The lagrangian of a particle in a plane in cartesian coordinates is given by
L=2mx˙2+2my˙2−V(x,y)
The action is thus given by
S=∫t0t1L(t,x,x˙,y,y˙)dt
Let us consider the variation of this action :
δS=δ∫Ldt=∫δLdt
Lagrangian variation is given by
δL=mx˙δx˙+my˙δy˙−δV(x,y)
Expanding δV gives
δV(x,y)=∂xVδx+∂yVδy
We also know that δx˙=(δx)˙ and thus the first two terms can be integrated by parts :
δS=[mx˙δx+my˙δy]t0t1−∫(mx¨δx+my¨δy+∂xVδx+∂yVδy)dt
We fix the trajectory at t0,t1 so the first term is zero. The second term should be zero for an arbitrary variation δx,δy. This gives us two Euler-Lagrange equations :
mx¨=−∂xVmy¨=−∂yV
which correspond exactly to Newton equations for a particle in a plane. These equations could be derived for the most general form of L by the same approach to obtain a system of Euler-Lagrange equations for a particle in a plane :
dtd∂q˙∂L−∂q∂L=0 for every coordinate q of the particle.
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