Show that the angular diameter distance has always a turning point

Question

I have the following cosmology exercise:

assume a flat Universe with a single component with an equation of state: \begin{equation} p = \omega \rho, \quad -1<\omega<1, \quad \omega = const. \end{equation} i) show that the angular diameter distance \begin{equation} d_A(z) = (1+z)^{-2} d_L(z) \end{equation} has always a turning point for $\omega >-1$.

My idea was to write the angular diameter distance as follow https://en.wikipedia.org/wiki/Angular_diameter_distance:

\begin{equation} d_A(z) = \frac{c}{H_0 q_0^2} \frac{zq_0 + (q_0-1)(\sqrt{2q_0+1} -1)}{(1+z)^2} \end{equation} and then replace for the deceleration parameter $q_0$ (choose for example $\Omega_m$ as the single component) (https://en.wikipedia.org/wiki/Deceleration_parameter)

\begin{eqnarray} q_0 &=& \frac{1}{2} \sum \Omega_i (1 + 3 \omega_i) \\ &=& \frac{1}{2} \Omega_m(1+3\omega_m) \end{eqnarray}.

Which would give:

\begin{equation} d_A(z) = \frac{c}{H_0 (\frac{1}{2} \Omega_m(1+3\omega_m))^2} \frac{z(\frac{1}{2} \Omega_m(1+3\omega_m)) + (\frac{1}{2} \Omega_m(1+3\omega_m)-1)(\sqrt{2 \cdot \frac{1}{2} \Omega_m(1+3\omega_m)+1} -1)}{(1+z)^2} \end{equation} where $\Omega_m = 1$ because it's a single component universe (is that correct?):

\begin{equation} d_A(z) = \frac{c}{H_0 (\frac{1}{2} 1(1+3\omega_m))^2} \frac{z(\frac{1}{2} 1(1+3\omega_m)) + (\frac{1}{2} 1(1+3\omega_m)-1)(\sqrt{2 \cdot \frac{1}{2} 1(1+3\omega_m)+1} -1)}{(1+z)^2} \end{equation}

and then derive this according to $z$ and replace the derivative with zero to find the solution. However I am not sure this is the right idea, The derivative will get really ugly, and I don't understand how the $\omega>-1$ conditions is relevant... And I don't know how to use the equation of state for ma calculation...Could anyone help me ?

Answer 1

In most general way, the equation can be written as

\begin{equation} \frac{H^2(z)}{H_0^2} = \sum_i \Omega_{i,0}~\text{exp}({3\int_0^z\frac{1+w_{i}(z)}{1+z}dz}) + \Omega_{k,0}a^{-2} \end{equation}

Let me call that fluid $g$ in this case you can write

\begin{equation} \frac{H^2(z)}{H_0^2} = \Omega_{\rm g,0}~e^{3(1+w_{\rm g})ln(1+z)} \end{equation}

\begin{equation} \frac{H^2(z)}{H_0^2} = \Omega_{\rm g,0}~(1+z)^{3(1+w_{\rm g})} \end{equation}

Thus

$$H(z) = H_0\sqrt{\Omega_{\rm g,0}~(1+z)^{3(1+w_{\rm g})}}$$

Thus

$$\chi = c\int_0^z \frac{dz}{H_0\sqrt{\Omega_{\rm g,0}~(1+z)^{3(1+w_{\rm g})}}}$$

$$\chi = \frac{c}{H_0}[\frac{2(1+z)^{(-3w-1)/2}}{-3w-1} - \frac{2}{-3w-1}]$$

$$\chi = \frac{c}{H_0}[m(1+z)^{1/m} - m]$$ for $m = \frac{2}{-3w-1}$

so

$$d_A = \frac{cH_0}{1+z}[m(1+z)^{1/m} - m]$$

From here I have plotted some graphs. Maybe you can take the derivative and do some mathematical steps from now on

Here the graphs for various results