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Sep 15 at 6:02 history edited Qmechanic CC BY-SA 4.0
Added explanation
Sep 15 at 4:25 comment added watahoo I understand that a conserved charge can be derived by an infinitesimal quasi-symmetry under which the Lagrangian is changed by $\frac{d}{dt}F(q,\dot{q},t)$. My question is different. I am asking why this can be understood as a symmetry even although the action is changed more than a constant.
Sep 15 at 4:15 comment added watahoo I of course agree with 1 & 2, but am not sure about 3. I looked at the link you shared and found that that question is similar to mine, although the OP cares more about the EOM rather than the action by itself. I read your answer but you did not explain why "quasisymmetry variations do not need to satisfy boundary conditions"
Sep 14 at 15:44 history answered Qmechanic CC BY-SA 4.0