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

6.5 Conformal form of the 3+1 Einstein system 97
Let us first transform Eq. (6.90). We can express the Laplacian of the lapse by applying the
divergence relation (6.37) to the vector v i = D i N = γ ij DjN = Ψ −4 ˜γ ij ˜ DjN = Ψ −4 ˜ D i N
DiD i N = Ψ −6
Di
˜ 6 i −6
Ψ D N = Ψ Di
˜ Ψ 2
D˜ i
N
= Ψ −4 Di
˜ ˜ D i N + 2 ˜ Di ln Ψ ˜ D i
N . (6.95)
Besides, from Eqs. (6.57), (6.72) and (6.76),
KijK ij =
Aij + K
3 γij
A ij + K
3 γij
= AijA ij + K2
In view of Eqs. (6.95) and (6.96), Eq. (6.90) becomes
3 = Ãij Ãij + K2
3
. (6.96)
∂
− Lβ K = −Ψ
∂t −4 Di
˜ ˜ D i N + 2 ˜ Di ln Ψ ˜ D i
N + N 4π(E + S) + Ãij Ãij + K2
. (6.97)
3
Let us now consider the traceless part, Eq. (6.94). We have, writing Aij = Ψ 4 Ãij and using
Eq. (6.70),
LmAij = Ψ 4 LmÃij + 4Ψ 3 LmΨ Ãij = Ψ 4
LmÃij + 2
˜Dkβ
3
k
− NK Ãij . (6.98)
Besides, from formulæ (6.29) and (6.33),
DiDjN = Di ˜ DjN = ˜ Di ˜ DjN − C k ij ˜ DkN
= ˜ Di ˜
DjN − 2
= ˜ Di ˜ DjN − 2
˜DkN
δ k i ˜ Dj ln Ψ + δ k j ˜ Di ln Ψ − ˜ D k ln Ψ ˜γij
˜Di ln Ψ ˜ DjN + ˜ Dj ln Ψ ˜ DiN − ˜ D k ln Ψ ˜ DkN ˜γij
. (6.99)
In Eq. (6.94), we can now substitute expression (6.98) for LmAij, (6.99) for DiDjN, (6.48) for
Rij, (6.95) for DkD k N and (6.49) for R. After some slight rearrangements, we get
∂
− Lβ Ãij = −
∂t 2
3 ˜ Dkβ k Ãij + N
+Ψ −4
KÃij − 2˜γ kl ÃikÃjl
− 8π Ψ −4 Sij − 1
3 S˜γij
− ˜ Di ˜ DjN + 2 ˜ Di ln Ψ ˜ DjN + 2 ˜ Dj ln Ψ ˜ DiN
+ 1
˜Dk
3
˜ D k N − 4 ˜ Dk ln Ψ ˜ D k
N ˜γij
+N ˜Rij − 1
3 ˜ R˜γij − 2 ˜ Di ˜ Dj ln Ψ + 4 ˜ Di ln Ψ ˜ Dj ln Ψ
+ 2
˜Dk
3
˜ D k ln Ψ − 2 ˜ Dk ln Ψ ˜ D k
ln Ψ ˜γij .
(6.100)