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This questions has been asked all over the net (here included) but I can't find a satisfactory answer or discussion. Some say it does not radiate if the acceleration is caused by a uniform gravity field. Some even say that it radiates in linear accelerators because of our imperfect technology. All this comes from university member (at least they claim to be). My head is spinning. I always thought (from CED) that any kind of acceleration causes the charge to radiate and loose energy, not that I say that this is the case, but the equivalence principle of relativity never crossed my mind while thinking about this puzzle. Today it did.

Can someone here give some good, kind of fresh, references (no ArXiV please!) or alternatively, try to explain better than what others have done?

Physics_maths
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    The "answer" is that it is still an open question whether uniformly accelerated charge radiates and this question is quite subtle and, for example, requires that one think very carefully about what "radiate" means. Also, there is the distinction between coordinate acceleration and proper acceleration. Now, think carefully about what that implies and you'll understand why you haven't found a "satisfactory answer" to your question. If you haven't read this already, try: http://www.mathpages.com/home/kmath528/kmath528.htm – Alfred Centauri Dec 05 '13 at 13:04
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    From the link above: "It's also possible to question whether the equations of electrodynamics really do imply that an accelerating charge necessarily radiates. Surprisingly, this is still an open question for the classical theory." – Alfred Centauri Dec 05 '13 at 13:06
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    Interesting. I guess then I should be happy that my head was spinning after all. – Physics_maths Dec 05 '13 at 13:09
  • Perhaps I should ask for a list of non-equivalent (to be defined) situations where charges DO radiate. – Physics_maths Dec 05 '13 at 13:19
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    No, I reckon you should read @AlfredCentauri's link and write what you understand from it up as an answer to your own question. That would be a great addition to this site. – Selene Routley Dec 05 '13 at 13:36
  • Yes, I didn't literally mean to ask that, it was just a way of saying. I agree with your suggestion. – Physics_maths Dec 05 '13 at 13:40
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    Why no ArXiV? Many/Most of the articles are simply preprints of published works. – Kyle Kanos Dec 05 '13 at 13:48
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    I guess preprints can be O.K. but I have a hard time to read 100 pages about something as controversial as this question/topic is in an arXiv article. Other stuff perhaps O.K. but this question seems very controversial to me. – Physics_maths Dec 05 '13 at 13:54
  • The link given by @AlfredCentauri quite interesting theory-wise, but are there any experimental tests on this topic? – Ruslan Dec 05 '13 at 15:53
  • Have a look also to this calculation http://physics.stackexchange.com/q/52764/ for radiation from a falling electron in the gravitational field of the earth. – anna v Dec 05 '13 at 20:44
  • http://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential Sounds like Liénard–Wiechert potential. –  Dec 05 '13 at 18:19
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    See Ben Crowell's answer here: http://physics.stackexchange.com/q/70915/. You may still be confused afterwards, but it will be a deeper confusion... – Art Brown Dec 06 '13 at 05:44
  • Has anyone studied "Classical Charged Particles" by Fritz Rohrlich and could comment on its application to this question? Is it worth reading? – JEM Apr 30 '14 at 13:46
  • Essentially a duplicate of http://physics.stackexchange.com/q/70915/2451 – Qmechanic Jun 02 '14 at 18:00

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This is an experimentalist's answer and yes, accelerated charged particles either in stable circular orbits or in linear acceleration do radiate.

Classically, any charged particle which moves in a curved path or is accelerated in a straight-line path will emit electromagnetic radiation. Various names are given to this radiation in different contexts. For example, when is occurs upon electron impact with a solid metal target in an x-ray tube, it is called "brehmsstrahlung" radiation.

So it is an experimental fact well known to particle physicists.

There exist classical electromagnetic calculations of this radiation, as seen in the link provided. The fact that the electrons did not fall into the nucleons from the loss of radiation was what prompted the quantized Bohr model and led to the discovery of quantum mechanics.

The controversy seems to be in introducing special relativity concepts for the motion of the charge and the observer, which I cannot really follow . I found this review though with references therein.

Edit after comments:

I would like to add to this "principle of equivalency" navel gazing that there also exists something called conservation of energy in a system. If a particle is radiating and it is not dissolving like the ISON comet recently, the energy in a system must come from somewhere. An elementary particle is intact through all the special relativity transformations, so the energy must be supplied by the potentials as perceived by the observer in whatever system he is set up in, as seen in this this answer to a similar question.

In the case of a charge at rest ( comoving system of observer and charge) there are no potentials to supply the energy. Approaching the situation differentially with acceleration going to zero, the radiated wavelength get a longer and longer wavelength, the energy taken from the acceleration of the observer, and when reaching at rest they become the static electric field .

Now it is well known that when reaching limits for electromagnetism and apparent paradoxes, these are resolved by going to the quantum mechanical framework. I suspect that until we get a unified quantized theory for general relativity and the three other forces the subject will remain at this level of a discrepancy/explanation as far as the equivalence principle is concerned.

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anna v
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  • so, if this is true, then charged objects at rest on a gravity well must radiate too. Do normal gravitational objects evaporate via this mechanism just as black holes? – lurscher Dec 05 '13 at 17:21
  • Thanks for your comment @anna. What are your thoughts about a charge at rest at the surface if the earth? Does it radiate? – Physics_maths Dec 05 '13 at 17:40
  • @lurscher: of course they won't radiate if they are at rest relative to the dragged aether ;) – Christoph Dec 05 '13 at 18:13
  • @Christoph what aether? :) According to equivalence principle, it's accelerated with $g$. – Ruslan Dec 05 '13 at 18:18
  • Your answer only considers cyclic motion (in synchrotron) and non-uniform acceleration (in Brehmsstrahlung). The main problem in this question as shown in the link given by AlfredCentauri's comment to question is uniformly accelerated (in comoving frame) motion - in this case theory doesn't give consistent answer. – Ruslan Dec 05 '13 at 18:23
  • @LoveLearning It has not been observed to radiate, and that is consistent with the fact that it is not accelerating. Acceleration means a dv/dt and it is not moving. The force from gravity is countered by the equal and opposite reactive force from the ground's atoms, and is at rest. – anna v Dec 05 '13 at 19:54
  • @anna so you're saying that the principle of equivalence is flawed? – Physics_maths Dec 05 '13 at 20:03
  • I am saying that for an experimentalist acceleration means dv/dt non zero, and if v is zero there is no acceleration. I have not delved into the fine points of the principle of equivalency to be able to give an opinion on it. Also @Ruslan there exists radiation for linear acceleration too http://en.wikipedia.org/wiki/Brehmstrahlung#Total_radiated_power – anna v Dec 05 '13 at 20:32
  • @Ruslan Have a look also to this calculation http://physics.stackexchange.com/questions/52764/falling-electron for radiation from a falling electron in the gravitational field of the earth. – anna v Dec 05 '13 at 20:45
  • @anna v But Experiments performed in a uniformly accelerating reference frame with acceleration a are indistinguishable from the same experiments performed in a non-accelerating reference frame which is situated in a gravitational field where the acceleration of gravity = g = -a = intensity of gravity field. http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/grel.html – Physics_maths Dec 06 '13 at 10:00
  • That is a correct statement, but it is not about a charge at rest, just that one cannot tell which is the source of the acceleration, gravitational or other( electrical attraction for example). – anna v Dec 06 '13 at 10:22
  • Let put this in another way: If I have a torchship able to accelerate to one g indefinitely. In the floor of the ship, I have one g environment, everything falls just like in earth's gravity. I have also a big charged ball with, I don't know, let's say a hundred Coulombs of charge (a lot of charge, if you do the math). Question: will this charge radiate electromagnetic radiation YES or NOT? I believe that it does, and doing so contributes to the total inertia of the ship: the mass inertia and the electromagnetic inertia – lurscher Dec 14 '13 at 02:26
  • Equivalence principle is valid only in a neighbourhood small enough so that tidal forces are small. But if the charge is condensed in a volume smaller than the typical equivalence neighbourhood, then the charge MUST radiate. – lurscher Dec 14 '13 at 02:28
  • quantum mechanics may explain where the energy comes from, but the classical limit should be describable by Maxwell electromagnetism and General relativity, and should match observations – lurscher Dec 14 '13 at 02:31
  • What I would like to understand is why hasn't this experiment been done? – lurscher Dec 14 '13 at 02:31
  • adding this link http://www.fisica.uniud.it/~deangeli/fismod/Radiation.pdf , i do not want to resurrect the question by editing http://www.fisica.uniud.it/~deangeli/fismod/Radiation.pdf – anna v May 23 '18 at 05:12
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A charge dropped in a uniform gravitational field will radiate, despite the equivalence principle suggesting that it won't. This is because the boundary conditions on the fields (asymptotic behavior for large distances) in the case of a uniform gravitational field are different from the case of a freely floating charge.

Count Iblis
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  • crazy that this answer seems to have been ignored, it says it all no? – ceillac Dec 05 '15 at 17:29
  • @borilla: This answer possibly was ignored, because it doesn't give any details or citations. The OP asked: "alternatively, try to explain better than what others have done?" This clearly doesn't meet that standard. – Peter Shor Jan 13 '17 at 16:26
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Do accelerated charges radiate or not?

It depends. A falling charged particle doesn't radiate. Nor does a charged particle accelerating because of a static electric field. Cyclotron radiation and synchrotron radiation are associated with acceleration due to a magnetic field.

This questions has been asked all over the net (here included) but I can't find a satisfactory answer or discussion. Some say it does not radiate if the acceleration is caused by a uniform gravity field.

A falling charge does not radiate. It doesn't matter whether the gravitational field is uniform. The particle is in free fall, and the principle of equivalence says standing on the ground is like accelerating through space. But note that a charged particle sitting on the ground doesn't radiate either. It only radiates when it stops falling because it hits the ground. Compare this with what happens to the electron when it's attracted to the proton to form a hydrogen atom.

Some even say that it radiates in linear accelerators because of our imperfect technology.

Can you give me a reference for that? See this: "While it is possible to accelerate charged particles using electrostatic fields, like in a Cockcroft-Walton voltage multiplier, this method has limits given by electrical breakdown at high voltages. Furthermore, due to electrostatic fields being conservative, the maximum voltage limits the kinetic energy that is applicable to the particles. To circumvent this problem, linear particle accelerators operate using time-varying fields..."

All this comes from university member (at least they claim to be). My head is spinning. I always thought (from CED) that any kind of acceleration causes the charge to radiate and lose energy

Not any kind, and not gravitational falling-down acceleration. That converts potential energy, which is mass-energy, which is internal kinetic energy, into external kinetic energy. It doesn't radiate it away whilst the particle is falling. In similar vein falling down doesn't hurt a man. It's the sudden cessation of falling down that causes his teeth to radiate across the pavement.

not that I say that this is the case, but the equivalence principle of relativity never crossed my mind while thinking about this puzzle. Today it did.

It shouldn't be a puzzle. IMHO it's only a puzzle because somebody has oversimplified some teaching. Or maybe confused two different fields of physics.

Can someone here give some good, kind of fresh, references (no ArXiV please!) or alternatively, try to explain better than what others have done?

I'm not sure I can. Have a look at this. And who's telling you that falling charged particles radiate. They don't. If they did Einstein's general relativity would be wrong, and it isn't. Instead it's one of the best-tested theories we've got. See The Confrontation between General Relativity and Experiment by Clifford M Will.

John Duffield
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