Degrees of freedom of neutrinos

In cosmology, in $g_{eff}$, the number of relativistic degrees of freedom, one finds that the contribution of neutrinos is 2 ($\frac{7}{8}\times 2$ more precisely) but what is due this 2 factor to ? Because of its spin, the neutrino has 2 degrees of freedom but if one considers a Dirac neutrino, there should also be another factor 2 due to its antiparticle, so why does one not consider a factor 4 (as for the electron-positron for instance) rather than a factor 2 in $g_{eff}$?

• Is it not the case that a neutrinos spin is always left-handed? – Thomas Russell May 24 '17 at 9:44
• I thought it was right only if the neutrinos are massless but we know today that is not the case anymore. – ketherok May 24 '17 at 10:15

The oscillation time for a traveling neutrino is $\sim (\gamma m)^{-1} \sim T/m^2$. One might think that the condition to check would be setting this equal to Hubble. This would be correct if the neutrino were by itself in free space. However, in practice the neutrino is constantly colliding with the rest of the thermal bath. These collisions are measuring the flavor of the neutrino and hence stop it from oscillating through the quantum Zeno effect. Therefore, the relevant quantity is actually the oscillating time over the mean free time, $(n \langle\sigma v\rangle)^{-1}$.
The temperature for which the neutrino begins to oscillate takes place below the electroweak phase transition and while neutrinos are relativistic, so we can write $n \langle\sigma v\rangle \sim G_F ^2 T ^5$. Setting this equal to the oscillation rate: $$\frac{m^2}{T} \sim G _F^2 T ^5$$ Sticking in $m\sim 0.1$eV I find, $$T\sim 10 \,{\rm MeV}$$ Thus, quite coincidentally, the right-handed neutrino can just begin to be thermally populated right before freezing out (which takes place at $\sim 1\, {\rm MeV}$). In practice, a serious computation is more involved as it needs to include all the different ways neutrinos interact. The net result seems to be that these constraints are mild.