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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.

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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.

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