Timeline for What is the definition of Curl in Mathematica?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 19, 2019 at 10:32 | comment | added | OA Fakinlede | @m_goldberg Thanks for your response. Don't know who did the editing. Will leave it that way - does not really hurt. My apologies. | |
| May 19, 2019 at 2:10 | comment | added | m_goldberg | @OAFakinlede. As far as I can tell, I didn't edit anything out your post; I only reformatted it. If I inadvertently did edit out something that you feel should be there, by all means edit the question to restore what is missing. | |
| May 18, 2019 at 8:29 | comment | added | OA Fakinlede | I apologize for my mistake. I misinterpreted your question. I know that you can have tensors of higher orders in 3D Euclidean Point Space. In fact, my present work is about fourth-order and higher-order tensors that appear in constitutive continuum modeling. I do not know practical uses for curls of tensors of higher orders than two. I have never seen one defined in the scope of my reading and research. I developed interest in the curl of order two tensors because I thought the definition was rather arbitrary and therefore tried to connect it to better-known constructs. | |
| May 18, 2019 at 7:08 | comment | added | Hosein Rahnama |
OK, that's fine. But one point I would like to mention is that you can have a tensor of arbitrary rank n in a 3D Euclidean space. Indeed, we encounter a lot of such creatures in mathematical physics. :)
|
|
| May 18, 2019 at 3:54 | comment | added | xzczd♦ |
@H.R. If I understand it correctly, Fakinlede's answer is just for the 2nd part of the question i.e. "How it can be related to the definition given in $(2)$?" Then I think the first step should be modified to Dot[Transpose@T, LeviCivitaTensor[3]]. ( @OAFakinlede Just compare your result to the CurlTen in the question. )
|
|
| May 17, 2019 at 20:04 | comment | added | OA Fakinlede | To H.R. 1. Formula tested correctly in Cylindrical and Spherical Polar coordinates using the same example problem in Mathematica 12, transformed to cylindrical and spherical polar coordinates; 2. Only interested in 3D Euclidean Point Space. 3. No, this method is different from xzczd. Here, we use transpose of the divergence of the contraction of the tensor argument with the LeviCivitaTensor[3]. I have proved elsewhere that it should give the same result but from different angles except for the scalar multiplier of 1/2. | |
| May 17, 2019 at 17:44 | comment | added | OA Fakinlede | Thanks for your question. | |
| May 17, 2019 at 14:36 | comment | added | Hosein Rahnama |
(+1) Hi there, and welcome to this community. :) Have you tested your formula against a tensor of an arbitrary rank n and in different curivilinear coordinates such as cylindrical or spherical? Your answer is somehow similar to that of xzczd.
|
|
| May 17, 2019 at 14:30 | history | edited | m_goldberg | CC BY-SA 4.0 |
Improved formatting
|
| May 17, 2019 at 12:28 | history | edited | OA Fakinlede | CC BY-SA 4.0 |
added 10 characters in body
|
| May 17, 2019 at 11:45 | review | Late answers | |||
| May 17, 2019 at 14:30 | |||||
| May 17, 2019 at 11:29 | history | answered | OA Fakinlede | CC BY-SA 4.0 |