Empty Universe with Curvature

• A static flat universe ($\mathrm{H(t)}=0$ and $k=0$)
• A negatively curved expanding/contracting Universe

We will now focus on the negatively curved expanding/contracting Universe, since we know the Hubble parameter is defined as: ${H=\frac{\dot{a}}{a}}^{}$ We obtain this equation: $\frac{da}{adt}=\pm \frac{c}{R_0}$
Solving this differential equation, we obtain the relation between scale factor and time (which is the ultimate goal of cosmology): $a\mathrm{(t)}=\frac{t}{t_0}$
Where $t_0=R_0/c$
This indicates that the universe expand linearly, in Newtonian terms, since there is nothing in the Universe to slow its expansion down, the rate of expansion will not change.
At first glance, this type of Universe sounds like only a math game, there is no possible application (since it is an empty Universe, nothing can be done inside), but if we take a step back and notice that if the density parameter is reasonable low (low enough to neglect its exsistance), we can apply this model to the Universe we want to describe and simplify calculations.

Matter Dominated Flat Universe

In this Universe, we can write the Friedmann Equation as: $H^2=\frac{8\pi G}{3}{\rho }_m$ Introducing the density parameter into this equation, we get: $\frac{H^2}{{H_0}^2}=\frac{{\Omega }_m\mathrm{(t)}}{{\Omega }_{\mathrm{m0}}}$

Also, it is trivial that the matter density is propotional to $a^{-3}$ (basically, the total matter of the Universe is conserved, and volume is propotional to $a^{-3}$) If we write ${\Omega }_m\mathrm{(a)}={\Omega }_{\mathrm{m0}}{a}^{-3}$ (Here density parameter is a function of scale factor, however scale factor is related to time, thus density parameter can also be expressed as a function of time, which is exactly what we will be doing now)
Then the rate of expansion can be written as: $\frac{da}{adt}=H_0\sqrt{{\Omega }_{\mathrm{m0}}{a}^{-3}}$
We take an educated guess that scale factor has a power law form: $a\propto t^q$
Thus on the left hand side of the equation, we have: $\frac{da}{adt}\propto t^{\mathrm{-1}}$
On the right hand side of the equation, we have: $\sqrt{a^{-3}}\propto t^{-\frac{3}{2}q}$

This indicates that in a matter dominated Universe, the expansion of the Universe follows: $a\propto t^{\frac{2}{3}}$
We can also write this equation like this: $a={\left(\frac{t}{t_0}\right)}^{\frac{2}{3}}$ If we rearrage the first equation, we will be able to obtain (remember that ${\Omega }_{\mathrm{m0}}$):
$dt=\frac{1}{H_0}{a}^{\frac{1}{2}}da$
If we integrate time from the start of Universe to time right now, we should integrate scale factor from 0 to 1 (remember that the present day value of scale factor is 1), solving the integral, we obtain the present age of the Universe in terms of Hubble constant: $t_0=\frac{2}{3H_0}$

Radiation Dominated Universe

Radiation behaves differently from matter, first, it is trivial that radiation can spread to all direction as the Universe expand, second, the wavelength of radiation is also affected by the expansion of the Universe, when the the Universe expand, the wavelength get streched, causing density to further decrease, to sum all the effects up, the density parameter of radiation is proportial to: $a^{-4}$ (3 from the spatial three dimension, 1 from the sterch of wavelength)

Just like before, we can write out the Friedmann Equation in a radiation dominated Universe: $\frac{da}{adt}=H_0\sqrt{{\Omega }_{\mathrm{r0}}{a}^{-4}}$
Repeating the exact same process, we shall see that: $a={\left(\frac{t}{t_0}\right)}^{\frac{1}{2}}$
The present age of Universe is: $t_0=\frac{1}{2H_0}$