Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 2 / 18D / d / Solution

For $E_0\ne 0$, put $\beta =v/c$. Part (c) givesThe vanishing of their cross product is equivalent toFor $0<\theta <\pi /2$ the two roots are $\tan (\theta /2)$ and $\cot (\theta /2)$. Only the first has $|\beta |<1$. The continuous endpoint $\theta =0$ gives $\beta =0$. Thus the physical boost isSubstitution gives ${\mathbf{E}}^{\prime }=c{\mathbf{B}}^{\prime }$, so the fields point in the same direction. At $\theta =\pi /2$ the formal root is $v=c$, which is not an inertial observer's velocity. The nonzero perpendicular equal-strength fields then have both electromagnetic field invariants zero. If they were parallel in a finite frame, those invariants would force both fields to vanish, impossible under an invertible Lorentz transformation. Hence no finite inertial boost makes them parallel in this case. If $E_0=0$, both fields are zero in every frame and no velocity is singled out.