Many thanks to Bart Andrews for this contribution!

### Question

Consider two systems \(I\) and \(II\) in contact with a common heat bath with temperature T and suppose that a mechanism exists which allows both systems to exchange particles. The probability that the composed system \(I + II\) is in a state for which system \(I\) has an energy between \(E_I\) and \(E_I+dE_I\) and a particle number \(N_I\), while \(II\) has an energy between \(E_{II}\) and \(E_{II} + dE_{II}\) with a particle number \(N_{II}\) is given by

\begin{equation}

p(E_I, N_I; E_{II}, N_{II})dE_I dE_{II} = \frac{\Omega_I(E_I,N_I)\Omega_{II}(E_{II},N_{II}) e^{-\beta(E_I+E_{II})}}{\sum_{N_I=0}^{N} Z_I(T,N_I)Z_{II}(T,N-N_I)}\,\,\,\,\,\,dE_IdE_{II}

\end{equation}

Here, \(Z_I(T, N_I)\) and \(Z_{II}(T, N_{II})\) are the partition sums of the single systems, \(\Omega_I(E_I, N_I)\) and \(\Omega_{II}(E_{II},N_{II})\) are the degeneracies of the states in the single systems and \(N = N_I + N_{II}\) is the total number of particles.

Show that for the most probable distribution \(\{N_I, N_{II}\}\) of particles, the sum of the free energies \(F_I + F_{II}\) is minimal and that the chemical potentials of both systems are equal.

### Solution

This question considers the most probable configuration of free particles and then asks for two things:

- Show that the total free energy of the system is minimal
- Show that the chemical potentials are equal

We shall start by showing that the chemical potentials are equal. Let the most probable value of \(N_I\) be called \(\tilde{N_I}\). We know that since the total number of particles stays constant, the particle number of the second system can be expressed as a function of the particle number of the first system.

\[N_{II} = N - N_I\]

Therefore, if we find the most probable distribution for \(N_I\) particles, then this will consequently be the most probable distribution of \(\{N_I,N_{II}\}\). So, we can proceed by finding the most probable distribution as follows:

Maximise the probability with respect to \(N_I\).

\[{\partial p \over \partial N_I} \stackrel{!}{=} 0\]

Use the product rule, set equal to zero and simplify.

\[\bigg({\Omega'_I}(E_I,N_I){\Omega_{II}}(E_{II},N-N_I)-{\Omega_I}(E_I,N_I){\Omega'_{II}}(E_{II},N-N_I)\bigg) \bigg|_{\tilde{N_I}} = 0\]

\[\frac{\Omega'_I(E_I,\tilde{N_I})}{\Omega_I(E_I,\tilde{N_I})} = \frac{\Omega'_{II}(E_{II},N-\tilde{N_I})}{\Omega_{II}(E_{II},N-\tilde{N_I})}\]

Multiply both sides by the Boltzmann constant.

\[k_B\frac{\Omega'_I(E_I,\tilde{N_I})}{\Omega_I(E_I,\tilde{N_I})} = k_B\frac{\Omega'_{II}(E_{II},N-\tilde{N_I})}{\Omega_{II}(E_{II},N-\tilde{N_I})}\]

Take the Boltzmann constant inside the derivatives.

\[{\partial (k_B \ln \Omega_I) \over \partial N_I} \bigg|_{\tilde{N_I}} = {\partial (k_B \ln \Omega_{II}) \over \partial N_{II}}\bigg|_{N-\tilde{N_I}}\]

Use the Boltzmann expression for the entropy.

\[{\partial S_I \over \partial N_I} \bigg|_{\tilde{N_I}} = {\partial S_{II} \over \partial N_{II}}\bigg|_{N-\tilde{N_I}}\]

Use the thermodynamic relation for the chemical potential.

\[\frac{\mu_I}{T} = \frac{\mu_{II}}{T} \]

Since \(T\) is constant for both systems due to the heat bath, it cancels.

\[\mu_I = \mu_{II} \; \blacksquare\]

Now we can proceed to the second part and show that the total free energy in this situation is minimal. From the second law of thermodynamics, we know that:

\(

\newcommand{\dslash}{\delta}

\)

\[dS\ge \frac{\dslash Q}{T}\]

\[T dS \ge \dslash Q\]

\[0 \ge \dslash Q - T dS\]

Additionally, from the first law of thermodynamics we know that:

\[dU = \dslash Q - P dV\]

Now taking the Legendre transform of the internal energy to Helmholtz free energy, \(F\equiv U – TS\), we obtain:

\begin{eqnarray}

dF & = & dU-TdS-SdT \nonumber \\

& = & \dslash Q-PdV-TdS-SdT \nonumber \\

& = & (\dslash Q-TdS) -PdV-SdT \nonumber \\

& = & \dslash Q-TdS \nonumber

\end{eqnarray}

Here, we have used the fact that the total volume and temperature of the two systems is constant and so \(dV=dT=0\). Combining this result with the second law of thermodynamics leads to the inequality:

\[dF \le 0\]

This implies that for a process at constant temperature and volume the Helmholtz free energy seeks a minimum and this minimum is achevied when \(dF=0\). So now we can look at our particular problem and calculate the total differential \(dF\).

\[dF = \left( \partial F \over \partial T \right) dT + \left( \partial F \over \partial V \right) {dV} + \left( \partial F \over \partial N_I \right) dN_I + \left( \partial F \over \partial N_{II} \right) dN_{II}\]

It is assumed that all other neccesary variables are kept constant when these partial derivatives are taken. Now we know that the temperature is constant because of mutual contact with a heat bath and so \(dT=0\). We also know that the total volume is kept constant because these systems are confined and so \(dV=0\). From the relation…

\[dN_I = -dN_{II}\]

…coupled with the thermodynamic definitions for chemical potential…

\[\mu_I = \left( \partial F \over \partial N_I \right) \; , \; \mu_{II} = \left( \partial F \over \partial N_{II} \right) \]

…it is clear that since the chemical potentials are equal, the last two terms cancel. This means that the total free energy is, in fact, at a minimum as required.

\[dF=0 \; \blacksquare \]