18.3: We Postulate That the Average Ensemble Energy Is Equal to the Observed Energy of a System

We will be restricting ourselves to the canonical ensemble (constant temperature and constant pressure). Consider a collection of \(N\) molecules. The probability of finding a molecule with energy \(E_i\) is equal to the fraction of the molecules with energy \(E_i\). That is, in a collection of \(N\) molecules, the probability of the molecules having energy \(E_i\):

\[P_i = \dfrac{n_i}{N} \nonumber \]

This is the directly obtained from the Boltzmann distribution, where the fraction of molecules \(n_i /N\) having energy \(E_i\) is:

\[P_i = \dfrac{n_i}{N} = \dfrac{e^{-E_i/kT}}{Q} \label{BD1} \]

The average energy is obtained by multiplying \(E_i\) with its probability and summing over all \(i\):

\[ \langle E \rangle = \sum_i E_i P_i \label{Mean1} \]

Equation \(\ref{Mean1}\) can be solved by plugging in the Boltzmann distribution (Equation \(\ref{BD1}\)):

\[ \langle E \rangle = \sum_i{ \dfrac{E_ie^{-E_i/ kT}}{Q}} \label{Eq1} \]

Where \(Q\) is the partition function:

\[ Q = \sum_i{e^{-\dfrac{E_i}{kT}}} \nonumber \]

We can take the derivative of \(\ln{Q}\) with respect to temperature, \(T\):

\[ \left(\dfrac{\partial \ln{Q}}{\partial T}\right) = \dfrac{1}{kT^2}\sum_i{\dfrac{E_i e^{-E_i/kT}}{Q}} \label{Eq2} \]

Comparing Equation \(\ref{Eq1}\) with \(\ref{Eq2}\), we obtain:

\[ \langle E \rangle = kT^2 \left(\dfrac{\partial \ln{Q}}{\partial T}\right) \nonumber \]

It is common to write these equations in terms of \(\beta\), where:

\[ \beta = \dfrac{1}{kT} \nonumber \]

The partition function becomes:

\[ Q = \sum_i{e^{-\beta E_i}} \nonumber \]

We can take the derivative of \(\ln{Q}\) with respect to \(\beta\):

\[ \left(\dfrac{\partial \ln{Q}}{\partial\beta}\right) = -\sum_i{\dfrac{E_i e^{-\beta E_i}}{Q}} \nonumber \]

And obtain:

\[ \langle E \rangle = -\left(\dfrac{\partial \ln{Q}}{\partial\beta}\right) \nonumber \]

Replacing \(1/kT\) with \(\beta\) often simplifies the math and is easier to use.