Hamilton: phase space, and flows that preserve volume

Idea. Use the definition, not the recipe: the recipe finds only the critical point, and two of the three regimes of part 1 have none.

1. \(p > 0\): \(px - e^x\) is strictly concave with its critical point at \(x = \ln p\), so \(f^*(p) = p\ln p - p\). \(p = 0\): \(\sup(-e^x) = 0\), approached as \(x\to-\infty\) and never attained; \(f^*(0) = 0\), which is also the limit of \(p\ln p - p\) as \(p\to0^+\). \(p < 0\): as \(x\to-\infty\), \(px\to+\infty\) and \(e^x\to0\), so \(f^*(p) = +\infty\). Geometrically: no line of negative slope lies below the exponential.

For \(f^{**}(x) = \sup_p\big(xp - f^*(p)\big)\), only \(p\ge0\) matter (the rest give \(-\infty\)). For \(p > 0\), \(\frac{\mathrm d}{\mathrm dp}\big(xp - p\ln p + p\big) = x - \ln p = 0\) at \(p = e^x\), a maximum since the bracket is concave in \(p\) (second derivative \(-1/p\)). The value is \(xe^x - xe^x + e^x = e^x\) ✓, and \(p = 0\) gives only \(0 < e^x\).

2. \(pv - \tfrac12mv^2 - cv\) is maximal at \(v = (p-c)/m\), so \[ H(p) = \frac{(p-c)^2}{2m}, \qquad p = \frac{\partial L}{\partial v} = mv + c. \] The momentum conjugate to \(x\) is not \(mv\) because the term \(cv\) in \(L\) depends on the velocity: \(p\) is whatever \(\partial L/\partial\dot x\) is. For a charge in a magnetic field this is the distinction between canonical momentum \(m\vec v + q\vec A\) and kinetic momentum \(m\vec v\). The Hamiltonian in terms of \(v\) is \((mv)^2/2m = \tfrac12 mv^2\): the linear term carries no energy.

3. \(b = a/(a-1) = 3/2\). At \((2,3)\): \(px = 6\) and \(x^3/3 + p^{3/2}/(3/2) = 8/3 + \tfrac23\cdot 5.196152 = 2.666667 + 3.464102 = 6.130768 \gt 6\) ✓. At \((2,4)\): \(px = 8\) and \(8/3 + \tfrac23\cdot8 = 8\): equality, because \(p = x^{a-1} = x^2 = 4\) — \(p\) is exactly the slope of \(f = x^3/3\) at \(x = 2\).

Check. Dimensions in part 2: \(c\) must be a momentum for \(cv\) to be an energy, so \((p-c)^2/2m\) is an energy ✓. Limit \(c\to0\): \(H = p^2/2m\) ✓. And the smooth-duality theorem: \(H'(p) = (p-c)/m = v\) ✓, \(H'' = 1/m = 1/L''\) ✓.

Idea. Degeneracy is a failure of the change of variables, not of the dynamics. The transform says precisely what goes wrong (a constraint on \(p\)); part 3 shows the dynamics can survive intact, in half the variables.

1. If \(\lVert\vec p\rVert\le1\), Cauchy–Schwarz gives \(\vec p\cdot\vec v\le \lVert\vec p\rVert\lVert\vec v\rVert\le\lVert\vec v\rVert\), so the bracket is \(\le0\), with \(0\) at \(\vec v = 0\). If \(\lVert\vec p\rVert > 1\), \(\vec v = t\vec p\) gives \(t\lVert\vec p\rVert(\lVert\vec p\rVert - 1)\to\infty\). So \(f^* = 0\) on the closed unit ball and \(+\infty\) outside. The recipe: \(\vec p = \vec v/\lVert\vec v\rVert\) is a unit vector for every \(\vec v\ne0\) and cannot be solved for \(\vec v\) (every positive multiple gives the same \(\vec p\)); plugging in anyway gives \(\vec p\cdot\vec v - \lVert\vec v\rVert = 0\) identically. So the recipe's "\(H = 0\)" is right on the sphere \(\lVert\vec p\rVert = 1\), and what it misses is that \(\vec p\) is confined there — a constraint.

2. \(f''(v) = (1+v^2)^{-3/2} > 0\). For \(|p| < 1\) the critical point of \(pv - \sqrt{1+v^2}\) is \(v = p/\sqrt{1-p^2}\), with value \(p^2/\sqrt{1-p^2} - 1/\sqrt{1-p^2} = -\sqrt{1-p^2}\). At \(|p| = 1\) the supremum is \(\lim_{v\to\pm\infty}(|v| - \sqrt{1+v^2}) = 0\), not attained. For \(|p|>1\) it is \(+\infty\). So \[ f^*(p) = -\sqrt{1-p^2}\ \ (|p|\le1),\qquad +\infty\ \ (|p| > 1). \] This is the deck's example \(f(v) = -\sqrt{1-v^2}\) with the roles exchanged: there a bounded velocity had an unbounded momentum; here an unbounded velocity has a bounded momentum. Transforming twice returns \(\sqrt{1+v^2}\), as involutivity promises.

3. \(p_x = -y/2\), \(p_y = x/2\); the velocity Hessian is the zero matrix. Euler–Lagrange: \[ x:\ \ \tfrac{\mathrm d}{\mathrm dt}\big(-\tfrac y2\big) = \tfrac{\dot y}2 - \partial_xV\ \Rightarrow\ \dot y = \partial_xV; \qquad y:\ \ \tfrac{\mathrm d}{\mathrm dt}\big(\tfrac x2\big) = -\tfrac{\dot x}2 - \partial_yV\ \Rightarrow\ \dot x = -\partial_yV. \] With \(q := y\), \(p := x\), \(H := V\) these read \(\dot q = \partial H/\partial p\), \(\dot p = -\partial H/\partial q\): Hamilton's equations on the two-dimensional \((y,x)\) plane. The naive recipe gives \(H = p_x\dot x + p_y\dot y - L = V\) on the four-dimensional space, and Hamilton's equations there would say \(\dot x = \partial V/\partial p_x = 0\) — false. The recipe's \(H\) is right; its phase space is wrong: the constraints \(p_x = -y/2\), \(p_y = x/2\) cut the four dimensions to two, and on those two \(x\) and \(y\) are conjugate.

This \(L\) is the phase-space action of the deck with \(y\) and \(x\) as \(q\) and \(p\), up to the total derivative \(\tfrac12\frac{\mathrm d}{\mathrm dt}(xy)\). It is also the \(m\to0\) limit of unit 12's charge in a uniform magnetic field (with \(qB = 1\)), where it gives the drift of the guiding centre along the level curves of \(V\); and it is Kirchhoff's Lagrangian for a point vortex in a two-dimensional fluid (cited: Saffman, Vortex Dynamics, the chapter on line vortices in two dimensions).

Check. \(\frac{\mathrm d}{\mathrm dt}V = \partial_xV\,\dot x + \partial_yV\,\dot y = -\partial_xV\partial_yV + \partial_yV\partial_xV = 0\): \(V\) is conserved, as the Hamiltonian must be ✓. Special case \(V = \tfrac12(x^2+y^2)\): \(\dot y = x\), \(\dot x = -y\), uniform circular motion ✓.

Idea. \(f^{**}\) is a supremum of affine functions lying below \(f\) — the supporting lines — so it is convex and below \(f\); and any convex differentiable \(g\) below \(f\) is itself a supremum of such lines.

1. Fenchel–Young: \(px - f^*(p)\le f(x)\) for all \(p\); take the supremum over \(p\) to get \(f^{**}(x)\le f(x)\). Convexity: \(f^{**}\) is a supremum of the affine functions \(x\mapsto px - f^*(p)\) (those with \(f^*(p)\) finite), and a supremum of affine functions is convex.

2. If \(g\le f\) then \(px - g(x)\ge px - f(x)\) for every \(x\), so \(g^*(p)\ge f^*(p)\). Then \(px - g^*(p)\le px - f^*(p)\), so \(g^{**}\le f^{**}\). By the involutivity theorem, \(g^{**} = g\). Hence \(g\le f^{**}\). So every convex differentiable minorant lies below \(f^{**}\), and \(f^{**}\) is a convex minorant by part 1; if it is differentiable it is the largest such.

3. Let \(h\) be the candidate of rung 3. It is \(C^1\): on \(|x| \gt 1\), \(h' = 4x(x^2-1)\), which tends to \(0\) at \(x = \pm1\), matching \(h' = 0\) inside. It is convex: \(h'\) is \(0\) on \([-1,1]\) and on \(|x|\gt1\) has derivative \(12x^2 - 4 \gt 0\), and it is increasing through \(\pm1\), so \(h'\) is nondecreasing everywhere. And \(h\le f\), since \(f\ge0\). By part 2, \(h\le f^{**}\). Conversely \(f^{**}\le f = h\) outside \([-1,1]\); and \(f^{**}(\pm1)\le f(\pm1) = 0\), so for \(x = (1-\lambda)(-1) + \lambda\cdot1\) convexity gives \(f^{**}(x)\le0 = h(x)\). So \(f^{**} = h\). \(\square\)

What it means. The bump between the wells is invisible to the transform: \(f^*\) is unchanged if \(f\) is replaced by \(h\). In thermodynamics this is Maxwell's construction — the free energy of a system with two coexisting phases is the convex envelope of the single-phase one — and in mechanics it is why the Legendre transform to a Hamiltonian needs \(L\) convex in \(\dot q\): if it is not, the transform silently replaces it by its envelope, and the Hamiltonian describes different dynamics.

Check. Numerically, \(f^*(0) = \sup(-f) = 0\) and \(h^*(0) = 0\) ✓; at \(x = 0\), \(f(0) = 1\) while \(f^{**}(0) = 0\) ✓. Limit: far from the wells \(f\) is convex and \(f^{**} = f\), as involutivity says it must be locally where no line has to bridge a gap ✓.

Idea. The Legendre transform does the frame change's bookkeeping: the Coriolis and centrifugal forces come out of one term, \(-\vec\Omega\cdot\vec L\), in \(H\).

1. \(\partial L/\partial\dot{\vec r} = m(\dot{\vec r} + \vec\Omega\times\vec r) = \vec p\), so \(\dot{\vec r} = \vec p/m - \vec\Omega\times\vec r\). Then \[ H = \vec p\cdot\dot{\vec r} - L = \frac{\lVert\vec p\rVert^2}{m} - \vec p\cdot(\vec\Omega\times\vec r) - \frac{\lVert\vec p\rVert^2}{2m} + V = \frac{\lVert\vec p\rVert^2}{2m} - \vec\Omega\cdot(\vec r\times\vec p) + V, \] using \(\dot{\vec r} + \vec\Omega\times\vec r = \vec p/m\) inside \(L\) and the triple product.

2. From rung 3, \(m\ddot{\vec r} = \dot{\vec p} - m\vec\Omega\times\dot{\vec r}\) \(= -\vec\Omega\times\vec p - \nabla V - m\vec\Omega\times\dot{\vec r}\). Substitute \(\vec p = m\dot{\vec r} + m\vec\Omega\times\vec r\): \[ m\ddot{\vec r} = -\nabla V - 2m\vec\Omega\times\dot{\vec r} - m\vec\Omega\times(\vec\Omega\times\vec r). \] Coriolis and centrifugal, as in unit 03 (no Euler term, since \(\vec\Omega\) is constant) ✓.

3. With \(\vec p = m(\dot{\vec r} + \vec\Omega\times\vec r)\): \(\lVert\vec p\rVert^2/2m = \tfrac12m\lVert\dot{\vec r}\rVert^2 + m\dot{\vec r}\cdot(\vec\Omega\times\vec r) + \tfrac12m\lVert\vec\Omega\times\vec r\rVert^2\) and \(\vec\Omega\cdot(\vec r\times\vec p) = m\dot{\vec r}\cdot(\vec\Omega\times\vec r) + m\lVert\vec\Omega\times\vec r\rVert^2\). So \[ H = \tfrac12m\lVert\dot{\vec r}\rVert^2 - \tfrac12m\lVert\vec\Omega\times\vec r\rVert^2 + V = E_{\text{inertial}} - \vec\Omega\cdot\vec L_{\text{inertial}}, \] since \(\vec p\) is the inertial momentum and \(\lVert\vec p\rVert^2/2m + V\) the inertial energy. \(H\) is conserved (no explicit \(t\)); it is the Jacobi integral of the restricted three-body problem. \(T + V\) is not conserved in general: \(H = T + V - \tfrac12m\lVert\vec\Omega\times\vec r\rVert^2\), and the last term changes whenever the distance from the axis does.

Check. Dimensions: \([\Omega\,rp] = \mathrm{s^{-1}\cdot m\cdot kg\,m\,s^{-1}} = \mathrm J\) ✓. Limit \(\vec\Omega = 0\): \(H = p^2/2m + V\) ✓. Special case: the hoop of the deck is this with the motion constrained to a circle through the axis, and its \(-\tfrac12mR^2\Omega^2\sin^2\theta\) is exactly \(-\tfrac12m\lVert\vec\Omega\times\vec r\rVert^2\) ✓.

Idea. \(D\) generates scaling — positions out, momenta in — and \(\dot D\) measures how the Hamiltonian responds to it. For a bound orbit \(D\) is bounded, so its derivative averages to zero.

1. \(\nabla\phi(g) = \phi'(g)\nabla g\) by the chain rule, so \(\{f,\phi(g)\} = \nabla f^\top J\,\phi'(g)\nabla g = \phi'(g)\{f,g\}\). \(\square\)

2. \(\{D,f\} = \sum_i(\partial_{r_i}D\,\partial_{p_i}f - \partial_{p_i}D\,\partial_{r_i}f) = \vec p\cdot\partial_{\vec p}f - \vec r\cdot\partial_{\vec r}f\). Hence \(\{D, r^2\} = -2r^2\), \(\{D, p^2\} = 2p^2\), and, with part 1 and \(\{D,r^2\}\), \(\{D, 1/r\} = \{D,(r^2)^{-1/2}\} = -\tfrac12(r^2)^{-3/2}(-2r^2) = 1/r\). The flow of \(D\): \(\vec r' = \{\vec r, D\} = \vec r\), \(\vec p' = \{\vec p, D\} = -\vec p\), so \(\vec r\mapsto e^s\vec r\), \(\vec p\mapsto e^{-s}\vec p\): dilation, with the momentum scaled inversely — canonical, as it must be, since \(\operatorname{diag}(e^s, e^{-s})\) has determinant \(1\) in each \((r_i,p_i)\) plane.

3. \(\dot D = \{D,H\} = \frac{2p^2}{2\mu} - \frac kr = 2T + V\). Over one period \(T_{\mathrm{orb}}\) of a bound orbit, \(\langle\dot D\rangle = (D(T_{\mathrm{orb}}) - D(0))/T_{\mathrm{orb}} = 0\), so \(2\langle T\rangle + \langle V\rangle = 0\). With \(\langle T\rangle + \langle V\rangle = E\): \(\langle T\rangle = -E\), \(\langle V\rangle = 2E\).

4. Per unit mass, \(\langle v^2\rangle = 2\langle T\rangle/\mu = -2\varepsilon = 4.97480\times10^{7}\ \mathrm{m^2\,s^{-2}}\), so \(v_{\mathrm{rms}} = 7053.2\ \mathrm{m\,s^{-1}} = 7.0532\ \mathrm{km\,s^{-1}}\) — less than \(13\%\) of the perihelion speed. The comet spends most of its \(75.3\) years slowly, far out.

Check. Dimensions: \([\{D,H\}] = [D][H]/\text{action} = \mathrm{J}\) since \([D]\) is an action ✓. Special case, a circular orbit: \(v^2 = GM/r\) and \(\varepsilon = -GM/2r\), so \(\langle v^2\rangle = GM/r = -2\varepsilon\) exactly ✓. For a general homogeneous potential of degree \(\kappa\), the same computation gives \(2\langle T\rangle = \kappa\langle V\rangle\); the oscillator, \(\kappa = 2\), has \(\langle T\rangle = \langle V\rangle\) ✓.

Idea. A map is canonical iff its Jacobian is symplectic; the other tests are the same condition read differently. A map that only rescales \(J\) keeps the form of the equations at the price of rescaling the Hamiltonian, which is exactly what "canonical" forbids.

1. \(M = \begin{pmatrix}0&1/m\omega\\-m\omega&0\end{pmatrix}\), \(\det M = 1\), so \(M^\top JM = J\) ✓. \(\{Q,P\} = \partial_qQ\,\partial_pP - \partial_pQ\,\partial_qP = 0 - \tfrac1{m\omega}(-m\omega) = 1\) ✓. \(F_1 = m\omega qQ\): \(p = \partial_qF_1 = m\omega Q\) and \(P = -\partial_QF_1 = -m\omega q\) ✓. The oscillator: \(p = m\omega Q\), \(q = -P/m\omega\), so \(K = \tfrac12m\omega^2Q^2 + P^2/2m\) — the same Hamiltonian with position and momentum exchanged.

2. \(M = \begin{pmatrix}1&0\\f''&1\end{pmatrix}\), \(\det = 1\) ✓. \(\{Q,P\} = 1\cdot1 - 0 = 1\) ✓. \(F_2(q,P) = qP - f(q)\): \(Q = \partial_PF_2 = q\), \(p = \partial_qF_2 = P - f'(q)\) ✓. This is unit 12's gauge freedom: \(L\to L + \frac{\mathrm d}{\mathrm dt}f(q)\) adds \(f'(q)\) to \(\partial L/\partial\dot q\). The change is canonical, so the gauge freedom of the Lagrangian is a canonical transformation of the Hamiltonian picture.

3. \(M = \lambda I\), \(M^\top JM = \lambda^2J\ne J\): not canonical. \(\{Q,P\} = \lambda^2\ne1\), the same verdict. No generating function: \(p\,\mathrm dq - P\,\mathrm dQ = (1-\lambda^2)p\,\mathrm dq\), whose coefficient vector \(((1-\lambda^2)p, 0)\) has antisymmetric derivative \((1-\lambda^2)J\ne0\), so it is not a gradient. Yet by rung 3, \(\dot w = J\nabla_w(\lambda^2K)\): Hamilton's equations hold with the Hamiltonian \(\lambda^2H(Q/\lambda, P/\lambda)\) — for every \(H\). So the three-tests theorem's condition (iii) must be read with \(K = H\circ\Phi^{-1}\) exactly; this map passes a weaker test (it is "conformally symplectic"). For the oscillator, \(\lambda^2H(Q/\lambda,P/\lambda) = H(Q,P)\): quadratic Hamiltonians hide the failure entirely.

4. \(M = \operatorname{diag}(A, B)\) with \(A = \begin{pmatrix}1&1\\1&-1\end{pmatrix}\), \(B = \tfrac12A\). Then \(M^\top JM = \begin{pmatrix}0&A^\top B\\-B^\top A&0\end{pmatrix}\) and \(A^\top B = \tfrac12A^2 = \tfrac12\cdot2I = I\) ✓. Brackets, from the formulas: \(\{Q_1,P_1\} = \tfrac12 + \tfrac12 = 1\), \(\{Q_1,P_2\} = \tfrac12 - \tfrac12 = 0\), \(\{Q_2,P_1\} = 0\), \(\{Q_2,P_2\} = 1\), and \(\{Q_i,Q_j\} = \{P_i,P_j\} = 0\) since each involves only positions or only momenta ✓. \(F_2\) of rung 2: \(Q_1 = q_1+q_2\), \(Q_2 = q_1-q_2\), \(p_1 = P_1+P_2\), \(p_2 = P_1 - P_2\), which inverts to the given \(P\)s ✓. It is a point transformation, and the rule "positions by \(A\), momenta by \(A^{-\top}\)" is what every point transformation does.

Check. Dimensions in part 1: \([p/m\omega] = \mathrm m\) and \([m\omega q] = \mathrm{kg\,m\,s^{-1}}\): \(Q\) is a length and \(P\) a momentum, so \(\{Q,P\}\) is dimensionless as it must be ✓. Limit in part 3: \(\lambda\to1\) is the identity, canonical ✓; \(\lambda = -1\) is canonical too (\(\lambda^2 = 1\)) ✓.

Idea. Pick coordinates in which the oscillator's orbits (ellipses) are the lines \(P =\) const and its motion is uniform in \(Q\). Canonicity is what guarantees the equations there are still Hamilton's.

1. \(\tan Q = m\omega q/p\) and \(2m\omega P = p^2 + (m\omega q)^2\): so \((p, m\omega q)\) has length \(\sqrt{2m\omega P}\) and angle \(Q\) from the \(p\)-axis, i.e. \(p = \sqrt{2m\omega P}\cos Q\), \(m\omega q = \sqrt{2m\omega P}\sin Q\), which is the claim.

2(a). With \(D = p^2 + m^2\omega^2q^2\): \(\partial_qQ = \frac{m\omega/p}{1 + m^2\omega^2q^2/p^2} = \frac{m\omega p}{D}\), \(\partial_pQ = -\frac{m\omega q}{D}\), \(\partial_qP = m\omega q\), \(\partial_pP = p/m\omega\). So \(\{Q,P\} = \frac{p^2}{D} + \frac{m^2\omega^2q^2}{D} = 1\) ✓.

2(b). Now treat \(Q,P\) as the canonical pair and compute \(\{q,p\}_{Q,P} = \partial_Qq\,\partial_Pp - \partial_Pq\,\partial_Qp\): \(\partial_Qq = \sqrt{2P/m\omega}\cos Q\), \(\partial_Pp = \sqrt{m\omega/2P}\cos Q\), \(\partial_Pq = \sin Q/\sqrt{2m\omega P}\), \(\partial_Qp = -\sqrt{2m\omega P}\sin Q\). So \(\{q,p\} = \cos^2Q + \sin^2Q = 1\) ✓. This is the determinant of the inverse Jacobian — the same theorem, but computed from different formulas, so an error in one does not reproduce in the other.

2(c). \(p = \partial_qF_1 = m\omega q\cot Q\), i.e. \(\tan Q = m\omega q/p\) ✓. And \(P = -\partial_QF_1 = \frac{m\omega q^2}{2\sin^2Q}\), with \(\sin^2Q = \frac{m^2\omega^2q^2}{D}\) from \(\tan Q = m\omega q/p\): \(P = \frac{m\omega q^2D}{2m^2\omega^2q^2} = \frac{D}{2m\omega}\) ✓. The condition \(\partial^2F_1/\partial q\partial Q = -m\omega q/\sin^2Q\ne0\) holds for \(q\ne0\).

3. \(H = D/2m = \omega P =: K(P)\). Then \(\dot Q = \partial K/\partial P = \omega\), \(\dot P = -\partial K/\partial Q = 0\): \(Q = \omega t + Q_0\), \(P = P_0\), and by part 1 \(q(t) = \sqrt{2P_0/m\omega}\,\sin(\omega t + Q_0)\) ✓.

4. \(E = \tfrac12m\omega^2A^2 = \tfrac12\cdot0.5\cdot16\cdot0.01 = 0.0400\) J, so \(P = E/\omega = 0.0100\ \mathrm{J\,s}\). The orbit is the ellipse with semi-axes \(A = 0.1\) m and \(m\omega A = 0.2\ \mathrm{kg\,m\,s^{-1}}\), of area \(\pi\cdot0.1\cdot0.2 = 0.0628319\ \mathrm{J\,s}\), and \(2\pi P = 0.0628319\ \mathrm{J\,s}\) ✓.

Check. Dimensions: \([P] = [p]^2/[m\omega] = \mathrm{J\,s}\), an action, and \(Q\) is dimensionless, so \(QP\) has the dimensions of \(qp\) as a canonical pair must ✓. The deck's theorem \(\mathrm dA/\mathrm dE = T\): \(A = 2\pi E/\omega\) gives \(T = 2\pi/\omega\) ✓.

Idea. Three computations that share no intermediate formula: the forward bracket, the backward bracket, and the generating function. If all three give "canonical", an algebra slip in any one of them would have shown.

1. With the rung-3 derivatives, \[ \{Q,P\} = \frac{c}{2uw}\big(2uc + 2u^2(c^2-s^2)\big) + \frac{us}{w}\Big(\frac su + 2cs\Big) = \frac{c^2 + uc^3 - ucs^2 + s^2 + 2ucs^2}{w} = \frac{1 + uc(c^2+s^2)}{w} = \frac{1+uc}{w} = 1. \]

2. From \(uc = e^Q - 1\) and \(us = Pe^{-Q}/2\): \[ q = u^2 = (e^Q-1)^2 + \tfrac14P^2e^{-2Q},\qquad \tan p = g := \frac{P}{2e^Q(e^Q-1)} . \] Then \(\partial_Qq = 2(e^Q-1)e^Q - \tfrac12P^2e^{-2Q}\), \(\partial_Pq = \tfrac12Pe^{-2Q}\), and with \(p = \arctan g\), \(\partial p = \partial g/(1+g^2)\), where \(1 + g^2 = q/(e^Q-1)^2\) (multiply out \(4e^{2Q}q\)), \(g_P = \frac1{2e^Q(e^Q-1)}\), \(g_Q = -\frac{P(2e^Q-1)}{2e^Q(e^Q-1)^2}\). Now \(\partial_Qq\,g_P = 1 - \frac{P^2e^{-3Q}}{4(e^Q-1)}\) and \(\partial_Pq\,g_Q = -\frac{P^2(2e^Q-1)e^{-3Q}}{4(e^Q-1)^2}\), so \[ \partial_Qq\,g_P - \partial_Pq\,g_Q = 1 + \frac{P^2e^{-3Q}\big(2e^Q - 1 - (e^Q-1)\big)}{4(e^Q-1)^2} = 1 + \frac{P^2e^{-2Q}}{4(e^Q-1)^2}, \] and dividing by \(1 + g^2\): \(\{q,p\} = \frac{(e^Q-1)^2 + P^2e^{-2Q}/4}{q} = \frac qq = 1\) ✓.

3. \(-\partial_pF_3 = (e^Q-1)^2\sec^2p = q\) gives \(\sqrt q\cos p = e^Q - 1\) (both sides positive on the region), i.e. the formula for \(Q\) ✓. \(-\partial_QF_3 = 2(e^Q-1)e^Q\tan p = 2\sqrt q\cos p\,(1 + \sqrt q\cos p)\tan p = 2(1 + \sqrt q\cos p)\sqrt q\sin p = P\) ✓. By the generating-function theorem the map is canonical.

4. For a \(2\times2\) Jacobian, \(M^\top JM = (\det M)J\), and \(\det M = \partial_qQ\,\partial_pP - \partial_pQ\,\partial_qP = \{Q,P\}\). So the symplectic condition and part 1 are literally the same number. In \(2n\ge4\) dimensions they are different computations (the entries of \(M^\top JM\) versus those of \(MJM^\top\)), equivalent by the lemma of the three-tests slide.

Check. A numerical spot check: at \(q = 1\), \(p = \pi/4\), \(Q = \ln(1 + 0.707107) = 0.534800\) and \(P = 2(1.707107)(0.707107) = 2.414214\); the inversion gives \(q = (0.707107)^2 + \tfrac14(2.414214)^2(0.343146) = 0.5 + 0.5 = 1\) ✓. Limit \(q\to0\): \(Q\to0\) and \(P\to0\), and the region shrinks to the origin, consistent with \(F_3\to0\) ✓.

Idea. Free flight is a pure shear: fast atoms get ahead. Area is fixed, so as the cloud lengthens it must thin — and "thin" means that atoms at one place have nearly the same velocity.

1. The flow matrix \(\begin{pmatrix}1&t/m\\0&1\end{pmatrix}\) has determinant \(1\). The corners \((0,0)\), \((\Delta x,0)\), \((\Delta x, m\Delta v)\), \((0, m\Delta v)\) go to \((0,0)\), \((\Delta x, 0)\), \((\Delta x + \Delta v\,t, m\Delta v)\), \((\Delta v\,t, m\Delta v)\): a parallelogram with base \(\Delta x\) and height \(m\Delta v\), area \(m\,\Delta x\,\Delta v = 1.443160\times10^{-25}\cdot10^{-4}\cdot10^{-2} = 1.44316\times10^{-31}\ \mathrm{J\,s}\), for all \(t\).

2. From \(0\) to \(\Delta x + \Delta v\,t = 100\ \mu\mathrm m + 200\ \mu\mathrm m = 300\ \mu\mathrm m\).

3. By rung 3 the allowed \(v\) form an interval of length \(\Delta x/t\), provided it lies inside \([0,\Delta v]\) — true for \(\Delta x\le x\le\Delta v\,t\), which is non-empty once \(t\ge\Delta x/\Delta v = 10\) ms. At \(t = 20\) ms, for \(100\ \mu\mathrm m\le x\le200\ \mu\mathrm m\), the local spread is \(\Delta x/t = 10^{-4}/0.02 = 5.00\ \mathrm{mm\,s^{-1}}\), half the initial \(1.00\ \mathrm{cm\,s^{-1}}\), and it falls as \(1/t\). No contradiction: the total velocity spread is still \(\Delta v\); what shrinks is the spread at fixed position, because the parallelogram tilts, and its vertical cross-section times its horizontal extent is (roughly) the conserved area.

Check. Dimensions: \([m\,\Delta x\,\Delta v] = \mathrm{kg\cdot m\cdot m\,s^{-1}} = \mathrm{J\,s}\) ✓. Limit \(t\to0\): the local spread formula \(\Delta x/t\) exceeds \(\Delta v\), so the cap \(\Delta v\) applies and nothing is lost ✓. Product check at \(t = 20\) ms: local velocity spread \(\times\) extent over which it holds, \(5\ \mathrm{mm\,s^{-1}}\times200\ \mu\mathrm m\), equals \(\Delta v\,\Delta x = 1\ \mathrm{cm\,s^{-1}}\times100\ \mu\mathrm m\) ✓.

Idea. Without friction the determinant is \(1\) and the shape can change but not the area; with friction the divergence \(-\gamma\) drains area at a fixed exponential rate, whatever the shape.

1. The flow is \(z\mapsto R(t)z\) with \(R = \begin{pmatrix}\cos\omega t&\sin\omega t/m\omega\\-m\omega\sin\omega t&\cos\omega t\end{pmatrix}\), \(\det R = 1\). The rectangle is \(z_0 + \{sa\,e_1 + ub\,e_2: s,u\in[0,1]\}\), so its image is \(Rz_0 + \{s\,aRe_1 + u\,bRe_2\}\): a parallelogram on the sides \(aRe_1\), \(bRe_2\), of area \(ab\,|\det R| = ab\). At \(t = \pi/2\omega\): sides \(a(0,-m\omega)\) and \(b(1/m\omega,0)\) — a rectangle again, with side \(b/m\omega\) along \(q\) and \(m\omega a\) along \(p\). It is congruent to the original (as a rectangle in these coordinates) iff \(b = m\omega a\). In between it is a genuine parallelogram. In the scaled coordinates \((\sqrt{m\omega}\,q,\ p/\sqrt{m\omega})\) the flow is a rotation and nothing deforms.

2. \(\operatorname{div} = -\gamma\), so by Liouville's formula the flow's determinant is \(e^{-\gamma t}\) and the area is \(ab\,e^{-\gamma t}\). Friction is exactly the term that breaks Liouville: it is not of the form \(-\partial H/\partial q\).

3. \(\gamma T_d = \frac{2\pi\gamma}{\omega_0\sqrt{1-1/4Q^2}} = \frac{2\pi}{Q\sqrt{1 - 1/4Q^2}} = 0.0628326\), so the factor per period is \(e^{-0.0628326} = 0.939101\), and after ten periods \(0.533484\): about half the area is gone in ten swings of a \(Q = 100\) oscillator.

Check. Dimensions: \(\gamma t\) dimensionless ✓. Limit \(\gamma\to0\) (\(Q\to\infty\)): factor \(\to1\), Liouville restored ✓. Unit 06's Abel identity, \(W(t) = W(0)e^{-\gamma t}\), is the same statement about the Wronskian's parallelogram ✓.

Idea. Volume preservation plus finite total volume forces returns. Nothing about the dynamics is used beyond those two facts.

1. Energy conservation gives \(\varphi_t(B)\subset B\) for all \(t\), positive and negative, so \(\phi\) is a bijection of \(B\). By Liouville, \(\operatorname{vol}\phi^k(U) = \operatorname{vol}U\) for every \(k\ge0\). If the sets \(\phi^k(U)\), \(k = 0,1,2,\dots\), were pairwise disjoint, then for every \(N\), \(\operatorname{vol}B\ge\operatorname{vol}\bigcup_{k \lt N}\phi^k(U) = N\operatorname{vol}U\), impossible since \(B\) is bounded. So \(\phi^i(U)\cap\phi^j(U)\ne\emptyset\) for some \(i \lt j\). Apply \(\phi^{-i}\): \(U\cap\phi^{j-i}(U)\ne\emptyset\). Take \(z\in U\) with \(\phi^{j-i}(z)\) in it... precisely: take \(w\in U\cap\phi^{n}(U)\), \(n = j - i\ge1\), and \(z = \phi^{-n}(w)\in U\); then \(\phi^n(z) = w\in U\). \(\square\)

2. Damped pendulum: the flow does not preserve volume (divergence \(-\gamma\)), so \(\operatorname{vol}\phi^k(U) = e^{-k\gamma\tau}\operatorname{vol}U\) and the pigeonhole fails; indeed almost every orbit spirals into the down position and never returns to a small set away from it. Free particle: the flow preserves volume but no energy shell is bounded (\(x\) is unrestricted), so the sets can march off disjointly for ever — and they do.

3. \(H = p_r^2/2\mu + V_{\mathrm{eff}}(r)\) with \(V_{\mathrm{eff}}\to+\infty\) as \(r\to0\) (because \(L\ne0\)) and \(V_{\mathrm{eff}}\to0\) as \(r\to\infty\). For \(E \lt 0\), \(H\le E\) forces \(V_{\mathrm{eff}}(r)\le E\), which confines \(r\) to \([r_-,r_+]\), and \(p_r^2\le2\mu(E - \min V_{\mathrm{eff}})\): bounded ✓. For Halley the theorem says that states near the present one recur — but you already know far more: every bound Kepler orbit is exactly periodic, with period \(75.3052\) yr. Recurrence is informative for systems too complicated to solve, such as a gas in a box, where Zermelo used it in 1896 against Boltzmann: the recurrence times are real but, for a macroscopic gas, vastly longer than the age of the universe.

Check. Special case, the harmonic oscillator: \(\{H\le E\}\) is an ellipse and \(\phi\) is a rotation in scaled coordinates; for \(\tau\) an irrational multiple of the period, points return arbitrarily close without ever repeating exactly — consistent with the theorem, which promises returns to \(U\), not to the point ✓.

Idea. Same cost, same accuracy per step; one map is canonical and one is not. Liouville turns that difference into bounded versus exponentially growing energy error.

1. \(M_E = \begin{pmatrix}1&h\\-h&1\end{pmatrix}\), \(\det = 1 + h^2\). \((q+hp)^2 + (p-hq)^2 = (1+h^2)(q^2+p^2)\), the cross terms cancelling. One period is \(2\pi/h = 628.32\) steps; \((1.0001)^{628.32} = 1.06484\): energy up \(6.48\%\) per period. After \(100\) periods, \((1.0001)^{62832} = 535\).

2. \(\det M = (1-h^2) + h^2 = 1\), so \(M^\top JM = J\): symplectic Euler is a canonical map. For the invariant, write \(z' = Mz\) and compute \(C(z') = q'^2 - hq'p' + p'^2\) with \(q' = (1-h^2)q + hp\), \(p' = p - hq\): \(q'^2 = (1-h^2)^2q^2 + 2h(1-h^2)qp + h^2p^2\); \(-hq'p' = -h\big[(1-h^2)qp - h(1-h^2)q^2 + hp^2 - h^2qp\big]\); \(p'^2 = p^2 - 2hqp + h^2q^2\). The coefficient of \(q^2\) is \((1-h^2)^2 + h^2(1-h^2) + h^2 = 1\); of \(p^2\), \(h^2 - h^2 + 1 = 1\); of \(qp\), \(2h(1-h^2) - h(1-h^2) + h^3 - 2h = -h\). So \(C(z') = C(z)\) ✓.

3. From \(|qp|\le\tfrac12(q^2+p^2)\): \((q^2+p^2)(1 - h/2)\le C\le(q^2+p^2)(1+h/2)\), which rearranges to the band. Its ends differ by the ratio \((1+h/2)/(1-h/2) = 1.005/0.995 = 1.01005\): the energy wobbles by about \(1\%\) at \(h = 0.01\), and that stays true after any number of periods.

Check. Limit \(h\to0\): both matrices tend to the identity, \(C\to q^2+p^2\), and the band closes onto the true energy ✓. The explicit method's growth per period, \((1+h^2)^{2\pi/h}\approx e^{2\pi h}\), tends to \(1\) as \(h\to0\) — so it is convergent; its defect is only that its error compounds, which is precisely what a non-unit determinant does ✓.

Idea. The oscillator separates into two identical one-dimensional oscillators rotating in step. Quadratic combinations that do not care about the common phase are conserved, and they close into \(\mathfrak{so}(3)\) — the analogue of Kepler's \(\mathfrak{so}(4)\), and the reason all orbits close.

1. \(\{H_x,H_y\} = 0\) (different variables), so \(H_x\) and \(H_y\) are separately conserved, and so is \(S_1\). For \(S_2\): \(\{xy, H\} = y\{x,H_x\} + x\{y,H_y\} = yp_x + xp_y\) and \(\{p_xp_y,H\} = p_y(-x) + p_x(-y)\); the sum is \(0\). For \(S_3\): it is \(\tfrac12L_z\), conserved because \(H\) is rotation-invariant (the lemma of the deck), or directly: \(\{xp_y - yp_x, H\} = p_xp_y + x(-y) - p_yp_x - y(-x) = 0\).

2. From rung 3, \(\{S_1,xy\} = \tfrac12(xp_y - yp_x)\). Similarly \(\{S_1,p_x\} = \partial S_1/\partial x = x/2\) and \(\{S_1,p_y\} = -y/2\), so \(\{S_1,p_xp_y\} = \tfrac12(xp_y - yp_x)\). Hence \(\{S_1,S_2\} = \tfrac12\cdot\tfrac12\cdot2(xp_y - yp_x) = S_3\) ✓. Next \(\{S_2,S_3\}\): with \(\{S_3,x\} = -\partial S_3/\partial p_x = y/2\), \(\{S_3,y\} = -x/2\), \(\{S_3,p_x\} = p_y/2\), \(\{S_3,p_y\} = -p_x/2\), one gets \(\{S_3, xy\} = \tfrac12(y^2 - x^2)\) and \(\{S_3,p_xp_y\} = \tfrac12(p_y^2 - p_x^2)\), so \(\{S_3,S_2\} = -\tfrac14(x^2 - y^2 + p_x^2 - p_y^2) = -S_1\), i.e. \(\{S_2,S_3\} = S_1\) ✓. Finally \(\{S_3,S_1\} = \tfrac14\big(\{S_3,p_x^2 + x^2\} - \{S_3,p_y^2 + y^2\}\big) = \tfrac14\big(p_xp_y + xy + p_xp_y + xy\big) = S_2\) ✓, using \(\{S_3,p_x^2\} = 2p_x\cdot p_y/2\), \(\{S_3,x^2\} = 2x\cdot y/2\), \(\{S_3,p_y^2\} = -p_xp_y\), \(\{S_3,y^2\} = -xy\).

3. \(4(S_1^2 + S_2^2) = \tfrac14(a - b)^2 + (xy + p_xp_y)^2\) with \(a = p_x^2 + x^2\), \(b = p_y^2 + y^2\), and \(4S_3^2 = (xp_y - yp_x)^2\). Now \((xy + p_xp_y)^2 + (xp_y - yp_x)^2 = x^2y^2 + p_x^2p_y^2 + x^2p_y^2 + y^2p_x^2 = ab\) (the cross terms \(\pm2xyp_xp_y\) cancel). So \(4\lVert\vec S\rVert^2 = \tfrac14(a-b)^2 + ab = \tfrac14(a+b)^2 = H^2\) ✓.

4. At fixed energy \(E\), \(\vec S\) lives on the sphere of radius \(E/2\), and each point of it is one orbit (all orbits are ellipses centred at the origin, and \(\vec S\) fixes the ellipse). Poles: \(S_3 = \pm E/2\), \(S_1 = S_2 = 0\) — maximal \(|L_z| = E\), the two circular orbits, one each way round. Equator: \(S_3 = 0\), no angular momentum — the straight-line orbits through the origin, with direction set by \((S_1,S_2)\): the angle of \((S_1,S_2)\) is twice the angle of the line. Rotation-invariance alone gives only \(L_z\), which does not explain why orbits close; the extra conserved \(S_1, S_2\) do, exactly as \(\vec A\) does for Kepler.

Check. Numbers: the orbit \(x = \cos t\), \(y = 0\) has \(E = \tfrac12\), \(S_1 = \tfrac14\), \(S_2 = S_3 = 0\): on the equator, radius \(\tfrac14 = E/2\) ✓. The circle \(x = \cos t\), \(y = \sin t\): \(E = 1\), \(S_3 = \tfrac12\cdot1 = E/2\), \(S_1 = \tfrac14(\sin^2t - \cos^2t + \cos^2t - \sin^2t) = 0\), \(S_2 = \tfrac12(\cos t\sin t - \sin t\cos t) = 0\): a pole ✓.