3+1 formalism and bases of numerical relativity - LUTh ...
3+1 formalism and bases of numerical relativity - LUTh ... 3+1 formalism and bases of numerical relativity - LUTh ...
48 Geometry of foliations Note that we have used K µ σn σ = 0, n σ ∇νnσ = 0, nσn σ = −1, n σ ∇σnν = Dν ln N and γ ν β nν = 0 to get the third equality. Let us now show that the term γ µ αγ ν β nσ ∇σKµν is related to LmK. Indeed, from the expression (A.8) of the Lie derivative: LmKαβ = m µ ∇µKαβ + Kµβ∇αm µ + Kαµ∇βm µ . (3.40) Substituting Eq. (3.22) for ∇αm µ and ∇βm µ leads to LmKαβ = Nn µ ∇µKαβ − 2NKαµK µ β − KαµD µ N nβ − KβµD µ N nα. (3.41) Let us project this equation onto Σt, i.e. apply the operator γ ∗ to both sides. Using the property γ ∗ LmK = LmK, which stems from the fact that LmK is tangent to Σt since K is [property (3.32)], we get LmKαβ = N γ µ αγ ν β nσ∇σKµν − 2NKαµK µ β . (3.42) Extracting γ µ αγ ν β nσ ∇σKµν from this relation and plugging it into Eq. (3.39) results in γαµ n ρ γ ν β nσ 4 R µ ρνσ 1 = N LmKαβ + 1 N DαDβN + KαµK µ β . (3.43) Note that we have written DβDαN = DαDβN (D has no torsion). Equation (3.43) is the relation we sought. It is sometimes called the Ricci equation [not to be confused with the Ricci identity (2.13)]. Together with the Gauss equation (2.92) and the Codazzi equation (2.101), it completes the 3+1 decomposition of the spacetime Riemann tensor. Indeed the part projected three times along n vanish identically, since 4 Riem(n,n,n,.) = 0 and 4 Riem(.,n,n,n) = 0 thanks to the partial antisymmetry of the Riemann tensor. Accordingly one can project 4 Riem at most twice along n to get some non-vanishing result. It is worth to note that the left-hand side of the Ricci equation (3.43) is a term which appears in the contracted Gauss equation (2.93). Therefore, by combining the two equations, we get a formula which does no longer contain the spacetime Riemann tensor, but only the spacetime Ricci tensor: γ µ αγ ν β 4 Rµν = − 1 N LmKαβ − 1 N DαDβN + Rαβ + KKαβ − 2KαµK µ β , (3.44) or in index-free notation: γ ∗ 4 R = − 1 N LmK − 1 N DDN + R + K K − 2K · K . (3.45) 3.4.2 3+1 expression of the spacetime scalar curvature Let us take the trace of Eq. (3.45) with respect to the metric γ. This amounts to contracting Eq. (3.44) with γ αβ . In the left-hand side, we have γ αβ γ µ αγ ν β = γµν and in the right-hand we
3.4 Last part of the 3+1 decomposition of the Riemann tensor 49 can limit the range of variation of the indices to {1,2,3} since all the involved tensors are spatial ones [including LmK, thanks to the property (3.32)] Hence γ µν4 Rµν = − 1 N γij LmKij − 1 N DiD i N + R + K 2 − 2KijK ij . (3.46) Now γ µν4 Rµν = (g µν + n µ n ν ) 4 Rµν = 4 R + 4 Rµνn µ n ν and − γ ij LmKij = −Lm(γ ij Kij with Lmγ ij evaluted from the very definition of the inverse 3-metric: γikγ kj = δ j i ⇒ Lmγik γ kj + γik Lmγ kj = 0 ⇒ γ il γ kj Lmγlk + γ il γlk ) + KijLmγ =K ij , (3.47) =δ i k ⇒ Lmγ ij = −γ ik γ jl Lmγkl ⇒ Lmγ ij = 2Nγ ik γ kl Kkl Lmγ lj = 0 ⇒ Lmγ ij = 2NK ij , (3.48) where we have used Eq. (3.24). Pluging Eq. (3.48) into Eq. (3.47) gives Consequently Eq. (3.46) becomes − γ ij LmKij = −LmK + 2NKijK ij . (3.49) 4 R + 4 Rµνn µ n ν = R + K 2 − 1 N LmK − 1 N DiD i N . (3.50) It is worth to combine with equation with the scalar Gauss relation (2.95) to get rid of the Ricci tensor term 4Rµνn µ nν and obtain an equation which involves only the spacetime scalar curvature 4R: 4 2 R = R + K + KijK ij − 2 N LmK − 2 N DiD i N . (3.51)
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3.4 Last part <strong>of</strong> the <strong>3+1</strong> decomposition <strong>of</strong> the Riemann tensor 49<br />
can limit the range <strong>of</strong> variation <strong>of</strong> the indices to {1,2,3} since all the involved tensors are spatial<br />
ones [including LmK, thanks to the property (3.32)] Hence<br />
γ µν4 Rµν = − 1<br />
N γij LmKij − 1<br />
N DiD i N + R + K 2 − 2KijK ij . (3.46)<br />
Now γ µν4 Rµν = (g µν + n µ n ν ) 4 Rµν = 4 R + 4 Rµνn µ n ν <strong>and</strong><br />
− γ ij LmKij = −Lm(γ ij Kij<br />
with Lmγ ij evaluted from the very definition <strong>of</strong> the inverse 3-metric:<br />
γikγ kj = δ j<br />
i<br />
⇒ Lmγik γ kj + γik Lmγ kj = 0<br />
⇒ γ il γ kj Lmγlk + γ il γlk<br />
) + KijLmγ<br />
<br />
=K<br />
ij , (3.47)<br />
<br />
=δ i k<br />
⇒ Lmγ ij = −γ ik γ jl Lmγkl<br />
⇒ Lmγ ij = 2Nγ ik γ kl Kkl<br />
Lmγ lj = 0<br />
⇒ Lmγ ij = 2NK ij , (3.48)<br />
where we have used Eq. (3.24). Pluging Eq. (3.48) into Eq. (3.47) gives<br />
Consequently Eq. (3.46) becomes<br />
− γ ij LmKij = −LmK + 2NKijK ij . (3.49)<br />
4 R + 4 Rµνn µ n ν = R + K 2 − 1<br />
N LmK − 1<br />
N DiD i N . (3.50)<br />
It is worth to combine with equation with the scalar Gauss relation (2.95) to get rid <strong>of</strong> the<br />
Ricci tensor term 4Rµνn µ nν <strong>and</strong> obtain an equation which involves only the spacetime scalar<br />
curvature 4R: 4 2<br />
R = R + K + KijK ij − 2<br />
N LmK − 2<br />
N DiD i N . (3.51)