2.4 Stationary solutions

When there is sufficient symmetry present, it is possible to construct initial data that are in true equilibrium. These solutions possess at least two Killing vectors, one that is timelike at large distances and one that is spatial, representing an azimuthal symmetry. When these symmetries are present, solving for the initial data produces a global solution of Einstein’s equations and the solution is said to be stationary. The familiar Kerr–Neumann solution for rotating black holes is an example of a stationary solution in vacuum. Stationary configurations supported by matter are also possible, but the matter sources must also satisfy the Killing symmetries, in which case the matter is said to be in hydrostatic equilibrium [8Jump To The Next Citation Point].

The basic approach for finding stationary solutions begins by simplifying the metric to take into account the symmetries. Many different forms have been used for the metric (cf. Refs. [8922Jump To The Next Citation Point34Jump To The Next Citation Point63Jump To The Next Citation Point21Jump To The Next Citation Point]). I will use a decomposition that makes comparison with the previous decompositions straightforward. First, define the interval as

[ ] ds2 = − ψ− 4dt2 + ψ4 A2 (dr2 + r2d𝜃2) + B2r2 sin2 𝜃(dϕ + βϕdt)2 . (53 )
This form of the metric can describe any stationary spacetime. Notice that the lapse is related to the conformal factor by
α = ψ− 2, (54 )
and that the shift vector has only one component
βi = (0,0,β ϕ). (55 )
I have used the usual conformal decomposition of the 3-metric (16View Equation) and have written the conformal 3-metric with two parameters as
( ) A2 0 0 &tidle;γij = ( 0 A2r2 0 ) . (56 ) 0 0 B2r2 sin2 𝜃
The four functions ψ, β ϕ, A, and B are functions of r and 𝜃 only.

The equations necessary to solve for these four functions are derived from the constraint equations (14View Equation) and (15View Equation), and the evolution equations (12View Equation) and (13View Equation). For the evolution equations, we use the fact that ∂tγij = 0 and ∂tKij = 0. The metric evolution equation (13View Equation) defines the extrinsic curvature in terms of derivatives of the shift

1-- ¯ ¯ Kij ≡ 2α(∇i βj + ∇j βi). (57 )
With the given metric and shift, we find that K = 0 and the divergence of the shift also vanishes. This means we can write the tracefree part of the extrinsic curvature as
ij − 10 ij − 2 ij A = ψ A&tidle; = 12ψ (𝕃&tidle;β) . (58 )
We find that the Hamiltonian and momentum constraints take on the forms given by the conformal thin-sandwich decomposition (51View Equation) with ij &tidle;u = K ≡ 0 and − 8 α&tidle; ≡ ψ. Only one of the momentum constraint equations is non-trivial, and we find that the constraints yield elliptic equations for ψ and β ϕ. What remains unspecified as yet are A and B (i.e., the conformal 3-metric).

The conformal 3-metric is determined by the evolution equations for the traceless part of the extrinsic curvature. Of these five equations, one can be written as an elliptic equation for B, and two yield complementary equations that can each be solved by quadrature for A. The remaining equations are redundant as a result of the Bianchi identities.

Of course, the clean separation of the equations I have suggested above is an illusion. All four equations must be solved simultaneously, and clever combinations of the four metric quantities can greatly simplify the task of solving the system of equations. This accounts for the numerous different systems used for solving for stationary solutions.

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