Timeline for Rigorous proof of Bertrand's Theorem for orbits under central force
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 1, 2021 at 8:26 | comment | added | Kashmiri | @mike stone, so basically we say if forces $f_i$ produce a closed orbit, then in the approximation they still would. Then we see that the approximation scheme gives us only two forces. Hence at best these two might be the ones producing closed orbits for any amount of deviations from circularity. Therefore we explicitly check them and find that indeed they are producing closed orbits. Hence we conclude that there are only two forces that can produce closed orbits for any amount of deviations from circularity. Is that what you meant? | |
| Nov 17, 2019 at 0:27 | comment | added | Ma Joad | Thank you, but I still have not got it :( I admit that I made a mistake in my last comment, but why is this true: "if the exact orbit is stably closed then so must the approximate orbit in the limit of very small perturbations"? How can I prove this? | |
| Nov 16, 2019 at 23:20 | history | edited | mike stone | CC BY-SA 4.0 |
added 1148 characters in body
|
| Nov 16, 2019 at 23:14 | comment | added | mike stone | I will respond by editing answer. | |
| Nov 16, 2019 at 22:11 | comment | added | Ma Joad | Even if the $O$ term is arbitrarily small, all we can say is that the exact solution and the approximate solution lies very close to each other. However, the author in the linked text is saying that if the approximate orbit is closed, then so is the exact orbit - this is NOT guaranteed by the fact that the approximate and exact solution lie very close to each other. | |
| Nov 16, 2019 at 14:38 | history | answered | mike stone | CC BY-SA 4.0 |