Hamiltonian Formulation Questions and Answers

Answer: A

The phase-space trajectory is sketch (A): smooth where it crosses the \(x\) axis and pointed where it crosses the \(p\) axis.

Energy conservation gives \(E = p^{2}/2m + \alpha|x|\), so

\[ |x| = \frac{1}{\alpha}\left(E - \frac{p^{2}}{2m}\right) \]

For \(x \gt 0\) and for \(x \lt 0\) the curve is a parabola in \(p\), opening toward the origin with its vertex on the \(x\) axis at \(x = \pm E/\alpha\), where it is smooth. The two parabolic arcs meet on the \(p\) axis at \(p = \pm\sqrt{2mE}\) at an angle, making corners there; this is because the force \(-\alpha\,\text{sgn}(x)\) jumps at \(x = 0\).

Sketch (B) has the corners on the \(x\) axis instead. The ellipse (C) is the harmonic oscillator, \(V \propto x^{2}\), and the rectangle (D) is a particle bouncing in an infinite square well, where \(|p|\) is constant.

Answer: D

The result is \(\frac{dp_{r}}{dt} = \frac{2p_{\theta}^{2} - mk}{2mr^{3}}\).

The force \(-\frac{k}{2r^{3}}\hat{r}\) comes from the potential \(V = -\frac{k}{4r^{2}}\). The Lagrangian in polar coordinates is \(\mathcal{L} = \frac{m}{2}(\dot{r}^{2} + r^{2}\dot{\theta}^{2}) - V\), with \(p_{\theta} = mr^{2}\dot{\theta}\). Lagrange's equation for \(r\) gives

\[ \begin{aligned} \frac{dp_{r}}{dt} &= \frac{\partial \mathcal{L}}{\partial r} = mr\dot{\theta}^{2} - \frac{k}{2r^{3}} \\ &= \frac{p_{\theta}^{2}}{mr^{3}} - \frac{k}{2r^{3}} = \frac{2p_{\theta}^{2} - mk}{2mr^{3}} \end{aligned} \]

The first term is the centrifugal term and the second is the applied force. The options with \(p_{\theta}^{2} - 2mk\) double the force term, and \(\frac{2p_{\theta}^{2} - mk}{mr^{3}}\) doubles both terms.

Answer: C

The canonical equations are ẋ = p_y/m, ṗ_x = -mω₀²y, ẏ = p_x/m, ṗ_y = -mω₀²x.

The Lagrangian couples the velocities crosswise, so each momentum involves the other velocity:

\[ \begin{aligned} p_{x} &= \frac{\partial L}{\partial \dot{x}} = m\dot{y}, \qquad p_{y} = \frac{\partial L}{\partial \dot{y}} = m\dot{x} \\ H &= p_{x}\dot{x} + p_{y}\dot{y} - L = \frac{p_{x}p_{y}}{m} + m\omega_{0}^{2}xy \end{aligned} \]

Hamilton's equations: \(\dot{x} = \partial H/\partial p_{x} = p_{y}/m\), \(\dot{y} = p_{x}/m\), \(\dot{p}_{x} = -\partial H/\partial x = -m\omega_{0}^{2}y\), \(\dot{p}_{y} = -m\omega_{0}^{2}x\).

The first two options pair ẋ with p_x, as for an ordinary kinetic term, which this Lagrangian does not have. The last option has the right pairing but the wrong sign in the force terms.

Answer: D

The correct statement is \(H\) is not the total energy of the system and is not conserved.

The bead's height is fixed by the rod, \(z = vt\), a time-dependent (rheonomic) constraint; only \(x\) is free. The Lagrangian and Hamiltonian are

\[ \begin{aligned} L &= \tfrac{1}{2}m(\dot{x}^{2} + v^{2}) - mgvt \\ H &= p_{x}\dot{x} - L = \frac{p_{x}^{2}}{2m} - \tfrac{1}{2}mv^{2} + mgvt \end{aligned} \]

The total energy would be \(\frac{p_{x}^{2}}{2m} + \frac{1}{2}mv^{2} + mgvt\), so \(H\) differs from it by \(mv^{2}\): with a moving constraint the Hamiltonian is not the energy. And \(H\) depends explicitly on time through \(mgvt\), so \(\frac{dH}{dt} = \frac{\partial H}{\partial t} = mgv \ne 0\): it is not conserved (the rod does work on the bead).

(This is a multiple-select question; the key accepts this option alone.)

Answer: B

The Hamiltonian is \(\dfrac{P_{\theta}^{2}}{2ml^{2}} + \dfrac{1}{2}mgl\theta^{2}\).

The generalised momentum is \(P_{\theta} = \partial L/\partial \dot{\theta} = ml^{2}\dot{\theta}\), so \(\dot{\theta} = P_{\theta}/(ml^{2})\). Then

\[ \begin{aligned} H &= P_{\theta}\dot{\theta} - L \\ &= \frac{P_{\theta}^{2}}{ml^{2}} - \frac{P_{\theta}^{2}}{2ml^{2}} + \frac{1}{2}mgl\theta^{2} \\ &= \frac{P_{\theta}^{2}}{2ml^{2}} + \frac{1}{2}mgl\theta^{2} \end{aligned} \]

As expected for a natural system, H is the total energy, kinetic plus potential, so the potential enters with a plus sign (option A has a minus) and keeps its factor 1/2 (options C and D drop it).

Answer: A

As \(t \to \infty\), \(q \to \infty\) and \(p \to 0\).

Hamilton's equations for \(H = pq\):

\[ \begin{aligned} \dot{q} &= \frac{\partial H}{\partial p} = q \\ \dot{p} &= -\frac{\partial H}{\partial q} = -p \end{aligned} \]

So \(q = q_{0}e^{t}\) and \(p = p_{0}e^{-t}\). For a non-zero starting \(q_{0}\) (positive, as the option assumes), \(q\) grows without bound, while \(p\) decays to 0. The product \(pq = H\) stays constant, as energy conservation requires.

Both tending to 0 or both to \(\infty\) would change \(pq\), and the last option reverses the two equations.

Answer: A

The general solution is \(q(t) = C_{1}\cos kt + C_{2}\sin kt\).

Hamilton's equations give

\[ \begin{aligned} \dot{q} &= \frac{\partial H}{\partial p} = p \\ \dot{p} &= -\frac{\partial H}{\partial q} = -k^{2}q \end{aligned} \]

so \(\ddot{q} = -k^{2}q\), simple harmonic motion with angular frequency \(k\). Its general solution is \(C_{1}\cos kt + C_{2}\sin kt\).

Exponentials \(e^{\pm kt}\) would solve \(\ddot{q} = +k^{2}q\), the sign of an inverted oscillator. Frequency \(k^{2}\) mistakes \(k^{2}\) in the Hamiltonian for the frequency, and \(kt + C_{1}\) is free motion.

Answer: B

The modified equations are \(\dot{q}_{i} = \frac{\partial H}{\partial p_{i}}\), \(\dot{p}_{i} = -\frac{\partial H}{\partial q_{i}} + Q_{i}\).

The conservative forces are contained in H. A non-conservative generalized force \(Q_{i}\) adds to the rate of change of momentum, as in Lagrange's equation, where \(\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_{i}}\) minus \(\frac{\partial L}{\partial q_{i}}\) equals \(Q_{i}\). Since \(p_{i} = \partial L/\partial \dot{q}_{i}\) and \(\partial L/\partial q_{i} = -\partial H/\partial q_{i}\), this gives \(\dot{p}_{i} = -\partial H/\partial q_{i} + Q_{i}\), while the first equation, which only defines the velocity, is unchanged.

Options A and C give \(\dot{p}_{i}\) the wrong sign of \(\partial H/\partial q_{i}\), and options C and D subtract the applied force instead of adding it.

Answer: C

The Lagrangian is \(L = \dfrac{\dot{q}_{1}^{2}}{4} + \dfrac{\dot{q}_{2}^{2}}{4} + kq_{1}^{2}\).

Hamilton's equation \(\dot{q}_{i} = \partial H / \partial p_{i} = 2p_{i}\) gives \(p_{i} = \dot{q}_{i}/2\). Then the Legendre transform \(L = \sum p_{i}\dot{q}_{i} - H\) gives

\[ \begin{aligned} L &= \frac{\dot{q}_{1}^{2} + \dot{q}_{2}^{2}}{2} - \frac{\dot{q}_{1}^{2} + \dot{q}_{2}^{2}}{4} + kq_{1}^{2} \\ &= \frac{\dot{q}_{1}^{2}}{4} + \frac{\dot{q}_{2}^{2}}{4} + kq_{1}^{2} \end{aligned} \]

The sign of \(kq_{1}^{2}\) flips because it enters H as part of the potential (\(H = T + V\), \(L = T - V\) with \(V = -kq_{1}^{2}\)). Options with \(-kq_{1}^{2}\) miss that, and those with denominators 2 or 1 use the wrong relation between \(p\) and \(\dot{q}\).

Answer: B

The area enclosed in phase space remains constant with time.

This is Liouville's theorem. For a Hamiltonian system the flow in phase space is incompressible: the divergence of the phase-space velocity vanishes,

\[ \begin{aligned} &\sum_{i}\left(\frac{\partial\dot{q}_{i}}{\partial q_{i}} + \frac{\partial\dot{p}_{i}}{\partial p_{i}}\right) \\ &= \sum_{i}\left(\frac{\partial^{2}H}{\partial q_{i}\partial p_{i}} - \frac{\partial^{2}H}{\partial p_{i}\partial q_{i}}\right) \\ &= 0 \end{aligned} \]

so a region of phase points may change shape as it moves but keeps its area (volume). Shrinking area belongs to dissipative systems, where trajectories crowd onto attractors, and growing or oscillating area does not occur in Hamiltonian flow.