3 ms·

L <> d/dt S, because dS/dt = 0 S is a functional of q which is a function of t, the t binding doesn't really exist on the left to take a derivative. every sys

L <> d/dt S, because dS/dt = 0 S is a functional of q which is a function of t, the t binding doesn't really exist on the left to take a derivative. every system minimizes the action. a subsystem is just some part of q (the coordinate of configuration space) Actual trajectories don't conserve momentum though. A pendulum in a potential has varying momentum.

hither_shores 4y ago
> every system minimizes the action Action is stationary, not necessarily minimal.
sixo 4y ago
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_mechanics 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.
- 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 ago
    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.

hither_shores 4y ago

> every system minimizes the action Action is stationary, not necessarily minimal.

sixo 4y ago

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_mechanics 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.

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 ago
  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.