Help in solving a cosmology exercise and explanation of a concept

Question

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?

Answer 1

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)

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

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

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