this article seems to refute some aspects of xkcd's answer:
https://physics.stackexchange.com/questions/370446/is-randal...
Can you use a magnifying glass and moonlight to light a fire? (2016)
181–190 of 277 posts
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#182Interesting 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…
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.
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?
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#183Earlier quoted context omitted.
>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 r…
>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 visible light and not hot enough. We can use scattered, since it is visible and hot enough.
When dealing with a gray-body, the equillibrium temperature of the body will be equal to the effective temperature of the incoming light, which will be equal to the effective temperature of the re-radiated + scattered light, since at equillbrium energy out equals energy in. So, assuming the moon is a graybody (and most objects tend to be roughly graybodies), we can use the surface temperature of the moon in our calculations instead of the effective temperature of the light that falls on it.
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#184Earlier quoted context omitted.
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.
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.
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#185Earlier 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.
Thanks, this is what I was missing and what I think is most obscure in the article!
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#186Earlier 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.
Perfect reflector exists only in theory. But we still put (imperfect) reflective coating on spacecrafts.
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#187I don’t buy the thermodynamic argument. Here’s a version I would believe: if you have a gadget that, exposed only to the sun and to empty space, heats some target hotter than the sun, then that gadget must not work if you take away the empty space part. This is because your gadget could be used to drive a heat engine, which is impossible without a temperature difference, and the sun is more or less a blackbody emitte…
> 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. With a simple lens you can focus hundreds of megawatts per square meter onto a surface.
But once you bounce that light off a diffuse surface, whatever concentration you had becomes the new maximum.
In your experiment, bathing something in moonlight would max out at 700 watts per square meter.
700 watts per square meter doesn't set things on fire. It can only heat a blackbody to 60 degrees C.
Even the full brunt of unaltered sunlight can only bring a blackbody up to 122C.
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Treating the moon as a blackbody or not doesn't actually change the equations. The important property is that it diffuses light. It resets your maximum concentration of light, because light that comes evenly from every direction can't be concentrated.
(I'm ignoring the part about wavelength filtering because it's confusing and would only make your piece of paper heat up less.)
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#188Earlier 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…
>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 reflector - it scatters most of the light that falls on it (and absorbs and re-emits most of the rest), so it is effectively a new light source.
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#189Earlier quoted context omitted.
> I redirect all the photons to a single point. For that to be possible with a blackbody light source, it has to itself be a single point. Which means a temperature of approximately infinity. For a real light source, one that has an area , you can at best focus it down to the same area. To get maximum light to a target, you either have to make the target almost touch the source, or you have to have it so no matter wh…
> For a real light source, one that has an area, you can at best focus it down to the same area. I'm not an optics expert, but this can't be true. You can clearly focus light emitting from an area into a smaller area (although probably not to an infinitesimally small area).
You can focus 1/100 of the light onto 1/100 the area. Or 1/1000 of the light onto 1/100 the area.
You can't increase the density of the light. You can't focus all of it onto a smaller area.
Is that clearer?
Re: Can you use a magnifying glass and moonlight to light a fire? (2016)
#190Earlier quoted context omitted.
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 r…
>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 visible light and not hot enough. We can use scattered, since it is visible and hot enough. When dealing with a gray-body, the equillibrium temperature of the body will be equal to the effective temperature of the incoming light, which will be equal to the effective t…
Suppose you are right. The effective temperature of the moon is 100 degrees Celsius. It is a grey-body, so its spectrum is dominated by black-body - e.g. it would be similar to spectrum of black-body with temperature lower than 100 degrees Celsius. So your argument is that the light we see from Moon is same as light we would see from a black body heated to less than 100 degrees C? Try putting a charcoal in boiling water and please report to us if you see it lighted up as white as the moon.