[Physics] Deriving Maxwell Equations in their covariant form

covarianceelectromagnetismhomework-and-exercisesmaxwell-equationsspecial-relativity

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?

Best Answer

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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