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The wavelength of light fermions is longer. Wouldn't this cause them to take up 'more space' so that they didn't overlap according to the Pauli exclusion principle? Am I totally misunderstanding something?

In a semi-related question... If this is true, what are the consequences for neutrinos?

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As asked your question is a little unclear. I assume you are talking about a states which have reason to be degenerate? And want to know how the size of the structure relates to the mass of the degenerate constituents.

Well, here are two examples. (Well, from some point of view they are two versions of the same example.):

  • The size of atoms (i.e. the electron clouds) of heavy elements and the size of the nucleus of the same.

  • White dwarf stars (held up by degenerate electron pressure) and neutron stars (help up by degenerate neutron pressure).

In both cases we have a structures whose size is determined by Fermi exclusion of some of the parts from the low level states. In both cases we compare structures of composed of electrons to structures composed of nucleons (and maybe some strange hadron in the neutron star---details still pending).

In both cases the electron-based structure is orders of magnitude larger than the nucleon-based structure of a similar number of components.

These comparisons are not perfect--the attractive potentials are different in both cases.

I find Fermi exclusion easier to visualize in momentum space, in which case the lighter fermions have higher velocities and larger orbital extents in any given attractive potential.

As for neutrinos they basically aren't any consequences: they only respond to the weak force and it's range is too short for neutrinos to form a degenerate gas. Nor can you put them in a box of any known material and attempt to squeeze them as they'll just fly through the walls.

The only place they could be made degenerate is very close to the event horizon of a black hole where from an outside perspective they will appear frozen in time in any case.

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Yes, this is basically what I was asking. And I'm curious about neutrinos from a theoretical perspective rather than a practical one. I mean, maybe someday we'll find a way to create a 'weak field' of appreciable strength and use that to trap neutrinos just like we use electromagnetic fields to trap electrons in quantum dots and the like. – Omnifarious Feb 17 '12 at 22:41
The range of a weak field is limited by the $W^{\pm}$ mass. – dmckee Feb 17 '12 at 23:32
Oh, the fact the W boson has mass has some bearing on the range of the weak force? Is this because the W boson will very quickly decay into other things? – Omnifarious Feb 18 '12 at 0:35

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