Physics 31600 R. Wald Classical Mechanics Autumn, 2002

Problem Set VII Solutions

1) (a) Show that the transformation on 2n-dimensional phase space associated with a coordinate transformation on configuration space, namely:

q_{i} → Q_{i}(q)

p_{i} → P_{i}(q, p) = ∑_{j} p_{j} \frac{∂q_{j}}{∂Q_{i}}

is a canonical transformation.

Solution: As was shown in class, in order to show that a transformation is canonical it is sufficient to demonstrate that the new coordinates satisfy the Poisson bracket as functions of the old coordinates. We first compute the easy bracket:

Ω^{(p,q)} (Q_{i}, Q_{j}) = ∑_{k=1}^{n} (\frac{∂Q_{i}}{∂q_{k}} \frac{∂Q_{j}}{∂p_{k}} − \frac{∂Q_{i}}{∂p_{k}} \frac{∂Q_{j}}{∂q_{k}} ) = 0

since the Q_{i} are independent of p_{i}. Next up are the mixed P Q brackets:

Ω^{(p,q)} (Q_{i}, P_{j}) = ∑_{k=1}^{n} (\frac{∂Q_{i}}{∂q_{k}} \frac{∂P_{j}}{∂p_{k}} − \frac{∂Q_{i}}{∂p_{k}} \frac{∂P_{j}}{∂q_{k}} ) = ∑_{k=1}^{n} \frac{∂Q_{i}}{∂q_{k}} \frac{∂q_{k}}{∂Q_{j}} − 0 = \frac{∂Q_{i}}{∂Q_{j}} = δ_{ij},

as desired. The last equality follows from the fact that the Q are good coordinates on configuration space so they must obey that relation. Finally, we need to do the P brackets, which are bit of mess:

Ω^{(p,q)} (P_{i}, P_{j}) = ∑_{k=1}^{n} (\frac{∂P_{i}}{∂q_{k}} \frac{∂P_{j}}{∂p_{k}} − \frac{∂P_{i}}{∂p_{k}} \frac{∂P_{j}}{∂q_{k}} ) = ∑_{k=1}^{n} ((∑_{l=1}^{n} p_{l} \frac{∂^{2}q_{l}}{∂q_{k}∂Q_{i}} ) \frac{∂q_{k}}{∂Q_{j}} − \frac{∂q_{k}}{∂Q_{i}} (∑_{l=1}^{n} p_{l} \frac{∂^{2}q_{l}}{∂q_{k}∂Q_{j}} )) .

Notice that in the terms involving second partial derivatives, we cannot blithely interchange order of differentiation because different things are being held fixed. For the q derivatives, the other q′s are being held fixed, while for the Q derivative it is the other Q′s. Now we interchange the order of summation and note that

∑_{k=1}^{n} \frac{∂^{2}q_{l}}{∂q_{k}∂Q_{i}} \frac{∂q_{k}}{∂Q_{j}} = \frac{∂^{2}q_{l}}{∂Q_{j}∂Q_{i}} .

Since these are partial derivatives where the same variables are being held fixed, we can just interchange order of differentiation to yield

∑_{l=1}^{n} p_{l} (\frac{∂^{2}q_{l}}{∂Q_{j}∂Q_{i}} − \frac{∂^{2}q_{l}}{∂Q_{i}∂Q_{j}} ) = 0,

as desired. ■

(b) On a 2-dimensional phase space, show that the transformation

q → Q = ln[\frac{1}{q} sin p]

p → P = q cot p

is canonical.

Solution: Again, we simply need to check that the new coordinates satisfy the canonical commutation relations as functions of the old coordinates. The QQ and P P brackets are trivial because the Poisson bracket is antisymmetric and so vanishes for any function of p and q. Thus, we just need to check the QP bracket, which is

Ω^{(p,q)} (Q, P) = \frac{∂Q}{∂q} \frac{∂P}{∂p} − \frac{∂Q}{∂p} \frac{∂P}{∂q} = [\frac{−1}{q}] [q(− csc^{2} p)] − [\frac{cos p}{sin p}] [cot q] = \frac{1 − cos^{2} p}{sin^{2} p} = 1,

as desired. ■

2) Give an “elementary” proof of Liouville’s theorem as follows: Introduce local canonical coordinates (q_{1}, p_{1}, q_{2}, p_{2}, ..., q_{n}, p_{n}) and pretend that these are Cartesian coordinates in 2ndimensional Euclidean space, R^{2n}. Consider a bounded region, R, of phase space covered by these coordinates. Let R_{t} denote the image of R under dynamical evolution by time t. Argue that the volume, V (R_{t}), of R_{t} must satisfy

\frac{dV}{dt} = ∫_{∂R_{t}} h⃗ · n̂dS

where ∂R_{t} denotes the boundary of R_{t}, h⃗ denotes the Hamiltonian vector field, and n̂ denotes the unit outward pointing normal to ∂R_{t}. Then show that dV/dt = 0.

Solution: Consider a volume R_{t} at a particular time in phase space. Corresponding to this volume we can construct an ensemble of particles wherein each point in phase space (q_{1}, p_{1}, q_{2}, p_{2}, ..., q_{n}, p_{n}) contained within the volume corresponds to the state of precisely one particle in the ensemble. In this case the volume is equal to the number of particles in the ensemble. If we wish to know the rate of change of the number of particles within this volume at time t, we must calculate the flux of the particles across the boundary of the volume of phase space. That is:

\frac{dV}{dt} = ∫_{∂R_{t}} ˙ x⃗ · n̂dS  (1)

where x⃗ is the coordinate vector in phase space.

But, Hamilton’s equations of motion are precisely ˙ x⃗ = h⃗, so we have the result:

\frac{dV}{dt} = ∫_{∂R_{t}} h⃗ · n̂dS  (2)

Furthermore, the divergence theorem allows us to write this as:

\frac{dV}{dt} = ∫_{∂R_{t}} h⃗ · n̂dS = ∫_{R_{t}} ∇ · h⃗dx⃗  (3)

However, we have:

∇ · h⃗ = ∑_{i} \frac{∂}{∂q_{i}} \frac{∂H}{∂p_{i}} − \frac{∂}{∂p_{i}} \frac{∂H}{∂q_{i}} = 0  (4)

Therefore \frac{dV}{dt} = 0

3) (a) Find the generating function, F_{2}(q, P), for the canonical transformation of problem

1(a) above.

Solution: F_{2} is given by F_{2} = F_{1} + ∑_{j} P_{j}Q_{j}. Thus, we first need to find F_{1} and then we can simply compute F_{2}. Now F_{1} satisfies

dF_{1} = ∑_{j} (p_{j}dq_{j} − P_{j}dQ_{j}) = ∑_{j} [p_{j}dq_{j} − (∑_{k} p_{k} (\frac{∂q_{k}}{∂Q_{j}} )_{\{Q,p\}}) (∑_{l} (\frac{∂Q_{j}}{∂q_{i}} )_{\{q,p\}} dq_{i})]

For fixed k and i, we have that

∑_{j} (\frac{∂Q_{j}(q)}{∂q_{i}})_{\{q,p\}} (\frac{∂q_{k}(q)}{∂Q_{j}})_{\{Q,p\}} = (\frac{∂q_{k}}{∂q_{i}} )_{\{q,p\}} = δ_{ik},

so that we get

dF_{1} = ∑_{j} [p_{j}dq_{j}] − ∑_{k,l} [p_{k}δ_{ik}dq_{i}] = ∑_{j} [p_{j}dq_{j}] − ∑_{k} [p_{k}dq_{k}] = 0.

Thus, we can take F_{1}(q, Q) = 0. Why can we take the generating function to be constant (which means that p and P are equal to zero)? Because q and Q do not form good coordinates on phase space; they only cover configuration space. Thus, we have the simple solution

F_{2}(q, P) = ∑_{j} Q_{j}(q)P_{j},

where we must explicitly insert the expression for Q_{j} in terms of q. ■

(b) Find the generating functions, F_{1}(q, Q) and F_{2}(q, P), for the canonical transformation of problem 1(b) above.

Solution: Again, we start we the formula

dF_{1}(q, Q) = pdq − P dQ.

Since we want F_{1} in terms of Q and q, we solve for p and P in terms of Q and q:

p = sin^{−1} (qe^{Q}) ; P = q cot p = \frac{\sqrt{1 − q^{2}e^{2Q}}}{e^{2Q}}

so that

dF_{1} = sin^{−1} (qe^{Q}) dq − \frac{\sqrt{1 − q^{2}e^{2Q}}}{e^{2Q}} dQ.

Integrating the coefficient of dq (which is \frac{∂F_{1}}{∂q}) with respect to q we get

F_{1} = q sin^{−1} (qe^{Q}) + \frac{\sqrt{1 − q^{2}e^{2Q}}}{e^{Q}} + f(Q).

Simply differentiating the above with respect to Q and comparing with the coefficient of dQ shows we can take f(Q) = 0, so we find that

F_{1} = q sin^{−1} (qe^{Q}) + \frac{\sqrt{1 − q^{2}e^{2Q}}}{e^{Q}} .

Now we know that F_{2}(q, P) = F_{1}(q, Q(q, P)) + P Q(q, P), so all we need to do is solve for Q in terms of q, P and substitute into the above expression. Using the above expression for P, we find that

P = \frac{\sqrt{1 − q^{2}e^{2Q}}}{e^{Q}} ⇒ Q = −\frac{1}{2} ln (P^{2} + q^{2})

which gives that

F_{2}(q, P) = q sin^{−1} [\frac{q}{\sqrt{P^{2} + q^{2}}} ] + P − \frac{P}{2} ln (P^{2} + q^{2}) = q cot^{−1} \frac{P}{q} + P − \frac{P}{2} ln(P^{2} + q^{2}) ■

(c) Explicitly choose a function of two variables, f(x, y). Then obtain the canonical transformations on a 2-dimensional phase space that it generates via the generating functions (i) F_{1}(q, Q) = f(q, Q) and F_{2}(q, P) = f(q, P).

Solution: Clearly, your results will depend on what function you choose. I will demonstrate with two different examples. I will start with my favourite function on ℝ^{2}, namely f(x, y) = x + y. Then

F_{1}(q, Q) = q + Q ⇒ p = \frac{∂F_{1}}{∂q} = 1 ; P = −\frac{∂F_{1}}{∂Q} = −1

F_{2}(q, P) = q + P ⇒ p = \frac{∂F_{1}}{∂q} = 1 ; Q = \frac{∂F_{1}}{∂P} = 1.

This illustrates an important point: even though any function will generate a “canonical transformation”, there is no guarantee that the resulting functions will be good coordinates on phase space even if the generating function satisfies every nice property of which you can think. Next up is slightly more complicated function, f(x, y) = (x + y)^{2}. Then

F_{1}(q, Q) = q^{2} + +2Qq + Q^{2} ⇒ p = \frac{∂F_{1}}{∂q} = 2(Q + q) ; P = −\frac{∂F_{1}}{∂Q} = 2(Q + q).

Solving for P and Q in terms of p and q gives q = Q and p = P. This is nothing but the identity transformation. For F_{2} we get

F_{2}(q, P) = q^{2} + 2qP + P^{2} ⇒ p = \frac{∂F_{1}}{∂q} = 2(q + P) ; Q = \frac{∂F_{1}}{∂P} = 2(q + P).

Solving for P and Q again gives Q = p and P = p − 2q. So this transformation turns the momenta into the coordinates and then does a little rearranging on the new momenta. ■