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\exnumber{3030}
\heading{Charged Bose gas in a divided box}
\auname{Or Dobkowski}
{\bf The problem:}
\Dn
Consider ${N}$ bosons with mass ${\mathsf{m}}$, positive charge ${e}$, and spin ${0}$.
The particles are in a box that is divided into two regions:
zero voltage region of volume $\Omega_0$, and voltage $V$ region of volume $\Omega_v$.
Assume that the bosons are condensed in the $\Omega_0$ region.
In items (3,5) assume that the gas in the $\Omega_v$ region
can be treated using the Boltzmann approximation.
\Dn
\begin {itemize}
\item[(1)]
Find the $V=\infty$ condensation temperature $T_c(\infty)$.
\item[(2)]
Find the $V=0$ condensation temperature $T_c(0)$.
\item[(3)]
Assuming an intermediate temperature,
find the critical voltage $V_c$ \\
below which the bosons are no longer condensed.
\item[(4)]
Write an exact expression for the energy $E(T,V)$ of the system
\item[(5)]
Write an expression for the heat capacity $C(T,V)$ of the system.\\
Keep only the leading correction in $V$.
\end {itemize}
Express the results using the thermal wavelength $\lambda_T$,
the variables ${\Omega_0, \Omega_v, N, T, eV}$, \\
and the functions ${L_{\alpha}(z)}$ and $\zeta(\alpha)$.
\begin{figure}[h!]
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\Dn\Dn
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In (1) \& (4) we will use two known results for a spinless non-relativistic bosons [given in the D.C lecture notes, and in K.Huang second edition p.289-290]
\eq{\frac{N}{\Omega}=\frac{1}{{\lambda_T}^3}Li_3/2(z)}{}
\eq{\frac{E}{\Omega}=\frac{3}{2}\frac{T}{{\lambda_T}^3}Li_5/2(z)}{}
($\Omega$ is the sign for volume in this exercise)\\
\noindent\rule{\textwidth}{1pt}
{\bf The solution:}
\Dn
(1) $V=\infty$ means there will be no particles in the high voltage part of the box B.
In that case we can consider just the $\Omega_0$ part. In this volume the chemical potential is $\mu=0 \rightarrow z=1$, we use eq.(1) and solve for $N$
\eq{N=n_0+\Omega\zeta\left(\frac{3}{2}\right)\left(\frac{m}{2\pi}\right)^\frac{3}{2}T^\frac{3}{2}}{}
to get the critical temperature one simply sets $n_0=0$ (meaning zero particles in the condensation), set $\Omega=\Omega_0$ and solve for $T_c$
\eq{T_c=\left(\frac{2\pi}{m}\right)\left(\frac{N}{\Omega_0}\frac{1}{\zeta\left(\frac{3}{2}\right)}\right)^\frac{2}{3}}{}
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\Dn
(2)
For $V=0$ we just have one box, with volume $\Omega_0+ \Omega_v$
\eq{T_c=\left(\frac{2\pi}{m}\right)\left(\frac{N}{\Omega_0+ \Omega_v}\frac{1}{\zeta\left(\frac{3}{2}\right)}\right)^\frac{2}{3}}{}
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\Dn
(3) Assuming $T_c(0)