Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.
Sign up to join this community
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?
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.
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.
