wu :: forums (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi) riddles >> hard >> Ellipsoid Power Generation (Message started by: SWF on Mar 24th, 2003, 7:45pm)

Title: Ellipsoid Power Generation
Post by SWF on Mar 24th, 2003, 7:45pm
A property of ellipsoids is that if the inside surface is a perfect reflector, a ray of light originating at one focus will reflect off of the ellipsoid and pass through the other focus.

In the figure below FC is a portion of ellipse with foci A and B.  DE is a portion of a larger ellipse also having foci A and B.  Arcs CD and EF are part of the circle with center B.  Points A, C, and D lie on the same line and A, E, and F are also collinear.

http://pages.prodigy.net/scottfr/Images/Ellipse.jpg

Think of a perfect reflector made by revolving this shape about line AB to form an axisymmetric shape. Any photon originating at A must strike one of the ellipsoid surfaces and will be reflected toward B. For photons orginating at B, some strike the ellipsoids and reflect toward A, but some also strike the spherical portion of the reflector and come right back to B.

Take two small identical objects of the same temperature and place one at A and the other at B inside the reflector.  Also the rest of the space inside the reflector is vacuum.  Objects emit thermal radiation based on temperature in the form of photons. All the radiation from A strikes B. Only a fraction of that emitted from B arrives at A. So with time B will become hotter than A.  In theory, this temperature difference can be used to generate power.

If this would not work, why not?  If it would work, why aren't ellipsoids being used to generate power?

(Edited to add the figure that had been deleted years ago. I just happened to run across the old figure, and remembered it had been deleted from here.)

Title: Re: Ellipsoid Power Generation
Post by Wacky on Mar 24th, 2003, 8:12pm

Quote:
 If this would not work, why not?  If it would work, why aren't ellipsoids being used to generate power?

Energy can't come from nowhere, therefore it can't possibly work. But where is the problem in the logic?

All the radiation from A strikes B. Only a fraction of that emitted from B arrives at A. So with time B will become hotter than A.  In theory, this temperature difference can be used to generate power.

I see several possible problems with this:

1. how are you going to find a perfect reflector?
2. How could the temperature difference be used to generate power when A and B are totally isolated and all
3. Wouldn't photons bounce off the objects as well?

Title: Re: Ellipsoid Power Generation
Post by wowbagger on Mar 25th, 2003, 2:36am

on 03/24/03 at 20:12:35, Wacky wrote:
 Energy can't come from nowhere, therefore it can't possibly work.

The energy doesn't come "from nowhere", it's just transferred from one object to the other.

Quote:
 I see several possible problems with this: 1. how are you going to find a perfect reflector?2. How could the temperature difference be used to generate power when A and B are totally isolated and all3. Wouldn't photons bounce off the objects as well?

1. & 2.: This is meant to be a "gedankenexperiment" (I suppose).
3.: I suggest we use perfectly absorbing black bodies.

on 03/24/03 at 19:45:12, SWF wrote:
 For photons orginating at B, some strike the ellipsoids and reflect toward A, but some also strike the spherical portion of the reflector and come right back to B.

Agreed.

Quote:
 Take two small identical objects of the same temperature and place one at A and the other at B inside the reflector. All the radiation from A strikes B. Only a fraction of that emitted from B arrives at A. So with time B will become hotter than A.

On condition that the amounts of radiation originating from the two bodies are the same. This, however, is not true, because when the object at B gets hotter, it will emit more radiation. I haven't done any calculations and generally don't trust my intuition with regard to thermodynamics, but this seems to be an important point. After all, this thing should eventually settle down in a dynamic equilibrium, shouldn't it?

Quote:
 If this would not work, why not?

Basically, it's because of the second law of thermodynamics, I'd say.

Title: Re: Ellipsoid Power Generation
Post by Icarus on Mar 25th, 2003, 4:42pm
Obviously, getting power from this situation would violate the 2nd Law of thermodynamics, but the 2nd law of thermodynamics is an observed principle - i.e. we believe it to be true simply because we have never seen it broken.

Because of the 2nd Law, we can be confident that there is a flaw in the scheme. The puzzle here, as with James Fingas' osmotic perpetual motion machines, is to figure out what that flaw is - if no flaw can be found, then it means that the 2nd law is false after all.

The flaw is not directly in the lack of perfect reflectors or blackbody radiators, since we can closely approximate both - close enough that losses would be less than the energy gain.

The increased radiation from B will cause it to eventually reach an equilibrium, but at a hotter temperature than A. The question is, why can we not harness that temperature difference to run a heat engine?

Presumably we could add heat conduits running from each focus to the apex of the ellipse on its side. Such conduits would not interfere with the transfer of heat between A and B, since no light path between the two foci crosses these lines. By means of these conduits we obtain the same temperature differential on the outside of the apparatus, where we can hook it up to our heat engine. As long as the heat flow through the engine does not exceed the ability of A to transfer heat to B, our engine runs for free!

So, where is the flaw?

Title: Re: Ellipsoid Power Generation
Post by towr on Mar 25th, 2003, 11:48pm
It would be a nice way to refridgerate things..

Title: Re: Ellipsoid Power Generation
Post by James Fingas on Mar 26th, 2003, 10:29am
A corrollary of the second law of thermodynamics is that no machine can be made which transfers thermal energy from a cooler object to a hotter object with no energy input.

Therefore, if the second law is indeed true, point B cannot become any hotter than point A. The flaw is not in how to get the energy out.
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The only flaw that I've thought of so far that could ruin this apparatus is the approximation of the objects as points. It seems at first that this wouldn't cause too much of a problem, but think of this:

1) I will talk about the "focal accuracy" of an ellipse. I admit that a photon coming exactly from one focus goes exactly to the other focus. However, a photon coming from near one focus will end up only near the other focus. Therefore, for finite-sized objects, not all photons from A will hit B. I will use the following approximation: a photon which passes by focus A at a distance dA will pass by focus B with a distance dB. For small dA, dB is roughly:

dB = dA * lB / lA

lA and lB are the distances to the ellipse from point A and from point B respectively.

2) If we choose the objects to be the same size, then object A is effectively larger compared to CF than object B is compared to DE.In this case, photons from A will have a much worse focal accuracy than photons from point B. dA will be roughly the radius of object A, but lA will be a lot smaller than lB for photons hitting CF, and will be fairly close to lB for photons hitting DE.

3) If a photon from A misses object B, then it reflects back towards object A, ideally hitting object A (okay, not always, but I'm sure not going to try the math!)

4) Because of this, some of the photons from object A end up returning to object A. You can see that the proportion is roughly equal to the size of DE over the size of CF, which also dictates the fraction of photons from object B that return to object B (by dictating the size of the circular arcs).

5) Now you say: what if we made object A smaller than B so that all of the photons from object A hit object B? The problem here is that the quantity of photons emitted is proportional to surface area. B will be emit more photons than A, exactly cancelling out any accuracy increases.

6) Now you say: what if we just shrank object A and object B until they were right at the focal points? The problem here is that no matter how small you make them to increase focal accuracy, the required focal accuracy decreases at the same time. In other words, you don't gain anything.
[/hide]

Title: Re: Ellipsoid Power Generation
Post by LZJ on Mar 30th, 2003, 11:42pm
Well, I guess James Fingas is right from a practical standpoint, but I think there is another flaw in the theoretical portion of this question (besides the fact that it violates the law of entropy).

Firstly, to address the other posts, the temperature difference could theoretically be used to power stuff and all, its just that the total thermal energy of A and B will drop when this is done. Therefore, I don't think that's the flaw.

What I noticed is that for this question, the radiation pressure will not be uniform, so one will notice a net force in the direction of AB. If the reflector isn't fixed in place, the drift will result in the photons near B losing KE, thereby 'correcting' the imbalance in radiation pressure, and these less energetic photons would in turn impart less KE to B compared to the ones near A, so the temperature may revert to its original value. If one takes the drift into account, the photons will no longer always pass through the focal points.

The above case assumes that the reflector isn't fixed in place. If it is fixed in place...well, I haven't really thought about it yet...

Title: Re: Ellipsoid Power Generation
Post by Chronos on Mar 31st, 2003, 2:53pm

Quote:
 Firstly, to address the other posts, the temperature difference could theoretically be used to power stuff and all, its just that the total thermal energy of A and B will drop when this is done. Therefore, I don't think that's the flaw.
OK, so that's not a problem with the First Law, but it's still a problem with the Second Law.  Getting work for no cost other than some objects getting colder is a decrease in entropy, which isn't allowed.  And if you're worried about the mirrors drifting relative to the blackbodies, then just put attachment struts along the axis where they won't interfere anyway.

I did actually figure out the same problem that James did, but unfortunately, he beat me to the keyboard.  For that matter, I didn't work out the math, either.  I was just relying on "well, the math has to  work, or the laws of thermodynamics wouldn't be valid!".

Title: Re: Ellipsoid Power Generation
Post by Icarus on Mar 31st, 2003, 4:36pm
The heat conduits along the axes necessary to get the supposed temperature difference out where it can be used could also double as supports to keep the radiators centered at the foci.

I don't see the size of the radiators as being a true flaw, anymore than the other abstractions are. If the device would work at the abstract limit (point radiators, true blackbodies, perfect reflectors, perfect heat conduits of zero radius), then it should be possible to approximate it near enough to still see an entropy decrease.

The flaw has to be in the argument that heat will transfer from A to B in the first place. That in and of itself is a violation of the 2nd law.

Note that every light path from A to B is also a path from B to A. Though I have not followed the idea far enough to see if it works, I am curious if perhaps B will radiate just as strongly back along these paths as A is radiating B, (with A and B at the same temperature).

Title: Re: Ellipsoid Power Generation
Post by LZJ on Mar 31st, 2003, 5:48pm
Its true that " every light path from A to B is also a path from B to A", but the converse isn't true: that is the crux of the problem.

As many of you have acknowledged, the size of the components isn't the main flaw: I still feel that the problem lies in the imbalance in radiation pressure within the system.

Title: Re: Ellipsoid Power Generation
Post by James Fingas on Apr 1st, 2003, 1:35pm
Icarus,

I really think that the size of the radiators is the key here. The problem has no meaning for zero-size radiators, because they can't radiate anything and they can't accept any radiation. For sizes larger than that, you have the radiator-size problem.

The reason the math works out that way for point radiators is that the behaviour of the mirrors has a discontinuity at that point. For any finite-sized region of space, there's no problem.

It's the same problem as the Carnot efficiency of a heat engine with a heat sink at absolute zero. Theoretically, you can get 100% efficient conversion of heat to energy. That's because the equation doesn't work at that one point. But for any finite range of heat source and heat sink temperatures, there's no problem.

Title: Re: Ellipsoid Power Generation
Post by LZJ on Apr 3rd, 2003, 5:23am
Hmm...James Fingas is right...the problem does lie with infinitely small points. Furthermore, it doesn't really violate the law of entropy in such a case: the photons, which all originate at 2 points, get increasingly disordered and scattered about throughout the interior of the reflector, so even if A does experience an increase in temperature, the change in entropy will likely >=0.

Title: Re: Ellipsoid Power Generation
Post by Icarus on Apr 3rd, 2003, 3:33pm
Sorry, but I still don't buy it. If you could design the radiators to emit photons only in an essentially radial direction, then the size of the radiator would not matter (almost anyway - it still gets in the way of some paths). Photons would take the same paths as if they were emitted from the foci.

Basically, my objection to this "solution" is that it is more of an engineering limitation, rather than anything fundamental. I believe that there is a deeper reason that this set up fails.

PS: James, your Carnot example doesn't work either. The only reasons we can't get perfect conversion of Heat into energy at absolute zero is that (1) by the 3rd LoT, we can't ever reach absolute zero, and (2) by the 2nd LoT, we can't ever reach the limiting efficiency of a Carnot cycle. The equation itself is accurate for T=0. And as we approach T=0, and reversability, effeciency does approach 100%.

Title: Re: Ellipsoid Power Generation
Post by Chronos on Apr 3rd, 2003, 6:56pm

Quote:
 Sorry, but I still don't buy it. If you could design the radiators to emit photons only in an essentially radial direction, then the size of the radiator would not matter (almost anyway - it still gets in the way of some paths). Photons would take the same paths as if they were emitted from the foci.
How do you restrict the radiators in this manner?  A hint for thinking about this:  What happens when a photon from elsewhere strikes a radiator obliquely?  If it gets absorbed, then the baffles or whatever you've installed must also be able to radiate away photons in that direction.  If it passes right through the baffles, then the baffles also can't restrict the outgoing photons.  If they reflect the photon, then you've essentially got absorbtion and instantaneous re-emission, from that illegal point.

Another way to think about this:  When you have an object in front of a curved mirror, you end up with a magnified image.  Optically, that magnified image will behave exactly like a real object of that size.  Put the object right at the focal point, and the magnification will be infinite.  When you have finite-sized objects, we know what this means for the images:  You'll effectively have two infinite blackbody plates parallel to each other, which will therefore be at the same temperature.  But what does infinite magnification do to point objects?

Title: Re: Ellipsoid Power Generation
Post by Icarus on Apr 5th, 2003, 8:49am
Actually, in the idealized case, that would not be a large problem, since the only source for photons would be the other radiator, and any rays from it that strike the reflectors would be coming in normal to the surface. The only case where photons would come in obliquely would be those that pass straight from one radiator to the other. The smaller the radiators are with respect to the overall size of the apparatus, the smaller the entropy gain from this is. So you still could have an overall entropy decrease.

But it is NOT my point here that the finite size problem is solvable. I don't know if my suggestion is workable, or if it can or cannot be solved by another means. What I am arguing is that the size problem is not the ultimate solution to this: I believe that there is a flaw in the analysis that has nothing to do with size. That if you could solve the size problem, you would discover that A and B still remain the same temperature.

Consider a variation of the arrangement: (I don't have a handy website set up to post pictures to, so you are just going to have to settle for a text drawing that only conveys vague generalities.)

-----------------------------------------------
|..................................|..........|
|....-----.........................-----......|
|....|.A.|.........................|.B.|......|
|....-----.........................-----......|
|..................................|..........|
-----------------------------------------------

With suitable design of the radiator and mirror shapes, you could ensure most of the photons leaving A arrive at B. And that those leaving B on the left arrive at A. Those emmited by B on the right are trapped behind the middle reflector and bounce around until they are eventually reabsorbed by B. The same situation as SWF describes in the original problem ensues: more photons going from A to B than from B to A.

Okay, I admit that is a little vague, but if the analysis SWF provides were accurate, and the only problem with the ellipsoidal generator were the finite size of A and B, then getting around the 2nd LoT would only require dreaming up a way to get object A to expose more radiating solid angle to B than B exposes to A. This does not strike me as an impossible task. The ellipsoidal method is just one of many possibilities for accomplishing this.

I have to believe that there is a deeper miscomprehesion in the analysis given, and that two objects A and B at the same temperature will radiate each other equally regardless of how much radiating solid angle each has exposed to the other.

Title: Re: Ellipsoid Power Generation
Post by LZJ on Apr 6th, 2003, 1:43am
Pardon me if I'm wrong, but I don't think there will be a violation of the 2nd LoT:

At the initial state, the energy will reside entirely in A and B. After that, however, there will be energy in the form of photons scattered about the container/reflector (radiation pressure). This would mean an increase in entropy that might negate the entropy loss due to the increase in temperature in B.

Title: Re: Ellipsoid Power Generation
Post by Icarus on Apr 6th, 2003, 11:41am
Actually, in SWF's original idealized scenario, if I may be allowed to ignore the size problem for a moment, this is not the case. Because we are assuming perfect reflectors, the photons are not scattered, but move along very definite paths. Information is not lost, so there is little entropy associated with the photons. My own suggestion is not well-enough defined to say one way or the other. But even if there is enough entropy in the photons to cover the entropy loss, you could "seed" the container with photons before your experiment begins, so any entropy associated with the photons is already present, and your experiment will show only the decrease.

Title: Re: Ellipsoid Power Generation
Post by LZJ on Apr 6th, 2003, 6:55pm
:-/ So has anyone come up with a reasonable explanation yet?

Title: Re: Ellipsoid Power Generation
Post by Chronos on Apr 6th, 2003, 7:17pm

Quote:
 Actually, in the idealized case, that would not be a large problem, since the only source for photons would be the other radiator, and any rays from it that strike the reflectors would be coming in normal to the surface.
You're missing the point.  Suppose you build your custom-made radiators, which are guaranteed to radiate only radially, and put them in SWF's device.  I then open up the device, take out one radiator, and put it out in the middle of a sunny field where light can hit it with any impact parameter.  What happens to the off-center photons?  If your device interacts with off-center photons in any way, then it can also interact with them in an exactly time-reversed manner:  If it can absorb an off-center photon, then it can emit a photon along the exact reversed trajectory, etc.  If your device does not interact at all with off-center photons, then you're effectively back to saying that it's zero size, which is unphysical.

Title: Re: Ellipsoid Power Generation
Post by towr on Apr 6th, 2003, 11:31pm
You can shield the radiator with some supercooled absorbent material, any photons that fall on it will be absorbed, turned to heat, removed, and thus no photons can be emited (since there simply is not enough energy for it)

Title: Re: Ellipsoid Power Generation
Post by Icarus on Apr 7th, 2003, 5:22pm

on 04/06/03 at 19:17:29, Chronos wrote:
 If your device interacts with off-center photons in any way, then it can also interact with them in an exactly time-reversed manner:  If it can absorb an off-center photon, then it can emit a photon along the exact reversed trajectory, etc.

So it reflects photons unless they come in at a vertical angle. Non-vertical photons reflected does not significantly interfere with the heat transfer mechanism, since in this scenario, they are extremely rare. And this is physically possible - every boundary has an angle outside of which all light is reflected, not absorbed.

But you seem to be missing MY point. Which as I said earlier is NOT that the size problem is solvable - I have admitted to not being sure of that - but rather that there must be something else wrong with the analysis SWF supplied. If SWF's analysis were correct other than accounting for the size, then getting around the 2nd LoT would still be possible. If not by this scenario, then by some other. Alll you have to do set up two objects so that one has a larger section of solid angle radiating at the other object, than the other object has radiating at it.

Title: Re: Ellipsoid Power Generation
Post by Chronos on Apr 9th, 2003, 8:07pm

Quote:
 If SWF's analysis were correct other than accounting for the size, then getting around the 2nd LoT would still be possible. If not by this scenario, then by some other. Alll you have to do set up two objects so that one has a larger section of solid angle radiating at the other object, than the other object has radiating at it.
This isn't a problem, unless you show that such a setup is possible.  Your text drawing won't do it, either:  It'll behave the same no matter where you put your "barrier mirror", in particular, what you have is the same as putting the barrier in the middle (with the same hole cut out of it).  Each blackbody receives as many photons as it emits.

Nor will towr's solution of supercooled superabsorbant material work.  A superabsorbant material is a blackbody, and it absolutely will radiate.  Decrease the temperature, and it'll radiate less, but it'll always be radiating some.  When one of these cold radiators absorbs a photon and heats up slightly, it'll also increase its radiation rate slightly until it next emits a photon, thereby staying in stable equilbrium.  All that making the objects cold accomplishes is that it means there will be less photons bouncing around inside.

Title: Re: Ellipsoid Power Generation
Post by Icarus on Apr 9th, 2003, 9:47pm
I'm sorry, but I don't feel free to glibly assume that because I haven't thought of a way to do something, it must not be possible. I have seen lots of variations of "Maxwell's Demon" before, and they all had a more fundamental flaw than the size problem is here. I still think that there must be something else wrong with this idea.

Title: Re: Ellipsoid Power Generation
Post by BNC on Apr 10th, 2003, 7:21am
“If this would not work, why not?  If it would work, why aren't ellipsoids being used to generate power?  “

It would work. It isn’t being used because no one thought about it. I just had a patent granted to me – I’m going to be rich beyond my wildest dreams. Muhahaha

Actually, two weeks later ( :-[) I got the same answer as did James Fingas, using much the same logic. I really DO think the size is the case here.
Coming from optics, I can tell you that “point sources” are used all the time – but only at a distance. Going “inward” to the point source will yield any number of “paradoxes” (e.g., we can “show” that light diffraction doesn’t exist). The point source black bodies used here may yield no energy – other wise they have infinite brightness (and the brightness conservation law will yield many more paradoxes). Approximating it with any finite size body (small as you wish) will have the problems mentioned by James (the magnification mismatch => losses are not symmetrical).

Title: Re: Ellipsoid Power Generation
Post by towr on Apr 10th, 2003, 10:45am
I think conservation of impuls might be the problem.. photons have impuls, and it needs to be conserved.

Photons veering of to the circle parts break that balance if the contraption doesn't move, and if it does move photons will be moving all over the place increasing entropy..

Or something..

Title: Re: Ellipsoid Power Generation
Post by LZJ on Apr 12th, 2003, 5:18am
Err...I've already mentioned that in my previous post...

Title: Re: Ellipsoid Power Generation
Post by James Fingas on Apr 15th, 2003, 5:46am
Sorry I've been gone from this thread for so long. I have three rebuttals for you, Icarus.

1) You say that your assumption of zero-volume light-emitting objects is more physically valid than my assumption of zero-temperature heat sinks. How is this so? Besides the fact that mine specifically has a law against it, and yours doesn't, of course...

2) Consider the following apparatus: construct two spheres, mirrored inside. One has internal radius A, and the other has internal radius B (A < B). Join the two spheres so that the distance from center to center is just slightly smaller than A+B. Then there will be a pinhole joining the interiors of the two spheres. Put a small ball in the center of each sphere, so that all light going through the pinhole is parallel. Now the amount of light passing through the pinhole from the ball in A will be larger than the amount of light passing through the pinhole from the ball in B (because ball A is closer to the pinhole). This will make the ball in sphere B become hotter than the ball in sphere A.

In case you don't see what I'm getting at, this is exactly analogous to the ellipsoidal reflector (for a very small pinhole).

3) Working from the other direction, if we assume that the second law of thermodynamics holds, then to avoid any power generators like the ellipsoid or these joined spheres, certain conditions must always hold. Here is my statement (I hope that this is clear enough for you to get the gist):

"Choose any two light-emitting surfaces A and B in space. Then, regardless of any distribution of mirrors, lenses, gravitational fields, etc. the 'optical coupling' from A to B must be the same as the optical coupling from B to A. That is to say, the probability that a light ray emitted from A will hit B is the inverse of the probability that a light ray emitted from B will hit A, and is equal to the ratio of the emitting surface areas of A and B. Because of this property, these two surfaces will approach thermal equilibrium with each other over time."

There is no problem if a single point in space does not satisfy this. It is sufficient that it be satisfied for any surface.

Now here are some more problems I don't have the answer to yet:

4) Consider my apparatus from point 2, and put a lens in the pinhole, constructed so that all light from ball A is focussed onto ball B, and all light from ball B is focussed onto ball A (i.e. ball B is the image of ball A). Now ball B will become hotter than ball A, right?

5) Or how about this one: Consider the ellipsoid power generator, but make the ball in the larger ellipse part a lot heavier, so that gravity will bend the light towards the ball. Now it will capture a greater portion of the other ball's light, and will become hotter than the ball in the smaller ellipse part, won't it?

Title: Re: Ellipsoid Power Generation
Post by BNC on Apr 15th, 2003, 10:49am
Point 4: No. Since A<B, to be an image of eachother, ball B must be larger than ball A, thus it emitts more energy => more energy reaches A.

Point 5: Without calculating, I assume that once an object heavy enough to bend light is placed in the generator, the path of many more light rays will be changed.. it will probably cancel out in the math.

Title: Re: Ellipsoid Power Generation
Post by James Fingas on Apr 15th, 2003, 2:07pm
BNC,

Ah, now I understand why I don't work with lenses ...

Title: Re: Ellipsoid Power Generation
Post by mike1102 on Jun 2nd, 2003, 5:52am
wow..... lots of confusion here about heat, temperature and heat transfer.

This "riddle" is really a heat transfer question. Heat transfer doesn't care about photons or black-body radiation/absorption or optics or most of the subjects brought up in the posts. Heat transfer cares about mass, temperature gradients, specific heat and thermal conductivity (see the heat equation). For heat to flow from one point to another, there must first be a temperature difference between those two points - no delta Temp - no heat flow!

Since we're starting with two identical objects at the same temperature and if the walls of the reflector can't absorb heat, there is no delta T, so there is no heat flow - end of story.

There is also an implication in the statement that heat must radiate uniformly into space - this is not the case in reality. Heat will flow in whatever direction there is a thermal gradient, according to a material's thermal conductivity.

My conclusion: This concept for power generation needs a major re-work. But keep trying...... ya just never know what the future holds.

Title: Re: Ellipsoid Power Generation
Post by towr on Jun 2nd, 2003, 7:42am
I think the point of this exercise was not to use the laws of thermodynaics to prove they apply to this special case..

Your story doesn't give any explanation of why there isn't any heat exchance (in this case) when there isn't a temperature difference.

Also when you have to objects of equal temperature that touch than their molecules do exchange kinetic energy = heat, but the exchange is equal in both direction (pretty much) giving a net result of no temprature change.
Here the construction of the problem suggest the exchange cannot be equal in both direction, and we have to figure out why it is nevertheless the case.

Title: Re: Ellipsoid Power Generation
Post by Rodrick Crider on Jun 3rd, 2003, 11:15am
I don't know anything about thermodynamics, so forgive me if this is completely wrong, but maybe the photons that are eminating from B and bouncing off of the spherical portion have a high probability of colliding with successive photons coming from B, thus causing those photons to scatter instead of returning to B?

If photons can pass through eachother, then this wouldn't make sense. I assume I am probably wrong since I have never seen two laser beams bounce off eachother. Oh well, just a thought...

Rod

Title: Re: Ellipsoid Power Generation
Post by Icarus on Jun 3rd, 2003, 6:07pm
Photons do not collide - they pass through each other. Otherwise sight would be useless because the thick sea of photons we live in would be scattering off each other, destroying the images they carry. Photons interact only with electromagnetically charged matter. Since photons themselves are neutral, they do not interact with each other.

Title: Re: Ellipsoid Power Generation
Post by kalie on Jun 3rd, 2003, 9:05pm
mike, i guess your the one whos confused. heat transfer does care about photons and radiation, your claim that these concepts are not involved is only valid for a high school student.

Title: Re: Ellipsoid Power Generation
Post by Rodrick Crider on Jun 4th, 2003, 9:57am
Thanks Icarus, the more I thought about I realized that that didn't make much sense...I better stick to the easies...

Title: Re: Ellipsoid Power Generation
Post by Icarus on Jun 4th, 2003, 3:28pm

on 06/04/03 at 09:57:01, Rodrick Crider wrote:
 Thanks Icarus, the more I thought about I realized that that didn't make much sense...I better stick to the easies...

Nobody always makes sense. About everyone else who has posted to this thread seems sure that I'm not making sense in it!  ;D

Title: Re: Ellipsoid Power Generation
Post by mike1102 on Jun 5th, 2003, 5:22am
I thought about it some more...... and I was wrong about heat transfer not caring about black body radiation. It certainly does. Black body radiation is the means by which the sun heats the earth and the earth re-radiates energy to keep from burning up.

But I know of no means to accomplish heat flow without a temperature difference. Just like in an electrical circuit, current will not flow without a potential difference; heat will not flow without a temperature difference.

Now.... back to the ellipsoid. If heat flows from A to B, then heat must also be able to flow from B to A. I don't think there is such a thing as a "heat diode" that would allow heat to flow in one direction and not in the opposite direction.

Title: Re: Ellipsoid Power Generation
Post by Icarus on Jun 5th, 2003, 5:07pm

on 06/05/03 at 05:22:53, mike1102 wrote:
 But I know of no means to accomplish heat flow without a temperature difference.

No one else seems to be able to figure it out either (unless this trick really does work ::)), that's why the 2nd law of thermodynamics is considered a "law"!

Title: Finite size of A and B
Post by Rujith de Silva on Oct 14th, 2003, 9:51am
I agree with Icarus that this puzzle should fail for a "robust"
reason.  Here's my idea of the "robust" reason:

Assume that A and B are spheres of equal finite size, but small
relative to the rest of the apparatus.  Would EVERY ray from A hit B?

Rays that emerge perpendicular to A's surface (i.e., that are
collinear with A's center) would always strike B, this being a basic
property of an ellipse (except for the arbitrarily small amount that
hit A again).

Now consider a nearly horizontal ray that emerges to the right from
the top of A, i.e., with its path as far away from A's center as
possible.  This goes to the right, strikes the far right wall, and
would always hit B.  Why?  Because the "inaccuracy" caused by its NOT
being from the center of A is reduced by the fact that the far right
wall is closer to B than to A.  Think about the angles of reflection.

However, consider a nearly horizontal ray that emerges to the LEFT
from A's top.  It strikes the far left wall, and then would MISS B,
because the "inaccuracy" is magnified by the far left wall being
closer to A than to B.

So all rays from A will NOT strike B, because of the effect of the
finite sizes of A and B.  Similarly, all rays from B will not hit A,
even ignoring those that hit the central "distortion."  This holds
true even if A and B are made arbitrarily small, because what matters
is the RELATIVE sizes of A and B.

(Note that all rays from B that hit the central "distortion" will hit
B again.)

What happens to all those rays from A that miss B, and B's rays that
miss A?  They will bounce around within the distorted ellipsoid and
hit A or B again after some number of trips.  Eventually, everything
will reach equilibrium when the rays permeating any unit area within
the ellipsoid's space are at the same temperature as A and B (sorry, I
don't know the exact thermodynamic terms).  At that point, A and B
will also be in equilibrium at the same temperature.  Proving this
requires more mathematics/thermodynamics, but I think the basic
argument is correct, and intuitively persuasive.

- Rujith.

Title: Re: Ellipsoid Power Generation
Post by Icarus on Oct 15th, 2003, 7:59pm
That is good solution, Rujith. It is I believe the one that James offered to which I objected as the sole cause, though.

However, recalling some of what I have read about other anti-entropic schemes makes me more doubtful that any other reason may exist. Scientific American once published an article about the 2nd Law that had several versions of Maxwell's demon in it. In particular, it examined some mechanical devices intended to move the hot particles only from one side to the other. They examined each and showed the flaw which would cause them to fail. But rather than deep physical principles that applied widely, what they found were simple mechanical problems that were different for each device, but ALWAYS present, no matter how hard you tried to remove them.

So, perhaps I am wrong about there being more.

Title: Re: Ellipsoid Power Generation
Post by Rujith de Silva on Jan 13th, 2004, 12:10pm
Icarus: sorry, I had seen the discussion on point-sized objects, but
had glazed over the stuff on finite-sized objects.  I agree that we're
looking for some simple, universal principle that cracks the problem,
but it's eluded us so far.  - Rujith.