= Solution
For $t\to+\infty$ with $qE>0$, an <asymptotic expansion> gives
$$
z(t)=ct-\frac{\mathcal E_0}{qE}+O(t^{-1}),
$$
so the longitudinal <velocity> tends to the speed of light. More generally, its limiting sign is the sign of $qE$.
For small $t$, the <taylor series> of the square root gives
$$
z(t)
=\frac{qEc^2}{2\mathcal E_0}t^2+O(t^4).
$$
In the nonrelativistic <limit> $\mathcal E_0\simeq mc^2$, this becomes
$$
z(t)\simeq\frac{qE}{2m}t^2
=\frac12at^2,
\qquad a=\frac{qE}{m}.
$$
The transverse motion is then uniform:
$$
\boxed{x(t)=\frac{p_0}{m}t}.
$$
Eliminating $t$ gives the nonrelativistic parabolic trajectory
$$
\boxed{
z(x)=\frac{qEm}{2p_0^2}x^2
}.
$$
Back to article page
