Timeline for Geometric Construct for Integrating Symmetric Tensors?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 18, 2016 at 5:09 | vote | accept | M. Pretko | ||
| Jul 17, 2016 at 9:58 | answer | added | Igor Khavkine | timeline score: 6 | |
| Jul 17, 2016 at 5:17 | comment | added | M. Pretko | To Peter, the integral you mentioned definitely exists as a valid subcase. My analysis so far indicates that this is just one degenerate subcase of a wider class of objects. I want to find a general scheme of integration which takes a symmetric tensor as input and then outputs a scalar quantity. Of course one can simply contract one index of the tensor with an arbitrary vector, $\tilde{A}_i = v_j A^{ij}$, reducing the problem to the vector case. But this does not appear to be the most general solution. | |
| Jul 17, 2016 at 5:08 | comment | added | M. Pretko | Thanks for the comments. To Willie, in general I know derivatives should have connection terms to be tensorial. For the time being, I'm only interested in the trivial connection. But I would also be interested to know the answer in the more general case. Simply replace the second derivative in the question above with the appropriate covariant derivatives. | |
| Jul 17, 2016 at 2:28 | comment | added | Peter Kravchuk | Why can't you just form $I_i=\int_C A_{ij} dx^j$? | |
| Jul 17, 2016 at 2:11 | comment | added | Willie Wong | The "second derivative" is not well defined until you specify a connection. (It is not tensorial/coordinate invariant). I am not sure how you want to think of $A$: is it an object subordinate to the chosen connection on $T^*R^{m}$. | |
| Jul 16, 2016 at 23:48 | history | edited | M. Pretko |
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| Jul 15, 2016 at 18:23 | review | First posts | |||
| Jul 15, 2016 at 18:53 | |||||
| Jul 15, 2016 at 18:21 | history | asked | M. Pretko | CC BY-SA 3.0 |