# How can I compute the scale factor for an open universe?

I'm struggling in obtaining an analytical expression for an open-universe $$a(t)$$. I know it is usually calculated from the second Friedmann equation, and with respect to the image I know how to compute it for a flat universe i.e. $$a(t)=(t/t_0)^{2/3}$$, for a $$\Lambda$$CDM universe and for the empty one is just linear in time (Milne, which is open but is also empty).

The more particular question is: why is the empty model (which should also be open) different from the open model in the image? Shouldn't a open universe evolve with a Milne scale factor ($$\propto t$$)?

The situation of a geometrically open universe (i.e. one with $$k=-1$$) is dealt with from p.24 of these lecture notes.

Fora "matter-dominated" universe (i.e. one where the energy density of matter is dominant, $$\Lambda=0$$), which appears to be what your sketch illustrates then we can use $$a^3 \rho = a_0^3 \rho_0 = \rho_0$$

The first of Friedmann's equations is $$\frac{\dot{a}^2 + kc^2}{a^2} = \frac{8\pi G\rho + \Lambda c^2}{3},$$ which becomes (using units where $$c=1$$) $$\frac{\dot{a}^2 - 1}{a^2} = \frac{8\pi G\rho}{3}$$ $$\frac{\dot{a}^2}{a^2} = \frac{8\pi G\rho}{3} + \frac{1}{a^2}$$ $$\dot{a} = \sqrt{\frac{8\pi G \rho_0 a_0^3}{3a} +1}$$

We can then make a change of variable to the conformal time, defined by $$\eta = \int^{t}_{0} \frac{dt'}{a(t')}$$ so that $$d\eta = dt/a$$ and $$\dot{a} = a^{-1} da/d\eta$$.

Thus we can write $$\frac{d\eta}{da} = \sqrt{\frac{1}{8\pi G \rho_0 a_0^3 a/3 +a^2}}$$

The solution to this differential equation is $$\eta = \cosh^{-1} \left(\frac{3a}{4\pi G \rho_0} + 1\right) + \eta_0 ,$$ but $$\eta = 0$$ when $$a=0$$, so $$\eta_0 = -\cosh^{-1}(1) = 0$$.

Finally then $$a = \frac{4\pi G \rho_0}{3} \left( \cosh \eta -1 \right),$$ where $$\eta$$ is the conformal time as described above.

Using $$a d\eta = dt$$, it can also be shown that $$t = \frac{4\pi G \rho_0}{3} \left( \sinh \eta - \eta \right)\ .$$

If $$\eta$$ is very large (i.e. at late times), then $$\cosh \eta\simeq \sinh \eta \simeq \exp(\eta)/2$$ and we can see that the behaviour is asymptotic to $$a = t$$.

Likewise, if the open universe is empty, with $$\rho=0$$ (and $$\Lambda=0$$) at all times, then the Friedmann equation becomes just $$\dot{a} = 1$$ and $$a = t$$.

• Thank you a lot, you have been very clear. – chiara iannetta Jul 23 '19 at 9:40