Can you use a magnifying glass and moonlight to light a fire? (2016) • Hacker News Board

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

#251

Post by sopooneo » Sat, Dec 22, 2018, 8:21 PM UTC

robbiep wrote: post #209

skolos wrote: post #197

Earlier quoted context omitted.

Looking through comments. There is still lots misunderstanding. So let's get deeper into the topic. In particular - how stuff gets heated by light and what light "temperature" is. Thermodynamics arguments are always good but you really need to understand that most of them describe closed systems under equilibrium. Real world is messier and if your understanding of thermodynamics is shaky its easy to get to wrong conc…

I am not an expert but on your oversimplification on the ‘High energy = higher wavelength’ Whilst this is true, it is also a bit irrelevant. Gamma rays are higher energy than visible light but you don’t need to focus gamma radiation (should that be easily possible) to start a fire, you can do it with visible light; or infrared light (which is lower wavelength again) at a high enough qanta

I don't have much to add, but I believe the poster you are replying to said that 'High energy = shorter wavelength'.

Equivalently, higher energy photons are higher frequency.

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

#252

Post by bscphil » Sat, Dec 22, 2018, 8:34 PM UTC

ummonk wrote: post #164

bscphil wrote: post #50

Earlier quoted context omitted.

The interesting thing to me is that Randall describes a very similar system just a few What-Ifs earlier: https://what-if.xkcd.com/141/ Here he says very clearly that if you "bundled" all the light from the sun and aim it at the earth it would heat the atmosphere to millions of degrees (the surface of the sun is much less than that). It's not at all clear to me what he means by "bundled" and why it's not contradictory…

Yeah, it does contradict that. You can't really "bundle" all the light from the sun like that and aim it at the Earth without violating thermodynamics. The way to do it would be something like wrapping the sun and the Earth together in a giant chamber made out of some perfect mirror (actually maybe the mirror isn't necessary), with the only exit being the dark side of the Earth. And that would heat up both the Earth…

Aha. I think you've helped me see the central issue with your point that it would also heat the sun to millions of degrees. Suppose there was some system that allowed you to collect the light of some source and dump it somewhere else: the drawing in the What-If seems to suggest a system of perfect mirrors with an "output tube" pointed at the earth, so that any photons leaving the tube hit the earth with a high degree of accuracy. What your point shows us is that even if we imagine an indestructible system of mirrors and lenses to do this, the result is that bottling up the energy that the source normally radiates away massively increases the temperature of the source (turning your mirrors into plasma and returning the system to normal, but we're ignoring that again). So entropy law isn't violated.

The reason I (and probably others) find Randall's explanation unhelpful is that obviously there's "enough" energy being reflected by the moon to start a fire (that's why people keep bringing up solar panels). The issue is that there's no way to optically redirect that energy into a small area without heating up your source to the same degree. Which is theoretically possible I suppose, but it's not the situation the What-If is talking about. Along with the issue that the light we see from the moon is mostly reflected rather than emitted (which changed the situation entirely), this makes the What-If explanation a little misleading.

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

#253

Post by sopooneo » Sat, Dec 22, 2018, 8:37 PM UTC

ummonk wrote: post #173

hinkley wrote: post #66

Isn't the bigger problem that sunlight is nearly parallel and moonlight is reflected off of a spherical surface? How are you violating conservation of energy if you're taking all of the light that would hit a square mile of the earth and concentrating it down to the size of a penny? If you can't concentrate light that way then how do focusing lenses on cutting lasers function? Makes no sense.

It's not because the moon is a spherical surface. If that were the issue, there would be a bright spot on the Moon where you can see the reflection of the Sun (you can see such specular reflection spots on e.g. many cars), and you could use that spot to light a fire. The issue is that the Moon barely reflects any of the light that falls on it. Most of the light is scattered, and most of the rest is absorbed and re-ra…

The scattering seems it should cancel out. As far as total energy reaching a point on earth (or the circular disk of a lens) just as much should get to you only due to scattering as failed to get to you due to scattering.

I don't think this changes the ultimate answer to the question.

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

#254

Post by AlexCoventry » Sat, Dec 22, 2018, 8:45 PM UTC

Dylan16807 wrote: post #249

AlexCoventry wrote: post #246

Earlier quoted context omitted.

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.

Thanks again for your help. I think that led me in the right direction.

This is the optical argument for maximum concentration given conservation of étendue (the same page has an optical argument for the conservation):

https://en.wikipedia.org/wiki/Etendue#Maximum_concentration

The angle subtended by the moon is approximately 0.54 degrees, or about 0.01 radians, so the maximum concentration factor is about 10,000. The moon provides about 0.1 lux of illumination, so the maximum illumination you can achieve by optics is about 1000 lux. The sun subtends about the same angle, and we receive 30,000-100,000 lux from it, and the maximum illumination you can achieve from concentrating it optically is about 1B lux.

I'm willing to believe that you can't light a fire from the moon, if the intensity's a million times lower.

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

#255

Post by AlexCoventry » Sat, Dec 22, 2018, 8:50 PM UTC

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…

What you say about blackbody being a red herring makes sense to me.

What do you think of the argument at this link, regarding maximum concentration achievable through optics?

https://en.wikipedia.org/wiki/Etendue#Maximum_concentration

I get that that implies a maximum concentration factor of about 10,000, for a light source subtending an angle of 0.54 degrees (the moon's angular size from Earth, and also approximately the sun's.)

Dylan16807 helped to prod me in this direction. https://news.ycombinator.com/item?id=18739190

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

#256

Post by ummonk » Sat, Dec 22, 2018, 9:45 PM UTC

nkurz wrote: post #196

ummonk wrote: post #188

Earlier quoted context omitted.

>Specifically, I feel certain that one can start a fire with sunlight reflected from a room temperature mirror, and don't understand the difference between a mirror and the moon within Munroe's argument. A mirror does specular reflection and thus conserves the etendue of the sunlight. You're concentrating the image of the sun in the mirror, not light from the mirror itself. The moon in contrast is mostly a diffuse re…

Yes, I can agree with that. The problem for me is fitting it into Munro's argument. He says (in boldface): "You can't use lenses and mirrors to make something hotter than the surface of the light source itself". This is true for a black body, but is it always true for an object being illuminated by another? I'm don't know that we can consider a diffuse reflector with a surface temperature of 100C as being equivalent…

An excellent point!

>Now change the moon to be more heat conductive (causing the surface temperature to drop due to more heat loss on the dark side),

Yeah, although this isn't too big in the case of the moon (unlike the Earth, it doesn't have an atmosphere and doesn't rotate rapidly, so there isn't too much redistribution of heat across its surface), it is definitely something that would confound the calculations. We'd still be able to just look at the effective temperature of light falling onto the moon and that would limit the temperature that we could light the object up to. But we wouldn't be able to use a direct measurement of the temperature of the surface of the moon.

> and more reflective (causing the surface temperature to drop further due to less absorption).

To the extent that it's a gray body (and most objects are approximately graybodies), this wouldn't actually lower the temperature. Absorptivity < 1 causes it to absorb less energy from the light, but for a gray body emmisivity equals absorptivity so it also radiates out less light too, and you actually end up reaching the same equillibrium temperature as fully absorptive black body.

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

#257

Post by username90 » Sat, Dec 22, 2018, 9:48 PM UTC

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…

Spectral composition is completely irrelevant for typical material and temperatures, it is just watt per square meter which matters. You can't concentrate watt per square meter to a higher level than the light had at its quadratic falloff source. You can show that this is true by using the blackbody example, concentrating its light would break the laws of thermodynamic. Moonlight has quadratic falloff from the moon and not the sun and is therefore impossible to concentrate further so at best you can use lenses and mirrors to get the same light intensity the moonlight have on the Moon which of course isn't enough to heat anything above the temperature of the Moon.

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

#258

Post by _0w8t » Sat, Dec 22, 2018, 9:56 PM UTC

nshepperd wrote: post #241

_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 ).

Yes, I stand corrected.

Essentially an optical system will bring Moon's surface closer, but even if it brings the surface within 1 cm from the wood, the defused Sun light scattered from that surface is not enough to ignite the fire.

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

#259

Post by ummonk » Sat, Dec 22, 2018, 9:58 PM UTC

kurthr wrote: post #243

ummonk wrote: post #184

Earlier quoted context omitted.

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 absorptio…

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

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

#260

Post by [deleted] » Sat, Dec 22, 2018, 10:02 PM UTC

[deleted]