We have a problem,
\(\cfrac { \partial \, B }{ \partial \, t } =-i\cfrac { \partial \, E }{ \partial x^{ ' } } \)
\( B=-i\int {\cfrac { \partial \, t }{ \partial x^{ ' } } \cfrac { \partial \, E }{ \partial x^{ ' } } } \cfrac { \partial \, x^{ ' } }{ \partial \,t }d\, t\)
\( B=-i\int { \cfrac { \partial \, t }{ \partial x^{ ' } }\partial \, E } \)
\( B=\cfrac { -i\,E }{ c } \) assuming \(B=0\) when \(E=0\)
where \(-i\) rotates \(E\) into the \(x^{'}\) direction which is the direction of \(B\).
Although, given,
\(\cfrac{1}{c}=\sqrt{\mu_o\varepsilon_o}\), we have
\( B^2=\cfrac { (-i)^2}{ c^2 }E^2 \)
\( B^2=-\mu_o\varepsilon_oE^2 \)
\( B^2=\mu_o\varepsilon_oE^2e^{i\pi} \)
or,
\( B^2=\mu_o\varepsilon_o(Ee^{i\pi/2})^2 \)
that \(B\) and \(E\) are \(\pi/2\) out of phase. And so,
\(\cfrac{1}{2\mu_o}\left|B\right|^2=\cfrac{1}{2}\varepsilon_o\left|E\right|^2 \)
which is consistent. Euler strikes again. \(e^{-i\pi/2}=-i\)
