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

where we have used the notation < y 2 >= ∫ y ′2 d 3 y ′ , to denote the average of the square of the y-coordinate
on the sphere and that ˆr = (cos φ sin θ, sin φ sin θ, cos θ). We have also used symmetry to deduce that the x ′ y ′
and z ′ y ′ terms in the integrand will not contribute. Similarly we deduce
We also know that because T 03 vanishes, h 03 = 0.
h 02 (r, θ, φ) = − 4Gρω sin θ cos φ < x2 >
r 2
Now let’s translate back to the metric perturbations, which gives us the following equations
h µµ = h µµ − h
h 01 = h 01
h 02 = h 02
h 03 = h 03
h = −h 00 + h 11 + h 22 + h 33
We can see readily that
h 01 = 4Gρω sin θ sin φ < y2 >
r 2
h 02 = − 4Gρω sin θ cos φ < x2 >
r 2
h 03 = 0
Because h ii = 0 for i = 1, 2, 3 we have that h ii = 1 2 h (which implies all hµµ are equal) so we can write the
trace entirely in terms of h 00 h = 2h 00
and we have h 00 = h 00 − 1 2 (2h00 ) which implies
h µµ = 1 2 h00 = 2GM
r
Now let’s transform from Cartesian to spherical coordinates using the transformation law for (0, 2)-tensors.
h 0r = ∂x
∂r h 01 + ∂y
∂r h 02 = (cos φ sin θ) ( −4Gρω sin θ sin φ < y 2 >
r 2 )
+ (sin φ sin θ)
(4Gρω sin θ cos φ < x 2 >
r 2 )
= 0
where we’ve used that by the spherical symmetry of the source distribution < x 2 >=< y 2 >.
h 0φ = ∂x
∂φ h 01 + ∂y
∂φ h 02
h 0φ = (−r sin φ sin θ) ( −4Gρω sin θ sin φ < y 2 > ) (4Gρω sin θ cos φ < x 2 > )
+ (r cos φ sin θ)
r 2 r 2
h 0φ = 4Gρω sin2 θ(< x 2 > + < y 2 >)
r
While they are equal we have kept the distinction between < x 2 > and < y 2 > so that we may make an
observation. In particular the combination ρ(< x 2 > + < y 2 >) = ∫ ρ(x ′2 + y ′2 )d 3 r ′ is equal to the moment
of inertia about the z-axis of the sphere (for a uniform sphere this is I = 2 5 MR2 = 8π 5 ρR5 ). We may then
rewrite this component as
h 0φ = 4G sin2 θL
r
where L = Iω is the angular momentum of the source matter.
Let’s focus on the remaining non-zero components of the metric,
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