# 3.3: Deriving the Boltzmann Equation I

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

In Sections 20-10 and 20-14, we develop the relationship between the system entropy and the probabilities of a microstate, $$\rho \left({\epsilon }_i\right)$$, and an energy level, $$P_i=g_i\rho \left({\epsilon }_i\right)$$, in our microscopic model. We find

\begin{align*} S &=-Nk\sum^{\infty }_{i=1}{P_i}{ \ln \rho \left({\epsilon }_i\right)\ } \\[4pt] &=-Nk\sum^{\infty }_{i=1}{g_i\rho \left({\epsilon }_i\right)}{ \ln \rho \left({\epsilon }_i\right)\ } \end{align*}

For an isolated system at equilibrium, the entropy must be a maximum, and hence

$-\sum^{\infty }_{i=1}{g_i\rho \left({\epsilon }_i\right)}{ \ln \rho \left({\epsilon }_i\right)} \label{maxentropy}$

must be a maximum. We can use Lagrange’s method to find the dependence of the quantum-state probability on its energy. The $$\rho \left({\epsilon }_i\right)$$ must be such as to maximize entropy (Equation \ref{maxentropy}) subject to the constraints

$1=\sum^{\infty }_{i=1}{P_i}=\sum^{\infty }_{i=1}{g_i\rho \left({\epsilon }_i\right)}$

and

$\left\langle \epsilon \right\rangle =\sum^{\infty }_{i=1}{P_i{\epsilon }_i}=\sum^{\infty }_{i=1}{g_i{\varepsilon }_i\rho \left({\epsilon }_i\right)}$

where $$\left\langle \epsilon \right\rangle$$ is the expected value of the energy of one molecule. The mnemonic function becomes

$F_{mn}=-\sum^{\infty }_{i=1}{g_i\rho \left({\epsilon }_i\right)}{ \ln \rho \left({\epsilon }_i\right)\ }+{\alpha }^*\left(1-\sum^{\infty }_{i=1}{g_i\rho \left({\epsilon }_i\right)}\right)+\beta \left(\left\langle \epsilon \right\rangle -\sum^{\infty }_{i=1}{g_i{\varepsilon }_i\rho \left({\epsilon }_i\right)}\right)$

Equating the partial derivative with respect to $$\rho \left({\epsilon }_i\right)$$ to zero, $\frac{\partial F_{mn}}{\partial \rho \left({\epsilon }_i\right)}=-g_i{ \ln \rho \left({\epsilon }_i\right)\ }-g_i-{\alpha }^*g_i-\beta g_i{\epsilon }_i=0$

so that

$\rho \left({\epsilon }_i\right)={\mathrm{exp} \left(-{\alpha }^*-1\right)\ }{\mathrm{exp} \left(-\beta {\epsilon }_i\right)\ }$

From

$1=\sum^{\infty }_{i=1}{P_i}=\sum^{\infty }_{i=1}{g_i\rho \left({\epsilon }_i\right)}$

the argument we use in Section 21.1 again leads to the partition function, $$z$$, and the Boltzmann equation

$P_i=g_i\rho \left({\epsilon }_i\right)=z^{-1}g_i\ \mathrm{exp}\left(-\beta {\epsilon }_i\right)$

3.3: Deriving the Boltzmann Equation I is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Paul Ellgen via source content that was edited to conform to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.