A How to prove this property of the Dual Strength Field Tensor?

  • Thread starter Gaussian97
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Hi, I've found this property of Strenght Field Tensors:

Where $$\tilde{F}_{\mu\nu}=\frac{1}{2}\varepsilon_{\mu\nu\alpha\beta}F^{\alpha\beta}, \qquad \varepsilon_{0123}=1$$

I've tried to prove this relation but I can't find a way to do it. I have done specific cases for 4x4 antisymmetric matrices and it seems to work, but I would appreciate having a formal proof.


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It's very dangerous to write the left-hand side of the 1st equation in this sloppy way. The ##F_{\mu \nu}## are antisymmetric and not symmetric. So it's important to keep the horizontal position of the indices carefully intact. So I can only gues, what you want to calculate (it will anyway only flip the sign, if you have a different contraction in mind):

$${F^{\nu}}_{\mu} {\tilde{F}_{\nu}}^{\lambda} = F_{\nu \mu} \tilde{F}^{\nu \lambda} =\frac{1}{2} F_{\nu \mu} \epsilon^{\alpha \beta \nu \lambda} F_{\alpha \beta}=-\frac{1}{4} \epsilon_{\nu \mu \gamma \delta} \epsilon^{\alpha \beta \nu \lambda} \bar{F}^{\gamma \delta} F_{\alpha \beta}=-\frac{1}{4} \delta_{\mu \gamma \delta}^{\lambda \alpha \beta}\bar{F}^{\gamma \delta} F_{\alpha \beta}=-\frac{1}{4} \delta_{\mu}^{\lambda} \tilde{F}^{\alpha \beta}F_{\alpha \beta}.$$
Ok, thanks! That is exactly what I want! Sorry for the notation 😅

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