Noether’s Theorem - Physics Courses

1 Chapter 7 noether s Continuous Symmetry Implies Conserved ChargesConsider a particle moving in two dimensions under the influence of an external potentialU(r). The potential is a function only of the magnitude of the vectorr. The Lagrangian isthenL=T U=12m r2+r2 2 U(r),( )where we have chosen generalized coordinates (r, ). The momentum conjugate to isp =mr2 . The generalized forceF clearly vanishes, sinceLdoes not depend on thecoordinate . (One says thatLis cyclic in .) Thus, althoughr=r(t) and = (t)will in general be time-dependent, the combinationp =mr2 is constant. This is theconserved angular momentum about the instead the particle moved in a potentialU(y), independent ofx, then writingL=12m x2+ y2 U(y),( )we have that the momentumpx= L/ x=m xis conserved, because the generalized forceFx= L/ x= 0 vanishes.

2 This situation pertains in a uniform gravitational field, withU(x,y) =mgy, independent ofx. The horizontal component of momentum is general, whenever the system exhibits acontinuous symmetry, there is an associatedconserved charge. (The terminology charge is from field theory.) Indeed, this is a rigorousresult, known asNoether s Theorem . Consider a one-parameter family of transformations,q q (q, ),( )where is the continuous parameter. Suppose further (without lossof generality) that at = 0 this transformation is the identity, q (q,0) =q . The transformation may benonlinear in the generalized coordinates. Suppose furtherthat the LagrangianLis invariant12 CHAPTER 7. noether S Theorem under the replacementq q. Then we must have0 =dd =0L( q, q,t) = L q q =0+ L q q =0=ddt L q q =0+ L q ddt q =0=ddt L q q =0.

3 ( )Thus, there is an associated conserved charge = L q q =0.( ) Examples of one-parameter families of transformationsConsider the LagrangianL=12m( x2+ y2) U px2+y2 .( )In two-dimensional polar coordinates, we haveL=12m( r2+r2 2) U(r),( )and we may now define r( ) =r( ) ( ) = + .( )Note that r(0) =rand (0) = , transformation is the identity when = 0. Wenow have =X L q q =0= L r r =0+ L =0=mr2 .( )Another way to derive the same result which is somewhat instructive is to work out thetransformation in Cartesian coordinates. We then have x( ) =xcos ysin ( ) y( ) =xsin +ycos .( )Thus, x = y , y = x( ) CONSERVATION OF LINEAR AND ANGULAR MOMENTUM3and = L x x =0+ L y y =0=m(x y y x).( )Butm(x y y x) =m z r r=mr2 .( )As another example, consider the potentialU( , ,z) =V( ,a +z),( )where ( , ,z) are cylindrical coordinates for a particle of massm, and whereais a constantwith dimensions of length.

4 The Lagrangian is12m 2+ 2 2+ z2 V( ,a +z).( )This model possesses a helical symmetry, with a one-parameter family ( ) = ( ) ( ) = + ( ) z( ) =z a .( )Note thata + z=a +z ,( )so the potential energy, and the Lagrangian as well, is invariant under this one-parameterfamily of transformations. The conserved charge for this symmetry is = L =0+ L =0+ L z z =0=m 2 ma z .( )We can check explicitly that is conserved, using the equations of motionddt L =ddt m 2 = L = a V z( )ddt L z =ddt(m z) = L z= V z.( )Thus, =ddt m 2 addt(m z) = 0.( ) Conservation of Linear and Angular MomentumSuppose that the Lagrangian of a mechanical system is invariant under a uniform translationof all particles in the ndirection. Then our one-parameter family of transformations is givenby xa=xa+ n,( )4 CHAPTER 7.

5 noether S THEOREMand the associated conserved noether charge is =Xa L xa n= n P,( )whereP=Papais thetotal momentumof the the Lagrangian of a mechanical system is invariant under rotations about an axis n, then xa=R( , n)xa=xa+ n xa+O( 2),( )where we have expanded the rotation matrixR( , n) in powers of . The conserved Noethercharge associated with this symmetry is =Xa L xa n xa= n Xaxa pa= n L,( )whereLis thetotal angular momentumof the Advanced Discussion : Invariance ofSObservant readers might object that demanding invariance ofLis too strict. We shouldinstead be demanding invariance of the actionS1. SupposeSis invariant undert t(q,t, )( )q (t) q (q,t, ).( )Then invariance ofSmeansS=tbZtadtL(q, q,t) = tbZ tadtL( q, q,t).( )Note thattis a dummy variable of integration, so it doesn t matter whether we call ittor t.

6 The endpoints of the integral, however, do change under thetransformation. Nowconsider an infinitesimal transformation, for which t= t tand q= q t q(t) are bothsmall. Thus,S=tbZtadtL(q, q,t) =tb+ tbZta+ tadtnL(q, q,t) + L q q + L q q +..o,( )1 Indeed, we should be demanding thatSonly change by a function of the endpoint ADVANCED DISCUSSION : INVARIANCE OFS5where q (t) q (t) q (t)= q t q t + q (t) q (t)= q q t+O( q t)( )Subtracting eqn. from eqn. , we obtain0 =Lb tb La ta+ L q b q ,b L q a q ,a+tb+ tbZta+ tadt( L q ddt L q ) q (t)=tbZtadtddt( L L q q t+ L q q ),( )whereLa,bisL(q, q,t) evaluated att=ta,b. Thus, if is infinitesimal, and t=A(q,t) ( ) q =B (q,t) ,( )then the conserved charge is = L L q q A(q,t) + L q B (q,t)= H(q,p,t)A(q,t) +p B (q,t).

7 ( )Thus, whenA= 0, we recover our earlier results, obtained by assuming invariance that conservation ofHfollows from time translation invariance:t t+ , for whichA= 1 andB = 0. Here we have writtenH=p q L ,( )and expressed it in terms of the momentap , the coordinatesq , and The HamiltonianThe Lagrangian is a function of generalized coordinates, velocities, and time. The canonicalmomentum conjugate to the generalized coordinateq isp = L q .( )6 CHAPTER 7. noether S THEOREMThe Hamiltonian is a function of coordinates,momenta, and time. It is defined as theLegendre transform ofL:H(q,p,t) =X p q L .( )Let s examine the differential ofH:dH=X q dp +p d q L q dq L q d q L tdt=X q dp L q dq L tdt ,( )where we have invoked the definition ofp to cancel the coefficients ofd q . Since p = L/ q , we haveHamilton s equations of motion, q = H p , p = H q.

8 ( )Thus, we can writedH=X q dp p dq L tdt .( )Dividing bydt, we obtaindHdt= L t,( )which says that the Hamiltonian isconserved( does not change with time) wheneverthere is noexplicittime dependence #1 : For a simpled= 1 system withL=12m x2 U(x), we havep=m xandH=p x L=12m x2+U(x) =p22m+U(x).( )Example #2 : Consider now the mass point wedge system analyzed above, withL=12(M+m) X2+m X x+12m(1 + tan2 ) x2 mg xtan ,( )The canonical momenta areP= L X= (M+m) X+m x( )p= L x=m X+m(1 + tan2 ) x .( )The Hamiltonian is given byH=P X+p x L=12(M+m) X2+m X x+12m(1 + tan2 ) x2+mgxtan .( ) ADVANCED DISCUSSION : INVARIANCE OFS7 However, this is not quiteH, sinceH=H(X,x,P,p,t) must be expressed in terms of thecoordinates and themomentaand not the coordinates and velocities. So we must eliminate Xand xin favor ofPandp.

9 We do this by inverting the relations Pp = M+mmm m(1 + tan2 ) X x ( )to obtain X x =1m M+ (M+m) tan2 m(1 + tan2 ) m mM+m Pp .( )Substituting into , we obtainH=M+m2mP2cos2 M+msin2 Ppcos2 M+msin2 +p22 (M+msin2 )+mg xtan .( )Notice that P= 0 since L X= the total horizontal momentum of the system (wedgeplus particle) and it is IsH=T+U?The most general form of the kinetic energy isT=T2+T1+T0=12T(2) (q,t) q q +T(1) (q,t) q +T(0)(q,t),( )whereT(n)(q, q,t) is homogeneous of degreenin the velocities2. We assume a potentialenergy of the formU=U1+U0=U(1) (q,t) q +U(0)(q,t),( )which allows for velocity-dependent forces, as we have withcharged particles moving in anelectromagnetic field. The Lagrangian is thenL=T U=12T(2) (q,t) q q +T(1) (q,t) q +T(0)(q,t) U(1) (q,t) q U(0)(q,t).( )The canonical momentum conjugate toq isp = L q =T(2) q +T(1) (q,t) U(1) (q,t)( )which is inverted to give q =T(2) 1 p T(1) +U(1).

10 ( )2A homogeneous function of degreeksatisfiesf( x1, .. , xn) = kf(x1, .. , xn). It is then easy to proveEuler s Theorem ,Pni=1xi f xi= 7. noether S THEOREMThe Hamiltonian is thenH=p q L=12T(2) 1 p T(1) +U(1) p T(1) +U(1) T0+U0( )=T2 T0+U0.( )IfT0,T1, andU1vanish, (q, q,t) is a homogeneous function of degree two in thegeneralized velocities, andU(q,t) is velocity-independent, thenH=T+U. But ifT0orT1is nonzero, or the potential is velocity-dependent, thenH6=T+ Example: A bead on a rotating hoopConsider a bead of massmconstrained to move along a hoop of radiusa. The hoop isfurther constrained to rotate with angular velocity = about the z-axis, as shown inFig. most convenient set of generalized coordinates is spherical polar (r, , ), in which caseT=12m r2+r2 2+r2sin2 2 =12ma2 2+ 2sin2 .( )Thus,T2=12ma2 2andT0=12ma2 2sin2.