\documentclass[12pt]{article}
\textheight 9.5in
\textwidth  6.3in
\topmargin -0.4in  
\headsep    0.4in  
\oddsidemargin  0.25in  
\evensidemargin 0.25in  

\def\pmbf#1{{\mbox{\boldmath${#1}$}}}
\def\undertext#1{$\underline{\smash{\hbox{#1}}}$}

\pagestyle{empty}
\begin{document}

%----------------------------------------------------%
%%%%%%%%%%%%%%%%% CLASSICAL MECHANICS %%%%%%%%%%%%%%%%
%----------------------------------------------------%
\begin{center}
{\bf ADVANCED DYNAMICS --- PHY--4241/5227}\\\vspace {2mm} 
{\bf HOMEWORK 5}\\%
\vspace {3mm} (\today) \\
\vspace {1mm}{Due on Monday, February 10, 2003} \\ 
\end{center}
\setcounter{equation}{0}
\bigskip

%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%
\centerline{\bf PROBLEM 13}
\medskip
By using the following definition for the antisymmetric
{\bf Levi-Civita} symbol:
$$
  \varepsilon_{ijk}=\cases{
    +1, & if $ijk$ is an even   permutation of 123; \cr
    -1, & if $ijk$ is an odd \  permutation of 123; \cr
   \phantom{+} 0, & otherwise;}
$$
show, in a line or two, that

\begin{description}
\item{a)} $\nabla\times\Big(\nabla\Phi({\bf x})\Big)
          \!=\!0\;.$

\item{b)} $\nabla\cdot\Big(\nabla\times{\bf A}({\bf x})
          \Big)\!=\!0\;.$

\item{c)} $\nabla\times\Big(\nabla\times{\bf A}({\bf x})
          \Big)\!=\!\nabla\Big(\nabla\cdot{\bf A}({\bf x})
          \Big) \!-\!\nabla^{2}{\bf A}({\bf x})\;.$
\end{description}
\medskip
\noindent
You may find useful the following two important identities:
\begin{eqnarray}
 \varepsilon_{ijk}\varepsilon_{ijl}&=&
  2\delta_{kl}\;, \nonumber \cr
 \varepsilon_{ijk}\varepsilon_{ilm}&=&
 \delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl} \;.
 \nonumber \cr
\end{eqnarray}

\bigskip
%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%
\centerline{\bf PROBLEM 14}
\medskip
The concept of a {\bf gauge transformation} emerges
from the observation that clearly distinct vector
potentials can generate the same exact magnetic
field.

\begin{description}
\item{a)} Show that if two vector potentials,
          ${\bf A}_{1}({\bf x})$ and
          ${\bf A}_{2}({\bf x})$, differ by
          the gradient of a scalar function
          $\Lambda({\bf x})$, then both
          potentials generate the same exact
          magnetic field ${\bf B}({\bf x})$.
          That is, if
          $
           {\bf A}_{2}({\bf x})=
           {\bf A}_{1}({\bf x})+
           \nabla\Lambda({\bf x})
          $
          and
          ${\bf B}({\bf x})\!=\!
           \nabla\times{\bf A}_{1}({\bf x})\;,$
          then
          ${\bf B}({\bf x})\!=\!
           \nabla\times{\bf A}_{2}({\bf x})\;.$

 \medskip
  Now consider the following three vector potentials
  with $B$ a constant:
   \begin{eqnarray}
    {\bf A}_{1}({\bf x})&=&+B
    \Big(x\hat{\bf y}-y\hat{\bf x}\Big)/2\;,
    \nonumber \cr
    {\bf A}_{2}({\bf x})&=&+Bx\hat{\bf y}\;,
    \nonumber \cr
    {\bf A}_{3}({\bf x})&=&-By\hat{\bf x}\;,
    \nonumber \cr
   \end{eqnarray}

\item{b)} Show that all three vector potentials
          satisfy:
          $\nabla\times{\bf A}=B\hat{\bf z}$
          and
          $\nabla\cdot{\bf A}=0$.

\item{c)} Find the three gauge transformations,
          {\it i.e.,} the  three scalar functions
          $\Lambda_{12}({\bf x})$,
          $\Lambda_{23}({\bf x})$, and
          $\Lambda_{31}({\bf x})$,
          that connect the above three gauge-equivalent
          vector potentials.
\end{description}
\bigskip
%%%%%%%%%%%%%%%%%%%%%%%%%%
\centerline{\bf PROBLEM 15}
\medskip

Show that the refraction of light as it passes from one 
medium (with $\varepsilon_{1}$ and $\mu_{1}$) to another 
one (with $\varepsilon_{2}$ and $\mu_{2}$) obeys Fermat's 
principle of least time.
%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%

\end{document}
