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Degeneracy pressure

  1. May 27, 2008 #1
    I was wondering something, a collection of fermions can resist compressing forces due to what is termed degeneracy pressure. I was wondering, which of the four fundamental interactions is this due to? Thanks.

    Molu
     
  2. jcsd
  3. May 27, 2008 #2
    To my knowledge, this is due to the Pauli exclusion principle, and not to force interaction between the electrons.
     
  4. May 29, 2008 #3
    as gendou2 said it's to do with Pauli exclusion, since

    [tex]\Delta{x}\Delta{p}\geq\frac{\hbar}{2}[/tex]

    when a material is compressed (such as the interior of a star under the effect of gravity) the uncertainty in x gets smaller, leading to less uncertainty in momentum. The fermions are called degenerate when the pressure due to this momentum equals or exceeds(?) the pressure due to the fermions thermal motion. So in a sense it's due to whatever of the fundamental forces is causing the compression, mainly gravity in a stellar core - which is what this problem is usually used for I think, although I'm sure it must be quite important in studying fusion.
     
  5. May 30, 2008 #4
    I understand that it's due to Pauli's exclusion principle. But how does that fit in with the Standard Model, where all forces can be classified as due to one of the four fundamental interactions? I mean, if a star is resisting a compressive force, that means an opposite force is acting. To which interaction can we attribute that force?

    Molu
     
  6. May 30, 2008 #5

    Dick

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    The 'force' is called 'degeneracy pressure'. As you probably know the fermions are excluded from occupying the same state, so they need a fixed number of states to exist in. As the star contracts the energy gap between those states increases, so the energy of the fermion population has to grow. And that increase is just due to the shrinking geometry they are confined to (as in the square well). So even if you could find a type of fermion that experienced no standard model interactions whatsoever and managed to confine them somehow, they would still exert this same 'pressure'.
     
  7. May 30, 2008 #6

    nrqed

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    The actual pressure is due to the usual forces of the Standard Model (In actuality, it will be mostly the electromagnetic force). If you put the fermions in a container at near zero temperature and you compress the container, the actual pressure exerted by the fermions on the box is simply the electromagnetic force between the fermions and the particles in the box. The reason resist compression is the degeneracy pressure but the actual force they exter on the container is electromagnetic (electric for the most part).

    It's not much different from heat pressure. The pressure is due to thermal agitation but thermal agitation is not a force. The actual pressure is due to the "collision" between the particles in the box and the container. But the collision occurs because of the electromagnetic force between the particles.


    At least, that's the way I think about it.
     
  8. May 30, 2008 #7

    Dick

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    You may think about it that way, but I don't think that's correct. Neutral fermions (like neutrinos) would exhibit the same degeneracy pressure. Are you going to say that's because of the weak force? Really, there is no interaction term necessary to compute it. You are right in that you need a force to confine them to begin with, but in this case that's gravity. And degeneracy pressure opposes gravity.
     
    Last edited: May 30, 2008
  9. May 31, 2008 #8
    So the degeneracy pressure is to be considered a characteristic of the interaction that is causing the compression?

    Molu
     
  10. May 31, 2008 #9

    Dick

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    As it is independent of the nature of the force causing the compression, I'd have a hard time think of it like that.
     
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