the dual of (28) is given by $ ^*\!{F}^{\mu\nu}\!=\!\varepsilon^{\mu\nu\alpha\beta}\partial_{\alpha}A_\beta$ and its divergence reads $\partial_\mu\,\!^*\!{F}^{\mu\nu}\!=\!\varepsilon^{\mu\nu\alpha\beta}\partial_\mu\partial_{\alpha}A_\beta,$ whose right-hand side identically vanishes because $\varepsilon^{\mu\nu\alpha\beta}$ is antisymmetric in the indices $\mu$ and $\alpha$ and the operator $\partial_\mu\partial_\alpha$ is symmetric in these indices. Therefore the additional required field equation is given by $$\begin{aligned}
\partial_\mu\,\!^*\!{F}^{\mu\nu}=0.\end{aligned}$$ The set formed by equations (24) and (25) is equivalent to the set formed by equations (29) and (31). Let us write (28) as $F^{\mu\nu} =(\delta^\nu_\lambda\partial^\mu-\delta^\mu_\lambda\partial^\nu) A^\lambda$, were $\delta^\nu_\lambda$ is the Kronecker delta. Using this expression for $F^{\mu\nu}$ together with (23) we obtain $$\begin{aligned}
F^{\mu\nu}=\mu_0 (\delta^\nu_\lambda\partial^\mu-\delta^\mu_\lambda\partial^\nu) \!\int\!\!G J^\lambda d^4x'.\end{aligned}$$ We now take the wave operator $\partial_\alpha\partial^\alpha$ to (32), use $\partial_\mu\partial^\mu G\!=\!\delta^{(4)}(x-x')$ and integrate over all spacetime, obtaining the wave equation $$\begin{aligned}
\partial_\alpha\partial^\alpha F^{\mu\nu}=\mu_0 (\partial^\mu J^\nu-\partial^\nu J^\mu).\end{aligned}$$ Our task will be complete if we appropriately specify the components of the four-current $J^{\mu}$, the four-potential $A^{\mu}$, the electromagnetic field $F^{\mu\nu}$ and its dual $^*\!{F}^{\mu\nu}$. The four-gradient is defined by $\partial_\mu=\big\{(1/c)\partial/\partial t, \nabla\big\}$. Therefore, if we write $$\begin{aligned}
  J^\nu\!=\!\{c\rho, \textbf{J}\}, \quad A^\nu\!=\!\{\Phi/c, \textbf{A}\},\end{aligned}$$ then (23)-(25) reproduce (5), (6) and (8) respectively. Similarly, if we write $$\begin{aligned}
F^{i0}\!=\!(\textbf{E})^i/c,\quad F^{ij}\!=\!-\varepsilon^{ijk}(\textbf{B})_k,\quad ^*\!{F}^{i 0}\!=\!(\textbf{B})^i,\quad ^*\!{F}^{i j}\!=\!\varepsilon^{ijk}( \textbf{E})_k/c.\end{aligned}$$ where $(\textbf{E})^i$ and $(\textbf{B})_k$ are the Cartesian components of the fields 