# Angular size of the largest region in causal contact at the moment of recombination

I want to calculate the angular size of the largest region in causal contact at the moment of recombination (z = 1100) to our universe, so we have: $$H_0 = 68$$ (km/s)/Mpc, $$\Omega_{m,0} = 0.31$$, $$\Omega_{\Lambda,0} = 0.69$$ and $$\Omega_{r,0}=9\times 10^{-5}$$. I know that:

$$\theta = \frac{d_{hor}(t_{ls})}{d_A}$$ where $$d_{hor}(t_{ls})$$ is the physical size of the largest region at the recombination and $$d_A$$ is the angular diameter distance.

My attempt: To find the distance $$d_{A}$$ at z = 1100, I can calculate the horizon distance today than use that $$d_A = \frac{d_{hor}(t_0)}{1+z}$$ To calculate this we note that:

$$d_{hor}(t_0) = c\int_{t_e}^{t_o}\frac{dt}{a(t)}$$ But you can write this equation in terms of z and (making limits from $$\infty$$ to $$0$$) to make calculations easier. Using that $$1 + z = a(t)^{-1}$$ we have:

$$\frac{dz}{dt} = \frac{dz} {da} \frac{da} {dt}$$ $$\frac{dz}{dt} = -\frac{1}{a^2} \dot{a}$$ $$dz = -\frac{H}{a}dt = -H(1+z)dt$$ $$dt = -\frac{dz}{H(1+z)}$$

Since at $$t_e$$ corresponds to $$z = \infty$$ and $$t_0 = 0$$ we have

$$d_{hor}(t_0) = c\int_{t_e}^{t_o}\frac{dt}{a(t)} = -c\int_{\infty}^{0}\frac{dz}{H}$$

Also we know using Friedmann equation that $$H = H_0\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}$$

So:

$$d_{hor}(t_0) = -\frac{c}{H_0}\int_{\infty}^{0}\frac{dz}{\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}}$$ Using the above given data, I can do this integral numerically and find that:

$$d_{hor} \approx 14121\; Mpc$$ and $$d_A = 14121/1101 \approx 12.943 \; Mpc$$ Until this point I think I'm right. Now the hard and dubious part:

To calculate $$d_{hor}(t_{ls})$$ I again tought about using the formula: $$d_{hor}(t_ls) = -\frac{c}{H_0}\int_{1100}^{0}\frac{dz}{\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}}$$ Since at $$t_e$$ corresponds to $$z = 1100$$ and $$t_0 = 0$$, but the result (13842 Mpc) here is very different from the one I expected (0.251 Mpc). Can someone give-me a help to calculate $$d_{hor}(t_{ls})$$?

1.For $$z=1100$$, $$Ω_r << 1$$, so it can be ignored.
2.For $$z>=9$$, $$Ω_Λ << Ω_m/(1+z)^3$$, so it can be ignored in this range.
1. When you integrate between z = 9 and z = 1100, the only integrand left is $$1/(Ω_m(1+z))^{3/2}$$, which is easily integrated.