The Gibbs distribution

Let us consider a system immersed in a heat bath that can exchange particles with the bath, which therefore acts also as a particle reservoir. We saw in Sec.  that the flow of particles is governed by the chemical potential, just as the flow of energy is determined by the temperature in an isolated system. Here we want to work out the conditions that establish equilibrium w.r.t the exchange of particles between the system and the reservoir.

Figure: A system exchanging energy and particles with a heat bath and particle reservoir. The Volume $V$ of thee system is kept constant.

Let the total number of particles be $N_0$, $N$ the number of particles of the system and $N_0 - N$ the number of particles in the reservoir. In analogy with the assumption that the heat bath is so large than exchange of energy with the system does no affect its temperature, we will make the assumption that the particle reservoir is so large that the chemical potential is unaffected by the exchange of particles with the system. Recall that the chemical potential measures the change of energy as one particle is added to the system, which in general may depend on how many particles the system already has. Indeed, in the previous section we have seen that two systems in contact with different chemical potentials will exchange particles until the two chemical potentials are equal. The above assumption then means that $N_0$ is so large compared to $N$ that a small change of $N_0$ causes a negligible change to the value of its chemical potential: the particle reservoir sets the chemical potential. Moreover, using relation , which establishes that in the absence of internal constraints two systems that can exchange particles have the same chemical potential, the chemical potential of the system is equal to that of the particle reservoir.

We now generalise Eq. , by including terms in the expansion that depend on the number of particles:

$$S_2(E_0 - E_{N,r}, N_0 - N) = \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$$

$$S_2(E_0,N_0) - E_{N,r} \left ( \frac{\partial S_2}{\partial E}\right )_{N,V} - N \left ( \frac{\partial S_2}{\partial N}\right )_{E,V} =$$

$$S_2(E_0,N_0) - \frac{E_{N,r}}{T} + N \frac{\mu}{T}. \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$$ (3.46)

We have not explicitly written the quadratic terms because they involve derivatives of the temperature and of the chemical potential, which because of our assumption of size of the heat bath and particle reservoir are constant. Note that now the energy has two subscripts, $N$ and $r$. This is because the ensemble of energy levels available to the system depends on the number of particles in it. The probability that the system is in one particular microstate $r$ with energy $E_{N,r}$ and number of particles $N$ is therefore a generalisation of :

$$p_{N,r} = \frac{1}{ \mathcal{Z} }\exp \{\beta (N\mu- E_{N,r})\},$$ (3.47)

which is known as the Gibbs distribution for the gran canonical ensemble, which is characterised by constant $V,T,\mu$. The gran partition function $\mathcal{Z}$ is the normalisation constant of $p_{N,r}$:

$$\mathcal{Z}(T,V,\mu) = \sum_N \sum_r \exp \{\beta (N\mu- E_{N,r})\},$$ (3.48)

which is a function of $V,T$ and $\mu$. Here $V$ and $N$ determine the energy levels $E_{N,r}$, and $T$ and $\mu$ the distribution of the energy and the number of particles between the system and the reservoir. Eq.  can also be rearranged as:

$$\mathcal{Z}(T,V,\mu) = \sum_N \exp \{\beta N\mu\} \sum_r \exp \{-\beta E_{N,r}\} = \sum_N \exp \{\beta N\mu\} Z(N,V,T),$$ (3.49)

with $Z(N,V,T)$ the partition function of the system with $N$ particles. We see that the grand partition function is a weighted average of the canonical partition functions over all possible number of particles.

As for the Boltzmann probability, if the energy levels $E_{N,r}$ have degeneracy $\Omega(E_{N,r})$, and $S(E_{N,r}) = k_{\rm B}\ln \Omega(E_{N,r})$, the probability for the system to have energy $E_{N,r}$ is:

$$p(N,E_{N,r}) = \frac{1}{ \mathcal{Z} }\exp \{\beta [N\mu- E_{N,r} + TS(E_{N,r})]\}.$$ (3.50)

The probability for the system to have $N$ particles, regardless of its energy, can be obtained by summing  over all microstates for each value of $N$:

$$p(N) = \sum_r \frac{1}{ \mathcal{Z} }\exp \{\beta [N\mu- E_{N,r}]\} = \frac{1}{ \mathcal{Z} } \exp \{\beta N\mu\} \sum_r \exp \{-\beta E_{N,r}\} =$$

$$\frac{1}{ \mathcal{Z} } \exp \{\beta N\mu\} Z(N). \quad \quad \qu... ...d \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad$$ (3.51)