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 \(\dot{q}\).