Since neutrinos interact constantly in the early universe, I assume that they are present as flavor eigenstates. However, they are Fermi-Dirac distributed, \begin{equation} f(E, T) = \frac{1}{e^{E/T} + 1}, \end{equation} where the energy $E = \sqrt{m^2 + p^2}$ depends on the mass (I know, they are ultra relativistic but that is not the point). But as flavor eigenstates, they do not have a specific mass. What mass goes the into the Fermi-Dirac distribution? Is it the "expected mass" weighting all masses by their probability?
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1$\begingroup$ You can't just plug in a value of $E$. You have to derive $f=\langle N\rangle$ from first principles from a partition function over all mass eigenstates. Presumably, since you can have neutrinos in all three of them, you should multiply three PFs, thus adding three $f$s obtained by log-differentiation. $\endgroup$– J.G.Commented Nov 30, 2023 at 17:36
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$\begingroup$ If you can elaborate on that it would be great, since it makes sense to tackle the question from first principles. But again, when computing the partition function one usually sums over energy states, so we have again the same problem... $\endgroup$– user268009Commented Dec 12, 2023 at 16:46
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$\begingroup$ We sum over energy states but, holding $p$ fixed, that's equivalent to mass states. Writing $E_i=\sqrt{m_i^2+p^2}$,$$Z(p,\,T)=\prod_i\left(1+e^{-E_i/T}\right)\implies f(p,\,T)=\sum_i\frac{1}{e^{E_i/T}+1}.$$ $\endgroup$– J.G.Commented Dec 12, 2023 at 17:06
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