0
$\begingroup$

I have a question regarding sources of electromagnetic potentials:

As far as I know, wave equations for the scalar and magnetic vector potentials can be derived from Maxwell equations employing the Lorenz gauge. If there are no charges then the wave equations are homogeneous. However in the presence of charges the wave equations are inhomogeneous and contain source terms. I haven't worked it out mathematically completely, but can't one show that charges and currents would then also be sources of electromagnetic energy, if one chooses vanishing initial conditions for the electromagnetic potentials? Because if the potentials are vanishing in the beginning, they would have to build up which basically means that electric fields would build up as well. However since the electric field has energy content then that would mean that the charges are "sources" of energy?

$\endgroup$
1
$\begingroup$

You cannot choose vanishing initial conditions, as the potentials must obey the inhomogeneous wave equation at all times and places. The charges can be the source of radiation energy. They can transform kinetic energy into radiation energy if they are accelerated.

$\endgroup$
  • $\begingroup$ I don't understand why one can't choose vanishing initial conditions. Isn't it a second order partial differential equation, which requires 2 initial conditions? $\endgroup$ – eeqesri Nov 17 at 12:04
  • $\begingroup$ They are inhomogenous equations. The source fixes the potentials up to some trivial integration constants. $\endgroup$ – my2cts Nov 17 at 12:18
  • $\begingroup$ I am not satisfied by this answer, because I read in Peskin Schröder that a Klein Gordon equation with a classical source exhibits particle creation by the source. I know that the question refers to classical fields, but the argument is very similar. So where does the energy of the created particles come from? $\endgroup$ – eeqesri Nov 17 at 16:57
  • $\begingroup$ If you are not satified because you have more questions than the one I answered, you may submit these one per post. $\endgroup$ – my2cts Nov 17 at 18:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.