- #1

TimeRip496

- 254

- 5

$$\frac{dA_j}{dt}=\frac{d}{dt}(\frac{\partial{}}{\partial{v_j}}(v.A)),$$

since the partial derivative will pick out only the j

^{th}component of the dot product. Now, since the scalar potential is independent of the velocity, we can add on a term containing it inside the partial derivative:

$$\frac{dA_j}{dt}=\frac{d}{dt}(\frac{\partial{}}{\partial{v_j}}(v.A-q∅)),$$ "

I don't understand the rationale behind these. It seems like the author is trying to get to the final equation so he guess the steps in order to get to the final equation.

Source: https://www.sccs.swarthmore.edu/users/02/no/pdfs/lorentz.pdf

Page 2 below eqn(13)