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My understanding is that the Pauli exclusion principle is always valid for fermions, as long as these fermions exist.

For example, when a star collapses under gravity, at first the Pauli exclusion applies to electrons until they combine with protons, at which point leptons leave as neutrinos and a neutron star forms.

From this point, the Pauli exclusion applies to neutrons until they break under gravity into quarks potentially forming a Quark Star:

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At this point the Pauli exclusion applies to quarks. The Up and Down quark are stable elementary particles that cannot break down into anything smaller like atoms or neutrons. Therefore, it would seem that the Pauli exclusion would still apply to quarks and prevent the collapse no matter how strong gravity is.

What is the current reasoning of how gravity overcomes the Pauli exclusion principle for quarks to form a black hole?

safesphere
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  • see https://physics.stackexchange.com/questions/352935/why-can-a-neutron-star-implode?rq=1 – ProfRob Nov 02 '17 at 00:47
  • "electrons until they combine with protons, at which point leptons leave as neutrinos" how exactly does an electron combine with a proton? – Árpád Szendrei Jan 25 '20 at 03:10
  • @ÁrpádSzendrei Here is the diagram: https://static.memrise.com/uploads/things/images/14977617_130404_1839_39.jpg - where time goes vertically from the bottom to the top. When gravity pushed the proton and electron together, the electron splits into a negative W-boson and a neutrino. The positive proton absorbs the negative W-boson and becomes a neutral neutron. The neutrino flies away. This interaction essentially is a part of the nuclear fusion process giving energy to the Sun to shine: 4 atoms of hydrogen convert to 1 atom of helium, so 2 protons become neutrons and 2 neutrinos fly away. – safesphere Jan 26 '20 at 10:32

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You may be misunderstanding the Pauli Exclusion Principle (PEP).The PEP states that no two fermions can occupy the same quantum state, not that they cannot occupy the same space or be compressed to whatever density you like. The quantum states here consist of two spin states for every possible momentum state. In a degenerate gas, all these states are filled up to the Fermi energy. All that happens when the neutron star (or quark star) gets smaller (or collapses), is that the Fermi energy just keeps increasing as the neutron (quark) density climbs, and the neutron (quark) degeneracy pressure just keeps increasing as a consequence.

However, in General Relativity, pressure (like mass/energy) is a source of gravitational curvature and actually increases the required pressure gradient needed to support the star. At a certain threshold radius - a small factor larger than the Schwarzschild radius, a point of instability is reached where increasing the pressure is actually counter-productive. Beyond this, you can make the pressure as large as you like and it will not prevent the formation of a black hole.

ProfRob
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  • So the momentum (pressure) increases and the kinetic energy contributes to gravity. I get it. The question remains, to what limit? It would seem that closer to the singularity the kinetic energy would increase without a limit, but this doesn't seen possible, as it would also make the total mass larger than it was. I asked this before, but the answer wasn't very clear to me: https://physics.stackexchange.com/questions/356029/energy-of-falling-into-a-black-hole I wonder if you have a better answer to that question. Thanks! – safesphere Nov 01 '17 at 23:55
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    It was the only one and I accepted it, as I was sure it wasn't incorrect, but you can see a long discussion there indicating the answer wasn't too clear. This doesn't prevent anyone from posting a better answer though, does it? In fact, I was surprised that the question didn't generate more interest, as I am sure most people have a little idea of what the relevant physics is. – safesphere Nov 02 '17 at 00:10
  • @safesphere To talk about anything near the singularity, you need a theory of quantum gravity. It's entirely natural that things seem non-physical if we try to apply theories without quantum gravity near a singularity. – Chris Nov 02 '17 at 00:22
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    The "gravitational mass" of a compact object is less than the sum of its baryonic components because it is bound and so the sum of its internal (kinetic) energy and potential energy is negative. @safesphere – ProfRob Nov 02 '17 at 00:23
  • @Chris How about a half way to the singularity or 90%... :) – safesphere Nov 02 '17 at 00:29
  • @safesphere and my comment above remains true at all stages of collapse. – ProfRob Nov 02 '17 at 00:34
  • @safesphere Rob Jeffries has the more applicable point here ;) PE decreases more than KE increases so the total mass goes down, not up – Chris Nov 02 '17 at 00:39
  • @RobJeffries "The "gravitational mass" of a compact object is less than the sum of its baryonic components because it is bound and so the sum of its internal (kinetic) energy and potential energy is negative." The rest mass of a baryon is 1% rest mass of the quarks, and 99% bonding energy (strong force). Are you talking about the same thing? – Árpád Szendrei Jan 25 '20 at 03:06
  • @RobJeffries Thanks again for your answer, it was very helpful! I also understand your comment on the balance of mass/energy, but I wonder if you could please clarify how it works with energy conservation. In a spherically symmetric collapse energy is defined and conserved. No matter how close to the singularity, the kinetic energy (and therefore the Fermi energy) cannot exceed the initial pre-collapse mass. The only way for this to work is if a limited Fermi energy has an unlimited number of levels (e.g. while approaching the speed of light. Is this the case or the explanation is different? – safesphere Jul 13 '20 at 21:21
  • @safesphere there are no limits to the Fermi energy. The number density of fermion energy states is $8\pi E_F^3/3h^3c^3$, or to put it another way, the Fermi energy goes up as number density to the power of one third. The total kinetic energy density is then $2 \pi E_F^4/h^3c^3$. Meanwhile, gravitational potential energy becomes more and more negative. – ProfRob Jul 13 '20 at 21:39
  • @RobJeffries Thank you Sir! You are a good man, one of a few true scientists spending time on this site to help others understand physics. Can you possibly please point me to a link or search on this topic? I’d like to understand how these densities are defined (e.g. per a 3D volume?) and why the Fermi energy there is in the forth power. Thanks again! – safesphere Jul 13 '20 at 23:16
  • @safesphere Density of states is $g(p)=8\pi p^2/h^3\ dp$ per unit volume ($p$ is momentum). If all states are filled up to the Fermi momentum ($E_F/c$) then you can integrate the density of states from zero to $E_F/c$ to find out the number of states per unit volume and integrate $E g(p)\ dp$ to find the total energy density. – ProfRob Jul 14 '20 at 06:55