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For a constant electromagnetic field tensor, the relativistic Lorentz force is a constant-coefficient linear system on four-velocity, with solutionThe electric-type blocks of $F$ generate hyperbolic rotations, while magnetic-type blocks generate ordinary rotations.

Use the Minkowski metric $\eta =\mathrm{diag}(-1,1,1,1)$ and define the electromagnetic field tensor byAs a rank-two Lorentz tensor, it transforms under a Lorentz transformation asTwo independent quadratic electromagnetic field invariants are

With $A^{\mu }=(\varphi /c,\mathbf{A})$, $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$, and $\mathbf{B}=\nabla \times \mathbf{A}$, the components areDirect contraction giveswhere the sign of $I_2$ changes if the opposite spacetime orientation is chosen. Since both are complete tensor contractions, $S$ and $T$ are Lorentz scalars.

For a constant uniform field, the relativistic Lorentz force is a constant-coefficient linear ordinary differential equation. Its solution is the matrix exponential

Now $S=T=0$ implies that the field is a null electromagnetic field:The stated axes allow us to choosePut $\Omega =qB/m$. On the $(ct,x,y)$ coordinates, the mixed tensor divided by $B$ isThe particle starts with four-velocity $u(0)=(c,0,0,0)$, so the exponential terminates after its quadratic term and givesIntegrating with the particle initially at the origin gives the relativistic trajectory in a constant null crossed fieldEliminating $\tau$ yieldsIf the original positive $x$ direction is opposite to $\mathbf{E}$, both the sign of $x$ and the displayed signed coefficient reverse.

For a constant electromagnetic field tensor, the relativistic Lorentz force is a linear system with constant coefficients, so

PutThe initial four-velocity is $(u^0,u^1,u^2,u^3)={\gamma }_0(c,0,v_0,0)$. For parallel fields along $x$, the field tensor has one hyperbolic block in the $(0,1)$ plane and one rotation block in the $(2,3)$ plane. With the orientation convention in the question,Taking the initial spacetime point to be the origin and integrating $u^{\mu }=d{x}^{\mu }/d\tau$ gives the relativistic motion in parallel electric and magnetic fieldsThe expressions have their continuous $a\to 0$ or $b\to 0$ limits when one field vanishes; reversing the sign of $q$ reverses the electric acceleration and magnetic rotation.

Eliminating proper time between the first two equations yieldsAt early coordinate time,which is uniformly accelerated motion with the transverse relativistic inertia factor ${\gamma }_0$. At late positive time,so the longitudinal speed approaches the speed of light.

The $y$-$z$ projection satisfiesIt is therefore a circle of radius and proper-time period

For the pitch magnitude assume $v_0>0$ and put $\alpha =|q|E/(mc)$, $\Omega =|q|B/m$, andNumber the $n$th proper-time period as $[(n-1)\Delta \tau ,n\Delta \tau ]$. Its longitudinal displacement isConsequently, as $n\to \infty$,where

Choose, for example,The fields are perpendicular and satisfy $E_0^2=c^2{B}_0^2$, so both invariants vanish. Every nonzero constant background with vanishing invariants can be brought to this form by a spatial rotation, up to signs.

In this frame, and with a convention consistent with the relativistic Lorentz force, the mixed tensor isCalling the displayed dimensionless matrix $N$, direct multiplication givesConsequentlyThis is a tensor equation. Since it holds in the canonical frame, it holds in every Lorentz frame for any null electromagnetic field.

Let $X^{\mu }(\tau )$ be the worldline, $u^{\mu }=d{X}^{\mu }/d\tau$ its four-velocity, andThe relativistic Lorentz force in the constant background isBecause $F^3=0$, its exponential terminates:The particle starts from rest, soUsing the matrix from part (c) givesIntegrating from the origin and using $X^0=ct$ yields the parametric trajectoryThis is the relativistic trajectory in a constant null crossed field. The real cubic relation for $t(\tau )$ is strictly increasing, so it determines $\tau$ uniquely as a function of coordinate time if an explicit $\mathbf{x}(t)$ is desired.

With $u^{\mu }=\gamma (c,\mathbf{v})$ and $d/d\tau =\gamma d/dt$, the spatial components of the relativistic Lorentz force giveThe time component gives $m{c}^{2}d\gamma /dt=q\mathbf{E}\cdot \mathbf{v}$. It says magnetic fields do no work and electric power changes relativistic energy.

When uniform $\mathbf{E}$ and $\mathbf{B}$ are parallel, the relativistic Lorentz force splits into independent hyperbolic motion along the fields and circular motion in the perpendicular plane. The circle is periodic in proper time, while electric acceleration makes the helix's longitudinal pitch grow exponentially from one proper-time cyclotron period to the next.