3
$\begingroup$

Essentially i have the same question as stated here:
What happens when an Electromagnetic wave encounters a conductor's surface?

I want a microsopically motivated answer, i.e. what the charges inside the conductor are doing. Alternatively i would also accept an explanation building on wave interference.

My perspective:
Assume an incoming plane electromagnetic wave from the negative x direction. When the electric field of the wave arrives at the surface of the conductor (x=0), the electric field will arrange the nearest free charges such that the net electric field inside the conductor will be annihilated. This is the microscopic reason why there is no field inside the conductor. The periodic acceleration of the charges induces dipole radiation which in combination with the incoming wave result in a standing wave on the negative x-axis.

My problem: $\quad$ What happens on the positive x-axis?
The dipole radiation emits waves also in the +x direction. The electric field will again be annihilated by the neighbouring free charges which again leads to dipole radiation. This "charge wave" will propagate until the end of the conductor (x=d). Now there are no charges left to annihililate the dipole radiation resulting in a "transmitted" electromagnetic wave (x>d).

My research:
I stumbled on the following theorem which sounds like an elegant explanation except that it only accounts for dieelectric materials and thus involves a transmitted (slower) wave which should not be the case for conductors.

The Ewald-Oseen Extinction Theroem says that the light emitted by the atoms has a component traveling at the speed of light in vacuum, which exactly cancels out ("extinguishes") the original light wave. Additionally, the light emitted by the atoms has a component which looks like a wave traveling at the slower speed of light in glass. Altogether, the only wave in the glass is the slow wave, consistent with what we expect from basic optics.

$\endgroup$

1 Answer 1

1
$\begingroup$

The radiating dipoles radiate in both directions. However, the radiated wave is $\pi$ out of phase with the incident wave. This is because the conductivity of a good metal is primarily imaginary, so the current is $\pi/2$ out of phase, and then the radiation is another $\pi/2$ delayed. Where the emitted wave is going in the same direction as the incident wave, they have interfere destructively at all points—the wave is zero everywhere inside the conductor. Where the emitted wave is traveling antiparallel to the incident wave, they interfere destructively at some places (the “nodes” of the standing wave), constructively other places (the “antinodes”), and somewhere in between everywhere else.

Of course, this isn’t the whole picture. Metals aren’t perfect, so there is some absorption of power into the metal, and consequently there is a finite “skin depth,” which describes the region of exponential decay of the field amplitude into the metal bulk. And then there is also the transfer of momentum (radiation pressure) as well as other more exotic phenomena.

$\endgroup$
3
  • $\begingroup$ Isn´t the energy of the incident wave already transferred to the kinetic energy of the charges? I thought that the reason for the transmission of the incident wave in dieelectrics is because of the bound charges (low imaginary refractive index). Whereas in conductors we have free charges and therefore the incident wave will fully transfer its energy to the charges (high imaginary refractive index). In my current view there should be nothing noteworthy of the incident wave left, only the secondary waves from the dipole radiation. $\endgroup$
    – hydra4jh
    Sep 18, 2022 at 10:02
  • $\begingroup$ I think i´ve found the answer: According to the Maxwell Equations the incident wave depends only on its source. When a charge moves driven by the incident wave it can only alter the total field (and its energy flux) by emitting a secondary wave. source (2nd answer) $\endgroup$
    – hydra4jh
    Sep 19, 2022 at 3:28
  • $\begingroup$ @hydra4jh physics often provides different ways to look at the same thing. In classical E&M, the incident wave can be taken to be constant, which then causes charges to move (then they emit waves of their own). This picture does not consider momentum and kinetic energy. However, of course, if you do consider momentum and energy, you’ll find that it all balances. But the momentum-energy picture often doesn’t give you details of how the interaction happens, only the overall accounting of the conserved quantities. $\endgroup$
    – Gilbert
    Sep 19, 2022 at 13:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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