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\begin{document}

\heading{E5969: 2D Coulomb gas} 

\auname{Yair Margalit} 

\textbf{The problem:}

\Dn

$N$ ions of positive charge $q$ and $N$ ions of negative charge
$-q$ are constrained to move in a two dimensional square of side
$L$ and area ${\mathsf{A}=L^{2}}$. The interaction energy of charge
$q_{i}$ at position $r_{i}$ with another charge $q_{j}$ at position
$r_{j}$ is $-q_{i}q_{j}ln|r_{i}-r_{j}|$, where $q_{i},q_{j}=\pm q$.
The mass of the ions is $m$.

\Dn

\begin{itemize}

\item [{{(a)}}]By rescaling space variables to $r_{i}:=r_{i}/L$,
the partition function can be written as $Z(L)=CL^{\alpha}$, where
$C$ does not depend on $L$. Find $\alpha$. Hint: $\sum_{ij}q_{i}q_{j}$
has a very simple dependence on $N$.

\item [{{(b)}}]Calculate the pressure, and show that for $T<T_{c}$
the system is unstable. Determine what is $T_{c}$. Comment on the
reason for this instability.

\item [{{(c)}}]Determine what is $C$ if the interaction can be neglected.
Handle the Gibbs factor correctly.

\item [{{(d)}}]Find the chemical potential $\mu(T,N,L)$, and solve
for $N(\mu,T,L)$. In particular write what is $N$ for a fixed $\mu$
in the limit $L\rightarrow\infty$ for both $T>T_{c}$ and $T<T_{c}$.

\end{itemize}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\textbf{The solution:}

\Dn

\begin{itemize}

\item [{{(a)}}]The system Hamiltonian is given by:
\begin{equation}
H=\sum_{i=1}^{2N}\frac{\mathbf{p}_{i}^{2}}{2m}+\sum_{\langle ij\rangle}^{2N}u(\mathbf{r}_{i}-\mathbf{r}_{j})
\end{equation}


Where the sum $\sum_{\langle ij\rangle}^{2N}$ is only over distinct
pairs. Thus the canonical partition function is given by:
\begin{equation}
Z_{2N}(\beta,V)=Z_{kin}\int_{0}^{L}d^{2}r_{1}\ldots dr_{2N}^{2}exp[-\beta\sum_{\langle ij\rangle}(-q_{i}q_{j})ln|\mathbf{r}_{i}-\mathbf{r}_{j}|]
\end{equation}


Where $Z_{kin}$ is the kinetic part of the partition function, including
the Gibbs factor. Using rescaling such that $r_{i}=\tilde{r}_{i}L$,
$dr_{i}=d\tilde{r}_{i}L$ we obtain:

\begin{align}
Z_{2N}(\beta,V) & =Z_{kin}\int_{0}^{1}d^{2}\tilde{r}_{1}\ldots d\tilde{r}_{2N}^{2}L^{4N}exp[\beta\sum_{\langle ij\rangle}q_{i}q_{j}ln(L|\mathbf{\tilde{r}}_{i}-\mathbf{\tilde{r}}_{j}|)]\label{eq:}\\
 & =Z_{kin}\int_{0}^{1}d^{2}\tilde{r}_{1}\ldots d\tilde{r}_{2N}^{2}L^{4N}exp[\beta\sum_{\langle ij\rangle}q_{i}q_{j}ln(|\mathbf{\tilde{r}}_{i}-\mathbf{\tilde{r}}_{j}|)]L^{\beta\sum_{\langle ij\rangle}q_{i}q_{j}}\label{eq:partition}
\end{align}


We now need to calculate the sum $\sum_{\langle ij\rangle}q_{i}q_{j}$.
Observing that there are in total $N(N-1)$ positive product pairs
and $N^{2}$ negative product pairs, by subtracting we get:
\begin{equation}
\sum_{\langle ij\rangle}q_{i}q_{j}=q^{2}(N_{pos}-N_{neg})=q^{2}[N(N-1)-N^{2}]=-q^{2}N
\end{equation}


Substituting into eq. \ref{eq:partition} we get:
\begin{equation}
Z_{2N}(\beta,V)=Z_{kin}L^{4N-\beta q^{2}N}\int_{0}^{1}d^{2}\tilde{r}_{1}\ldots d\tilde{r}_{2N}^{2}exp[\beta\sum_{\langle ij\rangle}q_{i}q_{j}ln(L|\mathbf{\tilde{r}}_{i}-\mathbf{\tilde{r}}_{j}|)]\equiv CL^{\alpha}
\end{equation}


\item [{{(b)}}]Since $C$ does not depend on $L$ calculating the
pressure is straightforward:

\begin{equation}
P=-\left(\frac{\partial F}{\partial A}\right){}_{\tau,N}=\tau\frac{\partial}{\partial A}ln\left(CA^{\alpha/2}\right)=\frac{\tau\alpha}{2A}
\end{equation}
The system will become unstable in the region where $\frac{\partial P}{\partial A}>0$:
\begin{align}
\frac{\partial P}{\partial A} & =-\frac{\tau\alpha}{2A^{2}}=\frac{N}{2A^{2}}(q^{2}-4\tau)>0\\
 & \Rightarrow T<\frac{q^{2}}{4k_{B}}\equiv T_{c}
\end{align}


This unphysical positive slope in which the pressure increases as
the area is increased indicates a phase transition (similar to the
Van Der Waals gas case) due to positive - negative ion pairing, and
in this regime both phases (ion pairs and free ions) coexist.

\item [{{(c)}}]If the interaction can be neglected then the partition
function takes the form:
\begin{align}
Z_{2N}(\beta,V) & =\frac{1}{(N!)^{2}}\left(\frac{1}{\lambda_{T}^{2}}\right)^{2N}\int_{0}^{L}d^{2}r_{1}\ldots dr_{2N}^{2}\\
 & =\frac{1}{(N!)^{2}}\left(\frac{L^{2}}{\lambda_{T}^{2}}\right)^{2N}\equiv CL^{4N}=\frac{\left(Z_{1}^{2D}\right)^{2N}}{(N!)^{2}}
\end{align}


Meaning the problem is reduced to the 2D ideal gas case where the
partition function can be factorized to single particle partition
functions. The only difference is the reduction of the Gibbs factor
which is due to the fact that we have two distinct groups of N ions
each.

\item [{{(d)}}]In order to find the chemical potential we will explicitly
write the partition function, including the kinetic part:

\begin{equation}
Z_{2N}(\beta,V)=\frac{L^{4N-\beta q^{2}N}}{(N!)^{2}\lambda_{T}^{4N}}\int_{0}^{1}d^{2}\tilde{r}_{1}\ldots d\tilde{r}_{2N}^{2}exp[\beta\sum_{\langle ij\rangle}q_{i}q_{j}ln(L|\mathbf{\tilde{r}}_{i}-\mathbf{\tilde{r}}_{j}|)]
\end{equation}


And the chemical potential of either the positive or the negative
ions:

\begin{align}
\mu_{\pm}(\tau,N,L) & =\left(\frac{\partial F}{\partial N_{\pm}}\right)_{\tau,A}\\
 & =-\tau\frac{\partial}{\partial N}ln\left[\frac{L^{4N-\beta q^{2}N}}{(N!)^{2}\lambda_{T}^{4N}}\int_{0}^{1}d^{2}\tilde{r}_{1}\ldots d\tilde{r}_{2N}^{2}exp[\beta\sum_{\langle ij\rangle}q_{i}q_{j}ln(L|\mathbf{\tilde{r}}_{i}-\mathbf{\tilde{r}}_{j}|)]\right]\\
 & =2\tau ln\left(N\lambda_{T}^{2}\right)-(4\tau-q^{2})ln(L)
\end{align}


Where the integral disappears being independent of N. In order to
get $N(\mu,\tau,L)$ we simply reverse the formula:

\begin{equation}
N(\mu,\tau,L)=\frac{e^{\frac{\beta\mu}{2}}}{\lambda_{T}^{2}}L^{2-\frac{\beta q^{2}}{2}}
\end{equation}


The limit of $L\rightarrow\infty$:

For $T<T_{c}$ $L$'s exponent is negative, we get $N(\mu,\tau,L)\rightarrow0$
indicating that the number of ions in the system goes to zero due
to pair creation and all the particles that remain are neutral.

For $T>T_{c}$ $L$'s exponent is positive and we get $N(\mu,\tau,L)\rightarrow\infty$
which is the same as the classical result, meaning interactions becomes
negligible compared to the temperature and we return to the ideal
gas case as in (c).

\end{itemize}
\end{document}
