 Solution

Lowering indices with the stated metric reverses the sign of each time-space component and leaves each space-space component unchanged. Contracting the two antisymmetric tensors, counting each unordered index pair twice, givesThese are independent Lorentz scalars; their signs follow from the explicit dual convention in the PDF.

Put $J=E\cdot B>0$ and $D=|E{|}^2-|B{|}^2$. In the parallel-field frame, invariance gives $E^{\prime }{B}^{\prime }=J$ and $E^{\prime 2}-B^{\prime 2}=D$. The positive product forces the two vectors to have the same orientation, so the use of positive magnitudes is legitimate. Substitution of $B^{\prime }=J/E^{\prime }$ gives $E^{\prime 4}-D{E}^{\prime 2}-J^2=0$. Its positive solution is $E^{\prime 2}=(D+\sqrt{D^2+4J^2})/2$. With $K=D/(2J)$,The denominator is positive for every real $K$. The other quadratic root is negative and cannot be a squared magnitude. No explicit boost transformation is needed because the two invariants determine the magnitudes uniquely in a frame where the fields are parallel.