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

34 Geometry of hypersurfaces
2.5 Gauss-Codazzi relations
We derive here equations that will constitute the basis of the 3+1 formalism for general relativity.
They are decompositions of the spacetime Riemann tensor, 4 Riem [Eq. (2.12)], in terms of
quantities relative to the spacelike hypersurface Σ, namely the Riemann tensor associated with
the induced metric γ, Riem [Eq. (2.34)] and the extrinsic curvature tensor of Σ, K.
2.5.1 Gauss relation
Let us consider the Ricci identity (2.34) defining the (three-dimensional) Riemann tensor Riem
as measuring the lack of commutation of two successive covariant derivatives with respect to the
connection D associated with Σ’s metric γ. The four-dimensional version of this identity is
DαDβv γ − DβDαv γ = R γ
µαβ vµ , (2.84)
where v is a generic vector field tangent to Σ. Let us use formula (2.79) which relates the
D-derivative to the ∇-derivative, to write
DαDβv γ = Dα(Dβv γ ) = γ µ αγ ν β γγ ρ∇µ(Dνv ρ ). (2.85)
Using again formula (2.79) to express Dνv ρ yields
DαDβv γ = γ µ α γν β γγ ρ ∇µ
γ σ νγρ λ∇σv λ
. (2.86)
Let us expand this formula by making use of Eq. (2.64) to write ∇µγ σ ν = ∇µ (δ σ ν + nσ nν) =
∇µn σ nν + n σ ∇µnν. Since γ ν β nν = 0, we get
DαDβv γ = γ µ α γν β γγ ρ
n σ ∇µnν γ ρ
λ∇σv λ + γ σ ν∇µn ρ nλ∇σv λ
=−vλ +γ
∇σnλ
σ νγρ
λ
λ∇µ∇σv = γ µ αγνβγγ λ∇µnν n σ ∇σv λ − γ µ αγσβγγ ρvλ∇µn ρ ∇σnλ + γ µ αγσβγγ λ
λ∇µ∇σv = −Kαβ γ γ
λ nσ ∇σv λ − K γ αKβλ v λ + γ µ αγ σ β γγ
λ ∇µ∇σv λ , (2.87)
where we have used the idempotence of the projection operator γ, i.e. γ γ ργ ρ
λ = γγ
λ to get
the second line and γ µ αγν β∇µnν = −Kβα [Eq. (2.76)] to get the third one. When we permute
the indices α and β and substract from Eq. (2.87) to form DαDβvγ − DβDγv γ , the first term
vanishes since Kαβ is symmetric in (α,β). There remains
DαDβv γ − DβDγv γ
=
v µ + γ ρ αγσβγγ
λ ∇ρ∇σv λ − ∇σ∇ρv λ
. (2.88)
KαµK γ
β − KβµK γ α
Now the Ricci identity (2.13) for the connection ∇ gives ∇ρ∇σvλ −∇σ∇ρv λ = 4Rλ µρσvµ . Therefore
DαDβv γ − DβDγv γ =
v µ + γ ρ αγ σ βγγ 4 λ
λ R µρσv µ . (2.89)
KαµK γ
β − KβµK γ α
Substituting this relation for the left-hand side of Eq. (2.84) results in
KαµK γ
β − KβµK γ α
v µ + γ ρ αγ σ βγγ 4 λ
λ R µρσv µ = R γ
µαβ vµ , (2.90)