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Classical Mechanics Examples (Canonical Transformation) Classical Mechanics Examples (Canonical Transformation) Dipan Kumar Ghosh Centre for Excellence in Basic Sciences Kalina, Mumbai 400098 November 10, 2016 1 Introduction In classical mechanics, there is no unique prescription for one to choose the generalized coordinates for a problem. As long as the coordinates and the corresponding momenta span the entire phase space, it becomes an acceptable set. However, it turns out in prac- tice that some choices are better than some others as they make a given problem simpler while still preserving the form of Hamilton's equations. Going over from one set of chosen coordinates and momenta to another set which satisfy Hamilton's equations is done by canonical transformation. For instance, if we consider the central force problem in two dimensions and choose Carte- k sian coordinate, the potential is − . However, if we choose (r; θ) coordinates, px2 + y2 k the potential is − and θ is cyclic, which greatly simplifies the problem. The number of r cyclic coordinates in a problem may depend on the choice of generalized coordinates. A cyclic coordinate results in a constant of motion which is its conjugate momentum. The example above where we replace one set of coordinates by another set is known as point transformation. In the Hamiltonian formalism, the coordinates and momenta are given equal status and the dynamics occurs in what we know as the phase space. In dealing with phase space dynamics, we need point transformations in the phase space where the new coordinates (Qi) and the new momenta (Pi) are functions of old coordinates (pi) and old momenta (pi): Qi = Qi(fqjg; fpjg; t) Pi = Pi(fqjg; fpjg; t) 1 c D. K. Ghosh, IIT Bombay 2 Such transformations are called contact transformations. However, in Hamiltonian mechanics, only those transformations are of interest for which the quantities Qi and Pi are canonically conjugate pairs. This means that there exists some Hamiltonian K(Q; P; t) with respect to which P and Q satisfy Hamilton's equations: @K _ Qi = @Pi @K _ Pi = − @Qi Note that Q and P are not specific to a particular mechanical system but are common to all systems having the same degree of freedom. We may, for instance, obtain P and Q for a plane harmonic oscillator and use the same set to solve Kepler's problem. Such transformations, which preserve the form of Hamilton's equations, are known as canon- ical transformation. Though the terms contact transformation is used synonymously with the term canonical transformation, not all contact transformations are canonical, as the following example shows. Let us consider a free particle system with the Hamiltonian p2 H = . Since q is cyclic, the conjugate momentum p is a constant of motion,p _ = 0, 2m p and we haveq _ = . Consider now, a contact transformation m P = pt; Q = qt (1) Does a Hamiltonian K(P; Q) exist for which P and Q satisfy the Hamilton's equations: @K P_ = − @Q @K Q_ = @P @P_ @Q_ Existence of such a Hamiltonian would imply = − as each of the expressions @P @Q @2K @P_ @Q_ 1 equals − . However, Since = = , this condition is not satisfied and the @P @Q @P @Q t modified Hamiltonian K does not exist. Thus the contact transformations (1) are non- canonical, they do not preserve the volume of the phase space. It may be noted that this does not imply that one cannot find an equation of motion using these variables. Indeed, since p is constant, dP=dt = p = P=t, which gives P = kt, where k is a constant. Likewise, we have, d Q dq p P t k = = = = dt t dt m m m c D. K. Ghosh, IIT Bombay 3 k so that Q = t2 + C, where C is a constant. m To illustrate a canonical transformation, consider the free particle Hamiltonian again for whichp _ = 0 andq _ = p=m, as before. Consider a linear transformation of (p; q) to a new set (P; Q), which is given by P = αp + βq Q = γp + δq (2) α β If the matrix of transformation is non-singular, i.e. if ∆ = 6= 0, then the above γ δ transformation is invertible and we have 1 p = (δP − βQ) ∆ 1 q = (−γP + αQ) (3) ∆ It can be checked that p β P_ = αp_ + βq_ = βq_ = β = (δP − βQ) m m∆ p δ Q_ = γp_ + δq_ = δq_ = δ = (δP − βQ) (4) m m∆ Is there a Hamiltonian K for which P and Q are canonical? It is seen that @P_ βδ = @P m∆ @Q_ βδ = − @Q m∆ @K so that K exists. To determine K note that P_ = − which gives @Q Z βδ β2 Q2 K = − P_ dQ = − PQ + + f(P ) m∆ m∆ 2 @K and Q_ = gives @P Z βδ δ2 P 2 K = − QdP_ = − PQ + + f(Q) m∆ m∆ 2 These two expressions give 1 K = (δP − βQ)2 2m∆ Except for the factor ∆, this is just the old Hamiltonian written in terms of the new vari- ables. For ∆ = 1, the two Hamiltonians are identical and the transformation corresponds c D. K. Ghosh, IIT Bombay 4 to a point transformation of Lagrangian mechanics. Consider a second example where, once again we take the free particle Hamiltonian of the previous example. Consider a transformation P = p cos q; Q = p sin q (5) The transformation is invertible and it is easily seen that p Q p = P 2 + Q2; q = tan−1 (6) P Using the same algebra as for the previous example, we get p2 Q p P_ =p _ cos q − qp_ sin q = − sin q = − P 2 + Q2 m m and P p Q_ = P 2 + Q2 m @P_ @Q_ It can be seen that = − , showing that the Hamiltonian exists. It is easy to show @P @Q 1 that the Hamiltonian is given by K = pP 2 + Q2. 3m However, instead of considering the free particle Hamiltonian, consider the Hamilto- nian of a particle in a uniform gravitational field. p2 H = + mgq 2m p In this casep _ = −mg andq _ = . We had seen that the transformation is invertible. In m this case, we get P_ =p _ cos q − qp_ sin q p2 = −mg cos q − sin q m P 1 Q = −mg − pP 2 + Q2 m pP 2 + Q2 and Q_ =p _ sin q +qp _ cos q Q p2 = −mg + cos q pP 2 + Q2 m Q 1 p = −mg − P P 2 + Q2 pP 2 + Q2 m c D. K. Ghosh, IIT Bombay 5 @P_ @Q_ One can check 6= − , so that the transformation is not canonical with respect to @P @Q the given Hamiltonian. This shows that (5) is not a valid canonical transformation. This is because, in order that a transformation may be canonical, it must satisfy Hamilton's equation for all systems having the same number of degrees of freedom. 2 Canonical Transformation & Generating Function We start with an observation that Hamilton's equations of motion can be derived from the Hamilton's principle by writing action written in the form S = R Ldt = R (pq_ − H)dt and optimizing the action. In deriving the Euler Lagrange equation, we had optimized action subject to the condition that δq(ti) = δq(tf ) = 0, i.e. there is no variation at the end points. We have Z tf δS = δ (piq_i − H)dt = 0 t0 with implied summation on repeated index and we have denoted the initial time as t0 so as not to confuse with the index i. The variation in the action is then given by Z tf @H @H δS = q_iδpi + piδq_i − δqi − δpi dt t0 @qi @pi Z tf @H d @H = q_i − δpi δpi + (piδqi) − p_iδqi − δqi dt t0 @pi dt @qi Z tf @H @H tf = q_i − δpi δpi − p_i + dt + [piδqi]t0 t0 @pi @qi We have seen earlier that setting δS = 0 for all variations of qi gave us Euler Lagrange equation and we expect that setting δS = 0 for all variations of qi and pi will lead to Hamilton's equations. However, there is a subtle difference between the two cases. In the Lagrangian formulation qi andq _i were not independent variables and hence we only needed variation with respect to qi. In the Hamiltonian formalism, however, we need to vary both qi and pi. THe last term of the above equation vanishes if δqi(t0) = δqi(tf ) = 0 and it does not requires the corresponding relation for vanishing of the momentum coordinates. Thus, we have an additional degree of freedom and we could define the action as Z dF (q; p) S = (p q_ − H + )dt i i dt so that for all variations of pi and qi, the Hamilton;'s equations would be valid. The variational principle requires the function pq_ − H to satisfy the Euler Lagrange equations. Thus we have d @ @ (pq_ − H) − (pq_ − H) = 0 dt @q_ @q d @ @ (pq_ − H) − (pq_ − H) = 0 dt @p_ @p c D. K. Ghosh, IIT Bombay 6 d @H @H Since H = H(p; q), we get from above, p + = 0, i.e.p _ = − and likewise, since dt @q @q @H @H pq_ − H no dependence onp _, −q_ + = 0, i.e.,q _ = . The variational porinciple @p @p satisfied by the old pair of variables (q; p) is Z δS = δ (pq_ − H)dt = 0 with no variation at the end points. With respect to the new variables, one then must have Z δS0 = δ (P Q_ − H)dt = 0 These two forms of S and S0 are completely equivalent if the two integrands differ either by a scale factor or by total time differential of a function F , i.e.
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