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

6.5 Conformal form of the 3+1 Einstein system 95
Since A is traceless, ˜γjkA jk = Ψ −4 γjkA jk = 0. Then the above equation reduces to DjA ij =
˜DjA ij + 10A ij ˜ Dj ln Ψ, which can be rewritten as
DjA ij = Ψ −10 ˜ Dj
Ψ 10 A ij . (6.81)
Notice that this identity is valid only because A ij is symmetric and traceless.
Equation (6.81) suggests to introduce the quantity 4
 ij := Ψ 10 A ij . (6.82)
This corresponds to the scaling factor α = −10 in Eq. (6.58). It has been first introduced by
Lichnerowicz in 1944 [177]. Thanks to it and Eq. (6.79), the momentum constraint equation
(4.66) can be rewritten as
˜Dj Âij − 2
3 Ψ6 ˜ D i K = 8πΨ 10 p i . (6.83)
As for Ãij, we define Âij as the tensor field deduced from Âij by lowering the indices with
the conformal metric:
Âij := ˜γik˜γjl Âkl
Taking into account Eq. (6.82) and ˜γij = Ψ −4 γij, we get
(6.84)
Âij = Ψ 2 Aij . (6.85)
6.5 Conformal form of the 3+1 Einstein system
Having performed a conformal decomposition of γ and of the traceless part of K, we are now
in position to rewrite the 3+1 Einstein system (4.63)-(4.66) in terms of conformal quantities.
6.5.1 Dynamical part of Einstein equation
Let us consider Eq. (4.64), i.e. the so-called dynamical equation in the 3+1 Einstein system:
LmKij = −DiDjN + N Rij + KKij − 2KikK k j + 4π [(S − E)γij
− 2Sij] . (6.86)
Let us substitute Aij + (K/3)γij for Kij [Eq. (6.57)]. The left-hand side of the above equation
becomes
LmKij = LmAij + 1
3 LmK γij + 1
3
K Lmγij . (6.87)
=−2NKij
In this equation appears LmK. We may express it by taking the trace of Eq. (6.86) and making
use of Eq. (3.49):
LmK = γ ij LmKij + 2NKijK ij , (6.88)
4 notice that we have used a hat, instead of a tilde, to distinguish this quantity from that defined by (6.76)