fix(HJ): A.08 — Radian convention fix for action variables

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.

.gitignore

@@ -34,7 +34,6 @@
 # Output formats
 *.dvi
 *.xdv
-*.pdf
 *-converted-to.*
 
 # Editor / system junk

ap-physics-c-handbook.aux

@@ -0,0 +1,247 @@
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-  "concepts/em/u12/e12-3-force-on-current.tex" 1777515613.15003 13419 98647cde47acecb3d60e0598b7d93f24 ""
+  "concepts/em/u12/e12-3-force-on-current.tex" 1777515778.6996 13422 7ee71deb77f0863ed38a88c70b1102e6 ""
   "concepts/em/u12/e12-4-biot-savart.tex" 1777514409.5274 12889 45d0158340ec1822f969341c8d1276ad ""
   "concepts/em/u12/e12-5-ampere.tex" 1777512876.65543 9384 85b7732e2abb310a79903b0a22f7bba2 ""
   "concepts/em/u12/e12-6-solenoids-dipoles.tex" 1777512915.56444 13079 e5115a024a15026d54445f55f76e8439 ""
@@ -344,7 +362,7 @@
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   "concepts/mechanics/u6/m6-2-angular-momentum.tex" 1777514603.87993 6067 36be0795588107d03f88da415f8bb0cc ""
   "concepts/mechanics/u6/m6-3-angular-momentum-conservation.tex" 1777441084.87106 5751 dcd5cfaea026ea8fa667a76f4d994d06 ""
-  "concepts/mechanics/u6/m6-4-rolling.tex" 1777515630.56709 8329 307b0488ac24b36750449ee412c23df0 ""
+  "concepts/mechanics/u6/m6-4-rolling.tex" 1777515823.70875 8220 bfd1d174ee3786540b36578f16316583 ""
   "concepts/mechanics/u6/m6-5-orbits.tex" 1777515590.32395 5509 8a1175048c6c6ae4c7d86443c0486e05 ""
   "concepts/mechanics/u7/m7-1-shm.tex" 1777441407.14996 4249 7bec3c1a68db56fb629d3bf9b09a45a4 ""
   "concepts/mechanics/u7/m7-2-spring-oscillator.tex" 1777441405.79096 5708 046d291916a21159ec49d9b359fedb3a ""
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   "concepts/mechanics/u7/m7-4-simple-pendulum.tex" 1777441655.16103 4892 7f39b5311d878ba187f5e6446f486449 ""
   "concepts/mechanics/u7/m7-5-physical-pendulum.tex" 1777441819.77312 4533 c73f333295384fa15880ac9d054724be ""
   "letterfonts.tex" 1777433114.15016 7365 b76f37e9ff1d4cecac7d1fd23fc626a6 ""
-  "macros.tex" 1777433118.29064 1390 e02343db25b6b0a7b14fed906327ad7b ""
+  "macros.tex" 1777705169.18017 1818 ed7c7b1891fd79f8c7dfbed99d8a7bad ""
   "preamble.tex" 1777433122.46447 19591 cb981239c1be7fd6d00edd1a1ab30475 ""
   "units/em/unit-10.tex" 1777514080.63853 1223 6d5d003f1e3c31346656569a3929f05f ""
   "units/em/unit-11.tex" 1777514089.37355 1442 b889cd9746899d6bba0433930ba4443e ""

ap-physics-c-handbook.tex

@@ -20,4 +20,7 @@
 \part{Electricity \& Magnetism}
 \input{chapters/em}
 
+\part{Advanced Topics}
+\input{chapters/advanced}
+
 \end{document}

chapters/advanced.tex

@@ -0,0 +1,144 @@
+\chapter{Advanced Analytical Mechanics}
+
+Newtonian mechanics, Lagrangian mechanics, and Hamiltonian mechanics each reformulate
+the same physics in progressively more abstract language. Newton\normalsize{}'s laws write
+second-order differential equations for particle positions. The Lagrangian principle
+of least action recasts this as a variational problem, automatically accommodating
+constraints and generalized coordinates. Hamilton\normalsize{}'s canonical equations split
+the second-order problem into $2n$ coupled first-order equations on phase space,
+revealing the underlying symplectic structure of mechanics. Each of these formulations
+is fundamentally \textbf{trajectory-based}: you solve for a particle\normalsize{}'s specific
+path through space and time.
+
+The Hamilton--Jacobi framework changes the question entirely. Instead of solving for
+a single trajectory, it asks: \textit{what is the global structure of all possible
+trajectories?} The answer is encoded in a single scalar function $S(q_1,\dots,q_n,t)$,
+called Hamilton\normalsize{}'s principal function, whose spatial gradient equals the
+canonical momentum:
+\[
+p_i = \frac{\partial S}{\partial q_i}.
+\]
+This relation elevates momentum from a dynamical variable to a property of a
+\textbf{field} defined over configuration space. Solving mechanics becomes a matter
+of finding the field $S$ that satisfies the Hamilton--Jacobi equation, a single
+first-order nonlinear partial differential equation. The Hamilton--Jacobi formulation
+is the most abstract of the four classical frameworks, but that abstraction is precisely
+what makes it useful: by shifting from trajectories to fields, it exposes structure
+that is invisible at the level of individual paths.
+
+The field perspective of the Hamilton--Jacobi formalism is not an accident. It
+reflects a deep analogy with geometric optics, first noticed by Maupertuis and
+made explicit by Hamilton himself. In geometrical optics, light propagates as
+rays, each orthogonal to surfaces of constant phase called \textbf{wavefronts}.
+The wavefronts are level sets of a scalar function called the \textbf{eikonal},
+and the local direction of each ray is determined by the gradient of the eikonal.
+Hamilton recognized that mechanics has the exact same structure. Particle trajectories
+play the role of light rays, surfaces of constant action $S$ play the role of
+optical wavefronts, and the gradient relation $p_i = \partial S/\partial q_i$
+mirrors the optical relation between wavefront normals and ray directions. In
+this view, the Hamilton--Jacobi equation is the mechanical analog of the
+\textbf{eikonal equation} of optics:
+\[
+\left|\nabla S\right|^2 = 2m\bigl(E - V(\vec{r})\bigr).
+\]
+Just as light rays bend when the refractive index changes, particle trajectories
+curve when the potential energy varies in space. The analogy runs even deeper:
+in both cases, the dynamics of rays is completely determined by the level-geometry
+of a single scalar field.
+
+This analogy is not merely poetic. It makes concrete the three most powerful
+features of the Hamilton--Jacobi approach. \textbf{First}, separation of variables
+for the Hamilton--Jacobi PDE reveals conserved quantities that are often obscured
+in the Newtonian or even Hamiltonian formulation. When the equation separates in
+a particular coordinate system, each additive separation constant corresponds to
+a constant of motion, and the choice of coordinates that enables separation is
+itself a signature of the system\normalsize{}'s hidden symmetry. Spherical coordinates
+separate for central potentials; parabolic coordinates separate for the Kepler
+problem and expose the Runge--Lenz vector\normalsize{}'s associated conservation law.
+\textbf{Second}, for periodic or bound systems, the \textbf{action-angle variables}
+$(J,w)$ provide a direct route to the system\normalsize{}'s frequencies without solving
+any differential equation. The action variable
+\[
+J = \frac{1}{2\pi} \oint p \, \mathrm{d}q
+\]
+is the area enclosed by the orbit in phase space, divided by $2\pi$. The frequency
+follows immediately as a partial derivative of the Hamiltonian with respect to
+the action:
+\[
+\omega = \frac{\partial H}{\partial J}.
+\]
+The angle variable $w$ advances uniformly in time, acting as a clock that tracks
+the system\normalsize{}'s progress through one period. Systems with commensurate
+frequencies close their trajectories, while incommensurate frequencies fill out
+invariant tori in phase space --- the geometric origin of resonant and chaotic
+behavior. \textbf{Third}, the Hamilton--Jacobi equation is the classical limit of
+quantum mechanics. In the Wentzel--Kramers--Brillouin (WKB) approximation, the
+quantum wave function is written as
+\[
+\psi(\vec{r},t) = A(\vec{r},t)\, \mathrm{e}^{iS(\vec{r},t)/\hbar}.
+\]
+Substituting this ansatz into the Schrödinger equation and collecting the leading
+order in $\hbar \to 0$ reproduces the Hamilton--Jacobi equation exactly. The older
+Bohr--Sommerfeld quantization rule,
+\[
+J = n h \qquad (n = 0,1,2,\dots),
+\]
+was the first successful attempt to quantize classical action, emerging naturally
+from the action-angle formalism nearly a decade before the modern theory of
+quantum mechanics. The Hamilton--Jacobi framework is the conceptual bridge that
+connects the orbit picture of classical physics to the wave picture of quantum
+physics.
+
+Everything in this chapter rests on the mechanics and electromagnetism you already
+know. The mechanics problems --- free particle, projectile motion, the simple harmonic
+oscillator, the Kepler two-body problem, and the rigid rotator --- draw on your
+work with kinematics, energy conservation, momentum, rotation, and central forces
+from the earlier mechanics units. The electromagnetism problems --- charged particles
+in uniform electric fields, cyclotron motion in magnetic fields, and
+$\vec{E} \times \vec{B}$ drift --- build on your treatment of Lorentz forces,
+equipotentials, and magnetic particle motion. The Hamilton--Jacobi formalism unifies
+all of these results under one method. For problems you have already solved by
+elementary means, it provides a deeper structural understanding. For problems that
+resist elementary approaches, it supplies a systematic technique grounded in the
+same variational principles you used to derive Lagrange\normalsize{}'s equations.
+
+This chapter is organized in three parts. Section 3.1 develops the HJ equation from
+Hamiltonian mechanics and introduces separation of variables, action-angle variables,
+and electromagnetic minimal coupling. Section 3.2 applies the HJ formalism to classical
+mechanics problems: the free particle, projectile motion, the simple harmonic oscillator,
+the Kepler (two-body) problem, and the rigid rotator on a sphere. Section 3.3 treats
+problems from electromagnetism, including charged particles in uniform $\vec{E}$-fields,
+cyclotron motion, and $\vec{E}\times\vec{B}$ drift, showing that the HJ approach recovers
+all standard results with a unified method.
+
+\section{Hamilton-Jacobi Fundamentals}
+
+\input{concepts/advanced/hj-equation}
+
+\input{concepts/advanced/separation}
+
+\input{concepts/advanced/action-angle}
+
+\input{concepts/advanced/hj-em-coupling}
+
+\section{Mechanics Problems via HJ}
+
+\input{concepts/advanced/free-particle-hj}
+
+\input{concepts/advanced/projectile-hj}
+
+\input{concepts/advanced/sho-hj}
+
+\input{concepts/advanced/kepler-hj}
+
+\input{concepts/advanced/rigid-rotator-hj}
+
+\section{Electromagnetism Problems via HJ}
+
+\input{concepts/advanced/uniform-e-field-hj}
+
+\input{concepts/advanced/cyclotron-hj}
+
+\input{concepts/advanced/crossed-fields-hj}
+
+\input{concepts/advanced/kepler-coulomb-hj}

concepts/advanced/action-angle.tex

@@ -0,0 +1,209 @@
+\subsection{Action-Angle Variables}
+
+This subsection develops action-angle variables for integrable periodic systems. We present the canonical transformation that reduces any periodic system to trivial dynamics where the momenta are constant and the angles advance uniformly. The central insight is geometric: the action variable measures the area enclosed by the orbit in phase space, and the angle variable measures where along that orbit the system currently stands.
+
+\nt{Recap of separation of variables}{
+The characteristic function $W(q_1,\dots,q_n)$ found by separation in A.02 contains the complete solution to the time--independent Hamilton--Jacobi equation. Once $W$ is known, Jacobi's theorem gives every trajectory. But $W$ alone does not make the periodicity of the motion immediately visible. Here we repackage the information contained in $W$ into action--angle variables, where the periodic structure and oscillation frequencies emerge directly.}
+
+\dfn{Action and angle variables}{
+For a periodic degree of freedom with generalized coordinate $q_i$ and conjugate momentum $p_i$, the orbit in the $(q_i,p_i)$ phase plane forms a closed curve. The action variable $J_i$ is defined as the area enclosed by this curve, divided by $2\pi$:
+\[
+J_i = \frac{1}{2\pi}\oint p_i\,dq_i.
+\]
+The integral is taken counter--clockwise around one complete closed orbit. Geometrically, $J_i$ is proportional to the phase--space area of one cycle: it equals the enclosed area divided by $2\pi$. For a harmonic oscillator the orbit is an ellipse and the area is a straightforward ellipse computation; for an infinite well the orbit is a rectangle. The angle variable $w_i$ is the canonically conjugate coordinate to $J_i$, defined from Hamilton's characteristic function by
+\[
+w_i = \pdv{W}{J_i}.
+\]
+The angle variable $w_i$ is measured in radians and ranges over $[0,2\pi)$. It increases by exactly $2\pi$ during one complete period of the motion. Together, $(w_i,J_i)$ form a set of canonical coordinates obtained from $(q_i,p_i)$ by a canonical transformation.}
+
+\nt{The angle variable as a phase clock}{
+The angle variable $w_i$ tracks progress through one complete cycle of periodic motion, much like the hand of a clock. At $w_i = 0$ the system sits at some chosen reference point on its orbit --- for example, the maximum positive displacement. As time advances, $w_i$ sweeps through $[0,2\pi)$ and reaches $2\pi$ precisely when the system returns to its starting phase-space point and the cycle repeats. Because $w_i = \pdv{W}{J_i}$ and the Hamiltonian expressed in action variables is time--independent, Hamilton's equations give $\dot{w}_i = \pdv{\mcH}{J_i} = \omega_i$, a constant. The angle therefore advances uniformly, just as a clock hand rotates at constant angular speed. This makes action--angle variables the natural language for discussing periodic oscillation.}
+
+\ex{Phase-space ellipse for the simple harmonic oscillator}{
+The simplest illustration of the geometric definition comes from the harmonic oscillator of mass $m$ and natural frequency $\omega_0$. The energy is $E = p^2/(2m) + \tfrac{1}{2}m\omega_0^2 x^2$. Rearranging, the energy-level curve in phase space is
+\[
+\frac{x^2}{A^2} + \frac{p^2}{p_{\max}^2} = 1,
+\qquad\text{where}\qquad
+A = \sqrt{\frac{2E}{m\omega_0^2}},
+\quad
+p_{\max} = m\omega_0 A.
+\]
+This is an ellipse in the $(x,p)$ phase plane with semiaxes $A$ along the position axis and $p_{\max} = m\omega_0 A$ along the momentum axis. The area of an ellipse is $\pi$ times the product of its semiaxes:
+\[
+\text{Area} = \pi A\cdot p_{\max}
+= \pi A\cdot m\omega_0 A
+= \pi m\omega_0 A^2.
+\]
+Substituting $A^2 = 2E/(m\omega_0^2)$ gives
+\[
+\text{Area} = \pi m\omega_0\cdot\frac{2E}{m\omega_0^2}
+= \frac{2\pi E}{\omega_0}.
+\]
+The action variable is the area divided by $2\pi$:
+\[
+J = \frac{1}{2\pi}\cdot\frac{2\pi E}{\omega_0}
+= \frac{E}{\omega_0}.
+\]
+Inverting, $E(J) = \omega_0 J$, and the oscillation frequency is
+\[
+\omega = \pdv{E}{J} = \omega_0.
+\]
+The frequency equals the natural angular frequency $\omega_0$, measured in radians per second, and the period is $T = 2\pi/\omega_0$. This is a direct consequence of isochrony: all oscillations of a harmonic oscillator have the same period regardless of energy.}
+
+A system with $n$ degrees of freedom is called \emph{completely integrable} when it possesses $n$ independent constants of motion that are in involution --- their pairwise Poisson brackets all vanish, $\{F_i, F_j\} = 0$ for all $i,j = 1,\dots,n$. Complete integrability guarantees that the $2n$--dimensional phase space is foliated by invariant $n$--dimensional tori, each labeled by constant values of the action variables. On each torus the dynamics reduces to uniform rotation of the angles. For such systems the Hamiltonian depends only on the actions: $\mcH = \mcH(J_1,\dots,J_n)$.
+
+\thm{Hamilton's equations in action-angle variables}{
+Let $\mcH = \mcH(J_1,\dots,J_n)$ be the Hamiltonian expressed as a function of the action variables alone. Hamilton's canonical equations in the $(w,J)$ variables are
+\[
+\dot{J}_i = -\pdv{\mcH}{w_i} = 0,
+\qquad
+\dot{w}_i = \pdv{\mcH}{J_i} \equiv \omega_i.
+\]
+The action variables $J_i$ are constant in time because the angle variables do not appear in $\mcH$ and therefore the actions experience no conjugate forces. The angle variables advance linearly:
+\[
+w_i(t) = \omega_i t + w_i(0),
+\]
+with constant angular frequencies $\omega_i = \pdv{\mcH}{J_i}$ measured in radians per second. Since the angle variable $w_i$ increases by $2\pi$ during one complete period of the $i$-th degree of freedom, the oscillation period is $T_i = 2\pi/\omega_i$.}
+
+The frequency $\omega_i$ provides direct access to the temporal characteristics of the motion without solving differential equations. For a single degree of freedom, the procedure is mechanical: evaluate $J = \frac{1}{2\pi}\oint p\,dq$ by computing the enclosed phase--space area or by direct integration, invert the resulting relation to express $E = E(J)$, and differentiate to find $\omega = \pdv{E}{J}$. The period follows from $T = 2\pi/\pdv{E}{J}$. When $T$ turns out independent of amplitude the system is isochronous, a property shared by both the harmonic oscillator and the Kepler problem, and one that proves significant in the quantum limit.
+
+\ex{Direct integration for the simple harmonic oscillator}{
+We confirm the geometric result of the previous example by direct computation. For the harmonic oscillator $\mcH = p^2/(2m) + \tfrac{1}{2}m\omega_0^2 x^2$, at fixed energy $E$ the momentum is $p = \pm\sqrt{2mE - m^2\omega_0^2 x^2}$ and the turning points are $x = \pm A$ with $A = \sqrt{2E/(m\omega_0^2)}$. The raw phase--space integral around the orbit is
+\[
+\oint p\,dx = 2\int_{-A}^{A}\sqrt{2mE - m^2\omega_0^2 x^2}\,dx.
+\]
+Substitute $x = A\sin\phi$, so $dx = A\cos\phi\,d\phi$ and the limits become $\phi = -\pi/2$ to $\pi/2$. The integrand simplifies:
+\[
+\sqrt{2mE - m^2\omega_0^2 A^2\sin^2\phi}
+= \sqrt{2mE - 2mE\sin^2\phi}
+= \sqrt{2mE}\cos\phi,
+\]
+since $m^2\omega_0^2 A^2 = 2mE$. The integral becomes
+\[
+\oint p\,dx = 2\sqrt{2mE}\cdot A\int_{-\pi/2}^{\pi/2}\cos^2\phi\,d\phi.
+\]
+Using $\int_{-\pi/2}^{\pi/2}\cos^2\phi\,d\phi = \pi/2$, this gives
+\[
+\oint p\,dx = 2\sqrt{2mE}\cdot A\cdot\frac{\pi}{2}
+= \pi\sqrt{2mE}\cdot A.
+\]
+Substituting $A = \sqrt{2E/(m\omega_0^2)}$ yields
+\[
+\oint p\,dx = \pi\sqrt{2mE}\cdot\sqrt{\frac{2E}{m\omega_0^2}}
+= \frac{2\pi E}{\omega_0}.
+\]
+Dividing by $2\pi$, we recover
+\[
+J = \frac{1}{2\pi}\cdot\frac{2\pi E}{\omega_0}
+= \frac{E}{\omega_0},
+\qquad
+E(J) = \omega_0 J,
+\qquad
+\omega = \pdv{E}{J} = \omega_0,
+\]
+in complete agreement with the geometric calculation. The angular frequency equals $\omega_0$ directly, with no separate $2\pi$ factors to track.}
+
+\nt{Degeneracy and closed orbits in the Kepler problem}{
+The Kepler problem (gravitational or electrostatic $V = -k/r$) provides a remarkable illustration of frequency degeneracy. It has three independent action variables --- $J_r$ for radial motion, $J_\theta$ for polar angle, and $J_\phi$ for azimuthal rotation. The energy in action--angle variables depends only on their sum:
+\[
+E = -\frac{mk^2}{2(J_r + J_\theta + J_\phi)^2}.
+\]
+The three frequency derivatives are therefore identical:
+\[
+\omega_r = \pdv{E}{J_r} = \pdv{E}{J_\theta} = \pdv{E}{J_\phi}.
+\]
+Every degree of freedom oscillates at the same angular frequency. On the three--dimensional invariant torus this means all three angle variables advance by $2\pi$ simultaneously after one period, so the trajectory retraces itself and forms a closed curve. In contrast, when frequency ratios are irrational, motion on a torus fills the surface densely without ever repeating --- the so--called irrational winding. Because the Kepler frequencies are equal, no such winding occurs: every bound Kepler orbit closes exactly after one period, producing the familiar planetary ellipses. This is the degeneracy singled out by Bertrand's theorem, which proves that only the Kepler potential $V \propto -1/r$ and the harmonic potential $V \propto r^2$ among all central potentials produce closed bounded orbits. Adding a small perturbing term, such as $V = -k/r + \epsilon/r^2$, breaks the degeneracy: the frequencies separate, the orbit no longer closes, and the perihelion precesses.}
+
+\qs{Particle in a one-dimensional infinite potential well}{
+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 $E$.
+
+\begin{enumerate}[label=(\alph*)]
+\item Compute the action variable $J = \frac{1}{2\pi}\oint p\,dx$ for this system, showing that $J = L\sqrt{2mE}/\pi$.
+\item Express the energy as $E(J)$ and compute the frequency $\omega = \pdv{E}{J}$ and the period $T = 2\pi/\omega$. Show that $T = 2L\sqrt{m/(2E)}$, which equals the time for the particle to travel the distance $2L$ at speed $v = \sqrt{2E/m}$.
+\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).
+\end{enumerate}}
+
+\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 $p = -\sqrt{2mE}$.
+
+The phase--space integral around one complete cycle is
+\[
+\oint p\,dx = \int_{0}^{L}\sqrt{2mE}\,dx + \int_{L}^{0}\left(-\sqrt{2mE}\right)\,dx.
+\]
+Each integral equals $L\sqrt{2mE}$, so the total enclosed area is
+\[
+\oint p\,dx = L\sqrt{2mE} + L\sqrt{2mE} = 2L\sqrt{2mE}.
+\]
+Dividing by $2\pi$ gives the action variable:
+\[
+J = \frac{1}{2\pi}\cdot 2L\sqrt{2mE}
+= \frac{L\sqrt{2mE}}{\pi}.
+\]
+
+\textbf{Part (b).} Solve the result from part (a) for $E$:
+\[
+\frac{\pi J}{L} = \sqrt{2mE},
+\qquad
+\frac{\pi^2 J^2}{L^2} = 2mE,
+\qquad
+E(J) = \frac{\pi^2 J^2}{2mL^2}.
+\]
+Differentiate with respect to $J$ to find the angular frequency:
+\[
+\omega = \pdv{E}{J} = \frac{\pi^2 J}{mL^2}.
+\]
+The period is $T = 2\pi/\omega$:
+\[
+T = \frac{2\pi}{\pi^2 J/(mL^2)}
+= \frac{2mL^2}{\pi J}.
+\]
+Substitute $J = L\sqrt{2mE}/\pi$ to express $T$ in terms of $E$:
+\[
+T = \frac{2mL^2}{\pi\cdot L\sqrt{2mE}/\pi}
+= \frac{2mL^2}{L\sqrt{2mE}}
+= \frac{2mL}{\sqrt{2mE}}
+= 2L\sqrt{\frac{m}{2E}}.
+\]
+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
+\[
+T = \frac{2L}{v} = 2L\sqrt{\frac{m}{2E}},
+\]
+which agrees exactly with the action--angle result.
+
+\textbf{Part (c).} The given values are $m = 9.11\times 10^{-31}\,\mathrm{kg}$, $L = 1.00\times 10^{-10}\,\mathrm{m}$, and $E = 1.60\times 10^{-19}\,\mathrm{J}$.
+
+First compute the product $2mE$:
+\[
+2mE = 2(9.11\times 10^{-31})(1.60\times 10^{-19})\,\mathrm{kg^2\!\cdot\!m^2/s^2}
+= 2.92\times 10^{-49}\,\mathrm{kg^2\!\cdot\!m^2/s^2}.
+\]
+Taking the square root:
+\[
+\sqrt{2mE} = 5.40\times 10^{-25}\,\mathrm{kg\!\cdot\!m/s}.
+\]
+Now compute the action variable $J = L\sqrt{2mE}/\pi$:
+\[
+J = \frac{(1.00\times 10^{-10})(5.40\times 10^{-25})}{\pi}\,\mathrm{kg\!\cdot\!m^2/s}
+= \frac{5.40\times 10^{-35}}{\pi}\,\mathrm{kg\!\cdot\!m^2/s}
+= 1.72\times 10^{-35}\,\mathrm{kg\!\cdot\!m^2/s}.
+\]
+Next, the period $T = 2L\sqrt{m/(2E)}$. Compute $m/(2E)$:
+\[
+\frac{m}{2E} = \frac{9.11\times 10^{-31}}{2(1.60\times 10^{-19})}\,\mathrm{kg/J}
+= 2.85\times 10^{-12}\,\mathrm{s^2/m^2}.
+\]
+Taking the square root:
+\[
+\sqrt{\frac{m}{2E}} = 1.69\times 10^{-6}\,\mathrm{s/m}.
+\]
+Multiplying by $2L$:
+\[
+T = 2(1.00\times 10^{-10})(1.69\times 10^{-6})\,\mathrm{s}
+= 3.37\times 10^{-16}\,\mathrm{s}.
+\]
+
+Therefore,
+\[
+J = 1.72\times 10^{-35}\,\mathrm{kg\!\cdot\!m^2/s},
+\qquad
+T = 3.37\times 10^{-16}\,\mathrm{s}.
+\]

concepts/advanced/crossed-fields-hj.tex

@@ -0,0 +1,251 @@
+\subsection{E × B Drift}
+
+This subsection shows how a uniform electric field crossed with a uniform magnetic field produces a constant guiding-centre drift, derivable from the Hamilton--Jacobi equation by recognizing the harmonic nature of the transverse motion.
+
+\dfn{Crossed-field Hamiltonian in the Landau gauge}{
+A particle of mass $m$ and charge $q$ moves in a uniform electric field $\vec{E} = E_0\,\hat{\bm{y}}$ and a uniform magnetic field $\vec{B} = B_0\,\hat{\bm{z}}$. We choose the Landau gauge for the vector potential, $\vec{A} = (-B_0 y, 0, 0)$, and the scalar potential $\varphi = -E_0 y$. The curl of the vector potential,
+\[
+\nabla\times\vec{A}
+= \left(\pdv{A_z}{y} - \pdv{A_y}{z}\right)\hat{\bm{x}}
++ \left(\pdv{A_x}{z} - \pdv{A_z}{x}\right)\hat{\bm{y}}
++ \left(\pdv{A_y}{x} - \pdv{A_x}{y}\right)\hat{\bm{z}}
+= B_0\,\hat{\bm{z}},
+\]
+reproduces the magnetic field, and the gradient of the scalar potential gives $\vec{E} = -\nabla\varphi = E_0\,\hat{\bm{y}}$. The electromagnetic Hamiltonian for a charged particle,
+\[
+\mcH = \frac{1}{2m}\bigl|\vec{p} - q\vec{A}\bigr|^2 + q\varphi,
+\]
+becomes explicitly
+\[
+\mcH = \frac{1}{2m}\Bigl[\bigl(p_x + qB_0 y\bigr)^2 + p_y^2 + p_z^2\Bigr] - qE_0 y.
+\]
+The coordinates $x$ and $z$ do not appear in $\mcH$, so they are cyclic and their conjugate momenta $p_x$ and $p_z$ are conserved.}
+
+\nt{Gauge choice for crossed fields. The Landau gauge $\vec{A} = (-B_0 y, 0, 0)$ differs from the gauge used in the pure magnetic-field problem (subsection A.11), where cylindrical symmetry guided the choice. Here, placing the electric field along the $y$-direction makes this particular Landau gauge algebraically convenient: because $\vec{A}$ depends only on $y$, the coordinate $x$ remains cyclic and $p_x$ is automatically conserved. This conserved momentum $\alpha_x$ couples directly into the guiding-centre position, allowing the drift velocity to emerge cleanly from the Hamilton--Jacobi formalism without solving coupled differential equations.}
+
+\thm{$E \times B$ drift velocity from the guiding centre}{
+For crossed fields $\vec{E} = E_0\,\hat{\bm{y}}$ and $\vec{B} = B_0\,\hat{\bm{z}}$, the guiding centre of the orbit lies at
+\[
+y_c = \frac{mE_0/B_0 - \alpha_x}{qB_0},
+\]
+where $\alpha_x$ is the conserved canonical $x$-momentum. The time-averaged $x$-velocity (drift velocity) is
+\[
+v_d = \frac{\alpha_x + qB_0 y_c}{m} = \frac{E_0}{B_0},
+\]
+pointing in the $\hat{\bm{x}} = \hat{\bm{E}} \times \hat{\bm{B}}$ direction. The drift velocity is universal: every particle, regardless of charge or mass, drifts at this same speed.}
+
+\pf{Derivation of the drift velocity from the Hamilton--Jacobi equation}{
+Because the Hamiltonian has no explicit time dependence, energy is conserved: $\mcH = E$. The time variable separates as $\mcS = W(x,y,z) - Et$. The coordinates $x$ and $z$ are cyclic in $\mcH$, so their conjugate momenta are constants:
+\[
+\pdv{W}{x} = \alpha_x,
+\qquad
+\pdv{W}{z} = \alpha_z.
+\]
+The full action takes the additive form
+\[
+\mcS(x,y,z,t) = \alpha_x x + W_y(y) + \alpha_z z - Et.
+\]
+The time-independent Hamilton--Jacobi equation $\mcH(\vec{r},\nabla\mcS) = E$ becomes
+\[
+\frac{1}{2m}\Bigl[\bigl(\alpha_x + qB_0 y\bigr)^2 + \bigl(\der{W_y}{y}\bigr)^2 + \alpha_z^2\Bigr] - qE_0 y = E.
+\]
+The term involving the only remaining unknown is isolated by solving for $\bigl(\der{W_y}{y}\bigr)^2$:
+\[
+\bigl(\der{W_y}{y}\bigr)^2 = 2mE - \alpha_z^2 + 2mqE_0 y - \bigl(\alpha_x + qB_0 y\bigr)^2.
+\]
+Expand the square $\bigl(\alpha_x + qB_0 y\bigr)^2 = \alpha_x^2 + 2\alpha_x qB_0 y + q^2B_0^2 y^2$ and collect terms by powers of $y$:
+\[
+\bigl(\der{W_y}{y}\bigr)^2
+= -q^2B_0^2 y^2 + 2mqE_0 y - 2\alpha_x qB_0 y + 2mE - \alpha_x^2 - \alpha_z^2.
+\]
+Factor the linear-$y$ terms:
+\[
+\bigl(\der{W_y}{y}\bigr)^2
+= -q^2B_0^2 y^2 + 2qB_0\bigl(mE_0/B_0 - \alpha_x\bigr)y + 2mE - \alpha_x^2 - \alpha_z^2.
+\]
+Complete the square on the right-hand side. Factor out $-q^2B_0^2$ from the quadratic and linear terms in $y$:
+\[
+-q^2B_0^2\Biggl[y^2 - \frac{2}{qB_0}\Bigl(\frac{mE_0}{B_0} - \alpha_x\Bigr)y\Biggr].
+\]
+Add and subtract the square of half the coefficient of $y$ inside the bracket:
+\[
+-q^2B_0^2\Biggl[y - \frac{1}{qB_0}\Bigl(\frac{mE_0}{B_0} - \alpha_x\Bigr)\Biggr]^2
++ q^2B_0^2\Biggl[\frac{1}{qB_0}\Bigl(\frac{mE_0}{B_0} - \alpha_x\Bigr)\Biggr]^2.
+\]
+The shift in brackets is the guiding-centre coordinate:
+\[
+y_c = \frac{mE_0/B_0 - \alpha_x}{qB_0}.
+\]
+Completing the square has a clear physical meaning: it reveals the equilibrium position $y_c$ at which the electric force $qE_0\,\hat{\bm{y}}$ is exactly balanced by the magnetic force arising from the guiding-centre's $x$-velocity. At $y = y_c$, the canonical momentum combination $\alpha_x + qB_0 y_c = mE_0/B_0$ gives precisely the velocity needed for the magnetic Lorentz force to cancel the electric force. Deviations from $y_c$ are therefore harmonic oscillations about this equilibrium.
+Substituting back, the full equation becomes
+\[
+\bigl(\der{W_y}{y}\bigr)^2 + q^2B_0^2\bigl(y - y_c\bigr)^2 = 2mE - \alpha_z^2 - \alpha_x^2 + q^2B_0^2 y_c^2.
+\]
+The right-hand side is a constant determined by the energy and separation constants. Define $C = 2mE - \alpha_z^2 - \alpha_x^2 + q^2B_0^2 y_c^2$. Then
+\[
+\bigl(\der{W_y}{y}\bigr)^2 + q^2B_0^2\bigl(y - y_c\bigr)^2 = C.
+\]
+This is precisely the Hamilton--Jacobi equation for a harmonic oscillator in the shifted variable $Y = y - y_c$, with frequency
+\[
+\omega_c = \frac{|q|B_0}{m}.
+\]
+The $y$-motion oscillates sinusoidally about $y_c$ with the cyclotron frequency. The explicit time dependence of the coordinate follows from inverting the generating function:
+\[
+y(t) = y_c + A\sin\bigl(\omega_c t + \varphi\bigr),
+\]
+where $A$ is the oscillation amplitude determined by the total energy and $\varphi$ is a phase set by initial conditions. Trigonometric averaging over one cyclotron period $T_c = 2\pi/\omega_c$ gives
+\[
+\langle y \rangle = \frac{1}{T_c}\int_0^{T_c} y(t)\,\dd t
+= y_c + \frac{A}{T_c}\int_0^{T_c} \sin\bigl(\omega_c t + \varphi\bigr)\,\dd t
+= y_c,
+\]
+because the sine function integrates to zero over a complete period. The guiding-centre position is thus the exact time-average of the $y$-coordinate.
+
+Now compute the kinematic $x$-velocity. From Hamilton's equation, $\dot{x} = \pdv{\mcH}{p_x}$:
+\[
+v_x = \pdv{\mcH}{p_x}
+= \frac{1}{m}\bigl(p_x + qB_0 y\bigr).
+\]
+The canonical momentum $p_x = \pdv{\mcS}{x} = \alpha_x$ is constant. With the trigonometric average $\langle y \rangle = y_c$ established, the drift velocity follows from substituting $y_c$ into the expression for $v_x$:
+\[
+\langle v_x \rangle = \frac{\alpha_x + qB_0 y_c}{m}.
+\]
+Substitute the explicit expression for the guiding centre:
+\[
+qB_0 y_c = qB_0\cdot\frac{mE_0/B_0 - \alpha_x}{qB_0}
+= \frac{mE_0}{B_0} - \alpha_x.
+\]
+The separation constant $\alpha_x$ cancels out:
+\[
+\alpha_x + qB_0 y_c = \alpha_x + \frac{mE_0}{B_0} - \alpha_x = \frac{mE_0}{B_0}.
+\]
+Dividing by $m$ gives the drift velocity:
+\[
+\langle v_x \rangle = \frac{E_0}{B_0}.
+\]
+The drift is positive in the $+x$ direction, equal to $(E_0/B_0)\,\hat{\bm{x}} = (\vec{E}\times\vec{B})/B_0^2$, and depends on neither the particle's mass nor its charge.}
+
+\cor{Comparison with the Lorentz-force prediction}{
+The steady-state solution of the Lorentz-force equation $m\dot{\vec{v}} = q(\vec{E} + \vec{v}\times\vec{B})$ for zero acceleration, $\dot{\vec{v}} = 0$, requires
+\[
+q\vec{E} + q\vec{v}\times\vec{B} = 0,
+\qquad\text{or}\qquad
+\vec{v}\times\vec{B} = -\vec{E}.
+\]
+Take the cross product of both sides with $\vec{B}$ from the right:
+\[
+(\vec{v}\times\vec{B})\times\vec{B} = -\vec{E}\times\vec{B}.
+\]
+Using the vector identity $(\vec{a}\times\vec{b})\times\vec{c} = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{b}\cdot\vec{c})\vec{a}$:
+\[
+(\vec{v}\cdot\vec{B})\vec{B} - B^2\vec{v} = -\vec{E}\times\vec{B}.
+\]
+Since $\vec{E} \perp \vec{B}$, the velocity component parallel to $\vec{B}$ does not contribute to the drift, and choosing $\vec{v}\cdot\vec{B} = 0$ gives
+\[
+\vec{v}_d = \frac{\vec{E}\times\vec{B}}{B^2}.
+\]
+With $\vec{B} = B_0\,\hat{\bm{z}}$ and $\vec{E} = E_0\,\hat{\bm{y}}$:
+\[
+\vec{v}_d = \frac{E_0 B_0\,\hat{\bm{x}}}{B_0^2}
+= \frac{E_0}{B_0}\,\hat{\bm{x}}.
+\]
+This matches the Hamilton--Jacobi result exactly. The cross product $\vec{E}\times\vec{B}$ determines the drift direction and division by $B^2$ converts the magnitude into a velocity. Both formalisms predict the same drift regardless of the particle's charge or mass.}
+
+\nt{Universality of the $E \times B$ drift velocity. The drift velocity $v_d = E_0/B_0$ is independent of both the particle's mass and its charge sign. A charge-sign reversal changes the sense of Larmor rotation and shifts the guiding centre $y_c$, but these two effects exactly compensate in the time-averaged $x$-velocity. An electron and a proton spiralling in the same crossed fields share the same guiding-centre drift, even though their gyroradii and cyclotron frequencies differ enormously. This universality is precisely why the result is called the $E\times B$ drift velocity: the expression $\vec{v}_d = \vec{E}\times\vec{B}/B^2$ contains no mass or charge, so every charged species drifts together. In plasma physics this means bulk plasma moves as a coherent fluid rather than separating into counter-streaming components.}
+
+\mprop{Properties of the $E \times B$ drift}{
+The drift velocity $\vec{v}_d = \vec{E}\times\vec{B}/B^2$ satisfies the following properties:
+
+\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
+\item Independence of charge sign. Positive and negative charges drift in the same direction at the same speed. The sign of $q$ cancels between the force $q\vec{E}$ and the Lorentz deflection $q\vec{v}\times\vec{B}$.
+\item Independence of mass. Heavy ions and light electrons drift side by side at the same velocity. The mass appears in neither $\vec{E}\times\vec{B}$ nor $B^2$.
+\item Direction perpendicular to both fields. The drift points along $\hat{\bm{E}}\times\hat{\bm{B}}$, orthogonal to the plane containing the two fields.
+\item Magnitude depends on the field ratio. $v_d = E_0/B_0$ grows with stronger electric field and weaker magnetic field. Doubling both fields leaves the drift unchanged.
+\item The guiding centre $y_c$ itself depends on $q$, $m$, and $\alpha_x$, but these dependencies cancel in the drift velocity $\langle v_x\rangle$.
+\item The transverse $y$-motion is a harmonic oscillation with cyclotron frequency $\omega_c = |q|B_0/m$ about $y_c$. The full trajectory is a trochoid: the superposition of circular Larmor motion and uniform drift.
+\end{enumerate}
+}
+
+\ex{Trochoidal orbit geometry}{
+When the transverse kinetic energy is large compared to the electric-field energy scale $E_0$ times the gyroradius, the particle traces a cycloid-like path in the $xy$ plane while drifting in $x$. If the drift speed exceeds the thermal speed, the orbit is a prolate trochoid with open loops; at equal speeds it is a common cycloid with cusps; and at slower drift the orbit folds back on itself as a curtate trochoid. In every case the guiding centre advances uniformly at $v_d = E_0/B_0$.}
+
+\qs{Electron in crossed electric and magnetic fields}{
+An electron travels through a region with crossed fields $\vec{E} = (500\,\mathrm{V/m})\,\hat{\bm{y}}$ and $\vec{B} = (0.01\,\mathrm{T})\,\hat{\bm{z}}$. The electron has mass $m_e = 9.11\times 10^{-31}\,\mathrm{kg}$ and charge $q_e = -e = -1.60\times 10^{-19}\,\mathrm{C}$.
+
+\begin{enumerate}[label=(\alph*)]
+\item Compute the drift velocity $\vec{v}_d = \vec{E}\times\vec{B}/B^2$ and show it equals $50\,000\,\mathrm{m/s}$ in the $\hat{\bm{x}}$ direction.
+\item Using the guiding-centre formula $y_c = (mE_0/B_0 - \alpha_x)/(qB_0)$, verify that the drift $\langle v_x\rangle = (\alpha_x + qB_0 y_c)/m$ is independent of both $m$ and $q$.
+\item For a proton ($m_p = 1.67\times 10^{-27}\,\mathrm{kg}$, $q_p = +e = +1.60\times 10^{-19}\,\mathrm{C}$) in the same fields, confirm that the drift velocity is identical to the electron's.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The cross product of the field vectors is
+\[
+\vec{E}\times\vec{B}
+= \bigl(500\,\mathrm{V/m}\bigr)\bigl(0.01\,\mathrm{T}\bigr)\,\hat{\bm{y}}\times\hat{\bm{z}}
+= 5.00\,\mathrm{V\!\cdot\!T/m}\,\hat{\bm{x}}.
+\]
+The unit check: $1\,\mathrm{V\!\cdot\!T/m} = 1\,(\mathrm{V/m})/(\mathrm{T}) = 1\,\mathrm{m/s}$, because $\mathrm{T} = \mathrm{V\!\cdot\!s/m^2}$. The squared magnetic field strength is
+\[
+B^2 = (0.01\,\mathrm{T})^2 = 1.00\times 10^{-4}\,\mathrm{T^2}.
+\]
+The drift velocity is
+\[
+\vec{v}_d = \frac{\vec{E}\times\vec{B}}{B^2}
+= \frac{5.00\,\mathrm{V\!\cdot\!T/m}}{1.00\times 10^{-4}\,\mathrm{T^2}}\,\hat{\bm{x}}.
+\]
+Equivalently, using the scalar ratio directly:
+\[
+\frac{E_0}{B_0} = \frac{500\,\mathrm{V/m}}{0.01\,\mathrm{T}}
+= 50\,000\,\mathrm{m/s}.
+\]
+Therefore,
+\[
+\vec{v}_d = 50\,000\,\mathrm{m/s}\,\hat{\bm{x}}.
+\]
+The drift is in the $\hat{\bm{x}}$ direction, perpendicular to both $\vec{E}$ ($\hat{\bm{y}}$) and $\vec{B}$ ($\hat{\bm{z}}$), consistent with the $\hat{\bm{E}}\times\hat{\bm{B}}$ rule.
+
+\textbf{Part (b).} The guiding-centre position is
+\[
+y_c = \frac{mE_0/B_0 - \alpha_x}{qB_0}.
+\]
+Substitute into the drift formula for the time-averaged $x$-velocity:
+\[
+\langle v_x\rangle = \frac{\alpha_x + qB_0 y_c}{m}.
+\]
+First evaluate the product $qB_0 y_c$:
+\[
+qB_0 y_c = qB_0\cdot\frac{mE_0/B_0 - \alpha_x}{qB_0}
+= \frac{mE_0}{B_0} - \alpha_x.
+\]
+The factors of $qB_0$ cancel cleanly. Now the numerator of the drift is
+\[
+\alpha_x + qB_0 y_c = \alpha_x + \Bigl(\frac{mE_0}{B_0} - \alpha_x\Bigr)
+= \frac{mE_0}{B_0}.
+\]
+Dividing by $m$:
+\[
+\langle v_x\rangle = \frac{mE_0/B_0}{m} = \frac{E_0}{B_0}.
+\]
+Both the charge $q$ and the mass $m$ have cancelled algebraically. The drift depends only on the field ratio $E_0/B_0$. For the numerical values of this problem:
+\[
+\frac{E_0}{B_0} = \frac{500\,\mathrm{V/m}}{0.01\,\mathrm{T}} = 50\,000\,\mathrm{m/s},
+\]
+which agrees with part~(a).
+
+\textbf{Part (c).} For the proton, the guiding centre is
+\[
+y_c^{\text{(p)}} = \frac{m_p E_0/B_0 - \alpha_x^{\text{(p)}}}{q_p B_0}.
+\]
+The proton mass $m_p = 1.67\times 10^{-27}\,\mathrm{kg}$ is about $1837$ times the electron mass, and the proton charge $q_p = +e$ has the opposite sign. The numerical value of $y_c^{\text{(p)}}$ therefore differs substantially from the electron's guiding centre. However, the cancellation in the drift formula is purely algebraic and does not depend on the numerical values of $m$ or $q$. The proton drift is
+\[
+\langle v_x^{(\text{p})}\rangle = \frac{E_0}{B_0}
+= \frac{500\,\mathrm{V/m}}{0.01\,\mathrm{T}}
+= 50\,000\,\mathrm{m/s}.
+\]
+This is identical to the electron drift velocity. Every charged particle in the same crossed fields drifts at the same speed in the same direction, as predicted by both the Hamilton--Jacobi and Lorentz-force formalisms.
+
+Therefore,
+\[
+\vec{v}_d = (50\,000\,\mathrm{m/s})\,\hat{\bm{x}}
+\qquad\text{for both electron and proton, independent of charge and mass.}
+\]

concepts/advanced/cyclotron-hj.tex

@@ -0,0 +1,342 @@
+\subsection{Cyclotron Motion}
+
+This subsection solves for a charged particle moving in a uniform magnetic field through the Hamilton--Jacobi equation, derives the helical trajectory by quadrature, and computes the action-angle variables that recover the cyclotron frequency.
+
+\dfn{Hamilton--Jacobi formulation of cyclotron motion}{
+A particle of mass $m$ and charge $q$ moves in a uniform magnetic field $\vec{B} = B_0\,\hat{\bm{z}}$. Choose the Landau gauge for the vector potential $\vec{A} = (0, B_0 x, 0)$. The curl verifies the field:
+\[
+\nabla\times(0, B_0 x, 0)
+= \left(0,\; 0,\; \pdv{(B_0 x)}{x}\right)
+= B_0\,\hat{\bm{z}}.
+\]
+Set the scalar potential $\varphi = 0$. The Hamiltonian is
+\[
+\mcH = \frac{1}{2m}\Bigl[p_x^2 + \bigl(p_y - q B_0 x\bigr)^2 + p_z^2\Bigr].
+\]
+In this gauge, the coordinate $x$ appears explicitly in $\mcH$ while $y$ and $z$ are absent, so $y$ and $z$ are cyclic: their conjugate momenta $p_y$ and $p_z$ are constants of the motion. The Hamilton--Jacobi equation for $\mcS(\vec{r},t)$ reads
+\[
+\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \bigl(\pdv{\mcS}{y} - q B_0 x\bigr)^2 + \left(\pdv{\mcS}{z}\right)^2\right] + \pdv{\mcS}{t} = 0.
+\]}
+
+\nt{Why Landau gauge?}{
+The Landau gauge $\vec{A} = (0, B_0 x, 0)$ deliberately breaks rotational symmetry in the $xy$-plane to make $y$ a cyclic coordinate. With $y$ cyclic, $p_y$ is conserved and the HJ equation separates. In the symmetric gauge $\vec{A} = \tfrac{B_0}{2}(-y, x, 0)$, neither $x$ nor $y$ is cyclic -- the Hamiltonian depends on both, preventing straightforward HJ separation. Even after rotating to $x', y'$ coordinates, the $y'$-coordinate is \textbf{not} cyclic in the symmetric gauge. Both gauges describe the same physics, but only the Landau gauge unlocks the separation ansatz $\mcS = W_x(x) + \alpha_y y + \alpha_z z - Et$.}
+
+\nt{Two momenta}{
+In a magnetic field two distinct notions of momentum appear. The \textbf{canonical momentum} $p_y = \pdv{\mcS}{y} = \alpha_y$ is conserved because $y$ is cyclic in the Landau gauge. The \textbf{kinetic momentum} $m v_y = p_y - q B_0 x = \alpha_y - q B_0 x$ is not conserved -- it rotates at the cyclotron frequency as $x(t)$ oscillates. Conservation of $p_y$ fixes the orbit guiding center at $X_c = \alpha_y/(q B_0)$, while the rotating kinetic momentum generates the circular gyration about that center.}
+
+\thm{Complete integral for cyclotron motion}{
+The cyclotron frequency is $\omega_c = q B_0/m$. The guiding-center $x$-coordinate is $X_c = \alpha_y/(q B_0)$ and the gyroradius is $R = \sqrt{2m E_\perp}/(q B_0)$, where $\alpha_y$ is the conserved canonical $y$-momentum and $E_\perp$ is the transverse energy. The complete integral of the Hamilton--Jacobi equation is
+\[
+\mcS = W_x(x) + \alpha_y y + \alpha_z z - Et,
+\]
+where the $x$-part of the characteristic function is
+\[
+W_x(x) = \frac{E_\perp}{\omega_c}\arcsin\!\left(\frac{x - X_c}{R}\right)
++ \frac{q B_0}{2}\,(x - X_c)\sqrt{R^2 - (x - X_c)^2}.
+\]
+Here $E_\perp = E - \alpha_z^2/(2m)$ is the energy of motion in the $xy$-plane alone. The complete integral is defined for $|x - X_c| < R$.}
+
+\pf{Separation of the HJ equation and integration of the x-dependent part}{
+Because the potentials are time-independent, separate as $\mcS(\vec{r},t) = W(\vec{r}) - Et$. Because $y$ and $z$ are cyclic coordinates, set $\pdv{\mcS}{y} = \alpha_y$ and $\pdv{\mcS}{z} = \alpha_z$. The remaining dependence on $x$ is carried by a single function $W_x(x)$, so $W = W_x(x) + \alpha_y y + \alpha_z z$ and the time-independent Hamilton--Jacobi equation reads
+\[
+\frac{1}{2m}\left[\left(\der{W_x}{x}\right)^2 + \bigl(\alpha_y - q B_0 x\bigr)^2 + \alpha_z^2\right] = E.
+\]
+Define the transverse energy $E_\perp = E - \alpha_z^2/(2m)$. The $x$-equation simplifies to
+\[
+\left(\der{W_x}{x}\right)^2 + \bigl(\alpha_y - q B_0 x\bigr)^2 = 2m E_\perp.
+\]
+Solve for the spatial derivative:
+\[
+\der{W_x}{x} = \pm\sqrt{2m E_\perp - \bigl(\alpha_y - q B_0 x\bigr)^2}.
+\]
+The square root is real when $\abs{\alpha_y - q B_0 x} \le \sqrt{2m E_\perp}$. Introduce the guiding-center coordinate
+\[
+X_c = \frac{\alpha_y}{q B_0}.
+\]
+Then $\alpha_y - q B_0 x = -q B_0(x - X_c)$, and the radicand factors as
+\[
+2m E_\perp - q^2 B_0^2(x - X_c)^2
+= q^2 B_0^2\left(\frac{2m E_\perp}{q^2 B_0^2} - (x - X_c)^2\right).
+\]
+Define the gyroradius $R = \sqrt{2m E_\perp}/(q B_0)$. The derivative of $W_x$ becomes
+\[
+\der{W_x}{x} = \pm q B_0\sqrt{R^2 - (x - X_c)^2}.
+\]
+This has the same square-root structure as the simple harmonic oscillator. Integrate by the trigonometric substitution $x - X_c = R\sin\theta$, giving $\dd x = R\cos\theta\,\mathrm{d}\theta$:
+\[
+W_x = \int q B_0\sqrt{R^2 - R^2\sin^2\theta}\cdot R\cos\theta\,\mathrm{d}\theta
+= q B_0 R^2\int \cos^2\theta\,\mathrm{d}\theta.
+\]
+The antiderivative of $\cos^2\theta$ is $\tfrac12(\theta + \sin\theta\cos\theta)$, so
+\[
+W_x = \frac{q B_0 R^2}{2}\bigl(\theta + \sin\theta\cos\theta\bigr).
+\]
+Evaluate the constant prefactor using $R^2 = 2m E_\perp/(q^2 B_0^2)$:
+\[
+\frac{q B_0 R^2}{2}
+= \frac{q B_0}{2}\cdot\frac{2m E_\perp}{q^2 B_0^2}
+= \frac{m E_\perp}{q B_0}
+= \frac{E_\perp}{\omega_c}.
+\]
+Now express the trigonometric factors in terms of $x$:
+\[
+\theta = \arcsin\!\left(\frac{x - X_c}{R}\right),
+\qquad
+\sin\theta = \frac{x - X_c}{R},
+\qquad
+\cos\theta = \frac{\sqrt{R^2 - (x - X_c)^2}}{R}.
+\]
+The product $\sin\theta\cos\theta$ is $(x - X_c)\sqrt{R^2 - (x - X_c)^2}/R^2$. Substituting back,
+\[
+W_x(x) = \frac{E_\perp}{\omega_c}\arcsin\!\left(\frac{x - X_c}{R}\right)
++ \frac{E_\perp}{\omega_c}\cdot\frac{(x - X_c)\sqrt{R^2 - (x - X_c)^2}}{R^2}.
+\]
+The coefficient of the second term simplifies as
+\[
+\frac{E_\perp}{\omega_c R^2}
+= \frac{E_\perp}{\omega_c}\cdot\frac{q^2 B_0^2}{2m E_\perp}
+= \frac{q^2 B_0^2}{2m\omega_c}
+= \frac{q B_0}{2},
+\]
+since $\omega_c = q B_0/m$. Therefore,
+\[
+W_x(x) = \frac{E_\perp}{\omega_c}\arcsin\!\left(\frac{x - X_c}{R}\right)
++ \frac{q B_0}{2}\,(x - X_c)\sqrt{R^2 - (x - X_c)^2}.
+\]
+The full complete integral is $\mcS = W_x(x) + \alpha_y y + \alpha_z z - Et$.}
+
+\cor{Helical trajectory from Jacobi's theorem}{
+Jacobi's theorem states that differentiating the complete integral with respect to each separation constant produces a constant fixed by the initial conditions. Differentiate $\mcS$ with respect to $E_\perp$ at fixed $x$:
+\[
+\pdv{\mcS}{E_\perp} = \pdv{W_x}{E_\perp} - t.
+\]
+Write $\chi = (x - X_c)/R$ and $U = \sqrt{R^2 - (x - X_c)^2}$. The partial derivative of $W_x$ with respect to $E_\perp$ is
+\[
+\pdv{W_x}{E_\perp} = \frac{1}{\omega_c}\arcsin\chi 
++ \chi\cos\chi\cdot\pdv{E_\perp/\omega_c}{E_\perp}\cdot\frac{1}{(E_\perp/\omega_c)}
++ \frac{q B_0}{2}(x - X_c)\cdot\frac{1}{2U}\pdv{R^2}{E_\perp}.
+\]
+Because $R^2 = 2m E_\perp/(q^2 B_0^2)$, one has $\pdv{R^2}{E_\perp} = 2m/(q^2 B_0^2)$. The last two terms are
+\[
+-\frac{1}{2\omega_c}\frac{\chi}{\sqrt{1-\chi^2}}
+= -\frac{x - X_c}{2\omega_c R\sqrt{1-\chi^2}},
+\]
+\[
+\frac{q B_0}{2}(x - X_c)\cdot\frac{m}{q^2 B_0^2 U}
+= \frac{m}{q B_0}\cdot\frac{x - X_c}{2U}
+= \frac{x - X_c}{2\omega_c R\sqrt{1-\chi^2}},
+\]
+which exactly cancel, as in the harmonic oscillator case. Hence,
+\[
+\pdv{W_x}{E_\perp} = \frac{1}{\omega_c}\arcsin\!\left(\frac{x - X_c}{R}\right).
+\]
+Set $\pdv{\mcS}{E_\perp} = \beta$ (constant):
+\[
+\frac{1}{\omega_c}\arcsin\!\left(\frac{x - X_c}{R}\right) - t = \beta,
+\]
+or equivalently,
+\[
+x(t) = X_c + R\sin\!\bigl(\omega_c(t + \beta)\bigr).
+\]
+Define the initial phase $\phi_0 = \omega_c\beta$, measured in radians.
+
+Differentiate $\mcS$ with respect to the separation constant $\alpha_y$:
+\[
+\pdv{\mcS}{\alpha_y} = \pdv{W_x}{\alpha_y} + y = \beta_y.
+\]
+Since $X_c = \alpha_y/(q B_0)$, the chain rule gives $\pdv{W_x}{\alpha_y} = -\pdv{W_x}{X_c}\cdot(1/q B_0)$. From the structure of $W_x$, this derivative evaluates to $-(x - X_c)/R\cdot(E_\perp/(\omega_c R)) - \tfrac12\sqrt{R^2 - (x - X_c)^2}\cdot(q B_0/q B_0)$, but the result is more easily found from the canonical relation $v_y = (\alpha_y - q B_0 x)/m$:
+\[
+v_y(t) = \frac{\alpha_y - q B_0 x(t)}{m}
+= \frac{q B_0\bigl(X_c - x(t)\bigr)}{m}
+= -\omega_c R\sin(\omega_c t + \phi_0).
+\]
+Integrating with respect to time,
+\[
+y(t) = Y_c + R\cos(\omega_c t + \phi_0),
+\]
+where $Y_c$ is an integration constant set by the initial conditions. Meanwhile, for the $z$-direction, $p_z = \alpha_z$ is constant, giving $z(t) = (\alpha_z/m)t + z_0 = v_z t + z_0$. The full trajectory is helical:
+\[
+x(t) = X_c + R\sin(\omega_c t + \phi_0),
+\qquad
+y(t) = Y_c + R\cos(\omega_c t + \phi_0),
+\qquad
+z(t) = v_z t + z_0.
+\]
+The projection onto the $xy$-plane is a circle of radius $R$ centered at $(X_c, Y_c)$, traversed at the constant angular speed $\omega_c$. Superimposed is uniform motion along the field direction at speed $v_z$.}
+
+\nt{Hamilton's equations shortcut for $y(t)$}{
+The same $y(t)$ result follows directly from Hamilton's equations without differentiating $\mcS$. Because $y$ is cyclic,
+\[
+\dot{p}_y = -\pdv{\mcH}{y} = 0,
+\]
+so $p_y = \alpha_y$ is constant. Hamilton's velocity equation gives
+\[
+\dot{y} = \pdv{\mcH}{p_y} = \frac{1}{m}\bigl(p_y - q B_0 x\bigr)
+= \frac{\alpha_y - q B_0 x(t)}{m}.
+\]
+Substituting $x(t) = X_c + R\sin(\omega_c t + \phi_0)$ with $X_c = \alpha_y/(q B_0)$:
+\[
+\dot{y} = \frac{q B_0(X_c - X_c - R\sin(\omega_c t + \phi_0))}{m}
+= -\omega_c R\sin(\omega_c t + \phi_0).
+\]
+Integrating once yields $y(t) = Y_c + R\cos(\omega_c t + \phi_0)$, matching the HJ-derived trajectory. All angle arguments are in radians.}
+
+\nt{Guiding-center independence}{
+The guiding center $X_c = \alpha_y/(q B_0)$ depends only on the conserved canonical momentum $\alpha_y$, not on the transverse energy $E_\perp$. Physically, increasing $E_\perp$ enlarges the gyroradius $R = \sqrt{2m E_\perp}/(q B_0)$ but does not shift the orbit center. Energy changes the orbit size, not the center. This reflects the fact that the magnetic force is always perpendicular to velocity: it does no work, cannot change the particle's speed, and merely redirects the motion into circles whose centers are determined by the initial momentum partition, not the total energy.}
+
+\mprop{Action-angle variables for cyclotron motion}{
+The action-angle formalism applied to cyclotron motion yields the following results:
+
+\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
+\item The action variable associated with the transverse motion, defined as one-over-$2\pi$ times the phase-space area enclosed by one gyration, is
+\[
+J = \frac{1}{2\pi}\oint p_x\,\dd x = \frac{1}{\pi}\int_{X_c - R}^{X_c + R} q B_0\sqrt{R^2 - (x - X_c)^2}\,\dd x.
+\]
+The integral is the area of a semicircle of radius $R$ multiplied by $q B_0$. With the $1/\pi$ factor:
+\[
+J = \frac{q B_0}{\pi}\cdot\frac{\pi R^2}{2}
+= \frac{q B_0 R^2}{2}
+= \frac{q B_0}{2}\cdot\frac{2m E_\perp}{q^2 B_0^2}
+= \frac{m E_\perp}{q B_0}
+= \frac{E_\perp}{\omega_c}.
+\]
+Geometrically, the full phase-space area $2\pi J/(q B_0) = \pi R^2$ is the area of the real-space circular orbit.
+
+\item Inverting the action--energy relation, the transverse energy as a function of the action is
+\[
+E_\perp(J) = \omega_c J.
+\]
+The Hamiltonian expressed in terms of the action variables is $E = \omega_c J + \alpha_z^2/(2m)$, linear in $J$ and quadratic in $\alpha_z$.
+
+\item The Hamilton--Jacobi frequency is $\pdv{E_\perp}{J} = \omega_c$, which already has units of angular frequency (rad/s). It depends only on the charge-to-mass ratio and the field strength, and is independent of the transverse energy $E_\perp$ and the gyroradius $R$. This amplitude independence is the hallmark of uniform circular motion in a constant magnetic field.
+
+\item The angle variable advances linearly in time: $w = \omega_c t + w_0$. The phase of the circular gyration, $\omega_c t + \phi_0$ (all angles in radians), equals $w$ up to a constant phase. The action-angle variables provide a clean canonical description even though the original Cartesian coordinates exhibit coupled oscillatory dynamics.
+\end{enumerate}
+}
+
+\nt{Comparison with the Lorentz force}{
+The Lorentz force law $m\,\ddot{\vec{r}} = q\,\dot{\vec{r}}\times\vec{B}$ with $\vec{B} = B_0\hat{\bm{z}}$ gives the component equations (see also e12-2 for cyclotron motion from Lorentz force):
+\[
+\ddot{x} = \frac{q B_0}{m}\,\dot{y},
+\qquad
+\ddot{y} = -\frac{q B_0}{m}\,\dot{x},
+\qquad
+\ddot{z} = 0.
+\]
+Introducing $v_x = \dot{x}$ and $v_y = \dot{y}$, these become $\dot{v}_x = \omega_c v_y$ and $\dot{v}_y = -\omega_c v_x$, whose solutions are
+\[
+v_x = v_\perp\cos(\omega_c t + \phi_0),
+\qquad
+v_y = -v_\perp\sin(\omega_c t + \phi_0).
+\]
+Integrating once more gives the same circular trajectory with radius $R = v_\perp/\omega_c = \sqrt{2m E_\perp}/(q B_0)$, and uniform $z$-motion. The Hamilton--Jacobi approach arrives at the identical values of $\omega_c$ and $R$ through a radically different route: solving a first-order nonlinear PDE by separation and quadrature, then differentiating the complete integral. The agreement reaffirms the consistency of the Hamiltonian and Newtonian formulations.}
+
+\qs{Proton cyclotron motion from the HJ complete integral}{
+A proton of mass $m = 1.67\times 10^{-27}\,\mathrm{kg}$ and charge $q = e = 1.60\times 10^{-19}\,\mathrm{C}$ moves in a uniform magnetic field $\vec{B} = (1.5\,\mathrm{T})\,\hat{\bm{z}}$. The transverse (perpendicular-to-field) kinetic energy is $E_\perp = 1.0\,\mathrm{keV} = 1.60\times 10^{-16}\,\mathrm{J}$.
+
+\begin{enumerate}[label=(\alph*)]
+\item Compute the cyclotron angular frequency $\omega_c = q B_0/m$ and the gyration period $T = 2\pi/\omega_c$.
+\item Find the gyroradius $R = \sqrt{2m E_\perp}/(q B_0)$ in meters.
+\item Compute the action variable $J = E_\perp/\omega_c$ in SI units and verify by differentiation that $\pdv{E_\perp}{J} = \omega_c$, recovering the cyclotron angular frequency.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The cyclotron angular frequency is
+\[
+\omega_c = \frac{q B_0}{m}.
+\]
+Substitute the given values:
+\[
+q = 1.60\times 10^{-19}\,\mathrm{C},
+\qquad
+B_0 = 1.5\,\mathrm{T},
+\qquad
+m = 1.67\times 10^{-27}\,\mathrm{kg}.
+\]
+Form the ratio:
+\[
+\omega_c = \frac{(1.60\times 10^{-19})(1.5)}{1.67\times 10^{-27}}\,\mathrm{rad/s}
+= \frac{2.40\times 10^{-19}}{1.67\times 10^{-27}}\,\mathrm{rad/s}.
+\]
+This gives
+\[
+\omega_c = 1.437\times 10^8\,\mathrm{rad/s}.
+\]
+Rounding to two significant figures (consistent with the field strength $1.5\,\mathrm{T}$),
+\[
+\omega_c = 1.4\times 10^8\,\mathrm{rad/s}.
+\]
+The gyration period is
+\[
+T = \frac{2\pi}{\omega_c}
+= \frac{2\pi}{1.437\times 10^8}\,\mathrm{s}
+= 4.37\times 10^{-8}\,\mathrm{s}.
+\]
+In more convenient units,
+\[
+T = 4.4\times 10^{-8}\,\mathrm{s} = 44\,\mathrm{ns}.
+\]
+
+\textbf{Part (b).} The gyroradius is
+\[
+R = \frac{\sqrt{2m E_\perp}}{q B_0}.
+\]
+Evaluate the numerator:
+\[
+2m E_\perp = 2(1.67\times 10^{-27}\,\mathrm{kg})(1.60\times 10^{-16}\,\mathrm{J})
+= 5.34\times 10^{-43}\,\mathrm{kg^2\,m^2/s^2}.
+\]
+Taking the square root:
+\[
+\sqrt{2m E_\perp} = \sqrt{5.34\times 10^{-43}}\,\mathrm{kg\,m/s}
+= 7.31\times 10^{-22}\,\mathrm{kg\,m/s}.
+\]
+The denominator is
+\[
+q B_0 = (1.60\times 10^{-19}\,\mathrm{C})(1.5\,\mathrm{T})
+= 2.40\times 10^{-19}\,\mathrm{C\,T}.
+\]
+Therefore,
+\[
+R = \frac{7.31\times 10^{-22}}{2.40\times 10^{-19}}\,\mathrm{m}
+= 3.05\times 10^{-3}\,\mathrm{m}.
+\]
+In more convenient units,
+\[
+R = 3.05\,\mathrm{mm}.
+\]
+
+\textbf{Part (c).} The action variable for the transverse cyclotron motion is
+\[
+J = \frac{E_\perp}{\omega_c}.
+\]
+Substitute the numerical values:
+\[
+J = \frac{1.60\times 10^{-16}\,\mathrm{J}}{1.437\times 10^8\,\mathrm{rad/s}}
+= 1.11\times 10^{-24}\,\mathrm{J\!\cdot\!s}.
+\]
+This gives
+\[
+J = 1.1\times 10^{-24}\,\mathrm{J\!\cdot\!s}.
+\]
+Now verify the energy--action relation $E_\perp(J) = \omega_c J$. Differentiating with respect to $J$:
+\[
+\pdv{E_\perp}{J} = \omega_c.
+\]
+The derivative directly equals the cyclotron angular frequency. For the numerical values,
+\[
+E_\perp(J) = \omega_c J
+= (1.437\times 10^8\,\mathrm{rad/s})(1.11\times 10^{-24}\,\mathrm{J\!\cdot\!s})
+= 1.60\times 10^{-16}\,\mathrm{J},
+\]
+which is exactly the original transverse energy of $1.0\,\mathrm{keV}$. The relation $\pdv{E_\perp}{J} = \omega_c$ holds both algebraically and numerically, confirming that the action-angle formalism reproduces the cyclotron angular frequency exactly.
+
+Therefore,
+\[
+\omega_c = 1.4\times 10^8\,\mathrm{rad/s},
+\qquad
+T = 44\,\mathrm{ns},
+\qquad
+R = 3.1\,\mathrm{mm},
+\qquad
+J = 1.1\times 10^{-24}\,\mathrm{J\!\cdot\!s}.
+\]

concepts/advanced/free-particle-hj.tex

@@ -0,0 +1,160 @@
+\subsection{Free Particle in 1D and 3D}
+
+The free particle is the simplest test of the Hamilton--Jacobi formalism, showing the full machinery in the least cluttered setting. With no potential to complicate the Hamiltonian, every step of the method -- separation of variables, identification of the complete integral, application of Jacobi\normalsize{}'s theorem -- can be seen clearly. Here we solve the equation in both one and three dimensions and recover the familiar result of uniform straight-line motion.
+
+\dfn{Free particle Hamiltonian and Hamilton--Jacobi equation}{
+For a free particle of mass $m$ the Hamiltonian is purely kinetic:
+\[
+\mcH = \frac{p^2}{2m}.
+\]
+In one dimension, substituting $p = \pdv{\mcS}{x}$ into the Hamilton--Jacobi equation $\mcH + \pdv{\mcS}{t} = 0$ yields
+\[
+\frac{1}{2m}\left(\pdv{\mcS}{x}\right)^2 + \pdv{\mcS}{t} = 0.
+\]
+The three-dimensional Hamiltonian is $\mcH = (p_x^2 + p_y^2 + p_z^2)/(2m)$ and the corresponding Hamilton--Jacobi equation is
+\[
+\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \left(\pdv{\mcS}{y}\right)^2 + \left(\pdv{\mcS}{z}\right)^2\right] + \pdv{\mcS}{t} = 0.
+\]}
+
+\thm{Complete integral for a 1D free particle}{
+The complete integral of the one-dimensional free-particle Hamilton--Jacobi equation is
+\[
+\mcS(x,t;E) = \pm\sqrt{2mE}\,x - Et,
+\]
+where $E > 0$ is the total mechanical energy. Jacobi's theorem $\pdv{\mcS}{E} = \beta$ gives the trajectory
+\[
+x(t) = \pm\sqrt{\frac{2E}{m}}\,(t+\beta) = v_0 t + x_0,
+\]
+with constant velocity $v_0 = \pm\sqrt{2E/m}$ and initial position $x_0 = v_0\beta$.}
+
+\pf{Derivation of the 1D and 3D free-particle action}{
+Because the free-particle Hamiltonian has no explicit time dependence, $\pdv{\mcH}{t} = 0$ and energy is conserved: $\mcH = E$. Use the time-independent reduction $\mcS(x,t) = W(x) - Et$. Substituting:
+\[
+\frac{1}{2m}\left(\der{W}{x}\right)^2 - E = 0.
+\]
+Solve for the derivative:
+\[
+\der{W}{x} = \pm\sqrt{2mE}.
+\]
+Integrate with respect to $x$ (absorbing the integration constant into the additive constant of $\mcS$):
+\[
+W(x) = \pm\sqrt{2mE}\,x.
+\]
+Reassemble the principal function:
+\[
+\mcS(x,t) = \pm\sqrt{2mE}\,x - Et.
+\]
+Jacobi's theorem requires $\pdv{\mcS}{E} = \beta$. Differentiate:
+\[
+\pdv{\mcS}{E} = \pm\frac{m}{\sqrt{2mE}}\,x - t = \beta.
+\]
+Solve for $x(t)$:
+\[
+x = \pm\frac{\sqrt{2mE}}{m}\,(t+\beta) = \pm\sqrt{\frac{2E}{m}}\,(t+\beta).
+\]
+Setting $v_0 = \pm\sqrt{2E/m}$ and $x_0 = v_0\beta$ gives $x(t) = v_0 t + x_0$.
+
+In three dimensions all Cartesian coordinates are cyclic, so each conjugate momentum is conserved. Setting $\pdv{\mcS}{x} = p_x$, $\pdv{\mcS}{y} = p_y$, $\pdv{\mcS}{z} = p_z$ as constants:
+\[
+W(x,y,z) = p_x x + p_y y + p_z z,
+\]
+with energy $E = (p_x^2 + p_y^2 + p_z^2)/(2m)$. The principal function is
+\[
+\mcS(x,y,z,t) = p_x x + p_y y + p_z z - Et.
+\]
+Treating $(p_x, p_y, p_z)$ as three independent separation constants, Jacobi's theorem gives
+\[
+\pdv{\mcS}{p_x} = x - \frac{p_x}{m}\,t = \beta_x,
+\qquad
+\pdv{\mcS}{p_y} = y - \frac{p_y}{m}\,t = \beta_y,
+\qquad
+\pdv{\mcS}{p_z} = z - \frac{p_z}{m}\,t = \beta_z.
+\]
+Each coordinate evolves linearly with time, confirming uniform straight-line motion in three dimensions.}
+
+\nt{Connection to earlier material}{Free-particle motion was already solved in Unit 1 (Kinematics), where constant velocity and linear position functions $x(t) = v_0 t + x_0$ were obtained directly from Newton\normalsize{}'s second law for zero net force. Unit 4 (Momentum and Impulse) used conservation of linear momentum to analyze collisions of free particles.
+
+Here we recover all of those results without writing a differential equation: each coordinate evolves as $q_i(t) = (p_i/m)t + \beta_i$, carrying a conserved momentum $p_i$ and zero acceleration $\ddot{q}_i = 0$. The Hamilton--Jacobi formalism reproduces familiar kinematics from the geometry of the action, treating the entire trajectory as a consequence of the action principle rather than Newton\normalsize{}'s second law.}
+
+\nt{Geometric picture of the action}{In three dimensions the principal function $\mcS(x,y,z,t) = p_x x + p_y y + p_z z - Et$ is a linear function, so its level sets are planes in configuration space. The surfaces $\mcS = \text{const}$ are parallel planes $p_x x + p_y y + p_z z = \text{const} + Et$, propagating along the direction of $\mathbf{p}$ with speed $|\mathbf{p}|/m$.
+
+This is the wavefront picture from the optics analogy discussed in A.01. The action $\mcS$ plays the role of an optical phase, and curves orthogonal to the wavefronts are the ray paths. Those ray paths coincide with the particle trajectories. For a free particle there is no potential to refract the rays, so the wavefronts remain perfectly planar and propagate without distortion. The constant velocity of the wavefronts reflects the constant speed of the particle itself.}
+
+As stated in Jacobi\normalsize{}'s theorem (A.01), setting $\pdv{\mcS}{\alpha_i} = \beta_i$ for each separation constant $\alpha_i$ gives the equations of motion.
+
+\qs{Free particle in three dimensions}{A free particle of mass $m = 2.0\,\mathrm{kg}$ passes through the origin at $t = 0$ with initial velocity $(3.0,\,4.0,\,0)\,\mathrm{m/s}$.
+
+\begin{enumerate}[label=(\alph*)]
+\item Write Hamilton's principal function $\mcS(x,y,z,t)$ using the additive separation ansatz $\mcS = p_x x + p_y y + p_z z - Et$, substituting numerical values for the momenta and energy.
+\item From Jacobi's theorem, $\pdv{\mcS}{p_x} = \beta_x$, $\pdv{\mcS}{p_y} = \beta_y$, $\pdv{\mcS}{p_z} = \beta_z$, find $x(t)$, $y(t)$, and $z(t)$ using the given initial conditions.
+\item At $t = 2.5\,\mathrm{s}$, compute the position vector $\mathbf{r}(t)$, the speed $|\mathbf{v}|$, and the kinetic energy $K$. Verify that $K$ equals the total energy $E$ found in Part (a).
+\end{enumerate}}
+
+\sol \textbf{Part (a).} Compute the canonical momenta from the initial velocity:
+\[
+p_x = m v_{x0} = (2.0)(3.0)\,\mathrm{kg\!\cdot\!m/s} = 6.0\,\mathrm{kg\!\cdot\!m/s},
+\]
+\[
+p_y = m v_{y0} = (2.0)(4.0)\,\mathrm{kg\!\cdot\!m/s} = 8.0\,\mathrm{kg\!\cdot\!m/s},
+\qquad
+p_z = 0.
+\]
+The total energy is
+\[
+E = \frac{p_x^2 + p_y^2 + p_z^2}{2m}
+= \frac{(6.0)^2 + (8.0)^2}{2(2.0)}\,\mathrm{J}
+= \frac{36 + 64}{4.0}\,\mathrm{J}
+= 25\,\mathrm{J}.
+\]
+Substitute these into the additive ansatz:
+\[
+\mcS(x,y,z,t) = 6.0\,x + 8.0\,y - 25\,t,
+\]
+where $x$ and $y$ are in metres, $t$ in seconds, and $\mcS$ in joule-seconds.
+
+\textbf{Part (b).} Apply Jacobi's theorem for each momentum component. For the $x$-coordinate:
+\[
+\pdv{\mcS}{p_x} = x - \frac{p_x}{m}\,t = \beta_x.
+\]
+At $t = 0$ the particle is at the origin, so $x(0) = 0$ and $\beta_x = 0$. Therefore,
+\[
+x(t) = \frac{p_x}{m}\,t = \frac{6.0}{2.0}\,t = 3.0\,t.
+\]
+For the $y$-coordinate:
+\[
+\pdv{\mcS}{p_y} = y - \frac{p_y}{m}\,t = \beta_y.
+\]
+With $y(0) = 0$, we have $\beta_y = 0$ and
+\[
+y(t) = \frac{p_y}{m}\,t = \frac{8.0}{2.0}\,t = 4.0\,t.
+\]
+For the $z$-coordinate:
+\[
+\pdv{\mcS}{p_z} = z - \frac{p_z}{m}\,t = \beta_z.
+\]
+Since $p_z = 0$ and $z(0) = 0$, we obtain $\beta_z = 0$ and
+\[
+z(t) = 0.
+\]
+
+\textbf{Part (c).} The position vector at $t = 2.5\,\mathrm{s}$ is
+\[
+\mathbf{r}(2.5\,\mathrm{s}) = \bigl(3.0(2.5),\,4.0(2.5),\,0\bigr)\,\mathrm{m}
+= (7.5,\,10.0,\,0)\,\mathrm{m}.
+\]
+The velocity is constant, $\mathbf{v} = (3.0,\,4.0,\,0)\,\mathrm{m/s}$, so the speed is
+\[
+|\mathbf{v}| = \sqrt{3.0^2 + 4.0^2} = 5.0\,\mathrm{m/s}.
+\]
+The kinetic energy at this instant is
+\[
+K = \tfrac{1}{2}\,m\,|\mathbf{v}|^2 = \tfrac{1}{2}(2.0)(5.0)^2\,\mathrm{J} = 25\,\mathrm{J}.
+\]
+This equals the total energy $E = 25\,\mathrm{J}$ from Part (a), confirming energy conservation for the free particle.
+
+Therefore,
+\[
+\mcS(x,y,z,t) = 6.0\,x + 8.0\,y - 25\,t,
+\qquad
+\mathbf{r}(t) = (3.0\,t,\,4.0\,t,\,0)\,\mathrm{m}.
+\]

concepts/advanced/hj-em-coupling.tex

@@ -0,0 +1,243 @@
+\subsection{Hamilton-Jacobi with Electromagnetic Fields}
+
+This subsection extends the Hamilton--Jacobi framework to a charged particle moving in electromagnetic fields by replacing the canonical momentum with the minimal-coupling substitution $p \to p - qA$.
+
+\nt{Where does the $q\,\vec{v}\cdot\vec{A}$ term come from?}{
+We can reverse-engineer the electromagnetic Lagrangian by demanding that the Euler--Lagrange equations reproduce the Lorentz force. Start from the ansatz $\mcL = \tfrac12 m v^2 - q\varphi + q\,\vec{v}\cdot\vec{A}$ and check the $x$-component. The Euler--Lagrange equation reads
+\[
+\frac{\dd}{\dd t}\!\left(\pdv{\mcL}{\dot{x}}\right) = \pdv{\mcL}{x}.
+\]
+The canonical momentum is $\pdv{\mcL}{\dot{x}} = m\dot{x} + qA_x$. Its total time derivative expands as
+\[
+\frac{\dd}{\dd t}(m\dot{x}+qA_x) = m\ddot{x} + q\pdv{A_x}{t} + q\dot{x}\pdv{A_x}{x} + q\dot{y}\pdv{A_x}{y} + q\dot{z}\pdv{A_x}{z}.
+\]
+The spatial derivative of the Lagrangian is
+\[
+\pdv{\mcL}{x} = -q\pdv{\varphi}{x} + q\dot{x}\pdv{A_x}{x} + q\dot{y}\pdv{A_y}{x} + q\dot{z}\pdv{A_z}{x}.
+\]
+Equating left and right sides and rearranging, the terms $q\dot{x}\pdv{A_x}{x}$ cancel, leaving
+\[
+m\ddot{x} = q\Biggl(-\pdv{\varphi}{x} - \pdv{A_x}{t}\Biggr) + q\dot{y}\Biggl(\pdv{A_y}{x} - \pdv{A_x}{y}\Biggr) + q\dot{z}\Biggl(\pdv{A_z}{x} - \pdv{A_x}{z}\Biggr).
+\]
+The first grouped term is $qE_x$ using $E_x = -\pdv{\varphi}{x} - \pdv{A_x}{t}$. The next two are the $x$-component of $q(\vec{v}\times\vec{B})_x = q(\dot{y}B_z - \dot{z}B_y)$ using $B_z = \pdv{A_y}{x} - \pdv{A_x}{y}$ and $B_y = \pdv{A_x}{z} - \pdv{A_z}{x}$. Thus $m\ddot{x} = q(E_x + (\vec{v}\times\vec{B})_x)$, matching the Lorentz force. The Lagrangian $\mcL = \tfrac12 mv^2 - q\varphi + q\,\vec{v}\cdot\vec{A}$ is not a guess --- it is uniquely fixed by the requirement that the variational principle yield $\vec{F} = q(\vec{E}+\vec{v}\times\vec{B})$.}
+
+\dfn{Lagrangian for a charged particle in electromagnetic fields}{
+Let a particle of mass $m$ and charge $q$ move with velocity $\vec{v}$ in an electromagnetic field described by the scalar potential $\varphi(\vec{r},t)$ and the vector potential $\vec{A}(\vec{r},t)$. The Lagrangian is
+\[
+\mcL = \tfrac12 m v^2 - q\varphi + q\,\vec{v}\cdot\vec{A}.
+\]
+The term $-q\varphi$ represents the electrostatic potential energy, while the velocity-dependent term $q\,\vec{v}\cdot\vec{A}$ is the magnetic interaction. Together they reproduce both the electric and magnetic parts of the Lorentz force when the Euler--Lagrange equations are applied. The canonical momentum conjugate to each spatial coordinate $r_i$ is
+\[
+p_i = \pdv{\mcL}{\dot{r}_i} = m v_i + q A_i,
+\]
+so that in vector notation,
+\[
+\vec{p} = m\vec{v} + q\vec{A}.
+\]
+The kinetic (mechanical) momentum is $m\vec{v} = \vec{p} - q\vec{A}$, and it is this combination that enters the kinetic-energy part of the Hamiltonian. The passage from a free-particle Hamiltonian to the electromagnetic one simply requires the substitutions $\vec{p} \to \vec{p} - q\vec{A}$ and $E \to E - q\varphi$. This procedure is called \emph{minimal coupling} because it represents the lowest-order way to couple electromagnetic potentials to particle motion: the potentials enter only through the simple shift of the canonical momentum, with no higher-order derivative couplings, no spin-dependent terms, and no direct coupling of the field-strength tensor to the particle coordinates.}
+
+\nt{Canonical versus kinetic momentum}{
+It is crucial to distinguish two notions of momentum for a charged particle. The canonical momentum $\vec{p} = m\vec{v} + q\vec{A}$ is the formal variable that appears in Hamilton's equations and the Hamilton--Jacobi equation. It is the quantity conserved when its corresponding coordinate is cyclic (absent from the Hamiltonian). The kinetic momentum $m\vec{v} = \vec{p} - q\vec{A}$, by contrast, is the physically measured momentum: it is the mass times the actual velocity of the particle, and its time derivative equals the mechanical Lorentz force $q(\vec{E}+\vec{v}\times\vec{B})$. These two momenta differ by $q\vec{A}$. Because $\vec{A}$ generally depends on position, the canonical momentum is not simply $m\vec{v}$, and its time derivative is not equal to the mechanical force. The canonical momentum is what Hamilton's equations govern; the kinetic momentum is what a detector would measure.}
+
+\thm{Hamiltonian for a charged particle in electromagnetic fields}{
+Let $m$ denote the mass, let $q$ denote the charge, let $\varphi(\vec{r},t)$ denote the scalar potential, and let $\vec{A}(\vec{r},t)$ denote the vector potential. The canonical momentum has components $p_i$. Then the Hamiltonian of the charged particle is
+\[
+\mcH(\vec{r},\vec{p},t) = \frac{1}{2m}\bigl|\vec{p} - q\vec{A}(\vec{r},t)\bigr|^2 + q\,\varphi(\vec{r},t).
+\]
+The corresponding Hamilton--Jacobi equation for the principal function $\mcS(\vec{r},t)$ follows by replacing $p_i$ with $\pdv{\mcS}{r_i}$:
+\[
+\frac{1}{2m}\left|\nabla\mcS - q\vec{A}\right|^2 + q\,\varphi + \pdv{\mcS}{t} = 0.
+\]}
+
+\pf{Derivation from the Lagrangian to the HJ equation}{
+Start from the Lagrangian
+\[
+\mcL = \tfrac12 m\,\dot{\vec{r}}\cdot\dot{\vec{r}} - q\varphi(\vec{r},t) + q\,\dot{\vec{r}}\cdot\vec{A}(\vec{r},t).
+\]
+Compute the canonical momentum by differentiating with respect to each velocity component:
+\[
+\vec{p} = \pdv{\mcL}{\dot{\vec{r}}} = m\dot{\vec{r}} + q\vec{A}.
+\]
+Invert this relation to express the velocity in terms of the canonical momentum:
+\[
+\dot{\vec{r}} = \frac{1}{m}\bigl(\vec{p} - q\vec{A}\bigr).
+\]
+Form the Legendre transform to obtain the Hamiltonian:
+\[
+\mcH = \vec{p}\cdot\dot{\vec{r}} - \mcL.
+\]
+Substitute the expressions for $\dot{\vec{r}}$ and $\mcL$. The kinetic-term piece gives
+\[
+\vec{p}\cdot\dot{\vec{r}} = \frac{1}{m}\,\vec{p}\cdot\bigl(\vec{p} - q\vec{A}\bigr) = \frac{1}{m}\bigl(p^2 - q\,\vec{p}\cdot\vec{A}\bigr),
+\]
+and the Lagrangian itself reads
+\[
+\mcL = \tfrac{1}{2m}\bigl|\vec{p} - q\vec{A}\bigr|^2 - q\varphi + \frac{q}{m}\bigl(\vec{p} - q\vec{A}\bigr)\cdot\vec{A}.
+\]
+The dot product in the last term of $\mcL$ expands as
+\[
+\frac{q}{m}\bigl(\vec{p} - q\vec{A}\bigr)\cdot\vec{A} = \frac{q}{m}\bigl(\vec{p}\cdot\vec{A} - qA^2\bigr).
+\]
+Subtracting $\mcL$ from $\vec{p}\cdot\dot{\vec{r}}$ gives
+\[
+\mcH = \frac{1}{m}\bigl(p^2 - q\,\vec{p}\cdot\vec{A}\bigr) - \Biggl[\frac{1}{2m}\bigl(p^2 - 2q\,\vec{p}\cdot\vec{A} + q^2A^2\bigr) - q\varphi + \frac{q}{m}\bigl(\vec{p}\cdot\vec{A} - qA^2\bigr)\Biggr].
+\]
+Combine the terms proportional to $\vec{p}\cdot\vec{A}$:
+\[
+-\frac{q}{m}\,\vec{p}\cdot\vec{A} - \frac{q}{m}\,\vec{p}\cdot\vec{A} + \frac{2q}{2m}\,\vec{p}\cdot\vec{A} = -\frac{q}{m}\,\vec{p}\cdot\vec{A}.
+\]
+Combine the terms proportional to $A^2$:
+\[
+-\frac{q^2}{2m}A^2 + \frac{q^2}{m}A^2 = \frac{q^2}{2m}A^2.
+\]
+Together with the $p^2$ terms, $\frac{p^2}{m} - \frac{p^2}{2m} = \frac{p^2}{2m}$. All terms assemble into the compact form
+\[
+\mcH = \frac{1}{2m}\bigl|\vec{p} - q\vec{A}\bigr|^2 + q\varphi.
+\]
+Now replace each component of the canonical momentum by the gradient of the action function, $p_i = \pdv{\mcS}{r_i}$, so that $\vec{p} \to \nabla\mcS$. The Hamilton--Jacobi equation $\mcH + \pdv{\mcS}{t} = 0$ then becomes
+\[
+\frac{1}{2m}\left|\nabla\mcS - q\vec{A}\right|^2 + q\,\varphi + \pdv{\mcS}{t} = 0.
+\]}
+
+\cor{Minimal coupling rule and gauge invariance}{
+The passage from a free particle to a charged particle in electromagnetic fields is effected by the minimal coupling substitutions
+\[
+\vec{p} \longrightarrow \vec{p} - q\vec{A},
+\qquad
+E \longrightarrow E - q\varphi.
+\]
+To understand why these substitutions are consistent, consider gauge transformations of the potentials. An arbitrary smooth scalar function $\chi(\vec{r},t)$ defines the gauge transformation
+\[
+\vec{A}' = \vec{A} + \nabla\chi,
+\qquad
+\varphi' = \varphi - \pdv{\chi}{t}.
+\]
+The physical electric and magnetic fields expressed in terms of potentials are $\vec{E} = -\nabla\varphi - \pdv{\vec{A}}{t}$ and $\vec{B} = \nabla\times\vec{A}$. Substituting the primed potentials,
+\[
+\vec{E}' = -\nabla\varphi' - \pdv{\vec{A}'}{t} = -\nabla\varphi + \nabla\pdv{\chi}{t} - \pdv{\vec{A}}{t} - \nabla\pdv{\chi}{t} = \vec{E},
+\]
+\[
+\vec{B}' = \nabla\times\vec{A}' = \nabla\times(\vec{A}+\nabla\chi) = \nabla\times\vec{A} + 0 = \vec{B}.
+\]
+Both $\vec{E}$ and $\vec{B}$ are unchanged, as required since potentials are a redundant mathematical description and only the fields are physically measurable. How does the Lagrangian behave? Under the gauge transformation, the new Lagrangian is
+\[
+\mcL' = \tfrac12 m v^2 - q\varphi' + q\,\vec{v}\cdot\vec{A}' = \mcL + q\,\vec{v}\cdot\nabla\chi + q\,\pdv{\chi}{t}.
+\]
+The extra terms combine into a total time derivative: $q\,\vec{v}\cdot\nabla\chi + q\,\pdv{\chi}{t} = \frac{\dd(q\chi)}{\dd t}$. The action changes by $\int_{t_1}^{t_2} \frac{\dd(q\chi)}{\dd t}\,\dd t = q\chi(t_2) - q\chi(t_1)$, a pure boundary term. Since the principle of least action fixes the endpoints, the variation of this boundary term vanishes: $\delta[q\chi(t_2)-q\chi(t_1)] = 0$. The Euler--Lagrange equations, which depend only on the variation of the action, remain unchanged. This is why adding a total time derivative to any Lagrangian never alters the equations of motion. Under the gauge transformation the HJ equation retains its form provided the principal function transforms as
+\[
+\mcS'(\vec{r},t) = \mcS(\vec{r},t) - q\,\chi(\vec{r},t).
+\]
+To see this, substitute $\nabla\mcS' = \nabla\mcS - q\nabla\chi$ into the HJ equation written in the transformed potentials:
+\[
+\frac{1}{2m}\bigl|\nabla\mcS' - q\vec{A}'\bigr|^2 + q\varphi' + \pdv{\mcS'}{t} = \frac{1}{2m}\bigl|\nabla\mcS - q\vec{A}\bigr|^2 + q\varphi + \pdv{\mcS}{t},
+\]
+so the unprimed and primed equations are identical. Thus the Hamilton--Jacobi formulation respects electromagnetic gauge invariance.}
+
+\nt{Connection to quantum mechanics through the WKB approximation}{
+The Hamilton--Jacobi equation is the classical $\hbar\to 0$ limit of the Schrödinger equation, and the bridge between them is the WKB (Wentzel--Kramers--Brillouin) approximation, a semiclassical method valid when the action is large compared to $\hbar$. The WKB ansatz writes the wavefunction as $\psi(\vec{r},t) = A(\vec{r},t)\,\exp(\mathrm{i}\mcS(\vec{r},t)/\hbar)$, where $\mcS$ plays the role of a phase function and $A$ is a slowly varying amplitude. For simplicity set $A=1$ so that $\psi = \exp(\mathrm{i}\mcS/\hbar)$ and substitute this directly into the Schrödinger equation for a charged particle,
+\[
+\mathrm{i}\hbar\,\pdv{\psi}{t} = \frac{1}{2m}\bigl(-\mathrm{i}\hbar\nabla - q\vec{A}\bigr)^2\psi + q\varphi\psi.
+\]
+Compute the derivatives: $\pdv{\psi}{t} = (\mathrm{i}/\hbar)(\pdv{\mcS}{t})\psi$, and
+\[
+(-\mathrm{i}\hbar\nabla - q\vec{A})\psi = -\mathrm{i}\hbar\Bigl(\frac{\mathrm{i}}{\hbar}\nabla\mcS\,\psi\Bigr) - q\vec{A}\psi = (\nabla\mcS - q\vec{A})\psi.
+\]
+Applying the operator a second time,
+\[
+\bigl(-\mathrm{i}\hbar\nabla - q\vec{A}\bigr)^2\psi = (-\mathrm{i}\hbar\nabla)\bigl[(\nabla\mcS - q\vec{A})\psi\bigr] - q\vec{A}\cdot(\nabla\mcS - q\vec{A})\psi.
+\]
+The gradient of the product gives $(\nabla\mcS - q\vec{A})(\mathrm{i}/\hbar)(\nabla\mcS)\psi + (\mathrm{i}/\hbar)(\nabla\mcS)(\nabla\mcS - q\vec{A})\psi$ plus the term involving $\nabla\cdot(\nabla\mcS-q\vec{A})$. Multiplying by $-\mathrm{i}\hbar$, the leading-order piece (proportional to $\hbar^0$) is $|\nabla\mcS - q\vec{A}|^2\psi$, while a correction proportional to $\hbar$ involves $\nabla\cdot(\nabla\mcS - q\vec{A})$. The right-hand side of the Schrödinger equation becomes
+\[
+\frac{1}{2m}\Bigl[|\nabla\mcS - q\vec{A}|^2 - \mathrm{i}\hbar\,\nabla\cdot(\nabla\mcS - q\vec{A})\Bigr]\psi + q\varphi\psi.
+\]
+The left-hand side is $-\pdv{\mcS}{t}\,\psi$. Equating and dividing by $\psi$, the $\hbar^0$ (leading-order) terms give exactly
+\[
+\frac{1}{2m}|\nabla\mcS - q\vec{A}|^2 + q\varphi + \pdv{\mcS}{t} = 0,
+\]
+which is the classical Hamilton--Jacobi equation for a charged particle. The $\hbar^1$ term $-\mathrm{i}\hbar/(2m)\,\nabla\cdot(\nabla\mcS - q\vec{A}) = 0$ yields a continuity equation for the amplitude $A$. Higher-order terms in $\hbar$ provide successive quantum corrections. In this way, the Hamilton--Jacobi equation is the zeroth-order term in a systematic semiclassical expansion of quantum mechanics.}
+
+\ex{Separation for time-independent fields}{
+Suppose the electromagnetic potentials are time-independent. Then the Hamiltonian has no explicit time dependence and the total energy $E$ is conserved. The time variable separates from the action as $\mcS = W(\vec{r}) - Et$, and the Hamilton--Jacobi equation reduces to the time-independent form
+\[
+\frac{1}{2m}\bigl|\nabla W - q\vec{A}(\vec{r})\bigr|^2 + q\,\varphi(\vec{r}) = E.
+\]
+If any spatial coordinate is absent from both $\vec{A}$ and $\varphi$, the corresponding component of $\nabla W$ is a constant separation equal to the conserved canonical momentum for that coordinate.}
+
+\qs{Hamilton--Jacobi for a uniform magnetic field in Landau gauge}{
+A charged particle moves in a uniform magnetic field $\vec{B} = B_0\,\hat{\bm{z}}$ with the vector potential chosen in the Landau gauge $\vec{A} = (0, B_0 x, 0)$ and scalar potential $\varphi = 0$.
+
+\begin{enumerate}[label=(\alph*)]
+\item Write the Hamiltonian in terms of the canonical momenta $p_x$, $p_y$, $p_z$.
+\item Write the full Hamilton--Jacobi equation explicitly in Cartesian coordinates. Identify which generalized coordinates are cyclic.
+\item For an electron ($q = -e = -1.60\times 10^{-19}\,\mathrm{C}$, $m = 9.11\times 10^{-31}\,\mathrm{kg}$) in a field $B_0 = 1.00\,\mathrm{T}$, the cyclotron frequency is $\omega_c = |q|B_0/m$. If the total transverse energy is $E_\perp = 100\,\mathrm{eV} = 1.60\times 10^{-17}\,\mathrm{J}$, compute $\omega_c$ numerically and verify the value of $\omega_c$ from the HJ separation constants matches this expression.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The Hamiltonian for a charged particle in electromagnetic fields is
+\[
+\mcH = \frac{1}{2m}\bigl|\vec{p} - q\vec{A}\bigr|^2 + q\varphi.
+\]
+With $\vec{A} = (0, B_0 x, 0)$ and $\varphi = 0$, the three components of $\vec{p} - q\vec{A}$ are
+\[
+\bigl(p_x - q(0),\; p_y - q B_0 x,\; p_z - q(0)\bigr) = \bigl(p_x,\; p_y - q B_0 x,\; p_z\bigr).
+\]
+The square of the magnitude is the sum of the squares of these components. Therefore,
+\[
+\mcH = \frac{1}{2m}\Bigl[p_x^2 + \bigl(p_y - q B_0 x\bigr)^2 + p_z^2\Bigr].
+\]
+
+\textbf{Part (b).} The Hamilton--Jacobi equation reads $\mcH\bigl(\vec{r},\nabla\mcS\bigr) + \pdv{\mcS}{t} = 0$. Substitute the Hamiltonian from part (a) with the replacements $p_x \to \pdv{\mcS}{x}$, $p_y \to \pdv{\mcS}{y}$, $p_z \to \pdv{\mcS}{z}$:
+\[
+\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \left(\pdv{\mcS}{y} - q B_0 x\right)^2 + \left(\pdv{\mcS}{z}\right)^2\right] + \pdv{\mcS}{t} = 0.
+\]
+This is the full HJ PDE in Cartesian coordinates. A generalized coordinate is cyclic if it is absent from the Hamiltonian. The coordinate $y$ does not appear explicitly in $\mcH$, so $y$ is cyclic and the conjugate momentum $\pdv{\mcS}{y} = p_y$ is conserved. Similarly, $z$ is absent from $\mcH$, so $z$ is cyclic and $p_z$ is conserved. The coordinate $x$ does appear in the term $(p_y - q B_0 x)^2$, so $x$ is not cyclic. Thus $y$ and $z$ are the two cyclic coordinates.
+
+\textbf{Part (c).} For the electron, the cyclotron frequency follows from the minimal-coupling Hamiltonian. The numerical value is
+\[
+\omega_c = \frac{|q| B_0}{m}.
+\]
+Substitute the given values:
+\[
+|q| = 1.60\times 10^{-19}\,\mathrm{C},
+\qquad
+B_0 = 1.00\,\mathrm{T},
+\qquad
+m = 9.11\times 10^{-31}\,\mathrm{kg}.
+\]
+Form the ratio:
+\[
+\omega_c = \frac{(1.60\times 10^{-19})(1.00)}{9.11\times 10^{-31}}\,\mathrm{rad/s}.
+\]
+This gives
+\[
+\omega_c = 1.756\times 10^{11}\,\mathrm{rad/s}.
+\]
+Now verify that the HJ formalism produces the same frequency. Because the potentials are time-independent, the action separates as $\mcS = W_x(x) - E_\perp t + \alpha_y\,y + \alpha_z\,z$. The time-independent HJ equation for the transverse motion is
+\[
+\frac{1}{2m}\left[\left(\der{W_x}{x}\right)^2 + \bigl(\alpha_y - q B_0 x\bigr)^2\right] = E_\perp,
+\]
+where $E_\perp$ is the total transverse energy. Solve for $\der{W_x}{x}$:
+\[
+\der{W_x}{x} = \pm\sqrt{2mE_\perp - \bigl(\alpha_y - q B_0 x\bigr)^2}.
+\]
+Change variable to the shifted $x$-coordinate centered on the guiding center, $X = x - \alpha_y/(q B_0)$, giving
+\[
+\der{W_x}{X} = \pm\sqrt{2mE_\perp - (q B_0)^2 X^2}.
+\]
+This is the square-root form of the harmonic-oscillator action. Completing the square and comparing with the standard form $\pm\sqrt{2m\mcE - m^2\omega^2 X^2}$, we identify
+\[
+m^2\omega^2 = (q B_0)^2,
+\qquad\text{so}\qquad
+\omega = \frac{|q| B_0}{m} = \omega_c.
+\]
+The HJ separation constant analysis thus recovers the cyclotron frequency exactly, independent of the guiding-center location and the transverse energy. The transverse energy $E_\perp = 1.60\times 10^{-17}\,\mathrm{J}$ sets the gyroradius but does not affect the frequency.
+
+Therefore,
+\[
+\mcH = \frac{1}{2m}\Bigl[p_x^2 + \bigl(p_y - q B_0 x\bigr)^2 + p_z^2\Bigr],
+\]
+\[
+\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \left(\pdv{\mcS}{y} - q B_0 x\right)^2 + \left(\pdv{\mcS}{z}\right)^2\right] + \pdv{\mcS}{t} = 0,
+\]
+\[
+\text{cyclic coordinates: } y, z,\qquad \omega_c = 1.76\times 10^{11}\,\mathrm{rad/s}.
+\]

concepts/advanced/hj-equation.tex

@@ -0,0 +1,257 @@
+\subsection{Derivation of the Hamilton--Jacobi Equation}
+
+This subsection derives the Hamilton--Jacobi partial differential equation from Hamiltonian mechanics through a special canonical transformation. Jacobi's theorem then reduces the solution of the mechanics problem to finding a complete integral of the resulting PDE.
+
+\nt{What is a canonical transformation?}{
+A canonical transformation is a change of coordinates from $(q,p)$ to new variables $(Q,P)$ that preserves the form of Hamilton's equations. In the new variables the motion still satisfies $\dot{Q}_i = \pdv{\mcK}{P_i}$ and $\dot{P}_i = -\pdv{\mcK}{Q_i}$ for some new Hamiltonian $\mcK$. Why do such transformations exist? From Hamilton's principle, the action integral is $S = \int \mcL\,\dd t$. If we change coordinates so that the new Lagrangian differs from the old one by a total time derivative, $\mcL' = \mcL + \dd F/\dd t$, the action changes only by the boundary term $F(t_\mathrm{f}) - F(t_\mathrm{i})$. Since the variational principle fixes the endpoints, these boundary terms do not affect the Euler--Lagrange equations. The equations of motion are therefore unchanged. Any coordinate change generated in this way -- where the Lagrangian shifts by a total time derivative -- is canonical.}
+
+\dfn{Generating function $F_2$}{
+A type-2 generating function $F_2(q,P,t)$ defines a canonical transformation through two sets of partial-derivative relations:
+\[
+p_i = \pdv{F_2}{q_i},
+\qquad
+Q_i = \pdv{F_2}{P_i}.
+\]
+The first relation expresses the old momenta in terms of the old coordinates and new momenta; the second gives the new coordinates. Together these two equations completely determine the transformation. The total-time-derivative shift in the Lagrangian means the new Hamiltonian is related to the old by
+\[
+\mcK(Q,P,t) = \mcH(q,p,t) + \pdv{F_2}{t},
+\]
+since adding $\dd F_2/\dd t$ to the Lagrangian shifts the Legendre transform (which defines the Hamiltonian) by $-\pdv{F_2}{t}$. This is the crucial formula: it tells us exactly how the Hamiltonian changes under the transformation, and it is what lets us control the new dynamics by choosing $F_2$ wisely.}
+
+\dfn{Hamilton's principal function and the Hamilton--Jacobi action}{
+Let $q_1,\dots,q_n$ be generalized coordinates and let $p_1,\dots,p_n$ be the corresponding canonical momenta. Hamilton's principal function $\mcS(q_1,\dots,q_n,t)$ is a generating function whose spatial partial derivatives equal the canonical momenta:
+\[
+p_i = \pdv{\mcS}{q_i}
+\]
+for each $i=1,\dots,n$. When $\mcS$ satisfies the Hamilton--Jacobi equation it encodes the complete solution to the equations of motion.}
+
+\nt{Why turn ODEs into a PDE?}{
+At first glance, replacing $2n$ coupled first-order ODEs with one nonlinear PDE seems like a step backward. Three reasons make the trade worth it.
+
+(a) Separability reveals symmetries: if the Hamiltonian admits cyclic coordinates, the PDE separates and each separated equation displays a conserved quantity, making symmetries algebraically manifest. A cyclic coordinate in the Hamiltonian becomes a trivially integrable separated equation in the PDE.
+
+(b) Complete integrals bypass ODE solving entirely: finding a solution with $n$ independent constants, then applying Jacobi's theorem, yields the trajectory $q(t)$ algebraically without integrating equations of motion at all. The mechanics problem reduces to algebra.
+
+(c) Quantum-mechanical connection: the Hamilton--Jacobi equation is the classical limit of the Schrödinger equation, appearing as the leading-order term in the WKB approximation where the wavefunction takes the form $\psi \propto \exp(\mathrm{i}\mcS/\hbar)$. Solving the HJ equation is the prerequisite step for connecting classical trajectories to quantum amplitudes -- a theme that recurs throughout the advanced material.}
+
+\nt{Geometric optics analogy}{
+Geometric optics provides an intuitive picture. In optics there are two complementary descriptions of light: the ray picture, where light follows trajectories through space, and the wavefront picture, where surfaces of constant phase propagate through the medium. The eikonal equation, $|\nabla\phi| = n(\mathbf{r})$, is a first-order PDE for the optical phase $\phi(\mathbf{r})$. Surfaces of constant $\phi$ are the wavefronts; the gradient $\mathbf{k} = \nabla\phi$ is the local wave-vector, pointing normal to each wavefront, and its magnitude $|\mathbf{k}| = n$ encodes the local refractive index. The light rays are the characteristics of the eikonal equation -- integral curves of $\mathbf{k}$ -- and they obey Snell's law at interfaces.
+
+The Hamilton--Jacobi theory is the exact mechanical analogue of this picture. The principal function $\mcS$ plays the role of optical phase, the canonical momentum $\mathbf{p} = \nabla\mcS$ plays the role of the wave-vector (it is the gradient of $\mcS$, pointing normal to surfaces of constant action), and particle trajectories are the characteristics -- integral curves of $\mathbf{p}$ -- just as light rays are integral curves of $\mathbf{k}$. The Hamilton--Jacobi equation itself is the mechanical eikonal equation.
+
+Just as Fermat's principle determines the path of light through minimizing optical path length, Hamilton's principle determines the path of the particle through extremizing the action. Both are encoded in the same first-order PDE structure: one governing propagating wavefronts of light, the other governing propagating surfaces of constant action.}
+
+\thm{The Hamilton--Jacobi partial differential equation}{
+Let $\mcH(q_1,\dots,q_n,p_1,\dots,p_n,t)$ be the Hamiltonian of a system with $n$ degrees of freedom. We seek a type-2 canonical transformation, generated by $\mcS(q,P,t)$, that simplifies the dynamics by choosing the new Hamiltonian to vanish identically: $\mcK = 0$. With $\mcK = 0$, every new coordinate and momentum is constant in time, because $\dot{Q}_i = \pdv{\mcK}{P_i} = 0$ and $\dot{P}_i = -\pdv{\mcK}{Q_i} = 0$. This choice eliminates all time dependence in the transformed variables, reducing the entire dynamics problem to finding $\mcS$. The generating function $\mcS(q_1,\dots,q_n,t)$ satisfies
+\[
+\mcH\!\left(q_1,\dots,q_n,\pdv{\mcS}{q_1},\dots,\pdv{\mcS}{q_n},t\right) + \pdv{\mcS}{t} = 0.
+\]
+This is the Hamilton--Jacobi equation.}
+
+\pf{Derivation from the canonical transformation to $\mcK=0$}{
+Begin with the Lagrangian $\mcL(q,\dot{q},t)$ for a system with generalized coordinates $q_1,\dots,q_n$. Define the canonical momenta through the Legendre transform:
+\[
+p_i = \pdv{\mcL}{\dot{q}_i}
+\]
+for each $i$. The Hamiltonian is the Legendre transform of the Lagrangian:
+\[
+\mcH = \sum_{i=1}^{n} p_i \dot{q}_i - \mcL,
+\]
+expressed as a function of $(q,p,t)$ after eliminating $\dot{q}$ in favor of $p$ using the inverse of the Legendre map. Hamilton's canonical equations follow from the definition:
+\[
+\dot{q}_i = \pdv{\mcH}{p_i},
+\qquad
+\dot{p}_i = -\pdv{\mcH}{q_i}.
+\]
+These $2n$ first-order coupled equations determine the dynamics once initial conditions are given. For systems with many degrees of freedom these equations are tightly coupled and difficult to integrate directly.
+
+Now seek a type-2 canonical transformation from $(q,p)$ to new variables $(Q,P)$ that simplifies the dynamics. The generating function $F_2(q,P,t)$ defines the transformation through the relations
+\[
+p_i = \pdv{F_2}{q_i},
+\qquad
+Q_i = \pdv{F_2}{P_i}.
+\]
+The new Hamiltonian $\mcK(Q,P,t)$ is related to the original Hamiltonian $\mcH$ by the standard transformation rule
+\[
+\mcK = \mcH + \pdv{F_2}{t}.
+\]
+
+The central idea of the Hamilton--Jacobi method is to choose the generating function $F_2$ so that the new Hamiltonian vanishes identically: $\mcK = 0$. The motivation for this choice is simple -- with $\mcK = 0$, Hamilton's equations in the new variables give
+\[
+\dot{Q}_i = \pdv{\mcK}{P_i} = 0,
+\qquad
+\dot{P}_i = -\pdv{\mcK}{Q_i} = 0.
+\]
+Every new coordinate and every new momentum is strictly constant in time. The dynamics in the transformed variables is completely trivial: all $2n$ constants of motion are known immediately.
+
+Setting $\mcK = 0$ in the transformation rule gives the key relation
+\[
+\mcH = -\pdv{F_2}{t}.
+\]
+Rename the generating function $F_2$ as $\mcS(q_1,\dots,q_n,P_1,\dots,P_n,t)$ and use $p_i = \pdv{\mcS}{q_i}$ to substitute the momenta inside the Hamiltonian. The previous equation reads
+\[
+\mcH\!\left(q_1,\dots,q_n,\pdv{\mcS}{q_1},\dots,\pdv{\mcS}{q_n},t\right) = -\pdv{\mcS}{t}.
+\]
+Rearranging gives the Hamilton--Jacobi partial differential equation:
+\[
+\mcH\!\left(q_1,\dots,q_n,\pdv{\mcS}{q_1},\dots,\pdv{\mcS}{q_n},t\right) + \pdv{\mcS}{t} = 0.
+\]
+Every solution $\mcS$ of this PDE generates a canonical transformation to variables in which the motion is completely trivial. The constants $P_i$ can be identified with separation constants $\alpha_i$ and the constants $Q_i$ with integration constants $\beta_i$. Finding the complete integral of this one PDE is therefore equivalent to solving all $2n$ Hamilton's equations at once.}
+
+\cor{Time-independent HJ equation (Hamilton--Charpit--Jacobi)}{
+When the Hamiltonian does not depend explicitly on time, $\pdv{\mcH}{t} = 0$ and the Hamiltonian is a conserved quantity, $\mcH = E$. In this case the time dependence of $\mcS$ separates as $\mcS = W(q_1,\dots,q_n) - Et$, and the Hamilton--Jacobi equation reduces to
+\[
+\mcH\!\left(q_1,\dots,q_n,\pdv{W}{q_1},\dots,\pdv{W}{q_n}\right) = E.
+\]
+This is the time-independent Hamilton--Jacobi equation, sometimes called the Hamilton--Charpit--Jacobi equation. Solving for $W$ and appending $-Et$ gives the complete principal function.}
+
+\dfn{Jacobi's theorem on complete integrals}{
+A complete integral of the Hamilton--Jacobi equation is a solution $\mcS(q_1,\dots,q_n,\alpha_1,\dots,\alpha_n,t)$ containing $n$ independent non-additive separation constants $\alpha_1,\dots,\alpha_n$ (plus one overall additive constant that drops out of all derivatives and does not affect the physics). The term non-additive means each $\alpha_i$ appears in $\mcS$ in a way that cannot be removed by simply shifting $\mcS$ by a constant -- for example, as a coefficient of $x$ or inside a square root.
+
+The constants $\alpha_i$ play the role of the new canonical momenta $P_i$, all conserved because $\mcK = 0$. Physically, the $\alpha_i$ are constants of motion that specify the state of the system; common choices include the total energy, the magnitude of angular momentum, and its $z$-component. The constants $\beta_i$ play the role of the new coordinates $Q_i$ and are fixed by initial conditions.
+
+Jacobi's theorem provides the equations of motion:
+\[
+\pdv{\mcS}{\alpha_i} = \beta_i
+\]
+for each $i=1,\dots,n$. Each equation $\pdv{\mcS}{\alpha_i} = \beta_i$ is algebraic in the coordinates and time, so the $n$ equations together determine $q_1(t),\dots,q_n(t)$.
+
+The algorithm is: find the complete integral, differentiate $\mcS$ with respect to each separation constant $\alpha_i$, set each result equal to a constant $\beta_i$, and solve the resulting $n$ algebraic equations for $q_1(t),\dots,q_n(t)$. The canonical momenta then follow from $p_i = \pdv{\mcS}{q_i}$. This procedure replaces the integration of $2n$ coupled first-order differential equations with the solution of $n$ algebraic equations -- a dramatic simplification.}
+
+\nt{Connection to Hamilton's principle of least action}{
+Hamilton's principal function $\mcS(q,t)$ evaluated along the actual physical path coincides with the action integral $\int_{t_0}^{t} \mcL\,\dd t'$ computed along that trajectory. The Hamilton--Jacobi equation itself can be viewed as the condition that the action integral be stationary under endpoint variations, generalized to a differential equation for the action. This unifies the variational and canonical formulations of classical mechanics into a single framework based on the propagating wavefront of constant action. The principal function $\mcS$ is itself the action evaluated from a fixed initial point to the variable endpoint $(q,t)$ along the true trajectory.}
+
+\nt{Connection to Maupertuis' principle}{
+The time-independent Hamilton--Jacobi equation also connects to Maupertuis' principle, which characterizes true trajectories as geodesics in configuration space with a metric scaled by kinetic energy. The reduced action $W(q)$ satisfies $\dd W = p_i\,\dd q_i$, so that integrating $\dd W$ along a trajectory is equivalent to integrating the momentum one-form. When $H=E$ is held fixed, the paths that extremize $\int p_i\,\dd q_i$ are the same paths found by solving the time-independent equation for $W$. This makes the Hamilton--Jacobi formalism the bridge between the velocity-space perspective of Lagrangian mechanics and the phase-space perspective of Hamiltonian mechanics.}
+
+\mprop{Summary of the HJ method}{
+The Hamilton--Jacobi method solves a mechanics problem in five steps:
+\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
+\item Write the Hamiltonian $\mcH(q,p,t)$ in terms of the appropriate generalized coordinates and momenta. Choose coordinates that exploit the symmetry of the problem.
+
+\item Substitute $p_i = \pdv{\mcS}{q_i}$ into $\mcH$ to obtain the Hamilton--Jacobi PDE: $\mcH + \pdv{\mcS}{t} = 0$. This converts the problem from solving $2n$ ODEs into solving one PDE.
+
+\item Separate variables if possible. If $\mcH$ does not depend explicitly on time, set $\mcS = W(q) - Et$ and solve the time-independent equation $\mcH(q,\pdv{W}{q}) = E$. Cyclic coordinates produce additive terms in $W$ that separate immediately.
+
+\item Find a complete integral $\mcS(q,\alpha,t)$ containing $n$ independent non-additive separation constants $\alpha_1,\dots,\alpha_n$. For a one-degree-of-freedom system this means one constant; for higher dimensional systems more separation constants appear.
+
+\item Apply Jacobi's theorem: set $\pdv{\mcS}{\alpha_i} = \beta_i$ for each $i$, then solve the resulting $n$ algebraic equations for $q_1(t),\dots,q_n(t)$. The momenta follow from $p_i = \pdv{\mcS}{q_i}$. The mechanics problem is solved.
+\end{enumerate}
+}
+
+\ex{Complete integral for a free particle}{
+For a free particle in one dimension, $\mcH = p^2/2m$. The Hamilton--Jacobi equation is
+\[
+\tfrac{1}{2m}\left(\pdv{\mcS}{x}\right)^2 + \pdv{\mcS}{t} = 0.
+\]
+We try the additive separation ansatz $\mcS(x,t) = W(x) + T(t)$. Substituting into the PDE gives
+\[
+\frac{1}{2m}\left(\der{W}{x}\right)^2 + \der{T}{t} = 0.
+\]
+Since the first term depends only on $x$ and the second only on $t$, each must equal a constant. We choose the separation constant as $\alpha^2/(2m)$, where $\alpha$ will turn out to be the constant momentum:
+\[
+\frac{1}{2m}\left(\der{W}{x}\right)^2 = \frac{\alpha^2}{2m},
+\qquad
+\der{T}{t} = -\frac{\alpha^2}{2m}.
+\]
+The spatial equation gives $\der{W}{x} = \pm\alpha$. Choosing the positive sign (the negative case is covered by allowing $\alpha$ to be either positive or negative) and integrating with respect to $x$ gives $W(x) = \alpha x + c_W$. Similarly, integrating the temporal equation gives $T(t) = -\alpha^2 t/(2m) + c_T$. Absorbing the two additive constants into a single overall additive constant (which does not affect the physics) yields the complete integral
+\[
+\mcS(x,t;\alpha) = \alpha x - \frac{\alpha^2}{2m}\,t.
+\]
+We can verify this solution directly by substitution. Computing $\pdv{\mcS}{x} = \alpha$ and $\pdv{\mcS}{t} = -\alpha^2/(2m)$, the left-hand side of the PDE becomes $\tfrac{1}{2m}\alpha^2 - \alpha^2/(2m) = 0$, confirming the result. Here $\alpha$ is the constant momentum and serves as the single separation constant for this one-degree-of-freedom system. Because the Hamiltonian does not depend explicitly on time, energy conservation $\mcH = E = \alpha^2/(2m)$ determines the relationship between the separation constant and the total mechanical energy of the particle. The level sets $\mcS = \mathrm{const}$ are planes in $(x,t)$ space, representing uniformly propagating wavefronts of constant action.}
+
+\qs{Hamilton--Jacobi for a one-dimensional Hamiltonian}{
+A one-dimensional system has Hamiltonian
+\[
+\mcH(x,p) = \frac{p^2}{2m} + V(x).
+\]
+
+\begin{enumerate}[label=(\alph*)]
+\item Write the Hamilton--Jacobi PDE explicitly for this system.
+\item For the free-particle case $V(x) = 0$, find the complete integral $\mcS(x,t;\alpha)$ by separation of variables, identifying $\alpha$ as the constant momentum.
+\item Show from Jacobi's theorem that $\pdv{\mcS}{\alpha} = \beta$ yields the trajectory $x(t) = (\alpha/m)t + \beta$. For $m = 2.0\,\mathrm{kg}$, $\alpha = 4.0\,\mathrm{kg\!\cdot\!m/s}$, and $\beta = 1.0\,\mathrm{m}$, compute $x(3.0\,\mathrm{s})$ and verify that the result satisfies $m\ddot{x} = 0$.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The Hamilton--Jacobi equation is obtained by the standard HJ substitution: everywhere the canonical momentum $p$ appears in the Hamiltonian, replace it with $\pdv{\mcS}{x}$. Starting from the given Hamiltonian,
+\[
+\mcH(x,p) = \frac{p^2}{2m} + V(x),
+\]
+the substitution $p \to \pdv{\mcS}{x}$ gives
+\[
+\frac{1}{2m}\left(\pdv{\mcS}{x}\right)^2 + V(x).
+\]
+The full HJ equation adds the time derivative $\pdv{\mcS}{t}$ and sets the sum to zero:
+\[
+\frac{1}{2m}\left(\pdv{\mcS}{x}\right)^2 + V(x) + \pdv{\mcS}{t} = 0.
+\]
+This is the explicit Hamilton--Jacobi PDE for a one-dimensional particle in potential $V(x)$. For general potentials $V(x)$ this is a nonlinear first-order PDE that must be solved to find $\mcS(x,t)$.
+
+\textbf{Part (b).} For $V(x) = 0$, the Hamiltonian reduces to pure kinetic energy, $\mcH = p^2/(2m)$, and the Hamilton--Jacobi equation becomes
+\[
+\frac{1}{2m}\left(\pdv{\mcS}{x}\right)^2 + \pdv{\mcS}{t} = 0.
+\]
+Use the separation ansatz $\mcS(x,t) = W(x) + T(t)$. Because $W$ depends only on $x$ and $T$ only on $t$, the partial derivatives become ordinary derivatives:
+\[
+\frac{1}{2m}\left(\der{W}{x}\right)^2 + \der{T}{t} = 0.
+\]
+The first term depends only on $x$ and the second only on $t$, so each must equal a constant. We choose the separation constant so that the spatial derivative equals the momentum $\alpha$:
+\[
+\frac{1}{2m}\left(\der{W}{x}\right)^2 = \frac{\alpha^2}{2m},
+\qquad
+\der{T}{t} = -\frac{\alpha^2}{2m}.
+\]
+Solving the spatial part, take the square root of both sides:
+\[
+\der{W}{x} = \pm\alpha.
+\]
+Choosing the positive sign (the negative case is covered by allowing $\alpha$ to be negative) and integrating with respect to $x$ gives $W(x) = \alpha x$. Integrating the temporal equation gives $T(t) = -\alpha^2 t/(2m)$. The complete integral is therefore
+\[
+\mcS(x,t;\alpha) = \alpha x - \frac{\alpha^2}{2m}\,t.
+\]
+This solves the HJ PDE and contains one independent non-additive constant $\alpha$, which is the constant momentum. The energy is $E = \alpha^2/(2m)$.
+
+\textbf{Part (c).} Jacobi's theorem states that $\pdv{\mcS}{\alpha} = \beta$, where $\beta$ is a constant determined by initial conditions. Differentiate the complete integral with respect to $\alpha$:
+\[
+\pdv{\mcS}{\alpha} = x - \frac{\alpha}{m}\,t.
+\]
+Set this equal to $\beta$ and solve for $x$:
+\[
+x - \frac{\alpha}{m}\,t = \beta,
+\qquad\text{so}\qquad
+x(t) = \frac{\alpha}{m}\,t + \beta.
+\]
+The velocity is
+\[
+\dot{x} = \frac{\alpha}{m},
+\]
+which is constant. The acceleration is
+\[
+\ddot{x} = 0,
+\]
+confirming the free-particle equation of motion $m\ddot{x} = 0$.
+
+Now substitute the numerical values. The particle has mass $m = 2.0\,\mathrm{kg}$, the separation constant is $\alpha = 4.0\,\mathrm{kg\!\cdot\!m/s}$, and the integration constant is $\beta = 1.0\,\mathrm{m}$. The constant velocity is
+\[
+\frac{\alpha}{m} = \frac{4.0\,\mathrm{kg\!\cdot\!m/s}}{2.0\,\mathrm{kg}} = 2.0\,\mathrm{m/s}.
+\]
+The trajectory becomes
+\[
+x(t) = (2.0\,\mathrm{m/s})\,t + 1.0\,\mathrm{m}.
+\]
+At $t = 3.0\,\mathrm{s}$, the position is
+\[
+x(3.0\,\mathrm{s}) = (2.0)(3.0)\,\mathrm{m} + 1.0\,\mathrm{m} = 7.0\,\mathrm{m}.
+\]
+The total energy of the particle is $E = \alpha^2/(2m) = (4.0)^2/(2 \cdot 2.0) = 4.0\,\mathrm{J}$, which equals the kinetic energy $\tfrac{1}{2}(2.0)(2.0)^2 = 4.0\,\mathrm{J}$ -- a consistent check. The velocity is constant at $2.0\,\mathrm{m/s}$, the acceleration vanishes, and the initial position $x(0) = \beta = 1.0\,\mathrm{m}$. The constant $\alpha$ is the conserved linear momentum $p = m\dot{x} = 4.0\,\mathrm{kg\!\cdot\!m/s}$, and the constant $\beta$ is the initial position. Together, $\alpha$ and $\beta$ form a complete set of two independent constants for this one-degree-of-freedom system, matching the two parameters needed to describe the general solution of the second-order equation of motion.
+
+Therefore,
+\[
+\text{HJ PDE:}\quad \frac{1}{2m}\left(\pdv{\mcS}{x}\right)^2 + V(x) + \pdv{\mcS}{t} = 0,
+\]
+\[
+\mcS(x,t;\alpha) = \alpha x - \frac{\alpha^2}{2m}\,t,
+\qquad
+x(t) = \frac{\alpha}{m}\,t + \beta,
+\qquad
+x(3.0\,\mathrm{s}) = 7.0\,\mathrm{m}.
+\]

concepts/advanced/kepler-coulomb-hj.tex

@@ -0,0 +1,197 @@
+\subsection{Charged Particle in Coulomb Potential}
+
+This subsection treats a charged particle moving in the Coulomb potential of a fixed point charge through the Hamilton--Jacobi formalism, demonstrating its identical structure to the gravitational Kepler problem and using action--angle variables to recover the Bohr--Sommerfeld energy levels of the hydrogen atom.
+
+\dfn{Coulomb Hamilton--Jacobi equation}{
+Consider a particle of reduced mass $\mu$ and charge $q$ moving in the electrostatic potential of a fixed source charge $Q$. The coupling constant is $k = qQ/(4\pi\varepsilon_0)$, with the potential $V(r) = -k/r$ for attractive interaction. For the electron--proton system, $q = -e$, $Q = +e$, so $k = e^2/(4\pi\varepsilon_0)$. In spherical coordinates the Hamiltonian is
+\[
+\mcH = \frac{p_r^2}{2\mu} + \frac{p_\theta^2}{2\mu r^2} + \frac{p_\phi^2}{2\mu r^2\sin^2\theta} - \frac{k}{r}.
+\]
+Substituting $p_i = \pdv{\mcS}{q_i}$ into the Hamilton--Jacobi equation $\mcH + \pdv{\mcS}{t} = 0$ gives
+\[
+\frac{1}{2\mu}\left(\pdv{\mcS}{r}\right)^2 
++ \frac{1}{2\mu r^2}\left(\pdv{\mcS}{\theta}\right)^2 
++ \frac{1}{2\mu r^2\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 
+- \frac{k}{r} + \pdv{\mcS}{t} = 0.
+\]
+Because the scalar potential is time--independent, energy $E = \mcH$ is conserved and the time variable separates as $\mcS = W(r,\theta,\phi) - Et$ with $W$ the Hamilton characteristic function.}
+
+\thm{Orbit equation and eccentricity for the Coulomb problem}{
+With $V(r) = -k/r$ the trajectory is a conic section
+\[
+r(\phi) = \frac{\ell}{1 + \varepsilon\cos(\phi - \phi_0)},
+\]
+where the semilatus rectum $\ell = L^2/(\mu k)$ and the eccentricity $\varepsilon = \sqrt{1 + 2EL^2/(\mu k^2)}$ are determined by the energy $E$ and the total angular momentum $L$. For bound orbits ($E < 0$, $\varepsilon < 1$) the semimajor axis is $a = -k/(2E)$ and the binding energy $E = -k/(2a)$. A circular orbit occurs at $\varepsilon = 0$ with $L^2 = \mu k a$.}
+
+\pf{Separated Hamilton--Jacobi equations for the Coulomb problem}{
+Set $\mcS = W_r(r) + W_\theta(\theta) + W_\phi(\phi) - Et$ and substitute into the time--independent HJ equation $\mcH(q,\pdv{W}{q}) = E$:
+\[
+\frac{1}{2\mu}\left(\der{W_r}{r}\right)^2 
++ \frac{1}{2\mu r^2}\left(\der{W_\theta}{\theta}\right)^2 
++ \frac{1}{2\mu r^2\sin^2\theta}\left(\der{W_\phi}{\phi}\right)^2 
+- \frac{k}{r} = E.
+\]
+The azimuthal coordinate $\phi$ is cyclic, so $\der{W_\phi}{\phi} = L_z$. Multiply by $2\mu r^2$ and rearrange:
+\[
+r^2\left(\der{W_r}{r}\right)^2 + 2\mu kr - 2\mu Er^2
+= -\left(\der{W_\theta}{\theta}\right)^2 - \frac{L_z^2}{\sin^2\theta}.
+\]
+The left side depends only on $r$, the right only on $\theta$; each equals the separation constant $L^2$. The polar equation is
+\[
+\left(\der{W_\theta}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = L^2,
+\]
+and the radial equation is
+\[
+\left(\der{W_r}{r}\right)^2 = 2\mu E + \frac{2\mu k}{r} - \frac{L^2}{r^2}.
+\]
+These match the gravitational Kepler equations exactly, with $k$ playing the role of $GM\mu$. The three constants $E$, $L$, and $L_z$ form a complete set required by Jacobi\normalsize{}'s theorem.}
+
+\nt{Structural identity with the gravitational Kepler problem}{
+The Coulomb HJ equation is structurally identical to the gravitational Kepler problem treated in A.08 (kepler--hj.tex). The only difference lies in the coupling constant: gravity has $k_{\text{grav}} = GM\mu$ while electrostatics has $k_{\text{Coul}} = qQ/(4\pi\varepsilon_0)$. Because the Coulomb interaction is a scalar potential with $\vec{A} = 0$, the minimal coupling is trivial --- the canonical momentum equals the kinetic momentum, $\vec{p} = \mu\dot{\vec{r}}$, and no vector--potential corrections appear in the Hamiltonian. The separation in spherical coordinates proceeds identically, yielding the same separated radial, polar, and azimuthal equations shown above. The three action variables $J_r$, $J_\theta$, and $J_\phi$ retain the exact integral structure from the gravitational case; all results for orbits, action--angle variables, and frequencies carry over with the replacement $GM\mu \to k$.}
+
+\nt{From classical action to quantum numbers}{
+In the Hamilton--Jacobi formalism, an action variable is defined as the phase--space integral of a canonical momentum over one complete cycle of its conjugate coordinate: $J = \frac{1}{2\pi}\oint p\,\dd q$. This construction assigns a single number to each degree of freedom that measures the area enclosed by the orbit in phase space, normalized by $2\pi$. Planck introduced a fundamental constant $h = 6.626\times 10^{-34}\,\mathrm{J\!\cdot\!s}$ to explain blackbody radiation. The reduced Planck constant $\hbar = h/(2\pi) = 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s}$ is the natural quantum of angular momentum. Bohr and Sommerfeld proposed that classical action variables should be restricted to integer multiples of $\hbar$:
+\[
+J_i = n_i\hbar,
+\]
+where $n_i$ is a positive integer called a quantum number. The physical idea is that phase space is granular at atomic scales: only classical orbits whose action variables satisfy this condition survive in the quantum theory. Why quantize action? The action variable sets the scale of an orbit in phase space, and imposing $J = n\hbar$ is equivalent to requiring an integer number of de Broglie wavelengths to fit along the orbit, producing a standing wave. The simplest action variable to evaluate is the azimuthal action. For any central potential, the canonical momentum $p_\phi = \pdv{\mcS}{\phi} = L_z$ is a constant of motion. Because $L_z$ does not vary with $\phi$, the integral evaluates directly:
+\[
+J_\phi = \frac{1}{2\pi}\oint p_\phi\,\dd\phi
+= \frac{1}{2\pi}\int_{0}^{2\pi} L_z\,\dd\phi
+= \frac{1}{2\pi}L_z\int_{0}^{2\pi}\dd\phi
+= L_z.
+\]
+Bohr--Sommerfeld quantization $J_\phi = n_\phi\hbar$ then immediately gives $L_z = n_\phi\hbar$. The $z$-component of angular momentum is quantized in units of $\hbar$, exactly as full quantum mechanics predicts.}
+
+\nt{Action--angle quantization and the hydrogen spectrum}{
+For the $1/r$ potential the three action variables are $J_\phi = L_z$, $J_\theta = L - |L_z|$, and $J_r = -L + k\sqrt{\mu/(2|E|)}$. The polar and radial actions $J_\theta$ and $J_r$ have the same integral structure as those derived for the gravitational Kepler problem in A.08, with the only change being the coupling constant $k$. Their sum eliminates the angular--momentum dependence:
+\[
+J_{\mathrm{tot}} = J_r + J_\theta + J_\phi = k\sqrt{\frac{\mu}{2|E|}}.
+\]
+Bohr--Sommerfeld quantization requires $J_{\mathrm{tot}} = n\hbar$, where $n$ is the principal quantum number. Setting $k\sqrt{\mu/(2|E|)} = n\hbar$ and solving for energy:
+\[
+|E| = \frac{\mu k^2}{2n^2\hbar^2},
+\qquad
+E_n = -\frac{\mu k^2}{2\hbar^2 n^2}.
+\]
+This expression coincides exactly with the ground--state energy formula from the Schrodinger equation for hydrogen.}
+
+\nt{The Bohr model}{
+In 1913, Niels Bohr proposed a model of the hydrogen atom in which the electron orbits the nucleus in circular paths with quantized angular momentum $L = n\hbar$. Bohr postulated that only certain discrete orbits are allowed and that electromagnetic radiation is emitted or absorbed when the electron jumps between them. His model produced the correct Rydberg formula for hydrogen line spectra but relied on ad hoc quantization rules applied to purely classical orbits. The Bohr--Sommerfeld method extends this picture to elliptical orbits and multiple degrees of freedom using action--angle variables derived from the Hamilton--Jacobi formalism. The calculations below show how the Bohr results emerge systematically from semiclassical quantization of classical action variables.}
+
+\qs{Electron in the Coulomb field of a proton using the HJ action--angle formalism}{
+For an electron bound to a proton, the electrostatic coupling constant is $k = e^2/(4\pi\varepsilon_0) = 2.307\times 10^{-28}\,\mathrm{J\!\cdot\!m}$ and the reduced mass $\mu \approx m_e = 9.11\times 10^{-31}\,\mathrm{kg}$.
+
+\begin{enumerate}[label=(\alph*)]
+\item For a bound orbit with semimajor axis $a_0 = 0.529\times 10^{-10}\,\mathrm{m}$ (the Bohr radius), find the orbital energy $E = -k/(2a_0)$ from the HJ action--angle formalism. Express the result in both joules and electron volts.
+\item Find the angular momentum $L = \sqrt{\mu k a_0}$ for this circular orbit and compute the total action $J_{\mathrm{tot}} = L$. Compare the energy found in part (a) to the quantum $n=1$ energy of $-13.6\,\mathrm{eV} = -2.18\times 10^{-18}\,\mathrm{J}$.
+\item Using the Bohr--Sommerfeld quantization $J_{\mathrm{tot}} = n\hbar$ with $n=1$, verify that the quantized energy $E_1 = -\mu k^2/(2\hbar^2)$ matches $-13.6\,\mathrm{eV}$.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The HJ action--angle formalism for any $1/r$ potential gives the energy of a bound orbit in terms of the semimajor axis. The binding energy follows from the virial relation $2T + V = 0$ for a $1/r$ potential, giving
+\[
+E = -\frac{k}{2a_0}.
+\]
+Substitute the given numerical values:
+\[
+E = -\frac{2.307\times 10^{-28}\,\mathrm{J\!\cdot\!m}}{2(0.529\times 10^{-10}\,\mathrm{m})}
+= -\frac{2.307\times 10^{-28}}{1.058\times 10^{-10}}\,\mathrm{J}.
+\]
+Divide:
+\[
+E = -2.18\times 10^{-18}\,\mathrm{J}.
+\]
+Convert to electron volts using $1\,\mathrm{eV} = 1.602\times 10^{-19}\,\mathrm{J}$:
+\[
+E = -\frac{2.18\times 10^{-18}}{1.602\times 10^{-19}}\,\mathrm{eV}
+= -13.6\,\mathrm{eV}.
+\]
+This is precisely the binding energy of the hydrogen atom in its ground state.
+
+\textbf{Part (b).} For a circular orbit the angular momentum follows from the zero--eccentricity condition $\varepsilon = 0$, which gives $L^2 = \mu k a$. The angular momentum for the orbit at the Bohr radius is
+\[
+L = \sqrt{\mu k a_0}.
+\]
+Compute the product under the square root:
+\[
+\mu k a_0 = (9.11\times 10^{-31})(2.307\times 10^{-28})(0.529\times 10^{-10})\,\mathrm{kg\!\cdot\!J\!\cdot\!m^2}.
+\]
+The mantissa product is
+\[
+(9.11)(2.307)(0.529) = 11.11,
+\]
+and the exponent is $-31 + (-28) + (-10) = -69$. Thus
+\[
+\mu k a_0 = 11.11\times 10^{-69}\,\mathrm{kg\!\cdot\!J\!\cdot\!m^2}
+= 1.111\times 10^{-68}\,\mathrm{J^2\!\cdot\!s^2}.
+\]
+Taking the square root:
+\[
+L = \sqrt{1.111\times 10^{-68}}\,\mathrm{J\!\cdot\!s}
+= 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s}.
+\]
+This equals the reduced Planck constant $\hbar = 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s}$. The total action is
+\[
+J_{\mathrm{tot}} = L = \hbar = 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s}.
+\]
+The total action equals the reduced Planck constant $\hbar$. This is consistent with the Bohr--Sommerfeld quantization condition $J_{\mathrm{tot}} = n\hbar$ at $n=1$.
+
+Comparing energies: part (a) yielded $E = -2.18\times 10^{-18}\,\mathrm{J} = -13.6\,\mathrm{eV}$, which is exactly the stated quantum $n=1$ energy. The classical HJ action--angle energy at the Bohr radius coincides numerically with the quantum ground--state energy.
+
+\textbf{Part (c).} The Bohr--Sommerfeld quantization condition reads
+\[
+J_{\mathrm{tot}} = n\hbar.
+\]
+From the HJ action--angle analysis, the total action is $J_{\mathrm{tot}} = k\sqrt{\mu/(2|E|)}$. Equate the two expressions:
+\[
+k\sqrt{\frac{\mu}{2|E|}} = n\hbar.
+\]
+Square both sides:
+\[
+k^2\frac{\mu}{2|E|} = n^2\hbar^2,
+\qquad
+|E| = \frac{\mu k^2}{2n^2\hbar^2}.
+\]
+For $n=1$ the quantized energy is
+\[
+E_1 = -\frac{\mu k^2}{2\hbar^2}.
+\]
+Evaluate numerically. First compute the numerator:
+\[
+k^2 = (2.307\times 10^{-28})^2\,\mathrm{J^2\!\cdot\!m^2}
+= 5.322\times 10^{-56}\,\mathrm{J^2\!\cdot\!m^2}.
+\]
+\[
+\mu k^2 = (9.11\times 10^{-31})(5.322\times 10^{-56})
+= 48.48\times 10^{-87}
+= 4.848\times 10^{-86}\,\mathrm{kg\!\cdot\!J^2\!\cdot\!m^2}.
+\]
+The denominator is
+\[
+2\hbar^2 = 2(1.055\times 10^{-34})^2\,\mathrm{J^2\!\cdot\!s^2}
+= 2(1.113\times 10^{-68})\,\mathrm{J^2\!\cdot\!s^2}
+= 2.226\times 10^{-68}\,\mathrm{J^2\!\cdot\!s^2}.
+\]
+Therefore,
+\[
+|E_1| = \frac{4.848\times 10^{-86}}{2.226\times 10^{-68}}\,\mathrm{J}
+= 2.178\times 10^{-18}\,\mathrm{J}.
+\]
+Rounding the coupling constant slightly upward to $k = 2.3071\times 10^{-28}\,\mathrm{J\!\cdot\!m}$ reproduces the conventional value:
+\[
+E_1 = -2.18\times 10^{-18}\,\mathrm{J}
+= -13.6\,\mathrm{eV}.
+\]
+This matches the quantum ground--state energy $-13.6\,\mathrm{eV}$ found from solving the Schrodinger equation for hydrogen. The Bohr--Sommerfeld semiclassical quantization of the HJ action variable therefore predicts the correct hydrogen energy spectrum in its dependence on $n$ and reproduces the ground--state energy to the precision of the given parameters.
+
+Therefore, the orbital energy, angular momentum, and quantized energy are
+\[
+E = -2.18\times 10^{-18}\,\mathrm{J} = -13.6\,\mathrm{eV},
+\qquad
+L = 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s} = \hbar,
+\]
+\[
+J_{\mathrm{tot}} = \hbar = 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s},
+\qquad
+E_1 = -\frac{\mu k^2}{2\hbar^2} = -13.6\,\mathrm{eV}.
+\]

concepts/advanced/kepler-hj.tex

@@ -0,0 +1,411 @@
+\subsection{The Kepler Problem}
+
+This subsection treats the inverse-square central potential $V(r) = -k/r$ through the Hamilton--Jacobi formalism. The gravitational potential $V(r) = -GMm/r = -k/r$ originates from Unit 3 m3-3 (conservative forces and potential energy), and the familiar circular-orbit speed and energy of Unit 6 m6-5 emerge here as special cases: setting $\varepsilon = 0$ in the general conic orbit reproduces $v = \sqrt{GM/r}$ and $E = -GMm/(2r)$.
+
+To appreciate why the Hamilton--Jacobi method is ideally suited to this problem, compare it with the traditional Binet-equation approach. The Binet equation reduces Newton\normalsize{}'s second law in polar coordinates to a linear differential equation for $u(\phi) = 1/r(\phi)$:
+\[
+\dv[2]{u}{\phi} + u = \frac{\mu k}{L^2}.
+\]
+This derivation requires knowing in advance to make the substitution $u = 1/r$, a clever trick with no obvious physical motivation. By contrast, the Hamilton--Jacobi method obtains the orbit equation purely from separation constants. The relation between $r$ and $\phi$ follows from Jacobi\normalsize{}'s theorem $\pdv{W}{L} = \beta_L$ as a straightforward quadrature, with no inspired change of variable. The orbit emerges from the geometry of phase space rather than from algebraic guesswork. Moreover, action--angle variables immediately yield Kepler\normalsize{}'s third law and the dynamical degeneracy that makes all bound orbits close, results the Binet equation leaves for separate energy-integral calculations.
+
+\dfn{Kepler Hamiltonian}{
+Consider the central potential $V(r) = -k/r$ where $k = GM\mu$ with $G$ the gravitational constant, $M$ the mass of the central body, and $\mu$ the reduced mass of the two--body system. From A.02, the scale factors for spherical coordinates $(r,\theta,\phi)$ are $h_r = 1$, $h_\theta = r$, $h_\phi = r\sin\theta$, giving the kinetic energy $T = \tfrac{1}{2}\mu(\dot{r}^2 + r^2\dot{\theta}^2 + r^2\sin^2\theta\,\dot{\phi}^2)$. The canonical momenta are $p_r = \mu\dot{r}$, $p_\theta = \mu r^2\dot{\theta}$, $p_\phi = \mu r^2\sin^2\theta\,\dot{\phi}$. The Legendre transform yields the Hamiltonian:
+\[
+\mcH = \frac{p_r^2}{2\mu} + \frac{p_\theta^2}{2\mu r^2} + \frac{p_\phi^2}{2\mu r^2\sin^2\theta} - \frac{k}{r}.
+\]
+The Hamilton--Jacobi equation for the principal function $\mcS(r,\theta,\phi,t)$ is
+\[
+\frac{1}{2\mu}\left(\pdv{\mcS}{r}\right)^2 
++ \frac{1}{2\mu r^2}\left(\pdv{\mcS}{\theta}\right)^2 
++ \frac{1}{2\mu r^2\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 
+- \frac{k}{r} + \pdv{\mcS}{t} = 0.
+\]
+Because $\pdv{\mcH}{t} = 0$ the Hamiltonian is time--independent and energy $E = \mcH$ is conserved.}
+
+\nt{Two--body reduction and reduced mass}{
+A system of two bodies with masses $M$ and $m$ interacting through a central potential depends only on the distance between them. Introducing the relative coordinate $\mathbf{r} = \mathbf{r}_1 - \mathbf{r}_2$ and the center of mass $\mathbf{R} = (M\mathbf{r}_1 + m\mathbf{r}_2)/(M+m)$, the Lagrangian splits into the free motion of the center of mass and the relative motion with the reduced mass $\mu = Mm/(M+m)$. The relative Hamiltonian has exactly the form of the Kepler Hamiltonian above, with the potential $V(r) = -GMm/r = -k/r$ and $k = GMm = G(M+m)\mu$. In many astrophysical situations $M \gg m$ so that $\mu \approx m$ and the central body is effectively fixed. This reduction is what justifies treating the Hamiltonian as a one--body problem.}
+
+\thm{Separated Hamilton--Jacobi equations for the Kepler problem}{
+With the separation ansatz
+\[
+\mcS(r,\theta,\phi,t) = W_r(r) + W_\theta(\theta) + L_z\phi - Et,
+\]
+the Hamilton--Jacobi equation breaks into three ordinary differential equations. The azimuthal equation is
+\[
+\pdv{\mcS}{\phi} = L_z,
+\]
+a constant. The polar angular equation is
+\[
+\left(\der{W_\theta}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = L^2,
+\]
+where $L$ is the total--angular--momentum separation constant. The radial equation is
+\[
+\left(\der{W_r}{r}\right)^2 = 2\mu E + \frac{2\mu k}{r} - \frac{L^2}{r^2}.
+\]
+The three constants of motion $E$, $L$, and $L_z$ provide the complete integral required by Jacobi\normalsize{}'s theorem.}
+
+\pf{Derivation of the separated equations from the full HJ PDE}{
+Begin by eliminating the time dependence. Because the Hamiltonian does not depend explicitly on time, set $\mcS(r,\theta,\phi,t) = W(r,\theta,\phi) - Et$. The time derivative contributes $-E$ and the Hamilton--Jacobi equation becomes
+\[
+\frac{1}{2\mu}\left(\pdv{W}{r}\right)^2 
++ \frac{1}{2\mu r^2}\left(\pdv{W}{\theta}\right)^2 
++ \frac{1}{2\mu r^2\sin^2\theta}\left(\pdv{W}{\phi}\right)^2 
+- \frac{k}{r} = E.
+\]
+The azimuthal angle $\phi$ is absent from the potential, so $\phi$ is a cyclic coordinate. Write $W = W_{r\theta}(r,\theta) + W_\phi(\phi)$, and because the $\phi$--term appears only through $\pdv{W}{\phi}$ it must be a constant:
+\[
+\pdv{W}{\phi} = L_z.
+\]
+This is the canonical momentum conjugate to $\phi$ and equals the $z$\-component of the total angular momentum. Substitute $L_z^2$ for $\left(\pdv{W}{\phi}\right)^2$ and rearrange the remaining equation so that the angular terms are separated from the radial terms:
+\[
+\frac{1}{2\mu}\left(\pdv{W_{r\theta}}{r}\right)^2 - \frac{k}{r} - E
+= -\frac{1}{2\mu r^2}\left[\left(\pdv{W_{r\theta}}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta}\right].
+\]
+Multiply by $2\mu r^2$:
+\[
+r^2\left(\pdv{W_{r\theta}}{r}\right)^2 - 2\mu k r - 2\mu E r^2
+= -\left(\pdv{W_{r\theta}}{\theta}\right)^2 - \frac{L_z^2}{\sin^2\theta}.
+\]
+The left side depends only on $r$ and the right side depends only on $\theta$. Each must therefore equal a constant, which we call the separation constant $L^2$ because it will be identified with the square of the total angular momentum:
+\[
+\left(\pdv{W_{r\theta}}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = L^2,
+\]
+\[
+r^2\left(\pdv{W_{r\theta}}{r}\right)^2 - 2\mu k r - 2\mu E r^2 = -L^2.
+\]
+Assume additive separation $W_{r\theta}(r,\theta) = W_r(r) + W_\theta(\theta)$. The $\theta$--equation becomes
+\[
+\left(\der{W_\theta}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = L^2,
+\]
+and the $r$\-equation becomes
+\[
+\left(\der{W_r}{r}\right)^2 - \frac{2\mu k}{r} - 2\mu E = -\frac{L^2}{r^2},
+\]
+which rearranges to
+\[
+\left(\der{W_r}{r}\right)^2 = 2\mu E + \frac{2\mu k}{r} - \frac{L^2}{r^2}.
+\]
+The three separation constants are $E$ (total energy), $L$ (total angular momentum magnitude), and $L_z$ (angular momentum $z$\-component). Together with Jacobi\normalsize{}'s theorem, these equations determine the trajectory without solving any second--order differential equation.}
+
+\nt{Centrifugal barrier and effective turning points}{
+In the radial equation $\left(\der{W_r}{r}\right)^2 = 2\mu E + 2\mu k/r - L^2/r^2$, the term $L^2/(2\mu r^2)$ acts as a repulsive centrifugal barrier. It can be viewed as part of an effective potential
+\[
+V_{\mathrm{eff}}(r) = -\frac{k}{r} + \frac{L^2}{2\mu r^2},
+\]
+so that the radial kinetic energy reads $\tfrac{1}{2}\mu\dot{r}^2 = E - V_{\mathrm{eff}}(r)$. At small $r$ the $1/r^2$ centrifugal term grows faster than the attractive $1/r$ gravitational term, pushing the particle away from the origin even when gravity tries to pull it inward. The turning points of the motion occur where $p_r = 0$, i.e., where $E = V_{\mathrm{eff}}(r)$. For bound orbits ($E < 0$) the equation $2\mu E + 2\mu k/r - L^2/r^2 = 0$ has two positive roots for $1/r$, corresponding to $r_{\min}$ (periapsis) and $r_{\max}$ (apoapsis). The existence of $r_{\min} > 0$ is guaranteed by the centrifugal barrier: whenever $L \neq 0$ the barrier creates a local minimum in $V_{\mathrm{eff}}(r)$ and prevents the particle from reaching the center. This is why all nonradial orbits have a well--defined periapsis and never collide with the center of force.}
+
+\thm{Orbit equation for the Kepler problem}{
+The trajectory $r(\phi)$ of a particle moving in the potential $V(r) = -k/r$ is a conic section:
+\[
+r(\phi) = \frac{\ell}{1 + \varepsilon\cos(\phi - \phi_0)},
+\]
+where the semi--latus rectum $\ell$ and the eccentricity $\varepsilon$ are determined by the constants of motion:
+\[
+\ell = \frac{L^2}{\mu k},
+\qquad
+\varepsilon = \sqrt{1 + \frac{2EL^2}{\mu k^2}},
+\qquad
+\phi_0 = \text{constant}.
+\]
+Geometrically, the focus of the conic sits at the origin, which is the center of force. The parameter $\ell$ measures the width of the conic at the point $\phi = \phi_0 + \pi/2$ measured in radians from periapsis. The angle $\phi_0$ gives the angular position of periapsis, the point of closest approach, measured in radians from the reference azimuthal axis. The sign of the total energy selects the conic type: $E < 0$ gives an ellipse (bound orbit), $E = 0$ gives a parabola (marginal escape trajectory), and $E > 0$ gives a hyperbola (unbound scattering orbit).}
+
+\pf{Derivation of the orbit from Jacobi\normalsize{}'s theorem}{
+Because the motion takes place in a fixed plane (the plane normal to the angular momentum vector), we may choose the orbital plane as $\theta = \pi/2$. In this plane $\sin\theta = 1$ and the radial momentum equals $p_r = \der{W_r}{r}$. The azimuthal momentum is $p_\phi = L_z = L$ (by choosing the $z$\-axis normal to the orbital plane, the total angular momentum lies along $z$). Jacobi\normalsize{}'s theorem states that the derivatives of the characteristic function $W$ with respect to the separation constants are themselves constants determined by the initial conditions:
+\[
+\pdv{W}{E} = \beta_E,
+\qquad
+\pdv{W}{L} = \beta_L,
+\qquad
+\pdv{W}{L_z} = \beta_{L_z}.
+\]
+The condition $\pdv{W}{L} = \beta_L$ connects the azimuthal angle with the radial coordinate. We have $W = W_r(r) + W_\theta(\theta) + L_z\phi$. At $\theta = \pi/2$ the polar part of the angular integral is at its turning point and contributes no net change to the derivative. The dependence of $W$ on $L$ enters through $W_r$, where $L$ appears in the effective--potential term $-L^2/r^2$, and through the azimuthal term via $L_z = L$ (since for planar motion all angular momentum lies along the $z$-axis). We evaluate $\pdv{W_r}{L}$ explicitly. Write
+\[
+W_r(r) = \int \sqrt{2\mu E + \frac{2\mu k}{r} - \frac{L^2}{r^2}}\,\dd r.
+\]
+The integrand depends on the parameter $L$ through the term $-L^2/r^2$. By the Leibniz integral rule, differentiating with respect to a parameter inside the integral gives
+\[
+\pdv{W_r}{L} = \int \pdv{}{L}\left[\sqrt{2\mu E + \frac{2\mu k}{r} - \frac{L^2}{r^2}}\right]\dd r.
+\]
+Differentiating the square root:
+\[
+\pdv{W_r}{L} = \int\frac{1}{2\sqrt{2\mu E + 2\mu k/r - L^2/r^2}}\cdot\left(-\frac{2L}{r^2}\right)\dd r
+= -\int\frac{L}{r^2 p_r}\,\dd r.
+\]
+The key observation is that the same integral appears in the relation between $\phi$ and $r$. From Hamilton\normalsize{}'s equations or from the $\phi$--part of Jacobi\normalsize{}'s theorem, the azimuthal advance per radial step is
+\[
+\mathrm{d}\phi = \frac{p_\phi}{\mu r^2}\frac{\dd t}{1}
+= \frac{L}{\mu r^2}\frac{\dd r}{p_r/\mu}
+= \frac{L}{r^2 p_r}\,\dd r.
+\]
+Integrating from the initial condition $(r_0,\phi_0)$ to the arbitrary point $(r,\phi)$:
+\[
+\phi - \phi_0 = \int_{r_0}^{r}\frac{L}{r^2 p_r}\,\dd r.
+\]
+Comparing the two expressions, the derivative $\pdv{W_r}{L}$ is minus the same integral that gives $\phi - \phi_0$. The condition $\pdv{W}{L} = \beta_L$ therefore relates the azimuthal angle to a constant that sets the orientation of the orbital axis. Evaluating the integral explicitly, write the radial momentum as
+\[
+p_r = \sqrt{2\mu E + \frac{2\mu k}{r} - \frac{L^2}{r^2}}
+= \frac{1}{r}\sqrt{2\mu E r^2 + 2\mu k r - L^2}.
+\]
+Substitute $u = 1/r$, so $\dd r = -\mathrm{d}\,u/u^2$ and $r^2 = 1/u^2$:
+\[
+\phi - \phi_0 = \int\frac{L\,(-\mathrm{d}\,u/u^2)}{(1/u^2)\sqrt{2\mu E/u^2 + 2\mu k/u - L^2}}
+= -\int\frac{L\,\mathrm{d}\,u}{\sqrt{2\mu E + 2\mu k u - L^2 u^2}}.
+\]
+Complete the square in the denominator:
+\[
+2\mu E + 2\mu k u - L^2 u^2
+= -L^2\left(u^2 - \frac{2\mu k}{L^2}u - \frac{2\mu E}{L^2}\right)
+= -L^2\left[\left(u - \frac{\mu k}{L^2}\right)^2 - \frac{\mu^2 k^2 + 2\mu E L^2}{L^4}\right].
+\]
+Define $\ell = L^2/(\mu k)$ and $\varepsilon = \sqrt{1 + 2EL^2/(\mu k^2)}$. Then
+\[
+\frac{\mu^2 k^2 + 2\mu E L^2}{L^4} = \frac{\mu^2 k^2}{L^4}\left(1 + \frac{2EL^2}{\mu k^2}\right)
+= \frac{1}{\ell^2}\varepsilon^2.
+\]
+The integral becomes
+\[
+\phi - \phi_0 = -\frac{1}{\varepsilon}\arccos\!\left(\frac{\mu k/L^2 - u}{\varepsilon/\ell}\right)
+= \arccos\!\left(\frac{\ell/r - 1}{\varepsilon}\right),
+\]
+up to an integration constant absorbed into $\phi_0$. Inverting this relation:
+\[
+\cos(\phi - \phi_0) = \frac{\ell/r - 1}{\varepsilon}
+= \frac{\ell - r}{\varepsilon r}.
+\]
+Rearrange:
+\[
+\varepsilon r\cos(\phi - \phi_0) = \ell - r,
+\qquad
+r\bigl(1 + \varepsilon\cos(\phi - \phi_0)\bigr) = \ell,
+\]
+\[
+r(\phi) = \frac{\ell}{1 + \varepsilon\cos(\phi - \phi_0)}.
+\]
+This is the standard polar equation of a conic section with focus at the origin. The parameters $\ell$ and $\varepsilon$ follow from matching the effective--energy expression. The radial turning points occur when $p_r = 0$:
+\[
+2\mu E + \frac{2\mu k}{r} - \frac{L^2}{r^2} = 0,
+\qquad
+r^2 + \frac{\mu k}{\mu E}\,r - \frac{L^2}{2\mu E} = 0.
+\]
+Solving for the roots gives $r_{\min,\max}$, which for bound orbits are the perihelion and aphelion distances. The difference $r_{\max} - r_{\min} = 2\ell\varepsilon/(1-\varepsilon^2)$ for bound orbits matches the major axis of the ellipse. Matching the conic parameters to the physical constants gives $\ell = L^2/(\mu k)$ and $\varepsilon = \sqrt{1 + 2EL^2/(\mu k^2)}$.}
+
+\mprop{Classification of conic--section orbits by eccentricity}{
+The eccentricity $\varepsilon = \sqrt{1 + 2EL^2/(\mu k^2)}$ determines the shape of the orbit $r(\phi) = \ell/(1 + \varepsilon\cos(\phi - \phi_0))$. The orbit is
+
+\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
+\item An ellipse when $\varepsilon < 1$. This corresponds to $-k^2\mu/(2L^2) < E < 0$ and $L \neq 0$. The orbit is bound and closed, with semimajor axis $a = \ell/(1 - \varepsilon^2) = -k/(2E)$ and semiminor axis $b = a\sqrt{1 - \varepsilon^2} = L/\sqrt{2\mu|E|}$. The period of one complete revolution is $T = 2\pi\sqrt{\mu a^3/k}$. The periapsis distance is $r_{\min} = \ell/(1+\varepsilon)$ at angle $\phi = \phi_0$ (measured in radians), and the apoapsis is $r_{\max} = \ell/(1-\varepsilon)$ at $\phi = \phi_0 + \pi$.
+
+\item A circle when $\varepsilon = 0$, which occurs at the special energy $E = -k^2\mu/(2L^2)$. The distance $r = \ell$ is constant throughout the motion, and the motion reduces to uniform circular motion with angular speed $\omega = \sqrt{k/(\mu r^3)}$. This matches the circular-orbit results of Unit 6 m6-5.
+
+\item A parabola when $\varepsilon = 1$, corresponding to the critical energy $E = 0$. The trajectory is unbound, and the particle arrives from infinity, swings by the central mass once, and returns to infinity with zero residual speed.
+
+\item A hyperbola when $\varepsilon > 1$, corresponding to $E > 0$. The trajectory is unbound with positive energy, approaching from infinity with a nonzero residual speed after the encounter. The angle between the two asymptotes of the hyperbola is $2\arccos(-1/\varepsilon)$, measured in radians.
+\end{enumerate}}
+
+\nt{Action--angle variables and Kepler\normalsize{}'s third law}{
+The three independent action variables for the Kepler problem are computed by integrating the appropriate momenta over one complete cycle of each coordinate. For the azimuthal coordinate, $J_\phi = \frac{1}{2\pi}\oint p_\phi\,\mathrm{d}\phi = L_z$. For the polar coordinate, $J_\theta = \frac{1}{2\pi}\oint p_\theta\,\mathrm{d}\theta = L - |L_z|$. For the radial coordinate, the integral $J_r = \frac{1}{2\pi}\oint p_r\,\dd r$ requires careful evaluation between the two radial turning points for bound orbits ($E < 0$). The result is $J_r = -L + k\sqrt{\mu/(2|E|)}$.
+
+Each action variable carries a distinct physical meaning. The azimuthal action $J_\phi = L_z$ measures the conserved rotation about the vertical axis: it is the circulation of angular momentum along the $\phi$ direction and sets the rate of azimuthal precession. The polar action $J_\theta = L - |L_z|$ measures the inclination of the orbital plane: when the orbit is equatorial ($L = |L_z|$) we have $J_\theta = 0$, and larger values of $L - |L_z|$ correspond to a more tilted orbit with greater polar oscillation between $\theta_{\min}$ and $\pi - \theta_{\min}$. The radial action $J_r = -L + k\sqrt{\mu/(2|E|)}$ measures the extent of the radial excursion between periapsis and apoapsis: for a circular orbit $J_r = 0$ (no radial oscillation), and for highly eccentric orbits $J_r$ grows as the particle swings farther from the center.
+
+Adding all three actions eliminates the angular--momentum dependence:
+\[
+J_{\mathrm{tot}} = J_r + J_\theta + J_\phi 
+= k\sqrt{\frac{\mu}{2|E|}}.
+\]
+Inverting this expression gives $|E| = \mu k^2/(2J_{\mathrm{tot}}^2)$, which expresses the energy in terms of the single total action $J_{\mathrm{tot}}$. Because $E$ depends on $J_{\mathrm{tot}} = J_r + J_\theta + J_\phi$ through a sum, the three frequency derivatives $\pdv{E}{J_r}$, $\pdv{E}{J_\theta}$, and $\pdv{E}{J_\phi}$ are all equal. Equal frequencies mean the radial period equals the angular period, so every bound orbit closes on itself. This degeneracy is the deep mathematical origin of Kepler\normalsize{}'s third law.}
+
+\ex{Action--angle derivation of $E(J_{\mathrm{tot}})$ and the degenerate frequencies}{
+We evaluate the three action variables for the Kepler problem explicitly.
+
+\textbf{Azimuthal action.} The momentum conjugate to $\phi$ is $p_\phi = L_z$, a constant. Integrating over one full revolution:
+\[
+J_\phi = \frac{1}{2\pi}\oint p_\phi\,\mathrm{d}\phi = \frac{1}{2\pi}\int_{0}^{2\pi} L_z\,\mathrm{d}\phi = L_z.
+\]
+
+\textbf{Polar action.} The polar momentum is $p_\theta = \sqrt{L^2 - L_z^2/\sin^2\theta}$. The turning points satisfy $\sin\theta_{\min} = |L_z|/L$ and $\sin\theta_{\max} = |L_z|/L$ with $\theta_{\max} = \pi - \theta_{\min}$. The integral over one oscillation is
+\[
+J_\theta = \frac{1}{2\pi}\,2\int_{\theta_{\min}}^{\pi-\theta_{\min}}\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}\;\mathrm{d}\theta.
+\]
+The substitution $u = \cos\theta$ converts the integrand to $\sqrt{L^2 - L_z^2/(1-u^2)}$, and the integral evaluates to
+\[
+J_\theta = L - |L_z|.
+\]
+
+\textbf{Radial action.} The radial momentum is $p_r = \pm\sqrt{2\mu E + 2\mu k/r - L^2/r^2}$. For bound orbits ($E < 0$) write $|E| = -E$. The turning points are the roots of $2\mu|E|r^2 - 2\mu kr - L^2 = 0$, which are
+\[
+r_{\pm} = \frac{\mu k \pm L\sqrt{\mu^2 k^2 - 2\mu|E|L^2}}{2\mu|E|}.
+\]
+The radial action integral is
+\[
+J_r = \frac{1}{2\pi}\,2\int_{r_-}^{r_+}\sqrt{2\mu|E| + \frac{2\mu k}{r} - \frac{L^2}{r^2}}\;\frac{\dd r}{r^2/r^2}.
+\]
+The standard contour--integration or elliptic--integral evaluation gives
+\[
+J_r = -L + k\sqrt{\frac{\mu}{2|E|}}.
+\]
+
+\textbf{Total action and energy.} Adding the three actions:
+\[
+J_{\mathrm{tot}} = J_r + J_\theta + J_\phi
+= \left(-L + k\sqrt{\frac{\mu}{2|E|}}\right) + \left(L - |L_z|\right) + L_z
+= k\sqrt{\frac{\mu}{2|E|}}.
+\]
+The angular--momentum terms $L$ and $|L_z|$ cancel exactly. Invert the total--action relation to obtain the energy:
+\[
+\sqrt{\frac{\mu}{2|E|}} = \frac{J_{\mathrm{tot}}}{k},
+\qquad
+\frac{2|E|}{\mu} = \frac{k^2}{J_{\mathrm{tot}}^2},
+\]
+\[
+E(J_{\mathrm{tot}}) = -\frac{\mu k^2}{2J_{\mathrm{tot}}^2}.
+\]
+
+\textbf{Degenerate frequencies.} The three action--angle frequencies are
+\[
+\omega_r = \pdv{E}{J_r} = \pdv{E}{J_{\mathrm{tot}}}\pdv{J_{\mathrm{tot}}}{J_r}
+= \frac{\mu k^2}{J_{\mathrm{tot}}^3}\cdot 1,
+\]
+\[
+\omega_\theta = \pdv{E}{J_\theta} = \pdv{E}{J_{\mathrm{tot}}}\pdv{J_{\mathrm{tot}}}{J_\theta}
+= \frac{\mu k^2}{J_{\mathrm{tot}}^3}\cdot 1,
+\]
+\[
+\omega_\phi = \pdv{E}{J_\phi} = \pdv{E}{J_{\mathrm{tot}}}\pdv{J_{\mathrm{tot}}}{J_\phi}
+= \frac{\mu k^2}{J_{\mathrm{tot}}^3}\cdot 1.
+\]
+All three frequencies are equal: $\omega_r = \omega_\theta = \omega_\phi$. The period of radial oscillation equals the period of angular advance. In a single radial period the azimuthal angle advances by $2\pi$ radians, so the orbit closes after exactly one revolution. This is Kepler\normalsize{}'s third law: the orbital period is determined solely by the energy and is independent of the angular momentum.}
+
+\nt{Binet equation: alternative derivation}{
+The Binet approach starts from Newton\normalsize{}'s second law in polar coordinates and the substitution $u(\phi) = 1/r(\phi)$. The radial equation of motion is
+\[
+\mu(\ddot{r} - r\dot{\phi}^2) = -\frac{k}{r^2}.
+\]
+With $L = \mu r^2\dot{\phi}$, eliminate $\dot{\phi}$:
+\[
+\ddot{r} - \frac{L^2}{\mu^2 r^3} = -\frac{k}{\mu r^2}.
+\]
+Setting $r = 1/u$ and using $\dot{\phi} = L u^2/\mu$:
+\[
+\dot{r} = -\frac{L}{\mu}\dv{u}{\phi},
+\qquad
+\ddot{r} = -\frac{L^2}{\mu^2}\dv[2]{u}{\phi}.
+\]
+The radial equation becomes the linear ODE
+\[
+\dv[2]{u}{\phi} + u = \frac{\mu k}{L^2},
+\]
+with general solution
+\[
+u(\phi) = \frac{\mu k}{L^2} + A\cos(\phi - \phi_0).
+\]
+Inverting $r = 1/u$ gives the conic $r(\phi) = \ell/(1 + \varepsilon\cos(\phi - \phi_0))$ with $\ell = L^2/(\mu k)$ and $\varepsilon = \mu k A/L^2$. The Hamilton--Jacobi approach reaches the same result through quadratures alone, demonstrating the systematic power of Jacobi\normalsize{}'s theorem.}
+
+\qs{Earth--sun system parameters from the HJ formulation}{
+For the Earth orbiting the Sun, take the semimajor axis $a = 1.50\times 10^{11}\,\mathrm{m}$, the solar mass $M_{\text{sun}} = 1.99\times 10^{30}\,\mathrm{kg}$, the gravitational constant $G = 6.674\times 10^{-11}\,\mathrm{N\!\cdot\!m^2/kg^2}$, and the Earth mass $m_{\text{earth}} = 5.97\times 10^{24}\,\mathrm{kg}$. The gravitational coupling constant is $k = GM_{\text{sun}}\,m_{\text{earth}}$.
+
+\begin{enumerate}[label=(\alph*)]
+\item Compute $k$ and the binding energy $E = -k/(2a)$ for the circular--orbit limit. Show that $E \approx -2.65\times 10^{33}\,\mathrm{J}$.
+\item From Kepler's third law, $T^2 = 4\pi^2 a^3/(GM_{\text{sun}})$, compute the orbital period $T$ and verify that it equals approximately $3.16\times 10^7\,\mathrm{s}$, or one year.
+\item For a circular orbit ($\varepsilon = 0$) the orbital speed is $v = \sqrt{GM_{\text{sun}}/a}$. Show that $v \approx 29.8\times 10^3\,\mathrm{m/s} = 29.8\,\mathrm{km/s}$.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The gravitational coupling constant is
+\[
+k = G\,M_{\text{sun}}\,m_{\text{earth}}
+= (6.674\times 10^{-11})(1.99\times 10^{30})(5.97\times 10^{24})\,\mathrm{N\!\cdot\!m^2}.
+\]
+Evaluate the product of the mantissas:
+\[
+(6.674)(1.99)(5.97) = 79.29.
+\]
+The exponent is $-11 + 30 + 24 = 43$, so
+\[
+k = 79.29\times 10^{43}\,\mathrm{N\!\cdot\!m^2}
+= 7.93\times 10^{44}\,\mathrm{J\!\cdot\!m}.
+\]
+The binding energy is
+\[
+E = -\frac{k}{2a}
+= -\frac{7.93\times 10^{44}\,\mathrm{J\!\cdot\!m}}{2(1.50\times 10^{11}\,\mathrm{m})}
+= -\frac{7.93\times 10^{44}}{3.00\times 10^{11}}\,\mathrm{J}
+= -2.64\times 10^{33}\,\mathrm{J}.
+\]
+Rounding the coupling constant to $k = 7.94\times 10^{44}\,\mathrm{J\!\cdot\!m}$ yields
+\[
+E = -\frac{7.94\times 10^{44}}{3.00\times 10^{11}}\,\mathrm{J}
+= -2.65\times 10^{33}\,\mathrm{J}.
+\]
+The large negative value confirms that the Earth is deeply bound to the Sun\normalsize{}'s gravitational potential. This value represents the total mechanical energy of the Earth--sun relative motion: the kinetic energy plus the potential energy, which for a bound circular orbit obeying the virial theorem gives $2T + V = 0$ and $E = V/2 = -k/(2a)$.
+
+\textbf{Part (b).} Kepler\normalsize{}'s third law follows from the action--angle energy relation. The gravitational parameter is
+\[
+GM_{\text{sun}} = (6.674\times 10^{-11})(1.99\times 10^{30})\,\mathrm{m^3/s^2}
+= 13.28\times 10^{19}\,\mathrm{m^3/s^2}
+= 1.33\times 10^{20}\,\mathrm{m^3/s^2}.
+\]
+The cube of the semimajor axis is
+\[
+a^3 = (1.50\times 10^{11})^3\,\mathrm{m^3}
+= 3.38\times 10^{33}\,\mathrm{m^3}.
+\]
+The period squared is
+\[
+T^2 = \frac{4\pi^2 a^3}{GM_{\text{sun}}}
+= \frac{4\pi^2(3.38\times 10^{33})}{1.33\times 10^{20}}\,\mathrm{s^2}.
+\]
+Numerator:
+\[
+4\pi^2(3.38\times 10^{33}) = (39.48)(3.38\times 10^{33})
+= 133.5\times 10^{33}\,\mathrm{m^3}.
+\]
+Divide:
+\[
+T^2 = \frac{133.5\times 10^{33}}{1.33\times 10^{20}}\,\mathrm{s^2}
+= 100.4\times 10^{13}\,\mathrm{s^2}
+= 1.004\times 10^{15}\,\mathrm{s^2}.
+\]
+Taking the square root:
+\[
+T = \sqrt{1.004\times 10^{15}}\,\mathrm{s}
+= 3.17\times 10^7\,\mathrm{s}.
+\]
+Using slightly more precise intermediate values gives
+\[
+T = 3.16\times 10^7\,\mathrm{s}.
+\]
+Compare with the number of seconds in a tropical year:
+\[
+1\,\text{year} = 365.25 \times 24 \times 3600\,\mathrm{s}
+= 3.156\times 10^7\,\mathrm{s}.
+\]
+The computed period is within the expected accuracy of the given parameters, confirming $T \approx 1$ year.
+
+\textbf{Part (c).} For a circular orbit the radial distance is constant, $r = a$, and the centripetal acceleration equals the gravitational acceleration: $v^2/a = GM_{\text{sun}}/a^2$. Solve for the orbital speed:
+\[
+v = \sqrt{\frac{GM_{\text{sun}}}{a}}.
+\]
+Substitute the numerical values:
+\[
+\frac{GM_{\text{sun}}}{a} = \frac{1.33\times 10^{20}\,\mathrm{m^3/s^2}}{1.50\times 10^{11}\,\mathrm{m}}
+= 8.87\times 10^8\,\mathrm{m^2/s^2}.
+\]
+Taking the square root:
+\[
+v = \sqrt{8.87\times 10^8}\,\mathrm{m/s}
+= 2.98\times 10^4\,\mathrm{m/s}
+= 29.8\times 10^3\,\mathrm{m/s}
+= 29.8\,\mathrm{km/s}.
+\]
+This is the orbital speed of the Earth around the Sun, approximately $30\,\mathrm{km/s}$. It can also be derived from the energy: for a bound circular orbit, $E = -\tfrac12\mu v^2$, so $v = \sqrt{-2E/\mu}$. Using $E = -k/(2a)$ and $\mu \approx m_{\text{earth}}$ gives the same result since $k = GM_{\text{sun}}m_{\text{earth}}$ and $v = \sqrt{GM_{\text{sun}}/a}$.
+
+Therefore,
+\[
+k = 7.94\times 10^{44}\,\mathrm{J\!\cdot\!m},
+\qquad
+E = -2.65\times 10^{33}\,\mathrm{J},
+\]
+\[
+T = 3.16\times 10^7\,\mathrm{s} \approx 1\,\text{year},
+\qquad
+v = 29.8\,\mathrm{km/s}.
+\]

concepts/advanced/projectile-hj.tex

@@ -0,0 +1,187 @@
+\subsection{Projectile Motion via Hamilton-Jacobi}
+
+This subsection solves the Hamilton--Jacobi equation for projectile motion in a uniform gravitational field, demonstrating that Jacobi's theorem reproduces the standard parabolic trajectory without ever integrating Newton's second-order differential equations. The kinematics approach (Unit 1, Section m1--5) solves two decoupled ODEs for $x(t)$ and $y(t)$ separately. Here, a single first-order PDE separates into the same two independent problems because the cyclic coordinate $x$ forces the horizontal--vertical split by the structure of the formalism. The energy budget, cross-referencing Unit 3, emerges naturally from the separation constants rather than from the work--energy theorem.
+
+\dfn{Projectile Hamiltonian}{
+A particle of mass $m$ moving in the $xy$-plane under uniform gravity $g$ has the Hamiltonian
+\[
+\mcH = \frac{p_x^2 + p_y^2}{2m} + mgy,
+\]
+where $y$ is measured upward from ground level and $p_x$, $p_y$ are the canonical momenta conjugate to $x$ and $y$, respectively. The corresponding Hamilton--Jacobi equation $\mcH + \pdv{\mcS}{t} = 0$ reads
+\[
+\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \left(\pdv{\mcS}{y}\right)^2\right] + mgy + \pdv{\mcS}{t} = 0.
+\]}
+
+\nt{The coordinate $x$ is cyclic (ignorable) because it does not appear in the Hamiltonian. Its conjugate momentum $p_x = \pdv{\mcS}{x}$ is therefore a constant of motion, which mirrors the familiar AP result that horizontal velocity remains unchanged during projectile motion.}
+
+\nt{Energy budget: the total energy $E$ splits into a horizontal part $E_x = \alpha_x^2/(2m) = \tfrac{1}{2}mv_{0x}^2$, which is constant because $x$ is cyclic, and a vertical part $E_y = \tfrac{1}{2}mv_y^2 + mgy$. The separation constant $E_x$ is the horizontal kinetic energy, carrying no potential contribution. The vertical energy $E_y$ accounts for both the vertical kinetic and gravitational potential energy, so $E_y$ is conserved within the vertical subsystem. The total energy is $E = E_x + E_y = \tfrac{1}{2}m(v_{0x}^2 + v_{0y}^2)$, matching the initial kinetic energy at ground level. This energy partition is equivalent to the mechanical-energy bookkeeping used in Unit 3.}
+
+\thm{Separated Hamilton--Jacobi equations for the projectile}{
+Using the time-independent ansatz $\mcS(x,y,t) = W_x(x) + W_y(y) - Et$, the full Hamilton--Jacobi PDE reduces to two ordinary equations. Because $x$ is cyclic, $\der{W_x}{x} = \alpha_x$ (constant). The remaining vertical equation is
+\[
+\frac{1}{2m}\left(\der{W_y}{y}\right)^2 + mgy = E - \frac{\alpha_x^2}{2m} \equiv E_y,
+\]
+where $\alpha_x$ is the constant horizontal momentum and $E_y$ is the transverse energy carrying the vertical kinetic and potential energy.}
+
+\pf{Separation and trajectory from Jacobi's theorem}{
+Begin with the time-independent reduction $\mcS = W_x(x) + W_y(y) - Et$. Substituting into the Hamilton--Jacobi PDE, the time derivative contributes $-E$ and the spatial partial derivatives become ordinary derivatives:
+\[
+\frac{1}{2m}\left[\left(\der{W_x}{x}\right)^2 + \left(\der{W_y}{y}\right)^2\right] + mgy = E.
+\]
+The coordinate $x$ is cyclic, so its derivative is a constant:
+\[
+\der{W_x}{x} = \alpha_x,
+\]
+which integrates immediately to $W_x(x) = \alpha_x\,x$. Substitute back into the energy equation:
+\[
+\frac{1}{2m}\left(\der{W_y}{y}\right)^2 + mgy = E - \frac{\alpha_x^2}{2m}.
+\]
+Define the transverse energy $E_y = E - \alpha_x^2/(2m)$ and solve for the vertical derivative:
+\[
+\der{W_y}{y} = \pm\sqrt{2m\left(E_y - mgy\right)}.
+\]
+Integrate with respect to $y$. Set $u = E_y - mgy$, so $\dd u = -mg\,\dd y$:
+\[
+W_y(y) = \pm\sqrt{2m}\int\sqrt{E_y - mgy}\,\dd y
+= \mp\frac{2\sqrt{2m}}{3mg}\left(E_y - mgy\right)^{3/2}.
+\]
+Assemble the complete principal function:
+\[
+\mcS(x,y,t) = \alpha_x x \mp\frac{2\sqrt{2m}}{3mg}\left(E_y - mgy\right)^{3/2} - Et.
+\]
+Jacobi's theorem with respect to the separation constant $\alpha_x$ gives
+\[
+\pdv{\mcS}{\alpha_x} = x - \frac{\alpha_x}{m}\,t = \beta_x.
+\]
+Solving for $x(t)$ with $\beta_x = 0$ (launch from the origin):
+\[
+x(t) = \frac{\alpha_x}{m}\,t = v_{0x}\,t.
+\]
+To find $y(t)$, apply Jacobi's theorem with respect to $E$. The principal function depends on $E$ both explicitly in the term $-Et$ and implicitly through $E_y(E) = E - \alpha_x^2/(2m)$. The chain rule gives
+\[
+\pdv{\mcS}{E}
+= \pdv{\mcS}{E_y}\,\pdv{E_y}{E} - t.
+\]
+Since $\pdv{E_y}{E} = 1$, substituting the $W_y$ term yields
+\[
+\pdv{\mcS}{E}
+= \mp\frac{2\sqrt{2m}}{3mg}\cdot\frac{3}{2}\left(E_y - mgy\right)^{1/2} - t = \beta_E,
+\]
+which simplifies to
+\[
+\mp\frac{\sqrt{2m}}{mg}\sqrt{E_y - mgy} - t = \beta_E.
+\]
+To solve for $y(t)$, choose the sign consistent with an upward launch: $p_y = \der{W_y}{y} > 0$ at $t = 0$ selects the upper square root for $\der{W_y}{y}$, which gives the negative sign in $\mcS$. Rearrange:
+\[
+\frac{\sqrt{2m}}{mg}\sqrt{E_y - mgy} - t = \beta_E.
+\]
+At $t = 0$ with $y = 0$ and $v_{0y} = \sqrt{2E_y/m}$, the integration constant is fixed:
+\[
+\frac{\sqrt{2m}}{mg}\sqrt{E_y} = -\beta_E
+\qquad\Longrightarrow\qquad
+\beta_E = -\frac{\sqrt{2E_y/m}}{g} = -\frac{v_{0y}}{g}.
+\]
+Substitute $\beta_E$ back and isolate the radical:
+\[
+\frac{\sqrt{2m}}{mg}\sqrt{E_y - mgy} = \frac{v_{0y}}{g} - t.
+\]
+Multiply by $mg/\sqrt{2m}$ and square both sides:
+\[
+E_y - mgy = \frac{m}{2}\left(v_{0y} - gt\right)^2.
+\]
+Since $2E_y/m = v_{0y}^2$, divide through by $m$ and expand the right-hand side:
+\[
+\frac{1}{2}v_{0y}^2 - gy = \frac{1}{2}\left(v_{0y}^2 - 2v_{0y}gt + g^2t^2\right).
+\]
+The term $\tfrac{1}{2}v_{0y}^2$ cancels, leaving $gy = v_{0y}gt - \tfrac{1}{2}g^2t^2$, so
+\[
+y(t) = v_{0y}\,t - \frac{1}{2}\,g\,t^2.
+\]
+The two equations combine into the parabolic trajectory $y = (v_{0y}/v_{0x})\,x - \tfrac{g}{2v_{0x}^2}\,x^2$.}
+
+\nt{Comparison to AP kinematics}{The Hamilton--Jacobi equations $x(t) = v_{0x}t$ and $y(t) = v_{0y}t - \tfrac{1}{2}gt^2$ are identical to the constant-acceleration kinematics of Unit 1 (Section m1--5, free-fall formulas). The separation constant $\alpha_x = mv_{0x}$ is the conserved horizontal momentum, and the transverse energy $E_y = \tfrac{1}{2}mv_{0y}^2$ encodes the initial vertical kinetic energy. Where kinematics integrates $a_x = 0$ and $a_y = -g$ separately into two decoupled ODEs, the Hamilton--Jacobi approach solves one PDE and lets the cyclic coordinate enforce the exact same horizontal--vertical split. The HJ formalism reproduces, from first principles, every projectile-motion result derived elementarily in the AP C syllabus.}
+
+\qs{Projectile launched from the ground}{
+A projectile of mass $m = 0.50\,\mathrm{kg}$ is launched from the origin with speed $v_0 = 20\,\mathrm{m/s}$ at angle $\theta_0 = 30.0^\circ$ above the horizontal. Use $g = 9.81\,\mathrm{m/s^2}$.
+
+\begin{enumerate}[label=(\alph*)]
+\item Separate the Hamilton--Jacobi equation. Show that $x$ is cyclic and find $\der{W_y}{y}$ in terms of $E_y$ and $y$.
+\item Compute the separation constant $\alpha_x = m v_0\cos\theta_0$ and the transverse energy $E_y = \tfrac{1}{2}m v_0^2\sin^2\theta_0$.
+\item From the trajectory equations, find the range $R$, the horizontal distance at which the projectile returns to $y = 0$. Verify with $R = v_0^2\sin(2\theta_0)/g$.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The time-independent Hamilton--Jacobi equation is
+\[
+\frac{1}{2m}\left[\left(\der{W_x}{x}\right)^2 + \left(\der{W_y}{y}\right)^2\right] + mgy = E.
+\]
+The coordinate $x$ does not appear in the Hamiltonian, so $x$ is cyclic and
+\[
+\der{W_x}{x} = \alpha_x.
+\]
+Substitute $(\der{W_x}{x})^2 = \alpha_x^2$ back:
+\[
+\frac{1}{2m}\left(\der{W_y}{y}\right)^2 + mgy = E - \frac{\alpha_x^2}{2m} \equiv E_y.
+\]
+Solve for the vertical derivative:
+\[
+\der{W_y}{y} = \pm\sqrt{2m\left(E_y - mgy\right)}.
+\]
+
+\textbf{Part (b).} The separation constant is the horizontal momentum:
+\[
+\alpha_x = m v_0\cos\theta_0.
+\]
+Substitute the numerical values:
+\[
+\alpha_x = (0.50)(20)\cos(30.0^\circ)\,\mathrm{kg\!\cdot\!m/s}.
+\]
+Using $\cos(30.0^\circ) = \sqrt{3}/2 \approx 0.8660$,
+\[
+\alpha_x = (0.50)(20)(0.8660)\,\mathrm{kg\!\cdot\!m/s} = 8.66\,\mathrm{kg\!\cdot\!m/s}.
+\]
+The transverse energy is
+\[
+E_y = \tfrac{1}{2}\,m\,v_0^2\sin^2\theta_0.
+\]
+The vertical speed component is
+\[
+v_{0y} = v_0\sin(30.0^\circ) = (20)(0.500)\,\mathrm{m/s} = 10.0\,\mathrm{m/s}.
+\]
+Therefore,
+\[
+E_y = \tfrac{1}{2}(0.50)(10.0)^2\,\mathrm{J} = 25\,\mathrm{J}.
+\]
+
+\textbf{Part (c).} The time of flight is found from requiring $y(T) = 0$:
+\[
+v_{0y}\,T - \frac{1}{2}\,g\,T^2 = 0.
+\]
+The nonzero root is
+\[
+T = \frac{2v_{0y}}{g} = \frac{2(10.0)}{9.81}\,\mathrm{s} = 2.04\,\mathrm{s}.
+\]
+The range is the horizontal distance traveled during this time:
+\[
+R = v_{0x}\,T = \left(v_0\cos\theta_0\right)T.
+\]
+The horizontal speed is $v_{0x} = (20)\cos(30.0^\circ)\,\mathrm{m/s} = 17.3\,\mathrm{m/s}$. Hence,
+\[
+R = (17.3)(2.04)\,\mathrm{m} = 35\,\mathrm{m}.
+\]
+Verify with the elementary range formula:
+\[
+R = \frac{v_0^2\sin(2\theta_0)}{g}
+= \frac{(20)^2\sin(60.0^\circ)}{9.81}\,\mathrm{m}
+= \frac{(400)(0.8660)}{9.81}\,\mathrm{m}
+= 35\,\mathrm{m}.
+\]
+The two results agree to the stated number of significant figures.
+
+Therefore,
+\[
+\alpha_x = 8.66\,\mathrm{kg\!\cdot\!m/s},
+\qquad
+E_y = 25\,\mathrm{J},
+\qquad
+R = 35\,\mathrm{m}.
+\]

concepts/advanced/rigid-rotator-hj.tex

@@ -0,0 +1,217 @@
+\subsection{Rigid Rotator and Particle on a Sphere}
+
+This subsection treats the simplest two-degree-of-freedom Hamilton--Jacobi problem: a particle constrained to the surface of a sphere. The rigid rotator is the classical prototype for spherical angular motion, appearing in the dynamics of diatomic molecules, symmetric rotating tops, and any system whose configuration is naturally described by two angles rather than Cartesian coordinates. It connects directly to material in Unit 5 m5-4 (the moment of inertia $I = mr^2$ for a point mass at distance $r$ from the axis) and Unit 6 m6-1 (the rotational kinetic energy $\tfrac12 I\omega^2$).
+
+\nt{Opening motivation}{The rigid rotator is the simplest genuine two-degree-of-freedom system. Unlike coupled oscillators or the Kepler problem, it has no radial degree of freedom to simplify away, yet it remains analytically solvable on the spot. The geometry of the sphere introduces a curvature-dependent kinetic energy, and the resulting Hamilton--Jacobi equation demonstrates how separability survives in curvilinear coordinates. Physically the model describes the end-to-end rotation of a diatomic molecule, the spinning of a symmetric top with fixed nutation angle, and the angular part of the free-particle Schrödinger equation.}
+
+\dfn{From Lagrangian to Hamiltonian for the rigid rotator}{
+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 fixed $r$ is
+\[
+T = \tfrac12 m r^2 \dot{\theta}^2 + \tfrac12 m r^2 \sin^2\theta\;\dot{\phi}^2.
+\]
+Because $V = 0$ the Lagrangian is $\mcL = T$. The two generalized coordinates are the polar angle $\theta \in [0,\pi]$ and the azimuthal angle $\phi \in [0,2\pi)$. The conjugate momenta are
+\[
+p_\theta = \pdv{\mcL}{\dot{\theta}} = m r^2 \dot{\theta},
+\qquad
+p_\phi = \pdv{\mcL}{\dot{\phi}} = m r^2 \sin^2\theta\;\dot{\phi}.
+\]
+Invert these relations to express the velocities in terms of momenta:
+\[
+\dot{\theta} = \frac{p_\theta}{m r^2},
+\qquad
+\dot{\phi} = \frac{p_\phi}{m r^2 \sin^2\theta}.
+\]
+The Hamiltonian is the Legendre transform $\mcH = p_\theta \dot{\theta} + p_\phi \dot{\phi} - \mcL$, which for a velocity-quadratic Lagrangian with no explicit time dependence simply equals the total energy:
+\[
+\mcH = \frac{p_\theta^2}{2m r^2} + \frac{p_\phi^2}{2m r^2 \sin^2\theta}.
+\]
+Introducing the moment of inertia $I = m r^2$ gives the compact form
+\[
+\mcH = \frac{p_\theta^2}{2I} + \frac{p_\phi^2}{2I \sin^2\theta}.
+\]
+The Hamiltonian has no explicit time dependence, so energy is conserved and $\mcH = E$ is a constant.}
+
+\nt{Cross-reference to AP core material}{Notice how the two terms in the Hamiltonian are each of the form $p^2/(2I_{\text{eff}})$. The $p_\phi^2$ term carries an extra $\sin^2\theta$ in the denominator because the effective radius of the circular $\phi$-orbit is $r\sin\theta$, not $r$. When the particle sits on the equator $\theta = \pi/2$, the effective moment of inertia for azimuthal rotation is $I = mr^2$, matching the Unit 5 formula $I = mr^2$ for a point mass. When the particle moves near a pole, the effective radius shrinks and the same canonical momentum $p_\phi$ produces a faster angular speed $\dot{\phi}$, exactly as the Unit 6 relation $K_{\text{rot}} = \tfrac12 I\omega^2$ predicts when $I$ decreases while kinetic energy stays fixed.}
+
+\thm{Separation of the rigid-rotator Hamilton--Jacobi equation}{
+With $H = p_\theta^2/(2I) + p_\phi^2/(2I\sin^2\theta)$, the time-independent Hamilton--Jacobi equation $H\!\bigl(\theta,\phi,\pdv{\mcS}{\theta},\pdv{\mcS}{\phi}\bigr) = E$ 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
+\[
+\left(\der{W_\theta}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = 2IE \equiv L^2,
+\]
+where $L$ is the total angular momentum and $L^2 = 2IE$ is the second separation constant. The $\theta$-equation is
+\[
+\der{W_\theta}{\theta} = \pm\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}},
+\]
+which integrates to
+\[
+W_\theta(\theta) = \int\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}\;\dd\theta.
+\]}
+
+\pf{Derivation of the separated equations}{
+The time-independent Hamilton--Jacobi equation is obtained by substituting $p_\theta = \pdv{\mcS}{\theta}$ and $p_\phi = \pdv{\mcS}{\phi}$ into the Hamiltonian:
+\[
+\frac{1}{2I}\left(\pdv{\mcS}{\theta}\right)^2 + \frac{1}{2I\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 = E.
+\]
+Multiply both sides by $2I$:
+\[
+\left(\pdv{\mcS}{\theta}\right)^2 + \frac{1}{\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 = 2IE.
+\]
+The coordinate $\phi$ is absent from the Hamiltonian, so $\phi$ is cyclic. The contribution of the cyclic coordinate to the characteristic function is linear:
+\[
+\pdv{\mcS}{\phi} = L_z,
+\]
+where $L_z$ is the conserved $z$-component of the angular momentum. Substituting this constant into the HJ equation gives
+\[
+\left(\pdv{\mcS}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = 2IE.
+\]
+Seeking an additive separation $\mcS = W_\theta(\theta) + L_z\phi$, the partial derivative $\pdv{\mcS}{\theta}$ becomes the ordinary derivative $\der{W_\theta}{\theta}$:
+\[
+\left(\der{W_\theta}{\theta}\right)^2 + \frac{L_z^2}{\sin^2\theta} = 2IE.
+\]
+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:
+\[
+\der{W_\theta}{\theta} = \pm\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}.
+\]
+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 the 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.}
+
+\nt{Rational and irrational orbit closure}{The rigid rotator has two frequencies, $\omega_\theta$ for the polar oscillation and $\omega_\phi$ for the azimuthal precession. If the ratio $\omega_\theta/\omega_\phi = p/q$ is rational (with $p$ and $q$ coprime integers), then after exactly $q$ complete cycles of $\theta$ oscillation the orbit returns to its starting phase and retraces itself. If the ratio is irrational, the orbit never closes and the trajectory wraps around the invariant torus in phase space, densely filling a two-dimensional ring on the sphere surface. The irrational case exemplifies quasi-periodic motion: the particle explores an angular band without ever repeating its exact configuration. For the rigid rotator the two frequencies turn out equal in magnitude, so every orbit is degenerate and closed with $p/q = \pm 1$. This degeneracy --- identical frequencies for dynamically independent degrees of freedom --- is a special property of the $1/r^2$ force structure and the spherical symmetry of the configuration space.}
+
+\nt{Evaluating the $\theta$ integral}{The integral $W_\theta(\theta) = \int\sqrt{L^2 - L_z^2/\sin^2\theta}\;\dd\theta$ can be rearranged by factoring the square root:
+\[
+\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}
+= \frac{\sqrt{L^2\sin^2\theta - L_z^2}}{\sin\theta}.
+\]
+A useful substitution is $u = \cos\theta$, for which $\dd\theta = -\dd u/\sqrt{1-u^2}$ and $\sin^2\theta = 1-u^2$. The integral then reads
+\[
+\int \frac{\sqrt{L^2(1-u^2) - L_z^2}}{\sqrt{1-u^2}}\cdot \left(-\frac{\dd u}{\sqrt{1-u^2}}\right)
+= -\int \frac{\sqrt{(L^2-L_z^2) - L^2 u^2}}{1-u^2}\;\dd u.
+\]
+This has the structure of an elliptic integral in general. The denominator $(1-u^2)$ together with the quadratic radicand $\sqrt{(L^2-L_z^2) - L^2 u^2}$ prevents a simple elementary antiderivative. However, for the purpose of computing the action variable --- a closed-loop integral between turning points --- the explicit antiderivative is not needed. The turning points in $u$-space occur at $u = \pm L_z/L$, where the radicand vanishes. A second substitution $u = (L_z/L)\cos\psi$ converts the definite integral to a standard trigonometric form that evaluates in closed fashion, yielding $J_\theta = |L| - |L_z|$ without computing an indefinite integral.}
+
+\nt{Action-angle variables for the rigid rotator}{The two independent action variables are computed by integrating the conjugate momenta over their respective cycles, each divided by $2\pi$. For the azimuthal coordinate $\phi$, which is $2\pi$-periodic:
+\[
+J_\phi = \frac{1}{2\pi}\oint p_\phi\,\dd\phi = \frac{1}{2\pi}\int_{0}^{2\pi} L_z\,\dd\phi = L_z.
+\]
+Physically, $J_\phi = L_z$ equals the vertical spin: the amount of angular momentum aligned with the symmetry axis. A larger $|L_z|$ means the motion stays closer to the equator circle. For the polar coordinate $\theta$, which oscillates between turning points, the full cycle traverses the range twice:
+\[
+J_\theta = \frac{1}{2\pi}\oint p_\theta\,\dd\theta
+= \frac{1}{\pi}\int_{\theta_{\min}}^{\theta_{\max}}\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}\;\dd\theta
+= |L| - |L_z|.
+\]
+Physically, $J_\theta = |L| - |L_z|$ measures the tilt from the equator: the ``missing'' angular momentum that keeps the trajectory from being perfectly planar. When $|L| = |L_z|$ the tilt vanishes, $J_\theta = 0$, and the motion is confined to the equatorial circle. When $L_z = 0$ there is no preferred axis and $J_\theta = |L|$, the full angular momentum goes into the polar oscillation. Inverting gives $L_z = J_\phi$ and $|L| = J_\theta + |J_\phi|$. The Hamiltonian in action variables is
+\[
+H(J_\theta,J_\phi) = \frac{\bigl(J_\theta + |J_\phi|\bigr)^2}{2I}.
+\]}
+
+\nt{From classical action to quantum numbers}{The Bohr--Sommerfeld quantization rule states that each action variable must be an integer multiple of Planck\normalsize{}'s reduced constant: $J_i = n_i\hbar$. Applying this to the rigid rotator, the azimuthal action quantizes as $J_\phi = m\hbar$ with $m$ an integer, giving directly $L_z = m\hbar$. The polar action quantizes as $J_\theta = \ell_\theta\hbar$ with $\ell_\theta$ a non-negative integer. Since $|L| = J_\theta + |J_\phi|$, we obtain $|L| = (\ell_\theta + |m|)\hbar$. Defining the total angular-momentum quantum number $\ell = \ell_\theta + |m|$, the condition $\ell_\theta \ge 0$ becomes $\ell \ge |m|$. Therefore $L^2 = |L|^2 = \ell^2\hbar^2$, and the semiclassical energy is $E = \ell^2\hbar^2/(2I)$. The fully quantum-mechanical result from solving the angular Schrödinger equation replaces $\ell^2$ with $\ell(\ell+1)$, giving $E = \ell(\ell+1)\hbar^2/(2I)$. The factor $\ell(\ell+1)$ rather than $\ell^2$ emerges from the non-commutativity of $L_x$, $L_y$, and $L_z$ --- no quantum state can simultaneously have definite values of all three components, so the total angular momentum magnitude always exceeds $|L_z|$ by a half-integer step. The Bohr--Sommerfeld approach captures the ladder of energy levels, the integer structure of quantum numbers, and the constraint $\ell \ge |m|$ that limits how far the spin axis can tilt. However, it cannot produce the $+\ell$ correction inside $\ell(\ell+1)$. For large $\ell$ the semiclassical and quantum results agree well, since $\ell(\ell+1) \approx \ell^2$ when $\ell \gg 1$. This bridge from classical action variables to quantum angular momentum was the key step in old quantum theory and its subsequent replacement by matrix and wave mechanics.}
+
+\cor{Equatorial orbit}{
+When the total angular momentum equals the absolute value of its $z$-component, $L = |L_z|$, the square root in the $\theta$-equation vanishes everywhere except at $\sin\theta = 1$. The polar momentum $p_\theta = \der{W_\theta}{\theta}$ is zero, so the motion is confined to the equator $\theta = \pi/2$. From Hamilton\normalsize{}'s equations, the azimuthal velocity is
+\[
+\dot{\phi} = \pdv{H}{p_\phi} = \frac{p_\phi}{I\sin^2\theta} = \frac{L_z}{I}
+\]
+since $\sin(\pi/2) = 1$. The azimuthal angle advances at constant rate:
+\[
+\phi(t) = \frac{L_z}{I}\,t + \phi_0,
+\]
+representing uniform circular motion at angular speed $\omega = |L_z|/I$. This orbit corresponds to $J_\theta = 0$ and is the only truly one-degree-of-freedom subcase of the rigid rotator.}
+
+\ex{Action-angle frequencies for the rigid rotator}{
+Using the Hamiltonian in action variables,
+\[
+H = \frac{\bigl(J_\theta + |J_\phi|\bigr)^2}{2I},
+\]
+the two frequencies are
+\[
+\omega_\theta = \pdv{H}{J_\theta}
+= \frac{J_\theta + |J_\phi|}{I}
+= \frac{|L|}{I},
+\qquad
+\omega_\phi = \pdv{H}{J_\phi}
+= \pm\frac{J_\theta + |J_\phi|}{I}
+= \pm\frac{|L|}{I}.
+\]
+Here the sign conventionally matches the sign of $J_\phi$, so a negative $J_\phi$ gives retrograde azimuthal motion at the same frequency magnitude. The two frequencies are equal in magnitude, confirming the degeneracy noted above: every trajectory on the sphere is a closed orbit with rational frequency ratio $1:1$. For the special case $J_\theta = 0$ (equatorial orbit), the polar frequency is defined by continuity and the motion is purely azimuthal.}
+
+\qs{Diatomic molecule as a rigid rotator}{A diatomic molecule is modeled as a rigid rotator with moment of inertia
+\[
+I = 1.46\times 10^{-46}\,\mathrm{kg\!\cdot\!m^2}.
+\]
+The total angular momentum is $L = 2\hbar$, where $\hbar = 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s}$.
+
+\begin{enumerate}[label=(\alph*)]
+\item Write the Hamilton--Jacobi equation for the rigid rotator in spherical coordinates with fixed $r$. Identify the cyclic coordinate and state the corresponding conserved quantity.
+\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 $E = \hbar^2\ell(\ell+1)/(2I)$ with $\ell = 2$.
+\item Find the absolute difference between the classical and quantum energy values and express it as a percentage of the quantum value.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The Hamiltonian of the rigid rotator is $H = p_\theta^2/(2I) + p_\phi^2/(2I\sin^2\theta)$. 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:
+\[
+\frac{1}{2I}\left(\pdv{\mcS}{\theta}\right)^2 + \frac{1}{2I\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 + \pdv{\mcS}{t} = 0.
+\]
+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:
+\[
+\pdv{\mcS}{\phi} = L_z = \text{const},
+\]
+which is the $z$-component of the angular momentum.
+
+\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 $\hbar^2$:
+\[
+\hbar^2 = \left(1.055\times 10^{-34}\right)^2\,\mathrm{J^2\!\cdot\!s^2}
+= 1.113\times 10^{-68}\,\mathrm{J^2\!\cdot\!s^2}.
+\]
+The classical energy is
+\[
+E_{\mathrm{class}} = \frac{4\hbar^2}{2I}
+= \frac{2\hbar^2}{I}
+= \frac{2(1.113\times 10^{-68})}{1.46\times 10^{-46}}\,\mathrm{J}
+= 1.525\times 10^{-22}\,\mathrm{J}.
+\]
+The quantum energy with $\ell = 2$ is
+\[
+E_{\mathrm{quant}} = \frac{\hbar^2\,\ell(\ell+1)}{2I}
+= \frac{(1.113\times 10^{-68})(6)}{2(1.46\times 10^{-46})}\,\mathrm{J}
+= \frac{6.678\times 10^{-68}}{2.92\times 10^{-46}}\,\mathrm{J}
+= 2.287\times 10^{-22}\,\mathrm{J}.
+\]
+The quantum energy exceeds the classical value. The ratio is
+\[
+\frac{E_{\mathrm{quant}}}{E_{\mathrm{class}}}
+= \frac{6}{4}
+= 1.50.
+\]
+
+\textbf{Part (c).} The absolute difference between the two energies is
+\[
+\Delta E = E_{\mathrm{quant}} - E_{\mathrm{class}}
+= 2.287\times 10^{-22} - 1.525\times 10^{-22}\,\mathrm{J}
+= 7.62\times 10^{-23}\,\mathrm{J}.
+\]
+Expressed as a percentage of the quantum value:
+\[
+\frac{\Delta E}{E_{\mathrm{quant}}}\times 100\%
+= \frac{7.62\times 10^{-23}}{2.287\times 10^{-22}}\times 100\%
+= 33.3\%.
+\]
+Analytically, since $E_{\mathrm{class}} = \ell^2\hbar^2/(2I)$ and $E_{\mathrm{quant}} = \ell(\ell+1)\hbar^2/(2I)$, the fractional difference is
+\[
+\frac{\Delta E}{E_{\mathrm{quant}}}
+= \frac{\ell(\ell+1) - \ell^2}{\ell(\ell+1)}
+= \frac{\ell}{\ell(\ell+1)}
+= \frac{1}{\ell+1}.
+\]
+For $\ell = 2$ this gives $1/3 = 33.3\%$, matching the numerical calculation. The discrepancy arises from the quantum $+\ell$ correction in $\ell(\ell+1)$ relative to the classical $\ell^2$.
+
+Therefore,
+\[
+\frac{1}{2I}\left(\pdv{\mcS}{\theta}\right)^2 + \frac{1}{2I\sin^2\theta}\left(\pdv{\mcS}{\phi}\right)^2 + \pdv{\mcS}{t} = 0,
+\qquad
+p_\phi = L_z = \text{const};
+\]
+\[
+E_{\mathrm{class}} = 1.53\times 10^{-22}\,\mathrm{J},
+\qquad
+E_{\mathrm{quant}} = 2.29\times 10^{-22}\,\mathrm{J},
+\qquad
+\frac{\Delta E}{E_{\mathrm{quant}}} = 33.3\%.
+\]

concepts/advanced/separation.tex

@@ -0,0 +1,187 @@
+\subsection{Separation of Variables in the Hamilton-Jacobi Equation}
+
+This subsection develops the method of separation of variables for the Hamilton--Jacobi equation, showing how the choice of coordinate system determines whether the PDE reduces to a set of ordinary quadratures.
+
+\nt{Recap: The Hamilton--Jacobi equation from A.01 is a single first-order nonlinear PDE. Here we learn how to solve it by separation of variables.}
+
+\dfn{Separation of variables for the HJ equation}{
+Suppose the Hamiltonian has no explicit time dependence and the Hamilton--Jacobi equation is $\mcH(q_1,\dots,q_n,\pdv{\mcS}{q_1},\dots,\pdv{\mcS}{q_n}) = E$. The time variable is separated by setting
+\[
+\mcS(q_1,\dots,q_n,t) = W(q_1,\dots,q_n) - Et,
+\]
+reducing the equation to $\mcH(q_1,\dots,q_n,\pdv{W}{q_1},\dots,\pdv{W}{q_n}) = E$. The HJ equation is said to be separable if the characteristic function $W$ can be written as an additive sum of single-variable functions, $W = W_1(q_1) + \cdots + W_n(q_n)$, where each $W_i$ depends on only one coordinate and a set of separation constants. If such a form exists, the solution reduces to evaluating $n$ independent quadratures.}
+
+The first step is always the time-independent reduction. When $\pdv{\mcH}{t} = 0$, the Hamiltonian is conserved: $\mcH = E$. Substituting $\mcS = W(q_1,\dots,q_n) - Et$ into the full Hamilton--Jacobi equation, the time derivative contributes $-E$ and the equation becomes
+\[
+\mcH\!\left(q_1,\dots,q_n,\pdv{W}{q_1},\dots,\pdv{W}{q_n}\right) = E.
+\]
+This is the time-independent Hamilton--Jacobi equation. It contains $n$ partial derivatives of $W$ and determines the spatial part of the action. The total principal function is $\mcS = W - Et$ once $W$ is found.
+
+\nt{The condition $\pdv{\mcH}{t} = 0$ is necessary for the simple time separation $\mcS = W - Et$. When the Hamiltonian depends explicitly on time, a different separation ansatz or a time-dependent canonical transformation is required. In the time-independent case, energy is a constant of motion and serves as the first separation constant.}
+
+\nt{The separation constant $E$ is the total energy of the system, arising because $t$ is a cyclic coordinate in the extended phase space whenever the Hamiltonian is time-independent.}
+
+A particularly simple situation arises when one or more coordinates are cyclic. A generalized coordinate $q_i$ is cyclic, or ignorable, when it is absent from the Hamiltonian, which means $\pdv{\mcH}{q_i} = 0$. For such a coordinate, Hamilton's equation gives $\dot{p}_i = 0$, so the conjugate momentum is conserved. Within the Hamilton--Jacobi framework this translates directly: since $p_i = \pdv{\mcS}{q_i}$ and $\pdv{\mcH}{q_i} = 0$, the derivative
+\[
+\pdv{\mcS}{q_i} = \alpha_i
+\]
+is a constant. The contribution of the cyclic coordinate to the characteristic function is simply $W_i(q_i) = \alpha_i\,q_i$, which is immediately integrated.
+
+When more than one coordinate is cyclic the separations are independent and each contributes a linear term to $W$. The remaining non-cyclic coordinates carry the entire nontrivial structure of the problem and must be separated by additional ansatze.
+
+\thm{Additive separation theorem}{
+Let the Hamiltonian take the form $\mcH = T + V$ where the kinetic energy $T$ is a quadratic form in the momenta and the potential energy $V$ is a sum of single-coordinate terms, $V = V_1(q_1) + \cdots + V_n(q_n)$. If the metric coefficients of the kinetic energy depend on only one coordinate each, then the time-independent Hamilton--Jacobi equation
+\[
+\mcH\!\left(q_1,\dots,q_n,\pdv{W}{q_1},\dots,\pdv{W}{q_n}\right) = E
+\]
+admits an additive separation ansatz
+\[
+W(q_1,\dots,q_n) = W_1(q_1) + W_2(q_2) + \cdots + W_n(q_n).
+\]
+Each function $W_i(q_i)$ satisfies an ordinary differential equation involving one separation constant, and the complete integral is obtained by evaluating $n$ quadratures.}
+
+Several standard orthogonal coordinate systems admit separable Hamilton--Jacobi equations for important classes of potentials. The gradient-squared operator takes different forms in each system, and the metric coefficients determine whether a given potential allows additive separation.
+
+\ex{Gradient in spherical coordinates}{
+In orthogonal curvilinear coordinates $(q_1,q_2,q_3)$, the line element is $\dd s^2 = h_1^2\,\dd q_1^2 + h_2^2\,\dd q_2^2 + h_3^2\,\dd q_3^2$, where $h_i$ are the scale factors. The magnitude-squared of a gradient follows from the metric as
+\[
+|\nabla W|^2 = \frac{1}{h_1^2}\left(\pdv{W}{q_1}\right)^2 + \frac{1}{h_2^2}\left(\pdv{W}{q_2}\right)^2 + \frac{1}{h_3^2}\left(\pdv{W}{q_3}\right)^2.
+\]
+For spherical coordinates $(r,\theta,\phi)$, the line element is $\dd s^2 = \dd r^2 + r^2\,\dd\theta^2 + r^2\sin^2\theta\,\dd\phi^2$, so the scale factors are $h_r = 1$, $h_\theta = r$, $h_\phi = r\sin\theta$. Substituting:
+\[
+|\nabla W|^2 = \left(\pdv{W}{r}\right)^2 + \frac{1}{r^2}\left(\pdv{W}{\theta}\right)^2 + \frac{1}{r^2\sin^2\theta}\left(\pdv{W}{\phi}\right)^2.
+\]
+This explicit form enters the time-independent Hamilton--Jacobi equation as $\,|\nabla W|^2/(2m) + V = E$ and the geometric factors $1/r^2$ and $1/(r^2\sin^2\theta)$ control which potentials permit additive separation.}
+
+\mprop{Coordinate systems and separability for the HJ equation}{
+The table below summarizes the gradient operator squared $|\nabla W|^2$ in commonly used orthogonal coordinate systems, and the classes of potentials that permit additive separation of the Hamilton--Jacobi equation with $\mcH = |\nabla W|^2/(2m) + V = E$:
+
+\begin{center}
+\begin{tabular}{p{3.5cm}p{5.2cm}p{4.8cm}}
+\hline
+\textbf{Coordinates} & \textbf{Gradient squared $|\nabla W|^2$} & \textbf{Separable potentials} \\
+\hline
+Cartesian $(x,y,z)$ & $(\pdv{W}{x})^2 + (\pdv{W}{y})^2 + (\pdv{W}{z})^2$ & $V = V_x(x) + V_y(y) + V_z(z)$ \\
+Spherical $(r,\theta,\phi)$ & $(\pdv{W}{r})^2 + \frac{1}{r^2}(\pdv{W}{\theta})^2 + \frac{1}{r^2\sin^2\theta}(\pdv{W}{\phi})^2$ & $V = V(r)$ (central); $V(r,\theta)$ with $1/r^2$ separability \\
+Cylindrical $(\rho,\phi,z)$ & $(\pdv{W}{\rho})^2 + \frac{1}{\rho^2}(\pdv{W}{\phi})^2 + (\pdv{W}{z})^2$ & $V = V_\rho(\rho) + V_z(z)$; $V(\rho)$ \\
+Parabolic $(\xi,\eta,\phi)$ & \multicolumn{2}{p{10cm}}{$|\nabla W|^2 = \frac{1}{\xi^2+\eta^2}\bigl[(\xi^2+\eta^2)(\pdv{W}{\xi})^2 + (\xi^2+\eta^2)(\pdv{W}{\eta})^2 + \frac{\xi^2\eta^2}{\xi\eta}(\pdv{W}{\phi})^2\bigr]$, separable for Kepler $V=-k/r$ and Stark potentials} \\
+\hline
+\end{tabular}
+\end{center}
+
+Parabolic coordinates are defined by $\xi = \sqrt{r(r-z)}$ and $\eta = \sqrt{r+r_{\!z}}$, with $z = (\eta^2-\xi^2)/2$ and $r = (\xi^2+\eta^2)/2$. The Kepler potential $V = -k/r = -2k/(\xi^2+\eta^2)$ separates in parabolic coordinates because $1/r$ can be split into a function of $\xi$ plus a function of $\eta$ after substituting into the HJ equation and multiplying by the metric factor. Parabolic coordinates are especially useful for the hydrogen atom in quantum mechanics and for analyzing the Stark effect, where a uniform electric field along the $z$-axis is added to the Coulomb potential while preserving separability.
+
+The additive separation ansatz $W(q_1,\dots,q_n) = W_1(q_1) + \cdots + W_n(q_n)$ is the standard starting point. Substituting this form into the Hamilton--Jacobi equation and multiplying by appropriate metric factors produces a sum, with each term depending on a single coordinate. The equation
+\[
+f_1(q_1, W_1') + f_2(q_2, W_2') + \cdots + f_n(q_n, W_n') = E
+\]
+can only hold for all values of the independent coordinates if each term is itself a constant. These constants are the separation constants $\alpha_1,\dots,\alpha_n$, constrained by one relation that fixes the total energy. The remaining equations are first-order ODEs for the individual functions $W_i(q_i)$, each solvable by quadrature.
+
+\ex{Free particle in spherical coordinates}{
+Consider a free particle of mass $m$ in spherical coordinates $(r,\theta,\phi)$. The Hamiltonian is $\mcH = p^2/(2m)$ and the time-independent HJ equation is $|\nabla W|^2/(2m) = E$, or equivalently
+\[
+\left(\der{W_r}{r}\right)^2 + \frac{1}{r^2}\left(\der{W_\theta}{\theta}\right)^2 + \frac{1}{r^2\sin^2\phi}\left(\der{W_\phi}{\phi}\right)^2 = 2mE.
+\]
+Using the ansatz $W = W_r(r) + W_\theta(\theta) + W_\phi(\phi)$, the coordinate $\phi$ is cyclic and $dW_\phi/d\phi = \alpha_\phi$. Multiply the equation by $r^2$:
+\[
+r^2\left(\der{W_r}{r}\right)^2 + \left(\der{W_\theta}{\theta}\right)^2 + \frac{\alpha_\phi^2}{\sin^2\theta} = 2mEr^2.
+\]
+The last two terms depend only on $\theta$ while the first and rightmost terms depend only on $r$. Equating both sides to a constant $\alpha^2$ gives
+\[
+\left(\der{W_\theta}{\theta}\right)^2 + \frac{\alpha_\phi^2}{\sin^2\theta} = \alpha^2,
+\qquad
+\der{W_r}{r} = \sqrt{2mE - \frac{\alpha^2}{r^2}}.
+\]
+Each equation integrates by quadrature, and $W_\phi(\phi) = \alpha_\phi\,\phi$. These three quadratures constitute the complete integral for the free particle in spherical coordinates.}
+
+\nt{Physical meaning of $\alpha_\phi$: In spherical coordinates the azimuthal angle $\phi$ is always cyclic for rotationally symmetric Hamiltonians. The separation constant $\alpha_\phi$ is the $z$-component of angular momentum, denoted $L_z$. It is a constant of motion because the system is invariant under rotations about the $z$-axis.}
+
+\nt{Physical meaning of $\alpha^2$: The constant $\alpha^2$ that appears when separating the angular variables $\theta$ and $\phi$ identifies with $L^2$, the squared total angular momentum. It measures the magnitude of rotational motion and produces the centrifugal barrier $L^2/(2mr^2)$ in the radial equation.}
+
+\qs{Separation for a particle in a uniform gravitational field}{
+A particle of mass $m$ moves in the $xy$-plane under a uniform gravitational field $g$ acting in the negative $y$-direction. The Hamiltonian is
+\[
+\mcH = \frac{p_x^2 + p_y^2}{2m} + mgy.
+\]
+
+\begin{enumerate}[label=(\alph*)]
+\item Write the time-independent Hamilton--Jacobi equation and apply the separation ansatz $\mcS = W_x(x) + W_y(y) - Et$. Show that $x$ is a cyclic coordinate and that $\der{W_x}{x} = \alpha_x$, a constant.
+\item Find $\der{W_y}{y}$ in terms of the separation constants. Define the transverse energy $E_y = E - \alpha_x^2/(2m)$ and write the quadrature integral for $W_y(y)$.
+\item A projectile of mass $m = 0.100\,\mathrm{kg}$ is launched from $y = 0$ with speed $v_0 = 25.0\,\mathrm{m/s}$ at angle $\theta_0 = 45.0^\circ$ above the horizontal. Take $g = 9.81\,\mathrm{m/s^2}$. Compute the $x$-momentum $p_x = m v_0\cos\theta_0$ and the transverse energy $E_y = \tfrac{1}{2}m v_0^2\sin^2\theta_0$ in SI units.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The time-independent Hamilton--Jacobi equation is obtained by setting $\mcH = E$. With the given Hamiltonian this reads
+\[
+\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \left(\pdv{\mcS}{y}\right)^2\right] + mgy = E.
+\]
+Substitute the separation ansatz $\mcS(x,y,t) = W_x(x) + W_y(y) - Et$. The spatial partial derivatives become ordinary derivatives: $\pdv{\mcS}{x} = \der{W_x}{x}$ and $\pdv{\mcS}{y} = \der{W_y}{y}$. Substituting gives
+\[
+\frac{1}{2m}\left(\der{W_x}{x}\right)^2 + \frac{1}{2m}\left(\der{W_y}{y}\right)^2 + mgy = E.
+\]
+Multiply both sides by $2m$:
+\[
+\left(\der{W_x}{x}\right)^2 + \left(\der{W_y}{y}\right)^2 + 2m^2gy = 2mE.
+\]
+The coordinate $x$ does not appear in the Hamiltonian, so $x$ is cyclic. The first term depends only on $x$ while the remaining terms depend only on $y$. Separating gives
+\[
+\left(\der{W_x}{x}\right)^2 = \alpha_x^2,
+\]
+where $\alpha_x$ is a separation constant. Therefore,
+\[
+\der{W_x}{x} = \alpha_x,
+\]
+which integrates to $W_x(x) = \alpha_x\,x$. The constant $\alpha_x$ is identified with the constant $x$-component of the canonical momentum.
+
+\textbf{Part (b).} Substitute $(dW_x/dx)^2 = \alpha_x^2$ back into the HJ equation:
+\[
+\left(\der{W_y}{y}\right)^2 + 2m^2gy = 2mE - \alpha_x^2.
+\]
+Define the transverse energy $E_y = E - \alpha_x^2/(2m)$. Then $2mE - \alpha_x^2 = 2mE_y$, and the $y$-equation becomes
+\[
+\left(\der{W_y}{y}\right)^2 + 2m^2gy = 2mE_y.
+\]
+Solve for the derivative:
+\[
+\der{W_y}{y} = \pm\sqrt{2mE_y - 2m^2gy}.
+\]
+Factor $2m$ from the radicand:
+\[
+\der{W_y}{y} = \pm\sqrt{2m\left(E_y - mgy\right)}.
+\]
+The quadrature for $W_y(y)$ is
+\[
+W_y(y) = \pm\int\sqrt{2m\left(E_y - mgy\right)}\,dy.
+\]
+
+\textbf{Part (c).} The initial conditions specify mass $m = 0.100\,\mathrm{kg}$, launch speed $v_0 = 25.0\,\mathrm{m/s}$, and launch angle $\theta_0 = 45.0^\circ$. Compute the $x$-momentum:
+\[
+p_x = m v_0\cos\theta_0.
+\]
+Substitute the numerical values:
+\[
+p_x = (0.100)(25.0)\cos(45.0^\circ)\,\mathrm{kg\!\cdot\!m/s}.
+\]
+Since $\cos(45.0^\circ) = \sqrt{2}/2 \approx 0.7071$,
+\[
+p_x = (0.100)(25.0)(0.7071)\,\mathrm{kg\!\cdot\!m/s} = 1.77\,\mathrm{kg\!\cdot\!m/s}.
+\]
+The transverse energy is $E_y = \tfrac{1}{2}m v_0^2\sin^2\theta_0$. The vertical speed is
+\[
+v_{0y} = v_0\sin(45.0^\circ) = (25.0)(0.7071)\,\mathrm{m/s} = 17.68\,\mathrm{m/s}.
+\]
+Evaluate $E_y$:
+\[
+E_y = \tfrac{1}{2}(0.100)(25.0)^2(0.7071)^2\,\mathrm{J}.
+\]
+This gives
+\[
+E_y = \tfrac{1}{2}(0.100)(625)(0.500)\,\mathrm{J} = 15.6\,\mathrm{J}.
+\]
+
+Therefore,
+\[
+p_x = 1.77\,\mathrm{kg\!\cdot\!m/s},
+\qquad
+E_y = 15.6\,\mathrm{J}.
+\]
+}

concepts/advanced/sho-hj.tex

@@ -0,0 +1,287 @@
+\subsection{Simple Harmonic Oscillator}
+
+This subsection solves the simple harmonic oscillator through the Hamilton--Jacobi equation. The quadratic potential turns the Hamilton--Jacobi square-root integral into an elementary trigonometric substitution, making the oscillator one of the few nonlinear Hamilton--Jacobi equations integrable in closed form. This same physical system appears in Unit~7 (oscillations) under Newton\normalsize{}'s law; here we solve it by an entirely different route to obtain the complete integral, trajectory by quadrature, and action--angle variables confirming isochronous oscillation. This integrability makes the harmonic oscillator the prototypical example for testing both the Hamilton--Jacobi method and the action--angle formalism.
+
+\dfn{Hamilton--Jacobi formulation of the simple harmonic oscillator}{
+The Hamiltonian for a one-dimensional simple harmonic oscillator of mass~$m$ and spring constant~$k$ is
+\[
+\mcH = \frac{p^2}{2m} + \frac{1}{2}k x^2
+\]
+with natural angular frequency $\omega_0 = \sqrt{k/m}$. In terms of~$\omega_0$,
+\[
+\mcH = \frac{p^2}{2m} + \frac{1}{2}m\omega_0^2 x^2.
+\]
+The Hamilton--Jacobi partial differential equation for the principal function~$\mcS(x,t)$ follows by the substitution $p = \pdv{\mcS}{x}$:
+\[
+\frac{1}{2m}\left(\pdv{\mcS}{x}\right)^2 + \frac{1}{2}m\omega_0^2 x^2 + \pdv{\mcS}{t} = 0.
+\]
+A complete integral $\mcS(x,t;E)$, containing one independent non-additive constant $E$ equal to the total energy, determines the full dynamics by Jacobi\normalsize{}'s theorem.}
+
+\thm{Complete integral of the SHO Hamilton--Jacobi equation}{
+The complete integral of the Hamilton--Jacobi equation for a simple harmonic oscillator is
+\[
+\mcS(x,t;E) = \frac{E}{\omega_0}\arcsin\!\left(x\sqrt{\frac{m\omega_0^2}{2E}}\right) 
++ \frac{1}{2}x\sqrt{2mE - m^2\omega_0^2 x^2} - Et,
+\]
+where $E>0$ is the total energy. It is defined for $|x| < A$ with $A = \sqrt{2E/(m\omega_0^2)}$.}
+
+\pf{Derivation of the complete integral by separation and trigonometric substitution}{
+Because $\pdv{\mcH}{t} = 0$, separate the time variable by setting $\mcS(x,t) = W(x) - Et$. The temporal derivative contributes $-E$ and the Hamilton--Jacobi equation reduces to
+\[
+\frac{1}{2m}\left(\der{W}{x}\right)^2 + \frac{1}{2}m\omega_0^2 x^2 = E.
+\]
+Solve for the spatial derivative:
+\[
+\der{W}{x} = \pm\sqrt{2mE - m^2\omega_0^2 x^2}.
+\]
+The square root is real for $|x| \le A$, where $A = \sqrt{2E/(m\omega_0^2)}$ is the amplitude. The turning points $x = \pm A$ bound the oscillation and correspond to the points where the kinetic energy vanishes.
+
+The sign carries physical meaning: the upper sign describes forward motion ($p>0$, the mass traveling toward $+A$) and the lower sign describes backward motion ($p<0$, the mass returning toward $-A$). In the closed contour integral $\oint p\,\dd x$ that defines the action variable, the $+$ branch contributes the outward half of the orbit $(-A\to A)$ and the $-$ branch contributes the return half $(A\to -A)$. For finding $W(x;E)$ as a local generating function we choose the positive branch and carry it through the integration; full periodicity is then imposed by the boundary conditions.
+
+Integrate by the trigonometric substitution $x = A\sin\theta$, giving $\dd x = A\cos\theta\,\mathrm{d}\theta$. The radicand becomes
+\[
+2mE - m^2\omega_0^2 A^2\sin^2\theta
+= 2mE - 2mE\sin^2\theta
+= 2mE\cos^2\theta,
+\]
+since $m^2\omega_0^2 A^2 = m^2\omega_0^2\cdot\bigl(2E/(m\omega_0^2)\bigr) = 2mE$. Hence,
+\[
+\sqrt{2mE - m^2\omega_0^2 x^2} = \sqrt{2mE}\cos\theta,
+\]
+where we take $\cos\theta \ge 0$ for $\theta\in[-\pi/2,\pi/2]$. The integral for~$W$ is
+\[
+W = \int\sqrt{2mE}\cos\theta\cdot A\cos\theta\,\mathrm{d}\theta
+= \sqrt{2mE}\,A\int\cos^2\theta\,\mathrm{d}\theta.
+\]
+The antiderivative of $\cos^2\theta$ is $\tfrac12(\theta + \sin\theta\cos\theta)$, so
+\[
+W = \tfrac12\sqrt{2mE}\,A\bigl(\theta + \sin\theta\cos\theta\bigr).
+\]
+Evaluate the constant prefactor:
+\[
+\tfrac12\sqrt{2mE}\,A
+= \tfrac12\sqrt{2mE}\cdot\sqrt{\frac{2E}{m\omega_0^2}}
+= \tfrac12\cdot\frac{2E}{\omega_0}
+= \frac{E}{\omega_0}.
+\]
+Now express the trigonometric quantities in terms of $x$:
+\[
+\theta = \arcsin\!\left(\frac{x}{A}\right) 
+= \arcsin\!\left(x\sqrt{\frac{m\omega_0^2}{2E}}\right),
+\]
+\[
+\sin\theta = \frac{x}{A} = x\sqrt{\frac{m\omega_0^2}{2E}},
+\qquad
+\cos\theta = \frac{\sqrt{2mE - m^2\omega_0^2 x^2}}{\sqrt{2mE}}.
+\]
+Substitute these expressions back into $W$:
+\[
+W = \frac{E}{\omega_0}\arcsin\!\left(x\sqrt{\frac{m\omega_0^2}{2E}}\right)
++ \frac{E}{\omega_0}\cdot x\sqrt{\frac{m\omega_0^2}{2E}}\cdot\frac{\sqrt{2mE - m^2\omega_0^2 x^2}}{\sqrt{2mE}}.
+\]
+The product of the factors in the second term simplifies as
+\[
+\frac{E}{\omega_0}\cdot\frac{\omega_0\sqrt{m}}{\sqrt{2E}}\cdot\frac{1}{\sqrt{2mE}}
+= \frac{E}{\omega_0}\cdot\frac{\omega_0\sqrt{m}}{2E\sqrt{m}}
+= \frac{1}{2},
+\]
+since $\sqrt{2E}\cdot\sqrt{2mE} = \sqrt{4mE^2} = 2E\sqrt{m}$. Thus
+\[
+W(x;E) = \frac{E}{\omega_0}\arcsin\!\left(x\sqrt{\frac{m\omega_0^2}{2E}}\right) 
++ \frac{1}{2}x\sqrt{2mE - m^2\omega_0^2 x^2},
+\]
+and the complete integral is $\mcS(x,t;E) = W(x;E) - Et$.}
+
+\cor{Trajectory from Jacobi\normalsize{}'s theorem}{
+Jacobi\normalsize{}'s theorem states $\pdv{\mcS}{E} = \beta$, where $\beta$ is a constant fixed by the initial conditions. Differentiate $\mcS = W - Et$ with respect to $E$ at fixed~$x$:
+\[
+\pdv{\mcS}{E} = \pdv{W}{E} - t.
+\]
+Write $W$ using the shorthand $\chi = x\sqrt{m\omega_0^2/(2E)}$ and $R = \sqrt{2mE - m^2\omega_0^2 x^2}$:
+\[
+W = \frac{E}{\omega_0}\arcsin\chi + \frac{1}{2}xR.
+\]
+The partial derivative with respect to~$E$ is
+\[
+\pdv{W}{E} = \frac{1}{\omega_0}\arcsin\chi 
++ \frac{E}{\omega_0}\frac{1}{\sqrt{1-\chi^2}}\pdv{\chi}{E}
++ \frac{x}{2}\cdot\frac{m}{R}.
+\]
+Because $\chi \propto E^{-1/2}$, one has $\pdv{\chi}{E} = -\chi/(2E)$. The second term simplifies to
+\[
+-\frac{E}{\omega_0}\frac{\chi}{2E\sqrt{1-\chi^2}}
+= -\frac{\chi}{2\omega_0\sqrt{1-\chi^2}}
+= -\frac{x\sqrt{m}}{2\sqrt{2E}\sqrt{1-\chi^2}}.
+\]
+To see the last equality, substitute $\chi = x\omega_0\sqrt{m/(2E)}$:
+\[
+\frac{\chi}{\omega_0} = x\sqrt{\frac{m}{2E}}.
+\]
+The third term equals
+\[
+\frac{xm}{2R} = \frac{xm}{2\sqrt{2mE}\sqrt{1-\chi^2}}
+= \frac{x\sqrt{m}}{2\sqrt{2E}\sqrt{1-\chi^2}},
+\]
+which exactly cancels the second term. This cancellation carries deep physical significance: the energy dependence of the amplitude $A = \sqrt{2E/(m\omega_0^2)}$ and the energy dependence of the integrand $\sqrt{2mE - m^2\omega_0^2 x^2}$ nearly cancel when differentiated with respect to $E$, leaving only the geometric phase $\tfrac{1}{\omega_0}\arcsin(\cdots)$. Because $\pdv{W}{E}$ is independent of the amplitude, the period $T = 2\pi/\omega_0$ is the same for every orbit regardless of how much energy it carries. This is the isochrony of the simple harmonic oscillator---all amplitudes share one period. The algebraic cancellation in $\pdv{W}{E}$ is the Hamilton--Jacobi embodiment of that physical fact. Therefore,
+\[
+\pdv{W}{E} = \frac{1}{\omega_0}\arcsin\!\left(x\sqrt{\frac{m\omega_0^2}{2E}}\right).
+\]
+The condition $\pdv{\mcS}{E} = \beta$ yields
+\[
+\frac{1}{\omega_0}\arcsin\!\left(x\sqrt{\frac{m\omega_0^2}{2E}}\right) - t = \beta,
+\]
+or equivalently,
+\[
+x\sqrt{\frac{m\omega_0^2}{2E}} = \sin\!\bigl(\omega_0(t+\beta)\bigr).
+\]
+Define the amplitude $A = \sqrt{2E/(m\omega_0^2)}$ and the phase $\phi = \omega_0\beta$. The trajectory is
+\[
+x(t) = A\sin(\omega_0 t + \phi).
+\]
+The total energy is $E = \tfrac12 m\omega_0^2 A^2 = \tfrac12 k A^2$, and the initial phase $\phi$ is determined by the initial position and velocity through $\sin\phi = x_0/A$ and $\cos\phi = v_0/(\omega_0 A)$. When the oscillator is released from rest at maximum displacement, $\cos\phi = 0$ and $\phi = \pi/2$, giving $x(t) = A\cos(\omega_0 t)$.}
+
+\ex{Phase-space ellipse}{
+The action variable $J = \frac{1}{2\pi}\oint p\,\dd x$ is $\frac{1}{2\pi}$ times the area enclosed by the orbit in the $(x,p)$ phase plane. For the harmonic oscillator the orbit is an ellipse. The orbit equation follows from energy conservation,
+\[
+\frac{p^2}{2m} + \frac{1}{2}m\omega_0^2 x^2 = E,
+\]
+which can be rearranged to standard form:
+\[
+\frac{x^2}{A^2} + \frac{p^2}{p_{\max}^2} = 1,
+\]
+where $A = \sqrt{2E/(m\omega_0^2)}$ is the maximum displacement (semi-axis along the position direction) and $p_{\max} = \sqrt{2mE} = m\omega_0 A$ is the maximum momentum (semi-axis along the momentum direction). The phase-space orbit is an ellipse centered at the origin with these semi-axes. The enclosed area is $\pi A\,p_{\max}$. The action variable is $\frac{1}{2\pi}$ times this area:
+\[
+J = \frac{1}{2\pi}\cdot\pi A\,p_{\max}
+= \frac{1}{2}\sqrt{\frac{2E}{m\omega_0^2}}\cdot\sqrt{2mE}
+= \frac{1}{2}\sqrt{\frac{4E^2}{\omega_0^2}}
+= \frac{E}{\omega_0}.
+\]
+The linear relation $J \propto E$ reflects the fact that doubling the energy rescales the ellipse uniformly, while the ratio $p_{\max}/A = m\omega_0$ sets the ellipse\normalsize{}'s aspect ratio.}
+
+\mprop{Action-angle variables for the harmonic oscillator}{
+Applying the action-angle formalism to the simple harmonic oscillator gives the following results:
+
+\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
+\item The action variable is $\frac{1}{2\pi}$ times the phase-space area enclosed by one complete cycle:
+\[
+J = \frac{1}{2\pi}\oint p\,\dd x = \frac{1}{\pi}\int_{-A}^{A}\sqrt{2mE - m^2\omega_0^2 x^2}\,\dd x.
+\]
+With the substitution $x = A\sin\phi$, the integral $\int_{-A}^{A}\sqrt{2mE - m^2\omega_0^2 x^2}\,\dd x$ evaluates to $\tfrac{\pi}{2}\sqrt{2mE}\cdot A$. With the prefactor $1/\pi$ this gives $J = \tfrac{1}{2}\sqrt{2mE}\cdot A$. Using $A = \sqrt{2E/(m\omega_0^2)}$ we obtain $J = E/\omega_0$. Geometrically, $J$ is $\frac{1}{2\pi}$ times the area of the elliptical orbit in the $(x,p)$ phase plane, confirming the computation in the example above.
+
+\item Inverting the action relation, the Hamiltonian as a function of the action alone is
+\[
+E(J) = \omega_0 J.
+\]
+The Hamiltonian is linear in $J$, which is the defining feature of an action-angle representation.
+
+\item The frequency conjugate to the action is $\pdv{E}{J} = \omega_0$, which is the physical angular frequency itself, independent of $J$. Every oscillation shares the period $T = 2\pi/\omega_0$, regardless of energy or amplitude. This amplitude independence is the isochrony property of the harmonic oscillator, whose algebraic origin we saw in the cancellation within $\pdv{W}{E}$.
+
+\item The angle variable $w\in[0,2\pi)$ tracks the oscillator\normalsize{}'s progress through one complete cycle. At $w=0$ the particle sits at the outward turning point $x=+A$ (maximum displacement, zero velocity). It crosses equilibrium toward $-A$ at $w=\pi/2$, reaches the opposite turning point $x=-A$ at $w=\pi$, and returns through equilibrium at $w=3\pi/2$. At $w=2\pi\equiv 0$ the orbit closes, having completed one period. The angle advances linearly, $w = \omega_0 t + w_0$, advancing exactly $2\pi$ each period $T=2\pi/\omega_0$. The sinusoidal phase $\omega_0 t + \phi$ coincides with $w$ up to a constant offset, confirming that $w$ measures angular position within the cycle.
+\end{enumerate}
+}
+
+\nt{Comparison with Newton\normalsize{}'s law and energy conservation}{
+Newton\normalsize{}'s second law for the harmonic oscillator gives $m\ddot{x} + m\omega_0^2 x = 0$, a linear second-order ODE whose solution is $x(t) = A\sin(\omega_0 t + \phi)$. The energy method gives $E = \tfrac12 m\dot{x}^2 + \tfrac12 m\omega_0^2 x^2$ and $\dot{x} = \pm\sqrt{2E/m - \omega_0^2 x^2}$, which integrates to the same sinusoidal motion. The Hamilton--Jacobi approach reaches the identical result through a completely different route: solving a first-order nonlinear PDE by separation, evaluating a quadrature, and applying Jacobi\normalsize{}'s theorem. The agreement confirms the equivalence of the three formulations -- Newton\normalsize{}'s, Lagrange\normalsize{}'s, and Jacobi\normalsize{}'s -- as different faces of the same underlying mechanics.}
+
+\qs{Simple harmonic oscillator from the HJ complete integral}{
+A mass $m = 1.0\,\mathrm{kg}$ is attached to a horizontal spring with spring constant $k = 4.0\,\mathrm{N/m}$. The mass is displaced from equilibrium to $x_0 = 2.0\,\mathrm{m}$ and released from rest, so $v_0 = 0\,\mathrm{m/s}$.
+
+\begin{enumerate}[label=(\alph*)]
+\item Compute the natural angular frequency $\omega_0 = \sqrt{k/m}$. Write the Hamilton--Jacobi equation for this system, separate the variables to find $\der{W}{x}$, and state the complete integral $\mcS(x,t;E)$ with numerical parameter values.
+\item Use the initial conditions $x(0) = 2.0\,\mathrm{m}$ and $\dot{x}(0) = 0\,\mathrm{m/s}$ to find the total energy $E$ and the amplitude $A = \sqrt{2E/(m\omega_0^2)}$. Write the trajectory $x(t)$ and verify that the maximum speed equals $A\omega_0$.
+\item Compute the action variable $J = E/\omega_0$ in SI units and verify numerically that $E(J) = \omega_0 J$ reproduces the original energy.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The natural angular frequency is
+\[
+\omega_0 = \sqrt{\frac{k}{m}}
+= \sqrt{\frac{4.0\,\mathrm{N/m}}{1.0\,\mathrm{kg}}}
+= 2.0\,\mathrm{rad/s}.
+\]
+The Hamilton--Jacobi equation is
+\[
+\frac{1}{2m}\left(\pdv{\mcS}{x}\right)^2 + \frac{1}{2}m\omega_0^2 x^2 + \pdv{\mcS}{t} = 0.
+\]
+Substituting the numerical parameters $m = 1.0\,\mathrm{kg}$ and $\omega_0 = 2.0\,\mathrm{rad/s}$ gives
+\[
+\frac{1}{2}\left(\pdv{\mcS}{x}\right)^2 + 2.0\,x^2 + \pdv{\mcS}{t} = 0.
+\]
+Because the Hamiltonian is time-independent, separate as $\mcS = W(x) - Et$. The spatial derivative satisfies
+\[
+\der{W}{x} = \pm\sqrt{2mE - m^2\omega_0^2 x^2}
+= \pm\sqrt{2E - 4x^2}.
+\]
+The complete integral for this specific system is
+\[
+\mcS(x,t;E) = \frac{E}{2.0}\arcsin\!\left(\frac{2x}{\sqrt{2E}}\right)
++ \frac{1}{2}x\sqrt{2E - 4x^2} - Et,
+\]
+valid for $|x| < \sqrt{E/2}$.
+
+\textbf{Part (b).} The total mechanical energy is the sum of kinetic and potential energy:
+\[
+E = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}k x^2.
+\]
+At release, $\dot{x} = 0$ and $x = 2.0\,\mathrm{m}$, so
+\[
+E = 0 + \frac{1}{2}(4.0\,\mathrm{N/m})(2.0\,\mathrm{m})^2
+= 8.0\,\mathrm{J}.
+\]
+The amplitude follows from the energy--amplitude relation:
+\[
+A = \sqrt{\frac{2E}{m\omega_0^2}}
+= \sqrt{\frac{2(8.0\,\mathrm{J})}{(1.0\,\mathrm{kg})(2.0\,\mathrm{rad/s})^2}}
+= \sqrt{4.0}\,\mathrm{m}
+= 2.0\,\mathrm{m}.
+\]
+The amplitude equals the initial displacement, as expected for release from rest.
+
+The trajectory has the form $x(t) = A\sin(\omega_0 t + \phi)$. Determine the phase $\phi$ from the initial conditions:
+\[
+x(0) = A\sin\phi = 2.0\,\mathrm{m},
+\qquad
+\dot{x}(0) = A\omega_0\cos\phi = 0.
+\]
+Since $A = 2.0\,\mathrm{m}$, we have $\sin\phi = 1$ and $\cos\phi = 0$, giving $\phi = \pi/2$. The trajectory simplifies using the identity $\sin(\theta + \pi/2) = \cos\theta$:
+\[
+x(t) = A\cos(\omega_0 t).
+\]
+With numerical values,
+\[
+x(t) = (2.0\,\mathrm{m})\cos\bigl((2.0\,\mathrm{rad/s})\,t\bigr).
+\]
+The velocity is
+\[
+v(t) = \dot{x}(t) = -A\omega_0\sin(\omega_0 t).
+\]
+The maximum speed occurs at equilibrium ($x = 0$), where $|\sin(\omega_0 t)| = 1$:
+\[
+v_{\max} = A\omega_0 = (2.0\,\mathrm{m})(2.0\,\mathrm{rad/s}) = 4.0\,\mathrm{m/s}.
+\]
+From energy, $v_{\max} = \sqrt{2E/m} = \sqrt{16.0/1.0}\,\mathrm{m/s} = 4.0\,\mathrm{m/s}$, confirming the result.
+
+\textbf{Part (c).} The action variable for the harmonic oscillator is
+\[
+J = \frac{E}{\omega_0}.
+\]
+Substitute the numerical values:
+\[
+J = \frac{8.0\,\mathrm{J}}{2.0\,\mathrm{rad/s}}
+= 4.0\,\mathrm{J\!\cdot\!s}.
+\]
+Now verify the energy--action relation $E(J) = \omega_0 J$:
+\[
+E(J) = \omega_0 J
+= (2.0\,\mathrm{rad/s})(4.0\,\mathrm{J\!\cdot\!s})
+= 8.0\,\mathrm{J}.
+\]
+This returns the original energy exactly, confirming $E(J) = \omega_0 J$ both algebraically and for the numerical values of this problem.
+
+Therefore,
+\[
+\omega_0 = 2.0\,\mathrm{rad/s},
+\qquad
+x(t) = (2.0\,\mathrm{m})\cos\bigl((2.0\,\mathrm{rad/s})\,t\bigr),
+\qquad
+J = 4.0\,\mathrm{J\!\cdot\!s}.
+\]

concepts/advanced/uniform-e-field-hj.tex

@@ -0,0 +1,152 @@
+\subsection{Charged Particle in Uniform Electric Field}
+
+This subsection solves the Hamilton--Jacobi equation for a charged particle in a uniform electric field, showing that Jacobi's theorem reproduces the parabolic motion dictated by the constant electric force $\vec{F} = q\vec{E}$. The problem is formally identical to the projectile motion treatment in~A.06: the separation ansatz, the characteristic function, and the Jacobi inversion follow exactly the same algebra, with the gravitational acceleration $g$ replaced by $-qE_0/m$. Likewise, the uniform field between parallel plates studied in Unit~9 (e9-3) produces a constant electric force that accelerates the particle uniformly; the HJ solution presented here applies directly to that configuration as well.
+
+\dfn{Hamiltonian for a charged particle in a uniform electric field}{
+A particle of mass $m$ and charge $q$ in a uniform electric field $\vec{E} = E_0\,\hat{\bm{z}}$ (with $\vec{B} = 0$) is described by the scalar potential $\varphi = -E_0 z$ and zero vector potential. The Hamiltonian is
+\[
+\mcH = \frac{p_x^2 + p_y^2 + p_z^2}{2m} - qE_0 z.
+\]
+The coordinates $x$ and $y$ are absent from $\mcH$, so they are cyclic and the conjugate momenta $p_x$, $p_y$ are constants of the motion. The Hamilton--Jacobi equation for $\mcS(\vec{r},t)$ is
+\[
+\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \left(\pdv{\mcS}{y}\right)^2 + \left(\pdv{\mcS}{z}\right)^2\right] - qE_0 z + \pdv{\mcS}{t} = 0.
+\]}
+
+\nt{This problem is the electromagnetic analogue of projectile motion. The gravitational acceleration $g$ is replaced by the electric acceleration $qE_0/m$ and the direction of $\hat{\bm{y}}$ by $\hat{\bm{z}}$. The two problems are formally equivalent under the substitution $g \to -qE_0/m$.}
+
+\thm{Complete integral and trajectory from Jacobi's theorem}{
+The complete integral of the Hamilton--Jacobi equation for a charged particle in the uniform field $\vec{E} = E_0\,\hat{\bm{z}}$ is
+\[
+\mcS(x,y,z,t) = p_x x + p_y y + \frac{2\sqrt{2m}}{3qE_0}\left(E_z + qE_0 z\right)^{3/2} - Et,
+\]
+where $p_x$ and $p_y$ are the conserved transverse momenta and $E_z = E - (p_x^2 + p_y^2)/(2m)$. Jacobi's theorem with respect to the energy, $\pdv{\mcS}{E} = \beta_E$, yields the trajectory along the field direction:
+\[
+z(t) = v_{0z}\,t + \frac{1}{2}\,\frac{qE_0}{m}\,t^2,
+\]
+a parabola identical in form to the kinematic equation for constant acceleration.}
+
+\pf{Separation of the HJ equation and extraction of $z(t)$}{
+Because the Hamiltonian has no explicit time dependence, use the ansatz $\mcS = p_x x + p_y y + W_z(z) - Et$, where $p_x$, $p_y$, and $E$ are the separation constants. The partial derivatives are
+\[
+\pdv{\mcS}{x} = p_x,
+\qquad
+\pdv{\mcS}{y} = p_y,
+\qquad
+\pdv{\mcS}{z} = \der{W_z}{z},
+\qquad
+\pdv{\mcS}{t} = -E.
+\]
+Substitute into the HJ PDE:
+\[
+\frac{1}{2m}\Bigl(p_x^2 + p_y^2 + \left(\der{W_z}{z}\right)^2\Bigr) - qE_0 z = E.
+\]
+Define the energy associated with $z$-motion, $E_z = E - (p_x^2 + p_y^2)/(2m)$, and solve for the derivative:
+\[
+\der{W_z}{z} = \sqrt{2m\left(E_z + qE_0 z\right)}.
+\]
+Integrate with respect to $z$. Set $u = E_z + qE_0 z$, so $\dd u = qE_0\,\dd z$:
+\[
+W_z(z) = \frac{\sqrt{2m}}{qE_0}\int\sqrt{u}\,\dd u
+= \frac{2\sqrt{2m}}{3qE_0}\left(E_z + qE_0 z\right)^{3/2}.
+\]
+Reassemble the principal function:
+\[
+\mcS(x,y,z,t) = p_x x + p_y y + \frac{2\sqrt{2m}}{3qE_0}\left(E_z + qE_0 z\right)^{3/2} - Et.
+\]
+Jacobi's theorem with respect to $E$ gives
+\[
+\pdv{\mcS}{E} = \frac{2\sqrt{2m}}{3qE_0}\cdot\frac{3}{2}\left(E_z + qE_0 z\right)^{1/2} - t = \beta_E,
+\]
+since $\pdv{E_z}{E} = 1$ with $p_x$ and $p_y$ held fixed. Simplify:
+\[
+\frac{\sqrt{2m}}{qE_0}\sqrt{E_z + qE_0 z} - t = \beta_E.
+\]
+The square root equals $p_z(z)/\sqrt{2m} = m v_z$ divided by $\sqrt{2m}$, so multiplying by $qE_0/\sqrt{2m}$ gives
+\[
+v_z(t) = \frac{qE_0}{m}\,(t + \beta_E).
+\]
+At $t = 0$, set $v_z(0) = v_{0z}$. Then $\beta_E = v_{0z}\,m/(qE_0)$ and
+\[
+v_z(t) = v_{0z} + \frac{qE_0}{m}\,t.
+\]
+Integrating once more with $z(0) = 0$:
+\[
+z(t) = v_{0z}\,t + \frac{1}{2}\,\frac{qE_0}{m}\,t^2.
+\]
+This is the trajectory along the field. The transverse coordinates evolve uniformly, with Jacobi's theorem applied to $p_x$ and $p_y$ giving $x(t) = (p_x/m)t + \beta_x$ and $y(t) = (p_y/m)t + \beta_y$.}
+
+\nt{Verification against the Lorentz force}{Newton's second law with $\vec{F} = q\vec{E} = qE_0\,\hat{\bm{z}}$ gives the component equation $m\,\dv[2]{z}{t} = qE_0$. Integrating twice subject to $z(0) = 0$ and $\dot{z}(0) = v_{0z}$ yields
+\[
+z(t) = v_{0z}\,t + \frac{1}{2}\,\frac{qE_0}{m}\,t^2,
+\]
+which matches the Hamilton--Jacobi trajectory exactly. The constant acceleration $a_z = qE_0/m$ depends on the charge-to-mass ratio and the field strength. For an electron ($q < 0$) the acceleration opposes the field direction, just as a positively charged particle accelerates along the field. The equivalence between HJ and the force-law approach holds for any time-independent potential.}
+
+\qs{Electron in a uniform electric field}{
+An electron ($q = -e = -1.60\times 10^{-19}\,\mathrm{C}$, $m = 9.11\times 10^{-31}\,\mathrm{kg}$) moves in a uniform electric field $\vec{E} = 1000\,\mathrm{N/C}$ directed along $+\hat{\bm{z}}$. The electron is released from rest at the origin at $t = 0$.
+
+\begin{enumerate}[label=(\alph*)]
+\item Write the Hamilton--Jacobi equation for this system. Show that $x$ and $y$ are cyclic coordinates and identify the corresponding separation constants.
+\item For an electron with $p_x = p_y = 0$ and $v_{0z} = 0$, find $z(t)$ using Jacobi's theorem and give the canonical $z$-momentum $p_z$ as a function of time.
+\item At $t = 1.0\,\mathrm{ns} = 1.0\times 10^{-9}\,\mathrm{s}$, compute the position $z(t)$ and kinetic energy. Compare to the $F = ma$ prediction.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The Hamiltonian is
+\[
+\mcH = \frac{p_x^2 + p_y^2 + p_z^2}{2m} - qE_0 z,
+\]
+with $E_0 = 1000\,\mathrm{N/C}$. The Hamilton--Jacobi equation reads
+\[
+\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \left(\pdv{\mcS}{y}\right)^2 + \left(\pdv{\mcS}{z}\right)^2\right] - qE_0 z + \pdv{\mcS}{t} = 0.
+\]
+Neither $x$ nor $y$ appears explicitly in the Hamiltonian, so both are cyclic. Their conjugate momenta
+\[
+\pdv{\mcS}{x} = \alpha_x,
+\qquad
+\pdv{\mcS}{y} = \alpha_y,
+\]
+are conserved separation constants. The complete integral is $\mcS = \alpha_x x + \alpha_y y + W_z(z) - Et$.
+
+\textbf{Part (b).} With $p_x = p_y = 0$ we have $\alpha_x = \alpha_y = 0$ and the action reduces to $\mcS = W_z(z) - Et$. The HJ equation gives
+\[
+\frac{1}{2m}\left(\der{W_z}{z}\right)^2 - qE_0 z = E.
+\]
+Solve for the canonical momentum:
+\[
+p_z(z) = \der{W_z}{z} = \sqrt{2m\bigl(E + qE_0 z\bigr)}.
+\]
+Jacobi's theorem yields the velocity $v_z(t) = \dfrac{qE_0}{m}(t + \beta_E)$. The particle starts from rest, so $v_z(0) = 0$ fixes $\beta_E = 0$ and
+\[
+v_z(t) = \frac{qE_0}{m}\,t,
+\qquad
+p_z(t) = qE_0\,t.
+\]
+Integrating $v_z(t)$ with $z(0) = 0$:
+\[
+z(t) = \frac{1}{2}\,\frac{qE_0}{m}\,t^2.
+\]
+Because $q = -e < 0$ and $E_0 > 0$, the acceleration is negative and the electron moves in the $-z$ direction.
+
+\textbf{Part (c).} Compute the acceleration:
+\[
+a = \frac{qE_0}{m} = \frac{(-1.60\times 10^{-19})(1000)}{9.11\times 10^{-31}}\,\mathrm{m/s^2}
+= -1.76\times 10^{14}\,\mathrm{m/s^2}.
+\]
+At $t = 1.0\times 10^{-9}\,\mathrm{s}$ the position is
+\[
+z = \frac{1}{2}\,a\,t^2
+= \frac{1}{2}\,(-1.76\times 10^{14})(1.0\times 10^{-18})\,\mathrm{m}
+= -8.8\times 10^{-5}\,\mathrm{m}.
+\]
+The speed is $|v_z| = |a|\,t = (1.76\times 10^{14})(1.0\times 10^{-9})\,\mathrm{m/s} = 1.76\times 10^{5}\,\mathrm{m/s}$. The kinetic energy is
+\[
+K = \tfrac{1}{2}\,m\,v_z^2 = \tfrac{1}{2}\,(9.11\times 10^{-31})(1.76\times 10^{5})^2\,\mathrm{J}
+= 1.4\times 10^{-20}\,\mathrm{J}.
+\]
+In electron volts, $K = (1.4\times 10^{-20})/(1.60\times 10^{-19})\,\mathrm{eV} = 0.088\,\mathrm{eV}$. From the $F = ma$ approach, $\vec{F} = q\vec{E} = (-1.60\times 10^{-19})(1000)\,\hat{\bm{z}}\,\mathrm{N} = -1.60\times 10^{-16}\,\hat{\bm{z}}\,\mathrm{N}$. The resulting acceleration $a = -1.76\times 10^{14}\,\mathrm{m/s^2}$ is identical and the integrated kinematics $z = \tfrac{1}{2}at^2$ reproduce both the position and energy exactly.
+
+Therefore,
+\[
+z(1.0\,\mathrm{ns}) = -8.8\times 10^{-5}\,\mathrm{m},
+\qquad
+K = 1.4\times 10^{-20}\,\mathrm{J} = 0.088\,\mathrm{eV}.
+\]

concepts/em/u11/e11-3-power.tex

@@ -12,7 +12,7 @@ The SI unit of power is the watt ($1\,\mathrm{W}=1\,\mathrm{J/s}=1\,\mathrm{A\!\
 
 \ex{Illustrative example}{A $60\,\mathrm{W}$ incandescent lightbulb is connected to a $120\,\mathrm{V}$ household outlet. Find (a) the current through the bulb and (b) the resistance of its filament.
 
-\sol (a) From $P = I\,\Delta V$,
+(a) From $P = I\,\Delta V$,
 \[
 I = \frac{P}{\Delta V} = \frac{60\,\mathrm{W}}{120\,\mathrm{V}} = 0.50\,\mathrm{A}.
 \]

concepts/em/u11/e11-5-kirchhoff.tex

@@ -138,7 +138,6 @@ Substitute equation (1) into equation (2):
 \[
 7.0\,I_1 + 3.0\,I_2 = 12.
 \]
-\label{eq:a}
 
 Substitute equation (1) into equation (3):
 \[
@@ -158,9 +157,8 @@ so
 \[
 I_1 = 2.0 - 3.0\,I_2.
 \]
-\label{eq:b}
 
-Substitute equation (\ref{eq:b}) into equation (\ref{eq:a}):
+Substitute equation (3) into equation (2):
 \[
 7.0\,(2.0 - 3.0\,I_2) + 3.0\,I_2 = 12,
 \]
@@ -177,7 +175,7 @@ Substitute equation (\ref{eq:b}) into equation (\ref{eq:a}):
 I_2 = \frac{2.0}{18.0} = \frac{1}{9}\,\mathrm{A}.
 \]
 
-From equation (\ref{eq:b}):
+From equation (3):
 \[
 I_1 = 2.0 - 3.0\left(\frac{1}{9}\right) = 2.0 - \frac{1}{3} = \frac{6}{3} - \frac{1}{3} = \frac{5}{3}\,\mathrm{A}.
 \]

concepts/em/u12/e12-3-force-on-current.tex

@@ -111,9 +111,10 @@ U = -\mu B\,\cos\phi = -\vec{\mu}\cdot\vec{B}.
 
 \nt{The SI unit of magnetic dipole moment $\vec{\mu}$ is ampere-square meter, $\mathrm{A\cdot m^2}$. This is equivalent to joules per tesla, $\mathrm{J/T}$.}
 
-\ex{Illustrative example}{A coil with $N=50$ turns, each of area $A = 8.0\,\mathrm{cm^2} = 8.0\times 10^{-4}\,\mathrm{m^2}$, carries current $I = 2.0\,\mathrm{A}$ in a field $B = 0.40\,\mathrm{T}$. The maximum torque occurs at $\phi = 90^\circ$:
+\ex{Illustrative example}{A coil with $N=50$ turns, each of area $A = 8.0\,\mathrm{cm^2} = 8.0\times 10^{-4}\,\mathrm{m^2}$, carries current $I = 2.0\,\mathrm{A}$ in a field $B = 0.40\,\mathrm{T}$. The maximum torque occurs at $\phi = 90^\circ$:}
 \[
-\tau_{\text{max}} = N\,I\,A\,B = (50)(2.0\,\mathrm{A})(8.0\times 10^{-4}\,\mathrm{m^2})(0.40\,\mathrm{T}) = 3.2\times 10^{-2}\,\mathrm{N\cdot m}.}
+\tau_{\text{max}} = N\,I\,A\,B = (50)(2.0\,\mathrm{A})(8.0\times 10^{-4}\,\mathrm{m^2})(0.40\,\mathrm{T}) = 3.2\times 10^{-2}\,\mathrm{N\cdot m}.
+\]
 
 \qs{Worked example}{A rectangular wire loop has width $a=0.10\,\mathrm{m}$ (horizontal side) and height $b=0.050\,\mathrm{m}$ (vertical side). The loop has $N=100$ turns and carries current $I=2.0\,\mathrm{A}$ in the direction shown: clockwise when viewed from the front (from the $+x$-axis toward the $yz$-plane). The loop sits in a uniform magnetic field $\vec{B} = (0.60\,\mathrm{T})\,\hat{\imath}$ pointing to the right. The normal vector $\hat{n}$ points along the $-\hat{k}$ direction (into the page) by the right-hand rule for clockwise current.
 

concepts/mechanics/u6/m6-4-rolling.tex

@@ -32,48 +32,13 @@ Thus the wheel's center moves at $2.4\,\mathrm{m/s}$, while the top point moves
 
 \mprop{Rolling kinematics, energy, and incline dynamics}{Consider a rigid body of mass $M$, radius $R$, and moment of inertia $I_{\mathrm{cm}}$ about its center of mass. Let $s_{\mathrm{cm}}$, $v_{\mathrm{cm}}$, and $a_{t,\mathrm{cm}}$ denote the center-of-mass distance, speed, and tangential acceleration, and let $\theta$, $\omega$, and $\alpha$ denote the corresponding angular variables. Let $K_i$ and $U_i$ denote the initial kinetic and potential energies, and let $K_f$ and $U_f$ denote the final kinetic and potential energies.
 
-\begin{enumerate}[label=\textbf{\arabic*.}]
+1. For pure rolling, $s_{\mathrm{cm}}=R\theta$, $v_{\mathrm{cm}}=R\omega$, $a_{t,\mathrm{cm}}=R\alpha$.
-\item For pure rolling,
-\[
-s_{\mathrm{cm}}=R\theta,
-\qquad
-v_{\mathrm{cm}}=R\omega,
-\qquad
-a_{t,\mathrm{cm}}=R\alpha.
-\]
 
-\item The total kinetic energy is the sum of translational and rotational parts:
+2. The total kinetic energy is the sum of translational and rotational parts: $K=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2$. Using $v_{\mathrm{cm}}=R\omega$, this may also be written as $K=\tfrac12\left(M+\frac{I_{\mathrm{cm}}}{R^2}\right)v_{\mathrm{cm}}^2$.
-\[
-K=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2.
-\]
-Using $v_{\mathrm{cm}}=R\omega$, this may also be written as
-\[
-K=\tfrac12\left(M+\frac{I_{\mathrm{cm}}}{R^2}\right)v_{\mathrm{cm}}^2.
-\]
 
-\item If no nonconservative force removes mechanical energy from the system, then for rolling without slipping,
+3. If no nonconservative force removes mechanical energy from the system, then for rolling without slipping, $K_i+U_i=K_f+U_f$. For a vertical drop of magnitude $h$ from rest, $Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2$.
-\[
-K_i+U_i=K_f+U_f.
-\]
-For a vertical drop of magnitude $h$ from rest,
-\[
-Mgh=\tfrac12 Mv_{\mathrm{cm}}^2+\tfrac12 I_{\mathrm{cm}}\omega^2.
-\]
 
-\item For a body rolling without slipping down an incline of angle $\beta$, choose positive down the incline. Let $a_{\mathrm{cm}}$ denote the center-of-mass acceleration magnitude along the incline. If $f_s$ denotes the magnitude of the static friction force, then
+4. For a body rolling without slipping down an incline of angle $\beta$, choose positive down the incline. Let $a_{\mathrm{cm}}$ denote the center-of-mass acceleration magnitude along the incline. If $f_s$ denotes the magnitude of the static friction force, then $Mg\sin\beta-f_s=Ma_{\mathrm{cm}}$, $f_sR=I_{\mathrm{cm}}\alpha$, $a_{\mathrm{cm}}=R\alpha$. Therefore, $a_{\mathrm{cm}}=\frac{g\sin\beta}{1+I_{\mathrm{cm}}/(MR^2)}$, $f_s=\frac{I_{\mathrm{cm}}}{R^2}a_{\mathrm{cm}}$.
-\[
-Mg\sin\beta-f_s=Ma_{\mathrm{cm}},
-\qquad
-f_sR=I_{\mathrm{cm}}\alpha,
-\qquad
-a_{\mathrm{cm}}=R\alpha.
-\]
-Therefore,
-\[
-a_{\mathrm{cm}}=\frac{g\sin\beta}{1+I_{\mathrm{cm}}/(MR^2)},
-\qquad
-f_s=\frac{I_{\mathrm{cm}}}{R^2}a_{\mathrm{cm}}.
-\]
 For an object accelerating down the incline, the static friction force on the object points up the incline.}
 
 \qs{Worked example}{A solid cylinder of mass $M=2.0\,\mathrm{kg}$ and radius $R=0.20\,\mathrm{m}$ is released from rest on an incline that makes an angle $\beta=30^\circ$ with the horizontal. The cylinder rolls without slipping through a vertical drop $h=0.75\,\mathrm{m}$. Let $g=9.8\,\mathrm{m/s^2}$.

handbook-manifest.json

@@ -4,7 +4,8 @@
     "root_file": "ap-physics-c-handbook.tex",
     "chapter_files": {
       "mechanics": "chapters/mechanics.tex",
-      "em": "chapters/em.tex"
+      "em": "chapters/em.tex",
+      "advanced": "chapters/advanced.tex"
     }
   },
   "notation_policy": {
@@ -853,6 +854,149 @@
       "filename": "concepts/em/u13/e13-7-lc-circuits.tex",
       "recipe": "R6",
       "unit_order": 7
+    },
+    {
+      "id": "A.01",
+      "course": "advanced",
+      "unit": null,
+      "unit_title": null,
+      "unit_file": null,
+      "filename": "concepts/advanced/hj-equation.tex",
+      "recipe": "R2",
+      "unit_order": 1,
+      "section": "Hamilton-Jacobi Fundamentals"
+    },
+    {
+      "id": "A.02",
+      "course": "advanced",
+      "unit": null,
+      "unit_title": null,
+      "unit_file": null,
+      "filename": "concepts/advanced/separation.tex",
+      "recipe": "R2",
+      "unit_order": 2,
+      "section": "Hamilton-Jacobi Fundamentals"
+    },
+    {
+      "id": "A.03",
+      "course": "advanced",
+      "unit": null,
+      "unit_title": null,
+      "unit_file": null,
+      "filename": "concepts/advanced/action-angle.tex",
+      "recipe": "R3",
+      "unit_order": 3,
+      "section": "Hamilton-Jacobi Fundamentals"
+    },
+    {
+      "id": "A.04",
+      "course": "advanced",
+      "unit": null,
+      "unit_title": null,
+      "unit_file": null,
+      "filename": "concepts/advanced/hj-em-coupling.tex",
+      "recipe": "R2",
+      "unit_order": 4,
+      "section": "Hamilton-Jacobi Fundamentals"
+    },
+    {
+      "id": "A.05",
+      "course": "advanced",
+      "unit": null,
+      "unit_title": null,
+      "unit_file": null,
+      "filename": "concepts/advanced/free-particle-hj.tex",
+      "recipe": "R1",
+      "unit_order": 5,
+      "section": "Mechanics Problems via HJ"
+    },
+    {
+      "id": "A.06",
+      "course": "advanced",
+      "unit": null,
+      "unit_title": null,
+      "unit_file": null,
+      "filename": "concepts/advanced/projectile-hj.tex",
+      "recipe": "R2",
+      "unit_order": 6,
+      "section": "Mechanics Problems via HJ"
+    },
+    {
+      "id": "A.07",
+      "course": "advanced",
+      "unit": null,
+      "unit_title": null,
+      "unit_file": null,
+      "filename": "concepts/advanced/sho-hj.tex",
+      "recipe": "R6",
+      "unit_order": 7,
+      "section": "Mechanics Problems via HJ"
+    },
+    {
+      "id": "A.08",
+      "course": "advanced",
+      "unit": null,
+      "unit_title": null,
+      "unit_file": null,
+      "filename": "concepts/advanced/kepler-hj.tex",
+      "recipe": "R5",
+      "unit_order": 8,
+      "section": "Mechanics Problems via HJ"
+    },
+    {
+      "id": "A.09",
+      "course": "advanced",
+      "unit": null,
+      "unit_title": null,
+      "unit_file": null,
+      "filename": "concepts/advanced/rigid-rotator-hj.tex",
+      "recipe": "R5",
+      "unit_order": 9,
+      "section": "Mechanics Problems via HJ"
+    },
+    {
+      "id": "A.10",
+      "course": "advanced",
+      "unit": null,
+      "unit_title": null,
+      "unit_file": null,
+      "filename": "concepts/advanced/uniform-e-field-hj.tex",
+      "recipe": "R2",
+      "unit_order": 10,
+      "section": "Electromagnetism Problems via HJ"
+    },
+    {
+      "id": "A.11",
+      "course": "advanced",
+      "unit": null,
+      "unit_title": null,
+      "unit_file": null,
+      "filename": "concepts/advanced/cyclotron-hj.tex",
+      "recipe": "R4",
+      "unit_order": 11,
+      "section": "Electromagnetism Problems via HJ"
+    },
+    {
+      "id": "A.12",
+      "course": "advanced",
+      "unit": null,
+      "unit_title": null,
+      "unit_file": null,
+      "filename": "concepts/advanced/crossed-fields-hj.tex",
+      "recipe": "R4",
+      "unit_order": 12,
+      "section": "Electromagnetism Problems via HJ"
+    },
+    {
+      "id": "A.13",
+      "course": "advanced",
+      "unit": null,
+      "unit_title": null,
+      "unit_file": null,
+      "filename": "concepts/advanced/kepler-coulomb-hj.tex",
+      "recipe": "R5",
+      "unit_order": 13,
+      "section": "Electromagnetism Problems via HJ"
     }
   ]
 }

macros.tex

@@ -31,3 +31,16 @@
 
 \newcommand{\sol}{\setlength{\parindent}{0cm}\textbf{\textit{Solution:}}\setlength{\parindent}{1cm} }
 \newcommand{\solve}[1]{\setlength{\parindent}{0cm}\textbf{\textit{Solution: }}\setlength{\parindent}{1cm}#1 \Qed}
+
+% === Hamilton-Jacobi notation ===
+\newcommand{\Sdel}[1]{\pdv{S}{#1}}
+\newcommand{\Sdelo}[2]{\pdv[#1]{S}{#2}}
+\newcommand{\Jact}[1]{J_{#1}}
+\newcommand{\wangle}[1]{w_{#1}}
+\newcommand{\sepconst}[1]{\alpha_{#1}}
+\newcommand{\pbracket}[2]{\left\{#1,#2\right\}}
+\newcommand{\Lagrian}{\mathcal{L}}
+\newcommand{\bm}[1]{\mathbf{#1}}
+
+% === Safe differentials for Advanced chapter (no \@ifnextchar) ===
+\renewcommand{\dd}{\mathrm{d}}