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

10.3 Free evolution schemes 179
Remark : We assume here specifically that Eq. (10.23) holds, because in the following we do
not demand that the whole Einstein equation is satisfied, but only its dynamical part, i.e.
Eq. (10.12).
As we have seen in Chap. 5, in order for Eq. (10.23) to be satisfied, the matter energy density
E and momentum density p (both relative to the Eulerian observer) must obey to the evolution
equations (5.12) and (5.23).
Thanks to the Bianchi identity (10.21) and to the energy-momentum conservation law
(10.23), the divergence of Eq. (10.19) leads to, successively,
∇µ (G µ α − 8πT µ α ) = 0
∇µ [F µ α + (H − F)γµ α + nµ Mα + M µ nα + Hn µ nα] = 0,
∇µF µ α + Dα(H − F) + (H − F)(∇µn µ nα + n µ ∇µnα) − KMα + n µ ∇µMα
+∇µM µ nα − M µ Kµα + n µ ∇µH nα − HKnα + HDα ln N = 0,
∇µF µ α + Dα(H − F) + (2H − F)(Dα lnN − Knα) − KMα + n µ ∇µMα,
+∇µM µ nα − KαµM µ + n µ ∇µH nα = 0, (10.24)
where we have used Eq. (3.20) to express the ∇n in terms of K and D ln N (in particular
∇µn µ = −K). Let us contract Eq. (10.24) with n: we get, successively,
n ν ∇µF µ ν + (2H − F)K + n ν n µ ∇µMν − ∇µM µ − n µ ∇µH = 0,
−F µ ν∇µn ν + (2H − F)K − Mνn µ ∇µn ν − ∇µM µ − n µ ∇µH = 0,
K µν Fµν + (2H − F)K − M ν Dν ln N − ∇µM µ − n µ ∇µH = 0. (10.25)
Now the ∇-divergence of M is related to the D-one by
Thus Eq. (10.25) can be written
Noticing that
DµM µ = γ ρ µ γσ ν ∇ρM σ = γ ρ σ ∇ρM σ = ∇ρM ρ + n ρ nσ∇ρM σ
= ∇µM µ − M µ Dµ ln N. (10.26)
n µ ∇µH = −DµM µ − 2M µ Dµ lnN + K(2H − F) + K µν Fµν. (10.27)
n µ ∇µH = 1
N mµ ∇µH = 1
N LmH = 1
∂
− Lβ H, (10.28)
N ∂t
where m is the normal evolution vector (cf. Sec. 3.3.2), we get the following evolution equation
for the Hamiltonian constraint violation
∂
− Lβ H = −Di(NM
∂t i ) − M i DiN + NK(2H − F) + NK ij Fij . (10.29)
Let us now project Eq. (10.24) onto Σt:
γ να ∇µF µ ν + D α (H − F) + (2H − F)D α ln N − KM α + γ α νn µ ∇µM ν − K α µM µ = 0. (10.30)