Homework 10 # 9.2, 9.6, 9.16, 9.31

Homework 10: # 9.2, 9.6, 9.16, 9.31 9.2 Show that the transformation for a system of one degree of freedom, Q = q cos α − p sin α P = q sin α + p cos α satisfies the symplectic condition for any value of the parameter α. Find a generating function for the transformation. What is the physical significance of the transformation for α = 0? For α = π/2? Does your generating function work for both of these cases? Answer: The symplectic condition is met if M J ˜M = J We can find M from ζ˙ = M η̇ = We know J to be P J= Solving M J M̃ we get 0 1 −1 0  q̇ ṗ  cos α − sin α sin α cos α  Q ˙˙  which is M) = M − sin α − cos α cos α − sin α − sin α − cos α 1 cos α − sin α  M (J ˜ M= = 0 1 −1 0   cos α − sin α sin α cos α  M J˜ Therefore M J ˜M = J and the symplectic condition is met for this transformation. To find the generating function, I will first attempt an F1 type and proceed to solve, and check at the end if there are problems with it. Rearranging to solve for p(Q,q) we have Q q cos α + sin α sin α p=− The related equation for F1 is p= ∂F1 ∂q Integrating for F1 yields q 2 cos α Qq + g(Q) + 2 sin α sin α Solve the other one, that is P (Q,q), it along with its relevant equation is F1 = − P = q sin α − q cos2 α Q cos α + sin α sin α P =− ∂F1 ∂Q Integrating −F1 = qQsin α − Q2 2 −F1 = − F1 = Q cot α + qQ( 2 2 cot α + 1 sin α − sin α) + h(q ) qQ + h(q ) sin α Q2 qQ cot α − + h(q ) 2 sin α Using both F1 ’s we find 1 Qq + (q 2 + Q2 ) cot α sin α 2 This has a problem. It blows up, sky high, when α = nπ. But otherwise its F1 = − ok, lets put the condition, α 6= nπ. If we solve for F2 we may be able to find out what the generating function is, and have it work for the holes, α = nπ . F2 (q,P,t)’s relevant equations are p= ∂F2 ∂q 2 q sin α P − cos α cos α q2 Pq − F2 = tan α + f (P ) cos α 2 p= and Q= ∂F2 ∂P Q = q cos α − (P − q sin α) tan α sin2 α P2 tan α + qP + g(q ) 2 cos α F2 = qPcos α − P2 1 − cos α) − tan α + g(q ) cos α 2 2 P qP tan α + g(q ) − F2 = 2 cos α F2 = qP(cos α + So therefore F2 = − 1 2 (q 2 + P 2 ) tan α + qP cos α This works for α = nπbut blows sky high for α = (n+ 21 )π. So I’ll put a condition on F2 that α 6= (n + 21 )π. The physical significance of this transformation for α = 0 is easy to see cause we get Q = q cos 0 − p sin 0 = q P = q sin 0 − p cos 0 = p This is just the identity transformation, or no rotation. For α = π/2 we get π π − p sin = −p 2 2 π π P = q sin − p cos = q 2 2 Where the p’s and q’s have been exchanged. Q = q cos 3 9.6 The transformation equations between two sets of coordinates are Q = log(1 + q 1/2 cos p) P = 2(1 + q 1/2 cos p)q 1/2 sin p • Show directly from these transformation equations that Q, P are canonical variables if q and p are. • Show that the function that generates this transformation is F3 = −(eQ − 1)2 tan p Answer: Q and P are considered canonical variables if these transformation equations satisfy the symplectic condition. M J ˜M = J Finding M : ζ˙ = M η̇ Q ˙˙  Mij= ∂ζi ∂ηj P ∂Q = ∂q =M q̇ ṗ  ∂q ∂Q∂P ∂q ∂Q ∂p ∂P ! M= ∂p q −1/2 cos p 2(1 + q 1/2 cos p) ∂Q −q 1/2 sin p = 1 + q 1/2 cos p ∂p ∂P = q −1/2 sin p + 2 cos p sin p ∂q ∂P = 2q 1/2 cos p + 2q cos2 p − 2q sin2 p ∂p Remembering cos2 A − sin2 A = cos 2A and 2 sin A cos A = sin 2A cos(A − B) = cos A cos B + sin A sin B 4 we can proceed with ease. JM= Now ˜ 0 1 −1 0 JM= −q 1/2 sinp 1+q 1/2 cosp 2(1+q 1/2 cos p) q −1/2 cosp  q −1/2 sin p + sin 2p q −1/2 cosp − 2(1+q q −1/2 sin p +1/2sin cos p2p ) 2q 1/2 ! cos p + 2q cos 2p q 1/2 sinp 2q 1/2 cos p + 2q cos 2p ! 1+q 1/2 cosp q −1/2 sin p + sin 2p 2(1+q 1/2 cosp) 1/2 cosp q −1/2 −q sinp M JM= ! q −1/2 cosp − 2(1+q q −1/2 sin p + 1/2 sincos2p p) q 1/2 sinp 2q 1/2 cos p + 2q cos 2p 1/2 1/2 2q cos p + 2q cos 2p cosp the diagonal terms disappear, and we are left with some1+q cosp You may 1+q see1/2 that algebra for the off-diagonal terms. ˜ Lets solve for ugly. M JM= 0 messy ugly 0  q −1/2 cos p −q 1/2 sin p (q −1/2 sin p+sin 2p)− (2q 1/2 cos p+2q cos 2p) ugly= 1 + 1/2 2(1 + q 1/2 cos p) q cos p ugly= − sin2 p − q 1/2 sin p sin 2p − cos2 p − q 1/2 cos p cos 2p 1 + q 1/2 cos p ugly= −(1 + q 1/2 (cos 2p cos p + sin 2p sin p)) 1 + q 1/2 cos p −(1 + q 1/2 cos p) = −1 1 + q 1/2 cos p Not so ugly anymore, eh? Suddenly ugly became pretty. The same works for messyexcept it becomes positive 1 because it has no negative terms out front. So finally we get ugly= ˜  0 1 M JM= =J −1 0 proves Q and P are canonical variwhich is the symplectic condition, which ables. To show that F3 = −(eQ − 1)2 tan p 5 ! generates this transformation we may take the relevant equations for F3 , solve them, and then solve for our transformation equations. q=− P =− ∂F3 = −[−(eQ − 1)2 sec2 p] ∂p ∂F3 = −[−2(eQ − 1) tan p]eQ ∂Q Solving for Q q = (eQ − 1)2 sec2 p √ q = eQ sec2 p 1/2 Q = ln(1 + q cos p) 1 +p This is one of our transformation equations, now lets plug this into the expression for P and put P in terms of q and p to get the other one. P = 2(1 + q 1/2 cos p − 1) tan p(1 + q 1/2 cos p) P = 2q 1/2 sin p(1 + q 1/2 cos p) Thus F3 is the generating function of our transformation equations. 9.16 For a symmetric rigid body, obtain formulas for evaluating the Poisson brackets ˙ (θ,φ,ψ)] [φ,f [ ˙ψ,f (θ,φ,ψ )] where θ, φ, and ψ are the Euler angles, and f is any arbitrary function of the Euler angles. Answer: Poisson brackets are defined by [u,v ]q,p= ∂u ∂v ∂u ∂v − ∂qi ∂pi ∂pi ∂qi From Goldstein’s section on Euler angles, we learned So calculating pφ − pψ cos θ ˙ I1 b − I1 a cos θ φ= = I1 sin2 θ I1 sin2 θ pφ − pψ cos θ ˙ [φ,f ] = [ I sin2 θ ,f ] 1 6 Note that f = f (θ,φ,ψ) and not of momenta. So our definition becomes ∂ ˙φ ∂f ∂pi ∂qi Taking only two derivatives because φ˙ doesn’t depend on pθ . We get [φ̇,f ] = − [φ̇,f ] = (− ˙ ] = [φ,f cos θ ∂f ∂f 1 ) + ( ) I1 sin2 θ ∂φ I1 sin2 θ ∂ψ ∂f 1 ∂f (− + cos θ) I1 sin2 θ ∂ψ ∂ψ For the next relation, ˙ [ψ,f] = − ∂ ψ̇ ∂f ∂pi ∂qi and ψ̇ = pψ I3 − pφ − pψ cos θ cos θ I1 sin2 θ This yields 2 ∂f ˙ = −( 1 + cos θ ) ∂f + −(− cos θ [ψ,f] ) I3 I1 sin2 θ ∂ψ I1 sin2 θ ∂ψ ˙ = −( [ψ,f] ˙ = [ψ,f] I3 cos2 θ ∂f I1 sin2 θ I3 cos θ ∂f + + ) I3 I1 sin2 θ ∂φ I3 I1 sin2 θ I3 I1 sin2 θ ∂ψ ∂f 1 ∂f (I3 cos θ − (I3 cos2 θ + I1 sin2 θ) ) I3 I1 sin2 θ ∂ψ ∂φ Both together, in final form ˙ ] = [φ,f ˙ = [ψ,f] ∂f ∂f 1 + cos θ) (− ∂ψ ∂ψ I1 sin2 θ ∂f 1 ∂f (I3 cos θ − (I3 cos2 θ + I1 sin2 θ) ) I3 I1 sin2 θ ∂ψ ∂φ 9.31 Show by the use of Poisson brackets that for one-dimensional harmonic oscillator there is a constant of the motion u defined as r k m u(q,p,t) = ln(p +ofimωq) iωt,ω = of motion? What is the physical significance this − constant 7 Answer: We have ∂u du = [u,H] + ∂t dt which we must prove equals zero if u is to be a constant of the motion. The Hamiltonian is H(q,p) = kq2 p2 + 2m 2 So we have du ∂u ∂H ∂u ∂H ∂u = − + dt ∂q ∂p ∂p ∂q ∂t du = dt imω p 1 ( )− (kq ) − iω p + imωq m p + imωq du iωp− kq iωp− mω2 q − iω= = − iω p + imωq p + imωq dt p + iωmq du = iω − iω= iω− iω dt p + imωq du = 0 dt Its physical significance relates to phase. Show Jacobi’s Identity holds. Show [f,gh] = g [f,h] + [f,g ]h where the brackets are Poisson. Answer: Goldstein verifies Jacobi’s identity [u,[v,w]] + [v,[w,u]] + [w,[u,v]] = 0 using an efficient notation. I will follow his lead. If we say ui ≡ ∂u ∂ηi vij≡ ∂v ∂ηi ∂ηj Then a simple way of expressing the Poisson bracket becomes apparent [u,v] = ui Jijvj 8 This notation becomes valuable when expressing the the double Poisson bracket. Here we have [u,[v,w]] = ui Jij[v,w ]j = ui Jij(vk Jkl wl )j Taking the partial with respect to ηj we use the product rule, remembering Jkl are just constants, [u,[v,w]] = ui Jij(vkjJkl wl + vk Jkl wlj) doing this for the other two double Poisson brackets, we get 4 more terms, for a total of 6. Looking at one double partial term, w we see there are only two terms that show up JijJkl ui vk wlj and Jji Jkl ui vk wjl The first from [u,[v,w]] and the second from [v,[w,u]]. Add them up, realizing order of partial is immaterial, and J is antisymmetric: (Jij+ Jji )Jkl ui vk wlj= 0 All the other terms are made of second partials of u or v and disappear in the same manner. Therefore [u,[v,w]] + [v,[w,u]] + [w,[u,v]] = 0 Its ok to do the second property the long way: [f,gh] = [f,gh] = ∂f ∂(gh) ∂f ∂(gh) − ∂qi ∂pi ∂pi ∂qi ∂f ∂g ∂h ∂f ∂h ∂g ( h+g )− (g + h) ∂qi ∂pi ∂pi ∂pi ∂qi ∂qi Grouping terms [f,gh] = ∂f ∂h ∂f ∂g ∂f ∂g ∂f ∂h −g h− h+g ∂qi ∂pi ∂pi ∂qi ∂qi ∂pi ∂pi ∂qi [f,gh] = [f,g ]h + g [f,h] 9