By CFRedDemon


2019-03-11 10:14:57 8 Comments

So I'm currently following a course in Cosmology and we're covering the densities of different species in the universe right now. Starting from the photon density $\rho_{\gamma}$ we need to derive the neutrino density. Apart from a different temperature and a factor 7/8 for the integral of the FD distribution I understand this. But for the photons, we have a degeneracy factor of 2 (two spin states). Then the book lists all characteristics for neutrinos (and the degeneracy factor):

  • spin degree
  • has antiparticles
  • has three generations (flavours)

So I would expect the degeneracy factor of neutrinos to be 2 (spin) x 2 (antiparticle) x 3 (flavour) => g = 12, so another factor of 6 in front of the photon density (g= 12 vs 2). But the book and other books all list a degeneracy factor of 2 x 3 = 6 for the neutrinos. So this results in factor 3 in front of the photon density (together with the other described differences). Why is this? Is the antiparticle part of the neutrinos not taken into account? Can someone please help me.

1 comments

@DomDoe 2019-03-11 11:12:41

In the SM, neutrinos are massless and therefore their helicity corresponds directly to their chirality. Also in the SM we find that neutrinos are always left-handed, right handed neutrinos do not exist. So if all neutrinos are left-handed, all anti-neutrinos must be right handed consequently. For that reason the spin states directly correspond to particle or anti-particle state. This eliminates a factor of 2 in the degeneracy chart.

Since we know the SM is imcomplete in this regard, this thread discusses the implications to the effective relativistic degrees of freedom if there was a right-handed neutrino: Degrees of freedom of neutrinos

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