Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 34D (Electrodynamics)

Combining the real transformations givesLet $v=\tanh \psi$, so $\gamma =\cosh \psi$ and $\gamma v=\sinh \psi$. The usual rotation about the positive $x$-axis has the $yz$ block $\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$. Taking $\boxed{n=(1,0,0),\theta =-i\psi }$ gives exactly the displayed transformation, because $\cos (-i\psi )=\cosh \psi$ and $\sin (-i\psi )=-i\sinh \psi$. This is a complex rotation, not a real spatial rotation.

Complex rotations preserve the bilinear dot product, without conjugation. Henceis invariant, giving the two independent real Lorentz invariantsTheir completeness refers to scalar invariants of the full Lorentz group. For a generic non-null field, a boost along $E\times B$ makes the fields parallel. Indeed with $s=|E\times B|$ and $A=E^2+B^2$, the required speed solves $s(1+v^2)-Av=0$; the smaller root is $v=[A-\sqrt{(E^2-B^2{)}^2+4(E\cdot B{)}^2}]/(2s)<1$. For $s=0$ the fields are already parallel. A real rotation then gives $E=en$, $B=bn$, leaving just $e^2-b^2$ and $eb$ as independent scalar parameters. Thus there are precisely two independent continuous scalar invariants on generic orbits; null fields have special orbits and the invariants do not distinguish zero from a nonzero null field.

If the printed phrase is read as invariance under only this single-axis boost, it admits extra fixed components $F_x$, so the assertion of precisely two must be understood in its Lorentz-scalar sense.