Three questions and explanations for the Lorentz invariant $E^2-c^2B^2$ It is demonstrated that the square trace of the electromagnetic tensor is nothing and it is valid:
$$
\mathrm{Tr}\,{F}^2_{\mu\nu}=\frac{2}{c^{2}}(E^2-c^2B^2).
$$
Proof: $F_{\mu\nu}=-F_{\nu\mu}$, hence
$$
\mathrm{Tr}\,{F}^2_{\mu\nu}=\sum_{\mu}\left(F^{2}\right)_{\mu\mu}=-\sum_{\mu\nu}F_{\mu\nu}F_{\nu\mu}=-\sum_{\mu\nu}F_{\mu\nu}^{2}=
$$
$$
=-2\left[B_{1}^{2}+B_{2}^{2}+B_{3}^{2}-\frac{1}{c^{2}}\left(E_{1}^{2}+E_{2}^{2}+E_{3}^{2}\right)\right]=
$$
$$=-\frac{2}{c^{2}}\left(B^2-\frac{E^2}{c^{2}}\right)=\frac{2}{c^{2}}\left(E^2-c^{2}B^2\right)$$
I have seen, also, this explanation of Lorentz invariant $E^2-c^2B^2$:

After, on the site Why is this invariant in Relativity: $E^2−c^2B^2$?
there are limited informations, mathematical and physical, for the following relationships:


*

*$E^2-c^2B^2=0$

*$E^2-c^2B^2>0$ 

*$E^2-c^2B^2<0$
For item 2.) $E^2-c^2B^2>0$ in $\Sigma$. Then there will be a reference system of $\Sigma'$ such that $\overline{B}'=\textbf{0}$ i.e. the interaction is purely electric. Why?
For item 1.) $E^2-c^2B^2=0$ in $\Sigma$ is the case with a plane wave: why? We can also say that if we have a plane wave in an inertial reference $\Sigma$ we will still find a plane wave in any other inertial reference $\Sigma'$.
For item 3.) $E^2-c^2B^2<0$ in $\Sigma$. Both $\overline{E}$ and $\overline{B}$ are different from zero in each reference system (otherwise both must be null and therefore there would be no electromagnetic wave). An example is a wire with current? It is correct and why?
 A: With tensor $F^{\mu \nu}$ we can make a Lorentz transformation (in the $x$-direction)
$$F'^{\mu \nu} = \Lambda^{\mu}_{\rho}\Lambda^{\nu}_{\sigma}F^{\rho \sigma}$$
then $F^{\mu \nu}$, its dual $\star F^{\mu \nu}$ and $\eta^{\mu \nu}$ are all tensors, so we can form scalars from them. The most obvious case is $F^{\mu \nu}\eta^{\mu \nu}=0$, but interestingly
\begin{align}
F^{\mu \nu}F_{\mu \nu} &= F^{0 \nu}F_{0 \nu} + F^{1 \nu}F_{1 \nu}  + F^{2 \nu}F_{2 \nu}  + F^{3 \nu}F_{3 \nu} \\
&= -F^{0 j}F^{0 j} - F^{1 0}F^{1 0}  + F^{1 2}F^{1 2}  + F^{1 3}F_{13}+ \ldots \\
&= \frac{1}{c^{2}}\left( -\overrightarrow{E^{2}}-E_{x}^{2}+c^{2}(B_{x}^{2}+B_{y}^{2})-E_{y}^{2}+c^{2}(B_{x}^{2}+B_{z}^{2})-E_{z}^{2}+c^{2}(B_{y}^{2}+B_{x}^{2})\right) \\
&= \frac{2}{c^{2}}\left( (c \overrightarrow{B^{2}})-\overrightarrow{E^{2}} \right)
\end{align}
So, if $E^{2} > cB^{2}$ we can choose a frame, $S'$ such that $B'=0$, if $E^{2} < cB^{2}$ we can choose a frame, $S'$ such that $E'=0$.
Alternatively, one could say that for $E^{2} > cB^{2}=E'^{2} > cB'^{2}$ so the frame with $B'=0$ gives $E^{2} > cB^{2}$ and the inequality is reversed if $B'=0$.
Using Hodge dual,
$$F^{\mu \nu}\star F^{\mu \nu} = \frac{-4}{c}\overrightarrow{E} \cdot \overrightarrow{B}$$
following the same rationale as above, one can show that $\overrightarrow{E} \cdot \overrightarrow{B}$ is invariant.
Note that a frame $S′$ can be chosen in which $\overrightarrow{E}′$ or $\overrightarrow{B}′$ is zero only if $\overrightarrow{E}′ \perp \overrightarrow{B}′$ in
S. 
