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According to the question Do positively charged particles exchange photons? there was an answer

Yes. Photons are the carriers for the electromagnetic force, regardless of the charges involved.

followed by a comment

Unless of course the charge is zero ;)

Now I'm getting curious how neutrons get scattered of each other?

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  • $\begingroup$ No charge so no electrostatic repulsion. But since the strong force plays a role in attracting them, they are bound in very short range but they stay apart due to pauli exclusion principle.....the strong force dominates in the range of $1.5 fm$ but the exclusion principle starts dominating from about $0.5 fm$ which is repulsive $\endgroup$ – Prasad Mani Oct 25 '16 at 5:40
  • $\begingroup$ @Prasad Mani What is about the neutrons magnetic dipole moments? How not to think about this? $\endgroup$ – HolgerFiedler Oct 25 '16 at 6:04
  • $\begingroup$ The spin magnetic dipole moment is what plays a role in the pauli exclusion principle.....for two identical fermions, it is more preferable for them to stay apart and have spins aligned than to stay closer and have spins antialigned $\endgroup$ – Prasad Mani Oct 25 '16 at 6:10
  • $\begingroup$ @Prasad Mani What intermediates the anti-alignment of the magnetic dipole moments? That seems to me to be a very interesting question. $\endgroup$ – HolgerFiedler Oct 25 '16 at 6:14
  • $\begingroup$ I dont really understand what you mean by 'intermediates' but i hope you got what i am saying..neutrons have spin $\frac{1}{2}$ and therefore obey the pauli exclusion principle, meaning two neutrons cannot occupy the same space at the same time. When two neutrons' wave functions overlap, they feel a strong repulsive force. $\endgroup$ – Prasad Mani Oct 25 '16 at 6:17
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The neutron is not an elementary particle. Its overall charge is zero, but depending on the energy available the fact that it is composed out of charged quarks becomes important. Thus both the strong force and the electromagnetic will play a role depending on the energies of the scatters and the targets.

For low energies , like the ones coming from fission reactions in reactors, matter is practically transparent to the neutrons due to their zero charge, and in order to absorb and contain them a lot of matter is needed. They can interact with the strong force with the nuclei of the atoms, be captured etc, generating secondary radiation centers.

Due to the quark structure the neutron has a magnetic moment that can interact with magnetic fields.

To interpret correctly the crossections the quark nature has to be taken into account where also the electromagnetic interactions will enter, example here., mathematical models are necessary.

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  • $\begingroup$ anna Does anything prevents you to make a clear statement that an EM interaction takes place? $\endgroup$ – HolgerFiedler Oct 25 '16 at 9:49
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    $\begingroup$ @HolgerFiedler I am saying it "Thus both the strong force and the electromagnetic will play a role" $\endgroup$ – anna v Oct 25 '16 at 11:14
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Neutrons scatter from other neutrons and protons through the nuclear force or "strong force", which is also what keeps neutrons within the atomic nucleus. The strong force is also what holds the constituents of nucleons (neutrons and protons) and other hadrons (the whole spectrum of mesons, baryons) together. The nuclear force which binds can also cause them to scatter. If you look at the diagram on Wikipedia, note that there's a very large repulsive force when the neutrons are too close.

Because of the repulsive part of the nuclear force, a neutron with a large kinetic energy will thus approach too close and bounce off another nucleon, whereas a slow neutron might bind with another nucleon.

Neutrons, although electrically neutral, consist of electrically charged particles (quarks) bound together as a neutral whole, and so they also have a magnetic moment.

(Aside: The references used in that Wikipedia article are all very obscure books. I was looking for something to read about this but was disappointed. I'll have another search on this site and then look it up again.)

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  • $\begingroup$ How about the scattering of two free neutrons (not bonded inside a nucleus)? $\endgroup$ – HolgerFiedler Oct 25 '16 at 6:38
  • $\begingroup$ It's the same force. If you look at the diagram on Wikipedia, note that it's very large repulsive force when the neutrons are too close. A neutron with a large kinetic energy will thus approach too close and bounce, whereas a slow neutron might bind with another nucleon. $\endgroup$ – Suzu Hirose Oct 25 '16 at 6:40
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    $\begingroup$ @PrasadMani - On reading it, I agree that it could be misleading. I might as well edit this I think. $\endgroup$ – Suzu Hirose Oct 25 '16 at 7:04
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    $\begingroup$ The strong nuclear force is repulsive at short ranges $<10^{-15}$ m. It is not due to the PEP between neutrons. $\endgroup$ – Rob Jeffries Oct 25 '16 at 8:18
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    $\begingroup$ You might attribute it to the PEP between quarks, but even neutrons with opposite spin are repelled at short separations. $\endgroup$ – Rob Jeffries Oct 25 '16 at 10:06
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In condensed matter research, neutron scattering experiments are very useful to study the magnetic structure of materials.

Neutron do indeed possess a magnetic moment, and thus interact with the local magnetic field. Electrons have a much larger magnetic moment, yet neutrons are used precisely because they have zero charge and do not interact electrically, but purely magnetically.

In the end it is still an EM interaction though. Microscopically, it is one of the charged quarks that interacts with the EM field.

At higher energies (read: particle accelerators), neutrons can also interact vi the strong nuclear force.

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  • $\begingroup$ polwel Neutrons interact "purely magnetically". How you couldn't call this interaction an EM interaction? $\endgroup$ – HolgerFiedler Oct 25 '16 at 7:25
  • $\begingroup$ In the end it is an EM interaction. Microscopically, it is one of the charged quarks that interacts with the EM field. $\endgroup$ – polwel Oct 25 '16 at 7:55
  • $\begingroup$ powel Coud you include the comment into your answer, please. $\endgroup$ – HolgerFiedler Oct 25 '16 at 7:59

protected by Qmechanic Oct 25 '16 at 12:15

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