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Note: I am NOT asking if moonlight can be used to start a fire. I am asking whether these particular arguments in support of the claim that it cannot are correct. i.e. I'm looking for an answer that addresses the physics of these arguments specifically.

On this page, the following claims are made:

  1. You can't start a fire with moonlight no matter how big your magnifying glass is.

  2. General rule of thumb: You can't use lenses and mirrors to make something hotter than the surface of the light source itself. In other words, you can't use sunlight to make something hotter than the surface of the Sun.

  3. Lenses and mirrors work for free; they don't take any energy to operate. If you could use lenses and mirrors to make heat flow from the Sun to a spot on the ground that's hotter than the Sun, you'd be making heat flow from a colder place to a hotter place without expending energy. The second law of thermodynamics says you can't do that. If you could, you could make a perpetual motion machine.

  4. (more claims which I'll omit here)

I neither believe the claim nor follow any of the reasoning.

First, I don't get the thing about the perpetual motion machine:

  • If this is regarding the first law of thermodynamics (conservation of energy), then it's perfectly possible to lose energy while still heating up an object, which would avoid perpetual motion, so I don't get the argument.

  • If this is regarding the second law of thermodynamics (increasing of entropy), then it's also invalid because the disorder in the system is still increasing.

  • If this is about something else, then I don't know what that is.

Second, here is a video of a guy using a mirror he can hold in his hand to light paper on fire.
Clearly the mirror itself isn't getting as hot as he's making the newspaper, and clearly the mirror is the one reflecting the sunlight.
So how can one claim that the paper fundamentally can't get hotter than the reflecting surface? Why can't the moon's surface fundamentally behave similarly (albeit with poorer reflectivity)?

Can someone explain? Is Randall confusing heat with temperature? Or maybe conduction with radiation? Or am I missing something subtle (or perhaps not so subtle)?

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user541686
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    You are missing at least two things. Firstly the temperature concerned is the temperature of the surface of the Sun, not the mirror: mirrors do not work by absorbing radiation and then reemitting it as a blackbody. Secondly, the actual argument is a little subtle: if you want to understand it look up 'conservation of etendue'. Finally this is probably a duplicate as this has been asked before here. –  Nov 23 '17 at 13:48
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    Your statement "it's also invalid because the disorder in the system is still increasing." is incorrect. If you use energy from an object at temperature $T_\mathrm{low}$ to heat an object at temperature $T_\mathrm{high}$, then yes, the entropy of the high-temperature object increases, but the entropy of the low-temperature object decreases more, which violates the Second Law. – Chemomechanics Nov 23 '17 at 19:27
  • @tfb: Wait, so you're saying the light from the moon absorbs radiation and then reemits it as a blackbody? That's why it looks white? (Feel free to mark as a duplicate if it is, I didn't find it after a quick search.) – user541686 Nov 23 '17 at 19:29
  • @Chemomechanics: I was talking about the entropy in the sun increasing, not in the paper. – user541686 Nov 23 '17 at 19:32
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    @Mehrdad: Ah, no, I hadn't realised that you were thinking of the moon as the mirror for the Sun. In that case I think it's more complicated. I think that if the moon was a mirror, then you could indeed make things much hotter than its surface in its reflection. But it's not, and I think the clue is that it's a diffuse surface, and then this comes down to conservation of etendue again. I don't completely understand this (which is why I didn't add an answer). –  Nov 23 '17 at 19:55
  • @tfb: But there's no fundamental distinction between diffuse and non-diffuse (specular?) reflection. It's a spectrum and nothing is perfectly specular. So it can't be the argument he's using. Also, my understanding is the diffuseness of the reflection is only relevant as far as how far you can be from the reflecting surface while still receiving a significant part of the energy, since changing the angle at each point doesn't change the total amount of energy reflected; the reflectivity (or albedo?) should be what matters, since that governs how much energy is absorbed. But that's secondary... – user541686 Nov 23 '17 at 20:34
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    Possible duplicate of Is it possible to start fire using moonlight? – sammy gerbil Nov 23 '17 at 21:06
  • @sammygerbil: It's not quite a duplicate. See my edit. – user541686 Nov 23 '17 at 21:16
  • @Mehrdad I think there is a difference, and I think that the difference is probably etendue. For instance, consider a very large white sheet of paper, lit by the Sun: it's albedo is close to 1, but I think you can't use it to start a fire. But you can start a fire with a mirror. –  Nov 23 '17 at 21:24
  • Like yours, the question by Calmarius is not asking for a Yes/No answer. It is asking if the thermodynamic argument is correct. It has 8 answers and numerous comments. Calmarius also posted Is it possible to focus the radiation from a black body to make something hotter than that black body?, which has 5 answers. Are you saying that none of these answers addresses what you are asking? – sammy gerbil Nov 23 '17 at 21:57
  • @sammygerbil: I didn't see the other question you just linked to until now. It'd take me a while to get through the answers, but even Randall himself points out that the moon's radiation is not blackbody radiation, so it's not clear to me that a blackbody radiation argument is correct for this question. A quick look over all the answers didn't help me understand anything here. In terms of the first link about starting fire with moonlight, they basically repeat Randall's arguments, which isn't helpful. – user541686 Nov 24 '17 at 03:16
  • Randall has to be wrong with this one. Even if you account for etendue, the moon reflects a huge amount of light energy, with a large enough aperture, it should easily overcome diffusion of light. The moon's brightness is 10^14 less than the sun so if you make a collector with area that much larger, I don't see why you wouldn't be able to do it. – A. C. A. C. Nov 24 '17 at 23:34

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So how can one claim that the paper fundamentally can't get hotter than the reflecting surface? Why can't the moon's surface fundamentally behave similarly (albeit with poorer reflectivity)?

Can someone explain? Is Randall confusing heat with temperature? Or maybe conduction with radiation? Or am I missing something subtle (or perhaps not so subtle)?

Munroe states that "You can't use lenses and mirrors to make something hotter than the surface of the light source itself" but this is not valid generally.

It can be derived under some assumptions about the light source and the irradiated body, some of which are:

  • the irradiated body is in energy flow equilibrium (energy that the body captures per some time interval, must be radiated away in the same time);

  • both light source and irradiated body behave as blackbodies.

If that was true for Moon and newspaper, the newspaper could not reach higher temperature than the Moon. This is because if the newspaper was hotter, it would send more radiation energy towards the Moon than it receives from the Moon and it would contradict the assumption that Moon is the source.

But even if we assume the energy flow is in equilibrium, real bodies such as Moon and newspaper are not blackbodies. They reflect part of the radiation that goes their way and their emission at some ranges of wavelengths may be lower than that of blackbody (quite common), or may be higher at some other ranges (if the material is fluorescent).

If the body irradiated by radiation coming from the Moon is made of material with low emissivity for Moon's thermal radiation spectrum peak and high emissivity for Sun's thermal radiation spectrum peak, the radiation reflected off Moon (with Sun-like spectrum) will have much stronger influence on the maximum temperature the body can reach.

Munroe wrote it himself:

"But wait," you might say. "The Moon's light isn't like the Sun's! The Sun is a blackbody—its light output is related to its high temperature. The Moon shines with reflected sunlight, which has a "temperature" of thousands of degrees—that argument doesn't work!"

It turns out it does work, for reasons we'll get to later.

But the rest of article repeats the result valid only under the blackbody assumption. He does not show anywhere in the article how the argument does work when Moon and the irradiated object are not blackbodies.

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Ján Lalinský
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  • Would you (or anyone else) know what happened to the other answer that used to be here? It was saying the claim was very close to being correct (within 3%) but now I'm confused why it's gone... was the answer wrong? – user541686 Nov 24 '17 at 21:59
  • It was deleted by its owner, probably because there was an issue and he wants to think about it before editing and making it visible again. – Ján Lalinský Nov 24 '17 at 22:07
  • In short, it assumed the bodies obey the Stefan-Boltzmann law, but this is not always true for real bodies. For any wavelength, real body will radiate less or more than blackbody, this is characterized by emissivity, which is a function of wavelength and temperature. A body with suitable emissivity function should be able to reach higher temperature than the temperature of the light source. – Ján Lalinský Nov 24 '17 at 22:08
  • Ah, okay thanks. Right, I understand that (and thanks, I've upvoted your answer), but after reading the other answer the part that I'm wondering is how much the temperature can go "higher" here. I imagined it could easily go high enough to make paper burn, but the answer was claiming in this case it'd only go up to 3% higher, which is pretty negligible and puts things in perspective. Now I'm not sure whether that claim was wrong or not... – user541686 Nov 24 '17 at 22:23
  • It cannot be said in general, it depends on the emissivity functions of Moon and the irradiated object. – Ján Lalinský Nov 24 '17 at 22:28
  • Yeah, for the 3% he/she was talking about the moon specifically. I believe he/she used the fact that the albedo was around 10-20%. It was definitely surprising to read, but very useful to know. – user541686 Nov 24 '17 at 22:32
  • @JanLalinsky "A body with suitable emissivity function should be able to reach higher temperature than the temperature of the light source." That can't be right. If that were the case it's trivial to build a heat engine that violates the second law of thermodynamics. – Chris Nov 25 '17 at 23:03
  • @Chris, by source I meant the Moon, since the radiation comes from the direction of the Moon, even if some part of it is not thermal emission of the Moon. This radiation has partially character of Sun's radiation, which has greater temperature than the Moon's surface. If there is a limit on achievable temperature, it should be Sun's temperature, not Moon's. – Ján Lalinský Nov 26 '17 at 01:22
  • @JánLalinský It seems pretty obvious to me that a low albedo object should act closer to a blackbody than to a perfect mirror. We agree that if the moon were a perfect blackbody than the highest achievable temperature would be the hottest surface temperature on the moon, yes? But the moon is much closer to a perfect blackbody (it has an albedo of .12) than a perfect mirror. – Chris Nov 26 '17 at 02:05
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    @Chris I am not sure what would happen if the bodies were blackbodies and we used lenses and mirrors to focus the radiation onto the heated object. But I do know that the limit of temperature the irradiated object can reach, if any, is determined by state of the EM radiation that it interacts with and not necessarily by temperature of the body it came from (since it is not a blackbody). Since this radiation is not equilibrium radiation, processes inside the heated body can probably achieve higher temperature than they would should the radiation be purely equilibrium at temp of the Moon. – Ján Lalinský Nov 26 '17 at 14:21
  • @JánLalinský It's at least pretty close to equilibrium radiation. A quick calculation shows that equilibrium is achieved when the temperature of the moon is about $120^\circ\rm C$, which is about the daytime high surface temperature of the moon. – Chris Nov 27 '17 at 00:14
  • @Chris, what is "pretty close" and what is your point? My point is the radiation coming from the Moon has no single temperature and for certain frequency bands it has stronger intensity (and for others lower) than that of a hypothetical blackbody of temperature and size equal to that of the surface of the Moon. This surplus is non-zero and can be used to heat another object to greater temperature than the blackbody would. The difference may be small, but it is hard to know without making detailed calculations. – Ján Lalinský Nov 27 '17 at 01:53
  • You say: "But the rest of article repeats the result valid only under the blackbody assumption." I don't think this isn't true Randall may have mangled the argument, but I think there is a real argument behind what Randall says. The basic idea: put a perfect blackbody on the surface of the moon. It will heat up to some temperature. Then you can never use moonlight to get anything hotter than that temperature – Peter Shor Nov 28 '17 at 14:02
  • @PeterShor, there may be a real argument but I did not see it there, and I am not sure about your claim either - why should temperature of a hypothetical blackbody in energy flux equilibrium with solar radiation limit temperature that can be reached with help of non-equilibrium radiation heating non-blackbody? You may be right, but I do not see the argument why. – Ján Lalinský Nov 29 '17 at 03:03
  • You're right about the blackbody argument. I had a sign backwards. But I still think the point is: Assuming the same intensity and spectrum of light is emitted from everywhere on the moon, you cannot heat up an object using moonlight and lenses to hotter than that object would become if it were surrounded by the light emitted from the moon's surface. – Peter Shor Nov 29 '17 at 13:21
  • That's possible but 1) I do not know why it should be true: albeit there may be some limitation to how well light can be focused and thus its intensity amplified, but how does this imply anything about temperature? The body is not blackbody and so there is no generally valid relation between spectral function of ambient radiation and temperature of the body; 2) it does not really give concrete limit to the reachable temperature, since it is not at all clear how great the temperature of an irradiated object would be if it were surrounded by the light emitted from the moon's surface. – Ján Lalinský Nov 29 '17 at 15:18
  • (1) You can't focus light to make it more intense than it is at the source. (Assuming the source isn't something like a laser or a mirror, that emits light preferentially in a single direction.) This argument is exactly the same as the argument that proves the same thing for blackbody radiation. (2) Randall did get it wrong, because different objects will heat up different amounts in the same (non-black-body) radiation. Think greenhouse effect. – Peter Shor Nov 29 '17 at 15:59
  • If we can focus light to make it more intense than at the visible source for laser light or light reflected off a mirror, why not for light coming from Moon's surface? Moon in non-blackbody non-Lambertian source of radiation, I doubt the same argument as for blackbody is valid for it. If you think it does, could you give a reference or post an answer where this would be explained? – Ján Lalinský Nov 29 '17 at 20:51
  • Reference. – Peter Shor Nov 30 '17 at 05:05
  • @PeterShor, the reference focuses on etendue in general, it says nothing about limiting the power per unit area to that near the body from which the radiation is coming. As you yourself have implied, if the divergence of incoming light rays is small, like for a mirror reflecting solar radiation, the intensity can be magnified beyond that near the mirror. The Moon is no mirror, but it does have very directional reflectance. https://en.wikipedia.org/wiki/Opposition_surge – Ján Lalinský Nov 30 '17 at 14:37
  • That article says that the directional reflectance effect for the moon is a 40% increase, probably from shadow hiding. Nowhere near enough to start a fire. – Peter Shor Nov 30 '17 at 17:06
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    The opposition effect of the Moon is just an example to show the problem with your general-sounding claim about impossibility to achieve higher light intensity. I do not claim anywhere that you can start a fire with Moon's light. This whole question is about something else - whether Munroe's argumentation is valid, and I think it is clear now that it is not. – Ján Lalinský Nov 30 '17 at 23:03