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I find this rather suspect. Write it as:

S[q(t), t_0, t] = int_{t_0}^{t} L(q(t'), q̇(t'), t') dt'

Formally, S is a function of the upper limit of integration, and dS/dt = L, yes? I don't see why we can't treat it this way. It is the arc-length formula of the space time metric, expressed as an integral in one privileged time coordinate ( https://en.wikipedia.org/wiki/Relativistic_Lagrangian_mechan... ). Though, it's Lorentz invariant, and we could express it as an integral along the trajectory of the particle, in which case it's just = a constant factor * the proper time. The whole idea of "varying q while keeping the boundaries q[t_0], q[t], t_0, and t fixed" is perfectly understandable as a condition on q, but doesn't stop us from using the formula for S in other ways--for one, to come up with a general principle behind the condition on q.

Actual trajectories of closed systems do conserve momentum. The earth is interacting with the pendulum.



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


There'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.




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