Question

For a system obeying Boltzmann statistics, we know what $\mu$is from Chapter 6. Suppose, though, that you knew the distribution function (equation $7.31$) but didn't know $\mu$. You could still determine $\mu$ by requiring that the total number of particles, summed over all single-particle states, equal N. Carry out this calculation, to rederive the formula $\mu =-kT\ln \left(Z_1/N\right)$. (This is normally how $\mu$ is determined in quantum statistics, although the math is usually more difficult.)

Short answer

The formula$\mu =-kT{\log }_e\left(\frac{Z_1}{N}\right)$is derived.

Step-by-step solution

Step 1. Given Information 

We are given a formula,

$\mu =-kT\ln \left(Z_1/N\right)$

Step 2. Deriving the formula

According to Boltzmann statistics, the average number of particles in the single state is determined by Boltzmann distribution function,

${\overline{n}}_{\text{Boltzmann}}=e^{\frac{-(\epsilon -\mu )}{kT}}$

N is the total number of particles, summed over all single-particle states,

$$\begin{gathered} N=\sum _s\exp \left[\frac{-\left(\epsilon \right(s)-\mu )}{kT}\right] \\ =\sum _s\exp \left(\frac{\mu }{kT}\right)\exp \left(\frac{-\epsilon \left(s\right)}{kT}\right) \\ =\exp \left(\frac{\mu }{kT}\right)\sum _s\exp \left(\frac{-\epsilon \left(s\right)}{kT}\right) \\ =\exp \left(\frac{\mu }{kT}\right)Z_1 \end{gathered}$$

As $Z_1=\sum \exp \left(\frac{-\epsilon \left(s\right)}{kT}\right)$,

Step 3. Simplifying

Simplifying, we get

$$\begin{gathered} \frac{N}{Z_1}=\frac{\mu }{kT} \\ \log \left(\frac{N}{Z_1}\right)=\frac{\mu }{kT} \\ -{\log }_e\left(\frac{Z_1}{N}\right)=\frac{\mu }{kT} \\ \mu =-kT{\log }_e\left(\frac{Z_1}{N}\right) \end{gathered}$$

Hence, the formula is derived.