= Solution
With Cartesian <Minkowski spacetime> coordinates $x^\mu=(ct,x,y,z)$, introduce the <four-current> $J^\mu=(c\rho,\mathbf J)$ and <electromagnetic four-potential> $\mathcal A^\mu=(\phi/c,\mathbf A)$. The antisymmetric <electromagnetic field tensor> $F^{\mu\nu}=\partial^\mu\mathcal A^\nu-\partial^\nu\mathcal A^\mu$ has components $F^{0i}=-E_i/c$ and $F^{ij}=-\epsilon^{ijk}B_k$. <Maxwell equations> become
$$
\boxed{\partial_\mu F^{\mu\nu}=\mu_0J^\nu,\qquad
\partial_\mu\widetilde F^{\mu\nu}=0,}
\qquad
\widetilde F^{\mu\nu}=\frac12\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma},
\quad \epsilon^{0123}=+1.
$$
The time and space components reproduce <Gauss's law> and the <Ampère-Maxwell equation>; the dual equation gives absence of <magnetic monopoles> and <Faraday's law of induction>. Antisymmetry gives <charge conservation> $\partial_\nu J^\nu=0$. The <relativistic Lorentz force> is $dp^\mu/d\tau=qF^{\mu\nu}u_\nu$. Every displayed equation transforms tensorially under a <Lorentz transformation>; the <electric field> and <magnetic field> are different components of the same object.
Two independent algebraic <electromagnetic field invariants> are
$$
\boxed{F_{\mu\nu}F^{\mu\nu}=2(B^2-E^2/c^2),\qquad
F_{\mu\nu}\widetilde F^{\mu\nu}=-4\mathbf E\cdot\mathbf B/c.}
$$
The second is a <pseudoscalar>: it is invariant under proper <Lorentz transformations> but changes sign under <spatial reflection>. For perpendicular fields, the sign of the first distinguishes a frame with purely <magnetic field> from one with purely <electric field>. Both invariants vanish for a <plane electromagnetic wave> with $\mathbf E\perp\mathbf B$ and $E=cB$, although the field itself is nonzero.
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