I can understand why 2 protons will repel each other, because they're both positive. But there isn't a neutral charge is there? So why do neutrons repel? (Do they, or have I been misinformed?)

The reason why I'm asking this is because, I've just been learning about neutron stars and how the neutrons are forced (as in, they repel) together according to my teacher (he's a great teacher btw, though what I just said doesn't make it seem so).

So I wondered, why they have to be forced by gravity and not just pushed?

  • $\begingroup$ Good old times, this was achieved by a principle, the "principle of inpenetratability of bodys" look for it in physics textbooks older than , say, 80 years. :=) $\endgroup$ – Georg Feb 3 '11 at 14:46
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    $\begingroup$ @Georg: This principle is gone because it isn't true. There is no inpenetrability in nature, that's just Aristotle gobbledygook. $\endgroup$ – Ron Maimon Apr 25 '12 at 23:09

Neutrons have spin 1/2 and therefore obey the pauli exclusion principle, meaning two neutrons cannot occupy the same space at the same time. When two neutrons' wavefunctions overlap, they feel a strong repulsive force. See http://en.wikipedia.org/wiki/Exchange_interaction .

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    $\begingroup$ Jonathan, this is the simple and most correct answer, but there is a lot to learn to really understand it. There is an explanation along these lines for sophomore college students in The Feynman Lectures on Physics, Vol III, lecture 6 (but read lectures 4 and 5 first). See e.g. amazon.com/Feynman-Lectures-Physics-Set/dp/0201021153 . Your teacher can probably help. $\endgroup$ – sigoldberg1 Nov 3 '10 at 3:50
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    $\begingroup$ While this answers the question in the body of the questions pretty well, it is incomplete with respect to the title as it does not explain why a bound state of two neutrons with opposite spins can't exist but the deuteron does. $\endgroup$ – dmckee Jun 9 '14 at 15:28
  • $\begingroup$ The PEP says that two fermions cannot occupy the same quantum state. There is no force involved here. Neutrons do repel each other but that is due to the strong nuclear force at small separations, not the PEP. $\endgroup$ – Rob Jeffries Oct 18 '14 at 19:43

Neutrons (and protons) being spin 1/2 fermions, must fit antisymmetric wavefunctions. This "wavefunction" doesn't always involve waves, though. For nucleons - the generic term for neutron or proton - this wavefunction for the pair is a produce of (1) a spatial part, (2) a spin part, and (3) an isospin part.

The isospin part is a clever way to describe charge possibilities of otherwise identical particles. We regard neutrons and protons as being in a sense identical. Just as a spin 1/2 particle can be "up" or "down" along some chosen axis, so is an isospin 1/2 particle can be "up" or "down" along an abstract mathematical axis - it's exactly the same SU(2) math as spin - but it plays out in physical reality as charge. For nucleons it's not +1/2 and -1/2 charge but with an offset, so we have +1 (proton) and 0 (neutron). This idea is from Heisenberg in 1932.

Now, how can the overall wavefunction of a pair of particles be antisymmetric? There are three factors - right away we can imagine three possibilites: any one factor being antisymmetric with the other two symmetric. We could also have all three antisymmetric at the same time.

An antisymmetric spatial wavefunction would have a node, like an atomic p orbital, like the electric potential around a dipole antenna. This is a higher energy state than a simple spherical blog, a Gaussian. Given the range of internuclear forces, this nodal antisymmetric wavefunction has more energy than if the two nucleons just stayed apart. This is a matter of radial or angular kinetic energy has to be either "zero" or some quantized value that exceeds "escape velocity" So forget that part of the system wavefunction being antisymmetric.

BTW, we don't have separate spatial wavefunctions for the two nucleons - whatever one does, the partner does the exact opposite, like a two-body celestial mechanics problem. They orbit a common barycenter.

The spin part could be antisymmetric. This is a bit tricky. If particle #1 is up and #2 is down, we can write "UD". There is also "DU". We form the spin part of the wavefunction for the pair as UD-DU. We could instead choose UD+DU but note this is symmetric. So are UU and DD. Just how UD-DU differs from UD+DU may mystify beginners in quantum mechanics, but it's important, and it's how physical matter works whether we humans like it or not. (You might also see where the 'u' and 'd' quarks got their names. The quark idea came along years after isospin.)

Neither D nor U is really a wave or a function; they're at most just rows and columns in matrices if you must represent them in familiar math. Otherwise quantum physicists just deal with these symbolically. Still the jargon is "wavefunction" - we silly humans and our primitive scientific language!

The same math applies for isospin. But the physics differs. We've concluded that the spatial part of the system wavefunction must be symmetric, so it's up to either the spin part or the isospin part to be antisymmetric. But not both! If spins are symmetric - they're parallel. This is experimentally the case - the Deuteron (obtain by distilling "heavy water" from water) - so we deduce that the isospin part is antisymmetric. That is, we must have one isospin "up" and one isospin "down" - neutron and a proton, not two neutrons or two protons.

Just why must the spins of the two nucleons be parallel? The strong force holding them together - the exchange of pions, kaons and other mesons - works better in that case. To explain that takes deeper analysis than I can go into here. When the spins are antiparallel, there's not enough force to keep the nucleons together.

This would be the case though, were to try pushing two neutrons together. They'd be both isospin "up" therefore symmetric isospin part of the wavefunction, therefore requiring an antisymmetric spin part, which leads to the pions and their buddies not getting as good a grip on the neutrons, which drift off going their separate ways.


I think there are two parts to this answer. The first is to do with an ensemble of neutrons in a dense fermion gas and the second is to do with the strong nuclear force between two neutrons (in a many-body nucleon system).

Neutrons in a dense gas will be degenerate. That is to say that the Pauli exclusion principle prevents more than two neutrons (spin up and spin down) occupying the same momentum eigenstate. This means that even at zero temperature, neutrons in the gas can have very large momenta up to the Fermi momentum. In fact in the centre of a typical neutron star, the neutron kinetic energy is becoming comparable with the neutron rest mass-energy even when it is "cold". The momentum of the neutrons leads to a degeneracy pressure that is able to (partially) support the weight of a neutron star.

However, pure degeneracy pressure actually ignores the interactions between the particles - it assumes they are ideal, non-interacting and point-like. From that point of view the neutrons can get very close together (and indeed @Marek they do "nestle" up to each other at densities approaching $10^{18}$ kg/m$^3$, the PEP does not forbid that because it is phase space [momentum x volume] that they have to be separated in, not just physical space).

But interactions do become important at these high densities. In a dense, many body nucleon gas (there are some protons in a neutron star too) the strong nuclear force is attractive at "long range" (where this means $>10^{-15}$m!) but becomes repulsive at shorter ranges. This latter hardens the equation of state (pressure increases more quickly with increasing density), allowing neutron stars to have much higher masses than would be allowed for ideal neutron degeneracy pressure alone. Ideal NDP would only allow neutron stars up 0.7Msun before collapse to a black hole, but with interactions they could become as high as 3Msun (a very uncertain number).

This doesn't really explain why n-n forces become repulsive at small separations (see DarenW's answer for that, though the situation becomes more complex and is less well understood in high density nucleon gases), but does I think clarify some confusion on this thread between the role of degeneracy pressure and the role of the strong nuclear force.

Edit: It is quite likely that at extremely high densities that the neutrons will undergo a phase transition. Firstly, their Fermi energies become large enough that the creation of other particles (e.g. hyperons or mesons) becomes possible. Secondly, the quarks in the neutrons may achieve asymptotic freedom and the neutrons "dissolve" into a quark-gluon plasma. Neither possibility is theoretically secure or has unambiguous observational evidence and it is unclear whether a stable neutron star can have a core featuring these phases or whether collapse to a black hole will already be in progress.

  • $\begingroup$ It seems pretty fundamental that the neutrons cannot be packed any tighter and retain their identity. Do you have any thoughts on what happens when a neutron star acquires enough mass to become a black hole? It seems like there must still be a solid surface of neutrons down there somewhere... $\endgroup$ – Brian Gordon Oct 7 '18 at 7:03

I think this may be a summary of other answers, but there are a couple things going on here. First, neutrons are electrically neutral, so an obvious force is the Van der Waals force. However, due to quantum mechanics and the Pauli exclusion principle (as noted above), neutrons' wave functions cannot (ish) overlap and so they are subject to the Neutron Degeneracy Pressure. I think if you understand these two concepts, then you should have a fairly good grasp on what is going on between neutrons in a neutron star.


Neutrons consist of quarks that are electrically charged, so when two neutrons get close enough to each other the higher electrical multipole moments will become relevant and cause repelling. But also note the magnetic moment and the strong force Cedric mentioned plus the Pauli exclusion mentioned by nibot.

  • $\begingroup$ As I understand it electrons do not consist of quarks, and that currently we don't know if or what they consist of? I don't really understand where the electrons in your answer came from? Sorry for all the questions, I'm just trying to learn as much as I can because the little amount I know I find fascinating. $\endgroup$ – Jonathan. Nov 2 '10 at 23:04
  • $\begingroup$ @Jonathan: I guess Tobias meant neutron instead of electron. $\endgroup$ – Cedric H. Nov 2 '10 at 23:17
  • $\begingroup$ @Cedric H.: yes that I meant. Thank you. (Please feel free to edit-correct such horrible errors if you spot them) $\endgroup$ – Tobias Kienzler Nov 3 '10 at 6:06
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    $\begingroup$ @Jonathan: Sorry for the confusion, as Cedric states I meant neutrons, not electrons. $\endgroup$ – Tobias Kienzler Nov 3 '10 at 6:07
  • $\begingroup$ Due to the Wigner-Eckart theorem a spin-half neutron may have a dipole moment, but no higher multipole moments. $\endgroup$ – rob Jun 9 '14 at 11:09

They repel because of the same reason why, say, atoms or molecules repel: because of Pauli principle which is the basis of elastiсity and solidity of matter. Why you cannot move through the walls? Why the dish holds food? Because all these objects are composed of fermions, the particles of substance.

This is opposed to say, light which can go through another light ray without any interaction. Because light is composed of bosons - the quants of radiation.


As they are not charged, they only interact through their magnetic moment (which at very low energy could lead to kind of Van der Waals forces, the strong force and the weak force.

If you try to create a nucleon-nucleon system and that you do some QM calculations using an Hamiltonian taking into account the nuclear force, the electric force and the spin 1/2 of the nucleons you will have different contribution to the total energy, some being positive, other negative.

For example, in the proton-proton case you cannot have a bound state because the electric repulsion is stronger.

You also have a contribution that depend on the [projection of the] spin of the nucleons. If they are equal the contribution lower the energy of the system and you can have a bound state.

That's the case of the deuton (a proton-neutron bound stage, very weak).

But in the neutron-neutron state, both being of spin 1/2, you cannot have the same projection, and thus you cannot obtain a bound state.

If you interpret that as "they repel each other" then you have your explanation.

  • $\begingroup$ -1, most if this talk is unimportant for the answer (only important point is the pressure of the degenerate Fermi gas) and parts about proton-proton and neutron-neutron are completely wrong. The only important things for discussing these interactions is the strong force and Pauli exclusion principle. Coulomb force is too weak on these scales to have any say (that's why all the rest of the nuclei are stable). Both for pp and nn what's important is that their spins are antialigned because of Pauli (this is the only correct part of this answer) but then nuclear interaction rules out bound states. $\endgroup$ – Marek Dec 26 '10 at 20:08
  • $\begingroup$ "parts about proton-proton and neutron-neutron are completely wrong" : "parts" like me answer is so damn long. @Marek altough you most of the time provides good answers, the way you interact is once again quite strange ... In addition you cannot say that coulomb force cannot be discussed in this case ... It plays a role in nuclear physics, even if in some cases it can be neglected (in any case one have to discuss why). $\endgroup$ – Cedric H. Dec 27 '10 at 12:05
  • $\begingroup$ sorry, I am just saying that this answer is bad because of the stated reason. E.g. you should not discuss Coulomb force because it is negligible when compared with the strong force which can keep huge number of protons together in nuclei. For the same reason van der Waals forces are unimportant. The only important effect is Pauli exclusion principle and this is almost not mentioned in your answer and is buried under irrelevant talk about negligible forces. That's why I think the answer is misleading, to say the least. $\endgroup$ – Marek Dec 28 '10 at 0:35
  • $\begingroup$ @Marek: You cannot really say that the Coulomb force don't play a role in nuclear physics ... And regarding the magnetic moment and VdW forces, it plays a role in experiments with neutrons at very low energies (almost at rest), so I think it is a good complement to other answers. That being said I can admit that it is not the most comprehensive answer I ever posted. $\endgroup$ – Cedric H. Jan 4 '11 at 11:54
  • $\begingroup$ I didn't say it doesn't play a role. But only a role of corrections. It's not the primary actor. Nothing wrong with complementary information but this answer doesn't present it that way. In any case, sorry if I offended you. But I still do think that the answer is quite confusing. $\endgroup$ – Marek Jan 4 '11 at 12:01

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