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White dwarfs and neutron stars
White dwarfs and neutron stars

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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 10 super 14 kg m super negative three , the most stable nuclei are actually far more neutron-rich, such as nickel-78 and iron-76. At still higher densities, above four multiplication 10 super 14 kg m super negative three , 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 two multiplication 10 super 17 kg m super negative three ), 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

cap e sub cap f of n equals cap e sub cap f of p plus cap e sub cap f of e full stop

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:

neutrons colon cap e sub cap f of n equals m sub n times c squared plus p sub cap f times left parenthesis n right parenthesis squared divided by two times m sub n
protons colon cap e sub cap f of p equals m sub p times c squared plus p sub cap f times left parenthesis p right parenthesis squared divided by two times m sub p comma

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

electrons colon cap e sub cap f of e equals p sub cap f of e times c comma

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

Now, because the Fermi momentum is p sub cap f equals left parenthesis three times n solidus eight times pi right parenthesis super one solidus three times h (Equation 8), we can write the equilibrium condition as

m sub n times c squared plus left parenthesis three times n sub n divided by eight times pi right parenthesis super two solidus three times h squared divided by two times m sub n equals sum with 3 summands m sub p times c squared plus left parenthesis three times n sub p divided by eight times pi right parenthesis super two solidus three times h squared divided by two times m sub p plus left parenthesis three times n sub e divided by eight times pi right parenthesis super one solidus three times h times c full stop

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 m sub n times c squared minus m sub p times c squared equals 1.29 MeV , so we can rearrange this as

left parenthesis three times n sub p divided by eight times pi right parenthesis super two solidus three times h squared divided by two times m sub p plus left parenthesis three times n sub p divided by eight times pi right parenthesis super one solidus three times h times c minus left parenthesis three times n sub n divided by eight times pi right parenthesis super two solidus three times h squared divided by two times m sub n equals 1.29 MeV full stop

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 2.5 multiplication 10 super 17 kg m super negative three , 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?

Comment

We again start by writing cap e sub cap f of n equals cap e sub cap f of p plus cap e sub cap f of e , but now we have EF = pFc for each type of particle where p sub cap f equals left parenthesis three times n solidus eight times pi right parenthesis super one solidus three times h . So we can write the equilibrium condition as

left parenthesis three times n sub n divided by eight times pi right parenthesis super one solidus three times h times c equals left parenthesis three times n sub p divided by eight times pi right parenthesis super one solidus three times h times c plus left parenthesis three times n sub e divided by eight times pi right parenthesis super one solidus three times h times c

and because np = ne in a neutral gas, this reduces to n sub n super one solidus three equals two times n sub p super one solidus three or equation sequence part 1 n sub n solidus n sub p equals part 2 two cubed equals part 3 eight . Thus, even in the ultra-relativistic case, neutrons far outnumber protons, by a factor of 8 to 1.