Chemical potential and fixed number of particles

[cryptist]

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?

 

[DrClaude]

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.

 

[jfizzix]

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.