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Suppose I have a charge moving back and forth above an infinite, grounded, conducting plane. Can I calculate the total radiated power by using image charges? That is, are the scalar and vector potentials the same in the upper-half space for all time for both the image charge "picture" and the standard picture?

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    $\begingroup$ No, the method of images is explicitly an electrostatic method. The reason is it relies on a uniqueness theorem for static electric fields with a given boundary condition, so that the actual distribution of charges can be replaced by a set of image charges that gaurantee the boundary condition. $\endgroup$
    – Michael
    Sep 11, 2013 at 5:58
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    $\begingroup$ In the non-static case this fails because the electric field is no longer determined by the boundary conditions alone: a wavepacket of EM radiation could be hiding out far away from the origin (without breaking $E,B\to 0$ as $r\to\infty$) and come in later (maybe an hour from now, maybe next year) and spoil your solution. It would be interesting to see in a detailed answer an example of how the method of images fails (or stops failing rather) in the limit of low frequencies... $\endgroup$
    – Michael
    Sep 11, 2013 at 5:59
  • $\begingroup$ BTW, you still might get away with it in some specific examples like the one you mentioned, but it won't work as a general method. $\endgroup$
    – Michael
    Sep 11, 2013 at 6:00

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Well the conducting ground plane (assumed perfect) simply reflects the radiation that comes from the moving charge. So based on your specification of the system, all the radiation from the moving charge (accelerating) either goes directly away from ground, or reflects off the infinite ground plane. This is in no way different from assuming no ground plane, and an image charge.

But of course, if you actually had two charges, well you would also have radiation into the other hemisphere, behind the ground plane.

So they are not quite identical, but the field pattern in space above the ground plane will be correct with either model.

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  • $\begingroup$ Transition radiation detectors work explicitly because of this as-if radiation pattern. And when the photon energies are high enough (as they can be when ultra-relativistic charges are involved, the radiation does go into the material as well). Fun stuff. $\endgroup$ Sep 14, 2013 at 0:16

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