Timeline for Does an on-shell symmetry necessarily change the Lagrangian by a total derivative?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
replaced http://physics.stackexchange.com/ with https://physics.stackexchange.com/
|
|
| Feb 16, 2015 at 1:41 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
Added the word on-shell in title so that title question is same as question 2. Dear Brian Bi, if u don't like my changes, please roll back or use the parts u like.
|
| Feb 16, 2015 at 1:08 | answer | added | Qmechanic♦ | timeline score: 2 | |
| Feb 16, 2015 at 0:18 | comment | added | Sofia | @BrianBi I saw explanation of the meaning of on and off-shell, and I saw their use in connection with Noether's theorem. | |
| Feb 16, 2015 at 0:06 | comment | added | Brian Bi | @ACuriousMind Actually Qmechanic says "(off-shell) quasisymmetry". My bad. | |
| Feb 16, 2015 at 0:03 | comment | added | ACuriousMind♦ | 1. Qmechanic nowhere uses the terms "off-shell symmetry", so I don't know what you're asking here? 2. Look at the derivation of the E-L equations. Their solutions are precisely the points where the infinitesimal variations do not change the action, hence only change the Lagrangian by a total derivative. | |
| Feb 15, 2015 at 23:56 | history | asked | Brian Bi | CC BY-SA 3.0 |