What happens to Protons and Electrons when a Neutron star forms? At some point gravity overcomes the Pauli Exclusion Principle ( I assume) and they are all forced together. What happens in the process?
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$\begingroup$ I would not state that gravity overcomes the Pauli Exclusion Principle since it is an idea. What gravity does do, is to squeeze all the particles together so that quantum states no longer exists. $\endgroup$– LDC3Commented Nov 30, 2014 at 14:23
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3$\begingroup$ @LDC3 quantum states always exist, different for different boundary conditions $\endgroup$– anna vCommented Nov 30, 2014 at 16:21
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$\begingroup$ @annav So the quantum states for normal matter are transformed into a different set of quantum states for compressed matter (where RobJeffries indicated that the electrons are degenerate)? Isn't that like saying the theories for normal matter no longer apply and a new set of theories are used? Oh, I should have stated that the quantum states for normal matter (and theories) are no longer viable. $\endgroup$– LDC3Commented Nov 30, 2014 at 16:31
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$\begingroup$ @LDC3 yes, normal matter would be correct. New solutions are needed but still the framework is quantum mechanical. $\endgroup$– anna vCommented Nov 30, 2014 at 16:45
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$\begingroup$ Indeed the quantum states are identical and they have a density of $8\pi p^2/h^3$, where $p$ is the momentum, whether we are talking about a perfect gas or a degenerate gas All that changes is the way they are filled. $\endgroup$– ProfRobCommented Nov 30, 2014 at 19:29
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2 Answers
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It is the Pauli Exclusion Principle that actually allows the formation of a "neutron" star.
In an "ordinary" gas of protons and electrons, nothing would happen - we call that ionized hydrogen! However, when you squeeze, lots of interesting things happen. The first is that the electrons become "degenerate". The Pauli exclusion principle forbids more than two electrons (one spin up the other spin down) from occupying the same momentum eigenstate (particles in a box occupy quantised momentum states).
In that case what happens is that the electrons "fill up" the low momentum/low energy states and are then forced to fill increasingly higher momentum/energy states. The electrons with large momentum consequently exert a degeneracy pressure, and it is this pressure that supports white dwarf stars.
If the density is increased even further - the energies of degenerate electrons at the top of the momentum/energy distribution get so large that they are capable of interacting with protons (via the weak nuclear force) in a process called inverse beta decay (sometimes referred to as electron capture when the proton is part of a nucleus) to produce a neutron and a neutrino. $$p + e \rightarrow n + \nu_e$$ Ordinarily, this endothermic process does not occur, or if it does, the free neutron decays back into a proton and electron. However at the densities in a neutron star, not only can the degenerate electrons have sufficient energy to instigate this reaction, their degeneracy also blocks neutrons from decaying back into electrons and protons. The same is also true of the protons (also fermions), which are also degenerate at neutron star densities.
The net result is an equilibrium between inverse beta decay and beta decay. If too many neutrons are produced, the drop in electron/proton densities leaves holes at the top of their respective energy distributions that can be filled by decaying neutrons. However if too many neutrons decay, the electrons and protons at the tops of their respective energy distributions have sufficient energies to create new neutrons.
Mathematically, this equilibrium is expressed as $$E_{F,p} + E_{F,e} = E_{F,n},$$ where these are the "Fermi energies" of the degenerate protons, electrons and neutrons respectively, and we have the additional constraint that the Fermi momenta of the electrons and protons are identical (since their number densities would be the same).
At neutron star densities (a few $\times 10^{17}$ kg/m$^{3}$) the ratio of neutrons to protons is of order 100. (The number of protons equals the number of electrons).
This calculation assumes ideal (non-interacting) fermion gases. At even higher densities (cores of neutron stars) the strong interaction between nucleons in the asymmetric nuclear matter alters the equilibrium above and causes the n/p ratio to decrease to closer to 10.
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2$\begingroup$ So the implication is that if we could remove a cubic metre of neutronium from a neutron star and put it in free space, we would get a very large explosion as neutrons decayed and released energy? $\endgroup$– user56903Commented Nov 30, 2014 at 17:15
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3$\begingroup$ @DirkBruere The explosion would only be caused by the release of pressure. Neutron decay is moderated by the weak force and is slow. The half life of a free neutron is 10 minutes! $\endgroup$– ProfRobCommented Nov 30, 2014 at 18:38
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1$\begingroup$ OK - let's add some numbers. We have 10^17 kg of neutrons, which means we get a decay of about 10^16 kg every 10 minutes or about 10^13 kg per second. That's about 10^40 neutrons/second. If each releases 0.7MeV we get about 10^45 eV which is around 10^26 J/s or roughly 10^11 megatonne equivalent nuclear explosion. Which is quite a lot. $\endgroup$– user56903Commented Nov 30, 2014 at 19:02
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1$\begingroup$ @DirkBruere Actually, it more like $5\times 10^{16}$ kg of neutrons in the first 10 minutes. Since this is an exponential decay, more than half of it ($2.5\times 10^{16}$ kg) will have decayed in the first 210 seconds (approximately). $\endgroup$– LDC3Commented Nov 30, 2014 at 19:14
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3$\begingroup$ @DirkBruere My mistake - the kinetic energy density of the neutrons is $3\times10^{32}$ Jm$^{-3}$. i.e. each neutron has $\sim 30$ MeV of KE and a speed of $\sim 0.2c$. Hence the ~1MeV released by beta decay and converted largely into KE of electrons and protons is a minor perturbation (impressive though it is). $\endgroup$– ProfRobCommented Nov 30, 2014 at 21:25
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Well let's talk about in a completely normal perspective. When a main sequence star loses its fuel it has to start combining heavier elements right? So hydrogen fuses to helium. Helium to carbon and oxygen. So for star like sun it does of have enough gravity to trigger helium fusion. As a result the core of the star gets crushed until the carbon is fused and electron degeneracy pressure kicks in to maintain the carbon core to form a white dwarf. In case of larger stars the gravity has enough power to overcome this electron degeneracy pressure and breaks their orbitals to push the electrons into the nucleus of the atom. Here the strong nuclear force takes over and the electrons and protons combine to give u neutrons, neutrinos and gamma rays. If the star is large enough it can even crush the neutrons and ultimately ends up as a black hole. However there is also another phenomenon known as a hypernova. Where a star is 200 times or so massive than the sun when it goes supernova there won't be a remnant left.
