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Most of us here would know that flames can be colored by adding an appropriate salt to the 'fuel' or by simply introducing it into a flame. Boron and Barium salts give the flame a green tinge, Strontium colors it red, Sodium does gold, copper a bluish-green, etc, etc.

Now from my (not-so-in-depth) high-schooler level of understanding, the energy released by whatever is being burnt to produce the flame in the first place (as is usually the case), excites electrons in metal ions in the salt to higher energy levels. And when these electrons 'de-excite' and drop back to the original energy levels, they emit that extra energy as photons (whose frequencies are determined by the Planck equation $E = h\nu$ ) and these photons are what we perceive as the colored flame.

If I'm not mistaken the book Harry Potter and the Philosopher's stone mentions a 'black flame' of sorts, being one of the obstacles Harry has to circumvent in order to reach the Stone. (Artist's depiction)

enter image description here

Now, as I understand it, for something to be 'black', it must absorb all ( or at least a fairly significant amount of) visible light, encompassing a wide range of frequencies.

Going back to colored flames, I mentioned that the ions must emit light energy, not absorb it, in order to color the flame.

Armed with all this information, would anyone know if it really is possible to create 'black colored flames'? Has it ever been achieved?

[I tried thinking it out myself.....just gave me a headache]

Edit 1: And yes, I've made note of what @Mithoron mentioned, but what I mentioned was just conventional logic, it isn't universal though.

...Newtonian mechanics was 'conventional logic', but we all saw how that failed in relativistic scenarios. So I didn't want to take the risk, hence I've asked If it actually is possible...

Edit 2: Just because someone mentioned otherwise in an answer... I emphasise that by black flames I don't mean invisible flames.

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    $\begingroup$ You basicly answered this - "black flame" is nonsense. $\endgroup$
    – Mithoron
    Commented Sep 24, 2016 at 17:04
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    $\begingroup$ Thinking outside the box here, vaporization of solid iodine may be the closest you can get to a "black flame". There is no chemical reaction involved, and the process consumes heat rather than releases it (so it cannot be self-sustaining), but it would generate dark purple wisps. $\endgroup$
    – Nicolau Saker Neto
    Commented Sep 25, 2016 at 3:39
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    $\begingroup$ Off course: it is called smoke $\endgroup$
    – Greg
    Commented Sep 26, 2016 at 12:56
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    $\begingroup$ The flame might of course give off radiation in a range not visible to the eye; one night even suppose that an exothermic reaction that heats the air around it and makes the air rise thereby makes a "flame" of sorts from the rising air. But that is invisible as such, not black. $\endgroup$
    – Oscar Lanzi
    Commented Nov 21, 2017 at 22:20
  • $\begingroup$ In a way, all flames are colored black. $\endgroup$
    – Ivan Neretin
    Commented Oct 14, 2018 at 9:00

5 Answers 5

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This video shows how a black flame is achieved. If you illuminate the fire with a monochromatic light source (sodium vapor lamp) and introduce a species in the fire that absorbs that wavelength (sodium ions) then the fire will in fact appear black under the illumination. A screen shot is shown below:

enter image description here
Click image for video.

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    $\begingroup$ This is an exceptionally interesting video, thanks very much for sharing it. I do feel, however, that a more proper description for that flame is "anti-yellow", or to be extremely precise, "anti-sodium D line", rather than black. It is not my intention to demote your contribution, nor to be needlessly nitpicky. I merely wish to state that there is a subtle distinction, which provides interesting food for thought and a nice stepping stone into further topics, such as atomic absorption spectroscopy. $\endgroup$
    – Nicolau Saker Neto
    Commented Oct 4, 2018 at 9:57
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    $\begingroup$ When you think of light, white is a mixture of all the colors, but paint and pigment black is a mixture of all the colors. Black to get in light is either the absence of light, or a white shade of black when compared to a white background. The parlor trick is maybe the only way to achieve the black flame. $\endgroup$
    – This random guy
    Commented Oct 4, 2018 at 21:12
  • $\begingroup$ Nonetheless it does change the proprities of the plasma. Where fire is a light source, the experiment also shows the differences between the flame compared to a lighter with an ordinary flame. The results depicted the lighter with no shadow and the black flame with a shadow. Like you said this does make an interesting question. I would also add that there are many theroys that darkness is faster then the speed of light. I’m sure I have a well appreciate video to correspond to that as well. $\endgroup$
    – This random guy
    Commented Oct 4, 2018 at 21:35
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If black means that the flame does not emit light that is visible to the human eye then there is a certain chance that black flames might exist.

Since hot flames emit black body radiation and the spectrum of visible light reaches from about $\pu{400nm}$ to $\pu{800nm}$, a black flame had to be cool enough to not emit too high intensities of light below $\pu{800nm}$. So we're searching for a cool flame.

While cool flames are hardly visible in daylight, it might be difficult to find one that is invisible in a dark environment. Here is a report of a cool flame that could only made visible in the dark by long photographic lighting.

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Prelude: I do not know this part of the story you refer to.

The picture you display here, however, contradicts the idea of "black flame", because if there were a black flame, it should absorb the light, and attenuate the sight across it. There is little evidence, however, that one of the seven containers is source of a "black flame" as the background is (still) clearly visible, and the blur likely more a question of "out of focus".

As @ZOZ correctly indicated, "black flames" you / the story refer(s) to do not exist as such. The black of a flame occasionally seen often is result of incomplete combustion in shortage of oxygen. To provide an example, starting a Diesel engine may generate soot.

A third cent. Your "Going back to colored flames, the ions must emit light energy, not absorb it, in order to color the flame." needs a revision. True, ions present in the flame may emit light due to their de-excitation; similar to additive color mixing. To enlarge the picture, however, absorption may still occur -- similar to subtractive color mixing. This is element specific (Fraunhofer lines in the solar spectrum, for example) and is used in analytical chemistry atomic absorption spectroscopy, too. The likelihood of the two processes is different, however. Hence to be efficient, AAS works on temperatures largely exceeding the ones provided by a normal fire, a candle, a medieval torch.

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There is no way to create black fire because when it comes to light black is the absence of light not the mixture of all the colors light can produce. Although pigments which refers to paint, mixing them would create black paint but there is no way to make black fire sorry.

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Using an (I think) hydrogen vapor lamp and salt water it is possible to make the fire absorb all of the light at the particular wavelength that the vapor lamp emits.

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    $\begingroup$ This answer has already been given. $\endgroup$
    – A.K.
    Commented Dec 8, 2018 at 20:18

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