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

36 Geometry of hypersurfaces
Example : We may check the Theorema Egregium (2.96) for the examples of Sec. 2.3.5. It
is trivial for the plane, since each term vanishes separately. For the cylinder of radius r,
R = 0, K = −1/r [Eq. (2.51)], KijK ij = 1/r 2 [Eq. (2.50)], so that Eq. (2.96) is satisfied.
For the sphere of radius r, R = 2/r 2 [Eq. (2.53)], K = −2/r [Eq. (2.59)], KijK ij = 2/r 2
[Eq. (2.58)], so that Eq. (2.96) is satisfied as well.
2.5.2 Codazzi relation
Let us at present apply the Ricci identity (2.13) to the normal vector n (or more precisely to
any extension of n around Σ, cf. Sec. 2.4.2):
If we project this relation onto Σ, we get
Now, from Eq. (2.75),
(∇α∇β − ∇β∇α) n γ = 4 R γ
µαβ nµ . (2.98)
γ µ αγ ν β γγ ρ 4 R ρ σµνn σ = γ µ αγ ν β γγ ρ (∇µ∇νn ρ − ∇ν∇µn ρ ) . (2.99)
γ µ α γν β γγ ρ ∇µ∇νn ρ = γ µ α γν β γγ ρ ∇µ (−K ρ ν − aρ nν)
= −γ µ α γν β γγ ρ (∇µK ρ ν + ∇µa ρ nν + a ρ ∇µnν)
= −DαK γ
β + aγ Kαβ, (2.100)
where we have used Eq. (2.79), as well as γ ν β nν = 0, γ γ ρa ρ = a γ , and γ µ αγ ν β ∇µnν = −Kαβ to
get the last line. After permutation of the indices α and β and substraction from Eq. (2.100),
taking into account the symmetry of Kαβ, we see that Eq. (2.99) becomes
γ γ ρ n σ γ µ αγ ν β 4 R ρ σµν = DβK γ α − DαK γ
β . (2.101)
This is the Codazzi relation, also called Codazzi-Mainardi relation in the mathematical
litterature [52].
Remark : Thanks to the symmetries of the Riemann tensor (cf. Sec. 2.2.3), changing the
index contracted with n in Eq. (2.101) (for instance considering nργ γσ γ µ αγ ν β 4 R ρ σµν or
γ γ ρ γ σ α n µ γ ν β 4 R ρ σµν ) would not give an independent relation: at most it would result in
a change of sign of the right-hand side.
Contracting the Codazzi relation on the indices α and γ yields to
γ µ ρ n σ γ ν β 4 R ρ σµν = DβK − DµK µ
β , (2.102)
with γ µ ρ n σ γ ν β 4 R ρ σµν = (δ µ ρ + n µ nρ)n σ γ ν β 4 R ρ σµν = n σ γ ν β 4 Rσν + γ ν β 4 R ρ σµνnρn σ n µ . Now,
from the antisymmetry of the Riemann tensor with respect to its first two indices [Eq. (2.14),
the last term vanishes, so that one is left with
γ µ α nν 4 Rµν = DαK − DµK µ α
We shall call this equation the contracted Codazzi relation.
. (2.103)