I'm looking at some solutions of the Friedmann equation at this website: Solutions to Friedmann equation
If I look at the first problem (Problem 1: matter or radiation in a flat Universe) and click on the "show" button, I get some solutions to some of this. Now, what I don't understand in this solution is when they get to this:
$$\frac{\dot{a}^{2}}{a^{2}}=\frac{8\pi G}{3}\cdot\frac{\rho_{r0}}{a^4}$$
Why have they done the substitution?: $$\rho_r=\rho_{r0}\left(\frac{a_0}{a}\right)^{4}$$ I know that the energy density, for radiation, is proportional to $a^{-4}$, but why do I have to include the zero version of $a$ and $\rho$?
Another thing is, where do they get the $\frac{32}{3}$ from in the square root? If I do the calculation (With $k=0$ and $\rho_r \propto a^{-4}$) I get (Please disregard the fact that I miss $\rho_{r0}$): \begin{align} \frac{\dot a^{2}}{a^{2}} &= \frac{8 \pi G}{3} \rho - \frac{kc^{2}}{a^{2}} \\ &\Updownarrow \\ \frac{\dot a^{2}}{a^{2}} &= \frac{8 \pi G}{3} \frac{1}{a^{4}} \\ &\Updownarrow \\ \left(\frac{da}{dt}\right)^{2} &= \frac{8 \pi G}{3} \frac{1}{a^{2}} \nonumber \\ &\Updownarrow \nonumber \\ \frac{da}{dt} &= \sqrt{\frac{8 \pi G}{3}} \frac{1}{a} \nonumber \\ &\Updownarrow \nonumber \\ da &= \sqrt{\frac{8 \pi G}{3}} \frac{1}{a} \,\, dt \nonumber \\ &\Updownarrow \nonumber \\ \int da &= \sqrt{\frac{8 \pi G}{3}} \frac{1}{a} \int dt \nonumber \\ &\Updownarrow \nonumber \\ a^{2} &= \sqrt{\frac{8 \pi G}{3}} t \nonumber \\ &\Updownarrow \nonumber \\ a &= \sqrt[4]{\frac{8 \pi G}{3}} t^{1/2} \nonumber \\ &\Updownarrow \nonumber \\ a &\propto t^{1/2} \end{align} Don't know if it is a flaw in my calculation, or I'm just missing something ordinary factor of some sort.
