## Jennifer Coopersmith

Print publication date: 2017

Print ISBN-13: 9780198743040

Published to Oxford Scholarship Online: June 2017

DOI: 10.1093/oso/9780198743040.001.0001

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

Source:
The Lazy Universe
Publisher:
Oxford University Press

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

(A6.3.1)
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($q˙$ means $d(q)dt$, $q′$ means $d(q)dτ$, and $Lˉ$ refers to the transformed Lagrangian).

Next we consider a more general time translation in which $ϵ$ itself is a function of time, $ϵ=ϵ(τ)$. We stipulate that $ϵ(τ)$ is infinitesimal, continuous, and that at the boundaries t is not transformed, in other words, $ta=τa$ and $tb=τb$, or $ϵ(τa)=ϵ(τa)=0$.

From $t=τ+ϵ(τ)$ we deduce that $d(t)dτ=1+d(ϵ)dτ$, and therefore that:

(A6.3.2)
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(remembering that $′$ is a shorthand for $ddτ$). Also, we have $q˙=d(q)dt=d(q)dτd(τ)dt=q′(1+ϵ′)−1$. Therefore, to first order, we have:

(A6.3.3)
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Substituting $τ$ for t, and $ddτ$ for $ddt$, we find that $L(qi,q˙i)$ becomes $Lˉ(qi,qi′(1−ϵ′))$. As $ϵ$ is small then $ϵ′$ also is small and $Lˉ$ may be expanded in a Taylor series expansion in $ϵ′$ which, to first order, gives:

(A6.3.4)
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(p.223) Also, in the action integral, dt becomes $(1+ϵ′)dτ$. Collecting all these parts together, and ignoring 2nd- and higher-order terms, we finally arrive at:

(A6.3.5)
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Now, from the Action Principle, we know that $δ∫tatbL(qi,q˙i)dt$ and $δ∫τaτbLˉ(qi,q′i)dτ$ are already guaranteed to be zero. So, we are left with a requirement just involving the last term:

(A6.3.6)
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We use our old trick1 - integration by parts - and arrive at:

(A6.3.7)
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The first integral contains nothing apart from a total differential, and so it becomes a boundary term:

(A6.3.8)
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However, because of our condition, $ϵ(τa)=ϵ(τa)=0$, then this boundary term is zero. Finally, we are left with just the second integral in (A6.3.7):

(A6.3.9)
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(p.224) The integral is preceded by the variation symbol, $δ$, but what is being varied? It is the infinitesimal function, $ϵ$, which acts as a ‘variation’, and because this variation is arbitrary (the function $ϵ$ can have any form - provided it is infinitesimal, continuous, and disappears at the boundaries), then it must be the coefficient of $ϵ$ which vanishes. In other words, in order that the integral is zero for arbitrary infinitesimal variations, it is necessary that:

(A6.3.10)
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which means that:

(A6.3.11)
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Without loss of generality, we may assert this same result in t coordinates:

(A6.3.12)
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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 $∑∂L∂q˙iq˙i−L=E$, 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.

## Notes:

(1) (Chapter 4 and the beginning of Chapter 6)