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

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

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

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

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

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

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.