Magnetic force term in Kobe's derivation of Maxwell's equations I'm trying to follow the paper "Derivation of Maxwell's equations from the gauge invariance of classical mechanics" by Donald Kobe, available here (American Journal of Physics, 1980): https://aapt.scitation.org/doi/10.1119/1.12094. There's a vector calc step, I think using integration by parts, that I'm having trouble following through to get a magnetic force term.
After a few pages (first column of p.350), the author gets a Lagrangian, equivalent to the following:
$$ L = \frac{1}{2} m\dot{\mathbf{r}}^2 - U(\mathbf{r}) + \frac{q}{c} \left( c A_0(\mathbf{r}, t)  + \dot{\mathbf{r}} \cdot \mathbf{A}(\mathbf{r},t) \right)  $$
The action is then $S= \int_{t_1}^{t_2} L \, dt$. To take the variation of this action, we start with
\begin{align} \delta S &= \int_{t_1}^{t_2} dt \left[ \frac{1}{2} m \, \delta (\dot{\mathbf{r}}^2 ) - \delta U(\mathbf{r}) + q \, \delta A_0(\mathbf{r}, t) + \frac{1}{c} \delta(\dot{\mathbf{r}} \cdot \mathbf{A}(\mathbf{r},t) )  \right] \\    
&= \int_{t_1}^{t_2} dt \, \left[ (\delta \mathbf{r}) \cdot \left[ -m\ddot{\mathbf{r}} - \nabla U + q \left( - \nabla A_0  - \frac{\partial \mathbf{A}}{\partial t} \right) \right] + \dot{\mathbf{r}}  \cdot ( \delta \mathbf{A}(\mathbf{r}, t) )        \right]
\end{align}
where the terms dotted against $\delta(\mathbf{r})$ come from integration by parts and the Leibniz rule. It's the last term, the magnetic force, that gives me grief. I need to show that
$$ \int_{t_1}^{t_2} dt \, [ \dot{\mathbf{r}}  \cdot ( \delta \mathbf{A}(\mathbf{r}, t) )   ] = \int_{t_1}^{t_2} dt \, [ (\delta \mathbf{r}) \cdot (\mathbf{r} \times (\nabla \times \mathbf{A})) ]  $$
However, when I expand the LHS integrand, I'm not sure how to rearrange the integral into the RHS form.
 A: Okay, I've realized this is actually very easy; I'm not 100% sure of the self-answer etiquette, so please let me know if I should do something different.
We expand the integrand as follows:
\begin{align}  \int_{t_1}^{t_2} dt \, [ \dot{\mathbf{r}} \cdot (\delta \mathbf{A}(\mathbf{r}) )  ]  &= \int_{t_1}^{t_2} dt \left[  \sum_{j,k=1}^3 \dot{r}_j \frac{\partial A_j}{\partial r_k} \delta (r_k)   \right] \\
&= \int_{t_1}^{t_2} dt \left[  \sum_{j=1}^3 \sum_{k \neq j} \dot{r}_j \frac{\partial A_j}{\partial r_k} \delta (r_k) + \sum_{\ell = 1}^3  \dot{r}_{\ell} \frac{\partial A_{\ell}}{\partial r_{\ell}} \delta (r_{\ell})  \right] \\
&= \int_{t_1}^{t_2} dt \left[  \sum_{j=1}^3 \sum_{k \neq j} \dot{r}_j \frac{\partial A_j}{\partial r_k} \delta (r_k) + \sum_{\ell = 1}^3 \delta (r_{\ell}) \left( \frac{dA_{\ell}}{dt}  - \sum_{m \neq \ell} \dot{r}_m \frac{\partial A_{\ell}}{\partial r_m}  \right)  \right] \\
&= \int_{t_1}^{t_2} (\delta \mathbf{r}) \cdot (\dot{\mathbf{r}} \times (\nabla \times \mathbf{A})) \, dt
\end{align}
as required since the variation of the integral of each total derivative is zero.
