# 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:

1. $$E^2-c^2B^2=0$$

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

3. $$E^2-c^2B^2<0$$

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?

• What sort of "explanation" are you looking for? They're just inequalities, it's not really clear to me what you want to know about them, could you elaborate? Apr 7 '17 at 10:57
• I just wanted to know what physically happens in these cases, and if you can give a more detailed mathematical justification of what is present on the site Why is this invariant in Relativity: $E^2−c^2B^2$? Apr 7 '17 at 20:09
• This seems a perfectly reasonable question to me. $E^2-c^2B^2=0$ implies no charge is present so the obvious example is an EM wave. The question is asking what sort of physical systems result in the two signs for the invariant. Apr 8 '17 at 8:28

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.