Thermal iimits of solar power.

[halfeye]

I've read in this forum that there is an upper limit of the heat achievable by solar power of the temperature of the sun's surface.

I'm not satisfied that that makes sense.

Thinking about it earlier today, the thought came to me that this is particularly debatable with reference to The Sun and Betelgeuse. The surface of the Sun is much hotter than the surface of Betelgeuse. However, Betelgeuse is brighter, mainly because it is much bigger. It is bigger because it is much more energetic.

The idea that the most energy that could possibly be extracted from Betelgeuse by a system of lenses and mirrors is less than the maximum energy that could be extracted from the Sun by a similar system seems odd to me.

[Bavarian itP]

Maybe start here. (https://what-if.xkcd.com/145/)

[gomipile]

I've read in this forum that there is an upper limit of the heat achievable by solar power of the temperature of the sun's surface.

I'm not satisfied that that makes sense.

Thinking about it earlier today, the thought came to me that this is particularly debatable with reference to The Sun and Betelgeuse. The surface of the Sun is much hotter than the surface of Betelgeuse. However, Betelgeuse is brighter, mainly because it is much bigger. It is bigger because it is much more energetic.

The idea that the most energy that could possibly be extracted from Betelgeuse by a system of lenses and mirrors is less than the maximum energy that could be extracted from the Sun by a similar system seems odd to me.

You're conflating energy, power, and temperature. These all have precise meanings distinct from each other. I could set 5 million 1500 Watt kettles to boil at the same time, and they'd heat the water to boiling at close to 100 Celsius. An oxygen-acetylene cutting torch would have much less overall power and use less energy than the system of 5 million kettles, yet it can get hot enough to cut through steel plate a couple of centimetres thick.

[crayzz]

The idea that the most energy that could possibly be extracted from Betelgeuse by a system of lenses and mirrors is less than the maximum energy that could be extracted from the Sun by a similar system seems odd to me.

That's not really an accurate summary of the issue. You can almost certainly extract more energy per unit time from Betelgeuse than you can from the sun; what you can't do is concentrate it in a system such that the system becomes hotter than the source.

[Tyndmyr]

The idea that the most energy that could possibly be extracted from Betelgeuse by a system of lenses and mirrors is less than the maximum energy that could be extracted from the Sun by a similar system seems odd to me.

It's a focusing problem. It's not that there isn't a ton of heat, it's that there's no way to concentrate it all in one spot perfectly.

That said, if we're working with insane megastructure level tech, there is a hack. If you put mirrors *all* around the sun(or any other source), you'll increase the heat of the overall system, and thus increase the heat you can focus on any point. For instance, visualize a spherical mirrored surface with a single pinhole. As the only point where light could escape, light striking anywhere else would be reflected and remain in the system. It doesn't truly break the rule, because the sun will heat up, and thus the pinpoint beam escaping will not actually be hotter than the surface, but it will be hotter than the surface would be without the system in place.

This is purely for illustration, the actual construction of such a thing would be more than a little insane for all kinds of practical reason. It's a spherical cow sort of example.

[Chronos]

First time I've ever seen a Dyson cow.

But back to the OP, one way I find helps to think of it is from the point of view of the ant being fried by the magnifying glass. When I look up in the sky, I see mostly cool darkness (well, a little scattered light, but mostly dark), plus a very small spot, a fraction of a square degree, that's at 6000 K. Average those out, and I end up at a temperature of a few hundred K, comfortable for humans (or ants). But when the ant looks up at the magnifying glass that's frying it, it sees the entire magnifying glass, which fills most of its sky, at 6000 K, and so it'll come to an equilibrium temperature that's close to that. Arrange enough mirrors and lenses, and you can make the ant completely surrounded by 6000 K solar surface, and so it'll come to an equilibrium temperature of exactly that. But you can't make the ant see any more than every direction at that temperature, so it can't get any hotter.

[halfeye]

It's a focusing problem. It's not that there isn't a ton of heat, it's that there's no way to concentrate it all in one spot perfectly.

That said, if we're working with insane megastructure level tech, there is a hack. If you put mirrors *all* around the sun(or any other source), you'll increase the heat of the overall system, and thus increase the heat you can focus on any point. For instance, visualize a spherical mirrored surface with a single pinhole. As the only point where light could escape, light striking anywhere else would be reflected and remain in the system. It doesn't truly break the rule, because the sun will heat up, and thus the pinpoint beam escaping will not actually be hotter than the surface, but it will be hotter than the surface would be without the system in place.

This is purely for illustration, the actual construction of such a thing would be more than a little insane for all kinds of practical reason. It's a spherical cow sort of example.

That's one of the ideas that's bugging me. If you put that spherical mirror around Betelgeuse, you'd get more heat out than you would if you put something similar around the sun.

Obviously with current tech we can't do it, but theoretically it seems potentially possible, and the theoretical results seem to be different than this theory of temperature limitation.

[Mechalich]

Note that the maximum temperature humans can currently engineer using concentrated solar is 1000 - 1500 C. This is an area of very active research as industrial groups are working hard to try and hit 1500 C and scale up, allowing solar to be used for things like cement production.

[gomipile]

That's one of the ideas that's bugging me. If you put that spherical mirror around Betelgeuse, you'd get more heat out than you would if you put something similar around the sun.

Obviously with current tech we can't do it, but theoretically it seems potentially possible, and the theoretical results seem to be different than this theory of temperature limitation.

A campfire is not an oven. An oven is not a campfire.

[Anymage]

That's one of the ideas that's bugging me. If you put that spherical mirror around Betelgeuse, you'd get more heat out than you would if you put something similar around the sun.

Obviously with current tech we can't do it, but theoretically it seems potentially possible, and the theoretical results seem to be different than this theory of temperature limitation.

If you put a dyson sphere around Betelguese and channeled all that energy into various contraptions, you could probably concentrate energy down to a point you could probably get that point to be hotter than a point on the surface. With that big an energy source you can do some nuts stuff.

As hinted at in the XKCD linked in the first comment, though, that requires gadgets that have their own energy requirements and failure points. Lenses and mirrors are passive and don't require any additional energy input. They also have various factors that limit their ability to concentrate energy past a certain point. If you leave optics and start using the energy to power different machines you can easily get different outcomes.

[halfeye]

The thing I come down to thinking in the end is how many variables does a photon carry? it obviously must have frequency or wavelength or energy, and it must have position and direction. Does it really have a lot of info about its origin encoded into it? that seems redundant and unnecessary.

[gomipile]

The thing I come down to thinking in the end is how many variables does a photon carry? it obviously must have frequency or wavelength or energy, and it must have position and direction. Does it really have a lot of info about its origin encoded into it? that seems redundant and unnecessary.

Did you read the XKCD What If linked near the beginning of the thread?

[halfeye]

Did you read the XKCD What If linked near the beginning of the thread?

Yes; I didn't get much from it. I don't recall anything about the information in a photon from it.

In theory, there could be anything in a photon, but there are so many photons that it almost has to be the case that there's as little information in there that's just enough to get the job done.

[gomipile]

Yes; I didn't get much from it. I don't recall anything about the information in a photon from it.

In theory, there could be anything in a photon, but there are so many photons that it almost has to be the case that there's as little information in there that's just enough to get the job done.

Re-read the part around the two diagrams of the Sun, a lens, and points A, B, and C.

[NichG]

Since the argument about these limits hinges on reversibility, I wonder if having accelerating lenses/mirrors is enough to beat it (of course at the cost of having to sustain those accelerations against forces exerted by the light).

The thought is based on whether you could e.g. chop a beam and restack it in an irreversible way to get around ettendue.

[factotum]

That's one of the ideas that's bugging me. If you put that spherical mirror around Betelgeuse, you'd get more heat out than you would if you put something similar around the sun.

You would get more *energy* out. The overall temperature would be lower, but Betelgeuse is far larger than the sun so even if the amount of heat per metre squared is lower, there's more energy. You're making the same error in your OP by assuming that something has to be hotter to emit more energy, which is manifestly not the case.

[Khedrac]

First time I've ever seen a Dyson cow.

But back to the OP, one way I find helps to think of it is from the point of view of the ant being fried by the magnifying glass. When I look up in the sky, I see mostly cool darkness (well, a little scattered light, but mostly dark), plus a very small spot, a fraction of a square degree, that's at 6000 K. Average those out, and I end up at a temperature of a few hundred K, comfortable for humans (or ants). But when the ant looks up at the magnifying glass that's frying it, it sees the entire magnifying glass, which fills most of its sky, at 6000 K, and so it'll come to an equilibrium temperature that's close to that. Arrange enough mirrors and lenses, and you can make the ant completely surrounded by 6000 K solar surface, and so it'll come to an equilibrium temperature of exactly that. But you can't make the ant see any more than every direction at that temperature, so it can't get any hotter.

This is the best summary explanation I have seen of this issue - very nice work!

[Mastikator]

In order to get all of the energy out of a star you need to get all the photons, not just the ones that are escaping at an angle perpendicular to the surface.

Using mirrors or lenses can not be used to make all the photons parallel so you can never make make a heated surface smaller than the surface area of the star.

Edit- the only way to use a lens to make an arbitrarily small area arbitrarily warm requires that you have actually parallel light. Only lasers are actually parallel and only lasers can be used to achieve arbitrarily high temperature.

[halfeye]

Re-read the part around the two diagrams of the Sun, a lens, and points A, B, and C.

As I read it that's saying you can't have infinite heat from a lens, which is not controversial at all. Lenses (except concave lenses) produce images, so you can backtrack from the image to the part a particular image came from, what we're hearing here is that you can't make an image smaller than 'x' where x is some proportion of the size of the lens.

Another thing I don't get is how one is supposed to make a perpetual motion device from being able to make an image that is hotter than its original. You need to shrink the size of the image, so you are not getting something as big as the original that is hotter, you are getting something much smaller that is hotter, which has less energy than the original, just in a different form.


You would get more *energy* out. The overall temperature would be lower, but Betelgeuse is far larger than the sun so even if the amount of heat per metre squared is lower, there's more energy. You're making the same error in your OP by assuming that something has to be hotter to emit more energy, which is manifestly not the case.

I thought I was aguing against that, apparently not so clearly as I would like. Betelgeuse is more energetic than the sun, and emits more light, from a much larger surface, so is cooler. What I find odd is the theory that you could get less hear from Betelgeuse, even though it is more energetic.

[Tyndmyr]

That's one of the ideas that's bugging me. If you put that spherical mirror around Betelgeuse, you'd get more heat out than you would if you put something similar around the sun.

While that's true, and you would get more heat out of Betelgeuse(I assume, I haven't actually looked at numbers for total heat output)....the sun is far smaller, and would thus have more heat/square foot, making for a more efficient use of dyson sphere materials.


Yes; I didn't get much from it. I don't recall anything about the information in a photon from it.

In theory, there could be anything in a photon, but there are so many photons that it almost has to be the case that there's as little information in there that's just enough to get the job done.

For this example, the information in a photon is largely irrelevant, we are only caring about the quantity and relative concentration.


Since the argument about these limits hinges on reversibility, I wonder if having accelerating lenses/mirrors is enough to beat it (of course at the cost of having to sustain those accelerations against forces exerted by the light).

The thought is based on whether you could e.g. chop a beam and restack it in an irreversible way to get around ettendue.

Yeah, the impossibility only holds true for an entirely passive system. Active systems can and do break this limit. In theory, you could have some clever device changing angle for every photon(again, wildly into the realm of sci fi) to *make* more parallel. Or absorbing a lot and re-emiting elsewhere, via a solar panel and laser setup.

An extremely rapidly accelerating panel could perhaps briefly reflect enough light to one point to exceed the limit, but the acceleration needed would be a ton of energy.

[gomipile]

Betelgeuse is more energetic than the sun, and emits more light, from a much larger surface, so is cooler. What I find odd is the theory that you could get less hear from Betelgeuse, even though it is more energetic.

You would get less temperature, but potentially more heat from Betelgeuse. Assuming a highly efficient Dyson shell or other heat engine that can process more power than the Sun's output, to make the comparison meaningful.

Heat is transferred energy, not temperature.

[halfeye]

You would get less temperature, but potentially more heat from Betelgeuse. Assuming a highly efficient Dyson shell or other heat engine that can process more power than the Sun's output, to make the comparison meaningful.

Heat is transferred energy, not temperature.

I was walking along the pavement thinking "I should have written "temperature" " after writing that.

I'm even thinking that that Dyson mirror might lower the Sun's temperature (and the equivalent one Betelgeuse's).


For this example, the information in a photon is largely irrelevant, we are only caring about the quantity and relative concentration.

I'm under the impression that with lasers, if you want more temperature, you just add more lasers.

If that is the case with the lasers, and you can't do the same with stars, then the photon has to carry information about the temperature of its source.

[Anymage]

You could certainly use a star to either power a laser directly, or store its energy for future use in lasers. You could then use those lasers to make a given spot as hot as your engineering allows. All those additional gadgets require energy to function and have their own wear and tear issues. Both the lasers themselves, and the energy infrastructure to get the energy from your dyson swarm to whatever devices you want to use.

Optics meanwhile are passive. Mirrors and lenses just sit there and let photons shift their paths. However, the same way that you can make a very good educated guess about where a ball came from if you have sufficient information about its position and momentum, the same could be said for photons. (Ignoring quantum weirdness, since optics cares about collective behavior over individual photons.) This photon here and that photon that landed a millimeter away took slightly different paths, and there's only so much that mirrors and lenses can do to monkey with that. Only being able to concentrate your image so much through optics and can only reasonably gather so much energy from the source, that limits how hot the concentrated image can wind up being.

[Tyndmyr]

I'm under the impression that with lasers, if you want more temperature, you just add more lasers.

If that is the case with the lasers, and you can't do the same with stars, then the photon has to carry information about the temperature of its source.

Nope. No information is being carried in any individual photon. It's not data that enforces this, it's distribution.

Lasers are an active, not a passive system, so they're not subject to this limit. You're putting energy into that system in the form of however you're powering the laser. They're not a good analogy for a star. You're producing a lot of light with very similar wavelengths and orientation with a laser. That's not true for a star, which is radiating light every which way.

That said, even lasers are not truly perfect, and you will hit some eventual limits. While they can be far more precisely focused, the light put out are usually not *exactly* identical in all respects, even if they are very close. Some dissipation will generally happen eventually, putting a hard cap on how many lasers you can add.

Think of it this way. If someone tosses you a pound of sand, you can theoretically catch it if it's in a bucket. If someone throws the same pound of sand into a strong wind, you will not catch it. Each grain of sand is scattered wildly. This is a (very rough) analogy for the photons. With a very large star, they are being scattered from a very large surface area. The more diffusely they are spread, the fewer any given area will catch.

[halfeye]

Optics meanwhile are passive. Mirrors and lenses just sit there and let photons shift their paths.

How does a photon know whether it arrived from a laser and is allowed to heat an area to a particular temperature, or arrived from the surface of a star and isn't allowed to heat that area?

[Tyndmyr]

How does a photon know whether it arrived from a laser and is allowed to heat an area to a particular temperature, or arrived from the surface of a star and isn't allowed to heat that area?

It doesn't know anything. This isn't a decision a photon makes. It's a law about how closely distributed the photons will be.

If you toss the bucket of sand into the wind, the sand doesn't decide to become widely distributed. It just is, because of the nature of wind and sand. This is the same.

[Chronos]

Lasers are subject to the same limitations as any other light source. They don't have a thermal distribution of wavelengths, but a laser of any given wavelength will have an equivalent temperature, and no matter how many of those lasers you have, and no matter what optics you shine them through, you'll never be able to use them to heat a spot to higher than that equivalent temperature. Laser beams are also never truly perfectly unidirectional: You can make them pretty tight, but there will always be some spread, if nothing else from the diffraction limit imposed by the wavelength and the aperture.

[Radar]

Another thing I don't get is how one is supposed to make a perpetual motion device from being able to make an image that is hotter than its original. You need to shrink the size of the image, so you are not getting something as big as the original that is hotter, you are getting something much smaller that is hotter, which has less energy than the original, just in a different form.
Lenses and mirrors do not use energy to do what they do nor do they (in theory) create any losses to the directed and shaped beams of light. This is the key point of the whole problem and it is indeed connected to perpetuum mobile.

If you were able to concentrate light (or anything really) without doing work (and that's what lenses and mirrors do) to obtain higher temperature than in your initial source, this creates temperature difference out of nothing lowering the universes entropy, which is not possible: you can obviously use that temperature difference to extract work out of the system using a standard heat engine. The residual heat from that engine could again be concentrated to the high temperature you need to keep the engine going. So in a closed cycle you convert thermal energy of a single source to useful work without resorting to using some lower temperature reservoir to dump the residual heat into.



I'm under the impression that with lasers, if you want more temperature, you just add more lasers.

If that is the case with the lasers, and you can't do the same with stars, then the photon has to carry information about the temperature of its source.
Even with lasers there will be limitation, but what counts here is the entropy of the light from a given source. Lasers produce by construction light of extremely low entropy so it is easier to obtain high temperatures using those. Stars are pretty much black bodies (as weird as it sounds), so the entropy of emitted light is actually very high - if I remember correctly emission of light with that distribution produces maximum of entropy for a given energy emitted. So if you want to concentrate it you are quite limited unless you put some work into it.

[Quizatzhaderac]

How does a photon know whether it arrived from a laser and is allowed to heat an area to a particular temperature, or arrived from the surface of a star and isn't allowed to heat that area?If the photon can get to the area, it can heat the area. The question is how many photon can be corralled into the area.

If the photons are well aligned (as in a laser) you can get them to travel together. If they can travel you get get a bunch of lasers to converge on the same spot.

If the photons aren't well aligned (like from a star) they won't travel together. If you try to get light from many points on Betelgeuse to the same spot, you'll lose most along the way since they didn't start out going in the same direction.

[NichG]

The issue isn't so much alignment as it is needing photons traveling on the same path to catch up with eachother. With reversible optics you'd have the problem that at some point on the reverse path, two photons at the same position and direction must spontaneously go in different directions. But with nonlinear optics or active optics (such as stimulated emission in a laser) you can have one point in space absorb two photons from different directions and times, and emit them in the same direction and time.

[Rydiro]

Does anyone have the optical laws/explanation for this? Maybe that would help halfeye.
My knowledge there ends at "image size is proportional to object size"

[halfeye]

Lasers are subject to the same limitations as any other light source. They don't have a thermal distribution of wavelengths, but a laser of any given wavelength will have an equivalent temperature, and no matter how many of those lasers you have, and no matter what optics you shine them through, you'll never be able to use them to heat a spot to higher than that equivalent temperature. Laser beams are also never truly perfectly unidirectional: You can make them pretty tight, but there will always be some spread, if nothing else from the diffraction limit imposed by the wavelength and the aperture.

Thanks for that, it reads more or less as expected, so that looks likely to be the orthodoxy.


Lenses and mirrors do not use energy to do what they do nor do they (in theory) create any losses to the directed and shaped beams of light. This is the key point of the whole problem and it is indeed connected to perpetuum mobile.

If you were able to concentrate light (or anything really) without doing work (and that's what lenses and mirrors do) to obtain higher temperature than in your initial source, this creates temperature difference out of nothing lowering the universes entropy, which is not possible: you can obviously use that temperature difference to extract work out of the system using a standard heat engine. The residual heat from that engine could again be concentrated to the high temperature you need to keep the engine going. So in a closed cycle you convert thermal energy of a single source to useful work without resorting to using some lower temperature reservoir to dump the residual heat into.

This reads very badly to me. Getting heat for free from nothing would indeed break entropy, but that isn't what I have been talking about at any time in this thread, though I may well have failed at times to express myself as clearly as I'd like. What I am talking about is getting a higher temperature in a smaller area. The total heat would be at worst the same, and in any real instantiation of the ideas almost certainly less.


Even with lasers there will be limitation, but what counts here is the entropy of the light from a given source. Lasers produce by construction light of extremely low entropy so it is easier to obtain high temperatures using those. Stars are pretty much black bodies (as weird as it sounds), so the entropy of emitted light is actually very high - if I remember correctly emission of light with that distribution produces maximum of entropy for a given energy emitted. So if you want to concentrate it you are quite limited unless you put some work into it.

There are problems with lenses and mirrors, it is relatively easy to make spherical surfaces, so lenses and mirrors are often made with such, but perfect optical surfaces would be parabolic, and sometimes the effort is made to produce parabolic surfaces, but it is much harder to make those. I presume parabolic surfaces are being used in this discussion, but if not that would explain everything, spherical aberation would limit any lens to images bigger than a certain fraction of the lens' diameter.


The issue isn't so much alignment as it is needing photons traveling on the same path to catch up with eachother. With reversible optics you'd have the problem that at some point on the reverse path, two photons at the same position and direction must spontaneously go in different directions. But with nonlinear optics or active optics (such as stimulated emission in a laser) you can have one point in space absorb two photons from different directions and times, and emit them in the same direction and time.

That reads like that same explanation from the link above with the wonky lens and the infinitely small image. I agree that an infinitely small image with infinite temperature is impossible, it's not what I've at any point been trying to suggest.

[NichG]

That reads like that same explanation from the link above with the wonky lens and the infinitely small image. I agree that an infinitely small image with infinite temperature is impossible, it's not what I've at any point been trying to suggest.

Eh, I don't think it has to do with infinity. Just, I was trying to imagine how one would even with arbitrary moving fiber optic cables make it so all of the light from two different point sources released at one moment in time would converge in the same location at the same time.

In order for that to happen, I either had to have one photon 'catch up' with another on the same path, or have the ability to inject a photon into a unidirectional beam from two directions at once.

So without something irreversible, I can't passively increase the peak photon density in a vacuum above the density of my sources.

[halfeye]

So without something irreversible, I can't passively increase the peak photon density in a vacuum above the density of my sources.

However, you can have sources with various densities? If the limit is the carrying capacity of the vacuum, how can you have different capacity sources?

[Captain Cap]

This reads very badly to me. Getting heat for free from nothing would indeed break entropy, but that isn't what I have been talking about at any time in this thread, though I may well have failed at times to express myself as clearly as I'd like. What I am talking about is getting a higher temperature in a smaller area. The total heat would be at worst the same, and in any real instantiation of the ideas almost certainly less.
You are mixing up the first principle of thermodynamics (conservation of energy, thus the impossibility to generate energy/heat from nothing) with the second one (entropy always increases in a closed system). Radar's point is relative to the latter, not the former.
Let's say that the entire universe consists only of the star (body A), the lenses apparatus (body B) and the "point" you want to heat up (body C), with the star having a fixed temperature Ta (heat bath) and the other two bodies temperatures Tb, Tc < Ta. Now let's assume that the apparatus is somehow perfectly efficient and doesn't heat up at all in the process, so that only Tc actually changes: the moment Tc reaches Ta, body C can't heat up any further without employing active mechanisms, this is because you now can't physically produce a non-zero gradient between A and C without external work, as it would violate the second principle of thermodynamics, in fact, defining with dS the total variation of entropy and with Q the heat exchanged after equilibrium, we would have dS = Q/Tx - Q/Ta < 0 (where Tx is some value such that Ta < Tx < Tf, with Tf the final temperature you want to reach).

[NichG]

However, you can have sources with various densities? If the limit is the carrying capacity of the vacuum, how can you have different capacity sources?

It has nothing to do with the carrying capacity of the vacuum though. It's that photons can't catch up with one another along the same path. You can of course irreversibly produce different numbers of photons at once, but that's not a passive optical process anymore.

[Radar]

This reads very badly to me. Getting heat for free from nothing would indeed break entropy, but that isn't what I have been talking about at any time in this thread, though I may well have failed at times to express myself as clearly as I'd like. What I am talking about is getting a higher temperature in a smaller area. The total heat would be at worst the same, and in any real instantiation of the ideas almost certainly less.
This was not about getting heat for free - nothing in the process would create new energy. It is all about the entropy of the energy and your ability to convert thermal energy to useful work. Getting higher temperature with the same energy means lower entropy. To use some energy for work, you need some energy with low entropy and a place to dump the high entropy leftover energy afterwards. Typically, this means two reservoirs: one of high temperature to take energy from and one with low temperature to dump the residual heat into - think of a typical heat engine (steam or internal combustion for example).

Now let's see, how increasing temperature without spending any extra energy would be a problem:
1. We have just one reservoir of whatever temperature.
2. By this hypothetical process, without putting in any extra work we take some energy from the reservoir and make a reservoir of higher energy with it (no new energy - just more "densely packed").
3. Since now we have two reservoirs of different temperatures, we can use them to do useful work.
4. If you continue this for long enough, you would fully convert the whole initial source of heat into useful work and that completely breaks the thermodynamics as it would allow you to effortlessly decrease entropy.

And breaking thermodynamics basically means breaking statistics as a whole.

[halfeye]

You are mixing up the first principle of thermodynamics (conservation of energy, thus the impossibility to generate energy/heat from nothing) with the second one (entropy always increases in a closed system). Radar's point is relative to the latter, not the former.

I believe you are mistaken about what I am thinking about.


Eh, I don't think it has to do with infinity. Just, I was trying to imagine how one would even with arbitrary moving fiber optic cables make it so all of the light from two different point sources released at one moment in time would converge in the same location at the same time.

What have point sources got to do with it? Photons start at a point, but any five or six start at five or six.

We can shine two torches at the same wall with no problem.


In order for that to happen, I either had to have one photon 'catch up' with another on the same path, or have the ability to inject a photon into a unidirectional beam from two directions at once.

So without something irreversible, I can't passively increase the peak photon density in a vacuum above the density of my sources.

I am sorry, I don't understand this at all.


It's that photons can't catch up with one another along the same path.

They travel at the same speed, so they wouldn't catch up? I'm definitely not understanding this.


It is all about the entropy of the energy and your ability to convert thermal energy to useful work.

Yeah, that's right. I am not in any way intentionally suggesting that that doesn't happen.


Getting higher temperature with the same energy means lower entropy.

Not necessarily. A higher temperature difference means more work can be done, a lower temperature difference means less work can be done, if the only energy you have is thermal. If everything you have is at a high temperature, you can do very little work with it. The heat death of the universe (if that's how it ends) is not at exactly zero Kelvin.


Now let's see, how increasing temperature without spending any extra energy would be a problem:

There is some energy put into the creation of lenses, trivial though it may be.


1. We have just one reservoir of whatever temperature.
2. By this hypothetical process, without putting in any extra work we take some energy from the reservoir and make a reservoir of higher energy with it (no new energy - just more "densely packed").

No, we do not take energy from the reservoir, the reservoir emits it while maintaining it's energy level.


3. Since now we have two reservoirs of different temperatures, we can use them to do useful work.

This needs thinking about, and might have something going for it, but I don't think I have so far read of a mechanism by which the universe might enforce this described.


4. If you continue this for long enough, you would fully convert the whole initial source of heat into useful work and that completely breaks the thermodynamics as it would allow you to effortlessly decrease entropy.

We are talking about a utterly miniscule fraction of the energy that the sun emits, and the timescale is billions of years.


The Sun's core fuses about 600 million tons of hydrogen into helium every second, converting 4 million tons of matter into energy every second as a result. This energy, which can take between 10,000 and 170,000 years to escape the core, is the source of the Sun's light and heat.


Core
Main article: Solar core

The core of the Sun extends from the center to about 20–25% of the solar radius.[70] It has a density of up to 150 g/cm3[71][72] (about 150 times the density of water) and a temperature of close to 15.7 million kelvins (K).

https://en.wikipedia.org/wiki/Sun


And breaking thermodynamics basically means breaking statistics as a whole.

Nope, breaking thermodynamics would break the universe, statistics can take its chances.

[NichG]

I believe you are mistaken about what I am thinking about.



What have point sources got to do with it? Photons start at a point, but any five or six start at five or six.

We can shine two torches at the same wall with no problem.


A torch doesn't fill every direction, so two torches each occupy a different range of incoming angles. If you had a solid sphere of torchfire surrounding a point and added an extra torch outside of that sphere, it's light would have to pass through another physical emitter in order to be able to add to the rest.

Now imagine rather than an emitter, we replace each torch with a fiberoptic cable carrying light from elsewhere. If that cable bends light into a particular direction there, then because of reversibility it cannot let light through from behind it in that same direction if it's a passive element.

For a passive optical element, every outgoing direction must correspond to exactly one incoming direction.



I am sorry, I don't understand this at all.

They travel at the same speed, so they wouldn't catch up? I'm definitely not understanding this.


Let's say you wanted to use the fiberoptics from the previous example to inject light on a path and then moved them out of the way so you could get light from behind as well. Because light travels at the same speed, you can't use this to increase the instantaneous intensity either. So merely allowing optical elements to move, even though it gives you something passively irreversible, isn't enough to let you stack light along a direction vector.

[Bohandas]

Maybe start here. (https://what-if.xkcd.com/145/)

But what if you had a bunch of lenses at different angles?

[Captain Cap]

Not necessarily. A higher temperature difference means more work can be done, a lower temperature difference means less work can be done, if the only energy you have is thermal. If everything you have is at a high temperature, you can do very little work with it. The heat death of the universe (if that's how it ends) is not at exactly zero Kelvin.
You are right, higher temperatures don't imply lower entropy per se, higher temperature gradients do. The point is that if you want to heat up a certain object to a higher temperature than the star, you must reach its temperature first: once that is done, any further increase of temperature would decrease entropy.


There is some energy put into the creation of lenses, trivial though it may be.
What matters is that our ideal lenses don't expend energy to do their job of redirecting light once there, they do it passively. The work you used to assemble the system is now irrelevant in the interaction between star and lenses; they may have very well been there since ever, for all it matters.


No, we do not take energy from the reservoir, the reservoir emits it while maintaining it's energy level.
The reservoir emits energy in "our" direction because there is a temperature gradient between the star and "us". If our apparatus had the same temperature as the star, it would irradiate as the star does and the net flux of photons would be zero.

[Radar]

We can shine two torches at the same wall with no problem.
And yet, we would not be able to make the wall hotter than a torch regardless of how many torches we use. In fact, we can explain the initial problem this way: if we view a star from any point, it takes some part of the sky (in other words, a given part of the whole spherical angle). What we can do with lenses and mirrors is to create a situation that at given point we see the star taking a bigger part of the sky. The absolute limit would be if a given point looked as if it is completely surrounded by the surface of the star. Such problems can be solved directly and you simply get at such a point temperature equal to that of the surface of the star.

You may wonder, why is that. Well, imagine a sphere as hot as the star surrounding some point. Each inside point of that sphere emits light in all directions - not just toward the center. This results in the radiation density being equal everywhere within that sphere and there is no particularly high heat concentration anywhere.


There is some energy put into the creation of lenses, trivial though it may be.
Irrelevant as lenses and mirrors in principle are not used up - they can work arbitrarily long.


No, we do not take energy from the reservoir, the reservoir emits it while maintaining it's energy level.
If some portion of energy is emitted, it is no longer at the star. Stars gradually lose energy due to radiation, so we actually do take the energy, if we try to use it for something.


This needs thinking about, and might have something going for it, but I don't think I have so far read of a mechanism by which the universe might enforce this described.
It is not about enforcing - it is just about allowing something. If you have two reservoirs of different temperature, you can use the thermal energy to perform useful work.


We are talking about a utterly miniscule fraction of the energy that the sun emits, and the timescale is billions of years.
Timescale does not matter, if we are talking about principles of physics. If something should cannot work in large scale due to fundamental reasons, it cannot work in a small scale for the same reasons.


Nope, breaking thermodynamics would break the universe, statistics can take its chances.
What I meant is that thermodynamics is just applied statistics - breaking it means that statistics is not working in our universe. And yes, it would break the universe as we know it.

[Lord Torath]

But what if you had a bunch of lenses at different angles?The best you can get is a bunch of mirrors and lenses that make it so that where ever you look, all you see is the sun. Which means you are effectively surrounded by the sun, and thus you will take the ambient temperature of the sun.

Again, see Randall Munroe's What If (https://what-if.xkcd.com/145/) article:

There's another way to think about this property of lenses: They only make light sources take up more of the sky; they can't make the light from any single spot brighter,[7] because it can be shown[8] that making the light from a given direction brighter would violate the rules of étendue.[9] In other words, all a lens system can do is make every line of sight end on the surface of a light source, which is equivalent to making the light source surround the target.
https://what-if.xkcd.com/imgs/a/145/linesight.png

If you're "surrounded" by the Sun's surface material, then you're effectively floating within the Sun, and will quickly reach the temperature of your surroundings.

[Bohandas]

I'm not entirely sure I understand why it would necessarily match the temperature. After all, it's not transmitting temperature, it's transmitting energy that excites the molecules of the target and thus raises its temperature

[crayzz]

I'm not entirely sure I understand why it would necessarily match the temperature. After all, it's not transmitting temperature, it's transmitting energy that excites the molecules of the target and thus raises its temperature

Thermal energy travels along temperature gradients.

That this holds true so consistently even for energy transfer through radiation is somewhat unintuitive; after all, we should be able to just get a bunch of mirrors and lenses and pile all these photons into a denser area, right?

But for a variety of reasons explained up thread, it doesn't actually work. Its like thinking you can stick a bunch of heat pipes into the surface of the ocean and expecting to boil tea with it: the energy is definitely there, but without doing some work yourself you have no way of moving the energy around beyond the point of thermal equilibrium.

[Lord Torath]

To put it a different way, light (and heat) can travel both ways along the mirrors and through the lenses, right? In the event that you managed to heat your target up hotter than the source of heat, light and heat would instantly start flowing the other direction, away from your target and back toward the source until their temperatures are in equilibrium.

[McGarnagle]

It might be helpful to think of things in terms of distributions of photon frequencies. Even if you collect all the photons from the surface of the Sun somehow, those photons will still have a frequency distribution governed by the Planck function for a 5800 K blackbody. Concentrating the radiant output of a blackbody source means an increase in brightness, not temperature.

[Bohandas]

Relating this back to the Cooking a chicken by slapping it (https://forums.giantitp.com/showthread.php?642796-Cooking-a-chicken-by-slapping-it) thread, could the chicken be heated to arbitrary temperatures by increasing the rate of the slaps without increasing their individual strength? I bring this up because it seems to be ultimately the same question

[Radar]

It might be helpful to think of things in terms of distributions of photon frequencies. Even if you collect all the photons from the surface of the Sun somehow, those photons will still have a frequency distribution governed by the Planck function for a 5800 K blackbody. Concentrating the radiant output of a blackbody source means an increase in brightness, not temperature.
That's not exactly accurate since temperature of a black body governs both the frequency distribution and radiation intensity from the surface of a given object. Besides, frequency filters could be used to shape the distribution but it would not help you in heating things beyond the temperature of the original source since the key point is how well can you concentrate the energy. To heat something up, means to push in more energy than it can radiate out. Form of the energy or in particular frequency of the incoming radiation does not matter.

[Lord Torath]

Relating this back to the Cooking a chicken by slapping it (https://forums.giantitp.com/showthread.php?642796-Cooking-a-chicken-by-slapping-it) thread, could the chicken be heated to arbitrary temperatures by increasing the rate of the slaps without increasing their individual strength? I bring this up because it seems to be ultimately the same questionThere is a difference there. In the chicken slapping, you are doing work. You are converting electrical energy into kinetic energy, and then transferring that kinetic energy into the chicken, where it is converted into thermal energy.

In the lenses-and-mirrors question, you are not doing work. The entire process relies on the natural flow of heat from a high-temperature object to a low temperature object. And as soon as the low temperature object is the same temperature as the high-temperature object, the flow of heat stops.

[halfeye]

I want to be very clear that I am not arguing against entropy, I am arguing that maybe entropy doesn't apply in the way some people think it does in this particular situation, however entropy always applies.

First off, how about transformation of electricity? The voltage can go up or down, the amperage can go up or down, and the only necessarily moving parts are switches to start and stop the process. This clearly doesn't break entropy.

Mirrors and lenses are not 100% efficient, there is loss of photons by reflection and absorbtion from lenses and absorbtion and transmission though mirrors.

Some people seem to be hung up on absolute temperature, I have a query, I would expect people not to go for this, but it seems worth asking because it's an odd one if people do, and the "why not" should clarify some things if as expected people don't. Suppose the target for the heating is cooled, does that change the amount of energy that can fall onto it?

This thing apparently gets up to 3.5k C, in Earth's atmosphere:

https://en.wikipedia.org/wiki/Odeillo_solar_furnace

https://upload.wikimedia.org/wikipedia/commons/thumb/a/ac/Four_solaire_001.jpg/440px-Four_solaire_001.jpg

I suggest something in space without Earth's atmosphere interfering should be able to get much hotter, particularly if the object was much nearer the sun. Though materials would be a problem, and we probably don't have a use for a temperature that high.

[warty goblin]

Mirrors and lenses are not 100% efficient, there is loss of photons by reflection and absorbtion from lenses and absorbtion and transmission though mirrors.

Which would mean you can heat your object to the temperature of the sun minus some factor depending on the efficiency of your mirrors.


Some people seem to be hung up on absolute temperature, I have a query, I would expect people not to go for this, but it seems worth asking because it's an odd one if people do, and the "why not" should clarify some things if as expected people don't. Suppose the target for the heating is cooled, does that change the amount of energy that can fall onto it?


The same amount of energy falls into the object no matter what. Say your focusing array can dump x joules/s into your object. If you don't cool the object, it's temperature will increase until some temperature T where it radiates x joules/s as a blackbody. If you cool the object, then it will not reach T because you are pumping some of the heat (hence energy) out of the object. This makes no difference to the amount of energy being put into the object.


This thing apparently gets up to 3.5k C, in Earth's atmosphere:

https://en.wikipedia.org/wiki/Odeillo_solar_furnace

The surface of the sun is about 5000 Kelvin. Nobody is saying you can't use the sun to get things very hot, they're saying you can't get it hotter than the sun.

[halfeye]

The surface of the sun is about 5000 Kelvin. Nobody is saying you can't use the sun to get things very hot, they're saying you can't get it hotter than the sun.

I am saying that doesn't make sense to me. Suppose you had a space capable version of that solar furnace, same specs, same sort of mirror, and you put it at 0.5 AUs from the sun. That means, each bit of mirror is getting 4 times the energy, and reflecting it with the same efficiency. How does that not produce at least twice the temperature?

[Captain Cap]

I am saying that doesn't make sense to me. Suppose you had a space capable version of that solar furnace, same specs, same sort of mirror, and you put it at 0.5 AUs from the sun. That means, each bit of mirror is getting 4 times the energy, and reflecting it with the same efficiency. How does that not produce at least twice the temperature?
You are considering only the energy input from the sun, but the higher the object's temperature becomes, the more energy it irradiates itself, and so the higher is the output. The maximum temperature you can reach without additional work is the equilibrium temperature for which input and output are the same.

Let's see this setup explicitly: we know that the power W irradiated (output) by a black body at temperature T is proportional to T^4, and so, adopting the appropriate unites of measures, we can write W = T^4 and Ws = Ts^4 (where s indicates the sun); if R is the distance from the sun and r its radius, the input power is Wi = Ws * (r/R)^2, therefore at the equilibrium we have T^4 = Ws * (r/R)^2 = Ts^4 * (r/R)^2 <= Ts^4.

[NichG]

I am saying that doesn't make sense to me. Suppose you had a space capable version of that solar furnace, same specs, same sort of mirror, and you put it at 0.5 AUs from the sun. That means, each bit of mirror is getting 4 times the energy, and reflecting it with the same efficiency. How does that not produce at least twice the temperature?

Lets say you have a lens system that makes the sun take up 1/2 of the sphere of view around a target point, and that gets you to, say, 2500K (for a 5000K source). If you move it closer, then depending on the optics, the sun may take up more or less of that sphere of view. If you can make it take up the full sphere of view, you could get it to 5000K. It's not physically possible for it to take up more than 100% of the sphere of view.

So the curve of temperature vs distance from the sun will depend on that optics calculation, and you may indeed find that its easy to get from something like 1% of field of view to 80% of field of view by moving it closer, but you will at some point start to get diminishing returns.

[Quizatzhaderac]

I am saying that doesn't make sense to me. Suppose you had a space capable version of that solar furnace, same specs, same sort of mirror, and you put it at 0.5 AUs from the sun. That means, each bit of mirror is getting 4 times the energy, and reflecting it with the same efficiency. How does that not produce at least twice the temperature?

If you move that whole apparatus twice as close to the sun, the first set of mirrors will receive 4 times as much light. However, the direction of travel for that light will very twice as much and less than 4 times the amount of light will following the complicated mirror and lenses path to the furnace core.

[halfeye]

Lets say you have a lens system that makes the sun take up 1/2 of the sphere of view around a target point, and that gets you to, say, 2500K (for a 5000K source). If you move it closer, then depending on the optics, the sun may take up more or less of that sphere of view. If you can make it take up the full sphere of view, you could get it to 5000K. It's not physically possible for it to take up more than 100% of the sphere of view.

So the curve of temperature vs distance from the sun will depend on that optics calculation, and you may indeed find that its easy to get from something like 1% of field of view to 80% of field of view by moving it closer, but you will at some point start to get diminishing returns.

At this point I disagree. What I suggested wasn't making the sun a bigger area of the object's sky, though coincidentally it would also do that to a small extent, but moving the object twice as close to the sun, where the radiation from the sun would be stronger by a factor of 4:

https://en.wikipedia.org/wiki/Inverse-square_law

I'm not sure how much energy is required for a given rise in temperature, but it seems to me that increasing the brightness of the source, ought to increase the temperature of the target.

We've discussed the permiability of space, and apparently it's not that which is the issue.

At all the ranges I am discussing, the sun would optically be at infinity.

[crayzz]

You can't apply the inverse square law and an optical infinity simplification at the same time. The inverse square law arises from the fact that your light source is not at optical infinity; that changing your distance from the light source increases or decreases the area of its image. If you're at optical infinity i.e. all your light rays are parallel, then you don't have that. It doesn't matter what your position is, or whether you move towards or away from the light source: the image of the light source has the same area, and thus the same wattage per area.

If you're not at optical infinity, then the surface area of the image grows quadratically as you move away, and so the wattage per area must decrease inverse to that because total power output must remain invariant.

If you're in a situation where moving towards or away from the light source substantially changes the power you can capture, you're in a position where the incident angle of the light rays change, and so you have to deal with the size of the image reflected off your solar furnace changing. Moving towards the light source increases the wattage you capture, but it also increases the area of the image you produce.

[halfeye]

You can't apply the inverse square law and an optical infinity simplification at the same time. The inverse square law arises from the fact that your light source is not at optical infinity; that changing your distance from the light source increases or decreases the area of its image. If you're at optical infinity i.e. all your light rays are parallel, then you don't have that. It doesn't matter what your position is, or whether you move towards or away from the light source: the image of the light source has the same area, and thus the same wattage per area.

If you're not at optical infinity, then the surface area of the image grows quadratically as you move away, and so the wattage per area must decrease inverse to that because total power output must remain invariant.

If you're in a situation where moving towards or away from the light source substantially changes the power you can capture, you're in a position where the incident angle of the light rays change, and so you have to deal with the size of the image reflected off your solar furnace changing. Moving towards the light source increases the wattage you capture, but it also increases the area of the image you produce.

I mean optical infinity in the camera/glasses/telescope sense, the focus is at infinity, the rays are parallel. By the time you are 3 metres from an object, you are more or less at infinity, the angle the rays arrive from is almost exactly 90 degrees. This is a peculiarity of cameras, the sort of glasses that are prescribed by opticians, and telescopes, it's really a bit odd, but it's only in really close to a subject that the angle of the light coming through the lens is significantly different from parallel. I am not suggesting that this is particularly significant to this debate, that was just a point of interest.

The inverse square law on the other hand, is in my opinion, entiely relevant to the debate.

[Captain Cap]

I'm not sure how much energy is required for a given rise in temperature, but it seems to me that increasing the brightness of the source, ought to increase the temperature of the target.
Technically, the brightness of the source is always the same, but I get what you mean: the higher is the power inputted by the sun (which in the setup considered can be described by an inverse-square law), the higher is the temperature.
That's correct, but a quantity that increases can have an upper bound, and in this case the upper bound is the superficial temperature of the sun, which you reach when your furnace system and the sun are at contact with each other.

[NichG]

I mean optical infinity in the camera/glasses/telescope sense, the focus is at infinity, the rays are parallel. By the time you are 3 metres from an object, you are more or less at infinity, the angle the rays arrive from is almost exactly 90 degrees. This is a peculiarity of cameras, the sort of glasses that are prescribed by opticians, and telescopes, it's really a bit odd, but it's only in really close to a subject that the angle of the light coming through the lens is significantly different from parallel. I am not suggesting that this is particularly significant to this debate, that was just a point of interest.

The inverse square law on the other hand, is in my opinion, entiely relevant to the debate.

They're not independent things, the inverse square law derives from the spreading of the light rays. If you assume parallel rays, you're already at infinity and the flux per unit area can't further decrease.

Think of it this way - if I put a Dyson sphere of radius r around a star, or a Dyson sphere of radius 2r, the energy collected over the entire surface of the sphere must be the same. But the area of the latter is 4 times higher, so the flux per unit area must by 4 times lower. It's not that the photons are becoming dimmer or losing energy, they're just spreading out. A lens can reverse that spread only to the level of the most concentrated the light has ever been, not beyond that point.

[Captain Cap]

They're not independent things, the inverse square law derives from the spreading of the light rays. If you assume parallel rays, you're already at infinity and the flux per unit area can't further decrease.
To be fair, the parallel rays approximation can be used also when not at infinity (if it wasn't the case, it would be useless), you just need to be aware of the error you are introducing by doing so: in this case, the relative error would go as L/R, with L the characteristic dimension of your apparatus and R the distance from the centre of the sun; if the radius r of the sun is far greater than the apparatus' size L, then you can reasonably adopt the parallel rays approximation throughout the range between surface of the sun and infinity.

Of course, for systems of the size of a Dyson sphere that envelope the sun the approximation always fails.

[Captain Cap]

(emphasis mine)

I... think you've got that backward? If you're much closer to the sun than it is wide, your object is going to be having light rays hitting it from all angles. If your object is very far from the sun, then the rays are closer to parallel.

If the sun takes up 30% of your field of view (ie you're very close to it), then its rays will span 54 degrees. If it takes up 1% of your field of view, it's rays are spread only across 1.8 degrees.
I was thinking about the net radiation flux (the quantity you actually apply the inverse-square law to) when I wrote the post, but yeah, the rays themselves are not parallel.

[halfeye]

Think of it this way - if I put a Dyson sphere of radius r around a star, or a Dyson sphere of radius 2r, the energy collected over the entire surface of the sphere must be the same. But the area of the latter is 4 times higher, so the flux per unit area must by 4 times lower. It's not that the photons are becoming dimmer or losing energy, they're just spreading out.

That's all true, but at this point I'm not talking about a Dyson sphere or anything like it.


A lens can reverse that spread only to the level of the most concentrated the light has ever been, not beyond that point.

That's the point of difference. Why not? How not?

The light is just a bunch of photons. If it's not going to converge beyond a certain point, something has to stop it. What is physically stopping that beam from converging?

We have a working solar furnace that gets up to 3,500 C / 3,200 K, if we replicate the relevant parts of that and put it half the distance to the sun of the original, it gets 4 times the number of photons. How does that not work out to four times the energy? I don't think the energy difference between temperatures is exponential, but I really don't remember, if it is we can move the replicated furnace even closer to the sun, until that exponential is reached.

[Captain Cap]

We have a working solar furnace that gets up to 3,500 C / 3,200 K, if we replicate the relevant parts of that and put it half the distance to the sun of the original, it gets 4 times the number of photons. How does that not work out to four times the energy?
Just because you input 4 times the energy it doesn't mean the object retains 4 times the energy.


I don't think the energy difference between temperatures is exponential, but I really don't remember, if it is we can move the replicated furnace even closer to the sun, until that exponential is reached.
The thermal energy of a body is directly proportional to its temperature.

[NichG]

That's all true, but at this point I'm not talking about a Dyson sphere or anything like it.

That's the point of difference. Why not? How not?

The light is just a bunch of photons. If it's not going to converge beyond a certain point, something has to stop it. What is physically stopping that beam from converging?


This all comes from reversibility, and basically corresponds to what that XKCD book called 'the law of etendue'. If I have some kind of propagating light wavefront, I can describe that as a distribution of energy over both position 'r' and direction 'k' that evolves with time. Passive optics apply reversible operations to that distribution, by which I mean that for the forward operation L, I can write down a unique inverse operation L^{-1}. That implies that I can never (with passive optics) perform any operation which would make it ambiguous where each parcel of energy came from originally - that would violate reversibility, and would correspond (thermodynamically) to performing work on the system - e.g. it would require energy or at least dissipation.

So imagine if I were to have an optical element which mapped a broad distribution in 'r' to a delta function in 'r'. That would in turn have to map a narrow distribution in 'k' to a broad distribution in 'k' for the operation to be reversible, because otherwise it would be ambiguous how much energy should go where on the reverse path. If you are 'out of space' to make the distribution in 'k' broader, then you can no longer make the distribution in 'r' more narrow. If you did, that would violate reversibility.

So its not like 'a new physical principle comes into play to stop you from making it coverge more', and rather that if you write down the equations for how optical elements interact with these distributions, you'll find that the full form of those equations always ends up having these diminishing returns built into them just because of the way that the optical elements behave *in detail*. The issue you're having is that you're combining a bunch of asymptotic approximations as if those approximations were the full story, but you're combining them in regimes where they no longer hold true, and as a result finding an outcome > 1.

See for example the Wikipedia page on Etendue: https://en.wikipedia.org/wiki/Etendue

Edit: Tried to plot it and this bit is wrong, I think I have a 1/x somewhere flipped. Will post again if I figure it out.
They have a formula for the 'maximum concentration': Cmax = n^2/sin^2(alpha), where n is the index of refraction of the medium in which the receiver is placed, and 2*alpha is the collection angle (the fraction of the incoming field of view along which light from the source is being captured). So if you set n=1 and consider something like a pinhole camera, 'alpha' is basically the angle subtended by the light source, which is proportional to 1/r with r being the distance, in which case you derive the 1/r^2 law in the limit that alpha << 1, because sin(alpha)~alpha when alpha is small. However, when you get close to the source, alpha approaches pi/2 and you're in a different limit entirely, so the 1/r^2 law cuts off.

[Quizatzhaderac]

I mean optical infinity in the camera/glasses/telescope sense, the focus is at infinity, the rays are parallel. By the time you are 3 metres from an object, you are more or less at infinity, the angle the rays arrive from is almost exactly 90 degrees.This is exactly the sense that matters and that you cannot simplify away. The imperfection doesn't matter much for a camera, but it matters for solar furnaces.

Let's say you have a 1 km radius reflecting dish heating a 1 meter radius solar furnace. If the light travels to the side one meter on the thousand meter journey between outer edge of the reflector dish and the furnace, it misses the solar furnace.

One meter out of a thousand is 3.5 arcminutes. Sunlight varies from parallel by 16 arcminutes, so the vast majority of the light that hits the rim of the reflector will miss the solar furnace.

[Radar]

That's the point of difference. Why not? How not?

The light is just a bunch of photons. If it's not going to converge beyond a certain point, something has to stop it. What is physically stopping that beam from converging?
Geometric argument: because the incoming photons are not going in parallel directions. A perfect lens can in theory (and ignoring more involved effects here) focus a parallel beam into a mathematical point. The thing is, ideally parallel beams do not and cannot exist. This approximation is especially true for black body radiation as every point of the source spreads the light equally in all directions. So instead of focusing a coherent beam, you try to focus light coming from many different direction with the same lens or mirror and if you try to filter it somehow, you are just losing energy in the process, which will not help you at all. Where the light will be directed by the lens or a mirror will obviously depend on the direction of the incoming light, so for larger angular size of the source you get proportionally larger focal area you can obtain even with perfect equipment.

[Chronos]

When you get twice as close to the Sun, the Sun is four times as bright, because it's taking up four times as much area in the sky. If you had a piece of paper held at arms' length, with a 30 arcsecond diameter hole cut in it, and looked at the Sun through that hole, you couldn't tell if you were at 1 AU and seeing the whole sun, or at 1/2 AU and seeing only a part of it. The "brightness density" of the Sun isn't changing, just its angular area.

And that's the same thing a collecting mirror does: It doesn't change the brightness density, just the effective area.

In either case, getting closer to the Sun or using a bigger mirror, the increase in total incoming energy is entirely due to the Sun (or equivalently, its image) taking up more angular area. And so, in both cases, the incoming energy maxes out when the angular area maxes out.

[Lord Torath]

I was thinking about the net radiation flux (the quantity you actually apply the inverse-square law to) when I wrote the post, but yeah, the rays themselves are not parallel.I misread your example, and realized you were correct. Then I deleted my post in embarrassment, hoping no one had seen it.

So much for hoping! :smallredface:

[Rockphed]

That's the point of difference. Why not? How not?

The light is just a bunch of photons. If it's not going to converge beyond a certain point, something has to stop it. What is physically stopping that beam from converging?.

It is called the diffraction limit. As you concentrate light it starts to interfere with itself. Any telescope (i.e. lens system) has a limit where it cannot make images smaller and still have them be images. At that point everything starts to blur together. I know it falls out of Maxwell's equations but I cannot remember exactly how to derive it.

[Radar]

It is called the diffraction limit. As you concentrate light it starts to interfere with itself. Any telescope (i.e. lens system) has a limit where it cannot make images smaller and still have them be images. At that point everything starts to blur together. I know it falls out of Maxwell's equations but I cannot remember exactly how to derive it.
In the case of concentrating Sun's energy things brake down long before the diffraction limit and the problem at hand is grounded in the optical geometry - there is no need to involve the wave equations here.

[Chronos]

And for what it's worth, the diffraction limit "falls out of Maxwell's equations" only in the sense that Maxwell's equations show that light is a wave. Diffraction works the same way for all waves, light, sound, ripples on the surface of water, whatever.

[Gnoman]

First off, how about transformation of electricity? The voltage can go up or down, the amperage can go up or down, and the only necessarily moving parts are switches to start and stop the process. This clearly doesn't break entropy.


Wattage=Voltage*Current

The total energy in a system is identical on both sides of a spherical transformer in a vacuum (practically, the output is slightly lower because some energy is wasted as heat within the transformer). For a given power source, it is impossible to get an output higher than the input, although you can add waste loads if you want to reduce the output for some reason. Wattage is constant. Thus the transformation of electricity does not have any unusual relationship to entropy.

Similarly, a star can only put out temperatures equal to what it itself is. You can get lower temperatures by adding waste load (all the areas around the star that aren't being illuminated), and you can shuffle some of the output around in the same conceptual way that you can shift voltages and currents around (in this case, you prevent solar energy from striking some areas in order to strike a different area more), but you can never, ever, ever get an output higher than the input.

There's a very simple (albeit potentially expensive) experiment that can be done to prove this, requiring a laser and a lens. If you take an industrial laser and shine it against different materials, using a lens will get a much thinner line on those materials that it is powerful enough to mark or cut, and it will produce those results significantly faster. What it will not do is mark or cut a material that the un-lensed beam cannot. This is because the final determiner of what effect the laser has is how hot it can make the target, and that is identical for both beams because they have the same source. The sole exception to this is if you use significantly different (but same cuttable/markable material) targets between the two beams - it is possible that if you try to use a larger sample for the wider beam it will transfer energy away faster than it can be applied, even if a thinner example would be affected.

I know this because I've done it - I spent most of a week back in high school trying to figure out why our lens didn't make the beam exponentially more powerful the way our simple area calculations suggested.

[halfeye]

Wattage=Voltage*Current

The total energy in a system is identical on both sides of a spherical transformer in a vacuum (practically, the output is slightly lower because some energy is wasted as heat within the transformer). For a given power source, it is impossible to get an output higher than the input, although you can add waste loads if you want to reduce the output for some reason. Wattage is constant. Thus the transformation of electricity does not have any unusual relationship to entropy.

That is exactly what I was saying. My point is that you can take either voltage or current to be equivalent to temperature, and overall wattage is exactly equivalent to overall wattage, but in the case of electricity there is not a hard limit on the output voltage or amperage provided you take the hit in the other aspect, so that overall output wattage falls due to the system not being 100% efficient..


Similarly, a star can only put out temperatures equal to what it itself is.

In the core of the sun that's millions of degrees Kelvin. The temperature at the surface is less because the heat is spread out and emitted.


You can get lower temperatures by adding waste load (all the areas around the star that aren't being illuminated), and you can shuffle some of the output around in the same conceptual way that you can shift voltages and currents around (in this case, you prevent solar energy from striking some areas in order to strike a different area more), but you can never, ever, ever get an output higher than the input.

I'm definitely not talking about getting more Watts out than were originally present.


There's a very simple (albeit potentially expensive) experiment that can be done to prove this, requiring a laser and a lens. If you take an industrial laser and shine it against different materials, using a lens will get a much thinner line on those materials that it is powerful enough to mark or cut, and it will produce those results significantly faster. What it will not do is mark or cut a material that the un-lensed beam cannot. This is because the final determiner of what effect the laser has is how hot it can make the target, and that is identical for both beams because they have the same source. The sole exception to this is if you use significantly different (but same cuttable/markable material) targets between the two beams - it is possible that if you try to use a larger sample for the wider beam it will transfer energy away faster than it can be applied, even if a thinner example would be affected.

I know this because I've done it - I spent most of a week back in high school trying to figure out why our lens didn't make the beam exponentially more powerful the way our simple area calculations suggested.

That is much better. Do you have any sources for that that aren't purely anecdotal? Is there any theory as to why that should be that isn't vague hand-waving?

[Captain Cap]

In the core of the sun that's millions of degrees Kelvin. The temperature at the surface is less because the heat is spread out and emitted.
Radiation wise, the sun interacts with the rest of the universe only through the energy irradiated by its surface, so its temperature anywhere else is irrelevant.

[Squire Doodad]

That is much better. Do you have any sources for that that aren't purely anecdotal? Is there any theory as to why that should be that isn't vague hand-waving?

I mean there's a What-If on the topic, concerning starting a fire with moonlight (https://what-if.xkcd.com/145/). Give it a read for a Munroesian Explanation.

[Radar]

That is exactly what I was saying. My point is that you can take either voltage or current to be equivalent to temperature, and overall wattage is exactly equivalent to overall wattage, but in the case of electricity there is not a hard limit on the output voltage or amperage provided you take the hit in the other aspect, so that overall output wattage falls due to the system not being 100% efficient..
The point is neither voltage nor current can be equivalent to temperature, since electrical current is the coherent motion of electrons, so its entropy is low and is not connected to the voltage or current values in any significant way - this is why you can transform the voltage almost however you like. With temperature and any kind of heat transfer (be it by radiation, convection or whatever else) it is completely different.

[Fat Rooster]

Also relevant is Maxwell's demon, and I'm surprised it hasn't been mentioned yet with regards to materials. The idea is that any sort of one way door for energy is prohibited, so that materials that absorb light at a particular frequency also emit it efficiently, and these two things are perfectly correlated. From that it might be easiest to consider each path individually. Lenses and mirrors can align beams quite closely, but they are always reversible. There is no possible way for two paths to be identical over some of their distance, and not over some other part. They can be close, but if they were aligned with mirrors they can be separated with mirrors.
If we consider the light entering and leaving a single path from a single point, at a single frequency, then we can say that either the material is perfectly reflecting or transmitting the light, or it is a good emitter at that frequency. If we follow the path back in time, and hit the sun, we can work out how much energy the material will be absorbing. If we know the temperature of the material, we can work out how much energy is going out at that frequency too*. If the material is hotter than the surface of the sun then we are guaranteed to be radiating more than we absorbing. While it might seem like we can do interesting things with manipulating the paths, this is true along every single individual paths. Managing to heat something that was already hotter than the surface of the sun would require adding up the effects of lots of paths, all of which are emitting more energy than they are absorbing, and coming up with a total where we are absorbing more than we are emitting**.

* You might then wonder if there is some way to trap the photons, such as by moving things so that our target will absorb light from the sun, but photons from it cannot escape and are redirected towards it. You can, but Maxwell's demon tells you that when those photons arrive back at our target they are on a path that could have lead back to the sun, so they are instead of energy from the sun, not in addition to it. In order to avoid energy escaping you need to close the door that lets energy get in.

**We cannot say where the energy will end up. Things move, light emitted back along a path that came from the sun is following a whole new path, but we can say that if any energy is getting in, then energy can get out.

That XKCD irks me somewhat, because it doesn't mention that it is possible to light a fire with a somewhat more complicated setup (though not with a magnifying glass), using the fact that we can filter by frequency. A material that does not absorb in the infrared spectrum but does absorb blue light can get extremely hot, even in moonlight. It only stops heating when it is emitting as much blue light as it is absorbing, and that means several hundred degrees! We can't heat a moon rock above 100C, but that doesn't actually tell us that we can't get a different material hotter than that. This is how greenhouses work, or in reverse, that cooling paint (https://www.parc.com/technologies/self-cooling-paint/).

[NichG]

That XKCD irks me somewhat, because it doesn't mention that it is possible to light a fire with a somewhat more complicated setup (though not with a magnifying glass), using the fact that we can filter by frequency. A material that does not absorb in the infrared spectrum but does absorb blue light can get extremely hot, even in moonlight. It only stops heating when it is emitting as much blue light as it is absorbing, and that means several hundred degrees! We can't heat a moon rock above 100C, but that doesn't actually tell us that we can't get a different material hotter than that. This is how greenhouses work, or in reverse, that cooling paint (https://www.parc.com/technologies/self-cooling-paint/).

Careful here, the math doesn't quite work out the way you're saying...

Constants aside, the power emitted per surface area and angle is ~ integral dw w^3/(exp(w/T)-1) where w is the frequency of light. Equilibrium occurs when the total power in = total power out. So now lets say you put a filtering function 0<=g(w)<=1 on that for a given target body, which applies equally for power absorbed and power emitted.

Total power in = integral dw g(w) input_spectrum(w)
Total power out = integral dw g(w) w^3/(exp(w/T)-1)

Lets say the input spectrum comes from a black body at temperature T'. That means that equilibrium occurs when:

integral dw g(w) w^3/(exp(w/T')-1) = integral dw g(w) w^3/(exp(w/T)-1)

That still occurs only when T=T'.

So regardless of a material's spectrum of absorbance and emittance (call this a transfer function so I don't have to keep writing that), it has the same equilibrium temperature with respect to an external black body source. Now, if the source has it's own filtering function 0<=h(w)<=0, it should be clear that it cannot increase the left side of that equation, only decrease it. So even if your emitter isn't a blackbody source, as long as its thermally induced emission there is no pair of transfer functions that can cause it to heat something else up to above its own temperature, only to a lower temperature at best.

[Fat Rooster]

Careful here, the math doesn't quite work out the way you're saying...

Constants aside, the power emitted per surface area and angle is ~ integral dw w^3/(exp(w/T)-1) where w is the frequency of light. Equilibrium occurs when the total power in = total power out. So now lets say you put a filtering function 0<=g(w)<=1 on that for a given target body, which applies equally for power absorbed and power emitted.

Total power in = integral dw g(w) input_spectrum(w)
Total power out = integral dw g(w) w^3/(exp(w/T)-1)

Lets say the input spectrum comes from a black body at temperature T'. That means that equilibrium occurs when:

integral dw g(w) w^3/(exp(w/T')-1) = integral dw g(w) w^3/(exp(w/T)-1)

That still occurs only when T=T'.

So regardless of a material's spectrum of absorbance and emittance (call this a transfer function so I don't have to keep writing that), it has the same equilibrium temperature with respect to an external black body source. Now, if the source has it's own filtering function 0<=h(w)<=0, it should be clear that it cannot increase the left side of that equation, only decrease it. So even if your emitter isn't a blackbody source, as long as its thermally induced emission there is no pair of transfer functions that can cause it to heat something else up to above its own temperature, only to a lower temperature at best.

Oh, sure. You absolutely cannot get hotter than the source of the light, but the moon is not the source of the light, and it irks me how he dismisses that and simplifies in a way that causes confusion because spectral effects are extremely important. You also could not use lenses to focus moonlight more effectively than if you just had a lump of your material sitting on the moon. You absolutely could craft a material that got hotter sitting on the moon than the moon rocks that surround it, even if it was not directly in sunlight. The factor of 400,000 that is lost to the whole reflecting off the moon part is more than made up for by the exponential in the equation and the fact we don't need to get that hot (250C is enough for paper). If our transfer function is 0 below the visible spectrum then the question becomes whether an object at 250C emits or absorbs more light in the visible spectrum when exposed to moonlight, or equivalently, whether we could see better looking into an oven at 250C with no other light sources, or a moonlit scene. Things don't even start to visibly glow until ~500C, and if we wanted to we could restrict ourselves further to the blue end of the spectrum.

I'm not saying any of this is practical (at best you could manage ~4mW/m2, so even if you could get it to work perfectly it would take a while), but it does not violate thermodynamics to use the visible part of moonlight to heat something above the flash point of paper despite the fact that moon rocks don't get that hot. It is impossible to use even the near infrared part, because then you would cool faster than you could possibly heat, but things need to get extremely hot before they start to emit any significant energy in the visible spectrum. Thermodynamics just means you cannot exceed the initial temperature, it doesn't say anything like "if you only get 1/4 of the light you can only heat to 1/4 of the temperature", or any function at all. Even if we get a millionth of the light, our limit is still the same temperature, because the exponential means that for any constant c and temperatures T > T' there is a frequency above which a body at temperature T will emit c times more energy at those frequencies than one at T'.

[NichG]

Oh, sure. You absolutely cannot get hotter than the source of the light, but the moon is not the source of the light, and it irks me how he dismisses that and simplifies in a way that causes confusion because spectral effects are extremely important. You also could not use lenses to focus moonlight more effectively than if you just had a lump of your material sitting on the moon. You absolutely could craft a material that got hotter sitting on the moon than the moon rocks that surround it, even if it was not directly in sunlight.

If it were in sunlight, you could do this. But you absolutely could not craft such a material that would get hotter in equilibrium than the moon rocks surrounding it when outside of sunlight. The equations in my last post are the proof of why that doesn't work.

[Fat Rooster]

If it were in sunlight, you could do this. But you absolutely could not craft such a material that would get hotter in equilibrium than the moon rocks surrounding it when outside of sunlight. The equations in my last post are the proof of why that doesn't work.

They are really not. Run the numbers and see. Set a g(w)=0 for low values of w, and g(w)=1 for high values. For low T the w/T is big before g(w) becomes 1, and that's an exponential term keeping power down. The emissions are small everywhere on the spectrum, either because of low transfer function, or low temperature. Power in meanwhile can be significant because the badly reflected sunlight can interact with the high part of the transfer function. Even the 20 fold reduction we eat from reflecting off moon rock is nothing on the power of that exponential term, meaning temperature has to be surprisingly high in order to be in equilibrium.
The point is that we are just using the moon rocks as (really bad) mirrors. Their temperature doesn't matter when used like that. The equilibrium temperature of a moon rock in that situation is going to be the same as any other moon rock, but that doesn't actually tell us anything about the equilibrium temperature of any other materials. It always depends on the transfer function. Normal materials are reasonably similar in having some IR response, but that doesn't tell us anything about the materials that can be produced, and definitely nothing about the thermodynamic limits of what a material can do. With enough layers of glass effective IR response can be brought down arbitrarily low, so it doesn't even need nanomaterial tricks. A good greenhouse design can do it. You can see that you can get to 250C, because objects at 250C are less bright in the visible spectrum than objects illuminated by moonlight. If you isolate a 250C object in all except the visible spectrum it will receive more light than it emits, even from moonlight. The power is miniscule, so your isolation needs to be extremely good, but it will be getting hotter if you manage that.

Maybe I'm misunderstanding, and you are just objecting to the practicality. I'm under no illusions that this is practical, just that thermodynamics has no beef with it.

Also, black body radiation for power in isn't really what you are looking for, because that implies you are completely surrounded by it. A <1 scalar multiple of black body is more appropriate, and when you do that it is rapidly pretty clear that power in and power out are rarely the same for particular values of w. In anything other than the completely surrounded case the specifics of g make a big difference.

[NichG]

They are really not. Run the numbers and see. Set a g(w)=0 for low values of w, and g(w)=1 for high values.

Sure. So lets take the source as a blackbody at T=90K (temperature of the dark side of the moon) and a threshold value of w at 10^15 Hz, well into the UV. Lets assume your material is completely buried in moon rock, so no losses to space or anything, just exchange between the sample and the moon rock.

The power absorbed by the sample per unit surface area is: 2h/c^2 * integral from 10^15 Hz to infinity [ w^3 / (exp(h w / k T) - 1)] dw. We can substitute u=(h/kT) w, and that becomes:

Power absorbed / m^2 ~= 2.78 * 10^-9 kg/(s^3 K^4) * T^4 * integral from 500 to infinity u^3/(exp(u)-1)

Now, at those values of u, the integral is basically u^3 exp(-u), which we can integrate in closed form, and it turns out to be 9 * 10^-210. So the total power absorbed per unit surface area is 1.64e-210 kg/s^3. A ridiculously small number because this material only emits or absorbs radiation in the high UV.

What's the temperature at which that is equal to the power emitted? Well, it's the same integral, but now with a temperature T'. Closed form here, approximating exp(big number)-1 ~ exp(big number), ends up being:

Power emitted / m^2 ~= T'^4 * 2.78 * 10^-9 * 9 * 10^-210 kg/(s^3 K^4). Notice something? It's the exact same equation, but with T' instead of T. So power emitted = power absorbed only when T' = T.

Now, what if we didn't consider the lunar rock to be a blackbody emitter? Well, that would strictly decrease the power absorbed at a given temperature, while leaving the power emitted to be the same. So in that case, you would have strictly T' < T.

[Bohandas]

What temperature, if any, would something have to be for a significant percentage of its blackbody radiation to emit mainly in the 300mhz-300-ghz range

EDIT:
And in particular, could frequencies like this dominate at some temperature or another that is less than or equal to 1000C?

Edit:
Because I read a Wired article (https://www.wired.com/2007/05/using-a-microwa/) about using radiation in this range to heat objects to 1000C

EDIT:
Tangential question. If you had unlimited time and resources, could you theoretically use the cosmic microwave background to cook chicken? Like if you had a parabolic reflector the size of a galaxy set up in intergalactic space or something like that

[NichG]

What temperature, if any, would something have to be for a significant percentage of its blackbody radiation to emit mainly in the 300mhz-300-ghz range

EDIT:
And in particular, could frequencies like this dominate at some temperature or another that is less than or equal to 1000C?

So one thing to keep in mind is that there is no frequency at which the total energy at that frequency decreases with increasing temperature. As you increase the temperature, the higher frequencies increase faster than the lower ones, but everything increases. That means that even if you have a frequency range that dominates at low temperature, the object is still emitting more energy in those frequencies at a higher temperature in which that range is not dominant anymore.

The temperature at which 300mhz is the peak frequency is ~0.003 Kelvin. The temperature at which 300ghz is the peak frequency is ~3 Kelvin. A 30 Kelvin object will emit more radiation at 300ghz than a 3 Kelvin one.

Edit: There's evidently a subtle thing about what 'peak' means for a density function, so what I did here (c/peak wavelength) isn't strictly correct, because dw and dlambda are different measures...

[Bohandas]

In any case, the point is that you can use electromagnetic radiation in that range to heat things to at least 100°C, and I've heard of it being used to heat objects to 1000°C (https://www.wired.com/2007/05/using-a-microwa/). You just need to have enough of it.

[NichG]

In any case, the point is that you can use electromagnetic radiation in that range to heat things to at least 100°C, and I've heard of it being used to heat objects to 1000°C (https://www.wired.com/2007/05/using-a-microwa/). You just need to have enough of it.

Yes, there's nothing about a certain frequency that inherently limits the temperature energy in that frequency can be used to heat something to. It's the balance of total power in versus total power out.

[Rockphed]

EDIT:
Tangential question. If you had unlimited time and resources, could you theoretically use the cosmic microwave background to cook chicken? Like if you had a parabolic reflector the size of a galaxy set up in intergalactic space or something like that

The short answer is no. The longer answer involves the effective temperature of the CMB (a few kelvin), its extent (4 pi steradians), and essentially everything else in this thread.

[Radar]

Yes, there's nothing about a certain frequency that inherently limits the temperature energy in that frequency can be used to heat something to. It's the balance of total power in versus total power out.
That and the ability to focus the energy well enough. If the source of radiation is directional, the light/microwaves/whatever can be focused far better than blackbody radiation.

[Fat Rooster]

Sure. So lets take the source as a blackbody at T=90K (temperature of the dark side of the moon) and a threshold value of w at 10^15 Hz, well into the UV. Lets assume your material is completely buried in moon rock, so no losses to space or anything, just exchange between the sample and the moon rock.

The power absorbed by the sample per unit surface area is: 2h/c^2 * integral from 10^15 Hz to infinity [ w^3 / (exp(h w / k T) - 1)] dw. We can substitute u=(h/kT) w, and that becomes:

Power absorbed / m^2 ~= 2.78 * 10^-9 kg/(s^3 K^4) * T^4 * integral from 500 to infinity u^3/(exp(u)-1)

Now, at those values of u, the integral is basically u^3 exp(-u), which we can integrate in closed form, and it turns out to be 9 * 10^-210. So the total power absorbed per unit surface area is 1.64e-210 kg/s^3. A ridiculously small number because this material only emits or absorbs radiation in the high UV.

What's the temperature at which that is equal to the power emitted? Well, it's the same integral, but now with a temperature T'. Closed form here, approximating exp(big number)-1 ~ exp(big number), ends up being:

Power emitted / m^2 ~= T'^4 * 2.78 * 10^-9 * 9 * 10^-210 kg/(s^3 K^4). Notice something? It's the exact same equation, but with T' instead of T. So power emitted = power absorbed only when T' = T.

Now, what if we didn't consider the lunar rock to be a blackbody emitter? Well, that would strictly decrease the power absorbed at a given temperature, while leaving the power emitted to be the same. So in that case, you would have strictly T' < T.

Oh, right. That wasn't what I meant at all. I meant on the day side of the moon, but under an umbrella so it wasn't exposed to direct sunlight (because then sunlight just dominates anyway). I was taking it as near the limit of what could be achieved with a magnifying glass and moonlight. I didn't at all intend to mean that the blackbody radiation from the moon rocks was in the least helpful, but the reflected sunlight certainly is, even massively weakened. The temperature of the moon rocks is irrelevant to that, and I took it as read that moonlight was mostly understood to be the reflected part, and not it's own blackbody part. A decent approximation is a blackbody at 6000K with all values divided by 400,000, and then adding on a blackbody at 350K. I don't think anybody was seriously suggesting they could get 90K moon rocks to do useful work, though I guess the natural conclusion to getting better than 5700C from solar power is that you could use a similar setup to do the same thing to get higher than 90K off a random moon rock (if I'm having to be clear, this is definitely impossible).

With regards to getting CMB to cook a chicken, certainly not directly, for the reasons outlined throughout this thread. If you had a lump of matter that was already colder than 2.7K though, you car create a heat engine between the CMB and your cold sink, and then use that useful work to do whatever you like. Sort of cheating though, as you 'expend' the cold sink to do it.

With regards to radio heating, it's called a microwave :smallsmile:*. The point is that microwaves are nowhere near black body radiation. They (similar to lasers, and LEDs, and many astronomical radio sources) rely on turning useful work into light directly, which means they do not have to care about the temperature of anything.** I've never actually run the numbers on how hot something would need to be to emit more microwave radiation than it absorbed inside a standard microwave, but I would not be surprised if it was in the millions of degrees***. Radio waves are used in fusion reactors for heating, which means we are capable of working with greater radio fluxes than a blackbody at 100 million K.

* There are some great designs out there for casting aluminium using an off the shelf microwave and a specially designed crucible. Not quite the thousands of degrees, but not exactly optimised either.

** I guess this means you might be able to use a pulsar to cook a chicken, on account of it being a non thermal emitter. Given it is also hot enough to do it thermally too, this isn't a big deal though.

*** The question might even be sort of meaningless, as it is highly dependent on how narrow your spectral peak is, and what frequency window we are actually talking about.

[Bohandas]

]With regards to radio heating, it's called a microwave :smallsmile:*. The point is that microwaves are nowhere near black body radiation.=

I don;t understand why that matters. Supposing we ran it all through an astronomical filter that only lets those wavelengths through, how could you then tell whether it came from a magnetron or somewhere else

[Captain Cap]

I don;t understand why that matters. Supposing we ran it all through an astronomical filter that only lets those wavelengths through, how could you then tell whether it came from a magnetron or somewhere else
By the intensity.

[Radar]

I don;t understand why that matters. Supposing we ran it all through an astronomical filter that only lets those wavelengths through, how could you then tell whether it came from a magnetron or somewhere else
If you have a magnetron or some other artificial source based on a resonant circuit, the wave is coherent, so it can be focused much better or accumulated in a resonance chamber. Black body radiation is very much not coherent: there are a lot of different incoming directions as was already discussed in this thread. No passive optical equipment can convert incoherent light into a coherent beam


By the intensity.
Not relevant actually. Or rather this is the effect of the generated wave being coherent. This is actually the reason why for the highest intensities people use lasers as they are the most coherent light sources we can make, so can be focused way better.

[halfeye]

With regards to radio heating, it's called a microwave :smallsmile:*. The point is that microwaves are nowhere near black body radiation. They (similar to lasers, and LEDs, and many astronomical radio sources) rely on turning useful work into light directly, which means they do not have to care about the temperature of anything.**

My question is how does the universe know, for any individual photon, whether it came from a source that had a particular temperature?

If we had that solar furnace set up half way to the sun, and we fired a laser at the target how would the universe differentiate the photons from that laser from the photons from the sun? I'm not suggesting that the photons from the laser would be enough to make a significant difference, I'm just absolutely fascinated that apparently they would be treated differently, and I want to know how and why.

[Gnoman]

It doesn't, and it doesn't have to. You can't have a reflection brighter than the source, and that's true for temperature as well.

[Anymage]

My question is how does the universe know, for any individual photon, whether it came from a source that had a particular temperature?

If we had that solar furnace set up half way to the sun, and we fired a laser at the target how would the universe differentiate the photons from that laser from the photons from the sun? I'm not suggesting that the photons from the laser would be enough to make a significant difference, I'm just absolutely fascinated that apparently they would be treated differently, and I want to know how and why.

The photons don't need to have any secret knowledge. If you were to stand off at the side and shine your laser at the solar furnace, you wouldn't have a good angle to have your laser hit a point at the focus. If you were to try injecting your laser at an angle where it could reach the focus, you'd be standing in the sunlight and thus blocking some of this. Even a fiber optic setup would cast some level of shadow here. If you were to cut a small hole in a mirror somewhere and shine your laser through, you'd similarly be preventing the photons that would have otherwise hit that bit of mirror from eventually reaching the focus. The optics simply don't allow for the direct injection of random extra photons into the beam.

[Fat Rooster]

I don;t understand why that matters. Supposing we ran it all through an astronomical filter that only lets those wavelengths through, how could you then tell whether it came from a magnetron or somewhere else

Well you couldn't be 100% sure that it wasn't a billion degree cloud of gas that was for some reason only emitting in the radio spectrum, but it is a pretty reasonable guess that it isn't, and some non thermal process is creating the light. While a black body can get to the intensity we observe, it has to be absurdly hot to do so. Non thermal processes can get to effectively arbitrary intensities without being hot. This doesn't violate thermodynamics because it relies on 'useful work', which can in some sense be regarded as a hot end with infinite temperature, but mostly just means the entropy has gone somewhere else.

[halfeye]

It doesn't, and it doesn't have to. You can't have a reflection brighter than the source, and that's true for temperature as well.

First off, I read your first sentence as contradicting your second.

Secondly, I am not talking about the surface of the mirror being hotter than the sun, which is where, as I understand it, the reflection is.


The photons don't need to have any secret knowledge.

If it were secret we woudn't be discussing it. Your side of this discussion seems to be saying that a if photon comes from a source with a particular temperature then it can't make its impact point have a higher temperature, while at the same time saying that the photon can't carry information about the temperature of the point it came from. I want to understand where that information is. That information surely has an entropy of its own.


If you were to stand off at the side and shine your laser at the solar furnace, you wouldn't have a good angle to have your laser hit a point at the focus. If you were to try injecting your laser at an angle where it could reach the focus, you'd be standing in the sunlight and thus blocking some of this. Even a fiber optic setup would cast some level of shadow here. If you were to cut a small hole in a mirror somewhere and shine your laser through, you'd similarly be preventing the photons that would have otherwise hit that bit of mirror from eventually reaching the focus. The optics simply don't allow for the direct injection of random extra photons into the beam.

I don't see that. There is clearly room at the side of the solar furnace where a laser could be sited, and a laser beam is relatively narrow, it ought to be possible to fire one through a small hole in the mirror that was brighter/hotter than the sunlight on the surface of the mirror at that point.

[NichG]

Oh, right. That wasn't what I meant at all. I meant on the day side of the moon, but under an umbrella so it wasn't exposed to direct sunlight (because then sunlight just dominates anyway). I was taking it as near the limit of what could be achieved with a magnifying glass and moonlight. I didn't at all intend to mean that the blackbody radiation from the moon rocks was in the least helpful, but the reflected sunlight certainly is, even massively weakened. The temperature of the moon rocks is irrelevant to that, and I took it as read that moonlight was mostly understood to be the reflected part, and not it's own blackbody part. A decent approximation is a blackbody at 6000K with all values divided by 400,000, and then adding on a blackbody at 350K. I don't think anybody was seriously suggesting they could get 90K moon rocks to do useful work, though I guess the natural conclusion to getting better than 5700C from solar power is that you could use a similar setup to do the same thing to get higher than 90K off a random moon rock (if I'm having to be clear, this is definitely impossible).


Okay, I do agree with this, and it seems to check out when I solve numerically for Tobj using step-function-like g(w). Basically the equilibrium temperature sits around 240K for both too-low cutoff and too-high cutoff, but there's a peak in the middle where I can get things to 1000K or so, for a cutoff around 1um. Gotta double check the numerics though because this is using really huge arguments to an exponential and I don't 100% trust that there isn't some issue with integration error.

[Radar]

My question is how does the universe know, for any individual photon, whether it came from a source that had a particular temperature?

If we had that solar furnace set up half way to the sun, and we fired a laser at the target how would the universe differentiate the photons from that laser from the photons from the sun? I'm not suggesting that the photons from the laser would be enough to make a significant difference, I'm just absolutely fascinated that apparently they would be treated differently, and I want to know how and why.
There is no secret knowledge. It is all about the coherence of the beam you want to focus. If you indeed have a parallel beam, you can focus it really well - up to the diffraction limit which is around a single wavelength in size. But any kind of blackbody radiation is not coherent - as each point of the star surface sends out radiation in all possible directions this is all simple geometrical optics. Even if you set your mirrors far enough to have almost parallel beams on each of them, that "almost" still ruins your efforts: light comes from different directions, so you cannot reflect it all in the same direction - the reflection will spread out as well. And if you had those mirrors far away from the star, they are also far away from the target, so any inaccuracy in directing the beam will create huge effects on the target. If on the other hand you place the mirrors close to the stars, the incoming beam will be decidedly less parallel, so the reflection will be spread around more and closer distance to the target cannot compensate that.

If I have some time, I will try to draw a few examples for this.

[halfeye]

And if you had those mirrors far away from the star, they are also far away from the target

This is mistaken. The solar furnace I have been talking about:

https://en.wikipedia.org/wiki/Odeillo_solar_furnace

is one AU from the sun, and the target is part of it, certainly within one kilometre, probably within 100 metres.

[Bohandas]

It's a focusing problem. It's not that there isn't a ton of heat, it's that there's no way to concentrate it all in one spot perfectly.

That said, if we're working with insane megastructure level tech, there is a hack. If you put mirrors *all* around the sun(or any other source), you'll increase the heat of the overall system, and thus increase the heat you can focus on any point. For instance, visualize a spherical mirrored surface with a single pinhole. As the only point where light could escape, light striking anywhere else would be reflected and remain in the system. It doesn't truly break the rule, because the sun will heat up, and thus the pinpoint beam escaping will not actually be hotter than the surface, but it will be hotter than the surface would be without the system in place.

This is purely for illustration, the actual construction of such a thing would be more than a little insane for all kinds of practical reason. It's a spherical cow sort of example.

You don't need the pinhole. If true as described this would already prove the point without qualification by having heated the sun using sunlight

[Anymage]

I don't see that. There is clearly room at the side of the solar furnace where a laser could be sited

Please show me how a laser could be fired at a paraboloid at the side in such a way that it still hits the focus.

Which is the main point. If you could inject extra photons from outside or concentrate them beyond what optics actually allows or get them to run more parallel than they actually do, you could get a given point to be hotter than the source. Just like how if you could tap into a parallel universe with differing natural constants, you could violate conservation of energy. The point of optics is that none of the things mentioned above are actually possible.


and a laser beam is relatively narrow, it ought to be possible to fire one through a small hole in the mirror that was brighter/hotter than the sunlight on the surface of the mirror at that point.

How hot is your laser? If it's hotter than the surface of the sun, of course it can make things hotter than the sun. Just shine the laser on the object directly.

If your laser is hotter than the small patch of sunlight that would have fallen on the hole you cut out, you just made your solar furnace slightly hotter. The Odelio furnace is roughly half as hot as the surface of the sun, so you're nowhere near the limit of being as hot as the sun.

Build a replica of the Odelio furnace on Mercury? That's roughly half as far so it'll receive more direct radiation, but the difference between incoming sunlight being nearly parallel and being actually parallel means that it's harder to focus the beams properly and you'll have to have a larger and thus more diffuse image at the focus.

Go full XKCD, say "we're doing the math, we can do whatever we like" and surround the sun and earth with a massive ellipsoid mirror with the sun at one focus and the earth at the other? If both the sun and the earth are point sources, you have perfect transfer of every photon and the two wind up the same temperature. Make them actual size, and the inability to get a perfect focus means that you won't get 100% energy transfer and wind up with just enough to equalize temperatures. And while if you could magically interject the beam of a laser pointer in somewhere you could wind up with a hotter temperature on earth, good luck doing so without having your laser cast a shadow due to being in the way of some solar photons. You're going to be hard pressed to have laser photons outweigh the value of the laser's shadow unless the laser is hotter than the surface of the sun.

[Quizatzhaderac]

If we had that solar furnace set up half way to the sun, and we fired a laser at the target how would the universe differentiate the photons from that laser from the photons from the sun? Well, you can't know for certain, but you could probably tell by the angle. The sunlight is coming in at a (relatively) broad range of angles and the laser light is coming in at a very narrow range of angles. So if the angle is too wide, you can be sure it's sunlight. If it's within the possible laser range you can do a calculation to determine the probability it came from the laser.

[Rockphed]

If it were secret we woudn't be discussing it. Your side of this discussion seems to be saying that a if photon comes from a source with a particular temperature then it can't make its impact point have a higher temperature, while at the same time saying that the photon can't carry information about the temperature of the point it came from. I want to understand where that information is. That information surely has an entropy of its own.

A photon has energy inversely proportional to its wavelength. How many photons at each wavelength something emits is determined by its absorption curve and its temperature. We normally talk about "black body" radiation. Any individual photon caries no information about where it came from, but the population of photons along a path carries that information. So we can look at the stars and tell how hot their surfaces are by looking at their spectrum. So we know that Sirius is super hot (I think) because we can tell what the population of photons it puts out looks like despite that it is giving us so little light that if Earth was only lit by Sirius at night it would be almost perfectly black.


You don't need the pinhole. If true as described this would already prove the point without qualification by having heated the sun using sunlight

The Sun is heated by fusion in its core. Tyndmere's experiment would change the rate at which energy leaves the sun and might result in the surface being hotter than in its current state but I am not prepared to state that for certain.

[Bohandas]

Ok, another idea. Does the double slit apparatus count as active or passive? The problem is something to do with photons interfering with other, right? So what if we ran them through something that forced them to act as a particle rather than a wave?

[halfeye]

Ok, another idea. Does the double slit apparatus count as active or passive? The problem is something to do with photons interfering with other, right? So what if we ran them through something that forced them to act as a particle rather than a wave?

The twin slits apparatus does that. They act as waves of particles. They are both waves and particles, all the time. I'm not comfortable with it, but that seems to be the way it is.

[Rockphed]

The two slit experiment can be used to find the relationship between uncertainty of momentum and position. I think it cones to something like h=delta p delta x, where h is the boltzmann constant, delta p is the uncertainty of momentum and delta x is the uncertainty of the position.

Two slits constitute a (really lossy) lens. Okay, they are a diffraction grating, but those have all the restrictions of a lens regarding ray tracing. You might be able to use a diffraction grating to add two rays, but only if they are already coherent and you probably scatter more than half the energy.

[Bohandas]

The implicztion was that we'd also be monitoring which slit each photon goes through. That kind of double slit apparatus. The kind that makes photons stop interfering with each other.

[Captain Cap]

The implicztion was that we'd also be monitoring which slit each photon goes through. That kind of double slit apparatus. The kind that makes photons stop interfering with each other.
While the double slit itself is passive, the action of monitoring the photons is not: to do so you have to inject stuff in the system to interact with the photons.

[Anymage]

First, the double slit experiment isn't about photons interacting with each other. It's about the photon self-interacting. Photons going through the slits can freely interact with other photons, and even with themselves over the distance between slit and detector.

Second and more importantly, the vast majority of optics focuses on light as rays and can quite happily ignore quantum effects. It's similar to Newtonian mechanics and moving objects. If you know an object's position, speed and direction, you can make very good guesses as to its past and future trajectory. And you aren't going to have two objects with the same position and momentum unless they're already on the same path. There is no way to have objects following different paths to spontaneously start following the same path. Optics says pretty much the same thing about light rays, only allowing for the effects of refraction and reflection.

[Quizatzhaderac]

The problem is something to do with photons interfering with other, right? So what if we ran them through something that forced them to act as a particle rather than a wave?The problem is that the same spot on the reflector dish will receive photons that are definitely coming in at different angles, and that many of those angles will be wrong to go on to hit the furnace. The photons are already acting like particles.

[halfeye]

Please show me how a laser could be fired at a paraboloid at the side in such a way that it still hits the focus.

My idea was to hit the focus from the side of the parabola without hitting the parabola in between.


How hot is your laser? If it's hotter than the surface of the sun, of course it can make things hotter than the sun. Just shine the laser on the object directly.

Yeah that's the plan. My point is there are photons coming from the laser, which are apparently allowed to be hotter than the sun, and photons from the sun which aren't allowed to be hotter than the sun, and I want to know how the two sets of photons are differentiated. They apparently aren't separated on the basis of frequency, so what does it?


If your laser is hotter than the small patch of sunlight that would have fallen on the hole you cut out, you just made your solar furnace slightly hotter. The Odelio furnace is roughly half as hot as the surface of the sun, so you're nowhere near the limit of being as hot as the sun.

Yeah, the heat isn't a significant factor, it's the fact that there are two populations of photons which are treated differently that I find astounding.

[NichG]

My idea was to hit the focus from the side of the parabola without hitting the parabola in between.

So, in total you have incoming directions which cover the surface of a sphere as your 'budget'. An infinite parabolic reflector takes up all angles except straight down the center, but in turn can extract nearly 100% of the possible power flux from a given light source at infinity. A finite parabolic reflector leaves an open circular patch, rather than an open point. Any extra light you want to pump in has to come along those open directions. So if you're using a finite reflector, you can supplement it, but you can only supplement it over a finite region - that's essentially your angular budget, and that angular budget + the power per solid angle coming off of your source determines the maximum possible power you can have entering the focal point.



Yeah that's the plan. My point is there are photons coming from the laser, which are apparently allowed to be hotter than the sun, and photons from the sun which aren't allowed to be hotter than the sun, and I want to know how the two sets of photons are differentiated. They apparently aren't separated on the basis of frequency, so what does it?


They're not 'differentiated' in the sense of the photons being fundamentally different. But different sources can have a different power per solid angle per m^2. The law of etendue implies you can't use passive optics to increase the power per solid angle per m^2, only decrease it. So you're stuck at whatever that value is for the brightest source you have access to. You can't 'layer' photons travelling in the same direction to e.g. superimpose two beams and get a beam twice as bright because photons can't catch up with each-other and because (for passive optics) paths are uniquely reversible, whereas a superimposed set of two parallel beams would imply that one destination had two source directions at one point.

[Radar]

This is mistaken. The solar furnace I have been talking about:

https://en.wikipedia.org/wiki/Odeillo_solar_furnace

is one AU from the sun, and the target is part of it, certainly within one kilometre, probably within 100 metres.
Same problems apply. The Sun as seen from Earth has a angular diameter of about 0.5 degree. Whatever mirrors you are using, the reflected light will have at least the same angle of spread which limits your ability to focus light.

Let's do some math to show how things play out: So you have a paraboloid mirror with a 100 m radius and let's assume a focal point that is on average those 100 m away from each point of that mirror. As seen from Earth, Sun has an angular diameter of about 0.5 degree. This is the spread of directions of the incoming light and by consequence the absolute minimum on the spread of light reflected from the mirror. So you will be able to get maximum focus of your light roughly in a ball 0.87 m in diameter (0.5 arc degree object 100 m away from the mirror) assuming a perfect mirror. So given sunlight's power density of 1370 W/m^2 you will push up to 4.3*10^2 W of power into your sphere. In thermal equilibrium your sphere will reach the temperature of about 4200 K. What prevents you from getting higher temperatures is the minimum size of the focal spot you can make.

Hypothetically with the same setup as before, if we used a source with better coherence which results in a spread of incoming light of let's say 0.1 degree, our focal spot would now have a diameter of just 0.17 m and given the same incoming light intensity of 1370 W/m^2 we end up with temperature of 9400 K. So better coherence of the source, higher possible temperature of the target.

Another hypothetical: we halve the distance of our solar furnace to the Sun. The Sun now has 1 degree angular diameter, so our focal spot is proportionally bigger with 1.75 m diameter, sunlight intensity increases to 5480 W/m^2 and the resulting temperature will be... 4200 K as all the effects of the distance to the Sun cancel each other out. More power, but larger focal spot.

Lasers produce coherent beams which have a spread very close to the diffraction limit, so the wider the aperture, the more parallel the beam is. This means you can focus laser light way better than sunlight. There is a bit more to it, but from the point of view of geometrical optics, they do allow as to get as parallel source as we want and the limit on the size of the focal spot is defined only by the wavelength of the produced light, which is typically many orders of magnitude below what we had for that solar furnace.

So the difference between various sources of light and how they can be used to heat something up is in how orderly the resulting light is. The more ordered the light, the more one can focus it and as a result the higher temperature we can get.

[halfeye]

So, in total you have incoming directions which cover the surface of a sphere as your 'budget'. An infinite parabolic reflector takes up all angles except straight down the center, but in turn can extract nearly 100% of the possible power flux from a given light source at infinity. A finite parabolic reflector leaves an open circular patch, rather than an open point. Any extra light you want to pump in has to come along those open directions. So if you're using a finite reflector, you can supplement it, but you can only supplement it over a finite region - that's essentially your angular budget, and that angular budget + the power per solid angle coming off of your source determines the maximum possible power you can have entering the focal point.

I do not understand this.


They're not 'differentiated' in the sense of the photons being fundamentally different. But different sources can have a different power per solid angle per m^2. The law of etendue implies you can't use passive optics to increase the power per solid angle per m^2, only decrease it. So you're stuck at whatever that value is for the brightest source you have access to. You can't 'layer' photons travelling in the same direction to e.g. superimpose two beams and get a beam twice as bright because photons can't catch up with each-other and because (for passive optics) paths are uniquely reversible, whereas a superimposed set of two parallel beams would imply that one destination had two source directions at one point.

I don't think it's useful in this context, but you can have two beams along the same path. You would need a pane of something reflective, that also allowed transmission from front to back, so one beam came in and reflected off the front, and another beam came from behind and transmitted straight through. You'd have losses on the tranmission, but severe losses on the reflection, so you probably couldn't get above 100%, but it is conceptually possible, the reversal splits the beams at the reflection.

[NichG]

I don't think it's useful in this context, but you can have two beams along the same path. You would need a pane of something reflective, that also allowed transmission from front to back, so one beam came in and reflected off the front, and another beam came from behind and transmitted straight through. You'd have losses on the tranmission, but severe losses on the reflection, so you probably couldn't get above 100%, but it is conceptually possible, the reversal splits the beams at the reflection.

You point it out yourself - that is not a reversible passive optical device, because there would be unavoidable losses involved in its operation. Basically, the device you describe is thermodynamically impossible as a passive system, at least for losses that added up to less than 50%.

[halfeye]

You point it out yourself - that is not a reversible passive optical device, because there would be unavoidable losses involved in its operation. Basically, the device you describe is thermodynamically impossible as a passive system, at least for losses that added up to less than 50%.

I do not understand your statement that the device is not reversible. It is not capable of producing more than one beam's worth of energy from its two combined beams, but it seems to me it clearly is reversible, if you fire the beams back where they came from, they are reversed and recombined with no problems at all. There are all sorts of losses in it, reflections from the front and back of the pane, absorbtion, reflections from the transmitted beam that don't line up with the transmissions from the reflected beam, and if you fire those beams back there will be more reflections and so on, but I can't see any way this is impossible, or threatens in any way to make a perpetual motion machine.


Same problems apply. The Sun as seen from Earth has a angular diameter of about 0.5 degree. Whatever mirrors you are using, the reflected light will have at least the same angle of spread which limits your ability to focus light.

Let's do some math to show how things play out: So you have a paraboloid mirror with a 100 m radius and let's assume a focal point that is on average those 100 m away from each point of that mirror.

I presume you are aware that with a paraboloid those averages would be considerably and consistently varied, unlike a spheroid, which would suffer spherical abberation. I'm also not at all sure that you'd want the focus that close to the surface.


As seen from Earth, Sun has an angular diameter of about 0.5 degree. This is the spread of directions of the incoming light and by consequence the absolute minimum on the spread of light reflected from the mirror. So you will be able to get maximum focus of your light roughly in a ball 0.87 m in diameter (0.5 arc degree object 100 m away from the mirror) assuming a perfect mirror. So given sunlight's power density of 1370 W/m^2 you will push up to 4.3*10^2 W of power into your sphere. In thermal equilibrium your sphere will reach the temperature of about 4200 K. What prevents you from getting higher temperatures is the minimum size of the focal spot you can make.

Hypothetically with the same setup as before, if we used a source with better coherence which results in a spread of incoming light of let's say 0.1 degree, our focal spot would now have a diameter of just 0.17 m and given the same incoming light intensity of 1370 W/m^2 we end up with temperature of 9400 K. So better coherence of the source, higher possible temperature of the target.

Another hypothetical: we halve the distance of our solar furnace to the Sun. The Sun now has 1 degree angular diameter, so our focal spot is proportionally bigger with 1.75 m diameter, sunlight intensity increases to 5480 W/m^2 and the resulting temperature will be... 4200 K as all the effects of the distance to the Sun cancel each other out. More power, but larger focal spot.

Lasers produce coherent beams which have a spread very close to the diffraction limit, so the wider the aperture, the more parallel the beam is. This means you can focus laser light way better than sunlight. There is a bit more to it, but from the point of view of geometrical optics, they do allow as to get as parallel source as we want and the limit on the size of the focal spot is defined only by the wavelength of the produced light, which is typically many orders of magnitude below what we had for that solar furnace.

So the difference between various sources of light and how they can be used to heat something up is in how orderly the resulting light is. The more ordered the light, the more one can focus it and as a result the higher temperature we can get.

I'm still not understanding why we can't use mirrors and lenses to make this incoming light more "ordered". We could theoretically boil water and make an electric light from that which could be as hot as we want, so I'm not seeing at all that that is significantly different from making temperature directly from the light. It's all energy, which can be measured in Watts, and more light is more Watts, so I don't see a thermodynamic problem.

[NichG]

I do not understand your statement that the device is not reversible. It is not capable of producing more than one beam's worth of energy from its two combined beams, but it seems to me it clearly is reversible, if you fire the beams back where they came from, they are reversed and recombined with no problems at all. There are all sorts of losses in it, reflections from the front and back of the pane, absorbtion, reflections from the transmitted beam that don't line up with the transmissions from the reflected beam, and if you fire those beams back there will be more reflections and so on, but I can't see any way this is impossible, or threatens in any way to make a perpetual motion machine.


Thermodynamically reversible doesn't just mean 'you can't make more than one beam worth of energy going backwards', it also means 'you can't make less than one beam worth of energy going backwards'. Idealized lenses and mirrors are reversible in that sense - its not just that they don't produce additional energy, they don't convert coherent energy to heat either.

If you're willing to suffer losses in energy, you absolutely could set up something with moonlight as an input and, say, produce a tiny speck at 1 billion Kelvin. For example, you could use a solar cell to charge a giant capacitor. The difference in target temperature and source temperature will determine the upper bound of the efficiency of the device that you could possibly build.

[Radar]

I presume you are aware that with a paraboloid those averages would be considerably and consistently varied, unlike a spheroid, which would suffer spherical abberation. I'm also not at all sure that you'd want the focus that close to the surface.
I simplified the math to make it tractable and get the key point across concerning target size and the resulting temperature.


I'm still not understanding why we can't use mirrors and lenses to make this incoming light more "ordered". We could theoretically boil water and make an electric light from that which could be as hot as we want, so I'm not seeing at all that that is significantly different from making temperature directly from the light. It's all energy, which can be measured in Watts, and more light is more Watts, so I don't see a thermodynamic problem.
You could for example produce electricity and use that to heat up your target arbitrarily high (within technological constrains), but this would incur inevitable losses that are rooted in thermodynamics: when you produce electricity, some of the source energy goes off as heat - this is inevitable as electrical current has very low entropy and sunlight (or any kind of blackbody radiation) has a comparatively higher entropy. So it has to go somewhere and take part of the energy with it. This way you do extract low entropy energy which you can use later to heat something how high our technology allows us regardless of the original source of energy. That being said, the lower the temperature of the original incoming radiation, the lower the efficiency of producing electricity as the entropy of light is related to its frequency distribution.

That being said, the initial problem was, if it is possible to use mirrors and lenses to heat something to a temperature higher than the source. Those are passive tools (as in they do not require any energy to operate) and the thermodynamic limits are applied directly. But even without involving thermodynamics, geometrical optics shows why it is not possible to do it as I tried to explain with my examples.

[halfeye]

Thermodynamically reversible doesn't just mean 'you can't make more than one beam worth of energy going backwards', it also means 'you can't make less than one beam worth of energy going backwards'. Idealized lenses and mirrors are reversible in that sense - its not just that they don't produce additional energy, they don't convert coherent energy to heat either.

Are any real world mirrors or lenses ideal in that sense? There is always some absorbtion as heat in either, and no lenses don't also reflect.

[Radar]

Are any real world mirrors or lenses ideal in that sense? There is always some absorbtion as heat in either, and no lenses don't also reflect.
That's not the point: they do not require energy to operate nor do they have to push out part of the incoming energy aside as a leftover the way heat engines, solar panels or other active devices do. There are always some losses beyond the ideal situation but it only means that in the the real world, you would get even worse results.

[halfeye]

That's not the point: they do not require energy to operate nor do they have to push out part of the incoming energy aside as a leftover the way heat engines, solar panels or other active devices do. There are always some losses beyond the ideal situation but it only means that in the the real world, you would get even worse results.

How is that not the point? They do absorb energy while operating, as heat perhaps, but how is absorbtion as heat different from other devices?

[Radar]

How is that not the point? They do absorb energy while operating, as heat perhaps, but how is absorbtion as heat different from other devices?
Because it is not inherent to their function - it is just a loss due to some imperfections. Moreover, those imperfection will not allow you to do something that an ideal mirror or lens could not do - they will only worsen the outcome and make the maximum temperature limit even lower.

On the other hand, the other devices even at their theoretical ideal will have inevitable losses that cannot be mitigated. Take Carnot cycle as one of the simplest examples: it has to dump heat into the cold reservoir in order to keep working and the ratio between energy used for work and that lost is determined by the temperatures of the reservoirs. Whatever you do, you cannot do better and actual engines obviously have even worse efficiency as nothing is ideal - especially since the isothermal compression and expansion would in theory take infinite time.

So the word "reversible" refers to processes that in ideal conditions are indeed reversible. In practice, you always have losses due to various imperfections, noise, etc. And if something is not possible in pure theory, it will not work in more realistic conditions either.

[halfeye]

Because it is not inherent to their function - it is just a loss due to some imperfections. Moreover, those imperfection will not allow you to do something that an ideal mirror or lens could not do - they will only worsen the outcome and make the maximum temperature limit even lower.

However, if the reason the theoretical items could not perform was because they were theoretical, the real items might be able to do what the theoretical ones couldn't?


On the other hand, the other devices even at their theoretical ideal will have inevitable losses that cannot be mitigated. Take Carnot cycle as one of the simplest examples: it has to dump heat into the cold reservoir in order to keep working and the ratio between energy used for work and that lost is determined by the temperatures of the reservoirs. Whatever you do, you cannot do better and actual engines obviously have even worse efficiency as nothing is ideal - especially since the isothermal compression and expansion would in theory take infinite time.

I don't understand that. I'm reminded of the real case of the wind powered vehicle that was faster than the wind driving it:

https://en.wikipedia.org/wiki/Blackbird_(wind-powered_vehicle)


So the word "reversible" refers to processes that in ideal conditions are indeed reversible. In practice, you always have losses due to various imperfections, noise, etc. And if something is not possible in pure theory, it will not work in more realistic conditions either.


In theory there is no difference between theory and practice. In practice there is.

[NichG]

However, if the reason the theoretical items could not perform was because they were theoretical, the real items might be able to do what the theoretical ones couldn't?


Well, if the reason you think it should be possible to heat something up above the source's temperature using optics doesn't actually make use of those specific dissipative interactions, it doesn't help resolve the state of your understanding I think... E.g. if the argument is 'I can just move closer', that argument doesn't actually depend on the degree of dissipation of your optics.

On the other hand, if you'd said something like 'I'm going to use a lens doped with a phosphorescent substance, so it retains and re-emits some photons at a different frequency' or 'I'm going to use a multi-photon process crystal to perform frequency doubling of the incoming sunlight' or things like that, those are mechanics which would explicitly make use of the dissipation to make a difference in the argument of how you could attain that higher temperature.

[Radar]

However, if the reason the theoretical items could not perform was because they were theoretical, the real items might be able to do what the theoretical ones couldn't?
No, at least not in this case. Let's get back to geometrical optics and start from the simplest possible case, which is a flat mirror (how it is relevant will be written later in this post):

1. Reflection means that the angle of reflected ray has the same angle with respect to the surface as the incident ray. That's an ideal mirror. A realistic mirror would scatter the incident light somewhat, so you get light going off in more or less a cone around that ideal reflection direction.

2. Let's take an ideal mirror and a light source that is not a single point (being infinitely far is the same thing as only the angular size matters), so if we look at a single point on the mirror surface, there is a whole cone of rays coming in from different spots on the source. When each ray is reflected ideally, the reflected light will form a cone with the same angular size as the incident one.

Now, what will happen with a realistic mirror? Each ray from the incoming cone of light will be reflected into many different direction roughly gathered in a cone as I wrote before. When you sum it all up, the incoming cone will be reflected into a wider cone (and without sharp edge, but that's not important) and the image of whatever is reflected becomes blurry.


So in my previous post I was working with ideal mirrors and the temperature at the focus was limited by the angular size of the light source, right? Once you account for the inaccuracies of your mirrors, the focal spot simply be even bigger as the angular size of the reflected beam will be inevitable larger than that of the incident one.

[georgie_leech]

No, at least not in this case. Let's get back to geometrical optics and start from the simplest possible case, which is a flat mirror (how it is relevant will be written later in this post):

1. Reflection means that the angle of reflected ray has the same angle with respect to the surface as the incident ray. That's an ideal mirror. A realistic mirror would scatter the incident light somewhat, so you get light going off in more or less a cone around that ideal reflection direction.

2. Let's take an ideal mirror and a light source that is not a single point (being infinitely far is the same thing as only the angular size matters), so if we look at a single point on the mirror surface, there is a whole cone of rays coming in from different spots on the source. When each ray is reflected ideally, the reflected light will form a cone with the same angular size as the incident one.

Now, what will happen with a realistic mirror? Each ray from the incoming cone of light will be reflected into many different direction roughly gathered in a cone as I wrote before. When you sum it all up, the incoming cone will be reflected into a wider cone (and without sharp edge, but that's not important) and the image of whatever is reflected becomes blurry.


So in my previous post I was working with ideal mirrors and the temperature at the focus was limited by the angular size of the light source, right? Once you account for the inaccuracies of your mirrors, the focal spot simply be even bigger as the angular size of the reflected beam will be inevitable larger than that of the incident one.

In other words, if a perfect mirror can only get you to a given temperature, an imperfect mirror that is inherently less efficient won't be able to do any better? And a similar argument applies for lenses?

[Radar]

In other words, if a perfect mirror can only get you to a given temperature, an imperfect mirror that is inherently less efficient won't be able to do any better? And a similar argument applies for lenses?
Yes, with trying to explain why is that.

[Lord Torath]

I'm still not understanding why we can't use mirrors and lenses to make this incoming light more "ordered". We could theoretically boil water and make an electric light from that which could be as hot as we want, so I'm not seeing at all that that is significantly different from making temperature directly from the light. It's all energy, which can be measured in Watts, and more light is more Watts, so I don't see a thermodynamic problem.Perhaps a visual indication?
You want a lens that does this:
https://cdn.discordapp.com/attachments/794253887686967337/981214076392833064/unknown.png

And that’s fine, it will take the rays coming from the indicated directions, and converge them into a beam. But the light from each point on the sun’s surface goes in all directions. The green lines show some other rays from the closest point of the star, and the lens will tend to scatter those rays, rather than aligning them with the others:
https://cdn.discordapp.com/attachments/794253887686967337/981213962169380924/unknown.png

There is no lens that can take two rays that hit the same point from different directions and direct them both in the same direction out the other side.

[halfeye]

In other words, if a perfect mirror can only get you to a given temperature, an imperfect mirror that is inherently less efficient won't be able to do any better? And a similar argument applies for lenses?

Except that an imperfect lens or mirror does not contradict the second law of thermodynamics which a perfect one might.

this is a diagram of the solar furnace I have been discussing, I cannot understand why this model would not work at half an AU, and if it would not as it stands, why it could not be modified (with for instance helistats altered to get more sunlight to the parabolic mirror)?

https://upload.wikimedia.org/wikipedia/commons/thumb/2/27/Four_en.svg/440px-Four_en.svg.png

[NichG]

Except that an imperfect lens or mirror does not contradict the second law of thermodynamics which a perfect one might.

this is a diagram of the solar furnace I have been discussing, I cannot understand why this model would not work at half an AU, and if it would not as it stands, why it could not be modified (with for instance helistats altered to get more sunlight to the parabolic mirror)?

https://upload.wikimedia.org/wikipedia/commons/thumb/2/27/Four_en.svg/440px-Four_en.svg.png

Well, as I said before, you're using things like a parallel rays approximation to understand how the device works. The reduction in intensity as you move away from a light source is a geometric consequence of the non-parallel nature of the rays from that source. So as you get closer to the sun, the parallel rays which all go perfectly into that parabolic focus will start to become more and more non-parallel, meaning that the size of the 'image' of those rays at the focus will grow and grow as you get closer. That's not to say you wouldn't get any improvement going from 1 AU to 0.5 AU, but as you get closer and closer, there are going to be diminishing returns. You won't just see a clean 1/r^2 scaling up of the power at the focus of the parabolic mirror. Instead, you'll see something that looks like 1/r^2 from far away, that eventually goes to a constant at r=0 rather than going to infinity.

Essentially, you're misinterpreting the 1/r^2 scaling as being something like 'the light itself is getting dimmer (intrinsic property)' when really it just comes from 'the same light is being spread out over a larger area (extrinsic property)'. Ultimately, the best you can do passively is take 'the entire output of the surface of the sun' and re-concentrate it down onto an area that is exactly the same as the surface area of the sun.

Here, this might be useful to get an intuition for these things. It's a web-based optics bench simulator: https://ricktu288.github.io/ray-optics/simulator/

[Radar]

Except that an imperfect lens or mirror does not contradict the second law of thermodynamics which a perfect one might.
None of them can contradict the second law of thermodynamics. The point is, an imperfect optical device obviously cannot focus things better than a perfect one - that's kind of there in the adjective and I think I described it in detail before.


this is a diagram of the solar furnace I have been discussing, I cannot understand why this model would not work at half an AU, and if it would not as it stands, why it could not be modified (with for instance helistats altered to get more sunlight to the parabolic mirror)?

https://upload.wikimedia.org/wikipedia/commons/thumb/2/27/Four_en.svg/440px-Four_en.svg.png
You will get higher power, but not temperature - the focus spot will simply get bigger. Why is that? That helistat setup makes the angular size of the incident beam substantially bigger, which inevitably cuts down your ability to focus it into a small area.

[DavidSh]

this is a diagram of the solar furnace I have been discussing, I cannot understand why this model would not work at half an AU, and if it would not as it stands, why it could not be modified (with for instance helistats altered to get more sunlight to the parabolic mirror)?

https://upload.wikimedia.org/wikipedia/commons/thumb/2/27/Four_en.svg/440px-Four_en.svg.png

Are the heliostats supposed to be flat mirrors? Then the angle of incidence equals the angle of reflection, and the width of the incoming beamlets should equal the width of the outgoing beamlets. Or is this just a drawing issue that shows them otherwise?

[Quizatzhaderac]

Are the heliostats supposed to be flat mirrors? Then the angle of incidence equals the angle of reflection, and the width of the incoming beamlets should equal the width of the outgoing beamlets. Or is this just a drawing issue that shows them otherwise?I'm pretty sure it's a simplification to make drawing easier.

this is a diagram of the solar furnace I have been discussing...THis diagram has some simpliciations and is analogous to Toarath's first diagram.

A more complete diagram would have two exta sets of lines.
1) Lines hitting the helostats but missing the parabolic mirror.
2) Lines hitting the helostats and parabolic mirrors, but missing the focus.

As you move the whole thing closer to the sun you get more of (1) and (2). Once you're close enough to the sun, moving the furnace closer just results in more light going to (1) and (2).
If you add more helostats and make the parabolic mirror bigger, you get a higher proportion of (2). With enough mirrors you eventually reach a point where additional mirrors are only increasing the amount of light going to (2).

[Mastikator]

Except that an imperfect lens or mirror does not contradict the second law of thermodynamics which a perfect one might.

this is a diagram of the solar furnace I have been discussing, I cannot understand why this model would not work at half an AU, and if it would not as it stands, why it could not be modified (with for instance helistats altered to get more sunlight to the parabolic mirror)?

https://upload.wikimedia.org/wikipedia/commons/thumb/2/27/Four_en.svg/440px-Four_en.svg.png

Sun rays are not parallel though. Not even to us on the surface of the Earth 149597871 kilometers away from the sun are they parallel. The sun occupies ~1/4 square degrees of the sky. If you want it to be more parallel it would have to be further away, which means you get less energy.

The way the math works out is that no matter how far away you are you can at theoretically best focus all of the sun's light into an area equally big as the surface of the sun. And when you do that you focus the sunlight into a temperature equal to the surface of the sun.

For sunlight to be parallel it would have to be observed at an infinite distance. And with an infinitely large lens at infinite distance you could focus the light to a mathematical point, however even then you couldn't get arbitrarily high temperature because at infinite distance it would take infinite time for the light to reach, meaning you'd end up with no energy actually. So no, not even then.

Edit-

Consider this. In this diagram you're showing the focus is only using the portion of the sun's light that happens to be parallel, which also happens to be only a fraction of the sun's emitted light. Light is emitted at 180 degrees squared from the surface and you're only getting the light that is traveling in parallel. Those foci never reach the surface temperature of the sun, at best they're a thousand degrees Celsius off.

[NichG]

(Bold formatting added by me)


Sun rays are not parallel though. Not even to us on the surface of the Earth 149597871 kilometers away from the sun are they parallel. The sun occupies ~1 square degrees of the sky. If you want it to be more parallel it would have to be further away, which means you get less energy.

The way the math works out is that no matter how far away you are you can at theoretically best focus all of the sun's light into an area equally big as the surface of the sun. And when you do that you focus the sunlight into a temperature equal to the surface of the sun.

For sunlight to be parallel it would have to be observed at an infinite distance. And with an infinitely large lens at infinite distance you could focus the light to a mathematical point, however even then you couldn't get arbitrarily high temperature because at infinite distance it would take infinite time for the light to reach, meaning you'd end up with no energy actually. So no, not even then.

Edit-

Consider this. In this diagram you're showing the focus is only using the portion of the sun's light that happens to be parallel, which also happens to be only a fraction of the sun's emitted light. Light is emitted at 180 degrees squared from the surface and you're only getting the light that is traveling in parallel. Those foci never reach the surface temperature of the sun, at best they're a thousand degrees Celsius off.

This is correct, except for the argument in the bold bit, which (whether it's true or not) is not the reason you can't get arbitrarily high temperature in that case.

When you have multiple things going to infinity, you have to be careful about how you take the limit. Here you've got a lens whose size grows to infinity, whose focal length grows to infinity, and whose distance from the source grows to infinity. However, you're taking the limit of 'the rays become parallel' first, then taking the infinite lens limit.

What you should instead do is look at the difference in angle between the upper-most ray impacting the lens and the lower-most ray impacting the lens. As you take both the distance and scale of the lens to infinity, that angular deviation can either go to zero or to 90 degrees depending on whether the lens gets larger faster than it gets further away, or slower. In the limit that the lens gets large slower than it becomes distant, the focus improves but the total amount of energy captured over the surface of the lens decreases with distance. In the limit that the lens gets large faster than it becomes distant, the quality of the focus decreases as you take the limit, so even though you eventually capture all of the energy coming out of one side of the sun, you end up concentrating it less and less.