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My previous question wasn't specific enough. I'll try to be more specific.

Let's imagine we have a hot body let's say 6000K hot that emits lots of thermal radiation. Let's assume 1kW of radiative power falls on on a m². Now if we use a lens with 1m² area, we can focus this 1kW into a smaller area which heats up until it radiates back the incoming power.

Now by using a lens with a smaller focal length, the image of the hot body can be smaller. So we can concentrate the radiation into a smaller area. The smaller the area you concentrate the power, the hotter that area becomes isn't it?

But I read in lot of places that this hot spot cannot be hotter than the body that that provides the energy, so in our example it cannot be hotter than 6000K, because it would violate the laws of thermodynamics. So what happens if you keep decreasing the focal length and size of the image? Maybe you cannot shrink the hot spot smaller than some size due to the wave interference and diffraction, thus ensuring this constraint?

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It is not possible because of conservation of etendue. This is based purely on geometry, not really a law of physics in that sense. No guarantees regarding quantum effects etc., but in the realm of ray optics it can't be done.

Basically, given any source of light radiating from finite surface to half space, you can never concentrate the entire emitted radiation to a smaller area than the original emitting area.

In your example of a focusing lens, you need to note that your black body is radiating at an angle of 180 degrees (full half space), and any lens you try to use is always capturing less than the total emitted energy.

See: http://en.wikipedia.org/wiki/Etendue

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    $\begingroup$ Here is a source which says that conservation of etendue means that in an optical system, luminance can't increase beyond the luminance at the source, at best it can remain constant (luminance is power per unit area delivered by light, and the Stefan-Boltzmann law says if power/area can't go up, blackbody temperature can't either). "Thermodynamics imposes this entropy-like property on etendue, ensuring the highest luminance is at the source, and conservation of etendue implies conservation of luminance in a lossless system." $\endgroup$ – Hypnosifl Oct 13 '14 at 14:05
  • $\begingroup$ As I read it, it is not "purely on geometry", it is a thermodynamics conservation expressed on a geometrical way. So it's a circular explanation by thermo, and it's still interesting to build this result via constructive physics. The point is that geometric approaches implicitly come with geometric optic, while to recover the thermodynamics property you have to consider the waviness of light (and its interference effects). $\endgroup$ – Fabrice NEYRET Jan 6 '17 at 18:31
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If you use complicated routes redefining "focusing" towards generating the temperature, yes.

Physicists at CERN's Large Hadron Collider have broken a record by achieving the hottest man-made temperatures ever - 100,000 times hotter than the interior of the Sun.

Scientists there collided lead ions to create a searingly hot sub-atomic soup known as quark-gluon plasma at about 5.5trillion C, the hottest temperature ever recorded in an experiment.

The LHC consumes 120 megawatts. This is something like the output of four Long Island solar farms . So, a small part of solar energy falling on earth has been used to create temperatures much higher than the sun's black body radiation temperature. I do not see any violation of thermodynamics.

Now if this energy that the solar panels gathered can be feasibly picked up by a huge optical system and focused to a point, I am in no position to guess. But if it can be done technically I do not again see any violation either of the first or of the second law:

The Clausius form :

Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time.

The change with LHC is obvious. I think the change with the parabolic mirror comes from the change in the light rays out of their free/non-interacting state.

EDIT

There have been lots of comments quoting that this cannot happen ( by authority not proof). The clearest in a comment by @Hypnosifl:

and here is another source which says "imagine some ideal solar concentrator that takes solar radiation with angular spread theta and accepts it from throughout a certain collector aperture area Ac, concentrating it onto a black-body receiver of some area Ar ... In the absence of any other heat losses, the black-body absorber will heat up until it reaches the same temperature as the source, and it will then be in equilibrium."

This states that when a specific area from the sun is focused on a specific area on the earth the maximum temperature reachable is the sun temperature and the two areas will be in thermodynamic equilibrium. In ideal conditions ( vacuum, no losses) it means that the earth area is reflecting back to the sun area as much energy as it is receiving. There is nothing in thermodynamics preventing the choice of a second set sun-area - earth-area, to get twice the energy. If one focuses from these different sun areas at the one (x,y,z) point on earth with two lenses, energy conservation says the temperature will go higher than the sun temperature. The new equilibrium will be reached by reflected rays going back to the sun in two paths, splitting the radiated back energy.

Thermodynamic laws work in isolated systems at equilibrium, afaik.

Edit after seeing the answer on solar furnaces by @BebopButUnsteady

It is the reversibility of the paths that brings entropy into question, and once a black body reaches the parent black body temperature equilibrium is reached. I think my statement of picking two different areas of the black body to focus on the same spot is possible: use a shade and take half of the sun image with one focus and the other with the other, and there will be two return paths, in effect two parent black bodies .

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    $\begingroup$ I suspect part of the idea behind the original question was why you couldn't passively focus the light in this way even if the whole system had reached equilibrium. The Earth's surface temperature is colder than the Sun's so the two are clearly not in thermal equilibrium (maybe they would be if you isolated them in a mirrored shell), and if the temperatures were the same there would be no way to generate power to create localized hotspots using solar panels (I believe all engines require some type of temperature differential to operate, even if they are not heat engines). $\endgroup$ – Hypnosifl Oct 12 '14 at 15:58
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    $\begingroup$ The question is about heating by radiation. Converting radiation energy into electricity and then using that one can do most anything, but this is irrelevant to the question. $\endgroup$ – Marc van Leeuwen Oct 12 '14 at 16:01
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    $\begingroup$ @Hypnosifl Suppose that one can focus with a parabolic lens enough to go to 2000K on a spot. All one has to do is to use four parabolic lenses on that spot to bring the temperature higher than the 6000. It is a matter of energy which is scalar and additive , it is algebra at this point. We are not talking of focusing the total energy of the sun, just some of it per meter square ( my solar panel example was used to show this, not clearly I guess :( ) $\endgroup$ – anna v Oct 12 '14 at 18:28
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    $\begingroup$ But the earth sun is not in a thermodynamic equilibrium. We are not at 6000K ! The problem is not an equilibrium problem. Everything would melt at that point ;). $\endgroup$ – anna v Oct 12 '14 at 18:35
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    $\begingroup$ I think if one set up such a system at 6000 equilibrium, a new temperature would be arrived at because the spot would be equilibrating with the rest of the earth/sun, so even if for a delta(t) it was higher the system will equilibrate. In addition I do not think you are correct that non equilibrium and equilibrium follow at the level you imply. There will be a limiting process in the variables. We would not have the CERN plasma either $\endgroup$ – anna v Oct 12 '14 at 18:39
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As mentioned in the other answers the etendue theorem rules this out for a system of mirrors and lenses. However, I think it's important to note that simple thermodynamic arguments are insufficient for the reasons given below.

I will answer the question using mirrors rather than lenses as it makes the physics clearer. Suppose we have a massive mirrored cavity at 0K:

Mirrored Cavity

At some time we switch on the sun:

Switch on the Sun

Now the cavity will slowly start to fill with radiation. Once the radiation density in the cavity has reached a certain level (which depends only on the sun's temperature) the sun will absorb just as much radiation as it emits and thermal equilibrium will be reached. Note that before equilibrium is reached heat is transfered from the sun into the cavity.

Now suppose we add another black body at the other end of the cavity, initially at lower temperature than the sun:

A second black body

Is it possible for the second black body to reach a temperature higher than the sun? Obviously not, for once it's temperature reaches that of the sun it will also absorb and emit radiation to/from the cavity at the same rate, and no more heat will be transfered (for simplicity, we are assuming that the sun's temperature is fixed).

However, this alone does not rule out the possibility to heat a second black body to a temperature higher than the sun's surface only by focussing the solar radiation. To understand why, it is useful to consider Anna's example of the LHC in more detail.

Assuming the LHC was powered only by solar energy (e.g. converted into electricity by photovoltic cells), this is what is happening. Heat $Q_{12}$ is transfered from the sun (temperature $T_1$) to the solar cells (temperature $T_2<T_1$) which act as heat engines converting part of the transfered heat into work $W<Q_{12}$. This work then powers a heat pump to transfer heat $Q_{23}$ to the quark-gluon plasma in the LHC (temperature $T_3>T_2$). The second law of thermodynamics puts some limits on the ratios of $Q_{12}:W$ and $W:Q_{23}$, but there is no problem with heat from the sun being transfered to the hotter plasma, due to the lower temperature of the intermediate system (the photovoltic cells).

So, we can use the sun's radiation to increase the temperature of a black body above that of the sun provided there is some intermediate system at lower temperature. In the OP's original question the space itself can in principle play the role of the intermediate system.

Consider the mirror cavity example again, but this time instead of allowing the sun to equilibriate with the cavity before adding the second body we add both to the 0K cavity at the same time, and assume that the temperature of the second body is equal to the temperature of the sun. Initially, the space inbetween the two black bodies is at a lower temperature to either and (as far as the second law of thermodynamics is concerned) there can be a net heat transfer from the sun to the second body provided there is a corresponding heat transfer from the sun into the cavity.

As mentioned above, the optics doesn't allow this, but I can't see any reason to rule it out on purely thermodynamic grounds.

Response to comments by Hypnosifl

Hypnosifl suggests that the non-equilibrium result can be derived from the equilibrium one with 'minimal assumptions' about the nature of the radiation. The gist of the argument (if I understand it correctly) is that since the radiation incident on the second black-body surface depends only on the source (the sun), this should be the same whether at equilibrium or not. I do not think this argument is sufficient for the following reason. In order for the the radiation field to ever reach equilibrium with the sun it is necessary to enclose it in a cavity of some kind (say a perfectly mirrored cavity for simplicity), in which case the radiation incident on the second body at equilibrium will include all sorts of complicated reflections of the cavity walls. Thus, until equilibrium is reached the incident radiation will be time dependent. (For example, suppose the second body is the Earth. Then the first radiation from the sun will reach the Earth after about eight minutes, and this will be radiation only directly from the sun. Some time later rays reflected once of the cavity will start to arrive, then double reflections, and so on.) Thus we cannot conclude that the luminance is the same in the non-equilibrium case as in the equilibrium case. You could maybe argue that it is impossible for the incident flux to decrease as a function of time (which is sufficient), but this is a more complicated argument.

[Edit:] It turns out that it is possible for the incident flux to decrease as a function of time after all, as the following example shows. Pass the radiation from the sun through a collimator into a large box. Then pass the collimated beam through a prism to separate different frequencies of light. Choose a particular frequency and pass that part of the separated light beam through another collimator. You now have a (reasonably) coherent beam of light. Now send this beam of light into a Michelson-Morley style interferometer with one arm significantly shorter than the other and place the test surface where the 'scope of the interferometer is. Light travelling the shorter path will reach the test surface first, giving a uniform intensity across the test surface. When the second beam arrives inteference fringes will be created, meaning that certain parts of the surface will see a decrease in intensity (while others will see an increase). Thus the argument at the end of the previous paragraph will not work. In any case, it is becoming clear that the thermodynamic argument requires a significant amount of extra work to complete it. [End edit]

Last night I thought of a similar argument from the equilibrium case to the non-equilibrium case, based upon the fact that at equilibrium thermodynamics tells us that the energy density of radiation is the same everywhere in the cavity (as long as you are in vacuum) and independent of the shape of the cavity and the direction of the radiation. By imagining how the radiation flows from one part of the cavity to another, this result seems to rule out the possibility of concentrating radiation from a uniform source (i.e. a source which emits equally in all directions) to an intensity higher than at the surface of the source in another part of the cavity, from which the non-equilibrium result follows. However, (i) I haven't made this argument rigorous and (ii) I suspect that the content of any rigorous argument along these lines would be similar to the conservation of etendue (i.e. it looks like some kind of conservation of volume in phase space). Thus this could hardly be called a 'minimal assumption' either.

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  • $\begingroup$ I agree it can't be ruled out on purely thermodynamic grounds, since an active concentrator like a laser powered by a solar engine can do it if the system is out of equilibrium. But consider the fact that the luminance a passive optical system concentrates on a surface depends only on the source of light rays (in this case the Sun), not the temperature of the surface where the rays are directed. Combined with the conclusion that you can't raise the surface temp. above 6000 K using the Sun at equilibrium, doesn't this imply you can't do it out-of-equilibrium with an optical system? $\endgroup$ – Hypnosifl Oct 13 '14 at 14:24
  • $\begingroup$ In other words, once you've ruled out the idea that it could work at equilibrium on purely thermodynamic grounds, I think you only need some very minimal assumptions about optics to conclude a passive optical system can't do it out-of-equilibrium either. Of course, if you want to rule it out on purely optical grounds rather than using a hybrid approach, then you have to get into more sophisticated ideas from optics like conservation of etendue. $\endgroup$ – Hypnosifl Oct 13 '14 at 14:34
  • $\begingroup$ I will reply to your comments in detail in the answer @Hypnosifl, to save this getting moved to chat... In short though, I think you can probably extend the equilibrium result to get the non-equilibrium one, but I don't think the extra work is as 'minimal' as you suggest. $\endgroup$ – Mark A Oct 14 '14 at 5:14
  • $\begingroup$ Thanks for addressing it. Since electromagnetic fields obey the superposition principle, I was thinking that if some luminance is being focused from one source (the sun) onto a surface by some system of lenses/mirrors, adding additional sources (the cavity walls) could only increase the luminance on that surface, not decrease it. It's true in some cases there can be destructive interference, but since we're talking about additional sources being added in the equilibrium case where all radiation is blackbody, I think one could add an assumption about how blackbody radiation is incoherent. $\endgroup$ – Hypnosifl Oct 14 '14 at 11:10
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    $\begingroup$ @Hypnosifl: it turns out you can get destructive inteference from blackbody radiation - I've added an example to the end of my answer. The discussion you cite does not consider the effects of optical elements being added to the system. I am becoming increasingly convinced that the thermodynamic argument is no simpler than the optical one. Any thoughts? $\endgroup$ – Mark A Oct 16 '14 at 10:11
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A quick back-of-the-envelope calculation using the Stefan-Boltzman law for a sphere with radius $r$

$$ P = \sigma T^4 4\pi r^2 $$

lets me conclude that you would need to focus $1\,\textrm{m}^2$ of sunlight onto a sphere with a radius smaller than $ 2.8\,\textrm{mm}$ to get its temperature above $6000\,\textrm{K}$.

How difficult is it to build such a focusing system? Thermodynamics tells us it's impossible, but I have not yet found an intuitive explanation using geometric optics. However, this explanation is possible!

There is a related question, where the one given answer recommends looking up the term Etendue. Unfortunately, this quickly leads to grimy math...

The gist is that it is the extent of the sun (or solid angle, as you view it from earth) that prevents you from building an imaging system that focuses the light well enough.

Edit: Also see this "what if" post, where Randall walks you right up to Etendue, and then skips past explaining why Etendue has to be conserved... Sigh.

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    $\begingroup$ Yes, I think a better understanding of etendue would probably be key to a detailed answer that's based on optics alone, but I don't understand it well enough to give such an answer myself. This article specifically cites conservation of etendue as a reason you can't violate the 2nd law by focusing light, though. $\endgroup$ – Hypnosifl Oct 12 '14 at 14:42

protected by Qmechanic Oct 12 '14 at 22:12

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