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

7.3 ADM mass 109
and
˜γij = fij + O(r −1 ) and
∂˜γij
∂xk = O(r−2 ). (7.35)
Thanks to the decomposition (7.32), the integrand of the ADM mass formula (7.14) is
D j γij − Di(f kl γkl) = 4 Ψ 3
D
∼1
j Ψ ˜γij
∼fij
+ Ψ
4
∼1
D j ˜γij − 4 Ψ 3
∼1
DiΨ f kl ˜γkl
∼3
− Ψ 4
∼1
Di(f kl ˜γkl), (7.36)
where the ∼’s denote values when r → ∞, taking into account (7.34) and (7.35). Thus we have
D j γij − Di(f kl γkl) ∼ −8DiΨ + D j ˜γij − Di(f kl ˜γkl). (7.37)
From (7.34) and (7.35), DiΨ = O(r −2 ) and D j ˜γij = O(r −2 ). Let us show that the unit determinant
condition (7.33) implies Di(f kl ˜γkl) = O(r −3 ) so that this term actually does not contribute
to the ADM mass integral. Let us write
with according to Eq. (7.35), εij = O(r −1 ). Then
and
Now the determinant of ˜γij is
⎛
det(˜γij) = det⎝
Requiring det(˜γij) = 1 implies then
˜γij =: fij + εij, (7.38)
f kl ˜γkl = 3 + εxx + εyy + εzz
(7.39)
Di(f kl ˜γkl) = ∂
∂xi(fkl ˜γkl) = ∂
∂xi (εxx + εyy + εzz). (7.40)
1 + εxx εxy εxz
εxy 1 + εyy εyz
εxz εyz 1 + εzz
= 1 + εxx + εyy + εzz + εxxεyy + εxxεzz + εyyεzz − ε 2 xy − ε2 xz − ε2 yz
⎞
⎠
+εxxεyyεzz + 2εxyεxzεyz − εxxε 2 yz − εyyε 2 xz − εzzε 2 xy. (7.41)
εxx + εyy + εzz = −εxxεyy − εxxεzz − εyyεzz + ε 2 xy + ε2xz + ε2yz −εxxεyyεzz − 2εxyεxzεyz + εxxε 2 yz + εyyε 2 xz + εzzε 2 xy . (7.42)
Since according to (7.35), εij = O(r −1 ) and ∂εij/∂x k = O(r −2 ), we conclude that
i.e. in view of (7.40),
∂
∂x i (εxx + εyy + εzz) = O(r −3 ), (7.43)
Di(f kl ˜γkl) = O(r −3 ). (7.44)
Thus in Eq. (7.37), only the first two terms in the right-hand side contribute to the ADM mass
integral, so that formula (7.14) becomes
MADM = − 1
2π lim
s
St→∞
i
DiΨ − 1
8 Dj
√q 2
˜γij d y . (7.45)
St