Timeline for Least Action in General Relativity
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 5, 2021 at 6:06 | comment | added | user1379857 | Sorry, I should have said extremize the action. You are right that the condition to extremize action is certainly not $d L/dt = 0$. The equation $d L/dt = 0$ is certainly not true for a general Lagrangian-- however it is true on shell for OP's example and for a free non relativistic particle. | |
| Mar 5, 2021 at 6:05 | history | edited | user1379857 | CC BY-SA 4.0 | deleted 4 characters in body |
| Mar 4, 2021 at 23:35 | comment | added | user92177 | Also, the lagrangian is not what's extremized. It's the action. "The Least Action Principle". Action is the time integral of L and it's extremized in terms of variations. | |
| Mar 4, 2021 at 23:31 | comment | added | user92177 | Perhaps I am misunderstanding, but it seemed that you were claiming $ \frac{d}{dt} L = 0 $ is the condition being extremizing the action. The condition is $ \delta \int dt L = 0 $. | |
| Mar 3, 2021 at 19:47 | comment | added | user1379857 | Yes, in the quantum path integral you sum over all virtual paths, not just the classical one | |
| Mar 3, 2021 at 13:43 | comment | added | Deschele Schilder | Like virtual particles in qft. They are off-shell too. | |
| Mar 3, 2021 at 0:49 | comment | added | user1379857 | That's right. It's a path which doesn't extremize the Lagrangian, and therefore is not "physical" (i.e. not a path the particle would take) but is a path nonetheless. It is sometimes called a "virtual path." | |
| Mar 2, 2021 at 23:19 | comment | added | Deschele Schilder | So the sin$(\omega t)$ trajectory is an imaginary one, in the light of the Lagrangian (which involves only a kinetic part, and no potential part corresponding to the sin$(\omega t)$-motion)? | |
| Mar 1, 2021 at 17:51 | history | edited | user1379857 | CC BY-SA 4.0 | typos |
| Mar 1, 2021 at 16:58 | comment | added | user92177 | $ L = H $ implies that $ \frac{d}{dt}L = 0 $ for on-shell which is true in your example. Claiming that $ \frac{d}{dt}L = 0 $ for all on-shell trajectories is not necessarily true. | |
| Mar 1, 2021 at 4:03 | history | answered | user1379857 | CC BY-SA 4.0 |