1. What happens to all of the electrons and protons in the material of a neutron star?

  2. Could there ever be an electron star or a proton star?

  • 1
    $\begingroup$ btw, there is "Electron stars for holographic metallic criticality": arxiv.org/abs/1008.2828; maybe some experts in the field can comment about this? $\endgroup$ – wonderich Jun 5 '14 at 2:46
  • 14
    $\begingroup$ I figure an electron star would probably better be referred to as a "Coulomb bomb"... $\endgroup$ – Johannes Jun 5 '14 at 2:48

If a dense, spherical star were made of uniformly charged matter, there'd be an attractive gravitational force and a repulsive electrical force. These would balance for a very small net charge: $$ dF = \frac1{r^2}\left( - GM_\text{inside} dm + \frac1{4\pi\epsilon_0}Q_\text{inside} dq \right) $$ which balances if $$ \frac{dq}{dm} = \frac{Q_\text{inside}}{M_\text{inside}} = \sqrt{G\cdot 4\pi\epsilon_0} \approx 10^{-18} \frac{e}{\mathrm{GeV}/c^2}. $$ This is approximately one extra fundamental charge per $10^{18}$ nucleons, or a million extra charges per mole — not much. Any more charge than this and the star would be unbound and fly apart.

What actually happens is that the protons and electrons undergo electron capture to produce neutrons and electron-type neutrinos.


The other answers cover your second question well enough, but there's some detail still missing on the first one - what happens to the protons and electrons in a star when it collapses into a neutron star. The basic answer is simple: they become those neutrons.

The reason that this happens is that, as it turns out, an {electron, proton} pair is sort of interchangeable with a neutron, or at least it's interchangeable given enough energy. The "more natural" version of the reaction, in fact, goes the other way: on its own, a neutron will in fact decay into a proton, emitting an electron in the process to keep the charge balance happy.

$$ n\to p^++e^-+\bar{\nu}_e $$

This is the most basic example of beta decay, and it has a half-life of about 15 minutes, which is fairly fast for a weak-force reaction.

The $\bar{\nu}_e$ thing is an antineutrino, which needs to be emitted to keep the lepton number constant. It has no mass, but it does carry energy, so what happens is that the neutron can turn into a proton and thereby lose a bit of mass, which becomes enough energy to materialize the electron and the antineutrino and accelerate them to keV energies.

Now, one of the cool things about particle physics is that it is all essentially time-reversible, which means that you can run any reaction in reverse. In this case, you can do something like $$p^++e^-\to n+\nu_e$$ if you have enough energy around to run it.

In any given star, you will have both reactions happening with some probability. You will have some amount of free neutrons sticking around, and these will decay into proton-electron pairs, but you will also have a lot of protons and electrons around in an energetic environment, so if two of them crash with enough energy, they will coalesce into a neutron for a bit.

The key word, though, is "enough" energy, and in a normal star the thermal energy - say, ~1keV for the 16 MK at the sun's core - is not enough to provide a significant fraction of the proton-electron collisions the ~780keV they need to produce a neutron. Nevertheless, in any thermal environment there will be some bits of the system that fluctuate to energies $E$ bigger than the thermal energy $k_B T$, with probability $e^{-E/k_B T}$. In this case, this gives a rough estimate that $e^{-1.35/780}\approx 0.1\%$ of proton-electron collisions produce a neutron, which is small but not completely negligible.

So much for normal stars in equilibrium. To make a neutron star, you need something else to break this equation, and this turns out to be an immense amount of pressure: the electron is essentially pushed into the proton by the surrounding plasma. Once nuclear fusion ceases to have fuel, the temperature can no longer keep up with the pressure and, at fairly constant temperatures*, the pressure rises to huge levels.

The reason the pressure changes the game is that the electron capture reaction significantly reduces the volume occupied by the system, which means that the environment performs work on the system by pushing it in, in exactly the same way that a piston performs work on a gas that's inside a box. It is this extra work, performed over a tiny volume by an absolutely humongous pressure, that provides the sizable >780 keV of energy required to make the electron capture reaction favourable.

* Or something like it. Experts, please correct me if I'm wrong.

  • $\begingroup$ The 780keV is the vacuum value for the threshold kinetic energy of the electron. In a neutron star, vastly more energy is required because the neutrons are degenerate. So more like tens of MeV. $\endgroup$ – Rob Jeffries Nov 30 '14 at 21:37
  • $\begingroup$ That's fair enough - I'm OK with that. The calculation with that number is for a normal star with few neutrons around. $\endgroup$ – Emilio Pisanty Dec 1 '14 at 10:16

The inner force of gravitation is so strong than outward pressure that the electron is forced inside the nucleus and fuses with the proton so become a neutral particle similar to neutron. In a sense , we can tell that the nuclues contains only neutron and thus called neutron star.

  • 2
    $\begingroup$ +1, although I think it would be correcter to say (I'm not a particle physicist) that the electron capture or "inverse beta decay" reaction, exactly the same as that undergone by a proton in certain unstable isotopes, EXACTLY yields a neutron. Also, I'm not sure that we can directly "tell" the neutron star is neutrons: my guess is that this is a theoretically inferred and we haven't confirmed it directly (since we haven't probed a neutron star with instruments), but I'd be interested to hear differently. $\endgroup$ – WetSavannaAnimal Jun 5 '14 at 3:36
  • $\begingroup$ Actually Johannes makes a good point in his comment above: "I figure an electron star would probably better be referred to as a "Coulomb bomb"" - this is another way we know that there can only be neutrons: the other kind would be unstable. $\endgroup$ – WetSavannaAnimal Jun 5 '14 at 3:41
  • 1
    $\begingroup$ Same comment as for Rob above: Neutron stars do not contain only neutrons. There are still plenty of protons and electrons (well ~10%). @WetSavannaAnimal one clue is that the direct URCA process could easily occur if the neutron star was not dominated by degenerate n. If the p/n ratio was high neutron stars would cool extremely fast. $\endgroup$ – Rob Jeffries Nov 21 '14 at 20:23

Addon to the present answers. They so far neglect the strong interaction, which keeps the known atom cores together, working "against" the mutual electric repulsion of the protons. But even $^2$He is not stable. Since the gravitational force is significantly weaker as the electromagnetic, proton stars are (as far as I know) not possible.

  • $\begingroup$ The strong force is a "contact force": the form of the potential is $V = \alpha_s r^{-1} \exp-m_\pi r$, where $m_\pi c^2 = \hbar c/\mathrm{1.2\,fm}$ is the mass of the pion. Nucleons separated by more than eight or ten femtometers don't feel the strong force, which is why there aren't any stable nuclei with masses greater than about 250 grams per mole. $\endgroup$ – rob Jun 5 '14 at 6:48
  • $\begingroup$ @rob I'd like to understand the relation of nucleon distance and mass per mol you're using - could you give me a hint where to look? (ie a related concept or so) $\endgroup$ – Volker Siegel Jun 5 '14 at 7:15
  • $\begingroup$ @rob I know, therefore it's just an addon to the prevoius answers $\endgroup$ – Lord_Gestalter Jun 5 '14 at 7:36
  • $\begingroup$ @VolkerSiegel Molar mass in grams per mole is approximately the number of nucleons in a nucleus (4 for helium, 27 for aluminum, 230-240 for uranium, usually written $A$). Nuclear matter has roughly constant density, so the radius of a nucleus is roughly $1.2\mathrm{fm}\,A^{1/3}$; even uranium only has radius $^3\sqrt{240}\approx6$. The Yukawa potential is important enough to find in any book on nuclear physics. $\endgroup$ – rob Jun 5 '14 at 12:46
  • $\begingroup$ Instability of ²He is irrelevant, because a hypothetical proton star is expected to be bound by its own gravity – a force that is negligible in ordinary nuclides. Is ²n stable, or do heavier isotopes of neutronium exist? $\endgroup$ – Incnis Mrsi Oct 24 '14 at 13:47

The answer to the main question is no. The repulsive force due to "like charges" is orders of magnitude larger than the attractive gravitational force, so it would be impossible to form a star. In the case of "opposite charges" the situation is now reversed, the opposite charges attract and Helium atoms are created. If enough of them are created, the attractive gravitational force increases until it is larger than the force separating the protons and the electrons and they "fuse," creating neutrons. This continues until all the electrons (or protons) are gone, thus a neutron star would be created.

  • $\begingroup$ And the same comment as above. Neutron stars are not just neutrons! $\endgroup$ – Rob Jeffries Nov 21 '14 at 20:24

Not the answer you're looking for? Browse other questions tagged or ask your own question.