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I was just thinking today that I usually see red flame, and have seen plenty of blue flame, but not green. My naive presumption for coloration of flames would suggest that I would see more green, so I'm trying to figure out if there is a flaw in my naive understanding.

My understanding/presumption was that the color of the fire was due to the heat. Something warm would produce inferred waves as a result of thermal radiation. If you heat an object up enough if will get 'red hot' because it's heat is so high that the object is more energetic and ultimately causes the thermal radiation to also produce higher energy waves, due to electrons getting excited to higher levels of the valent shell before the degrade back down to their 'normal' levels and thus releasing higher energy with the larger jump back to base energy state.

I further assumed the reason fire is usually red was that the temperature of typical fires we produce is at the right temperature to be emitting light in the red spectrum, and the blue flames we see on really hot fires is a sign that their temperature is high enough to cause the heated objects/air to emit thermal radiation in the blue spectrum etc.

However, Green is midway through red and blue on the light spectrum. If my naive assumption was right I would expect to see hot flames turn green before they turn blue. However, while blue flames aren't that uncommon to me I can barely recall seeing green flames, which I would have thought would require a lower heat and thus be easier to produce.

I assume I've over simplified the process with which light is being emitted, but I'm trying to figure out where my error is. Can anyone better explain to me why green flame seems so much less common, assuming the very premise is right.

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    $\begingroup$ To my knowledge, the blue color is not thermal radiation but results from the fluorescence of thermally excited molecules. Furthermore, you have to take into account that you have a rather broad distribution of different wavelengths in your thermal radiation, which eventually rather leads to "white" light for higher temperatures. $\endgroup$ – LLang Dec 8 '15 at 22:27
  • $\begingroup$ @LLang hmm. I think you just answered another question I was just thinking of, why white light seems to occur before blue in hot flames. White is when there are too many different visible wavelengths to distinguish, blue is when most of the wavelengths are ultraviolet so the only wavelengths left that I can actually see 'weaker' radiated length of blur purple on the lower end of the wavelength bell curve? $\endgroup$ – dsollen Dec 8 '15 at 22:36
  • $\begingroup$ @LLang so is it just that I go from red when most light is inferred and red is just the 'high' edge of the light belcurve, to white when multiple visible light spectrum are produced, to blue when I'm seeing the 'low' end of the belcurve when mostly ultraviolet is being produced? If so I fel silly because if I had just followed up on the white fire question I had I probably would have deduced the rest on my own; but was too distracted on the lack of green one lol $\endgroup$ – dsollen Dec 8 '15 at 22:39
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There are 2 different kind of flames, and thus of flame color mechanisms:

  • flames containing carbon, i.e., soot. Carbon is pretty opaque (it can create heavy smoke), and emit thermal blackbody spectrum. Then you don't have green for the same reason that you don't have green stars ( well explained in wikipedia : a blackbody spectrum is large, not peaky, and visual system integrates so that "max of bump in green" is seen as white).
  • "normals flames" (but carbon is so frequent on Earth flames that one often forgets all the other elements :-) ), where color is due to emissive peaks of the excited molecules (flame of sodium, etc). Then you can have green flames. For example, copper sulphate shows a green flame.
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  • $\begingroup$ A good answer. But can you elaborate briefly on why green doesn't show up in black body spectrum? $\endgroup$ – dsollen Dec 9 '15 at 14:31
  • $\begingroup$ edited. you can also google "green star" for more. $\endgroup$ – Fabrice NEYRET Dec 9 '15 at 22:55
  • $\begingroup$ By "carbone" do you mean "carbon"? $\endgroup$ – Stefan Monov Aug 6 '17 at 16:02
  • $\begingroup$ yes. (sorry, I'm French :-) ) edited; thanks. BTW, does "carbone" with "e" means something (different) in English ? $\endgroup$ – Fabrice NEYRET Aug 7 '17 at 13:54
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Green is roughly in the middle of the visible spectrum. So to see green light, you need not just energy at the "green" wavelength (ca 530nm), but also a relative lack of energy at other (red and blue) wavelengths, (say, 630 and 450nm).

Thus to see green rather than yellow or white, you need energy in a relatively narrow band. (You can achieve this by exciting relatively narrow bandwidth emitters in a flame, such as copper ions, but not normally by black body radiation.

A normal flame is a black body emitter - relatively broad band, thus more likely to emit white light, if it was at a high enough temperature (say, 6000K) Green does show up in this light, but because of the presence of red and blue in similar quantities, it doesn't appear green but white.

However at lower temperatures, its peak is at the red end of the spectrum or even well into the infra-red end. Then the visible part will appear yellow (3000K) (because the blue end of the spectrum is well down the emission curve) or even red, (1500K) as the peak is infra-red and only the upper end of the curve intersects the low (red) end of the visible band.

In theory, black body radiation at still higher temperatures (peaking in the UV region) would appear bluish, having less energy at the red end of the band. However "blue flame" temperatures simply aren't that high. Instead, there are no carbon particles to act as black body radiators, we are seeing something else. I'm not sure what, but I have heard that oxygen has emission lines in the blue region (as copper does in green) so it may be oxygen atoms acting as narrow band emitters.

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