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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!]
  \includegraphics[width=\linewidth]{ex3030.png}
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\Dn\Dn
\newpage

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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 $T_c(0)<T<T_c(\infty)$, the number of particles in region $\Omega_0$ is given by eq.(3) For the $\Omega_v$ region assuming the density is low, the Boltzmann approximation gives $\frac{N}{\Omega}=\frac{1}{{\lambda_T}^3}e^\frac{-\epsilon}{T}$, so that
\eq{N=n_0+\Omega_0\zeta\left(\frac{3}{2}\right)\frac{1}{{\lambda_T}^3}+\Omega_v\frac{1}{{\lambda_T}^3}e^\frac{-eV}{T}}{eq}

setting $n_0=0$ and solving for $eV_c$ one gets
\eq{eV_c=T\ln{\frac{\Omega_v}{N{\lambda_T}^3-\Omega_0\zeta\left(\frac{3}{2}\right)}}}{eq}

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\Dn

(4) To find the exact energy of the system one should go back to the basics, and start from the integral 
\eq{E=\int_{0}^{\infty} g(\epsilon)\epsilon f(\epsilon-\mu) d\epsilon}{eq}
where $ g(\epsilon)$ is the density of states, and $f(\epsilon-\mu)$ is the Bose occupation function.


\eq{g \left( \epsilon_k \right) = \Omega \frac{2 m^{3/2}}{(2 \pi )^2} \epsilon_k ^{1/2}}{eq}
\eq{f(\epsilon-\mu)=\frac{1}{e^{\beta(\epsilon-\mu)}-1}}{eq}

We have two regions to treat:
In both regions $\mu=0$ since we have chemical equilibrium, and the chemical potential the a condensation is zero.
In the $\Omega_0$ region, $\epsilon=\epsilon_k$, so we get $E_0=\int_{0}^{\infty} g(\epsilon_k)\epsilon_k f(\epsilon_k) d\epsilon_k$ which leads to eq(2) with $z=1$, thus we get

\eq{E_0=\Omega\frac{3}{2}\frac{T}{{\lambda_T}^3}Li_{5/2}(1)=\Omega_0\frac{3}{2}\frac{T}{{\lambda_T}^3}\zeta\left(\frac{5}{2}\right)}{eq}

In the $\Omega_v$ region, $\epsilon=\epsilon_k+eV$, so $g\left(\epsilon\right)=g\left(\epsilon_k+eV\right) =g\left(\epsilon_k\right)$
and $d\epsilon=d(\epsilon_k+eV)=d\epsilon_k$

\eq{E_v=\int_{0}^{\infty} g(\epsilon_k)(\epsilon_k+eV) f(\epsilon_k+eV) d\epsilon_k=\int_{0}^{\infty} g(\epsilon_k)\epsilon_k f(\epsilon_k+eV) d\epsilon_k+\int_{0}^{\infty} g(\epsilon_k)eV f(\epsilon_k+eV) d\epsilon_k}{eq}
The first integral is the same as eq (2) with $z=e^{\frac{-eV}{T}}$, the second integral is the same as eq(1) times a constant $eV$ and $z=e^{\frac{-eV}{T}}$
thus we get
\eq{E_v=\Omega_v\left(\frac{3}{2}\frac{T}{{\lambda_T}^3}Li_{(5/2)}\left(e^{\frac{-eV}{T}}\right)+eV\frac{1}{{\lambda_T}^3}Li_{(3/2)}\left(e^{\frac{-eV}{T}}\right)\right)}{eq}
\eq{E_{total}=\Omega_0\frac{3}{2}\frac{T}{{\lambda_T}^3}\zeta\left(\frac{5}{2}\right)+\Omega_v\left(\frac{3}{2}\frac{T}{{\lambda_T}^3}Li_{(5/2)}\left(e^{\frac{-eV}{T}}\right)+eV\frac{1}{{\lambda_T}^3}Li_{(3/2)}\left(e^{\frac{-eV}{T}}\right)\right)}{eq}
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\Dn

(5)
In the Boltzmann approximation $Li(z)\approx z$
\eq{E_{total}\approx\Omega_0\frac{3}{2}\frac{T}{{\lambda_T}^3}\zeta\left(\frac{5}{2}\right)+\Omega_v\left(\frac{3}{2}\frac{T}{{\lambda_T}^3}e^{\frac{-eV}{T}}+eV\frac{1}{{\lambda_T}^3}e^{\frac{-eV}{T}}\right)}{eq}
\eq{\frac{1}{{\lambda_T}^3}=\left(\frac{mT}{2\pi}\right)^{3/2}}{}
\eq{C=\frac{\partial E_{total}}{\partial T}}{}
Differentiating and keeping the leading order in $\frac{eV}{T}$ we get
\eq{C\approx \Omega_0\frac{15}{4}\frac{1}{{\lambda_T}^3}\zeta\left(\frac{5}{2}\right)+\frac{\Omega_v}{{\lambda_T}^3}\left(\frac{eV}{T}\right)^2e^{\frac{-eV}{T}}}{}
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