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I was recently told by a physics teacher that "any system of charges in which at least some of the charges are executing some sort of accelerated motion, will radiate and lose energy". This refers to classical electrodynamics (I'm not entering into quantum mechanics), that is, this statement should either be provable or disprovable from Maxwell's equations. I tried to think of a simple system of charges that would disproof this statement, since I intuitively think that it isn't true, but I haven't come up with any. The one example I could think of was a positive charge sitting on top of a negative charge, so that the net charge is zero. No matter how they move it is obvious that there will be no radiation. But my teachers did not accept this trivial example.

So my question is whether the statement is true or not, plus a proof or a counterexample of some charges executing accelerated motion without radiating.

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    $\begingroup$ Any DC circuit is a counterexample. This question is a pretty close duplicate of this one: physics.stackexchange.com/q/13361 $\endgroup$
    – user4552
    Commented Aug 12, 2011 at 3:16

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The claim that accelerated charges must radiate is simply false. There are very many simple situations in which they do, but in general things should be examined on a case-by-case basis; there is not simple thumb rule like "acceleration yields radiation."

The simplest way to see this is to consider a wire carrying a constant current. This situation is magnetostatic, with well-understood electromagnetic field given by Biot-Savart law... which trivially lacks any radiation, as the field is constant. This is independent of the shape of the wire, and yet, if the wire is not straight, the charges within it must accelerate at some point.

A more exotic could be a uniformly rotating ring of charge. Both charge density and current density will be constant, and will produce constant electric and magnetic fields. This can be experimentally realized with toroidal superconductors.

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  • $\begingroup$ You mean rotation around the ring axis perpendicular to the ring plane. It's true but for not radiating the charge must be uniformly distributed and the ring must be ideal in shape. $\endgroup$
    – Vladimir Kalitvianski
    Commented Aug 12, 2011 at 17:24
  • $\begingroup$ A superconducting ring doesn't need to be ideally-shaped to carry a persistent current. The end result is still rotating charge with no cyclotron-type radiation. $\endgroup$
    – Stan Liou
    Commented Aug 12, 2011 at 19:04
  • $\begingroup$ A stationary ring, superconducting or not, can be of any shape but a rotating one must be ideal. $\endgroup$
    – Vladimir Kalitvianski
    Commented Aug 12, 2011 at 19:10
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    $\begingroup$ If you keep the charge density and current constant, there cannot be radiation, whether in the context of non-relativistic electrodynamics or fully covariant Maxwell's equations. I'll edit in some details, which might also address Vladimir (not entirely sure what he's referring to). $\endgroup$
    – Stan Liou
    Commented Aug 15, 2011 at 18:15
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    $\begingroup$ @StanLiou Charges in a storage ring of a particle accelerator constitute a (constant, I assume) circular current and do radiate - this is often even their primary purpose (e.g. PETRA III). What is the fundamental difference between a ring carrying a constant current and electrons being being kept on a circular beam path? $\endgroup$
    – DK2AX
    Commented Apr 2, 2019 at 8:48
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The teacher is wrong. The simplest counterexample is the relativistic two-body problem solved in A. Schild, Phys. Rev., 131, 2762 (1963).

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    $\begingroup$ The Schild paper uses the "time-symmetric" Fokker-Tetrode form of E&M which is not equivalent to the standard E&M theory. In this "time-symmetric" theory an average of retarded and advanced fields is used. This suppresses radiation from accelerating charges, but is inconsistent with causality. Wheeler and Feynman restored causality in their "absorber" theory by the action of a distant absorbing medium. References are in the Schild paper. $\endgroup$
    – Mark Davidson
    Commented Mar 13, 2018 at 3:20

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