Boerseun's Perpetual Motion Machine...

[Boerseun]

Why would that be?

 

Consider:

 

A normal, humdrum point in the universe (say, P in your diagram) receives CMBR from the universe, as is. The black and red lines are equally likely. But all the red lines except R3 would have bypassed the point. Now, with a lens, they don't. All the red lines are focussed towards P. But your lens would then bend the black lines away, to be replaced by the red lines.... uh.... okeley dokely, I see where you're going with this... P would then receive the same amount of CMBR, regardless... doh!

 

Oh well, interesting thought, nonetheless.

[Qfwfq]

Pyro, that's clever! I'll have to consider some aspects carefully before saying for sure if it could make us rich! (Edit: but not with the spinning paddles, which wouldn't work, I'm thinking of a different idea.)

 

Ok, now just wondering, what if you replace CMB with cosmic rays?

Intuitively, I think it would amount to the same, since the latter have higher energy but lower intensity...

I'm not at all sure that cosmic rays can be considered thermalized, do you know?

 

But, I think you could avoid the black hole frequency converter by using a material in your heat engine which efficiently absorbs microwaves, like, water—rather than the fabric (or plastic??) of a paddle wheel.

The solid surface of highest emissivity and no spectral deviations is matt black. That's why they call it black body spectrum.

 

you could also use a material to convert microwaves to infrared (rather than using a black hole). I think, again, water would be best since it absorbs microwaves and emits infrared

It emits infrared according to how high its temperature is. If it's hot enough it even emits visible. If it isn't hot it emits more in the microwave range than infrared. So, inorder for it to convert low T black body radiation into infrared so as to heat something with it, you're hoping for it to be heated by the low T waves.

 

The temperature of space is not given by the wavelength alone. It's also given by the flux. Put another way: the temperature of a photon gas is a function of the density of photons (among other things).

Gosh, I doubt Boltzmann would agree with you there. It's like saying the temperature of a material body depends not only on the average kinetic energy of its particles but also on the density of them. Look it up in stat-mech.

 

So, have we defeated the 2nd principle of thermodynamics yet?

[modest]

But, I think you could avoid the black hole frequency converter by using a material in your heat engine which efficiently absorbs microwaves, like, water—rather than the fabric (or plastic??) of a paddle wheel.

The solid surface of highest emissivity and no spectral deviations is matt black. That's why they call it black body spectrum.

 

A blackbody is an idealized object so called because it absorbs all radiation that falls on it reflecting or letting pass none therefore appearing black if cold.

 

By "matt black" are you referring to a color or a material? The visible color of a material will not be a good indicator of the material's ability to absorb microwaves. Emissivity is wavelength-dependent in real objects. It is possible to plate, chemically treat, or paint the surface of a material to increase emissivity, but the effectiveness typically falls off > 9 μm.

 

The point I was making was that the plastic of a vacuum paddle wheel like Pyrotex references will not absorb microwaves efficiently, but other substances would. Water is an example of a substance which has a high (or relatively high) emissivity in the microwave range. This is evident if you put a microwave safe plastic plate in a microwave. you will notice the temperature doesn't rise. A dish of water will, however, rise in temperature very quickly despite its high specific heat capacity. Water is a good absorber of microwaves which a person would not know by the translucent nature of light in the visible spectrum.

 

This, by the way, is the reason CMBR imaging has been done with balloon-born instruments in Antarctica. The humidity in the atmosphere is very problematic in the microwave and far-infrared range. The balloons rise above the water vapor and Antarctica is one of the least humid places on earth.

 

you could also use a material to convert microwaves to infrared (rather than using a black hole). I think, again, water would be best since it absorbs microwaves and emits infrared

It emits infrared according to how high its temperature is. If it's hot enough it even emits visible. If it isn't hot it emits more in the microwave range than infrared. So, inorder for it to convert low T black body radiation into infrared so as to heat something with it, you're hoping for it to be heated by the low T waves.

 

If, by "low T waves" you mean microwaves then yes. It is, once again, completely possible to heat water with microwaves to boiling. It would radiate that heat in infrared.

Microwaves - Currently the basis of a widespread technology utilized in millions of households for heating food, microwave spectral wavelengths range from approximately one millimeter to thirty centimeters (or about one foot). The appeal of utilizing microwaves in food preparation results from the fortuitous circumstance that water molecules present in most foods have a rotational resonance frequency within the microwave range. At the frequency of 2.45 gigahertz (12.2 centimeter wavelength),

water molecules efficiently absorb microwave energy and subsequently dissipate radiation as heat (infrared)

. If containers composed of materials that do not contain water are utilized to hold food in the microwave oven, they will tend to remain cool, adding a significant additional convenience to microwave cooking.

Physics of Light and Color

 

The temperature of space is not given by the wavelength alone. It's also given by the flux. Put another way: the temperature of a photon gas is a function of the density of photons (among other things).

Gosh, I doubt Boltzmann would agree with you there. It's like saying the temperature of a material body depends not only on the average kinetic energy of its particles but also on the density of them. Look it up in stat-mech.

 

I understand what you're saying. I was thinking of boosting the luminosity while keeping the wavelength the same. That was the premise of the OP. The temperature of a body depends not only on the wavelength of light it receives, but its intensity. For example,

 

If you put a thermometer on a white piece of paper at noon in sunlight then it will absorb a certain amount of heat from the sun. The energy flux (or the power) available is around 600 watts per square meter. The thermometer will reach a certain temperature. A glass lens will not change the wavelength of light passing through it, but it can increase the luminosity or flux at the focal point. If you focus a magnifying lens on the thermometer the temperature will rise above its previous level—not because the wavelength of the light has changed, but because the intensity has changed.

 

So, have we defeated the 2nd principle of thermodynamics yet?

 

I believe the proof in my last post demonstrates that a lens or dish cannot increase the flux when the energy source is isotropic. Since there can be no power gain at the focal point this should prove that the lens will create no thermal gradient, and hence no perpetual motion machine.

 

~modest

[Boerseun]

Well, to b fair, even if it did work, it wouldn't be a "perpetual motion" machine, and the second law would nowhere be violated. It would merely be a concentration of energy that was released by the Big Bang and is now merely so diffuse as to be unusable.

 

But the Second Law would stay intact throughout the whole process - if it did work.

[Qfwfq]

OK fellas I'll try, but it seems it will take a mighty effort cuz it requires a thorough understanding. I only did the basics of the necessary topics (and long ago), so this discussion has been requiring me to think mighty carefully and shake the cobwebs out.

 

Well, to b fair, even if it did work, it wouldn't be a "perpetual motion" machine, and the second law would nowhere be violated.

I fully agree that it wouldn't be a perpetual motion machine (I think I so far forgot to say this) but I do say that any manner of doing it with the CMBR would be an exception of the 2nd principle of thermodynamics. You are, to all effect, trying to get a 2.71K "object" to heat another one which has a temperature slightly more than 100 times that. This is the basic fact upon which to seek out the detail that makes whatever idea or argument untenable.

 

A blackbody is an idealized object so called because it absorbs all radiation that falls on it reflecting or letting pass none therefore appearing black if cold.

Therefore appearing matt black, if cold.

 

This means that, as well as no chromatic deviations, it should also have perfect emissivity and this includes no reflectivity because it cannot have distinct values of emissivity and absorbtivity. Obviously, the colour it looks to the human eye is not sufficient, the semantics must be generalized to the whole EM spectrum or at least the range for which amplitude is relevant for the given temperature. A material surface that looks matt black isn't necessarily an excellent black body, but a perfect black body is necessarily as matt black as a terse night sky without the stars.

 

No material lacks chromatic deviations, though emissivity can be pretty nigh on total, when a previous job had me study optical pyrometers I learned a few things further to my uni courses; although written for his sales reps, Vern Lappe's tutorial was quite interesting. But that was just over 10 years ago, I can't remember the highest attained emissivity values but certainly in the 0.99 area and there is also a paint that gives a very precise value for calibration purposes, especially necessary with instruments of the kinds that rely more on intensity than on actual spectral form. The simplest kind of pyrometry, based only on intensity, has its shortcomings for the very reason that intensity doesn't really determine temperature.

 

An ideal black body is most excellently approximated not by a material surface but by... a hole. With a good design of the cavity, imperfect quality of the surfaces inside makes far less difference, the small aperture can be an excellent black body as far go emissivity and exact spectrum. This is how the phenomenology is studied in the lab; material property troubles get almost totally stamped out.

 

If, by "low T waves" you mean microwaves

I meant thermalized EM radiation at a low absolute temperature.

 

It is, once again, completely possible to heat water with microwaves to boiling.

I have used the odd microwave oven and I know how they work. Pay attention to what you quote. Straight before the words you bolded comes a mighty interesting thing:

At the frequency of 2.45 gigahertz (12.2 centimeter wavelength),

Why does it specify this frequency/wavelength? The range of wavelengths it gives shortly before that is that of what are considered microwaves, you can see that the high degree of absorption occurs at a specific peak of resonance; it is a chromatic deviation and in my understanding a quite tight one.

 

You can bring a hammer head and a piece of metal to a temperature higher than that of the warmest other object around; the mechanical energy moving the hammer isn't thermal and neither is the microwave radiation inside the oven.

 

I wasn't objecting that you propose to change the wavelength. I objected to the contrary that since you don't alter the spectrum, the temperature is still less than 3K.

If you put a thermometer on a white piece of paper at noon in sunlight then it will absorb a certain amount of heat from the sun. The energy flux (or the power) available is around 600 watts per square meter. The thermometer will reach a certain temperature. A glass lens will not change the wavelength of light passing through it, but it can increase the luminosity or flux at the focal point. If you focus a magnifying lens on the thermometer the temperature will rise above its previous level—not because the wavelength of the light has changed, but because the intensity has changed.

The solar surface is at about 6000K and this is the temperature of sunlight incident on this planet. A black body left in the sumer sun won't reach quite that temperature, but concentrating the rays can certainly bring it to a pretty high one. No conflict with thermodynamics.

 

I believe the proof in my last post demonstrates that a lens or dish cannot increase the flux when the energy source is isotropic. Since there can be no power gain at the focal point this should prove that the lens will create no thermal gradient, and hence no perpetual motion machine.

Are you sure the only problem is the isotropy?

 

Suppose you find a way of collecting all the EM radiation leaving the sun and focussing it onto some little excellent black body that could withstand arbitrarily high temperature, and as much as possible reduce dissipation from where the sun rays are not incident. What temperature do you reckon it would reach? What do you reckon would happen?

[Boerseun]

You are, to all effect, trying to get a 2.71K "object" to heat another one which has a temperature slightly more than 100 times that.

We're not discussing "objects" heating each other, we're discussing collecting and concentrating naturally occurring microwaves (albeit at very low energy). Let's say that the CMBR energy falling on a square meter per second equals a gnat's footstep. So, one m² won't help. But if you were to somehow concentrate the CMBR energy falling on a thousand km² onto a single m², then you'll have the energy of 1,000,000,000 gnats' footsteps available on a single square meter. There's your lens/microwave dish.

 

That is why we're discussing a way of concentrating it. Not a single ant in the history of the world has had his head explode because he was walking in the sun. But give a kid a magnifying glass and see what happens.

 

The point is, CMBR is radiation, it is there, it's photons bouncing around at a specific frequency like all other photons throughout the electromagnetic spectrum. If you can somehow concentrate it, you can potentially reach extremely high temperatures if your catchment area is big enough. Hence me wondering if such a device will fit on the planet - because it's so diffuse and of such low energy that any device big enough to catch enough of it will have to be ridiculously big.

Suppose you find a way of collecting all the EM radiation leaving the sun and focussing it onto some little excellent black body that could withstand arbitrarily high temperature, and as much as possible reduce dissipation from where the sun rays are not incident. What temperature do you reckon it would reach? What do you reckon would happen?

If you collect "all" the EM radiation coming from the sun and you focus all of it on an "excellent" or perfect black body that could withstand arbitrarily high temperatures, nothing will happen. You will not be able to detect anything coming from that body, seeing as it's a perfect black body. But then again, that is not what we're proposing.

[Jay-qu]

So you use one of these beasties:

First black hole for light created on Earth - physics-math - 14 October 2009 - New Scientist

 

Interesting thread you have going here..

 

To answer your question Q, you would simply have a replica of the sun

[Qfwfq]

To answer your question Q, you would simply have a replica of the sun

In a sense, yup, that was my drift. ;)

[modest]

Suppose you find a way of collecting all the EM radiation leaving the sun and focussing it onto some little excellent black body that could withstand arbitrarily high temperature, and as much as possible reduce dissipation from where the sun rays are not incident. What temperature do you reckon it would reach? What do you reckon would happen?

 

I believe your premise breaks the second law while my previous proof is the reason the law is upheld. The sun's temperature and total power output is related via the Stefan–Boltzmann law:

To find the total absolute power of energy radiated for an object we have to take into account the surface area' date=' A(in m[sup']2

[/sup]

):

 

[math] P= A j^{\star} = A \varepsilon\sigma T^{4}[/math]

This is because j^* in S&B's law has dimensionality of area^-1

If you propose that all of the sun's radiated energy is caught by a smaller object (a premise I think breaks the second law) then the smaller object must emit a similar amount of power. The temperature of the two bodies would then be:

[math]T_{sun} = \left( \frac{P}{\sigma A_{sun}} \right)^{1/4}[/math]

[math]T_2 = \left( \frac{P}{\sigma A_2} \right)^{1/4}[/math]

Power being constant for both, the temperature will be larger for the body with less surface area.

 

The answer to your question, then, is that the smaller body would quickly attain a higher temperature than the sun. They would have equivalent luminosities, but unequal temperatures. As an analogy, smaller stars that burn equally bright to larger stars are hotter^[1].

 

The scenario is, however, impossible geometrically. A focusing device (lens or dish) set adjacent to the sun can focus only parallel rays to the focal point. If you follow the ray back to the sun you would discover that the ray could just as easily have been pointed a slightly different direction and hit the point directly (i.e. with no lens or mirror). The only exception to this rule is if you had set a mirror behind the sun for the purpose of directing rays from the far side toward the focal point.

 

Drawing this out on a diagram I believe you will find that it is impossible to arrange focusing devices in a way which directs more light to their focal point than would be accomplish by substituting the sun with a hollow sphere containing the second body where the sphere has equal surface area to the sun.

 

Now consider the flux which the point in the hollow sphere receives. The problem is analogous to gravitational force in a hollow sphere. If the point is near the wall of the sphere in a certain direction then it will receive an equal flux from that direction that the wall emits (the flux is not lessened by distance). The surface area of the sphere creating flux increases in every direction proportionately to the square of the distance, yet luminosity falls off with the inverse square of the distance—thus flux absorbed is constant in every direction. The flux per unit area of the object in the hollow sphere is equal to the flux per unit area of the hollow sphere regardless of the object's location.

 

Therefore, focusing the light of one object onto another cannot increase the flux per unit area received by the second object beyond the flux per unit area emitted by the first.

 

I do not have access to the following paper, but I would guess it evokes a similar proof:

Abstract. The paradox proposed by Panse (1992)-that some reflector systems can focus sufficient radiation from a number of objects to heat another to a higher temperature-has been resolved for two specific cases. The problem highlights the fact that all focusing systems must be limited by the second law of thermodynamics, whence a corollary to the second law is proposed for focusing systems. The states that 'no system may focus from one object, through an aperture onto a second identical object, more radiation than may be emitted by the second object through the same aperture'. The application of this type of system to non-contact temperature measurement is discussed.

Focused radiation, the second law of thermodynamics and temperature measurements

 

To a proposition which can violate the second law it is easy to invoke the second law in rebuttal. It is harder, but perhaps more satisfying, to discover why the law is upheld against the example.

 

~modest

[Qfwfq]

Well, you seem to be getting there, so you agree that you were trying to do something that would violate the 2nd principle, which means isotropy wasn't the most fundamental objection.

 

The matter is complicated and certainly subtle:

To a proposition which can violate the second law it is easy to invoke the second law in rebuttal. It is harder, but perhaps more satisfying, to discover why the law is upheld against the example.

Agreed, I found the discussion interesting for this very reason, which you guys seemed both unaware of. The 2nd principle is a watchdog that you just can't disregard, when it growls you must sharpen your eyes and look out for the burglars. This is why I posed the question and, of course, when I had much earlier posed the one about a surface having distinct values of emissivity and absorbtivity, the answer is: even that would give us a way to solve the world's energy troubles. It is of course believed that no surface could have these distinct values.

 

There is however a fundamental reason why I don't think the 2nd principle can so easily imply there being a strict limitation on optical focussing, this is equivalent to simply invoking the 2nd principle and of course it remains interesting to examine the matter.

I believe your premise breaks the second law while my previous proof is the reason the law is upheld.

I don't think this is the way out of the tight spot and you make a tacit assumption about my premise.

 

The answer to your question, then, is that the smaller body would quickly attain a higher temperature than the sun.

I realize that the [imath]T^4[/imath] law suggests this, at least at first glance, but of course it cannot be so. I do differ about your reason:

The scenario is, however, impossible geometrically. A focusing device (lens or dish) set adjacent to the sun can focus only parallel rays to the focal point.

It is not geometrically impossible, you are assuming the specific case of a collimated beam, the limit case in which one of the two conjugate points is at infinity. A limiting factor is the wavelength, in ratio to the scale of the optical apparatus, but this can be arbitrarily reduced.

 

What I proposed was of course a gedankenexperiment but, to make a less unrealistic version, let's consider a spherical surface source in general. With a parabolic mirror sufficiently extended, you can get arbitrarily close to 50% (semispherical surface) onto a much smaller conjugate figure near the paraboloid's focal point. With a flat mirror enclosing the paraboloid, you get the equivalent of a source being two semispheres. A surface near the conjugate figure can be quite small and intercept almost all of the EM radiation form that source, except for the longest wavelength portion of the spectrum.

 

Therefore, focusing the light of one object onto another cannot increase the flux per unit area received by the second object beyond the flux per unit area emitted by the first.

You are attempting to generalize on the arguments in your preceding paragraphs, but you can't rule out there being counterexamples such as here above. I would be very wary of deducing a general implication on the optics.

[modest]

Well, you seem to be getting there, so you agree that you were trying to do something that would violate the 2nd principle, which means isotropy wasn't the most fundamental objection.

 

If a person could increase the flux of an isotropic source then it necessarily follows that the second law of thermodynamics is violated. I have demonstrated why that is and also why the flux cannot be increased and thus why the law is upheld.

 

I don't think this is the way out of the tight spot and you make a tacit assumption about my premise.

 

Your premise is well-stated. I'm not tacitly assuming that the premise is geometrically impossible with focusing devices. It's a proposition. One, which I might add, took quite a bit of work with a t-square, triangle, and compass before I had convinced myself.

 

I realize that the [imath]T^4[/imath] law suggests this, at least at first glance, but of course it cannot be so.

 

The results of the Stefan–Boltzmann law are as expected.

 

I do differ about your reason:It is not geometrically impossible, you are assuming the specific case of a collimated beam, the limit case in which one of the two conjugate points is at infinity. A limiting factor is the wavelength, in ratio to the scale of the optical apparatus, but this can be arbitrarily reduced.

 

The parallel rays emitted from the source form a single focal point. To consider non-parallel rays you must consider multiple focal points. Essentially, you are projecting an image of the sun's surface area onto some other surface area. With such a method I can find no way to increase the flux received per surface area of the target past the flux emitted per surface area of the source.

 

What I proposed was of course a gedankenexperiment but, to make a less unrealistic version, let's consider a spherical surface source in general. With a parabolic mirror sufficiently extended, you can get arbitrarily close to 50% (semispherical surface) onto a much smaller conjugate figure near the paraboloid's focal point. With a flat mirror enclosing the paraboloid, you get the equivalent of a source being two semispheres. A surface near the conjugate figure can be quite small and intercept almost all of the EM radiation form that source, except for the longest wavelength portion of the spectrum.

 

Prove it.

 

Consider that each point on the sun radiates in every possible direction. As that radiation propagates through space it becomes disperse giving the inverse square distance law.

 

A disk, for example, which is perfectly insulated on one side can have a given temperature and radiate energy. A second disk also insulated on one side of equal surface area as the first disk can act as a black body target set facing the first disk. If the target disk is going to attain an equal temperature as the radiating disk then there can be no distance between them. Each point on the radiating disk must have a full field of view of the target. If there is distance between them then only a portion of the radiating energy finds its way to the target.

 

If you were to draw a ray diagram of your example,

 

 

The two red lines are parallel rays. They hit a single focal point. That single focal point is not increased in flux/area beyond the sun's surface for reasons I gave in the previous proof. But, if you introduce non-parallel rays then you must increase the size of the target to some finite amount. Consider each point on the surface of the sun (I've drawn two). If your target is small then only a portion of the energy radiated from a point on the sun finds the target. This is analogous to the two disks described above. You cannot expect the sun to give its full compliment of radiated energy to something with a small apparent size.

 

In the above diagram the apparent size is only 6 degrees even with the help of the parabola. Regardless of the actual size of the target, the apparent size needs to be 180 degrees from any point on the surface of the sun to transfer all of the sun's energy. To transfer half to a target with half the radius you need 90 degrees... and so on.

 

You are attempting to generalize on the arguments in your preceding paragraphs, but you can't rule out there being counterexamples such as here above. I would be very wary of deducing a general implication on the optics.

 

I am at least giving a plausible reason that the second law is upheld. I obviously can't prove a negative, but until you give a better reason or show (actually show) that some mirror/lens system can increase the flux/area beyond the source then I feel no further need to prove that it can't.

 

~modest

[Qfwfq]

Modest, seriously, I have no time to get into an argument about the matter if you don't avoid misunderstandings. I tried to suggest a bit more thought on a few things but you skew my points, so it seems I have to let the cat out of the bag and be more explicit but I certainly couldn't get the content of entire textbooks into this thread. So avoid claims which aren't well grounded while challenging me to prove what I say; if you haven't been through the same courses I can't do much about it, it just takes me too much time.

 

If a person could increase the flux of an isotropic source then it necessarily follows that the second law of thermodynamics is violated. I have demonstrated why that is and also why the flux cannot be increased and thus why the law is upheld.

I did not disagree with your argument about isotropic flux --which is not to be confused with isotropic source-- and you have not effectively demonstrated the general case nor invalidated my argument about conjugate figures.

 

Your premise is well-stated. I'm not tacitly assuming that the premise is geometrically impossible with focusing devices. It's a proposition. One, which I might add, took quite a bit of work with a t-square, triangle, and compass before I had convinced myself.

Here you are making a further assumption about which assumption I meant, it wasn't that one:

The results of the Stefan–Boltzmann law are as expected.

Your assumption is about what it means and implies. Which is the cart and which is the horse?

 

My premise was in terms of the radiation being focussed onto a smaller body. You are tacitly assuming that all of it will be absorbed and that the total of reflected and emitted will be given exactly by [imath]T[/imath] and [imath]\epsilon[/imath] but you have no empirical nor theoretical evidence of this. Do you know of any? IMHO this is what is interesting, I have not yet found anything about reliable that decides the case; more below.

 

The parallel rays emitted from the source form a single focal point. To consider non-parallel rays you must consider multiple focal points. Essentially, you are projecting an image of the sun's surface area onto some other surface area. With such a method I can find no way to increase the flux received per surface area of the target past the flux emitted per surface area of the source.

If you cannot find the way, then you should improve your understanding of optics; you could also learn better about focal points and conjugate points and might even revise the opinions of which you have convinced yourself instead of telling me to prove what I said.

 

I am at least giving a plausible reason that the second law is upheld. I obviously can't prove a negative, but until you give a better reason or show (actually show) that some mirror/lens system can increase the flux/area beyond the source then I feel no further need to prove that it can't.

You did not show in general how the 2nd principle is not contradicted.

 

The Stefan Boltzmann law is a statement about the radiation emitted by a body and it does not consider incident radiation. It does not determine a temperature of radiation according to flux and doesn't constitute grounds for implying that concentrating the flux would be an exception to the 2nd principle of thermodynamics. Moreover, the theoretical grounds of it, given by Boltzmann, are based on the very same principle so it even smacks of circularity to imply an exception on the very grounds of the principle itself. Doesn't it?

 

Your conclusion therefore does not follow. To me it makes much more sense to conjecture that the target would reach the temperature of the source, with outgoing radiation flux equal to that which is incident according to Kirchhoff's thermal equilibrium law. I'm unaware of anything being published in support of this (which is why I consider it an interesting thing) but I hold this conjecture makes far more sense than assuming the validity of something in a manner whereupon it would imply an exception to the same principle on which it is argued.

 

The Panse paradox you mention was proposed in 1992 and it seems odd that nobody would have previously thought of it in a good century, if it really was unexplainable. I'm more inclined to think people simply reasoned as I did, rather than applying the Stefan-Boltzmann law out of context and even in a manner which turns out to be circular and drawing a conclusion that a principle which rests only on statistical considerations must pose a strict limitation on optics --which has a definitely non statistical nature.

 

I stated my example saying "all the solar radiation" to make it more illustrative but, according to your argument, a lot less than that would be sufficient to give an exception to thermodynamics including cases which certainly aren't optically inhibited.

[modest]

If you don't have an argument then please don't patronize mine. Your premise breaks the second law and my conclusion is the very reason the law is upheld.

As for focussing the light of the sun: Based on the discussion so far, you might think you could take all of the solar energy entering the lens and focus it on an arbitrarily small spot, thereby achieving an arbitrarily high temperature. Well, you can’t. Conservation of energy does not forbid it, but conservation of phase space (along with some other basic laws) does forbid it. Long before you achieved a spot that small, you would be violating the paraxial approximation. More to the point, if you do the full analysis, you would find that forming such a small spot would require impossibly large values of dX/dt (greater than c).

 

Not coincidentally, such a tight focus would also violate the second law of thermodynamics. If you could focus the sun’s rays more tightly than permitted by Liouville’s theorem, it would be possible to create a focal spot hotter than the surface of the sun. This is perfectly consistent with the first law of thermodynamics (conservation of energy), but would immediately violate the second law of thermodynamics, since you could in principle run a heat engine using the focal spot as the "hot" side of the engine, and the surface of the sun as the "cold" (!) side, thereby producing work using only one heat bath (the sun) rather than the conventional two.

 

To say the same thing in more detail: Let’s take a thermally non-conducting object and place it at the focal point of our system (point D in figure 1). A certain amount of energy falls on the image of the sun. The spot will heat up. Eventually it will become hot enough to glow. The temperature will stabilize at some temperature T such that the re-radiated power just matches the incident power. This temperature cannot be hotter than the surface of the sun; otherwise energy would be flowing from a (relatively!) cooler object to a hotter object, in violation of one of the corollaries of the second law of thermodynamics.

Phase Space of a Thin Lens

 

~modest

[Qfwfq]

If you don't have an argument then please don't patronize mine. Your premise breaks the second law and my conclusion is the very reason the law is upheld.

I did not patronize you, considering your replies of yesterday I said it's no use arguing, I don't have the time to go into the details that would be necessary.

 

What you quote is based mostly on the 2nd principle and so can't defeat my argument. The first paragraph might be a further consideration to my argument but it doesn't show that I was patronizing you, there were simply flaws in what you said. It is an interesting point but doesn't go much into detail, i would have to think better for it to convince me there's a strict and universal disequality on the temperature. It really doesn't support the things you said yesterday; you could have quoted that instead so it remains that you should be more reasonable in debate.

[modest]

I could not have quoted that as I had not yet read it. You should read the whole link.

 

~modest

[BrianG]

And then we could use the perpetual motion machine to provide green energy, a fuel without waste products. You must build a prototype immediately, to save the planet!

[Qfwfq]

I could not have quoted that as I had not yet read it.

Which does not justify the manner in which you did reply.

 

You should read the whole link.

I got back after having had a look at it, to say that it hasn't convinced me yet. Most of what it says has no bearing on my argument, as well as invoking the 2nd principle too, it talks essentially about a specific case in which there are obvious reasons for the limitation anyway. It is considering the flat lense approximation and the distant sun as a source. The only universal thing it says is conservation of phase space, but I had not neglected the matter of wavelength; I had even mentioned it though somewhat concisely. I would need to see a formal proof with integrals over the spectrum, else it seems to me that scaling up mirrors and/or raising temperature would get past the wavelength obstacle.

 

It also makes a quite preposterous statement:

It [Liouville’s theorem] is key to understanding the second law of thermodynamics; any system that violated Liouville’s theorem would immediately be usable as a perpetual motion machine.

The second sentence is logically equivalent to stating that Liouville’s theorem is implied by the 2nd principle, which I find absurd. To me it would seem more the other way around, Liouville’s theorem ought to be helpful in a formal demonstration of the 2nd principle, which however remains of a statistical nature so I waver even about the first sentence.

 

And then we could use the perpetual motion machine to provide green energy, a fuel without waste products. You must build a prototype immediately, to save the planet!

Quite right! :naughty: We agree optics can't be used to this end, we've just got into a tangle about exactly how and why it wouldn't work. :rolleyes: