 Solution

Use SI units and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. Write $x^{\mu }=(ct,x,y,z)$, ${\partial }_{\mu }=(c^{-1}{\partial }_t,\nabla )$ and the four-current $J^{\mu }=(c\rho ,\mathbf{J})$, with Greek indices from $0$ to $3$. The electromagnetic field tensor is the antisymmetric tensor whose components areIndices are raised and lowered with $\eta$. Define the dual electromagnetic field tensor by ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$, with ${ϵ}^{0123}=+1$. Then the covariant Maxwell equations areHere $\rho$ and $\mathbf{J}$ are charge and current density, ${\mu }_0$ is the vacuum magnetic permeability, and ${ϵ}_0{\mu }_0{c}^2=1$. For example the $\nu =0$ equation gives $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, and the spatial equations give $\nabla \times \mathbf{B}-c^{-2}{\partial }_t\mathbf{E}={\mu }_0\mathbf{J}$. With the stated dual convention, ${\tilde{F}}^{0i}=-B_i$ and ${\tilde{F}}^{ij}={ϵ}_{ijk}{E}_k/c$; the second tensor equation gives the two homogeneous Maxwell equations.