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Do black holes agree with the pauli exclusion?

  1. May 10, 2010 #1
    Is there agreement here or are all/most of the neutrons in the same state?
  2. jcsd
  3. May 10, 2010 #2
    One cannot apply such a principle in that case : the momenta are undefined, arbitrarily high, for such academic "particles confined to zero volume".
  4. May 10, 2010 #3
    So there is no quantum state for a black hole?
  5. May 10, 2010 #4
    Yes, there must certainly be a quantum state associated to a black hole in a quantum theory of gravity. This quantum state is just not suitably defined using the observables you refer to.

    About your first message, it is not correct to say that all particles which has fallen in a black hole are in the same quantum state only because they appear to us as occupying the same zero volume. There are many wrong things about this idea. The first thing that is wrong seems to me that those particles could all have different momenta. The next wrong thing would be that this "volume" is itself ill-defined in current theories, it is the volume of the singularity.

    About your second message, one should not assume that the quantum state of a black hole is defined by the quantum states of all particles which have fallen. A black hole carries as much entropy as possible, so it has "forgotten" as much as can be forgotten from the fallen material.
  6. May 10, 2010 #5
    What about when the star is collapsing and the volume is still very small but finite.

    I will try to explain what I was thinking when I posted.

    From what I understand the reason materials can only be compressed as much as they can (degeneracy pressure) is because of the pauli exclusion principle.

    If I have this right.. as electrons or any spin 1/2 particle must enter higher and higher energy levels the associated eigenfunction puts their position probability further and further away (in position space) from the central potential.

    I am wondering if when the "degeneracy pressure" is overcome if a particle of 1/2 spin is forced into an energy state already occupied by 2 particles.
  7. May 10, 2010 #6
    The concept of degeneracy pressure is useful to describe slow processes of collapse for moderate size stars. This degeneracy can never be overcome. Ever more energy must be put into the system to excite the fermions into higher momenta, until so much energy is stored that the BH forms.

    If you think in terms of a potential, the fermions trapped into it always have their wavefunction decaying exponentially beyond the surface of the potential, they remain trapped because there is no "valley" into which to tunnel (the gravitational potential is always below the vacuum). So as you add energy into the system, allowing ever higher harmonic levels inside the potential, general conservation of energy tells us that we build a depth of the gravitational potential increasing faster than higher harmonics are being occupied. I do not remember explicitly doing such a calculation, but if the depth would not grow fast enough, then our calculations would indicate that no black hole form, which is contrary to the general theorems and would only invalidate our approximations (the BH collapse is generic and does not require specific initial conditions such as spherical symmetry).
  8. May 11, 2010 #7


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    Quantum field theory in curved (here: singular) spacetime is afaik not well understood; there is not even a definition of "vacuum" on top of which a QFT with its particle-like excitations could be constructed. Therefore in a singular spacetime like a black hole there is no concept for "particles" in the usual sense.
  9. May 11, 2010 #8
    I agree with tom.stoer and think his point is quite relevant to go beyond what has been said. So to clarify my post above in 2 points, I only meant 1) specifically where the concept of pressure degeneracy is used (to compute stability condition when collapse does to a BH does not occur) 2) the limitation within the potential approach, if we contradicted the general GR theorems for BH formation, we would only invalidate this approach, not the theorems (potential approach is non-relativistic btw)
  10. May 11, 2010 #9
    So we don't know if 2 neutrons are occupying the same state?

    We just don't know if a neutron even exists anymore once a star of this magnitude starts to collapse?
  11. May 11, 2010 #10


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    GR predicts its own breakdown at singularities like a black hole or the big bang. QFT is no longer defined. Seriously we do not know if a neutron will exist at the singularity.
  12. May 11, 2010 #11
    I can only repeat that no fermion will ever occupy the same state in the collapse process. We expect all structures, including the hadronic structure of a neutron, to be torn apart under ever growing gravitational tidal forces. But if there is such a thing as a fundamental particle without structure, it could be a quark for instance, then still no two such fundamental particles will occupy the same state either. That remains valid at all times, that is to say when the available spatial volume shrinks to zero, the available momentum volume more than compensates to accommodate Heisenberg's inequality. The exact limit of zero volume, where your particle "sits on the singularity" is undefined as far as I know. I am unaware of any modification of GR or QM attempting to unify both where several fermions can occupy the same state.
  13. May 11, 2010 #12
  14. May 11, 2010 #13
    I was under the impression that the degenracy state was achieved in neutron stars of a certain mass, just microseconds before they collapse into a black hole. is that wrong??
  15. May 11, 2010 #14
    Which degeneracy? Neutron? Quark? The latter is yet to be observed, so you have degenerate electron matter forming White Dwarves, and DNM forming a neutron star, and maybe something between that and a BH. What happens when that is overcome is a singularity, as far as GR concerned and QM has nothing to say on the matter.
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