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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.

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.