Phys 231A Fall 2014Assignment 7 Solutions

Carroll, Chapter 7, Problem 2

(a) In Lorenz gauge, the linearized Einstein’s equations look like (Carroll (7.125))

hµν = −16πTµν

We can recover Maxwell’s equations1 Aµ = −Jµ from this if we make the sub-stitutions h0µ → Aµ as in Carroll’s (7.25), and 16πT0µ → Jµ. In this approximation,we are taking only parts of the stress tensor linear in the velocity (c.f. Wald p.78).We neglect the spacial components Tij, since they are higher than linear order in thevelocity.

Choosing the z axis as the axis of rotation, and using spherical coordinates, the stresstensor has (Cartesian) components

T00 =M

4πR2δ(r −R)

T01 = −MRΩ

4πR2sin θ sinϕ δ(r −R)

T02 =MRΩ

4πR2sin θ cosϕ δ(r −R)

T03 = 0

Now this is just an electrodynamics problem. We have a uniformly charged shellof charge density σ = −4M/R2 (since J0 = −ρ), hence by Gauss’s law ~E = 0 inside

the shell. There are a few approaches2 available to find ~B - the most elementary isjust to find ~A by integrating ~J against the Green’s function for Poisson’s equation:

Ai(x) =1

4π

∫d3x′

Ji(x′)

||x− x′||This is actually done as an example in Griffiths. The result is, inside the shell,

~A =RΩσ

3r sin θϕ =

RΩσ

3(xy − yx)

giving a uniform gravitomagnetic field

~B =2RΩσ

3z = −8MΩ

3Rz

(b) Let uµ be the tangent to the observer’s worldline. Take a purely spatial vectorSµ in the observer’s frame, i.e. a vector such that Sµuµ = 0. In the observer’s frame,

1Note that Carroll’s convention uses Heaviside-Lorentz units, unlike Wald.2c.f. Problem 5.13 of Jackson, or Example 5.11 of Griffiths.

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uµ = (1, 0, 0, 0) so that S0 = 0. Parallel transport of Sµ gives us uσ∇σSµ = 0. To

linear order in velocity, the parallel transport equation becomes

dSi

dt+ Γi0jS

j = 0 .

Using (7.19) of Carroll with h0i = Ai and (v ×B)i = (∂iAj − ∂jAi)vj, we have

dSi

dt= −Γi0jS

j = −1

2(∂jAi − ∂iAj)Sj

= δ[ki δ

l]j ∂kAlS

j

Carroll’s (2.71) gives

dSidt

=1

2(εijmε

mkl∂kAlSj) =

1

2εijm(∇× ~A)mSj

= −1

2((∇× ~A)× ~S)i = −1

2( ~B × ~S)i

=4M

3R(~Ω× ~S)i

We see that ~S changes in the plane of rotation - the shell drags the frame alongwith it as it rotates.

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Carroll, Chapter 7, Problem 4

Our goal is to show that the following gauge conditions are equivalent:

∂µhµν = 0 ⇐⇒ xµ = 0.

Let’s start with the second expression. As the problem suggests, it is beneficial tothink of xµ as scalar functions of spacetime. That is, µ is just a label for the particularscalar function. Their values are those of the coordinate x0, x1, etc. Thus, a definingcharacteristic is

∇νxµ = ∂νx

µ = δµν .

We then compute

xµ = ηαβ∇α∇βxµ

= ηαβ∇αδµβ

= ηαβ(∂αδ

µβ − Γσβαδ

µσ

)= −ηαβΓσβαδ

µσ

= −ηαβΓµβα

= −1

2ηαβηµρ (∂αhβρ + ∂βhαρ − ∂ρhαβ)

= −1

2ηµρ (2∂αhαρ − ∂ρh) .

Let’s now write this in terms of the trace-reversed perturbation hαβ = hαβ − hηαβ/2,

xµ = −1

2ηµρ(2∂αhαρ + ∂ρh− ∂ρh

)= −∂αhαµ.

Thus we see that if xµ = 0 it follows that ∂αhαβ = 0, so our two gauge conditions

are indeed equivalent.

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Carroll, Chapter 7, Problem 6

(a) In the Newtonian theory, we have

Mx = −GM2

(2x)2

Since we want x(t → −∞) = ±∞ and x(0) = 0, we assume a power-law depen-dence x(t) = atp. Plugging in this ansatz, we have

(p− 1)patp−2 = − GM

4a2t2p=⇒ t−2p−(p−2) = −4a3(p− 1)p

GM

which implies we must have p = 2/3. Plugging in this value of p gives a =(9GM/8)1/3.

(b) The Newtonian approximation is no longer reasonable when the objects’ speedsexceed c

x =2

3

(9GM

8t

)1/3

1 =⇒ x GM

2

Then it must be that the separation 2x = r GM . Similarly, the Newtonianlimit also breaks down at separation r = 2GM . Then we must likewise have r GM .

(c) From (7.140) of Carroll, we have

hij =2G

r

d2

dt2Iij(tr)

where tr is the retarded time, and Iij is given by (7.138):

Iij =

∫d3y yiyjT 00(t,y)

The 00 component of the stress tensor is just the energy density, given by

T 00 = M

[δ

(x− 9GMt2

8

)1/3

+ δ

(x+

9GMt2

8

)1/3]δ(y)δ(z)

The integrals of the δ-functions kill off all Iij except for Ixx, which is just

Ixx(tr) = 2M

(9GMt2r

8

)2/3

Now, tr = t −√R2 + (x(t))2, which we can approximate by t − R for large R.

Then the trace-reversed perturbation becomes

hxx =16GM

9r

(9GM

8tr

)2/3

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To express this in transverse-traceless gauge, we use (7.178)

hTTxx =

(P kx P l

x −1

2PxxP

kl

)hkl =

(P xx P x

x −1

2PxxP

xx

)hxx

Here, ni = [0, 1, 0], since the observer looks down at the collision along the y-axis.Then Pxx = 1, and we have

hTTxx =8GM

9R

(9GM

8tr

)2/3

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