# Time of deuterium bottleneck

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.

• 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. Apr 9 at 15:54