<s>[INST] Parameterizing a universe with non-zero curvature, some matter, no dark matter, and no dark energy Defining $\Omega_i$ by $\rho_i (t_0) = \Omega_i \rho_{c_0}$, we can obtain the below equality.
$$H^2 = H_0^2 \left(\frac{\Omega_r}{a^4} + \frac{\Omega_m}{a^3}+\frac{\Omega_k}{a^2} + \Omega_\Lambda\right)$$
What is the meaning of the $\Omega$ parameters? What do they sum up to?
Each of the parameters are used as coefficients to represent each of the partial densities in terms of the critical density of the universe. The sum is equal to $1$, which implies that we live in a flat universe. The density parameter $\Omega$ can be defined as the ratio of actual or observed density $\rho$ to the critical density $\rho_c$ of Friedmann's universe.
Let's set $\Omega_r=\Omega_\Lambda=0$, so we have non-zero curvature and some matter. Show that when $\Omega_k<0$ the solutions can be written in parametric form as
$$ 
\begin{align}
t (\theta) &= A \left(\sinh\theta - \theta\right) \\
a (\theta) &= B \left(\cosh \theta - 1 \right)
\end{align}
$$
I understand that using Friedmann's equations,
$$
    H^2 = \left( \frac{\dot{a}}{a}\right)^2 = \frac{8\pi G}{3}\rho - \frac{k}{a^2} + \frac{\Lambda}{3}
$$
might be helpful. I derived that for a normal Einstein-de Sitter universe $$\frac{\dot{a}^2}{a^2} = \frac{8\pi G\rho_0}{3a^3} \implies \dot{a}^2a = \frac{8\pi G \rho_0}{3} \implies \text{, with $\dot{a} = \frac{da}{dt}$, } \int \sqrt{a} da = \int \sqrt[3]{\frac{8\pi G \rho_0}{3}} dt \implies \frac{2a^{\frac{3}{2}}}{3} = 2t \sqrt[3]{\frac{\pi G \rho_0}{3}} \implies a(t) = t^\frac{2}{3} \sqrt[9]{81 \pi^2 G^2 \rho_0^2} \iff a(t) \propto t^{\frac{2}{3}} \implies H_0 = \frac{\dot{a}}{a}|t_0 = \frac{2t_0^{-\frac{1}{3}} }{3t_0^{\frac{2}{3}}} = \frac{2}{3t_0} \implies t_0 = \frac{2}{3H_0}$$
However, I don't understand how to parameterize the time and scale factor themselves as functions of $\theta$.
 [/INST] The Friedmann equation for these models can be written
$$ \dot{a}^2 = H_0^2(\frac{\Omega_{m}}{a} + 1 - \Omega_{m}) $$
For a universe with both matter and nonzero curvature, we have
$$ \frac{\dot{a}^2}{a^2} = H_0^2(\Omega_m a^{-3} + \Omega_k a^{-2}) \implies (\frac{da}{dt})^2 = H_0^2(\Omega_m a^{-1} + \Omega_k) $$
Therefore,
$$ H_0dt = \frac{da}{\sqrt{\Omega_ma^{-1} + \Omega_k}} = \frac{1}{\sqrt{\Omega_m}} \frac{a^{\frac{1}{2}}da}{\sqrt{1+a(\frac{\Omega_k}{\Omega_m})} } $$
It turns out that it is easier to first solve for the conformal time $d\eta = \frac{dt}{a}$. We have
$$ \eta = \int d \eta = \int \frac{dt}{a} = \frac{1}{H_0\sqrt{\Omega_m}} \int \frac{a^{-\frac{1}{2}} da}{\sqrt{1+a(\frac{\Omega_k}{\Omega_m})}} $$
Here begins our solution for positive curvature, i.e. $k > 0$ and therefore $\Omega_k = \frac{-k}{H_0^2} < 0$. Then, let $u^2 = \frac{-\Omega_k}{\Omega_m a}$, so that $u = \sqrt{\frac{-\Omega_k}{\Omega_m}}a^{\frac{1}{2}}$ and $du =  \frac{1}{2}\sqrt{\frac{-\Omega_k}{\Omega_m}}a^{-\frac{1}{2}}$.
We have
$$ \eta = \frac{2}{H_0\sqrt{\Omega_m}}\sqrt{\frac{\Omega_m}{-\Omega_k}} \int \frac{du}{\sqrt{1 - u^2}} = \frac{2}{H_0\sqrt{-k}} \sin^{-1} u $$
Inverting
$$ u = \sin\frac{\theta}{2}, \; \theta = H_0 \eta \sqrt{-\Omega_k} $$
under the same change of variables $a \to u$, $u^2 du = \frac{1}{2} \frac{-\Omega_k}{\Omega_m}^\frac{3}{2} a^\frac{1}{2} da$, the equation for $t$ becomes
$$ \frac{1}{H_0\sqrt{\Omega_m}} \int \frac{a^\frac{1}{2} da}{\sqrt{1+a(\frac{\Omega_k}{\Omega_m})}} = \frac{2}{H_0\sqrt{\Omega_m}} (\frac{\Omega_m}{-\Omega_k})^\frac{3}{2} \int \frac{u^2du}{\sqrt{1 - u^2}} $$
Now, changing $u = \sin \frac{\theta}{2}, du = \frac{1}{2} \cos \frac{\theta}{2} d\theta$, we have
$$ t = \frac{2 \Omega_m}{H_0(-\Omega_k)^\frac{3}{2}} \int \frac{\sin^2 \frac{\theta}{2} \cos \frac{\theta}{2} d \theta}{2\sqrt{1 - \sin^2 \frac{\theta}{2}}} = \frac{2 \Omega_m}{H_0(-\Omega_k)^\frac{3}{2}} \int \sin^2 \frac{\theta}{2} d\theta $$
Now, using $\cos \theta = \cos^2 \frac{\theta}{2} - \sin^2 \frac{\theta}{2} = 1 - \sin^2 \frac{\theta}{2}$, we find
$$ t = \frac{\Omega_m}{2H_0(-\Omega_m)^\frac{3}{2}} \int [1 - \cos \theta] d\theta = \frac{\Omega_m}{2H_0(-\Omega_k)^\frac{3}{2}}(\theta - \sin \theta) $$
Finally, recall that $a = -\frac{\Omega_m}{\Omega_k}u^2 = -\frac{\Omega_m}{\Omega_k}\sin^2 \frac{\theta}{2}$ so that we have a parametric solution for a cycloid, with $\theta = H_0 \sqrt{-\Omega_k}\eta$,
$$ t(\theta) = \frac{\Omega_m}{2H_0 (-\Omega_k)^\frac{3}{2}}(\theta - \sin \theta), a(\theta) = \frac{\Omega_m}{2H_0(-\Omega_k)^{\frac{3}{2}}}(1 - \cos \theta ), \textbf{ for positive curvature.} $$
Similarly, using $\theta$ as a dummy variable to parameterize this a negatively curved universe, we can make a substitution $Q$, where $Q = \frac{\sinh^2\frac{\theta}{2}}{a} = \frac{1 - \Omega_{m}}{\Omega_{m}}$, to obtain:
$$ t(\theta) = \frac{\Omega_{m}}{2H_0(-\Omega_k)^\frac{3}{2}}(\sinh \theta - \theta), a(\theta) = \frac{\Omega_{m}}{2(-\Omega_m)}(\cosh \theta - 1), \textbf{ for negative curvature.} $$
Take solutions from previous exercise and use perturbative expansion when $|\Omega_k|\ll 1$. How does the age of the universe vary with $H_0$ and $\Omega_k$, when $\Omega_k$ is small?
For the universe with a negative curvature, the time equation is as follows, knowing that $\Omega_m + \Omega_k = 1$:
$$ t(\theta) = \frac{1-\Omega_k}{2H_0(-\Omega_{k})^\frac{3}{2}}(\sinh \theta - \theta) $$
Since $\Omega_k = \frac{-k}{H_0}$, $\Omega_k < 0$ corresponds to a closed universe, where $k > 0$, which reaches a maximum turnaround scale factor $a_{ta} = \frac{\Omega_m}{-\Omega_k}$ at time $t_{ta}$, so $H_0 t_{ta} = \frac{\pi}{2}(\frac{\Omega_m}{-\Omega_k^{\frac{3}{2}}})$.
In similar fashion, for the universe with a positive curvature, the time equation is as follows, again knowing that $\Omega_m + \Omega_k = 1$:
$$ t(\theta) = \frac{1 - \Omega_k}{2H_0 (-\Omega_k)^\frac{3}{2}}(\theta - \sin \theta) $$
In both cases, as $\Omega_k \to 0$, both $t_{ta}, a_{ta} \to \infty$ and the solution approaches that of a flat universe without turnaround, i.e. $a(t) = (\frac{3}{2} \sqrt{\Omega_m} H_0 t)^{\frac{2}{3}}$, an asymptotically flat universe with an exceedingly large age.
$\textbf{Correction}$: As $\Omega_k \to 0$, the universe becomes primarily matter-dominated, implying that it asymptotically reaches a flat universe with the solution to elapsed time given by that of the Einstein-de Sitter model of the universe: $t = \frac{2}{3H_0}$.
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