If two neutrons were set to collide head-on, being electrically neutral:

  1. Will they feel each other's presence due to electrostatic interaction between quarks of approaching neutrons?
  2. When they approach very close enough, will they accelerate towards each other due to the effects of residual strong force?
  3. Will the acceleration experienced by each neutron be less than the acceleration experienced in a head-on collision between a proton and an antiproton?

I came across the following, but they did not address my question:

Note: I am not a physics student. I apologize for any improper terminology.

  • $\begingroup$ The biggest problem with this question is that you treat the idea of [X] and [Y] collide as if it is well defined, but it really isn't. You haven't even given the most fundamental piece of information which is what energy regime you're interested in. Nor does "head on" have a clear meaning in the quantum realm the way it does if you are envisioning these things as little billiard balls (which they aren't), because the more defined the impact parameter the less well known the transverse momentum and vice versa (this is the Heisenberg Uncertainty principle at work). $\endgroup$ Commented Dec 8, 2015 at 21:25
  • $\begingroup$ @dmckee, the reason I didn't mention energy because, I was not thinking about what come/result out of the collision. I now realize that the neutrons in question can not be treated as classical particles. $\endgroup$ Commented Dec 11, 2015 at 16:57

3 Answers 3


Neutrons are not classical particles as noted by @dmckee comment and their collision must be treated as a quantum phenomenon. With this proviso the answers to your questions are:

  1. Yes they will interact electromagnetically. Neutrons have an anomalous magnetic moment so they will behave like two magnets (attracting or repelling depending upon orientation). Since they are quantum objects, only two orientations (helicity states) are possible rather than many.

  2. Yes, they will be attracted via the strong force. A neutron and a proton will bind together to form a deuteron and since the forces between two neutrons are similar (though not identical) it is possible that they might bind. Since they are quantum objects it is not certain that they will bind (note the deuteron is weakly bound). Many experimental searches have been conducted for the two neutron bound state, but (as far as I know) none have been successful.

  3. Yes, the attraction between two neutrons is expected to be less than the attraction between a proton and an anti-proton both because of the smaller electromagnetic interaction and because of a smaller strong interaction.

  • $\begingroup$ Lewis Miller, thank you for the response. I did not understand what helicity states are and how neutrons get their magnetic moments from, however, your response satisfied my question very well. I will read more about those concepts. $\endgroup$ Commented Dec 11, 2015 at 17:02
  • $\begingroup$ Just to add some news on the bound neutron-neutron state: just recently researchers found that Tetraneutron (4n) might actual exist: asianscientist.com/2016/01/in-the-lab/tokyo-tetraneutron-proton $\endgroup$
    – Martin
    Commented Feb 27, 2016 at 0:55
  • $\begingroup$ Why is there a stronger strong interaction between antiproton & proton than neutron and neutron? $\endgroup$ Commented Jul 16, 2017 at 7:51
  • $\begingroup$ @Mockingbird See my answer to this question: physics.stackexchange.com/questions/288357/… $\endgroup$ Commented Jul 17, 2017 at 2:21

The answer by Lewis Miller is adequate qualitatively.

Please note though the very important fact that there exist no free neutrons to be set up as targets. Free neutrons have a lifetime of minutes and can be used as projectiles , but not as targets. Thus the neutron neutron scattering cross section, the scientific term for what you are asking , has to be inferred from neutron scattering on some target that has both protons and neutron and the contribution of neutrons to the scattering can be ascertained by subtracting the known proton crossections.

In this link you will find a lot of neutron scattering crossections on various targets . A search on google will give a number of books where the subject is treated. There is little interest in pure neutron neutron scattering, although the extraction has been made using deuterons as a target, since the neutron proton cross section is known form experiments on hydrogen. It is not simple:

The theory of scattering and production reactions on deuterium is developed with a view to the extraction of neutron target amplitudes from analysis of deuterium experiments. Corrections to the spectator model due to the following effects are examined: target particle binding corrections to the Impulse Approximation; (ii) the Pauli principle in charge exchange reactions; (iii) internal or „Fermi” motion of the target particles; (iv) multiple scattering and final state interactions. The closure approximation and the general analysis of single-arm spectrometer experiments are discussed. The connection at high energies and small momentum transfers between the general Watson Theory and the Glauber model is established, and the results of calculations of deuterium corrections for pion and rho photoproduction at high energies are reviewed. “Fresnel” and recoil corrections to Glauber theory are examined. An approximation scheme suitable to the analysis of high energy wide angle coincidence or bubble chamber experiments is developed. A special examination is made of the breakdown of the impulse approximation in the 3-3 resonance region, and of the applicability of lowest order binding and in multiple scattering corrections in this sensitive kinematical region.

  • $\begingroup$ You might also have added that my answer is applicable only for very low energy scattering (something that I forgot to mention). The situation becomes more complex and qualitatively different at high energies. Nevertheless, such high energy scattering should be quite rare in the natural world (and currently impossible in the Lab) for the reasons you outline. $\endgroup$ Commented Dec 9, 2015 at 14:35
  • $\begingroup$ @anna v, thanks a lot for posting your response to my question. I now understand that my question was based on incorrect assumptions. $\endgroup$ Commented Dec 11, 2015 at 16:47
  • $\begingroup$ @anna v, from your response I take away that neutrons can't be set up as targets and that their scattering cross section has to be calculated. Thank you for the links, but I was unable to understand much because of my lack of physics background. $\endgroup$ Commented Dec 11, 2015 at 16:54

The two neutrons would scatter from each other much like two protons, or a proton and a neutron, since the principal interaction between nucleons is the strong force and doesn't much care about electric charge. The parameter which describes the interaction is the "neutron-neutron scattering length." Here's a recent review on the subject, which seems to have been condensed by the author into a talk.

One of the first proposals to measure the neutron-neutron scattering length was to simultaneously detonate two nuclear bombs underground. Each bomb would liberate something like a mole of neutrons with a well-known energy spectrum, with enough flux that you could reasonably expect a few neutrons from the two sources to scatter from each other. I have heard that this experiment was actually performed in the 1980s, but the blast doors protecting the data acquisition equipment failed and the data were vaporized a moment after being collected. If a description of this experiment has actually been published it's in an obscure place that I haven't been able to track down.

To answer your questions directly:

  1. While neutrons do have a nonzero magnetic moment, the predominant interaction responsible for the measured neutron-neutron scattering length is the strong force. The magnetic interaction is a "dipole-dipole force" which decays like $r^3$ or $r^4$; even in the presence of a strong (Tesla) magnetic field, the magnetic force on a neutron is negligible.

  2. I suppose the two-neutron force is weakly attractive, but people don't usually talk about it that way because (a) it's such a short-range force that the neutrons must already be "touching" for it to matter, and (b) there isn't a dineutron bound state. (Because the strong force doesn't care about electromagnetism, the dineutron would have roughly the same energy as the diproton or the spin-zero state of the deuteron; both of those systems are unbound.)

  3. If the strong nucleon-nucleon interaction is attractive, I would expect the strong nucleon-antinucleon interaction to be repulsive. However a proton-antiproton pair has an electric attraction which is nonzero at large distances. That's a recipe for proton-antiproton pairs to have a stable bound state, held together by electromagnetism with annihilation delayed by the strong repulsion. This bound proton-antiproton "molecule" is called protonium.

    Compared to the p-p and n-n scattering lengths (about 18 fm) given in the link above, the proton-antiproton scattering length seems to be quite small, only about 1 fm. (Sources: 1992, 2006.) So the interaction between a pair of neutrons is much stronger than the interaction in protium. However, I haven't been careful in these few minutes of searching to keep track of the energies of the different collisions given for these data, nor to who is doing "Coulomb subtraction" and who isn't. Here's a fresh high-energy paper which reports a scattering length for antiproton-antiproton collisions; it's beautifully identical to the proton-proton scattering length, but the value for both is different from the p-p length given in my first links.

  • $\begingroup$ Your description of the "Two Bombs" experiment sounds like urban legend to me. I was a government employee during the 1980s and interacted with nuclear scientists from both Los Alamos and Livermore yet I never heard of it. $\endgroup$ Commented Dec 18, 2015 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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