Parameterizing a universe with non-zero curvature, some matter, no dark matter, and no dark energy

Question

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 $\theta$.

Answer 1

"What is the meaning of the Ω parameters? What do they sum up to?"

Ω_r is the current density (a=1) of radiation

Ω_m is the current density (a=1) of matter

Ω_Λ is a cosmological constant, also referred to as the dark energy density Ω_r+ Ω_m + Ω_Λ = 1

Ω_k is the current density (a=1) of curvature

If you assume Ω_r = Ω_Λ = 0, then Ω_m = 1, and Ω_k can be any real number.

Ω_k = 0 means the universe is flat.

Assuming the universe is isotropic and homogeneous, then Ω_k < 0 means the universe is finite and 3D-hyper-sperical, and assuming Ω_k > 0 means the universe is infinite and 3D-hyperbolic.

I also suggest you note that the assumptions Ω_r = Ω_Λ = 0 and Ω_k < 0 implies:

H^2 = (a˙/ a)^ 2 = (H_0)^2 x ((1/a^3) + ((Ω_k)/a^2)).

If we define B = -Ω_k, simplifying gives:

(da/dt)^2 = (H_0)^2 x (1/a - B).

I am sorry that this is as far as I can help you.

Answer 2

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: $t = \frac{2}{3H_0}$.