Timeline for Curl as a divergence... Is it possible?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Aug 3, 2018 at 21:59 | comment | added | nodarkside | Certainly the divergence decreases by one the order of the tensor over which operates. I need the divergence since I'm using the finite volume method to solve them; therefore, the divergence is transformed into a sum by use of the Gauss theorem. | |
| Aug 3, 2018 at 21:51 | comment | added | Qfwfq | I have the impression I answered the question for $\boldsymbol{\nabla}\times (\textrm{tensor})$ rather than $\boldsymbol{\nabla}\cdot (\textrm{tensor})$ (whatever those definitions are), cause the exterior differential $d$ keeps increasing the number of "indices" whereas your $\boldsymbol{\nabla}\cdot$ certainly decreases it... | |
| Aug 3, 2018 at 21:34 | comment | added | nodarkside | Thanks for your reply @Qfwfq ! Unfortunately I do not know very much about differential forms. Sorry... | |
| Aug 3, 2018 at 21:19 | comment | added | Qfwfq | Maybe an equivalent interpretation, closer to your question, is asking for a $1$-form $\beta$ on $M$ such that $d\phi\wedge d \alpha=d\ast \beta$ where $\ast$ is the Hodge star from a fixed Riemannian metric on $M$. | |
| Aug 3, 2018 at 21:13 | history | answered | Qfwfq | CC BY-SA 4.0 |