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A Lagrangian is like newtons equation, its a description of the system, you don't have any mathematical definition for it since it is a law of physics and not maths.

Edit:

> but each subsystem's world line curves to achieve SOME property. What is this property?

You have to provide that as a part of the explanation of the system. Lagrangians are just dumb functions, you have to be smart about how you choose them, there is no magic here you just have to understand the physics and construct the Lagrangian that has whatever properties you want.

I'm pretty sure you're answering a different question.

I'm saying:

- it appears that the "stationary action" is the same as "worldline follows a geodesic of the spacetime metric" (https://en.wikipedia.org/wiki/Relativistic_Lagrangian_mechan...). I first heard this from a particle physics professor in undergrad.

- for a compound system, the individual elements follow worldline geodesics, with the other particles factored out into a "potential" / as force terms in E-L equation. These represent how 1 particle, when distorted from its non-interacting trajectory, trades off against the other particles distorting from their OWN free trajectories. It's analogous to how heat energy flows between two systems to maximize their joint entropy; here, the particles' trajectories distort each other via forces to minimize their joint spacetime arc length. (As measured, presumably, in ANY reference frame)

- yes, in any specific context, the Lagrangian represents HOW those things trade off, in the same way that in any specific thermodynamics scenario the microstate structure can tell us d(entropy)/d(energy) and let us compute the equilibrium state.

- but how to complete the analogy to thermodynamic equilibrium--what is the "temperature", exactly, what is "heat"? In a QFT context, the interaction is ITSELF a particle, and has its own contribution to the Lagrangian and action, but how to think about it classically?

It has always seemed to me that this line of thinking is the most natural way to express "Stationary Action".

I have created a resource that I think addresses your dissatisfaction.

The information is available on physics.stackexchange https://physics.stackexchange.com/a/670705

I use 'Hamilton's stationary action' to refer to the action concept of Classical Mechanics.

For Hamilton's stationary action the standard presentation is that it is demonstrated that F=ma can be recovered from Hamilton's stationary action.

Here's the thing: in physics it is common that derivation can be performed in either direction, and that applies in this case too. Hamilton's stationary action can be derived from F=ma

The derivation proceeds in two stages: 1. Derivation of the Work-Energy theorem from F=ma 2. Demonstration that in all cases where the Work-Energy theorem holds good Hamilton's stationary action will hold good also

Importantly, it's not retracing of the steps. The from-F=ma-to-Hamilton derivation hinges on the Work-Energy theorem. It's a different path altogether.

The steps of the derivation show why Hamilton's stationary action holds good. It achieves the justification you are looking for.

The derivation that I present is for the case of Hamilton's stationary action specifically; I'm positive the reasoning generalizes to all areas where an action concept is applied.

The demonstration is illustrated with interactive diagrams. (On physics.stackexchange the diagrams are posted as animated GIFs, the frames of the GIF are successive screenshots of the interactive diagram.)

Each diagram has one or more sliders, to explore variation of a trial trajectory. The diagram shows how the kinetic energy and potential energy respond to variation sweep.

The interactive diagrams are on my own website: http://cleonis.nl/physics/phys256/energy_position_equation.p...

> S, action, is the arc-length of a world-line. L = dS/dt. It is a natural law that systems move along "straight" world-lines, geodesics, in the absence of interactions.

This is a roughly correct description of GR, where the curvature is given by the Einstein field equations. It's not true in classical electromagnetism, or in QM.

> The "future-looking" nature of the least action principle is unsatisfying: sure, the action is minimized over an interval in time. Since L = dS/dt, the E-L equations are that same principle expressed at a single moment in time--right?

Yes, although the global principle is that the action is stationary, not minimal.

> Is there some sense in which L is "orthogonal" to the space spanned by constant energy and momentum?

Not quite, but you're on the right track. Conserved quantities of integrable systems are "orthogonal" to the Hamiltonian H, in that the poisson bracket `{f, H} = -df/dt` and so `{f, H} = 0` for constant `f`. The Lagrangian arises as the Legendre transform of H.

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.

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.

> every system minimizes the action

Action is stationary, not necessarily minimal.