This is an interesting argument. Can I not reflect some sunlight off a mirror, then do the magnifying-glass-to-start-a-fire trick in daytime? Doesn't the mirror stay cool? Isn't the moon just a (poor) mirror for the sun's light?
Your mirror in sunlight works because the reflectivity or albedo of the mirror is very high relative to whatever target you're lighting on fire.
In a magical closed system where radiative heat transfer was the only factor, objects of differing reflectivity would eventually reach temperature equilibrium through black body radiation.
We aren't interested in closed systems though. The moon's temperature is set by the equilibrium between incident solar radiation and black body emission, most of which flies off into deep space making the system very open. Just like your mirror's temperature is the equilibrium of incident radiation, black body cooling, and convective cooling in Earth's atmosphere.
If the moon had a high reflectivity and/or a powerful cooling mechanism like convection, its equilibrium temperature would be far lower than the temperature of its emitted + reflected light. Unfortunately the moon's albedo is just 0.12 and black body radiation is all it's got, so the modest difference between its temperature and that of its light isn't enough to start a fire.
If this seems to fly in the face of all common experience, it's because things that look white aren't actually white in the infrared. Their emissivity is lower in the optical spectrum (where they absorb sunlight) than in the infrared (where they emit), so they are cooler than things that look black to us. In short, they aren't grey-bodies.
The situation is also more complicated because, as you say, that things we are used to also cool by other mechanisms.
According to the Stefan-Boltzmann Law, the energy put out in blackbody radiation is proportional to temperature to the 4th power. Therefore something that is 5000 degrees reflected off of an object with albedo 0.12 is putting out 0.12 times the energy it originally did, while something that is 2500 degrees only puts out (1/2)^4 = .0625 times as much. So the "temperature of the Moon's light" should be more than hot enough to light something on fire if it is focused right.
As a sanity check, compare how much light the moon puts out as a black body in shadow with what it reflects from the Sun. As another sanity check, compare how bright the Moon is versus a fire.
What am I missing here?
The moon is diffuse, so an incoming ray of sunshine is spread by the optically rough surface of the moon from an incident solid angle of 6 * 10^-5 steradians out into 2 pi steradians of the night sky, or a reduction in angular concentration by a factor of about 100,000 (totaling ~1 million after the albedo is accounted for).
It is like you are looking at the sun through a mirror so rough that the image of the sun is blurred over literally half the sky. Because this process does not create new photons, the blurred image must be far, far dimmer. This circumstance corresponds to the "most we could do" with lenses and mirrors focusing the moon, which is to fill the sky with an image of the moon/"blurred sun".
Unconcentrated moonlight corresponds to the same picture, except we only see a "cutout" disk of this blurred sun-image which is the size of the moon in the sky. Our crappy moon-mirror does not fill our vision, it is a porthole letting through only a tiny fraction of the blurry sun-image.
And of course if you imagine yourself as the ant under the magnifying glass, with your entire sky filled with moon, there is no way you could spontaneously become hotter than your moon-y surroundings.
However the key point is this. The Moon is in reality 400 times smaller than the Sun. Which means that an optimally placed lens can actually make the image of the Moon 400 times smaller, for 160,000 times as little area. If the lens is big enough that this target area is mostly losing heat through black body radiation, again it should wind up over half the temperature of the surface of the Sun. Which is more than hot enough to start a fire.
I don't understand this bit. I don't think there's any power law involved in reflection, if something has an albedo of 0.12 then it just reflects 0.12 times the incident radiation, doesn't it?
I think I agree with your overall point, though, which is that the moon isn't a black body radiator (well it is but only at a couple of hundred degrees C at most) but is just reflecting the sun's light (and those photons are hot enough to start fires).
Fiber optics could probably collect and point enough moonlight to light a match. If we call fiber a sort of flexible lens, then lenses can start fires.
If we imagine a full sphere of inward-pointing fibers, each one "looking" at the moon, then we see moon-surface in all directions from within this contraption. We are in a thermal bath of moon-temperature. We will not get hotter than the moon.
And fibers are not needed to create this circumstance. The same situation ("moon visible in all directions") could be created with a few lenses and mirrors.
The problem with sunlight reflected off the moon is that it's like having a very low transmittance ND filter in the optical path. There aren't a whole lot of photons entering the aperture, and as the material begins to heat, there are more photons and overall probably more energy going out. (I'm not giving an answer here, just some thoughts.)
Note too that i said match. Getting tiny bit of a matchhead hot enough to decompose isnt the same as burning an ant. A matchhead on the surface of the moon would probably ignite just fine (150*+o2).
(Well, the fiber you’re thinking of is. Fluorescent fiber is not, but that’s another story.)
> None of the fibers would "look at the moon". Fiber isnt the same as lenses.
I think you meant to ask "how is a fiber different from a lens?" instead of asserting "fiber isn't the same as lenses," which turns out to be completely wrong.
To a physicist, a lens is a shaped piece of refractive medium-- that is exactly what a fiber is. There is no magic inside the fiber; it does not add any photons. You are looking through a shaped piece glass, through which you will see a (possibly very distorted) image of what's on the other side.
In this case, it's the moon that is on the other side. The most you could do is surround your match/ant/whatever with fibers/lenses which are showing the moon on the other side of them. And from this fact it is unavoidable that you cannot make your surrounded subject hotter than the surface of the moon. If you are trying to say anything to the contrary, I'm sorry, but you are Flat Wrong.
Any other configuration of fiber would give you less moonlight than completely surrounding the subject, so it's not going to improve the odds of starting a fire.
> Note too that i said match. Getting tiny bit of a matchhead hot enough to decompose isnt the same as burning an ant.
I think you meant to ask "does the material or shape change anything?" instead of asserting "it isn't the same," which again, in the context of this question, turns out to be completely wrong.
The maxiumum temperature that can be imparted by a lens/mirror system doesn't depend on what it's focused on at all. In the long run, the ant/match/whatever will reach that maximum temperature and not get hotter. From a thermodynamics perspective, it is like putting an object in an oven of a particular temperature-- It doesn't matter if you put in a brick, or a cake, or match, or an ant; eventually they will all be 250 degrees if the oven is set to 250 degrees. In this case, the "oven" is the moon, and it's set to about 120C. A match combusts at about 600C. There will be no combusting of the match by putting it in the moon-oven. Sorry. End of story.
And if I can give some advice: If what you're hearing doesn't make sense to you, it is always safer to ask a question than to blindly assert your gut-guess of how you think it is. Someone who thinks they're an expert but spouts nonsense looks like a fool. Someone who asks a question, though, looks curious, which is smart. These two people have the same degree of knowledge, but one comes off looking much worse than the other. That's why asking questions is better, especially when you don't know the knowledge of the people who are listening.
Archers see this. Many archery sights use fiber to make nice illuminated dots, without batteries or leds. The light appearing out the end of the fiber is brighter than the skin, a rare practical use of "naked" optical fibers.
https://www.nanoptics.com/service/replacement-bowsight-fiber...
The above are junk fiber (little internal lensing) but you can see the effect.
No, reflection and black-body emission are different. The moon primarily produces the former.
Moonlight is not the result of the moon glowing incandescent.
The surface is very cold, its emissions as a black-body radiator are in the far infrared.
Moonlight is white light with a color temperature of several thousand Kelvin. This is reflected sunlight and has nothing to do with black-body radiators.
You could start a fire from the light of a single star if you had a big enough lens. You could also start a fire with the light from a hand mirror at the distance of the moon if it reflected the sun's light at you. But you'd need a very big lens.
That's my impression as well. A blackbody has albedo 0. The moon has an albedo of around 0.12. While I suspect you can't start a fire from moonlight in practice, I don't think the arguments in this article are correct.
I guess if you're reading that as some kind of absolutely logical argument. I read stuff like that as an abstraction which just kinda works in the messy real world. Pretty much like how the typical explanation for how a wing works turns out to be an over-simplification. It only partly works that way. Actual wings are complicated, but in the aggregate, they just get enough air molecules to go downward to net out the forces to keep the plane from going downward. It turns out that there are a lot of mechanisms contributing to this all at once. (Which is something else he discusses in that series.)
Ivanpah may be able to, I wonder if we can borrow it at night for an experiment - https://en.wikipedia.org/wiki/Ivanpah_Solar_Power_Facility
edit - And by whom. Should probably get it in early, just in case.
It's mostly not reflected. It's mostly scattered. If it were specularly reflected, there would be a bright spot on the moon (where you could see the reflection of the sun), and you'd be able to concertate light from that to light a fire.
The only reason the moon's surface doesn't reach such high temperatures is that the moon's surface is not thermally isolated from the rest of the night sky or from the body of the moon itself (which is also not isolated from the night sky).
The more general argument based on etendue, which applies to the moon, is: you can't make the incoming light any brighter than it is on the surface of the source (which doesn't have to be the original emitter, but can be any point along the path of the light). As a corollary this happens to mean you can't really make something hotter than a rock on the moon.
Suppose the moon was actually a flat mirror. Standing on the mirror-moon, you look at the ground. In most directions you would see the darkness of space (with a few reflected stars), but in one spot you would see about ~(0.5°)^2 solid angle of extreme brightness - the sun's reflection in the mirror. Standing on the mirror-moon, you could certainly use a magnifying glass to heat something to ignition temperature (ignoring the lack of oxygen) by making its environment that bright using the reflected light.
Similarly, if the moon was a mirror, you could certainly use the reflected light to start a fire standing on earth (provided you're lucky enough for the moon/earth/sun to line up just right so that the reflection of the sun is visible through small angle subtended by the moon from earth). It would basically appear as another sun in the sky when lined up properly.
In reality, instead of reflecting light like a mirror, the moon scatters light in all directions. Standing on the actual moon, looking at the ground, what you see is a lot of moon dirt, all of which about equally bright (ie. not very). The scattering smooshes the sun's light out in all directions, ensuring there's no visible "reflection" of the sun in any direction.
The most you can do with this scattered moon light is make the environment of an object as bright as the (mediocre) brightness of the moon rocks. But moon rocks already experience that environment of mediocre brightness, and reach only 100°C, so you won't be able to make your object much hotter than that.
I'm bothered by the modifer "much". If you are indeed talking about a physical principle, shouldn't this be an absolute limit rather than a suggestion? How much hotter does physics allow you to go? Are you sure it's not enough to allow ignition?
Along those lines, I'd assume that the surface temperature of the depends on the moon's shape and thermal conductivity. If I were to change the moon to be an ultra-thin and highly heat conductive hemispherical shell rather than a solid sphere, I'd assume the surface temperature would drop.
Assuming the amount of light reflected remains the same, does this imply that the maximum achievable temperature on earth with a magnifying glass drops as well? I don't see any physical reason that it should, but your logic would seem to imply that it must. Can you explain?
It's an absolute limit on the amount of incoming irradiance you can create to your object. The actual equilibrium temperature it reaches will depend on additional factors like how well your object loses heat (eg. by conduction) compared to a moon rock.
In this case, the temperature of moon rocks is probably a reasonable upper bound of the achievable temperature of an object on the earth:
- Moon rocks are in vacuum, while something on earth is in contact with air and dissipating heat by convection.
- Moon rocks are in contact with the surface of the moon (~100°C), whereas an earth object is in contact with the ground, or your hand, or whatever (~37°C, assuming your hand). So heat loss by conduction will be greater on the ground.
If you rigged up something to suspend your object in vacuum without touching anything so that conductive heat losses ~0, maybe you could get something slightly hotter than the average surface temperature of the moon. But not hotter than a well placed moon rock that already happens to be making near 0 contact with the moon's surface (due to standing on a point or something).
> If I were to change the moon to be an ultra-thin and highly heat conductive hemispherical shell rather than a solid sphere, I'd assume the surface temperature would drop.
In that case much more heat would escape around to the unlit side, and the moon's surface temperature would reach somewhere between the "day" (~100°C) and "night" (~-200°C) temperatures. Say around -50°C. In that case the surface temperature will be less representative of that achievable for an object on earth. A moon rock touching the ground would be in contact with -50°C, which is colder than the 37°C for an object held in your hand.
[The surface temperature of the moon is] an absolute limit on the amount of incoming irradiance you can create to your object.
This is true for a black body, but why are you convinced this is true for the actual moon? I think we agree that a more reflective moon could have a lower surface temperature while increasing incoming irradiance on the earth. And we both agree that the moon is partially reflective. Doesn't this mean that the surface temperature is not an absolute limit?
I think the correct statement is that the intensity of light from the sun to the moon gives a limit on both the surface temperature of the moon (highest if we assume the moon is a blackbody) and a limit on the amount of sunlight reflected toward the earth (highest if we assume the moon is a perfect reflector).
Since the moon absorbs about 90% of the light incident on it, we can assume that the surface temperature is lower than it would be if it was a perfect black body, presumably reaching a temperature corresponding to a sun that was about 10% less strong. The 10% of light that is reflected, although diffused in all directions, is much more intense when viewed from earth than low energy blackbody radiation that is also emitted. We know this intuitively because the sunlit moon is much brighter at night than the non-sunlit portion, and because the visible light is more energetic than the infrared, but could integrate across the energy spectrum to find an exact answer.
As such, unless we are willing to make some additional assumptions, I don't think we can make any firm claim about the the maximum temperature achievable on the earth using lunar reflected sunlight based only on knowledge of the surface temperature of the moon. In practice, the scattered sunlight doesn't provide a lot of energy, so heating with it will be difficult. But it's energy incident on the earth that matters, not the temperature of the lunar surface.
Would you agree with this summary? Are there additional assumptions that you think should be added that would provide the tighter limit you want? Alternatively, is there something other than "[The surface temperature of the moon is]" that you think I should have substituted for "It's"?
It's the radiance (brightness) of the moon (as perceived at the moon) that is an absolute limit on the incoming irradiance you can create, because of conservation of etendue.
Separately, the fact that moon rocks (which experience that exact amount of irradiance) reach some given temperature X°C while losing very little heat to conduction (certainly less than an object on earth would), shows that this level of irradiance is insufficient to heat up your kindling above X°C.
This argument has nothing to do with the moon being a black body, or an approximate black body, or any such thing. Just the fact that moon rocks, which are exposed to this light (with very little heat conductive losses), function as a kind of a thermometer which tells you how hot that light can make something[1]; and the answer is "a bit over 100°C"
> Since the moon absorbs about 90% of the light incident on it, we can assume that the surface temperature is lower than it would be if it was a perfect black body, presumably reaching a temperature corresponding to a sun that was about 10% less strong.
No... An object with 90% absorbance absorbs 10% less light energy, yes. But it also emits 10% less light energy, so the two effects cancel out and it reaches the same equilibrium temperature as a black body.
Well yes, this is what it should say! Is this our disagreement? Because for me (and I think for most other dissenters in this thread) the whole problem we have with Munroe's argument is that he keeps coming back to the surface temperature of the moon as the limiting factor. If he was simply to say that the moon is not bright enough, then we'd probably all agree.
the fact that moon rocks ... reach some given temperature X°C ... shows that this level of irradiance is insufficient to heat up your kindling above X°C.
I think this is the real point of dispute. We agree that this is true if we are only considering pure blackbody radiation. What's not clear (at least to me) is that this equivalence is still true when you include the directly reflected light. That is, no one thinks that you can start a fire using only the thermal infrared light from a dark moon. The question is whether it's hypothetically possible with a sufficiently bright sun and sufficiently reflective moon, without raising the surface temperature. Can you point to something that makes this argument more directly?[1]
the two effects cancel out and it reaches the same equilibrium temperature as a black body
I need to learn more about this. I have trouble thinking it applies correctly here, because it's assuming the moon is a perfect gray body. I think this assumption falls apart if it's actually reflecting light, which in fact we know it is. Or am I wrong? Does a silver mirror in space actually end up at the same equilibrium temperature as a lump of coal? I guess it could. This wouldn't harm much argument (the argument just requires that the temperature not increase), but would indicate that I'm not viewing things correctly.
Summarizing, I think the point of dispute is whether the surface temperature of an object in space can always be reasonably estimated from its brightness (and vice versa). We agree that it can be if it's a perfect black body. We agree that it's mathematically true if it's a "gray body". We disagree (I think) as to whether it's appropriate to make the simplification of assuming that all stellar objects are sufficiently close to "gray bodies" for the math to hold.
[1] Here's the outline of the counterargument. Start with a blackbody moon. Estimate that with perfect optics you can heat an object to X. Now increase the reflectivity of the no-longer-black-body, noting that the surface temperature does not increase. I'd argue that when you increase the reflectivity, the moon gets brighter, and thus you can heat your object to a higher temperature. You seem to be arguing that because the surface temperature remains the same, the attainable heat stays the same, even though you can collect more reflected energy.
Yes. It takes longer to reach the equilibrium, but it does reach the same equilibrium temperature.
> [1] Here's the outline of the counterargument. Start with a blackbody moon. Estimate that with perfect optics you can heat an object to X. Now increase the reflectivity of the no-longer-black-body, noting that the surface temperature does not increase.
Right, so, let's say the total solar power received by the blackbody moon is Ps. The moon increases in temperature until it reaches temperature Tb, at which the power emitted equals the power absorbed Pb = Ps. That is the equilibrium.
Now let's let the moon reflect (scattering) some light, by giving it a realistic albedo of 10%. Then of the solar power received at the moon, 0.9Ps is absorbed and 0.1Ps is scattered. But this albedo also causes the moon to emit 10% less light, so it now emits 0.9Pb at the same temperature Tb, hence it reaches equilibrium 0.9Ps = 0.9Pb at the same temperature.
Note that the total light energy leaving the moon is now: 0.9Pb (emitted thermal) + 0.1Ps (scattered) = 0.9Ps + 0.1Ps = Ps. Just the same as the amount that was emitted as a blackbody. So in fact it's not any brighter (in terms of power), just has a different spectrum.
The first is that the applicable "mirror" question here whether the surface temperature of the sunlit side of a slow rotating astronomical body is independent of reflectivity, not whether the eventual core temperature of a uniformly lit body is independent. Are you confident that the same reasoning applies? I'm not yet.
Next, if we are phrasing our question as to whether one can start a fire with a magnifying glass, we clearly do care about the difference between visible sunlight and low temp infrared. For Munroe's argument to really work, the surface temperature needs to be proxy for the collectible visible light that would be used by a magnifying glass.
There's also the directionality: none of the visible light is going to be reflected toward the "dark" side. The thermal radiation is also directional, but not to the same extent. As we move toward greater reflectivity, presuming a bright full moon, we do get more of the total energy available on earth. How much more? I don't know.
Lastly, which we haven't discussed, there are options for insulating the heated object on earth that are optically transparent (to allow the concentrated light in) but infrared reflective (to prevent thermal radiation from escaping). I think this "privileges" the reflected sunlight so that we might indeed be able to achieve a higher temperature than the reflective moon surface in vacuum.
I feel like you recognize these factors also, by your caveats that you might be able to get "a little bit" higher than the surface temperature. Without committing to a number or methodology, your implication is that this "little" must is small relative to the surface temperature of the moon, rather than small relative to the optical temperature of the sun. While I agree that the surface temperature is related to the achievable temperature for collected reflected light, I still don't think that the exact temperature is a hard limit.
I'll try to bow out here and not take up more of your time. You've definitely helped me to think through the issues here. If you happen to be interested in a somewhat parallel situation, you might enjoy this article that describes a cyclical water collection system based on collecting solar energy through an aerogel during the day, then using a condenser optically coupled to the dark sky at night: https://www.nature.com/articles/s41467-018-03162-7. Only loosely related to Munroe's hypothetical, but shows a real application of some of the same concepts.
Edit: I just noticed a link on the second page of this thread that had some useful discussion that might interest you: https://physics.stackexchange.com/questions/370446/is-randal.... If you expand out all the comments on the answer, Shor seems to be making the same argument you are, and Lalinský is making a better version of mine.
The mirror is a part of you hybrid reflective/refractive light concentration system. Of course it stays cool, otherwise your system is inefficient.
Doesn't the mirror stay cool? Isn't the moon just a (poor) mirror for the sun's light?
Precisely. So the maximum temperature that system can achieve is much less that your mirror/magnifying glass system. A perfect black body radiator is just about the poorest mirror there is. The moon's light isn't what you'd get from a perfect black body. However, it's a long, long ways away from being a perfect mirror and probably far closer to being a black body.
Why? If the moon were a perfect black body, but with a few small perfect mirrors on the surface to bounce a bit of sunlight towards Earth, couldn't we (in theory) focus that tiny bit of sunlight to heat something up to the temperature of the Sun? There may not be enough solar radiation to make it feasible in practice, but it seems like you need to actually do the math instead of just saying "the moon's surface temperature isn't high enough to start a fire, Q.E.D."
To get past the diffraction limits, those "few small perfect mirrors" on the Moon would have to be gargantuan. (If I'm remembering correctly, they'd be the size of the alien space ships in Star Trek that make the Enterprise-D look teeny.) Otherwise, viewed from the surface of the Earth, those mirrors would just constitute a minuscule brightening of the sky in the direction of the Moon.
Dandy.
Are you saying that mirrors do something other than absorbing and re-radiating light? In most explanations I've seen, this is exactly what they do:
How does the mirror reflect light? The silver atoms behind the glass absorb the photons of incoming light energy and become excited. But that makes them unstable, so they try to become stable again by getting rid of the extra energy—and they do that by giving off some more photons. (You can read about how atoms take in and give out photons in our article about light.) The back of a mirror is usually covered with some sort of darkly colored, protective material to stop the silver coating from getting scratched, and also to reduce the risk of any light seeping through from behind. Silver reflects light better than almost anything else and that's because it gives off almost as many photons of light as fall on it in the first place. The photons that come out of the mirror are pretty much the same as the ones that go into it.