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Aug 17, 2017 at 11:00 comment added user35305 I've updated my OP to reflect your suggested edits.
Aug 16, 2017 at 20:57 comment added user35305 Thanks for your response. Ideally I would like to prove it in general, but I wasn't quite sure how to do it, so for now I've restricted to no explicit time dependence in the Lagrangian and $\delta t$ time-independent. Also, why is it natural to try $q'(t')=q(t)$ in the case of time translation? Is it because one defines that the coordinates change as $q'(t)=q(t+\delta t)$ under time translation, and then one can simply make a change of variables, $t\rightarrow t-\delta t$ such that $q'(t')=q(t)$?
Aug 16, 2017 at 20:47 history edited Qmechanic CC BY-SA 3.0
Added explanation
Aug 16, 2017 at 19:26 history answered Qmechanic CC BY-SA 3.0