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In what way is degeneracy pressure related to the separation of fermions? Is there any influence at ranges like a meter and beyond. I expect the influence at those ranges to be immeasurably small. I just wonder if theory predicts an effect or if there is truly zero degeneracy pressure at long range.

I'm under the impression that if you have a block of iron that is a solid cubic meter, then each and every electron in that block will be found in a slightly different quantum state at any given moment. This being due to the PEP.

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  • $\begingroup$ Hi Alex, I saw your question above and it prompted me to ask a related one, I had always assumed degeneracy pressure did not apply in normal situations, physics.stackexchange.com/questions/197742/… $\endgroup$ – user81619 Aug 4 '15 at 23:27
  • $\begingroup$ Hey, hopefully we both get an answer. $\endgroup$ – Alex Aug 4 '15 at 23:46
  • $\begingroup$ I got this quote as a comment and it might apply to your question "At commonly encountered densities, this pressure is so low that it can be neglected." I would guess because it's not an electro-magnetic effect (afaik), but instead the PEP, it may have finite range and is zero at finite range, just a guess, best of luck with an answer from some who really knows. Also, if the PEP only applies within atoms, to prevent overfilling of energy levels, would that be affected by an atom relatively far away...bit out of my depth $\endgroup$ – user81619 Aug 5 '15 at 0:32
  • $\begingroup$ I'm under the impression that if you have a block of iron that is a solid cubic meter, then each and every electron in that block will be found in a slightly different quantum state at any given moment. This being due to the PEP. $\endgroup$ – Alex Aug 5 '15 at 0:42
  • $\begingroup$ Put that above comment in your post, definitely. It clarifies your question a lot. If I have you right, you are saying the PEP may still "push" the neighbouring atoms outer electrons a little bit from their normal orbital position? $\endgroup$ – user81619 Aug 5 '15 at 1:01
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If you are considering ideal fermion degeneracy pressure, then the answer is simple.

If the fermions are non-relativistic, the degeneracy pressure scales as $n^{5/3}$, where $n$ is the fermion number density. Thus for a fixed number of fermions the pressure decreases more rapidly than the volume increases.

If the fermions are not completely degenerate then the "contribution of degeneracy" to the pressure (it cannot really be separated as such) is lower, and of course the fermions do become less degenerate as you reduce their density.

So in principle there is always some degeneracy pressure, but it would quickly become negligible.

However, this discussion is about free fermions. Your question appears to be about atoms. I have never come across degeneracy pressure being explicitly discussed in atomic material, so am unable to properly address your question I think. The force between atoms is often represented by models such as the Lennard-Jones or Morse potentials. These are repulsive at short range, presumably as a result of overlapping electronic wavefunctions, but are small and attractive at long range. Of course all this means is that any repulsion due to the PEP at long range is overwhelmed by other (attractive) effects.

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