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I'm self-studying Modern Cosmology by Dodelson and Schmidt 2nd edition, and I've found a certain passage in Section 4.2.1 particularly interesting. It says,

Therefore, at high temperatures, there are as many neutrons as protons. As the temperature drops beneath 1 MeV, the neutron fraction goes down. If weak interactions operated efficiently enough to maintain equilibrium indefinitely, then it would drop to zero (even if free neutrons were stable).

Can anyone explain this part more thoroughly? Why would the weak interactions cause the neutron fraction to go to zero if the weak interaction could maintain equilibrium?

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The mention of the weak force is something of a red herring as any interaction that allows protons and neutrons to interconvert would give the same result.

Earlier in the chapter the authors describe how to calculate the relative number densities of two components that are in equilibrium. For neutrons and protons they assume that these can be treated as a gas of free neutrons and free protons, and they derive the equation 4.17:

$$ \frac{n_p^{(0)}}{n_n^{(0)}} = \frac{ e^{-m_p/T} \int dp ~ p^2 e^{-p^2/2m_pT}}{e^{-m_n/T} \int dp ~ p^2 e^{-p^2/2m_nT}} \tag{4.17} $$

And since the masses of the neutron and proton are very similar the integrals are very similar and can be set equal to a good approximation. Then we get equation 4.18 for the proton-neutron ratio:

$$ \frac{n_p^{(0)}}{n_n^{(0)}} \approx \exp\left(\frac{m_n - m_p}{T}\right) \tag{4.18}$$

Since $m_n - m_p > 0$ the ratio $n_p/n_n \to \infty$ as $T \to 0$ i.e. all the neutrons convert into protons. This result depends only on the fact the neutron is more massive that the proton and it applies as long as the two particles can freely interconvert.

The weak interaction is mentioned because it is the means by which neutrons and proton can interconvert. There is nothing special about the weak interaction in this context as any interaction that allowed the interconversion would have the same result.

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