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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2017 
Solutions} \medskip

\centerline{\bf Set 5} 

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\item{\bf 21. Electromagnetic field tensor in $\vec{E}$ and $\vec{B}$ 
fields.}
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\noindent A. From the $J^4$ component we get
$$ \partial_4E^{44}+\partial_1E^{14}+\partial_2E^{24}+\partial_3E^{34}=
   \frac{4\pi}{c}\,J^4 = 4\pi\,\rho = \nabla\cdot\vec{e} =
   \partial_1e^1+\partial_2e^2+\partial_3e^3\,. $$
Therefore, $E^{44}=0$ by anti-symmetry, $E^{14}=e^1$, $E^{24}=e^2$ and
$E^{34}=e^3$.

From the $J^1$ component we get
$$ \partial_4E^{41}+\partial_1E^{11}+\partial_2E^{21}+\partial_3E^{31}=
   \frac{4\pi}{c}\,J^1 = \partial_2b^3-\partial_3b^2 - \partial_4e^1\,.$$
Therefore, $E^{41}=-e^1$ consistent with $E^{14}=e^1$, $E^{11}=0$ by 
anti-symmetry, $E^{21}=b^3$ and $E^{31}=-b^2$.

From the $J^2$ component we get
$$ \partial_4E^{42}+\partial_1E^{12}+\partial_2E^{22}+\partial_3E^{32}=
   \frac{4\pi}{c}\,J^2 = \partial_3b^1-\partial_1b^3 - \partial_4e^2\,.$$
Therefore, $F^{42}=-e^2$, $E^{12}=-b^3$ both consistent with anti-symmetry,
$E^{22}=0$ by anti-symmetry and (new) $E^{32}=b^1$.

Using anti-symmetry all components are now determined and the $F^{\alpha\beta}$
field tensor reads
$$ \left(E^{\alpha\beta}\right)=\left(\begin{matrix} 
    0  & -b^3&  b^2&  e^1\cr
    b^3&  0  & -b^1&  e^2\cr
   -b^2&  b^1&  0  &  e^3\cr
   -e^1& -e^2& -e^3&  0  \cr\end{matrix}\right)\,.$$
The remaining equations from $\beta=3$ are not needed, but can be used 
for consistency checks.
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