5 ms·

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

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.

skolos 8y ago
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?
- ummonk 8y ago
  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.
  - lultimouomo 8y ago
    Thanks, this is what I was missing and what I think is most obscure in the article!
  - skolos 8y ago
    Perfect reflector exists only in theory. But we still put (imperfect) reflective coating on spacecrafts.
  - fpoling 8y ago
    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.
    - skykooler 8y ago
      The moon is very imperfect; it's darker than asphalt. http://wtamu.edu/~cbaird/sq/2015/08/06/why-is-the-moon-so-bright/ http://wtamu.edu/~cbaird/sq/2015/08/06/why-is-the-moon-so-br...
    - darkmighty 8y ago
      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.
  - kurthr 8y ago
    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.
    - ummonk 8y ago
      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 https://en.wikipedia.org/wiki/Opposition_surge
      - kurthr 8y ago
        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 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.
- klodolph 8y ago
  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.
smaddox 8y ago
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.
- klodolph 8y ago
  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”.
Retric 8y ago
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%3AAldrin_Apollo_11_original.jpg https://en.m.wikipedia.org/wiki/Apollo_11#/media/File%3AAldr...
deleted 8y ago
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skolos 8y ago

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?

ummonk 8y ago
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.
- lultimouomo 8y ago
  Thanks, this is what I was missing and what I think is most obscure in the article!
- skolos 8y ago
  Perfect reflector exists only in theory. But we still put (imperfect) reflective coating on spacecrafts.
- fpoling 8y ago
  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.
  - skykooler 8y ago
    The moon is very imperfect; it's darker than asphalt. http://wtamu.edu/~cbaird/sq/2015/08/06/why-is-the-moon-so-bright/ http://wtamu.edu/~cbaird/sq/2015/08/06/why-is-the-moon-so-br...
  - darkmighty 8y ago
    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.
- kurthr 8y ago
  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.
  - ummonk 8y ago
    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 https://en.wikipedia.org/wiki/Opposition_surge
    - kurthr 8y ago
      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 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.
klodolph 8y ago
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.

smaddox 8y ago

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.

klodolph 8y ago
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”.

Retric 8y ago

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%3AAldrin_Apollo_11_original.jpg https://en.m.wikipedia.org/wiki/Apollo_11#/media/File%3AAldr...

deleted 8y ago

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