sented by the number triple (a', a2, a'). Hence, without appeal-ing to geometry we could develop an algebraic theory of vectors.If b = (b', b2, b3), then a + b is defined by the number triple(a' + b', a2 + b2, a3 + b3), and xa = x(al, a2, a') is defined by thenumber triple (xa', xa2, xa3). From this the reader can provethat

(a', a2, a3) = a'(1, 0, 0) + a2(0, 1, 0) + a3(0, 0, 1)

The triples (1, 0, 0), (0, 1, 0), (0, 0, 1) form a basis for our linearvector space, that is, the space of number triples. We note thatthe determinant formed from these triples, namely,

1 0 0

0 1 0 =10 0 1

does not vanish. Any three triples whose determinant does notvanish can be used to form a basis. Let the reader prove thisresult. We can define the scalar product (inner product) oftwo triples by the law (a - b) = a'b' + a2b2 + a3b3.

9. Applications of the Scalar Product to Space Geometry(a) We define a plane as the locus of lines passing through a

fixed point perpendicular to a fixed direction. Let the fixedpoint be Po(xo, yo, zo) and let the fixed direction be given by thevector N = Ai + Bj + Ck. Let r be the position vector to any

-4

point P(x, y, z) on the plane (Fig. 18). Now POP = r - ro is per-pendicular to N so that

or[(x - xo)i + (y - yo)j + (z - zo)kl - (Ai + Bj + Ck) = 0

and

A(x - xo) + B(y - yo) + C(z - zo) = 0 (27)

This is the equation of the plane. The point Po(xo, yo, zo) obvi-ously lies in the plane since its coordinates satisfy (27). Equa-tion (27) is linear in x, y, Z.

(b) Consider the surface Ax + By + Cz + D = 0. Let P(xo,yo, zo) be any point on the surface. Of necessity,

Axo+Byo+Czo+D=0.

16 VECTOR AND TENSOR ANALYSIS [SEC.9

Subtracting we have

A(x - xo) + B(y - yo) + C(z - zo) = 0 (28)

Now consider the two vectors Ai + Bj + Ck and

(x - xo)i + (y - yo)j + (z - zo)k

Fia. 18.

Equation (28) shows that these two vectors are perpendicular.Hence the constant vector Ai + Bj + Ck is normal to the sur-face at every point so that the surface is a plane.

Fia. 19.

(c) Distance from a point to a plane. Let the equation of theplane be Ax + By + Cz + D = 0, and let P(E, % r) be any pointin space. We wish to determine the shortest distance from P


to the plane. Choose any point Po lying in the plane. It isapparent that the shortest dis-

will be the projection oftance

PoP on N, where N is a unitvector normal to the plane (seeFig. 19). Now

d = JPoP NJ

= IAZ+B, +Cr+DI (29)

(A2 + B2 + C2)f

where use has been made of thefact that

Axo+Byo+Czo+D=O. FIG. 20.

(d) Equation of a straight line through the point Po(xo, yo, zo)parallel to the vector T = li + mj + A. From Fig. 20 it is

z

Fia. 21.


apparent that r - ro is parallel to T so that r - ro = AT,

-- <A<+O0Hence (x - xo)i + (y - yo)j + (z - zo)k = A(li + mj + nk), sothat equating components yields

x xo y - yo -z - zo=Am n

(30)

By allowing A to vary from - oo to + oo we generate every pointon the line.

(e) Equation of a sphere with center at Po(xo, yo, zo) and radius a.In Fig. 21 obviously (r - ro) (r - ro) = a2, or

(x - xo) 2 + (y - yo) 2 + (z - zo) 2 = a2

Problems

1. Add and subtract the vectors a = 2i - 3j + 5k,

b = -2i+2j+2kShow that the vectors are perpendicular.

2. Find the cosine of the angle between the two vectorsa = 2i - 3j + k and b = 3i - j - 2k.

3. If c is normal to a and b, show that c is normal to a + b,a - b.

4. Let a and b be unit vectors in the x-y plane making anglesa and 6 with the x axis. Show that a = cos a i + sin a j,b = cos 9 i + sin 3 j, and prove that

cos (a - 6) = cos a cos S + sin a sin 8

5. Find the equation of the cone whose generators make anangle of 30° with the unit vector which makes equal angles withthe x, y, and z axes.

6. The position vectors of the foci of an ellipse are c and - c,and the length of the major axis is 2a. Show that the equationof the ellipse is a4 - a2(r2 + c2) + (c r)2 = O.

7. Prove that the altitudes of a triangle are concurrent.8. Find the shortest distance from the point A(1, 0, 1) to the

line through the points B(2, 3, 4) and C(-1, 1, -2).


9. Let a = 2i - j + k, b =i-3j - 5k. Find a vector cso that a, b, c form the sides of a right triangle.

10. Let r be the position vector of a point P(x, y, z), and let abe a constant vector. Interpret the equation (r - a) a = 0.

11. Given a = 2i - 3j + k, b = 3j - 4k, find the projectionof a along b.

12. Show that the line joining the end points of the vectorsa =2i -j - k, b = -i + 3j - k with common origin at O isparallel to the x-y plane, and find its length.

13. Prove that the sum of the squares of the diagonals of aparallelogram is equal to the sum of the squares of its sides.

14. Let a = a, + a2 where a, b = 0 and a2 is parallel to b.

Show that a2 = (iblb b, al = a - (a . b) b.JbJ2

15. Derive (25).16. Verify (26).17. Find a vector perpendicular to the vectors a = i - j + k,

b=2i+3j-k.18. Let a = f(t)i + g(t)j + h(t)k, and define

da = f'(t)i+ g'(t)j +

h'(t)kdt

Show thatda a = lal

daldt dt

19. Find the angle between the plane Ax + By + Cz + D = 0and the plane ax + by + cz + d = 0.

20, a, b, c are coplanar. If a is not parallel to b, show that

c=

ca ab aa caIcb bbi ab cb b

aa abab bb

21. For the an", Asa defined by (24) and (25), show that

a, -Ap r = Sp-y-1

where &0°= Iifa=l4,Se=Oif aPd f.22. If B1, B2, B3 are the components of a vector B, that is,

B = B'i + B2j + B3k (see Example 8 in regard to the super-


scripts), show that for a rotation of axes the components of the3

vector B become (BI, B2, B3) where B° _ I ayaB#, a = 1, 2, 3,Ps1

and B = BIi' + B2j' + Mk'. Read Example 8 carefully.23. Show that for a rotation of axes,

BICI + B2C2 + B3C3 = BICI + B2C2 +,&3103

This shows the invariance of the scalar product for rotations ofaxes. The invariance here refers to both the numerical invari-ance of the scalar product and the formal invariance,

3

BaC' = Z BaC'aaa1 amt

24. Prove the statements made in Example 9.25. Generalize the statements of Example 9 for n-tuples

(a', a2 a")

10. Vector, or Cross, Product. Given any two nonparallelvectors a and b, we may construct a third vector c as follows:

F ia. 22.

When translated so that theyhave a common origin, thetwo vectors a, b form twosides of a parallelogram. Wedefine c to be perpendicularto the plane of this parallelo-gram with magnitude equalto the area of the parallelo-gram. We choose that nor-mal obtained by the motionof a right-hand screw when ais rotated into b (angle of ro-

tation less than 180°) (see Fig. 22). A cross is placed between thevectors a and b to denote the vector c = a x b. The vector cis called the cross, or vector, product of a and b and is given by

c=axb=lallbl sin0E (31)

where JEl = 1. The area of the parallelogram is

A = !alibi sin 0

SEc. 121 TIIE ALGEBRA OF VECTORS 21

The cross product will occur frequently in mechanics and elec-tricity, but for the present Ave discuss its algebraic behavior. Itis obvious that a x b = -b x a, so that vector multiplicationis not commutative. If a and b are parallel, a x b = 0. Inparticular, a x a = 0.

11. The Distributive Law for the Vector Product. We desireto prove that a x (b + c) = a x b + a x c. Let

u= ax(b+c)-axb-axcand form the scalar product of this vector with an arbitraryvector v. We obtain

x(b+c)] - x b)

In Sec. 13 we shall show that a (b x c) = (a x b) c. Hence

This implies either that u = 0 or that v is perpendicular to u.Since v is arbitrary, we can choose it not perpendicular to u.Hence u = 0 and

ax(b+c) =a xb+a xc (32)

This proof is by Professor Morgan Ward of the California Insti-tute of Technology.

12. Examples of the Vector ProductExample 10

ixi=jxj=kxk=0ixj=k, jxk=i, kxl=j

For the vectors a = all + a2j + ask, b = b1i + b2j + b3k weobtain a x b = (a2b3 - a3b2)i + (a3b1 - alb3)j + (aib2 - a2b1)kby making use of the distributive law of Sec. 11. Symbolicallywe have

i j kaxb = a1 a2 as

bl b2 b3(33)

where (33) is to be expanded by the ordinary method of determi-nants.

22

Example 11.

Example 12.

or

c=b - ac xc=cx(b-a)

0=cxb-cxac xa = c xb

Fia. 23.

However, if two vectors are equal, their magnitudes are equal sothat

lellal sin I = lcilbl sin aand

a b c

sin a sin fl sin ti

Example 13. Rotation of a Particle. Assume that a particle isrotating about a fixed line L with angular speed w. We assumethat its distance from L remains constant. Let us define theangular velocity of the particle as the vector w, whose directionis along L and whose length is w. We choose the direction of win the usual sense of a right-hand screw advance (see Fig. 24).

VECTOR AND TENSOR ANALYSIS [SEC. 12

a = 2i - 3j + 5k, b = -i + 2j - 3k, so that

a xb =i j k

2 -3 5

-1 2 -3

Sine law of trigonometry

SEC. 13] THE ALGEBRA OF VECTORS 23

It is our aim to prove that the velocity vector v can be representedby w x r, where r is the position vector of P from any origintaken on the line L. Let the reader show that v and w x r areparallel. Now lw x rl = wa = speed ofP, so that L

aOP

%13. The Triple Scalar Product. Letus consider the scalar a (b x c). This vscalar represents the volume of the par-allelepiped formed by the coterminoussides a, b, c, since

a- (b xc) =fallbllclsin0cosa= hA = volume (35)

A being the area of the parallelogramwith sides b and c, the altitude of the parallelepiped beingdenoted by h (see Fig. 25).

Fia. 25.

Now

a (b x c) = (a,i + a2j + ask) .

= ai(b2ca - bac2) + a

i j kb, b2 bs

c1 C2 C3

(baci - bic3) + as(bic2 - b2c1)

so that

a1 a2 as

a (b x c) = b, b2 bs (36)C1 C2 C3


Notice that a (b x c) = (a x b) c since both terms representthe volume of the parallelepiped. It is also very easy to showthat the determinant of (36) represents (a x b) c. We usuallywrite a (b x c) = (abc) since there can be no confusion as towhere the dot and cross belong. We note that

(abc) = (cab) = (bca)

and that (abc) _ - (bac) = - (cba) = - (acb). These resultsfollow from elementary theorems on determinants. We are thusallowed to interchange the dot and the cross when working withthe triple scalar product. This result was used to prove (32).If the three vectors a, b, c are coplanar, no volume exists, and weat once have (abc) = 0. In particular, if two of the three vectorsare equal, the triple scalar product vanishes.

14. The Triple Vector Product. The triple vector producta x (b x c) plays an important role in the development of vectoranalysis and in its applications. The result is a vector sinceit is the vector product of a and (b x c). This vector is there-fore perpendicular to b x c so that it lies in the plane of b and c.If b is not parallel to c, a x (b x c) = xb + yc, from Sec. 6.Now dot both sides with a and obtain x(a b) + y(a c) = 0,

since a [a x (b x c)] = 0. Hence(a X = (a b) = X, where

X is a scalar, so that

a x (b x c) _ X[(a c)b - (a b)c] (37)

In the special case when b = a, we can quickly prove that X = 1.We dot (37) with c and obtain

c [a x (a x c)] = X[(a c)2 - a2c2Jor

-(a x c)2 = X[(a c)2 - a2c2J

by an interchange of dot and cross. Hence

-a2c' sine O= A(a'c2 cost 0- a2c2) = -Xa'c2 sin2 0

so that A = 1. Hence a x (a x c) = (a c)a - (a a)c. Fromthis it immediately follows that (a x b) x b = (a b)b - (b b)a.Now we prove that A = 1 for the general case. We dot (37)with b and obtain

SEC. 15] THE ALGEBRA OF VECTORS

x(b xc)]

-[(a xb)x b (a

x b) x (a b)(b c)

implying X = 1. Thus

a x (b x c) = (a c)b - (a b)c

We leave it to the reader to show that

25

(38)

(a x b) x c = (a c)b - (b c)a

Notice that a x (b x c) 5-1 (a x b) x c. If b is parallel to c,(38) reduces to the identity 0 = 0, so that (38) holds for any threevectors. The expansion (38) of a x (b x c) is often referred toas the rule of the middle factor.

More complicated products are simplified by use of the tripleproducts. For example, we can expand (a x b) x (c x d) byconsidering (a x b) as a single vector and applying (38).

(a x b) x (c x d) = (a x b d)c - (a x b c)d= (abd)c - (abc)d (39)

Also(c x d)

d) (b c) d) (40)

15. Applications to the Spherical Trigonometry. Consider thespherical triangle ABC (sides are arcs of great circles) (see Fig.26). Let the sphere be of radius 1. Now from (40) we see that

(a x b) (a x c) = (b c) - (a b) (a c)

The angle between a x b and a x c is the same as the dihedralangle A between the planes OAC and OAB, since a x b is per-pendicular to the plane of OAB and since a x C is perpendicularto the plane of OAC. Hence

sin y sin # cos A = cos a - cos y cos,6.


FIG. 26.

Problems

1. Show by two methods that the vectors a = 2i - 3j - k,b = -6i + 9j + 3k are parallel.

2. Find a unit vector perpendicular to the vectors

a=i-j+kb=i+i - k

3. A particle has an angular speed of 2 radians per second, andits axis of rotation passes through the points P(0, 1, 2),

Q(1, 3, -2)

Find the velocity of the particle when it is located at the pointR(3, 6, 4).

4. Find the equation of the plane passing through the endpoints of the vectors a = a1i + a j + ask, b = b1i + bd + bsk,c = c1i + cd + cak, all three vectors with origin at P(0, 0, 0).

5. Show that (a x b) x (c x d) = (acd)b - (bcd)a.6. Prove that d x (a x b) (a x c) = (abc) (a d).7. If a + b+ c= 0, prove that a x b= b x c= c x a,

and interpret this result trigonometrically.

SEC. 15] THE ALGEBRA OF VECTORS 21

8. Four vectors have directions which are outward perpen-diculars to the four faces of a tetrahedron, and their lengths areequal to the areas of the faces they represent. Show that thesum of these four vectors is the zero vector.

9. Prove that (a x b) (b x c) x (c x a) = (abc)2.10. If a, b, c are not coplanar, show that

d = (dbc) a + (adc) b + (abd)c

(abc) (abc) (abc)

for any vector d.11. If a, b, c are not coplanar, show that

(c d) (a d) (b d)d= a x b + - b x c -1- c x a(abc) (abc) (abc)

for any vector d.12. Prove that

a x(b xc) +b x(c x a)+ c x (a xb) =013. The four vectors a, b, c, d are coplanar. Show that

(a x b) x (c x d) = 0.14. By considering the expansion for (a x b) x (a x c), derive

a spherical trigonometric identity (see Sec. 15).15. Show that

(a x b) (c x d) x (e x f) = (abd) (cef) - (abc) (def)= (abe) (fcd) - (abf) (ecd)= (cda) (bef) - (cdb) (aef)

16. Find an expression for the shortest distance from the endpoint of the vector r, to the plane passing through the end pointsof the vectors r2, r8, r4. All four vectors have their origin atP(0, 0, 0).

17. Consider the system of equations

a1x + b,y + c1z = d,a2x + b2y + c2z = d2 (41)

asx + bsy + c3z = da

Let a = aui + a2j + aak, etc. Write (41) as a single vector equa-tion, and assuming (abc) F6 0, solve for x, y, z.

18. Find the shortest distance between two straight lines inspace.

28 VECTOR AND TENSOR ANALYSIS [Ssc. 15

19. Show directly that a x (b x c) _ (a c)b - (a b)c, wherea, b, c take on the values i, j, k in all possible ways. Now showthat, (38) is linear in a, b, c, that is,

a x [b x (ac + ad)] = as x (b x c) + #a x (b x d)

etc. Since any vector is a linear combination of i, j, k, explainwhy (38) holds for all vectors.

20. If a and b lie in a plane normal to a plane containing c andd, show that (a x b) (c x d) = 0.

21. If (a', a2, a3), (b', b2, b3) are the components of the vectorsa, b, show that three of the nine numbers Ca9 obtained by consid-ering cab = aab# - a8ba, a, a = 1, 2, 3, represent the componentsof a x b, that three others represent b x a, while the remainingthree vanish. Represent c,--P as a matrix and show that

cap = - G a

By considering a rotation of axes (see Example 8), show that theEaP in the new coordinate system are related to the caP in the old

a a

coordinate system by the equations &A = I I a aa,i9c°r,r-1 a=1

a, 6 = 1, 2, 3. Show that caP _ -Cfta.The numbers c' = a2b3 - a3b2, c2 = a3b' - a1b3,

c3 = a'b2 - a2b'

a

are the components of a x b. Show that ca = 1 apa#S under aB=1

rotation of axes (see Prob. 22, Sec. 9).22. We can construct a x b by three geometrical constructions.

We first construct a vector normal to b lying in the plane of aand b. We project a onto this vector, and finally we rotatethis new vector through an angle of 90° about the axis parallelto b, magnifying this newly constructed vector by the factor IblThe final result yields a x b. Use this to prove that

a x (b + c) = a x b + a x c

CHAPTER 2

DIFFERENTIAL VECTOR CALCULUS

16. Differentiation of Vectors. Let us consider the vectorfield

u = a(x,y,z,t)i+13(x,y,z,t)j+ '(x,y,z,t)k (42)

At any point P(x, y, z) and at any time t, (42) defines a vector.If we keep P fixed, the vector u can still change because of thetime dependence of its components a, S, and y. If we keep thetime fixed, we note that the vector at the point P(x, y, z) will, ingeneral, be different from that at the point

Q(x + dx, y + dy, z + dz)

Now, in the calculus, the student has learned how to find thechange in a single function of x, y, z, t. What difficulties do weencounter in the case of a vector? Actually none, since we easilynote that u will change if and only if its components change.Thus a change in a(x, y, z, t) produces a change in u in thex direction, and similarly changes in f3 and y produce changes in uin the y and z directions, respectively. We are thus led to thefollowing definition:

du = dai.+d,6i+dyk (43)

as as asdu (-aa

ax dx + ay dy + az dz + atd) i

( +(axdx+aydy+a-dz atdt)j

+ ax dx + dy -}- az dz }- at dt J k

For example, let r = xi + yj + zk be the position vector of amoving particle P(x, y, z) in three-space. Then

dr = dx i + dy j + dz k29


and

vdt dt 1 + dt + dt

k

d 2 z 2y , 2ra =

dt2 = dt2 I + dt2j + dt2 k

(44)

(45)

Equations (44) and (45) are, by definition, the velocity andacceleration of the particle. We have assumed that the vectorsi, j, k remain fixed in space.

If the vector u depends on a single variable t, we can define

dulim

u(t + At) - u(t)dt At-,o At

(46)

(see Fig. 27). It is easy to verify that (46) is equivalent to (43).Example 14. Consider a

particle P moving on a circle:Q/ C -A-. -44-1k -4- -

so that

Fio. 27.

angular speed w = - (Fig

28). We note that

r=rcos0i+rsin0j

v = dt = (-r sin 6i+rcos Bj) de

and

a = dt dt2 =(-r cos 0 i - r sin B j)

C

(dA2dt/

Therefore the acceleration is

a = -w2r (47)

The point P has an acceleration toward the origin of constantmagnitude w2r. This acceleration is due to the fact that the

SEC. 16] DIFFERENTIAL VECTOR CALCULUS 31

velocity vector is changing direction at a constant rate; it iscalled the centripetal acceleration.

z

Y

Fra. 28. FIG. 29.

Example 15. Let P be any point on the space curve (Fig. 29)

x=x(s)y = y(s)z = z(s)

where s is are length measured from some fixed point Q. Now

r = x(s)i + y(s)j + z(s)k

so thatdr _ dx , dy , dz

ds

and

dr dr dx 2 dy z (dz 2ds ds = ds + \dsI + \ds)

=dx2+dy2+dz2=1ds2

(48)

(49)

from the calculus. Hence ds is a unit vector. As As -> 0, the

position of Ar approaches the tangent line at P. Hence (49)

represents the unit tangent vector to the space curve (48).

32 VECTOR AND TENSOR ANALYSIS

17. Differentiation Rules. Consider

'p(t) = u(t) v(t)

[SEC. 17

'P(t + At) - ,P(t) = u(t + At) v(t + At) - u(t) o(t)

Now

(see Fig. 27), so that

u(t + At) = u(t) + Auv(t + At) = v(t) + AV

c(t+At At At At

and passing to the limit, we obtain

Similarly

d (u v) dv du_dt

- udt + dt'v

d(u x v) dv dudt -

ux dt + dt

v

d(fu) du d_f-dt f dt + dt

u

(50)

(51)

(52)

Notice how these formulas conform to the rules of the calculus.Example 16. Let u(t) be a vector of constant magnitude.

Therefore

u u = u2 = constant

By differentiating we obtain

du duu'dt+

u'dt=0

SEC. 17] DIFFERENTIAL VECTOR CALCULUS 3m

Hence either dt = 0 or dl is perpendicular to u. This is an

important result and should be fully understood by the student.The reader should give a geometric proof of this theorem.

Example 17. In all cases yu u = u2 where u is the lengthof u. Differentiation yields

2u ' dt = 2u d and

u'du

-du (53)auat

This result is not trivial, forIdul du.

FIG. 30.

Example 18. Motion in a Plane. Now r = rR, where R is aunit vector (see Fig. 30). Hence

-I-rV-dt dtR dR

NowR

is perpendicular to R (see Example 16). AlsoIdRI

= desince R is a unit vector. We can easily verify this by differ-

entiating R = cos B i + sin 9 j. Hence v =Wt-

R + r d8 P, where

P is a unit vector perpendicular to R. Differentiating again weobtain

_ dv d2r dr dR dr d6 d20 dB dP

a_

dt dt'R+dl dt +dtdt P+rdt2 P+rdt dtor

since

d2ra_dt R+2dtd6P+rd- P-r(d8)2R

P= -d6R(54)

34 VECTOR AND TENSOR ANALYSIS [Sic. 17

Thus

I d2r - (dO)2]Rid(odO\pdt r dt dt

(55)

Problems

1. Prove (51) and (52).2. Prove (54). /3. Differentiate l r - dr1 with respect to t.

4. Expand dt [p x (q x r)].

5. Show thatd r x dt) = r x

d2r

dr d2r6. Find the first and second derivatives of (r dt dt27. r = a cos wt + b sin wt; a, b, w are constants. Prove that

r xdt = wa xb andd +w2r = 0.

8. If r x dt = 0, show that r has a constant direction.

9. R is a unit vector in the direction r. Show that

r x drRxdR=7.2

10. If dt = w x a,d

= w x b, show that

dt(axb)=0x(axb)

11. If r = aew' + be-,", show that dt; -- w2r = 0. a, bare con-

start vectors.

12. Find a vector u which satisfies dtu = at + b. a, b are con-

stant vectors. Is u unique?d

13. Show thatdt()

r dt r dt r.


14. Let r° be the position vector to a fixed point P in space, andlet r be the position vector to a variable point Q lying on a spacecurve r = r($). Show that if the distance Pte(- is a minimum, thenr - r° is perpendicular to the tangent at Q. Show also that

z l2 d1r

r ds2 + ds, r0 ds2

15. If u = a(x, y, z, t)i + 13(x, y, z, t)j + y(x, y, z, t)k, showdu au au dx au dy au dzthat _dt at + ax dt + ay dt +

_az dt

z

Fia. 31.

16. The transformation between rectangular coordinates andspherical coordinates is given by

x=rsin0cos0y=rsin0sinrpz = r cos 0

where 0 is the colatitude, (p is the longitudinal or azimuthal angle,and r is the magnitude of the position vector r from the originto the particle in question. Find the components of the velocityand acceleration of the particle along the unit orthogonal vectorser, ee, e,, (see Fig. 31).

36 VECTOR AND TENSOR ANALYSIS 1SEc. 1b

17. Consider the differential equation

(i) du+2Adu+Bu=0

where A, B are constants. Assume a solution of the formu(t) = e'°'C, where C is a constant vector, and show thatu(t) = Clew,' + C2e'°2' is a solution of (i), w1, w2 being roots ofw2 + 2Aw + B = 0. Consider the cases for which A2 - B < 0,A 2 - B = 0, A2 - B > 0.

18. Find the vector u which satisfies

d3u d2u2du

0a dtedt3

such that u = i,du = j, d = k for t = 0.

19. If u, is a solution of

d8u+Ad2u+Bdu+Cu=0

and if u2 is a solution of

ddtu+Adtu+Bdt+Cu=F(t)

show that u1 + U2 is a solution of (ii) provided A, B, C areindependent of u. Why is this necessary?

20. A particle moving in the plane of (r, 8) has no transverse

acceleration, that is,

r

dtt r2 df) = 0. Show that the radius

vector from the origin to the particle sweeps out equal areas inequal intervals of time.

18. The Gradient. Let p(x, y, z) be any continuous differenti-able space function. From the calculus

dVa a a

=ax

dx +ay

dy +az

dz (56)

SEC. 181 DIFFERENTIAL VECTOR CALCULUS 37

Now let r be the position vector to the point P(x, y, z).

r = xi + yj + zk

If we move to the point Q(x + dx, y + dy, z + dz) (Fig. 32),

dr = dxi + dyj + dzk

Now notice that (56) contains the terms dx, dy, dz and the terms

aP, aSP,aP We define a new vector formed from gp by taking itsax 8y az

three partial derivatives. Let del rp = gradient p be defined by

Vsp=axi+ J+aZky

We immediately see that

d(p = dr VV (58)

We shall now give a geomet-rical interpretation of IV-p.

At the point P(xo, yo, zo), so hasthe value gp(xo, yo, zo) so that

go(x, y, z) = gp(xo, yo, zo)

represents a surface whichobviously contains the point

P(xo, yo, zo) FIG. 32.

(57)

As long as we move along this surface, (p has the constant value,p(xo, yo, zo) and d-r = 0. Consequently, from (58),

(59)

Now Vgp is a vector which is at once completely determined after,p has been differentiated, and Eq. (59) states that Vgp is perpen-dicular to dr as long as dr represents a change from P to Q, whereQ remains on the surface v = constant. Thus V(p is normal toall the possible tangents to the surface at P so that V(p mustnecessarily be normal to the surface (p(x, y, z) = constant (see

38 VECTOR AND TENSOR ANALYSIS [Sec. 18

Fig. 33). Let us now return to dip = dr Vv. The vector Vp isfixed at any point P(x, y, z), so that dip (the change in p) willdepend to a great extent on dr. Certainly dcp will be a maximumwhen dr is parallel to Vp, since dr VV = Idrl IV pl cos 0, and cos 8

VOis a maximum for 0 = 0°.Thus Vp is in the direction ofmaximum increase of p(x, y, z).Let ldrl = ds so that

d(P = u V(p (60)ds

Fia. 33.

where u is a unit vector inthe direction dr. Hence thechange of (p in any directionis the projection of Vp onthe unit vector having thisdirection.

Example 19. To find a unit vector normal to the surfacex2 + y2 - z = 1 at the point P(1, 1, 1). Here

,P(x,y,z) =x2+y2-zV(p=2xi+2yj-k

=2i-2j-katP(1,1,1)Thus

N=2i-2j-k3

Example 20. We find Vr if r = (x2 + y2 + Z2)1. The surfacer = constant is a sphere. Hence Vr is normal to the sphere andso is parallel to the position vector r. Thus Vr = kr. Now

dr = dr Vr = k dr r = kr dr from (53)Therefore

k = 1 and Vr = r = R (61)r r

Example 21Vf(u) = f'(u) Vu, u = u(x, y, z)


Proof:

Vf(u) =

ai+ayj+azk

= r(u) au i + f'(u)au j +

f'(u)au k

ax ay az

= f'(u) Vu

Example 22

Vf(ul, u2, . . . , un) =a i+a j+azky

_ of aua , of aua , of au--

1au, ax aua ay aua az )

a= Ifaua Vua (62)

Example 23. Consider the ellipse given by rl + r2 = constant(see Fig. 34). Now V(rl + r2) is normal to the ellipse. Let

Y

n

Fio. 34.

T be a unit tangent to the ellipse. Thus V(rl + r2) T = 0, and

W2 - T (63)

But from Example 20, Vrj is a unit vector parallel to the vector

AP, and Vr2 is a unit vector parallel to the vector BP. Equation


(63) shows that AP and BP make equal angles with the tangentto the ellipse.

19. The Vector Operator V. We define

v=ia +iay

a+kaz (64)

Notice that V is an operator, just as dx is an operator in the differ-

ential calculus. Thus

v - (1-+ja +kaz)'py

a vector operator because of its componentsa a a

It will help us in the future to keep in mind that Vax ay azacts both as a differential operator and as a vector.

Example 24

v (uv) =i a (uv)

+ Jq!!!) + k a (uv)

ax ay az

C'ax+j a+kc1v )u+Clax +jau+k )vy az y -az

V(uv) = U Vv + v Vu (65)

This result is easily remembered if we keep in mind that V is adifferential operator, so that we can apply the ordinary rules ofcalculus.

Problems

1. Find the equation of the tangent plane to the surfacexy - z = 1 at the point (2, 1, 1).

Sac. 191 DIFFERENTIAL VECTOR CALCULUS 41

2. Show that V(a r) = a, where a is a constant vector and ris the position vector.

3. If r = (x2 + y2 + Z2)'1, find Vr" by explicit use of (57).4. If So = (r x a) (r x b), show that

V,p = b x (r x a) + a x (r x b)

when a and b are constant vectors.5. Let gp = x2 + y2. Find IVpI and show that it is the maxi-

mum change of V.6. Find the cosine of the angle between the surfaces

x2y + z = 3

and x log z - y2 = -4 at the point of intersection P(- 1, 2, 1).7. What is the value of Vv(x, y, z) at a point that makes So a

maximum?8. The surfaces g(x, y, z) = constant and #(x, y, z) = con-

stant are normal along a curve of intersection. What is thevalue of V(p V4' along this curve?

9. What is the direction for the maximum change of the spacefunction gp(z, y, z) = x sin z - y cos z at the origin?

10. Expand V(u/v) where u = u(x, y, z), v = v(x, y, z).11. Let r and x be the distances from the focus and directrix

to any point on a parabola. We know that r = x. Show that(R - i) T = 0, where T is a unit tangent vector to the parabola,and interpret this equation.

12. Show that the ellipse r, + rz = c, and the hyperbolar, - ra = c2 intersect at right angles when they have the samefoci.

13. If Vv is always parallel to the position vector r, show thatrp=(p(r),r== x2+ysd-z2.

14. Find the change of g = xyz in the direction normal to thesurface yx2 + xy2 + z'y = 3 at the point P(1, 1, 1).

15. If f = f(xl, x', xa) (see Example 8), and if

x°t = x01(yl, y2, ya),

a = 1, 2, 3, show that

of of axea = 1, 2, 3aye aya


axa3

ax" aye 1 if« R

Using the fact that_ _

` , show thatax8 1 ay° ax8 0 if a 0 ,e

saof of ay '

a = 1, 2, 3.axa 9-1 ay8 axa

16. Apply the results of Prob. 15 above to the transformation

x = r cos 0y = r sin 0z = z

and show that af, 1 of , of are the components of Vf along theOr r ae az

three mutually orthogonal unit vectors e, es, e. which occur incylindrical coordinates.

17. If (_ v(x, y, z, t), show that d,P= ate +dr

V(p.

18. Ifr =xi+yj+zk,find

xyevzi:' + log(x + z

-X z+ y)

19. Ifu =u(x,y,z,t)showthat dt = au, +Cad V>u.

20. If u(tx, ty, tz) = t"u(x, y, z), show that (r V)u = nu.20. The Divergence of a Vector. Let us consider the motion

of a fluid of density p(x, y, z). We assume that the velocityfield is given by f = u(x, y, z)i + v(x, y, z)j + w(x, y, z)k. Thistype of motion is called steady motion because of the explicitindependence of p and f on the time. We now concentrate onthe flow through a small parallelepiped ABCDEFGH (Fig. 35),of dimensions dx, dy, dz.

Let us first calculate the amount of fluid passing through theface ABCD per unit time. The x and z components of thevelocity f contribute nothing to the flow through ABCD. Themass of fluid entering ABCD per unit time is given by pv dx dz.The mass of fluid leaving the face EFGH per unit, time is

r a(pv) 1pv +

aydy dx dz

SEC. 20) DIFFERENTIAL VECTOR CALCULUS 43

The loss of mass per unit time is thus seen to be equal to

a(pv)dx dy dz

ay

If we also take into consideration the other two faces, we findthat the total loss of mass per unit time is

[a(Pu)+o(Pv)+a(Pw)]dddz

z

A dy E

ry

Fia. 35.

so tha t

a(Pu) + a(pv) + a(Pw) (66)ax ay az

represents the loss of mass per unit time per unit volume. Thisquantity is called the divergence of the vector pf. We see atonce that

a(Pu) a(Pv)V V dif f

a(Pw) 1 dM67. (p ) = v (p ) =

ax ay ++=

az V dt( )

44 VECTOR AND TENSOR ANALYSIS [SEc. 20

since i, j, k are constant vectors. M and V are the mass andvolume of the fluid.

The divergence of any vector f is defined as V f. We nowcalculate the divergence of p(x, y, z)f.

Va(q.u) + a(te) + a(cyw)

ax ay az

_ (axau av aw arp a'p sip

x +ay+ az + uax +v ay +u'aa

(68)

We remember this result easily enough if we consider V as avector differential operator. Thus, when operating on spf, wefirst keep p fixed and let V operate on f, and then we keep f fixedand let V operate on V(V V is nonsense), and since f and Vp arevectors we complete their multiplication by taking their dotproduct.

Example 25. Compute V f if f = r/r' (inverse-square force).

V (r-'r) = r-3V r+r Vr-'

V (r- 3r) = 0 (69)

This is an important result. The divergence of an inverse-squareforce is zero. We note that

y

Example 26. What is the divergence of a gradient?

k/axe+ayz+az,

SEc.21] DIFFERENTIAL VECTOR CALCULUS

This important quantity is called the Laplacian of 0.

v = v2 =axe aye az2

45

(70)

21. The Curl of a Vector. We postpone the physical meaningof the curl and define

curl f=vxf=i j

a a a

ax ay az

u v w

Caw _ -J 1 _ awl+ k

1 _ auvxf = i +ay az az ax ax ay,

Example 27

V x r =

Example 28

1 i j k

v x (wf) =a a a

=0

(71)

ax ay azcpu vv cv[a(te) _ a()1 [a(sou) a(caw)

ay az JJ az ax

+ k [a(;) a(,Pu)]ax ay

aw avv x (cOf) = cO [i (ay az

+jau

az

Ow

ax

U v w

Vx('vf)=(Pvxf+Via xf (72)


This result is easily obtained by considering V as a vector differ-ential operator.

Example 29. To show that the curl of a gradient is zero.

V X (VP) =

i j ka a a

ax ay azav av avax ay az

a2 a2IPi -

ay az az ay

+ i

a2,p

a2, + k

02,p C1 2,p

az ax ax az/ \ax ay ay ax/\Hence

Vxvio =0 (73)

provided (p has continuous second derivatives.Example 30. To show that the divergence of a curl is zero.

V (v x f)a aw_av a au_aw' a(av_au)

= ax ay az + ay az ax / + az \ax ay/a2u a2ul a2V a2v a2w a2w

ay az az ay + az ax ax az + ax ay - ay axThus

(74)

Example 31. What does (u V)v mean? We first dot u withV. This yields the scalar differential operator

a a aux ax + ua ay + U. az

Then we operate on v obtaining

Thus

uxax+Uyay+u$az

df=axdx+af dy+azdzy

=dxa+dya+dzazy

SFC. 211 DIFFERENTIAL VECTOR CALCULUS 47

and

df = (dr V)f

since dr = dx i + dy j + dz k.If f = f(x, y, z, t),

df = (dr V)f + a dt

Example 32

+vsa= vJ + vyj + vZk

(75)

(76)

(v - V)r = v (77)

where r is the position vector xi + yj + zk.Example 33. Let us expand V(u v). Now

u x (V x v) = Vti(u v) - (u V)v

Here we have applied the rule of the middle factor, noting alsothat V operates only on v. Vn(u v) means that we keep thecomponents of u fixed and differentiate only the components of v.

Similarly, v x (V x u) = v) - (v V)u. Adding, weobtain

Vu(u v) + Vn(u v) = u x (V x v) + v x u)+ (u V)v + (v V)u

and

u x(V xv) + v x(V xu) + (78)

Example 34

V x (u x v) = (v V)u - v(V u) + u(V v) - (u V)v (79)


Example 35

V. (u xv) = Vu (u xv) +Vv (u xv)

(V (V (80)

Example 36

V X (V x v) = V(V v) - V2v (81)

Example 37. Let A = V x ((pi) where V2,p = 0. We now com-pute A V x A. Since A = V(p x i, we obtain

V x A = V x V xi = V iV2acix

from (7), Sec. 22. Thus

i j k

A V x A =app app app

ax ay az1 0 0

a2 a2y ()2

-P

ax2+j ay ax+kazax

app a2 app a2g= az ay ax ft azax

If also sp = X(x)Y(y)Z(z), we can immediately conclude that

22. Recapitulation. We relist the above results:1. V (UV) = u Vv + v Vu2.

V V xv+V(pxv4. V x (V(p) = 05. V. (Vxv)=06. V. (u x v) _ (V x u) - v - (V x v) - u7. V x (u x v) _ (v V)u - v(V u) + u(V v) - (u V)v8.

VV x r = 0

13. df = (dr V)f + a dt

Sxc. 22] DIFFERENTIAL VECTOR CALCULUS 49

14. dco = dr VV + LP dt

15. V- (r 3r) =0

Problems

1. Show that V2(1/r) = 0 where r = (x2 + y2 + z2)}.2. Compute V2 r, V2r2, V2(1/r2) where r = (x2 + y2 + z2)3.3. Expand V(uvw).4. Find the divergence and curl of (xi - yj)/(x + y); of

xcoszi+ylogxj -z2k.5. If a = axi + fiyj + yzk, show that V(a r) = 2a.6. Show that V x V(r)r] = 0 when r = (x2 + y2 + Z2) i and

r = xi + yj + zk.7. Let u = u(x, y, z), v = v(x, y, z). Suppose u and v satisfy

an equation of the form f(u, v) = 0. Show that Vu x Vv = 0.8. Assume Vu x Vv = 0 and assume that we move on the sur-

face u (x, y, z) = constant. Show that v remains constant andhence v = f(u) or F(u, v) = 0.

9. Prove that a necessary and sufficient condition that u, v, wsatisfy an equation f(u, v, w) = 0 is that Vu Vv x Vw = 0, or

au

ax

av

au

ay

ft

au

az

av0

ax ay az

aw aw awax ay az

This determinant is called the Jacobian of (u, v, w) with respectto (x, y, z), written J[(u, v, w)/(x, y, z)].

10. If w is a constant vector, prove that V x (w x r) = 2w,where r=xi+yJ+zk.

11. If pf = Vp, prove that f V x f = 0.12. Prove that (v V)v = 4 Vv2 - v x (V x v).13. If A is a constant unit vector, show that

V x (v xA)] =

14. If fl, f2, f3 are the components of the vector f in one set ofrectangular axes and 11, /, / are the components of f after a


rotation of axes (see Example 8), show thata fa

=aa_

aa=1 x a=1 x

so that V f is a scalar invariant under a rotation of axes. Alsosee Prob. 21, Sec. 9.

15. Prove (79), (80), (81).16. Let f = f1i + f2j + f3k and consider nine quantities

afi of,9is axi axis

i, j = 1, 2, 3

Show that gi; = -gi and that three of the nine quantities yieldthe three components of V x f. Use this result to show thatV x((pf) =('V xf+Vcpxf.

23. Curvilinear Coordinates. Often the mathematician, phys-icist, or engineer finds it convenient to use a coordinate systemother than the familiar rectangular cartesian coordinate system.If he is dealing with spheres, he will probably find it expedient todescribe the position of a point in space by the spherical coordi-nates r, 8, sp (see Fig. 31). Let us note the following: The spherex2 + y2 + z2 = r2, the cone z/(x2 + y2 + z2)'1 = cos 8, and theplane y/x = tan sp pass through the point P(r, 0, Sp). We mayconsider the transformations

r = (x2+y2+Z2)}z

'a = cos-(x2 + y2 + z2)

Sp =tan-1 y

X

as a change of coordinates from the x-y-z coordinate system tothe r-B-#p coordinate system. The surfaces

r = (x2 + y2 + z2)} = cl

8 = cos-' [z/(x2 + y2 + z2)}1 = c2, O = tan-' y/x = c3 are respecttively, a sphere, cone, and plane. Through any point P in space,except the origin, there will pass exactly one surface of each type,the coordinates of the point P determining the constants c1, c2, c3.

The intersection of the sphere and the cone is a circle, the circleof latitude, having e, as its unit tangent vector at P. This circleis called the (p-curve since r and 0 remain constant on this curve


so that only the coordinate p changes as we move along thiscurve. The intersection of the sphere and the plane yields the0-curve, the circle of longitude, while the intersection of the coneand plane yields the straight line from the origin through. P, ther curve. ee and er are the unit tangent vectors to the 0- andr curves, respectively. The three unit vectors at P, er, ee, e, aremutually perpendicular to each other and can be considered asforming a basis for a coordinate system in the neighborhood of P.Unlike i, j, k, they are not fixed, for as we move from point topoint their directions change. Thus we may expect to find morecomplicated formulas for the gradient, divergence, curl, andLaplacian when dealing with spherical coordinates.

Since Vv is perpendicular to the plane cp = constant, we musthave V(p parallel to e, Hence e9, = h3 Vsp, where hs is the scalarfactor of proportionality between e,, and Vp. If drs is a vectortangent to the p-curve, of length dss = jdr3j, we have from (58)

dip = drs VP = dr3e,p = ds3

so that ds3 = hs dip. Hence h3 ishs hs

that quantity which must be multiplied into the differential changeof coordinate gyp, namely, d-p, to yield arc length along the p-curve.Thus e,, = r sin 0 V(p, while similarly er = Vr and ee = r VB.We note that er = ee x e,, = r2 sin 0 VO X VV,

eB=e,xe,.=rsin 0V(pxVre,=erxee=rVrxVB

Any vector at P may be represented as f = flex + flee + f3ec.The scalars f1, f2, fs can be functions of r, 0, p. We may alsorepresent f as f = f1 Vr + f2r VO + far sin 0 VV and also byf = f1r2 sin O VO x V(p + f2r sin BV(p x Vr + far Vr x V8. We alsonote that the triple scalar product Vr V9 x Vtp is equal to(r2 sin 0)-1 and that dV = dsl ds2 ds3 = r2 sin 0 dr d9 dp.

Spherical coordinates are special cases of orthogonal curvilinearcoordinate systems so that we will proceed to discuss these moregeneral coordinate systems in order to obtain expressions for thegradient, divergence, curl, and Laplacian.

Let us make a change of coordinates from the x-y-z system to au1-u2-u3 system as given by the equations

Ul = u1(x, y, z)U2 = u2(x, y, z) (82)ug = u3(x, y, z)


We assume that the Jacobian J[(ul, u2, u3)/(x, y, z)] ;P,-, 0 so thatthe transformation (82) is one to one in the neighborhood of apoint. A point in space is determined when x, y, z are knownand hence when ui, u2, u3 are known. By considering

ui(x, y, z) = ci

u2(x, y, z) = C2, u3(x, y, z) = es, we obtain a family of surfaces.Through a point P(xo, yo, zo) will pass the three surfaces

ui(x, y, z) = ui(xo, yo, zo)

z

y

FIG. 36.

u2(x, y) z) = u2(xo, yo, Zo), and u3(x, y, z) = u8(xo, yo, zo). Let usassume that the three surfaces intersect one another orthogonally.The surfaces will intersect in pairs, yielding three curves whichintersect orthogonally at the point P(xo, yo, zo). The curve ofintersection of the surfaces uI = ci and u2 = C2 we shall call theus curve, since along this curve only the variable us is allowed tochange. Let u1, u2, us be three unit vectors issuing from Ptangent to the ui, us, us curves, respectively (see Fig. 36).

SEc. 231 DIFFERENTIAL VECTOR CALCULUS 53

Now Vu3 is perpendicular to the surface

u3(x, y, z) = u3(xo, yo, zo)

so that Vu3 is parallel to the unit vector u3. Hence u3 = h3 Vu3where h3 is the scalar factor of proportionality between us andVu3. Now let dr3 be a tangent vector along the us curve,dr3' = dss. Obviously dr3 us = dss, and

so that from (58)

dss = h3 du3 (83)

We see that hs is that quantity which must be multiplied intothe differential coordinate dus so that are length will result. Forexample, in polar coordinates ds = r do if we move on the 0-curve,so that r = h2.

Similarly, ul = h1 Vu1j u2 = h2 Du2, so that

u1 = U2 X u3 = h2h3 Vu2 X Vu3u2 = us x u1 = h3h1 Vu3 X Vu1 (84)113 = u1 x u2 = h1h2 Vu1 x out

and

Du1 Du2 X Vu3 =u1

h1 hU2

gx

Us

s= (h1h2ha)-1

Note that the differential of volume is

dV = ds1 ds2 ds3 = h1h2h3 du1 due du3

and making use of (85) as well as Prob. 9, Sec. 22,

(85)

dV = j (_x, y, z ) du1 due du3 (86)u1, U2, u3

Example 38. In cylindrical coordinates

ds2 = dr2 + r2 d02 + dz2

so that h1 = 1, h2 = r, h3 = 1.


Example 39. If f = f (U1, u2, u3), then from Example 21,

yf = ofVul + ` f vu2 + of vu3

au1 au2 au3

1 of 1 of 1 ofVf = u1+----u2+-''U3

h1 au1 h2 49u2 h3 au3

In cylindrical coordinates

of R+p+kar r a9 az

(87)

Our next attempt is to obtain an expression for the divergenceof a vector when its components are known in an orthogonalcurvilinear coordinate system. Now

f = flul + f2u2 + f3u3

= f lh2ha Vu2 x vu3 + f 2h3h1 Vu3 x Vul + fahlh2 Vu1 x vu2

from (84). Consequently

V f = V (f 1h2h3) vu2 x Vu3 + f 1h2h3 V (Vu2 X Vu3)

+ V(f 2h3h1) Dua x Vu1 + f 2h3hly (Vu3 X Vu1)+ V(f3h1h2) Vu1 X Vu2 + f3h1h2V (Vul x vu2) (88)

)Now V(f1h2h3) VU2 X Vu3 =

a(flh2ha Vu1 Vu2 x Vu3, andout

V (vu2 x vua) = 0

ao that (88) reduces to

a(h3h1f2) h,h2f3) )1 [o(h2h3fi)+h1h2h3 49U, au2 + 49u3

(89

If we apply (89) to the vector VV as given by (87), we obtain

21 a h2h3 a a (hAi a a a

+au2 h2 au2 + aus(h,h2

V V -h1h2h3 au1 { h1 49U, h3 au]

(90)


This is the Laplacian in any orthogonal curvilinear coordinatesystem.


l a a a 1a_ a aV2V r -I- + - r - (91)

r ar ar aB r aB az az

Example 41. Solve V2V = 0 assuming V = V(r),

r = (x2 + y2)iFrom (91)

1 d (LV = 0 or rdV

= c1r dr

rdr

J dand

V =cllogr+C2

Finally we obtain the curl of f.

f = f1u1 + f2u2 + faun= f 1h1 Vu1 + f2h2 Due + fshs Vua

and

V x f = V(f1hi) x Vul + V(f2h2) X Vu2 + V(faha) X Vus

since V x (Vul) = V X (Vu2) = V x (Vua) = 0. Now

a(f1h1)Vul

a(f1h1)Vu2 X Vu1V(f1h1) X VU, =

49U,X Vul +

49U2

a(flhl+ Vus X Vulaus

Replacing Vu2 x Vul by --us

etc., we obtainhlh2,

ul f a(hsf3) - a(h2f2J

)1 u2 ra(hlfl) ^ a(hsfa)V x f =

h2h3 L au2 aua + hsh1 L aua aulua La(h2f2) ` a(h1f1)

+ 1h2 49U1 49U2(92)


Problems

1. For spherical coordinates, ds2 = dr2 + r2 d92 + r2 sine 9 d(p2where B is the colatitude and (p the azimuthal angle. Show that

V Z V= 1 I a I r2 sin B a }-a

(sin 6a V

r2 sin 9 ar \\\ ar a9 ` aB/J

+ a (sin 9

2. Solve V2V = 0 in spherical coordinates if V = V(r).3. Express V - f and V x f in cylindrical coordinates.4. Express V f and V x f in spherical coordinates by letting

a, b, c be unit vectors in the r, 9, (p directions, respectively.5. Write Eq. (92) in terms of a determinant.6. Show that V x [(r V9)/sin 0) = V(p where r, 0, (p are spher-

ical coordinates.7. If a, b, c are the vectors of Prob. 4, show that

s =sin9car=0' a9= b, aab = 0 ab = -a ab

= cos 9 car a9 a(P

as as

ac ac _ar=0' a9=0'

ac = - sin0a - cos0b

8. If x = r sin 9 cos (p, y = r sin 9 sin (p, z = r cos 9, thenthe form ds2 = dx2 + dy2 + dz2 becomes

ds2 = dr2 + r2 d92 + r2 sin2 9 d(p23

Prove this. If, in general, ds2 = I (dxa) 2, and ifa-1

xa = xa(y', y2, y3)

a = 1, 2, 3, show that

ds2 =axa axa

dyo dyrY

a, ..ydyOdy

_ I go, dys dy''

X-f

SEC. 231 DIFFERENTIAL VECTOR CALCULUS 57

where3

ax" axa90Y = aI a" ay"

Check this result for the transformation to cylindrical coordi-nates:

x = r cos 0y = r sin 0z = z

and obtain ds2 = dr2 + r2 d02 + dz2.9. By making use of V2V = V(V V) - V x (V x V), find

V2V for V = v(r)e,, V being purely radial (spherical coordinates).Find V2V for V = f(r)e, + (p(z)e, in cylindrical coordinates.

10. Find V IV if V = w(r)k x r.11. Consider the equations

a2S

(X -f 'U)V(V s) + u V2s = p ate

X, µ, p constants. Assume s = eiP'sl, p constant, and show that

(X + p)V(V s1) + (A + pp2)sl = 0

Next show that [V2 + (µ + pp2)/(X + µ)](V s1) = 0,

X + µ 0.

12. If A = V x (¢r), V2¢ = 0, show that

1 a a2Y' a¢ a2Y'

sin 0 appaOar a0acpar

so that A V x A = 0 if, moreover, ¢ =13. Show that Cpi = Ae9 + Bey + Ce° satisfies V2sp1 = ci, and

show that if 402 satisfies V°02 = 0, then rp = Cpl + 4p2 also satisfiesV2ip = gyp. Find a solution of V2Sp = -(p.

CHAPTER 3

DIFFERENTIAL GEOMETRY

24. Frenet-Serret Formulas. A three-dimensional curve in aEuclidean space can be represented by the locus of the end pointof the position vector given by

r(t) = x(t)i + y(t)j + z(t)k (93)

where t is a parameter ranging over a set of values to < t < ti.We assume that x(t), y(t), z(t) have continuous derivatives of allorders and that they can be expanded in a Taylor series in theneighborhood of any point of the curve.

We have seen in Chap. 2, Sec. 16, that ds is the unit tangent

vector to the curve. Let t = ds- Now t is a unit vector so that

its derivative is perpendicular to t. Moreover, this derivative,

dse tells us how fast the unit tangent vector is changing direction

as we move along the curve. The principal normal to the curveis consequently defined by the equation

dt= Kn

ds(94)

where K is the magnitude of ds and is called the curvature. The

reciprocal of the curvature, p =I 1K, is called the radius of curva-ture. It is important to note that (94) defines both K and n,

K being the length of ds while n is the unit vector parallel to

dtds

At any point P of our curve we now have two vectors t, n at

right angles to each other (see Fig. 37). This enables us to set up58

SEC. 24] DIFFERENTIAL GEOMETRY 59

a local coordinate system at P by defining a third vector at rightangles to t and n. We define as the binormal the vector

b = t xn

All vectors associated with the curve at the point P can bewritten as a linear combination of the three fundamental vectorst, n, b, which form a trihedral at P.

z

0

xFia. 37.

Y

Let us now evaluate - -- anddn

Since b is a unit vector, its

derivative is perpendicular to b and so lies in the plane of t and n.Moreover, b t = 0 so that on differentiating we obtain

t = 0. Hence ` is also perpendicular to t so thatdd_b

must be parallel to n. Consequently, ds = rn, where r by defini-

tion is the magnitude of -. r is called the torsion of the curve.


Finally, to obtain dn, we note that n = b x t so that

ds b x ds +b x t = b x Kn + rn x t =- Kt- rb

The famous Frenet-Serret formulas are

2

1 =(dl (a2sin2t+a2cos2t+b2)

Successive derivatives are functions of t, n, b and the derivativesof K and r.

Example 42. The circular helix is given by

r = a cos ti+asintj+btkt =ds = (-a sin ti+acostj + bk) st

and

Hence

Now

so that

Also

t = (-a sin ti+acostj+bk)(a2+b2) 4

Kn = d = (-a cos t i - a sin t j)(a2 + b2)-t

dt

ds

do

Kn

- (Kt + Tb) 95)dsdb

dsrn

(a2 + b2)(.)2

K = a(a2 + b2)-i

b = t x n =i j k

-a sin t a cost b

-cos t -sin t 0

(a2 + b2)-1

= (b sin t i - b cos t j + ak) (a2 + b2)-+

SEC. 241 DIFFERENTIAL GEOMETRY

and

so that

dbds = rn = (b cos t i + b sin t j)(a2 + b2)-1

T = b(a2 + b2)-1

61

Problems

1. Show that the radius of curvature of the twisted curvex = log cos 0, y = log sin 0, z = V2 0 is p = csc 20.

2. Show that r = 0 is a necessary and sufficient condition thata curve be a plane curve.

3. Prove that T =

z(r'r"r"').

K

4. For the curve x = a(3t - te), y = 3at2, z = a(3t + t3),show that K = T = 1/3a(1 + t2)2.

AA _ do db dt dn5. Prove that

ds .=

ds0,

d8 .

_ds

= 0.ds

Kr,d8 -

=

6. Prove that r"' = -K2t + K'n - rKb, where the primesmean differentiation with respect to are length.

7. Prove that the shortest distance between the principalnormals at consecutive points at a distance ds apart (s measuredalong the arc) is ds p(p2 + r`2)_;

8. Find the curvature and torsion of the curvex = a(u - sin u), y = all - cos u), z = bu

9. For a plane curve given by r = x(t)i + y(t)j, show thatx,y - y,x

[(x')2 + (y1)2]1

10. Prove that (t't"t"') = K5ds (K)

11. Show that the line element ds2 = dx2 + dy2 + dz2 - c2 dt2remains invariant in form under the Lorentz transformation

- ytx = [1 - (V2/c2)1Iy = Iz=2

- (V/c2)xt =[1 - (V2/c2)J*


V, c are constants. The transformation ct = iT, ileads to the four-dimensional Euclidean line element

ds2=dx2+dy2+dz2+dr2

12. If xa = xa(s), a = 1, 2, . . . , n, represents a curve in ann-dimensional Euclidean space for which

d82 = (d_-1)2 + (dx2) 2 + . . + (dxn) 2

define the unit tangent vector to this curve, this definition being ageneralization of the definition of the tangent vector for the case

d2xan = 3. Show that the vector

ds2) a = 1, 2, . . . , n, is normal

to the tangent vector, and define the unit principal normal n,and curvature K, by the equations

d2xa dta

d82 - ds = Klnla, a = 1, 2,

a

Show thatds

todnla =

!Is

..,n

a = 1, 2, . . . , n, is normal to nl and that

Define the second curvature K2 and unit- K1.a-1

latib th dl a an2 y e equa ons = -Kltnorma + K2n2 , a = 1, 2,id.. . , n, and show that n2a is normal to to and nla if K2 ;P'- 0.

Continue in this manner and obtain the generalization of theFrenet-Serret formulas.

25. Fundamental Planes. The plane containing the tangentand principal normal is called the osculating plane. Let s be avariable vector to any point in this plane and let r be the vectorto the point P on the curve. s - r lies in the plane and is conse-quently perpendicular to the binormal. The equation of theosculating plane is

(s - r) b = 0 (96)

The normal plane to the curve at P is defined as the planethrough P perpendicular to the tangent vector. Its equation is

SEC. 26] DIFFERENTIAL GEOMETRY

easily seen to be

63

(s - r) t = 0 (97)

The third fundamental plane is the rectifying plane through Pperpendicular to the normal n. Its equation is

(s - r) n = 0 (98)

Problems

1. Find the equations of the three fundamental planes for thecurve

x = at, y=bt2, z=cta

2. Show that the limiting position of the line of intersectionof two adjacent normal planes is given by (s - r) n = p wheres is the vector to any point on the line.

26. Intrinsic Equations of a Curve. The curvature and torsionof a curve depend on the point P of the curve and consequentlyon the are parameter s. Let is = f(s), r = F(s). These twoequations are called the intrinsic equations of the curve. Theyowe their name to the fact that two curves with the same intrinsicequations are identical except possibly for orientation in space.Assume two curves with the same intrinsic equations. Let thetrihedrals at a corresponding point P coincide; this can be doneby a rigid motion.

Now

ds(t1. t2) = t1 ,cn2 + xni t2

T (nl n2) = n1 (-Kt - rb2) + n2 (--Kt, -- rbi) (99)

dAdding, we obtain

s 0

64 VECTOR AND TENSOR ANALYSIS (SEC. 27

so thatconstant = 3 (100)

since at Ptl=t2, n1=n2, b1=b2

Since (100) always maintains its maximum value, we must havedr, _ dr2

tl = t2, n1 n2, bi = b2 so thatds ds

or r1 = r2 locally.

Hence the two curves are identical in a small neighborhood ofP. Since we have assumed analyticity of the curves, they areidentical everywhere.

Problems

1. Show that the intrinsic equations of x = a(9 - sin 8),y = a(l - cos 8), z = 0 are p2 + s2 = 16a2, 7- = 0, where s ismeasured from the top of the are of the cycloid.

2. Show that the intrinsic equation for the catenary

y=a'(ex/a+e-(sla))2

is ap = s2 + a2, where 8 is measured from the vertex of thecatenary.

Fla. 38.

27. Involutes. Let us consider the space curve r. We con-struct the tangents to every point of r and define an involute

SEc. 271 DIFFERENTIAL GEOMETRY 65

as any curve which is normal to every tangent of r (see Fig. 38).From Fig. 39, it is evident that

r,=r+ut (101)

is the equation of the involute, u unknown. Differentiating(101), we obtain

tA d_u \ dsdr, Cdr

ds,^l ds+ u ds+dst ds,(102)

where s is are length along r and s, is arc length along r'. Using(95), (102) becomes r

ti = (t+uua+dut) d (103)ds ds,

Now t t, = 0 from the definition ofthe involute so that

du1 +- =0 and u=c - s(104) Fta. 39.

Therefore r, = r + (c - s)t, and there exists an infinite familyof involutes, one involute for each constant c. The distancebetween corresponding involutes remains a constant. An invo-

r

Fia. 40.

lute can be generated by unrolling a taut string of length c whichhas been wrapped along the curve. The end point of the stringgenerates the involute (see Fig. 40). What are some propertiesof the involute?


r1=r+(c-s)tdr, dr / dt dsti =ds, - Ids + lC - S) ds - t] dsl

(C - s)ds

K - nds1

Hence the tangent to the involute is parallel to the correspondingnormal of the curve. Since ti and n are unit vectors, we must

have (c - s) K dl

= 1. The curvature of the involute is

dt1 do ds (-Kt -7-b)obtained from - = Kin, = - - = Hencedsl ds ds1 (c - s)K

rK12 K2

+ r2(105)

K2(C - 8)2

Fla. 41.

28. Evolutes. The curve t'whose tangents are perpendicu-lar to a given curve is called theevolute of the curve. The tan-gent to r' must lie in the planeof b and n of r since it is perpen-dicular to t. Consequently

rl=r+un+vbis the equation of the evolute. Differentiating, we obtain

= dr1 = dr d_n db d_u dv

tl dsl - {ds + u ds + v ds + dsn + d8

bl Ads,

= It + u(-Kt -rb) +vTn+dsn +d8b]dsl

Now t t1 = 0, which implies I - uK = 0 or u = 1= p. Thus

K

dv d8tl=L(-ru+as), b+( +ds)n

Also t1 is parallel to r1 - r = un + vb (see Fig. 41). Therefore

(dv/ds) - UT (du/ds) + Pru V

SEC. 28] DIFFERENTIAL GEOMETRY

or

Therefore

uv' - vu' d vT = = - tan-' -u2+v2 ds u)

67

=r V

o

a Tds=tan--' -CU

--

and v = p tan (,p - c) since u = p. Therefore

r1 = r + pn + p tan (,p - c)b (106)

and again we have a one-parameter family of evolutes to thecurve T.

Problems

1. Show that the unit binormal to the involute is

b1 =Kb - Tt

(C - S)KK1

2. Show that the torsion of an involute has the value

T1= T - K

`dS

] [K(K2 + r2)(C - 8)I-1

3. Show that the principal normal to the evolute is parallelto the tangent of the curve 1'.

4. Show that the ratio of the torsion of the evolute to its curva-ture is tan (,p - c).

5. Show that if the principal normals of a curve are binormals(equal vectors not necessarily coincident) of another curve, thenc(K2 +,r2) = K where c is a constant.

6. On the binormal of a curve of constant torsion T, a point Qis taken at a constant distance c from the curve. Show that thebinormal to the locus of Q is inclined to the binormal of the givencurve at an angle

tan-'K(C2r2 + 1)}

CT2

7. Consider two curves which have the same principal normals(equal vectors not necessarily coincident). Show that the tan-gents to the two curves are inclined at a constant angle.


29. Spherical Indicatrices(a) When dealing with a family of unit vectors, it is often

convenient to give them a common origin and then to considerthe locus of their end points.This locus obviously lies on aunit sphere. Let us now considerthe spherical indicatrix of the tan-gent vectors to a curve r = r(s).The unit tangent vectors are

t(s) = ds- Let r, = t. Then

Fla. 42.

= dr, = dt ds ds

t' ds1 ds dal - ds,

Thus the tangent to the spheri-cal indicatrix P is parallel to the

normal of the curve at the corresponding point. Moreover,

1 = K d1' t, = n. Let us now find the curvature K, of the indi-

catrix. We obtain

dt, do ds 1=K,n,='--=-(-Kt-rn)as, ds ds, K

andK2 + r2

K12K

2

(b) The spherical indicatrix of the binormal, r1 = b.entiating,

dr,=

A ds = redst,=--- -

ds, ds ds1 ds1

Thereforerds=1

and t,=nds,

Differentiating,

dt, do ds 1 `-Kt - rn)dS1

= x,n1 =A dS;

_r

and

K12 =K2 + r2

T2

Differ-

SEC. 30J DIFFERENTIAL. GEOMETRY

Problems

1. Show that the torsion of the tangent indicatrix is

T(dK/ds) - K(dr/ds)Ti

K(K2 +T2)

2. Show that the torsion of the binormal indicatrix is

T(dK/ds) - K(dT/ds)Ti T(K2 + r2)

69

3. Find the curvature of the spherical indicatrix of the principalnormal of a given curve.

30. Envelopes. Consider the one-parameter family of sur-faces F(x, y, z, c) = 0. Two neighboring surfaces are

F(x, y, z, c) = 0

and F(x, y, z, c + Ac) = 0. These two surfaces will, in general,intersect in a curve. But these equations are equivalent to theequations F(x, y, z, c) = 0 and

F(x, y, z, c + Ac) - F(x, y, z, c) = 0Ac

where Ac # 0. As Ac --> 0, the curve of intersection approachesa limiting position, called the characteristic curve, given by

F(x, y, z, c) = 0aF(x, y, z, c) - 0 (107)

ac

Each c determines a characteristic curve. The locus of allthese curves [obtained by eliminating c from (107)] gives us asurface called the envelope of the one-parameter family. Nowconsider two neighboring characteristics

F(x,y,z,c) = 0aF(x,,z,c)

= 0ac

and (108)

F(x, y, z, c + Ac) = 0 aF(x, y, z, c + Ac) = 0ac

which, in general, intersect at a point. The locus of these pointsis the envelope of the characteristics and is called the edge of

70 VECTOR AND TENSOR ANALYSIS [SEc.31

regression. The edge of regression is given by the three simul-taneous equations

F(x, y, z, c) = 0aF(x, y, z, c) = 0

aca2F(x,y,z,c) = 0

ace

(109)

Example 43. Let us consider the osculating plane at a point P.From (96) we have [s - r(s)] b(s) = 0. If we let P vary, weobtain the one-parameter family of osculating planes given by

f(x, y, z, s) = [s - r(s)] b(s) = 0

where s is the parameter and s = xi + yj + A.

of dr ANow =

as dsb + (s -r).-

ds= (s - r) in, and setting

of= 0, we obtain (s - r) n = 0. This locus is the rectifying

as

plane. The intersection of f = 0 and as = 0 obviously yields

the tangent lines which are the characteristics. Now

a If

as2= -t n + (s - r) (-Kt - A) + (s - r) n ad

z

=It is easy to verify that s = r satisfies =a

=a

- 0, so thatY Y f as 4982

the edge of regression is the original curve r = r(s).A developable surface, by definition, is the envelope of a one-

parameter family of planes. The characteristics are straightlines, called generators. We have seen that the envelope of theosculating planes is the locus of the tangent line to the spacecurve P. In general, a developable surface is the tangent surfaceof a twisted curve. A contradiction to this is the case of acyiinder or cone.

31. Surfaces and Curvilinear Coordinates. Let us considerthe equations

x = x(u, v)y = y(u, v) (110)z = z(u, v)

SEc. 32] DIFFERENTIAL GEOMETRY 71

where u and v are parameters ranging over a certain set of values.If we keep v fixed, the locus of (110) is a space curve. For eachv, one such space curve exists, and if we let v vary, we shall obtaina locus of space curves which collectively form a surface. Weshall consider those surfaces (110) for which x, y, z have continu-ous second-order derivatives. Equation (110) may be written

r(u, v) = x(u, v)i + y(u, v)j + z(u, v)k (111)

where the end point of r generates the surface. The curvesobtained by setting v = constant are called the u curves, andsimilarly the v curves are obtained by setting u = constant.The parameters u and v are called curvilinear coordinates, andthe two curves are called the parametric curves.

32. Length of Arc on a Surface. If we move from the point rto the point r + dr on the surface, the distance ds is given by

N 2

ds2 =(arudu+-dv

or 9r o9r

J

2

due + 2au . av

du dv + (av)2 dv2C_

or

ds2 = E du2 + 2F du dv + G dv2 (112)

where\12 2

E au/ 'F

ao9r

u.

av'G

(Ov)

Equation (112) is called the first fundamental form for the surfacer = r(u, v). In particular, along the u curve, dv = 0, so that

(ds) = 1'E duand similarly (113)

(ds), = VG_ dv

NowOr

and av are tangent vectors to the u and v curves, so

that the parametric curves form an orthogonal system if andOr Or

only au av

72 VECTOR AND TENSOR ANALYSIS 1Szc. 33

Example 44. Consider the surface given by

r = r sin 8 cos v i+ r sin 8 sin V j+ r cos 8 k, r = constantDifferentiating,

ar = r cos 0 cos sP i + r cos 0 sin V j - r sin 8 kd0

ar = -r sin6sinpi+rsin8coscpja st

and

is 2c1r

E a8/ = r2,F=-e

c1r= 0, G = r2 Sin2 0

so that ds2 = r2 d82 + r2 sin2 0 dcp2 and the 0-curves are orthog-onal to the 9-curves. Of course the surface is a sphere.

33. Surface Curves. By letting u and v be functions of asingle variable t, we obtain

r = r[u(t), v(t)] (114)

which represents a curve on the surface (111). Along this curve,dvl

dt. dr is completely determined when dudr = (arau

dtdu + a

av

rWt

1ànd dv are specified, so that we will use the notation (du, dv) tospecify a given direction on the surface. Now consider another

curve such that ar =au

su + avav, where su and av are the

differential changes of u(t) and v(t) for this new curve. Now

dr or = E du au + F(du av + dv au) + G dv av (115)

so that two curves are orthogonal if and only if

Edu su +F(duav +dv &u) +Gdv av = 0or

6V dv

u+au/+Gd -=0 (116)E+FCa

If we have a system of curves on the surface given by the differ-ential equation P(u, v) su + Q(u, v) av = 0, the differential equa-

SEC. 341 DIFFERENTIAL GEOMETRY 73

tion for the orthogonal trajectories is given by

E+F,(_P+dv1 - GPdv =0 (117)

`` Q dull Q du

by Psince -bu Q

Problems

1. Find the envelope and edge of regression of the one-param-eter family of planes x sin c - y cos c + z tan 8 = c, where c isthe parameter and 0 is a constant.

2. Show that any two v curves on the surface

r = u cos v i + u sin v j + (v + log cos u)k

cut equal segments from all the u curves.3. Find the envelope and edge of regression of the family of

x2 y z2ellipsoids c2 (a2 +

b2+

c2= 1 where c is the parameter.

4. If 8 is the angle between the two directions given by

P due + Q du dv + R dv2 = 0

show that tan 0 = H(Q2 - 4PR)}/(ER - FQ + GP), where

Har 8

8uavl.

5. Provetvhl

at the differential equations of the curves whichbisect the angles between the parametric curves are

VEdu-VGdv=0and 1/E du + 1i dv - 0.

6. Given the curves uv = constant on the surface r = ui + vj,find the orthogonal trajectories.

7. Show that the area of a surface is given by

f f (EG - F2)1 du dv

34. Normal to a Surface. The vectorsor

and av are tangent

to the surface r(u, v) along the u and v curves, respectively.ar 8r ,

is a vector normal to the surface. NoteConsequently,au x 49V


that au need not be a unit tangent vector to the u curve since the

parameter u may not represent are length. Since

(ds),, = 1/E du

a necessary and sufficient condition for u to be arc length is thatE = 1. We define the unit normal to the surface as

n _ (Or/au) x (ar/av)(118)

(ar/au) x (ar/av)

35. The Second Fundamental Form. Consider all the planesthrough a point P of the surface r = r(u, v) which contain thenormal n. These planes intersect the surface in a family ofcurves, the normals to the curves being parallel to n. We nowcompute the curvature of any one of these curves in the direction(du, dv). Let ds be length of are along this curve. Now

dr Or du Or dvt=ds=auds+avdsTherefore

d2r dt 492r du 2 a2r du dv a2r dv 2ds2 ds

_ _K"n au2 (ds) + 2

au av A ds + av2 (ds)Or d2u Or d2v

+ + 119)

and

au ds2 av ds2

n)

since n Or-au

(n au2)(du)2ds+ 2 (n au2av) ds ds

)z

n-

= 0. Thereforeav

edue+2fdudv+gdv2Kn =

ds2

Knedue+2fdudv+gdv2

(120)Edue+2Fdudv+Gdv2


where we define

75

a2r a2r a2re=n-au2' f _n.

auav' (121)

The quantity e due + 2f du dv + g dv2 is called the second funda-mental form.

Now consider any curve t on the surface and let its normal ben, at a point P, the direction of r being (du, dv) at P. Let r' be

Fia. 43.

the normal curve in the same direction (du, dv) with normal n atP (Fig. 43). We have

r " r"=n -K K

since n rl" = r" n for two curves with the same (du, dv) [see(119)]. Therefore

COS0=Kn-K

so that

K = K,, sec 0 (122)

This is Meusnier's theorem.36. Geometrical Significance of the Second Fundamental

Form. We construct a tangent plane to the surface at the pointr(uo, vo). What is the distance D of a neighboring point

r(uo + Du, vo + Av)


on the surface, to the plane? It is D = Ar n. Now

r(uo + Au, vo + AV) = r(uo, vo) +arau

Au +arav

AV

1 a2r a2r

+ 2!

(a2r- Au2 + 2au av Au AV

+ av2 Av2 +

from the calculus. Consequently

1 / z z

D = Av2

except for infinitesimals of higher order. Thus

2D = e du2 + 2f du dv + g dv2 (123)

Problems

1. For the paraboloid of revolution

r=ucosvi+usinvj+u2kshow that E = 1 + 4u2, F = 0, G = u2, e = 2(1 + 4u)`}, f = 0,g = 2u2(1 + 4u2)-}, and find the normals to the surface and thenormal curvature for the direction (du, dv).

2. What are the normal curvatures for directions along theparametric curves?

3. Find the second fundamental form for the sphere

r=rsin0coscpi+rsin6sincpj+rcosOkr = constant.

4. Show that the curvature K at any point P of the curve ofintersection of two surfaces is given by

K2 Sln2 0 = K12 + K22 - 2K1K2 COS 0

where 9j, K2 are the normal curvatures of the surfaces in the direc-tion of the curve at P, and 0 is the angle between their normals.

5. Let us make a change of variable u = u(u, v), v = v(u, 1).Show that E, F, G transform according to the law

lz 2

ECaul

+ 2Fau au

+G(av\au


au au au av av au av avF =Eaudo +F (aft do +audo )+Gag do

UP-

au\2 au av (avl 2E Ii + 2F

avav + G àvl

and that E due + 2F du dv + G dv2 = E due + 2F du do + G W.Also show that

((au)2 +

au av (OV\2e = ± e Zf

au + g au Jau au1

au au au av av au av avau av au av au av au CIO]

- ± [e (au)22f

au

av

av v12

g + avav+ g(a

av J

37. Principal Directions. From (120) we have

(K .E - e) due + f) du dv + g) dv2 = 0 (124)or

A due + 2B du dv + C dv2 = 0

This quadratic equation has two directions (du, dv), (Su, Sv),which give the same value for x,,. These two directions willcoincide if the quadratic equation (124) has a double root. Thisis true if and only if

B2-AC= (K.F'-f)2- 0or

, 2(F2 - EG) + gE - 2fF) + (f2 - eg) = 0 (125)

Moreover, we havedu B

and d = - C if B2 - AC = 0,-Aso that

(K. E - e) du + (x F - f) dv = 06126)f) du + (xnG - g) dv = 0

The solutions of (124) give the two directions for a given x,,.When x is eliminated between (124) and (125), the two directionscoincide and satisfy

(Ef - Fe) due + (Eg - Ge) du dv + (Fg - Gf) dv2 = 0 (127)


The two directions, solutions of (127), are called principal direc-tions and are the only ones with a unique normal curvature, thatis, no other direction can have the same curvature. The normalcurvatures in these two directions are called the principal curva-tures at the point. The average of the two principal curvaturesis

HeG+gE-2fF2(EG - F2)

(128)

which is obtained by taking one-half of the sum of the roots of(125). The Gaussian curvature K is defined as the product ofthe curvatures, that is,

K= f2egF2EG (129)

A line of curvature is a curve whose tangent at any point has adirection coinciding with a principal direction at that point.The lines of curvature are obtained by solving the differentialequation (126). The curvature of a line of curvature is not aprincipal curvature since the line of curvature need not be anormal curve.

Example 45. Let us consider the right helicoid

r= ucospi+usine'j+cookWe have

= -u sinsau

=cosSPi+sinrpj,

Or

2 z r

a2G2-0' 8ua = -sin+Cos(pj,a2rQ= -ucosipi - usinrpj

Hence

(Or\2(ar)2 u2 + c2=au- 1, F =au - am 0' G = am


Also

79

n = (ar/au) x (,9r/app) = (c sin p i - c cos v j + uk) (C2 + u2)-}

(ar/au) x (ar/ap)j82r a2rf=n =

au ago2

-c(c2 + u2)-#

Equation (125) yields

- (u2 + C2)K,, + C2(C2 + u2)-1 = 0whence

C

u2+C2

The average curvature is H = 22+ C2) = 0, and

-the Gaussian curvature is K = (u2 C2)2' The differential

equation (126) for the lines of curvature becomes

-c(c2 + u2)-}dug + c(c2 + U2)Id tp2 = 0

so thatdu

d(p = ± (C2 + u2) b and = ± log (u + \/u2 + c2) + a

and the lines of curvature are given by

r = u cos [± log (u + u2 + c2) + aji + u sin -p j + cspk.

Referring to (126) for the two principal directions, we have

dv av Eg - Gedu + au Fg - Gf

dv av _ Ef - Fedu au Fg - Gf

Substituting (130) into (116) we obtain

E - F(Fg-Gf/ +G \F'g -Gf/ 0

so that the principal directions are orthogonal.

(130)


Now let us choose the principal curves as the parametric lines.Thus u = constant, v constant are to represent the principalcurves. These two curves satisfy the equation du dv = 0, sothat from (127) we must have

Ef - Fe = 0Fg-Gf=0Eg - Ge 0

From these equations we conclude that

f(Eg - Ge) = gfE - feG=Feg - eFg =0and F(Eg - Ge) = 0, so that f = F = 0. We have shown thata necessary and sufficient condition that the lines of curvature beparametric curves is that

f=F=0 (131)

Problems

1. Find the lines of curvature on the surface

x=a(u+v), y=b(u-v), z=uv2. Show that the principal radii of curvature of the right conoid

x = u cos v, y = u sin v, z = {f (v) are given by the roots of

f'=K2 - of"(u2 + fF )iK - (u2 + ft )2 = 0

3. The surface generated by the binormals of the curve r = r(s)is given by R = r + ub. Show that the Gauss curvature isK = -.r2/(1 + r2u2)2. Also show that the differential equationof the lines of curvature is

-T2 du2 - (K + Kr2u2 + d3 u) du ds + (1 + r2u2)T M = 0

38. Conjugate Directions. Let P and Q be neighboring pointson a surface. The tangent planes at P and Q will intersect in astraight line 1. Now let Q approach P along some fixed direction.The line 1 will approach a limiting position 1'. The directionsPQ and 1' are called conjugate directions.

We now compute the analytical expression for two directionsto be conjugate. Let n be the normal at P and n + do the

SFc. 38] DIFFERENTIAL GEOMETRY 81

Or Ornormal at Q, where dr = PQ =

andu + - dv. Let the direc-

tion of 1' be given by SrOr

=au

Su +arav

Sv. Since Sr lies in both

planes, we must have Sr n = 0 and Sr (n + dn) = 0. Thesetwo equations imply Sr do = 0, or

(auSu + av

Sv) Can du + a dv) = 0 (132)

Expanding, we obtain

far anav Or anau

auOr an

(_au-) du bu + I (-av -) Sv du + (au aU) au dvJ

ar an+ av av

Sv dv = 0 (133)

Now n au = 0, so that by differenuiating we see that

an ar a2r

which impliesan ar a2r-n- - = -eau au au,

Similarlyan Or an Or

av au au avan Or

-f

9

so that (133) becomes

e du Su + f (du Sv + dv Au) + g Sv dv = 0 (134)

If the direction (du, dv) is given, there is only one correspondingconjugate direction (Su, Sv), obtained by solving (134).

Now consider the lines of curvature taken as parametric curves.Their directions are (du, 0), (0, Sv). Equation (134) is satisfiedby these directions since f = 0 for lines of curvature, so that thelines of curvature are conjugate directions.

82 VECTOR AND TENSOR ANALYSIS [SFc. 39

39. Asymptotic Lines. The directions which are self-conjugateare called asymptotic directions. Those curves whose tangentsare asymptotic directions are called asymptotic lines. If a direc-

tion is self-conjugate,dv = Sv,

so that (134) becomesdu Su

e due + 2f du dv + g dv2 = 0 (135)

We see that the asymptotic directions are those for which thesecond fundamental form vanishes. Moreover, the normal curva-ture rc, vanishes for this direction.

If e = g = 0, f 0, the solution of (135) is u = constant,v = constant, so that the parametric curves are asymptotic linesif and only if e = g = 0,f3PK 0.

Example 46. Let us find the lines of curvature and asymptoticlines of the surface of revolution z = x2 + y2. Let x = u cos v,y = u sin v, z = u2, and

r=ucosvi+usinvj+u2k

We obtain

Or =cosvi-{-sinvj+2uk, ar= -u sinvi+ucosvjau av

azr =2k'

a2r

8u2 au av

n=

= -sinvi+cosvja2r - -ucosvi - usinvjav2

(ar/au) x (ar/av)(ar/au) x (ar/av)I

= (-2u2 cos v i - 2u2 sin v j + uk)u-1(1 + 4u2)'I

Thereforez =2(1+4u)-}, f=n a2r =0

au2 au avz

U2)-fg = n2

= 2u2(1 +4


ar arAlso F = -- - = 0, so that f = F = 0, and from (131) theau avparametric curves are the lines of curvature. The asymptoticlines are given by due + u2 dv2 = 0. These are imaginary, sothat the surface possesses no asymptotic lines.

Problems

1. Show that the asymptotic lines of the hyperboloid

r=acos0seeipi+bsin0sec4, j+ctan 4kare given by 0 ± ¢ = constant.

2. The parametric equations of the helicoid are

x=ucosv, y=usinv, z=cv

Show that the asymptotic lines are the parametric curves, andthat the lines of curvature are u + V 'u2+ c2 c2 = Ae}°. Showthat the principal radii of curvature are ± (u2 + c2)c-1.

3. Prove that, at any point of a surface, the sum of the normalcurvature in conjugate directions is constant.

4. Find the asymptotic lines on the surface z = y sin x.40. Geodesics. The distance between two points on a surface

(we are allowed to move only on the surface) is given by

8 = I.1 [E (dt/2 + 2F dt dt + G (i)]1 dt (136)

Among the many curves on the surface that join the two fixedpoints will be those that make (136) an extremal. Such curvesare called geodesics. We wish now to determine the geodesics.To do this, we require the use of the calculus of variations, andso we say a few words about this important method.

Let us first consider the integralfQ(xt,yi) (1 + y'dx (137)

(x,, VS)

We might ask what must be the function y = y(x) joining the twopoints P and Q which will make (137) a minimum. The readermight be tempted to say, y' = 0 or y = constant, since the inte-grand is then a minimum. But we find that y = constant will


not, in general, pass through the two fixed points. Hence thesolution to this problem is not trivial. We now formulate a

y more general problem: to findy = y(x) such that

j'f(x, y, y') dx (138)

is an extremal. The function'yI.-, B f(x, y,) is given. It is y(x)

ry and so also y' (x) that are,unknown Let y = (x) bey.

A r }' M It fthat unction which makes(138) an extremal Now 1 t

Fzo. 44.

erty that jp(a) = p(b) = 0 (see Fig. 44). Under our assumption,

J(a) = j,'f(x, Y, ?') dx (139)

is an extremal for a = 0. ConsequentlydJ

= 0 orda0

since

Y(x, a) = y(x) + arp(x), where

a b P.x a is arbitrary and independ-ent of x and jp(x) is any func-

dJ

da

as aY as + aY' as if +C7F

'p,

and for a=0,of of of ofa> r e Tay, ay'

We now integrate the right-hand term of (140) by parts andobtain

a-0

tion with continuous firstderivative having the prop-

=Ib\ay,+a-

(140)

of of 81' of 81" of of

bf b

aydx + Lay

P]

- Jab dx ay'/ `p dx = 0 (141)

SEc. 40] DIFFERENTIAL GEOMETRY

Now (p(a) = p(b) = 0 by construction of v(x), so that

85

(142)

Now let us assume thatay dx (ay') is continuous. If

ay a (-' is not identically zero on the interval (a, b), it

will be positive or negative at some point. If it is positive atx = c, it will be positive in a neighborhood of x = c from con-tinuity (see Sees. 42 and 43). We can construct (p to be positiveon this interval and zero elsewhere. Then

Jb[afd(of)]x ay`p dx > 0

y d

so that we have a contradiction to (142). Consequently, thefunction of y(x) must satisfy the Euler-Lagrange differentialequation

d of ofdx Cay' ay

0 (143)

If f = f(y, y'), we can immediately arrive at an integral of(143). Let us consider

(of d (af 11J, +

yy y

ay y/ dx TOdx f y yl aya

y [Of _ d (t\1 = 0 from (143)lay dx ay'

f - y aofy, = constant (144)

is an integral of (143) if f = f(y, y').Example 47. To extremalize (137), we have f = (1 + y'2)1,

which is independent of x. From (144),

1f - y

of= constant =

a

86

so that

and


(1 +y') - Y'(1 Jy'2), a

y'=and finally

y= ± av-1x+ 0 (145)

The constants a and fi are determined by noting that the straightline (145) passes through the two given fixed points.

Example 48. If f = f Cx',x 2, ... , x", dtl, dt2) ...dtn? t);

then fhfdt is an extremal when the xa(t) satisfy

d ofdt

(ftafa

axa = o (146)

for a = 1, 2, . .. , n with is = X.The superscripts are not

powers but labels that enable us to distinguish between the vari-ous variables. The formulas (146) are a consequence of the fact

that fo f dt must be an extremal when x`(1) is allowed to varywhile we keep all other x1 fixed, j = 1, 2, ... , i - 1, i + 1,

,n.Let us now try to find the differential equations that u(t) and

v(t) must satisfy to make (136) an extremal. We write

s = f " (E,42 + 2Fuv + Gv2)} dt

and apply (146), where x' = u and x2 = v. We thus obtain

dt \a4/ au - 0 (147)

d off`

dt avlofav

o (148)

where

f = (E,42+ 2Fuv + Gv2)i = dt' E = E(u, v), etc.

SEc. 401 DIFFERENTIAL GEOMETRY 87

Now

iceCIE

+ 2uvOF

+ v2aG---

af Eic + Fv of au au auau f au 2f


d Eu + Fvl _ 2t2 + 2"(W/au) + v2(aG/au)dt ds/dt / 2 ds/dt

(149)

and if we choose for the parameter t the are length s, then t = a

and dt = 1, so that (149) reduces to

opud(Eic + Fv) = 2(u2 aE + 2uv

OF+ 62

-n/A

ds

while similarly (148) yields\d (Fu+Gv) _2(u2av +24vav +v2av/

(150)

In Chap. 8 we shall derive by tensor methods a slightly differ-ent system of differential equations.

Example 49. Consider the sphere given by

r = a sin 0cos(pi+a sin 0 sin cpj+acosOkwhere ds2 = a2 do2 + a2 sin2 0 dp2 so that E = a, F = 0,

G = a2 Sin2 0,and

OE _ a E E _ OF _ OF _ aG _ 8G2

80 ac av a9 ap 0' aB

_- a sin 29

Hence (150) reduces to2

ds a ds/ 2 dssin 2B

d (a sin2 0 ds) = 0 (151)

Integrating (151) we have sine 0

dLIP = constant. We can choose

our coordinate system so that the coordinates of the fixed points


are a, 0, 0 and a, 0, 0. Hence sin' 8ds

= 0, andds

= 0, so that

0. Hence the geodesic is the are of the great circle joiningthe two fixed points.

Example 50. Let us find y(x) which extremalizes

f y(1 + y")} dx

Since f = y(1 + y")# = f(y, y'), we can apply (144) to obtaina first integral. We obtain y(1 + y'')t -- y''y(1 + y'')-l = a-1,and simplifying this expression yields y' = ± (a2y2 - 1)I. Afurther integration yields ay = cosh (0 ± ax). These are thecurves (catenoids) which have minimum surfaces of revolution.

Problems

1. Find the geodesics on the ellipsoid of revolutionx2+ z2 2

a2 + b2 = 1

Hint: Let x = u cos v, z = u sin v.2. Show that the differential equation of the geodesics for the

right helicoid x = u cos v, y = u sin v, z = cv is

du 1

TV+

-h

[(u2 + C2)(u2 + C2 h2)1], h = constant

3. Prove that the geodesics on a right circular cylinder arehelices.

4. Show that the perpendicular from the vertex of a rightcircular cone to the tangents of a given geodesic is of constantlength.

5. Find y(x) which extremalizes f'[(l + y')/y)]} dx.

CHAPTER 4

INTEGRATION

41. Point-set Theory. In geometry and analysis the studenthas frequently made use of the concept of a point and of thenotion of a set or collection of elements (objects, points, numbers,etc.). We shall not define these concepts, but shall take theirnotion as intuitive. We may be interested in the points subjectto the condition x2 + y2 < r2. These will be the points interiorto the circle of radius r with center at the origin. We might alsobe interested in the rational points of the one-dimensional con-tinuum. For convenience, we shall consider only points of thereal-number axis in what follows. Any set of real numbers willbe called a linear set. The integers form such a set, as do therationals and irrationals. All the definitions and theoremsproved for linear sets can easily be extended to any finite-dimensional space.

Closed Interval. The set of points { x j satisfying a < x S bwill be called a closed interval. If we omit the end points, thatis, consider those x that satisfy a < x < b, we say that theinterval is open (open at both ends). For example, 0 S x 5 1is a closed interval, while 0 < x < 1 is an open interval.

Bounded Set. A linear set of points will be said to be boundedif there exists an open interval containing the set. It must beemphasized that the ends of the interval are to be finite numbers,which thus excludes - oo, + oo.

An alternative definition would be the following: A set ofnumbers S is bounded if there exists a finite number N such that-N <x <NforallxinS.

The set of numbers whose squares are less than 3 is certainlybounded, for if x2 < 3 then obviously -2 < x < 2. However,the set of numbers whose cubes are less than 3 is unbounded, forx3 < 3 is at least satisfied by all the negative numbers. This setis, however, bounded above. By this we mean that there existsa finite number N such that x < N if x3 < 3. Certainly N = 2

89


does the trick. Specifically, a set of elements S is bounded aboveif a finite number N exists such that x < N for all x in S. Let thestudent frame a definition for sets bounded below.

We shall, in the main, be concerned with sets that contain aninfinite number of distinct points. The rational numbers in theinterval 0 < x < 1 form such a collection. Let the readerprove that between any two rationals there exists anotherrational.

Limit Point. A point P will be called a limit point of a set Sif every open interval containing P contains an infinite number ofdistinct elements of S. For example, let S be the set of numbers(1/2, 1/3, 1/4, . . . , 1/n, . . .). It is easy to verify that anyopen interval containing the origin, 0, contains an infinite num-ber of S. In this case the limit point 0 does not belong to S.It is at once apparent that a set S containing only a finite numberof points cannot have a limit point.

Neighborhood. A neighborhood of a point is any open intervalcontaining that point.

Interior Point. A point P is said to be an interior point of aset S if it belongs to S and if a neighborhood N of P exists, everyelement of N belonging to S. If S is the set of points 0 < x 5 1,then 0 and 1 are not interior points of S since every neighborhoodof 0 or 1 contains points that are not in S. However, all otherpoints of S are interior points.

Boundary Point. A point P is a boundary point of a set Sif every neighborhood of P contains points in S and points not inS. If S is the set 0 5 x <_ 1, then 0 and 1 are the only boundarypoints. A boundary point need not belong to the set. 1 is aboundary point of the set S for which x > 1, but 1 is not in Ssince 1 > 1.

Complement. The complement of a set S is the set of pointsnot in S. The complement C(S) has a relative meaning, for itdepends on the set T in which S is embedded. If S, for example,is the set of numbers -1 < x < 1, then the complement of Srelative to the real axis is the set of points lxl > 1. But thecomplement of -1 < x < 1 relative to the set -1 S x S 1 isthe null set (no elements). The complement of the set of nation-als relative to the reals is the set of irrationals, and conversely.

Open. Set. A set of points S is said to be an open set (not to beconfused with open interval) if every point of S is an interior

SEC. 411 IXTBGRATION 91

point of S. For example, the set S consisting of points whichsatisfy either 0 < x < 1 or 6 < x < 8 is open.

Closed Set. A set containing all its limit points is called aclosed set. For example, the set (0, 1/2, 1/3, 1/4, . . . , 1/n,

.) is closed, since its only limit point is 0, which it contains.

Problems

1. What are the limit points of the set 0 < x < 1? Is theset closed? Open? What are the boundary points?

2. Repeat Prob. 1 with the point x = removed.3. Show that the set of all boundary points (the boundary)

of a set S is closed.4. Prove that the set of all limit points of a set S is closed.5. Prove that the complement of an open set is closed, and

conversely.6. Why is every finite set closed?7. Prove that the set of points common to two closed sets is

closed. The set of points belonging to both Si and S2 is calledthe intersection of Sl and S2, written SI n82-

8. Prove that the set of points which belong to either S, or 82is open if S, and S2 are open. This set is called the union of S,and S2, written S, U S2-

9. An infinite union of closed sets is not necessarily closed.Give an example which verifies this.

10. An infinite intersection of open sets is not necessarily open.Give an example which verifies this.

Supremum. A number s is said to be the supremum of a set ofpoints S if

1. x in S implies x 5 s2. t < s implies an x in S such that x > tExample 51. Let S be the set of rationale less than 1. Then

1 is the supremum of S, for (1) obviously holds from the defini-tion of S, and if t < 1, it is possible to find a rational r < 1 suchthat t < r, so that (2) holds. We give a proof of this statementin a later paragraph.

Example 52. Let S be the set of rationals whose squares areless than 3, that is, SI x2 < 3. Certainly we expect the tobe the supremum of this set. However, we cannot prove thiswithout postulating the existence of irrationals. We overcome


this by postulating

Every nonempty set of points has a supremum (152)

Hence the rationals whose squares are less than 3 have a supre-mum. It is obvious that we should define this supremum as thesquare root of 3.

The supremum of a set may be + oo as in the case of the setof all integers, or it may be - oo as in the case of the null set.

The infemum of a set S is the number s such that1. x in S implies s < x2. t > s implies an x in S such that x < tExample 53. Let a > 0 and consider the sequence a, 2a, 3a,

... , na, ... . If this set is bounded, there exists a finitesupremum s. Hence an integer r exists such that ra > s - (a/2)so that (r + 1)a > s + (a/2) > s, a contradiction, since na 5 sfor all n. Hence the sequence Ina} is unbounded. This is theArchimedean ordering postulate.

Example 54. To prove that a rational exists between any twonumbers a, b. Assumed: a > b > 0, so that a - b > 0. Fromexample 53, an integer q exists such that q(a - b) > 1, orqa > qb + 1. Also an integer p exists such that p 1 = p > qb.Choose the smallest p. Thus p > qb ? p - 1. Hence

qa>gb+1zp>qb,and a > p/q > b. Q.E.D.

With the aid of (152) we are in a position to prove the well-known Weierstrass-Bolzano theorem.

"Every infinite bounded set of points S has a limit point." Theproof proceeds as follows: We construct a new set T. Into T weplace all points which are less than an infinite number of S. T isnot empty since S is bounded below. From (152), T has asupremum; call it s. We now show that s is a limit point of S.Consider any neighborhood N of s. The points in N which areless than s are less than an infinite number of points of S, whereasthose points in N which are greater than s are less than a finitenumber of points of S. Hence N contains an infinite number ofS, so that the theorem is proved. We have at the same timeproved that s is the greatest limit point of S. A limit point mayor may not belong to the set.

SEC. 41) INTEGRATION 93

Problems

1. The set 1/2, 1/3,. . . , 1/n, . . ., 1, 2, 3, . . . isunbounded. Does it have any limit points? Does this violatethe Weierstrass-Bolzano theorem?

2. Prove that every bounded monotonic (either decreasing orincreasing) sequence has a unique limit.

3. Prove that lim r" = 0 if Irl < 1. Hint: The sequencer, r2, . . . , r", . . . is bounded and monotonic decreasing forr > 0, and r"+1 = rr"

4. Show that if P is a limit point of a set S, we can pick out asubsequence of 8 which converges to P.

5. Show that (152) implies that every set has an infemum.6. Show that removing a finite number of elements from a set

cannot affect the limit points.7. Prove the Weierstrass-Bolzano theorem for a bounded set

of points lying in a two-dimensional plane.8. Let the sequence of numbers sl, 82, .. . , sn, . . . satisfy

the following criterion: for any e > 0 there exists an integer Nsuch that Is,,+F - snI < e for n ? N, p > 0. Prove that a uniquelimit point exists for the sequence.

9. A set of numbers is said to be countable if they can bewritten as a sequence, that is, if the set can be put into one-to-onecorrespondence with the positive integers. Show that a count-able collection of countable sets is countable. Prove that therationals are countable.

10. Show that the set S consisting of x satisfying 0 5 x 5 1is uncountable by assuming that the set S is countable, thenumbers x being written in decimal form.

Theorem of Nested Sets. Consider an infinite sequence ofnonempty closed and bounded sets S1, S2, . . . , Sn, . suchthat Sn contains that is, S1 M S2 D Sa M . Thereexists a point P which belongs to every Si, i = 1, 2, 3, . . . .

The proof is easy. Let P1 be any point of S1, P2 any point ofS2j etc. Now consider the sequence of points P1, P2, ... , P,,. . . . This infinite set belongs to S1 and has a limit point Pwhich belongs to 8, since Sl is closed. But P is also a limit ofP, Pn+1, ... , so that P belongs to Sn. Hence P is in all Sn,n = 1, 2, . . . .

Diameter of a Set. The diameter of a set S is the supremum ofall distances between points of the set. For example, if S is the


set of numbers x which satisfy I < x < 1, 3 < x < 7, thediameter of S is 7 - = 61. There are pairs of points in Swhose distance apart can be made as close to 6,91 as we please.

Problems

1. If a set is closed and bounded, the diameter is actuallyattained by the set. Prove this.

2. If, in the theorem of nested sets, the diameters of the Sapproach zero, then P is unique. Prove this.

The Heine-Borel Theorem. Let S be any closed and boundedset, and let T be any collection of open intervals having theproperty that if x is any element of S, then there exists an openinterval 7' of the collection T such that x is contained in T.The theorem states that there exists a subcollection T' of Twhich is finite in number and such that every element x of S iscontained in one of the finite collection of open intervals thatcomprise T'.

Before proceeding to the proof we point out that (1) both theset S and the collection T are given beforehand, since it is nogreat feat to pick out a single open interval which completelycovers a bounded set S; (2) S must be closed, for consider the setS(1, 1/2, 1/3, . . . , 1/n, . . .) and let T consist of the follow-ing set of open intervals:

TIIx such that Ix - 11 < 22

T21x such that Ix - _f < 32

T,,I x such that Ix - 1n

1

< (n + 1)2

It is very easy to see that we cannot reduce the covering of Sby eliminating any of the given T,,, for there is no overlappingof these open intervals. Each T,, is required to cover the point1/n of S contained in it.

Src. 421 INTEGRATION 95

Proof of the theorem: Let S be contained in the interval-N < x < N. This is possible since S is bounded. Now dividethis closed interval into two equal intervals (1) -N < x _:5- 0 and(2) 0 < x < N. Any element x of S will belong to either (1) or(2). Now if the theorem is false, it will not be possible to coverthe points of S in both (1) and (2) by a finite number of thegiven collection T, so that the points of S in either (1) or (2)require an infinite covering. Assume that the elements of S in(1) still require an infinite covering. We subdivide this intervalinto two equal parts and repeat the above argument. In thisway we construct a sequence of sets S, S2 S3 , suchthat each Si is closed and bounded and such that the diametersof the Si -* 0. From the theorem of nested sets there exists aunique point P which is contained in each Si. Since P is in S,one of the open intervals of T, say T, will cover P. rt'his Tphas a finite nonzero diameter so that eventually one of the Siwill be contained in TD, since the diameters of the Si --> 0. Butby assumption all the elements of this Si require an infinitenumber of the { T } to cover them. This is a direct contradictionto the fact that a single Tp covers them. Hence our originalassumption is wrong, and the theorem is proved.

42. Uniform Continuity. A real, single-valued function f (x)is said to be continuous at a point x = c if, given any positivenumber e > 0, there exists a positive number a > 0 such thatIf(x) - f(c)I < e whenever Ix - cI < S. The a may well dependon the a and the point x = c. The function f(x) is said to becontinuous over a set of points S if it is continuous at every pointof S.

We now prove that if f (x) is continuous on a closed andbounded interval, it is uniformly continuous. We define uni-form continuity as follows: If, for any e > 0 there exists a S > 0such that If(XI) - f(X2)1 < e whenever Ix1 - x21 < S, then f(x)is said to be uniformly continuous. It is important to noticethat 8 is independent of any x on the interval. We make useof the Heine-Borel theorem to prove uniform continuity. Choosean e/2 > 0. Then at every point c of our closed and boundedset there exists a S(c, a/2) such that If(s) - f (c)I < e/2 forc - a < x < c + S. Hence every point of S is covered by a26-neighborhood, and so also by a a-neighborhood. By theHeine-Borel theorem we can pick out a finite number of these


neighborhoods which will cover S. Let S, be the diameter of thesmallest of this finite collection of 6-neighborhoods. Now con-sider any two points x, and x2 of S whose distance apart is lessthan bl. Let xo be the center of the 6-neighborhood which coversx1. From continuity lf(xi) - f(xo)I < e/2. But also x2 differsfrom xo by less than 26, so that I f (x2) - f (xo) I < e/2. Conse-quently If(xl) - f(x2)1 < e. Q.E.D.

43. Some Properties of Continuous Functions(a) Assume f(x) continuous on the closed and bounded interval

a < x < b. As a consequence of uniform continuity, we canprove that f (x) is bounded. Choose any e > 0 and consider thecorresponding S > 0 such that If(XI) - f(X2)1 < e wheneverIx1 - x21 < S. Now subdivide the interval (a, b) into a finitenumber of S-intervals, say n of them. It is easily seen thatmax If(x) I < If (a) I + ne.

(b) If f (a) < 0 and f (b) > 0, there exists a c such that f (c) = 0,a < c < b. Let {x} be the set of all points on (a, b) for whichf (x) < 0. The set is bounded and nonvacuous since a belongsto {x}, for f(a) < 0. The set {x} will have a supremum; call it c.Assume f (c) > 0. Choose e = f (c) /2. From continuity, a 8 > 0exists such that If(x) - f (c) I < f(c)12 if Ix - cI < 6. Hencef(x) ? Jf(c) for all x in some neighborhood of x = c. Hence cis not the supremum of f x 1. Similarly f (c) < 0 is impossible, sothat f (c) = 0.

(c) We prove that f (x) attains its maximum. In (a) weshowed that f (x) was bounded. Let s be the supremum of f (x),a 5 x S b. As a consequence, f (x) <= s for all x on (a, b).Now consider the set s - e, s - e/2, . . . , s - e/n, ... ,where e > 0. Since s is a supremum, there exist x1, x2, ... ,x,,, ... such that f (xl) > s - e, f (x2) > s - e/2, ... ,

fxd } will have a limit point c. Let { xn' } be a subsequenceof f x,,) which converges to c. Then lim f (x.') z s, since

-s'-+ce/n -* 0 as n oo. But from continuity

lim f(x,') = .f(c).4C

Hence f (c) = s. Q.E.D.

SEC. 44] INTEGRATION

Problems

97

1. In the proof of (c) we exclude f(c) > s. Why?2. If f (x) is continuous on (a, b), show that the set of values

{f(x) } is closed.3. If f (x) has a derivative at every point of (a, b), show that

f (x) is continuous on (a, b).4. If f (x) has a derivative at every point of (a, b), show that a

c exists such that f(c) = 0, a < c < b, when f (a) = f (b) = 0.5. If f (x) has a derivative at every point of (a, b), show that

a c exists such that f(b) - f(a) _ (b - a)f'(c), a < c < b.6. Show that if two continuous functions f(x), g(x) exist such

that f (x) = g(x) for the rationals on (a, b), then f (x) = g(x) on(a,

7. Given the function f (x) = 0 when x is irrational, f (x) = 1/qwhen x is rational and equal to p/q (p, q integers and relativelyprime), prove that f (x) is continuous at the irrational points of(0, 1) and that f(x) is discontinuous at the rational points ofthis interval.

44. Cauchy Criterion for Sequences. Let x1, x2, ... ,xn, ... be a sequence of real numbers. We say that L is thelimit of this sequence, or that the sequence converges to L, if,given any e > 0, there exists an integer N depending on a suchthat IL - xnl < e whenever n > N(e). However, in most caseswe do not know L, so that we need the Cauchy convergencecriterion. This states that a necessary and sufficient conditionthat a sequence converge to a limit L is that given any e > 0,there exists an integer N such that Ixn - xml < e for n >_ N,m N. That the condition is necessary is obvious, for

IL - xnl < e/2,

IL xn,l < e/2 for m, n > N implies Ix,, - xml < e for m, n >_ N.The proof of the converse is not as trivial. Choose any e/2 > 0.Then we assume an N exists such that Ix,, -- xml < e/2 form, n ? N. Hence Ixn+ < I xNl + (e/2), n N, so that thesequence is bounded. We ignore xi, X2, . . . , xN_, since a finitenumber of elements cannot affect a limit point. From theWeierstrass-Bolzano theorem this infinite bounded set has atleast one limit point L. Hence, given an a/2, there exists an x.,with n > N, such that IL - xAi < e/2. But we also have


I xm - x,,l < e/2 for all m, n > N. Hence IL - xml < e for allm > N. Q.E.D.

Problems

1. Show how the convergence of a series can be transformedinto a problem involving the convergence of a sequence.

2. Show that the Cauchy criterion implies that the nth termof a convergent series must approach zero as n -- oo.

3. Show that the sequence 1, 1/2, 1/3, . . . , 1/n, . . . con-verges by applying the Cauchy test.

45. Regular Arcs in the Plane. Consider the set of pointsin the two-dimensional plane

P2 such that the set can be repre-sented in some coordinatesystem by x = f (t), y = sp(t),a S t < 0, where f(t) and ap(t)are continuous and have con-tinuous first derivatives. Suchcurves are called regular arcs.A regular curve is a set ofpoints consisting of a finitenumber of regular arcs joinedone after the other (see Fig.

Fia. 45.P4 45).

PoPI, P1P2, P2t 3, P3P4 arethe regular arcs joined at P1,

P2, P3. Notice that there are at most a finite number of dis-continuities of the first derivatives. In Fig. 45 the derivativesare discontinuous at P1j P2, P3.

46. Jordan Curves. The locus i x = f (t) a < t 5 fl, will be

l

y = P(t),called a Jordan curve provided that f (t) and (p(t) are continuousand that two distinct points on the curve correspond to twodistinct values of the parameter t (no multiple points).

A closed Jordan curve is a continuous curve having f(a),o(a) = 9(fl) but otherwise no multiple points.

From this we see that a Jordan curve is always "oriented,"that is, it is always clear which part of the curve lies between twopoints on the arc, and which points precede a given point. Weshall be interested in those curves which are rectifiable, or, in

Sr:c. 47] IN TkGRA TION 99

other words, we shall attempt to assign a definite length to agiven Jordan are or curve.

47. Functions of Bounded Variation. Let f(x) be defined onthe interval a < x < b. Subdivide this interval into a finitenumber of parts, say a = xo, x,, . . . x;, . . . , x,-1) x = b.Now consider the sum

I f(xi) - f(xo)I + I f(x2) - f(xi)1 + . . + l f(xn) - f(xn - 1)In

_ ± I f(xi) - f (xt-1)i_1

If the sums of this type for all possible finite subdivisions arebounded, that is, if

n

I f(xi) - f(xc-1)I < A < -o (153)s=x

we say that f(x) is of bounded variation on (a, b).A finite, monotonic nondecreasing function is always of

bounded variation since

n n

If(x=) - f(x=-,)1 = I If(x=) - f(xi-1)) = f(b) - f(a) = Ai=1 t=1

An example of a continuous function that is not of boundedvariation is the following:

f(x) = x sin 2x, 0< x 5 1

f (O) = 0

Let us subdivide 0 5 x <_ 1 into the intervals

(n+1) Sx<n,

n = 1, 2, .. . , N. Now f(1 /n) = (1/n) sin (irn/2) so that

N = 12 2 2

n If(I\n/ f \n } )I + 3

+ 5 + N

We cannot bound this sum for all finite N since the series divergesas N -> w. N was chosen as an odd integer.


48. Arc Length. Consider the curve given by x = f(t),y = c(t), and assume no multiple points. Divide the parametert in any manner into n parts, say

a = t0 < 11 < t2 < ' ' ' < to = ,8and consider

Sn = 1[x(ti) - x(ti-1)]2 + ly(ti) - y(tti-1)]2J}i=1

Y

F1o. 46.

This is the length of thestraight-line segments joiningthe points x(ti), y(ti), i = 0,1) 2, . . . , n (see Fig. 46).If the set of all such lengths,obtained by all finite meth-ods of subdivision, is bounded,we say that the curve is recti-fiable and define the length ofthe curve as the supremum ofthese lengths.

An important theorem is----x the following: A necessary

and sufficient condition thatthe curve described by

x = f(t), y = 'P(t)

be rectifiable is that f (t) and V(t) be of bounded variation. Letn

A = [I f (ti) - f(4.-1)12 + 14p(ti) - p(ti--1) J 2J1{=1

B = I IIf(t;) - f(4--1)I + Ip(k) - v(4-1)I If-l

so thatA==<Bs /2A

Consequently if the curve is rectifiable, f (t) and '(t) are ofbounded variation, and conversely, if f (t) and ap(t) are of boundedvariation, A is bounded, and hence the curve is rectifiable.

If f(t) and V'(t) are continuous, then from the law of the mean,I f(4) - f (ti--1) I = I f (ii) (ti - ti-1) I < A I ti - ti-il, where A is thesupremum of If(t)l, a 5 t =< # and 4_1 5 Ti 5 ti.

SEC. 49] INTEGRA TION

n n

E If() - f(t;-1)I < A 11(4- - till = A(# - a)i-1 i=1

101

Similarly, p(t) is of bounded variation, so that the curve is recti-fiable. Under these conditions it can easily be shown that thearc length is given by

s - f8 [(dt)2 + ( p)2J dt (154)

49. The Riemann Integral. We now develop the theory ofthe Riemann integral in connection with line integrals. Weneed a curve over which an integration can be performed and a

function to be integrated over this curve. Let r x

y=¢(t)a S t be a rectifiable are. Let f (x, y) be a function continu-ous at all points of the curve r. Subdivide the parameter tinto n parts, a = to < ti < t2 < < t, = 6. Let the coordi-nates of Pi be and let Osi be the length of are joiningPi-1 to Pi. Since f (x, y) is continuous, it will take on both itsminimum and maximum for each segment Pi_1P;. Multiply eachare length by the maximum value of f (x, y) on this are, say f v,and form the sum

n

J = A fii (xi, yi) A8i

Let .7 be the infemum of all such sums. Similarly, let K be thesupremum of all sums when the minimum value of f(x, y), say f,,,is used. If J = K, we say that f (z, y) is Riemann-integrableover the curve r, and write

J = K = fr f(x, y) ds = LfI(t), + * )} dt

whenever So' and J' are continuous.That J = K for a continuous function defined over a rectifiable

curve is not difficult to prove. For,

nJ-K=i-1

102 VECTOR AND TENSOR A V.I LYSIS [SEC. 50

and from the uniform continuity of f(x, y) we can subdivide thecurve r into arcs such that the difference between the maximumand minimum values of f (x, y) on any are is less than any givene > 0 so that

JJ - Kl < e11lAsil < eL (155)

We leave it as an exercise for the reader to prove that we can makeJ as close to J as we please and similarly that the difference

R=R1+R2(ii)

FIG. 47.

FIG. 48.

between K and K can be made arbitrarily small, for a sufficientlylarge number of subdivisions. Since the a of (155) is arbitrary,we can make the difference between J and K as small as weplease so that J = K since they are fixed numbers.

50. Connected and Simply Connected Regions. A region Ris said to be connected if, given any two points of the region, wecan join them by an arc, every point of the are belonging to theregion R. In Fig. 47, (i) is connected; (ii) is not connected. In(i), R is the nonshaded region.

Sic. 51] INTEGRATION 103

If every closed curve of a connected region R can be continu-ously shrunk to a point of R, we say that the region is simplyconnected. In Fig. 48, (i) is connected, but not simply con-nected; (ii) is simply connected. In (i) the curve r cannot becontinuously shrunk to a point. An analytic expression forsimple connectedness can be set up, but we shall omit this.

51. The Line Integral. Let

f = X(x, y, z)i + Y(x, y, z)j + Z(x, y, z)k

and consider the line integral f ( f ds) ds along a rectifiable

space curve r given by r = r(s). Since

dsds = dr = dx i + dy j + dz k

we have

Jr (f ds/ds = fr X dx + Y dy + Z dz (156)

We use (156) as a means of evaluating the line integral. Ifthe space curve is given by x = x(t), y = y(t), z = z(t), then(156) reduces to

f f dr = fo' [x(t)dt + Y(t) dt + Z(t) dtJ

dt (157)

In general, the line integral will depend on the path joiningthe two end points of integration. If f f dr does not dependon the curve r joining the end points of integration, we say that fis a conservative vector field. If f is a force field, we define

A

fBf - dr as the work done by f as the unit particle moves from

A to B. We shall now work out a few examples for the reader.Example 55. Let f = x2i + yaj, and let the path of integra-

tion be the parabola y = x2, the integration being performedfrom (0, 0) to (1, 1). We exhibit three methods of solution.

(a) Let x = t, so that y = t2. Thus

f = t2i + t6j, r = ti + t2j, dr = (i + 2tj) dt


and

ft t of f dr = f.' (t2 + 2t7) dt =

(b) Since y = x2 everywhere along r, f = x2i + x6j along r,and dr = dx i + dy j = (i + 2xj) dx, so that

Jr JO

1

(x2 + 2x') dx = A(c)

r (1, 1) ((1, 1) xaJ(o, 0) f dr = J(0, 0)

x2 dx + y$ dy - 31+y'1- 70 4 (0 12

(c) shows that the integral from (0, 0) to (1, 1) is independentof the path since x2 dx + y3 dy is a perfect differential, that is,x2 dx + ys dy = d[(xs/3) + (y4/4)].

Example 56. f = yi - xj, and let the path of integration bey = x2 from (0, 0) to (1, 1). Then

(1 f'f(1'1)ydx-xdy= x2dx-x(2xdx) _ -0,

, o) ro, o)

Next we compute the integral by moving along the x axis fromx = 0 to x = 1 and then along the line x = 1 from y = 0 toy = 1. We have

(f (0,0)) f("

o, o) 0, o) 1, o)

Along the first part of the path, f = -xj and dr = dx i, sincey = dy = 0. Along the second part of the path, f = yi - j,dr = dy j since x = 1 and dx = 0. Thus

f(111) x=1 v i

0) f.--=o - fv-of01-dy=-1

The line integral does depend on the path for the vector fieldf = yi - xj. We say that f is a nonconservative field.

52. Line Integral (Continued). Let us assume that f can bewritten as the gradient of a scalar (p(x, y, z), that is, f = 0p.Then the line integral JAB f dr is independent of the path of

SEC. 521 INTEGRATION

integration from A to B since

105

JAfdrJAVdrLd(B)(A) (158)Our final result in (158) depends only on the value of (p(x, y, z)when evaluated at the points A(xo, yo, zo), B(xi, yi, z,) and in noway depends on the path of integration. If our path is closed,then A = B and (p(B) = p(A), so that the line integral aroundany closed path vanishes if f = Vip. However, the region forwhich f = must be simply connected. Let us consider thefollowing example.

Example 57. Let- yi xif

=x2+y2LL

-rx2+y2Then f = VV where p = tan-' (y/x), and if we integrate f overthe unit circle with center at the origin, we have

d (tan_i )fo2T dO 2rxJ

so that our line integral does not vanish. The region for whichf = V(p is not simply connected since p is not defined at theorigin.

We now prove that if f f dr is independent of the path, thenf is the gradient of a scalar V. Let

P(x, y, z) =JP(z,y.z) f dr

o(xo, YO, zo)

Now(P(x + Ax, y, z) - (P(x, y, z)

Ax

JP(fdr)ds (159)

and since the line integral is independent of the path joining Pto Q, we choose the straight line from P to Q as our path ofintegration, that is, dr = dx i. Thus

lim cP(x+Ox,y,z) --P(x,y,z) =avAX-40 Ax ax

fx+Ax

X (x, y, z) dxiim z

= X (x, y, z)AX-0 Ox


from the calculus. We have assumed that f is continuous. Nowsimilarly

-y

= Y(x, y, z), az = Z(x, y, z)

so thata(P av , a,P

f =ax

i + a7 +az

k = VV (160)

If f has continuous derivatives, we can easily conclude whetherf is the gradient of a scalar or not. Assume f = VV, or

(1) X=-"`P, (2) (3)Z=az

Differentiating (1) with respect toy and (2) with respect to x,we see that

ax ayay ax

SimilarlyaYaZaz ay

azaxax az

(161)

This is the condition that V x f = 0. Conversely, assumeV x f = 0. Let

,P(x,y,z) = f:oX(x,y,z)dx+f,0 Y(xo, y, z) dy

+z,

Z(xo, yo, z) dz (162)

Nowa'P

= X (x, y, z)axaq

f.. -0Y

dx+Y(xo,y,z)

ay_ -L.

f ax dx + Y (x,), y, z)

= Y(x, y, z) - Y(xo, y, z) + Y(xo, y, z)= Y(x, y, z)

Sic. 53] INTEGRATION

Similarly 7.(x, y, z). Consequently

f axi+ayj+azk=Drp

107

We have proved that a necessary and sufficient condition for fto be the gradient of a scalar is that.

V xf =0 (163)

If f = V or V x f = 0, f is said to be an irrotational vector.Example 58. Let f = 2xyezi + x2ezj + x2yezk. Then

Vxf=

and

2xyez x2ez x2yeZ

i j k

a a a

ax ay az=0

=JX

2xy9 dx + 02ez dy + 02. 0 . 9 dz

= x2yeZ

so that f= V (x2yeZ + constant).

Problems

1. Given f = xyi - xj, evaluate f f dr along the curve y = x3from the origin to the point P(1, 1).

2. Show that if the line integral around every closed path iszero, that is, if jFf dr = 0, then f = V p.

3. Show that /r dr = 0.4. Show that the inverse-square force field f = -r/r3 is con-

servative. The origin is excepted.5. f = (y + sin z)i + xj + x cos z k. Show that f is conserva-

tive and find rp so that f = V p.6. Evaluate f x dy - y dx around the unit circle with center

at the origin.7. If A is a constant vector, show that /A dr = 0,

jrAxdr=053. Stokes's Theorem. We begin by considering a surface

of the type encountered in Chap. 3. Now consider a closed


rectifiable curve r that lies on the surface. As we move, alongthe curve r, keeping our head in the same direction as the normalcar ar

x ) we keep track of the area to our left. It is this surfaceau av

u curve

v curveFia. 49.

that we shall keep in touchwith, and r will be theboundary of this surface.We neglect the rest of thesurface r(u, v). We now con-sider a mesh of networks onthe surface formed by a col-lection of parametric curves.Of course, the boundary rwill not, in general, consistof arcs of these parametriccurves (see Fig. 49). Con-sider the mesh A BCD. Letthe surface coordinates of A

be (u, v), so that A(u, v), B(u + du, v),

C(u+du,v+dv)

D(u, v + dv) are the coordinates of A, B, C, D. We also assumethat the parametric curves are rectifiable. Now consider

ABCD f dr

The value of fat A is f(u, v); at B it is f(u + du, v); at D it isf(u, v + dv).Now

f (u + du, v) = f (u, v) + dfu= f(u, v) + (dru V)f

= f(u, v) + (au du v) f

except for infinitesimals of higher order. Similarly

f(u,v+dv) =f(u,v)+(-dvv)f

Six. 53] INTEGRATION 109

Hence, but for infinitesimals of higher order,

-auc1r

f +

av - fi au} dude

(164)

Nowfor ar an Or

(v x f) (au x av = (° x f) xau av

C\au °/ Jav ^ L \av

Hencee

v)fI arau

fiABCD f dr = (v x f) a x adu dv (165)

ar arNow d x a du dv = area of sector ABCD in magnitude, and

au avits direction is along the normal. We define

Or Orx du dvdd = au &

so that

(166)

IABCDfdr= (V

We now sum over the entire network. Interior line integrals

will cancel out in pairs leaving only fir f dr. Also

I (V f f (Vover surface 8 a

as the areas approach zero in size. We thus have Stokes'stheorem :

fir f dr = f f (V x f) dd (167)8


Comments1. The reader may well be aware that (165) does not hold for

a mesh that has r as part of its boundary. This is true, butfortunately we need not worry about the inequality. The lineintegrals cancel out no matter what subdivisions we use, and fora fine network the contributions of those areas next to r con-tribute little to f f (V x f) dd. The limiting process takes careof this apparent negligence.

2. We have proved Stokes's theorem for a surface of the typer(u, v) discussed in Chap. 3. The theorem is easily seen to betrue if we have a finite number of these surfaces connected con-tinuously (edges).

3. Stokes's theorem is also true for conical points, where nodd can be defined. We just neglect to integrate over a small areacovering this point. Since the area can be made arbitrarilysmall, it cannot affect the integral.

4. The reader is referred to the text of Kellogg, "Foundationsof Potential Theory," for a much more rigorous proof of Stokes'stheorem.

54. Examples of Stokes's TheoremExample 59. Let r be a closed Jordan curve in the x-y plane.

Let f = -yi + xj. Applying Stokes's theorem, we have

f f

f fS ax ay az

-y x 0

=2ffdydx=2AS

or

Area A =- xdy - ydx

For the ellipse x = a cos t, y = b sin t, dx = -a sin t dt,

dy = b cos t dt

(168)

and

A =f27

ab(cos2 t + sin 2 t) dt = grab

SEc. 541 INTEGRATION 111

Example 60. If f has continuous derivatives, then a necessaryand sufficient condition that ff dr = 0 around every closedpath is that V x f = 0.

If V x f = 0, then Cf U. dr = f f (V x f) dd = 0. Con-s

versely, assume f t dr = 0 for every closed path. If V x f p4 0,then V x f 0 at some point P. From continuity, V x f: 0in some region about P. Choose a small plane surface S in thisregion, the normal to the plane being parallel to V x f. Then

f- dr = f f V x f- dd > 0, a contradictions

Example 61. We see that an irrotational field is characterizedby any one of the three conditions:

(i) f = VP(ii) V x f = 0 (169)

(iii) f dr = 0 for every closed path

Any of these conditions implies the other two.Example 62. Assume f not irrotational. Perhaps a scalar

p(x, y, z) exists such that pf is irrotational, that is, V x (pf) = 0.If this is so,

pV xf+Vp xf = 0 (170)

and dotting (170) with f, we have, since f VA x f = 0, theequation

f- (V xf) = 0 (171)

If f = Xi + Yj + Zk, (171) may be written

X Y Za a a

= 0 (172)ax ay az

X Y ZIn texts on differential equations it is shown that (172) is alsosufficient for p(x, y, z) to exist. We call p(x, y, z) an integratingfactor.

Example 63. Let f = f(x, y, z)a, where a is any constant vec-tor. Applying Stokes's theorem, we have


f f x a) dds

f fs

a

f f dd x

f = a awe have

S

ff [a(V g) - (a V)gl dds

=a- f f (V.g)dd-a f f

V dd (summed). There-fore

xdr = a- f f (ddxv) xg

and

dr x g = f f (dd x V) x g (174)

S

We notice that in all cases

0 dr * f= Jf(doxv)*f (175)S

The star (*) can denote dot or cross or ordinary multiplication.In the latter case, f becomes a scalar f.

Sec. 541 INTEGRATION

Problems

113

1. Prove that fdr = 0 from (175).2. Show that L, 'dr x ri taken around a curve in thex-y plane

is twice the area enclosed by the curve.3. If f = cos y i + x(1 + sin y)j, find the value of ff - dr

around a circle of radius r in the x-y plane.4. Prove that fr dr = 0.5. Prove that ff dd x r = fir 2 dr.

s6. Prove that f u Vv dr = - ,f v V u dr.7. Prove that u Vv - dr = JJvu x Vv dd.

s8. If a vector is normal to a surface at each point, show that

its curl either is zero or is tangent to the surface at each point.9. If a vector is zero at each point of a surface, show that

its curl either is zero or is tangent to the surface.

10. Show that fi a x r dr = 2a. Jf dd, if a is constant.S

11. If fi E dr = -cat Jf B dd for all closed curves, shows

that V x E = ---C at12. By Stokes's theorem prove that V x (V(p) = 0.

13. Show that fi dr/r = f f (r/r3) x dd where r = Id.8

14. Find the vector f such that xy = f(O,(X' Y,) f dr.o, o)

15. If f = r/r3, show that V x f = 0 and find the potential yosuch that f = Vsp.

16. Show that the vector f = (- yi + xj)/(x2 + y2) is irrota-

tional and that (fr f dr = 2w, where r is a circle containing the

origin. Does this contradict Stokes's theorem? Explain.

17. Show that ff v x f dd = 0, where S is a closed surface.s

18. Let C1 and C2 be two closed curves bounding the surfacesS1 and S2. Show that


k. ki r122 dr1 dr2 = -4 fJ dal f f dd2St s2

f, fc, r122 dr1 x dr2 = - 2 f f dot x ff dd2Si s2

where r12 is the distance between points on the two curves.55. The Divergence Theorem (Gauss). Let us consider a

region V over which f and V f are continuous. We shall assumethat V is bounded by a finite number of surfaces such that ateach surface there is a well-defined continuous normal. We shallalso assume that f can be integrated over the total surface bound-ing V. Now, no matter what physical significance f has, if any,we can always imagine f to be the product of the density andvelocity of some fluid. We have seen previously (Sec. 20) thatthe net loss of fluid per unit volume per unit time is given byV f. Consequently, the total loss per unit time is given by

JfJ(v.f)dr (176)V

Now since f and V f are continuous, there cannot be any pointin the region V at which fluid is being manufactured or destroyed;that is, no sources or sinks appear. Consequently, the totalloss of fluid must be due to the flow of fluid through the boundaryS of the region V. We might station a great many observers onthe boundary S, let each observer measure the outward flow, andthen sum up each observer's recorded data. At a point on thesurface with normal vector area dd, the component of the velocityperpendicular to the surface is V N, where N is the unit outwardnormal vector. It is at once apparent that pV dd = f ddrepresents the outward flow of mass per unit time. Hence thetotal loss of mass per unit time is given by

Jff.dd (177)

Equating (176) and (177), we have the divergence theorem:

JJf.do=fJf(v.f)d.r (178)8 V

SEC. 55] INTEGRATION 115

For a more detailed and rigorous proof, see Kellogg, "Founda-tions of Potential Theory." We now derive Gauss's theorem by adifferent method. Let f be a differentiable vector inside a con-nected region R with rectifiable surface S. Surround any pointP of R by a small element of volume dr having a surface area AS.Form the surface integral JJf. dd and consider the limit,

As

fAs

liraAr-.0 DT

If this limit exists independent of the approach of Ar to zero, wedefine

Jff.dddiv f = lim As

AT-40 AT

We can write

(179)

tardivf= JJ (180)

where e -40 as Ar -* 0. If we now subdivide our region intomany elementary volumes, we obtain formulas of the type (180)for all of these regions. Summing up (180) for all volumes andthen passing to the limit, we have

fffdivfdr__ Jffdd (181)

Inthe derivation of (181) use has been made of the fact that foreach internal dd there is a -dd, so that all interior surfaceintegrals cancel in pairs, leaving only the boundary surface Sas a contributing factor. The sum of the ei Ar; vanishes in thelimit, for I Zet Aril 5 ZIel... OT S IEIo,.=V, and if div f is continu-ous, IeI,. - 0 as 2r --i 0.

By choosing rectangular parallelepipeds and using the methodof Sec. 20, we can show that

fff.doclu c3v aw

div f = slim AS(182)

AT ax + a + 49Zy


for f = ui + vj + wk, which corresponds to our original defini-tion of the divergence.

Example 65. Consider a sphere with center at 0 and radius a.Take f=r=xi+yj+zk,

xi+yj+zk)dS(dd =(r)

dS =(ai)a

Now V f =3, f dd = (1/a)(x2 + y2 + z2) dS = a dS on thesphere. Applying (178), f _f f 3 dr = f f a dS, or 3V = aS, where

Fzo. 50.

S is the surface area of thesphere. If V is known to be

g 47a8, then S = 4aa2.Example 6 6. Let f = qr/rs

and let V be a region surround-ing the origin and let S be itssurface. We cannot apply thedivergence theorem to this re-gion since f is discontinuous atr = 0. We overcome this diffi-culty by surrounding the originby a small sphere E of radius e

(see Fig. 50). The divergence theorem can be applied to theconnected region V'. The region V' has two boundaries, S andE. Applying (178),

ffJ V. () dr f rad+J1 r"d (183)8 E

In Example 25 we saw that V (r/r3) = 0. This implies that(183) reduces to

ff.!.do=(184)

Now for the sphere 1, dd = -r dS/e, since the outward normalto the region V' is directed toward the origin and is parallel tothe radius vector. Hence

JJ.dd= fff.dd= ffds=4fq

SFe. 551 INTEGRATION 117

since f f dS = The integral f f f dd is called the flux ofs

the vector field f over the surface S. We have, for an inverse-square force f = qr/r3,

fJ f dd = 4irq (185)S

Example 67. A vector field f whose flux over every closedsurface vanishes is called a solenoidal vector field. From (178)it is easy to verify that V f = 0 for such fields. Hence asolenoidal vector is characterized by either /f dd = 0 or

0.

Now assume f = V x g, which implies V f = V (V x g) = 0.Is the converse true? If V f = 0, can we write f as the curl ofsome vector g? The answer is "Yes"! We call g the vectorpotential of f. This theorem is of importance in electricitytheory, as we shall see later. Notice that g is not uniquelydetermined since V x (g + Vp) = V x g. We now show theexistence of g. Let f = Xi + Yj + Zk and assume

g=ai+ij+ykWe wish to find a, $, y such that f = V x g, and hence

x ay a#=---ay az

Yas ay

=az ax

Zas as=---ax ay

(186)

Now assume a = 0. Then X =ay - 8-, Y = - aye Z = as.ay 8z ax ax

Consequently if there is a solution with a = 0, of necessity

$6= fXzdx+o.(y,z)o

y = - xos Ydx+r(y,z)

Now V f = 0 orax aY + az) by assumption.

y


Hencef aY aZ ar _ aaJZoay az ay + az /

,"+ ay az

fx axdx

+ ar - as

ax ay az

= X (x, y, z) - X (xo, y, z) +oray -

ao

az

so that (186) is satisfied by a = 0, S = I o Z dx + or (y, z),

yjX

Y dx by choosing r = 0, o(y, z) = - JZX(xo,y,z) dz.

Hence f =Vxgwhereg=[jz

Z dx + c(y, z)] j - faVdxk.In general

g=[ faZdx+v(y,z)]j- foYdxk+Vp (187)

For example, if f = V(1/r), then V f = 0. Now

X=-- s, Y=-y, Z=-raz

where r2 = x2 + y2 + z2. Applying (187)

-z dx x y dxg= (x2+y2+z2), j+1 (x2+y2+z2)$k+V ,

andv arbitrary

g = (x2 + y2 + z2)#(y2 +z2)

(-zj + yk) + VSP

Example 68. Green's theorem. We apply (178) to fl = u Vvand f2 = v Vu and obtain

f f f V. (u Vv)dr= f f f ffR R S(188)

f f f V V. (v Vu) dr = f f f (v V2u + VV. VU) d7 = Jf v Vu ddR R S

Subtracting, we obtain

f f f (u V2v - v V2u) dr = f f (u Vv - v Vu) dd (189)R S


Example 69. Uniqueness theorem. Assume two functionswhich satisfy Laplace's equation everywhere inside a region andwhich take on the same values over the boundary surface S.The functions are identical. Let V2(P = 1724, = 0 inside R and(p ='p on S. Now from (188) we have

JJJov2odr+JfJvo.vodr=JJove.doR R

Define 0 - (p - ¢ so that 1720 = 0 over R and 0=0 on S.Hence f f f (V0)2 dT = 0, which implies V0 = 0 inside R.

R

Hence 0 = constant = p -'p, and since sc on S, we musthave cp ='G. We have assumed the existence of Vp, V¢ on S.

Example 70. Another uniqueness theorem. Let f be a vectorwhose curl and divergence are known in a simply connectedregion R, and whose normal components are given on the surfaceS which bounds R. We now prove that f is unique. Let f1 beanother vector such that V f = V f1, V x f = V x f1, andf dd = f, dd on S. We now construct the vector g = f - f1.We immediately have that V g = V x g = g dd = 0. InSec. 52 we saw that if V x g = 0, then g is the gradient of a scalarcp, g = V<p. Consequently, V g = V2ip = 0. Applying (188)with u = v = (p, we have

If! [V2cc+ (Vcc)21 dr = ff `p vv - dd = f 0R s

and hence fff (VV) 2 dr = 0 so that V p - 0 inside R, andR

g = Vp - 0, so that f = f 1 inside R.We can also prove that f is uniquely determined if its diver-

gence and curl are known throughout all of space, provided that ftends to zero like 1/r2 as r--, ao. We duplicate the aboveproof and need Jim f p VV dd = 0. If cp tends to zero like

1/r, then V<p tends to zero like 1/r2, and f f cc V(p V. dd tends tos

zero like 1 /r as r oo.

Example 71. Let f = f(x, y, z)a, where a is constant. Applying (178), we obtain


a- f ffdo= f f fv.(fa)dr=a f f fVfdrs V V

Hence

fffdd = fff 7fdr

s V

We leave it to the reader to prove that

JJdo*f = Jff(V*f)drs V

(190)

(191)

Problems

1. Prove that f f dd x f = f f f (V x f) dr.s y

2. Prove that ff dd = 0 over a closed surface S.s

3. If f = axi + byj + czk, a, b, c constants, show thatJff.do = 4(a + b + c), where S is the surface of a units

sphere.fffddAs4. By defining grad f = Jim

4,-+Osnow that

OT

gradf = afi+ afj+afkax ay az

fff.dd5. By defining div f = lim °3 , show that

A?--+O GT

div f =r2 sin 0 Lar

(r2 sin of,) + e (r sin 9 fe) + a (rf,)]

for spherical coordinates.6. Show that

f fJ JfJ (P 0fdr.

SEC. 55] INTEGRATION

7. If w = V x v, v = V x u, show that

'9v2dr = f f u xv.dd+ f f

R S R

8. Show that

121

f J f w Vu . Vv dr = f f uw Vv . dd - f if u V . (w Vv) dr

9. If v = Vp and V . v = 0, show that for a closed surface

f J f V2dr = Jf pv.dd.

10. Show that f f IrJ2r . da = 5 fJJ r2 dr.

11. If f = xi - yj + (z2 - 1)k, find the value of f f f . do

over the closed surface bounded by the planes z = 0, z = 1 andthe cylinder x2 + y2 = 1.

12. If f is directed along the normal at each point of theboundary of a region V, show that f f f (V x f) dr = 0.

V

13. Show that f f r x da = 0 over a closed surface.S

14. Given f = (xyez + log (z + 1) - sin x)k, find the value of

ffv x f . da over the part of the sphere x2 + y2 + z2 = 1 aboveS

the x-y plane.15. Show that (xi + yj)/(x2 + y2) is solenoidal.16. If f 1 and f2 are irrotational, show that f 1 x f2 is solenoidal.17. Find a vector A such that

f =- yzi - zxj + (x2+y2)k = V xA.

18. If r, 0, z are cylindrical coordinates, show that De andV log r are solenoidal vectors. Find the vector potentials.

19. Let S, and S2 be the surface boundaries of two regionsV1 and V2. Let r be the distance between two elementary


volumes d-r, and dT2 of V1 and V2. Show that

.fs, dii2 fsI rm dd1 = -m(mn + 1) fV (IT2 fv, rn:-2 dT,

and that

1s= dd2 fs, log r ddl = - fv,cIT2

dTlfV1 r

20. IfV - f = V xf ='1i+¢2j+4,ak,f = Xi + Yj + Zk,49 Oshow that V2X = - - a3 - z2' and find similar expressions

y

for V2Y, V2Z. Find a vector f such that V f = 2x + y - 1,Vxf=zi.

56. Conjugate Functions. Let us consider the two-dimen-sional vector field w = u(x, y)i + v(x, y)j and an orthogonalvector field wi = v(x, y)i - u(x, y)j. Obviously w - wl = 0.What are the conditions on u(x, y), v(x, y) which will make w andw1 irrotational? From Stokes's theorem

av auw dr = ffv x w dd = ffax ay

dydx

w1 dr =

s sau (192)

{ jJv = Jf ax-av

aydy dx

A necessary and sufficient condition that both w and w1 be

irrotational is thatav - au = 0 and - au -- av

= 0 (see Sec.ax ay ax a4

52). This yields

av auax

_ay

av auay ax

(193)

The reader who is familiar with complex-variable theory willimmediately recognize (193) as the Cauchy-Riemann equations,which must be satisfied for the analyticity of the complex func-tion w = v(x, y) + iu(x, y), i = .


On differentiating (193), we obtain

and

a2u a2u=V 2U = + 2

axe 49Y

a2v a2v=2V2v=+aaxey

au av au av _ax ax + ay ay

(194)

If functions u(x, y), v(x, y) satisfy Eqs. (193) we say thatthey are harmonic conjugates. The importance of such func-

y v

P (x, yl

Ix

Fia. 51.

tions is due to the fact that they satisfy the two-dimensionalLaplace's equation given by (194). If u satisfies V2u = 0, wesay that u(x, y) is harmonic.

Let us now consider two rectangular cartesian coordinatesystems, the x-y plane and the u-v plane (Fig. 51). Let

r=xi+yj.Now to every point P(x, y) there corresponds a point Q(u, v)given by the transformation u = u(x, y), v = v(x, y). Hence thevector w = u(x, y)i + v(x, y)j corresponds to the vector

r -- xi+yj.If now P(x, y) traverses a curve C in the x-y plane, Q(u, v) willtrace out a corresponding curve r in the u-v plane.

The curve u(x, y) = constant in the x-y plane transforms intothe straight line u = constant in the u-v plane. Similarly,

124 VECTOR AND TENSOR ANALYSIS [Ssc. 56

v(x, y) = constant transforms into the straight line v = constant.The two straight lines are orthogonal. Do the curves

u(x, y) = constant

v(x, y) = constant intersect orthogonally? The answer is "Yes"!The normal to the curve u(x, y) = constant is the vector

VU=au i+u

aay

and the normal to the curve v(x, y) = constant is Vv =av

ax i +49Vav

jyOu av au av

so that Vu Vv =ax ax + a = 0 from (194).

y ayExample 72. Consider the vector field w = 2xyi + (x2 - y2) j.

Here u = 2xy, v = x2 - y2, and

av = au= 2x

ax ay

av _ aur_ _ 2

ay ax y

so that u and v are conjugate harmonics. The curves

u = 2xy = constant

and v(x, y) = x2 --- y2 = constant are orthogonal hyperbolaswhich transform into the straight lines u = constant, v = con-stant, in the u-v plane (Fig. 52).

Example 73. Consider

w = ( tan-1 i + J log (x2 + y2)jx)

Here u(x, y) = tan-1 (y/x), v = I log (x2 + y2), and

au av y

ax ay x2 + y2

au av x

ay ax x2 + y2

so that u and v are conjugate harmonics.

SEc. 56] INTEGRATION 125

YI

Fia. 52.

Fra. 53.

0'

(3)(4)

------------

The circles x2 + y2 = constant transform into the straightlines v = J log c, while the straight lines y = mx transform intothe straight lines u = tan-' m (Fig. 53).

Example 74. If u(x, y) is given as harmonic, we can find itsconjugate v(x, y). If v(x, y) does exist satisfying (193), then

126 VECTOR AND TENSOR ANALYSIS f&c. 56

dv=axdx+-dy=aydx - auax dy

Now consider the vector field f =au i - au j.

We have thatay ax

V x f = - t x + a2 ) k = 0 by our assumption about u(x, y).y

Hence f is irrotational and so is the gradient of the scalar v,f = Vv, and

au jyOUv = fa a dx -8x

dy + c1!

(195)

As an example, consider u = x2 - y2, which satisfies V2u = 0.Hence

v = fox - 2y dx - f." 2 . 0 dy + c = - 2xy + c

Problems

1. Find the harmonic conjugate of xa - 3xy2, of ex coo y, ofx/(x2 + y2).

2. Show that u(x, y) = sin x cosh y and v(x, y) = cos x sinh yare conjugate harmonics and that the curves u(x, y) = constant,v(x, y) = constant are orthogonal. What do the straight linesy = constant transform into?

3. If u(x, y), v(x, y) are conjugate harmonica, show that theangle between any two curves in the x-y plane remains invariantunder the transformation u = u(x, y), v = v(x, y), that is, thetransformed curves have the same angle of intersection.

CHAPTER 5

STATIC AND DYNAMIC ELECTRICITY

57. Electrostatic Forces. We assume that the reader isfamiliar with the methods of generating electrostatic charges.It is found by experiment that the repulsion of two like pointcharges is inversely proportional to the square of the distancebetween the charges and directly proportional to the productof their charges. The forces act along the line joining the twocharges. We define the electrostatic unit of charge (e.s.u.) asthat charge which produces a force of one dyne on a like chargesituated one centimeter from it when both are placed in avacuum. The electrostatic intensity at a point P is the forcethat would act on a unit charge placed at P as a result of the restof the charges, provided that the unit test charge does not affectthe original distribution of charges. For a single charge q placedat the origin of our coordinate system, the electric intensity, orfield, is given by E = (q/r3)r. For many charges the field at Pis given by

E = - 4'a ra (196)s

a=1 ra

where ra represents the vector from P to the charge qa.We have seen in Example 25 that V V. (r/r3) = 0. Conse-

quently,

V.E=0 (197)

so that the divergence of the electrostatic-field vector is zero atany point in space where no charge exists. Hence E is solenoidalexcept where charges exist, for there E is discontinuous. If thecoordinates of P are t, n, t and the coordinates of qi are xi, yi, zi,then ri = I(rS - xi)2 + (,l - y,)2 + (pJ - z,)2]1, and

l a _ r;V1

ri o tr ri an rr;3

ri (xi - )i -I- (yi- 7/)i + (zi -t)k127


so that (196) reduces to

E = -Vp (198)

n

where P = I qa/ra. We call w the electrostatic potential.a-1

For any closed path which does not pass through a point charge,we have f E dr = - ^p dr = - f d(p = 0. Thus E is alsoirrotational, and

VxE=O (199)

We also note that f E dr = cp(P) - p(-o) = tp(P), since,p(oo) = 0. Hence the work done by the field in taking a unitcharge from P to oo is equal to the potential at P.

58. Gauss's Law. Let S be an imaginary closed surface thatdoes not intersect any charges. In Example 66 we saw thatJ f (qr/r3) dd = 4Tq. This is true for each charge q; inside S.S

Hence

J ! Laqara dd = 41r qa (200)

S a-1 raa a-1

For a charge outside 8,

Jf.do=JfJv.()dr=0 (201)

since there is no discontinuity in qr/r3, r > 0. Adding (200)and (201), we obtain Gauss's law,

JJE.dd=4irQ (202)s

where Q is the total charge inside S. The theorem in wordsis that the total electric flux over any closed surface equals 4wtimes the total charge inside the surface.

Example 75. We define a conductor as a body with no electricfield in its interior, for otherwise the "free" electrons would moveand the field would not be static. The charge on a conductormust reside on the surface, for consider any small volume con-tained in the conductor and apply Gauss's theorem.

SEC. 58] STATIC AND DYNAMIC ELECTRICITY 129

f f E dd = 4xq

and since E = 0, we must have q = 0. This is true for arbi-trarily small volumes, so that no excess of positive charges overnegative charges exists. Hence the total charge must exist onthe surface of the conductor.

If a body has the property that a charge placed on it continuesto reside where placed in the absence of an external electric field,

Fic. 54.

we call the body an insulator. Actually there is no sharp lineof demarcation between conductors and insulators. Every bodypossesses some ability in conducting electrons.

At the surface of a conductor the field is normal to the surface,for any component of the field tangent to the surface would causea flow of current in the conductor, this again being contrary tothe assumption that the field is static (no large-scale motion ofelectrons occurring). Such a surface is called an equipotentialsurface. The field is everywhere normal to an equipotentialsurface, for the vector E = -V(p is normal everywhere to thesurface V(x, y, z) = constant.

Example 76. Consider a uniformly charged hollow sphere E.We shall show that the field outside the sphere is the same as if


the total charge were concentrated at the center of the sphere

Fic. 55.

and that in the interior of thesphere there is no field.

Let P be any point outsidethe sphere with spherical coordi-nates r, 0, cp. Construct animaginary sphere through P con-centric with the sphere (seeFig. 54). From symmetry it isobvious that the intensity at anypoint of the sphere is the sameas that at P. Moreover, the

field is radial. Applying Gauss's law, we have f f E dd = 4,rQ,or

so thatf fEdS=Ef fdS=4irr2E=4aQ

E=Q

and E=Q

r (203)r

We leave it to the reader to show that E = 0 inside I.Example 77. Field w thin a parallel-plate condenser. Consider

two infinite parallel plates with surface densities a and -a.

Fic. 56.

From symmetry the field is normal to the plates. We applyGauss's law to the surface in Fig. 55 with unit cross-sectionalarea.

f f E do = E = 4ara (204)

so that the field is uniform.Example 78. We now determine the field in the neighborhood

of a conductor. We consider the cylindrical pillbox of Fig. 56and apply Gauss's law to obtain


EA = 4raAor

E = 47ro N (205)

where a is the charge per unit area and N is the unit normal vectorto the surface of the conductor.

Example 79. Force on the surface of a conductor. We considera small area on the surface of the conductor. The field at apoint outside this area is due to (1) charges distributed on therest of the conductor (call this field E1), and (2) the field due tothe charge resting on the area in question, say E2 (see Fig. 57).From Example 78, El + E2 = 41rv. Now the field inside the

Ei+E2

P

Fia. 57.

conductor at the point P' situated symmetrically opposite P iszero from Example 75. The field at P' is E1 - E2 = 0. ThusEl = 2o per unit charge. For an area dS the force is

dE = (2av) (v dS) = 27rv2 dS (206)

This force is normal to the surface. A charged soap film thustends to expand.

Problems

1. Two hollow concentric spheres have equal and oppositecharges Q and -Q. Find the work done in taking a unit testcharge from the sphere of radius a to the sphere of radius b, b > a.The outer sphere is negatively charged.

2. Find the field due to any infinite uniformly charged cylinder.3. Solve Prob. 1 for two infinite concentric cylinders.


4. Let ql, q2, . . . , q be a set of collinear electric chargesresiding on the line L. Let C be a circle whose plane is normalto L and whose center lies on L. Show that the electric flux

n

through this circle is N = 121rga(1 - cos Sa), where Na is thea-I

angle between L and any line from qa to the circumference of C.5. Let the line L of Prob. 4 be the x axis, and rotate a line of

force r in the x-y plane about the x axis (see Fig. 58). If noY

xa x2 xlx

q3 0 qa qi

I1

1

I

III `

/

Fia. 58.

charges exist between the planes x = A, x = B, show that theequation of a line of force is

n

qa(x - xa)[(x - xa)z + y2]i = constanta-1

6. Point charges +q, -q are placed at the points A, B. Theline of force that leaves A making an angle a with AB meets theplane that bisects AB at right angles in P. Show that

sin 2 = sin (+ PA B)

59. Poisson's Formula. In Sec. 57, we saw that

E_ -va-1 r,

SEC. 591 STATIC AND DYNAMIC ELECTRICITY 133

For a continuous distribution of charge density p, we postulatethat the potential is

= fff pdr

(207)

where the integration exists over all of space. At any point Pwhere no charges exist, r > 0, and we need not worry about theconvergence of the integral. Now let us consider what happensat a point P where charges exist, that is, r = 0. Let us surroundthe point P by a small sphere R of radius e. The integralf f f (p dr/r) exists if p is continuous. We define 9 at PY-R

as lim f f f (p dr/r). This limit exists, for using sphericaly-R

coordinates,If f f "d*T

= Ifo2r fo= 0 pr sin a dr d9 d<p` < MT2E2

where M is the bound of p in the neighborhood of P. Thus

I i l f pdT

_ fJf RpI

< M? (e2 + el )xV

where e' is the radius of the sphere R' surrounding P. TheCauchy criterion holds, so that the limit exists. In much thesame way we can show that

E = f f f Ta dr (208)

and that at a point P where a charge exists

E(P) = lim f uJ - dr+0 tr

converges.Now from Gauss's law

ff f f f pdr8 v


In order to apply the divergence theorem to the surface integral,we must be sure that V E is continuous at points where p iscontinuous. We assume this to be true, and the reader is referredto Kellogg's "Foundations of Potential Theory" for the proofof this. Thus

JJJ(v.E)dr = 47r f f f p drV V

Since (209) is true for all volumes, it is easy to see that

(209)

V E = 4irp (210)

provided V E and p are continuous. Since E _ - Vs,, we havePoisson's equation

V2,p = -4ap (211)

and at places where no charges exist, p = 0, so that Laplace'sequation, V2,p = 0, holds.


V 20r L or

(rOr + e r t90/ + 8z \r 8z/ J

Consider an infinite cylinder of radius a and charge q per unitlength. At points where no charge exists, we have V2,p = 0.Moreover, from symmetry, p depends only on r. Thus

r dr = constant = A

vAlogr+BE_ -Vip= -A r, r=xi -f- yl

Also 42rc = (Er)r_a = -A/a, so that q = 2,raa = -A/2, and

E=2r (212)r2


Example 81. To prove that the potential is constant inside aconductor. From Green's formula we have

Jfpvcc.ddJfJ(s7co)2dr+JJJrpv2codr

Inside the conductor no charge exists so that V 2(p = 0. Moreover,for any surface inside the conductor, E = -Vgp = 0 so thatJff (vp)2 dr = 0 for all volumes V inside the conductor.

V

Therefore (VV)2 = 0, andapp = app = app

_ 0, so that p = con-ax ay az

stant inside the conductor.

Problems

1. Solve Laplace's equation in spherical coordinates assumingthe potential V = V(r).

2. Find the field due to a two-dimensional infinite slab, ofwidth 2a, uniformly charged. Here we have p = p(x) and mustsolve Laplace's equation and Poisson's equation separately forfree space and for the slab, and we must satisfy the boundarycondition for the potential at the edge of the slab. The spaceoccupied by the slab is given by -a S x < a, - co < y < oo.

3. Solve Laplace's equation for two concentric spheres of radiia, b, with b > a, with charges q, Q, and find the field.

4. Solve Laplace's equation and find the field due to an infiniteuniformly charged plane.

5. Prove that two-dimensional lines of force also satisfyLaplace's equation.

6. Show that rp = (A cos nx + B sin nx) (Cell + De-Av) satis-

fies x + a2 $ = 0.49Y2

7. If Cpl and S02 satisfy Laplace's equation, show that cpl + 4o

and Ipl -- rp2 satisfy Laplace's equation. Does cl92 satisfyLaplace's equation?

8. If (pi satisfies Laplace's equation and cp2 satisfies Poisson'sequation, show that cpi + 4p2 satisfies Poisson's equation.

60. Dielectrics. If charges reside in a medium other than avacuum, it is found that the inverse-square force needs readjust-ment. That this is reasonable can be seen from the following


considerations. We consider a parallel-plate condenser sepa-rated by glass (Fig. 59). Assuming that the molecular structure

of glass consists of positive.+++++++++++.....++++ and negative particles, the

0/0/ electrons being bound to theGlassnucleus, we see that the fielddue to the oppositely charged---

-----j/Fm. 59 plates might well cause a dis-

placement of the electronsaway from the negative plate and toward the positive plate.This tends to weaken the field, so that E = 47rv/x, where x > 1.x is called the dielectric constant.

It is found experimentally that E -_ (qq'/Kr9)r for charges in adielectric. Applying this force, we see that Gauss's law is modi-fied to read

4 QfjE d. d=K

(213)

and if x is a constant,

f f 41rQ (214)

where D is defined as the displacement vector, D = xE _ -x V.Poisson's equation becomes V D = -V (K Vg) = 4rp, and forconstantx

4rpx

For p = 0 we still have Laplace's equation V2(p = 0.In the most general case, we have

3

D; = Z x;;E;, i = 1, 2, 3i=i

(215)

where D = Dli + D2j + Dak, E = E,i + E2j + Eak, and x;; = K;,.61. Energy of the Electrostatic Field. . Let us bring charges

q,, q2, . . . , q,. from infinity to positions P1, P2, . . . , P,., andcalculate the work done in bringing about this distribution. Ittakes no work to bring q, to P,, since there is no field. To bringq2 to P2, work must be done against the field set up by Q1. Thisamount of work is glg2/rl2, where r12 is the distance between Pland P2. In bringing q3 to P3, we do work against the separate


fields due to q, and q2. This work is gig3/r13 and g2g3/r23. Wecontinue this process and obtain for the total work

n

Wq;q'i=- -ri,

(216)

The J occurs because gig2/r12 occurs twice in the summationprocess, once as glg2/r12 and again as g2g1/r21. The quantity W

n

is called the electrostatic energy of the field. Since (pi = q;/rii,;al

n

we have W = gipi. For a continuous distribution of charge,i=1

we replace the summation by an integral, so that

W= jfff pv dr (217)

Now assume that all the charges are contained in some finitesphere. We have V D = 41rp so that

W f f f f f f

8- f f f

Applying the divergence theorem,

W f f f f87 S v

Now p,is of the order of 1/r for large r, and D is of the order of1/r2, while do is of the order of r2. We may take our volumeof integration as large as we please, since p = 0 outside a fixedsphere. Hence lim f f cpD dd = 0, so that

s

w= 8A f f f (E D) dr (218)


The energy density is w = (1/8ir)E D. For an isotropicmedium, D= KE and W = (1/87r) Jff KE2 dr.

Example 82. Let us compute the energy if our space containsa charge q distributed uniformly over the surface of a sphere ofradius a. We have

D = E = qr,r>>=ar

The total energy is

and D=E=O,r<a

22x xW=s- ffJ-sin O drdOdcp

q2

2a

62. Discontinuities of D and E at the Boundary of Two Dielec-trics. Let S be the surface of discontinuity between two mediawith dielectric constant K1 and K2. We apply Gauss's law to apillbox with a face in each medium (Fig. 60). Assuming no

K1

K2

charges exist on the surface ofdiscontinuity, we have

so that D2 n2 = 0.hSi = - n2, wence nl aven2=_nl

DN, = DN, (219)FIG. 60.

We have taken the pillbox veryflat so that the sides contribute a negligible amount to the flux.Equation (219) states that the normal component of the displace-ment vector D is continuous across a surface of discontinuitycontaining no charges.

We next consider a closed curve r with sides parallel to thesurface of discontinuity and ends negligible in size (Fig. 61).

Since the field is conservative,

ft dr = 0 or Er1 = Er1 (220)


In other words, the tangential component of the electric vectorE is continuous across a surface of discontinuity. Combining(219) and (220), we have

DN, DN,

ET, E,,K1EN, K2EN,ET, ET,

or

Fio. 62.

for isotropic media. Hence

tan01_Kltan 02 K2

(221)

which is the law of refraction (see Fig. 62).63. Green's Reciprocity Theorem. Let us consider any dis-

tribution of volume and surface charges, the surfaces being con-ductors. Let p be the volume density and a the surface density.If p is the potential function for this distribution of charges, then02p = -4up. We shall make use of the fact that E = -Vp andthat at the surface E. = 4arv, or E . dd = -4uo dS.

A new distribution of charges would yield a new potentialfunction cp' such that V2sp' = -41ro'. Our problem is to find

140 VECTOR AND TENSOR ANALYSIS 1SEC. 63

a relationship between the fundamental quantities p, o, p of theold distribution and p', o', of the new distribution. To do so,we apply Green's formula

JJJ (vV2V - ip'V',p) JJ(cV,' - v v) ddV S

which reduces to

-4v ii (,Pp '-V 'p)dr=4a f f('-rp'o)dSV s

or

f if rpp' dr + ff ' ds = f f f v p dr + jJ p dS (222)

This is Green's reciprocity theorem. It states that the poten-tial (p of a given distribution when multiplied by the correspond-ing charge (p', a,) in the new distribution and then summed overall of the space is equal to the sum of the products of the poten-tials (pp') in the new distribution by the charges (p, o) in the olddistribution, that is, a reciprocal property prevails.

Example 83. Let a sphere of radius a be grounded, that is, itspotential is zero, and place a charge q at a point P, b units fromthe center of the sphere, b > a. The charge q will induce acharge Q on the sphere. We desire to find Q. We construct anew distribution as follows: Place a unit charge on the sphere, andassume no other charges in space. The potential due to thischarged sphere is p' = 1/r. For the sphere we have initially

= 0, Q = ?, and afterward, cp' = 1/a, q' = 1. For the point Pwe have initially pp = ?, q = q, and afterward, V = 1/b, q' = 0.Applying the reciprocity theorem, we have

a b0.1+vp-0Q'+qso that Q = - (alb)q. This is the total charge induced on thesphere when it is grounded. Note that this method does nottell us the surface distribution of the induced charge. .

Problems

1. A conducting sphere of radius a is embedded in the centerof a sphere of radius b and dielectric constant K. The conductor

SFc.64] STATIC AND DYNAMIC ELECTRICITY 141

is grounded, and a point charge q is placed at a distance r fromits center, r > b > a. Show that the charge induced on thesphere is Q = -Kabq{r[b + (K - 1)a])-1.

2. A pair of concentric conductors of radii a and b are con-nected by a wire. A point charge q is detached from the innerone and moved radially with uniform speed V to the outer one.Show that the rate of transfer of the induced charge (due to q)from the inner to the outer sphere is

dQ=

dt-gab(b - a)-'V(a + Vt)-2

3. A spherical condenser with inner radius a and outer radius bis filled with two spherical layers of dielectrics Kl and K2, theboundary between being given by r = J(a + b). If, when bothshells are earthed, a point charge on the dielectric boundaryinduces equal charges on the inner and outer shells, show thatK1/K2 = b/a.

4. A conductor has a charge e, and V1, V2 are the potentialsof two equipotential surfaces which completely surround it(V1 > V2). The space between these two surfaces is now filledwith a dielectric of inductive capacity K. Show that the changein the energy of the system is - e(V l - V2) (K - 1 )K 1.

5. The inner sphere of a spherical condenser (radii a, b) has aconstant charge E, and the outer conductor is at zero potential.Under the internal forces, the outer conductor contracts fromradius b to radius b1. Prove that the work done by the electricforces is E2(b - b1)b-'b1-1.

64. Method of Images. We consider a charge q placed at apoint P(b, 0, 0) and ask if it is possible to find a point Q(z, 0, 0)such that a certain charge q' at Q will cause the potential overthe sphere x2 + y2 + z2 = a2, a < b, to vanish. The answer is"Yes"! We proceed as follows : From Fig. 63 we have

82 = z2 + a2 - 2az cos B

t== b2+a2-2abcos6

We choose z so that zb = a2, and call Q(a2/b, 0, 0) the imagepoint of P(b, 0, 0) with respect to the sphere. Thus

a2 a2s2 = a (a2 + b2 - 2ab cos 0) =

b2

0


and

The potential at S due to charges q and q' at P and Q is

P= s,-I- qt

=-t Cbq'+qJ

and p = 0 if we choose q' _ - (alb)q.

FIG. 63.

isThe potential at any point R with spherical coordinates r, 0, (p

_ q (a/b)q(r2 + b2 - 2rb cos 0)# [r2 + (a4/b2) - (2a2/b)r cos 0]1

(223)

with = 0 on S and V24 = 0 where no charges exist.Now let us consider the sphere of Example 83. The function

of (223) satisfies Laplace's equation and is zero on the sphere.From the uniqueness theorem of Example 69, F of (223) is thepotential function for the problem of Example 83. The radialfield is given by

8 q(r-bcos9)Er

or (r2 + b2 - 2rb cos B)'(a/b)q[r - (a2/b) cos 0]

[r2 + (a4/b2) - (2a2/b)r cos B];


and the surface distribution is given by

_ (Er)rs _ q b2 - a247r 4ir a(a2 + b2 - 2ab cos 9)'

Problems

1. A charge q is placed at a distance a from an infinite groundedplane. Find the image point, the field, and the induced surfacedensity.

2. Two semiinfinite grounded planes intersect at right angles.A charge q is placed on the bisector of the planes. What distri-bution of charges is equivalent to this system? Find the fieldand the surface distribution induced on the planes.

3. An infinite plate with a hemispherical boss of radius a isat zero potential under the influence of a point charge q on theaxis of the boss at a distance f from the plate. Find the surfacedensity at any point of the plate, and show that the charge isattracted toward the plate with a force

q2 4g2a3 fa

4f2 + (f' - a')2

65. Conjugate Harmonic Functions. If we are dealing witha two-dimensional problem in electrostatics, we look for a solu-tion of Laplace's equation V2V = 0. The curves V (x, y) = con-stant represent the equipotential lines. We know that thesecurves are orthogonal to the lines of force, so that the conjugatefunction U(x, y) (see Sec. 56) will represent the lines of force.We know that V2V = V2U = 0.

We now give an example of the use of conjugate harmonicfunctions. In Example 73 we saw that

U(x, y) = A tan-' yx

V (X, y) = 2 log (x2 + y2)

(224)

are conjugate functions satisfying Laplace's equation. If wetake V (x, y) as the potential function, then the equipotentialsare the circles (A/2) log (x2 + y2) = C, or x2 + y2 = e2cie.


Hence the potential due to an infinite charged conducting cylinderis

V (X, y) = 2 log (x2 + y2) =

2

log r2 = A log r, r2 = x2 + y2

since A log r satisfies Laplace's equation and satisfies the bound-ary condition that V = constant for r = a, the radius of thecharged cylinder. If q is the charge per unit length, then

_ aV

q (Er)r..a ar r-a A4ara24ra 41r 41r

so that A = -2q and V = -2q log r.

1e=Jr) U=U1 0 U=U0 (0=0)Frs. 64.

If we choose U(x, y) = A tan-1 (y/x) as our potential func-tion, then the equipotentials U = constant are the straight linesA tan-1 (y/x) = C, or y = x tan (C/A). As a special case wemay take the straight lines 0 = 0, 0 = v as conducting planesraised to different potentials (see Fig. 64). The lines of forceare the circles (A/2) log (x2 + y2) = V.

The theory of conjugate functions belongs properly to thetheory of functions of a complex variable. With the aid of theSchwarz transformation it is possible to find the conjugate func-tions associated with more difficult problems involving the two-dimensional Laplace equation.

Problems

1. By considering Example 72, find the potential functionand lines of force for two semiinfinite planes intersecting at rightangles.

SEC. 66J STATIC AND DYNAMIC ELECTRICITY 145

2. What physical problems can be solved by the transformationx = a cosh U cos V, y = a sinh U sin V? Show that

V2U = V2V = O

66. Integration of Laplace's Equation. Let S be the surfaceof a region R for which V24p = 0. Let P be any point of R, andlet r be the distance from P to any point of the surface S. Wemake use of Green's formula

fJf (,p 02' - ¢ V2V) dr = f f (9 Vi' DAP) ddR

We choose ¢ = 1/r, and this produces a discontinuity insideR, namely, at P, where r = 0. In order to overcome this diffi-culty, we proceed as in Example 66. Surround P by a sphereZ of radius e. Using the fact that V2,P = V24, = 0 inside R'(R minus the Z sphere), we obtain

0 = f f (9 V V-r) - dd + f f (p V 1 - Vp) - dd (225)8 r r // B r r //

Now V(1/r) = -r/r', and on the sphere X,

r r8 r e$

and (1/r) VV dd is of the order eIV(pf, so that by letting e --j 0,(225) reduces to

v(p) = 4 ffir VV - sa V r) dd (226)

This remarkable formula states that the value of Sp at any pointP is determined by the value of (p and V(p on the surface S.

Problems

1. If P is any point outside the closed surface S, show thatf f V(1/r)] dd = 0, where V2cp = 0 inside S and

r is the distance from P to any point of S.


a21P

2

2. Let rp satisfy V2cp = + 2 = 0. Let r be the closedaX2 y°boundary of a simply connected region in the x-y plane. If Pis an interior point of t, show that

1

1(1) app a[log (1/r)]

,P(P) - 2x `log r an an`0where use is made of the fact that

ds

f f (u V 2V - v V 2U) dA (u an -van) daA `

n being the normal to the curve.3. Let (p be harmonic outside the closed surface S and assume

that ip --> 0 and rI V pl -> 0 as r -' oo. If P is a point outside 8,show that

P(P) = 4x If (1r Vp - ip v T dds

where the normal dd is inward on S.4. Let so be harmonic and regular inside sphere 7. Show that

the value of p at the center of is the average of its values overthe surface of the sphere. Use (226).

67. Solution of Laplace's Equation in Spherical Coordinates.From Sec. 23, Prob. 1,

+ aB (sin o

+ 49(p (sin B0 (227)

To solve (227), we assume a solution of the form

V(r, 0, gyp) = R(r)6(0),P(,p) (228)

Substituting (228) into (227) and dividing by V, we obtain

sin 0 d( dR) 1 ( d8) 1 d2r2 - J +-A sin O - + -- = 0R dr dr 0 d9 d8 4) sin 0 d(p2

SEc. 67j STATIC AND DYNAMIC ELECTRICITY 147

Consequently

1 d dR\ 1 d d9 1 d4_ _ -2

r sin 0 - (229)R dr dr 0 sin 0 dB d9 sin 2 0 4, dp2

The left-hand side of (229) depends only on r, while the right-hand side of (229) depends on 0 and gyp. This is possible onlyif both quantities are constant, for on differentiating (229) with

respect to r, we obtaind

L Rd (r2i J

= 0. We choose as

the constant of integration c = -n(n + 1), so that

R dr (r2 dR) n(n + 1)

or

r2 4+24+n(n+1)R=0 (230)

It is easy to integrate (230), and we leave it to the reader to showthat R = Ar" + Br-n-1 is the most general solution of (230).Returning to (229), we have

41)

'e1d

= n(n -{- 1) sing 0 --sin

d0(sin 0 de) (231)

Since we have again separated the variables, both sides of(231) are constant. We choose the constant to be negative,-m2, m being an integer. This choice guarantees that the solu-tion of

d24 -I-mq =0d(p2 (232)

is single-valued when p is increased by tar. The solution of(232) is 4) = A cos mtp + B sin mcp.

Finally, we obtain that 0(0) satisfies

sin 0 d0 (sino)+[n(n+1)sin2o_m2Je=o (233)

We make a change of variable by letting µ = cos 0,

dµ= -sin0d0



(1 - µ2)dµ

[(1 - µ2) + [(1 µ2)n(n + 1) - m2]0

= 0 (234)

If we assume that V is independent of p (symmetry about thez axis), we have m = 0, so that (234) becomes

dIA(235)d

This is Legendre's differential equation.By the method of series solution, it can be shown that

0 = P"(14)-

satisfies (235); the P (µ) are called Legendre polynomials.Two important properties of Legendre polynomials are

following:

1 d"(µ2 - 1)nn! dµn2

f 11 P,n(;&)Pn(p) dµ = 0

J 11 P.'(µ) dµ = 2n + 1

the

if m : n (236)

(237)

We give a proof of (236). P. and P. satisfy

dµ[(1 - p!) dµn1

+ n(n + 1)Pn = 0 (238)

(239)

Multiplying (238) by P. and (239) by P. and subtracting, weobtain

P. d- [(1 - 2) a ,, - Pn d [(1 _ u2) d µ"'

+[n(n+1) -m(m+l)PPm=0

SFC. 671 STATIC AND DYNAMIC ELECTRICITY 149

or

µ[(1-M2)(Pmddn-PndPm)]

+ [n(n + 1) - m(m + 1)]P,aPm = 0 (240;

Integrating between the limits -1 and + 1, we obtain

[n(n + 1) - m(m + 1)J f i P,nPn dµ = 0

and ifm -d n,

1 PmPn dµ = 0

A particular solution of (227) which is independent of gyp, that

is, aV = 0, is given by V(r, 0) = (Anrn + Bnr-' 1)Pn(cos 0).aip

Now it is easy to show that any sum of solutions of (227) isalso a solution, since (227) is linear in V. Consequently a moregeneral solution is

V = (Anrn + Bn7rn-1)Pn(cos 0)n-0

(241)

provided that the series converges.If we wish to solve a problem involving V2V = 0 with spherical

boundaries, we try (241) as our solution. If we can find theconstants An, B. so that the boundary conditions are fulfilled,then (241) will represent the only solution, from our previousuniqueness theorems involving Laplace's equation.

We list a few Legendre polynomials:

Po(µ) = 1Pi(µ) = AP2(p) = . (3µs - 1)P,(µ) = 4(5Aa - 3µ)P,(µ) = 9(35."4 - 30u$ + 3) (242)

P.(0) = 0, n oddP-(0) = (_1)n/x 1-3-5 ... (n - 1)

2.4.6 ... n unevenPn(1) = 1

Pn(-µ) _ (-1)"Pn(µ)


Problems

1. Prove (237).2. Solve V2V = 0 for rectangular coordinates by the method

of Sec. 67, assuming V = X(x)Y(y)Z(z).3. Investigate the solution of VI V = 0 in cylindrical coordi-

nates.68. ApplicationsExample 84. A dielectric sphere of radius a is placed in a uni-

form field Eo = Eok. We calculate the field inside the sphere.The potential due to the uniform field is p = -Eoz = -Eor cos 0.There will be an additional potential due to the presence of thedielectric sphere. Assume it to be of the form ArPI = Ar cos 0inside the sphere and Br-2PI ° Br-2 cos 0 outside the sphere.We cannot have a term of the type Cr-2 cos 0 inside the sphere,for at the origin we would have an infinite field caused by thepresence of the dielectric. Similarly, if a term of the typeDr cos a occurred outside the sphere, we would have an infinitefield at infinity due to the presence of the sphere. If we let VIbe the potential inside and V11 the potential outside the sphere, wehave

VI = --Eor cos 0 + Ar cos 0

Vu = --Eor cos B + B cos 0r2

(243)

Notice that VI and Vn are special cases of (241). We have twounknown constants, A, B, and two boundary conditions,

VI= Vu at r=aaDN, = D.Y. or KI = a II at r = a (244)

(see Sec. 62). From (243) and (244) we obtain

A=K-'Eo, B_aaK-1Eox+2 K+ 2

so that

VI = - - 2 Eor cos 0 = -K

+

2 Eoz (245)I 'V+


We see that the field inside the dielectric sphere is

E_ -vVI=K+2Eo

and E is uniform of intensity less than E0 since K > 1. Outsidethe sphere

VII = -Eor cos B +K - 1 _ Eo cos 8 (246)K + 2 r2

The radial field outside the sphere is given by

E, aVII=Eocos0+2K + 2aa ocos9

For a given r the maximum E, is found at 0 = 0.Example 85. A conduct-

ing sphere of radius a andcharge Q is surrounded by aspherical dielectric layer upto r = b (Fig. 65). Let uscalculate the potential dis-tribution. From sphericalsymmetry V = V (r), so thatwe try

FlG. 65.

The boundary conditions are

(i) VI= VIIatr=baVI _ aVII at r = b

(11) Or =K

Or

(iii) Q =41r f f D dd = - J2rJT (a

T a2 sin 0 d9 dps 47 J

From (i) Alb = (B/b) + C; from (ii) -A/b2 = -KB/b2; from(iii) Q = (K/4x) (B/a2) f f dS = KB. Hence


VI = Q' VII =Q +90C 1

K(247)

Example 86. A conducting sphere of radius a and charge Q isplaced in a uniform field. We calculate the potential and thedistribution of charge on the sphere. We assume a solution

V = -Eor cos 0 +Br

The boundary condition is

Q = 4v Jf - avl dSr-a

so that

and

1 ,= 4

f2vf- (E cos 8 + a a2 sin 8 d8 dsp = BQ

V= -Eorcos8+QrFor the charge distribution

aVcIra-4"

4I(E0cos8+Q

Example 87. Consider a charge q placed at A (b, 0, 0). Letus compute the potential at any point P(r, 8, gyp) (see Fig. 66).The potential at P is

V = 4 = q(r2 + b$ - 2rb cos 8)_;

There are two cases to consider:(a) r < b. Let µ = cos 8, x = r/b, so that

V =

b

(1 - 2µx +2)-}

Now (1 - 2µx + x=)-} can be expanded in a Maclaurin series inpowers of x, yielding


m

n =_ m

V = b I b P.(--)()fl

(248)n0 n0

The proof is omitted here that the Pn(p) are actually the Legendrepolynomials. However, we might expect this, since V satisfiesLaplace's equation and P,,(µ)rn is a solution of VI V = 0.

z

Fia. 66.

(b) r > b. In this case

/b nV = q I Pn6u)

rn=0 r(249)

Notice that each term is of the form Pn(µ)r-n-1, which satisfiesLaplace's equation.

Example 88. A point charge +q is placed at a distance bfrom the center of two concentric, earthed, conducting spheresof radii a and c, a < b < c. We find the potential at a point Pfor a <r <b.

For r > b, V = (q/r) (b/r)'P,,(cos 0) due to the charge q;0

and for r < b, V = (q/b) 2 (r/b)'Pn(cos 0). Moreover, we have0


an induced potential of the form

V = q I (Anrn + Bnr n-1)Pn(cos 0) (250)0

which is due to the spheres, the An and Bn undetermined as yet.Hence

For r > b:

Vl = q I [Anrn + (Bn + bn)r-n-1]Pn(cos 6)0

For r < b:

V2 = q [(A,, + b-n-I)rn + Bnr-n-1]Pn(cos 00

The boundary conditions are

(i) V1=0atr=c(ii) V2 = O at r = a

These yield the equations

(i)(11)

so that

Ancn + (Bu + bn)c-n-1 = 0(An + b-n-1)an + Bna n-1 = 0

a2n+1 (C2n+l - b2n+l)'Bn T bn+l (a2n+l - C2n+1)'

(251)

(252)

b-n-1(C2n+1 - b2n+1)A. + b-(n+1) _

a2n+1 - C2n+l

Hencem b2n+1 - C2n+1 a2n+1

V2(P) = q j bn+l(a2n+l - C2n+1)(rn

- rn+1 Pn(cos 0) (253)-0n

Problems

1. Show that the force acting on the sphere of Example 86 isF = kQEok.

2. A charge q is placed at a distance c from the center of aspherical hollow of radius a in an infinite dielectric of constant K.Show that the force acting on the charge is


(K - 1)g2'' n(n + 1) 2n+1

c2 ,410n+K(n+ 1)(a/

3. A point charge q is placed a distance c from the center of anearthed conducting sphere of radius a, on which a dielectriclayer of outer radius b and constant K exists. Show that thepotential of this layer is

m

Vq (2n + 1 )b2n+1(rn - a2n+lr-n-1)

= -c n c0

cn { [(K +1)n + 1]b 2n+1 + (n +1)(K -1)a 2n+1 } Pn(cos e)

4. Show that the potential inside a dielectric shell of internaland external radii a and b, placed in a uniform field of strength E,is

= 9KEV

9K - 2(1 - K2)[(b/a)3 - 1]r cos 0

5. The walls of an earthed rectangular conducting tube ofinfinite length are given by x = 0, x = a, y = 0, y = b. A pointcharge is placed at x = xo, y = yo, z = zo inside the tube. Showthat the potential is given by

toV = 8q I (m2a2 +n2b2)-e-a-'b-'(mta'+n$t)i*(s-ss)

sinn1rx0

nil m-1 anrx mTryo miry

sina

sinb

sinb

69. Integration of Poisson's Equation. Instead of assumingthat p is harmonic, let us consider that 1p satisfies V2co = -47rp.By applying Green's formula as in Sec. 66, we immediately obtain

,P(P) = Jff rdr+Jf (rVp - jP 0 (254)

If we make the further assumption that rp is of the order of 1/rfor large r and that jVjpj - 1/r2, we see that by pushing S out toinfinity the surface integral will tend to zero. Our assumption isvalid, for if we assume the charge distribution to be bounded bysome sphere, then at large distances the potential will be of theorder of I /r, since we may consider all the charges as essentially


concentrated at a point. Thus

V(P) = JJf e dr (255)

70. Decomposition of a Vector into the Sum of Solenoidal andIrrotational Vectors. In Example 70, we saw that if Ifl tends tozero like 1/r2 as r --> oo, then f as uniquely determined by itscurl and divergence.

We now proceed to write f as the sum of irrotational andsolenoidal vectors. Let

W(P) = JfJ f dr(256)

where r is the distance from P to the element of integration dr.If we write f = fli + f j + f k, W = WA + W,,j + Wek, then

f ffrldr

W2 = f f f f 2dr (257)

M

W3=fffr$dr

00

We assume that the components of f are such that the integralsof (257) converge and that I Wl - 1/r, I V W,,l '' 1/r2, n = 1, 2, 3.From Sec. 69, V2W. = -4xfn, so that

V2W = -4,rf (258)

From (256)

VxW= JJJf xVdr40

Now

(259)

V x(V xW) = --V2W

SEc. 711 STATIC AND DYNAMIC ELECTRICITY 157

so that

and hence

f =4-V X X - 4

f = V x A + Vcp 260)where

A=4-VxW, =

Problems

1. Show that (256) is a special case of (254).2. Find an expression for (p(P) if V2(p = -41rp inside S and if

P is on the surface S.3. f = yzi + xzj + (xy - xz)k. Express f as the sum of an

irrotational and a solenoidal vector.71. Dipoles. Let us consider two neighboring charges -q and

+q situated at P(x, y, z) and Q(x + dx, y, z). The potential atthe origin 0(0, 0, 0) due to -q is -q/r, and that due to +q isq/(r + dr), where r = (x2 + y2 + Z2)I and

r+dr= [(x + dx)2 + y2 + Z2]1

The potential at 0(0, 0, 0) due to both charges is

q q g2drr + dr r r2

Now dr = x dx/r, so that rp - qx dx/r3. If we now let q -p coand dx -+ 0 in such a way that q dx remains finite, we have formedwhat is known as a dipole. Let r be the position vector from theorigin to the dipole, and let M = q dr, where dr is the vector fromthe negative charge to the positive charge; !drl = dx. We have(M r)/r8 = (qr dr)/ra = (qx dx)/r3, so that

M is called the strength or moment of the dipole. For morethan one dipole, the potential at a point P is given by


(Mi.ri)i= i ri3

where r; is the vector from P to the dipole having strength M.Example 89. The field strength due to a dipole is E _ --Vv

so that

E = Vs

r - ra (262)rs ) - r

Example 90. Potential energy of a dipole in a field of potential V.Let (p, be the potential at the charge q and c2 the potential atthe charge -q. The energy of the dipole is

W = spiq + IP2(-q) = q(spl - IP2) = g L ds = Mapp

as ao

where ds is the distance between the charges. Now

d,p = ds V p

so that W = M as VV = M V(p.

72. Electric Polarization. Let us consider a volume filledwith dipoles. The potential due to any single dipole is given by(261). If we let P be the dipole moment per unit volume, that is,P = lim (AM/or), then the total potential due to the dipoles is

A"o

= I f f r ra- dr (263)

Now V V. (P/r) = (1/r)V P - [(P r)/r3]. The reason that wehave taken V(1/r) t= r/r3 instead of -r/r3 is that

r = [( - x)2 + (n - y)2 + (J - z)2]'

and V performs the differentiations with respect to x, y, z, thecoordinates of the point P at which V is being evaluated. Thecoordinates t, ot, t belong to the region R and are the variables ofintegration, and r = (E - x)i + (n - y)j + (r - z)k. Hence(263) becomes

`P=fffv.(!)dr fffR R


and

(p - ff!.dd_ fff_!d r (264)

by applying the divergence theorem.Example 91. Let us find the electric intensity at the center

of a uniformly polarized sphere. Here P = Pok, so that V P = 0inside R. Hence (264) becomes

sp(x, y, z) = f f Pok dd

S x)2 + (71 - y)2 + ( - z) 2]iand

E= fj Po(k dd)[(E - x)i + (n - y)j + (1' - z)k]

S [(E-x)2+(7] -y)2+(J -z)2]$ni + 'k)Po(k dd)

E(0,0,0)ff (265)

Now for points on the sphere,

52 +' 72+-2 = a2,

and letting t = a sin B cos -o, n = a sin 0 sin tp, a cos 0, itis easily seen that (265) reduces to

E(0, 0, 0) = -frPok (266)

E is independent of the radius of the sphere. By superimposing(concentrically) a sphere with an equal but negative polariza-tion, we see that the field at the center of a uniformly polarizedshell is zero.

Problems1. Prove (262).2. Prove (266).3. If M1 and M2 are the vector moments of two dipoles at A

and B, and if r is the vector from A to B, show that the energyof the system is W = M1 Mgr-a - 3(M1 r)(M2 r)r-6.

4. The dipole-moment density is given by P = r over a sphereof radius a. Calculate the field at the center of the sphere.


73. Magnetostatics. The same laws that have held for elec-trostatics are true for magnetostatics with the exception thatOZcp.. = 0 always, since we cannot isolate a magnetic charge.We make the following correspondences, since all the laws ofelectrostatics were derived on the assumption of the inverse-square force law, which applies equally well for stationarymagnets.

Electrostatics MagnetostaticsEHq qm

D ---) B (magnetic induction)K <---- u (permeability) (267)

D = KE F----- B = µH

0

74. Solid Angle. Let r be the position vector from a point PN to a surface of area dS and unit normal N, that

is, do = N dS. We define the solid angle sub-tended at P by the surface dS to be (see Fig.67)

dct =r3

PFia. 67. The total solid angle of a surface is

J'r.do (268)S r3

Example 92. Let S be a sphere and P the origin so that

12(P)=fJrr4rdS=4u

Example 93. The magnetic dipole is the exact analogue of theelectric dipole. We consider a magnetic shell, that is, a thinsheet magnetized uniformly in a direction normal to its surface(Fig. 68). Let $ be the magnetic moment per unit area and


assume S = constant. The potential at P is given by

sff r,MdS=13 f f rras=On

Now let P and Q be opposite points onthe negative and positive sides of thesurface S. We have

,p(Q) = -P(4ir - 11)

so that the work done in taking a unitpositive pole from a point P on the nega-tive side of the shell to a point Q on thepositive side of the shell is given by

PFur. 68.

W = f," H H. dr = - f,' Vp . dr = *(P) - F(Q)=6U+$(4,r- 9)

W = 4x-8 (269)

75. Moving Charges, or Currents. If two conductors at differ-ent potentials are joined together by a metal wire, it is found thatcertain phenomena occur (heating of the wire, magnetic field), sothat one is led to believe that a flow of charge is taking place.Let v be the velocity of the charge and p the density of charge.We define current density by j = pv. The total charge passingthrough a surface per unit time is given by

ffpv.do_ Jfj.dd

Now the total charge inside a closed surface S is Q = f f f p dr.R

If there are no sources or sinks inside S, then the loss of charge

per unit time is given by - aQ = - fRf

atdr. Thus

ffpv.dd=- dr


Applying the divergence theorem, we have

or

0

= 0at

(270)

Equations (270) are the statement of conservation of electriccharge. We define a steady state as one for which p is independ-

ent of the time, a = 0, which implies V j = 0.

It has been found by experiment that if E is the electric field,then

j = XE = -A VM (271)

where X is the conductivity of the metal. This is Ohm's law.a

For the general case, j, = I Xa,gE#, and the simplest case,69=1

X = constant, so that V2(p = 0 for the steady state.We now compute the work done on a charge q as it moves

from a point of potential p, to one of potential 02, Ip2 > 92. Theenergy at (pi is qpl and at V2 is q(p2. The loss in energy is

W = (1v1 - Iv2)q

This loss in electrical energy does not go into mechanical energy,since the flow is assumed steady. Hence the electrical energyis converted into heat, Q = (,p1 - p2)q. The power loss is

P = dt = (1P1 - (P2) dq, and since 01 - 02 = RJ (another form

of Ohm's law, where R is resistance and J current), we have

P=RJ2 (272)

76. Magnetic Effect of Currents (Oersted). Experimentsshow that electric currents produce magnetic fields. Themathematical expression for the magnetic field is given by

dH = Jr x dr(273)

cra


where r is the vector from the point P; P is the point at which wecalculate the magnetic field dH due to the line current J in thatportion of the wire dr, and cis a constant, the ratio of the Q (tn'r)electrostatic to the electro-magnetic unit of charge (seeFig. 69). Biot and Savartestablished this law forstraight-line currents.

For a closed path

H= Jrxdr (274)cra

P (x,y, z)Fra. 69.

Now r = [(E - x)2 + (n - y)2 + ( - z)2];, and V(1/r) = r/ra,

where V = i + ja

+ k az. Henceclx y

H=1fiJVI x(Jdrc r c r

since V does not operate on dr and J is a constant. Thus

H = V x A, where A = (1/c) J dr/r = (11c) f f f i dr/r is

integrated over all space containing currents.

Now V A = fif V (j/r) dr f f (j/r) dd, so that if alls

currents lie within a given sphere, we may push the boundaryof R to infinity, since nothing new will be added to the integralyielding A. But when S is expanded to a great distance, j = 0on S, so that that V A = 0. Also

V x H = -VIA. Now since A = (1/c) f f f j dr/r, or0

A. = (1 /C) f f f is dr/r, A. = (1 /c) f f f j dr/r,

As fffis drr


we have from Sec. 69 that V2A = -- (47r/c)j. Thus

VxH=47jc

(275)

Example 94. The work done in taking a unit magnetic polearound a closed path r in a magnetic field due to electric currentsis

f f f fS c s

For an electric current J in a wire that loops r, we have

(276)

Example 95. The magnetic field at a point P, r units awayfrom an infinite straight-line wire carrying a current J, is obtainedby use of Example 94.

H dr = H(2irr) = 4a J so that H = 27c cr

Example 96. We compute the dimensions of c//. Nowf. = qaq,'/Kr2, and f. = qqm'/µr2 so that

[M][L] [g612 [q+x]2

[712 [K][L]2 [µ][L]2and

[J] [dq,/dt] [g] [M14[L][K]}

F (276)

[c] [c] [c][T] [c][T]2

rom ,

Work d 4'rJHso that

, r ==Unit pole c

[M][L]2 _ [J][gm][T]2 [c]

SEc. 77] STATIC AND DYNAMIC ELECTRICITY 165

and

yielding

[Ml}[L]; [M]}[Ll'[K]i

[7'][µl} [c][TJ2

L / i [TL]

We see that c// has the dimensions of speed. We shall soonsee the significance of this.

(1) (2)

Fca. 70.

77. Mutual Induction and Action of Two Circuits. Considertwo closed circuits with currents J, and J2 (Fig. 70). Themagnetic field at 0 due to J, is H, = V x A, where

A,=J, r drc (1)r

We define the mutual inductance of the two circuits as the mag-netic flux through the surface B due to a unit current in (1).This is

IM= f f L da= f fa a

1(2) A1. dr = c1(2) (1(1) drl) dr2

Hence

M = 1 dr, dr,C f(2)J(1) r (277)

The current element J2 dr2 seta up a magnetic field, so thatfrom Newton's third law of action and reaction, any magnetic


field will act on J2 dr2 with an equal and opposite force. Thus

2 drz x H, =J,cJ2

dr, x V r x dr2 (278)df = J (L)and integrating over (2) we obtain

f-J,J2f drzxf OZ xdr,C (2) (1) r

Now

dr2 x (TI 1 x dr,) = V 1 (dr, dr2) - (dr2 "V 1) dr,r r r

and J 2)[dr2 - v(1/r)] dr, =

I2)d(1/r) dr, = 0, so that

f = Jc2 J (2)1(1) (C'

1) (dr, dr2) (279)

This is the force of loop (1) on loop (2). It is equal andopposite to the force of loop (2) on loop (1), this being immedi-ately deducible from (279) when we keep in mind that

1 1v-=-V-r2, r12

In (279), r = r2,.Example 97. We find the force per unit length between two

long straight parallel wires carrying currents J, and J2. We use(278) and the result of Example 95. We have H, = (2J,/cd)iat right angles to the plane containing the wires. Hence

.df=Jzdr2x 2J,di 2J,J2= dr2xi

and the force per unit length is F = 2J,J2/cd. If the currentsare parallel, F is an attractive force; if the currents are opposite,F is a repulsive force.

Problems

1. From (278) show that f = J2 f f dd2 x (V X H,).a

2. Find the force between an infinite straight-line wire carryinga current J, and a square loop of side a with current J2, the


extended plane of the loop containing the straight-line wire, andthe shortest distance from the wire to the loop being d.

3. A current J flows around a circle of radius a, and a currentJ' flows in a very long straight wire in the same plane. Showthat the mutual attraction is 4irJJ'/c(sec a - 1), where 2a isthe angle subtended by the circle at the nearest point of thestraight wire.

4. Show that A = (J/c) f f dd x V (1 /r) for a current J in aS

closed loop bounding the area S. For a small circular loop, showthat A = (M x r/r$), where r is very much larger than the radiusof the loop and is the vector to the center of the circle, and where

M=c f f dd

78. Law of Induction (Faraday). It has been found by experi-ment that a changing magnetic field produces an electromotiveforce in a circuit. If B is the magnetic inductance, the fluxthrough a surface S with boundary curve r is given by f f B dd.

S

The law of induction states that

-ca f f

Applying Stokes's theorem, we have

_ - =VxEC at

(280)

The time rate of change of magnetic inductance is proportionalto the curl of the electric field. Equation (280) is a generaliza-tion of V x E = 0, which is true for the electrostatic case in which

B = 0 and for the steady state for which atB = 0.

79. Maxwell's Equations. Up to the present we have, for anelectrostatic field, V x E = 0, V D = 4,rp and, for stationarycurrents, V x H = (41r/c)j, V- B = 0.


Now V x E_- 1 aB is a generalization of V x E = 0.C at

Maxwell looked for a generalization of V x H = (41r/c)j. Hedecided to retain the two laws: (1) V D = 4rp as the definition

= 0 as the law of conservation ofof charge, and (2) V j +at

charge.Let us assume

VxH=4w(j+x)C

(281)

as a generalization of V x H = (47r/c) j. We take the divergenceof (281) and obtain

(282)

so that

V = - V j = at = Oar at (V . D)aD0

We can choose Z = so that

VxH= w-+--aD)

C (i

We call -aD

the displacement current.t

We rewrite Maxwell's equations

V D = 4irp

V X4c \1 + 41r aD1t

(283)

(284)


We have in addition the equation

(v) f=p(E+1vxBJ (285)\\ C

where f is the force on a charge p with velocity v moving in anelectric field E and magnetic inductance B. This result followsfrom Sec. 77.

Problems

1. Show that the equations of motion of a particle of mass mand charge e moving between the plates of a parallel-plate con-denser producing a constant field E and subjected to a constantmagnetic field H parallel to the plates are

-= Be - dymd d

ML = Hedx

dtz dt

Given that d dt = x = y = 0 when t = 0, show that

z = (E/wH) (1 - cos wt), y = (E/(X) (wt - sin wt), where

Hew = -m

2. From (iii) of (284) show that V - aB = 0.

3. From (i) and (iv) of (284) show that

D) -at

4. Write down Maxwell's equations for a vacuum wherej=p=O,D=E,B=H.

80. Solution of Maxwell's Equations for "Electrically" FreeSpace. We have p = j = 0 and c, u are constants. Equations(284) become


(i)

(ii)V E=0VH=0vxE= aH

K aEvxHcat(286)

We take the curl of (iii) and obtain

V x (V x E) = V(V E) - V2E _

or

V2E =µK a2E

C2 at2

by making use of (i) and (iv). Similarly

V2H - JLK a2H

C2 8t2

(287)

(287a)

Equation (287) represents a three-dimensional vector wave equa-tion. To illustrate, consider a wave traveling down the x axiswith velocity V and possessing the wave profile y = f(x) att = 0. At any time t it is easy to see that y = f(x - Vt). From

y = f(x - Vt) we have a2-y = f"(x - Vt) and

at= = V2f"(x - V0

so thata2y 1 a2yax2 V2 at2

(288)

Equation (287) represents three such equations, and µK/c2 playsthe same role as 1/V2, so that c/V has the dimensions of avelocity (see Example 96). c = 3 X 1010 by experiment.

z

Example 98. We solve the wave equation V2f = Y2 at2 in

spherical coordinates where f = f(r, t).

SEC. 80J STATIC AND DYNAMIC ELECTRICITY

Vf = f'(r, t) Vr = f' r

V2f =V.(rr/

= r V - r+V( )'r3f+rf" f f,

Our wave equation is19 2f 2 of 1 a2f

are+rar V28t2

Now let u(r, t) = rf(r, t); then

af_1au u 02fi0 2u2au 2

Or r Or r2 are r are r2 Or + r2

and substituting into (289), we obtain

82u a2u

are V2 at2

of which the most general solution is

u(r, t) = g(r - Vt) + h(r + Vt)and so

171

(289)

.f(r, t) = r [g(r - Vt) + h(r + Vt)] (290)

is the most general solution of (289).

Let us now try to determine a solution of Maxwell's equationsfor the case E = E(x, t), H = H(x, t).

aE=(x, t) aEy(x, t) 8Eg(x, t)Now V E = 0, so that

ax + ay + az0,

8E.(x, t)which implies ax = 0. We are not interested in a uniform

field in the x direction, so we choose E. = 0. Hence

E = EE(x, t)j + Es(x, t)kand similarly

H = Hy(x, t) j + H.(x, t)k


Now we use Eq. (iii) of (286), V x E _ - c axt , so that

i j ka a a

ax ay az

0 Ey E.

or

(i)

AaH, i LaH,k

c at c at

OE. ,.8H4ax c at

aE, aH,ax at

Similarly, on using (iv) of (286), we obtain

(i)aH, K aE8x c at

aH, K aE,

ax c at

(291)

(292)

The four unknowns are E., E,, H,,, H,, which must satisfy (291)and (292). If we choose H = E, = 0, we see that (i) of (291)and (ii) of (292) are satisfied. Differentiating (ii) of (291) withrespect to x and (i) of (292) with respect to t, we obtain

a2Ey UK 61%

axe C2 at2

We leave it to the reader to show that

82H. Ax a2H,

ax2 c2 at=

(293)

(294)

'These equations are of the type represented by (288). Hencea solution to Maxwell's equations is

E = [E,,(') (x - Vt) + Eyes) (x + Vt)JjH = [H,(')(x - Vt) + H,{2>(x + Vt)Jk (295)

where V = c(AK)"}.Both waves are transverse waves, that is, they travel down

the x axis but have components perpendicular to the x axis.


Also note that E H = 0, so that E and H are always at rightangles to each other.

By letting H. = Er, = 0, we can obtain another solution,E = E.(x, t)k, H = H (x, t)j. These two solutions are calledthe two states of polarization, the electric vector being alwaysoriented 90° with the magnetic vector.

Example 99. We compute the energy density.

We _ _ _2 2 2

B H µH2 µH.2w,n= = 2 2 = 2 =w.2

and w = w, + 2w, = 2 ,. = KE 2, and for both wavesw, = K(E,,2 + E.2). We have here used the fact that

(see Prob. 1).Example 100. Maxwell's equations in a homogeneous con-

ducting medium are

V.E=4pK

V.H = 0

7 E OH

c at

VxH=4c1uE+4v a/Assume a periodic solution of the form

E = Eo(x, y, z)e-;'e

H = Ho(x, y, z)e-'"e

Substituting into (iv), we obtain

e--;tee V x Ho = 4 t1oe.-;'eE0 - iKW

or

4a/VxHo

2Kw=-lo-- EoC 4r

(296)


This equation is the same as that which occurs for "electrically"free space with a complex dielectric coefficient.

Problems

1. By letting E = f(x - Vt), Hz = F(x -- Vt), V = c/,show that H. = VKIIA E.

2. Derive (287a).a2y 1 a2y3. Letr=x- Vt, s =x+ Vt, and show that-aX2 =

V2 at22

reduces to -ay = 0. Integrate this equation and show that theOr as

general solution of (288) is y = f(x - Vt) + F(x + Vt), wheref and F are arbitrary functions.

4. Prove that Maxwell's equations for insulators (a = 0) are

v x H=C

KaEat

and V x E_- c a (297)

5. Show that the solution of (297) can be expressed in termsof a single vector V, the Hertzian vector, where

2aV H=-vxat

and V satisfies V2V = c .at2

2W6. Prove thatE=-VxeH= -V(V.W)+Kµa isa

c at C2 at2

solution of (297), provided that W satisfies V2W = e192W

at2

7. Derive (294).8. Look up a proof of the laws of reflection and refraction.9. By considering (i) and (iv) of Example 100, show that

P = Poe-4x'ì`

81. Poynting's Theorem. Our starting point is Maxwell'sequations. Dot Eq. (iii) of (284) with H and Eq. (iv) with Eand subtract, obtaining

aBIDc(H-V xE - xH) = -H (298)

at at

SEc. 821 STATIC AND DYNAMIC ELECTRICITY 175

Now from (218)aw, 1 E aD aw,, _ 1 H aBat 4ir at at 4r at

Let us write j = jo + jc where jo represents the galvanic currentand j. = pv, the conduction current. Now

dr awmec,,,,a;c,, awM

at at

and from Sec. 75 it is easy to prove that E ja is Joule's power

loss = aQ Moreover, H V x E - E V x H = V (E x H), so

that we rewrite (298) asaw, aw, awm aQl

c V (E X H) = -41rat + at + at + at (299)

Integrating over a volume R and applying the divergencetheorem, we obtain

f if aQ dr + 4r f f E x H- dd fJfat

dr (300)

where w is the total energy density.We define s = (c/4x)E x H as Poynting's vector. Equation

(300) states that to determine the time rate of energy loss in agiven volume V, we may find the flux through the boundarysurface of the vector s = (c/4T)E x H and add to this the rateof generation of heat within the volume. It is natural tointerpret Poynting's vector as the density of energy flow.

Problems

1. Find the value of E and H on the surface of an infinitecylindrical wire carrying a current. Show that Poynting's vectorrepresents a flow of energy into the wire, and show that this flowis just enough to supply the energy which appears as heat.

2. Find the Poynting vector around a uniformly chargedsphere placed in a uniform magnetic field.

3. If E of Sec. 80 is sinusoidal, E = Eo sin co(x - Vt)k, findthe energy density after finding the magnetic wave H.

82. Lorentz's Electron Theory. For charges moving withvelocity v, j = pv, and Maxwell's equations become


(i)(ii)

(iii)

(iv)

4irp

V x H = 4c(pv + D)

(301)

These equations are due to Lorentz. From (ii) we can writeB = V x (Ao + Vx) = V x A. Substitute this value of B into

(iii) and obtain V x E C v x Af or its equivalent

vx(E+catA/ 0

Thus E +I A is irrotational, so that E + 1

A -vgp.

Let D= KE, B = pH, and substitute into (iv). We have

rIV x (V xA) = - Ipv+

z

4a( c

at2A-vaC1 SO

tµ

and since V x (V x A) = --V2A + V(V A), we obtain

z

v2Ac2 at2

cl-p v + V j, (302)

where

vA+c2at

Now

so that

K(--AKa2S0/= -K V2(p -

C at c at2

_ Kp, 12,p 4rp _ 14vg c2 812 K cat (303)


Equations (302) and (303) would be very much simplified if

we could makeKA a _V A +c- at

_ O. This is called the equa-

tion of gauge invariance. Let us see if this is possible.

Now B = V x A0 and Ei aA°

+=

C at-v4p° so that

I aAo l aAE = -cat - cat - vp

where A = A0 + V. Thus

and

Now we desire

or

1 (BA aA0 1 acat at -cat °x = v((P°

c°)

Iax _ _C at

(p + constant

1 a2x asoo aso

C ate at at

a'c2 at

ao z

C at C at21

2

vzxatz

-v A0cz

a °cz

(304)

The right-hand side of (304) is a known function of x, y, x, t.This equation is called the inhomogeneous wave equation, andif the equation of gauge invariance is to hold, we must be able tosolve it. If we can solve it, the Lorentz equations will reduceto four inhomogeneous wave equations and so will also be solv-able. They are

v2A-c2a2t2

= 4cµP,v,

v25p-=-=2

K a2c 4w- PC2 at K

(305)


Problems

1. For the Lorentz transformations (see Prob. 11, Sec. 24),show that

825, a2,p a24, 1 a2p a2tp a2,p a2(p 1 a2Vc2at2 c2ai2

We call the D'Alembertian.2. Consider the four-dimensional vector C; = (A1, A2, A3, -q'),

2

i = 1, 2, 3, 4, the A; satisfying V2A = C2 at2' while p satisfies2

V 2v = - 4, with H = V x A, and E= -cat - Let

X, = x) x2 = y, x3 = z, x' = ct, and show that

0 - H. H, - EsaC;_ 8C;

F C

Hr 0 - Hz - Es_" axi axf -HY Hx 0 --E,

E. EY E. 0

andShow that axr,s = 0, i = 1, 2, 3, 4, yields V x H =at-1

V E = 0, Fit = -F;;, i or j = 4, otherwise Fu = F;1. Alsoshow that

aF.# aFj- aF#,

axy + axp + ax.= ° a, P, y = 1,2,3,4

yields V x E _ -

c

aHand V H = 0.

4 4axa axs

3. If P;; show that for the Lorentz trans-

formations P12 = -171 =Ht

Complete the ma-ll - (y2/C2)]l

trix P 1.83. Retarded Potentials. Kirchhoff's Solution of

2172V 2 av -47rF(x, y, z, t)


To find the solution (p(P, t) of the inhomogeneous wave equa-tion at t = 0, we surround the point P by a small sphere of

Fia. 71.

radius e, and let S be the surface of a region R containing P (seeFig. 71). We apply Green's formula to this region.

f J f V2,p dT = f f Vp - V#) ddR B

+ f f (# V-p - p Vu') dd (306)9

We choose for ¢ a solution of 'V24, - V2 a = 0. We know

that 4, = f(r + Vt)/r is one such solution, where f is arbitrary.Equation (306) now becomes

V2fIf (4,,,2,p

t_V actd-r-4'rfff FOd-r= ff .. .R B

+ fJ . (307)a


Equation (307) is true for all values of t so that we may integrate(307) with respect to t between limits t = tl and t = t2. Weobtain

l dT - 4 j:2dt j f f F4, dTIII at at

1:2

rr

= f dt (if .. . +jf (308)

Now on E, ¢ _ (1/E)f( + Vt) and

v o da = - = 1 [f(E + Vt) + Vt)] dS2ar,._ E

so that (308) reduces to

12 f If [f(r + Vt) &p Vf'(r + Vt)1t2V2 r at r

fiFf(r+t,

dr

- 4w f if t)dt dr = f,t: dt f 1[f( Vt)

È

+ 91Ef'(E + Vt)

E2

f(E + Vt) R] , da .+ f f ...} (309)JJ

s

Let us now return to a consideration of f(r + Vt). Since fis arbitrary, let us choose f ° 0 for jr + VtI > 6, with the addi-tional restriction that f as

f(r + Vt) d(r + Vt) = 1, where 6 isarbitrary for the moment. Notice that f = 0 for Ir + Vti > a.

Now let us choose t2 > 0 and tl negatively large, so that forall values of r in the region R, jr + Vt[ > a. Hence

_[f(r + Vt) &p _ Vf'(r + Vt)ItL r at r t,

since Ir + V121 > a, I r + V4J > S.Moreover

ftF f(r + Vt) dt = 1 fttip f(r + Vt) d(r + Vt)

= V jda F

f (x) dx (310)

for a fixed r. Now if S is chosen very small, the value of (310)reduces to approximately

SBc. 831 STATIC AND DYNAMIC ELECTRICITY 181

1

ILdx = 1

V (F)r z=of(x) V (r t_-r/v

Hence the left-hand side of (3' 09) reduces to

4V()t_-r/vdr

Now considering the right-hand side of (309), we see that

lim f frf(E + Vt) Dtp +

(pf'(E + Vt) R] dd = 0e-.0 E L E E

(311)

since dd is of the order E2, and f, f , tp, Dtp are bounded for a fixed a.We also have that

t: f(E + Vt) 1lim -jI dt f f

'pE,- . dS = - v 47rcp(P) (312)

since f f dS = and for small 6,E

Finally,

dtlf(r+ Vt)D (rf'2 Dr] ddf f i:'s L r2

f= ii fl' dt{f(r+Vt)Dp-Vr].dd

s r r

- f f j,t dtr

f'(Dr . dd) = ff f .t dt [f(r + Vt)r

+ -f Dr] . dd + f f j i s rV f at dt (Dr dd) (313)s

on integrating by parts and noticing that f} = 0. Finally theright-hand side of (313) becomes equivalent to

V f f(rDs t - -r/V-,p 1 1 t3(o

t--r/V r rV at t- -r/V(314)


Combining (311), (312), and (314), we obtain

V(P) ` 1 JR JF=

-r/ vdr

Vr) dd (315)1

S t= -r/'V- i f J (v , ~ V r + rV at

Now let S recede to infinity and assume that (p, when

evaluated at t = -r/V on the surface S, have the value zerountil a definite time T. For large r, t = -r/V is negative andso is always less than T. Hence the surface element vanishes,and

(P)= d - (316)V JJJThe solutions to (305) are thus se

4PA f fI

t-

en

pv

i-r/V

to be

d( , t) = J eW

rr

t-(r/V)317)

,v(P,t)'JLI xwhe re V =

drt - (r/V)

Finally,B=VxA

l aA (318)E=-'cat-VThe physical interpretation of these results is simple. The

values of the magnetic and electric intensities at any particularpoint P at any instant t are, in general, determined not by thestate of the rest of the field (p, v) at that particular instant, butby its previous history. The effects at P, due to elements at adistance r from P, depend on the state of the element at a previ-ous time t -- (r/V). This is just the difference in time requiredfor the waves to travel from the element to P with the velocityV = c/ V AK, hence the name retarded potential. Had we con-sidered the function f(r - Vt), we should have obtained a solu-tion depending on the advanced potentials. Physically this isimpossible, since future events cannot affect past eventsl


Problem

1. A short length of wire carries an alternating current,j = pv = Io (sin wt)k, -1/2 5 z 5 1/2.

(a) At distances far removed from the wire, show that

A = j-o1 sin w (t - r J kcr c//

and that in spherical coordinates

A,. =jot sin wit - r1cos 0

cr \ /c/l

Ae= - Iotsinwlt- Isin0cr c//

A, =O

(b) Show that H, = He = 0, and that

Hw = cr sin 0IC

cos w Ct - c) + r sin w t - -)1

(c) Find p from the equation of gauge invariance, and then

E,, E8, E from E + c A - V o.

CHAPTER 6

MECHANICS

84. Kinematics of a Particle. We shall describe the motionof a particle relative to a cartesian coordinate system. Themotion of any particle is known when r = x(t)i + y(t)j + z(t)kis known, where t is the time. We have seen that the velocityand acceleration, relative to this frame of reference, will be givenby

i k+dt dtv' dtd2x d2y d zz

a =dtz 1 + dtz

j+ dt2 k

The velocity may also be given by v = vt, where v is the speedand t is the unit tangent vector to the curve r = r(t). Differ-entiating, we obtain

a=dt dtt+vdsdt =dtt+Kv2n (319)

by making use of (95). Analyzing (319), we see that the accelera-tion of the particle can be resolved into two components: a

tangential acceleration of magnitude dt, and a normal accelera-

tion of magnitude v2rc = v2/p. This latter acceleration is calledcentripetal acceleration and is due to the fact that the velocityvector is changing direction, and so we expect the curvature toplay a role here.

For a particle moving in a plane, we have seen in Sec. 17,Example 18, that the acceleration may be given by

ar

2r 2l

= [dt - r` de]R -}- r d (2 de P

184

SEC. 84J MECHANICS 185

Example 101. Let us assume that a particle moves in a planeand that its a celeration is only radial. In this case we mustchave r d [ r2 de1 = 0, and inte-

grating, +r2 dte = h = constant.

From the calculus we knowthat the sectoral area is givenby dA = }r2 dB (see Fig. 72).

Thus dA = constant, so that

equal areas are swept out inequal intervals of time.

Example 102. For a particlemoving around a circle r = b

with constant angular speed coo = we have dt = 0 and

d(r2wo) = 0, so that a = -bwo2R.

Fie. 72.

Example 103. To find the tangential and normal componentsof the acceleration if the velocity and acceleration are known.

and

v=vt,

a v = vat so that as =

Also

and

v

axv=va,nxt= -vanb

an = I$) i°IV,

Problems

1. A particle moves in a plane with no radial acceleration andconstant angular speed wo. Show that r = Ae"o' + Be at.

2. A particle moves according to the law

r = cos t i + sin t j + t2k

Find the tangential and normal components of the acceleration.

186 VECTOR AND TENSOR ANALYST ° [SEc. 85

3. A particle describes the circle r = a cos 0 with constantspeed. Show that the acceleration is constant in magnitude anddirected toward the center of the circle.

4. A particle P moves in a plane with constant angular speedw about 0. If the rate of increase of its acceleration is parallel

d2rto OP, prove that ate = 4rw2

5. If the tangential and normal components of the accelerationof a particle moving in a plane are constant, show that theparticle describes a spiral.

85. Motion about a Fixed Axis. In Sec. 10, Example 12, wesaw that the velocity is given by v = w x r. Differentiating, weobtain

dr dwa=wxa+dt xr

a = w x v + a x r

where a is the angular

(320)

acceleration --' Since v = w x r, we

wl

Fm. 73.

athave also

a=wx(wxr)+axr_ (w - r)w - w2r + a x r

If we take the origin on theline of w in the plane of themotion, then w is perpendiculartororor = 0, so that

a= -w2r+axra x r is the tangential accelera-tion, and w x (w x r) is thecentripetal acceleration.

If we assume that a particle P is rotating about two intersectinglines simultaneously, with angular velocities cal, w2 (Fig. 73), wecan choose our origin at the point of intersection so that

vi = wl x r, v2 = 632 x r

and the total velocity isV = V1 + V2 = (wi + (02) x r

SFC. 86] MECHANICS 187

A particle on a spinning top that is also precessing experiencessuch motion.

86. Relative Motion. Let A and B be two particles traversingcurves r, and r2 (Fig. 74). r, and r2 are the vectors from a point0 to A and B, respectively.

r2=r+rl (321)

Definition: dt is the relative velocity of B with respect to A,

written V4(B).

Fia. 74.

Differentiating (321), we have

dr2 dr dri

dt - dt + dt

or

Vo(B) = VA(B) + Vo(A) (322)

More generally, we have

Vo(A) = V4,(A) + VA,(A1) + VA.(A,) + .. . +Vo(A.)

It is important to note that V4(B) _ -VB(A).Example 104. A man walks eastward at 3 miles per hour, and

the wind appears to come from the north. He then decreaseshis speed to 1 mile per hour and notices that the wind comesfrom the northwest (Fig. 75). What is the velocity of the wind?

We have

V0(W) = VM(W) + V0(M) G(ground)

188 VECTOR AND TENSOR ANALYSIS ISEc. 86

In the first case

so thatVM(W) = -kj, V0(M) = 3i

VG(JV) = -kj + 3iIn the second case,

VM(W) = h(i - j), VG(M) = iso that

V0(W) = h(i - j) + i = (h + 1)i - hj,and

3=h+1, -k= -h, VG(W) =3i-2j

The speed of the wind is miles per hour, and its directionmakes an angle of tan-' I with the south line.

N

Ex.

SFia. 75. Fzo. 76.

Example 105. To find the relative motion of two particlesmoving with the same speed v, one of which describes a circle ofradius a while the other moves along the diameter (Fig. 76). Wehave

P=acos0i-}-asin0j, adze=v

Q = (a - vt)i

This assumes that both particles started together.

T -dQ=(-a sin0dO+v)i+acos0doi

VQ(P) = v(1 - sin 0)i + v cos 6 j

Sec. 87] MECHANICS 189

The relative speed is

IVQ(P)I = [v2(1 - sin 0)2 + v2 cos2 B]} = 2}v(1 - sin B)}

Maximum IVQ(P)I occurs at 0 = 3x/2, minimum at 0 = x/2.

Problems

1. A man traveling east at 8 miles per hour finds that the windseems to blow from the north. On doubling his speed, he findsthat it appears to come from the northeast. Find the velocityof the wind.

2. A, B, C are on a straight line, B midway between A and C.It then takes A 4 minutes to catch C, and B catches C in 6 min-utes. How long does it take A to catch up to B?

3. An airplane has a true course west and an air speed of200 miles per hour. The wind speed is 50 miles per hour from1300. Find the heading and ground speed of the plane.

87. Dynamics of a Particle. Up to the present, nothing hasbeen said of the forces that produce or cause the motion of aparticle. Experiment shows that for a particle to acquire anacceleration relative to certain types of reference frames, theremust be a force acting on the particle. The types of forcesencountered most frequently are (1) mechanical (push, pull), (2)gravitational, (3) electrical, (4) magnetic, (5) electromagnetic.We shall be chiefly concerned with forces of the types (1) and(2). For the present we shall assume Newton's laws of motionhold for motion relative to the earth. Afterward we shall modifythis. Newton's laws are:

(a) A particle free from the action of forces will remain fixedor will continue to move in a straight line with constant speed.

(b) Force is proportional to time rate of change of momentum,

that is, f = dt (mv). In general, m = constant, so that

The factor m is found by experiment to be an invariant for a givenparticle and is called the mass of the particle. In the theory ofrelativity, m is not a constant. my is called the momentum.

(c) If A exerts a force on B, then B exerts an equal and oppositeforce on A. This is the law of action and reaction: fAB = -fha


By a particle we mean a finite mass occupying a point in ourEuclidean space. This is a purely mathematical concept, andphysically we mean a mass occupying negligible volume as com-pared to the distance between masses. For example, the earthand sun may be thought of as particles in comparison to theirdistance apart, to a first approximation.

88. Equations of Motion for a Particle. Newton's second law

may be written f = m dt = ma. We postulate that the forces

acting on a particle behave as vectors. This is an experimentalfact. Hence if fl, f2j . . . , f act on m, its acceleration is givenby

fa1

(fl+f2+....+fn)_ 1 I f,_ fm mti_1 m

We may also write f = m d , where r is the position vector fromdt2

the origin of our coordinate system to the particle. If the particle

is at rest or is moving with constant velocity, thend

= 0, anddt2

so f = 0, and conversely. Hence a necessary and sufficient con-dition that a particle be in static equilibrium is that the vectorsum of the forces acting on it be zero.

A standard body is taken as the unit mass (pound mass). Apoundal is the force required to accelerate a one-pound mass onefoot per second per second. The mass of any other body can becompared with the unit mass by comparing the weights (force of

it lt l f th ty a mean sea eve ) of12/ m2 grav e wo

Fzo. 77.

objects. This assumes the equivalenceof gravitational mass and inertial mass.

Example 106. Newton's law of gravi-tation for two particles is that everypair of particles in the universe exertsa mutual attraction with a force directedalong the line joining the particles, themagnitude of the force being inversely

proportional to the square of the distance between themand directly proportional to the product of their masses.f12 = (Gmim2/r2)R (see Fig. 77). G is a universal constant. Let

SEC. 88] MECHANICS 191

the mass of the sun be M and that of the earth be m. Weshall assume that the sun is fixed at the origin of a given coordi-nate system (Fig. 78). The force act-ing on the earth due to the sun is

f = - (Gm.M/r3)rFrom the second law

GmM d2r dvrs r m dt2 dt

so that

Now

and hence

This implies

or

dv GMdt r3

d dv

dt (r x v) = r x dt

d dt(rxv)=rx(-GMr) =0

Fia. 78.

r x v = h = constant vector

(323)

M

drhrxa =

(324)

Since Ir x drl = twice sectoral area, we have 2 dA = Ihl, or equal

areas are swept out in equal intervals of time. This is Kepler's

first law of planetary motion. Moreover, r I r x dt ] = r h = 0,

so that r remains perpendicular to the fixed vector h, and themotion is planar. Now

a xh -GMrxh= -GMrx(rxv)

from (324), andd (v x h) = d x h, so that

d(v x h)

GMr x (r x v) (325)

192 VECTOR AND TENSOR ANALYSIS SEC. 88

Now r = rR, where R is a unit vector. Hence

v R


\d(vxh)=-GMrx /(rxrR)

= -GMRx(RxdR)

-GM K R . ddR

t / R - R2 dRJ

= GM ddR (326)

since R is a unit vector.Integrating (326), we obtain

vxh=GMR+kand

Thush' = GMr + rk cos (R, k) (327)

h2/GMr

1 + (k/GM) cos (r, k) (328)

We choose the direction of the constant vector k as the polaraxis, so that

h'/GMr1 + (k/GM) cos 8

(329)

This is the polar equation of a conic section. For the planetsthese conic sections are closed curves, so that we obtain Kepler'ssecond law, which states that the orbits of the planets are ellipseswith the sun at one of the foci.

Let us now write the ellipse in the formS

r 1+ e cos 9where e=

GM' p k


The curve ci osses the polar axis at 0 = 0, 0 = it so that the lengthof the major axis is

2, - ep + ep 2p 2h2

1+e 1-e 1-e2 GM(1-e2

For an ellipse, b2 = a2 - c2 = a2 - e2a2, orb = a(l - ,2)1. ThedA

area of the ellipse is A = Tab = ,ra2(1 - e2)1, and sincedt

= Jh,

the period for one complete revolution is

2A 2ara2(1 - e2)i ZralT - h - a1GiM1(1 - e2)1 = G1Mi

ThusT2 47r2 _0 GM = constant, for all planets (330)

This is Kepler's third law, which states that the squaresof the periods of revolution of the planets are proportional to thecubes of the mean distances from the sun.

Problems

1. A particle of mass m is attracted toward the origin with theforce f = - (k2m/r6)r. If it starts from the point (a, 0) with thespeed vo = k/21a2 perpendicular to the x axis, show that the pathis given by r = a cos 0.

2. A bead of mass m slides along a smooth rod which is rotatingwith constant angular speed w, the rod always lying in a hori-zontal plane. Find the reaction between bead and rod.

3. A particle of mass m is attracted toward the origin with aforce - (mk2/r3)R._ If it starts from the point (a, 0) with velocityvo > k/a perpendicular to the x axis, show that the equation ofthe path is

r = a sec C(OW

avo

- k2)i0I

4. In a uniform gravitational field (earth), a 16-pound shotleaves the putter's fingers 7 feet from the ground. At what angleshould the shot leave to attain a maximum horizontal distance?


5. Assume a comet starts from infinity at rest and is attractedtoward the sun. Let ro be its least distance to the sun. Showthat the motion of the comet is given by r = 2ro/(1 + cos 0).

89. System of Particles. Let us consider a system consistingof a finite number of particles moving under the action of variousforces. A given particle will be under the influence of two typesof forces: (1) internal forces, that is, forces due to the interactionof the particle with the other particles of the system, and (2) allother forces acting on the particle, said forces being calledexternal forces.

If r; is the position vector to the particle of mass m,, then weshall designate f,(e) as the sum of the external forces acting on thejth particle, and f,(i) as the sum of the internal forces acting onthis particle. Newton's second law becomes for this particle

f1( + f1( = m; d2r' (331)

Unfortunately, we do not know, in general, f;('), so that we shallnot try to find the motion of each particle but shall look ratherfor the motion of the system as a whole. Since Eq. (331) istrue for each j, we can sum up j for all the particles. This yields

n

f'c6> +f'(i)

d 2r,

m' dt2J=1 ;=1 1-1

From Newton's third law we know that for every internal forcen

there is an equal and opposite reaction, so that A f ®= 0.

This leavesn n

f, l m, dt2(332)

We now define a new vector, called the center-of-mass vector,by the equation

n

Im,r, n

ra =1' =n

(333)7-1

;11

SEC. 89) MECHANICS 195

The end point of r. is called the center of mass of the system.It is a geometric property and depends only on the position ofthe particles. Differentiating (333) twice with respect to time,we obtain

nMd2r` _ md?r,

dt2 ;=1 ' dt2


f f.(e)=M2_1°

z)ml

(334)

Equation (334) states that the center of mass of the systemaccelerates as if the total mass were concentrated there andall the external forces acted at that point.

Fia. 79.

Example 107. If our system is composed of two particles infree space and if they are originally at rest, then the center of

mass will always remain at rest, since f = 0 so that d2° = 0, and

r. = constant satisfies the equation of motion and the initial con-

dition 0. For the earth and sun we may choose the center

of mass as the origin of our coordinate system (Fig. 79). Theequations of motion for earth and sun are

d2r1 GmMR M d2r2 _ GmMRm

dt2= -

(rlr2)2f

dt2(rl - 12)2

Since r. = 0, we have mr, + Mr2 = 0, and

n

d2r1 _ -GM rld12 [1 + (m/M))2 rig


This shows that m is attracted toward the center of mass by aninverse-square force. The results of Example 106 hold by replac-ing M by M[1 + (m/M)1-2.

Problems

1. Show that the center of mass is independent of the originof our coordinate system.

2. Particles of masses 1, 2, 3, 4, 5, 6, 7, 8 are placed at thecorners of a unit cube. Find the center of mass.

3. Find the center of mass of a uniform hemisphere.4. Find the force of attraction of a hemisphere on another

hemisphere, the two hemispheres forming a full sphere.90. Momentum and Angular Momentum. The momentum of

a particle of mass m and velocity v is defined as M = mv. The

total momentum of a system of particles is given by M = j m;v1.

We have at once that

dM n dv1n

dt I mj dt = I f; (e) = fj-1 j-1

j-1

(335)

We emphasize again that the mass of each particle is assumedconstant throughout the motion.

The vector quantity r x my is defined as the angular momen-tum, or moment of momentum, of the particle about the origin 0.

f The total angular momentum is given

Fla. 80.

byn

H = E r; x mjvj (336)j-1

91. Torque, or Force Moment. Letf be a force acting in a given directionand let r be any vector from the originwhose end point lies on the line of

action of the force (see Fig. 80). The vector quantity r x fis defined as the force moment, or torque, of f about 0. For asystem of forces,

n

L = I r, x f, (337)j-1

SEC. 921 MECHANICS 197

We immediately ask if the torque is different if we use a differ-ent vector ri to the line of action of f. The answer is in thenegative, for

(r, - r) x f = 0

since ri - r is parallel to f. Hence rl x f = r x f.What of the torque due to two

equal and opposite forces bothacting along the same line? Itis zero, for

r1xf+r1x(-f)=r1x(f-f)=0

Two equal and opposite forceswith different lines of actionconstitute a couple (see Fig. 81).Let ri be a vector to f and r2 avector to -f. The torque dueto this couple is

L = r1xf+r2x(-f)= (r1 - r2) x f

Fio. 81.

The couple depends only on f and on any vector from the line ofaction of -f to the line of action of f.

Problems

1. Show that if the resultant of a system of forces is zero, thetotal torque about one point is the same as that about any otherpoint.

2. Show that the torques about two different points are equal,provided that the resultant of the forces is parallel to the vectorjoining the two origins.

3. Show that any set of forces acting on a body can bereplaced by a single force, acting at an arbitrary point, plus asuitable couple. Prove this first for a single force.

4. Prove that the torque due to internal forces vanishes.92. A Theorem Relating Angular Momentum with Torque.

We are now in a position to prove that the time rate of changeof angular momentum is equal to the sum of the external torquesfor a system of particles.


Sincen

H = Ir3xmjvj=1

we have on differentiating

j=1

drir, x

m'dt

n ndH _ d2r, dri dridt

Z r, x m,dt2 + j=1 dt x m' dt

n

I r' x (f1" + f, (') )j-1

naii = I r; xf;(e)

= Ldt j=1

(338)

93. Moment of Momentum (Continued). It is occasionallymore useful to choose a moving point Q as the origin of our

FIG. 82.

coordinate system. Let 0 be a fixed point and Q any point inspace. We define

n driHQa = I (r, -- rQ) x m;di

(339)j-1

The superscript a stands for absolute momentum, that is, thevelocity of m; is taken relative to 0, whereas the subscript Qstands for the fact that the lever arm is measured from Q to theparticle m; (Fig. 82). Differentiating (339), we obtain

SEc. 941 MECHANICS

dHQa _ (dri _drQl dr, nn` d2r;

dt (dt dt /X 7n,

dt + L (r. - rQ) x "nidt2

j=1 j=1

drQ n dr1 n

dl xLm'dt+Ij=1 j-1

n

(r1 - rQ) x (f.(e) + f .(i))

Now Mr = m;r,, so that M dt =j=1 l1

r1 x f;(') = 0 from Sec. 91, so thatj-1

199

dr, nm1 -, and f,(i)=0,

j=1

dHQa drQ dF n

dt = Mdt x dt + I (r, - rQ) x fj(e)

j-1

or

(e) - M (340)dtQa

LQdt

Qxdtc

We can simplify (340) under three conditions:

1. Q at rest, so thatdrQ

dt= 0

2. Center of mass at rest, dt` = 0

3. Velocity of Q is parallel to velocity of center of mass,drQ dre

dt dt0

In all three cases

dHe= LQ(e)

dt(341)

In particular, if LQ(e) = 0, then He - constant, and this is thelaw of conservation of angular momentum.

94. Moment of Relative Momentum about Q. In Sec. 93 weassumed that the absolute velocity of each particle was known.It is often more convenient to calculate the velocity of each

dr, - drQparticle relative to Q. This is

dt dtWe now define rela-


tive moment of momentum about Q asn

HQr = (rj - rQ) x mjdl

(rj - rQ) (342)j-1

Differentiating,

dHQ* (d2rj d2rQ

dt j-1(r' - rQ) x m'

dl2 dt2

= (r7 - rQ)

We see thatn

n

x I m, (rj - rQ)j-1

dtiQr 'rQLQ(' + x I mj (rj - rQ)a, d12

js1

Under what conditions does ddQ' = Lo ? We need

d2rQ n

x I mj(rj - rQ) = 0dt2

or

/ `x

(fj(ei d2rQ

f3(i))dt2

d2rQM

dt2x (r, - rQ) = 0

(343)

(344)

Now (344) holds if1. rQ = rQ, or Q is at the center of mass.

2. Q moves with constant velocity, dQ = 0.l

2

3. r. - rQ is parallel to2dt'

n

Problems

2a rQ d rQ1. Show that m4(r1 - rQ) x dt2 = M(r. - rQ) x

dt2jet2. A system of particles lies in a plane, and each particle

remains at a fixed distance from a point 0 in this plane, each


particle rotating about 0 with angular velocity w. Show thatn

Ho = Iw, where I = I m;r;2, and show that Lo = I atj=1

3. A hoop rolls down an inclined plane. What point can betaken as Q so that the equation of motion (343) would besimplified?

95. Kinetic Energy. We define the kinetic energy of a particleof mass m and velocity v as T = jmv v. C

MFor a system of particles, r.-rn 1 n 1 fdrs\ 2 `r P;

T 2 mv,22 m' dt 1 (345)

f=1 .1-1 rNow let r, be the vector to the center of 0mass C (Fig. 83). It is obvious that Fia. 83.

so thatr;=rr+(r,-r-)

dr; _ dr, ddt dt + dt

(r; - r-)

dr; dr;_ (dro)2

drd I d. 2

di dt dt+2

dt dt(r, - rc) -{- L dt (r; - rc) ]

Hence

1 (dr12 dr" n dT=2M dtJ +dt

Now Mr. =

2

+ 2 m' [dt (r' - r`)] (346)1j-

n n n

j-1 j-1 ;-1

m;r, r,, so that m1 alt (r; - r-) = 0,

and (346) reduces to

T = I M()2

-}- m; (r; - r.)]2a)(347)-1 2

[-ddt 1

This proves that the kinetic energy of a system of particles isequal to the kinetic energy of a particle having the total mass


of the system and moving with the center of mass, plus thekinetic energy of the particles in their motion relative to thecenter of mass.

96. Work. If a particle moves along a curve r with velocity vunder the action of a force f, we define the work done by thisforce as

W =

fr f fr f t

f acts at right angles to the path, no work is done.If the field is conservative, f = -V(p, the work done in taking

the particle from a point A to a point B is independent of thepath (see Sec. 52).Now

m,dvi = fj(a) + fi(.)dt

mivi .dvidt

= fi(e) . V1. + Vi

and integrating and summing over all particles,

n

Jòmivi dt' dt = La i:1

fi(a) . vi dt + fee:' fi('i . v; dt

or

n

lmi[vi2(ti) - vi2(to)l = W(e) W °i=1

(349)

This is the principle of work and energy. The change in thekinetic energy of a system of particles is equal to the total workdone by both the external and internal forces.

If the particles always remain at a constant distance apart,(ri - rk)2 = constant, the internal forces do no work. Let r,and r2 be the position vectors of two particles whose distanceapart remains constant, and let f and -f be the internal forcesof one particle on the other and conversely. Now

(r, - r2) (r, - r2) = constant

Sec. 97] MECHANICS

so that

(r, - r2) dr, - dr20

di dt/Also

W(o = ff f

f is parallel to r, - r2, Ave have f = a(r, - r2) and

f (v, - v2) = a(r, - r2) (v, - v2) = 0

from (350). Thus W(° = 0.

203

(350)

Problems

1. A system of particles has an angular velocity w. Show thatn

T = Jm;lw x r,J2.i-i

2. If to of Prob. 1 has a constant direction, show that T = }Iw2,n

where I = md;2, d; being the shortest distance from m; to

line of w.

3. Show that dT = w L, by using the fact that T = 4-mv;2

and that v, = to x r;.4. Show that the kinetic energy of a system of rotating par-

ticles is constant if the system is subjected to no torques. Whatif L is perpendicular to w?

5. A particle falls from infinity to the earth. Show that itstrikes the earth with a speed of approximately 7.0 miles persecond. Use the principle of work and energy.

97. Rigid Bodies. By a rigid body we mean a system ofparticles such that the relative distances between pairs of pointsremain constant during the discussion of our problem. Actuallyno such systems exist, but for practical purposes there do existsuch rigid bodies, at least to a first approximation. Moreover,the rigid body may not consist of a finite number of particles, butrather will have a continuous distribution, at least to the unaidedeye. We postulate that we can subdivide the body into a greatmany small parts so that we can apply our laws of motion forparticles to this system, this postulate implying that we can use


the integral calculus.the following form:

J f f f(e)R

Our laws of motion as derived above take

T = f f f -pv2 dr, p = densityR

ff pr dr

f f f p drR

2dr,dt2

H = f f f prxvdrR

-it ff r x f(e) dr

351)(

where f(.) is the external force per unit volume.98. Kinematics of a Rigid Body. Let 0 be a point of a rigid

P to P, for if r is the position vectorfrom 0 to P, we have r - r = con-stant throughout the motion so that

r dt = 0. Q.E.D.

We next prove that if two pointsof a rigid body are fixed, then allother particles of the body are rotat-ing around the line joining these twopoints. Let A and B be the fixedpoints and P any other point of the

body. From above we have

so that P is always moving perpendicular to the plane ABP.Moreover, since the body is rigid, the shortest distance from Pto the line A B remains constant, so that P moves in a circle

body for which 0 happens to be fixed.It is easy to prove that the velocityof any other point P of the body must

f V (P1 = yr be perpendicular to the line joining 0

Sec. 981 MECHANICS 205

around AB (Fig. 84). We saw in Sec. 10 that the velocity of Pcould be written

VP =wxrp

Is w the same for all particles? Yes! Assume Q is rotatingabout AB with angular velocity w,, so that vQ = to, x rQ. Now(rP - rQ)2 = constant, so that

(rP - rQ) (vp - vQ) = 0or

(rP - rQ) ((a x rp - w, x rQ) = 0Thus

x rP - xrQ = 0and

rQ x rp (w, - w) = 0

We leave it to the reader to conclude that w, = w.

VA

B

1

FIG 85.'

If one point of a rigid body is fixed, we cannot, in general, hopeto find a fixed line about which the body is rotating. However,there does exist a moving line passing through the fixed point sothat at any instant the body is actually rotating around this line.The proof proceeds as follows: Let 0 be the fixed point of ourrigid body and let r` be the position vector to a point A. Fromabove we know that the velocity of A, VA, is perpendicular torA. Construct the plane through 0 and A perpendicular to VA(Fig. 85). Now choose a point B not in the plane. We alsohave that vB rB = 0, so that we can construct the plane through0 and B perpendicular to vB. Both planes pass through 0, so


that their line of intersection, 1, passes through 0. Now con-sider any point C on this line. We have vc rc = 0. Moreover,(rc - rA) - (rc - rA) ° constant, so that

(re-rA).(vc-VA) =0

and (rc - rA) VC = 0, since vA is perpendicular to (rc - rA).

Similarly (re - rB) vc = 0. Hence the projections of vc inthree directions which are nonplanar are zero. This means that

FIG. 86.

Vc - 0, so that we have two fixed points at this particularinstant. Hence from the previous paragraph the motion is thatof a rotation about the line 1. If w is the angular-velocity vector,then v; = w x r;, where r; is the vector from 0 to the jth particle.

Now let us consider the most general type of motion of a rigidbody. Let -r represent a fixed coordinate system in space,and let 0-x-y-z represent a coordinate system fixed in the rigidbody (see Fig. 86). Let ,o; and r; represent the vectors from 0'and 0 to the jth particle, and let a be the vector from 0' to 0.We have ei = a + r,, and differentiating,

dLD; - da dr,dt dt + dt


Now l represents the velocity of Pi relative to 0. This means

0 is fixed as far as Pi is concerned, and from above we know thatdridt - w x ri. Thus

_dei = da

v' dt dt +wxr; (352)

that is, the most general type of motion of a rigid body is that of

a translationAdt

plus a rotation w x r;.

We next ask the following question: If we change our originfrom 0 to, say, 0" does w change? (Fig. 87.) The answer is"No"! Let b be the vectorfrom 0' to 0". Then 0 rf

=t= + w,xr;v; '

But

and

Thus

db da

dt dt+w x (b - a)

ri = (a -- b) + riFio. 87.

vi = d +w x(b - a)+wi x (a -b)+w, xri (353)

Subtracting (352) from (353), we obtain

(w - wi) x (b-a)+(w,-w) xr;=0or

(w - wl) x (b-a-r,)= xr;"=0We can certainly choose an r;" 0 and not parallel to the vectorw - wl, at any particular instant. Hence w, ° w.

Problems

1. Show that if r, and r2 are two position vectors from theorigin of the moving system of coordinates to two points in the

rigid body, then r, dt2 + r2 - dtl = 0.


2. A plane body is moving in its own plane. Find the pointin the body which is instantaneously at rest.

3. Show that the most gen-eral motion of a rigid body isa translation plus a rotationabout a line parallel to thetranslation.

99. Relative Time Rate ofChange of Vectors. Let S bean vector measured in theymoving system of coordinates(Fig. 88).

Fin. 88. S = S i + Sj + S,k (354)

To find out how S changes with time as measured by an observerat 0', we differentiate (354),

dS dS=. dS . dS, di dj dk

dt = dt 1+ dt'+ dt k+S=_dt+Sydt+S'dt (355)

We do not keep i, j, k fixed since i, j, k suffer motions relative to

0'. But we do know that dt is the velocity of a point one unit

along the x axis, relative to 0. Henced

= w x is dt = w x it

dkat

= w x k. Hence (355) becomes

dS dSs dS dS,

dt - dt i -}dt j + d k + w x (3j + S1t)

and

dS DS+ S

dt dt w x (356)

where DS represents the time rate of change of S relative to the

moving frame, for S= is measured in the moving frame and so d _t

is the time rate of change of S. as measured by an observer in themoving frame.


Intuitively, we expected the result of (356), for not only does Schange relative to 0, but to this change we must add the changein S because of the rotating frame. The reader might well ask,What of the motion of 0 itself? Will not this motion have to beconsidered? The answer is "No," for a translation of 0 onlypulls S along, that is, S does not change length or direction if 0is translated. It is the motion of S relative to the frame O-x-y-zand the rotation about 0 that produce changes in S.

Problems

1. Show that d dl

2. For a pure translation show thatdSdt =

DSdt

3. From (356) show that di = w x i.

100. Velocity. Let P be any point in space and let 9 and rbe the position vectors to Pfrom 0' and 0, respectively(see Fig. 89). Obviously

a + r, so that

dp A drv

_dt

_dt + dt

Now r is a vector measured inthe O-x-y-z system, so that(356) applies to r. This yieldsdr _ Dr7t dt + to x r and t

FIG. 89.

vdt dt --

xr+Dr

(357)

This result is expected.A

is the drag velocity of P, co x r is

the velocity due to the rotation of the 0-x-y-z frame, andDr

is

the velocity of P relative to the 0-x-y-z frame. The vector sumis the velocity of P relative to the frame 0'--i-r.

210 VECTOR AND TENSOR ANALYSIS (SEC. J 01

101. Acceleration. In Sec. 100 we saw that

A Drv=ac dt+wxr+dt

To find the acceleration, we differentiate (357) and obtain

dv_d2p_d2a d d(Drldt dt2 dt2 + dt (wxr) + dt dt

We apply (356) to w x r and obtain

Similarly

d (wxr) = w x (wxr) + D (w x r)

ddldtl xdt+d`dtlD

(358)

d2p d2a do Dr D2r_2 dt2

+ w x (wxr) +dt

x r + 2w xdt + dt2 (359)dt

Let us analyze each term of (359). If P were fixed relativeDr D2r

to the moving frame, we would have=

dt dt2 = 0 and conse-

quently P would still suffer the acceleration

d2a &dt2 +wx(wxr)+dt xr

This vector sum is appropriately called the drag accelerationof the particle. Now let us analyze each term of the drag accel-eration. If the moving frame were not rotating, we would have

2

0, and the drag acceleration reduces to the single term dta

This is the translational acceleration of 0 relative to 0'. Now inSec. 84 we saw that w x (w x r) represented the centripetal accel-

eration due to rotation and d x r represented the tangential

component of acceleration due to the angular-acceleration vector


d We easily explain the term D2ras the acceleration of P

relative to the O-x-y-z frame. What, then, of the term 2w x dr?

This term is called the Coriolis acceleration, named after its dis-coverer. We do not try to give a geometrical or physical reasonfor its existence. Suffice to say, it occurs in Eq. (359) and mustbe considered when we discuss the motion of bodies moving overthe earth's surface. Notice that the term disappears for par-

ticles at rest relative to the moving frame, for then dt = 0. Italso does not exist for nonrotating frames.

Now Newton's second law states that force is proportionalto the acceleration when the mass of the particle remains con-stant. It is found that the frame of reference for which this lawholds best is that of the so-called "fixed stars." We call such aframe of reference an inertial frame. Any other coordinatesystem moving relative to an inertial frame with constant velocity

2d Dis also an inertial frame, since from (359) we have d e = dt2r

d166 dabecause w = Of 1 0

dt, = constant, dta = 0.

Let us now consider the motion of a particle relative to theearth. If f is the vector sum of the external forces (real forces,

2

that is, gravitation, push, pull, etc.), then d p = m, and (359)

becomes

D2r d2a d4,3 Dr fdt2 dt2 - w x (w x r) - d x r - 2w x dt + m (360)

This is the differential equation of motion for a particle of massm with external force f applied to it.

Example 108. Let us consider the earth as our rotating frame.The quantity w x ((a x r) is small, since jwl 27r/86,164 rad/sec,and for a particle near the earth's surface, lrl , (4,000)(5,280)

2dfeet. Also

dt- 0 over a short time; da ,= 0 over a short time;



D2r

dt2-2w x dt + m (361)

Now consider a freely falling body starting from a point P atrest relative to the earth. Let the z axis be taken as the line

SFio. 90.

joining the center of the earthto P, and let the x axis betaken perpendicular to the zaxis in the eastward direction.We shall denote the latitudeof the place by A, assumingA > 0. The equation of mo-tion in the eastward directionis given by

d1xy- -2 (w xdj

, m_

dtt2 + Cf /Now f (force of attraction)

has no component eastward, so

that (f/ = 0. We do not know at but to a first approxima-m ,

tion it is -gtk + i. Moreover, w = w sin A k + w cos A j

(see Fig. 90). Hence (w x at)0 = -wgt cos A, and

d2x

dt' = 2wgt cos A (362)

If the particle remains in the vicinity of latitude A, we can keep Aconstant, so that on integrating (362), we obtain

dxdt

=wgt2cosA

X = 3ts cos A (363)

(363) is to a first approximation the eastward deflection of ashot if it is dropped in the Northern Hemisphere. If h is the

SEc. 1011 MECHANICS

distance the shot falls, then h = Jgt2 approximately, so that

wh2gh1}x = 2

3 cos X J(

213

Problems

1. Show that the winds in the Northern Hemisphere have ahorizontal deflecting Coriolis acceleration 2wv sin X at rightangles to v.

2. A body is thrown vertically upward. Show that it strikesthe ground 'jwh cos X (2h/g)1 to the west.

3. Choose the x axis east, y axis south, z axis along the plumbline, and show that the equations of motion for a freely fallingbody are

d2x +2wsinU z- 2wcos0 y=0W dt dt

dt +2w cos0dy =02dtz-g-2w sin0dx =0

where 0 is the colatitude.CO

Fio. 91.

4. Using the coordinate system of Prob. 3, let us consider themotion of the Foucault pendulum (see Fig. 91).

Let ii, i2, i$ be the unit tangent vectors to the spherical curvesr, 8, p. We leave it to the reader to show that the acceleration


along the is vector is 2 cos e .06 + sin 0 0 when the string is ofunit length. The two external forces are mgk along the z axis andthe tension in the string, T = - Tr = - Tit. We wish to findthe component of these forces along the i3 direction. T has nocomponent in the is direction. Now k is = 0, so that mgk hasno component along the is direction. Finally, we must compute

the is component of -2w xDr

The velocity vector is

Drdt

_ Bit + sin 6 rpis

Also w = w(- cos A j - sin X k), so that we must find the rela-tionship between i i2, is and i, j, k.Now

r = i, = sin0cosci+sin Bsin cpj+cos0kail

i2 = aB = cos 0 cos v i + cos 0 sin sin 0 k

_ 1 ail1s sin B ai - sin sp i + coo sp j

Thus

i = (i il)i, + (i i2)i2 + (i i3)is= sin 0 cos Tp it + cos 0 cos jP is -- sin rp is

j = sin 0 sin Sp i, + cos B sin 'P i2 + C0803k = cos 0 it - sin B i2

-2wx-=2w11

cos A sin 0 sin P

+ sin A cos 0

i2

-sinAsin0

is

0 6 sine c

and

r2w x -D = #(sin A sin 0 sin ip + sin A cos 0)(- dt

Equation (361) yields

2 cos 006+ sin 8;p = 2w(B sin A sin B sin rp + # sin A cos 0)(364)

SEC. 102] MECHANICS

For small oscillations, sin 0 0, and (364) reduces to

215

, = w sin X (365)

Hence the pendulum rotates about the vertical in the clock-wise sense when viewed from the point of suspension with anangular speed w sin X. At latitude 30° the time for one completeoscillation is 48 hours.

5. Find the equation of motion by considering the i2 compon-ents of (361) for the Foucault pendulum.

102. Motion of a Rigid Body with One Point Fixed. Themotion of a rigid body with one point fixed will depend on theforces acting on the body. Let O-x-y-z be a coordinate systemfixed in the moving body, and let O-- be the coordinate systemfixed in space. 0 is the fixed point of the body. In Sec. 94 we

saw that -Or = Lo. Now Ho* = f if r x p dt dr. We can

replace dt by to x r (w unknown). Thus

Let

H0? = f f f pr x (w x r) drR

= fJf p[r2w - (r w)r] dr (366)R

w=w=i+wyj+wr=xi+yj+zk

so that

r2w - (r w)r = (x2 + y2 + z2) (w i + wyj + W k)+ (xwx + ywy + zws) (xi + yj + zk)

= [(y2 + z2)wz - xywy - xzws]i

+ [-xyw. + (22 + x2)wy - yzw=lj+ [-xzwz - yzwy -- (x2 + y2)w]k

We thus obtain

H0? = i[w= f f f p(y2 + z2) dr - w f f f pxy dr - wz f f f pxz dr]

+ j[ - w=f f f pxy dr + wyf f f p(z2 + x2) dr - w: f f f pyz dr]+ k[ -w=f f f pxz dr - wyf f f pyz dr + w=f f f p(x2 + y2) dr] (367)


The quantities

A = f f f p(y2 + Z2) drB=fffp(z2+x2)drC = fffp(x2 + y2) drD = f f fpyzdrE = f f fpzxdrF = f f f pxy dr

(368)

are independent of the motion and are constants of the body.That they are independent of the motion is seen from the factthat for a particle with coordinates x, y, z, the scalars x, y, zremain invariant because the O-x-y-z frame is fixed in the body.The quantities A, B, C are the moments of inertia about thex, y, z axes, and D, E, F are called the products of inertia. Weassume the student has studied these integrals in the integralcalculus.

dHor _ DHorNow from Sec. 99 we have

dt dt+ w x Hor so that

DRO,Lo = dt + to x Hor

Hence

L=i+Lvj+LA=i(A ds-F dtEdts/

+

+j(-Fw

+Bdty-Ddt;/

+ k \-E dt - D dt + C dta/i k

wy WV wt

Aws - Fwv - Ewz, -Fws+Bwv - Dw=, -Ews - Dwv+Ccos(369)

In the special case when the axes are so chosen that theproducts of inertia vanish (see Sec. 107), we have Euler's cele-brated equations of motion :

SEc. 1031 MECHANICS

L. = A d[ + (C - B)wYwz

dwYL = B

dt+ (A - C)w,wZ

dw,L. = C

dt+ (B A)wzwy

217.

(370)

103. Applications. If no torques are applied to the body ofSec. 102, Euler's equations reduce to

(i)dwz

Adt-

+ (C B)wYw, = 0

B dt +(A-C)w,wx=0

C d,+(B-A}wZwY=0

(371)

Multiplying (i), (ii), (iii) by w,, respectively, and adding,we obtain

Awsdws + Bw

dwY+ Cw,

dwa= 0

dt dt dt

Integrating yields

Aws2 + Bw,,2 + Cw,2 = constant (372)

This is one of the integrals of the motion. We obtain anotherintegral by multiplying (i), (ii), (iii) by Aw,z, Bw,,, Cw,, and adding.This yields

dws 'dw'BZAZ C2

dw,+ wyw, + w,

dt dt=

dt0

so that

A 2w12 + B2wV2 + C2w,2 =constant (373)

If originally the motion was that of a rotation of angularvelocity w about a principal axis (x axis), then initially


wz(0) = coo

0

ws(0) = 0

(374)

and we notice that (371) and the boundary condition (374) aresatisfied by

wt(t) = -coo0

wZ(t) = 0

so that the motion continues to be one of constant angularvelocity about the x axis. Here we have used a theorem on theuniqueness of solutions for a system of differential equations.

Now suppose the body to be rotating this way and thenslightly disturbed, so that now the body has acquired the verysmall angular velocities w,,, wt. We can neglect ww: as comparedto and wswo. Euler's equations now become

B dty + (A - C)wswo = 0

Cdt +(B0wo = constant

(375)

Differentiating the first equation of (375) with respect to time

and eliminating d s' we obtain

zB dtzy + (- C)((A - B)w02w = 0 (376)

If A is greater than B and C or smaller than B and C, then(A - C) (A - B)

a2 =C

> 0, and the solution to (376) is

wy = L cos (at + a)

Also ws =aBL sin (at + a)

by replacing w,, in (375).wo(A - C)

SEC. 1041 MECHANICS

Problems

219

1. Solve the free body with A = B for wzj co,, co,.2. A disk (B = C) rotates about its x axis (perpendicular to

the plane of the disk) with constant angular speed wo. A con-stant torque Lo is applied constantly in the y direction. Findw and W..

3. Show that a necessary and sufficient condition that a rigidbody be in static equilibrium is that the sum of the externalforces and external torques vanish.

4. A sphere rotates about its fixed center. If the only forcesacting on the sphere are applied at the center, show that the initialmotion continues.

5. In Prob. 2 a constant torque Lo is also applied in the zdirection. Find w,, and ws.

104. Euler's Angular Coordinates. More complicated prob-lems can be solved by use of Euler's angular coordinates. LetO-x'-y'-z' be a cartesian coordinate system fixed in space, andlet O-x-y-z be fixed in the moving body (Fig. 92).

The x-y plane will intersect the x'-y' plane in a line, called thenodal line N. Let 0 be the angle between the z and z' axes,L' the angle between the x' and N axes, and (p the angle betweenthe nodal line and the x axis. The positive directions of theseangles are indicated in the figure.

The three angles &, 0, rp completely specify the configuration

of the body. Now d represents the rotation of the O-z'-N-T'

frame relative to the O-x'-y'-z' frame; de represents the rotation

of the O-z-N-T frame relative to the O-z'-N-T' frame, and finally,d(P

represents the rotation of the O-x-y-z frame relative to the

O-z-N-T frame. Therefore

d*

+ dO + dt gives us the angular

velocity of the O-x-y-z frame relative to the fixed O-x'-y'-z' frame,and

d1 + dO + d (377)


The three angular velocities are not mutually perpendicular.We now define i, j, k, i', j', k', N, T', T as unit vectors along the

x, y, z, x', y', z', N, T', T axes, respectively. Thus

w =d k

wzi + w,j + wLkwz'i' + wy j' + wZ k'

Fia. 92.

Now it is easy to verify that

(378)

i = cos rp N + sin cp Tj = - sinpN+cosjpTi' = cos 4, N - sin ' T'


so that

sin 4, N+cos#T'sin sin psin B

cosspsin 0

ws kiat at at

sin p sin 0 d + cos s dB

day dow = cos (p sin 8 --- - sin (p -

W: =

at atday dp

cos B d +dt

Rewriting this, we have

dtsin 0 sin (P +

docos <p

wy =d4,

sin B cos - dt8 sin So

wI = d cos 8 +- d

(379)

Also(d)2

wi=wss+w2+(O2= +(dB 2 d2dd +Cdt

dcp d#+ 2 cos 8

dt dt (380)

For the fixed frame

w=- ado dcp

wy sin>Gt - cos0sin8dt

d +cos6d

(381)

105. Motion of a Free Top about a Fixed Point. Let usassume that no torques exist and that the top is symmetric(A = B). Euler's equations become

222

(i)

VECTOR AND TENSOR ANALYSIS

A

[SEC. 106

(ii)

(iii)

A d' + (A - C)cvxwZ = 0

Cdts=O

(382)

Integrating (iii), we obtain wz = w, = constant. Multiplyby i = and add to (i). We obtain

A dt (w= + iwv) + (C - A)wo(wy - iwz) = 0

or

Integrating,

so that

A d (co., + iwy) = iwo(C - A)(wx + i )

wz + 2Wy = ae<'(C-A)/ALot

co. = a cos atwy = a sin at (383)

where a = [(C - A)/A]wo and a is a constant of integration.Now w2 = ws2 + Wye + ws2 = a2 + wp2 = constant, so that the

magnitude of the angular velocity remains constant during the

motion. Moreover,d

= 0, so that H is a constant vector in

fixed space. We choose the z' axis for the direction of H. Now

H = Awzi + Bw,j + Cwk= Aa cos at i + Aa sin at j + Cwok (384)

This shows that H rotates around the z axis (of the body) withconstant angular speed a = [(C - A)/A]wo, and since H is fixedin space, it is the z axis of the body which is rotating about thefixed z' axis with constant angular speed -a = [(A - C)/A]wo.Also H k = I HI cos 0 = Cwo, so that 9 is a constant sinceI HI = constant. We say that the top precesses about the z' axis.

106. The Top (Continued). We have assumed above that theweight of the top or gyroscope was negligible, or that the gyro-scope was balanced, that is, suspended with its center of massat the point of support, so that no torques were produced. We


shall now assume that the center of mass, while still located onthe axis of symmetry, is not at the point of support. We nowhave the following situation (Fig. 93) :

L = 1kx(.-Wk')= WI sin ON

The three components of the torque are

Lz = WI sin 0 cos 'vL" = - Wl sin 0 sin pL. = 0

ZI

Fro. 93.

Euler's equations become

Wl sin 0 cos p = Adw=

dt+ (C - A)w"ws

- WI sin 8 sin (p = Adwdt"

+ (A - C)wws (385)

0=CdsforA=B

Hence w: = coo. Multiplying Eqs. (385) by wx, w", Co., respec-tively, and adding, we obtain

2d (Awx2 + Bw"2 + Cw,2) = Wl sin 0(wy cos w" sin (p) (386)


From (379) we have wz cos P - w sin cp = de, so that (386)

becomes1 d (A w=2 + B

d8

2 dt dt

and integrating

Awx2 + Cwz2 = -2W1 cos 0 + k

or, again using (379),

(j)2

sin2 B +(dO)2

= a - a cos B (387)

a and a are constants.Now since Le = L k' = 0) we have HZ- = constant. Also

H = AwJ + Bw j + Cwsk, so that

sin 9+Cwocos0= constant

Replacing wr and w by their equals from (379), we have

A sin2 9 sine,p + d sin (p sin B cos + di cos2 V sin2 0

- dte cos V sin (p sin 9) + Cwo cos 8 = constant (388)

or A d sin2 9 + Cwo cos 9 = constant = He.

Let 16 = H,,/A, b = Cwo/A, so that (388) becomes

d' -bcos9dt - sin2 9

From (379)

(389)

d4'LIPw, = wo =

dtcos B +

dt(390)

Using (389), (387) becomesll2 x

= a - a cos 0( sin9 s 9/ +(de (391)

SEC. 107] MECHANICS

Let z cos 0, so that dt = - sin 6 -, and

2

( - bz)2 +Cdt = (a - az) (1 - z2)

Hence

225

t = fos [(a - az) (1 - z2) - (g - bz) 2]-} dz (392)

This integral belongs to the class of elliptic integrals. If we canintegrate and find z, then we shall know

d4, 9-bz dip d'y

dt - 1 - z2' dt= wa - dt

The reader should look up a complete discussion of ellipticintegrals in the literature.

Fm. 94.

107. Inertia Tensor. The moment of inertia of a rigid bodyabout a line through the origin may be computed as follows.Let the line L be given by the unit vector ro = li + mj + nk,and let r be the vector from 0 to any point P in the body,

r = xi + yj + zk

(see Fig. 94). The shortest distance from P to L is given by


D2 = r2 -- (r r0)2_ (x2 + y2 + z2) - (lx + my + nz)2_ (12 + m2 + n2)(x2 + y2 + z2) - (lx + my + nz)2=

1l2(y2

+ z2) + m2(z2 + x2) + n2(x2 + y2) - 2mnyz- 2lnzx - 2lmxy

Thus

I = f f fpD2dzdydx= Ale + Bm2 + Cn2 - 2mnD - 2n1E - 2lmF

Let us replace 1, m, n by the variables x, y, z, and let us considerthe surface

,p(x, y, z) = Axe + By2 + Cz2 --- 2Dyz - 2Ezx - 2Fxy= 1 (393)

A line L through the origin is given by the equation x = It,y = mt, z = nt. This line intersects the ellipsoid rp(x, y, z) = 1for t satisfying

(Al2 + Bm2 + Cn2 - 2Dmn - 2n1E - 2lmF)t2 = 1

or t2 = 1/I. The distance from the origin to this point of inter-section is given by

d = (1212 + m2t2 + n2t2)} = t = I-fso that

(394)

We know that a rotation of axes will keep I fixed, for the lineand the body will be similarly situated after the rotation. Wenow attempt to simplify the equation of the quadric surface,p(x, y, z) = 1. First, let us find a point P on this surface atwhich the normal will be parallel to the radius vector to thispoint. The normal to the surface is given by Vip, so that wedesire r parallel to Vv, which yields the equations

Ax - Ez - Fy By - Dz - Fx Cz - Dy - Ex(395)

x y z

Any orthogonal transformation (Example 8) will preserve theform of (393) and (395) with x, y, z replaced by x', y', z' andA, B, . . . , F replaced by A', B', .. . , P. Now choose the


z' axis through P so that x' = 0, y' = 0, z' _ satisfy (395).This yields - E'/O = - D'/0 = C', which means that

E'=D'=0and (393) reduces to

A'x'' + B'y'' + C'z'' - 2F'x'y' = 1 (396)

The rotationx"=x'cos0-y'sin0y"= x' sin0+y'cos0z" = z'

with tan 20 = F'/(B' - A') reduces (396) to

A"z" + D''y.,, + cf/z,F= = 1 (397)

This is the canonical form desired. We have thus proved theimportant theorem that a quadratic form of the type (393) canalways be reduced to a sum of squares of the form (397) by arotation of axes. In the proof we made the assumption thatthere was a point P such that r is parallel to V o, which yielded(395). We could have arrived at Eqs. (395) by asking at whatpoint on the sphere x2 + y2 + z2 = 1 is (p(x, y, z) a maximum.Since p(x, y, z) is continuous on the compact set

x2+y2+z2 = 1such a point always exists. Equations (395) are then easilydeduced by Lagrange's method of multipliers.

We can arrange the constants of inertia into a square matrix

- EI = -F B -D

-E -D C(398)

The elements of the matrix (an array of elements) are called thecomponents of I. Under a proper rotation we have shown thatwe can write

A" 0 0I= 0 B" 0

0 . 0 C"(399)


In general, under an orthogonal transformation, I will become

A' -F1 - E'I = -F' B' -D' (400)

- E' -D' C'

and the components of I in (400) will be related to the compon-ents of I in (398) according to a certain law. We shall see inChap. 8 that I is a tensor and so is called the inertia tensor.

Referring back to (367), we may write

H= A -F -E wZH _ -F B -D wb (401)Hz -E -D C ws

from the definition of multiplication of matrices, where

w = w1 + wyJ + w111 121 I31

If we write (398) as 1112 122 I32 and

w=w1i

113 123 I3

Ho' = H,i + H2J + H3k

+ w2j + w3k, then (367) may be written3

H, = I, I,-way j= 1, 2, 3 (402)aa1

which is equivalent to the matrix form (401).

Problems

1. Find the moments and products of inertia for a uniformcube, taking the cube edges as axes.

2. Show that the moment of inertia of a body about any lineis equal to its moment of inertia about a parallel line through thecenter of mass, plus the product of the total mass and the squareof the distance from the line to the center of mass.

3. Find the angular-momentum vector of a thin rectangularsheet rotating about one of its diagonals with constant angularspeed we.


4. If3 3

Hp = Ipawa, Hp = I Ipacoaa=1 a-1

3 3

lip = I ap'Ha, coo = apawa, 3 = 1, 2, 3a-1 a-1

for arbitrary wa, show that

3 3

1 Ipaaa = I Ia°apa, 6, a = 1, 2, 3a-1 amt

5. Let us consider the form

I = x2 + 9y2 + 18z2 - 2xy - 2xz + 18yz

We may write

I = (x2 - 2xy - 2xz) + 9y2 + 18yz + 18z2(x-y-z)2+8y2+16yz+1722(x y - z)2 + 8(y + z)2 + 9z2X2 +Y2 + Z2

where X = x - y - z, Y = V"8- (y + z), Z = 3z, a set of lineartransformations from x, y, z to X, Y, Z.

This method may be employed to reduce any quadratic formto normal form. However, the linear transformations may notbe a rotation of axes. Reduce I to normal form by a rotation ofaxes.

CHAPTER 7

HYDRODYNAMICS AND ELASTICITY

108. Pressure. The science of hydrodynamics deals with themotion of fluids. We shall be interested in liquids and gases, aliquid or gas being defined as a collection of molecules, which,when studied macroscopically, appear to be continuous in struc-

z ture. A liquid differs from a

FIG.

solid in that the liquid willyield to any shearing stress,however small, if the stressis continued long enough.

95.

All liquids are compressibleto a slight extent, but formany purposes it is simplerto consider the liquid as beingincompressible. We shallalso be highly interested inperfect fluids. These are

liquids which possess no shearing stresses.We now show that the pressure is the same in all directions

for a perfect fluid. Let us consider the motion of the tetra-hedron ORST (see Fig. 95). The face ORT has a force actingon it, since it is in contact with other parts of the liquid. Underthe above assumption, this force acts normal to the face. Call itAf,,. If we divide Af" by the area of the face ORT, AA,,, we

obtain the pressure on this face, P,, = f" The limit of thisAAy

quotient is called the pressure in the direction normal to the faceORT. The y component of the pressure on the face RST isP cos l4. Let f" be the y component of the external force perunit volume, and let p be the density of the fluid. The equationof motion in the y direction is given by

+f"oT=d

p A r y (403)= zdt°

230

SEC. 109] HYDRODYNAMICS AND ELASTICITY 231

since (p Ar) =dm

= 0. Now AA = AA,. cos 0, so that (403)dt dt

becomes

(Pv - Pn) + fv A oA dt Cp

Oz

i(404)

v

As A --- 0, we have 0, so that if we assume f,,, d pAA dt2finite, we must have P = P.. Similarly, P = P. = P. = p.Since the normal n for the tetrahedron can be chosen arbitrarily,the pressure is the same in all directions and p is a point function,p = p(x, y, z, t). We leave it to the student to prove that at theboundary of two perfect fluids the pressure is continuous.

109. The Equation of Continuity. Consider a surface S bound-ing a simply connected region lying entirely inside the liquid.Let p be the density of the fluid, so that the total mass of the fluidinside S is given by

M = ff p(x,y,z,t)d-rR

Differentiating with respect to time and remembering that x, y, zare variables of integration, we obtain

Mdd = fJ

fat dr (405)

Now there are only three ways in which the mass of the fluidinside S can change: (1) fluid may be entering or leaving thesurface. The contribution due to this effect is JJ vp dd.

s(2) matter may be created (source), or (3) matter may bedestroyed (sink). Let 4,(x, y, z, t) be the amount of mattercreated or destroyed per unit volume. For a source, 4, > 0, andfor a sink, 0 < 0. The net gain of fluid is therefore

ffJ1dT_ jfpv.dd (406)

Equating (405) and (406) and applying the divergence theorem,we obtain


ap + V (Pv) = #(x, y, z, t) (407)

This is the equation of continuity. For no source and sink,(407) reduces to

at + V - (Pv) = 0 (408)

If furthermore the liquid is incompressible, p = constant,aP

= 0, and (408) becomesat

V v = 0 (409)

If the motion is irrotational, that is, if f v dr = 0, thenV = so that the equation of continuity for an incompressiblefluid possessing no sources and sinks and having irrotationalmotion is given by

V2V = 0 (410)

We call p the velocity potential. We solve Laplace's equationfor rp, then compute the velocity from v = Vrp.

Problems

1. If the velocity of a fluid is radial, u = u(r, t), show that theequation of continuity is

ap LP+

P a

at + u ar r2 ar (r2u)

Solve this equation for an incompressible fluid, if '(r, t) = 1/r2.z - z

2. Showthaty=_2xyz i+ x 1<)zj+ kiss

(x2 + y2)2 (x2 + y2)2 x2 + y2

possible motion for an incompressible perfect fluid. Is thismotion irrotational?

3. Prove that, if the normal velocity is zero at every pointof the boundary of a liquid occupying a simply connected region,and moving irrotationally, rp is constant throughout the interiorof that region.

Sic. 110] HYDRODYNAMICS AND ELASTICITY 233

4. Prove that if v is constant over the boundary of any simplyconnected region, then (p has the same constant value throughoutthe interior.

5. Express (407) in cylindrical coordinates, spherical coordi-nates, rectangular coordinates.

110. Equations of Motion for a Perfect Fluid. Let us considerthe motion of a fluid inside a simply connected region of volumeV and boundary S. The forces acting on this volume are

(1) external forces (gravity, etc.), say, f per unit mass; (2)pressure thrust on the surface, - p dd, since dd points outward.The total force acting on V is

F = fff pfdr - f f pdd = f f f (pf - Vp)dr

The linear momentum of V is

M = J117 pvdr

and the time rate of change of linear momentum is

dMd fffpvdr

= I f pdT + f f f v(pdr)tfsince the volume V changes with time. However, p dr is themass of the volume dT, and this remains constant throughout

the motion, so thatd

(p dT) = 0. Since F = dd , we obtain

fff(pf-vp)drfffpdr

This equation is true for all V, so that

Ipf - VP=p dvdor

dt f - pVP (411)

This is Euler's equation of motion.

234 VECTOR AND TENSOR ANALYSIS SEC. 111

dvavFrom (76) we have that

dt at + (v V)v, so that an alterna-

tive form of (411) is

Vp (412)

Also from Eq. (9) of Sec. 22, Vv2 = 2v x (V x v) + 2(v - V)v, sothat (412) becomes

av 1

at+1VV2-Vx(VxV)=f-PVp (413)

2

111. Equations of Motion for an Incompressible Fluid underthe Action of a Conservative Field. If the external field is con-servative, f = -Vx, so that f - (1/p) Vp = -V[x + (p/p)] ifp = constan

(IV

t. Hence (413) becomes

\-

-vx(Vxv)=-V(x+P+2v2) (414)

We consider two special cases:(a) Irrotational motion. v = V\p and V x v = 0, so that (414)

becomes at = -v l x + +P

2 v2).\

(b) Steady motion. a = 0, so that (414) becomes

l 1vX(Vxv)=VlX++1v2P

For this case we immediately have that

v.[V(x+p+1v2)]=0

p 2

Hence V[x + (p/p) + +v2] is normal everywhere to the velocityfield v. Thus v is parallel to the surface X + (p/p) + j v2 =constant. The curve drawn in the fluid so that its tangentsare parallel to the velocity vectors at corresponding points iscalled a streamline. We have proved that for an incompressibleperfect fluid, which moves under the action of conservative

SEC. 111] HYDRODYNAMICS AND ELASTICITY 235

forces and whose motion is steady, the expression X + (p/P) +-v2 rernains constant along a streamline. This is the generalform of Bernoulli's theorem. If X remains essentially constant,then an increase of velocity demands a decrease of pressure, andconversely.

Problems

1. If the motion of a perfect incompressible fluid is both steadyand irrotational, show that x + (p/p) + V2 = constant.

2. If the fluid is at rest, dt = 0. Show that V x (pf) = 0,

and hence that f V x f = 0. This is a necessary condition forequilibrium of a fluid. Why must pf be the gradient of a scalarif equilibrium is to be possible?

3. If a liquid rotates like a rigid body with constant angularvelocity w = wk and if gravity is the only external force, provethat p/p = 1w2r2 - gz + constant, where r is the distance fromthe z axis.

4. Write (411) in rectangular, cylindrical, and sphericalcoordinates.

5. A liquid is in equilibrium under the action of an externalforce f = (y + z)i + (z + x)j + (x + y)k. Find the surfaces ofequal pressure.

6. If the motion of the fluid is referred to a moving frame ofreference which rotates with angular velocity w and has transla-tional velocity u, show that the equation of motion is

1 du dw Dr D2rf--Vp=dt+dt Xr+wx (wxr)+2wxdt+P dt2

and that the equation of continuity is

rat (P Rdt-)

7. The energy equation. For a simply connected region R withboundary S, the kinetic energy of R is

T=ifffpv2dr

R

Let the surface S move so that it always contains all the originalmass of R. Show that


dT = ffJ -dv2

dt R dt

= If! v.f!Vpdr

= f f f v.fdr- f f f f f pdt(dr) (415)R S R

Analyze each term of (415).8. For irrotational flow show that

at =- (x + P + 2 v2) + C(t),

z

and if p =p(p),

t-f- 2 + x + f-D(t).

112. The General Motion of a Fluid. Let us consider thevelocities of the particles occupying an element of volume of a

fluid. Let P be a point of thevolume or region, and let vprepresent the velocity of thefluid at P (Fig. 96). The veloc-ity at a nearby point Q is

vQ = vp + dvp= Vp + (dr . V)vp (416)

from (75). By (dr V)vp wemean that after differentiation,the partial derivatives of v arecalculated at P. We now re-place dr by r for convenience,so that r = xi .+ } + zk if wey

Fio. 96.consider P as the origin and x,

y, z large in comparison with x2, y2, z2, zy, etc. Equation (416)now becomes vQ = vp + (r V)vp. Now

r x (V x w) + (417)

from (9), (10), (12) of Sec. 22.Now let

w = (r V)vp = x av + y av + z avl (418)alp ay p azlp

SEC:. 1121 HYDRODYNAMICS AND ELASTICITY 237

and hence

ax av--ax ax

ay avP+yayay

az avI

+zP azazP=w

We did not differentiate the (' I IV' since they haveaxP ayP P

been evaluated at P and so are constants for the moment. Thus,using (417), we obtain

Moreover, v(418)], and

w = j V(r w) + J(V x w) x r (419)

=Vp+w, so thatV xv =V xw = (Vxv)p [see

VQ = VP+ -(V x v)P (420)

It is easy to verify that r w is a quadratic form, that is,

Axe+Bye+Cz2+2Dyz+2Ezx+2Fxy

and so by a rotation (Sec. 107), we can write

and

IV(r w) = axi + byj + czk

We may now write (420) as

VQ = VP + w x r + (axi + byj + czk) (421)

where w = J(V x v)p.Let us analyze (421), which states that the velocity of Q is the

sum of three parts:1. The velocity vp of P, which corresponds to a translation of

the element.2. w x r represents the velocity due to a rotation about a line

through P with angular velocity J(V x v) p.3. axi + byj + czk represents a velocity relative to P with

components ax, by, cz, respectively, along the x, y, z axes.The first two are rigid-body motions; they could still take place

if the fluid were a solid. The third term shows that particles at

238 VECTOR AND TENSOR ANALYSIS (SEc. 113

different distances from P move at different rates relative to P. Ifwe consider a sphere surrounding P, the spherical element is trans-lated, rotated, and stretched in the directions of the principalaxes by amounts proportional to a, b, c. Hence the sphere isdeformed into an ellipsoid. This third motion is called a purestrain and takes place only when a substance is deformable.Each point of the fluid will have the three principal directionsassociated with it. Unfortunately, these directions are not thesame at all points, so that no single coordinate system will sufficefor the complete fluid.

The most general motion of a fluid is that described above andis independent of the coordinate system used to describe themotion. It is therefore an intrinsic property of the fluid.

113. Vortex Motion. If at each point of a curve the tangentvector is parallel to the vector w = J(V x v), we say that the

curve is a vortex line. This implies thatdxdx =

dy=

dzwhere

wz wy w,dx, dy, dz are the components of the tangent vector and

w=wi+wyJ+w L

The integration of this system of differential equations yieldsthe vortex lines. The vortex lines may change as time goes on,since, in general, w will depend on the time.

Let us now calculate the circulation around any closed curvein the fluid.

C = ff (V xv) dd (422)r s

If V x v = 0, then C = 0. This is true while we keep the curver fixed in space. Let us now find out how the circulationchanges with time if we let the particles which comprise P moveaccording to the motion of the fluid. As time goes on, assumingcontinuity of flow, the closed curve will remain closed.Now

e = (423}r' r

where s is are length along the particular curve v, at some time t.At an instant later the curve t' has moved to a new position givenby the curve r". The velocity of the particles over this path is

SFC. 1141 HYDRODYNAMICS AND ELASTICITY 239

slightly different from that over r, and, moreover, the unit

tangents ds have changed. The parameter s is still a variable of

integration and has nothing to do with the time. Therefore

dt dt dsds + $6 v

d

Pdslds

dt

= T dt . dsEr- ds + v . ds \dt/

ds (424)

Euler's equation of motion (411) for a conservative field,

f = -VX

isdv = -VX - 1

dtVp = -VV, where V = x + f dp/p. There-

1

fore2

dt2I

_ --0d(V-J2) =0 (425)

We have arrived at a theorem by Lord Kelvin that the circu-lation around a closed curve composed of a given set of particlesremains constant if the field is con-servative, provided that the densityp is a function only of the pressure p.

If we now consider a closed curvelying on a tube made up of vortexlines, but not encircling the tube (seeFig. 97), then

C =

ff8

i Frdd i l t Vs norma o x v.s nce om Fio. 97.Kelvin's theorem, C = 0 for all time,so that the curve r always lies on the vortex tube.

114. ApplicationsExample 109. Let us consider the steady irrotational motion

of an incompressible fluid when a sphere moves through the fluid

240 VECTOR AND TENSOR ANALYSIS ISEC. 114

with constant velocity. Let the center of the sphere travel alongthe z axis with velocity vo. We choose the center of the sphereas the origin of our coordinate system. From Sec. 110, Prob. 6,we have

f - pop=at

and

0

HenceDr

= so that V2p = 0. Now at points on the surface

of the sphere we must have CYT)radWLY = 0, so that (a?Y-a = 0.Or

We look for a solution of Laplace's equation satisfying thisboundary condition, so that we try

'P = (Ar + !cos 8r/J

(see Sec. 67). We need

cosB=0-CA-

a3

B

(426)

so that B = a2A/2. Moreover, at infinity we expect the veloc-ity of the fluid to be zero, so that the velocity relative to thesphere should be -vo. Hence

a-= A = vVs - o

az z .and

s

P = -vo (r + 2r2 cos 0 (427)

The velocity of the fluid relative to the sphere is given by v = Vspand the velocity of the fluid is v = Vp + vok.

Example 110. Let us consider a fluid resting on a horizontalsurface (x-y plane) and take z vertical. Let us assume a trans-verse wave traveling in the x direction. For an incompressible

Sec. 114] HYDRODYNAMICS AND ELASTICITY

fluidz

2°2`P-ax+az?

0

241

(428)

We assume a solution of the form p = A (z)eaLT(X-'i)t.

Sub-stituting into (428), we obtain

2wtZ_voi f 41x2 d2Ae - 2 A (z) +

dx2 I = 0

so thatd2A _ 41r2

Adz2 \2 (429)

The solution to (429) is A = Ape<zr° + Boe (2r/1`)', and a realsolution to (428) is

(Aoe(2r/X)z + Boe-(zrn*)z) cosL

(x - vt)] (430)

The fluid has no vertical velocity at the bottom of the plane on

which it rests, so that v. =azL(P = 0 at z = 0. This yields

A 0 = B0, so that

= Ao(e(2r')= + ec2r/l.>=) cosI2r

(x - Vi)J

= 20 cosh (-- z) cos[xr (x - vt)] (431)

From Prob. 8, Sec. 110, we have

at -(x++2v2)+C(t)and for a gravitational potential, x = gz, so that

at= - (gz +

p+

2v2) + C(t) (432)

We now assume that the waves are restricted to small ampli-tudes and velocities, so that we neglect Jv2. Moreover, at the


surface, p, the atmospheric pressure, is essentially constant, so

that dp = 0. Differentiating (432), we obtain

a2( _ az dCate .9 at + dl

(433)

and again at the surface at = vZ = az , so that (433) becomes

a2V a_p dCate g az + dt

Substituting (431) into (434), we obtain

(434)

2

-v2 a2 A0 cosh zcosLA (x - Vt)

g 2 w -z cos

I2r (x - vt) ] + dt (435)

In order for C to be dependent only on t, we must have thecoefficient of cos [(27r/A) (x - vt) ]identically zero in (435). Hence

227r2 27r g 27rAo (- v cosh z + sink z = 0 (435a)

or

v2 = g tanh - z

In deep water z/X is large so that tank z 1, and the

velocity of the wave is v = (Ag/2,r)*.

Problems

1. Show that for steady motion of an incompressible fluidunder the action of conservative forces, (v O)w - (w V)v = 0,wherew = D x v.

2. Show that dt (Pl = \P of v for a conservative system.

SEc. 1151 HYDRODYNAMICS AND ELASTICITY 243

3. If C is the circulation around any closed circuit moving

with the fluid, prove that dC = p d (1) if the field is con-

servative and if the pressure depends only on the density.4. Show that v = 2axyi + a(x2 - y2)j is a possible velocity

of an incompressible fluid.5. Verify that the velocity potential (p = A[r + (a2/r)] cos 0

represents a stream motion past a fixed circular cylinder.115. Small Displacements. Strain Tensor. In the absence

of external forces, a solid body remains in equilibrium and theforces between the various particles of the solid are in equilibriumbecause of the configuration of the particles. If external forcesare added, the particles (atoms, molecules) tend to redistributethemselves so that equilibrium will occur again. Here we areinterested in the kinematic rela-tionship between the old positions p° r pof equilibrium and the new. We so Isshall assume that the deforma- po r ptions are small and continuous. Fia 98. .We expect, from Sec. 112, thatin the neighborhood of a given point Po, the remaining points willbe rotated about Po and will suffer a pure strain relative to Po.Let r be the position vector of P relative to Po, and let s be thedisplacement vector suffered by P, and so the displacement suf-fered by Po (Fig. 98). Then

S = so + ds = so + (r V)so (436)

Let s = u(x, y, z, t)i + v(x, y, z, t)j + w(x, y, z, t)k. Since wewill be dealing with static conditions,

s = u(x, y, z)i + v(x, y, z)j + w(x, y, z)k

From (420),S = so+J(V xs)P, (437)

whereasw=x-ax

as+yay

PO

as+z-Pc oz PO

since s = v At.We are interested in the position of P after the deformation

(now P') relative to the new position of Po (now Po ). This is


the vector r' = r `}- s - so, or

r' = r+4(D xs)po x r + (438)

Since J(V x s)P, x r represents a rigid-body rotation about P0,we ignore this nondeformation term and so are interested inr + 'I'p(r w). Now

1

D Cxauj a-l aw+2 tax +xyax +xz atPo P. 0

I r au av

+ 2D xy

y Po + y2 ay

+ Dazl a+ zyzx2 Po

and

awl+ yz -

PO ay Po

awP. az Po

r+ 1

aul _y au avl x (aw aux Ci + ax + 2 Cay + ax! + 2 \ax + az J ix (au av av _z (aw 8v

+yCl+ay 2y+x (aw au y aw av aw 1

az)+2Cay+az az JkThe partial derivatives are evaluated at the point Po.

Let us now consider the matrix

lIsr'II =

au 1 au av

i + ax 2 ay + Ox

2 Cay + ax/ + yi aw aul i law av

2 C 8x + azl 2 ày + az

(439)

(440)

The nine components of this matrix form the strain tensor. Ifwe write r = x'i + x2j + xak and r' = y'i + y2j + yak (see

SEC. 115) HYDRODYNAMICS AND ELASTICITY 245

Example 8), then r' = r + I V(r w) may be written

yi = sfix' + S2 'X2 + 83ix3, i = 1, 2,3or

3

y' = Sa'xa (441)a-1

We shall see in Chap. 8 that since r and r' are vectors, then, ofnecessity, the s/ are the components of a tensor. Notice thatSii = s; , so that the tensor is symmetric.

The ellipsoid which has the equation

( aul ( avl (awl (au avl1 + ax)xr+ 1 +a y2+ 1 + az)z2+ a +a xy

y yaw av aw au

+ (ay + az yz + (ax + azzx = 1 (442)

is called the strain ellipsoid. From Sec. 107 we know that wecan reduce the ellipsoid to the form

Ax" + By'2 + Cz'z = 1

by a proper rotation. The strain tensor becomes entirelydiagonal,

A 0 00 B 0

0 0 C

In the directions of the new x', y', z' axes, the deformation is apure translation, and these directions are called the principaldirections of the strain ellipsoid.

Let us now compute the change in the unit vectors, neglectingthe rotation term. The unit vector i has the components(1, 0, 0), so that from (438) and (439)

( au 1 au av 1 aw a_uli - r1 = \1 + ax) 1 + 2 (ay + ax 1 + 2 ax + az/ k

By neglecting higher terms such as(LU)

21 u - we havey

_ I au I avlIril 1 +

ax. Similarly j --> r2, and lr2l = 1

+ay

; k - rs,


and fr31 = 1 +awl

. The angle between r1 and r2 is given byI

au avcos B =

+r,jjr2+ v ay + ax'.

The terms of the strain tensor are now fully understood. Thevolume of the parallelepiped formed by r1, r2, r3 is

au av awV r2xrs = 1+ + +ax ay az

so that _V V _ au av aw

V ax+8y +az (443)

The left-hand side of (443) is independent of the coordinatesystem, so that V s is an invariant.

Finally, we see that the deformation tensor due to the tensor- V (r w) has the components

ax 2\ay+8x/ 2(ax+ az)1 au 8v'\ av 1 au av

2 (ay + ax/ ay 2 (ay + az1 aw au 1 (au av aw

2 \ ax + az 2 \ay + az az

2 (axl + axi/1I

aui

whereu1 =u, x1=xu2=v, x2=yu3=w, x3=z

(444)

116. The Stress Tensor. Corresponding to any strain in thebody must be an impressed force which produces this strain.Let us consider a cube with faces perpendicular to the coordinateaxes. In Sec. 108 we assumed no shearing stresses, but now weconsider all forces possible between two neighboring surfaces.

SF:C. 116] HYDRODYNAMICS AND ELASTICITY 247

Let us consider the face ABCD (Fig. 99). It is in immediatecontact with other particles of the body. As a consequence, theresultant force tx on the face ABCD can he decomposed into threeforces: txx, tyx, tzx, where txx is the component of tx in the x direc-tion, t,, is the component of tx in the y direction, and tzz is the

z

Fia. 99.

y

component of tt in the z direction. We have similar results forthe other two faces and so obtain the matrix

tzz tzy tzz

tyz tyy tyz

tzz tzy G.

(445)

These are the components of the stress tensor.By considering a tetrahedron as in Sec. 108, we immediately

see that if dd is the vectoral area of the slant face, then the com-ponents of the force f on this face are

f, = tzx dsz + tzy d s + tzz dszfy = tyx dsz + tyy dsy + tyz dszfz = tzz ds, + tzy dsy + tzz dsz

(446)

where dsz = i dd, dsy = j dd, dsz = k dd.


We immediately see that

tyx =Of.,

tx fx,as., asv

[SEC. 117

3

and that fi = I tia dsa, where f, = f,, f2 = f,,, f3 = f., t12 = t,,,a=1

. . . . We shall see laterdv

z

that this explains why the ti; arecalled the components of atensor.

Let us now consider theresultant force acting on avolume V with boundary 8(see Fig. 100). Wehavefrom(446)

fz = txx dsx + txv ds + tzx dst

so that

F. = fz _ JJtxzdSxY

+ t,, ds, + tx, dss

Fio. 100.

Applying the divergence theorem, we obtain

f r acZy4-9t"

acz=Fz - JI J ax + ay + az

dTV

with similar expressions for F,,, F.

s= ! J t.dd

where t = tz, + t,, k.

(447)

By letting V --p 0, we have that the x component of the forcetz

ax + qty" +per unit volume must be V -y

117. Relationship between the Strain and Stress Tensors. Inthe neighborhood of a point P in our region, let us choose thethree principal directions of the stress tensor for the axes of ourcartesian coordinate system. If we assume that the region isisotropic (only contractions and extensions exist), a cube withfaces normal to the principal directions will suffer distortions onlyalong the principal axes. Hence the principal directions of the

SEC. 1171 HYDRODYNAMICS AND ELASTICITY 249

strain ellipsoid will coincide with those of the stress ellipsoid. Inthis coordinate system

el 0 0 It, 0 0

ie17JJ = 00

000 t20 0 (448)

0

Our fundamental postulate relating the shear components withthose of the stress will be Hooke's law, which states that everytension produces an extension in the direction of the tension andis proportional to it. We let E (Young's modulus) be the factorof proportionality. Experiments also show that extensions infibers produce transverse contractions. The constant for thisphenomenon is called Poisson's ratio a. We thus obtain for therelative elongations of the cube in the three principal directionsthe following:

1 a 1 -SETel = ;ti --(t2+t3) _

e2

E E Eor 0,

t2 - (t3 + tl) = 1 r t2 -

1 a 1-r-a-e3 =

E3 - - (11 + t2) t3 "-

E(t1+t2+t3)

E (t1 + t2 + t3) (449)

(t1 + t2 + t3)

The formulas for el, e2, e3 apply only in the immediate neighbor-hood of a point P. Since points far removed from P will havedifferent stress ellipsoids, the principal directions will vary frompoint to point. Hence no single coordinate system will existthat would enable the stress and strain components to be relatedby the simple law of (449). Let us therefore transform thecomponents of the stress and strain tensors so that they may bereferred to a single coordinate system. The reader should readChap. 8 to understand what follows. If he desires not to breakthe continuity of the present paragraph, he may take formula(456) with a grain of salt, at least for the present. Example 8,Probs. 21 and 22 of Sec. 11, and Prob. 21 of Sec. 15 will aid thereader in what follows.

If x1, x2, x3 are the coordinates above, and if we change to anew coordinate system 11'. T2, 23 where

3

x' = I a'xx, i = 1, 2, 3 (450)a-1


then the transformation (450) is said to be linear. Notice thatthe origin (0, 0, 0) remains invariant.. If, furthermore, we desiredistance to be preserved, we must have

3 3

(.Ti), =(xi)

In Chap. 8 we shall easily show that this requires3 =0ifij

a.,aaia = S=i (451)a=1 = 1if ij

Equation (451) is the requirement that (450) be a rotation of axes.Moreover, since we are dealing with tensors, we shall see that thecomponents of the strain tensor in the x'-x2-x3 coordinate systemare related to the components in the x'-a 2-x3 system by thefollowing rule:

3 3

e;i = I I a,aa1 ea# (452)'6=1 a=1

If we now let i = j and sum on i, we obtain`3 3 3 3

L, ei, = I I Z aiaMeag:=1 i=1 0=1 a=1

3 3 3

Saisea# = I eaa

so that0-1 a-1 a-1

e11 + 922+933 = e11 + e22 + e33 = el + e2 + e3 (453)

This is an invariant obtained from the strain tensor [see (443)].A similar expression is obtained for the stress tensor; namely,that

111+122+133 = t11 +t22+133 = t1+t2+t3

Equations (449) may now be written as

1+0el - E t1+e2=1

e3=1Eor

ts+

(454)

SEC. 117] HYDRODYN.4.1IICS AND ELASTICITY

where 4, is the invariant

E (tl + t2 + t3) - (III + 122 + 133)

From Eq. (452) we have

and since eap

and

3 3

1 aiaaJaeaa = I aiaajaeaa=1 a=1

3

aiaaja (1 E to +)-l

a-1 a=1

1 + oeij = E of + 4,5ij

3 3 3

since 1j = I I aiaajalas = I aiaaala.6=1 a -l a=1

251

(455)

Equation (455) is the relationship between the components ofthe strain and stress tensors when referred to a single coordinatesystem. We have

ell = 1

E

Q111- E (tll + 122 + 133)

1 _ oT

922 =+E

Q122 - - (111 + t22 + 138)

e33 = 1 +O

133 - E, (111 + E22 + 133)

1 _912 = E

+ v112

1+0.l23923 = E

1 +Q_981 = E 131

3 3

eij = E aiaa,8easB=1 a-1

= 0 unless a = 6 [see (448)], we obtain

3 31+ aF, I aiaajala + Y' 1 aiaaja

a

(456)


Solving Eels. (456) for the t;; and removing the bars,

t11 =

we obtain

E au o (au av aw=i+o, axl-2vax+ay+az J

=E rav v au av awt22

1+aLay1-2v (ax+ay+az)]=

E taw vtss 1+aaz+1-2QV*s

E E (au avt12=t21= 2(1+o) ay+ax

E av awt23 = t32 = 2(1 + v) \az + ay

=E (aw au

t31 - t13 = 2(1+Q)ax +az

01

1 + [ell + 1 2- (ell + e22 + e33)

(457)

Equation (447) now becomes

E . (a2u a2v a2wF=

-}- v

[d?uax2 + 1 - 20 \49x2 + ax ay + ax az

E a2u a2v a2w a2u

+ 2(1 + u) aye ay ax + az ax + az2 1

_ E 1 a (au av aw-2(1+a)IV

zu+1 -2aaxax+ay+ az J

[vu+2(1+o) 1 1The forces per unit volume in the y and z directions are

Fy 2(1+a)[Dzv+1 12vay(V - s)IE {V2W+ 1 aI-2o-

az (0 s)

so that

f =2(1

+r)

[V2S+ 1 1 2cr

V(V s)] (458)

If we let R = RJ + R j + R2k be the external body force perunit volume, p the density of the medium, then Newton's second

SEc. 117] HYDRODYNAMICS AND ELASTICITY 253

law of motion yields

R+2(1

+o,)[V2S+-Lv(V.S)] =P

For the case R = 0, (459) reduces to

E 2

2(1 + a) [v2s + 1 -2a

°(° s) = p at2 (460)

In Sec. 70 we saw that a vector could be written as the sum ofa solenoidal and an irrotational vector. Let s= s1 + s2j whereV s, = 0 and V x s2 = 0. Since (460) is linear in s, we canconsider it as satisfied by s, and s2. This yields

E _ a2S,

2(1 + a)V2s1 - p

ate

and (461)E 1

°(° . s2)a2s2

2(1 +a,) [V2s2 + 1 -2a = p ate

However,

so thatV X (V X 62) = V(V S2) - V 'S2 = 0

E(1 T a)V2S2 = p - (462)(1 +a)(1 - 2a) at2

In Sec. 80, we saw that (461) leads to a transverse wave mov-ing with speed Vt = E/2(1 + a)p.

Equation (462) is also a wave equation, but the wave is nottransverse. Let us assume that the wave is traveling along thex axis. Then

s2=s2(X-Vt)= u(x - Vt)i + v(x - Vt)j + w(x - Vt)k

V X S2 = =0

a2s2

axJ+-k=0

clx


so that w and v are independent of x and therefore are independentof x - Vt. We are not interested in constant displacements, sothat s2 = u(x - Vt)i, and the di,,placement of s-2 is parallel tothe direction of propagation of the wave. The wave is thereforelongitudinal. The speed of the wave is

L'(1 - o.)V,=P(1 + v)(1 2v)

In general, both types of waves are produced, this result beinguseful in the study of earthquakes.

Problems1. Derive (451).2. If P, f--, f3 are the components of a vector for a cartesian

coordinate system, prove that the components P, P, j'3 of thisvector in a new cartesian coordinate system are related to the old

3

components by the rule J' = i = 1, 2, 3, using then=i

coordinate transformation (450).3. If the body forces are negligible and if the medium is in a

state of equilibrium, show that V2S +11 - V(V s) = 0.

4. If the strain of Prob. 3 is radial, that is, if s = s(r)r, find thedifferential equation satisfied by s(r).

5. Assuming a = 0 for a long thin bar, find the velocity ofpropagation of the longitudinal waves.

6. If µ = E/2(1 + v) (modulus of rigidity) and

A= Ea

(1 + a) (1 - 2Q)

show that Eq. (459) becomes's49R+AVss+ Pats

x x

7. Why do we use ats instead of in in (459)?

8. A coaxial cable is made by filling the space between a solidcore of radius a and a concentric cylindrical shell of internalradius b with rubber. If the core is displaced a small distance

Ssc. 118] HYDRODYNAMICS AND ELASTICITY 255

axially, find the displacement in the rubber. Assume that endeffects, gravity, and the distortion of the metal can be neglected.

118. Navier-Stokes Equation. We are now in a position toderive the equations of motion of a viscous fluid. In the case ofnonviscous fluids, we assumed no friction between adjacent layersof fluid. As a result of friction (viscosity), rapidly moving layerstend to drag along the slower layers of fluid, and, conversely, theslower layers tend to retard the motion of the faster layers. Itis found by experiment that the force of viscosity is directlyproportional to the common area A of the two layers and to thegradient of the velocity normal to the flow. If the fluid is mov-ing in the x-y plane with speed v, then the viscous force is

since av is the gradient of the speed normal to the direction of flow.

17 is called the coefficient of viscosity.We shall let Pt; be the stress tensor and u;; the strain tensor

for the fluid analogous to t; and e;; of the previous paragraphs.We have

_ E au Ov

P12 2(1 + v) ay + ax

where u, v, w are the components of the velocity vector v (seeSec. 112). For a fluid moving in the y direction with a gradient

in the x direction, we have u = 0 and au = 0, so thatay

_ E av _ avP12

2(1 + v) ax = '' ax

must be replaced by q.Hence the term2(1

Eor)

In addition to the stress components due to viscosity, we mustadd the stress components due to the pressure field, which weassume to be

I

-p 0 00 -p 0

0 0 -p


The equations of (457) become

P;i = 2rio;i + X(911 + 022 + r33)aii - pa+i (463)

where X is undetermined as yet. Now let i = j and sum on j.We see that

P11 + P22 + P33 = (2rj + 3X)(ail + 022 + 033) - 3p

We know that P11 + P22 + P33 is an invariant and that for thestatic case P11 + P22 + P33 = -3p. Consequently we choose2, + 3X = 0, so that (463) becomes

p;, = P;, = 2+10;1 - I V(0-11 + 022 + 033) - pa+i (464)

for small velocities. Moreover, the velocity vector is given by

v = u11 + 1627 + u3k

and 0;i =1

2 C

au;axi +

auiax:

' so that div v = V v = 011 + 022 + 033-

To obtain the equations of motion, we note that from (447)

ap11+

49p12+

49p13fl - axiax, axe

3

c3 apli= Gj=1 ax

3

I apt' Henceand in general f;

pF; + f1 = pdu;dt

-i

x1 = xwhere x2 = y

x3=z

becomes

pF. +3

ap;i du:

j =1axi P dt (465)

where F; is the external force per unit mass. From (464)

ap;i

axi

a0;i 2 a(diy v) ap17= 2,i axi - 3 axi aì - axi a+i

_ a au; au) 2 a(V v) ap'q -a; -

= axi axi + ax'-

3+1 axi axi

SEC. 1181 HYDRODYNAMICS AND ELASTICITY 257

and

3 api; 3 a au; au; 2 a (V v) apax? + axi/ 3 ' ax,1-1

The equations of motion (465) are

dui 3 a aui au; 2 a(V v) app - = pFi + v + -- - - 7 - (466)

dt JI1 ax' Cax' ax' 3 axi axi

or

pat = pf +,q V2v - - V(V v) - Vp (467)

For an incompressible fluid V v = 0, and

pd

Along with (467) we have the equation of continuity

at =0

Problems

1. Derive (467) from (466).2. Consider the steady flow of an incompressible fluid through

a small cylindrical tube of radius a in a nonexternal field. Let

v = vk and show that p = p(z) and +1 V2v = ap Show that the

boundary conditions are v = 0 when r = a, and v = v(r),

r2 = x2 + y2, and that = r dr (r dr) Hence show that

v = (A/4i7)(r2 - a2), where A is a constant and LP = A.

3. Consider a sphere moving with constant velocity vok (alongthe z axis) in an infinite mass of incompressible fluid. Choosethe center of the sphere as the origin of our coordinate system.

Show that the equation of motion is p aE = q V2v - Vp and that

the boundary conditions are v = 0 for r = a, v = -vok at r = oo


and that for steady motion a = 0 for any quantity 4, associated

with the motion. Moreover V v = 0. We shall assume that vand the partial derivatives of v are small. Show that this implies

(i) Vp = I V2v

Hence prove that V2p = 0. Now let v = -V(p + wlk and showthat

awi=v4o

49Z

z

Assuming p = -,1 V2p and V2w1 = 0, show that (i) and (ii)will be satisfied.

Let w, = 3voa/2r, rp = coz - (voa3z/4r3) + (3voaz/4r),

P=and show that

3,gvoaz

2r3

V2w1 = 0, V2p = 0, p = -n V2rpv= -V(p+wik=0forr=av = -vok for r = co

4. Solve for the steady motion of an incompressible viscousfluid between two parallel plates, one of the plates fixed, theother moving at a constant velocity, the distance between theplates remaining constant.

5. Find the steady motion of an incompressible, viscous fluidsurrounding a sphere rotating about a diameter with constantangular velocity. No external forces exist.

CHAPTER 8

TENSOR ANALYSIS AND RIEMANNIAN GEOMETRY

119. Summation Notation. We shall be interested in sumsof the type

S = alx, + a2x2 + . . . + anxn (468)

We can shorten the writing of (468) and write

n

S = Z aixi (469)i=i

Now it will be much more convenient to replace the subscriptsof the quantities x:, x2, . . . , x by superscripts, x', x2, . . . , X11.

The superscripts do not stand for powers but are labels thatallow us to distinguish between the various x's. Our sum S nowbecomes

n

S = aixi (470)

We can get rid of the summation sign and write

S = aixi(471)

where the repeated index i is to be summed from I to n. Thisnotation is due to Einstein.

Whenever a letter appears once as a subscript and once as asuperscript, we shall mean that a summation is to occur on thisletter. If we are dealing with n dimensions, we shall sum from1 to n. The index of summation is a dummy index since thefinal result is independent of the letter used. We can writeS = a;xi = a;x' = a,,xa = as0.

Example 111. If f = f(x', x2, , xn), we have from thecalculus

259

0,60 VECTOR AND TENSOR ANALYSIS [SEC. 120

df = ax dxl + ax dx2 + . + - dx"CIx"

of_ dx'ax'

= of dxa

axa

;.nddf

afdJxadt - axa dt

Example 112. Let S = gasxax#. The index a occurs both asa subscript and superscript. Hence we first sum on a, say from1 to 3. This yields

S = 9iax'x8 + 92ax'x' + g38xaxe

Now each term of S has the repeated index # summed, say, fromI to 3. Hence

S = g11x'x1 + 912x'x2 + 913x'x3 + 921x2x' + g22x2x2 + 923X2x3+ 931x3x' + g32x3x2 + 933x8x3

and S = gaoxaxa represents the double sum3 3

S = Z I g-Oxaxs0-1 a-1

We also notice that the gap can be thought of as elements of asquare matrix

gii 912 913

921 922 923

931 932 933

120. The Kronecker Deltas. We define the Kronecker o tobe equal to zero if i - j and to equal one if i = j:

a;=0,ipj1,i=j

We notice that Si = 52 _ - = SA = 1,1 1 rb_ = 03 = = +1 = u

(472)

SEC. 120) TENSOR ANALYSIS 261

ax'If xI, x2, ... , x" are n independent variables, then = S,axl

i

for if axe = 1, and if i j, there is no change in the vari-

able x' if we change x1 since they are independent variables, so

that ax! = 0.ax1

cisExample 113. Let S = aaxa. Then

ax

= as ax , and

as

ax" "- = as ga

Now Sµ = 0 except when a = µ, so that on summing on a wea(axa)

obtain = a".ax"

Example 114. Let S = apxax# = 0 for all values of the vari-ables x', x2, . . . , x^. We show that a,, + a;; = 0. Firstdifferentiate S with respect to x' and obtain

as =aa$xa

Taxp

ax' ax' + a,axa

xa = 0axi

= aaflxaaf + aa$67x8 = 0= aaixa + a;se = 0

Now differentiate with respect to xi so that

028 - a.v + aiob = 0axe ax'

and

a,,+a;, 0

We define the generalized Kronecker delta as follows:The superscripts and subscripts can have any value from 1 to n.If at least two superscripts or at least two subscripts have thesame value, or if the subscripts are not the same set of numbersas the superscripts, then we define the generalized Kroneckerdelta to be zero. If all the superscripts and subscripts are sep-arately distinct, and the subscripts are the same set of numbersQs the superscripts, the delta has the value of + 1, or --1, accord-


ing to whether it requires an even or odd number of permutationsto arrange the superscripts in the same order as the subscripts.

For example, 6x23 = 1, 6213 - - 1, 6123123 = 1, 6163 = 0,

a1438 - -11 a123 = 0221

It is convenient to defineeilh ... i.. =

= 6312323 0, 61213 = 0

and (473)Eiji, ... i. =

Problems

1. Write in full aaxa=bi,a,i= 1, 2, 3.2. If aae7xax8xY = 0, show that ai;k + aki; + a; + aJik +

akii + aiki = 0. 3i,ti:'412 ...n i.,i,...i.3. Show that 612..,, i. = 6;,i:...;..

4. If yi = aaxa, zi = by, show that zi = bas;x#.5. Prove that

bas = are - a8rbaBratYaapi = arst + airs + astr - sari - aisr - arts

6. Show that the determinant I at at can be written

1a1

zat

and that

2 2a1 a2

a2'= Etiiaia? = etiia1

2a2

1atat a2 a3

2 z za1 a2 a3

8 8 $a1 a2 a$

= Eiika;a;ak = eiikai4a8

7. Prove that ei1i, ... ie = e'li2 ' ' i*.

8. Show that aijaxe ax axi

9. If yi = yi(xl, x2) ... , xe), i = 1, 2, . . . , n, show thatayi ax"

a; =axa

ai assuming the existence of the derivatives. Alsoy

i e

show that ayi-

as = a?. Show thatyi

a2yi axa ax¢ ayi 492xa

axa 8x8 ayi 8yk + Oxa ayi ayk- 0

SEC. 1211 TENSOR ANALYSIS

10. If = cp(x', x 2 . . . . . . n),

xi = xi(y', y2' . , yn) = xi(y)

i = 1,2, n, and if

<P = 0(y1, y2, . . . , yn) = p[xl(y), x2(y), . . . , xn(y)]a C) p

show that y = Given 9,, = , show that

arpi aoi acs axe, axil

ayayi - (ax# _axa) ay, ayl

263

121. Determinants. We define the determinant lal by theequation

at at ata1 a2 an2 2 2

a1 a2 an

i,a1a2 an

n n na1 a2 an

l all = Erik ... i"ai,a?2 . . . a"n (474)

The reader should note that this definition agrees with thedefinition for the special case of second- and third-order deter-minants which he has encountered in elementary algebra. Thedefinition of a determinant as given by (474) shows that it con-sists of a sum of terms, nn in number. Of these, n! are, in gen-eral, different from zero. Each term consists of a product ofelements, one element from each row and column. The signof the term a a'a . a depends on whether it takes an evenor odd permutation to regroup i1i2 in into 12 . . n.

Since i1 and i2 are dummy indices, we can interchange themso that

fail = Ei,i,... ina,'a2 . . . ann = ei,i, ... inal=a2 . . , ann

An interchange of the subscripts 1 and 2, however, will meanthat an extra permutation will be needed on i2iii3 in. Thischanges the sign of the determinant. Hence interchanging twocolumns (or rows) changes the sign of the determinant. As an


immediate corollary, we have Ia;I = 0 if two rows (or columns)are identical.

Let us now examine the sumbill:... i. = Ei,i, ... a. (475)

If jl, j2, . . . , jn take on the values 1, 2, . . . , n, respectively,we know that (475) reduces to (474). If the ji, j2, .. . , j,.take on the values 1, 2, . . . , n, but not respectively, then wehave interchanged the rows. An even permutation reduces (475)to Ia;I, and an odd permutation of the i's reduces (475) to -Ia 1.If two of the j's have the same value, (475) is zero, since if tworows of a determinant are alike it has zero value. Hence

Es, ...,.a`;<1 . .. a;- (475a)and i i

t,44 ... ai.=

ai Ehj, ... i. (475b)

Example 115. We now derive the law of multiplication fortwo determinants of the same order. We have

I ail IbilIasIEC,;,... i.bi'b2 bri

Ei,i, ... i.a is = . . . a.,b b2 . . . b%

= Ei,i, ... i.(a1j1,bi') (ai,bi,) .s, 1 s, 2 { ,. x )

I ciI

where

abi (476)

Example 116. We now derive an expansion of a determinantin terms of the cofactors of the elements. We have

I41 = Ei,i,...i.aia8 . . . an

a; (Ei,i,... j.a2 . . . a'*)

= aIAQ

where Ao = AQ is called the cofactor of a,.In general,

a'0-b5-Pa'-#-K +,. . . an)= apAP. (f4 not summed)

Hence

aiAa = IaIB; (477)

SEC. 121] TENSOR ANALYSIS

Also

265

aaA7 = IaI (477a)

Example 117. Let us consider the n linear equations

yi = aQxa, a, i = 1, 2, .. , n (478)

Multiplying by Af, we obtain

A°y' =a' Asxa

so that summing on i, we have from (477)

A"y' = Ialaxa = Ialxx

If IaI 0 0, then

A°y' _ y'(cofactor of a' in IaI)IaI - IaI (479)

Example 118. Let yi = y'(x', x2, . , x"), i = 1, 2, . . . , n.i

In the calculus it is shown that if ayi) 96 0 at a point P and if the

partial derivatives are continuous, we can solve for the x' interms of the y's, that is, xi = xi(y', y2, ... , y") in a neighbor-hood of P.

Now we have identically

ayi ay' axa8i

ayi axa ayi

Forming the determinant of both sides,

ayi axaI O;I - lax- ayi

from (476). Henceaxi

ayl

ay

ax

The determinant II is called the Jacobian of the y's with respect

to the x's. We have shown that


J \xl,y2,. . . ,yry/Jfy',X2, . . . xn) - 1

', x2, , xn 1, y2,y (480)

Example 119. If the elements of the (leterminant lal are func-tions of the variables x', x22, . . . , x", we leave it to the studentto prove that

alai _ Aa aao

axp0

axµ

ay i

As a special case, suppose a = , where y' = y'(x', x2,49x1

yi axaNow a` =

at ax- ay,

ytLet Yi = 491,

ax

(481)

, xn).

Let us consider j to be fixed for the moment.

a , Xi = ax -, so that Yi = aQXa. Ify

l alax11

0

then

from (479), or

8x8

and so1

(cofactor of a? in

Ya(cotactor of a#" in lal)

lal

a

S; cofactor of aye in

cix1

1

(cofactor of Lyi in

ax

ay

ax

ay

ax

ay

8x

ay 8x8

TX 8y1

Applying this to (481), we have

Sic. 121 7'EXSOR ANALYSIS

a

or

a,y

8x

axµ

laa loglaxI axa a2ya

ft" aya axµ axa

We shall make use of this result later.

Example 120. Let yi = y'(xl, x2, , xn),

wish to prove that

a2xµ

ay

ax

267

(482)

3-1 0. We

axa ax'6 axA a2yi

- ayi ayk ay` axa axaayk aye

ayi axaNow S' = - -, so that upon differentiating with respect toax- a yi

yk, we obtainayi a2xa axa a2yi axa

0 - axaayk aye + ay; axa axa ayk

(483)

1+

Multiplying both sides of (483) by ayi and summing on i, we

obtaina2xµ axa axa axµ &2yi

o = + Q.E.D.ayk ayay' ayk ayi axa axa

As a special case, if y = f (x),d = - (dx3d2ydy

2x

2 dy dx2

Problems

1. What is the cofactor of each term of

i z ial a2 a3

2 2 2a1 a2 a3

as a3 a3l 2 3

ay axa a2ya

ax, ays ax), axa

2. Prove (481), (477).3. If JAI is the derminant of the cofactors of the determinant

IaI, show that I:4I = IaJ"-'.


4. If a = a, show that A; = A. Is laI = jail in all cases?aaxa

5. Find an expression forayi ayi ayk

6. If z' = z'(y1, y2, ... , yn), y' = yt(x1, x2, . . . , x"), showthat J(z/y)J(y/x) = J(z/x).

7. If 9i; = gap a i -, show that I#I = IgI 2a2i

_ a2{ _ axaI _I

8. If u' = ua axa, Vi = V. -, show that uaVU = uaVa, whereclt.

xi = x{(x1 x2 xn).{ i

9. If u{ = uaax

, show that u{ = uaax

ax"axa

10. If gi; = gaaa

, show that gi; = 9a$ax- 00

82i a2i axi axi{ axi azi

11. If u{ = uaa , ui = ua , show that u{ = uaaxa axa axa

12. Apply (476) for the product of two third-order determi-nants.

13. If A; = B,-s ayi ax , show that A{ = B, and that IAI = IBI,

A;A; = B )$B,!.14. If X is a root of the equation Iai; -- Xbifl = 0, show that X

axa axais also a root of Id;; - Ail = 0 provided that B;; = aaa aji

axa axa ax!ii = baa a axi a

0.

122. Arithmetic, or Vector, n-space. In the vector analysisstudied in the previous chapters, we set up a coordinate systemwith three independent variables x, y, z. We chose three mutu-ally perpendicular vectors i, j, k, and all other vectors could bewritten as a linear combination of these three vectors. Anyvector could have been represented by the number triple (x, y, z),where we imply that (x, y, z) -= xi + yj + zk. The unit vectorscould have been represented by (1, 0, 0), (0, 1, 0), and (0, 0, 1).A system of mathematics could have been derived solely bydefining relationships and operations for these triplets, and weneed never have introduced a geometrical picture of a vector.For example, two triplets (a, b, c), (a, #, y) are defined to be equal

SEC. 1221 TENSOR ANALYSIS 269

if and only if a = a, b = P, c = y. We define the scalar productas (a, b, c) (a, f4, -y) = as + bfi + cy. The vector product, dif-ferentiation, etc., can easily be defined. Addition of triples isdefined by (a,b,c)+(a,#,y)=(a+a,b+$,c+y). If Ais a real number, then A(a, b, c) is defined as (Aa, Ab, Ac). Theset of all triples obeying the rules

(i) (a, b, c)+(a,0,y) _ (a+a,b+i,c+y) (484)(ii) A(a, b, c) = (Aa, Ab, Ac)

is called a three-dimensional vector space, or the arithmetic spaceof three dimensions.

It is easy to generalize all this to obtain the arithmetic n-space.Elements of this space are of the form (x', x2, . . . , x"), the x'taken as real. In particular, the unit or basic vectors are(1,0,0, . . . ) 0), (0,1,0, . . . ,0), ... , (0, 0, . . .

We shall designate V. as the arithmetic n-space.By a space of n dimensions we mean any set of objects which

can be put in one-to-one reciprocal correspondence with thearithmetic n-space. We call the correspondence a coordinatesystem. The one-to-one correspondence between the elementsor points of the n-space and the arithmetic n-space can be chosenin many ways, and, in general, the choice depends on the natureof the physical problem.

Let the point P correspond to the n-tuple (x', x2, . . . , x^).We now consider the n equations

y' = y'(x', x2, ... , x"), i = 1, 2, . . . , n (485)

and assume that we can solve for the x', so that

x' = x'(y', y2, ... , y"), i = 1, 2, . . , n (486)

We assume (485) and (486) are single-valued. It is at onceobvious that the point P can be put into correspondence withthe n-tuple (y', y2, ... , y"). The n-space of which P is anelement is also in one-to-one correspondence with the set of(y', y2, . .. , y"), so that we have a new coordinate system.The point P has not changed, but we have a new method forattaching numbers to the points. We call (485) a transforma-tion of coordinates.


123. Contravariant Vectors. We consider the arithmeticn-space and define a space curve in this V by

x` = x`(t), = 1, 2, .. n (487)at<_/iNote the immediate generalization from the space curve x = x(t),y = y(t), z = z(t): In our new notation x = x', y = x2, z = x3.

We We remember thatdxdt

,ddt'

dz

dtare the components of a tangent

vector to this curve. Generalizing, we define a tangent vectorto the space curve (487) as having the components

d.ridt 'j = 1, 2, . . . , n (488)

Now let us consider an allowable (one-to-one and single-valued)coordinate transformation, of the type (485). We immediatelyhave that

y' = y'(x1, x2, . . . , x") = y`tx'(t), x2(t), . . . , x"(t)1 = y'(t)

as the equation of our space curve for observers using the ycoordinate system. The components of a tangent vector to thesame space curve (remember the points of the curve have notchanged; only the labels attached to these points have changed)are given by

dyi

dti = 1, 2, . . . , n (489)

Certainly the x coordinate system is no more important than

the y coordinate system. We cannot say that dt is the tangents

vector any more than we can say dt is the tangent vector. If

we considered all allowable coordinate transformations, we wouldobtain the whole class of tangent elements, each element claimingto be the tangent vector for that particular coordinate system.It is the abstract collection of all these elements that is said tobe the tangent vector. We now ask what relationship existsbetween the components of the tangent vector in the x coordinatesystem and the components of the tangent vector in the y coordi-

SF(,. 123] TENSOI? ANALYSIS

nate system. We can easily answer this question, for

dys 0y1 dxa

dt axa dt

271

(490)

sdyWe also notice that

dx1d =ay-

dl We leave it as an exercise

that this result follows from (490) as well as from (486).We now make the following generalization: Any set of numbers

Ai(xe, x2, . . . , xn), i = 1, 2, .. . , n, which transform accord-ing to the law

_ a

AL(P, 22, , xn) = Aa(xl, x2, . . . , x') ax(491)

axa

under the coordinate transformation x= = 21(x', x2, . . . , xn),are said to be the components of a contravariant vector. Thevector is not just the set of components in one coordinate systembut is rather the abstract quantity which is represented in eachcoordinate system x by the set of components Ai(x).

We immediately see that the law of transformation for acontravariant vector is transitive. Let

A'=Aaa2i, Ai=Aai.-OX* a2a

Then8A' _ A8 ax a = Aa ax ax _Aaax

a28 axa ate axa

which proves our statement.If the components of a contravariant vector are known in one

coordinate system, then the components are known in all otherallowable coordinate systems by (491). A coordinate trans-formation does not give a new vector; it merely changes thecomponents of the same vector. We thus say that a contravariant vector is an invariant under a coordinate transformation.An object of any sort which is not changed by transformations ofcoordinates is called an invariant.

Example 121. Let X, Y, Z be the components of a contra-variant vector in a Euclidean space, for an orthogonal coordinate


system, and let ds2 = dx2 + dy2 + dz2. The components of thisvector in a polar coordinate system -,re

ar ar arR=X- +Ya +Zaz= cos 0X+ sin 0Yy

ae+ Y

ae+ Z

CIO _ -sine X + cos eY0

X ax ya az^ r r

Z=Xa +Yaz+Zaz=Zy

where r = (x2 + y2)4, 0 = tan-' (y/x), z = z.The components R, 0, Z are not the projections of the vector

A = Xi + Yj + Zk on the r, 9, z directions. However, if the0-component is given the dimensions of a length by multiplyingby r, we obtain Or = - sin 0 X + cos 9 Y, which is the projec-tion of A in the 9-direction. We multiplied by r because r d8is arc length along the 0-curve. R, 0, Z are the vector com-ponents of the vector A in the r-e-z coordinate system, whereasR, re, Z are the physical components of the same vector.

Problems

1. If A'(x), B'(x) are components of two contravariant vec-tors, show that C'"(x) = A'(x)B'(x) transforms according to the

law C*' = Capax' araxa ate, where C" =

2. Show that if the components of a contravariant vectorvanish in one coordinate system, they vanish in all coordinatesystems. What can be said of two contravariant vectors whosecomponents are equal in one coordinate system?

3. Show that the sum and difference of two contravariantvectors of order n is another contravariant vector.

4. If X, Y, Z are the components of a contravariant vector inan orthogonal coordinate system, find the components'in a spher-ical coordinate system. By what must the 0 and Sp componentsbe multiplied so that we can obtain the projections of the vectoron the 0- and (p-directions?

'5. If A' = Aa a ' show that A' = Aa a.

SEC. 125] TENSOR ANALYSIS 273

6. Referring to Prob. 1, show that Cs, Caaax` ax'axa al$

N i N i

7. If Ai = ax1 Aa axa, show that Ati = l-A-

Iaz

124. Covariant Vectors. We consider the scalar point func-tion So = co(x', x2, . . . , x"), and form the n-tuple

(492)

Now under a coordinate transformation

av a(p axa

ay' axa ay'(493)

so that the elements of the n-tuple l 1 y2 , y"J are

related to the elements of (492) by (493). We say that the axi

are the components of a covariant vector, called the gradient of gyp.More generally, if

i=Aaa'J L' (494)

the A; are said to be the components of a covariant vector. Theremarks of Sec. 123 apply here. What is the difference betweena contravariant and a covariant vector? It is the law of trans-formation! The reader is asked to compare (494) with (491).We might ask why it was that no such distinction was made inthe elementary vector analysis. We shall answer this questionin a later paragraph.

125. Scalar Product of Two Vectors. Let A'(x) and &(x)be the components of a contravariant and a covariant vector.We form the scalar AaBa. What is the form of AaBa if we makea coordinate transformation?

Nowaax

c12'Aa=A8axe' Baaxa


so that

AaBa = AFB,axa axQ

axp axa

AaBa = A"B,a2s

= Asfl = AaBa

Hence AaBa is a scalar invariant under a coordinate trans-formation. The product (AaBa) is called the scalar, or dot,product, or inner product, of the two vectors.

Problems

1. If Ai and Bi are components of two covariant vectors, show

that Ci; = A;B, transforms according to the law Oil = Capaxa axe

axa ax?

2. Show that C = AB; transforms according to the law

C"i - C8axp

.V axa

a3. If Ai = A. axi show that A, = Aa

axi4. If p and J, are scalar invariants, show that

grad (vO) = c grad P + 0 grad cpgrad [F(op)] = F'(,p) grad rp

5. If A'Bi is a scalar invariant for all contravariant vectorsA', show that Bi is a covariant vector.

126. Tensors. The contravariant and covariant vectorsdefined above are special cases of differential invariants calledtensors. The components of the tensor are of the form Ta4b .. b;,where the indices a,, a2, . . . , a, b1, b2, . . . , b, run throughthe integers 1, 2, ... , n, and the components transform accord-ing to the rule

ax N alas axa* axle, axig1 b,b,... ax TaW*A:...,q, anal . . . aXa

alb,. . . a-. (495)


We call the exponent N of the Jacobian 1092X1 the weight of the

tensor field. If N = 0, we say that the tensor field is absolute;otherwise the tensor field is relative of weight N. A tensordensity occurs for N = 1. The vectors of Sees. 123 and 124 areabsolute tensors of order 1. The tensor of (495) is said to becontravariant of order r and covariant of order s. If s = 0, thetensor is purely contravariant, and if r = 0, purely covariant;otherwise it is called a mixed tensor.

Two tensors are said to be of the same kind if the tensors havethe same number of covariant indices and the same number ofcontravariant indices and are of the same weight. We can con-struct further tensors as follows:

(a) The sum of two tensors of the same kind is a tensor of thiskind. The proof is obvious, for if

a...b _ axN 8xrOxa albTc ... d - ax To ... r age ...

axd axa. . .

ax"

b = laXlN S-.. ax° . axr axa. . .

aA7c. d a ""T a atd axa ax0

then

Ua...b Fa...b) _ Ua...c...d c...d c...d c...z

ax

a2

N ax' axb

(b) The product of two tensors is a tensor. We showfor a special case. Let

Tba =

19i =

so that

a axQ a2"

alt T s axlaxa

Ox 3c

02i

a2S or,

ax(T bs')

=,ax

s ax8 a2a all(TWO) a26 ax* ax"

The new tensor is of weight N + N' = 3 + 2 = 5.(c) Contraction. Consider the absolute tensor

a ax0 axy axiAtk - AOY a2' axk axa

this


Replace k by i and sum. We obtain

43X'9 axy axiA,i - Aay

axi axi axaM ax, a 13x# y

= Aay ax axa= Aay axi sa

a axa= Apa axi

so that Ali are the components of an absolute contravariantvector. In general, we equate a certain covariant index to acontravariant index, sum on the repeated indices, and obtaina new tensor. We call this process a contraction.

(d) Quotient law. We illustrate the quotient law as follows:Assume that AiB,k is a tensor for all contravariant vectors Ai.We prove that B,k is a tensor, for

AiE1 = AQBayCIO axy 492i- x6axi ax; axa

axa axy= A'Bay

ax axkor

l axa axyAi ` $,k - Bay

axp avk= 0

Since A' is arbitrary, we must have f k = Bayaxa axk

a2 axthe desired

result.Example 122. The Kronecker delta, S, is a mixed absolute

tensor, foraxi axa = axi axa = 80 = as

' aga axi' axa axs a a

Example 123. If Ai and Bi are the components of a contra-variant and a covariant vector, then C = AiB, are the com-ponents of a mixed tensor, for

s axaAi = Aa ax , B' -_ B

axa a ali


CO that

C AaB#axa axe =

Csaxa Ox'

Example 124. Let gij be the components of a covariant tensor

so that gi, = gapax-,9x#- = Taking determinants and applyingaxt ox'

Example 115 twice, we obtain

laxIgl =1g1 lax or I#I} = IgIl

ax

a2

Now if Ai are the components of an absolute contravariant vec-

tor, then A{ = Aa ax Y so that

=ax

B, = IgI}A' aaxBa2i

axB.

ax-

OfIgI'Aa

axa

so that Bi ° IgI#Ai are the components of a vector density. Thismethod affords a means of changing absolute tensors into relativetensors.

Example 125. Assume gap dxa dxO an invariant, that is,

9aP d7a d2O = gp dxa dxfi

a

Now d2a = axu dxµ, so that

or

gapaxa a-

dxu dx = g,,, dxu dx'axu ax,

(9aO °1°a0

^ 9u dxu dx' = 0 (496)axu ax

If we assume gap = gaa, then since (496) is identically zero forarbitrary dxi, we must have (see Example 114)


axa axs9µY = 9a$

axµ axY(496a)

or the gµ, are components of a covariant tensor of rank 2.Example 126. If the components of a tensor are zero in one

coordinate system, it follows from the law of transformation(495) that the components are zero in all coordinate systems.This is an important result.

Example 127. Outer product of two vectors. Let A; and B; bethe components of two covariant vectors, so that

axa ax# axa a28C;; = A;B; = AaEp

axti axe =C°

axti axe

The C:; are the components of a covariant tensor of second order,the outer product of Ai and B.

Example 128. By the same reasoning as in Example 127, wehave that Cc; = A;B; - A;B, are the components of a covarianttensor of the second order. Notice that C,, is skew-symmetric,for C;; = -C;;. For a three-dimensional space

11C1;II =

0 A1B2 - A2B1 A1B8 - AaBI-(A1B2 - A2B1) 0 A2B3 - A3B2-(A1B8 - A3B1) -(A2B3 - A3B2) 0

The nonvanishing terms are similar to the components of thevector cross product.

Problems

axa ax0 axa axa1. If A = Aaoa i axe, show that A;; = Aaa

ax' axe2. Show that A,, can be written as the sum of a symmetric

and a skew-symmetric component.3. If A are the components of an absolute mixed tensor, show

that A; is a scalar invariant.4. If Aa# are the components of an absolute covariant tensor,

and if Aa#A$,, = 8 , show that the Aal are the components of anabsolute contravariant tensor. The two tensors are said to bereciprocal.


5. If Al, and A1, are reciprocal symmetric tensors, and if uiare components of a covariant vector, show that A,;uiul = Aiiu,u;,where ui = Ai°u,,.

6. Let Ail and Bit be symmetric tensors and let ui, vi be com-ponents of contravariant vectors satisfying

(A,,-KBi;)ui=0 i,j= 1, 2, . . . n(A,, -K'Bii)v' = 0, KX K

iProve that Ai,uiv' = Bi,uiv' = 0, and that K =

AiiutuWhyis KBiJuiui

an invariant?7. From the relative tensor A of weight N, derive a relative

scalar of weight N.8. If Ami is a mixed tensor of weight N, show that Amw is a

mixed tensor of weight N.9. Show that the cofactors of the determinant lair, are the

components of a relative tensor of weight 2 if ail is an absolutecovariant tensor.

10. If Ai are the components of an absolute contravariant

vector, show that axe are not the components of a mixed tensor.

127. The Line Element. In the Euclidean space of threedimensions we have assumed that

ds2 = dx2 + dye + dz2

In the Euclidean n-space we have

ds2 = (dx')2 + (dx2)2 + . .. + (dxn)2= bo dx° dxa

If we apply a transformation of coordinates

xi = xi(21, 22, . . . , 2' )

we have that dx' = axi d2a, so that (497) takes the form

ds2 = 6,0ax° axe

dxµ dxaxµ aV

We may write ds2 = vr d2" dz', where

axa 8x#n

ax° ax°9vr = a°a a _ ax+' any

(497)


Thus the most general form for the line element (ds)2 for aEuclidean space is the quadratic form

ds2 = gas dxa dx5 (498)

The gap are the components of the metric tensor (see Example125). The quadratic differential form (498) is called a Rie-mannian metric. Any space characterized by such a metric iscalled a Riemannian space. It does not follow that there existsa coordinate transformation which reduces (498) to a sum ofsquares. If there is a coordinate transformation

xi = xi(y', y2, . . . , y")

such that ds2 = Say dya dy5, we say that the Riemannian space isEuclidean. The y's will be called the components of a Euclideancoordinate system. Notice that gas = Sap. Any coordinate sys-tem for which the g;; are constants is called a Cartesian coordinatesystem.

We can choose the metric tensor symmetric, for

gii = Ij(gv + gig) + (gii - gii)

and the terms $(gt, - gii) dx' dx' contribute nothing to the sumds2. The terms J(g,i + gii) are symmetric in i and j.

Example 129. In a three-dimensional Euclidean spaceds2 = tax 1)2 + (dx2)2 + (dx3)2 for an orthogonal coordinate sys-tem, so that

1 0 00 1 0

0 0 1

Letx' = r sin 0 cos p = y' sin y2 cos y$x2 = r sin 0 sin rp = y' sin y2 sin y2x3=rcos9=y'cosy2

Now

axa 8x59ti(r, g, w) = gas ay; ayi

ax' ax' axe axe OxI 49x8

- ay: ayi + ay: ayi + ay' ayi


Hencegll = (sin y2 cos y3)2 + (sin y2 sin y3)2 + (cos y2)2

Similarly

922 = (y') 2,

so that

933 = (y')2(sin y2)2,

281

gi,=0 for i5j

ds2 = (dy')2 + (y')2(dy2)2 + (y' sin y2)2(dy3)2= dr2 + r2 do2 + r2 sin2 0

is the line element in spherical coordinates. Since the g's arenot constants, a spherical coordinate system is not a Cartesiancoordinate system.

Example 130. We define gii as the reciprocal tensor to gi,,that is, gi.2 gai = d,f (see Prob. 4, Sec. 126). The gii are the signedminors of the g;i divided by the determinant of the gi;. Forspherical coordinates in a Euclidean space

1 0 00 r2 00 0 r2 sin2 B

1 0 0

0 2 0r2

1

r2 sin2 90 0

Example 131. We define the length L of a vector Ai in aRiemannian space by the quadratic form

L2 = g pA°A# (499)

The associated vector of Ai is the covariant vector

i = gia

It is easily seen that Ai = gipAp, so that

L2 = gapg' g"'A"A.

We see that a vector and its associate have the same length.If L2 = 1, the vector is a unit vector.

Example 132. Angle between two vectors. Let Ai and B, beunit vectors. We define the cosine of the angle between these


two vectors by

cos 0 = A'Bi = Aig;;Bi = gi,AiB'= giiA,B1 = gi'A;B, (500)

If the vectors are not unit vectors,

gi,AiB;500cos 0 = a)(

If gi;A'B' = 0, the vectors are orthogonal. We must show thatI cos 01 < 1. Consider the vector AA' + 1Bi. Assuming a posi-tive definite form, that is, gaaza# > 0 unless z' = 0, we have

gas(XAa + µB")(XAa + kB") > 0or

y = A2(gaAaAa) + 2Aµ(gasA"B1) +µ2(gaaBaBa) > 0

This is a quadratic form in A2/µ2, so that the discriminant mustbe negative, for if it were nonnegative, y would vanish for somevalue of A/µ or µ/A. Hence

gaaAaBP < (gaaAaA8)4(gaaBaBa)}

or ,cos 01 < 1. Moreover, if Ai = kBi, it is easy to see thatcos 0 = ± 1. Hence I cos el < 1.

Example 133. A hypersurface in a Riemannian space is givenby x' = x'(ul, u2). If we keep ul fixed, ul = uo, we obtain thespace curve x' = x'(uo, u2), called the u2 curve. Similarly,x' = x'(ul, uo) represents a ul curve on the surface. These curvesare called the coordinate curves of the surface. We have atonce that on the surface

2axa axads2 = gad dxa dxa = I gap - -dui du'

ca-1 au' aui_ axa axa

dui du'g°a aui au

ds2 = h;; du' du' (501)

axa axawhere hi, = gaa au'aui


Example 134. The special theory of relativity. Let us considerthe one-parameter group of transformations

x = ,B(x - Vt)y=yz = 2

xJ

(502)

where,6 = [1 - (V2/c2)1-} and V is the parameter. c is the speedof light. These are the Einstein-Lorentz transformations (seeProb. 11, Sec. 24). The transformations form a group because(1) if we set V = 0, we obtain the identity transformationsx = 2, y = y, z = 2, t = t; (2) the inverse transformation existssince 2 _ #(x + Vt), y, t = z, t = O[t + (V/c')x], theinverse transformation obtained by replacing the parameter Vby - V; (3) the result of applying two such transformationsyields a new Lorentz transformation, for if

where _ [1 - (W'/c')]-;, then

x = (2- Ut)y=?!z

t=fi -Uxc

where

U` V-I-W :'^ r1 _ U2)-1

- 1 + (VW/c) `\ c IWe now assume that (x, y, z, t) represents an event in space

and time as observed by S and that (x, 9, 2, 1) represents thesame event observed by S (see Fig. 101).


The origin b has :f = y = 2 = 0' so that from (502)dx = - V'dt

showing that S moves with a constant speed - V relative to S.Similarly S moves with speed + V relative to S.

y

.7, 1,q'P1:Mf,714

-V

S

X

z zFia. 101.

S

X

From (502) we see that 0 and 0 coincide at t = 0. At thisinstant assume that an event is the sending forth of a light wave.The results of Prob. 11, Sec. 24 show that

x2 + y2 + z2 xa _'_2 + 22 - 4

t2 -12

- C

so that the speed of light is the same for both observers. Thisis one of the postulates of the special theory of relativity. Start-ing with this postulate and desiring the group property, we couldhave shown that the transformations (502) are the only trans-formations which keep dx2 + d y2 + dz2 - cz dt2 = 0 an invariant.

Let us now consider a clock fixed in the S frame. We havex = constant, so that dx = 0, and from (502) dt = 0 dt. Hencea unit of time as observed by S is not a unit of time as observedby S because of the factor 0 5-6 1. 8 remarks that S's clock isrunning slowly. The same is true for clocks fixed in the S frame.


We choose for the interval of our four-dimensional space theinvariant

ds2 = c2 dt2 - dx2 - d y2 -dz2 = (dx') 2 - (dx') 2 - (dx2) 2- (dxa) 2

where x' = x, x2 = y, x$ = z, x' = ct. The interval ds2 yieldstwo types of measurements, length and time, but takes care todistinguish between them. If we keep a clock fixed in the Sframe, then dx = dy = dz = 0, so that ds2 = c2 dt2, and themeasurement of interval ds is real and proportional to the timedt. Now if we keep t fixed, dt = 0, and

ds' = - (dx2 + dye + dz2)

so that ds is a pure imaginary, its absolute value denoting lengthas measured by meter sticks in a Euclidean space.

We shall describe the laws of physics by tensor equations, thecomponents of the tensors subject to the transformations (502).This will guarantee the invariance of our laws of physics.

The momentum of a particle of mass mo will be defined by

pa = mo d8 . If the speed of the particle is u,

u2()2

+()2

+ \dt/2

as measured by S, then

d82 = c2 dt2 - (dx2 + dye + dz2) = (c2 - u2) dt2so that

mo dxa I mo dxapa = (C2 - u2)} dt = C [1 - (u2/c2)]} dt

and

mo -- mp = [1 - (u2/C2)]}

We define the Minkowski force by the equations

_ d ( dxaf a - C2

ds m0 da )I d ( a

[1 - (u2/C2)]; dtm

dt ' a = 1, 2, 3, 4


The Minkowski force differs from the Newtonian force by thefactor [1 - (u2/C2))-1. The work done by the Newtonian force

Fa = d- (m ddt for a displacement dxa is

dE _ d(mxa) dxa

a-1

mou dua1I (mxa dxa + dd -4- dxa j

[1 - (u2/C2)11

and integrating, E = [1 - (u2/c2))-lmoc2 - moc2 = (m - mo)c2,with E = 0 for u = 0. Expanding [1 - (u2/c2)1-1 in a Maclaurinseries, we have E Jmou2 for (u2/c2) << 1.

The reader is referred to Probs. 1, 2, and 3 of Sec. 82 for theapplication of special relativity theory to electromagnetic theory.Let the reader derive (285) by use of this theory, choosing theframe S so that at a particular instant the charge p is fixed in thisframe. The force on the charge as measured by S is given by(285).

Problems

1. For paraboloidal coordinates

x1 = yly2 cos ysx2 = y1y2 sin yax8 = 3[(yl)2 - (y2)2]

show that ds2 = [(y')2 + (y2)21[(dyl)2 + (dy2)21 + (yly2)2(dy$)2.2. Show that for a hypersurface in a three-dimensional

8axa axa

Euclidean space, h1 =t9lis au?

a-13. Show that the unit vectors tangent to the ul and u2 curves

1

are given by1 ax

and18x

Vh11aul-V//-22aU2

4. If w is the angle between the coordinate curves, show thatcos w = his/.

SEc. 127J TENSOR ANALYSIS 287

5. (p(x', x2, . . . , x") = constant determines a hypersurfaceof a V,,. If dx` is any infinitesimal displacement on the hyper-

surface, we haveax

a dxa = 0. Why does this show that the axare the components of a covariant vector that is normal to thehypersurface?

6. If (p(x', x2, . . . , x") = constant, show that a unit vector

normal to the surface is given by (g°8a aP gra avp

ax, axo axa7. Consider the vector with components (dx', 0, 0, . . . , 0).

Under a coordinate transformation the components become1 2 "

Cax' dx', ax' dx', . . . , ax' dx' . Consider the new components

for the vectors with components (0, dx2, . . . , 0), . . . , (0, 0,. .. ) 0, dx"), and interpret

d2' dxa ... dx"= 1a1 dx' dx2 . . . dx"

Using the result of Prob. 7, Sec. 121, show that gl dx' dx2 dx"is an invariant. We define the volume by

V = f f . . . f dx' dxa . . . dx"

8. Show that ds is a unit vector for a V".

9. The surfaces xti = constant, i - 1, 2, ... , n, are calledthe coordinate surfaces of Riemannian space. On these surfacesall variables but one are allowed to vary. This determinessubspaces of dimensions (n - 1). If we let only x' vary, weobtain a coordinate curve. Show that the unit vectors to thecoordinate curves are given by a; = 1//, i = 1, 2, . . . , n,and that the angle of intersection between two coordinate curvesis given by r

gc,cosm;; _ 9g10. Show that a length observed by S appears to be longer as

observed by S. How does S compare lengths with S?


11. Let jx, jj,, jz, p be the components of a vector as measuredby S. What are the components of the same vector as measuredby S?

12. Letd2 x 72d 29 dz2

be the components of the acceleration of adie' die' die

2zdz 2y

particle as measured by S. Find dt2, dt2, dt2 from (502).

128. Geodesics in a Riemannian Space. If a space curve in aRiemannian space is given by xi = xi(t), we can compute thedistance between two points of the curve by the formula

r ( dxa dxO 4s - ` i \g°a dt dt

dt (503)

To find the geodesics we extremalize (503) (see Sec. 40). Thedifferential equations of the geodesics are [see Eq. (146)]

d of of

dt(of

ax' - 0

where f = (go '.#)} =Wt

Now

of 1 (aga9 zaxsax' 2f \ ax'

and

d (af d (ga;xa + gistdt \az' dt \ 2 ds/dt /

Fz2 ds/dt (g.a + g+sio + agxps xaxs + x° ve

(504)

2

2(ds/dt)2 dt2 (ga:xa +

2dIf we choose s for the parameter t, s = t, dt = 1, dtQ = 0, and

use the fact that gii = gii, (504) reduces to

a1 /agai ague agoal a o

g'ax + 2 610 + axa- -axti / x x = 0 (505)


Multiplying (505) by gri and summing on i, we obtain

xr + 2

gri`(89-i .99io

+ axa axi /or

d2xr , dxa dxP

dS2 + r°a ds ds -

where

9r r° (.9g.,a9oa ago

ra$ 2 \ ax$ + axa axQ

280

(506)

(507)

The functions r" are called the Christoffel symbols of the secondkind. Equations (506) are the differential equations of thegeodesics or paths.

Example 135. For a Euclidean space using orthogonal coordi-nates, we have ds2 = (dxl) 2 + + (dx") 2, so that gap = a.#

d1and ft"

0. Hence the geodesics are given by ds = 0 or

xr = ars + br, a linear path.Example 136. Assume that we live in a space for which

dal = (dx')2 + [(x1)2 + c2](dx2)2, the surface of a right helicoidimmersed in a Euclidean three-space. We have

{1gi?il =

1 0 1 0

0 (x1)2 + c2

Thus we have

rill =0,1r21 = r14 = 0,1 - -x1r22 - ,

r212

r2

01

(x')2 .+ C2

x1

(x') 2 + C2

so that the differential equations of the geodesics on the surfaceare

d2x' dx2 2- 0

ds2 -x 1

T8

d2x2 2x' dx' dx2d82 + (x')2 + C2 d8 ds =

0


Problems

1. Derive the r;k of Example 136.2. Find the differential equations of the geodesics for the line

element ds2 = (dxl)2 + (sin x')2(dx2)2.3. Show that rqn = r'a.4. For a Euclidean space using a cartesian coordinate system,

show that r, = 0.5. Obtain the Christoffel symbols and the equations of the

geodesics for the surface

xl = ul cos u2x2 = u' sin 412x3 = 0

This surface is the plane x3 = 0, and the coordinates are polarcoordinates.

+ g,arp,,.6. From (507) show that axµ = g,sroals

7. Obtain the Christoffel symbols for a Euclidean space usingcylindrical coordinates. Set up the equations of the geodesics.Do the same for spherical coordinates.

8. Write out the explicit form for the Christoffel symbols ofthe first kind : { i, jk I = g;,rlk.

9. Let ds2 = E due + 2F du dv + G dv2. Calculate IgJ, g'',i, j = 1, 2. Write out the rk.

10. If r = r;7 axs axk axe + a2x° axe show that r;,y - rrya. ax ax ax ax axare the components of a tensor.

129. Law of Transformation for the Christoffel Symbols.Let the equations of the geodesics be given by

d2xi d i J dxk

r_

0

and

ds2 + "k ds ds ^

AV d.V d2k -P°

(508)

0_2k ds dsds2 + (5 9)

for the two coordinate systems xi, xi in a Riemannian space. Wenow find the relationship between the r k and r; . Now

SEc. 129J TENSOR ANALYSIS

dxi _ a2' dxads

_axa ds

and

291

d2x` _ a2xi dxa dxa 9. d2x°

ds2 axa axa ds ds + axa ds2

Substituting into (509), we obtain

&V d2xa a22 i dxa dxa axi axk dxa dxaaxa ds' + ax# axa ds A +

rikaxe, ax" ds as

°We multiply (510) by axi and sum on i to obtain

0 (510)

d2x° axe axk a22i 49x°l dxa dxa

dx2+ (ri axgr +jk axa axa a2i axa axa a-V ds ds - 0

Comparing with (508), we see that (using the fact that rlk = rk;)

axa axy axi a2x° Oxirik aY axe axk axa + axi axk a2°

(511)

This is the law of transformation for the rk. We note that therk are not the components of a tensor, so that the rk may bezero in one coordinate system but not in all coordinate systems.

Example 137. From (481) we have

algl

8 = Iglgae ax'

and from Prob. 6, Sec. 128,

so that

= gas9°ar,, + gang°aFBp

= a;rap + a1r00=rap+r;p=2rµ

or

ax# = garap + g°arPA

C log _ r aaxp p (512)


Example 138. We may arrive at the Christoffel symbols andtheir law of transformation by another method. Differentiatingthe law of transformation

axe, ax8

g'' = ga$ a2i a2;with respect to xk, we have

ag;i agar axY axa ax8 axa 492X# ax0 aaxaaxk - axY a2k 49.t- __-

gay ak a + jI02i x2a2 a2k a.Ti (513)

If we now subtract (513) from the two equations obtained fromit by cyclic permutations of the indices i, j, k, we obtain

ti a axs axY axi a2xa axi

02i axk axa511a)r - rdy afj a2k axa +

where

agi, _ agile

r" = 2 g°' (agk,axj + axle ax"

Example 139. Let us consider a Euclidean space for which

ds2 = (dx 1) 2 + (dx2) 2 + . - . + (dx") 2

In this case the r (x) = 0. In any other coordinate system, wehave

a2x° a2iTV a2k ax°

If the new coordinate system is also Cartesian (the g,, = con-stants), then r k = 0, or

82x° =0aP a2k

x° = alga + b° (514)

where a', b° are constants of integration.Hence the coordinate transformation between two cartesian

coordinate systems is linear. If, furthermore, we desire thedistance between two points to be an invariant, we must have

n n n

dx° dx° _ d2° d2- _ a:a- dxa dx8ff-1 W-1 C-1


so that

293

n

I a,-,a; = Sas (515)V-1

A linear transformation such that (515) holds is called an orthog-onal transformation.

For orthogonal transformations,

ax. axe' =g"°a' axe

reduces to S;; =axa axe

We multiply both sides byamt

anda:' CIO ax),

sum on i, so thatazf axp 8x8 8x'axe

= aaa is a = #0 axe - iii (516)

Now let us compare the laws of transformation for covariantand contravariant vectors. We have

A' = Aaa2' axa_,axa

A; = Aa82'

Replacing621

by a from (516), we see that

n axa

Aaaa-1

(517)

(518)

so that orthogonal transformations affect contravariant vectorsin exactly the same way that covariant vectors are affected [com-pare (517) and (518)]. This is why there was no distinctionmade between covariant and contravariant vectors in theelementary treatment of vectors.

Problems

1. From (511) show that

_ a ax8 axy a2' 82xC, ofr'k

=ray

all To ax- + afj axk axe


2. By differentiating the identity gaga; = S , show that

ag{k

=_gkxr h4 - 9htr

axi h;

3. Derive (513) by performing the permutations.2 s0

4. If az.aQ

xkax = 0, show that

az axk = 0.

5. If090 My axi a21° axi

r;k = rPr +ax; 021; axa axi axk at°show that

a axA 8x' axi a2x' axir;k = r;, a.C axk axa + axi axk ax°

6. If xa = xa(ul, u2, . . . , u*), a = 1, 2, . . . , n, r < n, and

if h;; = gasaxa axs

and ifTauii 1 (ah ahk, ah,kl

{rik)h = 2 h.,auk + aui _ au°

show thataxa axs axy axa a2xs

hair k)k = gap(ra,)oaui aui auk + gas aut aui auk

7. Define gs(x) by the equation gas(x) = µ(x)gas(x). We seethat the metric tensor gas(x) is determined only up to a factor ofmultiplication u(x). In this space (conformal) we do not com-pare lengths at two different points, or, in other words, the unitof length changes from point to point. Show that

T;r(x) = r;, (x} + rPYaa + PSaY - ga°gstimo

1alogµwhere 'Po = 2 axe

I'sa, and ra sy are defined by (507) using

gas and gas.8. Prove that a geodesic of zero length (minimal geodesic)

is, xa(s) satisfies(506)

anddxa dxs

= 0] remains a[that gas ds Aminimal geodesic under a conformal transformation.

SEC. 130J TENSOR ANALYSIS 295

130. Covariant Differentiation. Let us differentiate the abso-lute covariant vector given by the transformation

At-Aaaxaaxi

We obtainaA; aAa axa axa a2xa

axe axa ax= ax' + Aa axi axi(519)

It is at once apparent thataA;axe are not the components of a

tensor. However, we can construct a tensor by the followingdevice: From (511a) (see page 292)

ax' axr axa 82x,7 axa

ax; axe axo + axi axi ax"(520)

Multiplying (520) by A. and subtracting from (519), we obtain

aA; Ox- axaA 1' rA( 521a _ O

axa-

&V- aa ax' axe

)(

so that if we define

49A;A;;=-Aa1 (522)8x1

we have thataxa axa

A..s = Aa.a axi axi

and A;,, is a covariant tensor of rank 2. The tensor is called thecovariant derivative of A; with respect to xi. The comma willdenote covariant differentiation. For a cartesian coordinate sys-

tem, r,,E = 0, so that A;,1 =aA;axe our ordinary derivative.

For a scalar of weight N we have

A=I-

A821


so that

axlN aA axa+ N

ax 8xa 02'

aax

ax

N-1

lax

and from (482),ax _ laxe a2xs

Henceaax'

ax ax i ax

aA ax N aA axa axaxi - latl axa axi + N H

N axa 82x0

(SEC. 130

A (523)axs &V C120

axa 2 , a

Multiplying r* = n., +a x ax

by NA and subtractingaxi axa Hi axQ

from (523), we have

OA - NAr; _ ax

ax

N (claAx

- NA r-.1, i aclx

(524)a

Hence A,, -= ax - NAr`,, is a relative covariant vector of weight

N. It is called the covariant derivative of the relative scalarA. For a cartesian coordinate system it reduces to the ordinaryderivative.

In general, it can be proved that if Tp,s,::: p; is a relative tensorof weight N, then

+ +axe

,,p a v,TW-a,ry" Ta,a,...a,

-A- Ain

(525)

is a relative tensor of weight N, of covariant order one greaterthan Ts,,:::a;, and it is called the covariant derivative of TB,A :::8.

Example 140. We have

g;;gii,k =

aaxk - gwl'.k - g+Nr k

so that from Prob. 6, Sec. 128,

g i.k = 0 (526)

SEC. 130J TENSOR ANALYSIS 297

Example 141. If p is an absolute scalar, _ gyp, we call

-p,; = ax the gradient of V.

Example 142. Curl of a vector. Let A; be an absolute covari-

ant vector. We have A;,; =aA;axj

- Aar . Similarly,

"'A,.; -axi - Ar,i

Hence Ai,5 - A;,; =aaxA;

-aA1

ax` is a covariant tensor of rank 2.

It is called the curl of the vector Ai. If the A; are the components

of the gradient of a scalar, Ai = a-P, thenaxi

a2ip 621P

curl A; == -ax1 ax' ax' ax'

= o

so that the curl of a gradient is zero. It can be shown that theconverse holds. If the curl is identically zero, the covariantvector is the gradient of a scalar.

Example 143. Intrinsic derivatives. Since Ai,1 anddx

are

.tensors, we know that Ai.;

dxi

dsis a covariant vector. We call it

the intrinsic derivative of A;. We have

dxi aAi dxi a dxfA;.,

ds a axe ds - A°`r`' ds

8A; dA; dx1

as ds - òar"ds(527)

and write the intrinsic derivative of A; asBA;

as

Example 144. The divergence of an absolute contravariantvector is defined as the contraction of its covariant derivative.


Hence

div A' = A' = ax -4- AarQ,

IgINow rQ; =a logx

from (512), so thatas

1 adiv A

_

t +Aa

ax- N/191 ax.

1

adiv As = aa (V f gj Aa)

In spherical coordinates, we have

`'9=so that

1 0 0

0 r2 00 0 r2 sin2 0

i= r2 sin 0

i adiv A' =

r2 sin 8 ar(r2 sin 0 A,)+

ad9

.(r2 sin 0 AB)

(528)

+ app (r2 sin 0 Ac)

and changing At and A" into physical components having thedimensions of A (see Example 121), we have

1div A' =

r2 sin oCar (r2 sin 0 A*) + a9 (r sin 0 A°) + a (rAc) J

Example 145. The Laplacian of a scalar invariant. If p is ascalar invariant, -r,; is the gradient of jp, and the div (p,,) is calledthe Laplacian of .p.

aLapcp=VZ(p=div(g,)=diva'

Thus

v 2(p (529)1,91

aaa (v' Fg19°°TX)

Sec. 1301 TENSOR ANALYSIS 299

We changed

a

into a contravariant vector so that we could

apply (528). The associate of81P

is g«i LIP.axe ax'

In spherical coordinates

1 0 0

0 r2 0

0 0 r2 sin2 B

so that

i

7 t g' l =

0 0

0

0

I

r2 sin 9

VF = 1 {

a1 r2 sin B aF -1- a (sin B

aF' + a / 1 aF'r2 sin 0 ar \\\ ar 00 \ 00a l\sin 9 app/)

Example 146. In Example 144 we defined the divergence ofMa

the vector A' as div A' = A + Aar' - For a EuclideanCti

space using cartesian coordinates, the rk = 0, so that

OA' 0A2 aA^div A'=axl+axe+ ... +axn

The quantity A* is a scalar invariant. If we let Ni be the com-ponents of the unit normal vector to the surface do, then AaNais also an invariant. In cartesian coordinates the divergencetheorem is

JJ divAdr = ffA.dd = JJA.N&r

In tensor form it becomes

fJfAT = JJA"Nad r (530)

We can obtain Green's formula by considering the covariantvectors 4O,; and y ,;. Now let

A, = kp,i - soO,c

300 VECTOR AND TENSOR ANALYSIS {SEC. 130

The associated vector of A; is A° = g'=A; = g°.i(4,v,s - 4,i).We easily see that A' = g°i(4y,; - Now g°'co,,Q is aninvariant and in cartesian coordinates reduces to

a29a2_ a24

(ax1)2+

(ax2)2+

(ax3)2_ Lap (p

Hence, using (530), we obtain

f if (t' Lap so - v Lap ¢) dr = f f g°iA;NQ doS

= f f Svf.,)N; dvS

Example 147. Let us consider the covariant vector Fa. Wemultiply it by the contravariant vector d--° and sum on a to

obtain the invariant F. dxa = F.ds

ds, which reduces to f dr

in cartesian coordinates. In Example 142 we constructed the curlof a vector, which turned out to be a tensor of rank 2. Wenow construct a vector whose components will also be those ofthe curl of a vector. We know that F.,s is a tensor. Now

define 4ah = 1 if a, ft, y is an even permutation of 1, 2, 3;w

1 '4°Ar = 0 otherwise. Thus

41st - ,is an odd permutation of 1, 2, 3;

211 - -LI

191

We obtain a new invariant

4132 =

G" = - 4"0''Fa,F

In cartesian coordinates Fa,P = ate, and

1 1411! = 0

1 = -EarilFR E2111F2 s + 021F3,2= aFs aF2

.a = -j axe - ax=


and similarly for G2 and G3. Hence Stokes's theorem in tensorform reads

a

f F. ds ds = f do (531)

Problems

a2i1. By starting with Ai = All , show that

Oza

A.1 axi + Aara1

is a mixed tensor.2. Prove that (g{aAa),, = g{aA'.3. Prove that (A-B.).1 = AaBa,, + A Ba.4. Prove that 14i = 0.5. Use (529) to find the Laplacian of F in cylindrical coordi-

nates.6. Prove that a; 0.

7. Prove that 1 axa (vflg g`-) + rQgas = 0.

8. As in Example 143, show that the intrinsic derivative

as ds + Aarocp ds is a contravariant vector.

9. Show that the intrinsic derivative of a scalar of weight N

is b = ds - NAr;, ds, so that if A is an absolute constant,

10. Show that (g,#AaAO) j = gapA ;Ap + 9apAaA,'

11. Show that A; , = aA' + r;,A' for an absoluteax°

mixed tensor A.12. Show that DZ(vo) V;k -I- 2 0v - DO -}- J, pz9.

1 013. Show that Aaa = gl

exa(VTg_j A") - Abr.

302 VECTOR AND TENSOR ANALYSIS [Sac. 130

14. If A; = A;(x, t) is a covariant vector, show that

oA; _ 8A; dx'at at + Ai,J dt

and hence that the acceleration fc °av; _ 8vi

at at + v ,,v'.15. Let X. be an arbitrary vector whose covariant derivative

vanishes; that is, Xa,$ = 0. Consider Jf Ta,' XaN,6 da, and applys

the divergence theorem to the vector TOXa. Hence show that

fJ T aaNN da = ff7 T dT.S

16. Let s; be the displacement vector of any particle from itsposition of equilibrium (see Sec. 115). We know that s;, is acovariant tensor. The relative displacements of the particlesare given by

as; = s,,J dx_ 4(s,,J + sJ,:) dx' + s,,.) dx'

Show that the term 4(8;,J - 8J,;) dxJ represents a rotation.We define the symmetric strain tensor E;J by the equation

Ei; = +(s.,J + ss,;)

The stress tensor T;, is defined by the equations

AF; = TJNJ Ao-

where AF; is the force acting on the element of area Aa withnormal vector Ni (see Sec. 116).

Let f, be the acceleration of the volume dT and F, be the forceper unit mass acting on the mass in question. Show that

f j f pF, dT + f f TJNJ da = f f f pf, dT

or using contravariant components,

fff pF'dr+ f f T*JNJda= fff pf'dTa s R


Now deduce the equations of motion

pFr + T = pfr

If Tn = pgri, show that

pFr - p.igr' = pfror

pFr --- p.r = pfr [see (411)]

303

131. Geodesic Coordinates. The equations of the geodesicsare given by

d2xi dxi dxk

as2+ r,k ds as

= 0

where the Ik transform according to the law

49x8 ax" axi 192xa axiask - ra + (532)sY ax axk axa 491i axk axa

We ask ourselves the following question : If the I k are differentfrom zero at a point x` = qi, can we find a coordinate systemsuch that rk 0 at the corresponding point? The answer is"Yes"!

Let(xi - q') + q«)(xd - 4s) (533)

so that a and I ax1i = 1, and moreover the transforma-a

tion (533) is nonsingular. The point xi = q' corresponds to thepoint x' = 0.

Now differentiating (533) with respect to x', we obtain

a =axii

+ (ri OPI

) Q(xa - qa) ate? (534)

because of the symmetry of ri,8. Hence 6z! = 6f.Q

Differentiating (534) with respect to xk, we obtain

a2x' axa axs a2x8

axk ax'+ (rae) a axk axi + (r10) a(x0 - q°`)

axk a21

304

so that


a2x'

axk axj

ax° axe

Q

= - (ray) 4 askQ

axe

-(raa)Q6-50 _ -k

Substituting into (532), we obtain

(rij k)o = (rfY)Qaj kaa - (rk)Qaa

= (r k), - (r;k)Q = o, Q.E.D.

Any system of coordinates for which (r7k)P = 0 at a pointP is called a geodesic coordinate system. In such a system,the covariant derivative, when evaluated at the origin, becomesthe ordinary derivative evaluated at the origin. For example,

(As')O _

\(Miaxi )0 + (ra;)o(Aa)o = ax, )0

since rq; = 0 at the origin.The covariant derivative of a sum or product of tensors must

obey the same rules that hold for ordinary derivatives of thecalculus, for at any point we can choose geodesic coordinates sothat

A'1OA' aBi _ a(Ai + Bi)

ax; +axi ax' (A'+B');

and A', + B`; - (A' + B'),, is a zero tensor for geodesic coordi-nates. Hence A + Bì - (Ai + B'),1 is zero in all coordinatesystems, so that

A` + B`; _ (A' + B').;

We leave it as an exercise for the reader to prove that

(A'B;).k = A'kB7 + A'B1,k

Equation (533) yields one geodesic coordinate system. Thereare infinitely many such systems, since we could have addedc,fiy(x) (x° - q') (xa - qP) (xr - q7) to the right-hand side of(533) and still have obtained (r k)o = 0.

A special type of geodesic coordinate is the following: Letx' = x'(s) be a geodesic passing through the point P, x' = xo, and

SEc. 131] TENSOR ANALYSIS 305

let xi Definer

X' = Vs (535)

where s is arc length along the geodesic. Each V determines ageodesic through P, and s determines a point on this geodesic.Hence every point in the neighborhood of P has the definitecoordinate xi attached to it. The equations of the geodesics inthis coordinate system are

d2z1 1 d21 dxk+ r'k ds ds = 0

dS2

2 i

But ds = V and ds = 0, so that

rikl:'k = 0 (536)

Since this equation holds at the point P for all directions t', wemust have r + r 1 = 21 k = 0, so that the x= are geodesiccoordinates. The X' = `s are called Riemannian coordinates.

Example 148. If ' is a unit vector, we have

gaa° = 1

The intrinsic derivative isa Q

gas s e + gaak°as = 0

since (g.p),1 = 0 (see Example 140). Hence gaat"as

0, and

gQs° Usp + -r"adx.

$) = 0. We see that the vector

dt1 dx°

is normal to the vector t'.

Problems

1. Show thatdx°, dx-

gab ds ds remains constant along a geodesic.

2. Show that for normal coordinates 2, r;2' k = 0.


3. If s is are length of the curve C, show that the intrinsici

derivative of the unit tangents

in the direction of the curved

has the components

pi -d2xs + . dx1 dxk

TSds2 r'k A

What are the components for a geodesic?a

4. Prove that bt (X"Ya) = sat Ya + Xa sat.

132. The Curvature Tensor. Let us consider the absolutecontravariant vector Vi. Its covariant derivative yields themixed tensor

v`; = a'ax; + V ar°`'

On again differentiating covariantly, we obtain

avt.v',k = axk' + v rak - viark

2 s a= a V + aV ri + Va arm, +axk 49x1 axk ' 49xk

a(a + VOr;, rak

Interchanging k and j and subtracting, we have

Vt;k - Vsk; = V"Ba;kwhere

(537)

Since V';k - Vik; and Vi are tensors, Va;k must be the componentsof a tensor, from the quotient law (Sec. 126). It is called thecurvature tensor. We can obtain two new tensors of the secondorder by contraction.


Let

ara ara

Ri'Ba;

axa axa + r ara; - r%raa (539)

This tensor is called the Ricci tensor and plays an important rolein the theory of relativity.

We obtain another tensor by defining

ara arai a;Si; = Be.. aa

ax1 axi

Evidently Si; = -8;i. and if we use the fact that

a log "ICI - raax" 1aµ.

we have that

a2 log a2 log -VIgl = 0

axi ax1 ax1 axi

(540)

Now Ri; - R;; = Si; = 0, so that the Ricci tensor is symmetricin its indices. We could have deduced this fact by examining(539) directly.

The invariant R = gi"Ri; is called the scalar curvature.133. Riemann-Christoffel Tensor. The tensor

Rhitk = gpaBk (541)

is called the Riemann-Christoffel, or covariant curvature, tensor.Let us note the following important result: Assume that theRiemannian space is Euclidean and that we are dealing with acartesian coordinate system. Since r;k(x) = 0, we have from(538)

B';ki = 0 (542)

in this coordinate system. But if B';k= = 0 in one coordinatesystem, the components are zero in all coordinate systems.Hence if a space is Euclidean, the curvature tensor must vanish.We shall show later that if B1,-k1 = 0, the space is Euclidean.

308 VECTOR AND TENSOR ANALYSIS [Sac. 134

If we differentiate (538) and evaluate at the origin of a geodesiccoordinate system, we obtain

a2raj _ a2rakBajk a- ax° axk 8x° axj

Permuting j, k, a and adding, we have the Bianchi identity

Bakj.o + Ba0j.k + Baka,1 = 0 (543)

134. Euclidean Space. We have seen that if a space isEuclidean, of necessity Bike = 0. We shall now prove that ifthe Bki = 0, the space is Euclidean. Now

axfi axy 00 a2xa ayirik(y) = rsY ayj ayk axa + ayj ayk aXa

If there is a coordinate system (x', x2, . . . , xn) for whichr;,.(x) = 0, then

a2xa ayir;k(y) = ayj ayk axa

(544)

and conversely, if (544) holds, the r;y(x) = 0. Now let us inves-tigate under what conditions (544) may result. We write (544)as

a2xa axeayj ayk = ayi rk(y1 (545)

which represents a system of second-order differential equations.Let us define

axau7-aycc

so that (545) becomesauk

ayj= u7Nkly)

(546)

(547)

For each a we have the first-order system of differential equationsgiven by (546) and (547), which are special cases of the moregeneral system


azk k=1, 2, ..,n+I= f (z', z2, . . . z", z"+1, y`, , yn)

ay' j=1,2,...,n(548)

If we let z' = x°, z2 = 211, z3 = 112i . . . , Zn+' = un, Eqs. (546)and (547) reduce to (548).

2 2

We certainly must have aye ayj = ayj yi, and this implies

afk afk az" _ aft af, av8y1 + az ay' - ay' + aZM ay'

or

afk afkfillay' + azi,

k fku

ay

+

az" fj

(549)

If the f; are analytic, it can be shown that the integrability con-ditions (549) are also sufficient that (548) have a solution satis-fying the initial conditions z1; = zo at yi = yo. The reader isreferred to advanced texts on differential equations and especiallyto the elegant proof found in Gaston Darboux, "Lecons sysOmesorthogonaux et les coordonnees curvilignes," pp. 325-336,Gauthier-Villars, Paris, 1910.

The integrability conditions (549), when referred to the system(546), (547), become

1jkua nklua

(550)aj"U'.=0

The first equation of (550) is satisfied from the symmetry of theI';k, and the second is satisfied if 19j"n = 0. Hence, if f"kj = 0,we can solve (545) for z' = x° in terms of y', y2, . . . , y". Forthe coordinate system (x', x2, . . , x"), we have T k(x) = 0.

Problems

1. Show that Rhijk = - Rihik = - Rhikj and that

Riijk = Rhikk = 0

2. Show that Rhijk + Rhkij + Rhiki = 0-3. If Rij = kgii, show that R = nk.


4. For a two-dimensional space for which g12 = 921 = 0,show that R12 = 0, 811922 = R22911 = R1221 and that

R _ R1221

911922

R:; = JRg.;.5. Show that B k1 76 0 for a space whose line element is given

by ds2 = (dx')2 + (sin xl)2(dx2)2.6. Derive (550) from (546), (547), (549).

7. If R = gtiaR,, show that (R'),;1 49R

= 2 ax;.

CHAPTER 9

FURTHER APPLICATIONS OF TENSOR ANALYSIS

i135. Frenet-Serret Formulas. Let Ai =

dsbe the unit tangent

vector to the space curve xi = xi(t), i = 1, 2, 3, in a Riemannianspace. In Example 148 we saw that the contravariant vector

t ° 0 }

asis normal to Xi. Let us define the curvature as K =

g°0 88 aasand the principal unit normal µi by

aai _as =

Kµi (551)

iSince pi is a unit vector, we know that aS is normal to pi. Now

g°0A°,us = 0, so that the intrinsic derivative yields

Sµa bA°g°'eA°

as+ g°0

bsµB = 0

since g°0,; = 0 or a3s 0. Hence

or

aµ0g°a A°

as+ Kg°aa°i8 = 0

µsg°aa°

(B68+ KAS = 0 (552)

since gp°µs = g°aX 'X 8 = 1.i

Equation (552) shows that Ss + KAi is normal to V, and since

5Fdi aµi

asand Ai are normal to µi,

as+ KAi is also normal to µi. We

311


define the binormal vi by the equation

v'- 1 Sµ + KAt

T (68or

aIAi

= -#c? - TY'as

-

where ,r is called the torsion and is the magnitude of(i)

(553)

(554)

Since v` is normal to both X and µ', we have

gaPvax$ = 0g«avO'µP = 0 (555)

By differentiating (555) and using (551), (554), (555), we leaveit to the reader to show that

/ avagaaµa (T/!a -

as= 0 (556)

The vector rµe - as is thus normal to all three vectors V, Af, Pi.

Since we are dealing in three-space, this is possible only ifa v

TEIZas

= 0, or

avi=

as(557)

Writing (551), (554), (557) in full, we have the Frenet-Serretformulas

& dxPKAi

ds ds

i

d + µa s = - (Ki + Tv`)

a

ds= Tµ

ds + Pa

(558)

SEC. 1361 FURTHER APPLICATIONS OF TENSOR ANALYSIS 313

For a Euclidean space using cartesian coordinates, the r = 0,and (558) reduces to the formulas encountered in Sec. 24.

Problems1. Derive (556).2. Using cylindrical coordinates,

ds2 = (dxl)2 + (x')2(dx2)2 + (dxa)2

and for a circle xl = a, x2 = t, x3 = 0. Expand (558) for thiscase, and show that K = 1/a, r = 0.

2i3. Show that Ss =

-K2X +as

U' - KTY'.

4. Since (558) is true for a Euclidean space using cartesiancoordinates, why would (558) hold for all other coordinate sys-tems in this Euclidean space?

2 . a Qd5. Since 'Xi = d3 , show that

d x + ra' dsdare thecom-d

ponents of a contravariant vector.136. Parallel Displacement of Vectors. Consider an absolute

contravariant vector Ai(xl, x2, . . . , xn) in a cartesian coordi-nate system. Let us assume that the components Ai are con-stants. Now

A' = Aaa2'

A' = Aaaxi

axa a2a

so that822' axfi

: - _ Aa dxydA8xs axa (921y

since dA' = 0. We thus obtain

dA = Aoa22i axe axa

d,yax' axa a2y 492'

From (511a) (see page 292)a2xa a2i- since r, = 0

r~"° = a2y a2° axa ;a2xi axO axa

ax8 axa axy axefrom (483)

so thatdAi = - A°1;y d2y (559)


In general, a Riemannian space is not Euclidean. We gener-alize (559) and define parallelism of a vector field Ai with respectto a curve C given by x' = x'(s) as follows: We say that Ai isparallelly displaced with respect to the Riemannian V,, along thecurve C, if

dAi dx*

ds °T ds

or

aas= d-si+r,,A°ds = 0 (560)

We say that the vector Ai suffers a parallel displacement alongthe curve. Notice that the intrinsic derivative of A' along thecurve xi(s) vanishes.

In particular, for a geodesic we haved2xi + ri dxa dxe = 0,ds2 °s ds ds

xiso that the unit tangent vector

dd suffers a parallel displacement

along the geodesic.Example 149. Let us consider two unit vectors Ai, Bi, which

undergo parallel displacements along a curve. We have

cos 0 = gapAaB$and

a(cos 0) SAa aBQ

as_ gas

asB$ + gaaAa

as= 0

so that 0 = constant. Hence, if two vectors of constant magni-tudes undergo parallel displacements along a given curve, theyare inclined at a constant angle.

Two vectors at a point are said to be parallel if their corre-sponding components are proportional. If A' is a vector ofconstant magnitude, the vector B' = WA', cp = scalar, is parallelto A'. If A' is also parallel with respect to the V. along a curve

txi = xi(s), we have Ss = 0. Now

aBi 0A' dc* dA'

1 dip d(log y)as - as + ds `4s ds - p ds r d8

SEC. 137] FURTHER APPLICATIONS OF TENSOR ANALYSIS 315

We desire B' to be parallel with respect to the V along the curve,so that a vector B' of variable magnitude must satisfy an equa-tion of the type

a13'

B = f(s)Bb (561)

if it is to be parallelly displaced along the curve.

Problems

1. Show that if the vector At of constant magnitude is par-allelly displaced along a geodesic, it makes a constant angle withthe geodesic.

2. If a vector A' satisfies (560), show that it is of constantmagnitude.

3. If a vector Bi satisfies (561) along a curve r, by lettingAt = #Bi show that it is possible to find ,y so that At suffers aparallel displacement along IF.

4. Let xi(t), 0 < t < 1, be an infinitesimal closed path. Thechange in the components of a contravariant vector on beingparallelly displaced along this closed path is 1 Ai = -jrra,Aa dxs,from (560). Expand Aa(x), ri (x) in Taylor series aboutxo = x4(0), and neglecting infinitesimals of higher order, show that

AA' = }RO,,A' xy dx# - x0 dxy

where Rte,, is the curvature tensor (see Sees. 132, 133, 134).137. Parallelism in a Subspace. We start with the Rie-

mannian space, V,,, ds2 = gab dxs dxs. If we consider thetransformation

xa = xa(ul, u2, . . . , u"), m < n (562)

we see that a point with coordinates ul, u2, . . . , u"' is a pointof V. and also a point of V,,. The converse is not true, for giventhe point with coordinates x', x2, . . . , x", there may not existU', u2, . . . , U. which satisfy xa = xa(ul, u2, . . , ur), sincem < n. Now

ds2 = 94 dxa dxs = gabaxa 8xs- - du' du;sub 8u1

= h;; du' du'


so that the fundamental metric tensor in the subspace, V,,,, isgiven by

axa MjE;j =gap

aui au1a

Now dxa = aua dui, so that if due are the components of a contra-

variant vector in the Vm, dxa are the components of the samevector in the V,,. In general, if ai(u', . . . , u»1), i = 1, 2, . . . ,

m, are the components of a contravariant vector in the V., wesay that

Aa = aui ai a = 1, 2, . .. , n (563)

are the components of the same vector in the V,,.

We now find a relationship between Maand as, where s is

arc length along the curve ui = ui(s) or the space curve

xa = xa[ui(s)]

Differentiating (563), we have

dAa axa dai a2xa duy ia

ds au' A + aui aui Tsand

6Aa dAaay

dxy _ ax,- dai a2xa du'as ds

+ (r *-yds aui ds + aui aui ds as

ax8 axy du1+ (rayaauui aui ds

Hence

ax' Ma9'a auk Ss

dai du' CIO axy ax' a2xa ax°l_ ;k ds + aids 9a«(ray>° aui au' auk + 9Qa aui aui auk

ax,S

9oa auk Ssa h`k dst + aids

h`k(r'i)h, (see Prob. 6, See. 129)

Sr;c. 1371 FURTHER APPLICATIONS OF TENSOR ANALYSIS 317

Hence

ax" SAa _ I dal dull9aa au as

hlkds + (r'i)ha dsk

y J

and

ax" SAa Sat9°a auk as hck as

(564)

From (564) we see that if ai is parallelly displaced alongi

xa[ui(s)], that is, ifas = 0, then Ss = 0. Thus the theorem:

If a curve C lies in a subspace V. of V,,, and a vector field in V.is parallel along C with respect to V,,, then it is also parallel alongC with respect to V.-

Problems

1. Prove that if a curve is a geodesic in a it is a geodesic inany subspace V. of V,,. Consider the unit tangents to thegeodesics.

2. By considering k fixed, show that auk is a contravariant

vector of V,, tangent to the uk curve, obtained by consideringul, u2, . . . uk-1, uk+', . . . u'" fixed in the equations

xa = xa(u' um)

3. If ai is parallel along C with respect to the Vm, show thatMa is normal to the space V, that is, normal to the ui curves,as

i = it 2, . . . , M.4. Under a coordinate transformation ui = ui(u', . .. , um),

i = 1, 2, . . . , m, the xa remain invariant. Hence show that

xa =axa

a au',agaa

9ad.i - ax° x.f

where the covariant derivatives are performed relative to theaai

metric hi;, that is, a1, = aa + ak(r kau'


a5. Show that ga$(x kxa + x {x k) + xaxIxk ax, = 0, where

covariant differentiation is with respect to ui and h;;.6. Show that A ; = x;ai + x°,as, for each a.138. Generalized Covariant Differentiation. The quantities

aui are contravariant vectors if we consider i fixed; for if

xa = xa(ul u'")

and ya = ya(xl, . . . , xn), a = 1, 2, . . . , n, then

aya _ aya axaaui CIO aui

ashowing that the aui transform like a contravariant vector.

axaHowever, if we consider a as fixed, the -- , i = 1, 2, . .. , m,

auiare covariant vectors in the V.; for if u' = ui(ul, .. . , u'n),

axa axa andwe have - = - - We propose to consider tensors of thisaui our au'

type, Latin indices indicating tensors of the V., and Greek indicesindicating tensors of the V..

Let us consider the tensor A7. We wish to derive a new tensorwhich will be a tensor in the V. for Greek indices and a tensor inV. for Latin indices. We consider a curve C in V. given byu` = ui(s) and by xa = xa(s) in V,,. Let bi be the components of avector field in V. parallel along C with respect to and let cabe the components of a vector field in V. parallel along C withrespect to V. We have

kdbi

ds +rrkb,dd

s= 0

dca dx#dsds=0

(565)

We now consider the product bicaA;. In V. this product isan invariant (scalar product) for each i, and in V. it is a scalarinvariant and is a function of are length s along C. Its derivativeis

SEc. 1381 FURTHER APPLICATIONS OF TENSOR ANALYSIS 319

i

ds(b°c«A;) bica

dsi +biAi"

ds + ds c°A;° d k\

= bica d' + A; ragd3

- Air,kds )

making use of (565). Since b1 and ca are arbitrary vectors, and

sinced

ds(b'c«A;) is a scalar invariant, it follows from the quotient

law thatdA°_

j + `4'r ,, dsdx#

- A`r'dull

ds k ds(566)

is a tensor of the same type as A7. We call it the intrinsic deriva-tive of A; with respect to s.

We may write (566) as

aA° axe \ duk(auk + A; r«a auk - Air;)

ds

k k

and since this is a tensor for all directions ds (the directions

of C are arbitrary), it follows from the quotient law that

axeAff:k =a

aAuk°

+ A;r auk - Air?k (567)

is the generalized covariant derivative of A7 with respect to theV,,,.

Problems

1. Why is A", a contravariant vector in V,i?ill

2. Show that

aA" ax* ax?Aa;:, = r,A,;

au,- r#rA

aut- rAflk

is a mixed tensor, by considering the scalar invariant b8c'daA';.a

3. Show that x = x`; = aus, and that


_ a2xa44 = au' aui

- r .xh + r;yx x

= xai + FRxx4.

Show that ga$xQ4 k = 0, and show by cyclicpermutations that gaaxa;x

k= 0.

5. The x°i of Prob. 4 are normal to the vectors 4k, the tangentvectors to the surface. Hence the x"i are components of a vectornormal to the subspace Vm. If Ni are the components of a unitnormal to Vm, we must have x = b;)Na. We call B = b;i du' duithe second fundamental form. Show that b;i = gapx" NP. Ifthe V is a Euclidean V3, gas = as, show that b11 = e, b12 = f,b22 = g (see Sec. 35) for the subspace r = r(u, v).

139. Riemannian Curvature. Schur's Theorem. Let us con-sider a point P of a Riemannian space. We associate with P twoindependent vectors X,, A2. These vectors determine a pencilof directions at P, given by

a1Al + a2A2 = aiX;

Every pair of numbers at, a2 determines a direction Ea. Sincethe geodesics are second-order differential equations, the point Pand the direction Sa at P determine a unique geodesic. The locusof all geodesics determined in this manner will yield a surface.In a Euclidean space the surface will be a plane, since the geo-desics are straight lines and two vectors determine a plane.

We now introduce normal coordinates ya with origin at P.The equations of the geodesics take the form ya = has, where

dyaka

= C dsp, and the geodesic surface is given by

ya = a'8A1 + a2sX2= ut?4 + u2A2 = uiX7 (568)

j summed from 1 to 2 and ul = a's, U2 = a2s.The element of distance on the surface is given by

ds2 = h;i du' dui (569)

and if ds2 = gab dya dyfi for the V, then

aya ayo= g X Ah = g (569a)asti ap au' aui


where the gas represent the components of the fundamentalmetric tensor in the system of normal coordinates.

Now let R;;kc be the components of the curvature tensor forthe surface S with coordinates ul, u2. Let us note the following:The gas of a Riemannian space completely determine the Chris-toffel symbols r;,, which in turn specify completely the Riemann-Christoffel tensor Ras,.a. Once the metric of a surface embeddedin a V. is determined, we can determine the F;k for this surface,and the R;;kc can then be determined. We need not make anyreference to the embedding space, V,,, to determine the R:;k1it is apparent that the ht; can be determined without leaving thesurface, so that all results and formulas derived from the h;; areintrinsic properties of the surface. All we are trying to say isthat ds2 = h21 dus du' is the fundamental metric tensor for aRiemannian space which happens to be a surface embedded in aRiemannian V,,. We shall use Latin indices for the space deter-mined by the metric h;; and Greek letters for the V,,.

The indices of R;;k1 take on the values 1 or 2, and from Prob. 1,Sec. 134, we have that

R1212 = R2121 = -R1221 = -R211281111 = 1?1122 = R1122 = . . = 81121 = . . = R2222 (570)

=0

If we make an analytic transformation, uti = uti(ul, u2), i = 1, 2,then

aua aub au` andR;;k1(u) = Rabcd(u) anti au' auk aul

and

au!, aub auc and81212 = Rabcd

aul au2 aa1 au2

au1 au2 au1 au2 2= R1212ka-lvau2

au2aul

by making use of (570). Thus

R 1212 =81212IJ1,LLaUa

aubMoreover, ds2 = h;; du= du', and h;; = hab -- --- so that

aut alai

(571)

(572)


IF = I hIJ 2. We rewrite (572) in the form

R1212 R1212

Ih IhI

(573)

Equation (573) shows that K is an invariant, and it is called theGaussian curvature. It is an intrinsic property of the surface.

We now determine an alternative form for K in terms of thedirections Ai and X and the curvature tensor for the V at thepoint P. The coordinate transformation between the Christoffelsymbols is given by (see Prob. 6, Sec. 129)

ay- 8y8 ayT a2yP ay°h;1rik(u) = gaprs.,(y) au' auj auk + gaa 8u1 auk aui

which reduces to

har3k = g.pr,,'X AgAk (574)

aya a2yasince aui - '' auk auj

0, from (568).

At the point P, rP1(y) = 0 (see Sec. 131), so that h;irik = oor himhiir;k = r, = 0. Hence the curvature tensor can bewritten

R1212(P) = h11Rg12(P)[ar1(P)

= h1iar'22(P)i

au2 au1 J(575)

from (538) and (541).From (574), h11r`21(u) = XRX , so that at the origin

of Riemannian coordinates

hitor,

= gaN y* (576)

since ag-0 = g,,911.,. + g,ar p = 0 at the origin (Prob. 6, See. 128),y

and from (569a) auk = X'XOXk agy = 0. SimilarlyY

"h1i 12 = gap ayr (577)

Sec. 139] FURTHER APPLICATIONS OF TENSOR ANALYSIS 323

Using (538), (541), (576), (577), it is easy to show that

R1212 = A72A1t2RapyrFinally,

and

so that

Thus

Ihl _hll

h2l

hi,

h12

h22

h12

h22hllh22 - hit

aya c3y9= =gap ga9Xx1

c3u1 c3u1

=gaAX2

Ihi = x7x2xip21'(gaagp. - ga,gpa)

7aps,X aj X 2'a1

2K

(579)

(580)

We are now in a position to prove Schur's theorem. If at eachpoint of a Riemannian space, K is independent of the orientation(X "j, as), K is constant throughout the space.

It follows at once if K is independent of X,, X7i that

Rap,,, = K(gaagp, - ga,gpe)and so

Rapa=, = K,N(gaegp, - ga,gp,,)Rap,, = K,,(gaagp, -- gaogpa)Raaµ.e = K,,(ga,gpµ - gw.gp,)

Adding and using Bianchi's identity, (543), we have

K.p(ga,gp - ga,gpa) + K.,(ga,,gsa -- ga,,gp,M) + K,n(ga,gp,, - gawgpr)=0

Multiplying the above equation by ga° and summing, we have

K, ,(ngp, - gp.) + K.,(gp - ngp,,) + K,,gp,, - K,,,gp, = 0or

(n - 2)gp,K,,, = (n - 2)K,,gpp

and gp,K,,, = gp K,,, or 5,'.K,,, = K,,, if n > 2. For n = 2 thereis no arbitrary orientation.


If we choose o = r X u, K, = 0. This is true for all µ since ILcan be chosen arbitrarily from I to n. Hence K = constantthroughout all of space. Such a space is said to be of constantcurvature.

Problems1. Derive (571).2. Derive (575).3. Derive (578).4. For a V3 for which gii = 0, i 54 j, show that if h, i, j are

unequal,

Rii =1

- Riaai9hh1 1

R, = - Raica + - Rhiihgii gii

s 1

R = Riiii9ii9ii

5. If R; = gaiRii, show that R"a =l aR2 axi

6. If Rii = kgii (an Einstein space), show that R = giiRii = nk,or Rii = (R/n)gii

7. Show that a space of constant curvature K is an Einsteinspace and that R = Kn(1 - n).

140. Lagrange's Equations. Let L be any scalar invariantfunction of the coordinates q', q2, . .. , qn, their time deriva-tives 41, 42, ... , 4n, and the time t:

L = L(qi, 4i, t) = L(qi, qi, t)

If we perform a transformation of coordinates,

qa = ga(gl, q2, . . . , qn)

agaa = 1, 2, . . . , n, then qa = aqa so that 4a is a function of

the Now

aL aL aqa aL 84aaqi = aqa aqi + a4a aqi

aL aqa aL a2ga41= aqa aqi + a4a a4i

aq0(581)


where we consider q' and q' as independent variables in L. Nowalso

aL _ aL a4 aL aqaaq' aqa aq; aqa aqr

so thatddt \agi1 - ala a \aq"I + aq q8 aqt (582)

Subtracting (581) from (582), we obtain

d d 8L,) c

dt(IL)

/ aq'aqaq-

L dt \aqa qla

dwhich shows that the - (aqa aqa

are the components of a

covariant vector.For a system of particles, let L = T - V, where T is the

1 (dI /s;12 _1

kinetic energy; T2

ma 2:=1 i1

V (1 2 s 1 2 s sxI x1 x1 x2, x2f x2f 7 xn)

is the potential function, - aV = (F*),. Thene

_d aL _ aL d _ n 1 8ga0 aV- (magma) _ m; x,a2 + -.dt ax; ax; dt 2 ax * ' ax;

=max;-(F,)a, ifgaQ=aaeand m,x; - (FT), = 0 for la Euclidean space and Newtonian

mechanics. HenceW \-x ax vanishes in all coordinate

systems. We replace the x; by any system of coordinates q',q2, . . . , qn which completely specify the configuration of thesystem of particles, and Lagrange's equations of motion are

d (;)aL _ aL 0, r= 1 2 n (583)aq=

' ' '

Example 150. In spherical coordinates, a particle has thesquare of the velocity v2 = t2 + r262 + r2 sin2 0 rp2, so that


L = T - V = 2 (t2 + r262 + r2 sin2 0 02) - V

- = m(r92 + r sing 0 #2) - aV

_d aL=

dt armr

and one of Lagrange's equations of motion is

mr - m(r62 + r sin 2 9 02) + aV = 0c1r

Since - aV represents the radial force, the quantity

r - (r62 + r sing 0 02)

must be the radial acceleration.If no potential function exists, we can modify Lagrange's equa-

tions as follows: We know that T = (m; 2)gapxaO is a scalarinvariant, so that

d aT aTQr = dt

(BT)axr

(584)

are the components of a covariant vector. In cartesian coordi-nates, the Q, are the components of the Newtonian force, so thatQ, is the generalized force vector. If f, are the components of theforce vector in a y'-yL - - -y" coordinate system, then

ayaQr - fa axf

anda

Qr dx r = fa 4 9 y dxr = fa dyaa

is a scalar invariant. The reader will immediately realize thatfa dya represents the differential of work, dW, so that

Q+aW

(585)= ax'

SEC. 1401 FURTHER APPLICATIONS OF TENSOR ANALYSIS 327

We obtain Qi by allowing xi to vary, keeping x', x2, , xi-',xi+', . . . , x% fixed, calculate the work A Wi done by the forces,and compute

d WiQi = lim -, i not summed..L- O AV

Example 151. A particle slides in a frictionless tube whichrotates in a horizontal plane with constant angular speed co.The only horizontal force is the reaction R of the tube on theparticle. We have T = (m/2) (j62 + r2A2), so that (584) becomes

mr" - mr02 = Q,

dt (mr2o) = Qe (586)

with Q, = 0, Qe =R(r de)

do= rR. The solution to (586) is

r = Aew' + Be", gR = 2mw dt, since 6 = w.

Problems

1. A particle slides in a frictionless tube which rotates in avertical plane with constant angular speed w. Set up the equa-tions of motion.

2. For a rigid body with one point fixed,

T =JA(w;+wy)+jCwa

Using Eulerian angles, show that if Q, = 0, then

C(O + cos 0 +') = RA¢sin20+Rcos0=S

A#-A,2sinocoso+R,1 sino =Qe

where R, S are constants of integration.aT

3. If T = q")gags, show that 2T =aq-

q°°.

4. Define p, =oL(q,

aandand assuming we can solve for

qq* = gr(gl) . . , q^, p,, show that the Hamiltonian


II defined by H = pages - L satisfies

II = T + V = It (a constant)

where V = V (q', . . , q"), T = aas(q', . . . , gn)gags. Alsoshow that

aHa ' pr

all = g.ap'.

These are Hamilton's equations of motion; the pr are called thegeneralized momentum coordinates. Show that they are thecomponents of a covariant vector.

5. By extremalizing the integral f " L(x1, x', t) dl, show thatLagrange's equations result.

6. If the action integral

A = f è [(h - '-6 dxs

]1

dXV )gas a aAis extremalized, show that the result yields

d aT IT aVdt Cair

- ax' = - ax,

where T + V = h, the constant of energy. The path of theparticle is the same as the geodesic of a space having the metric

ds2 = 2M(h - V)gas dxa dxs

141. Einstein's Law of Gravitation. We look for a law ofmotion, which will be independent of the coordinate system used,describing the gravitational field of a single particle. In thespecial theory of relativity, the line element for the space-timecoordinates is given by

ds2 = -dx2 - dy2 - dz2 + c2 dt2-dr2 - r2 do2 - r2 sin2 0 dp2 + c2 d12 (587)

In the space of x, y, z, t, the gas are constants and the space is


flat (Euclidean), so that BJk: = 0. For a gravitating particlewe postulate that the Ricci tensor Ri, vanish (see Probs. 5 and 6of this section). Since Ri; = R;;, a four-dimensional space yieldsn(n + 1)/2 = 10 equations involving the gi; and their deriva-

tives. From Prob. 7, Sec. 134, we have R ,i = 1 aR2 ax;

, where

R = gi°R°;, R = g°pRay, and for j = 1, 2, 3, 4 the 10 equationsare essentially reduced to 6 equations.

We assume the line element (due to Schwarzchild) to be ofthe form

ds2 = -e"''' dr2 - r2 d82 - r2 sin2 0 dp2 + e'(') dt2 (588)

so that our space is non-Euclidean. We do not include termsof the form dr do, etc., because we expect our space to be homo-geneous and isotropic. We have

911 = -ea,g22 -r2, g33 = -r2 sin2 0, 944 = e'

gi, = 0, i j (589)and

g11 = -e,

Now r 1',k - 2 9

i o e, we have

g22 -r-2' g33 =

g44 = e-', gii = 0,- (r2 sin2 6)-1

iPd j

ago; agke - ag;k

(axk + ax' ax°and since g'° = 0 for

rik n Caxk + ax's - agi) i not summed

If i, j, k are different, then rk = 0. We also see that

_ 1 ,i agiirii 2 g axi

rik =

2

9ii giikk

1 agkkrkk = - 2 9ii axi

(590)

Applying (590), we have


1

r22

1 _ 1 ,tag, _ 1dar11 2 g Or 2 dr

= _ 1g11 a92z =

1

1441 =

112 =

2 or

_ 1 g11 ag =aa _2 Or

-re a

-r sine 0 e-A

1 1 ag44 1 dv--g - _1 ag 1

z or

s zzs2

raas=

r1aa=

rt = - 94

2g

Or r_ 1 22 a9aa

2 9 00 =an..

2 dr

(591)

- sin 0 cos 0

2933

ae= cot 0

4 1 dv8g1

gsa2

Or r1 agsa

44 44

2 Or 2 dr

and all other r;k vanish.From (539)

R;; _

so that

ar:axe - axa + rar8, - r°raa

a 4

R11 - ar1R + art, + ar14 +rilril + risrs1 + ri3r 1ar Or Or

+ ri4ril - r1i(ri1 + rig + ria + r44)1 1 1 d2D 1 1 1(dv12 1 dar2 r2+2dr2+r2+r2+4`drl 2r dr

l da l da dv2r dr 4 dr dr

(592)

by making use of (591). Hence Einstein's law R;; = 0 yields

_ 1 d2v 1 (_f)2 1 A dv 1 AR11 2 dr2 + 4 Cdr 4 dr dr r dr - 0


Similarly

R22 = e-a C 1 - r (dr dr) I- 1= 02

1 (ddrv _ daR33 = sine 8 C 1 +

2 r dr I - sin2 0 = 0 (593)

r_ 1 dzv 1 (dv 2 1 dX dv 1 dvl _R44 =

e.-aL 2 dr2 4 dr + 4 dr dr r drJ - 0

Dividing R44 by e'-x and adding to R11, we obtain

d +dr =o

or

X + v = constant = co

We desire the form of (588), as r --f oo, to approach that of(587). This requires that X and v approach zero as r approachesco.. Hence \X + v = 0 or X P. From R22 = 0 we have

ev C1 + r dr) = 1. Let y = so that dr = 1 dy and'Y r

yCl+ydr)or

dy dr

1-y r

and

2m(594)r

where 2m is a constant of integration.The equations of the geodesics are

d2x' dx' dxka82 + r;k d ds = 0

which yield

d20 2 dr d_9

+2

CdLp)2

= 0Z-2 + 2T E ds A r33

Vs J


or

a dB aIf 0 = 2,ds

= 0 initially, then 0 =2

satisfies (595) and the

boundary conditions.We also obtain

ds2+1'11(ds/2

()2ds+ r44(d- 2

or

dt 1 da (dr)2a (thp)2 1 v-' dv (dt)2

ds2 + 2 dr ds - re-x

+ 2e

dr ds/

making use of 0 = 7r/2.Also

d29 a dr dcp

ds22I'19 =

ds dsd21 dt dr

0 or

ds2 + 2x14 ds ds -0 or

Integrating (597) and (598), we obtain

r2d(p

= hAdt c

ds = y

(599)

(600)

where h and c are constants of integration. Equation (588)becomes

d20 2 dr db (j2d,82

+r ds ds

- sin 0 cos 8ds

0 (595)

d2 2drdipds2 +r dsds

d2t dv dt dr

=0

= 0 (596)

=0 (597)

ds2 + dr ds ds- 0 (598)

dtlogd + v = log c or

ds2 = - 1 dr2 - r2 d(P 2 + y dt2y

or

1

1

\ds2 + y2-

\ds2 - r2yor

11

(r d(p)2 - r2 r4 + y (l zy


or

Ch drl 2 h.2 2m 2m h2

r2 dip/ + r2- c2 - 1 r + r

r2

and writing it = 1/r, we obtainz -()2

au + u2 =C

h21 + h2 u + 2mu8

and differentiating, we finally obtain

d 2

+ u = h2 + 3mu2d2U(601)

We obtain an approximate solution of (601) in the followingmanner: We first neglect the small term 3mu2 = 3m/r2, for large

z

r. The solution ofaddz + u = h2 is

T =u=-[l+eeos((p-w)jwhere e, w are constants of integration. This is Newton's solu-tion of planetary motion. We substitute this value of u in theterm 3mu2, and we obtain

d2u m 3m3 6m3W,p2+u =h2+h; + h4 ecos(cp-w)33

+ 23 h4 [1 + cos 2((p - co)]

We now neglect certain terms which yield little to our solutionand obtain

d z 3

d2U + u = h2 + 6m'e Cos (,p - w)

From the theory of differential equations the solution of our newequation is

z

U = h2 + e cos ((p - w) + h2 sin ((p - w)

= h2 [1 + e cos (rp - co - e)J, approximately

where e = (3m2,'h2)cp and e2 is neglected.


When the planet moves through one revolution, the advanceof the perihelion is given by 8(w + e) = (3m2/h2) 3 = 6irm2/h2.When numerical results are given to the constants, it is foundthat the discrepancy between observed and calculated results onthe advance of the perihelion of Mercury is removed.

Problems1. Derive (591).2. Derive (593).3. For motion with the speed of light, ds = 0, so that from

(599), h = oo, and (601) becomess

d2u + u = 3mu2 (602)

2

Integrate d Y + u = 0, replace this value of u in 3mu2, andP

obtain an approximate solution of (602) in the form

u = co`p+ 2(coal jp+2sin2(p)

where R is a constant of integration. Since u = 1/r, x = r coa rp,y = r sin tp, show that

m x2 + 2y2x = R + R(x2+y2)1

The term (m/R) (x2 + 2y2)/(x2 + y2)} is the small deviation of thepath of a light ray from the straight line x = R. The asymptotesare found by taking y large compared with x. Show that theyare x = R + (m/R) (± 2y) and that the angle (in radians)between the asymptotic lines is approximately 4m/R. This istwice the predicted value, on the basis of the Newtonian theory,for the deflection of light as it passes the sun and has been veri-fied during the total eclipse of the sun.

4. If R;; = ag;; is taken for the Einstein law, show that ify = el, then y = 1 - (2m/r) - }are and

d2Ud2 + U = h2 + 3mu2 -

3 h2 us

5. Assume the following: ds2 = gap dxa dxi, gap = 0 for a p&agaa dxa dx4

gae1, 8x4

= 0,ds ' 0, a = 1, 2, 3, %:W

ds1,


41 _ J944+ constant

xl = x, x2 = y, xa = z, x4 = ct. Show that the equations of thedzi

geodesics reduce to Newton's law of motion die + a ' = 0,

i = 1, 2, 3.

6. With the assumptions of Prob. 5 show that R44 = 0 yieldsLaplace's equation V24 = 0.

142. Two-pointTensors.

The tensors that we have studiedhave been functions of one point. Let us now consider thefunctions ga,s(xl, x2) which depend on the coordinates of twopoints. We now allow independent coordinate transformationsat the two points M1(x;, x;, . . . , x;), M2(x4i x2, xs, . . . ,If in the new coordinate systems xl, 22 we have

8xi 8xg2ga.0( 1, 22) = g, (xl, x2) 8-i - (603)

then the go.,A are the components of a two-point tensor, a covariantvector relative to Ml and a covariant vector relative to M2.Indices preceding the comma refer to the point M1, indicesfollowing the comma refer to M2. If we keep the coordinates ofM2 fixed, that is, if 2_ = x', then (603) reduces to

dxi

1

(604)xs) = gw.a(xl, x2) 8xso that relative to Ml, g,,s behaves like a covariant vector. Asimilar remark applies at the point M2.

We leave it to the reader to consider the most general type oftwo-point tensor fields. We could, indeed, consider a multipletensor field depending on a finite number of points. Whatdifficulties would one encounter for tensors depending on a count-able collection of points?

We may consider a two-point tensor field as special one-pointtensors of a 2n-dimensional space subject to a special group ofcoordinate transformations.

The scalar invariant

ds2 = ga,s(xl, x2) dxi dxs (605)

is an immediate generalization of the Riemann line element.Indeed, when x, and x2 coincide, we obtain the Riemann line


element. Assuming ds2 > 0 for a < t < 0, we can extremalize

fB(ga'odx" dx,f dsdt-dt dt

dt

and obtain a system of differential equations.

where

+ ra+.o,x-°ze1 + a,C..'6x1 axe2= 0z:1 1

x$ + r: xax2 + CO, 1 0 = 0

rye, =g ''° a x ' r,"# = g°'' axe

a9°.e _ a9°.,)C`

; a9µ.. _ ag°.

ax` axiz = 9"'ax1° axµ IF

9"'`9µ.i = a;7rdx;

x1 = -ds

(606)

(607)

(608)

The unique solutions of (606), xi(s), x4(s), subject to theinitial conditions x1(8o) = ao, X '(so) _ Oo, 1 '(so) = a', xs(so) = o;,are called dyodesice, or dyopaths.

Problems1. Derive (607).2. Show that the CQ,#(x1, X2) are the components of a two-point

tensor, a mixed tensor relative to M1, and a covariant vectorrelative to M2.

3. Show that the law of transformation for the linear connec-tion ri, is

ax; a2xi axiaxi'9X'(2 1, x2) = r°";,(xl, x2) azi awl ax; + ax; awl axi

4. Show that rip, = r , if and only if g.,p = axa, where1

,p,g(x x2) is a scalar relative to M1 and a covariant vector rela-tive to M2. If also r: = r' , show that of necessity

a29°`'0 = axi axQ

where 4, is a scalar relative to both M1 and M2.

SEc. 142] FURTHER APPLICATIONS OF TENSOR ANALYSIS 337

5. Ifds2 = -el, dr, dr2 - r,r2eM dpi er dt, dt2

- 1 1 11 - m2M 'e = [1mM2

(m + L JM)2 rl (m + r' M)2 r2

el=(1+rlC1+m/

ex = e2µ-r

show that the two-point tensors

ax aro, a,c T.a .+

1 2

ar:'ax- 9x° a,T

1 2

vanish identically (m, M are constants). Show that the dyo-desics satisfy

rir2 ds1 = he-,,

rir2 = he-a

d t 1 __ds

Cie_

dt2C2e=ds

d2v Mm M

d,P2 +v

L 1 + [1 + (m/M))'hi 11 + (m/M)]2h2l

+3Mv2r1+(mMM)JL

2

provided that Mr2 = mr1, v = 1/r1, hi = (M/m)h. For m << M,

Mm/hi << 1, we haved

+ vM

+ 3Mv2, the Einstein solu-

tion for the motion of an infinitesimal particle moving in the fieldof a point gravitational mass M.

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Brillouin, L. "Les Tenseurs," Dover Publications, New York1946.

Graustein, W. C. "Differential Geometry," The MacmillanCompany, New York. 1935.

Houston, W. V. "Principles of Mathematical Physics,"McGraw-Hill Book Company, New York. 1934.

Joos, G. "Theoretical Physics," G. E. Stechert & Company,New York. 1934.

Kellogg, O. D. "Foundations of Potential Theory," John Mur-ray, London. 1929.

McConnell, A. J. "Applications of the Absolute DifferentialCalculus," Blackie & Son Ltd., Glasgow. 1931.

Michal, A. D. "Matrix and Tensor Calculus," John Wiley &Sons, Inc., New York. 1947.

Milne-Thomson, L. M. "Theoretical Hydrodynamics," TheMacmillan Company, New York. 1938.

Page, L. "Introduction to Theoretical Physics," D. VanNostrand Company, New York. 1935.

Phillips, H. B. "Vector Analysis," John Wiley & Sons, Inc.,New York. 1933.

Smythe, W. R. "Static and Dynamic Electricity," McGraw-Hill Book Company, New York. 1939.

Thomas, T. Y. "Differential Invariants of Generalized Spaces,"Cambridge University Press, London. 1934.

Tolman, R. C. "Relativity Thermodynamics and Cosmology,"Oxford University Press, New York. 1934.

Veblen, O. "Invariants of Quadratic Differential Forms,"Cambridge University Press, London. 1933.

Weatherburn, C. E. "Elementary Vector Analysis," GeorgeBell & Sons, Ltd., London. 1921.

"Advanced Vector Analysis," George Bell & Sons, Ltd.,London, 1944.

339

3.10 VECTOR AND TENSOR ANALYSIS

"Differential Geometry," Cambridge University Press,London. 1927.

"Riemannian Geometry," Cambridge University Press,London. 1942.

Wilson, W. "Theoretical Physics," vols. I, II, III, Methuen& Co., Ltd, London. 1931, 1933, 1940.

INDEX

Acceleration, angular, 186centripetal, 30-31, 184Coriolis, 210-211linear, 30, 184, 210-211

Action integral, 328Addition, of tensors, 275

of vectors, 2

Angular momentum, 196-200Angular velocity, 22Are length, 71, 100Archimedean orderingArcs, 98

rectifiable, 98, 100regular, 98

Arithmetic n-space, 268-269Associated vector, 281Associative law of vector addition, 3Asymptotic directions, 81Asymptotic lines, 81Average curvature, 78

B

Bernoulli's theorem, 234-235Bianchi's identity, 308Binormal, 58, 312Biot-Savart law, 163Boundary point, 90Boundary of a set, 91Bounded set, 89Bounded variation, 99

C

Calculus of variations, 83-85Cartesian coordinate system, 280,

292Cauchy-Riemann equations, 122

Cauchy's criterion for sequences, 97Cauchy's inequality, 13Center of mass, 194Centripetal acceleration, 30-31, 184Ceva's theorem, 8Characteristic curves, 69Charges, 127

moving, 146Christoffel symbols, 289-293

law of transformation of, 290-291Circulation, 238Closed interval, 89

A

Continuity, 95equation of, 231-232uniform, 95

Contraction of a tensor, 275-276Contravariant tensor, 274-275Contravariant vector, 270-272Coordinate curves, 52Coordinate system, 9, 269Coordinates, geodesic, 303-305

Riemannian, 305transformation of, 269

Coriolis acceleration, 210-211

Components of a vector, 8, 270-271Conductivity, 162Conductor, 128

field in neighborhood of, 130-131force on the surface of, 131

Conformal space, 294Conjugate directions, 80Conjugate functions, 122, 143Connected region, 102-103Conservation of electric charge, 162Conservative field, 103

Closed set, 91Commutative law

tion, 3of vector addi-

Complement of a set, 90Components of a tensor, 274

postulate, 92

341


Couple, 197Covariant, differentiation, 295-296

generalized differentiation of, 318-319

Covariant tensor, 274-275Covariant vector, 273Curl, 45, 55, 297, 300

of a gradient, 46, 297Currents, displacement, 168

electric, 161-162Curvature, average, 78

of a curve, 58, 311Gaussian, 78, 322lines of, 78Riemannian, 307, 320-323tensor, 306-307

Curve (see Space curve)Curvilinear coordinates, 50, 70

curl, divergence, gradient, Lapla-cian in, 54-55

D

D'Alembertian, 178Del (v), 40Deflection of light, 334Deformation tensor, 246Desargues's theorem, 7Derivative, covariant, 295-296

intrinsic, 297-298of a vector, 29

Determinants, 263-267cofactors of, 264derivative of, 266multiplication of, 264

Developable surfaces, 70Diameter of a set, 93Dielectrics, 135-136Differentiation, covariant, 295-296

generalized covariant, 318-319

rules, 32of vectors, 29

Dipole, 157-158energy of, 158field of, 158magnetic, 160-161moment of, 157potential of, 157

Direction cosines, 13Directional derivative, 38, 297Discontinuities of D and E, 138-

139Displacement current, 168Displacement vector, 136Distributive law, 3, 11, 21Divergence, 42, 54, 120, 297-298

of a curl, 46of a gradient, 44

Divergence theorem of Gauss, 114-120, 299

Dot, or scalar, product, 10Dynamics of a particle, 189Dynamics of a system of particles,

194Dyodesics, 336

E

Edge of regression, 69-70Einstein, Albert, law of gravitation,

328-329space, 324special theory of relativity, 283-

286summation notation, 259

Einstein-Lorentz transformations,283

Electric field, 127discontinuity of, 138polarization of, 158-159

Electromagnetic wave equations,170-173

Electrostatic dipoles, 157-158Electrostatic energy, 136-138Electrostatic field, 127Electrostatic flux, 128Electrostatic forces, 127Electrostatic intensity, 127Electrostatic polarization, 158Electrostatic potential, 128Electrostatic unit of charge, 127Electrostatics, Gauss's law of, 128

Green's reciprocity theorem of,139-140

Ellipsoid of inertia, 226of strain, 245

INDEX

Energy, equation for a fluid, 235-236of electromagnetic field, 175of electrostatic field, 136-138kinetic, 201

Envelopes, 69Equation, of continuity, 231-232

of gauge invariance, 177of motion for a fluid, 233-236,

302-303Equipotential surfaces, 129Euclidean space, 8, 279, 308-309Euler's angular coordinates, 219-221

equation of motion, for a fluid, 233,302-303

for a rigid body, 216-217Euler-Lagrange equation, 85Evolutes, 66

F

Faraday's law of induction, 167Field, 9

conservative vector, 103nonconservative vector, 104solenoidal vector, 117steady, 9uniform, 10

Fluid, 230general motion of, 236-238

Force moment, 196-200Foucault pendulum, 213-215Frenet-Serret formulas, 60, 311-312Functions of bounded variation, 99-

100conjugate, 122continuous, 95

properties of, 96harmonic, 123

Fundamental forms, first, 71second, 74-75

Fundamental planes, 62-63normal, 62osculating, 62rectifying, 63

G

Gauge invariance, 177Gauss, curvature, 78, 322

343

Gauss, divergence theorem of, 114-120

electrostatic law of, 128Generalized force vector, 326

momentum, 328Geodesic coordinates, 303-305Geodesics, 83, 288-289

minimal, 294Gradient, 36, 120, 273, 297Gravitation, Einstein's law of, 328-

329Newton's law of, 190

Green's formula, 118, 299-300Green's reciprocity theorem, 139-

140Gyroscope, motion of, 222-225

H

Hamilton's equations of motion, 328Harmonic conjugates, 123

functions, 123, 143Heine-Borel theorem, 94Helix, 60Hooke's law, 249Hypersurfaces, 282

I

Images, method of, 141-143Induction, law of, 167Inertia, moment of, 216

product of, 216tensor, 225-228

Inertial frame, 211Infemum, 92Inhomogeneous wave equation, 177

solution of, 178-182Insulator, 129Integral, line, 101, 103-105

Riemann, 101Integrating factor, 111Integration, of Laplace's equation,

145of Poisson's equation, 155

Interior point, 90Interval, closed, 89

open, 89


Intrinsic equations of a curve, 63Invariant, 271Involutes, 64Irrotational motion, 232Irrotational vectors, 107, 111

J

Jacobian, 49, 265Jordan curves, 98

K

Kelvin's theorem, 239Kepler's laws of planetary motion,

191-193Kinematics, of a particle, 184

of a rigid body, 204-207Kinetic energy, 201Kirchhoff's solution of the inhomo-

geneous wave equation, 178-182Kronecker delta, 260-262

L

Lagrange's equations, 324-327Laplace's equation, 123

integration of, 145solution in spherical coordinates,

146-149uniqueness theorem, 119

Laplacian, 45, 298-299in cylindrical coordinates, 55in spherical coordinates, 56, 299

Law of induction, 167of refraction, 139

Legendre polynomials, 148-149Legendre's equation, 148Limit point, 90Line, of curvature, 78

element, 279of Schwarzchild, 329

of force, 132integral, 101, 103-105

Linear function, 3set, 89

Liquids, general motion of, 233-234Lorentz's electron theory, 175-177

transformations, 61, 283

M

Magnetic dipole, 160-161effect of currents, 162-164

Magnetostatics, 160Mass of a particle, 189, 285Maxwell's equations, 167-169

for a homogeneous conductingmedium, 173

solution of, 169-173Menelaus' theorem, 8Meusnier's theorem, 75Minkowski force, 285Moment of inertia, 216Momentum, 196

angular, 196-200generalized, 328relative angular, 199-200

Motion, in a plane, 33irrotational, 234relative, 187-188steady, 234vortex, 238-239

Moving charges, 161-162Mutual induction of two circuits,

165-166

N

Navier-Stokes equation, 255-257Neighborhood, 90Newton's law of gravitation, 190Newton's law of motion, 189, 211Nonconservative field, 104Normal acceleration, 184

plane, 62to a space curve, 58, 311to a surface, 73, 109

Number triples, 15, 268-269

O

Oersted, magnetic effect of currents,162-165

Ohm's law, 162Open interval, 89Open set, 90Orthogonal transformation 292-293Osculating plane, 62

INDEX 345

P

Parallel displacement, 313-315Parallelism in a subspace, 315-317Parametric lines or curves, 71Particle, acceleration of, 30, 210

angular momentum of, 196dynamics of, 189kinematics of, 184mass of, 189, 285momentum of, 196Newton's laws of motion for, 189,

211rotation of, 22velocity of, 30, 184, 209

Particles, system of, 194Perihelion of Mercury, 334Permeability, 160Planetary motion, 190-193Point, 89

boundary, 90interior, 90limit, 90neighborhood of, 90set theory, 89

Poisson's equation, 132-134integration of, 155

Poisson's ratio, 249Polarization, 158-159Potential, of a dipole, 157

electrostatic, 128vector, 117velocity, 232

Power, 162Poynting's theorem, 174-175Poynting's vector, 175Pressure, 230Principal directions, 77-78

Q

Quadratic differential form, 280Quotient law of tensors, 276

R

Radius of curvature, 24Recapitulation of differentiation

formulas, 48Reciprocal tensors, 281

Rectifying plane, 63Refraction, law of, 139Regions, connected, 102

simply connected, 102-103Regular arcs, 98Relative motion, 187-188

time rate of change of vectors, 208Resistance, electric, 162Retarded potentials, 178-182Ricci tensor, 307Riemann integral, 101Riemannian, coordinates, 305

curvature, 307, 320-323metric, 280space, 280

geodesics in, 288-289hypersurface in, 282

Riemann-Christoffel tensor, 307-308Rigid bodies, 203

motion of, 215-225

S

Scalar, 1curvature, 307gradient of, 36, 273, 297Laplacian of, 45, 298-299product of vectors, 10, 273-274

Schur's theorem, 323-324Schwarzchild line element, 329Second fundamental form, 74-75

geometrical significance, 75Sequence, 97

Cauchy criterion for convergenceof, 97

Set, 89bounded, 89closed, 91complement of, 90countable, 93diameter of, 93infemum of, 92limit point of, 90linear, 89open, 90supremum of, 91theorem of nested, 93

Simply connected region, 102


Sink, 231Solenoidal vector, 117Solid angle, 160Source, 231Space, conformal, 294Space curve, 31

arc length of, 100-101curvature of, 58, 311intrinsic equations of, 63Jordan, 98on a surface, 72radius of curvature of, 58tangent to, 31, 58, 311torsion of, 59, 312unit binormal of, 59, 312unit principal normal of, 58, 311

Space of n-dimensions, 268-269Special theory of relativity, 283-286Spherical coordinates, 35, 50

indicatrix, 68Steady field, 9Stokes's theorem, 107-112, 300-301Strain, ellipsoid, 245

tensor, 243-246Streamline, 234Stress tensor, 246-248Subtraction of vectors, 3Summation convention, 259Superscripts, 14, 259Supremum, 91-92Surface, 70

are length on, 71asymptotic curves on, 81average curvature of, 78conjugate directions on, 80curves on, 72developable, 70first fundamental form of, 71Gauss curvature of, 78geodescis of, 83normal to, 73, 109principal directions on, 77second fundamental form of, 74

T

Tensor, components of, 274contraction of, 275-276contravariant, 274-275covariant, 274-275curvature, 306-307deformation, 246inertia, 225-228mixed, 275Ricci, 307Riemann-Christoffel, 307-308strain, 243-246stress, 246-248two-point, 335weight of, 275

Tensors, 274-278absolute, 275addition of, 275cross product of, 278outer product of, 278product of, 275quotient law of, 276reciprocal, 281relative, 275

Theorem, of Ceva, 8of Desargue, 7of Menelaus, 8

Top, motion of, 222-225Torque, 196-200Torsion of a space curve, 59, 312Transformation of coordinates, 269Trihedral, 59Triple scalar product, 23Triple vector product, 24Two-point tensors, 335

U

Uniform continuity, 95Uniform vector field, 10Uniqueness theorems, 119Unit charge, 127

V

Vector, associated, 281basis, 3, 8center of mass, 194

Tangent to a space curve, 31, 58, 311 components of, 8-9Tangential acceleration, 184 conservative field, 108

INDEX 347

Vector, contravariant, 270-272covariant, 273curl of, 45, 55, 297, 300definition of, 1differentiation of, 29displacement, 136divergence of, 42, 54, 120, 297-298field, 9irrotational, 107, 111length of, 1, 281operator del (v), 40physical components of, 272potential, 117solenoidal, 117space, 268-269sum of a solenoidal and an irrota-

tional vector, 156-157unit, 1, 281zero, 1

Vectors, addition of, 2, 275angle between two, 10, 281-282differentiation of, 29equality of, 1fundamental unit, 8linear combination of, 3

Vectors, parallel, 2, 314parallel displacement of, 313-315scalar, or dot, product of, 10, 273-

274subtraction of, 3triple scalar product of, 23-24triple vector product of, 24-25vector, or cross, product of, 20-23

Velocity, angular, 22-23linear, 30, 184, 209potential, 232

Vortex motion, 238-239

W

Waves, equation of, 170inhomogeneous equation of, 177longitudinal, 253-254transverse, 172, 253

IN'eierstrass-Bolzano theorem, 92Work, 103, 202

Y

Young's modulus, 249

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Vector and Tensor Analysis