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Oct 23, 2023 at 9:32 history edited dennismoore94 CC BY-SA 4.0
replaced "orbital" with "path"
Oct 2, 2018 at 5:47 history edited Qmechanic CC BY-SA 4.0
Tried to make the title more descriptive
Oct 1, 2018 at 21:48 vote accept dennismoore94
Sep 30, 2018 at 22:45 comment added dennismoore94 OK, I think this is, what I actually don't understand: here $\delta p(t_1)=\delta p(t_2)=0$ is the same as $\delta \dot{q}(t_1)=\delta \dot{q}(t_2)=0$ would be in Lagrangian mechanics (again, thinking in cartesian coordinates), but in Lagrangian mechanics we don't have this kind of condition for the velocity. Or do we?
Sep 30, 2018 at 22:15 comment added Trevor Kafka I think you're confusing $\delta p$ and $p$. We're varying both $q$ and $p$ when we work in a variational principle but hold both $q$ and $p$ fixed at the endpoints of the path. $\delta q = \delta p = 0$ at the endpoints, even if the values of $q$ and $p$ are nonzero themselves.
Sep 30, 2018 at 22:12 history edited Qmechanic CC BY-SA 4.0
added 12 characters in body; edited tags
Sep 30, 2018 at 22:11 answer added Qmechanic timeline score: 5
Sep 30, 2018 at 21:54 comment added dennismoore94 Endpoints are fixed indeed ($\delta q(t_1)=\delta q(t_2)=0$), but for some function $f(\boldsymbol{q,p})$, $\delta f(\boldsymbol{q,p})=0$ would require $\delta p(t_1)=\delta p(t_2)=0$ too and I can't see how fixing the endpoints only (and not the derivatives!) guarantees this condition.
Sep 30, 2018 at 20:56 comment added Trevor Kafka Endpoints are held fixed during the path variation, so the variation of any function at the endpoints is zero.
Sep 30, 2018 at 20:40 history edited dennismoore94 CC BY-SA 4.0
edited title
Sep 30, 2018 at 16:33 history asked dennismoore94 CC BY-SA 4.0