Chemical potential and fixed number of particles

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The discussion addresses the chemical potential of a system with a fixed number of particles, questioning its physical significance. It is clarified that, similar to temperature at fixed entropy or pressure at fixed volume, one can indeed define the chemical potential in this context. The ideal gas is used as an example, where the occupancy is derived from classical Maxwell-Boltzmann statistics. The relationship between the total particle number and the partition function allows for the calculation of the chemical potential, leading to the expression μ = k_B T ln(η/η_q). This analysis demonstrates that discussing chemical potential with a fixed number of particles is both meaningful and mathematically valid.
cryptist
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Can we talk about the chemical potential of a system with fixed number of particles? Is this physically meaningful? Why/why not?

P.s: I know that chemical potential is the partial derivative of free energy with respect to number of particles. But in the formulation of grand canonical ensemble, we write N=Ʃf (for example f would be fermi-dirac distribution function) and can't we fix N in this formula, and solve chemical potential μ by changing energy ε?
 
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cryptist said:
Can we talk about the chemical potential of a system with fixed number of particles? Is this physically meaningful? Why/why not?

Yes, for the same reason you can talk about the temperature of a system with a fixed entropy or the pressure of a system with a fixed volume.
 
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You can indeed solve for the chemical potential \mu(N,V,T) in the way you suggest.

Consider the ideal gas as an example. Here, the occupancy comes from the classical maxwell-Boltzmann statistics f_{MB}(\epsilon) = e^{-(\frac{\epsilon-\mu}{k_{B}T})}.

The number of particles N can be expressed as

N=\sum_{j}g_{j}f_{MB}(\epsilon_{j})
where g_{j} is the number of states at energy level \epsilon_{j}

The partition function for a single particle Z_{1} is

Z_{1} = \sum_{j}g_{j} e^{-(\frac{\epsilon_{j}}{k_{B}T})}=\eta_{q}V : \eta_{q}=(\frac{m k_{B}T}{2 \pi \hbar^{2}})^{\frac{3}{2}}

Here \eta_{q} is a characteristic quantum concentration (you can see it has dimensions of \frac{N}{V}). It's a large concentration indicating when quantum effects (bose/Fermi statistics) must be taken into account. For ideal gases, the real concentration \eta=\frac{N}{V} is much less than \eta_{q}.

We can relate the total particle number N to the partition function Z_{1}, giving us

N=Z_{1}e^{-\frac{\mu}{k_{B}T}}

We can solve for \mu to find

\mu =k_{B}T \ln (\frac{N}{Z_{1}})

Then, substituting out expression for Z_{1}, we arrive at the final result

\mu =k_{B}T \ln (\frac{\eta}{\eta_{q}}) :\eta=\frac{N}{V}.

It's a total non-sequitur, but it seemed worth explaining since I had already written lecture notes on the subject.
 
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