Note on Electrodynamics
Particle Motion in Uniform Perpendicular Electromagnetic Fields
A test particle with mass $m$ and charge $q$ moving at velocity $\boldsymbol{v}$ in an electromagnetic field feels the Lorentz force
$$\frac{d }{d t} (\gamma m\boldsymbol{v}) = q\left(\boldsymbol{E} + \frac{\boldsymbol{v}}{c} \times \boldsymbol{B}\right).$$Since the magnetic force $\boldsymbol{v}\times\boldsymbol{B}/c$ does no work on the particle, the change of particle energy arises only from the electrical force
$$\frac{d \gamma}{d t} = \frac{q}{m c^2} \boldsymbol{E}\cdot\boldsymbol{v}.$$When the electromagnetic field is constant in time, analytical solution can be sought by changing the reference frame so that only the electrical or magnetic field exists.
$B>E$ If the magnetic field dominates, the electrical field vanishes in a reference frame which is moving with drift velocity
$$\boldsymbol{v}_d = \frac{\boldsymbol{E}\times\boldsymbol{B}}{B^2}c.$$In that reference frame, the test particle will gyrate around the constant magnetic field. Therefore, the particle’s motion in the lab frame will be a combination of gyration around the $\boldsymbol{B}$ and drift with $\boldsymbol{v}_d$. The energy of the test particle oscillates on the short timescale of the gyration period as $\boldsymbol{E}\cdot\boldsymbol{v}$ varies, but there is no net change of particle energy on the timescale much larger than the gyration period.
$E>B$ In case the electrical field dominates, the magnetic field will vanishes in a reference frame moving with drift velocity
$$\boldsymbol{v}_d = \frac{\boldsymbol{E}\times\boldsymbol{B}}{E^2}c.$$The test particle is accelerated in the new reference frame and the resulting motion in the lab frame will be a combination of acceleration along $\boldsymbol{E}$ and drift with $\boldsymbol{v}_d$. The particle is continuously gaining energy as $\boldsymbol{E}\cdot\boldsymbol{v}$ is always positive.
$E=B$
For the speical case where $E=B$, there is no special reference frame that only one field is present. The analytical solution is given by Landau and Lifshitz. Assume that the magnetic field is pointing in the $x$ direction and the electrical field in the $z$ direction, there are two constants of motion: the momentum in the $x$ direction $p_x=\gamma m v_x$ and $\alpha\equiv \gamma m c^2(1-v_y/c)$. The particle is accelerated in the direction of the electrical field following
$$2eEt = \left( 1+\frac{\varepsilon^2}{\alpha^2} \right)p_z +\frac{c^2}{3\alpha^2}p_z^3$$with $\varepsilon^2=m^2c^4+p_y^2 c^2$. But the particle is accelerated fastest in the $y$ direction (parallel to $\boldsymbol{E}\times\boldsymbol{B}$)
$$p_y = -\frac{\alpha}{2c} + \frac{p_z^2c^2+\varepsilon^2}{2\alpha c}.$$And the particle energy increases as
$$\gamma mc^2 = \frac{\alpha}{2} + \frac{p_z^2c^2+\varepsilon^2}{2\alpha}.$$Relativistic Plasma Oscillation
Consider plasma consisting of particles with charge $q$, mass $m$ and number density $n$ in an electrical field, the Lorentz force will accelerate the particle and induce a current density $\boldsymbol{j} = n q \boldsymbol{v}$,
$$\frac{d \boldsymbol{j}}{d t} = nq\frac{d \boldsymbol{v}}{d t} = \frac{n q^2}{\gamma m}\left(\boldsymbol{E}-\frac{\boldsymbol{V}}{c^2}\boldsymbol{E}\cdot\boldsymbol{V}\right).$$The nonzero charge density will then feedback to the electrical field
$$\frac{\partial \boldsymbol{E}}{\partial t} = -4\pi \boldsymbol{j}.$$Combining the equation together, we have
$$\frac{d^2 \boldsymbol{E}}{d t^2} = -\frac{4\pi n q^2}{\gamma m}\left(\boldsymbol{E}-\frac{\boldsymbol{V}}{c^2}\boldsymbol{E}\cdot\boldsymbol{V}\right).$$This equation exhibits the oscillatory behaviour for the electrical field at approximately the relativistic plasma frequency
$$\omega_{\rm p} = \sqrt{\frac{4\pi n q^2}{\gamma m}}.$$If the plasma is confined to move in the direction of the electrical field, $\boldsymbol{V}\parallel\boldsymbol{E}$, the oscillation frequency is modified as $\omega_{\rm p} = \sqrt{4\pi n q^2/\gamma^3 m}$.
Speiser Orbits
Let us consider the motion of a positron in a static electromagnetic field $\boldsymbol{B}=(B_0,B_y(x),0)$ and $\boldsymbol{E}=(0,0,E_z)$, where
$$ E_z=-aB_0, \qquad B_y=bB_0 h(x).$$Here $a>0$ and $b>0$ are constants, and $h(x)$ is a smooth function monotonically decreasing from $h\approx 1$ at $x<- \Delta$ to $h\approx -1$ at $x>\Delta$. This electromagnetic configuration describes the jump of $B_y(x)$ on the scale $\Delta$ that forms during the collision of two symmetric Alfven packets, as observed in our simulations (with $b=2A-1$). Non-relativistic particle motion in a magnetic jump with $h(x)=-x/\Delta$ at $|x|<\Delta$ was studied by Speiser 1965. We are interested here in the relativistic case where the particle is accelerated by $E_z$ to a high Lorentz factor $\gamma$. The constant $a$ is smaller than unity but close to it.
$\gamma'\sim\gamma_u\gamma_0\sim\gamma_u$ where $\gamma_0$ is the initial particle energy in the lab frame. We are interested in the case $a$ close to unity, so initially the particle has $v'_y\sim ac\gg |v'_x|=v_x/\gamma_u$, therefore we can approximate the equations of motion as
$$\dot{v'_y} = \frac{q}{\gamma' mc}B'_x v'_z,\qquad \dot{v'_z} = -\frac{q}{\gamma' mc}B'_x v'_y.$$The particle will undergo a gyration around $B_x$ with solutions given by (assuming initially $v'_z=0$, $v'_y=v'_0$)
$$ v'_y = v'_0\cos\left(\frac{qB'_x}{\gamma'mc}t'\right),\qquad v'_z = -v'_0\sin\left(\frac{qB'_x}{\gamma'mc}t'\right). $$$v'_z$ will develop values along $qE_z$ in the first half cycle of gyration before it reverses sign. Now consider the particle is inside the magnetic jump, the equation for $v'_x$ is
$$\ddot{x'} = \dot{v'_x} = -\frac{q B_0b v'_z}{\gamma' mc\Delta}x'.$$Before $v'_z$ reverses sign, the quantity $\omega^{'2}_{\rm osc}=q B_0bv'_z/\gamma'mc\Delta>0$, so the particle oscillates between magnetic jump with frequency $\omega'_{\rm osc}$. When $v'_z$ reverses sign, $|x|$ will increase exponentially and the particle will be ejected from the magnetic jump.
After the particle finishes half cycle of the gyration $v'_z$ reverses sign and $v'_y=-ac\approx -c$ is almost a constant, we can approximate the equation as
$$ \dot{v'_x} = -\frac{q}{\gamma' mc}B'_y v'_z,\qquad \dot{v'_z} = \frac{q}{\gamma' mc}(B'_y v'_x+B'_x c). $$Assuming the particle has reached the edge of the magnetic jump where $B'_y$ is also almost constant, the solution of the particle motion is
$$ v'_x = V'\cos\left(\frac{qB'_y}{\gamma'mc}t'+\phi\right) - c\frac{B'_x}{B'_y},\qquad v'_z = V'\sin\left(\frac{qB'_y}{\gamma'mc}t'+\phi\right) $$with $V'$ and $\phi$ being two constants for the amplitude and phase. Therefore, the particle will gyrate around $B_y$ with a net drift along $x$ away from the center.
In the lab frame, a particle initially with $|v_y|\ll c$ will be ejected with $|v_y|\sim 2u/(1+u^2/c^2)=2ac/(1+a^2)$ with energy gain
$$\gamma=\frac{1+a^2}{1-a^2}$$and its ejection velocity along $x$ is
$$ |v_x| \sim \frac{|B'_x/B'_y|c}{\gamma_u(1+u^2/c^2)} = \frac{(1-a^2)c}{(1+a^2)b}= \frac{c}{\gamma b}. $$The ejection velocity will decrease with $\gamma$.
