\documentclass[11pt,fleqn]{article}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% template that does not use Revtex4
%%% but allows special fonts
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%
%%% Please use this template.
%%% Edit it using e.g. Notepad
%%% Ignore the header (do not change it)
%%% Process the file in the Latex site 
%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


% Page setup
\topmargin -1.5cm      
\oddsidemargin -0.04cm   
\evensidemargin -0.04cm  
\textwidth 16.59cm
\textheight 24cm 
\setlength{\parindent}{0cm} 
\setlength{\parskip}{0cm} 


% Fonts
\usepackage{latexsym}
\usepackage{amsmath} 
\usepackage{amssymb} 
\usepackage{bm}


% Math symbols I
\newcommand{\sinc}{\mbox{sinc}}
\newcommand{\const}{\mbox{const}}
\newcommand{\trc}{\mbox{trace}}
\newcommand{\intt}{\int\!\!\!\!\int }
\newcommand{\ointt}{\int\!\!\!\!\int\!\!\!\!\!\circ\ }
\newcommand{\ar}{\mathsf r}
\newcommand{\im}{\mbox{Im}}
\newcommand{\re}{\mbox{Re}}

% Math symbols II
\newcommand{\eexp}{\mbox{e}^}
\newcommand{\bra}{\left\langle}
\newcommand{\ket}{\right\rangle}

% Mass symbol
\newcommand{\mass}{\mathsf{m}} 
\newcommand{\Mass}{\mathsf{M}} 

% More math commands
\newcommand{\tbox}[1]{\mbox{\tiny #1}}
\newcommand{\bmsf}[1]{\bm{\mathsf{#1}}} 
\newcommand{\amatrix}[1]{\begin{matrix} #1 \end{matrix}} 
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}

% Other commands
\newcommand{\hide}[1]{}
\newcommand{\drawline}{\begin{picture}(500,1)\line(1,0){500}\end{picture}}
\newcommand{\bitem}{$\bullet$ \ \ \ }
\newcommand{\Cn}[1]{\begin{center} #1 \end{center}}
\newcommand{\mpg}[2][1.0\hsize]{\begin{minipage}[b]{#1}{#2}\end{minipage}}
\newcommand{\Dn}{\vspace*{3mm}}


\def\Abs#1{\left| #1 \right|}
\def\Paren#1{\left( #1 \right)}		
\def\Brack#1{\left[ #1 \right]}		
\def\Seq#1{\left\langle #1 \right\rangle}

% Figures
\newcommand{\putgraph}[2][0.30\hsize]{\includegraphics[width=#1]{#2}}

% heading
\newcommand{\exnumber}[1]{\newcommand{\exnum}{#1}}
\newcommand{\heading}[1]{\begin{center} {\Large {\bf Ex\exnum:} #1} \end{center}}
\newcommand{\auname}[1]{\begin{center} {\bf Submitted by:} #1 \end{center}}

\begin{document} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\exnumber{3030}
\heading{Charged Bose gas in box with potential difference} 
\auname{Gilad Granit}


{\bf The problem:}

\Dn
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.


\begin{itemize}
\item[1)]
 Find the condition on ${N}$, so if ${V=0}$ then there's no
condensation, but if ${V=\infty}$ then there's condensation.


\item[2)]
 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}

\end {itemize}


\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


{\bf The solution:}

\Dn
 

\begin{enumerate}
\item[1)]
W
e start by calculating that the density of states 
\begin{align}
g \left( \epsilon \right) = c \cdot L^{3}   \epsilon ^{\alpha -1} = L^3 \frac{2 m^{3/2}}{(2 \pi )^2} \epsilon ^{1/2}
\end{align}

applying this to the term for the number of particles in the system
\begin{align}
N=\sum _r f\left( \epsilon _r -\mu \right) = \int ^{\infty} _{0} g\left( \epsilon \right) \cdot f\left( \epsilon _r -\mu \right) d \epsilon
\end{align}

the explicit solution for this integral is known, by defining the fugacity ${z=e^{\beta \mu}}$
\begin{align}
N=\frac{L^3}{\lambda_{T}^{3}} Li_{3/2} \left( z \right)
\end{align}

in order to get a BEC we require that the fugacity goes to unity, then we may use the following ${Li_{\alpha}(1)\sim \zeta(\alpha)}$
\begin{align}
N=n_0 +L^3 \zeta \left( \frac{3}{2} \right) \left( \frac{m}{2 \pi} \right)^{3/2} T^{3/2}
\end{align}

so we have a lower bound for the numberof particles when ${V\rightarrow \infty}$ in all the excited states of the gas.
\\
now let us find a higher bound when ${V= 0}$ this can be simply done by rewriting the density of states with the change ${L \rightarrow 2L}$ , and from it followes that 
\begin{align}
N=n_0 +2 \cdot L^3 \zeta \left( \frac{3}{2} \right) \left( \frac{m}{2 \pi} \right)^{3/2} T^{3/2}
\end{align}
finnaly arriving at 
\begin{align}
L^3 \zeta \left( \frac{3}{2} \right) \left( \frac{m}{2 \pi} \right)^{3/2} T^{3/2} < N <2 \cdot L^3 \zeta \left( \frac{3}{2} \right) \left( \frac{m}{2 \pi} \right)^{3/2} T^{3/2}
\end{align}

\Dn\Dn
%%%%%%%%%%%%%%%%%%%%%
\item[2)]

\begin{itemize}
\item[(a)]
We can start from equation (2)
\begin{align}
N_B= \int ^{\infty} _{eV} g\left( \epsilon \right) \cdot f\left( \epsilon _r -\mu \right) d \epsilon \sim \int ^{\infty} _{eV} g\left( \epsilon \right)  e^{- \beta \left( \epsilon _r -\mu \right)} d \epsilon
=\frac{L^3}{\lambda_{T}^{3}}  e^{- \beta \left( eV \right)}
\end{align}
from equation (7) we see that, as expected ,the battery effectivly raises the ground state in zone B so now if we want do not wish to create a BEC for which ${\mu = eV}$ ,  we have to take ${ T\gg eV}$

\Dn
%%%%%%%%%

\item[(b)]
we are given N particles in the system so ${N_A =N - N_B}$, in order to know how many particles are condenced we remmember that
 ${
N_A = n_0 + L^3  \zeta \left( \frac{3}{2} \right) \left( \frac{m}{2 \pi} \right)^{3/2} T^{3/2}
}$
, so the final
\begin{align}
n_0 = N - N_B - N_e = N - \frac{L^3}{\lambda_{T}^{3}} \left[  \zeta \left( \frac{3}{2} \right) +   e^{- \beta \left( eV \right)} \right]
\end{align}

\Dn

%%%%%%%%%

\item[(c)]
in order for there to be a BEC in zone A ${n_0 >0}$  ,starting from Eq (8)
\[
N - \frac{L^3}{\lambda_{T}^{3}} \left[  \zeta \left( \frac{3}{2} \right)+   e^{- \beta \left( eV \right)} \right]>0
\]

\[
 \frac{N\lambda_{T}^{3}}{L^3}> \zeta \left( \frac{3}{2} \right) +   e^{- \beta \left( eV \right)}
\]

\begin{align}
eV_c = -T \ln\left[ \frac{N\lambda_{T}^{3}}{L^3} -\zeta \left(  \frac{3}{2} \right) \right]
\end{align}

as can see we found a critical voltage, below it there will be no condensate state for the Bosons, that is ${V_c < V}$ is our condition.
%%%%%%%%%


\end{itemize}

\end{enumerate}
















%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document} 
