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Would it be correct to say that any charge oscillating in space (regardless of the spacial amplitude) at a given frequency will emit an EM wave of the same frequency?

related: What change in an EM field is required to create an EM wave?

Manu de Hanoi
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    this contains the "yes" answer and goes on https://www.toppr.com/guides/physics/electromagnetic-waves/introduction-electromagnetic-waves/ – anna v Jan 15 '19 at 13:46
  • you could have answered that to my comment to your answer in the related link provided. – Manu de Hanoi Jan 16 '19 at 02:59

2 Answers2

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Will any charge oscillating in space create an EM wave?

Almost always yes, but a counterexample was given by G.A. Schott, "The Electromagnetic Field of a Moving Uniformly and Rigidly Electrified Sphere and its Radiationless Orbits," Phil Mag Suppl 15 (1933) 752. For a uniformly charged spherical shell of radius $b$, considering only retarded solutions, the radiation reaction force at time $t$ is proportional to $v(t-2b)-v(t)$. You get no radiation if there's periodic motion with period $2b/n$.

It doesn't seem to be possible to access Schott's paper online, but there is a 1964 paper by Goedecke that has a discussion of this type of thing. Schott was opposed to quantum mechanics, and his work is often cited by kooks, such as Randell Mills, who claims to have disproved quantum mechanics and discovered an unlimited source of energy. However, there appears to be nothing wrong with Schott's actual work in the 1933 paper.

Reading between the lines, Schott's hope seems to have been that he could salvage classical physics by showing that a hydrogen atom would not have to collapse by radiation, since it might be possible for the electron's orbit to satisfy this condition for not radiating. But this would only seem to work if the electron was a rigid, extended body orbiting at some speed that happened to be exactly matched to its own dimensions, and the electron also can't satisfy the condition unless the diameter of the orbit is less than the radius of the sphere.

  • The answer above is an exception, most oscillating charges in general will generate EM waves of the same frequency. But it is a very interesting example. – PhysicsDave Jan 15 '19 at 14:53
  • @PhysicsDave: Good comment. I've edited the answer to make that more clear. –  Jan 15 '19 at 17:16
  • I had a look at the Schott 1933 article, the key paragraph is : "When the centre of a uniformly and rigidly charged sphere, with charge e and radius a, in purely translatory motion, describes a closed orbit periodically in a time 2a/cj, where j is any integer, the electromagnetic field at every outer point is a static field" I'd add that if you consider that the sphere has to move around the orbit at less than the speed of light, you get that the diameter of the orbit must be less than the radius of the sphere. – Manu de Hanoi Jan 16 '19 at 02:51
  • @ManudeHanoi: Good point, I'll add that. –  Jan 16 '19 at 17:05
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It depends on the available space. In a free space - yes. In a perfect cavity the proper cavity modes are discrete and start from some minimal frequency. So a "slow" charge oscillations may not excite even the lowest cavity mode. The corresponding EMF is then purely reactive and may not propagate.

Vladimir Kalitvianski
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  • do you mean a cavity like this : https://en.wikipedia.org/wiki/Optical_cavity ? These cavities seem to matter only for resonnance of standing waves. But my question is about what happens before that, wether it will resonate or not is another question right ? – Manu de Hanoi Jan 16 '19 at 01:34
  • No, I meant a "metallic" cavity, like a box with reflecting walls. – Vladimir Kalitvianski Jan 16 '19 at 04:49
  • i dont see how EM waves could escape a metallic box – Manu de Hanoi Jan 16 '19 at 06:00
  • In an ideal case, the EMV does not escape the box. The EMF may be very complicated time-dependent one, but it can be represented as a discrete Fourier sum over all possible eigen-frequencies, and what I am trying to say is that in this sum (superposition) will not be any term with the frequencies below "the ground state" ones. Thus, a "slowly accelerating" charge may not excite even the first standing mode. Instead, it will induce some currents in the walls, and all that, i.e., the charge and the walls, serving as a source, may not radiate even within the cavity. – Vladimir Kalitvianski Jan 16 '19 at 07:34
  • For a more realistic example, let us consider an LC circuit where the inductance is made in form of a torus so that no magnetic field exists outside it, and the capacity plates are at very short distance from each other so that the edge effects are negligeable. Then this sytem may have EMF inside it, but not outside it. – Vladimir Kalitvianski Jan 16 '19 at 07:49
  • in any case, we have EM waves, be it inside or outside right ? – Manu de Hanoi Jan 16 '19 at 08:14
  • Not really. Normally we call "waves" an EFM, that propagates eslwhere. An EMF that "stays" within a "circuit" has a constant energy (if there is no dissipation), but it does not propagate elswhere. – Vladimir Kalitvianski Jan 16 '19 at 09:16
  • This is interesting, +1, but I think it raises the question of how to define radiation. Normally we define radiation in terms of the asymptotic behavior of the fields at infinity, but that wouldn't apply here. –  Jan 16 '19 at 17:05
  • @BenCrowell: No, it is banal. The near time-dependent (and retarded too) field exists always, but it does not propagate to infinity. The radiation (the part of the total field propagating to infinity) may be suppressed in some cases. – Vladimir Kalitvianski Jan 16 '19 at 18:09