0

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?

Qmechanic
  • 220,844
Yolbeiker
  • 1
  • 2

1 Answers1

1

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.

fgoudra
  • 972