# Show that the angular diameter distance has always a turning point

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 ?

• (This is my idea) First you can find $H(z)$ in terms of $w$. And then you can use $$d_A = \frac{\chi}{1+z}$$ for $$\chi = c\int_0^z\frac{dz}{H(z)}$$ Dec 5, 2020 at 17:15
• Thanks for the reply. But how do I write $H(z)$ in terms of $\omega$ ? The definition of $H(z)$ is (en.wikipedia.org/wiki/Distance_measures_(cosmology)) : $$H(z)=H_0 \cdot E(z) = H_0 \cdot \sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_k (1+z)^2 + \Omega_\Lambda}$$. How to write the $\Omega$ in terms in $\omega$ ? Dec 6, 2020 at 17:38
• You do need to write to $\Omega$ in terms of the $w$. You need to start from the begining. Do you know about the fluid equations and obtain that equation for any parameter ? Dec 6, 2020 at 20:29
• The formula for $d_A$ that you found on wiki is valid for open/flat/closed models with only matter $(w=0)$. See Layla's answer for the model that you're looking for. Dec 6, 2020 at 21:58
• Pulsar really ? It's the Mattig formula in wiki it says that it's the formula for no cosmological constant but it does not say anything about radiation and matter. Why does an universe with only matter has $\omega = 0$ ? Dec 7, 2020 at 10:48

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 • For consistency, I'd recommend replacing your parameter $q$ with the deceleration parameter $q_0 = (1 + 3w)/2$. Dec 6, 2020 at 21:54
• I did not mean to use deceleration parameter..ıt was just a coincidence actually, Dec 6, 2020 at 22:28
• I know, but OP used it, and might get confused. Dec 6, 2020 at 22:38
• @Pulsar okay, thanks for the recommendation Dec 7, 2020 at 7:00
• Thank you this is very clear ! Where did you get the very first formula ? I have it neither in my lectures notes or in internet ( I haven't looked in book tho cause they are so many and our lecturer didn't give us one specifically) Dec 7, 2020 at 10:39