7.5: Entropy Can Be Expressed in Terms of a Partition Function

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We have seen that the partition function of a system gives us the key to calculate thermodynamic functions like energy or pressure as a moment of the energy distribution. We can extend this formulism to calculate the entropy of a system once its \(Q\) is known. We can start with Boltzmann's (statistical) definition of entropy:

\[S = k \ln(W) \label{Boltz} \]

with

\[W=\frac{A!}{\prod_j{a_j}!} \nonumber \]

Combining these equations, we obtain:

\[S_{ensemble} = k \ln\frac{A!}{\prod_j{a_j}!} \nonumber \]

Rearranging:

\[S_{ensemble} = k \ln{A!}-k \sum_j{\ln{a_j!}} \nonumber \]

Using Sterling's approximation:

\[\begin{split}S_{ensemble} &= k A\ln{A}-k A - k \sum_j{a_j\ln{a_j}} + k \sum_j{a_j} \\ &= k A\ln{A}- k \sum_j{a_j\ln{a_j}}\end{split} \nonumber \]

Since:

\[\sum_j{a_j}=A \nonumber \]

The probability of finding the system in state \(a_j\) is:

\[p_j=\frac{a_j}{A} \nonumber \]

Rearranging:

\[a_j = p_jA \nonumber \]

Plugging in:

\[S_{ensemble}=k A\ln{A}- k \sum_j{p_jA\ln{p_jA}} \nonumber \]

Rearranging:

\[S_{ensemble}=k A\ln{A}- k \sum_j{p_jA\ln{p_j}}- k \sum_j{p_jA\ln{A}} \nonumber \]

If \(A\) is constant, then:

\[k \sum_j{p_jA\ln{A}} = k A\ln{A}\sum_j{p_j} \nonumber \]

Since:

\(\sum_j{p_j} = 1\)

We get:

\[S_{ensemble}=k A\ln{A}- k \sum_j{p_jA\ln{p_j}}- k A\ln{A} \nonumber \]

The first and last term cancel out:

\[S_{ensemble}=- k \sum_j{p_jA\ln{p_j}} \nonumber \]

We can divide by \(A\) to get the entropy of the system:

\[S_{system}=- k \sum_j{p_j\ln{p_j}} \label{eq10}\]

If all the \(p_j\) are zero except for the for one, then the system is perfectly ordered and the entropy of the system is zero. The probability of being in state \(j\) is

\[p_j=\frac{e^{-\beta E_j}}{Q} \label{eq15}\]

Plugging Equation \ref{eq15} into Equation \ref{eq10} results in

\[\begin{align*}S &= - k \sum_j{\frac{e^{-\beta E_j}}{Q}\ln{\frac{e^{-\beta E_j}}{Q}}} \\[4pt] &= - k \sum_j{\frac{e^{-\beta E_j}}{Q}\left(-\beta E_j- \ln{Q}\right)} \\[4pt] &= - \beta k \sum_j{\frac{E_je^{-\beta E_j}}{Q}}+\frac{k\ln{Q}}{Q}\sum_j{e^{-\beta E_j}} \end{align*}\]

Making use of:

\[\beta k=\frac{1}{T} \nonumber \]

And:

\[\sum{\frac{e^{-\beta E_j}}{Q}}=\sum{p_j}=1 \nonumber \]

We obtain:

\[S= \dfrac{U}{T} + k\ln Q \label{20.42} \]

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