[NOTES/SM-06004] Grand Canonical Ensemble --- Pressure and Chemical Potential

$\newcommand{\pp}[2][]{\frac{\partial #1}{\partial #2}}\newcommand{\Zca}{\mathcal Z}$

Using thermodynamics to express TdS in terms of Gibbs energy and comparing TdS expression in terms of the grand canonical partition function we obtain pressure and chemical potential function in terms of grand canonical partition function.

Pressure and Chemical Potential

We will compute $TdS$ in two different ways. First we will use thermodynamic relations to express $TdS$ in terms of Gibbs energy $g=u-Ts+pv$. Next we find $TdS$ in terms of partition function. Comparing the two different expressions for $TdS$ obtained in two ways will lead us to the desired relations $ p= kT \dd{V}\ln \Zca$ and $\mu = (u-Ts+pv)$.

$TdS$ from thermodynamics
The second law of thermodynamics can be written as
\begin{equation}
du = T ds- p dv \label{EQ30}
\end{equation}
If $\bar{N}$ denotes the average number of particles
we have
\begin{equation}
U= \bar{N} u, \qquad S=\bar{N} s; \qquad V = \bar{N} v. \label{EQ12}
\end{equation}

Expressing $du, ds$ and $dv$ in terms of $dS, dU$ and $dV$,\label{EQ13}
\begin{eqnarray}
dU &=& d\bar{N} u + \bar{N} du\\
dV &=& d\bar{N} v + \bar{N} dv\\
dS &=& d\bar{N} s + \bar{N} ds
\end{eqnarray}
the second law,\eqref{EQ30}, gives
\begin{eqnarray}
dU&=& TdS -pdV + ( u-Ts+pv) d\bar{N} \\
&=& TdS -pdV + g d\bar{N}\\
TdS &=& dU + pdV - g d\bar{N} \label{EQ28}
\end{eqnarray}
where $g= u-Ts+pv $ is the Gibbs free energy per particle.

$TdS$ in terms of $d\Zca$:
Next wee compute $TdS$ in terms of partition function and compare with \eqref{EQ28}. Using \eqref{EQ11} we will compute $TdS$. For this purpose we need to compute change in $\ln \Zca$. Remembering that the grand canonical partition function is a function of $\beta, \mu$ and $V$, {\it i.e.} $\Zca= \Zca(\beta,\mu,V)$, therefore
\begin{equation}
d \ln \Zca = \Big(\dd{\beta}\ln \Zca\Big) d\beta + \Big(\dd{\mu}\ln \Zca
\Big) d\mu + \Big(\dd{V}\ln \Zca\Big) dV \label{EQ11A}.
\end{equation}
We note that
\begin{eqnarray}
d(\log \Zca)
&=& \frac{1}{\Zca} d\Zca \\
&=& \frac{1}{\Zca} \Big( \pp[\Zca]{\mu} d\mu + \pp[\Zca]{\beta} d\beta+
\pp[\Zca]{V} dV \Big)\\
&=& \beta \bar{N} d\mu + (\mu \bar{N} -U ) d\beta + \pp[\ln\Zca]{V} dV
\end{eqnarray}
We now use this answer for $d\ln \Zca$ to compute $TdS$. Taking $S$ from
\eqref{EQ11} we have
\begin{eqnarray}
TdS
&=& kT d [ \ln \Zca + \beta \bar{U} -\beta \mu \bar{N}] \\
&=& \beta^{-1}\Big[\big( \beta \bar{N} d\mu + (\mu \bar{N} -\bar{U} ) d\beta +
\pp[\ln\Zca]{V} dV \big) \\
&& ~~~~ + ( d\beta \bar{U} + \bar{U}d\beta ) \\
&& ~~~~ - ( \mu \bar{N} d\beta + \beta \bar{N} d\mu +\beta \mu d\bar{N})\Big]
\end{eqnarray}
On simplification we get
\begin{equation}
TdS = d\bar{E} + kT\frac{1}{\Zca}\pp[\ln\Zca]{V} - \mu
d\bar{N}.\label{EQ19}
\end{equation}
Comparing \eqref{EQ19} with \eqref{EQ28} we get the required relation.
\begin{equation}
p = kT \dd{V}\ln \Zca, \qquad \mu = g= (u-Ts+pv),
\end{equation}
where $u,s,v$ are specific internal energy, specific entropy and specific volume respectively. Thus the chemical potential is equal to the Gibbs free energy, $g$, per particle.

Expression for $\ln \Zca$ in terms of thermodynamic variables
Let $\Omega= k\ln \Zca$, then from \eqref{EQ11} we get
\begin{eqnarray}
\Omega = \bar{E}-TS - \mu \bar{N}\\
\end{eqnarray}
Gibbs energy $G$ is an extensive quantity and equals $ g \bar{N}=\mu \bar{N}$.
Therefore
\begin{eqnarray}
\Omega &=& \bar{E}-TS - G \\
&=& \bar{E}-TS -(\bar{E}-TS+pV),
\end{eqnarray}
and therefore we have
\begin{eqnarray}
\boxed{\Omega = - pV}.
\end{eqnarray}
Remembering that the independent variables for $\Omega$ are $T,V, \mu$
\begin{eqnarray}
d\Omega
&=& d( \bar{E} - TS - \mu \bar{N}) \\
&=& d\bar{E} - TdS -S dT -\mu d\bar{N} - \bar{N} d\mu\\
&=& -pdV -SdT -\bar{N} d\mu
\end{eqnarray}
Therefore we have
\begin{equation}
\boxed{ \pp[\Omega]{V}\Big|_{p,\mu} = -p;}\qquad
\boxed{\pp[\Omega]{T}\Big|_{\mu,p} = -S;}
\qquad \boxed{ \pp[\Omega]{\mu}\Big|_{T,V}=-\bar{N}.}
\end{equation}

Finally the Gibbs distribution can itself be written as
\begin{equation}
P(E_k,N) = \exp(-\beta(E_k-\mu N + pV))
\end{equation}