Timeline for Do geodesics avoid regions where the curvature diverges?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Jul 31, 2023 at 21:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jul 1, 2023 at 20:32 | comment | added | Daniel Asimov | Now that I've thought about it, it seems possible (but very counterintuitive to me): | |
| Jul 1, 2023 at 19:09 | comment | added | Daniel Asimov | I had never thought about this before, but am having trouble imagining a smooth 2D riemannian manifold M with boundary such that its gaussian curvature K(p) → ∞ as p → ∂M. Is that even possible? | |
| Jul 1, 2023 at 18:26 | answer | added | Anton Petrunin | timeline score: 2 | |
| Oct 13, 2022 at 22:31 | comment | added | Leo Moos | @WillJagy Sure, Clairaut's relation is essentially the heuristic which suggested to me that 'most' geodesics ought to avoid regions where the curvature is 'large'. However, I am not sure it gives much more here, as I don't want to impose rotational symmetry. | |
| Oct 13, 2022 at 21:14 | comment | added | Will Jagy | the thing that comes to mind is the Clairaut relation for surfaces of revolution such as $z = x^2 + y^2,$ with positive curvature near the origin. Every geodesic that does not pass through the origin (not a meridean) achieves a minimum distance from it, so every non-meridean geodesic can be produced by specifying a point where it is parallel to the $xy$ plane. en.wikipedia.org/wiki/… | |
| Oct 13, 2022 at 21:12 | history | edited | Leo Moos | CC BY-SA 4.0 |
rephrased question
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| Oct 13, 2022 at 15:22 | history | asked | Leo Moos | CC BY-SA 4.0 |