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DrClaude
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There were two interwoven discussions, so I split one off to https://www.physicsforums.com/threads/identical-and-indistinguishable-particles.939282/
This summarizes well how I always thought of the situation. So for me, Boltzmann follows from working with indistinguishable particles.One type is a gaseous assembly, in which the identical particles are the gas molecules themselves. In quantum mechanics one recognizes that the molecules are not only identical, but they are also (in principle as well as in practice) indistinguishable. It is not possible to ‘put a blob of red paint’ on one particular molecule and to follow its history. Hence the microstate description must take full account of the indistinguishability of the particles. Gaseous assemblies will be introduced later in Chapter 4.
In this chapter we shall treat the other type of assembly, in which the particles are distinguishable. The physical example is that of a solid rather than that of a gas. Consider a simple solid which is made up of N identical atoms. It remains true that the atoms themselves are indistinguishable. However, a good description of our assembly is to think about the solid as a set of N lattice sites, in which each lattice site contains an atom. A ‘particle’ of the assembly then becomes ‘the atom at lattice site 4357 (or whatever)’. (Which of the atoms is at this site is not specified.) The particle is distinguished not by the identity of the atom, but by the distinct location of each lattice site. A solid is an assembly of localized particles, and it is this locality which makes the particles distinguishable.
However, the derivations I've shared give Boltzmann only for distinguishable particles!DrClaude said:I am not sure I understand the original problem. To quote from T. Guénault, Statistical Physics (Springer):
This summarizes well how I always thought of the situation. So for me, Boltzmann follows from working with indistinguishable particles.
We know that the ##1/N!## is needed for an ensemble of identical and distinguishable (a better word might be classical) particles to make the entropy extensive. The factorial is not needed if the particles are not identical. Such is the case if two different gasses are allowed to mix. The entropy should increase when two subsystems containing different gasses are combined since mixing the gasses is an irreversible process.Philip Koeck said:I also noticed the opinion that the factor 1/N! is required even for distinguishable particles if they are identical, but I'm not sure what to think of that and it's not what textbooks say.
Again, its not that classical particles of a pure gas are indistinguishable but that they are identical. This means that permuting their positions in phase space on a constant energy manifold leads to the same microstate.Philip Koeck said:Textbooks claim that you need a factor 1/N! in the partition function to make sure that even "classical" particles are indistinguishable and the entropy becomes extensive.
Yes Boltzmann is for distinguishable particles and Bose-Einstein and Fermi-Dirac are for indistinguishable particles.Philip Koeck said:On the other hand the mentioned derivations give Boltzmann only for distinguishable and Bose-Einstein only for indistinguishable particles.
That would definitely solve my problem. So some textbooks simply got it wrong and the factor 1/N! does not account for industinguishability but only for identity.NFuller said:We know that the ##1/N!## is needed for an ensemble of identical and distinguishable (a better word might be classical) particles to make the entropy extensive. The factorial is not needed if the particles are not identical. Such is the case if two different gasses are allowed to mix. The entropy should increase when two subsystems containing different gasses are combined since mixing the gasses is an irreversible process.
As far as agreeing with what your textbooks say, I think the confusion is again coming from differences in opinion of how to use the words identical and indistinguishable. That topic is for another thread though.
Again, its not that classical particles of a pure gas are indistinguishable but that they are identical. This means that permuting their positions in phase space on a constant energy manifold leads to the same microstate.
Yes Boltzmann is for distinguishable particles and Bose-Einstein and Fermi-Dirac are for indistinguishable particles.
NFuller said:Yes Boltzmann is for distinguishable particles and Bose-Einstein and Fermi-Dirac are for indistinguishable particles.
If you have a system of identical or even slightly different, distinguishable particles, what distribution would you use to describe it?Lord Jestocost said:The Boltzmann statistics is the limiting form of the Bose-Einstein and Fermi-Dirac statistics for a system of identical, indistinguishable “particles” in case the number of energy states available to anyone “particle” is very much larger than the number of “particles” in the system.
Philip Koeck said:That would definitely solve my problem. So some textbooks simply got it wrong and the factor 1/N! does not account for industinguishability but only for identity.
Philip Koeck said:If you have a system of identical or even slightly different, distinguishable particles, what distribution would you use to describe it?
I didn't want to consider a system with a maximum of one particle per state. Let's make it very specific: What would be the most appropriate distribution function for a gas of Argon atoms at room temperature with a density low enough to use the ideal gas law?Lord Jestocost said:There are either identical or non-identical “particles” (slightly different is no criterion for statistical considerations). To find the statistical distribution describing the system, one starts to set up the appropriate partition function for the system in question.
Identical, distinguishable particles:
That can be the case when modeling the solid state. By assuming, for example, that the identical atoms are confined to lattice sites and that each site is occupied at most once, one “artificially” introduces distinguishability as the positions in the lattice can be considered as distinguishable labels.
Philip Koeck said:What would be the most appropriate distribution function for a gas of Argon atoms at room temperature with a density low enough to use the ideal gas law?
Philip Koeck said:If you have a system of identical or even slightly different, distinguishable particles, what distribution would you use to describe it?
I think I can make a summary that sort of works for me. Boltzmann is for distinguishable particles, and atoms in ideal gas are distinguishable since they are sufficiently far apart (far more states than particles). At very high density particles become indistinguishable and Bose-Einstein applies (maybe easiest for Helium). This means of course that the factor 1/N! in the N-particle partition function for an ideal gas has nothing to do with indistinguishability. Some testbooks simply get it wrong. In other words indistinguishability is not required to resolve Gibbs' paradox.Lord Jestocost said:The Boltzmann statistics is the limiting form of the Bose-Einstein and Fermi-Dirac statistics for a system of identical, indistinguishable “particles” in case the number of energy states available to anyone “particle” is very much larger than the number of “particles” in the system.
Philip Koeck said:...and atoms in ideal gas are distinguishable since they are sufficiently far apart
German is fine. Thanks for the suggestions.Lord Jestocost said:Identical atoms in an ideal gas are indistinguishable and everything can be derived by using quantum statistics. The classical approach (distinguishability) works under certain circumstances as a good approximation to derive the distribution functions, but is in principle “wrong.” There exists nothing like a gas of identical classical “particles” where the “particles” become distinguishable at low densities and indistinguishable at high densities.
Regarding the “Gibbs’ paradox”: This is nothing else but a superficial paradox.
To recommend textbooks in this context:
“Thermodynamik und Statistik” by Arnold Sommerfeld (maybe, there exists an English translation)
“Theorie der Wärme” by Richard Becker (maybe, there exists an English translation)
“An Introduction to Statistical Thermodynamics” by Terrell L. Hill
Philip Koeck said:I'll just repeat my problem (same as original question):
You state that atoms in an ideal gas are indistinguishable and obey Boltzmann.
A combinatorial derivation gives Boltzmann only for distinguishable and Bose-Einstein only for indistinguishable particles.
How is this possible?
Thanks, I'll have a look at that.Lord Jestocost said:You can find the lengthy math in Chapter 3 "General Relations for Independent Distinguishable and Indistinguishable Molecules or Subsystems" of Terrell L. Hill's textbook “An Introduction to Statistical Thermodynamics”.
What I've read in Hill so far seems to confirm the problem I see very nicely.Lord Jestocost said:You can find the lengthy math in Chapter 3 "General Relations for Independent Distinguishable and Indistinguishable Molecules or Subsystems" of Terrell L. Hill's textbook “An Introduction to Statistical Thermodynamics”.
Philip Koeck said:What I've read in Hill so far seems to confirm the problem I see very nicely.
I like the part where they write "... much argued over...".Lord Jestocost said:To my mind, there is no problem. According to Walter Guido Vincenti and Charles H. Krüger (see page 105 in “Introduction to Physical Gas Dynamics”, John Wiley and Sons, Inc., New York (1965)):
“The foregoing limiting results [for the BE and FD statistics, ed. LJ] can also be obtained directly on the basis of the so-called Boltzmann statistics. This approach, which was appropriate to the classical methods used before quantum statistical mechanics was known, assumes in counting up the microstates that the particles are truly distinguishable. This leads without difficulty to equation (6.1), but an approximate and much argued over correction to account for the actual indistinguishability is needed to arrive at equation (6.2). Because of this difficulty we have chosen to avoid the Boltzmann statistics here and obtain our results solely as the limit of the correct quantum statistics. For convenience we shall refer to these results as corresponding to the Boltzmann limit.”
Lord Jestocost said:“The foregoing limiting results [for the BE and FD statistics, ed. LJ] can also be obtained directly on the basis of the so-called Boltzmann statistics. This approach, which was appropriate to the classical methods used before quantum statistical mechanics was known, assumes in counting up the microstates that the particles are truly distinguishable. This leads without difficulty to equation (6.1), but an approximate and much argued over correction to account for the actual indistinguishability is needed to arrive at equation (6.2). Because of this difficulty we have chosen to avoid the Boltzmann statistics here and obtain our results solely as the limit of the correct quantum statistics. For convenience we shall refer to these results as corresponding to the Boltzmann limit.”
Remember, the ##1/N!## is not strictly a product of quantum mechanics. It is there because a permutation of the particles on a constant energy manifold in phase space does not alter the microstate. I think the text fumbles this point a bit and might be confusing you. Although all classical systems are in some way an approximation of a more fundamental quantum system, classical thermodynamics is still self consistent. Even if a system was truly classical, the ##1/N!## is still needed to give the correct counting.Philip Koeck said:The Boltzmann distribution is derived for distinguishable particles, the correction (1/N!, I assume) accounts for their "actual indistinguishability".
Philip Koeck said:I'm pretty sure I'm not. The argument in many/some textbooks is as follows: Distinguishable particles obeying Boltzmann give an entropy expression that is non-extensive and therefore wrong. If one includes a factor 1/N! to account for indistinguishability one arrives at the Sackur-Tetrode formula for entropy, which is extensive. Therefore particles obeying Boltzmann (such as atoms and molecules in a gas) have to be indistinguishable.
If this was only true in limiting cases it wouldn't be very helpful since the Gibbs paradox wouldn't be resolved in the general case.
Philip Koeck said:I'm pretty sure I'm not. The argument in many/some textbooks is as follows: Distinguishable particles obeying Boltzmann give an entropy expression that is non-extensive and therefore wrong. If one includes a factor 1/N! to account for indistinguishability one arrives at the Sackur-Tetrode formula for entropy, which is extensive. Therefore particles obeying Boltzmann (such as atoms and molecules in a gas) have to be indistinguishable.
If this was only true in limiting cases it wouldn't be very helpful since the Gibbs paradox wouldn't be resolved in the general case.
Strictly speaking this is true. The actual number of states valid for even small numbers of identical particles isstevendaryl said:So the heuristic of dividing by N! doesn't give the exact right answer.
Philip Koeck said:You can find my version of the mentioned derivations here:
https://www.researchgate.net/publication/322640913_A_microcanonical_derivation_gives_the_Boltzmann_for_distinguishable_and_the_Bose-Einstein_distribution_for_indistinguishable_particles
I will derive the most probable distribution of N particles among k energy levels from combinatorics and some classical thermodynamics.
NFuller said:A microstate is a unique distribution of particles in phase space. Swapping the position and momentum of two identical particles will give the same configuration in phase space and the same microstate. If we didn't get the same microstate, that would imply that some microstates with many possible permutations are much more likely than others. The problem is that such a system cannot be at equilibrium. At equilibrium, the system must be in a maximum entropy configuration which occurs when each microstate comprising the equilibrium macrostate is equally likely.
I agree with your rant that the definition of microstate, or whatever-state, is a bit sloppy. A microstate is generally interpreted to mean a specific configuration of the sub-units of the system. How to deal with those microstates, and what the physical meaning of those states is, depends on the type of ensemble used. In the micro-canonical ensemble, each microstate has an equal probability of being selected at equilibrium. In the canonical ensemble, the probability to select a microstate depends on the energy of the state and temperature of the system.Stephen Tashi said:The definition of microstate can be justified by "the voice of authority" or tradition. I assume it can also be justified by arguments about practicality along the lines of "It wouldn't make physical sense to define a microstate so it depended on which particular particles are in an energy level because ..." It isn't clear (to me) which type of justification is being used for item 1)
The justification used is the second law of thermodynamics which states that when the system reaches equilibrium, that is a maximum entropy state. Since we want statistical mechanics to reproduce classical thermodynamics, we impose the requirement that the equilibrium probability distributions maximize the entropy.Stephen Tashi said:It also isn't clear what justification is implied for item 2). It could be justified soley by empirical tests- or It might be justified by a Bayesian form of reasoning. If we take the Bayesian approach we face the usual criticism: You defined a particlar type of outcome (i.e. microstate) and as assumed a maximum entropy distribution for it, but why didn't you define a different type of outcome and assume a maximum entropy distribution for that type of outcome?
Philip Koeck said:Many textbooks claim that particles that obey Boltzmann statistics have to be indistinguishable in order to ensure an extensive expression for entropy. However, a first principle derivation using combinatorics gives the Boltzmann only for distinguishable and the Bose Einstein distribution for indistinguishable particles (see Beiser, Atkins or my own text on Research Gate). Is there any direct evidence that indistinguishable particles can obey Boltzmann statistics?
Do you have a pdf or a link for the approximation you could send easily?NFuller said:Strictly speaking this is true. The actual number of states valid for even small numbers of identical particles is
$$W=\frac{(N+g-1)!}{N!(g-1)!}$$
where ##g## is the number of states. For the case given above this means
$$W=\frac{(2+2-1)!}{2!(2-1)!}=3$$
In the high temperature limit where ##g>>1## and the low density limit where ##g>>N##, the expression above can be simplified to give the standard Boltzmann counting
$$W=\frac{g^{N}}{N!}$$
This wiki page actually goes through much of the derivation. https://en.wikipedia.org/wiki/Maxwell–Boltzmann_statisticsPhilip Koeck said:Do you have a pdf or a link for the approximation you could send easily?
What is the physical interpretation of "being selected"? Are we talking about picking a random time and taking the microstate of the system at that time to be the one that is selected?NFuller said:Ieach microstate has an equal probability of being selected at equilibrium.
The justification used is the second law of thermodynamics which states that when the system reaches equilibrium, that is a maximum entropy state. Since we want statistical mechanics to reproduce classical thermodynamics, we impose the requirement that the equilibrium probability distributions maximize the entropy.
NFuller said:This wiki page actually goes through much of the derivation. https://en.wikipedia.org/wiki/Maxwell–Boltzmann_statistics
Thus when we count the number of possible states of the system, we must count each and every microstate, and not just the possible sets of occupation numbers.
You seem to be saying that swapping two particles in different states does not lead to a different microsctate even if it's obvious that the particles have been swapped. My understanding was that swapping distinguishable particles in different states leads to a new micrstate even if the particles are identical.
You say in post #24
A microstate is a unique distribution of particles in phase space. Swapping the position and momentum of two identical particles will give the same configuration in phase space and the same microstate.
I've tried to fill in the gaps of this derivation (see appended pdf) and I don't get the same result. Am I making a mistake?NFuller said:This wiki page actually goes through much of the derivation. https://en.wikipedia.org/wiki/Maxwell–Boltzmann_statistics
YesStephen Tashi said:What is the physical interpretation of "being selected"? Are we talking about picking a random time and taking the microstate of the system at that time to be the one that is selected?
I'm sorry but I don't think I understand what you are asking. Can you rephrase this?Stephen Tashi said:I can understand that as a purely empirical claim. What I don't understand is an attempt to justify the definition of microstate by deductive logic - if that's what's being attempted.
If we are discussing Shannon entropy then when a system is at equilibrium some probability distributions of its properties may be maximum entropy distributions and others may not. How do we explain why assuming a maximum entropy for microstates (as defined by occupancy numbers) is a good idea - as opposed to assuming a maximum entropy distribution for selecting the type of state whose description includes which particular particles occupy various energy levels.
I think the closest I can give to a "consensus definition" is the one given in Kardar's statistical physics book. He saysStephen Tashi said:What is the consensus definition of a "microstate"?
At any time ##t##, the microstate of a system of ##N## particles is described by specifying the positions ##\vec{q}(t)## and momenta ##\vec{p}(t)## of all of the particles. The microstate thus corresponds to a point ##\mu(t)##, in the ##6N##-dimensional phase space ##\Gamma=\Pi_{i=1}^{N}\{\vec{q}_{i},\vec{p}_{i}\}##