Hamilton-Jacobi theory

Transcription

1 Hamilton-Jacobi theory November 9, 04 We conclude with the crowning theorem of Hamiltonian dynamics: a proof that for any Hamiltonian dynamical system there exists a canonical transformation to a set of variables on phase space such that the paths of motion reduce to single points. Clearly, this theorem shows the power of canonical transformations! The theorem relies on describing solutions to the Hamilton-Jacobi equation, which we introduce first. Integrability of the action We first define Hamilton s principal function. Let x i t and p i t satisfy Hamilton s equations of motion, and as for the integrability condition for the action. That is, we would lie to now when the action is a function and not a functional, S [ x i t ] S x i, t. The condition we need is just lie the vanishing curl of a force required for the existence of a potential function. Thining of the n + vectorp a p i, H integrated along a curve in dx a x i, t -space S p i dx i Hdt P a dx a the integrability condition is the vanishing of the higher-dimensional curl, Writing this in terms of p i, H, x j, t, P a X b P b X a p i x j p j x i H x i dp i dt dp i dt + H x i H + H The first is satisfied because x i and p j are independent, the middle two give one of Hamilton s equations, and the final equation is an identity. Therefore, S is a function if ṗ i H x. i The condition is not unique. Since we may integrate by parts, S p i dx i Hdt d p i x i x i dp i Hdt p i x i x t dp t i + Hdt 0

2 a similar argument applied to x i dp i + Hdt shows that ẋ i H p i gives integrability. Therefore, if the family of curves, x i t, p j t solve Hamilton s equations, then evaluating the action on those curves gives a function. Exercise: Carry out the demonstration that S becomes a function if ẋ i H p i The Hamilton-Jacobi equation Conversely, suppose we replace the action with a function, S [x] S x i, t. Then S x i, t p i dx i Hdt implies so that ds p i dx i Hdt x i dxi + dt p idx i Hdt p i x i H x i, p j, t If we replace p j in the Hamiltonian, we get a differential equation for Hamilton s principal function, H x i, x j, t This is the Hamilton-Jacobi equation. Example: Find the Hamilton-Jacobi equation for a simple harmonic oscillator Hamiltonian for the oscillator is H p m + x the Hamilton-Jacobi equation is + m x x Partial differential equations have free functions in their solutions. Thus, while Since the has the solution it has the general solution f x f f x, t ax + by + c f x, y g x + t + g + x t for any two functions g ±. For the Hamilton-Jacobi equation, canonical transformations can help introduce arbitrary functions. Suppose we have a solution for S of the form S g x i, α j, t

3 for some function g. Then a canonical transformation changes p i dx i Hdt by df, where f is the generating function. This means that S X i, P j, t g x i, α j, t + f, where the new coordinates are X i, P j. Now let Then we have so that α i dx i Hdt P i dx i H dt + df f f X i, α j x i α i P i dx i H dt x i dα i α i dx i + f X i dxi + f dα i + f α i dt P i + f f X i dx i + x i dα i + H H dt α i f X, α P i X i x i f X, α α i H H The Hamiltonian remains unchanged, the new coordinate X i is found by inverting x i fx,α α i to find x i as a function of the new coordinates and the constants α i. The new momentum P i is an arbitrary function of the new coordinates X i, and the principal function becomes S X i, t g x i, α j, t + f g x i X, α X, P, α X, P, t + f X i, α j X, P x i X, P α i X, P so the new form of the principal function depends on n arbitrary functions P i x, α. 3 The principal function as generator of a canonical transformation Suppose we find Hamilton s principal function, S x i, t. Its relationship to the momentum, p i x i, suggests that we may use it as a generating function for a canonical transformation, S x i, P j, t S x i, t. This turns out to be especially useful. We choose a generating function with independent variables x i, P j, so let f X i P i + S x i, t Then we have p i dx i Hdt P i dx i H dt + df P i dx i H dt X i dp i P i dx i + x i dxi + dp i + P i dt x i p i dx i + X i dp i + H + P i H dt so that the independent variables are now x i, P i, t, satisfying X i P i p i x i H H + 3

4 But we now that x i p i p i H so that H. The principal function has generated a transformation to a set of canonical variables for which the Hamiltonian vanishes! This maes Hamilton s equations trivial: so X i, P j simply stay at their initial values. 4 Examples 4. Example : Free particle Ẋ i P i The simplest example is the case of a free particle, for which the Hamiltonian is and the Hamilton-Jacobi equation is Let Then fx must satisfy and therefore H p m m S S fx Et df dx me fx mex c πx c where c is constant and we write the integration constant E in terms of the new constant momentum. Hamilton s principal function is therefore Then, for a generating function of this type we have S x, π, t πx π m t c p x π q π x π m t H H + H E 4

5 Because E H, the new Hamiltonian, H, is zero. This means that both q and π are constant. The solution for x and p follows immediately: x q + π m t p π We see that the new canonical variables q, π are just the initial position and momentum of the motion, and therefore do determine the motion. The fact that nowing q and π is equivalent to nowing the full motion rests here on the fact that S generates motion along the classical path. In fact, given initial conditions q, π, we can use Hamilton s principal function as a generating function but treat π as the old momentum and x as the new coordinate to reverse the process above and generate xt and p. 4. Example : Simple harmonic oscillator For the simple harmonic oscillator, the Hamiltonian becomes and the Hamilton-Jacobi equation is Setting S f x Et then m H p m + x + x x df + m dx x E df dx me mx and direct integration see examples, below give a solution for S, me f mx dx me E x dx so with x sin µ we have E f me E E sin µ cos µdµ E cos µdµ ω E + cos µ dµ ω E µ + ω sin µ E sin ω E x + sin sin E x cos sin E x E sin ω E x + E x E x 5

6 Letting as before, fx must satisfy and therefore fx m S + x S fx Et df dx m E x m E x dx π mx dx where we have set E π m. Now let mx π sin y. The integral is immediate: fx π mx dx π m cos ydy Hamilton s principal function is therefore S x, π, t π m π π y + sin y cos y m sin x m + m x m x π π π m t c π x m sin m + x π mx π π m t c and we may use it to generate the canonical change of variable. This time we have p x π + m x π π π mx + π mx + x π mx q π π sin x m + π m π m mx π mx π mx mx m x π m x π 6

7 + x π π mx π m t π sin x m π m π m t H H + H E The first equation relates p to the energy and position, the second gives the new position coordinate q, and third equation shows that the new Hamiltonian is zero. Hamilton s equations are trivial, so that π and q are constant, and we can invert the expression for q to give the solution. Setting ω, the solution is where x t π mω mω sin π q + ωt A sin ωt + φ q Aφ π Amω The new canonical coordinates therefore characterize the initial amplitude and phase of the oscillator. 4.3 Example 3: One dimensional particle motion Now suppose a particle with one degree of freedom moves in a potential Ux. Little is changed. The the Hamiltonian becomes H p m + U and the Hamilton-Jacobi equation is m S + Ux Letting S fx Et as before, fx must satisfy and therefore df dx m E U x m fx m E U xdx π mu xdx where we have set E π m. Hamilton s principal function is therefore π S x, π, t mu xdx π m t c and we may use it to generate the canonical change of variable. 7

8 This time we have p x π mu x q π x π mu xdx π π x 0 m t H H + H E The first and third equations are as expected, while for q we may interchange the order of differentiation and integration: q π mu xdx π m t π π π mu x dx π m t πdx π mu x π m t To complete the problem, we need to now the potential. However, even without nowing Ux we can mae sense of this result by combining the expression for q above to our previous solution to the same problem. There, conservation of energy gives a first integral to Newton s second law, E p m + U m dx dt so we arrive at the familiar quadrature t t 0 dt Substituting into the expression for q, q x x x0 x x 0 πdx π mu x π m + U mdx m E U x x 0 mdx m E U π m t 0 πdx x π mu x πdx π mu x π m t 0 x 0 πdx π mu x π m t 0 We once again find that q is a constant characterizing the initial configuration. Since t 0 is the time at which the position is x 0 and the momentum is p 0, we have the following relations: and which we may rewrite as t x p m + Ux p 0 m + Ux 0 E const. t t 0 x dx t 0 m E U x 0 x0 m dx E U dx m m E U π q const. 8

9 4.4 Example 4: Two dimensional oscillator Suppose we have a mass fastened to a spring, moving on a tabletop, so that action is S m ṙ + r ϕ r The canonical momenta are so the Hamiltonian is The Hamilton-Jacobi equation is m r p r mṙ p ϕ mr ϕ H m ṙ + r ϕ + r H p r + p ϕ m r + r + r + ϕ r Let S f r, ϕ Et so that m r + r + ϕ r E Now use separation of variables. Let Then f R r + Φ ϕ dr + dφ dr r + mr me dϕ dr + mr me dφ dr dr r mr 4 + mer dr r dφ dϕ Since the left side depends only on r and the right only on ϕ, each side must equal some constant, a : dr r mr 4 + mer a dr dφ dϕ ±a We immediately have and we may integrate to find R: dr dr R Φ ±aϕ + b mr + me a r me mr a r dr dϕ 9