4 ms·

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 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?
- nshepperd 8y ago
  > 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? 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.
  - nkurz 8y ago
    Thanks for the reply, and I think I agree with all the physical processes you describe, but I'm not convinced that your approximations are correct. I'm going to keep pushing a bit to see if we can resolve this as well. [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"?
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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?

nshepperd 8y ago
> 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? 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.
- nkurz 8y ago
  Thanks for the reply, and I think I agree with all the physical processes you describe, but I'm not convinced that your approximations are correct. I'm going to keep pushing a bit to see if we can resolve this as well. [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"?

deleted 8y ago

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