Saturday, August 2, 2014

electromagnetism - Deriving the Lorentz force from velocity dependent potential


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.



Answer



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.


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