Introduction to relativity

Introduction to Relativity


William D. McGlinn

The Johns Hopkins University PressBaltimore and London

© 2003 The Johns Hopkins University Press

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McGlinn, William D.Introduction to relativity/ William D. McGlinn.

p. cm.Includes bibliographical references and index.ISBN 0-8018-7047-X (hc. : alk. paper) -- ISBN 0-8018-7053-4

(pbk. : alk. paper)1. Relativity (Physics) I. Title.

QC173.55 .M38 2002530.11--dc21

2002070073

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To Louise

Contents

Preface xi

1 Foundations of Special Relativity 1

1.1 Introduction 11.2 Kinematics: The Description of Motion 11.3 Newtonian Mechanics and Galilean Relativity 31.4 Maxwell’s Equations and Light Propagation 71.5 Special Relativity: Einsteinian Relativity 10

1.5.1 Lorentz Transformation 101.5.2 Lorentz Transformation of Velocities 121.5.3 Lorentz Transformation with Arbitrary Relative

Velocity 13

1.6 Exercises 15

2 Geometry of Space-Time 17

2.1 Introduction 172.2 Invariant Length for Rotation and Euclidean

Transformations 172.3 Invariant Interval for Lorentz and Poincaré

Transformations 182.4 Space-Time Diagrams 19

2.4.1 Causality 212.4.2 Longest Elapsed Proper Time between

Two Events: The Twin Paradox 232.4.3 Length Contraction 262.4.4 Time Dilation 262.4.5 Doppler Shift 27

2.5 Vectors and Scalars 30

2.5.1 Euclidean Vectors and Scalars 302.5.2 Lorentzian Vectors and Scalars 312.5.3 The Doppler Shift Revisited 33

vii

2.6 Rotation and Lorentz Transformations as Groups 342.7 Exercises 35

3 Relativistic Dynamics 39

3.1 Introduction 393.2 Momentum in Galilean Relativity 403.3 Momentum-Energy in Einsteinian Relativity 413.4 The Geometry of the Energy-Momentum Four-Vector 44

3.4.1 “Elastic” Collisions 463.4.2 “Inelastic” Collisions 473.4.3 Particle Production 47

3.5 Relativistic Form of Newton’s Force Law 483.6 Dynamics of a Gyroscope 483.7 Exercises 50

4 Relativity of Tensor Fields 53

4.1 Introduction 534.2 Transformations of Tensors 53

4.2.1 Three-Tensors 534.2.2 Four-Tensors 55

4.3 Relativity of Maxwell’s Equations 584.4 Dynamics of a Charged Spinning Particle 614.5 Local Conservation and Gauss’s Theorem 634.6 Energy-Momentum Tensor 65

4.6.1 Energy-Momentum Tensor of Dust 674.6.2 Energy-Momentum Tensor of a Perfect Fluid 684.6.3 Energy-Momentum Tensor of the

Electromagnetic Field 724.6.4 Total Energy-Momentum Tensor of Charged

Dust and Electromagnetic Field 73

4.7 Exercises 74

5 Gravitation and Space-Time 77

5.1 Introduction 775.2 Gravitation and Light 775.3 Geometry Change in the Presence of Gravity 815.4 Deflection of Light in a Gravitational Field 83

6 General Relativity 87

6.1 Introduction 876.2 Tensors of General Coordinate Transformations 90

viii Contents

6.3 Path of Freely Falling Particles: Timelike Geodesics 926.4 Covariant Differentiation 946.5 Parallel Transport: Curvature Tensor 966.6 Bianchi Identity and Ricci and Einstein Tensors 1026.7 The Einstein Field Equations 1046.8 The Cosmological Constant 1076.9 Energy-Momentum Tensor of a Perfect Fluid in

General Relativity 1086.10 Exercises 110

7 Static Spherical Metrics and Their Applications 113

7.1 Introduction 1137.2 The Static Spherical Metric 1147.3 The Schwarzschild Solution 1167.4 Gravitational Redshift 1177.5 Conserved Quantities 1187.6 Geodesic Motion for a Schwarzschild Metric 120

7.6.1 Gravitational Deflection of Light 1237.6.2 Precession of the Perihelia of Orbits 125

7.7 Orbiting Gyroscopes in General Relativity 1287.8 Stellar Interiors 131

7.8.1 Constant Density Newtonian Star 1347.8.2 Constant Density Relativistic Star 135

7.9 Black Holes 1367.10 Exercises 139

8 Metrics with Symmetry 143

8.1 Introduction 1438.2 Metric Automorphisms 1438.3 Killing Vectors 145

8.3.1 Conserved Momentum 146

8.4 Maximally Symmetric Spaces 1468.5 Maximally Symmetric Two-Dimensional

Riemannian Spaces 153

8.5.1 Two-Dimensional Space Metric of PositiveCurvature 154

8.5.2 Two-Dimensional Space Metric with ZeroCurvature (Flat) 155

8.5.3 Two-Dimensional Space Metric with NegativeCurvature 155

Contents ix

8.6 Maximally Symmetric Three-DimensionalRiemannian Spaces 156

8.6.1 k 0= 1598.6.2 k 1= + 1598.6.3 k 1= - 159

8.7 Maximally Symmetric Four-DimensionalLorentzian Spaces 160

8.8 Exercises 161

9 Cosmology 163

9.1 Introduction 1639.2 The Robertson-Walker Metric 1659.3 Kinematics of the Robertson-Walker Metric 167

9.3.1 Proper Distance 1679.3.2 Particle Horizons 1689.3.3 Event Horizons 1709.3.4 Cosmological Redshift: Hubble’s

Constant 1719.3.5 Luminosity Distance 1739.3.6 Cosmological Redshift: Deceleration

Parameter 174

9.4 Dynamics of the Robertson-Walker Metric 175

9.4.1 Critical Density 1799.4.2 Cosmological Redshift: Distant Objects 1809.4.3 Cosmological Dynamics with � � 0 1829.4.4 Cosmological Dynamics with � ≠ 0 186

9.5 The Early Universe 187

9.5.1 The Cosmic Microwave BackgroundRadiation 188

9.5.2 Inflation 1899.5.3 Cosmic Microwave Background and

Cosmological Parameters 193

9.6 Exercises 196

Suggested Additional Reading 197

References 199

Index 201

x Contents

Preface

This book provides an introduction to the theory of relativity, bothspecial and general, to be used in a one-term course for undergradu-ate students, mainly physics and math majors, early in their studies. Itis important that students have a good understanding of relativity toappreciate the unifying principles and constraints it brings to bothclassical and quantum theories. The understanding of such a beauti-ful subject can bring a great deal of satisfaction and excitement to thestudent.

The history of both special and general relativity is given shortshrift in the book, however. I believe that an initial study of relativityis best done by a straightforward, linear development of the theory,without the twists and turns that are inevitably a part of a theory’shistory. With such an approach, students can understand and appre-ciate the structure of the resulting theory more quickly than if thehistorical path is followed. The historical path has an interest andvalue of its own, but the illumination of this path is best done by histo-rians of science. Therefore, for example, I have not discussed Mach’sprinciple—-the principle that inertial frames are determined by thedistribution of mass in the universe. Undoubtedly, Einstein was influ-enced by Mach, but in the end the answer to the question “DoesEinstein’s general relativity obey Mach’s principle?” is elusive.

The target audience imposes constraints on what material isincluded and on the level of sophistication, especially mathematicalsophistication, employed. I assume the student has had an introduc-tory course in physics, a knowledge of basic calculus, including simpledifferential equations and partial derivatives, and linear algebra,including vectors and matrices. In addition, it is useful, though notnecessary, that students be able to use a symbolic mathematicalprogram such as Mathematica or Maple. Most physics majors atAmerican universities have the required background at the begin-ning of their third year, many as much as a year earlier. With suchstudents in mind, I use the traditional tensor index notation and not

xi

the coordinate-free notation of modern differential geometry.Though the modern notation gives deeper and quicker insight intothe structure of the geometry of space-time, for a first introduction Ithink the traditional tensor index notation is preferable—-and thisnotation is almost always used in calculations. Also, by the use ofsymbolic programs students can do calculations that would be some-what prohibitive without them and, thus, develop understanding byway of calculation.

An author of such a textbook faces a choice as to how and when tointroduce the required new mathematics, the most important “new,”in this case, being tensor analysis (or differential geometry). Shouldone introduce the mathematics separately so that it stands alone, orshould one introduce it within the context of the physics to bedescribed, whereby it can be physically motivated? I do both to somedegree. In the study of special relativity, I introduce the concept ofthe space-time metric and associated tensors within the context of thephysics and in analogy with the comparable entities in three-dimen-sional Euclidean space of which students would be expected to havesome understanding. The concepts of a metric and associated tensorsunder the general coordinate transformations, required in the theoryof general relativity, are introduced somewhat abstractly but arequickly tied to the metric of space-time and physical tensors. Thispath entails a loss of rigor in the discussion. At times, an appeal ismade to the reasonableness of a result rather than giving rigorousarguments: it is clear in the context when this is done. One mathe-matical subject is presented in a “stand alone” manner: in Chapter 8,symmetries of a metric, called isometries, are characterized by the exis-tence of Killing vectors. This is a subject one might think is unneces-sary in such a book. However, symmetries of the metric are soimportant in studying dynamically conserved quantities and in thediscussion of cosmological models that I feel this introduction toKilling vectors is useful.

The book is roughly divided into two parts. The first part, Chapters1 through 4, concerns special relativity. General relativity, including achapter on cosmology, is covered in Chapters 6 through 9. Given theexciting theoretical and observational work being done at the presenttime, the student should find the study of the cosmology chapterparticularly satisfying. Chapter 5 is a reading of Einstein’s firstattempt at incorporating the equivalence principle into physics,thereby predicting the bending of light rays by the gravitational fieldof massive bodies, a reading in which the result is viewed as a changein the Minkowskian geometry of space-time. It serves as a physicalmotivation to changes introduced in his general theory of relativity

xii Preface

and is used to relate the coupling constant introduced in Einstein’sfield equations to Newton’s gravitational constant.

Except for Chapter 5, there are exercises associated with eachchapter. The number of exercises are few enough so that it is notunreasonable to expect the student to do all of them. The impor-tance of students doing these exercises cannot be emphasizedenough. Exercises are particularly important in the study of relativity.The student is susceptible to thinking the material is understoodbecause he or she can “read” the equations. The understandingcomes only through serious thought and by working out problems.Furthermore, some important results are contained only in the exer-cises.

A word about units. A unit of time is introduced so that the speedof light has value one. (In such a set of units, velocities are dimen-sionless.) This is the only change from standard units, such as MKSunits. I do not go “all the way” by defining the Planck units whereinthe velocity of light, Newton’s gravitational constant G, and Planck’sconstant h are all chosen to have value one. Such units are particu-larly useful when studying quantum gravity, a subject we are, mostdefinitely, not considering here.

I am grateful to my late friend, Jim Cushing, and to Steven Shore,who both took the time to read an early version of the notes uponwhich this book is based and to suggest improvements. Jim’s encour-agement was instrumental in my decision to put the notes in bookform suitable for publication.

Preface xiii


Chapter 1

Foundations of SpecialRelativity

1.1 Introduction

Before studying Einstein’s modification of the scientist’s view of spaceand time as held at the beginning of the last century, we will take aquick look at the theories of physical phenomena that were operativeat that time, for the space-time concepts of scientists were shaped bythese theories.

At that time, an overall aim of physicists was to give a mechanicalexplanation of physical phenomena based on Newton’s laws ofmotion. Theories existed for two basic forces: Newton’s theory for thegravitational force, and Maxwell’s theory for the electromagneticforce.

A relativity principle, recognized by Newton and, before him, byGalileo, applies in Newton’s mechanics. With the advent of electro-magnetic theory, as incorporated in Maxwell’s equations, it seemedthat the relativity of Newtonian mechanics could not be therelativity of Maxwell’s equations. We study this (Galilean) relativityof Newtonian mechanics to understand its conflict with the(Einsteinian) relativity of Maxwell’s equations.

1.2 Kinematics: The Description of Motion

The basic question answered (differently) by Galilean andEinsteinian relativity is, “How do different observers see the samemotion?” To address this question, one has to be able to describe the

1

motion of particles quantitatively. A notion prior to motion is that ofan event, that is, a position and a time, for motion is a sequence ofevents. To characterize the position of an event, an observer mustchoose a coordinate system. One can imagine constructing a latticeframe of orthogonal (micro) meter sticks so that the position of anevent that occurs at the intersection of three sticks has coordinates( , , ) ( , , )ml nl ol x y z= , where l is the length of the sticks, and ( , , )m n o areintegers. The position ( , , )0 0 0 is referred to as the origin of the coor-dinate system and is not a special point, if the space is homogeneous.We assume one can place the origin of the coordinates at any pointin space without prejudice and can pick the mutually orthogonal axispointing in any direction. This is clearly an assumption that space ishomogeneous and isotropic. Of course, an event may not occurprecisely at an intersection, but one can choose the length l as smallas one wishes for the accuracy required. Further, we assume that thespace is Euclidean (flat) and of infinite extent, that there is a distancebetween points in the space ( , , )x y z1 1 1 and ( , , )x y z2 2 2 given by( ) ( ) ( )d x x y y z z21 2

21 2

21 2

2= - + - + - , and that this distance does notdepend on the choice of origin or the direction of the chosen axis.This Euclidean nature of space is clearly an assumption, one that isrelaxed in Einstein’s theory of general relativity. (Imagine trying toset up a two-dimensional Euclidean lattice on a two-dimensionalsphere.)

The lattice permits one to record the position of an event. Howmight the time of an event be characterized? One would like to havea set of “synchronized” standard clocks at each lattice intersection sothat an event’s time measurement can be recorded locally. But howdoes one synchronize the clocks? Imagine firing “free” particles fromthe origin with a known velocity, in a direction of a lattice junction.Note that the velocity can be measured locally with nonsynchronizedclocks. Knowledge of the space coordinates of the junction, and thusthe distance from the origin, permits the clock at this junction to besynchronized with the origin clock, if the velocity of the free particleis constant and independent of direction. Frames relative to whichfree particles, particles that have no forces acting on them, move withconstant velocity are a special class called inertial frames. These areframes in which free particles obey Newton’s first law. In discussingspecial relativity, we restrict our description of motion to such inertialframes.

An event is then characterized by a set of space coordi-nates and a time coordinate. That is, we have four numbers( , , , ) ( , , , )t x y z x x x x0 1 2 3= , which we also write as ( , )t r . The motion ofa particle is described by the position r of the particle as a func-

2 Chapter 1. Foundations of Special Relativity

tion of the time t—-a sequence of events ( , ( ))t tr with the t rangingcontinuously over some interval. From this parameterization of themotion, we obtain the instantaneous velocity /d dtv r= or, in compo-nent form, ( , , ) ( / , / , / )v v v dx dt dy dt dz dtx y z = , and the acceleration

/ /d dt d dta v r2 2= = .

1.3 Newtonian Mechanics and Galilean Relativity

Newton’s force law relates the acceleration a of a particle with theforce F through the equation

mF a= . (1.1)

Here m is a fixed property of the particle, called the (inertial) mass.The value of the force Fmust be given by some force law if this equa-tion is not to be merely a definition of F in terms of m and a. This lawimplies that a particle under the influence of no force (a free parti-cle) has a 0= , and thus its velocity v is constant. The law is valid in aninertial frame.

How are these inertial frames related? What is the relation betweenthe descriptions of the motion of a particle referred to in one inertialframe and another? What is the relation between the force law asviewed in one inertial frame and another? These questions can beanswered by a knowledge of the transformation (translation) ofevents between inertial frames.

Consider two frames: an unprimed inertial frame S and a primedframe Sl moving with a constant velocity v0 relative to the unprimedframe. For simplicity we set up the coordinates of the frames in a par-ticular way. We choose the relative velocity of Sl in the x direction andhave the origin of the Sl frame move along the x axis, the origin of theS frame move along the xl axis, and set the time of the event when the

1.3 Newtonian Mechanics and Galilean Relativity 3

Figure 1.1 Two inertial frames.

origins agree to be zero for both sets of clocks. This is possible for ho-mogeneous and isotropic space-times. Furthermore, we will assumethat we can choose the yl and zl axes such that the the locus of allevents with y 0= (z 0= ) agree with that corresponding to y 0=l (z 0=l ),with the >z 0 ( >y 0) events agreeing with >z 0l ( >y 0l ). Figure 1.1 de-picts a single event “O” with space-time coordinates ( , , , )t x y z and( , , , )t x y zl l l l as measured in the unprimed and primed frames, respec-tively. From this figure we see that it is reasonable that events charac-terized by x 0=l are the set of events characterized by vx t 00- = interms of unprimed coordinates.

We might expect that the translation of event “O” between the twoframes is given by

.

v

t t

x x t

y y

z z

0

=

= -

=

=

l

l

l

l (1.2)

The first equation assumes that time is universal, that is, that all stan-dard clocks run synchronized. This translation of events is theGalilean transformation for the special coordinates chosen.

This derivation of the Galilean transformation is at best ad hoc. Wemerely wrote down the transformation equations as one might expectthem to be from viewing Figure 1.1. We now tighten up the derivationa bit in a manner that can be generalized to the Einsteinian case andto understand the assumptions hidden in this derivation.1 Forinstance, no mention was made about transforming between inertialframes—-ones for which a constant velocity observed in one frameimplies a constant velocity observed in the transformed frame. Itshould be reasonable to the reader, and it can be shown, that for thisto be so the transformation between the frames must be linear, thatis, of the form

n xx A Tv 0

3

= +=

v vn n!l , (1.3)

where Ano and Tn are constants and x t0/ . We assume this to be true.We can choose the space coordinates and the time coordinates sothat the event ( , , , )0 0 0 0 , referred to the primed coordinates, is the


1 The derivations of the Galilean and Lorentz transformations that follow are muchinfluenced by those in Rindler (1991).

event ( , , , )0 0 0 0 , referred to the unprimed coordinates, so that T 0=n .Eq. (1.3) then becomes

nx x xA A .v 0

3

= ==

v v v vn n!l (1.4)

In the last expression here, we have used the Einstein convention thatrepeated indexes indicate summation (a convention we will followhereafter).

What are these constants Ano for the transformation between thespecially chosen coordinates described above? Again, for our choiceof ,x yl l and zl axes, the events characterized by x 0=l are characterizedby vx t 00- = . Thus, we have

( ),vx A t A x A y A z A t A x A x t10 11 12 13 10 11 0= + + + = + = -l (1.5)

where A is some constant. If one considered the transformation fromthe primed coordinates to the unprimed, one would obtain

( ),vx A x t0= +l l l (1.6)

where Al is some constant. How are Al and A related? If one assumes an equality of the two

frames in the sense that no observation can be made that distin-guishes one frame as being different or special, then Al must equalA. For instance, consider an event O, say ( , , , )x L0 0 0=o , simultaneouswith the origin event in the unprimed frame. From Eq. (1.5), for thisevent x AL=l . Similarly, a different event Ol, say ( , , , ),x L0 0 0=ol simul-taneous with the origin event in the primed frame, from Eq. (1.6),has x A L= l . The assumed equality of the two frames requires A A=l

and thus Eq. (1.6) becomes

( )vx A x t0= +l l . (1.7)

Similarly, for our choice of axes, since events characterized by y 0=lare characterized by y 0= , the relation

y A t A x A y A z A y20 21 22 23 22= + + + =l (1.8)

follows. But, if we considered the transformation from the primed tothe unprimed frame, we would obtain

22 ,y A y= l l (1.9)

1.3 Newtonian Mechanics and Galilean Relativity 5

and, again with the assumed equality of the frames, A A22 22= l . Withthis, Eqs. (1.8) and (1.9) imply A 122

2 = . Since we chose our axis suchthat >y 0 corresponds to 22> ,y A A0 122= =l l . Similarly, we concludethat A 133= .

We see that a transformation between the inertial frames (linearityof transformation) with the axes as chosen and with the assumptionthat no inertial frame is special assumes the form

( )vx A x t0= -l

( )vx A x t0= +l l

y y=l

z z=l

.t A t A x A y A z00 01 02 03= + + +l (1.10)

Without further assumptions—-such as, perhaps, that space isisotropic—-these relations seem to be as far as we can go. Theassumption of a Newtonian universal time implies that A 100= and

, ,A i0 1 2 3i0 = = , and these, with the first two relations of Eq. (1.10),imply A 1= . The relations Eq. (1.10) reduce to

vx x t0= -l

y y=l

z z=l

,t t=l (1.11)

which are the Galilean transformation, Eq. (1.2), deduced before.The transformation of velocity components, defined by/ , , ,v dx dt i 1 2 3i i= = for an arbitrarily moving particle, immediately

obtains

xv v vx 0= -l

yv v y=l

z .v v z=l (1.12)

For the acceleration components, / , , ,va d dt i 1 2 3i i= = , we have

xa ax=l

ya ay=l

a az z=l . (1.13)

The two different frames record the same acceleration. If mF a=


is valid in the unprimed frame, then mF a=l l is valid in the primedframe if

.F F= l (1.14)

Thus, if Newton’s law of motion is valid in one frame, the lawis valid in all frames that move with constant velocity with respectto it if (1) the transformation of events is given by a Galilean trans-formation and if (2) all forces are the same in all such frames. Theforces cannot depend on the relative velocity. If the form of thelaw is not changed by certain coordinate transformations, the law issaid to be invariant with respect to the transformations considered.Newton’s law mF a= is invariant with respect to Galilean transforma-tions.

Newton believed in an absolute space or reference frame and thatinertial frames were those at rest or in uniform motion with respectto it. However, he recognized the difficulty in discovering which of allinertial frames is the absolute frame because his laws of motion areform invariant under change of frames if the transformation law ofevents is given by a Galilean transformation and if, under these trans-formations, the force doesn’t change. Such invariance of all physicallaws between all inertial frames is the principle of relativity. Note that ifthis principle is valid, then no experiment could be performed thatwould distinguish one particular frame as being special in any way.(Recall that we used this in the derivation of the Galilean transfor-mation.) The principle of relativity and the assumption of the exis-tence of a universal time give rise to Galilean relativity. Force laws, tobe consistent with this relativity, must be such that they give an iden-tical force in all inertial frames. Since the vector distance between twosimultaneous events is the same in all inertial frames, any law of forcebetween two bodies that depends only on the vector distance betweenthe bodies will satisfy this principle. Newton’s gravitational force issuch a law. However, a force law that depends upon both velocities ofthe two bodies (and not just on their relative velocity) would violatethis principle.

1.4 Maxwell’s Equations and Light Propagation

In 1861 James Clerk Maxwell, born in Edinburgh in 1831, formulatedmathematical equations that describe electromagnetic phenomena.These equations predict the existence of electromagnetic waves thattravel with a velocity that could be calculated from the theory—-calcu-lated in terms of a parameter of the theory that was experimentally

1.4 Maxwell’s Equations and Light Propagation 7

determined by measuring the electric force between charges (or themagnetic force between currents). The calculated velocity agreedwith that of light as determined by the Danish astronomer OlausRoemer in 1675 by observing the time lag of the eclipses of Jupiter’smoons—-a remarkable agreement that could only imply that light isan electromagnetic wave.

In spite of the great successes of Maxwell’s equations, most physi-cists (including Maxwell himself ) demanded a mechanical explana-tion of these waves and, thus, the existence of a medium, called“ether,” whose mechanical vibration constituted electromagneticwaves. The speed of light predicted by Maxwell’s equations would bethe speed in this all-encompassing ether. The principle of Galileanrelativity is clearly in trouble. If the ether were a valid concept, itwould provide the “absolute” reference frame, thus violating the prin-ciple; if Maxwell’s equations are valid in all inertial frames, and thusthe velocity of light is the same in all inertial frames, the principle ofGalilean relativity, which predicts the simple addition of velocities, isinvalid.

Irrespective of the existence of the ether, it was crucial to detect themotion of the earth with respect to the frame in which the speed oflight has a fixed standard value, independent of the direction ofpropagation. The most famous attempt to detect this motion was anexperiment performed by two Americans, Albert A. Michelson andEdward W. Morley. Michelson first performed the experiment in1881 and, in collaboration with Morley, again in 1887, with moresensitivity.

Michelson’s technique for measuring the motion of the earththrough the ether was to split a beam of light in two, send thesplit beams in two mutually perpendicular directions for approxi-mately equal distances (L in Fig. 1.2), reflect them back to acommon point, and measure the difference in travel times of the two.In the figure, the ether is assumed to have a velocity relative to theearth of magnitude v e, which is assumed to be perpendicular to thedirection traveled by the light in one path. One can show (seeExercise 1) that the expected difference in travel time td of the twobeams is

.v

t cLce

2

.d c m (1.15)

The difference in arrival times for a reasonable length L is very small—-it is proportional to the square of /v ce . If v e is taken to be the orbit


speed of the earth around the sun, about 30 km/s, /v c 10e4. - , which

when squared is a very small number indeed. Michelson did notmeasure the difference in arrival times directly. Rather, he measuredthe shift in the interference pattern as the apparatus was slowlyrotated. That is, as the apparatus was rotated the arms would inter-change the roles of lying parallel and perpendicular to the supposedflow of the ether, thus changing the difference in arrival times of thebeams traversing the two arms, resulting in a shift in the fringes of theinterference pattern. The total change in the lag time is twice that ofEq. (1.15). A change of one period x corresponds to a shift of onefringe! The number of fringes n shifted when the apparatus is rotatedby 90o is thus

n t c t2 2= =xd

md . .

vLc

2 e

2

mc m

Here m is the wavelength of the light used. The 1887 experiment ofMichelson and Morley had L 11. m, which with 5 10 7#=m - m, a typi-cal wavelength of visible light, yields .n 4. . This is quite detectable—Michelson and Morley had estimated they could detect .n 01= .However, the experiment gave a null result. No fringe shift wasdetected.

1.4 Maxwell’s Equations and Light Propagation 9

Figure 1.2 Michelson-Morley experimental setup.

1.5 Special Relativity: Einsteinian Relativity

It seems that Einstein was not influenced greatly by the Michelson-Morley experiment although he probably knew of the null result.Rather, he doubted the existence of an absolute frame of referencethat the Michelson-Morely experiment was attempting to detect. Inhis later years, Einstein recalled that as a boy of sixteen he wonderedhow a light wave would appear if one were moving along with it.2

Galilean relativity would predict a static, nonoscillatory wave. Butsuch a static wave does not satisfy Maxwell’s equations if the parametersthat define the theory do not change from one frame to another. After all,these parameters fix the speed of the wave. Thus, if there is a relativ-ity principle and Maxwell’s equations are correct, then Galilean rela-tivity and Newton’s mechanics are wrong.

In his remarkable paper of 1905 titled “On the Electrodynamics ofMoving Bodies,”3 Einstein broke with Galilean relativity and itsconcomitant view of space and time. In this paper he enunciated twopostulates:

1. The principle of relativity: All physical laws have the same form(i.e., they are invariant) in all inertial reference frames.

2. The speed of light is the same in all inertial reference frames.

Again, inertial frames are those in which isolated particles, thosewith no “force” acting on them, move with constant velocity. The firstpostulate implies that no measurement can be made that distin-guishes one frame as being special or different from any other frame.

The second postulate of Einstein, the invariance of the speed oflight to all inertial observers, immediately contradicts Galilean trans-formations of events, since such transformations imply that no speedis invariant. Special relativity is a study of transformations of events,called Lorentz transformations, that are consistent with the invariance ofthe speed of light.

1.5.1 Lorentz Transformation

We now derive the form of the transformation between the two iner-tial frames depicted in Figure 1.1. Because the speed of light is


2 See Autobiographical Notes in Schilpp (1949), p.53. For a view questioning the accuracyof Einstein’s recollection, see Bernstein (1973), p.38.3 Annalen der Physik 17, 891 (1905), reprinted (in English translation) in Einstein et al.(1923).

special—-we will see it is the only speed that has the same value for allinertial observers—-it is useful to introduce a time unit that reflectsthis special role. The time unit used is the amount of time it takeslight to travel a distance unit. Thus, if we measure distance in meters,the time unit will be “one meter of light travel time.” In such a systemof units, a velocity is dimensionless since time and distance have thesame units. When a velocity is expressed in terms of dimensionlessunits we will use the symbol b for velocity. | | 1=b for light. Note alsothat we usually indicate the time coordinate of an event by “x0” ratherthen “t.”

Again, the transformation of events between the two frames islinear, and we choose the space coordinates and the time coordinatesso that the “origin” events ( , , , )0 0 0 0 of the two coordinates are thesame event:

n .x A x= no ol (1.16)

As before, the events characterized by x 01 =l are characterized byx x 0r1 0- =b , where rb is the velocity of the primed coordinate withrespect to the unprimed. If one applies the principle of relativity asbefore, we again obtain Eqs. (1.10):

2

( )

( )

,

x A x x

x A x x

x x

x x

x A x A x A x A x

r

r

1 1 0

1 1 0

2

3 3

0 00 0 01 1 02 2 03 3

= -

= +

=

=

= + + +

b

b

l

l l

l

l

l (1.17)

where A is some constant dependent on rb .Now, we do not assume a Newtonian universal time x x0 0= l as we

did before, and do not obtain A 1= . However, consider a sequence ofevents of a light pulse traveling in the x1+ (and thus x1+ l ) directionwhose emission was the origin event ( , , , )0 0 0 0 . These events are char-acterized (for light 1=b ) by x x1 0= and x x1 0=l l . Using these relationsin the first two equations of Eqs. (1.17), we obtain

( ) ( )x A x x A x x1r r1 1 0 0 0= - = - =b bl l

0( ) ( ) .x A x x A x x1r r1 1 0 0= + = + =b bl l l (1.18)

From this it follows that

( ) .A 1 /

r

2 1 2= - b - (1.19)

1.5 Special Relativity: Einsteinian Relativity 11

The positive square root is chosen since we assumed the axes werechosen so that a light pulse traveling in the x1+ direction travels in thex1+ l direction. Furthermore, since the first two equations of Eqs.

(1.17) are valid for the transformation of any event , we can solve forx0l in terms of x0 and x1, and the last of Eqs. (1.17) becomes

( ) ( ).x x x1 /

r r0

2 1 21 0= - - +b b-l (1.20)

The Lorentz transformation of events between these two frames isgiven by

( )x x xr0 1 0= - +c bl

( )x x xr1 1 0= -c bl

2x x2=l

3 .x x3=l (1.21)

Here ( )1 /

r

2 1 2= -c b - , a standard notation. We refer to such a Lorentztransformation, that is Eq. (1.21), as canonical. Solving these equa-tions for the xn in terms of nxl , thus obtaining the inverse transfor-mation, one finds

0( )x x xr0 1= +c b l l

( )x x xr1 1 0= +c bl l

x x2 2= l

.3x x3 = l (1.22)

These equations can also be obtained from Eqs. (1.21) by substitutingnx x)n l and r r" -b b .

1.5.2 Lorentz Transformation of Velocities

First note, because the Lorentz transformations are linear, the differ-ence in the space-time coordinates of two events transforms in thesame way as the coordinates of a single event, that is,

( )x x xr0 1 0= - +c b� � �l

( )x x xr1 1 0= -c b� � �l

2x x2=� �l

.x x3 3=� �l (1.23)


Consider then two events, each corresponding to the position of aparticle and the time the particle is at the position, with the twoevents close in time. The above equations give the transformation ofthe difference in the space-time coordinates of two such events. If onedivides the last three of these equations by the first, one easily obtainsthe transformation equations that relate the components of the veloc-ity of the particle:

1

( )

( )

xx

x x

x x

1r

r

r

r1

0 1 0

1 0

1

1= =- +

-=- +

-b

b

b

b b

b b

��

� �

� �l

ll

0 ( )

( ) ( )

xx

x x

x 1

1

1/ /

r

r

r

r2

2

1 0

2

2 1 2

1

2

2 1 2

= =- +

-=- +

-b

b

b

b b

b b

��

� �

�l

ll

03 ( )

( ) ( ).

xx

x x

x 1

1

1/ /

r

r

r

r3

1 0

3

2 1 2

1

3

2 1 2

= =- +

-=- +

-b

b

b

b b

b b

��

� �

�l

ll

(1.24)

Two important observations should be made about these velocitytransformation equations:

1. It is rather easy to argue that if 11

2

2

2

3

2+ + =b b b then

1 2 3 12 2 2+ + =b b bl l l . Thus, if a particle is moving with a velocity of

light in the primed frame, it moves with the velocity of light in theunprimed frame. (See Exercise 2.)

2. If the particle is moving with a constant velocity in one frame, itmoves with a constant velocity in the other frame. Thus, if oneframe is inertial, the other is as well; these transformations relateinertial frames.

1.5.3 Lorentz Transformation with Arbitrary Relative Velocity

The generalization of the canonical Lorentz transformation Eq.(1.21) for which the relative velocity is in the direction of the x2(or x3) is clear. Thus,

2

3

( )

( )

.

x x x

x x

x x x

x x

r

r

0 2 0

1 1

2 0

3

= - +

=

= -

=

c

c

b

b

l

l

l

l (1.25)

1.5 Special Relativity: Einsteinian Relativity 13

It is useful to know the Lorentz transformation of events between“parallel” frames having arbitrarily directed relative velocity rb . Inorder to derive such transformations, we characterize properties ofthe canonical transformation with reference to the direction of therelative velocity b.

First, the equation for x0l involves only the component of r in thedirection of b and can be written in a rotation invariant form as

( ).x x rr0 0 $= -c bl

Second, the components of the space part of the event that areperpendicular to rb are unchanged. This can be assured by writing

( , , )xr r rS r r0= + b bl t .

Here S is a “scalar”— it does not change under a rotation of coordi-nates.

Finally, for the component of the event in the direction of rbt , r||l ,

r r|| $= cl xr r r0- cb b bt t .

These last two equations can be combined to yield a vector equa-tion for the transformed space components,

( )( ) ,xr r r1 r r r 0$= + - -c cb b bl t t (1.26)

which, with the “time” component equation from above,

( ),x x rr0 0 $= -c bl (1.27)

give a Lorentz transformation of events between “parallel” frameshaving arbitrarily directed relative velocity rb . These frames are“parallel” in the sense that the components of the relative velocity ofthe primed frame with respect to the unprimed are the negatives ofthe relative velocity of the unprimed with respect to the primed. Thisdoes not imply that the primed axis is parallel to the unprimed axis.Eq. (1.26), a vector expression, written in component form withrespect to any (space) coordinate system, gives, with Eq. (1.27), thetransformation of the coordinates of events. Note that if rb hascomponents ( , , )0 0rb , these expressions reduce to the canonicaltransformations.


We can easily deduce the transformation of velocities betweenthese “parallel” frames. From Eqs. (1.25) and (1.26), it follows that

0 ( )

( )( )x x

xrr

r r1

$

$= =

-

+ - -b

cc c

bb b b

��

� �� 0

r

r r r

0

lll t t

( )

( )

1

1

$

$=

-

+ - -

bcc c

bb b bb b

r

r r rt t

( )

( ) ( ) ( ).

1

1 r

$

# #=

-

- + -

cc c

b bb b b b b

r

r rt t

(1.28)

1.6 Exercises

1. Derive Eq. (1.15).

2. Show that if 11

2

2

2

3

2+ + =b b b in Eq. (1.24) then 1 2 3 .1

2 2 2+ + =b b bl l l

3. Show that if 1$ =b b in Eq. (1.28), then .1$ =b bl l

4. Show that for the transformation between frames given by Eqs.(1.26) and (1.27), the components of the relative velocity of theprimed frame with respect to the unprimed are the negatives ofthe relative velocity of the unprimed with respect to the primed.

5. In a given intertial frame, two particles are shot out simultaneouslyfrom a given point with equal speeds b and in orthogonal direc-tions, say x1(particle A) and x2 (particle B). (a) What are the speedsof each projectile relative to the other? (b) What angle does thevelocity of B make with the x1 axis of A as seen by an observermoving with A?

1.6 Exercises 15

Chapter 2

Geometry of Space-Time

2.1 Introduction

Einstein has changed immeasurably our concept of space-time. TheLorentz transformations imply that the time of an event as measuredin one frame depends on both the position and time of the event asmeasured in a second frame moving with respect to the first. Thus,both the time and the position of a single event are different in differ-ent frames. Is there some measure of an event (or difference betweentwo events) that remains invariant (i.e., is the same in all frames)?What characterizes the general transformation between inertialframes?

2.2 Invariant Length for Rotation and EuclideanTransformations

Turning to familiar ground, we study the invariant of the rotationtransformations in three space dimensions. Consider two space coor-dinate systems, relatively stationary and with the same origin, onerotated with respect to the other. The general form of the transfor-mation between such coordinates is given by

i ,x A xij jj 1

3

==

!l (2.1)

such that

ix xi

jj

2

1

32

1

3

== =

! !l (2.2)

17

That is, the squared length of the position vector is invariant underthe rotation.

One can consider a more general transformation between coordi-nate systems for which the origins do not agree, as in

i jA x Xx ij ij 1

3

+==

!l . (2.3)

Here X i are the components of some fixed translation vector. Forsuch a transformation,

i j ,x A xijj 1

3

=� �=

!l

and thus

x xi jji

2 2

1

3

1

3

=� �==

!! l (2.4)

Such transformations are referred to as Euclidean transformations.

2.3 Invariant Interval for Lorentz and PoincaréTransformations

Galilean transformations, including transformations to moving coor-dinates, are not characterized by a quadratic invariant such as Eq.(2.4). (There is a trivial quadratic invariant for Galilean transforma-tions, namely 0 .x x2

02=� �l ) However, there is an invariant, comparable

to that of Eq. (2.2), of the canonical Lorentz transformations given byEq. (1.21). The expression x x x x0

212

22

32- - - computed using the

primed and unprimed coordinates of an event is the same. Similarly,using the difference of the space-time coordinates of two events, wehave

0 1 2 3 .x x x x x x x x202

12

22

32 2 2 2 2= - - - = - - -x� � � � � � � � �l l l l (2.5)

We call 2x� the invariant interval (between the two events). The geom-etry of space-time defined by this invariant interval was introduced byH. Minkowski and is referred to as Minkowski space.1

We have, then, an invariant of the canonical transformationsbetween the specially related frames. But we can consider the trans-

18 Chapter 2. Geometry of Space-Time

1 Minkowski’s paper, given as an address in 1908, is reprinted (in English translation)in Einstein et al.(1923).

formations between frames whose relative motion is not along bothframes’ x1 axis, whose space axis directions do not agree or whoseorigin events are not the same. We expect that we could effect themost general transformation by performing a sequence of transfor-mations of the canonical type, space rotations, space translations, andtime translations. All leave 2x� invariant, since space rotations, spacetranslations, and time translations leave x0

2� and x x x12

22

32+ +� � � sepa-

rately invariant. Thus, these more general transformations are of theform

n ,x A x Xv 0

3

= +=

no o n!l (2.6)

such that 2x� is invariant. Here Xn is some constant translation inspace-time. Such transformations are referred to as Poincaré transfor-mations or inhomgeneous Lorentz transformations. The subset of suchtransformations with X 0=n is called the homogeneous Lorentz transfor-mations, or merely the Lorentz transformations.

Note that the invariant interval 2x� can be positive, negative, orzero in contrast to the rotation invariant distance that is positive forseparated points. A positive invariant interval is said to be “timelike,”a negative one “spacelike,” and a zero invariant interval is said to be“lightlike.” Two events corresponding to the position and time of aparticle moving with the velocity of light have 02=x� in all frames.Conversely, if a particle moves so that 02=x� , it is moving with thevelocity of light. The velocity of light is the same in all frames.

2.4 Space-Time Diagrams

It is sometimes useful, in the study of events and sequences of events,to plot events on a Cartesian space of two or three coordinate axes,with one axis representing time x0 and the other one or two axesrepresenting, say, x1 or x1 and x2 . These Cartesian coordinates are thespace and time coordinates of the event for a particular inertialframe. The set of events that are light-like with respect to the originevent, that is, those events which satisfy x x x 02

02

12

32= - - =x� , form

two cones, as depicted in Figure 2.1. These cones are the futureand the past light cones of the origin event. Events within these conesare timelike with respect to the origin event; that is, they satisfy

>x x x 0202

12

32= - -x� , whereas those outside the cones are spacelike,

with <x x x 0202

12

32= - -x� . The motion of a particle consists of a

sequence of events, and events are represented by points in the space-

2.4 Space-Time Diagrams 19

time coordinate system. Thus, the motion of a particle is representedby a line, called a world line. There is a restriction on the possibleworld lines of particles. A particle traveling faster than the speed oflight has never been observed and, as we will argue, “causality” wouldseem to dictate that such a velocity is not possible. The magnitude ofthe velocity of a particle is given by | | | / |xr 0=b � � , which is the inverseof the “slope” of the world line—-the “slope” of the world line mustbe greater than (or equal to ) one. For any event A that is on theworld line of a particle, the ensuing part of the world line must lie inthe future light cone of event A (see Fig. 2.1).

Also in the space-time coordinates of the unprimed frame, we canconsider the primed coordinates of events, which are related to theunprimed coordinates by the canonical Lorentz transformation Eq.(1.21). The world line of the origin of the primed coordinates,defined by x 01 =l , is given by the locus of events such that1 ( ) .x x x x xor0 r r1 0 1 0= = - =c b bl This line is depicted in Figure 2.2,

labeled as the axis 0xl . It makes an angle tan r1=i b- with the x0 axis.

All lines parallel to this axis are events of constant 1xl . Similarly,the locus of events simultaneous, in the primed frame, with theorigin event are characterized by 0 ( )x x x0 r0 1= = -c bl or x xr0 1= b ..This line is depicted in Figure 2.2, labeled as the axis 1xl . It makes anangle tan r

1=i b- with the x1 axis. All lines parallel to this axis areevents of constant 0xl , that is, they are loci of events that are simulta-


Figure 2.1 Space-time.

neous in the primed frame. This opposite direction of rotation ofthe time axis to that of the space axis reflects the difference betweenthe geometry of Minkowski space-time and space. In the figure, theprimed and unprimed coordinates of an event A are indicated. Acautionary note: the scales of the primed and unprimed coordinateaxes are not equal.

2.4.1 Causality

One can argue rather easily that, if two events, A and B, are separatedby a spacelike interval, there are inertial frames for which B occursafter, before, and simultaneous with A. Choose the A event to be theorigin event and the B event to have vanishing x2� and x3� compo-nents in some frame. One can always choose the space axes so thatthis is true. Thus, the B event has components ( , , , )x x 0 0B B

0 1 with( ) ( ) <l x x 0B B202

12- = - . Two such events are noted in Figure 2.3. The B

event lies on the hyperbola defined by < .l x x 0202

12- = - The interval’s

time component, that is, the time component of the B event, in aprimed frame moving with a velocity rb in the x1 direction, is given by

( ).x x xrB B

0 1 0= - +c bBl (2.7)


Figure 2.2 Space-time “rotation.”

The A and B events are simultaneous, in this frame, if /x xrB B0 1=b ,

which has a magnitude less than one since the interval was assumedto be spacelike. In fact, there exists an inertial frame, characterized bysome value for rb , for which x0

Bl takes on any given value. That is, inviewing the coordinates in Figure 2.3 as “primed” coordinates, thereexists a “primed” frame for which the event B has coordinates at anypoint on the hyperbola. Clearly, two events separated by a spacelikeinterval cannot be causally related. If, say, A caused (or effected)event B, for some observers B would occur before A and effect wouldprecede cause. This implies that particles cannot have a speed greaterthan that of light, for with such a particle, event A could be the emis-sion of such a particle and B could be its reception. Event B wouldthen be caused by event A. We conclude that a particle can onlyproceed into or on its own future light cone.

In a manner similar to the above, one can argue that, if the twoevents A and B are separated by a timelike interval, say with B in thefuture light cone of A, as shown in Figure 2.4, then there exists aninertial frame in which the interval has only a time component. Thatis, the two events occur at the same position. The value of this timecomponent ( ) /2 1 2=x x� � is called the elapsed proper time of the interval.It is the elapsed time of a clock stationary in the frame for which thetwo events occur at the same position.


Figure 2.3. Spacelike interval.

2.4.2 Longest Elapsed Proper Time between Two Events:The Twin Paradox

The familiar statement, “The shortest distance between two points isa straight line,” reflects the geometry of space for which the invariantdistance between two nearby points is ( )x y /2 2 1 2+� � . For consider thetwo points A and B and paths P1 and P2 depicted in Figure 2.5a. P2 hasa longer path length than the straight line path P1.

This can be shown in a rotated coordinate system (recall thatdistance is invariant under rotation) in which the path P1 lies only inthe y′−axis as in Figure 2.5b. The total length l2 of path P2 is

( ) .l x y /2

2 2 1 2= +� �! l l

But

( ) ( ) ,x y y l/ /2 2 1 2 2 1 21$+ =� � �!! l l l

where l1 is the length of path P1.


Figure 2.4. Timelike interval.

Figure 2.5. Shortest distance.

For a path in space-time—-a world line—-a comparable statementis: “The longest elapsed proper time of a world line between two time-like separated events is that of a straight world line.” This reflects thegeometry of space-time in which the invariant propertime betweentwo nearby timelike separated events is given by ( )x x /

02

12 1 2= -x� � � .

(We consider world lines for which x x 02 3= = , a restriction easilyrelaxed.) Consider timelike separated events A and B, with coordi-nates for some inertial frame as shown in Figure 2.6a and two worldlines between A and B, one “straight” P1 and the other not, P2. We willargue that the total propertime for P2 is smaller than that of P1. Thepath P1 represents the world line of the origin of an inertial framemoving with constant velocity /x x1 0=b , and thus we can imaginethese world lines viewed in such a frame, as in Figure 2.6b. Since thetotal elapsed propertime of a path is the sum of infinitesimal invari-ants, it can be computed in any inertial coordinate system. The totalproper time 2x of P2 is given by

0( )x x /2

212 1 2= -x � �! l l 0 .x x B0 1# = = x�! l l (2.8)

Note that we assume that path P2moves into its own future light cone.Thus, P1has the longest possible elapsed total proper time of any pathbetween A and B. If the paths P1 and P2 of Figure 2.6b were the worldlines of twins O1 and O2, respectively, Eq. (2.8) shows that between thetwo events A and B corresponding to the crossing of their world linesO1 has had a longer elapsed time than O2; O1 has aged more than O2..

(We assume the biological clock—-or any other clock—-runs synchro-nized with the proper time. The principle of relativity demands this.)This is the famous twin paradox—the paradox arises, supposedly,because if one considers the process from a coordinate system of O2


Figure 2.6. Longest proper time.

one might expect that O2 should have the longer elapsed time. Hewould not. There is no paradox—-there is no symmetry in the historyof the two twins. O1 stays in a fixed inertial coordinate system; O2changes inertial coordinate systems as he undergoes accelerationwith respect to O1.

We can illuminate the asymmetry in the twins’ history by consider-ing an example for which the world line of O2 is particularly simple,as depicted in Figure 2.7. The velocity of O2 is /3 4 for a time /T 2, asmeasured by O1, after first crossing the world line of O1. O2 thenchanges his inertial frame by changing his velocity to /3 4- relative toO1, who remains in a fixed inertial frame. O2 returns to intercept O1lsworld line at a time T after the first crossing. Two lines of events,simultaneous as observed by O2, are plotted. One line of events issimultaneous to an event on the world line of O2 immediately beforeO2 turns, the other line is simultaneous to an event on the world lineof O2 immediately after. The elapsed proper time 2x� for the twoparts of the world line for which O2 remains in an inertial frame isgiven by

( / ).T47

22=x�


Figure 2.7. Twins’ history asymmetry.

The elapsed proper time 1x� of the world line of O1 as observed byO2 for each of these periods 2x� is /T7 32. Thus,

.T327

47

1 2= =x x� �

There is complete symmetry in the behavior of moving clocks asseen from O1 and O2 while O2 remains in a fixed inertial frame.Nevertheless, the total elapsed time for O2’s clock between leavingand returning is /T7 4, whereas for O1’s it is T .

2.4.3 Length Contraction

We can use the Lorentz transformation of events (or the invariantinterval expression) to calculate the length of a moving stick, and wewill see that the stick is contracted. It is a matter of asking the correctquestion. Consider the world lines of the ends of a stick of length L0at rest in the primed frame and the two “events” ( , , , )0 0 0 0 and( , , , )L0 0 0c on the two world lines and simultaneous in the unprimedframe. (See Fig. 2.8.) Lc is the length “observed” by the unprimedobserver. From the Lorentz transformation equations, Eq. (1.21), theevent of the “leading” end of the stick, ( , , , )L0 0 0c , has primed coor-dinate 1x L Lc

00= = cl or ( )L L1 /

c r2 1 2

0= - b ; the stick is contracted!

2.4.4 Time Dilation

In a similar manner, we can use the Lorentz transformation of eventsto calculate what time interval is observed by the unprimed observer


Figure 2.8. Length contraction.

between two events that occur at the same position in the primedframe. The two events can be considered to be two successive ticksof a clock at rest, at 1x 0=l , in the primed frame. Consider then twosuch events, event A, the “origin” event, and B with primed coordi-nates ( , , , )T 0 0 0l . (See Fig. 2.9.) By use of the Lorentz transformationof events, Eq. (1.22), the event B has unprimed coordinates

( )x T T T x Tand1 /r r02 1 2

1/ = - = =b c b- l l . Time is dilated; moving clocksrun slowly! We have already seen this time dilation in the twin para-dox discussion.

The observation of time dilation, experienced by particles thatdecay, is made many times each day, thus confirming the predictionof special relativity. As an example, beams of rmesons are producedby bombarding nuclear targets with high-energy protons. These pionsleave the target with speeds up to .99=b . The lifetime of a pion in itsrest frame is about 5.4 meters. Thus, if time were not dilated, onewould expect the beam to drop in intensity by a factor of /e1 each 5.4meters that it travels. In fact, consistent with time dilation, it decaysmuch more slowly with distance. (See Exercise 6.)

2.4.5 Doppler Shift

Though the speed of light is the same in all inertial coordinatesystems, the frequency (and wavelength) changes from one frame toanother. This change is referred to as the Doppler shift. Space-timediagrams are useful in investigating this shift. Consider a light wavetraveling in the x+ direction with a period T as observed by the


Figure 2.9. Time dilation.

unprimed observer. What is the period T l in the primed framemoving in the x direction with velocity rb relative to the unprimedframe?

Figure 2.10 depicts the world lines of successive wave crests, onepassing x xat0 01 0= = and then the next passing x 01= at x T0 = . Since.

the crests travel with the velocity of light, their world lines haveslope 1. T l is the proper time of event A, with unprimed coordinates( , )x xA A0 1 corresponding to crest 2 passing x 01 =l . From the figure, we

see that

.x T x T xA ArA

0 1 0= + = + b

Thus,

/( ).x T 1Ar0 = - b

But

r( ) ( )( ) ( ) ( )

( ),T x x T T

T1 1 1

1A A

r r r

r202

12

2

2

2

2 22= - =

---

=-

+

b b

bbb

l

or

( )

( ).T T

1

1/

/

r

r1 2

1 2

=-

+

b

bl (2.9)

In terms of the frequency, /T1=o , we obtain

( )

( ).

1

1/

/

r

r1 2

1 2

=+

-o o

b

bl


Figure 2.10. Doppler shift.

If the primed frame is moving in the positive direction, ol is smallerthan o. If that frame is moving in the negative direction, ol is largerthan o.

It is instructive to derive the Doppler shift by considering how aplane light wave, traveling in the x+ direction, will look in the twoframes. Suppose in the unprimed frame the wave has a wave vector

/k 2= r m and a frequency 2=~ ro, with of course /k 1=~ , the velocityof light in our system of units. Let, then, the electric field in theunprimed frame be represented by

( ).cos kx xE E 1 00= - ~

The magnetic field of the wave has a similar form,

( ).cos kx xB B 1 00= - ~

Though not relevant to our considerations, E B 00 0$ = and, for ourwave, both E0 and B0 lie in the x x2 3- plane. The observant reader willnote that we are describing a plane-polarized electromagnetic wave.We might expect (see Sec. 4.5) that the electric field, in the primedframe, at a given event is a linear combination of the electric andmagnetic fields of the unprimed frame at the same event and thus isof the form

( ),cos kx xE E 1 00= - ~l l

which, when expressed in terms of the primed coordinates of theevent, becomes

1 00[ ( ) ( )]cos k x x x xE E r r 10= + - +c b ~c bl l l l l l

1 00 [ ( ) ( ) ].cos k x k xE r r= - - - +c ~b c b ~l ll


( ) ( ) ,k 1r r= - + = -~ c b ~ c b ~l

or

( )

( ).

1

1/

/

r

r1 2

1 2

=+

-o o

b

bl


2.5 Vectors and Scalars

2.5.1 Euclidean Vectors and Scalars

We saw that Newton’s equation mF a= has an invariant meaning,under Galilean transformations, if the components of F transformfrom one inertial frame to another in the same way as the componentsof a. This statement is of course true if the transformations arerestricted to the rotations. The equality of any two vectors has aninvariant meaning under rotation if indeed they transform in thesame way and we can thus attach an absolute meaning to a vectorindependent of the particular frame of reference. The componentsthen are the realization of the vector referred to a particular referenceframe. With this view, a three-vector is an object whose three compo-nents transform, under rotation, from one frame to another with thesame transformation as the difference in the coordinates of two posi-tions. Under a rotation, the coordinates of a position transform,

,x A xi ij j=l (2.10)

in such a way as to leave the distance squared between two positionsinvariant. We can write this invariance relation as

j ,r x x x x A x A xi ij j i ij ik k ij jm m2 = = =d d d� � � � � � �l l

where ijd is the Kronecker delta, defined by

, ,

, ,

i j

i j

1

0

if

ifij ==

d *Since this equality holds for all ,xk�

,A Aik ij jm km=d d (2.11)

or

.A Aik im km= d (2.12)

By definition, then, a three-vector v has components vi and vilrelated by

i .v A vij j=l

If one defines the inner product of two vectors a ba b i i$ = , then


�

i i .a b A a A b a b a ba b ij j ik k j jk k j j$ = = = =dl l (2.13)

That is, the definition of a b$ is such that if computed with compo-nents of vectors referred to different coordinate axes, the result isunchanged. Such objects are called three-scalars.

2.5.2 Lorentzian Vectors and Scalars

If an equation is to be valid in different inertial frames connected byLorentz transformations, the entities equated must transform in thesame way. Recall that the Lorentz transformations—-Eq. (2.6) withX 0=n —-are given by

n ,x A x= na al (2.14)

which leaves invariant 2x� , the invariant interval. We can write theinvariant interval of two events as

n o ,x n x x n x A x n A x2 = = =x� � � � �n no o no no o nb ba al l

with

, ,

, , , ,

, .

n

if

if or

if

1 0

1 1 2 3

0

=

= =

- = =

a b

a b

a b

ab

Z

[

\

]]

]](2.15)

Since this relation is true for any interval ,x A� n na satisfies

.A n A n=at ab bv tv (2.16)

nab is referred to as the Lorentz or Minkowski metric.2 Viewed as amatrix, n is diagonal with diagonal entries { , , , }1 1 1 1- - - . Incontrast, Euclidean space has a metric ijd .

In order to discuss form-invariant equations, it is useful todefine Lorentzian vectors, sometimes called four-vectors, as entitiesthat transform under Lorentz transformations as the difference


2 Loosely speaking, a space whose points are characterized by set of continuous vari-

ables gn , on which an infinitesimal “interval” is defined by a symmetric quadratic

expression ( )g2 = g g g� � �no n o is called a metric space with a metric ( )g gno .

�

between events x� n. Thus, a four-vector a has components an andnal related by

n .a A a= na al (2.17)

Now, defining the “inner” product of two four-vectors bya n ba b$ = o on n, we find3

o n .a n b A a n A b a n ba b$ = = =on oa a on nb b a ab bl l

The definition is invariant, that is, it is a four-scalar. Of course, anexample of such an inner product is the invariant interval.

The world line of a particle ( )x x0n gives rise to other useful four-vectors. The four-velocity U a is defined by

( ) .U xddx

0 / xaa (2.18)

Since dxa is a four-vector and dx is invariant (a scalar), it follows thatU a is a four-vector. We can see how U a is related to the three-velocityb by noting that

( ( ))d dx dx dx dx /02

12

22

32 1 2= - + +x (2.19)

implies

( ) ( )ddx

1 /0 2 1 2= - =x

b c b- , (2.20)

and thus

( ) ( )( , ).U xddx

dxdxddx

100

0/ = =x x

c b baa a (2.21)

The relativistic generalization of Newton’s force law uses anotherkinematic variable, the four-acceleration Aa defined by

AddU

/xaa . (2.22)


3 We use boldface to indicate four-vectors as we do three-vectors. The type should beclear in context. Actually, we usually indicate vectors by their components and we useGreek letters for four indices and Latin letters for three indices.

2.5.3 The Doppler Shift Revisited

In Section 2.3.5 we derived the Doppler shift for light traveling in thex+ direction between two frames in canonical relative motion, that is,

for the case where the direction of the light was in the direction ofrelative motion. It would not have been difficult to generalize thatargument to obtain the Doppler shift for any direction of propaga-tion of light. Now, however, we can use the properties of four-vectorsand four-scalars to simplify the argument and to illustrate the powerof invariance arguments. The same argument as before implies thatthe phase of the propagating wave is the same in all inertial frames;that is, it is a four-scalar. Let a wave propagating in a direction givenby the direction of the three-vector k be represented by

( )cos xE E k x 00 $= - ~ .

As noted, the phase of the wave is a scalar:

0 .x xk x k x0$ $- = -~ ~l l l l

Since this identity must be true for any event , ( , )x K k= ~n n musttransform as a four-vector (see Exercise 8), which implies, for acanonical transformation,

1

2

3

( )

( )

.

k

k k

k k

k k

r

r

1

1

2

3

= - +

= -

=

=

~ c b ~

c b ~

l

l

l

l (2.23)

With 1,cos cosk k k k1= =i il l l and, for light, k=~ and k=~l l, the first ofthese equations gives

( )cos1 r= -~ c b i ~l (2.24)

or

( ) .cos1 r= -o c b i ol (2.25)

Here ( )i il is the angle the direction of propagation of light makeswith the 1( )x x1 l axis. An expression for il in terms of i can be obtainedby dividing the second equation by the first equation of Eq. (2.23):


1 coscos

cos

cosk k1 r

r= = =-

-~ ~

i ib i

i bll

ll l l (2.26)

This direction relationship can also be obtained by transforming thevelocity of a pulse of light. (See Exercise 3.)

2.6 Rotation and Lorentz Transformations as Groups

Note that if the transformation equation Eq. (2.10) is written as amatrix equation,

,x Ax=l (2.27)

then Eq. (2.12) written in matrix form is

.AA I=u (2.28)

Any matrix transformation, A, that leaves the distance squared invari-ant has an inverse given by its transpose, Au . Au also leaves the distancesquared invariant, that is, AA I=u . The matrix product C of two suchmatrices, A and B, C BA= , leaves the distance squared invariant andhas as an inverse, C C AB1= =- u u u . Thus, the set of matrices that leavesthe distance squared invariant satisfies the group properties,4 withwith the group multiplication defined by matrix multiplication. Thisgroup is called the rotation group.

Similar to rotation transformations, Lorentz transformations canbe realized by matrices. If the transformation equation, Eq. (2.14), isviewed as a matrix equation,

,x Ax=l (2.29)

then Eq. (2.16), in matrix form, becomes

.AnA n=u (2.30)

From this it follows that DetA 1!= , where Det signifies the determi-nant. Thus, any matrix transformation, A, that leaves 2x� invariant,


4 A group G is a set such that (1) a multiplication that associates with each pair ofelements ,a b of G a third element c, written c ab= ; (2) the multiplication is associa-tive, ( ) ( )ab c a bc= ; (3) there exists an element e (the identity), such that ae ea=for all a !G ; and (4) for all a ! G there exists an element a 1- such thataa a a e1 1= =- - .

has an inverse A 1- . It is easy to show that A nA n1 1=- -u , that is, A 1-

leaves 2x� invariant. The matrix product C of two such matrices, Aand B, C BA= , leaves the x� invariant. Thus, the set of matrices thatleave x� invariant satisfies the group properties with the group multi-plication defined by matrix multiplication. The group is called theLorentz group.

2.7 Exercises

1. A stick of length Ll at rest in the primed frame makes an angle ilwith the xl axis, as shown in Figure 2.11. What angle does the stickmake with the x axis in the unprimed frame?

2. (A generalization of Exercise 1) A stick of “apparent” length Ll inthe primed frame moves with a velocity yx i j= +b bbl l l , making anangle il with the xl axis, as shown in Figure 2.12. (a) What angledoes it make with the x axis? (b) With what velocity is it moving inthe unprimed frame? (c) Does your answer to (a) reduce to thatof Exercise 1 if 0=bl ? (d) If x 0=bl and 0=il what is i?

3. A pulse of light is emitted at an angle il with respect to the xl axisin the primed frame. What angle does the direction of the light

2.7 Exercises 35

Figure 2.11. Moving stick.

Figure 2.12. Another moving stick.

make with the x axis in the unprimed frame? Why is this angledifferent from i in Exercise 1?

4. Argue that, if two events A and B are separated by a timelike inter-val, > 02x� , then there exists an inertial frame in which the inter-val AB has only a time component.

5. Express the four-acceleration A of a particle in terms of the three-acceleration a and the three-velocity b. What is the value of theinner product A UA U$ = ha ab b in the rest frame of the particle(and thus in any frame, since the inner product is a scalar)?

6. Two r+ mesons are created together. One moving with a velocityb is immediately inserted into a storage ring of radius R, as illus-trated in Figure 2.13. The second is created at rest. After one timearound the ring, thus rejoining its twin pion, what is the age of the

moving pion? What is the age of twin pion at rest at this time? Thepions have a lifetime of about 5.4 meters. If .98=b and R m30= ,.what is the probability that the moving pion survives one timearound the ring? And what is the probability that the stationarypion is still “alive” when the moving pion returns?

7. As shown in Section (2.6), Lorentz transformations can be imple-mented by matrix multiplication, that is, by a 4 4# matrix actingon a 4-column vector whose components are xn. (a) What is thematrix representing a canonical Lorentz transformation with rela-tive velocity rb ? (b) Space rotations are Lorentz transformations—they leave d 2x unchanged. What is the 4 4# matrix that representsrotation about the x3 axis by an angle i? (c) What is the matrix thatrepresents the Lorentz transformation that results if one firstimplements a rotation (b) with 0=i i , followed with a Lorentztransformation (a) and then a rotation (b) with 0=-i i ? (d) What


Figure 2.13. Twin pions.

is the result of (c) if /20=i r ? Might you have anticipated theresult? (It is advisable to use a symbolic program, Mathematica orMaple.)

8. By use of Eq. (2.30), AnA n=u , show that if Knx K nx= l l for all four-vectors x then K must transform as a four-vector.

9. Show that Eqs. (1.26) and (1.27) imply

0x xr r r r202$ $- = -l l l .

2.7 Exercises 37

Chapter 3

Relativistic Dynamics

3.1 Introduction

Newtonian mechanics satisfies the Galilean but not the Einsteinianrelativity principle. Clearly, then, the dynamical laws of Newton mustbe modified so that they are consistent with the Einsteinian relativityprinciple. First, we show how an important law of Newtonian mechan-ics—-the conservation of momentum—-is modified. Recall this law inNewtonian mechanics. If two particles collide, that is, interact, thenthe law states that the total momentum after the collision is the sameas before. In Newtonian mechanics the momentum p of a particle ofmass m, moving with velocity v, is given by

.mp v= (3.1)

That momentum is conserved and that the momentum of a particleis given by Eq. (3.1) follow from Newton’s third and second laws ofmotion. Rather than viewing conservation laws as resulting from aparticular form of a theory, one can take a more general view thatthey arise from assumed symmetries of the physical theory. Thus, theconservation of a vector, the momentum, follows from the homo-geneity of space. That is, the theory is invariant under the change offrames by translations. The particular form the momentum takes, forinstance, Eq. (3.1), results from invariance of the conservation lawsunder the transformations between different inertial frames. SinceGalilean relativity has a transformation law different than that ofEinsteinian relativity, it is not surprising that the form of the momen-tum for Newtonian mechanics must differ from that required by thespecial theory of relativity. In the following we assume that a vector

39

quantity associated with the colliding particles is conserved in all iner-tial frames and deduce the form this vector quantity must take forGalilean relativity and Einsteinian relativity.

We will see, of necessity, that the relativistic momentum of a parti-cle has an associated relativistic energy and that both are inexorablymixed by Lorentz transformations. This gives rise to an energy-momentum geometry, just as there is a space-time geometry as stud-ied in the last chapter.

Finally, in this chapter, the relativistic form of Newton’s force lawand the relativistic dynamics of the gyroscope will be studied.

3.2 Momentum in Galilean Relativity

Consider the simple case of the collision of two identical particles. Wewant to associate a vector quantity p with the particle moving withconstant velocity v. From the assumed isotropic property of space, itfollows that ( )fp vv 2= ; that is, p must point in the direction of v, theonly direction characterized by the motion.

We show that Galilean relativity implies ( )f v 2 is a constant. Thecollision we consider is that of two identical particles, A and B, whichbefore the collision move with velocities of magnitude v and reboundat right angles with velocity magnitude v f , as depicted in Figure 3.1a.For any ( )f v 2 , the total momentum is zero before the collision. Thus,the particles A and B must come off with velocities of the same magni-tude but of opposite directions. Now consider the same collision in aframe moving with velocity v in the negative x direction, as illustratedin Figure 3.1b. By the Galilean transformation of velocities, Eqs.

40 Chapter 3. Relativistic Dynamics

Figure 3.1. (a) Collision in zero momentum frame (b) Collision in lab frame.

(1.12), in this frame B is at rest and A is moving with velocity v2 in thepositive x direction before the collision, whereas after the collision

f vv i jvBf= +l

f .vv i jvAf= -l

Conservation of the x component of momentum gives

( ) ( ) ( ) .f v f fv v v v v v v4 2 f f2 2 2 2 2= + + +

This implies

( ) ( ),f fv v v4 f2 2 2= +

which must be true for any v and v f . Thus, ( )f v 2 is a constant m calledthe mass of the particle. The momentum of the particle is, as expected,

.mp v= (3.2)

3.3 Momentum-Energy in Einsteinian Relativity

The expression mp = b for the momentum is valid even in Einsteinianrelativity for particles with velocities much smaller than that of light( << 1b ).This is true because the Lorentz transformation of velocitiesis approximated by the Galilean transformations of velocities if therelative velocities of the two frames considered is small compared tothat of light. However, if the particle is moving with large velocity, weshould expect that mp = b cannot be the expression for the momen-tum, since transformations of velocities under Lorentz transforma-tions differ from those of Galilean transformations.

We want, then, to deduce the required form of the momentum byconsidering the collision of two identical particles using the Lorentztransformation between frames. The momentum is to be a three-vector, meaning it transforms as a vector under rotations. As before,the isotropic property of space implies ( )fp U= b . We use here

( )U / c b b, the space part of the four-velocity, rather then the veloc-ity b because of the simple transformation properties of the four-velocity. The space part of the four-velocity does point in thedirection of b. Consider again the collision of the identical particles,first in a frame in which they are moving with equal and oppositevelocities of magnitude ib and rebound at right angles with velocitiesof magnitude fb . (See Fig. 3.2a.) The geometry of the collision is the

3.3 Momentum-Energy in Einsteinian Relativity 41

same as that considered for the Galilean case. Clearly, the totalmomentum is zero, independent of the form of ( )f b , both before andafter the collision, and it is thus conserved. Now, view the collision ina frame moving, with respect to the first frame, in the negative x1direction with a velocity whose magnitude is ib . (See Fig. 3.2b.) By useof the transformation equations for velocity, Eqs. (1.24), and four-velocity—which of course transforms as a four-vector, Eq.(1.23)—-wecan calculate the total momentum before and after the collision inthe primed frame. For the velocity bl of the incoming particle and thex1l and 2xl components 1bl and 2bl of the outgoing particles, we obtain(refer to Fig 3.2 b)

(3.3)

The x2 component of the total momentum before and after thecollision is zero, independent of the function ( )f b . However, theconservation of the 1xl component of the momentum requires theequality

( )( ) .f U U f U

1

42 1

i

i iii

i f i if

2 2

2

1 0

2 2 2

0+

+ = + -b

bb b b b b

J

LKK `N

POO j8 B

(3.4)

For the case i f=b b , an elastic collision, Eq. (3.4) reduces to


Figure 3.2. Relativistic collision.

2 ( ) .

1

2

1 /

i

i

i

f i

2

1

2 1 2

=+

=

= -

bb

b

b b

b b b

l

l

l

( )( ) .f f

1

41

i

ii f i2 2

22 2 2

+= + -

b

bb b b

J

LKK `N

POO j (3.5)

For this to be true for any , ( )fi2b b must be a constant that is inde-

pendent of 2b , a scalar property of the particle called its (invariant)mass,m. Thus, by considering an elastic collision, we have determinedthe relativistic expression for the momentum, an expression thatmust be valid in inelastic collisions. With f set equal to m, Eq. (3.4)becomes

( ) ( ).

1

2

1

2/ /

i

i

f

i2 1 2 2 1 2-

=-b

b

b

b (3.6)

This implies ib must equal fb ; the collision must be elastic!The particle’s momentum is the mass times the proper time rate of

change of position;

.mdd mpr

= =x

cb (3.7)

For small velocities,

,mdxd mpr

v0

. =

the Galilean expression for the momentum.Note that the expression for p is the space-part of a four-vector. The

time component of this four-vector is

.p mddx

m00= =x

c (3.8)

This four-vector, with components ( , , , )p p p p0 1 2 3 , is referred to as theenergy-momentum four-vector, or merely the momentum four-vector. If oneknows the three-momentum p and mass m of a particle, and thusknows p0, in one frame, one may determine, by use of the transfor-mation properties of a four-vector, the value of the four-momentumin any frame. Using this one can easily argue that, if the total three-momentum is to be conserved in all frames, then the total timecomponent of the four-momentum must be conserved, as we deter-mined in our example of the collision of identical particles. This timecomponent p0 is called the energy of the particle.

The momentum four-vector has the components

.p mU=n n (3.9)

3.3 Momentum-Energy in Einsteinian Relativity 43

We see that the energy of a particle at rest is equal to its mass m.(Note that the energy has units of mass. If one wishes to express theenergy in conventional units, one would multiply by c2, givingp mc0

2= c, which for a particle at rest becomes E p mc02= = ).

It is reasonable to define the kinetic energy T as the energy a parti-cle has as a result of motion; thus,

,T p m m m0= - = -c

which for low velocities ( << 1b ) becomes

,Tm2

2

.b

the expression for kinetic energy in Newtonian mechanics.

3.4 The Geometry of the Energy-MomentumFour-Vector

Just as x x2 =x h� � �a ab b is invariant under Lorentz transformations,so too is the expression p n p m2=a ab b . The four-momentum for a.

massive particle is a timelike four-vector in the sense that its associ-ated invariant is greater than zero. If one considers a particle of amass m, the possible values of its energy-momentum components arerestricted by the relation

( ) .p p p p m02

12

22

32 2- + + = (3.10)

Using coordinate axes of ,p p0 1, and p2 (and setting p 03= ) we can plotthis equation and obtain a hyperbola, called the mass hyperbola,pictured in Figure 3.3. The values of the components of four-momen-tum of a particle of mass m must lie on this hyperbola. For instance,a particle at rest has p p 01 2= = and p m0 = . Note also that the four-momentum must lie in the “forward” cone since the energy p0 isgreater than zero. There is no mass hyperbola in the “backward”cone.

In the limit of m 0" , the mass hyperbolas approach the cone char-acterized by ( )p p p 00

212

22- + = , a mass hyperbola of zero mass. The

expression for the energy p m0 = c and the momentum mp = cbare both zero if m 0= , unless 1=b , in which case the expressionsare indeterminate. However, for very fast moving particles| |; | |p pforp p0 0. = the invariant mass is zero. For these reasons we

consider particles that travel with the speed of light to have invariant


mass zero. Their mass hyperbola is the cone depicted in Figure 3.4.Conversely, particles with zero invariant mass travel with the speed oflight. One can easily show (see Exercise 1) that, if the four-momentaof two particles are added (i.e., if the corresponding components areadded to obtain the components of the sum), the resulting four-momentum is timelike, the invariant being greater than zero, or lightlike, the invariant being zero. It is lightlike only if the two originalfour-momentum vectors are themselves lightlike, with their spacemomenta parallel. It follows that in adding the four-momenta of anynumber of particles, one always obtains a timelike four-vector (unless,of course, all the particles’ four-momenta that are added are lightlikewith all three-momentum vectors parallel). This four-momentum has

3.4 The Geometry of the Energy-Momentum Four-Vector 45

Figure 3.3. Finite mass hyperbola.

Figure 3.4. Zero mass hyperbola.

Image not available.

a nonzero invariant mass associated with it. Since this total four-vectoris always timelike, it follows that there exists a frame in which thespace components vanish. (Recall the argument that there alwaysexits a frame in which a timelike interval is a pure time interval.) Thesurviving time component is of course the invariant mass of the totalfour-momentum. Such a frame is referred to as the center of mass frame,or zero-momentum frame. It is the frame in which the total three-momentum is zero. As in Newtonian physics, it is quite often a usefulframe to use in the description of collisions. Of course, knowledge ofthe four-momenta of the particles before and after collision in oneframe implies knowledge of these four-momenta in any frame, sincethey transform as four-vectors.

It should be noted that the relativistic expression Eq. (3.10), whichrelates the energy and three-momentum of a particle, is confirmedevery day in many laboratories where relativistic collision experimentsare performed and analyzed, thus confirming the theory of specialrelativity.

3.4.1 “Elastic” Collisions

Consider, first, in the zero-momentum (unprimed) frame, the colli-sion of two identical particles, each of mass m, moving along the x1axis in opposite directions with velocity magnitude b. Without lossof generality, we can write that after collision the particles come offwith velocity components ( , , )0

f f1 2! b b with ( ) ( ) .

f f12

22 2+ =b b b This.

latter relation follows from the conservation of energy. Nowtransform these final momenta to the laboratory (primed) frame(where one particle is at rest before the collision), a frame thatmoves with speed b with respect to the zero-momentum frame.In this lab frame, one particle rebounds with three-momen-tum components ( ( ), , )m m m 0

f f1 2+c bc cb cb and the other with

( ( ), , )m m m 0f f1 2- -c bc cb cb . If i and z denote the angles that the

outgoing particles make with the incident path (i.e., in our case the1xl axis), we obtain

( ) ( ).tan tan1

f

f

f

f

1

2

1

22=

- +=i z

bc cb

b

bc cb

b

c

Since tan tani z �1, we see that the angle between the paths of the“rebounding” particles is not /2r . The particles always rebound inorthogonal directions in elastic Newtonian collisions of identicalparticles as viewed in the lab frame.


3.4.2 “Inelastic” Collisions

In Newtonian collisions (kinetic) energy need not be conserved. Sucha collision is referred to as a inelastic collision, and a particularlysimple highly inelastic collision is one in which the particles sticktogether after the collision occurs. However, in Einsteinian relativitythe conservation of three-momentum implies the conservation of thefourth component of the momentum, the energy. What does changein a collision in which the particles stick together?

Consider such a collision of two identical particles, each of massm0. We can view the collision in the zero-momentum frame in whichthe particles are moving toward each other with equal speed.(Consider drawing the four-momenta of the two particles in a coor-dinate system with axes p0 and p1.) Since there is only one particleafter the collision, the total four-momentum must be the four-momentum of the final particle. In the zero-momentum frame, thefour-momentum has only a time component and the final particle isat rest. It has a mass m given by

> .m p m m m m2 20 0 0 0 0= = + =c c c

The mass of the a final particle is greater then the sum of the massesof the two original particles.

3.4.3 Particle Production

Another interesting example of a collision is one involving particleproduction. In contrast to the “inelastic” collision in which a singleparticle emerges from the collision, in particle production more parti-cles emerge than collide. Consider the case in which an electron e-

collides with a positron e+, a particle with the same mass me as an elec-tron but with the opposite charge, and produces an additional elec-tron and an additional positron. Thus, we are considering the process

.e e e e2 2"+ +- + + -

In the zero-momentum frame, the two particles are traveling in oppo-site directions with equal speeds b. What is the minimum speed theymust have so that the process can occur? Again, the total four-momentum has only a time component p0 that must have a magni-tude of at least m4 e, the value if all of the particles are produced at

3.4 The Geometry of the Energy-Momentum Four-Vector 47

rest. Since the incoming particles each have a time component ofmomentum equal to mec, we have m m2 4e e=c if the particles areproduced at rest. Thus, /3 2=b is the minimum speed required toproduce the additional electron and positron.

3.5 Relativistic Form of Newton’s Force Law

Newton’s force law,

,dxd

fp

0= (3.11)

does not behave nicely under Lorentz transformations. Thus, f of oneframe is not equal to fl of another frame, in contrast to what obtainsunder Galilean transformations Eq. (1.14). An obvious relativisticgeneralization of Eq. (3.11) is

.Fddp

=xaa (3.12)

Fa is referred to as the four-force. The usefulness of such an entity is ofcourse manifest only if the four-force is given by some force law.However, the four-force can be related to the relativistic three-force fdefined by Eq. (3.11):

( ) , ( ) , .Fddxdxdp

dxdpdxdp

dxdpfi0

0 0

0

0 0

0= = =x

c b c baa d dn n

(3.13)Furthermore, we have

( ),

dxdp

dxd p m

pp f

f/

0

0

0

2 2 1 2

0

$$=

+= =b (3.14)

so that the rate at which the (relativistic) energy is changing in agiven frame is equal to the dot product of the velocity and the (rela-tivistic) three-force in that frame. Note also that in the instantaneousrest frame of the particle, Eq. (3.13) reduces to

( , ) , .Fdxd

fp

0 00

= =a d n (3.15)

3.6 Dynamics of a Gyroscope

In Newtonian mechanics, the rotational dynamics of a rigid body isgoverned by the equation


.dtdL

cmcm=x (3.16)

Here cmx is the total torque with respect to the center of mass, and L cmis the angular momentum of the body with respect to the center ofmass, that is, an angular momentum of the body with respect to thecenter of mass in an inertial frame moving with the center of mass. L cmtransforms as a three-vector under rotation. Under pure Galileantransformation—-no relative rotation of the two frames—-thecomponents of L cm do not change. A particularly simple but impor-tant case is 0cm=x , for which L cm is constant. It is constant irrespectiveof how the center of mass is moving! Such a spinning body, with

0cm=x , is called a gyroscope.What are the dynamics of a gyroscope in special relativity? In the

rest frame of the gyroscope, the angular momentum L cm is character-ized by a three-vector with some magnitude and direction, or a four-vector that in this frame has a vanishing time component, that is,( , )L L0 cm=a . The existence of such a spacelike four-vector describing

the intrinsic angular momentum of a rigid body can be made moreexplicit by considering the relativistic generalization of angularmomentum. Here we take it as a given. The relativistic dynamics of agyroscope is then easy to describe. Thus, the proper time depen-dence of ( )L xa is such that in the instantaneous inertial rest frame ofthe gyroscope,

.dxd

ddL L

0cm cm

0= =

x(3.17)

This is assured if

( ),

ddL

SU=xxa

a (3.18)

where S is a scalar and Ua is the instantaneous four-velocity of thegyroscope. This equation sets a four-vector equal to a four-vector.Thus, if correct in one inertial frame, it is correct in any other, and inthe gyroscope’s inertial frame the space components reduce to Eq.(3.17). What is the scalar S? We will see that it cannot generally bezero. Since L U 0=ha ab b is true in the gyroscope’s inertial frame, it is.

true in any frame. We have then

.ddL

U LddU

0+ =xh h

xa

ab b a abb

This, with Eq. (3.18) (recall that U U 1=ha ab b ), implies that

3.6 Dynamics of a Gyroscope 49

,S LddU

= - hxa abb (3.19)

which, used in Eq. (3.18), leads to

( ).

ddL

LddUU= -

xx

hx

av vb

ba (3.20)

A knowledge of the world line of the gyroscope, ( )x xi 0 , gives an equa-tion governing how La changes, and it does indeed change. If, forexample, the gyroscope moves in a circle, La will undergo a changewith each revolution. Even though in the instantaneous inertial frameof the gyroscope there is no rotation, we see that the “frame” of thegyroscope rotates. That is, if one considers a continuous set of iner-tial frames tangent to each point on the circle and moving with rela-tive velocity zero to the gyroscope such that nearby inertial frames areoriented without relative rotation (they are “parallel”), there is a rota-tion of the frame at the completion of the circle relative to the frameat the beginning.(See Exercise 7.) This effect is known as the Thomasprecession. A gyroscope moving in a circle with a constant speed bprecesses through an angle ( )2 1-r c in a direction opposite to thepath rotation. (See Exercise 6.)

3.7 Exercises

1. Show that for two massive particles, with masses m1 and m2 andfour-momenta p pand( ) ( )1 2

n n the sum of their four-momentum,P p p( ) ( )1 2= +n n n , is such that ( )P n P m m1 2

2$ +n no o , with equality.

obtaining only if the velocities of the two particles are equal. (Hint:Consider the invariant in the rest frame of one of the particles.)

2. Two particles with momenta (expressed in energy units of Mev)( , , , )3 1 1 0 and ( , , , )2 1 1 0- collide to produce two photons withmomenta ( , , , ), ( , , , )1 1 0 0 2 1 1 2 , and an unknown particle. (a)What is the momentum of the unknown particle? (b) What is itsmass? (c) What is the velocity of the center of momentum frame?

3. Compton scattering. A photon of frequency o scatters off acharged particle of mass m that is at rest, and the scattered photonmakes an angle i with the direction of the incoming photon.Show that

o( ),

cosh m

1 1 1= +

-o

il


where ol is the frequency of the outgoing photon. Use p h0 = o.

4. Show that the total momentum of two photons is timelike unlessthe photons’ directions are parallel.

5. Use the definition of f, Eq. (3.11), to show that

f f am$= + cbb .

Thus, generally, the three-acceleration is not parallel to the three-force. For what cases are they parallel?

6. A gyroscope moves counterclockwise in a circle, in the x x1 2-

plane, of radius R with a constant speed b. (a) Show that the four-angular momentum La of the gyroscope satisfies the equations

[ ]

[ ]( )

[ ]( ) .

sin cos

sin cos sin

sin cos cos

d

dLL L

ddL

L L

ddL

L L

01 2

2

11 2

2

21 2

2

= - +

= +

= - +

ii i bc

ii i bc i

ii i bc i

Here i is the angular position in the circle measured from thex1 axis.

Now let L L1 10= and L L 02 0= = at 0=i . (b) At =i i� , what is ( )L i�a

to first order in i� ? (c) What are the components a ( )L i�l in the“parallel” frame moving with the gyroscope at =i i� , again to firstorder in i� ? [Use Eqs. (1.26) and (1.27).] (d) From (c), findthrough what angle the gyroscope precessed while traveling i� .Through what angle will the gyroscope precess while traveling oneloop?

7. Consider two frames, ( )F( )1 b and ( )F( )2 +b b� moving parallelto the frame F with velocities b and +b b� , respectively, with9b b� . (a) What are the F( )2 coordinates of an event expressed in

terms of those of F to first order in b� ? Use Eqs. (1.26) and (1.27).(b) By use of the inverse of Eqs. (1.26) and (1.27), express the F( )2coordinates of an event in terms of those of F( )1 and show

0 ,xr r r( ) ( ) ( ) ( )2 1 2 1 1#= + -b

zb� l

3.7 Exercises 51

where ( )1 #= -cz b b� and b� l is the velocity of F( )2 with respectto F( )1 (as viewed from F( )1 ). Thus we see that, though the framesF( )2 and F( )1 are “parallel” to F, they are rotated with respect toeach other. (c) Relate this rotation to the precession of the gyro-scope in Exercise 6.


Chapter 4

Relativity of Tensor Fields

4.1 Introduction

So far we have dealt mainly with kinematic and dynamic quantitiesthat arise in the description of the motion of point particles.However, many quantities used in the description of physicalphenomena are “fields,” which are defined at all positions and for alltimes, that is, at all events. Examples are the electric field, a three-vector field, and pressure, a three-scalar field. The “three” here refersto the three space dimensions. We will characterize transformationproperties of such fields. By differentiating such fields, one formsother fields. As an example, the gradient of a scalar electric potentialgives the electric field. We generalize these three-field concepts tofour-fields and extend the idea of “gradient” to include its action onfour-fields. Here, of course, “four” refers to the four space-timedimensions. We then discuss some particularly important four-fields,namely electromagnetic fields and energy-momentum fields.

4.2 Transformations of Tensors

4.2.1 Three-Tensors

Recall that the rotation of the coordinate system is characterized by atransformation of the coordinates

x A xxxxi ij j

j

ij

j 1

3

22

= ==

!l l(4.1)

such that

53

.x xii

jj

2

1

32

1

3

== =

! !l (4.2)

We saw that Aij satisfies

A Amn jm kn jk=d d

.AA I=u (4.3)

We defined the elements of the rotation group to be any Aij that satis-fies Eq. (4.3), A three-vector a is defined by its transformation,

ii ,a A axxaij jjj

j 1

3

22

= ==

!l l

and, in analogy to a vector, a vector field ( )a r , is defined by the trans-formation

( ) ( ).a x A a xi ij j=l l (4.4)

One defines a three-tensor field of rank n as a field ( )T x...i k , with nindices, with the transformation

( ) ... ( ).T x A A T x... ...l n li nk i k=l l (4.5)

We see, from Eq. (4.3), that lmd , the metric, transforms as a three-tensor of rank 2. Viewed as a field, the metric has trivial “event”dependence. An important tensor, sometimes referred to as the Levi-Civita tensor, is the completely antisymmetric tensor of rank 3, ijke ,defined by

.

ijk

ijk

ijk

if is aneven permutationof

if is an odd permutationof

if any two of are equal

1 123

1 123

0

ijke = -

Z

[

\

]]

]](4.6)

That the transformed ijke is completely antisymmetric is easy tosee; that 1123e = is less so. In fact, ijk Det A ijke e=l . It follows from Eq.(4.3) that A 1Det != .1 (See Exercise 1.)

Several operations on tensors result in new tensors. The outerproduct of two tensors, for instance C A Bijk ij k= , form a tensor of a rankthat is the sum of the two tensors. Contraction—-summing over two

54 Chapter 4. Relativity of Tensor Fields

1 A “proper” rotation, one that can be effected by rotating about some directionthrough some finite angle, will satisfy A 1Det = . For an “improper” rotation, such asx xi i=-l , A 1Det =- .

repeated indices—-results in a tensor of rank 2 less. An example ofsuch an operation is the cross-product of two vectors A and B, result-ing in a vector C given by

.C A Bi ijk j ke=

An important operation performed on tensor fields is differentia-tion. We ask then how ( )/a x xi j2 2 transforms; that is, what is the rela-tion of j( )/a x xi2 2l l l to ( )/a x xi j2 2 ? This relation is determined by thetransformation properties of ( )a r and of xi. We have

k

( ) ( ) ( ),

xa x

Ax

a xAxx

x

a xiij

k

jij

k

m

m

j

22

2

2

22

22

= =ll l

l l(4.7)

where use has been made of the chain rule

k k

.x x

xx

m

m22

2222

=l l

But

,A x A A x xkm k km kj j m= =l

and thus

,xx

Ak

mkm2

2=l

which merely restates that the inverse of A is its transpose, Au . Usingthis result in Eq. (4.7), we obtain

k

( ) ( ).

xa x

A Ax

a xiij km

m

j

22

22

=ll l

(4.8)

We see that ( )/a x xj m2 2 transforms as a tensor of rank 2. The opera-tion / xm m2 2 2/ raises the rank of a tensor field by one, adding avector index m. A familiar example of this, the gradient of a scalarfield, results in a vector field.

This result, that operation by m2 on a tensor field produces a newtensor field of one higher rank, depends on the property of rotationsA Aij ik jk= d , which is not shared by Lorentz transformations.

4.2.2 Four-Tensors

Before we discuss the transformation properties of four-tensors andthe effect on transformation properties by action of partial deriva-

4.2 Transformations of Tensors 55

tives, we introduce an important change in notation that agrees withtraditional usage. Rather than indicating the space-time coordinatesof an event by xn, we will use xn. That is, we will use superscripts ratherthan subscripts. Thus, a Lorentz transformation will be written as

,x x= �nb

n bll (4.9)

where �bnl satisfies2

b tn n=� �n

ndd

btll

.n n=� �u (4.10)

Here nbt is the metric introduced earlier, which we write withsubindices. Note that the repeated indices are one subscript and onesuperscript. We will maintain this convention. Also, we have intro-duced a somewhat mixed notation; in writing primed event coordi-nates, the “xl”s are primed, not the index; the index merely takes onthe values , , ,0 1 2 3. However, in writing the coefficients � we primethe index to indicate the index of the primed coordinates. In thetransformation equation for coordinates, Eq. (4.9), the sum is overthe lower index of �, an unprimed coordinate. The coefficients �b

nl

implement the Lorentz transformation from the unprimed coordi-nates to the primed coordinates. The coefficients �

b

n

limplement the

transformation from the primed coordinates to the unprimed coor-dinates. The distinction is easily recognized by the notation.

The transformation of a contravariant (four) vector field withcomponents ( )V xn is given by

( ) ( ) ( ),V x V xx

x V x22

= =�nb

n b

b

nbll l l (4.11)

which gives the components of the vector in the primed frame at thesame event point. The designation “contravariant vector” is some-times used. We will see there is reason to introduce a different type of“vector” field called a covector—-sometimes a covariant vector field.

How does ( )/V x x2 2b n transform? That is, what is the relationbetween ( )/V x x2 2b nl l l and ( )/V x x2 2b n, something surely determinedby the transformations of V and x. We have


2 Similar to rotations, not all s�l satisfying Eq. (4.10) can be built up by a sequence ofsmall Lorentz boosts and “proper” rotations and thus are not all smoothly connectedto the identity. Again, 1Det !=� . In addition, from Eq. (4.10) it is easy to see that| | 100 $� . To be smoothly connected to the identity, then, 1Det = +� and 0

0� is posi-tive. The Lorentz group with such restrictions is referred to as the restricted Lorentzgroup.

n

( ).

xV x

xV

xx

xV

xV

22

22

22

22

22

= = =� � � �n

b

tb

n

t

tb

n

v

v

t

tb v

v

tl llll

l ll l

(4.12)

We have set n/x x2 2 = �v n v

ll where

n ,x x= �v v n

ll

the “inverse” of the original Lorentz transformation. Using the chainrule of differentiation, we find

tb b .xx

x

xxx

22

22

22

= = =d � �t

n

tn

n bn b

t ll

ll l

l(4.13)

Here t�b

l is the “matrix” inverse of �bnl. We can ask how b�

nl can beexpressed in terms of t�

b

l. First, define ntv so that n n = dtvvb b

t

( { , , , })n Diag 1 1 1 1= - - -tv . Then consider the Lorentz transforma-tion

.x x= �nb

n bll (4.14)

By use of Eq. (4.10), we can write

tbvn n = d� �tv a

ann

bl l

and thus obtain

v b v .x n n x n n x= =� � �t tv aan

n b tv aan

nl l l l


n .xx n nl 22

= =� �t

n

ttv

va

anl

l(4.15)

With

( )( ),A x

xV x22

/vt

v

t

the upper index of A transforms as dxt (a contravariant index),whereas the lower index transforms differently, as

x22 2/v v (covari-

ant index). Thus, we have

alat( ) ( ).A x A x= � �b b v

vt

ll l (4.16)

The definition of a mixed tensor of covariant rank n andcontravariant rank m is clear. A has covariant rank 1 and contravari-

4.2 Transformations of Tensors 57

ant rank 1. A particular example of a covariant vector (a covarianttensor of rank 1) is the “gradient” of a scalar field ( )xi ,

n nl( ) ( ).x x2 2=i i�b

bl l l

Note that, from Eq. (4.10), the metric nnd transforms as tensor ofcovariant rank 2; b�

nl effects the Lorentz transformation from x to xl.Furthermore, nbt transforms as a tensor of contravariant rank 2.

As for three-tensors, an important tensor, again sometimescalled the Levi-Civita tensor, is the completely antisymmetric tensorof contravariant rank 4 eabtv given by

.

if is an even permutationof

if is an odd permutationof

if any two of are equal

1 0123

1 0123

0

e = -

abtv

abtv

abtv

abtv

Z

[

\

]]

]](4.17)

And as for the completely antisymmetric three-tensor, there is acaveat to be added; actually,

.Dete e= �abtv abtvl

Just as in three-space, the outer product of two tensors, for instanceA Btb

v, forms a tensor of ranks which are the sums of the ranks of thetwo tensors.

Again, contraction can be used to form a new tensor—-here,however, the summing must be over an upper and lower index. Onecan see that such an operation produces a tensor of one lowercontravariant rank and one lower covariant rank. For example,consider

bn l l ln b b b( ) ( ) ( ) ( ).B x B x B x B x= = =d� � � � �nvn t a

av

vt a

av a

vav

t tlll l

Eq. (4.13) has been used. A particularly important contractionoperation is the raising (lowering) of indices by contraction withnat (with nat). As an example, given B B n Bwe define =t

a att. Note

dx dx dx n dx=nn a

ann is the invariant interval.

4.3 Relativity of Maxwell’s Equations

The Maxwell equations are invariant in form under Lorentz transfor-mations. This had been discovered by Lorentz, but he did not inter-pret their physical meaning in a way that gave them relevance outside


of electromagnetic theory. Thus, these equations need no modifica-tion to make them compatible with Einsteinian relativity. We nowexamine the Maxwell equations themselves to deduce the field trans-formation properties.

We begin with the continuity equation,

.t

J 0$22

d+ =t (4.18)

Here t is a electric charge density and J is the electric current density.It is natural to postulate that ( , )/cJt form a contravariant vector fieldJ a, for the continuity equation can then be written as

.J 02 =aa (4.19)

Consider now the inhomogeneous equations,

E 4$d = rt (4.20)

.c t cBE

J1 4

#22d - =

r (4.21)

We have postulated that the right-hand sides of these equationsconstitute the time and space components of a four-vector. If theseequations are to be true in all frames, the left-hand sides must form afour-vector too. But on the left-hand sides, the action by 2n, whichraises the covariant index by one unless contracted, is involved. Thus,these equations must be of the form

.F J42 = raab b (4.22)

This F ab, called the field-strength tensor, is a contravariant tensor ofrank 2 and linear in the B and E fields. Since there are but sixcomponents to these fields, one expects F ab is antisymmetric in thetwo indices, thus having six independent components. The timecomponent of Eq. (4.22) becomes Eq. (4.20) with the identificationF Ei i0 = . Eq. (4.21) implies that the components Bi must be in thespace-space parts F ij. It is easy to check that

FE

E

E

E

B

B

E

B

B

E

B

B

0

0

0

0

1

2

3

1

3

2

2

3

1

3

2

1=

-

-

-

-

-

-ab

p

r

qqqqqq

t

v

uuuuuu

(4.23)

4.3 Relativity of Maxwell’s Equations 59

satisfies Eq. (4.22). The components of the E and B fields are writtenwith upper indices to agree with the convention now being usedof writing the components of the position r and time x 0 with upperindices.

What about the homogeneous Maxwell’s equations,

B 0$d = (4.24)

?c tE

B10#d

22

+ = (4.25)

If F ab transforms as a contravariant tensor of rank 2, will these equa-tions be valid in all frames if valid in one? That is, can they be put intoa (Lorentz) invariant form? If one defines the antisymmetric tensorof contravariant rank 2 by

FB

B

B

B

E

E

B

E

E

B

E

BF

21

0

0

0

0

1

2

3

1

3

2

2

3

1

3

2

1= =

-

-

-

-

-

-fab abtv

tv

p

r

qqqqqq

t

v

uuuuuu

(4.26)

(said to be the “dual” tensor of F tv), one can see that the homoge-neous Maxwell equations can be written

,F 02 =aab (4.27)

which can also be expressed as

.F F F 02 2 2+ + =a bc c ab b ca (4.28)

The continuity equation and Maxwell’s equations can be writtenin invariant form. Can the Lorentz force law that gives the interactionof the electromagnetic field, a tensor of rank 2, and the chargedensity and three-current density, which taken together form a four-vector, be expressed invariantly? Considering that Newton’s forcelaw cannot be put in invariant form, one may expect some difficulty.First, consider the relativistic form of Newton’s force law for acharged particle with charge q in an electromagnetic field. In the restframe of the charged particle—-see Eq. (3.15)—-we have

( , ) , .F qdx

dE

p0 0 0= =a e o (4.29)


Any four-vector that reduces to this in the rest frame must be the four-force for a moving charged particle:

( )( , ).E BF qn U F x q q eE$ #= = +c b batb

t ab (4.30)

We deduce from this—-recall Eq. (3.13)—-that the three-force isgiven by

.f E Bq q #= + b (4.31)

This Lorentz force is determined by the transformation properties ofthe electromagnetic field and the form of the force in the rest frameof the particle. By writing the charge density and current density forpoint particles, one can deduce the Lorentz force per unit volume:

( ) ( ) ( ) ( ).x x x xE J Bf #= +t (4.32)

This is the space part of a four-vector field given by

( ) ( ) .f x F x J=b abb (4.33)

4.4 Dynamics of a Charged Spinning Particle

The study of the dynamics of a charged spinning particle is ofspecial interest since many charged “elementary” particles have a spin(intrinsic) angular momentum. Even though the spin is a quantummechanical property of the particle, some insight into the dynamicsof spin can be obtained by treating the spin as a classical angularmomentum of fixed magnitude—a gyroscope. (See Sec. 3.6.) In addi-tion, the power of invariance arguments is well illustrated by theirstudy. A rotating spherical charged particle has associated with it amagnetic moment n that points parallel/antiparallel to the “intrinsic”angular momentum S. When such a particle is at rest in the presenceof a magnetic field, a torque is exerted on the particle so that

.dxdS

B S Bmgq20 # #= =n (4.34)

Here q is the charge and m is mass of the “particle.” The constant g iscalled the gyromagnetic ratio. For a spinning body whose chargedensity to mass density ratio is constant, q 1= .

What is the dynamical equation for the spin of a charged particlein special relativity? That is, what form-invariant equation reduces to

4.4 Dynamics of a Charged Spinning Particle 61

Eq. (4.34) in the rest frame of the particle? As in the case of the angu-lar momentum vector of the gyroscope, we take it as given that thespin angular momentum S a is a four-vector that has a vanishing time-component in the rest frame of the particle and thus satisfies

.S U 0=aa (4.35)

For such a four-vector, in the rest frame of the particle, we have

,( ).mgqS F

mgqE S S B

2 2$ #=b

ab (4.36)

Consider the equation

,ddS CU

mgqS F

2= +x

aa

bab (4.37)

with C a scalar. (Recall Eq. (3.18) for the gyroscope.) In the restframe of the particle, the space components of Eq. (4.37) reduce toEq. (4.34).

By use of Eqs. (4.35) and (4.37) we find

.C SddU

mgqU F S

2=- +

xt t

ttb

b< F (4.38)

With this expression for C, Eq. (4.37) becomes

,ddS S

ddU

mgqU F S U

mgqS F

2 2=- + +

x x

at t

ttb

ba

bab< F (4.39)

If the electromagnetic field tensor vanishes, Eq. (4.39) becomesthe gyroscope equation, Eq. (3.20). If the electromagnetic field four-force, Eq. (4.30), alone acts on the charged particle, Eq. (4.39) canbe written as

.ddS

mq g

S F U UmgqS F1

2 2=- - +

x

a

ttb

ba

babd n

R

T

SSS

V

X

WWW (4.40)

In this case, the Thomas precession term ( / )S dU d Uxtt

a is partiallycanceled. If the particle has a gyromagnetic ratio of two, the Thomasprecession term is completely canceled, resulting in the equation

.ddS

mqS F=

x

a

bab (4.41)

In 1926, Uhlenbeck and Goudsmit introduced the idea of spin of anelementary particle and argued that if the electron had a gyromag-


netic ratio of two, various effects of line spectra could be explained.Finally, the quantum relativistic equation of Dirac results in a gyro-magnetic ratio of two for the particle it describes, thus giving a rela-tivistic basis for the value postulated by Uhlenbeck and Goudsmit toexplain line spectra.

4.5 Local Conservation and Gauss’s Theorem

The continuity equation, Eq. (4.18), implies the local conservation ofcharge, that is, that the rate of change charge in a three-volume V3equals the rate at which the charge current J brings charge into thevolume. This is seen by an application of the three-dimensionalGauss’s theorem:

.A A Sd dvV V3 3

$ $d =2##

Here V2 indicates the closed two-dimensional surface (boundary) ofthe volume, and dS is the outward directed element of the surface.Thus, Eq. (4.19) yields

.dtd d d SJ J ndSv

VVV 333

$ $=- =-t22###

The change Q� in the charge in the volume between t0 and t1 is givenby

( , ) ( , ) .Q t d t d Sr r J nd dtv vVt

t

VV1 0

30

1

33

$= - = -t t�2####

If V3 is all of three-space, and if J�0 only in a compact region ofspace, then Q 0=� . Local conservation of charge, Eq. (4.18), impliesglobal conservation of charge.

Let us now apply the four-dimensional Gauss’s theorem to theinvariant form of the continuity equation, Eq. (4.19):

J dx dx dx dx J dSV V

0 1 2 3

4 4

2 =2

aa a

a# #

( ) .J dS J dS J dS J dS 0V

00

11

22

33

4

= + + + =2# (4.43)

But what is the covariant vector dSa? Consider a four-volume,V x V x x xwith30

31 2 3=� � � � depicted in Figure 4.1. The three-dimen-

sional Gauss’s theorem applied to Eq. (4.19) for this four-volumegives

4.5 Local Conservation and Gauss’s Theorem 63

(4.42)

( , )J dx dx dx dx dS dx J x dx dx dxJ rV V Vx

x0 1 2 3 0 0

20 1 2 3

4 3 310

20

$2 = +2

aa# # ##

( , ) .J x r dx dx dx 0V

010 1 2 3

3

- =# (4.44)

By use of this and Eq. (4.43), for this four-volume, we can identify theparticular surface element dSa associated with the various parts of thethree-surface boundary:

( , , , )

( , , , )

( , )

d

dx dx dx

dx dx dx

dS x

S

0 0 0

0 0 0

0

1 2 3

1 2 3

0

= -b

Z

[

\

]]

]]

We can write these surface elements in a manner that illustratestheir covariant vector transformation properties. Consider, for exam-ple, the surface element ( , , , )dS dx dx dx 0 0 01 2 3=b . It is determined bythree four-vectors, namely ( , , , ), ( , , , )da dx db dx0 0 0 0 0 01 2= =n n , and

( , , , )dc dx0 0 0 3=n . The four-vector dSb must result from contraction ofthese three four-vectors with a tensor of covariant rank 4. Thus, wehave

.da db dcedS =b batva t v (4.46)

Here ebatv is the completely antisymmetric tensor of covariant rank-4. A general three-dimensional surface element dSb characterized bythree (ordered) four-vectors ,dx dx1 2

a t, and dx3v is given by


(4.45)

Figure 4.1. Four-volume V4.

for the surface at x20

for the surface at x10

for the four “vertical” sides of the surface.


.edS dx dx dx1 2 3=b batva t v (4.47)

As another example, a general surface element of the “vertical” sidesof our three-surface is characterized by the three four-vectors,

( , , , ), ( , )da dx db dB0 0 0 00= =n n , and ( , )dc dC0=n . This results in

( ,( ) ) ( , ).edS da db dc d d dx d dxC SB0 00 0#= = =b batva t v (4.48)

We can, of course, apply the four-dimensional Gauss’s theorem tothe four-divergence of any four-tensor field, ( )T x...ab v . Thus,

( ) ( ) .T x T x dS... ...

Vv 44

2 =2

vab v ab v

v## (4.49)

If ( ) ,T x 0...2 =vab v this gives

( ) ,T x dS 0...

V4=

2

ab vv#

which, if V4 is our volume of Figure 4.1, gives

( ) ( ) ( , )T x dS T x dS dx T x dx dx dxr... ... ...

V

ii

VVx

x0 0

20 1 2 3

4 3310

20

= +2 2

ab vv

ab ab# ###

( , )T x dx dx dxr 0...

V

010 1 2 3

3

- =ab# (4.50)

As in the case of the conservation of charge, if V3 is all of three-space,and if ( )T x...iab � 0 only in a compact region of space, then

( , ) ( , ) .T x dx dx dx T x dx dx dxr r... ...020 1 2 3 0

10 1 2 3=ab ab## (4.51)

( , )T x dx dx dxr...0 0 1 2 3ab# is conserved.

4.6 Energy-Momentum Tensor

One would expect that for physical phenomena described by fieldssuch as the electromagnetic fields, an energy and momentum shouldbe associated with the fields themselves. Similarly, we should expectthat any continuous distribution of matter, for example a gas, shouldhave an energy and momentum associated with it. Further, this fielddescription should permit us to associate an energy and momentumdensity with a particular point in space at a particular time, and notjust a total global energy and momentum. From the required rela-tivistic invariance we discuss what form this association must take.

4.6 Energy-Momentum Tensor 65

Consider, then, the momentum dpi and energy dp0 that arecontained in an infinitesimal volume element dx dx dx1 2 3. Heredp dpandi 0 are the space and time components of a single contravari-ant four-vector, which we write as dpa. Clearly, dpa is proportional todx dx dx1 2 3:

( ) ( )dp T x dx dx dx T x dS... ...1 2 30= =a a a .

Here we have written the three-surface element dx dx dx1 2 3 as dS0, thatis, as the time component of a covariant three-vector, as we learned itis in Section 4.5. The requirement that dpa transform as a vector inturn implies that ( )T x...a is a contravariant tensor of rank 2 and thus

( ) ( ) .dp T x dS T x dx dx dx0 1 2 3= =a abb

a

We see that ( )T x0a is the a component of the energy-momentumdensity at an event point x. From our discussion in Section 4.5, weknow that if T ab satisfies the local conservation equation,

,T 02 =bab (4.52)

then the total four-momentum,

( ) ( , , , ) ( , , , )p x T x x x x dS T x x x x dx dx dx0 0 1 2 3 0 0 1 2 3 1 2 3= =a abb

a##(4.53)

is conserved, that is it does not depend on x 0. All noninteracting orself-interacting fields should satisfy Eq. (4.52), the “local” conserva-tion law.

We know the physical meaning of ( )T x0a . What is the physical mean-ing of the remaining components, ( )T xja ? In order to answer this ques-tion, let us consider an infinitesimal four-volume defined by thefour-vectors ( , , , )dx 0 0 00 , ( , , , )dx0 0 01 , ( , , , )dx0 0 02 , and ( , , , )dx0 0 0 3 ..This volume has eight cubic three-surfaces each characterized by asurface element four-vector, Eq. (4.46). There are two timelike surfaceswith four-vectors ( , , , )dx dx dx 0 0 01 2 3 and its negative, and six spacelikesurfaces with four-vectors ( , , , ),dx dx dx0 0 02 3 0 ( , , , ),dx dx dx0 0 03 0 1

.

( , , , )dx dx dx0 0 0 0 1 2 , and their negatives. Now integrating Eq. (4.52)over the infinitesimal volume and applying the four-dimensionalGauss’s theorem, one obtains

[ ( ) ( )] [ ( ) ( )]

[ ( ) ( )]

[ ( ) ( )] .

T x dx T x dx dx dx T x T x dx dx dx dx

T x T x dx dx dx dx

T x T x dx dx dx dx

0 0 0 0 0 1 2 3 1 1 1 1 1 2 3 0

2 2 2 2 2 3 0 1

3 3 3 1 3 0 1 2

+ - = - +

+ - +

+ - +

a a a a

a a

a a

(4.54)


From this we see that in addition to T 00 being the energy density andT i0 the momentum density, T i0 is the energy flux across a two-surfaceof constant x i, and T ij is the momentum flux of component i across atwo-surface of constant x j. This can be due to fields carrying themomentum across and/or stress “forces” being exerted across thesurface. Since most readers would be more familiar with applyingGauss’s theorem in three dimensions than in four dimensions, theinterpretation of momentum flux is perhaps more easily seen by inte-grating Eq. (4.52) over some three -volume and applying Gauss’s theo-rem to the space divergence term:

.dxd T dV T dV T dSi

i ii

VVV0

0

333

2=- =-2

a a a### (4.55)

4.6.1 Energy-Momentum Tensor of Dust

Let us first discuss a rather trivial example of an energy-momentumtensor, that of “dust,” by which we mean a collection of particles thatat each point move together (not randomly) with some velocity. Inthe frame for which the particles are at rest (for a particular eventpoint) TD

000= t , and, in this frame, the three-momentum density is

zero, that is, T 0Di0 = . Similarly, since there is no energy flux in any

direction, T 0Di0 = . And there is no momentum flux (no stresses), since

the dust is noninteracting, that is, T 0Dij = . Here 0t is the mass density

in the rest frame of the dust and thus, by definition, is a scalar field.Note that m n0 0=t where m0 is the mass of the dust particles, and n isthe number density of dust particles. Then, with U a the four-velocityof the dust in any other frame, we have

( ) ( ) ( ),T x U x U xD 0= tab a b (4.56)

since this is tensor of rank 2 and reduces to the correct form in therest frame of the dust and is form invariant under Lorentz transfor-mations. From our characterization of dust, we have

.T 0D2 =bab (4.57)

The motion of the dust is governed by the (local) conservation ofenergy and momentum. That is the conservation of particles of dust,and zero acceleration of elements of the dust fluid is implied by Eq.(4.57). This is easily seen in the instantaneous rest frame of the dustat a particular event point. In such a frame, Eq. (4.57) becomes


( ( ) ( ) ( ))x U x U xx x

U

x

U0 0

0 00 0 0

022

2

22

22

= + +ttd t t db

a b aa

b

ba

( )

.x x

0

( )( )

jj

00

00

0 0$2

2

2

2d= + +

=

tt d t

bdb a a

In this equation ( )j indicates that the j index is not summed. The timecomponent of Eq. (4.58),

( ) ,x

x 000

0 $2

2d+ =

tt b (4.59)

implies the conservation of mass or the number of particles. Thespace components of Eq. (4.58),

( ) ,xx

0j

0 02

2=t

b (4.60)

shows the dust has zero acceleration.

4.6.2 Energy-Momentum Tensor of a Perfect Fluid

We will not consider in detail the energy-momentum tensor of ageneral fluid, but we find it useful to discuss some general propertiesin order to compare them with the properties particular to a perfectfluid. For a general fluid, then, consider the frame in which the mate-rial of the fluid is at rest at a particular event point. At this point, inthis particular frame, again T 00 0= t . However, though the energy fluxdue to the flow of the fluid is zero, heat flow can contribute to energyflux so that, generally, T i0 is not zero. Similarly, though the momen-tum density of the material of the fluid is zero, the heat flow cancontribute a momentum density with the result that, generally, T i0 isnot zero. A perfect fluid can be characterized as one such that at eachevent point in a frame moving with the fluid the nearby region is seento be isotropic—-all directions are equivalent. In such a frame,

, , ,T T j0 1 2 3Fj

Fj0 0= = = . If TF

i0 were not zero, then the direction of.

energy flux would be special, and if TFj0 were not zero, then the direc-

tion of the momentum density would be special. These results implythe there is no heat flow—-for a perfect fluid the changes occurringin the fluid are adiabatic. Since it is not assumed that the fluid isnoninteracting, we cannot conclude that T 0F

ij = . However, since alldirections are assumed equivalent, momentum flux and stresses mustbe radial. Thus T pF

ijij= d , where p is the pressure at the event point. The

shear forces are zero. Thus, in this frame TFab reduces to


(4.58)

.

T

T T

T

p

0

Fij

Fi i

F

ij

0 0

00

=

=

=

=

d

t (4.61)

Here t is the proper relativistic energy density, the energy density in therest frame of the fluid.

We can write TFab in invariant form, which at a given event point

reduces to Eq.(4.61) in the rest frame of the fluid. We have at ourdisposal the fluid’s four-velocity ( )U xa , the scalar fields ( )xt and ( ),p x

and the metric nab. Thus,

( ) ( ( ) ( )) ,T p x n x p x U UF =- + +tab ab a b (4.62)

which is easily seen to reduce to Eq. (4.61) in the rest frame of thefluid.

As for dust, the motion of the fluid is (partially) governed by thelocal conservation of energy momentum, Eq. (4.52). To see thephysics of this local conservation equation, we evaluate this equation,using the energy-momentum tensor given by Eq. (4.62), in the framemoving with the fluid at the particular event, with the result

( ) ( ( ) ( ))T

x

p xn

x

x p xF 0

022

2

2

2= - +

+tdb

ab

b

ab a

( ( ) ( )) ( ( ) ( ))x p xxU x p x

x

U0

0

22

22

+ + + +t t da

b

ba

( ( ) ) ( )x

p px x

pj

jj

00

0$2

2

2

2

2

2d= + + + + +

tt d t

bdb a a= G

.0= (4.63)

The time component of Eq. (4.63) can be written as

( ) (( ) ) .x

px

p 00 0$ $2

2

2

2d d+ + = + + =

tt

ttb b (4.64)

Integrating this equation over a small three-dimensional volumesurrounding the event and applying the three-dimensional Gauss’stheorem, we see that the rate of change of the energy in the smallvolume equals the rate at which the velocity field is carrying energyinto the volume plus the rate at which the pressure is doing work onthe volume. Note again that this is the only flow of energy into thevolume; the dynamics is describing a flow for which no heat is trans-mitted to a fluid element—-an adiabatic process.


The space component of Eq. (4.63) can be written as

( ( ) ( )) ( ),x p x p xa d+ =-t (4.65)

where a is the acceleration of the fluid element. Here ( )p xd- isthe force density exerted by the pressure. We see that the fluid inits rest frame has an “inertial mass density” that is equal to the( ) ( )x p x+t ! Except for a very dense or very relativistic fluid, p is much

less than t . For a nonrelativistic gas, mn 0c /t t , where m is the massof the particles and n is the number density. Thus, for a nonrelativis-tic gas, with pmuch less than t, ( ) ( )x p xa0 d=-t .

If the fluid consists of pointlike, nonrelativistic, noninteractingparticles, the energy of the nonrelativistic particles is /p m m 20 2. + b ..(There is no interaction energy or energy associated with rotation orvibration of the particles.) For such a nonrelativistic ideal monatomicgas (see Exercise 9),

.p23

0. +t t (4.66)

For an extremely relativistic gas of pointlike, noninteracting particles(see Exercise 9),

.p3.t (4.67)

Such relationships, between the pressure and energy density, arereferred to as equations of state.

The behavior of the fluid is then determined, in part, by the equa-tion of state of the fluid and the conservation of energy-momentumequations. In addition, in some cases, the number of particles isconserved (i.e., there is no creation or destruction of particles.) Forsuch cases, since a fluid element, defined as containing a fixedamount of particles, say N, undergoes only adiabatic changes, therelation between t, n and p is constrained, by the first law of thermo-dynamics, to satisfy

( ) .pd nN d n

NTd sN 0+ = =

tb cl mHere T is the temperature and s is the entropy per particle. The adia-batic (constant entropy) constraint can then be written as

p ndn

ndn d 0- - + =t t (4.68)

or


.n

pn

s constant22

= +t t=

(4.69)

An important physical process that occurs in fluids, including rela-tivistic fluids, is sound propagation. Sound propagation is governedby the dynamics resulting from the (local) conservation of theenergy-momentum equation, with the velocity of sound in the restframe of the fluid determined by the (adiabatic) equation of state.We can obtain the relation between the equation of state and thesound velocity by showing that energy density perturbations, in thisframe, satisfy the wave equation. Consider, then, small perturbations,pd dt, and db away from a uniform p0 and 0t and velocity zero, that

is, in the rest frame of the fluid. To lowest order in pd , dt, and db,.the time component of the conservation of the energy-momentumequation Eq. (4.64) becomes

( ) .x

p 00 0 0 $2

2d+ + =

dtt db (4.70)

Similarly, for the space components, Eq. (4.65)

( ) ( ) .p p xa 00 0 d+ + =t d (4.71)

With

( ) ,p xp

, s0 022=dt

dtt

Eq. (4.71) can be written

( ) ( ) ( ) .pp

xa 0, s

0 0

0 022

d+ + =tt

dtt

(4.72)

Finally, by subtracting the time derivative of Eq. (4.70) from thedivergence of Eq. (4.72), one obtains the wave equation satisfiedby dt:

( ),

xv 0s0 2

22 2

2

2d- =

dtdt (4.73)

with the speed of sound, v s, given by

( ) .p

v /

,s

s

1 2

0 022

=tt

(4.74)

A clear physical constraint on the (adiabatic) equation of state isthat the speed of sound is less than one. Note that, from Eq. (4.67),


the speed of sound of relativistic gas is ( / )1 3 /1 2. With the use of Eq.(4.69), for a nonrelativistic ideal gas, with an equation of state givenby Eq. (4.66), one finds that ( / )p mnv 5 3 /

s1 2. , which is much smaller

than the speed of light. (See Exercise 9.)

4.6.3 Energy-Momentum Tensor of the Electromagnetic Field

Elementary study of capacitors and inductors indicates that one canassociate an energy density equal to ( )/ )E E B B 8$ $+ r for electric andmagnetic fields. Similarly, a study of plane electromagnetic waves,shows that they carry a three-momentum density of ( )/E B 4# r. Onemight then expect that the energy-momentum tensor is quadratic inthe fields. Thus, two contractions are needed to reduce the outerproduct of the two tensor fields to a field of rank 2. Two such termsare n F Fab

vtvt and n F Fvt

av tb. There are other possible terms involv-ing the dual tensor F ab, Eq. (4.26). We do not use such terms butsee if we can write a candidate energy-momentum tensor for theelectromagnetic field as a linear combination of n F Fab

vtvt and

n F Fvtav tb. Of course, in the absence of charges and currents the

tensor must satisfy Eq. (4.52) and have an energy density, T 00, givenby ( )/E E B B 8$ $+ r. Such a tensor is

( ),T n F F n F F41

41

EM = +r

abvt

av tb abvt

vt (4.75)

which expressed in terms of the E and B fields becomes

TB E B E

B E B E

B E B E

B E B E B E B E

T

B E B EB B E E

41

21

4EM ij

3 2 2 3

3 1 1 3

2 1 1 2

3 2 2 3 3 1 1 3 2 1 1 2$ $

=-

- +

-

- - + -+

r r

ab

^ hp

r

qqqqqqq

t

v

uuuuuuu

(4.76)Here

( ( ))T E E B B E B41

21ij i j i j

ij2 2= - - + +

rd .

We see that ( )/T B B E E 8EM00 $ $= + r, the energy density, whereas

( )/S E B 4#= r is the momentum density as required.We now show that, in the absence of charges and currents, the

tensor TEMab satisfies the local conservation law. Thus, from Eq. (4.75),


.T n F F n F F n F F41

21

EM2 2 2 2= + +ra

abvt a

av tbvt

ava

tb abvt a

vt; EThe last two terms in the bracket can be written as

F n F n F F F F F21

212 2 2 2 2+ = + +vt

ava

tb aba

vtvt

t bv v tb b vt; 7E A .The factor in brackets on the right-hand side vanishes by Eq. (4.28).We then have

T n F F J F f41

EM2 2= = =-ra

abvt a

av tbt

tb b. (4.79)

Here f b is the four-force density (see Eq. (4.33)). In a region wherethe four current vanishes, T ab satisfies Eq. (4.52), as required.

4.6.4 Total Energy-Momentum Tensor of Charged Dust andElectromagnetic Field

Suppose the energy-momentum tensor of dust, Eq. (4.56), is notconserved; that is, it does not satisfy Eq. (4.52) because of interactionswith other fields. It remains true that Eq. (4.56) still defines a tensorof rank 2 and thus TD2b

ab is a four-vector. In the rest frame of the dustat a particular event point, this vector is given by Eq. (4.58):

( ( ) ) ( )Tx

x xxD

jj

00

00

0 0$22

2

2

2d= + +

tt d t

bdbb

ab a a .

The time component still must vanish—-the conservation of mass ornumber of particles. Even if some other field is interacting with theparticles, there cannot be a change in the energy density due to thisinteraction since, in this frame, 0=b . The space components are therate of change of the three-momentum density, which must be equalto the three-force density. If the dust particles are charged, each witha charge q, the dust interacts with the electromagnetic field and theforce density in this frame is Et , where the charge density can be writ-ten ( / ) ( / )qn q m q m U0 0

0= = =t t t . In an arbitrary frame, the forcedensity is given by the space part of

( ) ( ) ( ( ))f x F x J F mq

U x0= = ta abb

abb .

Thus, for charged dust we have

.T fD2 =aab b (4.80)


(4.77)

(4.78)

This, with Eq. (4.79), gives

( ) .T T 0D EM2 + =aab ab (4.81)

The sum of the energy-momentum of the charged dust and the elec-tomagnetic field is locally conserved.

4.7 Exercises

1. Space inversion, x xi i=-l , satisfies Eq. (4.2), and the A that imple-ments this transformation thus satisfies Eq. (4.3). (a) How does e ijktransform under space inversion? (b) What is the vector e x pijk j k inNewtonian mechanics? How does it transform under space inver-sion?

2. (a) Space inversion, ,x x x xi i0 0= =-l l , satisfies Eq. (4.10). Howdoes e abtv transform under space inversion? (b) Time inver-sion ,x x x xi i0 0=- =l l satisfies Eq. (4.10). How does e abtv trans-form under time inversion?

3. Show that E B$ and E E B B$ $- are scalar fields. (Hint: How canone form these scalars out of the electromagnetic field tensor?)

4. In a certain frame at a certain event point, E B 0$ = and<E E B B$ $ . Prove there exists a frame in which E 0= . Show that

such a frame is not unique. (Hint: Choose a special “axis” and usea canonical Lorentz transformation.)

5. In a certain frame at a certain event, E B$ �0. Prove that thereexists a frame in which E and B are parallel or antiparallel. Showthat such a frame is not unique.

6. (a) Transform, by a canonical Lorentz transformation, a generalelectromagnetic field tensor to the primed frame. (You are urgedto use Maple or Mathematica to perform matrix multiplication toeffect the Lorentz transformation.) (b) Show that the result can bewritten as

( )E EE

B12

$#= + - +c

bc c

bb bl

( ) .B BB

E12

$#= + - -c

bc c

bb bl


One can use these three-vector relationships to give the connec-tion between the fields as seen in frames with arbitrary relativevelocity.

7. By transforming the field-strength tensor due to a charge q at restin the primed frame, calculate the field in the unprimed frame inwhich the charge is moving in the x1 direction with constant veloc-ity b. Show that at the instant the charge is passing the origin

,sin

q

rB E E

r

1/

2 3 2 23 2#=

-=c b i

b 7 A ,

where i is the angle between the r and the x1- axis. Note that theelectric field and the magnetic field exist only in the x x2 3- planein the limit 1"b . You might find the following identity useful:

( )( )x y z r y z12 2 2 2 2 2 2 2 2+ + = - - +c c c .

8. Consider a charged spinning particle, with a gyromagnetic ratio oftwo, that is accelerated by a magnetic field only. Show that

ddS

00

=x

.ddS

0$=

xb

The latter result states that the component of the four-vector spinin the direction of b does not change. In turn, this implies in therest frame of the particle that the component of spin in the direc-tion tangent to the path doesn’t change.

9. Let n ba k be the number of molecules per unit volume with veloc-ity components between dandi i i+b b b . Thus, n d n3 =b b# a k ,.where n is the number per unit volume. Recall that for aNewtonian ideal gas the pressure p is related to the average valueof the kinetic energy of a molecule by

< >.p nm

32

2

2

=b

(a) By the usual argument, show more generally for a relativisticideal gas that

< >p n p31

$= b .

Here < > ( / )n n dp p1 3$ $= b bb b# a k is the average value of .p $b

4.7 Exercises 75

(b) Derive the “equation of state” Eq. (4.66) for a nonrelativisticgas and (c) Eq. (4.67) for an extreme relativistic gas. (d) For sucha nonrelativistic gas show that the velocity of sound is given by

( ) .mnp

v35 /

s1 2. (4.82)

10. In a frame F, a ring, formed by a large number of particles ofmass m rotates in the x y- plane about the origin with an angularvelocity ~. The ring has a radius r and a, small circular crosssection of area ad . The particles, of number N, are uniformlydistributed in the ring. (a) What is the number density in F? (b)What is the number density in the rest frame of the particles? (c)What is the energy-momentum tensor of this rotating “dust”? (d)What is the momentum flux at a given point in the torus? (e)How much momentum per unit time is passing a given crosssection? (f) Using (e), how many particles per unit time are pass-ing a given cross section? (g) Using (f), how many particles passa given cross section in the time / ?r r2r ~

11. Use the energy momentum tensor T ab of Exercise 10 to computeT 02aa and T i2a

a . Are the results what you would expect for thistensor? Why? (Note that yx = -b ~ and xy =b ~ .)


Chapter 5

Gravitation and Space-Time

5.1 Introduction

When Einstein published his special theory of relativity in 1905, therewere two well-established forces of nature for which there was a theo-retical description: the gravitational and the electromagnetic forces.Of course, as we have seen, the special theory of relativity fits well intothe theoretical framework of electromagnetic theory—recall thatEinstein’s 1905 paper was titled “The Electrodynamics of MovingBodies.” Such is not the case with Newton’s gravitational theory.Einstein’s consideration of the gravitational force led to extensions ofhis ideas on the structure of space-time and culminated in that mostbeautiful of theories, his general theory of relativity. In this chapterwe study ideas Einstein introduced in a paper published before hisfamous general theory of relativity paper. This study serves as a goodtransition for us—as it was perhaps for Einstein—between the specialtheory of relativity and the full-blown general theory.

5.2 Gravitation and Light

In 1911 Einstein published a paper titled “On the Influence ofGravitation on the Propagation of Light”1 in which he attempted tocalculate the effect of gravity on light propagation by applying an“equivalence principle,” a principle by which the equality of gravita-tional mass and inertial mass is made manifest. Before we consider his

1 Annalen der Physik 35 (1911), reprinted (in English translation) in Einstein et al.(1923).

77

arguments in this paper, let us recall some salient features ofNewton’s gravitational theory. Newton’s law of universal gravitationstates that the magnitude of the force of gravity that exists betweentwo bodies of gravitational masses m and M separated by a distance ris given by

.FrGmM

2= (5.1)

Here G is Newton’s universal gravitational constant. It was (and is)generally believed that the gravitational mass and the inertial mass ofa body are the same. This equality implies that the acceleration g of abody of mass m due to the gravitational attraction of a body of massM is independent of m and is given by

.grGM2=

If time is measured in distance units, this becomes

.gc rGM

rGM

2 2 2= =u

Here /G G c2/u . One might look at this equality of gravitational andinertial mass as an accident—Einstein did not. He replaced thisequality with a “hypothesis as to the nature of the gravitational field”2

that we will now discuss following Einstein’s original arguments.Consider a stationary system of coordinates, K (Fig. 5.1a) in a

region of space where there exists a homogeneous gravitational fieldwith a gravitational acceleration g taken to be in the negative x 3 direc-

78 Chapter 5. Gravitation and Space-time

2 Ibid.

Figure 5.1. Equivalent frames.

tion. Consider also, in a region of space free of gravitational fields, asecond system of coordinates K l (Fig. 5.1b) accelerating with aconstant acceleration g in the positive x 3l direction. If all relativevelocities are small, so that the Galilean transformations of velocitiesapply, the motion of a particle free of forces other than gravity are thesame in K and K l. Thus, as far as such motion is concerned, the twocoordinate systems are equivalent. One cannot distinguish betweenthe presence of a gravitational acceleration and an acceleration of thecoordinate system. Einstein hypothesized a more general equiva-lence, namely, that the two coordinates are equivalent with respect toall physical processes and thus can be used to deduce the effects ofgravity on light propagation.

We can infer what happens to a photon “rising” in coordinatesystem K by considering the photon in K l. In turn, we can concludewhat happens in K l by considering the photon in a coordinate systemK ll that is in the same region of space as K l but is not accelerating.(See Fig. 5.2.)

Let a photon of energy Ec be emitted at the origin along the x 3l axisat time x x 00 0= =l ll and received at a distance h from the origin. Sincethe K ll coordinate system is an inertial frame the photon proceeds upthe x 3ll axis without changing energy. Assume K ll is at rest withrespect to K l when the photon is emitted. For small relative velocity,the receiver’s position in K ll as a function of x 0ll is

( ) .x h g x213 0 2. +ll ll

5.2 Gravitation and Light 79

Figure 5.2. “Accelerated” photon.

The photon is received in a time x x h0 0. .ll l . At that time K l ismoving with velocity <<gh 1r.b with respect to .K ll When received,the photon has energy Ecl, in an inertial frame comoving with the K lframe, given by

( ) ( )E E ghE E gh1. .- -cc c c cl (5.2)

or

.E

E EEE

gh-

= = - = -d

d�c

c c

c

cl(5.3)

d� is the change in the potential energy per unit mass in Newton’sgravitational theory. From Einstein’s hypothesis, this is also the frac-tional change of the photon’s energy in a frame at rest in the pres-ence of a gravitational acceleration g. If one uses the quantumcondition E h= oc , Eq. (5.3) can be written

= -odo d�. (5.4)

This frequency shift can be argued from Einstein’s equivalencehypothesis without invoking the quantum condition. Imagine that asource at the origin of the K l inertial coordinate system emits a lightwave of fixed period Tsl in the x 3l direction. In the K ll inertial framethe source moves up the x 3ll axis with a velocity gxs

0.b ll . The periodof the wave emitted by the source, as measured in the K ll frame, is

s s s s( ) ( ) ( )T T T gx T1 1 1/s s2 1 2 0. .= - + +b bll l l ll l .

In the inertial frame K ll the period changes. We assume that g andsT l are small enough so that there is very little change in sb over

many periods. We want to know what period the detector at x h3=l

measures. Since K ll is an inertial frame, the wave proceeds up the x 3ll

axis with the velocity of light and with the (local) period unchanged.Thus, the light period, as observed in the K ll frame, at the detector atx h3=l is the period of the light that was emitted at an earlier timex h0.� ll . With a knowledge of the period of the emitted light we can

determine the period as observed in the K l frame using the Dopplershift to an inertial frame co-moving with the K l frame. We will use theinertial frame K ll that is moving with K lwhen the wave that is receivedat the detector is emitted. Thus, s sT T=ll l . When this is received atx h x3 3.=l ll , K l is moving with velocity ghr.b . Thus, the periodmeasured by the detector fixed in the K l frame, determined by theDoppler shift relation Eq. (2.9), is


d s s s( )

( )

( )

( )( )T T

gh

ghT gh T

1

1

1

11/

/

/

/

r

r1 2

1 2

1 2

1 2

. .=-

+

-

++

b

bl l l l .

With d/T1=ol l and sT=o l , this yields Eq. (5.4).In passing, we note a hypothesis that one can use in place of that

which Einstein assumed in his 1911 paper that is perhaps closer inspirit to the principle of equivalence of his later work and also resultsin Eq. (5.2). The result of an observation made by a freely falling observer isthe same as that of an observation made in an inertial frame in the absence ofgravity. The equivalence of this hypothesis to that of the 1911 paperis made manifest by viewing Figure 5.2 as depicting the freely fallingframe as K ll and K l as the frame ‘fixed’ in the gravitational field,which of course is accelerating with respect to K ll.

If one assumes that the expression for the change in frequency isvalid even when the gravitational field is not homogeneous, but forwhich d� is still small, one could calculate the frequency shift for aphoton emitted from the surface of a star of mass M and radius R andreceived far away (r 3. ). Then

RGM

=d�u

.

For the sun .2 12 10 6#.d� - so that / 2 10 6#. -do o - . This change infrequency of one part in a million is difficult to detect. A Dopplershift caused by velocity of 0.6 km/sec would be of the same magni-tude. Convective currents of hot gases on the surface of the sun easilyexceed this velocity, thus effectively masking the gravitational redshiftin frequency. It is referred to as a “redshift” because the frequency isshifted toward lower frequencies when the photon proceeds awayfrom a massive body.

The gravitational shift was first observed on earth by groups atHarvard University and Harwell, England. In the Harvard experimentthe photon “drops” (thus a blue shift) a distance of about 20 meters,resulting in a fractional frequency shift of about 10 14- , an incrediblysmall shift.”3

5.3 Geometry Change in the Presence of Gravity

We can regard the frequency change that occurs when a photonmoves into a region of different gravitational potential as an effect of

5.3 Geometry Change in the Presence of Gravity 81

3 See Pound and Rebka (1960).

a change of space-time geometry. These frequencies are thosemeasured by clocks (i.e., “atomic clocks”) at rest, located at the differ-ent positions. We refer to these clocks as standard clocks. The processwe are considering is essentially a static process—the gravitationalpotential is not changing. Thus, if we consider the light as wave prop-agating between two points S1 and S2, there is a constant number ofwavelengths between S1 and S2. Similarly, during the time, measuredby either standard clock, a wave at S1 goes through one cycle, and thewave at S2 goes through one cycle. Otherwise there will be a loss orgain of the number of wavelengths between S1 and S2. Since thefrequency at S2 is different than at S1, Eq. (5.4), as measured by stan-dard clocks, the wavelength 2m at S2 is different from the wavelength1m at S1. The relation between 1m and 2m is easily deduced by equating

the velocity of light at S2 (measured using a standard clock at S2) withthe velocity of light at S1 (measured using a standard clock at S1).Hence, we have

2 2 1 1=o m o m ,

and, by using Eq. (5.4) (with ,1 2 1= = -o o do o o , and )2 1= -d� � � ,

[ ( )]12 1 2 1= - -o o � �

we obtain

( ).

122 1

1=- -

mm

� �(5.5)

The wavelength increases as the wave goes “up.”In Figure 5.3 we have drawn this increase in wavelength for the case

of gz=� and /gh 1 2. . Of course 1/2 is not “small” and, thus, ourarguments for the shift are not valid, but we can at least depict theamount of the shift. We have drawn an x 0 axis, that is to be the timemeasured by a standard clock at S1. This implies the light cone, at S1,is a 45o cone (i.e., c 11= ).


Figure 5.3. Local wavelengths and local light cones.

However, since the wavelength at S2 is twice that at S1, the wavemoves twice the distance in one period, that is, with twice the velocity(using the standard clock of S1 for measuring time). Thus, the lightcone at S2 has a larger apex angle. In the absence of gravity the locallight cones would be identical; the local light cone at any point couldbe obtained by merely translating the light cones from any otherpoint. This is not true when gravitational fields are present.

Finally, we note that the velocity of light ( )c x at a point x, asmeasured using a standard clock at the point x0, is given by

( ) ( )( ( ) ( ))( )

( ( ) ( )).c x xx xx

x x1

100

0 00.= =

- -+ -m o

m o� �

� � (5.6)

Here we have assumed ( ) ( )<<x x 10-� � . The geometry of lightcones as depicted in Figure 5.3, with the velocity of light given by Eq.(5.6), is characterized by an invariant interval,

( ( ( ) ( )) ( ) ( ) ( ) ( ) .d x x dx dx dx dx1202 0 2 1 2 2 2 3 2= + - - - -x � � (5.7)

Here x 0 is the time as measured by a standard clock at point x0. It isclear the invariant interval represents a space-time geometry thatdiffers from that of the special theory of relativity.

5.4 Deflection of Light in a Gravitational Field

Having argued that the velocity of light (measured by a fixed stan-dard clock at one point in space) depends upon the position in thegravitational field, we can now calculate how much light is deflectedwhen the rays pass through such a field by applying Huygen’s princi-ple. Recall that Huygen’s principle states that to determine the posi-tion of wave front at some time x x0 0+ � knowing what the wave frontis at time x 0, one merely has to consider each point of the wave frontat x 0 to be a source of spherical waves. That is, let these waves propa-gate for a time x 0� , and the envelope of the spherical wave frontsform the wave front at x x0 0+ � .

Consider, then, a plane wave front of a light wave at time x 0. Let P1and P2 be two points on the wave front separated by a distance ldwhere the velocities are c1 and c2 respectively. We see from Figure 5.4that dz, the angle at which the wave is refracted in a time x 0d , isgiven by

( ).

lc c x1 2

0

.-

dzdd

5.4 Deflection of Light in a Gravitational Field 83

In a gravitational field c c1 2- = d�, where d� is the change in the grav-itational potential from point 1 to point 2. (See Eq. (5.6).) Thus, wehave

,lx 0=dz

dd d�

where / ld d� is the change per unit length of the gravitational poten-tial along the wave front. It is also the component of the gravitationalacceleration g along the wave front. Now consider a plane wavecoming from far off and passing by a body of mass M , as depicted inFigure 5.5. The total deflection z� is given by

.ddld dx 0= =z z� �##

Here the integral is taken over the full path, which is parameterizedby x 0, the standard clock time at the position at which the gravita-tional potential has zero value. For light passing near the surface ofthe sun, the case that Einstein treated (and, of course, the case ofmost interest), the deflection calculated is extremely small. Thus, it is


Figure 5.4. Bending of wave fronts.

a good approximation to merely sum up the infinitesimal deflectionsthat would occur if the ray passed along the undeflected path.Similarly, the velocity of light along the path is very nearly constant sothat ds dx0= , where s measures the distance along the path. (See Fig5.5.) Then we see that, when at a point characterized by ( , )r i , in anelapsed time dx 0 (distance ds dx 0= ), z changes by

( )( )

.cosddld ds

rGM ds

sGM ds

/2 2 2 3 2= = =+

z i��u u

Here � is the “impact parameter,” which is also, approximately, thedistance of closest approach to the mass M . The sum of dz over thepath, as the light comes in from far off to the left and proceeds outfar off to the right, is the total expected deflection z� :

( ).d

sGM ds GM2

/2 2 3 2= =+

=z z��

�3

3

-## u u

Using the mass of the sun for M and the radius of the sun for �, thepoint of closest approach, one obtains

. .0 875 arcsec.z�

This is the deflection that Einstein predicted in his 1911 paper forstarlight that passes near the sun. Of course, usually one cannotobserve starlight that passes near the sun—-one would be observingin daylight. However, as Einstein noted in his paper, “As the fixedstars in the parts of the sky near the sun are visible during a totaleclipse of the sun, this consequence of the theory may be comparedto experiment.”

We will see that this deflection is exactly half that predicted byEinstein’s general relativity theory.

5.4 Deflection of Light in a Gravitational Field 85

Figure 5.5. Gravitational deflection of a light ray.

Chapter 6

General Relativity

6.1 Introduction

In his 1911 paper, Einstein deduced the effect of gravity on light byapplying the principle of equivalence. However, the arguments wererestricted to nonrelativistic relative velocities of inertial coordinates.He used Newton’s theory of gravity in which effects are propagatedinstantly—the gravitational field due to a massive body depends onlyupon where it is at a given instant. This is incompatible with finitepropagation of effects as implied by Einsteinian relativity. In addition,the principle of equivalence was applied over space-time regions forwhich the gravitational effects were small.

But the principle of equivalence is a very attractive idea andEinstein built a theory of gravity with this principle as a cornerstone.

In the previous chapter we argued that some of the results of the1911 paper could be obtained by assuming a particular form of thespace-time metric gab that is used to form the invariant interval

,d g dx dx2 =x aba b

and that a world line of a photon is partially characterized by d 02=x .Though there exist locally inertial coordinate systems, that is, coordi-nate systems corresponding to freely falling frames, for which

( ) ( ) ( ) ( )d dx dx dx dx dx dx2 0 2 1 2 2 2 3 2= - - - =x haba b (6.1)

in some small space-time region around an event. However, a “globalinertial coordinate” system did not exist, meaning there do not existcoordinates xn for which Eq. (6.1) is valid for all regions. With this

87

observation in mind we can formulate the principle of equivalence asfollows:

At every space-time point in a gravitational field it is possible to choose a“locally inertial coordinate system” such that within a sufficiently small regionof the point the laws of physics take the same form as in an unaccelerated coor-dinate system in the absence of gravitation.

This principle implies that, given any point P in the four-dimen-sional space of events, one can find a set of four coordinates x o, whoseorigin is at P, in which the metric becomes locally Lorentzian; that is,

( ) (( ) ).g x O x 2= +hnoa

noa

Thus, in these coordinate systems, called local Lorentz frames or localinertial frames,

( )g P n=no no (6.2)

( )x

gP 0

2

2=a

no , (6.3)

but generally

( )x x

gP

2

2 2

2a b

no� .0

How to formally implement the principle of equivalence was theproblem facing Einstein. A theory of gravity becomes a theory as tohow the “local coordinate systems” are put together to form thegeometry of space-time in the large: What is the space-time metric( )?g xab

Einstein’s theory of how local inertial coordinate systems arepatched together was described in his paper titled “The Foundationof the General Theory of Relativity,” published in 1916 (Einstein1916). In this paper he developed equations for the space-time metricwhose source was the energy-momentum distribution.

Generally, there does not exist a global coordinate system forwhich the space-time metric is independent of the event point. Thus,it is reasonable that one must consider general coordinate systemsand the transformation of physical entities under general coordinatetransformations. To assure that the principle of equivalence is satisfied,one can impose the principle of general covariance, which we take toconsist of the following conditions (See Weinberg 1972, p. 72).

88 Chapter 6. General Relativity

1. All physical equations obey the laws of special relativity in theabsence of gravity.

2. The physical equations are generally covariant; that is, under ageneral change of coordinates the equations preserve their form.

3. The space-time metric is such that at any event point there exists a“locally inertial coordinate system.”

The principle of general covariance implies the principle of equiva-lence. Requirement (2) implies that the physical equations are validin locally inertial frames, and (1) implies that the equations in theseframes obey the laws of special relativity in the absence of gravity.Requirement (2) gives the resulting theory its name, general relativ-ity (GR). By itself, requirement (2) is somewhat empty. Many differ-ential equations, which might be candidates for physical laws, can bewritten in generally covariant form by the inclusion of terms whoseexpression depends on the particular coordinate system used, termsthat may in fact vanish in some particular coordinate system. It is thethree conditions taken together that give content to the principle. Inthe following, I present, a somewhat heuristic derivation of theEinstein equations, based on the principle of general covariance.

In the study of special relativity we were led to the study of tensors,entities that transform in particular ways under Lorentz transforma-tions. Special relativity invariance of equations was then assured ifphysical properties were expressed as tensors and physical equationswere equations between like tensors. To study covariance undergeneral coordinate transformations, one should understand tensors,entities that transform in particular ways under such general trans-formations.

In the following sections on tensors, geodesics, curvature, etc., thespace emphasized is the four-dimensional space of events. The discus-sion can be generalized, both to spaces of arbitrary dimension N andto a slightly more general metric. When a “squared distance” on aspace is defined by a quadratic differential form,

( ) ,dl g x dx dxiji j2 = (6.4)

such that dl 02$ and such that dl 02= implies that dx 0i= , then thespace is referred to as a Riemannian space. The metric is said to bepositive definite. A local frame theorem states that at any point P in thespace one can choose coordinates such that

( )g Pij ij= d (6.5)

6.1 Introduction 89

( ) ,x

gP 0i

ij

2

2= (6.6)

that is, there exist coordinates such that the metric is locallyEuclidean. The metric induced on a two-sphere by the imbedding inthe three-dimensional flat Euclidean space is an example of such aRiemannian metric:

( ) .sindl r d r d2 2 2 2 2 2= + +i z i (6.7)

If the condition that the metric be positive definite is relaxed to therequirement that it be nondegenerate, the space (or metric) is saidto be pseudo-Riemannian. “Non-degenerate” simply means thatg V V0 0implies= =ab

b b . That is, treated as a matrix, g has no zero.eigenvalue. For such a case, the local frame theorem states that at anypoint in the space one can choose coordinates such that

( )g P n=no no (6.8)

( ) ,x

gP 0

2

2=a

no (6.9)

where non is a diagonal matrix with 1+ ls and s1- l on the diagonal. Thenumber of s1+ l and s1- l on this diagonal matrix is characteristic ofthe metric and is referred to as the signature of the metric. The space-time metric is an example of a pseudo-Riemannian metric with butone 1+ and is referred to as a Lorentzian metric.

6.2 Tensors of General Coordinate Transformations

We need to consider tensors whose components can be defined withrespect to an arbitrary system of coordinates that label a particularevent (point) in our space. As in the case of tensors of special relativ-ity, the archetype of a contravariant rank 1 tensor is a differential ofevent space dxn. The (four) coordinates xn are not necessarily thespace-time of a standard clock and an orthogonal lattice. In fact, suchglobal coordinates in general do not exist. The transformation ofthese differentials from xn to coordinates ( )x xol is given by the chainrule,

.dxxx dx22

=o n

onl l (6.10)

(The transformations of the differential of space-time coordinatesunder Lorentz transformations are a particular subset of such trans-


formations.) The transformation of any contravariant rank 1 tensor( )A xn is given by

( ) ( ).A xxx A x22

=on

onl l l (6.11)

As under Lorentz transformations, the archetype of a covariantvector of rank 1 is the gradient of a scalar field ( )x2 in . Recall that ascalar field satisfies ( ) ( )x x=i il l . The chain rule in the form

xx222 2=o o

n

nl l

implies

( ) ( ).xxx x222 2=i io o

n

nl l l l(6.12)

The primed index on the symbol 2ol indicates a derivative with respectto a primed coordinate.

The meaning of mixed tensor of contravariant rank n and covari-ant rank m, defined by its transformation, is clear:

a ( ) ( ).A xxx

xx A x

22

22

=oa

b

n

o

bnl l

ll (6.13)

As for Lorentz tensors, the outer product of two tensors, for exampleA Btb

v, forms a tensor of ranks that are the sums of the ranks of thetwo tensors. Similarly, it is easily shown that the operation of contrac-tion of a covariant index with a contravariant index produces a tensorof one lower contravariant rank and one lower covariant rank.

Clearly, it is important to know how the metric ( )g xab transforms. Itstransformation is implied by the scalar character of the (local) invari-ant interval. That is, ( ) ( )d g x dx dx g x dx dx2 = =x ab

a bon

o nl l l l , or

( ) ( )( )( ) ( ) .d g x dx dx g xxx dx

xx dx g x dx dx2

22

22

= = =x aba b

ab o

ao

n

bn

ono n

ll

ll l l l l

We see that

( ) ( ) .g x g xxxxx

2222

=on ab o

a

n

b

l ll l

(6.14)

The metric transforms like a tensor of covariant rank 2. This mighthave been expected since dx dxa b transforms as a tensor ofcontravariant rank 2 and to form a scalar, d 2x , we would expect thesecontravariant indices to be contracted with two covariant indices.

As for Lorentz tensors, the “inverse” metric gab, defined byg g = dab

bt ta, is a contravariant rank 2 tensor, for we have

6.2 Tensors of General Coordinate Transformations 91

no( ) ( ) ( )

( )

.

g xx

xxx g x g x

x

xxx

xxxx g

g xxx

xx g

xx

xx

22

22

22

22

2222

22

22

22

22

=

=

= =

d

d d

ab

b

n

a

vab

b

n

a

v

n

t

o

h

th

aba

v

o

h

btth

ha

a

v

o

h

ov

l l l l l ll l

ll

ll

6.3 Path of Freely Falling Particles: Timelike Geodesics

Differentiation of a Lorentz tensor yields a Lorentz tensor of onehigher covariant rank. However, differentiation of these moregeneral tensors does not result in a tensor because the coordinatetransformations are not restricted to be linear like Lorentz transfor-mations. Before investigating how one might define a “differentia-tion” operation on general tensors that will result in a tensor, we willconsider a particular example that we might expect to result in aquantity that has tensorial transformation properties. Thus, we studythe equation satisfied by the motion of a particle that is moving underthe influence of gravity only—a freely falling particle. For any eventon the world line of the particle there exists a locally inertial frame,with coordinates gn, for which

( ) ( ).g n O 2= +g gno no

By the principle of general covariance, the equation satisfied by theparticle at 0=g is

.d

d02

2

=x

gn(6.15)

In local inertial coordinates the world line is a straight line if onerestricts to a small enough region about 0=g . Here dx is the invariantinterval along the world line of the particle. Note that there existsmore than one inertial frame at a point, for if ( ) ( )g n O 2= +g gno no ,.a change of coordinates n ( )O 2= +g g g�o o nll gives no ( )g n O 2= + gnol l ,.if n�

ol is a Lorentz transformation. We are parameterizing the worldline with the proper time measured along the world line, a parame-terization that must be modified when considering the motion of aparticle moving with the velocity of light. Eq. (6.15) is locally that ofa timelike geodesic, as we know that a free particle follows a path oflongest proper time between two points on its worldline. (There arepaths that in local coordinates are straight lines but which are space-


like, which we call spacelike geodesics.) Let us consider the form that Eq.(6.15) takes in an arbitrary coordinate system x to which the localcoordinate system g is related by the transformation ( ).xgn a

( ( ))xWriting g xa , we have

,dd

x ddx

2

2=

xg g

x

a

b

a b

(6.16)

from which

.d

d

dd x

x x x ddx

ddx

02

2

2

2 2

2

2

2 2

2= + =

x

g

x

g gx x

n b

b

a

t b

a b t

(6.17)

Multiplying Eq. (6.17) by /x2 2go a and summing over the repeatedindex a, we obtain

.dd x

x x

xddx

ddx

02

2 2

2 2

222

+ =x

g

g x x

o

t b

a

a

o b t

(6.18)

With a definition of Christoffel symbols ( )x�tbo as

( ) ,xx x

x2

2 2

222

=g

g�tbo

t b

a

a

o

(6.19)

Eq. (6.18) can be written

( ) .dd x x

ddx

ddx

02

2

+ =x x x

�o

tbo

b t

(6.20)

Thus, we have the equation for a freely falling particle in arbitrarycoordinates but the equation involves the Christoffel symbols that areexpressed in terms of the transformation equations between localinertial coordinates and the general coordinates. Eq. (6.20) is anequation for a geodesic, a path in space-time that has the longestproper time between any two points on the path, and thus oneexpects that the equation can be expressed in terms of the metric forarbitrary coordinates. (Note that under a change of parameterizationof the path from ato =x x xu , with a a constant, the geodesic equationis unchanged. Such parameters are called affine parameters.) We nowshow that ( )x�tb

o can be expressed in terms of the metric. First, notethat the transformation property of the metric implies that, near anevent point for which gb are local inertial coordinates,

( ) ( ) ( ( )).g xx x

gx x

n O 2

2222

2222

= = +g g

gg g

gno n

a

o

b

ab n

a

o

b

ab (6.21).

6.3 Path of Freely Falling Particles: Timelike Geodesics 93

The derivative of Eq. (6.21) with respect to xv, evaluated at 0=g , gives

( ) .x

gx n

x x x x x x

2 2

2

2

2 22

22

2 22

22

= +g g g g

vno

ac v n

a

o

c

v t

c

n

a= G (6.22)

From Eq. (6.19) we have

( ) .x x

xx

2

2 2

222

=g g

�t b

a

tbo

o

a

(6.23)

Using this and Eq. (6.21) in Eq. (6.22), we find

.x

gg g

2

2= +� �v

no

vnh

ho voh

hn (6.24)

Now add to Eq. (6.24) the same equation with n interchanged with v,and subtract the same equation with o interchanged with v to obtain

.x

g

x

g

x

gg2

2

2

22

2

2+ - = �v

nonvo

nnv

vnh

ho (6.25)

We have used the fact that �abo is symmetric under interchange of

anda b. By multiplying the above equation with got, we have

.gx

g

x

g

x

g21

2

2

22

2

2= + -�vn

t otvno

nvo

onv< F (6.26)

Having succeeded in expressing the Christoffel symbols in terms ofthe metric and its derivative evaluated in any coordinate system, wecan write the equation for a freely falling particle, Eq. (6.20), in termsof the metric in any coordinate system. In a locally inertial coordinatesystem, 0=�ab

t and Eq. (6.20) reduces to Eq. (6.15).Finally, note that the geodesics of Riemannian spaces are paths of

the shortest distance between two points, such as the great circles ona two-dimensional sphere, in contrast to timelike geodesics forLorentzian metrics that have the longest proper time between events.There is a caveat that should be added to the last statement: timelikegeodesics have a longer proper time than any “nearby” path betweentwo events.

6.4 Covariant Differentiation

Since Eq. (6.20) is valid in any coordinate system, one might expectthat, even though the two terms in the equation may not transformseparately as a contravariant vector of rank 1, the sum may. We canuse this observation to guess how to define a “differentiation” thatmight result in tensors. First note that /dx dxo is a vector since


.ddx

xxddx

22

=x x

n

o

n ol l

With /A dx d= xo o , Eq. (6.20) can be written as

( ) .ddA x A

ddx

0+ =x x

�o

tbo b

t

This suggests that for any contravariant vector ( )A xo , the object/DA dxo defined by

( )dDA

ddA x A

ddx

= +x x x

�o o

tbo b

t

(6.27)

is a contravariant vector. One can check that it is. It is called the direc-tional absolute derivative, or the directional covariant derivative, in thedirection /dx dxt . For any parameterized curve ( )x st ,

( )dsDA

dsdA x A

dsdx

= + �o o

tbo b

t

(6.28)

is a tensor, the absolute derivative of Ao in the direction of /dx dst .Furthermore, since

,dsdA

xAdsdx

22

=o

t

o t

we can write

( ) ,dsDA

xA x A

dsdx

22

= + �o

t

o

tbo b

t< Fwhich suggests that

( )AxA x A; 22/ + �t

ot

o

tbo b (6.29)

is a mixed tensor of covariant and contravariant rank 1. One can showthat it is. It is called the covariant derivative of ( )A x (note the notation“;t” for the covariant derivative). The connection between the abso-lute derivative /DA dso and the covariant derivative A ; t

o is

.dsDA A

dsdx

;=o

to

t

(6.30)

Of course, the equation for a geodesic that describes a freely fallingparticle, Eq. (6.20), can be written in terms of a covariant derivative.After all, we started our argument to suggest a definition of a covari-ant derivative with Eq. (6.20). First, note that the covariant definitionof the four-velocity Un is, of course, /U dx d= xn n ; this definition

6.4 Covariant Differentiation 95

reduces to the special relativity definition in a locally inertial frame.By use of the four-velocity, Eq. (6.20) can be written as

.dDU U U 0;= =x

o

to t (6.31)

Furthermore, by use of the covariant definition of the four-momen-tum, p mU=n n, Eq. (6.31) can be written as

,dDp

0=x

n

(6.32)

which in a locally inertial coordinate system reduces to /dp d 0=xn .The covariant derivative of an arbitrary tensor A ...

...ab is given by

A A A A......

......

......

......

; , f f= + + - -� �a tb

a tb

vtb

av

atv

vb (6.33)

Here we have introduced the notation “;t” for “ / x2 2 t.”It is important to note that g 0; =t

ab and g 0; =ab t . These follow easilyfrom the fact that tensor equations that are true in a local inertialcoordinate system are satisfied in all coordinate systems. gab and gabcan be taken in and out of covariant differentiation.

6.5 Parallel Transport: Curvature Tensor

Consider some path ( )x sn in event space parameterized by s and acontravariant vector Ao that satisfies

( ) ,dsDA

dsdA x A

dsdx

0= + =�o o

tbo b

t

or

( ) .dsdA x A

dsdx

= - �o

tbo b

t

(6.34)

By considering this equation in a locally inertial frame, we can give ita geometric interpretation. In such a frame, since 0=�tb

o , Eq. (6.34)becomes

.dsdA

0=o

That is, the components of the vector, in a locally inertial frame,do not change as the parameter changes—the vector is parallel-transported along the path. Note that if Ao satisfies Eq. (6.34), then

.dsdA A

dsDA g A

0= =o

ooon

n

(6.35)


This follows from g 0; =ab t . Similarly, if the path is a geodesic path,that is, it satisfies Eq. (6.31), then

.ddU A

0=x

aa

The scalar U Aa a is constant along the path. Loosely speaking, the“component” of A in the direction of the path doesn’t change.

A natural question to ask is, “If a vector, starting at a given point, isparallel-transported around a closed path, thus returning to the start-ing point, does the original vector result?” Consider parallel-trans-porting a two-dimensional vector on the surface of the sphere arounda close path starting at a pole, proceeding along a fixed longitude tothe equator, following the equator to some point, and then moveback to the pole along a fixed longitude, as depicted in Figure 6.1.The path chosen is a sequence of path sections each of which is ageodesic of the sphere. Thus, the angle the vector makes with eachpath section does not change as the vector is transported along thesection . We see that the resulting vector is generally not the originalvector, but has been rotated by an angle equal to the angle made bythe intersection of the two longitudes. The direction of rotation(counterclockwise) is the same as the path direction. Of course, if thetwo-dimensional surface is a flat surface, parallel transport of a vectoraround a closed path does not change the vector. A change in thevector results if the surface is curved.

6.5 Parallel Transport: Curvature Tensor 97

Figure 6.1. Parallel transport on a sphere.

But the characterizations of surfaces as being “flat” or “spherical”in a region is not exhaustive. Consider a two-dimensional surface thatcurves “up” in one direction and “down” in a perpendicular direc-tion—a “saddle” surface. Such a surface is depicted with a sphericalsurface and a plane (flat) surface in Figure 6.2. Note that the intrin-sic geometric properties of spherical and plane surfaces are every-where the same. Not so for the saddle surface—the point at the“center” of the saddle is the most symmetric point. Also depicted inthe figure are small “triangles” formed by geodesics for the threesurfaces. The triangle for the saddle is centered at the center of thesaddle. Of course, for the sphere and saddle triangle, the sides do notall lie in the same two-dimensional plane of the embedding three-space. Nevertheless, we see that the sum of the three angles made bythe intersections of the geodesics is >r for the sphere, <r for thesaddle, and =r for the plane surface. One can imagine that parallel-


Figure 6.2. Two-dimensional surfaces.

transporting a vector around the spherical triangle in a counter-clockwise direction as indicated in the figure results in a rotation ofthe vector in a counterclockwise direction, whereas for the saddletriangle the vector would be rotated in clockwise direction. Of course,for the plane triangle the vector is not rotated. The sphere is said tohave positive “curvature,” the saddle negative “curvature,” and theplane zero “curvature.”

To obtain a quantitative definition of curvature at a point P of atwo-dimensional surface that is intrinsic to the surface, consider thedistance C along a closed path that is the locus of all points that areat the same geodesic distance r from the point. For the case of thesphere (Fig. 6.3) for small /r R,,

( ) ( ) ( ).sinlC R

Rr r

Rr r r

2 26

262

3 3

.= - = -r r r

Here /l R1 2= , a reasonable definition for the curvature of a sphere.From this we obtain the following formula for this curvature:

,limlrr C3 2

>r 03=-

rr

-(6.36)

a result that depends only on the intrinsic properties of the metric ofthe two-dimensional surface with no reference to the embeddingspace. The sign of the curvature is positive if <C r2r , negative if>C r2r , and zero if C r2= r . With Eq. (6.36) as the definition of the

curvature at a point P of any two-dimensional surface, then

lR R11 2

= (6.37)


Figure 6.3. Curvature of sphere.

obtains, where R1 and R2 are the extrema of the radii of sectionsformed by planes normal to the surface at P with appropriate sign. (Asection is the curve formed by the intersection of the surface and aplane.) If the two extrema, which occur in perpendicular directions,are of the same sign, the curvature is positive, and if opposite it isnegative. For the saddle surface, one of the two radii is negative andthus it has a negative curvature.

To characterize the curvature properties in a small space-timeregion, we study the parallel transport of a contravariant vectoraround a “small” closed loop in our space-time manifold. The closedpath is given by ( )x sn , with , ( ) ( )s x x x0 1 0 1 0# # = =n n n. From Eq.(6.34) the change A Ain a vector� o n is given by

( ) ( ) .AdsdA ds x A x

dsdx ds

0

1

0

1

= = -� �oo

tbo b

t

## (6.38)

Expanding to lowest order around x0, we have

( ) ( )( )( ),x x

x

xx x0

0

02

2. + -� �

�tbo

tbo

atbo

a a (6.39)

and by use of Eq. (6.34),

( ) ( ) ( ) ( )[ ].A x A x x A x x x0 0 0 0. - -�b bavb v a a (6.40)

Using these last two approximations in Eq. (6.38) one obtains

Note that /dx dst is “small” and thus there is no zero-order contribu-tion to A� o, and since

,dsdx ds 0

0

1

=t

#

the first-order contribution vanishes. We see then, to second order,that

( )( ) ( ) ( ) [ ] .A

x

xx x A x x x

dsdx ds

0

0 0 0 00

1

2

2= - - -�

�� o

atbo

tho

abh b a a

t

#R

T

SSS

V

X

WWW

The integrals are antisymmetric in their indices. This can be seen byintegrating by parts as follows:


( )( )( )

( ) ( ) ( ) ( ) .

A xx

xx x

A x x A x x xdsdx ds

0

0

00

1

0 0 0 0

2

2= - + -

- -

� ��

�

otbo

atbo

a a

bavb v a a

t

#R

T

SSS

8

V

X

WWW

B

[ ] [ ] |[ ]

x xdsdx ds x x x

dsd x x

x dsss

00

1

0 01 0

0

1

- = - --

==a a

ta a t

a at# #

[ ] .dsdx x x ds0 0

0

1

= - -a

t t#


( ) ( )( ) ( ) ( ) ( )A

x

x

x

xx x x x

0 0

0 0 0 02

2

2

2= - + -�

� �� o

tbao

abto

tho

bah

aho

bth

R

T

SSS

V

X

WWW

( ) [ ] .A x x xdsdx ds

21

0 00

1

-b a at

#< F (6.41)

Since A� o is the difference of two vectors at the same event point, it is avector, as is Ab . Similarly the integral is a tensor of contravariant rank2. It consists of a sum of the product of two “small” changes in eventpoints. Thus, we might expect that Rbat

o , defined by

,R , ,= - + -� � � � � �bato

ba to

bt ao

tho

bah

aho

bth (6.42)

is a mixed tensor of contravariant rank 1 and covariant rank 3.One can show it is. It is called the Riemann curvature tensor. If at agiven space-time point, there exists a coordinate system such that( ) ( )g n O 3= +g gab ab , then R 0=bat

o .We were led to the Riemann curvature tensor by considering paral-

lel-transporting vectors around infinitesimal closed paths. By consid-ering a closed path defined by the sequence of four infinitesimals, , ,da db da db- -b v b v, one might expect that the Riemann tensor

would be involved if one considered the action on a vector field V a ofthe “commutator” of the two covariant derivatives in the directionsdab and dbv. “Commutator” means the difference between the actionsof taking the derivatives first in one order and then reversing theorder. That is,

.V V; ; ; ;-b va

v ba

This commutator is a tensor field of contravariant rank 1 and covari-ant rank 2. In a locally inertial frame, for which 0=�bt

a but for which,�bt v

a generally does not vanish, this commutator is easy to computewith the following result:

( ) .V V V; ; ; ; , ,- = -� �b va

v ba

tb va

tv ba t

But, since in a local inertial frame R , ,= -� �btva

bt va

bv ta (see Eq. (6.42)),

this can be written


.V V R V; ; ; ;- =b va

v ba

tbva t (6.43)

Note the commutators of ordinary derivatives vanish as would thecommutators of covariant derivatives at a point where the Riemanntensor vanishes.

The Riemann tensor has 4 2564= components; however, not all areindependent for the tensor possesses a large number of symmetriesbest expressed in terms of the completely covariant form obtainedfrom the above by “lowering” the o index :

.R g R/vbat vo bato

The following equalities are most easily derived in a local inertialframe (Exercise 3):

R R R R= - = - =vbat bvat vbta atvb

.R R R 0+ + =vbat vtba vatb (6.44)

From these it follows that

.R 0/vatv (6.45)

6.6 Bianchi Identity and Ricci and Einstein Tensors

Again, the following important identity, called the Bianchi identity, ismost easily shown in a local inertial coordinate system:

.R R R 0; ; ;+ + =vbat h vbha t vbth a (6.46)

An important tensor, called the Ricci tensor, is formed by a contrac-tion on the Riemann tensor:

.R R R/ =bt boto

tb (6.47)

Because of the symmetries of Rvbat this is the only independentcontraction on the Riemann tensor. Note that the contraction usedfor the definition of the Ricci tensor is a contraction of the contravari-ant index of the Riemann tensor with its middle covariant index.

Similarly, the Ricci scalar is defined as

.R g R/ btbt (6.48)


In order to see the implication of the Bianchi identity on the Riccitensor, we do a “Ricci”-like contraction on this identity;

[ ] .g R R R 0; ; ;+ + =vavbat h vbha t vbth a

As noted before, g 0; =tab and g 0; =ab t . Thus gab and gab can be taken in

and out of covariant differentiation, and, we have

.R R R 0; ; ;- + =bt h bh t bth aa (6.49)

An even more useful equation is obtained by a second contraction,

[ ] ,g R R R 0; ; ;- + =btbt h bh t bth a

a

or

.R R R 0; ; ;- - =h h tt

h aa (6.50)

This equation can be written in the form

( ) .R R2 0;- =dha

ha

a (6.51)

The completely contravariant form of this equation is

,G 0; =aat (6.52)

where

.G R g R21/ -at at at (6.53)

G at is called the Einstein tensor. We will see that it is of fundamentalimportance in the Einstein field equations for general relativity, as isthe identity Eq. (6.52), also referred to as the Bianchi identity.

Listed below are some differential geometry definitions and resultsthat will be used repeatedly in discussions that follow:

Christoffel symbols gx

g

x

g

x

g21

2

2

22

2

2= + -�vn

t otvno

nvo

onv< F

Geodesic ( )dd x x

ddx

ddx

02

2

+ =x x x

�o

tbo

b t

Directional covariant ( )dsDA

dsdA x A

dsdx

= + �o o

tbo b

t

derivative

6.6 Bianchi Identity and Ricci and Einstein Tensors 103

Covariant derivative ( )AxA x A; 22/ + �t

ot

o

tbo b

Riemann curvature tensor bat ,R , ,= - + -� � � � � �oba to

bt ao

tho

bah

aho

bth

Commutator of covariant V V R V; ; ; ;- =b va

v ba

tbva t

derivative

Ricci tensor R R R/ =bt boto

tb

Ricci scalar R g R/ btbt

Einstein tensor G R g R21/ -at at at

6.7 The Einstein Field Equations

In Chapter 5 we saw that the equivalence principle seems to implythat the metric is changed in the presence of matter. When restric-tion was made to low relative velocities and weak static fields, weargued that

.g 1 200. + � (6.54)

Here �, the Newtonian gravitational potential, satisfies the equation

,G42d = r t� u (6.55)

where t is the mass density. Thus, in a relativistic theory of gravity theequation

g G8200d = r tu (6.56)

must obtain in some low-speed and weak-field approximation. Sincet is the T 00 component of the energy-momentum tensor in the restframe of a fluid, we might expect that Eq. (6.56) should be replacedby a generally covariant equation with the energy-momentum tensoras the source of a tensor Oab of contravariant rank 2:

( ) ( ).O x kT x=ab ab

Oab should be expressible in terms of the metric and should involveup to second-order derivatives of the metric. Second-rank tensors that


might be used are , ,R g Rab ab and gab. Thus, we might expect as possi-bilities

,O R cg R g= + + mab ab ab ab

where c and m are constants. But the local conservation of energymomentum, Eq. (4.52), along with the principle of general covari-ance demands that

,T 0; =bab

and thus

O 0; =bab

must be generally true. We see, by use of Eq. (6.52), that this equa-tion is satisfied if /c 1 2=- and we are, plausibly, led to the followingfield equations for the metric:

( ) ( ) ( ),G x g x kT x+ =mab ab ab (6.57)

with the constants m and k to be determined. A restriction on theseconstants arises if Eq. (6.56) is to be satisfied in a weak-field, low-velocity limit. In the low-velocity limit <<T Tij 00 . (As an example,for a perfect fluid, see Eq. (4.46).) But this implies that

( ) ( ) ,G x g x 0ij ij .+ m (6.58)

or

( ).R g R g x21ij ij ij. - m (6.59)

We assume the meaning of the weak-field limit to be the existence ofcoordinates such that

, | |<< .g g g 1. +h d dab ab ab ab (6.60)

By use of such a metric and Eq. (6.59), we have

( ),R n R R R R R g x23ii ii00 00.= = - + + mab

ab (6.61)

6.7 The Einstein Field Equations 105

or

.R R g2 2 ii00. - - m (6.62)

With this, the { }0 0 component of Eq. (6.57) becomes

[ ] ,R g g g kT2 ii00 00 00 00.+ +m (6.63)

where we have used g R R00 00 00. . We need to know what R00 is in theweak-field approximation. We start with the Riemann tensor:

.R , ,. -� �batv

bt av

ba tv (6.64)

Terms quadratic in �s and thus quadratic in gd ab have been dropped.Using

R n R.obat ov batv

and Eq. (6.26), one obtains

[ ].R g g g g21

, , , ,. - - - +obat to ba tb oa ao bt ab ot (6.65)

Because we are dealing with static fields, all time derivatives vanish.We want an expression for R00. Within our approximation, we have

.R R R R g g21

21

,i i i i ii00

0000 0 0 0 0 00200d. . .- - - =-

Using this in Eq. (6.63), we obtain

[ ] .g g g g kTii200

00 00 00d .- + +m (6.66)

We see that Eq. (6.56) results if k G8=- r u and 0=m . The constant m,referred to as the cosmological constant, was not included at first inEinstein’s field equations. He added it later so that a static cosmolog-ical solution to the equations would exist. He discarded it still later.Observations of gravitating systems imply m is quite small. (See thenext section.) Except for some exercises and in Chapter 9, wherecosmology is discussed, we assume 0=m . With m included, Einstein’sfield equations become

( ) ( ) .G x R g R GT x g21

8/ - =- -r mab ab ab ab abu (6.67)

With m set equal to zero, the field equations are


( ) ( ).G x R g R GT x21

8/ - =- rab ab ab abu (6.68)

We find from this that

,R R GT GT2 8 8/- =- -r raau u (6.69)

and thus Eq. (6.68) can be written

( ) .R G T x g T821

=- -rab ab abu c m (6.70)

Note that in “vacuum,” ( )T x 0=ab and the field equations reduce to

.R 0=ab (6.71)

We will obtain a nontrivial solution to this equation in the nextchapter.

6.8 The Cosmological Constant

Eq. (6.67) can be rewritten as

( ) ( ( ) ).G x R g R G T xGg

21

88

/ - =- +rrmab ab ab ab abuu (6.72)

The cosmological term behaves as an effective energy-momentumdensity Tm

ab:

.TGg

8=rm

mab ab

u

It is present even where T 0=ab and is sometimes referred to as avacuum energy-momentum density.

With m sufficiently small, Eq. (6.66) would remain a good approx-imation to Eq. (6.56) and thus to Eq. (6.55) for the Newtonian grav-itational potential. We can ask how small mmust be so that it doesn’thave a significant effect on the dynamics of gravitating systems thatare known to be very well described by Newton’s gravitational theory.For m small, the gab in the second term of Eq. (6.66) can be replacedwith nab, and the equation can be then written as

,g GT8 2200

00d . -r mu (6.73)

or, in terms of the gravitational potential, �,

6.8 The Cosmological Constant 107

.G42d = - +m r t� u (6.74)

We see that m contributes an effective mass density, tm, given by

.G4

= -trm

m u

For a spherically symmetric system, this term gives a contribution tothe gravitational potential,

( ) .r r6

2

= -m�m (6.75)

Note we have set ( )0 0=�m and " 3�m as r " 3. Clearly, irrespectiveof how small m is, the approximation g n.ab ab breaks down for r large.

The potential ( )r�m gives a radial force per unit mass of /r 3m . Whena system is well described by a Newtonian potential ( )rN� , which givesa radial force per unit mass of ( )/d r drN- � , then

| | ( )rdrd r

3N%m � (6.76)

must be true. As an example, for a gravitational system such as oursolar system, whose dynamics is determined by a large mass M, with alargest observed orbit radius of rL, this inequality becomes

.rGM

L3%mu

(6.77)

With M the mass of the sun and rL the radius of the orbit of Jupiter,we have | | < m10 35 2m - - . The effect of the m term is larger the largerthe size of the system considered. That the Newtonian gravitationalforce law seems to be valid for systems of galaxies results in a morestringent bound, < m10 46 2m - - . But importantly, a much smaller | |mcan have a significant effect on systems of cosmological size.

6.9 Energy-Momentum Tensor of a Perfect Fluid inGeneral Relativity

The properties of energy-momentum tensor as the source of themetric are of course important in the applications of general relativ-ity. One of the most important is the local conservation law, which inspecial relativity, is expressed by Eq. (4.52)

.T 02 =bab (6.78)


The general covariant form of this equation is

.T 0; =bab (6.79)

That is, the ordinary derivative in Eq. (6.78) is replaced by the covari-ant derivative. Eq. (6.79) reduces to Eq. (6.78) in a local inertialframe, as it must to be consistent with the principle of general covari-ance.

In general relativity, as in special relativity, an important exampleof an energy-momentum tensor is that of a perfect fluid, which inspecial relativity has the form

( ) ( ( ) ( )) .T p x n x p x U U= - + +tab ab a b (6.80)

In general relativity the energy-momentum tensor of a perfect fluidmust then have the form

( ) ( ( ) ( )) ,T p x g x p x U U= - + +tab ab a b (6.81)

which is a rank 2 general covariant tensor and which reduces to Eq.(6.80) in a local inertial frame and is thus consistent with the princi-ple of general covariance. Recall that ( )p x and ( )xt are defined to bethe pressure and the energy density (at an event point x) as measuredby an observer in a locally inertial frame that is at rest with respect tothe fluid, and are thus scalar fields under general coordinate trans-formations.

For such a perfect fluid let us compute T ; bab to see what is implied by

energy-momentum conservation. Recall that g 0; =bab and, since ( )p x

and ( )xt are scalar fields, ( ) ( )p x p x; ,=b b and ( ) ( )x x; ,=t tb b. Thus, wehave

(( ) )T p U U p g; ; ,= + -tbab a b

b bab

( ) ( ( )p U U p U U U U U U, ,= + + + + +t t � �ba b a b

b vba v b

vbb v a9 C

.p g 0,- =bab (6.82)

In a local inertial frame, this becomes

( ) ( )( ) ,T p U U p U U p g 0; , , ,= + + + - =t tbab

ba b a b

b bab (6.83)

which in turn, in such a frame moving with the fluid, reduces to Eqs.(4.63) and (4.64).


6.10 Exercises

1. Develop your own Mathematica (or Maple) program that, given ametric, will calculate (a) the Christoffel symbols, �tb

o ; (b) thecurvature tensor, Rbat

o ; (c) the Ricci tensor, Rboto ; and (d) the Ricci

scalar, R.

2. For a closed path as indicated in Figure 6.4, evaluate

[ ]x xdsdx ds0

0

1

-a ab

# .

3. (a) Derive the equalities of the Riemann tensor, Eqs. (6.44) and(6.45). (b) Derive the Bianchi identity, Eq. (6.46).

4. Let the metric on the surface of a sphere be given by

( ) ( ) .sin sinds d d dx x dx2 2 2 2 1 2 2 1 2 2= + = +i i z

(a) What are the Christoffel symbols? (b) What are the nonvan-ishing components of the curvature tensor? (c) What is the Riccitensor? (d) What is the Ricci scalar? (e) What vector results if oneparallel transports a vector that starts with only a i component,A Ai i1= d , around a close path with da dai i1= d and db dbi i2= d in thenotation of Exercise 2? (f) Prove that the path defined by =zconstant is a geodesic.

5. (a) Calculate the curvature tensors and the Ricci scalars for themetrics

ds y dx x dyn n2 2 2= +


Figure 6.4. Closed path.

for , , ,n 1 2 3 4= . Are any of these metrics the metric of two-dimen-sional Euclidean space? (b) Calculate the curvature tensors andthe Ricci scalars for the metrics

ds x dx y dyn n2 2 2= +

for , , ,n 1 2 3 4= . Are any of these metrics the metric of two-dimen-sional Euclidean space? Why might you have anticipated theresult?

6. In the low-speed and weak-field approximation (Eq. (6.60)), showthat a freely falling particle of mass m satisfies the expected equa-tion

/ .ga 200d=-

Note in this approximation / /dx d dx di0 &x x. This result is valideven if m� 0, but is, of course, considered small.

6.10 Exercises 111

Chapter 7

Static Spherical Metrics andTheir Applications

7.1 Introduction

In the study of most theories, highly symmetric solutions play a defin-ing role. Compared to more general solutions, they are easier toobtain, and their implications for physical phenomena are moreapparent. They give insights into a theory that, quite often, havegeneral validity. And luckily nature is kind in providing examples thatare quite close to being symmetric. In this chapter we will study spher-ically symmetric solutions to Einstein’s field equations.

Einstein suggested three tests of his theory of general relativity:

1. The gravitational red shift of light.2. The deflection of light by the sun.3. The precession of the perihelia of the orbits of planets, in particu-

lar that of Mercury, the innermost planet.

These processes take place in empty space and for gravitational fieldsthat are almost static and, to a good approximation, sphericallysymmetric. For this reason, as well as for the intrinsic interest of anexact nontrivial solution to the field equations, we obtain a spheri-cally symmetric static solution of the vacuum field equations, theSchwarzschild metric. Then, though we have already derived the gravi-tational redshift for weak fields using the equivalence principle, wewill derive the magnitude of the effect for a static metric. We followthis with a discussion of geodesic motion for massive and masslessparticles in a Schwarzchild metric and apply the results to (2) and (3)above.

113

The dynamics of a gyroscope in general relativity is then consid-ered and specialized to a study of a gyroscope in orbit (i.e., one thatis freely falling), a generalization of the Thomas precession that occursin flat space. Measurement of this precession may prove to be one ofthe most sensitive tests of general relativity.

After studying the processes exterior to a massive body, we willderive the equations governing the dynamics of a fluid comprisingthe interior of such a body. For a particularly simple (and unphysical)fluid, we show that, in contrast to the Newtonian equations, the rela-tivistic equations limit the mass-radius ratio of a static massive body.

Consideration of the geometry of the Schwarzschild metric in theextreme gives rise to the concepts of event horizons and black holes,described in the last section of this chapter.

7.2 The Static Spherical Metric

The meaning of a static spherical metric can be characterized in a wayindependent of the coordinate system used (see Chapter 8). Here wetake the definition of such a metric to be one for which a set of coor-dinates { , }x x i0 exists for which the invariant interval is expressible interms of rotational scalars , , , ,dx x x r x dx dx dxi i i i i i0 = and which is x 0

independent. Thus, the invariant interval can be expressed in theform1

( )( ) ( ) ( )( ) ( ) ),d D r dx E r x dx dx F r x dx H r dx dx2 i i i i i i2 0 2 0 2= - - -x (7.1)

or in terms of the spherical coordinates, , ,r andi z

) .

( )( ) ( )

( ) ( )( sin

d D r dx E r rdrdx

F r r dr H r dr r d r d

22 0 2 0

2 2 2 2 2 2 2 2

= -

- - + +

x

i i i

One can transform to a new time variable so that a term like( )E r rdrdx2 0 does not appear in the metric. With the new time x 0l

defined by

( , ),x G x r0 0= l

the condition g 0r0 =l requires that

( ) ( ) .xG D r

rG rE r 002

222- =

lc m

114 Chapter 7. Static Spherical Metrics and Their Applications

1 This characterization of a static spherical metric follows Weinberg (1973).

This equation is satisfied if we choose ( )x x I r0 0= +l , with ( )I r given by

( )( )( ).

drdI r

D rrE r=

With this change of the time variable, the invariant interval becomes

( )( ) ( ) ( )( ),sind D r dx K r dr H r dr r d r d2 0 2 2 2 2 2 2 2 2= - - + +x i i il

where

( ) ( )( )( )

.K r r F rD rE r2

2

= -e oWe can further reduce the number of functions of r in the metric

to two, if we define a new “radial” variable

( ) .r H r r2 2=l

The invariant interval then becomes

( )( ) ( ) ,sind B r dx A r dr r d r d2 0 2 2 2 2 2 2 2= - - -x i i zl l l l l l

with( ) ( )B r D r/l

( )( )( )

( )( )

.A rH rK r

H rH r r

12

1

2

/ + +

-

ll= =G G

Finally, we drop the primes and write the invariant interval as

( )( ) ( ) .sind B r dx A r dr r d r d2 0 2 2 2 2 2 2 2= - - -x i i z (7.2)

Computing the Ricci tensor (say, by use of Mathematica), we find

( )( )

( )( )

( )( )

( )( )

( )( )

RrA rB r

A rB r

A rA r

B rB r

A rB r

41

200=- + + -l l l l lld dn n (7.3)

( )( )

( )( )

( )( )

( )( )

( )( )

RrA rA r

B rB r

A rA r

B rB r

B rB r

41

2rr =- - + +l l l l lld dn n (7.4)

( ) ( )( )

( )( )

( )R

A rr

A rA r

B rB r

A r12

1=- + - + +ii

l ld n (7.5)

( )sinR R2= zzz ii (7.6)

;R 0= nno � .o (7.7)

7.2 The Static Spherical Metric 115

7.3 The Schwarzschild Solution

We are now in a position to solve for the metric outside of a staticspherically symmetric body. In this region the metric satisfiesEinstein’s vacuum field equation, Eq. (6.71). We will impose thecondition that very far from the mass the metric becomes flat; that is,it is asymptotically flat. Note that this condition applied to the metricof the form Eq.(7.2) gives

( ) ( ) .lim limA r B r 1r r

= =" "3 3

(7.8)

In fact, more general considerations lead to Birkhoff’s theorem(Birkhoff 1923) that states that the spherically symmetric asymptoti-cally flat metric is unique. Thus, the static condition can be relaxedand the solution will apply outside of a spherically symmetricnonstatic mass. From Eqs. (7.3) to (7.6), the field equations Eq.(6.71) for the metric of Eq. (7.2) become

( )( )

( )( )

( )( )

( )( )

( )( )

RrA rB r

A rB r

A rA r

B rB r

A rB r

041

200 = =- + + -

l l l l lld dn n (7.9)

( )( )

( )( )

( )( )

( )( )

( )( )

RrA rA r

B rB r

A rA r

B rB r

B rB r

041

2rr = =- - + +l l l l lld dn n (7.10)

( ) ( )( )

( )( )

( )R

A rr

A rA r

B rB r

A r0 1

21

= =- + - + +ii

l ld n (7.11)

( ) .sinR R0 2= = zzz ii (7.12)

We see that there are three independent equations to be satisfied.From Eqs. (7.9) and (7.10), we have

( ) ( ) ( ) ( )( )

( )( )

.B rR

A rR

rA r B rB r

A rA r1

0rr00 + =- + =l ld n (7.13)

This, with the asymptotically flat condition, Eq. (7.8), implies

( )( ).A r

B r1

= (7.14)

Using this result in Eq. (7.11), we find

( ( )) ,rB r1 0- + =l (7.15)

which has a solution


( ) .B r rc

1= + (7.16)

Here c is some constant. Asymptotically, since the gravitational fieldis weak, Eq. (6.54) applies. This justifies writing

( ) ,g B r rGM rfor1 2 12

00 " 3.= + = -�u

(7.17)

and calling M the total gravitational mass. This parameter is thetotal mass of the solution in the sense that distant orbits areNewtonian orbits of a spherical mass M centered at r 0= . We see thatc of Eq. (7.16) is GM2- u and we have for the invariant interval

( ) ,sind rGM dx r

GM dr r d r d12

122 0 2

1

2 2 2 2 2 2= - - - - -x i i z

-u ud dn n(7.18)

a solution to the vacuum field equations found by Schwarzschild.The nonvanishing Christoffel symbols (see Exercise 2) for this metricare

=

=

( )

r

GM GM r

GMr rGM

2

2

r

r rr

r r

r r

r r r r

0 0

00

00

2

3

2

1

=

=

=

- +

-

-

i i z z

z zi

ii

ii

zz

zz

z iz

i zz

�

coti

cos sin

r

- i i

.

=

=- =-� � �

=

�

� �

GM r2 -

= = =� � � �

=� � / sin i

u u

uu

u

7.4 Gravitational Redshift

We return to the calculation of the change in frequency of a lightwave as it moves radially in the presence of a massive body whosemetric is stationary—-for instance, a Schwarzschild metric.Specifically, we ask, What is the relation between the frequency 1o

measured by an observer at “rest” at r1 and the frequency 2o measuredby an observer at “rest” at r2?

The proper time period of the wave measured by the observer fixedat r1 is

7.4 Gravitational Redshift 117

( ) ,d g r dx/1 00 1

1 2 0=x

where dx 0 is the change in coordinate time. The proper time periodof the wave measured by the observer fixed at r2 is

( ) ,d g r dx/2 00 2

1 2 0=x

where dx 0 is the same change in coordinate time as at r1. This is truebecause, as noted in Section 5.3, this is an essentially static processbecause the metric is static. The coordinate time period is the time ittakes one wavelength to pass the observer’s position and must be thesame for both observers. Thus, we can write

( )

( ),

dd

g r

g r/

/

2

1

00 21 2

00 11 2

=xx

from which follows

( )

( ).

g r

g r/

/

1

2

00 21 2

00 11 2

=oo (7.19)

How does this compare to the shift Einstein obtained in his 1911paper that we discussed in Chapter 5? That shift is given by Eq. (5.2),which written in terms of frequencies takes the form

( ( ) ( )).r r11

22 1= - -o

o� �

If we use the weak-field approximation g 1 200= + � in Eq. (7.19), wehave

( )

( )

( ( ))

( ( ))( ( ) ( )),

g r

g r

r

rr r

1 2

1 21/

/

/

/

1

2

00 21 2

00 11 2

21 2

11 2

2 1. .=+

+- -o

o

�

��

the same result Einstein obtained in his 1911 paper.

7.5 Conserved Quantities

Next we study the geodesic motion of both massive particles andmassless particles in the highly symmetric Schwarzschild metric.Realizing that symmetries give rise to conserved quantities, we wouldexpect to be able to find entities associated with the symmetries of themetric that are conserved as the particle moves along a geodesic.


Such conserved quantities will of course be useful in solving thegeodesic motion.

First, consider the motion for a massive particle. We subsequentlysee how our results need to be modified for a zero-mass particle. Thegeodesic equation can be written (Eq. (6.31))

,ddx U 0; =x

t

ta (7.20)

where /U dx d= xt t , the generalized “four-velocity”. In terms of thecovariant generalized “four-momentum” p mU=v v (recall that themetric can be taken inside the covariant derivative) this can beexpressed as

mU p 0; =tv t (7.21)

or as

.p p p p 0, - =�tv t vt

a ta

or as

( ) .

mddp

p p

g g g p p g p p21

21

, , , ,

=

= + - =

x�vvta t

a

gv t gt v vt gg t

gt vg t

(7.22)

We see that the geodesic equation can always be written as

.mddp

g p p21

,=xv

gt vg t (7.23)

From this equation follows a useful result:

If the metric gab is dependent of xv for some fixed v then pv is constantalong any geodesic.

Of course, the statement that the metric is independent of xv iscoordinate dependent; one would expect that the identification of aconserved quantity associated with a symmetry of the metric could bemade in a coordinate-independent way once the characterization ofthe symmetry of the metric is made in a coordinate-independentmanner. This will be done in some detail in Chapter 8.

How are these considerations modified for a zero-mass particle?Recall that in the geodesic equation, Eq. (7.23), if we parameterizethe geodesic path by an affine parameter s c= x for any constant c ,then the equation for the geodesic in terms of s is identical to that

7.5 Conserved Quantities 119

of x. We could, for instance, use /s m= x and for such we see/p dx ds=a a , and Eq. (7.23) becomes

.dsdp

g p p21

,=v gt vg t (7.24)

The generalization of these equations to zero-mass particles isclear. For a particle of nonzero mass we can use x to parameterize thegeodesic. For a zero-mass particle, we can take ( / )lims mm 0= x" . Ineither case, p p m2=a

a .

7.6 Geodesic Motion for a Schwarzschild Metric

The Schwarzschild metric is time independent and sphericallysymmetric. We would expect that, since it is spherically symmetric,geodesic motion is confined to a “plane”, as in the Newtonian case.We can easily see this is so. Consider the geodesic equation for pi:

.sin cosdsdp

g p p p p21

,= = i iigt i

g t z z

Since the metric is spherically symmetric for any point on thegeodesic, we can pick the spherical coordinates such that

/p and0 2= =i ri ; the particle is moving along the “equator” at thatinstant. But this equation shows that, at that instant, /dp ds 0=i . Thus,it continues to move along the equator, or in a plane. Without loss ofgenerality, we can assume the particle moves with /2=i r .

Since the metric is independent of x 0 and ,p0z and pz are constant.For a massive particle we define

, .p mE p mL0 = =-z

(Here E is a dimensionless constant, whereas the constant L has thedimension of length.) For a zero-mass particle we define

, .p E p L0 = =-z

(Here E has the dimension of mass, whereas L has the dimension oflength #mass.) Thus, for a massive particle,

p g p rMG mE120 00

0

1

= = -

-ud n (7.25)

p mddrr =x

(7.26)


,p g p r mL mdd2= = =xz-z zz

z (7.27)

whereas for a zero-mass particle,

p g p rMG E120 00

0

1

= = -

-ud n (7.28)

pdsdrr = (7.29)

.p g p r Ldsd2= = =z-z zz

z (7.30)

By use of Eqs. (7.25) — (7.27) , p p m2=aa becomes an equation for

/dr dx:

.rMG m E r

MG mddr r m L m1

212

1

2 2

1

22

2 2 2 2- - - - =x

- -

-u ud d cn n m

(7.31)

Thus the “radial” equation for a massive particle is

.ddr E

rL

rGM

1 12

22

2

2

= - + -x

uc d dm n n (7.32)

Similarly, the radial equation for a zero-mass particle is

.dsdr E

rL

rGM

12

22

2

2

= - -uc dm n (7.33)

We can obtain the orbit equation, an equation for /dr dz, by dividing/dr dx ( /dr ds) by /d dz x ( /d dsz ) for the massive particle (for the zero-

mass particle), with the result

/ddr

L r

ErL

rGM

1 12

2

2 4

22

2

=

- + -

z

u

ed d

on n

(7.34)

for the massive particle. With the substitution

,u r1= (7.35)

this orbit equation becomes

.ddu

LE MGu

Lu1 2

12

2

2

22= - - +

zue _ do i n (7.36)

7.6 Geodesic Motion for a Schwarzschild Metric 121

Proceeding similarly, the orbit equation for a zero-mass particle is

.ddu

LE MGu u1 2

2

2

22= - -

zue _o i (7.37)

We can get insight into the nature of the possible orbits of bothmassive and zero-mass particles by considering Eqs. (7.32) and (7.33)as “energy” conservation equations for one-dimensional motion witheffective potentials Vm and V0 for massive and zero-mass particles,respectively:

VrL

rGM

1 12

m 2

2

= + -ud dn n (7.38)

.VrL

rGM

12

0 2

2

= -ud n (7.39)

An effective potential Vm is shown in Figure 7.1 with L GM4= u . Notethat turning points of orbits occur at values of r such that ( )E V rm

2 = .We see that there are two orbits of fixed radius occurring at theextrema of ,V Vm m

a, and Vmb, the inner corresponding to an unstable

orbit of radius a GM4= u and the outer a stable orbit of radiusb GM12= u . For < <V E 1m

b 2 there exists an orbit with two turning pointsbetween r a= and r b= and an orbit with one turning point at a point<r a. For this case (L GM4= u ), all unbound orbits, which have >E 12 ,.

have no turning points. However, > >V if L GM1 4ma u and unbound

orbits with one turning point exist.


Figure 7.1. Effective potential for a massive particle.

The effective potential V0 is shown in Figure 7.2. One unstable orbitof fixed radius occurs at the single extremum of V0, which occurs atr GM3= u . For ( ) > >V GM E3 00

2u , there is an unbound orbit with oneturning point. If > ( )E V GM32

0u , the orbit has no turning points.

7.6.1 Gravitational Deflection of Light

In Chapter 5, we saw that Einstein, in his 1911 paper, predicted adeflection of .875 arcsec for light passing near the sun. The predic-tion arising from his general theory is not the same, as one canrealize by noting that both g gandrr 00 are changed from their flatvalue, not just g00, as is effectively assumed in the 1911 paper. Thedeflection is calculated by calculating the change in ,z dz, as thephoton comes in from r 3= , passes near the massive body at someclosest point, and then proceeds out to r 3= . The deflectionis -z dz r� . (See Fig. 7.3.) Equivalently, we could calculate the

change in z as the photon proceeds in from r 3= to the point ofclosest approach, subtract /2r , and double the result.

Note that the point of closest approach /r u10 0= is determined bysetting the right-hand side of the orbit equation, Eq. (7.37), equal tozero. This determines /E L2 2 in terms of u0 as

,LE MGu u1 22

2

0 02= - u_ i (7.40)


Figure 7.2. Effective potential for a zero–mass particle.

and, then, the orbit equation can be written in terms of u0. Since wewant to integrate to find the change in z, we write the resulting orbitequation as

.dud

MGu u MGu u1 2 1 2/

02 2

1 2

!= - - -z -

0u u_ _a i i k (7.41)

The plus obtains for an incoming photon with >L 0. Thus, we find

.

MGu u MG u

du2

1 2 1 22/

u

02 2

1 2

0

0

=

- - -

-z r�

u0

#u u_ _a i i k (7.42)

The integral to be evaluated is an elliptic integral.To obtain an analytic expression for the deflection angle in terms

of the point of closest approach u0, we do a weak field expansion ofEq. (7.42) using the small parameter GMuu . There are some subtletiesin such an expansion. The expression for solving /E L2 2 in terms of u0is linear in GMuu . Thus, in the integrand of Eq. (7.42) the small expan-sion parameter occurs in two forms, GMu0u and GMuu . Expansion of Eq.(7.42) to first order in these two parameters leads to

( ) ( ).

u udu MG

u u

u udu

2 2/ /uu

02 2 1 2

02 2 3 203 300

00

.-

+-

--

z r� ## u (7.43)

The first integral gives /2r . The second integral gives GMu2 0u . Thus,

we have

,GMu rGM

22

20

0. =

z� u u(7.44)

which yields a predicted deflection of .1 75=z� arcsec, twice thatpredicted by Einstein in his 1911 paper.

The deflection was measured by Dyson and Eddington (Dyson et al.1920) during a solar eclipse in 1919, yielding a result . .1 90 0 16!=z�arcsec, confirming the prediction within about 30% accuracy.


Figure 7.3. Light deflection.

7.6.2 Precession of the Perihelia of Orbits

In Newtonian dynamics all bound orbits for an attractive /r1 potentialand an isotropic simple harmonic oscillator close; that is, the periodof motion in the r variable is the same as that in the angle z variable.For other spherically symmetric potentials this is generally not thecase.

The “time” it takes for the orbiting particle to go from perihelionr- to aphelion r+ and back to perihelion is not the same as the time ittakes the particle’s angle position z to change by 2r. The orbitsprecess. (See Fig. 7.4.)

Now consider the orbit equations of a particle about a sphericalmass as given by Eqs. (7.27) and (7.32). The effective potential Vm ,Eq. (7.38), is not the effective potential of a /r1 potential (nor, forthat matter, is it one of an isotropic simple harmonic oscillator).Thus, one would expect that the orbits would precess. We will calcu-late the precession per orbit expressed in terms of r- and r+.

First, we show that the Newtonian orbit does not precess. Note thatsince the equivalent one-dimensional potential for the Kepler prob-lem does not contain a /r1 3 term, one would expect to obtain theNewtonian orbit equation if one neglects the /r1 3 term in Eq. (7.32).Indeed, if this term is dropped and the equation is multiplied by/m 2, the following equation results:


Figure 7.4. Perihelion precession.


.mddr E m

mrL m

rGmM

21

2

22

2

2 2

= - - -x

uc _ dm i nFor low velocity and a weak field d dt.x . Thus we see that the equiv-alent one-dimensional motion is that of the Kepler problem.Neglecting the corresponding u3 in Eq. (7.37), we obtain for the orbitequation

.ddu

LE MGu

Lu1 2

12

2

2

22= - - -

zue _o i (7.45)

The aphelion, /u r1=+ +, and perihelion, /u r1=- -, are determined bythe condition /du d 0=z . Solving the resulting two simultaneous equa-tions,

,LE MGu

Lu1 2

102

2

22- - - =! !

u_ ifor / /E L Land 12 2 2, one obtains

,L GM

u u GMu u1

2

22 =

+ -+ - + -

u

u(7.46)

and

.LE

GMu u22

2

=++ -

u (7.47)

With these expressions, Eq. (7.45) becomes

,ddu u u u u

2

= - -z - +e ^ _o h i (7.48)

and thus

.dud

u u u u

1/ /1 2 1 2=

- -

z

- +^ _h i (7.49)

As the particle proceeds from aphelion to perihelion, the change inz is thus

( ) ( )( ) ( )

.u uu u u u

du/ /

u

u

1 2 1 2- =- -

=z z r- +- ++

-# (7.50)

Since the change is the same as the particle proceeds from perihelionto aphelion, the total change in is2z r; the orbit closes.


It is clear how to calculate the precession using the orbit equationEq. (7.36). Express the right-hand side in terms of u uand+ -, obtainedby setting the right-hand side equal to zero; integrate the equation for/d duz from u+ to u- and double the result. The result is the precession

mod 2r. The integral to be evaluated is the elliptic integral

( ) ( ) .u u

LE MGu

Lu

du

1 21

/

2

2

22

1 2- =

- - +

z z- +

u_ di nR

T

SSS

V

X

WWW

(7.51)

As noted, the expressions for /E L2 2 and /L1 2 in terms of u+ and u- areobtained by solving the simultaneous equations that result fromsetting the right-hand side of Eq. (7.37) equal to zero:

.LE MGu

Lu1 2

102

2

22- - + =! !

u_ di n(If one views these equations as equations for u! in terms of/ / ,E L Land12 2 2 they are cubic and generally yield three values for>u 0. We expect this from the above discussion of the nature of orbits.

However, any two such solutions determine /E L2 2 and /L1 2.)We obtain

( ) ( ),LE

MG

u uu u u u MG u u u u

22 22

22 2 2 2=

+- + + + ++ -

+ - + - + - - +uu (7.52)

and

( ).L MG

u uu u u u1

22

2 2=+

- + ++ -+ - + -u (7.53)

The substitution of these expressions into Eq. (7.51) gives( ) ( )

( )( ) ( )( )( ).

u u

u u u u MG u u u u u u u

du

2/

u

u

1 2

- =

- - - - - + +

z z- +

+ - - + + -+

-#u_ i (7.54)

As expected, setting the term proportional to MG2 u to zero givesthe Newtonian result Eq. (7.49). Treating this term to first order,however, gives

( ) ( )u u .-z z- +

( ) ( ) ( ) ( )

( ).

u u u udu MG

u u u u

u u u du/ / / /

u

u

u

u

1 2 1 2 1 2 1 2- -+

- -

+ +

- + - +

+ ---

++

## u (7.55)


u-

u+

The first integral gives r, the Newtonian result, whereas the secondgives ( ) /MG u u3 2+ r+ -

u . The change in z as the particle moves fromaphelion to perihelion and back to aphelion is ( )GM u u2 3+ +r r + -

u ;the orbit precesses by an angle z� given by

( ) .GM u u GM r r3 31 1

1= + = +z r r� + -

+

u u b l (7.56)

The perihelion advances by this amount for each orbit as depicted inFigure 7.4.

For Mercury, the prediction is a perihelion advance of 43 arcsecper century, a prediction experimentally confirmed by observation.The actual observed advance must be corrected by effects of plane-tary perturbations to effect the agreement with the predictedadvance.

7.7 Orbiting Gyroscopes in General Relativity

We argued in Chapter 3, Eq. (3.20), that a gyroscope, in special rela-tivity, obeys the dynamical equation

( ).

ddL

LddU Un=-

xx

x

avvb

ba (7.57)

By the principle of general covariance, in curved space-time this equa-tion becomes

( ),

dDL

L gdDU U=-

xx

x

avvb

ba (7.58)

obtained merely by replacing the ordinary derivatives by covariantderivatives. For an orbiting gyroscope,

,dDU

0=x

b

(7.59)

which is the geodesic equation, and Eq. (7.56) becomes

( ),

dDL

0=xxa (7.60)

or, written out,

.ddL L U 0+ =x

�a

vba v b (7.61)


Note that this equation merely states that the four-vector La is paral-lel-transported (in space-time) along the orbit. We might then expectthat the value of the four-vector, when the gyroscope is at a point ona closed orbit, will not repeat when the gyroscope returns to thispoint. The gyroscope is expected to precess, meaning the “direction”of the spin of the gyroscope is different when it returns to its startingpoint from what it was at the start. One can observe the direction thegyroscope points with respect to the light received from fixed stars atthe beginning of an orbit and at the completion. A comparison wouldgive the angle of precession.

If we write ( )L 2ra for the value of La when the gyroscope completesthe orbit and L0

a for the value at the beginning of the orbit, theprecession angle, pd , after the completion of one orbit, satisfies therelation

( ),cos

L LL L 2

p0 0

0

$$

=dr

where L is the three-vector angular momentum in the rest frame ofthe gyroscope. Recall that in the rest frame of the gyroscope the time-component of the four-vector La is zero. Thus, we can rewrite thisrelation for pd in the covariant form,

p .=( )

cosL LL L 2

0 0

0dr

aa

aa (7.62)

We calculate this precession for a gyroscope in a circular orbit with/2=i r in the Schwarzschild metric, eventually restricting to weak

fields and small velocity. First, we write, in terms of z rather than xderivatives,

.ddL L

ddx

0+ =z z

�a

vba v

b

(7.63)

Furthermore, from Eqs. (7.25) and (7.27),

/, , , .

ddx

ddxdd

LE

GM rr

1 20 0 1

2

= =-z x z

xv v

ue o (7.64)

For a circular orbit, / ,u u r1= =+ - Eq. (7.52) gives

( ).

LE

GMrGM1 2/1 2=

-uu

(7.65)

With the Schwarzschild metric’s Christoffel symbols (Sec. 7.3), Eqs.(7.64 ) and (7.65), Eq. (7.63) becomes

7.7 Orbiting Gyroscopes in General Relativity 129

/, (

./ )

( / ) , ,ddL r

GMGM rL

rGM GM r L

GM r rL rL

1 21 2

1 2 0

/ /r

r

1 2 1 20

=- --

- -z

a

z

uu

u u

u

J

L

KKKKK

d d N

P

OOOOO

n n(7.66)

From this, we find that

.dd L

rGM L13r

r2

2

=- -z

ud n (7.67)

Let us consider the case for which ( ) ( , , , )L L0 0 0 0r0= =za . Clearly,

( ) ( )L U0 0 0=aa , as it must. Also, from Eq. (7.66), /dL d 0r =z at 0=z .

With these conditions the solution to Eq. (7.67) is

cosL L rGm

13

/

r r0

1 2

= - zud n

R

T

SSS

V

X

WWW, (7.68)

and thus

( ) .cosL L rGm

2 13

2

/

r r0

1 2

= -r rud n

R

T

SSS

V

X

WWW

(7.69)

One could use this solution in Eq. (7.65) and solve for the z depen-dence of L0 and Lz. However, to solve for the amount the gyroscopeprecesses in one orbit, we can use Eq. (7.69) in Eq. (7.62) with theresult

.rGM

rGM

13

2 23

/

p

1 2

.= - -d r r ru ud n (7.70)

This is referred to as the de Sitter precession. It is not the analog of theThomas precession—-rather, it is the precession of freely fallingframes, not frames that are accelerated, as is the case of the Thomasprecession. Note that the weak-field result, /GM r3 ru , is the precessionof the perihelion of an orbit in the limit of a circular orbit.

For a gyroscope orbiting around a spinning body, such as the earth,for which the metric is not quite spherically symmetric, there is anadditional contribution to the precession due to “frame dragging.”This additional precession is called the Lense-Thirring precession. Inorder to discuss this effect, we would have to derive the modificationof the metric due to rotation, something we will not do.

Both the de Sitter and the Lense-Thirring precession are to bemeasured by Stanford’s Gravity B Probe, scheduled for launch in 2003.


7.8 Stellar Interiors

We have obtained the metric outside of a static spherical distributionof matter and will now study the equations for a static sphericalmetric in the presence of matter. Such metrics are the metrics of theinterior of nonrotating stars and must smoothly go over to theSchwarzschild solution at the surface of the material.

The metric is determined, through the Einstein field equations, bythe properties of the matter as expressed by the energy-momentumtensor, which we assume to be that of a perfect fluid, and the equa-tion of state of the fluid, which relates the pressure of the fluid to itsenergy density.

Since the metric is assumed to be static and spherically symmetric,the invariant interval can again be put in the form of Eq. (7.2):

( )( ) ( ) .sind B r dx A r dr r d r d2 0 2 2 2 2 2 2 2= - - -x i i z (7.71)

This, of course, means that we again have

( )( )

( )( )

( )( )

( )( )

( )( )

RrA rB r

A rB r

A rA r

B rB r

A rB r

41

200=- + + -l l l l lld dn n (7.72)

( )( )

( )( )

( )( )

( )( )

( )( )

RrA rA r

B rB r

A rA r

B rB r

B rB r

41

2rr =- - + +l l l l lld dn n (7.73)

( ) ( )( )

( )( )

( )R

A rr

A rA r

B rB r

A r12

1=- + - + +ii

l ld n (7.74)

( )sinR R2= zzz ii (7.75)

;R 0= nno � .o (7.76)

Since the metric is static and spherically symmetric, the energy-momentum tensor must also be. Thus,

( ) ( ( ) ( )) ( ) ( ),T p r g p r r U r U r=- + + tab ab a b (7.77)

with ( )U U r0 0= da a. The four-velocity has a zero component only. (Onemight think it could have a nonvanishing r component, but if itdid T r0 would be nonvanishing, whereas R 0r0 = implies it vanishes.)Further, since ( ) ( ) ( )U U g U r U r g 1r0 0

00= =a bab and ( ) ( ) ( )U r U r B r0 0 1= - ,.

the energy-momentum tensor, in covariant form, becomes

7.8 Stellar Interiors 131

( ) ( ( ) ( )) ( ).T p r g p r r B r0 0=- + + t d dab ab a b (7.78)

For a perfect fluid ( ) ( )T T r p r3= = -tb

b . Thus, we have

/ ( ( ) ( )) ( ).T g Tp

g p r r B r22 0 0- =-

+ +t

t d dab ab ab a b (7.79)

With this and Eqs. (7.72) to (7.76), Einstein’s field equation, Eq.(6.70), becomes

( )( )

( )( )

( )( )

( )( )

( )( )

( )rA rB r

A rB r

A rA r

B rB r

A rB r

Gp

B r41

28

23

- + + - =-+

rtl l l l ll ud dn n (7.80)

( )( )

( )( )

( )( )

( )( )

( )( )

( )rA rA r

B rB r

A rA r

B rB r

B rB r

Gp

A r41

28

2- - + + =

-r

tl l l l ll ud dn n (7.81)

( ) ( )( )

( )( )

( ).

A rr

A rA r

B rB r

A rGp

r12

18

22- + - + + =

-r

tl l ud n (7.82)

The local conservation of energy-momentum equation, T 0; =bab ,

gives

( )( ) ( ) ( )

( )( )

.Tdrdp r

A r p A rB rB r2

0;r 1 1= + + =t- -bb l

(7.83)

(The other components of the equation are trivially satisfied.) Thecomputation of this covariant derivative requires knowledge of theChristoffel symbols for the metric of Eq. (7.71). (See Exercise 7.1.)Eq. (7.83) is not independent of Eqs. (7.80)–(7.82) but is satisfied asa result of the Bianchi identity. However, it has a particularly usefulform.

To solve for the metric, the procedure will be to use Eqs. (7.80)–(7.82) to obtain a solution for ( )A r and ( )B r in terms of a ( )p r and( )rt , and then Eq. (7.83) will be a first-order differential equation

relating ( )p r to ( )rt , to be solved consistent with the assumed equa-tion of state. To this end we see that a rather simple combination ofEqs. (7.80)—(7.82) yields an equation for ( )A r ,

( ) ( ) ( )

( )

( ),

B rR

A rR

r

R

r rA r

A r

r A rG

2 21 1

8rr002 2 2 2+ + =- - + =- r tii l u (7.84)

which can be rewritten as

( ( )).

drd rA r

G r8 11

2=- +r t-

u (7.85)

Equation (7.85) has a solution


( )( )

.A r rGm r

rc

12

1

= - +

-ue o (7.86)

Here

( ) ( ) ,m r r r dr4r

2

0= r t# l l l (7.87)

and c is an integration constant. Since the space-space part of themetric must become a three-dimensional Euclidean space for smallr, cmust be zero. With ( )A r known, Eq. (7.82) can be solved for /B Blwith the result

( )( )

( )

( ).

B rB r

r r Gm r

G pr m r

2

2 4 3

=-

+rlu

u

78

AB

(7.88)

This use in Eq. (7.83) results in the following equation, known as theOppenheimer-Volkoff (O-V)equation:

( )( )

( )

( ).

drdp r

pr r Gm r

G pr m r

2

4 3

=- +-

+t

r

u

u

78

AB

(7.89)

Integration of the O-V equation, consistent with the equation ofstate relating t to p, with some assumed value of ( ),p 0 will give ( )p rand ( )rt . Note that ( )p r is monotonically decreasing with increasingr if 0$t and ( ) <Gm r r2 u . Also note that the expression for ( )A rbecomes singular if ( )Gm r r2 =u . It is, thus, reasonable to assume that

( ) <Gm r r2 u in the interior of a static star. (See the next section.) Withthis condition, the pressure monotonically decreases from its centralvalue until it reaches zero at some value of r, denoted by rs, the“radius” of the star.2 The value of ( )m rs equals the “mass” parameterM of the Schwarzschild metric for the star’s exterior. We see that the“radius” and the “mass” of the star are determined by the centralpressure pc and, of course, the equation of state.

One should not make too much of the relation

( ) ( ) ,m r r r dr M4s

rs2

0= =r t#

a relation that is identical to that in Newtonian theory. Rememberthat ( )rt is the energy density in an inertial coordinate system at restwith respect to the fluid, and the coordinates { , , , }x r0 i z are not such


2 With the reasonable assumption that the equation of state is such that / >dp d 0t , theenergy density also monotonically decreases from its central value.

coordinates. More to the point, r dr4 2r is not a “proper” volumeelement. After all, the proper distance, at fixed coordinate time,between r and r dr+ is ( ( )/ )Gm r r dr1 /1 2- -u , not dr. Thus, a propervolume, at fixed coordinate time, is

( ),r

Gm rr drd1

/1 2

2- �

-ue onot r drd2 �. The integral cannot be identified with the total energy ofthe star, but only with the “mass” parameter of the exteriorSchwarzschild metric.

In the Newtonian limit with <<p t and ( ) <<Gm r ru , the O-V equa-tion becomes

( ),

drdp

r

Gm r2. -

u(7.90)

the Newtonian hydrostatic equilibrium equation. An interestingdifference in the stellar equilibrium equations of general relativityand Newtonian gravity is seen by comparing Eq. (7.90) with O-Vequation for a given density distribution ( )rt . It is easy to see that forthis case | / |dp dr is greater for the O-V equation than it is for theNewtonian Eq. (7.90). This in turn implies that the pressure at eachr is greater in the general relativistic star than in the Newtonian star.The relativistic star requires a higher pressure than a Newtonian starto keep the material from collapsing. A significant implication of thisis well illustrated by considering a star made of an incompressiblefluid, that is, one for which ( )rt is a constant 0t . In this case, for bothtype of stars

( ) ,m r r Mrr

34

s

30 3

3

= =r t (7.91)

where M is the “mass” parameter of the exterior Schwarzschild metricand rs is the “radius” of the star—the “radius” at which the pressurevanishes.

7.8.1 Constant Density Newtonian Star

The Newtonian equilibrium equation, Eq.(7.90), by use of Eq.(7.91)becomes

,drdp

rGM rs3=-u

(7.92)

which is easily integrated to yield


.prGM

rr

21

s s

2

= -u b l> H (7.93)

The condition ( )p r 0s = has been imposed. The central pressure ps is/GM r2 su and is finite for all M and rs.

7.8.2 Constant Density Relativistic Star

The O-V equation takes the form

( )( )( ) ,

drdp r

p pGrGr

33 8

40 0 2

0

= + +-

-t tr truu

(7.94)

or equivalently,

( )( ).

p pdp

GrGrdr

3 3 8

40 0

20

+ +=-

-t t r t

ruu

(7.95)

This last equation, upon integration, yields

.pp

G r

G r3

3 8

3 8/

s0

0

02

021 2

++

=-

-tt

r t

r tu

uJ

L

KK

N

P

OO (7.96)

Again, the condition ( )p r 0s = has been applied. Solving for ( )p r , wehave

( )( ) ( )

( ) ( ).p r

G r G r

G r G r

3 3 8 3 8

3 8 3 8/ /

/ /

s

s0

02 1 2

02 1 2

02 1 2

02 1 2

=-- - -

- - -t

r t r t

r t r tu u

u u(7.97)

This implies an upper limit on the density 0t for a star of a fixed-radius star. This upper-limit density is that for which the central pres-sure becomes infinite, which occurs when the denominator of theright-hand side of Eq. (7.97) vanishes. Thus,

< .Gr838

s20r tu (7.98)

Expressed in terms of the mass M of the star, this inequality becomes

< .rGM

94

s

u(7.99)

A static homogeneous star made of an incompressible fluid cannotexceed this mass -to- radius ratio. Since incompressible fluid has the“stiffest” equation of state possible, one would expect that theinequality Eq. (7.99) would hold for any equation of state. That suchis the case was proved by Buchdahl (Buchdahl 1959). It should be


noted that an incompressible fluid is unphysical. The velocity ofsound in a perfect fluid is given by /p2 2t, which for an incompress-ible fluid is infinite.

We can complete the solution for the metric of this relativistichomogeneous star by solving for ( )B r using Eq. (7.83) written in theform

( )( )

( )( )

.B rB r

p rp r2

0=-

+ tl l

(7.100)

Upon integration, this yields

( )( ( ))

/.B r

p r

GM r1 2 s

02

02

=+

-

t

t u_ i(7.101)

The boundary condition ( ) /B r GM r1 2s s= - u has been imposed. Thisassures that the internal metric matches the external metric at thestar’s surface. Finally, with ( )p r given by Eq. (7.97), we obtain

( ) .B r rGM

rGMr

413 1

212

//

ss

1 2

2

1 22

= - - -u uJ

LKKd N

POOn

R

T

SSS

V

X

WWW

(7.102)

It is noteworthy that the bound on /G rsu given by Eq. (7.99) restrictsthe size of the gravitational redshift of light emitted from the surfaceof a static spherical star. From Eq. (7.19), for the fractional increase inthe wavelength, denoted by z, we have

( )

( )( ) .z

g r

g rB r1 1 1/

//

1

2 1

1

2

00 11 2

00 21 2

11 2=

-= - = - = -

mm m

mm - (7.103)

Here r1 is the radial position at which the light is emitted and, it isassumed, the light is received at r 3= . Thus, a star satisfying thebound of Eq. (7.99) has a redshift parameter z less than two for lightoriginating at the surface. Of course, light coming from the interiorof the star, imagined opaque, does not respect this bound. In fact, fora star saturating the bound, ( )B 0 3= and thus z is unbounded.

7.9 Black Holes

It seems something peculiar occurs to the Schwarzschild metric, Eq.(7.18), at r r GM2H /= u . The coefficient of ( )dx 0 2 becomes zero—-theinvariant interval becomes independent of dx 0. In addition, the coef-ficient of dr 2 becomes infinite so that a small dr causes the dx to blowup. The metric has a very singular behavior here. Is this singular


behavior intrinsic to the space-time geometry or is it merely reflectingproperties of the space-time coordinates being used?

Consider the metric of a sphere expressed in terms of polar andazimuthal angles,

.sindl d d2 2 2 2= +i i z

At the pole, the coefficient of d 2z vanishes; however, we know thereis nothing peculiar about the geometry of the sphere at a pole—-thegeometry of a sphere is the same at all points. This singular behavioris a property of the coordinates, not an intrinsic property of thesphere.

Though the Schwarzschild metric is not well behaved at r rH= , andsome components of the Riemannian curvature tensor are singular atr rH= (see Exercise 2), the metric does satisfy the vacuum field equa-tions R 0=ab for both >r rH and <r rH. However, since <g 000 and >g 0rr

for <r rH in this region, x 0 is a spacelike coordinate and r is a timelikevariable. We obtain a good picture of this geometry by consideringthe light cones associated with “radially” traveling light defined by

( ) ( ) .d rrdx r

rdr1 1 0H H2 0 2

1

2= - - - =x-c cm m

These cones are depicted in Figure 7.5. Remembering that r is a“time” variable for <r rH the metric is not “stationary’’ that is, it is notr independent.

7.9 Black Holes 137

Figure 7.5. Schwarzschild light cones.

If we could perform a change of coordinates such that the metricand the Riemann tensor are well behaved at ,r rH= we would knowthat the singular behavior is not intrinsic to the metric. A manifesta-tion of the singular behavior of the metric is that the radial lightcones close up as r rH" . One can imagine keeping these cones openby choosing a new time variable x 0l that is r dependent, so that theingoing radial rays keep slope 1- . With a change of coordinates

( ), , ,x x f r r r0 0= + = = =i i z zl l l l l for >r rH, the condition that theingoing radial cone has slope 1- is

,dr

dfrr1

H

1

=- --

llb l

which gives

( ) , > .lnf r r rr r r1HH

H=- -l l lb lThe transformed Schwarzschild invariant interval in these coordi-nates, called the Eddington-Finkelstein coordinates, becomes

( ) .sind rrdx r

rdr r

rdx dr r d r d1 1 2H H H2 0 2 2 0 2 2 2 2 2= - - + - - -x i i zlc cm m

(7.104)Though the coordinate change was meaningful only for >r rH, theresulting metric is well behaved for >r 0, the vacuum field equationsare satisfied for >r 0, and the components of the Riemann tensor arewell behaved for >r 0 (Exercise 3). Using Eq. (7.104), we see that radi-ally “outgoing” light has a slope given by

.drdx

r rr r

H

H0

= -+l (7.105)

For >r rH this slope is in fact positive, but for <r rH it is negative. For<r rH, both radially proceeding light rays move inward. The resulting

light cones are depicted in Figure 7.6.Light cannot proceed outward originating at <r rH. Since the world

lines of particles of nonzero mass move within the future light cone,such particles cannot pass outward past rH, called the Schwarzschildradius. The “sphere” defined by r rH= is referred to as the black hole’sevent horizon. If the mass of an object such as a star becomes containedin this sphere, a black hole is formed. The ultimate fate of a stardepends on the dynamics of the material forming the star. Thoughthe exterior solution is a Schwarzschild metric, the interior solution,


and thus the development of the surface that, as described in the lastsection, is characterized by the radius at which the density and pres-sure vanish, depends on the details of the dynamics.

Nevertheless, we can question the ultimate fate of a spherical star ifits surface, moving inward, passes the radius of the Buchdal bound,Eq.(7.99), /r GM9 4s= u . Since the bound would be exceeded, the mate-rial cannot return to a static distribution. Rather, it must continue toflow inward. But to what end? The metric exterior to the matterremains Schwarzschild as follows from Birkhoff’s theorem. Thus,eventually the surface of the star must pass the Schwarzschild radius.A black hole, of necessity, forms.

A realistic study of the dynamics of a star is required to see underwhat conditions it would evolve so that a black hole is indeed formed.Such studies require knowledge of the equation of state of materialconsisting of nuclei and other elementary particles under conditionsfor which quantum effects are important. This we will not consider.

7.10 Exercises

1. Starting with the metric of the form Eq. (7.2), compute (a) theChristoffel symbols and (b) the Ricci tensor to obtain Eqs. (7.3)to (7.7).

7.10 Exercises 139

Figure 7.6. Future light cones in Eddington-Finkelstein coordinates.


2. (a) Using Schwarzschild coordinates, compute the Schwarzschildmetric’s Christoffel symbols. (b) Compute the metric’s Riemanntensor. Are any of the elements singular? If so, where?

3. (a) Using Eddington-Finkelstein coordinates, compute theSchwarzschild metric’s Christoffel symbols. (b) Compute themetric’s Riemann tensor. Are any of its elements singular? If so,where?

4. A photon travels in the unstable circular orbit at the radiusr GM3= u . What is the elapsed Schwarzschild coordinate time forone orbit?

5. A clock is in a circular orbit at r GM12= u . (a) How much timeelapses on the clock during one orbit? (b) A second clock is heldstationary at a point on the orbit of the first clock. What is theelapsed time the second clock reads between successive passes theof the first clock? (c) Is (a) or (b) greater? Is this what you expect?(Consider the twin “paradox’’.)

6. A particle is held at a fixed “radius’’ in Schwarzschild space. (a)What is the magnitude of the three-acceleration of the particle inan inertial frame instantaneously moving with the particle? (Hint:What is the covariant definition of the acceleration ( i.e., whattensor reduces to the four-acceleration in an inertial frame)? (b)In a weak-field region (i.e., where /GM r 1%u ), what is the accelera-tion? Is this what you would expect? (c) At the Schwarzschildradius, what is the acceleration?

7. A photon propagates radially outward in a Schwarzschild metricof mass M. (a) What is the elapsed coordinate time x 0 as it movesfrom r1 to r2 ? (b) In terms of p E0 = , what is the energy at r1 asmeasured by an observer fixed at r1? (Hint: The four-velocity ofthis observer /U dx d= xa a has “space’’ components zero. Also, inhis inertial frame /dx d 0i =x . But U U 1=a

a . So what is U 0? Thenconsider the invariant p Ua a.) (c) Show that (b) implies the gravi-tational redshift given by Eq. (7.19).

8. A particle of mass m is “freely falling” outward in a Schwarzschildmetric of mass M . (a) What is the value of the constant of motionE, in p mE0 = , such that the particle just reaches r 3= ? (b) Interms of this value of E, what is the energy of the particle at r r1= ,


as measured by on observer fixed at r1? (See the hint in Exercise3.) (c) In the weak-field approximation, /GM r 1%u , what is theapproximate energy of the particle at r1 as measured by anobserver fixed at r1? Interpret this expression in terms of theescape velocity one obtains in Newtonian theory.

9. Suppose the Einstein field equations have a nonvanishing cosmo-logical constant m. (a) For a static spherically symmetric metric ina vacuum, what are the equations for the metric components ( )A rand ( )B r . That is, what replaces the Eqs. (7.9) to (7.12)? (b) Solvethe equations of (a) and show there doesn’t exist an asymptoti-cally flat solution.

10. Derive the Eq. (7.66) for the precession of a gyroscope in circularorbit moving in a Schwarzschild metric. How does the de Sitterprecession for one orbit around the earth compare to theThomas precession one would calculate treating the gyroscope asan accelerating body moving in a circle?

7.10 Exercises 141

Chapter 8

Metrics with Symmetry

8.1 Introduction

The symmetries of a metric are important in the study of dynamicallyconserved quantities and, as we will see in Chapter 9, play a definingrole in the study of cosmology.1 And, of course, the assumption of asymmetry of the metric simplifies the Einstein equations, as we saw inthe discussion of the Schwarzschild metric and of the equations ofmotion for freely falling particles. We will formulate the meaning ofa symmetry of a metric without reference to a particular coordinatesystem which reflects the symmetry—-that is, we will characterize thethe symmetry in covariant language.

This characterization can be illustrated by the consideration of themetric of a two-dimensional sphere endowed with the metric inducedby imbedding in a three-dimensional Euclidean space. In the follow-ing we will use this case to illustrate symmetry concepts that have amore general applicability.

8.2 Metric Automorphisms

The two-dimensional sphere’s metric, using the usual polar coordi-nate system, is symmetric under rotation about the polar axis—-themetric is independent of the polar angle. But by considering themetric itself, how can we know it is symmetric under rotation aboutany direction in the embedding space? Indeed, what does it mean tosay that the metric is symmetric under these rotations?

1 Chapter 9 can, however, be understood without studying this chapter.

143

Under rotation each point of the sphere ( , )x i= i z moves to thepoint ( , )x j = i zl l l . Formally, we represent this as

( , , )=i i i z dl l (8.1)( , , )=z z i z dl l . (8.2)

Here the direction and magnitude of the rotation is indicated by d.Such a one-to-one mapping of a space to itself is called an automor-phism. There are many automorphisms of a space, and not all concerna symmetry of the metric. The definition of an automorphism makesno reference to the metric. What characterizes automorphisms thatimplement symmetry transformations of the metric? The rotations ofthe sphere carry any differential dx i to dx il in such a way that

( ) ( )g x dx dx g x dx dxiji j

klk l=l l l . (8.3)

That is, any “translated” local invariant interval is unchanged. This isclearly a property of the metric and the automorphism. Eq. (8.3) canbe written as

( ) ( ) .g x g xxxxx dx dx 0ij kl i

k

j

li j

2222

- =ll l

l le o (8.4)

Since this must be true for any interval dx jl , we have

( ) ( ) ,g x g xxxxx

0ij kl i

k

j

l

2222

- =ll l

(8.5)

or, equivalently,

( ) ( ) .g x g xxxxx

0kl ij k

i

l

i

2222

- =l l l (8.6)

The metric at the transformed point is related to the metric at theoriginal point by the coordinate transformation induced by the auto-morphism—-the automorphism is a symmetry transformation of themetric. Such an automorphism is called a metric automorphism or anisometry. We have our answer to the question of the meaning of asymmetry of a metric without reference to a particular coordinatesystem—-the existence of a metric automorphism.

If one views the automorphism as a coordinate change at the thepoint x, Eq. (8.6) can be written

( ) ( ), .g x g x xallij ij=l (8.7)

144 Chapter 8. Metrics with Symmetry

That is, the transformed metric is the same function of its argumentsas the original metric is of its arguments. The metric is said to be forminvariant.

8.3 Killing Vectors

Given a metric ( )g xij , a solution, ( )x xil , of the partial differential equa-tions Eq. (8.6) exhibits an isometry of the metric. Except for specialcases, these equations are complicated. (If the metric is indepen-dent of a coordinate, a simple solution of these equations can beobtained.) For this reason, it is well to consider infinitesimal isome-tries for which the motion of the points is small. Also, it seems reason-able that many, though perhaps not all, finite isometries could beimplemented by a sequence of infinitesimal ones. Consider such aninfintesimal transformation,

( ), << .x x x 1i i ie e= + gl (8.8)

To first order in e, Eq. (8.6) gives

( )( ) ( ) ( ) .

x

g xx g x

xg x

x0m

ij mil j

l

kj i

k

2

2

2

2

2

2+ + =g

g g (8.9)

Expressed in terms of the covariant components mg , this becomes

i j

x x x

g

x

g

x

g0 j i

mmij l

jil k

i

kj

2

2

2

2

2

2

2

2

2

2= + + - -

g gg g g

i j,

x x2j i l ij

l

2

2

2

2= + -

g gg �

or

;j i .0;i j + =g g (8.10)

Any vector field satisfying this equation is called a Killing vector. Theproblem of finding all of the infinitesimal symmetries of a metricreduces to the problem of finding all Killing vectors. It is interestingto note that the Killing vector equation of the covariant components,expressed in local inertial coordinates at a point, reduces to

,i j .0,j i+ =g g (8.11)

The form is independent of the signature of the metric. Killingvectors for all metrics satisfy the same set of equations at a point in

8.3 Killing Vectors 145

the coordinates of a local inertial frame. Of course, not every solutionof these (local) equations gives rise to a (global) Killing vector.However, if the metric is “flat” and, thus, there exists a global inertialcoordinate system, then in such a coordinate system, Eq. (8.11)applies everywhere. Then any solution of these equations is a Killingvector. The covariant Killing vectors of such metrics are all the same,independent of the signature of the metric.

8.3.1 Conserved Momentum

Now that we can characterize the existence of a symmetry of a metricin a way independent of the coordinate system used, we can answerthe question, asked in Chapter 7, of how to identify the conserved“momentum” associated with a symmetry. First, note that if a metricis independent of a coordinate x ( )b , then by Eq. (8.9) we have( ) ( )=g dba

ba as a Killing vector. (By enclosing an index in parentheses we

indicate a particular index in contrast to a generic one.) Also, weknow that p mU( ) ( )/b b is constant along a geodesic—-it is conserved.Furthermore, since ( ) ( )=g db

aba , mU m U( ) ( )= gb b

aa. Consider now

p m U m U( )/ =g gga

a aa, (8.12)

where U a is the generalized velocity of a geodesic and ga is any Killingvector. p( )g is a scalar, the generalized momentum in the direction ofthe Killing vector g. We can write

( ).

ddp

md

d Umddx U U

ddx( )

; ;= = +x x

gx

g gx

ga

av

va

aaa v

v

(8.13)

By use of Eqs. (7.20) and (8.10) we obtain

.ddp

U U 0( )

;= =x

gg a

a vv (8.14)

That is, p( )g

is constant along a geodesic. It is conserved.

8.4 Maximally Symmetric Spaces

Any linear combination of Killing vectors with constant coefficients isa Killing vector, and thus Killing vectors, form a linear vector space ofsome dimension. A natural question to ask is, What is, the maximumpossible dimension of this linear vector space—-that is, the dimension


for a maximally symmetric space.2 Consider, for instance, the two-dimen-sional sphere, a maximally symmetric space. It is reasonably clear thatthere exist three linearly independent infinitesimal metric automor-phisms corresponding to small rotations about three different axes ofthe embedding space. Similarly, for the two-dimensional Euclideanspace there are three: namely, two infinitesimal translations and asingle infinitesimal rotation. But what is the general result?

We have suggested that a knowledge of infinitesimal isometriesdetermines “finite” isometries by successive application. A moresurprising result is that a knowledge of the Killing vector field and itsfirst covariant derivative at a point determines the Killing vector every-where. The “commutator” of two covariant derivatives, for a generalvector field, is (the covariant form of Eq. (6.43))

.A A R A; ; ; ;- =-o t v o v t otva

a (8.15)

From this equation, with Eq. (6.44), we find that, for any vector field,

( ) ( ) .A A A A A A 0; ; ; ; ; ; ; ; ; ; ; ;- + - + - =o t v o v t v o t v t o t v o t o v

For a Killing vector ig , with the use of Eq. (8.10), this equation gives

.2 2 2 0; ; ; ; ; ;i j k i k j k i j- - =g g g

With this result and Eq. (8.15), we have

.R; ;i j k kjip

p=-g g (8.16)

This equation, with a knowledge of the Killing vector ( )Xig and itsderivative ( )X,i jg at a fixed point X, determines all of the third deriva-tives and, with the derivatives of Eq. (8.16), all of the higher deriva-tives. Thus, a power series in x X- can be developed giving ( )xig for xwithin the radius of convergence of the series.

The fact, that isometries are completely determined by their localbehavior can be used to determine the maximum possible number ofindependent Killing vectors ( )

ing in N dimensions. There are a maxi-

mum of N-independent vectors ( )X( )ing at a point and ( )/N N 1 2- inde-

pendent ( )X( );i jng (recall Eq. (8.10)) for a total maximum of ( )/N N 1 2+

independent Killing vectors. A maximally symmetric space has

8.4 Maximally Symmetric Spaces 147

2 The subsequent discussion of maximally symmetric spaces follows somewhat that ofWeinberg 1972.

( )/N N 1 2+ independent Killing vectors. For N 2= there are atmost three independent Killing vectors in agreement with our previ-ous observation for the two-sphere and two-dimensional Euclideanspace.

Consider these three infinitesimal isometries for the two- sphere atthe point ( , ) ( , / )X 0 2= =i z r . (One might think that it would be morenatural to consider the polar point. However, the metric, in terms ofpolar coordinates, is singular at that point—-some Christoffel symbolsdiverge—-whereas the polar coordinates are locally Euclidean coor-dinates at X.) Since the � vanish at the point, the covariant derivativesreduce to ordinary derivatives, and, thus, the Killing vectors satisfy( ) ( ), ,i jn

j in

=-g g . To first order in ( ),x X- we can realize the three inde-pendent Killing vectors by

( )x( )i i1

1=g d

( )x( )i i2

2=g d

( ) ( / )x 2( )i i i3

1 2= - -g zd i r d .

The first two are vectors whose first derivatives vanish at X, whereasfor the third, the vector vanishes at the point X. The correspondingisometries are locally two translations and a rotation about the pointX. Considering the action of an infinitesimal rotation about the point

( , ) ( / , / )X 2 2( )1 = =i z r r on the point X, we see it is the first translation,whereas the infinitesimal rotation about the point ( , ) ( , )X 0 0( )2 = =i z

is the second. Similarly, we can consider for an N-dimensional spacethe ( )/N N 1 2+ infinitesimal isometries in the vicinity of a point Xwhere the coordinates are chosen to be locally “flat”—-the �’s vanishat X. To first order in ( )x X- , we can realize the independent Killingvectors by

( ) , ,x n N1( )in

in f= =g d (8.17)

( ) ( ) ( ) , , ... , > .x x X x X l m N l m1( )ilm m m

il l l

im

= - - - =g d d (8.18)

(As before, the use of ( ) indicates that the enclosed indices areparticular in contrast to the generic, and their value characterizesa particular Killing vector.) The isometries are N “translations,”Eq. (8.17), and ( )/N N 1 2- “rotations,” Eq. (8.18). Note again thatthe form of these Killing vectors, expressed in covariant compo-nents, does not depend on the signature of the metric. These are the


(local) Killing vectors for any maximally symmetric N-dimensionalspace.

The “translations” of Eq. (8.17) can be characterized in a coordi-nate-independent manner as

( )X( )ing �0

( ) .X 0( );i jn

=g (8.19)

The existence of N such Killing vectors implies that the space is homo-geneous at the point X. There exists an isometry that will carry X intoany nearby point.

Similarly, the “rotations” of Eq (8.18) can be chosen so that theyare characterized as

( )X( );i jlmg �0

( ) .X 0( )ilm

=g (8.20)

The existence of ( )/N N 1 2- such Killing vectors implies that thespace is isotropic about the point X.

It is not too difficult to show that if a space is isotropic about allpoints, it is homogeneous at all points. (Just consider effecting thetranslation of the point X in some arbitrary direction by a rotationcentered about a suitable nearby point.) Thus, a space that isisotropic about all points is at any point isotropic and homogeneousand admits the maximum number of Killing vectors—-it is maximallysymmetric. Similarly, a space that is maximally symmetric is isotropic andhomogeneous, since at each point the ( )/N N 1 2+ independentKilling vectors are the N translations and ( )/N N 1 2- rotations, thelatter implying isotropy around the point.

It is reasonable that the information about the curvature tensorof a maximally symmetric space can be gained by considering the( )/N N 1 2- “rotation” Killing vectors, which are those associated

with rotation about a point. First, let us note that Eq. (8.16), evaluatedat this point for such a rotation Killing vector, provides no usefulinformation. One can see this by considering Eq. (8.16) in a local iner-tial coordinate system. It merely requires that the second derivatives ofthe Killing vectors vanish. We must look at higher covariant deriva-tives.

Note, as is easily shown in a local inertial coordinate system, that ageneralization of Eq. (8.15) for rank-2 tensors


,A A R A R A; ; ; ;ij k l ij l k iklm

mj jklm

im- =- - (8.21)

becomes, with A ( );ij i jmn

= g ,

.R R( ) ( ) ( ) ( ); ; ; ; ; ; ; ;i j k lmn

i j l kmn

iklp

p jmn

jklp

i pmn

- =- -g g g g (8.22)

Also, the covariant derivative of Eq. (8.16) becomes, with ( ); ;i j i j

mn=g g ,

( ) ( ) ( ),R X R X R X( ) ( ) ( ) ( ); ; ; ; ; ;i j k lmn

kij lp

pmn

kijp

p lmn

kijp

p lmn

=- - =-g g g g (8.23)

where ( )X 0( )pmn

=g has been used. With use of Eq. (8.23) and the formof ( )X

( );p lmng in local inertial coordinates given by Eq. (8.18), Eq. (8.22)

becomes

.R R R R R R R Rkijm

ln

kijn

lm

lijm

kn

lijn

km

iklm

jn

ikln

jm

jkln

im

jklm

in

- + + - =- + - +d d d d d d d d

(8.24)

As derived, this equation is valid only at the point X and only in localinertial coordinates. However, since it is assumed that all ( )/N N 1 2-

rotations give rise to a Killing vector, this equation is valid for allvalues of m and n. It follows that Eq. (8.24) is valid, at the point X, inany coordinate system, since it is covariant in form if m and n can takeon all values. Furthermore, if, as assumed, the space is isotropic aboutall points, then Eq. (8.24) is valid at all points. Contracting m with k,using the definition of the Ricci tensor, Eq. (6.47), R Rjil

nlijn=- and

R 0mijm = , Eq. (6.45), one obtains

( ) .N R R R1 lijn

il jn

jl in

- =- +d d (8.25)

This can be rewritten as

jl( ) .N R R g R g1 nlij il nj ni- =- + (8.26)

But R Rnlij lnij=- , so that we have

jn .R g R g R g R gil nj jl ni in lj li- + = - (8.27)

Now contracting i with n , after “raising” one of the indexes, we find

.NR R glj lj= (8.28)

Put into Eq. (8.26), this gives

( )( ).R

N NR g g g g1nlij lj ni il nj=-

- (8.29)


Eq. (8.29) is a necessary condition that a metric be maximallysymmetric. It is not sufficient. Consider, for example, a torus definedby a rectangle in a two-dimensional Euclidean space with oppositesides identified and with the metric dl dx dy2 2 2= + .(See Fig 8.1.)Such a space clearly satisfies Eq. (8.28) —-it is after all a flat space—-but it is not maximally symmetric. It is homogeneous but not iso-tropic about all points.

It is interesting and useful to determine what information aboutthe metric can be gained by assuming the metric is homogeneousabout each point, and, thus, satisfies Eq. (8.19), but not necessarilyisotropic. Of course, any result obtained would be valid for isotropicmetrics since isotropy about any point implies homogeneity aboutany point.

Eq. (8.22), with ( )A X 0( );ij i jn

= =g , becomes

,0( ) ( ); ; ; ; ; ;i j k ln

i j l kn

- =g g (8.30)

and the covariant derivative of Eq. (8.16) is

( )R X( ) ( ); ; ; ;i j k ln

kij lp

pn

=-g g . (8.31)

Here we have used Eq. (8.19). Together, Eqs. (8.30) and (8.31) imply

( ) ( ) .R X R X 0( ) ( )

; ;kij lp

pn

lij kp

pn

- + =g g (8.32)

Using Eq. (8.17), valid in a local inertial frame for the N-indepen-dent translations, we have


Figure 8.1. Torus.


( ) ( ) .R X R X 0( ) ( ); ;kij ln

lij kn- + = (8.33)

Similar to the above discussion of the implication of assumedisotropy of the metric, this equation is derived to be true only at thepoint X and only in local inertial coordinates. However, since it isassumed that every one of the N translations gives rise to a Killingvector, this equation is valid for all values of n. It follows that Eq.(8.33) is valid, at point X, in any coordinate system since it is forminvariant and true for any value of n. Since R 0pij

p / (Eq. (6.45)) andthus R 0;pij l

p / , then Eq. (8.33) gives ( )R X 0;lij pp = . Thus, if the metric is

homogeneous everywhere, we see that

( ) .R x 0;lij pp = (8.34)

This implies

( ) ( ( )) ( ) ,g R x g R x R x 0; ; ;li

lij p

p li

lij

pp j p

p= = =

which by use of Eq. (6.50) gives

.R R 0; ,i i= = (8.35)

The Ricci scalar is constant—-not a suprising result. This of coursemeans that the Ricci scalar in Eq. (8.29) for the Riemann tensor of amaximally symmetric space is a constant.

So, maximally symmetric spaces satisfy Eq. (8.29) with R a constant.Further, all such metrics with the same value of R (and the samesignature) are equivalent. That is, if there are two maximally symmet-ric metrics ( )g xkl and ( )g xijl l that satisfy Eq.(8.29) with the same valueof R, there exist functions ( )x x xr r=l l such that3

ij( ) ( ) .x

g x g x xxx

kl k

i

l

j

22

22

= l ll l (8.36)

This is a very useful result. The fact that all maximally symmetricmetrics, which satisfy Eq. (8.29) and have the same value for the(constant) Ricci scalar, are equivalent enables us to study thesemetrics by exhibiting any metric that satisfies Eq. (8.29) and studyingthe properties of the exhibited metric.


3 See Weinberg (1972) for a proof. In his proof, Weinberg exhibits a power seriesexpansion of the coordinate transformation that satisfies Eq. (8.36).

8.5 Maximally Symmetric Two-DimensionalRiemannian Spaces

We will see that a maximally symmetric three-dimensionalRiemannian space, as a three-dimensional subspace of four-dimen-sional space-time, is important in the study of cosmology. To that endwe study such a metric. To simplify the discussion, we first considermaximally symmetric two-dimensional Riemannian spaces. What arethe possible maximally symmetric two-dimensional space metrics?

Two possibilities are clear. One metric would be that of the surfaceof a sphere of radius K, embedded in a three-dimensional Euclideanspace, that is, a surface characterized by

.x y z K2 2 2 2+ + = (8.37)

That this two-dimensional space is maximally symmetric is clear. Itadmits three independent isometries corresponding to three rota-tions of the sphere. Furthermore, the metric (induced by the three-dimensional Euclidean metric) is locally Euclidean (spacelike). Thecircumference C of a “circle” is less then r2r , where r is its “radius,”the geodesic distance of the “circle” from its “center.” The space haspositive curvature. (See Eq. (6.36).) It is instructive to note that thethree-dimensional Euclidean embedding space is flat and, thus, ismaximally symmetric. It has six independent Killing vectors.However, the constraint surface, defined by Eq. (8.37), is not invari-ant under the three translations of the flat space. Thus, we mightexpect that the induced metric on the sphere might have 6 3 3- =

isometries, which it does.A second possibility is a flat two-dimensional surface. The three

independent isometries are the two translations and one rotation.This is a space of zero curvature for which C r2 0= r .

There exists a third type of two-dimensional maximally symmetricspace. The space can be visualized as a surface imbedded in a spaceendowed with a Minkowski metric of one “time” tr and two spacecoordinates ,x y with the surface characterized by

( ) .t x y K2 2 2 2- + =r (8.38)

The surface is depicted in Figure 8.2. It is a spacelike surface, whichis locally Euclidean. Any two nearby points are spacelike separatedwith the “distance” defined by that induced by the Minkowski metric.Note that the embedding three-dimensional Minkowski space is flatand thus maximally symmetric, possessing six independent Killing

8.5 Maximally Symmetric Two-Dimensional Riemannian Spaces 153

vectors. However, the constraint surface of Eq. (8.38) is not invariantunder the three translations of the flat space-time. Thus, we mightexpect that the induced metric on the sphere might have 6 3 3- =

isometries, and it does. The three independent isometries correspondto the two “Lorentz” boosts in the x and y directions and the rotationin the xy plane. This metric is a maximally symmetric metric of nega-tive curvature in the sense that the circumference C of a “circle” isgreater then r2r . It is instructive to show that this is the case for thissurface. Consider a circle defined by the locus of points x y r2 2

02+ = on

the surface. The circumference, using the metric induced on thesurface, is simply r2 0r . The points of this circle are all at minimumdistance (i.e., the geodesic distance) r from x y 0= = given by

( ) < .r x y t r/2 2 2 1 20= + -� � �! r (8.39)

C r2 0= r is greater then r2r .

It is useful to introduce two independent coordinates, one “radial”-like and the other “angle”-like, to characterize points in these spacesand to express the metric in terms of these coordinates. The use ofsuch coordinates will result in similar forms for the three metrics.

8.5.1 Two-Dimensional Space Metric of Positive Curvature

The metric of the imbedding space is

.ds dx dy dz2 2 2 2= + + (8.40)


Figure 8.2. Negative curvature surface.


With the differentials not independent but constrained by Eq. (8.37),this is also the metric induced on the sphere. The differential form ofthe constraint Eq. (8.37) is

( ) .zdz d x y2 02 2+ + = (8.41)

In terms of a “radial” coordinate r x y2 2 2= +l , this constraint equationcan be written as

.dzK rr dr22 2

2 2

=- ll l (8.42)

Thus, the invariant distance of points on the sphere, in terms of thepolar variables rl and ( / )tan x y1=i - , is given by

.dsK rK dr r d22 2

2 22 2=

-+ ill l (8.43)

By defining r Kr=l , which merely rescales the distance measurementof rl, we obtain

.ds Krdr r d1

2 22

22 2=

-+ id n (8.44)

Here r, which is dimensionless, ranges from 0 to 1 and to each valueof r and i there correspond two points on the sphere, one in the“north” hemisphere and one in the “south” hemisphere.

8.5.2 Two-Dimensional Space Metric with Zero Curvature (Flat)

This is easy.

( ) ( ).ds K dx dy K dr r d2 2 2 2 2 2 2= + = + i (8.45)

8.5.3 Two-Dimensional Space Metric with Negative Curvature

We proceed in a manner similar to that for the positive curvaturecase. The metric of the embedding space is

.ds dx dy dt2 2 2 2= + - r (8.46)

Again, for the metric on the surface the differentials are not inde-pendent but are constrained by Eq. (8.38). Thus, we have

8.5 Maximally Symmetric Two-Dimensional Riemannian Spaces 155

( ) .tdt d x y2 02 2- + + =r r (8.47)

In terms of a “radial” coordinate r x y2 2 2= +l , this constraint equationreads

dtK rr dr22 2

2 2

=+ ll lr . (8.48)

Thus, the invariant distance of points on the surface in terms of thepolar variables rl and i is given by

.dsK rK dr r d22 2

2 22 2=

++ ill l (8.49)

Again defining r Kr=l we obtain

.dt Krdr r d1

2 22

22 2=

++ ir d n (8.50)

Here r, which is dimensionless, ranges from 0 to 3 and to each valueof r and i there is but one point on the surface.

Combining the results of these three spaces we can write

,dt Kkrdr r d1

2 22

22 2=

-+ ir d n (8.51)

where k 1!= or 0. One may very well ask if these are the most generalmaximally symmetric metrics. By computing the curvature tensor andnoting that Eq. (8.29) is satisfied with R determined in magnitude byK and sign by k, using the uniqueness theorem, we can answer yes. Wewill return to this after we generalize the results to three space dimen-sions.

8.6 Maximally Symmetric Three-DimensionalRiemannian Spaces

We could obtain the maximally symmetric three-dimensionalRiemannian metric by enlarging the embedding space metrics of Eqs.(8.40) and (8.45) by one space dimension. However, to generalizethe results from two to three dimensions, let us merely consider themetric defined by

,sindl Kkrdr r d r d1

2 22

22 2 2 2 2=

-+ +i i zd n (8.52)


where K and k are constants. Here r is a radial-like variable and i andz are angle variables. The space is rotationally symmetric (isotropic)about r 0= and, for, k 0= , is the three-dimensional flat Euclideanspace. It is not clear that the space is isotropic about any point fork � 0. However, computing (say, by use of Mathematica), one obtainsfor the nonvanishing components of the curvature tensor

,

sin

R

R

R

R

R

R

R

R

R

R

R

R

kr

kr

krk

1

rr

rr

r r

rr

rr

rr r r rr

2

2 2

2

=

=

=

-

-

-

=

=

=

=

=

=

-

-

-

=

=

=

-

-

--

i

i i

z z

ii

ii

zz

ii

iziz

zizi

zz

iizz

zzii

zz (8.53)

and for the Ricci scalar

.RKk62=- (8.54)

From these results we see that the metrics defined by Eq. (8.52) satisfyEq. (8.29) if k 1!= or 0, a necessary condition when they are maxi-mally symmetric. (As noted, this is not a sufficient condition. Theembedding procedure would prove that they are indeed maximallysymmetric.) Again by the uniqueness theorem, are the most generalmaximally symmetric metrics. The coordinates used in Eq. (8.52)treat r 0= as special, but we now know all points are equivalent.

The nonvanishing Christoffel symbols are

( )

( )

.

sin

cos sin

cot

krkr

r kr

r kr

r

1

1

11

rrr

r

r

r r r r

2

2

2 2

=-

- -

- -

-

i

i i

i

�

ii

zz

ii

ii

zz

zz

zzi

izz

ziz

= = =

=

�

�

� � � �

�

� �

=

=

=

=

= (8.55)

The singular behavior of r r r r= = =� � � �ii

ii

zz

zz at r 0= and of rr

r� at r 1=for k 1=+ reflects the particular coordinate system and is not intrinsicto the metric. Clearly, for any value of k the metric for r small is thatof flat Euclidean space expressed in spherical coordinates.

Let us now investigate the properties of these spaces in the large.Since all points are equivalent, to discuss the properties of the mostgeneral geodesic we need consider only those passing through r 0= .The geodesic equation,

8.6 Maximally Symmetric Three-Dimensional Riemannian Spaces 157

,dld x

dldx

dldx

0i

jki

j k

2

2

+ =� (8.56)

for geodesics passing through r 0= , reduces to i and z beingconstant and to

.dld r

dldrdldr

0rrr

2

2

+ =� (8.57)

But since

,dldrdldr kr

Kgdldrdldr kr

K1 1

rrr

rr2 2= =� (8.58)

Eq. (8.57) has as solutions for k 1=+ , sinr lK= and for k 1=- ,.

sinhr Kl

= . For k 0= the solution is of course r Kl

= . In these solutionsl measures the total invariant distance from r 0= along the geodesic.A circular path is characterized by r r0= and by some fixed value forz. The invariant length circumference of these paths are

.C dl Kr d Kr20 00

2

= = =i rr

## (8.59)

These circles have invariant length radii l0, that is, an invariantdistance from r 0= , given by

,

,

, .

sin

sinh

l

K r

Kr

K r

k

k

k

1

0

1

0

10

0

10

=

=+

=

=-

-

-

Z

[

\

]]

]] (8.60)

Therefore, the circumference C is

,

,

, .

sin

sinh

C

KKl

l

KKl

k

k

k

2

2

2

1

0

1

0

0

0

=

=+

=

=-

r

r

r

Z

[

\

]]]

]]] (8.61)

What are the properties of these three generically different spaces? Inwhat respects do they differ?


8.6.1 k = 0

As noted, the metric is that of three-dimensional flat Euclidian space.The space is infinite in extent. The geodesics are “straight” lines andthe usual relationship between circumference and (invariant) radiusobtains, C l2 0= r .

8.6.2. k = +1

The equation for the geodesic, sinr lK= , tells us that, as we let the

geodesic distance l run to /K 2r , the coordinate r reaches its maxi-mum value of unity. This reflects the coordinate singularity in themetric at r 1= . However, we can continue the invariant distance to Krand the coordinate r returns to the value zero. One might expect thatthe r coordinate is one which, for fixed values of i and z, representsmore than one point in space, like a radial coordinate for a polarprojection of a sphere. Indeed, the metric for k 1=+ is the inducedmetric of a three-dimensional sphere imbedded in a four-dimen-sional Euclidean flat space. As l goes from 0 to /K 2r and on to Kr, thepoint on this sphere moves from the “north pole” to the “equator” onto the “south pole.” The geodesic continues with l going from Kr to/K3 2r to K2r, with the point going back to the “equator” and return-

ing to the “north pole.” So the invariant distance around the spacealong a geodesic is

.L K2= r (8.62)

One can say that the space has a radius K. The space is finite butwithout boundary. From Eq. (8.61) we see that the relationshipbetween the circumference of a circle and the invariant radius is notthat of flat space. Rather, the circumference increases more slowlythan the invariant radius, just as latitude circles on a two-dimensionalsphere. It is a space of positive curvature, as expected.

8.6.3 k = 1

The equation of the geodesic sinhrKl

= tells us that the geodesicdistance can increase without bound as does the coordinate r. Thespace is infinite. And again the relationship between the circumfer-

8.6 Maximally Symmetric Three-Dimensional Riemannian Spaces 159

ence of a circle and the invariant radius is not that of flat space. But,in contrast to the previous case, the circumference increases morerapidly than the invariant radius. This is a space of negative curvature.

8.7 Maximally Symmetric Four-DimensionalLorentzian Spaces

Though not having any direct application to the study of cosmology,it is of historical and formal interest to study maximally symmetricspace-time metrics, that is, Lorentzian spaces that have ten indepen-dent Killing vectors. One such space is of course Minkowskispace, which can be characterized as the maximally symmetric four-dimensional pseudo-Riemannian space with signature ( , , , )+ - - -

and with zero Ricci scalar. One might expect that one could obtainanother by considering a surface embedded in a five-dimensionalpseudo-Riemannian space with signature ( , , , , )+ + - - - —-thus, asurface

( ) ( ) (( ) ( ) ( ) ) ( ) ( )x x x x x r r a1 2 2 2 3 2 4 2 5 212

22 2+ - + + = - = (8.63)

in a space with a metric

( ) ( ) (( ) ( ) ( ) )d dx dx dx dx dx2 1 2 2 2 3 2 4 2 5 2= + - + +x

( ) .sindr r d dr r d r d12

12

12

22

22

22

22 2

22= + - - -i i i z (8.64)

The induced metric on the surface (after rescaling the r2 variable) is

( ) ( )sind a r drdr r d r d1

112 2 2

12

22 2

22 2 2

22= + -

+- -x i i i z . (8.65)

It is clear that the metric, called the anti–de Sitter metric, isLorentzian. The one timelike variable is 1i . A computation of theRicci scalar yields

,Ra122=- (8.66)

a negative R. A maximally symmetric four-dimensional Lorentzianspace with a positive R would complete the set of such metrics. (SeeExercise 2, Chapter 9.)


8.8 Exercises

1. The ten independent Poincaré transformations induce an isome-try of the Minkowski metric, and thus there are ten independentKilling vectors (the Minkowski metric is maximally symmetric) ofthe form given by Eqs. (8.17) and (8.18). (Let X 0= .) (a) What arethe contravariant components of the Killing vectors associatedwith the four infinitesimal coordinate translations? What are theconserved momentum components associated with these fourKilling vectors? (b) What is the Killing vector associated with aninfinitesimal Lorentz boost along the x1 axis? (c) What is theKilling vector associated with an infinitesimal rotation about the x 3

axis? What is the associated conserved (angular) momentumcomponent? (d) By considering the infinitesimal form of theLorentz transformation ( i.e., 1%b ), derive the contravariantcomponents of the resulting Killing vector. Do they agree with thatobtained in (b)?

2. The Schwarzschild metric is time independent and sphericallysymmetric. Because in Schwarzschild coordinates the metric is x 0

and z independent, p0 and pz are conserved. However, sphericalsymmetry implies there are two more independent conserved(angular) momentum components. (a) What is the Killing vectorassociated with the infinitesimal rotation about the x axis?(Consider the polar axis to be the z axis.) Check that this vectorsatisfies the Killing equation. (b) What is the conserved (angular)momentum component associated with this Killing vector?

3. (a) Derive Eq. (8.65) (b) Show that this metric is maximallysymmetric. (c) Obtain Eq. (8.66).

4. Show that the anti–deSitter metric satisfies Einstein’s vacuum fieldequations with a cosmological constant. What is the cosmologicalconstant in terms of the constant Ricci scalar?

8.8 Exercises 161

Chapter 9

Cosmology

9.1 Introduction

We have seen that Einstein’s replacement of the Newtonian gravita-tional force by a change in the geometry of space-time results inchanges in physical processes that take place in the vicinity of massivebodies. We now turn our attention to the effect of Einstein’s theoryon our view of the universe in the large. This study of the large-scalestructure and behavior of the universe constitutes the science ofcosmology. Prior to Einstein, most astronomers thought the universeto be infinite in extent, Euclidean, and with a uniform and staticdistribution of stars. Of course, one can see with the naked eye thatthe distribution of stars is not uniform. What was believed was that,on some large-scale average, the distribution was uniform. Even froma Newtonian standpoint there are difficulties with this view of theuniverse. First, one expects the gravitational force to be the onlysignificant force acting between stars, and it is rather easily arguedthat there cannot exist a static distribution of mass points withonly the force of gravity acting between them. Second, there isOlbers’ paradox, named after the German astronomer HeinrichOlbers (1758–1840). The paradox is between the existence of arather dark night sky and a static, uniform, and infinite universe.1

The paradox can be appreciated most simply by realizing that insuch a universe if one looked in any direction the line of sightwould eventually intercept a star. How bright would such a sky

1 For an extensive discussion of the history and resolution of Olbers’ paradox, seeHarrison (1987).

163

be? Consider the light that arrives from a portion of a star’s surfacethat is subtended by a viewing cone of some very small apex angle.This area is proportional to r 2 where r is the distance to the star.The intensity of light received is proportional to /r1 2. Thus, theamount of light energy received per unit time within the viewingcone is independent of the distance to the star: the brightnessdoes not depend on the distance to the star. The whole sky wouldbe as bright as a typical star’s surface. It is rather obvious that thenight sky is not that bright—nor the day sky, for that matter. Notealso that such a view of the universe implies a preferred referenceframe; namely, one in which the stars are seen, in the average, tobe at rest.

Most modern cosmological theories assume the cosmological princi-ple according to which all points and directions in the universeare essentially the same. More particularly, the assumption is madethat observations made in the universe are spatially homoge-neous and isotropic. Of course, the cosmological principle is notexact. There are clumps of matter, stars and galaxies, in the universe.Einstein has taught us that energy-momentum determines geom-etry. Thus, one implication of the cosmological principle is thatthe geometry of space, in the large, is homogeneous and isotropic.But to what observers is space homogeneous? And at what “times,”for the various observers distributed throughout space, does theuniverse appear the same? We address these questions in the nextsection.

The cosmological principle appeals to cosmologists for philosoph-ical and mathematical reasons. For most of the history of moderncosmology, it has been taken as an assumption. Indeed, it is difficultto imagine what form cosmological studies would take if some formof a cosmological principle were not operative. After all, there existsonly one universe to study and, if our observation point were notessentially like other points in the universe, what could we learnabout the universe in the large by studying it from our viewpoint?There are, however, philosophical and physical questions raised bythe special nature of universes that satisfy the cosmological principle,and these have given rise to attempts to derive the homogeneity andisotropy of space.

It may be that the universe does not satisfy the cosmological prin-ciple, or at least not exactly—or perhaps not for all time. But it is aunifying idea that makes a good starting point for organizing obser-vational data. The discovery and understanding of violations of theprinciple would be most interesting.

164 Chapter 9. Cosmology

9.2 The Robertson-Walker Metric

In order to use the cosmological principle we need to formulatemathematically what it says about the space-time metric. Having stud-ied the implications of symmetries on the metric in the last chapter,we are in a good position to do this. To define a time coordinate t, itis useful to use a local scalar property of the universe that is changing,such as the proper energy 0t density of the event point. (This, ofcourse, assumes that t is changing.) Here 0t is the energy density asseen by a local inertial observer for whom the local material of theuniverse is, on the average, not moving, that is, for whom the momen-tum density T io is zero. We can identify such a time coordinate at allpositions in the universe because, by the cosmological principle, allpoints are equivalent and experience the same history. We willfurther assume that such points follow space-time geodesics and useas the time coordinate the proper time of such an observer. This wewill call “cosmic” time. (The cosmological principle would then implythat the three-dimensional surfaces characterized by fixed cosmictime are maximally symmetric subspaces.)

In terms of the cosmic time t, we write the cosmological invariantinterval as

.d dt g dtdx g dx dx2 ii

iji j2 2

0= - -x (9.1)

The coefficient of dt 2 in this metric is unity because t is chosen to bethe proper time of a world line characterized by dx 0i= (i.e., U 0I = ).The assumption that such world lines are geodesics puts furtherrestrictions on the metric. The geodesic equation can always be writ-ten as (see Eq. (7.23))

.ddU

g U U21

,=xv

gt vg t

For a geodesic with U 0i= and with a metric for which g 100= , thisequation becomes

( ).

dtd g U

dtdg

0oi oi0

= = (9.2)

The gi0 are (cosmic) time independent. Further, the assumed homo-geneity and isotropy of space imply the existence of six independent(spacelike) isometries. Under the action of these isometries, cosmictime does not change, which implies, for the associated Killingvectors, 00=g . We can argue that there exists a change of coordinates

9.2 The Robertson-Walker Metric 165

( , )

,

x

t

X x x

x x t

i i i 0

0 0

=

= = =l

l l

l (9.3)

such that g 0i0 =l . Under this change of coordinates, we find

.Xg

xX g

xX

xX g

xg

xX gi i

k

k i

k j

kj i

k

k

j

kj0 0 0 0 022

22

22

22

22

= + = +ll l l l l

e o (9.4)

Thus, an X i that satisfies

( , ) ( , )g X xxX g X x 0k

ij

kji

00

00

22+ =ll

l (9.5)

induces a change of coordinates that results in g 0i0 =l . Eq. (9.5) canbe viewed as a coupled set of three ordinary differential equations(for fixed x il ). This set has a solution with arbitrary initial conditionsfor X i at some x x0

00=l , which can be chosen to be x il ; that is,

( , )X x x xi i i00 =l l l . With these initial conditions, we have x xi i= l on the

“surface” characterized by x x x0 000= =l . Of course, away from this

surface, that is, at x 0� x00, generally ( , )X x xi 0l � x il .

Thus, the invariant interval of Eq. (9.1) in the new coordinates, forwhich we drop the primes, becomes

( ) ( , ) .d dx g x x dx dxiji i j2 0 2 0= -x (9.6)

The six spacelike isometries still have 00=g , and they satisfy (see Eq.(8.9))

( , )( , ) ( , ) ( , ) .

x

g x xx x g x x

xg x x

x0m

ij mil j

l

kj i

k00 0 0

2

2

2

2

2

2+ + =g

g g (9.7)

Only the space-space part of the metric is involved. The metric( , )g x xij

0 is maximally symmetric for each value of x 0. Eq. (8.52) in theprevious chapter implies that space-space coordinates exist for whichthe space-space invariant interval has the form

( ) .sindl K xkrdr r d r d1

2 2 02

22 2 2 2 2=

-+ +i i zd n (9.8)

The constant k cannot change with time since it has discrete valuesand thus cannot change in a continuous manner. Finally, we have forthe general metric satisfying the cosmological principle,


( ) ,sind dt K tkrdr r d r d1

2 2 22

22 2 2 2 2= -

-+ +x i i zd n (9.9)

with ,k 0 1!= . This metric is known as the Robertson-Walker metric. The(large-scale average) metric history of a universe satisfying the cosmo-logical principle is contained in the function ( )K t and the constantk. Knowledge of ( )K t and the constant k must be gained by observa-tion and theoretical interpretation. Note that no dynamics, such asEinstein’s general relativity equations, have been used in determiningthe form of Eq. (9.9). Such dynamics can tell us about the possibletime dependence of ( )K t , its relation to k, and the relation to otherphysical entities such as the mass density of the universe. Beforemoving on to this, we will first study the kinematics of the Robertson-Walker metric.

9.3 Kinematics of the Robertson-Walker Metric

9.3.1 Proper Distance

Recall that a “typical” galaxy (or material) moves along a geodesiccharacterized by fixed r, i, and z with the cosmic time t being theproper time of the galaxy (or material). These coordinates arereferred to as co-moving coordinates. We assume r 0= is our galaxy’sposition. This is just a choice of coordinate system, not an observationthat our galaxy is special—recall the cosmological principle. Wewould like to know how far apart, at a given cosmic time t, are twogalaxies positioned at r 0= and r r1= . But first we will answer the ques-tion of how far apart, dl, two galaxies are at r and r dr+ (same anglecoordinates). We define this distance as the cosmic time it takes lightto travel from r to r dr+ . Since we have for the world line of a lightparticle d 02=x , we find

( )( )

dl K tkrdr

1 /2 1 2=-

, (9.10)

which changes with time as ( )K t changes. The proper distance dbetween the galaxies at r 0= and r r1= at a fixed cosmic time t is thendefined by

( , ) ( )( )

d t r K tkrdr

1 /

r

1 2 1 20

=-

1# . (9.11)

9.3 Kinematics of the Robertson-Walker Metric 167

As measured by proper distance, typical galaxies move apart ortogether as ( )K t increases or decreases. The quantity d is not the timeit takes light to travel from a typical galaxy at r1 to our galaxy. After all,d is a distance measurement at fixed cosmic time. But a photon whichreaches our galaxy travels “radially,” and, thus, satisfies

( )d dt

kr

K t dr

102 2

2

2 2

= --

=x .

From this we see that the cosmic time t1 at which light was emitted atr1 to arrive at our galaxy at the present time t0 is determined by therelation

( ) ( )K tdt

krdr

1 /

r

t

t

2 1 20

=-

0 1

1

## . (9.12)

Another useful “proper” distance is that which is perpendicular toa radial distance. Imagine that we observe an object, say a galaxy, thatsuspends an angle di (in some direction). What is the proper lengthof that object? The light signals we observe from the two ends of theobject, separated by a small angle di, are emitted at the same cosmictime t1 and at the same r value, say r1. From the operational definitionof proper distance given above, the proper (perpendicular) distancel9 between these positions at the cosmic time the light was emitted isgiven by

( ) .l K t r1 1= di9 (9.13)

Here ( )K t1 and r1 are related by Eq. (9.12).These considerations, particularly Eq. (9.12), can be used to illus-

trate the concept of horizons, that is, the limits of our view of theuniverse. In the next two sections we will consider two types of hori-zons, called particle horizons and event horizons by Rindler (1956).

9.3.2 Particle Horizons

Our particle horizon ( )r tPH 0is defined to be the largest value of r from

which we could have received a light signal up to the present. Moreaccurately, it is the position ( )r tPH 0

whose light signal, emitted at theearliest possible time, is just now reaching us. Of course, its valuedepends on the time dependence of ( )K t . Eq. (9.12) gives

( ) ( )K tdt

krdr

1 /

( )

,

r tt

2 1 200

=-3-

00 PH## . (9.14)


The lower limit of the time integral is 0 if ( )K t goes to zero in the pastat the cosmic time chosen to be zero (the “big bang” event), whereas,if such an event did not occur, the limit is 3- . A particle horizon willexist if

( )< .

K tdt

,

t

03

3-

0# (9.15)

For the big bang case, this inequality is satisfied if ( )K t vanishes moreslowly than t1 e- .

As an illustration of the particle horizon, Figure 9.1 shows theworld lines of light particles that were emitted toward us at time zeroat various r positions. (Remember r is a co-moving coordinate.) t0 ischosen to be equal to one. The lines are calculated for the k 0=universe and with the dynamic assumption that the universe can bedescribed as a dust field universe with zero pressure all the way backto time zero. (See Sec. 9.4.) The three world lines in the figure reacht 0= at , /r 1 3 4= , and /1 2. For this case ( )K t vanishes as t /2 3. We see


Figure 9.1. World lines of light emitted at the big bang.


that any light emitted at > ( )r r tPH 0at time t 0= has not yet reached us,

whereas light emitted from < ( )r r tPH 0has gone past us.

The integral on the right side of Eq. (9.14) can be evaluated toyield

( )

( ) ,

( ) ,

( ) , .

sin

sinhK tdt

r t

r t

r t

k

k

k

1

0

1,

t

1

10

=

=+

=

=-3

-

--

0PH

PH

PH

0

0

0

#

Z

[

\

]]

]]

(9.16)

For ,k 1= + if / ( ) >dt K t 2,

t

0r

3-

0# , the horizon photon will have passed

us at least once before. We will return to the discussion of particle

horizons after we study the cosmic dynamics of general relativity.

9.3.3 Event Horizons

Just as there may be particle horizons past which we have observed noevents up to the present time, there may be events that we can neversee. From Eq. (9.12) we see that the light signal of an event, occur-ring at r1 at cosmic time t1, reaches us at cosmic time t determined by

( ) ( )K tdt

krdr

1 /

r

t

t

2 1 201

=-

1## . (9.17)

Thus, there exists an event horizon ( )r tEH 1 for an event occurring atcosmic time t1 if

( )< .

K tdt

t

tmax

1

3# (9.18)

The event horizon ( )r tEH 1 is determined by

( ) ( )K tdt

krdr

1 /

r

t

t

2 1 20

EHmax

1

=-

## , (9.19)

where tmax is finite if at some time in the future ( )K t 0" , or infinity ifnot. As before, this can be written as

( )

( ),

( ),

( ), .

sin

sinhK tdt

r t

r t

r t

k

k

k

1

0

1

EH

EH

EH

t

t

11

1

11

max

1

=

=+

=

=-

-

-

#

Z

[

\

]]

]](9.20)

One should appreciate that such event horizons have no observa-tional consequence but might be of philosophical interest.


9.3.4 Cosmological Redshift: Hubble’s Constant

The time dependence of K has observational consequences, and thusobservational astronomy contributes to our knowledge of K. The bestand most direct information is obtained from the shift in thefrequency of light coming from distant galaxies. We can calculate thisfrequency by considering two successive crests of a light wave thattravels from a galaxy at r1 and is received by us (at r 0= ). Each crestsatisfies the equation

( )( )

.dt K tkrdr1

02 22

2

--

= (9.21)

If a crest is emitted at t1 from r1, it is received at the origin at t0,given by

( )( ) .

K tdt dr kr1 /

rt

t 02 1 2=- - -0

1 1

## (9.22)

If the next crest is emitted at time t t1 1+ d (from r1), it is received at theorigin at a time t t0 0+ d given by

( )( ) .

K tdt dr kr1 /

rt t

t t 02 1 2= - -

+

+

d

d

1 1

0 0

1

## (9.23)

Thus, we have

( ) ( ) ( ) ( ).

K tdt

K tdt

K tdt

K tdt

0t

t t

t

t t

t

t

t t

t t

- = - =++

+

+ dd

d

d

01 1

10 0

1

0

1

0 0 1#### (9.24)

Since ( )K t does not change significantly during the period of a typi-cal light wave, we find that

( ) ( ).

K tt

K tt

0

0

1

1=d d (9.25)

Of course >t t0 1. Since cosmic time is the proper time of typical galax-ies, t0d is the period of the light received, and t1d is the period emit-ted. The frequencies are related by

( )( ),

K tK t

1

0

0

1=oo (9.26)

where 0o is the frequency received and 1o is the frequency emitted.For the corresponding wavelengths, the relation is

( )( ).

K tK t

1

0

1

0=mm (9.27)


If ( )> ( )K t K t1 0 then >0 1o o and a “blue” shift results, whereas if( )< ( )K t K t1 0 then <0 1o o and a red shift obtains. To be able to

measure this shift, we must know the frequency emitted, which is notpossible for a single frequency. However, by identifying sets offrequencies as those of a particular emission spectrum, thus deter-mining sets of 1o ’s, the ratio ( )/ ( )K t K t0 1 can be determined.Observations from distant galaxies give a redshift indicating( )< ( )K t K t1 0 . The shift is usually expressed in terms of the redshift

parameter z defined as the fractional increase in the wavelength

,z1

0 1=-

mm m (9.28)

which, by Eq. (9.26), is

( )( )

.zK tK t

11

0= - (9.29)

This redshift is sometimes mistakenly referred to as a Doppler shift.It does have a Doppler component. For emission from nearby galax-ies, for which r1 and t t i0 - are small, we have

( )( )( )

( ) ( , ).zK t

K t t tr K t d t r

0

0 0 11 0 0 1. . .

-oo o (9.30)

The last equality follows from the definition of proper distance Eq.(9.11). This z is the result for the Doppler shift for low relative veloc-ity ( , ).d t r0 1o However, for light from far galaxies, for what cosmic time

would one calculate the “relative” velocity? Rather, the redshift is dueto the expansion (or contraction) of space itself during the transittime of the light.

Since, for nearby galaxies, the proper distance ( , )d t r0 1 is the timeof passage ( )t t0 1- , we can use Eq. (9.30) to obtain a proper distanced redshift z relationship:

( )( )( )

( )( )

.zK t

K t t tK tK t

d H d0

0 0 1

0

00.

-= =

o o(9.31)

The constant of proportionality H 0,

( )( ),H

K tK t

00

0=o

(9.32)

is called Hubble’s constant after the astronomer Edwin Hubble. It washe who first announced, in 1929, that observations “establish aroughly linear relation between velocities and distances among nebu-lae” (Hubble 1929). The most recent measurement of this constant ispresented in an article appropriately titled “Final Results from the


Hubble Space Telescope Key Project to Measure the HubbleConstant” (Freedman et al. 2001). The result is a value of about 70km s 1- Mpc 1- or a Hubble time H 0

1- of .1 3 1025# m or .13 8 109# years.2

One should appreciate that the Hubble time, H 01- , is a natural

timescale for the present universe. Similarly, ( )K t0 is a natural sizescale. We will at times find it useful to define a dimensionless timevariable t H t0=r and a dimensionless size variable ( ) ( )/ ( )K t K t K t0=r r

and write equations in terms of these variables.The measurement of the redshift parameter of close galaxies gives

us some information about K t^ h, namely, ( )/ ( )K t K t0 0o . But, if the

cosmologist is to obtain further knowledge of the time dependenceof ( )K t , measurements of redshifts of more distance galaxies have tobe related to the time dependence. Thus, it is necessary to definesome measurable “distance” that depends on ( )K t . The most usefuldistance is the luminosity distance.

9.3.5 Luminosity Distance

Suppose that astronomers have identified a standard type of galaxy(or star) and have data that indicates the total power P that it radiates.Then an observation of the luminosity (the power received per unitarea) of such a galaxy gives information about the distance of thegalaxy. If the space is Minkowskian and if the distant galaxy is at restwith respect to the observer, the luminosity L is given by

.LrP4 2=r

(9.33)

This is just the power divided by the area of a sphere at a distance r.Imagine that the galaxy is positioned at the center of a sphere ofradius r and is radiating uniformly in all directions. How is this rela-tion modified in a Robertson-Walker metric? What is the relationbetween the luminosity and the position of the standard typicalgalaxy at r? Two effects modify the relation. First, the photonsundergo a cosmological redshift, and second, the rate at which thephotons are received is not the same as the rate at which they wereemitted. We choose a coordinate system with the radiating galaxy atr 0= radiating at cosmic time t t1= , and our receiving galaxy is at r r1=and receiving at t t0= . By Eq. (9.26), a photon emitted with energyE he e= o is received with energy ( )/ ( )E h K t K tr e 1 0= o . Similarly, by Eq.


2 The Mpc, or megaparsec, is a unit of distance. 1 Mpc .3 26 106#= light years.3 08 1022#= m.

(9.25), if the time between emission of photons is t ed , then the timebetween reception at the “sphere” at r1 is ( )/ ( )t t K t K tr e 0 1=d d . Fromthese it follows that the power received on the sphere at r1 at time

( )/ ( )t PK t K tis021

20 . The area of the sphere at r1 at cosmic time t0

is ( )K t r4 20 1

2r , and thus the luminosity at r1 at t0 is

( ) ( )

( )

( )

( )L

K t rP

K t

K tP

K t r

K t

4 420 1

2 20

21

40 1

2

21= =

r r. (9.34)

If we define the luminosity distance dL1 for a typical galaxy at r1 to be

( )( )

( )

( ),d

K tK t r

K t

K t rL

1

20 1

1

0 11= = r r (9.35)

then

LdP4 L

2

1

=r

. (9.36)

By measuring the luminosity of a standard galaxy, that is, one whosepower is known, the astronomer can determine the luminositydistance. Note that d dL 11

" for small r1.

9.3.6 Cosmological Redshift: Deceleration Parameter

Measurement of the redshifts and luminosities for more distant galax-ies gives the astronomer more information about the time depen-dence of K than that given by H 0. So, we might consider a powerseries expansion in t t0- of K as

( ) ( ) ( ) ( ) .K t K t H t t q H t t121

0 0 0 0 02

02 f= + - - - +; E (9.37)

or, equivalently,

( ) ( ) ( ) ...K t t t q t t121

0 0 02= + - - - +r r r r r r . (9.38)

(Recall that we have defined ( ) ( )/ ( ))K t K t K t0=r r and t H t0=r .) Hubble’sconstant, H 0, has already been introduced, and q0 is known as thedeceleration parameter.

( )( )

( )( ).q K t

K t

K tK t0 0 2

0

0/ - =-po

rp r (9.39)

What does one measure to determine q0? One can measure (hope-fully) luminosity distances and redshifts for standard galaxies. What isneeded is a luminosity distance-redshift relationship that involves


both H 0 and q0 to replace the proper distance-redshift relationship,Eq. (9.31). One can obtain this relationship by first expanding Eq.(9.29) in a power series in t t0- and inverting it to obtain

.t tHz

qz1

120 1

0

0 2f- = - +d nR

T

SSS

V

X

WWW (9.40)

Using Eq. (9.12) to expand r1 to order ( )t t0 12- , we obtain

( )( )r

K tt t H t t1

21

10

0 1 0 0 12f= - + -; E. (9.41)

After expanding the time dependence of Eq. (9.35) and substitutingEqs. (9.41) and then (9.40) into that, one obtains the luminositydistance-redshift relationship, for small ,z

( ) .d H z q z211L 0

10

21

f= + -- ; E (9.42)

By measuring luminosity distance and redshifts of galaxies, theastronomer can determine H 0 and q0. The measurement of luminos-ity distance is, as one can imagine, fraught with error. How “standard”are standard galaxies? Do standard galaxies age? That is, do theychange so that a nearby standard galaxy differs from a distant stan-dard galaxy that is, after all, being viewed at an earlier cosmic time?

Of course, Eq. (9.42) is meaningful only for <z 1 and is merely aparameterization of the luminosity distance-redshift relationship.After studying the dynamics of the Robertson-Walker metric, we willobtain a dL1 versus z relationship valid for larger z under particulardynamical assumptions.

9.4 Dynamics of the Robertson-Walker Metric

The dynamics of the Robertson-Walker metric are governed by theEinstein field equation

.G g R Rg g GT21

8+ = - + =-m m rab ab ab ab ab abu (9.43)

We allow for the possibility of a nonvanishing cosmologicalconstant m.

Note that the cosmological principle applies to the energy-momen-tum tensor Tab. That is, it should be form invariant under the sixspacelike isometry transformations for fixed t. In the coordinates ofthe R-W metric, these isometry transformations transform only thespace-space part of the metric gij. But the space-space part of the


energy-momentum Tij transforms like the gij so that the form invari-ance of the Tij would be assured if

( )T f t gij ij1=- , (9.44)

where ( )f t1 is some function of t. Similarly, under these isometrytransformations, T00 transforms the same as g00, that is, it does notchange, and we might expect that

( ),T f t00 2= (9.45)

where ( )f t2 is some second function of t. Since the components gi0vanish and are not effected by the six isometries, if we assume that Ti0vanish and that Eqs. (9.44) and (9.45) are valid, the form invarianceof Tab would be assured. We can write such a tensor as

( ) ,T f f U U f g1 2 1= + -ab a b ab (9.46)

where Ua is the four-velocity of a typical galaxy which in Robertson-Walker coordinates becomes U 0= da a . This is the energy-momentumtensor of a perfect fluid with f p1= , the pressure, and f2 = t, theproper energy density. (See Sec. 4.4.2.) We expect the energy-momentum tensor of a universe satisfying the cosmological principleto be that of a perfect fluid, so that

( ) .T p U U pg= + -tab a b ab (9.47)

Indeed, if one computes the Einstein tensor Gab (using, say,Mathematica) for the Robertson-Walker metric, one finds that

( )G F t00 2=

( )G F t gij ij1=

,G 0i0 = (9.48)

with

( )

( ( ) )F

K t

k K t32 2

2

=-+ o

( )

( ) ( ) ( ).F

K t

k K t K t K t21 2

2

=-+ +o p

(9.49)

Eq. (9.48), along with Einstein’s field equations, implies that theenergy-momentum tensor of a universe satisfying the cosmological


principle must be that of a perfect fluid (Eq. (9.47)). With this resultand Eq. (9.49), the Einstein field equations can be written as

( )

( ( ) )( )

K t

k K tG t

382

2+- =m r t

ou (9.50)

and

( )

( ) ( ) ( )( ).

K t

k K t K t K tGp t

282

2

-+ +

+ =m ro p

u (9.51)

In addition to these equations relating the ( )tt and ( )p t to the metric,the motion of the fluid is governed by the conservation of the energy-momentum tensor T 0; =b

ab . For the case of an ideal fluid, these conser-vation equations become

( )T p U U p g; ; ,= + -tbab a b

bbab_ i

( ) ( ) ( )p U U p U U U U U U, ,= + + + + +t t � �ba b a b

b vba v b

vbb v a9 C

.p g 0,- =bab (9.52)

Here we have used g 0; =bab and the scalar character of p and t . For the

Robertson-Walker metric, U 0= da a , and thus Eq. (9.52) becomes

( ) ( )( ) .T p p p 0;0

00 00 0= + + + + - =t d t d d� �b

ab a abb a ao o o (9.53)

Furthermore, for the Robertson-Walker metric, we find that

KK3

0 =� bb

o

.000 =�a (9.54)

Putting these into Eq. (9.53), we obtain

( ) ( ) .T p pKK p3

0;0 0 0= + + + - =t d t d db

ab a a ao o o o (9.55)

The space components of this equation are trivially satisfied, whilethe time component is

( )pKK

3 0+ + =t to o

( ) ( ) ,t pKK

3 0+ + =t to rrro

(9.56)

which can also be written as


( )dt

d p KpK

33+

=t o

( ).

dt

d p KpK

33+

=tr

ro r (9.57)

With the time dependence of t and p viewed as given implicitly bytheir ( )K t dependence, Eq. (9.57) can finally be expressed as

( )dKd K

pK33

2= -t

( ).

dK

d KpK3

32= -

tr

rr (9.58)

This is a useful form of the conservation of energy-momentum equa-tion. Given the equation of state ( )p p= t , Eq. (9.58) presumably canbe integrated to give ( )Kt , with one integration constant ( )K0t . Thisresult can be inserted into Eq. (9.50) to be solved for ( )K t , again withone integration constant ( )K t0 . We will return to these considerationslater.

Note that if the two first-order differential equations, Eqs. (9.57)and (9.50), are satisfied, then the second-order equation, Eq. (9.51),is automatically satisfied. This is not surprising, since the localconservation of the energy-momentum tensor is implied by theBianchi identities. Nevertheless, the second-order equation obtainedby eliminating ( )K to from Eq. (9.51), with the aid of Eq. (9.50), provesuseful:

( )( )

( ) ( ) .K tK t G t p t

34

33

= - + +r t mp u _ i (9.59)

Furthermore, we can write the two controlling dynamical equa-tions, Eqs. (9.50) and (9.59), using the time variable tr and the “scalevariable” Kr . Note ( )K t 10 =r r and ( )/ ( )K t K t 10 0 =ro r r r . In terms of these.

variables, Eq. (9.50) becomes

( )

( ) ( )( ) ,

K t

K t tK tk

2

0

20

= + +tt

� � �-t mr r

ro r rr rf p (9.60)

and Eq. (9.59)

( )

( ) ( ) ( ).

K t

K t t p t21 3

00=-

++t

t� �t mr r

rp r r r(9.61)


Here we have defined the quantities

( )HG t

38

02 00

=r t�t

u

( )H K tk

k

02

02= -�

.H3 02=m�m (9.62)

These are related as

( )( )H

G tH K tk

H138

302 0

02

02

02= - +

r t mu

.k0= + +� � �t m (9.63)

The , k0� �t and �m can be viewed as the (dimensionless) energy

density contribution to H 0 from matter, curvature, and the cosmo-logical constant, respectively. Note that the value of k� , if not zero,determines the value of ( )H K t0 0 .

9.4.1 Critical Density

Astronomers can determine Hubble’s constant H 0, the decelerationparameter q0, by measuring the luminosity distance and the redshiftsof galaxies, and hopefully the present average density 0t . How doesthe dynamics relate these “kinematic” measurements to the constantk? From Eq. (9.63) we see that k is negative or positive as 1

0- -� �t m

is greater or less than zero or as ( )t0t is greater or less than a criticaldensity,

.GH

G8

3

8c 0

2= -tr r

mu u (9.64)

If ,H0 02=m and ( )t0t determine the value of k.

Now consider the second-order dynamical Eq. (9.61) evaluated atthe present time t0r for which ( )p t 00 .r . Using the definition of q0, Eq.(9.39), we find from Eq. (9.61) that

.q200= -

��

tm (9.65)

The deceleration parameter is positive if 0=m and negative only if> /2

0� �m t .


9.4.2 Cosmological Redshift: Distant Objects

As noted before, the luminosity distance versus redshift relation, Eq.(9.42), is merely a power-series development of the luminositydistance in terms of the redshift expected to be accurate only for <z 1at best. But astrophysical objects, quasars, have been observed atapparent redshifts up to 3 and the microwave background, discussedin the next section, has a redshift of about 103. Thus, having adistance-redshift relation for >z 1, or even one that is more exact for. ,z 5. is useful. Such a relation depends of course on the equation of

state governing the dynamics of ( )K tr r . A universe of “dust” is a goodapproximation for the present epoch. One expects that the dustapproximation should be valid back past the cosmic time at which weare observing quasars, but surely not back to the decoupling time ofthe microwave background radiation (see next section).

To obtain a relation applicable to >z 1, back during the timefor which the dust approximation is valid, we return to Eq. (9.12)and change the time integral to a Kr integral. Thus, Eq. (9.12)becomes

( ) ( ) ( ).

K t H K t K

dKkrdr1

1 /( )

( )

K t

K t r

0 02 1 2

0=

-1

0 1##ro r r

rr r

r r

(9.66)

After evaluating the right-hand integral, we can write this as

( ) ( ),

( ) ( ),

( ) ( ), .

sin

sinh

r

K t H K t K

dK

K t H K t K

dK

K t H K t K

dK

k

k

k

1

1

1

1

0

1

( )

( )

( )

( )

( )

( )

K t

K t

K t

K t

K t

K t

1

0 0

0 0

0 0

=

=+

=

=-

1

0

1

0

1

0

#

#

#

ro r r

r

ro r r

r

ro r r

r

r r

r r

r r

r r

r r

r r

J

L

KK

J

L

KK

N

P

OO

N

P

OO

Z

[

\

]]]]]

]]]]

(9.67)

For a dust-filled universe, we have p 0. . This fact and Eq. (9.58) give

.K03=t t -r (9.68)

Thus, for a dust-filled universe, Eq. (9.60) becomes

,K K Kk2 1 2

0= + +� � �-

t mro r r (9.69)

which, if used in Eq. (9.67), gives


( ),

( ),

( ), .

sin

sinh

r

K t HK K K

dK

K t HK K

dK

K t HK K K

dK

k

k

k

1

1

1

1

0

1

( )

( )

( )

( )

( )

( )

oK t

K t

oK t

K t

oK t

K t

1

0 0 2 41 2

0 0 41 2

0 0 2 41 2

=

+ +

+

+ +

=+

=

=-

� � �

� �

� � �

t l m

t m

t l m

1

0

1

0

1

0

#

#

#

r r r

r

r r

r

r r r

r

r r

r r

r r

r r

r r

r r

J

L

KKKK

J

L

KKKK

`

`

`

N

P

OOOO

N

P

OOOO

j

j

j

Z

[

\

]]]]]]]

]]]]]]

(9.70)

To obtain a distance redshift relation, we change the Kr integration toan integration over the redshift parameter z, using the relation

/( )K z1 1= +r , Eq. (9.29), to obtain

( )( ) ( )

,

( )( )

,

( )( ) ( )

, .

sin

sinh

r

H K tz z

dz

H K tz

dz

H K tz z

dz

k

k

k

1

1 1

1

1

1

1 1

1

0

1

o

z

o

z

o

z

1

0 0 3 21 2

0

0 0 31 2

0

0 0 3 21 2

0

=

+ + + +

+ +

+ + + +

=+

=

=-

� � �

� �

� � �

t l m

t m

t l m

1

1

1

#

#

#

J

L

KKKK

J

L

KKKK

`

`

`

N

P

OOOO

N

P

OOOO

j

j

j

Z

[

\

]]]]]]]

]]]]]]

(9.71)

The integrals can be performed, at least numerically, to give ( )r z1 1 .Finally, with these results and the relation /( ),K z1 1= +r Eq. (9.35) canbe used to obtain ( )d zL 11 . In Figure 9.2, ( )H d zL0 is plotted for modeluniverses with the representative values of s�l and q0 as listed in

Table 9.1. ModelUniverses

Table 9.1. Note that the curves for universes C and D cross at aboutz 4= . The luminosity distance is the same for both curves for this



value of z. One might appreciate from these curves how difficult itis to distinguish universes by measuring the luminosity redshift rela-tion.

9.4.3 Cosmological Dynamics with 0=m

The presence of a cosmological term has a significant effect on thedevelopment of the universe during an epoch when | |m is greaterthan or of the order of G8r tu or Gp8r u (see Eqs. (9.50) and (9.51)).We assume in this section that 0=m , and in the following section theeffect of a nonvanishing cosmological constant is considered. With0=m , if ( ( ) ( ))t p t3+t r r is positive, then Eq. (9.61) implies the rate of

change of the slope of the curve K versus t is negative; the curve isconcave downward. Furthermore, since at the present time K 1=ro , theslope at the present time is positive. Together these two propertiesof the curve imply that at some time t br in the past, ( ) .K t 0b =r r As( )K t 0b "r r the proper distance ( , )d t r approaches zero (Eq. (9.11)).

The cosmological metric becomes singular with the energy densitybecoming infinite. This singular event is sometimes referred to as the“big bang,” and t t b0 -r r would be considered the age of the universein Hubble time units. Note that if ( )K t 0/rp r in the past (i.e., for< )t t0r r , then t t 1b0 - =r r ; the age of the universe would be the Hubble

time. Furthermore, since ( )<K t 0rp r ,

< .t t 1b 0-r r (9.72)

That is, the age of the universe is less than the Hubble time.


Figure 9.2. Luminosity distance versus redshift.


In addition to postdicting the beginning of the universe, referredto as the big bang, we can predict possible behaviors of the universein the future. Of course, such predictions cannot be checked andthus are of less interest.

As noted above, at the present time p 0. ; that is, the equation ofstate of the universe is that of dust. The universe is expanding andthus will remain dustlike in the immediate future. We will firstconsider the universe during this expansion phase.

For a dust-filled universe with ,0=m Eq. (9.69) becomes

.K K k2 1

0= +� �-

tro r (9.73)

By use of this equation we can infer how the fate of the universedepends on the value of k.

1. k 0= ( 0k=� ): For this case, from Eq. (9.73), we see that Kro will notvanish in the future. Thus, the universe will remain dustlike andEq. (9.73) will remain valid. The solution to this equation is

( ) ( ).K t K t t23/ / /3 2

03 2 1 2

00= + -�t

r r r r r

From this we see that the “size” of the universe increases withoutlimit.

2. k 1=- ( > 0k� ): Again Kro will not vanish in the future. Thus, theuniverse will remain dustlike, Eq. (9.73) will remain valid, and Krwill continue to increase. Eventually the k� term will dominate theright-hand side of Eq. (9.73). Thus, as ,t K" 3r r increases withoutbound as tr.

3. k 1=+ ( < 0k� ): At some time in the future, Kro will vanish andbecome negative. The density will begin to increase and eventuallythe universe will cease being dust-like. However, with ( ( ) ( ))t p t3+t r r

being positive, the rate of change of the slope of the curve Kr versustr is negative, the curve is concave downward, and eventually Krvanishes—the universe ends.

Though we would not expect the universe to be one of dust backto the big bang and, for the case k 1= , forward to the end of theuniverse, it is instructive to see how a universe of dust would behave.In Figure 9.3 the solution ( )K tr r of Eq. (9.73) is graphed for threeuniverses (A, B, and C of Table 9.1), for which , ,k 1 0 1= - , respec-tively.


During the epoch for which p 0. , referred to as the matter domi-nated era, as K becomes small, t becomes large. Eq. (9.68) merelyreflects the fact that the mass contained in a coexpanding volume isfixed. We might expect that as t grows back in time, the equation ofstate would gradually switch over to that of a relativistic hot gas,namely,

.p3=t (9.74)

The epoch, for which Eq. (9.74) is a good approximation, is calledthe radiation-dominated era. In that case, Eq. (9.58) leads to

.K 4?t -r (9.75)

Using this in Eq. (9.50) we see that, at a time when ( )K t is small, forinstance, early in the life of the universe, then

( ) ,K t K 1? -ro r r (9.76)

and thus

.K t /1 2?r r (9.77)

The condition for a particle horizon exists. (See Eq. (9.15).) In fact,even if the matter-dominated era obtained back to the big bang, aparticle horizon would still exist.


Figure 9.3. ( )K tr r for a universe of dust.


In a universe with 0=m , from Eqs. (9.64) and (9.65),

q2c

00=t

t . (9.78)

Thus, we have the following possibilities:

> /

/

< / .

k

k

k

q

q

q

1

0

1

1 2

1 2

1 2

0

0

0

=

=

=-

= (9.79)

A direct observational determination of q0 is obtained by measur-ing how the luminosity distance dL varies with z, the redshift parame-ter. Figure 9.4, which graphs Eq. (9.42), shows the deviation of d HL 0

from linear for values of universes A, B, and C of Table 9.1, whichhave , / , /q 1 1 2 1 60= , respectively. We see then that a measurement ofH 0 determines the critical density ct , whereas a measurement of bothH 0 and the deceleration parameter q0 determines the present densityof the universe 0t . Conversely, a measurement of H 0 and a determi-nation of 0t determines q0. An independent determination of allthree is a consistency check on the theory. Unfortunately, the reason-able assumption that essentially all of the matter of the universe isconcentrated in the visible mass of galaxies gives rise to discrepanciesin the dynamics of galaxies and clusters of galaxies, perhaps indicat-ing the presence of a large amount of “dark matter.” The amount andnature of this dark matter are among the outstanding questionsfacing present-day cosmology.


Figure 9.4. Luminosity distance versus redshift for universes A, B, and C.


9.4.4 Cosmological Dynamics with m� 0

Admitting the possibility of a nonvanishing cosmological constantwidens the range of the possible dynamics of the universe. As notedbefore, observations imply that | |m is very small; | |< m10 46 2m - .Nevertheless, an even smaller | |m can have a significant effect oncosmological dynamics. If | |

0.� �m t then | | /G G c4 40 0

2. =m r t r tu .Since /Kg m100

26 3.t - , | | m10 54 2.m - . A universe even with | |�m

comparable to 0

�t has a very small m. One sure indication of a nonva-nishing m would be the observation of a negative value of q0. (See Eq.(9.65).) Evidence from observations of supernovae at up to a redshiftof 0.7 indicates that in fact <q 00 (Riess et al. 1998). Also, it is gener-ally believed that the universe is very nearly flat at the present time,or 0k.� (see Sec. 9.5), and thus Eq. (9.63) becomes

.10

. +� �t m (9.80)

Furthermore, the size of cosmic matter density, including the so-called dark matter inferred from galaxy dynamics, seems to imply that<. 50

�t . So, when 0k.� , then >. 5�m .3 Perhaps the best fit to data is/1 3

0.�t and thus /2 3.�m . Consider then, as a model of the

universe, a dust-filled universe with 0k=� , /1 30=�t and /2 3=�m . This

is model universe D of Table 9.1. Then Eq. (9.69) becomes

.K K K31

322 1 2= +-ro r r (9.81)

This equation can be integrated with the boundary condition( )K t 0 10= =r r . The resulting solution is plotted in Figure 9.5. The solu-

tion with 10=�t , corresponding to model universe C of Table 9.1, is

also included in the figure for comparison. Notice that the age of theuniverse for /2 3=� is approximately . H94 0

1- , whereas for the flatuniverse, with 0=m , it is smaller, about . H65 0

1- . This is not surprising,since the 0=m universe has much more mass for the same Hubbleconstant. Universe D has /q 1 20=- . (See Eq. (9.64).)

In Figure 9.6, Eq. (9.42), the variation of the luminosity-distancewith the redshift, is graphed for universe D. In the same figure, thegraph for a universe with 0=m and /q 1 60= , that is, universe C, isshown to give an idea of the measurements required to distinguishsuch cases.


3 See Section 9.5.2 for an argument as to why k� is approximately zero.

9.5 The Early Universe

The gross behavior of the universe at early times depended on theequation of state of the ideal fluid at the expected high densities andthe expected associated high temperatures. We will not discuss this inany detail as it would take us far afield into the discussion of thethermodynamics of the interaction of elementary particles and theirdescendants at these high densities and temperature. However,observations such as the relative abundance of nuclei and even thesize of the clustering of galaxies give the cosmologist insight into the


Figure 9.5. ( )K tr r for universes C and D.

Figure 9.6. Luminosity distance versus redshift for universes C and D.



correctness of the developing picture of the universe—or raise ques-tions about the consistency of the picture.

9.5.1 The Cosmic Microwave Background Radiation

The strongest evidence for the existence of this high density–hightemperature phase of the universe is the thermal cosmic backgroundradiation (CMB). At some early time, we expect that the electromag-netic radiation was in thermodynamic equilibrium with matterconsisting of electrons and nuclei, such as protons. This electromag-netic radiation would be black body in character. The energy densityper frequency interval ( , )Tn o , which is characteristic of the tempera-ture, is given by the Planck formula,

( , )( / )

.exp

Th kTh

18 3

=-

n oor o (9.82)

As the universe expanded and cooled, the radiation would remain inthermal equilibrium with matter by ionizing the atoms as they wereformed. This scenario would continue until the temperaturedropped to the point that the energy of a typical photon was too smallto ionize atoms—the radiation would thermally decouple from thematter. This would occur at a temperature of about 4000 K.4

Subsequent to this decoupling, the expansion of the universe causesthe radiation to undergo a redshift, which, with the concomitantexpansion of a volume element, maintains the black body characterof the radiation with a lower effective temperature. That the blackbody character is maintained is not too difficult to derive. In addition,the relation among the decoupling temperature, the present effectivetemperature, the size of the universe at decoupling and the presentsize is easily obtained.

In the following we denote the decoupling time as t d, the decou-pling temperature as Td, the present time as t0, and the present effec-tive temperature of the radiation as T0. It follows from the definitionof ( )n o and the quantum condition E h= o that the number ofphotons in a volume element dV in a range of frequencies do is givenby /hn o. With the assumption that the decoupled photons are notabsorbed, the number of photons of a fixed direction in a frequency


4 One would not expect a sharp decoupling. In fact, the high-frequency tail of theblack-body radiation would keep the radiation in thermal contact down to a lowertemperature. Thus, the effective decoupling temperature would be somewhat lowerthen 4000 K.

range d do and a propagating volume element dVd at the time ofdecoupling is equal to the number in a redshifted frequency ranged 0o and expanded volume element dV0 at the present time. Thus,

dd d 0( / )

( ).

exp h kTd dV

hd dV

18

d

d2

0

00-

=oro

oo

n oo (9.83)

The photons, in a frequency range d do at decoupling, would beredshifted to be in a frequency range d 0o at the present time, given by(see Eq (9.26))

( )( )

,dK tK t

ddd0

0=o o (9.84)

and the expanded volume element 0dV is related to the volumeelement ddV by

.d( )( )

dVK tK t

dVd

00

3

= d n (9.85)

Using Eqs. (9.84) and (9.85) in Eq. (9.83), we find that

( )( ( )/ ( ))

.exp h K t kT K t

h1

8

d d0

0 0

03

=-

n oo

r o (9.86)

Thus, the decoupled radiation maintains its black-body characterwith a present effective temperature T0 given by

( )( )( )

( ).T tK tK t

T tdd0

0= (9.87)

The observed value of ( )T t0 is about . K2 7 , which corresponds to aratio

( )( )

.K tK t

10d

0

3. - (9.88)

Recently, beautiful and sensitive measurements have been made ofthis CMB radiation. The radiation has been observed in differentdirections with surprisingly uniform results, with the effectivetemperature anisotropy, /T T� , being on the order of 10 6- , aftercorrection for the Doppler shift of the motion of our solar system inthe universe (Bennet et al. 1996).

9.5.2 Inflation

Why is it surprising that the thermal background radiation is souniformly the same when observed in different directions? Consider


the observations of radiation coming from one direction, and thenfrom the opposite direction. The radiation from one direction (say0=z ) was emitted at a radial position rd given by

( ) ( ).

K tdt

krdr

1 /

r

t

t

2 1 20

=-

0 d

d

## (9.89)

Here, as before, t d is the cosmic time at which decoupling occurred.The radiation from the opposite direction (say, =z r) was emitted atthe same radial position. But these are separated positions, and thequestion arises as to why they should have the same temperature tosuch a high degree of accuracy. The reader may of course say that thecosmological principle demands this. But should the isotropy of theuniverse at a given time hold over distances between points that werenever in causal contact? This would seem to indicate that the materialat these two positions were in thermal equilibrium at the (cosmic)time of decoupling, which in turn would seem to imply that one posi-tion was within the particle horizon (see Eq. (9.14)) of the other atthe decoupling time.

We can understand what is involved by studying Figure 9.7. Thefigure depicts world lines of photons that were emitted toward us at


Figure 9.7. World lines of light emitted at the “big bang.”


time zero at various r and from opposite directions, say,0 and= =i i r. The world lines are calculated for the same case as

that of Figure 9.1, that is, for a universe with ,k 0 0= =m , and dustfilled with the pressure assumed to be zero all the way back to t 0= .The set of four world lines for 0=i , originate at t 0= at , / , / ,r 1 3 4 1 2=

and /1 4; similarly, for the =i r world lines. We can read off the figureapproximately when a position of a given r with 0=i comes withinthe horizon of the position r with =i r, that is, when they come incausal contact. For example, the two positions for .r 1. clearly are incausal contact at .t 7= , the time at which the light emitted at .r 1.reaches us. In fact, they are in causal contact before .t 10= . The twopositions .r 3. have just come in causal contact at .t 3. , the time atwhich the light emitted at .r 3. reaches us, whereas the two positions.r 5. are clearly not in causal contact at .t 15= , the time at which the

light emitted at .r 5. reaches us.We see that the time t r

c at which the position ,r =i r comes in causalcontact with ,r 0=i is given by

( ) ( ) ( ).

K tdt

krdr

krdr

12

1/ /

r

r

rt

2 1 2 2 1 200

rc

=-

=--

### (9.90)

From this, the condition that the positions ,rd =i r and ,r 0d =i are incausal contact at decoupling time t d is

( )>

( ) ( ).

K tdt

krdr

K tdt

21

2/t

trt

2 1 200 -

=d

0dd ### (9.91)

The universe has undergone a large expansion by a factor of 103 sincedecoupling. The validity of this inequality is determined by how theexpansion from t 0= to t td= compares with this. We first change thet integration to a K integration. Eq. (9.91) becomes

> .KKdK

KKdK

2( )

( )( )

K t

K tK t

0 d

d 0## o o (9.92)

We consider a universe for which k 0= =m . Assume that the integralsare dominated by the contribution during the matter-dominated erafor which

( ) .KK H K t K CK/ / /0 0

3 2 1 2 1 2= =o (9.93)

Here we have used Eq. (9.69), with 0k= =� �m , and have set( )H K t C/

0 03 2 = . The required inequality, Eq. (9.92), becomes

( )>( ) ( )

CK t

CK t K t

2

/ / /d d1 2

01 2 1 2- (9.94)


or

( )

( )>

( )

( ).

K t

K t

K t

K t21

1/

/

/

/d d

01 2

1 2

01 2

1 2

- (9.95)

Clearly, this inequality is not satisfied since ( )/ ( )K t K t 10d 03. - . We

can also see this from Figure 9.7, since the figure is drawn for amatter-dominated universe extending back to t 0= . For such auniverse K t /2 3? and thus t 3 10d

5#. - . But we saw that .t 3. is theearliest time that light, emitted from opposite directions and just nowreaching us, came from causally connected sources. One might arguethat at least for the left integral of Eq. (9.92), a large part of thecontribution comes from the radiation-dominated era, that is, forwhich K 4?t - and KK constant.o . This makes the situation worse.The contribution to the integral during the radiation-dominantepoch becomes proportional to ( )K tr , the value at which the radia-tion epoch ends. Note that during a radiation-dominated eraK t /1 2? , whereas for the matter-dominated era, K t /2 3? , and a fasterexpansion results. If the inequality is to be satisfied, an expansionmuch faster than that of the matter-dominated era must occur forsome “inflationary” period before decoupling. An example of such aperiod would be one in which the expansion is exponential, to whit,

( ),expK at? (9.96)

with a positive constant .a K would then satisfy the equation

.KK aK 2=o (9.97)

For such a period beginning at t b and ending at t e, before decoupling,the contribution to the left-hand side of Eq. (9.90) is

( ) ( )( ) ( )

.aKdK

aK t K tK t K t

( )

( )

e b

e b

K t

K t

2 =-e

b

# (9.98)

If there is a large expansion during this period, that is,( ) >>> ( )K t K te b , then the inequality is surely satisfied if

( ) ( ) ( )( )>

( )

( ).

aK t K tC

aH

K tK t

K t

K t

4 41/ /

/

b b

d

01 2

0 0

01 2

1 2

= -c m (9.99)

If the inflationary expansion is large enough, this will hold. We cansee the effect of the inflationary expansion period on the worldlines of Figure 9.7. During the period that K satisfies Eq. (9.96)

( )expr at? - , t approaches zero logarithmically with increasing r. Theworld lines reach much farther out in r.


As noted by Guth (1981), existence of a period that gives rise to anexponential expansion is suggested by quantum theory models thatunify the weak electromagnetic and strong interactions and thatpredict a phase transition at high temperatures at which the systemmight attain a supercooled state whose vacuum would contribute tothe energy-momentum tensor a term ga ab, with a assuming a largepositive value for some time. During this time, the effect is the sameas that of a large positive cosmological constant of value G8=m r au .For a such an effective cosmological term, eventually the constant mterm dominates the k and t terms of Eq. (9.50), and Eq. (9.96) resultswith ( / )a G8 3 /1 2= r au .

A concomitant effect of the standard inflationary model is that,after inflation, that is, after the universe condenses from its super-cooled state, the universe is very nearly flat. Thus, | |k� is much smallerthan

0�t . Note that our model universes B and D of Table 9.1 satisfy

this condition.After condensation occurs, the universe would be in the very high

density and high temperature condition of the big bang.

9.5.3 Cosmic Microwave Background and CosmologicalParameters

We have seen that the determination of the cosmological parameters,the �’s, is a primary goal in cosmological studies. And, as we haveseen, a determination of the redshift-distance relation contributes tothe measurement of these parameters. Surprisingly, the study of thevery small anisotropy of the CMB is proving to be a marvelous secondtool in this study. As an illustration of the effect of geometry on theanisotropy of the CMB, we will discuss how geometry affects the posi-tion of the first “acoustic” peak. Imagine a small density perturbationpresent at the big bang or, equivalently, immediately after theuniverse condenses following the inflationary expansion. Such aperturbation will expand with the velocity of sound cs in the hot,dense relativistic fluid to the time of decoupling td out to a (radial)position rSH, the “sound horizon,” related by

( ).

krdr c

K tdt

1 / s

tr

2 1 200 -

=dSH ## ^ h (9.100)

The proper size dSH, Eq. (9.11), of this sound horizon at the time ofdecoupling is given by

( )( )

( )( )

d K tkrdr c K t

K tdt

1 /d s d

tr

SH 2 1 200

d

=-

=SH ##


( )( )

.K t Hc K t

KK

dK( )s d

K t

0 0 0=

d#ro r

rr

(9.101)

We have again changed the t integration to a Kr integration. Recallthat ( )K t 10d

3. -r . From Eq. (9.60), the denominator of the last inte-gral is

(( )

) .KKtK K K /

k0

4 2 4 1 20

= + +tt

� � �t mro r

rr r r (9.102)

With the assumptions that the relativistic equation of state obtainsfrom decoupling back to the big bang and a zero-pressure dust-filleduniverse back to decoupling, the energy density in this equation canbe written as

( ) ( )( )

( )

( ) ( )

( )

( )

( ).t t

K t

K t

K t K t

K t

K t

K td

d

d

d d4

4

30

4

4

0 4= = =t tt

trr r

r r

r r r r

r r

r r

r r

Here, the first equality results from applying Eq. (9.75), which is truefor a relativistic fluid, and the second from Eq. (9.68), which is validfor a dust-filled universe. With this expression for t substituted intoEq. (9.102) and the result substituted into Eq. (9.101), we obtain

( )( )

( ( ) ) ( )

( ) ( )d

K t Hc K t

K t K KdK

K t H

c K t K t/ /

/( )

SHs d

d k

s d dK t

0 02 4 1 2

0 01 2

1 2

00

.=+ +� � � �t m t

d# r r rr rrr

(9.103)The proper size of the sound horizon is a “perpendicular” proper

size as viewed by us. Thus, the angular size of this sound horizonSHdi , as viewed by us, is related to this proper size dSH by (see Eq.

(9.13))

( ) ,d K t rSH d d SH= di (9.104)with

( ) ( ),

( ) ( ),

( ) ( ), .

sin

sinh

r

K t H K K K

dK

K t H K K

dK

K t H K K K

dK

k

k

k

1

1

1

1

0

1

( )

( )

( )

d

oK t

oK t

oK t

0 02 4 1 2

1

0 04

1

0 02 4

1

1 2

1 2

=

+ +

+

+ +

=+

=

=-

� � �

� �

� � �

t l m

t m

t l m

d

d

d

#

#

#

r r rr

r r

r

r r r

r

r

r

r

J

L

KK

J

L

KK

N

P

OO

N

P

OO

Z

[

\

]]]]]

]]]]]

(9.105)

With rd known, Eqs. (9.103) and (9.104) determine the angular sizeof the sound horizon:


( )

( ).

K t H r

c K t/

/

SHd

s d

0 01 2

1 2

o

=di�t

r(9.106)

But what properties of the CMB are related to SHdi ? The sound-waves are density waves and thus are temperature waves. The temper-ature is higher (lower) than the ambient temperature where thedensity is larger (smaller) than the background density. Thus, atdecoupling the soundwaves leave a temperature variation imprint.These waves have an extent as large as the sound horizon—an angleextent as large as SHdi . Performing a harmonic analysis of theobserved temperature variation with angle—that is, expanding theobservations in terms of ( )cos ,i—one would expect the result wouldpeak at values of , such that an integer number of half-waves wouldcover the sound horizon. (One should perform a spherical harmonicanalysis—we are dealing with a two-dimensional surface.) The acous-tic peaks should occur at

( )

( ).

j

c K t

j K t H r/

/

jSH s d

d

1 2

0 01 2

0, = =di

r r �t

r (9.107)

The positions of the peaks clearly depend on the geometry throughthe terms /1 2

0�t and rd. The position of the first peak, 1, , calculated

using Eq. (9.107), is graphed in Figure 9.8 for three sets of �’s:(1)0=m , k 1=+ , (2) 0=m , k 1=- , and (3) . ( )k K t0 d= r is taken to be 10 3-

and c 3 /s

1 2= - .


Figure 9.8. First acoustic peak, l vs. o�t1 for three sets of �’s.


Recent analysis of the anisotropy of CMB data gives a strong indi-cation for a flat universe with /2 3.�m and /1 3

0.�t , the values of our

universe D (Melchorri et al. 2000). As we have seen, for such auniverse <q 00 , which agrees with the result of the high z redshiftobservations noted before.

Other questions need to be answered, supported by observations,that bear on the consistency of the emerging view of the big bangorigin of our universe. And there are apparent difficulties. But it mustbe said that Einstein’s general relativity theory, coupled with thecosmological principle, have been successful to a remarkable extentin developing a coherent picture of our universe.

9.6 Exercises

1. Consider Einstein’s field equations with a nonvanishing cosmo-logical constant m. (a) Show that there exists a static solution onlyif k 1= and > 0m . (b) For such a solution for a dust-filled universe,derive the relations of the “radius” K and the density t to thecosmological constant m.

2. Again, consider Einstein’s field equations for a ,k 0 1!= , homoge-neous, isotropic, and empty universe (i.e., p 0= =t ) and with apositive cosmological constant m. (a) What are the resulting space-time metrics? Use your program to calculate (b) the space-timecurvature tensors Rabvt and (c) the Ricci scalars for these metrics.(d) Show that these metrics are space-time homogeneous andisotropic. (Since all have the same constant Ricci scalar, they areequivalent—they are different forms of the de Sitter metric.)

3. For the metric of Eq. (8.65), which is called the anti–de Sittermetric, calculate (a) the space-time curvature tensors Rabvt and (c)the Ricci scalars. (d) Show that this metric is space-time homoge-neous and isotropic. (e) Show that this metric satisfies Einstein’sfield equations for an empty universe with a negative cosmologicalconstant.

4. Assume that after decoupling the photon gas maintains thermalequilibrium with itself. Using the knowledge that, for such aphoton gas, T 4?t , show that T K? , and thus Eq. (9.86) is valid.


Suggested Additional Reading

For a very readable introduction to special relativity and a source ofmany exercises, see Taylor and Wheeler (1966).

I just touched the surface of electromagnetic theory, using it as anexample of a relativistic field. An extensive treatment of the relativis-tic formulation of electromagnetic theory is given in the classic bookby J. D. Jackson (1975). For accessible discussions of relativistic fluidsand associated energy-momentum tensors, see Schutz (1985) andWeinberg (1972).

Our treatment of tensor analysis is based on coordinate transfor-mations. Students wishing to study the more modern coordinate-freeapproach to differential geometry applied to general relativity arereferred to Wald (1984). The mathematically inclined student mightwant to look at Nash and Sen (1983) for an introduction to mathe-matical constructs, such as manifolds, forms, connections, etc., usedin general relativity.

In the section on equilibrium stellar interiors, I did not coverreasonable equations of state for high density nor did I cover theevolution of stellar interiors. These topics must be studied beforequestions concerning the formation of black holes via stellar evolu-tion can be answered. A popular account of the science and historyof black holes is given in Thorne (1994).

Topics at the interface of astrophysics and cosmology are discussedin marvelous detail in Peebles (1993). Astrophysical observationshaving a defining effect on cosmological studies are being made atsuch a fast pace that many new results are not in older reviews. A goodsource for semitechnical articles on recent advances is the magazineScientific American.

197

References

Bennet, C. et al. (1996). Astrophysical Journal 464:1.Bernstein, J. (1973). Einstein, New York: Viking Press.Birkhoff, G. (1923). Relativity and Modern Physics. Cambridge, Mass.:

Harvard University Press.Buchdahl, H. A. (1959). Physical Review 116:1027.Dyson, F., Eddington, A., and Davidson, C. (1920). Phil. Trans. Roy.

Soc. 220:291.Einstein, A. (1916). Annalen der Physik 49:769.Einstein, A., Lorentz, H., and Minkowski, H. (1923). Principle of

Relativity. New york: Dover.Freedman, W., et al. (2001). Astrophysical Journal 553:47.Guth, A. (1981). Physical Review D 23:347.Harrison, E. R. (1987). Darkness at Night: A Riddle of the Universe.

Cambridge Mass.: Harvard University Press.Hubble, E. (1929). Proc. Nat. Acad. Sci. 15:168.Jackson, J.D. (1975). Classical Electrodynamics. New York: WileyMelchorri, A., et al. (2000). Astrophysical Journal Letters 536:63.Nash, C., and Sen, S. (1983). Topology and Geometry for Physicists.

London: Academic Press.Peebles, P.J.E. (1993). Principles of Physical Cosmology. Princeton,

N.J.:Princeton University Press.Pound, R. V., and Rebka, G.A. (1960). Physical Review Letters 4:337.Riess, A., et al. (1998). Astronomical Journal 116(3):1009.Rindler, W. (1956). Mon. Not. R. Astron. Soc. 116:662.Rindler, W. (1991). Introduction to Special Relativity. 2d ed. New York:

Oxford University Press. Schilpp, P.A., ed.(1949). Albert Einstein: Philosopher-Scientist. Evanston,

Ill.: Library of Living Philosophers.

199

Schutz, B.F. (1985). A First Course in General Relativity. Cambridge,U.K.: Cambridge University Press.

Thorne, K. S. (1994). Black Holes and Time Warps. New York: Norton.Taylor, E.F., and Wheeler, J. A. (1966) Spacetime Physics. San

Francisco; Freeman.Wald, R. M. (1984). General Relativity, Chicago: University of Chicago

Press.Weinberg, S. (1972). Gravitation and Cosmology. New York: Wiley.

200 References

Absolute derivative. See Directionalcovariant derivative

Acoustic peak, 193, 195Adiabatic, 68, 70Affine parameter, 93, 119Age of universe, 182Anti-de Sitter metric, 161Asymptotic flatness, 116

Bianchi identity, 102, 103, 132, 178Big bang, 169, 183, 193Birkhoff’s theorem, 116, 139Blackbody radiation. See Cosmic

microwave backgroundBlack hole, 136–39

event horizon, 138Buchdahl bound, 135, 139

Causality, 21–22Christoffel symbols, 93, 103

for maximally symmetric three-dimensional space, 157

and metric, 94for Schwarzschild metric, 117

Collisions, 39–47Commutator of covariant

derivatives, 101, 104, 147Co-moving coordinates, 167Conservation and symmetry, 39,

118, 146Conserved generalized momenta.

See Killing vectorsContinuity equation, 59

and local charge conservation, 63Contraction of tensor indices, 54,

58Contravariant index, 57Cosmic microwave background

(CMB), 188–89acoustic peak of, 193, 195anisotropy, 189, 193

Cosmic time, 165Cosmological constant, 106–08

and deceleration parameter,179

Cosmological distance. SeeLuminosity distance

Cosmological principle, 164and energy-momentum tensor,

175Covariant derivative, 95, 104, 109

of metric, 96Covariant index, 57Critical density, 179Curvature, 98–99

negative, 99, 154, 160positive, 99, 153, 159zero, 99, 153, 158

Curvature scalar. See Ricci scalarCurvature tensor. See Riemann

curvature tensor

Dark matter, 185Deceleration parameter, 174

and critical density, 185Decoupling of photons, 188Density

electric charge, 59electric current, 59energy-momentum, 66

de Sitter precession, 130Directional covariant derivative, 95,

96, 103Doppler shift, 27–29, 33, 172

Eddington-Finkelstein coordinates.See Schwarzschild metric

Effective potential, 122Einstein, A., 10, 77, 88, 113Einstein equations, 104–7, 116, 132

cosmological, 177Einsteinian relativity, 10, 39

Index

201

Einstein tensor, 103, 104for Robertson-Walker metric, 176

Electromagnetic field-strengthtensor, 55, 59

dual tensor of, 60Energy

density, 67, 69, 133, 176relativistic, 43kinetic, 44

Energy-momentum four-vector. SeeFour-momentum

Energy-momentum tensor, 65–67of charged dust and

electromagnetic field, 73cosmological, 175–177of dust, 67of electromagnetic field, 62local conservation of, 66of perfect fluid, 68–69, 108–9,

176Equality of gravitational and

inertial mass, 78Equation of state, 70, 133

for non-relativistic ideal gas, 65,70

for relativistic gas, 70and speed of sound, 71

Equivalence principle. See Principleof equivalence

Equivalent frames, 79Ether, 8Euclidean space, 2Euclidean transformations, 18Event, 2Expansion of the universe, 172,

183, 191and decoupling of photons, 188

Four-acceleration, 32Four-force, 43Four-momentum, 43–48

generalized, 119Four-scalar, 31Four-tensor, 55–58Four-vector, 31Four-velocity, 32

of a fluid, 69

generalized, 119Galilean relativity, 7, 40Galilean transformations, 4–6Gauss’s theorem 63

and local charge conservation,63–65

General relativity, 88, 89, 103, 113,134

General relativity field equations.See Einstein equations

Geodesic, 92–94, 103, 119, 165for Schwarzschild metric, 120–23in terms of momentum, 119of “typical” galaxy, 167

Gravitational constant, 78in Einstein’s field equations, 106

Gravitational potential, 80, 104, 108Group

definition, 34nLorentz, 35rotation, 34

Gyroscopein general relativity, 128–30in special relativity, 49–50

Homogeneous and isotropic space.See Maximally symmetricspaces

Horizonand causal contact, 190event, 170particle, 168–69, 190of Schwarzschild metric, 138

Hubble’s constant (H0), 172Hubble time, 173

and age of universe, 182Huygen’s principle, 83

IndicesLatin and Greek, 32nraising and lowering, 58repeated, 5, 55

Inertial frame, 2, 7, 10, 17, 25, 39,49, 80

global, 87local, 87–89, 92, 96, 109, 145

Inflation, 192

202 Index

Invariant interval, 18and causality, 21–22lightlike, 19and proper time, 22spacelike, 19timelike, 19

Invariant length, 18Isometries, 144

and Killing vectors, 145

Killing vectors, 145and conserved generalized

momenta, 146cosmological, 166and maximally symmetric spaces,

137Kronecker delta, 26

Length contraction, 26Lense-Thirring precession, 130Levi-Civita tensor, 54, 58Light cones, 19, 22, 82–83

for Schwarzschild metric, 137–39Light deflection, 83–85, 123–24Local frame theorem, 89, 90Local inertial coordinates. See Local

inertial frameLocal inertial frame, 87–89, 92, 96,

109, 145Local Lorentz frame. See Local

inertial frameLongest elapsed proper time, 24,

92, 93, 94Lorentz, H. 58Lorentz force, 61Lorentz transformations, 10–12, 31,

56for arbitrary relative velocity,

13–14canonical, 12as group elements, 34

Luminosity, 173Luminosity distance, 174

and redshift, 175, 180, 185, 186Mass, 41, 43

gravitational, 78, 117Mass hyperbola, 44

Matter-dominated era, 184, 191,192

Maximally symmetric spaces,146–52

condition for, 151for cosmology, 166four-dimensional Lorentzian, 160three-dimensional, 156–60two-dimensional, 153–56See also Killing vectors

Maxwell, J., 7Maxwell’s equations, 7, 58–61Metric, 31n, 87

and Christoffel symbols, 94covariant derivative of, 96Euclidean, 31and invariant interval, 31, 87inverse, 91isometry, 144locally Lorentzian, 88Lorentz, 31Minkowski, 31signature of, 90as tensor, 54, 58, 91of weak-field limit, 105

Metric space, 31nMichelson-Morley experiment, 8Minkowski space, 18Momentum, 39–48

conservation of, 39, 41, 42, 47

Newtonian mechanics, 3Newtonian universal time, 6Newton’s gravitational theory, 78

Olbers’ paradox, 163Oppenheimer-Volkoff equation,

133Orbits for Schwarzchild metric,

120–30effective potentials for, 122and light deflection, 123–24precession of, 125–28

Parallel transport, 92, 97, 99, 100on a saddle surface, 99on a sphere, 97, 99

Index 203

Perfect fluid, 68, 109, 176Poincaré transformations, 19Precession of perihelia, 125–28

lack of, for Newtonian orbit,125–26

Pressure, 68, 70, 176central; 133and equation of state, 70

Principle of equivalence, 81, 88Principle of general covariance, 88Principle of relativity, 7, 10Proper time, 22, 24, 165Pseudo-Riemannian space, 89

Radiation-dominated era, 184, 192Redshift

cosmological, 171, 174, 180and deceleration parameter,

174–75gravitational, 81, 117–18, 136and luminosity distance, 175,

180, 185, 186Relativistic three-force, 48Ricci scalar, 102, 104, 152Ricci tensor, 102, 104, 150Riemann curvature tensor, 101,

102, 104and commutator of covariant

derivatives, 102identities, 102

Riemannian space, 89Robertson-Walker metric, 167

dynamics of, 175–79energy-momentum tensor of, 176

Roemer, O., 7Rotation transformations, 17

as a group, 34

Scalars, 30–32four-scalar, 32three-scalar, 31

Schwarzschild metric, 113, 116–17Christoffel symbols for, 117in Eddington-Finkelstein

coordinates, 138event horizon for, 138light cones, 137, 138

See also Orbits for Schwarzschildmetric

Schwarzschild radius, 138Sound horizon, 193–94Space-time diagrams, 19–28Special relativity, 10, 27, 46, 49, 61,

89, 108Speed of sound, 71–72, 193Spin, 57Spin magnetic moment dynamics,

61–63Stanford’s Gravity B Probe, 130Static spherical metric, 104–15

Ricci tensor for, 105Stellar interior, 131–36

Newtonian, 134Newtonian constant density,

134–35Oppenheimer-Volkoff equation

for, 133relativistic constant density,

135–36

Tensorcontraction, 54, 58, 91contravariant, 56, 90covariant, 57, 91differentiation of, 55, 57of general coordinate

transformations, 86–92metric, 58, 91outer product, 54, 58, 91rank of, 54

Three-scalar, 31Three-vector, 30Thomas precession, 50, 82Time dilation, 26–27Torus, 151Twin paradox, 23–26

Universeage of, 182critical density of, 173of dust, 169, 180, 183energy-momentum tensor of, 176expansion of, 172, 182horizons, 168–70

204 Index

Robertson-Walker metric for, 167

Vector, 30–32contravariant, 56covariant, 58four-vector, 31inner product, 30, 32three-vector, 30

Velocity, 2Galilean transformation of, 6Lorentz transformation of, 13, 15units, 11

Volumefour-volume, 63–65, 66proper volume, 134three-volume, 63, 67

Wave equation for sound, 71World line, 20, 24–26, 28, 32, 50,

87, 92, 165, 167, 192

Zero momentum frame, 46, 47

Index 205

Science

Introduction to relativity