Violating the Second Law?

I'm guessing, but B radiate higher frequency (more energetic) light than A since it's a blackbody at a higher temperature, so could this extra radiation compensate for focal point effect you mention above? Or can someone else think of another effect that might be helping fix the Entropy problem?

Roger, when you said "I'm guessing, but B radiate higher frequency (more energetic) light than A since it's a blackbody at a higher temperature...", did you consider how B got hotter than A in the first place - in violation of the 2nd law?

The fact that B eventually radiates more than A is what is causing the equilibrium - but by then the 'crime' has been committed!

Can the reflectors heat up? In any real system I'm aware of, the mirrors will heat up slightly from incident radiation. Since the reflector is much closer to A than to B, would it heat up much more, reducing the energy focused on B?

Roger, you asked: "Can the reflectors heat up?"

The question stated "perfectly reflective cavity", ruling out absorption.

But, even slightly less than perfect reflectivity should make little difference, because of the relatively large difference between the radiation arriving at point A and B respectively. It is determined by the size of the spherical mid-section of the cavity.

It seems to me that the part of the radiation from point "B" that hits the inside of the circle will get relected right back at itself and take no further part in the game

Aurizon wrote: "It seems to me that the part of the radiation from point "B" that hits the inside of the circle will get reflected right back at itself and take no further part in the game."

This is exactly what is stated in the OP, but, per se, it makes no difference to the temperature of B. It does however mean that while all the radiation from point A reaches point B, not all radiation from point B reaches point A, causing more temperature at point B - in violation of the second law.

Interesting your problem!

Please visite also the adress:http://peswiki.com/index.php/Directory:Rauen_Environmental_Heat_Engine#Overview

Nic

I think the E-H Engine is as fundamentally flawed as the " heat engine" one can construct from the reflective vacuum vessel above! Nobody has ever demonstrated a violation of the 2nd law above reasonable doubt and probably nobody ever will. This does not stop people from trying, though. It is usually not easy to find the flaws, as this teaser shows!

I believe what is happening is that while all the radiation from A reaches B it is not evenly distributed as it approaches B but only arrives at B from the direction of the parabolic reflectors. As a result of this increased intensity the re-radiation from B in the direction of the parabolic reflectors would be proportionally higher would hence balance out. The net result would be not transfer of energy from A to B.

Masu, you said: "As a result of this increased intensity the re-radiation from B in the direction of the parabolic reflectors would be proportionally higher would hence balance out."

You mean the average temperature of B remains the same as A, while there are hot spots on B? Does this not still mean a transfer of heat from a colder to a hotter surface that violates the 2^nd law?

Yes it dose mean that doesn't it. Oh well scratch an I will go and procrastinate some more.

Jorrie, While I ponder your riddle, I thought I would pass along this example of violating the second law I found in an FAQ for the SimpleTech network server I just bought. It is excerpted from the answer to "Why does the outside ...(case)... get so hot?" "To ensure the hard drive operates within its temperature specifications, the metal case acts as a heatsink, pulling heat away from the hard drive inside. Due to this, the outside of the case is actually warmer than the drive itself." Amazing what you can do with market-speak, violate the laws of physics. While I'm at it, doesn't your riddle have something to do with the lensing effect of the mirrors equalizing the effective temperature that each point "sees" of the other?

rcapper, you wrote that the sales blurb said: "Due to this, the outside of the case is actually warmer than the drive itself." Amazing"

They could actually be right if they meant that the magnetic disc itself is cooler than the case. The heat surely comes mostly from the drive motor?

Turning to the teaser, point B sees all of point A's radiation and point A sees only part of B's, due to B seeing itself in the spherical part. But you are getting warmer (pun intended)!

Jorrie

You wrote: "They could actually be right if they meant that the magnetic disc itself is cooler than the case. The heat surely comes mostly from the drive motor?" I could see perhaps that that might be briefly after it was first powered, depending on the heat flow dynamics inside the device. However, since the outside of the case is the boundary at which all internal heat must find egress, once the temperature has reached an equilibrium, I don't think any point inside the case would be cooler than its surface. On your riddle, were B to rise above A wouldn't it simply re-radiate to A until there was equilibrium? And in fact, because the path between A and B is bidirectional, the points would never diverge?

Hi rcapper, you wrote: "However, since the outside of the case is the boundary at which all internal heat must find egress, once the temperature has reached an equilibrium, I don't think any point inside the case would be cooler than its surface."

Yep, I think you are right - my hasty remark was bad!

On the riddle, what you said last moves us further from the solution, not closer! One can't get away from the fact that point B receives more radiation than point A, unless B is hotter than A (so that B radiates at a faster rate than A).

The solution lurks in the lensing effect that you mentioned in post 12…

I had the sense that that was going in the wrong direction. How about this. Since only a portion of the energy radiated by B is incident upon all of A. the lensing effect redistributes the energy density in a manner such that each point effectively "sees" an equivalent temperature even though B may be at a higher temperature than A. It is frustrating to me because I remember a similar puzzle with the sun and a magnifying glass but can quite recall it.

rcapper, you are still violating the 2^nd law when you say: "...the lensing effect redistributes the energy density in a manner such that each point effectively "sees" an equivalent temperature even though B may be at a higher temperature than A."

The 2^nd thermo-law demands that the two masses must stay at identical temperatures. Otherwise, one can take them out and extract real work from them, then repeat the cycle, yielding 'free-energy'!

So the question is, what prevents B from getting hotter than A? It is in the 'lensing effect', but not quite as you described it.

I was, I am and I will be meticulous, hairsplitting, pedantic et cetera.

Do we wont (in 2006 !!!) distinguish between LAW (of Physics) and PRINCIPLES (of Thermodinamics) ?

Why not open a discussion on the differences between the two concepts ?

Please pardon any imprecision in my comments and please let us not corrupt another thread by diverging into philosophical ramblings that are irresolvable as anything more than personal opinion. It becomes so tiresome reading through perpetual silliness to extract some small gem of technical information. Surely there must be a site that would satisfy those who want to argue such matters somewhere else.

Hi Camillo, I think you are welcome to open such a discussion - "Do we wont (in 2006 !!!) distinguish between LAW (of Physics) and PRINCIPLES (of Thermodinamics) ?"

It will be interesting, but rather post it as a new thread, since it may be considered off-topic in this one.

Great suggestion, thanks.

Outstanding question Jorrie! I'm a first time responder, and found this very interesting, since reading it yesterday over coffee. I'll take a shot at the answer!

The flaw is in the statement "Since point B receives more radiation than point A...". The statement presumes that while all the energy radiated from A is received at B, not all the energy at B makes it back to A.

However, a careful analysis shows that every spot on A (every unit of area, dA, on the surface of the sphere at point A) can be mapped to a corresponding spot (differential unit of area, etc.) on B. So energy radiated off any arbitrary spot on A hits its corresponding mapped spot on B, (and only that spot on B). Energy radiated from any arbitrary spot on B hits either its corresponding spot on A (therby balancing the energy received from A), or, it reflects off the inner surface of the (spherical section) container and hits back where it started, also balancing.

JKS, you said "So energy radiated off any arbitrary spot on A hits its corresponding mapped spot on B, (and only that spot on B). Energy radiated from any arbitrary spot on B hits either its corresponding spot on A (thereby balancing the energy received from A), or, it reflects off the inner surface of the (spherical section) container and hits back where it started, also balancing."

If the radiation from every spot on A ends up on B, but not all spots on B ends up on A, it still means that B receives more energy from A than what A receives from B. So, B must get hotter than A, so that the lesser area on B radiate as much energy as A does, meaning energy flows from the cooler to the hotter body...

Jorrie,

Does some of the light focused on B arrive out of phase with other light arriving at B, cancelling out some of the ligh on B so that the incident energy is the same as received by A?

Great question by the way. I just wish I had a clue how to answer it.

Roger asked: "Does some of the light focused on B arrive out of phase with other light arriving at B, cancelling out some of the light on B so that the incident energy is the same as received by A?"

Sure, there will be out of phase pockets, but there should be as much constructive as destructive interference, averaging out, even over very small areas - I hope!

Jorrie, you objected to my earlier post, noting that: "If the radiation from every spot on A ends up on B, but not all spots on B ends up on A, it still means that B receives more energy from A ..."

Sorry, I should have been more precise - how about this?!

Any energy emitted from A, lands on B. Any energy emitted from B, that is not reflected directly back to B, lands on A. The energy transfer between the two bodies, A and B, balances, exactly.

The entire surface of A maps to a portion of the surface of B, and this same portion of B maps back to A. The remaining portion of the surface of B maps to itself, via reflection off the spherical reflective walls.

There is no mechanism for net heat transfer between A and B. A and B are initially at the same temperature. The radiant energy passing between them balances. The walls of the container neither emit, absorb, nor transmit (through) heat energy. Spheres A and B reside within a vacuum.

There is no net energy transfer between A and B, therefore there is no temperature change of either A or B. The second law of thermo remains intact.

I think rcapper really had the answer in post #14: without trying to analyse the shapes at all, every path of radiation involving reflections is bi-directional.

Quote: "I think rcapper really had the answer in post #14: without trying to analyse the shapes at all, every path of radiation involving reflections is bi-directional."

This is a very good argument, but it fails on these counts:

1) the absorption and radiation processes are not reflective by nature, so some of the absorbed energy will conduct as heat to other areas of mass B, which means...

2) some of the absorbed energy at B will be radiated in a direction that can only "see itself" in the spherical mid-section and will never reach A, but...

3) all of the absorbed energy at A will be radiated to B, on account of the geometry and the assumption that both A and B resemble point masses in relation to this vastness of the cavity.

I know I'm reaching here, but could the photons exert a pressure and move mass B to an equilibrium point in the chamber (since photons have momentum)?

You said

"1) the absorption and radiation processes are not reflective by nature"

Consider any infinitely small point on the surface of B "say": then any "ray" leaving it whatever route it follows and wherever it ends up is exactly matched by a ray in the opposite direction.

Just to consider one aspect of the shapes what happens to any ray from A which is reflected at some point(s) in its path from the spherical reflector.

Randall, you said: "Consider any infinitely small point on the surface of B "say": then any "ray" leaving it whatever route it follows and wherever it ends up is exactly matched by a ray in the opposite direction."

You are right as far as rays are concerned, but I do not think one can treat energy transfer simply on the principle of ray balancing. Say we started this riddle with mass A hotter than mass B. Then the balancing of rays would still hold, but heat will be transferred from A to B until they have equal temperatures.

As our 'resident physicist' has pointed out, most rays originating from A will miss B on the 'first pass'. They are then scattered throughout the cavity, until either A or B eventually absorbs them (or what is left of them after numerous partial absorptions by a less than perfect reflective cavity).

It seems to me that the problem lies largely (entirely?) in the false assumptions implied in the question. "Point sources" would have infinite temperatures for there to be ANY radiation... So in this form the problem amounts to hiding divisions by zero. Thus the only way to analyse point sources would be as the limit of finite-size ones.
If we therefore allow that the sources to have finite size (however small), we could see that the image of A from the reflector near A is larger than A, and the image of B due to this reflector is smaller than B. If the objects are the same size, energy from A will therefore miss B - in spite of the apparent "nominally" focus at its centre. This just covers one of the sets of light paths. However, a complete "ray analysis" would always show that the amount of energy received at B is the same as that leaving it (for the case of perfect reflectors and equal temperatures).
Incidentally, in the limit of a lossless cavity, the light intensity and spectral distribution throughout the enclosure will correspond to the radiation from the surface of a black body at the same temperature.

Physicist wrote: "Thus the only way to analyse point sources would be as the limit of finite-size ones.
If we therefore allow that the sources to have finite size (however small), we could see that the image of A from the reflector near A is larger than A, and the image of B due to this reflector is smaller than B."

Bingo!

Has the fact that a physicist (I presume) had to come to the rescue of us engineers got anything to due with the engineer's way of thinking (approximately, practical) vs. the more theoretical physics way of thinking?

Physicist,

I believe your answer is correct, but I have a question regarding something you said.

You said ""Point sources" would have infinite temperatures for there to be ANY radiation... "

Why would this be true? An answer in the form of an equation would be fine, or an explanation I can get my head around, I'm sure your not making this up but I had never heard that before and would like to know why.

Hi Roger, if Physicist will forgive for barging in here, the easiest way to get your head around it is to remember that point sources have zero volume, so any energy there will have infinite density - which is not on!

Perhaps physicist can expand on it, especially on what infinite temperature from a zero volume that radiates means? I guess it is just the same thing said differently!

That sounds reasonable. I guess my confusion comes from the fact that Temperature is average kinetic energy, but if the source is a point, there isn't any place for the object to move, so how can it have temperature? (Keep in mind, even as I ask this I'm sure there is a flaw in my logic here. An equation would probably help me see it)

Roger wrote: "An equation would probably help me see it"

I am sure physicist can provide us an equation that will illustrate this!

A very good analysis published on the web and titled: "An Optical Perpetual Motion Machine of the Second Kind" was written by Leigh Hunt Palmer of the Simon Frazer University, giving some equations applicable to this riddle.

I think your logic is pretty sound, Roger. It relates to an interesting point brought up earlier, re the difference in views (or perceptions) of engineers and physicists. The same differences show up in the perception of mathematics by mathematicians and engineers. All (E,P,&M) would agree easily on the temperature of an obviously finite object, such as a BB. All would agree that if we make that object smaller, and retain its energy, then its temperature must go up. Where the agreement can fall apart, is right at the transition from very, very small to infinitely small. Practically speaking, it makes a great deal of sense to say that an infinitely small particle has zero volume. But then, there is "nothing" to vibrate, so there can be no temperature. [Crudely speaking, it is a little like a light bulb being fed by increasing amounts of energy: it gets brighter and brighter, and then, poof, goes out. (Granted, the analogy is impossible loose.)]

Were we to consider A and B to be true points, then the problem could be viewed (at least from my perspective) in two very different but equally valid ways:

1. The points have no temperature at all: they can't… they have no volume and no mass. If they have no temperature, they also have no energy.
2. The points have infinite temperature, because their energy is infinitely concentrated. If they have infinite temperature, they also have infinite energy.

Yet another really interesting thread! Thanks, to all of you.

Isn't the question upside down. B can't be heated by A BECAUSE of the second law? A will always have to be at a higher level for the heat tranfer to happen. Sort of like putting a big reflector behind a fluro light and trying to focus it down to start a fire, can't be done

Jorrie I guess prcatical have the answer for the question.

think of two identical bodies A AND B of equal mass and volumes.

A is having temp of 80Deg and B is 100 Deg.

now if we keep them at a given distance, according to 2nd law heat should flow from B to A.

but now think of a lence that is having a focus distance of 1/4 of total distance and kept in beween closer to B.

all the heat wave Radiated from A will be focused on a very limited section of B this will start heating local area of B. despite of the fact its temp is more then A.

So in a way it is the energy density that is important.

rakesh_semwal, I'm afraid you are wrong when you say: "all the heat wave Radiated from A will be focused on a very limited section of B this will start heating local area of B. despite of the fact its temp is more then A."

In fact, a lot of the radiation from A misses B completely and is then scattered throughout the vessel. Read the reply from Physicist above!

Guru Ji

When I say all the heat means all the heat that is focused by lance,

What I am trying to say is, may be your system working like a heat pump. And that's why its not violation of 2nd law.

Theoretically we can increase the heat of local area of sun by putting a huge reflector on the earth and then focusing it back on sun just on area of say about 1 square meter. But it is not the violation of 2nd law. Despite of the fact sun is much hotter then earth.

Hi rakesh_semwal, you wrote: "What I am trying to say is, may be your system working like a heat pump. And that's why its not violation of 2nd law."

No, it's not working like a heat pump - it's not 'working' at all, because the two masses remain at the same temperature. That's why it's not violating the 2nd law!