Talk:Taken by Storm

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Gee, let me see if I understand this then - take 3 temperatures in a room: one in a freezer; one in an oven; and, one at some other randomly chosen location; add them together and divide by 3. That gives you the "average" temperature of the room. Is that right? Turn up the freezer and turn down the oven, the average might stay the same, but it wouldn't do today's or tomorrow's dinner much good! "If you connect two systems of different temperatures with a conductor, the temperatures of both systems will reach an equilibrium which is a weighted arithmetic mean" - well, you'd have to be dealing with two systems then would you not? Rather than just temperatures at a particular point that don't represent a system. Should a temperature taken at sea level on the coast be given the same weighting as one in the mountains thousands of miles away? If so, why? Or why not? What is the weighting? I'm no statistician, and I might not know much about thermodynamics, but I think that this is the point that Essex and McKitrick were trying to make. I think that this article is extemely pedantic, off the mark, and sounds very desperate.

No, you can't just add them together -- that's the whole point of it being weighted. You have to weight each temperature by how much of the room is that temperature.
If you want to understand the flaws in McKitrick's argument, Tim Lambert, a computer science professor in Australia, has written some more detailed critiques, one of which is referenced in the article already. Here's another that the article doesn't mention:
The most fundamental flaw is McKitrick's bizarre claim that there simply isn't any known way to calculate the average temperature of something by using sampling and statistics. Lambert disposes of this argument rather deftly (and with a bit of humor thrown in): "If you read the Wikipedia page on temperature you will discover that it does have a physical meaning and also that it is a physical quantity. And that there is the whole field of w:statistical mechanics based on the application of statistics to temperature. Go figure." I imagine that someone who isn't a statistician and doesn't know much about thermodynamics would find this comment "pedantic" and "off the mark," but you don't have to be a mathematical genius to figure out what Lambert is saying. It's pretty clear if you take a few minutes to read through what he has written.
With a little fleshing out, by the way, this article could be a worthy addition to our case studies section. If someone would like to do this, Lambert's critiques contain all of the information and analysis you'd need. (I particularly like the part when he points out that McKitrick skewed his mathematical model by dropping in temperatures of zero whenever he didn't have an actual temperature reading to stick in his spreadsheet.) --Sheldon Rampton 18:27, 10 Oct 2004 (EDT)
Unfortunately Sheldon, Tim Lambert's argument is completely wrong. It speaks volumes that you think it "worthy". See below for why.

The counter argument

I've moved the counter-argument offered by (really McKitrick?) to talk to respond to it --Tim Lambert 06:58, 9 May 2005 (EDT)

The Reply

Perhaps you'd like to move it back and stop trying to censor that which you can't answer.


"If you connect two systems of different temperatures with a conductor, the temperatures of both systems will reach an equilibrium which is a weighted arithmetic mean."

The problem with this statement is that either the person who wrote it is not a scientist or a mathematician or is just incompetent or did not read the book very carefully.

The point that McKitrick and Essex made in their book is that, at no point on any macroscopic scale is the atmosphere in thermal equilibrium, a key point that appears to be lost on the reviewer above. If the atmosphere was in thermal equilibrium then i) we would all be dead and ii) there would be no weather at all since weather is a function of the difference of thermodynamic states between one part of the atmosphere and another (for example, the tropics compared to the poles, or the sea-level troposphere compared to the stratosphere)

The key part of the argument given by the reviewer is simply lost. There is no theoretic reason why the "global temperature" cannot be calculated as a geometric mean rather than an arithmetic mean (or any other sort of mean), because temperature is an *intensive variable", that is, temperature is not per-something (like per square meter or per joule). Temperature is simply a numbering of thermodynamic states.

What the reviewer doesn't know is that taking the temperature of something which is not in thermal equilibrium has no theoretic underpinning in science because you're trying to convert an infinitely fine four dimensional temperature field (because it changes over time because its not in thermodynamic equilibrium) into a single number. Without a coherent theory that says how to do that, what you end up with has no physical meaning - it does not represent energy or anything else.

In the absence of a coherent theory of climate, the notion of a "global temperature" for the Earth is something which has no theoretic basis, and is further complicated by the fact that there no understood standard for how to measure the temperature of the earth.


The physical basis for average temperature is the equilibrium temperature. You don't have to actually allow the system to come into equilibrium to work out what the average will be -- it is just the weighted average, where the weights are the thermal mass (mass times specific heat) of each component.

Counter-counter-counter argument

Lambert simply doesn't know anything about thermodynamics. He even makes a classic mistake of claiming that if two systems in thermal equilibrium are brought together, the resulting equilibrium will be a weighted average. No, it won't. Guess why? I'll give you a hint

I agree with Lambert on this point- so long that the system is closed the equilibrium temperature will be an average weighted by the heat capacity of the two bodies, due to energy conservation. I don't see the point that you are trying to make in linking to an article on entropy- are you claiming that this would violate the second law of thermodynamics? If so, it should be easy to show the calculation which demonstrates this. As far as I can tell, so long as heat flows from the warmer to the cooler body (which is obviously the case here) entropy will increase and the second law obeyed.
Then you're simply as incompetent with regard to thermodynamics as Lambert. The point being, that even in the extremely simplified thought experiment given, that the temperature of the combined system at equilibrium would be less than the weighted average of the "thermal masses" because the total entropy must be greater than the sum total of the individual entropies. Read the passage on entropy again and then get back to me.
Insulting other contributors (such as calling them "incompetent") is not acceptable behavior on SourceWatch. If you wish to continue contributing, please show some manners. --Sheldon Rampton 12:18, 11 May 2005 (EDT)
You mean like describing John Brignell, a PhD and a Professor of electrical engineering as a "crank"? Or was this a mysterious term of endearment at SourceWatch? Be that as it may, I won't call anyone incompetent again. Hopefully Bob will be along shortly to explain the application of the 2nd Law of Thermodynamics to our simple thought experiment. Or maybe Tim will explain it himself.
I looked at your link to the entropy article- it was a general introduction and I didn't see where it discussed this particular problem. Please direct me to part which you feel supports your point. Or else show the calculation which demonstrates that "the temperature of the combined system at equilibrium would be less than the weighted average of the "thermal masses".
I'm sorry but I'm not going to quote you large parts of undergraduate texts on thermodynamics as it applies to closed systems where the energy is conserved. It has to do with the concept of Work and the quantity of disorder in a closed system called entropy. If you join two separate systems together then in order for the conductor to transmit energy it must do work. If it does work, the total amount of entropy must increase. In order for entropy to increase it must take thermal energy from the system. If it were not true then you could remove the conductor and the situation would be reversible. The result is that because entropy must increase, and total energy is conserved then the resultant temperature must be less than the "weighted average" of the two separate systems.
(Please sign your posts using four tildes.) You say: "If you join two separate systems together then in order for the conductor to transmit energy it must do work." This is wrong. A conductor of heat between two (otherwise isolated) systems does NO work. However, it does transfer energy. The transfer of energy without doing work is the DEFINITION of heat. Elementary example: Lets say I have two blocks of gold one of 1kg at T of 50degC, the other of 2kg at T of 25degC, that are both insulated on the outside except for one face which I now place in contact with each other. I allow the system to come to thermodynamic equilibrium. Both blocks now have a uniform temperature throughout which is 33.3degC. (Arithmetic: (1*(273+50) + 2*(273+25))/3 - 273. = 33.3 ). This is the properly weighted average. It is not something less that 33.3degC. The total energy before and after is the same, as you state. The entropy has increased, as you state. Where you go wrong again is in your statements "If it does work, the total amount of entropy must increase." The entropy increases even if NO work. "In order for entropy to increase it must take thermal energy from the system." Wrong. No thermal energy gets "taken" anywhere. Your next sentence is because you are thinking "work". Do you agree with my simple example, before we discuss the atmosphere? 05:06, 19 Jul 2005 (EDT)
That's why Lambert has run off like a little girl - because he's realised he's made a big mistake that could be spotted by any competent undergraduate of physics. The non-decreasing nature of entropy means that his simple argument falls to the ground. He delights to regale us with how he discovered that McKitrick made a trivial error in a calculation where he erroneously used degrees instead of radians, something that McKitrick fixed, re-calculated and submitted the paper with all of the conclusions intact, but when Lambert makes a basic screw-up in thermodynamics - guess what? He's nowhere to be found.
All of which has nothing to do with the Earth's atmospheric physics, because its not a thermodynamically closed system, never has been, never will be. But it does show that Lambert knows much less about basic science than he's letting on.

Please tell us all how to "weight" the atmosphere by "thermal mass"? Or to put it into mathematical form, how do you reduce a 4-dimensional vector quantity (like the temperature of the atmosphere) into a scalar quantity? The only way I know to reduce a vector to a scalar is by using another vector (which in this case would be dimensionless) in a dot product. But what is that function and where does it come from?

For someone who accuses others of not knowing anything about thermodynamics, this is a pretty trivial error to make. Temperature is certainly not a "4-dimensional vector quantity", but a scalar, as you should have learnt in high school. In this case it is a function of position, which makes it a scalar field. To obtain the mean temperature one can simply integrate the temperature function, multiplied by the weighting function (density multiplied by specific heat capacity) over position. No conversion between vectors and scalars is required.
The point that McKitrick and Essex make is that the temperature of the atmosphere is an infinitely fine 4-dimensional vector field - it varies both in position AND TIME because it is not, at any scale, in thermal equilibrium. All of the examples posted by Lambert (including the ones of statistical mechanics) assume a local thermodynamic equlibrium at some scale which does not exist anywhere in the Earth's atmosphere. There is no theory which says how to "add temperature multiplied by weighting function overposition" and then take a mean of such a calculation and expect it to make any sense from a physical point of view. It's a vector field and not a scalar, because it moves.
No, the temperature of the atmosphere is a scalar field. See here or here or here, or simply google "temperature scalar field". Your insistence that it is a vector field because "it moves" is complete nonsense. Learn the difference between a scalar and a vector.
From your linked articles:
1. "If the value is a scalar, the field is a scalar field; if the value is a vector, the field is a vector field. The temperature at all points in a room is an example of a scalar field. The velocity of the air at all points in a room is an example of a vector field."
2. "Examples of vector fields are wind velocity in a given region of the atmosphere, the gravitational field and the electric field, and the gradient of any scalar field."
3. "For example, temperature is a scalar quantity; by specifying the temperature for every point in the room, you have specified a scalar field."
But I said that the temperature of the atmosphere was a 4-dimensional vector field. The examples you gave were 3 dimensional, not 4. 4-dimensional (where one of those dimensions is time), specifies a vector field, not a scalar. The reason why they're using a room for a scalar field is that they're assuming that the system is thermodynamically isolated. something that is not true for the atmosphere.
In addition, your claim that local thermodynamic equilibrium does not exist anywhere in the Earth's atmosphere is contradicted by the AMS.
From your quote: "LTE occurs when the radiant energy absorbed by a molecule is distributed across other molecules by collisions before it is reradiated by emission." Unfortunately the assumption that LTE occurs like that is a classical assumption. Temperature has no meaning at the level of a single molecule absorbing and re-emitting single photons.
For pity's sake stop googling and start thinking.

The point which Lambert dances around is that the atmosphere is not in thermal equilibrium (which is why we have weather in the first place) and that the temperature of the atmosphere is not a scalar quantity.

Lambert isn't claiming that the atmosphere is in thermal equilibrium- he's saying that if it was then it would be at the average temperature (defined above). This is an important point since McKitrick claims that there is no physical basis for this quantity, which is clearly untrue.
Um hello? What are you and Lambert talking about then? McKitrick and Essex clearly make the point that if the atmosphere was in thermal equilibrium (other than the fact that there'd be no weather, natch) then the temperature of such an atmosphere would reduce to a scalar quantity. BUT IT ISN'T. It's no use saying "if the atmosphere was in thermal equilibrium then yadda yadda yadda". There is no constructive way in physics to add bits of the atmosphere in the way you describe in any manner. The earth's atmosphere is highly convective (that's why we have winds, storms and hurricanes). This form of energy is not represented in the construction of Lambert even though it dominates the energy balance of the atmosphere. When a box of the atmosphere is heated it doesn't just sit there and warm up, some of the energy is translated into motion.
It could be argued, that if you wanted to, there is no reason why you can't divide the atmosphere into tiny little boxes (assumed to be in thermodynamic equilibrium, take the temperature of each box, calculate the fourth power of the temperature of each box, add them all together, divide by the total number of boxes, take the fourth root, producing a single number (we'll call it the Mean Equilibrium Temperature) and then relate this to the total energy of the atmosphere via the Stefan-Boltzmann relation E=sigma x Temperature^4. I can completely justify that construction by reference to thermodynamic relations in equilibrium situations, but this isn't an equilibrium situation. Is there a theory that lets me add boxes, take the temperature, raise to the power of 4, take the mean, then take the fourth root and call this "Mean Equilibrium Temperature"? Nope. There is no theoretic reason why my construction is right or wrong compared to the "weighted arithmetic mean" that Lambert proposes.
This is bizarre. The Stefan-Boltzmann law gives the energy radiated from a blackbody per unit time per unit area. It has nothing to do with the "total energy" of the body, so I don't see how you can use it to relate the temperature to the total energy of the atmosphere. Your suggestion makes no physical sense.
Of course it doesn't. That's my (or rather McKitrick and Essex's) point. In order to make the calculation like that I had to assume that the atmosphere was made of something like "lucite", with uniform density and no convective motion. In effect, I reduced the vector field to a scalar. It doesn't make physical sense any more than Lambert's "thermal mass" idea. The atmosphere isn't like that, and there is no constructive way to average the atmosphere (not even to the level of molecules) by taking the temperature an infinitely dense 4-dimensional field.
"The Earth's Mean Temperature" is not a physically meaningful measure, it's (as McKitrick and Essex put it) a statistic with dimensions of temperature. It's A statistic, one of many possible statistics (and I've shown another). There's no theory that enables me to distinguish one from another.
So Lambert, and presumeably you following Lambert, are making suppositions about the physics of non-equilibrium systems by assuming them to be at local thermodynamic equilibrium, which they're not. There is no construction in physics which says how to take the temperature of the non-equilibrium atmosphere that is meaningful. That's the point of the passage regarding "Dr Thermos" in the book. Lambert doesn't understand the point.

Removal of links to Deltoid

Interesting...Tim Lambert seems to have put a bee in someone's bonnet. went through a number of articles taking out all links to Lambert's Deltoid weblog, including this one. I've reverted most of them.

Removal of links to Deltoid =

After castigating John Brignell as "a crank" for referring to his own book as a source for his statements, should Tim Lambert be spending what little credibility he has by referring to his own weblog? Or is it the case that John Brignell may not do what Tim Lambert does constantly?

And in the same spirit, perhaps you'd like to put back the "counter argument" rather than give in to Lambert's crude attempt at censorship?

whether the links were inserted by Lambert or not matters little. They are relevant and substantial contributions to the debate. There is no case for deleting them. The protocol is that if contributors delete material (other than minor changes) they add a comment to the talk page. That way other contributors can judge the reasons and decide whether they agree or not or suggest changes. In the absence of that it is liable to be interpreted as vandalism and reverted. Contributors who persistently attempt to vandalise pages are placed on the block list. --Bob Burton 16:14, 10 May 2005 (EDT)

My opinion... that Lambert is right and McKitrick and Essex are wrong. It was written:

The key part of the argument given by the reviewer is simply lost. There is no theoretic reason why the "global temperature" cannot be calculated as a geometric mean rather than an arithmetic mean (or any other sort of mean), because temperature is an *intensive variable", that is, temperature is not per-something (like per square meter or per joule). Temperature is simply a numbering of thermodynamic states.

This is somewhere between wrong and meaningless. In terms of bulk properties, temperature is a real physical thing and mass-weighting it is the obvious way to average. In terms of skin temperatures, then area-weighting is the obvious thing to do, and its what people do.

It is also said:

The point that McKitrick and Essex make is that the temperature of the atmosphere is an infinitely fine 4-dimensional vector field - it varies both in position AND TIME because it is not, at any scale, in thermal equilibrium

This isn't true: the atmos isn't infinitely fine. Atmos T is correlated on various scales. You don't need an indefinitely fine mesh of measurements (in space or in time) to get a good approx to the T elsewhere.

Its worth pointing out that McKitrick doesn't believe his own argument - he has devoted considerable time and effort (and failed...) to replicate the MBH temperature record of the northern hemisphere. If he really believed his own argument, he would be saying: the entire process is pointless.

BTW, I assume that taken by storm is about a bit more than glob-t-does-not-exist, so I suppose it would be nice of the page filled out a bit.

(William M. Connolley 12:12, 12 May 2005 (EDT)).

My opinion of Bill Connelley's opinion

In terms of bulk properties, temperature is a real physical thing and mass-weighting it is the obvious way to average. In terms of skin temperatures, then area-weighting is the obvious thing to do, and its what people do.:

It's always worth pointing out the boundary conditions under which such a construction should apply: "Mass-weighting" would be physically meaningful under the constraint that the mass of gas or area of skin can be considered to be at or near thermal equilibrium for the short time it is measured. That's why when the doctor takes your temperature he puts it somewhere where the sun don't shine, in order to get a reasonable mean internal temperature.

So the answer is that your reply is somewhere between wrong and meaningless. You 've obviously missed the point about entropy, but then there's no point dropping yourself in the same hole as Lambert, is there?

Its worth pointing out that McKitrick doesn't believe his own argument - he has devoted considerable time and effort (and failed...) to replicate the MBH temperature record of the northern hemisphere. If he really believed his own argument, he would be saying: the entire process is pointless.

Nobody has managed to replicate MBH98 completely because Michael Mann has refused to disclose his source code for the PCA process. He told Antonio Regalado of the Wall Street Journal that he "would not be intimidated" into handing over the code. Presumeably because it's something so powerful and dangerous it could be used by terrorists. Or more likely, because revealing that code would end any pretense that MBH98 was properly conducted science and would end Michael Mann's career.

So that failure is not for want of ability or trying. It's about stalling. McIntyre and McKitrick are not the only people waiting for the source code, Ulrich Cubasch is waiting as well (link to German article)

You're a big friend of Michael Mann, Bill. Perhaps you'd like to ask him for the source code. At least you won't intimidate him by asking, would you?

Oh and Wahl and Amman's replication fails multiple significance tests as well. What a shame!

As for the question of pointlessness, the other chapters in "Taken By Storm" (which Bill clearly hasn't read, so informed opinion it ain't) cover the issue of the meaningfulness (or not) of the concept of "global temperature".

You should try reading the book before replying. It's available on at a very reasonable price

Tim Lambert rides again

A new post on Tim Lambert's blog has appeared on this subject (since he's disappeared from this "discussion") repeating his failed argument about Ross McKitrick and Chris Essex's assertion that there is no physical meaning to the term "mean temperature" in a non-equlibrium system and apparently unable to rescue his basic mistake in thermodynamics that the average temperature of two systems joined together is just the "weighted average" of the two systems.

As proof of this stunning discovery he quotes...wait for it..a Flanders and Swann comic song about Thermodynamics! Call the Nobel Committee, we have a genius in our midst!

Wrong again, Tim!

Look up entropy and if you don't think that Total Energy=Thermal Energy + Entropy in a closed thermodynamic system, then you'll struggle with this really difficult article on the First Law of Thermodynamics where it explains that

The first law of thermodynamics is a consequence of conservation of energy and requires that a system may exchange energy with the surroundings strictly by heat flow or work. Therefore, for change in energy dE, heat change dQ , work done dW,
dE = dQ - dW

So if the total energy is constant ( ie dE=0 ) then the change in thermal energy is matched by the work done. Voila! if you join two systems that were in thermodynamic equilibrium by any means, then heat flows, work is done and the temperature of the resultant equilibrium sysem must be less than the "weighted average" of the temperatures of the original systems.


Now Tim, instead of quoting comic songwriters, would you care to quote a physics professor?

By the way, the notion of the non-physicality of the concept of "mean temperature" in non-equilibrium systems comes from Professor Chris Essex. But then, of course, the point was never about physics was it?

This post is completely out to lunch.
If the two bodies are considered to be one system, then
dE = dQ = dW = 0
There is no heat loss or gain, no energy change, and no work done.
If they are separate systems, the one loses energy (equal to the heat lost), and
the other gains it. The second law is completely independent from the first law
and only determines which direction the heat (equal to energy) will flow.
Sorry but the poster above keeps making basic mistakes in comprehension (and where is Lambert? Hiding from his own gross mistakes while criticising other's minor and corrected ones... ).
The reference to work comes from Lambert's thought experiment that if two independent closed thermodynamic systems are joined together by some means (Lambert specifies a conductor) then the resultant system's temperature at the new equilibrium would be the "weighted mean" of the original systems. Which is rubbish, since the conductor must do work, the entropy must rise and the resultant temperature will be less than the "weighted mean".
Here's a nice easy article on Entropy. Note the always increasing aspect of entropy when work is done. Happy reading!
Perhaps you could enlighten us about how the conductor must do work. This seems to be a rather novel idea. What is the potential that is being changed by the work? What is the force being acted against? Is this where you think my basic mistake in comprehension lies? Or someplace else?
It might help if you could explain your idea in terms of a practical example the rest of us could understand. For example, when an ice cube is dropped in a glass of water, my understanding is that heat flows freely from the water to the ice cube. (Or more accurately, over time more heat flows to the ice cube than back into the water, asymptotically approaching equal flows). The entropy is clearly increasing. But where exactly is the resulting work in this experiment?
Wrong. Heat energy only flows from hot to cold (Clausius' Theorem).
Clausius' Theorem does not contradict what I said. Is your understanding of thermodynamics so limited that it cannot be reconciled with black-body radiation? 04:27, 27 Aug 2005 (EDT)
Update: I see you have modified your last post to remove your contention that "dE <> 0". Is this a sign that you realise your theory is doomed?
Nope. It's that I inserted the wrong expression into the argument, thought about it, realised I'd have write 10 pages to explain it, then thought "why bother?"
You do have a kernel of truth in your argument. If work is done by the bodies, then their total energy, and thus temperature, would be reduced. Maybe you see now that work is not being done simply by connecting the bodies together.
Drivel. The total energy of the combined systems remains constant (Law of Conservation of Energy, remember that one?). Which part of this basic thermodynamic thought experiment do you not understand? The total energy of the combined system is
Etotal = E1 + E2
What don't I understand? I guess the part where if energy is removed from a system by work against a force outside the system, that it is "drivel" to think that energy remaining in the system is reduced. And no, the conservation law that I remember states that the energy is conserved. I don't remember the one you refer to. 19:09, 19 Jul 2005 (EDT)
There's a difference between the total energy of the combined system and the total thermal energy of the system.
You seem to have come to your misapprehension by an incorrect reading of the Entropy article you reference. It is true that if work is done, then the entropy increases. The converse is not necessarily true. It is not true that if the entropy has increased, then work has been done. You might want to read this article to get started.
We are not discussing the converse. If heat flows from one system to another then entropy must increase. If it does so, then the total thermal energy must be less that the sum of the thermal energies of the original systems. This is really basic Thermodynamics 101. There is a basic definition of the thermodynamic concept of work that you're missing.
I am afraid I only have the accepted definition of thermodynamic work to go on. If you have a different one from your Thermodynamics 101 course, you should share it with the rest of us. That would give us a better chance to understand what you are trying to tell us. Or are you sworn to secrecy? Is this also the reason why you won't tell us what work is being done by the conductor? It doesn't really help to enlighten us by referring to "knowledge" normal people don't have access to. 19:09, 19 Jul 2005 (EDT)
My analysis of this I inserted above when this came up earlier. Search for above. 05:25, 19 Jul 2005 (EDT)
Thanks. I didn't really mean to pose the question to people who actually understand it. I wanted to get this guy's explanation for such a wierd claim. But I am afraid we're just going to get another assertion that the conductor does "special" work, because he says so. 19:09, 19 Jul 2005 (EDT)
What's weird about it? Lambert proposes a simple scenario of an irreversible thermodynamic process and forgets about entropy, a fundamental universal phenomenon that is related to such concepts as the "Arrow of Time" (look it up). Is this an argument that because you don't understand it, therefore I must be wrong?
Lambert does not "forget" about entropy - he simply does not need it for his example. In order to calculate the final temperature when two blocks, initially at different temperatures, are connected by a thermal conductor, all you need is the First Law. When this calculation is done correctly, you do indeed find that the final temperature is a weighted average of the two initial temperatures, with the weights determined by the masses of the blocks and by their specific heats (if they are made of different materials.) Once you have found the final temperature, you can go ahead and calculate the entropy change if you wish, and you will find that it is positive in accordance with the Second Law. But you do not need to know the entropy change in order to calculate the final temperature. In fact, the opposite is true - you need to know the final temperature in order to calculate the entropy change. (unsigned)
It is not clear what you accuse us of not understanding. If you are implying that we don't understand your argument, then that is true. We have made guesses as to where the flaws in your thinking lie and tried to help you with them. But if you won't tell us your argument other than indirectly through misguided critiques of accepted thermodynamic theory, then we are only left with guessing.
If you suggest we don't understand the concept of entropy, then you are wrong. The unsigned post above this one demonstrates a clear understanding. What we are asking you to realise is that entropy and energy are different things. That is, "really basic Thermodynamics 101." 04:27, 27 Aug 2005 (EDT)