Timeline for Does an on-shell symmetry necessarily change the Lagrangian by a total derivative?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:40 | history | edited | CommunityBot |
replaced http://physics.stackexchange.com/ with https://physics.stackexchange.com/
|
|
| Feb 16, 2015 at 14:46 | comment | added | Qmechanic♦ | $\uparrow$ Yes. | |
| Feb 16, 2015 at 4:03 | comment | added | Brian Bi | So, basically, off-shell quasisymmetry means that the action always changes by a boundary term (or not at all) even if the original state was off-shell, is that right? | |
| Feb 16, 2015 at 1:49 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
Added explanation
|
| Feb 16, 2015 at 1:25 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
Added explanation
|
| Feb 16, 2015 at 1:08 | history | answered | Qmechanic♦ | CC BY-SA 3.0 |