Timeline for Physicist's request for intuition on covariant derivatives and Lie derivatives
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 16, 2020 at 18:59 | comment | added | Deane Yang | Looks better to me. | |
| Oct 16, 2020 at 18:18 | history | edited | Qfwfq | CC BY-SA 4.0 |
(I hope this time it's less wrong)
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| Oct 16, 2020 at 18:12 | history | edited | Qfwfq | CC BY-SA 4.0 |
(I hope this time it's less wrong)
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| Oct 16, 2020 at 18:06 | history | edited | Qfwfq | CC BY-SA 4.0 |
(I hope this time it's less wrong)
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| Sep 13, 2011 at 9:50 | comment | added | Deane Yang | As for the description of the Lie derivative, I should have been more careful. It is only the first sentence that I find misleading. The second sentence appears to be a correct description of the Lie derivative. | |
| Sep 13, 2011 at 9:49 | comment | added | Deane Yang | "I agree you don't have to compute it along a geodesic, but it's what is it morally supposed to mean": I don't really agree with the second half. It's just a directional derivative associated with the tangent vector. A geodesic works, but in this case plays no special role. So mentioning it is misleading. It is true that when you first learn about directional derivatives on $R^n$, you tend to define them in terms of straight lines. However, it is rather important in differential geometry to understand that straight lines are not special when computing or defining a directional derivative. | |
| Sep 12, 2011 at 21:45 | comment | added | Qfwfq | @DeanYang: your second remark make me think my interpretation of the Lie derivative is not quite correct as stated. Is there a way (given $T$ and $V$ in a neighborhood of $x$) to compute the Lie derivative $\mathcal{L}_VT$ within a fixed finite dimensional vector space (depending on $x$)? | |
| Sep 12, 2011 at 21:38 | comment | added | Qfwfq | @DeanYang: of course I agree the connection only depends on $v$, but he asked for an intuitive explanation, and I think the most intuitive meaning I can attach to the covariant derivative along a direction is: "directional derivative along the stright line (with respect to the connection, or metric if the connection is metric)". I agree you don't have to compute it along a geodesic, but it's what is it morally supposed to mean. | |
| Sep 12, 2011 at 17:27 | comment | added | Deane Yang | And the explanation of Lie derivative is quite misleading, because it makes it seem like the Lie derivative depends only on $T$ along the flow line of $V$. In fact, it depends on how both $T$ and $V$ behave in a neighborhood of the flowline. If you really want to use only data along the flow line, then you need to know their first order jets along the curve. | |
| Sep 12, 2011 at 16:18 | comment | added | Deane Yang | I don't see any reason why a covariant derivative has to be computed using a geodesic. You get the same answer using any curve with tangent vector $V$. | |
| Sep 12, 2011 at 15:50 | history | edited | Qfwfq | CC BY-SA 3.0 |
added 9 characters in body
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| Sep 12, 2011 at 15:44 | history | answered | Qfwfq | CC BY-SA 3.0 |