Timeline for Parity transformation property of $\epsilon^{\mu\nu\sigma\rho}$ and $F_{\mu\nu}$ (and $G_{\mu\nu}^a$)

11 events

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Feb 14, 2022 at 9:48	comment	added	Boson Bear		Although the question is bit old -- the transformation of non-Abelian gauge fields can be easily seen if you look at the covariant derivative. To make sure the covariant derivative to behave covariantly, $A_\mu$ should transform similarly as $\partial_\mu$.
S Jun 24, 2018 at 16:34	history	bounty ended	CommunityBot
S Jun 24, 2018 at 16:34	history	notice removed	CommunityBot
S Jun 16, 2018 at 14:51	history	bounty started	SRS
S Jun 16, 2018 at 14:51	history	notice added	SRS		Authoritative reference needed
Jun 15, 2018 at 5:14	history	edited	SRS	CC BY-SA 4.0	added 425 characters in body
Jun 14, 2018 at 14:46	answer	added	FrodCube		timeline score: 2
Jun 14, 2018 at 14:44	history	edited	AccidentalFourierTransform	CC BY-SA 4.0	deleted 6 characters in body
Jun 14, 2018 at 14:44	history	edited	SRS	CC BY-SA 4.0	added 351 characters in body; edited title
Jun 14, 2018 at 14:41	comment	added	gj255		Under a general transformation $A$, we have $\epsilon^{\mu \nu \rho \sigma } \mapsto A^\mu{}_\alpha A^\nu{}_\beta A^\rho{}_\gamma A^\sigma{}_\delta \epsilon^{\alpha \beta \gamma \delta} = \mathrm{det}(A) \epsilon^{\mu \nu \rho \sigma}$. If $A$ is a parity transformation then its determinant is $-1$.
Jun 14, 2018 at 14:35	history	asked	SRS	CC BY-SA 4.0