Talk:Thermodynamic temperature/Archive 2

Archive 1

Archive 2

Archive 3

July '06 comment

ACCOLADES ON THE WIKIPEDIA WAY OF COLLABORATIVE WRITING

The following e-mail was received on July, 18, 2006 from Dr. Martin Chaplin, BSc PhD CChem FIFST FRSC. He is the Professor of Applied Science, Water and Aqueous Systems Research, Head of the Food Research Centre, at London South Bank University. He wrote…

Hi

Thank you very much for your email.
I like your site very much and will link to it. It is great and shows excellent use for wikipedia.

Unfortunately I cannot help you as I do not know where that data can be found. I would work it out the same way you did (!!)

Best wishes
martin

Thanks to all who have been, and will continue to be, making contributions to this article (and for being patient with me when I am slow to understand your point). Greg L 22:36, 18 July 2006 (UTC)

Thanks to Dan

I just checked my “Wikipedia” archive in my e-mail program. This is where I store past correspondence with the various experts I’ve been in contact with while working on this article. I see that I’ve got 27 e-mails to and from Dan Cole. Dan is the über-expert in various thermodynamic intricacies like zero-point energy. He’s notable as the IBM researcher who has published many papers, including the one cited in note #1 of this article:

Derivation of the classical electromagnetic zero-point radiation spectrum via a classical thermodynamic operation involving van der Waals forces, Daniel C. Cole, Physical Review A, Third Series 42, Number 4, 15 August 1990, Pg. 1847–1862.

I wanted to express my appreciation as follows:

“ Thanks Dan, for all your time in patiently explaining things like absolute zero, zero-point energy, and heat energy to me. Contributors to Wikipedia don’t get to sign their name to articles. Behind the scenes, you’ve been helping to make this article what it is. All I can do is offer this humble acknowledgement and expression of appreciation. ”

Greg

Greg L 00:41, 22 February 2007 (UTC)

How do you link to a category page without putting linking page into category?

I don't think this article belongs in so many categories, but I wanted to link to the pages. Is there a way?

User:Pifvyubjwm 11:49, 30 December 2005 (UTC)

If you want to link to a category without including the page in the category, put a colon (:) in front of the word Category, as in this link to Category:Temperature, which should not put this talk page into that category. That link will be visible on the page, just not in the line of categories at the bottom of the page. Gene Nygaard 14:02, 22 September 2006 (UTC)

Heat Energy and Temperature and Proportionality

The article says: "Regardless of the substance though, a change in its heat energy produces a proportional change in its temperature on the thermodynamic (absolute) temperature scale" This statement is in general incorrect. The relationship between change in heat and change in temperature is given by the heat capacity. The heat capacity itself is, in general, a function of temperature. That is to say the amount of heat required for a given temperature change depends on the temperature. In general, there is no proportionality between change in heat and change in temperature, although they are related by the heat capacity. In some substances, over a limited range, there is a reasonable proportionality, eg water between 1C and 99C has an approximately linear relationship between change in heat and change in temperature. This breaks down significantly at phase changes, where heating a substance might not change its temperature at all. I don't want another edit war, so I won't correct the article on this issue, but hopefully someone else will. LeBofSportif 08:50, 8 June 2006 (UTC)

I generally agree with what you say above. I wanted to convey the concept without burdening the paragraphs with caveats like phase changes and volume changes. I was thinking about that sentence and came back to add the caveats anyway. What do you think of it as currently revised? Greg L 17:31, 8 June 2006 (UTC)

I still don't like proportional being in there. It misses the point that the heat capacity can vary significantly with temperature, even when one wouldn't expect it to. Explanation of anomalous heat capacities at low temperatures is one of the triumphs of the quantum theory of solids - the Dulong-Petit law doesn't hold at low T, and nobody knew why. The point, I suppose, is that Temperature and Energy are two very different things, which because of certain coincidental linearities often follow each other closely, eg water at nice temperatures. The article, especially since it is concerning "thermodyanic temperature" should strive to make the point that temperature and energy are different and in general ("in general", rather than "often") temperature and energy do not follow each other in a nice way at all. The only thing we can say heat capacity is never negative - you can never add heat and reduce the temperature, while keeping sensible things constant like volume - except for a few quirky experiments with only a few atoms. LeBofSportif 08:59, 9 June 2006 (UTC)

I appreciate your thoughts. Text mustn't be simplified to the point of making it incorrect. OK, revised again. It mentions that temperature is the “speed” of movement, and it expands on “specific heat.” What do you think?
Greg L 13:17, 9 June 2006 (UTC)

absolute zero and classical physics

I have rephrased the following sentence:

The thermodynamic temperature scale’s null point, absolute zero, is where all kinetic motion in the particles comprising matter ceases and they are at complete rest in the “classic” (non-quantum mechanical) sense.

There is no "classic" sense in which all motion ceases at absolute zero. Rather, classical physics predicts (incorrectly) that all motion ceases. --Trovatore 05:36, 11 July 2006 (UTC)

another try

My previous wording was a bit awkward; the parenthetical "incorrectly" kind of interrupted the flow. But the current version's still not good; it suggests that the difference between the classical and quantum accounts is one of interpretation, which it's not. How about this?

... absolute zero has traditionally been defined as the point at which all kinetic motion in the particles comprising matter ceases, and they are at complete rest. In fact, it is now known that, owing to quantum effects, some motion would remain even at absolute zero.

Of course, it's true that this version doesn't really say what absolute zero is, but that can come later; the lead section doesn't have to give all the details of a precise definition. --Trovatore 04:29, 12 July 2006 (UTC)

Classic Physics and absolute zero

• Trovatore. You recently made a contribution to an article on Thermodynamic temperature. You stated that absolute zero is "where classical physics (as opposed to quantum mechanics) predicts (incorrectly) that all kinetic motion in the particles comprising matter ceases." Please explain your reasoning. If one reduces the motion of particles to the point where all that remains are the fluxuations of random quantum uncertainty, then this is the point where it is no longer theoretically possible to remove any more kinetic energy from a substance. Isn't that right? This is the point in thermodynamic equations where absolute zero is defined to be precisely -273.15. Everything in the universe has random quantum uncertainty as to their position and this sets the baseline for "zero"; nothing can be more stationary that quantum uncertainty. Any motion beyond quantum uncertainty is kinetic motion. Is this not correct?

Expanding with a thought experiment: If there are two particles in a closed container and they have kinetic energy (move with respect to each other and/or the container), they must eventually collide; they have kinetic energy. At the classic definition of absolute zero, there is no particle motion as measured from particle-to-particle or particle-to-container. All that remains is quantum uncertainty as to their precise location at any one moment. All that remians is probabilities. According to quantum mechanics, there is an very remote chance a particle can briefly appear outside of the container. But on average, the particle's position remains at its original position and will never collide and recoil away from each other. Even if they momentarily appear at the same position due to quantum uncertainty, they will still reappear elsewhere (most probably near their "most probable" position again). If this is all true, then what is incorrect with the statement that "absolute zero is the point where all kinetic motion in particles (in the classic, non-quantum mechanical sense) ceases"?
Greg L 17:27, 11 July 2006 (UTC)

• Actually, the article has been reviewed by four Ph.D.s. One works for Argonne National Labs, one is a chemist, another works for the NIST (he was on the team that set a record cold temperature of 700 nK), and the fourth teaches second-year thermodyamics at a university. I've got two patents on new ways to calculate the equation of state of gases. A friend of mine who is an afficionado of quantum physics is currently reviewing the article (which is why I wanted to leave it as is until his review is done). The paragraph you have issue with is used througout Wikipedia in other articles like Kelvin and Absolute zero. It has withstood months of being read by others without being modified.

Perhaps there is another way to re-word this without leaving it like you had it "(incorrectly)". That's entirely unsatisfactory. If you want to use such a broad paint brush, you need to cite sources. I think it would also help if you could get the backing of some Ph.D.s to peer review what you are proposing.
Greg L 23:29, 11 July 2006 (UTC)

• Trovatore:
  

Here are some definitions I've found:
Meriam-Websters: a theoretical temperature characterized by complete absence of heat and motion and equivalent to exactly -273.15°C or -459.67°F

ChemiCool Dictionary: The zero point on the absolute temperature scale; -273.15°C or 0 K; theoretically, the temperature at which molecular motion ceases.
Greg L 00:03, 12 July 2006 (UTC)

Please point me to these Ph.D.'s. I freely admit I'm not a physicist (my PhD is in math) but the wording as it stands looks like nonsense to me. It would be like saying "classically, tunnel diodes don't work". If classical physics were correct, they wouldn't work. But the fact is that they do work, and the fact is that the particles do move at absolute zero (more precisely, the particles' state is a superposition of states with nonzero momentum). --Trovatore 00:44, 12 July 2006 (UTC)

• Trovatore:
  

You make an interesting point about tunneling diodes. The guy who really knows his quantum mechanics is the founder of the fuel cell company I worked at called Relion. He once went to a convention in Hawaii. While there in his spare time, he went to the beach, sat in a lawn chair, and read up on his favorite subject. I will forward your comment to him and will get back to you probably no later than Thursday. I shouldn't conjecture about the difference, but it could be that the huge de Broglie wavelength of electrons, coupled with the fact that they are under a huge voltage field (small voltage but really small joint) amounts to a different picture. Regards
Greg L 01:43, 12 July 2006 (UTC)

• All you can even hope to get from that is a quantitative difference, which is to say that the zero-point motion is small. Sure, it's small, but it has observable macroscopic effects (for example, it's what prevents helium from ever becoming a solid at atmospheric pressure, even at absolute zero). --Trovatore 02:37, 12 July 2006 (UTC)

Trovato: I see what you are saying. I Googled on your statement and see from an article (click here) about these effects. I think the existing statement needs revision. But it should be sweet and simple without a parenthetical caveat that leaves people dangling. Now the question becomes this: Is absolute zero the point where there would be zero atomic drift velocity if you release them? By this definition, if all the heat that "could" be removed were removed (a practical impossibility) and only the residual, unremovable effects of zero-point energy remains, then is that absolute zero? For helium (MW = 4.002602), a temperature of 1 pk equals an atom drift speed (not vector-isolated velocity) of 79.94 µm/s. Helium's vector-isolated velocity would be 45.58 µm/s @ 1 pk. Or is absolute zero where zero-point energy would also be absent? This isn't even a practical impossibility, it's a theoretical impossibility. My guess is that absolute zero is "yes" to the former question; that two particles with, for instance, 1 mm of separation and which theoretically have zero drift velocity with respect to each other would never collide because only zero-point motion remains. This answer is satisfying in one way because it follows the PV=nRT slope; that is, that classic velocity gives gases their pressure. It is unsatisfying because it doesn't account for the observable effects on helium. Still, it could be that zero-point's effect upon helium simply means that helium is another tool that opens a window into the quantum world and is distinct from thermodynamic issues. If the answer to the former conjecture is "yes", perhaps wording like this is in order: Absolute zero is the temperature where no more kinetic motion can theoretically be removed and all that remains is quantum mechanical zero-point energy. Upon reflection, that's not all that technically different from what's already in place. If the answer is "yes" to the second conjecture, then other wording is required. I think I need to contact the researcher in the here-cited article: P. G. Klemens at the University of Connecticut.Greg L 05:36, 12 July 2006 (UTC)

Done. I've e-mailed Dr. Klemens. Greg L 05:54, 12 July 2006 (UTC)

Are you guys arguing about what the proper definition of absolute temperature is? You don't have to email researchers in the field for the answer; it can be found in standard textbooks. In classical statistical mechanics, the parameter β is defined as a particular derivative of the number of states with respect to internal energy. Absolute zero occurs when this derivative goes to zero. It is true that for an ideal gas, classical statistical mechanics predicts that the absolute temperature goes to zero when the internal energy is removed. The appellation "incorrectly" doesn't seem to apply; this prediction is vacuously satisfied even by quantum systems. There is also a phenomenological definition of absolute temperature, which is nothing but an extrapolation of some curves. -lethe ^{talk +} 06:07, 12 July 2006 (UTC)

dU = T*dS - P*dV - other terms, where U is the internal energy, T is temperature, S is entropy, P is pressure, V is volume. Entropy, S = k*ln(#microstates consistent with the macrostate), where k is Boltzmann's constant. At absolute zero, you must consider the quantum mechanical effects. So with quantized levels (and holding volume, etc. constant), we get U1 - U0 = T0*(S1 - S0). Notice that temperature is not associated with a particular energy level, but with a transition between two neighboring levels. In reality, T0 can never be zero, because then the energy levels would be equal which they are not. JRSpriggs 06:28, 12 July 2006 (UTC)

Lethe, I think you're quibbling a bit on the "vacuous satisfaction" thing. First of all I'm not convinced it's vacuous; it seems to me that it is possible for a system to attain a temperature of absolute zero, if just by chance it radiates out its last quantum of heat and no more come in. Also ISTR that superfluid helium II is considered a mixture of two phases, and the actual superfluid part of it is at absolute zero (though the average temperature of the whole liquid, of course, is not). That may not be up-to-date.

But even if it were true that no system can attain absolute zero, it's still the case that quantum mechanics predicts a minimum kinetic energy per particle (in a given bound system) and classical mechanics does not. Classical mechanics is observed to be incorrect on this point. --Trovatore 08:16, 12 July 2006 (UTC)

(Edit conflict) I don't know anything about superfluid helium, so I can't comment on that. But for your final statement, let me make two comments. First, standard classical mechanics does indeed predict that a bound system can have no internal energy, which does violate the principles of quantum mechanics. However, the third law of thermodynamics states that absolute zero is unattainable, and I think this follows from the law of large numbers and the principles of energy transfer, and so applies even to classical systems. Thus classical statistical mechanics (the context in which temperature is defined) also seems to disallow absolute zero, so saying that classical mechanics makes incorrect predictions may not be accurate (at least in the thermodynamic limit). Secondly, the edit states that absolute zero is where all energy is lost. If the edit said that classical mechanics predicts this state is attainable, well we could argue about whether classical mechanics were correct or not. But the edit in question doesn't say that. It says there is no internal energy in a system at absolute zero. If classical mechanics also decrees that this state is not physically attainable (and there's no mention one way or the other than I can see), then the statement is not actually wrong. I do concede that this superfluid helium may be a counterexample though. If that is a system at absolute zero, then classical stat mech probably makes incorrect statements about it. -lethe ^{talk +} 08:47, 12 July 2006 (UTC)

I'm having a look at Kerson Huang's Stat Mech book. He shows that the first and second laws of thermodynamics can be derived from statistical methods, but is ominously silent about the third law. And in his treatment of the third law, he says it is the thermodynamic manifestation of quantum phenomena. So it may be the case that classical statistical mechanics alone does allow for violations of absolute zero. I might then modify my position to say that classical thermodynamics (including the third law as a postulate) does not allow systems to attain absolute zero, and this is not in disagreement with observation. -lethe ^{talk +} 08:59, 12 July 2006 (UTC)

Thanks to you all for weighing in on this subject. But the two answers seem somewhat contradictory to me. The only question at hand is whether this statement is true: Absolute zero, is where all kinetic motion in the particles comprising matter ceases and they are at complete rest in the “classic” (non-quantum mechanical) sense. This definition is located early in the article where it should be simple indeed. Whatever statement goes there for defining absolute zero, it shouldn't be burdened with formulas, constants, and caveats. It should be extremely concise, understandable to a 10th-grader (at this point in the article anyway), should avoid the standard weasely Wikipedia equivocations that occasionally result from “committee-itis”, and must be 100% scientifically true. The above definition doesn't say that absolute zero is achievable or not, it just says what it is. I believe the question boils down to whether absolute zero would theoretically be achieved if there was zero classic kinetic drift in the atoms and only zero-point energy remained. Is that absolute zero? Or is absolute zero where zero-point energy must also be absent? To this extent, I've e-mailed a Ph.D. researcher who's been studying zero-point effects on matter. I've already run this wording by a Ph.D. instructor who teaches 2nd-year college thermodynamics and he had no problem with it. If there is anyone who knows there stuff at this level, how say you? Greg L 08:44, 12 July 2006 (UTC)

I have edited the disputed sentence to something I think is closer to a good description, what do y'all think of my edits? -lethe ^{talk +} 09:09, 12 July 2006 (UTC)

Lethe: No, I don't think that does it. Quantum mechanics, per se, does not say absolute zero is not achievable (see my remarks above on that); rather, it says that even at absolute zero, some motion remains. Greg, Lethe's edit summary about there being "no such thing as 'at rest in the classic sense'" is right on target. The kinetic energy of the atoms at absolute zero is the actual state of affairs, not simply the prediction of quantum mechanics. What you're calling the "classic kinetic drift" is not separable from the quantum-mechanical motion; it's just the prediction of a different theory, classical mechanics; a prediction that happens to be wrong. --Trovatore 15:42, 12 July 2006 (UTC)

My understanding from some tengential remarks by Huang is that this is exactly what quantum mechanics says, though he wasn't entirely explicit about it. I can't go looking more carefully at the moment, but I guess what I want to find is some reference that says that absolute zero is attainable. My current understanding is that absolute zero is not attainable, precisely because of the residual motion of the ground state, and that this is codified in the third law. -lethe ^{talk +} 17:05, 12 July 2006 (UTC)

No, I don't think that's right. As I recall from Kittel and Kroemer, the precise definition of temperature is the reciprocal of the partial derivative of entropy with respect to internal energy, where entropy is defined as the logarithm of the degeneracy. This fudges over some points about discreteness; I've never been sure just how those are resolved. But it should work out that a system large enough for statistics to be applicable is at absolute zero when it's in the lowest-energy eigenstate of its overall Hamiltonian. --Trovatore 17:55, 12 July 2006 (UTC)

Yeah, that's the right definition of temperature. Does that number go to zero for quantum systems approaching their ground state? -lethe ^{talk +} 18:20, 12 July 2006 (UTC)

I believe so. Of course there's this quibble about what "derivative" means in the discrete case; technically you probably have to let the number of particles go to infinity as well, to make rigorous sense of it (and I'm not exactly sure what "degeneracy" means either, if energies can be close but not exactly the same; I think this is probably also a smoothing of a discrete picture). --Trovatore 18:24, 12 July 2006 (UTC)

In the thermodynamic limit (infinite number of particles), energy states form a continuum, and one speaks of density of states instead, and may use the methods of calculus with impunity. The thermodynamic limit is of course only an approximation of our discrete universe, but then so is everything in physics. -lethe ^{talk +} 18:32, 12 July 2006 (UTC)

• My ex-boss at a fuel cell company agrees with you guys. He's extremely technical and into the deep depths of quantum mechanics. He thinks absolute zero would include all motions, including zero-point energy. I think now that Trovatore was right all along: any definition must be consistent with the reality of zero-point energy because zero-point energy is actually able to make phonon waves in crystal latices. What Lethe has written—while a tad wordy—appears technically correct given what we believe at this point. I'm still expecting an answer from the Ph.D. who researched zero-point-induced phonon waves in helium to get his input too but I don't think we have to wait to revise what we have. I may also contact the Ph.D. instructor of college thermodynamics and see why he didn't flag this sentence. On a separate note: I think a detailed definition of absolute zero is probably best dealt with in the Wiki article: Absolute zero. When we're dealing with the introductory second paragraph in an article that doesn't deal directly with absolute zero, we should be able to make it simpler yet. I would propose something along the lines of:

The thermodynamic temperature scale’s null point, absolute zero, is a theoretical temperature characterized by complete absence of motion ^[1]

This definition has the twin virtues of being 1) suscinct, and 2) in harmony with Meriam-Webster's definition. The  [1] would be a reference stating that no motion includes quantum mechanical zero-point energy and that, while scientists can get close, absolute zero unachievable. The provided link to absolute zero would give the reader an opportunity to learn more. What do you think?
Greg L 19:37, 12 July 2006 (UTC)

The problem is that it's false; absolute zero is not characterized by the complete absence of motion. I agree with a point you made earlier in the talk page, that material should not be simplified at the expense of accuracy. That's what your proposed text above does. --Trovatore 19:43, 12 July 2006 (UTC)

latest version doesn't work

Greg, no, the latest version does not solve the problem. Absolute zero is not characterized by the complete absence of motion, and (this point refers to the footnote) the zero-point motion is not a barrier to attaining absolute zero, but rather is motion that would persist even at absolute zero. --Trovatore 20:36, 12 July 2006 (UTC)

Just to put my 2 cents in: Yeah, what Trovatore said. Thermodynamic temperature is defined as the partial derivative of the average interval energy with respect to entropy, and absolute zero is the point where that derivative is zero. In a classical universe all motion would stop, and in a quantum universe the system would be in the ground state, but that isn't how absolute zero is defined. The Meriam-Webster's definition is simple wrong. (Pilled Higher and Deeper) Nonsuch 20:55, 12 July 2006 (UTC)

So you are saying Meriam-Webster's definition is wrong. I still don't get it. Your logic seems circular. You stated that zero-point energy means that motion persists even at absolute zero. That suggests that absolute zero is where there is zero classic motion and all that remains is zero-point motion. But this is what you had a problem with. Your point is that helium can not crystalize at one standard atmoshpere because zero-point energy causes motions that prevent it. That objection is founded on the notions that 1) zero-point energy is motion, and 2) true absolute zero can not be obtained due to its effect. This reasoning is inescapably founded on the notion that absolute zero is a theortically unachievable point of zero motion where, even if one removes all classic kinetic motion, zero-point-induced motion still causes a background of motion. If this is true, then statement that "absolute zero is a theoretical temperature characterized by the complete absence of motion" is true. It also avoids having to declare that dictionaries and encyclopedias are wrong. Greg L 21:01, 12 July 2006 (UTC)

(1) is true, (2) is false. Even if you get helium to a temperature of absolute zero, at atmospheric pressure, it will still be a liquid. --Trovatore 21:07, 12 July 2006 (UTC)

A definition of absolute zero needs to cite an authoritative reference. Isn't it as simple as this: Either absolute zero is an unachievable state where all classic kinetic motion has ceased and all that remains is zero-point energy, or absolute zero is an unachievable state where threre is no motion whatsoever, not kinetic, and no zero-point energy. Use plain-speak. Absolute zero is where there is no what? Greg L 21:09, 12 July 2006 (UTC)

I think what Trovatore seemed to have been first saying makes the most sense. If helium can not freeze at room-temperature due to the kinetic effects of zero-point energy, then this is motion. I don't think the Meriam-Webster definition was poorly thought out and is incorrect; I belive it is spot-on precise. Absolute zero is where there is no atomic motion whatsover. It doesn't matter what causes that vibrational motion, if something makes it move (zero-point energy, “classic” kinetic translational motions) then it is motion and one has not achieved absolute zero. Greg L 21:26, 12 July 2006 (UTC)

There is no such thing as "classic kinetic motion". There's motion, period. Absolute zero is where classical physics predicts there would be no motion. Classical physics is wrong. I wonder if you've come to terms with that fact? --Trovatore 21:27, 12 July 2006 (UTC)

You don't have to be glib and patronizing. I've "come to terms" with the fact that classic physics breaks down when one presses sufficiently far into the realm of quantum mechanics. I also know that if one wants to be precise in the definition of absolute zero, then the definition must be consistent with quantum effects since these come into play when one tries to define someting with a "zero" in it. And I agree, when we are talking about "zero," that motion is motion. That's why absolute zero is correctly described as as state where there is none of it. It seems utterly ludicrous to suggest that absolute zero is where "there is no motion but there is still a little bit of motion due to zero-point energy which causes compresson waves in cystal lattices and prevents helium from freezing". Greg L 21:38, 12 July 2006 (UTC)

No one is claiming that there both is and isn't motion. Absolute zero is not correctly described as where there is no motion. I'm not being glib, just terse. --Trovatore 21:45, 12 July 2006 (UTC)

Ok. You don't think absolute zero is correctly described as where there is no motion. So you must believe absolue zero is where there is some motion. Is that right? Greg L 21:55, 12 July 2006 (UTC)

There is some motion at absolute zero, right. --Trovatore 22:07, 12 July 2006 (UTC)

Then doesn't that take us back to the beginning? Are you suggesting that absolute zero should be defined as a state where there is no motion except for quantum mechanical motion? Greg L 22:16, 12 July 2006 (UTC)

No, there is no distinction between "classical motion" and "quantum mechanical motion"; it's all just motion. Absolute zero is not defined in terms of motion. It's defined in terms of the derivative of entropy with respect to internal energy. (It is true, I think, that it's where there's the minimum possible motion, but I'm not 100% sure of that—as I said, I'm a mathematician, not a physiscist.) --Trovatore 22:20, 12 July 2006 (UTC)

Greg, are you perhaps taking the simplified definition near the top of the article to literarily? At its simplest, “temperature” is the measure of the kinetic energy resulting from the motions of matter’s particle constituents (molecules, atoms, and subatomic particles). The full variety of these motions comprises the total heat energy in a substance, which is a form of kinetic energy. The relationship of kinetic energy, mass, and velocity is given by the formula Ek = 1/2m • v 2. That is a reasonable, and relatively intuitive, approximation to the truth that needs to be modified, in particular, in the quantum realm. The true, but technical definition of thermodynamic temperature is the last equation in the article. Nonsuch 22:22, 12 July 2006 (UTC)

I don't pretend to have the expertise to decided for myself what the definition of absolute zero ought to be. I don't own a copy of Encyclopedia Britannica (its way too expensive). However, Encyclopedia Britannica Online directly addressed absolute zero and how it relates to zero-point energy in its truncated (but informative) explanation of zero-point energy. It defines zero-point energy as the “vibrational energy that molecules retain even at the absolute zero of temperature.” I trust this better than Merriam-Webster which was succinct but unfortunately didn't direct address the nuance of zero-point energy. Still, for confirmation, I e-mailed Dr. Beamish at the University of Alberta. They've performed recent work on Bose-Einstein condensates in helium and how zero-point energy affects things. Greg L 00:14, 13 July 2006 (UTC)

P.S. Don't beat up on me too hard. A lot of effort went into adding that table and all those notes. Most of all, I want everything there to be true. I also think really expansive discussions on the nature of absolute zero belong in Absolute zero. But this is a great forum for fleshing out its true definition because "they can read it here first." Greg L 01:02, 13 July 2006 (UTC)

No one wants to beat up on you. But the latest version still isn't true; there is no "non-quantum mechanical sense" in which the "particles are at complete rest" at absolute zero. --Trovatore 01:11, 13 July 2006 (UTC)

Oops. That wasn't what I intended to do. I think I've been staring at this for too long. Will fix. Greg L 01:40, 13 July 2006 (UTC)

I suggest "Absolute zero is the condition when the heat energy is at a minimum, that is, no more heat can be removed from the system.". This avoids all the theoretical stuff about zero-point motion and hypotheticals about what would be true if only classical physics were true. JRSpriggs 04:49, 13 July 2006 (UTC)

That's not bad at all! The current wording is sort of OK but I don't too much like the phrasing "except for quantum-mechanical motions", as though they were somehow a different kind of motion. --Trovatore 04:56, 13 July 2006 (UTC)

Thanks. “Quantum mechanical motions” refers more to what caused the motions than to what kind of motions they are. One can adiabatically cool atoms all day but if they get to the point where they're sucking on quantum mechanical vapors, it's time to quit. I think the tweaking of this section of the article over the last two days is a prime example of the Wikipedia way of collaboration. Greg L 05:28, 13 July 2006 (UTC)

Redundant With Temperature?

This article seems to have great overlap with the article on temperature. Indeed, since temperature is a thermodynamic concept, the title Thermodynamic temperature sounds a little like the Department of Redundancy Department. I guess I don't clearly see what the distinction between the two articles is meant to be? Perhaps this article is meant to contain the technical thermodynamic discussion, and the other one is meant to be the broad, intuitive introduction? In which case there is a lot of material that should switch places. For example, the section on Theoretical foundation of temperature should come over here, and some of the introductory material and the nice table of temperatures should migrate over there? And, perhaps this article should be renamed to make the distinction clearer? For example "Theory of temperature?" or "Foundations of temperature"? Nonsuch 22:43, 12 July 2006 (UTC)

No kidding. It's no fun trying to edit there. Greg L 00:16, 13 July 2006 (UTC)

I'll take that personally. LeBofSportif 21:05, 17 July 2006 (UTC)

;-) Greg L 07:17, 19 July 2006 (UTC)

Heat energy

• JRSpriggs: In thermodynamic temperature, are you sure about what you wrote about potential energy and the stretching and breaking of bonds? To break a bond (melting or vaporization), requires heat energy from an external source. So the potential energy would only be for the re-establishment of bonds (condensing and freezing) would it not? Do you have a text book definition that can be used here? When responding, please copy this and let's pick this up in the discussion page of thermodynamic temperature. Greg L 06:04, 14 July 2006 (UTC)

• I'm going to try my hand at crafting language that speaks to potential energy in the correct direction (condensing and freezing). If you can find better language yet, have at it. Greg L 06:08, 14 July 2006 (UTC)

• Very good point about heat. I've had this article reviewed by three Ph.D.s (one of whom teaches 2nd year thermodynamics at the university level) and they all missed that. There is a lot of information in the article and its hard to digest. I guess you get what you pay for (the reviews were free). I think I've captured your intent about heat but I changed potential energy to how it must be going in the cold direction (making of molecular bonds). Note that the whole article carefully builds upon the nature of the kinetic motions. The first few paragraphs deal with translational motion. Then the middle paragraphs introduce the concept of the degrees of freedom associated with the internal heat of molecules. Then it goes on to phase transitions. So I simplified your list of all the various types of motions to just “the full variety of these kinetic motions” so as to not hit the reader with too much too fast. I've found that the collaborative nature of Wikipedia tends to result in either equivocal statements that don't say much, or have so many technical caveats that it becomes too hard to read. It's a challenge writing concise introductory text that is both simple and still technically correct. Anyway, we've got kinetic energy, radiation energy, and potential energy that are "included" in heat. Are these all the contributors to heat? If so, we can say "is comprised of" rather than "includes." By the way, my ex-boss at a fuel cell company is reviewing this now. He's ultra, ultra-technical (he reads quantum physics books for recreation) and is taking his time on reviewing the article. We've been exchaning e-mails daily on it—most recently on getting "absolute zero" correct. I'm sure we'll be spending some time on heat now. Let me know if you find a text-book definition that rattles off all of heat's contributors with crisp language. Greg L 06:59, 14 July 2006 (UTC)

• P.S. At 30 GK, the Lorentz effect maxes out at 0.42% for atomic hydrogen; for other elements, it's less than that. Greg L 07:09, 14 July 2006 (UTC)

• This morning with a fresh perspective, I took a look at what I did last night with your work. Your point was well taken about heat. The article as it had originally been written had a few goofs where heat energy was used in place of kinetic and visa versa. No one caught it and it's now been corrected. The subject of temperature, heat, and kinetic energy is complex and needs to be built upon slowly. Otherwise one can undermine the objective of clearly communicating to the target audience (which for this article should be high schoolers in advanced science classes). So I moved the first mentioning of the potential energy of molecular bonds from the first paragraph (where the nature of kinetic energy is explained) to the paragraph that deals with phase transitions. Your correction to the first paragraph still stands: it no longer says the total heat energy is comprised of all the various motions, and now properly states that the full variety of kinetic motions contribute to the total heat energy. I left your caveat of "non-relativistic". I don't think "non-relativistic" and the "30 GK" is hitting the reader with too much too early. This morning I removed your reference to radiation being a form of heat energy (even though it's a true statement). But the issue isn't that it's a "form" of heat energy, it's whether it's a "contributor" to heat energy. Radiation is just a carrier of heat energy and derives its energy from the collisions due to kinetic motion. Black-body radiation is explained later in the article. Although it's tempting to do so, the nature of Wikipedia tends to slowly introduce too many new issues in introductory paragraphs even though the issues are addressed in later paragraphs. Accordingly, the challenge is to make sure what is written in introductory paragraphs is 1) technically true, 2) doesn't mislead, and 3) avoids sacrificing scientific rigor. As regards objective #3, I still must make a revision to a statement about the speed of electrons during cooling because zero-point energy disproportionally affects electrons. I recently received an e-mail from a Ph.D. department head at the University of Alberta. I had contacted them because they've done research on zero-point energy's effect on Bose-Einstein condensates. He wrote "{due to the complexity of the subject}, I really think you have to sacrifice rigor to produce easily understood (e.g. by a high school student) definitions of (thermodynamic) temperature." I'd like to avoid this if at all possible. Greg L 16:31, 14 July 2006 (UTC)

Thermodynamic temperature doesn't contribute to heat?

JRSpriggs: What was on your mind when you added the following:

“Notice that the part of the kinetic energy due to the average translational motion, i.e. the motion of the center of mass, does not contribute to the heat. Nor does the kinetic energy of the average rotational motion of the substance. Only random microscopic motions relative to the averages contribute.”(?)

…I wonder what your reasoning was. If you take helium at, say, 50 kelvin and add heat energy into it, only one thing happens, the translational motions of the helium atoms increases. Whether the helium is under constant pressure or at constant volume, the way you confirm that the translational motions have increased is by measuring the temperature of the gas. What you wrote had the force and effect of saying that the ideal gas law (for which helium is the best analog) is entirely wrong.

Anyway, thanks for the corrections to the "Derivations" section. I don't have sufficient math background to critically evaluate your changes. However, I see from your history of previous contributions as well as your user-talk that you have an interest in mathematics and I assume you are well versed in the discipline. Also, I added your caveat about tidal friction to the list of items contributing to Earth's core temperature. I also made Earth a link. As I stated in the "history" note, this whole caveat on gravitational friction and the decay of radioisotopes was already cited in the #19 footnote attached to that paragraph. I think a footnote is a much better place if one wants to add yet more mechanisms that contribute to the internal temperature of Earth. Greg L 20:02, 15 July 2006 (UTC)

You do not understand what I was saying. Did you not read "the motion of the center of mass"? If I put a block of ice in a rocket ship and accelerate it to a very high speed, does it melt because of that? Heat is random motions relative to the overall motion. The overall motion is not included. JRSpriggs 09:10, 16 July 2006 (UTC)

To use the language of statistics, it is not the MEAN of the velocity vectors of the various particles which contributes to heat; it is their VARIANCE which contributes to heat. Perhaps you are confusing "velocity" with "speed". JRSpriggs 09:15, 16 July 2006 (UTC)

JR: OK, now I understand what you meant. The statement can be easily interpreted to mean something entirely else. Often in discussions of temperature, "average translational motion" addresses the issue of how some molecules move faster than others in a substance. Also, when you say "center of mass", you were referring to the center of mass of an entire system, not the individual particles. Your statement really needed those two words. Just before what you had added, the preceding paragraph introduces the concept of translational motion. That paragraph conveys that the center of mass of an entire molecule moves (translational motion) due to heat energy, and that the average velocity (kinetic energy) of these translational motions defines the temperature of a substance. Pretty much any sentence which states that "average translational motion…does not contribute to heat" runs the risk of being interpreted as something other than what you intended.

I believe what you were referring to was the average motion of an entire system with respect to an observer, and that only random molecular motion with respect to the system’s averaged moving frame of reference (both linear and rotational) contributes to heat energy. With linear motion, this is true. With rotational motion in solids, this seems to be true. In fluids though, the rotating frames of reference you wrote about gets ugly to the extent I don't believe the statement is true. Getting this concept across with plain-speak could be a challenge and I doubt it is worth the effort. Intuitively, everyone knows that moving an object from here to there doesn't make it hotter while it is in motion. We intuitively know that Earth's motion around the Sun doesn't make everything hot. Rotation though, is another thing entirely. According to your suggestion, a rotating object must have the average thermal motion of its molecules measured with respect to a rotating frame of reference. But no matter how I look at the issue of rotating frames of reference, I come to the conclusion that your statement could only apply to a solid—or perhaps a superfluid. But for normal fluids, the average rotational motion of a system is a form of potential heat energy because if it is ever brought it to rest with respect to the observer (as in stopping the rotation of a box filled with a liquid for instance), the resulting turbulence would make the substance hotter. This is, of course, the definition of potential energy: movement with respect to an observer. If rotational motion in fluids can legitimately be considered a form of potential heat energy, then this would be in disagreement with what you wrote. So to this extent, I believe the portion of your proposed wording regarding rotating frames of reference was incorrect unless, possibly, it is buttressed with lots of further caveats and qualifications (such as rotational movement with respect to an observer’s fixed frame of reference and where it is assumed the rotational motion will never stop). One might correctly state that the average rotational motion doesn't contribute to temperature until the potential energy of rotation is brought to a stop. I doubt it is worthwhile to try to introduce such an intricacy so early in the article, especially when the linear portion is intuitively understood anyway. Greg L 20:16, 16 July 2006 (UTC)

P.S. Thanks for pointing out the potential energy of phase changes-to-come. The article is much better for that contribution. Greg L 01:37, 17 July 2006 (UTC)

Virtually any form of energy could be converted to heat. (As a last resort, you could drop it into a black hole and wait for the back hole to evaporate.) The question is "Is it heat NOW?". The overall rotational energy of a substance is not heat now. JRSpriggs 05:25, 17 July 2006 (UTC)

Is it heat now?(?) As you correctly pointed out, heat energy includes the potential energy of phase transitions towards a more ordered state. The potential energy of rotational motion (vorex-like action) in fluids is certainly a form of heat energy. Greg L 06:44, 17 July 2006 (UTC)

The reason that overall translational kinetic energy is not heat is that it is linked to a conserved quantity -- the linear momentum by K = p²/(2×m) where: K is that portion of the kinetic energy; p is the linear momentum vector; and m is the mass. Therefor, it cannot increase or decrease by exchanges with other forms of energy which are heat. Similarly, the kinetic energy of the overall rotation is linked to the angular momentum by K = J²/(2×I) where: J is the angular momentum; and I is the moment of inertia. So it cannot exchange energy with other forms without either a change in the moment of inertia of the object or a transfer of angular momentum to another entity. Of course, vortices within the substance have additional kinetic energy of their own which exceeds that required for the overall rotation and that can be considered to be heat.

Remember that heat is merely a way of giving an approximate description of forms of energy which we cannot describe exactly in terms of the macroscopic properties of the substance. But p, J, m, and I are macroscopic properties of the substance. These arguments do not apply to the potential energy of molecular bonds which can freely exchange with other forms of heat. JRSpriggs 04:03, 18 July 2006 (UTC)

JR: Everything you said above is true. In particular, the part where you say "so it {a system with overall angular momentum} cannot exchange energy with other forms without either a change in the moment of inertia of the object or a transfer of angular momentum to another entity" is also true. I would argue, as you correctly pointed out, that the overall translational velocity of an entire system is not heat energy; it is the potential energy of the overall mass and when it is brought to a stop, heat is generated in whatever braked it but is not technically generated within the moving system. The same situation applies to a rotating solid. Once again, this is potential energy that turns into heat in whatever braked it.

Angular momentum in a fluid within a container is another matter entirely. The question becomes this: in real-world (real-universe) situations, can cyclonic motion in fluids be considered potential heat energy given your just-quoted constraint? Let's take a simple analogy. On the one hand, we have a container of a fluid with an overall, average cyclonic action (angular momentum). On the other, we have a spinning gyroscope. Let's just assume that the source of energy that was required to get them spinning is unknown and that we simply accept that both are spinning. As you wrote, the gyroscope can not exchange energy without transferring momentum to another entity. And when this process finally occurs, the potential energy comes out of the gyroscope and is transfered elsewhere into whatever braked it. But in real life, with real containers, with real fluids that have any non-zero viscosity, the potential energy of gyroscope-like cyclonic motion (angular momentum) would almost immediately be converted within the fluid into molecular vibrations (increased its temperature).

This isn't merely the technicality of an abstract thought experiment; after all, isn't this pretty much what James Watt was doing? He was taking known amounts of mechanical energy (a dropping weight) and was coupling that into a spinning paddle that gave the fluid a cyclonic motion. Imagine doing this ourselves. Immediately after you stop the paddle, you've got a fluid with cyclonic motion. Just wait a half minute and all the energy of this cyclonic motion will have been converted into the kinetic energy of random molecular motion. For fluids with a non-zero viscosity, I think overall cyclonic motion in a system has to be considered as potential heat energy. In real life, one could mull over this issue for maybe a few minutes at most before the cyclonic motion stopped within a container and the substance simply got hotter. Greg L 06:17, 18 July 2006 (UTC)

Quick comment

User:Loom91, good work on removing that ridiculous header comment about high school students. Things like that are not appropriate. I just stumbled onto this page on doing a search for absolute temperature. The page needs a different and smaller image at the top of the page; the one presently there is too big and not relevant enough for the top of the article. I’ll try to clean the article a bit. To User:Greg L, you seem to be biasing this article to a slight degree. I would suggest that you loosen up a bit and try to seek conformity with the rest of the science articles in WP. Thanks:--Sadi Carnot 10:45, 7 August 2006 (UTC)

I cleaned up the article a little, added two sources, and some historical facts; one thing that needs to still be cleaned is the wordiness of the headers: they need to be shortened, more concise, and possibly reduced in amount. Adios:--Sadi Carnot 11:16, 7 August 2006 (UTC)

Sadi: "i.e." means "in other words." You're not using it correctly. It now effectively means "The null or zero point on the thermodynamic temperature scale, although technically impossible to reach, {in other words} according to the third law of thermodynamics, is the lowest possible temperature where nothing could be colder and no more heat energy can theoretically be removed from a substance." Of course, that doesn't make sense. Furthermore, the picture of the Z machine has an attribution copyright. The caption, like all the others instances of this picture in Wikipedia, is supposed to have “Courtesy Sandia National Laboratories” at the end of it. This was there before you deleted it. Greg L 19:27, 8 August 2006 (UTC)

Greg, "i.e." is the Latin abbreviation for id est and according to Merriam-Webster it means "that is"; yet it can mean "in other words" as well as other phrases, which are implied by the context of the sentence, for example: read here and read here. Bye: --Sadi Carnot 05:50, 13 August 2006 (UTC)

"Quantum mechanical motion/energy fluctuations"

Loom91, would you please explain your rationale for changing "quantum mechanical motion" to "quantum mechanical energy fluctuations"? There are no "energy fluctuations" at absolute zero; that's kind of the point: It's a pure eigenstate of the Hamiltonian, and the energy is precisely known. But there is motion; that is, the magnitude of the momentum has a nonzero expected value. --Trovatore 17:09, 7 August 2006 (UTC)

• Saying the molecules have motion will associate the classical concepts of motion with the term. I think it's better to say non-zero energy. How can the specification of a single parameter, the temperature, determine that the system is an eigenstate of the Hamiltonian? Loom91 06:57, 8 August 2006 (UTC)
  • But the particles do move in the classical sense. Or more to the point, they move, period; there's no "sense" involved; the prediction of classical phyisics that they don't move is simply wrong.
  • As to the second question: In general it wouldn't, but for absolute zero it does, because it means that the system is in the ground state, the lowest possible energy, and of course that's well-defined. --Trovatore 19:01, 8 August 2006 (UTC)
    • Of course they don't move in the classical sense. The classical concept of motion involves specifying the exact world-line of the particle, while in quantum mechanics we can't do that. What you mean is that the expectation value of linear momentum is non-zero, which is not the same thing.
      • Let me put it another way: There is no "classical" sense in which they "don't" move. Classical physics simply has no way of describing the actual state of affairs.
    • Specifying a system to be at absolute zero only determines that its thermal energy is minimum, but of course that's not the only contribution to the Hamiltonian, so how can you say it's the lowest energy eigenstate? Loom91 10:33, 10 August 2006 (UTC)
      • Well, so on this point I'll confess to not being 100% sure. Just the same, it's motion we're talking about, not "energy fluctuations". --Trovatore 15:34, 10 August 2006 (UTC)

triple point of water

Per Sadi Carnot's recent edits, thermodynamic temperature is the temperature "relative to the triple point of water". I don't think that's accurate; water doesn't have any distinguished role in the theory of temperature. The Kelvin scale might be defined in terms of the triple point of water, but that's just a unit of measurement, not thermodynamic temperature itself. --Trovatore 19:19, 7 August 2006 (UTC)

• I agree, triple point of water is only used to define scales, not temperature itself. Loom91 06:54, 8 August 2006 (UTC)

• I agree, the triple point of water is used only to define the kelvin and—exactly as Trovatore said—“water doesn't have any distinguished role in the theory of temperature.” And kelvin is but one scale used to delineate absolute temperature; Rankine is the other. Clearly, the attribute in common to both scales — and the unique characteristic of “Thermodynamic temperature” in general — is absolute zero. Measuring temperature relative to absolute zero is what makes "thermodynamic" temperature a key parameter of thermodynamics. Greg L 00:48, 13 August 2006 (UTC)

Did anyone happen to notice that this contribution was sourced by a "Dictionary" of Thermodynamics (from its two-page section on thermodynamic temperature)? --Sadi Carnot 05:58, 13 August 2006 (UTC)

• Would you care to give an exact quote from that text saying that the thermodynamic temperature in general (as opposed to the kelvin, a unit of measurement of thermodynamic temperature) is defined relative to the triple point of water?

I don't have that book, but Amazon lets you look at a few pages of it, and it so happens that the entry on "Absolute temperature" is among those pages. Here's what it says:

A temperature scale is said to be absolute when it is not related to the physical properties of any particular material. Expansion of mercury, electrical resistance of platinum or electromotive force of a thermocouple, for instance, give only a relative temperature scale. The thermodynamic temperature, defined by the second law, gives an absolute temperature scale.

Now frankly I find the passage a little dubious; the real distinction, as Greg says, is where the zero point is. But certainly it does not support the claim that thermodynamic temperature is defined in terms of water's triple point. Quite the opposite, in fact; it says that thermodynamic temperature gives an absolute temperature scale, which therefore (according to the quoted text) is not defined in terms of physical properties of a particular material (triple point is a physical property; water is a particular material). --Trovatore 06:20, 13 August 2006 (UTC)

• Sadi Carnot: If one carefully parses the first sentence you supposedly quoted from the Dictionary of Thermodynamics: (“In thermodynamics, the thermodynamic temperature, or absolute temperature, is the measure of temperature with reference to the triple point of water, which was fixed at exactly 273.16 K in 1954.”), one can see that each clause is based on a true fact. But taken as a whole, it is misleading to the point of being incorrect and is fatally flawed. It's precisely as useful as saying “the Celsius scale is defined with respect to boiling point of water.” The definition you had makes the error of assuming that the kelvin scale and thermodynamic temperature are one in the same. They are not. And even if they were one in the same, mentioning only one anchor point (the triple point of water) at the expense of omitting any mention of the principal one (absolute zero) makes it, effectively, totally wrong. In its broadest sense, thermodynamic temperature is simply the measurement of temperature with respect to absolute zero; that's what thermodynamic temperature is. Thermodynamic temperature is delineated (precisely represented via an artifact, protocol, or system) using either the kelvin or Rankine scales.

Now that I think about it, I realize that what you had was suitable for communicating what the kelvin scale is: “The kelvin scale, is the measure of thermodynamic temperature with reference to the triple point of water, which was fixed at exactly 273.16 K in 1954.” And this would still assume the reader is technically sophisticated, already understands thermodynamics, and already knows that all thermodynamic temperature scales are anchored at absolute zero.

Was that Dictionary of Thermodynamics book something college students were forced to pay for sight unseen? If your sentence really came out of that book, then that's the only way I can envision a publisher could have expected to make money with something like that in a free-market economy. Greg L 01:52, 14 August 2006 (UTC)

picture caption vs table

The caption says it reached 2 billion kelvin; the table says it hit 2 billion Celcius. Which is it?

Annonymous reader from Tokyo Japan: Both. The table actually says both 2 GK and 2 billion °C. The 273.15 deg C difference between the two values is too small to resolve for such a big value (which wasn't measured to an accuracy of a couple hundred degrees). This point is disclosed below the table in note E, which says "The 273.15 K difference between K and °C is ignored to avoid false precision in the Celsius value." Greg L 04:00, 14 December 2006 (UTC)

Coldest ocean water

I slightly change the note 24. Ocean water at the biggest depths is colder than 4 deg. C. Remember that ocean water is not distilled water. Ocean water don't have maximum density at 4 deg. 84.10.114.122 08:33, 6 November 2006 (UTC) a casual visitor, Poland.

Casual visitor from Poland: Thank you for pointing this out. The ocean bottom is indeed colder than 4 °C. I found this post by Rob Campbell, Ph.D Candidate, Oceanography, University of British Columbia. The post included an actual plot of temperature vs. depth. I updated the temperature range in the article's note accordingly. I also note and applaud your command of English. It's a heck of a lot better than my Polish. Greg L 04:09, 14 December 2006 (UTC)

“Comprised”/“composed”

With regard to my original caption, which read “[m]olecules have internal structure because they are comprised of atoms that…”, you are correct. I learned something new. Science interested me in school; grammar bored me. Wikipedia is a hell of way to learn from others. However, the word comprised has (arguably incorrectly) shifted meaning over time to the point that my original wording is becoming increasingly acceptable. I reworded the caption to avoid this issue. Free Dictionary.com says the following about comprise:

“ Usage Note: The traditional rule states that the whole comprises the parts and the parts compose the whole. In strict usage: The Union comprises 50 states. Fifty states compose (or constitute or make up) the Union. Even though careful writers often maintain this distinction, comprise is increasingly used in place of compose, especially in the passive: The Union is comprised of 50 states. Our surveys show that opposition to this usage is abating. In the 1960s, 53 percent of the Usage Panel found this usage unacceptable; in 1996, only 35 percent objected. ”

My 1976 edition of World Book Dictionary makes a similar statement. Note that I had used the word as if it meant composed, as in the increasingly acceptable (but classically incorrect) usage in “The Union is comprised of 50 states.” Still, roughly 35% of a dictionary’s “usage panel” (people who know proper grammar) would object to my original wording; you did. And now that I've studied the whole comprise/compose issue, I feel like you: “…are comprised of…” shouldn't be considered as acceptable. Accordingly, because the acceptable use of “comprise” is in a state of flux and means different things to different people, and because “…they comprise atoms…” is correct to purists but sounds odd to many plain folk, I've reworded the caption with “are composed of” to avoid this problem altogether. It's a passive-voice construction but reads smoothly.

By the way, I try to avoid passive-voice constructions but was amused when I Googled on {"are composed of" grammar} and the first hit was this grammar lesson site which says “Verb phrases are composed of the verbs of the sentence and any modifiers…” (my emphasis). Greg L 17:45, 27 January 2007 (UTC)

Overly broad and inaccurate statements

Enormousdude: With regard to these two changes, temperature is not simply the measure of “energy of random motion of particles.” That expression is overly broad and imprecise and reads as if one needs only to measure the energy one puts into a substance to determine its temperature. Those words don't account for why gasoline (C_p = 228) requires eleven times the energy to raise its temperature one kelvin as is required to raise helium’s temperature one kelvin (C_p = 20.8). Temperature is actually proportional to the kinetic energy of a specific kind of particle motion (the kind not absorbed internally to molecules). This issue is addressed soon enough (the very next paragraph actually) in the Overview section. Your edit also seemed also to have been written in great haste as it used the spelling “per perticle” and also relied upon rather unencyclopedic language “(say, of a piece of metal consisting of bunch of atoms).” Note too that the word “a” is missing before the word “bunch” (“…consisting of bunch…”). To properly address what you attempted to accomplish: establish the association between temperature and energy, one could simply transplant portions of the Overview section up to the introductory paragraph. However, this would defeat the purpose of keeping introductory paragraphs succinct and pithy — one of the qualities for which Wikipedia is famous. One can simply scroll down one paragraph to get onto that topic.

Also, the change of “vibrations” to “motion (vibration, rotations, etc)” uses imprecise, improper language. It has the effect of saying that "vibrational motion" is a special class of thermal motion distinct from "rotational motion". However, "vibrational" motion is actually the broadest way of describing all classes of thermal motions. If one wants more specificity in the type of vibration, one properly specifies "translational, "rotation," "bond length," and/or "bond angle" movements. These distinctions are addressed at the appropriate places in the article: in the Overview and The internal motions of molecules and specific heat sections. And once again, in your haste of writing even this short addition, you included no period at the end of "etc".

Occasionally, some scientific rigor must be sacrificed in order to explain complex scientific topics (often with obscure exceptions) in a clear manner. Many Web-based articles on temperature succumb too easily to this temptation in my opinion. By carefully addressing and building upon each new topic (such as the types of motion or the nature of kinetic energy), entirely true and completely true statements can be consistently made throughout the article.

Before making contributions to Wikipedia, I suggest that you carefully read entire articles in order to determine the terminology and conventions used throughout them. This will also give you the opportunity to understand what has been covered, where it is covered, and how articles progress from concept to concept. This will help ensure that articles are harmonious, flow well, and are consistent. This extra discipline might also diminish the haste with which you are making your edits. Although this might reduce the article-count that you’ve contributed to, you and I are after all, working on an encyclopedia. Greg L (my talk) 05:09, 20 March 2007 (UTC)

The heat of phase changes

I think that the following passage:

At one specific thermodynamic point, the melting point (which is 0 °C across a wide pressure range in the case of water), all the atoms or molecules are—on average—at the maximum energy threshold the lattice bonds can withstand without breaking and jumping to a higher quantum energy state. Quantum transitions entail a complete jump in the atoms’ electron configuration from one energy level to another; no intermediate values are possible. Consequently, when a substance is at its melting point, every joule of added heat energy only breaks the bonds of a specific quantity of its atoms or molecules..

is, not to put to fine a point upon it, a load of hooey. The phenomenon of latent heat has nothing to do with quantum mechanical restrictions on energy absorption - indeed, in most familiar examples of first-order phase transitions quantum effects are negligible as the intermolecular degrees of freedom are well within the classical regime at these temperatures. I don't believe that latent heat can be understood at all in terms of individual intermolecular "bonds" - it is a collective phenomenon, arising from the fact that the system has to reorganize itself on a large scale in order to move from one set of configurations to another set having a lower free energy. (Strictly speaking, latent heat only emerges in the limit of an infinite system; in practice, you can see its emergence in simulations of large, finite systems but you can't explain it at all on the basis of the properties of individual intermolecular bonds.)

Unless someone mounts a well-founded defense of this passage, I'm going to delete it.

Rparson 01:11, 2 August 2007 (UTC)

Rparson: I didn’t see your above alert until after you later deleted the text.

“[A] load of hooey”?!? Okaaaaay… That's fine; the text you deleted was flawed and, since it caused confusion with regard to “quantum,” merited a rewrite. Accordingly, I've revised the section you objected to. It no longer discusses "electron configuration" and related matters. However, all chemical bonding ("ionic," "covalent," or that catchall "molecular", etc.) are described by the laws of quantum electrodynamics (QED) and are quantized forces; there either is a bond, or there isn’t a bond; there is never “kinda sortuva bond.” That's why there are very precise valuse for enthalpies of fusion: a very precise amount of energy is required to break an individual bond (0.062284 eV per molecule in the case of water).

Given this reality, I can’t even begin to fathom your statement that “I don't believe that latent heat can be understood at all in terms of individual intermolecular "bonds" - it is a collective phenomenon” (my emphasis). Or this one: “Strictly speaking, latent heat only emerges in the limit of an infinite system.” Is this science or faith? I don't have much respect for any theory of latent heat that “strictly speaking” requires “infinite systems” since reality daily serves up abundant examples of melting (and specific heat) in systems quite a bit smaller than infinite—like an individual snowflake melting. Or an icecube, which always melts from the outside-in. It is inescapable that all melting, when considered at the molecular level, comprises the release of individual molecular bonds (albeit numerous ones). That much is just too obvious.

The section in question now (as of today anyway) reads as follows:

“ At one specific thermodynamic point, the melting point (which is 0 °C across a wide pressure range in the case of water), all the atoms or molecules are—on average—at the maximum energy threshold their chemical bonds can withstand without breaking away from the lattice. Chemical bonds are quantized forces: they either hold fast, or break; there is no in-between state. Consequently, when a substance is at its melting point, every joule of added heat energy only breaks the bonds of a specific quantity of its atoms or molecules… ”

You might enjoy this article: UCSD Scientists Use Laser And X-Ray Technologies To Watch A Different Way Of "Melting" Semiconducting Material …and this abstract: Ultrafast superheating and melting of bulk ice (also touched upon here in the Ice article).

Greg L (my talk) 04:09, 5 August 2007 (UTC)

You are mixing up several things (partly my fault, since I jumped around between several points that I should have made separately.) Sure, quantum mechanics is required to calculate the forces that act between atoms in a molecule. But once those forces are determined, you can often - not always, but often - calculate the motions of the atoms within the molecule by classical mechanics. As you put more and more vibrational energy into a molecule (or a solid) it behaves more and more classically. That's why heat capacities of solids approach the Dulong-Petit rule (i.e. obey the classical equipartition theorem) at high temperatures. Thus in the vast majority of cases, the motions of the atoms in a solid has already become essentially classical well before melting. And that's why you can perfectly readily simulate the melting transition using classical mechanics on a few thousand atoms interacting with Lennard-Jones potentials. All the quantum mechanics is buried in the LJ potentials.

You say that a chemical bond is "all or nothing" - sure, in the sense that as you put more vibrational energy into a molecule, you get wider and wider amplitude swings, but always a return - until you pass the dissociation energy and the molecule doesn't come back together again. But that' no different from escape velocity of a rocket - it's either in orbit, or it isn't. Classical mechanics in both cases.

As to the issue of "latent heat emerges in the limit of an infinite system", let me try to explain by means of a thought experiment. Let's start with a gas of Lithium molecules (diatomic, Li2). At low temperatures, the molecules are in their vibrational ground state. As you heat up the gas, they move to excited states; higher and higher temperatures lead to higher average vibrational energy and also a broader distribution over vibrational states, and soon enough you start to see some Li atoms. Kepp raising the temperature and the equilibrium shifts towards more atoms and fewer diatoms. Eventually you have no molecules left. Each individual molecule, of course, requires a well-defined minimum nergy to dissociate, but you don't see anything remotely like a "latent heat" - the energy increases smoothly with temperature.

Now repeat, with a Li nanocrystal - say 100 atoms or so - instead of diatomic molecule. This would be a tough experiment to do, but it's easy enough to simulate on a computer. In such simulations, what people see is a much narrower range of energies than in the single molecule as you "heat" the crystal. The vibrational amplitudes increase - but they don't get as big as in the isolated molecules, because the atoms get in each other's way. Then, over a narrow, but finite, range of temperatures, the crystal "melts" - the atoms now move around a lot more - and later on vaporizes. The curve of E vs T is continuous, but has definite step-like features in it. When you make the crystal larger, the range over which melting and boiling takes place gets narrower. Theoretically speaking, an actual discontinuity in the curve emerges as you take the limit # of particles going to infinity (the so-called thermodynamic limit but you are welcome to regard that as a mathematical convenience if you with - certainly 10^23 is close enough to infinity for all practical purposes.

So while I would not disagree that the existence of a bond dissociation energy - your "bond is either broken or it isn'") is required in order for there to be a latent heat, but I maintain that it is also necessary to have a large system - collective effects - in order for latent heat to show itself. Quantum mechanics only enters in so far as it determines what the interatomic forces are.

All that said, your current revision goes a long way to meeting my objections; I may massage the wording slightly. What bothered me about the earlier version was that it seemed to say that the system did some big quantum jump at the melting point - that implication is gone now.

--Rparson 20:47, 5 August 2007 (UTC)

Very well. Greg L (my talk) 20:53, 5 August 2007 (UTC)

Fig. 1

Fig. 1 states "slowed down here two trillion fold". Does that mean that atoms usually travel significantly faster than light?

• No. It just looks that way because the helium atoms are so small. Sped up two trillion fold (actual speed at 23 °C), they would be moving at a mean speed of 1359 m/s, which is 0.000453% the speed of light. Greg L (talk) 04:51, 29 June 2008 (UTC)

  P.S. The above was the mean “speed”; that is, velocity along any arbitrary vector at any instant. Note that the mean velocity of atoms and molecules of a gas is very, very close to the speed of sound in the gas. Take, for instance, air at median world-wide conditions at sea level. At 760 torr, 15 °C, and a dew point of 9 °C, the average molecular weight of air is 28.8413(4). Under these conditions, the mean molecular velocity (velocity vector component aligned with a particular direction) is 288.2 m/s, which is close to the actual speed of sound in air. Greg L (talk) 22:18, 29 June 2008 (UTC)

I think the objection of the original poster still stands though. The speed on screen is about 0.05m/s, and that is what people are thinking of when they read "slowed down here two trillion fold". (Implying speeds of 2 x 10^11 m/s !) It should be made clear what the length scale is in the diagram if the "two trillion" figure is to be used.LeBofSportif (talk) 09:03, 1 July 2008 (UTC)

Merge

Should probably be merged with temperature Headbomb {^{— Write so you cannot be misunderstood.}
_{— ταλκ / Wikiproject Physics: Projects of the Week} 04:30, 17 June 2008 (UTC)

• Ugh; I shudder at the thought of it. This article has been very stable for a long time. It covers the topic of both thermodynamic temperature and the thermodynamic principles underlying it. The Temperature article, IMO, needs to stay simpler so it is more accessible to a general-interest readership. The trouble with the Temperature article is precisely because it is more accessible to a general-interest readership, it tends to get revised with garbage and falsehoods and is in a continual state of flux as a result. If Temperature is getting remotely close to the technical level of this article, it should get reigned back in; not every article needs to be so technical. That’s why we have Quantum mechanics and Introduction to quantum mechanics. Greg L (talk) 04:58, 29 June 2008 (UTC)

• Support I read much of this article, and skimmed the rest, and there is nothing to separate this concept from temperature. A strange pretentious tone pervades this article, starting with the title, but there is much content that would substantially improve temperature, such as the history section. However, there is also some very bad science here, such as the assumption that degrees of freedom are somehow fundamentally limited to three. Also, an introduction to statistical mechanics is missing and necessary to explain eg the black body distribution or how to find the temperature of a small closed system. Merging will be tricky but is necessary as "thermodynamic" temperature is not a distinct concept, nor does this article deal more than the other one with the origins or application of thermodynamics. Potatoswatter (talk) 05:33, 6 July 2008 (UTC)

• There are three spatial degrees of freedom within which entire particles can move. That physical principal is a distinct concept from the internal degrees of freedom, which can number in the hundreds for complex molecules. Greg L (talk) 21:54, 4 September 2008 (UTC)

Three degrees of translation, two of rotation of the entire particle, and more of internal rotations and vibrations. Typically, far as I know, internal rotation or movement within a molecule does not contribute much to temperature, as limited energy can be stored. But the molecular rotation certainly does. Other degrees of freedom contributing to thermodynamics result from possible ionic and electronic interactions.

Anyway, looking again at section "The nature of kinetic energy, translational motion, and temperature", I see that it incorrectly defines temperature in terms of translational motion. Indeed, it totally misses the point of what a degree of freedom is, and why it is called that. Needs to mention that thermal energy is randomly exchanged between oscillators. Potatoswatter (talk) 02:13, 5 September 2008 (UTC)

• Crap, I knew I shouldn’t have answered you here and encouraged you. The article is correct. Please don’t screw with it unless you know what you are doing. The temperature of a substance depends only upon translational motion. There is nothing wrong with the Boltzmann constant. Any kinetic energy bound up in internal degrees of freedom—while it is transfered to and from and is in equilibrium with translational motion—only adds to a substance’s specific heat capacity, not to its temperature. Yes, there is some coupling going on if you look (theoretically) at an individual molecule: a long molecule spinning like a propeller can “thwack” another molecule (or container wall) with its advancing edge. But then, just as many molecules “anti-thwack” with their trailing edge. Averaged out across a statistically significant quantity of particles, the internal degrees of freedom of molecules are just heat sinks and the “temperature” of the molecular substance is simply a function of the kinetic energy of their translational motion. That’s why all the monatomic gases have a C_v of 12.4717 J mol⁻¹ K⁻¹ and any complex molecular gas with available internal degrees of freedom simply soaks up more heat energy per unit increment of temperature.

  And as far as your statement “[the article n]eeds to mention that thermal energy is randomly exchanged between oscillators.”, the article states

Accordingly, as heat is removed from molecules, both their kinetic temperature (the kinetic energy of translational motion) and their internal temperature simultaneously diminish in equal proportions. This phenomenon is described by the equipartition theorem, which states that for any bulk quantity of a substance in equilibrium, the kinetic energy of particle motion is evenly distributed among all the active degrees of freedom available to the particles.

Greg L (talk) 02:35, 5 September 2008 (UTC)

• P.S. You know that first footnote in the article? The one that cites Daniel C. Cole? He’s the guy who wrote a number of landmark papers on the goings on at absolute zero. I exchanged a dozen e-mails with him and he reviewed the article many time while I wrote it. I also had other Ph.D.'s who teach college thermodynamics review the article while I wrote it—many many times in fact. One of the professors (at London South Bank University) linked to this article for the benefit of his students after I contacted with him with a question. This article isn’t the result of some bullshit blown out of some random volunteer editor’s ass because he had some wild notion of how thermodynamics works; this article has been extraordinarily well researched, fully cited, and proof read by a half dozen pros, some of them world-class, published experts on the subject. It is correct. Greg L (talk) 02:46, 5 September 2008 (UTC)

• Lol, sounds like you're gettin' to know me. So, you know that kinetic energy is divided among all "degs of freedom" (aka harmonic oscillators) in a molecule, so why are translational oscillations more special than rotational ones? To condense your argument, "The temperature of a substance depends only upon translational motion. … Any kinetic energy bound up in internal degrees of freedom … only adds to a substance’s specific heat capacity, not to its temperature." You again neglected rigid rotation, so please review equipartition theorem. And checking specific heat capacity, note that d(temperature) = d(heat kinetic energy) / (specific heat capacity).

Yes, I’m familiar with Specific heat capacity, I helped to significantly expand the article. Greg L (talk) 17:36, 5 September 2008 (UTC)

• Certain oscillators require activation energy before they can represent a degree of freedom, if that is what you refer to, but that doesn't mean they don't contribute to temperature. Typically, heat capacity is approaches zero near absolute zero as matter becomes a Bose-Einstein condensate and acts like one big molecule with only five possible (big) oscillations. More heat activates more oscillators and specific heat capacity increases. This goes on until some internal oscillators get "filled up" with energy and fall apart, such as denaturing molecules and then ionizing atoms. This further increases the specific heat capacity by increasing the number of particles. It is true that activating more oscillators doesn't increase temperature, but false that additional oscillators don't affect temperature once activated.
• The equipartition theory states that all oscillators are equal. Each oscillator can be assigned a "temperature" based on its internal energy and the Boltzmann constant. This goes for translation and rotation alike. The difference is that translation is the lowest common denominator which applies to all systems. Potatoswatter (talk) 16:03, 5 September 2008 (UTC)

• I see you are a good sport Potatoswatter and aren’t trying to be an ass. But some of what you wrote above is beside the point and some seems to have been ralphed out onto this page for no particular reason other than to demonstrate that you are knowledgeable about this subject. I really don’t want to debate you here. I just don’t have the inclination you seem to have towards engaging in intellectual thrust & parry for the shear sport of it.

  You wrote “Each oscillator can be assigned a "temperature" based on its internal energy and the Boltzmann constant.” Well, you seemed to have missed the important point. For any statistically significant quantity of a substance in equilibrium, all available degrees of freedom have the same mean temperature. That goes for molecular substances. At room temperature, a nitrogen molecule has two internal degrees of freedom and the mean temperature of these internal degrees of freedom are the exact same temperature as that of its translational motion. So if a container of nitrogen is at room temperature (295 K), then its internal degrees of freedom are 295 K too. And since room-temperature nitrogen has two extra degrees of freedom, guess what(?): nitrogen has a C_v of 20.8 J mol⁻¹ K⁻¹, which is 166.7% that of monatomic substances such as helium (C_v of 12.4717 J mol⁻¹ K⁻¹).

• Exactly. Hence equipartition theorem and thermal equilibrium. Potatoswatter (talk) 23:35, 5 September 2008 (UTC)

• The article makes all this clear and your continued rankling and making observations on points that are clearly covered in the article makes it increasingly obvious that you really haven’t read it. If you think anything is incorrect with the article, first run your theories and understandings by an expert before you foul things up. I note that Dr. Kravitz in the below post seems to have some facility with the subject and he didn’t seem to key-in on any shortcomings in the article—he might even have floated a trial balloon below for mentioning his proposed temperature scale in the article.

  And finally, if you are trying to argue arcane nuances and exceptions to the general rule, all that is beyond the scope of the article. Greg L (talk) 17:26, 5 September 2008 (UTC)

• The only point we're arguing over is whether temperature is defined in terms of translational oscillation (current text) or all types of oscillation at once. From my thermo class, I don't recall an intermediate step of only thinking about translation, so your intro seems a little stilted. More generally, I simply think the article's tone differs from the WP norm, for better or worse, because of fewer authors and less linking. For example, the section on Thermodynamic_temperature#The_diffusion_of_heat_energy:_Black-body_radiation would gloss over less if you simply copy-pasted some relevant text from that article and used {{main}}. Perhaps that wouldn't particularly improve the article, but I think that reflects its status as a freestanding work, almost a textbook chapter, rather than an integral part of WP. All study of temperature involves thermodynamics. That is why I think it should be merged. Your arguments for separation (separatism?) sound too elitist. There are other places on the Web for works not intended for public modification.
• A separate article discussing phase changes, quantization, and general effects of low density of states might be nice, but this article doesn't go there: (Yes it does cover phase changes & quantization. You clearly haven’t read it. Here. Greg L (talk) 02:17, 6 September 2008 (UTC)) I see only phase changes in terms of quantity of heat needed, nothing about e^kT/E probabilities Potatoswatter (talk) 04:36, 6 September 2008 (UTC) the scope is exactly the same as temperature. Potatoswatter (talk) 23:35, 5 September 2008 (UTC)

← unindented

• The rigorous distinction of heat, temperature, internal degrees of freedom, and translational motion is to keep it clear that the temperature and volume of a gas and the pressure exerted by that gas on a container’s walls, the pV = nRT  business, are all intrinsically tied only to the particle kinetics of translational motion, as Ludwig Boltzmann figured out in the late 1870s. The pressure exerted by the gas can be perfectly calculated with classical mechanics calculations purely in terms of recoil kinetics involving perfectly elastic collisions of atoms and molecules due only to their translational motion. Indeed, the internal motions of molecules provide a splendid heat source to prevent the gas from cooling as rapidly as it would if it were monatomic, but these internal degrees of freedom do not directly contribute to the pressure (and temperature) of the gas—not on average. Said another way: the temperature and volume of any gas (monatomic or molecular) and the pressure it exerts on a container’s walls, can be perfectly modeled, on average, as simple billiard balls rattling about without concern for internal motion that might be within the billiard balls. Note the first animation. The caption says it’s helium. But it could just as well have been a complex molecule. In that animation do we need to see the internal degrees of freedom or know anything about the insides of these molecules in order to calculate the pressure the molecules are exerting on the container? No. All the information regarding pV = nRT   is satisfied with an external view of the recoil kinetics. Whether atoms or molecules, the speed of translational motions of plain old solid blue and red balls provides all the necessary information regarding pressure, temperature and volume.

• Hmm. I tend to overlook gases because I'm a solid state guy. Isn't it best to introduce heat and temperature as generally as possible before relating it to pressure? My thermo course just said, think of all these springs connected together, and there are several springs for each particle. So ideal gases are a little distracting. Potatoswatter (talk) 04:36, 6 September 2008 (UTC)

• Normally of course, all degrees of freedom—translational and internal—are in equilibrium; all temperatures are the same. But maintaining translational motion as distinct from internal degrees of freedom is useful when one starts examining non-equilibrium phenomenon. For instance, when scientists fire cold, large molecules through instruments, these distinctions are important when one ponders “what is the temperature of that particular molecule (that is to say, regardless of its velocity through the instrument, at what peak wavelength of blackbody radiation does it radiate as it flies through)” and then wonders “what will its kinetic impact be when it hits the instrument’s backstop,” or then considers “what would be the temperature of the gas be if one collected many of these molecules and allowed them to interact and reach equilibrium; what would the impact forces on the backstop then be?” Certainly these are advanced concepts, but you’d be amazed at the number of people who have a difficult time understanding the distinction of heat and temperature and specific heat capacity. To address the complexity of this issue, this article properly discusses the properties of matter that contribute to specific heat capacity, and what property is responsible for the pressure, temperature and volume of gases.

  Note too that the article was accurate enough that after I contacted the professor at London South Bank University for get a citation for something, he linked to this article within his curriculum Web page for the benefit of his students. He didn’t state a reason, other than he really liked it. Greg L (talk) 01:05, 6 September 2008 (UTC)

• Congratulations on the citation, but I'm pretty sure Jimbo doesn't want us to rest on our laurels. Um, I'm pretty sure blackbody radiation is quantized at the size of the emitting particle/system. Anyway, specific questions are often answered by specific research. Even I don't bother gaining general knowledge for expedient answers. Right now I'm supposed to be making a linear solver, as quickly as possible. I hate matrices and won't learn much by doing if I can help it. Rather than guess that our readers being "more technical" will be thinking about plasma guns or absolute zero or even ideal gases in particular, I think it's safest to stick with just plain elementary thermo at the intro. The number one question is "where does all the energy go?" and distinctions can be made later. Double anyway, it's not specifically my decision to make. It belongs to the community, which is why the contributions and merits of this article should be merged to the more prominently named one. Potatoswatter (talk) 04:36, 6 September 2008 (UTC)

• Incorrect regarding blackbody radiation being quantized at the size of the emitting particle/system. Yes, blackbody photons are quantized. But blackbody photons are always emitted from atom-size volumes: from perturbed electron orbitals that have been stretched or bumped (in this case, due to internal degrees of freedom and thermal agitation). You can sum the contributions from a hundred atoms comprising a single, big molecule, or you can sum the contributions from a trillion atoms or molecules. So in the case of something like a single, complex, 100+ dalton molecule, with very many internal degrees of freedom, the blackbody radiation being emitted also takes the form of a wide variety of quantized wave packets that sum to form a classic blackbody curve with a peak emittance wavelength that correlates to—and reveals—that particular molecule’s internal temperature.

  By the way, this property (blackbody radiation being emitted from individual molecules) was exploited when scientists in Austria did double-slit experiments with big molecules (C-70 buckeyballs). You know: those ‘interference pattern’ experiments that are classically done with photons and even electrons? They can make entire big-ass molecules interfere in double-slit experiments. They found that when the blackbody radiation of the emitting molecule achieved a sufficiently short wavelength (high temperature) that it could reveal “which slit” information, the interference pattern disappeared. They even could get C₄₄H₃₀N₄ molecules to interfere.

  Anyway, my point is that pV = nRT   is the product only of translational motion. Thought experiment: if one could magically get 100 ultra-cold molecules suspended in a chamber with near-zero motion in the three spatial degrees of freedom, and then rapidly heated them with a multi-direction flash of laser light (imagine one hundred molecules like that shown in the Fig. 3  animation, all static relative to each other): the pressure in the chamber wouldn’t rise whatsoever until two of those molecules drifted together and recoiled apart, setting off a cascade reaction. The kinetic energy bound in those molecules’ internal degrees of freedom would finally percolate into their three spatial degrees of freedom. For a brief moment, you’d have a “cold gas” of “hot molecules” before the system reached equilibrium. Greg L (talk) 18:34, 6 September 2008 (UTC)

• Yeah, you're right, they're quantized at the size of the enclosing system. That's the only way to set boundary conditions.
• I suspect you tend to focus on "interesting aspects" of science, and this might lead to most of what I find objectionable. Pedagogically, I tend to be a stickler for laying foundations. Still think that pressure should be in another section after the intro. Potatoswatter (talk) 04:20, 7 September 2008 (UTC)

• What better way to begin giving readers the “Ah haaa” of what temperature is (kinetics) than to first touch upon the relationship of pressure, temperature, and volume very early in the article? After all, it was the 1702–1703 work of Guillaume Amontons and his observation of how absolute temperature was related to absolute volume, and the 1777 work of Johann Heinrich Lambert in understanding how absolute temperature affected absolute pressure that laid the foundations of this science. The article starts with pV = nRT—exactly in accordance like the physical phenomena that were first discovered over one hundred years before Boltzmann finally explained the concept of temperature it in terms of the kinetics of bouncing billiard balls and their probabilities. Greg L (talk) 20:38, 7 September 2008 (UTC)

• Sometimes ontogeny recapitulates phylogeny, but it's not a hard rule. There are other ways to arrive at the question, "what is temperature," than wondering about ideal gases. If the presentation were more anecdotal, eg "let's step back and look at the guy who discovered temperature," I'd be more partial. Potatoswatter (talk) 21:25, 7 September 2008 (UTC)