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.
197 lines
12 KiB
TeX
197 lines
12 KiB
TeX
\subsection{Rigid Rotator and Particle on a Sphere}
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This subsection treats the motion of a particle constrained to a sphere of fixed radius using the Hamilton--Jacobi method, derives the separated equations for the two angular degrees of freedom, and connects the action-angle variables to rotational states of diatomic molecules.
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\dfn{Rigid rotator Hamiltonian}{
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Consider a particle of mass $m$ constrained to move on a sphere of fixed radius $R$ with no potential energy. The kinetic energy in spherical coordinates with $r = R$ is $T = \tfrac{1}{2}mR^2\dot{\theta}^2 + \tfrac{1}{2}mR^2\sin^2\theta\,\dot{\phi}^2$. Defining the moment of inertia $I = mR^2$, the canonical momenta are
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\[
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p_\theta = I\,\dot{\theta},
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\qquad
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p_\phi = I\sin^2\theta\,\dot{\phi}.
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\]
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The Hamiltonian is the Legendre transform of the Lagrangian:
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\[
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\mcH = \frac{p_\theta^2}{2I} + \frac{p_\phi^2}{2I\sin^2\theta}.
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\]
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Since the Hamiltonian has no explicit time dependence, energy is conserved and $\mcH = E$ is a constant.}
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\nt{The rigid rotator arises in molecular physics as the model for the rotation of diatomic molecules. The two nuclei are treated as point masses constrained to a fixed separation $R$ by a rigid bond, rotating freely about their center of mass. The moment of inertia $I = \mu R^2$ uses the reduced mass $\mu$ of the two-atom system. Because there is no potential energy, the problem is purely kinematic and governed by the geometry of the sphere.}
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The Hamiltonian depends only on $\theta$ and the two momenta, and it contains no explicit dependence on the azimuthal angle $\phi$. Therefore $\phi$ is a cyclic coordinate and its conjugate momentum is conserved.
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\thm{Separation of the rigid-rotator Hamilton-- Jacobi equation}{
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With $\mcH = p_\theta^2/(2I) + p_\phi^2/(2I\sin^2\theta)$, the time-independent Hamilton--Jacobi equation $\mcH\!\bigl(\theta,\phi,\pdv{\mcS}{\theta},\pdv{\mcS}{\phi}\bigr) = E$ separates as follows. Because $\phi$ is cyclic, $\pdv{\mcS}{\phi} = L_z$, a constant. Setting $\mcS = W_\theta(\theta) + L_z\phi - Et$ reduces the equation to
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\[
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\left(\der{W_\theta}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = 2IE \equiv L^2,
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\]
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where $L$ is the total angular momentum and $L^2 = 2IE$ is the second separation constant. The $\theta$-equation is
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\[
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\der{W_\theta}{\theta} = \pm\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}},
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\]
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which integrates to
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\[
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W_\theta(\theta) = \int\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}\;\mathrm{d}\theta.
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\]}
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\pf{Derivation of the separated equations}{
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The time-independent Hamilton--Jacobi equation is obtained by substituting $p_\theta = \pdv{\mcS}{\theta}$ and $p_\phi = \pdv{\mcS}{\phi}$ into the Hamiltonian:
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\[
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\frac{1}{2I}\left(\pdv{\mcS}{\theta}\right)^2 + \frac{1}{2I\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 = E.
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\]
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Multiply both sides by $2I$:
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\[
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\left(\pdv{\mcS}{\theta}\right)^2 + \frac{1}{\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 = 2IE.
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\]
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The coordinate $\phi$ is absent from the Hamiltonian, so $\phi$ is cyclic. The contribution of the cyclic coordinate to the characteristic function is linear:
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\[
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\pdv{\mcS}{\phi} = L_z,
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\]
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where $L_z$ is the conserved $z$-component of the angular momentum. Substituting this constant into the HJ equation gives
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\[
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\left(\pdv{\mcS}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = 2IE.
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\]
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Seeking an additive separation $\mcS = W_\theta(\theta) + L_z\phi$, the partial derivative $\pdv{\mcS}{\theta}$ becomes the ordinary derivative $\der{W_\theta}{\theta}$:
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\[
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\left(\der{W_\theta}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = 2IE.
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\]
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Define the separation constant $L^2 = 2IE$, which has the dimensions of angular-momentum squared and equals the square of the total angular momentum. Solving for the derivative:
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\[
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\der{W_\theta}{\theta} = \pm\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}.
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\]
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The right-hand side vanishes at the turning points where $L^2 = L_z^2/\sin^2\theta$, or equivalently $\sin\theta = |L_z|/L$. Between these turning points, the particle oscillates in $\theta$, tracing a cone on the surface of the sphere. The polar angle sweeps between $\theta_{\min} = \arcsin(|L_z|/L)$ and $\theta_{\max} = \pi - \theta_{\min}$, while $\phi$ advances monotonically. The trajectory is a closed orbit when the ratio of the azimuthal advance to the $\theta$-oscillation is rational.}
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\nt{Geometric interpretation of the orbit}{On the surface of the sphere, the angular momentum vector is fixed in space with magnitude $L$ and $z$-component $L_z$. The fixed polar angle that this vector makes with the $z$-axis is $\theta_{\text{cone}} = \arccos(|L_z|/L)$. The instantaneous position vector of the particle precesses around the angular-momentum vector, so the trajectory on the sphere is the intersection of the sphere with the cone defined by $\theta = \theta_{\text{cone}}$. In Hamilton's equations, the azimuthal rate is $\dot{\phi} = L_z/(I\sin^2\theta)$, which varies as $\theta$ oscillates. Near the turning points the denominator is small compared to the pole, so the particle slows in $\phi$ and spends more time near the maximum polar excursion.}
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\cor{Equatorial orbit}{
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When the total angular momentum equals the absolute value of its $z$-component, $L = |L_z|$, the square root in the $\theta$-equation vanishes identically except at $\sin\theta = 1$. The radial momentum $p_\theta = \der{W_\theta}{\theta}$ vanishes everywhere except on the equator $\theta = \pi/2$, where the denominator of $L_z^2/\sin^2\theta$ exactly matches the separation constant. The motion is therefore confined to the equatorial plane. From Hamilton's equations, the azimuthal velocity is
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\[
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\dot{\phi} = \pdv{\mcH}{p_\phi} = \frac{p_\phi}{I\sin^2\theta} = \frac{L_z}{I}
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\]
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on the equator where $\sin\theta = 1$. The azimuthal angle advances linearly in time:
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\[
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\phi(t) = \frac{L_z}{I}\,t + \phi_0,
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\]
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representing uniform circular motion at constant angular speed $\omega = |L_z|/I$.}
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\nt{Action-angle variables for the rigid rotator}{The two independent action variables are computed by integrating the conjugate momenta over their respective cycles. For the cyclic coordinate $\phi$:
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\[
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J_\phi = \oint p_\phi\,\mathrm{d}\phi = \int_{0}^{2\pi} L_z\,\mathrm{d}\phi = 2\pi L_z.
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\]
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For the oscillating coordinate $\theta$, the integral runs between the two turning points:
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\[
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J_\theta = \oint p_\theta\,\mathrm{d}\theta = 2\int_{\theta_{\min}}^{\theta_{\max}}\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}\;\mathrm{d}\theta = 2\pi\bigl(L - |L_z|\bigr).
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\]
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Inverting these relations gives $L_z = J_\phi/(2\pi)$ and $L = (J_\theta + |J_\phi|)/(2\pi)$. The Hamiltonian expressed in terms of actions is
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\[
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\mcH(J_\theta,J_\phi) = \frac{L^2}{2I} = \frac{\bigl(J_\theta + |J_\phi|\bigr)^2}{8\pi^2 I}.
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\]
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The frequencies follow from $\omega_i = \pdv{\mcH}{J_i}$. They are generally unequal, so the motion is quasiperiodic unless $L = |L_z|$ (equatorial orbit, $J_\theta = 0$).}
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\nt{Connection to quantum mechanics}{In the Bohr--Sommerfeld semiclassical quantization, the action variables are quantized as integer multiples of Planck's constant:
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\[
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J_\phi = m\,h,
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\qquad
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J_\theta = (l - |m|)\,h,
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\]
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where $l$ and $m$ are integers satisfying $l \ge |m| \ge 0$. Using $L_z = J_\phi/(2\pi) = m\hbar$ and $L = (J_\theta + |J_\phi|)/(2\pi) = l\hbar$, the semiclassical energy is $E = l^2\hbar^2/(2I)$. The fully quantized result from the Schrodinger equation is $E = \hbar^2 l(l+1)/(2I)$. The two agree in the limit of large $l$, since $l(l+1) \approx l^2$ for $l \gg 1$. For small $l$, the $+l$ correction in $l(l+1)$ represents a shift with no classical counterpart.}
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\ex{Action-angle frequencies for the rigid rotator}{
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Using the Hamiltonian in action variables,
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\[
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\mcH = \frac{\bigl(J_\theta + |J_\phi|\bigr)^2}{8\pi^2 I},
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\]
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the two frequencies are
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\[
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\omega_\theta = \pdv{\mcH}{J_\theta}
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= \frac{J_\theta + |J_\phi|}{4\pi^2 I}
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= \frac{L}{2\pi I},
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\qquad
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\omega_\phi = \pdv{\mcH}{J_\phi}
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= \pm\frac{J_\theta + |J_\phi|}{4\pi^2 I}
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= \pm\frac{L}{2\pi I}.
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\]
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Here the signs conventionally match the chosen signs of the actions, so $J_\phi$ can be negative and the $\pm$ sign on the right-hand side is the sign of $J_\phi$. These two frequencies are equal in magnitude, confirming the degeneracy. For the special case $J_\theta = 0$ (equatorial orbit), there is no $\theta$ oscillation and the motion is purely azimuthal at the single frequency $\omega = L/(2\pi I)$.
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}
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\qs{Diatomic molecule as a rigid rotator}{A diatomic molecule is modeled as a rigid rotator with moment of inertia
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\[
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I = 1.46\times 10^{-46}\,\mathrm{kg\!\cdot\!m^2}.
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\]
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The total angular momentum is $L = 2\hbar$, where $\hbar = 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s}$.
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\begin{enumerate}[label=(\alph*)]
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\item Write the Hamilton-- Jacobi equation for the rigid rotator in spherical coordinates with fixed $r = R$. Identify the cyclic coordinate and state the corresponding conserved quantity.
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\item Use the classical action-angle result $E = L^2/(2I)$ to compute the rotational energy of the molecule in SI units. Compare this to the quantum result $E = \hbar^2 l(l+1)/(2I)$ with $l = 2$.
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\item Find the absolute difference between the classical and quantum energy values and express it as a percentage of the quantum value.
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\end{enumerate}}
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\sol \textbf{Part (a).} The Hamiltonian of the rigid rotator is $\mcH = p_\theta^2/(2I) + p_\phi^2/(2I\sin^2\theta)$. The full Hamilton--Jacobi equation follows by replacing $p_\theta$ with $\pdv{\mcS}{\theta}$, $p_\phi$ with $\pdv{\mcS}{\phi}$, and appending the time derivative:
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\[
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\frac{1}{2I}\left(\pdv{\mcS}{\theta}\right)^2 + \frac{1}{2I\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 + \pdv{\mcS}{t} = 0.
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\]
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For time-independent motion, the action separates as $\mcS = W(\theta,\phi) - Et$. The Hamiltonian does not depend explicitly on $\phi$, so $\phi$ is the cyclic coordinate. Its conjugate momentum is conserved:
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\[
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\pdv{\mcS}{\phi} = L_z,
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\]
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which is the $z$-component of the angular momentum.
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\textbf{Part (b).} The classical action-angle result for the rigid rotator is $E = L^2/(2I)$. With $L = 2\hbar$, we have $L^2 = 4\hbar^2$. First compute $\hbar^2$:
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\[
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\hbar^2 = \left(1.055\times 10^{-34}\right)^2\,\mathrm{J^2\!\cdot\!s^2}
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= 1.113\times 10^{-68}\,\mathrm{J^2\!\cdot\!s^2}.
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\]
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The classical energy is
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\[
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E_{\mathrm{class}} = \frac{4\hbar^2}{2I}
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= \frac{2\hbar^2}{I}
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= \frac{2(1.113\times 10^{-68})}{1.46\times 10^{-46}}\,\mathrm{J}
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= 1.525\times 10^{-22}\,\mathrm{J}.
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\]
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The quantum energy with $l = 2$ is
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\[
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E_{\mathrm{quant}} = \frac{\hbar^2\,l(l+1)}{2I}
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= \frac{(1.113\times 10^{-68})(6)}{2(1.46\times 10^{-46})}\,\mathrm{J}
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= \frac{6.678\times 10^{-68}}{2.92\times 10^{-46}}\,\mathrm{J}
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= 2.287\times 10^{-22}\,\mathrm{J}.
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\]
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The quantum energy exceeds the classical value. The ratio is
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\[
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\frac{E_{\mathrm{quant}}}{E_{\mathrm{class}}}
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= \frac{6}{4}
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= 1.50.
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\]
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\textbf{Part (c).} The absolute difference between the two energies is
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\[
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\Delta E = E_{\mathrm{quant}} - E_{\mathrm{class}}
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= 2.287\times 10^{-22} - 1.525\times 10^{-22}\,\mathrm{J}
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= 7.62\times 10^{-23}\,\mathrm{J}.
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\]
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Expressed as a percentage of the quantum value:
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\[
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\frac{\Delta E}{E_{\mathrm{quant}}}\times 100\%
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= \frac{7.62\times 10^{-23}}{2.287\times 10^{-22}}\times 100\%
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= 33.3\%.
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\]
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Analytically, since $E_{\mathrm{class}} = l^2\hbar^2/(2I)$ and $E_{\mathrm{quant}} = l(l+1)\hbar^2/(2I)$, the fractional difference is
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\[
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\frac{\Delta E}{E_{\mathrm{quant}}}
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= \frac{l(l+1) - l^2}{l(l+1)}
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= \frac{l}{l(l+1)}
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= \frac{1}{l+1}.
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\]
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For $l = 2$ this gives $1/3 = 33.3\%$, which matches the numerical calculation. The discrepancy arises entirely from the quantum $+l$ correction in $l(l+1)$ relative to the classical $l^2$.
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Therefore, the Hamilton-- Jacobi equation is
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\[
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\frac{1}{2I}\left(\pdv{\mcS}{\theta}\right)^2 + \frac{1}{2I\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 + \pdv{\mcS}{t} = 0,
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\]
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the cyclic coordinate is $\phi$ with $p_\phi = L_z = \text{const}$, and the energies are
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\[
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E_{\mathrm{class}} = 1.53\times 10^{-22}\,\mathrm{J},
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\qquad
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E_{\mathrm{quant}} = 2.29\times 10^{-22}\,\mathrm{J},
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\qquad
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\frac{\Delta E}{E_{\mathrm{quant}}} = 33.3\%.
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\]
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