# Is Randall Munroe's “what-if” xkcd correct that magnified moonlight can't get things hotter than the moon? [duplicate]

### 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.

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)?

## marked as duplicate by sammy gerbil, stafusa, Jon Custer, Kyle Kanos, knzhouNov 24 '17 at 12:59

• 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. – tfb Nov 23 '17 at 13:48
• 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.) – Mehrdad Nov 23 '17 at 19:29
• @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). – tfb Nov 23 '17 at 19:55
• Possible duplicate of Is it possible to start fire using moonlight? – sammy gerbil Nov 23 '17 at 21:06

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.

• 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? – Mehrdad 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... – Mehrdad Nov 24 '17 at 22:23
• @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