# (p.222) Appendix A6.3 Energy conservation and the homogeneity of time

# (p.222) Appendix A6.3 Energy conservation and the homogeneity of time

We consider a case where the Lagrangian has no explicit dependence on time. Then we perform a translation of the time coordinate, $t\mapsto \tau $, where $\tau =t-\u03f5$, and where $\u03f5$ is some small constant. Invariance of the action principle before and after the translation implies that:

($\dot{q}$ means $\frac{d(q)}{dt}$, ${q}^{\prime}$ means $\frac{d(q)}{d\tau}$, and $\stackrel{\u02c9}{L}$ refers to the transformed Lagrangian).

Next we consider a more general time translation in which $\u03f5$ itself is a function of time, $\u03f5=\u03f5(\tau )$. We stipulate that $\u03f5(\tau )$ is infinitesimal, continuous, and that at the boundaries *t* is not transformed, in other words, ${t}_{a}={\tau}_{a}$ and ${t}_{b}={\tau}_{b}$, or $\u03f5({\tau}_{a})=\u03f5({\tau}_{a})=0$.

From $t=\tau +\u03f5(\tau )$ we deduce that $\frac{d(t)}{d\tau}=1+\frac{d(\u03f5)}{d\tau}$, and therefore that:

(remembering that ${}^{\prime}$ is a shorthand for $\frac{d}{d\tau}$). Also, we have $\dot{q}=\frac{d(q)}{dt}=\frac{d(q)}{d\tau}\frac{d(\tau )}{dt}={q}^{\prime}(1+{\u03f5}^{\prime}{)}^{-1}$. Therefore, to first order, we have:

Substituting $\tau $ for *t*, and $\frac{d}{d\tau}$ for $\frac{d}{dt}$, we find that $L({q}_{i},{\dot{q}}_{i})$ becomes $\stackrel{\u02c9}{L}({q}_{i},{q}_{i}^{\prime}(1-{\u03f5}^{\prime}))$. As $\u03f5$ is small then ${\u03f5}^{\prime}$ also is small and $\stackrel{\u02c9}{L}$ may be expanded in a Taylor series expansion in ${\u03f5}^{\prime}$ which, to first order, gives:

(p.223)
Also, in the action integral, *dt* becomes $(1+{\u03f5}^{\prime})d\tau $. Collecting all these parts together, and ignoring 2nd- and higher-order terms, we finally arrive at:

Now, from the Action Principle, we know that $\delta {\int}_{{t}_{a}}^{{t}_{b}}L({q}_{i},{\dot{q}}_{i})\phantom{\rule{thinmathspace}{0ex}}dt$ and $\delta {\int}_{{\tau}_{a}}^{{\tau}_{b}}\stackrel{\u02c9}{L}({q}_{i},{{q}^{\prime}}_{i})\phantom{\rule{thinmathspace}{0ex}}d\tau $ are already guaranteed to be zero. So, we are left with a requirement just involving the last term:

We use our old trick^{1} - integration by parts - and arrive at:

The first integral contains nothing apart from a total differential, and so it becomes a boundary term:

However, because of our condition, $\u03f5({\tau}_{a})=\u03f5({\tau}_{a})=0$, then this boundary term is zero. Finally, we are left with just the second integral in (A6.3.7):

(p.224)
The integral is preceded by the variation symbol, $\delta $, but what is being varied? It is the infinitesimal function, $\u03f5$, which acts as a ‘variation’, and because this variation is arbitrary (the function $\u03f5$ can have any form - provided it is infinitesimal, continuous, and disappears at the boundaries), then it must be the coefficient of $\u03f5$ which vanishes. In other words, in order that the integral is zero for *arbitrary* infinitesimal variations, it is necessary that:

which means that:

Without loss of generality, we may assert this same result in *t* coordinates:

This conserved quantity has units of energy. No assumptions have been made about *L* except that it is independent of the time. However, when we consider the simplest default form for *L* (*T* is quadratic in the speed coordinates, *V* depends only on the position coordinates, and *L*= $T-V$), then it turns out (see Section 6.7, Chapter 6) that $\left(\sum \left(\frac{\partial L}{\partial {\dot{q}}_{i}}\right){\dot{q}}_{i}-L\right)=E$, where *E* is the total energy.

Thus, assuming the validity of the Principle of Least Action, and assuming the homogeneity of time (*the requirement of invariance following a time transformation, even a time-dependent one*!), has resulted in the conservation of energy.