6.1.2 Angular momentum perturbations
Treating angular momentum perturbations is a bit more involved. One reason for
this is the fact that it inherently changes the form of the metric. From Equation 6-23, it
is clear that our metric perturbation will acquire an At4 component. Realizing this as a
perturbation towards the K~err spacetime, we will write it as
26aM~ sin2 8
has (6-51)
which is just the linearization about a = J/M~ of the corresponding component of
the (background) K~err metric. Because of this, there will be nonzero contributions to
him, hm and their complex conjugates which means that we must now take parity into
consideration. To that end we will introduce a gauge vector with components
it = P(t, 7) m(0, ) (6-52)
(r= Q(t, r) m,(0, ) (6-53)
1 i
le = [R(t, r)--(a +') +S(t, r) (a a')]em(e, ~)
2 2 sm0
=R(t, r) ~t(0, ~) + S(t, r) (6-54)
sin 8
i sin 0
(4= [R(t, r) (a a') S(t, r)> (a + a')]Nem(8, 4)
2 2
=R(t, r)Y m(0 ) Slt, r) sin 0@(0,4) (6-55)
where we've defined Q' =i (8T + T')L = (1m+-1m nd T- = ~(n 8') m =
~(1Ye -1 m,), where 1%m are the spin-weight +1 spherical harmonics discussed in
Appendix D. This form of the gauge vector was obtained by considering (a = (ma + (als
(mh (mm, and making use of the parity decomposition discussed in ChI Ilpter 3. This
makes it easy to see that P, Q and R represent the even-parity degrees of gauge freedom
and S represents the only odd-parity gauge freedom available. A natural question to ask is
what parity the perturbation in Equation 6-51 has. For an answer, we look to the source
terms. A quick computation reveals that m, = Im = m = -Im, from which it follows