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

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

Is this a question of the rhetorical type?

Twistar
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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; 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.

Not sure what you mean by "conventional" here. I'm taking thermal equilibrium to mean all subsystems of the system under consideration are at the same temperature. This is a pretty conventional definition of thermal equilibrium.
In the case we're talking about now the system consists of the sun, the laser, the laser gain mediums internal degrees of freedom, and the target.

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.

Hypnosifl
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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; 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.

As I said, you don't have to use the actual Sun, you could replace it with a blackbody at whatever temperature you like, one which wouldn't melt the laser's solid components if it was at the same temperature. The conclusion must be the same--even if the laser functions initially when it's at a different temperature of the blackbody, it must eventually stop working as its temperature comes to match the blackbody's.
Wolfkeeper wrote: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.

But the question was about whether there is some reasonable definition of "passive" such that a laser can be said to work in a passive way--surely no one would say that a refrigerator keeps things cool in a "passive" way, given that it requires a power supply.

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

Hypnosifl wrote:But the question was about whether there is some reasonable definition of "passive" such that a laser can be said to work in a passive way--surely no one would say that a refrigerator keeps things cool in a "passive" way, given that it requires a power supply.

Ah yeah, that's what this is about.

Yeah, so this is the distinction between thermal equilibrium and steady state. It sort of depends where you draw the box around your system. For example, if I draw a box around my refrigerator its not a purely isolated system. There is a wire carrying electrical power entering my box and maybe there is also energy leaving the box in the form of heat. Nonetheless, it is possible for all of the parameters which describe the inside of my box to constant. The temperature would be constant, the total energy would be constant etc. The energy fluxes would even be constant. This would be a system that is operating in steady state. However, I would be wary about saying it is in "thermal equilibrium". It is not the same temperature as its environment, and it is also being fed by a power supply. Both of those are tip offs that thermal equilibrium might not be the right way to describe this system. The idea here is that if I now make my box a little bigger and include all of the things with which the refrigerator is interacting, I find that my box now includes things which are at different temperatures than my refrigerator so I can't claim my total system is in thermal equilibrium.

So it goes for the laser. It is being pumped by the power supply (sun?) which means it might be operating in steady state, but we probably shouldn't think of it as being in thermal equilibrium.

And then I guess to tie this all back to the discussion at hand: The point is a laser is not a passive optical element, so it is possible to use a laser to take light from something at T1 and heat something else up to temperature T2 where T2>T1. Apparently this is something that CANNOT be said for passive optical elements.

I guess thermal equilibrium is really the key here. If your element comes to thermal equilibrium we can consider it to be passive. If it doesn't we have to consider it to be active. I think the non-linear crystals would exhibit similar out of equilibrium dynamics if you looked at the internal states of the crystal which give rise to the non-linearity. Not sure though. It's definitely an interesting question!

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

OK, so if we mean 'thermal equilibrium' in the sense of wikipedia:

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

then no, you can't heat stuff up, at all, without some source of low entropy/work etc.

But if we take 'passive' in the sense of what the questioner asked, where he specified that a transformer was 'passive', then yes, a laser is a pretty good approximation to that in the optical domain, and provided you're not at thermal equilibrium, which as I pointed out, is a stupid thing to apply, then yes you can heat an object hotter than a source.

Note that the questioner never specified or implied that it should all be at thermal equilibrium, this was an extra thing added by later posters; and I don't feel in any way that imposing it is a valid thing to do. In the electrical transformer analogy it's exactly like saying, 'well, if everything is at the same voltage you can't do it with a transformer, so a transformer can never give you higher voltages than you fed in'.

That wasn't the question; you've answered a different question.

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

Who was "imposing" thermal equilibrium on the original question? They were mostly just discussing a tangential question about what would happen under different circumstances.
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Tal-Seto
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### Re: Can you (passively) heat an object hotter than your source?

lgw wrote:Can you (passively) heat an object hotter than your source?

Imagine a rotational ellipsoid shaped mirror whith two spheres in the focuses. If the size of the spheres is not the same, the smaller sphere must radiate at a higher temperature.

The smaller spherre has a smaller surface and must do this at a higher temperature to transport the same amount of energy thru the smaller surface.
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Soupspoon
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### Re: Can you (passively) heat an object hotter than your source?

Tal-Seto wrote:
lgw wrote:Can you (passively) heat an object hotter than your source?

Imagine a rotational ellipsoid shaped mirror whith two spheres in the focuses. If the size of the spheres is not the same, the smaller sphere must radiate at a higher temperature.

The smaller spherre has a smaller surface and must do this at a higher temperature to transport the same amount of energy thru the smaller surface.

I'm not sure you're not confusing temperature and heat energy, but then many people do. I think you're answering another question, though.

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

Its just an old "paradox" to which the answer is the thermal radiation from each sphere is not radial, it's in all directions from each point on the surface, so it doesn't refocus onto the other one.

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

Soupspoon wrote:I'm not sure you're not confusing temperature and heat energy,

Can you explain that for me?
I can't see any flaw in my text.

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

Tal-Seto wrote: we simply ignore the facts.

After which, the spheres turn into unicorns and fly away.

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

Tal-Seto wrote:
Soupspoon wrote:I'm not sure you're not confusing temperature and heat energy,

Can you explain that for me?
I can't see any flaw in my text.

A luke-warm bath will melt a block of ice more surely than a single burning match. The match is at higher temperature but (even if you avoid extinguishing it in the attempt) probably does not have the total amount of energy necessary to melt the ice, yet the hypothetical bathwater does.

That's a very rough description for what I had (it turns out) misunderstood as your intended wording. But I hope it helps explain my meaning.

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

You talk about in a volume stored energy.
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gmalivuk
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### Re: Can you (passively) heat an object hotter than your source?

As already explained, it comes out every point of the surface in every direction. Each photon that doesn't come out perpendicular to the surface won't get reflected directly toward the other focus of the enclosure. The smaller the other object is, the fewer photons hit its surface.
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Eebster the Great
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### Re: Can you (passively) heat an object hotter than your source?

I have created a beautiful illustration of the distinction elliptic is making. You seem to think radiation looks like the illustration on the right, with every ray pointing radially away from the center of the sphere (that is, perpendicular to the surface). In fact, from any given point, an equal amount of radiation will be released in all possible directions, as on the left.

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

Eebster the Great wrote:Don't be discouraged if you can't match my MS Paint skills.
You're a Michelangelo, and way better than me. He could also ask himself why him and his 5 best friends all see a complete image of the moon when they sit on his patio at night?

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

Much apart from being so low as suggesting that there are at least four assumptions in that last statement, as a main point of order I'd like to suggest that you can generally see no more than half of a Moon, each, in that example. Although if it's a large patio...

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

Soupspoon wrote:as a main point of order I'd like to suggest that you can generally see no more than half of a Moon, each, in that example
Qué? Oh wait, the "generally" primed me. A very large patio would indeed be required.
Soupspoon wrote:Much apart from being so low as suggesting that there are at least four assumptions in that last statement
It's my story and I'm sticking with it. His five friends shoot a picture of the moon with a camera at the same moment. In my backwater part of the multiverse they get 5 pictures of the moon, each different only by the light lost to occlusion due to the different positions of the center line of the lenses at the moment the shutter triggers. As a side issue that is a naive view. Since light is a wave and a particle. And the eye doesn't see in all wavelengths and is itself composed of discreet light sensors.

On the original point, is the spectrum tied to the energy level at the moment of emission? It occurs to me, and I'm assuming this is faulty until told otherwise, but it might answer a question asked elsewhere. If the original spectrum reflects the max energy available from that spectrum then it explains why you can never get anything hotter than that temperature with a lens. And why you can light paper with starlight, and not moonlight.

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

Tal-Seto wrote:
Soupspoon wrote:I'm not sure you're not confusing temperature and heat energy,

Can you explain that for me?

Temperature is to thermal [heat] energy as speed is to kinetic [movement] energy.
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