> Lenses and mirrors work for free; they don't take any energy to operate. Wait a minute. Does that mean that I could get a tiny solar panel and light it up with a lens, instead of getting big panels? Energy output of a panel is proportional to the amount of light that hits it, right? I guess that lenses are ‘free’ only if they have no impurities, but even then, assuming solar panels are costlier than plastic lenses,…
Can you use a magnifying glass and moonlight to light a fire? (2016)
171–180 of 277 posts
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#172And what of materials that burn at a temperature below the temperature of the Moon? If 100C is the limit, there are materials that burn at much lower temperatures such as Phosphorous (34C).
I mean you can just take white phosphorous to a warm place and let it spontaneously combust.
The use case would be materials that don't burn at the Earth's surface temperature, but do burn at the Moon's peak surface temperature. But you could probably get those hot enough just by rubbing them or something.
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#173Isn'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-radiated.
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#174Oh god, can we please get a real physicist in here? This entire thread is a mess of computer programmers “well actually”ing other computer programmers and everyone being wrong.
Real physicist here. The linked article is correct.
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#175I love xkcd, but this is completely wrong. It is well known that the spectral temperature of the moon is about 4000K. See for example : http://www.lumec.com/newsletter/architect_06-10/the_sun_the_... or https://physics.stackexchange.com/questions/244922/why-does-... . That is the maximum temperature that you can achieve with light from the moon, no matter how concentrated. 4000 K is plenty hot enough to start a fire.
Spectral temperature is irrelevant. It's effective temperature that matters. https://en.wikipedia.org/wiki/Effective_temperature
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#176Earlier quoted context omitted.
The linked article is correct and the arguments are well known and accepted in the physics community and have been for a long time (much longer than Randall Munroe has been alive). A lot of the counter-arguments/speculation here in the comments is wrong. Reading this is is equivalent to reading a thread on a physics forum where with people arguing about an article saying that O(n*lgn) is really the best possible runt…
The linked article is correct and the arguments are well known and accepted in the physics community and have been for a long time (much longer than Randall Munroe has been alive). It's interesting what bothers different people. While many of the statements in this thread are probably wrong, not many of them bother me. But I find the lack-of-self-doubt and appeal-to-authority in your message to be genuinely offensive…
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#177Interesting 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…
>Moon surface temperature argument is incorrect To the extent that the moon acts as a greybody under sunlight, it is correct. And like most things, the moon will be close enough to a greybody that you could use the surface temperature as a first order approximation. It isn't necessary of course, since you can easily just directly estimate the amount of scattered / re-radiated energy from the amount of sunlight fallin…
Leaving this comment to make sure that others don't get confused by your response.
1. As you mentioned, the moon acting as greybody absorbs part of the radiation and scatters the rest. In the second sentence you mention that the moon's temperature is enough to describe this radiation. In the third sentence you mention that you don't need moon's temperature to do that. So which is which?
Correct answer - to describe re-radiated energy we need moon's temperature, but to describe scattered we don't. We can ignore re-radiated, since it is not visible light and not hot enough. We can use scattered, since it is visible and hot enough.
2. Not sure why it is "No" if you are agreeing with me. Also not sure what you mean by temperature represented by total radiation (as you mentioned in part 1 there are two parts - re-radiated - at temperature of the moon, and scattered - at temperature of Sun).
To see that there is a problem with your argument - first note that the spectrum of black-body radiation dominates spectrum of grey-body radiation [1] (i.e. if black-body radiation is not in visible spectrum, the grey body will not be in visible as well). Then think that at 100 degrees C there would be no visible radiation. So the light that we see from moon cannot be this cold. Its coming from sun and it is hot enough (since it is in visible spectrum).
[1] https://claesjohnsonmathscience.wordpress.com/2012/04/21/gre...
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#178Earlier quoted context omitted.
This works because you're just concentrating the image of the sun. It's the surface temperature of the sun that matters. Now if, instead, the mirror were scattering or absorbing and re-radiating the light (as the moon does), then its temperature would be limiting factor.
Now if, instead, the mirror were scattering or absorbing and re-radiating the light (as the moon does) Are you saying that mirrors do something other than absorbing and re-radiating light? In most explanations I've seen, this is exactly what they do: How does the mirror reflect light? The silver atoms behind the glass absorb the photons of incoming light energy and become excited. But that makes them unstable, so the…
I haven't done material physics in depth, so can't be 100% certain of this, but I don't see how that can be true. Electromagnetic reflection is what results in either specular reflection or diffuse reflection (i.e. scattering), not absorbing and re-radiating. Which is why the color of reflected light is dependent on the color of incoming light, and not on the spectral frequencies of the material (unlike re-radiated light).
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#179Earlier quoted context omitted.
You've found the right question to ask. Your mirror in sunlight works because the reflectivity or albedo of the mirror is very high relative to whatever target you're lighting on fire. In a magical closed system where radiative heat transfer was the only factor, objects of differing reflectivity would eventually reach temperature equilibrium through black body radiation. We aren't interested in closed systems though.…
The equilibrium temperature of a grey-body is actually independent of its emissivity. A black-blackbody absorbs more radiation than a white-blackbody, but it also emits more. The emissivity factors in both on the absorption and emission side, so the equilibrium temperature is independent of it. If this seems to fly in the face of all common experience, it's because things that look white aren't actually white in the…
Can you describe what black-blackbody and white-whitebody are?
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#180> You can't use lenses and mirrors to make something hotter than the surface of the light source itself. This is an interesting argument. Can I not reflect some sunlight off a mirror, then do the magnifying-glass-to-start-a-fire trick in daytime? Doesn't the mirror stay cool? Isn't the moon just a (poor) mirror for the sun's light?
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…
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?