Principle of least action

Nick R

What is the physically intuitive reason behind the fact that the action is stationary for the true path a particle takes?

I understand that a path that satisfies the Euler-Lagrange equation minimizes (or maximizes or "saddle-points") the result of the action functional.

I also understand that the euler-lagrange equation basically says F - F = 0 where F in one side is the force as it is defined (time change of momentum) and F on the other side is true simply by the mathematical definition of the potential energy (force integrated over distance).

Neither of these things explain why the quantity of action has any significance whatsoever. Is there a significance or did someone out there just make a very astute observation that this particular mathematical quantity does something under "true conditions"?

Explainations I have read of the lagrangian formulation of mechanics always simply state that the true path is that of least action and just show how the euler-lagrange equation arises from that. Sometimes they'll talk about the classic problem in calculus of variations that has to do with finding the shape of the ramp that will minimize the time taken for a marble to roll from starting point to ending point, but this appears to have no relation to the principle of least action, since most any ramp you can come up with will result in the marble going down a true, but different path. The marble could take any of those paths it has nothing to do with finding the marble's true path.

wywong

I am not sure if I fully understand your question which may be beyond my level. However I recalled something I read which might address your question. In one of his lectures, Richard Feynman explained why a particle always travel along a geodesic using the following argument (in my own wording):

The particle actually travels along different paths at the same time. When the times taken by adjacent paths are about the same, they are in phase and constructively interfere. When the times vary a lot, they tend to cancel each out. For a geodesic which is a local minimum and hence stationary, the former case applies.

I think the same argument (stationary->constructive inference of solutions) can be applied to your question.

Wai Wong

Andy Resnick

Because the action has units of energy*time, minimization of the action is similar to minimizing the time interval over which motion occurs- the Brachistochrone problem is usually stated as a time minimization.

Why the action is defined as the integral of the Lagrangian (as opposed to the Hamiltonian, for example), I am not sure.

http://en.wikipedia.org/wiki/Action_(physics)

The Lagrangian has many good properties, one of which is (I think) that the Lagrangian represents the energy available to perform useful work.

Nick R

Couldn't you also say the action has units of energy*time, so minimization of the action is similar to maximizing the potential energy over which motion occurs? (L = T - V)

But if were maximizing the potential energy, we're minimizing the kinetic energy (in a conserved system), so were minimizing the velocity, and thus maximizing the time...

Minimizing the time would likewise correspond to maximizing the kinetic energy, which increases the action...

Andy Resnick

I'm not sure what you are trying to say. The quantity being minimized is T-V (over the path), so it's not maximizing or minimizing either T or V, but of the excess T-V.

Nick R

The action is stationary for the true path: it could be at a minimum, maximum or a saddlepoint.