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> You can't use lenses and mirrors to make something hotter than the surface of the light source itself. This is an interesting argument. Can I not reflect som

> You can't use lenses and mirrors to make something hotter than the surface of the light source itself. 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?

nealabq 8y ago
If the light is being radiated from a black body surface, then you cannot make something hotter than that surface. But you're right, the light from the moon is reflected sun light, plenty hot to start a fire. The moon also radiates like a black body, but virtually all that light has longer wavelengths than the reflected visible light. 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.
- baddox 8y ago
  > But you're right, the light from the moon is reflected sun light, plenty hot to start a fire. 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.
  - stcredzero 8y ago
    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.)
    - baddox 8y ago
      It’s presented as an absolute logical argument of the form “you can’t use a lens to make anything hotter than the average surface of the moon.”
      - stcredzero 8y ago
        And it "kinda" works in the same way the simplified Bernoulli explanation of the wing works.
  - starbeast 8y ago
    >While I suspect you can't start a fire from moonlight in practice 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 https://en.wikipedia.org/wiki/Ivanpah_Solar_Power_Facility
    - alanh 8y ago
      now this is an idea i like.
      - starbeast 8y ago
        I wonder what the protocol is for 'borrowing' Ivanpah to point it at the moon at night and at what point in the exchange the phrase, 'well, you won't be using it', will finally occur. edit - And by whom. Should probably get it in early, just in case.
- heyjudy 8y ago
  Thunderf00t should totally make a lens to start a fire with starlight.
  - AlexCoventry 8y ago
    I think you'd need a different argument to show that's impossible, since the surfaces of stars are very hot, and their light is unmediated.
- ummonk 8y ago
  >But you're right, the light from the moon is reflected sun light, plenty hot to start a fire. 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.
  - smaddox 8y ago
    With a sufficiently isolated vacuum flask containing the tender and with an optical opening only large enough to view the moon, you wouldn't even need to concentrate the scattered light from the moon. The temperature of the inside of the flask would eventually reach the approximate blackbody temperature of the reflected light from the Sun (corrected for whatever frequencies are absorbed by the Moon). 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).
stcredzero 8y ago
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? 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.
- saagarjha 8y ago
  The moon is about as reflective as asphalt.
  - stcredzero 8y ago
    So it's approximately a black body. Even asphalt has tiny bits of specular reflection coming off of it.
- baddox 8y ago
  > So the maximum temperature that system can achieve is much less that your mirror/magnifying glass system. 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."
  - stcredzero 8y ago
    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? 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.
nickparker 8y ago
You've found the right question to ask. 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.
- btilly 8y ago
  Really? 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?
  - taneq 8y ago
    > 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 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).
deleted 8y ago
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ummonk 8y ago
This works because you're just concentrating the image of the sun. It's the surface temperature of the sun that matters. Now if, instead, the mirror were scattering or absorbing and re-radiating the light (as the moon does), then its temperature would be limiting factor.
- nkurz 8y ago
  Now if, instead, the mirror were scattering or absorbing and re-radiating the light (as the moon does) 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. https://www.explainthatstuff.com/howmirrorswork.html https://www.explainthatstuff.com/howmirrorswork.html
  - ummonk 8y ago
    I haven't done material physics in depth, so can't be 100% certain of this, but I don't see how that can be true. Electromagnetic reflection is what results in either specular reflection or diffuse reflection (i.e. scattering), not absorbing and re-radiating. Which is why the color of reflected light is dependent on the color of incoming light, and not on the spectral frequencies of the material (unlike re-radiated light).
nshepperd 8y ago
The thermodynamic argument applies to black bodies, and says that you can't make an object hotter than the surface of the emitter. That's pretty uncontroversial. 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.
- nkurz 8y ago
  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?
e12e 8y ago
The other obvious question is: could we mold one of the moon craters to a gigantic parabola and coat it to be (much more) reflective? And have the moon shine a concentrated death ray down to earth? We'd obviously have trouble aiming - but it should make for a spectacular experiment....

nealabq 8y ago

If the light is being radiated from a black body surface, then you cannot make something hotter than that surface. But you're right, the light from the moon is reflected sun light, plenty hot to start a fire. The moon also radiates like a black body, but virtually all that light has longer wavelengths than the reflected visible light. 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.

baddox 8y ago
> But you're right, the light from the moon is reflected sun light, plenty hot to start a fire. 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.
- stcredzero 8y ago
  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.)
  - baddox 8y ago
    It’s presented as an absolute logical argument of the form “you can’t use a lens to make anything hotter than the average surface of the moon.”
    - stcredzero 8y ago
      And it "kinda" works in the same way the simplified Bernoulli explanation of the wing works.
- starbeast 8y ago
  >While I suspect you can't start a fire from moonlight in practice 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 https://en.wikipedia.org/wiki/Ivanpah_Solar_Power_Facility
  - alanh 8y ago
    now this is an idea i like.
    - starbeast 8y ago
      I wonder what the protocol is for 'borrowing' Ivanpah to point it at the moon at night and at what point in the exchange the phrase, 'well, you won't be using it', will finally occur. edit - And by whom. Should probably get it in early, just in case.
heyjudy 8y ago
Thunderf00t should totally make a lens to start a fire with starlight.
- AlexCoventry 8y ago
  I think you'd need a different argument to show that's impossible, since the surfaces of stars are very hot, and their light is unmediated.
ummonk 8y ago
>But you're right, the light from the moon is reflected sun light, plenty hot to start a fire. 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.
- smaddox 8y ago
  With a sufficiently isolated vacuum flask containing the tender and with an optical opening only large enough to view the moon, you wouldn't even need to concentrate the scattered light from the moon. The temperature of the inside of the flask would eventually reach the approximate blackbody temperature of the reflected light from the Sun (corrected for whatever frequencies are absorbed by the Moon). 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).

stcredzero 8y ago

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? 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.

saagarjha 8y ago
The moon is about as reflective as asphalt.
- stcredzero 8y ago
  So it's approximately a black body. Even asphalt has tiny bits of specular reflection coming off of it.
baddox 8y ago
> So the maximum temperature that system can achieve is much less that your mirror/magnifying glass system. 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."
- stcredzero 8y ago
  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? 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.

nickparker 8y ago

You've found the right question to ask. 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.

btilly 8y ago
Really? 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?
- taneq 8y ago
  > 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 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).

deleted 8y ago

[deleted]

ummonk 8y ago

This works because you're just concentrating the image of the sun. It's the surface temperature of the sun that matters. Now if, instead, the mirror were scattering or absorbing and re-radiating the light (as the moon does), then its temperature would be limiting factor.

nkurz 8y ago
Now if, instead, the mirror were scattering or absorbing and re-radiating the light (as the moon does) 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. https://www.explainthatstuff.com/howmirrorswork.html https://www.explainthatstuff.com/howmirrorswork.html
- ummonk 8y ago
  I haven't done material physics in depth, so can't be 100% certain of this, but I don't see how that can be true. Electromagnetic reflection is what results in either specular reflection or diffuse reflection (i.e. scattering), not absorbing and re-radiating. Which is why the color of reflected light is dependent on the color of incoming light, and not on the spectral frequencies of the material (unlike re-radiated light).

nshepperd 8y ago

The thermodynamic argument applies to black bodies, and says that you can't make an object hotter than the surface of the emitter. That's pretty uncontroversial. 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.

nkurz 8y ago
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?

e12e 8y ago

The other obvious question is: could we mold one of the moon craters to a gigantic parabola and coat it to be (much more) reflective? And have the moon shine a concentrated death ray down to earth? We'd obviously have trouble aiming - but it should make for a spectacular experiment....