Canonical Transformations, Hamilton-Jacobi Equations, and Action-Angle Variables

Chapter 4 Canonical Transformations, Hamilton-Jacobi Equations, and Action-Angle Variables We’ve made good use of the Lagrangian formalism. Here we’ll study dynamics with the Hamiltonian formalism. Problems can be greatly simplified by a good choice of generalized coordinates. How far can we push this? Example: Let us imagine that we find coordinates qi that are all cyclic. Then ṗi = 0, so pi = αi are all constant. If H is conserved, then: H = H(α1 , . . . , αn ) (4.1) is also constant in time. In such a case the remaining equations of motion: q̇i = ∂H = ωi (α) ∂ αi ⇒ qi = ωi t + δi (4.2) All coordinates are linear in time and the motion becomes very simple. We might imagine searching for a variable transformation to make as many coordinates as possible cyclic. Before proceeding along this path, we must see what transformations are allowed. 4.1 Generating Functions for Canonical Transformations Recall the the Euler-Lagrange equations are invariant when: • a point transformation occurs Q = Q(q, t) with L[q, t] = L0 [Q, t]; • a total derivative is summed to the Lagrangian L0 = L + dF [q, t] . dt For H we consider point transformations in phase space: Qi = Qi (q, p, t) and Pi = Pi (q, p, t), (4.3) where the Hamilton’s equations for the evolution of the canonical variables (q, p) are satisfied: q̇i = ∂H ∂pi and ṗi = − ∂H . ∂qi (4.4) Generally, not all transformations preserve the equations of motion. However, the transformation is canonical if there exists a new Hamiltonian: K = K(Q, P, t), (4.5) where ∂K ∂K and P˙i = − . ∂Pi ∂Qi For notational purposes let repeated indices be summed over implicitly. Hamilton’s principle can be written as: Z t2 δ (pi q̇i − H(q, p, t)) dt = 0, Q̇i = (4.6) (4.7) t1 or in the new Hamiltonian as: Z t2 δ   Pi Q˙ i − K(Q, P, t) dt = 0. (4.8) t1 For the Eq.(4.7) to imply Eq.(4.8), then we need: λ(pi q̇i − H) = Pi Q˙ i − K + F˙ . (4.9) Since F˙ is a total derivative and the ends of the path are fixed: δq|tt21 = 0 and δp|tt21 = 0 ⇒ δF |tt21 = 0 (4.10) There are a few things to be said about transformations and λ. • If λ = 1 then the transformation is canonical, which is what we will study. • If λ = 6 1 then the transformation is extended canonical, and the results from λ = 1 can be recovered by rescaling q and p appropriately. • If Qi = Qi (q, p) and Pi = Pi (q, p) without explicit dependence on time, then the transformation is restricted canonical. We will always take transformations Qi = Qi (q, p, t) and Pi = Pi (q, p, t) to be invertible in any of the canonical variables. If F depends on a mix of old and new phase space variables, it is called a generating function of the canonical transformation. There are four important cases of this. 1. Let us take F = F1 (q, Q, t) (4.11) where the old coordinates qi and the new coordinates Qi are independent. Then: ∂F1 ∂F1 ∂F1 + q̇i + pi q̇i − H = Pi Q˙ i − K + F˙ 1 = Pi Q˙ i − K + Q̇i ∂t ∂qi ∂Qi (4.12) from this we see that Pi Q˙ i cancels and equating the terms with a q̇i , a Q˙ i and the remaining terms gives: pi = ∂F1 ∂Qi Pi = − , ∂F1 ∂Qi and K=H+ ∂F1 , ∂t (4.13) which gives us formula for a transformation: pi = pi (q, Q, t) and Pi = Pi (q, Q, t) (4.14) and connects K to an initial H. Example: if Q F1 = − , q then: p= Q ∂F1 = 2 ∂q q (4.15) P =− and ∂F1 1 = . ∂Q q (4.16) Writing the new coordinates as function of the old ones yields Q = pq 2 Example: Given the transformations   p Q = ln and q and P =  P =− 1 q q2 +1 2 (4.17)  p , q (4.18) we can prove they are canonical by finding a corresponding generating function. We know: ∂F1 = p = qeQ , (4.19) ∂q which gives us Z F1 = qeQ dq + g(Q) = q2 Q e + g(Q), 2 (4.20) and ∂F1 q2 dg P =− =− = − eQ − 2 dQ ∂Q ⇒ g(Q) = eQ .  q2 +1 2  p =− q   q2 + 1 eQ 2 (4.21) Thus F1 is given by:  F1 =  q2 + 1 eQ . 2 (4.22) 2. Let: F = F2 (q, P, t) − Qi Pi (4.23) where we wish to treat the old coordinates qi and new momenta Pi as independent variables. Then: ∂F2 ∂F2 ∂F2 + q̇i + Ṗi (4.24) q̇i pi − H = Q˙ i Pi − K + F˙ 2 − Q˙ i Pi − Qi P˙ i = −Qi P˙ i − K + ∂t ∂qi ∂Pi This corresponds to pi = ∂F2 ∂qi ; Qi = ∂F2 ∂Pi and K=H+ ∂F2 . ∂t (4.25) 3. We could also take F = F3 (p, Q, t) + qi pi (4.26) with the new coordinates Qi and the old momenta pi as independent variables. 4. Finally we could take F = F4 (p, P, t) + qi pi − Qi Pi with the old momenta pi and new momenta Pi as independent variables. This can be summarized in the table below. (4.27) Function F1 (q, Q, t) F2 (q, P, t) Transformations pi = ∂F1 ∂F1 , Pi = − ∂qi ∂Qi F1 = qi Qi Qi = pi , Pi = −qi ∂F2 ∂F2 , Qi = ∂qi ∂Pi F 2 = q i Pi Qi = qi , Pi = pi ∂F3 ∂F3 , Pi = − ∂pi ∂Qi F3 = pi Qi Qi = −qi , Pi = −pi pi = F3 (p, Q, t) qi = − F4 (p, P, t) Simplest case qi = − ∂F4 ∂F4 , Qi = ∂pi ∂Pi F 4 = p i Pi Qi = pi , Pi = −qi The simplest case of the 2nd (F2 ) transformation is just an identity transformation. For any of these Fi cases we also have: ∂Fi . (4.28) K=H+ ∂t If Fi is independent of time then this implies K=H (4.29) Mixed cases may also occur when more than two old canonical coordinates are present. (In this chapter we will be using Einstein’s repeated index notation for implicit summation, unless otherwise stated.) Example: consider F2 = fi (q, t)Pi (4.30) for some functions fi where i ∈ {1, . . . , n}. Then Qi = ∂F2 = fi (q, t) ∂Pi (4.31) is a coordinate point transformation. It is canonical with ∂fi Pj , ∂qj (4.32) p2 kq 2 + where k = mω 2 2m 2 (4.33) pi = which can be inverted to get Pj = Pj (q, p, t). Example: Consider the harmonic oscillator: H= Let us try the transformation: √ p = α 2mP cos(Q) α √ q= 2mP sin(Q) mω (4.34) for α constant. Then:
K = H = P α2 cos2 (Q) + sin2 (Q) = P α2 , (4.35) so the new momentum E (4.36) α2 is just proportional to the energy, while Q is a cyclic variable. Is this transformation canonical? We can find a generating function F = F1 (q, Q) by dividing the old variables: p = mω cot(Q). (4.37) q This gives us: Z ∂F1 1 p= ⇒ F1 = p(q, Q) dq + g(Q) = mωq 2 cot(Q) + g(Q) (4.38) ∂q 2 ∂F1 mωq 2 dg P =− = − 2 ∂Q 2 sin (Q) dQ Setting: dg 2P = 0 ⇒ q2 = sin2 (Q), (4.39) dQ mω √ which tells us the transformation is canonical if α = ω. This means: E (4.40) P = ω By Hamilton’s equations Eq.(4.4): ∂K Q̇ = = ω ⇒ Q = ωt + δ. (4.41) ∂P Putting this altogether, this gives the familiar results: r 2E q= sin(ωt + δ) (4.42) 2 √ mω p = 2mE cos(ωt + δ). P = Lets record for future use our final canonical transformation here: r √ 2P q= sin(Q) , p = 2mωP cos(Q) . mω So far, a transformation Q = Q(q, p, t) and P = P (q, p, t) is canonical if we can find a generating function F . This involves integration, which could be complicated, so it would be nice to have a test that only involves differentiation. There is one! 4.2 Poisson Brackets and the Symplectic Condition In Classical Mechanics II (8.223) the Poisson bracket of the quantities u and v was defined as  X  ∂ u ∂v ∂v ∂u {u, v}q,p = (4.43) − ∂qi ∂pi ∂qi ∂pi i It is easy to check that the following fundamental Poisson bracket relations are satisfied: {qi , qj }q,p = {pi , pj }q,p = 0 and {qi , pj }q,p = δij . (4.44) There are a few other properties of note. These include: {u, u} = 0, {u, v} = −{v, u}, {au + bv, w} = a{u, w} + b{v, w}, {uv, w} = u{v, w} + {u, w}v, {u, {v, w}} + {v, {w, u}} + {w, {u, v}} = 0, (4.45) (4.46) (4.47) (4.48) (4.49) for a, b constants. Eq.(4.49) is the Jacobi identity. The above looks a lot like the commutators of operators in quantum mechanics, such as: [x, ˆ p̂] = i~ (4.50) Indeed, quantizing a classical theory by replacing Poisson brackets with commutators through: [u, v] = i~{u, v} (4.51) is a popular approach (first studied by Dirac). It is also the root of the name “canonical quantization”. (Note that Eq.(4.48) was written in a manner to match the analogous formula in quantum mechanics where the operator ordering is important, just in case its familiar. Here we can multiply functions in either order.) Now we can state the desired criteria that only involves derivatives. Theorem: A transformation Qj = Qj (q, p, t) and Pj = Pj (q, p, t) is canonical if and only if: {Qi , Qj }q,p = {Pi , Pj }q,p = 0 and {Qi , Pj }q,p = δij . (4.52) To prove it, we’ll need some more notation. Let’s get serious about treating qi and pi on an equal footing together, defining the following two quantities:        ~η =       q1    ..  .    0n×n 1n×n  qn   and J =       p1 −1n×n 0n×n  ..  .  pn (4.53) where 0n×n is the n × n zero matrix, 1n×n is the n × n identity matrix. The following properties of J will be useful: J > = −J , J 2 = −12n×2n and J > J = JJ > = 12n×2n . (4.54) We also note that det(J) = 1. With this notation Hamilton’s equations, Eq.(4.4), can be rewritten as: ∂H ~η˙ = J ∂ ~η or ~η˙ = J ∇η~ H . (4.55) The notation ∇η~ H better emphasizes that this quantity is a vector, but we will stick to using the first notation for this vector, ∂H/∂~η , below. Although the Theorem is true for time dependent transformations, lets carry out the proof for the simpler case of time independent transformations Qi = Qi (q, p) and Pi = Pi (q, p). This implies K = H. Let us define:        ξ~ =       Q1  ..  .   Qn    P1  ..  .   Pn (4.56) which is a function of the original coordinates, so we can write: ~ η) ξ~ = ξ(~ (4.57) ~ Now consider the time derivative of ξ: ∂ξi ξ˙i = η̇j ∂ηj ⇔ ˙ ξ~ = M~η˙ where Mij = ∂ξi . ∂ηj (4.58) Here M corresponds to the Jacobian of the transformation. From the Hamilton’s equations, we know that ∂H ~η˙ = J . ∂~η (4.59) We want to show that : ∂H ˙ ξ~ = J ∂ξ~ for ~ η ) a canonical transformation. ξ~ = ξ(~ (4.60) Let us now consider: ∂ξi ∂ξi ∂H ∂ξi ∂ξl ∂H ξ˙i = η̇j = Jjk = Jjk ∂ηj ∂ηj ∂ηk ∂ηj ∂ηk ∂ξl ⇔ ∂H ˙ ξ~ = M JM > ∂ξ~ (4.61) for any H. Then ~ η ) is a canonical transformation ξ~ = ξ(~ iff M JM > = J (4.62) is satisfied. This is known as the “symplectic condition”. Moreover, since M J = J(M > )−1 and J 2 = −1, (4.63) we can write: J(M J)J = −JM = J(JM >−1 )J = −M >−1 J ⇒ JM = M >−1 J . (4.64) Thus we see that M JM > = J is equivalent to: M > JM = J . Now consider Poisson brackets in this matrix notation:  > ∂v ∂u J {u, v}q,p = {u, v}η~ = ∂~η ∂~η (4.65) (4.66) and the fundamental Poisson brackets are: {ηi , ηj }η~ = Jij (4.67) Then we can calculate the Poisson brackets that appeared in the theorem we are aiming to prove as  > ∂ξi ∂ξj {ξi , ξj }η~ = J = (M > JM )ij (4.68) ∂~η ∂~η This last equation is the same as Eq.(4.65). The new variables satisfy the Poisson bracket relationships Eq.(4.67): {ξi , ξj }η~ = Jij (4.69) if and only if M > JM = J (4.70) ~ η ) is canonical, Eq.(4.65), completing the proof. which itself is true if, and only if, ξ~ = ξ(~ There are two facts that arise from this. • Poisson brackets are canonical invariants {u, v}η~ = {u, v}ξ~ = {u, v}. (4.71) This is true because:  >   > ∂u > ∂v {u, v}η~ = M J M ∂ξ~ ∂ξ~  >  > ∂u ∂v ∂u ∂v = M JM > = J = {u, v }ξ~ ~ ~ ~ ∂ξ ∂ξ ∂ξ ∂ ξ~  ∂u ∂~η > ∂v J = ∂~η (4.72) (4.73) • Phase space volume elements are preserved by canonical transformations, as discussed in 8.223. Phase space volume is given by: Y Y Vξ~ = dQi dPi = |det(M )| dqj dpj = |det(M )|Vη~ . (4.74) i j However, we also have: det(M > JM ) = det(J) = (det(M ))2 det(J) ⇒ |det(M )| = 1. 4.3 (4.75) Equations of Motion & Conservation Theorems Let us consider a function: u = u(q, p, t) Then: u̇ = ∂u ∂u ∂u ∂H ∂u ∂H ∂u ∂u q̇i + ṗi + = − + , ∂qi ∂pi ∂t ∂qi ∂pi ∂pi ∂qi ∂t (4.76) (4.77) which can be written more concisely as u̇ = {u, H} + ∂u ∂t (4.78) for any canonical variables (q, p) and corresponding Hamiltonian H. Performing canonical quantization on this yields the Heisenberg equation of time evolution in quantum mechanics. There are a few easy cases to check. • If u = qi then: q̇i = {qi , H} + ∂qi ∂H = ∂t ∂pi (4.79) • If u = pi then: ṗi = {pi , H} + ∂pi ∂H =− ∂qi ∂t (4.80) Together the above two cases yield Hamilton’s equations of motion. • Also, if u = H then: Ḣ = {H, H} + ∂H ∂H = ∂t ∂t (4.81) Next consider what we can say if a quantity u is conserved. Here: u̇ = 0 = {u, H} + ∂u . ∂t (4.82) As a corollary, if ∂u = 0, ∂t (4.83) {u, H} = 0 ⇒ u is conserved. (4.84) then (In quantum mechanics this the analog of saying that u is conserved if u commutes with H.) Another fact, is that if u and v are conserved then so is {u, v}. This could potentially provide a way to compute a new constant of motion. To prove it, first consider the special case where: ∂v ∂u = =0 (4.85) ∂t ∂t then using the Jacobi identity we have: {H, {u, v}} = −{u, {v, H}} − {v, {H, u}} = −{u, 0} − {v, 0} = 0 For the most general case we proceed in a similar manner:     ∂v ∂u {{u, v}, H} = {u, {v, H}} + {v, {H, u}} = − u, + v, ∂t ∂t ∂ d = − {u, v} ⇒ {u, v } = 0 ∂t dt (4.86) (4.87) Infinitesimal Canonical Transformations Let us now consider the generating function: F2 (q, P, t) = qi Pi + G(q, P, t), (4.88) where F2 = qi Pi is an identity transformation, and ||  1 is infinitesimal. The function G(q, P, t) is known as the generating function of an infinitesimal canonical transformation. Using the properties of an F2 generating function we have: pj = ∂G ∂F2 = Pj +  ∂qj ∂qj ⇒ δ pj = Pj − pj = − ∂G ∂qj (4.89) giving the infinitesimal transformation in the momentum. Likewise: Qj = ∂G ∂F2 = qj +  ∂Pj ∂Pj (4.90) Since Pj = pj + O() and  is infinitesimal we can replace ∂G(q, P, t)/∂Pj = ∂G(q, p, t)/∂pj + O(). Therefore we have: Qj = qj +  ∂G + O(2 ) ∂pj ⇒ δqj = Qj − qj =  ∂G ∂pj (4.91) where now we note that we can consider G = G(q, p, t), a function of q and p, to this order. Returning to the combined notation of ~η > = (q1 , . . . , qn , p1 , . . . , pn ), Eq.(4.89) and Eq.(4.90) can be consisely written as the following Poisson bracket: δ~η = {~η , G} (4.92) Example: if G = pi then δpi = 0 and δqj = δij , which is why momentum is the generator of spatial translations. Example: if G is the z component of the angular momentum: X G = Lz = (xi piy − yi pix ) and  = δθ (4.93) i then the infinitesimal change correponds to a rotation δxi = −yi δθ , δyi = xi δθ , δzi = 0 δpix = −piy δθ , δpiy = pix δθ , δpiz = 0 (4.94) (4.95) which is why angular momentum is the generator of rotations. Important Example: if G = H and  = dt then {~η , G} = {~η , H }dt = ~η˙ dt = d~η On the left hand side we have the change to the phase space coordinates due to our transformation. On the right hand side we have the physical increment to the phase space variables that takes place in time dt. The fact that these are equivalent tells us that the Hamiltonian is the generator of time evolution. The infinitesimal transformation generated by the Hamiltonian is corresponds with the physical motion. Rather than trying to think of this as a passive coordinate change (q, p) → (Q, P ), it is useful if we can take an active view of the infinitesimal canonical transformation generated by H. Let the time t be a parameter for the family of transformations with  = dt: the initial conditions are: ~η0 (t0 ) = ~η0 (4.96) The result is a series of transformations of ~η that move us in a fixed set of phase space coordinates from one point to another: η(t2 ) η(t3 ) η(tn ) η(t1 ) ~η0 (t0 ) → ~η1 (~η0 , t1 ) → . . . → ~ηn (~ηn−1 , tn ) where tn = t is the final time (4.97) η(t0 ) Phase Space η All together, combining an infinite number of infinitesimal transformations allows us to make a finite transformation, resulting in: ~η = ~η (~η0 , t) or ~η0 = ~η0 (~η , t) (4.98) This is a canonical transformation that yields a solution for the motion! How could we directly find this transformation, without resorting to stringing together infinitesimal transformations? We can simply look for a canonical transformation with new coordinates Qi and new momenta Pi that are all constants, implying an equation of the type: ~η0 = ~η0 (~η , t) (4.99) Inverting this then gives the solution for the motion. This logic can be used to extend our proof of the Theorem in Section 4.2 to fully account for time dependent transformations. (see Goldstein). Using K = H +∂G/∂t, Goldstein also describes in some detail how the change to the Hamiltonian ∆H under an active infinitesimal canonical transformation satisfies: ∂G = −G˙ (4.100) ∆H = −{G, H} −  ∂t This says “the constants of motion are generating functions G of the infinitesimal canonical transformation that leave H invariant”; that is, G˙ = 0 if and only if ∆H = 0 under the transformation. Thus a conservation law exists if and only if there is a symmetry present. 4.4 Hamilton-Jacobi Equation Let us take the suggestion from the end of the previous section seriously and look for new canonical variables that are all cyclic, such that: Q˙ i = P˙ i = 0 ⇒ (Q, P ) are all constants . (4.101) If the new Hamiltonian K is independent of (Q, P ) then: Q̇i = ∂K =0 ∂Pi and ∂K P˙i = − = 0. ∂Qi (4.102) We could look for a constant K, but it it is simplest to simply look for K = 0. Using a generating function F = F2 (q, P, t), then we need K = H(q, p, t) + Because pi = ∂F2 = 0. ∂t ∂F2 , then we can rewrite this condition as ∂qi   ∂F2 ∂F2 ∂ F2 H q1 , . . . , q n , ,..., ,t + = 0, ∂q1 ∂qn ∂t (4.103) (4.104) which is the time dependent Hamilton-Jacobi equation (henceforth abbreviated as the H-J equation). This is a 1st order partial differential equation in n + 1 variables (q1 , . . . , qn , t) for F2 . The solution for F2 has n + 1 independent constants of integration, One of these constants is trivial (F2 → F2 + C for a pure constant C), so we’ll ignore this one. Hence, suppose the solution is: F2 ≡ S = S(q1 , . . . , qn , α1 , . . . αn , t), (4.105) where S is called Hamilton’s principal function and each αi is an independent constant. We can pick our new momenta to be the constants of integration Pi = αi for i ∈ {1, . . . , n} (so that P˙i = 0), thus specifying F2 = F2 (q, P, t) as desired. Then, using again the property of an F2 generating function (and K = 0), we have that the new constant variables are: Pi ≡ α i and Qi ≡ βi = ∂S(q, α, t) . ∂αi (4.106) We introduce the notation βi to emphasize that these are constants. From these results we can obtain a solution for the motion as follows. From the invertibility of our transformations we have: ∂S ∂αi ∂S pi (q, α, t) = ∂qi βi (q, α, t) = ⇒ qi = qi (α, β, t) , ⇒ pi = pi (q, α, t) = pi (q(α, β, t), α, t) = pi (α, β, t) . (4.107) (Note that function argument notation has been abused slightly here since pi (q, α, t) and pi (α, β, t) are technically different functions of their three arguments. Since we are always sticking explicit variables into the slots this should not cause confusion.) If desired, we can also swap our 2n constants αi and βi for 2n initial conditions qi0 and pi0 , to obtain a solution for the initial value problem. We obtain one set of constants in terms of the other set by solving the 2n equations obtained from the above results at t = t0 : qi0 = qi (α, β, t0 ) , pi0 = pi (α, β, t0 ) . (4.108) Thus we see that Hamilton’s principal function S is the generator of canonical transformations of constant (Q, P ), and provides a method of obtaining solutions to classical mechanics problems by way of finding a transformation. There are a few comments to be made about this. 1. The choice of constants αi is somewhat arbitrary, as any other independent choice γi = γi (α) is equally good. Thus, when solving the H-J equation, we introduce the constants αi in whatever way is most convenient. 2. What is S? We know that: Ṡ = ∂S ∂S ∂S Ṗi + , q̇i + ∂Pi ∂t ∂qi (4.109) but we also know that: ∂S = pi ∂qi , P˙i = 0 and ∂S = −H ∂t (4.110) Putting Eq.(4.109) and Eq.(4.110) together we have: Z Ṡ = pi q̇i − H = L ⇒ S= L dt (4.111) Thus S is the classical action which is an indefinite integral over time of the Lagrangian (so it is no coincidence that the same symbol is used). 3. The H-J equation is also the semiclassical limit of the quantum mechanical Schrödinger equation (0’th order term in the WKB approximation). To see this consider the Schrödinger equation i~   ∂ψ ~2 ∂ 2 = Hψ = − + V (q) ψ, ∂t 2m ∂q 2 (4.112) with the wavefunction ψ = exp(iS/~). At this point we are just making a change of variable, without loss of generality, and S(q, t) is complex. Plugging it in, and canceling an exponential after taking the derivative, we find ∂S i~ ∂ 2 S 1  ∂S 2 − =− + + V (q) . (4.113) 2m ∂q ∂t 2m ∂q 2 This equation is just another way of writing the Schr¨odinger equation, to solve for a complex S instead of ψ. If we now take ~ → 0 then we find that the imaginary term goes away leaving  ∂S  1  ∂ S 2 ∂S ∂S , (4.114) + + V (q) = + H q, 0= ∂q ∂t 2m ∂q ∂t which is the Hamilton-Jacobi equation for S with a standard p2 /2m kinetic term in H. Having set things up, it is always good for us to test a new formalism on an example where we know the solution. Example: let us consider the harmonic oscillator Eq.(4.33):
1 p2 + (mωq)2 = E (4.115) 2m Here we will look for one constant P = α and one constant Q = β. The H-J equation says !  2 1 ∂S ∂S 2 + (mωq) + = 0. (4.116) 2m ∂q ∂t H= In solving this, we note that the dependence of S on q and t is separable S(q, α, t) = W (q, α) + g(α, t), (4.117) which gives: 1 2m  ∂W ∂q 2 ! + (mωq)2 =− ∂g = α. ∂t (4.118) Since the left side is independent of t and the right hand side is independent of q, then the result must be equal to a separation constant α that is independent of q and t. We will choose our new P = α. Now we have ∂g = −α ⇒ g = −αt (4.119) ∂t where we have avoided the addition of a further additive constant (since our convention was to always drop an additive constant when determining S). To identify what α is note that H=− ∂S ∂g =− = α, ∂t ∂t (4.120) which corresponds to the constant energy, α = E. (4.121) The other equation we have to solve is 1 2m  ∂W ∂q 2 ! + (mωq)2 = α, so rearranging and integrating this gives the indefinite integral Z p W =± 2mα − (mωq)2 dq , (4.122) (4.123) which we will leave unintegrated until we must do so. The full solution is then given by: Z p 2mα − (mωq)2 dq. (4.124) S = −αt ± With this result for Hamilton’s Principal function in hand we can now solve for the equations of motion. The equations of motion come from (we now do the integral, after taking the partial derivative): ! r Z dq 1 mω 2 ∂S = −t ± m p ⇒ t + β = ± arcsin q . (4.125) β= ∂α ω 2α 2mα − (mωq)2 Inverting gives: r 2α sin(ω(t + β)), mω 2 so β is related to the phase. Next we consider p and use this result to obtain: q=± p= p √ ∂S = ± 2mα − (mωq)2 = ± 2mα cos(ω(t + β)) ∂q (4.126) (4.127) These results are as expected. We can trade (α, β) for the initial conditions (q0 , p0 ) at t = 0. The choice of phase (from shifting β so that ωβ → ωβ + π) allows taking the positive sign of each square root in the solutions above. Separation of variables is the main technique to solve the H-J equation. In particular, for a time independent H where ∂H Ḣ = =0 (4.128) ∂t we can always separate time by taking: S(q, α, t) = W (q, α) − α1 t, (4.129) where α1 has been chosen as the separation constant, then plugging this into the time dependent H-J equation yields (just as in our Harmonic Oscillator example):   ∂W ∂W H q1 , . . . , q n , ,..., = α1 . (4.130) ∂q1 ∂ qn This result is referred to as the time independent Hamilton-Jacobi equation. Since H˙ = 0, H is conserved, and equal to a constant α1 . If H = E then this constant is energy, α1 = E. The function W is called Hamilton’s characteristic function. The idea is now to solve the time independent H-J equation for W = W (q, α) where P = α still. If we follow the setup from our time dependent solution above then the equations of motion are obtained from the following prescription for identifying variables: ∂W for i ∈ {1, . . . , n} , ∂qi ∂S ∂W Q 1 = β1 = − t, = ∂α1 ∂α1 ∂W Qj = βj = for j ∈ {2, . . . , n} for n > 1 . ∂ αj pi = (4.131) Here all the Qi are constants. There is an alternative to the above setup, which allows us to not refer to the time dependent solution. The alternative is to consider W = F2 (q, P ) as the generating function, instead of S and only demand that all the new momenta Pi are constants with P1 = α1 = H for a time independent Hamiltonian H. At the start of chapter 4 we saw that this less restrictive scenario would lead to Qs that could have a linear time dependence, which is still pretty simple. This is almost identical to the above setup but we rename and reshuffle a few things. The following three equations are the same as before:   ∂W ∂W , Pi = αi and H q, = α1 (4.132) pi = ∂qi ∂q However, now we have a non-zero K and different equation for Q1 : K = H = α1 and Qi = ∂W for all i ∈ {1, . . . , n}. ∂αi (4.133) This means: ∂K ∂W = 1 ⇒ Q1 = t + β1 = (4.134) ∂α1 ∂α1 which is Eq. (4.131) but rearranged from the perspective of Q1 . For j > 1, the equations are the same as before Eq.(4.131): Q̇1 = Q̇j = ∂K =0 ∂αj ⇒ Qj = βj = ∂W ∂αj (4.135) In this language we do not need to mention S only W . There are a few comments to be made: 1. Again, the choice of α is arbitrary, and αi = αi (γ) is fine. If we do replace α1 = α1 (γ) ∂K = vi is a constant so that (potentially) all of the Qi become linear in then Q˙ i = ∂γi time: Qi = vi t + βi for all i ∈ {1, . . . , n} (4.136) 2. What is W ? We know that: ∂W q̇i = pi q̇i ⇒ W = Ẇ = ∂qi Z Z pi q̇i dt = pi dqi , (4.137) which is a different sort of “action”. 3. The time independent H-J equation has some similarity to the time-independent Schrödinger energy eigenvalue equation (both involve H and and constant E, but the former is a non-linear equation for W , while the latter is a linear equation for the wavefunction ψ). The H-J method is most useful when there is a separation of variables in H. Example: if H = h1 (q1 , q2 , p1 , p2 ) + h2 (q1 , q2 , p1 , p2 )f (q3 , p3 ) = α1 , (4.138) so that q3 is separable, then α1 − h 1 (4.139) h2 is a constant because the right hand side is independent of q3 and p3 . Thus we assign f (q3 , p3 ) = f (q3 , p3 ) = α2 (4.140) for convenience. We can then write:   ∂W3 W = W (q1 , q2 , α) + W3 (q3 , α) ⇒ f q3 , = α2 ∂q3     ∂ W 0 ∂W 0 ∂ W 0 ∂W 0 h1 q1 , q2 , , + α2 h2 q1 , q2 , , = α1 ∂q1 ∂q2 ∂q1 ∂ q2 0 Here, q1 and q2 may or may not be separable. If all variables are separable then we use the solution: X W = Wi (qi , α) i and (4.141) (4.142) (4.143) We can simply try a solution of this form to test for separability. Note that cyclic coordinates are always separable. Proof: let us say that q1 is cyclic. Then   ∂W ∂ W ∂W , ,..., = α1 , p1 ≡ γ and H q2 , . . . , qn , ∂q1 ∂q2 ∂ qn (4.144) where γ is constant. Let us now write W (q, α) = W1 (q1 , α) + W 0 (q2 , . . . , qn , α). (4.145) This gives us: p1 = ∂W1 = γ ⇒ W1 = γ q1 . ∂ q1 (4.146) Which gives us: W (q, α) = γq1 + W 0 (q2 , . . . , qn , α) (4.147) This procedure can be repeated for any remaining cyclic variables. Note that the choice of variables is often important in finding a result that separates. A problem with spherical symmetry may separate in spherical coordinates but not Cartesian coordinates. 4.5 Kepler Problem As an extended example, let us consider the Kepler problem of two masses m1 and m2 in a central potential (with the CM coordinate R = 0). The Lagrangian is: 1 L = mṙ2 − V (r) 2 where 1 1 1 ≡ + , m m1 m2 (4.148) and here m is the reduced mass. Any V (r) conserves L = r × p, so the motion of r and p is in a plane perpendicular to L. The coordinates in the plane can be taken as (r, ψ), so:  m 2 2 ˙2 L= ṙ + r ψ − V (r), (4.149) 2 with ψ being cyclic, which implies: pψ = mr2 ψ̇ is a constant. (4.150) In fact pψ = |L| ≡ `. Notationally, we use ` for the magnitude of the angular momentum L to distinguish it from the Lagrangian L. The energy is then: m `2 E = ṙ2 + + V (r), (4.151) 2 2mr2 which is constant, and this can be rewritten as: m 2 `2 ṙ + Veff (r) where Veff (r) = V (r) + , (4.152) 2 2mr2 is the effective potential, as plotted below for the gravitational potential. E= Veff(r) where Veff 0 r Figure 4.1: Plot of the effective potential Vef f along with the different qualitative orbits allowed in a gravity-like potential. The points correspond to turning points of the orbit. Writing the E-L equation for ṙ = dr/dt = . . . and then solving for it as dt = dr/(. . .), and integrating yields Z r dr0 q (4.153) t = t(r) =
2 `2 0 r0 E − V (r ) − 2mr02 m as an implicit solution to the radial motion. The orbital motion comes as r = r(ψ) or ψ = ψ(r) by using Eq.(4.150) and substituting, in Eq.(4.153). We have ψ˙ = dψ/dt = `/(mr2 ), so we can use this to replace dt by dψ in dt = dr/(. . .) to get an equation of the form dψ = dr/(. . .). The result is given by Z r dr0 q ψ − ψ0 = ` (4.154)
`2 r0 r 02 2m E − V (r0 ) − 2mr 02 k In the particular case of V (r) = − , the solution of the orbital equation is: r mk 1 = 2 (1 + ε cos(ψ − ψ 0 )) r(ψ) ` (4.155) where the eccentricity ε is given by: r 2E`2 (4.156) mk 2 Below are plotted the different qualitative orbits for the gravitic potential, with different ε or E (circular, elliptical, parabolic, and hyperbolic respectively). ε≡ 1+ ε =0 , E = Emin 0 <ε <1 , E <0 ε =1 , E =0 ε >1 , E >0 0 0 Figure 4.2: Different Orbits for the gravity-like potential. The orbits’ colors match those of Fig.(4.1). The unbounded orbits occur for E ≥ 0. The different curves correspond to the different possible conic sections. Consider solving this problem instead by the H-J method. Lets start by considering as the variables (r, ψ) so that we assume that the motion of the orbit is in a plane. Here   p2ψ 1 2 H= p + + V (r) = α1 = E . (4.157) 2m r r2 As ψ is cyclic, then pψ ≡ αψ is constant. Using: W = W1 (r) + αψ ψ, (4.158) then the time independent H-J equation is: !  2 αψ2 1 ∂ W1 + 2 + V (r) = α1 . 2m ∂r r (4.159) This is simplified to s ∂W1 = ∂r αψ2 2m(α1 − V (r)) − 2 r (4.160) and solved by Z s 2m(α1 − V (r)) − W = αψ ψ + αψ2 dr. r2 (4.161) The transformation equations are: ∂W t + β1 = =m ∂α1 Z dr q 2 αψ r2 , 2m(α1 − V (r)) − Z dr ∂W q = ψ − αψ β2 = ∂αψ r2 2m(α1 − V (r)) − 2 αψ r2 . (4.162) Thus we immediately get the radial equation t = t(r) and orbital equation ψ = ψ(r) from this, with αψ = ` and α1 = E, showing that the constants are physically relevant parameters. Let’s solve this problem again, but suppose the motion is in 3 dimensions (as if we did not know the plane of the orbit). Using spherical coordinates (r, θ, ϕ) this corresponds to   p2ϕ p2θ 1 2 p + + + V (r) = α1 . (4.163) H= 2m r r2 r2 sin2 θ Lets try a separable solution W = Wr (r) + Wθ (θ) + Wϕ (ϕ) . (4.164) Since ϕ is cyclic we know it is separable and that: Wϕ (ϕ) = αϕ ϕ . Together, this leaves;  2 ∂Wr 1 + 2 ∂r r  ∂ Wθ ∂θ 2 αϕ2 + sin2 (θ) (4.165) ! + 2mV (r) = 2mα1 . (4.166) Because the term: 2 αϕ2 (4.167) sin2 (θ) only depends on θ while the rest of the equation depends on r, it must be a constant so we can say:  2 αϕ2 ∂Wθ ≡ αθ2 (4.168) + ∂θ sin2 (θ) and the separation works. This then gives:  2 ∂Wr α2 + 2θ = 2m(α1 − V (r)), (4.169) r ∂r  ∂Wθ ∂θ + which is the same equation we considered before when assuming the motion was in a plane, with α1 = E and αθ = `. Eq. (4.168) says that p2θ + p2ϕ = `2 . 2 sin (θ) (4.170) Here pϕ is the constant angular momentum about the ẑ axis. 4.6 Action-Angle Variables For many problems, we may not be able to solve analytically for the exact motion or for orbital equations, but we can still characterize the motion. For periodic systems we can find the frequency by exploiting action-angle variables. The simplest case is for a single dimension of canonical coordinates (q, p). If H(q, p) = α1 then p = p(q, α1 ). There are two types of periodic motion to consider. 1. Libration (oscillation) is characterized by a closed phase space orbit, so that q and p evolve periodically with the same frequency. Figure 4.3: Phase space orbit of a libration (oscilation). The trajectory closes on itself, the state returns to the same position after some time τ . 2. Rotation is characterized by an open phase space path, so p is periodic while q evolves without bound. Figure 4.4: Phase space orbit of a rotation. Although the orbit is not closed, each period the evolution of the system is the same, leading to a orbit that repeats itself with a translation. Example: a pendulum of length a may be characterized by canonical coordinates (θ, pθ ), where: p2θ − mg a cos θ (4.171) E=H= 2ma2 This means: p (4.172) pθ = ± 2ma2 (E + mga cos θ) must be real. A rotation occurs when E > mga, and oscillations occur when E < mga. The critical point in between (when the pendulum just makes it to the top) is when E = mga exactly, and is depicted by a dashed line in the figure below. Rotation Libration/Oscilation −3.142 3.142 Figure 4.5: The pendulum exhibits both librations and rotations depending on the initial conditions. For either type of periodic motion, it is useful to replace P = α1 by a different (constant) choice called the action variable I J = p dq, (4.173) H where refers to a definite integral over one period in phase space. To see that J is constant, recall that p = p(q, α1 ), so plugging this into the definite integral we are left with J = J(α1 ). Also, we have the inverse, α1 = H = H(J), and can rewrite Hamilton’s characteristic function in terms of J by W = W (q, α1 ) = W (q, H(J)) = W (q, J) (where again the argument notation is abused slightly). The coordinate conjugate to J is the angle variable ∂W ∂J (where ω is not meant to imply an angular velocity). This means ω= ω̇ = ∂H(J) = ν(J) is a constant. ∂J (4.174) (4.175) As a result the angle variable has linear time dependence, ω = νt + β , (4.176) for some initial condition β. Dimensionally, J has units of angular momentum, while ω has no dimensions (like an angle or a phase). To see why it is useful to use the canonical variables (ω, J), let us consider the change in ω when q goes through a complete cycle. I I 2 I I ∂ W ∂ ∂W ∂ ∂ω p dq = 1 (4.177) dq = dq = dq = ∆ω = ∂J ∂q ∂q∂J ∂J ∂q where in the last equality we used the definition of J in Eq.(4.173). Also, we have ∆ω = ντ where τ is the period. Thus 1 ν= (4.178) τ is the frequency of periodic motion. If we find H = H(J) then ν= ∂H(J) ∂J (4.179) immediately gives the frequency ν = ν(J) for the system. Often, we then J = J(E) to get ν = ν(E) the frequency at a given energy. This is a very efficient way of finding the frequency of the motion without solving for extraneous information. Example: let us consider a pendulum with action-angle variables. We define: Ẽ = E mga (4.180) so that E˜ > 1 corresponds to rotation and E˜ < 1 corresponds to oscillation. This means p p pθ = ± 2m2 ga3 Ẽ + cos θ. (4.181) For E˜ > 1: Z π p p 2 3 J = 2m ga dθ Ẽ + cos θ , (4.182) −π which is an elliptic integral. For E˜ < 1: Z +θ0 p Z p p 2 3 2 3 J = 2m ga dθ Ẽ + cos θ + 2m ga p = 4 2m2 ga3 −θ0 Z θ0 −θ0 h dθ − p Ẽ + cos θ i θ0 dθ p Ẽ + cos θ , (4.183) 0 as the contributions from the four intervals that the pendulum swings through in one period are all equivalent. Here θ0 is the turning point of the oscillation, and E˜ = − cos(θ0 ). From this:  −1 ∂J ∂E = (4.184) ν= ∂J ∂E which we can solve graphically by making a plot of J vs E, then dJ/dE versus E, and finally the inverse ν = dE/dJ versus E. Figure 4.6: Plot of J(E) ˜ The discontinuity versus E. corresponds to the transition from Oscilation to Rotation. dJ Figure 4.7: Plot of . dE The discontinuity is logarithmic divergent so it is integrable. Figure 4.8: Plot of the frequency of oscillation ν(E) ˜ versus E. As E˜ → −1 we approach the small amplitudeplimit, where ν = (2π)−1 g/a. Example: let us consider the limit |θ|  1 of a pendulum, so: H= p2θ mga 2 + θ − mga 2 2ma 2 (4.185) We can actually consider this in the context of a general harmonic oscillator H= where: p2 mΩ2 2 + x, 2m 2 (4.186) r g pθ , x = aθ and p = (4.187) a a Notationally, Ω is used for the harmonic oscillator frequency to distinguish from the transformed angle variable ω. We then have: I I √ (4.188) J = p dq = ± 2mE − m2 Ω2 x2 dx Ω= Note that the coordinate does not need to be an angle, as may be the case for general x. This gives: √ Z x0 s √ x2 2mE 1 − 2 dx where x0 ≡ . (4.189) J = 4 2mE x0 mΩ 0 Solving the integral yields √ 2πmE 2πE J = π 2mEx0 = = , mΩ Ω (4.190) which gives us Ω ∂E = , ∂J 2π the expected cyclic frequency for the harmonic oscillator. (4.191) Multiple Variables: We can treat multiple variables (q1 , . . . , qn , p1 , . . . , pn ) with the actionangle formalism if each pair (qi , pi ) has an oscillatory or rotating orbit. Lets also assume that the H-J equation is completely separable into: X W = Wj (qj , α). (4.192) j Here we have ∂Wi pi = = pi (qi , α1 , . . . , αn ) ⇒ Ji = ∂qi I pi dqi = Ji (α1 , . . . , αn ) (4.193) where repeated indices do not correspond to implicit sums here. This implies that the inverse will be αj = αj (J1 , . . . , Jn ) and thus α1 = H = H(J1 , . . . , Jn ). Likewise: ωi = X ∂ Wj ∂W = = ωi (q1 , . . . , qn , J1 , . . . , Jn ) . ∂Ji ∂Ji j (4.194) Just as in the one dimensional case the time derivative of the angle variables is a constant ω̇i = ∂H = νi (J1 , . . . , Jn ) ∂Ji (4.195) which are the frequencies describing motion in this “multi-periodic” system. Due to the presence of multiple frequencies, the motion through the whole 2n-dimensional phase space need not be periodic in time. Example: in the 2-dimensional harmonic oscillator: x = A cos(2πν1 t) and y = B cos(2πν2 t) px = mẋ and py = mẏ (4.196) (4.197) ν1 is a rational number. ν2 The overall motion is not periodic in time unless Kepler Problem Example: Let us do a more extended and detailed example. Returning to the Kepler problem: V (r) = − k r (4.198) with its separable W : W = Wr (r, α) + Wθ (θ, α) + Wϕ (ϕ, α) . (4.199) If we take E < 0, we have oscillation in r and θ, along with a rotation in ϕ. In particular from solving our earlier differential equations for Wθ and Wr , we have W ϕ = αϕ ϕ Z s αθ2 − Wθ = ± αϕ2 dθ sin2 (θ) r Wr = ± Here we have I 2m(α1 − V (r)) − αθ2 dr r2 I I ∂W Jϕ = pϕ dϕ = dϕ = αϕ dϕ (4.200) ∂ϕ For the cyclic variable ϕ, we still call the constant pϕ periodic and will take the period to be 2π (arbitrarily since any period would work), which corresponds to particle returning to the original point in space. Thus Jϕ = 2παϕ , (4.201) where αϕ is the angular momentum about ẑ. Continuing, in a similar manner we have I Jθ = I pθ dθ = ∂W dθ = ∂θ s I ± αθ2 αϕ2 dθ − sin2 (θ) (4.202) Let us call: cos(γ) ≡ αϕ αθ (4.203) the angular momentum fraction. Then: s I Jθ = αθ ± 1− cos2 (γ) dθ sin2 (θ) (4.204) If we let sin(θ0 ) = cos(γ), then pθ = 0 at the turning points, θ ∈ {θ0 , π − θ0 }, as expected. Here one oscillator goes from π − θ0 → θ0 when pθ > 0, and in reverse for pθ < 0. Moreover, sin(θ)−2 is even about θ = π2 . This gives Z θ0 s cos2 (γ) Jθ = 4αθ 1− dθ. (4.205) 2 π sin (θ) 2 Making two more substitutions cos(θ) ≡ sin(γ) sin(ψ) , and then u ≡ tan(ψ) , after some work the expression becomes  Z ∞ 1 cos2 (γ) Jθ = 4αθ − du = 2παθ (1 − cos(γ)) 1 + u2 1 + u2 cos2 (γ) 0 = 2π(αθ − αϕ ), (4.206) (4.207) which gives Jθ + Jϕ = 2παθ . (4.208) Finally we can consider I r Jr = 2mE − 2mV (r) − (Jθ + Jϕ )2 dr 4π 2 r2 (4.209) We can immediately make some observations. We observe that Jr = Jr (E, Jθ + Jϕ ) is a function of two variables for any V = V (r), and thus if we invert E = E(Jr , Jθ + Jϕ ). This implies: ∂E ∂E = ⇒ νθ = νϕ (4.210) ∂Jθ ∂Jϕ The two frequencies are degenerate for any V = V (r). For the V (r) = −kr−1 potential, the integration can be performed (for example, by contour integration) to give (for E < 0) r r 2m 2m Jr = −(Jθ + Jϕ ) + πk ⇒ Jr + Jθ + Jϕ = πk . (4.211) −E −E This means: E=− 2π 2 k 2 m (Jr + Jθ + Jϕ )2 ⇒ νθ = νϕ = νr (4.212) In particular: ∂E 1 νr = = 4π 2 k 2 (Jr + Jθ + Jϕ )−3 = πk ∂Jr r −2E 3 m (4.213) which is the correct orbital frequency in a bound Kepler orbit. Using the relations between {α1 = E, αθ , αϕ } and {Jr , Jθ , Jϕ }, we can also get Hamilton’s characteristic function for this system as W = Wϕ + Wθ + Wr Z s Z s 2 J dθ −(2πmk)2 2mk (Jθ + Jϕ )2 ϕJϕ ϕ ± (Jθ + Jϕ )2 − ± = − dr. + 2 2π r (2πr)2 (Jr + Jθ + Jϕ )2 sin (θ) 2π This then gives the angle variables: ∂W = ωr (r, Jr + Jθ + Jϕ , Jθ + Jϕ ) ∂Jr ∂W = ωθ (r, θ, Jr + Jθ + Jϕ , Jθ + Jϕ , Jϕ ) ωθ = ∂Jθ ∂W ωϕ = = ωϕ (r, θ, ϕ, Jr + Jθ + Jϕ , Jθ + Jϕ , Jϕ ) ∂Jϕ ωr = (4.214) where ω̇r = νr , ω̇θ = νθ , and ω̇ϕ = νϕ . Of course, in this case, νr = νθ = νϕ . At this point we can identify five constants for the Kepler problem J1 J2 J3 ω1 ω2 = Jϕ = Jθ + Jϕ = Jr + Jθ + Jϕ = ωϕ − ωθ = ωθ − ωr . (4.215) ~ energy E, and (These 5 constants could also be identified from the angular momentum L, ~ Laplace-Runge-Lenz vector A.) What are they? There are two constants specifying the plane of the orbit (the x0 y 0 -plane), which are the inclination angle i and the longitude of the ascending node Ω. There are three constants specifying the form of the ellipse, which are the semi-major axis a (giving the size), the eccentricity ε (giving the shape), and the angle ω (giving the orientation within the plane). These are all shown in Fig. 4.9. Figure 4.9: Picture of an orbit in 3d and the five parameters necessary to fully specify it. The angles i, Ω and ω provide the orientation in space while a and ε provide the size and shape of the conic section. It can be shown that the relations between these constants and the ones above are s  2 2 J1 k J3 J2 = 1− cos(i) = a=− = 2 4π mk J2 2E J3 Ω = 2πω1 ω = 2πω2 providing a fairly simple physical interpretations to (J1 , J2 , J3 , ω1 , ω2 ). Also recall that J2 = 2παθ = 2π`. When the orbit is perturbed by additional forces (like a moon, another planet, general relativistic corrections, and so on), these action-angle variables provide a natural way to describe the modified orbit. We will soon see that they become functions of time. For example, from the above list of constants, the perturbed ω = ω(t) is the precession of the perihelion of an orbit. We will learn how to derive this time dependence next, in our chapter on Perturbation Theory.