in lectures we derived that the time for the deuterium bottleneck to pass (the time after which deuterium production is possible) is $t_D = (\frac{kT_D}{1MeV})^{-2} s$ (edited, used to lack s), where $kT_D \approx 0.05 MeV$ since only after the average temperature of the universe drops below value can deuterium be produced. This calculation gives about $t_D$ of about 200s or 3 minutes. However, I tried to calculate this time $t_D$ using the redshift integral equation:

$t_D = \frac{1}{H_0} \int_{z_D}^{\infty}\frac{dz}{(1+z)E(z)}$

Since at this time the universe was radiation dominated (RD), we can say $\Omega_r = 1$, $E(z)=\sqrt{\Omega_r(1+z)^4}=(1+z)^2$. Integrating I get:

$t_D = \frac{1}{H_0}\frac{1}{2(1+z_D)^2}$

I can relate this to temperature using the relation $T = T_0 (1+z)$, therefore:

$t_D = \frac{1}{2H_0}(\frac{T_0}{T})^2$

Using $T_0 = 2.73K$ and $H_0 = 100 h km/s/Mpc$, where I take $h=0.7$ I get that $t_D \approx 2.5s$ (edited, used to say 12000).

Why is there such a discrepancy between the answers?

Edit: I made a numerical mistake when in my final calculation for $t_D$, it should be about $t_D = 2.5$ seconds.

  • $\begingroup$ You are right, I made a calculator error when I calculated $t_D = 12000$, it should be about 5s like you say. My point still stands that this is a few orders of magnitude different from the 3 mins predicted using the first calculation. As for the ratio, my lecture notes use these types of ratios a lot and I am not quite used to them yet. There was a unit of $s$ missing from my question. I've made the edits. $\endgroup$
    – NX37B
    Apr 9 at 15:54


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