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

96 Conformal decomposition
hence
LmK = −DiD i N + N R + K 2 + 4π(S − 3E) . (6.89)
Let use the Hamiltonian constraint (4.65) to replace R + K 2 by 16πE + KijK ij . Then, writing
LmK = (∂/∂t − Lβ )K,
∂
− Lβ K = −DiD
∂t i N + N 4π(E + S) + KijK ij . (6.90)
Remark : At the Newtonian limit, as defined by Eqs. (5.15), (5.25) and (5.59), Eq. (6.90)
reduces to the Poisson equation for the gravitational potential Φ:
DiD i Φ = 4πρ0. (6.91)
Substituting Eq. (6.89) for LmK and Eq. (6.86) for LmKij into Eq. (6.87) yields
LmAij = −DiDjN + N Rij + 5
+ 1
DkD
3
k N − N(R + K 2
)
3 KKij − 2KikK k j
− 8π Sij − 1
3 Sγij
γij. (6.92)
Let us replace Kij by its expression in terms of Aij and K [Eq. (6.57)]: the terms in the
right-hand side of the above equation which involve K are then written
5K
3 Kij − 2KikK k j − K2
3 γij = 5K
Aij +
3
K
3 γij
− 2 Aik + K
= 5K
3 Aij + 5K2
9 γij − 2
3 γik
A k j + K
AikA k j + 2K
3 Aij + K2
9 γij
3 δk j
− K2
3 γij
− K2
3 γij
= 1
3 KAij − 2AikA k j . (6.93)
Accordingly Eq. (6.92) becomes
LmAij
= −DiDjN + N Rij + 1
3 KAij − 2AikA k
j − 8π Sij − 1
3 Sγij
+ 1
DkD
3
k
N − NR γij. (6.94)
Remark : Regarding the matter terms, this equation involves only the stress tensor S (more
precisely its traceless part) and not the energy density E, contrary to the evolution equation
(6.86) for K ij , which involves both.
At this stage, we may say that we have split the dynamical Einstein equation (6.86) in
two parts: a trace part: Eq. (6.90) and a traceless part: Eq. (6.94). Let us now perform the
conformal decomposition of these relations, by introducing Ãij. We consider Ãij and not Âij,
i.e. the scaling α = −4 and not α = −10, since we are discussing time evolution equations.