Lecture 20: Noether s Theorem

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1 Lecture 20: Noether s Theorem In our revew of Newtonan Mechancs, we were remnded that some quanttes (energy, lnear momentum, and angular momentum) are conserved That s, they are constant f no external nfluence acts on a system Snce we can thnk of the unverse as one (bg) system, ths means that energy and momentum can never be created or destroyed And we found that physcsts clng to ths dea of conserved quanttes, even when the expermental evdence seems to ndcate otherwse Now we are ready to learn why we feel these quanttes must be conserved

2 Symmetres Assume now that the Lagrangan s symmetrc under some transformaton of varables That s, all of the q s change accordng to some rule: q q s but the Lagrangan doesn t change, no matter what value of s s used: d L { q ( s ), q ( s ); t } 0 ds Noether clamed that for any such symmetry, the quantty must be conserved C p q ( ) dq( s) ds

3 Proof Let s see what happens when we take the dervatve of C wth respect to tme: dc ( ) ( ) p dq s q + p d dq s q dt ds dt ds Usng the defnton of p q, ths becomes: dc d L dq( s) L d dq( s) dt dt q + ds q dt ds We now assume that s s ndependent of tme, and use Lagrange s Equaton of Moton: dc L dq( s) L dq ( s) dt q + ds q ds

4 The rght-hand sde of the prevous equaton s equvalent to d L { q ( s ), q ( s ); t } ds But we requred ths dervatve to be 0! So we ve shown that: dc 0 dt In other words, C s conserved Ths result s one of the most mportant theorems n physcs. It holds not only for classcal mechancs, but also for quantum mechancs, relatvty, and quantum feld theores!

5 Examples of Conserved Quanttes If we express the Langrangan n rectangular coordnates, and fnd that t s nvarant under the transformaton x x + s what quantty s conserved? We know that the generalzed momentum assocated wth x s ust the lnear momentum So the conserved quantty s: dx C p p () 1 p ds For any Lagrangan symmetrc under spatal translatons, lnear momentum s conserved

6 Now suppose we wrte the Lagrangan n terms of angular varables: 1 2 L I θ U( θ) 2 In ths case, the generalzed momentum assocated wth θ s: L pθ I θ θ whch we recognze as Iω, the angular momentum For any Lagrangan symmetrc under rotatons, angular momentum s conserved

7 What About Symmetres n Tme? Ths s outsde the scope of our proof of Noether s Theorem though a more general and thorough proof would nclude t! Consder a Lagranan that has no explct tme dependence: L 0 t then the total tme dervatve s: {, } dl q q L L q + dt q q But Lagrange s Equaton of Moton requres that: L d L q dt q q

8 So we have: dl d L L q dt dt q + q q d L L q q dt q + q d L d L q q dt q dt q Whch can be rearranged to show: d dt q L L q 0 We call ths quantty the Hamltonan of the system (H)

9 Physcal Interpretaton of H We ve proven that f the Lagrangan doesn t depend explctly on the tme, the quantty H s conserved But the defnton of H doesn t gve us much ntuton nto what t represents And snce you ll be seeng Hamltonans for the rest of your lves as physcsts (especally n quantum mechancs ) we d better get an dea of what t s Let s start wth the usual case where the potental energy doesn t depend on velocty, and doesn t depend explctly on tme Also assume that the relatonshp between rectangular and generalzed coordnates has no explct tme dependence Systems wth ths property are called scleronomc

10 That means that: U U U q q ( ) and 0 In such cases, we have L ( T U) T + 0 q q q so we can rewrte the Hamltonan as: L T H q L q L q q

11 To see what the frst term represents, consder the total knetc energy of a system of N partcles: T 1 mα x 2 α, That s fne f we re usng rectangular coordnates, but what about generalzed coordnates? Let s substtute: x α, 2 (, ) x x q t α, α, x q + q x α, α, t 1 x T m α q + 2 α x α, α, q t 2

12 Expandng the squared term, we fnd: 1 x x x x x T m α q q + 2 q + 2 α α, α, α, α, α, k q k qk q t t 1 x x x x x qq + 2 q + 2 α, α, α, α, α, k α,, k q qk q t t 2 2 xα, For a scleronomc system, 0, so T reduces to: t 1 x x 1 x x qq a qq α, α, T mα qqk 2 α,, k q qk α, α, k, k k k, 2 α, q q k k,

13 We now recall our expresson for the Hamltonan: dt dq l d a q q T H q L q k, k k, ( δ l + δkl ) a, kq qk dql k, a q + a q k lk, k l, dt q a q q + a q q l l, k k l, l l l dql l, k l, l dt q 2 a q q 2T dq l l, k k l l l, k These two terms are dentcal the ndces are dfferent, but we sum over all of them

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