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In an electromagnetic waveguide, there is generally a "cutoff frequency." Electromagnetic waves with a frequency that is lower than this cutoff frequency will not propagate at all -- i.e., they will be exponentially attenuated.

Suppose that there is a source of electromagnetic radiation at the center of a waveguide (e.g., an oscillating electric dipole inside a rectangular waveguide), and that the emitted waves have a frequency below the cutoff frequency of the waveguide. All of the waves will be exponentially attenuated and thus will not propagate within/outside of the waveguide.

But, before the waves are sufficiently exponentially attenuated, they will be carrying away some energy from the radiating source in the form of Poynting flux. What happens to that radiation energy immediately before and after attenuation -- where does it go? Maybe the radiation is very quickly absorbed by the conducting boundary of the waveguide and thus heats up the waveguide's surface?

Most people would say that the Poynting flux is zero if the frequency of the waves is below the cutoff frequency of the waveguide, but the language of "exponential attenuation" suggests to me that the waves did travel for some short amount of time before being exponentially attenuated (and thus the Poynting flux was not zero before exponential attenuation).

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    $\begingroup$ It gets reflected towards the source. The phenomenon is equivalent to evanescent waves that can be observed near a surface with total internal reflection. $\endgroup$
    – CuriousOne
    Commented Mar 25, 2016 at 0:28
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    $\begingroup$ And does the source "re-absorb" the waves/radiated energy? Is that "re-absorption" the cause of the exponential attenuation? $\endgroup$
    – quantumflash
    Commented Mar 25, 2016 at 0:32
  • $\begingroup$ It doesn't matter where the energy goes, a traveling wave at that frequency is simply not a solution to the waveguide. Having said that, with most engineering setups the impedance of the waveguide will simply appear similar to a short (with some large imaginary component) to the transmitter. If you ever get involved with RF power circuits you will learn (probably the hard way for the first time), that measuring the impedance of an antenna (or a filter etc.) before attaching it to the transmitter is an excellent idea. :-) $\endgroup$
    – CuriousOne
    Commented Mar 25, 2016 at 0:36
  • $\begingroup$ Whether the source re-absorbs energy reflected back towards it depends entirely on the nature of the source. $\endgroup$
    – The Photon
    Commented Mar 25, 2016 at 14:51
  • $\begingroup$ I would disagree that a traveling wave below the cutoff frequency is not a solution to the waveguide. The language "exponentially attenuated" means that the wave was in fact emitted from the radiating source, but that very quickly it was attenuated. So it did travel spatially before being extinguished. So what would the Poynting flux in such a case be? Is the source radiating away / losing energy via Poynting flux? $\endgroup$
    – quantumflash
    Commented Mar 25, 2016 at 17:08

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I believe energy will be reflected from all sides in this case, which will not be much different from a case of a source in a cavity. The amplitude of the electromagnetic field will grow until losses in the walls of the waveguide become comparable to the power radiated by the source.

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    $\begingroup$ @quantumflash: You should understand that the exponential attenuation is spatial, not temporal (the amplitude decreases with distance from the source, not with time). With time, the amplitude will increase until saturation ( determined by the balance between energy input from the source and the losses). $\endgroup$
    – akhmeteli
    Commented Mar 25, 2016 at 0:50
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    $\begingroup$ @quantumflash: Energy is lost in the walls, as they are not perfect conductors (even if they are superconducting, there will be losses at nonzero frequency). $\endgroup$
    – akhmeteli
    Commented Mar 25, 2016 at 1:45
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    $\begingroup$ There needn't be much energy lost in the walls; an optical fiber is a waveguide, and transfers useful amounts of EM radiation for dozens of miles. The walls of such waveguides use no conduction and ohmic heating doesn't occur. $\endgroup$
    – Whit3rd
    Commented Mar 25, 2016 at 9:02
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    $\begingroup$ @Whit3rd: much energy - maybe not, but some energy is always lost, and these losses still provide saturation. $\endgroup$
    – akhmeteli
    Commented Mar 25, 2016 at 13:41
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    $\begingroup$ @quantumflash: I am not sure, but I suspect the Poynting flux is oscillating, probably, with a double frequency. There is also a small constant part of the flux related to the losses (and it should be directed towards the walls). $\endgroup$
    – akhmeteli
    Commented Mar 25, 2016 at 23:25

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