4 ms·

Reversible means you don't increase the entropy of the system. If you relax this restriction, you can do useful work, like powering an arc welder or growing car

by Gatsky 8y ago

Reversible means you don't increase the entropy of the system. If you relax this restriction, you can do useful work, like powering an arc welder or growing carrots to feed yourself and rubbing sticks together to start a fire. Merely making the lens non-reversible doesn't mean it will suddenly set things on fire.

DenisM 8y ago
How does it work in practice? I am dumping some amount of low-energy photons onto a target, the target is heated up and radiating out the same amount of energy it receives. As I increase the number of photons hitting the target its temperature raises and it radiates more heat outwards, shedding excess heat. Then as I keep increasing the number of photons, the temperature stops increasing at some point having reached the temperature of the photon source. I increase the number of photons a million times of the previous level, yet the object remains the same temperature. Where does the excess energy go? Is it not absorbed by the target? Is it reflected? Is it re-radiated?
- Gatsky 8y ago
  Well I think the thermodynamic analysis is useful in understanding things, that was my point. I’m not quite understanding the scenario you describe - are you asking how it is possible that you can’t heat up the target more than the origin? If so then take an example where the sun is a single point at 5000 degrees. It is easy to see that the most you could do with this point sun is heat up one point on your object to the same temperature, no more. Now what if there were two point suns next to each other? The discussion on xkcd about optics is saying you can’t superimpose the points on each other on the target under any circumstances, even with two separate lenses at the right angle. This doesn’t seem like it should be impossible though... What about two single photon sources, can’t they be pointed at exactly the same spot? Maybe the explanation here is that the target electron cannot interact with 2 photons at the same time, so you can’t ‘double heat’ a single particle. Or maybe that you can’t precisely target a single particle without decreasing the entropy of a closed system, which is impossible.
- Dylan16807 8y ago
  So you have a blackbody source, and a target. Initially it's in a steady state where the target is colder. You increase the number of photons hitting the target. The target gets hotter. All good so far. But we have to look at how you increased that number. You need to either move the source and target closer together, or you have to use lenses to simulate the same thing. You can keep hitting the target with photons from more and more directions, but no system of mirrors or lenses can increase the apparent brightness of the source. As you approach the point where the source fills the entire sky, the target will approach the same temperature. And then you can't increase the number of photons any more. Not without using a hotter source.
- URSpider94 8y ago
  The point is, you can never get to the point where you are dumping more photons in to make it hotter, because there are not more photons. There is a limit to how small you can focus a lens. Once the object you are focusing is in sharp focus, it will start to go out of focus again if you go past that point. So you can’t just arbitrarily increase the photon density further. If you make your lens bigger to gather more photons, then the focused image just grows by a corresponding amount. Optics (and thermodynamics) tell us that we can not generate a higher flux of photons into a target area, than left the equivalent area at the source. Assuming that both the source and target are black body radiators, that says the target can’t get hotter than the source.
  - Drury 8y ago
    Consider this: Say a light source has a T temperature resulting in X photons emitted. I redirect all the photons to a single point. I see arguments mentioning that that single point cannot be hotter than the source because there's no more photons to make it hotter. I now add a second light source of the same T temperature, that emits the same amount of photons and also focus all of them on that same point. I now have more photons, but temperature source of all photons is the same. How does adding more photons not make my point hotter?
    - URSpider94 8y ago
      In order to truly move all of the photons from the emitter to the target, you need to effectively surround the target with lenses, such that it looks from the surface of the target as if the emitter is filling the sky. If that is the case, then there is no extra room to fit the optics to focus another source onto the target. If you shrink the optics enough to make room for another source, then you also aren’t delivering all of the photons from the original source (you can’t be, since you aren’t covering all incoming angles), and therefore it’s not the same temperature as the source.
    - Dylan16807 8y ago
      > I redirect all the photons to a single point. For that to be possible with a blackbody light source, it has to itself be a single point. Which means a temperature of approximately infinity. For a real light source, one that has an area, you can at best focus it down to the same area. To get maximum light to a target, you either have to make the target almost touch the source, or you have to have it so no matter what direction you go from the target, you hit a lens that's bringing light from the source. Once you set either one of those up, there's nowhere to fit a new light source.

by Gatsky 8y ago

DenisM 8y ago

How does it work in practice? I am dumping some amount of low-energy photons onto a target, the target is heated up and radiating out the same amount of energy it receives. As I increase the number of photons hitting the target its temperature raises and it radiates more heat outwards, shedding excess heat. Then as I keep increasing the number of photons, the temperature stops increasing at some point having reached the temperature of the photon source. I increase the number of photons a million times of the previous level, yet the object remains the same temperature. Where does the excess energy go? Is it not absorbed by the target? Is it reflected? Is it re-radiated?

Gatsky 8y ago
Well I think the thermodynamic analysis is useful in understanding things, that was my point. I’m not quite understanding the scenario you describe - are you asking how it is possible that you can’t heat up the target more than the origin? If so then take an example where the sun is a single point at 5000 degrees. It is easy to see that the most you could do with this point sun is heat up one point on your object to the same temperature, no more. Now what if there were two point suns next to each other? The discussion on xkcd about optics is saying you can’t superimpose the points on each other on the target under any circumstances, even with two separate lenses at the right angle. This doesn’t seem like it should be impossible though... What about two single photon sources, can’t they be pointed at exactly the same spot? Maybe the explanation here is that the target electron cannot interact with 2 photons at the same time, so you can’t ‘double heat’ a single particle. Or maybe that you can’t precisely target a single particle without decreasing the entropy of a closed system, which is impossible.
Dylan16807 8y ago
So you have a blackbody source, and a target. Initially it's in a steady state where the target is colder. You increase the number of photons hitting the target. The target gets hotter. All good so far. But we have to look at how you increased that number. You need to either move the source and target closer together, or you have to use lenses to simulate the same thing. You can keep hitting the target with photons from more and more directions, but no system of mirrors or lenses can increase the apparent brightness of the source. As you approach the point where the source fills the entire sky, the target will approach the same temperature. And then you can't increase the number of photons any more. Not without using a hotter source.
URSpider94 8y ago
The point is, you can never get to the point where you are dumping more photons in to make it hotter, because there are not more photons. There is a limit to how small you can focus a lens. Once the object you are focusing is in sharp focus, it will start to go out of focus again if you go past that point. So you can’t just arbitrarily increase the photon density further. If you make your lens bigger to gather more photons, then the focused image just grows by a corresponding amount. Optics (and thermodynamics) tell us that we can not generate a higher flux of photons into a target area, than left the equivalent area at the source. Assuming that both the source and target are black body radiators, that says the target can’t get hotter than the source.
- Drury 8y ago
  Consider this: Say a light source has a T temperature resulting in X photons emitted. I redirect all the photons to a single point. I see arguments mentioning that that single point cannot be hotter than the source because there's no more photons to make it hotter. I now add a second light source of the same T temperature, that emits the same amount of photons and also focus all of them on that same point. I now have more photons, but temperature source of all photons is the same. How does adding more photons not make my point hotter?
  - URSpider94 8y ago
    In order to truly move all of the photons from the emitter to the target, you need to effectively surround the target with lenses, such that it looks from the surface of the target as if the emitter is filling the sky. If that is the case, then there is no extra room to fit the optics to focus another source onto the target. If you shrink the optics enough to make room for another source, then you also aren’t delivering all of the photons from the original source (you can’t be, since you aren’t covering all incoming angles), and therefore it’s not the same temperature as the source.
  - Dylan16807 8y ago
    > I redirect all the photons to a single point. For that to be possible with a blackbody light source, it has to itself be a single point. Which means a temperature of approximately infinity. For a real light source, one that has an area, you can at best focus it down to the same area. To get maximum light to a target, you either have to make the target almost touch the source, or you have to have it so no matter what direction you go from the target, you hit a lens that's bringing light from the source. Once you set either one of those up, there's nowhere to fit a new light source.