1 M3-4-5A16 Assessed Problems # 1: Do 4 out of 5 problems

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1 D. D. Holm M3-4-5A16 Assesse Problems # 1 Due 1 Nov M3-4-5A16 Assesse Problems # 1: Do 4 out of 5 problems Exercise 1.1 (Poisson brackets for the Hopf map) Figure 1: The Hopf map. In coorinates (a 1, a 2 ) C 2, the Hopf map C 2 /S 1 S 3 S 2 is obtaine b transforming to the four quaratic S 1 -invariant quantities Let the C 2 coorinates be expresse as (a 1, a 2 ) Q jk = a j a k, with j, k = 1, 2. a j = q j + ip j in terms of canonicall conjugate variables satisfing the funamental Poisson brackets {q k, p m } = δ km with k, m = 1, 2. (A) Compute the Poisson brackets {a j, a k } for j, k = 1, 2. The C 2 coorinates a j = q j + ip j satisf the Poisson bracket {a j, a k } = 2i δ jk, for j, k = 1, 2. Likewise a j a j = 2i q j p j (B) Is the transformation (q, p) (a, a ) canonical? Explain wh or wh not. Hint: a map (q, p) (Q, P ) whose Poisson bracket is {Q, P } = c{q, p} with a constant factor c is still regare as being canonical. The transformation (q, p) (a, a ) is inee canonical. The constant ( 2i) is inessential for Hamiltonian namics, because it can be absorbe into the efinition of time. (C) Compute the Poisson brackets among Q jk, with j, k = 1, 2.

2 D. D. Holm M3-4-5A16 Assesse Problems # 1 Due 1 Nov The quaratic S 1 invariants on C 2 given b Q jk = a j a k satisf the Poisson bracket relations, {Q jk, Q lm } = 2i (δ kl Q jm δ jm Q kl ), j, k, l, m = 1, 2. Thus, the o close among themselves, but the o not satisf canonical Poisson bracket relations. (D) Make the linear change of variables, X 0 = Q 11 + Q 22, X 1 + ix 2 = 2Q 12, X 3 = Q 11 Q 22, an compute the Poisson brackets among (X 0, X 1, X 2, X 3 ). The quaratic S 1 invariants (X 0, X 1, X 2, X 3 ) given b X 0 = Q 11 + Q 22, X 1 + ix 2 = 2Q 12, X 3 = Q 11 Q 22, ma be expresse in terms of the a j, j = 1, 2 as X 0 = a a 2 2, X 1 + ix 2 = 2a 1 a 2, X 3 = a 1 2 a 2 2. These satisf the Poisson bracket relations, {X 0, X k } = 0, {X j, X k } = 4ɛ jkl X l (E) Express the Poisson bracket {F (X), H(X)} in vector form among functions F an H of X = (X 1, X 2, X 3 ). The Poisson bracket {F (X), H(X)} is given in vector form as {F (X), H(X)} = 4X F X H X. It s the same as the Poisson bracket for the rigi bo. (F) Show that the quaratic invariants (X 0, X 1, X 2, X 3 ) themselves satisf a quaratic relation. How is this relevant to the Hopf map? The quaratic invariants (X 0, X 1, X 2, X 3 ) satisf the quaratic relation X 2 0(X) = X X X 2 3 = X 2. This relation is relevant. It completes the Hopf map, because level sets of X 0 are spheres S 2 S 3. Moreover, it is relevant to the Poisson bracket in vector form above, which ma be written using this relation as {F (X), H(X)} = 1 X0 2 2 X F X H X.

3 D. D. Holm M3-4-5A16 Assesse Problems # 1 Due 1 Nov Exercise 1.2 (Motion on a sphere) Figure 2: Motion on a sphere. Motion on a sphere: Part 1, the constraint Consier Hamilton s principle for the following constraine Lagrangian on T R 3, L(q, q) = 1 2 q 2 µ 2 ( 1 q 2 ). Here the quantit µ is calle a Lagrange multiplier an must be etermine as part of the solution. Provie a geometric mechanics escription of the namical sstem governe b this Lagrangian. In particular, compute the following for it. 1. Fibre erivative 2. Euler-Lagrange equations 3. Hamiltonian an canonical equations 4. Discussion of solutions. In particular, show that the solutions reall o escribe motion on a sphere. Do the initial conitions matter? (i) Fibre erivative This constraine Lagrangian is plainl hperregular, because which of course is an isomorphism. (ii) Euler-Lagrange equations p = L q = q q = µq The motion must preserve 1 q 2 = 0 an its time erivative q q = 0 for the initial conition q 0 v 0 = 0. Hence one requires (q q) = q 2 + q q = 0 b which q = µq implies µ = q 2 / q 2.

4 D. D. Holm M3-4-5A16 Assesse Problems # 1 Due 1 Nov Hence, the Euler-Lagrange equation for this constraine Lagrangian is q = q 2 q 2 q The right sie of this equation is the centripetal force, require for keeping the motion on the sphere, given b the constraint 1 q 2 = 0. (iii) Hamiltonian an canonical equations The corresponing Hamiltonian is obtaine b the Legenre transformation as, H(q, p) = 1 2 p 2 + µ 2 ( 1 q 2 ), (1) in which the variable p is the momentum canonicall conjugate to the position q. The canonical equations are q = {q, H} = H p = p, ṗ = {p, H} = H q = µq, an we must re-etermine µ = q 2 / q 2 as before. (iv) The energ corresponing to this Hamiltonian, when evaluate on the constraint is just the kinetic energ E = 1 2 q 2 + µ 2 ( 1 q 2 ) = 1 2 q 2 an it is conserve, as ma be checke from the equation of motion, which implies (log( q 2 q 2 ) = 0 an the result follows since q 2 = 1 on the constraint. The motion escribes a great circle on S 2, since that is, (q q) = q q = 0 q q(t) = q 0 v 0, which sas that q(t) is alwas in the plane perpenicular to q 0 v 0 an moving on the sphere if initiall q 0 v 0 = 0. Thus, since q(t) stas on the sphere in a plane passing through the origin, it is a great circle, which is a geoesic on S 2.

5 D. D. Holm M3-4-5A16 Assesse Problems # 1 Due 1 Nov Motion on a sphere: Part 2, the penalt Provie the same kin of geometric mechanics escription of the namical sstem governe b the Lagrangian L ɛ (q, q) = 1 2 q 2 1 4ɛ (1 q 2 ) 2 for a particle with coorinates q R 3 an constants ɛ > 0. For this, let γ ɛ (t) be the curve in R 3 obtaine b solving the Euler-Lagrange equations for L ɛ with the initial conitions q 0 = γ ɛ (0), v 0 = γ ɛ (0). Show that lim ɛ 0 γ ɛ(t) = γ 0 (t) traverses a great circle on the two-sphere S 2, provie that q 0 has unit length an that q 0 v 0 = 0. Hint: go through the same first three steps as in the constraine case. Then use the conserve energ to show that 1 γ ɛ (t) 2 0 as ɛ 0. (i) Fibre erivative The fibre erivative gives a linear relation p = L q = q This means the Lagrangian is hperregular. (ii) Euler-Lagrange equations q = 1 ɛ (1 q 2 )q. (2) Obviousl this equation can onl make sense if the ratio (1 q 2 )/ɛ converges as ɛ 0. The conserve energ will help us eal with this issue in a moment. (iii) Hamiltonian an canonical equations The corresponing Hamiltonian is obtaine b the Legenre transformation as, H(q, p) = 1 2 p ɛ (1 q 2 ) 2, (3) in which the variable p is the momentum canonicall conjugate to the position q. The canonical equations are q = {q, H} = H p = p, an the Hamiltonian H is conserve. ṗ = {p, H} = H q = 1 ɛ (1 q 2 )q, (iv) Discussion of motion an convergence in the limit as ɛ 0 The energ E ɛ for the Lagrangian L ɛ is conserve. In terms of the solution curve γ ɛ (t) R 3, this energ is expresse as 1 2 γ ɛ(t) ( 1 γɛ (t) 2) 2 1 = 4ɛ 2 v 0 2. In particular, this sas that γ ɛ (t) v 0 for all t. Also, from the above equation we conclue that ɛ γ ɛ (t) ( 1 γɛ (t) 2) 2 = ɛ v0 2. 2

6 D. D. Holm M3-4-5A16 Assesse Problems # 1 Due 1 Nov Therefore, γ ɛ (t) 2 v 0 2 is boune b its initial value, so taking the limit as ɛ 0 gives ɛ γ ɛ (t) 2 0. Consequentl, the expression for energ gives 1 γ ɛ (t) 2 0 as ɛ 0, that is, lim ɛ 0 γ ɛ (t) = 1. Thus, lim ɛ 0 γ ɛ (t) = γ 0 (t) exists at an fixe time t an lies on the unit sphere S 2. It remains to check that Lagrange s equations continue to make sense in the limit as ɛ 0. Inserting the conserve energ into Lagrange s equations for this problem iels γ ɛ = 1 ɛ (1 γ ɛ 2 )γ ɛ = ( v0 2 γ ɛ 2 ) γ ɛ =: F (γ ɛ, γ ɛ ). Since the equation γ ɛ = F (γ ɛ, γ ɛ ) is smooth in ɛ as ɛ 0, continuous epenence of solutions on parameters (a general fact from theor of orinar ifferential equations) shows that γ ɛ oes inee converge as ɛ 0. Observe that (γ ɛ γ ɛ ) = γ ɛ γ ɛ = 0 that is, Thus, γ ɛ (t) γ ɛ (t) = γ ɛ (0) γ ɛ (0) = q 0 v 0. γ 0 (t) γ 0 (t) = q 0 v 0, which sas that γ 0 (t) is alwas in the plane perpenicular to q 0 v 0. Thus, since γ 0 (t) stas on the sphere in a plane passing through the origin, it is a great circle, which is a geoesic on S 2.

7 D. D. Holm M3-4-5A16 Assesse Problems # 1 Due 1 Nov Exercise 1.3 (The free particle in H 2 : #1) Figure 3: Geoesics on the Lobachevsk half-plane. In Appenix I of Arnol s book, Mathematical Methos of Classical Mechanics, page 303, we rea. EXAMPLE. We consier the upper half-plane > 0 of the plane of complex numbers z = x + i with the metric s 2 = x It is eas to compute that the geoesics of this two-imensional riemannian manifol are circles an straight lines perpenicular to the x-axis. Linear fractional transformations with real coefficients z az + b (4) cz + are isometric transformations of our manifol (H 2 ), which is calle the Lobachevsk plane. 1 Consier a free particle of mass m moving on H 2. Its Lagrangian is the kinetic energ corresponing to the Lobachevsk metric Namel, L = m (ẋ2 + ẏ 2 ) 2 2. (5) (A) (1) Write the fibre erivatives of the Lagrangian (5) an (2) compute its Euler-Lagrange equations. These equations represent geoesic motion on H 2. (3) Evaluate the Christoffel smbols. Fibre erivatives: L ẋ = mẋ 2 =: p x an L ẏ = mẏ 2 Euler-Lagrange equations L ẋ = L x an L ẏ = L ) ( ẋ 2 = 0 an ( ẏ 2 =: p iel, respectivel: ) = ẋ2 + ẏ 2 3 (6) Expaning these equations iel the Christoffel smbols for the geoesic motion, ẍ 2 ẋẏ = 0 an ÿ + 1 ẋ2 1 ẏ2 = 0 Γ 1 12 = 2, Γ2 11 = 1, Γ2 22 = 1. 1 These isometric transformations of H 2 have eep significance in phsics. The correspon to the most general Lorentz transformation of space-time.

8 D. D. Holm M3-4-5A16 Assesse Problems # 1 Due 1 Nov (B) Hint: The Lagrangian in (5) is invariant uner the group of linear fractional transformations with real coefficients. These have an SL(2, R) matrix representation [ ] a b z = az + b (7) c cz + (1) Show that the quantities u = ẋ an v = ẏ are invariant uner a subgroup of these smmetr transformations. (2) Specif this subgroup in terms of the representation (7). (8) The quantities (8) are invariant uner a subgroup of translations an scalings. T τ : (x, ) (x + τ, ) Flow of X T = x, (δx, δ) = (1, 0), [X T, X S ] = X T. S σ : (x, ) (e σ x, e σ ) Flow of X S = x x +, (δx, δ) = (x, ). These transformations are translations T along the x axis an scalings S centere at (x, ) = (0, 0). The are represente b elements of (7) as [ ] [ ] 1 b a 0 T = an S = That is, the transformations T an S are isometries of the metric s 2 = (x )/ 2 on H 2 with T : a = 1 =, c = 0, b 0 an S : a 0, b = 0 = c, = 1. (C) (1) Use the invariant quantities (u, v) in (8) as new variables in Hamilton s principle. Hint: the transforme Lagrangian is l(u, v) = m 2 (u2 + v 2 ). (2) Fin the corresponing conserve Noether quantities. (1) The translations T along the x axis an scalings S centere at (x, ) = (0, 0) leave invariant the quantities u = ẋ an v = ẏ, in terms of which the Lagrangian L in (5) reuces to l(u, v) = m 2 (u2 + v 2 ). The reuce Hamilton s principle in the variables u an v iels, 0 = δs = δ = m b a b = m l(u, v) = a b b a m(uδu + vδv) u (δẋ uδ) + v (δẏ vδ) ( ) ( u v δx + + u2 + v 2 a ) δ + m [ u δx + v δ ] b a

9 D. D. Holm M3-4-5A16 Assesse Problems # 1 Due 1 Nov Thus, Hamilton s principle recovers equations (6) in the variables u an v. (2) Appling Noether s theorem to the enpoint term in these variables iels conservation of C T = u, for (δx, δ) = (1, 0) translations, an C S = ux + v for (δx, δ) = (x, ) scaling. (D) Transform the Euler-Lagrange equations from x an to the variables u an v that are invariant uner the smmetries of the Lagrangian. Then: (1) Show that the resulting sstem conserves the kinetic energ expresse in these variables. (2) Discuss its integral curves an critical points in the uv plane. (3) Show that the u an v equations can be integrate explicitl in terms of sech an tanh. Hint: In the uv variables, the Euler-Lagrange equations for the Lagrangian (5) are expresse as u = 0 an v + u2 + v 2 = 0. Expaning these equations using u = ẋ/ an v = ẏ/ iels u = uv, v = u 2 (9) (1) Equations (9) impl conservation of the kinetic energ l(u, v) = m 2 (u2 + v 2 ) = E (2) The integral curves of the sstem of equations (9) in the uv plane are either critical points along the axis u = 0, or the are heteroclinic connections between these points that are semi-circles aroun the origin on level sets of the energ E. The critical points at u = ẋ/ = 0 are relative equilibria of the sstem corresponing to vertical motion on the x plane. Those corresponing to upwar motion (ẏ > 0) are unstable an the ones corresponing to ownwar motion (ẏ < 0) are stable. (3) The trial solutions u = tanh an v = sech quickl converge to the exact solutions of the uv sstem. (E) (1) Legenre transform the Lagrangian (5) to the Hamiltonian sie, obtain the canonical equations an (2) erive the Poisson brackets for the variables u an v. Hint: {p x, p } = p x. The equations of motion on the Hamiltonian formulation are efine b introucing the momenta: p x = L ẋ = mẋ 2, p = L ẏ = mẏ 2,

10 D. D. Holm M3-4-5A16 Assesse Problems # 1 Due 1 Nov an the Hamiltonian One gets H = 2 ( p 2 2m x + p 2 ). B efining ẋ = 2 p x m ẏ = 2 p m the Hamiltonian can be written as u = p x /m, ṗ x = 0, ṗ = m H = h(u, v) = 1 2 (10) ( p 2 x + p 2 ). v = p /m, ( u 2 + v 2), an the equations of motion (10) become, using {p x, p } = p x, u = uv v = u 2. (11) These equations are Hamiltonian with respect to the Lie-Poisson bracket {u, v} = u, an the reuce Hamiltonian h(u, v) in terms of the invariant variables. Namel, u = {u, h} = uv v = {v, h} = u 2.

11 D. D. Holm M3-4-5A16 Assesse Problems # 1 Due 1 Nov Exercise 1.4 (The free particle in H 2 : #2) Figure 4: Geoesics on the Lobachevsk half-plane. Consier the following pair of ifferential equations for (u, v) R 2, u = uv, v = u 2. (12) These equations have iscrete smmetries uner combine reflection an time reversal, (u, t) ( u, t) an (v, t) ( v, t). (This is calle P T smmetr in the (u, v) plane.) (A) Fin 2 2 real matrices L an B for which the sstem (12) ma be written as a Lax pair, namel, as L = [L, B]. Hint: a basis of 2 2 real matrices is given b [ ] [ ] σ 1 =, σ = 1 0, σ 3 = [ ] Explain what the Lax pair relation means an etermine a constant of the motion from it. Hint: consier the similarit transformation L(t) = O 1 (t)l(0)o(t). We introuce two linear 2 2 matrices, one smmetric (L T = L) an the other skewsmmetric (B T = B), as require for the commutator [L, B] to be smmetric: [ ] [ ] [ ] v u L = = u v = uσ u v vσ 3, B = 1 [ ] 0 u = u [ ] 0 1 = u 2 u σ 2. Both matrices must be linear homogeneous, so that the commutator [L, B] an time erivative L can match powers using (12). The sl(2, R) σ-matrices satisf [σ 1, σ 2 ] = 2σ 3, [σ 2, σ 3 ] = 2σ 1, an [σ 3, σ 1 ] = 2σ 2. Thus, we fin the Lax pair relation, L [ = u2 σ 3 + uv σ 1 = [L, B] = uσ 1 vσ 3, u ] 2 σ 2. What the Lax pair relation means: isospectralit. The Lax pair relation implies that L(t) = O 1 (t)l(0)o(t), where B = O 1 Ȯ an O(t) SO(2).

12 D. D. Holm M3-4-5A16 Assesse Problems # 1 Due 1 Nov Thus, OL(t)O 1 = L(0) is conserve. That is, the flow generates a similarit transformation that conserves the initial spectrum of the 2 2 smmetric matrix L(0). Such a flow is sai to be isospectral. The traceless matrix L(t) has one inepenent eigenvalue an the sstem (12) has onl one conserve quantit. The conserve quantit is the eterminant, et L(t) = et L(0). (B) Write the sstem (12) as a ouble matrix commutator, L = [L, [L, N] ]. In particular, fin N explicitl an explains what this means for the solutions. Hint: compute tr (LN). Substituting N := example we ma set [ ] a 0 0 b N := into L [ ] = [L, [L, N] ] iels b a =, so for What this means for the solutions: Graient flow. The evolution b the ouble bracket relation L = [L, [L, N] ] is a graient flow that preserves the spectrum of L but ecreases the quantit tr (LN) accoring to tr (LN) = tr ([L, N]T [L, N]), until L becomes iagonal an hence [L, N] 0, because N is iagonal. Thus, the namics (12) becomes asmptoticall stea as L tens to a iagonal matrix. This means the sstem (12) must asmptoticall approach a stable equilibria that is consistent with its initial conitions an conservation laws. For the present case, substituting the explicit forms of L an N iels tr (LN) = 1 2 v = 1 2 u2 = tr ([L, N] T [L, N]), which hols b (12) an thus checks the previous calculation. In the present case, it will turn out that lim t u(t) = 0, which will verif [L, N] 0, as the off-iagonal parts of L will vanish asmptoticall. (C) Fin explicit solutions an iscuss their motion an asmptotic behaviour: (1) in time; an (2) in the (u, v) phase plane. Hint: keep the tanh function in min. Keeping the tanh function in min an recalling that tanh(ct) = c sech 2 (ct) we fin, for u(0) = c an v(0) = 0, sech(ct) = c sech(ct) tanh(ct), v(t) = c tanh(ct) an u(t) = c sech(ct), an of course we check, 2h = u 2 + v 2 = c 2 (tanh 2 + sech 2 ) = c 2. Motion an asmptotic behaviour. (a) In time: We have lim t (u(t), v(t)) = (0, c). Consequentl, the quantit u(t) falls exponentiall with time, from u(0) towar the line of fixe points at u = 0, while u(t) goes to a constant equal to u(0).

13 D. D. Holm M3-4-5A16 Assesse Problems # 1 Due 1 Nov (b) In the (u, v) phase plane: Since h is conserve, the motion is along a famil of semi-circles, each parameterise b its raius c = 2h, as u 2 + v 2 = c 2 for u > 0 an u < 0, ling in the upper an lower (u, v) half planes. These semi-circular motions are mirror images, reflecte across the line of fixe points at u = 0. The equations of motion are P T -smmetric, so the fixe points along u = 0 in the (u, v) plane are stable for v < 0, an unstable for v > 0. Thus, the two families of semi-circular motion both connect the line of fixe points at u = 0 to itself. One famil of semi-circles lies in the upper half (u, v) plane, an the other lies smmetricall place to complete the circles in the lower half (u, v) plane. The flows along each reflection-smmetric pair of semi-circles pass in the same (negative) v irection, from v = c to v = c. (D) Explain wh the solution behaviour foun in the previous part is consistent with the behaviour preicte b the ouble bracket relation. This analsis is consistent with the conclusion from the ouble-bracket relation L = [L, [L, N] ] that the namics of L-matrix [ ] v u L = u v asmptoticall becomes stea. In fact, since lim t u = 0 an lim t v(t) = c, the L-matrix asmptoticall iagonalises!

14 D. D. Holm M3-4-5A16 Assesse Problems # 1 Due 1 Nov Exercise 1.5 (Nambu Poisson brackets on R 3 ) Figure 5: Motion along intersections of surfaces in R 3. (A) Show that for smooth functions c, f, h : R 3 R, the R 3 -bracket efine b {f, h} = c f h satisfies the efining properties of a Poisson bracket. Is it also a erivation satisfing the Leibnitz relation for a prouct of functions on R 3? If so, wh? The R 3 -bracket is plainl a skew-smmetric bilinear Leibniz operator. Its Hamiltonian vector fiels are ivergence free vector fiels in R 3. These vector fiels in R 3 satisf the Jacobi ientit uner commutation. The ientification of the R 3 -bracket with its Hamiltonian vector fiels shows that it satisfies Jacobi. This will be mae clearer below. (B) How is the R 3 -bracket relate to the canonical Poisson bracket in the plane? choice of the R 3 -bracket The canonical Poisson bracket in the (x, )-plane is given b the particular {f, h} = z f h (C) The Casimirs (or istinguishe functions, as Lie calle them) of a Poisson bracket satisf {c, h}(x) = 0, for all h(x) Part (E) provies aitional hints to proving that the R 3 -bracket satisfies the efining properties of a Poisson bracket. What are the Casimirs for the R 3 bracket? Smooth functions of c are Casimirs for the R 3 -bracket given b {f, h} = c f h.

15 D. D. Holm M3-4-5A16 Assesse Problems # 1 Due 1 Nov (D) Write the motion equation for the R 3 -bracket ẋ = {x, h} in vector form using graients an cross proucts. vector fiel X h = {, h} has zero ivergence. Show that the corresponing Hamiltonian ẋ = {x, h} = c h The corresponing Hamiltonian vector fiel X h = {, h} has zero ivergence because the vector c h has zero ivergence, since it s a curl. (E) Show that uner the R 3 -bracket, the Hamiltonian vector fiels X f = {, f}, X h = {, h} satisf the following anti-homomorphism that relates the commutation of vector fiels to the R 3 -bracket operation between smooth functions on R 3, [X f, X h ] = X {f,h}. Hint: commutation of ivergenceless vector fiels oes satisf the Jacobi ientit an for the R 3 -bracket these vector fiels are relate to the Poisson bracket b [X G, X H ] = X G X H X H X G = {G, }{H, } {H, }{G, } = {G, {H, }} {H, {G, }}. Lemma. The R 3 -bracket efine on smooth functions (C, F, H) b {F, H} = C F H ma be ientifie with the ivergenceless vector fiels b [X G, X H ] = X {G,H}, (13) where [X G, X H ] is the Jacobi-Lie bracket of vector fiels X G an X H. Proof. Equation (13) ma be verifie b a irect calculation, [X G, X H ] = X G X H X H X G = {G, }{H, } {H, }{G, } = {G, {H, }} {H, {G, }} = {{G, H}, } = X {G,H}. Remark. The last step in the proof of the Lemma uses the Jacobi ientit for the R 3 -bracket, which follows from the Jacobi ientit for ivergenceless vector fiels, since X F X G X H = {F, {G, {H, }}} (F) Show that the motion equation for the R 3 -bracket is invariant uner a certain linear combination of the functions c an h. Interpret this invariance geometricall. (αc + βh) (γc + ɛh) = c h for constants satisfing αɛ βγ = 1. Uner such a (volume-preserving) transformation, the level sets change, but their intersections remain invariant.