3+1 formalism and bases of numerical relativity - LUTh ...
3+1 formalism and bases of numerical relativity - LUTh ...
3+1 formalism and bases of numerical relativity - LUTh ...
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118 Asymptotic flatness <strong>and</strong> global quantities<br />
Using this property, as well as expression (7.75) <strong>of</strong> dVµ with the components nµ = (−N,0,0,0)<br />
given by Eq. (4.38), we get<br />
<br />
Vt<br />
∇νA µν <br />
dVµ = −<br />
Vt<br />
∂<br />
∂xν √ µν<br />
−gA nµ<br />
√<br />
γ<br />
√ d<br />
−g 3 <br />
x =<br />
Vt<br />
∂<br />
∂xν √ 0ν<br />
γNA d 3 x, (7.78)<br />
where we have also invoked the relation (4.55) between the determinants <strong>of</strong> g <strong>and</strong> γ: √ −g =<br />
N √ γ. Now, since A αβ is antisymmetric, A 00 = 0 <strong>and</strong> we can write ∂/∂x ν √ γNA 0ν =<br />
∂/∂x i √ γ V i where V i = NA 0i are the components <strong>of</strong> the vector V ∈ T (Σt) defined by<br />
V := −γ(n · A). The above integral then becomes<br />
<br />
Vt<br />
∇νA µν <br />
dVµ =<br />
Vt<br />
1 ∂<br />
√<br />
γ ∂xi √ i<br />
γV <br />
√ 3<br />
γ d x = DiV<br />
Vt<br />
i√ γ d 3 x. (7.79)<br />
We can now use the Gauss-Ostrogradsky theorem to get<br />
<br />
∇νA µν <br />
dVµ = V i √ 2<br />
si q d y. (7.80)<br />
Vt<br />
Noticing that ∂Vt = Ht ∪ St (cf. Fig. 7.2) <strong>and</strong> (from the antisymmetry <strong>of</strong> A µν )<br />
we get the identity (7.74).<br />
∂Vt<br />
V i si = V ν sν = −nµA µν sν = 1<br />
2 Aµν (sµnν − nµsν), (7.81)<br />
Remark : Equation (7.74) can also be derived by applying Stokes’ theorem to the 2-form<br />
4 ǫαβµνA µν , where 4 ǫαβµν is the Levi-Civita alternating tensor (volume element) associated<br />
with the spacetime metric g (see e.g. derivation <strong>of</strong> Eq. (11.2.10) in Wald’s book [265]).<br />
where<br />
Applying formula (7.74) to A µν = ∇ µ k ν we get, in view <strong>of</strong> the definition (7.71),<br />
MK = − 1<br />
<br />
∇ν∇<br />
4π Vt<br />
µ k ν dVµ + M H K , (7.82)<br />
M H K := 1<br />
<br />
∇<br />
8π Ht<br />
µ k ν dS H µν<br />
will be called the Komar mass <strong>of</strong> the hole. Now, from the Ricci identity<br />
(7.83)<br />
∇ν∇ µ k ν − ∇ µ ∇ ν k ν<br />
=<br />
=0<br />
4 R µ νk ν , (7.84)<br />
where the “= 0” is a consequence <strong>of</strong> Killing’s equation (7.73). Equation (7.82) becomes then<br />
MK = − 1<br />
<br />
4 µ<br />
R<br />
4π<br />
νk<br />
Vt<br />
ν dVµ + M H K = 1<br />
<br />
4<br />
Rµνk<br />
4π Vt<br />
ν n µ √ γ d 3 x + M H K . (7.85)