
\item{\bf Electromagnetic field tensor in $\vec{E}$ and $\vec{B}$ fields.}

\noindent A. From the $J^0$ component we get
$$ \partial_0F^{00}+\partial_1F^{10}+\partial_2F^{20}+\partial_3F^{30}=
   \frac{4\pi}{c}\,J^0 = 4\pi\,\rho = \nabla\cdot\vec{E} =
   \partial_1E^1+\partial_2E^2+\partial_3E^3\,. $$
Therefore, $F^{00}=0$ by anti-symmetry, $F^{10}=E^1$, $F^{20}=E^2$ and
$F^{30}=E^3$.

From the $J^1$ component we get
$$ \partial_0F^{01}+\partial_1F^{11}+\partial_2F^{21}+\partial_3F^{31}=
   \frac{4\pi}{c}\,J^1 = \partial_2B^3-\partial_3B^2 - \partial_0E^1\,.$$
Therefore, $F^{01}=-E^1$ consistent with $F^{10}=E^1$, $F^{11}=0$ by 
anti-symmetry, $F^{21}=B^3$ and $F^{31}=-B^2$.

From the $J^2$ component we get
$$ \partial_0F^{02}+\partial_1F^{12}+\partial_2F^{22}+\partial_3F^{32}=
   \frac{4\pi}{c}\,J^2 = \partial_3B^1-\partial_1B^3 - \partial_0E^2\,.$$
Therefore, $F^{02}=-E^2$, $F^{12}=-B^3$ both consistent with anti-symmetry,
$F^{22}=0$ by anti-symmetry and (new) $F^{32}=B^1$.

Using anti-symmetry all components are now determined and the $F^{\alpha\beta}$
field tensor reads
$$ \left(F^{\alpha\beta}\right)=\left(\begin{matrix} 
    0  & -E^1& -E^2& -E^3\cr
    E^1&  0  & -B^3&  B^2\cr
    E^2&  B^3&  0  & -B^1\cr
    E^3& -B^2&  B^1&  0  \cr\end{matrix}\right)\,.$$
The remaining equations from $\beta=3$ can be used for consistency checks.
\bigskip

\noindent B. For $\beta=0$: 
$$\partial_{\alpha}\,^*F^{\alpha 0}=\partial_1\,B^1
+\partial_2\,B^2+\partial_2\,B^2=\nabla\cdot\vec{B}=0\,.$$
For $\beta=1$: 
$$\partial_{\alpha}\,^*F^{\alpha 1}=-\partial_0\,B^1-\partial_2\,E^3
+\partial_3\,E^2=-\partial_0\,B^1-(\nabla\times\vec{E}\,)^1=0\,.  $$
For $\beta=2$: 
$$\partial_{\alpha}\,^*F^{\alpha 2}=-\partial_0\,B^2+\partial_1\,E^3
-\partial_3\,E^1=-\partial_0\,B^2-(\nabla\times\vec{E}\,)^2=0\,.  $$
For $\beta=3$: 
$$\partial_{\alpha}\,^*F^{\alpha 3}=-\partial_0\,B^3-\partial_1\,E^2
+\partial_2\,E^1=-\partial_0\,B^3-(\nabla\times\vec{E}\,)^3=0\,.  $$
So,
$$ \nabla\cdot\vec{B}=0~~{\rm and}~~-\partial_0\vec{B}-\nabla\times
   \vec{E} = 0 \,.$$
\vfil

\end{enumerate}
\end{document}

\item{\bf Homogeneous Maxwell equations.}
\smallskip \noindent

We choose $\beta =0$ in 
$\partial_{\alpha}\,^*F^{\alpha\beta}=0$ and obtain:
$\partial_{\alpha}\,^*F^{\alpha 0}=0$. Using the definition of the 
dual tensor, this reads
$$ \frac{1}{2}\,\left(\partial_{\alpha}\,
   \epsilon^{\alpha 0\gamma\delta}\,F_{\gamma\delta}\right)=0\,$$
or
$$ \partial_{\alpha}\,F_{\gamma\delta} +
  \partial_{\gamma}\,F_{\delta\alpha} +
  \partial_{\delta}\,F_{\alpha\gamma}\ =\ 0 $$
for all three indices different and not 0. Repeating this for 
$\beta=1,\,2$ and 3, we find that the equation 
hold for all combinations of distinct indices $\alpha,\,\beta$
and $\gamma$. Due to the antisymmetry of electromagnetic 
tensor, the equation still holds when some of the indices
agree. Finally, all indices can of course be raised. This
proves the equivalence of the two versions of the homogeneous
Maxwell equations. More by brute force, one could just write
out all 64 equations.

More elegantly: Use
$$ \frac{1}{2}\,\partial_{\alpha}\,
   \epsilon_{\hat\alpha\beta\hat\gamma\hat\delta}\,
   \epsilon^{\alpha\beta\gamma\delta}\,F_{\gamma\delta} = 0 $$
and expand the contraction in $\beta$ of the two Levi-Civita symbols 
into Kronecker deltas.
