3
$\begingroup$

I see that Friedmann's first Equation (for flat space) is: $$\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G}{3}\rho.$$ And I know that Einstein's equation, just considering the time-time component is: $$R_{00}-\frac{1}{2}g_{00}R=8\pi G T_{00}.$$ And I know that $T_{00}$ in the tensor is $\rho$, so we get: $$R_{00}-\frac{1}{2}g_{00}R=8\pi G\rho.$$ Could someone please fill in the missing steps? How do we arrive at: $$R_{00}-\frac{1}{2}g_{00}R=3\left(\frac{\dot{a}}{a}\right)^2~?$$

$\endgroup$
2
  • $\begingroup$ I never found a solution filling in the missing steps. At best, textbooks (and I have at least five) just seem to leave this as an exercise. I have given a complete derivation of the Friedmann equation in my third book, for the case with curved space and cosmological constant, but it takes six pages of calculation, so I can't give an answer here. $\endgroup$ Jul 6, 2020 at 20:42
  • $\begingroup$ @CharlesFrancis - Are you available for a conversation in the h-bar? $\endgroup$
    – Gluon Soup
    Jul 6, 2020 at 21:25

2 Answers 2

3
$\begingroup$

This is a typical example where a Newtonian derivation is much simpler and quicker, and gives the same answer. Which you can easily find online.

But if you want to do this from within GR, then you have to work out the Ricci tensor entry $R_{00}$, the Ricci scalar $R$, and the metric entry $g_{00}$:

  • $g_{00} = 1$;

  • $R_{00}$: $$ R_{00} = R^m_{tmt} = R^r_{rtr} + R^\theta_{t\theta t} + R^\phi_{t\phi t} = -3 \frac{\ddot a}{a},$$ where each Riemann tensor depends on the Christoffel symbols (listed for instance in section C here);

  • $R$: $$R = g^{ik}R_{ik} = -6\frac{\ddot a}{a} - 6\left ( \frac{\dot a}{a} \right )^2 - 6\frac{1}{k^2a^2},$$ where $k^{-2}=0$ for flat space.

So putting it all together: $$ R_{00} -\frac{1}{2}Rg_{00} = -3\frac{\ddot a}{a}+3\frac{\ddot a}{a} + 3\left ( \frac{\dot a}{a} \right )^2.$$

Hence: $$3\left ( \frac{\dot a}{a} \right )^2 = 8\pi G\rho, $$ $$ \Rightarrow \left ( \frac{\dot a}{a} \right )^2 = \frac{8\pi G}{3}\rho. $$

$\endgroup$
9
  • $\begingroup$ Worth noting that the RHS follows from the stress energy tensor of a perfect fluid. $\endgroup$
    – bapowell
    Jul 7, 2020 at 0:11
  • $\begingroup$ Ah yes, good to point out that assumption. $\endgroup$
    – SuperCiocia
    Jul 7, 2020 at 0:30
  • $\begingroup$ Which you can easily find online. I have no idea even what search terms I would use. Any chance you could find one and post the link? It'd be educational to see how it's done with Newtonian physics. $\endgroup$
    – Gluon Soup
    Jul 7, 2020 at 0:52
  • $\begingroup$ @GluonSoup astronomy.nmsu.edu/aklypin/AST616/SimpleDerivation.pdf $\endgroup$
    – SuperCiocia
    Jul 7, 2020 at 1:05
  • $\begingroup$ By “easily” I meant that it’s the derivation you would usually find. Fewer sources do the actual GR one. Most university lecture notes always go classical. $\endgroup$
    – SuperCiocia
    Jul 7, 2020 at 2:58
1
$\begingroup$

$$\Gamma_{ab}{}^{c} = \frac{1}{2}g^{cd}\left(g_{ad, b} + g_{bd,a} - g_{ab,d}\right)$$

$$R_{ab} = \partial_{c}\Gamma_{ab}{}^{c} - \partial_{a}\Gamma_{bc}{}^{c} + \Gamma_{ab}{}^{c}\Gamma_{ce}{}^{e} - \Gamma_{ad}{}^{c}\Gamma_{bc}{}^{d}$$

So, given a metric, you can compute any christoffel symbol, and given any christoffel symbol, you can calculate the ricci tensor. Just turn the crank and compute $R_{00}$ and $R$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.