D.6 Lorentz force derivation

To derive the given Lorentz force from the given Lagrangian, plug the canonical momentum and the Lagrangian into the Lagrangian equation of motion. That gives

\frac{{\rm d}p_i}{{\rm d}t} + q \left(\frac{\partial A_i}{...
...hi}{\partial x_i}
+ q \frac{\partial A_j}{\partial x_i} v_j

This uses the Einstein convention that summation over $j$ is to be understood. Reorder to get

\frac{{\rm d}p_i}{{\rm d}t} =
q \left( - \frac{\partial ...
...al x_i} v_j
- \frac{\partial A_i}{\partial x_j} v_j \right)

The first parenthetical expression is the electric field as claimed. The quantity in the second parenthetical expression may be rewritten by expanding out the sums over $j$ to give

\frac{\partial A_i}{\partial x_i}v_i
- \frac{\partial A_...
...verline{\overline{\imath}}}} v_{\overline{\overline{\imath}}}

where ${\overline{\imath}}$ follows $i$ in the cyclic sequence $\ldots,1,2,3,1,2,3,\ldots$ and ${\overline{\overline{\imath}}}$ precedes it. The first two terms drop out and the others can be recognized as component number $i$ of $\vec{v}$ $\times$ $(\nabla\times\skew3\vec A)$. (For example, just write out the first component of $\vec{v}$ $\times$ $(\nabla\times\skew3\vec A)$ and compare it the expression above for ${\overline{\imath}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 and ${\overline{\overline{\imath}}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3.) Defining $\skew2\vec{\cal B}$ as $\nabla$ $\times$ $\skew3\vec A$, the Lorentz force law results.