0
$\begingroup$

Prompt radiation of a nuclear detonation in vacuum strips primary electrons from nuclei and radially drives them in an expanding sphere. The electrons are negatively charged and low mass while the nuclei are positively charged and high mass.

An idealized approximation of this, however rough, is a stationary positive spherical surface through which the negative spherical surface radially oscillates. Since there is no curl in the magnetic vector potential of the oscillating negative sphere, no photons can escape. Is there any energy loss mechanism in this idealized system?

If so, what is it?

If not, can a test charge outside of the maximum extent of the negative sphere sense any change in the electric field during an oscillation?

$\endgroup$
  • $\begingroup$ Related: physics.stackexchange.com/questions/59402/… $\endgroup$ – probably_someone May 3 '18 at 3:26
  • $\begingroup$ Thanks. That question and answer would seem to imply that outside the outermost and inside the innermost sphere, there is no electric field, and between them the electric field takes on the polarity of the innermost. So there is an E-field shell of varying thickness that oscillates in a square wave between positive and negative but has no affect on the environment. Right? $\endgroup$ – James Bowery May 3 '18 at 4:02
  • $\begingroup$ There can be no magnetic field without curl in the magnetic vector potential (by definition) and there is no curl in the magnetic vector potential even as the two spheres exchange inner vs outer position -- with the attendant reversal of E field in the shell between them. Part of the reason I asked this question is if there is any "physical reality" to the magnetic vector potential, its empirical effects should appear most prominently during a nuclear detonation in space due to the curl-free motion of the primary electrons. $\endgroup$ – James Bowery May 3 '18 at 16:28

Your Answer

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

Browse other questions tagged or ask your own question.