How is the first Friedmann equation derived from Einstein's field equations?

Question

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: $$R_{00}-\frac{1}{2}g_{00}R=3\left(\frac{\dot{a}}{a}\right)^2~?$$

Answer 1

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

Answer 2

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

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

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 $R$