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Sep 29, 2015 at 19:17 review Late answers
Sep 30, 2015 at 8:14
Apr 10, 2014 at 0:19 history edited Paul Masson CC BY-SA 3.0
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Jan 11, 2014 at 1:24 comment added Qmechanic @Dan: No, my answer seems to agree with Paul Masson's answer: It is not a symmetry of the Lagrangian. Rather it is a symmetry of the action (up to boundary terms). Phrased equivalently, it is a so-called quasisymmetry of the Lagrangian, cf. this Phys.SE answer.
Jan 11, 2014 at 0:32 review First posts
Jan 11, 2014 at 0:36
Jan 11, 2014 at 0:26 comment added Dan Qmechanic's answer seems to imply the presence of a transformation that leaves the Lagrangian constant by the inverse Noether's theorem. Does your answer disagree with theirs, or am I missing something?
Jan 11, 2014 at 0:16 history answered Paul Masson CC BY-SA 3.0