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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?

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  • $\begingroup$ link.springer.com/article/10.1007/BF00892879 $\endgroup$
    – G. Smith
    Commented Jul 2, 2019 at 22:09
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    $\begingroup$ Note that the neutron consists of three charged particles, one up and two down quarks. $\endgroup$
    – infinitezero
    Commented Jul 3, 2019 at 9:45
  • $\begingroup$ @infinitezero This note often appears, but is not necessary. My question aimed at the fact that not only charged particles radiate. $\endgroup$
    – HolgerFiedler
    Commented Jul 3, 2019 at 9:54
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    $\begingroup$ I think that is a matter of perspective. The net charge of a neutron is zero, but its inner structure leads to the radiation. $\endgroup$
    – infinitezero
    Commented Jul 3, 2019 at 10:00
  • $\begingroup$ There are two questions here, please note a scattering event is not the same as de/accelerating a neutron. Shaking a magnet will not result in EM radiation nor will starting or stopping one. Strongly suggest that the last sentence is revised or deleted. $\endgroup$
    – Jason
    Commented May 18, 2022 at 16:06

2 Answers 2

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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.

Edit

After a discussion with user Jason (see comments) I am now unsure whether what happens when a dipole accelerates along a line is correctly called 'radiation'. There is an outward-going change in the field, propagating at the speed of light, but there is some uncertainty in my mind now about whether it has enough energy and amplitude to be called 'radiation'. When a dipole oscillates in magnitude, on the other hand, there is certainly radiation, but that won't happen for a neutron. But when a dipole rotates the effect is the same as a pair of superposed oscillations in orthogonal directions, so that suggests a rotating dipole does radiate: see

https://physics.stackexchange.com/questions/158557/would-a-rotating-magnet-emit-photons-if-so-what-causes-the-torque-that-gradual#:~:text=This%20is%20simply%20a%20rotating,per%20standard%20M1%20radiation%20formulas.

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    $\begingroup$ 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. $\endgroup$
    – Gyromagnetic
    Commented Jul 3, 2019 at 13:30
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    $\begingroup$ @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. $\endgroup$
    – Andrew Steane
    Commented Jul 3, 2019 at 14:29
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    $\begingroup$ 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. $\endgroup$
    – rob ♦
    Commented Jul 3, 2019 at 16:36
  • $\begingroup$ It would be useful for this answer's completeness to calculate the acceleration needed to get RF/optical radiation from a neutron. $\endgroup$
    – KF Gauss
    Commented Jul 4, 2019 at 17:45
  • $\begingroup$ 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 ? $\endgroup$
    – Shashaank
    Commented Mar 10, 2021 at 14:50
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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.

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  • $\begingroup$ You need both fields to radiate, ie voltage and current in terms of antennas. Agree the neutron stopping will not radiate for the same reason any magnet cannot radiate but seems like that is not a part of the intended question. Suugest you restate your answer $\endgroup$
    – Jason
    Commented May 18, 2022 at 17:09

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