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I am trying to get a better understanding of the Lagrangian. From what I know, we say that each trajectory in physics must be a path that is at a minimum, which means that is satisfies the Euler-Lagrange equations. Thus, we define the Lagrangian, which is exactly this function.

Now, considering a non-relativistic newtonian problem, it is stated that the Lagrangian is the function $$ L = T - U $$ How does this function satisfy the Euler-Lagrange equations, and more importantly, how is it derived?

Qmechanic
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DLG03
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  • For a point mass https://basics.altervista.org/test/Physics/Me/dynamics_newton_lagrange_equivalence_point.html – basics Sep 14 '23 at 09:29
  • For a system of point masses https://basics.altervista.org/test/Physics/Me/dynamics_newton_lagrange_equivalence_points.html – basics Sep 14 '23 at 09:30
  • for a review of classical mechanics https://basics.altervista.org/test/Physics/Me/main.html you can find the notes on the equivalence of the Newtonian and Lagrangian formulation at the links (light blue) at the bottom of the page, in the section "Weak formulation" – basics Sep 14 '23 at 09:31
  • Thanks for your response, but in your notes, you define L = T - U. I want to know how you can derive these equations. – DLG03 Sep 14 '23 at 09:38
  • That's how things work in Physics. Sometimes you introduce definitions, especially when a quantity satisfies some conservation principle (or here a variational principle). It's possible to prove that the function defined as Lagrangian appears in the stationary action principle, that gives results that are equivalent to those of the Newtonian mechanics. – basics Sep 14 '23 at 10:05
  • Possible duplicates: https://physics.stackexchange.com/q/78138/2451 and links therein. – Qmechanic Sep 14 '23 at 10:19
  • David Tong: Lectures on Classical Dynamics – mmesser314 Sep 14 '23 at 13:54
  • While the usual approach is to posit The Lagrangian $(T-U)$ and then show that putting that Lagrangian into the Euler-Lagrange equation recovers F=ma, the relation is bi-directional; there is a valid way of starting with F=ma, and moving in all forward steps arrive at the Lagrangian $(T-U)$. This forward sequence of steps is laid out in an 2021 answer submitted by me, to a question about Lagrangian mechanics. The animated GIFs in that answer represent the content of interactive diagrams that are available on my own website. – Cleonis Sep 14 '23 at 19:14

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