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feat(HJ): Chapter 3 — Advanced Analytical Mechanics (Hamilton-Jacobi)

Browse Source 
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.
This commit is contained in:
Krishna Ayyalasomayajula
2026-05-02 02:21:53 -05:00
parent 7cf82f52da
commit 1c1a575f6e
22 changed files with 3048 additions and 7 deletions
Download Patch File Download Diff File Expand all files Collapse all files

43
ap-physics-c-handbook.aux
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\part{Electricity \& Magnetism}
\input{chapters/em}
\part{Advanced Topics}
\input{chapters/advanced}
\end{document}

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\chapter{Advanced Analytical Mechanics}
The Hamilton-Jacobi (HJ) formulation is the final reformulation of classical mechanics, expressing the entire dynamics of a system as a single first-order partial differential equation for a scalar function $S$, called the \textbf{principal function}. Solving the HJ equation by separation of variables often yields complete solutions more directly than the Lagrange or Hamilton equations -- especially for systems with symmetries and cyclic coordinates. The HJ framework also provides the classical foundation for the WKB approximation and connects to the Schrodinger equation in the $\hbar \to 0$ limit.
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}

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\subsection{Action-Angle Variables}
This subsection develops action-angle variables for integrable periodic systems, presenting the canonical transformation that reduces any periodic system to trivial dynamics where the momenta are constant and the angles advance uniformly.
\dfn{Action and angle variables}{
For a periodic degree of freedom with generalized coordinate $q_i$ and conjugate momentum $p_i$, the action variable $J_i$ is defined as the phase-space integral over one complete closed orbit:
\[
J_i = \oint p_i\,dq_i.
\]
The integral is taken over one full cycle of the periodic motion. The angle variable $w_i$ is the canonical coordinate conjugate to $J_i$, defined by differentiating Hamilton's characteristic function $W$ with respect to the action:
\[
w_i = \pdv{W}{J_i}.
\]
The angle variable $w_i$ increases by exactly one complete unit during one period of the associated periodic motion.}
\nt{The action variable $J_i$ equals the area enclosed by the orbit in the $(q_i,p_i)$ phase-space plane. This geometric interpretation makes it straightforward to evaluate $J_i$ for simple periodic systems: the integral reduces to computing the area of an ellipse (harmonic oscillator), a triangle plus its reflection (infinite well), or other phase-space shapes.}
In a completely integrable system with $n$ degrees of freedom, the Hamiltonian depends only on the action variables and not on the angle variables: $\mcH = \mcH(J_1,\ldots,J_n)$. Because the angles do not appear in $\mcH$, they are cyclic coordinates. This leads to the simplest possible Hamiltonian dynamics.
\thm{Hamilton's equations in action-angle variables}{
Let $\mcH = \mcH(J_1,\ldots,J_n)$ be the Hamiltonian expressed in action variables. Hamilton's canonical equations in the $(w,J)$ variables are
\[
\dot{J}_i = -\pdv{\mcH}{w_i} = 0,
\qquad
\dot{w}_i = \pdv{\mcH}{J_i} \equiv \omega_i.
\]
The action variables $J_i$ are constant in time, and the angle variables advance linearly:
\[
w_i(t) = \omega_i t + w_i(0).
\]
The frequency $\omega_i = \pdv{\mcH}{J_i}$ is independent of time. Since the angle variable $w_i$ increases by one unit over one complete cycle, the period of the $i$-th motion is $T_i = 1/\omega_i$.}
The frequency $\omega_i$ provides direct access to the temporal characteristics of the motion. When there is a single degree of freedom, the action is found by evaluating the integral $J = \oint p\,dq$, the Hamiltonian is inverted to give $E(J)$, and the period follows immediately from $T = 1/\pdv{E}{J}$. This procedure avoids solving the equations of motion directly. The physical angular frequency of the motion is $2\pi\omega_i$.
\ex{Simple harmonic oscillator in action-angle variables}{
The Hamiltonian for a one-dimensional simple harmonic oscillator of mass $m$ and natural angular frequency $\omega_0$ is
\[
\mcH = \frac{p^2}{2m} + \frac{1}{2}m\omega_0^2 x^2.
\]
At energy $E = \mcH$, the momentum is $p = \pm\sqrt{2mE - m^2\omega_0^2 x^2}$ and the turning points are at $x = \pm A$ with amplitude $A = \sqrt{2E/(m\omega_0^2)}$.
The action variable is evaluated by integrating over one complete oscillation:
\[
J = \oint p\,dx = 2\int_{-A}^{A}\sqrt{2mE - m^2\omega_0^2 x^2}\,dx.
\]
Substitute $x = A\sin\phi$, so $dx = A\cos\phi\,d\phi$ and the limits are $\phi = -\pi/2$ to $\pi/2$:
\[
\sqrt{2mE - m^2\omega_0^2 x^2}
= \sqrt{2mE - 2mE\sin^2\phi}
= \sqrt{2mE}\cos\phi.
\]
The integral becomes
\[
J = 2\sqrt{2mE}\cdot A\int_{-\pi/2}^{\pi/2}\cos^2\phi\,d\phi
= 2\sqrt{2mE}\cdot A\cdot\frac{\pi}{2}
= \pi\sqrt{2mE}\cdot A.
\]
Substituting $A = \sqrt{2E/(m\omega_0^2)}$ gives
\[
J = \pi\sqrt{2mE}\cdot\sqrt{\frac{2E}{m\omega_0^2}}
= 2\pi\,\frac{E}{\omega_0}.
\]
Inverting this relation expresses the energy as a function of the action:
\[
E(J) = \frac{\omega_0 J}{2\pi}.
\]
The frequency obtained from the action-angle formalism is the derivative of $E$ with respect to $J$:
\[
\omega = \pdv{E}{J} = \frac{\omega_0}{2\pi}.
\]
The period of oscillation is $T = 1/\omega = 2\pi/\omega_0$, and the physical angular frequency is $2\pi\omega = \omega_0$. Crucially, the frequency is independent of the energy $E$ and therefore independent of the amplitude $A$. This is the property of isochrony: all oscillations of a simple harmonic oscillator have the same period regardless of amplitude.}
\nt{The Kepler problem (gravitational or electrostatic $V = -k/r$) has three independent action variables $J_r$, $J_\theta$, and $J_\phi$. The energy depends on their sum:
\[
E = -\frac{2\pi^2 m k^2}{(J_r + J_\theta + J_\phi)^2}.
\]
The frequency derivatives $\pdv{E}{J_r}$, $\pdv{E}{J_\theta}$, $\pdv{E}{J_\phi}$ are all equal, so the three frequencies are degenerate. Degenerate frequencies mean every bound orbit closes on itself after one period. This degeneracy is the deep reason Kepler's ellipses are closed: the radial period equals the angular period. A small perturbation $V = -k/r + \epsilon/r^2$ breaks the degeneracy and produces precession.}
\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 $J = \oint p\,dx$ for this system, showing that $J = 2L\sqrt{2mE}$.
\item Express the energy as $E(J)$ and compute the frequency $\omega = \pdv{E}{J}$ and the period $T = 1/\omega$. Show that $T = 2L\sqrt{m/(2E)}$, which equals the time for the particle to travel the distance $2L$ at speed $v = \sqrt{2E/m}$.
\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 action integral over one complete cycle is
\[
J = \oint p\,dx = \int_{0}^{L}\sqrt{2mE}\,dx + \int_{L}^{0}\left(-\sqrt{2mE}\right)\,dx.
\]
Each integral equals $L\sqrt{2mE}$, so
\[
J = L\sqrt{2mE} + L\sqrt{2mE} = 2L\sqrt{2mE}.
\]
\textbf{Part (b).} Solve the result from part (a) for $E$:
\[
\frac{J}{2L} = \sqrt{2mE},
\qquad
\frac{J^2}{4L^2} = 2mE,
\qquad
E(J) = \frac{J^2}{8mL^2}.
\]
Differentiate with respect to $J$ to find the frequency:
\[
\omega = \pdv{E}{J} = \frac{J}{4mL^2}.
\]
The period is the reciprocal of the frequency:
\[
T = \frac{1}{\omega} = \frac{4mL^2}{J}.
\]
Substitute $J = 2L\sqrt{2mE}$ to express $T$ in terms of $E$:
\[
T = \frac{4mL^2}{2L\sqrt{2mE}}
= \frac{2mL}{\sqrt{2mE}}
= 2L\sqrt{\frac{m}{2E}}.
\]
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 action variable $J = 2L\sqrt{2mE}$:
\[
2mE = 2(9.11\times 10^{-31})(1.60\times 10^{-19})\,\mathrm{kg\!\cdot\!J}
= 2.92\times 10^{-49}\,\mathrm{kg^2\!\cdot\!m^2/s^2}.
\]
(The product $\mathrm{kg\!\cdot\!J}$ has the same dimensions as $\mathrm{kg^2\!\cdot\!m^2/s^2}$ since $1\,\mathrm{J} = 1\,\mathrm{kg\!\cdot\!m^2/s^2}$.) Taking the square root:
\[
\sqrt{2mE} = \sqrt{2.92\times 10^{-49}}\,\mathrm{kg\!\cdot\!m/s}
= 5.40\times 10^{-25}\,\mathrm{kg\!\cdot\!m/s}.
\]
Now multiply by $2L$:
\[
J = 2(1.00\times 10^{-10})(5.40\times 10^{-25})\,\mathrm{kg\!\cdot\!m^2/s}
= 1.08\times 10^{-34}\,\mathrm{kg\!\cdot\!m^2/s}.
\]
Next compute the period $T = 2L\sqrt{m/(2E)}$:
\[
\frac{m}{2E} = \frac{9.11\times 10^{-31}}{2(1.60\times 10^{-19})}\,\mathrm{kg/J}
= 2.85\times 10^{-12}\,\mathrm{s^2/m^2}.
\]
Taking the square root:
\[
\sqrt{\frac{m}{2E}} = \sqrt{2.85\times 10^{-12}}\,\mathrm{s/m}
= 1.69\times 10^{-6}\,\mathrm{s/m}.
\]
Multiply by $2L$:
\[
T = 2(1.00\times 10^{-10})(1.69\times 10^{-6})\,\mathrm{s}
= 3.37\times 10^{-16}\,\mathrm{s}.
\]
Therefore,
\[
J = 1.08\times 10^{-34}\,\mathrm{kg\!\cdot\!m^2/s},
\qquad
T = 3.37\times 10^{-16}\,\mathrm{s}.
\]

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\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{Choice of gauge does not affect physical observables. The Landau gauge $\vec{A} = (-B_0 y, 0, 0)$ breaks translational symmetry in $y$ but preserves it in $x$, making $p_x$ the conserved momentum. An alternative symmetric gauge $\vec{A} = \tfrac12 B_0(-y, x, 0)$ would be natural for purely rotational problems but obscures the drift structure we want to expose here.}
\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:
\[
y_c = \frac{mE_0/B_0 - \alpha_x}{qB_0}.
\]
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 time average of the position is the guiding centre, $\langle y \rangle = y_c$.
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. Averaging over one cyclotron orbit, $\langle y \rangle = y_c$, so the drift velocity is
\[
\langle v_x \rangle = \frac{\alpha_x + qB_0 y_c}{m}.
\]
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{A charge-sign reversal changes both the sense of Larmor rotation and the location of the guiding centre $y_c$, but these two effects exactly compensate in the averaged $x$-velocity. An electron and a proton spiralling in the same crossed fields therefore share the same guiding-centre drift, even though their individual gyroradii and rotation frequencies differ enormously. This universality is what makes the $E\times B$ drift so important in plasma physics: bulk plasma drifts as a coherent fluid.}
\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,
\[
\vec{v}_d = (50\,000\,\mathrm{m/s})\,\hat{\bm{x}}
\qquad\text{for both electron and proton, independent of charge and mass.}
\]

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\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 $\vec{A} = (0, B_0 x, 0)$ and 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{The Landau gauge $\vec{A} = (0, B_0 x, 0)$ breaks rotational symmetry and makes $y$ cyclic but $x$ not. Other gauges exist (for instance $A = (-B_0 y, 0, 0)$), but they lead to mathematically equivalent Hamiltonian systems. The physics of cyclotron motion -- circular gyration at the Larmor frequency -- is gauge-independent, as the magnetic field $\vec{B} = \nabla\times\vec{A} = B_0\hat{\bm{z}}$ is the same in every gauge.}
\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 $\phi_0 = \omega_c\beta$.
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 $v_z$.}
\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 is the phase-space area enclosed by one gyration:
\[
J = \oint p_x\,\dd x = 2\int_{X_c - R}^{X_c + R} q B_0\sqrt{R^2 - (x - X_c)^2}\,\dd x.
\]
The integral is twice the area of a semicircle of radius $R$ (multiplied by $q B_0$), so
\[
J = q B_0 \cdot \pi R^2
= q B_0 \cdot \pi\cdot\frac{2m E_\perp}{q^2 B_0^2}
= \frac{2\pi m E_\perp}{q B_0}
= \frac{2\pi E_\perp}{\omega_c}.
\]
Geometrically, $J/(q B_0)$ is the area of the circular orbit in the $xy$-plane.
\item Inverting the action--energy relation, the transverse energy as a function of the action is
\[
E_\perp(J) = \frac{\omega_c J}{2\pi}.
\]
The Hamiltonian expressed in terms of the action variables is $E = \omega_c J/(2\pi) + \alpha_z^2/(2m)$, linear in $J$ and quadratic in $\alpha_z$.
\item The Hamilton--Jacobi frequency is $\hat{\omega} = \pdv{E_\perp}{J} = \omega_c/(2\pi)$. The physical angular frequency is $\omega = 2\pi\hat{\omega} = \omega_c = q B_0/m$, which depends only on the charge-to-mass ratio and the field strength. It 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: $w = \hat{\omega}t + w_0 = (\omega_c/2\pi)t + w_0$. The phase of the circular gyration, $\omega_c t + \phi_0$, equals $2\pi w$ 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
\[
\ddot{x} = \frac{q B_0}{m}\,\dot{y},
\qquad
\ddot{y} = -\frac{q B_0}{m}\,\dot{x},
\qquad
\ddot{z} = 0.
\]
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 $J = 2\pi E_\perp/\omega_c$ in SI units and verify by differentiation that $\pdv{E_\perp}{J} = \omega_c/(2\pi)$, recovering the cyclotron 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
\[
J = \frac{2\pi E_\perp}{\omega_c}.
\]
Substitute the numerical values:
\[
J = \frac{2\pi(1.60\times 10^{-16}\,\mathrm{J})}{1.437\times 10^8\,\mathrm{rad/s}}
= \frac{1.005\times 10^{-15}}{1.437\times 10^8}\,\mathrm{J\!\cdot\!s}.
\]
This gives
\[
J = 7.0\times 10^{-24}\,\mathrm{J\!\cdot\!s}.
\]
Now verify the energy--action relation $E_\perp(J) = \omega_c J/(2\pi)$. Differentiating with respect to $J$:
\[
\pdv{E_\perp}{J} = \frac{\omega_c}{2\pi}.
\]
The physical angular frequency is recovered as $\omega = 2\pi(\pdv{E_\perp}{J}) = \omega_c$. For the numerical values,
\[
E_\perp(J) = \frac{\omega_c J}{2\pi}
= \frac{(1.437\times 10^8\,\mathrm{rad/s})(7.0\times 10^{-24}\,\mathrm{J\!\cdot\!s})}{2\pi}
= 1.60\times 10^{-16}\,\mathrm{J},
\]
which is exactly the original transverse energy of $1.0\,\mathrm{keV}$. The relation $\pdv{E_\perp}{J} = \omega_c/(2\pi)$ holds both algebraically and numerically, confirming that the action-angle formalism reproduces the cyclotron frequency exactly.
Therefore,
\[
\omega_c = 1.4\times 10^8\,\mathrm{rad/s},
\qquad
T = 44\,\mathrm{ns},
\qquad
R = 3.1\,\mathrm{mm},
\qquad
J = 7.0\times 10^{-24}\,\mathrm{J\!\cdot\!s}.
\]

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\subsection{Free Particle in 1D and 3D}
This subsection solves the Hamilton--Jacobi equation for a free particle in one and three dimensions, demonstrating that Jacobi's theorem reproduces 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 Newton's second law}{Each coordinate equation $q_i(t) = (p_i/m)t + \beta_i$ integrates a constant velocity $\dot{q}_i = p_i/m$. The acceleration vanishes, $\ddot{q}_i = 0$, which is precisely the result of Newton's second law for zero applied force. The Hamilton--Jacobi formalism therefore reproduces the familiar kinematic result of uniform motion along a straight line.}
\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 Verify that the trajectory matches $\mathbf{r}(t) = (3.0t,\,4.0t,\,0)\,\mathrm{m}$.
\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).} Assembling the three components:
\[
\mathbf{r}(t) = \bigl(x(t),\,y(t),\,z(t)\bigr) = (3.0\,t,\,4.0\,t,\,0).
\]
This matches the expected trajectory $\mathbf{r}(t) = (3.0t,\,4.0t,\,0)\,\mathrm{m}$, confirming the consistency of the Hamilton--Jacobi formalism for the free particle. The speed is $|\mathbf{v}| = \sqrt{3.0^2 + 4.0^2} = 5.0\,\mathrm{m/s}$, and the kinetic energy $E = \tfrac{1}{2}(2.0)(5.0)^2 = 25\,\mathrm{J}$ matches the energy from part (a).
Therefore,
\[
\mcS(x,y,z,t) = 6.0\,x + 8.0\,y - 25\,t,
\qquad
\mathbf{r}(t) = (3.0\,t,\,4.0\,t,\,0)\,\mathrm{m}.
\]

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\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 $p \to p - qA$.
\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.}
\nt{The canonical momentum differs from the kinetic momentum. The extra term $q\vec{A}$ in the canonical momentum is what distinguishes a charged particle's Hamiltonian dynamics from those of a free particle, even in the absence of a scalar potential. 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. Instead, Hamilton's equations for the canonical variables reproduce the full Lorentz-force law.}
\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
\[
\vec{p} \longrightarrow \vec{p} - q\vec{A},
\qquad
E \longrightarrow E - q\varphi.
\]
Under a gauge transformation of the potentials,
\[
\vec{A}' = \vec{A} + \nabla\Lambda,
\qquad
\varphi' = \varphi - \pdv{\Lambda}{t},
\]
the HJ equation retains its form provided the principal function transforms as
\[
\mcS'(\vec{r},t) = \mcS(\vec{r},t) - q\,\Lambda(\vec{r},t).
\]
To see this, substitute $\nabla\mcS' = \nabla\mcS - q\nabla\Lambda$ 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{A deep connection to quantum mechanics emerges from the WKB ansatz. Write the wavefunction as $\psi(\vec{r},t) = \exp\bigl(i\mcS(\vec{r},t)/\hbar\bigr)$. Substituting this form into the Schrodinger equation,
\[
i\hbar\,\pdv{\psi}{t} = \frac{1}{2m}\bigl(-i\hbar\nabla - q\vec{A}\bigr)^2\psi + q\varphi\psi,
\]
and collecting terms order by order in $\hbar$, the leading-order equation (proportional to $\hbar^0$) is exactly the classical Hamilton--Jacobi equation for a charged particle:
\[
\frac{1}{2m}\bigl|\nabla\mcS - q\vec{A}\bigr|^2 + q\varphi + \pdv{\mcS}{t} = 0.
\]
Higher-order terms in $\hbar$ account for quantum corrections. In this sense, the classical Hamilton--Jacobi PDE is the $\hbar\to 0$ limit of the Schrodinger equation.}
\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, the Hamiltonian is
\[
\mcH = \frac{1}{2m}\Bigl[p_x^2 + \bigl(p_y - q B_0 x\bigr)^2 + p_z^2\Bigr],
\]
the Hamilton--Jacobi equation is
\[
\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \left(\pdv{\mcS}{y} - q B_0 x\right)^2 + \left(\pdv{\mcS}{z}\right)^2\right] + \pdv{\mcS}{t} = 0,
\]
the cyclic coordinates are $y$ and $z$, and the cyclotron frequency is
\[
\omega_c = 1.76\times 10^{11}\,\mathrm{rad/s}.
\]

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\subsection{Derivation of the Hamilton-Jacobi Equation}
This subsection derives the Hamilton--Jacobi partial differential equation from the Lagrangian formulation through successive Legendre transforms and a special canonical transformation, and states Jacobi's theorem that reduces the solution of the mechanics problem to finding a complete integral of the resulting PDE.
\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{The Hamilton--Jacobi approach replaces the $2n$ coupled first-order Hamiltonian equations of motion with a single first-order nonlinear PDE for $\mcS$. The trade-off is between solving a system of coupled ODEs and solving a nonlinear PDE. In practice, the PDE is often separable, reducing to a set of ordinary equations that integrate more easily.}
\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. Then Hamilton's principal function $\mcS(q_1,\dots,q_n,t)$ satisfies
\[
\mcH\!\left(q_1,\dots,q_n,\pdv{\mcS}{q_1},\dots,\pdv{\mcS}{q_n},t\right) + \pdv{\mcS}{t} = 0.
\]}
\pf{Derivation from Lagrangian through canonical transformation to HJ}{
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 directly from this construction:
\[
\dot{q}_i = \pdv{\mcH}{p_i},
\qquad
\dot{p}_i = -\pdv{\mcH}{q_i}.
\]
These $2n$ first-order coupled equations fully determine the dynamics of the system once initial conditions are specified.
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
\[
\mcK = \mcH + \pdv{F_2}{t}.
\]
Choose the generating function so that the new Hamiltonian vanishes identically: $\mcK = 0$. With this choice, all new coordinates and momenta are constant in time by Hamilton's equations in the new variables:
\[
\dot{Q}_i = \pdv{\mcK}{P_i} = 0,
\qquad
\dot{P}_i = -\pdv{\mcK}{Q_i} = 0.
\]
This makes every $Q_i$ and $P_i$ a constant of motion, which trivializes the dynamics in the transformed variables. 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 action-angle-like variables in which the motion is completely trivial.}
\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.}
\mprop{Jacobi's theorem on complete integrals}{
Let $\mcS(q_1,\dots,q_n,\alpha_1,\dots,\alpha_n,t)$ denote a complete integral of the Hamilton--Jacobi equation, meaning it contains $n$ independent non-additive separation constants $\alpha_1,\dots,\alpha_n$ plus one true additive constant. Jacobi's theorem provides the equations of motion:
\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\item For each $i=1,\dots,n$, the equation
\[
\pdv{\mcS}{\alpha_i} = \beta_i
\]
determines a relation among the coordinates, time, and the two constants $\alpha_i$ and $\beta_i$. The constants $\beta_1,\dots,\beta_n$ are fixed by the initial conditions.
\item The $n$ equations from the previous point determine $q_1(t),\dots,q_n(t)$, thereby solving the mechanics problem entirely. The canonical momenta follow from
\[
p_i = \pdv{\mcS}{q_i}.
\]
\item In the language of canonical transformations, the separation constants $\alpha_i$ play the role of the new momenta $P_i$ and the constants $\beta_i$ play the role of the new coordinates $Q_i$. All are constant in time because the new Hamiltonian has been chosen to vanish.
\end{enumerate}
}
\nt{Connection to Hamilton's principle of least action}{The Hamilton--Jacobi formalism is deeply connected 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 $\int_{t_0}^{t} \mcL\,dt'$ computed along that trajectory. The Hamilton--Jacobi equation itself can be viewed as the condition that the action integral be stationary under variations of the endpoint, generalizing the Euler--Lagrange equations 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.}
\nt{Connection to Maupertuis' principle}{The time-independent Hamilton--Jacobi equation also connects to Maupertuis' principle of least action, which characterizes true trajectories as geodesics in configuration space with a metric scaled by kinetic energy. The reduced action $W(q)$ satisfies $dW = p_i\,dq_i$, so that integrating $dW$ along a trajectory is equivalent to integrating the momentum one-form. When $H=E$ is held fixed, the paths that extremize $\int p_i\,dq_i$ are the same paths found by solving the time-independent equation for $W$.}
\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.}
\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 $x(t) = (\alpha/m)t + \beta$, and verify that this satisfies the free-particle equation of motion $m\ddot{x} = 0$.
\end{enumerate}}
\sol \textbf{Part (a).} The Hamilton--Jacobi equation is obtained by replacing the canonical momentum $p$ with the partial derivative $\pdv{\mcS}{x}$. Substituting this into the given Hamiltonian gives
\[
\frac{1}{2m}\left(\pdv{\mcS}{x}\right)^2 + V(x) + \pdv{\mcS}{t} = 0.
\]
This is the explicit Hamilton--Jacobi PDE for a one-dimensional particle in potential $V(x)$.
\textbf{Part (b).} For $V(x) = 0$, the Hamilton--Jacobi equation reduces to
\[
\frac{1}{2m}\left(\pdv{\mcS}{x}\right)^2 + \pdv{\mcS}{t} = 0.
\]
Use the separation ansatz $\mcS(x,t) = W(x) + T(t)$. Substituting and using ordinary derivatives gives
\[
\frac{1}{2m}\left(\der{W}{x}\right)^2 + \der{T}{t} = 0.
\]
The first term depends only on $x$ and the second only on $t$, so each must equal a constant. 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 $T(t) = -\alpha^2 t/(2m)$. Therefore,
\[
\mcS(x,t;\alpha) = \alpha x - \frac{\alpha^2}{2m}\,t.
\]
This is the complete integral: it solves the HJ PDE and contains one independent non-additive constant $\alpha$.
\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 $x$:
\[
x - \frac{\alpha}{m}\,t = \beta,
\qquad\text{so}\qquad
x(t) = \frac{\alpha}{m}\,t + \beta.
\]
This is the trajectory. Differentiate once with respect to time to find the velocity:
\[
\dot{x} = \frac{\alpha}{m}.
\]
The velocity is constant, as expected for a free particle. Differentiate a second time:
\[
\ddot{x} = 0.
\]
Multiplying by the mass gives
\[
m\ddot{x} = 0,
\]
which is Newton's second law for a free particle. The trajectory is therefore verified. The constant $\alpha$ has the physical meaning $m\dot{x}$, confirming it is the conserved linear momentum. The constant $\beta$ represents the initial position of the particle at $t=0$, since $x(0) = \beta$. Together, $\alpha$ and $\beta$ form a complete set of independent constants for this one-degree-of-freedom system, providing the full two parameters needed to describe the general solution of the second-order equation of motion.
Therefore,
\[
\text{HJ PDE:}\quad \frac{1}{2m}\left(\pdv{\mcS}{x}\right)^2 + V(x) + \pdv{\mcS}{t} = 0,
\]
\[
\mcS(x,t;\alpha) = \alpha x - \frac{\alpha^2}{2m}\,t,
\qquad
x(t) = \frac{\alpha}{m}\,t + \beta.
\]

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\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's theorem.}
\nt{Structural identity with the gravitational Kepler problem}{
The Coulomb HJ equation is structurally identical to the gravitational Kepler problem. 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. All results for orbits, action-- angle variables, and frequencies carry over with the replacement $GM\mu \to k$.}
\nt{Action-- angle quantization and the hydrogen spectrum}{
For the $1/r$ potential the three action variables are $J_\phi = 2\pi L_z$, $J_\theta = 2\pi(L - |L_z|)$, and $J_r = 2\pi(-L + k\sqrt{\mu/(2|E|)})$. Their sum eliminates the angular-- momentum dependence:
\[
J_{\mathrm{tot}} = J_r + J_\theta + J_\phi = 2\pi k\sqrt{\frac{\mu}{2|E|}}.
\]
Bohr-- Sommerfeld quantization requires $J_{\mathrm{tot}}/2\pi = n\hbar$, 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. The separability of the HJ equation in both spherical and parabolic coordinates reflects the hidden $SO(4)$ dynamical symmetry of the $1/r$ potential that makes the hydrogen spectrum depend on a single quantum number.}
\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 $J_{\mathrm{tot}} = 2\pi L$. 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 $J_{\mathrm{tot}} = n h$ 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
\[
J_{\mathrm{tot}} = 2\pi L = 2\pi\hbar = h = 6.63\times 10^{-34}\,\mathrm{J\!\cdot\!s}.
\]
The total action equals Planck's constant $h$. This is consistent with the Bohr-- Sommerfeld quantization condition $J_{\mathrm{tot}} = n h$ 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
\[
J_{\mathrm{tot}} = n h = n\cdot 2\pi\hbar.
\]
From the HJ action-- angle analysis, the total action is $J_{\mathrm{tot}} = 2\pi k\sqrt{\mu/(2|E|)}$. Equate the two expressions:
\[
2\pi k\sqrt{\frac{\mu}{2|E|}} = 2\pi n\hbar,
\]
\[
k\sqrt{\frac{\mu}{2|E|}} = n\hbar.
\]
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
\[
E = -2.18\times 10^{-18}\,\mathrm{J} = -13.6\,\mathrm{eV},
\qquad
L = 1.055\times 10^{-34}\,\mathrm{J\!\cdot\!s} = \hbar,
\]
\[
J_{\mathrm{tot}} = h = 6.63\times 10^{-34}\,\mathrm{J\!\cdot\!s},
\qquad
E_1 = -\frac{\mu k^2}{2\hbar^2} = -13.6\,\mathrm{eV}.
\]

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\subsection{The Kepler Problem}
This subsection treats the inverse-square central potential $V(r) = -k/r$ through the Hamilton-- Jacobi formalism, derives conic-section orbits from Jacobi's theorem, and uses action-- angle variables to recover Kepler's third law and the degeneracy that makes bound orbits close.
\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. In spherical coordinates $(r,\theta,\phi)$ the kinetic energy is $T = \tfrac{1}{2}\mu(\dot{r}^2 + r^2\dot{\theta}^2 + r^2\sin^2\theta\,\dot{\phi}^2)$, and 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'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 $2\mu r^2$:
\[
r^2\left(\pdv{W_{r\theta}}{r}\right)^2 - \frac{2\mu k r}{1} - 2\mu E r^2
= -\left(\pdv{W_{r\theta}}{\theta}\right)^2 - \frac{L_z^2}{\sin^2\theta}.
\]
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's theorem, these equations determine the trajectory without solving any second-- order differential equation.}
\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:
\[
\ell = \frac{L^2}{\mu k},
\qquad
\varepsilon = \sqrt{1 + \frac{2EL^2}{\mu k^2}},
\qquad
\phi_0 = \text{constant}.
\]
The angle $\phi_0$ fixes the orientation of the conic in the orbital plane.}
\pf{Derivation of the orbit from Jacobi'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'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 in the $z$-direction). Differentiating $W_r$ with respect to $L$:
\[
\pdv{W_r}{L} = \int\frac{1}{2\der{W_r}{r}}\cdot\pdv{(\der{W_r}{r})^2}{L}\dd r
= \int\frac{1}{2p_r}\cdot\left(-\frac{2L}{r^2}\right)\dd r
= -\int\frac{L}{r^2 p_r}\,\dd r.
\]
The key observation is that the same integral appears in the relation between $\phi$ and $r$. From Hamilton's equations or from the $\phi$-- part of Jacobi'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 $T = 2\pi\sqrt{\mu a^3/k}$.
\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 $\omega = \sqrt{k/(\mu r^3)}$.
\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 $2\arccos(-1/\varepsilon)$.
\end{enumerate}}
\nt{Action-- angle variables and Kepler'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, $J_\phi = \oint p_\phi\,\mathrm{d}\phi = 2\pi L_z$. For the polar coordinate, $J_\theta = \oint p_\theta\,\mathrm{d}\theta = 2\pi(L - |L_z|)$. For the radial coordinate, the integral $J_r = \oint p_r\,\dd r$ requires careful evaluation between the two radial turning points for bound orbits ($E < 0$). The result is $J_r = 2\pi(-L + k\sqrt{\mu/(2|E|)})$. Adding all three actions eliminates the angular-- momentum dependence:
\[
J_{\mathrm{tot}} = J_r + J_\theta + J_\phi
= 2\pi k\sqrt{\frac{\mu}{2|E|}}.
\]
Inverting this expression gives $|E| = 2\pi^2\mu k^2/J_{\mathrm{tot}}^2$, 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'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:
\[
J_\phi = \oint p_\phi\,\mathrm{d}\phi = \int_{0}^{2\pi} L_z\,\mathrm{d}\phi = 2\pi L_z.
\]
\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
\[
J_\theta = 2\int_{\theta_{\min}}^{\pi-\theta_{\min}}\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}\;\mathrm{d}\theta.
\]
The substitution $u = \cos\theta$ converts the integrand to $\sqrt{L^2 - L_z^2/(1-u^2)}$, and the integral evaluates to
\[
J_\theta = 2\pi\bigl(L - |L_z|\bigr).
\]
\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
\[
J_r = 2\int_{r_-}^{r_+}\sqrt{2\mu|E| + \frac{2\mu k}{r} - \frac{L^2}{r^2}}\;\frac{\dd r}{r^2/r^2}.
\]
The standard contour-- integration or-- elliptic-- integral evaluation gives
\[
J_r = 2\pi\!\left(-L + k\sqrt{\frac{\mu}{2|E|}}\right).
\]
\textbf{Total action and energy.} Adding the three actions:
\[
J_{\mathrm{tot}} = J_r + J_\theta + J_\phi
= 2\pi\!\left(-L + k\sqrt{\frac{\mu}{2|E|}}\right) + 2\pi(L - |L_z|) + 2\pi L_z
= 2\pi k\sqrt{\frac{\mu}{2|E|}}.
\]
The angular-- momentum terms $L$ and $|L_z|$ cancel exactly. Invert the total-- action relation to obtain the energy:
\[
\sqrt{\frac{\mu}{2|E|}} = \frac{J_{\mathrm{tot}}}{2\pi k},
\qquad
\frac{2|E|}{\mu} = \frac{4\pi^2 k^2}{J_{\mathrm{tot}}^2},
\]
\[
E(J_{\mathrm{tot}}) = -\frac{2\pi^2\mu k^2}{J_{\mathrm{tot}}^2}.
\]
\textbf{Degenerate frequencies.} The three action-- angle frequencies are
\[
\omega_r = \pdv{E}{J_r} = \pdv{E}{J_{\mathrm{tot}}}\pdv{J_{\mathrm{tot}}}{J_r}
= \frac{4\pi^2\mu k^2}{J_{\mathrm{tot}}^3}\cdot 1,
\]
\[
\omega_\theta = \pdv{E}{J_\theta} = \pdv{E}{J_{\mathrm{tot}}}\pdv{J_{\mathrm{tot}}}{J_\theta}
= \frac{4\pi^2\mu k^2}{J_{\mathrm{tot}}^3}\cdot 1,
\]
\[
\omega_\phi = \pdv{E}{J_\phi} = \pdv{E}{J_{\mathrm{tot}}}\pdv{J_{\mathrm{tot}}}{J_\phi}
= \frac{4\pi^2\mu k^2}{J_{\mathrm{tot}}^3}\cdot 1.
\]
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 $2\pi$, so the orbit closes after exactly one revolution. This is Kepler's third law: the orbital period is determined solely by the energy and is independent of the angular momentum.}
\nt{Comparison with the Binet equation}{
The Binet equation is an alternative derivation of Kepler orbits that begins with Newton's second law and the substitution $u(\phi) = 1/r(\phi)$. The radial equation of motion in polar coordinates is
\[
\mu(\ddot{r} - r\dot{\phi}^2) = -\frac{k}{r^2}.
\]
With the angular momentum $L = \mu r^2\dot{\phi}$, eliminate $\dot{\phi}$ in favor of $L$:
\[
\ddot{r} - \frac{L^2}{\mu^2 r^3} = -\frac{k}{\mu r^2}.
\]
Write $r = 1/u(\phi)$ and use the chain rule to convert time derivatives into $\phi$ derivatives. Since $\dot{\phi} = L/(\mu r^2) = L u^2/\mu$:
\[
\dot{r} = \dv{r}{\phi}\dot{\phi} = -\dv{u}{\phi}\cdot\frac{1}{u^2}\cdot\frac{L u^2}{\mu}
= -\frac{L}{\mu}\dv{u}{\phi}.
\]
Differentiate once more:
\[
\ddot{r} = \dv{}{t}\!\left(-\frac{L}{\mu}\dv{u}{\phi}\right)
= -\frac{L}{\mu}\dv[2]{u}{\phi}\dot{\phi}
= -\frac{L^2}{\mu^2}\dv[2]{u}{\phi}.
\]
Substitute into the radial equation:
\[
-\frac{L^2}{\mu^2}\dv[2]{u}{\phi} - \frac{L^2}{\mu^2}\,u = -\frac{k}{\mu},
\]
which rearranges to the Binet equation:
\[
\dv[2]{u}{\phi} + u = \frac{\mu k}{L^2}.
\]
The general solution is
\[
u(\phi) = \frac{\mu k}{L^2} + A\cos(\phi - \phi_0),
\]
where $A$ is determined by the energy. Inverting $r = 1/u$ gives
\[
r(\phi) = \frac{1}{\mu k/L^2 + A\cos(\phi - \phi_0)}
= \frac{L^2/(\mu k)}{1 + \mu k A/L^2\cos(\phi - \phi_0)},
\]
which matches the form $\ell/(1 + \varepsilon\cos(\phi - \phi_0))$ with $\ell = L^2/(\mu k)$ and $\varepsilon = \mu k A/L^2$. This derivation is shorter but requires knowing the $u = 1/r$ substitution. The Hamilton-- Jacobi approach reaches the same result through quadratures without any clever change of variable, demonstrating the power of Jacobi's theorem as a unifying principle. Moreover, the action-- angle formalism provides immediate access to the energy-- period-- semimajor axis relations that the Binet equation leaves as an afterthought. The Binet method also cannot be extended to noncentral potentials or to higher dimensions without substantial modification, while the Hamilton-- Jacobi approach generalizes naturally to any separable system.}
\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'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'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,
\[
k = 7.94\times 10^{44}\,\mathrm{J\!\cdot\!m},
\qquad
E = -2.65\times 10^{33}\,\mathrm{J},
\]
\[
T = 3.16\times 10^7\,\mathrm{s} \approx 1\,\text{year},
\qquad
v = 29.8\,\mathrm{km/s}.
\]

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\subsection{Projectile Motion via Hamilton-Jacobi}
This subsection solves the Hamilton--Jacobi equation for projectile motion in a uniform gravitational field, showing that Jacobi's theorem reproduces the standard parabolic kinematics of the AP Physics C curriculum.
\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.}
\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):
\[
x(t) = \frac{\alpha_x}{m}\,t = v_{0x}\,t.
\]
From Jacobi's theorem with respect to $E$, using $\pdv{E_y}{E} = 1$:
\[
\pdv{\mcS}{E} = \mp\frac{2\sqrt{2m}}{3mg}\cdot\frac{3}{2}\left(E_y - mgy\right)^{1/2} - t = \beta_E,
\]
which simplifies to
\[
\mp\frac{\sqrt{2m}}{mg}\sqrt{E_y - mgy} - t = \beta_E.
\]
Solve the squared relation for $y(t)$. At $t = 0$ the particle is at $y = 0$ with vertical speed $v_{0y} = \sqrt{2E_y/m}$. The initial conditions fix $\beta_E$ and yield
\[
y(t) = v_{0y}\,t - \frac{1}{2}\,g\,t^2.
\]
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 standard AP C constant-acceleration kinematics. 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. The HJ approach thus 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,
\[
\alpha_x = 8.66\,\mathrm{kg\!\cdot\!m/s},
\qquad
E_y = 25\,\mathrm{J},
\qquad
R = 35\,\mathrm{m}.
\]

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

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\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.
\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.}
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.}
\nt{The additive separation theorem gives a sufficient condition for separability. A more general theory was developed by Levi-Civita and later refined by Stackel. The Levi-Civita separability conditions state that the Hamilton--Jacobi equation is separable in a given orthogonal coordinate system if and only if the Hamiltonian can be written as a sum, each term depending on only one coordinate and its conjugate momentum. Equivalently, the coefficient matrix of the quadratic kinetic-energy form, when written in these coordinates, must be a Stackel matrix. A Stackel matrix $S_{ij}$ is an $n\times n$ matrix whose $(i,j)$ entry depends only on the single coordinate $q_i$, and whose determinant is nonzero almost everywhere. The inverse of the Stackel matrix then relates the separation constants to the Hamiltonian components, producing the $n$ separated ODEs.}
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.
\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.}
\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,
\[
p_x = 1.77\,\mathrm{kg\!\cdot\!m/s},
\qquad
E_y = 15.6\,\mathrm{J}.
\]
}

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\subsection{Simple Harmonic Oscillator}
This subsection solves the simple harmonic oscillator through the Hamilton--Jacobi equation, obtains the complete integral and trajectory by quadrature, and computes the action-angle variables that confirm isochronous oscillation.
\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's theorem.}
\nt{The simple harmonic oscillator is one of the few nonlinear Hamilton--Jacobi equations that can be integrated in closed form. The quadratic potential turns the square-root integral into an elementary trigonometric substitution, and the resulting complete integral yields the standard sinusoidal trajectory. This integrability makes the harmonic oscillator the prototypical example for testing both the Hamilton--Jacobi method and the action-angle formalism.}
\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.
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's theorem}{
Jacobi'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 reflects the fact that the energy dependence of the amplitude and the energy dependence of the integrand conspire to leave only the angular part. 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 $x(t) = A\cos(\omega_0 t)$.}
\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 the phase-space area enclosed by one complete cycle:
\[
J = \oint p\,\dd x = 2\int_{-A}^{A}\sqrt{2mE - m^2\omega_0^2 x^2}\,\dd x.
\]
With the substitution $x = A\sin\phi$, the integral reduces to $J = \pi\sqrt{2mE}\cdot A$. Using $A = \sqrt{2E/(m\omega_0^2)}$ this becomes $J = 2\pi E/\omega_0$. Geometrically, $J$ is the area of the elliptical orbit in the $(x,p)$ phase plane.
\item Inverting the action relation, the Hamiltonian as a function of the action alone is
\[
E(J) = \frac{\omega_0 J}{2\pi}.
\]
The Hamiltonian is now linear in $J$, which is the defining feature of an action-angle representation.
\item The HJ frequency is $\hat{\omega} = \pdv{E}{J} = \omega_0/(2\pi)$. The physical angular frequency is $\omega = 2\pi\hat{\omega} = \omega_0$, which is independent of the action $J$. 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.
\item The angle variable advances linearly in time: $w = \hat{\omega}t + w_0 = \omega_0 t/(2\pi) + w_0$. The phase of the sinusoidal trajectory, $\omega_0 t + \phi$, equals $2\pi w$ up to a constant, matching the canonical construction.
\end{enumerate}
}
\nt{Comparison with Newton's law and energy conservation}{
Newton'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's theorem. The agreement confirms the equivalence of the three formulations -- Newton's, Lagrange's, and Jacobi'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 $J = 2\pi E/\omega_0$ in SI units and verify numerically that $E(J) = \omega_0 J/(2\pi)$ 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
\[
J = \frac{2\pi E}{\omega_0}.
\]
Substitute the numerical values:
\[
J = \frac{2\pi(8.0\,\mathrm{J})}{2.0\,\mathrm{rad/s}}
= 8\pi\,\mathrm{J\!\cdot\!s}.
\]
Evaluating numerically:
\[
J = 8\pi\,\mathrm{J\!\cdot\!s} \approx 25\,\mathrm{J\!\cdot\!s}.
\]
Now verify the energy--action relation $E(J) = \omega_0 J/(2\pi)$:
\[
E(J) = \frac{\omega_0 J}{2\pi}
= \frac{(2.0\,\mathrm{rad/s})(8\pi\,\mathrm{J\!\cdot\!s})}{2\pi}
= 8.0\,\mathrm{J}.
\]
This returns the original energy exactly, confirming $E(J) = \omega_0 J/(2\pi)$ both algebraically and for the numerical values of this problem.
Therefore,
\[
\omega_0 = 2.0\,\mathrm{rad/s},
\qquad
x(t) = (2.0\,\mathrm{m})\cos\bigl((2.0\,\mathrm{rad/s})\,t\bigr),
\qquad
J = 8\pi\,\mathrm{J\!\cdot\!s}.
\]

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\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 $\vec{F} = q\vec{E}$.
\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,
\[
z(1.0\,\mathrm{ns}) = -8.8\times 10^{-5}\,\mathrm{m},
\qquad
K = 1.4\times 10^{-20}\,\mathrm{J} = 0.088\,\mathrm{eV}.
\]

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13
macros.tex
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@@ -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}}