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

7.3 ADM mass 107
Example : Let us consider Schwarzschild spacetime and use the standard Schwarzschild coordinates
(x α ) = (t,r,θ,φ):
gµνdx µ dx ν
= − 1 − 2m
dt
r
2
+ 1 − 2m
−1 dr
r
2 + r 2 (dθ 2 + sin 2 θdϕ 2 ). (7.16)
Let us take for Σt the hypersurface of constant Schwarzschild coordinate time t. Then we
read on (7.16) the components of the induced metric in the coordinates (x i ) = (r,θ,ϕ):
γij = diag
1 − 2m
−1 , r
r
2 , r 2 sin 2
θ . (7.17)
On the other side, the components of the flat metric in the same coordinates are
fij = diag 1,r 2 ,r 2 sin 2 θ
and f ij = diag 1,r −2 ,r −2 sin −2 θ . (7.18)
Let us now evaluate MADM by means of the integral (7.14) (we cannot use formula (7.15)
because the coordinates (xi ) are not Cartesian-like). It is quite natural to take for St the
sphere r = const in the hypersurface Σt. Then ya = (θ,ϕ), √ q = r2 sin θ and, at spatial
infinity, si√q d2y = r2 sin θ dθ dϕ(∂r) i , where ∂r is the natural basis vector associated the
coordinate r: (∂r) i = (1,0,0). Consequently, Eq. (7.14) becomes
MADM = 1
16π lim
D
r→∞
r=const
j γrj − Dr(f kl
γkl) r 2 sin θ dθ dϕ, (7.19)
with
f kl γkl = γrr + 1
r2γθθ 1
+
r2 sin2 θ γϕϕ
= 1 − 2m
−1 + 2, (7.20)
r
and since f kl γkl is a scalar field,
Dr(f kl γkl) = ∂
∂r (fkl
γkl) = − 1 − 2m
−2 2m
. (7.21)
r r2 There remains to evaluate D j γrj. One has
D j γrj = f jk Dkγrj = Drγrr + 1
r 2 Dθγrθ +
1
r 2 sin 2 θ Dϕγrϕ, (7.22)
with the covariant derivatives given by (taking into account the form (7.17) of γij)
Drγrr = ∂γrr
∂r − 2¯ Γ i rrγir = ∂γrr
∂r − 2¯ Γ r rrγrr
Dθγrθ = ∂γrθ
∂θ − ¯ Γ i θr γiθ − ¯ Γ i θθ γri = − ¯ Γ θ θr γθθ − ¯ Γ r θθ γrr
(7.23)
(7.24)
Dϕγrϕ = ∂γrϕ
∂ϕ − ¯ Γ i ϕrγiϕ − ¯ Γ i ϕϕγri = − ¯ Γ ϕ ϕrγϕϕ − ¯ Γ r ϕϕγrr, (7.25)