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The $x$ component of the velocity isThusSolving this equation for $t$ givesSubstitution into the expression for $z(t)$ and the identity $1+{\sinh }^{2}u={\cosh }^{2}u$ produce the trajectoryThis is relativistic motion in a constant electric field with transverse momentum.

For a uniform constant electric field and zero magnetic field, the mechanical momentum changes linearly even though the velocity does not:Invert $\mathbf{p}=m\gamma \mathbf{v}$ using $\gamma =\sqrt{1+p^2/(m^2{c}^2)}$. The result for arbitrary initial direction isEquivalently, $\mathbf{v}=c\mathbf{p}/\sqrt{m^2{c}^2+p^2}$. 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.