In another forum, someone was explaining to me how to define a new time coordinate in a metric tensor to diagonalize it. They said that if I want spacelike hypersurfaces, the covariant derivative of $\tau$ should be timelike. Does $\tau$ count as a vector? If not, how would I take the covariant derivative?
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$\begingroup$ $\tau$ is a coordinate, meaning it is a function. It’s covariant derivative is by definition simply its exterior derivative/differential $d\tau$. CHecking whether this is timeline/space like amounts to looking at the sign of the coefficient of $d\tau^2$ in the metric. $\endgroup$– peek-a-booCommented Sep 1, 2022 at 19:57
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$\begingroup$ Would the covariant derivative with raised indice also just be the regular derivative in this case? $\endgroup$– user345249Commented Sep 1, 2022 at 20:47
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$\begingroup$ $\nabla^af=g^{ab}\nabla_bf=g^{ab}(df)_b=g^{ab}\frac{\partial f}{\partial x^b}$ if that’s what you want to know. This is what we’d call the ($a^{th}$ component of) gradient vector field of $f$, i.e $\text{grad}_g(f)$. $\endgroup$– peek-a-booCommented Sep 1, 2022 at 20:50
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$\begingroup$ Yes, that answers my question. Thank you. $\endgroup$– user345249Commented Sep 1, 2022 at 22:17
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