Reflection, Transmission, and Plasma Frequency

Question

Does anyone have a good, clear explanation of why and how this works? I don't understand the following.

Say you have a piece of metal with a plasma frequency $\omega_p$. This is like a resonant frequency, so I can believe that for frequencies $\omega \approx \omega_p$, we will have mostly absorption from the damping term.

Here is what I don't understand: we always say that we will have reflection for $\omega < \omega_p$ and transmission for $\omega > \omega_p$. Why is that? The resonance curve is symmetric, so it seems like the system would do more-or-less the same thing on either side of resonance.

Thanks!

Answer 1

If you have a very low frequency, the material behaves as a conductor: the electrons respond instantaneously to the excitation, and therefore the metal becomes a reflector (the boundary condition of "no E field parallel to the surface" is met).

If you have a very high frequency, the electrons don't have "time to react" at all - so the amplitude of their response is small (and shifted in phase). The incident EM wave will continue unmolested.

As you already said - at resonance, the electrons move just enough to absorb much of the energy of the incident wave.

Another way to put it: the resonance amplitude curve looks symmetrical - but the phase curve is not. Phase changes continuously as you go from below-resonance to above-resonance.