Derivation of $zz$-component of Einsteins Equations in AdS

Question

I am trying to understand how we get the Einsteins equations in here section 4.1 equation 4.2 where we use the metric
$$
ds^2 = a^2(z)(dz^2+dx^mu dx_mu)
$$
to derive the $zz$-component of Einstein's equation which should give
$$
3left(frac{a'}{a}right)^2-frac{3a^2}{L^2_{AdS}}=8pi G T_{zz}.
$$
I understand that the second term on the LHS comes from the fact that we have a negative curvature and the cosmological constant is given by $Lambda = frac{-(n-1)(n-2)}{2L^2_{AdS}}$ where $n=4$ in our case(the dimensions). When I try to compute this using
$$
R_{zz}-frac{1}{2}Rg_{zz}+Lambda g_{zz}=8pi G T_{zz}
$$
I don't get the desired result. Is there any special property of the metric(for starters it is symmetric) I can use to compute this without computing all Chrisstoffel symbols?

user142288 · Answer

There is a prefactor $3$ before $(a'/a)^2$, empirically, I can learn that this corresponds to $4$D geometry, because this prefactor depends on dimension by $n(n-1)/2$, where $n$ is number of dimension. It implies that $mu$ runs from $0$ to $2$ (not $3$);
It is not important what components of metric standing in $dx^mu dx_mu$, you can simply set it to be flat;
You had given the correct $Lambda$ term;
$R_{zz}=3 left[(a')^2-a a''right]/a^2$, $R=-6 a''/a^3$, $g_{zz}=a^2$;
Substituting item 4 into Einstein tensor, you can find the result.

Derivation of $zz$-component of Einsteins Equations in AdS

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