## Can you (passively) heat an object hotter than your source?

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thoughtfully
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### Re: Can you (passively) heat an object hotter than your source?

I am pretty sure that Thermodynamics requires a perfect absorber to also be a perfect emitter. This looks like a good link, but Hyperphysics isn't coming up for me right now. Ahh, here's a cached page:

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Derek
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### Re: Can you (passively) heat an object hotter than your source?

Reading this thread I had a question, but I think I have found the answer in the last few posts. I'll still present it for the sake of the discussion.

The emitted power of a black body is proportional to it's area. So say we took the entire radiation of the Sun and focused it on a much smaller black body. When they reach thermal equilibrium they will be the same temperature, but the Sun, with a much larger surface area, will be radiating far more power than the small black body. Energy obviously must be conserved, so where does this energy go?

From the last few posts I think the answer I have found is that you can't focus the radiation of the Sun that much. In fact, the smallest area onto which you could focus the Sun's radiation would have to be the area of the Sun (I think you could do this with an ellipsoid set of mirrors with the Sun at one focus, and a Sun-sized target at the other).

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### Re: Can you (passively) heat an object hotter than your source?

Xanthir wrote:Yes, the question is about passive lenses. Lasers or spacecraft are neither passive nor lenses.

It depends how you define passive. For me a sunlight pumped laser is passive, and can heat hotter than the source.

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### Re: Can you (passively) heat an object hotter than your source?

Wolfkeeper wrote:For me a sunlight pumped laser is passive
Sure, if you redefine "passive" to mean something no one else uses it to mean, then you can "passively" do all sorts of crazy things.
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### Re: Can you (passively) heat an object hotter than your source?

thoughtfully wrote:I am pretty sure that Thermodynamics requires a perfect absorber to also be a perfect emitter. This looks like a good link, but Hyperphysics isn't coming up for me right now. Ahh, here's a cached page:

Doesn't that just mean that a perfect absorber would also be an ideal Black body radiator?
Doesn't change the fact that additional light flux falling on the body must increase its temperature.
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### Re: Can you (passively) heat an object hotter than your source?

gmalivuk wrote:
Wolfkeeper wrote:For me a sunlight pumped laser is passive
Sure, if you redefine "passive" to mean something no one else uses it to mean, then you can "passively" do all sorts of crazy things.

The question is, what wouldn't be passive under this definition? I literally can't imagine a more active process short of explicitly violating thermodynamics because you're God.
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### Re: Can you (passively) heat an object hotter than your source?

jseah wrote:
thoughtfully wrote:I am pretty sure that Thermodynamics requires a perfect absorber to also be a perfect emitter. This looks like a good link, but Hyperphysics isn't coming up for me right now. Ahh, here's a cached page:

Doesn't that just mean that a perfect absorber would also be an ideal Black body radiator?
Doesn't change the fact that additional light flux falling on the body must increase its temperature.
Which increases the additional light flux leaving the body. And there's a limit to how much flux you can focus onto one body, which corresponds to the temperature of the source (assuming the source is itself a blackbody, at least).
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### Re: Can you (passively) heat an object hotter than your source?

gmalivuk:
It was written in response to this:
The non-intuitive part is that if you make a high-quality focusing mirror, which creates an image of the sun at nearly the temperature of the sun’s surface, and I do the same thing, and we adjust our mirrors so the focused images of the sun coincide, the resulting image will be about the same temperature as each individually, still not hotter than the sun’s surface.

Which would imply that the black body target would not increase in temperature as light flux increased.
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### Re: Can you (passively) heat an object hotter than your source?

jseah wrote:gmalivuk:
It was written in response to this:
The non-intuitive part is that if you make a high-quality focusing mirror, which creates an image of the sun at nearly the temperature of the sun’s surface, and I do the same thing, and we adjust our mirrors so the focused images of the sun coincide, the resulting image will be about the same temperature as each individually, still not hotter than the sun’s surface.

Which would imply that the black body target would not increase in temperature as light flux increased.
Ah, so you were responding to something written months ago on the last page, but quoting only something written days ago on this page.

It doesn't imply that, actually. If you have to turn or reflect the light to a different focus, the shape or size of the image will change. You cannot use lenses and mirrors to increase the flux from a given source to arbitrarily high levels.
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Wolfkeeper
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### Re: Can you (passively) heat an object hotter than your source?

Copper Bezel wrote:
gmalivuk wrote:
Wolfkeeper wrote:For me a sunlight pumped laser is passive
Sure, if you redefine "passive" to mean something no one else uses it to mean, then you can "passively" do all sorts of crazy things.

The question is, what wouldn't be passive under this definition? I literally can't imagine a more active process short of explicitly violating thermodynamics because you're God.

What definition of passive has anyone given? None!

We're talking about a rod, with silvered mirrors on each end and mirrors directing sunlight from the sides. There's no moving parts, no electricity, no external power source, other than the Sun. That's a passive system, sorry.

I don't think 'passive' is the correct term here, "linear optics" and "non linear optics" might be what you mean.

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### Re: Can you (passively) heat an object hotter than your source?

Well, electricity isn't magic, and you're still doing a transformation and venting waste heat and all of that. But given that everything is sort of still optically coupled - well, what happens in the gain medium is exactly the difference between a solar pumped laser and a passive optical device, and it's going to determine whether or not you can heat your target hotter than your source.
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jseah
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### Re: Can you (passively) heat an object hotter than your source?

Oh, a new idea!

The principle that prevents you from doing this is simple. For a given black body radiator, the maximum power a passive lens system can pack into a target's view of the sky is equal to the power output by the radiator at it's surface.

Therefore, any solution to doing so must necessarily exceed this amount in order to achieve a higher than source temperature. This is essentially only achievable by increasing the number or frequency of the photons.

Here's one way I could think of increasing the number, although converting this into a solution is still beyond me:

Basically, if your photon collection mirror is moving towards the target at 50% the speed of light, the photon 'density' of the space between the target and the mirror will necessarily double (assuming the mirror is perfect). Another way of looking at this is to say that the average distance between each successive photon is halved.

If the mirror is approaching at 2/3 the speed of light, the density increases by three times, etc.

Theoretically, this might let the target exceed the radiator's temperature. Temporarily, until the mirror passes the target. How you stop the mirror from hitting the target is better left unanswered (or just leave a target shaped hole and hope you aimed properly).
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### Re: Can you (passively) heat an object hotter than your source?

But then the system is not passive anymore. The photon's energy is increased by the energy they steal from the moving mirror. This slows the mirror down.
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### Re: Can you (passively) heat an object hotter than your source?

Neil_Boekend wrote:But then the system is not passive anymore. The photon's energy is increased by the energy they steal from the moving mirror. This slows the mirror down.

Hm, true. But the energy increase from the blue shifted photons will not account for all of the increased temperature. There is also an increased intensity due to the higher number of photons per second.

So you can account for the energy lost by the mirror, which will not fully account for the peak temperature achieved by the target.

Still, point taken.
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### Re: Can you (passively) heat an object hotter than your source?

Arbitrarily large sphere, completely reflective, enclosing both source and target.

That'll do it.

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### Re: Can you (passively) heat an object hotter than your source?

That'll heat the target hotter than the Sun was at first, but it'll heat the Sun by the same amount so the target will still not be hotter than the source.
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### Re: Can you (passively) heat an object hotter than your source?

Wolfkeeper wrote:We're talking about a rod, with silvered mirrors on each end and mirrors directing sunlight from the sides. There's no moving parts, no electricity, no external power source, other than the Sun. That's a passive system, sorry.

I don't think 'passive' is the correct term here, "linear optics" and "non linear optics" might be what you mean.

Doing a little googling, p. 71 of this book on the optics of solar cells gives an official-sounding definition:
A concentrator using only geometry, and not relying on a frequency shift, is called a geometrical, or passive, concentrator. A system that is concentrated only by a frequency shift is called a fluorescent, luminescent, or active system.

Incidentally, one point that occurs to me about the solar-pumped laser is that it would necessarily have to fail to heat a target to a temperature greater than the Sun in a hypothetical scenario where the Sun, the target, and the laser are all enclosed within a perfectly reflective chamber with no energy flowing in or out, and the whole system has had time to go to thermodynamic equilibrium. At equilibrium the temperature must be the same anywhere, and any local departure from this temperature would represent a reduction in entropy, so something must go wrong with trying to use the laser to heat the target to hotter than the sun (even if the target is placed a very short distance from the laser), but I wonder what the specific failing would be here--probably something to do with the fact that the lasing medium would also be the temperature of the Sun in this case (of course we are free to imagine replacing the Sun with some body at a cooler temperature in this hypothetical, though that would also mean the light being used to pump the laser would have a lower intensity)

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### Re: Can you (passively) heat an object hotter than your source?

Hypnosifl wrote:
Wolfkeeper wrote:We're talking about a rod, with silvered mirrors on each end and mirrors directing sunlight from the sides. There's no moving parts, no electricity, no external power source, other than the Sun. That's a passive system, sorry.

I don't think 'passive' is the correct term here, "linear optics" and "non linear optics" might be what you mean.

Doing a little googling, p. 71 of this book on the optics of solar cells gives an official-sounding definition:
A concentrator using only geometry, and not relying on a frequency shift, is called a geometrical, or passive, concentrator. A system that is concentrated only by a frequency shift is called a fluorescent, luminescent, or active system.

Incidentally, one point that occurs to me about the solar-pumped laser is that it would necessarily have to fail to heat a target to a temperature greater than the Sun in a hypothetical scenario where the Sun, the target, and the laser are all enclosed within a perfectly reflective chamber with no energy flowing in or out, and the whole system has had time to go to thermodynamic equilibrium. At equilibrium the temperature must be the same anywhere, and any local departure from this temperature would represent a reduction in entropy, so something must go wrong with trying to use the laser to heat the target to hotter than the sun (even if the target is placed a very short distance from the laser), but I wonder what the specific failing would be here--probably something to do with the fact that the lasing medium would also be the temperature of the Sun in this case (of course we are free to imagine replacing the Sun with some body at a cooler temperature in this hypothetical, though that would also mean the light being used to pump the laser would have a lower intensity)

Well, when the sun, the target, and the laser are in a closed system the target will heat to the temperature of the sun. Once it heats beyond that temperature it will radiate more energy, and that energy will go towards heating the sun. The system will end up in thermodynamic equilibrium eventually. The solar sail kinetic impact method has the same issue: it will heat the target object briefly, but that heat will quickly be radiated away and the object will cool. In a closed system that energy will go towards heating the sun and the object equally, and it will reach thermodynamic equilibrium eventually. That's just the nature of closed systems. Any method that seemingly breaks the laws of thermodynamics will do so for either a short time or require an open system.

Which brings me to a third method to passively heat an object hotter than the sun: wait. A very long time. At some point the sun will become a white dwarf, and that will eventually cool. Eventually the spontaneous movements of the particles making up the target object will be such that it has a temperature greater than that of the sun. Thermodynamics is just what happens in statistical mechanics when you have a lot of particles and a reasonable observation period. If you wait long enough there's a non-zero probability that a "big bang" event will occur at the location of your target object. Indeed, with this method the sun, the mirrors, and the lenses are irrelevant. Just immortality and a lot of patience.

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### Re: Can you (passively) heat an object hotter than your source?

SAI Peregrinus wrote:Well, when the sun, the target, and the laser are in a closed system the target will heat to the temperature of the sun. Once it heats beyond that temperature it will radiate more energy, and that energy will go towards heating the sun.

Yes, that was what I meant when I said the temperature should be the same everywhere when the whole system was at equilibrium. What I was thinking about was the fact that in a non-equilibrium situation, an "active" concentrator like a laser pumped by solar energy can potentially heat its target to a temperature hotter than the Sun (or whatever source of incoming light is being used to pump it). But thermodynamic arguments indicate this would fail to work if the whole system started out at equilibrium, so I was thinking about why the laser might fail to function in this situation. As I said, I would think it has something to do with the idea that in order for the laser to work in the non-equilibrium scenario, the lasing medium must be at a different temperature from the source of the light that's pumping it (or at least some vibrational modes in the lasing medium must be at a different temperature), so that's why it would cease to work at equilibrium when the lasing medium must be the same uniform temperature as the light source.

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### Re: Can you (passively) heat an object hotter than your source?

In fact, the black body spectrum, with emissivity 1, is the spectrum that corresponds to a perfect absorber (that is also the perfect emitter, as thoughtfully said).
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### Re: Can you (passively) heat an object hotter than your source?

Hypnosifl wrote:At equilibrium the temperature must be the same anywhere, and any local departure from this temperature would represent a reduction in entropy,

The question clearly is not about the heat death of the universe, so:
so something must go wrong with trying to use the laser to heat the target to hotter than the sun (even if the target is placed a very short distance from the laser),

is incorrect, and incredibly misplaced, at the heat death, the sun wouldn't be running, because if it was, the universe wouldn't be in thermal equilibrium.
but I wonder what the specific failing would be here--probably something to do with the fact that the lasing medium would also be the temperature of the Sun in this case

Pretty sure that there's no specific failing, and I think that the pumped laser photon frequency/temperature can even be higher than the sunlight photons doing the pumping (via multiphoton events). That doesn't mean that the net entropy is going down though(!)- there's going to be a lot of waste heat in the laser. Lasers can be very dangerous, specifically because they give much, much more intense light than their pumping, but they absolutely definitely obey the laws of thermodynamics!

As I say, these are nonlinear processes, and they are not considered in the The Brightness Theorem which applies only to linear processes and objects such as mirrors, lenses etc.

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### Re: Can you (passively) heat an object hotter than your source?

Wolfkeeper wrote:The question clearly is not about the heat death of the universe

I wasn't talking about the heat death of the universe, I said "a hypothetical scenario where the Sun, the target, and the laser are all enclosed within a perfectly reflective chamber with no energy flowing in or out, and the whole system has had time to go to thermodynamic equilibrium". In this case, the equilibrium temperature may be very high. And I realize that the original question wasn't about this, but I explained my reasons for musing about it in earlier comments (see my last one addressed to SAI Peregrinus).
Wolfkeeper wrote:is incorrect, and incredibly misplaced, at the heat death, the sun wouldn't be running, because if it was, the universe wouldn't be in thermal equilibrium.

See above. It's true you wouldn't have nuclear fusion at equilibrium, but in an isolated system of finite volume things can be hot at equilibrium, and hot objects will emit radiation at something close to the blackbody spectrum (the Sun's spectrum is already close to a blackbody spectrum for its surface temperature).
Wolfkeeper wrote:Pretty sure that there's no specific failing, and I think that the pumped laser photon frequency/temperature can even be higher than the sunlight photons doing the pumping (via multiphoton events).

But obviously that can't be true in an equilibrium situation, since the temperature must be uniform at equilibrium.
Wolfkeeper wrote:Lasers can be very dangerous, specifically because they give much, much more intense light than their pumping, but they absolutely definitely obey the laws of thermodynamics!

Sure they do, and that's exactly I conclude an optically pumped laser must ordinarily (in the non-equilibrium situation) rely on some sort of temperature differential to function; if they didn't require that then they could function at equilibrium when a hot body is emitting blackbody radiation towards them, but if they could function they could heat things to even hotter temperatures, which would contradict the original assumption that we were at equilibrium. So, it's a sort of physical proof-by-contradiction.

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### Re: Can you (passively) heat an object hotter than your source?

Hypnosifl wrote:
Wolfkeeper wrote:The question clearly is not about the heat death of the universe

I wasn't talking about the heat death of the universe, I said "a hypothetical scenario where the Sun, the target, and the laser are all enclosed within a perfectly reflective chamber with no energy flowing in or out, and the whole system has had time to go to thermodynamic equilibrium". In this case, the equilibrium temperature may be very high. And I realize that the original question wasn't about this, but I explained my reasons for musing about it in earlier comments (see my last one addressed to SAI Peregrinus).

Yes, you've answered a totally different question, and one in which the answer is that there's never, ever, any heat flow under any circumstances; and one where you can draw no conclusions with respect to the original question, nor with respect to lasers.

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### Re: Can you (passively) heat an object hotter than your source?

Wolfkeeper wrote:Yes, you've answered a totally different question, and one in which the answer is that there's never, ever, any heat flow under any circumstances; and one where you can draw no conclusions with respect to the original question, nor with respect to lasers.

Why do you think you can draw no conclusions about lasers? Do you agree that in a giant box containing the laser, the source of blackbody radiation that's pumping it, and the target (with only a photon gas in the space between them, say), if the contents of the box are in thermal equilibrium that implies the laser can't function? If so just consider a different situation where the system as a whole isn't at equilibrium, but the laser still has the same uniform internal temperature as in the first case and is still being pumped by light with the same spectrum and intensity. The only relevant physical input to the laser is the light hitting it, no other aspects of the system as a whole (like whether the source of radiation is internally powered by nuclear fusion, or whether the target the laser is aimed at is at the same temperature) should affect the laser's performance--do you disagree? If not then if the laser doesn't function in the equilibrium case, it shouldn't function in this case either.

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### Re: Can you (passively) heat an object hotter than your source?

I just mean it's a different scenario, you can't use it to prove that lasers wouldn't work.

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### Re: Can you (passively) heat an object hotter than your source?

Hypnosifl wrote:
Wolfkeeper wrote:Yes, you've answered a totally different question, and one in which the answer is that there's never, ever, any heat flow under any circumstances; and one where you can draw no conclusions with respect to the original question, nor with respect to lasers.

Why do you think you can draw no conclusions about lasers? Do you agree that in a giant box containing the laser, the source of blackbody radiation that's pumping it, and the target (with only a photon gas in the space between them, say), if the contents of the box are in thermal equilibrium that implies the laser can't function? If so just consider a different situation where the system as a whole isn't at equilibrium, but the laser still has the same uniform internal temperature as in the first case and is still being pumped by light with the same spectrum and intensity. The only relevant physical input to the laser is the light hitting it, no other aspects of the system as a whole (like whether the source of radiation is internally powered by nuclear fusion, or whether the target the laser is aimed at is at the same temperature) should affect the laser's performance--do you disagree? If not then if the laser doesn't function in the equilibrium case, it shouldn't function in this case either.

Not sure what your over arching point is but anyways:

When a laser is in operation the gain medium is in a distinctly NON-EQUILIBRIUM state. Namely, it has a population inversion. If you want to insist on using temperatures to describe things then you can say that the gain medium (or at least the system containing the degrees of freedom you use for lasing) are at a negative temperature. This is related to the fact that the HIGH energy states are more populated than the LOW energy states as would be expected for a positive temperature system.

When you pump a laser (with a positive temperature object) you drive a population inversion and thus create a non-equilibrium situation. Non-equilibrium by virtue of the fact that the TOTAL SYSTEM (pump + laser) now contains two sub-systems with different temperature.

In your situation that includes the sun the laser and the pump where thermal equilibrium has been reached. What would happen is roughly as follows: The laser would be on and lasing and the sun would be pumping it (but the system would not yet be at thermal equilibrium) then eventually the sun would use up all of its fuel and die and be unable to pump the laser any more so the laser would stop lasing. Then everything would come to thermal equilibrium.

edit: Oh, but in the intervening time before the sun died it would be possible for the laser to heat the target to hotter than the temperature of the sun.
The point is this isn't really an equilibrium situation and energy is flowing by avenues other than heat. There is non-thermal transfer of energy. You might call it coherent energy transfer.

TLDR: lasers are always out of equilibrium. You can't insist upon a system where a laser is lasing and the whole system is in thermal equilibrium.

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### Re: Can you (passively) heat an object hotter than your source?

Twistar wrote:TLDR: lasers are always out of equilibrium. You can't insist upon a system where a laser is lasing and the whole system is in thermal equilibrium.

Thanks for the explanation. As for the question about my overarching point, I was just using a simple thermodynamic thought-experiment to try to show that it must be the case that lasers can only function in an out-of-equilbrium state, without knowing the more technical details you supplied to verify this. So even aside from the technical meaning of a "passive concentrator" in optics which I cited in an earlier post, if we want to use some broader more intuitive notion of a "passive" device (as Wolfkeeper seemed to be doing with the comment 'For me a sunlight pumped laser is passive'), the relevance of this point is that if you have to periodically apply some work to the laser (distinct from the sunlight) in order to keep it from equilibrating, that might be seen as a reason not to call it a "passive" device.

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### Re: Can you (passively) heat an object hotter than your source?

Sorry, I thought I was interacting with people who had some understanding what they are on about; the last two posts have completely disproved that.

"TLDR: lasers are always out of equilibrium."

^ this is absolute, total bullshit, continuous wave lasers you just shine pumping light in the side and continuous laser light comes out one end.

"the relevance of this point is that if you have to periodically apply some work to the laser"

^ you don't. Yes, the laser dissipates heat energy, but that doesn't make it non passive. So does a mirror for example.

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### Re: Can you (passively) heat an object hotter than your source?

Wolfkeeper wrote:^ this is absolute, total bullshit, continuous wave lasers you just shine pumping light in the side and continuous laser light comes out one end.

But do you think they would continue to do this if the laser was at a uniform internal temperature before (or while) the pumping light hit the side, assuming the light was emitted by a blackbody at the same uniform temperature?
Wolfkeeper wrote:^ you don't. Yes, the laser dissipates heat energy, but that doesn't make it non passive. So does a mirror for example.

If you're answer to the above question is "no, the laser wouldn't work if it was at an uniform internal temperature before the light from the blackbody hit it", that implies if you want to get the laser to work, you have to do some work on it to create an internal temperature difference. If on the other hand you say "yes, the laser would work even if it was at an uniform internal temperature and the light came from a blackbody at the same temperature", then I would feel very confident in betting you were wrong, based on the argument I gave (but if you do say 'yes' and think I'm wrong about this, can you provide a source?)

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### Re: Can you (passively) heat an object hotter than your source?

Wolfkeeper wrote:Sorry, I thought I was interacting with people who had some understanding what they are on about; the last two posts have completely disproved that.

"TLDR: lasers are always out of equilibrium."

^ this is absolute, total bullshit, continuous wave lasers you just shine pumping light in the side and continuous laser light comes out one end.

"the relevance of this point is that if you have to periodically apply some work to the laser"

^ you don't. Yes, the laser dissipates heat energy, but that doesn't make it non passive. So does a mirror for example.

No reason to be rude. Very much unappreciated.

We're debating the definition of passive and equilibrium so everything I said was totally reasonable.

There is a distinction between thermal equilibrium and steady state. In my post I probably wasn't as clear as I could have been but I was talking about thermal equilibrium.

My point was as follows: a laser pumped by say a blackbody will not be in thermal equilibrium. The black body will be at some finite positive temperature. In steady state the laser will be a finite NEGATIVE temperature. The fact there is a temperature difference means that BY DEFINITION OF THERMAL EQUILIBRIUM the system is out of thermal equilibrium.

Your point seems to be that the system is in a steady state. In other words, the amount of energy in the gain medium is not changing and, for example, the amount of light intensity coming out of the laser is not changing. I would agree with all of these points. I would agree with the statement that a CW laser can operate in a steady state.

What I disagree with is the notion that the system is in thermal equilibrium. In the pumping state energy is flowing from the pump to gain medium. In thermal equilibrium there shouldn't be net energy flux. If you turn off the pump then the quantum systems in the gain medium will all drop to the ground state and thus the gain mediums temperature will go from finite and negative to finite and positive. The temperature will change when the pump is removed. Another argument that pumped system is not in thermal equilibrium.

Hypnosifl
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### Re: Can you (passively) heat an object hotter than your source?

Twistar wrote:My point was as follows: a laser pumped by say a blackbody will not be in thermal equilibrium. The black body will be at some finite positive temperature. In steady state the laser will be a finite NEGATIVE temperature. The fact there is a temperature difference means that BY DEFINITION OF THERMAL EQUILIBRIUM the system is out of thermal equilibrium.

But for an isolated system restricted to a finite volume (no energy or matter entering or leaving the volume in which it's enclosed), the maximum-entropy macrostate is always at thermal equilibrium, no? And in the thought-experiment I suggested where the laser and the blackbody are enclosed in a perfectly reflective chamber so the whole system is an isolated one, if we leave this system alone for enough time the 2nd law of thermodynamics indicates the entropy will increase until it reaches the maximum possible for this system, so the final "steady state" must be a thermal equilibrium.

Twistar
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### Re: Can you (passively) heat an object hotter than your source?

Hypnosifl wrote:
Twistar wrote:My point was as follows: a laser pumped by say a blackbody will not be in thermal equilibrium. The black body will be at some finite positive temperature. In steady state the laser will be a finite NEGATIVE temperature. The fact there is a temperature difference means that BY DEFINITION OF THERMAL EQUILIBRIUM the system is out of thermal equilibrium.

But for an isolated system restricted to a finite volume (no energy or matter entering or leaving the volume in which it's enclosed), the maximum-entropy macrostate is always at thermal equilibrium, no? And in the thought-experiment I suggested where the laser and the blackbody are enclosed in a perfectly reflective chamber so the whole system is an isolated one, if we leave this system alone for enough time the 2nd law of thermodynamics indicates the entropy will increase until it reaches the maximum possible for this system, so the final "steady state" must be a thermal equilibrium.

All correct. However, my point is that, vis-a-vis my post above, when the isolated system you've described reaches thermal equilibrium the laser must necessarily be not lasing.

1) If you leave isolated system alone long enough it reaches thermal equilibrium
2) If a laser is lasing it is not in thermal equilibrium
2a) Therefore if a laser is in thermal equilibrium it is not lasing

Therefore
3) If you leave an isolated system containing a laser alone long enough then eventually the laser will stop lasing.

Hypnosifl
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### Re: Can you (passively) heat an object hotter than your source?

Gotcha, I misunderstood what you meant by "steady state".

Twistar
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### Re: Can you (passively) heat an object hotter than your source?

Hypnosifl wrote:Gotcha, I misunderstood what you meant by "steady state".

Yeah I guess the point is it's not a permanent steady state. The laser will operate in steady state for a while, but properties of the pump (in this case the sun) will be changing as time goes on. In particular it will eventually use up all of its nuclear fuel so that the sun's internal degrees of freedom are all in thermal equilibrium etc and eventually it won't be pumping the laser anymore, it will just heat up the mechanical parts of the laser and it will heat up the gain medium but it won't optically pump it anymore and you'll just have a bunch of mass at some temperature, but no lasing. Very much like the heat death of the universe. You'll have a bunch of STUFF at some temperature but there will be maximum entropy which means it won't be doing anything interesting.

Put another way:

Another situation is as follows: Imagine you have some magical thing which will stay temperature T forever. Now put it in your big mirror sphere with a laser and target. The pump will pump the laser and the laser will have a population inversion and shine laser light onto the target. This will be the situation for all eternity. In other words the system will ALWAYS be out of equilibrium and it will never reach thermal equilibrium. That just isn't how the energetics work out in this particular (unphysical) thought experiment.

Hypnosifl
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### Re: Can you (passively) heat an object hotter than your source?

Twistar wrote:Another situation is as follows: Imagine you have some magical thing which will stay temperature T forever. Now put it in your big mirror sphere with a laser and target. The pump will pump the laser and the laser will have a population inversion and shine laser light onto the target. This will be the situation for all eternity. In other words the system will ALWAYS be out of equilibrium and it will never reach thermal equilibrium. That just isn't how the energetics work out in this particular (unphysical) thought experiment.

I don't understand why you say having a light source with constant temperature T would require "magic". In the confined mirrored chamber example where all emitted light eventually comes back to be re-absorbed by the bodies within the chamber, the object standing in for the Sun can remain at the equilibrium temperature T for "all eternity", which can be as high as we like depending on the total energy contained in the system, and even without any nuclear fusion it will continue to emit thermal radiation with a blackbody spectrum corresponding to that temperature T for all eternity (the Sun's spectrum is pretty close to the blackbody spectrum for its surface temperature anyway, the nuclear fusion just serves to keep the surface temperature approximately constant despite constantly losing thermal energy to radiation which isn't reflected back since we don't live in a convenient mirrored chamber). And we can set the intensity of the thermal radiation hitting the laser to whatever value we like by varying the size/distance of the blackbody, along with possibly putting some kind of passive lens between the blackbody and the laser.

So suppose we start the system in an out-of-equilibrium state, but in a state where the object we call the "Sun" is already at an internal equilibrium (no nuclear fusion happening inside, unlike the real Sun), and is shining purely due to blackbody radiation. Based on the total energy in the system we can predict what the final equilibrium temperature will be, even if the lasing medium starts at a lower temperature (and with excited states having a different collective temperature than unexcited states, I think) and the "Sun" starts at a higher temperature than the equilibrium temperature (but the Sun can be much larger than the lasing medium with many more degrees of freedom, so the difference between the Sun's initial temperature and the final equilibrium temperature can be made arbitrarily small even if the difference between the lasing medium's initial temperature(s) and the equilibrium temperature is large). Since there is no "magic" in this scenario the 2nd law must be obeyed, so even if the laser functions initially it must eventually stop as the whole system comes to equilibrium, even though there may have been very little change in the spectrum/intensity of the light from the "Sun".

Twistar
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### Re: Can you (passively) heat an object hotter than your source?

Hypnosifl wrote:
Twistar wrote:Another situation is as follows: Imagine you have some magical thing which will stay temperature T forever. Now put it in your big mirror sphere with a laser and target. The pump will pump the laser and the laser will have a population inversion and shine laser light onto the target. This will be the situation for all eternity. In other words the system will ALWAYS be out of equilibrium and it will never reach thermal equilibrium. That just isn't how the energetics work out in this particular (unphysical) thought experiment.

I don't understand why you say having a light source with constant temperature T would require "magic". In the confined mirrored chamber example where all emitted light eventually comes back to be re-absorbed by the bodies within the chamber, the object standing in for the Sun can remain at the equilibrium temperature T for "all eternity", which can be as high as we like depending on the total energy contained in the system, and even without any nuclear fusion it will continue to emit thermal radiation with a blackbody spectrum corresponding to that temperature T for all eternity (the Sun's spectrum is pretty close to the blackbody spectrum for its surface temperature anyway, the nuclear fusion just serves to keep the surface temperature approximately constant despite constantly losing thermal energy to radiation which isn't reflected back since we don't live in a convenient mirrored chamber). And we can set the intensity of the thermal radiation hitting the laser to whatever value we like by varying the size/distance of the blackbody, along with possibly putting some kind of passive lens between the blackbody and the laser.

So suppose we start the system in an out-of-equilibrium state, but in a state where the object we call the "Sun" is already at an internal equilibrium (no nuclear fusion happening inside, unlike the real Sun), and is shining purely due to blackbody radiation. Based on the total energy in the system we can predict what the final equilibrium temperature will be, even if the lasing medium starts at a lower temperature (and with excited states having a different collective temperature than unexcited states, I think) and the "Sun" starts at a higher temperature than the equilibrium temperature (but the Sun can be much larger than the lasing medium with many more degrees of freedom, so the difference between the Sun's initial temperature and the final equilibrium temperature can be made arbitrarily small even if the difference between the lasing medium's initial temperature(s) and the equilibrium temperature is large). Since there is no "magic" in this scenario the 2nd law must be obeyed, so even if the laser functions initially it must eventually stop as the whole system comes to equilibrium, even though there may have been very little change in the spectrum/intensity of the light from the "Sun".

I mean yeah that's the same as the first thing I said.

I'm not really sure what purpose the rest of this post I'm writing serves but I think it's sort of relevant regarding equilibrium vs out of equilibrium thermodynamics.

It depends how and which things you want to idealize. The thing is in the sort of system your describing I think one of the most likely outcomes is that everything sort of disintrigrates into fundamental particles or at least atoms pretty quickly and you don't have anything resembling a sun or target or laser. But you're imagining a system where that doesn't happen so I'm calling it magical.

Here's one important note about equilibrium thermodynamics. For two systems to come to thermal equilibrium they must be coupled somehow. This always KILLED me when I was first learning stat mech. If I have this black oven at some temperature why the hell does it just start spontaneously emitting light? Well the answer is because it has energy AND THE MEANS TO EMIT LIGHT. That last part is never explicitly stated. The walls of the chamber are coupled to the electromagnetic field. More explicitly, the walls of the chamber are made of atoms which contain protons and electrons. Some of the energy corresponding to the blackbody temperature is stored as electrons in excited electronic states. These electrons can relax into the ground state* and emit light. If your black body was just some black object not made out of electrons that COULDN'T emit light then you would NOT get blackbody radiation.

-Furthermore, people have done experiments showing that if you take some matter that can't emit at a certain range of frequencies (say something with a bandgap) and heat it up to a temperature where it would have blackbody radiation at those forbidden frequencies you can see the usual blackbody spectrum but with holes in it.

Anyways, the TLDR version of that is that the reason the lightfield comes into thermal equilibrium with the temperature of the matter object is because their degrees of freedom are coupled. The systems are coupled.

In our case maybe the story goes like this. Some particular frequency band pumps the gain medium (say the gain medium is a gas of atoms) to a negative temperature and the laser lases and hits the target. Maybe it heats it up hotter than the sun. Then as time goes on the IR and other frequencies of the sun heat up the outer casing of the laser, i.e. the box holding the gain medium. Then, via collisions with the wall, the atomic gas heats up until it is a Boltzmann gas. In other words the motional degrees of freedom have thermalized with the laser box which has thermalized with the sun. The kicker is still the electronic degrees of freedom. However, it is possible for the electronic degrees of freedom to thermalize with the motional degrees of freedom via atom-atom collisions). There is a question of whether this thermalization rate would be faster or slower than the pumping of the atoms by sunlight. I guess it would probably be faster. So in that situation the laser would stop lasing as soon as the electronic degrees of freedom thermalize with the rest of the sytem.

Again, TLDR: if the gain medium internal degrees of freedom of a laser are in thermal equilibrium with some positive temperature system then the laser will not be lasing.

*Through spontaneous emission via coupling to the vacuum electric field

Hypnosifl
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### Re: Can you (passively) heat an object hotter than your source?

Twistar wrote:I mean yeah that's the same as the first thing I said.

Well, your first comment said "properties of the pump (in this case the sun) will be changing as time goes on. In particular it will eventually use up all of its nuclear fuel", so I was making the point that if you start with the pump just being a blackbody at internal equilibrium, and made it very large compared to the laser, then both the change in its temperature and the change in its spectrum can be made arbitrarily small as the whole system including the laser goes to equilibrium.
Twistar wrote:It depends how and which things you want to idealize. The thing is in the sort of system your describing I think one of the most likely outcomes is that everything sort of disintrigrates into fundamental particles or at least atoms pretty quickly and you don't have anything resembling a sun or target or laser. But you're imagining a system where that doesn't happen so I'm calling it magical.

If you wait long enough all protons and neutrons in atomic nuclei recombine into iron nuclei because this is the most stable configuration, but in thermodynamics you are allowed to assume certain aspects of the system (like atomic nuclei other than iron) are stable on the time scale you're interested in, and hold them constant while allowing other aspects of the system to vary, finding the maximum-entropy state for the parts that can vary (see the discussion on p. 42 of this book). And I think in a case like the one I describe in the second paragraph of my last post, the time for the whole system to reach a uniform temperature should be very small compared to the time for the laser's physical structure to break down.

Another way of thinking about the scenario I was trying to describe is to imagine we start with a hollow spherical shell with a mirrored lining and a large hot blackbody at the center with a photon gas in the space between the body and the mirror, but at first there is no laser inside; then we can imagine that this system inside the spherical shell is already at thermal equilibrium to begin with. Now you can imagine briefly opening a trapdoor on the outside just long enough to bring in a laser which is at a different temperature, which is very small compared to the blackbody at the center (which again I'm imagining as Sun-like in size or even larger), and placing it at a distance where the intensity of light falling on it is similar to sunlight (the laser can be placed in orbit around the central blackbody, or perhaps just attached to an arm whose other end is attached to the inner surface of the spherical shell). Now this new system is not at equilibrium, but given that the laser is very small compared to the rest of the system and the rest was at equilibrium before the laser was brought in, I wouldn't expect it to take very long for the laser to be warmed up by radiative heat transfer and to reach the same temperature as the rest of the system, I think the time for this should be much less than the time for the laser to physically fall apart (unless the equilibrium temperature is enough to melt the laser's components, but there are solids with very high melting points).

So, we put the laser inside, and initially the combination of its lower internal temperature and the intensity/spectrum of the light pumping it is sufficient to produce a beam which can heat a nearby target to hotter than the blackbody's temperature. But if we leave the system sealed for say a year and come back to take a look, I'd expect that to be enough for the laser to heat up internally to match the temperature of the blackbody, at which point the broad thermodynamic argument suggests it should no longer be producing a beam that would heat anything to hotter than this temperature. And your argument below is giving a more detailed story of why this could be the case, correct?
Twistar wrote:In our case maybe the story goes like this. Some particular frequency band pumps the gain medium (say the gain medium is a gas of atoms) to a negative temperature and the laser lases and hits the target. Maybe it heats it up hotter than the sun. Then as time goes on the IR and other frequencies of the sun heat up the outer casing of the laser, i.e. the box holding the gain medium. Then, via collisions with the wall, the atomic gas heats up until it is a Boltzmann gas. In other words the motional degrees of freedom have thermalized with the laser box which has thermalized with the sun. The kicker is still the electronic degrees of freedom. However, it is possible for the electronic degrees of freedom to thermalize with the motional degrees of freedom via atom-atom collisions). There is a question of whether this thermalization rate would be faster or slower than the pumping of the atoms by sunlight. I guess it would probably be faster. So in that situation the laser would stop lasing as soon as the electronic degrees of freedom thermalize with the rest of the sytem.

Specifically, are you saying that if "electronic degrees of freedom thermalize with the motional degrees of freedom via atom-atom collisions" (with more energy in motional degrees of freedom due to heat transfer between the photon bath and the atoms the laser is made up of), this thermalization of the electronic degrees of freedom prevents it from lasing? (if so I'll take your word for it, I only know the broad outlines of how lasers work but this would presumably require some understanding of the detailed statistical mechanics)

Twistar
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### Re: Can you (passively) heat an object hotter than your source?

Hypnosifl wrote:Specifically, are you saying that if "electronic degrees of freedom thermalize with the motional degrees of freedom via atom-atom collisions" (with more energy in motional degrees of freedom due to heat transfer between the photon bath and the atoms the laser is made up of), this thermalization of the electronic degrees of freedom prevents it from lasing? (if so I'll take your word for it, I only know the broad outlines of how lasers work but this would presumably require some understanding of the detailed statistical mechanics)

That's right. You need a population inversion for a laser to function. A system which is at finite positive temperature does not exhibit a population inversion. Therefore if the gain medium has thermally equilibrated with the positive temperature heat source it will not have a population inversion so the laser will not lase.

"In other words, a population inversion (N2/N1 > 1) can never exist for a system at thermal equilibrium. To achieve population inversion therefore requires pushing the system into a non-equilibrated state."

edit: A system which exhibits a population inversion is at a negative temperature. This is related to the fact that as you add energy you actually DECREASE the entropy. Consider an ensemble of N two level systems. The lower energy level has energy 0. The higher one has energy E. If there were 0 energy then all of the systems would be in the ground state and there would be no entropy since there is just one configuration. Now lets add energy. Some of the systems will now be in the excited state. But which ones? There is some amount of entropy now. The maximum entropy occurs when the total energy is E*N/2. Exactly half the systems are excited. This is the maximum entropy state. If you add more energy you actually decrease the entropy. In the limit that the total energy is E*N you know that ALL of the systems are in the excited state. There is only one way for this to happen so again there is no entropy. Recalling the temperature is related to the derivative of entropy with respect to energy you can see how a negative temperature would arise.

edit2: And I guess just to clarify my thoughts: It's possible for a negative temperature system to be in thermal equilibrium so long as it is not coupled to any positive temperature system. The reason the gain medium can never be in thermal equilibrium if it has a population inversion is because it is ALWAYS coupled to vacuum electromagnetic field which is at positive temperature. Put another way, If you have a population inversion and you stop pumping the laser then the excited atoms will decay by spontaneous emission until it is back to a positive temperature state. This is what is meant by the notion that the gain medium can't be in thermal equilibrium at negative temperature.
Last edited by Twistar on Wed May 04, 2016 5:52 pm UTC, edited 1 time in total.

Wolfkeeper
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### Re: Can you (passively) heat an object hotter than your source?

But in a conventional sense, it will be in thermal equilibrium; the surface of the sun will be thousands of degrees, the laser will be at (say) 80C or whatever, and the object will be tens of thousands of degrees, and everything will be stable.

Sure, the heat from the laser, will be radiating to free space, as will the sun, and as will the object. And if you stick everything (including the sun) in a perfectly insulated box, it will suddenly not be in thermal equilibrium, it will warm up until everything melts/the laser stops lasing. But that's perfectly normal; you could say the same about a refrigerator at constant temperature; NOTHING around us is in perfect thermal equilibrium in a thermodynamic sense due to entropy considerations.

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### Re: Can you (passively) heat an object hotter than your source?

Wolfkeeper wrote:But in a conventional sense, it will be in thermal equilibrium
Is this the same "conventional" sense in which a laser is considered "passive"?
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