Polished surface of metal as a mirror

1. Apr 9, 2007

pixel01

Hi everybody
I am not so sure if this question is suitable to be in this box, but I hope it is. I know that all metals are opaque because they have lots of free electrons which absorb the energy of photons. But a polished surface of metal reflexes light other than absorbs it, especially copper, silver or gold. So why in this case, the free electrons do not absorb energy to jump to higher levels?
Thanx.

2. Apr 9, 2007

bdrosd

My understanding is that a free electron does absorb the photon by jumping to a higher level but then instantly jumps back down and re-emitts a photon so that for all practical purposes it is just reflected. This simple explaination has always bothered me since it is nearly the same explaination for how a non metallic opaque substance scatters light diffusely. In the latter case I suppose you could make an argument that there is a much greater period of time between when the incident photon was 1st absorbed and the 2nd photon emitted. This much longer time would then be responsible for the lack of phase correlation between the two and hence does not result in specular (mirror like) reflection. If anyone has some wisdom to offer I'd much appreciate it.

3. Apr 10, 2007

lpfr

Specular (mirror) reflection by metals of electromagnetic (EM) waves, including light, is not a quantum process. There are not absorbed and reemitted photons.
An EM wave has an alternating electric field that oscillates at the frequency of the wave. The limit condition of electric fields and perfect conductors is that the electric field in the perfect conductor must be zero. If not, the current should be infinite.
When free electrons in a metal "feel" an alternating electric field, they began to oscillate at the same frequency as the field. An oscillating electron radiates an EM wave polarized in the direction of its movement. That is, with the electric field parallel to the movement. This EM wave is radiates in all directions, with a maximum in the plane perpendicular to movement and a zero in the direction of movement.
In a perfect conductor, the amplitude of the emitted EM wave is identical to the incident wave. In a real metal the emitted wave is slightly smaller than the incident one.
The phase of the emitted EM wave is such that at the metal surface and in the metal side, the addition of the two waves is zero. In the incident wave side the two waves travel in opposite directions. Seen from outside one has the impression that the wave coming from the metal is the reflection of the incoming wave. It has the same frequency and amplitude. But, in reality, it is a wave that has been emitted by the metal electrons.
When the electric field is parallel to the surface of the mirror, the reemitted wave has the same polarization. When it is not the case the polarization of the reemitted wave is a little trickier.
There is not need of metals to obtain a "metallic" reflection: free electrons suffice. There are a lot of them in the ionospheric plasma. The ionosphere reflects radio waves at low radio frequencies (under a few MHz). And some metals can be transparent to light. The potassium is transparent to near UV.
Real metals are not perfect and their conductivity is frequency dependant. But except copper and gold, they reflect fairly uniformly in the visible band.

4. Apr 10, 2007

bdrosd

lpfr, Thanks for your insight. Your explaination is in terms of classical EM theory which is cetainly proper, but it should be possible to explain the same reflection phenomena via the quantum theory (i.e. photons). This was the question initially posed in the first post. I believe that we need to be able to explain the reflection of light from both approaches. Since incident photons exist, and in general they have no phase relationship to one another, their reflection must be explainable on a single photon basis. Fynmen's book "QED" does this nicely but I must admitt that I still struggle with understanding the difference between reflection from metals and smooth surfaces that scatter light diffusely.
P.S. another, more common, example of a "metallic" like reflection from a non metal is the low angle reflection of light from the boundary between hot and cooler air over the surface of a hot road- this change in index of refraction produces a reflection that we call a mirage.