Timeline for Transporting Tangent Vectors when Taking Lie Derivatives
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Aug 22, 2020 at 22:22 | history | bounty ended | JG123 | ||
| Aug 22, 2020 at 22:22 | vote | accept | JG123 | ||
| Aug 15, 2020 at 20:48 | history | edited | Deschele Schilder | CC BY-SA 4.0 |
deleted 1 character in body
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| Aug 15, 2020 at 15:19 | comment | added | JG123 | @ChiralAnomaly I'll take a look. | |
| Aug 13, 2020 at 16:54 | comment | added | mike stone | Just draw the pictures of the heads and tails of the vectors! | |
| Aug 13, 2020 at 16:27 | comment | added | JG123 | Forgive me for asking, but what is this "pictorial interpretation" precisely? Or would it be more rewarding to go through the computations myself and see what I can gain from that? | |
| Aug 13, 2020 at 13:53 | comment | added | mike stone | Yes. The $X$ the flow at at $x+\eta Y$ is $X(x)+\eta Y\partial X$, so each term in $X^\nu \partial_\nu Y^\mu -Y^\nu \partial_\nu X^\mu$ has a pictorial interpretation. | |
| Aug 13, 2020 at 13:22 | comment | added | JG123 | Is there a way to understand that expression pictorially? | |
| Aug 13, 2020 at 13:12 | comment | added | mike stone | I guess it is at $x^\mu +\epsilon X^\mu +\eta \epsilon Y^\nu \partial_\nu X^\mu$ because $({\mathcal L}_XY)^\mu = X^\nu \partial_\nu Y^\mu- Y^\nu \partial_\nu X^\mu$. | |
| Aug 13, 2020 at 13:08 | comment | added | JG123 | Thank you for your answer @mike stone and I apologize for my late response. I guess the crux of my issue is what happens to the "head" of the $Y$ arrow. I am struggling to visualize where it ends up landing when the $Y$ arrow is carried along by the fluid (whose velocity field is $X$) to $x + \epsilon X$. | |
| Aug 12, 2020 at 21:52 | history | answered | mike stone | CC BY-SA 4.0 |