The Magnetic Spherical Pendulum

Transcription

1 The Magnetic Spherical Pendulum R. Cushman L. Bates Abstract This article gives two formulae for the rotation number of the flow of the magnetic spherical pendulum on a torus corresponding to a regular value of the energy momentum mapping. One of these formulae is nonclassical and is based on an idea of Montgomery. In questo lavoro si danno due formule per il numero di rotazione del flusso del sferico pendolo magnetico su un toro che corrisponde a un valore regolare dell applicazione energia-momento. Una di queste formule è nonclassica ed è basata su un idea di Montgomery. In this article we look at the problem of computing a rotation number for the flow of the magnetic spherical pendulum. Mechanically, the magnetic spherical pendulum is the motion of a charged massive particle constrained to a sphere. The forces on the particle come from a gravitational field and a magnetic field due to a magnetic monopole located at the center of the sphere. At first sight this problem may seem obscure, if not baroque, but actually it is a mechanical model for the motion of the vertex of the Lagrange top (recall that the Lagrange top is the symmetrical top spinning in a gravitational field about a fixed point on its symmetry axis.) Geometrically, the magnetic spherical pendulum arises as a reduction of the Lagrange top by reducing out the symmetry about the body axis, and the value of angular momentum that we reduced by manifests itself as the strength of the magnetic field. We will give a fast heuristic derivation of this using Euler angles. It is interesting and more rigorous to do the calculation in an invariant manner, but we will not do so here for the sake of brevity. Keywords completely integrable system, particle mechanics

2 A Lagrangian for the top is L = 2 A( φ 2 + sin 2 φ θ 2 ) + 2 C( ψ + cos φ θ) 2 cos φ The notation is such that A and C are the moments of inertia, φ is the colatitude, θ is the longitude, and ψ is the rotation of the top about its symmetry axis. The Euler-Lagrange equations are d (A sin2 φ θ + C( ψ + cos φ θ) cos φ) = (A φ A sin φ cos φ θ 2 + C( ψ + cos φ θ) sin φ θ + sin φ = d (C( ψ + cos φ θ)) = Because of the third equation, set p ψ = L = j. Now j is a constant of ψ motion because ψ is an ignorable coordinate. Substitute this value of j into the first two Euler-Lagrange equations to obtain Note that the energy d (A sin2 φ θ + j cos φ) = A φ A sin φ cos φ θ 2 + j sin φ θ + sin φ = E = 2 A( φ 2 + sin 2 φ θ 2 ) + cos φ is a constant of motion for this reduced problem. Pushing everything over to the cotangent bundle by the Legendre transformation p θ = A sin 2 φ θ p φ = A φ we arrive directly at the Hamiltonian description with Hamiltonian h = 2A (p φ 2 + sin 2 φ p θ 2 ) + cos φ 2

3 and symplectic form ω = j sin φ dφ dθ + dθ dp θ + dφ dp φ. It is a simple calculation to check that the Hamiltonian equations of motion are what we get by pushing over the reduced Euler-Lagrange equations by the Legendre transformation. By inspection of the above two equations, one sees the validity of the mechanical interpretation of the motion of the vertex of the top which we gave at the beginning. Liouville integrability and reduction In this section we reduce out the axial symmetry of the magnetic spherical pendulum by using singular reduction []. We use this technique because it allows us to reduce at all values of the angular momentum. This means that our reduction captures those motions of the vertex of the top which pass through the north or the south pole as well as the motions corresponding to regular precession. These motions are physically interesting and are not susceptible to analysis using action angle variables. The reason for this is that the Liouville integrablility thorem, in its modern formulation by Arnol d [2], requires functional independence of the integrals. In addition, we explicitly realize each of the reduced spaces as an semialgebraic variety in R 3 with an explicit Poisson structure. As a byproduct of the invariant theoretic technique used in constructing the reduced space, we obtain a faithful geometric model of the singularities and how the reduced spaces fit together as the angular momentum is varied. We start by describing the magnetic spherical pendulum as a constrained Hamiltonian system. On R 3 let, be the Euclidean inner product and be the usual vector product. Define a 2-form Ω on the bundle π : T R 3 R 3 by Ω(x, y)((u, v), (r, s)) = dθ + µπt R3F () = v, r + s, u + µ x x 3, u r. Ω is the standard symplectic form dθ on T R 3 plus a term µf, which is the magnetic field of a monopole of strength µ located at the origin of R 3. It is 3

4 straightforward to check that Ω is symplectic. Suppose that H : T R 3 R is a smooth function. Then a short calculation shows that the Hamiltonian vector field X H associated to H is dx dy = H y = H x + µ x x 3 H y. (2) where On T R 3 define an S action Φ : S T R 3 T R 3 : (x, y) (R t x, R t y) (3) R t = cos t sin t sin t cos t. The infinitesimal generator of this S action is the vector field dx dy = x e 3 = y e 3. (4) Using (2), another short calculation shows that the Hamiltonian vector field X J corresponding to the Hamiltonian function J : T R 3 R : (x, y) x y, e 3 + µ x x, e 3 (5) is also given by (4). Thus J is the momentum mapping of the S action Φ. Consider the submanifold T S 2 T R 3 defined by F (x, y) = x, x = F 2 (x, y) = x, y =. Since the 2 2 matrix ({F i, F j } T S 2 ) = ( 2 2 ) 4

5 is invertible (see table for the structure matrix of {, } T R 3), T S 2 is a symplectic submanifold of T R 3. Therefore Ω T S 2 is a symplectic form on T S 2. The magnetic spherical pendulum is the Hamiltonian system (T R 3, Ω, H) with Hamiltonian H : T R 3 R : (x, y) 2 y, y + γ x, e 3 (6) constrained to (T S 2, Ω T S 2 ). In other words, it is the Hamiltonian system (T S 2, Ω T S 2, H T S 2 ). Using the Dirac prescription [3] we compute X H T S 2 as follows. Let H = H 2 {H, F 2} T R 3F + 2 {H, F } T R 3F 2 = 2 y, y + γ x, e 3 + 2( y, y γ x, e3 )( x, x ) x, y 2. Then using (2), X H dx dy on (T R 3, Ω) is = y + y( x, x ) 2 x, y x = γ e γ e 3( x, x ) ( y, y γ x, e 3 )x + 2 x, y y + µ x x 3 ( y + y( x, x ) 2 x, y x ). Since {H, F } T R 3 T S 2 = {H, F 2 } T R 3 T S 2 =, T S 2 is an invariant manifold of X H. Consequently, X H T S 2 = X H T S 2 is given by dx dy = y = γ e 3 + (γ x, e 3 y, y )x + µ x y. (7) Since H and T S 2 are invariant under Φ, L XJ T S 2 H T S 2 = {H, J } T R 3 T S 2 =, that is, J T S 2 is an integral of the magnetic spherical pendulum. Therefore, the magnetic spherical pendulum is Liouville integrable and has energy momentum mapping EM : T S 2 T R 3 R 2 : (x, y) (H T S 2 (x, y), J T S 2 (x, y)). 5

6 To remove the S symmetry defined by the S action Φ and reduce the magnetic spherical pendulum to a one degree of freedom Hamiltonian system, we use the technique of singular reduction []. To start with, the algebra of S invariant polynomials on T R 3 is generated by which satisfy the relation σ = x 3 σ 3 = y 2 + y2 2 + y3 2 σ 5 = x 2 + x 2 2 σ 2 = y 3 σ 4 = x y + x 2 y 2 σ 6 = x y 2 x 2 y σ σ 2 6 = σ 5 (σ 3 σ 2 2) σ 3 & σ 5. (8) Thus the space of S orbits T R 3 /S is the semialgebraic variety W defined by (8). Since T S 2 T R 3 is S invariant, the S orbit space T S 2 /S is the semialgebraic variety U defined by (8) together with σ 5 + σ 2 = σ 4 + σ σ 2 =. (9) Solving (9) for σ 4 and σ 5 and then substituting the result into (8) shows that the variety U is defined by σ 2 σ σ 2 6 = ( σ 2 )(σ 3 σ 2 2), σ & σ 3. () Since the S action has momentum mapping J T S 2, the reduced phase space M j = (J T S 2 ) (j)/s is the semialgebraic variety of U defined by () and σ 6 + µ σ = j. In other words, M j is the semialgebraic variety of R 3 defined by σ 3 ( σ 2 2 ) (j µσ ) 2 σ 2 2 =, σ & σ 3. () When j ±µ, the reduced space M j is diffeomorphic to R 2, being the graph of the function σ 3 = G(σ, σ 2 ) = (j µσ ) 2 + σ2 2, σ σ 2 <. When j = ±µ, M ±µ is not the graph of a function, because it contains a vertical line {(±,, σ 3 ) R 3 σ 3 } (see figure ). However M ±µ is still homeomorphic to R 2. 6

7 Figure. The reduced phase space M j. Since H T S 2 is S -invariant, it induces the Hamiltonian H j M j on the reduced space M j where H j : R 3 R : (σ, σ 2, σ 3 ) 2 σ 3 + γσ (2) This completes the reduction of the magnetic spherical pendulum to a one degree of freedom Hamiltonian system. What remains to be done is find Hamilton s equations for the reduced Hamiltonian H j M j. To do this we compute the Poisson structure on C (M j ) as follows. From the definition of the symplectic form Ω on T R 3 we obtain a Poisson bracket {, } T R 3 on C (T R 3 ), which is given by {f, g} T R 3 = i,j f g {ζ i, ζ j } T R 3, ζ i ζ j where (ζ,..., ζ 6 ) = (x, x 2, x 3, y, y 2, y 3 ) are coordinates on T R 3 and ({ζ i, ζ j }) is the skew symmetric structure matrix {, } T R 3 half of which is given in table. 7

8 {A, B} x x 2 x 3 y y 2 y 3 B x x 2 x 3 y µ x 3 x 3 µ x x 3 y 2 µ x 2 x 3 y 3 A Table. Structure matrix for {, } T R 3 Let (σ,..., σ 6 ) are coordinates on R 6. A straightforward calculation using table shows that C (R 6 ) has a Poisson bracket {, } R 6 half of whose skew symmetric structure matrix is given in table 2. {A, B} σ σ 2 σ 3 σ 4 σ 5 σ 6 B σ 2σ 2 2µ µ σ 2 σ x 3 6 σ x 3 5 2µ σ 3 σ x 3 σ 6 + 2σ2 2 2σ 3 4σ 4 2µ (σ x 3 σ 4 + σ 2 σ 5 ) σ 4 2σ 5 µ σ x 3 σ 5 σ 5 σ 6 A Table 2. Structure matrix for {, } R 6 Here x = σ 2 + σ5 2 Using table 2 it is easily shown that C = σ σ 2 6 σ 5 (σ 3 σ 2 2) and C 2 = σ 6 + µσ are Casimir elements of (C (R 6 ), {, } R 6), that is, for every f C (R 6 ), {C, f} R 6 = {C 2, f} R 6 =. Since W, the space of S orbits on T R 3, is the semialgebraic variety defined by C =, σ 3 & σ 5, the Poisson bracket on W has the same 8

9 structure matrix as the Poisson bracket on R 6. Because {σ 5 + σ 2, σ 4 + σ σ 2 } R 6 U = 2, U, the space of S orbits on T S 2, is a symplectic subvariety of W. Consequently, the Poisson bracket {, } U on U may be computed using the Dirac prescription: namely, for every f, g C (R 6 ) where {f U, g U} U = {f, g } R 6 U f = f 2 {f, σ 4 + σ σ 2 } R 6(σ 5 + σ 2 ) + 2 {f, σ 5 + σ 2 } R 6(σ 4 + σ σ 2 ) and similarly for g. Half of the skew symmetric structure matrix for {. } U is given in table 3. {A, B} σ σ 2 σ 3 σ 6 B σ σ 2 2σ 2 σ 2 2µσ 6 2σ σ 3 µσ 5 σ 3 2µσ 2 σ 6 A Table 3. The structure matrix for {, } U. The reduced space M j is the subvariety of U defined by C 2 = j. Since C 2 is a Casimir in (C (U), {, } U ), the Poisson bracket {, } Mj on M j has the same structure matrix as that give in table 3 with the last row and column deleted and σ 6 replaced by j µσ. This completes the construction of the Poisson structure on M j. It is easier to calculate in the ambient space R 3 with coordinates (σ, σ 2, σ 3 ) instead of on the reduced space M j. On C (R 3 ) define a Poisson bracket {, } R 3 half of whose skew symmetric structure matrix is given in table 4. {A, B} σ σ 2 σ 3 B σ σ 2 2σ 2 σ 2 2µ(j µσ ) 2σ σ 3 σ 3 A Table 4. The structure matrix for {, } R 3. 9

10 An inspection of table 4 shows that {σ i, σ j } R 3 = k ε ijk ψ σ k (3) where ψ(σ, σ 2, σ 3 ) = σ 3 ( σ 2 ) σ 2 2 (j µσ ) 2 and ψ = is the defining equation of M j. Since ψ is a Casimir element of (C (R 3 ), {, } R 3), the Poisson bracket {, } Mj on C (M j ) is obtained by restricting {, } R 3. From (3) it follows that for every f, g C (R 3 ) {f, g} R 3 = f g, ψ. Consequently, for h C (R 3 ) Hamilton s equations are dσ i = {σ i, h} R 3 = ( h ψ) i (4) for i =, 2, 3. Using (2) and (4), Hamilton s equations for the reduced Hamiltonian H j M j are dσ dσ 2 dσ 3 = σ 2 = γ( σ 2 ) + µ(j µσ ) σ σ 3 = 2γσ 2 (5) restricted to M j. Note that ψ is an integral of (5). We return to studying the energy momentum mapping EM. For fixed µ, the set Σ µ of critical values of EM is the same as the set of critical values of H j M j. These values (h, j) occur when the family of lines h = 2 σ 3 + γσ (6) in R 2 with coordinates (σ,, σ 3 ) intersects the curve {σ 2 = } M j given by = σ 3 ( σ 2 ) (j µσ ) 2 σ & σ 3 (7) at a point with multiplicity greater than one (see figure 2).

11 Figure 2. The critical points and values of H j {σ 2 = } M j. Solving (6) for σ 3 and then substituting the result into (7) shows that (h, j) Σ µ if and only if the polynomial V (σ ) = 2(h σ )( σ 2 ) (j µσ ) 2 (8) has a multiple root in [, ]. In other words, Σ µ is a µ = const. slice of a piece S of the discriminant locus of V. The piece S also occurs in the treatment of the Lagrange top [4], [5]. Now suppose that (h, j) is a regular value of EM and that EM (h, j) is nonempty. The line defined by (6) intersects M j {σ 2 = } in two geometrically distinct nonsingular points, whose abscissas are σ ± = x ± 3. Thus, in R 3 with coordinates (σ, σ 2, σ 3 ) the plane defined by (6) intersects M j in a smooth manifold (H j M j ) (h), which is diffeomorphic to a circle. From the reduction process we know that the (h, j)-level set of EM is an S bundle over the h-level set of H j M j. Because this level set bounds a 2-disc on M j, which is contractible, the bundle is trivial. Hence EM (h, j) is diffeomorphic to a 2-torus Th,j. 2 By construction, Th,j 2 is invariant under the flow of X H T S 2 and X J T S 2. A good way to vizualize Th,j 2 is to view it as some sort of bundle over its image under the projection π T S 2 : T S 2 T R 3 S 2 R 3 : (x, y) x

12 (see figure 3). In the following discussion we construct this bundle. By definition T 2 h,j T R 3 is given by the equations x 2 = x 2 + x x 2 3 = x y + x 2 y 2 + x 3 y 3 = 2 (y2 + y2 2 + y3) 2 + γx 3 = h x y 2 x 2 y + µ x 3 x Substituting the above equations into the identity gives = j. (9) (x y 2 x 2 y ) 2 + (x y + x 2 y 2 ) 2 = (x 2 + x 2 2)(y 2 + y 2 2) y 2 3 = V (x 3 ) = 2(h γx 3 )( x 2 3) (j µx 3 ) 2 (2) where x 3. Consequently, x 3 [x 3, x + 3 ] where V (x ± 3 ) = and < x 3 < x + 3 <, because (h, j) is a regular value of EM. Therefore the image of Th,j 2 under π T S 2 is contained in the annular piece A of S 2 defined by {x S 2 R 3 x 3 x 3 x + 3 }. Given x A, from (2) we find that y 3 = η V (x 3 ) where η 2 =. Solving for (y, y 2 ) gives x y 2 x 2 y = j µx 3 x y + x 2 y 2 = x 3 y 3 = ηx 3 V (x 3 ) y = [x x 2 2 (j µx 3 ) + ηx x 3 V (x 3 )] 3 y 2 = [ x x 2 2 (j µx 3 ) + ηx 2 x 3 V (x 3 )]. 3 Therefore for every x int A there are two points p ± T 2 h,j such that π T S 2(p ± ) = x; while for every x A, there is exactly one point p so that π T S 2(p ) = x. In other words, T 2 h,j is a pinched S bundle over A with bundle 2

13 projection π T S 2 Th,j. 2 The above argument shows that π T S 2 Th,j 2 has a fold singularity at every point on the two curves C ± = π T S ({x S 2 x 2 3 = x ± 3 }). Note that C + and C are each an orbit of X J T S 2 Th,j. 2 Figure 3. Image of T 2 h,j under π T S 2. 2 Two formulae for the rotation number In this section we derive two formulae for the rotation number of the flow of X H T S 2 on the 2-torus Th,j. 2 First we recall the definition of rotation number. Since (h, j) is a regular value of EM, d(h T S 2 ) and d(j T S 2 ) are linearly independent at every point of Th,j 2 = EM (h, j). Therefore X H T S 2 and X J T S 2 are linearly independent at every point of Th,j. 2 The curve C + on Th,j, 2 which is the image of the integral curve t ϕt (p) of X J T S 2 through p π T S ({x 2 3 = x + 3 }), is a cross section for the flow ϕt of X H T S 2 on Th,j 2 because 3

14 ) if q C +, then X H T S 2(q) is transverse to C + at q; and 2) since X H T S 2 does not vanish on Th,j, 2 there is a smallest T = T (p) > such that ϕt (p) C +. Let θ h,j be the smallest positive time such that ϕ θ h,j (p) = ϕ T (see figure 4). Then θ h,j is the rotation number of the flow of X H T S 2 on Th,j. 2 The rotation number does not depend on the choice of point p on C +. For if q C +, q p, then there is a smallest time S > such that ϕs (p) = q. Thus ϕ T (q) = ϕ = ϕ = ϕ T S S (ϕ S (p)) (p) (ϕt (p)), since [X H T S 2, X J T S 2] = (ϕ θ h,j (p)) = ϕ θ h,j which proves the claim. We now derive the classical formula for the rotation number of X H T S 2 on Th,j. 2 Using the map π T S 2 Th,j, 2 project the integral curve Γ : t ϕt (p) of X H T S 2 on Th,j 2 onto the curve γ : t γ(t) in the annular region A of S 2. Let x i A be coordinates on A with (x, x 2, x 3 ) being coordinates on R 3. Furthermore, let θ = tan x 2 and x 3 x (q), be coordinates on the universal covering space à of A. The following reasoning shows that a lift γ of γ to à satisfies the differential equations dθ dx 3 = j µx 3 (2) x 2 3 = ± V (x 3 ). (22) How the sign in (22) is chosen will be discussed below. By definition of Lie derivative, we find that dθ = L XH T S 2 θ 4

15 = x 2 + x 2 2 dx 2 (x x dx 2 ) = x y 2 x 2 y, using (7) x 2 + x 2 2 = j µx 3, using the definition of J T S 2 and the fact that x 2 3 Γ(t) (J T S 2 ) (j). Also, dx 3 = L XH T S 2 x 3 = π j (L XHj M j σ ) = πj (σ 2 ) using (5) = ±πj ( σ 3 ( σ) 2 (j µσ ) 2 ) since ψ is an integral of X Hj = ±πj ( 2(h γσ )( σ) 2 (j µσ ) 2 ), = η V (x 3 ), η 2 =. since Γ(t) (H T S 2 ) (h) The sign ambiguity in (22) is handled as follows. Suppose that γ(t ) = x ± 3 and at t ε > the value of η is known, then at time t + ε the value of η is the negative of η at t ε. Since Γ crosses C ± transversally at Γ(t ) and the mapping π T S 2 Th,j 2 has a fold singularity at Γ(t ), the curve γ has second order contact with C ± at γ(t ). The above sign convention ensures that the solutions of (2) and (22) in à are real analytic. Now consider the curve s ϕs (p) for s [, θ h,j ). Then the projected curve s π T S 2(ϕs (p)) is an arc of the small circle {x 3 = x + 3 } on S 2 which joins two successive points of intersection γ() and γ(t ) of γ [, T ] with {x 3 = x + 3 }. Let ϑ h,j be the angle between γ() and γ(t ) as measured from the center of the small circle {x 3 = x + 3 }. Then ϑ h,j is equal to the rotation number θ h,j, because ϑ h,j = = ϑh,j θh,j dθ L XJ T S 2 θ ds, by definition of rotation number 5

16 = θh,j ds, using (4) and the definition of θ = θ h,j. (23) Since V (x 3 ) > for x = (x, x 2, x 3 ) int A, from (22) we find that dx 3. Hence for t (, T/2) (T/2, T ) we may parametrize γ by x 3. Choose η = in (22) for t = ε. Then θ h,j = θh,j = = 2 Using (2) and (22) we get x 3 x + 3 x + 3 x 3 x + 3 θ h,j = 2 x 3 L XH T S 2 θ dθ x + 3 dx 3 + dx 3 x 3 dθ dx 3 dx 3. dθ dx 3 dx 3 j µx 3 ( x 2 3) V (x 3 ) dx 3, (24) which is the classical formula for the rotation number. To find a second formula for the rotation number we proceed as follows. Let C = {ϕs (p) s θ h,j } and C 2 = {ϕt (p) t T } be curves on Th,j 2 and let C be the curve C 2 C. Now θ T S 2 is the canonical -form on T S 2, θ being the canonical -form on T R 3. Then clearly we have C θ T S 2 = θ T S 2 C 2 θ T S 2. C (25) According to [6], calculating both sides of (25) will give a second formula for θ h,j. We shall do this, starting with evaluating the first term of the right hand side of (25). We find that C 2 θ T S 2 = = T T (θ T S 2 ) ( ϕ t (p) ) ( ( (θ T S 2 )(X H T S 2) )( ϕ dϕh T S2 t t (p) ) (p) ). 6

17 But (θ T S 2 )(X H T S 2)(x, y) = θ(x, y)(y, γe 3 + (γ x, e 3 y, y )x + +µx y) T S 2 ), using (7) = y, y T S 2, using the fact that θ(x, y)(u, v) = y, u = 2(H T S 2 )(x, y) 2(V S 2 )(x). where V(x) = γ x, e 3 is the potential energy of the unconstrained system (T R 3, Ω, H). Hence C 2 θ T S 2 = 2 T = 2hT 2γ = 2hT 2γ = 2hT 2γ (H T S 2 )(ϕt (p)) 2 T π j (σ M j ) ( ϕ t T (πt S (p) ) S2 2V)(ϕH T (p)) since t ϕt (p) lies in Th,j 2 EM (h, j) T since π j ϕ T σ (ϕ H j M j t (π j (p))) t (J T S 2 ) (j) = ϕ H j M j t π j σ (t) (26) where t (σ (t), σ 2 (t), σ 3 (t)) M j R 3 is the integral curve of the reduced vectorfield X Hj M j, which parameterizes the h-level set of the reduced Hamiltonian H j M j. Therefore we may write (26) as C 2 θ T S 2 = 2h = 4 = 4 = 4 T σ + σ σ + σ σ + σ 2γ T σ (t) h γσ dσ, parameterizing by σ instead of t σ h γσ σ 2 dσ, using (5) h γσ V (σ ) dσ, since ψ is an integral of (5) and σ(t) Hj (h). t (27) 7

18 We now compute C θ T S 2. As before we find that But C θ T S 2 = = θh,j θh,j ( θ T S 2 (ϕ s (p) ) ( ( (θ T S 2 )(X J T S 2) )( ϕ dϕj T S2 s (p) ) ds ds s (p) ) ds (θ T S 2 )(X J T S 2)(x, y) = (θ T S 2 )(x, y)( x e 3, y e 3 ), using (3) = y, x e 3 T S 2, since θ(x, y)(u, v) = y, u = (x y 2 y 2 x ) T S 2 = (σ 6 T S 2 )(x, y), which is invariant under the flow ϕ C θ T S 2 = θh,j s. Consequently, (σ 6 T S 2 )(p) = θ h,j ( σ6 T S 2 (p) ) = θ h,j (j µσ (p)), since J T S 2 = ( e 3, x y + µx 3 ) T S 2 and p C (J T S 2 ) (j) = θ h,j (j µσ + ) since σ = x 3 = x + 3 = σ + on C. (28) Finally we compute the left hand side of (25). Montgomery s idea [6] is to use Stokes theorem to evaluate C θ T S2. To carry this out we need to know that C bounds a suitable domain D. (This point seems to be incorrect in [6]). Certainly C is not null homologous on Th,j. 2 Therefore we must construct a suitable space D (J T S 2 ) (j) containing Th,j, 2 in which C is null homotopic and hence null homologous. To find D, we first fix a regular value j of J T S 2. Next observe that the image of C under the reduction map π j is the smooth closed curve (H j M j ) (h), which bounds a closed disc h = (H j M j ) (I h ) in M j. Here I h = [h min, h], where h min is the minimum value of H j M j. Since h is contractible, the S bundle πj ( h ) over h is trivial, that is, πj ( h ) is diffeomorphic to h S. Note that πj ( h ) = Th,j 2 and that πj ( h ) (J T S 2 ) (j). Finally put D = πj ( h ). To show that C is null homotopic in D, we note that D is a pinched closed -disc bundle over the annulus A S 2 with bundle projection π T S 2 D. The proof of this fact goes 8

19 as follows. Since D = {T 2 h,j h I h }, the space D T R 3 is defined by x 2 + x x 2 3 = x y + x 2 y 2 + x 3 y 3 = 2 (y2 + y2 2 + y3) 2 + γx 3 = h, h Ih x y 2 x 2 y + µ x 3 x = j, j µ. For each h I h, repeating the argument which lead to (2), we obtain where y 3 ± ( h) = ± V (x ± 3 ( h), h) Ṽ (x 3, h) = 2( h γx 3 )( x 2 3) (j µx 3 ) 2 and < x 3 ( h) x + 3 ( h) < are roots of Ṽ, that is, Ṽ (x± 3 ( h), h) =. From the definition of x ± 3 = σ ± ( h) as the abscissae of the intersection of the line h = σ σ with the graph of the function σ 3 = (j µσ ) 2, σ σ 2 <, (29) it follows from the strict convexity of (29) that the functions h σ + ( h) = x + 3 ( h) and h σ ( h) = x 3 ( h) are strictly monotonic decreasing and increasing, respectively, with x 3 (h min ) = x + 3 (h min ). Using the implicit function theorem we find that both of the functions h σ ± ( h) are smooth on (h min, h]. For h (h min, h], y 3 + ( h) > and y3 ( h) < ; while for h = h min, y3 (h min ) = = y 3 + (h min ). Thus for x A and h I h, we have (π T S 2 T 2 h,j ) (x) = {y 3 ± ( h)}, which implies that (π T S 2 D) (x) = [ min h I h y 3 ( h), max h I h y + 3 ( h) ] is a closed -disc containing. Hence D is a pinched closed -disc bundle over A. Viewing A as the zero section of the bundle D, we see that C is homotopic in D to the curve π T S 2(C) in A. Next we show that the curves γ 2 = π T S 2(C 2 ) and γ = π T S 2(C ) are homotopic in A leaving their end points fixed. Since γ and γ 2 both have the same beginning and end points, we can 9

20 lift them to curves γ and γ 2 in Ã, the universal covering space of A. In à the curves γ and γ have the same starting points. However, their end points can have θ coordinates differing by an integer multiple of 2π. From (2) it follows that dθ, because x 3 (, ) and j ±µ. Therefore γ 2 may be parameterized by θ instead of t. In fact, γ 2 is a function of θ where θ ranges in [, θ h,j ]. By the argument leading to (23) we see that γ is also a function of θ on [, θ h,j ]. Thus the end points of γ and γ 2 have the same θ coordinate and hence are the same point on Ã. Therefore γ and γ 2 are homotopic in A. Consequently, π T S 2(C) is null homotopic in A and thus in D, which is what we wanted to show. Let D be the domain in D bounded by C. Applying Stokes theorem, we find that C θ T S 2 = = = D (J T S 2 ) (j) D (J T S 2 ) (j) D (J T S 2 ) (j) dθ T S 2 ( Ω T S 2 µ π T S 2F ), using () πj Ω j + µ π D (J T S 2 ) T S 2F (j) using πj Ω j = Ω (J T S 2 ) (j) where Ω j is the symplectic form on M j and ( ) follows from the reduction theorem. = Ω j + µ F S 2. (3) π j (D) M j π T S 2 (D) Recall that for a regular value j ±µ, the reduced space M j R 3 is defined by ψ(σ) = σ 3 ( σ 2 ) (j µσ ) 2 = where σ 3 and σ <. A short calculation shows that the symplectic form Ω j on M j is given by ( ) Ω j (σ)(v, w) = ψ(σ), v w / ψ(σ) 2 (3) where v, w T σ M j = ker dψ(σ). Because µf is the magnetic field strength on S 2 of a monopole located at its center and π T S 2(D) is the region of S 2 bounded by π T S 2(C ) and π T S 2(C 2 ), we see that F h,j = F S 2. (32) π T S 2 (D) 2

21 is the magnetic flux through π T S 2(D) of the magnetic field of the monopole. Substituting (32) into (3) and then combining this result with (27) and (28), we see that (25) becomes where σ + (j µσ + h γσ )θ h,j = 4 σ V (σ ) dσ + A h,j µf h,j, (33) A h,j = Ω j. π j (D) M j We now compute the term A h,j. For j ±µ the reduced space M j is also defined by ϑ(σ) = σ 3 ( (j µσ ) 2 + σ 2 2 σ 2 ) = σ3 G(σ, σ 2 ), (34) since σ <. Suppose that σ M j, then ψ(σ) and ϑ(σ) are linearly dependent, because they are both normal to the surface M j R 3 at σ. From and we find that ψ(σ) = ( 2σ σ 3 + µ(j µσ ), 2σ 2, σ 2 ) Substituting (35) into (3) gives ϑ(σ) = ( G σ, G σ 2, ), ψ(σ) = ( σ 2 ) ϑ(σ). (35) Ω j (σ)(v, w) = ϑ(σ), v w /( σ 2 ) ϑ(σ) 2 (36) for v, w T σ M j = ker dϑ(σ). Pulling Ω j back by the mapping g : (, ) R R 3 : σ = (σ, σ 2 ) (σ, σ 2, G(σ, σ 2 )), which is a parametrization of M j, gives g Ω j ( σ)(, ) = σ σ 2 2

22 = Ω(g( σ)) ( = = G σ g( σ), G σ 2 g( σ) ) det ( ϑ(g( σ)), ( σ) ϑ(g( σ)) 2 2 σ 2 = dσ σ 2 dσ 2 (, ). σ σ 2 ϑ σ g( σ), ϑ σ 2 g( σ) ) Let D h,j be the projection of π j (D) along σ 3 -axis onto the σ -σ 2 plane. Then D h,j is the curve Therefore A h,j = = π j (D) D h,j σ + = 2 σ σ 2 2 = 2(h γσ )( σ 2 ) (j µσ ) 2 = V (σ ), for σ σ σ +. (37) Ω j = g Ω j, since g(d h,j ) = π j (D) D h,j σ 2 dσ σ 2 dσ 2 = dσ D h,j σ 2 by Stokes theorem V (σ ) σ 2 Substituting (38) into (33) gives σ + (j µσ + ) θ h,j = 2 σ dσ. (38) (j µσ ) 2 ( σ) V 2 (σ ) dσ µf h,j (39) which is the desired second formula for the rotation number θ h,j of the magnetic spherical pendulum. Substituting (25) into the right hand side of (39) gives the following quite surprising formula for the magnetic flux through π T S 2(D) σ + F h,j = 2 σ (σ + σ )(j µσ ) dσ. (4) ( σ) 2 V (σ ) 22

23 References [] Arms, J., Cushman, R., and Gotay, M., A Universal Reduction Procedure for Hamiltonian Group Actions, in: T. Ratiu (Ed.), Geometry of Hamiltonian Systems, Springer Verlag, New York, 99, pp [2] Arnol d, V., Mathematical Methods of Classical Mechanics, Springer Verlag, New York, 978. [3] Dirac, P.A.M., Lectures on Quantum Mechanics, Academic Press, New York, 964. [4] Cushman, R. and Knörrer, H., The momentum mapping of the Lagrange top, in: H. Doebner and J. Henning (Eds.), Differential Geometric Methods in Mathematical Physics, Lecture Notes in Mathematics, vol. 39, Springer Verlag, New York, 985, pp [5] Cushman, R. and van der Meer, J.-C., The Hamiltonian Hopf bifurcation in the Lagrange Top, in: Albert, C. (Ed.) Géométrie Symplectique et Mécanique, Lecture Notes in Mathematics, vol. 46, Springer Verlag, New York, 99, pp [6] Montgomery, R., How much does the rigid body rotate?, American Journal of Physics, 59 (99) R. Cushman L. Bates Mathematics Institute Department of Mathematics Rijksuniversiteit Utrecht University of Calgary Budapestlaan 6 25 University Drive, N.W. 358TA Utrecht Calgary, Alberta T2N N4 The Netherlands Canada 23