# Is it possible to start fire using moonlight?

You can start fire by focusing the sunlight using the magnifying glass.

I searched the web whether you can do the same using moonlight. And found this and this - the first two in Google search results.

What I found is the thermodynamics argument: you cannot heat anything to a higher temperature using black body radiation than the black body itself, and Moon isn't hot enough.

It may be true, but my gut feelings protest... The larger your aperture is, the more light you collect, also you have better focus because the airy disk is smaller. So if you have a really huge lens with a really short focus (to keep Moon's picture small), or in the extreme case you build a Dyson-sphere around the Moon (letting a small hole to the let the sunlight enter), and focusing all reflected light into a point it should be more than enough to ingnite a piece of paper isn't it?

I'm confused. So can you start fires using the Moon?

• You can focus moonlight on a small solar panel, accumulate the electricity in a small battery and use that to ignite whatever you like. None of this violates thermodynamics in the least, as long as the temperature of your solar cells is slightly lower than the temperature of the radiation (and the bandgap is chosen correctly), they will generate electricity. The "temperature" of moonlight is pretty close to that of sunlight because it's a reflected solar spectrum, not a thermal emission spectrum at the temperature of the lunar surface. Oct 12, 2014 at 8:51
• Your argument about black-body radiation from the moon holds for a new moon. I'll bet that agrees with your gut feeling. With a full moon, you're discussing reflected sunlight which isn't in thermal equilibrium with the moon surface. Oct 12, 2014 at 14:23
• If you are using hypergolics - yes, you can. No moon needed at all. Oct 12, 2014 at 21:03
• If the highest temperature you could achieve was determined by the temperature of the object reflecting the light, you wouldn't be able to start a fire with a laser reflected off a mirror. In fact, you wouldn't be able to start a fire with a laser period, since the operation of the laser depends on internal reflection. Oct 13, 2014 at 19:43
• Your sources make the crucial mistake by assuming that the only radiation from the moon is black-body radiation. This is false! Most of the radiation from the moon is reflected radiation from the Sun. So I think, yes you could given a large enough lens (see J's comment, about 17m). Black body radiation is what you see (or rather, don't see since the moon is not all that hot) during a lunar eclipse, Oct 13, 2014 at 20:20

Moonlight has a spectral peak around $$650\ \mathrm{nm}$$ (the sun peaks at around $$550\ \mathrm{nm}$$). Ordinary solar cells will work just fine to convert it into electricity. The power of moonlight is about $$500\,000$$ times less than that of sunlight, which for a solar constant of $$1000\ \mathrm{W/m^2}$$ leaves us with about $$2\ \mathrm{mW/m^2}$$. After accounting for optical losses and a typical solar cell efficiency of roughly $$20\ \%$$, we can probably hope to extract approx. $$0.1\ \mathrm{mW}$$ with a fairly simple foil mirror of $$1\ \mathrm{m^2}$$ surface area. Accumulated over the course of a whole night with a full moon, this leaves us with around $$6\ \mathrm h\times3600\ \mathrm{s/h}\times0.1\ \mathrm{mW}\approx2\ \mathrm J$$ of energy. That's plenty of energy to ignite a fire using the right chemicals and a thin filament as a heater.

• granted, if you have to wait a full night you might as well use the sun to do that! :D
– Ant
Oct 12, 2014 at 9:44
• To put it in the context of OP's magnifying glass, let's say you can start a fire by focusing daytime sunlight using a 25mm diameter lens. To gather the equivalent optical power using moonlight would require a lens over 17m in diameter (almost 60 feet). This would not fit in your camping rucksack.
– J...
Oct 12, 2014 at 10:07
• I'll accept this, because it answers my question. Although this isn't the answer I expected. So I will post another question soon that's more specific about the black bodies. Oct 12, 2014 at 11:02
• I thought you were making a point that this energy would not be enough to set anything on fire. Huh! Oct 14, 2014 at 12:37
• @Calmarius - if you were wondering about whether it could be done with a passive optical system involving lenses/mirrors, as opposed to something like a solar panel which functions as a type of engine to convert light to electricity, then the answer is no, conservation of etendue in optics implies a passive optical system can never focus light to a greater intensity than the intensity at the surface of the source--see some of the the answers to this question. Oct 20, 2014 at 22:33

At least one point in your favour is that the light we receive from the Moon has barely anything to do with its temperature. Instead it is mostly a secondary light source "reflecting" light from the Sun towards us.

The second point in your favour (I think) is that the thermodynamic argument seems pretty weak. We are not trying to make Earth as hot as the Sun or anything like that. The only thing we want is to gather enough energy in a sufficiently small volume with oxygen and some fuel to light a fire; hence most of the energy for the fire still comes from the enthalpy of the combustion reaction.

Overall, I would think this is not impossible but probably very inefficient because of the minute fraction of power we receive from the Sun's light scattered by the Moon.

• Well, the Moon's surface temperature during the day is about 123 celsius, which isn't hot enough to ignite paper or lighter fluid, so if we had to rely on black-body radiation from the Moon and pure optics then it would be impossible. But since as you say moonlight is reflected rather than re-emitted, it might be in-principle possible. I imagine you'd need a ridiculously large collection area though. Oct 12, 2014 at 9:27
• > "the light we receive from the Moon has barely anything to do with its temperature. Instead it is mostly a secondary light source "reflecting" light from the Sun towards us." I thought this too, but actually albedo of the Moon surface is quite low, around 0.1. This means only 10% of incident radiation energy is reflected by the surface. The remaining 90% is, (assuming energy equilibrium over many rotations of the Moon) radiated away as thermal radiation of the Moon. Nov 24, 2017 at 21:18
• It seems most of radiation energy Earth receives from the Moon is thermal. Still, the fact that neither the Moon nor the object to be heated are blackbodies means one could in principle heat to higher temperature than the surface of the Moon. See also my answer here: physics.stackexchange.com/questions/370446/… Nov 24, 2017 at 21:19

I am bumping this, because there still seems to be no consensus on the matter, even after Randall's What If post and the heated discussions that followed:

What If post

Reddit discussion, where people strongly disagree with Randall

My intuitive take (taking points from the discussions): Obviously, Randall's argument holds for blackbodies. However, a part of moonlight is diffuse reflected sunlight, and so should be able to start a fire.

@Marty Green: If my argument is correct, then "adding moons to the night sky" will certainly increase the temperature beyond the Moons' surface temperature (because all we do is add more mirrors).

• See my analysis of the contribution of the reflected sunlight which basically shows it's not significant. Sep 26, 2016 at 16:57
• Thermodynamic arguments kind-of work because moon is "100%" reflective (even though 88% of the light is "reflected" as infrared). But strictly speaking it isn't black body radiation: one could theoretically encase the oven with a material that reflects infrared but is transparent to visible light. However, this isn't the standard picture of a magnifying glass or other lens setup that doesn't care about wavelength. Jul 22, 2020 at 18:20
• @Floris thanks for that analysis. I think that analysis is the most complete one and nicely answers this question. I do think the way summarize it here is misleading though. By saying it is not significant you make it seem like the reflected light has no impact on the maximum temperature achievable. However the analysis there shows that the maximum temperature achievable is 198 celsius with reflection, vs 123 celsius without reflection. So it would be better summarized as: reflection contributes a significant amount of energy, but not enough to start a fire.
– M.D.
Dec 16, 2021 at 20:39
• @M.D. you are right I didn’t phrase that very well. It would be better to say that, based on the analysis I linked, even direct reflection off the moon cannot yield sufficient power to light a fire directly (if you collect over a long period into, say, a battery, all that changes of course). Dec 16, 2021 at 22:46

If you could fill the whole sky with moons you would not light a fire. It would be the same as looking up and seeing a wide expanse of bright shiny sand on a beach. What you can do with lenses and mirrors is no different than filling the sky with moons, so no: you cannot light a fire that way.

• This is actually a very good way of casting the problem in a way that it can be answered intuitively. Jan 6, 2016 at 23:45
• @Marty Green I love the analogy, but you know what? If we think of the Moon as a reflector, not a blackbody, I'm actually not sure that you're right - it doesn't seem obvious to me.
– Zubo
Jul 15, 2016 at 21:24
• If you fill the sky with mirrors, yes, you can start a fire. Jul 16, 2016 at 2:32
• This is not correct. Under your reasoning one cannot use lenses to start a fire even in sunlight. Nov 26, 2017 at 13:50
• The moon is not a mirror. Dec 3, 2017 at 13:11

Other answers here don't take into account two very important aspects. First, the heated point radiates too. Second, ideal lens with large diameter to focal length ratio don't exist. The latter can be proved with entropy.

Suppose the "magical" lighting (which might contain the ideal lens) system exists away from Earth. Suppose also we surround this system with a radiation bath of solid angle $4 \pi$ with temperature $T_0$. Then, due to the second law of thermodynamics, the system will be in equilibrium, when the temperature in the "magical" device is uniformly also $T_0$. Then, as to simulate some radiating object (such as Moon), we remove most of the radiation, and leave only a small proportion of the solid angle. Of course, if the "magical" lighting device consists only of lens and mirrors, the radiation towards the heated point can only decrease $\Rightarrow$ it's temperature $T \leq T_0$. Note that the temperature $T$ only depends on the power it is heated, not on the spectrum of radiation.

If arbitrarily large diameter to focal length lens existed, these could be used to focus light of a radiating body (such as Sun or Moon) to arbitrary intensities, giving rise to arbitrarily high temperatures (contradicting the proof).

Thus if the heated point is black, the maximum intensity of the black-body radiation is the intensity of the reflected light. Thus the maximum temperature might indeed be in the range of $0^{\circ} C$.

If the heated point is not black, but radiates only very high-frequency spectrum, the achievable temperatures would be higher, probably temperatures up to the temperature of the sun.

• It might seem that an arbitrarily large parabolic lens could be used to heat a point to arbitrary temperatures. However, it should be noted that the heating effect is zero, if we consider only a single direction of incident rays (solid angle is zero, temperature finite). Thus, we have to consider how the spot changes under small perturbations of incident angle. Result: very badly. Extreme coma effect is observed, which ruins our high intensity. Indeed, if we would be watching from the intense spot, we would observe just the reflected Moon, with the same solid angle intensity. Jan 4, 2015 at 18:55

It appears the question is specifically concerned with using a lens.

In the original question: "So if you have a really huge lens...and focusing all reflected light into a point it should be more than enough to ingnite a piece of paper isn't it?"

In which case using solar cells doesn't answer the question.

The answer is no. No matter the lens you cannot make the surface brighter than the surface of the moon. That's thermodynamics. see: second law of thermodynamics

The second law of thermodynamics states that the total entropy of an isolated system always increases over time

In other words energy cannot flow from a colder area to a hotter one.

It can also be explained using the optical calculations point of view. See: CuriousOne's comment about a passive optical system and conservation of etendue. You should see this post. Especially the answer by CountIbis which will explain the limits using optic calculations.

If the object radiates as a black body, has a radius of R, a temperature of TT and is a distance d away, then the flux of radiation reaching the lens is: $$F = \sigma T^4 \left(\frac{R}{d}\right)^2 = \sigma T^4 \alpha^2$$ The total power of the radiation entering the lens PP is the area of the lens opening times the flux: $$P = \pi r^2 F = \pi \sigma T^4\alpha^2r^2$$ This power ends up heating the area of the image in the focal plane. The flux of radiation there is: $$F_{\text{im}} = \frac{P}{\pi\alpha^2f^2} = \sigma T^4\frac{r^2}{f^2}$$ Suppose then that you put a black body in the image plane, then the temperature there would be $$T_{\text{im}}$$ where $$\sigma T_{\text{im}}^4 = F_{\text{im}}$$ therefore: $$T_{\text{im}} = \sqrt{\frac{r}{f}}T$$ The ratio of the focal length f and the lens diameter is called the F-number and this is always larger than 1. So, the factor multiplying TT in the above equation will always be smaller than 1, therefore you can never reach a higher temperature than the temperature of the object in this way.

You should also see @zubo 's links http://what-if.xkcd.com/145/

• If we assume that the moon radiates as a black body, yes. But if it behaves like a mirror, reflecting sunlight, then that might still work, no? That case isn't addressed here, I think.
– Zubo
Sep 29, 2016 at 2:57
• "No matter the lens you cannot make the surface brighter than the surface of the moon." Would using multiple lenses and mirrors to direct all of their own created points of light into a single point make a difference? Jul 10, 2017 at 21:31
• @Sean256 You cannot "smoosh" light beams into a single point. It violates the laws of étendue. Also an optical system has to always be reversible, so if the light was focused onto a single point the system could not be reversible since the light beams don't know where to go. The beams can only be focused onto an area not a point, which is very important. Jul 11, 2017 at 18:00
• If you have non-blackbody radiation, the amount it heats up an object can depend significantly on the optical absorbing/reflecting spectral properties of the object. [For example, the amount the sun heats up the Earth depends on the amount of CO2 in the atmosphere.] So your comment "you can never reach a higher temperature than the temperature of the object" is wrong. Dec 3, 2017 at 12:57

Since the question is not constrained in any way, the answer is yes, it is possible. Is it "realistically" possible, no.
To make it possible, all you have to do is concentrate the moonlight (optically,or electrically) until you have the energy required to reach the ignition point of the material. what makes it not realistic is the size, cost, and/or time involved in creating a "concentrator."

• This answer could be improved by reading up on the matter - the ideas you mention are in dispute and require arguments and explanation.
– Zubo
Jul 11, 2016 at 15:21
• The answer is constrained by “and follows the laws of thermodynamics”. A device such as you describe would violate those laws - if I can make A hotter than B with appropriate “mirrors” then I could make heat flow back from B to A through a heat pump - boom, perpetuum mobile. Only of your allow “concentration over time” (collect moonlight with a PV cell, charge battery) could you do this without violating laws of physics. Dec 18, 2020 at 12:06