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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.

Qmechanic
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idm
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1 Answers1

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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

$$\int_{t1}^{t2}ds=\int_{t1}^{t2}\left|\frac{d\mathbf{r}}{dt}\right|dt.$$

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.

J.G.
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G. Smith
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