I understand that two protons would repel due to them both being positively charged, however, wouldn't the strong force act on the two protons pulling them together? Would this mean that in this case the electromagnetic repulsive force is greater than the strong force? If so why? If not why would they repel?

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    $\begingroup$ The strong force has a very short reach. $\endgroup$
    – ACuriousMind
    Jan 29, 2015 at 19:38
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    $\begingroup$ Adding onto ACuriousMind's comment, think about nuclei. In the nucleus, the protons are sufficiently close together that the strong force attraction is stronger than the electromagnetic repulsion. $\endgroup$
    – Ryan Unger
    Jan 29, 2015 at 19:43
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    $\begingroup$ Both of the comments above miss the main point of the question, which is why, despite the strength of the nuclear force, is there not a proton-proton bound nucleus? $\endgroup$
    – ProfRob
    Jan 30, 2015 at 12:11

2 Answers 2


The electromagnetic repulsion between two protons is a long-range force, depending on $1/r^2$, where $r$ is the separation of the two protons. The electromagnetic repulsion between two protons is not the reason that they do not stick together; if they are forced together (or can tunnel through the Coulomb barrier) then short-range strong nuclear forces are much stronger than the electromagnetic force over separations $<1.7\times10^{-15}\ m$, yet they are unable to make a bound state consisting of two protons.

The reason for this is that although the nuclear interaction is symmetric to the isospin of the nucleons (i.e. to first order it does not depend on whether the nucleons are protons or neutrons) it does depend on the spins of the two particles. The attractive nature of the force is only strong enough to bind the two nucleons if they have aligned spins (as in the bound state of the deuteron which has a neutron and proton with aligned spins and total angular momentum 1). If the two nucleons were identical however, i.e. a p+p or n+n interaction, then a bound state state with aligned spins would be forbidden by the Pauli exclusion principle.

The deeper reasons behind this spin dependence will need an answer from someone with a much better understanding of these issues.


Yea, a proton and neutron stick together, but two of the sam kind don't. You don't get neutron balls even with only the interneucleon force and no electric repulsion.

I asked about it some years ago in a physics on-line forum, long before StackExchange. Ended up getting a textbook and eventually learning that "the force is largely insensitve to species (whether proton or neutron), but highly dependent on spin."

I got two different answers as to whether the diproton exists as a (fleeting) bound state.

The answer, relative to species and spin combinations of neucleons, has to do with "singlet state vs triplet state". That is a unique enough phrase that you ought to be able to search on that (e.g. class notes ...Hence the force is attractive for isospin singlets (T = 0) and repulsive for isospin triplets (T = 1).; more class notes in a nice PDF with illustrations.

Bottom line: the interneucleon force is not so simple as what you think of with gravity or electric or magnetic but with faster fall-off. It has complex behavior involving just how the particles are parked and what its neighbors are doing.

  • $\begingroup$ It is worth noting in this kind of context that the deuteron is stable by that Helium-2 is not. So one can not naive neglect the electromagnetic contribution to the energy. $\endgroup$ Jan 29, 2015 at 20:05
  • $\begingroup$ Otoh, textbooks say that the electric force is completely irrelevant. $\endgroup$
    – JDługosz
    Jan 29, 2015 at 20:54
  • $\begingroup$ Well, two protons would stick together, but the Helium-2 that you got out would beta-decay to a deuteron pretty quickly. $\endgroup$ Jan 29, 2015 at 21:18
  • $\begingroup$ Other sources state that a bound state never forms. From the class noted linked above (far easier than hittingnthe old dead-tree content!) The implication is that two nucleons are not bound together if their spins are antiparallel and this is consistent with there being no proton-proton or neutron-neutron bound states. In the case of identical particles the bound parallel spin state is forbidden by the Pauli exclusion principle. $\endgroup$
    – JDługosz
    Jan 29, 2015 at 21:25

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