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My professor said that it's easy to generalize to these definitions of Stoke's and Gauss' theorem from the 3 dimensional versions but didn't say much else. He threw the following on the chalk board:

$$ Gauss' \\ \\ \int{A^{\mu} \frac{1}{6} \epsilon^{\mu \nu \rho \sigma} dS_{\nu \rho \sigma}} = \int{\partial_{\mu} A^{\mu} d{\Omega}} $$

$$ Stoke's \\ \\ \frac{1}{2} \int{A^{\mu \nu} \frac{1}{2} \epsilon_{\mu \nu \rho \sigma} dF^{\rho \sigma}} = \int{\frac{\partial A^{\mu \nu}} {\partial{x^{\nu}}} \frac{1}{6} \epsilon_{\mu \nu \rho \sigma} dS^{\mu \rho \sigma}}$$

Where $d\Omega = dx^0 dx^1 dx^2 dx^3$ is the invariant spacetime volume and $dS_{\mu \rho \sigma} $ $($ and $\ dS^{\mu \rho \sigma}), \ dF^{\rho \sigma} $ are 3-surface and 2-surface elements, respectively.

For Gauss' theorem, it makes sense why the 1/6 is present, for the normalization of the rank-4 levi-cevita symbol, but why is there only a factor of 1/4 for Stoke's theorem? Also, what is the role of the levi-cevita symbol in both theorems, I'm unsure how to interpret the 2-surface and 3-surface elements. Some resources for an easy example using both theorems in this notation would be great. Thanks!

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    $\begingroup$ Not sure if this is correct but perhaps your professor is using a factor of 1/n! for each n form and the $\epsilon$ tensor is part of the definition of area and volume forms. Thus a factor of 1/6 is attached with the 3d volume form eds, factor of 1/2 with the 2d area form edf and a factor of 1/2 with 2 form A on the LHS of equation 2. On the RHS, the 2-form A is contracted with the derivative to give a 1 form so a factor of 1 for it. $\endgroup$
    – user10001
    Commented Feb 5 at 16:42

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