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?

• 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. Sep 11, 2013 at 5:58
• 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... Sep 11, 2013 at 5:59
• 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. Sep 11, 2013 at 6:00