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* There is a function L(q,q'), the Lagrangian. How do I compute this function? Does it have an analytical expression? How does it depend on p or F? I tend to t

* There is a function L(q,q'), the Lagrangian. How do I compute this function? Does it have an analytical expression? How does it depend on p or F? I tend to think of the Lagrangian as "something that someone defined that happens to be useful". In classical mechanics, the notion of total energy (kinetic plus potential) is something we can all understand. The Lagrangian, which is kinetic minus potential energy, looks strange but happens to observe unusually useful properties. If you integrate the Lagrangian of a point particle, it will always take the path that minimizes the integral. We call this integral the action and the universe seems to behave in a way that minimizes the action. I don't admit to have a deeper understanding of why this happens but every experiment ever done seems to observe this. The equation after Fig. 19-3 gives an analytic expression for the action: https://www.feynmanlectures.caltech.edu/II_19.html https://www.feynmanlectures.caltech.edu/II_19.html Different physical systems will have different Lagrangians but once formulated, the physical system will always behave in a way that minimizes the action. In principle, if you can formulate a Lagrangian for the universe, you can predict how it will behave because you know it will behave in a way that minimizes the action. EDIT: Added more details.

sixo 4y ago
I've always been unsatisfied with this description of the Lagrangian. I spent some time trying to figure out a better justification, and never was happy with it. Maybe someone reading this thread knows? A few things I have but could not fully translate into math: * 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. With an interaction the combined system (original + interactor) still moves in a "straight"/minimal-length world line (wrt to the proper time of the combined system) but each subsystem's world line curves to achieve SOME property. What is this property? * 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? * Is there some sense in which L is "orthogonal" to the space spanned by constant energy and momentum? Actual trajectories conserve E and p; their variation in E or p along the trajectory is 0. The actual trajectory is DEFINED by being the one where variation of L, _off the trajectory_, is 0. So lines of constant L are orthogonal to the subspace of constant E and p—is there more to it than that? Could we run that backwards to get a satisfactory derivation of L? Is this the same as saying that lines of constant L are lines of MAXIMAL variation in E and p (presumably, in 4-momentum)?
- Jensson 4y ago
  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.
  - sixo 4y ago
    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_mechanics 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".
- nh23423fefe 4y ago
  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.
eigenket 4y ago
Minor correction, but actually classical paths make the action stationary, they don't minimize it. All minima are stationary points but not all stationary points are minima. Sometimes the classical path maximises the action, and sometimes it is a saddle point.
deleted 4y ago
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sixo 4y ago

I've always been unsatisfied with this description of the Lagrangian. I spent some time trying to figure out a better justification, and never was happy with it. Maybe someone reading this thread knows? A few things I have but could not fully translate into math: * 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. With an interaction the combined system (original + interactor) still moves in a "straight"/minimal-length world line (wrt to the proper time of the combined system) but each subsystem's world line curves to achieve SOME property. What is this property? * 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? * Is there some sense in which L is "orthogonal" to the space spanned by constant energy and momentum? Actual trajectories conserve E and p; their variation in E or p along the trajectory is 0. The actual trajectory is DEFINED by being the one where variation of L, _off the trajectory_, is 0. So lines of constant L are orthogonal to the subspace of constant E and p—is there more to it than that? Could we run that backwards to get a satisfactory derivation of L? Is this the same as saying that lines of constant L are lines of MAXIMAL variation in E and p (presumably, in 4-momentum)?

Jensson 4y ago
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.
- sixo 4y ago
  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_mechanics 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".
nh23423fefe 4y ago
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.

eigenket 4y ago

Minor correction, but actually classical paths make the action stationary, they don't minimize it. All minima are stationary points but not all stationary points are minima. Sometimes the classical path maximises the action, and sometimes it is a saddle point.

deleted 4y ago

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