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import mmf_setup; mmf_setup.nbinit(quiet=True)
import logging;logging.getLogger("matplotlib").setLevel(logging.CRITICAL)
Assignment 6: Hamiltonian Dynamics#
Due: 11:59pm Friday 4 December 2023
Arbitrary Dispersion: Negative Mass and Special Relativity#
Consider a particle moving in 1D with kinetic energy \(E(p)\) moving under a constant force \(F\). Use Hamilton’s equations to find the general solution for the position \(x(t)\) of the particle. Check your answer with the familiar solution for \(E(p) = p^2/2m\). Discuss the physical meaning of \(E'(p)\) and \(E''(p)\) in terms of Newton’s law and the particle motion.
Note
This approach also works for \(E(p) = \sqrt{p^2c^2 + m^2c^4}\) where \(c\) is the speed of light. This gives the motion of a particle under constant force in special relativity The corresponding coordinate transformation into a co-moving constantly accelerating frame gives rise to Rindler coordinates, which are applicable close to the surface of the earth. These have interesting properties associated with general relativity, including time-dilation at different heights, and an event horizon at a distance \(d=mc^2/F\) below the observer. For the earth, this distance is about \(d\approx0.3\)pc, well past the limit where the approximation of a constant gravitational field breaks down.
Hamilton Jacobi Equation#
Your assignment is to analyze the motion of a harmonic oscillator using the Hamiltonian formalism. Please follow the outline given below for analyzing a free particle and complete the same type of analysis for the harmonic oscillator.
Free Particle (Sample Analysis)#
Lagrangian Analysis#
Analyze the problem using the Lagrangian formalism.
We use the generalized coordinate \(q = x\) and velocity \(\dot{q} = v\) so that the Lagrangian is:
\[ \mathcal{L}(q, \dot{q}, t) = \frac{m}{2}\dot{q}^2. \]The canonical momentum is:
\[ p = \pdiff{\mathcal{L}}{\dot{q}} = m\dot{q}. \]The Euler-Lagrange equation is:
\[ \dot{p} = m\ddot{q} = \pdiff{\mathcal{L}}{q} = 0 \]The general solution is
\[ \dot{q}(t) = v, \qquad q(t) = q_0 + v_0t. \]
Hamiltonian Analysis#
From above, we can form the Hamiltonian using the Legendre transform. First we invert \(p = m\dot{q}\) to find \(\dot{q} = p/m\), then we perform the Legendre transformation:
\[ H(q, p, t) = p \dot{q}(p) - \mathcal{L}(q, \dot{q}(p), t) = \frac{p^2}{2m}. \]Now we can express the problem in terms of Hamilton’s equation of motion:
\[\begin{align*} \begin{pmatrix} \dot{q}\\ \dot{p} \end{pmatrix} &= \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} \cdot \begin{pmatrix} \pdiff{H}{q}\\ \pdiff{H}{p} \end{pmatrix} = \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} \cdot \begin{pmatrix} 0\\ p/m \end{pmatrix}\\ &= \frac{1}{m} \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} q\\ p \end{pmatrix}. \end{align*}\]The solution can be expressed as:
\[\begin{align*} \begin{pmatrix} q\\ p \end{pmatrix} &= e^{\left(\begin{smallmatrix} 0 & 1\\ 0 & 0\end{smallmatrix}\right)\frac{t}{m}} \cdot \begin{pmatrix} q_0\\ p_0 \end{pmatrix} = \left( \mat{1} + \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}\frac{t}{m} \right) \cdot \begin{pmatrix} q_0\\ p_0 \end{pmatrix} = \begin{pmatrix} 1 & \frac{t}{m}\\ 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} q_0\\ p_0 \end{pmatrix}\\ &= \begin{pmatrix} q_0 + p_0t/m\\ p_0 \end{pmatrix} \end{align*}\]where we have used the fact that \(\mat{A}^n = \mat{0}\) for \(n>1\) where \(\mat{A}=\bigl(\begin{smallmatrix} 0 & 1\\ 0 & 0\end{smallmatrix}\bigr)\).
The Hamilton-Jacobi equation is:
\[ H\left(q, \pdiff{S}{q}, t\right) + \pdiff{S}{t} = \frac{1}{2m}\left(\pdiff{S}{q}\right)^2 + \pdiff{S}{t} = 0. = \frac{1}{2m}S_{,q}^2 + S_{,t} = 0. \]This equation is separable, and we may place the \(q\)’s on one side, and the \(t\)’s on the other to obtain:
\[ \frac{1}{2m}S_{,q}^2 = E = -S_{,t}. \]Integrating each side, we obtain:
\[ S(q, t) = \sqrt{2mE} q + f(t), \qquad S(q, t) = W(q) - Et, \]where \(f(t)\) and \(W(q)\) are the integration constants of each of the pieces. The solution is thus:
\[ S(q,t) = \sqrt{2mE}q - Et + S_0. \]Note that \(W(q) = \sqrt{wmE}q + S_0\) is called “Hamilton’s characteristic function” (i.e. in Fetter and Walecka) or sometimes the “abbreiviated action” (Landau and Lifshitz) and the form \(S(q,t) = W(q) - Et\) is always valid when \(H(q, p, t) = H(q, p)\) is independent of time.
We are free to choose any new coordinate \(Q\) as long as the invertability requirement still holds:
\[ \left(\frac{\partial^2 S}{\partial q \partial Q}\right) = \sqrt{\frac{m}{2E(Q)}}E'(Q) \neq 0. \]Since the new Hamiltonian \(H'(Q, P, t) = H(q, p, t) + S_{,t} = 0\) by construction, the equations of motion are \(\dot{Q} = \dot{P} = 0\) and \(P\) and \(Q=E\) are constants of motion.
A convenient choice is \(E(Q) = Q\): i.e. introducing the energy \(E\) as the new coordinate.
After this choice is made, we have the generating function for the canonical transformation:
\[ S(q, Q, t) = \sqrt{2mQ}q - Qt + S_0. \]The canonical momentum follows from the following, which may be inverted to express \(q(Q, P, t)\):
\[\begin{split} P = - \pdiff{S(q,Q, t)}{Q} = t - \sqrt{\frac{m}{2E}} q\\ q(Q, P, t) = \sqrt{\frac{2E}{m}}(t - P) = v_0(t - t_0) = q_0 + v_0t, \\ p(Q, P, t) = S_{,q} = \sqrt{2mE} = mv_0. \end{split}\]Thus, we see that the canonical momentum \(P\) to the energy \(Q=E\) is the initial time \(t_0\).
Inverting these, we have the explicit canonical transformation:
\[ Q(q, p, t) = E = \frac{p^2}{2m}, \qquad P(q, p, t) = t - \sqrt{\frac{m}{2E}}q = t - \frac{mq}{p}. \]We can now explicitly check that the Poisson bracket (I am using the convention of Landau and Lifshitz here) satisfies the canonical commutation relationships:
\[ [P, Q] = \pdiff{P}{p}\pdiff{Q}{q} - \pdiff{P}{q}\pdiff{Q}{p} = \frac{mq}{p^2}\cdot 0 - \left(-\frac{m}{p}\right)\cdot \frac{p}{m} = 1. \](This analysis is a little harder to do for the oscillator, so do not feel you have to do it.) Armed with the solutions, we may construct the action function for the path connecting \((q_0, t_0)\) to \((q_1, t_1)\):
\[ q(t) = q_0 + \frac{q_1 - q_0}{t_1 - t_0}(t-t_0), \qquad \dot{q}(t) = \frac{q_1 - q_0}{t_1 - t_0}. \]Hence, we have the action:
\[ \bar{S}(q_0, t_0; q, t) = \int_{t_0}^{t} \mathcal{L}\d{t} = \int_{t_0}^{t} \frac{m}{2}\frac{(q_1 - q_0)^2}{(t_1 - t_0)^2}\d{t} = \frac{m}{2}\frac{(q_1 - q_0)^2}{t - t_0}. \]This allows us to construct the general solution to any initial-value problem for the Hamilton-Jacobi equation:
\[ H(q, S_{,q}(q, t), t) + S_{,t}(q,t) = 0, \qquad S(q, t_0) = S_0(q) \]as
\[ S(q, t) = S_0(q_0) + \int_{t_0}^{t} L(q(t), \dot{q}(t), t)\;\d{t} \]where the action is computed over the trajectory starting from \(q(t_0) = q_0\) with initial momentum \(p_0 = S_0'(q_0)\) and ending at \(q(t) = q\) at time \(t\). For this problem \(p_0 = mv_0\) so we have
\[ q(t) = q_0 + v_0(t - t_0) = q_0 + \frac{S'_0(q_0)}{m}(t - t_0) \]which must be inverted to find \(q_0 = q_0(q, t, t_0)\). The explicit solution here is expressed in terms of this:
\[ S(q, t) = S_0(q_0) + \frac{1}{2m}[S'_0(q_0)]^2 (t - t_0). \]Since \(H\) is independent of time, we can take \(t_0 = 0\) without loss of generality. Now, consider an example problem from Arnold where he asks for the solution to this problem with initial conditions \(S_0(q) = \frac{mq^2}{2T}\) (though he chooses units where \(m=T=1\)). This can be explicity constructed:
\[ S'_0(q_0) = \frac{m q_0}{T} = mv_0, \qquad q = q_0 + \frac{q_0}{T}t, \qquad q_0 = \frac{q}{1 + \frac{t}{T}} = \frac{q T}{T + t}. \]The explicit solution is thus
\[ S(q, t) = \frac{m q^2 T}{2(T + t)^2} + \frac{1}{2m}\left(\frac{mq}{T + t}\right)^2 t = \frac{mq^2}{2(T + t)}. \]Note that this does not have the same form as the separable solution we constructed above. This is due to the choice of initial conditions \(S_0(q)\). In particular, our separable solution corresponds with the initial conditions \(S_0(q) = m v_0 q\) instead. Mathematically, however, the Hamilton-Jacobi equations can be solved with any arbitrary initial conditions.
Harmonic Oscillator#
Your assignment is to repeat a similar analysis with the harmonic oscillator.
Lagrangian Analysis#
Analyze the problem using the Lagrangian formalism.
Use the generalized coordinate \(q = x\) and velocity \(\dot{q} = v\) so that the Lagrangian is:
\[ \mathcal{L}(q, \dot{q}, t) = \frac{m}{2}\dot{q}^2 - \frac{k}{2}q^2. \]Find the the canonical momentum?
\[ p = ? \]Write the Euler-Lagrange equation:
\[ \dot{p} = ? \]Write down the general solution:
\[ q(t) = ? \]
Hamiltonian Analysis#
Analyze the problem using the Hamiltonian formalism.
Use the Legendre transform to write the Hamiltonian:
\[ H(q, p, t) = ? \]Now we can express the problem in terms of Hamilton’s equation of motion:
\[\begin{split} \begin{pmatrix} \dot{q}\\ \dot{p} \end{pmatrix} = ? = \mat{A}\cdot \begin{pmatrix} q\\ p \end{pmatrix}. \end{split}\]The solution can be expressed as:
\[\begin{split} \begin{pmatrix} q\\ p \end{pmatrix} = e^{\mat{A}t} \cdot \begin{pmatrix} q_0\\ p_0 \end{pmatrix} = ? \end{split}\]where you have used the properties of the matrix \(\mat{A}^n\) to explicitly compute the matrix exponential.
Write the Hamilton-Jacobi equation is:
\[ H\left(q, \pdiff{S}{q}, t\right) + \pdiff{S}{t} = ? = 0. \]This equation is separable, and we may place the \(q\)’s on one side, and the \(t\)’s on the other. Connect the two equations with the constant \(E\) and integrate each side to obtain both the *abbreviated action \(W(q)\) and the generating function \(S(q,t)\):
\[\begin{split} W(Q) = ?, \\ S(q, t) = W(q) - Et = ?. \end{split}\]Note: to complete the integrals here you will probably want to make a trignometric substitution \(\sqrt{k/2E}q = \sin\theta\).
You are free to choose any new coordinate \(Q\) as long as the invertability requirement still holds. Compute this a choose a reasonable coordinate \(Q\):
\[ \left(\frac{\partial^2 S}{\partial q \partial Q}\right) = ? \neq 0. \]Now express the generating function \(S(q,Q,t)\) for the canonical transformation in terms of your chosen coordinate:
\[ S(q, Q, t) = ?. \]Use this to determine the canonical momenta:
\[\begin{split} P = ?\\ q(Q, P, t) = ?\\ p(Q, P, t) = ?. \end{split}\]Invert these to get the explicit canonical transformation:
\[\begin{split} Q(q, p, t) = ?\\ P(q, p, t) = ?. \end{split}\]Explicitly check that the Poisson bracket (I am using the convention of Landau and Lifshitz here) satisfies the canonical commutation reationships:
\[ [P, Q] = ? = 1. \]Armed with the generating function designed so that \(H'(Q, P, t) = 0\), argue that \(Q\) and \(P\) are constant, and hence use your result from step 7. to write down the complete solution and compare with your previous results.
\[\begin{split} q(t) = ?\\ p(t) = ?. \end{split}\]Since this problem is both separable and periodic, we can express the solution in terms of action-angle variables. Using this formalism, compute the fundamental frequency of the system.
(Bonus) Use the same action-angle formalism to compute the fundamental frequencies for the simple pendulum of mass \(m\) on a massless rod of length \(l\), which has the following Hamiltonian expressed in terms of the angle \(q=\theta\) from the vertical:
Hints:
Don’t try to analytically evaluate the integrals – just get the answer expressed in terms of definite integrals.
Remember that there are two types of periodic orbits here: librations (if \(1 - \cos q < 1\) always) and rotations where \(q\) takes on all values. These will have different endpoints in your integrals.