% Bernd Berg Oct 07 2017.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2017 }
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\centerline{\bf Set 6} \smallskip

19. Dual Tensor.  \smallskip

From $^*E^{\alpha\beta}=\frac{1}{2}\,\epsilon^{\alpha\beta\gamma\delta}
E_{\gamma\delta}$ we obtain for the dual tensor
$$  ^*E^{12}=E_{34},~^*E^{13}=-E_{24},~^*E^{14}=E_{23}\,,$$
$$  ^*E^{23}=E_{14},~^*E^{24}=-E_{13},~^*E^{34}=E_{12}\,.$$
In matrix form the dual tensor reads then
$$ \left(^*E^{\alpha\beta}\right)=\left(\begin{matrix} 
   0        & ^*E^{12} & ^*E^{13} & ^*E^{14} \cr
   ^*E^{21} &  0       & ^*E^{23} & ^*E^{24} \cr
   ^*E^{31} & ^*E^{32} &  0       & ^*E^{34}\cr
   ^*E^{41} & ^*E^{42} & ^*E^{43} &  0\end{matrix}\right) =
    \left(\begin{matrix} 
    0      &  E_{34} & -E_{24} &  E_{23}\cr
   -E_{34} &  0      &  E_{14} & -E_{13}\cr
    E_{24} & -E_{14} &  0      &  E_{12}\cr
   -E_{23} &  E_{13} & -E_{12} &  0\cr\end{matrix}\right)\,.$$
Using the $E_{\mu\nu}$ equations of (7.32) this becomes  
$$ \left(^*E^{\alpha\beta}\right)=\left(\begin{matrix} 
    0   & -e_3 &  e_2 & -b_1 \cr
    e_3 &  0   & -e_1 & -b_2 \cr
   -e_2 &  e_1 &  0   & -b_3 \cr
    b_1 &  b_2 &  b_3 &  0 \cr\end{matrix}\right)\,.$$
The equation $\left.^*E^{\alpha\,4}\right._{,\alpha}=0$ is
the homogeneous Maxwell equation 
$$\nabla\cdot\vec{b}=0\,.$$
The other three equations $\left.^*E^{\alpha\,i}\right._{,\alpha}=0$,
$i=1,2,3$ yield the homogeneous Maxwell equation
$$ \nabla\times\vec{e}+\frac{1}{c}\,\frac{\partial\vec{b}}{\partial t}
   =0\,.$$
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