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

5.2 Energy and momentum conservation 73
Remark : We have written the derivative of E along n as a Lie derivative. E being a scalar
field, we have of course the alternative expressions
Ln E = ∇nE = n · ∇E = n µ µ ∂E
∇µE = n = 〈dE,n〉. (5.10)
∂x µ
Ln E is the derivative of E with respect to the proper time of the Eulerian observers: Ln E =
dE/dτ, for n is the 4-velocity of these observers. It is easy to let appear the derivative with
respect to the coordinate time t instead, thanks to the relation n = N −1 (∂t −β) [cf. Eq. (4.31)]:
Then
in components:
Ln E = 1
∂
− Lβ E. (5.11)
N ∂t
∂
− Lβ E + N
∂t D · p − KE − KijS ij + 2p · DN = 0 , (5.12)
∂ ∂
− βi
∂t ∂xi
E + N Dip i − KE − KijS ij + 2p i DiN = 0. (5.13)
This equation has been obtained by York (1979) in his seminal article [276].
5.2.3 Newtonian limit
As a check, let us consider the Newtonian limit of Eq. (5.12). For this purpose let us assume
that the gravitational field is weak and static. It is then always possible to find a coordinate
system (x α ) = (x 0 = ct,x i ) such that the metric components take the form (cf. N. Deruelle’s
lectures [108])
gµνdx µ dx ν = − (1 + 2Φ) dt 2 + (1 − 2Φ)fij dx i dx j , (5.14)
where Φ is the Newtonian gravitational potential (solution of Poisson equation ∆Φ = 4πGρ)
and fij are the components the flat Euclidean metric f in the 3-dimensional space. For a weak
gravitational field (Newtonian limit), |Φ| ≪ 1 (in units where the light velocity is not one, this
should read |Φ|/c 2 ≪ 1). Comparing Eq. (5.14) with (4.48), we get N = √ 1 + 2Φ ≃ 1 + Φ,
β = 0 and γ = (1 − 2Φ)f. From Eq. (4.63), we then obtain immediately that K = 0. To
summarize:
Newtonian limit: N = 1 + Φ, β = 0, γ = (1 − 2Φ)f, K = 0, |Φ| ≪ 1. (5.15)
Notice that the Eulerian observer becomes a Galilean (inertial) observer for he is non-rotating
(cf. remark page 44).
Taking into account the limits (5.15), Eq. (5.12) reduces to
∂E
∂t
+ D · p = −2p · DΦ. (5.16)