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# 6.1 Neutron star composition

For normal matter, the most stable configurations, with the lowest binding energy per nucleon, are nuclei around iron-56. However, when the density approaches , the most stable nuclei are actually far more neutron-rich, such as nickel-78 and iron-76. At still higher densities, above , a phenomenon called neutron drip occurs, whereby neutrons ‘leak out’ of nuclei and an equilibrium mixture of nuclei, neutrons and electrons exists. When the density exceeds that of nuclear matter (at about ), the nuclei begin to merge and a dense gas of protons, neutrons and electrons is produced. Under these conditions there are complicated and uncertain interactions between nuclei, so the equation of state for neutron stars is not known.

Despite this, a reasonable idea of the composition of a neutron star can be gained by considering a crude model in which the neutron star is treated as an ideal gas of degenerate electrons, protons and neutrons. In this situation, the normal beta decay of free neutrons is blocked because of the Pauli exclusion principle: there are no available quantum states left for the protons and electrons to occupy, so the neutrons are not able to decay. In particular, neutrons cannot decay if their Fermi energy is less than the sum of the Fermi energies of the electrons and protons, but they can decay if their Fermi energy is greater than that of the protons plus the electrons. An equilibrium will therefore exist when

The neutrons and protons can be considered to be non-relativistic; hence their Fermi energies are given in each case by the sum of a rest-mass energy term and a kinetic energy term:

where mn and mp are the masses of the neutron and proton, and pF(n) and pF(p) are their Fermi momenta. Conversely, the electrons are much less massive than the protons and neutrons, so they can be considered ultra-relativistic. Their Fermi energy is simply

where pF(e) is the electron’s Fermi momentum.

Now, because the Fermi momentum is (Equation 8), we can write the equilibrium condition as

Because the gas will be electrically neutral, the number densities of the protons and electrons must be equal: np = ne. Furthermore, the mass-energy difference between the neutron and proton is , so we can rearrange this as

This equation can be solved numerically to find the ratio of neutrons to protons (nn / np) at a given density. For instance, for a typical neutron star density of , the equation predicts a neutron-to-proton ratio of about 200 to 1. So, as long as there is one proton and one electron for every 200 neutrons, the neutrons are prevented from decaying, and the neutron star remains supported, essentially by the pressure of degenerate neutrons. Whereas white dwarfs are supported by electron degeneracy pressure, neutron stars are supported against further collapse by neutron degeneracy pressure.

## Activity 10

Suppose that the neutron star is so dense that the protons and neutrons (as well as the electrons) are ultra-relativistic. What is the ratio of neutrons to protons in this case?

### Discussion

We again start by writing , but now we have EF = pFc for each type of particle where . So we can write the equilibrium condition as

and because np = ne in a neutral gas, this reduces to or . Thus, even in the ultra-relativistic case, neutrons far outnumber protons, by a factor of 8 to 1.

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