Age of universe vs Hubble time in Milne universe

Question

Consider an empty universe where energy density $\varepsilon = 0$, thus the Friedmann Equation can be reduced into:

$\dot a^2= -\frac{kc^2}{R_O^2}$

$k$ is the curvature of space, $R_0$ is the radius of curvature. Here $k = -1$ or $k = 0$ since positive $k$ gives an imaginary value to $\dot a$.

Now we consider a universe where $k = -1$, also called a Milne universe. By Friedmann equation, if one integrate on both side, the result is:

$a(t) = \frac{c}{R_0}t$

Where $t$ is the time since the beginning of universe.

From here my textbook defines $t_0 = \frac{R_0}{c}$, so that $a(t) = \frac{t}{t_0}$

I have three questions:

1. What is $t_0$ actually referring here, is it the current age of universe or is it Hubble time of the universe.
2. If $t_0$ referring to the current time of universe, doesn't that mean the universe is expanding at the speed of light?
3. Why would this definition of $t_0 = \frac{R_0}{c}$ be legit?

Answer 1

At $t = t_0, \, a(t_0) =1$. Thus, $t_0$ is the time when the scale factor is unity. If you want to assume that the scale factor is one right now, then $t_0$ is today.

To measure expansion rate, I would consider looking at the change in proper distance $d$. Its rate of change can be written as $$ \dot{d}(t) = H(t) d(t) = \frac{d(t)}{t} \,$$

where the last RHS bit was calculated using the provided information. Thus we see that for some $d(t)$, the expansion rate will indeed become larger than $c$ but not for all $d$'s. You are using $R_0$ instead.