Chapter 6 Lecture 1 Canonical Transformations Akhlaq Hussain 1 6.1 Canonical Transformations 푁 Hamiltonian formulation 퐻 푞푖, 푝푖 = σ푖=1 푝푖 푞푖ሶ − 퐿 (Hamiltonian) 휕퐻 푝푖ሶ = − 휕푞푖 (푯풂풎풊풕풐풏′풔 푬풒풖풂풕풊풐풏풔) 휕퐻 푞푖ሶ = 휕푝푖 one can get the same differential equations to be solved as are provided by the Lagrangian procedure. 퐿 푞푖ሶ , 푞푖 = 푇 − 푉 (Lagrangian) 푑 휕퐿 휕퐿 − = 0 (Lagrange’s equation) 푑푡 휕푞푖ሶ 휕푞푖 Therefore, the Hamiltonian formulation does not decrease the difficulty of solving Problems. The advantages of Hamiltonian formulation is not its use as a calculation tool, but rather in deeper insight it offers into the formal structure of the mechanics. 2 6.1 Canonical Transformations ➢ In Lagrangian mechanics 퐿 푞푖ሶ , 푞푖 system is described by “푞푖” and velocities” 푞푖ሶ ” in configurational space, ➢ The parameters that define the configuration of a system are called generalized coordinates and the vector space defined by these coordinates is called configuration space. ➢ The position of a single particle moving in ordinary Euclidean Space (3D) is defined by the vector 푞 = 푞(푥, 푦, 푧) and therefore its configuration space is 푄 = ℝ3 ➢ For n disconnected, non-interacting particles, the configuration space is 3푛 ℝ . 3 6.1 Canonical Transformations ➢ In Hamiltonian 퐻 푞푖, 푝푖 we describe the state of the system in Phase space by generalized coordinates and momenta. ➢ In dynamical system theory, a Phase space is a space in which all possible states of a system are represented with each possible state corresponding to one unique point in the phase space. ➢ There exist different momenta for particles with same position and vice versa. 4 6.1 Canonical Transformations ➢ To understand the importance of Hamiltonian let us consider a problem for which solution of Hamilton’s equations are trivial (simple) and Hamiltonian is constant of motion. ➢ For this case all the coordinates “푞푖” of the problem will be cyclic and all conjugate “푝푖” momenta will be constant. Since 푝푖 = 훼푖 = 퐶표푛푠푡푎푛푡 휕퐻 휕퐻 And 푞푖ሶ = = = 휔푖 휕푝푖 휕훼푖 푞푖 = 휔푖푡 + 훽푖 훽푖 is constant and can be find by the initial conditions. But in real problem it is not necessary that all the coordinates are cyclic. 5 6.1 Canonical Transformations ➢ Practically, it rarely happens that all the coordinates are cyclic. ➢ However a system can be described by more than one set of generalized coordinates. ➢ The motion of particle in plane is described by generalized coordinates either the cartesian coordinates. In cartesian coordinates 푞1 = 푥, & 푞1 = 푦 In polar coordinates 푞1 = 푟, & 푞1 = 휃 Both choices are equally valid, but one of the set may be more convenient for the problem under the consideration. Not that for the central force neither x, nor y is cyclic while the second set does contain a cyclic coordinate 휃 6 6.1 Canonical Transformations ➢ The number of cyclic coordinates thus depend on choice of generalized coordinates, and for each problem there may be one choice for which all the coordinates are cyclic. ➢ Since the generalized coordinates suggested by the problem will not be cyclic normally, we must first derive a specific procedure for transforming from one set of variables to some other set that may be more suitable. 7 6.1 Canonical Transformations ➢ Let us consider transformation equations 푄푖 = 푄푖 푞푖, 푝푖, 푡 , & 푃푖 = 푃푖 푞푖, 푝푖, 푡 ➢ Such that the general dynamical theory is invariant under these transformations. ➢ Let us consider a function 퐾 푄푖, 푃푖, 푡 such that 휕퐾 휕퐾 푃푖ሶ = − & 푄ሶ푖 = 휕푄푖 휕푃푖 푄푖 & 푃푖 are called canonical coordinates and transformation 푞푖 → 푄푖 & 푝푖 → 푃푖 푄푖 = 푄푖 푞푖, 푝푖, 푡 , & 푃푖 = 푃푖 푞푖, 푝푖, 푡 are known as canonical transformations. 8 6.1 Canonical Transformations Here “퐾 푄푖, 푃푖, 푡 ” play role of Hamiltonian and 푄푖 & 푃푖 must satisfy Hamilton’s principle. 푡2 (훿 ׬ σ 푃푖 푄ሶ푖 − 퐾 푄푖, 푃푖, 푡 푑푡 (1 푡1 푡2 (훿 ׬ σ 푝푖 푞푖ሶ − 퐻 푞푖, 푝푖, 푡 푑푡 = 0 (2 푡1 Equation (1) and Equation (2) may not be qual, therefore we can find a function “F” such that 푡2 푑퐹 ׬ 푑푡 = 퐹 푡2 − 퐹 푡1 푡1 푑푡 푡2 푑퐹 and δ ׬ 푑푡 = 0 where δ퐹 푡2 = δ퐹 푡1 푡1 푑푡 푑퐹 and σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + (3) 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 9 Function “퐹” is called generating function. 6.1 Canonical Transformations ➢ There are four different possibilities for “퐹” 1) 퐹1 푞푖, 푄푖, 푡 Provided that 푞푖, 푄푖 are treated as independent 2) 퐹2 푞푖, 푃푖, 푡 Provided that 푞푖, 푃푖 are treated as independent 3) 퐹3 푝푖, 푄푖, 푡 Provided that 푝푖, 푄푖 are treated as independent 4) 퐹4 푝푖, 푃푖, 푡 Provided that 푝푖, 푃푖 are treated as independent 10 6.1 Canonical Transformations Case I: Eq. 3 can be written as 푑퐹 σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 1 (4) 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 푑퐹1 푞푖,푄푖,푡 휕퐹1 휕퐹1 휕퐹1 Since = σ 푞푖ሶ + σ 푄ሶ푖 + Therefore Eq. (4)… 푑푡 휕푞푖 휕푄푖 휕푡 휕퐹1 휕퐹1 휕퐹1 σ 푝푖 푞푖ሶ − 퐻 푞푖, 푝푖, 푡 = σ 푃푖 푄ሶ푖 − 퐾 푄푖, 푃푖, 푡 + σ 푞푖ሶ + σ 푄ሶ푖 + 휕푞푖 푑푄푖 휕푡 Comparing coefficient of “푞푖ሶ ” & “푄ሶ푖” on both sides 휕퐹1 푝푖 = (5)a 휕푞푖 휕퐹1 푃푖 = − (5)b 휕푄푖 휕퐹1 And 퐾 푄푖, 푃푖, 푡 = 퐻 푞푖, 푝푖, 푡 + (5)c 휕푡 11 6.1 Canonical Transformations ➢ From Eq. (5a) we can determine “푝푖” in terms of 푞푖, 푄푖 and t and the inverse transformation 푄푖 in terms of 푞푖, 푝푖 푎푛푑 푡 Eq. (5)a ⇒ 푄푖 = 푄푖 푞푖, 푝푖, 푡 Eq. (5)b ⇒ 푃푖 = 푃푖 푞푖, 푝푖, 푡 & Eq. (5)c provide connection between new and old Hamiltonian 12 6.1 Canonical Transformations Case II: For 퐹2 푞푖, 푃푖, 푡 generating function 휕퐹1 Since 푃푖 = − 휕푄푖 Therefore, 퐹1 푞푖, 푄푖, 푡 + σ 푃푖 푄푖 = 퐹2 푞푖, 푃푖, 푡 푑퐹 Since σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 1 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 푑 ⇨ σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 퐹 푞 , 푃 , 푡 − σ 푃 푄 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 2 푖 푖 푖 푖 푑퐹 ⇨ σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 2 − σ 푃 푄ሶ − σ 푃ሶ 푄 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 푖 푖 푖 푖 푑퐹 ⇨ σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = −퐾 푄 , 푃 , 푡 + 2 − σ 푃ሶ 푄 푖 푖 푖 푖 푖 푖 푑푡 푖 푖 푑 휕퐹2 휕퐹2 휕퐹2 Since 퐹2 푞푖, 푃푖, 푡 = σ 푞푖ሶ + σ 푃푖ሶ + 푑푡 휕푞푖 휕푃푖 휕푡 13 6.1 Canonical Transformations Putting in previous equation 휕퐹2 휕퐹2 휕퐹2 ෍ 푝푖 푞푖ሶ − 퐻 푞푖, 푝푖, 푡 = −퐾 푄푖, 푃푖, 푡 + ෍ 푞푖ሶ + ෍ 푃푖ሶ + − ෍ 푃푖ሶ 푄푖 휕푞푖 휕푃푖 휕푡 Comparing coefficient of “푞푖ሶ ” & “푃푖ሶ ” on both sides 휕퐹2 푝푖 = (6)a 휕푞푖 휕퐹1 푄푖 = (6)b 휕푃푖 휕퐹 And 퐾 푄 , 푃 , 푡 = 퐻 푞 , 푝 , 푡 + 2 (6)c 푖 푖 푖 푖 휕푡 14 6.1 Canonical Transformations Case III: For 퐹3 푝푖, 푄푖, 푡 generating function 휕퐹1 Since 푝푖 = 휕푞푖 Therefore, we can write 퐹1 푞푖, 푄푖, 푡 − σ 푝푖 푞푖 = 퐹3 푝푖, 푄푖, 푡 푑퐹 Since σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 1 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 푑 ⇨ σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 퐹 푝 , 푄 , 푡 + σ 푝 푞 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 3 푖 푖 푖 푖 푑퐹 ⇨ σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 3 + σ 푝 푞ሶ + σ 푝ሶ 푞 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 푖 푖 푖 푖 푑퐹 ⇨ −퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 3 + σ 푝ሶ 푞 푖 푖 푖 푖 푖 푖 푑푡 푖 푖 푑 휕퐹3 휕퐹3 휕퐹3 Since 퐹3 푝푖, 푄푖, 푡 = σ 푝푖ሶ + σ 푄ሶ푖 + , putting in above equation. 푑푡 휕푝푖 휕푄푖 휕푡 15 6.1 Canonical Transformations Putting in previous equation 휕퐹3 휕퐹3 휕퐹3 −퐻 푞푖, 푝푖, 푡 = ෍ 푃푖 푄ሶ푖 − 퐾 푄푖, 푃푖, 푡 + ෍ 푝푖ሶ + ෍ 푄ሶ푖 + + ෍ 푝푖ሶ 푞푖 휕푝푖 휕푄푖 휕푡 Comparing coefficient of “푝푖ሶ ” & “푄ሶ푖” on both sides 휕퐹3 푞푖 = − (7)a 휕푝푖 휕퐹3 푃푖 = − (7)b 휕푄푖 휕퐹 And 퐾 푄 , 푃 , 푡 = 퐻 푞 , 푝 , 푡 + 3 (7)c 푖 푖 푖 푖 휕푡 16 6.1 Canonical Transformations Case IV: For 퐹4 푝푖, 푃푖, 푡 generating function 휕퐹1 휕퐹1 Since 푝푖 = & 푃푖 = − 휕푞푖 휕푄푖 Therefore, we can write 퐹1 푞푖, 푄푖, 푡 − σ 푝푖 푞푖 + σ 푃푖 푄푖 = 퐹4 푝푖, 푄푖, 푡 푑퐹 Since σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 1 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 푑 ⇨ σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 퐹 푝 , 푄 , 푡 + σ 푝 푞 − σ 푃 푄 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 4 푖 푖 푖 푖 푖 푖 푑퐹 ⇨ σ 푝 푞ሶ − 퐻 푞 , 푝 , 푡 = σ 푃 푄ሶ − 퐾 푄 , 푃 , 푡 + 4 + σ 푝 푞ሶ + σ 푝ሶ 푞 − σ 푃 푄ሶ − σ 푃ሶ 푄 푖 푖 푖 푖 푖 푖 푖 푖 푑푡 푖 푖 푖 푖 푖 푖 푖 푖 푑퐹 −퐻 푞 , 푝 , 푡 = −퐾 푄 , 푃 , 푡 + 4 + σ 푝ሶ 푞 − σ 푃ሶ 푄 푖 푖 푖 푖 푑푡 푖 푖 푖 푖 푑 휕퐹4 휕퐹4 휕퐹4 Since 퐹4 푝푖, 푃푖, 푡 = σ 푝푖ሶ + σ 푃푖ሶ + , putting in above equation.