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

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:

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

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

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

1. 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.

## 1 Answer

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).

• Ah, that makes things a lot clearer from a general standpoint. Thanks for the great answer, but I'd like to know how to incorporate the transformations of the differential and derivative directly into the action, as done in the book to obtain the result. Particularly, I'm curious about which step(s) from my question is/are logically correct, if any. – GodotMisogi Sep 30 '18 at 14:56