A First Course in General Relativity

Bernard F Schutz

(2nd Edition, Cambridge University Press, 2009)

Solutions to Selected Exercises(Version 1.0, November 2009)

To the user of these solutions:

This document contains solutions to many of the Exercises in the secondedition of A First Course in General Relativity. The textbook oers an ex-tensive collection of exercises, some of which prove results omitted from thetext, or form the basis for later exercises, or provide the foundation for re-sults in later chapters. Doing exercises is integral to the process of learning asubject as complex and conceptually challenging as general relativity. Thesesolutions, therefore, are meant to help users of this book master exercisesthat they might have had diculty with. It is assumed that an attempt hasalready been made at solving the problem!

Solutions are provided to those problems which might present particularconceptual challenges; those which require just routine algebra, and thosewhich require computer solutions, are generally not included.

I hope you nd the solutions helpful!

Bernard SchutzPotsdam, Germany

1

Chapter 1

1 3.71024 kgm1; 3.51043 kgm; 107; 104 kg; 1.11012 kgm3;103 kgm3; 3.7 1016 kgm3.

2 3106 ms1; 91035 Nm2; 3.3109 s; 91016 Jm3; 91017 ms2.3 (a)(k) See the top panel in Figure 1. Note that most of the items

continue into other quadrants (not shown). (l) See the bottom panelin the same gure. Note that light always travels at speed 1 in thisframe even after reecting o of a moving mirror.

tO

x (m)

(m)

1

2

3

321

(l)

(a)(b)tO

xO

(c)

(c)(d)

(e)(f )

(f )

(g)

(h)

(i)

(j)

(k)

tO

xO (m)

(m)

1

2

3

321

tO

xO

reflection

absorption

mirror

detector

Figure 1: Solution to Ex. 3 of Chapter 1. See the solution text for explana-tion.

2

5 (a) See the top panel in Figure 2. The particles (short dashed worldlines) are emitted at event A and reach the detectors at events B and C.The detectors send out their signals (more short dashed world lines)at events D and E , which arrive back at the spatial origin at eventF . The line BC joining the two reception events (long dashed) at thedetectors is parallel to the x=axis in our diagram, which means thatthey occur at the same time in this frame. Note that the lines DFand EF are tilted over more than the lines AB and AC because thereturning signals go at speed 0.75 while the particles travel only atspeed 0.5.

tO

x (m)

(m)

-2

2

4

42-2

tO

xO

O

-4

6(a)

tO

x

(m)

(m)

-2

2

4

42-2

O

-4

6(c)

A

A

B

B

C

C

D

D

E

E

F

F

Figure 2: Solution to Ex. 5 of Chapter 1. See the solution text for explana-tion.

3

(b) The experimenter knows that the detectors are equidistant fromx = 0, and that the signals they send out travel at equal speeds.Therefore they have equal travel times in this frame. Since they arriveat the same time, they must have been sent out at the same time.This conclusion depends on observer-dependent things, such as thefact that the signals travel at equal speeds. Therefore the conclusion,while valid in this particular frame, might not be valid in others.

(c) See the bottom panel in Figure 2. This is drawn using the axes offrame O, which we draw in the usual horizontal and vertical directions,since any observer would normally draw his/her axes this way. In thisframe, the frame O moves forwards at speed 0.75. These axes aredrawn in gray, and calibrated using invariant hyperbolae (not shown),just as in the solution to Exercise 1.3. Then the events are locatedaccording to their coordinate locations on the axes of O. Note thatto do this, lines of constant tO must be drawn parallel to the xO axis,and lines of constant xO must be drawn parallel to the tO axis. Onesuch line is the long-dashed line BC. The detectors (heavy lines) mustpass through the points xO = 2 m on the xO-axis. The two signal-emission events D and E are clearly not simultaneous in this frame,although they are simultaneous in O: D occurs much earlier than E .Note also that the signal sent at the event E and received at F remainsat rest in O, since it was sent backwards at speed 0.75 in frame O,exactly the same speed as the frame O.(d) The interval is easily computed in frame O because the events Dand E have zero separation in time and are separated by 4 m in x.So the squared interval is 16 m2. To compute it in frame O, measureas carefully as you can in the diagram the coordinates tO and xO forboth emission events. Given the thickness of the lines representing thedetectors, you will not get exactly 16, but you should come close.

6 Write out all the terms.

7 M00 = 2 2,M01 = ,M11 = 2 2,M22 = a2,M33 =b2,M02 = M03 = M12 = M13 = M23 = 0.

8 (c) Use various specic choices of xi; e.g. x = 1,y = 0,z =0M11 = M00.

10 Null; spacelike; timelike; null.

4

11 Asymptotic refers to the behavior of the curves for large values of xand t. But when these variables are suciently large, one can neglecta and b, and then one has approximately t2 + x2 = 0, leading tot = x. This approximation is better and better as t and x get largerand larger.

13 The principle of relativity implies that if time dilation applies to oneclock (like one based on light travel times over known distances, whichis eectively the sort we use to calibrate our time coordinate), then itapplies to all (like the pion half-life). Algebra gives the result.

14 (d) For (a), 3.7 105.16 (a) In Fig. 1.14, we want the ratio t/t for event B (these are its time-

coordinate dierences from A). The coordinates of B in O are (t, v t).The rst line of Eq. (1.12) implies t = t(1 v2)1/2.

17 (a) 12m.

(b) 1.25 108 s = 3.75m; 211m2: spacelike.(c) 9m; 20m. (d) No: spacelike separated events have no unique timeordering. (e) If one observer saw the door close, all observers musthave seen it close. The nite speed of transmission (< c) of the shockwave along the pole prevents it behaving rigidly. The front of the polemay stop when it hits the wall, but the back keeps moving and cantbe stopped until after the door has shut behind it. Many apparentparadoxes in special relativity are resolved by allowing for the nitespeed of transmission of pressure waves: for a body to be perfectlyrigid would violate special relativity, since the communication amongits dierent parts would have to happen instantaneously.

18 (b) tanh[N tanh1(0.9)] 1 2(19)N .19 (c) The analog of the interval is the Euclidean distance x2 + y2. The

analog of the invariant hyperbola is the circle x2 + y2 = a2. The circleallows one to transfer the measure of length along, say, the x-axis upto the y-axis.

21 The easiest way to demonstrate these theorems is graphically: put twoarbitrary events onto a spacetime diagram, and join them. If they aretime-like separated, then the line joining them will be more verticalthan a light-line, and so there is a frame in which it is parallel tothe time-axis, in which frame it will represent a line of xed spatial

5

position: the two events will be at the same spatial point in this newframe. This proves (a). The proof of (b) is similar.

6

Chapter 2

1 (a) 4; (b) 7, 1, 26, 17; (c) same as (b); (d) 15, 27, 30, 2 (dierentfrom (b) and (c) because the sum is on the second index, not the rst);(e) A0 B0 = 0, A2B3 = 0, A3B1 = 12, etc.; (f) 4; (g) the subset of(e) with indices drawn from (1, 2, 3) only.

2 (c) , free; , dummy; 16 equations.

5 (b) No.

6 To get the basis vectors of the second frame, compose velocities to getv = 0.882.

8 (b) The dierence of the two vectors is the zero vector.

10 Choose each of the basis vectors for A, getting the four equations inEq. (2.13) in turn.

12 (b) (35/6, 37/6, 3, 5).14 (a) (0.75/1.25) = 0.6 in the z direction.15 (a) (, v, 0, 0); (b) (, vx, vy, vz) with = (1 v v)1/2. (d)

vx = Ux/U0 = 0.5, etc.

18 (a) Given a a > 0, b b > 0, a b = 0; then (a + b) (a + b) =a a + 2a b+b b > 0.(b) If a is timelike, use the frame in which a (a, 0, 0, 0).

19 (b) v = t(1 +2t2)1/2, = 1.1 1016 m1; t = 2.0 1017 m =6.7 108 s = 22 yr.(c) v = tanh (), x = 1[(1 + 2t2)1/2 1], 10 yr.

22 (a) 4 kg, 3.7 kg, 0.25 (ex+ey); (b) (3, 12 , 1, 0) kg, 3 kg, 2.8 kg, 16ex+13ey, 0.2 ey.

23 E = m + 12m|v |2 + 38m|v |4 + . . .; |v |2 = 2/3.24 Work in the CM frame.

25 (a) Lorentz transformation of p. (b) v = 2cos /(1 + cos2 ).

27 The cooler, because ratio of the rest masses is 1 + 1.1 1016.

7

31 2h cos ; h cos .

33 Emax = 8 105 mp, above the -ray band.

8

Chapter 3

3 (b) (i) 1; (ii) 2; (iii) 7; (iv) same.4 (b) p (1/4,3/8, 15/8,23/8); (b) 5/8; (d) yes.5 The order of the multiplication of numbers does not matter, nor does

the order in which the sums are done; summation over produces theunit matrix because the two matrices are inverse to one another;summation over is just multiplication by the unit identity matrix.

6 (a) Consider p = 0, an element of the basis dual to {e}; (b) (1, 0,0, 1).

8 See Figure 3 for the diagram. Note that in the representation of a one-form as a set of surfaces, the extent of the surfaces and the numberof them is not important. The critical property is their orientationand spacing. The one-form dt has surfaces separated by one unit inthe t-direction so that, when the basis vector e0 is drawn on top ofit, it crosses exactly one surface, and when the other basis vectors areplaced on it they cross no surfaces (they lie in the surfaces).

tO

x (m)

(m)

1

2

3

321O

} dttO

x (m)

(m)

1

2

3

321O

}

dx

Figure 3: Solution to Ex. 8 of Chapter 3. See the solution text for explana-tion.

9

9 dT (P) (15,15); dT (L) (0, 0).10 (a) A partial derivative wrt x holds all x xed for = , so

x/x = 0 if = . Of course, if = then x/x = 1.12 (a) By denition, n(V ) = 0 if V is tangent to S, so if V is not tangent,

then n(V ) = 0.(b) If V and W point to the same side of S then there exists a positivenumber such that W = V + T , where T is tangent to S. Thenn( W ) = n(V ) and both have the same sign.

(c) On the Cartesian basis, the components of n are (, 0, 0) for some. Thus any n is a multiple of any other.

16 (e) 10 and 6 in four dimensions.

18 (b) q (1,1, 1, 1).20 (a) In matrix language, A is the product of the matrix with

the column vector A, while p is the product of the transpose of with the column vector p. Since

is inverse to

, these are

the same transformation if equals the transpose of its inverse.

21 (a) The associated vectors for t = 0 and t = 1 point inwards.

24 (b) No. The arguments are dierent geometrical objects (one vector,one one-form) so they cannot be exchanged.

25 Use the inverse property of and

.

26 (a) AB = AB = AB = AB = AB .Therefore 2AB = 0. The justication for each of the above steps:antisymmetry, symmetry, relabel dummy indices, relabel dummy indices.

28 Arbitrariness of U .

30 (a) Since D D = x2 + 25t2x2 + 2t2 = 1, D is not a four-velocityeld.

(f) 5t.

(h) 5t because of (e).(i) the vector gradient of has components {,} = (2t, 2x,2y, 0).(j) U D [t2, 5t3 + 5x(1 + t2),

2(1 + t2), 0].

10

33 (d) Given any matrix (A) in 0(3), let () be the 4 4 matrix1 0 0 000 (A)0

.

Show that this is in L(4) and that the product of any two such matricesis one of the same type, so that they form a subgroup. These arepure rotations of the spatial axes (relative velocities of the two framesare zero). Transformations like Eq. (1.12) are pure boosts (spatialaxes aligned, relative velocity nonzero). The most general Lorentztransformation involves both boost and rotation.

34 (c) guu = gvv = 0, guv = 1/2, guy = guz = gyz = 0, gyy = gzz = 1.(e) du = dt dx, dv = dt + dx,g(eu, ) = dv/2,g(ev,) = du/2.Notice that the basis dual to {eu, ev}, which is {du, dv}, is not thesame as the set of one-forms mapped by the metric from the basisvectors.

11

Chapter 4

1 (a) No. (b) Yes. (c) Dense trac can indeed resemble a continuum,and trac congestion on highways often shows wave-like behavior,with cars moving through a compression zone (slow speed) into a rar-efaction zone (fast speed) and back into a compression zone. See, forreferences, the Wikipedia article on Trac Flow. (d) This is onesituation where cars are not a continuum: they move through the in-tersection one by one. (e) If the plasma is dense enough then it is acontinuum, but very raried plasmas, particularly in astronomy, canbe dicult to describe fully this way.

2 Particles contributing to this ux need not be moving exclusively inthe x direction. Moreover, consider a change to the non-orthogonalcoordinates (t, x, y =: x+ y, z). A surface of constant x is unchanged,so the ux across it is unchanged, but the x direction, which is thedirection in which now t, y, and z are constant, is in the old ex +ey direction. Is the unchanged ux now to be regarded as a ux inthis new direction as well? The loose language carries an implicitassumption of orthogonality of the coordinates in it.

3 (a) In Galilean physics p changes when we change frames, but inrelativity p does not: only its components change.

(b) This is because the usual Galilean momentum is only a three-vector. However, if in Galilean spacetime we dene a four-vector(m,p) then the Galilean transformation changes this to (m,p mv),where v is the relative velocity of the two frames. This is an approxi-mation to the relativistic one (see Eq. (2.21)) in which terms of orderv2 are neglected.

4 The required density is, by denition, N0 in the frame in which U (1, 0, 0, 0). In this frame N U = N0.

8 (a) Consider a two-dimensional space whose coordinates are, say, pand T , each point of which represents an equilibrium state of theuid for that p and T . In such a space the d of Eq. (4.25) is justd, where is whatever vector points from the old state to thenew one, the change in state contemplated in Eq. (4.25). Since wewant Eq. (4.25) to hold for arbitrary , it must hold in the one-form version. See B.F. Schutz, Geometrical Methods of MathematicalPhysics (Cambridge University Press, Cambridge, 1980).

12

(b) If q = dq, then TS/xi = q/xi, where xi is either p or T . Theidentity 2q/Tp = 2q/pT implies (T/p)(S/T ) = S/p,which will almost never be true.

9 Follow the steps leading to Eq. 4.33 but changing the rst index 0to, say, x. Use the form of Newtons law that says that the force isthe rate of change of the momentum. Interpret the terms analogousto those on the left-hand-side of Eq. 4.30 as uxes of momentum.

11 (a) By denition of rotation: see Exer. 20b, 3.10.(b) Suppose M has the property OTMO = M for any orthogonalmatrix O. Consider the special case of a rotation about x3, whereO11 = cos ,O12 = sin ,O21 = sin ,O22 = cos ,O33 = 1, all otherelements zero. Then OTMO = M for arbitrary implies M13 =M23 = M31 = M32 = 0,M11 = M22, and M21 = M12. By relabeling,a rotation about x implies M12 = M21 = 0,M33 = M22. Therefore Mis proportional to I.

13 U = U ; is constant, so , = 0;U,U = U,u(relabeling) = U ,U (symmetry of ).

14 Since U MCRF (1, 0, 0, 0), multiplying by U and summing on picks out the zero component in the MCRF.

16 There is no guarantee that the MCRF of one element is the same asthat of its neighbor.

20 (a) In Eq. 4.58, let V be N but now show that the expression doesnot equal zero, but instead equals the integral of over the four-dimensional volume. Interpret the result as the dierence betweenthe change in the number of particles and the number that have en-tered over the boundaries. Show that this means that is the rate ofcreation of particles per unit volume per unit time.

(b) F 0 is the rate of generation of energy per unit volume, and F i isthe ith component of the force. F is the only self-consistent general-ization of the concept of force to relativity.

21 (a) T = 0UU , U (1, , 0, 0), = (1 2)1/2.(b) At any point on the ring, the particles have speed a. At position(x, y) we have U (1,y, x, 0), = (1 2a2)1/2. In theinertial frame their number density is N [22a(a)2]1, which equals

13

nU0, where n is the number density in their rest frame. Thereforen = N [22a(a)2]1 and T = mnUU.

(c) Add (b) to itself with . For example, T 0x = 0 and T xx =2mn2y22.

22 No bias means that T ij is invariant under rotations. By Exer. 11T ij = pij for some p. Since T 0i = 0 in the MCRF, Eq. (4.36) holds.Clearly = nm, where = (1 v2)1/2. The contributions of eachparticle to T zz, say, will be the momentum ux it represents. For aparticle with speed v in the direction (, ), this is a z component ofmomentum mv cos carried across a z = const. surface at a speedv cos . If there are n particles per unit volume, with random velocities,then T zz = n(mv)(v) times the average value of cos2 over the unitsphere. This is 13 , so T

zz = p = nmv2/3. Thus, p/ = v2/3 13 asv 1. (In this limit m remains nite, the energy of each photon.)

23 This exercise prepares us for computing the quadrupole radiation ofgravitational waves in Ch. 9. Note that, although the stress-energytensors components vanish outside a bounded region of space, thedomain of integration in the spatial integrals in this exercise can stillbe taken to be all of space: since the divergence vanishes outside thesystem, integrating over the exterior does not add anything. Thisseemingly small detail will be used in all parts of this exercise.

(a) The integral is over spatial variables, so the partial derivative withrespect to time may be brought inside, where it operates only on T 0.Then use the identity T , = 0 to replace T 0,0 by T j,j. Thisis a spatial divergence, so its integral over d3x converts to a surfaceintegral, by Gauss law in three dimensions. Since the region of in-tegration in space is unbounded, this surface can be taken anywhereoutside the domain of the body, where the stress-energy componentsall vanish. That means that the surface integral vanishes, and thisproves the result.

(b) The computation follows that in (a) closely. First bring one time-derivative inside the integral, replace T 00,0 by T 0k,k, and this timeintegrate by parts on xk. The integrated term is a full divergence,and vanishes as above. But the integration by parts leaves anotherterm involving (xixj)/xk. Since partial derivatives of one coordinatewith respect to another are zero, the only terms that survive thisdierentiation are where the indices i or j equal k. We can writethe result using Kronecker deltas: (xixj)/xk = ikxj + jkxi. The

14

integral involves this expression multiplied by T 0k and summed on k.This summation leaves a relatively simple integrand: T 0ixj T 0jxi.But this integral still has one further time-derivative outside it. Bringthis one in now, let it operate on the components of T (it doesnot aect the coordinates xi and xj), and again replace these time-derivatives by the spatial divergences. Again integrate by parts andevaluate the full divergence on a surface outside the body. There willbe more Kronecker deltas, but when summed they will give the simpleresult.

(c) This is a variation on the procedure in (b). Follow the same steps,but there are now more factors of xk so the result will be the one givenin the problem.

24 (c) R/x = [(1 v)/(1 + v)]1/2.25 (h) Ex = Ex; let Eyey+Ezez be called E, the part of E perpendicu-

lar to v . Similarly, let E = Eye +Eze z. Then E = (E+vB).

15

Chapter 5

3 (b) (i) Good except at origin x = y = 0: usual polar coordinates. (ii)Undened for x < 0, fails at x = 0, good for x > 0: maps the right-hand plane of (x, y) onto the whole plane of (, ). (iii) Good exceptat origin and innity: an inversion of the plane through the unit circle.

4 A vector has a slope which is the ratio of its y- and x-components. Forthe given vector this is (dy/d)/(dx/d) = dy/dx, which is the slopeof the curve.

5 (a) and (b) have same path, the unit circle x2 + y2 = 1. But theirtangent vectors are dierent even at the same point, because theparametrization is dierent. Additionally, for this problem the points = 0 in (a) and t = 0 in (b) are dierent.

6 Recall the solution to Exer. 3.8.

7

11 = x/r = cos ;

21 = y/r2 = 1r sin ;

12 = y/r = sin ;

22 = x/r2 = 1r cos ;

11 = x/r = cos ;

21 = y/r = sin ;

12 = y = r sin ;22 = x = r cos .

8 (a)

f = r2(1 + sin 2)

V r = r2(cos3 + sin3 ) + 6r(sin cos )

V = r sin cos (sin cos ) + 3(cos2 sin2 )W r = cos + sin

W = (cos sin )/r

16

(b)

(df)x = 2x + 2y; (df)y = 2x+ 2y

(df)r = 2r(1 + sin 2) = f/r

(df) = 2r2 cos 2 = 12(df)x + 22(df)y

(c)(i)

(V )r = V r = r2(cos3 + sin3 ) + 6r(sin cos )

(V ) = r2V = r3 sin cos (sin cos ) + 3r2(cos2 sin2 )(W )r = W r = cos + sin

(W ) = r2W = r(cos sin )

(c)(ii) Same result by a dierent method, e.g.:

(W )r = 11(W )x + 21(W )y = cos + sin

10 The key thing is to prove linearity. The lower index, associated withthe derivative, is linear in whatever vector we give it, just as is thederivative of a scalar function. So if we double the vector argument,we get a derivative twice as large, since it has to approximate thechange in the function when we go twice as far. Similarly, the one-form argument is linear (associated with the upper index) because ofEq. 5.52.

11 (a) The Christoel symbols are zero in Cartesian coordinates so theresult is:

V x,x = V x;x = V x/x = 2x; V x,y = 3; V y,x = 3; V y,y = 2y.

(b) Although it is possible to do this using matrices, the straightfor-ward expansion of the summations in the transformation equation isless error prone. Thus, the r r component of v is

V r ;r = 11V ,

= 1111V 1,1 + 1

211V 2,1 + 1

121V 1,2 +1

221V 2,2

= 2r(cos3 + sin3 ) + 6 sin cos .

17

Other components are:

V r ; = 2r2 sin cos (sin cos ) + 3r(cos2 sin2 )V ;r = 2 sin cos (sin cos ) + 3(cos2 sin2 )/rV ; = 2r sin cos (sin + cos ) 6 sin cos

(c) This gives the same as (b).

(d) 2(x + y).

(e) 2r(sin + cos ), same as (d).

(f) Same as (d).

12 (a) Same components as in Exer. 11a.

(b),(c) These components are the same for both (b) and (c) and arerelated to the answers given for Exer. 11c as follows: pr;r = V r ;r, pr; =V r ;, p;r = r2V ;r, p; = r2V ;. It happens that p;r = pr; for thisone-form eld. This is not generally true, but happens in this casebecause p is the gradient of a function.

14 Two selected results: Ar ; = r(1 + cos tan ); Arr ;r = 2r.15 Of the rst-derivative components, only V ; = 1/r is nonzero. Of

the second-derivative components, the only nonzero ones are V ;r; =1/r2, V r ;; = 1, and V ;;r = 1/r2. Note that this vector eld isjust the unit radial vector of polar coordinates.

17

e

x=[

x(e)

] =

ex

+ ,e

= +,

20 = 12g(g, + g, g, + c + c c).

21 (a) Compute the vectors (dt/d, dx/d) and (dt/da, dx/da) and showthey are orthogonal.

(b) For arbitrary a and , x and t obey the restriction |x| > |t|. Thelines {x > 0, t = x} are the limit of the = const. hyperbolae asa 0+, but for any nite the limit a 0+ takes an event to theorigin x = t = 0. To reach x = t = 1, for example, one can set = ln(2/a), which sends as a 0.

18

(c) g = a2, gaa = 1, ga = 0, a = 1/a, a = a, all otherChristoel symbols zero. Note the close analogy to Eqs. (5.3), (5.31),and (5.45). Note also that g (the time-time component of the metric)vanishes on the null lines |x| = |t|, another property we will see againwhen we study black holes.

22 Use Eq. (5.68).

19

Chapter 6

1 (a) Yes, singular points depend on the system.

(b) Yes: if we dene r = (x2 + y2)1/2, the map {X = (x/r) tan(r/2),Y = (y/r) tan(r/2)} shows that, as a manifold, the interior of theunit circle (r < 1) is indistinguishable from the whole plane (X,Yarbitrary). This map distorts distances, but the metric is not part ofthe denition of the manifold.

(c) No, discrete.

(d) This consists of the unit circle and the coordinate axes. It has thestructure of a one-dimensional manifold everywhere except at the veintersection points, such as (1, 0).

2 (a) Normally no metric is used on this manifold: since the axes repre-sent physically dierent quantities, with dierent units, no combina-tion like p2 + q2 is normally useful or meaningful.

(b) The usual Euclidean metric is normally used here.

(d) On the dierent segments the one-dimensional Euclidean metric(length) is used.

4 (a) Symmetry on (, ) means there are 12n(n+1) = ten independentpairs (, ). Since can assume four values independently, there are40 coecients.

(b) The number of symmetric combinations (, , ) is 16n(n + 1)(n + 2) = 20, times four for , gives 80.

(c) Two symmetric pairs of ten independent combinations each give 100.

5 Carefully repeat the at-space argument in a local inertial frame.

6 gg, = gg, (symmetry of metric) = gg, (relabelingdummy indices) = gg, (symmetry of metric). This cancels thesecond term in brackets.

9 For polar coordinates in two-dimensional Euclidean space, g = r2.In three dimensions it is g = r4 sin2 . Notice in these cases g > 0,so formulae like Eqs. (6.40) and (6.42), which are derived from Eq.(6.39), should have g replaced by g.

10 The vector maintains the angle it makes with the side of the triangleas it moves along. On going around a corner the angle with the new

20

side exceeds that with the old by an amount which depends only onthe interior angle at that corner. Summing these changes gives theresult.

15 Consider a spacelike curve {x()} parametrized by , and going fromx(a) to x(b). Its length is

ba [g(dx

/d)(dx/d)]1/2/d. Assumethat is chosen so that the integrand is constant along the given curve.Now change the curve to {x() + x()} with x(a) = x(b) = 0.The rst-order change in the length is, after an integration by parts, ba [

12g,U

U d(gU)/d]x d, where U dx/d. For ageodesic, the term in square brackets vanishes.

18 (b) Each pair or is antisymmetric, so has six independent com-binations for which the component need not vanish. Each pair canbe chosen independently from among these six, but the componentis symmetric under the exchange of one pair with the other. (c) Eq.(6.69) allows us to write Eq. (6.70) as R[] = 0. There are onlyn(n 1) (n 2)/6 = four independent choices for the combination, by antisymmetry. In principle is independent, but if equalsany one of , , and then Eq. (6.70) reduces to one of Eq. (6.69).So Eq. (6.70) is at most four equations determined only by, say, thevalue . However, Eq. (6.69) allows us also to change Eq. (6.70) toR[] = 0. Thus, if we had earlier taken, say, = 1 and = 2, thenthat equation would have been equivalent to the one with = 2 and = 1: all values of give the same equation.

25 Use Eqs. (6.69) and (6.70).

28 (a) We desire the components g in the new (spherical) coordinates,so we need to compute derivatives of the Cartesian with respect to thespherical, e.g.

x/r = sin cos,

and so on. Then we compute the new components from the transfor-mation law, e.g.

gr =x

r

x

gxx +

x

r

y

gxy + . . . ,

summing over all components of the metric in Cartesian coordinates.Since most of the components are zero the algebra is not very long.

21

(b) If one moves on the surface of the sphere then one is at constantr, so that dr = 0. With this in the line element, only the second andthird rows and columns of the matrix are relevant.

(c) g = r2, g = (r sin )2, g = 0.

29 R = sin2 .

30 The most sensible coordinates to use are Euclidean: unwrap the cylin-der so that it lies at and put a at coordinate system on it. To makea cylinder you just have to remember that when you come to the edgeof the paper you jump to the other side, where the join was. Butlocally, this makes no dierence to the geometry. Since the metric isEuclidean, its derivatives all vanish, and the Riemann tensor vanishes.

32 Compare with Exer. 34, 3.10.35 The nonvanishing algebraically independent Christoel symbols are:

ttr = , rtt = exp(2 2),rrr = ,r = rexp(2),r = r sin2 exp(2), r = r = r1, = sin cos , = cot . Here primes denote r derivatives. (Compare these withthe Christoel symbols you calculated in Exer. 29.) As explained inthe solution to Exer. 18, in calculating R we should concentrateon the pairs () and (), choosing each from the six possibilities(tr, t, t, r, r, ). Because R = R , we do not need tocalculate, say, Rr after having calculated Rr. This gives 21 in-dependent components. Again following Exer. 18, one of the com-ponents with four distinct indices, say Rtr, can be calculated fromothers. We catalog, therefore, the following 20 algebraically indepen-

22

dent components:

Rtrtr = [ ()2 ] exp(2),

Rtrt = Rtrt = Rtrr = Rtrr = 0,Rtt = r exp(2 2),Rtt = Rtr = Rtr = Rt = 0,

Rtt = r sin2 exp(2 2),Rtr = Rtr = Rt = 0,Rrr = r,Rrr = Rr = Rr = 0,

Rrr = r sin2 ,

R = r2 sin2 [1 exp(2)].

See if you can use the spherical symmetry and time independence ofthe metric to explain why certain of these components vanish. Alsocompare R with the answer to Exer.29 and see if you can explainwhy they are dierent.

36 Since = 0 is already inertial, we can look for a coordinate trans-formation of the form x

= ( + L)x, where L is of order .

The solution to Exer. 17, 5.9, gives , which must vanish at P .Since

= +L +L,x, we nd L(,) = 12

at P . The

antisymmetric part, L[,], is undetermined, and represents a Lorentztransformation of order . Since we are only looking for an inertialsystem, we can set L[,] = 0. Calculating at P (as in Exer. 3, 7.6, below) gives the new coordinates. In particular, the equationxi

= 0 gives the motion of the origin of the new frame, whose acceler-

ation is d2xi/dt2 = itt = ,i. We shall interpret this in the nextchapter, where we identify as the Newtonian gravitational potentialand see that this acceleration expresses the equivalence principle.

37 (b) The ranges of the coordinates must be deduced: 0 < < , 0 < < , 0 < < 2. Then the volume is 22r3.

38 2r sin .

23

Chapter 7

1 Consider a uid at rest, where U i = 0 and U0 = 1 in a local inertialframe. Then n/t = R, so Eq. (7.3) implies creation (or destruction)of particles by the curvature.

2 g00 = (1 2), gij = ij(1 + 2).3 000 = , 00i = ,i, 0ij = ij , i00 = ,i, i0j = ij ,

ijk = jk,i ij,k ik,j.7 (a)(i) In the Minkowski metric, all metric components are indepen-

dent of the given coordinates (t, x, y, z), so the momentum components(pt, px, py, pz) are conserved.

(a)(ii) The Schwarzschild metric components are independent of timet, so the component pt is conserved. Also, the metric does not dependon the coordinate , so that the associated momentum p is conserved.In part (b)(ii) below we will see that this is an angular momentum.

(a)(iii) The Kerr metric is a somewhat more complicated generalizationof Schwarzschild. Like Schwarzschild, however, the components are allindependent of the two coordinates t and . Therefore there are twoconserved momentum components, pt and p.

(a)(iv) The Robertson-Walker metric does depend on time t, but likethe previous two it is independent of the coordinate . Therefore pis conserved.

(b)(i) This coordinate transformation is to spherical coordinates, andit produces a metric whose components are independent of the angle. This means that p is conserved, the angular momentum aboutthe axis of our coordinates. But this axis could have been chosento point in any direction, so there are three independent conservedcomponents of angular momentum, usually referred to as (Jx, Jy, Jz).This exhausts the set of conserved quantities.

(b)(ii) Now we can see that there are more conserved quantities forSchwarzschild. The metric is written in coordinates that look like thespherical coordinates of Minkowski spacetime that we have just de-rived, and indeed the way it depends on the angles and is identicalto Minkowski spacetime in spherical coordinates. This means that themetric is in fact spherically symmetric. Now, since no metric compo-nent depends on , then certainly p is conserved. But we can alsoimagine performing exactly the kinds of coordinate transformations

24

(rotations) on these coordinates as one would perform in Minkowskispacetime if one wanted to re-orient the symmetry axis of the coordi-nates in a dierent direction, and that would again produce a metricindependent of the new angle . So, just as in at spacetime, thereare actually three conserved quantities associated with angles, threeangular momentum components. There are no other conservation lawsfor Schwarzschild.

(b)(iii) There is no spherical symmetry for the Kerr metric, so we donot nd any further conserved quantities.

(b)(iv) The Robertson-Walker metric is spherically symmetric in thesame way as Schwarzschild and Minkowski are. Therefore there arethree conserved angular momentum components. But linear momen-tum is not conserved.

9 (a)

R0i0j = ,ij + ij,00,R0ijk = ,0jik ,0kij ,Rijkl = ik,jl + jl,ik jl,ik jk,il.

(c) The acceleration, in Newtonian language, is ,i. The dier-ence between the accelerations of nearby particles separated by j istherefore j.ij.

10 (b) In terms of a Lorentz basis, the following vector elds are Killingelds: et, ex, ey, ez , xeyyex, yezzey, zexxez, tex+xet, tey+yet, tez+zet. Any linear combination with constant coecients of solutions toEq. (7.45) is a solution to Eq. (7.45), which means that the analogouselds to these in another Lorentz frame are derivable from these.

25

Chapter 8

2 (c) (i) 7.4251011 m2; (ii) 8.2611012 m2; (iii) 1.0901016 m1;(iv) 2.756 1012.(d)mPL = 2.176108 kg; tPL = 5.3901044 s. Typical elementary-particle lifetimes are 1024 s or greater. The heaviest known particlesare less than 1023 kg.

3 (a)(i) 2.122 106; (ii) 9.873 109; (iii) 6.961 1010; (iv)9.936 105.(b) We and everything on Earth are in free-fall around the Sun, so thevalue of the Suns potential does not matter, nor even the value of itsgradient (the gravitational acceleration the Sun produces on us). Whatwe do experience is the tidal force of the Sun, which is comparable insize to that of the Moon, and which raises ocean tides and distortsEarths shape. The tidal forces arise in the second derivatives of theSuns potential, i.e. the dierences in the Suns acceleration acrossEarth. We experience Earths gravity, even though its potential ismuch smaller than the Suns, because we are not in free-fall on Earth.The forces we feel from our weight are actually the forces of the groundor oor pushing up on us, stopping us from falling.

5 (a) This is an exercise in index gymnastics. Make sure that youuse all possible symmetries, in this case the freedom to exchange theorder of partial derivatives. (b) Since the expression is linear, a gaugetransformation is just the same as adding a pure-gauge value, and wehave just shown that this vanishes. Therefore the Riemann tensor atlinear order is gauge-invariant.

9 (b) No contradiction because this expression was derived by applyingthe gauge condition Eq. 8.33, as is computed explicitly in the nextexercise, Ex. 10. One could use this condition to remove the time-derivatives from the time-components of Eq. 8.42, at the expense ofmaking them more complicated to write down. The important thingis that the gauge condition Eq. 8.33 creates a relationship among thedierential equations in Eq. 8.42, so that there are still only 6 inde-pendent equations with second-time derivatives.

11 Gauge transformation: A A + f,. Lorentz gauge condition:A, = 0. These are very similar to Eq. 8.24 and Eq. 8.33. The dif-ferences are all to do with the extra index needed in general relativity

26

because our fundamental eld is a tensor (the metric) rather than avector.

13 The ratio of momentum to mass is velocity, so the ratio of the corre-sponding densities, T 0i/T 00, will be of order v, where v is a typicalvelocity. In the low-velocity limit this will be small. The argument forthe stresses is similar but more subtle. In a uid the stress componentsare just pressures, and when one uses statistical mechanics to derivepressure from the random motions of particles of the gas, one ndsp v2. Therefore in this case T ij/T 0i 1. In more general mate-rials, the stresses can be made of direct forces between particles, buteven here they must be of the same size: too great a stress would cre-ate forces that would accelerate a body to relativistic speeds, violatingthe low-velocity assumption.

17 (a) M = Rv2 = (C3/2P 2) 1.3 km 1M. (b) 100 km 68M.18 (a)|| 4 1035 m2. (b) This value of represents a mass density

that one would get by spreading the mass of the Sun over the wholeSolar System. This is very much bigger than the average density ofthe universe, since the nearest star is much further away than Pluto,and on top of that there are vast empty spaces between galaxies. Soa cosmological constant of this size would have enormous and easilyobservable consequences for the evolution of the universe. It followsthat a of the same size as the cosmological mass density would notbe observable in the Solar System.

19 (a) T 00 = , T 01 = x2, T 02 = x1, T 03 = 0. The componentsT ij are not fully determined by the given information, but they mustbe of order vivj , i.e. of order 2R2.

(b) Since 2h00 = 16, h00 is just minus four times the Newtonianpotential, h00 = 4M/r exactly. For h0i we have

h0i = 4

y2|x y|1d3y.

Use the binomial expansion

|x y|1 = r1[1 + x y/r2 + 0(R/r)2].By symmetry,

yid3y = 0,

yiyjd3y = 0 if i = j, and

27

y1y1d3y =

y2y2d3y =

y3y3d3y = (4/15)R5.

This implies h01 = (16/15)R5x2/r3. In terms of the angularmomentum J , we nd

h01 = 2Jx2/r3, h02 = 2Jx1/r3, h03 = 0.

These fall o as r2 and are correct to order r3. A more careful studyof the properties of the solutions of 2f = g would show that these arein fact exact: the higher-order terms all vanish in this simple situation.The components hij are small compared to h00 and h0i because T ij issmall. Therefore Eq. (8.31) gives h and the metric

g00 = 1 + 2M/r + 0(2R2),g01 = 2Jx2/r3 + 0(3R3), g02 = 2Jx1/r3 + 0(3R3), g03 = 0,gij = ij(1 + 2M/r) + 0(2R2).

Compare this with Eq. (7.8). In standard spherical coordinates, g00is the same, g0r = g0 = 0, g0 = 2J sin2 /r, and the spatial lineelement is

dl2 = [1 + 2M/r 0(2R2)](dr2 + r2d2 + r2 sin2 d2).

(c) Such a particle obeys the geodesic equation with p0 := E andp := L constant, pr and p zero. To this order, the normaliza-tion p p = m2 implies E = m(1 M/r + L2/2m2r2), just as inNewtonian theory (except for the rest mass). The r component ofthe geodesic equation implies, again to lowest order, L = m(Mr)1/2,again as in Newtonian theory. One orbit, = 2, will take a timet = (dt/d). Now dt/d = (dt/d)/(d/d) = U0/U = p0/p,and this can be expressed in terms of E, L, and the metric; a straight-forward calculation gives (t)prograde (t)retrograde = 8J/M , in-dependently of r. In principle, this allows measurement of a bodysangular momentum by the study of particle orbits far from it.

(d) 0.16ms.

28

Chapter 9

7 A solution of Eq. (9.22) is uniquely determined by the initial positionand U. The function U = 0 satises Eq. (9.22) for all time (byvirtue of Eq. (9.23)), and so must be the unique solution for initialdata in which U = 0.

8 No: the equivalence principle.

9 For the light beam, ds2 = 0 dt/dx = (gxx/|gtt|)1/2 1 + 12hTTxx (t).Therefore, if the remote particle is at coordinate location x = , thetime elapsed for a round trip of light is t (2+hTTxx ), where hTTxx is some mean value of hTTxx during the time of ight of the photon.Since hTTxx changes with time while does not, free particles do seeaccelerations relative to their neighbors.

10 (a) For example, Rtyxz = 122hyz ; (d) t = 12B(x t)2, i = 0.13 One way to do this is with rotation matrices. But consider a more

direct way. Write down the new basis vectors in terms of the old:

ex = (ex + ey)/2, ey = (ey ex)/

2.

The components of the tensor hTT are its values on the basis vec-tors, so its components in the new frame are, for example, hTTxx =hTT (ex , ex). Putting the expressions for the new basis vectors intothis gives hTTxx = h

TTxy +(h

TTxx + h

TTyy )/2. The vanishing trace gives the

desired result. (Note the sign errors in the statement of the problem!)

20 In the TT coordinates of the wave, choose the waves direction to be z,the ellipses principal axes to be x and y, and the masses separation tobe along the unit vector s (sin cos, sin sin, cos ) in the usualspherical coordinates of the frame. Then in Eq. (9.45) we replace hTTxxby hTTss sshTT = sin2 (cos 2 + ia sin 2)hTTxx . There will be nodriving term if = 0, i.e. if the masses lie along the direction of thewaves propagation (compare with Exer. ??).

26 (a) Does not violate relativity: this is just a coordinate speed, de-pendent on the coordinate system. Relativity insists that the properdistance divided by proper time along a light ray should always be 1,but there is no constraint on the coordinate speed. (c) The statementis correct as long as the distance is small enough that there is a localinertial frame covering both ends of the light path. If the distance is

29

larger, and the geometry is time-dependent, then there is not a uniquemeaning to the distance to a remote object: it depends on time andthe path being measured. (e) This is a coordinate-independent result,since it is a proper time as measured on a single clock. So it would bethe same in any coordinate system.

30 I lm =

A m(A)xi(A)x

i(A).

31 (a) Iij = (4/3)ijr4dr; Iij = 0.

(b) Iij =: Result of (a) +Maiaj ; Iij = M(aiaj 13 ija2).(c)

Ixx = a2M/5, Iyy = b2M/5, Izz = c2M/5,

Ixx = (2a2 b2 c2)M/15,Iyy = (2b2 a2 c2)M/15,Izz = (2c2 a2 b2)M/15,

all other components zero.

(d)

Ixx = (a2 cos2 t + b2 sin2 t)M/5,

Iyy = (b2 cos2 t + a2 sin2 t)M/5,

Izz = c2M/5,

Ixy = cos t sin t(a2 b2)M/5,Ixx = [a2(3 cos2 t 1) + b2(3 sin2 t 1) c2]M/15,Iyy = [b2(3 cos2 t 1) + a2(3 sin2 t 1) c2]M/15,Izz = (2c2 a2 b2)M/15,Ixy = Ixy,

others zero.

(e) Ixx = 2ma2 = Iyy, others zero;Ixx = 2ma2/3 = Iyy, Izz = 4ma2/3, others zero.(f) Same as (e).

(g) Ixx = 2m(A2 cos2 t +Al0 cost + l20/4), others zero;Ixx = 2Ixx/3, Iyy = Ixx/3 = Izz, others zero.(h) Ixx = (m + M)(A2 cos2 t + Al0 cost + l20/4), other zero;Ixx = 2Ixx/3, Iyy = Ixx/3 = Izz, and 2 = k/, where =mM/(M +m) is the reduced mass.

30

35 There is no radiation in (a)(c) and (e), and no quadrupole radiationin (f).

For (d), rst put the time-dependence of Iij from Exer. 31 into complexform, e.g., Ixx = 110M(a

2 b2) exp(2it). Then use Eqs. (9.84)(9.86) with = 2 and the correct permutations of the various indices.Results:

Along x-axis: hTTzz = hTTyy = 25M2(a2 b2) exp[2i(t r)]/r,hTTzy = 0.

Along y-axis: hTTzz = hTTxx = 25M2(a2 b2) exp[2i(t r)]/r,hTTxz = 0.

Along z-axis: hTTxx = hTTyy = 45M2(a2 b2) exp[2i(t r)]/r,hTTxy = ih

TTxx .

This means that the radiation is linearly polarized in the equatorialplane, circularly polarized along the z-axis. Notice there is no radiationif the body is axially symmetric, i.e. if a = b.

For (h), there is no radiation on the x-axis. On the y-axis:

hTTzz = (m + M)2{2A2 exp[2i(t r)] + Al0 exp[i(t r)]}/r,hTTyy = hTTzz ,hTTxy = 0.

36 Without loss of generality choose the wave to be moving in the x yplane. The unit vector along its direction has components nx =cos , ny = sin .Then Pxx = sin2 , Pyy = cos2 , Pzz = 1, Pxy = sin cos , others zero. By matrix multiplication or by just writ-ing out all the summation terms, we nd hTTxx = f sin

2 , hTTyy =f cos2 , hTTxy = f sin cos , hTTzz = f , others zero, where f =m2 sin2 {2A2 exp[2i(t r)] + Al0 exp[i(t r)]}/r. If we letli be ( sin , cos , 0), then l and ez are two orthogonal vectors per-pendicular to the motion of the wave. We nd hTTll := l

ilj hTTij = f .Recalling hTTzz = f , we see that the wave is always 100% linearlypolarized with the ellipses axes in the x y plane and parallel to ez:the component rotated by 45 is absent.

38 Let the wave be traveling in the x z plane at an angle with thez axis. Let ey and l (cos , 0, sin ) be orthogonal vectors in the

31

plane of polarization. Then hTTyy = (1 12 sin2 )f, hTTly := lihTTiy =i cos f, hTTll := lilj hTTij = hTTyy , where f = 2ml202 exp[2i(t r)]/r. From Exer. ??b we see that the wave is elliptically polarizedwith principal axes l and ey. The percentage of circular polariza-tion is |hTTly /hTTyy |2, which ranges from zero at the equator to 100% atthe poles.

39 (a) Let the stars orbit in the x-y plane, with the center of mass atthe origin. If the polar coordinates of m are (r1, ), both functions oftime, and if those of M are (r2, + ), then their total separation isr = r1 + r2, with r1/r = M/(m + M), r2/r = m/(m + M). In termsof these variables the quadrupole tensor follows straightforwardly:

Ixx = r2 cos2 ,

Iyy = r2 sin2 ,

Ixy = Iyx = r2 sin cos ,

with = mM/(m+M), which is called the reduced mass of the binarysystem.

Let the reduced orbit have eccentricity e and semi-major axis a. (Notethat the statement of the problem incorrectly called a the distance ofclosest approach). The Newtonian equations imply

r(t) =a(1 e2)

1 + e cos[(t) 0] ,

where 0 is the orientation of the major axis of the ellipse, the direc-tion of their separation when they are closest. Similarly, the angularvelocity of the orbit is

(t) =ddt

= r(t)2[(m + M)a(1 e2)]1/2.

The orbital period is

P = 2[

a3

m + M

]1/2.

The total energy E and angular momentum L, being conserved, canbe computed at any point; the point of closest approach = 0 is

32

convenient. The results are E = mM/2a, L2 = 2(m+M)a(1 e2).From these follow the desired relations:

P 2 = 2(M + m)23/2E3,e2 = 1 + 2EL2/(M + m)23,a = mM/2E.

(Recall that E < 0.)

(b) The trace of Ijk is r2, which is not a constant (unlike the case ofcircular orbits). The components of Ijk are:

Ixx = r2(cos2 13),Iyy = r2(sin2 13),Ixy = Iyx = r2 sin cos ,

Izz = 13r2.

(c) The radiation requires two time-derivatives of Ijk.

40 Bring out the R1 as r1, as in Eq. (9.103), but expand t R abouttr. Eq. (9.104) and its consequences still follow. The rst term is Eq.(9.105) but higher terms depend on (d/dt)n

yiyk . . . ylym d3y, where

there are n factors of yj. If is spherical, then the integrations give in-dices of the form jk lm (see Exer. 42) and permutations. The TTprojection eliminates traces and so completely eliminates these terms.

33

Chapter 10

1 This is essentially a repeat of Exer. 28, except with the time-dimensionpresent. However, the transformation aects only the spatial coordi-nates, so the solution follows in simple way.

3 The factor e in Eq. 10.11 expands in the weak-eld approximationto 1, and in this limit the relativistic function just becomes theNewtonian potential . When multiplied by the expression for E =p0 given in Eq. ch07:eqn7.34, this factor cancels the gravitationalpotential energy, leaving just the rest-mass and kinetic energy.

4 First calculate the components of the Ricci tensor R = gR .This would give, for example,

Rtt = e2Rtrtr + r2Rtt + (r sin )2Rtt= [ ()2 2/r] exp(2 2).

Then, from all the components of the Ricci tensor one forms the Ricciscalar, R = gR . Finally one forms the Einstein tensor, e.g.

Gtt = Rtt 12gttR.

5 The denition of a static metric is that this transformation must leaveall quantities invariant, but it sends t t and hence dr/dt dr/dt. Only if this vanishes can it be invariant.

7 The construction proceeds as described in Sec. 10.5, except that tointegrate equations with p one has to specify the function S in theequation of state. It would be simple to give a function S(r), butnot very physical (unless S is constant through the star), because onecant know ahead of time how the stars structure will come out of theequations. It is more physical to give S(m(r)), so that the entropy is afunction of the mass interior to a radius r. This might approximatelydescribe a star formed by spherical collapse that conserves entropy, sothat the rest-mass inside a given shell would be constant during thecollapse and so would be the entropy, leading to a xed relationshipbetween entropy and interior rest-mass. Now, m(r) is not exactlyproportional to the rest mass, so giving S(m(r)) only approximatesthis situation. It would be possible, but more complicated, to computethe rest-mass interior to r and take the entropy to be a function of

34

that. Since there is no dierential equation for S one does not changethe manner of integration; one just adds an auxiliary function or look-up table to determine the entropy at any radius and hence allow thepressure there to be computed.

8 (a) Use reasoning similar to that in Exer. 5. (b) Make sure to includethe Christoel symbols in the computation of the covariant derivative!These can be found in Exer. 35 of Sec. 6.9.

10 (b) exp() = 0.999 997, 0.760; z = 3 106, 0.315.15 Putting the power series into the dierential equations and matching

up powers of r shows that 1 = 0 and

2 = 2c(c + pc)(c + 3pc)/(3cpc).

The power series solution for the three functions then becomes

= c + 2r2,p = pc(1 + 2c/cr2),m = 4r3c/3 + 4r52/5.

Estimate the error in neglecting (uncalculated) the next term of eachequation to be about equal to the square of the last calculated term,i.e., choose r such that the calculated terms are smaller than

(0.01)

3%. This means thatr2 |c/301c|.

19 (a) 1.121042 kgm2 s1; (b) 1.4104 s1; yes, by 50%; (c) 2.7104;(d) 5 109 Gauss.

35

Chapter 11

1 In Eq. 11.9, nd the minimum approach distance by setting drd = 0,and calling this radial distance b. If M = 0 solving for b gives theresult. Note that if m = 0 the result is the same as if one had startedwith Eq. 11.10. The impact parameter is dened for any orbit (evenif M is not zero) by the equation

b = L/[E2 m2]1/2.It can be thought of as the oset of the aim of the trajectory fromthe center of the metric, when the trajectory is far away. If M is non-zero then the trajectory will approach the center more closely thanb. For a photon orbit with m = 0, again even when M is not zero,one can replace in Eq. 11.12 L by bE. Then if one re-scales the aneparameter to a new one = E, the resulting equation dependsonly on one parameter, b. Re-scaling the ane parameter is alwaysallowed and does not change the path of the geodesic through thespacetime. So we learn that photons follow geodesics that dependonly on their aim, or oset, from the center. Massive particles donot: the degree to which they approach more closely than b dependson their velocity far away, which (unlike that of a photon) depends onthe initial conditions.

4 The key radii are 3M , the location of the unstable photon circularorbit, and 6M , the location of the last stable circular orbit for massiveparticles. So if a star has a radius of 2.5M , as in (a), then all possibleorbits exist outside it, even photon circular orbits. If a star, as in (b),has a radius of 4M , then it will not have any photon orbits, but thelast stable massive-particle orbit will still be outside its surface, so allstable massive-particle orbits will exist. Finally, in case (c), if the starhas a radius of 10M then there will be stable circular orbits downto its surface, and of course any quasi-elliptical and quasi-hyperbolicorbits that approach no closer than the surface radius.

5 (a) The question is badly worded, since the value 12M/R is the rela-tivistic value of g00, not its Newtonian approximation. The functionthat is to be compared with the Newtonian potential is

(g00)1 =

M/rM2/r2. . .. This diers from the Newtonian potential M/rby the term M2/r2, which makes a 1% change when M/r = 0.01.

(b) For a 106M black hole, this is at a distance of about 1.51011 m,just 15 time larger than the radius of a normal star. So nor more than

36

of order 1000 normal stars can be entirely inside the highly relativisticzone around a black hole of that mass. But if the mass is 109M, thenthe radius is 1000 times larger, and the volume is 109 times larger, sothat billions of normal stars could be in the highly relativistic regionnear such a mega-massive black hole.

7 (a) Use Eq. (11.24) in ds2 = g00dt2 + gd2 to get = 207M .

(b) Same as coordinate time interval, Eq. (11.25): 2010M .

(c) Integrate ds2 = g00dt2 over the time in (b): 402M .

(d) 6.4ms: innermost stable orbit.

(e) 40/2 yr, independent of M .

9 (a) Turning points are at r = 12.5M and 5.47M . (Notice that thehigh L enables the particle to turn around inside 6M . This is not acontradiction with is radius being the last stable orbit, because thatapplies only to circular orbits.) The orbit changes by = 7.4 radbetween these points. A full orbit has = 14.8 rad, or a perihelionshift of 8.5 rad. The approximation Eq.11.37) gives only 2.3 rad. Thisshows that highly noncircular orbits can have much greater shifts.

10 (a) (E/L)2 = [1 + 36M2/L2 + (1 12M2/L2)3/2]/(54M2).14 In arc sec per orbit and per year: Venus (0.052, 0.085), Earth (0.038,

0.038), Mars (0.025, 0.013).

20 (a) The Christoel symbols may be computed as a special case of thosein the solution of Ex. 35 of Sec. 6.9.

(d) The quantity computed here is called the Riemann scalar. Notethat it is perfectly well-behaved at the horizon and only singular atthe center r = 0.

22 Approximately 105 M.

24 The point u = v = 0 has r = 2M , so is located on the horizon.We expect it to be locally at, but this can also be easily shown byexpanding the metric in a Taylor series in u and v. It is clear fromEq. 11.68 that the expansion does not contain terms linear in u andv, and so with some rescaling of the coordinates by constant factors itcan be brought into the form of a Minkowski metric with correctionsof order u2 and v2. Of course, the metric is locally at everywhereexcept at the singularity at r = 0, but we have chosen to look at the

37

origin because it belongs to the extended Kruskal-Szekeres metric andis not easily examined in Schwarzschild coordinates, and also becauseit is relatively easy to do the computation there!

25 Estimate the total breakup force to be M/R2, giving RT (MH/)1/3.For the given density we nd that the disruption mass is about 106 M.This is an interesting mass considering that there is strong evidencethat a black hole of about this mass resides in the center of our Galaxy.

31 The ZAMO has angular velocity so that U/U0 = . Denoting U0

by the variable A, one can compute A from the normalization conditionUUg = 1 in the Kerr metric.

34 (a) (22)M , (b) E m1 +m2 (m21 +m

22) m1(1 12m1/m2).

37 (a) The photon follows a null line along which only the coordinates tand change. So the relevant interval is

ds2 = 0 = gttdt2 + 2gtdtd+ gd2.

This can be solved to give the photons angular velocity:

ddt

=

2 + 2,

where is the frame-dragging angular velocity dened in Eq. 11.90,whose value in the equatorial plane ( = /2) is

=gtg

=2Ma/r

r2 + a+2Ma2/r,

and where is the basic photon circular angular velocity and is dened(again in the equatorial plane) by

2 =gttg

=1 2M/r

r2 + a2 + 2Ma2/r.

The forward-going photons angular velocity is (2+2)1/2 +, whilethe backward-going angular velocity is [(2 + 2)1/2 ]. Thebackward-going photon travels more slowly than the forward-goingone. This is an illustration of the action of the dragging of inertialframes, which drags in the forward direction.

Notice that, as r gets large, approaches 1/r while falls o as 1/r3.So dominates the angular velocity and the frame-dragging makes

38

a correction. The limiting value of = 1/r is just what one needs inorder for the photon to travel around a distant circle of radius r in atime equal to the circumference of the circle, 2r.

The coordinate time that elapses is 2 divided by the absolute valueof the angular velocity. In the forward direction this is

t+ =2

(1 +

2

2

).

In the backward direction this is

t =2

(1 +

2

2+

).

The dierence in the round-trip times is

t t+ = 42 .

There is no redshift or blueshift of the photons: the components ofthe photons 4-momentum are constant in time as it moves aroundthe circle, since the geometry is the same everywhere along the path.

(b) The ZAMO observer orbits with angular velocity . It follows thatthe angular velocity of a photons relative to him is d/dt , whichevaluates simply to (2 + 2)1/2: for this observer the two photonshave the same (absolute value) angular velocity, so they take the sametime to go once around and meet him again! If he launches the twophotons simultaneously, they arrive back and meet him at the sametime. From the point of view of the rst observer at rest, the forward-going photon is going faster but has to travel further in order to meethim again, since he moves around the black hole in the meantime,while the backward-going photon is going slower but has less far to gosince the moving observers motion takes it towards this photon.

Notice that all the speeds of the photons we are talking about arecoordinate angular velocities. The photons always travel at speed 1relative to any local inertial frame, but the curvature of space and timemakes them take dierent amounts of time to traverse their circularpaths.

39

Chapter 12

8 (a) Use d2 = dr2/(1 + r2). Integrating leads to r = sinh.

(c) A Lorentz transformation leaves the metric and hence the hyper-bola unchanged, but changes the origin of spatial coordinates. Anypoint on the hyperbola can be made into this origin by choosing thecorrect transformation.

9 (a) In the text it is pointed out that p is conserved due to the homo-geneity of the universe. A radially propagating photon has just twonon-zero four-momentum components, p0 and p, so we have

0 = g00(p0)2 + g(p)2,

which by the diagonality of the metric can also be written as

0 = g00(p0)2 + g(p)2.

For each kind of universe g = R2(t), so that g = 1/R2(t). Sinceg00 = 1, we nally get that

p0 = p/R(t).

The constancy of p establishes the fact that the locally measuredenergy of the photon, p0, decreases with time as 1/R(t). Notice that,since the location of the origin of coordinates in a homogeneous spaceis arbitrary, the radially-moving photon really can be any photon, sothis law is quite general.

(b) The redshift relation follows from the fact that the wavelength ofa photon is inversely proportional to its energy.

10 In Eq. 12.67, which is the result of the previous Exercise, expand R(tr)about R(te):

R(tr) = R(te) + R(te)(tr te) + . . . .Now, for a photon traveling at the speed of light, tr te = d0, thedistance to the emitter as dened in Eq. 12.24. So we get

z = (R/R)(te)d0.

Now, to the accuracy we are working (rst order in time intervals ord0) we can replace te in this expression with tr, which is the same astodays time t0 as used in Eq. 12.24. This makes z from Eq. 12.67identical to v in Eq. 12.24 at this order.

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13 Make sure you use the denition of the luminosity distance dL inEq. 12.34.

15 The result follows from the fact that the energy density of black-bodyradiation is proportional to T 4.

18 Pressure forces depend on pressure gradients, and in a homogeneousuniverse the gradient is zero. So a cosmological tension does not leadto a pulling inward. Rather, the entire eect of the negative pressureis in the active gravitational mass per unit volume term + 3p inEq. 12.55, where the large negative pressure overwhelms the densityand causes gravity itself to become repulsive.

22 (a) The energy density of black-body radiation is r = 4T 4/c, where is the Stefan-Boltzmann constant, whose value in SI units is 5.67108Wm2K4. Divide by another factor of c2 to get this in mass-density units. When divided by m we get

=rm

4m1 104.

Then we get the number of photons per baryon to be about 2m1109.In the early universe, if there were no baryon-antibaryon asymmetry,then one would have expected of order one baryon and one antibaryonper photon, because the baryons would have been in thermal equilib-rium with the photons at temperatures where the photons had enoughenergy to create baryon-antibaryon pairs. What this large number tellsus is that the excess of baryons over antibaryons is of order one partin 109.

23 z = 3 103.24 The density must exceed = 3H20/8 = 10

26kgm3. This is calledthe closure density or critical density and is denoted by c.

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