We define the action of a system as $$S(q)=\int_{t_1}^{t_2}L(t,q(t),q'(t))dt,$$ where $q(t)$ is the evolution of the system and $L$ is the Lagrangien. How can a stationary point of $S$ can describe the mouvement of a system ? So, how did we arrive at :

if $q$ is a stationary point of $$\int_{t_1}^{t_2}(T-V)(t,q(t),q'(t))dt,$$ then $q(t)$ describe the mouvement... (where $T$ is the kinetic energy and $V$ the potential). How did we arrived to that ? It looks so magic for me, I would like to understand a bit better the motivation behind.


marked as duplicate by Qmechanic lagrangian-formalism Dec 30 '18 at 19:51

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Consider the simplest system of a force-free particle in flat 3D space. It travels in a straight line with constant velocity.

But geometrically, a straight line is the shortest path between two points, so the trajectory minimizes the length along the curve


In other words, in the case of a free particle, its motion minimizes (makes stationary) a particular time-integral over the path. So it shouldn’t be too surprising that this same kind of stationary-integral thing works in more general cases.

For example, you could notice that uniform motion also makes stationary the integral of the velocity squared, which is proportional to the kinetic energy. And then, to introduce forces, you could look at whether the integral of the total energy $T+V$ is stationary. When you try this using the calculus of variations, you discover that it isn’t, but it becomes obvious that the integral of $T-V$ is!

So, to discover the Principle of Stationary Action, all you really need to know to get started is that a straight line is the shortest distance between two points.

  • $\begingroup$ It implies that the future position is known, which is not the case in Classical Mechanics. The knowns are the initial data and they determine where the particle will move. $\endgroup$ – Vladimir Kalitvianski Dec 30 '18 at 18:38
  • $\begingroup$ The Principle of Stationary Action is very much part of classical mechanics. It is simply a different way to formulate it. It does not imply that the future position is known. It simply says that the future position can be found from the initial data by requiring that the action to get to the future position be stationary. $\endgroup$ – G. Smith Dec 30 '18 at 19:37
  • $\begingroup$ Wrong. The future position is found from the initial data (or from the current position) by integrating the equation of motion: $x(t+dt)=x(t)+v(t)dt$, $v(t+dt)=v(t)+a(t)dt$. $\endgroup$ – Vladimir Kalitvianski Dec 30 '18 at 19:45
  • $\begingroup$ I didn’t dispute that. I’m sorry that you don’t understand the Principle of Stationary Action. I suggest reading Goldstein’s “Classical Mechanics”, which will clarify that they are equivalent approaches. $\endgroup$ – G. Smith Dec 30 '18 at 19:48

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