 Solution

For a constant electromagnetic field tensor, the relativistic Lorentz force is a linear system with constant coefficients, so

PutThe initial four-velocity is $(u^0,u^1,u^2,u^3)={\gamma }_0(c,0,v_0,0)$. For parallel fields along $x$, the field tensor has one hyperbolic block in the $(0,1)$ plane and one rotation block in the $(2,3)$ plane. With the orientation convention in the question,Taking the initial spacetime point to be the origin and integrating $u^{\mu }=d{x}^{\mu }/d\tau$ gives the relativistic motion in parallel electric and magnetic fieldsThe expressions have their continuous $a\to 0$ or $b\to 0$ limits when one field vanishes; reversing the sign of $q$ reverses the electric acceleration and magnetic rotation.

Eliminating proper time between the first two equations yieldsAt early coordinate time,which is uniformly accelerated motion with the transverse relativistic inertia factor ${\gamma }_0$. At late positive time,so the longitudinal speed approaches the speed of light.

The $y$-$z$ projection satisfiesIt is therefore a circle of radius and proper-time period

For the pitch magnitude assume $v_0>0$ and put $\alpha =|q|E/(mc)$, $\Omega =|q|B/m$, andNumber the $n$th proper-time period as $[(n-1)\Delta \tau ,n\Delta \tau ]$. Its longitudinal displacement isConsequently, as $n\to \infty$,where