Timeline for Geodesics and potential function
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 24, 2020 at 3:09 | comment | added | Deane Yang | I was referring to the second equation, which has no $R$. | |
| Aug 23, 2020 at 23:15 | comment | added | Bruno Peixoto | Does the last paragraph clarify my intentions regarding my doubt? I enjoy José Figueroa's answer. Michael provides some intuition, but it lacks rigor. Deane, R regards the Rayleigh function and not the curvature notation. | |
| Aug 23, 2020 at 23:12 | history | edited | Bruno Peixoto | CC BY-SA 4.0 |
added 311 characters in body
|
| Aug 23, 2020 at 17:01 | comment | added | Deane Yang | Perhaps you could revise what you wrote and be more specific and precise about what you want to know? I would observe, however, if there is a non constant potential, then there is a force acting on the particle’s motion and therefore the path with least energy is no longer a geodesic. The second equation can be viewed geometrically as saying something about the curvature of the path. | |
| Aug 23, 2020 at 14:32 | comment | added | Michael Engelhardt | I'm also not sure I really understand the issue, but my reaction is that the geodesic equation and its generalizations on the one side, and the boundary conditions on the other side are separate ingredients. Ideally, you would solve the geodesic equation in general, leaving you with integration constants, and then you fix the integration constants to the boundary data, in whatever way they're given. So the geodesic equation, and additional potential, etc. terms don't depend on the form of the boundary data, these are separate pieces of information. Maybe I'm missing something. | |
| Aug 23, 2020 at 12:50 | comment | added | José Figueroa-O'Farrill | @BrunoPeixoto I'm not sure what the question actually is. If I re-interpret it as asking for a geometrical set-up to discuss lagrangian/hamiltonian systems with potentials, then I have two comments: (1) you may be able to rewrite the equation as a geodesic equatoin for an affine connection which is not necessarily metric (cf. Newton-Cartan theory) and (2) you may be able to view it as a geodesic equation in an auxiliary space relative to the Levi-Civita connection (cf. the Eisenhart lift). | |
| Aug 23, 2020 at 12:16 | comment | added | Bruno Peixoto | Why doesn't it have a mathematical answer? The physical statement relies on the mathematical toolset to perform its calculation. | |
| Aug 23, 2020 at 12:14 | comment | added | Ben McKay | The role of those terms is the same as their role in Newtonian classical mechanics, where the Riemannian manifold is flat Euclidean space. So you should ask a physicist or look in a classical mechanics textbook to see how such terms appear. Your questions have no mathematical answer, only an answer in physical intuition. | |
| Aug 23, 2020 at 12:10 | comment | added | Bruno Peixoto | The questions are: What is the geodesic equation when the boudary values are not points A and B but initial position and velocity? What is the influence of the potential term on the geodesic equation when employed as a Point A B condition? The same for with Rayleigh i.e . dissipative function? | |
| Aug 23, 2020 at 12:10 | history | edited | Bruno Peixoto | CC BY-SA 4.0 |
added 6 characters in body
|
| Aug 23, 2020 at 11:35 | comment | added | Jaap Eldering | What is your question here? From the looks of it, $R$ is just another potential term just like $V$, that could be absorbed into it. | |
| Aug 23, 2020 at 6:30 | history | edited | YCor |
edited tags
|
|
| Aug 23, 2020 at 1:05 | history | edited | Bruno Peixoto | CC BY-SA 4.0 |
added 2 characters in body
|
| Aug 14, 2020 at 18:04 | review | Close votes | |||
| Aug 30, 2020 at 3:02 | |||||
| Aug 14, 2020 at 17:02 | review | First posts | |||
| Aug 14, 2020 at 17:55 | |||||
| Aug 14, 2020 at 16:54 | history | asked | Bruno Peixoto | CC BY-SA 4.0 |