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\begin{document} \huge

\begin{center}
{Mathematical Physics --- PHZ 3113}\\\vspace {2mm} 
Curl; Vector Integration Homework (February 7, 2013) \\
\end{center}
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The dual of the Euclidean electromagnetic field tensor
is defined by
\begin{eqnarray} \label{dFmn}
   ^*F_{\alpha\beta} = \frac{1}{2}\,\epsilon_{\alpha\beta\mu\nu}
     F_{\mu\nu}
\end{eqnarray} 
where the indices run from 1 to 4, the Einstein convention is used 
and $F_{\mu\nu}$ is an anti-symmetric (i.e., $F_{\mu\nu}=-F_{\mu\nu}$) 
rank two tensor
\begin{eqnarray} \nonumber
  \left(F_{\mu\nu}\right)\ =\ \left( \matrix{ 
           0   & F_{12} & F_{13} & F_{14} \cr
        -F_{12}&   0    & F_{23} & F_{24} \cr
        -F_{13}& -F_{23}&   0    & F_{34} \cr
        -F_{14}& -F_{24}& -F_{34}& 0    } \right)\ .
\end{eqnarray} 

\begin{enumerate}

\item Calculate the elements of $^*F_{\alpha\beta}$ in terms of the 
independent elements of $F_{\mu\nu}$ and write $^*F_{\alpha\beta}$
as a matrix.

Solution:
\begin{eqnarray} \nonumber
  ^*F_{12}&=&\frac{1}{2}(\epsilon_{1234}F_{34}+\epsilon_{1243}F_{43})
  \ =\ +F_{34} \\ \nonumber
  ^*F_{13}&=&\frac{1}{2}(\epsilon_{1324}F_{24}+\epsilon_{1342}F_{42})
  \ =\ -F_{24} \\ \nonumber
  ^*F_{14}&=&\frac{1}{2}(\epsilon_{1423}F_{23}+\epsilon_{1432}F_{32})
  \ =\ +F_{23} \\ \nonumber
  ^*F_{23}&=&\frac{1}{2}(\epsilon_{2314}F_{14}+\epsilon_{2341}F_{41})
  \ =\ +F_{14} \\ \nonumber
  ^*F_{24}&=&\frac{1}{2}(\epsilon_{2413}F_{13}+\epsilon_{2431}F_{31})
  \ =\ -F_{13} \\ \nonumber
  ^*F_{34}&=&\frac{1}{2}(\epsilon_{3412}F_{12}+\epsilon_{3421}F_{21})
  \ =\ +F_{12} 
\end{eqnarray} 
\begin{eqnarray} \nonumber
  \left( ^*F_{\alpha\beta} \right) &=& \left( \matrix{ 
           0   & +F_{34}& -F_{24} & +F_{23} \cr
        -F_{34}&   0    & +F_{14} & -F_{13} \cr
        +F_{24}& -F_{14}&    0    & +F_{12} \cr
        -F_{23}& +F_{13}& -F_{12}& 0    } \right)\ .
\end{eqnarray} \bigskip

\item The electromagnetic field tensor can be written in form of
derivatives of a 4-potential 
\begin{eqnarray} \label{Fmn}
  F_{\mu\nu}\ =\ \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}\,.
\end{eqnarray} 
Use this expression to proof the homogeneous Maxwell equations 
in their form
\begin{eqnarray} \label{ddFmn}
  \partial_{\alpha} ^*F_{\alpha\beta}\ =\ 0\,.
\end{eqnarray} 

Solution:
\begin{eqnarray} \nonumber
  \partial_{\alpha} ^*F_{\alpha\beta} &=& \frac{1}{2}\,
  \epsilon_{\alpha\beta\mu\nu}\,\partial_{\alpha}\,F_{\mu\nu}
  \\ \nonumber &=& \frac{1}{2}\,\epsilon_{\alpha\beta\mu\nu}\,
  \partial_{\alpha}\,\left(\partial_{\mu} A_{\nu} -
                           \partial_{\nu} A_{\mu} \right)
\end{eqnarray} 
Now
\begin{eqnarray} \nonumber
  \epsilon_{\alpha\beta\mu\nu}\,\partial_{\alpha}\,\partial_{\mu} 
  \,A_{\nu} &=& 0 \\ \nonumber
  \epsilon_{\alpha\beta\mu\nu}\,\partial_{\alpha}\,\partial_{\nu} 
  \,A_{\mu} &=& 0
\end{eqnarray} 
because each time an antisymmetric tensor is contracted with 
a symmetric tensor ($\partial_{\alpha}\,\partial_{\mu}$ and 
$\partial_{\alpha}\,\partial_{\mu}$). This proofs (\ref{ddFmn}).
\bigskip

Remark: Therefore (\ref{Fmn}) implies that no magnetic monopoles 
exist, because (\ref{ddFmn}) includes (upon identification of the
$\vec{E}$ and $\vec{B}$ fields) the relation $\nabla\cdot\vec{B}
=0$.

\end{enumerate}

\end{document}
