# Modes inside a cavity and black body radiation

Consider a perfect conductor that encloses a spatial volume such as a parallelepiped or cylinder. If we solve Maxwell's equations inside that volume, seeking solutions that depends on time with a dependency of the form $e^{-i\omega t}$, we find that only TE and TM modes can exist inside of that volume (and no TEM modes). However both TE and TM modes have a cutoff frequency. This seems to imply that any EM wave inside the cavity cannot have any frequency and that it should be greater than a threshold.

However if we look at the problem from another perspective, the one of a black/grey body at a temperature $T > 0K$, we'd think that the walls are emitting EM waves without any cutoff frequency (and with a continuous spectrum).

I understand that the sum of two solutions to Maxwell's equations in the cavity is also a solution and I think that I could write any allowed EM wave as a sum of TE and TM modes, but if both TM and TE modes have a cutoff frequency, I don't see how I could obtain an EM wave with a lower frequency that the cutoff one.

Hence I don't see how to reconciliate the blackbody radiation with TE and TM modes inside a cavity. Where do I go wrong?

• Compare the energy of photons at the cut-off frequency with typical temperatures. Oct 21, 2015 at 23:21
• If I consider a circular cylindrical cavity with radius equal to 0.5 m, I get a cuttoff frequency of about 1.4 GHz for TM modes. That would be photons corresponding to about $1.52 \times 10^{-25}$ J. Or wavelength of about 3.02 meters... which is bigger than the radius of the cavity. Assuming the above is right, the main idea is that almost any wavelength produced by a black body at ~290 K (which has a peak at around $10 \mu m$) is allowed to exist in the cavity. Is this correct? (more to come) Oct 22, 2015 at 0:16
• Then what would happen if I reduce the size of the cavity enough so that microwaves can't exist there due to the cutoff frequency? The black body radiation wouldn't occur? Oct 22, 2015 at 0:16
• The confusion here comes from assuming that a hollow conductor is a black body. It's not. If that answers your question I'm happy to copy it into an answer, but I suspect you'll want more information. Please let me know. Oct 23, 2015 at 20:26
• @SebastianRiese it's not hard at all to make a cavity small enough that the cutoff frequency $\nu$ corresponds to a temperature $T = h \nu / k_b$ which is easily achieved in cryogenic systems. Oct 23, 2015 at 20:28

It's true that a hollow conductor has a minimum cutoff frequency. However, a hollow conductor is not a black body. A black body has perfect absorption of radiation at all frequencies, while a perfect conductor perfectly reflects all radiation.

A black body emits radiation according to the Planck law. Since the black body absorbs all incoming radiation, but then re-emits it according to the Planck law, it must be capable of converting from e.g. a single high energy photon to several lower energy ones. The hollow perfect conductor can't do this because it reflects all radiation without ever absorbing and "processing" it.

For more details I strongly recommend reading the other Physics Stack Exchange post Why is black the best emitter? and the associated answer.

• I am extremely satisfied with your answer. Now I have a related question. I am wondering why metals seem to behave more like black bodies than to perfect conductors when they are heated up, since they become red/white as if they emitted closely to black bodies. Am I correct in thinking that metals stop to become conductors at high temperature (say over 1500 K)? Oct 24, 2015 at 15:05
• @no_choice99 You should ask this as another question, but I'll give you a hint. Metals are poorer conductors as you go to higher frequency. Also, the frequency at which the blackbody spectrum is largest goes up as you raise the temperature. Oct 24, 2015 at 15:16