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\sect{Charged Bose gas in a divided box}

Consider ${N}$ bozons with mass ${\mathsf{m}}$, positive charge
${e}$ and spin ${0}$. The particles are in a tank in thermic
equilibrium, and temperature ${T}$. The tank has two zones ${A}$
and ${B}$, The volume of each zone is ${L^{3}}$.A battery creates
potential difference  ${V}$ between the zones. The potential in
every zone is homogenous.


Find the condition on ${N}$, so if ${V=0}$ then there's no
condensation, but if ${V=\infty}$ then there's condensation.

answer: ${L^{3}\zeta\left(\frac{3}{2}\right)\cdot
\left(\frac{mT}{2\pi}\right)^{\frac{3}{2}}<N<2L^{3}\zeta\left(\frac{3}{2}\right)\cdot
\left(\frac{mT}{2\pi}\right)^{\frac{3}{2}}}$

Below, assume that the particles in zone ${A}$ are in a
condensation state and the particles in zone ${B}$ can be described
in the Boltzman proximity frame.

\begin {itemize}
\item[(a)]

What is the number of the particles in zone ${B}$.
What is the condition for ${V}$, so that Boltzman proximity will be
valid
\item[(b)]

What is the number of the particles in zone ${A}$. How many
of them are in condensation state?


\item[(c)]
Show that the condensation in zone ${A}$ as long as
${V_{c}<V}$. Find an explicit expression for ${V_{c}}$.

\end {itemize}

remark: This problem is formally identical to the 'bozons' problem
with spin ${\frac{1}{2}}$ in magnetic field.

[zone ${B}$] down, [zone ${A}$] up, and potential difference ${eV
\gamma B}$.




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