I understand that the EM field has an EM charge and that a nucleus made of neutrons and protons has an opposite EM charge, and this attracts.

And because the electron's kinetic energy (that would make the electron fly away) repels, so the attraction and the repelling equal out.

So the electron is in a stable quantized energy level in the EM field (it will not change its energy level without for example a photon absorption), and it also makes the nucleus stable.

I understand the neutron does not have EM charge and that strong force and residual strong force (nuclear force) works between two neutrons, keeps them stick together.


  1. Can a nucleus of only neutrons exist, and be stable? Can the nuclear force work between two neutrons even without protons next to them?

  2. Can this nucleus exist and be stable (since it does not have EM charge) without an EM field?

  • $\begingroup$ you should look up neutrium - yes its a thing $\endgroup$ – Cursed Apr 1 '18 at 9:43
  • $\begingroup$ Try reading this and see if some of your doubt is resolved. en.wikipedia.org/wiki/Neutronium $\endgroup$ – Netravat Pendsey Apr 1 '18 at 9:45
  • $\begingroup$ As a nucleus is a group of particles that electrons can orbit around, I'd say no to this as without an EM field the electrons cannot orbit. $\endgroup$ – StephenG Apr 1 '18 at 10:31
  • $\begingroup$ A short summary on the search of trineutron and tetraneutron Roman Ya. Kezerashvili 2016 "In light of a new experiment which claims an identification of tetraneutron [3], we discuss the results of experimental search of trineutron and tetraneutron in different nuclear reactions. A summary of theoretical studies for trineutron and tetraneutron within variety of approaches such as variational methods, the method of Faddeev and Faddeev-Yakubovsky equations, and the method of hyperspherical harmonics are presented." $\endgroup$ – Keith McClary Apr 1 '18 at 17:13
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    $\begingroup$ The part of your question relating to electromagnetic fields is unclear. EM fields are not usually described as "stable" or "unstable", nor as being charged themselves. Rather, charged particles always have EM fields, and uncharged particles may also (for instance, the neutron has a magnetic field). I especially can't make any sense at all of your first paragraph. $\endgroup$ – zwol Apr 1 '18 at 19:43

The two neutron state (the dineutron) is known to be unbound. As far as I know no definitive calculations have been done for the trineutron, quadneutron,etc states but they are expected to be unbound as well.

A couple of the comments have mentioned neutronium, but I'm not sure that counts as a bound state. We expect it to exist only where gravitational forces hold the neutrons together.

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    $\begingroup$ Maybe we can see them as gravitationally bound states. $\endgroup$ – peterh Apr 2 '18 at 0:52
  • $\begingroup$ John, neutronium is made up science fiction. Neutrons cannot exist without at least some of them decaying into protons and electrons. $\endgroup$ – Rob Jeffries Apr 2 '18 at 10:22
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    $\begingroup$ @RobJeffries I think we can afford to be generous with the terminology and accept neutronium to mean a form of degenerate matter that is mostly neutrons. In any case I don't think the OP had it in mind when asking the question. $\endgroup$ – John Rennie Apr 2 '18 at 11:03
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    $\begingroup$ Given that the question is specifically about the absence of charged particles, my pedantry on this point is entirely justified. $\endgroup$ – Rob Jeffries Apr 2 '18 at 11:49

The closest you will ever come in nature to pure neutron matter is the nuclear matter in neutron stars, but it is not pure either, being suffused with protons and electrons. Neutrons are 1.4 MeV heavier than protons and tend to undergo beta decay to $pe\bar{\nu }$, unless stabilized by the exclusion principle and an abundance of protons, electrons, or antineutrinos in the vicinity. Were it not for the Coulomb repulsion between protons, ordinary nuclear matter would have more protons than heavier neutrons!

Let us concentrate on nuclear matter. Recall that in a non-interacting Fermi gas, all momentum states below the spherical Fermi surface of radius ${{p}_{F}}$ are filled. The condition for overall electrical neutrality is $p_{p}^{3}=p_{e}^{3}$, based on the volume inside the Fermi sphere. At zero temperature, we may identify the chemical potential with the corresponding energy, $\mu ={{E}_{F}}=\sqrt{p_{F}^{2}+{{m}^{2}}}$. The condition for chemical equilibrium is ${{\mu }_{n}}={{\mu }_{p}}+{{\mu }_{e}}+{{\mu }_{{\bar{\nu }}}}$. Interactions roil the Fermi surface but don’t actually change the equilibrium significantly. Since nuclear forces are independent of charge, they do not distort the difference ${{\mu }_{n}}-{{\mu }_{p}}$.

When a neutron star forms, most protons undergo inverse beta decay, and it seems safe to assume that the neutrinos escape, hence ${{\mu }_{{\bar{\nu }}}}=-{{\mu }_{\nu }}=0$. We now have enough conditions to determine the proportions of neutrons, protons, and electrons at any given density.

The chemical potential of neutrinos is more relevant cosmologically. If God had assigned the creation of the universe to his most incompetent archangel-trainee, who then poured too many antineutrinos into the mix, you might very well have nothing but pure neutron matter, and you wouldn’t be here to ask about it. (Hmmm ... neutrons instead of neurons ... even worse than having rocks in one’s head.)

At densities well above what is expected in neutron stars, nucleons would overlap, so it would be more accurate to think in terms of quark matter, comprising u and d quarks in three colors. The conditions become ${{\mu }_{d}}={{\mu }_{u}}+{{\mu }_{e}}+{{\mu }_{{\bar{\nu }}}}$ and $-p_{e}^{3}+\tfrac{2}{3}(3)p_{u}^{3}-\tfrac{1}{3}(3)p_{d}^{3}=0$.


protected by Qmechanic Apr 2 '18 at 14:18

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