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Instructor Solutions Manual: Modern General Relativity Mike Guidry This document gives the solutions for all problems at the ends of chapters for the first edition of Modern General Relativity: Black Holes, Gravitational Waves, and Cosmology by Mike Guidry (Cambridge University Press, 2019). Unless otherwise indicated, literature references, equation numbers, figure references, table references, and section numbers refer to the print version of that book.

Introduction 1.1 From Eq. (1.2), the value of γ is infinite if v = c, so there is no Lorentz transformation to an inertial frame corresponding to a rest frame for light. 1.2 Since E = mγ , for a 7 TeV proton,

γ= Then from the definition of γ ,

E 7 × 1012 eV = = 7460. m 938.3 × 106 eV

v = c

1−

1 = 0.999999991. γ2

This is a speed that is only about 3 meters per second less than that of light. 1.3 This question is ambiguous, since it does not specify whether the curvature is that of the surface itself (which is called intrinsic curvature) or whether it is the apparent curvature of the surface seen embedded in a higher-dimensional euclidean space (which is called the extrinsic curvature). In general relativity the curvature of interest is usually intrinsic curvature. Then the sheet of paper can be laid out flat and is not curved, the cylinder is also flat, with no intrinsic curvature, because one can imagine cutting it longitudinally and rolling it out into a flat surface, but the sphere has finite intrinsic curvature because it cannot be cut and rolled out flat without distortion. The reason that the cylinder seems to be curved is because the 2D surface is being viewed embedded in 3D space, which gives a non-zero extrinsic curvature, but if attention is confined only to the 2D surface it has no intrinsic curvature. This is a rather qualitative discussion but in later chapters methods will be developed to quantify the amount of intrinsic curvature for a surface.

Coordinate Systems and Transformations 2.1 Utilizing Eq. (2.31) to integrate around the circumference of the circle, s 2 I I Z +R dy 2 2 1/2 dx 1 + C = ds = (dx + dy ) = 2 , dx −R subject to the constraint R2 = x2 + y2 , where the factor of two and the limits are because x ranges from −R to +R over half a circle. The constraints yield dy/dx = −(R2 − x2 )−1/2 x, which permits the integral to be written as s Z R R2 . dx C=2 R 2 − x2 −R Introducing a new integration variable a through a ≡ x/R then gives C = 2R

Z +1

da √ = 2π R, −1 1 − a2

since the integral is sin−1 a. In plane polar coordinates the line element is given by Eq. (2.32) and proceeding as above the circumference is I

(dr2 + r2 d ϕ 2 )1/2 s 2 Z 2π Z 2π dr =R = d ϕ r2 + d ϕ = 2π R, dϕ 0 0

ds =

where r = R has been used, implying that dr/d ϕ = 0. 2.2 Under a Galilean transformation x ′ = x − vt and t ′ = t it is clear that the acceleration a and the separation vector r = ∆xx between two masses are unchanged. Thus the second law F = maa and the gravitational law F = Gm1 m2 r̂/r2 are invariant under Galilean transformations. 2.3 Our solution follows Example 1.2.1 of Foster and Nightingale [88]. The tangent and dual basis vectors, and the products for gi j = g ji = e i ·ee j , were worked out in Example 2.3. The elements for gi j = g ji = e i ·ee j can be obtained in a similar fashion. For example, g12 = g21 = ( 21 i + 12 j )·( 12 i − 12 j ) = 41 − 14 = 0,

where the orthonormality of the cartesian basis vectors has been used. Summarizing the results,  2    1 4v + 2 4uv 2v 0 −v 2 1  gi j =  4uv −u 4u2 + 2 2u gi j =  0 2

−v −u 2u2 + 2v2 + 1

Coordinate Systems and Transformations

θ x

Fig. 2.1

Figure for Problem 2.5.

By direct multiplication the product of these two matrices is the unit matrix, verifying Eq. (2.26) explicitly for this case. Utilizing Eq. (2.29), the line element is ds2 = gi j dui du j = guu du2 + 2guvdudv + 2guwdudw + gvvdv2 + 2gvw dvdw + gwwdw2 = (4v2 + 2)du2 + 8uvdudv + 4vdudw + (4u2 + 2)dv2 + 4udvdw + dw2 where gi j = g ji has been used and no summation is implied by repeated indices. 2.4 Using the spherical coordinates u1 = r

u2 = θ

u3 = ϕ

defined through Eq. (2.2) and the results of Example 2.2, e1 ·ee1 = 1

e2 ·ee2 = r2

e3 ·ee3 = r2 sin2 θ ,

while all non-diagonal components vanish. Thus the metric tensor is   1 0 0 . gi j =  0 r 2 0 2 2 0 0 r sin θ

The corresponding line element is

ds2 = dr2 + r2 d θ 2 + r2 sin2 θ d ϕ 2 , where Eq. (2.29) has been used. 2.5 This solution is based on Problem 1.2 in Ref. [88]. From the parameterization r = xii + y j + zkk with x = (a + b cos ϕ ) cos θ

y = (a + b cos ϕ ) sin θ

z = b sin ϕ ,

where the radius of the doughnut a and radius of the circle b are defined in Fig. 2.1 [this document], the tangent basis vectors are

Coordinate Systems and Transformations

∂r = − sin θ (a + b cos ϕ ) i + (a + b cos ϕ ) cos θ j ∂θ ∂r eϕ = = −(b sin ϕ cos θ ) i − (b sin ϕ sin θ ) j + (b cos ϕ ) k . ∂ϕ eθ =

The corresponding elements of the metric tensor gi j = e i · e j are gϕϕ = b2

gϕθ = gθ ϕ = 0

gθ θ = (a + b cos ϕ )2 .

2.6 The tangent basis vectors and metric tensor gi j were given in Example 2.4. Since gi j is the matrix inverse of gi j , which is diagonal, 1 0 1 0 ij gi j = −→ g = 0 r2 0 1/r2 Then the dual basis may be obtained by raising indices with the metric tensor: e i = gi j e j , giving 1 e 1 = g11 e 1 + g12e 2 = e 1 e 2 = g21 e 1 + g22e 2 = 2 e 2 r for the elements of the dual basis. 2.7 For a constant displacement d in the x direction x′ = x − d

y′ = y

z′ = z.

dy′ = dy

dz′ = dz

Since d is constant dx′ = dx

and therefore ds′ 2 = ds2 . From Eq. (2.41), a rotation in the x − y plane may be written x′ = x cos θ + y sin θ

y′ = −x sin θ + y cos θ

z′ = z,

which gives the transformed line element 2

ds′ = (dx′ )2 + (dy′ )2 + (dz′ )2 = (cos θ dx + sin θ dy)2 + (− sin θ dx + cos θ dy)2 + dz2 = (cos2 θ + sin2 θ )dx2 + (cos2 θ + sin2 θ )dy2 + dz2 = dx2 + dy2 + dz2 = ds2 . Therefore the euclidean spatial line element is invariant under displacements by a constant amount and under rotations. 2.8 Taking the scalar products using Eqs. (2.8), (2.9), and (2.20) gives V = e i ·(V j e j ) = V j e i ·ee j = V j δ ji = V i , e i ·V

ei ·V V = ei ·(V j e j ) = V j ei ·ee j = V j δij = Vi , which is Eq. (2.22).

Coordinate Systems and Transformations

2.9 Utilizing that the angle θ between the basis vectors is determined by cos θ = e 1 · e 2 /|ee1 ||ee2 |, the area of the parallelogram is dA = |ee1 ||ee2 | sin θ dx1 dx2

= |ee1 ||ee2 |(1 − cos2 θ )1/2 dx1 dx2 1/2 1 2 dx dx . = |ee1 |2 |ee2 |2 − (ee1 ·ee2 )2

The components of the metric tensor gi j are e 1 ·ee2 = g12 = g21

|ee1 ||ee1 | = e 1 ·ee1 = g11

|ee2 ||ee2 | = e 2 ·ee2 = g22 ,

so the area of the parallelogram may be expressed as 1/2 1 2 p dx dx = det g dx1 dx2 , dA = g11 g22 − g212

where det g is the determinant of the metric tensor. This is the 2D version of the invariant 4D volume element given in Eq. (3.48).

Tensors and Covariance 3.1 For the three cases

∂ x′ µ ∂ x′ ν α β ∂ x′ µ ∂ x′ ν αβ V V = T ∂ xα ∂ xβ ∂ xα ∂ xβ ∂ xα ∂ xβ ∂ xα ∂ xβ ′ Tµν = Vµ′ Vν′ = ′ µ ′ ν Vα Vβ = ′ µ ′ ν Tαβ ∂x ∂x ∂x ∂x ∂ xα ∂ x′ ν ∂ xα ∂ x′ ν β ′ν ′ ′ν β Tµ = Vµ V = ′ µ V V = T . α ∂ x ∂ xβ ∂ x′ µ ∂ xβ α

T′

µν

= V′ V′ =

3.2 From Eqs. (3.50) and (3.51) with indices suitably relabeled

∂ xβ ∂ xα ∂ 2 xα + A α ν µ ∂ x′ ∂ x′ ∂ x′ ν ∂ x′ µ ! ′λ α β ∂ 2 xα ∂ x′ λ ∂ xγ κ ∂x ∂x ∂x − Γαβ ′ µ ′ ν + ′ µ ′ν Aγ ∂ x ∂ x ∂ xκ ∂ x ∂ x ∂ xα ∂ x′ λ

A′µ ,ν − Γ′ µν A′ λ = Aα ,β

∂ xβ ∂ xα ∂ 2 xα + A α ∂ x′ ν ∂ x′ µ ∂ x′ ν ∂ x′ µ λ ′ α β ∂ x ∂ x ∂ x ∂ xγ ∂ 2 xα ∂ x′ λ ∂ xγ − Γκαβ ′ µ ′ ν A − Aγ γ ∂ x ∂ x ∂ xκ ∂ x′ λ ∂ x′ µ ∂ x′ ν ∂ xα ∂ x′ λ ∂ xβ ∂ xα ∂ 2 xα = Aα ,β ′ ν ′ µ + Aα ′ ν ′ µ ∂x ∂x ∂x ∂x α ∂ xβ ∂ x ∂ 2 xα − Γκαβ ′ µ ′ ν Aκ − Aα ′ ν ′ µ ∂x ∂x ∂x ∂x β α α ∂x ∂x ∂ x ∂ xβ = Aα ,β ′ ν ′ µ − Γκαβ ′ µ ′ ν Aκ ∂x ∂x ∂x ∂x ∂ xα ∂ xβ = Aα ,β − Γκαβ Aκ , ∂ x′ µ ∂ x′ ν = Aα ,β

which is Eq. (3.52).

3.3 (a) Since δµν is a rank-2 tensor with the same components in all coordinate systems (see Section 3.8), under a coordinate transformation gµα gαν = δµν becomes g′µα g′αν = δµν . Since gµν is a tensor, if we assume g µν is also a tensor then g′µα =

∂ xκ ∂ xη gκη . ∂ x ′ µ ∂ x ′α

g′αν =

∂ x′α ∂ x′ν ρσ g . ∂ xρ ∂ xσ

Then evaluating g′µα g′αν , g′µα 6

∂ x′α ∂ x′ν ρσ ∂ xκ ∂ xη ∂ x′α ∂ x′ν ρσ ∂ x σ ∂ x ′ν g = g g = = δµν , κη ∂ xρ ∂ xσ ∂ x ′ µ ∂ x ′α ∂ xρ ∂ xσ ∂ x′ µ ∂ xσ

Tensors and Covariance

where we have used

∂ xη ∂ x′α = δρη ∂ x′α ∂ xρ

gκρ gρσ = δκσ .

Comparing the result g′µα

∂ x′α ∂ x′ν ρσ g = δµν ∂ xρ ∂ xσ

with g′µα g′αν = δµν requires that

∂ x′α ∂ x′ν ρσ g . ∂ xρ ∂ xσ which is the transformation law for a rank-2 contravariant tensor. Note that this result is an example of the quotient theorem described in Problem 3.13. Since gµα gαν = δµν and gµν and δµν are known to be tensors, g µν must also be a tensor. g′αν =

(b) From Eq. (3.44) an arbitrary rank-2 tensor can be decomposed into a symmetric and antisymmetric part, gµν = 21 (g µν + gν µ ) + 21 (g µν − gν µ ). Inserting this in the line element gives ds2 = g µν dxµ dxν = 21 (g µν + gν µ )dxµ dxν + 12 (gµν − gν µ )dxµ dxν = gµν + 12 (gν µ − gν µ ) dxµ dxν = gµν dxµ dxν .

Thus only the symmetric part of gµν contributes to the line element. 3.4 Under the transformation x → x′ ,

∂ xα ∂ xβ ∂ x′µ ∂ x′α γδ g T αβ ∂ x ′ν ∂ x ′α ∂ xγ ∂ xσ α ′µ ∂ xα ∂ x′ µ γ ∂x ∂x = gαδ T γδ ′ν = T α ′ν , γ ∂x ∂x ∂ x ∂ xγ where in going from the first line to the second line ′µ

T ν = g′να T ′µα =

∂ xβ ∂ x′α β = δδ ∂ x ′α ∂ x δ has been used. This is a tensor transformation law so it is valid in all frames. Proceeding in similar fashion, ′ ′ Tµν = g′µα gνβ T ′αβ =

∂ xε ∂ xγ Tεγ , ∂ x ′ µ ∂ x ′ν where in the last step

∂ xε ∂ xλ ∂ xγ ∂ xδ ∂ x′α ∂ x′β τθ g g T ελ γδ ∂ x ′ µ ∂ x ′α ∂ x ′ν ∂ x ′β ∂ xτ ∂ xθ

∂ xλ ∂ x′α = δτλ ∂ x ′α ∂ x τ

∂ x δ ∂ x ′β = δθδ ∂ x ′β ∂ x θ

gελ gγδ T λ δ = Tεγ

Tensors and Covariance

have been used. This is a tensor transformation law so it is valid in all frames. 3.5 (a) For example, consider a rank-4 tensor Tβ ′ µνα

Tβ

µνα

. Its transformation law is

∂ x ′ µ ∂ x ′ν ∂ x ′α ∂ x η γδ ε T . ∂ x γ ∂ x δ ∂ x ε ∂ x ′β η

Now set α = β for this tensor (implying a sum on this index). The resulting quantity must have two upper indices by the summation convention, so define it to be T µν : β

T µν ≡ δα Tβ

µνα

= Tα

µνα

Is T µν a tensor? From the preceding equations, its transformation law is T′

µν

′ µνα

≡ Tα

′ µνα

= δα Tβ

∂ xη ∂ x ′ µ ∂ x ′ν ∂ x ′α ∂ x η γδ ε γδ ε Tη = T ε ′ δ β ∂ xγ ∂ xδ ∂ xε ∂ x′α η ∂x ∂x ∂x ∂ x′µ ∂ x′ν η γδ ε ∂ x′µ ∂ x′ν γδ η ∂ x′µ ∂ x′ν γδ = δ T = T = T , ∂ xγ ∂ xδ ε η ∂ xγ ∂ xδ η ∂ xγ ∂ xδ β ∂x

= δα

′ µ ∂ x ′ν ∂ x ′α

∂ xγ

which is the transformation law for a contravariant rank-2 tensor. Similar proofs can be carried out for tensors of any order. Thus, setting an upper and lower index equal on a rank-N tensor and summing yields a tensor of rank N − 2. ν

(b) For example, consider the linear combination of two rank-2 tensors, Tµ = aAµ + ν bBµ . The transformation law is

∂ x ′ν ∂ x β α ∂ x′ν ∂ xβ α A + b B ∂ xα ∂ x′ µ β ∂ xα ∂ x′ µ β ∂ x ′ν ∂ x β ∂ x ′ν ∂ x β α α α aAβ + bBβ = α = α T . µ ′ ∂x ∂x ∂ x ∂ x′ µ β A similar proof holds for any such linear combination of tensors. Tµ′ ν = a A′µν + bB′µν = a

3.6 The line element is ds2 = −dt 2 + dr2 + r2 d θ 2 + r2 sin2 θ d ϕ 2 , so the non-zero components of the metric are g00 = gtt = −1 g11 = grr = 1 g22 = gθ θ = r2

g33 = gϕϕ = r2 sin2 θ

and det g µν = −r4 sin2 θ . Then from Eq. (3.48) the invariant volume element is dV = (−detg µν )1/2 dr d θ d ϕ = r2 dr sin θ d θ d ϕ , which gives a volume V=

dV =

Z R 0

r2 dr

Z π 0

sin θ d θ

Z 2π 0

d ϕ = 34 π R3 ,

as expected. 3.7 Since A · B = Aµ B µ is a scalar it is unchanged by a coordinate transformation. Thus from the vector transformation law for Bµ ∂ xν µ ∂ xν µ µ A′µ B′ = Aµ Bµ = Aν ′ µ B′ −→ Aν ′ µ − A′µ B′ = 0. ∂x ∂x

Tensors and Covariance

But B′ µ is an arbitrary vector that does not generally vanish. Thus the quantity in parenthe′ ν ′µ ses must be equal to zero, implying that Aµ = ∂ x /∂ x Aν , which is the transformation law for a dual vector. 3.8 This problem is adapted from an example in Ref. [88]. From the transformation equations between spherical and cylindrical coordinates assuming u = (r, θ , ϕ ) and u′ = (ρ , ϕ , z), 1

u′ = ρ = r sin θ = u1 sin u2 2

u′ = ϕ = u3 3

u′ = z = r cos θ = u1 cos u2 and the inverse transformations are q q u1 = r = ρ 2 + z2 = (u′ 1 )2 + (u′ 3 )2 ! ′1 −1 ρ −1 u 2 = tan u = θ = tan z u′ 3 2

u3 = ϕ = u′ . From these the partial derivative entries in the matrices U and Û defined in Example 3.7 may be computed directly. For example,

∂ u′ 1 ∂ = 2 (u1 sin u2 ) = u1 cos u2 = r cos θ ∂ u2 ∂u " !# ′1 ∂ u2 ∂ u′ 3 cos θ 2 −1 u . Û1 = tan = = = 1 1 3 1 2 3 2 ′ ′ ′ ′ ′ r ∂u ∂u u (u ) + (u ) U21 =

Computing all the derivatives and assembling them gives    sin θ sin θ r cos θ 0  cos θ   U = 0 Û =  0 1  r cos θ −r sin θ 0 0

0 0 1

 cos θ sin θ  , − r  0

and by explicit matrix multiplication, ÛU = I. 3.9 From Eqs. (3.45) and (3.46), 1 Tαβ (γδ ) − Tβ α (γδ ) 2 1 1 (Tαβ γδ + Tαβ δ γ ) − 12 (Tβ αγδ + Tβ αδ γ ) = 2 2 1 Tαβ γδ + Tαβ δ γ − Tβ αγδ − Tβ αδ γ . = 4

T[αβ ](γδ ) =