# Deriving Maxwell Equations in their covariant form

Mawell Equations, in a particular unit system, are: \begin{eqnarray} \nabla \cdot \vec{E} &=& \rho &(1)\\ \nabla \times \vec{B} &=& \frac{\partial \vec{E}}{\partial t} + \vec{J}&(2)\\ \nabla \cdot \vec{B} &=& 0&(3)\\ \nabla \times \vec{E} &=& -\frac{\partial \vec{B}}{\partial t}&(4)\\ \end{eqnarray}

If we introduce the matriz $$F^{\alpha\beta} = \left( \begin{array}{cccc} 0 & E_x & E_y & E_z \\ -E_x & 0 & B_z & -B_y \\ -E_y & -B_z & 0 & B_x \\ -E_z & B_y & -B_x & 0\\ \end{array}\right)~(5)$$

I derived that, Maxwell's equations (1) and (2) are simplified in $\partial_\alpha F^{\alpha\beta} = -J^\beta$. But, S. Weinberg also say that, Maxwell's equations (3) and (4) are simplified in the form $$\epsilon^{\alpha\beta\gamma\delta} \partial_\beta F_{\gamma\delta} = 0$$

How can I derived this last equation?

• Hint: Choose $\alpha = 0$ in $\epsilon^{\alpha \beta \gamma \delta}$ and verify that this equation is $\nabla . B = 0$. In the same way, choose $\alpha = i$ for $i=1,2,3$ and show that you get $\nabla \times E = \dot{B}$. – OkThen May 11 '17 at 20:09
• Corresponding Lagrangian story: physics.stackexchange.com/q/71611/2451 – Qmechanic Mar 16 '18 at 19:19

$G^{\mu\nu} = \frac{1}{2}\epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}$
Where $\epsilon^{\mu\nu\alpha\beta}$ is the totally anti symmetric tensor.