6 ms·

Really? According to the Stefan-Boltzmann Law, the energy put out in blackbody radiation is proportional to temperature to the 4th power. Therefore something

by 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).
- sandworm101 8y ago
  All photons can start fires. They arent hotter or colder individual photons, just photons at different energy levels/wavelengths/relative velocity. Heat happens once photons collide with stuff. Heat is a group effort. Get enough photons to hit something and it will warm. Photon colour, and the reflectivity of the struck object, alters the needed number but with infinite photons fire (300*?) is always possible. 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.
  - gowld 8y ago
    If your fiber network could heat the match hotter than the moon, why wouldn't the match heat the moon instead? How do you cram all the ends of those fibers (with total surface area equal to the moon) into a target smaller than the moon? Fiber optics aren't lasers.
    - sandworm101 8y ago
      Because reversing the system isnt straitforwards. Fiber tends to bend light back towards the middle of the fiber. Shining light back down the middle doesnt mean it will reappear at the same point, which is important for many types of lasers. 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/ https://www.nanoptics.com/service/replacement-bowsight-fiber... The above are junk fiber (little internal lensing) but you can see the effect.
  - tbabb 8y ago
    Sorry, this is wrong. Please don't "explain" your misunderstanding as if you know the answer; ask questions instead. 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.
    - sandworm101 8y ago
      Not the system i was talking about. None of the fibers would "look at the moon". Fiber isnt the same as lenses. 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).
gnopgnip 8y ago
As a comparison the moon and sun are roughly the same size in the sky from earth. The sun puts out about 1000 watts per square meter on a clear day at noon in summer, the moon puts out about 0.0025 watts per square meter during full moon on a clear day, 1/400000 as much.
- btilly 8y ago
  Your figure for the energy from the Moon seems overstated. https://education.seattlepi.com/moonlight-strong-enough-power-solar-panels-4884.html https://education.seattlepi.com/moonlight-strong-enough-powe... quotes a ratio of 2.3 million to one. 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.
tbabb 8y ago
The angular concentration of the light. 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.
- colanderman 8y ago
  This is the primary reason. You can no more start a fire with sunlight reflected off the moon at night than with sunlight reflected off a sheet of paper during the day. (In fact you can do better with the sheet of paper, because you could in theory surround it with lenses to recapture the diffuse light.)

by btilly 8y ago

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).

sandworm101 8y ago
All photons can start fires. They arent hotter or colder individual photons, just photons at different energy levels/wavelengths/relative velocity. Heat happens once photons collide with stuff. Heat is a group effort. Get enough photons to hit something and it will warm. Photon colour, and the reflectivity of the struck object, alters the needed number but with infinite photons fire (300*?) is always possible. 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.
- gowld 8y ago
  If your fiber network could heat the match hotter than the moon, why wouldn't the match heat the moon instead? How do you cram all the ends of those fibers (with total surface area equal to the moon) into a target smaller than the moon? Fiber optics aren't lasers.
  - sandworm101 8y ago
    Because reversing the system isnt straitforwards. Fiber tends to bend light back towards the middle of the fiber. Shining light back down the middle doesnt mean it will reappear at the same point, which is important for many types of lasers. 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/ https://www.nanoptics.com/service/replacement-bowsight-fiber... The above are junk fiber (little internal lensing) but you can see the effect.
- tbabb 8y ago
  Sorry, this is wrong. Please don't "explain" your misunderstanding as if you know the answer; ask questions instead. 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.
  - sandworm101 8y ago
    Not the system i was talking about. None of the fibers would "look at the moon". Fiber isnt the same as lenses. 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).

gnopgnip 8y ago

As a comparison the moon and sun are roughly the same size in the sky from earth. The sun puts out about 1000 watts per square meter on a clear day at noon in summer, the moon puts out about 0.0025 watts per square meter during full moon on a clear day, 1/400000 as much.

btilly 8y ago
Your figure for the energy from the Moon seems overstated. https://education.seattlepi.com/moonlight-strong-enough-power-solar-panels-4884.html https://education.seattlepi.com/moonlight-strong-enough-powe... quotes a ratio of 2.3 million to one. 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.

tbabb 8y ago

The angular concentration of the light. 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.

colanderman 8y ago
This is the primary reason. You can no more start a fire with sunlight reflected off the moon at night than with sunlight reflected off a sheet of paper during the day. (In fact you can do better with the sheet of paper, because you could in theory surround it with lenses to recapture the diffuse light.)