The rest of the argument seems to be that light cannot be optically condensed to a single point as there will always be some dispersion due to diffraction, and the size and shape of the dispersion is dictated by that of the source. That is you can't make the target denser then source, hence the temperature must be less.
EDIT: Several commenters submitted that lenses are reversible while solar panels are not, and this makes all the difference. My retort is that I can make non-reversible lenses by covering them in a thin layer of dust. Since the lense system is now non-reversible can I use these sub-par lenses to create higher temperature than I could with clear lenses?
There's a difference. By going through solar panels you would increase entropy because the conversion of sunlight to electricity can't be 100% efficient. And if you allow the solar panels to heat up too much because of that waste heat, then that efficiency will drop further.
If a lens could heat a point to a temperature higher than the sun, then there'd be no such loss and you'd be magically decreasing entropy.
Were is this reasoning wrong?
This boils down to a moderately heated BB receiving an large stream of moderately powered photons and either rejecting them or first absorbing and then radiating them away at the same pace without changing its own temperature, regardless of how many photons are coming in.
I think that this issue is addressed by footnote 2
> And, more specifically, everything [lenses and mirrors] do is fully reversible—which means you can add them in without increasing the entropy of the system.
Presumably we can't say the same of the electrical devices that we use to collect, store, and transmit sunlight, or to create high temperatures from this stored energy. For example, if you use a "lunar panel" to store energy in a battery and then heat something on an electric stove, many of the components in this process will not be reversible, differently from mirrors and lenses. (Putting a hot object on top of the stove won't cause the lunar panel to emit light back in the direction of the moon!)
So I think footnote 2 is actually very important, because it's not that we can never use any energy source to create something hotter than that source, it's that we can never do so using only reversible processes, including purely passive optics.
I guarantee you this can be done. Arc welding  goes to many thousands of degrees, up to 20,000C. A square mile of solar panels will most certainly give me enough energy to power an arc welder. In fact I will probably will get away with just 100kw of pwoer, so 500 panels give or take.
The solar panel system is different - it's a heat engine that takes Q units of heat energy from one body, transfers a portion of it Q₁ to a higher temperature body and transfers the remainder of it (Q - Q₁) to a lower temperature body (the immediate surroundings). That's OK with thermodynamics, subject to upper bounds on the ratio Q₁/Q that depend on the absolute temperatures of the bodies involved.
This suffices to show that the perfect lens system can't work. The dusty lens system can't work because adding dust to a perfect lens doesn't turn it into a heat engine.
Except it's not "heat energy", it's electromagnetic energy that's getting transferred. If the target object is smaller than the source object, you can have an amount of energy that will increase the target's heat above that of the source.
Thermal radiation == electromagnetic radiation.
It's okay because solar panels are not perfectly efficient while lenses are, in theory, perfectly efficient. That efficiency loss is in essence the "cost" of moving heat from a colder to a hotter place. I'm guessing, but don't quote me on it, that solar panel efficiency is related to the temperature of the sun and the local environment just like any other heat engine.
You are correct,
(1) can't heat an object to a higher temperature than a source object using lenses - you can't transfer heat from a cooler object to a warmer object without work and lenses don't provide a mechanism for work. This still leaves room for reflected sunlight from moon albedo getting something hot enough to burn, at least from a thermodynamic perspective.
(2) you can get something arbitrarily hot using lunar panels, this is just a bad heat pump (max COP 1) where you feed work generated by the solar panels into a perfectly thermally isolated system with a resistive heater, and we could do this a bit more efficiently with a true heat pump.
I am dumping some amount of low-energy photons onto a target, the target is heated up and radiating out the same amount of energy it receives. As I increase the number of photons hitting the target its temperature raises and it radiates more heat outwards, shedding excess heat.
Then as I keep increasing the number of photons, the temperature stops increasing at some point having reached the temperature of the photon source. I increase the number of photons a million times of the previous level, yet the object remains the same temperature.
Where does the excess energy go? Is it not absorbed by the target? Is it reflected? Is it re-radiated?
Initially it's in a steady state where the target is colder.
You increase the number of photons hitting the target. The target gets hotter. All good so far.
But we have to look at how you increased that number. You need to either move the source and target closer together, or you have to use lenses to simulate the same thing.
You can keep hitting the target with photons from more and more directions, but no system of mirrors or lenses can increase the apparent brightness of the source.
As you approach the point where the source fills the entire sky, the target will approach the same temperature.
And then you can't increase the number of photons any more. Not without using a hotter source.
There is a limit to how small you can focus a lens. Once the object you are focusing is in sharp focus, it will start to go out of focus again if you go past that point. So you can’t just arbitrarily increase the photon density further.
If you make your lens bigger to gather more photons, then the focused image just grows by a corresponding amount.
Optics (and thermodynamics) tell us that we can not generate a higher flux of photons into a target area, than left the equivalent area at the source. Assuming that both the source and target are black body radiators, that says the target can’t get hotter than the source.
Say a light source has a T temperature resulting in X photons emitted. I redirect all the photons to a single point. I see arguments mentioning that that single point cannot be hotter than the source because there's no more photons to make it hotter.
I now add a second light source of the same T temperature, that emits the same amount of photons and also focus all of them on that same point. I now have more photons, but temperature source of all photons is the same. How does adding more photons not make my point hotter?
For that to be possible with a blackbody light source, it has to itself be a single point. Which means a temperature of approximately infinity.
For a real light source, one that has an area, you can at best focus it down to the same area. To get maximum light to a target, you either have to make the target almost touch the source, or you have to have it so no matter what direction you go from the target, you hit a lens that's bringing light from the source. Once you set either one of those up, there's nowhere to fit a new light source.
I'm not an optics expert, but this can't be true. You can clearly focus light emitting from an area into a smaller area (although probably not to an infinitesimally small area).
You can't increase the density of the light. You can't focus all of it onto a smaller area.
Is that clearer?
If that is the case, then there is no extra room to fit the optics to focus another source onto the target.
If you shrink the optics enough to make room for another source, then you also aren’t delivering all of the photons from the original source (you can’t be, since you aren’t covering all incoming angles), and therefore it’s not the same temperature as the source.
What about two single photon sources, can’t they be pointed at exactly the same spot? Maybe the explanation here is that the target electron cannot interact with 2 photons at the same time, so you can’t ‘double heat’ a single particle. Or maybe that you can’t precisely target a single particle without decreasing the entropy of a closed system, which is impossible.
Here he says very clearly that if you "bundled" all the light from the sun and aim it at the earth it would heat the atmosphere to millions of degrees (the surface of the sun is much less than that). It's not at all clear to me what he means by "bundled" and why it's not contradictory to what he says in the article here. Presumably some kind of lens / mirror system could be used?
It seems to me that in this article he has in mind some highly abstract system that's fully reversible. Of course, in that case, once the target object gets hot enough it will start emitting light and result in equilibrium. But it's not clear to me that this describes what would actually happen with a real optical system! E.g. (a) much of the light the target receives is going to be absorbed and reemitted away from the lens, (b) what if you removed the lens targeting system at the precise moment the light impacted the target, so that the system couldn't be reversed, etc.
Edit: one more thing. The surface of the lit side of the moon can reach 260 degrees F, and dry wood can potentially catch fire as low as 300 degrees F. And the moon has some reflectivity as well. So even taking Randall's claims on their face, I'm skeptical that you could not start a fire (in some materials at least) using moonlight.
The reason I (and probably others) find Randall's explanation unhelpful is that obviously there's "enough" energy being reflected by the moon to start a fire (that's why people keep bringing up solar panels). The issue is that there's no way to optically redirect that energy into a small area without heating up your source to the same degree. Which is theoretically possible I suppose, but it's not the situation the What-If is talking about. Along with the issue that the light we see from the moon is mostly reflected rather than emitted (which changed the situation entirely), this makes the What-If explanation a little misleading.
Yes, yes I can. 1000 solar panels (200 watts each) will power a an electric arc welder up to 20,000 C continuously for as along as the sun shines in the sky.
There is nothing here that collects energy across time, all energy is immediately imparted onto the electric arc.
The solar panel itself is already non-reversible, and so is the arc welder, right? That seems to be the important difference from the lens.
In that sense, we don't run either of them "for free" in the way that the xkcd piece describes.
Are you sure about that? First off, take an off-the-shelf solar panel, point it at the moon in the middle of the night. You get a grand total of nothing. Okay, it was a cheap panel, that might not generalize to anything.
But more importantly, by the argument laid out in the article, your solar panels cannot work in moon light. (Or maybe they work at horrible efficiency, because the moon is a bit warmer than the earth.) I'm not sure I buy that argument; maybe you should run the experiment.
Yes, this is what would happen.
Solar panels will work in moonlight if and only if you can make fire from moonlight with a magnifying glass.
The answer depends on how good a mirror the moon is. It calls for a real experiment, not a thought experiment. I don't really know which way it will go.
I put in place 1000 solar panels and aim them at the sun thus procuring 200kw of power. Next I use it to power an arc welder, producing 10,000 C of heat. Therefore I have used a 6,000C sun to produce 10,000C on earth.
Would a similar setup work with the moon? I don't know. That was not my point, my point was that it certainly is possible to produce higher temperature at the target than it was at source.