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Is the principle of least action sort of a greedy algorithm that all mechanical systems follow?, sometimes to minimise and sometimes to maximise the quantity we call action, at each individual step.

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To echo ACuriousMinds answer, there is a difference between formulating a boundary value problem and an initial value problem. Optimisation procedures in my mind are initial value problems. The principle of stationary action is a well-posed boundary condition problem. The equivalence of the two is not guaranteed and I am not confident that the stationary action principle can be formulated as an initial value problem.

I would welcome informed comments or corrections to this. For further reading please see the following post.

Is the principle of least action a boundary value or initial condition problem?

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  • $\begingroup$ However, we end up using the solution for initial value problems as well. Thank you for linking the question. $\endgroup$ – Viesr Dec 13 '15 at 18:18
  • $\begingroup$ So, I can take the principle of least action to be sort of a 'guiding principle', sort of an intelligent guesswork principle for deriving the true equations of motion that we really require and who are actually causing the principle to be true? $\endgroup$ – Viesr Dec 13 '15 at 18:22
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No, the principle of least action (more properly, the principle of stationary action) is not an algorithm. In particular, it doesn't have steps.

It just states that the physically realized path of any system is a critical point of the action functional, which equivalently means its Euler-Lagrange equations are the equations of motion.

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  • $\begingroup$ Yes, but it is a critical point for each infinitesimal displacement. $\endgroup$ – Viesr Dec 13 '15 at 17:57
  • $\begingroup$ @Viesr: That depends on your action functional, which is typically local since it is almost always the integral of a local Lagrangian. I don't see what that would have to do with it being an "algorithm", though - it just isn't. That you can use algorithms to solve the equations of motion obtained through the principle has nothing to do with the principle itself. (This was a reply to your now-gone comment whether it is local) $\endgroup$ – ACuriousMind Dec 13 '15 at 17:57
  • $\begingroup$ @Viesr: "it is a critical point for each infinitesimal displacement" - I don't know what that means, or what it has to do with the question. $\endgroup$ – ACuriousMind Dec 13 '15 at 17:58
  • $\begingroup$ The way I am thinking about this is, at each step, the particle moves infinitesimally such that the action is stationary. Also the action is stationary for the complete path. So, it is achieved sort of in terms of these small small action stationarisation in some form, that leads to the whole action being stationary as a whole. $\endgroup$ – Viesr Dec 13 '15 at 18:02
  • $\begingroup$ In greedy algorithms to solve optimisation problems, what one does is choose the best option available at each individual step. So, the final optimal state is achieved as you solve little little optimisation problems to the best of your ability at each step leading to the whole solution being optimised. I am thinking that each infinitesimal step, the particle is doing sort of a similar thing and the final minimisation/extremisation of the action is ultimately coming out of the fact that the particle is extremising every infintesimal movement it makes. $\endgroup$ – Viesr Dec 13 '15 at 18:04

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