# Why the Fermi's energies of the proton, neutron and electron are related in this way in a neutron star?

I'm referring to this answer made by ProfRob about why neutrons are stable against beta decay in neutron stars.

I've partially understood the answer: when the Fermi's momentum of the electron $$p_f = (3\pi^2n_e\hbar^3)^{1/3}$$ increases above the maximum momentum $$p_{max}$$ produced by beta decay, the process stops. I calculated $$q_{max}$$ by imposing the conservation of energy and momentum in the center of mass frame : $$m_nc^2 = \sqrt{p_e^2c^2 + m_e^2c^4} + \sqrt{p_p^2c^2 + m_p^2c^4}$$ and $$\vec{p_e} = -\vec{p_p}$$ and found that $$p_{max}$$ is about $$1.2 \rm\, MeV$$. The notation is self-explanatory: $$n_e$$ represents the number density of electrons, $$m$$ is the mass and $$\vec{q}$$ is the momentum.

I don't get why, at equilibrium, the Fermi energies must be in this particular relation: $$E_{F,n} = E_{F,p} + E_{F,e}$$

• Thanks, I corrected that. Jun 20, 2023 at 15:10

The total energy density of the gas is given by $$u = u_n + u_p + u_e\ , \tag*{(1)}$$ assuming that neutrons, protons and electrons are the only species present and $$u_n$$, $$u_p$$ and $$u_e$$ are the energy densities of the neutrons, protons and electrons, given by a well-known expression that depends only on the number densities of those species in degenerate fermion gases.
However, we know that by charge conservation and conservation of baryon number that $$n_p = n_e \tag*{(2)}\ ,$$ $$n_n + n_p = {\rm constant}\ , \tag*{(3)}$$ where $$n_n$$, $$n_p$$ and $$n_e$$ are the number densities of neutrons, protons and electrons.
If we allow the system to reach equilibrium it will minimise the total energy density. If we differentiate eqn. (1) with respect to say the neutron number density and equate this to zero (for a minimum), then we can also use equations (2) and (3) to says that $$dn_p = dn_e$$ and that $$dn_p = - dn_n$$. Thus $$\frac{du}{dn_n} = \frac{du_n}{dn_n} + \frac{du_p}{dn_n} + \frac{du_e}{dn_n} = \frac{du_n}{dn_n} - \frac{du_p}{dn_p} - \frac{du_e}{dn_e} = 0 \ .$$
But for a degenerate gas, $$du/dn$$ is the Fermi energy and so we can say that $$E_{F,n} - E_{F,p} - E_{F,e} = 0\ .$$
Note that $$p_{\rm max}$$ in your calculation of 1.2 MeV/$$c$$ (in the rest frame of the neutron) cannot be directly compared with the Fermi momenta of the degenerate electrons, which are defined in a laboratory rest frame. $$p_{\rm max}$$ (in the lab frame) increases with the Fermi energy of the neutrons.