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Thermodynamics argument seems iffy to me. I can cover the entire surface of the earth with solar panels to harvest the moon light and use the combined electricity to melt iron. If this is ok with thermodynamics then so is using lenses.

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?




> I can cover the entire surface of the earth with solar panels to harvest the moon light and use the combined electricity to melt iron. If this is ok with thermodynamics then so is using 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[1]. 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.

[1] https://en.wikipedia.org/wiki/Thermodynamic_efficiency_limit


The region of higher temperature is much smaller. Entropy is not a pure measure of energy or randomness, but a measure of that within a volume. You'd have smaller entropy in a small volume, but higher around it because you would have diverted the ray that would have hit outside the concentrated region.

Were is this reasoning wrong?


Yeah this is all rather confusing...

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.


Yeah I don't see how this argument holds water at all. If you have 10 kW worth of photons focused onto a square inch that absorbs 90% of those photons then that's 9 kW of power that isn't just going to magically disappear and it's not going to reach equilibrium until it's emitting 9 kW of power itself which is certainly going to be thousands of degrees. While the entropy argument is certainly an interesting thought experiment all that means is that the global entropy must somehow still be increasing.


I was wondering about this too. After all, human beings have produced temperatures that were hotter than the sun's surface and almost all of the energy that we used to do that ultimately came from the sun (except for very small components that came from nuclear reactions on Earth and geothermal energy, which also comes from nuclear reactions inside Earth).

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 can convert the optics to non-reversible by spreading a thin layer of dust on top. Now both systems are non-reversible and both allow higher temperatures? Or are we looking at a particular kind of non-reversibility?


I think the thermodynamics argument is that only non-reversible systems cause energy to flow from a lower-temperature source to a higher-temperature destination, not that all non-reversible systems do so. For example, if you set up a solar panel that directly feeds a resistor, you have a dissipative system that's unlikely to get hotter than the sun. However, a solar panel array and/or battery could conceivably power a fusion reaction that does get hotter than the sun.


>if you set up a solar panel that directly feeds a resistor, you have a dissipative system that's unlikely to get hotter than the sun.

I guarantee you this can be done. Arc welding [1] 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.

[1] https://hypertextbook.com/facts/2003/EstherDorzin.shtml


Okay, sure, as long as you're not trying to say this contradicts the post. "A resistor" was just an example of how you could waste the power and not get much heat.


Covering the lens with dust would let you light a fire. The dust will warm slightly above ambient temperature in the lens' shadow, so you can use it to power a Carnot engine that winds up a spring that is then released to rub a twig against a log.


This is genius.


Wouldn't this layer of dust also stop the lense from working normally? If I understand the other commenters here, it's not just about reversibility, but "useful energy" loss - with lenses you are not paying the needed "efficiency tax" to achieve higher temperatures than the source.


What the thermodynamic principle says you can't do is to take Q units of heat energy from one body and transfer it all to a higher temperature body.

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.


> you can't do is to take Q units of heat energy from one body and transfer it all to a higher temperature body

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.


> Except it's not "heat energy", it's electromagnetic energy that's getting transferred.

Thermal radiation == electromagnetic radiation.


I don't think it could remain above the source temperature in equilibrium, because it will radiate back toward the source!


It's impossible to aim the source at a smaller target because the source dissipates in all directions, unless you make something like a laser.


>If this is ok with thermodynamics then so is using lenses.

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.


> solar panel efficiency is related to the temperature of the sun and the local environment just like any other heat engine

You are correct, https://en.wikipedia.org/wiki/Solar_cell_efficiency#Thermody...


You are correct about the solar panels. Solar panels are more efficient when they are colder, i.e. the temperature delta between the sun and the panel is larger.


The difference is that a lens is a reversible system, whereas using energy from solar panels to melt iron is not. If you don't have a reversible system it is perfectly possible to take energy from a cold object and make a hot object hotter, but you cannot do this reversibly.


Both these thermodynamic arguments look good, I don't see any contradiction though

(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.


The entropy argument is a red herring for your system (and wrong). The difference is that you would have to store your energy from the solar panel for time t1 and then expected it in time t2 much less than t1 to achieve your desired effect. The system in question doesn't allow storing the energy for later quicker dispersal


No, I do not store any energy. My solar panels outputs are wired directly to the input of an arc welder. Powered from the sun light (surface temperature 6,000C) this system can easily give me an arc of 10,000 C.


Reversible means you don't increase the entropy of the system. If you relax this restriction, you can do useful work, like powering an arc welder or growing carrots to feed yourself and rubbing sticks together to start a fire. Merely making the lens non-reversible doesn't mean it will suddenly set things on fire.


How does it work in practice?

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?


So you have a blackbody source, and a target.

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.


The point is, you can never get to the point where you are dumping more photons in to make it hotter, because there are not more photons.

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.


Consider this:

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?


> I redirect all the photons to a single point.

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.


> For a real light source, one that has an area, you can at best focus it down to the same area.

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 focus 1/100 of the light onto 1/100 the area. Or 1/1000 of the light onto 1/100 the area.

You can't increase the density of the light. You can't focus all of it onto a smaller area.

Is that clearer?


In order to truly move all of the photons from the emitter to the target, you need to effectively surround the target with lenses, such that it looks from the surface of the target as if the emitter is filling the sky.

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.


Well I think the thermodynamic analysis is useful in understanding things, that was my point. I’m not quite understanding the scenario you describe - are you asking how it is possible that you can’t heat up the target more than the origin? If so then take an example where the sun is a single point at 5000 degrees. It is easy to see that the most you could do with this point sun is heat up one point on your object to the same temperature, no more. Now what if there were two point suns next to each other? The discussion on xkcd about optics is saying you can’t superimpose the points on each other on the target under any circumstances, even with two separate lenses at the right angle. This doesn’t seem like it should be impossible though...

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.


The interesting thing to me is that Randall describes a very similar system just a few What-Ifs earlier: https://what-if.xkcd.com/141/

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.


Yeah, it does contradict that. You can't really "bundle" all the light from the sun like that and aim it at the Earth without violating thermodynamics. The way to do it would be something like wrapping the sun and the Earth together in a giant chamber made out of some perfect mirror (actually maybe the mirror isn't necessary), with the only exit being the dark side of the Earth. And that would heat up both the Earth and the Sun to millions of degrees.


Aha. I think you've helped me see the central issue with your point that it would also heat the sun to millions of degrees. Suppose there was some system that allowed you to collect the light of some source and dump it somewhere else: the drawing in the What-If seems to suggest a system of perfect mirrors with an "output tube" pointed at the earth, so that any photons leaving the tube hit the earth with a high degree of accuracy. What your point shows us is that even if we imagine an indestructible system of mirrors and lenses to do this, the result is that bottling up the energy that the source normally radiates away massively increases the temperature of the source (turning your mirrors into plasma and returning the system to normal, but we're ignoring that again). So entropy law isn't violated.

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.


Solar panels lose energy to heat in this case, so entropy does not decrease.


I was wrong about what I wrote here and am removing it.


>The analogous problem for you would be that you couldn't maintain your electric heat ray (or whatever) for a time equal to or greater than the time you spent collecting.

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.


Yeah, maybe it was a red herring (a black-body herring?) to bring up batteries here.

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.


> I can cover the entire surface of the earth with solar panels to harvest the moon light

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.


There is no need for an experiment, the moonlight does produce energy in a solar panel. Obviously the power produced will be much much lower than with the sun. I remember that as a child I was able to use a solar calculator indoor with just the power gathered from an incandescent light.


>(Or maybe they work at horrible efficiency, because the moon is a bit warmer than the earth.)

Yes, this is what would happen.


I'll repeat for the reflexive downvoters:

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.


You're not addressing the essential part of my comment. Let me clarify:

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.


Monroe’s explanation only applies to optical systems, ie reflecting and focusing light, which is a thermodynamically reversible process (yes, even with dust. Dust doesn’t make the lens thermodynamically irreversible, it just scatters and absorbs some of the light so it doesn’t reach the lens. The irreversibility is a property of the lens). Solar panels are converting light energy into electrical energy, which is not an irreversible process - it creates entropy, and therefore the rules are different.


Sure it is, just not in real-time. The lense fails because there is no storage mechanism.




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