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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?

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  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. $R_{zz}=3 \left[(a')^2-a a''\right]/a^2$, $R=-6 a''/a^3$, $g_{zz}=a^2$;
  5. Substituting item 4 into Einstein tensor, you can find the result.
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  • $\begingroup$ 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 $R$? $\endgroup$ Jul 4 '21 at 15:04
  • $\begingroup$ Also I find $R_{zz}=3(\frac{a''}{a}-(\frac{a'}{a})^2)$. So we have a sign difference as well. $\endgroup$ Jul 4 '21 at 17:52
  • $\begingroup$ @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. $\endgroup$
    – user142288
    Jul 5 '21 at 1:12

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