# Will a free neutron radiate if it is decelerated?

In this answer it is said (and I fully agree):

Yes, a ... photon can accelerate a lone neutron. The kinetic energy imparted to the neutron reduces the photon's wavelength (redshifts it) by the same amount, so the total energy of the system remains the same.

In turn, the opposite process has to be possible too. Neutrons are able to radiate. This usually is said only for charged particles.

Electromagnetic waves are emitted by electrically charged particles undergoing acceleration, and these waves can subsequently interact with other charged particles Electromagnetic radiation

Will a free neutron radiate if it is decelerated?

• link.springer.com/article/10.1007/BF00892879 – G. Smith Jul 2 '19 at 22:09
• Note that the neutron consists of three charged particles, one up and two down quarks. – infinitezero Jul 3 '19 at 9:45
• @infinitezero This note often appears, but is not necessary. My question aimed at the fact that not only charged particles radiate. – HolgerFiedler Jul 3 '19 at 9:54
• I think that is a matter of perspective. The net charge of a neutron is zero, but its inner structure leads to the radiation. – infinitezero Jul 3 '19 at 10:00

The neutron is magnetic. It is a tiny little magnet. In more formal language, it carries a magnetic dipole moment of size $$\mu_n = −9.6623647(23) \times 10^{−27} {\rm J\,T}^{−1}.$$ This is what allows it to interact with electromagnetic waves---or, to say the same thing another way, with photons. This also means that when accelerated then yes, it will generate electromagnetic radiation. This effect is much smaller that the radiation of a charged particle with the same acceleration.

You can associate this radiation with the presence of accelerating charge, indirectly, by noting that the neutron has charged quarks inside it, but strictly speaking those components have to be treated by quantum theory so they shouldn't be thought of as little separate charges. The calculation in terms of magnetic dipole moment is more appropriate.

• Nice Answer! Don't wanna be a killjoy but this is only true for a spin-polarized neutron, right? If you catch a neutron out in the wild (no magnetic fields) it's very likely its quantum-states will be thermalized, in which case, decelerating it would not radiate. – Gyromagnetic Jul 3 '19 at 13:30
• @Gyromagnetic A thermal ensemble is equivalent to a mixture with each neutron in a pure state, so each neutron does emit, but if it is close to other neutrons (within of order a wavelength) then I guess the interferometric cancellation will surpress the net effect. – Andrew Steane Jul 3 '19 at 14:29
• Two notes. (1) The neutron magnetic moment is more conveniently written in different units, as 50 nano-eV per tesla. Repolarizing in a magnetic field has a big effect on the motion of "ultra-cold" neutrons, with kinetic energies below 100 neV, but not so much on thermal neutrons with milli-eV energies. (2) A second neutron isn't required for interference effects. In single-crystal neutron interferometer experiments, it's not uncommon to obtain an interference pattern even though the number of events where two neutrons were present in the interferometer at the same time was zero. – rob Jul 3 '19 at 16:36
• It would be useful for this answer's completeness to calculate the acceleration needed to get RF/optical radiation from a neutron. – KF Gauss Jul 4 '19 at 17:45
• But the net charge of the quarks inside the neutron cancel. Would there still be an acceleration? Do you mean that charges inside the neutron will form a kind of dipole and that will radiate ? – Shashaank Mar 10 at 14:50

We take a charged ball and shake it. There must be electromagnetic radiation from the shake. When we shake an electrically neutral object, it does not emit electromagnetic radiation from the shaking.

This situation cannot be explained by photons. Otherwise, an electrically neutral object will radiate when it is shaken. Because charged particles (proton, electrons) inside object radiate photons, objects radiate. But this is clearly not the case.