Null crossed electromagnetic field

Perpendicular fields with $\mathbf{E}\cdot \mathbf{B}=0$ and $E^2-c^2{B}^2=0$ form a null electromagnetic field. The constant-field Lorentz-force generator is nilpotent, so its exponential terminates as a polynomial.

ChooseA particle released from rest at the origin has the proper-time trajectoryEliminating proper time gives the semicubical parabolaThis signed coefficient assumes that the positive $x$ direction is chosen along $\mathbf{E}$; reversing that axis changes the signs of both $x^3$ and its coefficient.