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From this previous question Charge, velocity-dependent potentials and Lagrangian where the citation is shown at the page 22, §1.5 of the book Classical Mechanics of Goldstein, we read that

"an electric charge $q$ of mass $m$ moving at a velocity $\mathbf{v}$ in a region containing both electric field $\mathbf E(t,x,y,z)$ and magnetic field $\mathbf B(t,x,y,z)$ ($\mathbf B$ and $\mathbf E$ are derivable from a scalar potential $\phi(t, x, y, z) $ and a vector potential $\mathbf A(t,x,y,z)$), nowing that $$\mathbf E=- \nabla \phi - \frac{\partial {\mathbf A}} {\partial t}, \quad \mathbf B= \nabla \times {\mathbf A}. "\tag{1.61}$$

Why must be

$$\color{red}{\large \mathcal U=q \phi - q {\mathbf A} \cdot{\mathbf v} \quad ?}\tag{1.62}$$

Exist a physics proof of this equality?

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Because that combination, after use of the Euler-Lagrange equation, gives you the Lorentz force.

Lagrangians can never really be "proved". Once you know the equation of motion (be it classical mechanics, quantum or electrodynamics), you can "come up" with a Lagrangian that gives the correct answer. Lagrangian are better for they are scalars, manifestly symmetric, and more compact.

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  • $\begingroup$ In the meantime, I would like to thank you very much for your reply. Could you do me a kindness, if it is possible, to have a simple proof...because that combination, after use of the Euler-Lagrange equation, gives you the Lorentz force. Thank you very much. $\endgroup$ – Sebastiano Sep 6 at 21:15
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    $\begingroup$ phys.ufl.edu/~pjh/teaching/phy4605/notes/… $\endgroup$ – SuperCiocia Sep 6 at 21:16
  • $\begingroup$ Is it also the answer of Claudius' user? I have understood 50% the method of Claudius and 50% your notes. I'm not gonna make a mixture and just get confused, am I? $\endgroup$ – Sebastiano Sep 6 at 21:18
  • $\begingroup$ ... I beg your pardon? Who's Claudius? $\endgroup$ – SuperCiocia Sep 6 at 21:19
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    $\begingroup$ Ah I see, on your other question. I think he does the beginning, whereas the link goes all the way. I'm happy to discuss steps of the proof in the chat you'd like. $\endgroup$ – SuperCiocia Sep 6 at 21:20

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