3+1 formalism and bases of numerical relativity - LUTh ...

134 The initial data problem
Figure 8.2: Same hypersurface Σ0 as in Fig. 8.1 but displayed via an embedding diagram based on the metric
γ instead of ˜γ. The unit normal of the inner boundary S with respect to that metric is s. Notice that D · s = 0,
which means that S is a minimal surface of (Σ0, γ).
since ˜γ(˜s, ˜s) = Ψ −4 ˜γ(s,s) = γ(s,s) = 1. Thus Eq. (8.55) becomes
Di(Ψ 4 ˜s i ) = S 1 ∂
√
f ∂xi
4 i
fΨ ˜s S = 0. (8.57)
Let us introduce on Σ0 a coordinate system of spherical type, (x i ) = (r,θ,ϕ), such that (i)
fij = diag(1,r 2 ,r 2 sin 2 θ) and (ii) S is the sphere r = a, where a is some positive constant. Since
in these coordinates √ f = r 2 sin θ and ˜s i = (1,0,0), the minimal surface condition (8.57) is
written as
i.e. ∂Ψ
∂r
1
r2 ∂ 4 2
Ψ r
∂r
r=a = 0, (8.58)
Ψ r=a
+ = 0 (8.59)
2r
This is a boundary condition of mixed Newmann/Dirichlet type for Ψ. The unique solution of
the Laplace equation (8.53) which satisfies boundary conditions (8.43) and (8.59) is
Ψ = 1 + a
. (8.60)
r
The parameter a is then easily related to the ADM mass m of the hypersurface Σ0. Indeed using
formula (7.66), m is evaluated as
m = − 1
2π lim
∂Ψ
r→∞ ∂r r2 sin θ dθ dϕ = − 1
2π lim
∂
4πr2 1 +
r→∞ ∂r
a
= 2a. (8.61)
r
r=const
Hence a = m/2 and we may write
Ψ = 1 + m
2r
. (8.62)