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Jul 21, 2017 at 11:15 comment added AccidentalFourierTransform @bianchira fair enough.
Jul 21, 2017 at 2:57 comment added tparker The same ambiguity can come up even if you only have one derivative. Suppose you have a one-form $A_\mu$. If we were to interpret $\partial_\mu A^\mu$ as the four-divergence of the corresponding four-vector $A^\mu$, then we would have $\partial_\mu A^\mu \neq \partial^\mu A_\mu$, because the former would equal $\partial_\mu (g^{\mu \nu} A_\nu) = A_\nu \partial_\mu g^{\mu \nu} + g^{\mu \nu} \partial_\mu A_\nu = A_\nu \partial_\mu g^{\mu \nu} + \partial^\nu A_\nu$. In practice, we usually adopt the notational convention that all derivatives are taken before raising or lowering any indices.
Jul 20, 2017 at 17:43 comment added zzz @AccidentalFourierTransform no this should stay on physics SE because it's a notation issue, and the notation is one invented and popularized and abused by physicists.
Jul 20, 2017 at 16:28 comment added tparker Your same objection also applies to two contravariant partial derivatives, which look even more like they "should" commute.
Jul 20, 2017 at 16:24 comment added Prahar No. Your feeling is wrong. You cannot commute raised partial derivatives.
Jul 20, 2017 at 16:12 answer added zzz timeline score: 3
Jul 19, 2017 at 21:53 comment added Prof. Legolasov What is your definition of $\partial^{\mu}$?
Jul 18, 2017 at 18:46 comment added Kyle Kanos Probably related: physics.stackexchange.com/q/187590/25301
Jul 18, 2017 at 18:26 answer added J.G. timeline score: 9
Jul 18, 2017 at 18:23 review Close votes
Jul 19, 2017 at 19:02
Jul 18, 2017 at 18:15 history edited Qmechanic CC BY-SA 3.0
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Jul 18, 2017 at 17:55 review First posts
Jul 18, 2017 at 18:00
Jul 18, 2017 at 17:55 history asked user163020 CC BY-SA 3.0