Special Relativity by A. P. French

Problem 1-23 A charged particle of rest mass $m_0$, charge $q$ and initial velocity $v_0$ enters a uniform electric field of strength $\mathscr{E}$ perpendicular to $v_0$. Find the resulting trajectory.

Solution We will set up our coordinate system so that the initial velocity is $v_0 \hat{\mathbf{i}}$ and that the electric field is $\mathbf{E}=\mathscr{E}\hat{\mathbf{j}}$. The main differential equation which describes the trajectory is:

$$ (1)\qquad \mathbf{F} = \mathrm{d}\mathbf{p}/\mathrm{d}t $$

Breaking down by components and using the definition $\mathbf{p}(t)=m(t)\mathbf{v}(t)$, we have

$$ \begin{aligned} \cfrac{\mathrm{d}}{\mathrm{d}t}[m(t)x'(t)] &= 0 \\ \cfrac{\mathrm{d}}{\mathrm{d}t}[m(t)y'(t)] &= \mathscr{E}q \end{aligned} $$

We can immediately integrate to find

$$ \begin{aligned} m(t)x'(t) &= m(0)v_0 \\ m(t)y'(t) &= \mathscr{E}qt \end{aligned} $$

Recall from basic principles of relativity that $m(t)=(1-v^2/c^2)^{-1/2}m_0$. So these equations become

$$ \begin{aligned} (1-v_0^2/c^2)^{1/2}x'(t) &= (1-v^2/c^2)^{1/2}v_0 \\ m_0 y'(t) &= (1-v^2/c^2)^{1/2} \mathscr{E}qt \end{aligned} $$

Square both sides and then multiply both sides by $c^2$; also replace $v^2 = \dot{x}^2+\dot{y}^2$.

$$ \begin{aligned} (c^2-v_0^2)\dot{x}^2 &= (c^2 - \dot{x}^2 - \dot{y}^2)v_0^2 \\

m_0^2 c^2 \dot{y}^2 &= (c^2-\dot{x}^2 - \dot{y}^2) \mathscr{E}^2q^2t^2 \end{aligned} $$

Conveniently, this is an algebraically linear system of equations in $\dot{x}^2$ and $\dot{y}^2$. Solving for these as unknowns we find (letting $P_0=\sqrt{c^2-v_0^2} \mathscr{E}q/E_0$ for convenience)

$$ \begin{aligned} \dot{x} &= \frac{ v_0 }{ \sqrt{1 + P_0^2 t^2} } \\ \dot{y} &= \frac{ c P_0 t }{ \sqrt{1 +P_0^2t^2} } \end{aligned} $$

We can readily integrate both of these (with initial condition $x(0)=y(0)=0$). The first succumbs to the substitution $P_0t = \sinh(s)$ while the second is calculated by u-substitution $u=1+P_0^2t^2$.

$$ \begin{aligned} x(t) &= \frac{v_0}{P_0} \log\left( P_0t + \sqrt{1+P_0^2t^2} \right) \\

y(t) &= \frac{c}{P_0}\left(-1+ \sqrt{ 1 + P_0^2 t^2 }\right) \end{aligned} $$

The relativistic effect is that the particle slows down somewhat in the horizontal direction as it gains inertial mass due to its increasing velocity. So, at low speeds the path is nearly parabolic as in classical physics, but over time it becomes more like an exponential graph shape, with $x(t) \approx \frac{v_0}{P_0} \log t$ and $y(t) \approx ct$. An interesting consequence of the analysis is that there is no vertical asymptote. The particle travels arbitrarily far in the x-direction, albeit slowing down over time, as it speeds up to light-speed in the y-direction.