How does a Faraday Cage work in terms of photons?

Question

High frequency radiation is able to penetrate solid material by virtue of the wavelength being far smaller than the interatomic distances of the lattice. In order for the radiation to be stopped, a photon needs a direct collision with an atom. In such instances the thickness and density of the material determine how effective a shield it is.

Yet, for EM radiation with higher wavelength, very thin conductive surfaces can act as a shield because of Gauss' Law which states that fields can't exist within an equipotential.

So my question is - what is the difference between these two scenarios in terms of photon absorption? Is there a different mechanism of absorption? Is there are different property of long wavelength photons? Why is a direct atomic collision not needed?

Thanks Mark, after re-reading my question perhaps I didn't phrase it right.

I understand the difference beween dielectrics and conductors, and also quantized energy states. What I'm trying to get at is if a gamma ray were to hit a conductive shield it would still pass right through UNLESS the shield was thick enough and dense enough. ie. a gamma ray is unaffected by conductive materials BUT lower frequency photons are stopped by the electron cloud.

Now this can all be well understood by classical field and wave theory. My question is what is the difference between the PHOTONS that means that gamma rays can ONLY be stopped by direct atomic collisions whereas lower frequency radiation can interact with the electron cloud?

I know that the frequency is different, but what does that mean in terms of photons as discrete entities?

Answer 1

You need to understand the difference between a dielectric and a conductor.
In a dielectric it is the intermolecular bond that absorbs the photon energy - this is limited by QM ie it can only absorb certain frequencies.

In a conductor the electrons form an "electron" gas they are not bound and hence are not subject to QM limits. This is why metal appear "metallic"

Hope this helps

I did my udergrad electromag exam a few months ago so here goes.

The interaction of the electron and the photon can be considered as a Simple Harmoic Motion but with a driver and a damper, whereby the electron vibrates due to the electromagnetic field of the photon. I can't write the equation here but in effect it's a second order differential equation that can have a complex solution (as in square root of minus 1 type complex). Because of the resistance to oscillation of the electron the electromagnetic waves attenuates which result it it penetrating to a certain depth, the skin depth. The skin depth is proportional to the angular frequency of the EM wave (w) raised to the power of -1/2
At high frequencies the electron cannot respond fast enough to vibrate in sympathy with the electromagnetic wave so the wave passes through "mostly" unhidered. I guess that it might hit a nucleus in the end.

The Open University Book SMT359 Electromagnetic Waves - Covers this quite well, If you want I can send you a pdf of the section you require, although you will need at least pretty good maths (say 1st year degree), which given the question, I guess you have.
Hope this helps.

Answer 2

Dear Andy,

I haven't had time to give a lot of thought to this question BUT:

At low frequencies the EM waves interact with the 'sea of electrons' in a conducting medium. By LeChatelier's theorem the electrons move in such a way as to negate the EM wave. (Same as Lenzes Law)

At high frequencies, the electrons (in the sea) have too much inertia to react to the EM waves and the waves pass through. At high enough EM frequencies even good conductors such as Cu and Ag are di-electrics. ( I seem to remember calculating the transition frequency in my Physics Course. It was a mass on a spring type of calculation I think)
At these high frequencies the photons pass through the material unless they hit the atomic nucleus field and undergo scattering that way.

The answer is that even metals do not act as conductors at very high frequencies!

Look up Quantum Electrodynamics. Feynmann wrote a good little book on this. The mathematics is a fearsome thing to behold!

. CopyLeft RC

Answer 3

Closed conductors are not equipotentials internally. They are equipotentials at the closed surface, but there may be internal field structure depending on interanl charge distribution.
EM radiation is waves and not charge, so an EM wave can enter a closed conductor without creating a net field which would change the potential at the surface.
The energy transfer from photons to the electron cloud around a metal is a process know as evanescence, which cause waves to attenuate and lose energy. It is a wave theory and quantum mechanical effect and can't be explained using a particle model. Essentially in your example the energy is transfered to the material not because it is a Gaussian surface but because of the intrinsic properties of the material to demonstrate evanescence at particular frequencies.

Answer 4

Faraday cages can be used to block out electromagnetic radiation which come in the form of waves. The lower the frequency the larger the grid, the higher the frequency the smaller the grid. Light waves could theoretically be blocked, but their frequency is so high that you'd have to use an impossibly small mesh. Gamma ray frequency is much higher. Faraday cages are really only good for very low frequency waves such as radio. Even microwaves are too high for practical usage of a a Faraday cage. Additional info: Alien Steve is correct ... I wasn't thinking of solid metal as being a Faraday cage. His answer is on the money (he gets my vote).