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\begin{document}
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\heading{E2060: Particle on a ring with electric field} 
\auname{Rotem Kupfer}

\textbf{The problem:}

\Dn

Particle of mass ${\mathsf{m}}$ and charge ${e}$ is free to move
on a ring of radius ${R}$. The ring is located in the ${x-y}$ plan.
The position of the particle on the ring is ${x=R\cos\left(\theta\right)}$
and ${y=R\sin\left(\theta\right)}$. There is an electric field ${E}$
in the $x$ direction. The temperature is ${T}$.

\begin{itemize}
\item [(1)] Write the Hamiltonian ${H\left(\theta,p\right)}$ of the particle. 
\item [(2)] Calculate the partition function ${Z\left(\beta,{\cal E}\right)}$. 
\item [(3)] Write an expression for the probability distribution ${\rho\left(\theta\right)}$. 
\item [(4)] Calculate the mean position ${\langle x\rangle}$ and ${\langle y\rangle}$. 
\item [(5)] Write an expression for the probability distribution ${\rho\left(x\right)}$.
Attach a schematic plot. 
\item [(6)] Write an expression for the polarization. Expand it up to
first order in ${E}$, and determine the susceptibility. 
\end{itemize}
\Dn Use the following identities: \[
\frac{1}{2\pi}\int_{0}^{2\pi}\exp\left(z\cos\left(\theta\right)\right)d\theta=I_{0}\left(z\right)\]
 \[
I_{0}'\left(z\right)=I_{1}\left(z\right)\]
 \[
I_{0}\left(z\right)=1+\left(\frac{1}{4}\right)z^{2}+\left(\frac{1}{64}\right)z^{4+...}\]


\Dn\Dn

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\textbf{The solution:}

\Dn

(1) The Hamiltonian is:

\[
\mathcal{H}=\frac{P_{\theta}^{2}}{2mR^{2}}-eER\cos\theta\]


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\Dn

(2) The partition function is therefor:

\[
Z=\frac{1}{2\pi\hbar}\intop_{-\infty}^{\infty}\eexp{-\frac{\beta P_{\theta}^{2}}{2mR^{2}}}dP_{\theta}\intop_{0}^{2\pi}\eexp{\beta eER\cos\theta}d\theta=\sqrt{\frac{mR^{2}}{2\pi\beta}}\frac{2\pi}{2\pi}\intop_{0}^{2\pi}\eexp{\beta eER\cos\theta}d\theta=\sqrt{\frac{mR^{2}}{2\pi\beta}}2\pi I_{0}(\beta eER)\]


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\Dn

(3) By definition:

\[
\rho\left(\theta\right)=\intop_{-\infty}^{\infty}\rho\left(\theta,P_{\theta}\right)dP_{\theta}=\intop_{-\infty}^{\infty}\frac{\eexp{-\frac{\beta P_{\theta}^{2}}{2mR^{2}}+\beta eER\cos\theta}}{Z}dP_{\theta}=\frac{\sqrt{\frac{mR^{2}}{2\pi\beta}}\eexp{\beta eER\cos\theta}}{\sqrt{\frac{mR^{2}}{2\pi\beta}}2\pi I_{0}(\beta eER)}=\frac{\eexp{\beta eER\cos\theta}}{2\pi I_{0}(\beta eER)}\]


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\Dn

(4) The mean ${\langle x\rangle}$ is given by:

\[
<x>=<R\cos\theta>=\intop_{0}^{2\pi}R\cos\theta\rho\left(\theta\right)d\theta=\intop_{0}^{2\pi}R\cos\theta\frac{\eexp{\beta eER\cos\theta}}{2\pi I_{0}(\beta eER)}d\theta=...\]


\[
...=\frac{R}{2\pi I_{0}(\beta eER)}\intop_{0}^{2\pi}\cos\theta\eexp{\beta eER\cos\theta}d\theta=\frac{RI_{1}(\beta eER)}{I_{0}(\beta eER)}\]


The mean ${\langle y\rangle}$ is given similarly:

\[
<y>=<R\sin\theta>=\intop_{0}^{2\pi}R\sin\theta\rho\left(\theta\right)d\theta=\intop_{0}^{2\pi}R\sin\theta\frac{\eexp{\beta eER\cos\theta}}{2\pi I_{0}(\beta eER)}d\theta=0\]


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\Dn

(5) The probability distribution ${\rho\left(x\right)}$ is given
by a simple substitution of random variable. It is important to notice
that ${\theta(x)}$ is not injective so there is a factor of 2:

\[
x=R\cos\theta\rightarrow\theta=\arccos\frac{x}{R}\]


\[
\rho\left(x\right)=2\rho\left(\theta(x)\right)\left|\frac{d\theta}{dx}\right|=\frac{\eexp{\beta eEx}}{\pi I_{0}(\beta eER)}\frac{1}{\sqrt{R^{2}-x^{2}}}\]


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(6) The polarization is the logarithmic derivation of the partition
function by the external electric field:

\[
P(E)=\frac{1}{\beta}\frac{\partial\ln Z}{\partial E}=\frac{1}{\beta}\frac{\partial}{\partial E}\ln(\sqrt{\frac{mR^{2}}{2\pi\beta}}2\pi I_{0}(\beta eER))=\frac{1}{\beta}\frac{\partial\ln2\pi I_{0}(\beta eER))}{\partial E}=eR\frac{I_{0}^{'}(\beta eER)}{I_{0}(\beta eER)}\]


By expanding to a Taylor series near ${E=0}$:

\[
P(E)=eR\frac{\frac{\beta eER}{2}+o(E^{3})}{1+\frac{(\beta eER)^{2}}{4}+o(E^{4})}\]


The linear susceptibility is therefor:

\[
\chi^{(1)}=\frac{\beta e^{2}R^{2}}{2}\]

\end{document}
