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The Lazy UniverseAn Introduction to the Principle of Least Action$

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

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

Source:
The Lazy Universe
Author(s):

Jennifer Coopersmith

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, tτ, where τ=tϵ, and where ϵ is some small constant. Invariance of the action principle before and after the translation implies that:

(A6.3.1)
δtatbL(qi(t),q˙i(t))dt=δτaτbLˉ(qi(τ),qi(τ))dτ=0

(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)
dt=(1+ϵ)dτ

(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)
q˙=q(1ϵ)

Substituting τ for t, and ddτ for ddt, we find that L(qi,q˙i) becomes Lˉ(qi,qi(1ϵ)). 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)
Lˉ(qi,qi(1ϵ))=Lˉ(qi,qi)iLˉqiqiϵ

(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)
δA=δtatbL(qi,q˙i)dt=δτaτbLˉ(qi,qi)dτδτaτbiLˉqiqiLˉϵdτ=0

Now, from the Action Principle, we know that δtatbL(qi,q˙i)dt and δτaτbLˉ(qi,qi)dτ are already guaranteed to be zero. So, we are left with a requirement just involving the last term:

(A6.3.6)
δτaτbiLˉqiqiLˉϵdτ=0

We use our old trick1 - integration by parts - and arrive at:

(A6.3.7)
δA=δτaτbddτiLˉqiqiLˉϵdτδτaτbddτiLˉqiqiLˉϵdτ=0

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

(A6.3.8)
iLˉqiqiLˉϵτaτb

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)
δτaτbddτiLˉqiqiLˉϵdτ=0

(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)
ddτiLˉqiqiLˉ=0

which means that:

(A6.3.11)
iLˉqiqiLˉ=constant

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

(A6.3.12)
iLq˙iq˙iL=constant

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= TV), then it turns out (see Section 6.7, Chapter 6) that Lq˙iq˙iL=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)