0
$\begingroup$

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 $d_{hor}(t_{ls})$?

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.