Answer by Raziel for Gauss' theorem for null boundaries
Gauss' Theorem has nothing to do with the (pseudo-)metric. Is just a consequence of Stokes' theorem.Stokes's theorem says that, for any $n-1$ form $\omega$,$$ \int_M d\omega = \int_{\partial M} \omega....
View ArticleAnswer by Bence Racskó for Gauss' theorem for null boundaries
Okay, as Futurologist pointed out, the decomposition $X=\langle X,N\rangle N+Y$ is incorrect. In fact, this was quite a rookie mistake.Instead, let $\{e_1,...,e_{n-1}\}=\{\partial/\partial...
View ArticleGauss' theorem for null boundaries
Prenote: I have asked this question first on math stackexhange, but a user suggested that mathoverflow might be a better place for this question. Upon thinking about it I have agreed with him and...
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