# Help in solving a cosmology exercise and explanation of a concept

I'm trying to solve an exercise which should be easy but I'm very rusty and I don't really know how to do it.

The exercise is the first one in this link, I have problems with the point 3 of this exercise. I'll present here a short summary of what one need to know to help me, and my attempt

Consider the following Friedmann and continuity equation

$$H^2=8 \pi G \rho_T -\frac{k}{a^2}$$ $$\dot{\rho}_T= - 3 H (\rho_T+p_T)$$

The possibility for the equation of state are 2:

matter: $$p_m=0$$

some fluid with negative pressure: $$p_s=\gamma\rho_s$$ with $$-1\leq \gamma \leq -1/3$$

After asking to find the evolution of the energy densities in the case only one of them was present, and after asking to derive the equation for $$\ddot{a}$$ (and both can be used for the next points) the exercise says

Having defined $$\Omega_i(a)=\rho_i(a)/\rho_{cr}(a)$$, $$\rho_{cr}=3H^2(a)/(8\pi G)$$ for $$i=(m,s)$$ and $$1 + z = 1/a$$, find an expression for the parameter $$\Omega_{m,0}=\Omega_m(a=1)$$ as a function of the two following quantities only:

1) $$z_\star$$, the redshift at which the condition $$\dot{a} = \ddot{a}=0$$ is satisfied (the relation between redshift and scale factor is $$1 + z = 1/a$$);

2) $$n = 3 + 3\gamma$$.

My (wrong) attempt: from the Friedmann equation and from the fact that for a general equation of state $$p=w \rho$$ we have $$\rho \propto a^{-3(1+w)}$$ I can obtain

$$\Omega_{m,0} = (1+z)^3(1-\Omega_k) - (1+z)^{3-n} \Omega_{s,0}$$

only as a function of $$n$$ and not as a function of $$z_\star$$

So now I have basically 2 questions:

1) Can you help me solve the exercise?

2) Simultaneously with (1) can you explain a bit the physical meaning of $$z_\star$$ since I have never encoutered, I mean it is a redshift at which the Hubble factor is zero (since $$\dot{a}=0$$), what does it mean?

• Can you please link/quote where you found this exercise? Commented Mar 4, 2020 at 1:16
• @magma It's an old exercise in an admission test for the PhD in trieste. I have it on my pc I think I can find it from their site Commented Mar 4, 2020 at 15:32
• Should we assume $\kappa=0$ ? Commented Mar 5, 2020 at 10:20
• I am thinking that we can use $q=(\Omega_i(1+3w_i))/2$. But I am not exactly sure what should we get. $\dot{a}$ means that the expansion of the umiverse pauses at that instant. Commented Mar 5, 2020 at 10:21
• @Reign Hi, I edited the question with a link to the full exercise Commented Mar 5, 2020 at 21:06

Let me share my solution.

So by using Friedmann equation we can write,

$$H^2 = \frac{8\pi G}{3}(\rho_m + \rho_s) - \frac{k}{a^2}$$

since we are looking for $$a_{*}$$ such that, $$\dot{a}=\ddot{a}=0$$, we can set $$H=0$$.

$$\frac{8\pi G}{3}(\rho_m + \rho_s) = \frac{k}{a_{*}^2}$$

or

$$\frac{8\pi G}{3}(\rho_{m,0}a_{*}^{-3} + \rho_{s,0}a_{*}^{-3(1+\gamma)}) = \frac{k}{a_{*}^2}$$

Let us divide it by $$\rho_{crit,0}=\frac{3H_0^2}{8\pi G}$$

$$\frac{8\pi G}{3}(\Omega_{m,0}a_{*}^{-3} + \Omega_{s,0}a_{*}^{-3(1+\gamma)}) = \frac{k8\pi G}{a_{*}^23H_0^2}$$

Hence,

$$(\Omega_{m,0}a_{*}^{-3} + \Omega_{s,0}a_{*}^{-3(1+\gamma)}) = \frac{k}{a_{*}^2H_0^2}~~(1)$$

Now let us look the acceleration equation

$$\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}(\rho_m + \rho_s(1+3\gamma))$$

Set $$\ddot{a}=0$$ for $$a_{*}$$. So we can write

$$\rho_m = -\rho_s(1+3\gamma)$$

or

$$\Omega_{m,0}a_{*}^{-3}=-\Omega_{s,0}a_{*}^{-3(1+\gamma)}(1+3\gamma)$$ $$\Omega_{m,0}=-\Omega_{s,0}a_{*}^{-3\gamma}(1+3\gamma)~~(2)$$

Inserting (2) in (1)

$$\Omega_{m,0}a_{*}^{-3}-\frac{\Omega_{m,0}}{(1+3\gamma)a_{*}^{-3\gamma}}a_{*}^{-3(1+\gamma)} = \frac{k}{a_{*}^2H_0^2}$$

$$\Omega_{m,0}a_{*}^{-3}[1-\frac{1}{1+3\gamma}] = \frac{k}{a_{*}^2H_0^2}$$

$$\Omega_{m,0} = a_{*}\frac{k}{H_0}\frac{1+3\gamma}{3\gamma}$$ $$\Omega_{m,0} = (1+z_{*})^{-1}\frac{k}{H_0}\frac{n-2}{n-3}$$

• Hi, sorry for my late answer. I think this solution is ok and I'd check it. But can you expand a bit on the meaning of $dot{a}=\ddot{a}=0$? Commented Mar 9, 2020 at 15:33
• @AnOrAn Hey, hmm I mean what kind of meaning are you referring to ? It represents that the expansion of the universe stops. I am not sure what else we can say Commented Mar 9, 2020 at 18:37