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    $\begingroup$ I don't know the answer, but it's probably a truism that if the electrons can move around well enough to emit photons, they can equally well absorb similar photons. I assume this is true of a plasma, for example. I suppose different frequencies (energies) could be involved. In which case, maybe red-hot glass is still transparent to UV? $\endgroup$
    – Roger Wood
    Commented Jan 9, 2021 at 5:52
  • $\begingroup$ @RogerWood Yes, that must be right. I think there is an experiment to be done at a glassblower's shop with a photodiode, a meter, and a red, green and blue laser pointer and some kind of pyrometer. Plotting the attenuation at three wavelengths as a function of temperature might show the red attenuation start to increase before noticeable in the blue. $\endgroup$
    – uhoh
    Commented Jan 9, 2021 at 6:01
  • $\begingroup$ I have a hunch that thermal excitation first moves electrons to higher bands allowing them to radiate - no. In solids and liquids the light is emitted by black body emission not by transitions between excited states. $\endgroup$
    – John Rennie
    Commented Jan 9, 2021 at 7:32
  • $\begingroup$ See What are the various physical mechanisms for energy transfer to the photon during blackbody emission?. I won't close your question, but I do think it's basically a duplicate of this. $\endgroup$
    – John Rennie
    Commented Jan 9, 2021 at 7:34
  • $\begingroup$ @JohnRennie blackbody emission is not a mechanism. The emission process is the opposite of whatever absorption proceses stop you seeing through the glass. $\endgroup$
    – ProfRob
    Commented Jan 9, 2021 at 10:08