Schwinger's variation of the action of point particle with *both* time and position as independent variables

Question

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:

$$ \mathrm dt \to d(t+\delta t) = \left(1 + \frac{\mathrm d\delta t}{\mathrm d t}\right)\mathrm dt,\tag{8.8}$$ $$ \frac{\mathrm d}{\mathrm dt} \to \left(1 - \frac{\mathrm d \delta t}{\mathrm d t}\right)\frac{\mathrm d}{\mathrm dt}.\tag{8.9}$$

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:

1. Start with the variation of the action defined as an integral over the Lagrangian and vary sensibly:

$$ \delta W_{12} = \delta \left(\int_2^1 L(\mathbf r, \dot{\mathbf r}, t)\right) = \int^1_2 \left[\delta L\,\mathrm dt + L\,\delta(\mathrm dt)\right] $$

2. Treat variation as a "transformation", i.e. substitute the transformed parameters in the Lagrangian:

$$ \int_2^1 L\left[\mathbf r + \delta\mathbf r, \left(1 - \frac{\mathrm d\delta t}{\mathrm dt}\right)\frac{\mathrm d}{\mathrm dt}\left(\mathbf r + \delta\mathbf r(t)\right), t + \delta t\right]\left[1 + \frac{\mathrm d\delta t}{\mathrm dt}\right]\mathrm dt $$

3. 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. is correct.

Answer 1

Here is perhaps a clearer approach. OP essentially already mentions that Schwinger et al. are allowing time reparametrizations on a fixed parameter interval $$[\lambda_i,\lambda_f]~\ni~\lambda~~\mapsto ~~x^{\mu}(\lambda)~\in~\mathbb{R}^4, \qquad {\bf x}~\equiv~{\bf r}\circ t, \qquad x^0~\equiv~t.\tag{A}$$

The action is $$ S[x^{\mu}]~=~\int_{t_i}^{t_f} \!dt ~L~=~\int_{\lambda_i}^{\lambda_f} \!d\lambda ~\dot{t}L, \qquad L~=~L(x^{\mu},{\bf v}), \qquad {\bf v} ~:=~\frac{d{\bf r}}{d t}~=~\frac{\dot{\bf x}}{\dot{t}},\tag{B}$$

where dot means differentiation wrt. $\lambda$. Define momentum and energy as $${\bf p}~:=~\frac{\partial L}{\partial {\bf v}}\qquad\text{and}\qquad h~:=~{\bf v}\cdot{\bf p}-L .\tag{C}$$

Infinitesimal variation of the action yields

$$\delta S ~:=~S[x^{\prime \mu}]- S[x^{\mu}] ~=~\ldots~=~\int_{\lambda_i}^{\lambda_f} \!d\lambda \sum_{\mu=0}^3\frac{\delta S}{\delta x^{\mu}}\delta x^{\mu} + \left[\sum_{\mu=0}^3\frac{\partial [\dot{t}L]}{\partial \dot{x}^\mu}\delta x^{\mu} \right]_{\lambda=\lambda_i}^{\lambda=\lambda_f} $$ $$~=~\ldots~=~\int_{t_i}^{t_f} \!dt\left(\frac{\partial L}{\partial {\bf r}}-\frac{d{\bf p}}{dt}\right)\cdot \delta {\bf r} + \int_{t_i}^{t_f} \!dt\left(\frac{\partial L}{\partial t}+\frac{dh}{dt}\right)\cdot \delta t + \left[{\bf p}\cdot\delta{\bf r} -h \delta t\right]_{t=t_i}^{t=t_f}, \tag{D}$$ which is equivalent to the last expression in eq. (8.11).