Timeline for Tying knots via gravity-assisted spaceship trajectories
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Jun 3, 2018 at 6:41 | comment | added | Alec Rhea | @JosephO'Rourke Thank you for the clarification, I meant that my statement about realizing differentiable paths using a Lagrangian needs the paths to be twice differentiable at least for it to be true. I think fedja's intuition is more precise, and correct -- to make the observation slightly more explicit, we can note that the contribution to the spaceship Lagrangian for each planet is of the form $G\frac{m_i}{r_i},$ so as suggested we should be able to place sufficiently small masses sufficiently close to the unperturbed path to reproduce any loop. | |
| Jun 2, 2018 at 23:18 | comment | added | Geoffrey Irving | I would expect fedja’s method to be extendable to produce exactly periodic orbits, assuming you close off the induction with some sort of fixpoint theorem. | |
| Jun 2, 2018 at 22:08 | history | edited | Joseph O'Rourke | CC BY-SA 4.0 |
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| Jun 2, 2018 at 22:03 | comment | added | Joseph O'Rourke | Just a minor comment re @fedja's "Now tie the knot with a broken line ": That is a stick knot. | |
| Jun 2, 2018 at 0:20 | comment | added | fedja | You can make a hyperbolic orbit around a single planet, which on large scale is just a turn. Now tie the knot with a broken line and place masses close to the turns. Note that you can use an arbitrarily small mass to make any turn in a small external field if you place it strategically near the empty space trajectory (this requires proof, of course, but intuitively it is clear). Thus, adding new turns won't spoil the existing ones and you should be able to happily do the induction. I'll try to make some rigorous sense of it later. | |
| Jun 1, 2018 at 23:51 | answer | added | Piyush Grover | timeline score: 8 | |
| Jun 1, 2018 at 23:29 | history | edited | Joseph O'Rourke | CC BY-SA 4.0 |
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| Jun 1, 2018 at 23:26 | comment | added | Joseph O'Rourke | @AlecRhea: I don't think the smoothness of the path matters, as a knot is a topological object. If we assume the trajectory never directly hits a point-planet, then I believe the path is at least differentiable. | |
| Jun 1, 2018 at 22:54 | comment | added | Alec Rhea | I think I meant ‘smooth’ instead of ‘differentiable’ above. | |
| Jun 1, 2018 at 22:49 | comment | added | Alec Rhea | If they are effectively the same then my training as a physicist tells me that the answer is ‘yes up to ambient isotropy’ since differentiable paths through $\mathbb{R}^3$ can be realized as the trajectory of a particle (or spaceship) subject to some Lagrangian and initial conditions, and we should be able to generate this Lagrangian with a collection of point masses (planets) in $\mathbb{R}^3$. This comes from physical intuition though, so a grain of salt is appropriate. | |
| Jun 1, 2018 at 21:37 | comment | added | Alec Rhea | Does the differentiability of $K$ as a path matter? For instance, if you were to imagine a knot with sharp corners between some planets, is that effectively the same as (ambiently isotropic to?) a knot following ‘almost’ the same path but with smooth corners? | |
| Jun 1, 2018 at 12:11 | history | edited | Joseph O'Rourke | CC BY-SA 4.0 |
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| Jun 1, 2018 at 11:59 | comment | added | Joseph O'Rourke | @ThomasRot: Yes. I suspect that is considerably more difficult to achieve. | |
| Jun 1, 2018 at 11:33 | comment | added | Thomas Rot | You can also ask if every knot type occurs as a periodic orbit in the system. | |
| Jun 1, 2018 at 11:26 | history | asked | Joseph O'Rourke | CC BY-SA 4.0 |