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Solar Fusion Magnifying Glass

Make It Big, Focus Sunlight on a Stream of Heavy Water
 (+2, -3) [vote for, against]

To create the extremes of high temperature and pressure needed for nuclear fusion, a solar collector would be used to focus a large amount of energy on a small point. The goal is to reproduce the temperature within the sun (~15E6 K) by concentrating sufficient incoming solar light (~15E2 K apparent temperature). This means a 1E4 concentration is required to recreate the Sun's apparent core temperature over a solid angle equal to that of the Sun. The Sun's is about 33 arc-minutes across, so its solid angle is about 2.8E-4, so an additional concentration of about 4E4 is required. So, as a rough estimate, we could create Sun- core temperatures if we could concentrate light by a factor of about 1E9.

This is difficult because the diffraction limit means that large lenses are required to focus light to a small point. To get this concentration (spot area of 1E-9 of the lens area), assuming an f/1 lens (let's keep it practical here) and assuming we want a 1mm^2 spot area, we would need a lens diameter of 31 meters.

The largest telescope currently in existence has an effective aperture (~lens diameter) of 11.9 meters.

To be conservative, I propose building a reflecting telescope with a 100m lens diameter, placing a target of fissile material at the focus, then aiming the telescope at the Sun.

No one has done this yet (e.g. with a solar collector array) because the collectors are deliberately designed not to fuse/vaporize the equipment where the light is focused, and building an optically accurate reflecting telescope of that size becomes very expensive (i.e. if special steps are not taken, the enormous lens will focus the light to a blurry spot a few feet across, and no fusion will occur).

A linear accelerator would be required to inject additional fissile material into the beam focus. Once fusion occurs, additional material could be added to power-up the reaction to the desired size.

 — sninctown, Dec 12 2012

Equation: diameter of Airy disc http://www.wolframa...28100m%29%29%29*50m
for 100m aperture, 800um light [sninctown, Dec 12 2012]

Solar thermal power plants http://www.volker-q...amentals2/index.php
[spidermother, Dec 12 2012]

Harnessing light from hotter stars Moon_20mirrors_2e_2...by_20bright_20stars
[spidermother, Dec 12 2012]

Non Imaging Optics http://www.nature.c...1/abs/339198a0.html
[MechE, Dec 13 2012]

More non-imaging optics http://en.wikipedia...nergy_concentration
Wikipedia -- Nonimaging Optics [sninctown, Dec 13 2012]

One I prepared earlier... Increase_20solar_20...20max_20temperature
How to increase the max temp... [Ling, Dec 13 2012]

 //The goal is to reproduce the temperature within the sun (~15E6 K) by concentrating sufficient incoming solar light (~15E2 K apparent temperature).//

A solar concentrator can create temperatures arbitrarily close to, but not exceeding, the temperature of the radiation source - in this case, the surface of the sun. To achieve a higher temperature would be to break the laws of thermodynamics, as energy would be flowing from a cooler body to a hotter one.
 — spidermother, Dec 12 2012

There is a difference between heat and temperature though is there not? If all the heat passing through a 100m diameter lens converged onto a tiny spot, wouldn't the temperature potentially rise above that of the heat source? It is sort of like hydraulics - other than being completely different.
 — AusCan531, Dec 12 2012

 Well, it's not heat passing through the lens, it's radiation ... but that's not really the point. But no, it's temperature that's strictly limited. There are all sorts of ways to prove this - geometrically, thermodynamically, etc. - and the answer comes out the same every time.

 For instance, it would only be possible to focus the sun's energy into an arbitrarily small area if the sun were infinitely small, which would mean that it would be infinitely hot.

Alternatively, no optical system can do better than reaching out and smooshing your collector right onto the surface of the sun, like a cosmic pie-to-the-face; and that would also result in a temperature, at best, equal to that of the surface of the sun.
 — spidermother, Dec 12 2012

 The real problem is that there are two theoretical upper limits to concentration.

 The first, as [spider] said is thermodynamic, and Treceiver is less than or equal to Tsource.

The second is optical, and depends on the materials involved, and has a theoretical upper limit somewhere around 100,000 in glass, and a real world limit around half that. You might do a little better with higher index of refraction materials, but not enough to reach fusion energies even if it was thermodynamicly possible.
 — MechE, Dec 12 2012

I only like to resort to the real world when something is not provably impossible even in theory ;-).
 — spidermother, Dec 12 2012

 That's known as a solar thermal power plant or a solar furnace and is WKTE.

 To satisfy my curiousity, I ran the numbers for diamond lenses (highest index of refraction i could find for a clear material) on a 1mm square target, and you could concentrate a theoretical limit of 300 watts on said target. (On the earths surface, you could maybe double that in orbit without the atmosphere in the way).

The calculation of the resulting temperature is left as an excersize for the student. Assume a perfect black body.
 — MechE, Dec 12 2012

Again using diamond lenses, a 100m telescope would have a minimum spot diameter of somewhere around 8 cm.
 — MechE, Dec 12 2012

For spot size I get about half a millimeter with a 100m optic-- see link to equation. I used the Airy disk formula: for a 100m disc, the Airy disc formula gives the angle from the center to the first minimum of the diffraction pattern.
 — sninctown, Dec 12 2012

 Again, the temperature is easy to find - for a perfect optical system, it is the same as the effective temperature of the radiation. The ratio of the distance of the image and the object from the lens is equal to the ratio of the size of the image and the size of the object. Thus, because emissivity equals absorbtivity for any object (including a perfect black body), at all wavelengths, the equilibrium temperature of the image on an insulated black body is equal to the temperature of the object multiplied by its effective emissivity (which is close to 1 in the case of the sun).

 //if it really worked like this, you could hook up a heat engine from the hot object to the cold object and achieve perpetual motion// Yes, that is precisely why the image cannot be hotter than the source (the sun). It would, indeed, allow perpetual motion.

(It's not really my argument, it's just basic physics).
 — spidermother, Dec 12 2012

I agree that the Sun would see a 100m image of the hot object (the temperature of the object across the entire 100m telescope aperture, if looking into the wrong end), and the object would see the temperature of the Sun (if looking from the focus toward the optic). What confuses me is that for energy balance, there would have to be a huge number of photons per area coming from the hot object, or a smaller number of higher-energy photons, to match the energy coming in from the Sun. An example of this is a Dyson sphere around the Sun-- at equilibrium, the energy radiated from the outside of the Dyson sphere would be the same as the energy radiated from the Sun. Since the Dyson sphere has more area, it would be cooler than the surface of the Sun. The telescope/object setup seems analogous-- if the small object is the same temperature as the Sun, then it's not emitting as much energy out of the telescope as the Sun is putting into it. Does this make sense?
 — sninctown, Dec 12 2012

Yes, it makes sense. And at equilibrium, the image is not only radiating the exact same amount of energy as it receives (assuming no other losses), it is (statistically) radiating the exact same per-photon energy distribution (or, equivalently, the same wavelength distribution) as it receives.
 — spidermother, Dec 12 2012

 Sninc- the airy disk defines how tightly a given wavelength of light can be focused, but we aren't focusing light from a point source. The sun has a disk of radius .25 degrees as seen from earth, this means that a lens (or reflector) will produce an image of the source, with size dependent on the index of refraction. That's where my number comes from.

For a 2d concentrator, it's index of refraction^2/sin^2 apparent angle of source.
 — MechE, Dec 12 2012

 As for the Dyson sphere, at equilibrium it also radiates the same amount of energy as it receives, but because it is at a lower temperature, it radiates longer wavelengths (less energetic photons, but more of them (not //a smaller number of higher-energy photons//, you got that the wrong way around)) than it receives. This is possible because it has a much greater surface area than that of the sun.

The large number of low energy photons has more entropy than the small number of high energy photons. Free lunch averted.
 — spidermother, Dec 12 2012

Yes, I'm pretty sure the Airy disk can be ignored in this context; it is one of the factors that make the focus less than perfect, and therefore the achievable temperature less than the surface of the sun, but it only makes a tiny difference if the aperture is large.
 — spidermother, Dec 12 2012

Wouldn't a fresnel lens do the trick equally well in a smaller diameter?
 — zen_tom, Dec 12 2012

 Even if the magnifying glass technique could reproduce the temperature of the Sun's core, you'd still need to reproduce the pressure in order to get fusion.

 The core of the Sun is at about 15 million degrees, but tokamaks run at about 150 million degrees. This is because we can't build a pressure vessel to contain 340 billion atmospheres of pressure, so we have to turn the temperature up instead to get a useful reaction rate.

This is probably a good thing, from a safety point of view (minimising the mass of material that has to be heated to the working temperature).
 — Wrongfellow, Dec 12 2012

 'k, obviously this is an off-Earth installation because you won't be able to get any where near 1mm2 before the air distorts too much to hold a focal area.

Re: temperature limitations... hmm. Obviously true in the case of conduction. Obviously not true in the case of the average temperature of an area, eg: if you make your apparatus cover an angular area larger than that of the Sun then your average temperature will drop while the temperature at the target remains the same.
 — FlyingToaster, Dec 12 2012

 //you won't be able to get any where near 1mm2 before the air distorts too much to hold a focal area// I disagree. Air distortion is not the limiting factor; if it were, then you could not resolve detail much smaller than the diameter of the sun with a terrestrial telescope, whereas clearly you can. The size of an image of the sun directly relates to the sun's apparent size.

Well, of course you can get a temperature greater than the average temperature of the sun and some arbitrary region of space surrounding it, but that's not really helpful, is it? The point is that you can't, by focussing light, exceed (or even quite reach) the temperature of the hottest thing available; which in this case, is the surface of the sun.
 — spidermother, Dec 12 2012

No, I mean the air would get hot and turbulent on the way to the focal point, throwing the focus off. I realize there's an argument coming up for "well all those frequencies have already been absorbed by the atmosphere", but I think concentrating 10,000 m2 into 1mm2 is gonna be pretty rough on the air near the focal point.
 — FlyingToaster, Dec 12 2012

^so put the focal point in a vacuum. It will probably need to be in a vacuum anyway to prevent the containment vessel from vaporizing in an unholy inferno.
 — DIYMatt, Dec 12 2012

 This gives me such a cool idea! (+)

...I think
hmmmmmm
 — 2 fries shy of a happy meal, Dec 12 2012

 //the air would get hot and turbulent on the way to the focal point// Good point; and there will be plasma, too. I've heard that if a large solar concentrator is focussed in mid air, you see a bright point hanging there.

 While I'm still confident that you can't exceed the temperature of the surface of the sun, I can' t find any information on the theoretical maximum temperature or concentration; the search space is dominated by practical energy harvesting concentrators.

 For example:

 "A point-focus system such as a dish can intensify sunlight to several thousand times its normal level and achieve temperatures of well over 1000°C" How many thousand times? How much over 1000°C?

 "Although simple, these solar concentrators are quite far from the theoretical maximum concentration." Which is...?

Frustrating.
 — spidermother, Dec 12 2012

 See my numbers above for the theoretical maximum for lens style concentrators, about 350,000 for theoretically perfect diamond optics, with about half that for real world optics. I included the equation for lenses in my immediatly prior anno.

I'll double check the textbook for the maximum for reflective parabolic concentrators when I get home tonight, I think it's lower.
 — MechE, Dec 12 2012

Another prospective, at JET, we use about 50 MW of power and won't get ignition. JET is a stepping stone and wasn't supposed to get ignition... That would equate to roughly a mirror with 50,000 square metres, roughly 250 metres across. Then you have to consider how long you can keep hold of the heat. At JET it's about 0.6 seconds, without magnetic confinement, the particle accelerator would zip the fuel right past the hot bit far too fast. So the solution presented would work even less well than a proper fusion experiment that isn't supposed to work...
 — saedi, Dec 12 2012

Another prospective, at JET, we use about 50 MW of power and won't get ignition. JET is a stepping stone and wasn't supposed to get ignition... That would equate to roughly a mirror with 50,000 square metres, roughly 250 metres across. Then you have to consider how long you can keep hold of the heat. At JET it's about 0.6 seconds, without magnetic confinement, the particle accelerator would zip the fuel right past the hot bit far too fast. So the solution presented would work even less well than a proper fusion experiment that isn't supposed to work...
 — saedi, Dec 12 2012

 If my reasoning above is correct, the theoretical maximum concentration is equal to the ratio of the surface area of a sphere with radius equal to the earth's distance from the sun, to the surface area of the sun. This is 11,5328 for the semimajor axis; far from 1E9. That does not account for efficiency; it is simply the concentration that produces a flux equal to that at the sun's surface.

 Sorry, that should be the apparent area of the sun's disk, not the surface area of the sun. Multiplying by 4 gives 46,154. See below.
 — spidermother, Dec 12 2012

"The theoretical maximum concentration factor is 46,211. It is finite because the sun is not really a point radiation source. The maximum theoretical concentration temperature that can be achieved is the sun’s surface temperature of 5500°C" (Link).
 — spidermother, Dec 12 2012

//Another prospective// Another prospective what?
 — MaxwellBuchanan, Dec 12 2012

 Alright, I agree with spidermother now...(area)* (solid angle) is the same in both cases. The hot object has a small area but large solid angle, while the sun has a large area but small solid angle. So, I agree that a solar concentrator can only make something as hot as the Sun's surface.

The small spot size I came up with is the image of only a small patch on the Sun's surface. I agree the minimum image size of the sun using a 100m diameter lens would be much larger.
 — sninctown, Dec 12 2012

 Hang on.

 So, the problem is that you can't focus sunlight to produce heat hotter than the surface of the sun.

However, there are other stars with much higher surface temperatures. We're just going to require a bigger telescope.
 — MaxwellBuchanan, Dec 12 2012

 Just for fun, here's an extreme illustration that this is the theoretical limit:

 Construct a perfect, spherical mirror, completely enclosing the sun, with the sun at its exact centre. (Anyone need more glue?) Now, any ray leaving the surface of the sun will be reflected back to the surface. Clearly, the radiation flux reflected back to the surface exactly equals the flux leaving the surface.

This is the Ultimate Solar Concentrator; no improvement is possible. Any change to this design (for example, an an attempt to concentrate the output onto a smaller region of the sun) will fail, and instead result in some rays missing altogether.
 — spidermother, Dec 12 2012

 — spidermother, Dec 12 2012

How about if we hold the fusion fuel and run really, really fast towards the sun, thereby blueshifting the light?
 — MaxwellBuchanan, Dec 12 2012

 I really, really, really do know better than to even mildly question [spidermother] on this technical stuff but what the hell, I will learn something....

 Surely if one thought in terms of heat energy (or radiative energy or whatever) entropy will always have its way. But isn't temperature just a measure of how much heat energy is present within a mass? If one collected all the heat energy present within a swimming pool of water at, say, 30C then transferred that energy into a teacup of water the temperature of that water would sky-rocket - even if the whole process is massively lossy?

In the poster's example, the heat energy into the focal point is continually being added into a tiny spot from a 7,800M2 lens. In the thought experiment of smooshing the collector onto the face of the sun the temperature of a 7,800M2 collector obviously couldn't be hotter than the sun's surface, but if that continual outpouring of heat energy from 7,800M2 was focussed onto 1M2 behind the collector wouldn't the temperature exceed the sun's temperature even though there would be inevitible energy losses along the way. There would be 7,799M2 behind the collector which would be commensurately cooler than they would be otherwise be with that energy going into the single square metre. If temperature is a function of heat energy / mass (not sure of this!) then reducing the mass increases the temperature.
 — AusCan531, Dec 12 2012

 Okay, the formula for paraboloid solar concentration is "sin^2(phi)/(4sin^2[theta])" where phi is the half angle of the parabola from the point of view of the concentrator (maximum 90, obviously before it starts self occluding) and theta is the half angle of the source, m so .25 for the sun.

Net result, ~13,000 for an ideal and perfect optical concentrator.
 — MechE, Dec 12 2012

 Thanks, [MechE], that seems right.

 //(maximum 90, obviously before it starts self occluding)//

While that's a practical limitation, it's not a theoretical one. The ideal concentrator is an ellipsoid (with the sun at one focus and the collector at the other), not a paraboloid; although the distinction is only important for rather large collectors. An ellipsoidal concentrator can achieve a concentration of 46,000 odd. (But only by completely enclosing the sun and the collector. And the number is meaningless anyway at that scale; it's better to speak of a concentration of 1 relative to the sun's surface.)
 — spidermother, Dec 12 2012

Which leads naturally to a solution. We just have to build the ellipsoidal reflector, and only harvest some of the energy at the focus. The remaining light will feed back into the sun, effectively insulating it, until the temperature of the sun's surface approaches the current temperature of its core.
 — spidermother, Dec 13 2012

 [AusCan] //In the poster's example, the heat energy into the focal point is continually being added into a tiny spot from a 7,800M2 lens// But the problem is that you cannot, using any system of optics, achieve that level of concentration.

You can achieve greater concentration (and temperatures hotter than the surface of the sun) in other ways - for example, you could use concentrated solar radiation to power a laser. But the power in the laser light will be less than the power in the sunlight harvested to power the laser. Most of the power will be lost as low grade (= low temperature) heat.
 — spidermother, Dec 13 2012

 Yes, less power, less heat, etcetera BUT greater temperature. Losses all over the place but it still achieves a greater temperature within a small receiving area than the temperature at the large emitting area.

//But the problem is that you cannot, using any system of optics, achieve that level of concentration.// What level? It doesn't have to be focussed to a single molecule just to an area a few thousand times smaller than the incoming lens area.
 — AusCan531, Dec 13 2012

 //What level?// The level required to achieve a temperature greater than that of the surface of the sun, or (equivalently) a radiant flux density greater than that leaving the surface of the sun. I know that's not really an answer - just a restatement of the above - but it's the reality, and optics, thermodynamics, etc. are all in agreement on that.

 So, a real-world 7,800 m² lens can at best produce an image of the sun on the order of a square metre or so.

 To do better than that, you need to harness some high-temperature energy (sunlight) to drive a separate process that produces a smaller amount of even higher temperature energy. (The system also produces a quantity of lower temperature energy).

Similarly, if you had a swimming pool of water at 30ºC, and another swimming pool of water at 0ºC, you could use that temperature difference to drive a stirling engine, whose output drives a smaller stirling engine, to produce a teacup of boiling water. But you can't directly syphon off even a thimble-full of water hotter than 30ºC - which would be (roughly) equivalent to using optics to direct even a portion of the sun's light into an area small enough to create a temperature hotter than the surface of the sun, which you also can't do.
 — spidermother, Dec 13 2012

 PAH! Next you'll be trying to tell me that light is both a wave and a particle at the same time. Nice try Poindexter.

Just to prove I'm right, I've pointed a temperature gauge at the sun from Perth and got a reading of 22C and have quickly constructed a giant lens over Adelaide, which by my calculations, should reach a temperature of around 38C today. Please monitor the result and report back with the empirical results to prove who is right. If you don't, I shall sharpen the focus considerably. Howdya like THAT image!
 — AusCan531, Dec 13 2012

[Aus] You did see my above anno on the maximum possible optical concentration on a 1mm^2 target, right? 300 Watts. No matter how big your collector, that 300 watts/mm is a pretty hard limit.
 — MechE, Dec 13 2012

 Yes [MechE] just because I'm arguing doesn't mean I think I'm right. Fools rush in and all that.

But, for my own understanding, if you continually poured 300 watts into that mm (presupposing the mentioned perfect black body) for a very long time and it wasn't emitting energy faster than receiving it wouldn't the temperature continue to climb?
 — AusCan531, Dec 13 2012

 But it _would_ emit energy as fast as it received it; not for any boring practical reason, but because the absorption and emission of radiation from a black body (or anything else) is perfectly symmetrical.

Otherwise, you could simply place a material that had a greater absorbtivity than emissivity (for any or all wavelengths) in a vacuum in sunlight, and it would get hotter than the sun.
 — spidermother, Dec 13 2012

 //maximum possible optical concentration on a 1mm^2 target, right? 300 Watts//

There must be something wrong there; the maximum concentrated flux for any optical system should be about 1,368 W/m² * 46,211 = 6.3E7 W/m² = 63 watts/mm².
 — spidermother, Dec 13 2012

 [Spider] That's for a mirror type concentrator, you can do better with a TIR/lens concentrator, with the upper limit being dependent on the index of refraction. Given perfect diamond optics, the concentration ratio reaches approximately 305,000.

 If you can find something with an index of refraction higher than 2.41, it could go higher, although I suspect (although can't prove it) that there is a point of diminishing returns as more energy is lost internal to the optic.

The highest reported actual concentration is 56,000 in glass optics, as of the publishing of my old textbook (Principals of Solar Engineering, Goswami et al, 1999). I suspect the target has to be index matched, or you'll lose most of the benefit at the lens exit.
 — MechE, Dec 13 2012

//305,000// //56,000// Are you sure those numbers are the ratio of irradiance at the focus to irradiance of the unfocussed sunlight, and not based on some other way of defining 'concentration'?
 — spidermother, Dec 13 2012

Yup, I'm sure. See link. The relevant factor is non- imaging optics.
 — MechE, Dec 13 2012

"We have used a non-imaging design to concentrate terrestrial sunlight by a factor of 56,000, producing an irradiance that could exceed that of the solar surface." <Eats hat>.
 — spidermother, Dec 13 2012

I call shenanigans on the factor of 56,000...I think etendue/throughput is still conserved regardless of whatever non-imaging funky optic is concentrating the light. See wikipedia link for examples of non-imaging optics.
 — sninctown, Dec 13 2012

[sninctown] That's what I was thinking, too. Notice that they maintain that it's impossible to exceed the temperature of the surface of the sun. This may be one of those tricksy exceptions that prove the rule, like group velocities (but not information) propagating faster than C. In other words, while that flux density may technically exist, it's not possible to capture real energy at that power density.
 — spidermother, Dec 13 2012

 I'm gonna go out on a limb and figure that "can't exceed" means that, in the case of the Sun, you couldn't exceed X-rays' black-body temp at the target no matter what kind of perfect Dyson reflector you built.

 Likewise, if we were trying to focus from Jupiter, the theoretical cap would be a bbt in the high-radio or IR range (or whatever the highest freq it emits is).

 So it may actually be impossible to jumpstart fusion using concentrated solar energy.

Of course then again if you filtered out all the ghetto frequencies...
 — FlyingToaster, Dec 13 2012

This would not work at all for a variety of reasons as stated above.
 — Kansan101, Dec 13 2012

 — Ling, Dec 13 2012

//I propose some speed. Blue shifted radiation? Link.// If only someone had suggested that a couple of screenfulls ago.
 — MaxwellBuchanan, Dec 13 2012

hmmm, does a solar eclipse create a gravitational lens? and if so is it possible to use that lensing to generate temperatures hotter than the surface of the sun?
 — 2 fries shy of a happy meal, Dec 13 2012

 //Construct a perfect, spherical mirror, completely enclosing the sun, with the sun at its exact centre. (Anyone need more glue?) Now, any ray leaving the surface of the sun will be reflected back to the surface. Clearly, the radiation flux reflected back to the surface exactly equals the flux leaving the surface.//

 I like this idea. Here's how I see it playing out:

 In the mean-time the sun is still producing energy. Therefore the temperature of the sun rises. The sun will tend to expand as it heats up, which will affect the rate at which it produces more heat. However, I imagine it will still be net positive for heat.

Now, depending on the size of the perfect mirror sphere, at some point the sun's outer layer will start to impact it. What happens then depends on the nature of said hypothetical material and whether it absorbs heat by convection.
If it does then the entire sphere would start to heat up, and potentially radiate heat into the universe. If it's perfectly reflective on the outside nothing obvious will happen until the material is affected by its temperature
If it fails at a single point we then have a leak. If the heat loss stabilises the internal temperature then the entire thing will whizz off a little like a deflating balloon, squirting out a powerful jet of plasma. If it fails progressively then all the contained matter would be released essentially at once and the whole thing would brew up, at something up to the power of a supernova.
 — Loris, Dec 13 2012

Put a frequency doubler - say, Nd:YVO maybe - ahead of your final concentrator. I believe the temperature maximum is related to the wavelength of the energy input, is it not? This won't get to solar-core temps, obviously, but I think you could crack the sun-surface-temperature barrier.
 — lurch, Dec 15 2012

 No, the attainable temperature is not directly related to wavelength; it has more to do with power density (how much the light can be concentrated).

 For example, it's possible to generate temperatures on the order of the sun's surface temperature in a kitchen microwave oven, despite the long wavelength.

Also, frequency doubling only works on coherent (e.g. laser) light.
 — spidermother, Dec 15 2012

[Lurch] You may be thinking of the quantum limit to focusing (the Airy disk mentioned above), which does improve with shorter wavelengths, but that's only for relatively small light sources, so again, typically lasers.
 — MechE, Dec 15 2012

 It all comes down to entropy.

 Sunlight has a relatively high entropy (compared to laser light, for example), which directly reflects the temperature of the surface. Being concentrated (point source), coherent, collimated, high-frequency, or monochromatic all equate to low entropy, and therefore to higher attainable temperature; but there is no way to decrease the total entropy of sunlight.

It is, of course, possible to produce a small amount of low entropy light using the energy of a large amount of high entropy light, but not (AFAIK) by any purely optical means.
 — spidermother, Dec 15 2012

 //You may be thinking of the quantum limit to focusing

No.
 — lurch, Dec 16 2012

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