Difference between Urca process and $\beta$-decay in neutron stars cooling

Question

I was reading about the Urca process and its importance in the cooling of astrophysical compact objects. Indeed, it is supposed to be one of the major contribution to the cooling of neutron stars (please, correct this sentence if it is not actually supported). Now, the general expression for the Urca process is:

$B_1\rightarrow B_2 + l + \bar{\nu_1}$

Where $B_1$ and $B_2$ are baryons and $l$ is a lepton. This is, in my understanding, the general expression for $\beta$-decay:

$n\rightarrow p + e^- + \bar{\nu_e}$

Since the neutron star is made mostly by neutrons, why is the Urca process indicated instead of the $\beta$-decay ? In other words, what else is going to contribute to neutrino emission and so to the neutron star cooling by neutrinos?

Relating to this question, also muons can be produced, but this asks for more energetic neutrons, which is a limited condition, since the particles are tightly packed in the neutron star. So, what am I missing?

Answer 1

A lot of what is written in the question is not correct.

The URCA process would be cycles of beta decay followed by inverse beta decay as follows.

$$n \rightarrow p + e + \bar{\nu_e}$$ $$ p + e \rightarrow n + \nu_e$$

The neutrinos escape from the star, carrying away energy.

The URCA process is very important during the collapse to a neutron star state, during the initial cooling of a neutron star and possibly in the most extreme density cores of neutron stars at later times.

The direct URCA process can be be blocked by degeneracy. In the neutron fluid, there must be a small number (of order 1 part in 100) protons and electrons, because neutrons are unstable and undergo beta decay. However, once the electron Fermi energy reaches the maximum possible decay energy of a beta electron, then the reaction ceases because there are no free energy states for it to occupy. Instead, inverse beta decay (or neutronisation) becomes possible, if the electron Fermi energy + proton Fermi energy equals or exceeds the neutron Fermi energy. An equilibrium is set up so that $$E_{F,n} = E_{F,p} + E_{F,e}$$.

Ok, having got this you should then look at my answer to this question, where I explain why the direct URCA process is blocked unless densities are very high ($>8\times10^{17}$ kg/m$^3)$, because it is impossible for the reactions to simultaneously conserve energy and momentum. At lower densities, the modified URCA process is inefficient, yet more effective. This is similar to the URCA process but includes a bystander particle (a neutron) to enable simultaneous conservation of energy and momentum, at the expense of requiring an extra particle to be within $kT$ of it's Fermi surface.

I do not really understand what you are asking in the last part of your question. Muons are produced at high densities. The higher density leads to higher neutron Fermi energies, and once this Fermi energy exceeds the Fermi energy of the protons plus the rest mass energy of the muons (105 Mev) then they can emerge. Almost equivalently, one can demand that the Fermi energy of the electrons in the gas reach 105 MeV and then muons can be created. It is the appearance of muons at $\sim 8\times10^{17}$ kg/m$^3$ that increases the proton fraction, reduces the difference between the neutron and proton Fermi momenta and makes the direct URCA process possible.