Krishna Ayyalasomayajula
1c1a575f6e
Add Chapter 3 with 13 concept files covering: - HJ Fundamentals: derivation, separation, action-angle, EM coupling - Mechanics problems: free particle, projectile, SHO, Kepler, rigid rotator - EM problems: uniform E-field, cyclotron, E×B drift, Coulomb Also: manifest update (13 entries), macro additions (HJ + \bm + \dd override), unicode cleanup, compilation fixes.
167 lines
8.4 KiB
TeX
167 lines
8.4 KiB
TeX
\subsection{Action-Angle Variables}
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This subsection develops action-angle variables for integrable periodic systems, presenting the canonical transformation that reduces any periodic system to trivial dynamics where the momenta are constant and the angles advance uniformly.
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\dfn{Action and angle variables}{
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For a periodic degree of freedom with generalized coordinate $q_i$ and conjugate momentum $p_i$, the action variable $J_i$ is defined as the phase-space integral over one complete closed orbit:
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\[
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J_i = \oint p_i\,dq_i.
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\]
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The integral is taken over one full cycle of the periodic motion. The angle variable $w_i$ is the canonical coordinate conjugate to $J_i$, defined by differentiating Hamilton's characteristic function $W$ with respect to the action:
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\[
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w_i = \pdv{W}{J_i}.
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\]
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The angle variable $w_i$ increases by exactly one complete unit during one period of the associated periodic motion.}
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\nt{The action variable $J_i$ equals the area enclosed by the orbit in the $(q_i,p_i)$ phase-space plane. This geometric interpretation makes it straightforward to evaluate $J_i$ for simple periodic systems: the integral reduces to computing the area of an ellipse (harmonic oscillator), a triangle plus its reflection (infinite well), or other phase-space shapes.}
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In a completely integrable system with $n$ degrees of freedom, the Hamiltonian depends only on the action variables and not on the angle variables: $\mcH = \mcH(J_1,\ldots,J_n)$. Because the angles do not appear in $\mcH$, they are cyclic coordinates. This leads to the simplest possible Hamiltonian dynamics.
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\thm{Hamilton's equations in action-angle variables}{
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Let $\mcH = \mcH(J_1,\ldots,J_n)$ be the Hamiltonian expressed in action variables. Hamilton's canonical equations in the $(w,J)$ variables are
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\[
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\dot{J}_i = -\pdv{\mcH}{w_i} = 0,
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\qquad
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\dot{w}_i = \pdv{\mcH}{J_i} \equiv \omega_i.
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\]
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The action variables $J_i$ are constant in time, and the angle variables advance linearly:
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\[
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w_i(t) = \omega_i t + w_i(0).
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\]
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The frequency $\omega_i = \pdv{\mcH}{J_i}$ is independent of time. Since the angle variable $w_i$ increases by one unit over one complete cycle, the period of the $i$-th motion is $T_i = 1/\omega_i$.}
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The frequency $\omega_i$ provides direct access to the temporal characteristics of the motion. When there is a single degree of freedom, the action is found by evaluating the integral $J = \oint p\,dq$, the Hamiltonian is inverted to give $E(J)$, and the period follows immediately from $T = 1/\pdv{E}{J}$. This procedure avoids solving the equations of motion directly. The physical angular frequency of the motion is $2\pi\omega_i$.
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\ex{Simple harmonic oscillator in action-angle variables}{
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The Hamiltonian for a one-dimensional simple harmonic oscillator of mass $m$ and natural angular frequency $\omega_0$ is
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\[
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\mcH = \frac{p^2}{2m} + \frac{1}{2}m\omega_0^2 x^2.
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\]
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At energy $E = \mcH$, the momentum is $p = \pm\sqrt{2mE - m^2\omega_0^2 x^2}$ and the turning points are at $x = \pm A$ with amplitude $A = \sqrt{2E/(m\omega_0^2)}$.
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The action variable is evaluated by integrating over one complete oscillation:
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\[
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J = \oint p\,dx = 2\int_{-A}^{A}\sqrt{2mE - m^2\omega_0^2 x^2}\,dx.
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\]
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Substitute $x = A\sin\phi$, so $dx = A\cos\phi\,d\phi$ and the limits are $\phi = -\pi/2$ to $\pi/2$:
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\[
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\sqrt{2mE - m^2\omega_0^2 x^2}
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= \sqrt{2mE - 2mE\sin^2\phi}
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= \sqrt{2mE}\cos\phi.
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\]
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The integral becomes
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\[
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J = 2\sqrt{2mE}\cdot A\int_{-\pi/2}^{\pi/2}\cos^2\phi\,d\phi
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= 2\sqrt{2mE}\cdot A\cdot\frac{\pi}{2}
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= \pi\sqrt{2mE}\cdot A.
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\]
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Substituting $A = \sqrt{2E/(m\omega_0^2)}$ gives
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\[
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J = \pi\sqrt{2mE}\cdot\sqrt{\frac{2E}{m\omega_0^2}}
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= 2\pi\,\frac{E}{\omega_0}.
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\]
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Inverting this relation expresses the energy as a function of the action:
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\[
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E(J) = \frac{\omega_0 J}{2\pi}.
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\]
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The frequency obtained from the action-angle formalism is the derivative of $E$ with respect to $J$:
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\[
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\omega = \pdv{E}{J} = \frac{\omega_0}{2\pi}.
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\]
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The period of oscillation is $T = 1/\omega = 2\pi/\omega_0$, and the physical angular frequency is $2\pi\omega = \omega_0$. Crucially, the frequency is independent of the energy $E$ and therefore independent of the amplitude $A$. This is the property of isochrony: all oscillations of a simple harmonic oscillator have the same period regardless of amplitude.}
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\nt{The Kepler problem (gravitational or electrostatic $V = -k/r$) has three independent action variables $J_r$, $J_\theta$, and $J_\phi$. The energy depends on their sum:
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\[
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E = -\frac{2\pi^2 m k^2}{(J_r + J_\theta + J_\phi)^2}.
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\]
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The frequency derivatives $\pdv{E}{J_r}$, $\pdv{E}{J_\theta}$, $\pdv{E}{J_\phi}$ are all equal, so the three frequencies are degenerate. Degenerate frequencies mean every bound orbit closes on itself after one period. This degeneracy is the deep reason Kepler's ellipses are closed: the radial period equals the angular period. A small perturbation $V = -k/r + \epsilon/r^2$ breaks the degeneracy and produces precession.}
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\qs{Particle in a one-dimensional infinite potential well}{
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A particle of mass $m$ is confined to a region $0 < x < L$ by infinite potential walls, so $V(x) = 0$ for $0 < x < L$ and $V = \infty$ elsewhere. Inside the well the Hamiltonian is $\mcH = p^2/(2m)$ and the total energy is $E$.
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\begin{enumerate}[label=(\alph*)]
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\item Compute the action variable $J = \oint p\,dx$ for this system, showing that $J = 2L\sqrt{2mE}$.
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\item Express the energy as $E(J)$ and compute the frequency $\omega = \pdv{E}{J}$ and the period $T = 1/\omega$. Show that $T = 2L\sqrt{m/(2E)}$, which equals the time for the particle to travel the distance $2L$ at speed $v = \sqrt{2E/m}$.
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\item For an electron with mass $m = 9.11\times 10^{-31}\,\mathrm{kg}$ confined to a region of width $L = 1.00\times 10^{-10}\,\mathrm{m}$ with total energy $E = 1.00\,\mathrm{eV} = 1.60\times 10^{-19}\,\mathrm{J}$, compute the numerical values of $J$ (in $\mathrm{kg\!\cdot\!m^2/s}$) and $T$ (in seconds).
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\end{enumerate}}
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\sol \textbf{Part (a).} Inside the well the particle has kinetic energy $E = p^2/(2m)$, so the magnitude of momentum is $|p| = \sqrt{2mE}$ and is independent of position. The particle travels back and forth between the walls at $x = 0$ and $x = L$. During the forward leg the momentum is $p = +\sqrt{2mE}$ and during the return leg $p = -\sqrt{2mE}$.
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The action integral over one complete cycle is
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\[
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J = \oint p\,dx = \int_{0}^{L}\sqrt{2mE}\,dx + \int_{L}^{0}\left(-\sqrt{2mE}\right)\,dx.
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\]
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Each integral equals $L\sqrt{2mE}$, so
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\[
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J = L\sqrt{2mE} + L\sqrt{2mE} = 2L\sqrt{2mE}.
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\]
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\textbf{Part (b).} Solve the result from part (a) for $E$:
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\[
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\frac{J}{2L} = \sqrt{2mE},
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\qquad
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\frac{J^2}{4L^2} = 2mE,
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\qquad
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E(J) = \frac{J^2}{8mL^2}.
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\]
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Differentiate with respect to $J$ to find the frequency:
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\[
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\omega = \pdv{E}{J} = \frac{J}{4mL^2}.
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\]
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The period is the reciprocal of the frequency:
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\[
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T = \frac{1}{\omega} = \frac{4mL^2}{J}.
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\]
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Substitute $J = 2L\sqrt{2mE}$ to express $T$ in terms of $E$:
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\[
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T = \frac{4mL^2}{2L\sqrt{2mE}}
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= \frac{2mL}{\sqrt{2mE}}
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= 2L\sqrt{\frac{m}{2E}}.
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\]
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Independently, the particle's speed inside the well is $v = \sqrt{2E/m}$, and the round-trip distance is $2L$. The travel time for one complete cycle is
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\[
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T = \frac{2L}{v} = 2L\sqrt{\frac{m}{2E}},
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\]
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which agrees exactly with the action-angle result.
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\textbf{Part (c).} The given values are $m = 9.11\times 10^{-31}\,\mathrm{kg}$, $L = 1.00\times 10^{-10}\,\mathrm{m}$, and $E = 1.60\times 10^{-19}\,\mathrm{J}$.
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First compute the action variable $J = 2L\sqrt{2mE}$:
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\[
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2mE = 2(9.11\times 10^{-31})(1.60\times 10^{-19})\,\mathrm{kg\!\cdot\!J}
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= 2.92\times 10^{-49}\,\mathrm{kg^2\!\cdot\!m^2/s^2}.
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\]
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(The product $\mathrm{kg\!\cdot\!J}$ has the same dimensions as $\mathrm{kg^2\!\cdot\!m^2/s^2}$ since $1\,\mathrm{J} = 1\,\mathrm{kg\!\cdot\!m^2/s^2}$.) Taking the square root:
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\[
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\sqrt{2mE} = \sqrt{2.92\times 10^{-49}}\,\mathrm{kg\!\cdot\!m/s}
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= 5.40\times 10^{-25}\,\mathrm{kg\!\cdot\!m/s}.
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\]
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Now multiply by $2L$:
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\[
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J = 2(1.00\times 10^{-10})(5.40\times 10^{-25})\,\mathrm{kg\!\cdot\!m^2/s}
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= 1.08\times 10^{-34}\,\mathrm{kg\!\cdot\!m^2/s}.
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\]
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Next compute the period $T = 2L\sqrt{m/(2E)}$:
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\[
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\frac{m}{2E} = \frac{9.11\times 10^{-31}}{2(1.60\times 10^{-19})}\,\mathrm{kg/J}
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= 2.85\times 10^{-12}\,\mathrm{s^2/m^2}.
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\]
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Taking the square root:
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\[
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\sqrt{\frac{m}{2E}} = \sqrt{2.85\times 10^{-12}}\,\mathrm{s/m}
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= 1.69\times 10^{-6}\,\mathrm{s/m}.
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\]
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Multiply by $2L$:
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\[
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T = 2(1.00\times 10^{-10})(1.69\times 10^{-6})\,\mathrm{s}
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= 3.37\times 10^{-16}\,\mathrm{s}.
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\]
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Therefore,
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\[
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J = 1.08\times 10^{-34}\,\mathrm{kg\!\cdot\!m^2/s},
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\qquad
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T = 3.37\times 10^{-16}\,\mathrm{s}.
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\]
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