\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}
\usepackage{graphicx}


% 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}}

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

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


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



\heading{Ex 2040: An ideal gas in a box with a gravitation field } 
\auname{Dotan Davidovich} 

{\bf The problem:}

\Dn
An ideal gas in a 3D box: $V=L\cdot L\cdot (Z_b-Z_a)$ is placed in an external gravitational field that points along $ -\hat{z}$.
\Dn


a)Find the 1-particle partition function $Z_1(\beta,Z_a,Z_b)$.
\Dn

b)What is the N-particle partition function $Z_N(\beta,Z_a,Z_b)$.
\Dn

c)What are the forces $F_a$ and $F_b$ acting on the floor and on the ceiling ?
\Dn

d)What is the difference between these forces? explain your  result.
\Dn

[[FIGURE]] %\includegraphics[scale = 0.6]{2040pic}

\Dn

{\bf The solution:}
\Dn

a)The one-particle partition function is : 
\[\tag{1} Z_1(\beta,Z_a,Z_b)=\iiiint e^{-\beta(\frac{p^2}{2m} + mgz)}\frac{d^{3}pdxdydz}{(2\pi)^{3}}=\frac{L^{2}}{(2\pi \lambda_T)^{3}\beta mg} (e^{-\beta mgZ_a} - e^{-\beta mgZ_b } )  \]

   Where $\lambda_T=\sqrt{\frac{\beta}{2\pi m}}$ is the thermal wave length.
\Dn

b)There are no interactions so the partition function gets factorized for each degree of freedom :
\[\tag{2}  Z_N(\beta,Z_a,Z_b)=\frac{1}{N!}[Z_1(\beta,Z_a,Z_b)]^N      \]

c)We can think of the floor and the ceiling as pistons and we can define the forces $F_a$ and $ F_b$ from the work :

\[\tag{3} dW=F_b\cdot dZ_b + F_a (-dZ_a)\]

Which means the work done by the system is positive for lifting the ceiling and for lowering the floor.
\Dn
The forces are :
\[\tag{4}  F_b=\frac{1}{\beta}\frac{\partial ln(Z_N)}{\partial Z_b}=Nmg\Big (\frac{e^{-\beta mg Z_b }}{e^{-\beta mg Z_a } -e^{-\beta mg Z_b }}  \Big)    \]

\[\tag{5}  -F_a=\frac{1}{\beta}\frac{\partial ln(Z_N)}{\partial Z_a}=-Nmg\Big (\frac{e^{-\beta mg Z_a }}{e^{-\beta mg Z_a } -e^{-\beta mg Z_b }}  \Big)    \]

d) The difference is :

\[\tag{6}  F_a - F_b = Nmg  \] 

This result should be intuitive since for $ dZ_a=dZ_b=dZ $ we get $ -dW=Nmg\cdot dZ  $ which is the work done on the system when lifting the box in $dZ$ ,or the gravitational potential  energy gained by each particle.  



\end{document} ‎‎‎ 
