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Please try to provide a sufficient answer, and when it is just „because it satisfies Newton‘s equations“, please try to give an example or explain it. If you know it, I would be very happy if you could tell me how Lagrange himself came up with this.

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  • $\begingroup$ Does this answer your question? Deeper Meaning to the Nature of Lagrangian $\endgroup$ – Davide Morgante Jul 1 at 19:30
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/78138/2451 $\endgroup$ – Qmechanic Jul 1 at 22:34
  • $\begingroup$ Here is how I understand your question: we have the two forms: Newton's second law on one hand and variational calculus with the Lagrangian $T - V$ on the other hand, and they are mathematically equivalent; proofs are abundant, and you accept these proofs. The nagging question is: how can two formulations that look so different be mathematically equivalent? How is that possible? The following visual demonstration is designed specifically to explain that: physics.stackexchange.com/a/556160/17198 $\endgroup$ – Cleonis Jul 3 at 13:39
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The Lagrangian isn't defined as $T-V$, it just turns out that way in a lot of situations. The electromagnetic Lagrangian, for example, is not $T-V$. The Shankar quantum mechanics book states (pg 84):

$L$ is

that function $L(q, \dot{q}, t)$ which, when fed into the Euler-Lagrange equations, reproduces the correct dynamics. The rule $L=T-U$ becomes just a useful mnemonic for the case of conservative forces.

So if it is a function that fits the Euler-Lagrange equations and reproduces the correct physics, then it's a Lagrangian.

A very simple example for when it is $L=T-U$ is whenever an object is falling. $T=\frac{1}{2}m\dot{y}^2$ and $U=mgy$, so the Lagrangian is $L=T-U=\frac{1}{2}m\dot{y}^ 2 - mgy$. Feeding these into $$\frac{d}{dt}\frac{\partial L}{\partial \dot{y}}=\frac{\partial L}{\partial y} \ \ \ \text{(Euler-Lagrange equation)}$$ comes out to $$\ddot{y}=g$$ telling you that the acceleration is just $g$. You could ge this form Newton's second law also $$F=ma=m\ddot{y}$$ setting this equal to the force of gravity (since we're talking about a falling object) $$m\ddot{y} = mg$$ $$\ddot{y} = g$$ So I got the same thing from Newton and Lagrange

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