[NOTES/SM-04015] Equilibrium of a System with a Heat Reservoir

For an isolated system the energy is constant and the equilibrium states are characterized by \(E,V,N\). The postulate of equal a priori probability then means that all micro states are equally probable.

A system \(\mathcal S\), when in contact with a heat bath \(\mathcal S_0\), keeps exchanging the energy continuously and and therefore \(E\) of the system is not a constant of motion. In this case postulate of equal a priori probability does not apply. and the different microstates are not equally probable. The probability \(p_r\) now depends on the energy \(E_r\) of the microstate and the temperature of the heat bath \(T\). Here we wish to find this probability using the Boltzmann relation
\begin{equation}\label{BolF} S = k\log \Omega. \end{equation}

The postulate of equal a priori probability, though not applicable to the system \(\mathcal S\), but it can be applied to the composite system consisting of \(\mathcal S, \mathcal S_0\).

The condition of thermal equilibrium in this case requires that
\begin{equation}
\pp[S_0]{E_0} = \pp[S]{E} = \frac{1}{T}.
\end{equation}
be satisfied. Here \(S, S_0\) are the entropies of the system and the heat bath respectively. The entropies \(S, S_0\) are related to the number of corresponding micro states by the Boltzmann formula \eqref{BolF}. Therefore, the number of micro states of the heat bath with energy \(E_0\) is given by
\begin{equation}
\Omega_0(E_0)= \exp(S(E_0)/k).
\end{equation}
Similarly, the number of micro states of the system \(\mathscr S\) with energy \(E_r\) is
\begin{equation}
\Omega(E_r)= \exp(S(E_r)/k).
\end{equation}

Hence the probability that the system \(\mathcal S\) is in state \(E_r\) is \begin{equation}
p_r = C \Omega(E_r)= C\exp(S(E_r)/k)
\end{equation}
where \(C\) is a normalization to be determined by the requirement that \(\sum_r p_r=1\) may be satisfied.

Since the combined system is isolated, the total energy \(E_t=E+E_0\) is a constant. Therefore, the number of microstates,\(\Omega(E_r)\), of \(\mathcal S\) with energy \(E_r\) is equal to the number of microstates of the heat bath \(\Omega_0(E_t - E_r)\). The probability \(p_r\) can therefore be written as
\begin{equation}
p_r = C \Omega_0(E_t-E_r) = C \exp(S_0 (E_t-E_r)/k).
\end{equation}
Expanding the entropy \(S_0\) in powers of \(E_r\) we get
\begin{eqnarray}
S_0(E_t - E_r) \\&=& S_0(E_t) - \frac{E_r}{k} \pp[S_0 (E_t)]{E_t} + \frac{1}{2}\PP[S_0(E_t)]{E_t} + \ldots.\end{eqnarray}
The third terms is related to the variation in temperature of the heat bath and can be neglected. We will skip details which can be found in Mandl, cited at the end. Thus we get
\begin{eqnarray}
S_0(E_t - E_r)
&=& S_0(E_t) - \frac{E_r}{k} \pp[S_0 (E_t)]{E_t} \\
&=& S_0(E_t) -\beta E_r,
\end{eqnarray}
where \(\beta =kT\) and \(T\) is the temperature of the heat bath.

Thus we get the final expression for the probability \(p_r\)
\begin{equation}
p_r =C e^{-\beta E_r}.
\end{equation}
Normalizing the sum of probabilities to unity
fixes the constant \(C\) to be
\begin{eqnarray}
C \sum_\text{All MS} e^{-\beta E_r}=1 \Rightarrow
C=\frac{1}{Z}
\end{eqnarray}
Thus
\begin{equation}
p_r =\frac{1}{Z} e^{-\beta E_r},
\end{equation}
where \(Z\)is given by
\begin{equation}
Z =\sum_\text{All MS} e^{-\beta E_r}
\end{equation}
and is known as {\tt canonical partition function}. The sum over \(\text{All MS }\) here means a sum over all possible microstates of the system. In general there will be several microstates with the same energy. Therefore the sum can be done by summing over all states with fixed energy \(E_r\), followed by sum over all different energies counted once. If the number of microstates with a given energy \(E_r\) is \(g(E_r)\), then we get
\begin{eqnarray}
Z &=& \sum_\text{All MS} e^{-\beta E_r}\\
&=& \sum_{E_r} g(E_r) e^{-\beta E_r}.
\end{eqnarray}

The final expression for probability of a microstate to have energy \(E_r\) is \begin{equation}
p_r = \frac{1}{Z} e^{-\beta E_r}.
\end{equation}
This result is known as the {\it Boltzmann distribution}.

The macrostates of a system in contact with a heat bath is specified by \(T,V,N\) and not by \(E,V,N\) because the energy is not constant. For a macrostate of a system we can only talk about average energy \(\bar E\) given by
\begin{equation}
\bar E = \sum_{E_r} g(E_r) E_r p_r = \sum_{E_r} g(E_r) E_r e^{-\beta E_r}.
\end{equation}
The average energy is a function of \(T,V,N\). The expression for the average energy can also be written as

\begin{equation}
\bar E = - \pp[\ln Z]{\beta}.
\end{equation}
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It turns out that the fluctuations, in energy for a large number of particles is very small compared to the energy and \(\bar E\) is the energy of the system for all practical purposes. To verify this statement we compute the standard deviation \(\Delta E\) defined by
\begin{equation}
(\Delta E)^2 = \overline{(E-\bar E)^2} = \overline{E^2} - \bar E ^2
\end{equation}
Differentiating \(\ln Z\) w.r.t. \(\beta\), we get
\begin{eqnarray}
\DD[\ln Z]{E} = -\dd[\bar E]{E} = -\dd[T]{\beta} . \pp[\bar E]{T}=-kT^2 C.
\end{eqnarray}
where \(C= \pp[\bar E]{T}\) is the heat capacity of the system at constant external parameters, \(V\) in the present case. Using the expressions of \(\bar E\) and \(\Delta E\), we get \begin{equation}
\frac{\Delta E}{\bar E} = \frac{(kT^2C)^{1/2}}{E}. \end{equation}
In the above expression \(kT^2\) is independent of \(N\) and \(C, \bar E\) are intensive quantities proportional to \(N\). Therefore
\begin{equation}
\frac{\Delta E}{\bar E} \sim \frac{1}{\sqrt{N}}.
\end{equation}
This equation shows that the relative fluctuation of energy of system in a heat bath is proportional to \(N^{_1/2}.\) For a macroscopic system \(N\sim 10^{23}\) and \(\frac{\Delta E}{\bar E} \sim 10^{-11}.\) Hence the average energy of macroscopic system in equilibrium with a heat bath is determined accurately. This enables us to talk about \(\bar E\) as the energy of the system.