05.02.E1 · symplectic / hamiltonian

Hamiltonian mechanics and canonical transformations exercise pack (Arnold Part III supplement)

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Formal definition of the pack Intermediate

Arnold's Part III recasts mechanics on the cotangent bundle $T^{*} M$ with its canonical symplectic form $ω = d p \land d q = - d θ$ , $θ = p d q$ . A Hamiltonian $H : T^{*} M \to R$ defines the vector field $X_{H}$ by $ι_{X_{H}} ω = d H$ , whose flow is Hamilton's equations $\overset{q}{˙} = \partial_{p} H$ , $\overset{p}{˙} = - \partial_{q} H$ . The Poisson bracket ${f, g} = ω (X_{f}, X_{g})$ controls the time evolution of observables; a transformation is canonical exactly when it preserves $ω$ , and generating functions parametrise such transformations. The Hamilton-Jacobi equation $\partial_{t} S + H (q, \partial_{q} S, t) = 0$ uses a single generating function to integrate the whole flow.

This pack collects nine exercises — two easy, four medium, three hard — each with a hint and a full solution. It is meant to be read alongside its prerequisite units 05.02.01, 05.02.02, 05.02.05, 05.05.03, 05.05.04, and 05.02.07, not as a standalone development. The problems are grouped by Arnold section: Hamilton's equations and Poisson-bracket computations (easy/medium), canonicity tests and the four generating-function types (medium), and Hamilton-Jacobi separation of variables (hard).

The conventions throughout are Arnold's: $ω = d p \land d q$ (so $\overset{q}{˙} = \partial_{p} H$ , $\overset{p}{˙} = - \partial_{q} H$ ), ${f, g} = \partial_{p} f \partial_{q} g - \partial_{q} f \partial_{p} g$ in one degree of freedom with the sign giving $\dot{f} = {H, f}$ , and the four generating functions $S_{1} (q, Q)$ , $S_{2} (q, P)$ , $S_{3} (p, Q)$ , $S_{4} (p, P)$ in Goldstein's labelling.

Key theorem with full solution Intermediate

Before the pack proper, we work one exercise in full as an exemplar of the format. The remaining eight follow the same structure (problem, hint, full answer in <details> blocks).

Lead exercise. Show that a transformation $(q, p) \mapsto (Q, P)$ generated by $S_{2} (q, P)$ via $p = \partial_{q} S_{2}$ , $Q = \partial_{P} S_{2}$ is canonical, and verify on $S_{2} = q P$ (the identity) and $S_{2} = \frac{1}{2} (q^{2} + P^{2}) cot$ ... — take $S_{2} = f (q) P$ .

Solution. A type-2 generating function $S_{2} (q, P)$ defines the transformation implicitly by

p = \frac{\partial S _{2}}{\partial q}, Q = \frac{\partial S _{2}}{\partial P} .

To see canonicity, recall the defining identity of generating functions: $p d q - P d Q = d S_{2} - d (P Q)$ , equivalently $p d q + Q d P = d S_{2}$ . Taking the exterior derivative of $p d q + Q d P = d S_{2}$ and using $d^{2} = 0$ :

d p \land d q + d Q \land d P = 0 ⟹ d p \land d q = d P \land d Q .

That is exactly $ω = d p \land d q = d P \land d Q$ — the transformation pulls the canonical 2-form back to itself, so it is a symplectomorphism, i.e. canonical.

Now $S_{2} = f (q) P$ for a diffeomorphism $f$ . Then $p = \partial_{q} S_{2} = f^{'} (q) P$ and $Q = \partial_{P} S_{2} = f (q)$ . So the point transformation is $Q = f (q)$ on the base, with momentum $P = p / f^{'} (q)$ (solving $p = f^{'} (q) P$ ). Check directly:

d Q = f^{'} (q) d q, d P = d (\frac{p}{f ^{'} ( q )}) = \frac{d p}{f ^{'} ( q )} - \frac{p f ^{''} ( q )}{f ^{'} ( q ) ^{2}} d q .

Then $d P \land d Q = (\frac{d p}{f ^{'}} - \frac{p f ^{''}}{( f ^{'} ) ^{2}} d q) \land f^{'} d q = d p \land d q$ (the $d q \land d q$ term vanishes). So $S_{2} = q P$ gives the identity ( $f = id$ ), and any base diffeomorphism $f$ lifts canonically to $T^{*} M$ — the cotangent lift. $□$

This is the operational meaning of "generating function": it produces a canonical transformation for free, with canonicity guaranteed by $d^{2} = 0$ . Every generating-function exercise downstream runs this argument with a different $S$ .

Exercises Intermediate

Exercise 3 (medium, proof). Jacobi identity gives a conserved bracket.

Suppose $f$ and $g$ are both conserved along the Hamiltonian flow of $H$ (so ${f, H} = {g, H} = 0$ ). Use the Jacobi identity to show ${f, g}$ is also conserved. This is Poisson's theorem.

Hint

The Jacobi identity is ${{f, g}, H} + {{g, H}, f} + {{H, f}, g} = 0$ .

Answer

A function $u$ is conserved iff ${u, H} = 0$ (since $\overset{u}{˙} = {u, H}$ ). Apply the Jacobi identity to $f, g, H$ :

{{f, g}, H} + {{g, H}, f} + {{H, f}, g} = 0.

By hypothesis ${g, H} = 0$ and ${H, f} = - {f, H} = 0$ , so the last two terms vanish, leaving

{{f, g}, H} = 0.

Therefore ${f, g}$ is conserved. This is Poisson's theorem: the Poisson bracket of two integrals of motion is again an integral of motion. It is the mechanism by which a symmetry algebra (e.g. the $so (3)$ angular-momentum brackets ${L_{i}, L_{j}} = ϵ_{ij k} L_{k}$ ) closes, and it sometimes manufactures new conserved quantities. Arnold §40.

Exercise 4 (medium, symbolic). Test a transformation for canonicity.

Decide whether $(q, p) \mapsto (Q, P) = (q cos α - p sin α, q sin α + p cos α)$ — rotation by a fixed angle $α$ in phase space — is canonical.

Hint

Compute $d P \land d Q$ and compare to $d p \land d q$ . Equivalently check ${Q, P} = 1$ .

Answer

Compute the Poisson bracket ${Q, P}$ in the original coordinates:

{Q, P} = \partial_{q} Q \partial_{p} P - \partial_{p} Q \partial_{q} P .

Here $\partial_{q} Q = cos α$ , $\partial_{p} Q = - sin α$ , $\partial_{q} P = sin α$ , $\partial_{p} P = cos α$ . So

{Q, P} = cos α \cdot cos α - (- sin α) \cdot sin α = cos^{2} α + sin^{2} α = 1.

Since ${Q, P} = 1 = {q, p}$ , the transformation preserves the fundamental bracket, hence is canonical. Equivalently $d P \land d Q = d p \land d q$ because the Jacobian is a rotation, with determinant $1$ . Phase-space rotations are canonical — they are the time- $α$ flow of the harmonic-oscillator Hamiltonian $\frac{1}{2} (q^{2} + p^{2})$ . Arnold §45.

Exercise 5 (medium, symbolic). The $S_{1}$ generating function $q Q$ swaps coordinate and momentum.

For the type-1 generating function $S_{1} (q, Q) = q Q$ , with relations $p = \partial_{q} S_{1}$ , $P = - \partial_{Q} S_{1}$ , find the resulting canonical transformation and interpret it.

Hint

Compute $p$ and $P$ from $S_{1} = q Q$ , then solve for $(Q, P)$ in terms of $(q, p)$ .

Answer

For $S_{1} (q, Q) = q Q$ :

p = \frac{\partial S _{1}}{\partial q} = Q, P = - \frac{\partial S _{1}}{\partial Q} = - q .

So $Q = p$ and $P = - q$ . The transformation is

(q, p) ⟼ (Q, P) = (p, - q) .

It interchanges coordinate and momentum (up to a sign), demonstrating that in Hamiltonian mechanics positions and momenta are on equal footing — there is no intrinsic distinction, only a symplectic pairing. Canonicity check: $d P \land d Q = d (- q) \land d (p) = - d q \land d p = d p \land d q = ω$ . This $S_{1} = q Q$ is the standard example showing that type-1 generators can produce transformations no point (configuration-space) transformation could. Arnold §45-§46.

Exercise 6 (medium, numeric). Liouville volume preservation for a linear flow.

The Hamiltonian flow of $H = \frac{1}{2} (p^{2} + ω_{0}^{2} q^{2})$ is linear in $(q, p)$ . Without integrating, argue that it preserves phase-space area, and confirm by computing the trace of the linearised generator.

Hint

Liouville's theorem: every Hamiltonian flow preserves $ω^{n} / n!$ . For a linear flow $\overset{z}{˙} = A z$ the volume is preserved iff $tr A = 0$ .

Answer

Hamilton's equations give $\overset{q}{˙} = p$ , $\overset{p}{˙} = - ω_{0}^{2} q$ , i.e. $\overset{z}{˙} = A z$ with $z = (q, p)^{⊤}$ and

A = (0 - ω_{0}^{2} 10) .

A linear flow $e^{t A}$ has Jacobian determinant $det e^{t A} = e^{t tr A}$ . Here $tr A = 0 + 0 = 0$ , so $det e^{t A} = 1$ for all $t$ : the flow preserves area. This is the special case of Liouville's theorem 05.02.07 — every Hamiltonian generator has trace zero because $A = J Hess (H)$ with $J$ the symplectic matrix, and $tr (J Hess (H)) = 0$ by antisymmetry of $J$ against the symmetric Hessian. Hamiltonian flows are incompressible. Arnold §38.

Exercise 7 (hard, symbolic). Harmonic oscillator by Hamilton-Jacobi.

Solve the harmonic oscillator $H = \frac{1}{2} (p^{2} + ω_{0}^{2} q^{2})$ by the time-independent Hamilton-Jacobi equation: find a complete integral $W (q; α)$ of $H (q, \partial_{q} W) = α$ , and identify the resulting action variable.

Hint

Set $p = \partial_{q} W$ and solve $\frac{1}{2} ((\partial_{q} W)^{2} + ω_{0}^{2} q^{2}) = α$ for $\partial_{q} W$ , then integrate. The action is $I = \frac{1}{2 π} \oint p d q$ .

Answer

The time-independent Hamilton-Jacobi equation is $H (q, \partial_{q} W) = α$ , i.e.

\frac{1}{2} ((\partial_{q} W)^{2} + ω_{0}^{2} q^{2}) = α ⟹ \partial_{q} W = 2 α - ω_{0}^{2} q^{2} .

Integrating gives the complete integral $W (q; α) = \int 2 α - ω_{0}^{2} q^{2} d q$ , depending on the single parameter $α$ (the energy). The full action is $S = W - α t$ ; the new coordinate $β = \partial_{α} W - t$ is constant along the motion, which integrates the flow by quadrature.

The action variable is the loop integral over one period, where $q$ runs between the turning points $\pm 2 α / ω_{0}$ :

I = \frac{1}{2 π} \oint p d q = \frac{1}{2 π} \oint 2 α - ω_{0}^{2} q^{2} d q = \frac{1}{2 π} \cdot \frac{2 π α}{ω _{0}} = \frac{α}{ω _{0}} .

(The loop integral is the area $π ab$ of the ellipse with semi-axes $a = 2 α / ω_{0}$ , $b = 2 α$ , giving $π \cdot 2 α / ω_{0}$ .) So $α = ω_{0} I$ , that is $H = ω_{0} I$ in action-angle variables: the oscillator energy is linear in the action with frequency $ω_{0} = \partial H / \partial I$ . Arnold §47-§50, units 05.05.04, 05.02.04.

Exercise 8 (hard, symbolic). Separation of variables for the Kepler problem.

For the planar Kepler Hamiltonian $H = \frac{1}{2} (p_{r}^{2} + p_{φ}^{2} / r^{2}) - k / r$ , show that the Hamilton-Jacobi equation separates as $W = W_{r} (r) + p_{φ} φ$ , and write the radial quadrature.

Hint

$φ$ is cyclic, so $\partial_{φ} W$ is a constant $= p_{φ}$ . Substitute the ansatz $W = W_{r} (r) + p_{φ} φ$ into $H (q, \partial_{q} W) = E$ .

Answer

The Hamilton-Jacobi equation $H (q, \partial_{q} W) = E$ is

\frac{1}{2} ((\partial_{r} W)^{2} + \frac{( \partial _{φ} W ) ^{2}}{r ^{2}}) - \frac{k}{r} = E .

Since $φ$ is cyclic ( $H$ has no explicit $φ$ ), try the additively separated ansatz $W = W_{r} (r) + W_{φ} (φ)$ . The $φ$ -dependence enters only through $(\partial_{φ} W)^{2} / r^{2}$ , so $\partial_{φ} W$ must be a constant; call it $p_{φ}$ (the conserved angular momentum). Then $W_{φ} = p_{φ} φ$ and the radial equation is

\frac{1}{2} ((\partial_{r} W_{r})^{2} + \frac{p _{φ}^{2}}{r ^{2}}) - \frac{k}{r} = E .

Solving for $\partial_{r} W_{r}$ and integrating gives the radial quadrature:

W_{r} (r) = \int 2 E + \frac{2 k}{r} - \frac{p _{φ}^{2}}{r ^{2}} d r .

So $W (r, φ; E, p_{φ}) = W_{r} (r) + p_{φ} φ$ is a complete integral depending on the two parameters $(E, p_{φ})$ — exactly $n = 2$ for two degrees of freedom, as Jacobi's theorem requires. The Kepler problem is integrated by quadratures; the integrals are the action variables yielding the orbit equation and Kepler's third law. Arnold §47-§48.

Exercise 9 (hard, proof). Time evolution is a canonical transformation.

Let $g^{t}$ be the time- $t$ flow of a Hamiltonian $H$ on $(T^{*} M, ω)$ . Prove that each $g^{t}$ is a symplectomorphism, $(g^{t})^{*} ω = ω$ , using Cartan's magic formula.

Hint

Show $\frac{d}{d t} (g^{t})^{*} ω = (g^{t})^{*} L_{X_{H}} ω$ , then compute $L_{X_{H}} ω$ with Cartan's formula $L_{X} = d ι_{X} + ι_{X} d$ .

Answer

The flow of a vector field satisfies $\frac{d}{d t} (g^{t})^{*} α = (g^{t})^{*} L_{X_{H}} α$ for any form $α$ . Apply this to $ω$ :

\frac{d}{d t} (g^{t})^{*} ω = (g^{t})^{*} L_{X_{H}} ω .

Compute $L_{X_{H}} ω$ by Cartan's magic formula $L_{X} = d ι_{X} + ι_{X} d$ :

L_{X_{H}} ω = d ι_{X_{H}} ω + ι_{X_{H}} d ω .

Now $d ω = 0$ ( $ω$ is closed), so the second term vanishes. The defining property of the Hamiltonian vector field is $ι_{X_{H}} ω = d H$ , so the first term is $d (d H) = d^{2} H = 0$ . Therefore $L_{X_{H}} ω = 0$ .

Hence $\frac{d}{d t} (g^{t})^{*} ω = 0$ , so $(g^{t})^{*} ω$ is constant in $t$ ; at $t = 0$ it equals $ω$ , giving $(g^{t})^{*} ω = ω$ for all $t$ . The Hamiltonian flow is a one-parameter family of canonical transformations. Wedging $n$ times, it also preserves $ω^{n} / n!$ — that is Liouville's theorem 05.02.07 as a corollary. Arnold §38, §45.

Exercise pack EP. Arnold Mathematical Methods of Classical Mechanics Part III supplement: Hamilton's equations, Poisson brackets, canonical transformations, the four generating functions, and Hamilton-Jacobi theory across §37-§48.

Prerequisites

05.02.01 pending
05.02.02 pending
05.02.05 pending
05.05.03 pending
05.05.04 pending
05.02.07 pending

Tier anchors

beginner
intermediate: Arnold *Mathematical Methods of Classical Mechanics* Part III exercises (§37-§48: Hamilton's equations, Poisson brackets, canonical transformations, generating functions, Hamilton-Jacobi); Goldstein *Classical Mechanics* Ch. 8-10 problems
master

References

Arnold — Mathematical Methods of Classical Mechanics (2nd ed., GTM 60) · Part III (Ch. 6-9) exercises: §37-§38 (Hamilton's equations, Liouville), §40 (Poisson brackets), §45-§46 (canonical transformations, generating functions), §47-§48 (Hamilton-Jacobi, separation of variables)
Goldstein, Poole, Safko — Classical Mechanics (3rd ed.) · Ch. 8-10 — Hamilton's equations, canonical transformations, the four generating functions, Hamilton-Jacobi theory
Cannas da Silva — Lectures on Symplectic Geometry · §18-§22 — Poisson brackets, Hamiltonian vector fields, symplectomorphisms

Estimated time

intermediate: 5h