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\begin{document}
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\exnumber{2170}
\heading{Polarization of two-spheres system inside a tube}
\auname{Ido Moskovich}
{\bf The problem:}
\Dn
Given two balls in a very long, hollow tube, with length ${L}$.
The mass of each ball is ${\mathsf {m}}$, The charge of one ball is
${-q}$ and the charge of the other one is ${+q}$. The ball's radius
is negligible, and the electrostatic attraction between the balls is
also negligible. The balls are rigid and can't pass through each
other. The balls are attached to a drop, whose surface tension
causes it's gravity constant ${\gamma}$ to work on the balls toward
each other (The force does not depend on the distance between the
balls). The system is in an external electric uniform field
${\bar{\varepsilon}=\varepsilon\hat{x}}$ and in thermal equilibrium in
temperature ${T}$.
\begin {itemize}
\item[(a)]
Write the hamiltonian of the system
${H\left(p_{1},p_{2},x_{1},x_{2}\right)=E_{k}+V\left(x\right)}$ when
${E_{k}}$ is the kinetic energy. Define properly ${V\left(x\right)}$
when ${x=x_{2}-x_{1}}$ an--d write a diagram of ${V\left(x\right)}$.
\item[(b)]
Calculate the partition function ${Z\left(\beta,
\varepsilon\right)}$.
\item[(c)]
Find the probability function of $x, \, {\rho\left(x\right)}$ and
the average distance ${\langle x\rangle}$ between the balls. Express
again ${\rho\left(x\right)}$ by ${\langle x\rangle}$.
\item[(d)]
Find the polarization ${p}$ as a function of ${\varepsilon}$.
Use the partition function.
\item[(e)]
Develop ${P\left (\varepsilon\right)}$ up to first order in
the field: ${P\left(\varepsilon\right)=P_{0}+\chi\varepsilon+O
\left(\varepsilon^{2}\right)}$.
\end {itemize}
This development is valid in a weak field, Define what is a weak
field. Express your answers with ${L, \mathsf{m}, q, \gamma, T,
\varepsilon}$.
\Dn\Dn
\begin{figure}[h!]
\advance\leftskip-0.4cm
\includegraphics[width=180mm]{Capture}
\label{fig:method}
\end{figure}
\Dn\Dn
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{\bf The solution:}
\Dn
\begin{itemize}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item [(a)]
Since we assume $L\gg \langle x \rangle$ and we derive in the continuing that $\langle x \rangle = \frac{1}{\beta \left( \gamma - q \varepsilon \right)}$, this derivation is valid in a weak field with respect to the mechanical attraction force $\gamma$,
\begin{equation}
\varepsilon\ll \gamma/q
\end{equation}
The hamiltonian of the system is:
\begin{equation} \label{Hamiltonian}
\mathcal{H} = E_k + V(x)
\end{equation}
Where the kinetic energy is:
\begin{equation} \label {potential}
E_k=\frac{p_1 ^2}{2\mathsf {m}} + \frac{p_2 ^2}{2\mathsf {m}} \\
\end{equation}
\Dn
The electrostatic potential is $q \varepsilon \left( x_1-x_2 \right)$. The surface tension potential is $\gamma \left( x_2-x_1 \right)$. Thus, the potential energy is,
\begin{align} \label {kinetic}
V(x) = \begin{cases} \left( \gamma -q \varepsilon \right)x & \:\:\: 0 < x < L \\
\infty & \:\:\: else \end{cases}
\end{align}
\Dn
\begin{figure}[h!]
\center
\includegraphics[width=100mm]{diagram}
\label{fig:method}
\\{Graph 1: Potential energy vs. distance between the particles. Both axes are normalized with respect to its maximum values.}
\end{figure}
\Dn
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\Dn
\item [(b)]
Let us define $\lambda^{-1}_T=\sqrt{\frac{mT}{2 \pi \hbar^2} }$. Due to the assumption that $L\gg \frac{1}{{\beta \left( \gamma - q \varepsilon \right)}}$, it is a good approximation to integrate the center of mass, $X$, from $0$ to $L$ while calculating the partition function,
\begin{equation}
Z = \frac{1}{\left( 2 \pi \hbar \right)^2}\int e^{-\beta \left( \frac{p_1 ^2}{2m} + \frac{p_2 ^2}{2m} \right) } dp_1 \, dp_2
\int_{0}^{L} dX \int_{0}^{L} e^{-{\beta \left( \gamma - q \varepsilon \right)}x } dx =
\lambda_T^{-2} \cdot L \cdot \frac{\left(1- e^{-{\beta \left( \gamma - q \varepsilon \right)}L} \right)}{{\beta \left( \gamma - q \varepsilon \right)}}
\end{equation}
\Dn
Thus, the partition function is approximately,
\begin{equation}
Z \approx \frac{L}{\lambda_T^{2}} \frac{1}{\beta \left( \gamma - q \varepsilon \right)}
\end{equation}
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\Dn
\item [(c)]
The probability density function of $x$ is,
\begin{equation}
\rho(x) = \frac{ e^{-{\beta \left( \gamma - q \varepsilon \right)}x }}{ \int_{0}^{L} e^{-{\beta \left( \gamma - q \varepsilon \right)}x } dx}
= {\beta \left( \gamma - q \varepsilon \right)}\cdot e^{-{\beta \left( \gamma - q \varepsilon \right)}x }
\end{equation}
The mean value of $x$ is,
\begin{align}
\langle x \rangle &= \int_{0}^{L} x \cdot\beta \left( \gamma - q \varepsilon \right) e^{- \beta \left( \gamma - q \varepsilon \right)x } \, dx
= \left. -x e^{-\beta \left( \gamma - q \varepsilon \right)x } \right|^{L}_{0} + \int_{0}^{L} e^{-\beta \left( \gamma - q
\varepsilon \right)x } \, dx \\
&= -L e^{-\beta \left( \gamma - q \varepsilon \right)L } - \frac{\left( e^{-\beta \left( \gamma - q \varepsilon \right)L } -1 \right)}{\beta \left( \gamma - q \varepsilon \right)}
\end{align}
Therefore,
\begin{equation}
\langle x \rangle \approx \frac{1}{ \beta \left( \gamma - q \varepsilon \right) }
\end{equation}
\Dn
Using $\langle x \rangle$, the probability density function is,
\begin{equation}
\rho_{(x)} \approx \frac{1}{\langle x \rangle} \cdot e^{-{x}/{\langle x \rangle} }
\end{equation}
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\Dn
\item [(d)]
The polarization is,
\begin{equation}
\langle P \rangle=\frac{1}{\beta} \frac{\partial \ln{Z}}{\partial \varepsilon} =
\frac{1}{\beta} \frac{\partial \left(-\ln{\beta \left( \gamma - q \varepsilon \right)}\right)}{\partial \varepsilon} = \frac{q}{\beta \left( \gamma - q \varepsilon \right) } = q \langle x \rangle
\end{equation}
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\Dn
\item [(e)]
The Taylor expansion of the polarization is,
\begin{equation}
\langle P \rangle= \frac{q}{\beta \gamma \left( 1 - \frac{\varepsilon}{(\gamma/q)} \right) } =
\frac{q}{\beta \gamma }\left( 1 + \frac{\varepsilon}{(\gamma/q)} \right) +O_{(\varepsilon^2)}=
\frac{q}{\gamma }T+ \left(\frac{q}{ \gamma }\right)^2 T\varepsilon +O_{(\varepsilon^2)}
\end{equation}
Thus,
\begin{equation}
P_0=\frac{q}{\gamma }T \\
\chi=\left(\frac{q}{ \gamma }\right)^2 T
\end{equation}
\end{itemize}
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\end{document}