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I'm attempting to work through the third chapter of Dodelson's Modern Cosmology. He gives an approximation for the neutron-proton conversion rate, $\lambda_{\rm np}$ that he lifted from Bernstein, but doesn't give any of the science behind it.

Digging further, I found a paper that derives the rate by first calculating the number densities for neutrinos and positrons:$$n_{\nu_e}=\frac{3\zeta(3)}{4\pi^2}T^3;\quad n_{e^+}=\frac{3\zeta(3)}{2\pi^2}T^3$$ where $\zeta$ is the Riemann zeta function and $\zeta(3)$ is 1.20. This paper then gives examples at $1\,\rm MeV$:$$n_{\nu_e}=1.2\times 10^{31}\, {\rm cm}^{-3};\quad n_{e^+}=2.4\times 10^{31}\, {\rm cm}^{-3}$$ But I can't get the formulas to work. Plugging in $T=1\,{\rm MeV}=1.16\times 10^{10}\,{\rm K}$, I get $$n_{\nu_e}=1.4\times 10^{29}\, {\rm cm}^{-3};\quad n_{e^+}=2.86\times 10^{29}\, {\rm cm}^{-3}.$$ They're off by more than a factor of 100. Are these formulas right? How do I produce the correct number densities of these particles as a function of temperature?

Quark Soup
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  • The units don't seem to be correct. Temperature does not have the dimensions of inverse length. There must be factors of $k_B$ if you use Kelvin and $\hbar$'s and $c$'s are needed somewhere as well. – mike stone Jan 01 '22 at 14:28
  • I agree. Still trying to work this out. 'Natural Units' are the bane of my existence. Even if you wanted to be lazy and select some unit where the speed of light is 1, you should still include 'c' (or some placeholder) in the formula for the dimensional analysis. – Quark Soup Jan 01 '22 at 16:46
  • I recommend you to check out the book of Kolb-Turner : Early Universe, i think they also discuss these things there. At least the lecturer of mine was going trough that book when I studied the BBN. – Kregnach Jan 01 '22 at 22:48

1 Answers1

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You need to include a factor of $(\hbar c)^{-3}$ (that has been set to unity in the paper) and then measure the temperature in Joules.

Notes added after the remark abut units: I think my units are correct: $\hbar c= 3.16152677...×10^{−26}$ J⋅m so $kT/\hbar c$ has units of inverse meters.

Including a factor of 2 for the two spin directions and using $p=\hbar k$, we have in detail $$ N/V=n= 2\times \int \frac{d^3 k}{(2\pi)^3} \frac{1}{e^{|p|c/kT}+1}\\ = 2 \frac 1{(2\pi \hbar )^3} \int_0^\infty \frac{4\pi p^2dp} {e^{|p|c/kT}+1}\\ = 2 \frac {(kT)^3}{(2\pi \hbar c )^3} 4\pi \int_0^\infty \frac{x^2 dx}{e^{x}+1}. $$ I havent worked out the integral for a while, but I think that it is proprtional to $(7/8)\zeta(3)$.

mike stone
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    Indeed. But then you get the density in $m^{-3}$, not $cm^{-3}$, so you have to divide by $10^6$ to compare to the values quoted in the paper. – Alfred Jan 01 '22 at 18:28
  • Got it working, thanks! – Quark Soup Jan 01 '22 at 19:37
  • Alright, this checks out, except your final expression has an extra factor of 2. It should be $$\frac {(kT)^3}{(2\pi \hbar c )^3} 4\pi \int_0^\infty \frac{x^2 dx}{e^{x}+1}$$ – Quark Soup Jan 01 '22 at 22:38
  • @Gluon Soup. Is my factor of 2 for spin up/spin down not correct? – mike stone Jan 01 '22 at 22:41
  • I can't say whether your last formula is correct or not, but it doesn't agree with the paper I cited and once I got the density formulas working, they appear to be giving the right neutron-proton reaction rates. – Quark Soup Jan 01 '22 at 23:58