= Solution
For a uniform constant <electric field> and zero <magnetic field>, the <mechanical momentum> changes linearly even though the velocity does not:
$$
\mathbf p(t)=\mathbf p_0+e\mathbf E t,\qquad \mathbf p_0=m\gamma_0\mathbf v_0,\qquad
\gamma_0=(1-v_0^2/c^2)^{-1/2}.
$$
Invert $\mathbf p=m\gamma\mathbf v$ using $\gamma=\sqrt{1+p^2/(m^2c^2)}$. The result for arbitrary initial direction is
$$
\boxed{\mathbf v(t)=\frac{\gamma_0\mathbf v_0+(e\mathbf E/m)t}
{\sqrt{1+\left|\gamma_0\mathbf v_0+(e\mathbf E/m)t\right|^2/c^2}}.}
$$
Equivalently, $\mathbf v=c\mathbf p/\sqrt{m^2c^2+p^2}$. This gives $\mathbf v(0)=\mathbf v_0$ and guarantees $|\mathbf v|<c$ at every finite momentum. The transverse momentum stays constant, but its velocity component generally decreases because the <Lorentz factor> increases. Thus an initial transverse velocity must not be treated as constant during the electric acceleration; this is the <relativistic motion in a constant electric field with transverse momentum> effect.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2018-q1-electric-velocity.png]
{title=Constant electric force changes momentum linearly while speed stays below light speed}
{description=Original example with initial velocity perpendicular to the electric force and magnitude $0.6c$. The longitudinal velocity tends to $c$, while the transverse velocity tends to zero. Time is scaled by $mc/(|e|E)$, and the positive longitudinal direction is chosen along the force.}
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