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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 ($\propto t$)?

enter image description here

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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 $a = t$.

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

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