We can achieve a simplified version of the Lorentz force by $$F=q\bigg[-\nabla(\phi-\mathbf{A}\cdot\mathbf{v})-\frac{d\mathbf{A}}{dt}\bigg],$$ where $\mathbf{A}$ is the magnetic vector potential and the scalar $\phi$ the electrostatic potential.

How is this derivable from a velocity-dependent potential $$U=q\phi-q\mathbf{A}\cdot\mathbf{v}?$$

I fail to see how the total derivative of $\mathbf{A}$ can be disposed of and the signs partially reversed. I'm obviously missing something.


2 Answers 2


Velocity-dependent potential is not strictly a potential. Lagrange equations say that

$$\frac{d}{dt}\frac{\partial L}{\partial \bf{v}} = \frac{\partial L}{\partial \bf{r}}$$

You have $L = L_0 - U$ where $L_0$ corresponds to free motion (e.g. $L_0 = mv^2/2$ or $L_0 = -mc^2\sqrt{1-(v/c)^2}$).

If $U$ does not contain $\bf{v}$ you have ${\partial L}/{\partial \bf{v}} = \bf{p}$ and so $\dot{\bf{p}} = -\nabla U$.

In this case, however $U$ contains $\bf{v}$ so on the you have

$$\frac{d}{dt}\left({\bf{p}} + q\bf{A}\right) = -\nabla U$$


Hints: Use

$$\frac{\partial U}{\partial {\bf v}}= -q{\bf A}, $$

and the defining property of a velocity-dependent potential:

$${\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}.$$

See e.g. Herbert Goldstein, Classical Mechanics and Wikipedia for more details.

  • $\begingroup$ Thank you Qmechanic, for your gracious hint, better than an answer. :) Actually the first relation is the one I missed. $\endgroup$
    – Valentina
    Commented Sep 14, 2013 at 16:15

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