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What is "degeneracy pressure"?

I know there are 4 fundamental forces- EM, gravity, weak and strong. But then degeneracy comes along ubiquitously in everything right from neutron star to the electronic configuration. What are the causes of degeneracy in fermions??

Questions in SE suggests EM and degeneracy pressure are entirely different. But are they really? I mean what's the origin of degeneracy pressure among the fundamental forces?

Vineet Menon
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2 Answers2

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The use of the term "force" in quantum mechanics can be misleading, since the macroscopic classical force does not directly translate at the quantum level. That's why I prefer to speak about the 4 fundamental interactions rather than about forces.

And the answer to your question is: the degeneracy pressure is not linked to any of the four fundamental interaction. This pressure directly comes from the Pauli exclusion principle and kinetic energy (see below). Your question is similar to question in classical mechanics, where one would ask for the force responsible of the pressure of a perfect gas.

Now a rough calculation. Suppose you have a bunch of $n$ non-interacting spin-1/2 fermions confined in a volume $V$. Due to the Pauli exclusion principle, each of them is confined in a volume $\sim\frac{V}{2n}$, so we can imagine it in a cell of lateral dimension $$\Delta x\sim\left(\frac{V}{2n}\right)^{\frac13}.$$ The Heisenberg uncertainty principle then states that $\Delta x \Delta p \ge \frac\hbar2.$ We have therefore $$\Delta p \ge \frac{\hbar}{2\Delta x} \sim \frac{\hbar n^{1/3}}{2^{2/3}V^{1/3}}.$$ We will from now on be on the low temperature limit, where the Heisenberg uncertainty will be supposed to be saturated.

The average kinetic energy of each of these fermion is then given by $$E=\frac{\Delta p^2}{2m}\sim \frac{\hbar^2n^{2/3}}{2^{7/3}V^{2/3}m},$$ and the total internal energy by $$U=nE=\frac{\Delta p^2}{2m}\sim \frac{\hbar^2n^{5/3}}{2^{7/3}V^{2/3}m}.$$

Standard thermodynamics tels us how to compute the pressure from internal energy : $$P=-\left(\frac{\partial U}{\partial V}\right)_{S}\sim \frac{\hbar^2n^{5/3}}{2^{4/3}3V^{5/3}m} .$$ Since this is derived without any interaction, it is clearly independent of them.

Edited to add numerical evaluation:

If one wants to express this in terms of macroscopic quantities like the molar mass $M$ and the density $\rho$, one has $$P\sim\frac{\hbar^2\mathcal N^{5/3}}{2^{4/3}3m} \left(\frac{\rho}{M}\right)^{5/3},$$ where $m$ is the electron mass. The left-hand fraction is 6.9 SI units. If we suppose that a typical condensed matter has $M\sim 10 \mathrm g/\mathrm{mol}$ and $\rho\sim 10^3 \mathrm{kg}/\mathrm{m}^3$, $\rho/M\sim 10^5 \mathrm{mol}/\mathrm{m^3}$ and $P\sim 1.5 \mathrm{GPa}$ which is the correct order of magnitude.

Chris
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Frédéric Grosshans
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  • In classical mech, the pressure force of a gas is EM in nature. I know it's an analogy, but it's better if you explicitly state that given the context of the question. – Manishearth Mar 12 '12 at 15:01
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    @Manishearth No, there is no EM used to derive the pressure of a gas. Escept, of course if it is a plasma. If the atoms were little hard sphere, the formulas from thermodynamics would give the same results. And actually, when Boltzmann derived the perfect gas law from statistical mechanics, there was no possible link with EM. – Frédéric Grosshans Mar 12 '12 at 15:08
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    It's not used to derive the pressure, true, as we can do this by using momentum. But where do you think the force involved when a gas molecule knocks against a wall comes from? It is the EM repulsion of electron clouds, isn't it? Or it may even be the PEP... This sort of confusion is best avoided. – Manishearth Mar 12 '12 at 15:11
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    In that case, it's the PEP or a combnination of PEP and EM. But the actual nature of this force is irrelevant. I agree it can be confusing. I first thought of another example : "... where one would ask the force responsible for inertia." But I think the perfect gas example is closer to the degeneracy pressure. – Frédéric Grosshans Mar 12 '12 at 15:45
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    "But I think the perfect gas example is closer to the degeneracy pressure" -> Yeah kinda true in a way, because it is entirely caused by thermal fluctuation. PV=NkT seems to be the "heisenberg uncertainty" equivalent of statistical mechanics. – pcr Mar 12 '12 at 16:14
  • Not sure: both PEP and EM are considered exchange interaction. Photons are exchanged in EM so it is possible to go to the case when the photon itself is classical (indirect interaction, and it is local all the time classically). While in the PEP case, the two fermions are exchanging among themselves directly. The classical version of this exchange, if exists, would have been nonlocal. – pcr Mar 12 '12 at 16:21
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    Aah, I see.. Analogifying it with inertia makes sense.. Though I wouldn't call it irrelevant..No issue though, its a small point :) – Manishearth Mar 12 '12 at 16:49
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    @FrédéricGrosshans: so to conclude, PEP pressure cannot be attributed to any of the 4 interactions. But then doesn't that makes it 5th fundamental interaction? – Vineet Menon Mar 12 '12 at 17:47
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    @VineetMenon : No it doesn't make it an interaction. One can draw a parallelism with inertia in classical mechanics : in some sense, it resists movement like friction, but it does not make it a force like friction at all. – Frédéric Grosshans Mar 12 '12 at 17:53
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    But isn't it the coulomb force that causes each particle to be confined to a finite volume? Each electron is going to have a quantized wavefunction because of the potential of each electron's coulomb interaction.

    (And I would think that maybe you could obtain higher-order energy wavefunction solutions to schrodinger's equation as well)

     – Steven Sagona Jul 18 '17 at 03:26
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    +1 "Your question is similar to question in classical mechanics, where one would ask for the force responsible of the pressure of a perfect gas." this analogy is exactly the one I happenned upon when I was struggling to understand all this. I think the misconception arises because we think of neutron stars and the like as "solid", and our everyday experience of solids is that they're, well, solid i.e. we think of them as nondynamic entities and we thus lose sight of the dynamics and inertial effects. At least, this was what was going on inside my head before I felt I had a grip on this issue – Selene Routley Jul 18 '17 at 04:25
  • @StevenSagona: it can be the Coulomb Force (e.g. for delocalized electrons in metals and most ordinary solids and liquids), but also the strong nuclear interaction (in nuclei) or gravitation (in neutron stars or white dwarfs). In all these case, the degeneracy pressure is the same – Frédéric Grosshans Jul 18 '17 at 05:55
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    "The use of the term "force" in quantum mechanics can be misleading, since the macroscopical classical force does not directly translate at the quantum level. That's why I prefer to speak about the 4 fundamental interactions than about the force." I sympathize with this concern, but relative to classical particles the PEP is also sort of like an interaction, so I'm not sure this change in terminology actually helps anything. – Rococo Jan 19 '18 at 00:16
  • @Rococo Ultimately, it's just nomenclature. I think it's quite likely that, if degeneracy pressure had been experimentally observed before the development of quantum mechanics, it would be labeled as a fundamental force. Sure, it's different in nature than the strong, weak, and electromagnetic forces, but then so is gravity. – Chris Jan 19 '18 at 02:35
  • @Chris this isn't really the place for a detailed discussion, but I disagree with this statement entirely. Pauli exclusion effects cannot change the energy or momentum of a particle, so they really are quite different than the four 'gauge-like' forces. – Rococo Jan 19 '18 at 16:43
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    @Rococo I'm curious what you mean- if Pauli exclusion effects are responsible for halting the collapse of star, it's hard to see how you can say they can't effect the momentum of a particle. The evolution of a given particle's momentum is certainly quite different than it would be if there were no exclusion. – Chris Jan 19 '18 at 22:32
  • @Chris I mean, more or less, literally what I said- that PEP effects cannot make a particle transition from state (p,E) to (p',E'). This is completely consistent with degeneracy pressure and similar effects. I think that the ability to do this is a reasonable criterion for something to be a 'force'. – Rococo Jan 20 '18 at 05:57
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    @Rococo Fair enough. I'm not really saying it should be labeled a force, anyway. – Chris Jan 20 '18 at 06:11
  • Basically what Is called degenerate gas or matter is "where" the discussed fermions actually escape degeneration. Right? – Alchimista Dec 21 '20 at 13:37
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Its EM that keeps electrons from occupying the same space and manifests the pauli exclusion principle. Its that simple. The are only four forces and not a fifth pauli exclusion force. In a gas its also the EM force that manifest the pressure. Don't over complicate things.

Albert Smith
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    I agree there is no fifth force ; but it is NOT EM which prevents electrons from occupyig the same space. Otherwise, what prevents neutrons to occupy the same space ? – Frédéric Grosshans Sep 12 '16 at 10:28
  • @FrédéricGrosshans basically we are back with the fact that matter can't superimposed on "itself". It might lack formalism but it makes sense and I have no problem to see it as a fundamental properties, without requiring forces or labels. – Alchimista Jan 10 '21 at 17:55
  • @FrédéricGrosshans I would assume for neutrons it would be the color force (or strong nuclear force, if you prefer that name) which keeps neutrons from occupying the same space. I'd imagine electromagnetism (between the individual quarks) would play a much smaller role as well, but perhaps one so negligible that it can be ignored. I don't know enough to agree or disagree with Albert here, but if Albert's conceptualization is right, then I imagine what I just said is the answer. – mtraceur Jan 23 '23 at 06:07