Geodesics of the Kerr metric

Therefore, once we have solved (4.97), we have the expressions of
the conjugate momenta (and then of ẋ µ ) in terms of four constants,
which allow to write the solutions of the geodesic equation in a
closed form, in terms of integrals.
In general, it is more difficult to solve (4.97), which is a partial
differential equation, than to solve thegeodesicequation,butinthis
case we can find such a solution.
First of all, we can use what we already know, i.e.
These conditions require that
H = 1 2 gµν p µ p ν = 1 2 κ
p t = −E constant
p φ = L constant . (4.99)
S = − 1 2 κλ − Et + Lφ + S(rθ) (r, θ) (4.100)
where S (rθ) is a function of r and θ to be determined.
Furthermore, we look for a separable solution, by making the
ansatz
S = − 1 2 κλ − Et + Lφ + S(r) (r)+S (θ) (θ) . (4.101)
Substituting (4.101) into the Hamilton-Jacobi equation (4.97), and
using the expression (3.16) for the inverse metric, we find
−κ + ∆ ( ) dS
(r) 2
+ 1 ( ) dS
(θ) 2
Σ dr Σ dθ
− 1 ]
[r 2 + a 2 + 2Mra2
∆
Σ
sin2 θ E 2 + 4Mra
Σ∆ EL + ∆ − a2 sin 2 θ
Σ∆ sin 2 θ L2 =0.
(4.102)
Using the relation (3.12)
(r 2 + a 2 )+ 2Mra2
Σ
sin2 θ = 1 Σ
and multiplying by Σ = r 2 + a 2 cos 2 θ,weget
( ) dS
−κ(r 2 + a 2 cos 2 (r) 2 ( dS
(θ)
θ)+∆ +
dr dθ
[
(r 2 + a 2 ) 2 − a 2 sin 2 θ∆ ] (4.103)
86
) 2