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\begin{document} \Large

\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
{Solutions Final (May 2, 2013)}  % \\
\end{center}
\setcounter{equation}{0}

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

\bigskip

Einstein's convention is used in the following problems and
$\epsilon_{ijk}$ is the Levi-Civita tensor. 

\begin{enumerate}

\item Calculate
\begin{eqnarray} 
  \epsilon_{12k}\,\epsilon_{21k} &=& ~~~~~~~~~~~~~~\\
  \epsilon_{ij3}\,\epsilon_{ij3} &=& \\
  \epsilon_{ijk}\,\epsilon_{ijk} &=& 
\end{eqnarray} 
Solution:
\begin{eqnarray} \nonumber
  \epsilon_{12k}\,\epsilon_{21k} &=& \epsilon_{123}\,\epsilon_{213}\ 
  =\ -1\,,\\ \nonumber
  \epsilon_{ij3}\,\epsilon_{ij3} &=& \epsilon_{123}\,\epsilon_{123} 
  + \epsilon_{213}\,\epsilon_{213}\ =\ +2\,,\\ \nonumber
  \epsilon_{ijk}\,\epsilon_{ijk} &=& 3!\ =\ +6\,.
\end{eqnarray} 

\item Rewrite the expression
\begin{eqnarray} \label{curlAxB}
  &\epsilon_{ijk}\,\epsilon_{klm}\,\hat{x}_i\,\partial_j\,(A_l\,B_m)\,,&
  \\ \nonumber &{\rm where\ the}\ \hat{x}_i\ {\rm are\ Cartesian\ unit\
  vectors\ and}\ \partial_j=\frac{\partial~}{\partial x_j}\,,\ \rm into&
\end{eqnarray} 

(A) A vector product. \medskip

(B) A sum of scalar products. \bigskip

Solution: \medskip

(A) The expression (\ref{curlAxB}) is the definition of
\begin{eqnarray} \nonumber
  \nabla\times \left(\vec{A}\times\vec{B}\right)\ .
\end{eqnarray} 

(B) It can be transformed into a sum of scalar products:
\begin{eqnarray} \nonumber
  \epsilon_{ijk}\,\epsilon_{klm}\,\hat{x}_i\,\partial_j\,(A_l\,B_m)
  &=& \left(\delta_{il}\,\delta_{jm}-\delta_{im}\,\delta_{jl}\right)\,
  \hat{x}_i\,\partial_j\,(A_l\,B_m)\\ \nonumber &=&
  \hat{x}_i\,\partial_j\,(A_i\,B_j)-\hat{x}_i\,\partial_j\,(A_j\,B_l)
  \\ \nonumber =\
  \left(\vec{B}\cdot\nabla\right)\,\vec{A}+
  \vec{A}\,\left(\nabla\cdot\vec{B}\right)&-&
  \left(\vec{A}\cdot\nabla\right)\,\vec{B}-
  \vec{B}\,\left(\nabla\cdot\vec{A}\right)\,.
\end{eqnarray} 
\bigskip

\bigskip
\item Consider spherical coordinates. Calculate the time derivative
$$ \dot{\hat{\theta}}\ =\ \frac{d\,\hat{\theta}}{d\,t}$$
of the unit vector $\hat{\theta}$ and express the result in spherical 
coordinates. 
\bigskip

\noindent Solution: From the definitions
\begin{eqnarray} \nonumber
  \hat{r}&=&\sin(\theta)\,\cos(\phi)\,\hat{x} +
            \sin(\theta)\,\sin(\phi)\,\hat{y} +
            \cos(\theta)\,\hat{z}\,, \\ \nonumber
  \hat{\theta}&=&\cos(\theta)\,\cos(\phi)\,\hat{x} +
                 \cos(\theta)\,\sin(\phi)\,\hat{y} -
                 \sin(\theta)\,\hat{z}\,, \\ \nonumber
  \hat{\phi}&=& -\sin(\phi)\,\hat{x}+\cos(\phi)\,\hat{y}
\end{eqnarray} 
we get
\begin{eqnarray} \nonumber
  \frac{\partial\,\hat{\theta}}{\partial\,r} &=& 0\,, \\ \nonumber
  \frac{\partial\,\hat{\theta}}{\partial\,\theta}
   &=&-\sin(\theta) \cos(\phi)\,\hat{x} -
        \sin(\theta) \sin(\phi)\,\hat{y} \\ \nonumber
   &~&-\cos(\theta)\,\hat{z}\ =\ -\hat{r}\,,\\ \nonumber
  \frac{\partial\,\hat{\theta}}{\partial\,\phi} 
  &=&-\cos(\theta)\,\sin(\phi)\,\hat{x} +
      \cos(\theta)\,\cos(\phi)\,\hat{y} \\ \nonumber
  &=& \cos(\theta)\,\hat{\phi}\,.
\end{eqnarray} 
Therefore, 
\begin{eqnarray} \nonumber
  \dot{\hat{\theta}}\ =\ 
  \frac{\partial\,\hat{\theta}}{\partial\,\theta}\,\dot{\theta} 
  + \frac{\partial\,\hat{\theta}}{\partial\,\phi}\,\dot{\phi}\ =\ 
  - \dot{\theta}\,\hat{r} + \cos(\theta)\,\dot{\phi}\,\hat{\phi}\,.
\end{eqnarray} % \vfil\eject 
\bigskip

\bigskip
\item Find the eigenvalues and (normalized) eigenvectors of the matrix
\begin{eqnarray} \nonumber
  \left( \begin{matrix} +1  &  -1  &   0  \cr
                  -1  &  +2  &  -1  \cr
                   0  &  -1  &  +1  \end{matrix} \right)\ .
\end{eqnarray} % \vfil\eject 
\bigskip

\noindent Solution: The characteristic equation of the eigenvalue 
problem is
\begin{eqnarray} \nonumber % \label{det}
  {\rm det}\, \left| \begin{matrix} +1-\lambda  &  -1  &   0  \cr
                  -1  &  +2-\lambda  &  -1  \cr
                   0  &  -1  &  +1-\lambda  \end{matrix} \right|
  &=& -\lambda^3+4\,\lambda^2-3\,\lambda\ =\ 0\
\end{eqnarray} % \vfil\eject 
and the first eigenvalue is seen to be
\begin{eqnarray} \label{la1}
  \lambda_1 &=& 0\,.
\end{eqnarray} % \vfil\eject 
Dividing $\lambda$ out gives the quadratic equation
\begin{eqnarray} \nonumber
  \lambda^2-4\,\lambda+3 &=& 0
\end{eqnarray} % \vfil\eject 
from which the eigenvalues
\begin{eqnarray} \label{la23}
  \lambda_{2,3} &=& 2 \pm \sqrt{4-3}\ =\ 2\pm 1\ =\
  \begin{cases} 1\,,\cr 3\,.\end{cases}
\end{eqnarray} % \vfil\eject 
follow.
\bigskip

The eigenvector for $\lambda_1=0$ follows from
\begin{eqnarray} \nonumber
  \left( \begin{matrix} +1  &  -1  &   0  \cr
                  -1  &  +2  &  -1  \cr
                   0  &  -1  &  +1  \end{matrix} \right)\, \left( 
  \begin{matrix} v_1^1\cr v_1^2\cr v_1^3  \end{matrix} \right) &=& 0\,.
\end{eqnarray} % \vfil\eject 
and
\begin{eqnarray} \nonumber
   v_1^1  -v_1^2        &=& 0\ \Rightarrow\ v_1^2=v_2^1\,,\\ \nonumber
  -v_1^2 +2v_1^2 -v_1^3 &=& 0\\ \nonumber
          -v_1^2 +v_1^3 &=& 0\ \Rightarrow v_1^3=v_1^1
\end{eqnarray} % \vfil\eject 
and the middle equation gives then $v_1^2=v_1^1$. Up to a $\pm$ 
convention the first eigenvector becomes
\begin{eqnarray} \nonumber
  \hat{v}_1 &=& \frac{1}{\sqrt{3}}\,\left(
  \begin{matrix} 1\cr 1\cr 1\end{matrix} \right)\,.
\end{eqnarray} % \vfil\eject 
\bigskip

The eigenvector for $\lambda_2=1$ follows from
\begin{eqnarray} \nonumber
  \left( \begin{matrix}  0  &  -1  &   0  \cr
                        -1  &  +1  &  -1  \cr
                         0  &  -1  &   0  \end{matrix} \right)\, \left( 
  \begin{matrix} v_2^1\cr v_2^2\cr v_2^3  \end{matrix} \right) &=& 0\,.
\end{eqnarray} % \vfil\eject 
and
\begin{eqnarray} \nonumber
          -v_2^2        &=& 0\ \Rightarrow\ v_2^2=0\,,\\ \nonumber
  -v_2^2 + 0     -v_2^3 &=& 0\ \Rightarrow\ v_2^3=-v_2^1\\ \nonumber
          -v_2^2        &=& 0\,.
\end{eqnarray} % \vfil\eject 
Up to a $\pm$ convention the second eigenvector becomes
\begin{eqnarray} \nonumber
  \hat{v}_2 &=& \frac{1}{\sqrt{2}}\,\left(
  \begin{matrix} +1\cr 0\cr -1\end{matrix} \right)\,.
\end{eqnarray} % \vfil\eject 
\bigskip

The eigenvector for $\lambda_3=3$ follows from
\begin{eqnarray} \nonumber
  \left( \begin{matrix} -2  &  -1  &   0  \cr
                        -1  &  -1  &  -1  \cr
                         0  &  -1  &  -2  \end{matrix} \right)\, \left( 
  \begin{matrix} v_3^1\cr v_3^2\cr v_3^3  \end{matrix} \right) &=& 0\,.
\end{eqnarray} % \vfil\eject 
and
\begin{eqnarray} \nonumber
  -2v_3^1-v_3^2        &=& 0\ \Rightarrow\ v_3^2=-2v_3^1\,,\\ \nonumber
   -v_3^1-v_3^2 -v_3^3 &=& 0\ \Rightarrow\ v_3^3=-v_3^1-v_3^2=
   -v_3^1+2v_3^1=v_3^1\\ \nonumber -v_3^2 -2v_3^3 &=& 0\,.
\end{eqnarray} % \vfil\eject 
Up to a $\pm$ convention the third eigenvector becomes
\begin{eqnarray} \nonumber
  \hat{v}_3 &=& \frac{1}{\sqrt{6}}\,\left(
  \begin{matrix} +1\cr -2\cr +1\end{matrix} \right)\,.
\end{eqnarray} % \vfil\eject 
\bigskip

Consistency check: $\hat{v_i}\cdot\hat{v}_j = \delta_{ij}$.
\end{enumerate}

\end{document}

