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While reading this article I got stuck with Eq.$(54)$. I've been trying to derive it but I can't get their result. I believe my problem is in understanding their hints. They say that they get the result from the Gauss embedding equation and the Ricci identities for the 4-velocity, $u^a$. Is the Gauss equation they refer the one in the wiki article?

Looking at the terms that appear in their equation it looks like the Raychaudhuri equation or the Einstein field equations are to be used in the derivation in order to get the density and the cosmological constant, but even though I realize this I can't really get their result.

Can anyone point me in the right direction?

$Note:$The reason why I'm trying so hard to prove their result is because I wanted to know if it would still be valid if the orthogonal space were 2 dimensional (aside some constants). It appears to be the case but to be sure I needed to be able to prove it.

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Is the Gauss equation they refer the one in the wiki article?: I think it is, according to page 610 (or 30 of the file) of link.springer.com/article/10.1007%2Fs10714-009-0760-7. –  Taiki May 28 '13 at 16:14
    
@Taiki nice reference. It appears to be so. I think I might be on the right track, after several pages of calculations. Your reference has been very helpful. Thank you. –  PML May 29 '13 at 21:35

1 Answer 1

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+50

I don't have time to do the full calculation (it will be rather long!), but I'll indicate what I think are the steps that get you there:

We start with our congruence given by the normalized vector field $u^a$, $u_au^a=-1$

The covariant derivative of $u^a$ splits into a part parallel to the congruence and a part orthogonal to it: $$\nabla_au_b=-u_a\dot{u_b}+{\tilde{\nabla}}_au_b $$ Where the tilde derivative is defined by projecting orthogonal to $u^a$ $${\tilde{\nabla}}_au_b=h_a^ch_b^d\nabla_cu_d $$ $$h_a^b=\delta_a^b+u_au^b $$Now we can decompose ${\tilde{\nabla}}_au_b$ into its irreducible parts $$ {\tilde{\nabla}}_au_b = \omega_{ab}+\frac{1}{3}\Theta h_{ab}+\sigma_{ab}$$ Where $\omega_{ab}$ is the antisymmetric part, $\Theta$ is the trace part, and $\sigma_{ab}$ is the trace-free symmetric part.

In most derivations of the Gauss-Codazzi equations, they assume that $u_a$ is vorticity-free ($\omega$ is the vorticity). Here we can't make that assumption. We wish to investigate curvature orthogonal to the congruence so we want to calculate $$({\tilde{\nabla}}_a{\tilde{\nabla}}_b-{\tilde{\nabla}}_b{\tilde{\nabla}}_a)X_c $$ where X is a vector field orthogonal to the congruence. Directly substituting for the ${\tilde{\nabla}}$ factors a couple of pages of calculation got me to $$ ({\tilde{\nabla}}_a{\tilde{\nabla}}_b-{\tilde{\nabla}}_b{\tilde{\nabla}}_a)X_c$$ $$=2\omega_{ab}{\dot{X}}_{<c>}+(^{\perp}R_{abcd})X_d+(K_{cb}K_{da}-K_{ca}K_{db})X^d $$where$$K_{ab}={\tilde{\nabla}}_bu_a $$ (I'm using the angle brackets and time derivatives defined in eqns (9) and (10) of your reference and the perp just means project all free indices using the $h$'s. Also the Gauss Codazzi section in Wald is useful here. Oh BTW I can't guarantee signs and factors of two!).

I believe the next step would be to contact this equation to obtain the desired three-Ricci tensor. It contains all the ingredients in your desired equation (54). The only problem is that you still have the (projected) Riemann tensor involved. To get rid of that you would have to use the field equations - this will bring in ingredients like $\pi_{ab}$

Sorry - it's more of a long hint than an answer, but it is a rather messy calculation! (you may have already completed it yourself now...)

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Ah, your answer/hint (=)) is spot on. I wasn't sure of one definition and an incorrect use was making me run in circles. Thank you very much for your effort. I just checked Wald's book and indeed is very (very!) useful. Thank you once again. I didn't know that the calculation was so messy or I should have given a greater bounty...=/ Thank you once again. –  PML Jun 1 '13 at 14:53

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