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In the case of classical mechanics, when describing the movement of a particle that is subject to a force, Newton's law gives us the law of motion of the particle: $$F(t)=m\cdot a(t)$$

We can also apply the Lagrangian instead: $$L=\frac 1 2 m\cdot v^2 - V(x)$$ Applying the principle of least action by the Euler-Lagrange equation we get: $$\frac d {dt}\frac {dL}{dv}=\frac {dL}{dx}$$ Which simply reduces to Newton's equation.

My question is: If minimizing the action simply reduces to finding Newton's equation, then what is the added value of the lagrangian in the first place? Why not just start with Newton's equations?

Qmechanic
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user56834
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  • Possible duplicates: https://physics.stackexchange.com/q/254266/2451 , https://physics.stackexchange.com/q/15899/2451 , https://physics.stackexchange.com/q/8903/2451 and links therein. – Qmechanic Mar 20 '18 at 05:41
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    As you solve more and more complicated mechanics questions, the utility of the lagrangian approach will be become evident! :) – CAF Mar 20 '18 at 07:54

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