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comment Covariant derivative and Leibniz rule
OK thank you, last thing if I may, I don't really understand why $u^j$ here is a scalar, from what I know scalar are objects which do not transform when changing coordinates. doesn't $u^j$ transform as a contravariant vector?
Jul
31
comment Covariant derivative and Leibniz rule
Another reason why that definition makes sense to me is that on the first derivation I displayed, it seems like the covariant derivative acts like the regular derivative when acting on $u^j$, just like you wrote in answer 1, I don't understand why this is, I don't understand what you mean by $u^j$ is a "non-specific component"
Jul
31
comment Covariant derivative and Leibniz rule
about question 2, it was online on a video lecture, which I won't make you sit through unless you're interested, by I also saw it on a forum post on Physics forum while searching for an answer for this on google, here's the post: physicsforums.com/showpost.php?p=3991829&postcount=3
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revised Covariant derivative and Leibniz rule
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