# On generalised potential in Electrodynamics

I'm studying Lagrangian Mechanics from Goldstein's Classical Mechanics. My question concerns Section 1.5 which talks about velocity-dependent potentials. I am actually unsure about how Equation 1-64' was derived: $$F_x = g \left[ - \frac{\partial}{\partial x} \left( \phi - v \cdot A \right) - \frac{d}{dt} \left( \frac{\partial A \cdot v}{\partial v_x} \right) \right]$$ Using equation 1-64: $$F = g \left[ - \nabla \phi - \frac{\partial A}{\partial t} + \left( v \times \left( \nabla \times A \right) \right) \right]$$

If the x-component of 1-64 is to be the same as 1-64', some algebra tells us that the following must be true:

$$\frac{dA_x}{dt}=\frac{d}{dt} \left( \frac{\partial A \cdot v}{\partial v_x} \right)$$

Why is this so? Does this have something to do with the definition of $$A$$, because I haven't done my Electrodynamics course yet. I'd be grateful for any help.

• Yes, the vector potential is encoding information about the electromagnetic field and you are supposed to treat the charge in question as only just measuring the value of the field at the point it is residing. i.e. that the charge in question is so weak as to negligibly affect the field it is at, so that the measurement is of the background field. Jun 27, 2023 at 9:23
• I suspect this has to do with relativity and how Lorentz boosts affect 4-vectors. The components of a 4-vector don't vary in spacial directions perpendicular to the boost direction, and after all, classical electromagnetism is a covariant theory. Jun 27, 2023 at 9:26

This is because the magnetic gauge potential $${\bf A}({\bf r},t)$$ does not depend on the velocity $${\bf v}$$. Note that the position $${\bf r}$$ and the velocity $${\bf v}$$ are here independent variables, cf. e.g. this Phys.SE post.