7 ms·

Interesting article. Right in many places. Wrong (possibly) in main conclusion. Entropy argument - correct in the sense that using radiation from black body we

by skolos 8y ago

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 radiation's temperature. Also the argument about impossibility of concentrating light into a dot is correct (although even if it were possible we still would not be able to get higher temperature - light would not be energetic enough for that). The important part is - we could concentrate light into a dot only if it consist of parallel rays - i.e. only for an object that is infinitely far away. Moon surface temperature argument is incorrect. A body at 100 degrees Celsius does not radiate in visible spectrum, so the light we see is not produced by Moon's temperature. It is reflected Sun light. So Moon's temperature doesn't matter. Moon surface does absorbs some light, changing spectral composition from about 5.7kK (Sun's surface temperature) to about 4kK. So we should consider moon to be a part of optics not emitter. Hence the question is now - can we concentrate moon light enough so that intensity at the concentration point is higher than thermal loss into environment (only then we will be able to raise temperature in the concentration area enough for combustion - remember that light is "hot" enough for this)? I don't have answer for that - need to do calculations. What can be a deal breaker? Remember that Moon is much closer than Sun, so rays come to us even less parallel, so the area into which we can concentrate light reflected from the Moon is even larger than the Sun's, so together with lower intensity of light from Moon we might have trouble achieving the necessary intensity for combustion. However big enough lens probably will work. And yes - I'm a physicist by training.

neltnerb 8y ago
Yeah, this was my immediate thought as well. Yes, you can't concentrate solar energy to heat something hotter than the surface of the sun. But the emission of a 100C blackbody is not visible either, so it's clearly an incorrect hypothesis on it's face that this temperature causes moonlight. It's a bad mirror, not a radiator.
- tbabb 8y ago
  And it's a mirror that subtends a tiny fraction of the sky, and a tiny fraction of the sun's blurred virtual image.
  - neltnerb 8y ago
    For sure, the light is super dim. I'm not sure it's possible to focus the energy enough to light a fire. I'm just sure the reason is not because moonlight is entirely composed of radiation from a 100C blackbody. This is just immediately obvious because we can see it, so this whole article is a little weird.
skolos 8y ago
As I thought more of why one might get confused with this question is because one might miss that there are two effects here, not one. To have a combustion we need two things - light of high enough temperature and light of high enough energy concentration. The two are not the same. All visible light has high enough temperature. However the concentration is the problem Why do we need high concentration for combustion? If we don't supply enough energy to compensate for heat loss, the area will never get hot enough. By the way - interesting corollary of this is that even with sun lights - if we have a mechanism that takes away heat fast enough (say by blowing cold air at the area) we would not be able to reach combustion with large mirror in direct sunlight.
- njarboe 8y ago
  "if we have a mechanism that takes away heat fast enough" An interesting demonstration of this idea is putting a paper cup full of water into a fire. It won't burn where the water is touching it and eventually the water will start to boil.
- darkmighty 8y ago
  Correct! Concentration is technically called energy flux, and is measured in W/m².
Robotbeat 8y ago
Right. We can't concentrate light brighter than it is on the surface of the Moon HOWEVER, because the light's effective temperature is much higher than the temperature of the Moon itself, we CAN use frequency-selective filters to heat up a sample, in principle to the point of combustion. Thermodynamics allows this. Basically, the greenhouse effect.
ummonk 8y ago
>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 falling on it, without needing to use the temperature. >Hence the question is now - can we concentrate moon light enough so that intensity at the concentration point is higher than thermal loss into environment (only then we will be able to raise temperature in the concentration area enough for combustion - remember that light is "hot" enough for this)? No, because the body you're trying to heat up will act as an approximate greybody, so (neglecting conduction and convection which could further lower the temperature) its temperature has to max out at the Stefan-Boltzmann temperature represented by the total radiation.
- skolos 8y ago
  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/grey-vs-black-bodies/ https://claesjohnsonmathscience.wordpress.com/2012/04/21/gre...
nshepperd 8y ago
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.
skolos 8y ago
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 conclusions. Black and grey body discussion is in same realm, so we will avoid it here. Let's consider piece of wood and try to see what it takes to combust it. We need two things - oxygen and high enough temperature. We are keeping this piece of wood in air, so we have oxygen. The temperature is how fast particles that compose piece of wood are moving. The wood consists of molecules, which are in turn consist of atoms. You can imagine a molecule is a bunch of atoms connected with each other by springs (the springs are created by electromagnetic forces when atoms lend and borrow electrons). The higher the temperature - the larger oscillations of these springs. When the temperature is high enough some of these springs can break and combine with oxygen to release energy - combustion. Suppose light strikes some piece of wood. What happens? Photons hit molecules and they can interact with an electron or proton - they exchange momentum (and energy), which means that one of the atoms in a molecule gets a bump - springs would start oscillating harder. Note, low energy photons cannot swing springs a lot. We need photons of high energy to swing spring to high temperature (ok - this part is too oversimplified, I can extend on this if there are questions). The energy of photons depends on their waivelength - the shorter waivelength - the more energetic photons. So we need photons with energy that could break molecular bonds (springs) to be able to heat up to combustion temperature. Visible light definitely has photons of such energy. So we start sending light down our piece of wood, it starts heating up. But it sits in air and probably fixed by some supporting stand. This piece of wood starts exchanging heat with surrounding materials. To be able to overcome this we need to send lots of photons on this piece of wood. How do we do this? We use lens to collect the photons from broader area and send it down to the wood. Note that it doesn't matter where photons came from, were they scattered or produced by sun, what temperature of scattering surface was, not even what temperature of sun was. All what is matter - can we collect enough high energy photons. So we would need to calculate what is energy flux of visible light photons at the surface earth and see if we could come up with a lens to focus these photons on piece of wood to create sufficient intensity. We know that we can see Moon in visible light, so it does sends bunch of energetic photons to us. All these photon are coming from moon direction (so we don't care that moon scatters them in other directions as well). So the task is to find what is the energy flux of these photons at earth surface and what size of lens we would need to get sufficient intensity.
ufmace 8y ago
Trying to think about this some. If the Moon was actually a giant mirror-lens thing that was the size of the Moon, but perfectly shaped to reflect and concentrate sunlight shining on it into as small an area as possible on the Earth's surface, you'd start a fire with that very easily. Probably turn a county-sized area of the crust into lava or something actually. On the other hand, if the Moon was a perfect reflector, but a big sphere/hemisphere pointed at the sun, then it would receive a lot of high-energy radiation, but reflect it back out in every direction all over the universe. I'm not sure of the math for that offhand, but it sounds difficult to concentrate it back to something even as concentrated as direct sunlight. I suppose it would involve considering the fraction of the total solid angle the reflecting area of the Moon would shine at. Even the entire Earth's surface would be a really tiny fraction of that area, so sounds impossible to concentrate things back to fire-starting intensity. Assuming these scenarios are basically right, then the actual Moon is a lot closer to the second than the first. So starting a fire isn't practical with any lens system you could build on Earth. Maybe you could build some massive mega-lens thing in space near the moon that concentrated enough of it back on the Earth to be near-sunlight brightness. Or just set up a more manageable sized reflector on the Moon itself, or in space, to reflect sunlight in a more concentrated way to a small area on the Earth.
kurthr 8y ago
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.
AlexCoventry 8y ago
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 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 https://news.ycombinator.com/item?id=18739190
username90 8y ago
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.

by skolos 8y ago

neltnerb 8y ago

Yeah, this was my immediate thought as well. Yes, you can't concentrate solar energy to heat something hotter than the surface of the sun. But the emission of a 100C blackbody is not visible either, so it's clearly an incorrect hypothesis on it's face that this temperature causes moonlight. It's a bad mirror, not a radiator.

tbabb 8y ago
And it's a mirror that subtends a tiny fraction of the sky, and a tiny fraction of the sun's blurred virtual image.
- neltnerb 8y ago
  For sure, the light is super dim. I'm not sure it's possible to focus the energy enough to light a fire. I'm just sure the reason is not because moonlight is entirely composed of radiation from a 100C blackbody. This is just immediately obvious because we can see it, so this whole article is a little weird.

skolos 8y ago

As I thought more of why one might get confused with this question is because one might miss that there are two effects here, not one. To have a combustion we need two things - light of high enough temperature and light of high enough energy concentration. The two are not the same. All visible light has high enough temperature. However the concentration is the problem Why do we need high concentration for combustion? If we don't supply enough energy to compensate for heat loss, the area will never get hot enough. By the way - interesting corollary of this is that even with sun lights - if we have a mechanism that takes away heat fast enough (say by blowing cold air at the area) we would not be able to reach combustion with large mirror in direct sunlight.

njarboe 8y ago
"if we have a mechanism that takes away heat fast enough" An interesting demonstration of this idea is putting a paper cup full of water into a fire. It won't burn where the water is touching it and eventually the water will start to boil.
darkmighty 8y ago
Correct! Concentration is technically called energy flux, and is measured in W/m².

Robotbeat 8y ago

Right. We can't concentrate light brighter than it is on the surface of the Moon HOWEVER, because the light's effective temperature is much higher than the temperature of the Moon itself, we CAN use frequency-selective filters to heat up a sample, in principle to the point of combustion. Thermodynamics allows this. Basically, the greenhouse effect.

ummonk 8y ago

>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 falling on it, without needing to use the temperature. >Hence the question is now - can we concentrate moon light enough so that intensity at the concentration point is higher than thermal loss into environment (only then we will be able to raise temperature in the concentration area enough for combustion - remember that light is "hot" enough for this)? No, because the body you're trying to heat up will act as an approximate greybody, so (neglecting conduction and convection which could further lower the temperature) its temperature has to max out at the Stefan-Boltzmann temperature represented by the total radiation.

skolos 8y ago
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/grey-vs-black-bodies/ https://claesjohnsonmathscience.wordpress.com/2012/04/21/gre...

nshepperd 8y ago

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.

skolos 8y ago

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 conclusions. Black and grey body discussion is in same realm, so we will avoid it here. Let's consider piece of wood and try to see what it takes to combust it. We need two things - oxygen and high enough temperature. We are keeping this piece of wood in air, so we have oxygen. The temperature is how fast particles that compose piece of wood are moving. The wood consists of molecules, which are in turn consist of atoms. You can imagine a molecule is a bunch of atoms connected with each other by springs (the springs are created by electromagnetic forces when atoms lend and borrow electrons). The higher the temperature - the larger oscillations of these springs. When the temperature is high enough some of these springs can break and combine with oxygen to release energy - combustion. Suppose light strikes some piece of wood. What happens? Photons hit molecules and they can interact with an electron or proton - they exchange momentum (and energy), which means that one of the atoms in a molecule gets a bump - springs would start oscillating harder. Note, low energy photons cannot swing springs a lot. We need photons of high energy to swing spring to high temperature (ok - this part is too oversimplified, I can extend on this if there are questions). The energy of photons depends on their waivelength - the shorter waivelength - the more energetic photons. So we need photons with energy that could break molecular bonds (springs) to be able to heat up to combustion temperature. Visible light definitely has photons of such energy. So we start sending light down our piece of wood, it starts heating up. But it sits in air and probably fixed by some supporting stand. This piece of wood starts exchanging heat with surrounding materials. To be able to overcome this we need to send lots of photons on this piece of wood. How do we do this? We use lens to collect the photons from broader area and send it down to the wood. Note that it doesn't matter where photons came from, were they scattered or produced by sun, what temperature of scattering surface was, not even what temperature of sun was. All what is matter - can we collect enough high energy photons. So we would need to calculate what is energy flux of visible light photons at the surface earth and see if we could come up with a lens to focus these photons on piece of wood to create sufficient intensity. We know that we can see Moon in visible light, so it does sends bunch of energetic photons to us. All these photon are coming from moon direction (so we don't care that moon scatters them in other directions as well). So the task is to find what is the energy flux of these photons at earth surface and what size of lens we would need to get sufficient intensity.

ufmace 8y ago

Trying to think about this some. If the Moon was actually a giant mirror-lens thing that was the size of the Moon, but perfectly shaped to reflect and concentrate sunlight shining on it into as small an area as possible on the Earth's surface, you'd start a fire with that very easily. Probably turn a county-sized area of the crust into lava or something actually. On the other hand, if the Moon was a perfect reflector, but a big sphere/hemisphere pointed at the sun, then it would receive a lot of high-energy radiation, but reflect it back out in every direction all over the universe. I'm not sure of the math for that offhand, but it sounds difficult to concentrate it back to something even as concentrated as direct sunlight. I suppose it would involve considering the fraction of the total solid angle the reflecting area of the Moon would shine at. Even the entire Earth's surface would be a really tiny fraction of that area, so sounds impossible to concentrate things back to fire-starting intensity. Assuming these scenarios are basically right, then the actual Moon is a lot closer to the second than the first. So starting a fire isn't practical with any lens system you could build on Earth. Maybe you could build some massive mega-lens thing in space near the moon that concentrated enough of it back on the Earth to be near-sunlight brightness. Or just set up a more manageable sized reflector on the Moon itself, or in space, to reflect sunlight in a more concentrated way to a small area on the Earth.

kurthr 8y ago

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.

AlexCoventry 8y ago

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 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 https://news.ycombinator.com/item?id=18739190

username90 8y ago

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.