\documentclass[12pt]{article} % BB APr 01 2010.
\usepackage{epsfig}
\textheight 9.0in \textwidth  6.0in \topmargin -0.4in \headsep 0.4in  
\oddsidemargin  0.25in  \evensidemargin 0.25in  

\def\sqr#1#2{{\vcenter{\hrule height.#2pt
   \hbox{\vrule width.#2pt height#1pt \kern#1pt
   \vrule width.#2pt} \hrule height.#2pt}}}
\def\square{{\mathchoice\sqr64\sqr64\sqr{4.2}3\sqr{3.0}3}}
\def\undertext#1{$\underline{\smash{\hbox{#1}}}$}
\pagestyle{empty}

\begin{document}

\begin{center}
{\bf ADVANCED DYNAMICS --- PHY 4241/5227}\\\vspace {2mm} 
{\bf SOLUTIONS -- SET 13}\\%
\end{center}
\setcounter{equation}{0}

\noindent
Problem 46: The equation for the redshift is
\begin{eqnarray} \nonumber
  \lambda' &=& \lambda\,\sqrt{\frac{1+\beta}{1-\beta}}\\ \nonumber
  \left(\frac{\lambda'}{\lambda}\right)^2 &=& {\frac{1+\beta}{1-\beta}}
  \\ \nonumber
  \left(\frac{\lambda'}{\lambda}\right)^2 -\beta\,
  \left(\frac{\lambda'}{\lambda}\right)^2 &=& 1+\beta
  \\ \nonumber
  \left(\frac{\lambda'}{\lambda}\right)^2 - 1 &=& \beta\,\left[ 1
  + \left(\frac{\lambda'}{\lambda}\right)^2 \right]
  \\ \nonumber
  \beta &=& \frac{(\lambda'/\lambda)^2-1}{(\lambda'/\lambda)^2+1}
\end{eqnarray}
With $\lambda'=729.2$ and $\lambda=364.56$ we find
$$ \beta\ =\ \frac{v}{c}\ =\ 0.6\,.$$

% \begin{figure}[tb] \vspace{5pc}
% \centerline{\hbox{ \psfig{figure=figs/ladder.eps,width=7.5cm} }}
% \caption{Ladder for assignment~14.} \label{fig_lader} \end{figure}

\end{document}
