Example

Consider a system of uncorrelated particles distributed among $m$ boxes. Assume that the average number of particles per box $\overline{n_j} << 1$, so that there is either one or none particle per box --i.e., $n_j=1,0$ with $n_j$ the number of particles associated with box $j$. The goal of this section is to compute the size of the fluctuations in the total number of particles in the system and to show that the size of such fluctuations is much smaller than the average number of particles $\overline{N}$, when $\overline{N}$ is sufficiently large. The ensemble average square fluctuation in the number of particles

$$\overline{(\delta N)^2} = \overline{(N - \overline{N})^2} = \overline{N^2} - \overline{N}^2,$$ (90)

can be computed as follows

$$\overline{(\delta N)^2} = \sum_{j=1}^m \sum_{k=1}^m <n_j n_k> -(\sum_{j=1}^m <n_j>)^2.$$ (91)

Factorizing the sums we obtain,

$$\overline{(\delta N)^2} = \sum_j <n_j^2> + \sum_{j=1}^m \sum_{k\n... ...^m <n_j n_k> - \sum_{j=1}^m <n_j>^2 -\sum_{j=1}^m \sum_{k \neq j}^m <n_j><n_k>.$$ (92)

Note that since $n_j=0, 1$, $<n_j^2>=<n_j>$. Furthermore, $<n_j n_i>=<n_j><n_i>$ when $j \neq i$ simply because the particles are uncorrelated. Therefore,

$$\overline{(\delta N)^2} = \sum_{j=1}^m <n_j> - \sum_{j=1}^m <n_j>^2 =\sum_{j=1}^m <n_j>(1-<n_j>).$$ (93)

Considering that $<n_j> << 1$,

$$\overline{(\delta N)^2} = m <n_1> = \overline{N}.$$ (94)

Therefore,

$$\frac{\sqrt{\overline{(\delta N)^2}}}{\overline{N}} = \frac{1}{\sqrt{\overline{N}}},$$ (95)

i.e., the size of the fluctuations in the number of particles in the system becomes negligible small when $\overline{N} \sim$   10$^{23}$. As a by-product of this derivation we can obtain the thermodynamic equation that establishes the relationship between the number of particles, the temperature and the chemical potential. Substituting Eq. (89) into Eq. (94) we obtain,

$$\overline{N} = \frac{\partial^2 \text{ln} \Xi}{\partial (\beta \mu)^2} \Bigg )_V = \frac{\partial \overline{N}}{\partial (\beta \mu)}.$$ (96)

Therefore,

$$\frac{\partial \text{ln} \overline{N}}{\partial (\beta \mu)} \Bigg )_V = 1,$$ (97)

or dividing both sides of Eq. (96) by V,

$$\left( \frac{N}{V} \right) = \beta \mu + c,$$ (98)

where $c$ is a constant.