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In the lectures, we have defined the generalized momentum in the Lagrangian to be:

$$p_i=\frac{\partial L}{\partial\dot q_i}.$$

But with this definition, if we do not make any assumptions about the system, the momentum becomes:

$$p_i=\frac {\partial T}{\partial\dot q_i} -\frac {\partial U}{\partial \dot q_i}.$$

As far as I know, the definition for momentum in the Newtonian mechanics is: $$p= \frac {\partial T}{\partial \dot r}= \frac {\partial\ \frac 12 m \dot r^2}{\partial\dot r} = m\dot r$$

Then what is the physical meaning of the potential term $$-\frac {\partial U}{\partial \dot q}$$ in the Lagrangian formalism and where does it come from?

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  • $\begingroup$ This can be answered by asking what $U$ is a function of. You should think of $U$ as being a scalar field, which means that it is independent of the speed moving through the medium. Because $U$ isn't a function of $\dot q_i$, this derivative is always $0$. $\endgroup$
    – Joshua G-F
    Commented Jun 28, 2023 at 21:24
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    $\begingroup$ check out en.wikipedia.org/wiki/Rayleigh_dissipation_function $\endgroup$
    – hyportnex
    Commented Jun 28, 2023 at 21:27
  • $\begingroup$ At first I thought like that too, but apparently the potential for the magnetic force is dependent on the velocity, thus the existence of velocity-dependent potentials cannot be excluded. That is exactly what made me confused actually $\endgroup$
    – gluon
    Commented Jun 28, 2023 at 21:29
  • $\begingroup$ The magnetic potential isn't really a "magnetic potential energy," but rather a different mathematical object. The magnetic potential is the vector $A$ such that $\nabla\times A=B$. This is similar to how the electric potential is the scalar $V$ such that $-\nabla V=E$. The thing about Lagrangians is they start as being defined as $L=K-U$, but later you define an interaction by the Lagrangian and not the potential and kinetic energies. $\endgroup$
    – Joshua G-F
    Commented Jun 28, 2023 at 21:32

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