An Introduction toTensor Calculus,

Relativity and Cosmology

Third Edition

,Y

D. F. LawdenDepartment ol Mathematics

The Unil'ersity (f Aston in Birminyham

• 'I ~ "f.( "

1807~19ll2fill<..,H\-

JOHN WILEY & SONSChichester' New York' Brisbane' Toronto' Singapore

First puhli.\hed 1962hy Methuen & Co Ltd

Second edition 1968Reprinted twice

First puhlished as a S,'ience Paperhack 1967hy Chapman and Hall Ltd

II New Feller Lane, London EC4P 4EEReprinted 1975

(Cd 962, 1967, 1975 D. F. LllWden

Copyright « 1982 by John Wiley & Sons Ltd.

All rights reserved.

No part of this book may be reproduced by any means, nortransmitted. nor translated into a machine language without thewritten permission of the publisher.

Library oj ConKre",. ('utah'KinK in Publication Data:

Lawden, Oerek F.An Introduction to tensor calculus. relativity and cosmology.- Jrd ed.

Rev. cd. of: An Introduction to tensor calculus and relativity. 2nd ed.Bibliography: p.Includes index.I. Relativity (PhySICS) 2. Calculus of tensors. J. Cosmology. 1. Title

QCI7.1.55.U8 1982 5JOI'1 81 14801ISBN 0471 HXl82 X (cloth) AACR2ISBN 0471 IlXl96 X (paper)

Briti..h Library Catal"Kuing in Publication Data:

Lawdcn. D. F.An introduction to tensor calculus, relatIvItylllld cO'll1ology. .lId cd.I. RelatIvity (Physics)2. Calculus of tcnsorsJ. Cosmology-- Mathcmatics1. Title5.10.1'1 QCl7.155

ISBN 0471 10082 X (cloth)ISBN 0471 10096 X (paper)

Typeset by Macmillan India Ltd, Bangalore, and printed in the United States of America,by Vad-Ballou Press Inc. Binghamton, N.Y.

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Contents

Preface, .... ,

List of Constants, , . , .

ix

xiii

Chapter I Special Principle of Relativity. Lorentz Transformations II. Ncwton's laws of motion ... , . I2. Covariance of the laws of motion 33. Special principle of relativity . . . 44. Lorentz transformations. Minkowski space- time 65. The special Lorent" transformation . . . . . 96. Fitzgerald contraction. Time dilation . . . . 127. Spacelike and timelike intervals. Light cone 14Exercises I 17

Chapter 2 Orthogonal Transformations. Cartesian Tensors 21K Orthogonal transformations. . 219. Repeated-index summation convention. . 23

10, Rectangular Cartesian tensors. . 24II. InvarIants. Gradients. Derivatives of tcnsors . 2712. Contraction. Scalar product. Divergencc 2813. Pseudotcnsors. . . 2914. Vector products. Curl 30Excrcises 2 . 31

Chapter 3 Special Relativity Mechanics 3915. The velocity vector. . . . 3916. Mass and momentum .. , 4117. The force vcctor, Energy. . 4418. Lorentz tra nsforma tion eq ua tions for force. 4619. Fundamental particles. Photon and neutrino. 4720. Lagrange's and HamIlton's equations 4821. Energy momentum tensor. .... , . 5022, Energy-momentum tensor for a fluid 5323. Angular momentum 57Exercises 3 . 59

VI

Chapter 4 Special Relativity Electrodynamics. 7324. 4-Current density. 7325. 4-Vector potential. . . . 7426. The field tensor. . . . . 75

I 27. 'Lorentz transformations of electric and magnetic vectors 772R. The Lorentz force. . . . . . . . 7929. The energy- momentum tensor for an electromagnctic field 79Exercises 4 . . . . . . . . . . . . . . . . . . X2

Chapter 5 General Tensor Calculus, Riemannian Space. X630. Generali~ed N-dimensional spaces. X631. Contravariant and covariant tensors. X932. The quotient theorem. Conjugate tensors. 9433. Covariant derivatives. Parallel displacement. Alline connection l)'i

34. Transformation of an alnnity . 9X35. Covariant derivatives of tensors . . . . . . . 10036. The Riemann Christolfel curvature tensor. 10237. Metrical connection. Raising and lowering indices. 1053X. Scalar products. Magnitudes of vectors. 10739. Geodesic frame. Christollcl symhob . lOR40. Bianchi identity. . . . III41. The covariant curvature tensor. . . II I42. Divergence. The Laplacian. Einstein's tensor 11243. Geodesics 114Exerciscs 5 . . II"!

Chapter 6 General Theory of Relativity 12744. Principle of equivalence 12745. Metnc in a gravitational field. 13046. Motion of a free particle in a gravitational field. 13347. Einstein's law or gravitation. 1354X. Acceleration of a particle in a weak gravitational Ileid 13749. Newton's law of gravitation. . . . 13950. Freely falling dust cloud. . . . . . 14051. Metrics with spherical symmetry. 14252. Schwarzschild's solution. . . . . . 14553. Planetary orbits. . . . . . . . . . . 14754. Gravitational deflection of a light ray. 15055. Gravitational displacement of spectral lines. 15256. Maxwell's equations in a gravitational field. 15457. Black holes. . . . . . 155'IX. Gravitational waves. 159Exercises 6. 163

Chapler 7 Cosmology59. Cosmological principle. Cosmical time60. Spaces of constant curvature61. The Rohertson Walker metric.62. H uhble's constant and Ihe deccleral ion parameter.6:\. Red shIft "I galaxies64. Lunllnosity dIstance65. CosmIC dynamICs.66. Model universes of l'instein and de Sitter67. Fnedmann lllllverscs6X. Radlilt IOn model69. Particle and event horiIons .Exercises 7

References

Bibliol:raphy.

Index.

VII

174174176IXOIXIIX2IX3IX5IXXIX9193195197

199

200

201

Preface

The revolt against the ancient world view of a universe centred upon the earth,which was initiated by Copernicus and further developed by Kepler, Galileo andNewton. reached its natural termination in Einstein's theories of relativity.Starting from the concept that there exists a unique privileged observer of thecosmos. namely man himself, natural philosophy has journeycd to the oppositepole and now accepts as a fundamental principle tha t all observers are equivalent.III the sense that each can explain the behaviour of the cosmos by application ofthe same sct of natural laws. Another line of thought whose completedevelopment lakes placc within the context or special relativity is that pioneeredby Maxwell. electromagnetic field theory. Indeed. smce the Lorent? transf.)rm­ation equations upon which the speCIal theory is based constitute none other thanthe transformation group under which Maxwell's equations remain of invariantform. the relatIvistic expression of these equations discovered by Minkowski ismore natural than Maxwell's. In the history of natural philosophy. therefore,relativity theory represcnls the culmlllati,)H of thrce centuries of mathematicalmodellIng of the macroscopic physical world; it stands at the end ofan era and is amagnificent and tittlng memorial to the golden age of mathematical physics whichcame to an end at llIe tUlle of the First World War. Eiw.tein's triumph was also histragedy: although he was inspired to c,'eatc a masterpiecc. this proved to be amonument to the past and its very perfection a barrier to future development.Thw;, although all the implications of the general lheory have not yet beenuncovered, the barrenness of Emstein's later explorations indicates that thegrowth areas of mathematical physics lic elsewhere. presumably in the fecund soilof quantum and elementary-particle theory.

Nevertheless, relativity theory, especially the special form, provides a found­ation upon which all later developments have been constructed and it seemsdestined to continue in this role for a long time yet. A thorough knowledge of itsclements is accordingly a prereqUIsite for all students who wish to understandcontemporary theorics of the phySIcal world and possibly to contribute to theirexpansion. This being universally recognized, university courses in appliedmathematics and mathematical physics commonly include an introductorycourse in the subject at the undergraduate level, usually in the second and thirdyears, but occasionally even III the tirst year. This book has been written toprovide a suitable supporting text for such courses. The author has taught thistype ofelass for the past twenty-live years and has become very familiar with thedifliculties regularly experienced by students when they first study this subject:

IX

the identification of these perplexities and their careful resolution has thereforeheen one of my main aims when preparing this account. To assist the studentfurther in m,lsterillg the subject, I huve collected together a lal'ge numher ofexercises and these will he found at the end of each chapter: most h<lve been set uscourse work or In examinatIOns for my own classes <md, I think, cover almost allaspect> normally treated at this level. It is hoped, therefore, that the book will alsopwvc helpful to lecturers as a source of prohlems for setting in exercise classes.

When preparing my plan for the development of the subject, I decided todisregard completely the historical order of evolution of the ideas and to presentthese in the most natural logical and didactic manner possible. In the case of afully established (and, indeed, venerable) theory, any other arrangement for anIntroductory text is unjustifiable. As a conseqnence, many facets of the suhjectwhich were at the centre of attention during the early years of its evolution haveheen relegated to the exercises or omItted entirely. For example, details of thesemlIwl Michelson Morley experiment and its associated calculations have notheeo included. Although this event was the spark which ignited the relativistictmder, It is now apparent that this was an historical accident and that, beingImplicit in Maxwell's principles of electromagnetism, it was inevitable that thespecial theory would be formulated near the turn of the century. Neither is theexperiment any longer to he regarded as a crucial test of the theory, since thetheory's manifold implication" for all branches of physics have providedcOllntless ot her checks, all of which ha ve told 111 its favour. The early controversiesattending the birth of relativity theory are, however, of great human interest andstudents who WIsh to follow these arc referred to the books hy Clark, HolTmannand Lanczos listed in the Bibliography at the end of this book.

A curious feature of the history of the special theory is the persistence ofcertainparadoxes which arose shortly after it was first propounded by Einstein andwhich wcre largely disposed of at that time. In spite of this, they are rediscoveredevery decade or so and editors of popular scientific periodicals (and occasionally,and more reprehensibly, serious research journals) seem happy to provide spacein which these old battles can be refought, thus generatIl1g a good deal ofacrimony on all sides (and, presumahly, improving circulation). The source of theparadoxes is invariably a failure to appreciate that the ,special theory is restrictedin its validity to inertial frames of reference or an inahility to jettison theNewtonian concept ofa unique ordering ofevents in time. Complete books basedon these misconceptions have been published by authors who should knowbetter, thus giving students the unfortunate impression that the consistency ofthis system of ideas is still in douht. I have therefore felt it necessary to mentionsome of these 'paradoxes' at appropriate points in the text and to indicate howthey are resolved: others have been used as a basis for exercises, providingexcellent practice for the student to train himself to think relativistically.

Much of the text was originally published in 1962 under the title AnIntroduction to J ensor Calru/us and Re/atiFity, All these sections have beenthoroughly revised in the light of my teaching ex perience, one or two sections

XI

have heen discarded as containing material which has proved to be of littleimportance for an understanding of the hasi"s (e.g. relative tensors)and a numberof ncw s(:ctions have heen added «(~.g. equa tions of motion of an clastic fluid, blackhoJes, gravItatIOnal waves, and a more detailed account of the relationshipbetween the metnc and amne connections). But the main improvement is theaddition ofa chaplercovering the application of the general theory to cosmology.As a result of the great strides made in the development of optical and,particularly, radio astronomy during the last twenty years, cosmological sciencehas moved towards the centre of interest for physics and very few universitycourses 111 the general theory now fail to include lectures in this area,

It is a common (and dcslrable) practice to provide separate courses in the specialand general theories, the special being covered in the second or third under­grndua te year and the general in the linal year of the undergraduate course or thefirst year ofa postgraduate course. The book has been arranged with this in mindand the first four chapters form a complete unit, suitable for reading by studentswho may not progress to the geFieral theory. Such students need not be burdenedwith the general theory of tensors and Riemannian spaces. hut can acquire amastery of the prInciples of tlte speCial theory using only the unsophisticated toolof Cartesian lcns()rs in Euclidean (or quasi-EuclIdean) space. In my experience.even students who intend to take a course in the general theory also benefit fromexposure to the special theory in this form. since it enables them to concentrateupon the dilliculties of the relatiVity prinCiples and not to be distracted byavoidahle complexities of notation. I have no sympathy with the teacher who.encouraged hy the shallow values of the times. regards it as a virtue that hislectures exhibit his own present mastery of the subject rather than hisappreciation of his students' bewilderment on being led into unfamiliar territory.All students should. 111 any case. he a ware of the simpler form the theory of tensorsassumes when the transformation group is restricted to be orthogonal.

As a consequence of my decision to develop the special theory within thecontext of Cartesian tensors. it was necessary to reduce the special relativisticmetric to Pythagorean form by the introduction of either purely imaginaryspatial coordinates or a purely imaginary time coordinate for an event. I havefollowed Minkowski and put X4 = ict; thus. the metric has necessarily been takenin the form

ds 2 = dx l2 +dx/ +dx/+d'\4 2 = dx 2 +dy 2 +d;:2_('2d/ 2

and ds has the dimension of length. I have retained this definition of the intervalbet ween two events ob~rved from a freely falling frame in the general theory; thisnot only avoids confusion but, in the weak-field approximation. permits thedistinction between covariant and contravariant components of a tensor to beeliminated hy the introduction of an Imaginary time. A disadvantage is that ds isimaginary for timeltke intervals and the interval parameter s accordIngly takesimaginary values along the world-line of any material body. Thus, when writingdown the equations for the geodesic world-line of a freely falling body, it is

XII

usually convenient to replace s by r, defined by the equation s = in, r being calledthe proper time and dr the proper time interval. However, it is understoodthroughout the exposition of the general theory that the metric tensor for space­time gij is such that ds 2 = gij dx' dxJ : a consequence is that the cosmical constantterm in Einstein's equation of gravitation has a sign opposite to that taken bysome authors.

References in the text are made by author and year and have been collectedtogether at the end of the book.

I), 1-, L.AwmN

Department oj Mathematics.{'he University of Aston in Birmingham.May. 1981.

List of Constants

In the 51 system of units:Gravitational constant = G = 6.673 x lO" m' kg I S 2

Velocity of light = ,. = 2.99X X 10" ms 'Mass of sun = 1.991 x 10,10 kgMass of earth = 5.979 x 1024 kgMean radius of earth = 6.371 x 10" mPermittivity of free space = "0 = KX54 x 10 '2

Permeability of free space = 110 = 1.257 x 10 6

xiii

20

frame 5' relative to S il~ observed from S. the event B occurs a time T before A,What is the distance between the events as observed from 57 (Assume D > c1',)

(Ans. u = 2e 2 DT/ (D2 + c2T 2), 15 = D.)

18. In the frame S, a particle is projected from 0 at t = 0 with the velocitycomponents (le, k, 0) and thereafter moves so that its acceleration is constant ofmagnitude 9 and is directed along the y-axis in the negative sense. Write down thex and y coordinates of the particle a time 1 later. The particle's motion is observedfrom the 5 frame. If u = k, using the transformation equations (5.X), obtainequations for its coordinates (.X, J')at timeTin this frame. Deduce that the anglemade by the velocity of projection with the x-axis is 60' and that the particle'sacceleration is 4Xg/25.

19. The clocks at the origins of the frames S, S are synchronized to read zerowhen they pass one another and the clocks stationary in either frame aresynchronized with the clock at the origin of the frame. At time 1 in S, the clock at 0passes a clock C fixed to the x-axis of S. Show that C registers 1 and 0 registersJ (I - 1/2/(,2)t at this instant. Observed from S. clock C is slowed by a factorJ (I - u 2

/(2

); in this frame, therefore, it might be expected that when C registers I,

o would register t/ J (I - u2 /c 2). Resolve the paradox by showing that, in the S

frame, the clock C is not synchronized with 0, but that C is always U2

1/(,2 ahead ofO.

20. In the frame S, at t = I, a particle leaves the origin 0 and moves withconstant velocity in the xy-plane having components Vx = 5('/6, vy = c/3. Whatare the coordinates (x, y) of the particle at any later time I? If the velocity of Srelative to Sis u = 3('/5, calculate the coordinates (x, y)ofthe particle at time 1 in Sand deduce that the particle makes its closest approach to 0 at time T = 220/113.

21. S, Sare the usual parallel frames with the origin 0 of5 moving along the x­axis of S with velocity u, An observer A is stationed on the x-axis at x = a and anobserver If is stationed on the x-axis at x = a. Show that, in both the frames, theevents (i) 0 passes 0 and (ii) A passes If, are separated by a time T, but that theorder of occurrence of these events is dillerenL Calculate the value of T. If T= a/3(', show that u = 3c/5.

CHAPTER 2

Orthogonal Transformations. CartesianTensors

8. Orthogonal transformations

In section 4 events have been represented by points in a space t!4' The resultingdistribution of points was described in terms of their coordinates relative to a setof rectangular Cartesian axes. Each such set ofaxes was shown to correspond toan observer employing a rectangular Cartesian inertial frame in ordinary 1,('3­

space and clocks which are stationary in this frame. In this representation, thedescriptions of physical phenomena given by two such inertial observers arerelated by a transformation in 1,('4 from one set of rectangular axes to another.Such a transformation has been given at equation (4.9) and is called an orthogonaltransformation. In general. If Xi. X,(i = 1.2..... N)are two sets of N quantitieswhich are related by a linear transformation

N

Xi = L uijxJ+h j

j ~ 1

and. if the coefficients a'l of this transformation are such that

N N

L (x, - YY = L (Xi - VYj", I 1::0;- 1

(8.1)

(8.2)

is an identity for all corresponding sets x" Xi and Yi. y" then the tramformation issaid to be orthogonal. It is clear that the X" ,Xi may be thought of as thecoordinates ofa point in I,('N referred to two different sets ofrectangular Cartesianaxes and then equation (8.2) states that the square of the distance bet ween twopoints is an invariant. independent of the Cartesian frame.

Writing Zi = X, - y" Zi = .x, - J". it follows from equation (8.1) that

N

i l = L uijz)j-=:- I

(R.3)

Let Z denote the column matrix with elements Zi' i the column matrix withelements i, and A the N x N matrix with elemenb aij' Then the set of equations

21

22

(8.3) is equivalent to the matrix equation

z = Az

Also. if z' is the transpose of z.

N

" ,2L. "i = I

and thus the identity (R.2l may be written

But. from equatit)n (H.4l.

;;' = ;;'A'

(8.4)

(1'.5)

(8.6)

(H.7)

Substituting ill the left-hand member ofequation (1'.6) from equations (RA). (1'.7).

it will be found that

This can only be true for all z if

;;' A' Az = z'z

A' A = I

(1'.8)

(1'.9)

where I is the unit N x N matrix.Taking determinant,; of both members of the matrix equation (8.9), we tind that

IAI 2 = I and hence

IAI = ± I (IUO)

A is accordingly regular. Let A ~ I be its inverse. Multiplication on the right byA - I of both members of equation (8.!)) then yields

It now I(lllows that

A' ~~ A I

AA' = AA I = I

(H.II)

(IU2)

(IU3)

Let ()'j be the i;th element of l. so that

bij:~: ;~~}The symbols t),j are referred to as the K ron('ckcr deltas. Equations (8.9). (8.12) arenow seen to be equivalent to

N

L ajju jk = ()jk

i == I

(8.14)

(8.15)

(91)

23

respectively. These conditions are necessarily satisfied by the coefficients auofttetransformation (8.1) if it is orthogonal. Conversely, if either of these conditions IS

satisfied, it is easy to prove that equation (X.6) follows and hence that tbetransformation is orthogonal.

9. Repeated-index summation convention

At this point it is convenient to introduce a notation which will greatly abbreviatefuture manipulative work. It will be understood that, wherever in any term of mexpression a literal index occurs tWIce, this term is to be summed over all possiblevalues of the index. For example, we shall ahbreviate by writing

N

L. urh, = a,h,r -= I

The index must be a literal one and we shall further stipulate that it must bl asmall letter. Thus a2h2' aNhN are individual terms of the expression a,h" and nosummation is intended in these cases.

Employing this convention, equations (8.14) and (8.15) can be written

£ljl(1ki = li jk(~2)

respectively. Again, with Zi = Xi - Y.. equation (8.2) may be written

(U)

More than one index may be repeated in the same term, in which case morethan one summation is intended. Thus

N N

ulJhjk£"k = L. L. U1jhjk('kj - 1 k:::; I

flA)

It is permissihle to replace a repeated index hy any other small letter, provided thereplacement index does not occur elsewhere in the same term. Thus

19.5)

hut

[9.6)

irrespective of whether the right-hand member is summed with respect to j arnot.A repeated index shares this property with the variable of integration in a de1initeintegral. Thus

b b

lJ (x)dx = U(y)dy (9.7)

A repeated index is accordingly referred to as a dummy index. Any other inde, willbe called a free index. The same free index mu~t appear in every term of anequation, hut a dummy index may only appear in a single term.

24

The reader should note carefully the identity

(9.8)

for it will be of frequent application. (jij is often called a suhsriwt;on operator,since when it multiplies a symbol such as a j , its etlect IS to replace the index} by i.

10. Reetangular Cartesian tensors

Let Xi' y, be rectangular Cartesian coordinates of two points Q, P respectively in""N' Writing Zi = Xi - Y" the Zi are termed the components of the di,\placeml'/Itvector PQ relative to the axes being used. If ,X;, .Vi are the coordinates ofQ, P withrespect to another set of rectangular axes, the new coordinates will be related tothe old by the transformation equations (8.1). Then, it';:i arc the components ofPQ in the new frame, it follows (equation (8.3)) that

(10,1)

Any physical or geometrical quantity A having N components Ai defined in thex-frame and N components A, similarly defined in the .x-frame, the two sets ofcomponents heing related in the same manner as the components Zi' Zi of adisplacement vector, i.e. such that

(10.2)

is said to he a (lector in 1\N relative to rectangular Cartesian reference frames. Weshall frequently abbreviate 'the vector whose components are A,' to 'the vectorA,'.

If A" Biare two vectors, consider the N 2 quantities Ai Bj • Upon transformationof axes, these quantities transform thus:

(10.3)

Any quantily having N 2 components C j defined in the x-frame and N 2

components (',j defined similarly in the x-frame, the two sets of componentshemg related by a transformation equation

(10.4)

is said to be a tensor of the second rank. We shall speak of 'thc tensor C i/. Such atcnsor is not, necessarily, representable as the product of two vectors.

A set of N 3 quantities Dij, which transform in the same mannt:r as the productof three vectors AiBl'., form a tensor of the third rank. The transformation law is

(10.5)

The generalization to a tensor of any rank should now be obvious. Vectors are,of course, tensors of the tirst rank.

If Air B,} are tensors, the sums Aij + Bij are N 2 quantities which transformaccording to the same law as the Ai} and Bij• The sum of two tensors of the second

25

rank is accordingly also a tensor of thts rank. This result can be generalizedimmediately to the sum of any two tensors of identical rank. Similarly, theditTerence of two tensors of t he same rank is also a tensor.

Our method of intwducing a tensor implies that the product of any number ofvectors is a tensor. Quite generally, if A'j ,.' B;j , arc tensors of any ranks(which may he different), then the product A,j B., is a tensor whose rank isthe sum oftlie ranks of the two I'actors. The reader should prove this formally fora product such as A,j B"m, by writing down the transformation equations. (N.B.the indices in the two factors must he kept distinct, for otherwise a summation isimplied and this complicates matters; see section 12.)

The components ofa tensor may be chosen arbitrarily relative to anyone set ofaxes. The components of the tensor relative to any other set are then fixed by thetransformation equations. Consider the tensor of the second rank whosecomponents relative to the x,-axes are the Kronecker deltas ''iii" In the ,xi-frame,the components are

(10.6)

by equations (9.2). Thus this tensor has the same components relative to all sets ofaxes. It is termed the fundamell/al /ell.\or of the second rank.

If, to take the particular case of a third rank tensor as an example,

(10.7)

for all values of i, j, k, A;j' is said to he ,\Y'llllle/r;( with respect to its indices i,.i­Symmetry may he with respect to any pair of indices. If A'j' is a tensor, itsproperty of symmetry with respect to two indices is preserved upon transform­ation, for

Ii-jill. = LJ jl(J,mLJli.n Aim,.

= LJ1m(JJ101i." Amlll

= ;(jA: (lO.H)

where, in the second line, we have rearranged and put A'm. = AmI.' Unless aproperty is preserved upon transformation, it will be of little importance to liS, forwe shall later employ tensors to express relationships which are valid for allobservers and a chance relationship, true in one frame alone, will he of nofundamental significance.

Similarly, if

Aijk = - AJik (10.9)

for all values of i. j, k. A;j, is said to be skew-symmetric or an/i-symme/ric withrespect to its first two indices. This property also is preserved upon tranAorm­ation. Since A tI. = - A tlk' A tlk = O. All components of A'j' with the first twoindices the same are clearly zero.

A tensor whose components are all zero in one frame, has lero components inevery frame. A corollary to this result is that if A'j, " B'j arc two tensors of

26

the same rank whose corresponding components are equal in one frame. thenthey are equal in every frame. This follows because Ai; . - B'j . is a tensorwhose components are all zero in the first frame and hence in every frame. Thus. aten50r equation

(10.10)

is valid for all choices of axes.This explains the importance of tensors for our purpose. By expressing a

physical law as a tensor equation in 6'4. we shall guarantee its covariance withrespect to a change of inertial frame. A further advantage is ihat such anexpression of the law also implies that it is covariant under a rotation and atranslation of axes in $ 3. thus ensuring that the law conforms to the principlrs ofisotropy and homoqeneity o!, 5pace.

The tirst principle states that all directions in space ar,~ equivalent in regard tothe formulation of fundamental physical laws. Examples arc that the inertia of abody in classical mechanics is independent of its direction of motion and that thepower ofattract ion ofan electric charge is the same in all directions. In the vicinityof the earth. the presence of the gravitational fIeld tends 10 cloud our perceptionof the validity of this principle and the vertical direction at any point on thesurface is sharply distinguished from any horizontal dIrection. But this is a purelylocal feature and the crew of a spaceship have no ditliculty in accepting theprinciple. Mathematically, the principle requires that the equation expressing abasic physical law must not change its form when the reference frame is rotated.Laplace's equation V2 V = 0 is well known to possess this property, whneas theequation (l V/(lX + (l V/(ly + (I V/ilz = 0 does not; thIS explains why Llplace'sequation occurs so frequently in mathematical physics. whereas the otherequation docs not.

The second principle affirms that all regions in space are also equivalent. i.e.that physical laws are the same in all parts of the cosmos. Covariance under atranslation of axes is the mathematical expression of this requirement.

Both principles are almost certainly a consequence of the uniformity withwhich matter and radiation are distributed over the universe. It is doubtfulwhether either would be valid in a cosmos not possessing this property.

A less well-established third principle is that of spllllUl parity. This requires thatphysical laws should be impartial as between left- and right-handedness. Themathematical formulation ofa law obeying this principle will be covariant undera transformation from a right-handed 10 a left-handed Cartesian frame or viceversa. Another way of expressing this principle is that, if the universe wereobserved in a mirror. the laws which would appear to govern its behaviour wouldbe identical with the actual laws. For example. observation of the planetarymotions in a mirror would alter their senses of rotation about the sun, but the lawof gravitation would be unaffected. Although the more familiar laws are inconformity with this principle, those governing the behaviour of some funda­mental particles do not appear 10 have this simple symmetry. Provided that the

27

orthogonal transformations upon which our tensor calculus is being built are notrestricted to be such that IA I = + 1 (i.e. transformations between frames ofopposite handedness are permitted), all tensor (and pseudotensor) equations willbe in conformity with the principle of spatial parity.

From what has just been said, it is evident that the calculus of tensors is thenatural language of mathematical physics, relativistic or non-relativistic. Itguarantees that the equations being considered are of the type which canrepresent physicallaws. However, in classical physics, a three-dimensional theorywas adequate to ensure conformity with the principles of isotropy, homogeneityand parity of space. In relativistic physics, a four-dimensional theory is needed toincorporate the additional special principle of relativity.

11. Invariants. Gradients. Derivatives of tensors

Suppose that Vis a quantity which is unaffected by any change of axes. Then V iscalled a .\Ctllar invariant or simply an inr'llrionl. Its transformation equation issimply

(11.1 )

As will be proved later (section 24), the charge ofan electron is independent of theinertial frame from which it is measured and is, therefore, the type of quantity weare considering.

Ifa value of V is associated with each point of a region of <I' N, an invariant fieldis defined over this region. In this case V will be a function of the coordinates Xi'

Upon transformation to new axes, V will be expressed in terms of the newcoordinates Xi; when so expressed, it is denoted by V. Thus

(lUi

is an identity. The reader should, perhaps, be warned that Vis not, necessarily, thesame function of the Xi that V is of the Xi'

If Ai) is a tensor, it is obvious that V Ai} is also a tensor of the second rank. It i!therefore convenient to regard an invariant as a tensor of zero rank.

Consider the N partial derivatives DV/Dx i• These transform as a vector. T~

prove this it will be necessary to examine the transformation inverse to (8.1). hthe matrix notation of section 8, this may be written

X = A '(x-h)= A'(x-h)

having made use of equation (lU 1). Equation (11.3) is equivalent to

Xi = a;j(xj-h j )

where a;j is the ijth element of A'. But a;j = aJi and hence

Xi = llJi(Xj -hj)

(lU

(11.4)

(I U)

28

It now follows that

and hence that

ilV aV (lx) DV---- = -.----- = a··i1.\"j (lXj (lx, I) (1-'\ J

(11.6)

(11.7)

proving that (1 V / (IXi is a vector. It is called the W(/(Ji,,"t of V and is denoted hygrad V or V V.

If a tensor Ai) ... is defined at every point of some region of i'N, the result is atensorfield. The partial derivatives (I Ai) .. ./(lS, can now he formed and constitutea tensor whose rank is one greater than that PI' Ai)' We shall prove this for asecond rank tensor field A,j' The argument is easily made general. We have

hy e4uation (11.6).

('if. tll-,j = ~l= (a;r u /"A,,)(x. (·x.

(I tlX,= 1- (u"llj,A,,)

(.x, (lx.

(lA,.,~

= Ujrll j.,Ukt -~l~~~--(11.8)

12. Contraction. Scalar product. Divergence

If two indices arc made identical, a summation is implied. Thus, consider Aij,'

Then

(12.1 )

There are N" quantities A;j" However, of the indices in A;jj' only i remains free torange over the integers 1,2, ... , N, and hence there are hut N quantities Aij} andwe could put B; = AU}. The rank has heen reduced hy two and the process isaccordingly referred to as contraction.

Contraction ofa tensor yields another tensor. For example, if Bi = Ai}} then,employing equations (9.2),

(12.2)

Thus B; is a vector. The argument is easily generalized.In the special case of a tensor of rank two, e.g. Ai}, it follows that Xi; = A;;, i.e.

Ai; is an invariant. Now, if A;, B; arc vectors, A;B} is a tensor. Hence, AiB; is aninvariant. This contracted product is called the inner product or the scalar product

2

of the two vectors. We shall write

A,B,=A'B ( 12.:

In particular, the scalar product ofa vector with itself is an invariant. The positivsquare root of this invariant will he called the m(/!Jnitude of the vector. Thus, if Aithe magnitude of A" then

A 2 = A;A, = A'A = A2

In 1)3, if II is the angle between two vectors A and B, then

ABcosli = A·B

In I)N, this equation is used to define /I. Hence, if

A·B= 0

(12.'

(12

(12.<

then 1/ = 1rr and the vectors A, B are said to be ortllO!Jon(/l.If A; is a vectorfield, ?Aji'x j is a tensor. Bycontraction it follows thatt1A;/I'x i

an invariant. This invariant is called the div"'!Jcl!ce of A and is denote~ by div ,Thus

('AdivA = .. '

(lX.(12.

More generally, if A,j is a tensor field, (1 A'j II'x, is a tensor. This tens,derivative can now be cOlltracted with respect to the index r and any other indto yield another tenSt>r, e.g. l'A u /1" )' This contraction is also referred to as tdivergence of A,j with respect to the index j and we shall write

(12

13. Pseudotensors

'.!I,; is a pseudo tensor if, when the coordinates arc subjected to the tramformati(X. I), its components transform according to the law

'.ITij = IAI(/"(/JI '.!l"r (13

I A 1 being the determinant of the transformation matrix A. Since for orthogoltransformations IAI = ± I (equation (8.10)). relative to rectangular Cartesifmmes, tensors and pseudotensors are identical except that, for certain changesaxes, all the components of a pscudotem.or will be reversed in sign. For examlif in Iff 3 a change is made from the right-handed system of axes to a left-hanesystem, the determmant of the transformation will be - 1and the cornponent~

a pseudotensor will then be subject to this additional sign change.Let l'u. • be a pseudotem.or of the Nth rank which is skew-symmetric.",

respect to every pair of indices. Then all its components are zero, except thosewhich the indices i, j, ... , n are all different and form a permutation of

30

numbers 1,2, ... , N. The effect of transposing any pair of indices in cij ... n is tochange its sign. It follows that if the arrangement i,j, . .. , n can be obtained fromI, 2, ... , N by an even number of transpositions, then cij n = + 1"2. N'

whereas if it can be obtained by an odd number cij .. n = - 1'12 N' Relative tothe x;-axes let c l2 N = 1. Then, in this frame, cij .. n is 0 if i,j, '.' . , n is not apermutation 01 1, 2, ... , N, is + I if it is an even permutation and is - I Ilit is anodd permutation. Transforming to the .x',-axes, we tlnd that

(13.2)

But cij • is also skew-symmetric with respect to all its indices, since this is aproperty preserved by the transformation. Its components are also 0, ± Itherefore and cij .. • is a pseudotensor with the same components in all frames.It is called the fA'"i Cir'ila pseudotensor.

It may be shown without dilliculty that:(i) the sum or difference of two pseudotensors of the same rank is a

pseudotensor.(ii) the product of a tensor and a pseudotensor is a pseudotensor.

(iii) the product of two pseudotensors is a tensor.(iv) the partial derivative of a pseudotensor with respect to Xi is a pseudotensor.(v) a contracted pseudotensor is a pseudotensor.

Thus, to prove (iii), let '.!Ii , 'Bi be two pseudovectors. Then

'lI, 'Bj = I A 12ai,ajl '.!l" 'B1 = (li,ajl '.!l" 'B1 (13.3)

The method is clearly quite general. The remaining results will be left as exercisesfor the reader.

14. Vector products. Curl

Throughout this section we shall be assuming that N = 3, i.e. the space will beordinary Euclidean space.

Let A;, B; be two vectors. Then cu,A,B, is a pseudotensor of rank 5.Contracting twice, we get the pseudovector

(14.1)

(14.2)

whose components are

<£,: A2 B)-A)B2 }

<£2 - A)B, -A,B)<£) = A,B2 -A 2 B.

Provided we employ only right-handed systems of axes or only left-handedsystems in tff), <£; is indistinguishable from a vector. If, however, a change is madefrom a left-handed system to a right-handed system, or vice versa, thecomponents of <£i are multiplied by -1 in addition to the usual vectortransformation. Since it is usual to employ only right-handed frames, <£; is often

31

referred to as a vector (or an axial ('eC/ur) and treated as such. It is then called thevee tor produ('/ of A and B and we write

(t=AxB

Vector multiplication is non-commutative, for

(14.3)

(14.4)

having made usc of r,., = -- r'1" However, vector multiplication obeys thedistributive law, f()r

Ax(B+C)= r,j.A)(B.+C.) = r'1.AjB.+r,j.A/'.

=AxB+AxC (14.5)

We now introduce the abbreviated notation (IAd(l x ) = Ai. r Any index after acomma will hereafter indicate a partial differentiation with respect to thecorresponding coordinate; thus, A,. j' is a second denvative.

Now suppose A, is a vcdor field. We can first construct a pseudotensor of rank5 r,j.A" S' Contracting twice, we get the pseudovector

(14.6)

This has components

~,(lA 3 (lA 2

(lX2 (lX,

~2=(lA, (lA 3

(lX 3 11X,

~.,=(IA

2 (lA,

(lX, (lX2

(14.7)

and is denoted by curl A. It, also, is an axial vector.Equation (14.1) can still be employed to detine a vcctor product when either or

both of the vectors A, B arc replaced by pseudotensors. If only one is replaced by apseudovector the right-hand member or equation (14.1) will involve the productof two pseudovectors and a vector. The resulting vector product will then be avector. Similarly, by replacing A in equation (14.6) by a pseudovector, the curl orapseudovector is defined as an ordinary vector.

E:xercises 2

I. Show that, in two dimensions, the general orthogonal transformation hasmatrix A given by

(COS /I sin II)

A=- sin II cos II

32

Verify that IA I = 1 and that A - I = A', Tij is a ten~or in thi~ space. Write down infull the transformation equatIOns for all its components and deduce that 7;; is aninvariant.

2. x= Ax, x= Bx are two successive orthogonal transformations relative to

each of which T,j transforms as a ten.~or. Show that the resultant transformation x= BAx is orthogonal and that Tij transforms as a tensor with respect to it.

3. If A;. lJ, are vectors and X,jA,B j is an invariant. prove that X,j is a tcnsor.4. Vcrify that the transformation

IXI = 15(5x, -·14x l + 2x,)

IXl = -3(2x I +X2 +2x,)

is orthogonal. A vector field is defined in thc x-frame by the equations A, = xi,A l = x~, A, = x~. Calculate the field in the i-frame and verify that divA is aninvariant.

5. AUk is a tensor, all of whose components are zero, except for the following:A III = A 212 = 1, Alil = - 2. Calculate the components of the vector A,j;' Showthat the transformation

XI = ~(-3xI -6x l ·-2x,)

Xl = +( -2x I +3x l -6x,)

x,=~( 6x l -2x 2 --3x,)

is orthogonal and calculate the component AlB of the tensor in .x-frame. Writedown the equations of the inverse transformation. If B;j is a tensor whosecomponents in the .x-frame all vanish except that li 13 = 1, calculate Bil .

(Ans. (-1, 1,0); 120/343;6/49.)6. If A = (l - B) (l + 8)- I, where B is a skcw-symmetric matrix. show that A is

orthogonal. Taking

B=I-~ ~ ~I-2 0 0

calculate A and write down the rectangular Cartesian coordinate transformationequations x = Ax. In the x-frame, the tensor C;J is skew-symmetric and C 12

= C13 = I,C23 = O. Calculate thecomponentC12 in the x-frame. In the x-frame,all the components of the tensor D;jk vanish except the following D121' = - 1,j)'122 = 2, iil2 , = 5. Calculate the component Dill in the x-frame. Calculate thecomponents of the vectors D;jj and C,jD;jk in the x-frame. (Ans. CI2 = 1; Dill =- 980/729; (- 14/9, - 8/9, - 8/9); (35/9, - 32/9, -7/9).)

33

7. A,j is a tem.or field defined in the x-frame by the equation A.) = X,Xj'

Calculate its components at the point I' where XI = 0, X2 = X3 = I. Thecoordinates X; of a point in the x-frame arc related to the coordinates x, of thesame point in the x-frame by the equati'Hls

.~, =~(-3xl-6x2-2xd

X 2 = ~(-2x, +3\1 ··6xd

x, = ?(6x, . 2X2 -<h.d

Calculate the componcnt A" of the tensor field at P. In the x-frame, prove thatA;J'j == 4x" A, J") = 12. (Ans. A 'I = 64/49.)

X. x is the positipn vector ofa point P with respect to an origin O. 01' is rotatedthrough an angle 1/ about an axis whose direction is determined by the unit vectoru. If the new position vector of P is X, prove tltat

x = xcos 1/ + (I - cos 1/) (x' u)u + u x x sin 1/

Deduce th.lt tlte coordinate transformation generated when a rectangularCartesian frame 0 XIX1\] is rotated tltwugh an acute angle sin '(4/5) about anaxis through 0 having direction ratios (1, 2, 2) to give a new frame 0 X, .x 2 X 3 is

45 x, = 29xI + 2Xx l -- 20X3

45'~1 = -20x I +35x l +20x,

45x 3 = 2Xx, -4xl + 35x-,

9. Verify that the transformation

x, = ~(XI -Xx l +4x])

X 2 = ~ (4xI + 4Xl + 7x])

x, = !(Xx, -Xl -4x.d

is orthogonal. In the x-frame, the tensor A,j is skew-symmetric and All = A,]

= I, A13 = O. Calculate the component All in the x-frame. In the x-frame, all thecomponents of the tensor B,). vanish except tlte following: B"'ll = - I, B'll = 2,j3"'23 = 5. Calculnte the component B'I' in tlte x-frame. Calculate the com­ponents of tlte vectors B;JJ and Ai} B,j' in the x-frame. (Ans. Xil = 1/3; B'I'= IXX/729; 2/9, - 16/9, X/9; 47/27. 11/27, -10/27.)

10. In the x-frame in ,g"., a tensor field is defined by the equation A,j' = x?+2xj +x1. Calculate the divergence of the vector field A,ji' Also, calculate thecurl of the vector field Ai)j' (Am.. 16 (x, -+ Xl + x 3 ); 6(x l - x 3 ), etc.)

II. A pair of rectangular Cartesian frames are related by the equations

.x, = h (5xI -14X l + 2.x])

'\1= -.\(2\,+x l +2xd

x] = 1\ (10x, +2Xl -IIX.d

34

A'i' is a tensor, all of whose components v<tnish in the x-frame except thefollowing: A II I = A 222 = 2, A 112 = 4, Al33 = 13. Calculate (a) the componentsof the vector A;}) in the x-frame, and (b) it 123. If Bi} is a tensor whose componentsin the x-frame all vanish except 812 , 82." which are both unity, calculate Bll .

If Vis an invariant field given in the x-frame by V = xi, calculate the field in the x­frame and the components of grad Vin this frame, where .x, = x2 = .x-, = 9. (Ans.(a) (-12, -9,6); (b) -1396/225; B ll = -2/3; VV= (2, -4,4).)

12. IntheframeOxlx2x-"allcomponentsofA,jareleroexceptAI2 = -All= I. If the transformation equations to the .x-frame are

.XI =xlcosrx+x2 sinrx

X2 = -XI sinrx+x2cos rx

.\\ = XJ

prove that II 12 = I.13. If A'j = x~ + x; (i,j = 1,2,3), prove that (a) A;j'j = 2 (XI + X2 + X-' + x,)

(bl A,l''i = 12.14. Verify that the coordinate transformation

25.\1 = 9x I +20x l + 12x-,

25.X2 = 12x I -15x2 + 16x-,

5x-, = -4xI + 3x-,

is orthogol1dl and calculate the component 11-, II of the tensor A'j' in the x-frameif all its components in the x-frame vanish except for the following: A 123 = 25,A222 = - 6. All components of the tensor B,) in the .x-frame are zero, except forthe components B" (j = 1,2,3). Show that, in the x-frame, all the componentsH'l (j = 1,2,3) vanish. (Ans. ~ 192/25.)

15. (i) x is a column matrix such that x'x = I. If A = I - 2xx', where I is a unitmatrix, prove that A is orthogonal. If x' = rx(l, -2,3) where rx is a scalarmultiplier, calculate rx and hence find A. Written in matrix form, the coordinatetransformation betwccn two rectangular Cartesian frames is .x = Ax (A is thematrix just calculated). In the x-frame, all the components of the tensor B,j" arezero, except that B 1133 = - 20, B, 232 = 29. Calculate the component 82221 inthe x-frame. What is the value of B;;j) '? If <l:, is a pseudotensor with components(2,3,6) in the x-frame, lind its components in the .x-frame. (Ans. 82221 = 72/49;Bi,jj = -20; (0, -7,0).)

(ii) A,j(x l , x 2 , xd is a tensor field. If A'j = (i'ixL where i\, is the Kroneckerdelta, list the non-Lero components of ;)Aii/(h•. Deduce the value of iJA II /(lX I atthe point where .x, = I, X2 = 2, X-' = I. (Take the frames to be related as in (i).)(Ans. (\411 /(l.X I = 4/7.)

16. Verify that thc transformation

7x I = 3xI +6x 2 -2x-,

7X2 = 2x I - 3X2 - 6x-,

7.\-, = 6x I -2X2 + 3x-,

35

is orthogonal and calculate the component .432 of the tensor Aij in the x-frameifall its components in the x-frame are zero except for the following: All = 10,An = 29. If the only non-zero component of the tensor Bijk in the x-frame isBB2 = 343, calculate the component Bill in the x-frame. If <£'j is a pseudotensorand <£33 = 98, all other components being zero, calculate [12' (Ans. .432 = 6;Bill = 24; ['12 = -24.)

17. If A is a square skew-symmetric matrix, show that A 2 is symmetric. If,also,A3 = - A, show that the matrix B given by B = 1+ 2A 2 is orthogonal. If

A = ('-~-h

U h)° c-c °

prove that A3 = -- A provided u2 + h2 + c2 = 1. Taking u = 1/3, h = c = 2/3,calculate the orthogonal matrix B. Written in matrix form, the coordinatetransformation between two rectangular cartesian frames is .x = Bx. In the x­frame, the tensor eij has all its components equal to 1. Calculate the component(:23 in the x-frame. In the x-frame. a vector field Ai has components A I = xi + x~

+ xL A2 = A 3 = 0. Obtain formulae for the field's components in the .x-frame.Calculate the divergence of the field in both frames and show that the results areequal. (Ans. 132.1 = 91/81.)

18. (i) The equations

'X I = ~(UXI + 8X2 +4x,)

x2 = ~(4xI + 11X2 + 7x 3 )

.x3 = ~(8xI + X2 + ex,)

represent a transformation between rectangular Cartesian axes. Calculate thevalues of u, h, c. In the x-frame, all the components of the tensor A ij are zeroexcept An = 9. Calculate the component If31 in the x-frame.(Ans. U = 1, h = -4, C = -4; .431 = 4/9.)

(ii) If A,jkl is a tensor, prove that A,jk' is also a tensor. In the x-frame referred toin (i),all the components of the tensor A"k1 vanish except for the following: A 1131

= 4, A3133 = 5, A2222 = 18. Calculate all the components of the tensor Aijk , inthe x-frame and the (1,1 )-component in the x-frame. (Ans. 44/3).

19. Verify thal the transformation

,X I = i l\(9x I +20x2+ 12x3)

X2 = 21~(12xI -15x2 + 16x3)

x 3 = ~(4xI ·-3xd

is orthogonal and write down the inverse transformation. In the x-frame, all thecomponents of the tensor A,jk vanish except A III = 1. Calculate the componentif12 .1 in the x-frame. In the x-frame, all the components of the tensor B'jk vanishexcept that 8212 = 1,8."3 = B112 = 13121 = 2. Calculate the components of thevector B'j; in the x-frame. (Ans. AI23 = 432/3125; (3,0,4).)

36

20. Show that, for all angles rx, Ii, the transformation

\, = x, cos rx cos {i -+ X2 cos rx sin {i ~ .X" sin ,Cf.

\2 = -x,sin{i+x2eos{i

x, = x, sin Cf. cos {i + x 2 sin rx sin {i +x, cos rx

is orthogonal. Obtain the form taken by the tramformation when rx = {I = rr/4and. in this case, calculate the component If, '1.' of the tensor A i]" in the Y-frame,if the only non-..:cro components in the >;-framc are A ,,2.\ = All 1., = I. (Ans, 0,)

21. If A. B arc orthogonal matrices of the ,arne 01 der. prove that A B is

orthogonal. If

A = ~ L~ ~ :;) ,\ 0 0 5

B=A ( ~ ~ ~\-·4 0 3)

calculate the orthogonal matrix e = A B, The coordinates x, and .x; in tworectangular Cartesian frames are relatcd by the transformation x = ex, All thccomponents of the tensor A'l' vanish in the ,\-(rame except All' = I, A '11 = 25.Calculate thc component 11'.'2' in the x-frame. In the x-frame, the only non-/,ero

components of the tensor B'1kl arc H.\211 = -5. BU'l = 10. B.\ll,' = 15,C:lIeulate the components of the vcctor il'I.B,]" in the x-frame ami deduce itscomponents in the ,-frame. (Ans, A'l' ~. 4~/5; 4464/25, - 96/), - 4~/2S,)

22, 11''11,; is a pseudotensor, 'H, is a psnldnVl'ctor and E, is a vector, provc that'1Iil 'B, E] IS an invariant.

2.1, A vector field has components in the x-frame given by

A, = sin(x]x,), A l = n'sl\,x,). A, = tanC".x l )

Calculate curl A at the point x, = Xl = \\ = ~ >./rr. If the .x-fr'jme is obtainedfrom the x-frame by reversing the sense "fthe Xl-axis, what arc the componentsof curl A in the ,x-frame') (Ans, >./ rr(\ -+ 1/2 )2), -)rr(\ - 1/2 )2), - ~ ) (2rr);- ,/rr(\ + 1/2 )2), - )rr(\ - 1/2 J2). 1V (lit).)

24. x',. x, (i = 1,2,3) an: coordinates of the same point with respect to twodifferent rectangular Cartesian frames, H

X, = rx(ax, +2x l +5x,)

Y l = {i(x, -+ /lXl +2x,)

X.\ = ,'(2x, ··Ilxl +cx,l

where rx, /I, " are all positive, calculate the values of a, /I. y, lJ, h. ('. If A'lU is a tensorwhose only non-/,ero components in the x-frame are AI.II,' = I, AU]} = 20,calculate its component AJ 11' in the x-frame, In the ,x-frame, the only non-zerocomponents of the tensors A" Bi] are as follows: if, = 3, III = I, 8" = 5, Bll

= 3, Calculate the components of the tensor A;B;i in the x-frame. (Ans a = - 14,h = 2. ( = 10. rx = 1/15, {i = 113. " =.1/15; ,-1 111 , = 2/5: (3. -9.12),)

25. A is an anti-symmetric matrix such that A' = A. Prove that the matrix B

37

= I + A + A 2 is orthogonal. Show that the matrix

I 2)o 2

-2 °satisfics thc statcd conditions ,1I1d hencc calculate the orthogon<ll matrix B. Thccoordinatcs of a point relative to a pair of rectangular Cartesian frame, arerelated by the matrix transformation x = Bx. A tensor A,j has all its compon:ntszero in the ~-frame except that A]2 = 81. Calculate the component X12 in th: .x­frame. If the tensor Bij • has all its components zero in the x-frame except 11 1 2.\

= 729. calculate the component B.\21 in the x-frame. (Ans. Al2 = 32;

B,I21 = --121'.)26. In $.1' prove that

curl grad V = 0, div curl A = 0.

27. In If." prove that

(i) ~ik/ ~i.," = (},mbl. - ()•• il1m

(ii) ~,k/ ~i'm ~ 2b1m

28. In If]. show that

29. In ~.I' prove that

curl curl A = grad div A - V2 A

(Hill/: Employ ExerCise 27(i).)30. In $], prove that

(i) A x (B x C) = A' CB - A· BC

(id

31. In <1I"N. prove that

div VA = V div A + A . grad V

32. In $]. prove that

(i) curl V A = V curl A - A x grad V

(ii) div(A x B) = B'curl A - A 'curl B

(iii) curl(AxB)= B'YA-A'YB+AdivB-BdivA

(iv) grat!(A' 8) = 8·VA t A'V" t A XCllrl" -I" xcuriA

38

where

:n. If Au is a tensor and Bij = Ajh prove that Bu is a tensor. Deduce that if Auis symmetric in one frame, it is so in all.

34. Prove that bij'l;. = bj •

and that eu• e'mn has the value + I if i, j, k are all different and (/mll) is an evenpermutation of (ijk), -I if i,j, k are all different and (/mll) is an odd permutation of(ijk), and 0 otherwise. Deduce that

e ij• elmn = bj/,5 jm b.n + bimb jn "" + b,n,)jlb'm

- ,)in(ijm,)" - ,5j/(jjn".m - "im,)jlb'n

Hence prove that

35. In <53 , prove that

(i)

(ii)

(axb)'(cxd)= a'cb·d-a,db·c

(a x b) x (c x d) = [acdJb - [bedJa

= [abdJc-[abe]d

where [abel = a'bxc.

CHAPTER 3

Special Relativity Mechanics

15. The velocity vector

Suppose that a point P is in motion relative to an inertial frame S. Let ds ~e thedistance between successive positions of P which it occupies at times I, r + dlrespectively. Then, by equation (7.4). if dI is the proper time interval belweenthese two events,

(15.1)

where II = ds/dl is the speed of P as measured in S. Now, as shown in section 7, dIis the time interval between the two events as measured in a frame for whi;h theevents occur at the same point. Thus dr is the time interval measured by adockmoving with P. til is the time interval measured by clocks stationary in S.Equation (15.1) indicates that, as observed from S, the rate of the dock rrovingwith P is slow by a fador (I _1,2 /e l )1 /2. This is the phenomenon of time dilationalready commented upon m section 6. If P leaves a point A at I = 11 and arnves ata point Bat 1= 12 , the time of transit as registered by a clm:k moving withP willbe

"I 2 -I, = f (I_1'2/e 2)1'2dl (152)

The successive positions of P together with the times it occupies these pmitionswnstitute a series of events which will lie on the point's world-line in Minlowskispace time. Erecting rectangular axes in space time corresponding 10 therectangular Cartesian frame S, let x" Xi + dx, be the coordinates of adjacentpoints on the world-line. These points will represent the events (x. y, Z, I), (s + dx,

y +dy, Z+ dz, 1+ dl) in S. If (1:" "y' ",) are the components of the velocity vector vof P relative to S, then

dxp =-

x dl'dl'

tJ = ~

Y dl'

dzI' =, dl

(15.3)

v does not possess thc transformation properties of a vcctor relativc toorthogl)nal transformations (I.c. Lorcnl/. transformations) in space time It is a

39

(15.4)

40

vector relative to rectangular axes ~tationary in S only. However, we can define a4-ve/ociIY vee/or which does possess such properties as follows: dx; is adisplacement vector relative to rectangular axes in space- time and dT is aninvariant. It follows that dx,/dT is a vector relative to Lorentz transformationsexpressed as orthogonal transformations in space, time. It is called the 4-velocityvector of P and will be denoted by V.

V ;;an be expressed in terms of y thus:

dx; dx; d/ 1 1 . I 1.. = =(I-·"/e) lX;dT dl dT

by equation (IS.\). Also, from equations (4.4) we obtain

It now follows from these equations that

V = (I-v l /c l r 111(IJ" "y' "" ie) = (I -"l/clr 1/1(y, it")

(15.5)

(15.6)

where the notation should be clear without further explanation.Knowing the manner in whi;;h the components of V transform when new axes

are ;;hosen in spa;;e time, equation (15.6) enables us to ;;alculate how thecomponents of y tmnsform when S is replaced by a new inertial frame S. Thus,consider the orthogonal transformation (5.1) which has been interpreted as achange from an inertial frame S to another ,I{related to the first as shown in Fig. 2.The corresponding transformation equations for V arc

v, = VI cosrx+ V4 sinrx

~ = - VI sin rx + ~4 cos rx(15.7)

(15.8)

By equation (15.6), these equations are equivalent to

(l_Vljcl) 111V, = (l-I,ljcl)-111(Il,eosrx+lesmrx) }

(I -i//elr 111 Vy = (I _I,l/e l ) Ill",

(I --li/el)-lll l" = (I _IJ1/el)--1,11J

(I _Vl/,l)- Illk = (I - "l/e l )' 111( '-I'x sin rx +1(" cos rx)

where v is the velo;;ity of the point as measured in the frame S: Substituting forcos rx. sin rx from equations (5.7), equations (15.8) can be written

Vx = Q(vx -u) ')

vy = Q(I _Ul/Cl)1111'yj (15.9)

Ii, = Q(I_Ul/Cl)111 1J,

I = Q(I - UIJx/c l )

[I_i)l/("l JIll

where Q = (I =~2kl)( ,'_ ul7cl) (15.10)

41

Dividing the lirst three cqlIations (159) by the fourth, we obtain the specialLorentz transformal ion equations for v in their linal form, viz.

!' = ''x -u ..x I -.. III'..Ie2

(I - tl 2/(

2)' .2",r, = I ~u;-../e2

(I - I/ 2/( 2)'121',!'- =

I - IIl'x/e2

(15.11)

If II and " are small by comparison with e, equations (15.11) can be replaced bythe approximate equations

(15 2)

These arc equivalent to the vector equation (1.1) relating velocity measurernellsin two inertial frames according to Newtonian mechanics.

Since, by the fourth of equations (15.9), Qmust be real. equation (\ 5.10) imphcsthat ifii < e then" < c. Thus, If a point is moving with a velocity approaching (inSand S is moving relative to S with a velocity of the same order, the poirt'svelocity relativc to S will still be less than c. Such a result is, ofcourse,complctdyat variance with das.'tcal ideas. In particular, if a light pulse is being propag<Jledalong Ox so that r. = c,'',. ~ I', = 0, then it will be found that ,-" = e, ,',. = 1', = O.This conlirrm that light is propag<Jted with speed e in all inertial frames.

The transformation inverse to (15.11) can be found by exchanging 'barred' 'lI1d'unbarred' velOCity components and replacing 1/ by -'1/.

Suppose th<lt particles A and lJ movc along 1he x-axis of a frame S with speeds3e/4 in opposite directIons, both k<lvrng 0 at time I = O. At ,my later time I, thcirx-coordin<ltes will he x.4 = - 3et /4, x/j = 3el/4 and their distance apa!'t will bcx/j- XA = .lel/2. Clearly, this distance inneases at a rate 3e/2. However, this is lOttheir rclativc velocity and the fact that the rate excecds e docs not conflict with :herestIlt alrcady dcrived that no material hody can he ohserved from any inerrialframe to h<Jve a speed greater than e. There is no special-relativity prohibitionagainst the d,stance hetween two bodies increasing at a rate greater than c. To tlndthe velocity of IJ relative to A, It IS necessary to introdnee a second inertial frarre ,..with its orrgin at A; the velocity of H in S is then the velocity of B relative to A.Since the veloeity of S relative to S is tI = - 3e/4 and the velocity of B in S i, I'x

= 3("/4, the tlrst of equations (15.11) gives i\ = 24e/25 « c) as the velocity of Brelative to A.

16. Mass and momentum

In section 2 it was shown that Newton's laws of motion conform to the specialprinciple of rcla tivIty. Ilowever, the argUlTlcnt involved classical ideas concerning

42

space-time relationships between two inertial frames and these have since beenreplaced by relationships based upon the Lorentz transformation. The wholequestion must therefore be re-examined and this we shall do in this and thefollowing section.

We shall begin by considering the conservation of momentum, equation (1.:1),for the impact of two particles by which mass is defined in classical mechanics.Since the velocity vectors u" etc. are not vectors relative to orthogonaltransformations in space time, and indeed transform hetween inertial frames in avery complex manner, it is at once evident that equation (1.3) is not covariant withrespect to transformations het ween inertial frames. It will accordingly hereplaced, tentatively, by another equation, viz.

(16.1 )

where U" etc. are the 4-velocities of the particles and M" M 2 are invariantsassociated with the particles which will correspond to their classical masses. Thisis a vector equation and hence is covariant with respect to orthogonaltransformations in space time as we require. Equation (16.1) will be abhreviatedto the statement

EM V is conserved (16.2)

and then, hy equation (15.6), this implies th<Jt

Em(v, ie) is conserved ( 16.3)

(16.4)whereM

m = .(l_l,2/e 2)'12

By consideration of the first three (or space) components of (16.3), it will he clearthat

E mv is conserved (16.5)

and, by consideration of the fourth (or time) component that

Em is conserved (16.6)

[f, therefore, m is identified as the quantity which will play the role of theNewtonian mass in special relativity mechanics, our tentative conservation law(16.1) is seen to incorporate both the principles of conservation of momentumand of mass from Newtonian mechanics. The principle (16.1) is accordinglyeminently reasonable. However, our ultimate justifkation for accepting it is, ofcourse, that its consequences are verified experimentally. We shall refer to suchchecks at appropriate points in the later development.

It appears from equation (16.4) that the mass of a porticle must now beregarded as being dependent upon its speed v. [f!i = 0, then m = M. Thus M is themass of the p<Jrticle when me<Jsured in an inertial frame in which it is stationary.M will be referred to as the rest mass or proper rna.,., and will, in future, be denoted

(In)

43

by m". To dIstinguish It from m". m is often called the inertial ma8.'. Tocn

mom = (I '=;,ipjh2

Clearly m -+ u: as v -+ c. implying that inertia effects become increasingly seri~us

as the velot.:lty of light is approached and prevent this velocity being attained byany material particle. This is in agreement with our earlier observations. Form~la(16.7) has been verified by observation of collisions between atomic nuclei andcosmic ray particles (e.g., see Exercise 27 at the end of this chapter).

We shall define the 4-momentllm vector P of a particle whose rest mass is moandwhose 4-velocity IS V, by the equation

P= mo V

Since m" i.- an invariant and V is a vector in space time, P is a vector. Byequaton(156),

P = m" (I - lJ2 /( 2) . 1/2 (v,ie) = (mv, imc) = (p, im(') (U.9)

where p = mv is the classical momentum.Relative to the special orthogonal transformation (5.1), the transformatIOn

equations for the components of Pare

1"2 = 1'2}

p.l = 1\(16.10)

Substituting for the components of P from equation (16.9) and similarly for P,and employing equations (5.7), it will be found that

.. px-· mll

Px = (l-~2;;jF'

Py = Py

pz = P,

. m - PxU/C2m = ..._ .. -- ... 'O.

(I _u 2/( 2)1/2

(16.11 )

(16.12)

(1613)

Equations (16.11) constitute the special Lorentz transformation equations for Ihecomponents of the momentum p and equation (16.12) the correspondingtransformation equation for mass. Since Px = mv" this equation can also bewritten

_ I - UI'x/C2m = _. m

(I_U 2 /C 2 )1'2

This reduces to the classical form of equation (2.4) if u, Vx are negligible bycomparison with c.

44

17. The force vector. Energy

We have seen that in classical mech,mics, when the mass of a particle has heendetermined, the force acting upon it at any im;tant is specified hy Newton's secondlaw. Force receives a similar defintion ill special rclativity mechanics. The mass of,I particle with a given velocity can be determined by permitting it to colhde with astandard particle and 3pplying the principle of momentum conservation.Fquation (16.7) thcn gives Its mass at any velocity. The force f acting upon aparticlc having mass III 3nd vel(lcity v rel3tive tp some inertial frame is thendefined by the equation

d dpf = (mv) =

dl dl(17.1 )

where p is the particle's momcntum. Clearly f""ill be dependent upon the inertialIi'ame employed. a departure from d,l>;sical mechanics.

Definition (17.1) implies that, if equal and opposite forces act upon twocolliding particles, momentum is conserved. However, although experimentconfirms that momentum is indeed conserved, Newton's third law cannot beincorpor,jled in Ibc new mecbanics, for it will appear later that, if the forces areequal and oppo.sitc for one inertial observer, ill veneral they are not so for all suchobservers. Equatioll (16.1) therefore replaces Ihis law III the new mechanics.

f is not a vector with respect to Lorenl/. transformations in space time.However, a 4:!o,-ec F call he defined which has tillS property. The naturaldefinition is clearly

tiP dVF= = "'"dr or ( 17.2)

P being the 4-momentum and r the proper time I"l' the partiele. F is immediatelyexpressible in terms of f for, by equation (16.9),

dF = (p, ime)

dr

d dl=. (p, ime)

dl dr

= (I _1· 2 /( 2 ) -1/2 (p, iri1(')

= (I _1,2 /( 2 ) 1/2 (f. imc) (17..1)

The vectors V, F are orthogonal. This is proved as follows: From equation( 15.6),

(17.4)

Differentiating with respect to r.

I.e. V' F = 0

45

(17.5)

as dated. This result has very important consequences. Substituting for Vand Ffrom equations (15.6) and (17.1) respectively, it is clear that

ThIs is equivalent to

(I -- ,,2 /e') I (v, ic)' (f, i,;le) = O.

v·f--e'ril = O.

(176)

(17.7)

But, bydclinition, v·fisthe rate at which fisdolllg work. It follows that the workdone by the force actillg Oil the particle during a time interval (II' /2) is

(17.I:n

"

The dassical equation of work is

work done = increase in kinetic energy (17.9)

where T = !mv2 is th(' kinetic energy (KE). Equation (l7.R) indicates that inspecial-relativity mechanics we must define T by a lormula of the type

T = me' -+ constant (17.10)

When I' = 0, r = 0 and this determines the unknown constant to be _. moe2• Thus

(17.11 )

If I'/e is small (I - ,";e') 1/, = 1-+ ("/2e' approximately and the above equ­ation reduces io T= !m()I", in agreement with classical theory.

According to equation (17.10), any increase in the kinetic energy of a particlewill r"esult in a proportional increase in its mass. Thus, if a body is heated so thatthe thermal agitation of its molecules is increased, the masses of these particles,and benee the total body mass, will increase in proportion to the heat energywhich has been communicated.

Again, suppose two equal clastic particles approach one another along thesame straight line with equal speeds r·. II" their rest masses are both mo, the netmass in the system before collision is

It has been accepted as a fundamental principle that this mass will be conservedduring the collision. However, from considerations of symmetry, it is obvious thatat some instant during the impact both particles will be brought to rest and theirmasses at this instant will be proper masses m~. By our principle,

(17.12)

(17.13)

46

[t follows, therefore, that, at this instant, the rest mass of each partit.:le hasincreased by

.__~_1110._ _ m = T/e 2

(I - v2 /( 2 ) I /2 0

where T is the original KE of the particle and use has been made of equation(17.11). Now, in losing this KE, the particle has had an equ<JI amount of workdone upon it by the force of interaction and this has resulted in a distortion of theelastic material of which it is made. At the inst<Jnt each panicle is [nought to rest.this distortion is at a m<Jximum and the elastic potential energy as measured hythe work done will he eX<Jctly T. [f we assume th<Jt this increase in the intern<JIenergy of the p<Jrtlt.:le leads to a proportional increase in mass. the increment ofrest mass (17.13) is explained. [f the particles <Ire not perfectly elastiC. the workdone in hringing them to rest will not only increase the internal elastic energy, hutwill also generate heat. Both forms of energy will then contribute to increase theres t rna sses.

Such consider;llions ;IS these suggest very strongly that mass and energy areequivalent, being two dill"erent measures of the same physic;d quantity. Thus, thedistinction hetwecn mass and energy which was maintained in claSSical physicaltheories. has now heen abandoned. All forms of energy E, mechanical. thermal,electromagnetic, are now taken to possess inertia of mass m, according to

Ein,\lein\ equi/lion, viz.

(1714)

(17.15)

Conversely. any particle whose mass is m. has associated energy F and. hyequation (17.11),

moc 2 is interpreted <Js the internal energy of the particle when stationary. If thcp<Jrticle were converted completely into electromagnetic radiation, moc 2 wouldbe the energy released. This is the source of the energy released in an atomicexplosion. The mass of the material products of the explosion is slightly less thanthe net mass present hefore the explosion. the difference heing accounted for hythe mass of the energy released. Even a small mass deficiency implies that­an immense quantity of energy has been released. Thus, if m = I kg, c = 3x 10R ms ", then E = 9 X 10 16 J = 2.5 X 1010 kWh.

The principle of conservation of mass, which has heen incorporated into thenew mechanics, is now seen to be identical with the pnnciple of conservation ofenergy, which is accordingly also regarded as valid in the new mechanics.However, the distinction between the two principles. which was a feattHt of theolder mechanics, has disappeared.

18, Lorentz transformation equations for force

By equ,ltion (17.7),

il;,ec

f-v (I X.I)

(18.2)

(llU)

(18.4)

47

Referring to equation (17.3), F can now he completcly expressed in terms ofr;thus

F = (I _1,1/(2

) III (r,;. f-V)

Relative to the special Lorentz transformation, the transformation equa ionsfor the components of F arc

j~ = 1-'1 coS::1. I- /<"4 sin (f. r 2~" I-'l}/<"4 = - I-' I SIll (f. -t 1-'4 cos rx F, I· .\

Substituting from equation (18.2) into the first three of these equations andemploying equations (5.7), it follows that

T" = Q (I, -- (u] f-v )

}: = Q(I -- u 2 /1'2 ) I /11,I: Q(I _ ,,] /el)1 /2/,

whcre Q is given by equation (15.10). Substituting for Q from the four~l ofequations (15.9), it will be found that

(18.5)

These are the special Lorentz transformation equations for f. [I' u, r, are negli~ible

by comparison with 1', these equations reduce to the classical form of equ;\tion(26)

It is clear from equations (\ 8.5) that, if equal and opposite forces are observedfrom S to act upon two particles, the forces observed from oS'will not be so relatedunless the particles' velocities are the same.

19, Fundamental particles, Photon and neutrino

By eliminating mand v between equations (\6.7), (\ 7.14)and the equation p = mvgiving the linear momentum of a part ide, it will be found that

(19.1)

This useful equation relates the total energy F: ofa particle (including its inta-nalenergy) with its momentum p. A special case of great importance is when mo = 0,and then

E = cp (\9.2)

48

For such a particle, m = Ele2 = pic; i.e. its rest mass vanishes, but its inertial massis non-zero. This result is inconsistent with equation (16.7), unless v = c (in whichcase the right-hand member becomes indeterminate). We conclude that anyparticle having zero rest masS must always move with the speed of light.

Two such particles are known, the photon and the neutrino. The former is aquantum of electromagnelic energy and the latter is a particle which is generatedin some interactions between fundamental particles governed by the weakinteraction .force (e.g. the decay of a neutron into a proton). Neither particleexhibits any electric charge. [I' either particle is absorbed by other matter, it losesits identity and delivers its energy and momentum to the absorbing body -- theheating of a metal plate placed in sunlight is an example of the absorption ofphotons. Neutrinos arc exceptionally difficult to ahsorb and hence todetect there is a high probability that a neutrino from the sun will pass rightthrough the earth without interaction with a single one of its atoms.

As an example ora particle interaction in which a neutrino is involved, considerthe decay ofa negative pion (a meson) into a muon (heavy electron)and a neutrino.Assuming that the pion is at rest in the laboratory frame, the momenta orthe muonand neutrino will have equal magnitudes p, but will be in opposite senses. [I'm.. mp

are the proper masses of the pion and muon respectively, the principle ofconservation of energy leads to the equation

m,e2 = ep+c J(p2 + m;(2) (19.3)

having used equations (19.1) and (19.2). Solving for p, we tind

p = elm; - mN2m,

The energy of the muon is now found to be

Ep = m.e 2- ep = c2(m; + m,:)/2m.

and the KE of this particle is accordingly

T,. = Ep - mpe2 = c2 (m, - mp )2 12m,

(19.4)

119.5)

(19.6)

From tables, we find that the rest masses in atomic energy units are m,c2 = 140MeV, m.c 2 = 106 MeV (mega-electron volts). Hence, Tp = 4.1 MeV. This valuehas been checked experimentally by observing the length of the path of the muOnin the resistive medium in which it is generated (usually liquid hydrogen).

20. Lagrange's and Hamilton's equations

Suppose that a particle having constant rest mass mo is in motion relative to aninertial frame under the action of a force derivable from a potential V. Then itsequations of motion are

etc. (20.1)

49

Expressed in Lagrange form, these equations must be

d (?f-) (~L~ , ' etc.

dr ?x I ('X

and hence I. must be a function of x, Y. =, .x, .i'. Z. such that

(, I. mo'\' ('L ?V(l_X (I .:: l~f/('2)1 ,2 ('X ?x

etc,

Since v2 = .x 2 + .i.2 + Z2, these equations can be validated by taking

L= -moe2(1~1'2/(.2)1,2_V

which is accordingly the Lagrangian for the particle.Now

(20.2)

(20.3)

(20.4)

(, I,..... = P.\,,'x

etc. (20.5)

and it follows exactly as in classical theory that, if the Hamiltonian H is defined bythe equation

(20.6)

and is then expressed as a I'tInction of the quantities \', y, =, fI" fly. fI, alone, theLagrange equations (20.2) are equivalent to Hamilton's equations

etc. (20.7)

Nownlor1

fI,I" + fI,·I., + fI,I', = 1 2 212. ( -I' /e )

and hence

,tlo 1,1 .1 ,1/. 2 I//= 2'21/2+m,,( (I-It' )'2+V

(I --I' Ie )

nJoe l

=- " + V(I - 1. 2/e 2 )1.2

=E+V

the total energy. precisely as for classical theory.

But

2 2 2 m~ {'2

fix + fly + fI, ="\ _1,21,,2

2 1 nl(~('2-moe + - "-""

I - 1,2 leI

_ m;',2 + 1,2/e2

(20.8)

(20.9)

(20.10)

(20.11)

50

and it follows that

e = ('2(p~ + p; +p; + 1115 (2 )

Substituting in equation (20.9),

H = c(p; +p; + p; + 1115('2)1/2 + V (20.12)

expressing H as a function ofx, y, z, Px. Py' p,. The reader is now left to verify thatHamilton's equations are equivalent to the equations of motion (20.1).

21. Ener~y- momentum tensor

Suppose that there is acontinuous distribution of mass over some region ofspace.In this section, we shall suppose this to take any physical form what~oever. Forexample, the distribution may be in the form of the molecules of an elastic bodyand, in this case, the mass must include a component corresponding to the mass ofthe potential energy of the field ofstress, in addition to the inertia of the particlesthemselves. Such a field will be electromagnetic in nature, the electric chargespresent in the molecules being ultimately responsible for its presence; we shallnot, therefore, exclude a further contribution to till: maSs energy di,tributionfrom any other electromagnetic field which may happen to be present. Anyrandom motion of particles exhibiting itself as heat energy will also make acontribtltion. Equations governing this combined mass flow will now be derived.

Let S be an inertial frame Ox 1 x 2 Xl and let 11', 11", etc., be the densities ofinertial mass at a point of the frame due to the various contributors. Then, ifv', v",etc., arc the respective velocities of flow of these components, the net density oflinear momentum g will be given by

where

g =Il'V' + Il"v" + ., . = IlV, (21.1)

(21.2)

is the net density of inertial mass. Equation (21.1) defines the mean velocity ofmass flow v. In time dl, the mass flowing acroSs an area dA having unit normal n is

11' v' . n d A dl + 11" v" . n d A dl + ... = 11 v. n d A dl (21.3)

thus, the rate of mass flow across unit area is 11 v . n = g' n, implying that g is alsothe current density vector for the mass flow. The components of g will be writtengo (Greek indices will range over values 1,2,3).. Let g"1 be the current density vector for the flow of the x.-component ofmomentum, i.e. the rate of flow of this component of momentum across unit areawith unit normal n is g'o/. n. The xp-component of ~,ol will be denoted by gop. In thespecial case ofa cloud of non-interacting particles (no stress field), whose velocityof flow is v, since the density of the xo-component of momentum is go, the quantityof this component passing over the unit area in unit time is go v· n. It follows that

51

~(') = fI, v and hence,

(21.4)

A simple distribution of this type will be referred to as an incoherent cloud.For any distribution which includes material particles, in addition to the

internal forces of interaction between the particles, there may be other forcesacting upon them due to agents which arc regarded as external to the system; suchforces will be termed extemu/forces. Let dw be the volume occupied by a smallclement of the fluid or ~olid which is formed from these particles; if v is the flowvelocity of the clement and dmo is its proper volume (i.e. volume measured in aframe in which the clement is momentarily stationary), then

(21.5)

since all lengths parallel to the flow will be subject to a Fitzgerald contraction. Ifdf is the resultant external 3-force acting on the clement, we shall define the 3­f()fce density d at the clement to be such that df = d dw. Similarly, if dF is theexternal4-force on the clement, the 4-foree density will be D, where dF = D dwo;since dWI) is it 4-invariant, this defines D as a 4-vector. Refercm:e to equation (18.2)shows that we can write

and this is equivalent to the equation

Ddwo = (I-t,2/c2r- 1 /2 (d,id'v/c)dw

It now follows from equation (21.5) that

D = (d. id'v/c)

(21,6)

(21.7)

(21.8)

which relates 3- and 4-force densities.Now suppose L is a closed surface which is stationary relative to the frame S

and let dl1 be the area ofa surface clement whose outwardly directed unit normalis n. Then the rate of increase of the total mass inside L must equal the rate ofinflow of mass across L, plus the rate at which external forces acting on anyparticles inside L generate energy (and hence mass) by performing work, Let rdenote the interior of Land dw a volume clement of C Then conservation ofinertial mass is expressed by the equation,

d f, Jldw = - rg' n dl1 + ~\ f, d· vdwdt I 1 (I

(21.9)

Converting the surface integral into a volume integral over r by the divergencetheorem, this equation is seen to be equivalent to

f {('III}I tit +div~-c2d'v dw=O (21.1 0)

52

Since r is arbitrary, this implies that

(111 1~1; + g,., = ("i d, v" (21,11 )

where the summation convention is being applied to the Greek indices.Since the x,-{;omponent of the linear momentum of the particles within the

volume element dw will be increased at a r,He d, dw by the external forces and g(,1is the current density vcctor for the flow of this component of momentum, theequation corresponding to (21,9) expressin~ the conservation of linear momen­tum is

d f, y,dw =dT I'

(21,12)

Again, by application of the divergence theorem, we can show that this impliesthat

(21,13)

These equations (21.1 \), (21,13), of conservation of inertial mass (or energy)and linear momentum can be expressed in four-dimensional form by theintroduction of Minkowski coordinates Xi and by the definition "f a 4-tensor T'Jaccording to the equations

(21,14)

It may be verified that, with this notation, the equations reduce (0 the form

(21,15)

where the 4-force density D; is given by equation (21.8). T,] is called theenergy momenTum tensor for the mass -energy distribution. By assuming T;Jbehaves like a 4-tensor on transformation between inertial frames, equation(21.15) is guaranteed to be valid in all such frames and the special principle ofrelativity is satisfied.

In the special case ofan incoherent cloud ofparlicles flowing with velocity v, wehave y, = JIL', and g,p = g,vp = IIv,vp. Equations (21,14) now yield

(21.16)

where V = (1_V 2 /C 2 )-112(V, ic) is the 4- velocity of !low and

1100 = (1 - v2 /("2) Ji (21,1 7)

Since the density of rest mass of the cloud observed from S is II J (1 - v2/(

2), the

rest mass of the particles in the volume element dw is IIJ(1-v2 /c 2 )dw.Observed from a frame So in which the particles in this element are momentarilyslationary, the volume of the element will be dwo; hence, the density of propermass in S" is II J (1 _,,2 /e 2 )t!w/dw" = JI(1 ,,2/(,2) = JI"", having used equation

53

(21,5). 1100 is accordingly referred to a~ the pruper demir y uf proper ma.\S of thecloud; it is a 4-invariant and equation (21.16) clearly expresses T" as a 4-tensor.

In many circumslances, I he 4-force demity D, of the external force Iield can beexpressed as the dIvergence of a second rank tensor, i.e. we can write

Equalion (21.15) then reduces to

(Til + .'I"),, = 0

(21.1 R)

(21.19)

This shows that the external force field can be treated as an additional componentof the original mass energy dislribution, contributing its own energy­momentum lensor S,j' The Iield is then regarded as pos~essingenergy of density(', 11 = - .'144 , x,-component of momentum of density g, = S,./ie, and themomentum flow within lhe field is described byg,p = S,p, There being no externalforces operating on the enlarged system. it is said to be iso/ared. and ifT" is takento denote the com bined energy momen tum tenSor, the equations ofconservationof energy and momentum for the overall distribution take the form

(21.20)

i.e. T" has a va nishing divergence.

22. Enerl:Y momentum tensor for a fluid

In this section we will calculate the equatiom of motion for an clastic fluid movingunder the action of an external force field of density iI,. From these, theenergy momentum tensor f(lr the tluid. including its internal stress field, can bederived.

Let T/" be the stress tensor, i,e. (T ,a• TIa' T3a ) arc the components of the forceexerted across unit area, whose normal is parallcl to the x,-axis. by the particles onthe side for which X a takes lesser values upon the particles on the side for which X a

takes greater values. Consider a tlllid clement in the shape ofa small tetrahedron,three of whose faces arc normal to the axes and whose fourth face has unit normalII, (Fig. 4). If Ox" bx lo bX3 are the respective lengths of the edges parallel to theaxes, the stress force acting on the face parallel to the coordinate plane OX, XI willhave components ~(T", T,,, T3.lbx,OX3; the forces acting on the two facesparallel to the other two coordinate planes can be calculated similarly. Let sbA bethe force applied to the sloping face. (,A being its area, Then the x,-component ofthe equation of motion of lhe element is

j ".' I '.' , ., 'A"" J _ dr,2T"(J\,()X1 +2 T.,,bx,b·\'+2 T,.Ii'l.\,OX, t.\,O +,;i'lx,i'lx,()X 3(,- dl

(22.1)

where r, is the momentum of the clemen!. Since the face of area ~('X,bX3 is theprojectIOn of lhc area ()A Oll 10 the COOl(flllHlt: plant: 0\1 \ .. IIlbA ~o\,(h ,;

54

X3

~)

x~1

FIG. 4

~x?

similarly, 11 2<'lA = ~O\.,(lx" ".,oA = ~OXIOX1. Thus,

S, = - (.,,11, + T'2"2 + T,.\n,)+ (terms of third order in oX,)/oA. (22.2)

In the limit as (lx, -> 0, this gives

(22.3)

Now con:-;idef the motion ofa small clement of fluid, of any ,hape, hounded hya surf:lce L (2: move~ with the fluid). If da is an element of L (Fig. 5) whose unitnormal is II" t he force exerted on it hy t he neigh houring fluid is - "lllIpda and theresultant stress foree on the t.:0ll1r1cte element is Iherdore

o

x,

I'ru. 5

X2

55

where r is the interior of L and we have used thc divergence theorem. Thus, the(orce density for the stress field is - T,p. p.

Let fI, be the total momentum density for the fluid. including the elasticpotential energy generated by the stress field. If ,)W now represents the volume ofthe fluid element inside L, the momentum of the element is fI,"'w and the rate ofchange of momentum is

d . d~ d.(fI,()W) = bw + fI, (hw)

dl dl dl

where the derivatives are calculated following the fluid motion. Thus.

df!, (lfI"dl (II + (lpfl,. p

(22.5)

(22.6)

(22.7)

where v, is the velocity of flow. During a short time aI, the surface element dl1

traces out a volume y' n dl1 ill and the increase in the volume of (jw is accordingly

,)1 r. y' n dl1 = 1)1 f. div y dw = ,)1 ,)(t) '·P. P

Thus,d .

(il(l)) = ,)Wl" "dl "."

Equati,>ns (22.5). (22.6) now givc for the rate of momentum change

(1

1;1, + r'pfl,. p + fl,(lp. p)dW = ("y, + i' (o,IJP))dWl'I (11 (lXp

We can now writc down the equation of motion of the element. viz.

or

(22.8)

(22.9)

(22.10)

(22.11)

(22.12)

At this stage it should be noted that we are disregarding any flow of heat whichmay take place by conduction within the fluid. Such a flow of energy wouldcontribute its own momentum and further terms would need to be included inequation (22.11) to allow for this.

We next calculate the equation of energy for the fluid element. The rate atwhIch the stress force acting upon dl1 docs work is - T.flllpV.dl1 and the total rateof doing work by these forces on the element is therefore

--f r',T,pllfJdl1 = --f (1 (", T,p )dwr r~xp

56

by the divergence theorem. The rate of doing work by the external forces on theelement is Jat,.d(l). [I' IJ is the density of the total inertial mass of the fluid(including the elastic potential energy and any heat generated by compression),the energy of the clement is e21/d(l). Thus, the equation of energy is

(2213)

Using equation (22.X) and

(22.14)

equation (22.13) gives

(22.15)

We now compare the equations (22.11), (22.15) with the equations (21.13),(21.11), for 11 general mass energy flow. It is seen that, for the fluid, we must take

IiI. = Ill'a + (.2 I'pT p. (22.16)

(2217)

Substituting for Ya from equation (22.16) into (22.17), we get the alternativeformula

Iqap ~ Ill'al'p + Tap + ('2 T ,a1'yl'p (22.IX)

(22.19)

Equations (21.14) now yield the components of the energy momentum tensor,VIZ.

Tap = Il!'al'p + Tap + 12 T y.I',.l·,1 jl('

I1~4 = 1'4a = ic(IJVa + -2 VpT pa )

('

1'44 = _e211

Alternative forms for these components involving the 4-velocity of flow V, Gtn

also be found (sec Exercise 6X at the end of this chapter).Two special cases of these results are of great importance. The tirst is that of an

incoherent cloud for which we have Tap = O. Then l~p = 1J1'.1'p, T a4 = i£'lJl'a,

1'44 = -e21J, and using equation (21.17), we derive equation (21.16) again. Itshould, however, be noted that equation (21.17) is not valid in the general case ofan elastic fluid, since the inertial mass density IJ includes a component due to the

57

elastic energy and the vallie or this component in the rest frame cannot be roundby simple multipltcation by (I-1'1/C1).

The second special case is tha t of a rerfecl fluid. This is defined to be a fluid inwhich there arc no sheartng stresses in a frame S" relative to which it is at rest.Thus. in S", r;'t/ ,-~ ",I,p where" is the pl""\.\l"'''. Equations (22.19) n,>w give for thecomponents of the energy momentum tensor in SO the values

(1:~) = i"0 0(l " 0001'

000

o j(l

o-- ("1 1/00

(22.20)

(22.22)

II' Vi is the 4-velocity of flow of the fluid, it is now easy to veriry that the tensorequation

TiJ=(1100+p/c1P;Vj+rl)iJ (22.21)

is valid in the frame S°(jl"o, I' heing invariants) and, hence, is valid III all frames.Thus, if II is the demity of inertial mass in a frame S,

Jl = -. I~4ic1 c, 1100 t 1'/("1 I'

. 1 - 1,1 /e 1 ,,1

or (2223)

It now follows that

1~4/i(" = (1100 +r/(1) - 1" __I _1,1/(1

= (II + 1'/,,1)1', (22.24)

Comparing this last equation with equation (22.16), we deduce that 1'1', = 1'pr p,

identically, which implies that

(22.25)

i.e. there is no shearing stress in any frame and the pressure is the same in allrrames.

23, An~ular momt'nlum

A particle having momentum 1', = ml', at a point x, relative to arrame S is definedto have mlgulal" momenlum

(23.1)

about the fixed point a,. Clearly, this delines the angular momentum as an anti­symmetric 3-tensor. [n elementary mechanics, a more usual definition is by thevector product (x -. a) >< p. i.e. the pseudovector ~,tl, (\'p _.. lip )1',.; however, thecomponents or this pseudovector are found to be (Ir B , 1r 3 !> Ir ll ) and these are

58

three of the non-vanishing components of h.p(the other three are h l2 '" - h 23 ,

etc.), so that the definitions are essentially equivalent.Iff. is the 3-force acting on the particle. then

dh.p dx. dx p dpp dp.dl dl Pp - d"1 P. +(x. -(/.) dl -(xp -ap) dl"'

(23.2)

since P. = ml,. = In.x. and r. = .I~. The right-hand member of the last equation isanother anti-symmetric 3-tensor called the mumenl of the force/~about the point(/a' Denoting this mom~nt by map, we have derived the equation of angularmomentum, viz.

(23.3)

[I' maP = 0, then h.p is constant and the angular momentum is conserved.[n the case of the continuous distribution of mass energy considered in

section 21, the momentum or an element of volume dw is gadw and the angularmomentum of the whole system about a. is defined by the equation

h.p = r {(x.-aa)gp-(xp-ap)ga}dwJr= I r {(x. - a.)Tp4 - (x p - (/p )T'4 }dw

Ie JI (23.4)

where r is the region occupied by the system and Ti) is the energy momentumtensor. [I' the system is imagined to be situated in otherwise empty space, so thatthere is no container exerting forces on its bounding surface, the region r can beextended to include the rest of space. In these circumstances, ditTerentiatingequation (23.4) with respect to I, we find

dlr.p rdl J, {(x. - (/.)Tp4 .4 - (x p- ap)Tp4 .4 }dw

i ((xa - a.) (D p- Tp,.y) - (x p- ap)(D. -." 7~y.,)} dw,

1{(xa-aa)Jp-(xp-ap)J.}dw

- r {[(x.-a.)Tp,J.,-[(xp-ap)T.,l,-x•. yTpy+xp.,Ta,}dw (23.5)J,having used equations (21.8) and (21.15). Applying the divergence theorem to thetirst two terms involving the energy momentum tensor, since T,) vanishes at agreat distance from the distribution, these terms make zero contribution. Thus,

59

the equation reduces to the form

dhap i{ i----- = (x -a)d -(x -a)J \dw+ (Tpa-Tap)dwdl I a • p . p p. J r (23.6)

(Note that x•. , = bar') The first integral in the right-hand member of equation(23.6) is the moment map of the external field forces about a. and it follows that theequation of angular momentum (23.3) is valid for the distribution if T.p issymmetric. Since 1~4 = T4 ., this condition is equivalent to the requirement thatthe energy momentum tensor should be symmetric.

The assumption that T'j is symmetric, and hence that the angular momentum ofa continuous distribution not acted upon by external forces is conserved, isalways made and is, indeed, necessary for the development of the general theoryof relativity (see section 47). Reference to equations (22.19) indicates that thisassumption impltes that the stress tensor r. p for an ela-;tic fluid cannot besymmetric as was always assumed in the classical theory; It will, however, be verynearly symmetric in the case when its components and the flow velocity aresufliciently small.

Exercises 3

I. Obtain the transformation equations for v by differentiating the Lorentztransformation.

2. Obtain the transformation equations for the acceleration a by diflerentiat­ing the transformation equations for v and express them in the form

_ (l __ u2 /e 2 ).112a = - ---- a

x (I _ 1'xll/(2).\ x

__ 1 - u2 /e 2( ,'yu/e 2

)a = --------- ---- ---. - a + ----------- ---- -- ay (l-v

xu/e 2 )2 ... I -v

xu/c 2 x

a, = (l1_~;~~~:~;)2 ((/, + "i~:~~~~;;;l ax)Deduce that a point which hIlS uniform acceleration in one inertial frame has not,in general, uniform acceleration in another.

3. A nucleus is moving along a straight line when it emits an electron. As seenfrom the nucleus, the electron's velocity is 61'/7 making an angle of 60" with itsdirection ofmotiol1_ A stationary observer measures the angle between the lines ofmotion of nuclcus and electron to be 30'. Calculate the speed of the nucleus.(Ans. 3c/5.)

4. A nucleus is moving with velocity 3c/5 when it emits a /i-particle withvelocity 3('/4 relative to itself in a direction perpendicular to its line of motion.Calculate the velocity and direction of motion of the /i-particle as seen by astationary observer. If the II-particle is emitted with velocity 3e/4 in such a

60

direction that the stationary observer sees its line of motion to be perpendicular tothat of the nucleus, calculate the direction of emission as seen from that nucleusand the velocity of the {I-particle as seen by the stationary observer. (Ans. ]('/5 at45" to line of motion; TC - a to line of motion where cos a = 4/5; ge/I6.)

5. Show that the 4-velocity V is of constant magnitude ic6. A beam of light is being propagated in the xy-plane of S at an angle a to the

x-axis. Relative to ,;;; it is observed to make an angle ix with 0.>:. Prove theaberration of light formula, viz.

.. cot a - (u/cJcosec acot a = ' ,

(l_u 2/e 2)112

Deduce that, if u ~ e, then

u~a = rt - a = sin a

approximately.7. A particle of rest mass m" is moving under the action of a force f with

velocity v. Show that

1ft ov mol'i,/c2f =' _.," +. .....' V

(I-1'2/e2)111dl (I_,,2/e 2)31 2

Hence, if the acceleration dv/dl is parallel to v, show that

/tlo dvf - ....- (I __ ,.I/( 2)3 /2dl'

and if the acceleration is perpendicular to v, then

/tl" dvf = -.---'.-......-(I-v 2 /e 2 )1/2 01

8. Two particles are moving along the x-axis ofa frame S with velocities t'" ('2'

Calculate the velocity u with which a parallel frame S must move parallel to the x­axis ofS, if the particles have equal and opposite velocities relative to S. Show thatthe magnitude of these velocities is

e 2_ (',"2 - (('2 _ vi)' 12 (e 2

_ tJ~)' 12

assuming v, > 1'2 > O.9. A bullet of length J is moving with velocity tJ. The line of sight from a camera

makes an angle a with the bullet's velocity. Behind the bullet and parallel to itsaxis is a stationary measuring scale. [fthecamera takes a photograph of the bulletagainst the background provided by the scale, show that the bullet's length as itappears on the scale is

(I - 1'2 /e 2)' 12J

~~---~- -~------

I + I' COS a/c

61

10. A cart rolls on a tahle with velocity kc. A smaller cart rolls on the first inthe same diret:tion with velocity he relative to the first cart. A third cart rolls onthe second wtth relative velocity he and so on up to n carts. If ('(', is the velocity ofthe rth cart relative to the tahlc, prove that

1', + kt' +1 =, I +k,',

Deduce that

(I +/..)" .- (I - k )"!' = ." (I + k)" + (I - k)"

What is the limit of ,," as n -+ (I? (Ans. ('.)II. A nucleus disintegrates into two parts, A and B which move with equal and

opposite velocities of magnitude V. A then ejects an electron whose velocityohserved from A has magnitude V and direction perpendicular to the direction ofA's motion. Show that, as ohserved from B, the electron's velocity makes an angle(j with the direction of A's mOlion, where tan (j = ~(I - VI /1'2) and calculate themagnitude of the velocity of the elet:tron relative to B. (Ans. V {4 +(I _. V I /C 2 )2) '12/(1.- V 2 /( 2 ).)

12. A rocket moves along the x-axis in S, commencing its motion with velocity"0 and ending it with velocity 1",. [f ", is the jet velocity as measured by the crew(assumed constant), show that the mass ratio of the manoeuvre (i.e. initialmass/final mass) as measured hy the crew is

[(c + ",)(e - (10) -j"/2"(c-I'I)(C+"o)_

What does this reduce to as (' -+ 'Y '! Deduce that, if the rocket starts from rest in Sand its jet is a stream of photons, the mass ratio to velocity v is

)(C+I')C-t'

Show that, with a mass ratIo of 6, the rocket can attain 35/37 of the velocity oflight in S. _ _

IJ. S, S~ Sarc inertial frames with their axes parallel. fhas a velocity u relativeto:\ and S'has a velocity I' relative to S, both velocities being parallel to the x-axes.If transformation from Sto Sinvolvcsa rotation through an angle (j of the axes inspa~-time and transformation from S to 5a rotation fI, a transf()fmation from Sto S involves a rotation l' where y = (j + II. Deduce from this equation therelativistic law for the composition of velocities, viz.

11+"W=

I + "I'/e 2

62

14. A force f acts upon a particle of mass m whose velocity is v. Show that

dv f· vf=nr +2V

dt c

15. An electrified particle having charge e and rest mass m" moves in a uniformelectric ficld of intensity E parallel to the x-axis. II' it is inilially at rest at the origin,show that it moves along the x-axis so that at time t

where k = eE/mQ• Show that this motion approaches that predicted by classicalmechanics as c -+ rf'. (I t may be assumed that the force acting upon the particle iseE in the direction of the field at all times.)

16. A tachyon transmitter always emits a tachyon at a speed" > c relative toitself. Observers A, Bare e4uipped with such transmitters and B is moving awayfrom A with velocity u < c. A transmits a tachyon towards B, who is at a distanced as measured from A when he receives it. B immediately transmits a tachyon backtowards A. Show that A receives this tachyon a time

d 2 2- ,- (2u - " - U 1'/e )v(r-u)

after transmitting his own. Deduce that A receives the reply before (I) his act ofInInsmission if

" e+J(c2- u 2)> - -- ------e u

17. v, Yare the 3-and4-velocitiesofa point. If a = dvjdt is the 3-accclerationand A = dV/dr is the 4-acceleration, prove that

A2 = (I - 1'2 /(2)' 3 {(('2 - v2 )a 2 + 2vva'v --lh/ }/e 2

18. A mirror moves perpendicular to its plane with velocity I' and away from asource of light. A ray from the source is reflected by the mirror. If () is the ray'sangle of incidence, show that the angle of reflection is </>. where

(ll2 + ('2 )cos() - 2n,cos </> = --i'- ---.. '-

II + ('2 - 2vecos(}

19. Two trains, each having the same rest length L, arc moving in oppositedirections with equal speeds U on parallel tracks. State the time Tthey take 10 passone another according to a classiral, non-relativistic, calculation. Show that thetime taken, as measured by a driver of one of the trains and using a relativisticcalculation. is also T.

20. v, v are the velocities of a point relalive to the inertial frames S. grespeclively. RepresentIng these veClOrs as position vectors ill an independent ,1,('.\,

6J

show that

where {I = (I -ul/el) Ijl and Q ~c 1/(1 + u·Y/e l ).Show further that

u2{1 v = QI (I - {I)u x (y xu) + {II/ (u + Y)]

and hence veri fy tha t

21. A luminous disc of radius a has its centre fixed at the point (x, 0, 0)01' the S·frame and its plane is perpendicular to the x-axis. It is observed from the origin ifI

the S-frame at the instant the origins of the two frames coincide and is measurecto subtend an angle 20:. Prove that, if a ~ x, then

0J(e+lI)tano: = __ ---X e-li

(Hint: employ the aberration of light formula, exercise 6 above.)22. A panicle moves along the .v-axis of the frame S with velocity v and

acceleralion a. Show thaI the panicle's acceleration in ,.:;- is

. (I -- 1/2/( 2).1/1il = - -, - -, a

(I - 1ll'/e2)'

If the particle always has constant acceleration 0: relative to an inertial frame inwhich it is instantaneously at rest, prove that

d(/iI') = 0:

dl

where II = (I - 1,2 /( 2), 1/2 and I is time in S.

Assuming that the particle is at rest at the origin of S at I = 0, show that its x­coordinate at time ( is given by

o:x = e1 [(1 +0:2(2/e2)1/1_1]

23. Three rectangular Cartesian inertial frames S, S, Sare initially coincident.As seen from S, S moves WIth velocity u parallel to Ox and, as seen from S, 'Smoves with velocity I' parallel to O\" If the direction of S's motion as seen from Smakes an angle O_,:"ith Ox and the'direction of S's motion as seen from S makesan angle </> with Ox, prove that

I' ( 112)1/2tanO = 1- 1

U ('

I' ( 1,1) Ij2tan¢= 1- 2

U ('

64

Deduce that, if u, v ~ e, then

approximately.24. The inertial frames S, S have their ,Ixes parallel and the origin of S moves

along the x-axis of S with velocity II. A rigid rod lies along the .x-axis of ,C;; and isattached to it. II'Tis the rod's length as measured in Sand I is its length measured inS, show that 1= T(l - u2 /("2)1 il. S' is a third parallel inertial frame whose originalso moves along the x-axis of S. Observed from S', the origins of Sand Shaveequal and opposite velocities. Show that the velocity of S' ohserved from S is

A rod, identical to the one already referred to, lies along the x'-axis of S' andmoves with this frame. Its length observed from S is L. Show that

25. Two particles, each having rest mass 1110 , are moving in perpendiculardirections with the same speed 1e. They collide and cohere to form a singleparticle. Show that its rest mass is .)(14/3)1110 , (Assume there is no radiation ofenergy.)

26. A particle of rest mass m l and speed v collides with a particle of rest massm1 which is stationary. Arter collision the two particles coalesce. Assuming thatthere is no radiation ofenergy, show that the rcst mass of the combined particle isM, where

and find its speed.27. A particle is moving with velocity u when it collides with a sta tionary

particle having the same rest mass. After the collision the particles are moving atangles 0, </> with the direction of mot ion of the first particle before collision. Showthat

2tanOtan</> = .----

y+1

wherey = (I _u2/( 2 )-lil. (lfe --+ cr., y --+ I and 0+ </> = !rr. This isthe prediction

of classical mechanics. However, if the particles arc electrons and u is near to C' invalue, 0 + </> < !rr. This effect has been observed in a Wilson cloud chamber.)(Hint: Refer the collision to an inertial frame in which both particles have equaland opposite velocities prior to collision.)

65

28. A body of mass M disintegrates while at rest into two parts of rest mlssesM, and M2' Show that the energies f .. 1':2 of the parts are given hy

M; I M;2M

2'.1. Two partidcs having rcst masscs "'" "'2 arc moving with velocities !J" U2

respeclivcly, when thcy collillc and cohcrc. 1f!1 is Ihe angle hetwecn their li[]~s ofmotion hefore collision, shnw that llic resl mass of the combined paniclels m,where

Show that, for all values of 11, m .:> "'I + m] and explain Ihe increase in rest nass.30. A photon having energy 1'; collides with a stationary clectron whose resl

mass is mo. As a result of the colliSion Ihe direction of the photon's mOli,n isdeflected through an angle (! and its energy is reduced to E. Prove that

moe](1 _1)= I-cosOE f

Deduce that the wavelength), of the photon is increased hy

211 ] I I!'.). = sin if,nloc

where h is Planck's constant. (This is the COlllpTOn eflee(. For a photon, take.Ie = he/E.)

31. A particle P having resl mass 2m" collides wilh a stalionary partirle Qhaving rest mass mo. Ahcr Ihc collision, Ihe resl mass of Q is unchanged, hll therest mass of P lias heen reduced to 11I0' If Ihc lines of motion of the two pari idesafter the collision both make an angle of 30 with the original line of motim P,calculate Ihe onginal velocity of P and the momentum acquired by Q. IAllS.3J5e/7; J 15mo<"')

32. A particle is moving with velocity I' when it disintegrates into two ph~tons

having energies f::", El , moving in directions making angles 11,11 with the onginaldirection of motion and on opposite sides of this direction. Show that

e -- t'tan~l1tan~/1 =

c+ "Deduce that, if a photon disintegrates into two photons. they must both m~ve inthe same direction as the original photon.

33. A stationary particle having rest mass 3m" disintegrates into a plir ofparticles, each of rest mass mo, and a neutrino. The directions of motion ~f theparticle pair are at an angle 20, where ms(! = 1/3. Calculate the energy cf theneutrino and show that the speed of each of the other two particles is 3e/5.(Ans.Il>lo( 2

)

66

34. A cosmic ray particle has rest mass mo and is moving with velocity 3c/5relative to a stationary observer. It is seen by this observer to emit a gamma rayphoton with energy moc 2/4 in a direction making an angle 01'60" with its originalline or motion. Show that the rest mass or the particle is reduced by a quarter andcalculate the angle through which its velocity is deflected and its new speed. (Ans.tan -I (J3/5); J7c/4.)

35. A neutron having rest mass mN is stationary when it disintegrates into aproton (rest mass mp). an electron (rest mass mEl and a neutrino. The protonmoves in the opposite direction to the other two particles. which move along thesame straight line. IrTis the kinetic energy or the proton, prove that the kineticenergy or the electron is c(mEc - k)2 /2k, where

Tk"" (mN-mp)c---- J(2m pT+T 2/c 2

)('

36. A particle having rest mass mo is at rest when it emits two photons, each orenergy imoc 2 The particle recoils with rest mass lmo along a line bisecting theangle between the tracks or the photons. Calculate the angle between these tracksand the particle's velocity or recoiL II' the photons are observed by an observermoving with the particle, show that the angle between their tracks is seen to be 20:,where sin 0: "" 1j7. (Ans. 30"; J3c/2.)

37. A nucleus has rest mass M. Whilst at rest, it emits a photon. II' the internalenergy or the nucleus is reduced by Eo in the process, show that the energy or thephoton is E, where E = Eo(l - Eo/2M( 2

).

38. The lines or motion or a particle having rest mass mo and a photon areperpendicular to one another. The total energies or the particle and photon are E,£ respectively. II' the particle absorbs the photon, show that its rest mass isincreased to Mo, where M5 = m5 + 2E£/c4

.

39. A photon having energy E is moving along the x-axis when it encounters astationary particle having rest mass mo. The particle absorbs the photon and thenemits another photon having the same energy in a direction parallel to the y-axis.Calculate the direction and magnitude or the final momentum or the particle andshow that its rest mass is reduced to the value J (m~ - 2t? /( 4

).

40. A mass 3.l.m(.l. > I) at rest disintegrates into three fragments (each or restmass m) which move apart in directions making equal angles with each other.Show that, in a rrame or rererence in which one or the rragments is at rest, theangle between the directions or motion or the other two rragments is2 cot 1 (J3.l.).

41. A positron travelling with velocity 3c/5 is annihilated in a collision with astationary electron, yielding two photons which emerge in opposite dire;.;tionsalong the track or the incoming particle. II' m is the rest mass or the electron andpositron, show that the photons have energies 3mc2 /4 and 3mc 2/2.

42. A positron having momentum p collides with a stationary electron. Bothparticles are annihilated and two photons are generated whose lines or motionmake equal angles 0: on opposite sides of the line or motion orthe positron. Prove

67

that p siniX taniX "" 2mc where m is the rest mass of both the positron and electron,If iX = 60", calculate the velocity of the positron. (Ans. 4c/5.)

43. A particle of rest mass mo collides elastically with an identical stationaryparticle and, as a result, its motion is deflected through an angle O. If T is its KEbefore the collision and T' is its KE afterwards, show that

44. A particle of rest mass m. collides elastically with a stationary particle ofrest mass m2 « m.) and, as a result, is deflected through an angle iI. If E, E' are thetotal energies of the particle m I before and after collision respectively, prove tha t

(E +m1(1)E' -ml c2 E -mic4

cosO = -- .... - ..~ .---.-.- "'-[(e -mic4 )(E'l -mic4 )]1/2

45. A pion having rest mass mo is moving along the x-axis of an inertial frameOxyz with speed 4c/5, when it disintegrates into a muon having rest mass 2mo/3and a neutrino. The neutrino moves parallel to the y-axis. Prove that the anglemade by the muon's velocity with the x-axis is tan - I (l/8)and calculate the energyof the neutrino. (Ans. moe l /6.)

46. A nucleus having rest mass mo disintegrates when at rest into a pair ofidentical fragments of rest mass pmo(A < I). Show that the speed ofone particlerelative to the other is 2 J(I -A2 )C/(2 _A l

). If A is small, show that this speed isless than c by a fraction A4 /8.

47. A positron collides with a stationary electron and the two particles areannihilated. Two photons are genera ted, the lines of motion of which make anglesof 30" and 9tr with the original line of motion of the positron. Calculate theoriginal velocity of the positron and show that the energy of one of the photons i~

equal to the in ternal energy of the electron. (The rest masses of an electron and 2

positron are equal.) (Ans. J3c/2.)48, A moving positron collides with a stationary electron, Both particles an

annihilated and two photons are generated, the lines of motion of which bot~

make angles of 60" with the original line of mOl ion of the positron, Prove that thetotal energy of each photon is 4moc213, where mo is the rest mass of the positronand of the electron.

49. A nucleus having rest mass mo is moving with velocity 4('/5 when it emits aphoton having energy moe l /3 in a direction making an angle of 60" with the line ofmotion of the nucleus. Show that the subsequent direction of motion of thenucleus makes an angle tan' I (J3/7) with its initial direction of motion anJcalculate the new rest mass of the nucleus, Show that, relative to an inertial framein which the nucleus was initially at rest, the line of motion of the photon makesan angle of 120' with thc original direction of motion of the nucleus. (Am.-J 13mo/6.)

68

50. A pion has rest mass mo and momentum P when it disintegrates into a pairof photons having energies E and E'. The directions of motion of the photons areperpendicular and that of the photon having energy E makes an angle rx with theoriginal direction of motion of the pion. Prove that

P = moc(sin 2rx) - If], E = moc l J dcot rx),

£' = moe l J(~tanrx).

51. A particle having rest mass mo is moving with an unknown velocity when itabsorbs a neutrino whose energy is 3moc l /2. The angle between the paths of theparticle and neutrino is rx, where cosrx = 1j3_ After absorption, the rest mass of theparticle is 2mo- Calculate the original velocity of the particle and show that itspath is deflected through an angle fI, where tanli = 4 J2/5, as a result of theencounter_ (AIlS_ 3c/5.)

52. A particle whose rest mass is mo moves along the x-axis ofan inertial frameunder the action of a lorce

, 2m oc l uj = (~-=-~l

AI time t = 0, the particle is at rest at the origin 0, Show that the time Iaken forthe particle to move from 0 to a point x ( < 0) is given by

I [XJ' /]( = - (x + 3u)3c (/

53. Oxr are rectangular axes of an inertial frame. A particle having rest massmo is projected from the origin with momentum Po along Ox. II is acted upon by aconstant force/parallel to Or. Show that its path is the catenary

Wo ( Ix )y = - ," cosh ' - I.f CPo

where w5 = m5c4 + p{,c 2

54, A particle having rest mass mo moves along a straight line under Ihe actionofa frictional force of magnitude mou/k opposing its motion; I' is the speed of theparticle and k is a constant. Show that the time which elapses whilst the particle'svelocity IS reduced from 41'/5 to 3c/5 is [log (3/2) + 5/12Jk.

55. A particle having rest mass mo moves on the x-axis under an attractive forceto the origin of magnitude 2moc l /x l

. Initially It is at rest at x = 2. Show that itsmotion is simple harmonic with period 471:/(.

56. A space ship, with its motors closed down. is moving at high velocity (J

lhrough stationary interstellar gas which causes a retardation as measured by thecrew of magnitude rxl'], Show that the distance it moves through the gas whih;t itsvelocity is reduced from V to U is

I II _!log 1+ x IVrx x I -x U

57. A particle has rest mass moand 4-momentum P. An observer has 4-velocit)Y in the same frame. Show that, for this observer, the particle's:

(i) energy is - p. Y;(ii) momentum is of magnitude J[P2 + (p.y)1/c2];

(iii) velocity is of magnitude J[I +e2p 2/(p.y)2J.(HilI(: All these expressions are invariant.)

5R. A particle has rest massmu and moves along the x-axis under the action of~

force given at any point having coordinate x by

, muc 1w2xf = -- i;T=- (,l~]+ ~/~])3/]

wand a being constants. It is projected from the origin with velocity wa. Shoythat its velocity at any later time is given by 1,2 = (li(a 2 -- x 2

). What does thi,imply for the particle's motion?

59. A particle of rest mass mu moves along the x-axis of an inertial frame unde­the action of a force

, moe 2

f = 2(1 + 2X[/2),1/2

At time 1 = 0, the particle is at rest at the origin. Show that, at any later time I, it,coordinate is given by

x = 2 + el ~ 2 J (I + et)

60. A particle of rest mass /flo moves under the action of a central f01'ce. (r, l:)are its polar coordinates in its plane of motion relative to the force centre as poll.VIr) is its potential energy when at a distance r from the centre. Obtain Lagrange'sequations for the motion in the form

d. )2 I d 2 .d

(yr)~yrl + .. V' = 0, ., (yr 0) = 01 /flo dl

where 1'= [1_(r 2 +r2()2)/e 2] [/2 Write down the energy equation for themotion and obtain the differential equation lor the orbit in the form

22(d2

tl ) C~V,h II dlf + U = m~(:2 V

where u = I/r and It, C are constants. In the inverse square law case when V ~

~ p/r, deduce that the polar equation of the orbit can be written

lu = 1+ /,COSI/O

where /12 = I _p2/m~h2e2. If p/molJ( is small, show that the orbit is ap­proximately an ellipse whose major axis rotates through an angle Trp2/m~h2 (2 purevolution.

61. A particle, having rest mass mo, is at rest a; the origin of the x-axis at time1 = O. It is acted upon by a force f: directed along the posItive x-axis, whOle

70

magnitude when the particle's velocity is I' is given byf = moke 2Ill. Show that attime I( > 0), I' = e sinO, where 0 is positive acute and sec/! = I + kl. Deduce that,at t he same time, the coordinate of t he particle is given by x = e(tan/!- O)/k.

62. If v is the 3-velocity ofa particle and {i = (1- ,,2I(2 )- [/2, prove that v'v= I,i, and

V' d (flv) = (h'Ddl

If mo is t he particle's rest mass. define the 3-force f acting upon it and deduce fromthe above result that v'f= rile2

, where m is the inertial mass.63. A particle having rest mass mo moves along the x-axis under a force of

attraction towards the origin - mooix. It is initi>llly at rest at the point x = a..Show that the velocity with which it passes through the origin is

wae J (4e 2 + oill 2)~--2e2+ (~2~2-

64. If the force f always acts along a normal to a particle's path, show that thesreed I' of the panicle is constant. Write down the equation of motion of therarticle and deduce Ihat till" curvature of Ihe rath is given by K =.Ilml,2 If theparticle moves in a eircle 01 radius a under a COnSlal1l radial fon;e.l; show that itsspeed I' is given by

where A=.IiJl2moe2 and mo is the particle's rest mass.65. A nucleus is moving along a straight hIll' when it ejects an electron. As

measured by a stationary observer, the speed of the electron is !c and the anglebetween the lines of motion of the nucleus and electron is 60". If the speed of theelectron relative to the nli(!cus is also ie, calculate the speed of the nucleus. (Ans.8cI17.)

66. A particle having rest mass mo. initially at reSl at the origin of an inertialframe. moves along its x-axis under t he act ion of a variable force fdirected alongthe axis and given by the formula f = moe l l2 J (I + X). Show that the particle'svelocity I' is given by I' = Xl /2 el(l + X)I /2. Putting x = sinh 20, if ( is the time and (= 0 at O. prove that c( = /I + sinhllcoshll.

67. A particle having rest mass mo is moving with speed ie when it is subjectedto a retarding force. When the particle's inertial mass is m. the magnitude of theretarding force is IXm

2 (IX is constant). Show that the time needed by the force tobring the particle to rest is rcc/6molX.

68. If Tu is the energy--momentum tensor for an elastic fluid and V, is its 4­velocity of flow, by verifying the equation TuVi = -C

2/100 V; in a frame in which

the fluid is momentarily at rest, prove it in any frame. Deduce the equations

II. = (/loov. + r.pt'pl( 2)1 (! _v2

Ie]), /l = 1100 + 1I.1'.le]

71

Hence derive the following formulae for the clements of l~j:

T,/I = lion V, VII + r,p + T" V, V/de2

7:'4 = T4 , = II"" V, V. + T,/I Vp V4 /e 2

1~4 = /1"0 V4 V4 - T,p V, Vp/e 2

69. A perfed fluid is streaming radially outwards across the surface ofaspherewith radius H and centre O. If the motion is steady and there IS no exterml forcefield, show that equation (21.20) leads to the equations

d 3 2dp d).(r ('A) -I r = 0, r'-r + 3A = 0

dr dr u

where r is radial distance from 0, p is the pressure, v is the speed of flow and A= (/loo+p/e 2)r/r(l-t,2/e2). If p vanishes over the sphere r = Rand p= Patgreat distances, and if /10" is constant outsidc the sphere, show that in thi\ region

p = (P + e2/1(0) J (I -- (,2 /(

2) - e21/oo

r11'(I-t,2/e2) \/2 = R2 J(p2 +2Pe2/1o,,)/e/l,,0

70. A straight rod has cross-sectional area A and mass m per unit length. It liesalong the x-axis of an inertial frame in a state of tcnsion F. Show Ihat theencrgy momcntum tensor has components whIch art: Ihe clements of a 4 x 4diagonal matrix. with diagonal elements ( - F/ A, 0, 0, - me 2

/ A). Deducr that arobserver moving along the x-axis with speed u, "ees the inertial mass per uni,length of the rod to be

('2/100 + I? \ iuT - ------- '--\4- l_u2 /e 2 e

T22 = I~2' T13 = r~_"

m -- FU 2/c4-- I .:.: ~i/~i

Deduce that F cannot exceed me 2•

71. Assuming that the energy momentum tensor T i} is a tensor with respect t,a general Lorentz transforma tion Xi = ai}x} + hi, write down the transfclrmatioequations for Tu in the special case where a4, = a,4 = 0, a44 = I. Deduce th2/I, g" g,P are 3-tensors with respect to a simple rotation of the frame Ox,x 2 -,

without relative motion.72. Relative to a frame S, a fluid has flow velocity (u, 0, 0) at a certain point. I

the frame SO rela tive to which the fluid is s ta tionary at the point, the stress tenSehas components I~pand the fluid density is /100' Show that the energy- mDmentultensor in the frame S has components

o 2T _ 1'1 +l/ooU

1 I -- I =_ u2 /(:2 '

72

-

e2/IO O + T~ 1112("2

T44 = _.- -1= utleY -

and dcdul:c that

T31 = (I _u 2(e 2)- 1!2'~1'

T\4 = (I -u2(e 2r 1!2'~3iu/e.

T I I r\) 1- TIl = (I _1I 2(e 2) 1/2 rill, TIJ = (I _I//(2) 1 I] r\I.1'

T21 = (l _u2/(2)112T~I. T21 = T~2' '2.\ = '~3'

T.n = (I _U2/('2)112<~I' T.n = T~l' TlJ = '~J.

CHAPTER 4

Special Relativity Electrodynamics

24. 4-Current density

In this chapter we shall study the electromagnetic field due to a flow of chargewhich will be assumed known. Relative to an inertial frame S, let p be the chargedensity and v its velocity of lh)w. Then, if j is the current density,

j = pv (24.1)

Assuming that charge can neither be crea ted nor destroyed, the equa tion ofcontinuity

,'pdivj+ - =0

1"/(24,2)

will be valid for the charge flow in S. This equation must be valid in every inertialframe and henee must be expressible rn a form which is covariant with respect toorthogonal transf(lrmations in space tIme. Introducing the coordinates Xi byequations (4.4) and employing equation (24.1), equation (24,2) is seen to beelJuivalent to

(24.3)

This equation is covariant as required if (PI'" pI',., pI'" ;cp) are the fourcomponents of a vector in space time. For, if J is this vector, equation (24,3) eanbe written

J,., = 0 (24.4)

and this is covariant with respect to orthogonal transformations, Now, byequation (15.6),

(24.5)

where V is the 4-velocity of flow and hence J is a vector if p( I _1·2

/("2)I,J is an

invariant. Denoting the invariant by Po' we have

(24.6)

73

74

It follows that p = Po if v = 0 and hence that Po is the charge density as measuredfrom an inertial frame relative to which the charge being considered isinstantaneously at rest. Po is called the proper charge density.

J is called the 4-current density and it is clear from equation (24.5) that

J = Po V = (j, icp) (24.7)

It is now clear that, when J has been specified throughout space time, the chargeflow is completely determined, for the space components of J fix the currentdensity and the time component fixes the charge density. Hence, given J, theelectromagnetic field must be calculable. The equations which form the basis forthis calculation will be derived in the next two sections.

Let dwo be the volume of a small element of charge as measured from aninertial frame So relative to which the charge is instantaneously at rest. The totalcharge within the element is Podwo. Due to the Fitzgerald contraction, the volumeof this element as measured from S will be d{/), where

dw = (1 -V2/Cl)1/2dwo

The total charge within the element as measured from S is therefore

pdw = p(1 - V2/( 2)1 12 do>o = Podwo

(24.8)

(24.9)

(25.1)

(25.2)

by equation (24.6). It follows that the electric charge on a body is invariant for allinertial observers.

25. 4- Vector potential

In classical theory, the equations determining the electromagnetic field due to agiven charge flow are Maxwell's equations (3.1)- (3.4). To ensure covariance of thelaws of mechanics with respect to Lorentz transformations, it proved necessary tomodify classical Newtonian theory slightly. However, it will be shown thatMaxwell's equations are covariant without any adjustment being necessary.Indeed, the Lorentz transformation equations were first noticed as the transform­ation equations which leave Maxwell's equations unaltered in form.

To prove this, it will be convenient to introduce the scalar and vector potentials,tj> and A respectively, of the field. It is proved in textbooks devoted to the classicaltheory (Coulson and Boyd, 1979) that A satisfies the equations

. I iltj>dlv A +~--- = 0

(,2ilt

V2A-~ il2~ =('1 (ltl - Jloj

and tj> satisfies the equation

(25.3)

75

where ('2 = 1/J10/;0' We now define a 4-vector potential 0 in any inertial frane Sby the equation

0= (A, iNc) 125.4)

25.5)

It is easily verified that equations (25.2), (25.3) are together equivalent ~) theequation

where the operator [J2 is defined by

[25.6)

The space components or equation (25.5) yield equation (25.2) and the timecomponent, equation (25.3). If Q" J i are the components or nand J respecively,equation (25.5) can be written

[25.7)

in which form it i~ clearly covariant with respect to Lorentz transformllionsprovided n is a vector. This confirms that equation (25.4) does, in fact, ddine aquantity with the transformation properties of a vector in space· time.

Next, it is necessary to show that equation (25.1) IS also covariant with nspeelto orthogonal transformations in space time. It is clearly equivalent 1D theequation

divO=Q'.i=O

which is in the required form.J heing given, n is now determined hy equations (25.7) and (25.R).

26. The field tensor

(25.R)

When A and 4> are known in an inertial frame, the electric and magnetic intelsitiesE and B respectively at any point in the electromagnetic field follow from theequations

?AE = - grad 4) - ..

(It

B = curl A

(26.1)

(26.2)

Making use of equations (4.4) and (25.4), these equations are easily shown to beequivalent to the set

_ i E = ~!.?4 DQ 1

C x (lX, (lX4

i. (lQ4 (lQ2- E = -. - .

(' y (lX2 (lX4

i. (lQ4 DQ,1'['" =. (lX-, (lX

4

(26.3)

76

(1Q (lQ2

B, = -1--'(Ix]( X 2

(lQ, i'QB =.1 (26.4)

.1' (lXJ i"'\ I

(lQ (lQ\B, =.

).

(1.\ I (1\2

Equations (26.3) and (26.4) indicate that the six components of the vector~ i E/,.,B with respect to the rectangular Cartesian inertial frame S are the six distInctnon-J:ero components in space time of the skew-symmetric tensor n

J. , -- n,. i'

We have proved, therefore, that equations (26.1), (26.2) are valid in all inertialframes if

( 0

B, -- Bv . IE'/')-- B, 0 Bx - if~·\,j,.

(265)(I'i;! =BI' - Bx 0 -iE),.

iE),. if),. iE),. ()

is assumed to transform as a tensor with respect to Or! hogonal transforma tion~ inspace time. The equations (26.3) and (26.4) can then be summarized in the tensorequation

(266)

F,j is called the electromugrwti,.jidd II'fJ.\or_ The close relationship between theelectric and magnetic aspects of an electromagnetic field is now revealed as beingdue to their both contributing as (':Jmponents to the field tensor which serves tounite them.

Consider now equations (3.2) and (3.3). Employing the field tensor defined byequation (26.5)and theeurrent density given by equation (24.7),and recalling thatB = J10 H, D = Do E, these equations are seen to be equivalent to

(26.7)

or, in short,

(26.8)

an equation which is covariant with respect to Lorentz transformatIOns.Finally, consider equations (3.1) and (3.4). These can be written

(~FI4 (IF ?F+ 42 + 2,1 =0(1.\2 (lx.., (lX4

(lF41 (~FI 1 (11"'4-

=()+ . + .tl \ ., (1 \4- ('XI

(' I- 12 i'I' ("F+ 24 + 41 =0

(1'\4 (lXt (lX 2

(lFB (IF\! (IF

+. '. + , 12 =0;"":Xl (lX

2 (·X.\

(2t

These equations arc summari)'ed thus:

(26.

(26.

lfany pair from i, i, J" arc equal, since F'j IS skew-symmetric, the left-hatld memlof this equation is identically )'efll and the equation is trivial. The fOIl' possicases when i, i, J,; arc dIstinct are the equations (26.9). Equation (26.10)is a len:equal ion and is therefore also covariant With respect to Lorentz transformatio

To sum lip, Maxwell's equations in 4-dimensional covariant form are:

F,j. j = It"J, lF'j. , + Fj" , + F". j = 0 .r

Given J, at all pomts in space time, these equations determine the iield tensorThe solution can be found in terms of a vector potential Q, which satisfiesfollowing eq ua tions:

Q", =0 }Q,.)j = - /loJ,

n, being determined, F'j follows from the equation

F'j = Qj., -Q" j

(26.

(26.

(2

(2

27. Lorentz transformations of electric and ma~netic vectors

Since F'j is a tensor, relative to the special Lorentz transformation (~.1) its n,zero components transform thus:

F2 ., = 1-'2.1

F.11 = F.1lcoslX + FJ4sin IX

FI2 = Fl 2coslX+F42 sinlX

FI4 = 1-'14

(24= -F2,sinlX+F14 coslX

P.14 = - F.11 sin IX + F.14COS IX

78

Substituting for the components of Fij from equation (26.5) and for sin IX, cos IX

from equations (5.7), the above eqnations (27. f) yield the special LorenlJ:transformation equations for B, viz.

Similarly, equations (27.2) yield the transformation equations for E, viz.

Ex=f;" E,.={I(Ev-uB,), [,={I([-;,+uBv)

(27.3)

(274)

The inverse equations can he written down hy exchanging 'harred' and 'unbarred'symhols and replacing 1/ by -II.

As an example of the usc to which these transformation formulae may he put,consider the electromagnetic field due to an infinitely long, uniformly chargedwire lying at rest a long the .x-axis of the inertial frame ,'i. I I' lJ is the charge per unitlength, it is well known that the electric intensity is everywhere perpendicular tothe wire and is of magnitude i//(2m:oi'), where r is the perpendicular distance fromthe wire. Thus, at the point (.x,)'. z), the components of E are given by

Ex = 0, (27.5)

The magnetic induction vanishes,The electromagnetic field observed from the parallel inertial frame S (relative

to which S has velocity (u, 0, 0)) is given by the inverses of equations (27.3) and(27.4) to have components

Bx = 0, By ={Iuliz

2~;~C2 (y2+ Z2) ,

(276)

at the point (x, y, z) (having used the transformation equations y = }'. z = z). Asegment of the wire having unit length in S' will appear in S to have lengthJ (1 - 11 2/(.2); however, the charge on the segment must be the same in bothframes, viz. q. It follows that the charge per unit length as observed in S is if = {Ii].Thus, the charge which flows past a fixed point on the x-axis of S in unit time willbe {lUi] = i; i therefore measures the current flowing along this axis. Since c2

= l/p",;", equations (27.6) can now be written

13, = 0, By =Poiz

2IT(y2 +Z2)'

(27.7)

These equations imply that the magnetic induction is of magnitude p oi/2lTr and

that the B-lines are circles with centres on Ox and planes parallel to Ovz. Th sresult f(lr a long straight current is a well-known one in the classical theory. Ticelectric intensity is of magnitude qj2nfor and is directed radially from the wire;however, in the case ofa current due to the flow of negatively charged electrons na stationary wire, this field IS cancelled hy the contrary field due to the positi1ccharges on the atomic nuclei.

28. The Lorentz force

We shall now calculate the force exerted upon a point charge e in motion in <Inelectromagnetic field.

At any instant, we can choose an inertial frame relative to which the poiltcharge is instantaneously at rest. Let Eo he the electric intensity at the poiltcharge relative to this frame. Then, by the physical definition of electric intensilyas the force exerted upon unit stationary charge, the force exerted upon e will teeEo. II follows from equation (I ~.2) that the 4-force acting upon the charge in thiSframe is given hy

The 4-vclocity of the charge in this frame is also given by

v = (0, it')

(28. )

(28.:)

and hence, hy equation (26.5),

eF i ) II) = e(Hxo, £,0, E,o, 0) = (eEo,O) (28,J)

It has accordingly been shown that, in an inertial frame relative to which thecharge is instantaneously stationary,

(28.~)

But this IS an equation between tensors and is therefore trne for all inertial frame!.Substituting in equation (2~.4) for the components Fi, Fir Vi from equatioll5

(18.2), (26.5) and (15.6) respectively, the following equations are obtained:

f~ = e(B,I',- -- Byl', + Ex)

Iv = e(Bx,., - B,"x + Ev)f~ = dByl'x - Bx"v + E,)

These equations are equivalent to the 3-vector equation

f = dE + v x B)

f is called the Lorentz loree acting upon the charged particle.

(28.5)

(21:1.1:)

29. The energy momentum tensor for an electromagnetic field

Suppose that a charge distribution is specified by a 4-current density vector J. IId(J)o is the proper volume of any small element of the distribution and Po is the

80

proper density of the charge, the charge within the element will be Podwo. Itfollows from equation (28.4) that the 4-force exerted lIpon the clement by theelectromagnetic field is given by

(29.1)

V being the 4-velocity offlow for the clement. Employing equation (24.7), this lastequation can be written

(29.2)

and it follows from the definition given in section 21 that the 4-force density forthe electromagnetic field is given by

(29.3)

(29.4)

Substituting for J, from the first ofequations (26,11), we can express D i in termsof the field tensor thus:

D i = - FijF jk . kJ10

We will now prove that the right-hand member of this equation is, apart fromsign, the divergence of a certain symmetric tensor Sij given by the equation

J10Sij = Fik/;jk -ll5ij F u F U

and called the energy-momentum tensor of the electromagnetic field.Taking the divergence of Sij, we have

/IOSij.j = F ik . j fjk + FikFjk.j - ~bijFu Fu,j

Now

since Fij is skew-symmetric. Thus

Fik.jFjk 0" ~ (Fik,j + Fji.dfjk

Also

(29.5)

(29.6)

(29.7)

(29.8)

(29,9)

and it follows from these results that the first and last terms of the right-handmember of equation (29.6) can he combined to yield

1(Fik• j + Fji,k + Fkj.i)Fjk

and this is zero by the second of equations (26.11),Hence

(29.10)

(29.11)

8\

Substituting for the components of the field tensor from equation (26J), thecomponents of Sij are calculable from equation (29.5) as follows: If IX, fJ take anyof the values 1,2,3, then writing £1 for EX' £0 2 for By, etc.,

IXffJ (~9.12)

If i =; = I,(t9.13)

S2" S-,-, may be expressed similarly and therefore, in general, if IX, fJ, = I 2,3,

(!9.14)

Apart from sign, this is Maxwell's stress tensor /,j' /oj is only a tensor with nspectto rectangular frames stationary in the inertial frame being employed.

Also, if IX = 1,2,3,

i i= ExH= S

c c

where S is Poynting's vee/or.Finally,

(29.15)

129.16)

where U is the energy density in the electromagnetic field.These results may be summarized conveniently by exhibiting the comp~nents

of Sij in a matrix thus:

We can now write down the equation of motion for a charge cloud movingunder the action of the electromagnetic field it generates. If T;j is the lineticenergy momentum tensor for the cloud, equations (21.\5) and (29.11) show thatthe equation of motion can be written

or

Tij,j= -Sij,j 129.18)

i.e. the divergence of the total energy- momentum tensor vanishes. If the chargedparticles forming the cloud do not interact except via the electromagnetic field, i.e.the cloud is incoherent, T'1 is given by equation (21.16). If, however, the plftic1esconstitute an ionized flUid, equations (22.\9) or (22.21) must be used to cQlculateTij.

It was shown in section 21 that T••lie equals the density of the x.-component ofthe linear momentum of a system. Since S••lie = SJc 2, the density of till linear

82

momentum of an electromagnetic field is g = 8/c 2, where 8 is Poynting's vector.

Alternatively, as explained in section 21, g can be interpreted as the currentdensity vector for the inertial mass flow and thus, ("2 g = 8 gives the rate ofenergyflow across unit area placed perpendicular to the direction of this flow; this is theusual significance attached to Poynting's vector.

According to the theory in section 21, - S44/("2 should equal the density ofinertial mass for the field. We have found that - S44/c2 = U/c 2 and, since U is theenergy density, our result is in conformity with expectations.

The results which have been obtained may be summarized as follows: Ifmomentum of density 8/("2 and energy of density U are ascribed to theelectromagnetic field, equation (29.19) shows that the net momentum and energyof the field and charge will be conserved.

Exercises 4

I. Write down the special Lorentz transformation equations for J and deducethe transformation equations for j, p, viz.

T. = (I - U2/C 2)-1/2(jx -pu), }y =}y

p=(I-u2/c2)-1/2(p-}xu/c2), T,=},

2. Deduce from the Maxwell equation Fij.} = J1. oJ; that div J = O.3. Verify that the field tensor defined in terms of the 4-potential Q i by equation

(26.13) satisfies Maxwell's equations (26.11) provided Q; satisfies the equations(26.12).

4. (i) Prove that

FijFij = 2J1.0(J1.0112 - Co £2)

and deduce that J1. oH 2- 60 £2 is invariant with respect to Lorentz

transformations.(ii) Prove that

CijklFijFk1 = - 8il<:' R/c

and deduce that E· B is an invariant density with respect to Lorentztransformations.

5. ] f U is the energy density and S is the Poynting vector for an electromagneticfield, prove that U2 - 8 2 is invariant.

6. An observer 0 at rest in an inertial frame Ox)';;( finds himself to be in anelectric field E = (0, E, 0), with no magnetic field. Show that an observer (Jmoving according to 0 with uniform velocity Vat right angles to E, finds electricand magnetic fields II, 8 connected by the relation

(,28+ V x II = 0

7. If Si) is the energy-momentum tensor for an electromagnetic field, provethat its trace, viz. Si/, is zero.

83

8. A plane monochromatic electromagnetic wave is being propagated in adirection parallel to the x-axis in the inertial frame S. Its electric and magneticfield components are given by

E = [0, asinm(t - x/t'), 0]

aB = [0,0, sinlt* - x/c)]

c

Show that, when observed from the inertial frame S. it appears as the planemonochromatic wave

E = [0, AasinAw(l- x/c), 0]_. a _8 = [0,0, A sinAw(t - x/c)J

c

where A = )('_- u/c.)1-1- u/c

u being the velocity of ~r relative to S (i.e. both the amplitude and frequency arereduced by a factor A. The reduction in frequency is the Doppler effect.)

9. Show that the Hamiltonian for the motion of a particle with charge e andmass m in an electromagnetic field (A, tj» is

H = c[ (p -eAy+ m2 c2J/2 + etj>

(Hint: Show that Hamilton's equations yield the equation

d(II (mv) = e(E + v x B).)

10. Verify that, in a region devoid of charge, equations (26.12) are satisfied by

provided Ai, kp are constants such that

Aik, = 0, kpk p = 0

By considering the 4-vector property of Q" deduce that Ai must transform as a 4­vector under Lorentz transformations. Deduce also that krXp is a scalar undersuch transformations and hence that kp is a 4-vector.

A plane electromagnetic wave, whose direction of propagation is parallel to theplane Oxy and makes an angle ex with Ox, is given by

where v is the frequency. The same wave observed from a parallel frame Oxyzmoving with velocity Ii along Ox, has frequency Ii and direction of propagationmaking an angle IX with Ox. By writing down the transformation equations for the

84

vector kp , prove that

u1- --CO,IX

CIi - - ----v

- (I _ U2/(2)L2 ' cosiX =

UCOSIX- -­

('

U1- cos IX

C

II. Ox}': is an inertial frame S, A particle having rest mass mo and electriccharge q moves in the xv-plane under the action of a uniform magnetic field Bdirected along the z-axis, Show that the particle's speed t' is constant and that,with a suitahle choice of coordinates, its trajectory is the circle

x = Rsinwl, y'= Rcoswl

where

Sis an inertial frame 0xyz parallel to Sand 0 moves along Ox with speed u.Calculate u and Bso that uniform fields E = (0, £0,0), B = (0,0, Bo)are observedin S. Hence descrihe the motion of a charged particle released in these fields andshow that its average velocity is £01Bo along the x-axis.

12. The frame S is parallel to the frame S and is moving along the x-axis withspeed u. In the frame S, there is a uniform electric field (0, E, 0) and a uniformmagnetic field (0,0, B). Show that it is possible to choose the value of u so thatthe field in the frame S is entirely magnetic and that its magnitude is thenJ(B2

- £2/(,2). What is the direction of this field? (Ans. Parallel to z-axiS-)13. A charge q has rest mass mo and is moving in the positive sense along a

negative x-axis with speed u, when it enters a magnetic held, having componentsBx = By = 0, Bz = B (constant), confined to the region 0,,:; x ~ ll. There IS nofield in the regions x < 0, x > ll. Explain why the inertial mass of the chargeremains constant during its motion through the field and show that its path is thecircle

x2 + y2 + 2ky = 0, z = 0

where k = mou/qB(l-u2/c 2)1 /2. What is the condition that the charge will beturned back hy the field? (Ans. k < ll.)

14. A plane electromagnetic wave of frequency f is heing propagated in adirection making an angle 0 with the x-axis. Its electric and magnetic fieldcomponents are given by

E = (- AXsinO, A Xc()sli, 0)

B = (0,0, AX/c)

where

. 2 .( xcosu+ysinO)X = Sin rr! I - --- .-. c

Show that, when ohserved from a frame S which is parallel to OXI': and moveswith a velocity (('eosll, 0, 0) relative to O'F, the wave has components

f: .~ ( - A X sinO, 0, 0)

B= (0,0, A ,X'smll/e)

where

x = sin [2rr/sinll(l-- i'/e) Iand (,\, I', ::, I ) are space time wordinates III ,'l, What is the direction ofpropag,ltion in S and what is thc ohserved frequency') (Ans, Parallel to x-axis ,IIfrequency I sinO)

J 5. V; is the 4-velocity of flow of a conducting medium and .I, is the 4-currenldensity ofa charge flow in the medium. Ohm's law IS vahd for the medium, rr heingits conductivity, Prove that

(lIilll: Verify this equation in a frame for whieh the medium is at rcst using Ohm'slaw j = rrE.)

16, A uniform magnetic Iield of Induction 13 is directed along the z-axis of aninertial frame. Show that the energy momentum tensor f()r the field h<Jscomponents which arc the elements of a 4 x 4 diagonal IHatrix, with diagonalclements 8 1 /2/Jo( I, I, - I, - I).

17, A point charge (' moves along the z-axis ofan inertial frame S with constantvelocity v, Calculate the electromagnetic field in a parallel inertial frame whoseorigin moves with the charge and deduce the field in S, Hence show that, at theinstant I ~, 0, when the eha rge passes through the origin 0 of S, the electric field isdirected radially from 0 and its magnitude at thc point having spherical polarcoordinates (f, II, (f» is givcn hy

Show, also, that the magnetic field in S at this instant is given hy B = (v X E)/(''.

(.10.1 )

CHAPTER 5

General Tensor Calculus. Riemannian Space

30. Generalized N-dimensional spaces

In Chapter 2 the theory of tensors was developed in an N-dimensional Euclideanspaceon the understanding that the coordinate frame heingemployed w,ts alwaysrectangular Cartesian. If Xi, Xi + dXi are the coordinates of two neighhouringpoints relative to such a frame, the 'distance' ds hetween them is given hy theequation

If .'C" .\:, + dX i are the coordinates of the same points with respect to anotherrectangular Cartesian frame, then

(30.2)

and it follows that the expression dxidx; is invariant with respect to atransformation of coordinates from one rectangular Cartesian frame to another.Such a transformation was termed orthogonal.

Now, even inlr" it is very often convenient to employ a coordinate frame whichis not Cartesian. For example, spherical polar coordinates (I', 0, tj» are frequentlyintroduced, these heing related to rectangular Cartesian coordinates (x, y. z) hythe equations

X = rsinOcostj>, y = rsinO sintj>, z = rcosO (10.3)

In such coordinates, the expression for ds 2 will he found to he

ds2 = dx2 + dy 2 + dz2

= d,.2 + 1'2 do 2 + ,.2 sin 2 (I dtj>2 (30.4)

and this is no longer of the simple form of equation (30.1). The .;oordinatetransformation (30.3) is accordingly not orthogonal. In fact, it is not even linear,as was the most general coordinate transformation (8.1) considered in Chapter 2.

The spherical polar coordinate system is an exam pie ofa ('urlJilinear coordinateframe in $,. Let (u. v, w) he quantities related to rectangular Cartesian coordinates(x, y, z) hy equations

86

IJ == Il(X, .1', Z), lJ = I'(X, .1', Z), Ii' = W(X, .1', Z) (30.5)

87

such that, to each point there corresponds a unique triad of values of (u, v, w) aoJto each such triad thcre corresponds a unique point. Then a set of values cf(u, t', w) will serve to identify a POint in 11'.1 and (u, P, w) can he employed oS

coordinates. Such generali7ed coordinates are called erJrl'i!inear coordinates.The equation

u(x, y, z) = u"

where Uo is some constant. defines a surface in 6'.1 over which u takes the constantvalue u". Similarly, the equations

(30.i)

define a pair of surfaces on which I' takes the value Po and w the value w)respectively. These three surfaces will all pas.s through the point Po havin~

coordinates (/1o, 1'0, wo)as shown in hg. 6. They are called the coordinate s/lr{at'e,through 1'0' The surfaces r = "1), W ~. W'o will inten,eel in a curve P"U along whidI' and w will he constanl in value and only II will vary. PoU is a ('oordinate IiII<'through Po. All('gether, three coordtl1atc lines pass through Po. The equations ~

= constant, r' = constant, W = constant define three families of coordinate sur·faces corresponding to the three f~lmilies of pl,1I1es parallel to the coordinateplanes x = 0, y = 0, Z = 0 of a rectangular Cartesian frame. Pairs of thesesurfaces intcrscct in coordinate lincs which correspond to the parallels to thtcoordinate axes in a Cartesian frame.

w

FiG. 6

Solving equations (30.5) for (x, y, z) in terms of (II, I', w), we obtain the inversetransformation

x = X(II, t', w), Y = y(u, r, W), z = z(u, v, w) (30.8)

Let (x, y, z), (x + dx, y +- dt" z + dz) he the rectangular Cartesian coordinates oftwo neighhouring points and let (u, 1', wi, (/I + du, I' + dr', W + dw) he their

88

respective curvilinear coordinates. Differentiating equations (30.8), we obtain

etc. (30.9)

Thus, if ds is the distance between these points,

d.\2 = dx 2 +d.l'l +dz'

= AdIJ 2 + Bdr,2 +Cdw2 +2Fdl'dw+2Gdliuu+211dudr' (30.10)

giving the appropriate expression for ds' in curvilinear c00rdinates. It will benoted that the coeflicients A, B, etc., are, in general, functions of (u, r', w).

If, therefore, curvilinear coordinate frames arc to be permitted, the theory oftensors developed in Chapter 2 must be modified to make it independent of the.~pccial orthogonal transrormations for which ds' " always expressible in thesimple form of equation (30.1). The necessary modiHcations will be described inthe later sections of this chapter. However, these modifications prove to be orsucha nature that the amcnded thcory makes no ,lprcal to the special mctric;)1properties of Euclidean spaee, i.e. the theory proves to be applicable in morcgcneral spaces for which Euclidcan space is a particular case. This we shall nowexplain further.

Let (x l, x', ... , x N ) be curvilinear coordinates in IJ'N'* Then, by analogy withequation (H). Ill), if us is the distance between two ncighbouring pl,inls, it can beshown that

(30.11 )

wherc tile coeflicicnts Y'} of the quadratic form in tile -.;j will, in gencral. befunctions of these coordinates. Since the spacc is Euclidcan, it is possible totransform from the curvilinear coordinates x j to Cartesian coordinates / so that

(30.12)

Clearly. the reduction of ds' to this simple fonn is only possible because thefunctions Y'1 satisfy certain conditions. Convc"dy. the satishlction of theseconditions by the Y>j will guarantee that coordinates y' exist for which ds' takesthe simple form (30.12) and hence that the space is Euclidean. However, inextending the theory of tensors to be applica ble to curvilinear coo I'd ll1a te frames.we shall, a t a certain stage, make use of the fact that ds' is expressible in the form(30.11), but no use will be made of the conditions satisfied by the wcflieients q,}

wllich arc a consequence of the space being Euclidean. It follows that theextended theory will be applicable in a hypothetical N-dimensional space forwhich the 'distance' ds between neighbouring points Xi, Xi + dx i is given by anequation (30.11) in which the Yi} are arhirraryful/('rions of the x'. t Such a space is

'" The coordJnate~are here distinguished hy superscripts instead of ~ubscflp{s for a reason which willbe given later.+ rxccrt Iha t p:1I tlal derl\:\llvcs or the fl,; will he 1.Issumcd to CXJsl and to he l'on(inllOUS (0 allY orderrc:qUll"cd hy the dh.:ory,

89

said to be Riemannian and will be denoted by !if N'/}N is a particular :ifN for whichthe !/ij satisfy certain conditions. The fight-hand member or equation (30.11) istcrmed the metrlc of the Riemannian spacc.

The surface of the Earth provides an example of an ,YI2' II' II is the co-latitudeand tj> is the longit ude of any point on the Earth's surface, the distancc d,Y betweenthe points (Ii, (Pl, (II + dll, </1 + d,/!) is given by

ds 2 = R 2 (d1l 2 +sin 2I1d</>2) (30.13)

where R is the earth's radius. For this space and coordinate frame, the !/'J take theform

(30.14)

It is not possible to define other coordinatcs (x, y) in terms or which

ds 2 = dx 2 +d y 2 (30.15)

over the whole surh\(;e, i.e. this .if2 is not Euclidcan. Howcver, thc surraces or aright circular cylinder and cone arc Fuclidean; the proor is lert as an exercise rorthe reader.

It will be proved in Chapter IJ that, in the presence or a gravitational field,space time ceases to be Euclidean in Minkowski's sense and becomes an ,if4' Thisis our chief reason for considering such spaces. However. we can generalize theconcept of the space in which our tensors arc to be defined yet further. Untilsection 37 is reached, we shall make no funher rererence to the metric. Thisimplies that thc theory of tensors, as developed thus hlf, is applicable in a verygeneral N-dimcnsional space in whidl it is assumed it is possible to set up acoordinate frame but which IS not assumed to posse,ss .1 metric. In such ahypothetical space, thc distance bctwcen two points is not even defined. It will bereferred to as .'I·N. ,ifN is a particular fl' N for which a metric is specified.

31. Contravariant and covariant tensors

Let x' be the coordinates ofa point Pin:l'N relative to a coordinate frame which isspecified In some manner which does not concern us here. Let x' be thecoordinates of the same point with respect to another reference frame and letthese two systems of coordinates be related by equations

(J 1.1)

Consider the neighbouring point P' having coordinates Xi + dx' in the fIrst frame.Its coordinates in the secl)nd frame will be Xi + dx" where

(31.2)

and summation with respect to the index i is understood. The N quantities dx i aretaken to he the componcnts of the di.YI'/acl'lII,'1II ,'"c/or PP' referred to the first

90

frame. The components of this, vector referred to the second frame are,correspondingly. the dx l and these are rela ted to the components in the first frameby the transformation equation (31.2). Such a displacement vector is taken to bethe prototype for all ('o/llrQlJariant lleC/ors.

Thus. Ai are said to be the components of a contravariant vector located at thepoint x" if the components of the vector in the 'barred' frame are given by theequation

(31.3)

It is important to observe that. whereas in Chapter 2 the coefJicients alj occurringin the transformation equation (10.2) were not functions of the Cartesiancoordinates Xi so that the vector A was not, neccssarily, located at a definite pointin r! N' the coeflicicnts (l'(i/<lx i in the corresponding equation (31,3) are functionsof the Xi and the precise location of the vector A i must be known before itstransformation equations are determinate. Thl!; can be expressed by saying thatthere are no free vectors in :/'N'

The form of the transformation equation (31.3) should be studied carefully. Itwill be observed that the dummy index} occurs once as a superscript and once as asubscript (i.e. in the dcnominator of thc partial derivative). Dummy indices willinvariably occupy such positions in all exprcssions with which we shall heconcerned, Again. the free index i occurs as a superscript on both sides of theequation. This rule will be followed in all later developments. i.e. a free index willalways occur in the same position (upper or lower) in each term of an equation.Finally, it will assist the reader to memori:t:e this transformation if he notes thatthe free index is associated with the 'baITed' symbol on both sides of the equ'ltion.

A contravariant vector AI may be defined at onc point of.'l'N only. However, if itis defined at every point of a certain region, so that the A I are functions of the x" aCO/llral'ariant veuor field is said to exist in the region.

If V is a quantity which is unaltered in value when the reference frame ischanged, it is said to be a scalar or an i/ll>ariant in Y'N' Its transformation equationis simply

(31.4)

Since this equation involves no coefJicients dependent upon the Xi, the possibilitythat Vmay be a free invariant exists. However, Vis more often associated with aspecific point in Y'N and may be defined at all points of a region of Y'N, in whichcase it defines an invariant field. In the latter case

(31.5)

j7 will then. in general, be a quite distinct function of the Xi. If, however. in thisfunction we substitute for the Xi in terms of the Xi from equation (3 Ll), byequation (31.4) the right-hand member of equation (31.5) must result. Thus

V(x',x 2•••.• X N )= V(x'.x 2

, •.• ,XN ) (31.6)

(3IJ)

V being an invariant field. consider the N derivatives cV/rix i. In the xi-fraIre.

the corresponding quantities are ilV/ili' and wc have

{l V ri V {lX j iJx j() V

sincc, by cquation (31.6), whcn Vis cxpressed as a function of the Xi it rcduces 10

V. As in eha pter 2, the {l V/,lX i arc taken to be the componen ts of a vector calledthe wudielll or Vand denoted by grad V. However, its transformation law (31.7)ls

not the same as that for a contravariant vcctor, viz. (31.3). and it is taken to be tIeprototype for another species of vectors called ('ovariant l'eClors.

Thus, Bi is a covariant vector if

(31.l)

(3U,

Covariant vectors will be distinguished from contravariant vectors by writingtheir components with subscripts instead of superscripts. This notation sappropria te, for (I V/{lX i is a covariant vector and the index i occurs in th~

de.l0minator of this partial derivative. The vector dx', on the other hand, has beenshown to be contravariant in its transformation properties and this is correctlyindicated by the upper position of the index. This is the reason for denoting trecoordinates by Xi instead 01'\" although it must be clearly understood that the ;'alone arc not the components of a vector at all.

The reader should check that the three rules formulated above in relation to tretransformation equation (31.3), apply equally to the equation (31.8).

The generali)'.ation from vectors to tensors now proceeds along the same lim;as in section 10. Thus. if Ai. BJ are two contravariant vectors, the N 2 quantitie;A'IJj arc taken as the components ofa contravariant tensor of the second rank. It;transform<llion equation is f(lUnd to be

-._. (Jx i i'J.xjAt l

A'BJ=-- -- A B(lX k ,lxi

Any set of N 2 quantities Cij transforming in this way is a romravariant tensOl.Again, if Ai, Bj arc vectors, the first contravariant and the second covarianl,

then the N 2 quantities A' B j transform thus:

(31.1 0

Any set of N 2 quantities C; transforming in this fashion is a mixed tensor, i.e. itpossesses both contravariant and covariant propert ies as is indicated by the tw~

positions of its indices.Similarly, the transformation law for a covariant tensor of rank 2 can b<

assembled from the law for covariant vectors.The further gencrali),'<ltion to tensors of higher ran k should now be an obvioUi

step. It will be suflicient to give one example. Ajk is a mixed tensor of rank 3

92

having hoth the covariant and contravariant properties indicated by the positionsof its indices. if it transforms according to the equation

(31.11 )

The components ofa tensor can be given arbitrary values in anyone frame andtheir values in any other frame arc then uniquely determined by the transform­ation law. (\lnsider the mixed second rank tensor whose components in thexi-frame are h;. the Kronecker deltas (,l; = O. i f j and ,l; = I. i = j). Thecomponcnts in the x'-frarne are ;) ;. where

- t1x' i1.\/ -k

,l' = .... ,lJ (lXk p\) I

Px' (1.'(1<.

;lxk

(31.12)

Thus this tensor has the same components in all frames and is called the!ulldamell/ol mixed (enSOf. However, a second rank ..01'(1/"1011/ tensor whosecomponents in the x'-frame are the Kronecker deltas (in tillS case denoted by J,,j.has diflerent components in other frames and is accordingly of no special interest.

It is reasonable to enquire at this stage why the distinction between covariantand contravariant tensors did not arise when the coordinate transformationswere restricted to be orthogonal. Thus. SUPP"Se that Ai. H, are contravari,1I1t andcovariant vectors WIth respect to the orthogonal transformation (H. I ). Thc inversetransformation has been shown to be equation (11.5) and it follows from thesetwo equations that

(31.13)

For the particular case of orthogonal transformations. therefore. equations (31.3)and (3I.H) take the form

(31.14)

It is clear that both types of vector transform in an identical manner and thedistinction hetween them cannot. therefore. be maintained.

As in the case of the Cartesian tensors of Chapter 2. new tensors may be formedfrom known tensors by addition (or subtraction) and multiplication. Onlytensors of the same rank and type may be added to yield new tensors. Thlls. if A)k'B)k are components of tensors and we define the quantities C)k by the equation

(31.15)

91

then Cik arc the components of a tensor having the covariant and contravarianlproperties indicated by the position of its indices. However, Ai, B,) cannot bladded in this way to yield a tensor. Any two tensors may be multiplied to yield ~

new tensor. Thus, if A~, B~ arc tensors and we define N 5 quantities C~:,n by thlequation

(31.16)

these are the components of a fifth rank tensor having the covariant andcontravariant properties indicated by its indices. The proofs of these statementlare left for the reader to provide.

If a temor is symmetric (or skew-symmetric) with respect to two of illsuperscripts or to two of Its subscripts in anyone frame, then it possesses thilproperty in every frame. The method of proof is identical with that of thlcorresponding statement fi)r Cartcslan tensors given in section 10. However, ii'A~ = A{ is true for all i, i when one refCrence frame is being employed, thilequation will not. in general, be valid in any other frame. Thus, symmetry (Oi

Skew-symmetry) ofa tensor wilh respect to a superscript and a subscript is not, ilgeneral. a covariant property. The lensor h~ is exceptional in this respect.

Another result of great importance which may be established by the samlargument we employed in the particular case of Cartesian tensors, is that alequation between tensors of the same type and rank is valid in all frames if it i\valid in one. Tlus implies th,lt such tensor equations arc wvariant (i.e. are of invariable form) wilh respect to transformations between reference frames. The useful·ness of tensors for our later work will be found to depend chiefly upon this propert}

A symbol such as A~k can be ('Oil/roC/cd by selting a superscript and a subscriplto be the same letter. Thus Ai" A:karc the possible contractions of A~k and each, bJthe repeated index summation convention, represents a sum. Since in the symbolAi"j alone is a free index, this entity has only N components. Similarly A:k has !'Icomponents. It will now be proved that, if Aik is a tensor, its contractions are aiscitensors. Specifically. we shall prove that B) = A~i is a covariant vector. For

(31.17,

94

estahlishing the result. This argument can ohviously he generalized to yield theresul t tha t any con tracted tensor is itself a tensor of ran k two less than the tensorfrom which it has heen derived and of the type indicated hy the positions of itsremaining free indices. In this connection it should be noted that, if Ajk is a tensor,A;j is not, in general, a tensor; it is essential that the contraction be with respect toa superscript and a suhscript and not with respect to two indices of the same kind.

If Alk , B~ are tensors, the tensor Ajk B'; is called the outer product of these twotensors. If this product is now contracted with respect to a superscript of one{actor and a subscript of the other, e.g. Ajk B;, the result is a tensor called an i,tnerproduct.

32, The quotient theorem, Conjugate tensors

It has been remarked in the previous section that both the outer and innerproducts of two tensors are themselves tensors. Suppose, however, that it isknown that a prod uct of two factors is a tensor and that one of the factors is atensor, can it be concluded thaI the other factor is also a ten,or'? We shall provethe following quutietll theorem:

lIthe result oftakitl!ftlte product (outer or inner) ofa qillen set o!elt'ments with atensor o!any spe('ified type and arhilrary ('ompmlelll.\' is knowlltl) he a tensor, thenthe given elements are tire components of a tellsor.

It will be sullicient to prove the theorem true for a particular case, since theargument will easily be seen to be ofgenel1tl application. Thus. suppose the Ajk areN J quantities and it is to be established that these are the components ofa tensorof the type indicated by the positions of 1he indices. Let B~ hc a mixed tensor 01rank 2 whose components can be chosen arbitrarily (in anyone frame only 01cour,e) and suppose it is given that the inlier product

(32.1 )

is a tensor for all such B~. All componcnb have been assumed calculated withrespect to the x-frame. Transforming to the .x-frame, the inner product is given totransform as a tensor and hence we have

(32.2)

where Aj: are the actual components replacing the A;k when the reference frame ischanged. Let A;k be a set of elements defined in the x-frame by equation (31.11).Since this is a tensor transformation equation, we know that the elements sodefined will satisfy

Subtracting equation (32.3) from (32.2), we obtain

(A;: -;r;dB~ = 0

(32.3)

(32.4)

Since B~ has arbitrary components in the x-frame, its components in the x-frame

95

(32.5)or

are also arbitrary and the components R~ eHn assume any convenient values.Thus. taking R~ = I when k = K and B~ = 0 otherwise. equation (32.4) yields

A;~ -,4;1< = ()

A;~ = A;KThis being true for K 1.2....• N. we have quite generally

(32.6)

This impliel; that A jk does transform al; a tenwr.We will fir.,t give a very simple example of the ,lpplication of this theorem. Let

A' be an arbitrary contravariant vedor. Then

c\jAI = A' (32.7)

and since the right-hand member of this equation is certainly a vector. by thequotielll theorem 6j is a tensor (as we have proved earlier).

As a second example, let fI'l be a symmetric covariant tenwr and let g= Ig,jl bethe determinant whose clements are the tensor" components. We shall denote'byGil the co-factor in this determinant of the clement gil" Then, although Gij is not atensor. if!/ f 0, Gijlg = yij is a symmetric contravariant tensor. To prove this. wefirst observe that

(32.8)

and hence. dividing by !I,

(32.9)

Now let A' be an arbitrary contravariant vector and define the covariant vector Bi

by the equation

(32.10)

Since g of O. when the components of Hi arc chosen arbitrarily, the correspondingcomponents 01" A' can always be calculated from this last equation. i.e. B, isarbitrary with A'. But

(32.11)

having employed the second identity (32.9). It now follows by the quotienttheorem that gil is a contravariant tensor. That it is symmetric follows from thecircumstance that (i'l possesses this property, gt). 0'1 are said to be cO/ljugate toone another.

33. Covariant derivatives. Parallel displacement. Affine connection

In the earlier sections of this chapter. the algebra of tensors was established and itis now time to explain how the concepts of analysis can be introduced into thetheory. Our space i/N has N dimensions, but is otherwise almost devoid of special

characteristics. Nonetheless, it has so far been able to provide all the facilitiesrequired of a stage upon which the tensors are to play their roles. It will now hedemonstrated, however, that additiomtl features must be built into the structureof .IN' before it can function as a suitable environment for the operations oftensor analysis.

It has been proved that. if tj> is an invariant field, (!tj>/(lX i is a covariant vector,But, if a covariant vector is differentiated, the result is not a tensor. For, let A, hesuch a vector, so that

(33.1)

Differentiating both sides of this equation with respect to xi, we obtain

(l Xk (lX' (lA k n2 x k

iJxi ix1 iJ~T + ~xi~j] Ak (33.2)

The presence of the second term of the right-hand member of this equationreveals that DA,/(!xi does not transform as a tensor. However, this fact can bearrived at in a more revcaling manner as follows:

Let P, p' be the neighbouring points Xi, Xi + dx i and let A" Ai + dA; be thevectors of a covariant vector field associated with these points respectively. Thetransformation laws for these two vectors will be different, since the coeflicicnts ofa tensor transformation law vary from point to point in .'I'N. It follows that thedifference of these two vectors, namely dA i, is not a vcctor. However,

(33.3)

and, since dx i is a vector, if Ai, i were a tensor, dA, would be a vector. A" I cannotbe a tensor, therefore. The source of the ditliculty is now apparent. To define Ai, i

it is nccessary to compare the values assumed by the vector field Ai at twoneighbouring, but distinct, points and such a comparison cannot lead to a tensor.If, however, this procedure could be replaced by another, involving thecomparison of two vectors defined at the same point. the modifIed equation (33.3)would be expected to be a tensor equation featuring a new form of derivativewhich is a tensor. This leads us quite naturally to the concept of paralleldisplacement.

Suppose that the vector Ai is displaced from the point P, at which it is defined,to the neighbouring point P', without change in magnitude or direetiUll, so that itmay be thought of as being the same vector now defined at the neighbouringpoint. The phrase in italics has no precise meaning in.'l'N as yet, for we have notjcfined the magnitude or the direction of a vector in this space. However, in theparticular case when ,'j'N is Euclidean and rectangular axes are being employed,this phrase is, of course, interpreted as requiring that the displaced vector shallpossess the same components as the original vector. But even in trN' if curvilinear:oordinates are being used, the directions of the curvilinear axes at the point P'

97

will, in gencral, he diflerent from their direction~at P and, as a consequence, thecomponents of the displaced vector will not be identical with its componentsbefore the displacement. In :/','1' thercforc, componcnb of thc displaced vectorwill b.e denoted by A, + ,lA" This vector can now bc comparcd with the ticld vcctorA, +dA, at the same point p', Since the two vcctors are defined at the same point,thcir ditlercnce is a vector at this pOllll, i,e, dA; (ill" is a vector. The modifiedequation Ln.3) is accordingly expected to be of the form

(33.4)

wherc A", is the appropriate replacemcnt for ii" r Since dx' is an arbitrary vectorand the left-hand member of equation (.HA) is known to be a vector, A", will, bythe quotient theorem, be a covariant tensor. It will be termed the C()(Iarian(deriI'Mil'/' of A,. Thus, the problem of defining a tensor derivative has been re­expressed as the problem of defining parallel displacement (infinitesimal) of avector.

We are at liberty to define the parallel di~plaeementof A, from P to P' in anyway we shall lind convenient. However, to avoid confusion, it is necessary that thedefinition we accept shall he in conformity with that adopted in <fiN, which is 1I

special casc of .'1'" Suppose. thercf(lre. that (lUl'.'I'N is Euclidean and that y' arcrectangular Cartesian coordinates in this space, Let B, be the components of thevector field A, with re~pect to thesc rectangular axes, Then

(33.5)

If the parallel displacement ofthc vector A, to the point P' is now carried out, it~

CartesIan components n, Will not change. i,c. (iB, = 0, Hence, from the tirst ofequations (33,5), we ohtain

(33.6)

Substituting for B) into this equation from the second of equations (33.5), we findthat

where

(lA, f':kA,dx'

1" _ i ,2y) j' x'IA; - ;\.';;x" pyJ

(33.7)

(33.8)

This shows that. in .-:1','1, the (ill, are bilinear fonns in the A,and dx', In.'l 1<' we shallaccordlllgly (lIji/l1' the oA, by the equation (33,7), determining the N" quantitiesr~k arhitrarily at every POint 01'.'1 N' This set of quantitlcs 1':. is called an affi/lity

... Suh.lcct to the rcqultcment that the r:~ art: l:\lhtlnuow, lunctlor\\ 01 the,' and PO\5>C,o;;S continuous

parllal dcthi.ltlvc'\ W the older nL'n'~"";lty to hthdate all laler ;11~lInH:IlI\,

(33.9)

98

and specifies an affine connection between the points of Y'N' A space which isaffinely connected possesses sutlicient structure to permit the operations of tensoranalysis to be carried out within it.

For we can now write

d - < - iJA'd 1 _ ['k d 1Ai vA i - 0 • X ijA k Xox)

((lA, k ) .

= '~j - fijA k dx l

(~x

But. as we have already explained. the left-hand member of this equation is avector for arbitrary dx1 and hence it follows that

(33.10)

(34.1)

is a covariant tensor, the covariant derivative of Ai'

It will be observed from equation (33.10) that, if the components of the atlinityall vanish over some region of .'I'N. the covariant and partial derivatives areidentical over this region. However, this will only be the case in the particularreference frame being employed. In any other frame the components of theaffinity will, in general, be non-zero and the distinction between the twoderivatives will be maintained. In tensor equations which are to be valid in everyframe. therefore, only covariant derivatives may appear, even if it is possible tofind a frame relative to which the affinity vanishes.

We have stated earlier that. when defining an affine connection, the com­ponents ofan aflinity may be chosen arbitrarily. To be precise, a coordinate framemust first be selected in .'I'N and the choice of the components of the affinity is thenarbitrary within this frame. However, when these have been determined, thecomponents of the affinity with respect to any other frame are, as for tensors,completely fixed by a transformation law. We now proceed to obtain thistransformation law for affinities.

34. Transformation of an affinity

The manner in which each of the quantities occurring in equation (33.10)transforms is known, with the exception of the aflinity n1. The transformationlaw for this affinity can accordingly be deduced by transformation of thisequation. Relative to the x-frame, the equation is written

- (lAiAi;] = 75ii r~lAk

Since Ai, A,;j are tensors,

(34.2)

~. ox' ox'A=----A

••J O.Xi oi] ,.t

Substituting in equation (34.1). we obtain

(34.3)

(34.4)

Employing equation (D.IO) to substitute for A and cancelling a pair of identicalterms from the two members of equation (34'.4), this equation reduces to

(34.5)

(34.6)

Since A, is an arbitrary vector, we can equate coefficients of A, from the twomembers of this equation to obtain

Dx' (Ix' Dx' iJ2 x'r~ ilik - (Sii ii:pr :, + ;j.iiiJ~j

Multiplying both sides of this equation by Dx'jiJx' and using the result

ilY{ ax'i"--x' ;iik

yields finally

(34.7)

(348)

(34.9)

which is the transformation law for nn affinity.It should be noted that, were it not for the presence of the second term in the

right-hnnd member of equation (34.R), I'~ would transform as a tensor of thethird rank having the covarinnt and contravariant characteristics suggestedby the positions of its indices. Thus, the transformation law is linear in thecomponents of an atJinity but is not homogeneous like a tensor transformationlaw. This has the consequence that, if all the components of an affinity are zerorelative to one frame, they are not necess;lrily zero relative to another frame.However, in general, there will be no frame in which the components ofan affinityvanish over a region of.'!'N, though it will be proved that, provided the affinity issymmetric, it is always possible to find a frame in which the components all vanishat some particular point (see section 39).

Suppose r~, r~* are two allinities defined over a region of .'I'N' Writing downtheir transformation laws and subtracting one from the other, it is immediate that

-k -k (l:e (lX' iJx'r-r*=---· ··(f'-P*)

I) rJ i)x" (lxi t:x.) Sf $1

i.e. the difference of two aflinities is a tensor. However, the sum of two allinities isneither a tensor nor an aflinity. It is left as an exercise for the reader to show,

100

similarly. thai the sum or an atlinity r~i and a tensor A~J is an aflinity.If nj is symmetric with respect to its subscripts in one frame, it is symmetric in

every frame. For, from equation (34.8),

r~ = (1xk

(lX' ~x' P + (lXk

(12X'

)' ()x" P.x) (lx' ,~t Px' trx.1iJ.x i

llxk (lX' (lX' (lXk (1 2 x'r;,+ ....itx' i).x' (l.x) (lX' (1;(';(;,\';

= r~j

where, at the first step, we have put r;, = r~,.

35. Covariant derivatives of tensors

(34.10)

In this section, we shall extend the process of covariant dilferentiation to tensorsof all ranks and types.

Consider first an invariant field V. When V suffers parallel displacement from Pto P', its value will be taken to be unaltered, i.e. ii V = 0 m all frames. Hence

is the counterpart for an invariant of equation (.BA). It follows that

V,= Vi

135.1)

(35.2)

i.e. the covariant derivative ofan invariant is identical with its partial derivative orgradient.

Now let Bi be a contravariant vector field and Ai an arbllrary covariant vector,Then Ai B' is an invariant and, when parallel displaced from P to P', remainsunchanged in value. Thus

or

o(A,Bi) = 0

oA i Hi + A,bB' = 0

and hence, by equation (33.7),

AkDBk = - r'~ Akdx j 11' (35.3)

But, since the Ak are arbitrary, their coeflieients 111 the two members of thisequation can be equated to yield

(35A)

This equation defines the parallel displacement of a contravariant vector. Thecovariant derivative of the vector is now deduced as before: thus

(35.5)

101

and since d x' is an arbitrary vector and dBk _liB' is therl known to be a vector,

I'Bk

B'= +rkB'I P,:1 I)

(35.6)

is a temor called the covarialll derivative of Bk.

Similarly, if A~ is a tensor field, we consider the parallel displacement of thtinvariant A~B,Ci. where B"C' arc arbItrary vectors. Then, from

/)(A~B,C') = ()

and equations (3.17) and (,Vi.4), we deduce Ihat

IIA~ ~ r,~A:dxk .. r,: A~ d x'

It now follows that

(35.7)

(35.HI

(35.9)

is the covariant derivative required.The rule for linding the covartant derivative of any tensor will now be plain

from examination ofequatiorl D5.9), ViI., the appropriate partial derivative is lirstwritten down and this is therl followed by 'aflinity terms; the 'aflinity terms arcobtained by writing down an lI1r1er product of the allilllty and the tensor withrespect to each of its indices in turn, prefixll1g a positive sign when the index i.contravariant and a negatIve sign when it is covariant.

Applying this rule to the tcnsor field whose components at every point arc thos1of the fun,lamerltal tensor II;. It will be found that

(35 IG)

Thus. the fundamental tensor behaves like a constant in covariant differentiatior"FInally. In tlw, section, we shall dcmonSlrate that the ordinary rules for th:

diflcrenllation of sums and products apply to the process of covariantdlll"crentiallon.

The rlght-han,1 member of equatwn (35'» bcing linear 111 the tensor A~, itfollows immedIately that If

then

Now suppose that

Then

Ci=A~j-B~

C;, = A;, + 13;,(35.11)

(,151 =)

(35.1 ~)

. I~Ci

C', = " k + r:kC'('.x-

,"(1

Xk (AiB')+ r:kAjB'

102

((lA;. ) . (llBj .).

= 2.x:,,+f;,Aj-f;,A; B'+ ~ii.x:' +f:,B' A;. . .

= A;,Bj + Bj,Aj (35.14)

which is the ordinary rule for the difTerentiatlOn of a product.

36. The Riemann Christoffel curvature tensor

[fa rectangular Cartesian coordinate frame is chosen in a Euclidean space $ Nandif A' are the components ofa vector defined at a point Q with respect to this frame,then 6A; = 0 for an arhitrary small parallel displacement of the vector from Q.This being true for arbitrary A', it follows from equation (35.4) Ihat fi, = 0 withrespect to this frame at every point of $ N' Suppose C is a closed curve passingthrough Q and that A' makes one complete circuit of C. being parallel displacedover each clement of Ihe palh. Then its components remain unchangedthroughout the motion and hence, if A' + i\A i denotes the vector upon its returnto Q,

(36.1)

Since i\A i is the difference between two \e..:tors both defined at Q, it is itself avector and equation (36.1) will therefore be a vector equation true in all frames.Thus, in t.'N. parallel displacement ofa vector through one circuit ofa closed curveleaves the vector unchanged.

If. however, Ai is defined at a point Q in an amnely connected Spilce .'I'N, notnecessarily Euclidean, it will no longer be possible, in general, to choose acoordinate frame for which the components of the aflinity vanish at every point.As a consequence, if A i is parallcl displaced around C. its components will varyand it is no longer permissible to suppose that upon its retum to Q it will beunchanged, i.e. 1\.4 itO. We shall now calculate i\A' when Ai is parallel displacedaround a small circlIlt C endosing the poinr P having coordinates Xi (Fig. 7) atwhich it is initially defined.

uv

p

c

FlU. 7

103

Let U be any point on this curve and let Xi + ~i be its coordinates, the ~i beingsmall quantities. V is a point on C ncar to U and having coordinates Xi + ~i + d~i.

When A' is displaced from U to V, its components undergo a change

(36,2)

(36.3)

where r;k and Aj are to be computed at lJ. Considering the small displacementfrom P to U and employing Taylor's theorem, the value of rik at U is seen to be

:lr i

r' + { ik "jk Px' ~

to the first order in the ~/. In this expression, the affinity and its derivative are to becomputed at P. Aj in equalion (36.2) represents the vector after its paralleldisplacement from P to U, i.e. it is

(36.4)

(365)

(366)

(36.10)

where A< A' and ['/, arc all to be cakulated at the point P. To the tirst order in ~'

therefore, equation (36.2) may be written

)A' = "-l r! AJ + (Aj ,0 [';k - ['/ r- j A')~/Jd ,.( Jk (l.'(/;k rl ~

Integrating around C. it will be found that

~A' ~ - ri AJ ld~k +- (r' f"' _ iWik )Aj l "d~kJk:r rk j/ (lx' :r~

( (

where the dummy indicesj, r have been interchanged in the final term of the right­hand member of equation (36.5).

Now fd~k = ~~k = 0 (36.7)

Also fd(('~k) = A(~'~k) = 0 (36.H)

c

so that f~ld~k = f'kdel (36.9)- ~ ~

( (

implying that the left-hand member of this equation is skew-symmetric in Iand k.Since ~', d~k arc vectors, it is also a tensor. Denoting it by ex kl , we have

Nkl _ ,l(dd'k 'kd")

~ - 2j ~ ~ -~ ~

\04

and equation (36.6) then reduces to the form

(36.11)

Apart from its property of skew-symmetry, ex" is arbitrary. Nonetheless, since itis not completely arbitrary, the quotient the(Hem (section 32) cannot be applieddirectly to deduce that the contents of the bracket in equation (36.11) constitute atensor. [n fact, this expression is not a tensor However, it is easy to prove that, ifX~, is skew-symmetric with respect to k, 1 ilnd if Y', defined by the equation

(36.12)

is a vector for arbitrary skew-symmetric tensors 'Xkl

, then X~, IS also a tensor.To prove this. let II" be an arbitrary symmetric tensor. Then the components of

the tensor

y" = ex kl + II"

are completely arbitrary, for, assuming k < I,

)," = ex kl + II", 1"k = - ex kl + Il kl

(36.1J)

(36.14)

and it follows that the values of l', y'k can be chosen arbitrarily and then

(3615)

i.e., it is only necessary to fix the value~; of ex".llkl in the cases" < 1in order that theykl shall assume any specified values oler the complelc range of its superscripts,with the exception ofthe cases when twn superscripts are equal. II' the superscriptsarc equal, ex kl = 0 and ykl = II". But thesl~ Il kl are also arbitrary lwd hence so againare the y" with equal superscripts.

Since {ikl is symmetric and X~, is skew-symmetric,

X~, {ikl = 0

Adding equations (36.12) and (36.16), we obtain therefore

(36.16)

(36.\7)

But ).kl is an arbitrary tensor and hence, by the quoticnt theorem, Xii is a tensor.The multiplier ofex kl in equation (36.11) is not skew-symmetric in k, I. However,

it can be made so as follows: Interchange the dummy indices k, 1in this equationto obtain

(36.18)

Adding equations (36.11) and (36.18) and noting that ex" = - ex'k , it will be foundthat

(36.19)

105

The bracketed expression is now ,kew-symmetric in k,l and hence

(?f' (~ri)r i [" _ r' r" + 1

1-- l' 4'

rk j/ rl;k (".xk ('lx' '

is a tensor. Al bcing arbitrary, it follows that

11"W,'I = 1",,1",1 - I" r' +( jl

... ... rl Jk t i '(k

is a tensor, [t is the Riemat/t/ Chrisroffel ('Im'(ltllre ret/"or,

E4uation (36,19) can now be written

~A'~, ~Hj.,,41():kI

(36.20)

(36.21 )

(36,22)

II' IJ;., is contracted With respect to the indices i and I. the resulting tensor iscalled the RiClI rell,\or and is delloled hy 1<,., Thus

Ri• = B;., (36,23)

This tensor has an important role to play in Einstein's theory of gravitation, SinceBju is skew-symmetric with respect I(l the indices k and I. its l'ontractioll with

respeel to i and k yields only the Ricci lensor agatn in the form - R "' However,eontmctlon with respecl In the indices i and i yields anOlher second rank lensor,\'U,

(36,24)

37, Metrieal connection, Raising and lowering indices

In this sectIOn we shall further particularize our space .'I'y by supposing it to beRiemannian, That IS, we shall suppose that a 'distance' or illrer('(ll ds between twoneighbouring points x', x' +d x' is defined by the e4ua tion

d,1 2oc !i'Jdx'dx' (37,1)

where the N 2 coeflicients!i.; are specified in some coordinate frame at every pointOfY'N' It will be assumed, without loss of generality, that the !iil are symmetric,Such a relationship between all pairs of adjacent points is called a merril'al

CO/lt/('crwn and the expression (37,1) for ds 2 is termed the nwrrj(',

F Of any two neighbouring points, d,1 will be regarded as an invariant associatedwith them and thc !i'l must accordingly transform so that this shall be so, Sincedxidxl is an arbitrary symmetric tensor, !i'l is symmctric and ds 2 is an invariant, itfollows by a modified 4uotient theorem ~imilar to the one proved in section 36that !iil is a tcnsor. It IS called the fllt/damenral cOI'(lI'iwl/ lenIOr, The contravarianttensor which is conjugate to !i'l (see section 32), viz. !i'l, is termed the!ulldamenral

fO/ltrlllJuriat/r remor, This exists only if !i = 1!i,jl1' 0, which we accordinglyassume to be the easc,

106

[n the case when !}iN is Euclidean, rectangular Cartesian coordinates yi can bedefined and, in such a frame, g,} = iii}' Consider a contravariant vector havingcomponents Ai in a general curvilinear x-frame and components Bi in the y­frame. [n the Cartesian frame, covariant and contravariant vectors are indis­tinguishable, so that it is natural to define covariant components for the vector bythe equation

(37.2)

[n the x-frame, let Ai be the components of the covariant vector B,. Then

(37,3)

This follows since (i) it is a tensor equation and (ii) it is valid in the y-frame inwhich it takes the form

(37.4)

[I' 9fN is not Euclidean, equation (37,3) is taken to define the covariantcomponents ofa vector whose contravariant components are Ai, This process ofconverting the contravariant components of a vector into covariant componentsis termed lowering the index.

[I' B, is a covariant vector, its contravariant expression is determined by raisingthe itidex with the aid of the fundamental contravariant vector. Thus

(.17.5)

For the notation to be consistent, it is necessary that if an index is first loweredand then raised, the original vector should again be obtained, This is seen to be thecase for, if Ai is formed from Ai (equation (37,3)), the result of raising its subscriptis (equation (37.5»)

(37,6)

where equations (32,9) have been used in the reduction. Similarly, if an index isfirst raised and then lowered, the original covariant vector is reproduced.

Any index ofa tensor can now be raised or lowered in the obvious way. Thus, ifAit is a tensor, we define

(37,7)

To allow for the possibility that indices may be raised or lowered during acalculation, it will be convenient to displace the subscripts to the right of thesuperscripts, It is also often helpful to keep a record of these operatior\s by placinga dot in the gap resulting from the raising or lowering of an index, Theseconventions are illustrated in equation (37,7).

Suppose an index of the fundamental tensor gil is raised. The result is

(37.8)

I.e. the mixed fundamental tensor. The same tensor results when an index of gil is

107

lowered. [I' both subscripts of Y,j are rai~ed, the result is

(37.9)

Our notation is elllirely consi~tent, therefore, and !J", !J,j, b~ are taken to be thecovariant, contravariant and mixr.:d components respectivr.:ly of a single rUllda­mental tr.:nsor,

Consider thr.: inner product of two vectors A', B,. We have

A'B, = 9';A;lJ,

= A;!J';R,

= AjBj

= A,B'

It is dear that the dummy index occurring in the expression for an inner productcan be raised in one factor and lowered In the other without affecting the result.This is obviously valid for the inner product of any pair of tensors.

38. Scalar products. Magnitudes of vectors

[n iiP N the magnitude of the displacement vector dx' is taken to be ds as given b~

equation (37.1). [f A' is any other contravariant vector, it may be represented as <I

displacement vector and then Its ma!Jllitude is thr.: invariant A, where

(3R.I

This equation is accordingly taken to define the magnitude of A'.Raising and lowering the dummy indices in equation (38.1), we obtain tht

equivalent result

(38.21

It is natural to assume that the associated vectors A" A' have equal magnitudeland hence A is also taken to he the magnitude of A,. Equation (3R.2) indicates ho...this can be calculated directly from A,.

Since !JdA I = A, and i/; A; = A', equations (3R.I), (3R.2) are also both seen to beequiVdlcnt to the equation

A 2 = A,A'

The scalar product of two vectors A, B is defined to be the invariant

It will be noted that

(301

(38.41

(38.51

By analogy with <5.\0 we now define the angle Ii between two vectors A, B to besuch that

ABcosli = A· B (38.6)

i.e.

108

A'BcosO = ~__ '_

J[(AJA}(B'B,)]

If 0 = -!rr, the vectors are said to be orrhoyorl(ll and

AiB, = ()

(38.7)

(388)

39. Geodesic frame. Christoffel symbols

It is always possible to choose a coordinate frame in which the components of themetric tensor are stationary at an assigned point, i.e. (lad(lX' all vanish at thepoint. Such a frame is said to be yeoJesic at the point.

For, if Xi = u' are the coordinates of a point A ill an x-fnlme, suppose wetransform to a new frame by the quadratic transformation

(39.1)

where x' = 0 at A and the constant coefficients a~, a re symmetric in the indicesi. "(no loss of generality). Differentiation yields the equations

(':2 X i

/'ljjPxk :-:7 U;k (39.2)

In particular, at the point A

(lX i .

1 j = 0;,(.x

(12 X i

p~j~)Xk = ajk (39.3)

The transformation equation for the metric tensor is

(lX' (JX'gij = (l.xi (lXJf.I,.s

Differentiating with respeet to X', we find that

(l!l'J (J2 x' iJx' iJx' (12 x' (lX' iJx' fly" flx'flX' = (7X'(lxi iJXJ r/-., + (10 flx')]X) !I" + (lX' ,7XJ iJx' (li'

Substituting from equations (39.3), it follows that at the point A,

(lY;j, , (lY;jiJ'x' = a"y,j + aj-,y;, t (1;;'

(39.4)

(39.5)

(39.6)

Suppose, if possible, the coeflicients a~, are chosen so that flgd(lX' = O. Then,writing a;,o,j = a"j, it is necessary that

(397)

Cyclically permuting the indices i, j, k, two further equations are obtained, viz.

aj,,+akij= -(JYj";flx i

a'ji + U;j' = - (ly,./i1x j

(39.8)

(39.9)

109

Since a,jk is symmetric in the indices i,j, hy adding the last pair of equations andsuhtracting equation (39.7), we get

139.10)

[ii, kJ is called the Chri.ltoffel s \'mho/ of the first k i/l(/. It is now easily verifitd thatthe conditIOn (39.7) IS satisfied provided fI'Jk is defined hy equation (39.1 1)) and,hence, that ?y,/(lXk = () at A.

[ii. kJ is not a ten.sor, hut its indices invariahly hehave like subseriptsin anyformula in which it occurs. It is symmetric in the indices i, i.

We now deduce that

where {, k j} is Christoffel's s \'mho/"lt!le second kind. It is clearly obtained fr'lm thesymbol of the first kind [ii, kl hy raising the final Index k. It, also, is not a tensor,but i,j always behave as subscripts and k behaves as a superscript; it is symmetricin i and j.

Suppose we next transform ff()m the .x-frame to a y-frame hy a lineartransformation

[39.12)

where the h; are constants. Then d \' = hj dyj and it is a well known resu t fromalgebra that the quadratic form iJ,jd.x'd.x l can he reduced to the diagond form

(dyl)' + (dy')' + .. , + (dyN)' (39.13)

at the point A by proper choice of the h; (some of Ihese coeflicients may hale to hegiven imaginary values). Let h'l he the metric tensor in the v-frame. Thm

(39.14)

(39.15)

and, in particular, h;j = (l'l al A. Differentiating the last equation, we obtain

'h ,--( ;j = h' h'· h~ ( {/"(lyk I J (lX'

showing that (lh,/"'l vanishes at A. Thus, in the y-frame, h'j is stationa:y withvalue i5,j at A. The implication is that, in a small neighbourhood of A, thecoordinates y' will behave like rectangular Cartesian coordinates, The y-frame isthe closest approximation to a rectangular Cartesian frame that can be flled tothe YP N in the neighbourhood of A,

If the y-frame were exactly rectangular Cartesian (i,e, the space wereEuclidean), in order that the law for parallel displacement of a vector, equation(33,7), should agree with the one usually adopted, it would be necessary lO:ake allthe components of the aflinity to be zero in this frame. It is natural, thcrcl'ore, todefine the aflinity at the point A of ,if N so that, in the v-frame, its eomponwts arc

110

zero, With this choice of aflinity, in the y-frame covanant derivatives will reduceto ordinary partial derivatives at the point A: in particular,

(31),16)

at A. But this equation is a tensor equation and, being valid in one frame, isaccordingly valid in all frames. Thus, with this choice of afllnity

!I",. =0 (39,17)

Assuming that the affinity is defined at all points of :1P N in this manner, the lastequation will be valid throughout the space and in all frames. Since this aflinity isclearly symmetric in the y-frame, it will he symmetric in all frames.

Writing equation (39.17) out at length, we have

(W.1K)

Cyclically permuting the indices i, j, k to ohtain two further equations. it nowfollows as from equations (39.7} (39.9) that

(39.19)

Raising the index k, this gives

(39,20)

The affinity determined by this equation in any frame will be called the metricaffinity and will invariably be assumed in all later developments,

Since equation (39,17) is valid using the metric affinity, the mctri(; tensorbehaves like a constant with respect to covariant differentiation. Further. since

(39,21)

by taking the covariant derivative of both members of this equation, we 'lbtain

!I;im!lkj c= 0

Multiplying by {/h and summing with respect to k, we then find that

(39,22)

(39.23)

Thus, all forms of the metric tensor behave like constants under covariantdifferentiation.

It is now clear that

(Yu Al ).• = !I'jA' k (39.24)

This shows that the lowering of an index followed by a covariant diflcrentiationyields the same result as when these two processes are reversed. Similarly, it can beshown that the raising of an index and covariant differentiation are two processeswhich commute.

III

40. Bianchi identity

If we choose a frame which is geodesic at a point x', j)g'Jj (lXk will vanish at the

point and the two Christoffel symhols will therefore also vanish at the point.Thus, all the components of the aflinlly will vanish at the point and covariantderivatives will reduce to partIal derivatives there. It then follows from equation(39.23) that the partIal derivatives (lgiJj(lX k all vanish at the point, also.

In such a geodesIc frame, therefore.

_ (121"~1 (12 r~k

- i'lx m px1< - tlxm(l,\,J(40.1)

since the r~k (hut not their derivatives necessarily) all vanish at this point.Cyclically permuting the indices /.;,/. mill equation (40.1). we obtain

(12 r~,

{lx"t'lx m

{'l2 r~m

(lX' (lX k

(40.2)

(40.3)

Addition of equations (40.1). (40.2). (40.3). yields the following identity

(40.4)

But this is a tensor equation and, having been proved true in the geodesic frame,must be true in all frames. Also, since the chosen point can be any point of.'JP N' it isvalid at all points of the space. It is the Bianchi identity.

41. The covariant curvature tensor

The components of B';k/ are not all independent since the tensor is skew­symmetric in the indices k,/. In addition, however, if the affinity is symmetric, it iseasily verified from equation (36.21) that

(41.1 )

If the affinity is metrical, by lowering the contravariant index of theRiemann- Christoffel tensor, a completely covariant curvature tensor Bijk , isderived. This has a number ofsymmetry properties, one of which is obtained fromour last equation immediately by lowering the index i throughout to give

(41.2)

Further such properties can be established by first calculating an expression forthe tensor in a geodesic frame. Thus, in such a frame

112

(41.3)

(41.4)

since (\/'/?x k = 0 in a geodesic frame. Using the result y,,!!" = (I; and substitut­ing for the ChristoflCI symbols, we now find from equation (41.3) that

. I ( (12 gil (12 gJk (1 1. f/;k (ll.gJI)l1/ ikl = 2 ii.~i(lXk + ()x/(l X l .-. (lXi(lX' -. (lX'(lX k

The following equations are now easily verified:

BI .ikl = --B/ lkl

B iikl = ~B'ilk

Bilkl = Bk11.i

(41.5)

(41.6)

(41.7)

Being true in the geodesic frame, these tensor equations must be valid in allframes. N()le that the tensor is skew-symmdric with n:spccl 10 its firsl pair ofindices and its last pair.

Also lowering the superscript i throughout tbe Bianchi identity (40.4), weobtain

(4I.X)

42. Divergence. The Laplacian. Einstein's tensor

If the covariant derivative of a tensor lield is found and then contracted withrespect to the index of differentiation and any superscript, the result is called adit'('rgenCl' of the tensor. With respect to orthogonal coordinate transformationsin f,,,, the partial and co variant derivatives are identical and then this definition ofdivergem:e agrees with that given in section 12.

From the tensor AkJ, two divergences can be formed. viz.

(42.1)

A contra variant vector possesses one divergence only, which is an in variant. 1ft heaflinity is the metrical one, such a divergence is simply expressed in terms ofordinary partial derivatives thus: since a deriva tive ofa determinant can be foundby diflerentiating each row separately and snmming the results, we deduce that

(42.2)

Since

113

(42.3)

equation (42.2) reduces to

,1 11 " [... , J l' .J) ~ I , I1.J = 'HI ( '/,"; + ,/,1 "= '-0"", .\

(42.4)

Helice I , I I ,1 J"JI = J ' ,( 0)

II l'X(42.5)

(426)

Now let .4; he a vector field. Its divergence is

I I" ,111' ,1 'JJ

JII 1 ,+.4 Jl .,(JO)

II (.X 'X

I ,1

J1 ,( JO.4')

ill \

which is the expre.'>sion required.In partin"ar, if the vector field is ohtained from an invariant V hy taking its

gradicnt, WC have

and hcnce

,01'11. "';;: (1:(1

," V.4' = II')

, t'\" I

(42.7)

(42.K)

From equation (42.6), it now follows that the divergence of this vector is

(42.9)

The right-hand member of this equation represents the form taken by theLaplacian of Vin a general Riemannian space. In I,' N' employing rectangular axes,'I'i = ,);i, (/ = I and thus

(42.10)

which is its famIliar form.We shall now calculate the divergence of the Ricci tensor Rj , (equation (36.23)).If the metric aflinity is heing employed, this tcnsor is symmetric, for

Rk} ~~ lJ~JI -= y.r RrkJ1 -= yl' BWk = yl' Bl)kr

c= B;" =c R" (42.11 )

having employed equations (41.5)--(41.7). Raising either index accordingly yieldsthe same mixed tensor Ri. If this is contracted, an invariant

R = Rj (42.12)

is obtained. R is called the ClJr!Jature s('{dar of .9P N •

The skew-symmetry of Bilkl with respect to its first two and its last two indicesmeans that equation (41.8) can be writen

(42.13)

Multiplying through by gOgjk, we then get

and this is equivalent to

(42.14)

or

Thus

9 lk R)k;m - 9 jkR)m.k - gil R,m.1 = 0

Rm-2R~;k = 0

R~.k = trJR/Dx'"

(42.15)

(42.16)

(42.17)

is the divergence of the Ricci tensor.Consider now the mixed tensor

Its di vergence is

(42.18)

=0 (42.19)

This is Einstein's Tens(lr. Its covariant and contravariant components are

(42.20)

(43.1)

respecti vely.

43. Geodesics

Let C be any curve constructed in a space :JfN having metric (37.1) and let s be aparameter defined on C such that, if s, s + ds are its values at the respectiveneighbouring points P, P' on C, then ds is the interval between these two points. IfXi are the coordinates of any point P on C, then the curve will be defined byparametric equations

Since dx i are the components of a vector and ds is an invariant, dxi/ds is acontravariant vector at P. Its magnitude is, by equation (38.1),

(g .dxi d_~:)112. '1 ds ds (43.2)

115

and this is unity by equation (37.1). dx'jds is termed the unit tangell/to the curveat P, its direction being that of the displacement dx' along the curve from P.

Suppose C possesses the property that the tangents at all its points are parallel,I.e. the curve's direction is constant over its whole length. This property is clearlyquite independent of the coordinate frame being employed. In $ 3, such a curvewould, of course, be a straight line. In :#,\, the curve will be called a qeoJes;c. Ageodesic is accordingly thc counterpart of the Euclidean straight line in aRiemannian space. Suppose P, p' are ncighhollring points on a geodcsic havingcoordinates x', x' -+- d \' resrecti vely. II' the ul1ltlangent ill P is parallel displaced toP', it will then be idcntieal with Ihe actual unit tangent at this point. Now, bycquation (35.4), after parallel displacement from P to p', the unit tangent hascomponents

dxi (dX') dx'. +c1 =

ds ds ds(433)

But the actual unit tangent for the point P' has components

The vectors (43.3) and (43.4) are identical provided

(43.4)

(435)

If these equations are satisfied at every point of the curve (43.1), it is a geodesic.The N equal ions (43.5) are sL'Cond-order differential equations for the

functions xi(s) and their solution will involve 2N arhitrary constants. If A, B arctwo given points having coordinates x' = a', x' = h' respectively, the 2Nconditions thaI the geodesic must contain these points will, in general, determinethe arbitrary constants. Hence there is, in general, a unique geodesic connectingevery pair of points. However, in somr cases, this will not be so. For example,thegeodesics ()n the surface ofa sphere ( .if 2) are great circles and, in general, there aretwo great circle arcs joining two given points, a major arc and a minor arc. Also, ifthese points are diamterically opposed to one another,there is an infinity of great­circle arcs connecting them.

Since dx'jd.\ is everywhere a ullit vector, on a geodesic

dx'dx J

g . - I. ,j ds ds - (43.6)

This must, accordingly, bea first integral of the equations (43.5). To show that thisis the casr, multiply equations (43.5) through by 2g"dx'jds and sum with respectto i 10 obtain

(43.7)

116

Now

Also

dx'd 2 xi d ( dxidX') dOi,dxidx'2i/i'd:~ ds i = ds lJi' ds ds - ds ds <Is

d jdkd' djdkd''i x X X [ '. ] x X X21Ji' r· k ·· - - = 2 }k, r ..} ds ds ds ds d., ds

dx j dx k dx'=(Uk,r]+[rk,ij)ds ds d':

(43.8)

(43.9)

("/i' dx k dx) dx'

('Xk <Is ds ds

dy), d x} dx'

d., ds ds

By addition of equations (43.H)and (43.9), it will be seen that equation (43.7) canbe expressed in the form

d(1J dxidXj)ooOds ." ds ds /

Upon integration, there results the first integral

dx'dx'lJij ds ds = comtant

(43.10)

(43.11 )

The constant of integration must, of course, he taken to be unity.The definition of a geodesic which has been given at the beginning of this

section cannot be applied to the class of curves for which the interval ds betweenadjacent points vanishes. For such a curve, the parametric representat ion (43.1) isnot appropriate and a unit tangent cannot be defined. Instead, suppose that a(1 I )c(lrrespondence is set up between the points of the curve and the val lies ofaninvariant) in some interval )n ~ Ie ~ ic I, so that parametric equations I'm thecurve can be writtcn

Xi = x i (';') (4312)

It will be assumed that the derivatives dxijd). all exist at each point of the curve.These derivative., constitute a contravariant vector and this has I.ero magnumfefor, since d.\ ~ () along the eurw,

d x' d.x j

Oi} dA dA() (43.13)

This vector will be in the direction of the displacement vector along the curve dx i

and WIll be called a zero {ani/nil to the curve. The curve will be termed a n,,11

r/('ot!esic if the zero tangents at all points of the curve are parallel. This impliesthat, when the zero tangent at P is parallel displaced lO the adjacent point P', itmust be parallellO the .r.ero tangent at this latler point, and since the magnitudesof these two vectors at P' are the same, they will be taken to be idelltical. The

condition for this to be so is found, as before, to be

d 2 x; dxid\'1" - 0

d A2 + j' dAd;' -

117

(43.14)

These are, therefore, the equations of the null geodesics. It may now be shown, byan argument simdar to tha t culminating in equation (43.11), tlla t a first integral ofthese equations IS

(43.15)

(43.16)

In this case the constant must be zero.Equation (43.5) may he put in an alternative form which is more convenient for

particular calculations, as follows: Multiply through hy 29" and sum with respectto;; the resulting equation is equivalent to

d(7 dX;) dlf"dx' .dx)dx'~lf" - 2 + 29" r;, d I. 0'= 0

ds ds d., ds ., (.\

Now

(43.17)

and

(43.IK)

Equation (43.16) accordingly reduces to

(43.19)

Eqnation (43.14) for a null geodesic may he expressed similarly.

Exercises 5

I. A ij is a covariant tensor. If B,; = A j" prove that R" is a covariant tensor.Deduce lhat, if A,) is symmetric (l1f skcw.symmetric) in one frame, it is symmetric(or skew-symmetric) in all. (/lim: The equations A,) = A j" A,) = - A j; are tensorequations.)

2. (x, y, z)are rectangular Cartesian c()ordinates ofa point Pin<l".1and (r, Ii, 4))

life the corresponding spherical polars related to the Cartesians hy equations(30.3). A is a contravariant vector defined at P having components (A" A V

, A') in

118

the Cartesian frame and components (A', AO, A"') in the spheric1l1 polar frame,Express the polar components in terms of the Cartesian components. 0 1,02,03are rectangular Cartesian axes such that P lies on Oland OJ lies in the plane Oxy.If (A " A 2

, ;1") are the components of A in this Cartesian frame, show that

A' = A', A 2 = rAil, A J = r sinO A4>

(Noll': Assume the Carlesian axes are right-handed.)3. If A, is a covanant vector, verify that R" = A,,} - A}., transforms like a

covariant tensor. (This is curl A.) [I' A is the gradient ofa scalar, verify that itscurlvanishes.

4. If A,} is a skew-symmetric covariant tensor, verify that

transforms as a tensor.5. Assuming the transfl)/mation inverse to (31.1) exists, prove that each

determinant l(lx'll"x'l, l(lx'111.VI, is the reciprocal of the other. A~ is a mixedtensor with respe<:t to this transformarion. Show that the determinant IAil is aninvariant.

6. If r~J is an amnity. show that the torsion defined hy

nl = ~(r~j - r~,)

is a tensor. !I" is a symmetric tensor. Write down an expression flH its covariantderivative II".,' By considering this equntil1n and two similar equations ohtainedby ,-,yelic permutation of the indices i,j, k, show that if the covariant derivative ofY'l is to vanish identically, the aHinity must he given hy

7. Show that

A,., - A,., = A',j - Aj "

provided the afTini ty is symmetric.8. Show that

A." -- A .." = B'l' A, +(rk) -1;,),.1.,

and deduce that B,}, is a tensor and that covariant dillcrentiations arecommutative in a space for which Bi" = (} and the allinity is symmetric. Obtainthe corresponding result for a contravariant vector A'

9. A, is defined at the point x' and is parallel displaced around a small contourenclosing the point. Prove that the increment in A, resulting from one circuit isgiven by

~Ai = -~B~j,Alexl'

where ex}' is defined by equa tion (36.10).I (}. The parametric equations of a curve in .'/'N are

Xi = x'(r)

119

1 is an invariant parameter. A tensor A j is defined over a region containing thecurve. P, P' are neighhouring points I, 1 + /1.1 on the curve and /I. A j is defined to bethe difference between the actual value of the tensor at P' and the value of thetensor at P after it has been parallel displaced to P'. Prove that

DAj . /l.Aj I dx k

= lim = A I01 A, • ., /1.( /. dl

(DA i/D1 is called the inlrinsit' Jeri(ll/live of the tensor along the curve.)II. Verify that { j kI transforms as an atJinity.12. [I' A,} is symmetric, prove that Ai} k is symmetric in i and}.13. Show that the number of the components of B~k1 which may be assigned

values arbitrarily is, in general, 1N '(N - I). [I' the affinity is symmetric, show thatthis number is ~ N 2(N 2 - I). (Hint. Usc equation (41.1 ).)

14. Show that the number of the compollcnts of B'jk/ which may be assignedvalues arhitrarily is N 2 (N 2 -1)/12. (flilll: Usc equations (41.2), (41.5), (41.6),(41.7).)

15. By differentiating the equation

with re~rect to Xl, show that

and hence that

Deduce that 9 'i = O.

16. If the affinity is the metric one, prove that

. (7. (J2 . rJR jk = Bjkl = - 1 i { /k} + ;;-TJ k log J 9 + { ,'d (/d - [/k) ." ,log J (/

ex (IX (X (·X

(Hint: Employ equation (42.5).) Deduce that R jk is symmetric.17. [I' 0, tj> are eo-Iat itude and longitude respectively on the surface of a sphere

of unit radius, ohtain the metric

d.l"2 = d0 2 + sin 2 /idtj>2

for the surface. Show that tire only non-vanishing three index symbols for this Yf 2

are

Show also that the only non-vanishing components of B iikl are

B 1212 = -B 1221 = B2121 = -B2112 = sin 2 0

(r > (1)

120

and that t he components of the Ricci tensor are given by

R 12 =R 2 ,=0, R11=-I, R 22 =-sin 2 1i.

Prove that the curvature scalar is given by R = - 2.18. Employing equation (42.9), obtain expressions for V 2 V in cylindrical and

spherical polars.19. [n a certain coordinate system

where 4>, r/J are functions of position. Prove that Btu is a function of r/J only. [I'r/J = -log (o,x') prove that

RJk = B~k'= 0

20. [n the :Il 2 whose metric is

dr 2 + r2dli 2 r2 dr 2d,·2 = __ _

. r2 _ (12 - (--':2-~ ;,2 )2

prove that the differential equation of thc gcodesics may be written

where k 2 is a constant such that k 2 = I if, and only if, the geodesic is null. Byputting r dli/dr = tan 4>, show that if the space is mapped on a Euclidcan planc inwhich r, Ii <lre taken as polar coordinates, the geodesics are mapped as straightlines, the null geodesics being tangents to the circle r = a.

21. A 2-space has metric

ds 2 = gl,(dx'f +g22(dx 2)2

where 111" g22 are functions of x I and x 2B,jkl is its covariant curvature tensor

and R,j is its Ricci tensor. Prove that

R l2 = 0, R"g22 = R 22 g" = B'22'

[fR = 9'iRij' show that R = 2B'22'/(9, ,922)' Deduce that R'J = 1R9".

22. Prove that

(i)

(ii)

.. I iJ ).. 'k'A'J = .( gA'J)+A' J J 1.' )g(lX" l,k!

X'i =0.'J

provided Xij is skew-symmetric. Hence prove that, for any tensor A,j

121

23. A l:l!fve C has parametric l:quatlOlls

x' .= x/(r)

and' joins two points A and B. The length of the curvc is defincd to he

f" f" J( dx

i dX))L = ds = q,) dr dr dr

A A

Write down the Euler conditions that }, should be stationary with respect to allsmall variations from C and by changing the independent variahle in theseconditions from r to .\, show that they arc idcntil:al with equatiofts (43.19). (Thisprovides an alternative definition for a geodesic.)

24. If r;, is a symmetrll: aflinity, show that

ri~ = ri, + ,ljA, + ,lkA I

is also a symmetric aflinity.[f Bi", Hi:, arc the Riemann Christoffel curvature tensors relative to the

aflinilies r;" ri~ respectively, prove that

Bi~, = Bi" .~ ,I kA j/ -- ,IiA i' + ,Ii( A'I -- A,,)

whefe A,) = A,Aj-A,.,.Hence show that if A, is the gradicnt of a scalar, thcll

25. Provc that the aflinity transformations form a group.26. Prove that

27. Two metrics arc defined in .if,. viz.

ds 2 = q'Jdx'dxJ, d.\2 = e"q'jdx'dx j

where (J is a function of the x'. [f ri" ri, arc metric aflinities constructed fromthese metrics, prove that

where

Curvature tensors Bi'" Bi'/ arc constructed from these aflinities. Prove that

B;,/ = Bj,/ + Aii, - Aikl + A;, A j/ - A;/ A j,

[22

Deduce that

R~, = RI' + A;J.' - A;" + A~.A;j - A;,Aj,

and show also that A:j = !NCf,j'28. Oblique Cartesian axes are taken in a plane. Show that the contra variant

components of a vector A can be obtained by projecting a certain displacementvector on to the axes by parallels to the axes and the covariant compollents byprojecting by perpendiculars to the axes.

29. Define coordinates (r, tj» on a right circular cone having semi-vert ical angleex so that the metric for the ~urface is

ds 2 = dr 2 +,.2 sin 2 r1. dq,2.

Show that the family of geodesics is given by

r = usec(,psillex -{i)

where u, {i arc arbitrary constants. Explain this result by developing the cone intoa plane.

30. An :iP N has metric

where ,,\ is a function of the Xi. Show that Ihe only non-vanishing Chrbtoflclsymbols of the second kind arc

where "\, = l7,,\/l7x'. Deduce that

{ripH;i} = !(N + 2)"\;' - P,"\,

and that the scalar curvature of this space is given by

R = (N -IV A["\,, + i(N - 2)"\,"\,J

where "\" = 11 2,,\/<"x' ,lx'.31. II is the co-latitude and tj> is the longitude on a unit sphere, so that the metric

for the surface is

The covariant vector Ai is taken with initial components (X, Y) and is carried, byparallel displacement, along an arc of length tj> sin ex of the circle 0 :, ex, Show thatthe components of Ai altain the final value~

A I = X cos (tj>eosex)+ Y cosec ex sin (tj>cos ex)

A 2 = --X sin ex sin (tj>cosex)+Ycos(tj>cosex)

123

Verify that the magnitude of the vector A, is unaltered by the displacement.320 An "if J has metric

ds 2 = icdr 2 +r 2(d0 2 +sin 2 0d(p2)

where A is a function ofralone" Show that, along the geodesic for which 0 = ~I,

dO/ds = 0 at s = 0,

where r = h sec r/J" Interpret this result geometrically when ic = I.33. i (i = 1,2,3,4) arc rectangular Cartesian coordimlles in <f 4 , Show that

J' I ~, R cos 0

y2 = R sinOcos¢

yJ ,= R sin Ii sin 1/ cos r/J

y 4 = R sin Ii sin tjI si n r/J

are parametric equations ofa hypersphere of radius R" If (0, (P, r/J) are taken a,coordinates 011 the hypcrsphere, show that the metric for this ,if] is

ds 2 = R 2[dIi 2 +sin 21i(dtjl2 +-sin 2 (pdr/J2)]

Deduce that in this ,if),

13 1212 = R 2 sin 21i, BBV = R 2 sin 4 lisin 2 tjI,

IJ]I.Jl = R' sin' Osin 2 tjI,

all other distinct components being zero" Hence show that

B'i" = K (y,,!!), -iJill/i')

where K = I;R 2 (This IS the condition for the space to be of constalllRiemannian l:urvature K")

34. An :#2 has metric

d"\2 = sech 2 r(dx 2 +d y 2)

Find the equation of the family of geodesics"Ii, tjI are co-tati tude and longitude respectively on the surface of a sphere of unit

radius, Mercator's projection is obtamed by plotting x, j' as rectangular Cartesiancoordinates in a plane, taking

x = (P, y = tog cOt !Ii

Calculate the metric for the spherical surface in terms of x and y and deduce thatthe great l:ircles are represented by the curves

sinh y = ex sin (x + {i),

124

-g"1 /k H/,}36. An .!II, has metric ds' = 2</Jdxdy, where ¢ = (p(x, y). Calculate the

component R"" of the curvature tensor and state the values of the remaining 15componcnts. Deduce that the space is nat provided

("'lji {1¢ {1¢tj! =

{lX{ly {lX {1 Y

Putting ¢ = e", obtain the general form for c/> satisfying this condition. Deducethat coordinates ( '1 can be found such that the metric takes the formd.s' = 2d~ d'1.

37. [I' A, is such that A..J + A" = 0, by cyclically permuting the indices i,j, kinA.. ,k - Au, = A,B;Jk to give two further equations, prove thar A,.'k = .- A,B kij .

3X. c/i( /I, I") and l~ (u, t') arc the real and imaginary parts of an analytic functionf (w) of the complex variable w = 1/ + il'. Show that the equations x = tj!(u, 0),.\' = t/J (II, I') transform the Pythagoreun metric d.s' = dx 2 + dy' int0 the metric

By takingf(w) = l/w,explain how it is possible to write down the equati0n of thefamily of geodesics in a space whose metric is

dr,J .\ dr"ds 2

=. (1/1 1-1/)1

Also obtain this equation by transforming the metric uSing the equationsu = I' cos 0, V = I' sin (I, and writing down the difrerential equations for thegeodesics in terms of I' and O.

39. (x, y )are rectangular Cartesian coon.linates and (I', II )are polar coordinatesin a Euclidean plane. A ij is a symmetric tensor field defined in the plane by itscomponents A xx = A yy = 0, A xy = A yx '" x/.r+ y/x. Calculate the contravariantpolar components of the tleld in terms of I' and (I,and deduce that A" + 1'1 A"" = O.(Ans. A" = 2, A'" = 2cot20/r, A"" = -2/1".)

40. X, y, z are rectangular Cartesian coordinates in tIJ .j. Parametric equationsfor a hyperbolic paraboloid are taken in the form x = u + v, y = u - v, z = uv, Acovariant tensor field on the surface is defined by the equations A•• = 11

2, A."

= A,.• = - Ill" A,,,. = v2 Show that the contravariant components arc one­quarter the covariant components.

125

41. x, yare rectdngular Cartesian coordinates in a Euclidean plane and u, v arecurvilinear coordinates detined by x = II cosh II cos r, y = II sinh II sin t'. A ccvari­ant vector has components A" Ayat the point (x, .1') and curvilinear compo~entsA", A,.. Show that

2 .A,.o (A" smh /1 cos /... it, C\lsh /1 sin ,.)/ (cosh 211 - cos 21')

"42. x, J' are rectangular Cartesian coordiantes in a plane. Curvilinear

coordinates 1/, I' are defined by the transformation equations I< = 1(.x 2 _ .1'2),

I' =\\'. Sketch the families of coordinate lines I< = cons!., t' = eonst. alld showthat the metric in the fir-frame is

A covariant vector has Cartesian components (Ax, A y ) and curvilinear com­ponents (A", A,.). Show that

A" = (xA, - yA y )/(x 2 + 1'2)

and derive the corresponding formula for A".

43. (x, r) are rectangular Cartesian coordinates in a Euclidean plane and (I<, I')

arc curvilinear cuordinates defined hy the equations x = ; (u 2 + 1,2), Y = HI'. Acovariant vector field A, is defined over the plane in the uv-frame by the equationsA" = A" = (11 2 - ,,2)2 Show that its divergence is equal to 2(u -1,)2/(1<+ v).Cakulate the contravariant vector at the point I< = 0, V = I and use the law ofparallel displacement along the curve u = () to calculate the parallel dispacedvectOf at the point I< = 0, I' = 2. (Ans. A" = A" = ;.)

44. An ,if 2 has metric ds 2 = d\2 + x 2d y 2 Calculate the components of itsmetric aflinity. Deduce that the divergence of the vector field whose covariantcomponents are given by A x = x cos 2.1', A y = - x 2sin 2y, vanishes.

45. If the x-frame is geodesic at the point x', prove that

and deduce that R;, = 1R.,.46. An ,if], has metrIC d.1 2 = y 2d\ 2+ d y 2 Deduce the parallel tra~sfer

equationsI

Axdy .. yA"dx, ,\/1,.I'

126

Using these equations, parallel transfer the vector along the curve y = sec x fromthe point x = O. y = I at which its components are A, = O. Ay = I, to the point x= rr/3. y = 2.

47. If Ii, 4> are latitude and longitude respectively on the surface of a globe ofunit radius, show that the geodesics on the globe have equations tan 0= tan ex sin (4) + Ii), where ex, {J are constants.

48. A space (if 2 has metric ds 2 = sech 1 y (dx 2 + dl' 2). A vector A; is paralleldisplaced from the point x = 0, y = h to the point x = u, y = h along the liney = h. Its initial components are (X, Yj. Show that its final components are givenby

AI = Xcos(utanhh)+Ysin(a tanhh)

A 2 = -Xsin(utanhh)+Ycos(utanhh)

49. Replacing A; in the argument of section 34 by A;, obtain the transform,ation law for an affinity in the form

i'l2 Xi i'lx" (1x.''*

(lX r nX s i'l.x~j i)iik

Prove that this is equivalent to equation (34.R). (Hillt: Differentiate(IX i (1.'(" •.

- = ,)'.)nx r i'lj) J"

50. Using the transformation law for an affinity in the form given in the lastexercise. if r; = yik r'~k show that an .x,frame call always be found sllch thatF' = 0 everywhere. Show that this frame is deteflllllled by the equations

and that these equations can be written V2 x' = o. (The coordinates x; are said tobe hurl/(()/Iir.)

CHAPTER 6

General Theory of Relativity

44. Principle of equiva[enee

The special theory of relativity rejects the Newtonian concept of a privilegedobserver, at rest in absolute space. and for whom physical laws assume theirsimplest form. and assumes instead that these laws will be identical for aUmembers of a class of inertial observers in uniform translatory motion relative tl)one another. Thus. although the existence ofa silly Ie privileged observer is deniedthe existence of a clas,\ of such observers is accepted, This seems to imply that, ii'all matter in the universe were annihilated except for a single experimenter and hi;laboratory, thIS observer would.nonctheless, beable to distinguish inertial frame;from non-inertial frames by the special simplicity which thc descriptions ofphysical phenomena take with respect to the former. The further implication is,therefore. that physiwl space IS not simply a mathematical abstraction which it i~

convenient to employ when considering distal]l"e relationships between mlilerialbodies, but exists in its own right as a separate entity with suflicient internalstructure to permit the definition of inertial frames. However, all the availableevidence suggests that physical space cannot be defined except in terms ofdistance measurements between physical bodies, For example, such a space can beconstructed by setting up a rectangular Cartesian coordinate frame comprisingthree mutually perpendicular rigid rods and then defining the coordinates of thepoint occupied by a material particle by distance measurements from these rods inthe usual way, Physical space is. then, nothing more than the aggregate of allpossible coordinate frames, A claim that physical space exists independently ofdistance measurements between matcrial bodies. can only be substantiated if aprecise statement is given of the manner in which its existence can be detectedwithout carrying out such measurements, This has never been done and we shallassume, therefore, that the special properties possessed by inertial frames must berelated in some way to the distribution of rna tter within the universe and that theyarc not an indication ofan inherent structure possessed by physical space when itis considered apart from the matter it contains, This line of argument encouragesus to expect, therefore, that, ultimately, all physical laws will be expressible informs which are quite independent of any coordinate frame by which physicalspace is defined, i,e. that physicallaw<; arc identkal fOI' all ohscrvers, This is the

[27

[28

gelleral principle o!relaril'iry, This does not mean that. when account is taken ofthe actual distribution of matter within the universe. certain frames will not proveto be more convenient than ()thers. When caleulating the field due to adistribution of electric charge. it simplifies the calculations enormously if areference frame can be employed relative to which the charge is wholly at rest.However, this does not mean that the laws of electromagnetism are expressiblemore simply in this frame. but only that this particular charge distribution is thendescribed more simply. Similarly. we shall attribute the simpler forms taken bysome calculatIons when carriell out in inertial frames. to the special relationshipthese fwmes hear to the mailer present in thc universe. Fundamentally. thcrefore.all observers WIll be regarded as equivalent and. by employing the same physicallaws. will arrive at identical conclusions con.:erning the development of anyphysical system.

The main difliculty which arises when we try to express physical laws so that theyare valid for all observers is that, if test particles are released and their motionsstudied from a frame which is being accelerated with respect to an inertial frame.these motions will not be uniform and this fact appears to set such frames apartfrom inertial frames as a special class for which the ordinary laws of motion donot apply. However. by a well-known device of Newtonian mechanics, viz. theintroduction of inerriulfim·es. accelerated frames .:an bc treated as though theywere inertial and this suggests a way out of l)lIr difllculty. Thus suppose a spacerocket. moving ill roC/IO, is being accelerated unifolrnly by the action of itsmotors. An observer inside the rocket will note that unsupported particlesexperiem:e an acceleration parallel to the axis of the rocket. Knowing that themotors a re operating. he will attri bute this acceleration to the fact that his naturalreference frame is being accelerated relative to an inert ial frame. However he may,ifhe prefers. treat his reference frame as inertial and suppose that all bodies withinthe rocket are being subjected to inertial forces acting parallel to the rocket's axis.11' a is the acceleration of the rocket, the appropriate inell ial force to be a pplied toa partide llf mass m is rna. Similarly. if the rocket's ml'tors are shut down bUI therocket is spinning about its axis, an ohserver within the rocket will again note thatfree particles do not move uniformly relative to hiS "urroundings and he mayagain avoid attributing this phenomenon to the fact that his frame is not mertial,by supposing certain inertial forces (viz. centrifugal and Coriolis forccsl to actupon the particles. Now it is an obvious property of each such inertial force that itmust cause an acceleration which is independent of the mass of the bOdy uponwhich it acts, for the force is always obtained by multiplying the body's mass by anacceleration independent of the mass. This property it shares with a gravitatt•.>nalforce, for this also is proportional to the mass of the particle being attracted andhence induces an acceleration which is independent of this mass. This indeprn­dence of the gravitational acceleration ofa particle and its mass has been checkedexperimentally with great accuracy by E6tv6s. If, therefore. we regard theequivalence of inertial and gravitational forces as having been esta blished, inertialforces l'an be thought of as arising from the presence of gravitational fields. This is

129

the principle (If equivalence. By this principle, in the case of the uniformlyaccelerated rocket, the observer is entitled to neglect his acceleration relative to aninertial frame, provided he accepts the existence ofa uniform gravitational field ofintensity a parallel to the axis of the rocket. Similarly, the observer in therotating rocket may disregard his motion and accept, instead, thc existence of agravitational field having such a nature as to account for thc centrifugal andCoriolis fo rccs.

By appeal to the principle of equivalence, therefore, an observer employing areference frame in arhitrary motion with respect to an incrtial frame, maydisregard this motion and assumc, instead, the existencc of a gravitational field.Thc intensity of this field at any point within the frame will be equal to the inertialforce per unit mass at the point. By thIS device, every observer becomes entitled totreat his reference frame as being at rest and all observcrs accordingly becomeequivalent. However, the reader is probably still not convinced that thedistinction between accelerated and inertial frames has been effectively elim­inated, but only that it has been concealed by means of a mathematical devicehaving no physical sigmficance. Thus, he may point out that the gravitationalfields which have been introduced to account for the inertial forces are 'fictitious'fields, which may be completely removed by choosing an inertial frame forreference purposes, whereas 'real' fields, such as those due to the earth and sun,cannot be so removed. He may further object that no physical agency can be heldresponsible for the presence of a 'fictitious' field, whereas a 'real' field is caused bythe presence of a massive body. These objections may be met by attributing such'fictitious' fields to thc motions of distant masses within the universe. Thus, if anobserver within the uniformly accelerated rodet takes himself to he at rest, hemust accept as an obser vable fact that all bodics within the universe, including thegalaxies, possess an additional acceleration of a relative to him and to thismotion he will be able to attribute the presence of the uniform gravitational fieldwhich is affecting his test particles. Again, the whole universe will be in rotationabout the observer who regards himself and his space-ship as stationary when it isin rotation relativc to an inertial frame. It is this rotation of the masses of theuniverse which we shall hold responsible for the Corio lis and centrifugalgravitational fields within the rocket. But, in addition, these 'inertial' gravitationalfields will account for the motions of the galaxies as observed from the non­inertial frame. Thus, for the observer within the uniformly accelerated rocket auniform gravitational field of intemity a cxtends over the whole of space and isthe cause of the acceleration of the galaxies; flJr the observer within the rotatingrocket, the resultant of the centrifugal and Coriolis ficlds ading upon thc galaxiesIS just sufJicient to account for their accelerations in their circular orbits abouthimself as centre (the reader should verify this, employing the results of ExercisesI, No. I). On this view, therefore, inertial frames possess particularly simpleproperties only because of their special relationship to the distribution of masswithin the universe. In much the same way, the electromagnetic field due to adistribution of electric charge takes an especially simplc form when dcscribed

130

relative to a frame in which all the charges are at rest (assuming such exists). If anyother frame is employed. the field will be complicated by the presence of amagnetic component arising from the motions of the charges. However. thismagnetic field is not considered imaginary because a frame can be found in whichit vanishes. whereas for certain magnetic fields such a frame cannot be found. Thelaws of electromagnetism are taken to be valid in all frames. though it is concededthat, for solving particular problems, a certain frame may prove to be pre­eminently more convenient than any other. Neither. therefore. should thecentrifugal and Corio lis fields be dismissed as imaginary solely because they canbe removed by proper choice of a reference frame. although it may be convenientto make such a choice of frame when carrying out particular computations. Inshort, the general principle of relativity can be accepted as valid and. at the sametime, the cxistence of the inertial frames accounted for by the simplicity of themotions of the galactic masses with respect to these particular frames.

The notion that the existence of inertial frames is bound up with the large-scaledistribution of matter within the universe is referred to as Mach's principle.Although Einstein was powerfully influenced by the principle when developinghis general theory, he was disappointed to discover that it still permits theexistence of universes in which local inertial frames are not in uniform non­rotatory motion relative to the overall matter distribution. The complet~

integration of Mach's principle into the theory is yet to be accomplished.The previously unexplained identity of inertial and gravitational ma.ses is

easily deduced as a consequence of the principle of equivalence. For, consider aparticle of mass m which is being observed from a non-inertial frame. Agravitational force equal to the inertial force will be observed to act upon thisbody. This force is directly proportional to the inertial mass m. But, by theprinciple of equivalence, all gravitational forces are of the same nature as thisparticular force and will, accordingly, be directly proportional to the inertialmasses of the bodies upon which they act. Thus the gravitational 'charge' of aparticle. measuring its susceptibility to the influences of gravitational fields, isidentical with its inertial mass and the identity of inertial and gravitational masseshas been explained in a straightforward and convincing manner.

45. Metrie in a gravitational field

Suppose that a space-station in the shape of a wheel ha. been constructed in aregion of space far from other attracting bodies and that it is set rotating in itsplane about its centre with angular velocity w. An observer 0, wearing a space­suit, is located outside the station and does not participate in the rotary motion;his frame of reference is therefore inertial. °watches C, a member of the station'screw, measuring the dimensions of the station using a metre rule. C first measure.the radius of the station from its centre to its outer wall by laying his rule alongone of the corridors forming a .poke of the wheel. °notes that the rule is movinglaterally throughout the measuring process, but this motion does not afTect its

131

length ill his frame and he will accordingly agree with the radius r recorded by C.C next lays his rule around the outer wall of the station and records a perimeter p.During this process, however, 0 sees the rule moving longitudinally with velocitywr and its length will be reduced by a factor J (I - (1)2 r2

/(2

). He will accorcinglycorrect the length of the perimeter found by C to the value p J (I - w2 r! /( 2

).

Since O's frame is inertial, Euclidean geometry is valid for all space measurementsreferred to the frame and he must find that

pJ(I-w2 r 2 /c 2 )=2rrr [45.1)

Thus p = 2rrr( I - (1)2 r2 /( 2 ) 1/2 [45.2)

This last equation indicates that C will discover that the Euclidean formulap = 2rrr is not valid for measurements made in the rotating frame of the space­station. But C is entitled to regard the station frame as being at rest, provided heaccepts the existence ofa gravitational field which will al'Count for the centflfugaland Coriolis forces he experiences. We conclude that, relative to a frame at lest insuch a gravitational field, spatial measurements will not be in conformity withEuclidean geometry.

By the pnnciple of equivalence. the conclusion which has just been reachedconcerning the non-Euclidean nature of space in which there is present agravitational field of the centrifugal Coriolis type, must be extended 1.0 allgravitational fields. However. in the case of a field such as that which surroundsthe earth. it will not be possible (as it is for the centrifugal Corio lis field) to find aninertial frame of reference relative to which the field vanishes and for whi,h thespatial geometry is Euclidean. Such a field will be termed irreducihle. Evenin anirreducible field. however. a frame can always be found which is inertial for asufficiently small region of space and a sufJiciently small time duration. Thus.within a space-ship which is not rotating relative to the extragalactic nebulae andwhich is falling freely in the earth's gravitational field. free particles will f0110wstraight-line paths at constant speed for considerable periods of time and theconditions will be inertial. A coordinate frame fixed in the ship will accordinglysimulate an inertial frame over a restricted region of space and time and itsgeometry will be approximately Euclidean.

Since a rectangular Cartesian coordlllate frame can be set up only in a ,pacepossessing a Euclidean metric. this method of specifying the relative positions ofevents must be abandoned in an irreducible gravitational field (except over smallregions as has just been explained). Instead. the positions and times of all eventswill be specified by reference to a very general type of frame which wecan supposeconstructed as follows: Imagine the whole of the cosmos is filled by a fluid whosemotion is arbitrary but non-turbulent (i.e. particles of the fluid which arc initiallyclose together. remain in proximity to one another). Let each molecule of the fluidbe a clock which runs smoothly, but not necessarily at a constant rate as judged bya standard atomic clock. No attempt will be made to synchronize clocks whi,hareseparated by a finite distance. but it will be assumed that, as this distance tends tozero. the readings of the Clocks will always approach one 'Inother. Each clock will

132

be allocated three spatial coordinates ~ 1. ~2. e according to any scheme whichensures that the coordinates of adjacent clocks only differ infinitesimally. Thecoordinates ~> ofa clock will be supposed never to change. Any event taking placeanywhere in the cosmos can now be allocated unique space- time coordinates~i (i = 1.2. 3.4)as follows: (~I.e,C)are Ihe spatial coordinates belonging to theclock which happens to be adjacent to the event when it occurs. and ~4 is the timeshown on this clock at this instant.

We shall now further generalize the coordinates allocated to an event. Letx; (i = I. 2, 3.4) be any functions of the ~; such that, to each set of vulues of the ~;

there corresponds one set of values of the Xi. and conversely_ We shall write

(45.3)

Then the x" also. will be accepted as coordinates. with respect to a new frame ofreference, of the event whose coordinates were previously taken to be the ~'. Itshould be noted that, in general. each of the new coordinate., x' will depend uponboth the time and the position of the evenl, i.e.. it will not necessarily be the casethat three of the coordinates x' arc spatial in nalure and one is temporal. Allpossible events will now be mapped upon a space .'/4' so tha! each event isrepresented by a point of the space and the y' will be the coordinates of this poin!with respect !o a coordmate frame. .'1'4 will be referred to as the .'!lU('(' lillie

cOnlintlllnJ.II has becn remarked that, in any gravitational field. it is always possible to

define a frame relative to which the field vunishes over a restricted rcgion andwhieh behaves as an inertial frame for events occurring in this region andextending over a small interval or time. Sueh a frame will be falling freely in thegravitational field and will accordingly be referred to as a !o('ul/i'ee-/IlIl/i·lInJe.Suppose, then, that such an inertial frame S is found for two contiguous events.Any other frame in uniform motion relative to S will also be inertial for theseevents. Observers at rest in all such frames will be able to construct rectangularCartesian axes and synchronize their standard atomic clocks in the m<lnnerdescribed in Cltapter I and hence measure the proper time interval dr between theevents. If, for one such observer. the events at the points having r('clangularCartesian coordinates (x, .1', ~), (x + dx, .I' + d y, ~ +- d;) occur at the times i. I + dlrespectively, then

dr2 = dl 2 - 12(dx 2 +dy 2 + d~2)C

The itllervu! between the events ds will be defined by

ds 2 = _c 2 dr2 ~, dx2 + d y 2 + d~2 - ("2 dt 2

(45.4)

(45.5)

The coordinates (y• .1',;, I )01' an event in this quasi-inertial frame will be related tothe coordinates Xi defined earlier, by equations

etc. (45.6)

and hence(IX

dx = dx',Px'

etc.

133

(45.7)

Substituting for l/x. dr, <I:. dl in equation (45.5), we obtain the result

lis' = Y'jdx' dxj (45.8)

determining the interva I lis between two events contiguous in space time, relaliveto a general coordinate frame valid for the whole of space-time. The space-limecontinuum can accordingly be trea ted as a Riemannian space with metric givem byequation (45.8).

As explained in section 7. <lr can be time like or spacclike according as it is 'eal

or imaginary respectively. [fit is real. it will be possible ror a standard clock to bepresent at both the events x', Xi + dx' and the time which elapses between them asmeasured by this clock will be dr. Alternatively. if ds is imaginary. ds/ic can beinterpreted as the time between two contiguous events as measured by a stand~lrd

elock present at both.

46. Motion of a free particle in a \:ravitationa[ field

[n a region of space which is at a great distance from material bodies. rectang~larCartesian axes Ox .1': can be found constituting an inertial frame. [I' time ismeasured by clocks synchroniled within this frame and moving with it. themotion ofa freely moving test particle relative to the frame will be uniform. Thus,if (x• .1'. z) is the position of such a particle at time I. its equations of motion can bewritten

(46.1 )

Let ds be the interval between the event of the particle arriving at the point (X, J. z)

at time 1and the contiguous event of the particle arriving at (x + dx. y + dy.: +d:)at 1 + dl. Then ds is given hy equation (45.5) and. If I' is the speed of the particle. itfollows from this equation that

(46.2)

Since IJ is constant. it HOW follows that equalions (46.1) can be expressed in theform

(46.3)

Also, from equation (46.2) it may be deduced that

d'l

d ~' = 0.\

(46.4)

Equations (46.3) and (46.4) determine the family of world-lines offree particles inspace time relative to an inertial frame.

134

Now suppose that any other reference frame and procedure for measuring timeis adopted in this region of space. e.g. a frame which is in uniform rotation withrespect to an inertial frame might he employed. Let (Xl. x 2

, X". x 4) be the

coordinates ofan event in this framc. The interval between two contiguous eventswill (hen he given by equation (45.l\). [I' an ohserver using this frame releases a testparticle and observes its motion relalive to the frame. he will note that it is notuniform or even rectilinear and will he ahle to account for this fact hyassumingthe prcsence of a gravitational field. He will Hnd that Ihc particle's cquations ofmotion arc

(46.5)

This must be the case for. as shown in section 43. this is a tensor equation defininga geodesic and valid in every frame if it is valid in one. But. in the xyzt-frame, the!1O) arc all con,;tant and thc three indcx symhols vanish. Hencc, iu this frame. theequations (46.5) reduce to the equations (46.3) and (46.4) and these are known tohe true for the partiele's motion. We havc shown, therefore, that thc efTect of agravitational field of the redueible variety upon the motion ofa tcst particle can beallowed for when the form taken by the metric tensor !I,j of the space timemanifold is known relative to the frame being employed. This means that the !Ii}determine. and are determined hy, the gravitational field.

The ideas of the previous paragraph will now he extended to regions of spacewhere irreducihle gravitational fiekb are present. It has been pointed out that. forany sufliciently small region of suclt space and interval of time. an inertial framecan he found and consequently the paths of frcely moving particles will hegovcrned in such a small region hy equations (46.5). II will now be assumed thatthese arc t he equations of motion of free particles without any restriction, i.e. thatIhe world-line of a free particle is a gcodesic for the space time mani fold or thatthe world-line of a free particle has col1'otant direction. This appears to be thenatural generalillition of the Galilean law of inerlia whereby, even in anirreducihle gravitational field, a particle's trajectory through space- time is thestraightest possible after consideration has heen given to Ille intrinsic curva ture ofthe continuum. It will then follow that the motions of particles falling freely inany gravitational field can he determined relative 10 any frame whcn thecomponents gij of the metric tensor for this frame are known. Thus the !lij willalways specify the gravitational fil'ld ohserved to be present in a frame and theonly distinction between irreducihle and reducible fields will be that. for the latterit will he possible to find a coordinate frame in space- lime for which the metrictensor has all its comjX)nents leI(l except

!Ill = g22 = g,,,, = I 944 = _c 2 (46.6)

whereas for the former this will not be possible.It will be proved in section 50 that the assumption we are making can be derived

from Einstein's law of gravitation and hence docs not constitute an additionalhasic hypothc>;is of thc theory.

(47.2)

135

Since the Christoffel symbols vanish in a frame which is geodesic at some pointof space time, in such a frame equations (46.5) reduce to d 2 x i jds 2 = ,) over asmall neighbourhood of the point. II; in addition, the flame is chosen to be quasi­Euclidean with metric (45.51, equations (46.1) will be valid over the neighbour­hood and a freely falling body will have very nearly uniform motion. Sud a framecan therefore be identified with a local freely falling frame.

47. Einstein's law of J:ravitation

According to Newtonian ideas. the gravita tiona I field which exists in any region ofspace is determined by the dtstribution of matter. This suggests that the metrictensor of the space time manifold. which has been shown to be closely related tothe observed graVItational tield.shOlrid becalclliable when the matterdistcibutionthnHlghollt space ttme is known. We first look, therefore, for a tensor quantitydescribing this matter distrihutton with respect to any frame in space time andthen attempt to relate this to the metric tensor. The energy momentum tensor J:

j,

defined in section 21 with respect ttl an inertial frame. immediately suggelts itself.Both matter and electromagnetic energy contribute to the componenll of thistensor but since, according to the special theory. mass and energy are basicallyidentical, it is to be expected that all forms of energy, including tle elec­tromagnetic variety. will contribute to the gra vitational field.

Since the energy momentum tensor has hcen defined in inertial frames only.this definition must now he extended ttl apply to a general courdinate frame inspace time. This can be carried out thus' In the neighbourhood of a point P ofspace time, a frame with coordinates \" can be defined which is geodesic at P andwhose metric reduces ttl the Euclidean form (39. D) at P. As explained in the lastsection, this frame will correspond to a local freely falling quasi-inertial frame inwhich the yi will behave like Mmkowski coordinates; we shall assume lhal theequations of the special theory are valid in this frame at P. The transformationequations relating the coordinates y' ofan event to its coordinates x' with respecttoany other coordinate frame can now be found. Then, ifTi,O) are the eomilonentsof the energy momenlum tensor in the .v-frame at the point P, its compollents inthe x-frame at this point can be determined from the appropriate tensortransf(mnation equaltons. Thus, the covariant energy- momentum tensor willhave components T'j in I he x-frame given by

(lV' ;;V"~

Tij = iO-; 'I' j T,~O) (47.1)IX IX

Since covariant and contravariant tensors are indistinguishable with respect torectangular Cartesian axes, T,\(l) can also be taken to be the components of aeontravariant tensor in the y-frame and the components of this tensor in the x­frame will then be given by the equation

1 .i 1 jT'} = (.X IX TIO)

,1 r' (lr·. r,'

136

Similarly, the components of the mixed ener~y momentum tensor are given by

(47.3)

These transformations can be carried Oil! at every point of space time, thusgenera tmg for the x-frame an energy momentum tensor field throll~hollt thecontinuum. It is left as Hn exercise for the reader to show that the last threeequatIons are consistent, i.e. raising the incbees of T,j as givcn by equation (47. J)

leads to T'J as given by equation (47.2) (sec Exercise I at the cnd of this chapter).Consider the tensor equation

r J .-~' {)., (47A)

Expressed in terms of the coordinates \" at any point of space time. this simplifiesto

T'O) =0"~,I

(47.5)

which is equation (21.20). Being valid i.n olle frnme, therefore, equation (47A) istrue for all frames. Thus, the divergence of the energy momentum tensorvanishes. If, therefore. this tensor is to be rdated to the me:ric tensor (I,l' therelationship should bc of such a f(1I m that it implies equation (47A). Now

y': = 0

by equation (3\).23) and hence. a{Orliori,

(47.6)

The law

~(: 0

'I'll l fJ d

(47.7)

(47.H)

where A is a univcrsal constant, would accordingly be satisf'lctory in this respect.However. ovcr a region in which matter and em:rgy wcre absent so that T,j = 0,this would imply that

(47.9)

whidl is clearly incorrect. Further. according to Newtonian theory. if II is thedensity of maller, the gravitHtional field can be dcrived from a pl)lential func!JonU satisfying the equation

(47.10)

where G is the gravitational constant. The ncw law of gravitation which is beingsought must include equation (47.10) as an approximation. But, as appears fromequation (21.14), T44 involves Jl and it seems reasomlt>ie, therefore, to expect thatthe other member of the equation expressing the new law of gravitation willprovide terms which can receive an approximate interpretation as V 2 U. Thisimplies that second-order derivatives of the metric tensor components willprobably be present. We therefore have a requirement for a second rankcontravariant symmetric tensor involving second-order derivatives of the Y;j and

137

of vani,hing divergence to which T'J can be assumed proportional. Einsll:in'stensor (42.20) possesses these characteristics and consequently we shall put

(47.11 )

where" is a constant of proportionality which must be related to G and which weshall later prove to be positive. Fquation (47.11) expresses bn5lein's 1m\' o{grl11'ilf/l;on; by lowering the indices <;uccessively, it may be expressed in the twoalternative forms

R;-1b;R = -hi;

R" 1Y'JR ~ - h'J:)

If equation (47.12) is contracted, it is found that

R ~ hT

(47.12)

(47U)

(47.14)

where T= J:'. It now follows that Einstein's law of gravitation can alse) beexpressed in the form

(47.15)

with two other forms obtained hy raising subscripts.Since the divergence of y'j vani.shes, a possible alterna tive to the law (47.11) is

(47.16)

where 1\ is a constant. The law (47.11) gives results which agree with observationover regions of space of galactic dimensions, so that It is certain that, even il'l\ isnot zero, it is exceedingly small. However. the extra term has entered into wmecosmological investigations (sec Ch"rter 7).

48. Acceleration of a particle in a weak gravitational field

In a gravitational field, such as the one due to the earth, the geometry of space isnot Euclidean and no truly inertial frame exists. In spite of this, we experieme nopractical difliculty in estahlishing rectangular Cartesian axes Oxyz at the eurth'ssurface relative to which for all practical purposes the geometry is Euclidean andthe behaviour of electromagnetic systems is indistinguishable from theirbehaviour in an inertial frame. It must be concluded, therefore, that such agravitational field is compmatively weak and hence that, with respect to such axesand their associated clocks, the space -time metric will not differ greatly from tha tgiven by equation (45,5), Putting

in terms of the x; the metric will be given by

ds 2 = dx'dx' 148,2)

138

approxima tely. With respect to the xi-frame, it will accordingly be assumed tha t

(48.3)

(48.4)

where the ('iij afe Kronecker deltas and the h'i are small by comparison.Consider a particle moving in a weak gra vitational field whose metric tensor is

given by equation (48.3). The contravariant metric tensor will be given by anequation of the form

where the k ii are of the same order of smallness as the h'i' Then, since

I (llh'k (lh'k llh .. )[ij,k] = J. +.!. _ ...'J2 llx' ax J ux k

(48.5)

it follows that, to a first approximation,

V} = Oh[ij, r] = I (fJh ik + alJ,k _ <lh ii )J 2 ax' fJXJ (lx k (48.6)

The equations of motion of the particle can now be written down as at (46.5).By equation (46.2),

dx i dx i dr= """ = (v1- _( 2 )-1/2 (v, ie)ds dr ds

(4H.7)

(48.8)

where v is the particle's velocity in the quasi-inertial frame. Hence, if the particle isstationary in the frame at the instant under consideration,

dx i

- = (0, I)ds

and the equations of motion (46.5) reduce to the form

d1x i .

'dsT + {4'4} = 0 (48.9)

(48.10)1iJh44 llh 4i2 fix i - iJ~4

d 2x i

d~;2

correct to the first order in thehij' Substituting from equation (48.6), this is seen tobe equivalent to

(48.11)

Differentiating equation (48.7) with respect to s and making use of equation(46.2), we obtain

d2Xi 2 2". 1 (dV ) dv 2 '2

d'~2 = (v -c) d;'O -vdr

(p -e")' (v,ic)

and, when v = 0, this reduces to

d 2x i

ds 2(48.12)

(48.13)

139

From equations (4X.IO) and (4X.12), we deduce that the components of theacceleration of the stationary particle in the directions of the rectangular axes are

_ ('2 ( lr'h.4~ _ ?h4 • )

2 (lX' (h 4

for i = 1,2,3. Reverting to the original coordinates (x, y, z, r), these componentsare written

etc (48.14)

Hence, if the field docs not vary with the time, the acceleration vector is

(48.15)

But, if U is the Newtonian potential function for the field, this acceleration will be- gmdU. It follows that, for a weak field. a Newtonian scalar potential U existsand i; related to the space time metric by the equation

Alternatively, we can write

2UY44 = 1+ ­

('2

(48.16)

(4X.17)

49. Newton's law of gravitation

In this section it will be shown that Newton's law of gravitation may be deducedfrom Einstein's law in the normal case when the gravitational field's intensity isweak and the matter distribution is static.

First consider the form taken by the Riemann Christoffel tensor in thespace time of a weak field. In the x'-frame, the metric tensor is given by equation(48.3) and the Christoffel three-index symbols by equation (4X.6). If products ofthe hi) are to be neglected, equation (36.21) shows that

(49.1)

approximately. Hence the Ricci tensor is given by

140

(49.2)

(49.3)

If the matter distribution is static in the quasi-inertial frame being employed, theh,j will be independent of f and equation (49.3) reduces to

R 44 = {V 2 h44 (49.4)

where V2 = (1 2/ IIX 2 + ,,2/1,

/ + ,,2/I' Z2 If U is the Newtonian potential for thefield, equation (4H.16) now shows that

(49.5)

Assuming that no electromagnetic field is present and that the contribution tothe energy- momentum tensor of any stress forces wIthin the matter distnbutionresponsible for the gravitational field IS negli!!lble, T~j will be determined byequation (21.16). But. since the distribution is static, its 4-velocity of flow Vatevery point IS (0, ;c)and hence all components otT,)' with the exception ofT44 , arezero. In this case,

(49.6)

where /1"", for ,ero velocity of matter, is the ordinary mass density. Also

(49.7)

The 44-compollent of Einstein's gravitatiolllaw in the form of equation (47.15)can now be expressed approximately

or (49.H)

This is the Poisson equation (47.10) of clas;,ical Newtonian theory, provided weaccept

(49.9)

This specifies K in terms of the gravitational constant.

SO. Freely falling dust cloud

Consider the case ofa cloud of particles falling freely in the field of the cloud itsell~

there being no other forces present in the system. The energy -momentum tensor

141

for such an incoherent cloud has been calculated relative to an inertial frame asequation (21.10). This equatIOn will be taken to provide a definition of 7ii in thefreely falling frame at any poinl of the cloud.

In an arbitrary x-frame. let x' = .\ '(r) be parametric equations of the wcrld-lincof some particle of the cloud. r being the proper time measured by a sandardclock moving with the parllcle. Then the 4-vdocity offlow of the particle 'I time ris defined by the equatloll

dx' dx'V' = = ic

dr ds(50.1 )

where s is the interval parameter measured along the world-line. The squaleofthemagnitude of the 4-veloeity of flow is

This equation can also be written

Vi V' = .- c 1

Now consider the tensor equation

T'i = /100 V' V}

(502)

(50.3)

(511.4)

where /loo is the mass density as measured in a freely falling frame movi~g withthe cloud; /100 is clearly a 4-illvariant. In any frecly I~tlling frame. this equationreduces to equation (21.16) and is accordingly valid; this establishes its va.ldity inall frames.

Equation (47.4) is known to follow from Einstein's equation of gravitation and.in this case. takes the form

(JIOO V' V})., = (Jloo Vi).} V' + /1(10 Vi V'} = 0

Multiplication by Vi now gives

(/1 00 VI)., V' V, + /100 V', V, VI = 0

Diflercntiating equation (50.3) with respect to xi, we find

(50.5)

(506)

(50.7)V V'+VV' =01,/ I ,/

Raising and lowering the index i in the two 1~letors of the tirst term, this equa tionis seen to be equivalent to

V' V = ().' 'Equatioll (50.6) accordingly reduces to

(JloO VI)., = 0

Equation (50.5) now gives

V' Vi = 0.'

(50.R)

(50.9)

(50.10)

142

or

Hence

or

This can also be written

( (~.v' + r~ v.) vj = 0('Xl )

d 2x i. dx' d.x j

- + r~. ... = 0d.1 2

I ds ds

(50.11)

(50.12)

(50.13)

(50.14)

a result which proves that the world-lines of the particles of the cloud aregeodesics.

That the world-lines of freely falling particles in general circumstances aregeodesics can be derived from equation (47.4), proving that Einstein's lawincludes its own law of motion for a particle in a gravitational field.

51. Metrics with spherical symmetry

When a change is made in the space time coordinate frame from coordinates Xi tocoordinates x'. the metric tensor (};) will change to g,; hy the law of transformationof a covariant tensor. In generaL the iJij will be functions of the x' and the ?i'l willbe functions of the .x', but it will not usually be the case that the !I,j are the same

functions of the 'barred' coordinates that the iJijare of the 'unbarred' coordinates,i.e., the functions (jijexk) are not .tilrm il/varjanT under general coordinatetransformations. Howcver, in some special cases, it is possible for these functionsto be form invariant under a whole group of transformations, and we shall studysuch a case in this section.

In a gravitational field, the geometry can only be quasi-Fuclidean andconsequently rectangular Cartesian axes do not exist. Nevertheless. no dillicultyis experienced in practice in defining such axes approximately and we shallsuppose, therefore, that the coordinates x, y. z, r of an event in the gwvitationalfield about to be considered are interpreted physically as rectangul:ll' Cartesiancoordinates and time. We shall now search for a metric which, when expressed inthose coordinates, is form invariant with respect to the group of coordinatetransformations which will be interpreted physically as rotations of therectangular axes Oxyz (r is to remain unaltered). To be precise. it will he supposedthat spatial coordinates (x, y, z) can be defined 'wcl! that the metric y,j(x,.r, z, r) is

143

form invariant under the group of orthogonal transformations x = Ax, wherex = (x, J', zjI, x = Lx, )'. ijl and AAI = I. Such a metric will be said to bespherically symmefric about O.

Invariants for this group ofcoordinate trall'formations. which are ofdegree nohigher than the second in the coordinate dillcrentials dx, dy, dz. are

(51.1)

Introducing spherical polar coordinates (I', II. (M. which will be d(:fined by theequations (30.31, these invanants may be written

It follows that

1'2, I'dI', dr2+r2dIl2+r2sin2I1d¢2

r. dr. dll 2 + sin 211 d¢2

(51.2)

(51.3)

are invariants. The most general metric with spherical symmetry can now be builtup in the form

d.\2 = A(r,l)dr' + B(r,f)(dll' +sin'lIdq)2)

+C(r,f)drdf +D(r,f)df 2 (51.4)

(51.6)

(51.5)1'" = B(r. I)

ds' = E(r'. I)dr" + r" (dll' + sin'lId¢')

+ F(r', f )dr' df + C(r', f )df 2Then

We now replace I' by a new coordinate r' according to the transformationequation

(51.8)

In a truly inertial frame. spherical polar coordinates can be defined exactly and themetric will, by equation (45.5). be expressed in the form

ds 2 = dr 2 +r2(dIl 2 +sin 2I1d¢2)-c 2df2 (51.7)

Comparing equations (51.6)and (51.7). it is clear that in a region for which (51.6)is the metric. r' will behave approximately like a true spherical polar coordinate r.We shall accordingly drop the primes and wnk

ds 2 = E(r, f )dr' + 1" (dll' + sin'lId1)')

+F'(r.f)drdf +C(r.f)df 2

If our frame is quasi-inertial. equation (51.7) must be an approximation forequation (51.8) and the following equations must therefore be trueapproximately:

E(r,f) = 1. F(r,f) = 0, C(r,f) = _(,2 (51.9)

Consider now the special case when the gravitational field is static in the quasi­inertial frame for which (1',0, ¢ )areapproximate spherical polar coordinates and I

is the time. The functions E, F. G will then be independent of f. Also, space timewill be symmetric as regards past and future senses of the time variable and thisimplies that ds 2 is unaltered when df is replaced by - df. Thus F = 0 and we have

(51.10)

(51.12)

144

where (1, h are functions of I' hoth approximating unity in a weak field.At any fixed instant T, the metric of spacc in the presence of this gravitational

field can he ohtained from the last equation hy putting dT = O. Thus, it is

ds 2 = lIdr 2 + 1'2 (d1l 2 + sin 2 I1d<p 2) (51.11)

Consider the 'circle' ,. = 1'" in the 'plane' II = ~ 1[, The length of the element of thecircle with end points (1'0, <p). (1'", <p + dq,), has length d., = 1'" d<p. Thus the totallength of the circle is 2nro' However, 1'" will not be the length ofa radius <p = <Po ofthis circle for, if an element of such a radi us has end points (I', <Po), (I' + dr, <Po ), thelength of the element is ds = a l

-2 dr and the total length of the radius isaccordingly

[11 1/2dr

Clearly, the Euclidean formula for the circumlCrence of a circle of giwn radiusdoes not apply.

For the metric (51.1 0), taking

we have

(5113)

411 = a, 422 = r'. 4.L1 = 1'2 sin 2 11, 444 = - hl'2

all other 4'J being zero. Thus

and hence

(51.14)

(51.15)

II 14 =

a(5116)

all other 4" being zero. The three,index symbols call now he calculated and,putting a = 1'0, h = 1'1', those which do not vanish an: listed helow:

(5117)

145

{1 I {1}S

= = cot /I2 3 3 2

{ I} -1'1' 'sin2/11 3 =

J 2 } _ _ sin /I cos /I)3 3 --

primes denoting differentiations with respect to r.The non-Lero comp0ncnts of the Ricci tensor are now calculated to he:

R 11 = 1{1" + l{1'2 -la' {I' - I a'I'

R22 = c'(ir{1' -ira' + 1)­

R.u = Rn sin 2 {}

1R44 = e 2 ell

'( -111" -i{I'2 + la' {I' - {i')I'

(5118)

52. Schwarzschild's solution

The static. spherically symmetrical metric (51.10) will determine the gravitationalfield of a static distrihution of matter also having spherical symmetry, provided itsatisfies Einstein's equations (47.15). We shall consider the special case when :hewhole of space is devoid of matter. apart from a spherical hody with its centre atthe centre of symmetry O. Then 7:) =, 0, T = 0 at all points outside the hody andEinstein's equations reduce in this region to

R'J = 0

By equations (51.18), these are satisfied hy the metric (51.10). provided

{1 "+'{i'2 I '{i' 2, 02 -'2a - a ~=

I'

1I'{I' - ~rex' + 1 = 1"

fi" + 1{1' 2 - ~ ex' {I' + 2{1' = 0I'

Subtracting equation (52.2) from (52.4), it follows that

ex + {i = constant

(5~.1)

(52.2)

(52.1)

(524)

(52.5)

But. as 1' ..... OC', we shall assume that our metric approaches that given by equation(51.7). valid in the absence ofa gravitational field. Thus, at infinity, ex = {i = Oa~d

146

henceex.+/J=O

Eliminating fJ from equation (52.3), it will be found that

rex.' = 1- e"

(52.6)

(52.7)

The variables are separable and this equation is easily integrated to yield

a = e" = (1-2m/r)-1

where m is a constant of integration. Then

h = eP = I -2m/r

(52.8)

(52.9)

and it may be verified that each of the equations (52.2H52.4) is satisfied by theseexpressions for ex. and {i.

We have accordingly arrived at a mNric

2 dr2

2 'l2 . 2 'l 2 ( 2m) 2ds = _ .. --..- + r (d/, + sm Oa</J ) -c 1---- dl1- (2m/r) r

(52.10)

which is spherically symmetrical and can represent the gravitationallield outsidea spherical body with its centre at the pole of spherical polar coordinates (r, 0, 4».This was first obtained by Schwarzschild. It will be proved in the next section thatthe constant m is proportional to the mass of the body. This may also be deducedfrom equation (4X.17), for the potential U at a distar1l'c r from a spherical body ofmass M is given by

and hence

GMU = ..--

2GM94..' = I -- "2--'

• C r

(52.11)

(52.12)

Now Y4.4 is the coeflicient of (Jx 4)2 = - c2 d1 2 in the metric and hence

Comparing equations (52.9) and (52.13), it will be seen that

(1M

(52.13)

(52,14)

It is clear from equation (52.10) that the metric is not valid for r = 2m= 2GM/c2

, This is the Schwarz,w'hild radius. In SI units, c = 3 X 108 and for theearth GM = 3,991 x 10'4, so that the radius for this body is about <) mm; since themetric is only applicable in the region outside the earth, no difficulty isencountered in this case, However, for an exceptionally dense body, the radius

147

may extend into the surrounding space and, for values ofr Ie,s than the radius, themetric needs special consideration (sec section 57).

53. Planetary orbits

The attractions of the planets upon the sun cause this body to have a smallacceleration relative to an inert ial frame. If, therefore, a coordinate frame movingwith the sun is constructed, relatiVt' to this frame there will be a gravitational fieldcorresponding to this acceleration in additIOn to that of the sun and planets.However, for the purpose of the following analy,is, this lield and the lields of thcplanets will bc neglectcd. Thus, rclative to spherical pol;lr coordinate, havingtheir pole at the centre of the sun, thc gravitational lield will be assumeddetermined by the Sl:hwarz,child mctric (52.10). The planets will be tre;lted asparticles possessing negligible graviUltional held" whose world-lines arcgeodesics in space time. We proceed to calculate the,e geodesic,.

Since the intervals betwecn adjacent pl1ints on the world-line Ill' a particle areneccS!>arily timelike, .\ will be purely mlaginary along such a curve. Whencalculating geodesics it is usually mOrc convenient, therefore, tl) replace.\ by randto work from the metric cxprc"ion for dr'. Thll" in thi, ,cction, theSchwarzschild metric will be takell iu the form

dr' = _ 1,( dr' +r'(dll'+sin'lId1"))+(I.-2rn/r)dt 2

c~ I - 2111/r

and equations (43.19) for the gelHle,ics bel:ome

(531)

(535)

(534)

(53.3)

d ( I'dI') rn (dr)l (dll)' ., (d 4»' me2(dt)2+ --I' -I' sin II -- t-_· - -0

dr r-2rndr (r-2nl)' dr dr . dr 1'2 dr -

(5.12)

d (1'2 dll') .. 1" sin II cos II (d4))2 = 0dr dr. dr

d (1'2 sin 2 l1d4

') oc 0dr dr

d ('r ·2/11 dt) =0dr I' dr

The lirst of this set of eqll'ltions will be replaced by the first integral (43.6), viz.

1', (ldlr)2+1',{(dll)2 +sin'lI(d(!»'}_C'(r_2 t1l )(dt)' = -c'

I' - dll r d r d r I' d r(53.6)

(5l9)

148

We now choose the spherirul polar coordinates so that the planet is movinginitially in the plane II = til. Then dll/dr = 0 initially and hence, by equation(53.3), d 2 O/dr 2 = Oat this instant. By repeated dilferentiation of this equation andsubstitution of initial values, it is found that all derivatives of II vanish inilially.Hence, by Maclaurin's theorem, 0 = ~Il for all values ofr, proving that the planetcontinues to move in the 'plane' 0 = in indefinitely.

Integrating equations (5~.4) and (53.5), and putting II = ~Il, we get

dql h

dr 1'2(53.7)

dr kr(53.8)

dr I' - 2m

where h and I.. are constants of integration.SUbstituting for d<p/dr, dr/dr from the last two equations and putting II = ~Il in

equation (53.6), it follows that

(dr)2 h2 2m('2_.. + ,(r-2m)=('2(k1 -1)+dr 1" I'

Then, eliminating dr between this equation and equation (5 1.7), we obtain theequation for the orbit, viL.

(h dr)2 h 2 2 2 21111'2 2mh 2

. + = I' (k - I) + +r 2 d</J r2 r r J

With 11 = 1/1', this reduces to the form

( dU)2 + u1= c: (k 2 _ I) + 2m~.2 U+ 2m1l"d</J h h

(5~.IO)

(53.11 )

(53.12)

(53.13)

Dillcrentlating through with respect to (p, this equation takes a form which is!;lIlJihar 11\ the theory of orbits. viL.

d 2 u me 2

I- II 'r ~mu2d¢2' = h2

The corresponding equation governing the orbit according tv classicalmechanics is

d 2 u GMd</J2 +u = ,;2

where M is the mass of theattracting body and h is the constant velOCity momentof the planet about the centre of attraction, i.e.

(53,14)

149

If we identify the time variable t of classical theory with the proper time r in therelativistic theNY, equations (53. 7)and (5.1.14) become identical and our ehoi(e ofIt for the constant in equation (53.7) is justified. Also, provided we take

(5315)

(5316)

(confirming equation (52.14)), equation (53.12) corresponds to the c1asscalequation (53.13), although there is now an additional term 3»1u 2

• The ratio 01 theadditional term 3mu 2 to the 'inverse square law' term me 2/h 2 is

3lt2

,,1 = 3 r2 J>1(.1 (.2

by equation (53.14). rJ> is the transverse component or the pillnet's velocity and,for the planets 01' the solar system, takes its largest value in the case of MerClry.viz. 4.H x 104 m/s. Since c = 3 x 10" mis, the ratio of the terms is in this easc i· 7x 10 -". which is vcry small. However, the clfect 01' the additional term provel tobe cumulative, as wIll now be proved, and Il)r this reason an observational chickcan be made.

The solution of the classical equation (53.13), vi/.

U - II J I + (' cos( '" - 6)) I- 1t 1 I .~, J (53. 7)

(53.JS)

(53.1~)

where 11= GM = mel, l' is the eccentricity 01' the orbit and Ii) is the longitudeofperihelion. will be an approximate, though highly accurate, solution of equati~n(53.12). Hence the error involved in taking

3ml11

3mu 2 = :, {1+ecos(</J-U))}2It

will be absolutely inappreciable, since this term is vcry small in any ca~e. EqUllticn(53.12) can accordingly be replaced by

d111 II 3mJ1l 2

d</J2 +U = Itl +h4 {1 +('COS(</J-U))}

This equation will posscss a solution of the form (53.17) with additioOiI'particular integral' terms corresponding to the new term (53.18). These prove tobe as follows:

(53.2~)

The constant term cannot be observationally separated from that alreadvoccurring in equation (53.17). The term in cos 2 (</J - (1)) has amplitude too smailfor detection. However, the remaining term has an amplitude which increaseswith </Jand its ellcct is accordingly cumulative. Adding this to the solution (53.17 1

,

(53.21 )

150

we obtain

11 { 3mJ1c }[/= 2 l+eeos(<!J-'I))+i <!Jsin(<!J-(IJ)

h h

11 (I • ".,=h2 ' +ecos(<!J-it)-lJw),

where h(l) = 3mW/J/h2 and we have neglected terms O(h(j)2).

Equation (53.21) indicates that the longitude or perihelion should steadilyincrease according to the cquati(\n

(53.22)

where I = hl /ll is the semi-latus redum or the orbit. Taking JI = I . 33 X 1020 51units for the sun. c = 3 x 10" and 1= 5·79 X 1010 for Mercury. it will be foundthat the predicted angular advance of perihelion per century for this planet's orbitis 43". This is in agreement wIth the observed value. The advances predicted forthe other planets are too small to be observable at the present time.

54. GravillUional deflcNioll of a lil:hl ray

In seetion 7 it was shown that the proper time interval between the tr'lnsmissionof a light signal and its reception at a distant point is zero. It was there assumedthat the signal was being propagated 111 an incrtial frame and hence that nogravitational lield was present. This result can be expressed by saying that

d., = () (54.1)

for any two neighbouring points on the world-line of a light signal. Now. nullgeodesics in the space time having metric (45.5) arc delined by equation (54.1)and the equations

(54.2)

for the three index symbols are all zero. Equations (54.2) imply that along a nullgeodesic x. y. z are linearly dependent upon f. But this is certainly tllle I()r thecoordinates of a light signal being propagated in an inertial frame. We concludethat the world-lines of light signals are null geodesics in space time. in this case.

Since an inertial frame can always be found for a sulliciently small space timeregion even in the presence 01''1 gravitationallield. it follows that the world-lineofa light signal in any such region is a null geodesic. We shall accept the obviousgeneralization of this result. viz. that the world-lines of light Signals over anunlimited region of spaee time are null geodesies.

We shall now employ this prineiple to calculate the path 01''1 light ray in thegravitational field of a spherical body. Taking the space- time metric in the5chwarl:schild form (equati(\n (52.10)). the equations governing a null geodesic are

151

identical with theequations (53.2) (53.5)arter r has been replaced by A. The firstintegral (43.13) takes the form

r (dr)2 {(dO)2 (d</l )2 1 ("2 (dt )2--- 1 +r2

. + sin 2() . (- (r-2m) . = ()

r -2m dA dll dll) r ,df.(54.3)

Without loss of generality, we shall again put II -= ~1!, so that a ray in theequatorial plane is heingconsidered and then proceed exactly as in the last sectionto derive the equatIOn

(54.4)

where u = I/r. This equation determines the I;imily of light rays in the equatori;ilplane.

As a first approximation to the solution of equation (54.4), we shall neglect theright-hand memher. Then

(54.5)

where R, ex <Ire constants of integratlDn. This is the polar equation ofa straight linewhose perpendicular dl';[ance from the centre of attractIon IS R. As might havebeen expected, therefore, provided the gravitational held is not too III tense, thelight rays will he straight hnes. This deduction is, of course, confirmed byobservation. Thus, as the moon's motion causes its disc to approach the positionofa star on the celestial sphere and ultimately to occult this body, no appreciabledeflection of the position of the star <\11 the celestial sphere can he detected.

Again, without loss of generality, we shall put ex = 0 so that the light ray,as given hy equation (54.5), is parallel to the Y-;lxis (</I = ± ~1!). Then. puttingu = cos<p!R in the right-hand memher of equation (54.4). this heeomes

d 211 3m 2d</l2+ U = R2 cos <I> (54.6)

The additional 'particular integral' term is now found to he

m2 (2 -cos 2 </I)

R(54.7)

(54.8)

and hence the second approximation to the polar equation of the light ray is

I lit1/ = R cos </I + R2 (2 --cos2 </I)

At each end of the ray II = () and hence

nl 2 2m- cos </I - cos </I - - = ()R R

(54.9)

152

Assuming m/R to be small, this quadratic equation has a small root and a largeroot. The small root is approximately

cos <P =2m

R(54.10)

and hence (n 2m)<P=±2+R (54.11)

(5412)R

at the two ends of the ray. The angul:lr detlectiol1 in the ray caused by its passagethrough the gravitational licld is accordingly

4m

approximately.For a light ray grazing the sun's sllrlilce.

R = sun's radius = 6·95 x 10" m and m = 1'5 x 10 3 m

Thus the predicted deflection is 8·(,: x 10- 6 radians. or about 1·77'". Thisprediction has been checked by observing a star close to the sun's disc during atotal eclipse. The experimental findings arc in accord with the theoretical result

55. Gravitational displacement of spectral lines

A standard clock will be taken to be any device which experiences a periodicmotion. each cycle of which is indistinguishable from every other cycle. Thepassage of time between two events which occur in the neighbourhood of theclock is then measured by the n umber of cycles and fraction of a cycle which thedevice completes between these two instants. The clocks employed to determinethe time coordinate ~4 of an event ill section 45 were not, necessarily, standardclocks. Such coordinate cloch can have arbitrary variable rates, the onlyrequirement being that, if A, B arc two cvents in the vicinity of a coordinate clockand B occurs aner A, then the coordinate-time lor 13 must bc greater than thecoordinate-time for A.

The successive oscillations of atoms governing the ml,tion of a modern atomicclock arc indistinguishable from onc another and it has heen assumed that such aclock is being used whenever standard time is measured. The constancy of the rateof this fundamental physical process is not susceptible to experimental check,since it is the standard against which all other rates (e.g. the rate of rotation of theearth) arc measured. By international agreement, one second is the time whichelapses when a specifiC type of atomic system perlorms a specilic num bel' 01oscillations and this definition applies in all regions of the cosmos and at allepochs; this fact should be borne in mind when phrases such as 'the IIrst secondafter the big bang' occur in cosmological studies.

153

As explained in section 45, if x', x' + dx' arc lhe space time coordinates of twoadjacent events, then dr = ds/ie is the time separating the events as measured byastandard clock which is presenl at hOlh events. It is assumed that the intelvalbetween the events is timclike (i.e, d r real) and that the clock is in a state offrce fallduring its passage fwm one event to the olher; alternatively, if the clock is notfreely t~dling. it is assumed that any effect on its rate of the gravitational lieU itexperiences is corrected for.

Let x' (i = 1.2.3.4) he the coordinates of an event with respect to semespace time reference frame, x' ..\2, x' being interpreted physically as spatialcoordinates rela tive to a slatic frame and \4/1l" as lime. Ifa standard clock is at restrelative to this trame, 1'01 adjacent poinb on its world-line dx l = dx 2 = (k' = 0and hence

(55.2)

where we have put x 4 = ief. The time r measured hy the standard clod istherefore related to the time f shown on the coordinate clock at (Xl, x 2

, Xl) bylheequation

r = fJ «I44)df

In the special case of the coordinate frame employed in section 48 which wasstationary in a relatively weak static gravitationaltield, it was proved that 944 isgiven in terms of the Newtonian scalar potential U for the field by theapproximate equation (48.17). Thus

(2U )1/2

dr = 1+ e2 df (553)

relates time intervals measured hy a stationary standard clock and a coordirateclock at a point in a gravilationalfield where the potential is U. Now, when :t isemitting its characteristic spectrum. an atom is operating as a standard clock.Consider, therefore, an atom for which the period (from standard tables) of onecomplete cycle of radiation corresponding to a certain spectral line is r. If such anatom is stationary in the frame at a point P where the potential is U I' the time forone complete cycle of the radiation as measured by a standard clock at the pointwill be r and the coordinate-time for the cycle will be f, where

(55.4)

Suppose this radiation is received at another fixed point Q where the potential isU2' Let Tbe the difference between the coordinate-time of emission of light fromP and the coordinate-time of its reception at Q; since the gravitational field isbeing assumed static and P, Q are t1xed, T will be a fixed constant. Thm, itsuccessive crests of the light wave are emitted from P at coordinate-times fo,10 + I, these crests will be received at Q at coordinate-times 10 +T, fo +T + I. Itfollows that the period of the radiation as measured hy the coordinate clock at Q

154

will also be 1. Htlwever. a standard clock at Q will measure the period to be r',where

(55.5)

(55.7)

Hence, if \. is the standard frequency of the spectral line being observed and v' isthe observed frequency of the line at Q. then

I,' = r = (I +2(11/<,2 )1/21 (55.6)

r' I + 2V z/c

In particular, jf V I < V 2 , the observed light will have its frequency shiftedtowards the red.

In the case of an atom nn the surfilce of the sun observed from a point on theearth\ surface. it will be found that, in SI units.

VI = -1.914x 1011. V 2 = -9.512x 108

v' = 0.99999791'and thus

This effect is so small that it is very difficult to measure. i1owever. in thecasc of thecompanion of Sirius. the predicted c!kct is 30 times larger and has beenconfirmed by observation.

56. Maxwell's equations in a gravitational field

Over any sufficiently small region of space and restricted interval of time it ispossible to define a rectangular Cartesian inertial frame. i.e. the frame in 'free fall'in the gravitational field. If the electric and magnetic components of theelectromagnetic field arc measured in this frame. the field tensor Fii defined byequation (26.5) can be found. Employing the appropriate transformationequations. the components of this tcnsor relative to general coordinates Xi in thegravitational field can be computed. No distinction is made between covariantand contravariant properties relative to the original inertial frame so that. whentransforming, F iJ may be treated as a covariant contravariant or mixed tensor. Ifit is treated as a covariant tensor. the covariant components Fu in the general x'·frame will be generated. Ifit is treated as a contraval iant or as a mixed tensor. thecontravariant or mixed eomponents [-'J, F; respectively will be generated. In thisway, the field tensor is defined at every point of space- time. Similarly. a current·density vector with covariant componellts J, and cOl1lravariant components J'isdefined relative to the xi·frame.

Consider the equations

F~; = 110 Ji

F'l.k + F,k" + Fk,.i = 0

(56.1)

(56.2)

These are tensor equations and hence are valid in every space -time frame if theyare valid in anyone. But. relative to the inertial coordinate frame (x, Y. Z, icc)

155

which can be fonnd for any sutliciently small space time region, these equationsleduce to equations (26.11) and hcnce arc valid over such a region. Regardin!! thewhole of space time (1$ an aggregate of such small clements, it follows thatequations (56.1). (56.2) arc universally tlue.

Since F,j is skew-symmetric,

. ~Fu . .F'J = + I , 'F" + I J I F"./ t1xJ (r J' Ir J'

(56.3)

(56AI

by equation (42.5) (4 has been replaced by - 4, since 4 is always negative for a realgravitational field). Equation (56. f) is accordingly eqUivalent to

I (' .

J ( ) ' .1 : J (- 4 )F'1: = 110J'-- 4 I X

Also, in view of the skew-symmetry of the tleld tensor, it follows that equation(56.2) is equivalent to

(56.5)

Theenergy momentum tensor for the field is found from equation (29.5) to begiven by

Jlo S; = F;k F 1k - ~ 6; F kI F kl

It now follows from Maxwell's equations that

S(.J = - F;J P = - D,

where D, is the 4-force density acting upon the charge distribution.

57. Black holes

(56.6)

(56.7)

The Schwarzschild metric is only valid in the region outside a sphericallysymmetric attracting body. Thus, if the radius of this body exceeds 2m, thecircumstance that the component 411 of the metric tensor becomes infinite atr = 2m creates no difficulty. If, however, the body's radius is less than 2m, thesphere r = 2m lies in empty space and the nature of the field in the vicinity of thiSsphere needs careful study.

Although the metric is clearly invalid over the sphere r = 2m, it is an acceptablesolution of the Einstein equation in the region 0 < r < 2m. Consider a bodymoving radially in this region, not necessarily in a statc of free fall. Then {land q,

156

will both be constant and the metric equation reduces to

c2dr2 = ~- 1 (dr2 _ ('2 ~2 dt2) (57.1)

where ~ = 2m/r - I > 0, along the body's world-line. Given the equation ofmotion I' = r(t), this equation determines the proper time r shown on a standardclock moving with the body. But dr must be real and it follows that either (i) dr/dt> ca or (ii) dr/dt < - ca. These inequalities show that it is impossible for a bodyto be stationary relative to our coordinate frame in this region. This implies thatour picture of the framc as a set of coordinate clocks measuring the time t andstationary at the points (I'. O. 4» ceases to be applicable. Evidently, the staticconditions we have been envisaging in the neighbourhood of the attracting bodyarc not present in this region.

Next, consider a body tailing freely along a radius towards the centre ofattraction in thc region I' > 2m. Taking as initial conditions t = 0, I' = R,dr/dt = O. equations (53.5) and (53.6) lc,td to the equation of motion

(57.3)

(57.2)

Thus

(dr)2 = 2mc2 (I _2m ') I (I _~m)2 (1- I)dt R ) I' V R

(R )11

2 f" rJI2

dl'ct = 2~ - I .' (;=2~~)(R__ ~)TI2

and it is clear that this integral diverges to + (XJ as I' -> 2m. This means that, in theSchwarzschild frame, the body willnccJ an intlnite coordinate time to reach thesphere I' = 2m. If the body is obscrved optically by an observer stationed at aconsiderable distance from the centre ofattraction, since allowance must be madefor the coordinate time needed for photons leaving the body to reach histelescope. the observed motion of the body, as measured hy his coordinate clock,will be further retarded. But his coordinate clock will be almost indistinguishablefrom a standard clock and it follows th,lt the apparent time of fall of the body tothe Schwarzschild radius according to an external obserwr uSlllg an atomic clockwill also be infinite.

If. however, instead of eliminating the proper time r bl-'tween equations (53.5)and (53.6), the coordinate time t is eliminated, the resulting equation is

(~:y= 2mc2C-~)

After integration with r = 0 at I' = R, this gives

er = J(R 3 /2m)[ J(p _1'2) + j cos- l (21' -I)J

(57.4)

(57.5)

where p = r/R and the inverse cosine is taken in the first or second quadrants. rwill be the time recorded by a clock moving with the body and equation (57.5)shows that this remains tinite for values of I' through the value 2m to zero.

It is now evident that the reference frame we have been using is unacceptable ifmotions across the Schwarzschild sphere arc to be studied and that, in particular,

157

the coordinate time r becomes infinite at r = 2m for some events which can occurin the experience ofcertain observers. I t appears, therefore, that it is a deficiency inthe reference frame which is re.'iponsible for the anomaly in the metric and our.expectation is that the infinity can be removed by transformation to a new frame.This view of the matter is supported hy the fact that (} is finite at r = 2m, indicatingthat there is no singulilfity of space time in thIs region.

The suggestion arising from our calculations is that r sh0uld be replaced by anew coordinate time /I defined by a transformation equation

/I = r +/(r) (57.6)

where! (r) hecomes negatively infinite at r = 2m in such a way as to cancel theinfinity which we have seen to arise in r for certain events taking place on r = 2m.Substituting

dr = du -J' (r)dr

in the Schwarzschild metric. this transforms to

(57.7)

where

rF =_..r - 2m

( 2 /2(r - 2m)/

r(57.9)

We can now remove the inl1nity by choosingJ(r) such that

c/'=r/(r-2m)

Thus, we take

e/(r) = r + 2m log (r - 2m)

and the metric then assumes the form

(57.10)

(57.11)

(57.12)

This metric must clearly satisfy Einstein's equation in l'uello. However, the fieldin the new frame is no longer static in the sense assumed in section 51; the presenceofa term involving the first power of dll shows that the field is not symmetric withrespect to the paq arid future, i.e. the sense of description of its trajectory by afreely falling particle cannot be reversed with impunity.

Let us study. once again, a body moving radially, but not necessarily fallingfreely. Along its world-line, we have

d.~2 = 2edrdu - el (I - 2m/ r) dul

Since ds l must be negative for any possible motion,

dr< k(1 -2m/r)

dll

(57.13)

(57.14)

158

If I' < 2m, this implies that dr/du is negative and that the hody must move towards0; in particular, it cannot remain stationary. Thus, this is a region of irresistiblecollapse towards the centre of attraction for all physical bodies. It will be notedthat the transformation has eliminated the possibility of outwards radial motionwhich existed when the Schwarzschild form of the metric was taken; thispossihility can be recovered by changing the sign of u.

Now consider the motion of a hody filliing freely along a radius from an initialstate of rest dr/du = 0 at I' = R > 2m. Since i141 = C, 1144 = - c2 (I - 2m/r),equation (53.5) must be replaced by

d (dr dU)-,(I .. 2m/I') = 0dI dI dI

Together with the lirst integral

drdu > (dU)22e- - c" (I -·2m/r) --dI dI dI

this leads to the following quadratic f(lr dr/du:

( dr )2 +4mc(I_- 1 Jdr +2/1/C2(1-.2rm)(RI _ 1'1)=0

du I' R I tiu

The roots are

(57.15)

(57.16)

(57.17)

(57IH)

If r > 2m, one root is positive and onc is negative. However, I' must dccreaseiniltally (otherwise the square root in (57.1 H) hec"IlWs imaginary) and so thenegative root is taken. I' then decreases steadily to I' n, iU, passage through theSchwarzschild radius being unremarkable. Once inside the Schwarzschild sphere,as already proved, the possibility of escape from the attr;H:tion no longer exists.

The world-lines of photom moving radially arc null geodesics governed hy theequation

2cdrdu -e2(1-2m/r)du 1 = 0

There are two families of such geodesics. viz.

(57.19)

and~.u = 0dr '

For the first family, equation (57.7) gives

du('

dr

21'

r-2m(57.20)

dlc

dr I' -2m(57.21)

provided I' > 2m. This corresponds to a photon moving towards the centre of

159

attraction. h)r the second family. we find

dlc

dr r -2m(57.22 1

in the same region; i.e. a photon moving away from the ccntre of attraetion. Aphoton hdonging to the first family crosses the Schwarzschild sphere and thellfalls into O. Insidc thi, sphere. the photons can he separated into two classes(i) those for which U IS constant along a world-line these photons could havetheir source ou tSlde the sphere: (ii) thme f()l which the sccond ofequations (57.201is valid and, hence,

C1J = 21' + 4/11 log (2/11- r) + constant (57.231

Asr ..... 2m, u ..... - '""j and these photons cannot have had an external source. SincedlJ/dr < 0, these photons also 11111 Into O.

It is now clcar that, In the field dc,crrhed hy the metric (57.12), no photon OtpartIcle can cross the Schwarzsclllid sphere in the sense I' increasing. On the othelhand, any photon or partIcle which crosses the sphere in the reverse sense i~

ahsorhed and cannot return to the external world. The conditions inside thesphere <Ire accordingly relCrred to as a Mack holc. It is thought possihle h)astrophysicists that some stars may have collapsed under their own gravitationaattraction to a radius less than theIr Schwarlschild radius. In such a case. al

explained ahove, further contraction would hewmc irresistihle and the stalwould collapse to a singular point having infinite density. Such a collapse woul~

require an infil1lte time hy terrestrial clocks so that. as,uming the age of thecosmos to he limtc, It might he ohjected that no such objccts can yet have comeinto existence. However, the idea "l' a cosmos of present events, all happenin~

simultaneously relative to some universal time scale, is quite foreign to relativit)theory, so that the ohJect 1<\11 i, mcanlllgle;,;,. The hard fact is that the possibIlity O'a space,hip hIlling into the hlaek hole created hy such an ohJect, in a Ilme whIch nlinite measured hy an on··hoard dock, is a real one. A few cases of ohJects whidappear to he in the early ,t<lge, of gravitational collapse have already beendetected.

If the metric (57.12) is transformed by changing the sign of lJ, another metricsatisfying Einstein's equation is generated. A similar analysis shows that thilgoverns the field in the vicinity ofa while hole, where matter and photons can onl~

cross the Schwarlschild sphere in an outgoing sen'e. Thus, a white hole behavelas an irresistible source and a blaek hole as an irresistihle sink. Being invarianlunder a sign reversal of I, the Schwar/.schild metric permits a black and a whitehole to exist together.

SIl. Gravitational waves

Thf<\ughout this sect ion it is assumed that the gra vitationailleid is weak and thaithe coordinates x' are quasi-Minkowskian, as explained in section 4X. Thus. th(

160

metric tensor is given by equation (4K.3) and terms of second or higher degree inthe hi) or their derivatives will be negleeted. We shall further suppose that thecoordinate frame i~ harmonic (see Exercises 5, No. 50). so that the metric tensorsatisl\cs the condition

(58.1 )

It can be proved that a transformation of the form Xi = Xi + ~i(X), where thefunctions ~' are small with the hi}' can always be made so that the x-frame isharmonic (see Exercises 6. No. 37); this means that t he harmonic coordinates willalso be quasi-Minkowskian.

To the first order, equation (58.1) reduces l()

[ii,k] = h'k.i -1h,;" = 0

DifTerentiation leads to

Exchanging indices j, k and adding the new equation to (58.3), we i-tet

hij, ik + hik . ij - h", )k = 0

(5K.2)

(5K.3)

(58.4)

The Ricci tensor has already been calculated to the first order ofapproximationat equation (49,2). Using the last result, thiS gives

Also

(5K.5)

R = Rjj = 11r,;."

Thus. Einstein's tensor is givcn by

R jk -hljk R = 1hjk ,ii --!'ljkh"'ii = ihJk,ii

where

(5K.6)

(5K.7)

(5K.K)

Einstein's equation of gravitation is now expressible in the form

02hjk = hjk' ii = - 2KTjk (5K.9)

The harmonic condition (5K.2) can also be written

h;k.i = 0

In empty Sp;\l'C, equation (5K.9) reduces to

(5K.IO)

(5K.ll )

which is the wave equation, showing that gravitational waves are propagated in"/lClIO wit h the velocity of light.

In the case of a plane wa ve, we can write

h;k = AJk exp Uk,x,)

161

(58.12)

where it is understood that the real part of the complex exponential is to be taken.Equation (58.11) is satisfied provided

k,k, = 0

and the condition (58.10) also requires that

k,A'k = 0

(58.13)

(58.14)

Since A'k is symmetric, the last equation shows that the A i4 can be expressed illterms of the A,'II (ex, If = 1,2, 3). By further transformation of coordinates, it callbe shown that all amplitudes can he expressed in terms of two parameters onlyand hence that gravitational waves have, esselllially, only two modes ofpolari72t ion.

The solution of the wave equation (58.4) with source term -2hTJk is wellknown to he given by Kirehholl"s formula (Bateman, 1952):

(58.15)

where Xo = (xl" x~, v,:), x = (x' , x 2, Xl) are posit ion veet ors wit h respect to the

ongin 0 of the frame in use, and /. ,~ IXo _. x I is the distance bet ween these points;Vis the region of space over which 7~k is non-vanishing. Note that '0 is retarded inthe integrand hya time,Ie, since the cllect of the source at x will not he felt at Xountil the time f(lr its transmIssion over the distance,. has c1apsed. In the case ofasource which is contlned to a small region of space including the origin 0, if'0 ~·Ixol is large compared with the dimensions of this region, equation (58.15)

can be approximated by

(5816)

But, as explained above, the components h;4 can be obtained easily once the iI;~

have been found. We shall now show that a further simplification of the lastformula is possible in these cases.

hrst note that

Since the divergence of T,j vanishes, we ha ve

7~i.' + T,4.4 = 0

and thus

(58.17)

(58.18)

(58.19)

162

Integrating over the region V, the integral of the left-hand member is seen to bezero by application of Gauss's divergence theorem (assuming T,r vani"hes overthe bounding surface); we accordingly obtain the result

(58.20)

(58.21)

Exchanging the indices a, If and adding the new identity to (58.20), we lind

f i d fT,Jld V = - (7~4XJI -+ 7ji4x')d V2cdT

We next integrate the identity

(Ty4 x'x Jl )." = T4y"x'x P+T'4XJl +7p4 x' (58.22)

over V. By the divergence theorem, t he integral of the lert-hand member vanishes.Hence

f(T,4XP+7p4X')dV= - fT4/.rX'XlldV

Equations (58.21) and (58.23) now yield the resnit

(58.23)

(58.24)

where we have again made use of the equation 7;j.) = O. But, equation (21.14)shows that T~4 = - )1('2 and equations (58.16) and (58.24) thercrore lead to thefinal result

. _ 2(; d 2 f _I. '.ph.p(Xo,T)- 4 2 )1(X,T 'o,<)x X dV(' 'II liT

(58.25)

It should be noted that it is the second time derivative of the .'('(oml moment ofthe mass distribution which is responsible for the gravitational wave. In thecorresponding electromagnetic situation, it is the second time derivative of thefinT moment of the eharge distribution which is responsible for the elec­tromagnetic wave.

In~truments have been devised to detect the small variations in the gravi­tational field caused by waves proceeding from possible sources within the galaxy(e.g. pulsatIng neutron stars, binary stars or Sl1pernova explosions), but no clearlyunambiguous results have yet been obtained. Such instruments allempt tomeasure the small strains ind uced in very large masses of metal by the tidal forcescaused by the passage through them of gravitational waves.

163

Exercises 6

I. Ifthc v-frame is defined a~ in section 47, show that the metric tensor in thc x­frame is given by

i'\,k(~Vk

q'j = ,<\1 '-<Xl

Hence lower the index) in T'! defined by equation (47.2) and show thatt he result ISr; as defined in equation (47.3).

2. Given th,lt space time has the metric

d.,2 = d x2+ d y 2+ ,,2t'd: 2 _ ,,24>d,2

where /I, q, arc function" pf: only, prove that the Riemann Christofrel tensorvanishes if, and only rt,

where dashes denote drtrercntiations with respect to :.1fql = - 0, prove that thespace time is flat provided (I' j log (/ + h:), where (/, h are constant,.

3. If space titHe ha, the metrIc

d.,' = ,,'(d,-' f d: 1)+ ,-',. "d</I' ._,.1'£1,'

where), I' arc functiol1s of,- and: only, ,how that the tiekl equatIOns in emptyspace R'j 0.0 require that A and I' should satIsfy the equatlOm

) I -j 1'1 '0 )r(l'; -- I'i)

;., + 1', '0. '-1',1',

I'll +-1'22 + ~ 1','= ()

)'1 , + ) 12 +- 1'1' f I'll + I (I'; -+- II}) ,~ 0

where subscript, I and 2 denote partial difl'erentiations .... ith re,pect to rand:respectively.

4. H space time has the mctric

d,,2 = (' 2" (d \ 2+ £I y 2 -+ d: 2 _ d, ')

where" is constant, and r' = X' + .1,2 -+- :", dots denoting difrerentiations withrespect to " show that for a freely falling body

I -I'" = (1·- V1je'"

where I' = V at x = 0,5. If space time has the metric

ds 2 = a 2 (dx' -+ dv 2 -+ d: 2) c'ad,2

where a = 1/ (\ - kx) and k is eon,tan!. and I' is as defined in the previous exercise,

164

prove t hat for a freely falling body

V2 _r 2 =kc 2 x

where l' = Vat x = 0-fl, The space time metric over a certain region of empty space is

ds ' == c"(dx 2 + dv 2 + dz') ePdr'

wherr a.11 8re functions of z alone. Show that Einstein's equation is satisfiedprovided

a" +-la".f ~a'll' = ()

a" + ~/I" +-l/l" -la'li' = 0

II" + SlI'2 + la'll' = 0

Deduce that c" = A (I<. - Z)4. ell == B(k -:) '. where A. B. k are constants.7. Show thai the space time metric

ds ' == c"dr' + r'dO' ·t cPdz' - cl'dr 2

where r. O. : are quasi-cylind rical polar coordinates and r is the time and a.11. yarefunctions of,. alone. sat is lies FInstein's equat ion ill l'aCl/II. provided

fI' +- y" + 1/1'2 + h,l - ~a'-1a'll' - ky' == ()

a' = II' + 1"II" + 1/1" - ~a' II' +Wy' + ~ II' = 0

y" + }y'2 - !a'y' + !/1'y' .f ~ y' = 0

dashes indicating diiTerentiations with respect to r. Deduce that

1'" == AI" () f "I. eP == HI' - ". cY = Cr "

where A. B. C. ,l.11 arc constants and ,lJl = 2(,l + p).

X. A certain I~gion of space time has metric

d.\' == dx' +dy' + d;;' - x 2 dr 2

A particle is stationary at the point x = I . .I' = z = 0 at r = O. If the particle isreleased at this instant and falls freely, show that it moves along the x-axIs withequation of motion x = sechr. A photon is emilled from the point (1,0, O)atl = 0In the direction of the positive y-axis. Show that at this instant x == z == 0, .I' = Iand that the path of the photon is the circle x' + .1" =,. 1.

9, De Sitter's universe has metric

ds 2 = A - Idr' + r 2(d1l 2+ sin'lIdl{;2) _ A,,2dr'

where A == I - 1'2/ R2, R being constant. At r = 0, a photon leaves the origin I' == 0and travels outwards along the straight line 0 == constant, <p == constant. Find itscoordinate I' at time ( and show that I' = ! R when ( = R (log 3 )/2l' and that I' -+ Ras ( -+ rx).

16)

10. 1', /I, z arc quasi-cylindrical coordinates in a gravitational field determinedby the metric

ds' = 1'2(d1'2 + d/l 2) + l'(d;;2 _ d(')

A particle is projected from the point I' = 1, /I = 0, Z = 0 in the field with suchvelocity that r = i = 0, /I = j3/2 (dots denote differentiations with respect to I).Prove that. if the particle falls freely. it moves in the plane;; = 0 between thecircles I' == 1, ,. = 3, first touching the outer circle where () = j 3n. A photon isemilled from the pornt I' = 1.0 = 0,;; = 0 and moves initially so that r = i = 0,Prove that its path is the spmtl I' =- 1+101 in tile plane;; = 0.

11. The metrIC for de Siller's universe can he cxpressed in t he form

d.,2 = 1'1""(<lx2 +dy'+d;;2)_("d('

where R is a constant and x, .1'.;; can be treated as rectangular Cartesiancoordinates. Show that the trajectories of freely falling particles and photons arc,traight lines. A particle is prolet:ted from the origin at I = °with a velocity Ialong thc positIve x-axis. Prove that lis x-coordinate at time I is given by

Vx = R [(' - Vi (1" -- V' + 1'1' '''")]

A body at the point x = X Oil the x-axis emits a photon wwards the 0rigin atR

I = O. Show that the pl10t011 arrives at 0 at time I = - log (I - X / R). Discussc

the case where X > R.12. ,.. /I, </J are quasi-spherical polar coordinates in a gravitational field which is

spherically symmetric about a centre ofallraction I' = O. The spaee--time metric is

ds' = ( I' __ )' dr' + r'd/l' + r ' sin1/1d1)1 __!.. dl 11'+1 1'+2

A particle is projected from the point I' = I. /I = !n, </J = 0 at I = °with velocitysuch that': = 0, () = O.;p = l/j6 and falls freely. Show that the particle'strajectory lies in the plane /I = !n and has polar equation

5 - cos(u</J)1'=

3 + cos (u</J)

where u = j(R/3). Deduce that the particle moves between two circles of radii 1and 3 and calculate the increment III </J between two successive contacts with oneof these circles. (Ans. 2n/1l.)

13_ Taking the metric for de Siller's universe in the form stated in exercise 9.filld equations of motion for a particle proj<"'l:ted from the point I' = !R.

I ) . J3 C • I'/I = 2n, </J = °with such velocity that r = I = o. </J = 2 R and thereafter ails

under gravity. Show that its trajectory lies in the plane /I = !n and that its polar

166

equatioll is

,. = R;(5cos1rj> _ 1)112.

14. Oxyz is a quasi-rectangular Cartesian coordinate frame comtructed in.acertain gravitational field. If t is the time measured by a system of clocksstationary in the frame, the space time metric isds1 = z(dx 2 + dy' + dz 2

-- dt 2). A

particle is pl(ljected from the point (0, O. I) at t = 0 with velocity components.X = 1'( < I), .i' = z= 0 and thereafter flllis freely. Show that its trajectory is aparabola and hes in the .\Z-plalle and that the particle arrives at the xv-plane attillie t = 2/jll - 1,2).

15. Show that Einstein's equatioll (47.10) can he written in the form

16. Inside a static gravitating homogeneous sphere of liquid. the properdensity is jJ (a constant) and the pressure is 1'. The energy momentum tensor Itaszero components except for r: = I ~ = r ~ = p. I ~ = - ,.'!'- Assuming that themetric of the field inside the sphere is givcn hy equation 151.10) with (/ = ,,'

and h = "P, show that Einstein's equation (47.15) can he satisfied by making (f.. IIand I' satisfy the equations

I11'= (,,"--1)+/\1'''"1'

I'

dashes denoting differentiations with respect to 1'. A~suming (J. = 0 at I' = 0 andI' = 0 at I' = (/ (the surface), deduce that

where 4 = /\C'II/:', and that

(I -- qr')112 - (I _ (fa 1 )'-'I' = c2JJ .-- -.--.--- - -

3( I -- 4(/1)1 11 - (I _ 41'2)1 /i

17. Obtain the equations of motion of photons Inovll1g radially inside theSchwarzschild sphere and deduce that a photon moving away from the eentrc 0takes an infinite coordinate time t to reach the sphere and a photon movingtowards the centre from I' = R ( < 2m) takes a time t = I given by

c1 = - R - 2mlog( 1- R;2m)

to reach O.ilL Obtain equations (57.2) and (.q.:.) for a body falling freely towards thc

centre of attraction in the region I' > 2m and hence prove that

(R )112[ J (I-T\1'f = - I .J lr(R - I'll + (R + 4m)cos- I (r/R)112 - 2mlog ---

2m _ _ 1+1')

where

)' = [2m(R -1')1 112

r(R -2m)

Deduce that f -+ ex, as I' -+ 2m.19. Obtain equations (57.4) and (57.5) and deduce that the time recorded 0

standard dock attached to a freely tailing body as it falls from the Schwarnchsphere to the centre ofattraction is 71m/c. Calculate this t)me in the case ofa bluhole having solar mass. (Ans. 16 ).Is.)

20. Verify that the metric tensor given by equation (57.12) satisfb Einsteiequation in empty space.

21. Show that the Kruskal Szekcres transformation

11 = (r/2m-ljI12e,,4mcosh(cf/4m)

I' = (r/2m- ljI 12e'14m sinh(u/4m)

converts the Schwarzschild metric to the form

12m]ds 2 =: (' ,,2m(du2 _ dl,2) + r 2(d1l2+ sin 2I1d<1J2)

I'

where I' is given in terms of 11, I' by the equation

u 2 - 1,2 = (r/2m - I )e'12m

Deduce that the world-lines of radially moving photons are u ± v = constan22. Show that the transformation

2u = {, + ' __ 1']1 2

3a

1/2 2 1'1 12 + av = f + 2ar - a log --_.~

1'11 2 - a'

where a2 = 2m, puts the Schwarlschild metric into the form

where ).13 = 9a 2/4.23. A photon is emitted from the point r = m, /I == 171, <1J == 0, inside a blae

hole (Schwarzschild coordinates) with angular velocities () = O,;P = 3J3c/IShow that f = ± 2,/71' initially. In the case when the initial value of f is negativ

168

show that the photon moves in the plane {J = 171 and falls into the centre along thetrajectory

6m 21-- = :1coth }(ex.--</J)-II'

where ex. = logcl(5+ y '21).24. Show that the only possible circular orbits for a photon in a Schwarzschild

Held all have radius I' = 3m and that their period in coordinate time is 6v /:171mjc.Show that these orbits are unstable.

25 A body moves in a circular orbit of radius I' in the plane /I = 171 in aSchwarzschild field. Show that I' > 3m and that the angular velocity d</J/dt isrelated to ,. in the same way as in classical theory. Show that the period of themotion as measured by a standard clock attached to the body is

2711'('-_:1)1c m

Show, also, that the period as measured by an observer using a standard clockwho is stationary at some point on the orbit is

2nr (I' _ 2)~c m

Show that the orbit is unstable if 3m < I' < 6m, but is stable otherwise.26. 1',/1, </J are Schwarzschild coordinates. A fixed observer at the point R, /I, 1>

transmits a wireless signal radially towards the attracting body. The signal isreflected by asmall body at the point 1',/1, </Jand returns to the observer. Show thatthe time elapsing between transmission and reception as measured by theobserver's standard clock is

2 12( R-2m)-(I-2m/R)/ R-r+2nlklg··--- ---c I' -··2m

Calculate the distance covered by the signal and deduce that, according toclassical theory, the time for the double journey would be

2[ 2 "2 2 12 R I/2+ (R -2m)I/2](R -2mR) . - (I' -·2mr) , + 2mlog-- ,--- ----.-

C ,.1/- + (I' --2m)I/2

Show that, to the Ilrst order in m, the diflerence between these times is

~!ll(IOgR+ I' -I)(' I' R

(Note: This result suggests a method of checking the general theory using theSun's field and Mercury or Venus as the reflector.)

IJIf'~

*

1.

I~.

I

27. An atom, which is statinnary at a Schwarlschild conrdinatc disunce r 1'1the ccntre of a sphcrically symmetric body, emits light of frequenCj I' whicobserved by a stationary observer at a coordinate distance R ( >") fromcentre. Show that the obscrved frcquency is I' - ,h" where

'I I)M/l' = mC - R

to thc first order in m.

2R. By replacing the spherical polar cnordinate r oceurrilg inSchwarlSehiid metric (52.10) by a ncw coordinate r' where

( m)'r =,.' I +. 2(

obtain this metric in 'isotropic' form, vi?

ds 2 = (I + /11.)4 (d(2 + ,.'2d/l2 +(2Sin2/1d</>1)_(I-m/2r:)"(2dt22,. l+m/2r

29. Employing a ccrtain frame, an evcnt is specified by spatial roordin<(x, y, z) and a time t. The corresponding space time manifold has t1etric

ds l = dx' +- dr' +- elz 2 + 2atdxdt - (1'2 -- a 2 t 2 )dt 2

Show that a particle blling freely in the gravitatiouallield observed ill the frahas equations of motion

x=A+fJt-iat 2, y=C+Dt, z=E+Ft,

where A, fJ, C, D, 1:, F are constants. By transforming to coordinates (x', .1', z

where x' = x + jat 2, and recalculating the metric, explain this result

30. (x I, x 2, x-') are spatial coordinates of an event relative to a frameS and x

the time of the event measured by a clock in S. A second frame 1 is litlli~g freel)the neighbourhood of the event and may be regarded as inertial. 0 1,1// .rectangular Cartesian axes in 1and 1,4/ie represents the time within 1 a~ meaSllIbysynchroni/.ed clocks attached to the frame. Show that f/'j' the metric lensor iris given by

(1 Vk (ll,k

!l1/ = ~1-\"1 ,<;r

P is a point, fixed in S, having coordinates (Xl, Xl, x J). At the instant x 4

, 1chosen so that P IS instantaneously at re.st in I. Deduce that

(Cy4 ~li4

(,,(1 J(!J44)

at x'. dl is the dIstance betwecn P and a neighbouring point

P' ( \ I +- d\ I, .\2 +d\' ,\,\ +- dx 3 )

170

as measured by a standard rod in 1 at the instant x·. Prove that

dt> = dy'dy' = )'J"dxJdx"

where ct, J., J1 range over the values 1,2, :1, and

~,_ _ ?/)4!1 JI4J-I/I-(.JA/I

!/44

(i'", i, the metric tensor for the ,if, which is S at the instant x•. ):11. Oxyz is a rectangular Cartesian inertial frame I. A rigid disc rotates in the

x\,-plane about its centre 0 with angular velocity w. Polar coordinates (I', II) in aframe R rotating with the disl: are defined by the equations

x = reos(1I +wr), .I' = rsin(1I + wI)

where r is the time measured by synchronized clocks in the inertial frame. If thetime o I' an event in R is taken to be the time shown by an adjacent clock in I, showthat the spal:e-time metric associated with R is

ds 2 = dr 2+ r2dll 2+ 2wr2dlldr - (c2 -r2o/)dr 2

Deduce that the metric for geometry in R is given by

r2dll 2dt 2 = dr2+ -. - -

l-o/r2Jc2

(Hilll: employ the result of the previous exercise.) Hcnce show that the family ofgcodcsics on the disc is determined by the equation

. (11) 11II =consl.-sm' , I' -r2J(r2-112)

where 1', = c/wand lal < 1'1' Sketch this family. What is the physical significanceof I',?

:12, x'(i = 1,2,3,4) are three space coordinates and time relative to a referenceframe S. A test particle is momentarily at rcst in S at the point (Xl, x 2

, x 3) at the

time x.. If .iJ'l is the metric tensor for the gravitationallidd in S, write down theconditions that the world-line of the particle is a geodesic and deduce that

!/ .el 2x' = I ((~iJ.4 + !Ii". ()•• ) _ DY;4

. "(dx·)2 2 h:' II.. ?x· ,]x.at the point Xi, Hence show that the covariant components of the particle'sacceleration in S are given by

d 2x P au 2 1/2 (ly,

Y'P ld';.j2 = '. ;l~-' - (c + 2U) i1x.

where y,p is defined in exercise 30 and

!/44 = - (('2 + 2U), y, = !/,./ ,,/ ( - !/•• )

171

(U, )', are the gravitational scalar and vector potentials respectively.)Show that, in the case of the space- time metric appropriate tn the rotati~g

frame of exercise ~I, the gravitational vector pntential vanishes and the scalelrpotential is given hy U = ;(1)2 r 2. Interpret this result 111 terms of the centrifu~al

force.~~. De Sitler's universe has metric

d.,l = A Idr 2 +r'd0 2 + r'sin 20dlj)2 -- Ac2 dl 2

where A = I _. r2 /R 2, R heing eonstill1t. Ohtain the differential equations satIS­

fied hy the null geodesics and show that along null gcodesics in the plane 0 = ~n,

dl" 221.2l/ = r(r -l/ )d¢

where II is a constant. Deduce that, if r, ¢ arc taken to he polar coordinatcs in tllisplane, the paths of light rays in this universe are straIght lines.

34. EinsteIn's llniver,e has the metnc

where (1",0, ql) are spherical polar coordinates. Ohtain the equations governingthe null geodeSICS and show tlwt, in thc plane II = ;n, these curvcs satisfy tileequation

(dl" )2 r2 (I _ ;"'2) (Jlr2 _ I)dill

wherc J1 is a constant. Puttlllg r200 1/1', integratc this equatIOn and hcnce dcduce

that the paths (If hght rays III thc planc () = in arc thc ellipses

;'~'+J1y2=1

where (~, .\'larc rectangular Cartcsian cO(lrdinates. Show, also, that the time takmhya photon tI' make onc complete circuit of an lOll ipse IS 2n/((';.I").

:l5. If the metric of spacc' time is

ds2 = 112(dx2 +dv 2 + d;2) - bdl l

where a is a function of x alone and" is a CIlmtant, ohtain the dillCrenti;tlequations governing thc world-lines of frcely falling particles. If x, .1',; afeinterpreted as rectangular Cartesian coordinates by an ohserver and I is his tilllevariable, show that there is an energy equation for the particles in the form

k11" - = constant21l

~6. (r, 0, rjJ, I) arc interpreted as spherIcal polar cQordinates and time. ~

wavitationallidd is Glllsed hy a pOint elect I ic chall(e at Ihe pole. ASSllmilll( tlt"t

172

the space- time metric is givel] by equation (51.10) and that the 4-vector potentialfor the electromagnetic field of the e!lar)!e is gIven by D, = (0,0,0.,), whereX = x(r), calculate the covanant components of the field tensor Foj from equation(26.6)and deduce the contravariant components V'l. Assuming that J' = 0, provethat Maxwell's equations (56.4) and (56.5) arc all satisfied if

dX if= .,j(ah)

dr 47!1;or 2

where if is a constant.Calculate the clcments of the mixed energy momentum tensor from equation

(56.6) and whte down Einstein's equations (47.15) for the gravitational ficle!'Show that these arc sal1sfied provided

where m is a constant.37. If the coordinates Xi arc quasi-Minkowskillll so that the metric tensor is

giVl'n by equation (4IU), show that the transformation x' = Xi + ~'(x) makes thex-framc harmonic prewided the ~' satisfy the conditions

(Neglect second order terms in the hil and usc the conditilln given III Fxcrcises 5,No. 50.) Show also that h~i = 0 provided the functions ¢i satisfy the additionalcondit ion ¢:, = iii ... If the v-frame is harmonic before transf6rmation, show thatthe v-frame is also harmonic pl'Ovided ¢: Ii 'C.' O.

3~. A sphere of mass M IS expanding in SUch Ii manner that its dells!ly remainsuniform. I I' 1I(1) is its radius at time I. show that, at a large distance r I'Will Ib (·entre.the gravitational wave generated has components

h' I' I' 4GM'2 ..II = 122 = I JJ = "4 (ll + Oll)

5«

the components h'12' h~J' hJ I being lero. (The: bracketed expreSSi,)H is to becalculated at the appropriate retarded time.)

39. A uniform rod of mass M and length 211 is pivoted with its eemre at theorigin of the x-frame and rotates in the x 2xrplane with angular vclo(:ity w. Showthat, at a large distance r from the rod, the gravitational wave generated has non­zero components

h~J = A sin2wI

wherc A = 4GMa 2 ,,)2/.1rc' and the instant 1=0 has been chosen appropriately.40. Show that, if t he cosmical constant term is retained in Einstein's equation,

it reduces in empty space to R'j + !\y,j = O. Deduce that the spherically symmctnc

173

Schwarzschild solution (cf. equation (52.9)) IS given by

Using the approximate equation (4H.17), show that this implies the existence ofanadditional force 01 repubio!l fronl the celltre proportional to the radius r.

CHAPTER 7

Cosmology

59, Cosmolo2ical principle, Cosmical time

Cosmology is the study of the large-scale features of the universe, such as thedistribution and motions of the galaxies and the density of radiation and dustthrough intergalactIc space. It is also concerned with the manner in which thesefeat ures can be expected to change over very long periods of time measured inbillions (10") of years. i.e, with the evolution of t he cosmos. Such calculations alsothrow light on the stages through which the cosmos has passed to arrive at itspresent state, and allempt to answer the question. did thc universe have abeginIlwg in time or has it always existed') If the universe had a beginning. as theevidence now ,trongly suggests. the study of its ,tatc dunn!! the very early stagesof lis cvolution is called Cl)sn!I)!Iony.

Since the galaxies arc electrically uncharged. thc only force influencing theirmot ion is graVJlY. Thus. cosmology is necessarily founded on a theory ofgravitation. It has been shown (sec. e.g., Bondi, 1(60) that the Newtolliantheory isquitc capable of generating models for the cosmos which provide cxplanationsfor ma ny of Its observed feat ures. Howevel, these models necessarily a,sume thatspace is Euclidean. whereas Einstein's theory indicates that, in the presence of agravitational field, space becomes curved and its geometry is then Riemannian.The curvature generated by the gravitational allraetion of a galaxy is inap­preciable and may be disregarded so long as we coniine our allentiOllto regions ofspace whose dimensions arc comparable With those ofa galaxy, but this elTect hasmajor consequenecs when the spatial extension of the whole cosmos iscollsidered; in particular, as we shall sec. a possible consequence is that the totalvolume of space is finite and, therefore, that the universe is not potentiallY infinitein extent as a Newtonian cosmology must assume. Only cosmic models which arcin accord wit h general relativity theory will accordingly be st udied.

The reader will be presumed familiar with the basic facts relating to thedistribution of malleI' and radiation over the cosmos as it is observed in thepresent epoch (Rowan-Robinson, 1(79). The mass of the radiation IS roughlyone-thousandth of the mass of the galactic malleI' and its gravitational etlcet istherefore negligible by comparison with the allraetion of the galaxies. However,at earlier epochs, the contribution to the gravitational field of the radiation was

174

probably much more cunsiderable and, during the tirst million years after the 'bigbang', it is thought that the cosmos was dominated by its radiation; this radiati()nis assumed to have been In equilibrium with the mailer and hence to ha'eacquired a black-body frequency distribution. The remnant of this black-bodyradiation In the present epoch was detected by Penzias and Wilson in 1965 andthis still forms the major part of the total cosmic radiation. It is not known whatproportion of the mailer in the universe has been allracted into the galaxies; t~e

density of mailer in intergalactic space is certainly so small as to have noobservable elTect on the light transmilled through these regions from the mOltdistant sources, but the volume of space is so large that this observation is n(ltinconsistent with the hypothesis that the net mass of intergalactic mailer is manynmes that of the mailer present in the galaXIes. As will be seen, our ignorance Inregard to this datum prevents our reaching a firm conclusion whether the cosmosis tinite or IntinIlc in extent. Although the galaxies often occur in cluster" frol!1our viewpoint their overall distribution ,lppears to be Isotropic and homoge­neous. At very great distances, the galactic density is observed to increase, but itmust be remembered that such observations arc carried out by light which w.semitted at a much earlier epoch when all the galaxies arc thought to havc bel'l1closer together; it is assumed tha t, at the 'present cosmicalt ime' (precise detinitimfollows later), the density of galactic mass is uniform throughout the cosmos.

That the galactic density is decreasing as the universe evolves is in accordamewith the observed recession of the galaxies. To be more precise, what is observed IS

that the spectrum of the light from a distant galaxy is shifted towards the red endof the spectrum by an amount which is approximately proportional to i:sdistance. This is Hllhhll''s IIII\'. The reduction in frequency is interpreted as aDoppler elTect ca used by the motion of the galaxy a way from the observer alongthe line of sight. Since it is supposed that the whole universe is in a state ofexpansion, each galactic observer will experience a recession of all the othergalaxies in accordance with Hubble's law. Clearly, if matter is conserved duringthis expansion, a steady reduction in its density is inevitable and there is now anaccumulation of evidence that the matter density was indeed greater in the distantpast than it is today. However, a steady-state cosmology (Bondi and Gold, 194K,Hoyle, 194R) has been proposcd in which the galactic density remains constantdue to the continuous crcation of matter in intergalactic space; this mallercondenses into new galaxies and so maintains a steady-state distribution.

As a first step towards the construction ofa mathematical model of the cosmos,we shall treat the galaxies as point masses or molecules forming a galactic gas andfurther assume that this gas behaves like thc perfect fluid studied in section 22. Inparticular, its energy momentum tensor will be supposed given by equatim(22.21). At the present epoch,t his gas is exceedingly rarefied and, since the randommotions of the galaxies relative to the background black-body radiation (whichprovides a natural frame of reference) arc of relatively small magnitude, thepressure associated with the gas is very low; thus, at this and later epochs, it will bepermissible to neglect the pressure and to treat the gas as an incoherent dUll

176

cloud. However, during the early phases ofcosmic evolution, the temperat ure andpressure arc believed to have been very high indeed and the pressure terms cannotthell be neglected; further, during these phases the contribution of thebackground radiation becomcs significant and terms representing this contri­bution must also be included in the energy momentum tensor.

We next assume that there arc no privileged galactic observers, i.e. all observersmoving with the galact ic gas will be assumed to see the same large-scale process 01

evolution of the cosmos. This is the cosmoloqicul principle. The steady-statetheory is based upon an extension called the perfect cosmoloqical principle; thisassert s tha t a II galactic observers scc the same large-sC'.J Ie state of the cos mos <It <III

times. Observation supports the tirst principle bllt not the second.If the cosmological principle is accepted and the perfect principle rejected, it is

possible to define an absolute cosmiclIl time, i.e. a way of assigning times to cosmicevents which is independent of the observer. For all galactic observers willexperience the same process of cosmic evolution and the various characteristicstages of this process can be alloca ted times according to some agreed s<;ale. It isnot necessary at this point in the argument to tic the scale to time measured bystandard clocks; we only require that the later stages 01 an observer's experiencebe allocated times which arc greater than the times allocated to earlier stages.Then, the cosmicaltime of any event can be defined unambiguously as the timerecorded for the event by an adjacent galactic observer using the agreed timescale. Thus, the state of the COSl1l0S at any epoch is now defined to be the set 01

events whose cosmicaltimes arc ,Ill equal to the cosmical time of the epoch. Since,at a given epoch, all galactic observers will be experiencmg similar processes, thelarge-scale state of the C()smos at a given epoch must be homogeneous andisotropic for each such observer.

60, Spaces of constant curvature

At a given epoch, as we have just seen, the state of our cosmological model mustbe homogeneous and isotropic. In particular, the three-dimensional space inwhich the model is constructed must have these properties. The surface of asphere is a two-dimensional space of this type embedded in "'.J and it is obviousthat a three-dimensional hypersphere embedded in J'4 will have all thecharacteristics we need for our purpose. Also,just as the ordinary sphere includesthe tfJ 2 plane as a special ease when its radius becomes infinite, a hypersphere ofinfinite radius will correspond to II j , which is an especially simple case of ahomogeneous and isotropic space. In addition, we shall be led quite naturally toconsider a third class of such spaces which, like tfJ j, but unlike the hypersphere,have infinite volume. These three types of space all have constant curvaturescalars R and arc therefore called spaces of constant curl'ature. All these spaceshave positive-definite metrics, as they must have if their geometry is to beEuclidean over sutliciently small regions. It may be proved that there arc no otherRiemannian spaces having such metrics which arc homogeneous and isotropic.

177

Let (x, \', =, II) be rectangular Cartesian coordina tes in ~ 4. Then, a hypersrhereof radius S has equation

(60 I)

Since II is determmed in terms of x, y. = by this equation, a coordinate frame forpoints 011 this hypersllrlace can be constructed by first allocating coordimltesex, .1', =) to t he pomt having coordinates (\ . .1', =. II) int he ( 'artesian frame. I'rlnidedx, ,', = arc small by comparison wllh S, they will behave approximatel) likerectangular Cartesian coordlllates in f, \ and we shall therefore deline quasi­spherical rolar coordinates (,.. II. <I») by the usual transf(,rmation equations

X c: ,. sin II cos eI),

Equations (60.1) and (60.2) give

.1'= ,. Sill II sin <I), = =- ,.cos II (60.2)

(60.3)

from which we lind by dill'crentiation that

(60.4)

(60.6)

Hence. the distance d.\ between the points (\, .1', =, II) and (x + dx, .\' + dy, = t· d=,II + dll) on the hypersphere is given by

d,v 2 ~c d \2 + d y2 + d=2 + dll 2

= d,.2 + ,.2 (d1l2 + sin 2l1drjJ2) +,.2 d,.2 / (S2 _ ,.21

)2. d,.2 +,.2 (d1l 2 + siu 2 IIdrjJ2) (60.5)52 _,.2

This is a metric for the hyperspherical.!IP) Clearly, by taking I/S2 = 0, the metricreduces to that for $.\ in ordinary spherical polars.

II the curvature scalar is calculated from this metric, it will be found that itequals .- hiS2, i.e. is constant. It is now evident that if S2 is replaced by - S" thecurvature scalar will still be cOllstant with value 6/.\'2 and the space will remainhomogCllcous and ISotropic. This is the third type of such a space; its metric is

<;2d,v" ~ . . d,.2 + ,.2 (d1l 2 + sin 2 IIdrjJ")

S2 +,.2

Since these spaces arc homogeneous and isotropic, the pole" .CO 0 call be tJkcnto be any point and the axes from which II and rjJ arc measured can be taken in anypair of perpenlhcular directions. If,. is small compared with S, both metrics (60.5)and (6().6) approximate to the spherical polar metric for <1), implying that thespaces are Fuchdean over small regions.

Consider the circle r ~"' constant, II = !n, ill the space with metric (60.5) Thedistance between nC'ighbouring points rjJ, </) + drjJ on the eircle is given by the

(flO.7)

17K

metric to be d., = ,.dcp and the circumference of the circle is accordingly 2n,..Along a radius of this circle dl! = d¢ -~ 0 and the distance between points r, r + dris given to he

I ntegrating from,. = 0 to ,., the length of the radius is found to bc

I' = S sin j (,./S) (flO.H)

Thus, r = S sin (p/S) and the circumference c is given in terms of the radius I' hy

c= 2nS sin (p/S) (60.9)

Since sin (p/S) < pIS, c is smaller than 2np, which is the Euclidean result.The formula (60.9) receives a simple interprctatlOn in the allied case of the .'112

which is the surface of an ordinary sphere of radius S. The quantities S, r, I' areindicated in Fig. H, from which the relatIOnships just found are readily secn to bevalid. Clearly, as I' increases, r first incrcases until it achieves a maximum valuc S,and thereafter decreases until I' = nS, when r becomes lero. It now appears thatour coordinate system is ambiguous, in that two dillcrent points can have thesame coordinates; this deficiency can be rectified by replacing r by I' using thetransformation equation (60.X), giving a new metric

(6010)

This transformation has also eliminated the singularity in the metric (60.5) at

r = S. Like the Schwarzschild singularity, this is a property of the coordinateframc and evidently docs not correspond to a singularity in the space itself.

Putting piS = t/J, the metric (flO.IO) can also be exprcssed III the convenicntf(lrllJ

d.,2 = S2 idt/J2 + sin 2t/J(dll l + sin2I1d¢2):

(Sec Exercises 5, No. :\3.)

FIG. H

(60.11)

179

In the case of the space with metric (60.6),r can assume all positive valuesand isnot ambIguous. By pUlling r = S SInh t/J. this metric can be transformed to

(50.12)

Another form for these mctrics which will be specially important IHer isobtained by putting r = Sa. The metries (60.S) and (60.6) can then beth bcexpressed by

d.\l=sl da2

+al (dll l +sin l lld4>l)] (50.13)LI-ka l

where k = I in the casc (hO.5)and k = .. I in the case (60.6). The special Eudideancase can also he accommodated by permitting k to be lcro. The new coordillate ais dimensionlcss and. if k = I. is restricted to values satisfying 0 S a S I. FCJr theother values of k. a takes all positive values.

To calculate the volume of some region of an :!IP ~.Iet y. (1X = 1.2.3) he ge~desic

rectangular Cartcslan coordInatcs in the neighbourhood ofsome point P (section39). Assuming that thc metric is positive definitc (as in the present case), the ", willall be real. Then. if.\· are coordInates with respcct to any other frame. the netrictensor in thc x-frame will be givcn by

,11'" (1 1,b (1.1" (ly"

!/./I = (1~. (1.~11 b,b = /1x • ('x#(~.14)

But. thc volume () V of a small rcgion A in the neighbourhood of P is gi\en by

bV = Iff dVldl'ldl'~ = ffI (l(y' • .I'l.J"') dx'dxldx-' (>0.16)A' '. AI1(xl, Xl. x-')

Thus. the volume enclosed by the coordInate surfaces Xl, Xl, x J• X1_ dx '.

Xl +dx l , \-' +dx-' is

(~0.17)

The formula for the volume Vol' a finite rcgion F of :!IP, now follows. ViZ.

(~0.18)

In the casc of the space with metric (60.5). q = Slr 4 sinlO/(Sl _rl ) anJ thewhole space has volume

fR f' fl' Srl

sin II2 dr dll "'2" . 2 d4>o 0 0 J (5 -- r )

(~O, 19)

180

where the factor 2 is needed since the range 0 < r < S covers only half the sphere(see above). Performing the integrations, this volume is found to be 21(2.'13.

It will be found that thc IOtal volume of the space with mctric (60.6) is infinite,as in the Eudidean case.

61. The Robertson Walker metric

In this section, we shall cakulate a space time metric for the cosmos in a frameformed by clocks moving with the galactic gas. all readin!! cosmicaltime x 4 Asexplained in section 45, the spatial coordinates x" of each clock never change andthe frame is therefore said to be co-moving with the gas.

Viewed from anyone of the clocks, the cosmos is isotropic, i.e. it is impossibleto specify any direction having special properties. Let !f,j be the metric tensor inthis x-frame. Suppose we carry out a spatial coordinate transformation byrelabelling the clocks. leaving their readings unchan!!ed; such a transformationwill take the form

Thc transformed metric tensor is !f,j and we shall have

~l\1 {~I.\J {1 x iJ

Y.4 = (""..x::t (;X~4 Ylj -= (l.xa YJl4

(61.1 )

(61.2)

This equation shows that ii" bellaH's as a covariant .I-vector with respect tospatial coordinate transformations and hence determines a special direction atevery point of 3-space. This is COllt rary to our assumption of isotropy for galacticobservers and we conclude that !/,4 = 0 throughout sp<Jce time. Thus.

(61.3)

Now consider thc cvents llf a coordinate dock indicating the times x4•

x4 + dx4. Let dT be the proper time interval between these events. Since thespatial coordinates of the clock never change. the metric (613) shows that

(61.4)

This equation determines the relationship between the cosmical time x4 and thestandard time Tshown on an atomic clock moving with the galactic observer. Butthis relationship must be independent of the galactic observer, since all arecquivalent, and it follows that !/44 can only dep~nd on x4. We can accordinglytransform from x4 to a new cosmical standard time 1 by a tran\,formatioll

('I = JJ( -Y44)dx4

so that the metric (61.3) reduces to

ds 2 = !/"'idx"dx~ - c 2 dl 2 (61.6)

Taking a section 1 = constant of space time at a particular cosmical time. an

.!lP J wIth metric

ds1 = !I,pdx'dxD

i, obtained. This is our model for the cosmos at this cosmic'll instant.cosmologica I principle, any galactic observer will find this :!IP J to behomog(and isotropIc. Choosing himself as pole, he will therefore be lble tocoordinates (IT, II, Ij») for which the metric (61.7) takes the form (60.13). Usiframe and the new l'Osnlical lime I, the metric (61.6) finally assumRobertson Walker form

S is a constant flH the cosmos at allY given time I. but will in genera! vary wwill be referred to as the cosmic .\Cale/acTor; only when k = I can cQsmie sppictured as a hypcrsphere of radius S in c'4.

It remain" to chcck that the Irame of reference is co-moving. ,IS assumedoutset. The galaxies will be fallmg freely in the gravitationallield a,sociatelthe metric (h I.K) and their wurld-lines must therefore be geodesics Since,supposing the spatial coordinates of a galaxy remalll constant. " .~ COlalong a galactic world-line. Suhstituting in the geodesic equations (43.~

condition they arc satisfied is found to be r'~4 = O. For the met lie (61.l)condition reduces to the requirement C.1J44/(~X' = O. which is clea-Iy true.

62. Hubble's constant and the deceleration parameter

The behaviour of the cosmic model derived from the Robertson Walker I'

will be determincd when the value of k and the dependence of the cosmicfactor S on the cosmiealtimc I are known. From the physical data a\ailable(19K I l. neither of these pieces of informa tion can be derived with my deglaccuracy. It seems likely (see section 65) that k = + I and that tk univeclosed, i.e. of finite volume. In regard to S(I). the value of its tirst derivatknown roughly. but even the sign of its second derivative is in doubt.althou~

general consensus of opinion is that it is negative, i.e. the cosmic expanslslowing down. Instead of quoting values of these two derivatives, it isconvenient to work with the parameters

fI = S/S4= -5S/52

H is called fluhhll''s con."anl and has reciprocal time dimension; 4 is calle.deceleralio/l par(lmeler and is dimensionless. In the present epoch, the val.1/fI is often quoted to be abou t 1.8 x 10' 0 years, whereas the value of 4 is thoby some cosmologists to be about unity. although others would not extnegative values.

(623)

182

At a fixed cosmic'll time I, the Robertson - Walker metric requires that ordinaryspace should have the metric (60.13). Any galaxy can be thoughtofas being placedat the pole a = 0 and then the radial (or proper) distance of any other galaxy(a, 0, 4» is given by

d = S f: 7ild~ k(2 ) = exS

where ex = sin - la if k = I, ex. = a if k = 0, and ex = sll1h la if J" = - I. Thus, therate of recession of this galaxy from the galaxy at the origin is givcn by

d = exS = lid ((,24)

(63.1 )

(63.2)

This is Huhhle's law that, at a given cosmical time, the rate of recession of anygalaxy is proportional to its distance. H has the same value for all galacticobservers at the time I, but will, in general, itself vary with f. Clearly, this lawcannot be verified directly, since neither the distance d nor its rate of change aredirectly observable; in the next two sections, we shall derive an alternativerelationship between associated quantities which can be measured.

63. Red shift of galaxies

Suppose that a galaxy G (aI, 01,4> I) is being observed through a telescope fromthe pole O. Successivecrestsofa light wave emitted by G at times II, II + dl l arereceived at 0 at times ' 0 ,'0 + dl o respectively. The world-line of each crest is aradial null geodesic along which fI and rjJ remain constant (the reader should checkthat the equations of a null geodesic can he satisfied with 0 and (p constant). TheRobertson- Walker metric shows that along such a world-line,

da c~--- -- -~ - dtJ (I _ka l

) S

Integration of this equation for the motion of each crest yields the equations

r7i/~-k~2) = ct ~ = c r++d~"o ~~Since dt o, dt l will be small, the last equation implies that

dIu dl l

S(lo) S(t l )(633)

de I is the period of the emitted waveas measured by a standard clock at G,and dl ois the period of the received wave as measured by a similar clock at O. Sincewa velength is proportional to period, if Ao, Al are the wavelengths of the receivedand emitted light respectively,

AO/A, = SolS,

Where So = S(lo) and SI = S(lI)'

(63.4)

IB

Thus, if the received light is redder than the emilled light, Ao = 1.1 + AA I , whereL'.A I is positive. The red-sluft t'lctOl" : is detined hy the equation

AAI i o ISo

I: = - -i' l ,i. I .'II

((,35)

Clearly: IS positive if So > .'II, i.e. the universe is expanding.If the observed galaxy is nllttoo distant,'o -'1 will be relatively small and 'We

can expand .'1(11) in a Taylor expansion thus:

.'I (I I ) = .'I (10) - (10 - " ) S(,0) + ! (10 - , I )2 .~ (10) + .= .'1'(10) ~ I -flo (To -(1).- ~qoFl~(lo -(1)2 + } (636)

where Flo. '/0 arc the values of H uhble's constant and the deceleration parameterat the instant,o ofohservation. Substituting from the last equation in (63.5), we

derive the result

: = flo (10 .. , I ) + (1'/0 + I) fI 1; (T 0 - , I )2 + ... (637)

By observing: for a number of galaxies and calculating (To - , I) for each, "'isexpansion provides a means of estimating the values of fI and q at the preseltepoch. The calculation llf (10·",) is considered in the next section.

64. Luminosity distance

If all galaxies possessed the same intrinsic luminosity, i.e. emilled light energy 1tthe same rate. and if thiS luminosity were independent of the time, the ohserved I)rapparent luminosities of galaxies would depend upon their distances according '0

a cakulahle formula and an observation of the apparent luminosity would thmprovide us with a measure of the distance. Although there is considerablevariatIon in the strengths of the galaxies as light sources, by continilgobservations to galaxies of a particular type and stage of evolution, this variationcan be reduced and thus estimates of their distances can be obtained. Sincegalaxies tend to occur in clusters, once t he distance of one member ofa cluster hiSbeen found, the intrinsic luminosity of the other members can be determined. thisproviding further useful infoflJ1ation in regard to the probable intrimicbrightness of galaxies of other types; this information can then be utili/ed asabasis for later distance determinatilllls.

Suppose we take the pole 0 of coordinates (a. Ii, ¢) at some distant galaxyGwhose luminosity IS to be observed and let our point of observation A harecoordinates (a. !rr, 0). Photons emilled by (j will have null geodesics as worlt­lines, along which Ii and 1> will heconstanl. Consider a photon which travels alorgthe ray Ii = ~ rr. 4' = I:, where I: is very smaiL This photon will ultimately arrive It

the point (a, ~rr, I:) ofclosest approach to A when its distance from A will be givmby equation (60.13) to be

(64.1)

184

where So is calculated at the time of arrival to of the photon in the vicinity of A. Itnow follows that, if the telescope at A has aperture of radius a, the photon will becollected by the telescope provided ,: < alSoa. Thus, the telescope will collect allphotons which left G along paths enclosed within a right circular cone of semi­vertical angle alSoa. This cone embraces a solid angle

(64.2)

where ex = na l is the telescope's aperture; it follows that the proportion ofphotons leaving G which are collected at A is ex/(4nS~al).

According to cquati(ln (63.4), if 1'1 is the frequency ofa photon when it leaves G

at time t I, its frequency Vo on arrival at A at lime to is given by 1'0 = .'II l'l/So. Butthe energy ofa pho10n b rela ted to its frequency by the formula E = hI', where hisPlanck's constant. Thus, the energy of the photon is also reduced hy a factor.'I I ISo. Further, equation (63.3) indicates that the photons emitted from G over atime interval dt I arriw at A over the longer lime interval dlo = So dt I 1.'1'1' Hence,the rate of reception of light energy is additionally reduced by a factor SI ISo. If,therefore, L is the inrl'il/sic llimillosity of (;. i.e. the total rate at which it emitsradiation, the rate at which light energy is collected by the telescope at A is givenhy

(64.3)

The apparenr llimillosi! I' i ..f a celestial ohjc,:t is defined to be the rate at whichlight energy from the ohJe<:t flows across unit area normal to the line of,ight at thepoint of ohservation. The last result shows that for G,

(64.4)

(64.5)

If space were Euclidean and G were stationary at a distance d from A, we shouldhave

Substituting the actual value of I from (64.4) and solving for d, we lind

d = aS~/SI (64.6)

(64.8)

This result is termed the IlImnll/sily disTallce of G.Returning to equation (64.4l, we shall obtain an expansion for I in powers of

(10-11)' Equation (63.2) shows that

t7(\d~k;'i)= cL~~ (64.7)

Assuming a is relatively small (i.e. G is not too distant), the integrand of the left­hand member of this equation can be expanded in powers of a to give

f: J (1~~7~';i) = a + : kaJ+ ...

, dx'V' = _

dr

Also. equation (63.6) leads to the result

I I f 1

). \1 +Ho(ro-r)+ ... j

S(t) SUo

Hence. the tight-hand member of equation (64.7) can be exparided in

f'" dr cc ,= , i(r,,-r l )+iH,,(r,,-ttl2 + ... :

,. S So

Equating the expansions (64.~nand (64.10). we sec that. to the tir,t orderqllantities. (f = c(to -tl)/S", It follows that. to the second order in (to

Substitution from equations (63.6) and (64.11) in equation (64.':) now yexpansion

Finally. equation (63.7) shows that to the tirst order (t" - rd =2/ H o. Itfrom the same equation. therefore. that

Hf)U" -rtl = z - dqf) + 1)22 +

Suhsiitutlllg for (tf) -rl) from this equation into equation (6412) acc,leads to t he expansion

The importance of this last equation is that it relates two obsenable qu,and z. By litting it as closely <Il, possible to the available data. estmates otq(l have been obtained. A more precise relationship follows from dyconsiderations (sec Exercises 7. No.5).

65. Cosmic dynamics

To gain further information in regard to the probable values ofth: functi,and q(t) at cosmical times in the remote past and future. it is nece,determine the equations of motion of the cosmos. These will be prmEinstein's equations of gravitation. It is therefore necessary tirst to calclenergy momentum tensor for the galactic gas.

The 4-velocity of a particle whose world-line in the x-frame has e(

Xi = x'(r) (r = "/ie belllg proper time measured hya standard clo:k movthe particle) has been defined (section 50) to be the contravariant vectorhy

186

If the frame is inertial and Minkowski coordinates are used, this definition is inagreement with that adopted in the special theory (section 15).

Now consider a perfect fluid. At any poin t in the fluid, we can construct a freelyfalling frame which is locally inertial and in which Minkowski coordinates can bedefined. The special theory is valid in any such frame and the fluid's properdensity of proper mass 1100 and pressure p can be defined as 4-invariants (section22). The energy- momentum tensor in the frame then follow. from equation(22.21 ). In the general x-frame, this tensor must therefore be given by the eq uation

(65.2)

for this is a tensor equation and reduces to the valid equation (22.21) in the freelyfalling Minkowski frame.

Along a world-line of a particle of the galactic gas, the coordinates (a, fJ, 1»remain constant and hence, from the metric equation (61.8), we deduce that r = f.

Thus, the 4-velocity of this particle has component. (V;) = (0,0,0, 1) in theRobertson Walker frame. The non-zero contra variant components of the metrictensor are

qll = (1 _k(J2)/Sl, q22 = I/S2a 2, !I'D = coseclO/S 2(Jl, !/44 = -I/el

(65.3)

We can now calculate the non-zero components of the energy--momentum tensorfor the gala(:tic gas from equation (65.2); they are

1''' = p(l -kal)/Sl )1'11 = p/ Slal

1'33= pcoseclO/Slal

T 44 = 11

(65.4)

(65.5)

where we have deleted the subscripts in 1100 for convenience. The covariantcomponents now follow immediately, viz.

1'" = pSl/(1 -kal )

1'11 = pSlal

1'33 = pSlal sinlfJ

1'44 = p.c4

The non-zero components of the Ricci tensor for the Robertson-Walkermetric may be verified to be

Rll = -P/(l-kal ), R11 = _Pa2, R]3 = -PalsinlO, R44 = 3.~!S

(65.6)

where P=2k+(SS+2S1)/Cl (65,7)

dots denoting differentiations with respect to f.

The reader may also verify that the curvature scalar is given by

R = R: = -6(SS +,~l +kc')/C'Sl

187

(65.8)

We can now constrw:t Einstein's equations (47.16) (covariant form); only Iwodistinct equatiom emerge, viz.

2SS + .Sl + kc l _ (l/\S' = _ 1((1 pSl

3(,S2 -j_k(2) _(2/\S' = I(C 4 j1S2

(659)

(6510)

Together with an equation of state p = p(j1), these equations arc sufficienl todetermine the unknown function,; S, p and Ii.

Before attempting to integrate these equations, certain important conclusiDnscan be reached by a dire<:t study.

Equati()n (65.10) can be written in the form

(6511 )

from which it follows that k = + I if

(6512)

and k = - I if the reven;e inequality is true. p,. is therefore a critical density (theclosure density) for the gala<:tic gas, determining whether the universe is of finiteor infinite volume. Assuming /\ = 0 and taking fI = 1.8 x 10' I" (at the prelentepoch), 1(4 = 8rrG = 1.67 x 10- 0 (all SI units), we calculate that

(65.13)

If the matter in the galaxies were spread uniformly over the whole of space at thepresent epoch, it is estimated that its density would be j1 = 3 x IO-l"kgm- J

.

Thus j1 < j1, and this datum suggests that the universe is open. However,thepresence of a very tenuous intergalactic dust or gas, far below the limit of possibleobservation, could easily reverse this conclusion. We accordingly seck furtherinformation from equatIOn (65.9).

This equation can be written

(1

5' k = (2q -. I) H l + c l/\ _ /((1 P (6514)

Again laking /\ = 0 and assuming p to be negligible, we conclude that k = +1 ifq > ~. Unfortunately, the red shift luminosity relationship, from which the v21ueof q is derivable (sec section 64). has not been established with sufficient certaintyto de<:ide for or against this inequality. However, the dala available tendl tosupport it.

188

66. Model universes of Einstein and de Sitter

The fIrst solution of the dynamical equations (65.9) and (65.10) was suggested byEinstein himself. His proposal was made some years before Hubblr published hisobservations relating to the recession of the galaxies, and the possibility thai theuniverse was in a state of expansion was not considered by astronomers at thattime. Einstein's universe was therefore slatic and the equations were satisfied bytaking S to be eonslanl. Pressure and denSIty had then 10 satisfy the equations

kKp = /\ -. S2

lkKc

2Ii =' -/\S2

(66.1)

(66.2)

Cle'lrly. if /\ = O. since p and /l cannot be negative. the only possible solution isfor p. Ii and k all to be zero, i.e. an empty. infinile. Euclidean cosmos. It was inorder to be able to reject this solution that Einstein introduced the cosmicalconstant term into his equation of gravitation.

Adding equations (66.1) and (66.2). we lInd

(66.3)

(66.4)

proving that k must be positive. i.e. k = + I and the universe is closed. Equations(66.1) and (66.2) now show that /\ must satisfy the inequalilies

I 352 ~ /\ ,,; 52

5 is the radius of the universe, which is certainly very large. indicating that /\ willbe quite inappreciable except for phenomena on the cosmic se<t1e. Following uponHubble's discovery. Einstein abandoned his model and with it. the cosmicalconstant. Since then. /\ has been put to zero in most cosmological investigationsand we shall follow this praetice in the remaining sections of this chapter (exceptin certain exercises at the end).

The necessity for including the cosmical constant term if the universe is to bestatic. follows from very elementary considerations. Without it. an initially staticcosmos will collapse under the gravitational attractions of its constituent galaxies.If /\ is positive. the term implies the existence of a counterbalancing long-rangerepulsive force between the galaxies. which increases with their distance ofseparalion (Exercises 6, No. 40). However. although these two opposinggravitational forces can result in eq uilibrium, putting p = 0 to match the presentstate of the cosmos and hence /\ = 1/52 • it may be proved I hat the eq uilibri urn isunstable (Exercises 7. No.8).

Following upon Einstein's proposed model, de Sitter derived another which isof some historical interest. In this context. his model is most conveniently

con~tructed by adopting as our basil: assumption that Hubble's constdnt doeschange with time, i.e.

Integrating, we get

SIS = H = constant

S = A exp HI

(t

(f

where A is the value of S at some arhitrarily chosen origin of cmmical tlSllhstitllting in eljuations (65.9) and (h5.10), we lind that

(f

(f:

(h

Neither of these limits can he negative, so we must reljuire that

A=3H'lc'

Thus, eljuations (M.7) and (hh.X) lead to the result

31'+c')1=0

from which we conclude that

I' = 0, II = °

(M

(66

(66

and, therefore, that k = O.Thus, de Sitter's universe is Euclidean, hut heing empty is only of acade

interest. Although the model appears to he in a state ofexpansion, since no rnais present, this phenomenon has no physical basis; indeed, it is possible to deristatic metrie for the model by carrying out certain tranformations ona and I

Exercjse~ 7, No.3). If the galaxies are represented hy freely falling particles halnegligible mass and constallt spatial coordinates, an expanding infinile cosm(ohtained; however, each galactic observer is limited to a finite universe, sinc(cannot penetrate beyond a certain distance called his Cl'etll horizon (see Exerc7, No.4)

67. Friedmann universes

These models are constructed from eljuatiom (65.9) and (65.10) by puttingI' ,i.e. by treating the galactic gas as an incoherent dust cloud. We shall also put I

zero.

190

We first note that equations (65.11) and (65.14) reduce 10 the forms

/(c 4 J1 = 3H' + 3kc' / 5'

(2(/-I)H' = kc 2/S'

It follows that

/(c4 J1 = hqH'

and hence that q can never be negative.

Equation (65.9) gives

Writing this equation in the form

d' 2 , •

j(55 ) = -kc 5

(f

we can integrate and obtain

D being constant.Substitution in equation (h5.10) now yields

liS] = 3D//(c' = constant

(67.1)

(h72)

(h7.3)

(67.4)

(h7.5)

(67.6)

(67.7)

showing that the constant D must he positive. This is the equation ofconservalionof mass (see Exercises 7. No.2). III the l:ase of a tinite cosmos. the IOtal volume isknown to he 211:' 5J and the total proper mass is accordingly 271 2 ~JI1. whil:h isconserved. This is to be ex peeted. since there is no interael ion bet ween Iheparticles of the dust cloud and Iheir proper masses therefore never change.

Wc shall ncxt integrate equation (67.6). with k taking its three possible valucs.separately.

(i) If k = I. the equation is

SS2 = c' (D - 5)

Changing the dependent variable from S to u by the tram,formation

5 = ~D(l-cosu)

we have S = ! Dli sinu and equation (67.8) leads to the equal ion

tD(I-cosu)1i = c

This integrates immediately to the form

Cf = tD(u -sinu)

(67.8)

(67.9)

(67.10)

(67.11)

choosing f to be I:ero when u = 0 (i.e. the origin of cosmieal time is laken 10 be theinstant when S = 0). Equations (67.9) and (67.11) determine 5 as a functIOn of f.

5

o rrDI2c ."Olc

since these equations are the well-known parametric equations 1'01 a eycloplot of S against cf must take this form (see Fig. 9).

Our conclusion is that the tinite universe will expand from a singularity ato a maximum radius D when 1/ = nand t = nDl2e,and will then cGntraet bthe singular ~tatc at II = 2n, t = n DIe. The actual behaviour in thevicinitysingularity is, of course, not predicted hy our analysis. since p has been ne~

and the contrrhution of the radiation ignored.(il) If k = O. equation (67.6) gives

S.~l = e'D

The variables separate and integration yields

S3 = ge' Dr ' 14

again taking r = () at S = O. This universe expands from a singularitcontinues to do so indefinitely (Fig. 9). Space is Euclidean and the cosmos i~

(iii) If k = - I, equation (67.6) wkes the form

S~;, = e'(D + S)

As in the case k = I, we change the variable to 1/, this time by the t,ansforr

S = !D(coshll - I)

and integrate to lind that

et = iD (sinhll - II)

taking II and t to be zero at S = O. As in the previous case, the initial expannever reversed and the universe is open (Fig. 9).

(h717)

192

It remains to calculate two constants of integration, viz. D and the value to of tin the present epoch, in terms of flo and '10'

Putting present values into equation (67.2), if I.. +0 we get

., ke 'S" = ",

(2ilo - l)fI 0

Equation (h7.h) can be written in the form

fll = e' (/) - k8)/S3

Thus,

(HIX)

(67.19)

If I.. +0, equation (67.17) now shows that

21'D = -- '10(2'10 _Ill', '10 >..l. I.. =0

flo

21' (h720)= '1o(l-2ilo) 3",0«/o<tJ..=-1

flo

If I.. = 0, equation (67.2) requires that 'I = ~ and equatIOn (67.19) gives

(67.21 )

(67.22)

So, flo are independent parameters for this model. H0wcver. So is superllu0us andmay be eliminated by transformation of the coordinatc a to a' = S"a; So thendisappears from the metric. EqUIvalently, we may take So c_ I and write equatIOn(67.13) in the final form

The present age to of the universe is now calculable for the three types of model.

(i) If I.. = I, '10 >!. equation (67.9) shows that

eosuo=I-·2So/D= -Iqo

(67.23)

having used equations (h7.17) and (67.20). Equation (67.11) now yields the result

(6724)

the imcrsc cosine being taken in the lirst or second qWldrant If values (/0 Cc L1/11" = 1.X x 10 10 ycars arc substituted, we lind to = 10'" )CaIS very ncarly. i.e.ten billion years.(ii) I1'1.. = - I. 0 < ilo < L a similar calculation leads to the result

Ito = - - [( I - 2'10) 1_ '10(1 - 2'10) -J 1 cosh -I ('1() I - I)J

I1 0(67.25)

19

Suppose we calculate (/0 from equation (67.3), taking II = 3 x 10- 28 kgll I (i.eassuming all the matter has been attracted into the galaxies) and ifu = 1.lx 10 18 S l Then '/0 .= 0.025 and the f(lrmula (67.24) gives 10 = 1.6 X IJ)IO year(16 billion years).

(iii) Joinally. if k -~ O. '/ =, L putting S = I into equation (67.22) gives

(672f

With 1/1/ 0 = LX x 10 10 years. this makes the present age of the cosrr.os to b12 billion years.

For the early stage of cosmic expansion, the model's failure to indude thcontribution of the radiation renders it invalid. However, since this stage iexpected to be short labout 10" years), the corrections which need to bemade t,the values of ' 0 calculated above to allow for this failure are negligible.

The stage of radiation dominance is studied in the next section.

6R. Radiation model

During the early stages of cosmic expansion, it is believed that the dominatinfactor was the electromagnelIc radi'llion. In this section, we sh'lll studysimplified cosmological model, containing radiat ion alone, which will providelirst approximation to this early state of the universe.

We shall assume that the radiation is isotropic for each galactic observer anhas energy density U (the same for all observers at a cosmiealtime I). Such aobserver is in a statc of free filII and can be equipped with rectangular axes anassociated standard clocks, which behave locally like an inertial fr<lIl)(M IIlkowski coordinates .1" can be defined in this frame and, with respect to theS'the observer's 4-velocity is V = (0, ic) (sect ion IS).

In tillS y-frame. the results of the specIal theory are applicable and, in plrticulaequation (29.5) defines the energy momentum tensor .'I,) for the radimion. Wsh'lll assume that the time variations of the field components Ea , lJa of thlr;ldiatlon are qUlle random and that there is no correlation between the~

components. Thus. if CI. l' If.

I/1(H a H p ) = 0 (6X.

where m(D ) denotes the mean value. Also, since the radiation is isotrope, m( E;and m(I1;) will be independent of CI.. Hence, taking mean values in cquatio(29.16). we lind

(6X.:

U being the mean energy dellf,ity of the radiation.We call now calculatc the mean values of the components of S,j' IfCl. +If,

follows from equations (29.14) and (6X.I) that m(SaP) = O. Also, since t~

radiation is in a ste,ldy state, the rate of energy flow in any direction is ocro an,

194

hence, the mean values of the components of the Poynting vector all vanish; i.e.m(S.4) = O. It is clear, therefore, that only the nleans of the diagonal componentsS I I' S22' 5 33 , ,'144 are non-zero and, for these, eq ua tions (29.13) and (29.16) give

m(SII) = m(S22) = m(S,.,) = ~I;om( Fo';) + I/lom(H; ) = ~ V (68.3)

m(S44)= -u (68.4)

A tensor equation for S'1, valid in any fr,lme, can now he wrillen down, viz.

(68.5)

(68.6)

where V' is the 4-velocity of the local galaetic observer. V is uniquely defmed andis thus a 4-invarianl. Hence, this equatiolI cert,ilnly de tines a cont ravariant tensor.The equation is easily verified to give the mean components just calculated inthe ,v-frame. It is therefore valid in all frames. If the equation is compared withequation (65.2) for a perfect fluid, it wiil be seen that the radIation beh<tves like aperfect fluid 01 density V Ie' and pre~sure V;3.

In the Rohertson Walker frame, (V'\ 1,0,0,0, I). Hence, the non-zerocomponents of the- energy momentum (en,or in this frame arc:

,'1'1 = V(I -J,(T')/3S' ),'122 = VIOS'(T')

SB = V cosec' O/(S~(T')

5 44= Vic'

Using the components of the Ricci tcnsM aln:ady calcu"lled In section 65, theEinstein equations I'M the model can now bl' calculated. They pro"e to he

2,'1.\' + 5' + kc' -('-~ i\S' = - .\I\C' us'3(S' +kC')_(·1i\S' = /(c'VS'

(68.7)

(68.8)

(68.9)

Since S will be comparatIvely sm<tll durilig this pha~e, even if i\ is l10n-zero, theterms containing this conslHnt will b(' negligible. We accordmgly put i\ = O. Also,ifk *0, the term kc' will he negligible by e•.>mparison with the terms 2SS and .~1.

For, during the maller-dominated phase, equation (67.6) shl''-'' that

.~' = c' (1) __ k1\ s )

and kc' is small by comparison with 5;1 prOVIded DIS ~ I. Bill

D

,'I

D So

So S

2q(l So

riq;~~ 11','1(68.10)

having used equations (67.17) and (67.20). At the beginning of this phase, SolS islarge and it follows that DI ,'I is large. Th us, kc' is negligible by comparison with

(6Xl 1)

195

,~2 and, since ,~ takes even larger values during the earlier radiation dominatedphase, we shall neglect hC 2 during this phase. We have proved, therefore, tha inthis early stage of the cosmic expansion, the behaviour of the cosmm isindependent of the values of 1\ and k.

Eliminating (J between equations (6X.7) and (68.8), it is now found that

,', "2 d .'.SSt,) =dl(SS)=O

Two integrations now yield the results

,\. = A/S, (6K12)

where A is constant. Substituting in equation (6X.8), we lind

KC2 U = 3,~2 /.'1'2 = 3/(41 2) (6813)

If we assume that during this phase the radiation is in thermal equilibrium \liththe matter present, then it will po.,sess a black-body spectrum and thetemperature Twill be given in terms of the energy density U by Stefan's law, liz.

(6XI4)

where (1= 7.5x IO'- 16 Jm OK 4 is the Stefan-Boltzmann constant. TIIUS,equation (6X.13) leads to the following formula forT:

(6XI5)

One second aftcr the inception of the expansion, this formula indicate' atemperature of 1.52 x 1()10 K for the cosmos.

69. Particle and event hori.wns

Equation (64.7) detennines the coordinate IT of an object which is observed in Qurtelescopes at the present epoch ' 0 , by light which was emitted by the object at timeI,. Since I I cannot be less than the time I = 0 at which the cosmic expansIOncommenced (accepting the big-bang hypothesis), the most distant object whichwe can observe today has coordinate IT given by

f" dlT f'" dlo J{i kIT 2)="C 0 S

The proper distance of such an object IS thereillre

(6~.1 )

(6U)f" dlT f'" dld,,=So oJ(I_kIT 2 )= cSo oS

d" is said to be the proper distance of the parlicle horizon.In the case of the Friedmann model with h = + I, equations (67.9) and (67.11)

196

give

k=O

dl !D(I -cosll)duc = - ...-..- -- ... -.. = duS !D(1 -·cos II)

and, thus,

il,,=SOIIO= _I' (2,/0_1)-112 CO,-I(1 -I)flo ~

by equations (67.17) and (67.2.\1.The reader is left to obtain th~ r"sliits

21'd ll =­

flo

J,,= ~~(1-2'/0)-I!lCOShl(±-I)­if k = I

(69.3)

(69.4)

(69.5)

(69.6)

for thc other values of k, in a similar manner.In particular, if k = + I, '/0 = I, 1/ flo = I.X x J()' 0 years, we calculate that

il" = 2.X X 10 10 light years.Equation (64.7) can be utilized in all alternative manner. If an event occurs at

the point with coordinate a at the present epoch '0' we shall observe it at time I I

provided

f"- _.da_. = I' C'dl (69.7)o J(l- k(

2) J", S

In the case of the Friedmann model with k = + I, Wto' must have I I < nD/c, smcethe cosmos collapses to a point "t tinw 11 IJlc; for k = () or - I, I I can be arbit rarilylarge. Thus, the proper distance Illthc present epoch "fihe most distant evellt wecan ever hope to see is given hy

il, = cSo f'O"' d: (69.X)In ~

where ' m" = nD/c if k = + I, and ' m" = (£, if k = 0 or _.1. If il,. = ex.', th...n allevents happening in the present epoch will ultimately be observed in ourteles...opes. ill, is tcrmed the proper distance of the el'f'1lI lIoriZllIl.

In the case k = + 1, we cakulate that

(69.9)

provided the inverse cosine is taken in the range (n, 2n). With '/0 = I, 1/ flo = 1.8X 10'0 years, it will be found that il,. = 8.4 X 10 10 light years. Light from events

occurring at a greater proper distance will not have reached us befon t he emcollapses to a sll1gulamy.

The cases k = °and -·1 ooth give an infinite value of d} am all C\

happening today will, in these models, ultimately be observable m the cal

Exercises 7

1. Transforming to a new radialeoordinate r by theequation a = 1/( 1 + ~I

obtain the Robertson Walker metric in the form

( .., )2d.I" = . ,,'. 'dr' +r'(dli 2 +sin'lidd,2)' -e'dr'

I + lkr 2 I 'I' I

2. From the conservation equation 1 4'., = 0, obtain the equatiOI

d , 3 '2;(liS) + ,S Sp = 0

dr ('

Deduce that, if p = 0. then II 'X I/S"­3. Show that the de Siller metric

ds 2 = A' exp(2HT)(da' + a 2 dll 2 -+- a' sin 2 11d rjJ') - ('2dl

can be tran.,formed to t he static form

d.I'~ dr' +r'(dli'+sin 2 I1drjJ')-e'(l lI'r'/e')L1T)1 _11 2

,.' /e'

by the translilflnation equations

r exp (- liT)

a = A J(1 ·-II",'/e')'

4. A photon is eInllled from thc point (a, II, qi) along the radius to the origiltime r in the de Siller universe whose metric is given in the previous exc'cise. Silthat the lillie taken to reach the origin is

1 l IiA J... log 1-· aexp(IIr)./I e

lienee show that if the proper di.,tance of the point from the origin at the timegreater than e/ H, the photon will never arrive at O.

5. Show that, for all the Friedmann models with 1\ = 0, p = 0, thelumino~distance d I of a g,tiaxy whose red shift is ~ " given by

cd I = - 2 l q,,: + ('I" ~ 1): "I (2'/0: + 1) - I : ].

/1"'1"

If z is small, verify the expansion (64.14).

198

6. Show that if 1\ is not assumed to vanish in the Friedmann model, then 5(1)satisfies

5S 2 = ('2(D - k5 + !I\SI)

where D is a mailer density parameter defined by the equation Ke2j15 1 = 3D.

Show that the spceial case k = 0, D = °yields the de Siller universe.7. Sketch the graph of 5S' against S for the Friedmann model with cosmical

constant (previous exercise) and deduce that (i) S increases from lcro to amaximum value and then decreases hack to zero in the cases: (a) k = + I, 1\

< 4/(9D 2), (h) k = 0,1\ < 0, (c) k = ·--1,1\ < 0; (ii) S increases steadily from

zero and tends to infinity in the cases: (a) k = + 1,1\ > 4/(9D 2

). (b) k = 0,1\ ? 0. (c) k '" - 1.1\ ? 0. In case (i) (a), if 1\ is posllive,show that there is also a sol Ut ion in which S decreases to a nOll-zero minim urn andthereafter steadily increases towards infinity.

8. If k = + I and 1\ = 4/(9D 2), show that the radius 5 of the Friedmann model

can first increase from zero to a value I/.J 1\, when the cosmos allains the statICEinstein state (section 66) and thcn, if slightly disturbed from this state, S mayeither decrease back to zero or continue to increase indefinitely. (This shows thatthc statIc Einstein universe is unstable.)

9. A cosmos containing radiation, but no maller, is governed by the equations(68.7) and (68.8). Show that

5 2S2 = e 2 (D - k5 2 + ~1\54)

where [) is an energy density parameter defined by the equation 3D = KUS 4

10. Sketch the graph l)f S2.~2 against S fllr the universe descrihed in theprevious cxercise and dcdlH:e that all the condusions Ii.sted in exercise 7 are valid it4/(9[)2) is replaced hy 3j(4D).

11. For the universe deserihed in exercise 9, if k ~. 1,1\ = 3j(4D), and 5 = 0 at I

= 0, prove that at any later time I,

S2 = 2D{ 1 --exp( -('1/ JD)}

If S = J (2D) at I = 0, prove that thc universe is static hut unsLible.

References

Bateman. H. (1952). Pi/rl/ill !JI//Ncmial FifUlJliolls 0/ Malhm/llfical Physicl. ClmbridgtUniversity Pre". p IH4.

Bondi. H. (1960). Co.lnfoloq... Cambridge University Press.Bondi. H .. and (;old. T (194H) Mon. Nol. ROI· . .4.l/r. Soc.. lOll, p. 252.Couboll. C A., and Boyd. T J. M. (1979). Flc('lril'il\', Longman, p. J()9Hoyle. F (J94H). MOil. NOI. Ro.\'. A.\lr. Soc., lOll. p. J72.Rowan·Robmsoll. M. (1979) CO.ll/1I1' I oml.lco!'e. Oxford University Press.

199

Bibliography

Adler. R., Bazin. M .. and Shiffer. M. (1965). Infrodu('/lOf! 10 General Rl'ial;l·iIY. McGraw-Hill.

Aharonl, J. (1959). 'lIw Special I1lCorc o! RcIOli/'iIY, Oxford University Press.Angel. R. B. (19HO). Relalwily: The Theory and il.l· Pllllosopl1\'. Pergamon.Berry, M. ( 1976). Principles o! ('osmolol/l' alld Grarila/loll, Cambndge University Press.Clark, R. W. (197.'). Fin.Hcin. Ihe Li!e and TIIII<'s. Hodder and Stoughton.Dirac. P. A. M. (1975). General 7heorl' 01 Re/(Jli,';ly. Wiley.Dixon. W. G. (197H). Special ReIOliI'IfY. Cambridge University Press.Einstein. A. (1967). The Meallinl/ o! Reldlwill', Chapman and Hall.Fock, Y. (1964). Iheon' o! Space. nme alld (jral'//d/(oll, Pergamon.Frankel. T. (1979). Grarillllionul CUrl'olure, Freeman.Graves. J. C (1971). nIl' ConeeplUal FoundaliiJlls o! ('olllell/pOmn' Rl'iOli"itl, Theo"y.

M. L T. Pre~s.

Hawking. S. W., and Ellis. G. F. R. (1973). Ihe I,a,,!!e Scale S,,,uc(u,,e oj Space 71me.Cambndge University Pre~s.

Heidmann, J. (19HO). Relil/II'lS/lI' ('oslI/olo!!.,'. Springcr.Hotl'nHlnn, Il. (197.')' Alhen Fin.Hem. ('''<'<110'' and Rchel, Ullrt-Davis and MaeGlbbon.Hoyle, F., and Narlikar, J. Y. (1974). A"lion ala lii.IIIlIII·C o! /'hy.I'lCs ami CO.I·IIlIIlol/l·,

Freeman.Kilmister, C. W. (1970). Sllecial Theon' iii Relillirily. Pergamon.Kilmlster. C. W. (1973). (ieneral 7helir\' iii Neld/(I'ily. Pergamon.Lancl,os, C (1974). Ihe Dnsle;n /)cmde (1'I1i.' 1'115). lick SCIence.Landau, L. D., and Lif~hit7. E. M. (1975). 'lJ1C C/<J.Isical 'Thco"y o! Fieldl. Pergnmon.Lnndsberg, P. T .. and Evalll;. D. A. (1977). MOIhemalical ('o-",wlo!!y, Oxford University

Prcss.Misner. C. W .. Thorne, K. S. and Wheeler. J. A. (197:\). (i,.,I/·tlalion. Freeman.Ml<iller. C. (1952). I hf'OlY o! Rclwil'ltl'. Oxfold University Press.Narlikar, J. (1979). I.CI·lu"es on (jenaol Rclalu·ily I/nd ('o.""IOI",fY. Macmillan.Papapetrou, A. (1974). l,eclun·.1 on Genl'/ol Rl'iaIJl'/ly, Reidel.Petrov, A. Z. (1969). Ei,lS/em Spaces. 1',·r,,"l11on.Raychaudhuri. A. K. (1979). 7heor('/lml ('osmolo!!y, Oxford University Press.Rindler, W. (1966). Sl,ecial ReIUl;,'itl'. Oliver and lloyd.Rowan-Robinson. M. (1977). ('o.I'IIIOI"i1c. Oxford University Press.Schrildinger. E. (1950). Space Time Simi/urI', Cambridge lJniverSlty Press.Scmma, D. W. (1971). Modem Cosmoltll/Y, Cambridge lJmvcrsity Press.Synge, J. L (1960). RelUlil'l/y The (leneral 111,,01'.1', North-Holland.Synge, J. L. (1964). Relali"itv - Ih" SpcClal 1 h"or.\', North-Holland.Tolman, R. C. (1934). R"lali"ily. TIt"mlOd.\'namics and ('osnlOlo!!.I', Oxford University

Pres,.Weinberg. S. (1972). Gr£/"ilalion ami COJl1lOlo!!,l'. Wiley.Weinberg. S. (1977). 7h" I'/rsl 7h,.,·(· MinU/".I. Deutsch.

200

Index

Aberration of light. 60AcceleratIon. motion with c()n~tant, 63

transformation of. 59Adler. R.. 200Aether.5Affine connection. 9HArtI'lily,97

metric. 110symnletric. 100transformation of. 98

Affinities. difference of. 99Aharoni. J., 200Angel, R. H.. 200Angle hctweell vectors. 29, 107Angular nHlIIlc'ltum, 57Axial vector,.' I

Hateman, II" 161. 199Hazin. M., 200Hcrry. M., 200Bialll;hi identity. I I IBig hang. 175Black body radiation of cosmos, 175Bllndi. II" 174, 175, 199Boyd.T.J.M., 74,199

Cartcsian tellsor, 24Charge. conservation of, 73

field of moving, 85Charge d~nsity, propcr, 74

transformation of. 82Christo/Tel symbols. 109Clarkc, R. W.. 200Clock. coordinate, IJ I. 152

standard, 'I, 1.'1. 152Clock paradox. 1\ 14.20Clocks. synchroni~,ationof. 7Cloud. incoherent, 51, 140

Components of vector. 24Compton efTect. 65Conjugate tensors, 95Coordinate hnes, 87Coordinates. curvilinear. 86

geodesic, 108spherical polar, 86

Coordinates of an ewnt, 'I, 132Coordinak surfaces, 87Copernicus, 5('osmical constant, JJ7, 188.198Cosmical time. 176Cosmic ray particles, 43

decay of, 13Cosmic scale factor. 181Cosmogony. 174Cllsmological principle, 176

perfect, 176Cosmology. steady state, 175Cosmos, radiation dominated, 175, 193Coulson, C. A.. 74, 199Covariant derivative, 97, 98. 100Covariam drllcrentiations. commutativity

of. 118Curl. 31.118Current density, 73,154

transl'ormntion of. 82Curvature. space of constant, 123, 176Curvature scalar. 114Curvalure tcnsor. covariant, III

Riemann -Christoffel. 105symmetry of. I 12

Curvature tensor for a weak freid, 139

Deceleration parameter. 181De Sit tcr's universe. 164, 165, I'll, 188.

197Dilation of time. 13. 39

J01

202

Dirac, P. A. M., 200Displacement vector, 24, 89Divergence, 29, I 12Dixon, W.G., 200Doppler effect. 83

Earth's surface, metric of, 89Einstein. A., 200Finstein's equation. 46Einstein's law of gravitation. 137Einstein's tensor, 114, lJ7Einstein's universe, 171, 188

stability of, InElectric charge, conservation of. 73

invariance of, 74proper density of. 74

Eleetric charge density, 5, 73transformation of. 82

Electric current, field of straight, 78Electric current density, 5, 73, 154

trausformation of, 82Electrie displacement, 5Electric intensity. 5, 75

transformation of. 77Electroma~netie field tensor, 76,154Ellis, G. F. R., 200Energy, conservation of, 46, 53, 82

equivalence of mass and, 46kinetic, 45particle's internal, 46

Energy density, electromagnetic. 81,82Energy··momentum tensor, 52, 1.'5

divergence of, 53, 136c1eetromagnetic, 80, 155fluid's, 5l>, 70symmetry of, 59

Eotvos, L., 128Equivalence, principle of, 129Euclidean geometry, 6Euclidean spaee, 88,96, 106, 109Evans, D. A., 200Events, 7

causal relation between, 16coordinates of, 7, 132

Fall, particle in free, 133, 140Fit zgerald contraction, 12, 17Pluid, EM tensor of, 56

perfect, 57Fock, V., 200Force, 2, 44

centrifugal, 3. 17, 128

Coriolis, 3, 17, 128fictitious, 2, 129inertial. 3,128rate of doing work by, 45, 51transformation of, 47

Force density, 51Frame, comllVing reference. 180

g.eneral reference, 13 Iinertial. I, 127local free·fall, 132

Frankel, T., 200hiedmann univen;es, 189, 198Fundamental tensor, 25, 105

covariant derivative of. 101Future, absolute, 16

Galactic gas, 175closure density for, 187EM tensor of, 186

Galactic observers, 176Galaxies, distrihution of. 175

gravitational field of, 129propel' distance of, 182red shift of, 175, 182

GalilelLn law of inertia, 134Galilean transformation. IIGeneral principle of relativity, 128Geodesic, I 14, 121, 134

equations of, 115, 117null, 116, 150

Geodesic frame, 108Gold, T., 175,199Gradient, 28,91Graves, J. C, 200Gravitation, Einstein's law of, 137, 166

Newton's law of, 139, 140Gravitational constant, 136, 140Gravitational field, irreducible, 13 I

weak,137Gravitational field of a point charge, 172Gravitational field of a sphere, 145Gravitational waves, 160

polarization of, 161

Hamiltonian, 49electromagnetic, 83

Hamilton's equations, 49Harmonic coordinates, 126, 160, 172Hawking, S. W., 200Heat, mass of, 45lIeidmann, J., 200Hoffmann, B., 200

Hole, hlack, 159white, 159

Homogeneity (If space, 26Horizon. event. I H<.I, 196

particle, 195Hoyle, F., 175. 199,200Huhhle's constant, lH I, 192flul>ble's law, 175. IH2Hypcrsphere, 12.'. 176

metric of, I 23, 177

Index, dummy, 23free, 23raising and lowering, 106

Inertial fe)rces, 12HInertial frame. I, 127

local, 132l)uasi , 13H

lnterVlI!. 105. 132proper time. 14, 132

lnkrval hetween evellts, 132Intrinsic derivative, 119Invariant. 27.90

covariant derivative of, 100parallel displacemcnt of, 100

Inverse square law, Einstein's correction to,14H

Isotropy of space, 26

Kilmister, C. W., 200Kinetic energy, 45Krunecker deltas, 22, 25,92Kruskal·· Slekeres transformation. 167

Lagrange's el)uations, 49Lagrangian, 49Lanuos, c., 200Landau, L. D., 200Landsberg. P. T., 200Laplace\ el)uation, 26La plaei,ln, I 13Length. 12Levi-Civita pseudotensor, 30l.ifshit7, E. M., 200Light, linllting velocity of, I ILight cone. 16Light pulse, wllvefront of, 7Light ray, gravitatiollal dellection 01',152Light signal, world· line 01',150Light waves, velocity of, 5Lorentz force, 79

Lorentz transformation, general, 9inverse, I Ispecial,9, 17

Luminosity, apparent, 183, 184intrinsic, 183, 184

Luminosity and red shift, 185, 197Luminosity distance, 184, 197

Mach's principle, 130Magnetic induction, 5, 75

transformation of, 77Magnetic intensity, 5Mass, 3

conservation of, 42, 45,51, 190el)uivalencc of energy and, 46inc rtia I, 43inertia I and gravitational. 130invariance of Newtonian, 4IHelper,42proper density of proper, 53rest, 42transllmnation of, 43

Maxwell's equations, 5, 77, 154Maxwell's stress tensor, 81Mercator's projection, 123Mercury. advance of perihelion 01',1;0Metric, 89, 105

form invariant. 142spl.erically symmetric. 143

Metric aflinity, 110Metrical connection, 105Metric of a gravitational field, 13JMetric of a sphere, 119Metric tensor, 105

covariant derivative of, 110Michelson-Morley experiment, 5Minkowski coordinates. 8Minkowski space-time, 8, 15Misner,C. W., 200Moller. C., 200Moment of force, 58Momentum, 3, 43

conservation 01',42,52, 82density of, 50trallsformation of, 43

Momentum density, 50electromagnetic, 82

MUOll,48

Narlikar,J. V., 200Neutrino, 48

203

204

Newtonian potential, 136, 139, 140, 146.153

Newton's first law, INewton's law of gravitation, 139Newton's laws, covariance of, 4Newton's second law, 2, 44Newton's third law, 3, 44Non·Euclidean space, 88

Ohm's law, 85Orbit, planetary, 69, 147Orthogonal transformation, 21, 86,92Orthogonal vectors, 29, 108

Papapetrou, A., 200Parallel displacement of vectors, 96Parity of space, 26Particle. Hamiltonian for, 49. 83

internal energy 01',46Lagrangian for, 49

Particles. collision 01',3,42Past, absolute, 16Penzias, A. A.. 175Perihelion, advance of, 150Petrov, A. Z., 200Photon, 48Pion, 48Planetary orbits, 147Poisson's equation, 140Potentials, electromagnetic, 74Poynting's vector, 81, 82Present. conditional, 16Privileged observers, 127Pressure, 57Product, inner, 28. 94

outer, 94scalar. 28, 107vector, 3 I

Proper mass, 42proper density of, 53

Pseudotensor,29

Quotient theorem for tensors, 94

Radiation, EM tensor of isotropic, 194Raychaudhuri, A. K., 200Relativity, general principle of, 128

special principle of, 4Ricci tensor, 105

divergence of, I 14symmetry of, 113

Riemann-Christoffel curvature tcnsor,102.105, III

Riemannian space, 89Rindler, W., 200Robertson-Walker metric, 181, 197

Ricci tensor for, 186Rocket, motion 01',61Rowan-Robinson, M., 174, 199,200

Scalar, 27,90Scalar product, 28, 107Schwarzschild radius, 146, 156Schwarzschild's metric, 146, 147

anomaly in, 155Schr6dinger, E., 200Sciama, D. W., 200Shiffer, M., 200Shift factor, red, 183Simultaneity, 13Sirius, companion of, 154Space, gravitational curvature of, 13 ISpace, physical, 127Spaeelike interval, 15Space-time, Min kowski, 8, 15Space-timc continuum, 132Special principle of relativity, 4Spectral lines. gravitational shift of, 154Spherical polar coordinates, 86Stefan's law, 195Stress tensor, 53, 7 I

Maxwell's. 81Summation convention, 23Synchronization of clocks, 7, ! 31Syngc, J. L.. 200

Tachyon, 62Tangent, unit, 115

zero, 11(\Tensor, Cartesian, 24

contrauion of, 28, 93contravariant, 90covariant, 9 Icovariant derivative 01',97,98,100fundamental, 25, 92, 106mixed,91parallel displacement of, 10 Irank of, 24skew-symmetric, 25,93symmctrie, 25,93

Tcnsor equation, 26, 93

Tensor field, 28derivative of, 28

Tensor product, 25,93,94covariant derivative of, 101

Tensor sum, 24, 92covariant derivative of, 101

Thorne, K. S., 200Time interval, proper, 14, 132Timelike interval, 15Tolman, R. C. 200Torsion, 118

Universe, present age of, 192

Vector, axial, 31Cartesian, 24contravariant, 90covariant, 9 Icovariant and contravariant

components of, 106displacement, 24, 89

free, 90magnitude of, 29, 107

Vector identities, 37Vector product, 31Vectors, angle between, 29, 107

orthogonal, 29, 108scalar product of, 28, 107

Velocities, composition of,61Velocity, relative, 41Velocity vector, 39

transformation of, 41Volume of space, 179

Weak gravitational field, 137Weak interaction, 48Weinberg, S., 200Wheeler.J. A., 200Wilson, R. W., 175Work,45World-line, 15World-lines of free particles, 134, 142

205