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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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HolgerFiedler
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  • https://link.springer.com/article/10.1007/BF00892879 – G. Smith Jul 02 '19 at 22:09
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    Note that the neutron consists of three charged particles, one up and two down quarks. – infinitezero Jul 03 '19 at 09:45
  • @infinitezero This note often appears, but is not necessary. My question aimed at the fact that not only charged particles radiate. – HolgerFiedler Jul 03 '19 at 09:54
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    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 03 '19 at 10:00
  • 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. – Jason May 18 '22 at 16:06

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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.

Andrew Steane
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    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 03 '19 at 13:30
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    @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 03 '19 at 14:29
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    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 03 '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 04 '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 '21 at 14:50
  • @Shashaank It is a magnetic dipole as I stated. The charge is only indirectly involved, in that in order to have non-zero magnetic dipole I think there has to be some charge somewhere. But magnetic dipoles radiate when accelerated. – Andrew Steane Mar 10 '21 at 15:58
  • @AndrewSteane Yes, I totally agree with you are saying. I get it perfectly. What I am saying is that just as you say that to have some non zero magnetic dipole moment some charge is needed in the first place, similarly won’t those 3 quarks inside the neutron also give rise to an electric dipole moment which should also radiate on being accelerated? Am I missing something? Do the quarks don’t give rise to an electric dipole moment? – Shashaank Mar 10 '21 at 16:04
  • @Shashaank ah-ha, now you have the famous electric dipole moment question. It is an open question whether the neutron has a non-zero electric dipole moment. If it did then it would violate CP symmetry so the weak and/or strong force would presumably be involved. You can look up the upper bounds on this e.d.m. and find out whether, if non-zero, it would compete with m.d.m. for radiative coupling; I have not done that calculation. – Andrew Steane Mar 10 '21 at 16:17
  • Accelerating a neutron will not make it radiate. The opposite of a photon scattering off of a neutron is a neutron scattering off of a photon which will cause it to change velocity and momentum but this is very different from radiation from an accelerating neutron. The last two sentances in your answer should be remove or revised and this would be more in line with the QM statements in the last part – Jason May 18 '22 at 17:14
  • @Jason I claim that, according to classical electromagnetism, accelerating a magnetic dipole will cause radiation (just as an oscillating magnetic dipole also produces radiation, called magnetic dipole radiation). If you are telling me this is wrong, then could you explain a bit further what is my mistake? – Andrew Steane May 18 '22 at 20:06
  • @AndrewSteane, I can try, regardless of the internal makeup the neutron has no net charge and is only a magnetic dipole due to QM spin and an electrical dipole due to internal makeup. Classically, you need both changing current and voltage to radiate, each EM wave has both E and H compoents. With no net charge overall, there is no E field component (1/r terms) needed to radiate. Accelerating magnetic dipole may induce e fields but these have no 1/r terms. The radiator has to have both. – Jason May 19 '22 at 22:08
  • Many ways to look at it, do hydrogen atoms (not ions) radiate? Does dropping magnets from the sky produce EM pulses once they hit the ground? Most metorites are magnetized right? Please keep in mind Im not talking about the internal charges moving relative to each other because that would/should radiate. How convincing is this quick burlb? – Jason May 19 '22 at 22:08
  • Pardon these quick blurbs – Jason May 19 '22 at 22:09
  • @Jason what about the 21cm line in hydrogen (an example of radiation owing to change in orientation of a magnetic dipole) – Andrew Steane May 20 '22 at 18:35
  • Very nice example but that is literally QM. Classically, that is rotation, not linear acceleration. If you model it as a current loop then there would be a change of E & H. Very interesting tho, kudos and thanks for bringing that up. – Jason May 20 '22 at 21:24
  • @AndrewSteane, see respone above, but also consider the fact that there is no neutron collider – Jason May 20 '22 at 21:58
  • @Jason and by the way, I think that a magnet bouncing off the ground would emit bremsstrahlung (very weakly of course). – Andrew Steane May 21 '22 at 12:59
  • @AndrewSteane, are you gaslighting me, lol, accelerating neutrons and bouncing magnets wont radiate. Accelerating electric dipoles wont radiate if the charges dont move relative to each other, and magnetic ones certainly dont. An electron's spin flipping without any proton nearby wont result in any radiation. Since everything is linear then there should be a companion radiation spectra for hydrogen's 21cm line that is due to the zero solution where proton distance is at infinity. Go to a EE department and talk to a EM waves specialist, trust me, worth the time. – Jason May 21 '22 at 16:50
  • @Jason I will think about this some more. I accept that I have not done a full calculation and my intuition may be out. Suppose we take a dipole (either magnetic or electric) located at the origin, then push it to one side, making it move to some distance away and then stop. There will be a change in the field. That change will propagate outwards at the speed of light (of this, at least, I am certain). My intuition wants to call the propagating change "radiation". Are you saying that that would be a misuse of the terminology? (I also think there would be a 'radiation reaction' force.) – Andrew Steane May 21 '22 at 21:23
  • @AndrewSteane, yes since radiation is used for desribing propagation of energy and photons. Non radiating fields dont propagate and are equivalent to converging series, nonzero at infinity but finite in nature. Example being integral of 1/r being infinite and non converging while integral of 1/r^2 is finite and converging, well, never starting at zero of course. The EM fields that radiate have 1/r terms that carry energy away and all other terms represent reactive energy stored inside induced magnetic or electric fields aka inductors and capacitors. Does this help? – Jason May 21 '22 at 22:30
  • Btw, there was at least three posts that Ive seen with someone saying what you are. Worst, most confused scattering with other things – Jason May 21 '22 at 22:40
  • Looks like pulsars a good example of what I was saying. Current theory is that the magnetic field induces electric fields on local charges which are the source of radiation. Dead pulsars are still spinning but cannot be detected. In fact the number of dead pulsars is an open question in astrophysics. This is wiki sourced but appears pretty clear. Nonetheless everyone on that link you added is saying aomething different. Thanks for the conversation, I am a bit smarter or more knowledgeable, at least temporarily, thanks to you. – Jason May 29 '22 at 20:56
  • Finally proved this to myself, E=0 for every pure magnet. See this news article below and let me know if you update your answer so that I can delete these comments. – Jason Jun 07 '22 at 18:01
  • https://www.sciencenews.org/article/pulsar-radio-waves-neutron-star-astronomy – Jason Jun 07 '22 at 18:01
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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.

Cang Ye
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  • 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 – Jason May 18 '22 at 17:09