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

3.4 Last part of the 3+1 decomposition of the Riemann tensor 47
Remark : An illustration of property (3.32) is provided by Eq. (3.24), which says that Lmγ
is −2NK: K being tangent to Σt, we have immediately that Lmγ is tangent to Σt.
Remark : Contrary to Ln γ and Lmγ, which are related by Eq. (3.26), Ln γ and Lmγ are
not proportional. Indeed a calculation similar to that which lead to Eq. (3.26) gives
Therefore the property Lmγ = 0 implies
Ln γ = 1
N Lmγ + n ⊗ D ln N. (3.36)
Ln γ = n ⊗ D ln N = 0. (3.37)
Hence the privileged role played by m regarding the evolution of the hypersurfaces Σt is
not shared by n; this merely reflects that the hypersurfaces are Lie dragged by m, not by
n.
3.4 Last part of the 3+1 decomposition of the Riemann tensor
3.4.1 Last non trivial projection of the spacetime Riemann tensor
In Chap. 2, we have formed the fully projected part of the spacetime Riemann tensor, i.e.
γ ∗ 4 Riem, yielding the Gauss equation [Eq. (2.92)], as well as the part projected three times
onto Σt and once along the normal n, yielding the Codazzi equation [Eq. (2.101)]. These two
decompositions involve only fields tangents to Σt and their derivatives in directions parallel to
Σt, namely γ, K, Riem and DK. This is why they could be defined for a single hypersurface.
In the present section, we form the projection of the spacetime Riemann tensor twice onto Σt
and twice along n. As we shall see, this involves a derivative in the direction normal to the
hypersurface.
As for the Codazzi equation, the starting point of the calculation is the Ricci identity applied
to the vector n, i.e. Eq. (2.98). But instead of projecting it totally onto Σt, let us project it
only twice onto Σt and once along n:
γαµn σ γ ν β (∇ν∇σn µ − ∇σ∇νn µ ) = γαµn σ γ ν β 4 R µ ρνσn ρ . (3.38)
By means of Eq. (3.20), we get successively
γαµ n ρ γ ν β nσ 4 R µ ρνσ = γαµn σ γ ν β [−∇ν(K µ σ + D µ ln N nσ) + ∇σ(K µ ν + D µ ln N nν)]
= γαµn σ γ ν β [ − ∇νK µ σ − ∇νnσ D µ ln N − nσ∇νD µ ln N
+∇σK µ ν + ∇σnν D µ ln N + nν∇σD µ ln N ]
= γαµγ ν β [Kµ σ ∇νn σ + ∇νD µ ln N + n σ ∇σK µ ν + Dν ln N D µ ln N]
= −KασK σ β + DβDα ln N + γ µ αγ ν β nσ ∇σKµν + Dα ln NDβ ln N
= −KασK σ β
+ 1
N DβDαN + γ µ α γν β nσ ∇σKµν. (3.39)