I have a follow-up on this post. The way I understand it, if one generally has a velocity-dependent potential $U(q, \dot q, t)$, then we can derive/define a generalized force $$Q_k = \frac{d}{dt}\frac{\partial U}{\partial \dot q_k} - \frac{\partial U}{\partial q_k}.$$ Now, this has a practical application when it comes to deriving the Lorentz force, as is done in the post I referenced to, but I don't understand yet why the generalized force that is derived there is the Lorentz force. I mean, when we take the generalized potential $U\left(q, \dot q, t\right) = q\left( \phi - \mathbf{\dot{\vec{r}}} \cdot \mathbf{\vec{A}} \right)$ and calculate the generalized force, why does it have to be the Lorentz force?
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1$\begingroup$ Because the potential you are using is for Lorentz force. $\endgroup$– HimanshuCommented Oct 28, 2020 at 11:31
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$\begingroup$ But if you want to re check, you can solve Lagrange's equation to derive lorentz force. $\endgroup$– HimanshuCommented Oct 28, 2020 at 11:33
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$\begingroup$ Related : Deriving Lagrangian density for electromagnetic field. $\endgroup$– VoulkosCommented Oct 28, 2020 at 19:45
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I recommend the book by Landau Lifshitz volume 2 paragraph 17 formula (17.3), $$\frac{d\vec p}{dt}=-\frac{1}{c}\frac{\partial \vec A}{\partial t}-\nabla \phi+\frac{e}{c}[\vec v,\vec H]$$ which describes the equation of motion of a particle in an electromagnetic field. This chapter has a derivation of this formula.
answered Oct 28, 2020 at 17:37