This question about whether or not it is possible to focus black-body radiation to make something hotter than the radiation's source was answered mostly negative: the second law of thermodynamics and/or the fact that etendue cannot be reduced are the reasons named.

Now, consider the following scenario:

Take an infinitely (or sufficiently) large plate as black-body radiator at temperature $T_r$. You have a power source available which is able to hold the plate's temperature constantly at this arbitrary chosen temperature.

Place a perfect parabolic mirror with its symmetry axis parallel to the plate's normal, opening toward the radiator, so that much light is focused in the focal point. If you now place a perfect black-body sphere with temperature $T_s<T_r$ at the focal point, it will absorb the energy and heat up.

enter image description here

Let's do an net-energy analysis. The photonic energy hitting the sphere is proportional to the geometrical cross section $A_m$ of the mirror: if the mirror is twice as large, it collects twice as much light and the sphere absorbs twice as many photons per unit time.

Since the sphere is a black-body, it emits photons in a radiation according to the Planck spectrum of $T_s$. As far as I understand, the total energy output of that radiation depends on temperature and surface area only. Since the surface is constant, $T_s$ is the only parameter.

In thermal equilibrium, the sphere will emit as much energy as it receives, and we have

$$ E_{in}(A_m) = E_{out}(T_s) $$

We can make the left hand side arbitrarily large since we have an infinitely large plate and we can use a broader or further extended mirror.

The right hand side apparently has a limit: the energy output of the maximally reachable temperature $E_{out}(T_r)$. So, if it is really true, that the sphere cannot get hotter than $T_r$, where does the excess energy go if $E_{in}(A_m) > E_{out}(T_r)$?

Or is that a thing that can never happen? Did I make some other mistake?

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    $\begingroup$ The radiation from a black body radiator is not parallel rays. If you were to make the radiation parallel with a collimator, you would end up with a much lower total intensity than given by the black body formula. Rays that are not parallel, however, do not get focused on the sphere by the parabolic mirror. $\endgroup$
    – CuriousOne
    Commented Oct 14, 2014 at 14:56
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    $\begingroup$ @CuriousOne: If the radiation is isotropic, most of the light will be radiated orthogonally of the plate, since $p(\theta)\propto cos(\theta)$. In my argumentation I don't care about the photons that are emitted in another direction, because one can increase the size of the mirror to compensate. Important is, that photons hitting the sphere are distributed according to the Planck spectrum. $\endgroup$
    – M.Herzkamp
    Commented Oct 14, 2014 at 17:05
  • $\begingroup$ Your argument is based on a misunderstanding of the amount of light that can be focused with a parabolic mirror. Every optical path is completely reversible. Imagine a 1W source of 0.01m diameter at the center of a 1m mirror. How much power is absorbed by a black plate in front of the mirror? Now make that mirror 10m in diameter. Does the power increase by a factor of 100? Why not? What does that mean for the reverted light path? $\endgroup$
    – CuriousOne
    Commented Oct 15, 2014 at 0:34
  • $\begingroup$ @CuriousOne: In your case, the power does not increase by a factor of 100, because the sphere radiates equally strong in all directions. So the power density absorbed by the plate is lower on the outside than on the inside. In my case, the output of the plate is uniform. That means, the sphere is not irradiated equally from all sides. So it still might be possible to arbitrarily increase the power input by scaling the mirror. $\endgroup$
    – M.Herzkamp
    Commented Oct 15, 2014 at 5:15
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    $\begingroup$ @annav: No, T_r is the temperature of the black body radiator. If we use its radiation, we cannot heat something up to a higher temperature that T_r. that is why it is the maximum temperature. It may be well below any critical temperature of the material. $\endgroup$
    – M.Herzkamp
    Commented Nov 30, 2015 at 11:37

2 Answers 2


UPDATED: I now think my previous answer was wrong, because the set up would be equivalent to the following question: Is a black body sphere inside a black body shell hotter than the shell?

Just change the question to add a carefully crafted lens that focuses all the radiation into the sphere (you could make the shell as large as you want), which of course is impossible to make or it would violate the second law.

  • $\begingroup$ Thanks. That was what I suspected. Now someone should post an answer to the linked question. $\endgroup$
    – M.Herzkamp
    Commented Oct 15, 2014 at 5:18
  • $\begingroup$ This is not a second law violation. Of course one cam heat things above the temperature of the hottest radiation source, but not with a mirror. This is a case of ettendue: en.wikipedia.org/wiki/Etendue. The phase space volume element occupied by photons can not be compressed with optical means alone. $\endgroup$
    – CuriousOne
    Commented Oct 15, 2014 at 5:39
  • $\begingroup$ @CuriousOne can you send me a link or give me a reference with the demostration? it is not clear to me from the wikipedia article that the conservation of etendue prevents the sphere to become hotter, as it only applies to refractions and reflections, and the light path is not reversible, it is absorbed at the sphere surface. The light path is not reversible. Also, could you point out what is wrong in the argument of the original OP? you might be right, but prove it or send a link to a proof, please. $\endgroup$
    – user65081
    Commented Oct 15, 2014 at 8:28
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    $\begingroup$ It's the conservation of etendue together with the characteristics of black body radiation. One can, of course, imagine a light source that has a black body spectrum but a much higher intensity (or a much narrower angular distribution) than a black body with the same temperature would have, but that can't be a black body for thermodynamic reasons. $\endgroup$
    – CuriousOne
    Commented Oct 15, 2014 at 17:52

I will answer this from the OP, presupposing that there is excess energy according to the arguments offered in the comments:

So, if it is really true, that the sphere cannot get hotter than $T_r$, where does the excess energy go if $E_{in}(A_m) > E_{out}(T_r)$?

Conservation of energy holds for an isolated system. Your system is an infinite hot plate at an initial temperature $T_r$, a small sphere at the focus of the parabolic mirror, and an electromagnetic field covering the vacuum ( where the masses are placed). The system is not isolated when you keep the temperature by outside energy input at $T_r$ .

Let us see what happens when there is only an initial temperature $T_r$ , an isolated system.

You have ignored the ambient electromagnetic field , it carries energy (and there is where any excess energy can be stored) going back to the infinite plane and it will be reabsorbed, in the thermodynamic process of equilibrating the temperatures.The plane and the ball will reach a temperature lower than $T_r$.

If you enter a source keeping the temperature of the infinite plane fixed, once the temperature of the ball and mirror are at $T_r$ , any excess will be stored initially in the electromagnetic field, with continuous reflections raising the poynting vector of the electromagnetic field. Please note that the system trying to keep the temperature of the plane at $T_r$ will need to cool it so that it stays at that value.

The moral of the example is that energy conservation holds for isolated systems, in this case the source/sink keeping the temperature fixed has to be considered in the isolation, so excess energy can be extracted to the sink.


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