# Why can't I build a non-reversible optic setup which focuses the sun to higher temperatures?

Following this question I would like to challenge one of the assumptions.

The standard answer is that thermodynamics prohibits focusing the sun to a spot such that the spot reaches a higher temperature than the sun itself, because lenses and mirrors are reversible machines and the argument goes from there.

I don't see why the reversibility should hold for composite machines. As in, if I have one lens/mirror apparatus here which focuses the sun to, say, 5,500K, and a duplicate apparatus there which creates another such spot...

...and I tilt them so that the spots overlap...

...then intuitively the overlapping spot should have a temperature significantly higher than 5,500K...

...and the machine isn't reversible because a photon striking any given point could have taken either path, so the thermodynamic argument against the above result doesn't apply.

What's wrong with this reasoning?

• Could you please explain your argument? I can't understand what you are trying to say. You can use energy from the sun to build an apparatus that generates temperatures higher than the sun; you could probably set up a solar power plant and drive a high power laser using it to generate high temperatures (much higher than the sun), but I am getting the feeling that this is not what you are asking. – Harsha Sep 11 '16 at 10:46
• Ideal mirrors and lenses do not change the wavelength/frequency of the incident light, just the area over which they are collected/distributed. It might be possible to get an object "hotter" than the incident light if that object cannot radiate away energy fast enough (not sure if any material would actually satisfy this, but I was just thinking about energy balance... keep dumping energy in without letting much leave). – honeste_vivere Sep 11 '16 at 18:29
• Optics is reversible (you can prove this just by looking at the laws for reflection and refraction, or at Maxwell's equations), so the premise doesn't make sense. – knzhou Sep 11 '16 at 22:53
• Nope, I was wrong... see comments and answers for http://physics.stackexchange.com/q/279146/59023. – honeste_vivere Sep 16 '16 at 22:31

The ideal 5500K apparatus must produce perfect optical coupling between the spot and the sun. This will mean that all light emitted by the spot must be channeled back in the direction of the sun. No light emitted by the spot is allowed to be absorbed or escape into other directions in space.

With this in mind, the apparatus must surround the spot from all sides, so you cannot simply bring two such machines together.

I would explain this as follows:

The (hypothetical) apparatus that focuses sunlight to "5500K" will illuminate the target from the full half-space. So it cannot be combined with itself to generate higher photon flux.

Another way to put it is that you need many focusing apparatuses that illuminate the target from all sides to reach 5500 K in the first place.

I would relate this to the fact that the smallest spot size can only be achieved if the illumination is from the full half-space, or the optics has a maximum numerical aperture.

Does this make sense to you?

• Not really. If I understand "half space" correctly, focusing the entire half space onto a sun-diameter spot would yield an equal surface temperature. To make a spot 1% of the area come to the same temperature I should only need to focus 1% of the light. So I'll make two 5,500K spots which each capture 1% of the half-space light. – spraff Sep 11 '16 at 11:07
• If you have a focusing apparatus which illuminates the target with a small angle (low numerical aperture), the diffraction limited spot size is large. The angle is given by the aperture of the incoming light and the distance of the lens to the target. But small spot size is essential for high power density. This means large aperture (lens) and lens close to the sample. for high power density. Ideally the illumination is from the full half space. Imagine a very large lens very close to the target surface (i.e. very low focal length). – Andreas H. Sep 11 '16 at 11:29
• My argument is that having the smallest spot size possible is incompatible with multiple illumination – Andreas H. Sep 11 '16 at 11:29
• I am not saying that we can reach 5500K with this apparatus at all. I am not sure of this. But if we have a "5500K apparatus" it needs to have maximum NA, because otherwise is could be improved by making the NA higher or, equivalently, decrease the spot size. – Andreas H. Sep 11 '16 at 11:33
• It sounds like you're saying "you can't build a lens which collects $x\%$ of the half-space light and focus it to a spot smaller than $x\%$ of the source area". Is that essentially it? – spraff Sep 11 '16 at 13:43

If one (let's call it a lens) optic focuses an image of the sun onto a target, then it does so, not from all directions (4pi steradians), but only from the circle of the lens. It is then possible to position another lens with a clear line-of-sight to the object, and focus more sunlight from the second lens. The limit is, after all 4pi steradians of the surroundings of the target are filled with lenses focusing sunlight, you have finally coupled your target with ONLY ONE object, the sun. Until you completely surround the target, the blank directions are presumably colder than the target and it is going to radiate away in those directions, and not achieve solar equality.

After you surround the target, you can no longer add one more lens.

One way a composite machine can exceed this limit, is to run a solar steam engine, which drives a generator, which fires a spark plug.
The spark plug can achieve a higher temperature than the sun.

Don't forget the Lagrange Invariant. That's what it's all about. The sun has a finite size in object space and therefore has a field. Faster optical systems will provide smaller spot sizes of the sun but the angles will be larger. Longer focal length systems will provide larger images but the angles are smaller. As you squeeze the size down the angles go up, just like in diffraction. alpha = 2.44 lambda/diameter. Like in a projector, if you have a dim image you can't just put in a higher Wattage lamp because to get more Wattage, you must increase the size of the filament and a larger filament means a larger field of view for the optical system and won't get through the field stop.