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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)
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)
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 AN; 25% of AN or even 0.000001% of AN 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)
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 = kAN. 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. S B H arris 22:41, 27 October 2011 (UTC)
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, AN 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 AN.
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)
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. S B H arris 21:32, 31 October 2011 (UTC)
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)
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. S B H arris 23:49, 31 October 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 = kBT is somehow imprecise, and that the various parts of the equation, E, kB and T are somehow incomplete or imprecise? E = kBT 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)
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. Spinning Spark 13:02, 1 November 2011 (UTC)
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. 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)
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)
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 kB) 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)
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 = kBT connect that energy and temperature? Are you saying that E = kBT is wrong? -- Damorbel ( talk) 21:34, 1 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)
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. S B H arris 16:15, 2 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. S B H arris 16:13, 3 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)
![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 | Archive 3 | Archive 4 |
![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 | Archive 2 | Archive 3 | Archive 4 |
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)
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)
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 AN; 25% of AN or even 0.000001% of AN 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)
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 = kAN. 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. S B H arris 22:41, 27 October 2011 (UTC)
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, AN 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 AN.
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)
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. S B H arris 21:32, 31 October 2011 (UTC)
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)
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. S B H arris 23:49, 31 October 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 = kBT is somehow imprecise, and that the various parts of the equation, E, kB and T are somehow incomplete or imprecise? E = kBT 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)
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. Spinning Spark 13:02, 1 November 2011 (UTC)
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. 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)
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)
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 kB) 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)
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 = kBT connect that energy and temperature? Are you saying that E = kBT is wrong? -- Damorbel ( talk) 21:34, 1 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)
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. S B H arris 16:15, 2 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. S B H arris 16:13, 3 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)
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