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Should the energy distribution of neutrinos be affected by Fermi-Dirac statistics? And if so, what would the consequences be? Could this locally cause weak interaction because of the Pauli Exclusion Principle?

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  • $\begingroup$ NB: When you say "weak interaction", I assume you mean the pseudo-interaction (no forces are involved) in degenerate fermion matter. It is certainly not a weak interaction in the sense of exchanging virtual W and Z particles. $\endgroup$
    – ProfRob
    Jul 20, 2015 at 13:42
  • $\begingroup$ @RobJeffries. Weak interaction might not be the right term, but I mean beta decay etc. $\endgroup$
    – Lehs
    Jul 20, 2015 at 13:52
  • $\begingroup$ But that is indeed a "weak interaction" and has nothing to do with the Pauli Exclusion Principle between identical neutrinos. Do you mean could the neutrinos be dense enough to prevent beta decay? $\endgroup$
    – ProfRob
    Jul 20, 2015 at 13:58
  • $\begingroup$ @RobJeffries. What I'm wondering is if there might be an explanation to the decay as caused by "neutrino crowding" among termic neutrinos - if there are such neutrinos in the neutrons? $\endgroup$
    – Lehs
    Jul 20, 2015 at 14:30
  • $\begingroup$ "neutrino crowding" would prevent beta decay if the neutrino Fermi energy (in a degenerate neutrino gas) exceeds the maximum possible energy of the decay neutrino. At lower levels of degeneracy you would be reducing the available phase space for the decay neutrino so I presume the reaction would be somewhat inhibited. $\endgroup$
    – ProfRob
    Jul 20, 2015 at 15:35

1 Answer 1

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Yes, neutrinos should obey Fermi-Dirac statistics and yes, the Pauli Exclusion Principle should operate for neutrinos.

But let's examine how dense the neutrino population has to be for this to be important.

The Fermi momentum is given by $$ p_F = \left( \frac{3}{8\pi}\right) h n_{\nu}^{1/3} $$ where $n_{\nu}$ is the neutrino number density.

In order to be degenerate (and hence strongly affected by the PEP), then the following criterion can be used $$ \frac{E_{F}}{kT} \gg 1,$$ where $E_{F}$ is the kinetic energy associated with particles with the Fermi momentum.

Neutrinos have very small rest masses and so would usually be considered relativistic, so we have $$\frac{p_F c}{kT} \gg 1$$ which leads to $$ n_{\nu} > \frac{8\pi}{3}\left(\frac{kT}{hc}\right)^3 = 2.8 \times 10^{6} T^3$$ in units of m$^{-3}$.

So, if this inequality is satisfied then neutrino degeneracy starts to become important.

Some examples:

The cosmic neutrino background has $T=1.95$ K, and a predicted number density of (for each neutrino species) of around $5.65 \times 10^{7}$ m$^{-3}$. Thus $E_{F}/kT = 1.4$. Not enough for the PEP to be super-important, but it is not completely negligible either. Actually, for finite neutrino masses it turns out that the neutrinos are probably non-relativistic at the present day resulting in the PEP being rather more important, because the neutrino temperature is lower than the 1.95K calculated for massless neutrinos. As a result there is a neutrino degeneracy pressure that may influence cosmic clustering of neutrinos (for instance see Pfenniger & Muccione 2006).

In the centre of a core-collapse supernova, the temperatures can reach $\sim 10^{11}$ K and the neutrino density could be as much as $10^{44}$ m$^{-3}$ per species. In this case $E_{F}/kT \sim 50-100$ and neutrino degeneracy is important. The resultant degeneracy pressure is an important component of the dynamics of supernovae explosions. It is partly neutrino degeneracy pressure that prevents the rest of the star collapsing onto the central proto-neutron star core.

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