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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)

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