Timeline for Why do we need a metric to define gradient?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Feb 4, 2021 at 9:33 | answer | added | Andrew Steane | timeline score: 1 | |
| Dec 31, 2020 at 1:52 | comment | added | MaximusIdeal | I think a simple observation can clarify a lot: Note that the formula for the gradient in Cartesian coordinates vs spherical coordinates is completely different. If you were told to calculate the gradient, how would you know what coordinates you were working in to use the correct formula? The metric is what would tell you. | |
| Dec 31, 2020 at 1:35 | answer | added | MathFoliage | timeline score: 2 | |
| Feb 24, 2019 at 19:57 | history | edited | Qmechanic♦ |
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| Feb 23, 2019 at 14:54 | answer | added | Manza | timeline score: 11 | |
| Jul 18, 2017 at 13:20 | answer | added | James Baugh | timeline score: 2 | |
| Jun 22, 2014 at 21:34 | history | edited | Qmechanic♦ |
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| Jun 20, 2014 at 21:43 | comment | added | joshphysics | @Christoph I completely agree that if we want to define the gradient as a vector field, then we need the tangent-cotangent isomorphism to do so and that the metric provides a natural method for generating it. I am, however, used to thinking of the gradient as the differential itself, not its dual. Having said this, I did some literature searching, and I think it's more common for the gradient to be defined as the corresponding vector field, so I'm inclined to agree that my comment is a bit misleading. | |
| Jun 20, 2014 at 20:38 | comment | added | Christoph | @joshphysics: your comment is misleading: if we want to define a vector field dual to the differential (which is what the gradient is), we need to specify an isomorphism between the tangent and cotangent spaces because there's no canonical one; a metric (or more generally, any non-degenerate bilinear form) does just that; covariant derivatives do not enter the picture: the covariant derivative of a function is the plain old differential | |
| Jun 20, 2014 at 19:03 | answer | added | Christoph | timeline score: 5 | |
| Jun 19, 2014 at 23:10 | comment | added | joshphysics | I'd say it's a bit of a stretch to say that you need a metric to define the gradient of a scalar. Even if you're working with a smooth manifold that has no defined notion of metric, the partial derivatives $\partial_i$ are well-defined; at each point they form the basis for the tangent space generated by the given system of local coordinates. However, if you want the notion of a covariant derivative operator that can act not just on scalar fields, but also on vector fields and tensor fields of higher rank, then you need a connection, and one can generate such a connection via a metric. | |
| Jun 19, 2014 at 21:23 | answer | added | Jordan | timeline score: 3 | |
| Jun 19, 2014 at 21:20 | answer | added | Robin Ekman | timeline score: 16 | |
| Jun 19, 2014 at 21:12 | history | asked | Jiang-min Zhang | CC BY-SA 3.0 |