I was thinking the other day and it looks like I need feedback if I am being stupid or crazy.
My question is quite simple - does Poynting's theorem cover the situation of charged particles escaping through surface (just like radiation part does) of the volume over which we calculate?
I honestly don't see it through some simple argument and I would like to avoid the necessity of boring calculations.
For example:
Two point charges
[itex]T<0[/itex]: I have two standing particles with the same charge and a distance [itex]d[/itex] between them. No mechanical work being done, no change in fields inside chosen volume (with some big radius [itex]R[/itex]), no [itex]E\times B[/itex] escaping through surface.
[itex]T>0[/itex]: I let them free and they start to move away from each other. Now the situation will be more interesting (and pain to calculate, at least for me). Particles accelerate and the bubble of changing fields will propagate from particles to the surface of our chosen volume. I really don't want to integrate fields of moving particles.
[itex]T=+\infty[/itex]: Particles are in infinity and all the energy which was originally hidden in [itex]\approx \int_V E^2[/itex] have left our volume. Some of it changed into kinetic energy of those particles.
And the rest must have crossed the surface as [itex]E\times B[/itex]. Well, maybe it did, it just doesn't "feel" right with my current state of intuition. Second example might be better.
One point charge
[itex]T<0[/itex]: Chunk of EM radiation goes in (it's not exactly classical electrodynamics but lets take photon). Photon is in [itex]T=0[/itex] completely absorbed by so far standing particle.
[itex]T>0[/itex]: Particle moves with constant speed, so the shape of EM fields will be a bit easier.
If I oversimplify this example and say that electron moves very very slow, then I could neglect [itex]B[/itex] field. And we can see where would such oversimplification lead.
[itex]\approx j\cdot E[/itex] is no concern, after photon absorption there is no mechanical work to be done. All the original energy in [itex]\approx \int_V E^2[/itex] must leave through surface as [itex]E\times B[/itex]. And yet, is quite hard to do with negligible [itex]B[/itex].
Two possibilities comes to my mind.
My question is quite simple - does Poynting's theorem cover the situation of charged particles escaping through surface (just like radiation part does) of the volume over which we calculate?
I honestly don't see it through some simple argument and I would like to avoid the necessity of boring calculations.
For example:
Two point charges
[itex]T<0[/itex]: I have two standing particles with the same charge and a distance [itex]d[/itex] between them. No mechanical work being done, no change in fields inside chosen volume (with some big radius [itex]R[/itex]), no [itex]E\times B[/itex] escaping through surface.
[itex]T>0[/itex]: I let them free and they start to move away from each other. Now the situation will be more interesting (and pain to calculate, at least for me). Particles accelerate and the bubble of changing fields will propagate from particles to the surface of our chosen volume. I really don't want to integrate fields of moving particles.
[itex]T=+\infty[/itex]: Particles are in infinity and all the energy which was originally hidden in [itex]\approx \int_V E^2[/itex] have left our volume. Some of it changed into kinetic energy of those particles.
And the rest must have crossed the surface as [itex]E\times B[/itex]. Well, maybe it did, it just doesn't "feel" right with my current state of intuition. Second example might be better.
One point charge
[itex]T<0[/itex]: Chunk of EM radiation goes in (it's not exactly classical electrodynamics but lets take photon). Photon is in [itex]T=0[/itex] completely absorbed by so far standing particle.
[itex]T>0[/itex]: Particle moves with constant speed, so the shape of EM fields will be a bit easier.
If I oversimplify this example and say that electron moves very very slow, then I could neglect [itex]B[/itex] field. And we can see where would such oversimplification lead.
[itex]\approx j\cdot E[/itex] is no concern, after photon absorption there is no mechanical work to be done. All the original energy in [itex]\approx \int_V E^2[/itex] must leave through surface as [itex]E\times B[/itex]. And yet, is quite hard to do with negligible [itex]B[/itex].
Two possibilities comes to my mind.
- Poynting's theorem covers a lots of situations, but it just isn't build for this. And we can avoid such situation all together if we choose infinite area over which we calculate (particles never reach surface).
- My intuition is wrong and all the energy from static fields really goes through surface as [itex]E\times B[/itex].