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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?

untreated_paramediensis_karnik
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  • Compare the energy of photons at the cut-off frequency with typical temperatures. – Sebastian Riese Oct 21 '15 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)

     – untreated_paramediensis_karnik Oct 22 '15 at 00:16
  • 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? – untreated_paramediensis_karnik Oct 22 '15 at 00:16
  • 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. – DanielSank Oct 23 '15 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. – DanielSank Oct 23 '15 at 20:28
  • Yes I would appreciate more information. For example even if the walls are a grey body (they don't emit as much as a black body would for a same temperature), I am still in the same trouble.

    I think we could consider an idealized microwave oven as a real world example of such a cavity at T > 0K.

     – untreated_paramediensis_karnik Oct 23 '15 at 20:30
  • Can you explain why you want to think about a hollow conductor in the first place? It's not a black body. If you couple it to a black body then you can make interesting arguments, but by itself it's not a black body. Also, please use @DanielSank in your comments so I know you've responded. – DanielSank Oct 23 '15 at 21:15
  • @DanielSank I know (but this explains why it doesn't matter for a practical hollow body at the photon energies the radiation laws were first discovered). The other thing is, that the Planck distribution still holds for the available modes. – Sebastian Riese Oct 23 '15 at 21:39
  • @DanielSank I understand now. I chose a hollow conductor because I know that inside those there would be some cutoff frequency for the EM waves. A perfect conductor would reflect all light and therefore cannot be a black body. So he answer to my question is that with the set up I've described there would be no black body radiation at all. I'd appreciate if you could write an answer and add any relevant remark if you have some. – untreated_paramediensis_karnik Oct 23 '15 at 22:45

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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.

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DanielSank
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  • 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)? – untreated_paramediensis_karnik Oct 24 '15 at 15:05
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    @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. – DanielSank Oct 24 '15 at 15:16