Together, the current answers to Is the visible light spectrum from "red-hot glass" at least close to Blackbody Radiation? explain that while we can not necessarily call a heated sample of good quality transparent glass a blackbody emitter, the shape of the thermal emission spectrum may still approximately follow Planck's_law if the emissivity is nonzero and relatively constant over the visible wavelength range.

That question shows several examples of apparently transparent glass (at least when cool) glowing brightly when it is hot enough.

I have two interrelated questions about this:

  1. Hot glass during glass blowing gets quite bright. I have no measurements but it certainly seems brighter than one would expect if the emissivity were as low as the absorptivity. Does heated glass or do transparent covalent crystals (e.g. quartz, diamond) glowing yellow-white become less transparent by the same mechanism that allows it to radiate light, or does it remain just as transparent and the visible radiation is actually as low as one would calculate from Kirchhoff's_law $a = \varepsilon$?
  2. What electronic mechanisms produce visible thermal radiation transparent covalent glasses and crystals? Reading @JohnRennie's answer I have a hunch that thermal excitation first moves electrons to higher bands allowing them to radiate, which suggests that yellow-white hot glass would also be darker (more absorbing) than room temperature glass, but I'm just grasping at straws.
  • 1
    $\begingroup$ 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
    Jan 9 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
    Jan 9 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. $\endgroup$ Jan 9 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. $\endgroup$ Jan 9 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
    Jan 9 at 10:08

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