# Heat Equilibrium in elliptical cavities, and the second Law

1. Feb 16, 2016

### jfizzix

I read this what-if XKCD article recently:

(Fire from Moonlight)
http://what-if.xkcd.com/145/

..so here's a nice little brain-teaser about radiating bodies in thermal equilibrium!

Let's say you have a large, perfectly reflecting elliptical cavity (a prolate ellipsoid). At one focus, you place a golf-ball sized blackbody (ball A) that's at 2000 Kelvin, and at the other focus, you place another golf ball blackbody (ball B) at the same temperature (2000 Kelvin). You seal up the cavity, and let it come to equilibrium.

At equilibrium, it makes sense that both balls will have equal temperature (i.e., the same temperatures they started with). The power absorbed and emitted by each ball will be identical,

However, the challenge comes from the case where ball B is made smaller, say, a marble-sized blackbody.

With the elliptical geometry of the cavity, and the spherical geometry of balls A and B, it seems that all light emitted by A, would be absorbed by B, being a blackbody and all.
However, the power emitted by a blackbody per unit surface area is proportional to the fourth power of its temperature (the Stefan-Boltzmann law).
What this would mean, is that at thermal equilibrium, the marble blackbody would have to have a higher temperature than the golf-ball blackbody, on account of its smaller surface area absorbing the same amount of radiation.

The second law of Thermodynamics forbids heat from flowing from cold to hot without extra work.
The question is, how are we to understand Balls A and B, remaining at the same temperature given their geometries, and the shape of the elliptical cavity?

2. Feb 17, 2016

### Staff: Mentor

Isn't the solution the same as in the XKCD situation? Namely, that your statement
is incorrect, that if ball B is made smaller, some of the light emitted by A will be reflected back to A?

3. Feb 17, 2016

### Daz

What this boils down to is the fact that both black bodies are extended sources and radiate with a Lambertian distribution. What that article calls conservation of étendue is what most of us call the Lagrange invariant. Very crudely, it says that in a first-order optical system the product of surface area times solid angle subtended by the optical system is a constant: the Lagrange invariant.

So, the marble has a smaller surface area but it also subtends a smaller solid angle at the surface of the golf ball. Remembering that both black bodies radiate in all directions with a distribution of angles given by Lambert’s law, the energy received by an element of surface is proportional to its area times the solid angle it subtends at the source. The Lagrange invariant says that if you use an optical system to increase one, the other goes down in equal measure. Ergo: no net transfer of energy.

4. Feb 17, 2016

### Daz

As an interesting aside, that article about fire from the moon isn’t technically accurate, because the radiation we receive from the moon isn’t completely thermal. If it were, it too would have a Lambertian distribution and the moon would look like a Lambertian sphere. It would look like this:

In fact, when we look at the moon we see a more-or-less uniform disc, like this:

The reason (which puzzled early stargazers for hundreds of years) is that the surface of the moon is a powdery crystalline texture. The crystallites in the moon’s surface act as microscopic retroreflectors (like cats eyes) giving it the appearance of a uniform disc.

So, in principle, it ought to be possible to get fire from the moon because the moon is acting more like a diffuse mirror (reflecting thermal radiation from the sun) than a thermal source.

5. Feb 17, 2016

### Staff: Mentor

Right, and that specific example of two balls in an ellipsoid has been discussed so often that it becomes really boring.

@Daz: the effect is even more complicated. A full moon is much brighter than you would expect, because a significant part of the light is reflected back under a small angle relative to the sunlight direction.

And yes, getting fire should be possible. While the moon just reflects ~10% of the sunlight, and only ~50% of the sunlight is visible light, this can be used in materials that absorb and emit visible light, but do not absorb/emit too much infrared.

6. Feb 17, 2016

### jfizzix

Yes, that's correct, though on the face of it it still feels weird that spherically symmetric light entered at one focus, will not converge completely on the second focus.

The weirdness goes away when you consider it like a gas of photons.
Photons emitted from any point on a blackbody could be emitted in any direction above the tangent plane defined by that point. Then, a good fraction of the light emitted by the golf ball body, will not be heading in the direction of the second focus (including reflections), and so could miss the marble blackbody altogether, getting reabsorbed in the golf ball blackbody.

7. Feb 17, 2016

### Andy Resnick

It's always amusing when someone re-discovers a long-known result. In this case, it can be easily found in "On the Possible and Impossible in Optics", by Slyusarev:

https://www.researchgate.net/publication/235161170_ON_THE_POSSIBLE_AND_IMPOSSIBLE_IN_OPTICS_O_VOZMOZHNOM_I_NEVOZMOZHNOM_V_OPTIKE [Broken]

A marvelous 1963-era document that demolishes both Archimedes' death ray and optical concentrators by page 5.

Last edited by a moderator: May 7, 2017
8. Feb 18, 2016

### DrDu

The situation is easiest understood assuming one of the two balls is a point. Only those rays emitted by the extended ball will end up in the other focus which are emitted perpendicularly to the balls surface. The ones emitted at other angles will miss the focus and be reflected back on the ball.

9. Feb 18, 2016

### Jano L.

I can't read the paper. How do you mean "demolishes Archimedes' death ray" ? Are you referring to the story about setting ships on fire by concentrating solar radiation?

Last edited by a moderator: May 7, 2017
10. Feb 18, 2016

### Andy Resnick

Yes- setting fire at a distance. It fails for the same reason this 2 objects in an ellipse 'paradox' fails, why you can't make a thin collimated pencil of intense light, and other related 'myths and legends'. The underlying conceptual flaw is the same- not understanding that extended luminous objects cannot be replaced by an equivalent point source, which can be done in mechanics (center of mass) and electrostatics (Gauss's law derivations of Coulomb's law).

11. Feb 19, 2016

### Randy Beikmann

My thermo professor asked why you could never take the rays of the Sun, whose surface is at 5800 K, and focus the rays to raise the temperature of a spot on the Earth's surface to more than 5800 K (which would obviously break the 2nd law of thermodynamics). As in the above, it's the finite width of the radiating surface of the Sun that sets a minimum to the area of the focused beam, and limits the surface temperature.

12. Feb 19, 2016

### Ken G

Yes, this is an example of what gets called "the brightness theorem", which says that the specific intensity along a ray is not affected by a lens. This means you cannot use a lens to "brighten" any given ray that leads back to the Sun, all you can do is make more rays from the target connect to the Sun. That also implies the surface of the target cannot be raised above about 5800 K.

13. Feb 20, 2016

### DrDu

You could use the light of the sun to pump a laser.

14. Feb 23, 2016

### Warp

The xkcd article says that you cannot get a spot of sunlight that's hotter than the surface of the Sun, because thermodynamics. However, I'm not convinced it's that simple.

The Sun has a surface area A. Every point on that surface as a certain temperature, and thus emits a certain amount of energy. If we could take all that energy and concentrate it on a surface of area B, which is smaller than A, then each point in B would be hotter than each point in A, because there is more energy per square unit in B than there is in A. I don't think thermodynamics is broken because the total amount of energy is still the same. It's just that it's now concentrated in a smaller area.

However, perhaps it's physically impossible to gather all the energy from a larger area A, and concentrate it on a smaller area B, at least not with an optical lens (or even a paraboloid mirror)?

Maybe that's the case with the Sun. The Sun is big, the lens is small. But what happens if the source of light is much smaller than the lens (or the mirror)? Could the lens concentrate the light coming from the small source of light, and make the spot hotter than the surface of that light? I don't know.

15. Feb 23, 2016

### DrDu

That's exactly why people here were considering two hot bodies inside an ellipsoid. The ellipsoid will act as a mirror which reflects all of the light from one body. However it is also not able to concentrate it in a region which is smaller than the source.

16. Feb 23, 2016

### Staff: Mentor

You cannot.
And the xkcd article explains why.
No.

17. Feb 23, 2016

### Warp

The thought came to mind (although I'm certainly not the first to think about it, because it seems a rather obvious thought experiment) that what would happen if you were to surround a black-body radiator with a spherical mirror (so that the radiator is in the center of it).

Wouldn't the surface get hotter because the radiation it's emitting is coming back? How does that work?

18. Feb 23, 2016

### Staff: Mentor

It would be in perfect equilibrium and emit as much radiation as it absorbs. If you do that to a star, the surface would indeed get hotter, because it is heated from the even hotter interior, but that is a different topic.