Velocity-dependent potentials and the dissipation function

Question

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?

Answer 1

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.

Answer 2

If you use that definition for a potential then the generalised force Qi( see Goldstein equation 1.53) can be replaced by a velocity dependent potential( equation 1.58) that leads to a beautiful equation called Lagrange equation( equation 1.57)