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When defining $L=T-V$, and using Euler-Lagrange equations ($\partial_x L = \frac{d}{dt} \partial_{\dot{x}} L$), we get back $m \ddot{x} = - \frac{dV}{dx}$ ONLY WHEN ASSUMING that $V(x, \dot{x}) = V(x)$ and $T(x, \dot{x}) = T(\dot{x})$.

However, these assumtions aren't always true (if I'm not mistaken, it may be the case in the electromagnetic potential).

Thus, my question is : why is $L=T-V$ even when those assumptions break?

Qmechanic
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Dor Harpaz
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    Because what doesn't go in the Lagrangian through the potential, i.e. the non-conservative actions, can be represented as generalized forces – basics Oct 25 '22 at 16:56
  • Possible duplicates: https://physics.stackexchange.com/q/78138/2451 and links therein. – Qmechanic Oct 25 '22 at 16:57
  • Best paper imho on that subject. – Kurt G. Oct 25 '22 at 17:11
  • Indeed those assumptions are not always applicable, and when they aren't Hamilton's action is not applicable. A necessary condition is that there is a well-defined potential energy. In order to have a well-defined potential energy: the force that is involved must be a conservative force, and generalized coordinates, when used, must be holonomic. In classical mechanics: when those conditions are met: $L=T-V$. To understand why for classical mechanics $L=T-V$ see my discussion of Hamilton's stationary action. (Demonstration with diagrams) – Cleonis Oct 25 '22 at 20:48

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