\documentclass[12pt]{article} % BB Jan 11 2013.
\textheight 9.0in
\textwidth  6.0in
\topmargin -1.0in  
\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} \huge

\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
{\bf Einstein Convention Homework 1} \\ 
\end{center}
\setcounter{equation}{0}

\begin{enumerate}

\item Use the $3D$ identity
\begin{eqnarray} \label{epsdel}
  \epsilon_{ijk}\epsilon_{ilm} =
  \delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}
\end{eqnarray} 
to calculate
\begin{eqnarray} 
  (\hat{a}\times\hat{b})\cdot(\hat{a}\times\hat{b})
\end{eqnarray} 
where $\hat{a}$ and $\hat{b}$ are unit vectors. Eliminate all 
$\hat{a}\cdot\hat{b}$ in favor of $\cos\theta$. 

Solution: Let
\begin{eqnarray} \label{hataxhatb}
  \hat{a}=\left(\matrix{\tilde{a}_1\cr\tilde{a}_2\cr\tilde{a}_3}\right)
  ~~{\rm and}~~ \hat{b}=
  \left(\matrix{\tilde{b}_1\cr\tilde{b}_2\cr\tilde{b}_3}\right)\,.
\end{eqnarray} 
Then
\begin{eqnarray} \nonumber
  &~& (\hat{a}\times\hat{b})\cdot(\hat{a}\times\hat{b})\ =\ 
  \epsilon_{ijk}\epsilon_{ilm}\tilde{a}_j\tilde{b}_k
  \tilde{a}_l\tilde{b}_m \\ \nonumber \\ \nonumber
  &=& \left(\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}\right)\,
  \tilde{a}_j\tilde{b}_k\tilde{a}_l\tilde{b}_m \\ \nonumber
  \\ \nonumber
  &=& \left(\tilde{a}_l\tilde{b}_m\tilde{a}_l\tilde{b}_m -
  \tilde{a}_m\tilde{b}_l\tilde{a}_l\tilde{b}_m\right)\\ \nonumber
  \\ \nonumber &=& 
  \left(\hat{a}\cdot\hat{a}\right)\,\left(\hat{b}\cdot\hat{b}\right)-
  \left(\hat{a}\cdot\hat{b}\right)\,\left(\hat{b}\cdot\hat{a}\right)\\
  &=& 1 - \cos^2(\theta)\ =\ \sin^2(\theta)\,.
\end{eqnarray} 

\end{enumerate}

\begin{center}
{\bf Einstein Convention Homework 2} \\ 
\end{center}

\noindent 1. Use the identity (\ref{epsdel}) to eliminate the vector 
products from the expression
\begin{eqnarray} 
  \vec{a}\times\left(\vec{b}\times\vec{c}\right)
\end{eqnarray} 
Solution (compare book p.33):
\begin{eqnarray} \nonumber
  &~&\vec{a}\times\left(\vec{b}\times\vec{c}\right)  \\ \nonumber
  &=&\epsilon_{ijk} \hat{x}_i a_j\epsilon_{klm} b_l c_m \\ \nonumber
  &=&\epsilon_{kij} \epsilon_{klm} \hat{x}_i a_j b_l c_m \\ \nonumber
  \\ \nonumber
  &=&\left(\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}\right)
     \hat{x}_ia_jb_lc_m \\ \nonumber \\ \nonumber
  &=& \hat{x}_ia_jb_ic_j - \hat{x}_ia_jb_jc_i 
      \\ \nonumber \\ \nonumber
  &=&\vec{b}\,\left(\vec{a}\cdot\vec{c}\right) -
     \vec{c}\,\left(\vec{a}\cdot\vec{b}\right)\,. 
\end{eqnarray} 

\noindent 2. Use the definition
\begin{eqnarray} 
  \left(\vec{b}\times\vec{c}\right)_i\ =\ \epsilon_{ijk} b_j c_k 
\end{eqnarray} 
of the $i^{th}$ component of the vector product $\vec{b}\times\vec{c}$
to proof
\begin{eqnarray} 
  \vec{a}\cdot\left(\vec{b}\times\vec{c}\right) =
  \vec{b}\cdot\left(\vec{c}\times\vec{a}\right) =
  \vec{c}\cdot\left(\vec{a}\times\vec{b}\right)\,.
\end{eqnarray} 
Solution (compare book p.30):
\begin{eqnarray} \nonumber
 &~&\vec{a}\cdot\left(\vec{b}\times\vec{c}\right) =
    \epsilon_{ijk}a_ib_jc_k \\ \nonumber 
    \\ \nonumber \\ \nonumber
 &=& \epsilon_{jki}b_jc_ka_i =
     \vec{b}\cdot\left(\vec{c}\times\vec{a}\right) \\ \nonumber
    \\ \nonumber \\ \nonumber
 &=& \epsilon_{kij}c_ka_ib_j =
     \vec{c}\cdot\left(\vec{a}\times\vec{b}\right)\,.
\end{eqnarray} 

\end{document}
