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

72 3+1 equations for matter and electromagnetic field
5.2 Energy and momentum conservation
5.2.1 3+1 decomposition of the 4-dimensional equation
Let us replace T in Eq. (5.1) by its 3+1 expression (4.10) in terms of the energy density E, the
momentum density p and the stress tensor S, all of them as measured by the Eulerian observer.
We get, successively,
∇µT µ α
= 0
∇µ (S µ α + nµ pα + p µ nα + En µ nα) = 0
∇µS µ α − Kpα + n µ ∇µpα + ∇µp µ nα − p µ Kµα − KEnα + EDα ln N
+n µ ∇µE nα = 0, (5.2)
where we have used Eq. (3.20) to express the ∇n in terms of K and D ln N.
5.2.2 Energy conservation
Let us project Eq. (5.2) along the normal to the hypersurfaces Σt, i.e. contract Eq. (5.2) with
n α . We get, since p, K and D ln N are all orthogonal to n:
Now, since n · S = 0,
Similarly
n ν ∇µS µ ν + nµ n ν ∇µpν − ∇µp µ + KE − n µ ∇µE = 0. (5.3)
n ν ∇µS µ ν = −S µ ν∇µn ν = S µ ν(K ν µ + D ν ln N nµ) = KµνS µν . (5.4)
n µ n ν ∇µpν = −pνn µ ∇µn ν = −pνD ν ln N. (5.5)
Besides, let us express the 4-dimensional divergence ∇µp µ is terms of the 3-dimensional one,
Dµp µ . For any vector v tangent to Σt, like p, Eq. (2.79) gives
Dµv µ = γ ρ µ γµ σ ∇ρv σ = γ ρ σ ∇ρv σ = (δ ρ σ +nρ nσ)∇ρv σ = ∇ρv ρ −v σ n ρ ∇ρnσ = ∇ρv ρ −v σ Dσ ln N
(5.6)
Hence the usefull relation between the two divergences
or in terms of components,
∀v ∈ T (Σt), ∇·v = D·v + v · D ln N , (5.7)
∀v ∈ T (Σt), ∇µv µ = Div i + v i Di ln N. (5.8)
Applying this relation to v = p and taking into account Eqs. (5.4) and (5.5), Eq. (5.3) becomes
Ln E + D · p + 2p · D ln N − KE − KijS ij = 0. (5.9)