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 $$H = H_0\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}$$


$$ d_{hor}(t_0) = -\frac{c}{H_0}\int_{\infty}^{0}\frac{dz}{\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}}$$ 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 Answer 1


I offer several suggestions.

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.

3.I have used a spreadsheet to numerically calculate this integral with respect to a (not z) for a=0.1 to a=1. I used da = 0.01, da = 0.005, and da = 0.0025, and the differences were very small.

  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.

Hope this helps.


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