Instructor Solution Manual for Einstein Gravity in a Nutshell 1st edition A. Zee

SOLUTION MANUAL

Solutions Manual to Einstein Gravity in a Nutshell A. Zee

PRINCETON UNIVERSITY PRESS PRINCETON AND OXFORD

c 2013 by Princeton University Press Copyright Published by Princeton University Press, 41 William Street, Princeton, New Jersey 08540 In the United Kingdom: Princeton University Press, 6 Oxford Street, Woodstock, Oxfordshire OX20 1TW press.princeton.edu All Rights Reserved ISBN (pbk.) 978-0-691-16035-1 This manual has been composed in Computer Modern with LATEX by Princeton Editorial Associates Inc., Scottsdale, Arizona Printed in the United States of America

Acknowledgments I am grateful to Joshua Feinberg for looking at some of these solutions as well as the solutions in the text. I also thank Sebastian Fischetti for providing the solutions to some of the exercises.

iii

0

Setting the Stage

Prologue:

Three Stories

1 Solution in back of book.

2 Denote the position of the ant by A and the point on the edge halfway between the ant and the drop of honey by M. The shortest path between A and M is the great circle joining the two points. By symmetry, the second half of the trek from M to the honey is also determined.

3 Use the same “cut the cylinder and unroll it” trick as in the text, and follow the discussion in the text allowing for the difference in speed. Evidently, the ant now spends more time on the outside of the glass than on the inside.

I

From Newton to Riemann: Coordinates to Curvature

I.1

Newton’s Laws

1 Solution in back of book.

2 Solution in back of book.

3 Solution in back of book.

4, 5 Solution in back of book.

6 Let R denote the radius of the earth, d the distance from the midpoint of the tunnel to the center of the earth, r the distance from an arbitrary point along the tunnel to the center of the earth, and w the linear coordinate measuring the distance along the tunnel joining city A and city B. (You need to draw a figure to follow this discussion.)

One check to see if you drew the figure correctly:√r ranges between √ d and R. Let the w coordinate be set up so that it ranges from − R2 − d2 to + R2 − d2 , vanishing at the midpoint of the tunnel. According to Newton’s two superb theorems, the force acting on the railroad car (of mass m) when it is at the point specified by r is given by Fc = −G(M r3 /R3 )m/r2 = −GM mr/R3 . But this force is pointed toward the center of the earth. What we want is the force F along the tunnel, and so we have to multiply Fc by the cosine of some angle; you can see from your figure that the cosine equals w/r. Thus, we have F = Fc w/r = −GM mw/R3 . Now Newton’s ma = F tells us that ẅ = −GM w/R3 = −(g/R)w with g the acceleration due to gravity at the surface of 2 2 ). the earth (which you learned in school is given by ∼32 feet/sec or ∼9.8 meters/sec √ This equation, supplemented by the boundary condition w(t = −T /2) = − R2 − d2 and ẇ(t = −T /2) = 0 (in light of the symmetry of we set up coordinates √the situation, − d2 cos[ g/R(t + (T /2))]. The so that w(t = 0) = 0), has the solution w(t) = − R2 condition w(t = 0) = 0 then fixes the period T = π R/g. The remarkable result is that the transit time, which works out to be about 42 minutes, does not depend on d (and hence the location of the two cities.)

1

I.2

Conservation Is Good

∂V (x) dxia a ma dt = − a ∂xia = 0. If you are confused about the last equality, ∂ ∂ consider the N = 2 case in which ( ∂x i + ∂xi )V (x1 − x2 ) = 0.

d 1 We have dt

1

I.3

2

Rotation: Invariance and Infinitesimal Transformation

1 Consider an infinitesimal rotation by an angle δθ in the (1-2) or (x-y) plane. Then since p is a vector, its components transform like (using (14)) ⎛ 1 ⎞ ⎛ 1⎞ p + δθ p2 p ⎝p2 ⎠ → ⎝p2 − δθ p1 ⎠ p3 p3 and likewise for q. The array of three numbers then transforms like 2 ⎛ 2 3⎞ ⎛ ⎞ p − δθ p1 q 3 p q 3 1 ⎝ p3 q 1 ⎠ → ⎝ ⎠ q2 1 p q 2 + δθ 1 2 2 1 p q p + δθ p q − δθ q But if this array were a vector, it should have transformed like ⎛ 2 3⎞ ⎛ 2 3 ⎞ p q p q + δθ p3 q 1 ⎝p3 q 1 ⎠ → ⎝p3 q 1 − δθ p2 q 3 ⎠ p1 q 2 p1 q 2 Since the array does not transform as it should under this rotation, it does not transform like a vector under rotations and is therefore not a vector.

For the other array, let us define the coefficients ai as ⎛ 1⎞ ⎛ 2 3 ⎞ a p q − p3 q 2 ⎝a2 ⎠ = ⎝p3 q 1 − p1 q 3 ⎠ a3 p1 q 2 − p2 q 1 Then we note that in index notation, we can write ai = εijk pj q k Under an infinitesimal rotation, pj → (δ jk + Ajk )pk , where by (14) the matrix A is antisymmetric and traceless. Then the ai transform like

ai → εijk δ j + Aj δ km + Akm p q m

= εijk pj q k + εi k Akm + εijm Aj p q m

= εijk pj q k + εinj Ajp δ n δ mp − εinj Ajp δ p δ mn p q m

= εijk pj q k + δ n δ mp − δ p δ mn εinj Ajp p q m To get to the second line, we only kept terms linear in A (since this is supposed to be an infinitesimal rotation), and to get to the third line, we used the antisymmetry 2

of εijk , relabeled indices, and explicitly included the Kronecker deltas. Now, we can use the identity εijk εk m = δ i δ jm − δ im δ j to write ai → εijk pj q k + ε mk εknp εinj Ajp p q m

= εijk pj q k + ε mk δ pj δ ik − δ pi δ jk Ajp p q m

= εijk pj q k + ε mk Ajj δ ik − Aki p q m = εijk pj q k + Aik εk m p q m

= δ ik + Aik εk m p q m = δ ik + Aik ak To get to the second line, we used our Levi-Civita identity again. To get to the fourth line, we used that fact that Ajj = 0, since A is traceless. Finally, to get to the fifth line, we used the antisymmetry of A and ε and relabeled some indices. Thus the components ak transform correctly under infinitesimal rotations, and so they do form a vector.

2 Solution in back of book.

3 The infinitesimal generator of rotations about the x-axis is Jx given in (14); the generator of finite rotations is Rx (θx ) = eθx Jx . Note that for integer n, we have ⎛ ⎞ ⎛ ⎞ 0 0 0 0 0 0 J 2n = ⎝0 1 0⎠ J 2n+1 = ⎝0 0 1⎠ 0 −1 0 0 0 1 Then in the power series expansion for Rx (θx ), we can separate even and odd powers: ∞

Rx (θx ) =

1 n n θx Jx n! n=0 ∞

∞

1 1 2n 2n θx2n+1 Jx2n+1 θx Jx + (2n + 1)!

(2n)! n=0 n=1 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 1 0 0 0 0 0 ∞ 0 0 0 ∞ 1 1 θx2n + ⎝0 0 1⎠ θx2n+1 = ⎝0 0 0⎠ + ⎝0 1 0⎠ (2n)!

(2n + 1)!

0 0 0 0 0 1 n=0 0 −1 0 n=0 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 1 0 0 0 0 0 0 0 0 = ⎝0 0 0⎠ + ⎝0 1 0⎠ cos θx + ⎝0 0 1⎠ sin θx 0 0 0 0 0 1 0 −1 0 ⎞ ⎛ 1 0 0 sin θx ⎠ = ⎝0 cos θx 0 − sin θx cos θx =I+

For Ry (θy ), note that we can simply permute the axes x → y, y → z, and z → x, which corresponds to just reshuffling the rows and columns of the matrix Rx (θx ), so 3

we can immediately write ⎛

cos θy Ry (θy ) = ⎝ 0 − sin θy

0 1 0

⎞ sin θy 0 ⎠ cos θy

By explicit computation, note that ⎞⎛ ⎞ ⎛ cos θy 0 sin θy 1 0 0 sin θx ⎠ ⎝ 0 1 0 ⎠ Rx (θx )Ry (θy ) = ⎝0 cos θx 0 − sin θx cos θx − sin θy 0 cos θy ⎛ ⎞ cos θy 0 sin θy cos θx sin θx cos θy ⎠ = ⎝ − sin θx sin θy − sin θy cos θx − sin θx cos θx cos θy ⎛ ⎞ cos θy − sin θx sin θy cos θx sin θy cos θx sin θx ⎠ = Ry (θy )Rx (θx ) = ⎝ 0 − sin θy − sin θx cos θy cos θx cos θy 4 Let’s start with (19):

ij J(mn) = −i δ mi δ nj − δ mj δ ni

Then we have J(mn) , J(pq)

ij

kj ik = J(mn) J(pq) − (m ↔ p, n ↔ q)

mi nk = − δ δ − δ mk δ ni δ pk δ qj − δ pj δ qk − (m ↔ p, n ↔ q)

= −δ np δ mi δ qj + δ nq δ mi δ pj + δ mp δ ni δ qj − δ mq δ ni δ pj − (m ↔ p, n ↔ q)

= δ np δ mj δ qi − δ mi δ qj + δ nq δ mi δ pj − δ pi δ mj (1) pi nj ni qj qi nj mq ni pj mp +δ +δ δ δ −δ δ δ δ −δ δ ij ij ij ij − iδ mq J(np) + iδ nq J(mp) + iδ mp J(nq) = −iδ np J(mq)

and thus ij ij ij ij + δ mp J(nq) − δ np J(mq) − δ mq J(np) J(mn) , J(pq) = i δ nq J(mp)

5 Solution in back of book.

I.4

Who Is Afraid of Tensors?

transforms like a vector, as suggested by the 1 This follows from the observation that ∇ notation.

2 Just change the sign in (5): S ij → S ij

= =

T ij + T ji = Rik Rjl T kl + Rjk Ril T kl Rik Rjl T kl + Rjl Rik T lk = Rik Rjl (T kl + T lk ) = Rik Rjl S kl 4

(2)

i

3 d3 x = d3 xJ, where the Jacobian J = det ∂x ∂xj = det R = 1.

4 Solution in back of book.

5 Solution in back of book.

6 The first index i can take on D values. Then for a specific i, j can take on D − 1 values, and k only D − 2 values. Since T ijk is totally antisymmetric, we see that T 1 D(D − 1)(D − 2) components. For D = 3, there is only one component, namely has 3!

123 T .

1 ijk ijk ε T . Under a rotation, 7 It has only one component, which we can write as T 123 = 3!

ijk ijk ijk ii jj kk i j k i j k i j k ε T → ε R R R T = (det R)ε T = εi j k T i j k , where we used the definition of the determinant.

8 Solution in back of book.

9 Solution in back of book.

I.5

From Change of Coordinates to Curved Spaces

1 Solution in back of book.

2 We have x = θ cos ϕ and y = θ sin ϕ, so that dx = cos ϕ dθ − θ sin ϕ dϕ

dy = sin ϕ dθ + θ cos ϕ dϕ

dθ = cos ϕ dx + sin ϕ dy

θ dϕ = − sin ϕ dx + cos ϕ dy

Small x and y implies small θ, so in the metric ds2 = dθ2 + sin2 θ dϕ2 , we expand sin2 θ = θ2 − θ4 /3 + O(θ6 ). The metric then becomes

2 ds2 = dθ2 + 1 − 13 θ2 + · · · (θ dϕ)

2 2 = (cos ϕ dx + sin ϕ dy) + 1 − 13 θ2 + · · · (− sin ϕ dx + cos ϕ dy) 2

= dx2 + dy 2 − 13 (−θ sin ϕ dx + θ cos ϕ dy) + · · · 2

= dx2 + dy 2 − 13 (−y dx + x dy) + · · ·

x2 y2 2 dx + 1 − dy 2 + 23 xy dx dy + · · · = 1− 3 3 3 Solution in back of book.

4 Solution in back of book.

5

5 Solution in back of book.

6 Recall that the curvature of a space can be evaluated by considering the ratio of the circumference to the ratio of infinitesimally small circles. If the radius of the circles is ε, the curvature is

6 C R = lim 2 1 − ε→0 ε 2πε where C is the circumference of the circles. We are told that ε = θ and C = 2π sin θ, so we have

6 C R = lim 2 1 − ε→0 ε 2πε

6 2π sin θ = lim 2 1 − θ→0 θ 2πθ

6 = lim 2 1 − 1 − 16 θ2 + O(θ4 ) θ→0 θ 6 = lim 2 16 θ2 + O(θ4 ) θ→0 θ =1

ds2 = dx2 + dy 2 = 7 Define the circle by x2 + y 2 = 1. First, obtain the metric: √ 2 2 dx /(1 − x ). In polar coordinates, θ√= arctan y/x = arctan √1 − x2 /x. By direct differentiation, we obtain dθ = −dx/ 1 − x2 . We recognize 1/ 1 − x2 as precisely √ g.

8 The actual length width and height of the volume element √ are given by adx, bdy, and √ cdz, and so the volume is indeed given by d3 x g = d3 x a2 b2 c2 .

9 Our coordinate transformation between the Cartesian coordinates X i and spherical coordinates θi is X n = cos θn

n−1

sin θi for n ≤ d,

i=1

X d+1 =

d

sin θi

i=1

To prove the Pythagorean relation, we proceed by induction. For the base case of the S 1 , it is trivial: we have X 1 = cos θ1 , X 2 = sin θ2 , so clearly (X 1 )2 + (X 2 )2 = 1.

For the inductive hypothesis, assume the Pythagorean relation is satisfied for S d−1 ;

6

then we show that it is satisfied for S d as well:

2 2 (X n ) + X d+1 = 1

d

n=1 d

cos2 θn

n−1

n=1 d−1

cos2 θn n=1

n−1

d

sin2 θi +

i=1

sin2 θi + cos2 θd

d−1

i=1 d−1

sin2 θi + sin2 θd

i=1

i=1 d−1

cos2 θn n=1

n−1

sin2 θi = 1 sin2 θi = 1

i=1 d−1

sin2 θi + cos2 θd + sin2 θd

i=1

sin2 θi = 1

i=1 d−1

cos2 θn n=1

n−1

d−1

sin2 θi +

i=1

sin2 θi = 1

i=1

To get to the third line, we explicitly removed the last term from the summation and from the second product. Now we recognize the final expression as the Pythagorean relation for S d−1 , which is satisfied by our inductive hypothesis, and so the Pythagorean relationship is satisfied for S d .

To find the metric on S d , we proceed recursively. Consider the coordinates X i , i = 1, . . . , d in which we embed an S d−1 :

X n = cos θn

n−1

sin θi for n ≤ d − 1,

Xd =

i=1

d−1

sin θi

i=1

i , i = 1, . . . , d + 1 in which we embed an S d : and the coordinates X

n = cos θn X

n−1

sin θi for n ≤ d,

d+1 = X

i=1

d

sin θi

i=1

We can relate these coordinates via d = X d cos θd , X

i = X i for i = 1, . . . , d − 1, X

7

d+1 = X d sin θd X

Then the metric on S d will be d+1

i dX

dΩ2d =

2

i=1 d−1

i dX

=

2

d + dX

2

d+1 + dX

2

i=1 d−1

2

2 2 + cos θd dX d − X d sin θd dθd + sin θd dX d + X d cos θd dθd

2

2 2 + dX d + X d dθd2

2

2 + X d dθd2

dX i

= i=1 d−1

dX i

= i=1 d

dX i

= i=1

The first term is just the metric on an S d−1 , so we have

2 dΩ2d = dΩ2d−1 + X d dθd2 The metric on an S 1 is just dΩ21 = dθ12 , so with our expression for X d , we can iteratively construct the metric on an arbitrary S d : dΩ2d = dθ12 + sin2 θ1 dθ22 + · · · + sin2 θ1 · · · sin2 θd−1 dθd2 10 Solution in back of book. √ √ 11 We have g = g11 g22 · · · gdd = (sin θ1 ) · · · (sin θ1 · · · sin θd−1 ), where in this product, evidently sin θ1 appears (d − 1) times, sin θ2 appears (d − 3) times, and so on. It is easier for you to grasp what is going on than for me to say it! The area element is given by this times dθ1 dθ2 · · · dθd and thus is equal to dθ for d = 2, sin θ1 dθ1 dθ2 for d = 3, and sin2 θ1 sin θ2 dθ1 dθ2 dθ3 for d = 4. (You can easily keep going.) π Integrating over 0 ≤ θi < π for 1 ≤ i < d and over 0 ≤ θd < 2π, and using 0 dθ sin θ = 2 π and 0 dθ sin2 θ = π2 , we obtain that the generalized area of (2π), 2(2π) = 4π, and ( π2 )2(2π) = 2π 2 for S 1 , S 2 , and S 3 , respectively.

2 2 12 Distance around the equator: √ 2π(b +a )/b. Distance around a circle of fixed longitude π through the poles: 2 0 dθ b2 + a2 cos2 θ, which can be evaluated in terms of elliptic functions, but for a = b, this gives 7.64a. The area is 4π(b2 + a2 ). √ 13 Comparing two similar triangles in figure 3 gives us (L− L2 − r2 )/r = 2L/ρ. Solving, we obtain the stated result.

14 The demonstration of the result, that angles are the same as calculated with the two different metrics, is given in the solution I.6.10.

8

15 To compare the expression for the divergence given in appendix 3, ∂r W r + ∂θ W θ + θ θ ∂ϕ W ϕ + 2r W r + cos = r12 ∂r (r2 W r ) + sin1 θ ∂θ (sin θW θ ) + ∂ϕ W ϕ (notice that sin θ W we have merely gone back to the first form given in (26)) with the expression ·E = 12 ∂r (r2 Er ) + 1 ∂θ (sin θEθ ) + 1 ∂ϕ Eϕ given in elementary texts, we see ∇ r r sin θ r sin θ that we have to lower indices and to make a trivial change in normalization. We have W μ = g μν Wν with g rr = 1, g θθ = 1/r2 , and g ϕϕ = 1/(r2 sin2 θ). Furthermore, from the combination W μ ∂μ in the expression we started the discussion with in appendix 3, we see that W θ and W ϕ have an extra dimension of inverse length compared to W r , while Er , Eθ , and Eϕ in the elementary discussion all have the same dimension. Thus, we define Wr = Er , Wθ = rEθ , Wϕ = r sin θEϕ . We recover the usual expression.

16 Solution in back of book.

17 Solution in back of book.

18 Solution in back of book.

I.6

Curved Spaces: Gauss and Riemann

1 The coordinates (x, y) that the Eskimo mites use correspond to distance traveled from the north pole along the sphere. The coordinates (x̃, ỹ) used in (2), however, correspond to distance traveled from the north pole along a tangent plane. If θ is the polar angle of the sphere, we should therefore expect x2 + y 2 = θ2 , while x̃2 + ỹ 2 = sin2 θ. We can then write the two coordinates as follows: x = θ cos ϕ x̃ = sin θ cos ϕ

y = θ sin ϕ ỹ = sin θ sin ϕ

We can rearrange these into the explicit coordinate transformation

arcsin x̃2 + ỹ 2

x̃ = x̃ + 16 x̃ỹ 2 + 16 x̃3 + · · · x= x̃2 + ỹ 2

arcsin x̃2 + ỹ 2

y= ỹ = ỹ + 16 x̃2 ỹ + 16 ỹ 3 + · · · x̃2 + ỹ 2 Plugging this into the Eskimo mites’ metric from exercise I.5.2, we get

x2 dx + 1 − dy 2 + 23 xy dx dy + · · · ds = 3

= 1 + x̃2 dx̃2 + 1 + ỹ 2 dỹ 2 + 2 x̃ỹ dx̃ dỹ + · · ·

2

y2 1− 3

2

This is clearly of the form given in (2), with a2 = b2 = 1 and c = 0.

2 Let us take the torus to be a tube of radius a coiled into a circle of radius L. Embedding the torus in usual 3-dimensional Euclidean space, we take its axis to be the z-axis, 9

so if we introduce the usual azimuthal angle ϕ in the (x-y) plane and an additional angle θ that parametrizes the circles of constant φ, we can parametrize the torus as X = (L + a sin θ) cos ϕ Y = (L + a sin θ) sin ϕ Z = a cos θ Now, consider some point on the torus at coordinates (ϕ, θ). The tangent plane at this point is defined by the unit normal to the torus, which is just = (sin θ cos ϕ, sin θ sin ϕ, cos θ) N This plane is also spanned by the two tangent vectors vϕ = ∂ϕ (X, Y, Z) = (−(L + a sin θ) sin ϕ, (L + a sin θ) cos ϕ, 0) vθ = ∂θ (X, Y, Z) = (a cos θ cos ϕ, a cos θ sin ϕ, −a sin θ) To use the tangent plane method, we want to express the derivatives of the normal in the basis { vϕ , vψ }: vector N sin θ vϕ L + a sin θ = (cos θ cos ϕ, cos θ sin ϕ, − sin θ) = 1 vθ ∂θ N a

= (− sin θ sin ϕ, sin θ cos ϕ, 0) = ∂ϕ N

The shape operator (that is, the matrix M ) is then

sin θ 0 M = L+a sin θ 1 0 a so the intrinsic curvature of the torus is detM =

sin θ a(L + a sin θ)

3 (a) The metric in the new coordinates becomes ds2 = dκ2 + (1 + 2κ) dζ 2 + · · ·

2 2 = (dω + φdφ) + 1 + 2ω + φ2 ((1 − ω)dφ − φdω) + · · · = dω 2 + dφ2 + · · · All linear terms have now canceled. (b) If we take ζ to be the angular coordinate of the polar coordinates scaled by r∗ and κ to be the normalized deviation of the radial coordinate from r∗ , we can write a coordinate transformation as κ = r − r∗ 10

ζ = r∗ θ

Then starting from the metric in polar coordinates, the metric becomes ds2 = dr2 + r2 dθ2

κ 2 2 dζ = dκ2 + 1 + ∗ r

2κ = dκ2 + 1 + ∗ dζ 2 + · · · r

which for r∗ = 1 is exactly the metric we started with. The civilization has thus been living in a neighborhood of the point (r, θ) = (1, 0).

4 Solution in back of book.

5 Solution in back of book.

6 Let’s express the metric in exercise 3 in the locally flat coordinates (ω, φ). We know that the space is really flat, and that we can obtain the metric in the coordinates (ω, φ) from the metric in polar coordinates via r = 1 + ω + 21 φ2 θ = φ − φω Thus we have ds2 = dr2 + r2 dθ2 2

(3)

2 2

2

(−φ dω + (1 − ω) dφ) = (dω + φ dφ) + 1 + ω + φ

2 2 = 1 + φ dω − 2φω dφ dω + 1 + 2φ2 − 2ω 2 dφ2 + · · · 1 2

where the dots represent terms cubic or higher in ω and φ. We can now read off the coefficients Bμν,λσ : Bωω,φφ = 1 Bφφ,ωω = −2 Bωφ,ωφ = − 21 (4) The intrinsic curvature is then 2Bωφ,ωφ − Bωω,φφ − Bφφ,ωω = 2 (− 12 ) + 2 − 1 = 0 as we should expect, since we know the space is just flat space.

2 2 7 Any metric in D = 1 will take the form ds = g11 (x) dx . But if we change coordinates to y = g11 (x) dx (which will always be well defined at least in some region of the space), the metric becomes ds2 = dy 2 . In this new coordinate, all derivatives of the metric are zero, so curves have no intrinsic curvature.

8 Solution in back of book.

11

9 As is often the case, it is simpler to work out things in general than this specific example. Suppose we have arrived at gμν (x) = δμν + 2Aμν,λ xλ + · · · . (We define Aμν,λ here with an extra 2 for convenience.) Now let xμ = x μ + Lμνλ x ν x λ + · · · . Then

gρσ

=

(δμν + 2Aμν,λ xλ )(δρμ + 2Lμρψ x ψ )(δσν + 2Lνσφ x φ ) + · · ·

= δρσ + 2(Aρσ,λ + Lσ,ρλ + Lρ,σλ )xλ + · · · where the first index in Lσ,ρλ is lowered with the Euclidean metric δμν and I have introduced a comma to emphasize its symmetry properties. Thus, we could remove the linear terms in gμν by Lσ,ρλ + Lρ,σλ = −Aρσ,λ for L. Using an identity similar to that in the appendix to chapter I.4, or simply by inspection, we find Lρ,σλ = − 12 (Aρσ,λ + Aρλ,σ − Aλσ,ρ ). Note that, not surprisingly, this is closely related to the formula for the Christoffel symbol in terms of the derivatives of the metric.

10 Solution in back of book.

11 Solution in back of book.

12 We immediately see with no extra work that the Poincaré half plane is conformally flat:

dx2 + dy 2 ds2 = = Ω2 (y) dx2 + dy 2 y2 where Ω(y) = 1/y. For the sphere, we need to do a little more work. Start with the metric on S 2 in standard coordinates:

dθ2 2 + dφ ds2 = dθ2 + sin2 θ dφ2 = sin2 θ sin2 θ

Define a new coordinate ψ = dθ/ sin θ = ln tan(θ/2); then sin2 θ = sech2 ψ, so that we have

ds2 = Ω2 (ψ) dψ 2 + dφ2 with Ω(ψ) = sechψ. Thus the sphere is also conformally flat.

13 Solution in back of book.

14 According to appendix 3, the n-dimensional hyperbolic space is described by ds2 = dr 2 2 2 equal to ds2 = Ω2 (dρ2 + ρ2 dΩ2n−1 ), and we have the two 1+r 2 + r dΩn−1 . Set this √ conditions r = Ωρ and dr/ 1 + r2 = Ωdρ = rdρ/ρ. Integrating this equation, we find r = 2ρ/(1 − ρ2 ) and Ω = 2/(1 − ρ2 ).

I.7

Differential Geometry Made Easy, but Not Any Easier!

√

dϕ dϕ X −aϕ 2 1 We easily calculate t = ddlX = ddϕ / a2 + 1. We dl . From t = 1, we obtain dl = e √

d t dϕ −aϕ / a2 + 1. Notice then calculate ddlt = dϕ dl , the magnitude of which gives κ = e that κ → 0 as ϕ → ∞, but κ → ∞ as ϕ → −∞; that is, as the spiral spirals in toward the origin, it winds itself tighter and tighter. We next find b, which turns out to be just (0, 0, 1). As might be expected, τ = 0.

12

2 Solution in back of book.

3 Solution in back of book.

4 A simple calculation shows that κ = a and τ = b, in complete accordance with our geometric intuition. For b = 0, the curve does not twist.

5 Evaluating Gauss’s equation, we find that Kμν = −gμν and thus G = 1.

6 Solution in back of book.

II

Action, Symmetry, and Conservation

II.1

The Hanging String and Variational Calculus

√ 1 We have F (u) = u/ 1 + u2 . Thus, instead of the first term in (3), we would obtain dφ d 1 T dx ( ). dφ 2 dx 1+( dx )

2 Solution in back of book.

3 Solution in back of book. ∂φ 2 4 The metric is given by ds2 = dx2 + dy 2 + dφ2 = dx2 + dy 2 + ( ∂φ ∂x dx + ∂y dy) = √ ∂φ 2 ∂φ 2 ∂φ ∂φ 2 2 (1 + ( ∂x ) )dx + (1 + ( ∂y ) )dy + 2 ∂x ∂y dxdy. The area element dxdy g is then up to

∂φ 2 2 quadratic order in φ given by dxdy 1 + ( ∂φ ∂x ) + ( ∂y ) . Thus, the amount by which

∂φ 2 2 the area is stretched is given by dxdy 12 [( ∂φ ∂x ) + ( ∂y ) ].

5 Solution in back of book.

II.2

The Shortest Distance between Two Points

1 From (11) we obtain sin2 θ dϕ dl = K ≤ 1, which when plugged into (12) gives us dθ 2 K2 ( dl ) + sin2 θ = 1. The potential has a minimum value of K 2 ≤ 1 at θ = π/2. The particle oscillates between two values of θ determined by sin2 θ = K 2 .

2 We need to consider only three cases (up to permutations of the indices) in which the Christoffel symbol does not vanish: (1) Γ111 = − 12 g 11 ∂1 g11 , (2) Γ211 = 12 g 22 ∂2 g11 , and (3) Γ112 = 12 g 11 ∂2 g11 . Note that Γ123 = 0.

3 The solution is actually given in the text. dy 2 2 4 The definition of length l gives us ( dx = y 2 . The geodesic equation dl ) + ( dl ) dy 2 d dx dx 2 2 2 4 2 dl ( dl /y ) = 0 gives dl = y /b. Plugging this in, we obtain ( dl ) = y − y /b .

With the change of variable y = b sin θ, we find x = ∓b cos θ + x∗ , in agreement with the result in appendix 4.

13

5 The Christoffel symbol Γηωσ vanishes in Cartesian coordinates. Thus, the Christoffel ∂x λ ∂ 2 xη r symbol in polar coordinates is given by Γ λ μν = ∂xη ∂x μ ∂x ν . We obtain Γθθ = 2 2 ∂r ∂ 2 x ∂r ∂ y ∂θ ∂ 2 x ∂θ ∂ y 1 θ ∂x ∂θ 2 + ∂y ∂θ 2 = −r. Similarly, Γrθ = ∂x ∂r∂θ + ∂y ∂r∂θ = r .

6 Solution in back of book. λ λ λ λ 7 Plugging into (30), we obtain Γ λ μν = Γμν + Lμν , and so set Lμν = −Γμν .

8 Solution in back of book.

II.3

Physics Is Where the Action Is

T 1 Set m = 2, so that S = 0 (q̇ 2 − ω 2 q 2 ). Write the solution of the equation of motion q̈ = −ω 2 q as q0 (t) = a cos ωt + b sin ωt with a, b fixed by some boundary conditions (with q(0) = q(T ).) Let q = q0 + δq. Then the second order variation of the action is T δS (2) = 0 ((δ q̇)2 − ω 2 δq 2 ) with the boundary conditions δq(0) = δq(T ) = 0. We can (2) already see that we could < 0 if δq varies slowly ∞ enough in time. More ∞ make δS explicitly, write δq = 1 cn sin(nπt/T ). Then δS (2) = 12 1 ((nπ/T )2 − ω 2 )c2n . For T large enough such that π/T < ω, we could, by choosing cn appropriately, get either sign for δS (2) .

2

2

3

h hT T hT 2 We find S(q2 ) = 2h 3T − 3 , and so S(q2 ) − Smin = 6T + 6 + 24 > 0.

3 We simply take the answer given in the text, set h = 0 and restore g.

4 Solution in back of book. +

∂ + ∂ 5 dx+ ∂x∂+ = dx+ ∂x ∂x+ ∂x + = dx ∂x + .

6 Use Cartesian coordinates. Let a string of length 2L be tied down a at x = ±a.

Call the area enclosed between the string and the x-axis 2A = −a dx y(x). The

a dy 2 ) = 2L. (The various factors of 2 are for constraint reads 2P ≡ −a dx 1 + ( dx convenience.) Thanks to Lagrange, we simply plug and chug. We have the obvious

a dy 2 symmetry condition y(x) = y(−x). Extremizing A + λP = 0 dx y + λ 1 + ( dx ) with the boundary condition y(a) = 0 and the Lagrange multiplier λ, we obtain dy 2 dy 2 d √ λ dx the Euler-Lagrange equation dx = 1, which implies ( dx ) = λ2x−x2 . We dy 2 1+( dx ) √ √ integrate to obtain y = λ2 − x2 = a2 − x2 with the second equality fixed by the boundary condition.

We find the circle as expected. Imposing the constraint, we find

√ a a dy 2 2L = 2 0 dx 1 + ( dx ) = 2a 0 dx 1/ a2 − x2 = πa. By the way, the geometrical proof I learned as a kid is given here: www.math.upenn.edu/˜shiydong/Math501X-5Isoperimetric.pdf.

14

II.4

Symmetry and Conservation

1 We now have rotational invariance around the x-, y-, and z-axes. Simply repeat k what we did to get to (10). The 3 conserved quantities are Li = m ijk xj dx dt , where ijk is the totally antisymmetric symbol defined in chapter I.3. For example,

3 3 dx2

x × p from L1 = m x2 dx dt − x dt . This is of course just the familiar formula L = elementary physics.

III

Space and Time Unified

III.1

Galileo versus Maxwell

1 Solution in back of book.

III.2

Einstein’s Clock and Lorentz’s Transformation 2

2

(Δt) and thus the relation between Δt and Δt.

1 Squaring, we find (Δt )2 = c24L u2 −u2 = 1− c2

The rest follows.

2 You can simply plug in and verify that one matrix is the inverse of the other. The easiest way to show this is to use the form of the transformation in light cone coordinates (5). Then u → −u corresponds to φ → −φ and eφ going into its inverse.

3 With no loss of generality, let the object be moving in the (x-y) plane. Then √ the differential form of the Lorentz transformation gives dt = dt−udx , dx = 1−u2

vx +u dx−udt √ √ , dy = dy. Dividing dx = dx−udt , dy by dt , we obtain vx = 1+uv , vy = x 1−u2 √1−u2 v y 1 − u2 1+uvx . If we insist on writing this using vectors, we obtain the rather unwieldy expression

v = [ 1 − u2 u2 v + ((1 − 1 − u2 ) u · v + u2 ) u]/(u2 (1 + u · v ))

This exercise reinforces the point, to be made in chapter III.6, that it is advisable to deal with objects that transform “naturally.” v∗ +u 4 The limiting value v∗ is determined by v∗ = 1+uv . Solving, we obtain v∗2 = 1, as ∗ expected. Note that v0 does not enter, also as expected.

III.3

Minkowski and the Geometry of Spacetime

1 Solution in back of book.

2 We have dt = ρ cosh T dT + sinh T dρ, dr = ρ sinh T dT + cosh T dρ. Thus, dt2 − dr2 = ρ2 dT 2 − dρ2 .

3 Solution in back of book.

15

4 Solution in back of book. μ

μν 5 Let ω μν ≡ aμ V ν − V μ aν so that dW dτ = −ω Wν . Then

d dW μ = −2ω μν Wμ Wν = 2ω νμ Wν Wμ = 0 (Wμ W μ ) = 2Wμ dτ dτ where the last step follows from the fact that something equal to minus itself vanishes.

This is just an example of the general result we learned in chapter I.4 that the product of an antisymmetric tenor (namely ω μν ) and a symmetric tensor (namely μ d μ Wμ Wν ) vanishes. Next, (aμ V ν − V μ aν )Vν = −aμ = − dV dτ . Finally, − dτ (Vμ W ) = μν μν Vμ ω Wν + Wμ ω Vν = 0.

6 Solution in back of book.

7 In the context of appendix 1, the statement is true by definition. The metric ημν regarded as a matrix is left invariant; in particular, the number of eigenvalues equal to +1 or −1 cannot change. This statement has more content for curved spaces or spacetimes. It amounts to the intuitive statement that a space cannot be turned into a spacetime by a coordinate transformation, and vice versa. In the context of chapter I.6, when we go to locally flat coordinates, we could scale the diagonal elements of gμν (0) to ±1, but we cannot change their signs.

8 Solution in back of book.

9 In the experimentalist’s frame, uμ = (1, 0) and hence the energy of the particle is given by E = −uμ pμ . Since (in any frame) p 2 = E 2 + p2 , plugging in what we have for E, the stated result follows.

√ 10 We have d(t, x) ≡ dAB + dBC = t2 − x2 + (2 − t)2 − x2 . By symmetry, take 0 < t < 1. Since

−1

∂d(t, x) −1 2 2 2 2 = −x <0 t − x ) + ( (2 − t) − x ∂x the distance d(t, x) reaches it maximum at x = 0, where d(t, 0) = 2.

III.5 The Worldline Action and the Unification of Material Particles with Light 1 Solution in back of book.

2 Under a Weyl transformation, we have γ → e4ω(τ,σ) γ and γ αβ (τ, σ) → e−2ω(τ,σ) γ αβ (τ, σ).

1 1 1 4ω(τ,σ) ) 2 γ αβ e−2ω(τ,σ) (∂α X μ ∂β Xμ ) = S is left invariant.

Thus S = 2 T dτ dσγ 2 (e

III.6 Completion, Promotion, and the Nature of the Gravitational Field 1 Solution in back of book.

16

2 Solution in back of book.

3 Solution in back of book.

4 Solution in back of book.

5 Solution in back of book.

6 The string current found in the text takes the form

∂τ q μ ∂τ q ν μν J (x) = dτ dσ det δ (d) (x − q(τ, σ)) ∂σ q μ ∂σ q ν Now, consider making a change of variables to τ = τ (τ, σ) and σ = σ (τ, σ). Then the measure in the integral changes by a Jacobian: ∂τ dτ dσ = det

∂τ ∂τ ∂σ

∂σ ∂τ ∂σ ∂σ

dτ dσ = J dτ dσ

Meanwhile, the partial derivatives transform like ∂σ ∂τ ∂σ q μ ∂τ q μ + ∂τ ∂τ ∂σ ∂τ ∂σ q μ ∂τ q μ + ∂σ q μ = ∂σ ∂σ ∂τ q μ =

so that we have

∂ qμ det τ μ ∂σ q

∂τ q ν ∂σ q ν

= det

∂τ ∂σ μ μ

∂τ ∂τ q + ∂τ ∂σ q ∂σ ∂τ μ μ

∂σ ∂τ q + ∂σ ∂σ q

= det

= det

=J

−1

∂τ ∂τ ∂τ ∂σ

∂τ ∂τ ∂τ ∂σ

∂σ ∂τ ∂σ ∂σ ∂σ ∂τ ∂σ ∂σ

∂τ q μ ∂σ q μ

∂ qμ det τ μ ∂σ q

det

∂τ q μ ∂σ q μ

∂τ q ν ∂σ q ν

∂σ ∂τ ν ν

∂τ ∂τ q + ∂τ ∂σ q ∂σ ∂τ ν ν

∂σ ∂τ q + ∂σ ∂σ q

∂τ q ν ∂σ q ν

∂τ q ν ∂σ q ν

Putting these back together, we see that the factors of the Jacobian J cancel precisely, and we get

∂ q μ ∂τ q ν J μν (x) = dτ dσ det τ μ δ (d) (x − q(τ , σ )) ∂σ q ∂σ q ν so the string current is indeed invariant under transformations of string coordinates.

17

7 Recall that angular momentum can be written as J = r × p , so we have, for instance, Jz = xpy − ypx , and so forth. Of course, we really should think of rotations in general dimensions, in which case we want to label them with two indices, not one, so we instead consider J ij = xi pj − xj pi . Now, the components of momentum are related to the components of the stress tensor as pi = T 0i , so we might be tempted to say that the angular momentum is J ij = xi T 0j − xj T 0i . However, this doesn’t transform like a tensor; we really need to integrate it over some 3-volume, so we write

Mij = d3 x xi T 0j − xj T 0i We conclude that the Mij describe the angular momentum of a system. If we upgrade from only spatial indices to Lorentz indices, and define

Mμν = d3 x xμ T 0ν − xν T 0μ we get an object that describes both angular momentum (Mij ) and conserved quantities associated with Lorentz boosts (M0i ).

Note that the 12-component M12 = d3 x x1 T 02 − x2 T 01 is precisely the elementary definition of angular momentum around the 3rd axis for an extended body. It seemingly could depend on the time t when the integral is evaluated. Let us check that in fact it does not (namely that angular momentum is conserved):

i

d ij 3 0j j 0i M = d x x ∂0 T − x ∂0 T = d3 x T ij − T ji = 0 dt In the second equality we used ∂0 T 0ν = −∂i T iν , and in the third equality, the fact that T μν is symmetric.

IV

Electromagnetism and Gravity

IV.1

You Discover Electromagnetism and Gravity!

2

μ

dp 1 We have dp dτ = 2pμ dτ μν antisymmetry of F .

IV.2

= m−1 F μν pμ pν = 0. The last equality is due to the

Electromagnetism Goes Live

2 2 ).

2 − B + Fij2 ) = 21 (E 1 L = − 14 Fμν F μν = − 14 (−2F0i

2 (a) ∂μ T μν = (∂μ F μλ )F νλ +F μλ ∂μ F νλ − 12 Fσρ ∂ ν F σρ . In the absence of charged particles, the first term vanishes by Maxwell’s equation (13). Next, (16) tells us that the sum of ∂ ν F σρ and the two terms obtained from it by cyclically permuting the indices νσρ vanish. Hence the third term in the expression above for ∂μ T μν can be rewritten as + 12 Fσρ (∂ σ F ρν + ∂ ρ F νσ ) = Fσρ ∂ σ F ρν . In contrast, the second term in ∂μ T μν , 18

namely F μλ ∂μ F νλ , can be written as −Fμλ ∂ μ F λν . We thus obtained the desired result. Throughout, we raise and lower indices with the Minkowski metric and use the antisymmetry of Fμν . (b) In the presence of charged particles, the first term in the expression in part a μν for ∂μ T μν does not vanish; instead, Maxwell’s equation (13) gives ∂μ Telectromagnetic = μν νλ −Jλ F . We have to add to this ∂μ Tparticles . We simply repeat the calculation in exercise III.6.3 except that in the last step, we have to use the equation of motion in the presence of the electromagnetic field, namely (IV.1.23) from the preceding chapter: dpμ dXaν μν μν μ νλ a Jλ = −∂μ Telectromagnetic . dτa = ea F ν (Xa (τa )) dτa . We obtain ∂μ Tparticles = F Hence the desired result.

3 We have T 00 = F 0i F 0i + 14 Fσρ F σρ . Plugging in (IV.1.17), we obtain the stated result.

4 Solution in back of book.

5 Solution in back of book.

6 Use the definition F̃μν = − 12 μνλσ F λσ and plug the identity ηλσ λμρω σνζψ = −(η μν η ρζ η ωψ + . . .) into ηλσ F̃ μλ F̃ νσ . . . .

7 Solution in back of book.

8 Solution in back of book.

9 This follows almost instantly: T ≡ ημν T μν = Fμλ F μλ − ( 14 )4Fσρ F σρ = 0.

10 Solution in back of book.

Prologue to Book Two:

The Happiest Thought

1 Solution in back of book.

V

Equivalence Principle and Curved Spacetime

V.1

Spacetime Becomes Curved

1 Solution in back of book.

V.2

The Power of the Equivalence Principle

1 Solution in back of book.

19

V.3

The Universe as a Curved Spacetime

d dt 2 x 2 dt d2 t dt x 2 x d2 x 2 d ( dτ = 2( dτ 1 We find dτ ) −a2 ( d ) ) dτ 2 −2aȧ( dτ )( d dτ dτ ) −2a ( dτ ) dτ 2 = 0 upon using (4) and (5).

2 With g00 = −1, g01 = − sin x, g11 = cos2 x, we evaluate (9) to find dl2 = dx2 . To understand this simple result, note that if we replace t by t̃ = t−cos x, the spacetime is in fact flat.

3 For a(t) = βtα , we obtain 2β(1 − α)R = (t1−α − t1−α ) and thus D(R; tS ) = R S 1 1−α 1−α 1 ) − tS ]. For α < 1, (1 − α) > 0, so that as R tends infinity, so 2 [(2β(1 − α)R + tS 1 does tR . In the limit R → ∞, we obtain D(R; tS ) ∝ R 1−α , namely ∝ R3 for a matter 2 dominated universe and ∝ R for a radiation dominated universe. In the opposite limit, for 2β(1 − α)R t1−α , we find that D(R; tS ) βtα S R = a(tS )R, as expected.

S t dt In general, for R small, we have tR tS , and thus 2R = tSR a(t) (tR − tS )/a(tS ) 1 and so D(R; tS ) = 2 (tR − tS ) a(tS )R. a(T ) with T = t(R; tS ) determined 4 It is merely a matter of evaluating 1 + z ≡ ωωre = a(t S) R t(R;tS ) dt

by R = 0 dr = tS a(t) .

For a(t) = et we obtain (1 + z) = 1/(1 − RetS ). In the R → 0 limit, z

RetS .

1−α 1−α For a(t) = βtα we obtain (1 + z) = t−α ) . In the R → 0 limit, S (β(1 − α)R + tS we have z αβR/tS 1−α .

1

5 Solution in back of book.

6 Solution in back of book.

7 Solution in back of book.

8 Consider the discussion in the text about sending a message carried by electromagnetic wave at time tS to our friend located at r = R and getting a response back at time dt = dr r2 . Integrating this along the outbound trip and the tR . We now have a(t) 1−k L2

return trip, we obtain tR tS

dt =2 a(t)

dr

= 2L(sin−1 (R/L)

or

sinh−1 (R/L))

2 1 − k Lr 2

for a closed and open universe, respectively. For a given a(t), this determines tR = tR (R, tS ), which in turn fixes the distance between us and our friend D(R; tS ) = 1 2 (tR − tS ).

20

V.4

Motion in Curved Spacetime

1 Simply plug (23) into (17).

2 Solution in back of book.

3 Simply vary to obtain the equations of motion with A and B dependent on t as well as r.

4 Solution in back of book.

5 Solution in back of book.

6 Solution in back of book.

V.6

Covariant Differentiation

1 You can show this either by verifying that the divergence as stated transforms correctly or by pretending that T μν is given by W μ U ν .

2 Simply plug in and do an ordinary integration by parts.

3 For the last part of the exercise: Dλ g μν = ∂λ g μν + Γμλρ gρν + Γνλρ gμρ = −g μσ (∂λ gσω − Γηλσ gηω − Γηλω gση )g ων = −g μσ (Dλ gσω )g ων = 0, where we used (7) in the second equality.

4 Solution in back of book.

5 Solution in back of book. dS(ζ) 6 For a scalar S define DS(ζ) Dζ = dζ . This transforms like a scalar.

7 In exercise 6, let S = Uμ W μ . Then as we would expect, the covariant derivaDUμ (ζ) dUμ (ζ) tive of a vector with a lower index along a curve is given by Dζ = dζ − ν λ Γμλ (X(ζ))V (ζ)Wν (ζ). Given how tensors transform as if they were made of products of vectors with an upper index and of vectors with a lower index, we can now immediately write down the covariant derivative of tensors along a curve. For example, DT μν dT μν = + Γμλρ V λ T ρν − Γρλν V λ T μρ Dζ dζ

VI

Einstein’s Field Equation Derived and Put to Work

VI.1

To Einstein’s Field Equation as Quickly as Possible

1 Note: A solution was already provided in the back of the book, but here is a more detailed solution.

21

This is just plug-and-chug, using the Christoffel symbols we already know: Dθ Wθ = ∂θ Wθ − Γθθθ Wθ − Γϕ θθ Wϕ = ∂θ Wθ Dθ Wϕ = ∂θ Wϕ − Γθθϕ Wθ − Γϕ θϕ Wϕ = ∂θ Wϕ − cot θ Wϕ Dϕ Wθ = ∂ϕ Wθ − Γθϕθ Wθ − Γϕ ϕθ Wϕ = ∂ϕ Wθ − cot θ Wϕ Dϕ Wϕ = ∂ϕ Wϕ − Γθϕϕ Wθ − Γϕ ϕϕ Wϕ = ∂ϕ Wϕ + sin θ cos θ Wθ Dθ U θ = ∂θ U θ + Γθθθ U θ + Γθθϕ U ϕ = ∂θ U θ ϕ θ ϕ Dθ U ϕ = ∂θ U ϕ + Γϕ θθ U + Γθϕ U

= ∂θ U ϕ + cot θ U ϕ Dϕ U θ = ∂ϕ U θ + Γθϕθ U θ + Γθϕϕ U ϕ = ∂ϕ U θ − sin θ cos θ U ϕ ϕ ϕ θ Dϕ U ϕ = ∂ϕ U ϕ + Γϕ ϕθ U + Γϕϕ U

= ∂ϕ U ϕ + cot θ U θ Next, note that

(Dθ Wρ ) U ρ + Wρ (Dθ U ρ ) = (Dθ Wθ ) U θ + (Dθ Wϕ ) U ϕ + Wθ Dθ U θ + Wϕ (Dθ U ϕ ) = (∂θ Wθ ) U θ + (∂θ Wϕ − cot θ Wϕ ) U ϕ

+ Wθ ∂θ U θ + Wϕ (∂θ U ϕ + cot θ U ϕ )

= (∂θ Wθ ) U θ + Wθ ∂θ U θ + (∂θ Wϕ ) U ϕ + Wϕ (∂θ U ϕ )

= ∂θ Wθ U θ + Wϕ U ϕ = Dθ (Wρ U ρ )

(Dϕ Wρ ) U ρ + Wρ (Dϕ U ρ ) = (Dϕ Wθ ) U θ + (Dϕ Wϕ ) U ϕ + Wθ Dϕ U θ + Wϕ (Dϕ U ϕ ) = (∂ϕ Wθ − cot θ Wϕ ) U θ + (∂ϕ Wϕ + sin θ cos θ Wθ ) U ϕ

+ Wθ ∂ϕ U θ − sin θ cos θ U ϕ + Wϕ ∂ϕ U ϕ + cot θ U θ

= (∂ϕ Wθ ) U θ + Wθ ∂ϕ U θ + (∂ϕ Wϕ ) U ϕ + Wϕ (∂ϕ U ϕ )

= ∂ϕ Wθ U θ + Wϕ U ϕ = Dϕ (Wρ U ρ ) and thus we have shown explicitly that (Dμ Wν ) U ν + Wν (Dμ U ν ) = Dμ (Wν U ν ).

Solution in back of book.

22

3 Simply plug in and verify.

4 Again, simply plug in and verify.

5 In proving relations like this, we can pretend, once again, that Tμρ is equal to the product of two vectors Wμ Yρ .

6–7 Solution in back of book.

8 Solution in back of book.

9 The only nonzero Christoffel symbols are Γyyμ = (∂μ φ)/φ and Γμyy = −η μν ∂ν φ. Then the nonzero components of the Ricci tensor are Rμν = Rσ μσν + Ry μyν = −∂ν Γyyμ − Γyνy Γyyμ ∂μ ∂ν φ =− φ Ryy = Rμ yμy + Ry yyy = ∂μ Γμyy − Γyyμ Γμyy = −φ φ √ √ where φ = η μν ∂μ ∂ν φ = ∂μ ( −η η μν ∂ν φ)/ −η. The Ricci scalar is then R = η μν Rμν + g yy Ryy = −

2 φ φ

10 Solution in back of book.

11 Solution in back of book.

12 Go to a locally flat system of coordinates to verify the stated relation.

13 Solution in back of book.

ρ2 −2 14 We are given ds2 = 1 + 4L (dx2 + dy 2 ) with ρ2 = x2 + y 2 , that is gij = Ω2 δij . To 2 calculate the scalar curvature using (32), we merely have to set d = 2 and to calculate

ρ2 −1 ∂i Ω and ∂i ∂j Ω with Ω = 1 + 4L . We find that R = 2.

2 15 (a) Since the Weyl tensor has all the same symmetries as the Riemann tensor, the only trace we need to evaluate is g μρ Cμνρσ = Rνσ + (d − 2)−1 (Rνσ (2 − d) − gνσ R) + ((d − 1)(d − 2))−1 (d − 1)gσν R = 0. (b) Simply evaluate C̃μνρσ by plugging the result of exercise 12 into (33).

16 DU DV W λ = DU V ν W λ;ν = U μ (V ν W λ;ν );μ = U μ V ν;μ W λ;ν + U μ V ν W λ;ν;μ . Interchange U and V and subtract. The stated result follows.

23

17 (a) By direct calculation: the metric gxx = y 2 , gyy = x2 , gxy = 0 leads to Γxxy = 1/y, Γxyy = −x/y 2 , with the other components either vanishing or given by x ↔ y. Then Rxyxy = ∂x Γxyy − ∂y Γxxy − Γxxy Γxxy − Γxyy Γyxy = 0. (2) Let x = eu , y = ev .

Then ds2 = e2(u+v) (du2 + dv 2 ). So Ω = e(u+v) = xy. ∂ ∂ Ω

(b) From exercise 13: we have R̃ = 0 − 2δ ρω ρΩ3ω − (−2)δ ρω

VI.2

(∂ρ Ω)(∂ω Ω) = 0. Ω4

To Cosmology as Quickly as Possible

1 Solution in back of book.

2 Again, Einstein’s equation is solved if p + q + r + s = p2 + q 2 + r2 + s2 = 1. This has the interesting solution p = q = r = 21 , s = − 12 so that the 5th dimension (for the meaning of 5, see chapter X.1) contracts.

3 Calculate the Christoffel symbol, then the Ricci tensor. Then solve.

VI.3 The Schwarzschild-Droste Metric and Solar System Tests of Einstein Gravity 1 Just continue the computation started in the text.

2 Solution in back of book.

3 Solution in back of book.

4 Solution in back of book.

5 Solution in back of book.

6 The solution is given in the appendix to chapter VII.4.

7 This involves a straightforward computation best performed with the help of a computer.

VI.4 Energy Momentum Distribution Tells Spacetime How to Curve 1 The electromagnetic stress tensor was given in (8); since we’re working in flat spacetime here, we’ll go ahead and replace gμν → ημν : Tμν = Fμλ Fν λ − 14 ημν Fσρ F σρ Getting Tij is now simply a matter of breaking up the index summations into sums over time and space components (using the fact that Fμν is antisymmetric in its indices), 24

making the substitutions F0i = Ei and Fij = εijk Bk , and using some identities involving the Levi-Civita symbol: Tij = Fiλ Fj λ − 14 ηij Fσρ F σρ

= Fi0 Fj 0 + Fik Fj k − 14 δij 2F0k F 0k + Fk F k = −Ei Ej + εik εjkm B Bm − δij (−2Ek Ek + 14 εk m εk n Bm Bn ) 2 + 2δμν Bm Bn = −Ei Ej + (δij δ m − δim δ j ) B Bm − 41 δij −2E

2 + 2Bm Bm = −Ei Ej + δij B B − Bj Bi − 14 δij −2E

2 + B 2 = −Ei Ej − Bi Bj + 12 δij E

The trace of the stress tensor is thus T = η μν Tμν = −T00 + δ ij Tij

2 − B 2 + B 2 + B 2 2 − E 2 + 32 E = − 12 E =0 as it should.

2 The divergence of the stress energy tensor of a gas of particles is

dXaν dXaμ 1 μν Dμ Tparticles Dμ δ 4 (x − Xa ) =√ ma dτa dτa dτa −g a

dXaν d 4 1 ma dτa δ (x − Xa ) =√ dτa dτa −g a

d2 Xaν 4 1 ma dτa δ (x − Xa ) = −√ −g a dτa2 To get to the second line, we used the chain rule to write (dXaμ /dτa )Dμ = d/dτa , and to get to the last line, we integrated by parts. But since no external forces act on the particles, we have by the geodesic equation that ma

dXaμ dXaν d2 Xaν Dμ = ma =0 dτa dτa dτa2

μν and thus Dμ Tparticles = 0.

3 The stress energy tensor is

dXaμ dXaν 4 1 μν T =√ ma dτa δ (x − Xa ) − gσλ F μσ F νλ + 14 g μν Fσλ F σλ −g a dτa dτa 25

Let’s take the divergence term-by-term, beginning with the last two terms (that is, the stess energy tensor of the electromagnetic field):

μν = Dμ −gσλ F μσ F νλ + 14 g μν Fσλ F σλ Dμ TEM

= −gσλ F μσ Dμ F νλ + F νλ Dμ F μσ + 14 g μν Dμ Fσλ F σλ Maxwell’s equations with a source give us that Dμ F μσ = −J σ , and noting that

σλ = 2F σλ Dμ Fσλ , we are left with Dμ Fσλ F

μν = −gσλ F μσ Dμ F νλ − F νλ J σ + 12 g μν F σλ Dμ Fσλ Dμ TEM = F ν σ J σ − g σλ F μ σ Dμ F ν λ + 12 g μν F σλ Dμ Fσλ = F ν σ J σ − g σν F μλ Dμ Fσλ + 12 g μν F σλ Dμ Fσλ Relabeling indices in the last two terms allows us to rewrite them as μν = F ν σ J σ − 12 g μν F σλ (2Dσ Fμλ − Dμ Fσλ ) Dμ TEM

= F ν σ J σ − 12 g μν F σλ (2Dσ Fμλ + Dμ Fλσ ) The first term in the parentheses can be rewritten as 2F σλ Dσ Fμλ = F σλ Dσ Fμλ + F σλ Dσ Fμλ = F σλ Dσ Fμλ + F λσ Dσ Fλμ = F σλ Dσ Fμλ + F σλ Dλ Fσμ = F σλ (Dσ Fμλ + Dλ Fσμ ) This leaves us with μν = F ν σ J σ − 12 g μν F σλ (Dσ Fμλ + Dλ Fσμ + Dμ Fλσ ) Dμ TEM = F νσJσ

where we used the Bianchi identity to kill the quantity in parentheses. The current of a collection of charged particles is

dXaσ 4 1 Jσ = √ ea dτa δ (x − Xa ) −g a dτa Therefore we have 1 μν Dμ TEM =√ −g 1 =√ −g

dτa F ν σ

ea a

ma

a

dτa

dXaσ 4 δ (x − Xa ) dτa

d2 Xaν 4 δ (x − Xa ) dτa2

where we used the Lorentz force law ma

dXaμ dXaν d2 Xaν dXaσ Dμ = ma = ea F ν σ 2 dτa dτa dτa dτa 26

Next, we need to find the divergence of the stress energy tensor of the particles. Since all the x dependence is in the delta function, we have

dXaν dXaμ 1 μν √ Dμ Tparticles = ma dτa Dμ δ 4 (x − Xa ) −g a dτa dτa

dXaν d 4 1 ma dτa δ (x − Xa ) =√ −g a dτa dτa

d2 Xaν 4 1 ma dτa δ (x − Xa ) = −√ −g a dτa2 To get to the second line, we used the chain rule to write (dXaμ /dτa )Dμ = d/dτa , and to get to the last line, we integrated by parts. This result is precisely the opposite of the divergence of the stress energy tensor of the electromagnetic field, so we have found that μν μν Dμ T μν = Dμ Tparticles + Dμ TEM =0 4 Solution in back of book.

5 Solution in back of book.

6 Just follow the same steps leading to (18).

7 Solution in back of book.

8 Solution in back of book.

VI.5

Gravity Goes Live

1 The Palatini action is S=

1 16πG

√ d4 x −g g μν Rμν (Γ) + Smatter

The variation with respect to g μν is easy:

√ √ √ 1 δS = d4 x −g δg μν + δ −g g μν Rμν (Γ) + −g g μν δRμν (Γ) + δSmatter 16πG

√ 1 d4 x −g Rμν − 12 gμν R δg μν + g μν δRμν (Γ) + δSmatter = 16πG The quantity in brackets being contracted with δg μν gives us the Einstein field equations, as usual: Rμν − 12 gμν R = 8πGTμν Next, we need to vary the piece containing δRμν (Γ): √ √ −g g μν δRμν (Γ) = −g g μν δ ∂σ Γσμν − ∂μ Γσσν + Γλμν Γσλσ − Γλσν Γσλμ √ = −g g μν ∂σ δΓσμν − ∂μ δΓσσν + δΓλμν Γσλσ + Γλμν δΓσλσ − 2δΓλσν Γσλμ 27

Now, let’s work on integrating the first two terms by parts. Using our handy identities ∂σ g μν = −g μλ g νρ ∂σ gλρ √ 1√ ∂σ −g = −g g λρ ∂σ gλρ 2 we can write (dropping total derivatives, since they’ll drop out when we integrate by parts) √ √ −g g μν ∂σ δΓσμν = −∂σ −g g μν δΓσμν

√ = −g g μλ g ρν ∂σ gλρ − 12 g μν g λρ ∂σ gλρ δΓσμν

√ √ − −g g μν ∂μ δΓσσν = − −g g μλ g ρν ∂μ gλρ − 12 g μν g λρ ∂μ gλρ δΓσσν As a result, we find √

−g g μν δRμν (Γ) =

√ μλ ρν −g g g ∂σ gλρ − 12 g μν g λρ ∂σ gλρ δΓσμν − g μλ g ρν ∂μ gλρ − 12 g μν g λρ ∂μ gλρ δΓσσν

+ g μν δΓλμν Γσλσ + Γλμν δΓσλσ − 2δΓλσν Γσλμ √

−g g μλ g ρν ∂σ gλρ − 12 g μν g λρ ∂σ gλρ

− g δλ g ρν ∂δ gλρ − 12 g δν g λρ ∂δ gλρ δσμ + g μν Γλσλ + g λρ Γμλρ δσν − 2g λν Γμσλ δΓσμν

−g g μν δRμν (Γ) =

√

Now, we must assume that the connection Γ is symmetric in its two lower indices, so we really need to symmetrize the expression in brackets above in the indices μ and ν.

Then assuming the connection Γ doesn’t appear in Smatter , the equations of motion for the connection are g λ(μ g ν)ρ ∂σ gλρ − 12 g μν g λρ ∂σ gλρ

(μ μ) − g δλ g ρ(ν ∂δ gλρ − 12 g λρ g δ(ν ∂δ gλρ δσμ) + g μν Γλσλ + g λρ Γλρ δσν) − 2g λ(ν Γσλ = 0 μ)

The parenthesis notation used here means symmetrization: for example, g λ(ν Γσλ = 1 λν μ λμ ν 2 (g Γσλ + g Γσλ ).

By contracting μ and ν, we get a formula for Γλσλ in terms of g λρ Γμλρ . Plugging back in and contracting σ and ν then gives a formula for g λρ Γμλρ . Plugging that back in μ)

gives a formula for g λ(ν Γσλ . The intermediate results are: Γμσμ = 14 g μν ∂σ gμν + 12 g μν ∂μ gνσ − 12 gμσ g νλ Γμνλ g λρ Γμλρ = g λν g μρ ∂ν gλρ − 12 g λρ g σμ ∂σ gλρ 1 μ) g λ(ν Γσλ = g λ(μ g ν)ρ ∂σ gλρ 2 28

We can use this last result to reconstruct Γμνσ by themselves by using the symmetry of the connections. Indeed, first let us remove the symmetrization parentheses by writing g λν Γμσλ = 12 g λν g μρ ∂σ gλρ + C μν σ where C μν σ are constants (to be determined) antisymmetric in μ and ν. Lowering the index ν gives Γμσν = 12 g μρ ∂σ gνρ + C μ νσ Since the right-hand side must be symmetric in σ and ν, we conclude that C μ νσ must provide a contribution of (1/2)g μρ ∂ν gσρ . Then the antisymmetry of C μ νσ in μ and ν requires another term −(1/2)g μρ ∂ρ gσν , so that we have C μ νσ = 21 g μρ ∂ν gσρ − 12 g μρ ∂ρ gσν These are all the terms that the coefficients C may contain, subject to the symmetry requirements and the fact that the C contains only first derivatives of the metric.

Thus we have specified the C uniquely, and we are left with Γμσν = 21 g μρ (∂σ gνρ + ∂ν gσρ − ∂ρ gσν ) as expected.

2 Solution in back of book.

VI.6

Initial Value Problems and Numerical Relativity

1 Solution in back of book.

2 Solution in back of book.

VII

Black Holes

VII.1

Particles and Light around a Black Hole

1 Plugging in the values of various quantities given in the text, we obtain gtt (V t )2 + S −1 S −1 gϕϕ (V ϕ )2 = − 2 /(1 − rrS ) + l2 /r2 = −(1 − rrS )(1 − 3r + r2rS (1 − 3r = −1.

2r ) 2r ) 2 The desired function r(ϕ) is readily obtained by integrating (10).

3 Solution in back of book.

4 Solution in back of book.

29

VII.2

Black Holes and the Causal Structure of Spacetime

1 The flat space metric in spherical coordinates is ds2 = dr2 + r2 dφ2 + dz 2 We wish to find a curve z = f (r) so that the induced metric on the curve reproduces the metric of a constant-time slice of Schwarzschild, (12). Thus dz = f (r) dr, and the induced metric becomes ds2 = (1 + f (r)2 )dr2 + r2 dφ2 For this to agree with a constant-time slice of Schwarzschild, we require 1 + f (r)2 = 1/(1 − rS /r), so we have

rS ⇒ f (r) = 2 rS (r − rS ) f (r) = r − rS

VII.3

Hawking Radiation

1 Solution in back of book.

VII.4

Relativistic Stellar Interiors

1 Plugging M(r) = (4π/3)r3 ρ = (r/R)3 M into (13), we see that the expression in the

3 second parentheses on the right hand side becomes 1+ 4πRMP (r) , so that the equation simplifies to dP rdr GMρ

=− 3 4πR3 P r2 P R 1+ ρ 1+ M 1 − 2GM R3 which can now be integrated to give the stated result for the pressure profile.

2 Look at Tolman’s paper.

VII.5

Rotating Black Holes

1 Solution in back of book.

2 Solution in back of book.

3 This is totally straightforward. Since ρ2 = r2 + O(a2 ) and Δ = r2 + O(a2 ), we have ds2Kerr = ds2Schwarzschild − (2rS a sin2 θ/r)dtdϕ + O(a2 ).

4 This is also straightforward. We read off from the Kerr metric that Σ2 = ρ2 (r2 + a2 ) + rS a2 r sin2 θ = (r2 + a2 − a2 sin2 θ)(r2 + a2 ) + rS a2 r sin2 θ = (r2 + a2 )2 − Δa2 sin2 θ.

Substituting, we obtained the stated form.

5 Expand out (23) and compare with the result of the previous exercise.

30

6 We find that ω(r, θ) =

rS ar 2 2 2 (r + a cos θ)(r2 + a2 ) + rS a2 r sin2 θ

For large r, ω → rS a/r3 independent of θ, while for small r, ω → rS r/(a3 cos2 θ), which → 0 except for θ = π/2, that is, in the equatorial plane. Thus, for θ = π/2, ω(r, θ) attains a maximum at a value of r determined by solving a quadratic equation.

You could verify that ω(r, θ) does not depend on θ for r satisfying r 2 + a2 − rS r = 0, that is, on the outer and inner horizons. When plotted as a function of r, the curves ω(r, θ) for various values of θ intersect for r on the outer and inner horizons. We plot ω(r, π/2), ω(r, π/4), ω(r, 0) in two figures, VII.5.6.a for rS = 3, a = 1, and VII.5.6.b for rS = 3, a = 3/2 (an extremal case). I intentionally left the curves unlabeled. ω 1.0 0.8 0.6 0.4 0.2

1

2

3

4

r

3

4

r

Figure VII.5.6.a ω 0.6 0.5 0.4 0.3 0.2 0.1 1

2

Figure VII.5.6.b 31

7 We have from the text that

Ω± = ω ±

gtt

gϕϕ

ω2 +

where ω = −gtϕ /gϕϕ . At θ = π/2, we have 1 Δ − a2 rS =1− =1− 2 r r x rS g arS −gtϕ = = r x

1 rS (r2 + a2 )2 − Δa2 2 2 2 2 2 = rS x + g 1 + =r +a 1+ gϕϕ = x r2 r −gtt =

Then it follows that rS Ω± = rs

−gtϕ ±

2 +g gtϕ ϕϕ |gtt |

gϕϕ

g/x ± g 2 /x2 + (x2 + g 2 (1 + 1/x))(1 − 1/x) = x2 + g 2 (1 + 1/x)

g ± g 2 + (x3 + g 2 (x + 1))(x − 1) = x3 + g 2 (x + 1)

g ± x g 2 + x(x − 1) = x3 + g 2 (x + 1)

8 By now, you should find this straightforward to work out, following the general treatment first given in chapter V.4. Allow me to point out though, that you could save considerable labor by realizing that you can simply make some minor adjustments to the same problem worked out for a massive particle in appendix 3.

So, start from (46). We have the same equation except that we should now interpret dt and ζ̇ = dϕ Newton’s dot notation as ṫ = dζ dτ , with ζ the affine parameter. Now, after setting θ̇ = 0, we have gtt ṫ2 + 2gtϕ ṫϕ̇ + gϕϕ ϕ̇2 + grr ṙ2 = −x, with the parameter x equal to 0 rather than 1. Thus, with the proper understanding of what Newton’s dot means, we can now treat the massive and the massless cases at the same time. After a few arithmetic steps, we arrive at an equation of the form 2 ṙ2 + · · · = −x/grr = −x((r2 + a2 − rrS )/r2 ) = −x(1 + ar2 − rrS ).

But we already did the arithmetic in appendix 3! All we have to do now is to replace the 0 on the right hand side of (47) by the round parentheses in the preceding equation.

Thus, we have

l2 + a2 1 − 2 rS rS (l − a )2 a2 rS 2 2 + − −

+ 1 = 1 + − ṙ − 2 3 2 r r r r r We see that several terms cancel between the left and the right hand sides, and we end up with ṙ2 + r12 {l2 − a2 2 − (l − a )2 rrS } = 2 . As was explained in chapter VI.3, the 32

freedom to scale ζ means that the physics can only depend on the impact parameter b ≡ |l|/ , and not on l and separately. Dividing by l2 (which we can absorb into ṙ2 ), we obtain the stated result.

Note that b is defined to be positive. Taking r → ∞ in (4) and (5) in the text (with τ replaced by ζ), we see that > 0, while l can take on either sign. This explains the appearance of sign(l) in the equation of motion.

9 To transform Kerr I to Boyer-Lindquist, set du = dT + dr + rS rdr/(r2 + a2 − rS r) and dϕ = −dΦ − adr/(r2 + a2 − rS r). After some arithmetic, we recover the BL form (15) of the Kerr metric (with t renamed T , and ϕ renamed Φ to avoid obvious confusion.)

To transform Kerr II to Kerr I, set t = u − r, x = sin θ(a sin ϕ + r cos ϕ), y = sin θ(r sin ϕ − a cos ϕ), z = r cos θ. Again, after some arithmetic, we obtain Kerr I. 10 In contrast to the Boyer-Lindquist form used in the text, the Kerr I form of the metric mentioned in exercise VII.5.9 depends on rS in only one place, and in a very simple way. Indeed, we have ds2

=

−(du + a sin2 θdϕ)2 + 2(du + a sin2 θdϕ)(dr + a sin2 θdϕ) rrS +ρ2 (dθ2 + sin2 θdϕ2 ) + 2 (du + a sin2 θdϕ)2 ρ

For a given a, the M dependence is isolated in the fourth term. Write du + a sin2 θdϕ as lμ dxμ , which defines the vector lμ . As we will soon see, it is advantageous to depart from tradition and order the coordinates as xμ = (u, r, ϕ, θ). With this convention, we have lμ = (1, 0, a sin2 θ, 0). The first three terms define ⎞ ⎛ −1 1 0 0 ⎟ ⎜ 1 0 0 a sin2 θ 0 ⎟ gμν =⎜ 2 2 ⎠ ⎝ 0 a sin θ (r + a2 ) sin2 θ 0 0 0 0 r2 + a2 cos2 θ With our unusual ordering of coordinates, the 4 × 4 block that is the metric breaks up into a 3 × 3 block and a 1 × 1 block.

0 Next, we have to invert gμν = gμν + (rrS /ρ2 )lμ lν to obtain lμ = g μν lν . But we can save ourselves some arithmetical labor by noticing that lμ = (1, 0, a sin2 θ, 0) is just the 0 ν 0 l , and thus we see that lμ = (0, 1, 0, 0). This solves lμ = gμν lν = gμν second row of gμν ν μ since manifestly lν l = 0. (Now that we have found l by eyeball so to speak, we can easily argue that the solution is unique, since gμν is not singular at a generic point in spacetime.)

0 μ ν Now we obtain immediately gμν lμ lν = 0 and gμν l l = 0, as desired.

11 The calculation is straightforward but tedious and is perhaps best done by a computer.

Satisfyingly, the Riemann curvature tensor turns out to vanish. For the reader who wishes to do a few terms by hand, here are a couple of Christoffel symbols to provide a check: Γuθθ = −2r r2 + a2 /(2r2 + a2 + a2 cos 2θ) and Γrrϕ = −2ar sin2 θ/(2r2 + a2 + a2 cos 2θ).

33

VII.6

Charged Black Holes

1 The Reissner-Nordström metric is ds2 = −A(r) dt2 +

1 dr2 + r2 dΩ2 A(r)

where A(r) = (r −r+ )(r −r− )/r2 . A photon moving in a radial direction has 4-velocity uμ = (ut , ur , 0, 0), and normalization requires 2 uμ uμ = −A(r) ut +

1 2 (ur ) = 0 A(r)

Thus, since dt/dr = ut /ur , we obtain

dt dr

2 =

1 1 dt =± ⇒ A(r)2 dr A(r)

Integrating, we get

±(t − t0 ) =

dr A(r)

r − r+ r − r− 1 2 2 r ln − r− ln = r − r0 + r+ − r− + r0 − r+ r0 − r− where we introduced constants of integration so that the photon is at r = r0 at time t = t0 .

2 Setting dt = dt̄ − (A−1 − 1)dr, the metric becomes

ds2 = −A

dt̄ −

2 1 1 − 1 dr + dr2 + r2 dΩ2 A A

= −A dt̄2 + 2(1 − A) dt̄ dr + (2 − A) dr2 + r2 dΩ2 → −B, B → 3 Maxwell’s equations are invariant under the duality transformation E μ μ μ μ μ μ

E, Je → −Jm , Jm → Je , where Je and Jm are the electric and magnetic 4-currents, respectively. We can therefore take the Reissner-Norström solution and replace the electric charge Qe with a magnetic charge −Qm to obtain the solutions for a black hole with magnetic charge. We get the same metric again: ds2 = −

(r − r+ )(r − r− ) 2 r2 dr2 + r2 dΩ2 dt + 2 (r − r+ )(r − r− ) r

except that now, r± is given by r± = M ±

M 2 − Q2m

In general, for a solution containing both electric and magnetic charge, we would replace Q2 → Q2e + Q2m .

34

VIII

Introduction to Our Universe

VIII.1

The Dynamic Universe

1 Solution in back of book.

2

T 2 The corresponding equation is now Ṙ2 = 4R 2 +1, with the solution R(t) = The small t behavior is the same whether the universe is closed or open.

t(T + t).

3 Solution in back of book.

4 Set up the problem as in the preceding exercise. In this case, the cosmological equation reads Ṙ2 = 2T R + 1. The solution is given parametrically by R(η) = T (cosh η − 1) and t(η) = T (sinh η − η). Again, check the small t behavior of R(t).

5 Simply solve (26) and (27). It is instructive to plot R(t) for the closed, flat, and open universes and compare. For large t, the curvature term does not matter.

VIII.2

Cosmic Struggle between Dark Matter and Dark Energy

1 Solution in back of book.

2 Solution in back of book.

3 Solution in back of book.

4 Solution in back of book.

5 Solution in back of book.

IX

Aspects of Gravity

IX.1

Parallel Transport

1 Solution in back of book.

2 Let the circle be described by (x, y) = (L + ε cos η, ε sin η). Then r = L + ε cos η and θ = arctan(y/x) ε sin η/L. We find

L+ε

π dη sin2 η = πε2 /L

2

dr θ = 2

x2 + y 2

drθ = (2ε /L) L−ε

0

As a check, dθ r = πε2 /L. This example shows that unless the coordinates all have dimensions of length, we cannot expect aσλ to have dimensions of length squared, of course.

3 Solution in back of book.

35

IX.3

Geodesic Deviation

1 On the sphere, we know that geodesics follow great circles. Consider two nearby great circles of constant longitude; we can parametrize the geodesics following these circles as xμ (τ ) = (τ, 0), so that the velocity of each geodesic is uμ = dxμ /dτ = (1, 0), which is properly normalized. We take τ = 0 at the pole, so τ = π/2 at the equator. If the geodesics are separated by a latitude of ϕ0 at the equator, we can then write the separation vector as εμ = (0, ϕ0 sin τ ). Finally, the only nonzero Christoffel symbols are Γθϕϕ = − cos θ sin θ, Γϕ θϕ = cot θ, and the only independent component of the Riemann tensor is Rθϕθϕ = sin2 θ. We are now ready to verify (6). First, note that Rϕ σρλ uσ uρ ελ = −ϕ0 sin τ

Rθ σρλ uσ uρ ελ = 0, Next, note that D 2 εθ = 0, Dτ 2

D 2 εϕ = −ϕ0 sin τ Dτ 2

thus verifying (6) for the sphere.

2 Solution in back of book.

2 μ

σ

ρ

dx λ 3 It’s just a matter of repackaging (6) DDτε2 = Rμσρλ dx dτ dτ ε and evaluating the left hand side as in (34) and (35).

4 Solution in back of book.

IX.4 Linearized Gravity, Gravitational Waves, and the Angular Momentum of Rotating Bodies 1 Solution in back of book.

3 2 As indicated in the hint, d y T ij (t, y ) = − d3 y y j ∂k T ik = d 3 y y j ∂0 T i0 =

d d d3 y y j T i0 = ( 12 dt d3 y (y j T i0 + y j T i0 )). Repeat the same trick: d3 y (y j T i0 + dt

3 i j 00 d y j T i0 ) = − d3 y y i y j ∂k T 0k = ( dt d y y y T ). We obtain the stated result.

3 Solution in back of book.

IX.6 Isometry, Killing Vector Fields, and Maximally Symmetric Spaces 1 The first equation in (5) gives ξ θ = F (ϕ), the second then gives ξ ϕ = − cot θF (ϕ) + g(θ), so that the third gives F (ϕ) + F (ϕ) + g (θ) = 0. This forces g (θ) to be a constant, and so g(θ) = c + c1 θ, but periodicity implies that c1 = 0. The rest is elementary.

2 Our three Killing vector fields are ξ(1) = sin ϕ ∂θ + cot θ cos ϕ ∂ϕ ,

ξ(2) = cos ϕ ∂θ − cot θ sin ϕ ∂ϕ , 36

ξ(3) = ∂ϕ

On the equator, θ = π/2, we get ξ(1) = sin ϕ ∂θ ,

ξ(2) = cos ϕ ∂θ

so ξ(1) and ξ(2) point “up” and “down” the sphere, depending on the value of ϕ.

Near the north pole of the sphere, where the coordinates fail to be well defined, cot θ = 1/θ + · · · , so we have ξ(1) = sin ϕ ∂θ +

1 cos ϕ ∂ϕ + · · · , θ

ξ(2) = cos ϕ ∂θ −

1 sin ϕ ∂ϕ + · · · θ

and thus ξ(1) and ξ(2) are dominated by a component in the ϕ direction, unless cos ϕ = 0 or sin ϕ = 0.

3 Solution in back of book.

IX.7

Differential Forms and Vielbein

1 Solution in back of book.

2 Solution in back of book.

3 The transformation from spherical to Cartesian coordinates is tan ϕ =

y x

cos θ =

It then follows that dϕ =

z z = 2 r x + y2 + z2

1 (x dy − y dx) x2 + y 2

and d cos θ

= =

dz z 1 − ( 2 ) (x dx + y dy + z dz) r r r dz z 2 (x + y 2 ) 3 − 3 (x dx + y dy) r r

Note that (x dx + y dy)(x dy − y dx) = (x2 + y 2 )dxdy and hence d cos θ dϕ

=

(xdzdy − ydzdx)/r3 − zdxdy/r3

=

−(xdydz + ydzdx + zdxdy)/r3 = − ijk xi dxj dxk /r3

g Thus, F = 4π d cos θ dϕ clearly represents a radial magnetic field. We can see this by writing F in terms of components: ⎛ ⎞ 0 0 0 0 ⎜0 0 z −y ⎟ ⎟ Fμν ∝ ⎜ ⎝0 −z 0 x⎠ 0 y −x 0

37

But the components of the field strength tensor are related to the electric and magnetic fields by ⎛ ⎞ Ey Ez 0 Ex ⎜−Ex 0 −Bz By ⎟ ⎟ Fμν = ⎜ ⎝−Ey Bz 0 −Bx ⎠ −Ez −By Bx 0 Thus for our field strength tensor, the electric field is zero, and components of the magnetic field go like Bx ∝ x, By ∝ y, Bz ∝ z with the same proportionality constant, indicating that the magnetic field is radial.

4 Solution in back of book.

5 Solution in back of book.

6 Solution in back of book.

IX.8

Differential Forms Applied

1 Solution in back of book.

2 Solution in back of book.

3 Solution in back of book.

4 Solution in back of book.

IX.9

Conformal Algebra

1 Solution in back of book.

2 A straightforward calculation gives ds2 = ημν dxμ dxν = (f 8 /(y 2 )5 )(y 2 ημν + 8yλ yρ )dy μ dy ν , which is assuredly not conformally flat.

3 Solution in back of book.

IX.10

De Sitter Spacetime

1 This is just a matter of straightforward calculation.

2 After working through a few metrics, you will realize that it would be easiest to write a computer program to do the actual calculation 3 We begin with ds2 = −(1 − r2 )dt2 + 38

dr2 + r2 dΩ2 1 − r2

Defining dp = dt+dr/(1−r2 ) and dq = −dt+dr/(1−r2 ), we find that dt = (dp−dq)/2 and dr = (1 − r2 )(dp + dq)/2, so that the metric becomes (1 − r2 )2 (dp + dq)2 2 ds2 = − 14 (1 − r2 ) (dp − dq) + + r2 dΩ2 4 1 − r2

= 14 (1 − r2 ) −(dp − dq)2 + (dp + dq)2 + r2 dΩ2 = (1 − r2 ) dp dq + r2 dΩ2 4 This is also just a matter of straightforward calculation.

5 Solution in back of book.

6 Again, this is just a matter of straightforward calculation.

7 de Sitter’s original metric is

ds2 = − cos2 χ dt2 + dχ2 + sin2 χ dΩ2 The 4-velocity of a line of constant χ, θ, ϕ is uμ = (ut , 0, 0, 0), so the geodesic equation becomes 0 = uμ Dμ uν ∂uν + Γνμλ uμ uλ = ∂τ 2 ∂uν + Γνtt ut = ∂τ The ν = t component is satisfied automatically; to satisfy the others, we need Γνtt = 0 for ν = χ, θ, ϕ. We have Γχtt = − 12 g χχ ∂χ gtt = − sin χ cos χ Γθtt = 0 = Γϕ tt So indeed, the geodesic equation can only be satisfied if χ = 0, so that we have Γχtt = 0.

X

Gravity Past, Present, and Future

X.3

Effective Field Theory Approach to Gravity

1 Solution in back of book.

X.4

Finite Sized Objects and Tidal Forces in Einstein Gravity

1 The gravitational field we feel every day is given, to first approximation, by the Schwarzschild metric produced by the earth. For an everyday slowly moving object, with 3-velocity essentially zero, so its 4-velocity is Ẋ μ = (1/ 1 − rS /r, 0, 0, 0), 39

where rS is the Schwarzschild radius of the earth. (Note that gμν Ẋ μ Ẋ ν = −1.) As per the definition of the gravitational electric field given in the text, Eμν = Rμρνσ Ẋ ρ Ẋ σ , we need rS 1 ) Err = Rrtrt (V t )2 = grr Rrtrt /(1 − rS /r) = − 3 ( r 1 − rrS Similarly, we obtain Eθθ = Eϕϕ = − 12 Err . For the gravitational magnetic field Bμν , we need all components of the dual of the curvature tensor with two t indices, which in turn requires all components of the Riemann tensor with only one t index. The gravitational magnetic field is zero (this isn’t so surprising: we expect magnetic fields to be produced by moving charges, but relative to us standing on the surface of the earth, the earth isn’t moving).

X.5

Topological Field Theory

1 We have E = 20 × 3/2 = 30, and so, using Euler’s theorem, V = 2 + E − F = 2 + 30 − 20 = 12. How many triangles meet at each vertex? The number is 3 × 20/V = 60/12 = 5. The angular deficit at each vertex is 2π − 5(π/3) = π/3. The total angular deficit is 12 × π/3 = 4π. Descartes is some kind of genius!

√ √ 2 We have, from exercise I.5.16, g = a(L + a sin θ), and hence d2 x gR = 2π 2π 2 0 dθ 0 dϕ sin θ = 0.

3 Solution in back of book.

X.6

A Brief Introduction to Twistors

1 We have

p · q = (εαβ λα μβ ) εα̇β̇ λ̃α̇ μ̃β̇ ≡ λ, μ [λ, μ]

Note the appearance of the two Lorentz invariants λ, μ and [λ, μ] defined in the text.

2 Solution in back of book.

3 Solution in back of book.

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