I was wondering whether the decay of neutrons and protons (if they happen to be able to decay, as it is predicted by some GUTs) could be avoided in some cases.

Let's begin with neutrons:

In principle neutrons have a very short time when they are isolated (around 10 minutes) and they suffer beta decay, but because of electron degeneracy pressure, it is heavily supressed in a neutron star (What stabilizes neutrons against beta decay in a neutron star?).

So, if this happens, then, shouldn't neutron stars be "safe" from the decay of neutrons and protons (in case there is proton decay, as there are almost no protons but mainly neutrons)?

I read an article, A Dying Universe: The Long Term Fate and Evolution of Astrophysical Objects, where they describe that as neutron stars would have protons (for example, they would have ordinary matter in their surface) the protons would decay, decreasing the degeneracy pressure on the neutrons and allowing bets decay to occur. But, I was thinking: Is it impossible for a pure neutron star to exist? Even if initially they had protons, couldn't there be big-enough neutron stars with enough mass that, even if their protons decayed, the gravity of the neutron star itself would maintain the degeneracty pressure? And if they could, would the surface suffer beta decay?

Then we go for the protons:

Could there be some situations in which, even if protons could decay, this could be somehow avoided? For example, if there was a white dwarf or a neutron star (like in the previous case) with a very high angular momentum, could it cause some kind of force or pressure (like a high centripetal force) that would keep a similar degeneracy pressure that would keep protons from decaying?

If not, can you think of any other ways? Or proton decay, if it occurs, is unavoidable and therefore neutron decay as well?

  • $\begingroup$ Your "some" GUTs predict nucleon decay at large, not just proton decay.... $\endgroup$ Commented Apr 20 at 21:28
  • $\begingroup$ Related physics.stackexchange.com/q/307885/226902 $\endgroup$
    – Quillo
    Commented Apr 20 at 22:02
  • $\begingroup$ @CosmasZachos including electrons? $\endgroup$
    – vengaq
    Commented Apr 21 at 15:15
  • $\begingroup$ Positrons; absolutely! See Fig 4 of this: an elementary process underlying proton decay is $ud\to u^c e^+$, but it obviously underlies $n\to e^+ \pi^-$ as well! $\endgroup$ Commented Apr 21 at 19:54
  • $\begingroup$ @CosmasZachos but this shows that protons can decay into positrons, I was asking whether electrons/positrons are also predicted to decay in GUTs $\endgroup$
    – vengaq
    Commented Apr 21 at 20:08

1 Answer 1


Neutron decay is blocked because the possible decay products (protons and electrons) are already present as degenerate gases and will not admit the creation of more of them at energies below their respective Fermi energies.

In other words it isn't the pressure of the gas that is important here - that is simply a consequence of the high number densities of the particles present.

If protons were to decay into say neutral pions and positrons, well those particles do not exist in great numbers in neutron star interiors - degeneracy would not be an issue, the positrons will just annihilate with electrons and pions are not fermions in any case. There will therefore not be any mechanism to block proton decay in a manner analogous to the way neutron decay is blocked in neutron star interiors.

If protons do decay, and electrons are also annihilated, preserving charge neutrality, then both the Fermi energies of the protons and electrons will get smaller. This allows room at the top of their Fermi distributions to allow the decay of neutrons (into protons and electrons). The neutrons that (unavoidably) decay will come from the top of their Fermi distribution and so the overall degeneracy pressure will decrease.

And to be clear: it is not possible to have a stable neutron star without the appropriate fraction of accompanying protons and electrons, so as to satisfy the equilibrium condition $$E_{F,n} = E_{F,p} + E_{F,E}\ . $$

  • $\begingroup$ Is the proton gas also degenerate? I would have expected the proton number density to be equal to the (degenerate) electron number density, and therefore much lower than the neutron density, even though the positions should have the same temperature as the neutrons ans their similar wavelengths/effective volumes. $\endgroup$
    – rob
    Commented Apr 20 at 21:35
  • 1
    $\begingroup$ @rob "Is the proton gas also degenerate? " Yes. $n_n/n_p \sim 10-50$ and $E_F/kT \propto n^{2/3}$. For neutrons $E_F \sim 100$s of MeV, so $E_F \sim 10$s of MeV for protons and they are completely degenerate for $T< 10^{10}$ K. $\endgroup$
    – ProfRob
    Commented Apr 21 at 0:13
  • $\begingroup$ @ProfRob wouldn't it be possible to have a neutron star with a unusual low number of protons (lost by some other mechanism than proton decay), so that even if the few protons remaining decay it would not be sufficient to low the degeneracy pressure (or the density of the particles) enough to decay the entire star? Or perhaps the gravity of a neutron star with a very large mass (maybe on the brink of becoming a black hole) could maintain the degeneracy even if the protons decay? $\endgroup$
    – vengaq
    Commented Apr 21 at 16:28
  • $\begingroup$ @vengaq as I've said in my answer, the pressure, or the gravity, is not what prevents the neutrons decaying. Higher mass neutron stars have a greater fraction of protons. $\endgroup$
    – ProfRob
    Commented Apr 21 at 16:32
  • 1
    $\begingroup$ @vengaq If there were no protons, then neutrons would decay and create them. You can't have a neutron star with no protons. $\endgroup$
    – ProfRob
    Commented Apr 21 at 17:29

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