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I was wondering about hydrostatic equilibrium (the balance of radially inward and outward forces) in large (i.e. above the Chandrasekhar limit) stars. It is often said that when the gravitational force exceeds any outward forces or pressures, mainly the electron degeneracy pressure I'm thinking, the star collapses into a black hole. But how can this happen without the Pauli exclusion principle being violated?

A similar question: Does black hole formation contradict the Pauli exclusion principle?

The answer by @Siva in that question makes some sense to me, but I don't understand at what point we start our counting of states from the object being a star to a black hole. What happens in the middle? Is there a sharp change?

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  • $\begingroup$ Related: How strong is electron degeneracy pressure? $\endgroup$
    – rob ♦
    Commented Jul 6, 2014 at 3:52
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    $\begingroup$ There's more than one possible way to construe "right before an object turns into a black hole." There's the formation of the event horizon, which is usually taken to be the defining property of a black hole. There is also the formation of the singularity, or the point at which a particular infalling particle reaches the singularity. Which of these did you have in mind? But how can this happen without the Pauli exclusion principle being violated? What makes you think that the exclusion principle would be violated? $\endgroup$
    – user4552
    Commented Aug 31, 2017 at 3:41
  • $\begingroup$ See physics.stackexchange.com/questions/352935/… $\endgroup$
    – ProfRob
    Commented Aug 31, 2017 at 7:05
  • $\begingroup$ and physics.stackexchange.com/questions/173107/… $\endgroup$
    – ProfRob
    Commented Aug 31, 2017 at 7:11

2 Answers 2

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When the force of gravity overcomes electron degeneracy, the electrons are effectively compressed into the nucleus where they are captured and bound by the nucleus (primarily protons) to form a large neutron mass. At this point, the star is even more dense and the process repeats as more electron-degenerate matter falls in and is bound. Meanwhile, energy is given off and this propels an inflationary pressure that can keep the star from collapsing instantly. Over time, the star cools and all remaining matter is bound by gravitation into a neutron star. This object is composed primarily of degenerate neutron matter. [Should the mass exceed 2-3 times the mass of our sun, this star will collapse into a black hole because gravity will overcome the neutron degeneracy pressure. Recall that neutrons, being fermions, cannot occupy the same energy state. Such a collapse is sudden and produces a supernova explosion from the energy transferred to outer layers by the energy released during collapse. This material can include heavier atoms than iron and seeds the surrounding space with the same stuff of which we and our rich planet are made.]

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    $\begingroup$ Hi Rex, you've really simplified here. It's not completely wrong but more details would be nice. Also, the Chandrasekhar limit is 1.4 solar masses. $\endgroup$
    – Brandon Enright
    Commented Apr 30, 2014 at 3:50
  • $\begingroup$ Fermions can and do occupy the same energy state! $\endgroup$
    – ProfRob
    Commented Aug 31, 2017 at 6:45
  • $\begingroup$ Neutron stars are really not supported by neutron degeneracy pressure. $\endgroup$
    – ProfRob
    Commented Aug 31, 2017 at 7:16
  • $\begingroup$ @RobJeffries "Fermions can and do occupy the same energy state" How so? hyperphysics.phy-astr.gsu.edu/hbase/pauli.html " . Link please? $\endgroup$
    – anna v
    Commented Aug 31, 2017 at 14:02
  • $\begingroup$ @annav That is appalling. The PEP forbids fermions occupying the same quantum state. Two fermions can have identical energy but different quantum numbers. That is what degeneracy means. $\endgroup$
    – ProfRob
    Commented Sep 1, 2017 at 16:14
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This answer expands on @Rex' s answer, so please read it to get the complete picture. It expands on the part elementary particles have in a black hole creation.

It is often said that when the gravitational force exceeds any outward forces or pressures, mainly the electron degeneracy pressure I'm thinking, the star collapses into a black hole. But how can this happen without the Pauli exclusion principle being violated? . . .

When the simple hydrogen equation does not hold because the potential has been distorted by the gravitational one, an electron can be captured by a proton. This makes a neutron and an electron neutrino. Neutrinos being weakly interacting escape and the neutrons make a neutron star which continues to collapse towards a black hole , if the mass is large enough. There is no problem with the Pauli exclusion or lepton number at this level. Neutrons are composed of quarks which are charged and also obey the Pauli exclusion principle. When the density due to the gravitational collapse becomes large then the whole will turn into a quark gluon plasma. That is as far as elementary particle interactions have taken us. Research is ongoing.

but I don't understand at what point we start our counting of states from the object being a star to a black hole. What happens in the middle? Is there a sharp change?

The point about a black hole is the total mass, such that it does not allow anything to escape from a certain radius. The quantum mechanical behavior from a certain point on is an effective theory joining quantum mechanics and gravitation, a process that is at the frontier of research. It depends on your definition of sharp. Supernovas are sharp.

Supernovae can be triggered in one of two ways: by the sudden reignition of nuclear fusion in a degenerate star; or by the gravitational collapse of the core of a massive star. In the first case, a degenerate white dwarf may accumulate sufficient material from a companion, either through accretion or via a merger, to raise its core temperature, ignite carbon fusion, and trigger runaway nuclear fusion, completely disrupting the star. In the second case, the core of a massive star may undergo sudden gravitational collapse, releasing gravitational potential energy that can create a supernova explosion.

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    $\begingroup$ This reads as a random mishmash of thoughts without any logical structure, and it doesn't seem to answer the question. $\endgroup$
    – user4552
    Commented Aug 31, 2017 at 3:44
  • $\begingroup$ @BenCrowell Maybe my english is at fault. This is not supposed to be a complete answer . but a partial one, to address the highlighted parts of the question because it would be too long for a comment. I had the impression my first sentence says this. It is not my fault it is upvoted in front of the more general answer. It is the peculiar voting system of this site . $\endgroup$
    – anna v
    Commented Aug 31, 2017 at 4:03
  • $\begingroup$ This doesn't address the fundamental misunderstanding of the PEP. $\endgroup$
    – ProfRob
    Commented Aug 31, 2017 at 7:16

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