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Using the Friedmann metric I've been trying to calculate how long the universe was radiation energy dominated. I've reduced the metric to:

$c.dt$ = $a(t).dr$

where a is the scale factor. I can work out the this since I know the redshift then was ~3000. But how do get rid of the dr and get an equation that relates $dt$ to $a(t)$ ?

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My answer to How does the Hubble parameter change with the age of the universe? (which is itself adapted from Equation for Hubble Value as a function of time) explains how to calculate the scale factor. In fact we calculate the time as a function of the scale factor rather than the other way around. The equation we use is:

$$ t(a) = \frac{1}{H_0}\int_0^a \frac{a'\,\text{d}a'}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a' + \Omega_{K,0}\,a'^2 + \Omega_{\Lambda,0}\,a'^4}} $$

where $H_0$ is the current value of the Hubble parameter, and the $\Omega$s are the current fraction of the critical mass.

If we consider a flat universe then the curvature contibution is zero, $\Omega_{K,0} = 0$, and if we're interested in the handover between radiation and matter domination we working in a regime where dark energy can be ignored. That means our equation simplies to:

$$ t(a) = \frac{1}{H_0}\int_0^a \frac{a'\,\text{d}a'}{\sqrt{\Omega_{R,0} + \Omega_{M,0}\,a'}} $$

The expansion is dominated by radiation when $\Omega_{R,0} \gg \Omega_{M,0}\,a$ and dominated by matter when $\Omega_{R,0} \ll \Omega_{M,0}\,a$, so we can conveniently define the handover point when $\Omega_{R,0} = \Omega_{M,0}\,a$. So this happens when:

$$ a(t) = \frac{\Omega_{R,0}}{\Omega_{M,0}} $$

From the Planck experiment we have:

$$\begin{align} \Omega_{R,0} &= 9.24\times 10^{-5}\\ \Omega_{M,0} &= 0.315 \end{align}$$

So the crossover happened when $a(t) \approx 0.00029$. Using the spreadsheet linked in my answer I calculate this happened at about $t = 50000$ years.

Wikipedia says it happened at about 47000 years, which seems close enough to my answer given the vagueness of the term dominated.

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  • $\begingroup$ Okay I see where the derivation of the first equation comes from and how to get too the second equation but do you actually integrate the equation at all ? and substitute the value of $a(t)$ in ? $\endgroup$ – AlwaysStuck Nov 27 '15 at 17:32
  • $\begingroup$ @AlwaysStuck: the integral doesn't have a simple closed form so I did the integration numerically. The spreadsheet I used is here. It'll be read only for you, but you can take a copy and play with changing the parameters. $\endgroup$ – John Rennie Nov 27 '15 at 17:36

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