\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}}

% 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{E010: Average distance between two particles in a box} 
\auname{Elkana Porat}


{\bf The problem: }

\Dn

In a one dimensional box with length ${L}$, two particles have
random positions $x_1$, $x_2$ . The particles don't know about each other. The
probability function for finding a particle in a specific place in
the box is uniform.
Let ${r = x_1 - x_2}$ be the relative location of the particles. 
Find ${\langle\hat{r}\rangle}$ and the dispertion relation ${\sigma_{r}}$:

\Dn

(1) By using the adding rules for the expectation values and variances of the independent positions.

(2) By calculating the probability function: ${f\left(r\right)dr={P}\left(r<\hat{r}<r+dr\right)}$.

\Dn
\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn

(1) The probability function for each particle is given by:
$${f\left(x\right) = \left\{\begin{array}{ccc}
	1/L &, & 0<x<L\\
	0 &, &o.w. \end{array}\right.}$$
Thus:
\begin{eqnarray}
   \left<\hat{x}_{1,2}\right> & = & \int^{L}_{0}{\frac{xdx}{L}} = L/2\nonumber\\
	\left<\hat{x}_{1,2}^2\right> & = & \int^{L}_{0}{\frac{x^2dx}{L}} = L^2/3\nonumber\\
	\sigma^2_{1,2} & = & \left<x_{1,2}^2\right> - \left<x_{1,2}\right>^2 = L^2/12\nonumber
\end{eqnarray}

Now, the relative location is defined as $\hat{r} = \hat{x}_1 - \hat{x}_2$, so by using the rules for adding expectation values and variances we get:
$$ \left<\hat{r}\right> = \left<\hat{x}_1\right> - \left<\hat{x}_2\right> = 0 $$
$$ \sigma_{r}^2 = \sigma^2_1 + \sigma^2_2 = L^2/6 $$

\Dn\Dn

(2) We need to calculate $P\left(r<\hat{r}<r+dr\right)$:
$$P\left(r<\hat{r}<r+dr\right) = \iint_{r<\left(x_1-x_2\right)<r+dr}{f\left(x_1\right)f\left(x_2\right)dx_1dx_2} \Dn = \frac{1}{L^2} \iint_{\begin{array}{c}
    _{r<\left(x_1-x_2\right)<r+dr} \\ 
	 _{0<{x_1,x_2}<L}
\end{array}} {dx_1dx_2} $$

To write the integration limits explicitly we divide to two cases:

\Dn

\ (a) \underline{$r > 0$}:
\ \ $ x_2 + r = x_1 < L\ \Rightarrow\ x_2 < L - r$
\begin{eqnarray}
&&\ \ \ \ P\left(r<\hat{r}<r+dr\right) = \frac{1}{L^2} \int_{0}^{L-r}{dx_2} \int_{x_2+r}^{x_2+r+dr}{dx_1} = \frac{dr}{L^2}\left(L-r\right) \nonumber
\end{eqnarray}

\Dn

\ (b) \underline{$r < 0$}:
\ \ $ x_2 + r = x_1 > 0\ \Rightarrow\ x_2 > -r$
\begin{eqnarray}
&&\ \ \ \  P\left(r<\hat{r}<r+dr\right) = \frac{1}{L^2} \int_{-r}^{L}{dx_2} \int_{x_2+r}^{x_2+r+dr}{dx_1} = \frac{dr}{L^2}\left(L+r\right) \nonumber
\end{eqnarray}

\Dn
Combining the results we get:
$$f\left(r\right)dr = \frac{dr}{L^2}\left(L - \left|r\right|\right)
\ ;\ r\in[-L,L]$$

And finaly we can calculate:

$$\left<\hat{r}\right> = \int_{-L}^{L}{\left(L-\left|r\right|\right)r\frac{dr}{L^2}} = 0 $$
$$\sigma^2_r =\left<\hat{r}^2\right> = 2\int_{0}^{L}{\left(L - r\right)r^2\frac{dr}{L^2}} = L^2/6 $$


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





