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\begin{document} 
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\heading{E5969: 2D Coulomb gas} \auname{Michael Epshtein}

\textbf{The problem:}\\


\begin{flushleft}
N ions of positive charge q and N ions of negative charge -q are constrained
to move in a two dimensional squre of side L and area $L^{2}.$ the
interaction energy of charge $q_{i}$ at position r$_{i}$ with another
charge $q_{j}$ at position r$_{j}$ is $-q_{_{i}}q_{_{j}}\; ln\left|r_{i}-r_{j}\right|$
where $q_{_{i}},q_{_{j}}=\pm q$ . mass of the ions is $m$.
\par\end{flushleft}

\Dn 
\begin{description}
\item [{{(a)}}] By rescaling space variables to $r_{i}=\frac{1}{L}r_{i}$,
The partition function can be written as $Z(L)=CL^{\alpha}$ find
$\alpha$ where C dont depend on L. Note that: $\underset{i<j}{\overset{2N}{\sum}}q_{_{i}}q_{_{j}}$has
a 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 Gibbs 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 $A\rightarrow\infty$ for both $T>T_{c}$ and $T<T_{c}$. 
\end{description}
\Dn

\textbf{The solution:}

\Dn

(a) we can assume that$\underset{j=1}{\overset{N}{\sum}}q_{_{j}}=Nq$
and $\underset{j=N+1}{\overset{2N}{\sum}}q_{_{j}}=-Nq$

\[
\underset{i<j}{\overset{2N}{\sum}}q_{_{i}}q_{_{j}}=\underset{j=2}{\overset{2N}{\sum}}q_{_{1}}q_{_{j}}+\underset{j=3}{\overset{2N}{\sum}}q_{_{2}}q_{_{j}}+...+\underset{j=N}{\overset{2N}{\sum}}q_{_{N-1}}q_{_{j}}+\underset{j=N+1}{\overset{2N}{\sum}}q_{_{N}}q_{_{j}}+\underset{j=N+2}{\overset{2N}{\sum}}q_{_{N+1}}q_{_{j}}+...+\underset{j=2N}{\overset{2N}{\sum}}q_{_{2N-1}}q_{_{j}}{\normalcolor =}\underset{j=N+1}{\overset{2N}{\sum}}q_{_{N}}q_{_{j}}=-Nq^{2}\]


\begin{flushleft}
For understanding this result let us look at $\underset{j=N}{\overset{2N}{\sum}}q_{_{N-1}}q_{_{j}}$
, here $q_{_{N-1}}$ is a positive charge that interact with $N$
negative charges from $q_{_{N+1}}$ to $q_{_{2N}}$ and one positive
charge $q_{_{N}}$therefore $\underset{j=N}{\overset{2N}{\sum}}q_{_{N-1}}q_{_{j}}$$=-(N-1)q^{2}$ 
\par\end{flushleft}

\begin{flushleft}
$\underset{j=N+2}{\overset{2N}{\sum}}q_{_{N+1}}q_{_{j}}$ here q$_{N+1}$is
a negative charge that interact with only negative charges from $q_{_{N+2}}$
to $q_{_{2N}}$ therefore $\underset{j=N+2}{\overset{2N}{\sum}}q_{_{N+1}}q_{_{j}}$$=+(N-1)q^{2}$,
all other parts will reduced in same way 
\par\end{flushleft}

\begin{flushleft}
only $\underset{j=N+1}{\overset{2N}{\sum}}q_{_{N}}q_{j}$ remains
while all $N$ ions in this summation are negative . 
\par\end{flushleft}

\[
H=\underset{i=1}{\overset{2N}{\sum}}\frac{{\displaystyle p_{i}^{2}}}{2m}-\underset{i<j}{\overset{2N}{\sum}}q_{_{i}}q_{_{j}}\; ln\left|r_{i}-r_{j}\right|\]


\begin{flushleft}
\[
Z_{(L)}=\underset{i}{\overset{2N}{\prod}}\underset{-\infty}{\overset{+\infty}{\int}}exp(-\beta\frac{{\displaystyle p_{i}^{2}}}{2m})\; dp_{i}\overset{2N}{\underset{i}{\prod}}\underset{L^{2}}{\int}exp\;(\beta\underset{i<j}{\overset{2N}{\sum}}q_{_{1}}q_{_{j}}\;\ln\left|r_{i}-r_{j}\right|)\; dr_{i}=\frac{1}{\lambda^{4N}}\overset{2N}{\underset{i}{\prod}}\underset{L^{2}}{\int}exp\;(\beta\underset{i<j}{\overset{2N}{\sum}}q_{_{1}}q_{_{j}}\;\ln\left|r_{i}-r_{j}\right|)\; dr_{i}\]

\par\end{flushleft}

\begin{flushleft}
for $r_{i}=\frac{1}{L}r_{i}$ 
\par\end{flushleft}

\[
Z_{(L)}=\frac{1}{\lambda^{4N}}\overset{2N}{\underset{i}{\prod}}\overset{1}{\underset{0}{\int}}exp\;(\beta\underset{i<j}{\overset{2N}{\sum}}q_{_{i}}q_{_{j}}\;\ln\frac{1}{L}\left|r_{i}-r_{j}\right|)\;(\frac{1}{L})^{2}dr_{i}=\frac{1}{\lambda^{4N}}\; L^{\beta Nq^{2}}\overset{2N}{\underset{i}{\prod}}\int exp\;(\beta\underset{i<j}{\overset{2N}{\sum}}q_{_{i}}q_{_{j}}\;\ln\left|r_{i}-r_{j}\right|)\;(\frac{1}{L})^{2}dr_{i}\]


\[
=C\; L^{(\beta Nq^{2}-4N)}\]


\begin{flushleft}
Finelly we get $\;\;\alpha=\beta Nq^{2}-4N$ 
\par\end{flushleft}

(b) $P={\normalcolor \left(\cfrac{\partial F}{\partial L^{2}}\right)_{T,N}=\tau\cfrac{\partial}{\partial L^{2}}\ln Z(L)}=\tau\cfrac{\partial}{\partial L^{2}}\ln\;(C\; L^{(4N-\beta Nq^{2})})=\tau\;\frac{N}{L^{2}}\;(\;2-\frac{1}{2}\beta q^{2}\;)$

\begin{flushleft}
The system is unstable when $(\frac{\partial P}{\partial L^{2}})_{T,N}>0$ 
\par\end{flushleft}

\[
\left(\cfrac{\partial P}{\partial L^{2}}\right)_{T,N}=\frac{N}{L^{2}}(\frac{1}{2}q^{2}-2\tau)>0\]


\begin{flushleft}
we define low $T$ by $k_{B}T<k_{B}T_{c}=\frac{1}{4}q^{2}$ 
\par\end{flushleft}

\begin{flushleft}
the reason for the unstabillity is that in low $T$ the ions behavior
will depend mostly from their charges and not $T$ therefore the distance
between oposite charged ions will be shorter. 
\par\end{flushleft}

\begin{flushleft}
for $T\gg q^{2}$we get a pressure of ideal gas. 
\par\end{flushleft}

(c) C depend only from momentum integrals and from the Gibbs factors
therefore

\[
C=\frac{1}{\lambda^{4N}}(\frac{1}{N!})^{2}\]


(d) $\;\; Z(L)=\frac{1}{\lambda^{4N}}(\frac{1}{N!})^{2}L^{2(2N-\frac{1}{2}\beta Nq^{2})}$

\[
F=-\tau\; lnZ(L)=2N\tau\;[\;\ln(\frac{\lambda^{2}N}{L^{2}})-1\;]+\frac{1}{2}Nq^{2}\;\ln L^{2}\]


\[
\mu=\left(\frac{\partial F}{\partial N}\right){}_{L^{2},T}=2\tau\;\ln(\frac{\lambda^{2}N}{L^{2}})+\frac{1}{2}q^{2}\;2\ln L\]


\[
N=\frac{1}{\lambda^{2}}L^{2(1-\frac{1}{4}q^{2}\beta)}\exp(\frac{1}{2}\beta\mu)\]


\begin{flushleft}
for A$\rightarrow\infty$ 
\par\end{flushleft}

\begin{flushleft}
$T>T_{c}\;,\;\; N\rightarrow\infty$ the system acting as ideal gas
$PV=NTk_{B}$ 
\par\end{flushleft}

\begin{flushleft}
$T<T_{c}\;,\;\; N\rightarrow0$ the distance between the particles
will go to zero ratio to the area. 
\par\end{flushleft}

\Dn \Dn

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\end{document}
