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

68 3+1 decomposition of Einstein equation
Using Eqs. (4.104), (4.108) and (4.105), we have
H = √ γ Kγ ij − K ij (−2NKij + Diβj + Djβi) − N √ γ(R + KijK ij − K 2 )
= √
γ −N(R + K 2 − KijK ij
) + 2 Kγ j
i − Kj
i Djβ i
= − √
γ N(R + K 2 − KijK ij
i
) + 2β DiK − DjK j
i
+2 √
γDj Kβ j − K j
iβi . (4.110)
The corresponding Hamiltonian is
H =
Σt
H d 3 x. (4.111)
Noticing that the last term in Eq. (4.110) is a divergence and therefore does not contribute to
the integral, we get
H = −
NC0 − 2β i √ 3
Ci γd x , (4.112)
where
Σt
C0 := R + K 2 − KijK ij , (4.113)
Ci := DjK j
i − DiK (4.114)
are the left-hand sides of the constraint equations (4.65) and (4.66) respectively.
The Hamiltonian H is a functional of the configuration variables (γij,N,β i ) and their con-
), the last two ones being identically zero since
jugate momenta (π ij ,π N ,π β
i
π N := ∂L
∂ ˙
N
= 0 and πβ
i := ∂L
∂ ˙ = 0. (4.115)
βi The scalar curvature R which appears in H via C0 is a function of γij and its spatial derivatives,
via Eqs. (4.72)-(4.74), whereas Kij which appears in both C0 and Ci is a function of γij and πij ,
obtained by “inverting” relation (4.108):
Kij = Kij[γ,π] = 1
√ γ
1
2 γklπ kl γij − γikγjlπ kl
. (4.116)
The minimization of the Hilbert action is equivalent to the Hamilton equations
δH
= ˙γij
(4.117)
δπij δH
δγij
= − ˙π ij
(4.118)
δH
δN = − ˙πN = 0 (4.119)
δH
= − ˙πβ
δβi i = 0. (4.120)