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Contractions of Lorentz tensors have no free indices and are Lorentz scalars. The antisymmetry and the components in part (a) giveThus $\mathbf{E}\cdot \mathbf{B}$ and $|\mathbf{B}{|}^2-|\mathbf{E}{|}^2/c^2$ are the two electromagnetic field invariants under proper Lorentz transformations. The second contraction uses the orientation-dependent epsilon tensor: strictly, $\mathbf{E}\cdot \mathbf{B}$ is a pseudoscalar, changing sign under an improper transformation such as spatial parity. Under the boosts in this question it is invariant.

For $E_0\ne 0$, put $\beta =v/c$. Part (c) givesThe vanishing of their cross product is equivalent toFor $0<\theta <\pi /2$ the two roots are $\tan (\theta /2)$ and $\cot (\theta /2)$. Only the first has $|\beta |<1$. The continuous endpoint $\theta =0$ gives $\beta =0$. Thus the physical boost isSubstitution gives ${\mathbf{E}}^{\prime }=c{\mathbf{B}}^{\prime }$, so the fields point in the same direction. At $\theta =\pi /2$ the formal root is $v=c$, which is not an inertial observer's velocity. The nonzero perpendicular equal-strength fields then have both electromagnetic field invariants zero. If they were parallel in a finite frame, those invariants would force both fields to vanish, impossible under an invertible Lorentz transformation. Hence no finite inertial boost makes them parallel in this case. If $E_0=0$, both fields are zero in every frame and no velocity is singled out.

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.

Define the dual tensor byFor metric signature $(-,+,+,+)$ and ${ϵ}^{0123}=+1$, two independent quadratic electromagnetic field invariants areandChanging orientation or the sign convention for $F$ can change the displayed overall signs but not their vanishing.

For a vacuum plane electromagnetic wave with wavevector $\mathbf{k}$,Hencefor every transverse polarization. Both Lorentz scalars therefore vanish.