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According to everything I've been taught about incandescence and black-body radiation, and some quick Googling to confirm I'm not crazy, just about everything, regardless of composition, should start glowing red at about the same temperature- 798K, the Draper point, where sufficient power in the black-body radiation curve crosses into the visible spectrum to be visible.

I have just been informed by a metallurgist friend, however, that different metals in his experience begin to glow red at wildly different temperatures; typically, just below their melting points. For example, apparently aluminum glows red at much lower temperatures than steel.

My hypothesis so far: The metals in question are far from perfect black bodies (reasonable, since most metals are shiny), and differing levels of emissivity in the low end of the visible spectrum require different temperatures to raise total emission in that range to visible levels. This, however, does not explain why there should be any connection between glow-point and melting point.

Am I close to correct? Is there another better explanation? Or is my friend simply crazy?

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    $\begingroup$ Most interesting. How does your friend explain how a pyrometer works? The modern multicolour pyrometer is error prone: variations in surface emissivity with frequency will trick it. But it is not THAT unreliable: it works pretty well for blooms of iron from a steelworks (I had quite a bit of experience with this kind of thing as a very young engineer). The point is: pyrometers assume a scaled (by emissivity) version of Planck's law to calculate the temperature from the spectrum, so their reasonable reliability means that the theory must work pretty well in many case. $\endgroup$
    – Selene Routley
    Commented Nov 12, 2014 at 12:47
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    $\begingroup$ See also my answer here. The main part is Kirchhoff's law of thermal radiation $\endgroup$
    – Ruslan
    Commented Nov 12, 2014 at 13:34
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    $\begingroup$ As common counterexamples, neither mercury nor gallium glow red anywhere near their melting points. They will near 798K though. $\endgroup$
    – Asher
    Commented Oct 11, 2015 at 15:12
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    $\begingroup$ @Aabaakawad I was commenting on the posted question, but thank you for the clarification. $\endgroup$
    – Asher
    Commented Oct 11, 2015 at 16:44
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    $\begingroup$ @Aabaakawad Of course: that is why they are multicolor pyrometers. But the shape of the emission spectrum still stands pretty well, albeit with a different, temperature-dependent, roughly wavelength independent multiplier $\epsilon(T)$. But the OP's question seems to be denying the shape of the spectrum itself. Witness that a material with a temperature-dependent, roughly wavelength independent emissivity will still glow red at the same temperature as a black body - only the intensity will be different $\endgroup$
    – Selene Routley
    Commented Oct 12, 2015 at 5:52

3 Answers 3

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All materials have some color (i.e. are not perfectly black or white bodies). So, even at incandescent temperatures, if they are illuminated, there is reflected light with color. A good example of a non-black-body would be glass; softening a glass rod in a Bunsen burner flame, it does glow red, but there's little optical coupling of the transparent material with visible light, so it's likely you will see a yellow sodium glow in the Bunsen burner's flame when the glass rod starts to melt, rather than the red of the hot glass. Optical pyrometers work best when peeking through a hole into a dark oven chamber. No matter how inefficient the light emission is, when the oven's content is at a uniform high temperature, the black-body curve is the color emitted, because non-blackness means you see reflection or transmission, of light from other (equally hot) surfaces in the oven.

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  • $\begingroup$ "(...)little optical coupling of the transparent material with visible light(...)". I don't understand this. Could someone elaborate? $\endgroup$
    – Mads Skjern
    Commented Jul 26, 2019 at 22:27
  • $\begingroup$ @MadsSkjern A reflector or transparent object has little color of its own, and cannot easily emit light because of that (it is a truism that 'white' objects are poor thermal emitters as well). The familiar color of red-hot steel comes from the (thin) black oxide on the steel surface, but silver doesn't show 'redness' when at the same temperature. $\endgroup$
    – Whit3rd
    Commented Jul 27, 2019 at 3:47
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Different metals would to glow at different temperatures because of their different abilities to hold on to electrons. Some metals hold their electrons very weakly and some hold it very tightly. The temperature at which it glows is dependent on the strength of this force. Things glow either because of absorption or emission spectrum. The lesser the force with which the nucleus holds the electron, the lesser is the energy required to make it glow. Actually, to get a more satisfying answer, you must post this question in chemistry stack exchange.

Edit: The ability to remain solid is dependent on this force too, hence the connection between melting point and glowing red.

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  • $\begingroup$ This is also covered in physics.stackexchange.com/q/64088 $\endgroup$
    – Eubie Drew
    Commented Oct 12, 2015 at 6:44
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    $\begingroup$ I'm unaware of any connection between electron binding energy and blackbody radiation, could you provide a source? $\endgroup$
    – pentane
    Commented Oct 12, 2015 at 18:17
  • $\begingroup$ Photoelectric effect clearly explained black body radiation. Einstein proposed the connection between frequency of light and work function. You can refer several sources for the same. $\endgroup$
    – ShankRam
    Commented Oct 14, 2015 at 5:57
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There is no direct relation between melting point and colour of light produced, it's just that some heat energy is used in breaking intermolecular forces and a part of it is transferred to atoms. So for a higher melting point, a bigger part of energy is used in breaking intermolecular attraction and to change its state of matter.

The rest is explained by black body radiation's explanation by Max Plank. When energy is given to atom, it's valence electron gets excited and jumps to higher energy levels and return back to original shell by emitting electromagnetic rays of different wavelength like first low energy radiations like that of infrared, red and so on as the heat increases.

And your question of different light in different elements is explained by law of conservation of energy. neglecting energy lost in breaking intermolecular forces, same energy given to two different atoms produce same light no matter which element that it is.

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    $\begingroup$ I think you got it wrong: black body radiation has nothing to do with electronic transitions in atoms.Instead, the radiation is explained if we treat the particles that makes up the black body as oscillators. Your penultimate sentence is also wrong. $\endgroup$
    – untreated_paramediensis_karnik
    Commented Nov 20, 2015 at 10:47
  • $\begingroup$ Bohr's atomic model can explain black body radiation in which it's clearly explained atoms jump orbits by taking quanta of energy and return back by emitting EM Rays of certain wavelength . And by the way there can be more than one explanation to single phenomenon. $\endgroup$
    – Shehbaj Singh
    Commented Nov 20, 2015 at 12:30
  • $\begingroup$ The blackbody spectrum is smooth/continuous over all wavelengths, it is not discrete as it would be if (as you claim) blackbody radiation was caused by electronic transitions. $\endgroup$
    – pentane
    Commented Nov 20, 2015 at 15:22
  • $\begingroup$ @pentane Even if a material was atomic with only two possible energy levels, it could still emit blackbody radiation so long as there was some opacity (perhaps due to natural, pressure or thermal broadening) in the line wings. However you are correct in that it would be difficult to make such an object as a blackbody - it would need to be physically very thick in order to be optically thick at all wavelengths. $\endgroup$
    – ProfRob
    Commented Nov 21, 2015 at 12:21

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