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Can you use a magnifying glass and moonlight to light a fire? (2016)

what-if.xkcd.com

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Re: Can you use a magnifying glass and moonlight to light a fire? (2016)

#241

Post by nshepperd »

_0w8t wrote: post #217

The article reasoning can be shortened to observation that passive optical system does not change the wavelength of photons and to trigger a fire the wavelength has to be short enough. But the conclusion of the article is wrong. The surface temperature of the Moon has very little to do with the wavelength of the reflected photons. Consider a surface covered with ideal tiny mirrors each pointing to random direction. O…

That's incorrect. It has nothing to do with the wavelength of the light.

> Still the reflected light has original wavelength of the light of the Sun. Collect enough of it and that triggers fire.

You can't collect enough of it into one place with a passive optical system, because it's been irreversibly scattered by the moon's surface (Read: irreversible increase of https://en.wikipedia.org/wiki/Etendue).

Re: Can you use a magnifying glass and moonlight to light a fire? (2016)

#242

Post by Dylan16807 »

amluto wrote: post #218
Dylan16807 wrote: post #187

Earlier quoted context omitted.

> The target will be in a bath of 499-501nm light at 1/2 the intensity (energy density per unit volume) of the sun, which is far more than half the temperature of the sun. It’ll catch fire after a while. Unconcentrated sunlight is slightly under 1400 watts per square meter. It's equivalent to a temperature of 122C. You can concentrate sunlight coming from the sun, because the sun only fills five millionths of the sky…

> 700 watts per square meter doesn't set things on fire. It can only heat a blackbody to 60 degrees C. > (I'm ignoring the part about wavelength filtering because it's confusing and would only make your piece of paper heat up less.) I’m afraid you’re ignoring the interesting bit. You say that 700 W/m^2 doesn’t set things on fire. This is not true. Sure, 700 W/m^2 applied to some target that is allowed to radiate its…

I see. I couldn't quite follow which parts you were saying to insulate, and your mention of "half the temperature of the sun" lead me to misinterpret. That's fine then, I think. Not an expert on wavelength filters.

Re: Can you use a magnifying glass and moonlight to light a fire? (2016)

#243

Post by kurthr »

ummonk wrote: post #184
skolos wrote: post #182

Earlier quoted context omitted.

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?

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.

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.

Re: Can you use a magnifying glass and moonlight to light a fire? (2016)

#244

Post by kurthr »

skolos wrote: post #145

Interesting article. Right in many places. Wrong (possibly) in main conclusion. Entropy argument - correct in the sense that using radiation from black body we cannot use lenses to heat another body to the temperature higher than original. Easy to understand why - the first body has a temperature, radiation has the same temperature, if we apply the radiation to another object it will not heat up more than the radiati…

I think there may be an astronomer or two (at Lick or McDonnald?) which have 30"+ telescopes and actual human usable eye pieces (most only have CCD mounts), which could actually run this experiment.

I'm sure they don't want smoke in the observatory, but you could do it in a box. Then you could fill the box with variable oxygen levels (to make up for altitude) and substitute N2 with Ar/Kr/Xe to change the heat dissipation.

Re: Can you use a magnifying glass and moonlight to light a fire? (2016)

#245

Post by MertsA »

DenisM wrote: post #96
pierrebai wrote: post #37

Earlier quoted context omitted.

The region of higher temperature is much smaller. Entropy is not a pure measure of energy or randomness, but a measure of that within a volume . You'd have smaller entropy in a small volume, but higher around it because you would have diverted the ray that would have hit outside the concentrated region. Were is this reasoning wrong?

Yeah this is all rather confusing... This boils down to a moderately heated BB receiving an large stream of moderately powered photons and either rejecting them or first absorbing and then radiating them away at the same pace without changing its own temperature, regardless of how many photons are coming in .

Yeah I don't see how this argument holds water at all. If you have 10 kW worth of photons focused onto a square inch that absorbs 90% of those photons then that's 9 kW of power that isn't just going to magically disappear and it's not going to reach equilibrium until it's emitting 9 kW of power itself which is certainly going to be thousands of degrees. While the entropy argument is certainly an interesting thought experiment all that means is that the global entropy must somehow still be increasing.

Re: Can you use a magnifying glass and moonlight to light a fire? (2016)

#246

Post by AlexCoventry »

Dylan16807 wrote: post #240
AlexCoventry wrote: post #230

Earlier quoted context omitted.

From an optical perspective, how does the argument dismissed at the start of the OP break down? Is there some reason the moon's light, gathered from a lens covering hundreds of acres, carries insufficient energy to light a fire in concentrated form?

The optical argument goes like this: Measure the brightness (in energy per square meter) right at the moon's surface. No matter what you do with lenses, you can't concentrate moonlight beyond this level. You can make an enormous lens where all the energy coming off a particular acre of moon goes through it. But you can't focus all of it onto a spot smaller than an acre, no matter what you do.

Thanks for the explanations.

The current top comment here seems to agree with my original intuition, though.

https://news.ycombinator.com/item?id=18739120

Re: Can you use a magnifying glass and moonlight to light a fire? (2016)

#247

Post by empiricus »

darkmighty wrote: post #224
skolos wrote: post #201

Earlier quoted context omitted.

Black body photons are coming at random times. Laser photons are coherent, so are timed. Think of swing - if you try to push it randomly you'll get it swing as far as hardest push. But if you push periodically, you can swing it very far with small pushes.

That's not important at all; what's important is that black body radiation has a fixed maximum flux -- its spectrum or lack of coherence isn't why you can't heat another body to a greater temperature. It comes back to etendue, or if you prefer the 2nd law. You could reproduce any fixed black body spectrum (to arbitrary accuracy) from a set of thermal sources and filters (or a set of lasers, LEDs, etc. with random pha…

Interesting, this makes more sense. Now I have to read more on the conservation of entendue.

Re: Can you use a magnifying glass and moonlight to light a fire? (2016)

#248

Post by nkurz »

nshepperd wrote: post #239
nkurz wrote: post #180

Earlier quoted context omitted.

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 conduct…

> 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,…

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"?

Re: Can you use a magnifying glass and moonlight to light a fire? (2016)

#249

Post by Dylan16807 »

AlexCoventry wrote: post #246
Dylan16807 wrote: post #240

Earlier quoted context omitted.

The optical argument goes like this: Measure the brightness (in energy per square meter) right at the moon's surface. No matter what you do with lenses, you can't concentrate moonlight beyond this level. You can make an enormous lens where all the energy coming off a particular acre of moon goes through it. But you can't focus all of it onto a spot smaller than an acre, no matter what you do.

Thanks for the explanations. The current top comment here seems to agree with my original intuition, though. https://news.ycombinator.com/item?id=18739120

As a few people pointed out, the reflection off the moon ruins the étendue, so there's no concentrating it beyond what all the rocks on the surface already experience.
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