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Does black hole formation contradict the Pauli exclusion principle?

+ 7 like - 0 dislike
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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?

This post imported from StackExchange Physics at 2014-04-24 02:34 (UCT), posted by SE-user jk88
asked Jan 16, 2014 in Theoretical Physics by jk88 (35 points) [ no revision ]
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!

This post imported from StackExchange Physics at 2014-04-24 02:34 (UCT), posted by SE-user Brandon Enright
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.

This post imported from StackExchange Physics at 2014-04-24 02:34 (UCT), posted by SE-user jk88
There are probably lots of other options besides that. I really don't know.

This post imported from StackExchange Physics at 2014-04-24 02:34 (UCT), posted by SE-user Brandon Enright

2 Answers

+ 6 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-04-24 02:34 (UCT), posted by SE-user Siva
answered Jan 17, 2014 by Siva (710 points) [ no revision ]
+ 1 like - 1 dislike

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.

Also see: http://www.physicsforums.com/showthread.php?t=600360

This post imported from StackExchange Physics at 2014-04-24 02:34 (UCT), posted by SE-user Simon Bridge
answered Jan 17, 2014 by Simon Bridge (0 points) [ no revision ]
+1 for pointing out that "what happens in the event horizon stays in the event horizon"

This post imported from StackExchange Physics at 2014-04-24 02:34 (UCT), posted by SE-user jk88

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