In Chapter 8, pages 86-87, equations (8.5)-(8.11) of Julian Schwinger et al., Classical Electrodynamics, the equations of motion for the following action principle of a point particle in an external potential are derived in an unconventional manner, from what I've seen.
The action is given as:
$$ W_{12} = \int_2^1 \left[\frac{1}{2}m\frac{(\mathrm d\bf r)^2}{\mathrm dt} - \mathrm dt\, V(\mathbf r,t)\right]\tag{8.10} $$
with the variations on $$\mathbf r(t) \to \mathbf r(t) + \delta\mathbf r(t)\tag{8.5}$$ and $$ t \to t + \delta t(t),\tag{8.6}$$ such that at the endpoints $$\delta t(t_1) = \delta t_1, \qquad \delta t(t_2) = \delta t_2,\tag{8.7}$$ so that the limits of integration are not changed. Thus time is a function of the parameter time $t$ (what I assume is notational abuse). This is different from the usual approach in which $\mathbf r$ and $\dot{\mathbf r}$ are considered as independent variables and varied accordingly.
The corresponding changes in the time differential and time derivative are:
$$ \mathrm dt \to \mathrm d(t+\delta t) = \left(1 + \frac{\mathrm d\delta t}{\mathrm d t}\right)\mathrm dt,\tag{8.8}$$ $$ \frac{\mathrm d}{\mathrm dt} \to \left(1 - \frac{\mathrm d \delta t}{\mathrm d t}\right)\frac{\mathrm d}{\mathrm dt}.\tag{8.9}$$
The following variation is then presented:
$$\delta W_{12} = \int^1_2 \mathrm dt\left\{m\frac{\mathrm d\bf r}{\mathrm dt}\cdot\frac{\mathrm d}{\mathrm dt}\delta\mathbf r - \delta\mathbf r \cdot \nabla V - \frac{\mathrm d\delta t}{\mathrm dt}\left[\frac{1}{2}m\left(\frac{\mathrm d\bf r}{\mathrm dt}\right)^2 + V \right] - \delta t \frac{\partial}{\partial t}V\right\}. \tag{8.11}$$
I've attempted to figure out how to derive this, but keep getting stuck at the first step itself. Here are some possibilities I considered:
- Start with the variation of the action defined as an integral over the Lagrangian and vary sensibly:
$$ \delta W_{12} = \delta \left(\int_2^1 L(\mathbf r, \dot{\mathbf r}, t)\right) = \int^1_2 \left[\delta L\,\mathrm dt + L\,\delta(\mathrm dt)\right] $$
- Treat variation as a "transformation", i.e. substitute the transformed parameters in the Lagrangian:
$$ \int_2^1 L\left[\mathbf r + \delta\mathbf r, \left(1 - \frac{\mathrm d\delta t}{\mathrm dt}\right)\frac{\mathrm d}{\mathrm dt}\left(\mathbf r + \delta\mathbf r(t)\right), t + \delta t\right]\left[1 + \frac{\mathrm d\delta t}{\mathrm dt}\right]\mathrm dt $$
- Consider the variation of the free particle's kinetic energy term, which leads to the following expression as I see it:
\begin{align*} \delta\left(\frac{\mathrm d\mathbf r}{\mathrm dt}\right)^2 & = 2\left(\frac{\mathrm d\mathbf r}{\mathrm dt}\right)\left[\delta\left(\frac{\mathrm d}{\mathrm dt}\right)\mathbf r + \frac{\mathrm d}{\mathrm dt}\delta\mathbf r\right] \\ & = 2\left(\frac{\mathrm d\mathbf r}{\mathrm dt}\right)\left[\left(1-\frac{\mathrm d\delta t}{\mathrm dt}\right)\frac{\mathrm d\mathbf r}{\mathrm dt} + \frac{\mathrm d}{\mathrm dt}\delta\mathbf r\right] \end{align*}
After the usual Taylor expansion up to first order, I can't see any of these leading to the variation given in the book. Which method, if any, is correct? I'm also not particularly sure if 3.'s approach via variation of an operator is correct.
Note: I posted this question on Physics SE before and received a more general answer which is useful, but I would still like to know which of the steps would be the mathematically correct way in this approach, if any.