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

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

Can you (passively) heat an object hotter than your source? OK, that's awkwardly worded, let me try to explain.

Different forms of power have different sorts of "potential difference" (is there a more general term for this??) that can each be increased "passively" while conserving power. A transformer can give you more voltage, a gear can give you more torque, a couple of pistons in a hydraulic system can give you more pressure, all while conserving the input power/energy (modulo losses) making the laws of thermodynamics happy without the need for separate input power. For heat this "potential difference" is temperature, but I can't think of an analogous "transformer" for temperature.

I vaguely remember from lessons about black bodies that you can't do this with thermal radiation: for example, no matter how many mirrors you use in your solar furnace, you can't heat something above the surface temperature of the Sun. I don't see an obvious way to do a passive "higher temperature out than in" device without changing the thermal energy to some other sort along the way (obviously, you could use some of the power to drive a compressor or heater). Am I just wrong about this? Forgetting something obvious and everyday? Or is there a fundamental reason there's no "transformer" for temperature.

I guess "can you passively increase the frequency of light" is a similar question. I see something about this on MadSci.org, but I'm not sure I believe it.
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untitled
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### Re: Can you (passively) heat an object hotter than your sour

You need to trade off something for some other thing... a transformer trades current for voltage, a gear trades angular speed for torque moment, the piston trades length for volume... for the black body, you could probably trade spectrum width for spectrum peak power (a single frequency with high energy instead of multiple frequencies with the energy distributed among them). But then again, I am not an engineer.

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

lgw wrote:for example, no matter how many mirrors you use in your solar furnace, you can't heat something above the surface temperature of the Sun.
Really? Why is that true? Or are you just speculating?
I guess "can you passively increase the frequency of light" is a similar question.
Yes. It is called upconversion and there are lots of materials that do this. Two photons are consecutively absorbed by an electron, which then relaxes back to it's ground state and releases a single, higher energy photon. Many people are working on potential uses for this effect to produce more efficient photoelectric cells. For example, if your cell absorbs best in UV and if you step up visible light to UV energies, then more will be absorbed and converted to electricity.
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### Re: Can you (passively) heat an object hotter than your sour

You can heat something to a higher temperature than the Sun. The system won't be in thermal equilibrium for a long while.
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### Re: Can you (passively) heat an object hotter than your sour

eternauta3k wrote:You can heat something to a higher temperature than the Sun. The system won't be in thermal equilibrium for a long while.

Not just using lenses & mirrors [citation needed ].

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

"Can you passively heat an object hotter than your source"

From another forum:

Every optical ray that you can trace which goes from the surface of the sun to the target also has an opposite optical ray which you can trace which goes from the target to the surface of the sun. When the target is hotter than the sun then along each ray there will be a net transfer of energy from the target to the sun. Integrate over all possible rays and you get heat transfer only from hot to cold, never the other way around. That is regardless of the details of your optical device.

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

That seems to be making a basic fallacy along the lines of being naive about the statement "there are as many numbers in [0,1] as in [0,2]". Yes, you can make a 1-1 map between them, but there are still important ways in which the former is half the latter.

In this case, yes, there's technically a ray going in each direction, but if there are giant mirrors at the sun, bouncing and focusing light to your second source, the mirrors near your second source can easily take up much less angular diameter than the ones at the sun. Most of the second point's light will radiate away from the mirror, and not reflect back to heat the sun up.

I'd believe someone who made a reasonable-sounding argument with math about this, because I've been tricked by my intuition in things like this before, but it appears obviously fallacious that you can't passively heat something hotter than the sun, simply because heat is about energy density, and I can passively raise the energy density of a small region arbitrarily high with focusing lenses and mirrors.

I clearly can't passively heat something *sun-sized* to be hotter than the sun, however.
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### Re: Can you (passively) heat an object hotter than your sour

Your second source isn't hotter than the sun until it starts to heat the sun up more than the sun does it.
That's what it means to be hotter.

stianhat
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### Re: Can you (passively) heat an object hotter than your sour

Depends what you mean by passively.

A heat pump can heat something hot by cooling something cold. Basically all it takes is manipulating the internal energy. It is theoretically possible to do this completely passively, but not particularily effective.

lgw
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### Re: Can you (passively) heat an object hotter than your sour

So all of this is really about black body-shaped radiation curves, right? What happens if we step up the frequency? Is the maximum temperature you can radiatively heat something to just a function of the frequency (distribution) of the light? If I increased the frequency of the heating element in my stove to look like sunlight could I then make a (very tiny) solar furnace? (Obviously net heat transfer couldn't go up.)

I hadn't really understood passive frequency-increasing materials existed before this thread - still seems weird.
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Qaanol
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### Re: Can you (passively) heat an object hotter than your sour

It is called the brightness theorem.

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.

And of course that must be the case, for if we had a passive optical system that could use radiation from one blackbody to heat another blackbody to a higher temperature, then we could use it between two objects in thermal equilibrium to heat one and cool the other without adding energy. It would be Maxwell’s demon, which the 2nd law disallows.

So this is not a property of lenses or optics per se, but is even more fundamental than that. The exact mechanism by which the temperature is limited in this case, however, is non-obvious. Is it diffraction limits of lenses? Destructive interference of a non-coherent light source? Something else?

Turns out it’s actually simple geometry: since each point on an extended object radiates in all directions, no matter what you do with your lenses the best you can get is a sharp image of the object that is the same brightness as the original. Just trace the rays through any lens configuration you like, and calculate the radiance.

Conversely, you can also view it as the defining fact of temperature: heat flows from hot objects to cold objects. For one object to heat another, the second object must be cooler. It doesn’t matter how the heat flows, just that it flows from hot to cold.
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### Re: Can you (passively) heat an object hotter than your sour

Xenomortis wrote:Your second source isn't hotter than the sun until it starts to heat the sun up more than the sun does it.
That's what it means to be hotter.

Ok, I think this is the source of my confusion/disagreement, then. I'm defining it as "temperature", rather than "total energy", or something. Does that make a difference, when the thing I'm heating is way smaller than the sun?

But Qaanol's talk about the brightness theorem seems to say that, no, it doesn't make any difference, and it's just an unintuitive result. I'm fine with that, as long as I'm completely sure that this isn't just a weird interpretation thing.
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### Re: Can you (passively) heat an object hotter than your sour

And of course that must be the case, for if we had a passive optical system that could use radiation from one blackbody to heat another blackbody to a higher temperature, then we could use it between two objects in thermal equilibrium to heat one and cool the other without adding energy. It would be Maxwell’s demon, which the 2nd law disallows.

This would only be relevant for an optical system that transfers all heat trough it from a cold to a hot source. But there's nothing wrong, thermodynamically speaking, with a system that sends some heat from cold to hottter, as long as there is another heat flow from cold to even colder to dump the entropy. For example, you can power a laser (or just a resistive heater, or a heat pump) from a photoelectric cell, and create a hot spot that is hotter than the sun. As long as you can dump the inevitable waste heat.

I am no expert on optics, but surely this can be done through purely optical means? Suppose you split the sunlight by wavelength, and you focus only the short-wavelength beam on a spot. That should be able to generate a hot spot hotter than the sun, or am I mistaken?

Qaanol
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### Re: Can you (passively) heat an object hotter than your sour

Right you are, Zamfir. A laser powered by a solar panel cooled by, let’s say a really long metal rod pointing away from the sun, should work.

With passive optics though, mirrors, lenses, prisms, and so forth, any path for light of a certain frequency to go from the sun to the target, is equally valid as a path for light of that same frequency to go from the target to the sun. So the brightness law will hold for each frequency independently.

If the target is hotter than the sun, then it radiates more intensely at all frequencies than the sun. Any frequency of light that can get from the sun to the target, can travel the same path in reverse to get from the target to the sun. And the intensity will be greater going from the hotter object to the cooler object, so there will be a net energy flux out of the target, toward the sun.

In other words, if the target somehow gets hotter than the sun, then it will cool down. Sunlight cannot provide enough energy flux to maintain the elevated temperature of the target.

Actually, there’s probably a caveat that this only applies to linear optics. If you get nonlinear optics you can do things like frequency doubling. Combine that with a lens that reflects all but a narrow range of frequencies, and you can pump your target with a higher intensity of high-frequency light than the sun naturally emits, which means you could indeed get it to a higher temperature than the sun: namely to the temperature where the target radiates in that high frequency at the elevated level.

And…if you had nonlinear optics on the target-side of the lens, you could frequency-shift any outgoing light of a wavelength that would pass through the lens to a different wavelength that will reflect off it, which would mean you could get your target arbitrarily hot.
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### Re: Can you (passively) heat an object hotter than your sour

Qaanol wrote:With passive optics though, mirrors, lenses, prisms, and so forth, any path for light of a certain frequency to go from the sun to the target, is equally valid as a path for light of that same frequency to go from the target to the sun. So the brightness law will hold for each frequency independently.

But the target doesn't necessarily emit the same spectrum as the source. You can heat a spot with hard UV (separated out from the solar spectrum), but the spot itself will radiate approximately as a black body. The UV part of that black body spectrum will symmetrically return to the sun, the rest could in principle be captured, until the outgoing UV component alone balances the incoming UV light. At that point, the spot might well be hotter than the sun.

At least, that's what my gut feeling says to me. I could be wildly wrong here. The main idea is that if you split black-body radiation by wavelength, then the short-wavelength beam will behave roughly like a beam from a hotter source, while the long-wavelength part behaves like a beam from a colder source. With the hot beam you can heat a place to a temperature beyond the original source, but the "cold" beam can only heat another place to a lower temperature.

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

Nope, think about geometrically. Surround your target with a perfectly reflective spherical mirror, so all outgoing radiation from it is recaptured. Make one panel of that mirror be transparent to just one specific frequency, say blue. Set up a lens (or another mirror) to focus sunlight through that panel onto the target.

The target heats up as it absorbs blue light. The target starts radiating as a blackbody. All the radiation from the target is reflected back on the target by the spherical mirror to heat it further, except for the blue light that falls on that one panel.

Blue light radiating from the target in the direction of the special panel passes out of the sphere, hits the lens, and gets focused onto the sun. As long as the target is cooler than the sun, the blue light along that same path from the sun to the target is more intense, so energy flows from the sun through the lens into the sphere and heats the target.

But when the target is the same temperature as the sun, it radiates blue light at the same intensity (amount per solid angle) as the sun. The brightness theorem then guarantees, for purely geometric reasons, that the blue light from the sun shining on the lens is equally bright as the blue light from the target shining on the lens from the other direction. So there is no net energy flux of blue light between the sun and the target.

If the target were somehow made hotter than the sun, then there would be an energy flux in the blue light from the target toward the sun and the target would cool down.
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### Re: Can you (passively) heat an object hotter than your sour

This isn't quite the same thing I don't think, but it seems similar enough to add here.

My intuition keeps on wanting to tell me that if you had one object with a massive heat capacity, one object with a tiny heat capacity, and heated them both to the same temperature, the high heat capacity one would have way more energy in it. If theres way more energy in the high heat capacity object than the low capacity object at equal temperatures, then there should be a way for the energy to be shared equally by the two objects resulting in the low capacity object to be heated to a higher temperature than the high capacity one was originally.

Now I'm pretty sure that reasoning is flawed, and indeed I'm pretty sure I've proven to myself it doesn't work quite like that especially because I'm pretty sure temperature itself is just a measure of the kinetic energy of the particles so equal temperature should mean equal energy, but since differing heat capacities are a thing I'm not sure I grok what the deal is here.

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

Sure, you could put the same amount of heat into the low- and high-capacity objects. But the question is, when you put them in thermal contact (physically or by your clever optical system), which way does heat flow? What we know from thermodynamics is that heat flows in such a way to increase entropy. Entropy will be maximized when all degrees of freedom have the same energy in them. Your high-capacity object has more degrees of freedom than the low-capacity one (that's what heat capacity is measuring), so the energy per degree of freedom is unequal. When brought into thermal contact, heat will flow from the low-capacity hot object to the high-capacity cold object until the energy per degree of freedom, or temperature, is the same in both.

You can move heat from the cold object to the hot object, but you have to discard entropy somehow. If you're keeping the energy of the system constant, this means you'll have to input some low-entropy energy and export some high-entropy energy. Whether you can set up a system that does this "passively" depends on what your definition of "passively" is.
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### Re: Can you (passively) heat an object hotter than your sour

Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time. Heat cannot spontaneously flow from cold regions to hot regions without external work being performed on the system.

That's the second law of thermodynamics, in the form in which it was first published by Clausius in 1854. And in this house we obey.

"I guess "can you passively increase the frequency of light" is a similar question."

Well, you've got frequency multiplication with nonlinear optics, where n multiple photons of wavelength lambda are converted into a single photon of wavelength lambda / n, for example two 1064nm photons from an infrared Nd:YAG laser into a single visible green 532 nm photon in the visible green.
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lgw
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### Re: Can you (passively) heat an object hotter than your sour

Minerva wrote:Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time. Heat cannot spontaneously flow from cold regions to hot regions without external work being performed on the system.

That's the second law of thermodynamics, in the form in which it was first published by Clausius in 1854. And in this house we obey.

"I guess "can you passively increase the frequency of light" is a similar question."

Well, you've got frequency multiplication with nonlinear optics, where n multiple photons of wavelength lambda are converted into a single photon of wavelength lambda / n, for example two 1064nm photons from an infrared Nd:YAG laser into a single visible green 532 nm photon in the visible green.

Doesn't the latter let you do the former? Again, the Laws of Thermodynamics are happy as long as energy is conserved, and there's some loss associated with the transform. The question is, is there some passive mechanism that would give you the same total heat (modulo losses) at twice the temperature, much as a gear can give you twice the torque at the same power.

Would passively stepping up the frequency of the emitted light let you raise the temperature of the target above the source? What about the clever suggestion above of just blocking the low-frequency light so that only the higher-frequencies illuminate the target? Is there anything beyond frequency that sets a theoretical limit on the temperature of the target object?
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### Re: Can you (passively) heat an object hotter than your sour

lgw wrote:Doesn't the latter let you do the former? Again, the Laws of Thermodynamics are happy as long as energy is conserved, and there's some loss associated with the transform. The question is, is there some passive mechanism that would give you the same total heat (modulo losses) at twice the temperature, much as a gear can give you twice the torque at the same power.

The same heat at twice the temperature has less entropy, so no. You can put heat into something hot by simultaneously dumping some into something cold. That is called a heat engine and is not considered a passive process.

The laws of thermodynamics are satisfied as long as energy in conserved and entropy is not decreased. "And there is some loss" is too vague.

lgw wrote:Would passively stepping up the frequency of the emitted light let you raise the temperature of the target above the source?

Some of the light is fundamentally lost in a frequency doubler. The resulting light will not be able to raise the target higher.

lgw wrote:What about the clever suggestion above of just blocking the low-frequency light so that only the higher-frequencies illuminate the target? Is there anything beyond frequency that sets a theoretical limit on the temperature of the target object?

It was already explained by someone else. As the temperature of something increases it increases the emition of all wavelengths, the short ones are just increased more than the long ones. So whichever frequencies you let through and which you reflect, the hot source is going to radiate more of those through the window and therefore decrease in temperature while the cold one gets hotter.

The limit to temperature depends on the ratio of entropy to energy in the light (although that of course can be hard to calculate). A perfect laser beam is, like a voltage source, in principle entropy-free and both can (be used to increase temperature without limit. Their entropy-free nature is in the end much of what makes lasers and electric power so versatile.

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

Does passively igniting a fire count?^^

Could someone explain the "When the target is hotter than the sun then along each ray there will be a net transfer of energy from the target to the sun. " reasoning to me? Sure the object is hotter but the sun is significantly bigger and would still put out more energy. Is it just that optics don't work that way, and you can't make a focus spot so small that the sun being bigger matters?

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

Heat flows (passively) from high temps to low ones, so long as there's a conduit. That's fundemental. The sun soaks it up. It just doesn't notice, is all.

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lgw
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### Re: Can you (passively) heat an object hotter than your sour

Tass wrote:The laws of thermodynamics are satisfied as long as energy in conserved and entropy is not decreased. "And there is some loss" is too vague.

That's oversimplified. Many systems decrease entropy "locally", it just takes power to do so (increasing entropy somewhere like the Sun). I don't think that's a fundamental bar to a system that raised temperature, locally decreasing entropy, as long as the energy lost in the transform was appropriate (adequate to cover the decrease in entropy, to spell it out).

Tass wrote:
lgw wrote:Would passively stepping up the frequency of the emitted light let you raise the temperature of the target above the source?

Some of the light is fundamentally lost in a frequency doubler. The resulting light will not be able to raise the target higher.

No, that's doesn't even make sense. The "amount of light" isn't relevant here - perhaps we're capturing the entire light output of the Sun to raise the temperature of a 1 mg target. We can afford some losses in power along the way, I think.

If you focus the entire light output of the Sun on a 1 mg sample with just lenses and mirrors, the sample will quickly reach (nearly) the temperature of the Sun's surface, even if those mirrors and lenses are quite inefficient indeed. So what about adding a frequency doubler to the mix?

Tass wrote:...
The limit to temperature depends on the ratio of entropy to energy in the light (although that of course can be hard to calculate). A perfect laser beam is, like a voltage source, in principle entropy-free and both can (be used to increase temperature without limit. Their entropy-free nature is in the end much of what makes lasers and electric power so versatile.

Can anyone add to this? Give a better idea of what "ratio of entropy to energy in the light " actually means? I don't remember anything about "entropy of light" from studying black-body radiation - seemed to me mostly about frequency. If you're just talking about the fact that irradiance normally falls off with the square of distance (but doesn't with a laser) that's not really relevant here, I don't think? The amount of w/m2 can be increased trivially with a lens, but that doesn't change the max temp of the target.

thoughtfully wrote:Heat flows (passively) from high temps to low ones, so long as there's a conduit. That's fundemental. The sun soaks it up. It just doesn't notice, is all.

For radiative transfer, isn't it the case that heat is always flowing in both directions (so long as there's a symmetrical conduit)? It's just that the net heat flow will be from high temp to low, even if the low temp source is a far more powerful emitter?
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### Re: Can you (passively) heat an object hotter than your sour

lgw wrote:
Tass wrote:...
The limit to temperature depends on the ratio of entropy to energy in the light (although that of course can be hard to calculate). A perfect laser beam is, like a voltage source, in principle entropy-free and both can (be used to increase temperature without limit. Their entropy-free nature is in the end much of what makes lasers and electric power so versatile.

Can anyone add to this? Give a better idea of what "ratio of entropy to energy in the light " actually means? I don't remember anything about "entropy of light" from studying black-body radiation - seemed to me mostly about frequency. If you're just talking about the fact that irradiance normally falls off with the square of distance (but doesn't with a laser) that's not really relevant here, I don't think? The amount of w/m2 can be increased trivially with a lens, but that doesn't change the max temp of the target.

You want to look for the thermodynamics of a photon gas. In a perfect laser, all the photons are in the same mode -- they have the same frequency and the same phase. Naturally, the entropy is zero. On the other extreme, a thermal photon gas (that from a black body) has maximal entropy. Wikipedia cites an entropy of S = 4U/3T, which seems reasonable.

Note that this means the entropy per unit energy goes like 1/T. Low-temperature black-body radiation carries more entropy than the same amount of energy as high-temperature radiation. This is very handy for us. The Earth's energy balance is dominated by the 6000 K black-body radiation received from the sun and the 300 K black-body radiation emitted into space. The energy being absorbed is about equal to the energy being emitted, but the radiated energy carries away much more entropy than the absorbed radiation delivers. This is what makes vacuum cleaners possible.

Circling back to your original question, any process that moves heat from a cool reservoir to a warm reservoir must have at least one of the following:
1. An input of (low-entropy) energy.
2. An output of (high-entropy) energy.
Most would agree that any process using (1) isn't passive, but (2) offers room for argument over semantics. The simplest example of (2) would be to add a third, cold reservoir (say, the icy blackness of space). Setting up a heat-engine driven by the temperature difference between the cool and cold reservoirs, we use the energy produced to drive a heat pump to move heat from the cool to the warm reservoir. Does this count as "passive"?

Now you can set up a complicated system with lenses and mirrors and prisms and non-linear optics, but you can't get around those restrictions above. If you manage to transfer heat from the cool reservoir to the warm reservoir without inputting any energy, you must be dumping energy and entropy somewhere else. Whether this disqualifies the mechanism from being "passive" depends on your definition. But it is certainly unlike the examples of a transformer or gearbox, in that some energy loss (from our two reservoirs) is fundamental to the process.
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### Re: Can you (passively) heat an object hotter than your sour

lgw wrote:
Tass wrote:The laws of thermodynamics are satisfied as long as energy in conserved and entropy is not decreased. "And there is some loss" is too vague.

That's oversimplified. Many systems decrease entropy "locally", it just takes power to do so (increasing entropy somewhere like the Sun). I don't think that's a fundamental bar to a system that raised temperature, locally decreasing entropy, as long as the energy lost in the transform was appropriate (adequate to cover the decrease in entropy, to spell it out).

Global entropy, obviously. I assumed that was understood. That is not oversimplified, that is the second law.

When you raise the temperature somewhere you actually increase entropy there. You just get less entropy per energy the hotter it is.

And no it is obviously not fundamentally impossible to to take heat energy from somewhere at intermediate temperature and move it to somewhere with a high temperature, as long as you also dump some energy somewhere with a low temperature. That is called a heat engine it is just not passive.

Energy is never lost, that's the first law. But it can be converted to heat at a low temperature, and yes if you make enough of that then you can make the rest into very ordered energy: High temperature or even "infinite" temperature (electricity, laser light, mechanical energy etc.)

lgw wrote:No, that's doesn't even make sense. The "amount of light" isn't relevant here - perhaps we're capturing the entire light output of the Sun to raise the temperature of a 1 mg target. We can afford some losses in power along the way, I think

The point is you can't do that. You can't focus the entire output of the sun onto a milligram. The sun is big, its rays are going in many directions at once. The light has disorder it has a profile corresponding to about 5500K. Not just in spectrum but in spatial movement too. At the surface of the sun itself this is obvious, light is coming from every direction and it has the spectrum of a 5500K blackbody. Anything put there will heat up to 5500K and then achieve equilibrium. Further away the light is obviously dimmer, so something at the earths distance only achieves an equilibrium temperature of a few hundred K, but the light is still as ordered as 5500K light. The light we recieve is more parallel because the sun is far away, and it is shorter wavelength than a 300K body emits. We can use this to get higher temperature. With a magnifying glass you can focus the light. It gets brighter, but it also becomes less parallel. With a perfect arrangement of mirrors and lenses you can get sunlight converging from every side, in the center you'll then have conditions similar to those at the sun. Likewise we can use the wavelength. A blackbody at 300K emits at much lower frequencies. We can let the shortwave sunlight through a window while trapping the emitted infrared. The temperature will gradually rise until the insides start emiting at frequencies that are lost through the window.

I am saying that combining the two ideas won't help. You can focus 5500K light onto a frequency doubler, but the losses in one area will beat the gains in another and you'll still have light at less than 5500K.

Of course the sun has power to spare. You can put a PV-panel out in the sun 80% of the light hitting it becomes heat at 300K, the other 20% becomes entropy-free electricity. You can the n use this electricity to power the LHC if you will, heating tiny amounts of matter to trillions of degrees. It is just that no one would consider this a passive process.

Edit: Maybe I should clarify, when I say "coming from all direction" it is not that light is converging - it obviously is if you focus it, but this is an ordered form of "from all directions". It is about the correlation between spatial position and the directions of the rays. At any place on earth the direction of sun rays vary by about half a degree, because that is the sun discs extend in the sky. As you focus it this variance will increase. Eventually you have a place where the variance is 360 degrees - there is simply no correlation in the directions of the rays in this place. At this point you cannot focus it any tighter, because sending one ray towards a specific smaller target will divert another one away from it.

Notice it you take a magnifying glass out into the sun. Depending on the size of the glass and its focal length, you can't seem to make the spot smaller than a few millimeters. If you did the same with a laser then you could make it converge to the point where diffraction of light sets the limit at some hundred nanometers, if the lens is good enough.
Last edited by Tass on Thu Feb 06, 2014 12:47 pm UTC, edited 1 time in total.

zukenft
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### Re: Can you (passively) heat an object hotter than your sour

so being "passively" means that storing the energy is not allowed?
what if you first focus the sunlight into a distance a few light-seconds away, then rotate the mirror and make them converge with a newly-focused sunlight?

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

zukenft wrote:so being "passively" means that storing the energy is not allowed?
what if you first focus the sunlight into a distance a few light-seconds away, then rotate the mirror and make them converge with a newly-focused sunlight?

Sending them far away makes them more parallel, but dimmer. Focusing them again returns you to where you were (barring losses).

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

PeteP wrote:Is it just that optics don't work that way, and you can't make a focus spot so small that the sun being bigger matters?
Yes, basically. The size onto which a perfect lens can focus the sun's image is still relatively large, and corresponds to the maximum limit of the sun's own surface temperature.
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### Re: Can you (passively) heat an object hotter than your sour

Wow, that's the weirdest thing I've read on these forums, and I read the OTT. So you're saying that if I try to redirect the all light from the Sun onto a small target, some of the light will, what, push the other light out of the way? How does that even? Destructive interference? I'm at a loss here (assuming the target is at least several wavelengths of that light across, of course).

More generally, could someone spell out just what fundamental property of physical law limits the temperature of the target given which properties of some light? Here I was thinking it had something to do with frequency, seems I'm way off ...
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### Re: Can you (passively) heat an object hotter than your sour

lgw wrote:So you're saying that if I try to redirect the all light from the Sun onto a small target, some of the light will, what, push the other light out of the way? How does that even? Destructive interference? I'm at a loss here (assuming the target is at least several wavelengths of that light across, of course).

Nothing as weird as that... it's just that you can't redirect "all" of the light from the Sun onto a small target. There's a nonzero minimum size that you can focus it all onto.

Remember, the Sun isn't a point source, and the light from the Sun is coming from a lot of different points in space, in a lot of different directions. Say you stick a big lens/mirror/whatever in front of the sun. Each point on the surface of that lens/mirror is receiving light from a number of different directions from the sun, and they'll all refract/reflect differently, and won't be all focused on the same point. There's only so much you can do to pull them in. If you move the lens/mirror further away from the sun, then the incoming light is less spread out, it looks more parallel, but the intensity is also reduced, since you're further away... so while you can focus the light more effectively, you have less to work with, and these two effects cancel each other out... and the more your lens focuses the light, the more chaotic it will be within that focus, so there's a limit to how focused you can make it.

The Brightness Theorem says that, even if there are no losses anywhere in the system, then the absolute maximum you can get with passive optics is that the focused image has the same brightness as the surface of the sun - no further. Obviously, if there are losses, you can't even reach that.

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

Tass wrote:
zukenft wrote:so being "passively" means that storing the energy is not allowed?
what if you first focus the sunlight into a distance a few light-seconds away, then rotate the mirror and make them converge with a newly-focused sunlight?

Sending them far away makes them more parallel, but dimmer. Focusing them again returns you to where you were (barring losses).

I think he was talking about using the time delay for light to travel to distant mirrors to temporarily spike the amount of energy being focused. e.g. focus the sun's light on a mirror 5 light-seconds away for a second, then on one 4 light seconds away for a second, etc. such that the light from all of them hits the object in question all at the same time.

And a separate stretching "passively" situation: What if the object you're focusing the sun's light on was moving in the direction of the sun at relativistic speeds, such that the sun's light was heavily blueshifted.

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

Soralin wrote:
Tass wrote:
zukenft wrote:so being "passively" means that storing the energy is not allowed?
what if you first focus the sunlight into a distance a few light-seconds away, then rotate the mirror and make them converge with a newly-focused sunlight?

Sending them far away makes them more parallel, but dimmer. Focusing them again returns you to where you were (barring losses).

I think he was talking about using the time delay for light to travel to distant mirrors to temporarily spike the amount of energy being focused. e.g. focus the sun's light on a mirror 5 light-seconds away for a second, then on one 4 light seconds away for a second, etc. such that the light from all of them hits the object in question all at the same time.

Well the sun has been a continuous source for billions of years. The light from a second ago is just as good as that from four seconds ago. Unless you do something active you will lose ome of one when you gain some of the other.

Soralin wrote:And a separate stretching "passively" situation: What if the object you're focusing the sun's light on was moving in the direction of the sun at relativistic speeds, such that the sun's light was heavily blueshifted.

Yes that is perfectly doable. Now you are transfering energy from the objects motion into heating it. You could also just let it crash into something stationary.

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

The argument against the optical heating mechanism seems to be that, when the object's surface temperature reaches the source's temperature, outgoing radiation balances incoming radiation and the temperature stops rising.

My question is, what happens in very short timescales before you reach thermal equilibrium, and therefore don't have a well-defined temperature.

I guess in that case the original question doesn't make sense because hotter = has a greater temperature => has a well-defined temperature.
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### Re: Can you (passively) heat an object hotter than your sour

Sorry to necro, but there was no good final answer on this, so I'll provide one.

If you are able to focus the light well enough, you can make the object hotter than the sun, frequency of the light isn't a restriction; an infrared laser can make something glow much more than red-hot, and easily cut through metal! Something that the things people have been saying should rule out.

But... You can't focus the light well enough!
If you decrease the angular size of the sun without reducing the radiation (squashing it), you can focus the light more and thus make the object hotter, but you've made the sun hotter since it's smaller but just as emissive. So the object is still no hotter than the sun.
If you decrease the angular size of the sun by moving away from it, you're getting more focus but less light unless you scale the size of the lens with the distance (in which case, focus stays the same I think? The focal area is definitely smaller relative to the lens though.).

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

Well, you could also theoretically pump a continuous-wave laser with sunlight. You could pump a lot of them, and aim them all at a pellet of lithium deuteride or similar in a hohlraum in such a way that the light causes an inertial confinement fusion reaction in the pellet, a la the NIF. That would momentarily be hotter than the photosphere and chromosphere of the sun, but it relies on more than just the transfer of heat energy. It also utilizes the momentum of the incoming light. Likewise one could simply launch a cold spacecraft with a solar sail on a trajectory such that the impact of that craft on a target body will produce temperatures hotter than the sun which powered the craft's acceleration. These methods are cheating by using more than just the thermal energy of the emitted light.

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

Yes, the question is about passive lenses. Lasers or spacecraft are neither passive nor lenses.
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### Re: Can you (passively) heat an object hotter than your source?

There was one bit earlier in the thread that I have a question about:
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.

If at the focus point a perfect absorber was placed, the temperature of the absorber must necessarily increase for each additional image.

Each of the images consists of a set of photons carrying an amount of energy. The absorber's energy in is equal to the total energy of the photons incoming.
Assuming the absorber's own black body radiation isn't reflected back onto the absorber by the lens system (we'll assume the absorber's radiated energy goes back to the sun), energy out is equal to it's own black body radiation. Since the energy out increases only with increasing temperature, increased energy in must necessarily increase temperature.
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### Re: Can you (passively) heat an object hotter than your source?

A perfect absorber doesn't emit any heat, and thus is not anywhere near the temperature of the sun's surface. It either doesn't have a temperature, or is absolute zero, whichever makes more sense in your context.
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### Re: Can you (passively) heat an object hotter than your source?

Why would it not emit heart? We are assuming it gives off black body radiation, no?
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