What exactly is the action?
For a wide class of systems the Lagrangian is the difference between kinetic and potential energy, L=T-V. (Not their sum — that is the energy; the difference is what least action minimizes.) The action is L integrated over the time of the motion.
The action is a functional: feed it an entire path \mathbf{q}(t) and it returns a number.
The calculus of variations
To find the stationary path, we use the calculus of variations. Take the true path q(t) and compare it with a nearby wiggled path q(t)+\delta q(t), where the wiggle \delta q vanishes at the two fixed endpoints. Demand that the resulting change in the action vanish to first order.
The first variation of the action, from perturbing both q and its velocity \dot q.
- Note that \delta\dot q = \tfrac{d}{dt}\delta q — varying the path also varies its slope.
- Integrate the second term by parts to move the time-derivative off \delta q.
- The boundary term \left[\tfrac{\partial L}{\partial \dot q}\delta q\right]_{t_1}^{t_2} vanishes because \delta q=0 at both endpoints.
- What remains is \int(\cdots)\,\delta q\,dt = 0 for every choice of \delta q.
- By the fundamental lemma of the calculus of variations, the bracket itself must be zero everywhere.
The Euler-Lagrange equation
The bracket that must vanish is the Euler-Lagrange equation — the beating heart of Lagrangian mechanics. There is one such equation for each generalized coordinate, and each is a second-order differential equation for the motion.
One equation per coordinate q_i — the same form in Cartesian, polar or any coordinates you like.
Newton's second law recovered as a special case of Euler-Lagrange.
What we assumed — and its limits
The great payoff is coordinate independence: the Euler-Lagrange equation has the identical form no matter which coordinates you chose, so you write L in whatever variables are convenient and turn the crank. That is why a hard Newtonian problem often becomes a short Lagrangian one.