An Introduction to Tensors and Relativity

An Introduction to Tensors and Relativity Peter K. S. Dunsby Department of Mathematics and Applied Mathematics, University of Cape Town, South Africa. March 21, 2000

To my students.

Poets say science takes away from the beauty of the stars { mere globs of gas atoms. Nothing is \mere". I too can see the stars on a desert night, and feel them. But do I see less or more? The vastness of the heavens stretches my imagination { stuck on this carousel my little eye can catch one - million - year - old light. A vast pattern { of which I am a part.... What is the pattern, or the meaning, or the why? It does not do harm to the mystery to know a little about it. For far more marvelous is the truth than any artists of the past imagined it. Why do the poets of the present not speak of it? What men are poets who can speak of Jupiter if he were a man, but if he is an immense spinning sphere of methane and ammonia must be silent? Richard Feynman

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Contents List of gures Preface 1 The Special Theory of relativity 1.1 1.2 1.3 1.4 1.5 1.6

Pre - relativistic physics . . . . . . . . . . . . . . . . . The equations of electromagnetism . . . . . . . . . . The principle of Special Relativity . . . . . . . . . . . Spacetime diagrams and the Lorentz transformations The spacetime interval . . . . . . . . . . . . . . . . . Minkowski spacetime . . . . . . . . . . . . . . . . . . 1.6.1 The null cone . . . . . . . . . . . . . . . . . . 1.7 Consequences of the Einstein postulates . . . . . . . 1.7.1 Time dilation . . . . . . . . . . . . . . . . . . 1.7.2 Length contraction . . . . . . . . . . . . . . . 1.7.3 The twin paradox . . . . . . . . . . . . . . . . 1.8 Velocity composition law . . . . . . . . . . . . . . . .

2 Vectors in Special Relativity

2.1 Four - vectors . . . . . . . . . . . . . . . . . 2.2 Four - velocity, momentum and acceleration . 2.2.1 Relativistic Doppler shift . . . . . . . 2.2.2 Four - acceleration . . . . . . . . . . . 2.3 Relativistic dynamics . . . . . . . . . . . . . 2.3.1 Example 2.1 . . . . . . . . . . . . . . 2.3.2 Example 2.2 . . . . . . . . . . . . . .

3 Tensors in Special Relativity

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3.1 Metrics and forms . . . . . . . . . . . . . . . . . . . . . . . 3.2 One - forms . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Gradients . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 The metric as a mapping of vectors onto one - forms 3.3 More general tensors . . . . . . . . . . . . . . . . . . . . . iii

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vii viii 1 1 3 5 6 11 11 12 14 14 15 16 16

18 18 22 24 25 25 26 27

29 29 30 32 33 35

3.3.1 Tensors of type 0=2 . . . . . . . 3.3.2 Tensors of type 0=N . . . . . . 3.3.3 Tensors of type 2=0 . . . . . . . 3.3.4 Tensors of type M=0 . . . . . . 3.3.5 Tensors of type M=N . . . . . . 3.3.6 Index \raising" and \lowering" 3.4 General properties of tensors . . . . . . 3.5 Tensor derivatives and gradients . . . .

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The story so far . . . . . . . . . . . . . . . . . . . . The gravitational redshift experiment . . . . . . . . Non - existence of an inertial frame at rest on earth Mass in Newtonian theory . . . . . . . . . . . . . . The principle of equivalence . . . . . . . . . . . . . The principle of equivalence in action . . . . . . . . 5.6.1 Eect of gravity on light . . . . . . . . . . . 5.6.2 Eect of gravity on time . . . . . . . . . . . 5.6.3 Towards spacetime curvature . . . . . . . .

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4 Application of Tensors in Special Relativity 4.1 The energy - momentum tensor . . . . . . . 4.1.1 General uids . . . . . . . . . . . . 4.1.2 Conservation of energy - momentum 4.1.3 Conservation of particles . . . . . . 4.1.4 Perfect uids . . . . . . . . . . . . 4.1.5 The conservation equations . . . . 4.2 The Electromagnetic tensor . . . . . . . .

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5 Conceptual Basis of General Relativity 5.1 5.2 5.3 5.4 5.5 5.6

6 Curved spacetime and General Relativity 6.1 6.2 6.3 6.4 6.5

Manifolds, tangent spaces and local inertial frames . Covariant derivatives and Christoel symbols . . . Calculating from the metric . . . . . . . . . . Tensors in polar coordinates . . . . . . . . . . . . . Parallel transport and geodesics . . . . . . . . . . . 6.5.1 The variational method for geodesics . . . . 6.5.2 The principle of equivalence again . . . . . . 6.6 The curvature tensor and geodesic deviation . . . . 6.6.1 The curvature tensor . . . . . . . . . . . . . 6.6.2 Properties of the Riemann curvature tensor 6.7 Geodesic deviation . . . . . . . . . . . . . . . . . . 6.8 The Bianchi identities; Ricci and Einstein tensors . 6.8.1 The Ricci tensor . . . . . . . . . . . . . . . iv

35 36 37 37 38 38 39 40

42 42 45 45 47 47 47 49

51 51 52 54 55 58 61 61 62 63

65 65 67 70 71 74 75 78 79 80 82 85 87 88

6.8.2 The Einstein Tensor . . . . . . . . . . . . . . . . . . . . . . 88

7 Einstein's Field Equations - The Main Goal of The Course

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7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 7.2 The non - vacuum eld equations . . . . . . . . . . . . . . . . . . . 91 7.3 The weak eld approximation . . . . . . . . . . . . . . . . . . . . . 93

8 The Schwarzschild Solution 8.1 8.2 8.3 8.4 8.5 8.6

The line element . . . . . . . . . . . . . . . . . . The Christoel symbols . . . . . . . . . . . . . . The Ricci Tensor . . . . . . . . . . . . . . . . . . The vacuum eld equations . . . . . . . . . . . . The Schwartzschild solution . . . . . . . . . . . . General discussion of the Schwartzschild solution 8.6.1 Time dilation in a gravitational eld . . . 8.6.2 Length contraction in a gravitational eld 8.7 Orbits in Schwartzschild spacetime . . . . . . . . 8.7.1 Solution for timelike orbits and precession 8.7.2 The bending of light . . . . . . . . . . . .

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98 99 100 101 102 103 104 104 104 106 107

List of Figures 1.1 I do not know what I may appear to the world; but to myself I seem to have been only like a boy playing on the sea - shore, and diverting myself in now and then nding a smoother pebble or a prettier shell than ordinary, whilst the great ocean of truth lay all undiscovered before me. Isaac Newton . . . . . . . . . . . . . . . . . . . . . . . . 1.2 James Clerk Maxwell . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Hendrik Antoon Lorentz . . . . . . . . . . . . . . . . . . . . . . . . 1.4 In the light of knowledge attained, the happy achievement seems almost a matter of course, and any intelligent student can grasp it without too much trouble. But the years of anxious searching in the dark, with their intense longing, their alternations of con dence and exhaustion, ant the nal emergence into the light - only those who have themselves experienced it can understand that. Albert Einstein. 1.5 A simple spacetime diagram . . . . . . . . . . . . . . . . . . . . . . 1.6 The ct axis of a frame whose velocity is v relative to O. . . . . . . . 1.7 Light re ection in O . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8 Light re ection in O as measured by O. . . . . . . . . . . . . . . . 1.9 The views of space and time that I wish to lay before you have sprung from the soil of experimental physics, and therein lies their strength. They are radical. Henceforth space by itself and time by itself are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality. Hermann Minkowski . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.10 The lightcone. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.11 ict space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.12 Length contraction . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 The worldline of a particle with four - velocity U . . . . . . . . . . . 2.2 The colliding particles in Example 1. . . . . . . . . . . . . . . . . . 2.3 Sciamatic diagram illustrating Compton scattering. . . . . . . . . . 3.1 p~(A) is the number of surfaces the vector A pierces. . . . . . . . . . 3.2 World line of particle with four - velocity U. . . . . . . . . . . . . . vi

2 4 5

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4.1 5.1 5.2 5.3 5.4 5.5 5.6 5.7 6.1 6.2 6.3 6.4 7.1 8.1 8.2 8.3 8.4 8.5

Energy ow across a uid element. . . . . . . . . . . . . . . . . . . The Gravitational redshift experiment. . . . . . . . . . . . . . . . . Minkowski geometry for the Pound - Snider Experiment. . . . . . . . The Galileo Piza experiment. . . . . . . . . . . . . . . . . . . . . . The lift experiments . . . . . . . . . . . . . . . . . . . . . . . . . . The bending of light . . . . . . . . . . . . . . . . . . . . . . . . . . Space time diagram of rocket undergoing uniform acceleration g . . The lift experiments again . . . . . . . . . . . . . . . . . . . . . . . Parallel transport of a vector V along a timelike curve with tangent U. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Parallel transport around a closed loop in at space. . . . . . . . . . Parallel transport around a closed loop ABCD. . . . . . . . . . . . Geodesic deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . A SUMMARY OF WHAT WE HAVE DONE : ) = . . . . . . . . Schwarzschild . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . De ection of light ray . . . . . . . . . . . . . . . . . . . . . . . . . . Diagram showing the total de ection . . . . . . . . . . . . . . . . . Einstein ring. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Einstein ring lensing event . . . . . . . . . . . . . . . . . . . . . . .

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45 53 54 56 60 61 63 64 75 80 81 85 97 103 108 109 109 110

Preface This set of lecture notes has evolved from a third year undergraduate course on Tensor Calculus and General Relativity which I have taught at the University of Cape Town since 1993. An online version can be found at http://shiva.mth.uct.ac.za/gr96/gr96.html Enjoy.....

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Chapter 1 The Special Theory of relativity 1.1 Pre - relativistic physics The starting point for our work are Newtons laws of motion. These can be stated as follows:

Free particles move with constant velocity. The vector force F is proportional to the rate of change of momentum i.e. F = dtd (mv). To every action there is an equal and opposite reaction. The rst of these laws singles out inertial frames as the non - accelerating ones. Consider now a frame O [ i.e. a set of spatial coordinates (x; y; z) and a time coordinate t ] and another frame O with coordinates (x; y; z; t) which moves in the x direction with uniform speed v relative to the frame O. Common sense suggests that the two sets of coordinates are related by

x = x vt ; y = y ; z = z ; t = t : These are the Galelian transformations . 1

(1.1)

CHAPTER 1. THE SPECIAL THEORY OF RELATIVITY

2

Figure 1.1: I do not know what I may appear to the world; but to myself I seem to have been only like a boy playing on the sea - shore, and diverting myself in now and then nding a smoother pebble or a prettier shell than ordinary, whilst the great ocean of truth lay all undiscovered before me. Isaac Newton If the particle has a velocity u with components (u1; u2; u3) in O, its velocity in O is:

u1 = u1 v ; u2 = u2 ; u3 = u3 ; or

u = u v ;

(1.2) (1.3)

where

1 ; u = dx2 ; u = dx3 : u1 = dx (1.4) 2 dt 3 dt dt More generally if the coordinate axes and the origins of O and O dier then: X = RX vt d ; (1.5)

where X has components (x; y; z). Here R is a rotation matrix aligning O and O , v is the relative velocity of O with respect to O and d is the displacement of the origin from O . Since the transformation is linear [ constant velocity in O ) constant velocity in O ], O is inertial if O is. Thus there are an in nite set of inertial frames, all moving uniformly with respect to each other.


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All of Newtons laws apply in any inertial frame since a = ddtu = ddtu ; (1.6) and F is invariant. Thus we have Newtonian [ Galelian ] Relativity. The laws of mechanics do not allow measurement of absolute velocity, however one can measure absolute acceleration. Newton explained inertial frames in terms of absolute space identi ed with the center of mass of the solar system or a frame of \ xed stars". However this is unsatisfactory because:

There is no unique identi cation [ many inertial frames ]. Philosophically unappealing since absolute space aects everything but is aected by nothing.

1.2 The equations of electromagnetism Maxwell's equations for the electromagnetic eld [ in units with 0 = 0 = c = 1 ] are: r B @@tE = 4j ; r E + @@tB = 0 ; r E = 4 ; r B = 0 : (1.7) These equations show that one can have electromagnetic waves in a vacuum which travel with a speed c = 3 108 ms 1 , Thus light is a form of electromagnetic radiation. An Ether was postulated as the medium which carries these waves [ i.e. the speed c is the speed with respect to the ether ] and naturally this was identi ed with the absolute space of Newton. Attempts to measure the ether drift by Michelson and Morley in 1887 gave a null result, so eects like stellar aberration could not be due to the earth dragging the ether. Another great problem was that Maxwell's equations did not appear to obey the principle of Galelian Relativity i.e. they were not invariant under the Galelian


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Figure 1.2: James Clerk Maxwell transformations. This means that in a moving space ship the electric and optical phenomena should be dierent from those in a stationary ship!!! Thus one could use for example optical phenomena to determine the speed of the ship. One of the consequences of Maxwell's equations is that if there is a disturbance in the eld such that light is generated, these electromagnetic waves go out in all directions equally and at the same speed c. Another consequence of the equations is that if the source of the disturbance is moving, the light emitted goes through space at the same speed c. This is analogous to the case of sound being likewise independent of the motion of the source. This independence of the motion of the source in the case of light brings up an interesting problem. Suppose we are riding in a car that is going at a speed u, and light from the rear is going past the car with speed c. According to the Galelian transformations, the apparent speed of the passing light, as we measure it in the car, should not be c, but c u. This means that by measuring the speed of the light going past the car, one could determine the absolute speed of the car. This is crazy because we already know that the laws of mechanics do not permit the measurement of absolute velocity.

Something is wrong with the laws of Physics!! At rst people thought that it was Maxwell's equations that were at fault and they modi ed them so that they were invariant under the Galelian transformations; but this led to the prediction of new electrical phenomena which could not be found experimentally.


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Figure 1.3: Hendrik Antoon Lorentz Then in 1903 Lorentz made a remarkable and curious discovery. He found that when he applied the following transformations to Maxwell's equations 2 t = qt vx=c ; 1 v2=c2 x = q x vt2 2 ; 1 v =c y = y ;

z = z ;

(1.8)

they remained invariant. These are the famous Lorentz transformations and we will discuss them in more detail in section 1.4.

1.3 The principle of Special Relativity In 1905 Einstein generalized the Galelian relativity principle [ applicable only to Newtons laws ] to the whole of Physics by postulating that:

All inertial frames are equivalent for all experiments i.e. no experiment can measure absolute velocity. Maxwell's equations and the speed of light must be the same for all observers. Einstein's motivation was to avoid inconsistencies between Maxwell's equations and Galelian relativity. The Lorentz transformations must relate to actual space and time measurements.


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Figure 1.4: In the light of knowledge attained, the happy achievement seems almost a matter of course, and any intelligent student can grasp it without too much trouble. But the years of anxious searching in the dark, with their intense longing, their alternations of con dence and exhaustion, ant the nal emergence into the light - only those who have themselves experienced it can understand that. Albert Einstein. Special Relativity abolishes the idea of absolute space [ for example the ether ] and absolute time, but it leaves unexplained the origin of inertial frames. Mach's principle says that the inertial frames are determined be the rest of the matter in the universe [ i.e. those that are non - accelerating with respect to the rest of the universe ]. Einstein tried later to incorporate this idea into General Relativity.

1.4 Spacetime diagrams and the Lorentz transformations Probably the easiest way to understand what the physical consequences of the above postulates are, is through the use of simple spacetime diagrams. In Figure 1.5 we illustrate some of the basic concepts using a two dimensional slice of spacetime. A single point, of xed x and t is called an event. A particle or observer moving through spacetime maps out a curve x = x(ct), and so represents the position of the particle at dierent times. This curve is called the particles world - line . The gradient of the world - line is related to the particle's velocity, d(ct) = c ; (1.9) dx v so light rays [ v = c ] move on 45o lines on this diagram.

CHAPTER 1. THE SPECIAL THEORY OF RELATIVITY ct

v dx2 + dy2 + dz2 ; (1.25) the curve is timelike in that interval;

ds2 > 0 ) c2dt2 < dx2 + dy2 + dz2 ;

(1.26)

the curve is spacelike in that interval; and

ds2 = 0 ) c2dt2 = dx2 + dy2 + dz2 ;

(1.27)

the curve is null in that interval. Note that the sign of ds2 is a matter of convention, invariance between frames is what is important.

1.6.1 The null cone Consider a null ray [ for example a light ray propagating in the x direction: then dx = c, so the null ray represents a particle moving at the speed of light. If we dt

CHAPTER 1. THE SPECIAL THEORY OF RELATIVITY cdt

Future

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Path of massive particle (timelike)

dx Elsewhere

dy

Past

Figure 1.10: The lightcone. consider 3D Minkowski spacetime then the null rays lie on the surface of a cone [ see Figure 1.9 ]: Before we look at some of the wonderful consequences of the Einstein postulates, let us return brie y to the Lorentz transformations. If we de ne tanh = vc we can write them in the following way:

x = x cosh ct sinh ; ct =

x sinh + ct cosh :

(1.28)

ict

ict iφ

x iΦ x

Figure 1.11: ict space


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Formally this corresponds to a rotation through an angle i in (ict; x) space [ see Figure 1.11 ].

1.7 Consequences of the Einstein postulates 1.7.1 Time dilation

Suppose we have someone in a spaceship O moving at some velocity v relative to an observer O. The guy in the spaceship turns his stereo on and starts listening to some rap music!! With a clock he measures the time interval between two beats t = tB tA. what time interval does the observer O measure? To calculate this it is easier to use the inverse Lorentz transformations. By the Relativity principle we have: t = t + vx=c2 ; (1.29) so

h i tB tA = tB tA + cv2 (xB xA ) : (1.30) But the stereo is stationary relative to O so xA = xB , so we end up with: t = tB tA = t = t = q t v2 : 1 c2

(1.31)

Since t > t, time in the spaceship slows down! That is, when the clock in theqspaceship records 1 second elapsed, as seen be the man in the ship, it shows 1= 1 v2=c2 seconds to the man outside. The slowing of clocks in a moving system is a real eect and it applies equally to all kinds of time for example biological, chemical reaction rates, even to the rate of growth of a cancer in a cancer patient! How is this so you may ask? If the rate of growth of the cancer was the same for a stationary patient as for a moving one, it would be possible to use the rate of cancer development to determine the speed of the ship! A very interesting example of the slowing down of time is that of the cosmic ray muons. These are particles that disintegrate spontaneously after an average


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Frame O x

x

A

B

Frame O

Figure 1.12: Length contraction lifetime of about 2:2 10 6 seconds. It is clear that in its short lifetime a muon cannot, even at the speed of light, travel more than 600 m. But although the muons are created at the top of the atmosphere, some 10 km up, we can detect them down here on earth. How can that be!!? From the muon point of view (i.e. their frame of reference) they only live about 2 s. However from our point of view they live considerably indeed long enough to reach the surface of the earth q longer, (by a factor of 1= 1 v2=c2).

1.7.2 Length contraction Now suppose the observer O wants to measure the length of the spaceship. He can only do this by making an instantaneous measurement of the spatial coordinates of the end of the ship i.e. xA and xB . The Lorentz transformations give you:

x1 = (x1 vt1) x2 = (x2 vt2) ;

(1.32)

so the length is

L = x2 x1 = 1 (x2 x1) = 1L ;

(1.33)


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since t1 = t2 [ instantaneous measurement by O ]. Writing this result in terms of v and c we have: q (1.34) L = 1 v2=c2 L : Note that this is not a physical eect on the rod but an eect of spacetime itself.

1.7.3 The twin paradox To continue our discussion of the Lorentz transformations and relativistic eects, we consider the famous \twin paradox" of Peter and Paul. When they are old enough to drive a spaceship, Paul ies away from earth at very high speed. Because Peter who is left on the ground, sees Paul going so fast, all of Pauls' clocks appear to go slower , from Peter's point of view. Of course, Paul notices nothing unusual. After a while he returns and nds that he is younger that Peter! Now some people might say \heh, heh, heh, from the point of view of Paul, can't we say that Peter was moving and should therefore appear to age more slowly? By symmetry, the only possible result is that they are both the same age when they meet". But in order for them to come back together and make a comparison, Paul must turn around which involves decelerating and accelerating, and during that period he is not in an inertial frame. This breaks the apparent symmetry and so resolves the paradox.

1.8 Velocity composition law Suppose a particle has a velocity u = (u1; u2; u3) in O, then at time t its spacetime coordinates are x = (ct; u1t; u2t; u3t). In a frame O moving at a speed v relative to O , its spacetime coordinates will be:

h

x = c t

vu1 c2

; (u1t vt) ; u2t ; u3t

i

(1.35)

The velocity in O simply comes from dividing the space parts by the time part i.e. u1 = ddxt1 = 1u1 vuv1 ; c2

CHAPTER 1. THE SPECIAL THEORY OF RELATIVITY u2 = = ddxt2 = u2 vu1 ;

1 c2 u3 = ddxt3 = u3 vu1 :

1 c2

17

(1.36)

If u1 = u and u2 = u3 = 0 then

u = 1u uvv : c2

(1.37)

This looks like the Galelian relative velocity if juj ; jvj c. If u = c, u = c, so we con rm the universality of the the speed of light.

Chapter 2 Vectors in Special Relativity 2.1 Four - vectors A four - vector is a quantity with four components which changes like spacetime coordinates under a coordinate transformation. We will write the displacement four - vector as: x !O (ct ; x ; y ; z) = f g ; = 0 ; 1 ; 2 ; 3 :

(2.1)

x is a frame independent vector joining near by points in spacetime. !O means: has components in frame O. x are the coordinates themselves [ which are coordinate dependent ]. In frame O , the coordinates are fx g so: x !O fx g

= 0 ; 1 ; 2 ; 3 :

(2.2)

The Lorentz transformations can be written as x

3 X = x ; =0

(2.3)

where is the [4 4] Lorentz transformation matrix

0 v 0 0 1 BB v c 0 0 CC c =B @ 0 0 1 0 CA 0

0 0 1

where = (1 v2 =c2) 1=2 and x , x can be regarded as column vectors. 18

(2.4)

CHAPTER 2. VECTORS IN SPECIAL RELATIVITY

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The positioning of the indices is explained later but indicates that we can use the Einstein summation convention: sum over repeated indices, if one index is up and one index down. Thus we can write: x = x :

(2.5)

is a dummy index which can be replaced by any other index; is a free index, so the above equation is equivalent to four equations. For a general four - vector (2.6) A !O A0 ; A1 ; A2 ; A3 ; we can write

A = A :

(2.7)

If A and B are two four - vectors, clearly C = A + B and D = A are also four - vectors with obvious components

C !O A0 + B 0 ; A1 + B 1 ; A2 + B 2 ; A3 + B 3 ; and

D !O A0 ; A1 ; A2 ; A3 :

(2.8) (2.9)

In any frame O we can de ne a set of four - basis vectors:

and

e0 !O (1 ; 0 ; 0 ; 0) ;

(2.10)

e1 !O (0 ; 1 ; 0 ; 0) ; e2 !O (0 ; 0 ; 1 ; 0) ;

(2.11) (2.12)

e3 !O (0 ; 0 ; 0 ; 1) :

(2.13)

In general we can write (e ) = = 1; = = 0 ; 6= ;

(2.14)


20

where labels the basis vector and labels the coordinate. Any four - vector can be expressed as a sum of four - vectors parallel to the basis vectors i.e.

A = Ae = A e :

(2.15)

The last equality re ects the fact that four - vectors are frame independent. Writing: (2.16) A e = A e = A e = A e ; where we have exchanged the dummy indices and and and , we see that this equals A e for all A if and only if

e = e :

(2.17)

This gives the transformation law for basis vectors:

e0 e1 e2 e3

= e0 =

v e c 1

;

v e + e 1 c 0

;

= e2 ; = e3 :

(2.18)

Note that the basis transformation law is dierent from the transformation law for the components since A takes one from frame O to O. So in summary for vector basis and vector components we have:

e = e ; A = A : Since O has a velocity v relative to O , we have:

e = ( v)e ) e = (v)e = (v) ( v)e ) (v) ( v) = :

(2.19)

(2.20)


21

So ( v) is the inverse of (v). Likewise

A = ( v)A = ( v) (v)A = A :

(2.21)

It follows that the Lorentz transformations with v gives the components of a four - vector in O from those in O . The magnitude of a four - vector is de ned as A2 = A A:

A2 = A0 + A1 + A2 + A3 ; 2

2

2

2

(2.22)

in analogy with the line element

ds2 = c2 dt2 + dx2 + dy2 + dz2 :

(2.23)

The sign on the A0 will be explained later. This is a frame invariant scalar. A is spacelike if A2 > 0, timelike if A2 < 0 and null if A2 = 0. The scalar product of two four - vectors A and B is:

Since

A B = A0 B 0 + A1 B 1 + A2B 2 + A3B 3 :

(2.24)

(A + B) (A + B) = A2 + B2 + 2A B ;

(2.25)

A B is frame independent. A and B are orthogonal if A B = 0; they are not necessarily perpendicular in the spacetime diagram [ for example a null vector is orthogonal to itself ], but must make equal angles with the 45o line. Basis vectors form an orthonormal tetrad since they are orthogonal: e e = 0 if 6= , normalized to unit magnitude: e e = 1:

0 1 0 0 01 C B e e = BB@ 00 01 10 00 CCA 0 0 0 1 = :

We will see later what the geometrical signi cance of is.

(2.26)


22

ct U

α

x

Figure 2.1: The worldline of a particle with four - velocity U .

2.2 Four - velocity, momentum and acceleration We must express Newton's laws of motion in terms of four - vectors so that they are frame invariant and consistent with Special Relativity. The four - velocity of a particle is the tangent to its worldline of length c [ see Figure 2.1 ]: (2.27) U = ddx ; U U = c2 ; where is the proper time [ the time measured in the particle's own rest frame ]. This is the most natural analogue of the three - velocity. It is clearly a four - vector since both dx and d are invariant. In the particle's own rest frame O , the four - velocity is

U = c e0 = (c ; 0 ; 0 ; 0) :

(2.28)

It follows that in a general frame O:

U = (c e0) = c0 = (c ; v ; 0 ; 0) ;

(2.29)

or

U = (c ; v ) ; (2.30) where v is the particle's three - velocity. For low velocities v c, U 0 c and the spatial part is nearly the same as v. If an observers own four - velocity is written as W = (c ; 0 ; 0 ; 0) in O we have

= U c2W :

(2.31)

CHAPTER 2. VECTORS IN SPECIAL RELATIVITY We can then write the particle's three - velocity as [ EXERCISE 2.2 ]: v = (W W ) U W :

UW

23

(2.32)

This is a spacelike vector expressed in coordinate - independent form [ although W singles out a particular observer ]. For photons the four - velocity is not de ned since dx dx = 0, i.e. there is no frame in which a photon is at rest. The four - momentum is de ned by

P = m0 U ; P P = m20 c2 :

(2.33)

where m0 is the rest mass of the particle i.e. the mass in its own rest frame. The spatial part of P [ the three - momentum ] is m0 v so the apparent mass exceeds the rest mass. m = m0 = q m0 2 2 : (2.34) 1 v =c The time part of the four - momentum is the energy of the particle E divided by c: (2.35) P 0 = Ec = 1c P W = m0 c = mc ; so we have E = m0 c2 = mc2 : (2.36) For v c E = m0 c2(1 v2=c2) 1=2 m0 c2 + 12 m0 v2 : (2.37) Since the second term is the kinetic energy, we interpret the rst term as the rest - mass energy of the particle. In general P P = m20 c2 implies that

E 2 = m20 c4 + c2p p ; (2.38) where p is the particle's three - momentum. Since m ! 1 as v ! c, requiring in nite energy, we infer that the particles with non - zero rest mass, can never reach the speed of light. For photons traveling in the x - direction E = cp1 , P:P = 0 , m0 = 0 ; (2.39) hence photons have zero rest mass.


24

2.2.1 Relativistic Doppler shift Consider a photon moving at an angle with respect to the x - axis. Its three velocity is (c cos ; c sin ; 0), so its four - momentum is

! h h h (2.40) P = c ; c cos ; c sin ; 0 ; where h is Planck's constant and the frequency. In a frame O with three - velocity (v; 0; 0) relative to O the frequency is . Using the Lorentz transformations we get P 0 = 0 P = 00P 0 + 01 P 1 ;

therefore

" # h = h v h cos ; c c c2 so we get the following result: = 1q v=c cos : 1 v2 =c2

(2.41) (2.42) (2.43)

If = 0, so that the photon moves in the same direction as O , we have

v u = u t 1 v=c : 1 + v=c

(2.44)

For low velocities v c this reduces to = 1 v ; (2.45) c This is the usual Doppler shift = vc , modi ed at large v. If = =2, so the photon moves perpendicular to O, we have = : (2.46) This is the transverse Doppler shift and is a consequence of time dilation.


25

2.2.2 Four - acceleration The four - acceleration of a particle is de ned as dU = dt d [ (c; v)] d d dt d (c; v) : = dt since ! d = d 1 v2 1=2 dt dt c2 3 = c2 v dv dt 3 = c2 v ddtv ; we have 4 A = 2 (0; v_ ) + c2 v v_ (c; v) :

(2.47)

(2.48) (2.49)

For speeds much less than the speed of light v c, we obtain v v_ (2.50) A = c ; v_ ; therefore spatial part approximates the usual three - acceleration at low velocities. Note that U U = c2 ) U ddU = 0, so the four - acceleration is orthogonal to the four - velocity. Finally Newton's second law requires that we de ne the four - force as

F = m0 A :

(2.51)

2.3 Relativistic dynamics For interacting particles we postulate that

X i

Pi

(2.52)

is conserved [ where i labels the particle ] since this is the natural analogue of Newton's law.

CHAPTER 2. VECTORS IN SPECIAL RELATIVITY u1

26 u2

Before U After

Figure 2.2: The colliding particles in Example 1. Conservation of the time component corresponds to energy conservation [ with rest mass being included in this ]. Conservation of spatial components corresponds to conservation of three - momentum. Note that Pi Pi is evaluated at a particular time by dierent observers and they will see dierent events as simultaneous. However Pi Pi will be conserved in all frames. The center of momentum frame for a particular system of particles is the one in which Etotal X P i = c ; 0; 0; 0 : (2.53) i

2.3.1 Example 2.1 As a simple example, consider two particles with rest masses m1 and m2 , moving with speeds u1 and u2 (u1 > u2), which collide and coalesce [ see Figure 2.2 ]. What is the mass M and speed U of the resulting particles. The initial four - momenta are Pi = mi i (c ; ui ; 0 ; 0) ; (2.54) where i = 1 : : : 2 and The nal four - momentum is

2 ! 1=2 u

i = 1 c2i :

(2.55)

P = M (c ; U ; 0 ; 0) :

(2.56)

Conservation of four - momentum gives

P = P1 + P2 ;

(2.57)


27

and squaring this relation gives Now and This gives

P2 = P21 + P22 + 2P1 P2 :

(2.58)

P2i = m2i c2 ; P2 = M 2 c2 ;

(2.59)

P1 P2 = m1 m2 1 2 u1u2 c2 : s

(2.60)

u u 1 2 M (2.61) c2 : For u1 c and u2 c, M m1 + m2 which is the usual Newtonian result. From the x and t components of the four - momentum equation we have

= m21 + m2 + 2m1m2 1 2 1

M U = m1 1u1 + m2 2u2 ; M c = m1 1c + m2 2c : Dividing we get

+ m2 2u2 : U = m1m 1u 1 + 1 1 m2 2 For u1 c and u2 c we obtain m2 u2 : U = m1mu1 + 1 + m2

(2.62) (2.63) (2.64)

2.3.2 Example 2.2 Consider a collision between a photon with frequency moving in the x direction, with an electron of rest mass me initially at rest [ see Figure 2.3 ]. Before the collision, the four - momenta of the photon and electron are ! h h Q1 = c ; c ; 0; 0 ; P1 = me (c; 0; 0; 0) : (2.65) After the collision we have

! h h h Q2 = c ; c cos ; c sin ; 0 ; P2 = me (c; v sin ; v cos ; 0) ;

(2.66)


28 Q

2

Θ Q

P1

1

Ψ

Before

After P2

Figure 2.3: Sciamatic diagram illustrating Compton scattering. where is the photon's frequency after the collision, v is the speed of the electron and , are shown in Figure 2.3. Conservation of four - momentum gives

P1 + Q1 = P2 + Q2 :

(2.67)

Rearranging and squaring we get

P22 = (P1 + Q1 Q2)2 :

(2.68)

Using P21 = P22 = m2e c2 and Q21 = Q22 = 0 we obtain

P1 Q1 P1 Q2 Q1 Q2 = 0 ; which gives

(2.69)

2 h me h + me h = c2 (cos 1) : (2.70) Finally after simplifying this expression we arrive at the famous Compton scattering formula ! m e c2 1 1 2 = : (2.71) sin 2 2h

Chapter 3 Tensors in Special Relativity We need to express the equations of physics in a frame independent way. This can be done for some equations by using four - vectors [ e.g. Newton's Laws ] but what about the rest of physics!!? We now extend this idea by introducing more general coordinate independent quantities called Tensors. In this section we will con ne our discussion to Minkowski spacetime, although the concepts will apply to curved spacetime and General Relativity.

3.1 Metrics and forms Recall that the scalar product of two four - vectors is A B:

so where

A B = (A e ) B e = A B e e ;

(3.1)

A B = A B ;

(3.2)

0 1 0 0 01 B 0 1 0 0 CC = B B@ 0 0 1 0 CA

(3.3)

0 0 0 1 are the components of the metric tensor. We can think of the metric tensor as a map which takes two four - vectors A and B into the reals: g(A; B) = A B. 29

CHAPTER 3: TENSORS IN SPECIAL RELATIVITY

30

The map is linear in the arguments, in the sense that

g(A + B; C) = g(A; C) + g(B; C) :

(3.4)

We can generalize this by de ning a tensor of type 0=N as a map which takes N four - vectors into the reals which is linear in all its arguments, for example the metric tensor is a type 0=2 tensor.

3.2 One - forms These are tensors of type 0=1 and they map four - vectors into the reals. They are denoted by p~, so p~(A) is a real number. One - forms form a vector space since (p~ + q~) A = p~(A) + q~(A) ; (p~) A = p~(A) :

(3.5)

It is called the dual vector space to distinguish it from the space of four - vectors. The components of p~ are p = p~(e). We can write

p~ !O fp g :

(3.6)

Generally we have

p~(A) = p~(A e) = A p~(e ) = A p = A0 p0 + A1 p1 + A2p2 + A3p3 :

(3.7)

Note that the signs are positive [ c.f A B = A0 B 0 + A1B 1 + A2B 2 + A3B 3 ]. We can think of vectors as columns and one - forms as rows: 0 A0 1 B 1C (3.8) p~(A) = (p0 p1 p2 p3 ) BB@ AA2 CCA : A3

p~(A) is more fundamental than A B since the latter is de ned only if there is a metric i.e. A B = A B . Now let us look at how one - forms transform:

p = p~(e ) = p~ e = p :

(3.9)


31

A ~ p

Figure 3.1: p~(A) is the number of surfaces the vector A pierces. So one - forms transform like basis vectors, not like vector components. Now since

A p = A ( p)

we have

= A p ;

(3.10)

A p = A p ;

(3.11)

so p~(A) is frame independent. We can introduce a one - form basis fw~ g, so that Then

p~ = pw~ :

(3.12)

p~(A) = p w~ (A) = p w~ (A e ) = pA w~ (e ) = pA ;

(3.13)

so we must have

w~ e = :

This gives the basis for one - forms. It is said to be dual to fe g. One can show that w~ = w~ ;

(3.14)

(3.15)

so the basis one - forms transform like vector components [ as required notationally ]. Both vectors and one - forms have four components but they have dierent geometrical interpretation. Vectors are like arrows but one - forms can be thought


32

ct U

α

φ=φ( ct,x,y,x)

x

Figure 3.2: World line of particle with four - velocity U. of as like three dimensional surfaces with the spacing between the surfaces de ning the magnitude of p~ [ see Figure 3.1 ].

3.2.1 Gradients For a scalar eld (x) and a world line of some particle x( ), we have

( ) = (ct( ); x( ); y( ); z( )) :

(3.16)

If U is the tangent to the curve [ the four - velocity of the particle, see Figure 3.2 ] then: d = c @ dt + @ dx + @ dy + @ dz (3.17) d @ct d @x d @y d @z d so d = @ U 0 + @ U 1 + @ U 2 + @ U 3 ; (3.18) d @ct @x @y @z since U = ddx [ see section 2.2 ]. This de nes a one - form since it maps U into real numbers and represents the rate of change of along a curve with tangent U. ! @ @ @ @ ~ !O (3.19) d @ct ; @x ; @y ; @z : In three dimensions one thinks of a gradient as a vector [ normal to surfaces of constant ] but d~ is a one - form and speci es a vector only if there is a metric. Now how do the components of d~ transform?

~ d = d~ :

(3.20)

CHAPTER 3: TENSORS IN SPECIAL RELATIVITY But we also have by the chain rule: @ = @ @x ; @x @x @x which means that ~ @x ~ d = @x d ;

so

and since x ; = we have

@x ; = @x

@ d~x : d~x = w~ ; d~ = @x This is a useful result, that the basis one - form is just d~x .

33

(3.21) (3.22) (3.23) (3.24)

3.2.2 The metric as a mapping of vectors onto one - forms We now introduce the idea that the metric acts as a mapping between vectors and one - forms. To see how this works, consider g and a single vector V. Since the metric g requires two vectorial arguments, the expression g (V; ) still lacks one: when another one is supplied, it becomes a number. Therefore g (V; ) considered is a linear function of vectors producing real numbers: a one - form. We call it V~ ( ). So we have g (V; ) = V~ ( ) : (3.25) Then V~ is a one - form whose value on a vector A is V A:

V~ (A) = g (V; A) = V A : ~ . They are What are the components of V

(3.26)

V = V~ (e) = V e = e V = e V e = V (e e ) = V :

(3.27)

CHAPTER 3: TENSORS IN SPECIAL RELATIVITY Thus

34

V0 = V 0 = V 000 = V 0

(3.28)

V1 = V 1 = V 111 = V1 : This can be summarized as follows: If

(3.29)

and

then

V !O (a; b; c; d) ;

(3.30)

V~ !O ( a; b; c; d) :

(3.31)

A = A :

(3.32)

~ are obtained from those of V by changing the sign of the The components of V time component. Since det = 1 [ i.e. non zero ], there exists an inverse metric which we can write as [ In Special Relativity the components of are the same as , but this will not be true in the curved spacetime of General Relativity ]. The inverse metric de nes a map from one - forms to vectors

In particular, we can map the gradient one - form d~ into a vector gradient: ! @ @ @ @ d !O @ct ; @x ; @y ; @z : (3.33) We can regard a vector as a 1=0 tensor i.e. a map from one - forms into the reals, so A (p~) = Ae p w~ = Ap ew~ = A p = p~ (A) (3.34) and

A = A (w~ ) : (3.35) The inverse metric can be used to de ne the magnitude of a one - form:

p~2 = pp = p20 + p21 + p22 + p23 ;

(3.36)

and the scalar product of two one - forms:

p~ q~ = pq = p0 q0 + p1q1 + p2 q2 + p3 q3 :

(3.37)


35

These are identical to the corresponding quantities for vectors, i.e.

p~2 = p2 ; p~ q~ = p q :

(3.38)

Thus one - forms are timelike/spacelike/null if the corresponding vectors are.

3.3 More general tensors 3.3.1 Tensors of type 0=2

The product of two one - forms, written as p~ q~ de nes a linear map which takes two vectors into the reals:

p~ q~ (A; B) = p~ (A) q~ (B) :

(3.39)

It is therefore a 0=2 tensor. denotes the outer product. It is the formal notation to show how the 0=2 tensor is formed from two one - forms. Note that this product is non - commutative since p~(B)q~(A) gives a dierent result [ EXERCISE 3.4 ] i.e. p~ q~ 6= q~ p~ : (3.40) The most general 0=2 tensor is a linear sum of such outer products. So

f (A; B) = f (Ae ; B e ) = A B f (ee ) = A B f ;

(3.41)

where f are the components of the map f and we have used linearity. If we take a basis for f as w~ [ 16 components ], then

f = f w~ :

(3.42)

But

f = f (e; e ) = f w~ (e ; e )

) w~ (e; e ) = ;

(3.43)

CHAPTER 3: TENSORS IN SPECIAL RELATIVITY so we have

w~ = w~ w~ :

36 (3.44)

Under a Lorentz transformation, the components of f become:

f = f :

(3.45)

A 0=2 tensor is symmetric if

f (A; B) = f (B; A) , f = f ;

(3.46)

f (A; B) = f (B; A) , f = f :

(3.47)

and anti - symmetric if

It follows that any 0=2 tensor can be uniquely decomposed into a symmetric and anti - symmetric part. f = f( ) + f[ ] ; (3.48) with the symmetric and anti - symmetric parts given by

and

f( ) = 21 (f + f )

(3.49)

f[ ] = 12 (f f ) :

(3.50)

3.3.2 Tensors of type 0=N We de ne a tensor of type 0=N as a linear map from N vectors into the reals:

T = T :::w~ w~ : : : ;

(3.51)

and it's components transform according to

T ::: = : : : T:::

(3.52)

It can be split into symmetric and anti - symmetric parts on any two indices.


37

3.3.3 Tensors of type 2=0 Recall that a vector can be regarded as a 1=0 tensor [ i.e. a map of one - forms into the reals ]. Therefore the outer product of two vectors de nes a 2=0 tensor, i.e. it is a linear map from two one - forms into the reals:

V W(p~; q~) = V(p~)W(q~) :

(3.53)

It follows that the most general 2=0 tensor is a linear sum of such outer products:

f (p~; q~) = p q f ; where

f = f (w~ ; w~ ) ;

f = f e e :

(3.54) (3.55)

Under a Lorentz transformation, the components of f are:

f = f :

(3.56)

As before we can separate f into symmetric and anti - symmetric parts:

and

f ( ) = 12 f + f ;

(3.57)

f [ ] = 21 f f :

(3.58)

3.3.4 Tensors of type M=0 We de ne a tensor of type M=0 as a linear map that takes M one - forms into the reals. So we can write T = T :::e e : : : ; (3.59) whose components transform according to

T ::: = : : : T ::: :

(3.60)


38

3.3.5 Tensors of type M=N Finally we de ne a tensor of type M=N as a linear function that maps M one - forms and N vectors into the reals:

T = T ::: :::e : : : w~ : : : :

(3.61)

The components of T transform according to

T ::: ::: = : : : : : : T :::::: :

(3.62)

These results just follow from linearity.

3.3.6 Index \raising" and \lowering" In the same way that the metric maps a vector V into a one - form, it maps a M=N tensor into a M 1=N + 1 tensor, i.e. it lowers an index. Similarly the inverse metric maps a M=N tensor into a M + 1=N 1 tensor, i.e. it raises an index. So for example

T = T ; T = T ; T = T :

(3.63)

In Special Relativity, raising or lowering a 0 component changes the sign of the component; raising or lowering 1, 2, or 3 components has no eect. We can operate the inverse metric on the metric to get the Kroneker delta [ EXERCISE 3.5 ]: = = : (3.64) So far we have con ned our attention to Lorentz frames [ i.e. inertial frames ]. We can also allow more general coordinate transformations in a more general space i.e. x = x (x ). We then de ne

= @x : @x

(3.65)


39

Tensors will then transform as before, for example

A::: ::: = : : : A:::::: :

(3.66)

Old fashioned texts regard the above as the de nition of a tensor. Raised indices are called contravariant because they transform \contrary" to basis vectors: A = @x (3.67) @x A ; while lowered indices are called covariant:

@x : p = @x

(3.68)

In particular one - forms are sometimes called covariant vectors, while ordinary vectors are called contravariant vectors.

3.4 General properties of tensors Tensors have the following general properties:

If A::: ::: and B ::: ::: are M=N tensors, so are A::: :::, B ::: ::: and A::: ::: + B ::: ::: for any and . Thus tensors of a given type form a vector space.

If A::: ::: is a M=N tensor and B :::::: is of type M 0 =N 0, then the outer product A::: :::B :::::: is a tensor of type M + M 0 =N + N 0 .

For any tensor of type M=N , one can construct a tensor of type M 1=N

1 by contracting an upper index with a lower index for example A is constructed from A [ ! or ! ]. Note however that there may be several ways of construction and they give dierent tensors.

Tensors can be symmetric on pairs of upper or lower indices for example

T ::: ::: = T ::: and T = T ::: , or anti - symmetric. However symmetric or anti - symmetric pairs of indices with one index up and the other down cannot be de ned since the relationship need not hold in all frames.


40

3.5 Tensor derivatives and gradients The partial derivative of a tensor is not generally itself a tensor although it is for a scalar: @x@ is a 0=1 tensor. For a vector A, we have ! @A = @ @x A @x @x @x @x @A 2 x @x @x @ (3.69) = + A : @x @x @x @x @x @x It follows that @A @x transforms like a 1=1 tensor only if the second term is zero, i.e.

@ 2 x @x = 0 @x @x @x ) x = B x + C ;

(3.70)

where B and C are constants. This is true for at space with coordinates (ct; x; y; z) under the Poincare transformations and in this case B may be interpreted as a 4D rotation matrix.

Note however that it is not true for at space with a general coordinate system or in the curved spacetime of General Relativity. We can de ne the derivative of a general M=N tensor T along a curve parameterized by the proper time as follows: # " dT = lim T ( + ) T ( ) : (3.71) d !0 If the basis vectors and one - forms are the same everywhere then: ! dT = dT ::: e : : : w~ : : : (3.72) d d where dT ::: ::: = @T ::: ::: U ; (3.73) d @x and U is the tangent to the curve. Thus dT d is also like a M=N tensor, written as

rUT !O fT ::: :::; U g ;

(3.74)


41

where ; means @=@x . We can then de ne a M=N + 1 tensor rT:

rT T ::: :::; e : : : w~ : : : w~ : This is the tensor gradient [ remember the gradient of a scalar d~ ].

(3.75)

Chapter 4 Application of Tensors in Special Relativity 4.1 The energy - momentum tensor Consider a pressure - less distribution of non - interacting particles [ called dust ], with rest mass m and number density n in the momentarily comoving reference frame [ MCRF ] . The density in this frame is = nm : (4.1) In a general frame the number density will go up by a factor

and so

n ! n

(4.2)

m ! m ;

(4.3)

! 1

v2 c2

1

:

(4.4)

Thus the density is not a component of a four - vector. We will see that it is a component of a 2=0 tensor. We can introduce a number ux four - vector N:

N = nU !O (nc; nv) = (nc; nvx; nvy ; nvz ) ; 42

(4.5)

CHAPTER 4: APPLICATION OF TENSORS IN SPECIAL RELATIVITY 43 where nvx is the ux per unit area across a surface with normals in the x direction etc, and n can be interpreted as the ux across a constant t surface. Thus N combines the ux and the number density in a single four - dimensional quantity. Note that N N = n2 c2 : (4.6) The most convenient de nition of the energy - momentum tensor is in terms of its components in some arbitrary frame.

T d~x ; d~x = T ;

(4.7)

where T is the ux of - momentum across a surface of constant x . By momentum we mean the component of the four - momentum p. Let us see how this de nition ts in with the discussion above. Consider rst T 00. This is de ned as the ux of 0 - momentum [ energy divided by c ] across a surface of constant t. This is just the energy density.

T 00 = energy density :

(4.8)

Similarly, T 0i is the ux of energy divided by c across a surface of constant xi :

T 0i = (energy flux=c) across a surface xi = const :

(4.9)

Then T i0 is the ux of i - momentum across a surface of constant t: the density of i - momentum multiplied by c:

T i0 = i momentum density c :

(4.10)

Finally T ij is the j - ux of i - momentum:

T ij = flux of i momentum across a surface xj :

(4.11)

For any particular system, giving the components of T in some frame, de nes it completely. For dust, the components of T in the MCRF are particularly simple. There is no motion of the particles, so all i - momenta are zero and all spatial uxes are

CHAPTER 4: APPLICATION OF TENSORS IN SPECIAL RELATIVITY 44 zero. Therefore:

00 T MCRF = c2 = mnc2 ; 0i T MCRF = T i0 MCRF = T ij MCRF = 0 :

(4.12)

It is easy to see that the tensor P N has exactly these components in the MCRF, where P = mU is the four - momentum of a particle. It follows that, for dust we have T = P N = mnU U = U U : (4.13) From this we conclude that the components of T are:

T = U U ; or in matrix form:

0 c2 0 0 0 1 B 0 0 0 0 CC T = B B@ 0 0 0 0 CA : 0

(4.14)

(4.15)

0 0 0

In a frame O with U !O (c; v), we therefore have

T 00 = U 0 U 0 = 2c2 ; T 0i = U 0 U i = 2 cvi ; T i0 = U i U 0 = 2 cvi ; T ij = U i U j = 2 vivj :

(4.16)

These are exactly what we would calculate from rst principles, for the energy density, energy ux, momentum density and momentum ux respectively. Notice one important property of T: it is symmetric:

T = T : This will turn out to be true in general, not just for dust.

(4.17)

CHAPTER 4: APPLICATION OF TENSORS IN SPECIAL RELATIVITY 45 x

a 3 4

2

1 y a

Figure 4.1: Energy ow across a uid element.

4.1.1 General uids A general uid has pressure p [ random velocities ] and viscosity [ interaction between particles ]. T has the same interpretion as for dust except that T 0i and T ij need no longer be zero in the MCRF. In particular T 0i represents heat conduction and T ij represents the stress forces between adjacent uid elements [ i'th component of force per area across a surface of constant xj ].

4.1.2 Conservation of energy - momentum Since T represents the energy and momentum content of the uid, there must be some way of using it to express the law of conservation of energy and momentum. In fact it is reasonably easy. Consider a cubical uid element [ see Figure 4.1 ] of side a, seen only in cross section below [ z direction suppressed ]. Energy can ow across all sides. The rate of ow across face (4) is ca2T 0x(x = 0), and across (2) is ca2 T 0x(x = a); the second term has a minus sign because T 0x represents energy owing in the positive x - direction, which is out of the volume across face (2). Similarly, the

CHAPTER 4: APPLICATION OF TENSORS IN SPECIAL RELATIVITY 46 energy owing in the y direction is a2 cT 0y (y = 0) a2cT 0y (y = a). The sum of these rates across each face must be equal to the rate of increase of energy inside the cube @=@t(a3 T 00): This is the statement of conservation of energy. Therefore we have: @ (a3 T 00) = a2 c hT 0x(x = 0) T 0x(x = a) + T 0y (y = 0) T 0y (y = a) @t i + T 0z (z = 0) T 0z (z = a) : (4.18) Dividing by a3 and taking the limit a ! 0 gives

# " @ T 00 = c @ T 0x + @ T 0y + @ T 0z : @t @x @y @z Dividing by c we get @ T 00 + @ T 0x + @ T 0y + @ T 0z : @ (ct) @x @y @z Since x0 = ct, x1 = x, x2 = y and x3 = z, we can write this as or

(4.19)

(4.20)

T 00;0 + T 01 ;1 + T 02 ;2 + T 03;3 = 0 ;

(4.21)

T 0; = 0 :

(4.22)

This is the statement of the law of conservation of energy. Similarly momentum is conserved. The same mathematics applies. with the index 0 changed to i [ the spatial components ] i.e.

T i ; = 0 :

(4.23)

The general conservation law of energy and momentum is therefore

T ; = 0 : This applies to any material in Special Relativity.

(4.24)

CHAPTER 4: APPLICATION OF TENSORS IN SPECIAL RELATIVITY 47

4.1.3 Conservation of particles It may also happen, that in a uid ow, the total number of particles does not change. This conservation law is derivable in the same way as the conservation of energy and momentum [ EXERCISE 4.1 ]: @ N0 + @ Nx + @ Ny + @ Nz = 0 ; (4.25) @ (ct) @x @y @z or N ; = 0 , (nU ); : (4.26) We will con ne ourselves to discussing uids which obey this conservation law. This is hardly any restriction since we can always take n to be the number density of baryons.

4.1.4 Perfect uids Finally we come to the type of uid which will be of most interest to us. A perfect uid in Relativity is de ned to be a uid that has no heat conduction in the MCRF. It is the generalization of the ideal gas in ordinary thermodynamics. In the MCRF the components of T are 0 c2 0 0 0 1 B 0 p 0 0 CC (4.27) T = B B@ 0 0 p 0 CA : 0 0 0 p It is easy to show that in a general frame we have [ EXERCISE 4.2 ]: p (4.28) T = + c2 U U + p : This is the energy - momentum tensor of a perfect uid.

4.1.5 The conservation equations We can now apply the conservation laws T ; = 0 and N ; = 0 to nd the conservation equations. We have: p T ; = + c2 U U + p ; = 0 ;

CHAPTER 4: APPLICATION OF TENSORS IN SPECIAL RELATIVITY 48 1 p = n + c2 nU U + p ; + p; = 0 ; 1 p (4.29) = nU n + c2 U + p; = 0 : ; If we multiply by U and use U = U ; and we get which gives

U U = c2 ;

1 @ (U U ) = 0 , U U = 0 ; ; 2 @x 1 + p n U U = 0 ; ; n c2 ;

! d = + p=c2 dn ; d n d where dd U @x@ is the derivative along the world line of the uid. The i components of T ; = 0 give p 2! + p=c i i i U ; U + c2 p; + nU U =0; n ;

(4.30) (4.31) (4.32) (4.33)

(4.34)

so in the MCRF [ U i = 0 ] this equation is dU i + p + i r p = 0 : (4.35) d c2 Equation (4.33) is the energy conservation equation and equation (4.35) is the momentum conservation equation [ or acceleration equation ]. In the Newtonian limit v c and p c2 so in this case the conservations equations become [ EXERCISE 4.2 ] d = m dn ; (4.36) dt dt and # " @ v (4.37) dt + (v r) v = rp ; where we have used

U i ; U = @dtv + (v r) v :

(4.38)

CHAPTER 4: APPLICATION OF TENSORS IN SPECIAL RELATIVITY 49

4.2 The Electromagnetic tensor Maxwell's equations for the electromagnetic eld [ in units with 0 = 0 = c = 1 ] are: r B @@tE = 4j ; r E + @@tB = 0 ; r E = 4 ; r B = 0 : (4.39) De ning the anti - symmetric tensor F with components: 0 0 Ex Ey Ez 1 B Ex 0 Bz By C C B@ E y F = B A Bz 0 Bx C z y x E B B 0 the electric and magnetic elds are given by

(4.40)

E = (E x; E y ; E z ) = F 01 ; F 02; F 03 ; B = (B x; B y ; B z ) = F 23 ; F 31; F 12 :

(4.41) (4.42)

If we also de ne a current four - vector:

J = (; j x; j y ; j z ) ;

(4.43)

Maxwell's equations can be written as [ EXERCISE 4.3 ]

F ; = 4J ;

(4.44)

F; + F; + F; = 0 ;

(4.45)

where F = F . We have now expressed Maxwell's equations in tensor form as required by Special Relativity. The rst of these equations implies charge conservation

Proof:

J ; = 0 , @ @t + r:j = 0 : 4J ; = F ; = F ;

(4.46)

CHAPTER 4: APPLICATION OF TENSORS IN SPECIAL RELATIVITY 50 =

F ;

) F ; = 0 ) J ; = 0 :

(4.47)

By performing a Lorentz transformation to a frame moving with speed v in the x direction, one can calculate how the electric and magnetic elds change:

F = F :

(4.48)

We nd [ EXERCISE 4.3 ] that Ejj = E x is unchanged, while

E? = (E + v B) ;

(4.49)

where Ejj and E? is the electric eld parallel and perpendicular to x^ . Thus E and B get mixed. The four - force on a particle of charge q and velocity U in an electromagnetic eld is [ EXERCISE 4.4 ]:

K = qF U = q (E:v; E + v B) :

(4.50)

The spatial part of K is the Lorentz Force and the time part is the rate of work by this force. By writing J = qU, Maxwell's equations give [ EXERCISE 4.4 ]:

where

K = T ; ;

(4.51)

T = 41 F F 41 F F :

(4.52)

This is the energy momentum tensor of the electromagnetic eld. Note that T is symmetric as required and the energy density is [ EXERCISE 4.4 ]

T 00 = 81 (E2 + B2 ) :

(4.53)

Chapter 5 Conceptual Basis of General Relativity 5.1 The story so far We are at last ready to embark on our central task, namely, that of extending Special Relativity to a theory which incorporates gravitation. In this section we will consider the physical principles which guided Einstein in his search for the general theory. There is a school of thought that considers this an unnecessary process, but rather argues that it is sucient to rst state the theory and then investigate its consequences. There seems little doubt, however, that consideration of these physical principles helps gives insight into the theory and promotes understanding. The mere fact that they were important to Einstein would seem sucient to justify their inclusion. If nothing else, it will help us understand how one of the greatest achievements of the human mind came about. Many physical theories today start by specifying a Lagrangian from which everything ows and we could adopt the same attitude with General Relativity. Although this is a very beautiful way about going about things, in taking that approach we would miss out on gaining some understanding of how the framework of General Relativity is dierent from that of Newtonian theory and Special Relativity. Moreover if we discover limitations in the theory, then there is more chance of rescuing it by investigating the physical basis of the theory rather than simply tinkering with the mathematics - an unfortunate trait of much of modern 51

CHAPTER 5. CONCEPTUAL BASIS OF GR

52

theoretical physics these days! But before we embark on this exciting journey of discovery, we must rst remind ourselves of where we have got to. So far we have only discussed Special Relativity. Here forces have only played a background role and we have never introduced gravitation explicitly as a possible force. One aspect of Special Relativity is the existence of a global inertial frame, all of whose coordinate points are always at rest relative to the origin, and all of whose clocks run at the same rate relative to the origin's clock. From Einstein's postulates we were led to the idea of the spacetime interval ds2 which gave an invariant geometrical meaning to certain physical statements. We discovered that the mathematical function that calculates the spacetime interval is the metric, and so the metric of Special Relativity is de ned physically by lengths of rods and the readings of clocks. This closeness to experiment is of course its strength. Let us now ask the following question:

Is is possible to construct a frame in which all clocks run at the same rate? This is a crucial question and we will show that in a non - uniform gravitational eld the answer, experimentally , is NO.

So: Gravitational elds are incompatible with global Special Relativity i.e. it is impossible to construct a global inertial frame. We shall see that in small regions of spacetime [ regions small enough that non - uniformities of the gravitational eld are too small to measure] one can always construct a \Local Inertial Frame" [LIF]. In this sense we will have to build Special Relativity into a more general theory.

5.2 The gravitational redshift experiment This \gedanken" experiment was rst suggested by Einstein. Suppose we have a tower of height h on the surface of the earth. A particle of rest mass m is dropped from the top of the tower and falls freely with the acceleration g [ see Figure 5.1 ].

CHAPTER 5. CONCEPTUAL BASIS OF GR m

53 γ

Figure 5.1: The Gravitational redshift experiment.

p

It reaches the ground with a velocity v = 2gh, so its total energy E , as measured by an observer at the foot of the tower is E = mc2 + 12 mv2 + O(v4) = mc2 + mgh + O(v4) : (5.1) Suppose the observer has some magical method of converting all this energy into a photon of the same energy [ this is a thought experiment after all! ]. Upon its arrival at the top of the tower with energy E the photon is again magically 2. It must be that m m; otherwise, changed into a particle of rest mass m = E=c perpetual motion could result, so E = mc2 . We therefore obtain: E = mc2 gh + O(v4) ; = 1 (5.2) E mc2 + mgh + O(v4) c2 and since E = h and E = h we nd:

! gh = 1 c2 : (5.3) We therefore predict that a photon climbing in the earths gravitational eld will lose energy and will consequently be redshifted. the redshift is: Zg = = gh (5.4) c2 : This was tested by Pound and Snider in 1965 using the Mossbauer eect [ photons from atomic decay peak sharply at a particular frequency ]. They measured the redshift experienced by a 14:4 Kev rays from the decay of 57 Fe in climbing up


54

t

γ ∆t

∆t

t

b

Bottom

Top

x

Figure 5.2: Minkowski geometry for the Pound - Snider Experiment. a 20 m tower by determining the speed at which a detector at the top must be moved in order to maximize the detection rate i.e. the velocity blueshift balances the gravitational redshift. They found:

Zg = 2:57 0:26 10 15 :

(5.5)

This experimental veri cation of Einstein's thought experiment is a death - blow of one's chances of nding a simple special relativistic theory of gravity!

5.3 Non - existence of an inertial frame at rest on earth If Special Relativity is to be valid in a gravitational eld, it is a natural rst guess to assume that the \laboratory" frame at rest on earth is an inertial frame. Let us draw a spacetime diagram representing the above experiment [ see Figure 5.2 ]. We consider light as a wave, and look at two successive crests of the wave as they move upward in the gravitational eld. The top and bottom of the tower have vertical world lines in this diagram since they are at rest. Light is shown moving on a curved line, to allow for the possibility that gravity may act on on light in an unknown way, de ecting it from a null path. But no matter how light is aected by gravity the eect must be the same on both wave crests, since the gravitational eld is not time dependent. Therefore the two crests ' paths are congruent, and


55

we conclude from this hypothetical Minkowski geometry that tt = tb :

(5.6)

But we know that t = 1 , and since the Pound - Snider experiment tells us that b > T we know that tt > b . Therefore we have to conclude that our answer using Minkowski geometry is wrong!

So the reference frame at rest on earth is not inertial!! Is this the end of Special Relativity!!?..... Not quite. We have shown that a particular frame is not inertial, not that there are no inertial frames. We will nd that there are certain frames which are inertial in a restricted sense, but before we de ne these frames let us rst consider what we mean by the mass of a body.

5.4 Mass in Newtonian theory So far we have been rather vague about what we mean by the mass of a body. Even in Newtonian theory we can ascribe three masses to any body which describe quite dierent properties:

Inertial mass mI : This is a measure of its resistance to change in motion or inertia. Passive gravitational mass mP : This is a measure of its reaction to a gravitational eld. Active gravitational mass mA : This is a measure of its source strength for producing a gravitational eld. Let us discuss each of these in turn. Inertial mass mI is the quantity occurring in Newton's second law [ F = mI a ]. It is a measure of a body's inertia. Note that as far as Newtonian theory, this mass has nothing to do with gravitation. The other two masses however do.


56

Splat

After

Before

Figure 5.3: The Galileo Piza experiment. Passive gravitational mass mP measures a body's response to being placed in a gravitational eld. Let the gravitational potential at some point be , then if mP is placed at this point, it will experience a force on it given by

F = mP r :

(5.7)

On the other hand active gravitational mass mA measures the strength of the gravitational eld produced by the body itself. If mA is placed at the origin, then the gravitational potential at any point a distance r from the origin is given by A = Gm (5.8) r : We will now see how these three masses are related in the Newtonian framework. Galileo discovered in his famous Pisa experiments [ see Figure 5.3 ] that when two bodies are dropped from the same height, they reach the ground together irrespective of their internal composition. Let's assume that two particles of inertial mass mI1 and mI2 and passive gravitational mass mP1 and mP2 are dropped from the same height in a gravitational eld. We have:

mI1 a1 = F1 = mP1 r ; mI2 a2 = F2 = mP2 r :

(5.9)


57

The observational result is a1 = a2 from which we get on dividing

mI1 = mI2 : (5.10) mP1 mP2 Repeating this experiment with other bodies, we see that this ratio is equal to a universal constant say. By a suitable choice of units we can take = 1, from which we obtain the result:

inertial mass = passive gravitational mass. This equality is one of the best tested results in physics and has been veri ed to 1 part in 1012. In order to relate passive gravitational mass to active gravitational mass, we make use of the observation that nothing can be shielded from a gravitational eld. Consider two isolated bodies situated at points Q and R moving under their mutual gravitational attraction. The gravitational potential due to each body is A A 1 ; 2 = Gm2 : 1 = Gm r r The force which each body experiences is

F1 = mP1 rQ2

F2 = mP2 rR 1 :

If we taken the origin to be Q then the gradient operators are rR = ^r @r@ = rQ ; so that P A P A F1 = Gmr12m2 ^r ; F2 = Gmr22m1 ^r :

(5.11)

(5.12)

(5.13) (5.14)

But by Newton's third law F1 = F2 , and so we conclude that

mP1 = mP2 ; mA1 mA2 and using the same argument as before, we see that

Passive gravitational mass = active gravitational mass.

(5.15)


58

That is why in Newtonian theory we can simply refer to the mass m of a body where m = mI = mP = mA : (5.16) This may see obvious to you, but it has very deep signi cance and Einstein used it as the central pillar for his equivalence principle.

5.5 The principle of equivalence One important property of an inertial frame is that a particle stays at rest or moves with uniform velocity unless it is acted on by a force [ Statement of Newton's rst law ]. In the last section we discovered that bodies freely falling in a gravitational eld all accelerate at the same rate regardless of their internal composition. It follows that relative to a non - rotating freely falling frame, at least locally, particles remain at rest or move in straight lines with uniform velocity, since this frame accelerates at the same rate as particles do. By locally we mean that observations are con ned to a region over which the variation of the gravitational eld is un - observably small. This leads to the following statement of the principle of equivalence.

POE 1: There are no local experiments which can distinguish non rotating free fall in a gravitational eld from uniform motion in space in the absence of a gravitational eld. We conclude that a non - rotating freely falling frame is a local inertial frame. Let's check this by viewing the Pound - Snider experiment from the view point of a freely falling frame. Let us take the particular frame to be at rest when the photon begins its upward journey and falls freely after that. Since the photon rises a distance h, it takes a time t = hc to arrive at the top. In this time, the frame has acquired a velocity v = gt = ghc downward relative to the tower.


59

We can now use the redshift formula from section 2.2 to calculate the photon's frequency relative to the freely falling frame:

v u ff = u t 1 + gh=c2 1 + gh ; t 1 gh=c2 c2

(5.17)

so

! gh ff = t 1 + c2 ! ! gh gh (5.18) 1 + c2 1 c2 ; since we neglect terms of higher order [ as we did in section 5.2 ]. So there is no redshift in a freely falling frame thus con rming that it is a local inertial frame. Einstein noted one other coincidence in Newtonian theory which proved to be of great importance in formulating the principle of equivalence. All inertial forces are proportional to the mass of the body experiencing the force. There is one other force which behaves in the same way, that is the force of gravitation. For, if we drop two bodies in the earths gravitational eld, they experience forces m1 g and m2 g respectively. This coincidence suggested to Einstein that the two eects should be considered as arising from the same origin. Thus he suggested that we treat gravitation as as inertial eect as well, in other words as an eect which arises from not using an inertial frame. Comparing the force mg of a falling body with the inertial force ma suggests the following version of the principle of equivalence.

POE 2: A frame linearly accelerated relative to an inertial frame in special relativity is locally identical to a frame at rest in a gravitational eld. These two versions of the principle of equivalence [ POE 1, POE 2 ] can be vividly clari ed by considering the famous \gedanken" experiments of Einstein called the lift experiments. We consider an observer con ned in a lift with no windows in it or other methods of communication with the outside world. The observer is allowed equipment to carry out simple dynamical experiments. The object of the exercise is to try and


60

g

1

2 mg

4

3 mg

g

Figure 5.4: The lift experiments determine the observers state of motion. Let us consider four cases [ see Figure 5.4 ].

Case 1: The lift is placed in a rocket ship in a part of the universe far removed from gravitating bodies. The rocket is accelerated forward with a constant acceleration g relative to an inertial observer. The observer releases a body from rest and sees it fall to the oor with acceleration g. Case 2: The rocket motor is switched o so that the lift undergoes uniform motion relative to the inertial observer. A released body is found to remain at rest relative to the observer in the lift. Case 3: The lift is next placed on the surface of the earth, whose rotational and orbital motions are ignored. A released body is found to fall to the oor with acceleration g. Case 4: Finally, the lift is placed in an evacuated lift shaft and allowed to fall freely towards the center of the earth. A released body is found to remain at rest relative to the observer.


61

g

Figure 5.5: The bending of light Clearly, from the point of view of the observer in the lift, cases 1 and 3 are indistinguishable, as required by POE 2, and cases 2 and 4 are indistinguishable, as required by POE 1. This is the principle of equivalence between gravity and acceleration, and is a corner stone of the theory of General Relativity. In more modern terminology, what we have described is called the weak equivalence principle, \weak" because it refers only to gravity. We shall later use the strong equivalence principle, which says that one can discover how all the laws of physics behave in a gravitational eld by postulating that their laws in a freely falling inertial frame are identical to their laws in Special Relativity i.e. when there are no gravitational elds.

5.6 The principle of equivalence in action Let us now look at two examples of how we use the principle of equivalence.

5.6.1 Eect of gravity on light Since we can think of light as having a mass h c2 , relative to an observer in an accelerated lift, light will curve downwards [ see Figure 5.5 ]. Therefore by the principle of equivalence, light will also be bent in a gravitational eld [ examples: light de ection due to the sun's gravitational eld; observed during a total solar eclipse. Lensing of cosmological sources. ].


62

5.6.2 Eect of gravity on time Consider a rocket of height h undergoing acceleration g relative to an outside observer. Let a light ray be emitted from the top (B) at time t = 0 and be received at the bottom (A) at time t = t0 in the frame of the outside observer [ see Figure 5.6 ]. A second ray is emitted at t = and received at t = t0 + t. One can show [ EXERCISE 5.1 ] that ! gh = t 1 + 2 ; (5.19) c where we have assumed that gh c2 1 (i.e non - relativistic motion). Using the equivalence principle we know that the same relation must apply of there is a dierence in gravitational potential B A between two points B and A in a gravitational eld. i.e. B A = t 1 + : (5.20) c2 If A ! 1 where = 0 [ no gravitational eld ] and B is taken to be a general point with position vector r, we expect ! ( r ) = t 1 + c2 : (5.21) Since (r) is negative, the time measured on B 0s clock [ as seen by A at in nity ] is less than the time measured on A0 s clock, i.e. clocks run slow in a gravitational eld. One can interpret this by imagining that the spacetime metric has the non Minkowski form: 2 2 2 2 2 ds2 = 1 + 2 (5.22) c2 c dt + dx + dy + dz : Then the proper time measured by a clock at xed (x; y; z) in a time t measured at in nity is p (5.23) d = 1c ds2 ; therefore (5.24) = 1 + c2 t ; for c2 1. This corresponds to spacetime curvature.

CHAPTER 5. CONCEPTUAL BASIS OF GR t

63

A B

t0 +∆t ∆τ t0

h

x

Figure 5.6: Space time diagram of rocket undergoing uniform acceleration g

5.6.3 Towards spacetime curvature The world lines of free particles have been our probe of the possibility of constructing inertial frames. In Special Relativity two such world lines which begin parallel to each other remain parallel, no matter how far extended. This is exactly the property that straight lines have in Euclidian geometry. It is natural, therefore, to discuss the geometry of spacetime as de ned by the worldlines of free particles. In these terms Minkowski space is a at space [ it is not Euclidian space because the metric is inde nite ( + ++) ]. To discuss non - uniform gravitational elds let us consider the lift experiments again, this time making them big enough so that variations of the gravitational eld can be measured [ see Figure 5.7 ].

Case 1: From the point of view of the observer in the lift, the two bodies fall to the ground parallel to each other. Case 2: The bodies remain at rest relative to the observer. Case 3: The two bodies fall towards the center of the earth and hence fall on paths which converge.


64

g

1

2

3

4

g

Figure 5.7: The lift experiments again

Case 4: The bodies appear to the observer to move closer together because they are falling on lines that converge towards the center of the earth. We thus conclude that gravitational spacetime is therefore not at - it is curved. A classic example, which we will us many times to illustrate our ideas, is the surface of a sphere or balloon. Locally straight lines on a sphere extend to great circles and two great circles always intersect [ at the poles ]. Nevertheless, close to any point, we can pretend the geometry is at. This is true also for Riemannian spaces: they all are locally at, but the locally straight lines [ geodesics ] do not usually remain parallel. Einstein's important advance was to see the similarity between Riemannian spaces and gravitational physics. He identi ed the trajectories of freely falling particles with the geodesics of a curved geometry: They are locally straight since spacetime admits local inertial frames, but globally they do not remain parallel.

Chapter 6 Curved spacetime and General Relativity 6.1 Manifolds, tangent spaces and local inertial frames A manifold is a continuous space whose points can be assigned coordinates, the number of coordinates being the dimension of the manifold [ for example a surface of a sphere is 2D, spacetime is 4D ]. A manifold is dierentiable if we can de ne a scalar eld at each point which can be dierentiated everywhere. This is always true in Special Relativity and General Relativity. We can then de ne one - forms d~ as having components f; @x@ g and vectors V as linear functions which take d~ into the derivative of along a curve with tangent V: (6.1) V d~ = rV = ;V = d d :

Tensors can then be de ned as maps from one - forms and vectors into the reals [ see chapter 3 ]. A Riemannian manifold is a dierentiable manifold with a symmetric metric tensor g at each point such that

g (V ; V ) > 0 for any vector V, for example Euclidian 3D space. 65

(6.2)

CHAPTER 6. CURVED SPACETIME AND GR

66

If however g (V; V) is of inde nite sign as it is in Special and General Relativity it is called Pseudo - Riemannian. For a general spacetime with coordinates fxg, the interval between two neighboring points is ds2 = g dxdx : (6.3) In Special Relativity we can choose Minkowski coordinates such that g = everywhere. This will not be true for a general curved manifold. Since g is a symmetric matrix, we can always choose a coordinate system at each point x0 in which it is transformed to the diagonal Minkowski form, i.e. there is a transformation @x (6.4) @x such that

0 1 0 0 01 B 0 1 0 0 CC B@ 0 0 1 0 CA : g (x0) = = B

(6.5)

0 0 0 1 Note that the sum of the diagonal elements is conserved; this is the signature of the metric [ +2 ]. In general will not diagonalize g at every point since there are ten functions g (x) and only four transformation functions X (x ). We can also choose so that the rst derivatives of the metric vanishes at x0 i.e. @g = 0 (6.6) @x for all , and . This implies

g (x ) = + O[(x )2] :

(6.7)

That is, the metric near x0 is approximately that of Special Relativity, dierences being of second order in the coordinates. This corresponds to the local inertial frame whose existence was deduced from the equivalence principle. In summary we can de ne a local inertial frame to be one where

g (x0) =

(6.8)

CHAPTER 6. CURVED SPACETIME AND GR for all , , and for all , , . However

67

g ; (x0) = 0

(6.9)

g ; (x0) 6= 0

(6.10)

for at least some values of . , and . It re ects the fact that any curved space has a at tangent space at every point, although these tangent spaces cannot be meshed together into a global at space. Recall that straight lines in a at spacetime are the worldlines of free particles; the absence of rst derivative terms in the metric of a curved spacetime will mean that free particles are moving on lines that are locally straight in that coordinate system. This makes such coordinates very useful for us, since the equations of physics will be nearly as simple as they are in at spacetime, and if they are tensor equations they will be valid in every coordinate system.

6.2 Covariant derivatives and Christoel symbols In Minkowski spacetime with Minkowski coordinates (t; x; y; z) the derivative of a vector V = V e is just @ V = @V e ; (6.11) @x @x

since the basis vectors do not vary. In a general spacetime with arbitrary coordinates, e which vary from point to point so @ V = @V e + V @ e : (6.12) @x @x @x Since @ e =@x is itself a vector for a given it can be written as a linear combination of the bases vectors: @ e = e : (6.13) @x The 's are called Christoel symbols [ or the metric connection ]. Thus we have: @ V = @V e + V e ; (6.14) @x @x

CHAPTER 6. CURVED SPACETIME AND GR so Thus we can write where

@ V = @V e + V @x @x

68

!

e :

@V = V e ; ; @x V ; = V ; + V

(6.15) (6.16)

:

(6.17)

Let us now prove that V ; are the components of a 1=1 tensor. Remember in section 3.5 we found that V ; was only a tensor under Poincare transformations in Minkowski space with Minkowski coordinates. V ; is the natural generalization for a general coordinate transformation. Writing @x @x , we have:

V ; = V ; + V = @x@ ( V ) + V @ + V : + V = @V @x @x Now therefore so we obtain:

(6.18)

@ e = e ; @x

(6.19)

@ e w~ = e w~ = = ; @x

(6.20)

@ + V @ e w ~ V ; = @V + V @x @x @x @ + V w @ e + V = @V

~ @x @x @x = @V + V @ + V w~ @ ( e ) @x @x @x @ + V w @ e + V = @V

~ @x @x @x + V w~ e @@x (6.21)


69

= @ we obtain: Now using = , w~ e = and @@x @x

@ + V V ; = @V + V @x @x = @V @x + V ! @V = @x + V ;

so

V ; = V ; :

V @@x (6.22) (6.23)

We have shown that V ; are indeed the components of a 1=1 tensor. We write this tensor as rV = V ; e w~ : (6.24) It is called the covariant derivative of V. Using a Cartesian basis, the components are just V ; , but this is not true in general; however for a scalar we have: @ ; r = d~ ; (6.25) r ; = @x since scalars do not depend on basis vectors. @ e w Writing = @x ~ , we can nd the transformation law for the components of the Christoel symbols. @ e w~ = w~ @ e =

@x @x @ e + w~ e @ = w~ @x

@x = + @@x (6.26) This is just 2 = @x @x @x + @x @ x : (6.27) @x @x @x @x @x @x

We can calculate the covariant derivative of a one - form p~ by using the fact that p~(V) is a scalar for any vector V:

= p V :

(6.28)


70

We have @p @V r = ; = @x V + p @x @p V + p V p V = @x ; ! @p = @x p V + pV ; :

(6.29)

Since r and V ; are tensors, the term in the parenthesis is a tensor with components: p; = p; p : (6.30) We can extend this argument to show that

r T T; = T; T

r T T ; = T ; + T

r T T ; = T ; + T

6.3 Calculating

+ T

T

;

; :

(6.31)

from the metric

Since V ; is a tensor we can lower the index using the metric tensor:

V; = gV ; :

(6.32)

V; = (gV ); = g; V + g V ; :

(6.33)

But by linearity, we have: So consistency requires g; V = 0. Since V is arbitrary this implies that

g; = 0 :

(6.34)

Thus the covariant derivative of the metric is zero in every frame. We next prove that = [ i.e. symmetric in and ]. In a general frame we have for a scalar eld :

; = ; ; = ;

;

:

(6.35)


71

in a local inertial frame, this is just ; @x@ 2@x , which is symmetric in and . Thus it must also be symmetric in a general frame. Hence is symmetric in and :

= : (6.36)

We now use this to express

in terms of the metric. Since g ; = 0, we have:

g ; =

g +

g

:

(6.37)

By writing dierent permutations of the indices and using the symmetry of , we get g ; + g; g ; = 2g : (6.38) Multiplying by 21 g and using g g = gives

= 12 g (g ; + g; g ; ) :

(6.39)

Note that is not a tensor since it is de ned in terms of partial derivatives. In a local inertial frame = 0 since g ; = 0. We will see later the signi cance of this result.

6.4 Tensors in polar coordinates The covariant derivative diers from partial derivatives even in at spacetime if one uses non - Cartesian coordinates. This corresponds to going to a non - inertial frame. To illustrate this we will focus on two dimensional Euclidian space with Cartesian coordinates (x1 = x, x2 = y) and polar coordinates (x1 = r, x2 = ). The coordinates are related by

1=2 x = r cos ; y = r sin ; = tan 1 xy ; r = x2 + y2 :

For neighboring points we have @r dx + @r dy = cos dx + sin dy ; dr = @x @y and @ dx + @ dy = 1 sin dx + 1 cos dy : d = @x @y r r

(6.40) (6.41) (6.42)


72

We can represent this by a transformation matrix :

dx = dx ; where

cos sin 1 sin 1 cos r r

=

(6.43)

!

:

(6.44)

Any vector components must transform in the same way. For any scalar eld , we can de ne a one - form:

! @ @ d~ ! @r ; @ :

(6.45)

We have

@ = @ @x + @ @y @ @x @ @y @ @ ; + r cos = r sin @ @x @y

(6.46)

and

@ = @ @x + @ @y @r @x @r @y @r @ ; + sin = cos @ @x @y This transformation can be represented by another matrix : @ = @ ; @x @x where ! cos r sin = sin r cos : Any one - form components must transform in the same way. The matrices and are dierent but related:

=

cos sin 1 sin 1 cos r r

!

cos r sin sin r cos

!

= 10 01

(6.47)

(6.48)

(6.49)

!

:

(6.50)


73

This is just what you would expect since in general = . The basis vectors and basis one - forms are

er = r e = xr ex + y r ey = cos ex + sin ey ; e = e = x ex + y ey = r sin ex + r cos ey ;

(6.51)

and

w~ r = dr~ = r w~ = r xw~ x + r y w~ y = cos w~ x + sin w~ y ; ~ = w~ = xw~ x + y w~ y = 1r sin w~ x + 1r cos w~ y : (6.52) w~ = d Note that the basis vectors change from point to point in polar coordinates and need not have unit length so they do not form an orthonormal basis:

jer j = 1 ; je j = r ; jw~ r j = 1 ; jw~ j = 1r : The inverse metric tensor is:

g =

1 0 0 r 2

!

(6.53)

(6.54)

so the components of the vector gradient d of a scalar eld are: (d)r = gr ; = grr ;r + gr ; = @ @r ; (d) = g ; = gr ;r + g ; = r12 @ (6.55) @ : This is exactly what we would expect from our understanding of normal vector calculus. We also have: @ er = @ (cos e + sin e ) = 0 ; (6.56) x y @r @r and

Since

@ er = 1 e ; @ e = 1 e ; @ e = re : r @ r @r r @

(6.57)

@ e = e ; @x

(6.58)


74

we can work out all the components of the Christoel symbols: =1; =1; r = r; (6.59) r r r r and all other components are zero. Alternatively, we can work out these components from the metric [ EXERCISE ]: i h = 1 g g + g g : (6.60) ; ;

; 2 In fact this is the best way of working out the components of , and it is the way we will adopt in General Relativity. Finally we can check that all the components of g ; = 0 as requited. For example

g;r = g;r = (r2);r

r g

2 (r 2 ) = 0 : r

r g

(6.61)

6.5 Parallel transport and geodesics A vector eld V is parallel transported along a curve with tangent U = ddx ; (6.62) where is the parameter along the curve [ usually taken to be the proper time if the curve is timelike ] if and only if dV = 0 (6.63) d in an inertial frame. Since dV = @V dx = U V ; (6.64) ; d @x d in a general frame the condition becomes:

U V ; = 0 ;

(6.65)

i.e. we just replace the partial derivatives (;) with a covariant derivative (;). This is called \the comma goes to semicolon" rule, i.e. work things out in a local inertial


V

75

Timelike curve U

Figure 6.1: Parallel transport of a vector V along a timelike curve with tangent

U.

frame and if it is a tensor equation it will be valid in all frames. The curve is a geodesic if it parallel transports its own tangent vector:

U U ; = 0 :

(6.66)

This is the closest we can get to de ning a straight line in a curved space. In at space a tangent vector is everywhere tangent only for a straight line. Now

U U ; = 0 ) U U ; + Since U = dxd and

d d

U

U

=0:

(6.67)

= U @x@ we can write this as

d2 x + dx dx = 0 : (6.68) d2 d d This is the geodesic equation. It is a second order dierential equation for x (), so one gets a unique solution by specifying an initial position x0 and velocity U0.

6.5.1 The variational method for geodesics We now apply the variational techniques to compute the geodesics for a given metric. For a curved spacetime, the proper time d is de ned to be (6.69) d 2 = c12 ds2 = c12 g dx dx :


76

Remember in at spacetime it was just d 2 = c12 dxdx : (6.70) Therefore the proper time between two points A and B along an arbitrary timelike curve is ZB Z B d AB = d = d d A A " ZB1 dx #1=2 dx g (x) d d d ; = (6.71) A c so we can write the Lagrangian as

L(x ; x_ ; ) = 1c and the action becomes

A = AB =

"

ZB A

dx #1=2 dx g (x) d d ;

(6.72)

L(x; x_ ; )d :

(6.73)

Extremizing the action we get the Euler - Lagrange equations:

! @L = d @L : @x d @ x_

Now and Since we get and

(6.74)

dx ! 1=2 @g dx dx dx g d d @x d d ! @ L = 1 g dx dx 1=2 (2)g dx : @ x_ 2c d d d ! 1 g dx dx 1=2 = d c d d d

(6.76)

@ L = 1 d @g dx dx @x 2 d @x d d

(6.78)

@ L = d g dx = g dx ; @ x_ d d d

(6.79)

@L = 1 @x 2c

(6.75)

(6.77)


77

so the Euler - Lagrange equations become: " # 1 d @g dx dx = d g dx : 2 d @x d d d d Multiplying by

Using we get

d d

we obtain

"

#

(6.80)

1 @g dx dx = d g dx : 2 @x d d d d

(6.81)

dg = @g dx ; d @x d

(6.82)

1 @g dx dx = g d2x + @g dx dx : (6.83) 2 2 @x d d d @x d d Multiplying by g gives " # d2x = g @g 1 @g dx dx : (6.84) d 2 @x 2 @x d d Now " # @g dx dx = 1 @g dx dx + @g dx dx @x d d 2 @x d d @x d d " # 1 @g @g dx dx = 2 @x + @x d d : (6.85) Using the above result gives us dx dx d2x = 1 g [g + g g ] ; ; ; d 2 2 d d

dx dx ; = (6.86) d d so we get the geodesic equation again d2x + dx dx = 0 : (6.87) d 2 d d This is the equation of motion for a particle moving on a timelike geodesic in curved spacetime. Note that in a local inertial frame i.e. where = 0, the equation reduces to d2x = 0 ; (6.88) d 2


78

which is the equation of motion for a free particle. The geodesic equation preserves its form if we parameterize the curve by any other parameter such that d2 = 0 ) = a + b (6.89) d 2 for constants a and b. A parameter which satis es this condition is said to be ane.

6.5.2 The principle of equivalence again In Special Relativity, in a coordinate system adapted for an inertial frame, namely Minkowski coordinates, the equation for a test particle is: d2 x = 0 : (6.90) d 2 If we use a non - inertial frame of reference, then this is equivalent to using a more general coordinate system [x = x (x )]. In this case, the equation becomes d2x + dx dx = 0 ; (6.91) d 2 d d where is the metric connection of g , which is still a at metric but not the Minkowski metric . The additional terms involving which appear, are inertial forces. The principle of equivalence requires that gravitational forces, a well as inertial forces, should be given by an appropriate . In this case we can no longer take the spacetime to be at. The simplest generalization is to keep as the metric connection, but now take it to be the metric connection of a non - at metric. If we are to interpret the as force terms, then it follows that we should regard the g as potentials. The eld equations of Newtonian gravitation consist of second - order partial dierential equations in the potential . In an analogous manner, we would expect that General Relativity also to involve second order partial dierential equations in the potentials g . The remaining task which will allow us to build a relativistic theory of gravitation is to construct this set of partial dierential equations. We will do this shortly but rst we must de ne a quantity that quanti es spacetime curvature.


79

6.6 The curvature tensor and geodesic deviation So far we have used the local - atness theorem to develop as much mathematics on curved manifolds as possible without having to consider curvature explicitly. In this section we will make a precise mathematical de nition of curvature and discuss the remaining tools needed to derive the Einstein eld equations. It is important to distinguish two dierent kinds of curvature: intrinsic and extrinsic. Consider for example a cylinder. Since a cylinder is round in one direction, one thinks of it as curved. It is its extrinsic curvature i.e. the curvature it has in relation to the at three - dimensional space it is part of. On the other hand, a cylinder can be made by rolling a at piece of paper without tearing or crumpling it, so the intrinsic geometry is that of the original paper i.e. it is at. This means that the distance between any two points is the same as it was in the original paper. Also parallel lines remain parallel when continued; in fact all of Euclid's axioms hold for the surface of a cylinder. A 2D ant con ned to that surface would decide that it was at; only that the global topology is funny! It is clear therefore that when we talk of the curvature of spacetime, we talk of its intrinsic curvature since the worldlines [ geodesics ] of particles are con ned to remain in spacetime. The cylinder, as we have just seen is intrinsically at; a sphere, on the other hand, has an intrinsically curved surface. To see this, consider the surface of a sphere [ or balloon ] in which two neighboring lines begin at A and B perpendicular to the equator, and hence are parallel. When continued as locally straight lines they follow the arc of great circles, and the two lines meet at the pole P. So parallel lines, when continued, do not remain parallel, so the space is not at. There is an even more striking illustration of the curvature of the surface of a sphere. Consider, rst, a at space. Let us take a closed path starting at A then going to B and C and then back to A. Let's parallel transport a vector around this loop. The vector nally drawn at A is, of course, parallel to the original one [ see Figure 6.2 ]. A completely dierent thing happens on the surface of a sphere! This time the vector is rotated through 90 degrees! [ see the balloon example ]. We will


80

B

A C

Figure 6.2: Parallel transport around a closed loop in at space. now use parallel transport around a closed loop in curved space to de ne curvature in that space.

6.6.1 The curvature tensor Imagine in our manifold a very small closed loop whose four sides are the coordinate lines x1 = a, x1 = a + a, x2 = b, x2 = b + b. A vector V de ned at A is parallel transported to B . From the parallel transport law V ; re V = 0 ) @V (6.92) 1 @x1 =

it follows that at B the vector has components

Z B @V 1 = dx 1 A @x Z ) V (B ) = V (A) x2=b 1 V dx1 ; V (B )

V (A) +

(6.93)

where the notation \x2 = b" under the integral denotes the path AB . Similar transport from B to C to D gives

V (C ) = V (B ) and

V (D ) = V (C ) +

Z

x =a+a

Z x =b+b 1

2 V

dx2

;

(6.94)

1 V

dx1

:

(6.95)

1

The integral in the last equation has a dierent sign because of the direction of transport from C to D is in the negative x1 direction.


81

B

^ e 1

x 1 =a

C

^ e 2

A

x 1 =a+ δa

x 2 =b D x 2 =b + δb

Figure 6.3: Parallel transport around a closed loop ABCD. Similarly, the completion of the loop gives

V (Afinal ) = V (D) +

Z

x1 =a

2 V

dx2

:

(6.96)

The net change in V (A) is a vector V , found by adding (6.93) - (6.96).

V = V (Afinal ) V (Ainitial ) Z Z 2 = 1 2 V dx x1 =a+a x =a Z Z 1 + 2 1 V dx 2 x =b+b

To lowest order we get

V

= +

Z b+b b

Z a+a a

x =b

a @x@ 1 (

b @x@ 2 (

2 V

1 V

2 V

1 V

dx2

dx1

:

(6.97)

mu ) dx2

) dx1

# @ @ (6.98) ab @x1 ( 2 V ) + @x2 ( 1 V ) : This involves derivatives of 's and of V . The derivatives of V can be eliminated "

CHAPTER 6. CURVED SPACETIME AND GR using for example

@V = @x1

This gives

V = ab [

1;2

2;1 +

1 V

2

82

:

(6.99)

1

1

2 ] V

:

(6.100)

To obtain this, one needs to relabel dummy indices in the terms quadratic in . Notice that this just turns out to be a number times V summed on . Now the indices 1 and 2 appear because the path was chosen to go along those coordinates. It is antisymmetric in 1 and 2 because the change V would have the opposite sign if one went around the loop in the opposite direction. If we use general coordinate lines x and x , we nd

V = ab [ De ning

;

R

we can write

; +

;

; +

] V

:

(6.101) (6.102)

V = abR V w~ w~ :

(6.103)

R are the components of a 1=3 tensor. This tensor is called the Riemann curvature tensor.

6.6.2 Properties of the Riemann curvature tensor Recall that the Riemann tensor is

R

;

In a local inertial frame we have

R = Now

; +

:

(6.104)

= 0, so in this frame

;

;

:

= 21 g (g ; + g; g ; )

(6.105) (6.106)

CHAPTER 6. CURVED SPACETIME AND GR so

;

= 12 g (g ; + g; g ; )

83 (6.107)

since g ; = 0 i.e the rst derivative of the metric vanishes in a local inertial frame. Hence R = 21 g (g ; + g; g ; g ; g; + g ; ) : (6.108) Using the fact that partial derivatives always commute so that g ; = g ; , we get (6.109) R = 21 g (g; g; + g ; g ; ) in a local inertial frame. Lowering the index with the metric we get

R = g R (6.110) = 21 (g; g; + g ; g ; ) : So in a local inertial frame the result is R = 12 (g; g; + g ; g ; ) : (6.111) We can use this result to discover what the symmetries of R are. It is easy to show from the above result that and

R = R = R = R

(6.112)

R + R + R = 0 :

(6.113)

Thus R is antisymmetric on the nal pair and second pair of indices, and symmetric on exchange of the two pairs. Since these last two equations are valid tensor equations, although they were derived in a local inertial frame, they are valid in all coordinate systems. We can use these two identities to reduce the number of independent components of R from 256 to just 20.


84

A at manifold is one which has a global de nition of parallelism: i.e. a vector can be moved around parallel to itself on an arbitrary curve and will return to its starting point unchanged. This clearly means that

R = 0 ;

(6.114)

i.e. the manifold is at [ EXERCISE 6.5: try a cylinder! ]. An important use of the curvature tensor comes when we examine the consequences of taking two covariant derivatives of a vector eld V:

rr V = r (V ; ) = (V ; ); +

V ;

V ;

:

(6.115)

As usual we can simplify things by working in a local inertial frame. So in this frame we get

rr V = (V ; ); = (V ; +

V );

= V ; +

; V

+

V ;

:

(6.116)

The second term of this is zero in a local inertial frame, so we obtain

rr V = V ; +

; V

:

(6.117)

Consider the same formula with the and interchanged:

r rV = V ; +

; V

:

(6.118)

If we subtract these we get the commutator of the covariant derivative operators r and r : [r ; r ] V = r r V r r V = (

;

; ) V

:

(6.119)

The terms involving the second derivatives of V drop out because V ; = V ; [ partial derivatives commute ].


85

V/ A/ ξ

a

V

A

Figure 6.4: Geodesic deviation Since in a local inertial frame the Riemann tensor takes the form

R = we get

;

(6.120)

[r; r ] V = R V :

(6.121)

;

;

This is closely related to our original derivation of the Riemann tensor from parallel transport around loops, because the parallel transport problem can be thought of as computing, rst the change of V in one direction, and then in another, followed by subtracting changes in the reverse order.

6.7 Geodesic deviation We have shown that in a curved space [ for example on the surface of a balloon ] parallel lines when extended do not remain parallel. We will now formulate this mathematically in terms of the Riemann tensor. Consider two geodesics with tangents V and V0 that begin parallel and near each other at points A and A0 [ see Figure 6.4 ]. Let the ane parameter on the geodesics be . We de ne a connecting vector which reaches from one geodesic to another, connecting points at equal intervals in . To simplify things, let's adopt a local inertial frame at A, in which the coordinate x0 points along the geodesics. Thus at A we have V = 0.


86

The geodesic equation at A is

d2x j = 0 (6.122) d2 A since all the connection coecients vanish at A. The connection does not vanish at A0 , so the equation of the geodesic at A0 is d2x j + (A0) = 0 ; (6.123) 00 d2 A where again at A0 we have arranged the coordinates so that V = 0. But, since A and A0 are separated by the connecting vector we have 0

00 (A0 ) = 00; ;

(6.124)

the right hand side being evaluated at A, so d2 x j = : 00; d2 A Now x (A0) x (A) = so

(6.125)

0

d2 = d2x j d2x j = : (6.126) 00; d2 d2 A d2 A This then gives us an expression telling us how the components of change. Consider now the full second covariant derivative rV rV . Now 0

rV rV = rV (rV )

= d (rV ) + 0 rV : d In a local inertial frame this is d (r ) rV rV = d V ! d d = d d + 0 2 = dd + 0;0 ;

(6.127)

(6.128)


87

where everything is again evaluated at A. Using the result for

rV rV = (

0;0

d2 d2

we get

00; )

= R 00 = R V V ;

(6.129)

since we have chosen V = 0 . The nal expression is frame invariant, so we have in any basis

rV rV = R V V :

(6.130)

So geodesics in at space maintain their separation; those in curved space don't. This is called the equation of geodesic deviation and it shows mathematically that the tidal forces of a gravitational eld can be represented by the curvature of spacetime in which particles follow geodesics.

6.8 The Bianchi identities; Ricci and Einstein tensors In the last section we found that in a local inertial frame the Riemann tensor could be written as R = 12 (g; g; + g ; g ; ) : (6.131) Dierentiating with respect to x we get

R ; = 12 (g; g; + g ; g ; ) :

(6.132)

From this equation, the symmetry g = g and the fact that partial derivatives commute, it is easy to show that

R ; + R ; + R ; = 0 :

(6.133)

This equation is valid in a local inertial frame, therefore in a general frame we get

R ; + R ; + R ; = 0 :

(6.134)

This is a tensor equation, therefore valid in any coordinate system. It is called the Bianchi identities, and will be very important for our work.


88

6.8.1 The Ricci tensor Before looking at the consequences of the Bianchi identities, we need to de ne the Ricci tensor R : R = R = R : (6.135) It is the contraction of R on the rst and third indices. Other contractions would in principle also be possible: on the rst and second, the rst and fourth, etc. But because R is antisymmetric on and and on and , all these contractions either vanish or reduce to R . Therefore the Ricci tensor is essentially the only contraction of the Riemann tensor. Similarly, the Ricci scalar is de ned as

R = g R = g g R :

(6.136)

6.8.2 The Einstein Tensor Let us apply the Ricci contraction to the Bianchi identities

g [R ; + R ; + R ; ] = 0 :

(6.137)

Since g ; = 0 and g ; = 0, we can take g in and out of covariant derivatives at will: We get: R ; + R ; + R ; = 0 : (6.138) Using the antisymmetry on the indices and we get

R ; R ; + R ; ; so

R ; R ; + R ; = 0 : These equations are called the contracted Bianchi identities. Let us now contract a second time on the indices and :

This gives

(6.139) (6.140)

g [R ; R ; + R ;] = 0 :

(6.141)

R ; R ; + R ; = 0 :

(6.142)

CHAPTER 6. CURVED SPACETIME AND GR so

R; 2R; = 0 ;

(6.143)

2R; R; = 0 :

(6.144)

i 2R 12 gR ; = 0 :

(6.145)

or Since R; = gR; , we get

h

89

Raising the index with g we get

h

De ning we get

i R 12 g R ; = 0 :

(6.146)

G = R 21 g R

(6.147)

G ; = 0 :

(6.148)

The tensor G is constructed only from the Riemann tensor and the metric, and it is automatically divergence free as an identity. It is called the Einstein tensor, since its importance for gravity was rst understood by Einstein. We will see in the next chapter that Einstein's eld equations for General Relativity are G = 8G (6.149) c4 T ; where T is the stress - energy tensor. The Bianchi Identities then imply

T ; = 0 ; which is the conservation of energy and momentum.

(6.150)

Chapter 7 Einstein's Field Equations - The Main Goal of The Course 7.1 Introduction In order to have a complete theory of gravity, we need to know

How particles behave in curved spacetime. How matter curves spacetime. The rst question is answered by postulating that free particles [ i.e. no force other than gravity ] follow timelike or null geodesics. We will see later that this is equivalent to Newton's law F = mr in the week eld limit [ =c2 1 ]. The second requires the analogue of r2 = 4G. We rst consider the vacuum case [ = 0 ] ) r2 = 0. The easiest way to do this is to compare the geodesic deviation equation derived in the last section with its Newtonian analogue. In Newtonian theory the acceleration of two neighboring particles with position vectors x and x + are: d2xi = @ (x) ; d2(xi + i) = @ (x + ) ; (7.1) dt2 @xi dt2 @xi so the separation evolves according to: d2 i = @ (x) @ (x + ) dt2 @xi @xi @ ((x) (x + )) : = @x (7.2) i 90

CHAPTER 7. Einstein's Field Equations This gives us since

91

d2 i = @ 2 j ; dt2 @xi @xj

@ : (x + ) (x) = j @x j This clearly is analogous to the geodesic deviation equation

rV rV = R V V ;

(7.3) (7.4) (7.5)

provided we relate the quantities @x@i2@x j and R V V Both quantities have two free indices, although the Newtonian index runs from 1 to 3 while in the General Relativity case it runs from 0 to 3. The Newtonian vacuum equation is r2 = 0 which implies that @2 @xi @xi

=0;

so we can write

R V V = 0 : Since V is arbitrary we end up with R = 0 :

(7.6) (7.7) (7.8)

These are the vacuum Einstein eld equations.

7.2 The non - vacuum eld equations The General Relativity version of r2 = 4G must contain T rather than since we saw in Special Relativity that c2 is just the 00 component of the energy momentum tensor. This is expected anyway since in General Relativity all forms of energy [ not just rest mass ] should be a source of gravity. To get the General Relativity version of the equations involving T we just replace the Minkowski metric by g and the partial derivative (; ) by the covariant derivative (; ). For example the energy - momentum tensor for a perfect

uid in curved space time is p T = + c2 U U + pg ; (7.9)

CHAPTER 7. Einstein's Field Equations

92

and the conservation equations become

T ; = 0 :

(7.10)

In the above we have just used the strong form of the Equivalence Principle, which says that any non - gravitational law expressible in tensor notation in Special Relativity has exactly the same form in a local inertial frame of curved spacetime. We expect the full [ non - vacuum ] eld equations to be of the form

O(g) = T ;

(7.11)

where O is a second order dierential operator which is a 0=2 tensor [ since T is the stress energy tensor ] and is a constant. The simplest operator that reduces to the vacuum eld equations when T = 0 takes the form

O = R + g R :

(7.12)

Now since T ; = 0 [ T ; = 0 in Special Relativity ], we require O ; = 0. Using g ; = 0 gives R + g R ; = 0 : (7.13)

Comparing this with the double contracted Bianchi identities

G ; = 0 ;

(7.14)

we see that the constant has to be = 21 . Thus we are led to the eld equations of General Relativity: (7.15) R 12 g R = T ; or

G = T :

(7.16)

In general we can add a constant so the eld equations become

R 21 g R + g = T :

(7.17)

In a vacuum T = 0, so taking the trace of the eld equations we get

R 12 Rg + g = 0 :

(7.18)


93

Since g = 4 and the Ricci scalar R = R we nd that R = 4, and substituting this back into the eld equations leads to

R = g :

(7.19)

We recover the previous vacuum equations if = 0. Sometimes is called the vacuum energy density. We have ten equations [ since R is symmetric ] for the ten metric components. Note that there are four degrees of freedom in choosing coordinates so only six metric components are really determinable. This corresponds to the four conditions

G ; = 0 ;

(7.20)

which reduces the eective number of equations to six. It is very important to realize that although Einstein's eld equations look very simple, they in fact correspond in general to six coupled non - linear partial dierential equations.

7.3 The weak eld approximation We have to check that the appropriate limit, General Relativity leads to Newton's theory. The limit we shall use will be that of small velocities vc 1 and that time derivatives are much smaller than spatial derivatives. There are two things we must do:

We have to relate the geodesic equation to Newton's law of motion [ i.e. the second law ]. and

Relate Einstein's eld equations to the Newton - Poisson equation. Let's assume that we can nd a coordinate system which is locally Minkowski [ as demanded by the Equivalence Principle ] and that deviations from at spacetime are small. This means we can write

g = + h ;

(7.21)


94

where is small. Since we require that g g = , the inverse metric is given by

g = h :

(7.22)

To work out the geodesic equations we need to work out what the components of the Christoel symbols are:

= 21 g (g ; + g; g ; ) :

(7.23)

Substituting for g etc. in terms of h we obtain

= 12 (h ; + h; h ; ) :

(7.24)

The geodesic equations are

d2x + dx dx = 0 : (7.25) d 2 d d But for a slowly moving particle t so d2x + dx dx = 0 : (7.26) dt dt dt Also dxdti = O(), so we can neglect terms like ij dxdti dxdtj . The geodesic equation reduces to d2 x + dx0 dx0 = 0 ; (7.27) 00 dt dt dt2 so the \space" equation (three - acceleration) is

Since

dx0 dt

= c we get

Now i

d2xi + i dx0 dx0 = 0 : 00 dt dt dt2

(7.28)

d2xi = c2 i : 00 dt2

(7.29)

00 = 12 (hi0;0 + hi0;0 21 h00;i ;

h00;i) (7.30)


95

where we have neglected time derivatives over space derivatives. The spatial geodesic equation then becomes d2xi = c2 h = c2 r h : (7.31) dt2 2 00;i 2 i 00 But Newtonian theory has d2xi = r ; (7.32) i dt2 where is the gravitational potential. So we make the identi cation ! 2 g00 = 1 + c2 : This is equivalent to having spacetime with the line element ! ! 2 2 2 2 2 ds = 1 + c2 c dt + 1 c2 dx2 + dy2 + dz2 : This is what we deduced using the Equivalence Principle in section 4.5. Let's now look at the eld equations [ with = 0 ]:

R 12 g R = T :

(7.33)

(7.34)

(7.35)

Taking the trace we get

R 2R = T = T

)R =

T :

(7.36)

This allows us to write the eld equations as

i h R = T 21 g T :

(7.37)

Let us assume that the matter takes the form of a perfect uid, so the stress - energy tensor takes the form: p T = + c2 U U + pg : (7.38) Taking the trace gives p 2 (7.39) T T = c + c2 + 4p ;


96

so the eld equations become

p p 1 R = + c2 U U + 2 c2 c2g : The Newtonian limit is >> cp2 . This gives R = U U + 21 c2 g :

(7.40) (7.41)

Look at the 00 component of these equations:

R00 = c2 21 c2 = 21 c2 to rst order in . Now

(7.42)

R = ; ; to rst order in . The (0; 0) component of this equation is

(7.43)

R00 = 00; 0;0 ; and since spatial derivatives dominate over time derivatives, we get

(7.44)

R00 = i00;i :

(7.45)

R00 = i00;i = 12 h00;ii = 12 c2 :

(7.46)

r2 = 12 c4 :

(7.47)

r2 = 4G ;

(7.48)

So the eld equations are This is just

Comparing this with Poisson's equation:

we see that we get the same result if the constant is = 8G (7.49) c4 : We can now use this result to write down the full Einstein eld equations:

R 21 g R + g = 8cG4 T :

(7.50)


SPECIAL RELATIVITY

97

GENERAL RELATIVITY

FLAT

Add

CURVED

SPACETIME

mass/energy

SPACETIME

Minkowski Coordinates

Γ

α

General Coordinates

Metric is η αβ

α

βγ = 0 , R βγµ = 0

Transform to frame or generalized

an accelerating choose some coordinates.

Metric is g

Γ

α

αβ α βγ =/= 0 , R βγδ =/= 0

Transform to

a non-rotating

freely falling

frame and take

Minkowski

coordinates.

A Non - Inertial Frame

Γ

α

βγ =/= 0 α But R βγδ = 0

A Local Inertial Frame Metric is η αβ + Ο( x µ ) 2

Γ

α

so still flat spacetime

Γα

βγ are inertial forces

α βγ = 0 , R βγδ =/= 0

Transform to frame wrt

Γ

α

an accelerating free fall.

α βγ =/= 0 , R βγδ =/= 0

Equivalence Principle - gravity also an inertial force

Figure 7.1: A SUMMARY OF WHAT WE HAVE DONE : ) =

Chapter 8 The Schwarzschild Solution The gravitational elds which are most important in our daily lives, namely that of the earth, moon and sun, are produced by slowly rotating, nearly spherical mass distributions, i.e. they are approximately spherically symmetric. Because of their importance we will try and solve Einstein's equations for such bodies.

8.1 The line element We shall try and introduce coordinates which are appropriate to the problem. Since our choice of coordinates always lead to requirements on the metric functions, we must proceed carefully in order not to lose solutions by making the restrictions too strong. Spherical symmetry evidently signi es that in the three dimensional space T = constant, all radial coordinates are equivalent and no perpendicular direction is singled out; in spherical polar coordinates we have [ for constant T ]

i h d(3) s2 = g11 (R; cT )dR2 + f (R; cT ) d2 + sin2 d2 :

(8.1)

The most general ansatz for the line element in the full four - dimensional spacetime is therefore

ds2 =

g00 (R:cT )c2dT 2 + g11 (R; cT )dR2 i h = f (R; cT ) d2 + sin2 d2 + 2g01(R; cT )dRdcT : 98

(8.2)

CHAPTER 8. THE SCHWARZSCHILD SOLUTION

99

We can simplify this line element further using the coordinate transformation so that and

r2 = f (R; cT ) ;

(8.3)

@f d(cT ) ; 2rdr = @F dR + @R @ (cT )

(8.4)

! ! ! ! @f dRd(cT ) : (8.5) @f 2 dR2 + @f 2 d(cT )2 + 2 @f @R @ (cT ) @R @ (cT ) Using these results we can be bring the line element into the form 4r2dr2 =

ds2 =

b(r; T )c2 dT 2 + h2 (r; T )dr2 2a(r; T )b(r; T )cdTdr + r2 d2 + sin2 d2 :

(8.6)

Notice that this already contains the usual two dimensional spherical surface element. A further transformation

e=2 d(cT ) = bd(cT ) + adr

(8.7)

eliminates the undesired non - diagonal terms. Thus we arrive at the Schwartzschild form of the line element of a spherically symmetric non - rotating body:

ds2 = e(r;t) c2 dt2 + e(r;t) dr2 + r2 d2 + sin2 d2 :

(8.8)

8.2 The Christoel symbols The next step on our journey towards the solution of the eld equations is to calculate the components of the Christoel symbols. We have

= 1 (g (8.9) 2 ; + g; g ; ) : Of the total of forty independent components, only twelve are non - zero, for example r

rr

= 21 gr (gr;r + rr;r grr;) = 21 grr grr;r = 21 e

@ e @r

;

(8.10)

CHAPTER 8. THE SCHWARZSCHILD SOLUTION so

r

rr

100

= 21 0 ;

(8.11)

where @=@r is denoted by a prime. The rest of the components are [ EXERCISE 7.1] r r tt

rt

= 12 _ ;

r

= 12 e( ) 0 ;

r

re ;

=

r = 1 r

= 1r ;

t

r

= 21 0 ;

= r sin2 e ;

=

;

t

= cot ;

rt

t tt

rr

sin cos ;

_ ( ) ; = 21 e

= 21 _ ;

(8.12)

where the dot denotes @=@t.

8.3 The Ricci Tensor From the Riemann tensor:

R =

;

; +

r

r

;

(8.13)

we obtain:

Rr r =

r

;r

R =

;

R =

Rt t =

t

r

rr

+

r

t rt

; + r r

; +

;t

r; +

t

t

r

r r +

t; +

r

r t t rt + tt

t

t

:

(8.14)

Unless = or ( ; ) = (r; t) these components vanish. Also Rrt = 0. Using the above expressions let us work out the non - zero components of the Ricci tensor R . For example

Rrr = Rr rrr + R rr + R rr + Rt rtr


101

= Rrr + Rrr + Rtrtr r;r + r r rr

=

+

t

+

t

r;r + t

rr;t t tt rr

r r rr

r r

r

r

t r rt;r + rt rr t

t rt rt

t

rr

r

rt

;

(8.15)

so

Rrr = r12 + 21 1r 0 r12 + r12 + 12 1r 0 r12 + 12 e( ) + 21 _ _ _ e( ) 1 00 + 1 0 0 + 1 _ e _ ( ) 1 02 + 1 _ 2e( ) : (8.16) 2 4 4 4 4 This simpli es to " _2 _ # 00 02 0 0 0 + _ ( ) (8.17) Rrr = 2 4 + 4 + r + e 2 4 4 The other components of the Ricci tensor are [ EXERCISE 7.2 ]: " 00 02 0 0 0 # _ 2 _ Rtt = e( ) 2 + 4 4 + r 2 4 + 4_ ; _ Rrt = r ; r 0 0 R = e 1 + 2 ( ) + 1 ; R = sin2 R :

(8.18) (8.19) (8.20) (8.21)

8.4 The vacuum eld equations Outside the eld producing mass the energy - momentum tensor vanishes i.e. T = 0. The eld equations are therefore

R = 0 : It follows that all the components of R vanish.

(8.22)


102

From Rrt = 0 we have immediately that _ = 0; thus depends only on the radial coordinate r. It follows that R = 0 can then only be satis ed if 0 is also independent of time, i.e. = (r) + f (t) : (8.23) Since occurs in the line element in the combination e(r) ef (t) c2dt2 , one can always make the term involving f (t) vanish by the coordinate transformation

dt0 = ef=2 dt ;

(8.24)

so that in the new coordinates = (r) and = (r). That is if the metric components no longer depend on time. We have proved Birkho's theorem: every spherically symmetric vacuum solution is independent of time, i.e. the solution is static. If one considers the vacuum gravitational eld produced by a spherically symmetric star, then the eld remains static even if the material in the star experiences a spherically symmetric radial displacement [ explosion ]. Thus Birkho's theorem is the analogue of the statement in electrodynamics that a spherically symmetric distribution of charges and currents does not radiate.

8.5 The Schwartzschild solution For static vacuum elds, the eld equations reduce to 00 + 02 0 0 0 = 0 ; 2 4 4 r 1 ( 0 + 0) = 0 ; r e (1 r0) 1 = 0 :

(8.25) (8.26) (8.27)

The second of these equations is equivalent to

(r) = (r) :

(8.28)

Under the substitution = e , the third equation transforms into the dierential equation (8.29) 0 + 1r = 1r ;


103

Figure 8.1: Schwarzschild whose general solution is

= e = e = 1 2GM (8.30) r ; where 2GM is a constant of integration. Thus, the spherically symmetric vacuum solution, rst found in 1916 by Schwartzschild, has the line element 1 2 dt2 + 1 2GM 2 + r2 d2 + sin2 d2 : (8.31) c dr ds2 = 1 2GM c2 r c2r

8.6 General discussion of the Schwartzschild solution In order to understand the physical properties of the Schwartzschild solution we rst have to clarify the physical signi cance of the integration constant 2GM c2 . This is best done through a comparison with Newtonian theory. For large values of the coordinate r, the line element deviates only a little from that of at space, and from our analysis of the Newtonian limit we see that 2GM ; 1 + 2 1 (8.32) 2 c c2 r so (8.33) = GM r : We thus have to interpret the Schwarzschild solution as the gravitational eld outside a spherically symmetric mass distribution whose [ Newtonian ] mass is M .


104

Notice that the line element is singular when r = 2GM c2 . This is the Schwartzschild radius or the gravitational radius of the source. For example for the Sun rG = 2:96 km and for the earth it is rG = 8:8 mm. From the line element, we can immediately deduce some of the eects we derived using the equivalence principle.

8.6.1 Time dilation in a gravitational eld For a clock at xed (r; ; ) ie dr = d = d = 0 we have 2 2 ds2 = 1 2GM c2 r c dt : The proper time is d 2 = c12 ds2, so 2GM 1=2 d = 1 c2r dt ; so clocks run slow in a gravitational eld.

(8.34) (8.35)

8.6.2 Length contraction in a gravitational eld For a \ruler" at xed (t; ; ) we have

1=2 2 GM ds = 1 c2r dr ; (8.36) so rulers \shrink" in a gravitational eld. Remember that this eect could not be deduced from the equivalence principle.

8.7 Orbits in Schwartzschild spacetime To nd the motion of planets and light rays in a Schwartzschild spacetime we must rst nd the geodesic equations. This is best done by working from the Lagrangian " dx #1=2 dx 1 : (8.37) L = c g d d Assuming that the orbits remain permanently in the equatorial plane [ as in Newtonian theory ] i.e. = 2 , the Lagrangian is:

8 !2 39 dt !2 1 2 2GM 1 dr !2 =1=2 < 2 GM d 2 5 4 L = : 1 c2 r d c2 1 c2 r d + r d ; :

(8.38)

CHAPTER 8. THE SCHWARZSCHILD SOLUTION The Euler - Lagrange equations are ! @L = d @L : @x d @ x_ Now @ L = 0 ) @ L = const ; @t @ t_ so 2GM dt 1 c2r d = k ; k_ = 0 :

This is just energy conservation. Also @L = 0 ) @ so r2_ = h ;

@ L = const ; @ _

105

(8.39) (8.40) (8.41) (8.42)

h_ = 0 : (8.43) This is conservation of angular momentum. Remember that L = with = 1 for timelike orbits and = 0 for null orbits, so !23 2GM dt !2 1 2 2GM 1 dr !2 d 2 5 : (8.44) 4 2 = 1 c2r d c2 1 c2r d + r d We have dt = 1 2GM 1 k ; d = h : (8.45) d c2 r d r2 Substituting these into the above, we get 1 r_ 2 + r2_ 2 1 2GM GM2 = 1 c2 k2 2 : (8.46) 2 c2 r r 2 This is the Newtonian energy equation with a modi cation to the _ 2 term. Using dr and putting U = 1 we get r_ = _ d r !2 dU + U 2 1 2GM U 2GM2 U = c2 (k2 2 ) : (8.47) h2 d c2 h2 This equation can in fact be integrated immediately, but it leads to elliptical integrals, which are awkward to handle. We therefor dierentiate to obtain the equation: d2U + U = GM2 + 3GM U 2 : (8.48) d2 h2 c2


106

For timelike orbits [ = 1 ] this is just Newton's equation

d2U + U = GM2 ; d2 h2

(8.49)

apart from the last term.

8.7.1 Solution for timelike orbits and precession The solution to the Newtonian equation give the well known conics

U0 = ` 1 (1 + e cos ) ;

(8.50)

h2 and e is the eccentricity of the orbit [ if e = 0 it is circular ]. where ` = GM We can obtain an approximate solution U1 to the exact orbit equation [ valid for r >> GM c2 ], if we substitute the Newtonian solution into the term quadratic in U , that is if we solve d2U1 + U = 1 + 3h2 (1 + e cos )2 ; (8.51) 1 ` c2 `3 d2 where we have put = 1 [ for timelike orbits ]. This equation is solved by choosing a particular integral of the form [ EXERCISE 7.3]:

U1 = A + B sin + C cos 2 : This gives

!2 h 3 h U = U0 + ` `c 1 + e sin + e2 12

(8.52)

i 1 6 cos 2

:

(8.53)

The most important term is the one that is linear in e because it is the only one which in the course of time [ with many revolutions of the planet ] becomes larger and larger. We therefore ignore the other corrections to U0 and obtain

Now

2 3 !2 1 h U = ` 41 + e cos + 3e `c sin 5 :

(8.54)

h2 = GM` = GM 1 = r : `2c2 `2c2 c2 ` `

(8.55)

CHAPTER 8. THE SCHWARZSCHILD SOLUTION But we have ` >> rG so

h2 `2 c 2

107

1, and this simpli es our solution to

8 2 !2 3 9 < = h 1 U ` :1 + e cos 41 3 `c 5 ; :

(8.56)

The orbit of the planet is thus only approximately an ellipse. The solution for U is still a periodic function, but no longer with a period 2. The point at which the orbit is closest to the sun is reached again only after an additional rotation through the angle

!

2 = 2 `32hc2 = 6GM (8.57) `c2 : This is the famous perihelion precession of planetary orbits. For Mercury we obtain 43:0300 per century.

8.7.2 The bending of light We next consider the case of the trajectory of a light ray in a spherically symmetric static gravitational eld. The calculation is essentially the same as that given in the last section, except that light rays travel on null geodesics, so that = 0. The dierential equation we need to solve is therefore d2U + U = 3GM U 2 : (8.58) d2 c2 In the limit of Special Relativity, the last term vanishes and the equation becomes d2 U + U = 0 : (8.59) d2 The general solution can be written in the form U = 1b sin ( 0) ; (8.60) where b is the closest approach to the origin [ or impact parameter, see Figure 8.2 ]. This is the equation of a straight line as goes from 0 to 0 + . The straight line motion is the same as predicted by Newtonian theory.


108

Light ray P φ

φ

o

x

O

Figure 8.2: De ection of light ray We again solve the General Relativity problem by taking the general solution to be a perturbation of the Newtonian solution: U = U0 + U1 ; U0 = 1b sin ; (8.61) where we have taken 0 = 0 for convenience. It follows that the equation for U1 is: d2U1 + U = 3GM sin2 : (8.62) 1 d2 c2 b2 This equation can be solved by trying a particular integral of the form

U1 = A + B sin2 : This gives [ EXERCISE 7.3 ] so the full solution is

U1 = 2cGM 2 b2

GM sin2 ; c2b2

U = 1b sin + 2cGM 2 b2

(8.63) (8.64)

GM sin2 : (8.65) c2b2 Let us now calculate the de ection of a light ray from a star which just grazes the sun [ see Figure 8.3 ]. When r ! 1 U ! 0, so GM sin2 : 0 = 1b sin + 2cGM (8.66) 2 b2 c2b2 at the asymptotes = 1 and = 2 + , and taking 1 we get: (8.67) 0 = 1b 1 + 2cGM 2 b2 (8.68) 0 = 1b 2 + 2cGM 2 b2 :


109

Aparent star

δ

r

b ψ 1

φ ψ

2 Observer

Sun Star

Figure 8.3: Diagram showing the total de ection δ=β+α/2 β Quasar

D

b

α/2

Galaxy

d

Earth

Figure 8.4: Einstein ring. The total de ection is = 1 + 2 :

= 4GM bc2 :

(8.69)

This works out to be about 1:7500 and was con rmed by Eddington in 1919 during a solar eclipse. Another beautiful example of the bending of light is the gravitational lens. Take the example of a Quasar directly behind a galaxy in our line of sight. The distance of closest approach corresponds to an angle = 4GM (8.70) bc2 : Now from the diagram [ see Figure 8.4 ] above we have (8.71) = 2 + = db + Db :


110

Figure 8.5: Einstein ring lensing event since both and are small. It follows that the impact parameter can be written as ( !)1=2 4 GM Dd b = c2 d + D : (8.72) So the image of the quasar appears as a ring which subtends an angle

" #1=2 2 b GMD 4 = d = c d(d + D) :

(8.73)

Problems 1 1.1 Show that the electromagnetic wave equation,

@2 + @2 + @2 1 @2 = 0 @x2 @y2 @z2 c2 @t2 is invariant under a Lorentz transformation.

1.2 The equation for a spherical pulse of light starting from the origin at t = t = 0 is,

x2 + y2 + z2 c2t2 = 0: Show from the Lorentz transformations that observer O will also measure this same pulse to be spherical, in accord with Einstein's second postulate stating that the velocity of light is the same for all observers.

1.3 A - meson with an average lifetime of 2 10 6 sec is created in the upper

atmosphere at an altitude of 6000 m. When it is created it has a velocity of 0:998c in a direction towards the earth. What is the average distance that it will travel before decaying, as determined by an observer on earth? Consider an observer at rest with respect to the - meson. What is the distance he measures from the point of creation of the - meson to the earth? Comment on the physical signi cance of these results.

1.4 Consider a radioactive nucleus that moves with a constant speed 0:5c

relative to the laboratory. The nucleus decays and emits an electron with a speed 0:9c in a direction perpendicular to the direction of [ the laboratory's ] motion as determined by an observer at rest with respect to the nucleus. 111

PROBLEMS 1

112

Find the velocity and the direction of motion of the electron as measured by an observer in the laboratory frame.

1.5 The speed of light in still water is c=n, where the index of refraction for water is approximately n = 4=3. Fizeau, in 1851, found that the speed relative to the laboratory) of light in the water moving with a speed V [ relative to the laboratory ] could be expressed as u = nc + kV; where the \dragging coecient" was measured by him to be k = 0:44. Determine the value of k predicted by the Lorentz velocity transformations.

** If you have any problems please come and see me **

Problems 2 2.1 A car 5 m long tries to get into a garage 3 m long by driving into it at a

speed of 0:8c. Show that in the frame of the garage, the whole car can indeed enter the garage before its front strikes the wall. Also calculate the length of the garage as seen by the driver and prove that he expects to strike the wall 4=c seconds before the back of the car gets in the garage. Recalling that the maximum speed of propagation of information is c, explain how the car ts into the garage before the news that the front of the car has hit the garage wall reaches the back of the car.

2.2 In section 2.2 we showed that the four - velocity of a particle in a frame O is given by

U = (c; v) : Show that for an observer at rest relative to O [ i.e. with a four - velocity W = (c; 0; 0; 0) ], the particle's three - velocity v can be written as

W) U W : v = (W UW 2.3 A particle of rest mass m1 moving with velocity u1 along the x - axis

collides elastically with a stationary particle of rest mass m2 and as a result m1 and m2 are de ected through angles and respectively. If E and E 0 are the total energies of the particle m1 before and after the collision respectively, show that 2 0 2 22 cos = (Eq+ m22 c )E2 4 m202c E 2m4 c1 : (E m1 c )(E m1 c ) 113

PROBLEMS 2

114

HINT:Try looking at the square of the four - momentum of the re ected m2 and use E 2 p pc2 = m2 c4 to eliminate u1 and u01 . 2.4 De ne the four - velocity U and the four - acceleration A of a particle in Special Relativity and specify how they relate to the Newtonian velocity v and Newtonian acceleration dv=dt. If the rest mass of the particle is m0 , what is its four - momentum P? A particle moves with variable velocity v relative to some inertial frame, under the action of a force f . Show that m0 v dv v ; f = m0 dv + h

1

i v2 1=2 2 c

dt

h c2 1

i v2 3=2 2 c

dt

where v is the magnitude of v. Infer that f = h mv02 i3=2 ddtv 1 c2 if the acceleration is parallel to v, while f = h mv02 i1=2 ddtv 1 c2 if the acceleration is perpendicular to v. Suppose the particle moves along the x - axis under a force of magnitude 2 f = (2am0cx)a2 ; being at rest at t = 0. Show that the time taken to move to the point with coordinate x (< a) is

1=2 t = 31c xa (x + 3a) :


Problems 3 3.1 Prove, by writing out all the terms, the validity of the following

p~ (Ae ) = A p~ (e) : Let the components of p~ be ( 1; 1; 2; 0), those of A be (2; 1; 0; 1) and those of B be (0; 2; 0; 0). Find (i) p~(A); (ii) p~(B); (iii) p~(A 3B); (iv) p~(A) 3p~(B).

3.2 Given the following vectors in O:

A !O (2; 1; 1; 0) ; B !O (1; 2; 0; 0) ; C !O (0; 0; 1; 1) ; D !O ( 3; 2; 0; 0) ; (a) show that they are linearly independent. (b) Find the components of p~ if

p~(A) = 1 ; p~(B) = 1 ; p~(C) = 1 ; p~(D) = 0 : (c) Find the value of p~(E) for

E !O (1; 1; 0; 0) : (d) Determine whether the one - forms p~, q~, ~r and ~s are linearly independent if q~(A) = q~(B) = 0, q~(C) = 1, q~(D) = 1, ~r(A) = 2, ~r(B) = ~r(C) = ~r(D) = 0, ~s(A) = 1, ~s(B) = 1, ~s(C) = ~s(D) = 0.

3.3 Let p~ !O (1; 1; 0; 0) and q~ !O ( 1; 0; 1; 0) be two one forms. Prove by trying two vectors A and B as arguments, that p~ q~ 6= q~ p~. Then nd the components of p~ q~. 115

PROBLEMS 3

116

3.4 (a) Given the components of a tensor M as the matrix 00 1 0 01 BB 1 1 0 2 CC B@ 2 0 0 1 CA : 1 0 2 0

nd (i) the components of the symmetric tensor M ( ) and the antisymmetric tensor M [ ] ; (ii) the components of M ; (iii) the components of M ; and (iv) the components of M . (b) For the tensor whose components are M , does it make sense to speak of its symmetric and antisymmetric parts? If so, de ne them. If not, say why. (c) Raise an index of the metric tensor to prove that

= :

3.5 Suppose A is an antisymmetric 2=0 tensor, B is a symmetric 0=2 tensor, C is an arbitrary 0=2 tensor, and D is an arbitrary 2=0 tensor. Prove: (a) A B = 0; (b) A C = A C[ ]; (c) B D = B D( ) .

3.6 In some frame O, the four - vector elds U and D have the components [ with units where c = 1 ]:

p U ! 1 + t2; t2 ; 2t; 0 ; p D ! x; 5tx; 2t; 0 ;

and the scalar has the value

= x2 + t2 y2 : (a) Find U U, U D, D D. Is U suitable as a 4 - velocity eld? Is D?

PROBLEMS 3

117

(b) Find the spatial velocity v of a particle whose four - velocity is U, for arbitrary t. What happens to it in the limits t ! 0, t ! 1? (c) Find U for all . (d) Find U ; for all and . (e) Show that U U ; = 0 for all . Show that U U; = 0 for all . (f) Find D;.

(g) Find U D

for all .

; (h) Find U U D ;

similar?

and compare with (f) above. Why are the two answers

(i) Find ; for all . Find ; for all . What are the numbers f;g the components of? (j) Find rU, rU D. rD and rDU.


Problems 4 4.1 By considering the ux of particle number across the surface of a cube of side a and letting a ! 0, derive the conservation law: N ; = 0 :

4.2 The energy - momentum tensor for a perfect uid is: 0 c2 0 0 0 1 BB 0 p 0 0 CC B@ 0 0 p 0 CA 0 0 0 p

in the momentarily comoving reference frame [ MCRF ]. By applying a Lorentz transformation , show that in a general frame it is given by p T = + c2 U U + p : Prove that the time and space parts of the conservation equations T ; = 0 in the MCRF are d = + p=c2 dn dt n dt and p + c2 U i ; U + i p; = 0 ; and derive the Newtonian limit of these equations. [ This is just lling in

the algebra of the derivation in the lectures.] 118

PROBLEMS 4

119

4.3 Using the de nition of the electromagnetic tensor F and the current four - vector J given in the lectures, show that Maxwell's equations can be written as

F ; = 4J and

F; + F; + F; = 0 : Perform a Lorentz transformation on F to a frame O with velocity v in the x - direction to prove that the part of the electric eld perpendicular to x^ changes to (E? + v B), while the part along x^ is unchanged.

4.4 The four - force on a particle of charge q and four - velocity U is K = qF U : Express its components in terms of E, B and the three - velocity v. By writing J = qU and using Maxwell's equations, show that

K = T ; ; where

T = 41 F F 14 F F is the energy - momentum tensor of the electromagnetic eld. Infer that the energy density is T 00 = 81 E2 + B2 :

NOTE: This question is worth a bottle of wine!!! ** If you have any problems please come and see me **

Problems 5 5.1 A rocket of height h undergoes a constant acceleration g. Light rays

emitted from the top at t = 0 and t = reach the bottom at times t = t0 and t = t0 + t. Prove that, so long as gh

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