# Degeneracy Pressure, What is it?

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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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.

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.