# Does black hole formation contradict the Pauli exclusion principle?

A star's collapse can be halted by the degeneracy pressure of electrons or neutrons due to the Pauli exclusion principle. In extreme relativistic conditions, a star will continue to collapse regardless of the degeneracy pressure to form a black hole. Does this violate the Pauli exclusion principle? If so, are theorists ok with that? And if it doesn't violate the Pauli exclusion principle, why not?

• Take the case of a star -> neutron star first. There is enough pressure (energy density) to convert protons into neutrons and radiate the leptons away as neutrinos. I'm not sure what happens in a collapse to a black hole but my guess is that something similar happens with the quarks. Good question! – Brandon Enright Jan 16 '14 at 21:03
• Then perhaps analogously to your example of $e + p \rightarrow n + \nu$ there could be some unknown process that allows $u + d \rightarrow X$ where $X$ is either bosonic or somehow free to propagate away. – kd88 Jan 16 '14 at 21:17
• There are probably lots of other options besides that. I really don't know. – Brandon Enright Jan 16 '14 at 21:21
• – user4552 Aug 31 '17 at 3:21

I don't have a very satisfactory description of the microscopic picture, but let me share my thoughts.

The Pauli exclusion doesn't quite say that fermions can't be squeezed together in space. It says that two fermions can't share the same quantum state (spin included). A black hole has an enormous amount of entropy (proportional to its area, from the famous Bekenstein-Hawking formula $S = \frac{A}{4}$) and hence, its state count is $\sim e^A$.

Now, this might not seem like a big deal since usual matter has entropy proportional to volume. However, volume of such collections is also proportional to the mass. This means that a counting of the number of states goes as $e^M$

For a black hole, it's Schwarzschild radius is proportional to the mass, hence $A \sim M^2$. So, the number of states scales as $e^{M^2}$ which is much much more than ordinary matter, especially if the mass is "not small". So there seem to be a lot of quantum states into which one can shove the fermions.

So it seems like the fermions should have an easier time in a black hole than in (say) a neutron star.

• There may be two different reasons why people would imagine the exclusion principle to be violated in gravitational collapse. One is the one given in the question, and other has to do with what happens below the Planck scale, at the formation of the singularity. This answer is interesting, but it seems to have more to do with the latter question, not the one asked by the OP. – user4552 Aug 31 '17 at 3:25
• So it seems like the fermions should have an easier time in a black hole than in (say) a neutron star. The Bekenstein-Hawking entropy doesn't represent the microstates of the infalling matter, since, e.g., an eternal black hole has the same entropy. AFAICT semiclassical gravity doesn't give a satisfactory answer as to what microscopic degrees of freedom are represented by black hole entropy, but presumably they would be gravitational in nature. – user4552 Aug 31 '17 at 4:03
• @ Ben Crowell-I had thought the NON-eternality of ANY black hole was why Hawking radiation (whose existence seems to be very widely accepted, in spite of the trans-Planckian problem you mention) had been hypothesized, so please correct me if I'm wrong. Among cosmologies, the trans-Planckian problem is dodged by Nikodem J. Poplawski's "Cosmology with torsion" (the subject of many papers by him between 2010 and 2019), because it uses fermionic interactions of Einstein-Cartan gravity, whose Cartan radius is much larger than the Planck length. – Edouard Jun 4 '19 at 3:12
• "an easier time" from outside observers, which is (at least partially) what matters for answering this question. I don't think the infalling fermions share this point of view... – Kevin Kostlan Sep 19 '19 at 3:21

All the known laws of physics, including the exclusion principle, are believed to be valid at all times during the collapse, up until the matter that you're talking is just about to hit the singularity. ("Just about to hit" may mean when the density reaches the Planck density, so that quantum gravity effects become important, or it may be a little earlier, if there is other physics beyond the standard model that we don't know about.)

Note that just because an event horizon has formed, and some matter has fallen past the event horizon, that doesn't mean that the conditions experienced by that matter are extreme. They needn't be extreme at all. The equivalence principle says that the laws of physics are always locally the same, because spacetime can always locally be approximated as flat, so that special relativity applies. Actually the conditions at the formation of the event horizon are pretty solidly within the range of conditions (temperature, pressure) that can be described by the standard model of particle physics.

Degenerate matter is simply matter in a state where the degeneracy pressure is significant. In a white dwarf or neutron star, for example, the degeneracy pressure happens to be in equilibrium with gravity. The matter can be compressed further without violating the exclusion principle. The exclusion principle essentially has the effect of imposing a maximum wavelength on each particle, which goes roughly like $$\lambda \sim (V/N)^{1/3}$$, where $$V$$ is the total volume and $$N$$ is the number of identical fermions. If $$V$$ gets smaller, you get a smaller wavelength, therefore bigger momenta and higher pressure.

In stars that aren't massive enough to form black holes, you reach a point where this higher pressure gets big enough to give an equilibrium with gravity. If the star is more massive, so that we are going to form a black hole, then we just don't reach such an equilibrium. The wavelengths of the fermions simply get very short, and their momenta very high, as we approach the formation of the singularity.

We don't try to say anything about the singularity itself using the presently known laws of physics. In pure classical general relativity, the singularity is not even considered to be part of the spacetime.

• Does it mean that Pauli's principle does not prohibit particles occupying the same state, but rather it says that there will be a resistance when this is forced upon the particles (a kind of repulsion/pressure). – Jus12 May 15 '15 at 11:13
• "singularity is not even considered to be part of the spacetime." - Can you elaborate further on this? – SurpriseDog Jul 19 '19 at 16:14
• @SurpriseDog An good introduction to the challenge of defining and understanding GR singularities can be found at god-does-not-play-dice.net/Geroch_1968.pdf – jawheele Dec 29 '20 at 23:00

Does this violate the Pauli exclusion principle? If so, are theorists ok with that?

The short answers are "yes" and "yes". Recall that we are talking about what happens inside the event horizon ...

Perhaps the density of states diverges as volume decreases. However iirc most thinking is around the idea that there is a quark degeneracy limit that has to be overcome like the neutron degeneracy limit.

i.e. those people speculating some process that combines quarks into some boson can pat themselves on the back.

The bottom line is that we don't know enough about how matter/energy behaves in such extreme conditions to be able to do more than speculate.

• +1 for pointing out that "what happens in the event horizon stays in the event horizon" – kd88 Jan 17 '14 at 10:40
• Recall that we are talking about what happens inside the event horizon ... That's irrelevant. The equivalence principle says that the laws of physics are always locally the same, because spacetime can always locally be approximated as flat, so that special relativity applies. The bottom line is that we don't know enough about how matter/energy behaves in such extreme conditions to be able to do more than speculate. What extreme conditions do you have in mind? The conditions are not necessarily extreme just because you're inside the event horizon. Do you mean at the singularity? – user4552 Oct 17 '14 at 21:35