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

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 $$3\left(\frac{a'}{a}\right)^2-\frac{3a^2}{L^2_{AdS}}=8\pi G T_{zz}.$$ 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 $$R_{zz}-\frac{1}{2}Rg_{zz}+\Lambda g_{zz}=8\pi G T_{zz}$$

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?

1. 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 $$3$$);
2. It is not important what components of metric standing in $$dx^\mu dx_\mu$$, you can simply set it to be flat;
3. You had given the correct $$\Lambda$$ term;
4. $$R_{zz}=3 \left[(a')^2-a a''\right]/a^2$$, $$R=-6 a''/a^3$$, $$g_{zz}=a^2$$;
• I calculate the same expression for $R_{zz}$ but for some reason I don't get the same result for $R$ when I use $R = g_{\mu\nu}R^{\mu\nu}$(which means I probably made a mistake for $R_{00},R_{11}$ and $R_{22}$). Can you give me some hints on how to calculate $R$? Jul 4 '21 at 15:04
• Also I find $R_{zz}=3(\frac{a''}{a}-(\frac{a'}{a})^2)$. So we have a sign difference as well. Jul 4 '21 at 17:52
• @chillyspangko The calculation of $R$ is standard, I just picked up its definition, substituted the metric and got the results; The only trick what I used is included in item 2, it helped me reduce a lot of workload. As to the different sign, did you employ an opposite convention of signature? Sorry, I can not guess how you get that result. Jul 5 '21 at 1:12