\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{2050}
\heading{Pressure by a particle in a spring-box system}
\auname{Eli Gudinetsky}
\Dn
A spring that has an elastic constant ${K}$ and
natural length ${L}$ is connected between a wall
at $x=0$ and a piston at~${x=X}$.
Consequently the force that acts of the piston is ${F_0=-K(X-L)}$.
A classical particle of mass ${\mathsf{m}}$ is attached
to the middle point of the spring.
The system is at equilibrium, the temperature is~${T}$.
\begin {itemize}
\item[(1)]
Write the Hamiltonian (be careful).
\item[(2)]
Write an expression for the partition function ${Z\left(\beta,X\right)}$.
The answer is an expression that may contain a definite integral.
\item[(3)]
Write an expression for the force ${F}$ on the piston.
The answer is an expression that may contain a definite integral.
\item[(4)]
Find a leading order (non-zero) expression for ${F-F_0}$
in the limit of high temperature.
\item[(5)]
Find a leading order (non-zero) expression for ${F-F_0}$
in the limit of low temperature.
\end {itemize}
Your answers should not involve exotic functions,
and should be expressed using $(X,L,K,\mass,T)$.
\Dn\Dn
{\bf The Solution:}
\Dn
\begin{enumerate}
\item
When cutting a spring in half its spring constant is doubled. This is understood by viewing the whole spring as 2 identical halves connected in series, so
\begin{align*} K^{-1}_{whole}=K^{-1}_{half}+K^{-1}_{half}\Rightarrow K_{half}=2K_{whole}\, .
\end{align*}
The Hamiltonian is therefore,
\begin{align}
\mathcal{H}=\frac{p^2}{2m}+\frac{2K\Paren{x-L/2}^2}{2}+\frac{2K\Brack{\Paren{X-x}-L/2}^2}{2}\, .
\end{align}
Here, $x$ is the position of the particle, $p$ is its momentum, the second term is due to the left spring and the third term is due to the right spring.
\item
Writing the partition function we have,
\begin{align}
Z=\frac{1}{2\pi}\int_{-\infty}^{\infty}{\mathrm{d}p\int_{0}^{X}{\mathrm{d}x\, e^{-\beta \Brack{p^2/2m+{2K\Paren{x-L/2}^2/}{2}+{2K\Paren{{X-x}-L/2}^2}/{2}}}}} \, .
\end{align}
Let us rewrite the argument of the spatial exponent,
\begin{align*}
&\Paren{x-L/2}^2-\Paren{X-x-L/2}^2=2x^2+L^2/2+X^2-2Xx-LX=\\
&=2\Paren{x^2-xX+X^2/4}+\Paren{L^2-2LX+X^2}/2=2\Paren{x-X/2}^2+\Paren{L-X}^2/2 \, .
\end{align*}
Substituting the above and solving the kinetic part we get,
\begin{align}
Z=\frac{1}{\lambda_T}e^{-\beta K\Paren{L-X}^2/2}{\int_{0}^{X}{\mathrm{d}x\, e^{-2\beta K \Paren{x-X/2}^2}}} \, .
\end{align}
\item
The generalized force associated with the parameter $X$ is the force $F$ on the piston. By definition it is,
\begin{align}
\Seq{F}&=\frac{1}{\beta}\frac{\partial \ln{Z}}{\partial X}=\frac{1}{\beta}\Brack{\beta K \Paren{L-X}+\frac{\partial}{\partial X}\ln\Paren{\int_{0}^{X}{\mathrm{d}x\, e^{-2\beta K \Paren{x-X/2}^2}}}}\\
&=F_0+\frac{1}{\beta}\frac{\partial}{\partial X}\ln\Paren{\int_{0}^{X}{\mathrm{d}x\, e^{-2\beta K \Paren{x-X/2}^2}}} \, .
\end{align}
\item
In the limit of high temperatures, or $\beta\to0$, we can expand the exponent in the intergal and have:
\begin{align}
\Seq{F}-F_0&\approx\frac{1}{\beta}\frac{\partial}{\partial X}\ln\Paren{\int_{0}^{X}{\mathrm{d}x\, 1}}=\frac{1}{\beta}\frac{\partial}{\partial X}\ln{X}=\frac{1}{\beta X} \, .
\end{align}
\item
Let us denote
\begin{align}
I\Paren{X,\beta}=\int_{0}^{X}{\mathrm{d}x\, e^{-2\beta K \Paren{x-X/2}^2}}=\frac{2}{\sqrt{2\beta K}}\int_{0}^{\sqrt{\frac{\beta K}{2}}X}{\mathrm{d}y\, e^{-y^2}}\, ,
\end{align}
(in the last equality we just changed variables) giving us:
\begin{align}
\Seq{F}-F_0=\frac{1}{\beta I\Paren{X,\beta}}\frac{\partial}{\partial X}I\Paren{X,\beta}\, .
\end{align}
Notice that in the limit $\beta\to\infty$ we have $I\Paren{X,\beta}\to\sqrt{\pi}/2$ becase of the upper bound of the integral. This gives as the approximated value:
\begin{align}
\Seq{F}-F_0\approx\sqrt{\frac{2K}{\pi\beta}}e^{-\beta K \frac{X^2}{2}}\, ,
\end{align}
where the last exponent came from the fundamental theorem of calculus.
\end{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document}