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\begin{document} \huge

\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
{\bf Levi-Cevita Tensor 2 Applications} \\ (January 14, 2013) \\
\end{center}
\setcounter{equation}{0}

\rightline{Group \#~~~~~~~~}

\leftline{Participating students (print):} \vskip 60pt

\centerline{------------------------------------------------------}

\bigskip

\begin{enumerate}

\item Write down the values of the cyclic permutations of 
$\epsilon_{123}$ and then of $\epsilon_{213}$. Do you get
all $3D$ values this way? Which are positive and which 
are negative? \bigskip
\begin{eqnarray}
   \epsilon_{123} = +1\,,~ \epsilon_{231} = +1\,,~
   \epsilon_{312} = +1\,,~\\
   \epsilon_{132} = -1\,,~ \epsilon_{321} = -1\,,~
   \epsilon_{213} = -1\,.~
\end{eqnarray}
We get all 6 non-zero values. The cyclic permutations of
$\epsilon_{123}$ are $+1$ and the cyclic permutations of
$\epsilon_{213}$ are $-1$.

\item In $3D$
\begin{eqnarray}
   \sum_{j=1}^3 \sum_{k=1}^3 \epsilon_{1jk}\,a_j\,b_k 
   &=& a_2b_3-a_3b_2\,,~~~\\
   \sum_{j=1}^3 \sum_{k=1}^3 \epsilon_{2jk}\,a_j\,b_k
   &=& a_3b_1-a_1b_3\,,~~~\\
   \sum_{j=1}^3 \sum_{k=1}^3 \epsilon_{3jk}\,a_j\,b_k 
   &=& a_1b_2-a_2b_1~~~\,.
\end{eqnarray}
\item In $3D$ 
\begin{eqnarray}
   \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \epsilon_{ijk}\,
   \hat{x}_i\,\,a_j\,b_k &=& \\ \nonumber
   \hat{x}_1\,(a_2b_3-a_3b_2) &+& \\ \nonumber
   \hat{x}_2\,(a_3b_1-a_1b_3) &+& \\ \nonumber
   \hat{x}_3\,(a_1b_2-a_2b_1) &=& \\ \nonumber
   \vec{a}\times\vec{b}\,. 
\end{eqnarray} \smallskip

{\bf Definition of the determinant of a $nD$ matrix:} \smallskip
\begin{eqnarray} \nonumber
   \left|\matrix{a_{11} & \dots & a_{1n} \cr
                   .    &       &   .    \cr
                   .    &       &   .    \cr
                   .    &       &   .    \cr
                 a_{n1} & \dots & a_{nn}}\right|\ =
   ~~~~~~~~~~~~~~~~~~~~~~~~~\\ \\ \nonumber 
   \sum_{i_1=1}^n \dots \sum_{i_n=1}^n 
   \epsilon_{i_1\dots i_n}\,a_{1i_1}\dots a_{ni_n}\,.
\end{eqnarray} \bigskip

\item In $2D$
\begin{eqnarray}
   \sum_{i=1}^2 \sum_{j=1}^2 \epsilon_{ij}\,a_{1i}a_{2j}
    = \epsilon_{12}a_{11}a_{22} + \\ \nonumber
    \epsilon_{21}a_{12}a_{21} = a_{11}a_{22} - a_{12}a_{21}\,.
\end{eqnarray} 

\item In $3D$ (using cyclic permutations)
\begin{eqnarray}
   \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \epsilon_{ijk}\,
   a_{1i} a_{2j} a_{3k}\ =~~~~~~\\ \nonumber
   +a_{11}a_{22}a_{33}+a_{13}a_{21}a_{32}+a_{12}a_{23}a_{31}
   \\ \nonumber
   -a_{11}a_{23}a_{32}-a_{13}a_{22}a_{31}-a_{12}a_{21}a_{33}
\end{eqnarray} 
Notice that this corresponds to a well known rule for 
evaluating the determinant of a $3\times 3$ matrix.
\bigskip

\item Substitute in the previous expression 
\begin{eqnarray} 
  a_{11}=\hat{x}_1,\ a_{12}=\hat{x}_2,\ a_{13}=\hat{x}_3\,.
\end{eqnarray} 
It becomes
\begin{eqnarray}
   \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \epsilon_{ijk}\,
   \hat{x}_i a_{2j} a_{3k} &=&~~~~~\\ \nonumber
   \hat{x}_1\,(a_{22}a_{33}-a_{23}a_{32}) &+&\\ \nonumber
   \hat{x}_2\,(a_{23}a_{31}-a_{21}a_{33}) &+&\\ \nonumber
   \hat{x}_3\,(a_{21}a_{32}-a_{22}a_{31}) &.&
\end{eqnarray} 

\item Substitute in the previous expression 
\begin{eqnarray} 
  &~& a_{21}=a_1\,,\ a_{22}=a_2\,,\ a_{23}=a_3\,,~~~~~\\ 
  &~& a_{31}=b_1\,,\ a_{32}=b_2\,,\ a_{33}=b_3\,.~~~~~
\end{eqnarray}
It becomes
\begin{eqnarray}
   \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \epsilon_{ijk}\,
   \hat{x}_i a_j a_k &=&~~~~~\\ \nonumber
   \hat{x}_1\,(a_2b_3-a_3b_2) &+&\\ \nonumber
   \hat{x}_2\,(a_3b_1-a_1b_3) &+&\\ \nonumber
   \hat{x}_3\,(a_1b_2-a_2b_1) &.&
\end{eqnarray} 

\item In $3D$, write $\vec{a}\times\vec{b}$ as determinant (book p.23).
\bigskip
\begin{eqnarray} 
  \vec{a}\times\vec{b}=\left|
  \matrix{\hat{x}_1&\hat{x}_2&\hat{x}_3\cr
          a_1      & a_2     & a_3     \cr
          b_1      & b_2     & b_3     } \right|
\end{eqnarray} \bigskip

\item {\bf Proof the $3D$ identity}
\begin{eqnarray} \label{epsdel}
  \sum_{i=1}^3 \epsilon_{ijk}\epsilon_{ilm} =
  \delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}
\end{eqnarray} 
by calculating the expression for the nine possibilites
of values for $jk$, i.e., 11, 12, 13, 21, 22, 23, 31, 
32, 33.
\begin{eqnarray} \nonumber
  \sum_{i=1}^3 \epsilon_{i11}\epsilon_{ilm}&=&0 =
  \delta_{1l}\delta_{1m}-\delta_{1m}\delta_{1l}\,,
  \\ \nonumber
  \sum_{i=1}^3 \epsilon_{i12}\epsilon_{ilm}&=&
  \epsilon_{312}\epsilon_{3lm} = 
  \delta_{1l}\delta_{2m}-\delta_{1m}\delta_{2l}\,.
\end{eqnarray} 
Now there are only two non-zero cases: $lm=12$ and $lm=21$.
One sees immediately that left and right side of the last 
equal sign agree in each case. Similarly, this holds for 
the other values discussed below.
\begin{eqnarray} \nonumber
  \sum_{i=1}^3 \epsilon_{i13}\epsilon_{ilm}&=&
  \epsilon_{213}\epsilon_{2lm} = 
  \delta_{1l}\delta_{3m}-\delta_{1m}\delta_{3l}\,,
  \\ \nonumber
  \sum_{i=1}^3 \epsilon_{i21}\epsilon_{ilm}&=&
  \epsilon_{321}\epsilon_{3lm} = 
  \delta_{2l}\delta_{1m}-\delta_{2m}\delta_{1l}\,,
  \\ \nonumber
  \sum_{i=1}^3 \epsilon_{i22}\epsilon_{ilm}&=&0 =
  \delta_{2l}\delta_{2m}-\delta_{2m}\delta_{2l}\,,
  \\ \nonumber
  \sum_{i=1}^3 \epsilon_{i23}\epsilon_{ilm}&=&
  \epsilon_{123}\epsilon_{1lm} = 
  \delta_{2l}\delta_{3m}-\delta_{2m}\delta_{3l}\,,
  \\ \nonumber
  \sum_{i=1}^3 \epsilon_{i31}\epsilon_{ilm}&=&
  \epsilon_{231}\epsilon_{2lm} = 
  \delta_{3l}\delta_{1m}-\delta_{3m}\delta_{1l}\,,
  \\ \nonumber
  \sum_{i=1}^3 \epsilon_{i32}\epsilon_{ilm}&=&
  \epsilon_{132}\epsilon_{1lm} = 
  \delta_{3l}\delta_{2m}-\delta_{3m}\delta_{2l}\,,
  \\ \nonumber
  \sum_{i=1}^3 \epsilon_{i33}\epsilon_{ilm}&=&0 =
  \delta_{3l}\delta_{3m}-\delta_{3m}\delta_{3l}\,.
\end{eqnarray} 

\end{enumerate}

\end{document}

\item Use the identity (\ref{epsdel}) 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 $\cos\theta$.

Calculation: 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} 
  &~& (\hat{a}\times\hat{b})\cdot(\hat{a}\times\hat{b})\ = \\ \nonumber
  &~& \sum_{i=1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \sum_{l=1}^3
  \sum_{m=1}^3 \epsilon_{ijk}\epsilon_{ilm}
  \tilde{a}_j\tilde{b}_k\tilde{a}_l\tilde{b}_m\ = 
\end{eqnarray} 
\begin{eqnarray} \nonumber
  \sum_{j=1}^3 \sum_{k=1}^3 \sum_{l=1}^3 \sum_{m=1}^3 \left(
  \delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}\right)\,
  \tilde{a}_j\tilde{b}_k\tilde{a}_l\tilde{b}_m\ = \\ \nonumber
\end{eqnarray} 
\begin{eqnarray} \nonumber
  \sum_{l=1}^3 \sum_{m=1}^3 \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
  \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)\
  &=&\\ \nonumber 1 - \cos^2(\theta)\ =\ \sin^2(\theta)\,.&~&
\end{eqnarray} 

\item Show $\vec{a}\times\vec{b}=a\,b\,\sin\theta$ (book p.21).

From (\ref{hataxhatb}) we have $\hat{a}\times\hat{b}=\sin\theta$.
Hence,
\begin{eqnarray} 
  \vec{a}\times\vec{b}\ =\ a\,b\,\hat{a}\times\hat{b}\ =\
  a\,b\,\sin\theta\,.
\end{eqnarray} 
