2 ms·
Hmm I still don't see why you can make that argument? I'm thinking that by chain rule d/dt S = dq/dt dS/dq. But we assert that the variation of S is zero for r
by nh23423fefe 4y ago
Hmm I still don't see why you can make that argument?
I'm thinking that by chain rule d/dt S = dq/dt dS/dq. But we assert that the variation of S is zero for real trajectories. so dS/dq = 0
- sixo 4y agoThere's no way that's right. It's mixing up "t" as the upper limit of integration (S[q] is really S[q, t_0, t]) vs t as the argument of q; dS/dt in the latter sense doesn't even make sense because the argument of q is "integrated out" in the expression of S.