General Relativity: Homework 4 Solutions - Department of Physics ...

equations, we are free to choose our coordinates. There are four coordinates, so we can equivalently enforce
four conditions on the solution to our metric. Before we decide what conditions to impose, let’s write the
Einstein tensor in terms of the trace-reversed perturbation h µν = h µν − 1 2 hηµν
G µν = 1 (
∂ µ ∂ λ (h νλ + 1 2
2 hηνλ )+∂ ν ∂ λ (h µλ + 1 2 hηµλ )−∂ λ ∂ λ (h µν + 1 2 hηµν )−∂ µ ∂ ν h+η µν ∂ λ ∂ λ h−η µν ∂ λ ∂ ρ (h λρ + 1 2 hηλρ ) )
which simplifies to
G µν = 1 (
∂ µ ∂ λ h νλ + ∂ ν ∂ λ h µλ − η µν ∂ λ ∂ ρ h λρ − ∂ λ ∂ λ h µν )
2
Now we can decide what conditions we would like to put on the perturbation terms in order to simplify the
calculations. Noting that the four-divergence of the perturbation comes up in three of the four terms, if we
enforced the four equations
∂ λ h µλ = 0
then, after neglecting the time derivatives, we are led to the tidy expression
G µν = − 1 2 ∇2 h µν
and now Einstein’s equation reads
− 1 2 ∇2 h µν = 8πGT µν
The only non-zero terms in the energy-momentum tensor are those with at least one time component. Therefore
we can set h ij = 0 for i, j = 1, 2, 3.
Consider the first equation
Let’s use the general form of the solution to solve
∫
h 00 (r) = 4Gρ
∇ 2 h 00 = −16πGρ
1
|r − r ′ | d3 r ′ ≈ 4Gρ 1 ∫
r
d 3 r ′ = 16πGρR3
3r
= 4GM
r
1
where we have just taken the first order term of the Taylor expansion
|r−r ′ | = 1 r′·ˆr
r
(1 +
r
+ ...) for r >> r ′ .
Now consider
∇ 2 h 01 = −16πGρωy
the solution to which is
∫
h 01 y ′ d 3 r ′
(r, θ, φ) = 4Gρω
|r − r ′ |
1
where the integral is taken over the source matter. Now expand
|r−r ′ |
to the first two orders, this will be
necessary because we will see that the first order term will be zero.
∫
h 01 (r, θ, φ) = 4Gρω
( 1 r + r′ · ˆr
r 2 )y′ d 3 r ′ = 4Gρω
∫ ( 1
r y′ + r′ · ˆr
r 2 y′) d 3 r ′
The source matter has a spherical distribution, and as such the region of integration is symmetric under sign
changes of the coordinates. We may then see that the first term in the integral, being odd in y ′ , will yield no
contribution.
h 01 (r, θ, φ) = 4Gρω 1 ∫
r 2 (r ′ · ˆr)y ′ d 3 r ′ = 4Gρω 1 ∫
r 2 (x ′ cos φ sin θ + y ′ sin φ sin θ + z ′ cos θ)y ′ d 3 r ′
h 01 (r, θ, φ) = 4Gρω sin θ sin φ < y2 >
r 2
6