Non-Noether symmetries in singular dynamical systems

George Chavchanidze
Department of Theoretical Physics,A. Razmadze Institute of Mathematics,1 Aleksidze Street, Tbilisi 0193, Georgia
abstract. In the present paper geometric aspects of relationshipbetween non-Noether symmetries and conservation laws in Hamiltoniansystems is discussed. Case of irregular/constrained dynamical systemson presymplectic and Poisson manifolds is considered.
keywords. Non-Noether symmetry; Conservation laws; Constrained dynamics;
msc. 70H33, 70H06, 53Z05
Georgian Math. J. 10 (2003) 057-061


Noether's theorem associates conservation laws with particular continuous symmetries ofthe Lagrangian. According to the Hojman's theorem [1]-[3] there exists the definite correspondence betweennon-Noether symmetries and conserved quantities. In 1998 M. Lutzky showed that several integrals ofmotion might correspond to a single one-parameter group of non-Noether transformations[4]. In the present paper, the extension of Hojman-Lutzky theorem to singular dynamical systems is considered.
First of all let us recall some basic knowledge of description of the regular dynamical systems(see, e. g. [5]).In this case time evolution is governed by Hamilton's equationiXhω + dh = 0,where ω is the closed(dω = 0) and non-degenerate(iXω = 0 ⇒ X = 0) 2-form,h is the Hamiltonian andiXω denotes contraction ofX with ω.Since ω is non-degenerate, this gives rise to an isomorphism between the vectorfields and 1-forms given by iXω + α= 0.The vector field is said to be Hamiltonian if it corresponds to exact formiXfω + df = 0.The Poisson bracket is defined as follows:{f , g} = Xf g = − Xg f = iXfiXgω.By introducing a bivector field W satisfyingiXiYω = iW iXω ∧ iYω,Poisson bracket can be rewritten as{f , g} = iW df ∧ dg.It's easy to show thatiXiYLZω =i[Z,W] iXω ∧ iYω,where the bracket [ · , · ] is actually a supercommutator,for an arbitrary bivector field W = s Vs ∧ Us we have[X,W] = s[X,Vs] ∧ Us+ sVs ∧ [X,Us]Equation (6) is based on the following useful property of the Lie derivativeLXiWω = i[X,W]ω +iWLXω.Indeed, for an arbitrary bivector fieldW = s Vs ∧ Us we haveLXiWω = LXsiVs ∧ Usω =LXs iUsiVsω= s i[X,Us]iVsω +s iUsi[X,Vs]ω +siUsiVsLXω =i[X,W]ω + iWLXωwhere LZ denotes the Lie derivative along the vector field Z.According to Liouville's theorem Hamiltonian vector fieldpreserves ωLXfω = 0;therefore it commutes with W:[Xf ,W] = 0.In the local coordinates zs whereω = rsωrsdzr ∧ zs bivector fieldW has the following formW = rsWrs∂zr∂zs whereWrs is matrix inverted to ωrs.

Case of regular Lagrangian systems

We can say that a group of transformationsg(z) = ezLE generated by the vectorfield E maps the space of solutions of equation onto itself ifiXhg*(ω) + g*(dh) = 0For Xh satisfyingiXhω + dh = 0Hamilton's equation.It's easy to show that the vector field E should satisfy[E , Xh] = 0Indeed,iXhLEω + dLEh =LE(iXhω + dh) = 0since [E,Xh] = 0. When E is not Hamiltonian,the group of transformations g(z) = ezLE is non-Noethersymmetry (in a sense that it maps solutions onto solutions but does not preserve action).
theorem. (Lutzky, 1998) If the vector field E generates non-Noether symmetry, then the following functions are constant along solutions:I(k) = iWk ωEk       k = 1...n,where Wk and ωEk are outerpowers of W and LEω.
proof. We have to prove that I(k) is constant alongthe flow generated by the Hamiltonian. In other words, we should find thatLXhI(k) = 0 isfulfilled. Let us considerLXhI(1)LXhI(1)= LXh(iWωE) =i[Xh , W]ωE+ iWLXhωE,where according to Liouville's theorem both terms[Xh , W] = 0 andiWLXhLEω =iWLELXhω =0 since [E , Xh] = 0 and LXhω = 0 vanish.In the same manner one can verify thatLXhI(k) = 0
remark. Theorem is valid for a larger class of generators E .Namely, if [E , Xh] = Xf where Xf isan arbitrary Hamiltonian vector field, then I(k) is still conserved. Such asymmetries map the solutions of the equationiXhω + dh = 0on solutions ofiXhg*(ω) +d(g*h + f) = 0
remark. Discrete non-Noether symmetries give rise to the conservation ofI(k) = iWkg*(ω)kwhere g*(ω) is transformed ω.
remark. If I(k) is a set of conserved quantitiesassociated with E and f is any conserved quantity, then the set of functions{I(k) , f} (which due to the Poisson theorem are integrals of motion) is associated with[Xh , E]. Namely it is easy to show by taking the Liederivative of (15) along vector field E that{I(k) , f} = iWkωk[Xf , E] is fulfilled.As a result conserved quantities associated with Non-Noether symmetries form Lie algebra underthe Poisson bracket.
remark. If generator of symmetry satisfies Yang-Baxter equation[[E[E , W]]W] = 0 Lutzky's conservation laws are in involution [7]{Y(l) , Y(k)} = 0

Case of irregular Lagrangian systems

The singular Lagrangian (Lagrangian with vanishing Hessian) leads to degenerate 2-formω and we no longer have isomorphism between vector fields and 1-forms.Since there exists a set of "null vectors" us such thatiusω = 0       s = 1,2 ... n − rank(ω),every Hamiltonian vector field isdefined up to linear combination of vectors us. By identifying Xfwith Xf + sCsus, we can introduce equivalence class Xf (then all us belong to0 ).The bivector field W is also far from being unique, but ifW1 and W2 both satisfyiXiY ω =iW1,2 iXω ∧ iYω,theni(W1 − W2) iXω ∧ iYω = 0       ∀X,Yis fulfilled. It is possible only whenW1 − W 2 = svs ∧ uswhere vs are some vector fields andiusω = 0(in other words when W1 − W2 belongs to the class0)
theorem. If the non-Hamiltonian vector field Esatisfies [E , Xh] = 0 commutationrelation (generates non-Noether symmetry), then the functionsI (k)= iWkωEk        k = 1...rank(ω)(where ω E = LEω) are constant along trajectories.
proof. Let's consider I(1)LXhI(1)= LXh(iWωE)= i[Xh , W]ωE +iWLXhωE = 0The second term vanishes since [E , Xh] = 0 andLXhω = 0. The first one iszero as far as [Xh , W] = 0 and[E , 0] = 0 are satisfied. SoI (1) is conserved.Similarly one can show that LXhI(k) = 0 isfulfilled.
remark. W is not unique, but I(k) doesn't dependon choosing representative from the class W.
remark. Theorem is also valid for generators E satisfying [E , Xh] = Xf
example. Hamiltonian description of the relativistic particle leads to the following actionA = p0dx0 + spsdxswherep0 = (p2 + m2)1/2with vanishing canonical Hamiltonian and degenerate 2-form defined byp0ω = s(psdps ∧ dx0 + p0dps ∧ dxs).ω possesses the "null vector field"iuω = 0u = p0∂x0 + sps∂xs.One can check that the following non- Hamiltonian vector fieldE =p0x0∂x0+ p1x1∂x1 + ⋯ + pnxn∂xngenerates non-Noether symmetry. Indeed, E satisfies[E , Xh] = 0 because ofXh = 0 and [E,u] = u.Corresponding integrals of motion are combinations of momenta:I(1) = spsI(2) = r > s prpsI(n) = spsThis example shows that the set of conserved quantities can be obtained from a singleone-parameter group of non-Noether transformations.
acknowledgements. Author is grateful to Z. Giunashvili and M. Maziashvili forconstructive discussions and particularly grateful to George Jorjadze for invaluable help.This work was supported by INTAS (00-00561)and Scholarship from World Federation of Scientists.


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