Poincaré (non-holonomic Lagrange) Equations

Transcription

1 Department of Theoretical Physics Comenius University Bratislava Student Colloqium and School on Mathematical Physics, Stará Lesná, Slovakia, August 23-29, 2010

2 We will learn: In which direction Poincaré generalized Lagrange equations

3 We will learn: In which direction Poincaré generalized Lagrange equations How can one derive them from the least action principle

4 We will learn: In which direction Poincaré generalized Lagrange equations How can one derive them from the least action principle What is interesting about their structure

5 We will learn: In which direction Poincaré generalized Lagrange equations How can one derive them from the least action principle What is interesting about their structure What happens, in particular, on Lie groups

6 We will learn: In which direction Poincaré generalized Lagrange equations How can one derive them from the least action principle What is interesting about their structure What happens, in particular, on Lie groups How they are related to Euler (rigid body motion) equations

7 We will learn: In which direction Poincaré generalized Lagrange equations How can one derive them from the least action principle What is interesting about their structure What happens, in particular, on Lie groups How they are related to Euler (rigid body motion) equations How Četajev repeated the story with Hamilton equations

8 Contents 1 2 Frame components of velocity and momentum Variational field W and velocity field V How to express δx i and δv a in terms of W Variation of S - Poincaré equations 3 4 Frame components with respect to left invariant vector fields Particular case - Euler dynamical and kinematical equations 5 Variation of the action pdq Hdt in terms of (x i, p a ) Četajev equations

9 Frame components of velocity and momentum Variational field W and velocity field V How to express δx i and δv a in terms of W Variation of S - Poincaré equations Coordinate components of the velocity vector In standard Lagrange equations d L dt ẋ i L x i = 0 (1) we encounter variables x i (generalized coordinates) and ẋ i (generalized velocities). From the geometrical point of view the coordinates x i represent the curve γ(t) whereas v i = ẋ i are the coordinate components of its velocity vector: x i (t) γ(t) v i (t) ẋ i (t) γ(t)

10 Frame components of velocity and momentum Variational field W and velocity field V How to express δx i and δv a in terms of W Variation of S - Poincaré equations Frame components of the velocity vector Consider a frame field e a e a (x) = e i a(x) i i = e a i (x)e a (x) (2) (any - orhonormal, left invariant, even the original coordinate one). Then the velocity vector may be expressed in two ways γ(t) = v i (t) i = v a (t)e a (x(t)) (3) The variables v a are frame ("vielbein") components of the velocity vector ("pseudovelocities"). So, we can also write x i (t) γ(t) v a (t) γ(t)

11 Frame components of velocity and momentum Variational field W and velocity field V How to express δx i and δv a in terms of W Variation of S - Poincaré equations Relation between the two kinds of components From (39) we see that v i i = v a e a v a = e a i v i v i = e i av a (4) This means that Lagrangian may also be expressed in terms of the mixed variables (x i, v a ) L = L(x i, v a ) rather than L = L(x i, v i ) (5)

12 Frame components of velocity and momentum Variational field W and velocity field V How to express δx i and δv a in terms of W Variation of S - Poincaré equations Variation of Lagrangian In order to derive equations of motion via the least action principle in terms of (x i, v a ), we need the variation δl(x i, v a ) = L x i δx i + L v a δv a (6) As is the case for δẋ i, the variation δv a is related to δx i. The explicit formula is, however, more subtle. Therefore we prefer to gain some geometrical insight into the variational procedure. (In his paper, Poincaré simply states the result with laconic comment "Or on trouve aisément" (one easily finds)).

13 Frame components of velocity and momentum Variational field W and velocity field V How to express δx i and δv a in terms of W Variation of S - Poincaré equations Variational field W We want to perform a small variation of the curve γ(t). Consider a vector field W an let Φ ɛ denote its flow. Then the variation may be realized as γ(t) Φ ɛ (γ(t)) γ ɛ (t) Φ s W (7) In this way the flow produces a two-dimensional (narrow) surface S. The field W is tangent to the surface.

14 Frame components of velocity and momentum Variational field W and velocity field V How to express δx i and δv a in terms of W Variation of S - Poincaré equations Velocity field V There is another vector field living on the surface S. Each curve γ ɛ (t) induces in all its points velocity vector γ ɛ (t). Altogether, collection of all velocity vectors of all curves results in a vector field on S - the velocity field V. In each point of the surface S, the two fields, V and W, span the tangent space to the surface.

15 Frame components of velocity and momentum Variational field W and velocity field V How to express δx i and δv a in terms of W Variation of S - Poincaré equations Commutator of V and W vanishes The field V is Lie-dragged with respect to W. Then the Lie derivative vanishes, L W V = 0, or, put another way, the two fields commute [W, V ] = 0 (8) In terms of frame components of the fields this reads 0 = [W, V ] a = WV a VW a + c a bc W b V c (9) where c c ab (x) are anholonomy coefficients of the frame field e a [e a, e b ] =: c c ab (x)e c (10)

16 Frame components of velocity and momentum Variational field W and velocity field V How to express δx i and δv a in terms of W Variation of S - Poincaré equations Variation δx i in terms of W In coordinates, the variation of x i (t) reads x i (t) x i (t) + δx i (t) But we know (see (7)) that the new curve arises by means of the flow of the original curve along W. So abstractly γ(t) Φ W ɛ (γ(t)) (11) in coordinates x i (t) x i (t) + ɛw i (t) (12) and, consequently δx i = ɛw i (13)

17 Frame components of velocity and momentum Variational field W and velocity field V How to express δx i and δv a in terms of W Variation of S - Poincaré equations Variation δv a in terms of W (1) We have v a e a = γ (14) (v a + δv a )e a = γ ɛ = V (γ ɛ (t)) V (Φ ɛ (γ(t))) (15) (the two e a s on the left sit in different points) hence δv a = ɛwv a (16) V (Φ ɛ (γ(t))) = V a (Φ ɛ (γ(t)))e a = (Φ ɛv a )(γ(t))e a = (ˆ1 + ɛl W )V a )(γ(t))e a = V a (γ(t)) + ɛ(wv a )(γ(t))e a v a e a + ɛ(wv a )e a

18 Frame components of velocity and momentum Variational field W and velocity field V How to express δx i and δv a in terms of W Variation of S - Poincaré equations Variation δv a in terms of W (2) Now the vanishing of the commutator [V, W ] (8, 9) comes in handy and we get δv a = ɛ(ẇ a + c a bc v b W c ) (17) WV a = VW a c a bc W b V c = VW a + c a bc V b W c = Ẇ a + c a bc v b W c Everything is evaluated at the original curve γ(t), so that Vf = γf = ḟ and also V a reduce to v a.

19 Frame components of velocity and momentum Variational field W and velocity field V How to express δx i and δv a in terms of W Variation of S - Poincaré equations Using of the results for δx i, δv a for performing δl Using results (13) and (17), we get δl(x i, v a ) = L x( δx i + L i v δv a a ) = ɛ L W i + L x i v (Ẇ a + c a a bc v b W c ) ( ) = ɛ W a e a L + L v (Ẇ c + c c c ba v b W a ) = ɛw a ( e a L + cba c v b L = ɛw a ( e a L + c c ba v b L v c ) v + ɛẇ a L c v a d dt ) L v + d a dt ( ɛw a L v a )

20 Frame components of velocity and momentum Variational field W and velocity field V How to express δx i and δv a in terms of W Variation of S - Poincaré equations Variation δs of the action functional S[γ] Then δs = ɛ t2 t 1 dt ( e a L + c c ba v b L v c d dt ) [ ] L L t2 v a W a + ɛ v a W a t 1 (18) Using standard arguments (δx i vanishing at t 1 and t 2 and "arbitrary" inside the interval) we at last get d dt L v a e al + cab c v b L v c = 0 Poincaré equations (19)

21 Just n first-order equations For L(x i, v a ), Poincaré equations d dt L v a e al + cab c v b L v c = 0 (20) represent just n first-order equations. We know that Lagrange equations represent n second-order equations. Where is the origin of the discrepancy? Don t panic. The remaining information sits in (4) ẋ i = e i a(x)v a (21)

22 Complete set of equations to be solved In order to use the equations for actual computation one has to write down the complete set of equations (Poincaré, 1901) d dt L v a e al + c c ab v b L v c = 0 (22) v a = e a i (x)ẋ i (23) Note that in general they form a coupled system of 2n first-order differential equations. In this respect they resemble standard Hamilton, rather then standard Lagrange, equations.

23 Particular case - holonomic (coordinate) frame e a = a In the particular case of holonomic (coordinate) frame e a = a 1. ebc a 0 2. e a L ea(x) i i L i L L x i 3. L v a L v i The complete system reduces to d dt equivalent to standard Lagrange equations L v i L x i = 0 (24) v i = ẋ i (25)

24 Elementary example - free motion in Euclidean plane In polar orthonormal frame we have e r = r e ϕ = (1/r) ϕ γ = ṙ r + ϕ ϕ = v r e r + v ϕ e ϕ (26) [e r, e ϕ ] = ( 1/r)e ϕ crϕ ϕ = cϕr ϕ = 1/r (27) L(r, ϕ, v r, v ϕ ) = 1 2 ((v r ) 2 + (v ϕ ) 2 ) (1/2)(ṙ 2 + r 2 ϕ 2 ) (28) v r 1 r v ϕ v ϕ = 0 v ϕ + 1 r v r v ϕ = 0 (29) v r = ṙ v ϕ = r ϕ (30) (True, but no profit here :-)

25 Frame components with respect to left invariant vector fields Particular case - Euler dynamical and kinematical equations Specific features on Lie groups There are important examples when 1. the configuration space has the structure of a Lie group Then it is natural to choose left invariant frame field e a. Position-dependent anholonomy coefficients c c ab (x) reduce to position-independent structure constants c c ab of the corresponding Lie algebra.

26 Frame components with respect to left invariant vector fields Particular case - Euler dynamical and kinematical equations Left invariant kinetic energy 2. Kinetic energy is left invariant Then e a L = e a (T U) = e a U. Moreover, left invariant metric tensor reads g = I ab e a e b I ab = const. (31) Therefore T = 1 2 g( γ, γ) = 1 2 I abv a v b (32)

27 Frame components with respect to left invariant vector fields Particular case - Euler dynamical and kinematical equations Resulting complete set of equations Under these conditions, the complete set (22) and (23) reduces to I ab v b + c dab v b v d = e a U (33) v a = e a i (x)ẋ i (34) where c dab := I dc c c ab (35) Note the structure similarity of this set of equations to Euler (rigid body motion) equations.

28 Frame components with respect to left invariant vector fields Particular case - Euler dynamical and kinematical equations Euler dynamical and kinematical equations And indeed, for G = SO(3) = configuration space of rigid body and I ab = diag (I 1, I 2, I 3 ) = moments of inertia of the body, the set of equations I ab v b + c dab v b v d = e a U (36) v a = e a i (x)ẋ i (37) is nothing but well-known Euler dynamical (36) and kinematical (37) equations. Here, frame components v a play the role of "body components" of angular momentum of the rotating body.

29 Četajev equations Further Reading Coframe components of the momentum vector In (39) we introduced frame components of the velocity vector. In the same way one can define coframe components of the momentum covector. Consider a coframe field e a (dual to e a ) e a (x) = e a i (x)dx i dx i = e i a(x)e a (x) (38) Then the momentum covector may be expressed in two ways p(t) = p i (t)dx i = p a (t)e a (x(t)) (39) The variables p a are coframe components of the momentum covector.

30 Četajev equations Further Reading Frame coordinates on TM and T M Consider a frame field e a and the dual coframe field e a. Then any vector on M may be expressed as v = v i i = v a e a and any covector on M may be expressed as p = p i dx i = p a e a. This means that one can use as local coordinates either (x i, v i ) on TM (x i, p i ) on T M (40) or Clearly (x i, v a ) on TM (x i, p a ) on T M (41) v a = e a i v i v i = e i av a p a = e i ap i p i = e a i p a (42)

31 Četajev equations Further Reading Expressing p i dx i Hdt in frame components In order to derive Hamilton-like equations from the least action principle we need to express the term p i dx i in terms of frame components. We have Then the action functional reads S[ γ] = p i ẋ i = p i e i av a = p a v a (43) t2 t 1 (p a v a H(x i, p a ))dt (44) Here γ(t) (x i (t), p a (t)) is a curve in phase space.

32 Četajev equations Further Reading Variation of p i dx i Hdt in frame components (1) We have δs = t2 t 1 (δp a v a + p a δv a δh(x i, p a ))dt (45) where δh = H x i δx i + H p a δp a (46) δx i = ɛw i (47) δv a = ɛ(ẇ a + c a bc v b W c ) (48) Since the curve (x i (t), p a (t)) is regarded as a curve in phase space, variations δx i and δp a are to be treated as independent.

33 Četajev equations Further Reading Variation of p i dx i Hdt in frame components (2) After usual per partes we get δs = + t2 t 1 t2 t 1 δp a ɛw a ( v a H p a ) dt (49) ( ṗ a e a H c c ab v b p c ) dt (50) + ɛ (p a W a ) t 2 t1 (51) Remember standard variational Hamiltonian folklore: δx i vanishing and no restrictions on δp a at t 1 and t 2 ; both "arbitrary" inside the interval. (52)

34 Četajev equations Further Reading Variation of p i dx i Hdt in frame components (3) Then one obtains v a = H p a (53) ṗ a = e a H c c ab v b p c (54) or, using the kinematical equations ẋ i = e i a(x)v a, (Četajev, 1927). ẋ i = ea(x) i H (55) p a ṗ a = e a H cab c H p c (56) p b

35 Četajev equations Further Reading How Poisson bracket looks in frame components If f (x i, p a ) is any function on phase space, then ḟ = f x i = H p a {H, f } ẋ i + f p a ṗ a (e a f ) f p a So, Poisson bracket explicitly reads (e a H) + p c cab c H (x) p a f p b {F, G} = F (e a G) G (e a F ) + p c c c F G ab (x) (57) p a p a p a p b

36 Četajev equations Further Reading Example - Hamilton equations for Dido problem (1) Finding subriemannian geodesics needs solving Hamilton equations, often fairly complicated (see my Stará Lesná 2009 lectures). It might be instructive to confront both pictures - standard holonomic versus non-holonomic. Easy example - Dido problem. The frame is e 1 = x (y/2) z e 2 = y + (x/2) z e 3 = z so the frame components of momentum read p 1 = p x (y/2)p z p 2 = p y + (x/2)p z p 3 = p z

37 Četajev equations Further Reading Example - Hamilton equations for Dido problem (2) Resulting equations: Hamilton equations ẋ = p x y(p z /2) ẏ = p y + x(p z /2) ż = (xẏ yẋ)/2 ṗ x = (p y + xp z /2)(p z /2) ṗ y = (p x yp z /2)(p z /2) ṗ z = 0 Četajev equations ẋ = p 1 ẏ = p 2 ż = (xẏ yẋ)/2 ṗ 1 = p 2 p 3 ṗ 2 = +p 1 p 3 ṗ 3 = 0 Which system would you prefer to solve?

38 Četajev equations Further Reading For Further Reading H. Poincaré. Sur une forme nouvelle des équations de la méchnique. C.R.Acad.Sci.Paris, v.132, p , 1901

39 Četajev equations Further Reading For Further Reading H. Poincaré. Sur une forme nouvelle des équations de la méchnique. C.R.Acad.Sci.Paris, v.132, p , 1901 M. Četajev. Sur les équations de Poincaré. C.R.Acad.Sci.Paris, v.185, p , 1927

40 Četajev equations Further Reading For Further Reading H. Poincaré. Sur une forme nouvelle des équations de la méchnique. C.R.Acad.Sci.Paris, v.132, p , 1901 M. Četajev. Sur les équations de Poincaré. C.R.Acad.Sci.Paris, v.185, p , 1927 V.I. Arnold, V.V. Kozlov, A.I. Neishtadt Mathematical Aspects of Classical and Celestial Mechanics. Springer, Berlin Heidelberg, 2009

41 Četajev equations Further Reading For Further Reading H. Poincaré. Sur une forme nouvelle des équations de la méchnique. C.R.Acad.Sci.Paris, v.132, p , 1901 M. Četajev. Sur les équations de Poincaré. C.R.Acad.Sci.Paris, v.185, p , 1927 V.I. Arnold, V.V. Kozlov, A.I. Neishtadt Mathematical Aspects of Classical and Celestial Mechanics. Springer, Berlin Heidelberg, 2009