# Age of universe vs Hubble time in Milne universe

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?

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.

• My interpretation on universe expanding in light speed is this: since the size of universe is $R_0$ and the age of universe is $t_0$, from Friedmann equation it can be given that $c= \frac{R_0}{t_0}$, $\frac{R_0}{t_0}$ is the expansion rate of universe, which is light speed, where did I mess up? Commented Jun 9 at 23:42
• To measure expansion, I would consider looking at the change in proper distance $d$. Its rate of change can be written as $\dot{d}(t)=H(t) d(t) = d(t)/t$. For some $d(t)$, the expansion rate will become larger than $c$ but not for all $d$'s. You are using $R_0$ instead.
– S.G
Commented Jun 10 at 4:44
• Oh, I see. So I am using the wrong measurement here. Commented Jun 10 at 5:20
• Ok, I have modified my response to accommodate the above comment into it and also removed the bit that did not help you last time.
– S.G
Commented Jun 12 at 23:00