What Is This?

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.  $e^{-i\pi/2}=-i$