StateOfTheEqn said:
I didn't actually say the critical density is constant. I said it varies as ##a^{-3}##.
Sorry, I missed this; still, it is not true in general that critical density "varies as ##a^{-3}##."
StateOfTheEqn said:
For the cosmological equation with ##k=0 ##, ##(\dot a/a)^2=8 \pi G \rho /3 ##. ... But in this case, ##\rho=\rho_{critical} ##.
Exactly. This gives that the general form for critical density is
$$\rho_{critical} = \frac{3}{8\pi G} \left(\frac{\dot{a}}{a} \right)^2 .$$
StateOfTheEqn said:
For the cosmological equation with ##k=0 ##, ##(\dot a/a)^2=8 \pi G \rho /3 ##. The solution of this differential equation is ##a=a_0 t^{2/3} ## if ##t=(6 \pi G \rho)^{-1/2} ##.
This is true for pressure-free matter (dust), but ##t=(6 \pi G \rho)^{-1/2}## is not true for radiation, or for a mixture of radiation and dust, or when the cosmological constant is non-zero, ... .
Assuming that a continuum approximation is valid on a large scale, matter in the universe is modeled by a perfect fluid with density ##\rho\left(t\right)## pressure ##p\left(t\right)##. An assumption of spatial homogeneity and spatial isotropy leads to an FLRW metric with scale factor ##a\left(t\right)##.
To solve for the three functions ##\rho\left(t\right)##, ##p\left(t\right)##, and ##a\left(t\right)##, three equations are needed. One equation is the equation that, above, you called "the cosmological equation", which follows from Einstein's equation with a FLRW used for the left side, and the stress-energy tensor for a perfect fluid used for the right side. Einstein's equation leads also to a second independent equation, but, here, it is more useful to use an equivalent independent equation,
$$\dot{\rho} + \left(\rho + \frac{p}{c^2}\right) \frac{3\dot{a}}{a} = 0.$$
The left side of EInstein's equation is divergence-free, and, consequently, so is the right. Taking the divergence of the stress-energy tensor for a perfect fluid leads to the second equation. The third equation is an equation of state that relates density and pressure. An often used equation of state is (with ##c## restored) ##p = w\rho c^2##, with ##w## constant. As I said in my first post, radiation has ##w = 1/3##.
Using the equation of state ##p = w\rho c^2## in the second equation gives ##\rho \propto a^{-3\left(1 + w\right)}##. Using this and ##w = 1/3## in the second equation leads, for radiation, to ##a \propto t^{1/2}## and ##\rho = \rho_{critical} \propto a^{-4}##.
Thus, for radiation (and in general)
George Jones said:
It is easy to show that the critical density evolves in time in such a way that ##\left( \rho +\rho_r \right) - \rho_{critical}## never changes sign.