# How do you find potential by Lagrangian formalism?

Suppose a ball is falling towards earth and hence by Lagrange equation we can find $T$ and $V$ where $V$ is $mgh$. But we know $V$ only because we know $F = mg$. Now since Lagrange equation doesn't require the concept of force (correct me if I am wrong), how do you find potential in a place where we have no intuition of force and are not allow to find it.

• You don't "find" the Lagrangian or potentials in the Lagrangian formalism. The Lagrangian formalism uses a given Lagrangian as its starting point. You need to use an educated guess (like guessing the potential from the experimentally observed force) for it and see whether it predicts the physics correctly. – ACuriousMind Sep 26 '15 at 20:47
• Related: physics.stackexchange.com/q/20298/2451 and links therein. – Qmechanic Sep 26 '15 at 22:03

Well I think this might be your problem; I've certainly never heard it said that you are not allowed to find forces. The Euler-Lagrange equations are simply another tool to finding the dynamics, but that doesn't mean you have to start from scratch and cannot use knowledge gained from other courses (e.g. that gravitational potentials go as $mgh$) for use in solving your problems.
In a general Lagrangian formalism, $L$ doesn't equal $T - V$. Rather it is a function of some field (be it scalar field, vector field, or whatever other field that is useful...) $\phi$, derivatives of $\phi$ and spacetime (x,y,z,t). This function is chosen so that the equations of motion produce the correct physical phenomena. In general, canonical momentum conjugate $\pi$ is defined by $\frac{\partial L}{\partial (\frac{\partial\phi}{\partial t})} = \pi$ And this is the most obvious bridge between the language of Lagrangian Field theory and classical mechanics. Other than that simple expressions in mechanics don't carry over to complicated fields...