Timeline for Chain rule for covariant derivative?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Mar 22, 2021 at 5:25 | vote | accept | TaeNyFan | ||
| Mar 21, 2021 at 20:21 | comment | added | octonion | @BenceRacskó, Ok yes searching the topic I see you yourself have been interested in an extension of the definition of the covariant derivative to densities in the past: (math.stackexchange.com/questions/2267059/…). | |
| Mar 21, 2021 at 20:05 | comment | added | Bence Racskó | @octonion it's a density, which can be covariantly differentiated | |
| Mar 21, 2021 at 19:31 | comment | added | octonion | @BenceRacskó, How? It's not a scalar | |
| Mar 21, 2021 at 10:58 | comment | added | Bence Racskó | @octonion But you can define a covariant derivative on $\sqrt{g}$. | |
| Mar 20, 2021 at 21:14 | comment | added | G. Smith | Yes. The chain rule works for covariant derivatives. As far as I know, it works for any kind of derivative. | |
| Mar 20, 2021 at 11:15 | comment | added | TaeNyFan | @G.Smith I'm thinking of taking the covariant derivative $\nabla_c \sqrt{g}$ where $g$ is the metric determinant. How can I use some sort of chain rule to take care of the square root? | |
| Mar 19, 2021 at 20:34 | comment | added | G. Smith | @Shashaank It doesn’t satisfy the transformation rule for a tensor with two covariant indices. | |
| Mar 19, 2021 at 19:10 | comment | added | Shashaank | @G.Smith why exactly is it not a tensor | |
| Mar 19, 2021 at 17:57 | comment | added | G. Smith | $\sqrt{t_{ab}}$ are not components of a tensor so your expression is physically meaningless. | |
| Mar 19, 2021 at 17:20 | answer | added | octonion | timeline score: 1 | |
| Mar 19, 2021 at 16:55 | history | edited | Qmechanic♦ |
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| Mar 19, 2021 at 16:35 | history | asked | TaeNyFan | CC BY-SA 4.0 |