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I was wondering about the nature of object's colour. I know that an object get its colour from the absorption of visible electromagnetic radiation, reflecting all the other wavelength. But if we take the case of a black object that absorbs every visible light, I know that photons will be absorbed by some molecules, then will be re-emitted with less energy because some of the energy has been "passed on" the molecule, in which it goes faster, thus, giving us heat. So, if we put a very intense light, does it simply change the amplitude and the object is still black or does the wavelength shift and gives a different result?

My guess would be that no matter the amplitude, the wavelength are the same and thus the black object will still appear black but I want to be sure with, maybe, a more scientific explanation? If you have also links of some sort, I would gladly appreciate it!

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    $\begingroup$ Every material reflects some amount of light. The one that reflects the least is here en.m.wikipedia.org/wiki/Vantablack $\endgroup$ – safesphere Oct 20 '17 at 17:24
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    $\begingroup$ Black isn't a wavelength. $\endgroup$ – JMac Oct 20 '17 at 17:30
  • $\begingroup$ If it gets very hot, it won't reflect, but will emit blackbody radiation at a certain point. $\endgroup$ – Vendetta Oct 20 '17 at 17:54
  • $\begingroup$ @safesphere Woah. Vantablack is creepy looking. $\endgroup$ – Chris Oct 20 '17 at 20:21
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As I understand your question, you are asking about an ideal black body. Keep in mind that such a thing does not exist in nature. But as long as we know we are talking about an ideal, first, the body will never reflect any light. What will happen is that the intense light falling on it will drive up the temperature of the black body, and the light emitted will depend, both in amplitude and wavelength, on that temperature. The more intense the light, the higher the black body temperature will become. The resulting spectrum will become both more intense and more blue, following the standard black body spectrum, which you can read about in many sources.

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    $\begingroup$ Assuming the blackbody has a positive heat capacity. It turns out that adding energy to a star (which is pretty close to an ideal blackbody) actually decreases the temperature. $\endgroup$ – Chris Oct 20 '17 at 18:41
  • $\begingroup$ @chris Good point. $\endgroup$ – bob.sacamento Oct 20 '17 at 18:43
  • $\begingroup$ @Chris "adding energy to a star [...] decreases the temperature". This seems counterintuitive. Could you please elaborate? Are you referring to Fermi temperature or actual temperature? $\endgroup$ – safesphere Oct 20 '17 at 20:39
  • $\begingroup$ @safesphere I agree! It is counterintuitive. I am referring to the actual surface temperature of the star. This is the case for any gravitationally bound system- the virial theorem implies that increasing the total energy of a gravitationally bound system actually decreases the kinetic energy. And since temperature is proportional to the average kinetic energy, this decreases the temperature of the star. $\endgroup$ – Chris Oct 20 '17 at 20:50
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Yes if it's a perfect black body then it would be black. You should read about https://en.m.wikipedia.org/wiki/Black_body

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  • $\begingroup$ This is dubious. Blackbodies don't look black except at low temperatures. If you shine a sufficiently intense light on a blackbody, it will no longer look black. $\endgroup$ – Chris Oct 20 '17 at 18:45
  • $\begingroup$ The OP said visible light. It wouldn't matter how intense it was it wouldn't raise the temperature of the black body enough to see a difference in visible color. So it would still appear black $\endgroup$ – Bill Alsept Oct 20 '17 at 19:04
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    $\begingroup$ How do you figure? Visible light can have an arbitrarily large intensity, which would heat the blackbody up to an arbitrarily large temperature. $\endgroup$ – Chris Oct 20 '17 at 19:12
  • $\begingroup$ I don't think so but I could be wrong. The visible light will of course add energy to the black body but it will radiate at low frequencies way before it gets to the visible red frequency range. $\endgroup$ – Bill Alsept Oct 20 '17 at 19:29
  • $\begingroup$ That's definitely wrong. The power radiated by a blackbody is given by $P=\sigma A T^4$, where $A$ is the surface area and $\sigma$ is a constant. If the blackbody has a light of irradiance $I$ shone on it uniformly over it's surface, it will rise in temperature until $\frac{P}{A}=I$, and the temperature it reaches will be $T=\left(\frac{I}{\sigma}\right)^{\left(\frac{1}{4}\right)}$. This formula is unbounded: arbitrarily large light intensities lead to arbitrarily large temperatures. $\endgroup$ – Chris Oct 20 '17 at 20:19

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