By the second law of thermodynamics, you shouldn't be able to use any amount of mirrors/lenses to focus sunlight onto an object and heat it past the surface temperature of the sun (approximately 5800K).

In principle this makes sense to me, but I'm more concerned with the mathematics behind this. Whenever I stop to think about energy, things seem to break down in my mind. I just can't get past the fact that the more rays of light your focus (ie, the more photons), the more energy that would be passing into your system. If you increase the energy of your system, is seems like the average energy of the molecules in your system will continue to increase.

Can anybody show me a mathematical counterexample? I guess I just don't understand all of the equations that necessarily go into this calculation, because I keep getting that the energy of your system increases with each photon that passes into your system. Thanks!

If you believe my statement is incorrect, please look at problem 91 here and explain why I'm incorrect and how the answer is E http://www.physics.ohio-state.edu/undergrad/greStuff/exam_GR9677.pdf

  • $\begingroup$ I'd like to point out that the fusion experiments at NIF (National Ignition Facility at Lawrence Livermore) use mirrors and lenses (and nonlinear optical crystals) to make a small target very hot. Explain that! $\endgroup$
    – DarenW
    Oct 17, 2013 at 19:09
  • 1
    $\begingroup$ Laser light is coherent, it's not at a certain temperature. $\endgroup$
    – Bubble
    Oct 17, 2013 at 19:14
  • 2
    $\begingroup$ The reason is quite simple: Absorbing bodies also emit thermal radiation. Once the object is hot, it shines back at the sun. To prove the symmetry microscopically you need to consider the entendue of the light path, the object's spectral characteristics, and so on. Or, you can simply appeal to the second law of thermodynamics that if the object were to get any hotter than the sun, you could build a heat engine using the object as heat source and the sun as the cold sink, and extract energy for free. $\endgroup$
    – Nanite
    Oct 17, 2013 at 19:19
  • $\begingroup$ I have come to agree that E is the correct answer. Second law dictates that you cannot transfer heat from a colder to a hotter body, but the other way round. $\endgroup$
    – mcodesmart
    Oct 18, 2013 at 0:21

1 Answer 1


The picture in the GRE problem is misleading (probably deliberately) as it shows all rays coming out of the oven as being parallel. Light coming from a thermal source will be distributed in direction as well as frequency, so only a fraction gets focused at the desired spot. There is something in optics called the "etendue" (look it up) which describes how light is distributed in angle as well as in space. The fact that the etendue is conserved in optics (basically Louiville's theorem, conservation of volume in phase space) means that a passive optical system cannot increase the temperature of light. So E is the correct answer.

Let me explain further, since I know this will be controversial. When you go outdoors you are exposed to a 5000 degree thermal radiation source, however that source is limited in solid angle (the solid angle subtended by the sun) which is why you are not burnt to a crisp. If you set up a bunch of parabolic mirrors or lenses or whatever, you can increase the solid angle subtended, but only up to a point. There are only 4 pi sterradians available. If you manage 4 pi sterradians then you have the full 5000 degree thermal distribution, but you can't go beyond that.

What the etendue tells us is that if we want to squeeze the radiation in space (focusing) we spread it out in angle. It is the distribution in both space and angle which characterizes the thermal distribution, not simply the intensity.

  • $\begingroup$ Two things. First off, e&m waves from the sun are pretty much parallel on a cloudless day to any lens you set up since the sun is so far away. Second off, if the waves were scattered in random directions, it would be even harder to focus the light. So I don't see what you're getting at here, and it doesn't really give much of a mathematical arguement $\endgroup$ Oct 17, 2013 at 18:29
  • $\begingroup$ No, light from the sun is distributed in solid angle. The solid angle distribution is important to the thermodynamic characteristics. $\endgroup$
    – Horatio
    Oct 17, 2013 at 18:54
  • $\begingroup$ Horatio has it correct. The sun is NOT a point source; it has an apparent angular diameter of about 30 minutes of arc; same as the moon. No optical system can form an image that is higher "brightness" than an Aplanatic system can (corrected for both coma and spherical aberration). Aplanatic systems obey the optical sine theorem, which says the quantity N.H. Sin(U) is invariant. Clausius derived this from the second law of thermodynamics. $\endgroup$
    – user26165
    Oct 18, 2013 at 19:49
  • $\begingroup$ Hi, does all heat just stop flowing at thermal equilibrium? My book says that if the environment chooses to interfere with a system, it must expend more energy than the energy naturally generated by the system? (Like carrying a bucket of water against the waterfall will take more energy than the amount of energy that will be released when water falls on its own) $\endgroup$ Aug 5, 2015 at 18:20
  • $\begingroup$ And the same with heat pumps, why? $\endgroup$ Aug 5, 2015 at 18:20

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