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As far as I understand, lowering the temperature of a bunch of fermions can let them reach their zero-point energies and subsequent lowest energy levels. That would make sense, since you'd be taking energy away from the fermions.

However, why does density lead to degeneracy? Is it for a similar reason?

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One of the several ways to understand the Pauli Exclusion principle is in terms of the number of fermions that can fit into a particular volume of Hamiltonian phase space. Say the volume we are interested in is a hyper-rectangle from $(x,y,z,p_x, p_y, p_z)$ to $(x+\Delta x,y+\Delta y,z+\Delta z,p_x+\Delta p_x, p_y+\Delta p_y, p_z+\Delta p_z)$. It has volume

$$V = V_s \cdot V_p = \left[(\Delta x)(\Delta y)(\Delta z)\right]\left[(\Delta p_x)(\Delta p_y)(\Delta p_z)\right]\;.$$

The number is that will fit is $\mathcal{N} = V/\hbar^3$ where $\hbar$ is the reduced Planck's constant.

Now, let us assume that we have $N$ atoms (I'm going to pretend the fermions are weakly interacting like an approximately ideal gas just to keep the language easy, so I'll call them 'atoms') of diffuse gas in a non-porous container of volume $V_s$ but in contact with a thermal reservoir of temperature $T$. The average momentum of the gas particles will be $\bar{p} = \sqrt{2 m k_b T}$, where $k_b$ is Boltzmann's constant, and $m$ is the mass of a single atom. Keep in mind that the momentum distribution will have tails that run from zero up to several times the $\bar{p}$. Let's say that they occupy a total momentum volume of $V_p \approx 400\,\bar{p}^3$ (the four hundred comes from assuming a sphere of radius $3\bar{p}$). By assumption that gas is diffuse, so $N \ll \mathcal{N}$.

Now we start reducing $V_s$, but slowly so the gas retains it's original temperature. At first all is well, but keep in mind that $\mathcal{N}$ is dropping proportionally. Eventually the number of atoms with nearly zero momentum gets to be too large to fit in the remaining spacial volume while keeping the same momentum.

That is, there is a limit $\mathcal{N}_l$ to the number of atoms with very low momentum that will fit in that volume and you're trying to force extras in there. No problem, though: you're doing work on the gas as you compress it, you'll just force a few to higher momentum as you squeeze. At this point the gas in decidedly not "diffuse" anymore, and the momentum distribution is no longer thermal at the low end but it is still thermal at the high end. You're just starting to have to do more work to compress the gas than you expected: the gentle on-set of degeneracy pressure.

Keep pushing and the product of the much reduced spacial volume and the volume of momentum space below $\bar{p}$ will start to approach $\hbar N$: you're out of space to for atoms with momentum comparable to the free average value at the temperature of the reservoir, and the degeneracy pressure is really setting in as you have to do more work to push atoms up to higher momentum states to squeeze them any closer together. The gas is significantly degenerate.

Keep pushing (much harder now) and the momentum volume needed to keep $V_s \cdot V_p$ above $\hbar^3 N$ will demand momentum far about the free thermal value. Every low lying level will be full and the gas will be fully degenerate.

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Also, one aspect of degeneracy that is often overlooked is that whether or not it is to be associated with high density depends on the other constraints on the system. The first answer above gave a complete accounting of the situation when the temperature T is being held fixed. An astrophysically important case is when you have a self-gravitating ball of gas, so T rises as the density rises. But there's another situation you rarely see, which is a box of constant volume V, and heat is just leaking out of the box. That situation will not change the density at all if we neglect forces between the particles (in reality, we'd have phase changes). So it's kind of an accident of the contexts in which we find degeneracy that it can be associated with high density. It's really net loss of heat that brings on degeneracy, you can squeeze a gas adiabatically all you want and it will not go degenerate at any density.

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Degeneracy, in this context, is defined as a deviation from a classical Boltzmann energy distribution of the fermions. It is known that fermions are governed by Fermi-Dirac statistics due to the Pauli exclusion principle. https://en.wikipedia.org/wiki/Fermi–Dirac_statistics At low densities and high temperatures this distribution can be approximated by the Boltzmann distribution. Deviation from the Boltzmann distribution occurs first at the low energy states which cannot be filled with more than 2 electrons due to the Pauli principle. https://en.wikipedia.org/wiki/Pauli_exclusion_principle

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  • $\begingroup$ The guy who downvoted this answer obviously didn't have the guts to give any reason for it. $\endgroup$
    – freecharly
    Nov 12, 2016 at 20:23

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