Timeline for How do we calculate the directional derivative of a vector field? (If there is such a thing.)
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 23, 2020 at 13:19 | comment | added | Vercassivelaunos | If you really have to use coordinates, you should at least use index notation, because otherwise it will be atrocious to read. Then you can write $\mathrm dT_j=\mathrm dx_i\partial_i T_j=\mathrm JT\cdot\mathrm d\vec l$, where $\mathrm JT=(\partial_i T_j)_{ij}$ is the Jacobian. So $\frac{\mathrm d\vec T}{\vert\mathrm d\vec l\vert}=\mathrm JT\cdot \vec e_l$ with $\vec e_l$ the unit vector in direction $\vec l$. | |
| Aug 23, 2020 at 12:36 | comment | added | Elon Tusk | How do I do the same for a vector field? | |
| Aug 23, 2020 at 12:36 | comment | added | Elon Tusk | I arrived at $\frac {dT}{|d\vec l|} = |\vec \nabla T| \cos\theta$ from the expression: $dT=\frac {\partial T}{\partial x} dx +\frac {\partial T}{\partial y} dy+\frac {\partial T}{\partial z} dz$, which I believe is the total differential. I wrote the same equation as $dT=\biggr{(}\frac {\partial T}{\partial x} e_{x} +\frac {\partial T}{\partial y} e_{y}+\frac {\partial T}{\partial z} e_{z}\biggr{)}.(dx\space e_x+dy\space e_y+dz\space e_z)$ where $e_x,e_y,e_z$ are the unit vectors in the x,y,z directions respectively. This is the same as $dT=\vec \nabla T.d\vec l$ | |
| Aug 23, 2020 at 10:52 | comment | added | Vercassivelaunos | @enzotib: Thanks for the heads up. Should be better now. | |
| Aug 23, 2020 at 10:50 | history | undeleted | Vercassivelaunos | ||
| Aug 23, 2020 at 10:50 | history | edited | Vercassivelaunos | CC BY-SA 4.0 | deleted 659 characters in body |
| Aug 23, 2020 at 10:43 | history | deleted | Vercassivelaunos | via Vote | |
| Aug 23, 2020 at 10:43 | comment | added | Vincenzo Tibullo | The question was about the directional derivative of a vector field | |
| Aug 23, 2020 at 10:41 | history | answered | Vercassivelaunos | CC BY-SA 4.0 |