Answer to Question #294224 in Mechanics | Relativity for Delightful

Question

 Derive Lagrange’s equation for a particle in a plane.

Answer 1

The lagrangian of a particle in a plane in cartesian coordinates is given by

$\mathcal{L} = \frac{m\dot{x}^2}{2} + \frac{m\dot{y}^2}{2}- V(x,y)$

The action is thus given by

$S = \int_{t_0}^{t_1} \mathcal{L}(t, x, \dot x, y, \dot y) dt$

Let us consider the variation of this action :

$\delta S = \delta \int \mathcal{L} dt = \int \delta \mathcal{L} dt$

Lagrangian variation is given by

$\delta \mathcal L = m\dot x \delta \dot x + m \dot y \delta \dot y - \delta V(x,y)$

Expanding $\delta V$ gives

$\delta V(x,y) = \partial _x V \delta x + \partial _y V \delta y$

We also know that $\delta \dot x = \dot{(\delta x)}$ and thus the first two terms can be integrated by parts :

$\delta S = [m \dot x \delta x+m\dot y \delta y]_{t_0}^{t_1} - \int (m\ddot x \delta x + m\ddot y \delta y + \partial_x V \delta x + \partial_y V \delta y)dt$

We fix the trajectory at $t_0, t_1$ so the first term is zero. The second term should be zero for an arbitrary variation $\delta x, \delta y$. This gives us two Euler-Lagrange equations :

$m\ddot x = -\partial_x V \\ m\ddot y = -\partial_y V$

which correspond exactly to Newton equations for a particle in a plane. These equations could be derived for the most general form of $\mathcal{L}$ by the same approach to obtain a system of Euler-Lagrange equations for a particle in a plane :

$\frac{d}{dt}\frac{\partial \mathcal L}{\partial \dot q}-\frac{\partial \mathcal L}{\partial q}=0$ for every coordinate $q$ of the particle.