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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?

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    $\begingroup$ 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 $\nabla \times E = \dot{B}$. $\endgroup$
    – OkThen
    Commented May 11, 2017 at 20:09
  • $\begingroup$ Corresponding Lagrangian story: physics.stackexchange.com/q/71611/2451 $\endgroup$
    – Qmechanic
    Commented Mar 16, 2018 at 19:19

1 Answer 1

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This equation is derived the same way as the first one but considering instead the Dual Electromagnetic tensor:

$G^{\mu\nu} = \frac{1}{2}\epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta}$

Where $\epsilon^{\mu\nu\alpha\beta}$ is the totally anti symmetric tensor.

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