Why is the number density of neutrons much larger in neutron stars?

Question

There is a particular argument given in Concepts In Thermal Physics by Blundell that I'm not able to understand:

A free neutron can decay with a mean life of about $15$ minutes but in stars, we have to consider the equilibrium $$n \rightleftharpoons p^++e^-+\nu_e$$ Because the electrons are relativistic, their Fermi energy is proportional to $p_F\propto n^{1/3}$, while the neutrons are non-relativistic and so their Fermi energy is proportional to $p^2_F\propto n^{2/3}$. Thus at high density, an equilibrium can be established in the reaction.

This implies that the Fermi momentum of the electrons is much smaller than that of neutrons and hence the number density of electrons will be much smaller than that of the neutrons. This moves the equilibrium of the reaction to the left side.

First, why do we need high density for equilibrium to be established? Once this is true, I think the other things are quite easy to see.

Answer 1

The neutrons are surrounded by a dense, degenerate electron (and proton) gas, which is produced by neutron decay. The degeneracy means that all energy states are totally filled by electrons up to a maximum value called the Fermi energy. This Fermi energy will increase if you make the electron gas denser.

Neutrons are stabilised if the electron gas density is high enough. The high density is required to push the Fermi energy of the degenerate electrons high enough that it blocks the usual neutron beta decay. That is, any newly created electron would be likely to have an energy lower than the electron Fermi energy, and since all available energy states up to the Fermi energy are already filled, then the new electron cannot be created (due to the Pauli Exclusion Principle, which forbids two electrons occupying the same quantum state).

But as the density increases so does the neutron Fermi energy and this means the decay electron can also be more energetic. An equilibrium is setup whereby the Fermi energies are related by $$E_{F,n}= E_{F,p} + E_{F,e}$$

When the above equation is solved at neutron star densities, one finds that the equilibrium is reached at about 1 electron (and proton) for every 50 neutrons, but the ratio gets smaller at higher densities.

Answer 2

In nature, a lone neutron is not stable. It will, as you mentioned about it’s half life, breakdown pretty quickly. This is because the neutron is more massive and thus energetic than the sum of its constituents; nature always leans toward the minimum energetic level, and for a lone neutron that is achieved by decay. However in a hyper dense environment, like in a neutron star, you’ve got gravity pushing all these particles with such force that they went from a white dwarf or a stellar remnant and recombination of protons and electrons into a more compact form, namely, neutrons.

To answer your question, high density allows an otherwise unstable particle to remain stable, since the decaying of which no longer is a means of an energetic release. There may be a bit more nuance here in the details, but that’s the general idea.