Suppose the light source is a black body with average temperature 2000K, and I want to use it to heat another black body to 2500K, via passive devices only. Here passive device means lens, mirror, medium... anything that doesn't actively do work.

I understand I can't focus the light beam unlimitedly small under geometrical optics' framework, my question is actually regarding thermodynamics. The second law of thermodynamics said that this is not possible, I cannot create temperature higher than the light source itself, otherwise this is a second-kind perpetual motion machine. The best I can do is a temperature equal to 2000K.

Now here is my question: can I use an absorbing medium, cut down low frequency and keep the high frequency part of the spectrum only, to make the spectrum that just shapes like 2500K? (but less total energy, of course) Can I use this "fake" 2500K spectrum heat the blak body to higher than 2000K? If the answear is yes, does this violates the second law of thermodynamics?

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  • $\begingroup$ The answer must be no, but I don't understand why. The conservation of etendue thing means you can't just focus things arbitrarily, but I don't see what stops you using a passive filter to make the spectrum of the source look hotter than it is (and, obviously, much less bright). There must be something that makes this not work. So this is a really interesting question. $\endgroup$
    – user107153
    Apr 12 '16 at 13:41
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    $\begingroup$ @tfb As I write below, the center wavelength of the spectrum does not matter much; you need also high radiance (W/m²/sr²) for high equilibrium temperatures. For instance of a non-thermal source, the microwave oven generates a very low frequency of 2.5 GHz, yet its radiance is high and it can create high temperatures when focused. $\endgroup$
    – dominecf
    Apr 12 '16 at 13:49
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    $\begingroup$ I think that you are overthinking this. How exactly do you think that blocking a large fraction of your light is going to help collect more light from the source? $\endgroup$
    – Rococo
    Apr 12 '16 at 14:09
  • $\begingroup$ @Rococo Good point. $\endgroup$
    – dominecf
    Apr 12 '16 at 14:13
  • $\begingroup$ I think you have your answer, which is as simple as "second law of thermodynamics". As for the nomenclature, what about "solar cells and an arc discharge" is not "passive"? There is no additional energy source in either. It doesn't invalidate your question, of course, but one has to be careful with the usual meaning of "passive", which does not always lead to the desired result of its use. $\endgroup$
    – CuriousOne
    Apr 12 '16 at 20:00

You propose to use a color filter to change the effective color temperature of a blackbody source. This however never allows to reach higher light intensity, or more exactly radiance, which is the decisive factor for the maximum temperatures higher than the source.

I do not know the exact definition of a passive device. Provided the device can not use the outer space as a cold bath for conversion of the light energy into, e.g., electricity, the answer is that reaching higher-than-source temperatures is simply not possible.


Here's a thought experiment, together with why it will not work. This is a very hand-waving argument, for which I apologise (I would add it as a comment but it is too long).

Imagine a black-body source which completely surrounds some object. That object will clearly end up at the same temperature as the source (for the sake of concreteness the source could be the cosmic microwave background).

Now insert a shell of filters between the source and the object, arranged so that the spectrum is filtered to make it look hotter.

So the object should now end up hotter, right, as everywhere it looks is now a hotter source?

Well, no. Because the filters must be doing something with the radiation they are not transmitting: in particular they must reradiate it. The end result of that will be that the spectrum the object sees will be no hotter than the one it originally saw.

  • $\begingroup$ Oh I think I get it, the absorber will heat up, if I use something else to cool it down, then that 's where I have to do work, thus second law of thermodynamics is not violated. $\endgroup$
    – LePtC
    Apr 12 '16 at 14:55

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