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({qj}, {pj}, t)

Pi = Pi({qj}, {pj}, 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 canonical 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 variables. 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. if X ˙ X PiQi − K = λ (piq˙i − H) (7) i i known as the scale transformation or, X dF X P Q˙ − K + = p q˙ − H (8) i i dt i i i i known as the canonical transformation or by a combination of both X dF X P Q˙ − K + = λ (p q˙ − H) (9) i i dt i i i i which is usually referred to as extended canonical transformation If we rewrite (8) as X X pidqi − Hdt − PidQi + Kdt = dF (10) i i we notice that the difference between the two differential forms must be an exact differential. Analogous statement can be made regarding (9) as well. It turns out that this condition of exact differentiability is both necessary and sufficient condition for a transformation to be canonical.

2.1 Scale Transformation Scale transformation is achieved by multiplying the old coordinates and momenta with some scale factors to get the corresponding new quantities, Pi = νpi and Qi = µqi. The scale factors can be diﬀerent for momenta and the coordinates but must be the same for all coordinates (and all momenta). P and Q are obtainable from the new Hamiltonian K through Hamilton’s equations

˙ ∂K ˙ ∂K Pi = − Qi = ∂Qi ∂Pi c D. K. Ghosh, IIT Bombay 7

Thus we have, using Pi = νpi and Qi = µqi, ∂K νp˙i = − ∂µqi which gives 1 ∂K p˙i = − µν ∂qi ∂H Comparing this with the equationp ˙i = , we get K = µνH. This gives ∂qi X ˙ X PiQi − K(P, Q, t) = µν (piq˙i − H(p, q, t)) i i Comparing this with (7) we get λ = µν.

2.2 Generating Function

The function F in (8) or (9) is, in principle, a function of pi, qi,Pi,Qi and t. It is known as the generating function of canonical transformation. We will illustrate it with a couple of examples. We will use Einstein summation convention according to which a repeated P index in a single term implies a sum, i.e. figi = i figi. Example : Identity Transformation:

Consider a transformation F = qiPi − QiPi in which the dependence on time is implicit. We have dF P Q − K + = P Q − K +q ˙ P + q P˙ − Q˙ P − Q P˙ i i dt i i i i i i i i i i ˙ = −K + (qi − Qi)Pi + Piq˙i

≡ piq˙i − H

The equation is satisﬁed by identity transformation pi = Pi, qi = Qi and K = H. Example: Swapping Coordinates and Momenta We can generalize the above example a consider a generating function of the following type (summation convention used)

F = fi(q1, q2, . . . , qn; t)Pi − QiPi where fi is a function of the arguments q1, q2, . . . , qn and t. We then have ˙ dF ˙ ∂fi ∂fi ˙ ˙ ˙ PiQi − K + = PiQi − K + q˙jPi + Pi + fiPi − (QiPi + QiPi) dt ∂qj ∂t ˙ ∂fi ∂fi = −K + (fi − Qi)Pi + q˙jPi + Pi ∂qj ∂t

≡ piq˙i − H c D. K. Ghosh, IIT Bombay 8

∂f The set of equations above can be satisﬁed by taking Q = f , K = H + i P and i i ∂t i ∂fi requiring piq˙i = q˙jPi which can be rearranged as follows: ∂qj

∂fi ∂fj piq˙i = q˙jPi = q˙iPj ∂qj ∂qi where in the last term we have interchanged the sum over i and j. This gives

X ∂fj p = P i ∂q j j i where we have explicitly reintroduced the sum for clarity. These are n equations which must be inverted to get Pi. It may be noted that the choice of F is not unique corresponding to a particular canonical transformation. For instance, if we add any g(t) to dF dF dg F it dopes not aﬀect the action integral, because → + does not change the dt dt dt Lagrangian equations. Let us illustrate how to ﬁnd generator. Let K(Q, P, t) = H(q, p, t). Plugging this into (8), we get dF X = (p q˙ − P Q˙ ) dt i i i i i This is not very interesting because there is no time dependence. The simplest way to satisfy it is F = F (q, Q) We have dF X ∂F ∂F = q˙ + Q˙ dt ∂q i ∂Q i i i i X ˙ = (piq˙i − PiQi) i ∂F ∂F This gives with = p ; = −P . A trivial way to satisfy the equations is to take ∂q i ∂Q i P i i F (qi,Qi) = i qiQi so that ∂F = Qi = pi ∂qi ∂F = qi = −Pi ∂Qi which simply requires us to swap the coordinates and the momenta in the Hamiltonian formalism (but for a sign). c D. K. Ghosh, IIT Bombay 9

3 Diﬀerent types of Generating Functions

We have remarked that the generating function F is a function of the old coordinates (and momenta) q, p, the new coordinates Q, P and time t. However, since Q and P themselves are functions of q and p (and vice-versa) , only two of the four variables are needed in the description of the generating function. We need one each from the old pair and the new pair to describe the generating function, in addition to time. There are four possibilities and depending on the combination that we choose, the generating functions are classiﬁed as type 1, 2, 3 or 4.

3.1 Type 1 Generator

Type 1 generator, denoted by F1 is a function of qi and Qi. We denote

F = F1(qi,Qi, t) (11)

Since F1 = F1(q, Q, t), we can write

dF ∂F1 ∂F1 ˙ ∂F1 = q˙i + Qi + dt ∂qi ∂Qi ∂t X X ˙ ≡ piq˙i − PiQi + K − H (12) i dF where the last expression is from (8). We can compare the last two expressions for 1 dt to get

∂F1 pi = ∂qi ∂F1 Pi = − ∂Qi ∂F K = H + 1 (13) ∂t Note that (12) can be rewritten as

dF1 = pidqi − PidQi + (K − H)dt (14) If the Hamiltonian has no explicit time dependence, the generating function will also not have an explicit time dependence and in such a case K = H. In such a situation (8) can be written as (summation convention used) dF = p q˙ − P Q˙ dt i i i i For instantaneous transformation, we can write the above as

dF = piδqi − PiδQi c D. K. Ghosh, IIT Bombay 10

Since the left hand side is a perfect diﬀerential, so is the right hand side. In order that this may be so, we require ∂p ∂P i = i (15) ∂Qi ∂qi When the functional dependence are given, one can always use (15) to test whether the transformation is canonical.

As shown earlier, one of the simplest examples of a F1 type transformation is F1(qi,Qi) = qiQi, which gives the swapping transformation pi = Qi and Pi = −qi. Let us see what it p2 1 does to the Hamiltonian of the Harmonic oscillator for which H = + kq2. Since the 2m 2 Hamiltonian has no explicit time dependence, we have K = H. In such a case

Q2 1 K = + kP 2 2m 2 The new equations of motion are ∂K Q˙ = = kP ∂P ∂K Q P˙ = − = − ∂Q m Thus we get Q Q¨ = kP¨ = − m Q˙ which has the solution Q = Q cos(ωt + δ) . We can obtain P from the equation P = 0 k Q ω which gives P = − 0 sin(ωt+δ). Since this is equal to −q, we can identify q = Q ω/k. k 0 0 We then can write 0 q = q0 cos(ωt + δ ) where δ0 − δ = π/2.

3.2 Type 2 Generator

Type 2 generators, donoted by F2(q, P, t) can be obtained from the type 1 generator F1(q, Q, t) by means of a Legendre transformation. Since we have, in view of (13)

∂F1 Pi = − ∂Qi and we wish to replace the variable Qi by the new variable Pi, we deﬁne

X F2(qi,Pi, t) = F1(qi,Qi, t) + PiQi (16) i c D. K. Ghosh, IIT Bombay 11

Note that starting with the equation (14) for F1 can be written (use summation convention) by observing dF 1 = p q˙ − P Q˙ + K − H (17) dt i i i i Using (16), we have dF dF 2 = 1 + P Q˙ + P˙ Q dt dt i i i i ˙ = piq˙i + PiQi + (K − H) (18)

Since F2 depends on q and P , we can write

dF2 ∂F2 ∂F2 ˙ ∂F2 = q˙i + Pi + dt dt ∂qi ∂Pi ∂t Comparing it with (17), we get

∂F2 = pi ∂qi ∂F2 = Qi ∂Pi ∂F 2 = K − H (19) ∂t 3.3 Type 3 Generator

In a similar way as above, we can deﬁne F3(p, Q, t) through a Legendre transform on F1. ∂F1 Referring to (13), we have, since the variable qi is being replaced by pi, and, pi = ∂qi

F3(p, Q, t) = F1(q, Q, t) − piqi We then have dF dF 3 = 1 − p q˙ − q p˙ dt dt i i i i ˙ = −qip˙i − PiQi + (K − H) (20)

On the other hand, since F3 depends on (pi,Qi, t), we can write,

dF3 ∂F3 ∂F3 ˙ ∂F3 = p˙i + Qi + dt ∂pi ∂Qi ∂t Comparing this with (20) we get

∂F3 = −Pi ∂Qi ∂F3 = −qi ∂pi ∂F 3 = K − H (21) ∂t c D. K. Ghosh, IIT Bombay 12

3.4 Type 4 Generator

Finally we would like to deﬁne F4(p, P, t). In this case it is better to start with F2 or F3 and do a Legendre transformation. Starting with F3, we have to replace Qi with Pi. In view of (21), we have

F4(p, P, t) = F3(p, Q, t) − QiPi We then have dF dF 4 = 3 + Q P˙ + P Q˙ dt dt i i i i ˙ = −qip˙i + QiPi + (K − H)

Since F4 = F4(p, P, t), we have

dF4 ∂F4 ∂F4 ˙ ∂F4 = p˙i + Pi + dt ∂pi ∂Pi ∂t Comparing with the preceding relation, we have we get

∂F4 = −qi ∂pi ∂F4 = Qi ∂Pi ∂F 4 = K − H (22) ∂t

3.5 Summary of the diﬀerent types of Generators The following table gives a summary of relationships for diﬀerent types of generators.

Generators Derivatives Relationship

∂F1 ∂F1 ∂pi ∂Pj F1(q, Q, t) pi = ; Pi = − = − ∂qi ∂Qi ∂Qj ∂qi

∂F2 ∂F2 ∂pi ∂Qj F2(q, P, t) pi = ; Qi = = ∂qi ∂Pi ∂Pj ∂qi

∂F3 ∂F3 ∂Pi ∂qj F3(p, Q, t) Pi = − ; qi = − = ∂Qi ∂pi ∂pj ∂Qi

∂F4 ∂F4 ∂Qi ∂qj F4(p, P, t) qi = − ; Qi = = − ∂pi ∂Pi ∂pj ∂pi c D. K. Ghosh, IIT Bombay 13

4 Invariance of Poisson’s Bracket

A canonical transformation preserves the Poisson’s bracket. This property can be used to verify whether a transformation is indeed canonical. Let (q, p) → (Q, P ) be a transformation. The set (q, p) satisﬁes Poisson’s bracket relationship, i.e.

{qi, pj} = δij (23)

What we need to show is that Poisson’s bracket calculated using either basis is the same, i.e.,

{Qi,Pj}QP = {Qi,Pj}qp (24)

Clearly, since {Qi}, {Pi} are canonically conjugate pairs, they satisfy

{Qi,Pj}QP = δij (25)

To prove (24), we note X ∂Qi ∂Pj ∂Qi ∂Pj {Qi,Pj}qp = · − · ∂qk ∂pk ∂pk ∂qk k X ∂Qi ∂qk ∂Qi ∂pk ∂Qi = · − · − = = δij (26) ∂qk ∂Qj ∂pk ∂Qj ∂Qj k In proving the above, we have used (15). In a similar way, we can prove that the Poisson’s brackets of the old variables are preserved in the new basis,

{qi, pj}qp = {qi, pj}QP In general, for any two dynamical variables X and Y , one can show that the Poisson’s bracket relationship remains the same in either basis

{X,Y }QP = {X,Y }qp (27)

5 Illustrative Problems

Example 1: Harmonic Oscillator The Hamiltonian for the Harmonic oscillator is p2 1 H(p, q) = + mω2q2. Let us carry out a transformation 2m 2 p = f(P ) cos Q f(P q = sin Q (28) mω f(P )2 This converts the Hamiltonian to a function of P alone with K = H = . Note that 2m the Hamiltonian is cyclic in Q so that P is constant. The problem is to ﬁnd f(P ) such c D. K. Ghosh, IIT Bombay 14 that the transformation in canonical.

Let us try an F1 type transformation for which,as per eqn. (13) ∂F p = 1 ∂q ∂F P = − 1 ∂Q ∂F K = H + 1 = H (29) ∂t the last equality because the Hamiltonian is time independent. From (28), we get,

∂F p = mωq cot Q = 1 ∂q Integrating, we get mωq2 F = cot Q 1 2 The momenta conjugate to Q is then given by ∂F P = − 1 ∂Q mωq2 = cosec2Q 2 p2 1 = 2mω cos2 Q which gives √ p = 2P mω cos Q √ f(P )2 so that f(P ) = 2P mω. This in turn gives the new Hamiltonian to be K = = ωP . 2m Thus the transformation (28) is equivalent to √ p = 2P mω cos Q r 2 q = sin Q mω Inverting, we get

1 P = (q2m2ω2 + p2) 2mω mωq Q = tan−1 p c D. K. Ghosh, IIT Bombay 15

Note that ∂Q mωp = ∂q p2 + m2ω2q2 ∂P p = ∂p mω ∂Q mωq = − ∂p p2 + m2ω2q2 ∂P = qmω ∂q which shows that

{Q, P }qp = 1 i.e. as expected, the transformation preserves Poisson’s bracket. E As the energy is constant we have P = . The Hamilton’s equations for the new ω variables then gives ∂K Q˙ = = ω ∂P yielding Q = ωt + α The original problem is now solved, √ √ p = 2mE cos Q = 2mE cos(ωt + α) r r 2P 2E q = sin Q = sin(ωt + α) mω mω2 Example 2: 1 1 Consider the Hamiltonian H = (p2 +q2). Determine if the transformation Q = (q2 +p2) 2 2 −1 and P = − tan (q/p) is canonical. If so, ﬁnd a generating function of type F1.

In order that the given transformation is canonical, pδq −P δQ = dF must be an exact diﬀerential. Note that q q2 + p2 pδq − P δQ = pδq + tan−1 δ( ) p 2 q q = p + q tan−1 δq + p tan−1 δp p p

In order that the above is an exact diﬀerential,

∂ q ∂ q p + q tan−1 = p tan−1 ∂p p ∂q p

p2 It can be easily seen that both sides of the above condition equals so that the p2 + q2 transformation is canonical. c D. K. Ghosh, IIT Bombay 16

The Poisson’s brackets can be shown to be preserved:

∂ 1 ∂ q ∂ 1 ∂ q {Q, P } = (q2 + p2) − tan−1 − (q2 + p2) − tan−1 ∂q 2 ∂p p ∂p 2 ∂q p (q/p2) 1/p = q − p − = 1 1 + q2/p2 1 + q2/p2

We can obtain F by integrating either of the factors. For instance, a bit of exercise in integration gives

Z q F = p + q tan−1 dq p pq q2 + p2 q = + tan−1 2 2 p However, the above form cannot be a generating function as it has no dependence on the new variables. Since we wish it to be of form F , we need to eliminate p. This is done q 1 q q by substitution p = p2Q − q2 so that tan−1 = tan−1 = sin−1 √ . With p p2Q − q2 2Q these substitutions, we get

q 1 p F = Q sin−1 √ + q 2Q − q2 2Q 2 1 ∂K The Hamiltonian K = (q2 + p2) = Q, so that Q˙ = = 0, i.e. Q is constant, and 2 ∂P ∂K P˙ = − = −1. Since Q is constant, so is the Hamiltonian and P decreases with time. ∂Q Example 3: sin p Prove that the transformation Q = ln and P = q cot p is canonical and derive all the q diﬀerent forms of generating functions. We have, for the given teansformation 1 δQ = cot pδp − δq q Thus 1 pδq − P δQ = pδq − q cot p cot p δp − δq q = (p + cot p) δq − q cot2 p δp

This is an exact diﬀerential if ∂ ∂ (p + cot p) = (−q cot2 p) ∂p ∂q c D. K. Ghosh, IIT Bombay 17

It can be seen that both sides equal − cot2 p and hence the transformation is canonical. It is also easy to see that Poisson’s bracket is preserved.

The structure of generating function is Z F = (p + cot p)dq = pq + q cot p

(We can also integrate the other factor over p and show that both the forms are the same if the constant integration is taken to be zero). We will now express this in terms of diﬀerent forms of the generating function. Note that for the given transformation, sin p = qeQ which gives cos p = p1 − q2e2Q and q cot p = pe−2Q − q2. Substituting these, we get −1 p 2 2Q p −2Q 2 F1 = q cos 1 − q e + e − q

Suppose we wish to ﬁnd the generator form F2. In such a case, we need to replace p by P , which is quite straightforward. We can ﬁnd F2 by a Legendre transformation. Since ∂F P = − 1 , we have F (q, P ) = F (q, Q) + PQ. Thus ∂Q 2 1

1 p F (q, P ) = F (q, Q) = q sin−1(qeQ) + 1 − q2e2Q + PQ 2 1 eQ

However, as F2 must be expressed as a function of q and P , we must accordingly express p1 − q2e2Q 1 the variables. We had shown that P = , which gives eQ = . Thus eQ pP 2 + q2 ! −1 q P 2 2 F2(q, P ) = q sin + P − ln(P + q ) pP 2 + q2 2