I was at a museum recently, and there was a display on neutron stars. It said that neutron stars are made only of neutrons, which honestly didn't make much sense to me - neutrons decay very quickly on their own, so how do neutron stars "last", so to speak?

So naturally, I checked wikipedia, which provides this diagram:

enter image description here

Electrons and protons do seem to be present. Nowhere is there a "layer" that's only neutrons.

This leads to two questions:

  1. Is Wikipedia wrong, or the museum (normally I'd trust wikipedia, but this is a not-insignificant museum that I went to)?
  2. Is there anything that could explain the museum display if wikipedia is right?
  • $\begingroup$ Wikipedia is right. I'm very sure. Check this out: physics.stackexchange.com/q/206856 $\endgroup$ – Hritik Narayan Jun 27 '17 at 16:22
  • $\begingroup$ Closely related: physics.stackexchange.com/q/63383 physics.stackexchange.com/q/9098 and links therein, plus I thought there was a "composition of neutron stars" question, but I haven't yet found it. $\endgroup$ – dmckee --- ex-moderator kitten Jun 27 '17 at 16:47
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    $\begingroup$ As an aside, the museum display may be knowingly simplifying for an unsophisticated audience. The neutrons are roughly 1800 times as numerous as the electrons and protons so why quibble? $\endgroup$ – dmckee --- ex-moderator kitten Jun 27 '17 at 17:07
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    $\begingroup$ Yes, mostly neutrons are the key. AFAIK, a NS is so named b/c the process $p+e^-\to n+\nu_e$, which can't happen w/o protons & electrons. $\endgroup$ – Kyle Kanos Jun 27 '17 at 18:43
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    $\begingroup$ @dmckee You are quantitatively way off. $\endgroup$ – ProfRob Jun 27 '17 at 21:59

The picture from Wikipedia is (qualitatively) correct and so is your intuition.

Neutrons are unstable and decay unless the decay is blocked by the presence of a degenerate electron gas with a Fermi energy that is as large as the maximum possible energy of the electron produced in a beta decay.

If all fermion species are degenerate, and they are at neutron star densities, then the Fermi energies of neutrons, protons and electrons come into an equilibrium where $$E_{f,n} = E_{f,p} + E_{f,e}$$ This, combined with charge conservation, leads to the calculation that there are about ten to a hundred times as many neutrons than protons/electrons in what started as a pure neutron gas; the exact ratio being density dependent (smaller at higher densities).

Aside from this, the equilibrium state of "cold" neutron star matter varies with density, as shown in the Wikipedia diagram. The n,p,e fluid probably makes up the greater part of the mass of a neutron star, but by no means can it ever be said that a neutron star is made up only of neutrons, and in fact free neutrons only appear at densities above a few $10^{14}$ kg/m$^3$, some way in to the neutron star.

NB This is where the diagram is quantitatively incorrect. $\rho_0$ is supposed to be the nuclear saturation density of about $2.3\times 10^{17}$ kg/m$^3$, so free neutrons appear at just over $10^{-3}\rho_0$. In addition, nuclei lose their identity and become an n,p,e gas at about $0.2\rho_0$.


Neutrons are stable in the neutron star because there is such high pressures that neutrons are unable to decay. The states for protons and electrons make sure that the neutrons do not decay. Neutrons do decay but not enough to make a dent on the neutron star. Also any neutrons that do manage to decay to protons most likely convert back to neutrons. There is no complete layer with neutrons. Just more neutrons the further down you go in a neutron star. Both the museum and Wikipedia are right. When they say composed of "all neutrons" they do not literally mean made of completely neutrons. Instead they mean neutron stars have tremendous amount of neutrons but still have a small amount of other stuff.

  • $\begingroup$ Also the core of a neutron star may contain quark gluon plasma. They may also contain strange quarks. This is because the pressure is so great that the individual neutrons break into their quarks and gluons. Some of those quarks gets converted to strange quarks. $\endgroup$ – Roghan Arun May 6 '20 at 14:14

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