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# Second Law of Thermodynamics Temporal Violator

Oopsy.
 (+4, -3) [vote for, against]

This will probably serve to reveal my deep-seated misunderhension of thermodynamics, but hey.

So, the second law of thermodynamics states that heat will not flow from a cooler to a hotter body. This has been used elsewhere to argue that solar ovens cannot generate temperatures greater than that of the surface of the sun. OK, this I believe.

But what happens if you take the sun's energy and concentrate in _time_ rather than just in _space_?

Suppose we build a fancy solar furnace, and some lenses and whatnot which focus sunlight into a tight beam (it needn't be quite parallel, and I realize that it can't be arbitrarily narrow). Let's also assume that our system is at the limits of efficiency, so that it can create a temperature (when it hits a target) equal to that of the sun's surface.

So far, so fine and dandy. Now we set up some mirrors, and a method for tilting those mirrors. Some of these mirrors are off in space, several light-seconds away from Earth.

To begin with, we bounce our beam of light off a mirror here on Earth, out into space, where it hits another mirror that sends it back to a target here on Earth. The round trip takes 60 seconds.

After one second, we tilt the Earthbound mirror to direct the beam of light out into space towards a different mirror, which again bounces the light back to the target here on Earth; however, this different mirror is closer, so the light only takes 59 seconds to make the round trip.

One second later, we tilt the Earthbound mirror again, this time sending the light on a 58 second trip into space, off a mirror, and back to the target.

And so on.

After we've been doing this for a minute, the first light beam will just be hitting the target. As will the second light beam, and the third, and the fourth...

The result is that we have 60 seconds' worth of light beam all hitting our target over a 1 second interval.

If the original beam was capable of heating the target to the same temperature as the surface of the sun, and if we're delivering 60 times as much energy during that 1 second, then...

I know I'm wrong, but I don't see how.

 — MaxwellBuchanan, Sep 04 2013

Related Portable_20solar_20...igh_20concentration
[MaxwellBuchanan, Sep 04 2013]

F&S First and Second Laws http://www.youtube....watch?v=VnbiVw_1FNs
Thermodynamics Described [csea, Sep 05 2013]

Optical non-reciprocity http://arxiv.org/abs/1305.1793
Maxwell's Demon? [csea, Sep 05 2013]

Maxwell's demon http://en.wikipedia...iki/Maxwell's_demon
Historical Energy Diode [csea, Sep 05 2013]

short section from one of my favourite books featuring a short explanation by De Selby [xenzag, Sep 08 2013]

Fresnel lens solar laser http://www.technolo...olar-powered-laser/
[not_morrison_rm, Sep 08 2013]

Light Measurement units and terminology http://en.wikipedia...Photometry_(optics)
Note section on Photometric versus Radiometric quantities [csea, Sep 10 2013]

Prior Discussion Solar_20Fusion_20Magnifying_20Glass
[MechE, Sep 13 2013]

Feynman Lecture on the Topic. http://www.youtube....watch?v=EyssfKRsgMU
The best explanation I can offer. [WcW, Sep 19 2013]

Bankrupting the coal industry without having to violate the laws of nature http://www.dogpile....krupt+coal+industry
[Sunstone, Sep 20 2013]

Fenniman Lecture on the topic, pt2 http://www.youtube....watch?v=kMSgE62S6oo
Ok so we come into this in the second part. [WcW, Sep 20 2013]

Not the most detailed, but http://www.scientif...actly-does-light-tr
I've read several sources, but it's difficult to refind them. [MechE, Sep 23 2013]

Several mechanisms discussed. [MechE, Sep 23 2013]

Black bodies and the failure of the classical model http://hyperphysics.../hbase/mod2.html#c3
Here it explains how intensity cannot increase temperature, no matter how high the intensity is. [WcW, Sep 25 2013]

Etendue http://en.wikipedia.org/wiki/Etendue
So it turns out there's a word for what I've been saying. [MechE, Sep 26 2013]

South Australia becomes hotter than the surface of the sun solely through solar radiation. http://www.news.com...rfku9-1226800307918
Adelaide one day into 5 day heatwave [AusCan531, Jan 13 2014]

Argument http://xkcd.com/1166/
[spidermother, Jan 15 2014]

The second law, explained somewhat. http://en.wikipedia...energy_dispersal%29

Yet another very simple explanation. http://www.animatio...ectrum.html#entropy
It's VERY SHORT. read the last two parts. [WcW, Feb 20 2014]

Photon Gas http://en.wikipedia.org/wiki/Photon_gas
Order in Light Density [MechE, Feb 22 2014]

solar furnace http://en.wikipedia.../wiki/Solar_furnace
Wouldn't you know it? It is in France. [popbottle, Feb 27 2014]

Maximum Concentration http://www.eolss.ne...08/E6-106-06-00.pdf
Directly under the third equation. Notice that the solution involves the sun's half angle, which increases the closer you get. [MechE, Feb 27 2014]

Talmudic debate - according to request http://www.debate.o...-their-frequency/1/
[pashute, Apr 23 2014]

 What if we make this array of yours but instead of sending the beams into space mirrors we simply build 60 of them and target the same, uhm, target. Same thing yes?

In other words, if we add a gallon of water with a temprature of 80 degress fahrenheit to a gallon of water of 80 degrees fahrenheit, we have two gallons of water of 80 degrees fahrenheit and no water of 160 degrees or no cloud of steam.
 — zeno, Sep 04 2013

Apparently Piko, god of thermodynamics will hit your experimental device with his big hammer and wander off, cackling maniacally.
 — Custardguts, Sep 04 2013

 Yes, there is that. But what I don't get (with either temporal summing of sunlight, or summing of multiple beams at the same time) is this: if 1 beam can heat a target to temperature T, then two beams, delivering twice the power to the same target, should heat it to >T.

This is clearly not the case, but I don't see why not.
 — MaxwellBuchanan, Sep 04 2013

 //This has been used elsewhere to argue that solar ovens cannot generate temperatures greater than that of the surface of the sun.//

 Do you happen to have a citation for this? I can't find anything about it. According to Wikipedia, however, “[t]he chromosphere, transition region, and corona are much hotter than the surface of the Sun.” So I'm not really sure your premise is correct.

 But let's say for the moment that it is. The target body, being nearly as hot as the sun, would itself be radiating heat energy so rapidly that it would be impossible to ever get it to the temperature of the sun. As the target's temperature goes up, the rate at which it absorbs heat goes down, and the rate at which it emits heat goes up.

 So what happens to the additional light that strikes the body? It's effectively reflected and not converted into heat. Since the object is nearly the same temperature as the Sun, it would itself be emitting black-body radiation in the form of light. Additional heat would simply cause it to radiate more light, the net effect being that the light (or some percentage of it) is reflected.

That's my bullshit theory, anyway.
 — ytk, Sep 04 2013

 //Do you happen to have a citation for this?// I was referring to the linked idea.

 //As the target's temperature goes up, the rate at which it absorbs heat goes down// Why? I'm not saying you're wrong, I'm just asking why.

 At solar temperatures, it all gets a bit complicated. Maybe something strange happens to heat absorbtion at high temperatures.

So, if you prefer, imagine that the "sun" is at 300K (slightly warm). We focus its energy (infrared, I guess) onto a target and manage it to warm the target to 300K. Now I use my temporal summation method to deliver twice the power to the target in the same time. The heat absorbtion by the target surely will not change much if its temperature rises just above 300K. So why doesn't it get hotter, even if only to 301K?
 — MaxwellBuchanan, Sep 04 2013

 Well, the target can't absorb energy without also being able to emit energy (barring an endothermic reaction, of course, but then it still wouldn't be getting hotter). If the target were to get hotter than the Sun, it would stop receiving heat energy and start emitting it, making the Sun hotter.

 I think you're hung up on the idea that the time- delayed light is coming from the Sun, when that's not really relevant. An object as hot as the Sun would be emitting its own light. Could you take that light and reflect it back onto itself in order to make it even hotter? Of course not— remember that the reason the object is emitting light in the first place is because it has become sufficiently hot.

 At some point, though, the Sun and the target would have to be in thermal equilibrium (again, if the object were hotter than the Sun, it would be heating the Sun, so the cooler object is always being heated by the hotter object). Being in thermal equilibrium, they essentially become part of the same system. So your idea is equivalent to reflecting sunlight back into the Sun in order to make it hotter. It just won't work, since the more heat you put into the Sun the more heat and light it emits.

(Should reiterate that I have no formal training in any of this whatsoever, so don't assume anything I assert is necessarily true or entirely accurate.)
 — ytk, Sep 04 2013

 Yes, all that you say makes sense.

 Howevertheless. Suppose I'm shining my beam (just one beam) on the target. Suppose the power of this beam is P1. And suppose my mirrors and everything else are sufficiently perfect that the temperature of the target becomes equal to that of the sun.

 At this point, we can assume (if everything is in steady state) that the target is emitting as much energy as it's receiving (P1), so that its temperature stays constant.

 If I now add a second beam, I am delivering twice as much energy to the target (P2). If its temperature is not allowed to increase, it must now emit twice as much energy (ie, P2) as before. Yes?

But then we have a paradox. The target stays at the same temperature, yet it changes from emitting P1 to emitting P2. How can the same target emit twice as much energy, when its temperature stays the same?
 — MaxwellBuchanan, Sep 04 2013

I think that it is a hypothetical example.
 — rcarty, Sep 04 2013

//Yes, all that you say makes sense. Howevertheless...// [Marked for Tagline]
 — AusCan531, Sep 04 2013

 //But then we have a paradox. The target stays at the same temperature, yet it changes from emitting P1 to emitting P2. How can the same target emit twice as much energy, when its temperature stays the same?//

 Think of the additional energy (P2 - P1) as simply being reflected off the target, rather than being absorbed and re-emitted. The system itself stays in equilibrium.

Remember that the system includes both the target, and the energy being absorbed and emitted. The overall temperature of the /system/ can increase, even if one component of that system remains at a constant temperature.
 — ytk, Sep 04 2013

Reflection is a good way to think of it. But can you explain this: Light with a shorter wavelength has more energy, right? So, let's say we get our solar reflectors and hurtle them (albeit for a short while) towards the Sun and Blue shift the light. Does the Sun appear hotter?
 — Ling, Sep 04 2013

 I have no idea, but I'm (+)ing this on spec.

Hey... would combining the focussed light from multiple stars generate more heat than any single one?
 — 2 fries shy of a happy meal, Sep 04 2013

 During the time we are moving towards the Sun, it does. The blueshift is due to the Doppler effect. As we move towards the sun, we plow into more photons over that time period than if we were standing still. The rate of energy accumulation increases while we're moving, thus the Sun appears hotter to us.

 Interesting point. For that brief time period, it would seem that, since the Sun is hotter, we can get hotter as well. The only problem is that, from the perspective of the Sun, the reverse is apparently true—the body moving towards the Sun is also blueshifted, and thus hotter.

Each body would observe an apparent increase in its own, as well as the other body's temperature for the duration of the movement. To an outside observer, however, it would make no difference. Each body is absorbing heat at the same rate it is emitting it.
 — ytk, Sep 04 2013

 //Could you take that light and reflect it back onto itself in order to make it even hotter? Of course not— remember that the reason the object is emitting light in the first place is because it has become sufficiently hot.//

 This is categorically wrong in the case of a sun or other "heat source". If you're talking about a hot body which is cooling down by radiating, yes of course. But if the sun were at temperature x, radiating heat at wattage y, and you enclosed it in a perfectly internally reflecting dyson sphere, it would heat up - absolutely. That's because the radiation wattage y is linked to the temperature x - if you add more heat, the temperature has to increase in order for the radiation to increase and reach equilibrium. Of course, over the course of time the heating would then cause the sun to expand, etc etc and we'd need an astrophysicist to tell us what the overall effect would be on the solar chemistry - but it would definitely heat up until the other effects came into play.

 Another way to look at this problem is that there's nothing special about the solar surface temperature - it's not an absolute in nature or anything. So your example of the target object somehow changing it's absorbtion rate so as to only absorb enough incident radiation so it stays at or below solar temperature isn't valid, because we could easily* focus light from a different heat source, or indeed change the spectra of the sun, which in your line of reasoning means the target somehow "knows" something about the sun and can change it's absorbtion coefficient depending on what your original source of light is. That doesn't read as true.

 *(actually not easily. how about perceivably)

Once again I'll ask if anyone has a link to an informed debate (no offence, I just think no one here including me is qualified to really say categorically what the right answer is) on this topic?
 — Custardguts, Sep 04 2013

[MaxwellBuchanan] I think I see how you are wrong. Your premise of getting to solar temperature with a single concentrated beam is flawed. If there is space to shine a second beam onto your target, then before the mirrors are adjusted to "turn on" that second beam, the target sees a bit of cold sky in that direction and a lot of energy gets radiated in that direction with none returning. You can of course fix that by putting a perfect heat mirror all around your target with a hole just large enough to let your beam in, but then to add another beam, you need to cut a hole in your heat mirror. Assuming your heat mirror is perfect and there are no other loss mechanisms, adding the second beam doesn't change the amount of energy traveling in both directions from your target once it reaches the temperature of the sun, it just changes how fast you can approach that temperature.
 — scad mientist, Sep 05 2013

If you could combine the beams of light to come from the same direction (for example with a hypothetical true one-way mirror), then you could theoretically exceed the temperature of the sun.
 — scad mientist, Sep 05 2013

 //scad mientist// That is the standard proof that such a mirror cannot exist - it would allow the second law to be violated.

Likewise, the thing about reflection is a red herring. While the emissivity of real substances can change with temperature, that is not relevant here, since we can postulate any hypothetical emissivity profile. And the (instantaneous) emissivity of any substance at any wavelength is always exactly equal to its absorptivity at that wavelength, so an object cannot becoming more reflective while maintaining the same emissivity.
 — spidermother, Sep 05 2013

 //Your premise of getting to solar temperature with a single concentrated beam is flawed.//

Indeed. To put it another way, the point at which all possible "beams" have been deployed is precisely the point at which the theoretical temperature reaches that of the surface of the sun. (Technically, we should be speaking of the _effective_ temperature of the sun, since, as [ytk] pointed out, there is not a single temperature involved).
 — spidermother, Sep 05 2013

 Oh, and since each reflector has to occupy its own segment of space (so as not to shade any other reflector), and since the sun's output is effectively constant, the time element is also a red herring. Moving the mirrors doesn't help, either; the moment a mirror needs to move out of the way of the light from a more distant mirror is precisely the moment it needs to be in that very spot, to reflect its own light.

 A Maxwell's demon mirror - one that positions itself to reflect individual photons from the front, and nimbly jumps out of the way to allow others to pass through from behind - would unfortunately generate more entropy than it saved; in other words, it could theoretically work, but it would need a power source at least large enough to generate the increased temperature directly.

If the sun's output were pulsed, it should be possible to use this type of delay to synchronise the pulses, such that the temperature reached is closer to the hottest effective sun temperature, rather than the average.
 — spidermother, Sep 05 2013

 That [spidermother] is a cagey one guys. I don't think we'll ever pin him down because he's probably right. It's just like him you know.

ANYway, another way to build a temporal violator would be to use solar energy charge a bank of batteries over a long period of time then use that energy to focus a beam of energy onto a small point for a short period of time. Hopefully, resulting in a temperature higher than Sol's.
 — AusCan531, Sep 05 2013

//he's probably right. It's just like him you know.// Not only is he being correct, but he's doing it on purpose.
 — pocmloc, Sep 05 2013

 I'm missing something about your original concern with the 2nd Law, MB.

 If you have a pound of coal, it has a finite amount of energy. It also has, given the amount of available oxygen, an amount of energy it can produce by burning, per second.

 You can alter how quickly or hotly it burns. But you cannot make it produce more energy than is in the carbon bonds, ultimately.

By collecting energy over time you will still not exceed the total amount of initial available energy, which is all that's truly required to maintain the law. Put another way, you do not have a closed system.
 — theircompetitor, Sep 05 2013

I don't think you're addressing MB's design. The mirrors aren't moving out of the way, instead they are being used as a directional lag-time inducer depending on their angle.
 — RayfordSteele, Sep 05 2013

 I'm failing to see how temperature is important. If you accumulate solar energy in a humongous battery via a PV cell over several days, you could release that energy in seconds by shorting the battery. And you could probably jimmy up an electrical system to attain a temperature higher than that of the sun's surface (corona discharge, lightening, etc).

The 2nd law concerns spontaneity, a.k.a., a hot bun left in a cool room will heat the room up a very small amount, yet a cold bun left in a warm room will never draw heat out of a room to become a hot bun.
 — the porpoise, Sep 05 2013

yes, Ray, just as porpoise is, that it is time is effectively providing storage, not really changing the thermodynamic properties of the system
 — theircompetitor, Sep 05 2013

 Those 'bakers incorporating energy storage are on the right track. It should be quite possible to use solar cells to charge a capacitor over some relatively long period of time, to accumulate (n) Joules of energy, then discharge the capacitor with a very low resistance/inductive secondary path to produce much higher temperatures that the sun's surface for a much shorter time.

e.g. 1kJ of energy (1000 Watt-seconds) may be stored in a capacitor at 1Watt for 1000 seconds, then discharged in under a microsecond, with a net power of 1GW (10^9 W.) This should be enough to exceed solar surface temperature.
 — csea, Sep 05 2013

Also note (humorous) [link] which includes the punchline of the second law, "Heat won't pass from a cooler to a hotter; you can try it if you like, but you far better not-ta!
 — csea, Sep 05 2013

 Hang on a second. I missed something.

 Up there ^ it was suggested that you can only focus sunlight to generate Sun-surface temperatures if you capture _all_ of the light coming out of the Sun.

 I don't think that can be right. Suppose we have a mirror system which captures all (or 99.99%) of the Sun's light, and focusses it down to a spot where it generates solar-surface temperatures.

 Now I add a second, identical sun next to the first one, but the mirrors don't capture this second sun's light at all. The temperature of my spot should stay the same as it was in the last paragraph. So, I am now capturing only 50% of the light of my "double sun", yet the temperature of the spot is still the same as the surface temperature of either part of this 'double sun'.

Clearly, therefore, I can't need to focus _all_ of the sun's light onto a target to raise that target to solar-surface temperatures.
 — MaxwellBuchanan, Sep 05 2013

 //Clearly, therefore, I can't need to focus _all_ of the sun's light onto a target to raise that target to solar-surface temperatures.//

 Well, yes. Temperature doesn't really matter. It's energy that matters.

 Your sixty 1-second-long light bursts together carry the same energy as the one 60-second long light burst that you started with. The fact that they all arrive at the target at the same time is neat, but hardly material to conservation of energy (1st law).

As for the 2nd law, well, light arrives at Earth at about 100 W/m^2, which is piddly. The fact that it originated from a surface that might be 1,000,000 K means what? CERN can apparently produce temperatures upwards of 1,000,000,000,000 K and they are not using all of the sun's energy to do it. Not sure how the 2nd law applies to your system of even if your system is a closed one.
 — the porpoise, Sep 05 2013

 //Seems you *do* need to get the light to do some work to transfer it to a hotter body.//

 I'm fine with that. So (if I understand, which happens by chance now and again), this is equivalent to saying that I can build a machine which uses the falling of a ton of water to shoot an ounce of water to a much greater height.

 But by that analogy, I can use sunlight to create an arbitrarily high temperature in a small target; and it's simple conservation of energy that says that the target has to be much smaller than the solar area whose radiation I'm collecting (or has to be heated for only a fraction of the time).

 Which brings me back to the question: given that I'm happy to 'waste' most of the solar energy in doing work, can I then create an arbitrarily high temperature, given an arbitrarily small target and/or a means of compressing the light flux in time?

And yes, clearly I _can_ in the case of a solar panel running a laser; so the real question is can I build something which does the same job as the solar-panel- plus- laser setup, but using only things like reflection, refraction and suchlike? And if not, why not?
 — MaxwellBuchanan, Sep 05 2013

No, because the men in black will come and get you if you do.
 — RayfordSteele, Sep 05 2013

They've been. We had tea. They left.
 — MaxwellBuchanan, Sep 05 2013

 I think there must be a fundamental mistake with the interpretation of the 2nd law in the original idea here, otherwise there would be no heat pumps. In them, energy is added to the system and overall efficiencies are considerably less than 100%, with greater entropy, but heat is clearly being transferred from cooler to warmer, albeit not through radiation.

 My first thought was to ignore the sun and substitute a 100W light bulb. I figured it must be possible to focus the reflected light to a point and raise it above the temperature of the light bulb. But then I realised that the filament of a light bulb is pretty hot. Then I thought that the wires aren't hot and they transfer heat to the filament? Is radiation a special case and you can transfer heat in any other way, like rubbing sticks together?

Or is it that a "black body" assumes a naturally- occurring object with no energy passing through it by exotic means?
 — marklar, Sep 05 2013

 //a fundamental mistake with the interpretation//

Welcome to my world.
 — MaxwellBuchanan, Sep 05 2013

 //Temperature doesn't really matter. It's energy that matters.//

Yes!
 — csea, Sep 05 2013

 Yes, I'm happy with that concept.

But an assertion was made on the linked idea that no amount of lensing and mirroring will allow a solar furnace to create temperatures higher than that of the surface of the sun.
 — MaxwellBuchanan, Sep 05 2013

So... what you are all saying is that we need to build a dyson sphere mirror just for the heck of it?
 — zeno, Sep 05 2013

 ///so the real question is can I build something which does the same job as the solar-panel- plus- laser setup, but using only things like reflection, refraction and suchlike? And if not, why not?//

 I don't see why not. I don't think the photons leaving the sun would remember the temperature of their origin. If you reflected every single photon leaving the sun onto the head of a pin, it would be hot. I'm not sure how hot, but let's assume hotter than the sun proper.

 Now you build a big perfectly insulating box around the sun, your optics, and the head of the pin (see Sky Tube Ring for instructions). The sun burns and the head of the pin gets really hot. But eventually the nuclear fuel of the sun runs out and all the matter within the box all settles to temperature T. Is there any reason to believe that temperature T would be different if your optics were installed misaligned and did not actually heat the pin?

 Now let's assume that while we're waiting for the system to reach temperature T, we measure the temperature of the head of the pin and wonkers it's hotter than the sun. So, we slowly move the pin into contact with the sun in the hope of creating some kind of perpetual feedback loop. The pin, being hotter, is now actually heating the sun.

 What we have in effect is an optical system that creates a localized hot spot on the sun. Clearly all energy is coming from the sun, so we haven't violated 1st law. My understanding of 2nd law is that this tends not to happen spontaneously. The sun does not just suddenly decide to have a localized hot spot (aside from when the nuclear fuel gets sloshed around by comet strikes etc).

But hey, I may be complete wrong.
 — the porpoise, Sep 05 2013

 This is an interesting topic, and clearly, I'm no expert on the subject.

 But I find it amusing to see that the classical "one- way energy diode" is called "Maxwell's daemon." Perhaps [MB] is distantly related to a scientist who could propose an experiment to elucidate this matter.

 — csea, Sep 05 2013

My intercalary twin is a sixth cousin twice removed, but he kept coming back.
 — MaxwellBuchanan, Sep 05 2013

 // Up there ^ it was suggested that you can only focus sunlight to generate Sun-surface temperatures if you capture _all_ of the light coming out of the Sun. // Who said that? I saw a statement from [spidermother] and I thought I made it clear as well that your TARGET must be getting light from all directions, but that light could be a very small fraction of the light for the sun. If you capture twice the light, your target will approach it's final temperature twice as fast.

 Here's a conceptually simple way to focus light on a point and get it to the sun's temperature (obviously limited somewhat with non-perfect mirrors and other hardware): Create a deep parabolic reflector such that when standing at the focus, the round opening of the dish is so far away that if it is pointing at the sun, only the sun is visible. This will look like a very long narrow tube that tapers to a point at one end right near the focus. From the focus, the sun is the only thing visible out the end of the pipe, and in all other directions, the sun is reflected off the inner wall of the pipe. Since the only thing visible in every direction is the sun, material at the focus will absorb heat through radiation until it is radiating the same amount of heat that it is absorbing.

This is somewhat impractical to build because if the focus is 1mm from the bottom of the parabola, the opening would be 1.8m in diameter, but it would be about 210m tall. Not to mention the difficulties involving in suspending your target at the focus in a perfect vacuum.
 — scad mientist, Sep 05 2013

Tried to follow this with limited resources and may be thoroughly confused so can only regurgitate some stuff I've read somewhere- A. You don't get something for nothing in physics. B. The duality of light, particle or energy? C. You can try to pull yourself up by your bootstraps but good luck.
 — cudgel, Sep 05 2013

 Thanks!Oh right, you, were talking to... never mind, carry on then.

 But just so you know, I'm borrowing your good-luck-bootstrap-pull-wish unless claimed by another.

True story...
 — 2 fries shy of a happy meal, Sep 06 2013

Pardon me...does this have something to do with simulated heat?
 — normzone, Sep 06 2013

 //Temperature doesn't really matter. It's energy that matters.//

 Temperature _does_ matter.

 //Your sixty 1-second-long light bursts together carry the same energy as the one 60-second long light burst that you started with.//

 No. The sixty 1-second-long light bursts together carry (at best) the same energy as one 1-second-long light burst (from an optimal Perfectly Ordinary Reflector).

 //If you reflected every single photon leaving the sun onto the head of a pin, it would be hot. I'm not sure how hot, but let's assume hotter than the sun proper.//

Unfortunately, that is not possible, even in theory. It would require another kind of Maxwell's daemon.
 — spidermother, Sep 06 2013

 Well, now I'm more confused.

Why don't 60 one-second bursts of sunlight carry 60 times the energy of a single one-second burst?
 — MaxwellBuchanan, Sep 06 2013

Because each of the 60 optical pathways can, on average, deliver at most 1/60 of the sunlight that can be delivered to the target by an ordinary (non-time-delayed) mirror.
 — spidermother, Sep 06 2013

 Ah, right, with you.

 I suspect I am floundering a little. So, let me get back to the thing that started all of this, with three questions:

 (1) Is it fair to say that, using a parabolic mirror (or lens) of any size, you cannot focus sunlight onto a target in such a way as to raise its temperature above that of the sun?

 (2) And is it also fair to say that, given a good enough mirror/lens and an ideal target, you could heat the target until it was arbitrarily close in temperature to the sun?

(3) And if the foregoing answers are both "yes", is it also true that you could use mirrors/lenses to capture light from only a fraction of the sun (a hemisphere, say, or a smaller part), and heat a [smaller] target arbitrarily close to the temperature of the sun?
 — MaxwellBuchanan, Sep 06 2013

 4) is the "nay" based on geometry rather than physics ?

 If you have 2 points at the focii of a large ellipsoid mirror, the radiator and the radiatee then, if you collapse the ellipsoid in a rather precise fashion, the radiatee will be receiving a certain amount of radiation and should, at least until the radiation it reemits reaches the radiator, be hotter.

Likewise if your radiatee is one of the focal points of two or more intersecting ellipsoids, all of which having a radiator as the other focal point, then it will be hotter, at least until such time as equilibrium equilibriates.
 — FlyingToaster, Sep 06 2013

 (1) Yes (2) Yes (3) Yes

What you cannot do is to use any optical method to raise an object of any size _above_ the surface temperature of the sun without capture/storage of energy.
 — csea, Sep 06 2013

 Thinking about the original time delay mirror system, it seems to me a parabolic mirror is already this. If one assumes a photon to bounce off the mirror at a right angle, any reflected light travels a path equal to sides a and b of a right triangle. Light directly hitting the target travels a path equal to the long side c of that right triangle. If the two photons hit at the same time, the reflected photon must be from an earlier graduating class because it has travelled farther before impact on the target. The larger the triangle the earlier the graduating class.

It seems to me that this in one way is analogous to the capture / storage of energy - this energy is "stored" because it is traveling through space and so available for redirection.
 — bungston, Sep 06 2013

 (4) Geometry AND Physics - If the heated object came to have a higher temperature than the Sun, it would radiate energy towards the sun, and therefore cool until it reached thermal equilibrium.

Any linear optical system intended to circumvent this would fail due to symmetry. A nonlinear system, including one with energy storage, could well do the job.
 — csea, Sep 07 2013

 Erm, there is this mysterious black stuff found underground, which can create temperatures hotter that the surface of the sun.

 I call this substance "coal". Admittedly you do also need the power station and a laser.

 At which point I can't work out who has won, is this Maxwell's "concentrating in time" or not?

More to the point, why would anyone want to make part of the earth hotter than the surface of the sun for considerable periods of time? Is it just cutting out the middleman for global warming.
 — not_morrison_rm, Sep 07 2013

 //(1) Yes (2) Yes (3) Yes//

Excellent - binary answers are what I was lookng for. Are all the physicists agreed on those three, before I move on to (4), (5)...?
 — MaxwellBuchanan, Sep 07 2013

 No matter how many '9' waves hit a target you won't be able to bring it to a '10'.

It's like tipping a waiter in dollar bills using quantum theory. No matter how many dollars you throw at him, you will never tip more than a dollar.
 — daseva, Sep 07 2013

 //(1) Yes (2) Yes (3) Yes//

Agreed.
 — spidermother, Sep 07 2013

 Further to what [bungston] said, you can design a reflector system with (a) only short paths, (b) only long paths, or (c) some combination of long and short paths.

The problem for this idea is that they are all equivalent. It doesn't matter how long ago the photons arriving right now left the sun or left a mirror, because the sun was approximately the same brightness 20 minutes ago as it was 10 minutes ago.
 — spidermother, Sep 07 2013

 //More to the point, why would anyone want to make part of the earth hotter than the surface of the sun for considerable periods of time? Is it just cutting out the middleman for global warming.//

Why would anyone _not_ want to make part of the earth hotter than the surface of the sun for considerable periods of time?
 — spidermother, Sep 07 2013

 //Erm, there is this mysterious black stuff found underground, which can create temperatures hotter that the surface of the sun.

 I call this substance "coal". Admittedly you do also need the power station and a laser.

 At which point I can't work out who has won, is this Maxwell's "concentrating in time" or not?//

 The ancient plants harvested sunlight, and used it to do work - admittedly, rather fiddly work, involving plucking bits off some molecules and squishing them onto others, like setting little molecular jack-in-the-boxes. The power station converted that molecular work into heat, then finally converted that into electrical work. The laser converted electrical work into a low-entropy form of energy (the excited state of the laser), then emitted low-entropy light, which was focussed to produce low-entropy heat, hotter than the surface of the sun.

 Total entropy increased at every step. It's quite permissible to use work to move heat from a cooler to a hotter object.

Concentration of energy over time is not required, although it does help. A very large area of solar panels, or a smaller area with a large capacitor bank to store work energy over time, could be used as coal substitutes.
 — spidermother, Sep 07 2013

 // It's quite permissible to use work to move heat from a cooler to a hotter object. //

 OK, I'm happy with that too.

 So then the question really boils down to this:

"Given that a solar cell, accumulator and laser can do solar-powered work to create a temperature higher than that of the sun (inefficiently, and increasing entropy overall); can a system made only of lenses and mirrors achieve the same end _in principle_?"
 — MaxwellBuchanan, Sep 07 2013

Certainly not a _static_ system. I can think of a couple of scenarios involving moving mirrors, but they involve the production and consumption of work and/or information.
 — spidermother, Sep 07 2013

Given that the sun's temperature was roughly the same in the carboniferous period, I conclude that it's not cheating to use this "coal" stuff. Getting coal from the far future would be, as the sun would be hotter then.
 — not_morrison_rm, Sep 07 2013

In the end, seeing as we have exceeded the sun's temperature using science, the only answer is a definitive YES: we can concentrate energy to create higher frequencies. My contention is that it requires strategic use of stored chemical and nuclear energy and the subsequent release to achieve higher radiation frequencies than are witnessed emanating from the sun. Mirrors and 'hold the gates until the horses are mad' type strategies will not prevail.
 — daseva, Sep 07 2013

 I wonder if one problem with these ponderments is that the sun is considered a point, with the heated object in question a point. If both are dimensionless points then the second law holds.

If there are dimensions, though, it seems silly. Assume a parabolic mirror near the sun with focal point on the sun's surface. This reflects the sunlight emitted by area x^2 back onto an area of the sun measuring 0.01x^2. Would that area not warm up? Even if not reflected back onto the sun, but onto some durable object smaller than x^2 it seems that this object would become warmer than any same-sized area on the surface of the sun.
 — bungston, Sep 07 2013

 //Assume a parabolic mirror near the sun with focal point on the sun's surface. This reflects the sunlight emitted by area x^2 back onto an area of the sun measuring 0.01x^2.//

That would be the case if the light leaving the sun were parallel (like a laser), but it's not, so the sunlight emitted by area x^2 is reflected back onto an area of the sun measuring x^2. Try it with light from any large, diffuse source, such as a computer monitor or a fluorescent tube. It's not possible to focus such light into a tiny, hot spot.
 — spidermother, Sep 07 2013

 //Certainly not a _static_ system.//

 OK, so what is the fundamental difference between:

 (a) a conventional "static" optical system (which can't raise temperatures above solar surface) and

 (b) a solar cell+laser system.

In other words, what is it about the solar cell+laser that makes it capable of doing what a static system of mirrors/lenses cannot do?
 — MaxwellBuchanan, Sep 07 2013

In the second case, work is done. In the first case, it isn’t.
 — pocmloc, Sep 07 2013

OK, so then is it possible to devise a system of mirrors and lenses which will do work?
 — MaxwellBuchanan, Sep 07 2013

 Hmm.

 OK, I get the argument about work being needed to move heat against a gradient.

What happens if I put a frequency-doubling crystal in the path of the light? Such things exist, but maybe they only work with coherent light. If such a thing did work, then I would presumably get out a [smaller amount of] light which, being of shorter wavelength, would be as if it had come from a hotter source, no?
 — MaxwellBuchanan, Sep 07 2013

 {lambda} = h {nu} to you, my good sir!

Thanks for the review of basic physics!
 — csea, Sep 07 2013

The way I compute this is to imagine if a complex system of inclined planes can make the water, from a waterfall, spout higher than the lip of the waterfall. Of course with tech (generators and pumps), the height can be reached.
 — wjt, Sep 08 2013

you could make _some_ of the water pouring over the edge go higher than the waterfall, without importing work to the system.
 — FlyingToaster, Sep 08 2013

We seem to have the makings of the world's largest Turkish bath, what with a high-temperature heat source and an airborne waterfall.
 — not_morrison_rm, Sep 08 2013

Yes [FT] but then the bulk of the water would be doing work on the small amount to shoot it higher? Not something that can be done with light.
 — pocmloc, Sep 08 2013

//solar cell, laser// umm... wait, why the intermediate step ? how'bout Sun>lasing-material>lens.
 — FlyingToaster, Sep 08 2013

Hey, where are my patent rights on the solar powered laser? I can do you a good deal on coal,if that doesn't work out.
 — not_morrison_rm, Sep 08 2013

 Among the 125,000 Google hits for "solar powered laser" there's a youtube vid "Building a Solar Powered Laser - Part 1".

There's no Part 2. So he was either unsuccessful or successful.
 — FlyingToaster, Sep 08 2013

 The thing that I have trouble with is this:

 1. The temperature of an object is the average kinetic energy of all the particles. If we talk about averages, then some must be more, some less. So do we day that the Sun is an average temperature? Some particles must be more?

 2. The radiation from an object is broadband. Energy of the radiation is related to the frequency.

 3. Kirchhoff's law says absorbtivity and emmisivity of an object is the same for each wavelength, so as the object heats up, it finally reaches equilibrium of energy received and energy released (and there are some who say that Kirchhoff's law is not quite obeyed in some cases with metal flakes and so on).

4. So, if the flux is high enough, and the broadband radiation is filtered so that only higher frequencies reach the object, then what does that mean? Does it mean equilibrium is reached, but with a higher energy received and released?
 — Ling, Sep 08 2013

 //Among the 125,000 Google hits for "solar powered laser"

Hmm, google. First on the list is a company called "solar lasers" who have laser cutters, but they are mains powered, then some guy with a magnifying glass with the old 2D computer plotter mechanism. Best one is the Fresnel lens laser, linked.
 — not_morrison_rm, Sep 08 2013

 I had no idea I knew so little about the second law.

 [FT]'s suggestion of a laser driven directly by sunlight (as opposed to via a solar cell) is interesting, as is [Ling]'s idea of selecting only the shortest wavelengths from the spectrum of solar radiation. And there's also the unanswered question of whether you can use a frequency- doubling nonlinear crystal to make sunlight "bluer".

Would any of these work? If not, why not?
 — MaxwellBuchanan, Sep 08 2013

If not then "why", so simply "yes".
 — rcarty, Sep 08 2013

 //OK, so then is it possible to devise a system of mirrors and lenses which will do work?//

 A solar sail is nothing more than a mirror that uses light to do work.

Use photon pressure (solar sail effect) to accelerate a mirror towards the sun. The light reflected off the moving mirror will be blue shifted, and therefore more energetic (as mentioned above). In theory, such a system could generate temperatures hotter than the effective sun temperature, using only mirrors, powered only by sunlight - a Second Law of Thermodynamics Violeter
 — spidermother, Sep 08 2013

Actually, I don't think selecting the higher frequencies to get higher temperatures is violetting anything...as the Sun's particles that are radiating those higher frequencies are already at the higher temperature...we're just making the average different?
 — Ling, Sep 08 2013

 // And there's also the unanswered question of whether you can use a frequency- doubling nonlinear crystal to make sunlight "bluer".//

 Frequency-doublers (obviously) don't increase the energy content of the light - the higher frequency photons have twice the energy, but are half as abundant. Nor do they allow it to be concentrated further. The power per unit area is therefore the same (at best). It is power density that determines the temperature of the target, not wavelength per se.

(I don't know to what extent frequency doublers can be used for sunlight. I think they only work well with lasers. But that is besides the point.)
 — spidermother, Sep 08 2013

I'm sorry I came to this party so late, but ytk, way up there near the top, is correct. That pesky concept of thermal equilibrium comes to save the day for the 2nd Law. Firstly, equilibrium means constant in time, so your little scheme of moving mirrors kind of violates that condition. To get around that, you could just put a mirror in every location of interest, and not have to move or tilt a single mirror. Now you have a time-constant system that can do no worse - only better - than your original proposal. Secondly, your hot spot on Earth will get so hot when it reaches equilibrium as to radiate back toward the sun, establishing this "thermal equilibrium" of which scientists speak.
 — sqeaketh the wheel, Sep 08 2013

[Ling] Selecting frequencies using filters isn't a problem, but it can only decrease the temperature of the target. Again, anything that reduces the power density reduces the temperature.
 — spidermother, Sep 08 2013

But, don't we start with a higher power density by concentrating the light? Use enough density, and only use the high energy/frequency light which makes the Sunlight appear to originate from a higher average temperature. OK, you might say: the power density could not be high enough to compensate for the loss in filtration...so let's consider two mirrors very close to the Sun, and each will somehow filter out the low frequencies and half of the power. Together, they feed the same power to an object, but with only high frequency light.
 — Ling, Sep 08 2013

[squeketh] A solar concentrator is not an equilibrium situation, except insofar as the target's emission balances its absorption when it is at its maximum temperature. The sun's temperature and output can be assumed to be constant, and to be unaffected by the temperature of the concentrator.
 — spidermother, Sep 08 2013

[Ling] In this context, the temperature that the sun appears to be is not determined by its relative spectral distribution, but by the power density. If you are paid in \$50 and \$100 notes, you can't increase your spending power by burning all the \$50s - even if you have two jobs.
 — spidermother, Sep 08 2013

 I'm with [Ling] in not understanding this.

 I've got a light beam, corresponding to a small fraction of the sun's surface focussed onto as small an area as possible, and which can heat a small target to (say) 6000K, or whatever temperature the surface of the sun is. That light beam has a given total power, and a given mean wavelength (energy per photon).

 First puzzlement: By applying two such beams to the same target, I can't raise its temperature any higher - and I don't see why. What is the mechanism by which it fails to get hotter?

Second puzzlement: I filter out the 50% of the power that comes from the reddest photons, leaving blue beams with only half the original power of each beam. Yet if I now apply four of these beams to the same target (so, a total of twice the original power that raised the target to 6000K; _and_ the photons are bluer), its temperature still doesn't rise above 6000K. Again, I believe it but I don't see why.
 — MaxwellBuchanan, Sep 08 2013

 //I've got a light beam, corresponding to a small fraction of the sun's surface focussed onto as small an area as possible, and which can heat a small target to (say) 6000K, or whatever temperature the surface of the sun is.//

 No, you haven't. It is not possible to produce such a beam. Any system that approaches the temperature of the sun means that the target sees nothing but sun in all directions. There is simply no room to squeeze in additional 'beams' (quotes added because it is not possible to direct sunlight into narrow, laser-like, sun-surface-intensity beams at all).

 //By applying two such beams to the same target, I can't raise its temperature any higher - and I don't see why.//

 You can't raise its temperature by applying two such beams, because you can't apply two such beams. See above.

 // Yet if I now apply four of these beams to the same target ...//

I'll leave that as an exercise for the reader.
 — spidermother, Sep 08 2013

I'm with De Selby as always.
 — xenzag, Sep 08 2013

 OK, sorry to labour this, but I really want to get my head around it.

 A solar furnace in Odeillo can generate temperatures of up to 3,500K on a target described as being "the size of a cooking pot" (whatever that means). The furnace mirror is big - looks like maybe 100m on a side. The focal length looks also to be something like 100m

 Now, it would presumably be possible to build such a furnace with a longer focal length (say, 1km), and achieve more or less the same result, no?

 And if so, it would presumably be possible to build ten such furnaces, arranged and shaped such that their foci were all at the same place (I think).

 So, if I did that, would the target not get hotter than the 3500K produced by a single furnace? If I kept adding more furnaces, would its temperature asymptote closer and closer to that of the sun's surface?

(I have another question that follows on from the answer to this one.)
 — MaxwellBuchanan, Sep 08 2013

As I understand it in my most simple terms: Radiation is only heat in the sense that absorbing radiation (p) causes an individual nucleus to be raised to the frequency of that radiation. Since the frequency (vibration) of radiant energy is much higher than what we regard as "normal" we say that the interaction with radiation "heats". But by the same action radiation may also "cool" by forcing a nucleus into a lower energy state (wave-particle mass duality) while still respecting conservation of energy. So the most that any radiation source can do is raise or lower the frequency of the surface that it hits to the highest frequency that is present in the radiation (or as we see it the "temperature" of that radiation). We don't need a complicated apparatus to do this and it is easily achieved with a simple magnifying glass. At that point the body is at the excitation temperature of the radiation source. No matter how much more radiation that source absorbs and re-emits it will not vibrate any faster and as of yet we have found no way to exceed this capacity of materials to re-emit radiation. For reference, please regard the use of lasers for cooling. Cheers and to all a wonder filled day.
 — WcW, Sep 08 2013

Far be it for me to suggest empiricism, but maybe get a candle and a load of mirrors and see if you can make a higher temperature?
 — not_morrison_rm, Sep 08 2013

I'm happy to believe that I can't create a higher temperature - I just want to understand the mechanism.
 — MaxwellBuchanan, Sep 08 2013

I have nothing of value to contribute at this time.
 — Alterother, Sep 08 2013

 Good analogy [n_m_r]. Puts the project in perspective. Like the old saying goes : 'Tis better to light one small candle than to curse the 2nd law of Thermodynamics.'.

Keep us posted [Alter]
 — AusCan531, Sep 08 2013

[WcW] Some of what you said may be correct, but a solar furnace (and heat itself, for that matter) is a macroscopic, statistical thing. The details of quantum interactions can be safely ignored.
 — spidermother, Sep 09 2013

 //Now, it would presumably be possible to build such a furnace with a longer focal length (say, 1km), and achieve more or less the same result, no?//

 No. Increasing the focal length (without correspondingly increasing the diameter of the mirror) decreases the temperature at the target. Adding more mirrors merely brings the temperature back towards that of the original (short focal length) mirror.

Think of it this way. At the point where you have squeezed in every possible extra mirror, the result resembles one single, large mirror (Fresnel-like or not).
 — spidermother, Sep 09 2013

 Measure radiant heat of candle flame, insert candle into a mini-dyson sphere of bacofoil, with a long tube, measure radiant heat at the end of the tube.

My money is still either on 1) coal or 2) some kind of quantum effect that any part of the sun's surface being concentrated in a chrono-battery will instantly drop in temperature, because life is like that.
 — not_morrison_rm, Sep 09 2013

 I presume it is not linear. e.g. a mirror of area x heating a target to temperature 90% of solar surface temperature. Double the size of the mirror, will the temperature of the target surface increase to 91%? i.e every doubling of the size of the mirror gets you incrementally closer to 100% without ever quite reaching it. Eventually the sun and target are entirely surrounded by mirrors, and you have (to within a gnat's nadger) reached 100%.

And I'm going to guess that it's not the amount of radiation reaching the target per se, but the balance between radiation in and radiation out.
 — pocmloc, Sep 09 2013

 //I'm going to guess that it's not the amount of radiation reaching the target per se, but the balance between radiation in and radiation out//

Both are true. Assuming that the target is a perfect black body, and radiation is the only means of heat transfer, the energy absorbed is precisely equal to the amount of radiation reaching the target. The temperature reached is precisely the temperature at which radiation in balances (equals) radiation out. That temperature is determined by the power density of the incoming radiation, and nothing else.
 — spidermother, Sep 09 2013

 Of course, radiation obeys quantum laws. But the scale (moles of atoms and photons) means that an attempt to analyse a solar concentrator at a quantum scale is unnecessary, and potentially misleading. The second law, after all, is a statistical law - it describes what is so overwhelmingly probable as to be effectively certain, especially for large numbers of particles.

For instance, the sun emits gamma rays, which can produce temperatures in the millions of degrees - momentarily, for one or a few particles. But that's not relevant here.
 — spidermother, Sep 09 2013

 //I presume it is not linear. e.g. a mirror of area x heating a target to temperature 90% of solar surface temperature. Double the size of the mirror, will the temperature of the target surface increase to 91%? i.e every doubling of the size of the mirror gets you incrementally closer to 100% without ever quite reaching it. Eventually the sun and target are entirely surrounded by mirrors, and you have (to within a gnat's nadger) reached 100%.//

 I don't see why a theoretical concentrator needs to be especially large.

For example, consider a fairly deep dish, which produces a crisp image of the sun just inside the plane of the dish's edge. Place at the focus a small disk-shaped target, perfectly black on the side facing the dish, and perfectly reflective on the side facing the sun. Apart from the small region of dish shadowed by the target, which reduces the flux by about 1/45,000, the black side of the target receives the full sun-surface intensity, and therefore gets extremely close to the theoretical maximum temperature - regardless of the size of the mirror.
 — spidermother, Sep 09 2013

I always get the feeling, at this point in such discussions, that nature has just decided on these laws, and then thrashes around wildly looking for reasons why they can't be broken.
 — MaxwellBuchanan, Sep 09 2013

Thus, we again demonstrate that the halfbakery is a great place to get together with people of a similar inclination to collaborate on reinventing the wheel. No for reals, my peeps, the reason you are limited particular temperatures when you heat a body with radiation IS due to the "temperature" of that radiation. Exposing a hotter body to "colder" radiation does cool that body and this has been worked out for, I daresay, a hundred years or so. Sure, it's a "quantum" affect, every damn effect in this universe is a quantum effect. Its the same effect that sets the temperature of the surface of the sun, the very best reference for a place where solar radiation is concentrated. If it were possible to "heat" things hotter with solar radiation, then the one place it would happen would be on the sun.
 — WcW, Sep 09 2013

 [WcW] What do you mean by the "temperature" of radiation? It's not a concept I'm familiar with*.

 (No-one's denying that quantum mechanics is behind this, it's just an inappropriate level to explain a macroscopic event. It's like trying to explain why a computer game is difficult to beat by discussing the physics of logic gates).

 I know that a black body of a particular temperature emits a characteristic spectral distribution, and you could loosely refer to the temperature of that radiation. But you seem to mean something else. For example, can you explain why a microwave oven (with a radiation temperature, in that sense, of a few Kelvins) can make things hotter than pointing your finger at them (a radiation temperature of about 300 kelvin)?

Alternatively, one can measure the effective temperature of a radiation field at a particular point - it is the temperature that a black body at that point would reach in the absence of conduction and convection. But that depends on flux (the effective radiation temperature in unfocussed sunlight is less than that in focussed sunlight).
 — spidermother, Sep 09 2013

 In the simplest terms it is the measurable temperature that a body will achieve when exposed only to that specific radiation, in the case of light it was established that each emissions band in the spectrum had a specific temperature a long time ago. It actually takes very little light to figure this out, no massive solar collectors, no giant space mirrors. A prism, a filament and a platinum thermocouple in a vacuum is all that is really required. Apply light until the terminal temperature is reached, turn off light, measure the rate of cooling, add that to the final temperature, and voila, the "temperature" of that emissions band is known to you. Doubling, tripping, quadrupling the intensity of the light source does nothing but double triple and quadruple the amount of energy that the platinum thermocouple re-emits. From there it is plain to see that the whole system is kept in balance by this mechanism, the surface of the sun is at the "temperature" that it is at because it is in a near perfect balance between the various radiations produced by fusion, not at the frequency (heat) of the highest energy photons (though you could say that it passes through this state) and not at the "heat" of the lowest photons but somewhere in between, and emitting photons constantly to keep itself in this state. Bad science education (as reflected in this forum) means that people misunderstand how light becomes heat and how the system remains balanced. Radiation isn't a question about the thermodynamic relationship between the sun and the earth, it's a question of the balance between a single photon and a single atom.

A body exposed to one frequency of light will eventually reach an equilibrium between that frequency and it's own ability to radiate or otherwise loose energy. It will never go to a temperature higher than the frequency of that light [radiation] (unless fusion or fission is being caused.)
 — WcW, Sep 09 2013

 I respectfully disagree.

 It is easy to demonstrate that a high intensity of low-frequency light can heat an object to a higher temperature than a lower intensity of high frequency light.

 Light *in equilibrium* (which implies light that is in equilibrium with some matter, since photons do not interact directly with each other) has a characteristic frequency *distribution* (not a particular single frequency) that relates to its effective temperature. (But the systems we are discussing are not in equilibrium. Nonetheless, it happens that sunlight is a reasonable approximation of black-body radiation).

 //the surface of the sun is at the "temperature" that it is at because it is in a near perfect balance between the various radiations produced by fusion// is incorrect. The temperature of the surface of the sun, the temperature of objects exposed to sunlight, and the frequency spectrum of the sun's radiation are vastly different from those derived from fusion events. Last I checked, I do not explode into a giant ball of x-ray emitting plasma when I step outside.

 Most of the radiation leaving the sun represents ordinary thermal radiation. For the purposes of this discussion, it is useful to think of the sun as a big hot glowing ball, and wildly inaccurate to think of it as a series of photon-emitting fusion events. Those events occur deep in the sun, and very few of the photons produced by them directly leave the sun.

 //A body exposed to one frequency of light will eventually reach an equilibrium between that frequency and it's own ability to radiate or otherwise loose energy. It will never go to a temperature higher than the frequency of that light [radiation]// is nonsense. Read, for instance, about the use of lasers to produce staggeringly high temperatures in fusion research. It is the intensity, not the frequency, that allows those temperatures to be reached.

Unless I've missed something, you seem to be implying that the question "What is the temperature, in Kelvins, of 300 nm light?" has a precise answer. As far as I know, it doesn't.
 — spidermother, Sep 09 2013

 At this point, I am prompted to ask the question which was first asked by my Great Aunt Agapantha shortly before she was rescued from the Titanic's starboard cocktail lounge: "Would somebody be so kind" she asked, peering over her pince-nez, "as to explain to me, calmly, simply and in terms readily comprehensible to the lay person, what precisely the fuck is going on?"

 I am still confused, and I sense that my confusion runs deep - possibly as deep as others' understanding of this matter.

 I thought I was seeing the light with the idea that radiation has a sort of an effective temperature, dependent on its wavelength, and that when a body was heated to that temperature it would start giving out as much radiation as it absorbed. That, I could handle. Yet clearly that is not the case, as the microwave example and various statements show.

 So, if I could capture enough sunlight and 'magically' focus it on a small enough target, I _could_ heat that target above the sun's temperature. But geometry means that I can't actually focus enough sunlight on a small enough spot, so I can never do better than solar temperatures.

 It just strikes me as a very weasily way for nature to enforce the second law of thermodynamics. In short, I am not impressed by the way the universe organises itself - it seems far too heavily dependent on fine print and lawyer's clauses.

 I am tempted to ask what would happen if I made a really huge fibreoptic bundle, and stretched one end of it to narrow all the fibres down, so that 1 billion metre-diameter fibres tapered down to become 1 billion micron-diameter fibres. I'd point the big end at the sun, and the small end at an ant, and see what happened.

 However, I suspect that nature is going to tell me that the light will escape as the fibres taper, or that the ant will wander away, or some other such weaselry. Or antelry.

I'm prepared to put up with quantum mechanics - that's just nature saying it hasn't got time to do all the calculations and will tell me the answer when it's got a moment. Fair enough. But when thermodynamic laws are found to be upheld only by the power of geometry I think that's pretty cheap.
 — MaxwellBuchanan, Sep 09 2013

<starts working on a machine to cool the surface of the sun to something like 12 degrees c>
 — not_morrison_rm, Sep 09 2013

 Confusing, isn't it. If you do not believe in the experiments that established the principal that each emission band had a specific frequency and temperature and that light sources produced light that could only raise the temperature of an object to the temperature of that light, no matter how much of it you focused on them then the whole freaking thing is a special sort of mystery. That there are exceptions, such as the introduced electromagnetism of the microwave, and the confusing fact that the lower energy photons mixed in with the higher energy photons actually lower the potential terminal temperature, only further demonstrates the phenomenon.

That, when bombarded with a specific frequency of photons, barring other phenomena, an object can never exceed the "temperature" of those photons not matter how many of them there happen to be is a very well established scientific fact.
 — WcW, Sep 09 2013

 It may be well established, but a great many things are well established on the basis of very poor practices. It may be the law, but it seems to me to depend on a very ad hoc collection of supports.

 Actually I take that back. Something cannot depend on supports. It can depend from them, or rest on them. But in any case I don't like it. If I had a truck, I would have it no further with the second law of thermodynamics.

And hang on a second. If I take enough infra-red lasers and point them at something black, I bet I can raise its temperature to orange-hot.
 — MaxwellBuchanan, Sep 09 2013

yep. luckily an atom is not limited in the ways that it may express it's excitement at being raised to a higher frequency, and in so doing. it may well do it by producing photons of a higher temperature. Kinda boggles the mind really.
 — WcW, Sep 09 2013

but not a paradox. the second law still applies it is fortunate that materials can emit radiation at high energies without actually getting really hot in a general way.
 — WcW, Sep 09 2013

So we're back the point that the only reason I can't heat something to 10,000K with sunlight is that geometry stops me from focussing enough of the sun's light onto a small enough spot.
 — MaxwellBuchanan, Sep 09 2013

nope, you aren't going to exceed 5,900k. Filter the light and you might get up to 8500k, but there is very little radiation at that frequency.
 — WcW, Sep 09 2013

 //nope, you aren't going to exceed 5,900k. Filter the light and you might get up to 8500k, but there is very little radiation at that frequency.//

Hang on. I thought the point was that there ain't no way to focus sunlight to get above solar temperatures, and filtering out part of it will only make things worse. Or did I miss something, again?
 — MaxwellBuchanan, Sep 09 2013

 the sun is a bazillion degrees but who cares? the light is all that matters, and a photon of light can only raise an atom to its own frequency, no more.

 I know it's frustrating to think that you can make something as hot as that and all it will do is emit some pathetic radiation. I mean really, the sun is freaking hot, but all it can do is put out tons of low energy photons.

I like it because the alternative is the immediate heat death of the universe. And we're going to live with it.
 — WcW, Sep 09 2013

 No but, hang on. The surface of the sun is at 6000K. And the whole argument, above, was telling me that no amount of focussing, mirroring or filtering of sunlight will heat a target above that temperature.

 But you're saying that with enough focussing and some filtering, I can.

So, what gives?
 — MaxwellBuchanan, Sep 09 2013

here's what gives, when you expose an object to a mixture of radiation it actually can only arrive at an average temperature between the energies of the photons. the sun provides a healthy spread of radiation, and if you filter out the lower frequency photons there is a small quantity in the dangerous ultraviolet and above range that is actually higher than 8000k but thankfully for us the quantity is vanishingly small, even before the ozone layer.
 — WcW, Sep 09 2013

 And yet, according [Spidermother]: //Selecting frequencies using filters isn't a problem, but it can only decrease the temperature of the target. Again, anything that reduces the power density reduces the temperature.//

Between you, me and [Spidermother] we seem to have arrived at three diametrically opposed points of view.
 — MaxwellBuchanan, Sep 09 2013

 ...And I think some of us might be claiming some authority or expertise that evidently, isn't real.

Damn you anonymous internet forums !
 — Custardguts, Sep 09 2013

 I suppose you have your choice of realities, I generally trust the one that scientists come up with, but you can believe in something that makes more sense to you if you want. I've railed against some windmills in my day, maybe this your windmill, man, have a good go at it.

Spider has a theory: that energy is added by every photon regardless of its frequency. I have mentioned a few examples that contradict this. The theory of every photon adds more energy and energy is heat thus every photon makes a thing hotter is simply contradicted by observations and easily performed tests. It is based on a flawed understanding of how photons interact with atoms and was completely discarded ages ago because it was wrong.
 — WcW, Sep 10 2013

 -{What is the temperature, in Kelvins, of 300 nm light?" has a precise answer. As far as I know, it doesn't.}-

 The "temperature" of light at 300nm is 10,000 Kelvin. works out easily since the math is 3,000,000/(nm) =K

A body bombarded with 300nm photons will reach a terminal temperature of 10,000K. If it is hotter, it will be cooled to that temperature.
 — WcW, Sep 10 2013

 {anyone ready to second that statement}

>?
 — Custardguts, Sep 10 2013

Feel free to look it up. Wein's law, I believe.
 — WcW, Sep 10 2013

//if it is hotter it will be cooled to that temperature// wait, what ?
 — FlyingToaster, Sep 10 2013

 //3,000,000/(nm) =K//

 That gives the frequency in gigahertz, not the temperature in Kelvins.

 //300nm photons will reach a terminal temperature of 10,000K//

 Then how do you explain temperatures of millions of Kelvin, produced using 351 nm laser light?

I will put my reputation on the line and say that you are simply wrong, [WcW].
 — spidermother, Sep 10 2013

 People here have reputations? Sheesh, who knew! Must get one myself...

 This is all getting messy. I am sure that it's all very well-known physics, but there seems to be a lot of disagreement over the well-known physics.

I am not on any side - I just want to understand (in a graspable, palpable way) whether (or why) shining more light on something does or doesn't make it hotter. Or not.
 — MaxwellBuchanan, Sep 10 2013

 Is it easier to recouch the question in terms of lightbulbs, ovens or chickens, in such a way as to make it easier to think about?

 Is there some ideal object, say some special kind of lightbulb object, that we can make some general assumptions about.

All this talk about suns and atmosphere etc is clouding what ought to be a relatively simple issue. i.e. what's the nature of the transmission of "temperature" through the means of "radiation". It might make sense to define what we mean by those terms first, and get some agreement on that before proceeding.
 — zen_tom, Sep 10 2013

Objects get hotter when you shine more light on them for a perfectly straightforward reason: more light means more energy means more heat.
 — spidermother, Sep 10 2013

I've included a [link] up there ^ to some of the more common Light Measurement units and terminology. Be prepared to be overwhelmed!
 — csea, Sep 10 2013

 //Objects get hotter when you shine more light on them for a perfectly straightforward reason: more light means more energy means more heat.//

 OK, so we're now back to the point that you can't collect all the sunlight originating from a given area of the sun's surface, and focus it onto a target smaller than that original area, therefore the target can't get hotter than the sun's surface.

Is that a fair summary so far?
 — MaxwellBuchanan, Sep 10 2013

Only if the heat is collected in the form of heat. Heat doesn't get any hotter once you convert it into heat, but the heat source can be concentrated for greater effect. For instance, if you collect the hydrogen plasma ejected from the surface of the sun, compress it into a smaller volume and let it out through a little tiny hole, the heat then produced will be hotter than the plasma was originally. So for purposes of pedantry, I think it important that we differentiate 'heat' from 'heat source'.
 — Alterother, Sep 10 2013

Basically everything works pretty much exactly the way Spider says until you start to get really hot, then the law of diminishing returns takes over and no matter how many photons you pump in they can't excite the atoms more than their own frequency allows. It's damn hot, but not anywhere near the amount of excitation that an atomic nucleus can reach. There's all sorts of parts to it, Quanta, the radius of the electron orbitals, and then the whole bit about the lower quanta actually acting against the higher quanta when you get above their temperature. It's really fascinating reading, and a lot of stuff that was only theory in the 1920's is now substantiated in the laboratory. The clearest implications are in the area of photovoltaics where sorting the quanta and harvesting them to maximum effect is critical. Then the notion of the electron volt, rather than the temperature is the more applicable concept. Gamma Radiation is the hottest common form of radiation in our universe and each photon is carrying a significant amount of energy, atoms bombarded with gamma rays reach energies higher than a billion kelvin and a photon beam of one Plank frequency would cause excitation in the range of 10^17K. There is no notion of how radiation hotter than gamma radiation might be produced though.
 — WcW, Sep 10 2013

 //for purposes of pedantry, I think it important that we differentiate 'heat' from 'heat source'.// Quite so, but I specifically referred to 'sunlight'.

[spidermother] is my previous annotation essentially corrrectish?
 — MaxwellBuchanan, Sep 10 2013

Let me also just clarify that this limit applies only to black body behavior, nothing stops you from using the energy garnered to power an arc welder, or a microwave generator which can produce much higher temperatures.
 — WcW, Sep 10 2013

 //nothing stops you from using the energy garnered to power an arc welder//

 I'm afraid I'm going to have to correct you on that one. I cannot garner solar energy and power an arc welder, even if I have an immense parabolic mirror with a 100% efficient photovoltaic cell at its focus. I have been banned from any sort of welding since the fire. At least in the billiards room.

 But I di very gress. I am still intrigued by three things:

 (1) Can my tapering bundle of optical fibres not condense sunlight more than can be done by mirrors and lenses alone, to create higher temperatures?

 (2) Why do standard wine bottles contain precisely one glassfull less than one requires?

(3) Is there no way in which purely optical components can do "work", and thereby concentrate solar energy in the same way that a solar cell plus laser can?
 — MaxwellBuchanan, Sep 10 2013

 4) There's nothing in anything I read on Wein's law that describes the affect that you [WCW] stated above, ie that an object will adjust to meet the "temperature" of the light incident on it. From what I can tell, you're the only one making this assertion, and I can't find any support for it. Now I really have no idea either way, but I have at least moderate reading comprehension.

???
 — Custardguts, Sep 10 2013

 hmm, I'm wondering how something gets to be a black-body radiator in the first place. the Sun is mostly just H and He, but even combined, the emission/absorption spectra of each doesn't even remotely resemble the Sun's spectral output.

 //1// A mess of optical fibres isn't going to be near as efficient as a mess of lenses and mirrors.

 //2// That's because of the sediment trap in the bottom. Wines with a flat-bottomed bottle you often find you have much much more than needed.

 //3// You might be able to do something neat with Bose-Einstein Concentrate.

//4// I think that was just a generic remark where "cooler" means "heats less".
 — FlyingToaster, Sep 10 2013

 //  hmm, I'm wondering how something gets to be a black-body radiator in the first place. the Sun is mostly just H and He, but even combined, the emission/absorption spectra of each doesn't even remotely resemble the Sun's spectral output//

Just think how black The Sun would be if it wasn't glowing so brightly...
 — Ling, Sep 10 2013

 // OK, so we're now back to the point that you can't collect all the sunlight originating from a given area of the sun's surface, and focus it onto a target smaller than that original area, therefore the target can't get hotter than the sun's surface.

 Is that a fair summary so far?//

 Yes. And more generally, whatever fraction of the sun's output you collect, you can't focus it onto less than that fraction of the sun's area.

 //(3) Is there no way in which purely optical components can do "work", and thereby concentrate solar energy in the same way that a solar cell plus laser can?//

Sun-pumped lasers might qualify.
 — spidermother, Sep 11 2013

I don't suppose we can expect everyone to accept quantum electrodynamics at the same time. Spider's postulation does not explain why we can't get incredibly high temperatures in a solar collector, it doesn't explain why lasers of different colors have different features, it isn't compatible with photovoltaic observations, it's the classical model and it has been superseded.
 — WcW, Sep 11 2013

 //Yes. And more generally, whatever fraction of the sun's output you collect, you can't focus it onto less than that fraction of the sun's area.//

 Is that an existing law, or can we claim it?

"The Second Law of Thermodynamics cannot be fooled, no matter how much smoke and how many mirrors are used"
 — Ling, Sep 11 2013

[csea] Photometry relates to brightness as perceived by the human eye. Radiometry covers units relevant to this discussion.
 — spidermother, Sep 11 2013

 // hmm, I'm wondering how something gets to be a black-body radiator in the first place. the Sun is mostly just H and He, but even combined, the emission/absorption spectra of each doesn't even remotely resemble the Sun's spectral output//

The light-emitting region of the sun is largely plasma, which does not have discrete spectral lines. Atoms and molecules are found further out, and contribute absorption lines to sunlight.
 — spidermother, Sep 11 2013

 //does not explain why we can't get incredibly high temperatures in a solar collector//

I thought the whole point was that you _can't_ get incredibly high temperatures in a solar collector - at least, not above the temperature of the solar surface. Record is something like 3,500 or 4000K, as compared to a solar surface temperature of around 6,000K
 — MaxwellBuchanan, Sep 11 2013

//I don't suppose we can expect everyone to accept quantum electrodynamics at the same time// I think there are plenty of people here ready to accept anything but you have to explain it coherently and convincingly first, not just point out their ignorance to them.
 — pocmloc, Sep 11 2013

 Maybe, in the far flung future, where control of individual photons, by optics, is like placing Lego blocks, a dynamic structure/environment could be designed that could create any temperature desired.

This act would be technology and not the natural order.
 — wjt, Sep 11 2013

Well said, [poc].
 — Alterother, Sep 11 2013

 // (1) Can my tapering bundle of optical fibres not condense sunlight more than can be done by mirrors and lenses alone //

I would think that if the fiber tapers, the reflecting walls are no longer parallel so some of the light light coming in the fat end will reverse direction and go back out the way it came in.
 — scad mientist, Sep 11 2013

nothing limits the amount of sunlight you can force into a tiny location. Take the surface of the sun itself. Pretty much a lazy worst case scenario, right: you're sitting there, right on top of a massive emitter, photons pumping through you like grand central station, and what, what is this? You only measure 5600 kelvin?!? WHY??? Because at that point even a giant concave death ray mirror Isn't going to get you above 5600K. The sun's core is at a bazillion degrees, but the radiation escaping it is only 5600K.
 — WcW, Sep 11 2013

 //photons pumping through you like grand central station, and what, what is this? You only measure 5600 kelvin?!? WHY??? Because at that point even a giant concave death ray mirror Isn't going to get you above 5600K. //

 Mate that doesn't even make sense. The first instance has nothing do do with the second, and you're not describing the mechanism at all. And what does the solar core temperature have to do with the plasma temperature at the periphery (some millions of kilometers distant), aside from a simple heat balance relationship. Nothing, that's what.

I have to be honest here, I'm still waiting on a second, convincing opinion in support of your statements about incident light "forcing" objects to adopt the representative temperature.
 — Custardguts, Sep 11 2013

 We've had much of this discussion before, see my link. What it comes down to is that an object that is heated by concentrated sunlight, in equilibrium, radiates out as much heat as it receives. With reversible objects, much of this heat is returned to the sun. In equilibrium, the sun gets a tiny bit hotter, depending on how much total energy is intercepted for your heating. In the case of a dyson sphere, with 100% interception, the sun gets significantly hotter, otherwise, generally not.

 As far as the "temporal violation", it's relatively simple. A passive system can't improve on the version produced by simple concentration, since the energy flow is in equilibrium.

An active system could (so you move optics as appropriate to correct for the speed of light), but there is in input in energy and entropy that offsets your "violation".
 — MechE, Sep 13 2013

 //An active system could (so you move optics as appropriate to correct for the speed of light), but there is in input in energy and entropy that offsets your "violation"//

But, in sort of theory, the energy needed to move the optics could be arbitrarily small. A mirror, moved against a spring, stores the energy used as spring compression available for later use. However, this then brings one back to Maxwell's Demon, which the second law again seems to weasel out of by means of the small print.
 — MaxwellBuchanan, Sep 13 2013

 The energy to move the optics cannot be arbitrarily small, because the more energy you want to introduce, the more complex or massive the system becomes.

Say, hypothetically, you have two focusing mirrors, one at distance of 1 light second and one at a distance of 2. You start with the 2 second mirror, move the 1 second mirror in place. In order to move the second mirror in place in much less than a second, you are going to have an extremely high acceleration/deceleration.
 — MechE, Sep 13 2013

 If we were dealing with collimated light, it would be no problem. Just use a tilting mirror to send the beam through either of two different fixed paths - one path being longer than the other; the amount of tilt needed can be arbitrarily small.

 Even with moving (as opposed to tilting) mirrors, the energy used to move them (however large) can be recovered in principle, simply by decelerating them after the move.

However (a) we're not dealing with collimated light and (b) there's always a subclause.
 — MaxwellBuchanan, Sep 13 2013

 If we were dealing with collimated light, it would be no problem. Just use a tilting mirror to send the beam through either of two different fixed paths - one path being longer than the other; the amount of tilt needed can be arbitrarily small.

 Even with moving (as opposed to tilting) mirrors, the energy used to move them (however large) can be recovered in principle, simply by decelerating them after the move.

However (a) we're not dealing with collimated light and (b) there's always a subclause.
 — MaxwellBuchanan, Sep 13 2013

This is about the fundamental property of photons, not optics, mirrors, space, the sun, total energy in systems, or even thermodynamic rules.
 — WcW, Sep 13 2013

 It shouldn't be. There is nothing to prevent a photon from adding energy to an object, no matter how energetic the object is.

 The fundamental matter is black body radiation. For any given level of radiometric power input, an isolated object in a vaccum will reach a temperature such that the object is emitting the same amount of radiometric energy. Typically, this will be in the form of a larger number of lower energy photons (Infrared, most commonly). A small percentage will actually be higher energy due to anti-Stokes Raman scattering, but that's not really relevant to the discussion, and the energy is still conserved, since it's complemented by Stokes Raman scattering.

 Since the object cannot dissipate more energy than it is absorbing, and since all objects closely approximate a black body with regards to the wavelengths they absorb (we can ignore reflection completely), the hottest an object (without a fuel source) can be is the temperature of the object heating it. At that point the object is emitting exactly the same, and the same wavelength photons it is taking in, resulting in no temperature increase.

The fact that some of these photons are returning to the source instead of continuing out into space results in an increase in the temperature of the source, which in turn allows the the object to get slightly hotter as well. However, if the object is planetary sized or smaller, and the source is the sun, this effect is so negligible as to be irrelevant.
 — MechE, Sep 13 2013

 As far as I can tell from all that has been said, countersaid, unsaid, resaid and then said again the other way around, the whole second law business is just based on the fact that you can't catch all the light from a given area of a heat source and focus it onto a smaller area than that from which it came.

 As I've mentioned, I'm rather disappointed that this thermodynamic law relies on cheap geometry tricks in order to uphold itself.

 It's a bit like saying "You can't exceed the speed of light, because there's a tree in the way."

Which, incidentally, I would like to shamelessly [marked- for-tagline] to express my disappointment with the universe in general.
 — MaxwellBuchanan, Sep 13 2013

 Well yes. It's fair to say you could do the physically impossible, if only you could do the physically impossible.

However, it's more like saying that you can't exceed the speed of light because it would require an infinite amount of energy. It's an inherent property of the system, not a really a cheap trick.
 — MechE, Sep 13 2013

 //an inherent property of the system, not a really a cheap trick//

 Sounds cheap to me.

 "I'd like to violate the second law of thermodynamics, please." "Ah, I'm afraid you can't do that." "Why not? Is it because, at a deep relativistic level, the very fabric of spacetime itself warps in such a way as to nullify any attempt to concentrate light beyond a certain limit?" "Ah, no, not really, and in any case the fabric of spacetime itself isn't 'very'." "Well, then, is it because at a quantum level the inherent and mysterious unpredictability of microscopic events leads to a statistical diffusion of matter and energy in such a way as to ineffably confound our efforts to concentrate energy." "Ah, no, that's not what it says in my manual. And what's 'ineffably' mean?" "Well then, does the existence of the second law give us insight into the mysterious interplay between dark energy and dark matter, and their role in the continued and hitherto unexplained accelerating expansion of the universe?" "I can ask my line manager, but as far as I know, no." "Well then, why on earth not?" "It's because the mirrors won't fit."

See what I mean? Cheap tricks of geometry.
 — MaxwellBuchanan, Sep 13 2013

Well if you had a four-dimensional periscope that could peer over the edge of all those edge-to-edge mirrors, you’d be sorted.
 — pocmloc, Sep 13 2013

The issue will never be the optics of mirrors or lenses, or the fact that some of the energy goes back to the sun or any such gobbledygook. Imagine for a moment, a perfect mirrored sphere, with a black body mass in the middle. Through a tiny hole we lase in light of a specific frequency, lets say 300nm, which is said to have a specific "temperature". All of this lased light hits the black body and it heats up, emitting more and more photons as it's effective temperature rises. For a moment we discard all questions about wavelengths and orbitals and just use the average of the distributions as have been extensively studied. As the black body becomes hotter and hotter there is a continuous equalibrium between admission and emission of photons, taking the "pure" lased light and distributing across a curve of higher and lower emission spectra. As the black body nears the temperature of the lased light (in this case ~6000k) the black body begins to emit photons that are at higher energies than 6000k, and a still equal number of photons distributed below that, in a perfectly even distribution around the 6000K point. Gradually the space inside the perfect sphere fills with photons in this perfect distribution, the black body emitting and absorbing all of them at an equal rate. This body is at equilibrium with the photons bombarding it; it's not that the system doesn't contain more and more energy, as we pump into it, but the fraction of that energy that is in the excitation of the atoms has been saturated and all of the additional energy is in the form of photons bouncing around. Another way to look at it is that there is an abrupt, brick wall like diminishing in returns when the black body approaches the temperature of the light, such that massive additions of photons make no appreciable difference in its excitation.
 — WcW, Sep 13 2013

 Yes indeed, but [spidermother] seemed quite adamant that radiation doesn't have a "temperature" (of 6000K or anything else) in that sense, and argued that the equilibrium temperature reached depends just on the total power input, not on the wavelength of the light.

I have no idea if this is true or not, but I assume it is a simple and well-understood matter. Problem is that either [spidermother] or you is wrong, and I do not know which.
 — MaxwellBuchanan, Sep 13 2013

photons of different frequencies carry different amounts of energy. A gamma ray has so much energy that individual absorptions are observable events. A photon in the IR range has very little energy, per article. While we might not agree about using the word "temperature" to describe this energy, nevertheless, the value that we assign is based on the above described experiment. Each frequency of light will excite a black body to a terminal temperature if given the chance. Thus we acquired this "thermal" understanding of light as represented by "color temperature" (see wiki article of same name). Spider is wrong.
 — WcW, Sep 13 2013

 //in a perfectly even distribution around the 6000K point//

 Not true. The distribution of photons emitted by a black body in no way resembles a normal distribution. It is a mostly one sided distribution with a very short tail into the shorter (higher energy) wavelengths from the peak. You seem to be confusing "color temperature" with temperature. Light does not have a temperature. It can be described by the temperature that an equivalent black body would give. Since incandescents worked by heating an element to that temperature, that was the historical way to do it, but even then it describes only the spectrum, not a real temperature.

 But since most newer or narrower light (including laser) sources do not use pure heat (instead relying on fluorescence or electroluminescence), it is more than a little outdated.

In the case you described, a black body inside a perfect spherical mirror, it would continue heating up, and continue emitting more and more light until such time as the emission back through your laser input point was equal to the energy of the laser going in. This has nothing to do with any inherent properties of the photon or the material at the center of the sphere.
 — MechE, Sep 13 2013

 // Each frequency of light will excite a black body to a terminal temperature if given the chance.//

And we cross commented here. The color temperature is the spectrum that a black body would give off, not the other way around.
 — MechE, Sep 13 2013

 "Temperature" is a measure of the average photonic energy being emitted per of an object.

 At maximum flux density of a black body, the peak frequency (and I'll assume that the sum of the energy emitted by higher-than-the-peak frequencies equals the sum of the energy emitted by lower-than-the-peak frequencies) is the temperature.

 At non-maximum flux density, for instance an LED emitting a green light, the average flux density makes it a lower temperature emitter. Of course you could say that the green LED's temperature is somewhere in the radio spectrum. But people would throw things at you.

 ----

 I think the only application of the SLoT is that if you have a hot body and a cold body then eventually the temperatures will equilibriate if there's a conduit between them, Maxwell's Daemon notwithstanding. It's not particularly sequiturous to the problem at hand.

 ----

 The problem at hand seems to be "Can you focus light from an object, for which every point on the surface radiates in all directions, onto something to make it hotter than the original object". (This rules out lasers)

 ----

 Obviously, in the case of a magnetron, a parabolic'ish reflector and a turkey, the answer is yes.

So now does the question need to be redefined to include "distance from the Sun" andor "blackbody" ?
 — FlyingToaster, Sep 13 2013

 If I may quote my late uncle Thaprick, "aaargh!".

 So, if I shine _enough_ light onto a finite target, I can raise it to any temperature, regardless of the colour of the light?

 [WcW] says "no - the light has an effective temperature related to its wavelength, and the target can never get hotter than that, which is why you can't focus sunlight to give arbitrarily high temperatures."

 [spidermother] and others say "yes - all that matters is the total energy flux, so enough red light will heat something to blue hotness; but geometric optics prevent you from delivering more energy per unit area than the sun gives out per unit area, which is why you can't focus sunlight to give arbitrarily high temperatures."

Two alternative explanations of the second law. Which is it?
 — MaxwellBuchanan, Sep 13 2013

 [Spider] is absolutely correct. If you can produce enough light, it is possible to reach any temperature.

Specific case in point. CO2 lasers used for machining emit in the infrared, at a wavelength of 9.4 microns, they have a black body color temperature of 308.3K. But the cut edges of metal are (briefly) red hot, and thus must have a color (and black body) temperature around 1500K.
 — MechE, Sep 13 2013

 Yes, I was wondering about laser cutters but hadn't found the wavelength of CO2 lasers - thankyou.

Therefore, we are left with "It's because the mirrors won't fit" as the only thing holding the second law together, which seems very insatisfactory to me.
 — MaxwellBuchanan, Sep 13 2013

 It's not that the mirrors won't fit, it's that equilibrium is maintained.

 If you took a dyson ellipsoid mirror, such that your target was at one focus and the sun was at the other, 100% of the sun's emitted energy would be focused on your target. But, and this is the critical bit, 100% of your target's energy is focused back on the sun.

 Why this is critical is when your target approaches the sun's surface temperature. At that point, the energy emitted by your target is approaching the energy emitted by the sun. Since minimal energy is escaping the system, the temperature of both will continue to increase. In this case, equilibrium would be the point at which the energy escaping from your dyson ellipsoid is equal to the output of the sun, and both the target and the sun would be at a higher temperature than the sun started.

 Even with a little tiny earth based parabolic mirror that in no way approaches the sun's temperature, you are doing the same thing. It's just that the fraction of the sun's energy you are returning to it is so vanishingly small that the temperature increase is well below any existing instrument's ability to pick it up.

So, in summary, you can increase a target's temperature above the default solar surface temperature, but not above it's current surface temperature, whatever that happens to be.
 — MechE, Sep 13 2013

 // the energy emitted by your target is approaching the energy emitted by the sun.//

Yes, but imagine if the target were the size of a baseball. Then, at equilibrium it would be receiving all the sun's energy, and emitting the same; but it would have to be much hotter than the [much bigger] sun to emit the same amount of energy. So it comes back to the purely geometric problem that you can't focus the light from a big sun onto a much smaller target.
 — MaxwellBuchanan, Sep 13 2013

Ah, but thats a rather neat function of snell's law. Your disappointment is my amazement on how all the little bits of the universe function together.
 — MechE, Sep 13 2013

^^^ So what's the difference between that and a globular magnetron, parabolic mirror and turkey ? Even though the turkey will be getting a fraction of the microwaves emitted by the spherical emitter, it will heat up far faster and maintain a higher temperature than the emitter will from reradiated warmth.
 — FlyingToaster, Sep 13 2013

 Equilibrium again. The turkey never reaches equilibrium, and thus never begins to emit the same level of radiation back to the magnetron as the magnetron does in the first place.

And remember, the magnetron is not a black body radiator, so it's emissions are not dependent on it's temperature, whereas the turkey could be approximated as one, but as it cooks, much of the heat is removed in the form of flowing steam and juices/drippings. And if you go to far, as gaseous combustion products. That is why the turkey can maintain a high temperature despite never reaching equilibrium with the magnetron.
 — MechE, Sep 13 2013

 //thats a rather neat function of snell's law//

 Yes, you see, that's just the sort of thing I'd expect.

"Snell's Law". The name alone suggests that this is some picker-snip's, quell-tenter's addendum to patch up a glaring loophole (or, if not a loophole, then certainly a phole of some kind) in a badly- drafted thermodynamic law. "Snell's Law" is just another way of saying "because the mirrors won't fit".
 — MaxwellBuchanan, Sep 13 2013

 //in a perfectly even distribution around the 6000K point//

 Not true. The distribution of photons emitted by a black body in no way resembles a normal distribution.

 -Even, in that there is equal area underneath the curves on each side. -

 It is a mostly one sided distribution with a very short tail into the shorter (higher energy) wavelengths from the peak. You seem to be confusing "color temperature" with temperature.

 -No, science is confusing you by using the same word for two different things.-

 Light does not have a temperature. It can be described by the temperature that an equivalent black body would give. Since incandescents worked by heating an element to that temperature, that was the historical way to do it, but even then it describes only the spectrum, not a real temperature.

 -nay, it does describe both the light and the temperature of a black body-

 But since most newer or narrower light (including laser) sources do not use pure heat (instead relying on fluorescence or electroluminescence), it is more than a little outdated.

 -Utter poppycock. It is not dated, you are simply observing that most things to not behave as black bodies do. Black bodies are the concept that allows us to understand how radiation works-

 In the case you described, a black body inside a perfect spherical mirror, it would continue heating up, and continue emitting more and more light until such time as the emission back through your laser input point was equal to the energy of the laser going in.

 -true, the system cannot be made a perfect store for photons, such a thing is impossible. But long before we reach the point that the leakage is equal to the input the black body will be almost perfectly in agreement with the energy of the lased photons, the leakage is a rather negligible factor a perfect sphere the size of a grape with a hole of one Plank diameter would hold all of the energy of the universe before it was "full" and emitting as much as it was absorbing. The Plank is that small, the grape is that large.-

 This has nothing to do with any inherent properties of the photon or the material at the center of the sphere.

-What you are proposing is the ultraviolet catastrophe, a concept that is fundamentally disproved by observation. the object in the sphere does stabilize in temperature, otherwise, as is observed, the surface of the sun should emit massive quantities of very high energy photons, which it does not.-
 — WcW, Sep 13 2013

 [WcW] Please respond specifically to my comment regarding CO2 laser cutting then. Your understanding here is fundamentally wrong.

 I don't argue that the object at the center of the sphere doesn't stabilize at a given temperature, but that temperature is in no way related to your input light's "temperature". It is instead related specifically to the equilibrium point. And since in the case you mentioned your input is a narrow frequency source, and your output is a blackbody source, the two would not even match in color/wavelength distribution.

 And to clarify, nothing I have mentioned in any way leads to the "ultraviolet catastrophe" issue. All I have said is that energy in equals energy out, regardless of the form of the energy.

//Even, in that there is equal area underneath the curves on each side.// Have you seen a black body curve? Specifically because of Planck's law, which prevents the existence of those "many high energy photons", the fall off at shorter wavelengths is extremely sharp, with no long tail, vs a broad fall off with a long tail towards the longer wavelengths.
 — MechE, Sep 13 2013

I like symmetry: is the energy equal on both sides ?
 — FlyingToaster, Sep 13 2013

 Nope, the energy curve is what I've been describing. The photon count curve would be even more extreme.

There's also the problem of describing this as "both sides". A normal distribution occurs when you have a central point that things can fall on either side of. The radiation curve simply happens to have a peak, not really the same thing.
 — MechE, Sep 13 2013

My use of lasing was a screw up, coherent light behaves differently from thermal light and is not limited to black body behavior. It isn't nearly the only exception but I was careless in my use of the word. If you use coherent light, then the black body will get much hotter than the color temperature of the light.
 — WcW, Sep 14 2013

 This idea has evolved from 'violating the second law of thermodynamics', and has become 'heating a target hotter than the sun using only optical means of concentrating sunlight'. Which is Fine, but title should change.

 I think that the parameter of interest in this problem is etendue. It is a measure of the disorder in an amount of light. It is similar to entropy.

If you take all the light that enters your optical system and concentrate it to the limit, you will not exceed the sun's temperature. But if you devise a system to select out the lowest-etendue rays, then concentrate those to limit, you may exceed the sun's surface temperature. A pair of distantly-seperated stops would work to select the most collimated rays, but the stop pair and concentrator would have to track the sun's path.
 — afinehowdoyoudo, Sep 14 2013

 Coherent light does not "behave differently" than black body light. EM radiation is EM radiation.

 Of course by making this claim, you have changed your terms somewhat. It is not possible for a black body source to heat a black body target beyond itself, but that is the point we've been discussing, and has to do with the energy limits, still not anything inherent in the photons.

Let me make this clear, energy flux is all that matters. If energy in is greater than energy out, the object gets hotter until the two terms equalize. The case of source and target being black body means they will equalize at the same point, but if it were theoretically possible to input more black body radiation, the object would get hotter, without regard to the color temperature of the source.
 — MechE, Sep 14 2013

 // if it were theoretically possible to input more black body radiation, the object would get hotter, without regard to the color temperature of the source.//

 Precisely. So, can I devise a system of tiltable mirrors by which I can focus sunlight onto a target via either of two different paths - one path being 1 light-hour long, the other being only 30 light- minutes long? If so, then by tilting the mirrors at the right time, I can deliver double the intensity of light to the target, for a 30 minute period, during which time the target should heat up above solar temperatures. Note that the energy needed to tilt the mirrors can be recovered, if they move against springs.

My guess is that the answer is going to be no, because of some geometric small print which says that there's only one possible path that delivers sunlight efficiently, or whatever.
 — MaxwellBuchanan, Sep 14 2013

 You are correct that the answer is no.

 For a given target size, there is a finite area of mirror that can be focused on it from a given source. This area is a function of the target size and the apparent size of the source, as well as the mirror distance from the target (it's actually simpler if you think of it in radians, because then it's size instead of distance). In order to approach solar temperatures (in orbit) your mirrors would need to approximate perfect and still would have to occupy 100% of this space.

 In order for your timed effect to work, both mirror sets would have to be in this space. Therefore the nearer (shorter) path mirror would have to occlude the larger mirror, eliminating the doubling effect.

Of course the same effect could be achieved by using the second mirror to power a laser, which is then directed at the target. Since the laser has a very small apparent size at the time it reaches the target, it can be targeted on a much smaller area than the reflector. Doing this, however, will still increase entropy by way of heat losses in the steps from sunlight to laser light.
 — MechE, Sep 14 2013

And pivoting or otherwise moving the nearer mirror out of the way doesn't help. The nearer mirror can either reflect light towards the target, or allow light from a more distant mirror to pass through the same space, but not both at the same time. The two are equivalent - either way, the same amount of light passes the same point at the same time, and reaches the target at the same time. There is no way to superimpose the light from different mirrors unless they occupy non-conflicting pathways in space.
 — spidermother, Sep 14 2013

 Erm, black holes?

 If you were to point your sunlight-collecting mirrors at a black hole, the sunlight passes beyond the Schwarzschild radius* the light can't bounce back...but don't ask me how you'd get any power out of it.

 * Not to be confused with the Schwarzenegger** radius (being the infinitesimal difference between plots of two films he has appeared in in the 1980's)

*** Not to be confused with the Schwarzkopf radius, being the figure derived from the waist measurement of the General.
 — not_morrison_rm, Sep 14 2013

 Apparently there's a theoretical relationship between Hawking radiation and the temperature of a black hole that follows , so if it's absorbing more light, it heats up, it radiates out more.

Regardless, I don't believe (less than certain here) that gravitational lensing allows any more concentration than optical. Since physics kind of breaks down around singularities, I won't swear either way.
 — MechE, Sep 14 2013

 //The nearer mirror can either reflect light towards the target, or allow light from a more distant mirror to pass through the same space, but not both at the same time.//

 But that's my point - we make the distances large enough that the two paths happen at different times.

 Consider two consecutive one-second-long "bursts" of light from the sun, with half an hour in between them.

 The first lot of light (pulse A) heads out into space. It encounters the first mirror, but that first mirror is turned edgeways on, and thus does not significantly block the light. Pulse A therefore continues to a far- distant mirror, which sends it back the way it came.

One Pulse A has passed that first mirror, that mirror rotates, so that the second pulse (B) hits it face-on and rebounds. After that, the first mirror again tilts out of the way before....ah, hang on. I see where this is going to go bear-shaped.
 — MaxwellBuchanan, Sep 14 2013

 MechE's point about the ratio of target size to collector area is quantified in terms of etendue. If you hate frogs then call it geometric extent. Light with a low ge is more focussed or more collimated and is capable of heating a target to higher temperature (in the focussed configuration). Lasers.

 Selecting out the 'best' rays from a column of sunlight, geometrically allows greater concentration and hence higher temperature at target, while entropy is preserved or increased.

Problem solved, and I feel confident that my esteemed peers at the hb will be struck silent by the need to say no more, and I will score a 'last comment'
 — afinehowdoyoudo, Sep 14 2013

// I will score a 'last comment'// Undoubtedly.
 — MaxwellBuchanan, Sep 14 2013

 // I see where this is going to go bear-shaped.// Finally!

 //Selecting out the 'best' rays from a column of sunlight, geometrically allows greater concentration and hence higher temperature at target, while entropy is preserved or increased.//

I think that that suffers the same drawback as filtering out the lowest frequencies. The 'best' rays would make it to the target anyway (without any selection), so removing the less-than-best rays can only decrease the temperature.
 — spidermother, Sep 15 2013

I thought photon energy is not the same as heat. The complex interaction that is the transfer of a photon to heat is at the heart of this. heat -> all photons -> heat (naturally downhill) . If a photon sits in the exact electromagnetic volume of another photon, doesn't the molecule/atom just see one photon?
 — wjt, Sep 15 2013

Lasers totally confuse the issue because their co-ordination allows them to excite atoms more than the pattern of incoherent light can. by acting in the same plane, rather from a random distribution they do not act against one another. This means the nucleus reaches a higher level of excitation without emitting the higher energy photons that would allow it to return to a lower energy state.
 — WcW, Sep 16 2013

No. Just no.
 — MechE, Sep 16 2013

Really man, really dude? You want to have a go at explaining coherent/incoherent light? Be my guest.
 — WcW, Sep 16 2013

 Not really, because it's not relevant in this case. All that matters is energy balance. A black body source can't heat a black body target above it's own temperature simply because the energy transmitted at a given color temperature cannot exceed the energy that produces said color temperature.

Focused LEDs, which are non-coherent, can provide any amount of energy at any given wavelength (color temperature). Thus they can heat an object to much higher than their apparent color temperature. Lasers are the same. Coherence doesn't matter, except that coherent light can be focused extremely tightly, and maintain that focus over long distances allowing it to be even more concentrated than narrow spectrum LEDs.
 — MechE, Sep 16 2013

 oops, I was wrong. The setup with two widely separated apertures will produce a column of rays that are more collimated than regular sunlight, and can hence be focussed to a smaller spot, but the intensity of the 'improved' light will be reduced by the same factor as the minimum spot area, so there is no net gain. Geometry got the better of me there.

 There is a book by Roland Winston that delves into this in great detail, I think its called Non-Imaging Optics.

*backs away slowly, smiling and nodding*
 — afinehowdoyoudo, Sep 17 2013

you have made an assertion that I was not able to find evidence for. Can you point to an example of an incoherent light source heating something to a temperature in kelvin greater than the frequency as given by Weins law?
 — WcW, Sep 17 2013

 Wein's law does not give "the" frequency associated with a temperature; it describes how the frequency distribution changes with temperature.

 Instead of graciously admitting that you were wrong in your assertion that light of a particular frequency always produces a particular temperature, you have instead changed to a new (bogus) theory in which coherent light is the only exception. Have you already forgotten that your example to back up your first version was laser cooling?

The best I can say is that there is a tiny grain of truth in all WcW's trollish waffle, which is that in the special case of perfect black bodies in thermal equilibrium there is a direct correlation between frequency *distribution* (not frequency per se) and temperature. But there is no law directly linking the frequency of light - coherent or otherwise - to the temperature of an object illuminated by that light.
 — spidermother, Sep 17 2013

 Microwave oven?

Black body temperature for microwaves is at or below room temperature. It can boil water.
 — MechE, Sep 17 2013

 Microwaves correspond to temperatures just a few degrees above absolute zero, and can melt steel. Strictly speaking, magnetrons produce fairly cohent light, but that can be fixed by doing the experiment in a large, irregularly shaped chamber.

Anyway, there's no need to provide such examples. WcW is wrongly asserting a causality from a partial correlation. One might as well say "Cows have non-cubical udders because grass is green. You don't believe me? Give one example of a cubical udder caused by green grass."
 — spidermother, Sep 19 2013

<clicks on idea, starts skimming through replies, slowly backs away>
 — DIYMatt, Sep 19 2013

So, why do rooms painted blue look colder than those painted red? And why do people look cooler wearing light-absorbing shades?
 — MaxwellBuchanan, Sep 19 2013

 If there is a misunderstanding in my knowledge it is not as profound as your misunderstand of how a microwave works. We all recognized the fact that there are exceptions to the rules that guide the relationship between photon interactions and and temperature, and that the dielectric action of microwaves are one example. There are other exceptions. They do not help us understand why when intense sources of incoherent light are focused on small areas those areas are not raised above specific temperatures no matter how large the mirrors and the intensity of the light source. This phenomenon is observed, your explanation does not fit it.

 I have posted a link to the foremost mind on the topic. I would invite you to watch the entire series, but this lecture does cover this specific question.

This is the best I can do.
 — WcW, Sep 19 2013

 What misunderstanding? First, I provided an example of EM radiation producing a greater temperature than it's color temperature (CO2 laser). You then, incorrectly, claimed that coherent light was an exception, using a rationale that makes no sense whatsoever.

 You asked for a case where non-coherent EM radiation produces a temperature greater than it's "color temperature" and I provided one. The mechanism of heating is not relevant. If you are going to except all cases where the mechanism of heat transfer is not black body radiation, then you and I are in agreement, but only because it's a very specific case.

 As far as other cases where narrow wavelength incoherent light is used for heating, it's not a very common application, simply because black body sources are relatively efficient. I did the theoretical research for a design that used focused infra-red LEDs as a narrow wavelength heating application, and it was entirely feasible. For other reasons (the cost of LEDs required to reach our target temperature), it was not practical.

 I do not have time to listen to the Feynman lecture at this point, but since I'm pretty certain he never made such a fundamental error in physics as you are, I am okay in stating that it is still you who are in error.

 Let me try a few simple statements and you tell me which one you disagree with:

 1) LEDs produce incoherent light whose wavelength distribution (apparent color temperature) is not related to their energy output. (This is in contrast to black bodies where Wein's law defines this relation).

 2) It is possible to produce an LED that outputs more energy than an identically sized black body at it's apparent color temperature.

 3) It is possible to focus the light from said LED to the same degree you could focus the light from an identically sized black body. (This is a little untrue, the narrower wavelength distribution improves focus, but I'll go with it as a first approximation).

 3) This allows more energy to be focused on the target from the LEDs than could be achieved by the identically sized black body.

 4) An object heats as a function of the energy it absorbs less the energy it re-radiates.

 5) Radiation by a black body object is purely a function of temperature.

 6) Absorption by a black body object is constant.

 7) Energy is conserved.

If you take the sum of these statements, you realize that by putting sufficient EM energy onto a target, it is possible to heat an object to greater than the "color temperature" would indicate, and that doing so is possible with non-black body light sources.
 — MechE, Sep 19 2013

 //I have posted a link to the foremost mind on the topic. I would invite you to watch the entire series, but this lecture does cover this specific question.//

 Did you post the wrong link? It's a good lecture (I love Feynman), but does not touch on the present question whatsoever. He is talking about QED, and reflection from pairs of surfaces, and how these things can be explained by probability amplitudes. (I did watch the whole lecture.)

If you have a Feynman lecture that's related to the second law, please post it.
 — MaxwellBuchanan, Sep 19 2013

No problem here on violating the Second Law of Thermodynamics. Any US president can use the power of "Executive Order" to override the Congress, Senate and the laws of physics if need be. The perpetual motion community has been lobbying the current US president to strike down the first law of thermodynamics as well. So far they have had no success due to the president's ties to big oil and big oil's resistance to any physics law changes that would bankrupt the oil industry. The coal industry opposes any changes in the laws of thermodynamics as well, but they have no pull in a government who wishes to bankrupt them already through legal means.
 — Sunstone, Sep 20 2013

 [WcW] have you actually watched either of the Feynman/"Fenniman" lectures you've posted as links? The second one, like the first, is wonderful but doesn't touch on the second law of thermodynamics. It's a continuation of QED and, in particular, the way it relates to classical optics.

If I have been dumb and have missed a passing remark on the subject of the second law, please feel free to correct me - and tell me where in the hour-long lecture it is.
 — MaxwellBuchanan, Sep 20 2013

 Maxwell, this isn't a question of thermodynamics. It has nothing to do with the second law in any other way that that photons do no violate thermodynamics. Though there is a strong desire within this group to see heat and energy as material that can be collected in an unlimited fashion, the reality is that photons are not effectively absorbed unless they can move an electron from one specific energy state to another, which has a specific impact on the excitation of the nucleus. If the photon does not interact in this way then it cannot be absorbed, and due to a symmetrical nature to the electron/photon relationship the tendency to absorb photons and thus arrive at that excited state is perfectly mirrored by the ability of a nucleus at that state to produce identical photons and thus return to a less excited state. As the material approaches the energy of the radiation the more and more perfect this symmetry of radiation is, every photon is driving that nucleus to vibrate in close phase with it's electrons, and due to the fact that every photon is coming at a different phase, the ability to dissipate that energy comes in the form of an increasing level of identical emissions, and due to the fact that there is a perfect symmetry function between absorption and emission; the number of times that the electrons will rise in energy and fall in energy is apparently limitless. You could say that when you reach the "temperature" of the color of the light, the nucleus no longer participates in absorption, simply letting the electrons do all of the dancing.

 Then we start to look for exceptions, and there are many, especially with the confusing behavior of electrons and dipoles to move rather than absorb the electron wave form in an isolated fashion. The most fascinating observation is that photons may force dipoles to switch their polarity, which means that rather than rising and falling in orbit, electrons move displacing the nucleus. The electrons are not displaced to positions where they can cycle back down to produce a photon and thus do not cause the electrons to re-emit in a reciprocal fashion. In this case there is no symmetry, the material cannot re-emit microwaves because dipole switching does not produce photons.

 Lasing also presents a special challenge, since lased light is in phase with itself it monopolizes the action of the electrons. Rather than moving in every direction, the electrons must now move in one direction (polarity?) only. This means that while the nucleus is still excited in all directions, it may only dissipate energy in one phase, which means that the excitation of the nucleus in the other phases becomes increasingly unlimited. If all of the electrons are forced into the energy level and phase of the lased photons, they are not available to the nucleus to dissipate energy in the other angles. In this way the nucleus may become very excited, but be denied the ability to symmetrically dissipate that energy.

So, there you have it, Thermal radiation in a nutshell. If it's incoherent light, this is how it interacts with electrons. We know that this is true by observation and the exceptions prove, rather than disprove this conclusion.
 — WcW, Sep 20 2013

 OK, but why pick two random Feynman lectures on QED?

 Also, I wasn't aware of the nucleus's involvement in all this, but that could be my ignorance.

 So, two plausible arguments, equally strongly held by their respective advocates:

 (a) The colour of light corresponds to a temperature; no amount of light can heat a target above that temperature. When the target is at that temperature, it can no longer absorb light of that colour.

 (b) Light is just energy; if you increase the intensity, the target will get hotter until it starts to radiate as much power as it's receiving. If you increase the intensity again, the target will get hotter still.

 and each of these two answers provides a different way for the second law of thermodynamics to be adhered to, which was the original question in all this.

 Now, I know that this is not a difficult physics question, and I know that it's like physics 101 (or at most 102). So it's frustrating that there are two answers.

 Incidentally, for the record, I'm happy to be convinced of either of the two arguments. I'm also sure that the second law of thermodynamics is right. I'm just trying to understand how, in the special case of lenses and mirrors focussing sunlight, the mechanisms of nature preserve the second law.

(To take an analogy: I'm happy that perpetual motion is impossible; but if I come across something that claims to produce perpetual motion, I want to be able to see where and how it fails, like there are frictional losses here, or it takes energy to cut a magnetic flux line, or whatever.)
 — MaxwellBuchanan, Sep 20 2013

 The exceptions do not prove this, because you fail to suggest how the exceptions exclude the possibility of my version still being correct. You also lack observation that in any way proves that this is impossible.

 Also, you appear to be confusing fluorescence and raman photon emission with black body radiation. The former are discrete emissions that relate to bound electrons moving between orbitals. This is why they produce photons at very specific energy levels and produce the emission lines of a spectrograph. (Or in the case of raman scattering, discrete energy shifts rather than levels, but that's not really an absoprtion).

 That latter is a wide band emission that relates to the excitation of free electrons and does not have the same discrete energy levels. This is why black body radiation is a frequency distribution, not a single frequency.

 Since these free electrons, by way of their interactions with nuclei, add kinetic energy to the atoms in their material, they heat the object in question without being concerned about the specific frequencies involved.

Again, please review my earlier statements and tell me which one, specifically you disagree with. Very specifically, I would like to hear your explanation of how your approach gets away with allowing a violation of the law of conservation of energy. (Specifically, if more radiometric energy is added to the system, what happens to it such that the target object doesn't gain any energy).
 — MechE, Sep 20 2013

 The Feynman lectures explain how it is possible for a the polarity, the coherence or incoherence of the light to interfere with the process of photon emission that allows the nucleus to reach symmetry when bombarded with radiation from an incoherent source, while gaining energy in a rather unlimited fashion from excitation from an coherent source.

 There is only one way to produce a photon, by the motion of an electron from a higher energy position into a lower energy position. The mechanisms that cause this action are diverse; in this case we are speaking only of causes originating from the nucleus or caused by the action of a photon on that nucleus and complimenting electrons.

The photons, while they could impart incredible energy to the material they bombard do not simply because all materials that can efficiently receive radiation can also efficiently produce it. The more of that radiation you bombard with, the more closely the nucleus is to the frequency which causes it to emit that specific radiation the more efficiently it does so. If this were not the case, the very heavily radiated sources, such as those near stars would emit large quantities of exotic high frequency high energy particles (the violet catastrophe theory) the simple fact that they are not, even though they experience insane intensities of radiation.
 — WcW, Sep 20 2013

 Your physics gets worse as time goes on. Your comments about particles near stars are irrelevant, because all visible parts if a star approximate black bodies, and as a result are subject to the limitations of black bodies, to which we have already agreed, although we remain divided on the mechanism.

However, You have again failed to answer a very simple question. With non black body light sources, it is possible to concentrate more energy on a target at a given wavelength than can be achieved from a black body source. Given that, in your conception, where does the energy go?
 — MechE, Sep 21 2013

 //So, two plausible arguments, equally strongly held by their respective advocates:

 (a) The colour of light corresponds to a temperature; no amount of light can heat a target above that temperature. When the target is at that temperature, it can no longer absorb light of that colour.

 (b) Light is just energy; if you increase the intensity, the target will get hotter until it starts to radiate as much power as it's receiving. If you increase the intensity again, the target will get hotter still.//

 [MaxwellBuchanan] I hope you have noticed that exactly one halfbaker is proposing argument (a), while several others are in general agreement that (b) is correct (at least when discussing large-scale interactions involving near black bodies), and that (a), and nearly everything said halfbaker has said in support of (a), is a load of old dingo's kidneys.

Perhaps it's time to stop feeding the troll.
 — spidermother, Sep 21 2013

 As an original proponent of "a)" a few posts ago, it was a wild but neat sounding guess which easily disposes of the problem "why can't you make something hotter than the sun using solar rays ?"

 Due to the blinding obvious disexample of a microwave oven, I think it could be rephrased as

 "A body under the influence of only black body radiation will not be able to increase its physical temperature past the black body temperature of the radiant source".

That still seems to make sense in some manner.
 — FlyingToaster, Sep 21 2013

 //I hope you have noticed that exactly one halfbaker is proposing argument (a), while several others are in general agreement that (b) is correct//

 Yes, I had noticed. I also took the liberty of asking a physicist at work (who happens to be CTO at a photonics company), who agreed with (b), but he had to think about it.

 There is a certain attractiveness (to a non- physicist) about (a), in as much as when a body reaches a certain temperature, it is as likely to give out a (say) red photon as it is to receive energy from a (say) red photon. It's not immediately and obviously wrong.

 So, as a non-physicist, there's no killer argument that makes me say "ah, of course.", and I have no objective evidence to say that person X knows more about it than person Y, and I was hoping for something clearer than a majority decision.

I'm not trying to feed trolls (and I am sure [WcW] is not trolling, but holds his belief strongly and is arguing for it) - I'm just trying to understand and hoping that someone can present an irrefutable argument which is simple enough for me to follow.
 — MaxwellBuchanan, Sep 21 2013

 Well, if we're talking about acoherent stationary black bodies then still sounds true but that's due to geometry: you simply can't focus more energy onto a target than a radiator(s) emit per unit of surface area unless you have angular coherency from the radiator.

 A sphere radiator that rapidly closes in on the target ball inside of it would circumvent that due to doppler shift. Sort of. Blueshifting the radiator means that its apparent temperature rises on one side and cools (redshifts) on the other.

This post would circumvent it also. If the radiator was active like a sun then having it's outward light reflected back on it would raise its temperature, thus potentially raising the temperature of the ball inside the sphere.
 — FlyingToaster, Sep 21 2013

 // still sounds true but that's due to geometry://

 Yes, but let me make it clear - I accept that you can't focus (all the) light onto a smaller area than it was emitted from. But the question - the (a) vs (b) - has been reduced to a simpler one:

Will an unlimited amount of incoherent light, of a given colour, hitting a target, raise that target's temperature arbitrarily high (answer b); or can it only raise it to a temperature limited by the colour of the light (answer a)?
 — MaxwellBuchanan, Sep 21 2013

 well, concerns BB radiation which I think is produced by free electrons running around body slamming each other, whereas the emission/absorption spectra of a substance concerns shells and nuclei and molecular bonds and more macro constructs.

 How the hell a radio is tuneable is beyond me. [edit: oh, yeah, duh]

Anyways, BB radiation isn't a resizeable distribution curve, it's fixed: temperature <x> always equals energy per area <y> at frequency curve <z>. A blackbody microwave oven would only be useful in warming stuff up a few degrees from absolute zero.
 — FlyingToaster, Sep 21 2013

 //BB radiation isn't a resizeable distribution curve//

Yes, but suppose (for the sake of argument) that all of the light coming from 1000 black body sources could be collected and focussed on one black body target (equal in size to each of the sources), would the target's temperature exceed that of any of the sources?
 — MaxwellBuchanan, Sep 21 2013

 well, with magic optics, perhaps warped spacetime, I assume that a bb target would end up radiating a bbr temperature corresponding to all the energy hitting it, since the definition is "absorbs all the radiation that hits it".

Likewise a bb target which is receiving bb radiation attenuated by a factor of 1,000 would radiate at a lower bbr temperature corresponding to 1/1000 the energy.
 — FlyingToaster, Sep 21 2013

 //temperature always equals energy per area at frequency curve //

 Indeed (although it should be power per unit area). If we know or or we can derive the other two. That is why there is a perfect correlation between frequency distribution and temperature in the special case of black bodies at thermal equilibrium.

 But neither frequency nor frequency distribution is the *cause* of the temperature reached by the target.

 If you shine an incandescent flashlight at a distant wall, the wall never becomes white hot - you can change the temperature of the target without changing the frequency distribution of the light.

 Many real-world objects (white paint, metals) have emissivity spectra wildly different from black bodies, but are nonetheless able to reach thermal equilibrium via radiation - you can change the frequency distribution of the light without changing the temperature of the target.

Finally (to the best of my knowledge), you can simulate a given emission spectrum - black-body or otherwise - at an intensity greater than that capable of being emitted thermally, and heat a target to a temperature exceeding that achievable with thermal radiation - you can heat an object to temperatures exceeding the so-called colour temperature of the light. For example, a large bank of de-tuned lasers collectively simulating the frequency distribution of sunlight could heat a target hotter than the surface of the sun (even if coherency were eliminated).
 — spidermother, Sep 21 2013

 //Yes, but suppose (for the sake of argument) that all of the light coming from 1000 black body sources could be collected and focussed on one black body target (equal in size to each of the sources), would the target's temperature exceed that of any of the sources?//

Yes, it would. (But of course, it can't be done without using magic or consuming additional energy or information.)
 — spidermother, Sep 21 2013

 and as long as we're on the subject of thermal radiation...

If you magically float a piece of glass with a high transmittance in the infrared, inside a (perfectly evacuated) Bell jar inside a classroom... will that piece of glass, which isn't being affected by conduction or convection, equilibriate at room temperature ?
 — FlyingToaster, Sep 21 2013

 I think that question has too many uncertainties for a concise answer. Instead:

 Q. Imagine a room at thermal equilibrium, at temperature x. We place inside that room an evacuated bell-jar, containing a levitating piece of special glass, at temperature y, perfectly transparent* to all frequencies emitted at temperatures from x to y. Will the special glass reach thermal equilibrium with the rest of the room?

 A. No.

(It needs to be perfectly transparent (or perfectly reflective, or some combination - that is, have zero emissivity) to all frequencies present, right down to the longest radio waves. Otherwise, it will reach equilibrium, albeit slowly.)
 — spidermother, Sep 21 2013

If all the electrons that would be able to make a particular photon admission are already in the excited state then and a new photon arrives at that energy the photon will not be absorbed. This means that a nucleus becomes saturated for that frequency of light. Beyond this level of interactions it cannot absorb more photons. If the nucleus is at a temperature to cause it's own electrons to rise to that level naturally then it is naturally already saturated. With no ability to move electrons the photon cannot impart energy to the nucleus. For any given frequency of light there is an intensity that will saturate a given element. This also represents a temperature. It's not trolling to point out this simple principal which is fundamental to optics is also critical to maintain the balance of radiation that we experience. That photons interact with electrons is not a matter of dispute. I am simply adding the observation that when an nucleus becomes hot enough it will naturally move it's own electrons into the same excited states that a photon would, and thus become a better emitter than receiver. This simple observation is the reciprocal of the quantized nature of light.
 — WcW, Sep 21 2013

//This means that a nucleus becomes saturated for that frequency of light.// Isn't it the electrons (or the atom as a whole) that would become saturated? Surely the nucleus is a red goose, rather like discussing the Earth's in the context of weather - there'd be no weather if there weren't an Earth to hold the atmosphere down, but the Earth itself doesn't do anything very weathery.
 — MaxwellBuchanan, Sep 21 2013

It's the only thing I know how to do.
 — MaxwellBuchanan, Sep 21 2013

I think that that is a fascinating question. Clearly the nucleus is the place where the property which we refer to as "temperature" is present. This motion or excitation of the massive particles occurs independent of the electrons, and while it affects them, it occurs even when they are not present, or when their number changes. You could imagine that the nucleus "pushes" the electrons out as it becomes hotter but this is not literally true. Figuratively it is true. A very hot atom has forced all of its low lying electrons into higher energy states and they are no longer rising and falling at those energy levels. At this temperature you could say that a low energy level photon will not be able to raise any electrons to higher energy states because none are available. So yes, the action of the nucleus in changing the states of the electrons is critical to this process, and that in effect it is the energy present in the nucleus limits the electrons available for interactions that absorb energy. There are other potential interactions that a atom might have with that light (Feynman Lectures) that do not involve admission but admission is the mechanism by which photons "heat".
 — WcW, Sep 21 2013

I get the impression we're drifting away from the mainstream a little here.
 — MaxwellBuchanan, Sep 21 2013

not at all actually. I admit that I am more than a little surprised by the general lack of enthusiasm for this topic here. The question of how a photon becomes heat seems a crucial one, and yet, "it just happens because photons are energy" isn't an adequate tool.
 — WcW, Sep 21 2013

 soooo... we're back to "water off a duck's back" again ?

'Slike lookin' at one of those pitchers, which pixels are made from smaller pitchers, which pixels...
 — FlyingToaster, Sep 22 2013

 [MaxwellBuchanan] //Will an unlimited amount of incoherent light, of a given colour, hitting a target, raise that target's temperature arbitrarily high (answer b); or can it only raise it to a temperature limited by the colour of the light (answer a)?//

 (b) for example, suppose you wanted to raise the temperature of an object to be almost as hot as the sun using only infrared light, you could covered it with a reflective filter that reflects visible light but transmits infrared, and then use mirrors so all that you could see (if you could see in infrared) from its location is the sun (and by symmetry, all the infrared radiation it emits is reflected into the sun). Then it would be absorbing more infrared than it emitted until it's about the temperature of the sun. The same filter that prevented it from absorbing visible light would reflect back any visible light it emitted,

 I think of the light being as intense/dense because it comes from a given temperature source. This density is includes: the amount of light of a given frequency range, in a given size area, over a given range of angles.

Light from a clear daytime sky is a mix of temperatures - mostly hot in the direction of the sun, mostly cooler in other directions, (though not so much cooler in the blue light frequency range)
 — caspian, Sep 22 2013

WcW- You are, still, confusing fluorescence with thermal radiation. Until you understand the difference, there is no point in continuing this discussion.
 — MechE, Sep 22 2013

 What a perfectly sensible thing to say. Of course I am confusing those things, and of course I am speaking to you.

 Essentially we are having the same debate that was happening around 1917 when people were grappling with photon model for light.

 So maybe you can follow my logic, and in reflection, the logic that some people much smarter than me were using at the time.

 1) light is observed to be absorbed and emitted from materials in discrete bands.

 2) these bands are found to be the the transitions between electron orbital positions

 3) If there is no available electron for the photon to excite, admission will not occur.

 4) admission is the only way for a photon to directly excite a nucleus, that is, to heat it.

 5) as a nucleus grows hotter we observe that it emits radiation of higher frequencies and stops emitting radiation at lower frequencies. By inference we assume that this means that electrons are no longer present in lower energy positions; no longer cycling between lower energy levels.

 6) because we have learned that there is a reciprocal relationship between electron orbit movement and the ability of an atom to absorb or emit photons of the same frequency we know that an excited atom that is not emitting radiation at a specific frequency will not admit the same radiation.

 7) we deduct that an atom that is exposed to a specific frequency of radiation heat until all of the electrons that respond to that frequency are no longer available.

 8) We get the 1921 Nobel Prize for physics.

9) We begin working on why there appear to be exceptions to these very simple observations, including lasers, microwaves, etc.
 — WcW, Sep 23 2013

 //1) light is observed to be absorbed and emitted from materials in discrete bands.//

 Not exclusively true. It is true for isolated atoms (fluorescence), but once you get to molecules, the light is absorbed and emitted elsewhere in the system. Free electrons. Latice Vibrations. Etc. All of those little kinetic actions that collectively make up heat. That is why Wien's Law produces a broad curve rather than discrete lines, a frequency distribution, not a single frequency.

 //5) as a nucleus grows hotter we observe that it emits radiation of higher frequencies and stops emitting radiation at lower frequencies. By inference we assume that this means that electrons are no longer present in lower energy positions; no longer cycling between lower energy levels.//

Also not true. The atom continues to emit radiation at the lower energy levels, it's just that the peak shifts to the higher energy levels. The total energy emitted at the lower levels still continues to rise.
 — MechE, Sep 23 2013

 There are a lot of exceptions. The photo receptivity of chlorophyll has a higher than expected admission of light due to a quantum mechanical feature we are only beginning to understand. But this principal, this essential action of heat and electrons is the basic standard.

The exceptions are interesting but they do not apply to the basic question which is about collecting heat from photons. Essentially speaking nuclear absorption, via electron excitement is the mechanism and this mechanism is clearly self limiting. That there might be exceptions where photons are accepted, but not absorbed and not transformed into heat it is interesting but not applicable to the question at hand.
 — WcW, Sep 23 2013

Please look up "phonons". There are other mechanisms at work as part of the routine, not as exceptions.
 — MechE, Sep 23 2013

we may infer that the greater the rate of admission the greater the time that the electrons are spending in the higher energy position, the less they are available for excitation in the lower range, essentially speaking the lower frequency is never completely saturated, there is always a chance that a photon might sneak in an excitation between emissions but that chance is diminishing with the square of the temperature. I'm not suggesting that increasing the intensity doesn't have an impact, i'm asserting that at infinite intensity you will not exceed a particular temperature. At this point the electrons are cycling at their frequency (in the pictoseconds) with no time in the rest state at all.
 — WcW, Sep 23 2013

 Simply put, you still have not answered where the energy goes, so you claim it cannot be absorbed.

 You also have not answered why Planck's Law (more accurate than Wein's) produces a continuous wide distribution of wavelengths, whereas your model should seem to produce a distribution of spectral lines.

 — MechE, Sep 23 2013

possibly you can illuminate us all with the other mechanisms at work. Please explain another mechanism whereby a photon may impart energy to a material outside of the already well established dielectric effect, and the special interactions of coherent light.
 — WcW, Sep 23 2013

The energy is not absorbed, it remains in the photon, which must go elsewhere. It is observed that elements do have specific lines of emission and admission. Molecules have different lines, materials even more. The hypothetical "black body" has such a wide range of bonds and electron configurations that a wide range of frequencies of light quickly find a home on the immediate surface. But even in this case, even when you allow for a material with perfect absorption, it is recognized that that body will act as similarly perfect emitter, that is, that the self same electrons will emit the same frequencies of light when excited.
 — WcW, Sep 23 2013

 I'll freely admit I am not perfectly clear on the entire thing, myself, but the major issue appears to be that dipole heating (as is found in microwaves) is present in pretty much all cases. This results in kinetic transfer to the atoms within a material, without the mediation of electron band excitation.

I have also mentioned the excitation of free electrons in a medium, which then transfer some of their energy to atoms in the material through electromagnetic interaction.
 — MechE, Sep 23 2013

So you are saying that black bodies get more reflective as they get hotter. This is not consistent with theory.
 — MechE, Sep 23 2013

You could say that a hot black body becomes reflective to photons that strike it, but the interaction is not that simple. it is simpler to say that such a body no longer absorbs energy from those photon interactions.
 — WcW, Sep 23 2013

If a body does not absorb a photon, it either reflects it or transmits it. There is no other possible interaction. Since I am choosing to believe you are not claiming the material becomes transparent, reflection is the only option.
 — MechE, Sep 23 2013

This is where the feynman part comes in. In this interaction where the photon does not loose energy it can change properties.
 — WcW, Sep 23 2013

 — MechE, Sep 23 2013

refraction is the more likely case when we regard what happens with a perfect black body, when it is hot it becomes a perfect refractor.
 — WcW, Sep 23 2013

So you ARE claiming the object becomes transparent. Since refraction only occurs when the photon transitions between two transparent media.
 — MechE, Sep 23 2013

I read your link. Yes, we are allowing for a perfect black body, one that can absorb AND emit radiation at a very wide range of energies. how this material achieves this property is not in debate, just that this process is reversible, that the electrons that perform it do so with perfect reversibility.
 — WcW, Sep 23 2013

diffraction is the more correct word. my apologies.
 — WcW, Sep 23 2013

 Except that real, pure materials can approximate black bodies in terms of their Planck's law spectrum, without regard to their specific electron configurations.

So it's not some special property of a theoretical black body, it's inherent in all materials.
 — MechE, Sep 23 2013

Again, diffraction occurs when a wave passes through gaps in an object. So now your target is becoming porous (a lattice?) at high temperatures.
 — MechE, Sep 23 2013

net effect is that the photons would be radiated from the surface as if the material was emitting them in the usual chaotic fashion.
 — WcW, Sep 23 2013

along with a compliment of lower and higher energy radiation distributed above and below the frequency we are adding.
 — WcW, Sep 23 2013

And again, I'm reduced to "no, just no". You just invented a mechanism in an attempt to justify your approach.
 — MechE, Sep 23 2013

 I have shown the mechanism whereby materials absorb energy other than at their fluorescent excitation energies. I have shown that materials produce light outside of their fluorescent emission spectra. Your theory allows for neither.

 Your theory also does not allow for exceptions for coherent light, because as far as the electrons in their orbitals are concerned, one light source is identical to another. Coherent light does not somehow have more energy than incoherent light at a given radiometric power and identical wavelength, such that it could more readily or severely excite electrons. In fact, since coherent light is mono-chromatic, it doesn't even have the advantage of the few higher energy photons that black body radiation does.

I don't know what else I can say, but that you are, simply, guilty of a basic misunderstanding of the mechanisms involved, and wrong in your extrapolation therefrom.
 — MechE, Sep 23 2013

 Possibly so. I'm not sure precisely why that is so upsetting to you personally but I am very interested in this process and I am not deterred from exploring it simply because it seems to frustrate you. I am looking for examples that contradict my postulate, an example of a material being heated to a temperature higher than the temperature of the highest radiation that they are exposed by simple admission of incoherent light. Your postulation is that this is common, and that no special mechanism is required to achieve it. If you feel so passionately, please just show me an example so that I may also be informed. Microwaves, dipoles, and lasers are out.

p.s. your link to the 'Scientific American' Q&A seems to support my view.
 — WcW, Sep 23 2013

 No the SA link describes direct transfer of photon energy into molecules, not through electron intermediaries.

 And I do not postulate that such heating is common, just that it is possible. It would not be common because sources of non-black body radiation are a relatively new phenomenon (from a human point of view). And, in the temperature range humans use, such heating would generally not be as efficient as simply raising the temperature of your black body source.

 Also, given that, to the best of my understanding, and based on my second link, dipole heating is the primary mechanism for radiation heat transfer to atoms/molecules as opposed to electrons, any attempt to exclude it is going to prevent me from making my case. If this is, in fact, the case, then the microwave oven is a clear example, which you are unfairly excluding.

Therefore let me, turn that back on you. What makes microwave radiation (which we know heats primarily by dipole excitation) different from all other forms of radiation? I say that nothing does, except that the energy level of microwave radiation is so low that no fluorescent excitation occurs, and, as such, it heats only through dipole excitation. Once the energy level comes up into the IR, a mix of the two starts coming into play.
 — MechE, Sep 23 2013

mech, my man, we're talking about the sun. You believe that a prominent concept in physics that has massive implications for the way that stars work can't be demonstrated?
 — WcW, Sep 23 2013

 I'm muddling through, not working against you MechE. You have to agree that this is very important, nay, critical factor in the balance of radiation that would populate the universe. It's not a trivial matter and I am not cranking you here. A very few times while exploring in this discussion I felt like I was confident that I had misunderstood the science, that I didn't comprehend it, but the shoe never dropped, I never found the Aha! moment that would prove to me that material could be superheated by conventional radiation that did not induce a di-electric effect, and that wasn't otherwise coherent or current inducing.

If I am correct then no light source of any intensity can heat beyond its temperature without resorting to other phenomena, and if I am wrong it should be easy to focus light to a higher temperature than the source.
 — WcW, Sep 23 2013

 Again, you are misunderstanding what I have said. I did not say it can't be demonstrated, I said is not routinely demonstrated. And even that may be incorrect if I am right in saying that you have been unfair to exclude microwaves and dielectric heating, as I my current understanding is that dielectric heating is a mechanism that is found in all EM/material interactions.

 I also believe you have been unfair to exclude coherent light, since it is functionally no different than any other source.

 The problem is that I am not a specialist in this, I do not have access to a fully equipped lab to provide a demonstration. Nor can I provide (or in many cases truly comprehend) the math that would be required to prove what I am saying conclusively. I have provided every bit of evidence I can scrape up to prove that this is, in fact, the case, and I know it, both instinctively and through my understanding of basic physics to be the case, but I cannot point to a specific example other than those I have provided, simply because I am uncertain as to what has been done.

 Therefore, the fact that you have declined or been slow to answer some critical questions is, in fact frustrating. The biggest one outstanding is the question of why black bodies produce a continuous spectrum, instead of the discrete emission lines that a purely fluorescent behavior would produce.

 I also question where the statement that this would have a critical affect "balance of radiation that would populate the universe" comes from, because that is not really under discussion. Almost all radiation in the universe is black body or unconcentrated fluorescent in nature, neither of which is relevant to this discussion.

Electroluminescence and similar affects are not found to any significant extent in nature, and they are required to develop a case where the total radiative energy is greater than the equivalent black body would produce. Therefore it is only when LEDs or other non-thermal lighting are developed that this discussion begins to have real world relevance.
 — MechE, Sep 24 2013

 black bodies produce and absorb diverse (albeit imperfectly) wavelengths of light because they contain a diversity of electron positions and relationships. One example would be carbon black, which contains thousands of distinct bond and shell electron configurations, bond and angles and combinations of the two. As a result, very little of the light that hits it does not find a applicable home. Why do I have to explain this?

 Finally, we are talking about black body radiation, or "thermal radiation" as it is also frequently called.

 If heat concentration is possible, it should be possible on even a small scale, and if it isn't possible, then it should also be possible to demonstrate that on a small scale.

For instance it should be possible to focus a source of incandescent light on a sufficiently small space to demonstrate that something can be heated to hotter than the color of that light.
 — WcW, Sep 24 2013

 Please revisit the earlier part of this discussion where I, and others, state that it is not possible to achieve this with a thermal light source (such as an incandescent bulb), because it is not geometrically possible to concentrate the light from your source onto a target smaller than your source.

It only applies when you have a non-thermal light source, such that the total energy at a given wavelength can be greater than the energy that would be available from a black body source at that wavelength.
 — MechE, Sep 24 2013

 Well, [MechE]'s and [spidermother]'s argument makes more sense to me than [WcW]'s, but aesthetically I prefer [WcW]'s.

I accept, but don't like, the fact that the second law of thermodynamics is only upheld by the geometrical constraints on focussing and concentrating light. I'd be happier if there were some fundamental property of light which, like hot water, couldn't heat an object above its own temperature.
 — MaxwellBuchanan, Sep 24 2013

But hot water could heat something above its own temperature, if you first concentrated the hotter molecules in one area...
 — Ling, Sep 24 2013

Realistically, this is more like expecting a limitation on electricity or kinetic energy being able to produce heat. Radiometric energy is not heat.
 — MechE, Sep 24 2013

I was pointing out the analogue between concentrating light too much as being somehow banned by Thermodynamics, and concentrating hot water molecules also seemingly banned by Thermodynamics.
 — Ling, Sep 24 2013

That's sophistry. I don't limit you to one source, have as many as you like. One, ten, a thousand and focus them on one spot. This isn't an optics problem,its a property of light problem. Its lazy to say that somehow intensity is heat, but that it cannot be demonstrated as such no mater the wattage at our disposal.
 — WcW, Sep 24 2013

 It doesn't matter how many sources you have, each source occupies a limited portion of the apparent "sky" of your target. No matter what you do, the best you can achieve is the entire "sky" occupied by your source(s), and that becomes my earlier comment on a dyson ellipsoid. You can do this with any target size, but size of the sky (area focused on the target) is proportional.

 As far as demonstrating it, the critical item is not the wattage, it's the light source. The energy of black body radiation is defined exclusively by the source's temperature. Since this also defines the "color temperature" of the light, the two are correlated.

 The energy of electro-luminescence (such as LEDs) is not defined by it's temperature Therefore it can generate more energy, and the color temperature is not correlated with it. Therefore a demonstration is possible with LEDs. If it happened that my work used IR rather than Blue LEDs, I could probably rig up that demonstration for you. Since it doesn't, and I don't want to know what a 10,000 K spot looks like, I'm not going to try.

This is not sophistry, it is the fundamental limiting factor on why it is impossible to violate the second law of thermodynamics with an optical approach. Apparently you have never actually understood the position you are arguing against.
 — MechE, Sep 24 2013

It's ironic that a discussion on thermodynamics becomes heated.
 — MaxwellBuchanan, Sep 24 2013

^ Generating more and more heat while simultaneously becoming more and more opaque (at least to the lesser lights like myself). "Myself" - also known as the "Pasty White Body" problem.
 — AusCan531, Sep 24 2013

 Wow, chill with the ad hominem. Are you really saying that intensity doesn't matter? Can we just put non thermal light aside?

If we double the intensity, what happens to the temperature? What happens if we multiply it by 10000? Where is the point where thermal radiation does what you say it can do and actually heats to a higher temperature? Also, how are you not my case now?
 — WcW, Sep 24 2013

 //If we double the intensity, what happens to the temperature? What happens if we multiply it by 10000?//

The temperature goes up. I think almost everybody here was saying that it goes apart from you.
 — MaxwellBuchanan, Sep 24 2013

 What ad hominem attack? I stated a fact. You made it clear by your arguments that you didn't understand ours. I did not proceed to attack your position as an outgrowth of that.

 And, also, if you read my statement in context, it does not say that wattage doesn't matter, just that it's not the critical factor.

 From your question it appears that you are asking what happens if you increase the amount of thermal radiation hitting the target. If it were possible to do so, the target temperature would increase. However, if you have a thermal source, the only way to increase the amount of radiation is by increasing it's temperature. There is no other way, so the temperature of the target never exceeds the source.

 As far as why I am not arguing your case, it's because you claimed that there was no condition under which light (later modified to non-coherent light) can result in a higher temperature than it's apparent color temperature. However, your argument would mean that increasing the intensity, however it might be done, would have no effect.

 Our argument is that increasing the intensity would produce an increase in temperature, regardless of the color temperature, it's just that such an increase in intensity is not possible for black body radiation. That is why non-thermal light sources are required.

P.S. This is now the idea where my screen name appears the largest number of times (if you don't group the three ATTR ideas together).
 — MechE, Sep 24 2013

look there are ways to increase intensity without increasing temperature. turning on a second identical filament for example. Just explain to me what happens to the temperature of a small spot when we focus multiple beams of incandescent light on it, do we ever go above the temperature of the light? this seems like a good first step in grappling with this problem. If I give you an infinitely intense flashlight bulb, can you heat an object above the temperature of that light with it or not. Perfect everything. Yes, or no.
 — WcW, Sep 25 2013

 As I have said multiple times, if such a thing as an infinitely intense incandescent flashlight could exist, then yes. But turning on a second filament, or any of the other tricks you seem to think could work, won't. Period.

In order for them to work, you would be required to focus the light from multiple filaments onto a single filament, and you can't do that.
 — MechE, Sep 25 2013

 Hmm, why would we not focus the numerous filaments on our hypothetical black body? Also, what prevents us from focusing the filaments upon one another? What happens when you focus the light from twin filaments on one another?

Also I would like for you to re-read that last statement and tell me if you would stand behind it. It's a strong assertion, and one that I seem to have no problem violating with my limited resources of imagination, not to mention flashlights.
 — WcW, Sep 25 2013

Can you explain the scenario where it is possible to do something with thermal radiation but somehow impossible to replicate it with any human means at any scale? So impossible to replicate that you can't propose a scenario where it could be done. That doesn't sound like possible, it sounds like impossible, and I'm not buying this notion that all that holds us back from demonstrating a universally applicable concept is bad optics and not enough D cell batteries.
 — WcW, Sep 25 2013

 And I repeat, again, it is not possible to do this with thermal radiation. I agreed to a hypothetical you proposed, but stated that said hypothetical is impossible. How many times do I have to say that?

 In order for one active filament to raise another, identical and passive, filament to it's own temperature the entire energy output of the first filament must be directed to the second filament. This can, theoretically, be achieved with an elliptical (perfect) mirror, with the filaments at the foci.

 If you add a second active filament, it will not be at the focus of the mirror, so it's radiation will not be directed at the passive filament. No matter what you do with this system, you can't change that.

 If you double the size of your active filament, only a portion of it's energy will focus on the passive filament, producing the same effect.

If you make your passive target larger, you can focus the light from both active filaments on it, but only in proportion to the size increase.
 — MechE, Sep 25 2013

 As far as focusing two active filaments on each other, of course the temperature of both will increase. But that is not heating a target element hotter than the source element. The two will increase in perfect lockstep, and in no way violates anything I said.

Heck, you would get the same result under your hypothesis, because the exchanged light would bring the element up to temperature, and then the additional electrical heating would produce an increase in temperature.
 — MechE, Sep 25 2013

 dude, you have my permission to use more than one mirror and more than one filament. use as many as you need. I'm glad we are on the same page.

 I'm going to go ahead and assume that you agree with me that if had a large parabolic mirror here on earth and we focused the light of the sun on a small spot that small spot would not exceed the temperature of the sunlight.

 What if we towed that mirror in space, in the direction of the sun. As we grew closer and closer would the temperature ever exceed the temperature of the light? What if we towed it all the way to the surface of the sun and aimed it at a spot in the admittedly diffuse gas cloud, would that spot grow hotter?

When do we ever get to push our black body to a temperature higher than that of the light it is exposed to? What level of intensity will achieve the affect you assert can happen?
 — WcW, Sep 25 2013

 Your permission doesn't mean anything. It's the universe's that I would need. You can't just add more mirrors or more filaments. This is the fundamental point of my argument, and until you understand that, this discussion isn't doing any good.

 What's important here is that the sun (or any black body source) is not a point source. As you move your focusing mirror closer to the sun, the size of the focal spot your mirror can produce gets larger (because the diameter of the sun, in steradians, increases).

 The percentage of the sun's energy your mirror intercepts goes up, but the percentage of the intercepted light that is focused on your target goes down. The net result is completely neutral.

I will say this one more time. It is impossible to focus or direct the light from a black body source to achieve this effect. It only becomes possible when you have light from a non-black body source.
 — MechE, Sep 25 2013

 Then it must be due to some fundamental relationship between the photons and the atoms they interact with. The surface of the sun outputs some 63 million watts of energy per square meter. Plenty enough that if this was an optics problem all you need to do is get close enough and your object will start out pretty damn near the temperature we want to exceed.

 What I am hearing is that you agree with me that intensity, no matter what the level, will not elevate a black body to temperatures higher than the frequency of that light.

 And now all we need to do is figure out why not.

 And also why some kinds of light seem to violate this rule even though they are far less intense.

 And then why you think we disagree about something.

And then why my model, which uses electron dynamics is wrong.
 — WcW, Sep 25 2013

 No, I do not agree with you.

 A black body cannot be elevated above the source temperature (or color temperature, since they are correlated) by black body radiation.

 A black body can be elevated above the source "color temperature" with non-black body radiation.

 The reason it cannot be achieved with black body radiation is because light sources are not point sources, and as a result, there is an upper limit on how much you can focus the light with optics. That limit means it is not possible to focus the energy from the source more tightly than it is at the source.

 This is actually true regardless of the source of the radiation. But, for non-black body sources, the temperature, spectrum, and intensity are not correlated. Therefore it is possible to generate more intense light per unit area at a given spectrum from non-black body sources than it is from black body sources. Therefore, focusing this light will back to it's source strength will result in higher temperatures than the color temperature would indicate.

 Intensity is all that matters for temperature, it's just that you cannot generate higher intensities with a black body source.

This is an optical problem, pure and simple.
 — MechE, Sep 25 2013

 surely the completely spherical perfect mirror can overcome that objection, what happens to the energy there?

Also I disagree that aiming a mirror at the sun does much of anything to the temperature. It seems clear to me that no amount of additional radiation of the same temperature would do anything noticeable to the material there.
 — WcW, Sep 25 2013

 No it can't.

If the energy is completely captured, then the temperature of the source will go up as well.
 — MechE, Sep 25 2013

 That is why I discussed the ellipsoidal mirror, above. This allows you to direct 100% of the energy from your source to your target. It only works if the target is at least as big as the source. If it is smaller, some of the light will miss the target, and the net result will still produce an equal temperature.

 In practice, in this case, the source will also increase in temperature, until they reach a new equilibrium where the energy loss at the mirror is equal to the output of the source, but that's not real relevant here.

If a perfect mirror were possible, you would get infinite temperature, which tells us that such a mirror is not possible in the real world.
 — MechE, Sep 25 2013

Aiming a mirror at the sun does essentially nothing to the temperature because you are returning some vanishingly insignificant portion of it's energy to it. If you constructed a mirrored dyson sphere, it most definitely would heat up.
 — MechE, Sep 25 2013

whoa, whoa, who said anything about energy? the energy is inside the sphere, no doubt, but remains as radiation and never becomes higher in average energy value.
 — WcW, Sep 25 2013

good, now we have something. So if we had such a sphere, and it did not become hotter than the radiation source you would accept that your theory was incorrect, right?
 — WcW, Sep 25 2013

 No, the sphere wouldn't become hotter than the radiation source. The radiation source would become hotter.

Again, "my" theory includes the fact that you cannot achieve the affect under discussion with black body radiation.
 — MechE, Sep 25 2013

Radiation is a form of energy, and energy is conserved. Your source is pumping more energy into the system (from chemical, electrical or nuclear sources). It has to go somewhere.
 — MechE, Sep 25 2013

 Let me try to state how this works in clear, simple sentences:

 1) EM Radiation cannot be concentrated to a higher intensity than it's source.

 2) The intensity of a black body source is directly correlated with it's temperature.

 3) Therefore, black body radiation cannot be concentrated on a target such that the temperature of the target exceeds that of the source.

 4) The intensity of a non-black body source is not correlated with the temperature that would generate black body radiation of the same spectrum.

 5) Therefore it is possible to have a non-black body source that is more intense per unit area than would be produced by a black body producing the same spectrum.

6) This extra intensity can be used, with the same limits on focusing, to produce a higher temperature than would be found in a black body source producing the same spectrum.
 — MechE, Sep 25 2013

 These are not related to the fundamental principles, but explain my answers to some of your questions:

 7) A black body target at the same temperature as the black body source will radiate energy of the same intensity as the source.

 8) This energy will be returned, by way of the focusing optics, to the source.

9) This will cause both objects to heat up until such time as a new equilibrium is reached, either through increased radiation that is not captured by the optics, or by losses through the optics (heat off the back of a mirror, for instance).
 — MechE, Sep 25 2013

 can we agree that that equilibrium is never higher than the temperature of the source radiation, and that there must be a mechanism that explains this?

 In effect if we do not allow the source to heat, that the system contains more and more photons, more and more energy and no element of it is getting hotter than those photons.

Once the source and the sink reach the same temperature the system may contain more energy but that energy is not in the form of thermal energy, this is true no matter what the intensity of the source for thermal radiation.
 — WcW, Sep 25 2013

 You cannot prevent the source from heating. The object being heated returns energy (in the form of photons) to the source, which heats the source. There is no way to avoid this. The effect may be absolutely minuscule if your source is the sun and your target is a marble, but it still happens.

The only way to avoid this would be to cool the system, and this extracts energy, which means that there are not "more and more photons".
 — MechE, Sep 25 2013

who cares what happens to the source? We can ignore it, lets say that it is regulated to a specific intensity. You keep getting hung up on that, and it doesn't matter. More energy will go into the perfect dyson ball than comes out. It takes the energy in the sun a hundred thousand years to get from the core to the edge of the sun, in effect the sun "contains" one hundred thousand times its annual output, all in photons, all trapped inside it, and all without becoming any hotter than they are. Photons have a specific energy. That energy can only be absorbed by nuclei that are colder (on average) than they are unless some additional process occurs. And for black bodies it does not, so it isn't an optics problem, its a light frequency and electrons problem as a result, no condensation of light, not increase in intensity or optical apparatus or change in the intensity or number of sources, number of photons, will have any impact. It doesn't in the sun, it won't here on earth.
 — WcW, Sep 25 2013

 Actually the energy in the sun is generally not in photons. Lots of thermal energy being conducted and convected.

 And more energy will not go into a dyson sphere then comes out. The total energy off the back side of a dyson sphere is equal to the output of the source enclosed. It may be lower quality (more photons at longer wavelengths), but the amount is the same. The exception being if some of that energy is converted into stored electricity, chemical energy, or other non-thermal varieties.

 However, there's a reason I isolated those three propositions, because they are not directly relevant to the thesis, merely an effect of trying to demonstrate it.

 Anyway, I have made the statement of my thesis. You have made the statement of yours. I have stated a way they can be tested (with non-black body radiation of sufficient intensity and/or sufficiently low color temperature).

 I have pointed out two cases where they are routinely tested (cutting lasers and microwaves), and you have rejected them.

 I know for certain that I am right and you are wrong. You apparently believe the opposite.

At this point, I don't believe we have anything else to discuss.
 — MechE, Sep 25 2013

Valid. I really wasn't clear on what you were saying, and I still feel that it is paradoxical and not supported by observation. I am glad that we have both agreed that the limit that may be achieved in solar collection is the temperature of the apparent source.
 — WcW, Sep 25 2013

 Hang on a mo.

 Is that the faint whiff of a consensus that I see staggering toward me across the desert of confusion? Is it possible to see a whiff? And why do I ask so many rhetorical questions? Who knows?

 Annnnnnnyway. I *think* the consensus that's emerging (with various degrees of consensus), is as follows:

 (1) Geometric optics mean that I can't take all of the sun's light and focus it onto a target smaller than the sun.

 (2) More generally, I can't take all the light from any given area of the sun's surface, and focus it onto a target of a smaller area.

 (3) And further, if I have many identical suns, I can't capture light from each of them and combine it onto a single target, such that the target receives more light per unit area than any of the suns is giving out, per unit area.

 (4) The net result of (1) through (3) is that I cannot, in crude terms, "concentrate" the light from the sun or any other black-body source. The best I can do, in a perfect system, is to make the light hitting the target as dense as the light leaving the source(s).

 (5) Because of (4), I cannot heat a black-body target above the temperature of the black-body source. This is a consequence of geometry.

 (6) If some impossible lens/mirror system _did_ allow me to violate (1) through (3), so that I _could_ concentrate light to an unlimited extent, then I _would_ be able to heat the target above the temperature of the source; but no such impossible system can exist. And therefore the second law of thermodynamics holds.

Before moving on, who's an aye and who's a nae to the foregoing?
 — MaxwellBuchanan, Sep 25 2013

I'm the aye on every statement there.
 — MechE, Sep 25 2013

 1) optics says that you can focus light to any intensity that you desire given a source. The sun is a great source.

 2) that this increase in intensity simply means more photons per unit area

 3) that no matter how many photons hit an area, if the radiation is incoherent the body will never rise above the temperature of the radiation.

4) energy is conserved because radiation is not absorbed beyond this thermal point but is re-emitted limitlessly.
 — WcW, Sep 25 2013

geometry has nothing to do with it. Geometry does not limit the intensity of light that can be achieved.
 — WcW, Sep 25 2013

 [WcW] I want to believe you, because if I did, the second law would seem more fundamental to me.

 But I am beginning not to believe you, and it's your statement (1) that I don't believe.

 So, here's an experiment. Take a positive lens, and use it to focus the image of a table lamp onto a piece of paper (I'm actually doing that now - well, in between typing).

 On the one hand, the image is smaller than the actual table light. In my case, the lamp is about 20cm long, and the focussed image is about 1.5cm long. In area terms, that's about a 180 fold reduction. I can't make the image any smaller than that.

 On the other hand, the lens I'm using is only about 10cm in diameter, and is about 2m from the light. So I'm only focussing about 1/6400th of the light. So, the "density" of light hitting the paper is only (1/6400)*180 = 0.03 compared to the "density" of light coming from the lamp.

 I can imagine that, if I use a bigger lens, I can catch more of the lamp's light. But I can also half-see that, whatever I do, I'm not going to be able to capture _all_ the lamp's output and focus it onto a target smaller than the lamp itself.

I don't like this fact, but I now accept that it's true.
 — MaxwellBuchanan, Sep 25 2013

maybe not, why are you trying to solve this problem with optics? why not get a more intense source of the same light? for instance why not two, ten, or a thousand sources, all focused on the same round object from different angles. How do optical properties actually prevent you from delivering whatever number of photons you desire?
 — WcW, Sep 25 2013

 That's exactly what I thought to begin with - give me multiple table lamps, and lens focussing each of them onto the same area of paper.

 But then do the math(s).

 Imagine that, instead of one lens focussing the image on one lamp onto the paper, I instead have a large number of lenses arranged on the surface of a sphere, each focussing the light from its own lamp onto the surface of a central (spherical) target. For the lens I'm currently holding, the focal length is something like 20cm, which means I can only fit a few tens of such lenses around my target. The target needs to be a sphere of about 1.5-2cm diameter to accommodate the superimposed images (each image being 1.5cm long).

 This means that each of maybe 100 lenses can focus 1/6400th of a lamp's light onto the target, and the target has something like 1/90th the area of a lamp. So, I'm still only "concentrating" the light to 100*90/6400 of its intensity at the surface of the lamp. [EDIT - if it's 100*90/6400 then the concentration factor is 9000/6400 or about 1.5; this was my bad. I think my guesstimated numbers are slightly out, and the true value is 1 or less than 1. My bad - apologies.]

Now, I know you're going to say something like "make the focal length of the lenses longer, so they can be moved further from the target, so you can fit more lenses and more lamps around the target". I almost believe that. But I actually believe that the simple laws of geometry mean that you'll never deliver more light per square centimetre than leaves any one lamp (per square centimetre). I don't like it at all, but I believe it.
 — MaxwellBuchanan, Sep 25 2013

 Because, in order to focus all of the energy from a single source on the target (as can only be achieved by the elliptical mirror previously mentioned), you will have to distribute it over all surfaces of the target.

 If you are focusing light from n sources on an indentically sized target, you are only going to be able to direct, at most, 1/n of the energy from each source into the target.

 Let me try to explain this another way. If you were standing on the surface of a spherical target at the focus of an elliptical mirror, and you looked up, every bit of "sky" would look like the object at the other focus of the mirror. This shows that light from every portion of that object would be reaching a corresponding portion of the target, and, if the targets were the same size (or if the target is larger), all of the light from the source object would be hitting the target.

 If the source is larger, then some light would miss the target, and the target will not reach the same temperature.

 If the target is larger, it will heat to some lower temperature where the radiation off of it's larger surface is sufficient to balance the incoming radiation.

 If they are the same size, the target will heat to the same temperature as the source.

 On the other hand, if you try to focus any light from a second source on the object, this can only be done by removing some portion of the elliptical reflector, to let the light through. This results in some of the energy from the original source being lost. This loss, and the gain from the second source will at a theoretical maximum, balance.

This repeats no matter how many different sources you try to use. Each one simply replaces the energy lost by reduction of that original elliptical mirror.
 — MechE, Sep 25 2013

I hate to admit this, but I'm with [MechE]. I see how geometry has this all sewn up.
 — MaxwellBuchanan, Sep 25 2013

 consider the experience of the H2 in the sun itself. it is bathed in radiation from every direction all the time. Why is it the specific temperature that it is? Is it that it would be hotter if it just received more radiation? Fewer places can be conceived of as containing more radiation than the material of the sun itself and yet it remains at a paltry 5300 degrees. Yes there are optical barriers to concentrating light from weak sources. But when given a strong source you would anticipate it to follow whatever rule we were describing. How is it possible for the sun to remain so cold, and yet emit and contain so much energy?

If the whole thing is an optics problem, why is my microwave able to heat hotter than the sun with a paltry 1500w element, while the massive nuclear core of the sun cannot heat the entire mass to a higher level using gamma waves.
 — WcW, Sep 25 2013

 //Yes there are optical barriers to concentrating light from weak sources. But when given a strong source you would anticipate it to follow whatever rule we were describing.//

 If there are barriers to concentrating light from weak sources, why would there not be barriers to concentrating light from a strong source? How does optics change?

[WcW], I'm with you in spirit - I don't like the idea that I can't concentrate light, using optics, indefinitely. Yet that seems to me to be the case. I accept that no amount of optics will result in a net concentration of the light relative to its [black body] source. That sucks, and it seems to me like a very small-print-ish way to enforce the Second Law, but there you go.
 — MaxwellBuchanan, Sep 25 2013

no, optics are limited, they can only focus light so much, but they do not change the fundamental properties of light. We have to look to areas where the intensity of radiation is very high, high enough that the properties of the radiation itself become apparent.
 — WcW, Sep 25 2013

 I'm missing something here, and I apologize for not grasping your argument.

But, on a post-two-bottles- of-Pinot-Grigio note: if we accept that lenses can't concentrate "weak" light beyond the intensity of the source, I am pretty sure that "strong" light will either follow the same rules or (possibly) worse.
 — MaxwellBuchanan, Sep 25 2013

 Only the outer layers remain at 5800 degrees. For the most part (and ignoring fluctuations), it gets much hotter as you go in.

However, at 5800 degrees, the energy emitted by a black body is 63x10^6 W/m^2. That's a lot of energy. Enough, in fact, over the total area of the sun, to completely balance the energy of fusion occurring inside the star.
 — MechE, Sep 25 2013

Incidentally, are there any experts on the Talmud here? I get the feeling that we would benefit from some system of organizing comments, countercomments, and metacomments.
 — MaxwellBuchanan, Sep 25 2013

We are grappling with two entirely different scales of concept. If your focused light never comes anywhere near heating an object to the temperature of the light itself, how could you observe this phenomenon? And if you proscribe that this is impossible then we are on the same page, because I would argue that you cannot heat an object to a higher temperature than the source even if it is actually submerged in that source.
 — WcW, Sep 25 2013

 //We have to look to areas where the intensity of radiation is very high, high enough that the properties of the radiation itself become apparent.//

 While I agree that MBs table lamp might not be ideal for this (assuming it's a frosted bulb), the filament inside it is. Or a candle flame. Or any other source of black body radiation.

 Black body radiation never gets more intense than it's temperature would indicate. That is, any 5800 K object radiates at the exact same intensity. A 273 K object radiates at a different intensity, but again, it doesn't matter what 273 K object it is, it will be the same.

 A higher temperature gives a higher radiation level, and a higher total intensity, but the three do not vary with respect to each other.

Thus it doesn't matter what the scale is, the properties of radiation are always apparent.
 — MechE, Sep 25 2013

 // I *think* the consensus that's emerging (with various degrees of consensus), is as follows: //

 ...

 //    Before moving on, who's an aye and who's a nae to the foregoing? //

 Aye.

 // (6) If some impossible lens/mirror system _did_ allow me to violate (1) through (3), so that I _could_ concentrate light to an unlimited extent, then I _would_ be able to heat the target above the temperature of the source; but no such impossible system can exist. And therefore the second law of thermodynamics holds //

Yes. For example a system with a one way mirror is one of those impossible systems. Minor quibble with number six though - the system could just expend energy rather than be impossible. For example a fast moving, blue shifting mirror.
 — caspian, Sep 25 2013

right, so we can agree that actually sitting inside the surface of the sun is as hot as the sun is going to get you no matter what. total and maximal exposure to the radiation from the sun.
 — WcW, Sep 25 2013

 If we ignore the insulating properties of the outer layer of the sun. And if we assume you mean sitting in the outermost layers of the photosphere.

 Then yes. Again, there has never been any argument on whether a black body source has an upper limit on the temperature an object heated by it can reach.

The debate is on the reason for that limit.
 — MechE, Sep 25 2013

so, even though we are completely bombarded by radiation from every direction we will never go above 5300K. the temperature and frequency of the light that we are now receiving from all sides at a very high intensity. You guys understand intensity, right? It's pretty high in the sun.
 — WcW, Sep 25 2013

 what if it was half as intense, would we be half as hot?

Same radiation, from all directions, just half as intense, does our temperature fall? How much?
 — WcW, Sep 25 2013

Ive posted a link that I feel finally explains my view of the position. It clearly states that matter which no longer contains any cold nuclei becomes transparent to lower energy radiation, which is hard for me to imagine, and explains how the lack of electrons at a particular energy level allows radiation to ignore the nucleus entirely. It is as weird as anything in quantum physics.
 — WcW, Sep 25 2013

 "if there are no available quantized energy levels with spacings which match the quantum energy of the incident radiation, then the material will be transparent to that radiation, and it will pass through."

Which is precisely what happens when a an atom is hot enough to auto-excite it's own electrons out of the range of a received photon.
 — WcW, Sep 25 2013

So the real practical limit the temperature you can achieve with a incoherent thermal of radiation isn't the intensity that optics limits you to its the properties of the radiation itself. An object that is at the temperature of the light no longer absorbs it and becomes transparent to it. At first this notion of transparency was cryptic to me, but on further exploration I find that experiments demonstrate it very clearly.
 — WcW, Sep 25 2013

This was my Aha! moment. The part I was missing before. A body that would absorb thermal radiation at a frequency of 3000K, will stop absorbing it when it reaches 3000K. At that temperature it will only interact with the fraction of the radiation that is required to keep it at 3000K otherwise becoming completely transparent to it. This explains how so much low energy light escapes the sun, and why you don't violate thermodynamics with intense sources of light.
 — WcW, Sep 25 2013

//half as intense// okay, that one I can answer: BBR power is proportional to T**4, so half the power is about 0.84 the original temperature.
 — FlyingToaster, Sep 25 2013

 Where in that link is the text you cited? I cannot find it on a first pass.

 Edit- Never mind, found it.

 It doesn't say what you drew from it at all. It says that the reason that some materials are transparent to some radiation is because they don't have an appropriately sized quantum gap. However, by definition, a black body does not have this problem.

 And, since excitation of an electron is not heat, the simple fact that the incoming photons excite an electron does not mean it stays excited. In fact, in the case of fluorescence, where the photon is emitted at lower energy, the remaining energy remains in the atom as vibrational energy (heat). So the atom heats up, but the electron is back in it's original position, it is ready to be excited again.

 If this didn't happen, over and over again, the atom would not heat at all.

In addition, the incoming photons can establish resonance with all sorts of different things in the material, including vibrational and torsional modes, that provide a wide range of energy bands, regardless of the how excited the atom already is.
 — MechE, Sep 26 2013

 And again, you misunderstand. The black body radiation in the sun is no more intense than the radiation coming from any other 5800 K object.

Since the sun is larger, there is more total, but whether the target is exposed to it from sources in all directions, or whether it is exposed to it from a single source by focusing optics in all directions, the intensity is the same.
 — MechE, Sep 26 2013

 I just had a thought which might help tie together some of what [wcw] is saying with how I understand some of the physics.

 The emission spectra of an object at a given temperature - is that tied to the red-and-blue shift resulting from the atoms vibrating around in the hot substance - ie is it really possible that *in the frame of reference* of an atom, the photon emitted is purely defined by the temperature of that atom, but due to the fact that at that temperature the atom has a given velocity, that there is a statistical spread of the atom's velocity with reference to an external observer, yielding an emmission continuum rather than an emmission line? When taken in bulk form, this would mean there is a (probably skewed) normal curve of emmitted photon wavelengths.

Is there any truth to the above? If so, it could be that some of what [wcw] is saying about fixed temperature-frequency relationships is true - but that on the atomic scale, motion of the atoms "smears" the distribution.
 — Custardguts, Sep 26 2013

 Yes and no. The sheer number of different possible energy states, combined with blurring, is what produces the broad spectrum.

 But it is not a limiting factor in how much photonic energy the material can absorb.

*And this now beats out any two of the ATTR ideas for appearances of my screen name.
 — MechE, Sep 26 2013

 "The sun puts out as much radiance as simple black body of the same area and temperature."

 Not true, the sun puts out substantially more energy than a perfect black body of the same temperature and surface area would. I invite you to do the maths on this one.

 The excitation level of all electrons within a material is are in complete agreement with the temperature of that material.

 As temperatures rise there are logarithmically fewer electrons in any given lower energy state.

 Not I am going to ask you to take a leap of imagination with me. Admittedly it may seem a leap to far:

 What If I am correct: A photon cannot interact with a material unless it finds a receptive electron, it may have numerous interactions, but in the cases we are caring about here, it must find an electron it can excite or it will not become heat. Other things might happen, but since they will not affect the temperature of the matter, we will discard these things. (-all interactions between photons and matter are mediated by electrons-)

In this way a hot material has fewer and fewer places that a cool photon may hit and at a certain point, as we observe, very low energy radiation may pass quite a distance through a material before it has that chance interaction that lets it become heat. In this way, while the material may be very dense, from the perspective of infrared radiation it is very transparent, and becomes increasingly so as the material gets hotter.
 — WcW, Sep 26 2013

 //Not true, the sun puts out substantially more energy than a perfect black body of the same temperature and surface area would. I invite you to do the maths on this one.//

 A black body at 5800 K emits 63x10^6 W/m^2. The sun emits 63x10^6 W/m^2. 63x10^6=63*10^6. That's the math, what else did you want me to do?

 (Okay, this isn't quite true. The sun does emit a little more, because some of the sun's emissions come from deeper layers with higher temperatures, but they are only more intense because the temperature is higher).

 //The excitation level of all electrons within a material is are in complete agreement with the temperature of that material. As temperatures rise there are logarithmically fewer electrons in any given lower energy state.//

 False. Electrons move up and down without much regard to temperature.

 They are more likely to be excited in a hot material, simply because they are more likely to have encountered a photon coming from another particle, or have captured energy from atomic vibration, but they still step down to emit the actual photon, and don't step back up until they are re-excited.

 //all interactions between photons and matter are mediated by electrons//

 False. Per your own link "Electron energy levels have been used as the example here, but quantized energy levels for molecular vibration and rotation also exist."

 Furthermore, from the same source "transitions between rotational quantum states are typically in the microwave region of the electromagnetic spectrum." Thus proving that microwave heating is one of the regular processes found in black body radiation, and a microwave oven is a clear case of such radiation heating above it's "color" temperature.

Finally. Even if any of what you have said were true, you have not, in any way, demonstrated that this is the limiting factor on the temperature of a black body receiving electromagnetic radiation. I, on the other hand have made a definite argument, which I realize you may not understand, on why the temperature is limited by optics. Said argument backed up tentatively by [MB]'s experiment.
 — MechE, Sep 26 2013

 It is significant, but it might not be important for the wider point. That the sun may be a superior radiator to perfect black body is a complicated thing to explain, but this may also be a compromise between flux temperature and radiance. The total energy fits very well, but there is even less high energy radiation than you would expect.

Therein lies the crux of the issue Mech. If hot material could be excited by photons of a lower energy then you would expect the surface of the sun to be emitting at least some high energy radiation to dissipate this effect, and you would expect filaments focused on each other here on earth to heat one another. (such that high energy photons would be produced) and they do not. This property, of systems to accept radiation in a reciprocal way is absolutely critical, it cannot be ignored simply because it is hard to accept because amongst other things, it is why the laws of thermodynamics work the way that they do and why a large hot object cannot heat a small object to a higher temperature than it is itself. By focusing on optics and images, and insisting that cold radiation can act on a hot object you are asking to violate this essential reciprocity that is definitive and not limited by the number of photons, but by their very character.
 — WcW, Sep 26 2013

the optics part is really interesting, but it doesn't matter. Even if we collected all the light coming out of the sun, every single photon, we could not heat an object to a higher temperature than that 5778 peak through thermal heating. Theoretically it is possible to do this as hard as it might be to imagine it.
 — WcW, Sep 26 2013

 //If hot material could be excited by photons of a lower energy then you would expect the surface of the sun to be emitting at least some high energy radiation to dissipate this effect, and you would expect filaments focused on each other here on earth to heat one another. (such that high energy photons would be produced) and they do not.//

 As I have clearly and repeatedly stated, it is not possible for a black body source to excite a target above it's own temperature. Therefore, there is no reason to expect the sun to emit such high energy photons.

And filaments focused on each other do, indeed, heat each other up. Where is your evidence that they don't?
 — MechE, Sep 26 2013

 //Even if we collected all the light coming out of the sun, every single photon, we could not heat an object to a higher temperature than that 5778 peak through thermal heating. Theoretically it is possible to do this as hard as it might be to imagine it.//

I have never said otherwise. But it is an energy balance and optical problem, not some limit on how much energy light can add to a system. You cannot focus all of that energy onto a target less than the size of the sun. As long as your target is as big as the sun, you can focus all of the energy on it, but it will take all of that energy simply to bring the target up to the temperature of the sun.
 — MechE, Sep 26 2013

once you reach the temperature of the light it doesn't add any more energy to the system because the light no longer interacts with the system, the light passes through it as if it was not there at all. If a photon does not "see" an electron at the right energy level to excite, it has no way to add energy to the material at all. You do not believe that hot materials also have uniformly excited electrons but that's how it works.
 — WcW, Sep 26 2013

 And I tell you that is not true.

 Unless you can point me to an outside source that agrees with you (and no source you linked to to date does), I will continue to consider that argument false.

 In fact, I can think of a relatively simple experiment that will prove it either way (although not one I have the equipment to do).

 Take two equal temperature near black body sources. Put them next to each other. Take a flat plate light power meter. Measure the light coming from one of the objects. Do this with the meter between them, facing only one of the objects. Then do it with the meter outside so the farther object is eclipsed by the nearer.

 If the nearer object is transparent to light from the farther object, you will get a higher reading from the second measurement (as you will see light from both objects). If it is not, the readings will be (nearly) identical.

I can promise you, you will get the latter result.
 — MechE, Sep 26 2013

 //You do not believe that hot materials also have uniformly excited electrons but that's how it works.//

If that were true, the object in question would only emit in the wavelengths between the highest excitations. Instead, it is monotonically increasing at all wavelengths. This indicates that those electrons are still returning to the ground state on a regular basis.
 — MechE, Sep 26 2013

You seem very sure that you can heat one black body with another of the same temperature. This would be over unity. The second body will perfectly transmit all of the received radiation, absorbing none of it or you would violate thermodynamics and logic. Both will cool more slowly because they are a less efficient shape for radiating given their volume.
 — WcW, Sep 26 2013

 //You seem very sure that you can heat one black body with another of the same temperature. This would be over unity.//

 No, it most definitely would not be, this is not a closed system. In all cases I have described there is a "source". Meaning that the object in question is adding energy to the system, whether electrical to electromagnetic in the case of a filament, or nuclear to EM in the case of the sun.

 By directing all of the energy from that source to a target, you will heat the target. But whatever is doing that directing also does the reverse. This means that the energy is trapped in the system, traveling back and forth between the two objects while the source continues to add more.

And to clarify, you can't heat one with the other, you can only heat the system, which means both increase in temperature in lock step.
 — MechE, Sep 26 2013

Imagine now, that one of the black bodies in an optical apparatus that directs all of its radiation onto a spot on the other body. can we in this way heat the second body? No, we never arrive at a higher temperature, even though we are able to bathe that one spot with more radiation.
 — WcW, Sep 26 2013

That is not possible. Please see the wikipedia article on etendue that I just linked. It's exactly what I've been trying to explain in that regard.
 — MechE, Sep 26 2013

sophistry, we aren't limited to simple optical mechanisms. Lets pretend we have perfect fiber optical transmission. If all of the surface radiation from one body is directed onto part of the area of a second body that part does not rise to a higher temperature even though it is bathed in radiation.
 — WcW, Sep 26 2013

if it did you would have constructed an over-unity machine. it would be a practical Maxwell's Demon.
 — WcW, Sep 26 2013

The classical model had numerous problems. This was only one of them.
 — WcW, Sep 26 2013

It doesn't matter how complex an optical mechanism you use, etendue is conserved. You simply cannot increase the basic radiance of a system through optical elements (and fiber optics are still optical elements). The only problem here is that you do not understand that.
 — MechE, Sep 26 2013

 "You simply cannot increase the basic radiance of a system through optical elements"

 Yep. And even when you focus that radiance on a small area, that area does not rise to a higher temperature than the source, no matter how large that source was.

So we are in total agreement, light from a black body like the sun cannot heat an object to a higher temperature than the radiation itself, no matter how much of it we collect and focus. Thus any photons that enter beyond that point cannot impart any energy. (for some reason that we do not agree about)
 — WcW, Sep 26 2013

 No! No! No!

 Have you understood anything I've said? Did you bother to read the etendue article at all?

It's nothing to do with the ability of photons to impart their energy. They could if you could get them there. It's the fact that you cannot focus the light any more tightly than the source.
 — MechE, Sep 26 2013

oh, I missed that point. Who cares. I can take light from multiple places on the same source and focus them on a surface are smaller than the surface area collected from.if I do this I direct more radiation from a source onto a spot. That spot is then receiving more radiation than the source is producing from any given equivalent area.
 — WcW, Sep 26 2013

No you can't.
 — MechE, Sep 26 2013

 //Lets pretend we have perfect fiber optical transmission.//

 OK, here's something where I can give a definite answer. If you pump enough red laser light down a multimode fibre, you can melt aluminium (I can vouch for this, having done so), which means the target is definitely being heated above "red heat".

 [WcW] will argue that that's because it's laser light. I would argue that after light has bounced around in a fibre it doesn't really matter whether it started out from a laser or from a black body radiator.

 So, sufficient red light can heat a target beyond "red" heat.

 Now, I do fully appreciate that the laws of optics mean you can't do this with a red-hot black-body source, because you can't concentrate its "red" light down to a higher flux density than that of the source.

 But that was the whole point of the argument. Sufficiently concentrated light can heat a target to any temperature; it's just the geometry of concentrating light that prevents you from violating the second law.

[OK, I admit to some residual doubt, and the possibility that [WcW] may be right. Nobody here is an idiot.]
 — MaxwellBuchanan, Sep 26 2013

// I can take light from multiple places on the same source and focus them on a surface are smaller than the surface area collected from.// I think the point is that you *can't* do that - the geometry of the multiple mirrors and lenses means that you can only collect a fraction of the light which is being emitted from any one place on the source. So I can focus the light from ten different square metres of the sun's surface onto one square metre, but I'll only be focussing *part* (at most 1/10th) of the light from each of those ten square metres. This is sadly true.
 — MaxwellBuchanan, Sep 26 2013

Laser light is tricky, I myself cannot explain. There is only one property that I can see that makes lased photons unique, and that is that materials that are heated by them are not able to re-emit them as a recourse. In this way when you make lased light you may actually be paying some entrophy penalty to produce pairs of photons that can act as a waveform of twice the frequency forcing an electron interaction where one was not immediately apparent to a single photon. This is one area where the wave/particle thing gets really crazy. It is obvious that interactions of this sort are also common in incoherent light and do allow the occasional high energy excitation to occur, but the lower energy interactions are far more statistically likely in incoherent light. So it's muddy, but the statistical properties always keep the average at unity. The higher level of synchronization in lased light greatly increased the chance that these exotic interactions will occur, lattice excitation, and dual (possibly trebble) photon excitations included. In thermal light they also occur, but in tandem with an even number of equally exotic low energy format admissions. In this way you could say that material properties are still symmetrical in realtionship to radiation, and that materials that respond to synchronous double photon excitations are equally likely to unstack a waveform of the double frequency into two low energy interactions.
 — WcW, Sep 26 2013

So, given two suns, both in the sky at the same time, and two magnifying glasses, and one ant, it takes the same amount of time to cook the ant? The ant never gets hotter than if you dipped him in whichever sun was hotter, but he sure gets there faster.
 — WcW, Sep 26 2013

 No, because neither magnifying glass is focusing anywhere near the full intensity of either sun (and the ant never reaches solar temperatures, because neither is focusing even half).

Each glass is occupying only a fraction of the possible optical/focus area available, but there is a hard upper limit on that area. In order to raise the temperature of your target to the temperature of your source, you would have to occupy that full area, and once you do that, there is no space available to introduce a second source.
 — MechE, Sep 26 2013

 OK, [WcW] I'll give you a thought experiment to explain why you must be wrong on at least one of these two points: (a) that you can focus light down as much as you like and (b) that the colour of the light sets a limit on the temperature you can produce.

 Here's the experiment.

 Take one sun, and suppose its yellow-hot. Its light contains a distribution of wavelengths, peaking in yellow but with a tail out to the blue. Let's suppose that 1/1000th of its light is in this blue tail. Put a blue-pass filter in the way, and focus the remaining 1/1000th blue light onto a small target.

 Now take a million such suns, all with the same blue filter, and focus all of their light onto a small target.

 The target is now being hit by 1000 times (one million / 1000) as much light as the total light from any one sun, and that light is exclusively blue.

 By your own arguments, that must heat the target up to greater than the temperature of any one sun.

So one of your points - (a) or (b) must be wrong. I think that (a) is absolutely wrong and it's easy to prove it. I also think (b) is wrong, but that's harder to prove.
 — MaxwellBuchanan, Sep 26 2013

 I have read that in every single case, when you tow a car behind another car using a rope, the equilibrium speed of the towed car never exceeds the speed of the towing car, and the colour of the rope doing the towing is the same as the colour of the rope being towed. Therefore there must be some fundamental property of ropes of a given colour that determines the speed of the cars. If a towed car were travelling faster than blue, and were attached to a leading car by a blue rope, it would be slowed down to the "speed" of blue rope. It has noting to do with geometry, or Newton's laws, or engine displacements, or the speed of the leading car. The speed of the towed car is determined purely by the "colour speed" of the rope towing it. Each of the individual wheel bearings can only absorb a particular colour of rope. Once saturated with that colour, the "speed" of the bearings cannot be increased any more.

That is honestly how absurd [WcW]'s comments appear to me, which is why I have stopped attempting to address them. [MechE], I bow to your superior patience. You are right. [WcW] is wrong.
 — spidermother, Sep 26 2013

 Can I respectfully direct [WcW] to the foregoing thought experiment? I think [WcW] is wrong, but it's not as clear-cut to me (a non-physicist) as it is to [spidermother] that his argument is absurd.

The reason I don't think [WcW] is being absurd is that, if someone told me "a red hot object can't absorb red light to become hotter", I would have to stop and think about it. I think he's wrong, but not absurd, which is why it's an interesting discussion.
 — MaxwellBuchanan, Sep 26 2013

 If I weren't rather bored at work right now, or if I weren't learning things in attempting to fully support my arguments or counter his, I probably would not be this patient.

I'll admit, I'd be happier about it if I thought he was honestly attempting to understand my arguments rather than glossing over them to the point where I have to repeat them multiple times.
 — MechE, Sep 26 2013

 One small quibble, [MechE]. It is possible to focus *laser* light to a higher intensity than the source (if you consider the aperture of the laser to be the source), because the light is nearly parallel - it behaves like light from a more distant, higher intensity source.

Thermal emission, on the other hand, is scattered - which is part of the reason it can't be concentrated beyond the intensity of the source.
 — spidermother, Sep 26 2013

I am very sure that filtering out the small fraction of high energy radiation can heat a small area to a higher temperature. Just in the same way that if we pass an intense source of florescent light through a prism and place a thermometer in each band, each one will have a different temperature.
 — WcW, Sep 27 2013

(.) (.)
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- -
_ _ ZZzzzz...
 — rcarty, Sep 27 2013

The saturation-based transparency thing was mentioned by Feynman in one of those lectures (I think: the unsynched audio trashed my attention span).
 — FlyingToaster, Sep 27 2013

 //I am very sure that filtering out the small fraction of high energy radiation can heat a small area to a higher temperature.//

So, then, according to your idea, if we use filters in addition to lenses and mirrors, and use the light from multiple identical suns, we can heat a target above the temperature of any one of those suns? Which would be a second-law violation? That's why I think you're wrong.
 — MaxwellBuchanan, Sep 27 2013

yes, in my theory the small fraction of high energy radiation can be separated and that fraction can be used to heat a small mass to the energy level of that radiation, but not higher. You could also infer that this was already happening no matter if you separate the radiation or not, that those high energy interactions are already "part of the blend" and that they are far more likely to be translated into photons of lower energy than they are to excite the emission of even higher energies, and to the extent that they do they actually enable the nucleus to cool below the energy of the "highball" photon.
 — WcW, Sep 27 2013

 So, you're saying that the small blue fraction of sunlight could heat a sufficiently small target to temperatures higher than that of the sun.

Which is why I think you must be wrong.
 — MaxwellBuchanan, Sep 27 2013

the nucleus that emitted that photon was no hotter (more excited) than the nucleus that receives it MB. If the two are treated as the "sender" and the "receiver" then we cannot say that the law has been violated can we. Touche. I say it proves that I am correct.
 — WcW, Sep 27 2013

 What are nuclei doing emitting photons??

If you mean that the blue part of sunlight comes from the much hotter core, I don't think you're right. My (limited) understanding of solar dynamics is that photons generatd in the core are absorbed and re- emitted countless trillions of times on their way to the sun's surface, and that the light we see has all been emitted from very close to the sun's surface.
 — MaxwellBuchanan, Sep 27 2013

We are talking about heat, right? Heat, temperature, is a property that bodies with mass have. The nucleus contains energy in the form of "heat" which it may dissipate by exciting available electrons. In so doing the nucleus causes the production of an electron. Conversely, if available electrons are excited near the nucleus by a photon, through a process that is equally hard to explain, one possible result is that the energy of the photon becomes heat in the nucleus. So, heat from a nucleus becomes a photon. A nucleus "emits" a photon.
 — WcW, Sep 27 2013

 MB, I I just don't know were you are starting from. Due to certain factors of uncertainty, within a given mass there is a random distribution of thermal energies. As with lasers there are also occasional incidents of photon stacking and other "exotic" events. These factors result in a distinctive bell curve of energy distributions, roughly similar to the bell curve of the radiation histogram. So, for any material at 3000K there is a even distribution above and below that temperature, which means some fraction of the material is emitting radiation at 3200k, and some is below 3000K and can absorb radiation at that level.

Given this, for twin materials we will always see some of these interactions.
 — WcW, Sep 27 2013

Basically we blew off the classical demonstration of this which is the "flashlight facing flashlight" model. Light from one flashlight is directed at the reflector of the other flashlight in a relatively uniform fashion but neither element grows hotter (although the non-thermal elements of the bulbs will fail if the intensity is great enough) this demonstrates that simple increases in intensity do not result in increases in temperature for radiating bodies. We do not, as a result, get a pair of flashlights that are now producing UV light.
 — WcW, Sep 27 2013

 Hang on a mo. I was pretty sure that photons are given off by electrons hopping between orbitals. I guess nuclei give off gamma rays during nuclear transmutations, but they're the exception.

 So, either I've horribly misconstrued all the (finite amount of) physics I ever learned; or you're talking about something else. (Or both.)

 Howevertheless, regardless of whether the photons come from the nucleus, the electrons, or Wallmart, I think my objection stands, so I'll re-iterate it:

 (1) You're saying that a hot object gives off a broad spectrum of light, the distribution of which is characteristic of the temperature of the object, yes?

 (2) This light always has a tail, however small, at the upper end (so a red-hot object gives off a tiny amount of blue light), yes?

 (3) I can filter out the redder end of the spectrum to leave only a small amount of blue light, yes?

 (4) I can then concentrate that light down as far as I like, onto as small a target as I like, yes?

 (5) Which, in turn, means that I can heat the target to a temperature corresponding to "blue hot", yes?

 (6) Which means that the target winds up hotter than the source, yes?

 If you're saying (1) to (6) are all true, then that's a violation of the second law, as far as I understand it.

[EDIT: [WcW] I'm not trying to be argumentative and I think this is a great discussion - I just really really want to get my head around this.]
 — MaxwellBuchanan, Sep 27 2013

I know what you are thinking, "lets build a cascade, and as a result produce a few exotic radiation, and a few exotic hot nuclei" but don't waste your time, to the extent to which that is possible, it is already happening in the unfiltered reciever, there is already a 1/1000000 hot photon/nucleus happening in the receiver. With filtering you can sort the piles of particles, but you cannot change the overall balance of energies.
 — WcW, Sep 27 2013

 I wasn't thinking of a cascade.

 (1) Given enough suns and enough mirrors, filters and lenses, I could heat a golfball above the temperature of any one of those suns (according to your reasoning).

 (2) By the same token, I should therefore be able to use the cosmic microwave background to heat that golfball to any temperature I choose.

This seems, to me, to be incorrect.
 — MaxwellBuchanan, Sep 27 2013

And I still state that you are wrong about the flashlight experiment. Both elements will heat up, and as a result emit higher energy light.
 — MechE, Sep 27 2013

 nope, what I'm saying is that the transmitter contains a small number of nuclei that are blue hot. The radiation that they produce could produce a twin someplace else. To the extent that there are diverse photons there are diverse nuclei, and to the extent there are diverse nuclei they cannot be ignored. Temperature is like average height, there is a population distribution, and within that population some people are much taller, and some people are much shorter. Nothing prevents us from putting all the tall people on a basketball team.

Making a small national basket ball team doesn't make the country better at basket ball, it just means that instead of the occasional dunk distributed across the population, many dunks are happening in one spot. And just because more dunks are happening in one spot than average doesn't mean that your team is getting any taller.
 — WcW, Sep 27 2013

prove it.
 — WcW, Sep 27 2013

if that were the case then you should be able to turn one of the flashlights off and achieve the same result with optics and mirrors, or as as we call it here "a temporal light collection device".
 — WcW, Sep 27 2013

 //Making a small national basket ball team doesn't make the country better at basket ball//

 On the other hand, if I could take the best basketball players from every planet in the universe, and assemble them into one team, I would have a pretty good team.

 [WcW] I hear what you are saying (actually no, I read what you are writing) about temperature being an average, and the photons therefore having a distribution around an average.

But your proposed mechanism still allows me to heat a golfball to white heat, using nothing but cosmic microwave background. Maybe I can actually do that - I just find it hard to believe.
 — MaxwellBuchanan, Sep 27 2013

Given a perfect apparatus to collect it, yes. A perfect pinhole cavity collector would reach the temperature of any source that it was directed at, no matter the intensity, if given enough time. This means that a lossless cavity collector would reach the opacity temperature of the background radiation, somewhere around 3000K if directed into deep space. eventually after the the cavity was at this temperature the radiation in and out of it would equalize and the pinhole would begin to radiate output equal to the input.
 — WcW, Sep 27 2013

mind you, the final temperature you achieve is going to be lower based on a realtionship between the intensity and the aperture. A very small aperture (though obviously large enough to let the light in) means a very slow rate of ingress, and a larger aperture means a lower terminal temperature. The "less perfect" black cavity heats faster and is lossier, the most perfect black body reaches the perfect temperature, but takes an infinite amount of time. The this affect falls logarithmically with the intensity.
 — WcW, Sep 27 2013

 even if you assemble all of the best basketball players in the world none of them will dunk a 30ft basket, and to the extent that any of them could dunk an amazing 16ft basket, they were doing it before you collected them together, it's just that now you know where to look for it, and before, you didn't.

 The exotic events that you find in the exotic sorted categories of light were happening in the "general population" of interactions, you just didn't know where to look for them.

Also, though it is neat that we can do this light sorting, we are reducing entropy so we must pay the entropy tax someplace in the system, likely at the filter. Pika gets his, if not by hammer, then by more subtle means.
 — WcW, Sep 27 2013

also, friends, lets recognize that we are replicating the 1913 debate on this issue, a noble thing to do. I feel honored that we have done this a hundred years later. It brings into stark relief the desperate need for more, rather than less discussion of scientific principals. More rather than less patience in those discussions. And more respect for the notion of rational, rather than personal methods of dialogue.
 — WcW, Sep 27 2013

We can discuss lasers next; how the heat waste at the lased surface is the entropic penalty for the higher incidence of high energy ordered interactions at the target. And how it is impossible to get away from paying this penalty.
 — WcW, Sep 27 2013

 // if that were the case then you should be able to turn one of the flashlights off and achieve the same result with optics and mirrors, or as as we call it here "a temporal light collection device".//

 You can. Energy trapped in the system increases the temperature of the system. It's functionally no different than wrapping an insulating layer around a convectively cooled object. It will increase in temperature until the energy coming of the backside of the mirror/optics balances the energy the source is putting in.

Heck, under your theory, the same thing happens, except the energy is all light instead of thermal.
 — MechE, Sep 27 2013

 So, the heat-death of the universe is avoidable?

 I take your point about selecting the extremes from a distribution, but I don't believe the conclusion. Given a region of space, starting at a uniform temperature, I do not believe that a target placed in front of any arrangement of mirrors, filters and lenses will spontaneously heat up. Are you convinced that it will?

And also, could you clarify why you talked about nuclei emitting photons? I had believed that photons were emitted by electrons hopping up and down (well, down at least) in their orbits; I've never heard "hot nuclei" being mentioned in discussions of light before.
 — MaxwellBuchanan, Sep 27 2013

 an object placed inside a universe that consists solely of uniform background radiation will not heat up more than the temperature of the background radiation no matter what you do with lenses and mirrors for the simple reason that while an optical mechanism may be able to make radiation that wouldn't normally hit the target hit the target, it will at the same time deflect at least as much other radiation that would normally hit the target away from hitting the target.

 Of course this could just be me being pessimistic.

I thought I was onto something with fisheye lenses for a bit.
 — FlyingToaster, Sep 27 2013

Thermal energy in the nucleus excites electrons, and electron excitations may become thermal energy. The nucleus never produces a photon, but the energy in the nucleus can cause "spontaneous" emission of photons. This is thermal radiation, in a nutshell.
 — WcW, Sep 27 2013

 MB- Surprisingly enough, I actually learned during this discussion that sources other than electrons can emit photons. This has nothing to do with what [WcW] is saying, other than invalidating his claim that microwave ovens are somehow an exception to his arguments, but it is interesting.

Molecular vibration and rotation have quantized energy levels, and are generally responsible for (and excited by) emissions in the Infrared and Microwave range respectively.
 — MechE, Sep 27 2013

 Well, then I've learned something too. Thank you, [WcW].

But it still leaves me struggling to believe that mirrors, filters and lenses can create a hot-spot in what is initially an environment at a uniform temperature.
 — MaxwellBuchanan, Sep 27 2013

 What you are saying is true FT, an object that can act as a radiator will stay damn cold in deep space. A cavity black body is the worst possible radiator, radiation that goes into it is very unlikely to come back out and it has a perfectly non emissive outside surface so other losses are eliminated. given such a perfect construction, little light escapes until the entire interior is as hot as the light, then the interior acts as as if it were a perfect reflector, so it takes even longer before the light in and out are equal.

We don't build perfect black body collectors, we build awful ones and correct our temperatures with math.
 — WcW, Sep 27 2013

 Ackshully what I was sayin' was that in a perfectly uniform environment (no stars, just a constant level of radiation coming in from all directions at any point), you can't get any warmer than the fog no matter what you do with lenses or mirrors. Again this has nothing to do with physics, just geometry.

(is "background radiation" what you get from focusing on nothing ? or is it the average radiation from all sources)
 — FlyingToaster, Sep 27 2013

the uniformity is the illusion MB, there is no thermal uniformity, which is why thermal sources of light produce distributed bands of radiation. They are diverse populations. In non thermal radiation the light is in very specific because it isn't coming from a diverse population of distributed thermal energies.
 — WcW, Sep 27 2013

 Yes, and Maxwell's Demon is a thought experiment on using that variation to violate the second law of thermodynamics. But it's also been proven, conceptually and mathematically to be impossible.

 Yet you are proposing that it is possible to passively heat a target beyond the source, therefore allowing the extraction of work from a uniform starting temperature, in direct violation of the second law.

Simply put, that can't happen.
 — MechE, Sep 27 2013

That's precisely what you are wrong about. On a one molecule to another basis we aren't heating above the source. The source molecule for the higher energy photon WAS hotter than the median temperature, and in that way the best that molecule can do is deliver its own temperature to another molecule, no better. It's 1/1. one nucleus in the source one nucleus in the receiver. One blue pixel. we can cluster all the blue pixels from the source into one side of the screen, but we can't defeat the fact that it's 1/1. You never end up with more total blue pixels (hot nuclei) than you had in the source.
 — WcW, Sep 27 2013

Hmmmm.
 — MaxwellBuchanan, Sep 27 2013

 Except you can redirect the light from your target back to one of the sources. Since, according to your theory, that will heat the source, the source will no emit more blue photons. And you, again, filter those blue back to your target, while it's still generating some additional random blue photons.

And eventually, after many exchanges, both are blue hot.
 — MechE, Sep 27 2013

not so. the chances of a blue nucleus emitting a blue photon are still very low. those blue photon receivers are still going to emit the same random distribution of radiation as a result, just on the upper end. and yes, this means a few excitations above the blue range BUT those were going to occur in the random mix anyway. Shift the filter up to include only them and the ones above them are similarly rarified. Energy is also order; Entropy is fighting you at every step, driving order down, the more steps the more entropic action will take your orderly blue pixels and turn them into red pixels.
 — WcW, Sep 27 2013

the sun has the same problem. Every step that the radiation takes increases the tendency for it to become less ordered, cooler, two hundred red photons for one gamma ray. The eventual fate of the universe is to be populated with lots and lots of cold matter and cold radiation.
 — WcW, Sep 27 2013

 At this point, and until you can link to a source that actually supports your argument, I give up.

 I have spent more time/effort on this anno thread than any other (and I am grouping all three ATTR ideas at this point). You are wrong. Your basic idea is wrong. Your contention that EM radiation cannot heat an object to a higher temperature is wrong, and microwave ovens prove that. Your belief that your argument is supported by a quantum understanding of black body radiation is wrong. Your persistent statements that photon emission is exclusively mediated by excited electrons are wrong, and your own link proves that. Your contention that light can be focused without limit is wrong, and a basic understanding of optics proves that. This latest argument that it is possible to cause heat to flow from a colder point to a hotter without energy input is wrong. Your contention that placing a spherical mirror around an illumination source will not make it heat up is wrong.

You are wrong. Cite a source that proves otherwise.
 — MechE, Sep 27 2013

Yes, the perfect black cavity seems to fight this process, but still the best it can do is steal a little bit of the order from another source. Once the source grows dim, the black cavity will also get cold.
 — WcW, Sep 27 2013

 Oh, and to clarify, //The eventual fate of the universe is to be populated with lots and lots of cold matter and cold radiation.// Is one of the statements I agree with, but your understanding of how quickly it happens is still wrong, and an atom/molecule excited by a blue photon will preferentially emit a blue or green one, not red. The net result will still be a hotter source, and this is contrary to that exact statement.

Oh and look up anti-stokes scattering to see a case in which an electron mediated photon emission can actually produce a higher energy than the input. It's rare, but if you are able to select for them, you will, slowly gain in temperature.
 — MechE, Sep 27 2013

 For those with limited physics backgrounds (my formal physics studies only went to second-year university level, but everything discussed here is far more elementary than that) who may still be wondering whether [WcW] is contributing anything useful to this discussion (beyond the intellectual exercise of contradicting him) I will borrow from Richard Dawkins and say that [WcW] is either ignorant, insane, or wicked. (Those aren't mutually exclusive, of course.)

I could go on, but I think my point is made. If the rest of you are still having fun, fine, but I don't intend to respond to another of [WcW]'s comments.
 — spidermother, Sep 28 2013

 //sources other than electrons can emit photons//

Simply put, charges create electric fields, moving charges create magnetic fields, and accelerating charges (positive or negative) create electromagnetic fields (and photons; you can't have one without the other).
 — spidermother, Sep 28 2013

This is all getting a bit ad hominemy...
 — MaxwellBuchanan, Sep 28 2013

Personal, yes. Ad hominem, no. I'm not attempting to use a person's character to prove an argument, but rather using a person's argument to draw inferences about his character.
 — spidermother, Sep 28 2013

 I'm not really interested in [WcW]'s character, I just want to get my head around the physics.

 I think [WcW] is wrong, but it's still an enlightening discussion.

 I think that thinking about the distribution of photon energies from a black body source is interesting and, since it's clearly true that there are some "hotter" photons in there, it must mean that the second law is upheld only due to the geometric impossibility of concentrating light far enough. (I know [WcW] disagrees with me on that one.)

So, from my perspective, it's been interesting. I am still disappointed, though, by the fact that the second law relies on what I see as nit-picking (though insuperable) geometry problems.
 — MaxwellBuchanan, Sep 28 2013

 You want a solar collector that makes things hotter than the heat of the sun. I have made a effort to explain why this will never be a problem of intensity. It is possible that the universe is playing a jest on us, that a series of mirrors could cause a lot of light to arrive at one spot at the same time and that you would have achieved something that simple optics cannot. As you intuit, this does not seem to be the case. Forcing more photons to arrive at the same time by sending them through a temporal delay circuit works fine, the intensity goes up. In fact, there isn't any apparent material limit on the number of photons that can live in a given space by optics or by temporal cadence. The end result is still the same.

We can take each exception to this an examine it, each one can be explained, and when examined they all make sense, but principally that would require less shouting and some sort of establishment of basic principals.
 — WcW, Sep 28 2013

 //Forcing more photons to arrive at the same time by sending them through a temporal delay circuit works fine, the intensity goes up//

 Alas, it doesn't - geometry has it all sewn up. I thought about a system which guided light down an optical fibre, with a switch that could send packets of light via progressively shorter shunts so that they'd all hit the target at the same time. But that can't work either - all the light packets have to go through the "output" switch of the shunt at the same time, in order to hit the target at the same time, and they can't.

However you try to figure the geometry, it prevents you from concentrating light in this way, either in space or in time. It's a dirty trick to hold the Second Law together, I say.
 — MaxwellBuchanan, Sep 28 2013

 What happens to a resistive filament if it is placed in spherical mirror? Does it get infinitely hot, does the sphere fill with gamma rays?

 Does one filament sitting right next to another cause either one of them to be hotter?

If optics fails us, why cannot we resort to simply heating the target to the right temperature, and then bombarding it with radiation and seeing what happens? Does the cooler radiation, or the equally hot radiation heat the target when we cheat, and preheat the target, which should allow for greater than 100% efficiency?
 — WcW, Sep 28 2013

 We've been calling it "geometry" for convenience, but I'd be interested in what the actual term is for the twig of maths that includes radiation (in a mathy sense) and flux; the latter seems a calculus-y thing.

 Note that a certain part of my brain is still holding out hope for using fisheye lenses.

All of which doesn't explain why my 'fridge won't work.
 — FlyingToaster, Sep 28 2013

 figgering round filaments, right next to each other, that's a temperature increase of about 4.6%.

 (The filaments block(ie: absorb) 1/6 of the other's output, so the total radiative area (that isn't being reabsorbed) has decreased by 1/6. But the absorbed radiation is being reradiated, so that's a radiative density increase of 1/5th, or a factor of 1.2, the 4th root of which is 1.046...)

 Which doesn't count the possibility of the absorbed power recharging the battery which would be just annoying.

The other thing, same theory, ignoring backtalk (which you can't of course in reality), the initial temperature would be doubling every 16 seconds.
 — FlyingToaster, Sep 28 2013

logically the filament in the center of a ring of other filaments should be able to arrive at a significantly higher temperature, even though they are all identical.
 — WcW, Sep 28 2013

yeah, but you can do the math on that one: I did the other two.
 — FlyingToaster, Sep 28 2013

my math teaches me that ten filaments behave exactly as one large one does. That removing the intervening material does nothing of note. That thermally two neighboring filaments behave the same as a single filament of the same shape touching. The gap between them, having no meaning.
 — WcW, Sep 28 2013

right, so you can't pin it down any more than I can eh ?
 — FlyingToaster, Sep 28 2013

sure, I just did. the gap between two filaments doesn't let them heat each other at all. They are just as able to do it as they were when they were touching, which wasn't at all. They may insulate each other, but insulation doesn't let them achieve a higher temperature.
 — WcW, Sep 28 2013

I'm pretty sure that's 'cuz your filaments are 1 dimensional lines, whereas mine are projections of a 2-dimensional circle (a rod)... which is less wrong.
 — FlyingToaster, Sep 28 2013

no, what I am saying is that there is no math. no math. a hot rod standing next to an identical hot rod does not get any hotter on the side facing than the side away. not 4% not .004%. The only action is insulative, that side simply cools more slowly. so there is no math for them getting hotter, only getting colder more slowly.
 — WcW, Sep 28 2013

 Yes, I see where you're coming from. But the difference is "temperature" and "power". Currently, during this post we're defining temperature as being the peak frequency in a black-body radiation, ie: a red peak is 4,000K, a blue peak is 5,500K (both figures are pulled out of my butt), stuff like that.

 The actual black-body graph which peaks at frequency indicates, not only "temperature" but how much power is being tossed out. It's totally proportionate: "red" is always (say) a billion watts per square meter (I'm not looking up the actual figures, feel free).

 Actually there is one figure I remember: room temperature (300 Kelvin) is 459W per square meter. Mind you it's 459 W that is in the form of a radiation spectrum that goes from a frequency of 0 to somewhere in the infrared.

Hang on, let me catch my breath, see if I remember where I was going with this.
 — FlyingToaster, Sep 28 2013

nothing magical or new happens when keep the rods warm with electricity, or with conduction, or with angry words, the same principals apply. so as the rods are the same temperature they can only insulate each other, with an insulating action that is most efficient when they are actually touching and acting as if they were a solid, single, larger, rod. Where does the math help this?
 — WcW, Sep 28 2013

 Right, anyways so 27deg Celsius, 300deg Kelvin (okay I'll grant that's a warm room) is define as a power emission of 459 Watts per square meter. Now the power to temperature relationship is a 4th power thing, so if you want to double the temperature you have to 16x's the power, which means if you want 600degK the power output is gonna be like 7.3 kilowatts per square meter.

 Okay, so back to the Sun. At the source (ie: the Sun's surface) it's power output is a godzillion watts per square meter, but by the time it gets to the Earth it's been diluted to 1kW/m2. If you were to shift all the frequencies around so that amount of power looked like a black-body curve, then the temperature would be about 86 degrees Celsius. So space is hot when the sun's shining but not too hot in the grand scheme of things.

 Anyways, now I remember where I was; that's what I was trying to do to calculate the new temperature of the rod.

 ------

 But the first two problems are sort of easy in that respect.

 You can surround a circular rod with 6 similar rods, just touching, to occlude all the radiation from that rod, right ? So one rod next to each other will block 1/6 of the light.

 But all that absorbed radiation is being (conducted to the rest of the rod but not back to the power source, and) reradiated.

 Which means that even though the same amount of power is being pumped through that it's radiating from 5/6 the surface. Which means that the power per area is 6/5 of what it was.

 The 4th root of 6/5 is 1.046...

So that's an equivalent temperature increase of 4.6%
 — FlyingToaster, Sep 28 2013

 I think the major difference of opinion here is that a black body will absorb everything that's thrown at it and (assumedly) reradiate everything in a black-body manner, by which I mean if you shoot a very small amount of gamma rays at a black body you'll get back a bunch of radio and microwaves and infrared in the shape of a black body spectrum.

Whereas in reality with real materials... no clue.
 — FlyingToaster, Sep 28 2013

 This idea starts out like the Book of Job, with [MB] in the title role, raging against Nature for its shabby behaviour. Various half-bakers then line up as Job's Comforters.

 It departs from the script when our own [Eliphaz the Temanite] grabs someone's beard, and [Bildad the Shuhite] starts beating him over the head with a stick.

This means we never get to the part when [MB] spends enough time by himself to arrive at the thought that Nature might have got it right after all, what with [MB] himself being a product of Nature.
 — pertinax, Sep 29 2013

 The Book of Elephant by a Dozen Irate Blind Men.

I think I ended up learning something... I'm not entirely sure what it was and I'm pretty sure it will be of no practical use whatsoever, but that just puts it in the same category as Humanities' courses, which people apparently sign up for all the time.
 — FlyingToaster, Sep 29 2013

 //I think I ended up learning something//

I think this discussion has brought us to the very edge of reason. However, I'm not sure which side we started from.
 — MaxwellBuchanan, Sep 29 2013

I agree, I learned a lot.
 — WcW, Sep 29 2013

 At the risk of awakening the sleeping dog after the horse has bolted, I asked two physicists, independently, about this. I asked them "if a black body target is hit with an indefinitely increasing amount of incoherent red light, will it continue to heat up indefinitely, or will it reach a limit determined by the colour of the light?"

 Both of them said that, given enough red light (or light of any colour), the black body target could be heated indefinitely. They pointed out also that a black body is a theoretical abstraction, but that it could be approximated fairly well, and that a real object could be heated by red light to whatever temperature caused it to melt or vaporize.

 These two physicists are from the Cavendish lab here in Cambridge, and are working at (and in one case being tech. director of) the optics-oriented company which I'm currently working with on a different project. So I'm inclined to believe them, with the caveat that there is always the chance they're both mistaken in the same way.

So, this leaves simple geometry (and hence the inability to focus light from a finite source to an infinitesimal point) as the only thing holding the second law together.
 — MaxwellBuchanan, Oct 02 2013

I'm pretty comfortable with the notion that the dipole action of the infrared light is non reciprocal and that the material cannot reciprocate by producing a perfect analogy, in the way that the electron can produce a near perfect "reflection" photon. We have no sources of intense incoherent light, and as such it isn't easy to test this premise. If infrared light falls into a similar behavior with microwaves, in that the objects that are bombarded with it are unable to produce it reciprocally, the materials that receive infrared radiation are still able to produce their own spontaneous radiation at higher energies once they are hot enough to do so, and may do so so long as they have electrons available to do so. To the extent that they do not do this, they are not true black bodies.
 — WcW, Oct 02 2013

of course I would be fascinated to take a look at examples of materials better able to absorb incoherent infrared radiation than they are able to produce it. Possibly your friends can produce a brief explanation of this in the context of a temperature above 3500k, which is the theoretical temperature of infrared light and is perfectly capable of destroying most terrestrial materials.
 — WcW, Oct 02 2013

 Just a quick interruption: Is it possible that this is the longest singularly-directed science thread on the website?

Carry on.
 — shapu, Oct 02 2013

I think it's approaching its own personal heat-death. I'm not completely happy, but all I can do is weigh the opinion of two professional physics/optics people against that of one esteemed halfbaker.
 — MaxwellBuchanan, Oct 02 2013

 // materials better able to absorb incoherent infrared radiation than they are able to produce it//

 No such material exists, and that wasn't what was being said here at all.

 //We have no sources of intense incoherent light, and as such it isn't easy to test this premise.//

 I have indicated that a test is possible with high power IR LEDs, which I don't happen to have, but they do exist. It would also theoretically be possible to test this with sufficiently high power LEDs of any wavelength, but since the target temperature goes up so rapidly with color, the physical testing apparatus would be difficult.

Thus a real world test is most definitely possible.
 — MechE, Oct 02 2013

I long ago agreed that infrared light seems to buck the paradigm, I'm curious to see the notion established. Any form of radiation which causes a material to heat but which it cannot produce as a result has unique potential.
 — WcW, Oct 02 2013

 //I long ago agreed that infrared light seems to buck the paradigm//

 Not that I ever saw. You agreed that coherent light did, and microwave, but never IR. Also, let's be clear, I am not saying there is anything special about IR, just that it happens to generate a peak in a temperature range that can be worked with easily.

 //I'm curious to see the notion established. Any form of radiation which causes a material to heat but which it cannot produce as a result has unique potential.//

 No one has said that is happening. Please explain where you are getting that interpretation. Any material heats, and the produces (an approximation of) black body radiation based on that heat.

And [Shapu], I believe the answer is yes based on word count, with a caveat. ATTRR split into two ideas. The original has 41,000 works, "Non- Working Reactionless Drive" has 20,000ish. This idea is only at ~43,000, so it's got a ways to go to reach the combined total.
 — MechE, Oct 02 2013

Any theory that depends on infrared radiation being qualitatively different from, say, blue light seems wrong to me.
 — MaxwellBuchanan, Oct 02 2013

infrared and microwaves have a unique interaction mode with molecules in that microwaves can twist them, and infrared waves can cause them to oscillate, blue radiation is incapable of doing either of these things, the frequency being to great to resonate with either function. In this way microwaves and infrared are kin in that they may interact with the nucleus within a single waveform.
 — WcW, Oct 02 2013

 So, if infrared, then what about far visible red? Orangey-red? Yellow? And if some substances can be stimulated in one way by infrared, aren't there others that would respond similarly to red? Or orange? Or yellow?

I truly cannot believe that something fundamental about light applies only to some wavelengths.
 — MaxwellBuchanan, Oct 02 2013

It's true though, and it explains why we can get specific materials so hot with these forms of radiation, also why the lasers used to cool are in this range.
 — WcW, Oct 02 2013

 So, if I had a target like (guessing here) diamond or graphene, where bonds are strong and atoms vibrate at higher frequencies, they'll respond to blue light in the same way as other materials to do red light?

 Are you saying that we can violate the second law of thermodynamics using lenses and mirrors for some combinations of wavelength and target material?

 I can follow your reasoning, and I don't think it's stupid, but I do think it's wrong. I can imagine the second law being upheld by either the limiting "temperature" of a given colour of light; or by the geometric limits of light focussing. But not sometimes by one and sometimes by the other.

(Hang on - someone at the door.)
 — MaxwellBuchanan, Oct 02 2013

(Back. It was Mr. Occam wanting his razor back.)
 — MaxwellBuchanan, Oct 02 2013

I don't see what this has to do with thermodynamics, order declines and energy is conserved. Nobody is yet looking to push the rock uphill, we're just looking to understand why it keeps rolling downhill. Yes, there are materials that are resonated by higher and lower wavelengths but the chance of hitting this interaction declines with the increase in size.
 — WcW, Oct 02 2013

 And we add another thing you apparently don't understand. Laser cooling is tuned to the atom in question, and it does take advantage of a specific transition in the atom. But it has nothing to do with the wavelength of the light as such.

Laser cooling works by hitting the atom with an energy just below an excitation energy. The energy for the transition to actually happen is taken from the kinetic energy of the atom. Thus what's critical is the relation between the transitions in the atom and the energy of the light, not anything inherent in the light itself.
 — MechE, Oct 03 2013

 And infrared and microwave do not have a "unique" interaction. They primarily interact with rotational and vibrational modes, because these are typically lower energy, and thus at the optimal level for these forms of energy. However, visible light, UV, etc, can and do still have these interactions, it's just that they can also trigger the higher energy interactions of electron excitation (and will often do so in combination with one of the former).

I have to say the "theory of the gaps" doesn't work any better than the "god of the gaps".
 — MechE, Oct 03 2013

Still have that unshakeable belief in the Second Law of Thermodynamics now [spidermother]? If so, we'll give it another couple of days. [link]
 — AusCan531, Jan 13 2014

 I don't have an unshakeable belief in anything. The closest I come to belief is not to bother questioning things that are firmly established, and beyond my ability to question as deeply as the experts have. The laws of thermodynamics could be loosely restated as 'things that are overwhelmingly improbable don't tend to happen'.

 I once tried to teach powers and logs to a friend. At one point, he said "How do I know that you're not just making this up?" I replied that (a) I wasn't, and (b) it doesn't matter; once you understand, you can see for yourself that powers and logs make sense in their own right. It's like that with thermodynamics. A system involving very, very large numbers of interactions - such as large volumes of gas, or large solar-thermal collectors, obeys the laws of thermodynamics simply because an increase in information (or a decrease in entropy) is statistically impossible. Once you understand that, belief is not necessary; it's so obvious as to be tautological.

 — spidermother, Jan 13 2014

 This is no different to any complex parabolic reflector. What this idea loses being different to the parabolic side geometrically, it gains on the complexity side, temporally. There is no real need to delve into complicated entropy, enthalpy, black body whadda whadda. Photons have a horrible habit of not wearing a watch. For all we know there may only be one photon in the universe, at any given point "in time". Wheeler would say electron, but I defer.

Simple parabolic reflectors have a one bounce focal point. Complex parabolic reflectors have a multiple bounce focal point. There are those that say there is no such thing as a "one bounce", and light just exhibits its highest probability outcome over many "reflections" (incidence = reflection) so that in essence the entire reflective array produces a high probability that the photon will arrive at a particular point. In a perfectly elastic world this means all complex reflections can be reduced to simple (one bounce) reflections. That is all that is really happening here, except your array is not static. By swapping a static, in time, array for a static, in time, focal point, you are not really changing anything. I think that is where the majority of confusion is coming in.
 — 4whom, Jan 14 2014

 //the time element is also a red herring.// [Spidermother], way up there ^.

I'm not disagreeing with you, [4whom]; but I felt that the points needed to be made that (1) there is a theoretical maximum temperature for all solar- thermal collectors, which can be determined in a straightforward way by assuming that they obey the laws of thermodynamics, and (2) altering the path lengths does not change this. My unnecessary whadda whadda was intended to counter a higher level of unnecessary whadda whadda. Thus, you don't need to talk about emissivities, wavelengths, 'temperatures' of radiation; the laws of thermodynamics have all that covered. And, as you say, you don't need to talk about thermodynamics; the lack of wristwatces on photons has that covered (Link).
 — spidermother, Jan 15 2014

 //So, the second law of thermodynamics states that heat will not flow from a cooler to a hotter body. This has been used elsewhere to argue that solar ovens cannot generate temperatures greater than that of the surface of the sun. OK, this I believe.//

 You shouldn’t believe this argument, because the “heat flows from hot to cold” consequence of the 2nd LoT applies to bodies that are thermally – which is essentially the same as physically – connected, not bodies thermally insulated from one another where one heats the other via radiation, as is the case with the surface of the sun and a solar oven.

 It’s easy, in principle to construct a solar oven hotter than the surface of the Sun: 1st, assure the oven is well-insulated from its surroundings, so it loses heat only by radiation, not conduction or convection; 2nd, make the oven black enough that it doesn’t reflect much sunlight; 3nd, make the oven’s inside small enough that its radiative power at a temperature greater than the surface of the Sun’s is small enough; 4rd, focus sunlight on it equal to that radiative power.

For example, let’s say the oven’s inside is a 1 m radius sphere orbiting the Sun at the same distance as the earth. Using the good ‘ole Stefan– Boltzmann law and some simple geometry and astronomy facts, I get that it would need an efficient circular lens with radius about 435 m.
 — CraigD, Jan 15 2014

[Craig], you might want to check the foregoing annotations a bit more deeply before getting yourself in this one up to the philtrum.
 — MaxwellBuchanan, Jan 15 2014

 [Craig]- There's a limit to how much you can focus light with optics. That limit prevents radiatively coupled objects from heating up to a higher temperature than the source. A simple experiment to prove it, take a decent sized magnifying glass outside, and look at the spot it produces. No matter what you do with it, the spot will never truly become a point. That is because the sun itself is not a point, so the light encountering the optics does so from multiple different angles.

 Also a factor is that //3nd, make the oven’s inside small enough that its radiative power at a temperature greater than the surface of the Sun’s// doesn't work. The radiative power at a given temperature is defined by Planck's law, and approximates a black body curve, no matter what the size.

So yes, it does follow the second law of thermodynamics, is the moment the two bodies equilibrate, they are radiating the same amount of energy from sink to source as from source to sink.
 — MechE, Jan 16 2014

 [MechE] //There's a limit to how much you can focus light with optics//

 True, but this limit is a function (the "Abbe diffraction limit", in a vacuum, diameter of spot = wavelength / (2 sin(angle)) ) of the angle the light and the wavelength of the light. Calculate this for an angle near 90 deg and wavelength in the visible light range (.0000004 to .0000008 m, where most of the Sun’s power is), and you’ll get a value of about half the wavelength of the light, much smaller than the 1 m sphere in my example. Were these values not in this range, it wouldn't be possible to burn microscopic features using focused light, like the optical disk burner in a typical computer does.

 //Also a factor is that //3nd, make the oven’s inside small enough that its radiative power at a temperature greater than the surface of the Sun’s// doesn't work. The radiative power at a given temperature is defined by Planck's law, and approximates a black body curve, no matter what the size.//

 A typical arrangement and integration of Planck’s law gives not the total power radiated by a body of a given temperature, but the power per area (eg: W/m^2). The total power radiated by a body – its luminosity – depends both on its temperature (per Planck’s law) and its surface area.

 Thus the Sun, with a surface temperature of about 5780 K, has a luminosity of about 3.8 x 10^24 W, while a small blob of melted tungsten (1 ATM melting point 3683 K, boiling 5933) filament heated to the same temperature radiates only on the order of 100 W.

 //So yes, it does follow the second law of thermodynamics, is the moment the two bodies equilibrate, they are radiating the same amount of energy from sink to source as from source to sink.//

 Yes, but the Sun and a solar oven are not a closed system in thermal equilibrium.

 My last ano wasn’t intended as any sort of poke at the laws of thermodynamics, but to refute the claim that “solar ovens cannot generate temperatures greater than that of the surface of the sun“.

Ignoring material limitations (ie: a small body of ordinary materials at 5800 K would be hard to hold together!), in principle, the temperature to which a body can be heated by the light of the Sun is limited only by the total power (luminosity) of the Sun. As the total entropy (disorder) of the universe isn't decreased by such an arrangement, it doesn't violate the 2nd law of thermodynamics, or any other physical law.
 — CraigD, Jan 16 2014

 It doesn't matter whether the bodies are in 'physical' contact, or linked by radiation. If you think about it, most ordinary objects can never touch each other directly; energy can only pass from one to the other via fields, particularly electromagnetic ones. Increase the separation, and the very best you can do with optics is to make them behave as if they are in intimate contact.

Lasers are limited by the Abbe diffraction limit, but the concentration of sunlight is limited by the size of the sun. You can't concentrate light from a diffuse source to an intensity greater than that of the source.
 — spidermother, Jan 16 2014

 See my simple experiment. The limit is not the diffraction limit, it is the conservation of etendue.

 Simply put, rays coming from the left side of the sun hit any given point on your optics at one angle, and rays from the right side of the sun hit at a different angle. The result is that those rays are directed differently, and the amount of that difference at the focal plane is the minimum spot size.

 And you're right to say that the total power is a function of temperature and size, but the power per unit area is purely temperature. With light from a black body source (such as the sun), and per my above anno, you cannot focus more area on the target than it has. Therefore, what matters is power per unit area.

And no, the target and the sun are not a closed system in thermal equilibrium, but since there is not a source of power other than the sun with regards to a solar furnace, they can be treated as such, with any imperfections in that assumption only serving to degrade the temperature of the target, not increase it.
 — MechE, Jan 16 2014

I believe that it is hard to grapple with this question unless you are willing to actually calculate the number of photons involved in the communication. If the number of photons passing entering an area goes up by 10x then people are right to ask "Where did all those photons go, why didn't they become heat?". Questions of optics are very interesting but they aren't actually critical, a sheet hung between two suns gets no hotter than a sheet dipped in one sun.
 — WcW, Jan 16 2014

I'm getting deja vu all over again.
 — MaxwellBuchanan, Jan 16 2014

 A sheet placed between two suns will not get as hot as a sheet dipped into a sun, because the total impinging energy is less. Power falls off with distance.

With ideal optics, you could bring it back up to the same level, but no farther than that. Optics do matter.
 — MechE, Jan 16 2014

What? I thought the whole point was that they did scream, that space was filled with screaming people. Otherwise how would you know you couldn’t hear them?
 — pocmloc, Jan 16 2014

If you were close enough to a person screaming at you, you'd hear it: the air molecules would bounce off your helmet.
 — FlyingToaster, Jan 16 2014

True, but only if you were not also screaming.
 — pocmloc, Jan 17 2014

Who said you were getting a helmet [FT]?
 — AusCan531, Jan 17 2014

'tis in my contract, near the bits about fresh grapefruit juice and decent coffee.
 — FlyingToaster, Jan 17 2014

 There’ve been no new annotations in this idea for a while, but the principle at issue concerning the claim that the 2nd law of thermodynamics prohibits a solar oven – a body heated by radiation from the Sun – from having a temperature higher than the surface of the Sun’s –is important enough, I think, to warrant an attempt to revive it.

 It’s correct to read the 2nd law as stating that a closed thermodynamic system consisting of the surface of the sun and a smaller body – the solar oven – will eventually have the same temperature – that is, that the system will reach thermal equilibrium.

 Were we by some means able to close off the system from the rest of the universe – say enclose the Sun and the furnace perfectly reflective vessel, and stop the Sun’s conversion of mass into energy (the nuclear fusion that occurs in its core) without changing its size, the Sun (from corona to surface to core) and the solar oven would reach equilibrium, all with the same temperature. No trickery, such as changing the shape of the vessel or adding lenses to the system, would change this

The mistake in concluding this applies to the actual Sun and a solar oven is that such a system is not closed. Energy is continuously added to the system by the fusion within the Sun (counterintuitively, although this occurs within the geometric inside of the system), and removed from the system by radiation leaving it. An open system is not required by the laws of thermodynamics to reach thermal equilibrium, as witnessed by the obvious fact that we and everything on Earth is not the same temperature as the surface of the Sun.
 — CraigD, Feb 08 2014

So if you put a thermonuclear device into your solar oven, you could get your food hotter than the sun?
 — pocmloc, Feb 08 2014

 Yes, [Craig].

 But I believe (reluctantly) the geometric argument made extensively in the foregoing (and going and going) annos.

I am not happy about it. I don't like the fact that the second law of thermodynamics depends on some seemingly unrelated failings in geometry, but that appears to be the way the universe works.
 — MaxwellBuchanan, Feb 08 2014

 [CraigD] You are right; but I don't think it makes any difference. What matters is that, for the purpose of this discussion, the sun has a temperature.

 I don't see a problem with talking about a solar collector at thermal equilibrium with the sun (or at least, arbitrarily close to thermal equilibrium), even though the system as a whole (sun and collector) is neither a closed system nor at thermal equilibrium.

Of course, unlike a theoretically perfect solar collector, the Earth is far from thermal equilibrium with the sun; it is linked by radiation both to the sun and to the rest of the universe, so its temperature is somewhere in between.
 — spidermother, Feb 08 2014

 [Craig] Yes, it's possible to increase the total heat in the system, but doing so increases the temperature of the sun as well. (e. g. simply wrapping an insulating blanket around the sun).

It's still not possible for the solar oven to be hotter than the current temperature of the sun.
 — MechE, Feb 09 2014

That's not a valid argument, [bigs]. The person outside the sphere will be in darkness, so they won't be able to find the window handles. I'm surprised you overlooked that.
 — MaxwellBuchanan, Feb 10 2014

 //All the energy of the sun hitting the target.//

 You've got the basic concept, but the above isn't quite right.

 In your design, assuming perfect mirrors, more of the energy would be returned to the star in question, and only some fraction would go down the mirrored passage to the target. The exact amount would depend on the size of the mirrored passage. In the extreme case, where the passage is essentially the diameter of the spheres, and you form an ellipsoid, then 100% of the sun's energy heads towards the target.

 This point is critical, because it is what lead's to the conclusion you eventually reach. That is, if you open a window, and direct a tunnel from a second star into it, that tunnel can funnel, at most, the same fraction of the second star's energy that the window is no longer directing from the first sun, thus exactly replacing it if the two stars are identical.

(P.S. Now 100 posts or mentions of my SN in this idea, can we just let it die already?)
 — MechE, Feb 10 2014

Don't the stars cast a little bit of light? At the very least you could see the handles silhouetted against them.
 — pocmloc, Feb 10 2014

 Problem is it's not really intuitive; hold your hand up to the sun and it feels warm; use a magnifying glass and it's warmer.

 Look at it this way: if the sun were inside out, ie: a spherical shell with vacuum inside, and you put a target inside that radiating shell... would it get hotter than the shell ? _Could_ it ? even if you employed all sorts of optics. Can you make the food inside an oven hotter than the oven ?

Then why would you expect a sphere 93,000,000 miles away to do that for you ?
 — FlyingToaster, Feb 10 2014

Of course, that being said, a 20,000 km span parabolic mirror could dump enough energy into a 1cm ball bearing to be the equivalent. :)
 — FlyingToaster, Feb 11 2014

 Math for trying to heat up a 1cm ball-bearing of black-body unobtanium to the temperature of the surface of the Sun, at the distance of the Earth:

 - Power output at the surface of the Sun is 6,250 W/cm² (stated like that it almost seems reasonable, doesn't it)

 - Power received at the Earth is 0.136 W/cm²

 6,250/0.136 = 45,955.9

 To keep it straight up we'll calculate using the entire surface of the ball-bearing, which is 3.14 square cm.

 So we need 144,374.7 square centimeters of sunlight.

 That works out to a parabolic mirror of 428.75 cm diameter or roughly 4 and a quarter metres.

This seems to slightly contradict my estimate of the previous annotation, which was based on temperature at the Earth and Sun with a 4th power in there somewhere.
 — FlyingToaster, Feb 14 2014

Please check the maximum possible concentration for your parabolic mirror, remembering that the sun is 0.25 degrees as seen from the earth. That's going to limit the maximum size of your mirror relative to the size of your target.
 — MechE, Feb 14 2014

If the object follows black body behavior, no matter how little or much radiation it receives, given perfect insulation it will eventually reach a photonic equilibrium. That is to say that it will eventually be the same temperature as the the frequency of the radiation allows. This means that radiation has a specific "temperature" a fact established by this very test. In this way it matters not at all how the radiation is collected, or even how much of it you can collect, the terminal temperature is determined by how closely your system retains the energy. If you are confused about where the extra energy and information goes when you pour it into a black box that never gets any hotter, fear not, the box still contains all that you put into it, only eventually every photon you add to the box is translated into a distribution of photons in the box in which the bounce around until they escape the inlet. Eventually given a perfect black cavity the light escaping is equal to the light entering, the contents of the box are at the "temperature" (frequency) of the light and the system is in photonic equilibrium, just as the areas deep in the photosphere of the sun are.
 — WcW, Feb 14 2014

 No. Just no. Yes, the black body will reach equilibrium. No, it has nothing to do with the frequency of the radiation involved.

We've been over this. The more radiation you put in, the hotter it gets. This is a function of the energy balance, not a property of the radiation.
 — MechE, Feb 14 2014

We can and have demonstrated that that isn't the case. Pumping red photons into a perfect cavity does not produce violet photos, and the temperature inside said box does not rise infinitely, it rises to a specific temperature then stops.
 — WcW, Feb 15 2014

 I thought we'd agreed that a black body was a magical object that reradiates the power being tossed at it vs. its surface area. Okay, I agreed... I may have also been the only one in the conversation. But it makes sense.

 //recalculate... 0.25 degrees//

 Estimated (finely) as [ tan 0.25º x (428.75cm / 2) ], the maximum spread is 0.935cm on a 1cm target. We're good.

 There's more legroom than what it looks like too, since that's the edges of the parabola: the middle sections are much closer to the target.

Though it is interesting to note that apparently you couldn't get *that* much hotter.
 — FlyingToaster, Feb 15 2014

Yep, now take that property (not magical, simply never prefect) and turn it inside out to produce a trap for light, an absorbent cavity. Now, with proper insulation, you no longer need to worry about collecting a lot of light at once, you can wait as the trap gradually comes to perfect equilibrium with the source. In this way you could achieve the temperature of the sun without any mirrors or lenses.
 — WcW, Feb 15 2014

This is pushing a child on a swing, with the same energy, at random intervals. There would have to be some seriously clever timings ( and a special swing set) for the child to do a three-sixty. Can a black body 'ignite' ?
 — wjt, Feb 15 2014

 //We can and have demonstrated that that isn't the case.//

 Please show me where that was demonstrated, because it wasn't anywhere in this conversation. In fact, I can clearly point out where the contrary was demonstrated in at least two cases (microwave and laser).

 I can also point out that if you put in exclusively red photons, even if all it does is produce the same temperature that produced the peak in those red photons, you will get out some violet out, since while the upper end of a black body emission curve trails off quickly, it still does have an upper end. Therefore, your basic argument fails.

 The extension also fails, in that, unless there is a geometric limit, I could select for those violet photons, concentrate them in a new black body, and get a higher temperature. Repeat indefinitely, and you've got gamma rays shooting around.

Energy matter. Wavelength doesn't. It's that simple.
 — MechE, Feb 15 2014

Why does everything come down to 'significant degree' ?
 — wjt, Feb 16 2014

MechE- Your assertion is that thermodynamics doesn't apply to microwave sources or lasers and that we can easily build a Maxwell's demon that will shine out blue light if we shine red light in the other end? The first part seems very much as impossible as the second part and I am very interested in understanding more about both. Specifically, describe how a passive device can cascade low frequency light into high frequency light, achieving a net loss of entropy. Such a device would certainly win you at least one Nobel prize.
 — WcW, Feb 16 2014

 I'm trying to figure out how you possibly get that from anything I said?

 In thermodynamics, energy is what matters, not wavelength. It's possible to put more energy in at a given wavelength than would be produced by a black body at that wavelength. The net result is a target hotter than your input temperature. As I have said repeatedly, this cannot be done with a black body source, since said source outputs in proportion to it's input temperature, but it can be done with a non-black body source such as a magnetron or a laser (or even an LED).

 And no, I'm not claiming maxwell's demon is possible. Your assertion that wavelength is what matters implies that it is possible. I'm arguing that it isn't, and therefore your assertion is wrong

 Given your assertion, a (perfectly insulated) black body fed only red light will eventually heat to red hot. However, a black body radiates a spectrum, including some light that is shifted about the peak temperature. If you place a band pass filter in the output of your black body, you can isolate that higher energy light. Then, according to your statements, you can use a series of mirrors and lenses to isolate that higher energy light, and direct it into another black body. Repeat, as needed, and you get a higher wavelength output than you used as an input.

 Since your supposition allows this, there must be something wrong with your supposition. There are two things wrong. The first is your assertion that wavelength is what matters. Since energy is what matters, the first portion of this is possible. It is possible to isolate a (very limited) amount of violet light from a red hot source. The return on energy will be very small. If this isn't immediately obvious, simply put a blue optical filter in the sunlight. The total energy that hits the ground under the filter will be lower than full sunlight, but the energy per photon will be higher. This is no different than filling a higher pond through wave action. The very tops of the waves can flow over a dam and end up in a body of water that has an average height higher than the average height of the ocean it came from. The total wave energy expended to do this, however is greatly in excess of the gravitational potential energy of the higher pool.

The second error is your belief that light from a non-point source can be infinitely concentrated. This is what prevents you from reconcentrating the filtered blue light so it is more energy dense than the input source. Since etendue is conserved, you are stuck with the reduced intensity, higher energy light, and nothing you can do will reconcentrate it to a higher energy than your input source.
 — MechE, Feb 16 2014

Why can't I just use solar power to heat something to a higher temperature than the sun? What's the difference?
 — Voice, Feb 16 2014

 The point that you seem intent on ignoring is that there is no limit on the number of photons you can put into a box. There is no limit to the number of photons that empty space can contain, and there is no limit on the number of photons that "material" can contin. You cannot pump order into a system using disordered information. A box does not become infinitely hot when you pump low ordered light into it, it simply has more and protons with the same level of order. Similarly an object within that box cannot arrive at a higher level of order (become infinitely hot) because it is similarly limited. The only way to force a higher level of order using radiation is to "order" it by arranging it by wavelength into more ordered forms. In this way the low ordered microwave information from a low ordered source cannot be used to produce highly ordered (infinitely hot) atoms. Similarly the descent into entrohpy means that the black cavity filter arrangement you propose cannot translate disorganized ifrared light into disorganized ultraviolet light.

We seem to agree on the basic premise, but disagree about the implications of entropy. Photons do not have an inevitable terminal relationship with mater that they pass through. No matter how many photons you put into matter or space you cannot increase their total order. This is the fundamental limit of thermodynamics, not entendue. Entendue is made completely moot by the black cavity experiment.
 — WcW, Feb 16 2014

 //Why can't I just use solar power to heat something to a higher temperature than the sun?// Because there's no material that can stand up to that temperature without blowing apart long before it gets there.

 A few annos ago I calc'd that (if such a material or containment method existed) you could do it in Earth orbit, given the amount of energy available, but not by much. I didn't run it for the Earth's surface (where most of the energy has been absorbed by the atmosphere) because that would have involved calculus which is something I prefer to admire from a distance. Feel free; the important bits are insolation and divergence angle, plugged into a paraboloid equation.

 / red in, blue out / again, you can, with some extreme codicils. If you pump a frequency line of red into a black-body-box, what will come back out the hole is a whole spectrum shaped like a black-body curve, based on the amount of energy you're pumping in: nothing to do with the frequency being pumped in, just the overall amount of energy. This will include, but is not limited to, blue.

 If you put a red-pass filter on the pinhole then only red will come back out, until such time as the box runs out of energy. If you keep pumping in red then eventually the amount of red coming out will equal the amount of red going in and you won't be able to get the inside any more excited.

 If you put another hole in it with a blue-pass filter, then red will come out one hole and blue out the other. The stuff inside will continue to bounce around until it all seeps out the red and blue holes.

 //describe how a passive device can cascade low frequency light into high frequency light// Two low-frequency photons hit a molecule which says "thankyou" and spits out one higher-frequency photon. Nothing to do with black-body radiation, it's called "asymmetrical optics". Some laser pointers use it to bump up their output into the visible range. One Nobel please.

[ edit: it's actually called "non-linear optics" ]
 — FlyingToaster, Feb 16 2014

yes, for the thousandth time laser light is coherent. It is more ordered than light from incoherent sources and thus is able to be translated into other forms of energy that are less ordered. What matters is the overall level of order, much like stacked waves coherent photons carry more information and energy than their incoherent bretherin.
 — WcW, Feb 16 2014

They've done it with incoherent light as well.
 — FlyingToaster, Feb 16 2014

"they" have?
 — WcW, Feb 16 2014

 I did have to look it up to justify the principle of "Maxwell's Demon's Refrigerator" where some of the BB radiation inside the box is upconverted and escapes through a highpass mirror, making the insides colder over time.

I don't recall which of the articles I read (skimmed).
 — FlyingToaster, Feb 16 2014

I suppose my inability to google it explains why the universe isn't full of UV radiation.
 — WcW, Feb 16 2014

"Up-conversion with non-coherent excitation by sunlight with wavelengths between 550 and 700nm is demonstrated." [S Baluschev - 2008]
 — FlyingToaster, Feb 16 2014

 Because energy is conserved, and down conversion is more probable. Contrary to what you appear to think, the fact that light can increase in energy when interacting with matter does not result in a UV catastrophe, since there are other mechanisms in play. Please look up Raman scattering, and the relative values of stokes and anti-stokes scattering, and how they vary with temperature, which is one such mechanism.

 Again, you exclude laser light for no valid reason, you exclude microwaves for no valid reason, and you've failed to explain how a monochromatic (or narrow) input spectrum produces the full spectrum a black body source would output.

Let me put this very simply. Your basic claim that matter stops absorbing photons of a given wavelength as soon as it reaches a certain temperature is false. It has been repeatedly shown to be false in this discussion, and you have failed to provide any evidence that supports it. Do so, and there may be some point in continuing this discussion.
 — MechE, Feb 16 2014

We seem to agree about everything but the results. And when it comes down to that suddenly you say "the proof is there" but you can't point to it. I go and look for what you are talking about and all I can find agrees with my postulation, Every paper, every measurement. Thermodynamics is sound, the information model for energy is sound. I'm not the one shouting case closed, I'm the one pointing to a large body of observations and pointing out consistency. The trend is always to descend in order. Radiation that exhibits lower levels of order may be scattered translated into a mixture of photons of higher and lower levels of disorder, but the trend is always biased to more photons and less order.
 — WcW, Feb 17 2014

Feynman said something about transparency in a lecture... I'm guessing its WcW's link. (warning: the audio isn't synched)
 — FlyingToaster, Feb 17 2014

 The proof is a microwave producing objects heated to the near infrared. The proof is in anti-stokes raman scattering. The proof is in the fact that black body radiation is a distribution, not monochromatic.

 There may come a point where an object can't absorb more energy in the form of photons, but it has nothing to do with wavelength. I'm not entirely clear on that, but if it occurs, it occurs in the range of plasmas, not our red hot black body.

 And you are applying thermodynamics incorrectly. You are claiming a general trend to disorder in a pumped system. If more energy is pumped in, order can increase. This can't happen with a black body source because more energy can't be pumped in, the optics don't allow it. It can happen with a laser, or a microwave and so on. Your attempts to exclude microwave ovens, in particular, are in the form of special pleading, always suspect in a physics argument.

And as far as all of these articles you claim support your information, you have failed to link to a single one. None of the ones you have linked to actually state any such thing.
 — MechE, Feb 17 2014

I am applying thermodynamics perfectly. Thermodynamics states that there is no way to increase the overall level of order in a closed system without adding more ordered ingredients. The contents of our closed box photon trap can never become more ordered than the input radiation because in every interaction the statistical result is a lower level of order. The contents of the box are moot, the space inside the box is moot. No matter how many photons it contains it will always be able to contain more and those more photons will always be of a lower average order than their precursors. (though there will be more of them, in complete respect to the conservation of energy and information) your argument is undermined by every observation that you add, because you choose to see only the fraction that appears more ordered, ignoring the inescapable fact that in every case the observation includes a decline in order. The coffin nails, as it were on your case are in the reciprocity of photon interactions and if, in fact the property of entropy and order is conserved between the energy of photons and the "temperature" of atoms. Since you appear to agree with observation that mater does not absorb photons above a certain energy (plasma! not plasma! moot point) you are in complete agreement with the second law, my points and the whole scheme and you don't even realize it.
 — WcW, Feb 17 2014

 It is not a closed system. More energy is being added. And added energy allows for a local decrease in entropy.

 And it is not a moot point, since you are arguing that it is a function of wavelength, not the complete ionization of the atoms in question. Ordinary matter emitting infrared is never saturated.

Let me repeat. Explain microwaves heating something to a red heat. If you can explain that, we are getting somewhere.
 — MechE, Feb 18 2014

If you put metal into a microwave it is clearly able to produce much more exotic and highly organized radiation than "red heat". It is indisputable that the entropic properties of radiation are not set by the frequency of individual photons, but by the order across the system. When that spoon you left in the soup is making a crazy light show inside the microwave, producing UV, and god knows what else, the level of order and entrophy contained within the system is always in the decline, always descending from the order of the input. The output of a magnetron is not incoherent radiation thus it carries an order of magnitude more energy (order) than microwave emissions from a hot body. This is why a small number of microwaves from an oven can cause burns. The sun is not a giant microwave, it is a black body, emitting incoherent radiation.
 — WcW, Feb 19 2014

 All the waves from a magnetron (once they actually get into the chamber) carry is more energy, which is exactly the argument I'm making. Each individual photon doesn't have more, it's just a higher photon flux. Your argument that coherent radiation is somehow different is simply wrong. The energy per photon is exactly the same of the wavelength is the same. That's sort of inherent in the definition of a photon. Therefore the energy of each photon interaction is the same. The only advantage of coherence is that it allows more energy to be directed into a finite point, because the source is effectively at infinity, allowing higher concentration (see my link on conservation of etendue).

Also, entropy increases in a closed system. If energy is coming from the outside, it is not a closed system. If that weren't the case, it would be impossible to synthesize complex molecules by adding energy. Therefore, if additional energy is available you can get spontaneous decreases in entropy. Again, look up what happens to the relative rates of stokes and anti-stokes Raman scattering when the temperature of the target increases.
 — MechE, Feb 19 2014

In effect we cannot build a system that passively increases order, the contents of our reflective box can never allow it to translate 100% of the input radiation from a lower order to a higher order. As a result the "temperature" of the contents is limited to the median order of the input radiation. If it were otherwise a reflective ball with a bead of carbon in it would be able to translate red light into violet light perfectly.
 — WcW, Feb 19 2014

 No. No. No.

Are you even reading my comments? Once energy is added, it is not passive. A heat pump decreases entropy locally through the use of outside energy. Complex chemical synthesis does the same. Heck, evolution does the same.
 — MechE, Feb 19 2014

You can't pump order into a system. The photons (energy) you you add to the system don't add order. Every time you force order into one part of the system it comes at the cost of greater entropy elsewhere. Inside the box the balance is always identical to the order (entropy) of the ingredients you put into the box. The energy of the system rises, but the order of the ingredients is perfectly conserved. We cannot design a system which will defeat this. We cannot produce a system that upconverts photons. Your model would allow for this by allowing radiation to freely heat to any temperature. This would clearly violate the second and third laws.
 — WcW, Feb 19 2014

The system cannot become more ordered overall.
 — WcW, Feb 19 2014

unless the universe is contracting.
 — FlyingToaster, Feb 19 2014

"Second Law of Thermodynamics Temporal Violetter"
 — Ling, Feb 19 2014

 Again, you are defining your system to narrowly. Yes, the universe has to get less ordered, but the box can get more ordered. If you put two red photons worth of energy in a molecule, you can get a single violet photon back out. (colors used as an example, exact energy not calculated) This is not likely to happen, but it does happen sometimes. If the total energy in the molecule is extremely high, the probability increases. Therefore, if you pump enough energy into the system, you can drive up the probability of this occurring.

This only works if you have some component actively pumping more energy into the system, but it does work if that is happening. The act of introducing more energy (against entropy) is the step you are missing. And yes, it requires that total entropy increases elsewhere, but so does a generator.
 — MechE, Feb 19 2014

You know that isn't true because it causes a conservation paradox. The penalty for the higher level of order has to be paid inside the box. For every high energy particle you produce you have to dump some entropy into the box and that entropy has to take some form that can never be translated into higher order so it can't be heat. That means that the entropy that you are dumping into the box to let you produce the higher ordered photons has a defined thermal limit. You cannot get all the energy back out of the box in a higher form. Sorry, that's clearly impossible.
 — WcW, Feb 19 2014

 No, because the entropy penalty is paid to bring the box up to it's high energy state. The net entropy of the system can be increased by the production of a high energy photon if the energy in the atom is higher quality than the photon being produced. That is, if the atom is sitting there with most of it's electrons kicked up a band and it's vibrational modes saturated, and a low energy photon comes in, the energy it takes when re-emitted by the atom decreases the energy level of the atom, resulting in a more uniform energy profile. And the effect that puts the atom into that high energy state is pumping in energy against the gradient.

And I never said you can get all the energy back out in a more ordered state. Some of the energy will be absorbed, and is used to maintain the atoms in their excited state. Losses will exist in the form of escaping lower energy photons as well. But a significant percentage of the photons can be returned in a higher energy state.
 — MechE, Feb 19 2014

 Now you know you are wrong and you just won't admit it.

 "Losses will exist in the form of escaping lower energy photons as well. But a significant percentage of the photons can be returned in a higher energy state."

 We aren't letting those photons get away. They stay in the box. Their presence in the box is moot because at their less ordered state they can no longer add any order to the atoms in the box because those atoms as ordered than they are. The matter is as hot as they can make it; It is at the temperature (order) of the input light.

 Ferme.

The excitation of the nucleus (temperature) by a photons is limited to the degree of order found in those photons. This limit cannot be exceeded.
 — WcW, Feb 19 2014

 And again, no. There are at least two (and probably more, but I know these two) mechanisms where by multiple low energy photons become fewer high energy photons.

 One is "two photon absorption" where two photons strike an atom at the same time (well, within the decay period), exciting a single electron two quanta, and then are re-emitted as a single photon. Obviously the more photons in the box, the more frequently this will occur.

 Another way they do it is by anti-stokes raman scatering. This is the result of reflection (not absorption) that happens to interact with an electron naturally moving down between energy levels (typically from a previously absorbed photon), or with a vibrational mode of the molecule (related to temperature/kinetic energy), increasing the energy of the reflected photon. Again, this increases greatly with the temperature of the atom, which increases with the energy trapped in the box.

 Let me make this clear. These are both real things that result in the emission of photons at higher energies than the accepted photon. The are both statistically rare, but both are subject to increase in more energetic systems.

 Therefore, your argument that low energy photons cannot become high energy photons is invalid. They can only do so because some other photon somewhere gave up it's energy, either completely (to heat the molecule in question) or partially (resulting in a lower energy photon). In energetic equilibrium, the latter is more common, so you'll get a very few high energy photons and lots of lower energy photons. In the case of a superexcited environment, where more and more photons are being pumped in, the net result will be lots of high energy photons. Again, this is not a violation of the second law, because the entropy is spent to pump photons into the box against the energy gradient.

Also, I asked for a link that supports your argument. I understand the second law, I would guess better than you do. I also understand it is A) statistical in nature, not absolute, and B) not strictly applicable to a system that has source of outside energy.
 — MechE, Feb 19 2014

Now you are just being coy. You just admitted that we don't get all of the energy out of the box that we put in. What is the fate of the remaining energy? (hint: it can't be heat)
 — WcW, Feb 19 2014

 Actually, you've managed to twist me up, so I've been combining to separate issues. The entropy cost to raise the temperature of the box is paid by the mechanism pumping in photons.

 The entropy cost for the higher energy photons is paid by increased numbers of lower energy photons, or motion in the molecules. Again, the second law is a statistical certainty, not an absolute one on a case by case basis. That means there are lots of low energy photons floating around, but given that we are continuously, pumping in more photons, there are always a few more getting kicked up to a higher energy.

 Your argument is that the energy of a photon is fixed, and can only go down. I have described two mechanisms whereby it can go up. Unless you prove those mechanisms don't exist, then that portion of your argument is false.

 Therefore, according to your argument, we should be able to screen out and concentrate those photons, which by your definition, should allow a higher temperature.

 I counter that by saying that those higher energy photons exist, but it is impossible to concentrate them sufficiently to produce the higher temperature. Therefore, my arguments are consistent with the mechanisms described.

And why can't it be heat? Uniform heat in an isolated system is not convertible to any higher energy form, exactly what the second law requires.
 — MechE, Feb 20 2014

 So we're now up to four things your theory can't account for.

 anti-Stokes Raman scattering

 two photon absorption

 microwave heating to red hot

 IR laser cutting of metal

My stated arguments account for all of these, and you have yet to bring up a single real world example that supports your theory.
 — MechE, Feb 20 2014

You are still dodging the question. If the black box is putting out some higher energy radiation then the entropy piper has to be paid. The fraction that does not become high energy radiation has to go somewhere, but IT CANNOT BECOME HEAT, because then we would be able to get it out as high energy radiation thus it must stay in the box in some form. This was observed by Plank about a hundred years ago. Changing the subject is undermining your case we can, and have, addressed every one of your points and I will again if you will just address this one fundamental observation.
 — WcW, Feb 20 2014

 It is heat. Uniform heat is the highest entropy state in existence. You cannot extract energy from uniform heat. It may generate high energy photons, but those photons cannot be isolated to produce a higher energy result. That is a function of optics, and again, the basis of my argument.

 You can only extract energy from this system because it is hot relative to the outside world, and that's not relevant if we look at the box in isolation. And if we do look at the box in terms of the total system, the cost for the low entropic, high temperature box is paid for by the source pumping in photons. There is no entropy debt here.

I have not dodged a single question. I may have, at times, been unclear about an explanation, but I have always attempted to clarify it. You, on the other hand, have specifically ignored the four issues I mentioned above.
 — MechE, Feb 20 2014

 The underlying question of if a single photon can be made to heat an already hot material to a higher temperature without limit is clearly, unequivocally and resolutely no. And thus an infinite number of photons of identical energy and disorder similarly cannot do so.

 All observations of this rule being violated come in systems where radiation is more ordered than the radiation from a perfect black body.

Radiation can be made to carry more information (energy) than you would calculate from its frequency alone. Ordered radiation can do more work per photon than disordered radiation. The question of how hot a thing will get when exposed to radiation has two factors, the first is the degree of order found in the radiation and the second is the efficiency with which the atoms heated convert this order into order in their own excitation. The baseline energy of a photon is set by its frequency, but the amount of information that photons may carry when ordered is many times greater. four photons cohesive photons of a particular frequency may carry far more that 4x the energy of their frequency and thus we may build cutting lasers, microwaves, etc.
 — WcW, Feb 20 2014

 //The underlying question of if a single photon can be made to heat an already hot material to a higher temperature without limit is clearly, unequivocally and resolutely no.//

No, it is not. That is your argument, but you have not supported it. If you have an atom at thermal equilibrium, and you hit it with a photon, it will heat up. If you hit that material with a steady stream of photons, it will heat up until it reaches a new thermal equilibrium where it is emitting as much energy as it is absorbing.
 — MechE, Feb 20 2014

Yeah, I have to say I agree with [MechE] on that one. If I have a target and hit it with 100 red photons, it'll heat up by the same amount as if I hit it with 40 (or however many) blue photons with the same total energy.
 — MaxwellBuchanan, Feb 20 2014

No, it cannot. A photon has energy equal to Planck's constant times the speed of light over it's wavelength. It does not matter if it is coherent or incoherent, that is a fixed value. There is no increase, singly or in aggregate from coherence. The only thing coherence does is eliminate interference and dispersion. This means that a coherent source maintains it's structure over a (nominally) infinite length, has an effective aperture at infinity, and thus can be much more tightly focused than a regular source. Again, if this is not correct, cite a source.
 — MechE, Feb 20 2014

 — WcW, Feb 20 2014

Your ability to hit an atom with a photon declines the hotter it becomes. At a certain energy it becomes practically impossible, unless you are able to dramatically increase the likeliness of an "exotic event" such as two photons hitting it in the same phase in rapid succession. Cohesive radiation can do more work, given the correct receiver, because the interaction is statistically more likely to cause "exotic events". Thus it is more ordered, thus it may deliver more energy per photon.
 — WcW, Feb 20 2014

 No, you did not, or not exactly. That that is saying is that multiple photons, acting together, can have effects that a single photon cannot. The advantage of a coherent beam is that multiple photons will, inherently and consistently, act together.

 //a sufficiently intense but low frequency electric field could produce an electric field of magnitude 100 MV/m, which is enough to ionise atoms, even though one photon might not have nearly enough energy for ionisation.//

 Note the reference to the consistent energy of the single photon at the end. Whats important to realize here is that even with incoherent radiation, it will sometimes be the case that the same sort of field can be produced in a localized area. By definition, incoherent radiation will have randomized peaks and valleys in it's field strength. This is what makes something like two photon absorption occur.

 Also relevant is the line:

 //The point here is that the intensity of the radiation produced by the magnetron or klystron in the microwave oven is much greater than that of its thermal radiation.//

It is in fact possible to have have EM radiation of a greater intensity (meaning more photons in the same space) than that from a black body, and that is exactly what is needed to heat a body to a greater temperature than it's color temeperature. However, it is not inherently neccesary that the energy be coherent, and LEDs are a source of such higher intensity, non-coherent light.
 — MechE, Feb 20 2014

 //unless you are able to dramatically increase the likeliness of an "exotic event"//

 Such as by increasing the flux of photons through the space of the atom? Also, as long as the atom is emitting light at a given wavelength, it can absorb a photon at that wavelength, and once a photon is absorbed, there is a chance of it's energy (or some of it) being transferred to the atom rather than re-emitted. If it is absorbed, then that same quanta is available to accept another photon of the same wavelength, just as if it was re-remitted.

 Please note that the reverse of this is also true, and energy can be added from the atom to a photon. This is (one of) the mechanisms that produces higher energy photons. It will continue to happen until the atom is in equilibrium with the high photonic flux.

 //thus it may deliver more energy per photon.//

No, see my above statements. It can more consistently deliver multiple photons.
 — MechE, Feb 20 2014

No, ordered photons from lower entropy sources contain more total information, they are more ordered and thus they can do more work. This is a conservation principal; they have to be able to do more work.
 — WcW, Feb 20 2014

 An individual photon cannot, because it is not "ordered" in and of itself. I agree, you've made your point that the coherent beam can do more work, simply because an atoms is statistically more likely to encounter multiple photons acting together.

 However, that doesn't prevent a disordered source from doing the same thing, just not as frequently. The same mechanism still exists, and still acts the same way, it just does so only when a given alignment occurs, which it will on some occasions (see "two photon absorption). Thus the total intensity of the disordered light must be higher to get the same effect, but it can get the same effect.

And that still leaves Raman scattering, which doesn't require multiple photon interactions at the same time.
 — MechE, Feb 20 2014

Well then I am going to agree with you that otherwise incohesive photons of light packed very close together and traveling in the same direction are more ordered than black body radiation. Thus these "intense" or collumated sources of light have more information and thus more heat than black body radiation because they are inherently less entropic than the same number of photons coming from all angles, in this way a cutting beam of radiation may reach a higher temperature than the same material in a photon flux of the same intensity. In each and every example however the radiation carries a strict limit to the order it delivers to the system and this is the fundamental limit that ties together optics and quantum mechanics. It limits how much up scattering can occur, and it prevents the overall order of the system from increasing above that of the sum of the ingredients added to it.
 — WcW, Feb 20 2014

 We are so close to accord, so close that I will try to tie raman scattering in. Raman scattering is the up-conversion or down-conversion of a photon through interaction with an atom. In this case the photon steals from or deposits some order from the heat of the nucleus.

If we think of the black body as a system that takes ordered radiation and distributes it into a bell curve of disordered radiation above and below the average order, then a Raman Scatterer would do a very similar thing, except the scatterer would resist certain energies, displacing radiation from those energies to higher and lower states but still producing the same area above and below the average order of the input radiation.
 — WcW, Feb 20 2014

 I'm tempted to leave it there, but the underlying argument is still wrong. Focused black body radiation and focused LED radiation are identical, except that LED radiation can be more intense than black body at a given color. It's not that the photons are more ordered, it's simply that there are more of them.

 And there is no inherent property in photon/atom interaction that limits it to a set temperature. More photons will increase the temperature, partly through higher multiple photon interactions, and partly simply through an increase in partial or complete photon absorption. Once that happens, the atom will increase in temperature, and it will emit higher energy photons.

 This is where the fact that we are pumping more energy into the system matters. Entropy is conserved because the generation and concentration of intense light is a low entropy process, so even if the individual photons are low energy/high entropy, the total focused amount is high energy/low entropy.

The basic mechanism you are arguing, that light can be infinitely concentrated, but not heat infinitely is still wrong. Intensity/energy, not wavelength, is still what matters.
 — MechE, Feb 20 2014

 //Your ability to hit an atom with a photon declines the hotter it becomes.//

Why? If I fire a rifle at randome at a herd of bison (yes, of course I've herd of bison), I'm equally likely to hit one whether they're standing still or milling around chaotically.
 — MaxwellBuchanan, Feb 20 2014

Because very excited buffaloes become transparent to bullets.You get that things can be transparent, right?
 — WcW, Feb 20 2014

I think that any argument that hinges on the existence of transparent buffalo (especially when I specified bison) is doomed to collapse. But do carry on.
 — MaxwellBuchanan, Feb 20 2014

 Again we are very close but there is a thought experiment that clarifies the matter. Consider a perfectly reflective ball containing a photon flux of X representing an energy of Y. The photons are all of an identical frequency and they have a perfect distribution of angles, phases, orders and modes. The space is large enough relative to their frequency that they are not limited in degrees of freedom and there are no optical modes or localizations. Let us also say that there are quite a few, as many as you want but not so many that two photon physics is a significant factor. f phot Into the magic globe we insert a very hot bead of carbon, carbon so hot that it is producing black body radiation of a frequency higher than the radiation bouncing about in the magic globe.

 What happens then?

 In your theory the carbon should be a net absorber of photons, and as a result it should heat up even further, emitting more and more high energy radiation as it translates the high concentration of photons into a lower concentration of high energy photons and a much hotter bead of carbon

 In your theory the hot carbon bead is able to "order" the photons.

 In reality the hot carbon bead is not able to order the photons and it actually cools down to the temperature (order) that they exhibit.

It does not matter what the bead is made of, or how many identical photons are in the magic globe, make it as many or as few as you want. The presence of many disordered photons does not enable a material to increase their order.
 — WcW, Feb 20 2014

 //In your theory the carbon should be a net absorber of photons//

 IFF the photon density is such that the energetic balance of the system favors it. That is if the light intensity is so high that the carbon atom is impacted by more energy than it gives off.

 //The presence of many disordered photons does not enable a material to increase their order.// Yes it can. There is an energy gradient between the photons the carbon atom is "holding" as thermal energy, and the photons present in the sphere. This gradient disappears when both equilibrate, even if it is at a higher than starting temperature. The disappearance of that gradient increases total entropy, even if the photons in question end up at a lower entropy.

Your argument is no different than me claiming the same thing for a cold (below color temperature) bead in the sphere. We both agree that the bead will heat up to a given temperature (although we disagree where). Even if you were correct, and the limit is the color temperature, the entropy of the bead has decreased by being raised to that temperature. This is permitted because the gradient between the photons and the bead been eliminated, providing a surplus of entropy.
 — MechE, Feb 20 2014

 I'm lost at where there can be any argument then. The equilibrium point has to be at the "light temperature" for photons that are otherwise completely disorganized, and at some higher temperature for photons that have any other order, and all the material can do is express that higher order (temperature).

The proximity of photons to one another is not a form of order so long as they do not have other common features. No matter what their density, atoms immersed in them will equiabrate at or below their "light temperature". Freaky exotic packed photon events are equally likely to cool as they are to heat.
 — WcW, Feb 20 2014

 Proximity of photons is not a form of order, but energy density is. Concentrated energy is ordered energy. In the situation you suggested, the concentration of energy (in photons) is more dense than the concentration in the carbon atom. It is sufficiently so that by the time it equilibrates, the temperature of the atom is higher than it's initial temperature.

There is no "light temperature". Light to temperature is an energy transfer and nothing more. No conditions exist on it.
 — MechE, Feb 20 2014

 nature does seem to abhor photon over-population and the solution seems to be that photons get together and from mater. This rather than the up-conversion of photons seems to be the predominant mechanism to maintain conservation.

When I use the word "light temperature" what I am referring to is the black body temperature at which this frequency of light is average.
 — WcW, Feb 20 2014

 There are conditions (limits) on the transformation of light energy to heat. First, statistically, it cannot increase order. Our hot bead of carbon reaches equalibrium when all the radiation is entirely incoherent and it is at the temperature that would cause it to emit the same distribution of radiation.

At this point the volume of radiation is relatively moot, it could increase 10x 100x 1000x and the carbon ball would remain essentially static. So long as the average order of the radiation remains stable the order "temperature" of the carbon bulb remains the same.
 — WcW, Feb 21 2014

 Your first statement is mostly correct. Your second is not. It gives no credit to the fact that increasing the light energy in the system brings the two components of the system out of equilibrium again.

Do you agree with the statement that a system at equilibrium is less ordered (higher entropy) than a system with an energy gradient in it?
 — MechE, Feb 21 2014

 You know what, forget that question. Let me, again, make a series of statements, and you tell me which one(s) you disagree with.

 1. One of the ways photons interact with atoms/molecules is by exciting their electrons a given quanta of energy.

 2. Typically these photons are reemitted with the same energy (rayleigh scattering), however, they can be completely (absorption) or partially (fluorescence, stokes Raman scattering) converted to kinetic energy in the atom (heat).

 3. When any of the three above (re-emission, absorption, or partial absorption) happens, the electron is returned to it's prior energy state, priming the molecule to interact with another photon of the same energy.

 4. Most electrons in a material spend most of their time in their lower energy state, simply because electron/photon interactions are relatively rare.

 5. Increasing the number of photons in an area increases the chance of electron/photon interactions.

 6. The number of such interactions that can occur is therefore dependent on the intensity of the light in the range of that quanta.

 7. The reverse of step two can also happen, wherein thermal energy is converted to photons, or to higher energy photons.

 8. The probability of this occurring is dependent on the kinetic energy (temperature) of the molecule in question.

 9. Thermal equilibrium is the point at which the energy being converted to kinetic energy per 6 is the same as the energy being converted from kinetic energy per 8.

10. Given 9, if the number of photons in the system increases and behave as per 6, 8 requires that the temperature also increase in order to maintain equilibrium.
 — MechE, Feb 21 2014

 1) True

 2) True, but only if the atom has an electron energy state available for one of these processes.

 3) False. If any energy (order) is deposited the excitation states of the electron fields are changed. The excitation of the nucleus forces the electrons into higher energy states where they remain until the nucleus cools . The atom thus less receptive to the same admission.

 4) False. Most electrons spend their time in whatever state is dictated by the excitation (heat) of the nucleus. (This is chem 101 stuff)

 5) False. Electron/Photon interactions are only more likely when there are photons AND appropriate electron energy states available. Since the process takes time a material may easily be saturated with photon adsorptions.

 6) False; The number of electron/photon interactions is limited by the rate at which the electron can move from state to state. Since the number of electrons and potential states is limited this, rather than the number of photons, is the limiting factor.

 7) True, with the caveat that the tendency is always in the direction of greater entropy.

 8) True, spontaneous emission, up-conversion, down-conversion, and Rayleigh scattering all have thermal components. Exotic events need not take any energy though.

 9) True by definition.

 10) False. Once the electron modes at that energy range are saturated the system is at (very near) thermal equilibrium and further radiation simply passes through the material as if it were not present. It it will never reach perfect thermal equilibrium as the material will always "cool" itself slightly by disputatious increasing the entropy of the radiation that it emits over what it receives.

See what you think.
 — WcW, Feb 21 2014

 3. Your response strongly implies my point 8, that re-excitation of the electrons can be driven by thermal energy, but ignores the fact that emission of a photon (even a reduced energy photon) resets the electron energy level to it's ground state. That extra energy is transferred to the nucleus, and no longer effects the electron except as mentioned in 8.

 Since electrons only move in quanta, no intermediate higher state is possible.

 4. Your own theory requires that molecules have significant spare capacity in this regard. Remember that a blue hot black body is not only emitting a lot of blue light (and UV), it is still emitting more red light than a red hot object (the emissions are monotonically increasing at all wavelengths with temperature). It therefore must, when red hot, have extra absorption capacity in that wavelength, which is required to maintain equilibrium if it is heated to blue light.

 In summary, your saturation argument can be proved wrong by the monotonically increasing intensity of all wavelengths of black body radiation as temperature increases.

All remaining objections are derived from that misunderstanding, so I'll leave it there.
 — MechE, Feb 21 2014

 -"3. Your response strongly implies my point 8, that re-excitation of the electrons can be driven by thermal energy, but ignores the fact that emission of a photon (even a reduced energy photon) resets the electron energy level to it's your earlier ground state. That extra energy is transferred to the nucleus, and no longer effects the electron except as mentioned in 8."-

 >>Correct except for one problem, the photon falling to a lower than "normal" energy state has two ways to return to normal. One way is the adsorbtion of a photon, and the other is to do it by taking energy out of the neucleus. In the first case, yes, this would add energy to the neucleus, but in the second case the energy is simply subtracted. The hotter the neucleus the more the balance shifts to net energy loss. <<

 -"4. Your own theory requires that molecules have significant spare capacity in this regard. Remember that a blue hot black body is not only emitting a lot of blue light (and UV), it is still emitting more red light than a red hot object (the emissions are monotonically increasing at all wavelengths with temperature). It therefore must, when red hot, have extra absorption capacity in that wavelength, which is required to maintain equilibrium if it is heated to blue light."-

 >> This nearly unlimited capacity is certainly why you cannot heat beyond the order of light using entirely in cohesive photons. The fact that electrons are producing photons from energy in the nucleus makes them unavailable, rather than more available for photon adsorption. This is the self limiting aspect of the system and what perpetually pushes entropy up. The atom can always say "no, I'm to busy increasing disorder to adsorb another photon of that sort" and there is nothing the photon can do about it. <<

 -"In summary, your saturation argument can be proved wrong by the monotonically increasing intensity of all wavelengths of black body radiation as temperature increases"-

 >>Actually quite the opposite, the monotonicaly increasing emission actually shows that the black body is always going to be at a slightly lower level of order than the surrounding system because it will always be more able to shift photons to higher level of entrophy than it is to adsorb them and shift them to a higher level of order (temperature)>>

You clearly recognize the players here but they all play for the other team. In every case you are pointing out a mechanism heavily biased against your model. Even under the most saturated conditions statistics still favors entropy and the closeness in time for each incident in-cohesive photon can only move the nuclear energy closer to becoming a perfect black body.
 — WcW, Feb 21 2014

There is no correction in your response to my response to point 3. I have clearly stated that emission of energy (which is not the photon falling to "below normal energy level", it is the photon being excited to above normal energy level due to the transfer of thermal energy) is dependent on temperature. But it is only dependent on the kinetic energy/temperature of the atom, and thus is independent of the status of incoming light. Exactly my original point 8.
 — MechE, Feb 21 2014

 And as far as your response to point 4, it is completely wrong. The only mechanism you have proposed for the limit of an atom to absorb energy from incoming photons of a given wavelength is saturation of that energy band. I have just conclusively proven that the atoms are not saturated at, say, a red heat.

 Therefore, if the atom is at equilibrium when it is absorbing the same amount and spectrum of black body radiation as it is emitting, it still has the capacity to absorb more photons of the same wavelength. You have proposed no other mechanism to prevent this. Thus, if those additional photons are provided, the atom will absorb them.

Once it absorbs them, it does contain the additional energy from them. Thus it is no longer in equilibrium. The only possible way for it to return to equilibrium is to increase the energy of the radiation it emits, and the only way that happens, per my original point 8, is if it is at a higher temperature.
 — MechE, Feb 21 2014

 Again I don't need a mechanism, the only way to achieve photon densities high enough to defeat Planks constant is with an ordered source of radiation. Our black body collector is a measure of the order found in the radiation. It turns that order into a temperature, an absolute measure of order. If shining a red laser into it produces a temperature of 12000* then that source of light was very much more ordered than a black body source of the same number of photons at thermal flux. One can produce the thermal energy of 12000* the other cannot, no matter what mechanism we throw at it. That's order, and energy and it takes many forms in the light itself.

The only real disagreement between the systems is that one system(the classical) fails to accurately predict how much energy will be found in radiation and the other provides tools to do so. The classical system also seemed to allow for up-conversion of photons which was in conflict with observation. This was resolved when the additional modes of order in radiation were established.
 — WcW, Feb 22 2014

 Yes you do need a mechanism. Simply saying something can't happen, when they way it does happen has been shown is not a valid argument.

 As far as why you are wrong with regards to entropy, I think I have found the answer, and it is in the fact that your earlier statement //The proximity of photons to one another is not a form of order// is incorrect. I hadn't though out the implications of this until last night, but photons do exert a pressure, and increasing the number of photons increases it. Increased pressure can do work, so it represents an additional form of order not accounted for in your model.

See my latest link on "photon gas". There is a relationship between energy, pressure, and volume and temperature of a black body chamber. If you increase the energy (by adding photons) in a chamber of a constant size, pressure and temperature both go up to reach a new equilibrium.
 — MechE, Feb 22 2014

Looks like we are in total agreement then. When we compress photons in a black body we can hide energy in them without increasing the temperature and then get it out again either by letting them into another space where they "become" hotter or letting them do some work and remain at the same temperature.
 — WcW, Feb 22 2014

 We can also get it out by letting the temperature increase.

And compression is the same as pumping in more photons, what I was saying before.
 — MechE, Feb 22 2014

The only way to pump the photons is to give them more order than thermal radiation. This is the most minimal level of order.
 — WcW, Feb 22 2014

Only to the extent you need to give them more intensity than thermal radiation, e.g. if your black body chamber has a small hole in it, black body radiation will come out at a certain intensity. If you shine a more intense, but otherwise equally chaotic light of the same peak wavelength (such as can be generated by an LED) back through that hole, that has the effect of pumping energy into the system.
 — MechE, Feb 22 2014

 that's not how thermodynamics works, it doesn't necessitate that the sun can't heat something to a higher temperature than itself. The sun's heat comes from the nuclear reactions inside it, which are constantly PRODUCING and radiating energy. So it's constantly producing new heat energy and buttloads of radiation. That radiant energy can heat things when it strikes them. Focusing the radiation with lens-like devices could easily heat them to temperatures highers than the sun. Currently we do not have the technology to build giant mirror arrays in space near the sun, but if you could you could easily be close by, have a really big array bounce energy onto a small point, and make it super hot. Lord knows you could get it so hot as to cause said target to start undergoing nuclear reactions.

The only thing thermodynamics would actually dictate is that if the sun were just a giant hot ball, that was not producing energy and just happened to be a hot giant ball of gas, and nothing was making it hotter, then THAT ball couldn't make anything hotter than it already is. This is obvious even before you ever take a physics class
 — EdwinBakery, Feb 23 2014

 Edwin- Please read this entire conversation. We've covered this in, in detail. You definitely cannot make it work the way you describe, we've spent half a year trying to convince each other on the details of why.

That is exactly how thermodynamics works, you cannot make heat flow passively from one temperature (the surface of the sun) to a higher temperature. The basis that I believe is correct is conservation of etendue, which prevents you from concentrating sunlight in the way you describe. You can get most of that from the relevant link.
 — MechE, Feb 23 2014

All sources of radiation that deviate from a black body distribution also contain more order per photon than thermal radiation does. The classical model that you find so irresistible fails to to predict exactly how much order and energy will be in a given source of non thermal radiation. Energy in radiation is clearly more than a simple count of photons at frequencies: that can only give us the minimum energy. It also gives us the minimum temperature. You cannot pump light. It can only be produced by thermal systems at an absolute minimum of order, or by more ordered systems resulting in light of a higher order and delivering more energy.
 — WcW, Feb 23 2014

 That's an interesting concept...that ordered radiation has more energy, or more efficient transfer of energy. But where does that energy come from?

 Stupid analogy: a swing is pushed by many individual forces at random times. Maybe it mostly swings to a certain angle. Occasionally it might swing a little. Compare to those same forces timed to the swing, and the swing mightily swinging back and forth.

In both cases the energy transfer efficiency is different...just like impedance matching.
 — Ling, Feb 23 2014

 So basically you're saying that my original contention, that non black body radiation can heat an object up to higher than it's apparent color temperature, was right, but you're disagreeing with why?

 Of course, you are still making a major mistake. Order and energy are not the same thing. Two equal bodies at 10 degrees have the same energy as the same two if one is 15 and one is at 5. They do not, however, have the same order. There is a temperature gradient between the two, which allows that energy to be used to do work. Denser or more organized photons may have more order, and thus be able to do more work, but they have no more energy than if they are completely chaotic.

Also, and I just realized this, you are mistaken in thinking that I am working from the classical model. However, the quantum mechanical model does not ignore light intensity. For instance, if you have a PV cell that is being struck by a given wavelength of light, the wavelength determines whether electrons are moved at all, but the intensity determines how many electrons are moved. At no point have I been claiming that the major action is that photons will move an electron if they aren't of a sufficient wavelength to do so (although obviously two photon absorption does allow this on rare occasions). However, that doesn't prevent multiple interactions at one energy level causing enough thermal energy to accumulate in the nucleus to produce an interaction at a higher energy level, if the intensity is such that net energy is transferred to the nucleus.
 — MechE, Feb 23 2014

 A black body can (by definition) absorb or emit a photon of any wavelength. That point is missing from the above few screensful. Individual atoms, and some simple molecules, have discrete energy levels, but most other types of matter don't.

For example, as has been mentioned waaay up there^, the glowey part of the sun consists mainly of plasma, not uncharged atoms - which is why it approximates a black body. To expand on the swing-set analogy, the ropes are broken and the swings and swing-sets are careering all over the park. Any push, regardless of its strength or timing, can involve an exchange of energy with a swing.
 — spidermother, Feb 24 2014

 That has been touched on, repeatedly.

 However, almost all matter has discrete energy levels, plasma being the exception, touched on in a bit. There may be hundreds of energy levels or just a few dozen (hydrogen). What's important is that many emitted photons are smeared by the inclusion or omission of multiple different energy levels (commonly vibrational) in a single photon emission. That produces a spectrum of emissions for a single primary band. Pure hyrdrogen does not approximate a black body, because it has too few energy levels and too few of these secondary interactions, and thus produces clean, narrow emission lines. Pure carbon does approximate a black body, because it has many more primary and secondary emission bands.

And plasma closely approximates a black body because it includes a cloud of free electrons, which means that they truly can interact at just about any energy level.
 — MechE, Feb 24 2014

 I stand corrected - most types of matter, particularly those approximating black bodies, *effectively* don't have discrete energy levels, as far as their emissivity spectrum is concerned. I felt that atomic energy levels and black body emission and absorption were being conflated.

Liquid water, glasses, wood, metals, plastics, rocks, chlorophyll - and most other amorphous or complex substances - have broad, or no, emission / absorption peaks across large parts of the electromagnetic spectrum.
 — spidermother, Feb 24 2014

Hang on - I thought chlorophylls (of which there are several) _did_ have fairly well-defined absorbtion spectra?
 — MaxwellBuchanan, Feb 24 2014

Well-defined absorption spectra are not the same as discrete energy levels. (I realise there are several chlorophylls; "chlorophyll" is a common, but lazy, shorthand.)
 — spidermother, Feb 24 2014

Actually, they are. A discrete absorption spectra is due to a discrete energy level in the absorbing compound. However, clorophyll peaks are still incredibly broad when compared to something like hydrogen lines, which are essentially monochromatic.
 — MechE, Feb 24 2014

If we can collect weak sources light and arrive at 90% of temperature X and then collect ten times as much radiation and arrive at 95% of temperature X, and collect 100 times as much light and arrive at 99% of temperature X then it is only best to describe temperature X as the terminal temperature of that light from that source; the terminal level of order that that light can deliver.
 — WcW, Feb 25 2014

 Who said that's all we can do? That statement is not supported by any prior made in this discussion, not even your own.

If the source we can collect from is x, and we collect 5% per unit area of the target, we'll arrive at 5% of X. If we can collect 100%, we'll arrive at 100%. And if, somehow, we could collect 110%, we'll arrive at a temperature that is 110%.
 — MechE, Feb 25 2014

Not true mech. Not true at all. That's a tragic misunderstanding. A black cavity collector can achieve 90% of the radiator temperature even at distances as great as those between the earth and the sun.
 — WcW, Feb 25 2014

Once again, please support that statement. Unless you are talking about concentrated light, the energy simply isn't there to do so. If you are talking about concentrated light, then of course, but only if the concentrator is sufficient to bring the incoming light density up to 90% of what's found at the surface of the sun.
 — MechE, Feb 25 2014

Read the idealizations section on the wiki about black bodies. If given enough time a back body will come to thermal equilibrium with the source of radiation even if the number of photons is very small. There are factors in efficiency, but it is exactly as I have described. 90% of T at this distance. 95% of T at half the distance, 97% at half again, so on, so forth.
 — WcW, Feb 25 2014

 Oh, an idealized black body, yes. Because it doesn't lose any energy to the outside. Therefore, the only effect of getting closer is increasing the percentage of the "sky" that is filled with your source. The closer you get, the higher the percentage, the closer you get to unity.

Again, it's a limit in the etendue of the system, not an energetic one.
 — MechE, Feb 25 2014

Call it what you like. My description was accurate and yours was not. x, 10x 100x the number of photons and the results are only marginally higher. The property of the terminal temperature is in the photons not in their number.
 — WcW, Feb 25 2014

 Your reference justifies the eventual thermal equilibrium of an ideal black body in isolation. It does not say anything about the number of photons intercepted. Thats the point I was actually asking for a reference on.

 What you are describing is a reasonable, but not exact approximation of what would happen if you moved a black body further and further from a star. The intensity of radiation received by the object falls by the square of the radius, starting from unity.

 At distance X, you get n photons. At distance 2x, you get n/4 photons, at distance 3x, you get n/9. And so on. The number of photons the target is receiving is what falls off with distance, not the temperature per photon.

 This has nothing to do with the ability of the photons to achieve a certain temperature, and exclusively depends on the geometry of the problem.

Remember, that we both agree a black body can't increase the temperature of another black body over the color temperature. But you have failed to indicate why this relatively simple geometry problem in any way prevents a non-black body source from doing so, since it can have an intensity greater than the black body spectrum for a given temperature.
 — MechE, Feb 25 2014

Starlight is hardly black body radiation. All cases of thermal radiation follow the same rules. With an efficient collector there is only a marginal difference between x, 10x and 100x intensity. The terminal temperature is the "light temperature" which is the actual temperature of the source. When you provide for perfect insulation optics become completely moot, unity occurs when the receptor is exposed to photons. The black cavity receptor is the most efficient way of finding the "temperature" of the light. The source of thermal radiation cannot be more efficient than a black body as a source of photons.
 — WcW, Feb 25 2014

 All thermal radiation is approximating black body radiation, and that's how we've been referring to it. When I say non-black body, I mean non- thermal. And barring hydrogen lines, stars more closely approximate black bodies than just about anything else.

 And there definitely is a difference between 1x, 10x, 100x. Energy in equals energy out. And in further double checking the wiki entry, because something didn't seem right, it did not say that the black body reaches any given percentage of the source temperature at a given distance. It says it does reach equilibrium, but not what temperature that equilibrium is.

In fact it will be colder by a factor of the photon flux involved. Energy in equals energy out. That's what matters. If one gamma ray comes in, you'll get a nice distribution of low energy photons out. If a blue laser of the same total energy goes in, you'll get the same distribution out, and if an infrared laser of the same energy goes in, you'll get the same distribution out.
 — MechE, Feb 25 2014

The temperature in the black cavity collector is only slightly below what you find at the source of thermal radiation. For non thermal sources the same rules apply; a black body collector doesn't care, 1x 10x 100x the black body translates it into a final temperature. That means that intensity is, in fact, meaningless given proper insulation. Put in a red laser beam, get out the equivalent black body radiation. The inside will be at "the temperature" of the ordered red photons.
 — WcW, Feb 26 2014

 No it won't. It will be at the spectrum that would be produced by that energy of black body radiation.

 Under your latest theory, if I shine a 1W blue laser into an approximate black body, with an opening of 1cm area, I will somehow, eventually, manage to get out a continuous 7059W. That number is fixed. A black body at 5940 K produces 7059 W/cm^2, with it's peak at 488 nm, the same as a particular variety of Argon laser.

 (And before you say anything about lasers, I can easily focus a 1W blue LED into a 1cm^2 target, same result).

Therefore, you are claiming that, despite conservation of energy, somehow, after sufficient time to reach equilibrium, a constant 1W input will output over 7000 times as much. Congrats, you've completely solved global warming.
 — MechE, Feb 26 2014

Of course I suppose that's offset somewhat by the fact we could put in a 1W maser, and only get out 0.46 x 10^-11 watts. Just wondering where all that extra energy goes.
 — MechE, Feb 26 2014

That's just dumb. If you put 1w of light of any source in you will get 1w out eventually. The higher the order of that light source the higher the order of the output light. It's really simple. Admittedly with very weak sources leakiness and taking a long time both hit you very hard but eventually you will get your 1w out with the spectral distribution from the terminal temperature of your source radiation. The same photons just really far apart ....
 — WcW, Feb 26 2014

 Thermal radiation doesn't work that way.

A given temperature produces a given spectrum with a given power output. If you change any one of the three, the other two change as well. That's the only way black body radiation works.
 — MechE, Feb 26 2014

That's the way it works when you force it to pass through a pinhole mech. Weak tea. The restriction on light getting in and out is perfectly reciprocal. Shine some really ordered light in, and, eventually, you get the same energy out, translated into thermal radiation. This translation process lets you find the absolute order of the inlet light. Hint: It's a limited temperature in every case. Yes the box gets very hot, yes it's a lot of energy to get it there. Once it gets hot though the energy that is leaking out of the pinhole will never be greater than the energy we are feeding in, and at unity, never any less (minus inevitable leakiness).
 — WcW, Feb 26 2014

 Nope. Nope. Nope. Nope. Nope.

 Yes, energy in equals energy out. But that's the last thing you said that was right. And in case it wasn't clear, my prior statements regarding an imbalance were to point out the absurdity of your apparent position.

 So now, your current argument is that the spectrum of the outgoing light is the same, based on the spectrum of the incoming light, regardless of the energy involved, and that is just plain wrong.

 Both the energy output and the spectrum output are completely and exclusively defined by the temperature of the black body. Please look up the Stefan-Boltzman law, which states that energy out is equal to the fourth power of the temperature times a constant. If you then move on to Planck's law, and plug in that temperature, you can find the energy at any given wavelength. Once again, this is a fixed value at a given wavelength for any particular temperature.

 It doesn't matter if your black body is a pinhole or a sphere. I picked the size I did because W/cm^2 is a convenient unit. If I shine a 2W laser in a cavity type black body approximation with a 2cm^2 opening, the numbers are the same as the 1W and 1cm^2 set up. If you shine the 1W laser in, the total power out is the same, the power per unit area is 1/2, and the equilibrium temperature of the black body is 1/16. (that fourth power in the Stefan-Boltzmann law), and the spectrum shifts longer (lower energy) appropriately.

This is one of the fundamental concepts of black body radiation.
 — MechE, Feb 26 2014

It's a pinhole for a reason. A square centimeter hole would be far to large unless the cavity was truly immense. No, think of a real pinhole, a hole the size of the head of a pin. Your argument that a black cavity collector cannot work 'because math' flies in the face of very practical observation. And the spectrum of light is exactly correct for the temperature, it's the intensity that makes it equal to the input. The structure allows us to produce outputs of any intensity/temperature combination and thus produce accurate translations of radiation into energy and thus temperature, no matter how weak the source is.
 — WcW, Feb 26 2014

 When did I say a black cavity collector couldn't work? All I said was that a black cavity collector with opening size x or a sphere with a surface area x behave the same.

And there is no requirement that a black cavity collector use a pinhole, just that the opening be significantly smaller than the chamber of the collector. However, if you insist on a pinhole, it doesn't matter. Take the size of the opening down to 0.1mm^2. I can still focus my incoming 0.1mW laser down to pass in through that opening, and I will still only get 0.1mW out. The total power going in per unit area of of output is the same. The energy flux is the same. The final temperature of the black body is the same, and the spectrum of the emitted light is the same. And it remains the same whether I do this with gamma rays, a blue laser, or an infrared laser.
 — MechE, Feb 26 2014

 //produce outputs of any intensity/temperature combination//

 No. It doesn't. I agree that by varying either the energy input or the size of the output you can get the black body to any temperature, and therefore any intensity, but the two are absolutely correlated. Please read the second sentence of the wiki on on Black-body Radiation.

Again. A fundamental principle of thermal radiation is that temperature, intensity (power/unit area) and emission spectrum are all perfectly correlated.
 — MechE, Feb 26 2014

Yes, but just as with the thermal sources the 1x 10x 100x observation applies. Each source no matter how weak may be collected until the inlet and outlet intensities are equal, in this way any source may be treated as if it were a thermal source. The shift from 1x to 10x to 100x in the balance of the photons is only due to the system being leaky, it's still 95%T, 97%T, 99%T, but the intensity of the output goes up equivalent to the input. And yes, this would mean 10K temperatures from a red light laser, etc. Mostly we don't try to make cavities this efficient because they are VERY SLOW and the very high temperatures tend to destroy the media. Instead it is easy to use math to determine the rate of heating, and by that, what the final temperature would be.
 — WcW, Feb 26 2014

 What "1x 10x 100x observation". Your entire observation there was based on a misunderstanding of how radiation falls off with distance.

 And you somehow still seem to be making the claim that the output wavelength is somehow dependent on the input wavelength.

 And "in this way any source may be treated as if it were a thermal source" makes no sense in this context. Any energy source is equivalent, yes, but that is contradictory to the rest of your arguments that somehow wavelength matters. And it also ignores the fact that non-thermal radiation sources can be more concentrated than thermal energy sources at a given wavelength.

 "Mostly we don't try to make cavities this efficient because they are VERY SLOW and the very high temperatures tend to destroy the media". This is nonsensical. We don't make black body cavities, as a rule, because they serve no practical purpose. And the efficiency of a black body cavity would have nothing to do with the final temperature, it would solely have to do with how much energy is lost through points other than the emission area. Only the energy input determines the final temperature, which is usually not all that high, since the energy input is usually only a few kilowatts per square meter at most.

"Instead it is easy to use math to determine the rate of heating, and by that, what the final temperature would be." Yes it is. The rate of heating doesn't matter, at all. The final temperature will be the fourth root of the energy input over the area of the output times the Steffn-Boltzmann constant, and the radiation distribution would be according to Planck's law.
 — MechE, Feb 27 2014

 Are you saying that a black cavity collector does not work as described? That such a device would not closely approximate the temperature (unity) of a (thermal) source at even a great distance (low intensity)? And that such a device wouldn't exceed unity even if very close to a (thermal) source (high intensity)?

What the hell man. I'm entering into this discussion in good faith.
 — WcW, Feb 27 2014

At which point the geometrical properties of the pinhole need verification: does it pass radiation only as a one-dimensional line, or hemispherically ?
 — FlyingToaster, Feb 27 2014

A pinhole in a flat surface would result in a Lambertian distribution (180 degrees, but strongest in the centre and approaching zero at the edges). This follows from simple geometry: the intensity is proportional to the apparent area of the hole.
 — spidermother, Feb 27 2014

 I'm not sure where you got the impression that a black cavity collector (or any black body) will equilibrate at the temperature of the source. That is only true if the light from the source is focused to the same intensity per unit area on the target that it leaves the source. Yes, black bodies do reach equilibrium, but equilibrium is about energy balance, not equal temperature. And the spectrum and intensity of black body radiation is defined by the temperature of the source. As the radiation spreads out with distance, the intensity decreases, but the spectrum doesn't change. Therefore a black body further away receives the spectrum of the source, but does so at a reduced intensity.

 A black cavity collector at the surface of the sun will equilibrate at 5778K, with an energy flux in and out of 6.319x10^7 W/m^2. The same collector, at earth's orbit, collecting unconcentrated sunlight only receives 1336 W/m^2, and will equilibrate at 391.8K.

 That same collector, still in earth's orbit, but now equipped with a concentrator, such that it is once again receiving energy at the intensity of 6.319x10^7 W/m^2 will once again reach the temperature of the sun. That is the theoretical upper limit of concentration, at no point can black body radiation be concentrated to a higher level than the source. Equilibrium is energy balance, not temperature.

I literally could not make sense of your previous post, the assumptions in it did not make coherent sense. I now understand that you are holding to a false understanding of how black bodies function, and of how the intensity of radiation falls off with distance, and that helps somewhat.
 — MechE, Feb 27 2014

Perhaps it's time to ask someone with a magnifying glass and a thermometer to step outside and settle this.
 — MaxwellBuchanan, Feb 27 2014

Fascinating. The thermalization of radiation by a black cavity is impossible in your model even though it is clearly observed in experiments. Also, as your model allows we reach unity with a mirror here on earth, and we reach unity without a mirror at the surface of the sun but somehow that same mirror does not allow us to exceed unity at the sun. For awhile I though that we were having a regular, sane, disputation about facts, but in my research I happened across a couple of sites that basically represented a community of die hard people just like you who refuse to let go of the notion that "heat transfer" is unlimited. The case against a quantized model, against the notion of order being conserved is made far more eloquently than you make it. It is like the case for a flat earth or perpetual motion, for some reason it has an addictive appeal. Either the observations about what happens in a pinhole cavity with thermal radiation are accurate (that direct thermal radiation allows the radiator and the receiver to come to the same temperature no matter how far apart they are given perfect insulation) or the evidence for that has been faked for over a century.
 — WcW, Feb 27 2014

 Yes, it does prohibit reaching greater than unity at the sun, because the surface area of the sun that can be focused into the target is no greater than the open area of the target. This holds true regardless of how far you are from the sun. Please, again, read the link on etendue, particularly the portion towards the bottom on "maximum concentration".

 And can you please point to a single experiment that shows the temperature of a black body at a distance from the source approaches the temperature of the source, without concentrating light? I would really like to see that, because in my own research, I have seen no such thing. Thermal equilibrium is reached when the energy passing into and out of the black body equalizes, not when the temperatures are equal. So no, thermalization is not impossible under my theory, it just doesn't happen where you think it does. (To clarify, this is for the case where you have a source black body, such as the sun, that gets it's energy elsewhere, that is the steady state equilibrium of the second black body, it does not apply to the case where you start with one hot and one cold black body, wherein they would equalize in temperature, but at a lower temperature than the starting point)

 And I stand by my statement that doing so would violate conservation of energy. Again, a list of statements:

 1. The intensity (energy per unit area) of radiation from a black body is a solely a function of it's temperature. (Steffan-Boltzman Law)

 2. The intensity of radiation falls off as the square of the distance from a black body. (surface area of a sphere)

 3. Therefore, a black body sitting further away from the source black body receives less intense radiation. (1 and 2)

 4. Equlibrium is the point at which energy in is equal to energy out (definition)

 5. The temperature at which the target black body will emit the same amount of energy it is receiving is lower than the temperature of the source black body (1, 3, and 4).

 Which of these is incorrect, and why?

I submit I have been just as patient with your irrational arguments as you claim to have been with mine, and dismissing mine as irrational without supporting your arguments is not to your benefit.
 — MechE, Feb 27 2014

And again, to clarify, I am not now, and never have been basing any of my arguments on the classical model. I feel justified, however, in basing them on the fact that I've already proved that a black body over a range of temperatures is not saturated at any wavelength that peaks at the lower end of that range (Planck's law, and monotonically increasing energy at any given wavelength), and therefore a true black body (which would be at any temperature) is never saturated.
 — MechE, Feb 27 2014

 You know what. It's simple. Your way requires a violation of either Stefan-Boltzmann's Law (The energy out of a black body is solely a function of it's temperature), Planck's Law (The spectrum of a black body is solely a function of it's temeprature), or the law of conservation of energy (the energy into a black body at equilibrium must be equal to the energy out).

 My way requires that you misunderstand how black bodies work, and how optics work.

Which of those two do you think is more likely?
 — MechE, Feb 27 2014

 Talmudic debate - upon author's request.

 — pashute, Apr 23 2014