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There has been numerous question, some violent even in physics@SE regarding PEP and EM forces.

But what baffles me is what is degeneracy pressure? I know there are 4 fundamental forces- EM, gravity, weak and strong.

But then degeneracy comes along in everywhere, from neutron star to the electronic configuration. What causes this 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?

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up vote 7 down vote accepted

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.

And the answer to your question is : the degeneracy pressure is not linked to any of the four fundamental interaction. This pressure directly com, 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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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
@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
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
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
@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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