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The conclusion is correct. It doesn't rely on the moon being a black body radiator, but on etendue, which shows that the most you can do with a system of lenses and mirrors is to create an environment (on earth) as bright as the environment on the moon. The fact that rocks on the moon's surface only reach 100°C shows that an environment like that is not bright enough to light a fire.

Trying to follow your argument. Suppose we put absolutely reflective mirror on the orbit. Since its absolutely reflective its temperature will be 0 Kelvin (in theory, very low in practice - there is a reason solar bound spacecrafts are covered with reflective surfaces). So we would not be able to use the light it reflects to heat anything?

Specular reflector preserves the etendue of the light rays that fall on it. Diffuse reflector (such as the moon) does not; it becomes a new light source instead, resulting in higher etendue.

In short - perfect reflector preserves etendue, but imperfect does not.

But the light coming from the moon is a combination of diffuse and specular reflection... and again (just like for the black body argument) the fraction of light which is specular is most important. The fact that the moon subtends 0.5deg out of 180, while reflecting more than 10% of light (12% bond albedo) suggests to me that what we see is not only lambertian (or cos2) diffuse reflection, but it depends on absorption so I don't know enough to calculate.

Whoa. I don't think most of the rest is specular reflection, since we never see any specular highlights on the moon, but if it is non-Lambertian, it might be possible to heat it up better than if the moon were actually a black-body or pefect Lambertian scatterer.

Thanks for pointing this out; I seem to have learnt of a new phenomenon I wasn't previously aware of: https://en.wikipedia.org/wiki/Opposition_surge

Actually, I was reading a very old article that some astronomers (lunonomers?) had seen intermittant specular reflections of small patches on the moon, with some calculations estimating size. https://doi.org/10.1016/0019-1035(68)90077-8

At this point I almost think an experimental measurement of temperature equilibrium would be publishable from one of the old re/frac/flectors you can put your head into.

Thanks, this is what I was missing and what I think is most obscure in the article!

The question is to what extend the moon surface is imperfect reflector. For example, if the mirror has a hole, it will be an imperfect reflector, but you can compensate for it using a bigger lense.

By imperfect it's sufficient to be diffuse. You can verify the Moon is diffusive to a good approximation because of the approximately uniform appearance of the full Moon.

The idea is that you can "organize" or "revert" any ray bundle from a system of non-absorbing lenses and specular reflectors, but if your reflector has billions of tiny irregularities it's not viable to build such a system (it's equivalent to an ideal diffuser, in which light is isotropically reflected). The ideal diffusion process is clearly not reversible by itself: if you shine a beam onto a diffuser it spreads the light; if you expose it to the same light (with reversed directions), it again diffuses it instead of reverting to the original beam. In theory again the physical laws of electromagnetism are time reversible, but in practice the effort to revert some systems might be too demanding (you can even do better -- see Maxwell's demon); manipulation of physical apparatus and information acquisition/manipulation itself has a cost that surpasses any gains.

The moon is very imperfect; it's darker than asphalt. http://wtamu.edu/~cbaird/sq/2015/08/06/why-is-the-moon-so-br...

Perfect reflector exists only in theory. But we still put (imperfect) reflective coating on spacecrafts.

Just a minor correction—I don’t see any reason why a perfect reflector would happen to be 0K, but since it has zero emissivity, you wouldn’t be able to measure the temperature. Which is not quite the same thing as 0K.

In practice, any reflector is imperfect, and therefore has nonzero emissivity, and therefore goes towards thermal equilibrium with its environment.

Except the moon is acting as a (partial) reflector, and so is part of the optical system. So, really, you're limited to the temperature of the Sun, not to that of the moon. So the conclusion is still potentially incorrect.

That’s really stretching “optical system” if you include the moon.

Sure, if we reshaped the moon into a giant mirror, we could use a lens light a fire using the “moon”.

First some light is reflected from the moon without increasing the temperature of moon rocks. Second, the moon landings avoided peak moon temperatures making the fact astronots survived irrelevant. Just look at how long their shadows where, it was a long way from noon and 127c surface temperatures.

Finally paper burns at ~233c but plenty of things ignite at lower temperatures. Hay’s ignition temperature is ~130c which considering the moon is 127c and you will concentrate extra reflected sunlight seems very viable.

However, the earth’s atmosphere blocks a lot of moonlight, so it’s more likely to work in a pressurized cabin at high altitude or spacecraft. Even your lenses are going to be an issue.

PS: A lunar day is almost a month long. They are casting long shadows which means they are a long way from noon. https://en.m.wikipedia.org/wiki/Apollo_11#/media/File%3AAldr...

Yeah, this was my immediate thought as well. Yes, you can't concentrate solar energy to heat something hotter than the surface of the sun. But the emission of a 100C blackbody is not visible either, so it's clearly an incorrect hypothesis on it's face that this temperature causes moonlight.

It's a bad mirror, not a radiator.

And it's a mirror that subtends a tiny fraction of the sky, and a tiny fraction of the sun's blurred virtual image.

For sure, the light is super dim. I'm not sure it's possible to focus the energy enough to light a fire. I'm just sure the reason is not because moonlight is entirely composed of radiation from a 100C blackbody. This is just immediately obvious because we can see it, so this whole article is a little weird.

>Moon surface temperature argument is incorrect

To the extent that the moon acts as a greybody under sunlight, it is correct. And like most things, the moon will be close enough to a greybody that you could use the surface temperature as a first order approximation. It isn't necessary of course, since you can easily just directly estimate the amount of scattered / re-radiated energy from the amount of sunlight falling on it, without needing to use the temperature.

>Hence the question is now - can we concentrate moon light enough so that intensity at the concentration point is higher than thermal loss into environment (only then we will be able to raise temperature in the concentration area enough for combustion - remember that light is "hot" enough for this)?

No, because the body you're trying to heat up will act as an approximate greybody, so (neglecting conduction and convection which could further lower the temperature) its temperature has to max out at the Stefan-Boltzmann temperature represented by the total radiation.

Leaving this comment to make sure that others don't get confused by your response.

1. As you mentioned, the moon acting as greybody absorbs part of the radiation and scatters the rest. In the second sentence you mention that the moon's temperature is enough to describe this radiation. In the third sentence you mention that you don't need moon's temperature to do that. So which is which?

Correct answer - to describe re-radiated energy we need moon's temperature, but to describe scattered we don't. We can ignore re-radiated, since it is not visible light and not hot enough. We can use scattered, since it is visible and hot enough.

2. Not sure why it is "No" if you are agreeing with me. Also not sure what you mean by temperature represented by total radiation (as you mentioned in part 1 there are two parts - re-radiated - at temperature of the moon, and scattered - at temperature of Sun).

To see that there is a problem with your argument - first note that the spectrum of black-body radiation dominates spectrum of grey-body radiation [1] (i.e. if black-body radiation is not in visible spectrum, the grey body will not be in visible as well). Then think that at 100 degrees C there would be no visible radiation. So the light that we see from moon cannot be this cold. Its coming from sun and it is hot enough (since it is in visible spectrum).

[1] https://claesjohnsonmathscience.wordpress.com/2012/04/21/gre...

>Correct answer - to describe re-radiated energy we need moon's temperature, but to describe scattered we don't. We can ignore re-radiated, since it is visible light and not hot enough. We can use scattered, since it is visible and hot enough.

When dealing with a gray-body, the equillibrium temperature of the body will be equal to the effective temperature of the incoming light, which will be equal to the effective temperature of the re-radiated + scattered light, since at equillbrium energy out equals energy in. So, assuming the moon is a graybody (and most objects tend to be roughly graybodies), we can use the surface temperature of the moon in our calculations instead of the effective temperature of the light that falls on it.

Suppose you are right. The effective temperature of the moon is 100 degrees Celsius. It is a grey-body, so its spectrum is dominated by black-body - e.g. it would be similar to spectrum of black-body with temperature lower than 100 degrees Celsius. So your argument is that the light we see from Moon is same as light we would see from a black body heated to less than 100 degrees C? Try putting a charcoal in boiling water and please report to us if you see it lighted up as white as the moon.

I think you're mixing up spectral temperature with effective temperature.

Effective temperature is about total power (per the Stefan Boltzmann law), not about the color of the light (per Wien's law).

Compare https://en.wikipedia.org/wiki/Effective_temperature with https://en.wikipedia.org/wiki/Color_temperature

The moon surface is not in equilibrium as there is thermal flow towards moon interior.

> And like most things, the moon will be close enough to a greybody that you could use the surface temperature as a first order approximation.

No, it won't, and no you can't. The OP already pointed out that the Moon is too cold for its blackbody radiation to reach the visible. All the visible light from the Moon is reflected from the Sun. The Sun's radiation's blackbody temperature is the ultimate limit, here, not the Moon's.

As I thought more of why one might get confused with this question is because one might miss that there are two effects here, not one.

To have a combustion we need two things - light of high enough temperature and light of high enough energy concentration. The two are not the same. All visible light has high enough temperature. However the concentration is the problem

Why do we need high concentration for combustion? If we don't supply enough energy to compensate for heat loss, the area will never get hot enough.

By the way - interesting corollary of this is that even with sun lights - if we have a mechanism that takes away heat fast enough (say by blowing cold air at the area) we would not be able to reach combustion with large mirror in direct sunlight.

"if we have a mechanism that takes away heat fast enough"

An interesting demonstration of this idea is putting a paper cup full of water into a fire. It won't burn where the water is touching it and eventually the water will start to boil.

Correct! Concentration is technically called energy flux, and is measured in W/m².

What you say about blackbody being a red herring makes sense to me.

What do you think of the argument at this link, regarding maximum concentration achievable through optics?

https://en.wikipedia.org/wiki/Etendue#Maximum_concentration

I get that that implies a maximum concentration factor of about 10,000, for a light source subtending an angle of 0.54 degrees (the moon's angular size from Earth, and also approximately the sun's.)

Dylan16807 helped to prod me in this direction. https://news.ycombinator.com/item?id=18739190

Right. We can't concentrate light brighter than it is on the surface of the Moon HOWEVER, because the light's effective temperature is much higher than the temperature of the Moon itself, we CAN use frequency-selective filters to heat up a sample, in principle to the point of combustion. Thermodynamics allows this. Basically, the greenhouse effect.

Looking through comments. There is still lots misunderstanding. So let's get deeper into the topic. In particular - how stuff gets heated by light and what light "temperature" is.

Thermodynamics arguments are always good but you really need to understand that most of them describe closed systems under equilibrium. Real world is messier and if your understanding of thermodynamics is shaky its easy to get to wrong conclusions. Black and grey body discussion is in same realm, so we will avoid it here.

Let's consider piece of wood and try to see what it takes to combust it. We need two things - oxygen and high enough temperature. We are keeping this piece of wood in air, so we have oxygen. The temperature is how fast particles that compose piece of wood are moving. The wood consists of molecules, which are in turn consist of atoms. You can imagine a molecule is a bunch of atoms connected with each other by springs (the springs are created by electromagnetic forces when atoms lend and borrow electrons). The higher the temperature - the larger oscillations of these springs. When the temperature is high enough some of these springs can break and combine with oxygen to release energy - combustion.

Suppose light strikes some piece of wood. What happens? Photons hit molecules and they can interact with an electron or proton - they exchange momentum (and energy), which means that one of the atoms in a molecule gets a bump - springs would start oscillating harder.

Note, low energy photons cannot swing springs a lot. We need photons of high energy to swing spring to high temperature (ok - this part is too oversimplified, I can extend on this if there are questions). The energy of photons depends on their waivelength - the shorter waivelength - the more energetic photons. So we need photons with energy that could break molecular bonds (springs) to be able to heat up to combustion temperature. Visible light definitely has photons of such energy.

So we start sending light down our piece of wood, it starts heating up. But it sits in air and probably fixed by some supporting stand. This piece of wood starts exchanging heat with surrounding materials. To be able to overcome this we need to send lots of photons on this piece of wood. How do we do this? We use lens to collect the photons from broader area and send it down to the wood. Note that it doesn't matter where photons came from, were they scattered or produced by sun, what temperature of scattering surface was, not even what temperature of sun was. All what is matter - can we collect enough high energy photons. So we would need to calculate what is energy flux of visible light photons at the surface earth and see if we could come up with a lens to focus these photons on piece of wood to create sufficient intensity.

We know that we can see Moon in visible light, so it does sends bunch of energetic photons to us. All these photon are coming from moon direction (so we don't care that moon scatters them in other directions as well). So the task is to find what is the energy flux of these photons at earth surface and what size of lens we would need to get sufficient intensity.

Your argument about low energy photons being capped at a black body level for heating is nonsense, a system can absorb all photons it can emit so typical matter can more or less absorb everything. Once energy from a photon has been absorbed it will diffuse in the material, increasing its temperature, energy quanta doesn't come into consideration here.

I am not an expert but on your oversimplification on the ‘High energy = higher wavelength’

Whilst this is true, it is also a bit irrelevant. Gamma rays are higher energy than visible light but you don’t need to focus gamma radiation (should that be easily possible) to start a fire, you can do it with visible light; or infrared light (which is lower wavelength again) at a high enough qanta

I don't have much to add, but I believe the poster you are replying to said that 'High energy = shorter wavelength'.

Equivalently, higher energy photons are higher frequency.

Higher energy = lower wavelength is not oversimplification. The formula is: photon energy is plank constan times speed of light divided by wavelength. You right - you don’t need gamma rays, but you cannot do it with low frequency radio waves. You need photons that energetic enough to break chemical bonds.

Depends on what you consider low frequency. Radio waves can be used to excite vibrational modes. The resultant thermal energy can be used to break chemical bonds.

Trying to think about this some.

If the Moon was actually a giant mirror-lens thing that was the size of the Moon, but perfectly shaped to reflect and concentrate sunlight shining on it into as small an area as possible on the Earth's surface, you'd start a fire with that very easily. Probably turn a county-sized area of the crust into lava or something actually.

On the other hand, if the Moon was a perfect reflector, but a big sphere/hemisphere pointed at the sun, then it would receive a lot of high-energy radiation, but reflect it back out in every direction all over the universe. I'm not sure of the math for that offhand, but it sounds difficult to concentrate it back to something even as concentrated as direct sunlight. I suppose it would involve considering the fraction of the total solid angle the reflecting area of the Moon would shine at. Even the entire Earth's surface would be a really tiny fraction of that area, so sounds impossible to concentrate things back to fire-starting intensity.

Assuming these scenarios are basically right, then the actual Moon is a lot closer to the second than the first. So starting a fire isn't practical with any lens system you could build on Earth. Maybe you could build some massive mega-lens thing in space near the moon that concentrated enough of it back on the Earth to be near-sunlight brightness. Or just set up a more manageable sized reflector on the Moon itself, or in space, to reflect sunlight in a more concentrated way to a small area on the Earth.

I think there may be an astronomer or two (at Lick or McDonnald?) which have 30"+ telescopes and actual human usable eye pieces (most only have CCD mounts), which could actually run this experiment.

I'm sure they don't want smoke in the observatory, but you could do it in a box. Then you could fill the box with variable oxygen levels (to make up for altitude) and substitute N2 with Ar/Kr/Xe to change the heat dissipation.

Spectral composition is completely irrelevant for typical material and temperatures, it is just watt per square meter which matters. You can't concentrate watt per square meter to a higher level than the light had at its quadratic falloff source. You can show that this is true by using the blackbody example, concentrating its light would break the laws of thermodynamic. Moonlight has quadratic falloff from the moon and not the sun and is therefore impossible to concentrate further so at best you can use lenses and mirrors to get the same light intensity the moonlight have on the Moon which of course isn't enough to heat anything above the temperature of the Moon.