I How to picture the vector potential?

[Rick16]

TL;DR Summary

developing the Lorentz force law from the Lagrangian for the magnetic field

Susskind (in The Theoretical Minimum, volume 1, pages 203-205) writes the Lagrangian for the magnetic field as $L=\frac m 2(\dot x^2+\dot y^2 + \dot z^2)+ \frac e c (\dot x A_x +\dot y A_y +\dot z A_z)$ and then calculates $\dot p_x =ma_x + \frac e c \frac d {dt} A_x=ma_x + \frac e c(\frac {\partial A_x} {\partial x}\dot x + \frac {\partial A_x} {\partial y}\dot y + \frac {\partial A_x} {\partial z}\dot z)$.

I have problems with the last step. I might have written $\frac {dA_x} {dt} =\frac {dA_x} {dx} \frac {dx} {dt}$, ignoring the other terms. How can I know that each component of the vector potential depends on all the coordinates? I would have to somehow picture the vector potential in order to come to this conclusion, but the vector potential seems to be a purely theoretical construct. How can I know how to treat it in situations like this one?

 

[ergospherical]

Yes, in the general case $\mathbf{A} = \mathbf{A}(x,y,z,t)$ is some vector field that depends on position (and time), and you'll need to use the full total derivative, e.g.

$\dot{A}_x = (\partial_x A_x) \dot{x} + (\partial_y A_x) \dot{y} + (\partial_z A_x) \dot{z} + (\partial_t A_x)$

If the field changes with time, then $\partial_t \mathbf{A}$ will also be non-zero.

Why don't you have a look at a few concrete examples, e.g. wires, solenoids...:
https://www.feynmanlectures.caltech.edu/II_14.html

 

[Rick16]

Thank you, I think I got it. If I have any vector field $\vec v=\vec v(x,y,z)$, then in general the componentes would be $v_x =v_x(x,y,z), v_y=v_y(x,y,z), v_z=v_z(x,y,z)$, and only in a special case would I have $v_x=v_x(x), v_y=v_y(y), v_z=v_z(z)$. If I don't know what a vector field actually looks like, I have to assume the general case. For some reason, I assumed $v_x=v_x(x), v_y=v_y(y), v_z=v_z(z)$, and it already seems strange to me why I assumed that. Sorry to have bothered you with this.

 

[ergospherical]

No problem!