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?

  • $\begingroup$ Compare the energy of photons at the cut-off frequency with typical temperatures. $\endgroup$ Oct 21, 2015 at 23:21
  • $\begingroup$ 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) $\endgroup$ Oct 22, 2015 at 0:16
  • $\begingroup$ 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? $\endgroup$ Oct 22, 2015 at 0:16
  • $\begingroup$ 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. $\endgroup$
    – DanielSank
    Oct 23, 2015 at 20:26
  • $\begingroup$ @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. $\endgroup$
    – DanielSank
    Oct 23, 2015 at 20:28

1 Answer 1


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.

  • $\begingroup$ 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)? $\endgroup$ Oct 24, 2015 at 15:05
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    $\begingroup$ @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. $\endgroup$
    – DanielSank
    Oct 24, 2015 at 15:16

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