Question #294224

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


1
Expert's answer
2022-02-08T08:16:11-0500

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

L=mx˙22+my˙22V(x,y)\mathcal{L} = \frac{m\dot{x}^2}{2} + \frac{m\dot{y}^2}{2}- V(x,y)

The action is thus given by

S=t0t1L(t,x,x˙,y,y˙)dtS = \int_{t_0}^{t_1} \mathcal{L}(t, x, \dot x, y, \dot y) dt

Let us consider the variation of this action :

δS=δLdt=δLdt\delta S = \delta \int \mathcal{L} dt = \int \delta \mathcal{L} dt

Lagrangian variation is given by

δL=mx˙δx˙+my˙δy˙δV(x,y)\delta \mathcal L = m\dot x \delta \dot x + m \dot y \delta \dot y - \delta V(x,y)

Expanding δV\delta V gives

δV(x,y)=xVδx+yVδy\delta V(x,y) = \partial _x V \delta x + \partial _y V \delta y

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

δS=[mx˙δx+my˙δy]t0t1(mx¨δx+my¨δy+xVδx+yVδy)dt\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 t0,t1t_0, t_1 so the first term is zero. The second term should be zero for an arbitrary variation δx,δy\delta x, \delta y. This gives us two Euler-Lagrange equations :

mx¨=xVmy¨=yVm\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 L\mathcal{L} by the same approach to obtain a system of Euler-Lagrange equations for a particle in a plane :

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


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