# Why neutron degeneracy pressure is much stronger than electron degeneracy pressure?

I am asking about Pauli Exclusion Principle stating that no two fermions can have the same 4 quantum numbers which is why we have periodic table of elements, but straight to my question: despite Pauli Exclusion Principle is not an actual force then why neutron degeneracy pressure is much stronger than electron degeneracy pressure?

• relevant physicsforums.com/threads/… Jan 18, 2022 at 7:23
• At same density, electron degeneracy pressure is stronger than neutron d.p., since neutrons are heavier. The second answer in @annav 's link is probably what you are looking for. Jan 18, 2022 at 9:24
• relevant: physics.stackexchange.com/a/484517/226902 and physics.stackexchange.com/q/141865/226902 (in short: Pauli Exclusion -> need to stack fermions into states higher and higher amounts of momentum, because the the low-lying states are filled -> even at $T=0$, we have a lot of kinetic energy that is responsible for the degeneracy pressure -> the higher the mass of the fermion, the larger this kinetic energy and pressure). Jan 18, 2022 at 10:11
• @Koschi yes ,mfb's answer the "At the same density electron degeneracy pressure is larger. That's why white dwarfs are supported by it. If the electron degeneracy pressure gets too large then electron capture (proton+electron -> neutron+neutrino) become energetically favorable and the electron degeneracy pressure cannot increase more. If the mass is too large the object collapses. The created neutrons don't have a mechanism where they could disappear, their degeneracy pressure can grow to much larger values in a more compact object.", italics mine, reconciles it with the formulae found Jan 18, 2022 at 10:13

Let $$\sim$$ denote equality to within dimensionless factors dimensional analysis can't determine. Consider a star where the degeneracy pressure is due to $$N$$ fermions of a mass-$$m$$ species comprising a proportion $$p$$ of the stellar mass; for a neutron star, we can take $$p=1$$, whereas $$p$$ is very small when the degeneracy pressure is due to electrons.
A gravity pressure $$\sim G(Nm/p)^2/R^4$$ balances a degeneracy pressure $$\sim\hbar^2n^{5/3}/m$$, with $$n:=N/R^3$$ the number density of the relevant species. Hence $$R^4\sim\frac{\hbar^2p^2}{Gm^3N^{1/3}}$$, and the pressure $$\sim G^5N^{10/3}m^{14}/(\hbar^8p^{10})$$.
So it's a matter of comparing values of $$N^{10/3}m^{14}/p^{10}$$. The values of $$m/p$$ are similar in the two star types, so the neutron's greater mass is crucial. (Also, the neutron star will have greater $$N$$.)