In an answer to this SE question, the respondent explains that heating a perfect diamond will not cause it to glow with thermal blackbody radiation. I don't quite follow his explanation. I think it comes down to: there is no mechanism for diamond to generate light in the visible region of the spectrum.
He mentions that interband transitions are well out of the visual range, so there will be no contribution from that.
He mentions that the Debye temperature for diamond is > 2000 K. I presume that the argument here is that optical phonons will be frozen out, too. (But diamond doesn't have infrared-active phonons, does it?)
So is that why hot diamond doesn't glow?
I suppose that if one considers real (not ideal) crystals, imperfections, impurities, and the existence of surfaces lead to the possibility of emission mechanisms, and thus glow. In fact it might be the case that a finite but otherwise perfect crystal might have an extremely faint glow.
Is this basically the reason that hot diamond does not glow? Further elucidation welcome.
Best Answer
This is because of Kirchhoff's law of thermal radiation. The corollary from it is that emissivity of a material is equal to its absorptivity.
As diamond is transparent even at large temperatures, which can be seen in this answer, its absorption coefficient is very low. Thus its thermal radiation in that spectral region is also very low.