Source: D. Tong, Electromagnetism , Problem Sheet 3, Question 9.
A particle of rest mass $m$ and charge $q$ moves in a constant
uniform electric field $\mathbf{E} = (E,0,0)$. It starts from the origin with
initial momentum $\mathbf{p} = (0,p_0,0)$. Show that the particle traces out
a path in the $(x,y)$ plane given by
\[x = \frac{\mathcal{E}_0}{qE}\left(\cosh\!\left(\frac{qEy}{p_0 c}\right)-1\right),\tag{1.1}\]
where $\mathcal{E}_0 = \sqrt{p_0^2c^2 + m^2c^4}$ is the initial kinetic energy
of the particle.
We start with the covariant form of the Lorentz force:
\[m\frac{dU^\mu}{d\tau} = qF^{\mu\nu}U_\nu.\tag{1.2}\]
Given $U^\mu = (\gamma c, \mathbf{p})$, it becomes
\[\frac{d\mathbf{p}}{dt} = q\mathbf{E},\tag{1.3}\]
where $\mathbf{p}$ is the relativistic 3-momentum, namely $\mathbf{p} = \gamma m \dot{\mathbf{x}}$. Thus, we can write
\[\frac{d\mathbf{x}}{dt} = \frac{\mathbf{p}}{\gamma m} = \frac{\mathbf{p}c^2}{\mathcal{E}}.\tag{1.4}\]
This gives
\[\begin{cases}
\dot{x} = \dfrac{p_xc^2}{\sqrt{\mathcal{E}_0^2+p_x^2c^2}},\\
\\
\dot{y} = \dfrac{p_0c^2}{\sqrt{\mathcal{E}_0^2+p_x^2c^2}}.
\end{cases}
\tag{1.5}\]
Since we can easily solve for $p_x = qEt$, we can solve the two equations and find
\[y = \frac{p_0c}{eE}\sinh^{-1}\left(\frac{cqEt}{\mathcal{E}_0}\right),\tag{1.6}\]
and
\[x = \frac{c}{qE}\left(\sqrt{\mathcal{E}_0^2+q^2E^2t^2c^2}-\mathcal{E}_0\right).\tag{1.7}\]
Putting these together, we find
\[x = \frac{\mathcal{E}_0}{qE}\left(\cosh\!\left(\frac{qEy}{p_0 c}\right)-1\right).\tag{1.8}\]
Figure 1. Trajectory of a relativistic charged particle in a uniform electric field $\mathbf{E}=(E,0,0)$, in natural units $\mathcal{E}_0/qE = p_0c/qE = 1$. The particle starts at the origin with momentum $\mathbf{p}_0$ in the $y$-direction and follows $x = (\mathcal{E}_0/qE)(\cosh(qEy/p_0c)-1)$.