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?


2 Answers 2


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.

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    $\begingroup$ @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. $\endgroup$ Commented Mar 12, 2012 at 15:08
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    $\begingroup$ 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. $\endgroup$ Commented Mar 12, 2012 at 15:11
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    $\begingroup$ 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. $\endgroup$ Commented Mar 12, 2012 at 15:45
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    $\begingroup$ "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. $\endgroup$
    – pcr
    Commented Mar 12, 2012 at 16:14
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    $\begingroup$ +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 $\endgroup$ Commented Jul 18, 2017 at 4:25

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.

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    $\begingroup$ 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 ? $\endgroup$ Commented Sep 12, 2016 at 10:28
  • $\begingroup$ @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. $\endgroup$
    – Alchimista
    Commented Jan 10, 2021 at 17:55
  • $\begingroup$ @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. $\endgroup$
    – mtraceur
    Commented Jan 23, 2023 at 6:07

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