# The problematic 1/N in canonical partition function

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• ShayanJ
In summary: It could be that quantum mechanics is necessary to fully understand and describe these types of systems. In summary, there is a paradox in the classical statistical mechanics calculation of the canonical partition function for a system of particles with no interaction and only external potentials. The factor introduced to solve the Gibbs paradox seems to be causing a paradox itself. When considering particles with different parameters, the limit of these parameters being equal results in a different partition function than expected. This may be due to the difficulty of classical statistical mechanics to account for indistinguishability, and quantum mechanics may be necessary to fully understand and describe these systems.
ShayanJ
Gold Member
The canonical partition function in classical statistical mechanics is calculated by ## Q_N(V,T)=\frac 1 {N! h^{3N}}\int e^{-\beta H(\mathbf q,\mathbf p)}d^{3N}q \ d^{3N}p ##. The ## \frac 1 {N!} ## is there to prevent the Gibbs paradox. But now consider a system of N particles that have no interaction with each other and only interact with external potentials. This means that their canonical partition function can be written as ## Q_N(V,T)=\frac{[Q_1(V,T)]^N}{N! h^{3N}} ## where ## Q_1(V,T)=\int e^{-\beta H(q,p)}d^3 q \ d^3p ##. So for ## N=N_1+N_2 ## particles of the same type, I'll have ## Q_{N_1+N_2}(V,T)=\frac{[Q_1(V,T)]^{N_1+N_2}}{(N_1+N_2)! h^{3(N_1+N_2)}} ##.
Now suppose there are two types of particles differing in a parameter which appears in their Hamiltonians. So for a system consisting of ## N_1 ## particles of the first kind and ## N_2 ## particles of the second kind, the partition function is ## Q^{(1)}_{N_1}(V,T)Q^{(2)}_{N_2}(V,T)=\frac{[Q^{(1)}_1(V,T)]^{N_1}[Q^{(2)}_1(V,T)]^{N_2}}{N_1!N_2! h^{3(N_1+N_2)}} ##. Now if I take the limit that the parameter is the same for the two kinds of particles, I expect to get ##Q_{N_1+N_2}(V,T)##, but instead, I get ##Q^{(1)}_{N_1}(V,T)Q^{(2\to 1)}_{N_2}(V,T)=\frac{[Q^{(1)}_1(V,T)]^{N_1}[Q^{(2\to 1)}_1(V,T)]^{N_2}}{N_1!N_2! h^{3(N_1+N_2)}} ## which is different from what I expect because instead of ## (N_1+N_2)! ## in the denominator, it has ## N_1!N_2! ##. It seems that the factor introduced to solve the Gibbs paradox, is itself causing a paradox. What is wrong here? What is it that I don't see?
Thanks

ShayanJ said:
Now if I take the limit that the parameter is the same for the two kinds of particles,
I'm pretty sure this is your problem right here. The particles are only indistinguishable when the parameters are identical. You can't continuously tune into or out of indistinguishability. So if the first kind of particle is indexed by a parameter ##a## like so: ##Q^{(a)}##, and the second kind of particle is indexed like so: ##Q^{(b)}##, you can't continuously walk ##b## into ##a##. In other words, the condition for indistinguishability (where the additivity condition ##(N_1+N_2)!## would apply) is ##\delta_{ab}##.

ShayanJ
Interesting point. But now consider a system whose behavior can be described using a model with free electrons and some free quasi-particles with electric charge equal to -e and effective mass that can be tuned using e.g. an applied electric field. You're saying that if I tune the mass of the quasi-particles to be exactly equal to the mass of the electrons, these quasi-particles can still be distinguished from the electrons? Is there such a model?

ShayanJ said:
Interesting point. But now consider a system whose behavior can be described using a model with free electrons and some free quasi-particles with electric charge equal to -e and effective mass that can be tuned using e.g. an applied electric field. You're saying that if I tune the mass of the quasi-particles to be exactly equal to the mass of the electrons, these quasi-particles can still be distinguished from the electrons? Is there such a model?
In this hypothetical scenario, if the only observables were mass and charge, and a quasiparticle had the same mass and charge as an electron, then I don't think they'd be distinguishable.

TeethWhitener said:
In this hypothetical scenario, if the only observables were mass and charge, and a quasiparticle had the same mass and charge as an electron, then I don't think they'd be distinguishable.
So just imagine my first post is about such a model and I'm tuning the mass of the quasi-particles to be equal to the mass of the electrons and there is no other difference between them. So now you also seem to agree that I should get ## Q_{N_1+N_2}(V,T) ## in the limit. And we're back to the problem!

ShayanJ said:
So just imagine my first post is about such a model and I'm tuning the mass of the quasi-particles to be equal to the mass of the electrons and there is no other difference between them. So now you also seem to agree that I should get ## Q_{N_1+N_2}(V,T) ## in the limit. And we're back to the problem!
Yeah that's a pickle. I'll have to think a little harder about it. Maybe they are distinguishable. If you're able to tune the mass, then wouldn't that qualify as a measurement?
Edit: I'm not sure how much of this is applicable to classical statistical mechanics.

TeethWhitener said:
If you're able to tune the mass, then wouldn't that qualify as a measurement?
Well, people do such things in labs all the time and don't have to deal with measurements. So I guess its not. But I also somehow understand why its not a measurement.
TeethWhitener said:
I'm not sure how much of this is applicable to classical statistical mechanics.
Good point. Maybe classical statistical mechanics is simply not able to take into account indistinguishability satisfactorily.

## 1. What is the problematic 1/N in the canonical partition function?

The 1/N term in the canonical partition function arises from the quantum mechanical nature of particles and represents the degeneracy of states. This term becomes problematic when dealing with systems with a large number of particles, as it can lead to diverging values for the partition function.

## 2. How does the problematic 1/N affect calculations in statistical mechanics?

The problematic 1/N can lead to inaccurate results in statistical mechanics calculations, particularly in systems with a large number of particles. It can also make it difficult to apply statistical mechanics to real-world systems, as the assumptions made for simplification may not hold for large systems.

## 3. Can the problematic 1/N be avoided in statistical mechanics calculations?

While it is not possible to completely avoid the 1/N term in the canonical partition function, there are methods and approximations that can be used to mitigate its effects. This includes using different ensembles, such as the grand canonical ensemble, or applying perturbation theory to the partition function.

## 4. How does the problematic 1/N relate to the thermodynamic limit?

The thermodynamic limit is the concept of taking the limit of an infinitely large system. In this limit, the 1/N term becomes negligible and does not affect the thermodynamic properties of the system. However, for systems with a finite number of particles, the 1/N term can significantly impact the results.

## 5. Are there any real-world applications for the problematic 1/N in statistical mechanics?

While the problematic 1/N can make it challenging to apply statistical mechanics to real-world systems, it also has important applications in understanding and predicting the behavior of materials at the nanoscale. This is because in this scale, the number of particles is relatively small, and the 1/N term becomes significant in determining the properties of the system.

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