\subsection{Cyclotron Motion}

This subsection solves for a charged particle moving in a uniform magnetic field through the Hamilton--Jacobi equation, derives the helical trajectory by quadrature, and computes the action-angle variables that recover the cyclotron frequency.

\dfn{Hamilton--Jacobi formulation of cyclotron motion}{
A particle of mass $m$ and charge $q$ moves in a uniform magnetic field $\vec{B} = B_0\,\hat{\bm{z}}$. Choose the Landau gauge for the vector potential $\vec{A} = (0, B_0 x, 0)$. The curl verifies the field:
\[
\nabla\times(0, B_0 x, 0)
= \left(0,\; 0,\; \pdv{(B_0 x)}{x}\right)
= B_0\,\hat{\bm{z}}.
\]
Set the scalar potential $\varphi = 0$. The Hamiltonian is
\[
\mcH = \frac{1}{2m}\Bigl[p_x^2 + \bigl(p_y - q B_0 x\bigr)^2 + p_z^2\Bigr].
\]
In this gauge, the coordinate $x$ appears explicitly in $\mcH$ while $y$ and $z$ are absent, so $y$ and $z$ are cyclic: their conjugate momenta $p_y$ and $p_z$ are constants of the motion. The Hamilton--Jacobi equation for $\mcS(\vec{r},t)$ reads
\[
\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \bigl(\pdv{\mcS}{y} - q B_0 x\bigr)^2 + \left(\pdv{\mcS}{z}\right)^2\right] + \pdv{\mcS}{t} = 0.
\]}

\nt{Why Landau gauge?}{
The Landau gauge $\vec{A} = (0, B_0 x, 0)$ deliberately breaks rotational symmetry in the $xy$-plane to make $y$ a cyclic coordinate. With $y$ cyclic, $p_y$ is conserved and the HJ equation separates. In the symmetric gauge $\vec{A} = \tfrac{B_0}{2}(-y, x, 0)$, neither $x$ nor $y$ is cyclic -- the Hamiltonian depends on both, preventing straightforward HJ separation. Even after rotating to $x', y'$ coordinates, the $y'$-coordinate is \textbf{not} cyclic in the symmetric gauge. Both gauges describe the same physics, but only the Landau gauge unlocks the separation ansatz $\mcS = W_x(x) + \alpha_y y + \alpha_z z - Et$.}

\nt{Two momenta}{
In a magnetic field two distinct notions of momentum appear. The \textbf{canonical momentum} $p_y = \pdv{\mcS}{y} = \alpha_y$ is conserved because $y$ is cyclic in the Landau gauge. The \textbf{kinetic momentum} $m v_y = p_y - q B_0 x = \alpha_y - q B_0 x$ is not conserved -- it rotates at the cyclotron frequency as $x(t)$ oscillates. Conservation of $p_y$ fixes the orbit guiding center at $X_c = \alpha_y/(q B_0)$, while the rotating kinetic momentum generates the circular gyration about that center.}

\thm{Complete integral for cyclotron motion}{
The cyclotron frequency is $\omega_c = q B_0/m$. The guiding-center $x$-coordinate is $X_c = \alpha_y/(q B_0)$ and the gyroradius is $R = \sqrt{2m E_\perp}/(q B_0)$, where $\alpha_y$ is the conserved canonical $y$-momentum and $E_\perp$ is the transverse energy. The complete integral of the Hamilton--Jacobi equation is
\[
\mcS = W_x(x) + \alpha_y y + \alpha_z z - Et,
\]
where the $x$-part of the characteristic function is
\[
W_x(x) = \frac{E_\perp}{\omega_c}\arcsin\!\left(\frac{x - X_c}{R}\right)
+ \frac{q B_0}{2}\,(x - X_c)\sqrt{R^2 - (x - X_c)^2}.
\]
Here $E_\perp = E - \alpha_z^2/(2m)$ is the energy of motion in the $xy$-plane alone. The complete integral is defined for $|x - X_c| < R$.}

\pf{Separation of the HJ equation and integration of the x-dependent part}{
Because the potentials are time-independent, separate as $\mcS(\vec{r},t) = W(\vec{r}) - Et$. Because $y$ and $z$ are cyclic coordinates, set $\pdv{\mcS}{y} = \alpha_y$ and $\pdv{\mcS}{z} = \alpha_z$. The remaining dependence on $x$ is carried by a single function $W_x(x)$, so $W = W_x(x) + \alpha_y y + \alpha_z z$ and the time-independent Hamilton--Jacobi equation reads
\[
\frac{1}{2m}\left[\left(\der{W_x}{x}\right)^2 + \bigl(\alpha_y - q B_0 x\bigr)^2 + \alpha_z^2\right] = E.
\]
Define the transverse energy $E_\perp = E - \alpha_z^2/(2m)$. The $x$-equation simplifies to
\[
\left(\der{W_x}{x}\right)^2 + \bigl(\alpha_y - q B_0 x\bigr)^2 = 2m E_\perp.
\]
Solve for the spatial derivative:
\[
\der{W_x}{x} = \pm\sqrt{2m E_\perp - \bigl(\alpha_y - q B_0 x\bigr)^2}.
\]
The square root is real when $\abs{\alpha_y - q B_0 x} \le \sqrt{2m E_\perp}$. Introduce the guiding-center coordinate
\[
X_c = \frac{\alpha_y}{q B_0}.
\]
Then $\alpha_y - q B_0 x = -q B_0(x - X_c)$, and the radicand factors as
\[
2m E_\perp - q^2 B_0^2(x - X_c)^2
= q^2 B_0^2\left(\frac{2m E_\perp}{q^2 B_0^2} - (x - X_c)^2\right).
\]
Define the gyroradius $R = \sqrt{2m E_\perp}/(q B_0)$. The derivative of $W_x$ becomes
\[
\der{W_x}{x} = \pm q B_0\sqrt{R^2 - (x - X_c)^2}.
\]
This has the same square-root structure as the simple harmonic oscillator. Integrate by the trigonometric substitution $x - X_c = R\sin\theta$, giving $\dd x = R\cos\theta\,\mathrm{d}\theta$:
\[
W_x = \int q B_0\sqrt{R^2 - R^2\sin^2\theta}\cdot R\cos\theta\,\mathrm{d}\theta
= q B_0 R^2\int \cos^2\theta\,\mathrm{d}\theta.
\]
The antiderivative of $\cos^2\theta$ is $\tfrac12(\theta + \sin\theta\cos\theta)$, so
\[
W_x = \frac{q B_0 R^2}{2}\bigl(\theta + \sin\theta\cos\theta\bigr).
\]
Evaluate the constant prefactor using $R^2 = 2m E_\perp/(q^2 B_0^2)$:
\[
\frac{q B_0 R^2}{2}
= \frac{q B_0}{2}\cdot\frac{2m E_\perp}{q^2 B_0^2}
= \frac{m E_\perp}{q B_0}
= \frac{E_\perp}{\omega_c}.
\]
Now express the trigonometric factors in terms of $x$:
\[
\theta = \arcsin\!\left(\frac{x - X_c}{R}\right),
\qquad
\sin\theta = \frac{x - X_c}{R},
\qquad
\cos\theta = \frac{\sqrt{R^2 - (x - X_c)^2}}{R}.
\]
The product $\sin\theta\cos\theta$ is $(x - X_c)\sqrt{R^2 - (x - X_c)^2}/R^2$. Substituting back,
\[
W_x(x) = \frac{E_\perp}{\omega_c}\arcsin\!\left(\frac{x - X_c}{R}\right)
+ \frac{E_\perp}{\omega_c}\cdot\frac{(x - X_c)\sqrt{R^2 - (x - X_c)^2}}{R^2}.
\]
The coefficient of the second term simplifies as
\[
\frac{E_\perp}{\omega_c R^2}
= \frac{E_\perp}{\omega_c}\cdot\frac{q^2 B_0^2}{2m E_\perp}
= \frac{q^2 B_0^2}{2m\omega_c}
= \frac{q B_0}{2},
\]
since $\omega_c = q B_0/m$. Therefore,
\[
W_x(x) = \frac{E_\perp}{\omega_c}\arcsin\!\left(\frac{x - X_c}{R}\right)
+ \frac{q B_0}{2}\,(x - X_c)\sqrt{R^2 - (x - X_c)^2}.
\]
The full complete integral is $\mcS = W_x(x) + \alpha_y y + \alpha_z z - Et$.}

\cor{Helical trajectory from Jacobi's theorem}{
Jacobi's theorem states that differentiating the complete integral with respect to each separation constant produces a constant fixed by the initial conditions. Differentiate $\mcS$ with respect to $E_\perp$ at fixed $x$:
\[
\pdv{\mcS}{E_\perp} = \pdv{W_x}{E_\perp} - t.
\]
Write $\chi = (x - X_c)/R$ and $U = \sqrt{R^2 - (x - X_c)^2}$. The partial derivative of $W_x$ with respect to $E_\perp$ is
\[
\pdv{W_x}{E_\perp} = \frac{1}{\omega_c}\arcsin\chi 
+ \chi\cos\chi\cdot\pdv{E_\perp/\omega_c}{E_\perp}\cdot\frac{1}{(E_\perp/\omega_c)}
+ \frac{q B_0}{2}(x - X_c)\cdot\frac{1}{2U}\pdv{R^2}{E_\perp}.
\]
Because $R^2 = 2m E_\perp/(q^2 B_0^2)$, one has $\pdv{R^2}{E_\perp} = 2m/(q^2 B_0^2)$. The last two terms are
\[
-\frac{1}{2\omega_c}\frac{\chi}{\sqrt{1-\chi^2}}
= -\frac{x - X_c}{2\omega_c R\sqrt{1-\chi^2}},
\]
\[
\frac{q B_0}{2}(x - X_c)\cdot\frac{m}{q^2 B_0^2 U}
= \frac{m}{q B_0}\cdot\frac{x - X_c}{2U}
= \frac{x - X_c}{2\omega_c R\sqrt{1-\chi^2}},
\]
which exactly cancel, as in the harmonic oscillator case. Hence,
\[
\pdv{W_x}{E_\perp} = \frac{1}{\omega_c}\arcsin\!\left(\frac{x - X_c}{R}\right).
\]
Set $\pdv{\mcS}{E_\perp} = \beta$ (constant):
\[
\frac{1}{\omega_c}\arcsin\!\left(\frac{x - X_c}{R}\right) - t = \beta,
\]
or equivalently,
\[
x(t) = X_c + R\sin\!\bigl(\omega_c(t + \beta)\bigr).
\]
Define the initial phase $\phi_0 = \omega_c\beta$, measured in radians.

Differentiate $\mcS$ with respect to the separation constant $\alpha_y$:
\[
\pdv{\mcS}{\alpha_y} = \pdv{W_x}{\alpha_y} + y = \beta_y.
\]
Since $X_c = \alpha_y/(q B_0)$, the chain rule gives $\pdv{W_x}{\alpha_y} = -\pdv{W_x}{X_c}\cdot(1/q B_0)$. From the structure of $W_x$, this derivative evaluates to $-(x - X_c)/R\cdot(E_\perp/(\omega_c R)) - \tfrac12\sqrt{R^2 - (x - X_c)^2}\cdot(q B_0/q B_0)$, but the result is more easily found from the canonical relation $v_y = (\alpha_y - q B_0 x)/m$:
\[
v_y(t) = \frac{\alpha_y - q B_0 x(t)}{m}
= \frac{q B_0\bigl(X_c - x(t)\bigr)}{m}
= -\omega_c R\sin(\omega_c t + \phi_0).
\]
Integrating with respect to time,
\[
y(t) = Y_c + R\cos(\omega_c t + \phi_0),
\]
where $Y_c$ is an integration constant set by the initial conditions. Meanwhile, for the $z$-direction, $p_z = \alpha_z$ is constant, giving $z(t) = (\alpha_z/m)t + z_0 = v_z t + z_0$. The full trajectory is helical:
\[
x(t) = X_c + R\sin(\omega_c t + \phi_0),
\qquad
y(t) = Y_c + R\cos(\omega_c t + \phi_0),
\qquad
z(t) = v_z t + z_0.
\]
The projection onto the $xy$-plane is a circle of radius $R$ centered at $(X_c, Y_c)$, traversed at the constant angular speed $\omega_c$. Superimposed is uniform motion along the field direction at speed $v_z$.}

\nt{Hamilton's equations shortcut for $y(t)$}{
The same $y(t)$ result follows directly from Hamilton's equations without differentiating $\mcS$. Because $y$ is cyclic,
\[
\dot{p}_y = -\pdv{\mcH}{y} = 0,
\]
so $p_y = \alpha_y$ is constant. Hamilton's velocity equation gives
\[
\dot{y} = \pdv{\mcH}{p_y} = \frac{1}{m}\bigl(p_y - q B_0 x\bigr)
= \frac{\alpha_y - q B_0 x(t)}{m}.
\]
Substituting $x(t) = X_c + R\sin(\omega_c t + \phi_0)$ with $X_c = \alpha_y/(q B_0)$:
\[
\dot{y} = \frac{q B_0(X_c - X_c - R\sin(\omega_c t + \phi_0))}{m}
= -\omega_c R\sin(\omega_c t + \phi_0).
\]
Integrating once yields $y(t) = Y_c + R\cos(\omega_c t + \phi_0)$, matching the HJ-derived trajectory. All angle arguments are in radians.}

\nt{Guiding-center independence}{
The guiding center $X_c = \alpha_y/(q B_0)$ depends only on the conserved canonical momentum $\alpha_y$, not on the transverse energy $E_\perp$. Physically, increasing $E_\perp$ enlarges the gyroradius $R = \sqrt{2m E_\perp}/(q B_0)$ but does not shift the orbit center. Energy changes the orbit size, not the center. This reflects the fact that the magnetic force is always perpendicular to velocity: it does no work, cannot change the particle's speed, and merely redirects the motion into circles whose centers are determined by the initial momentum partition, not the total energy.}

\mprop{Action-angle variables for cyclotron motion}{
The action-angle formalism applied to cyclotron motion yields the following results:

\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\item The action variable associated with the transverse motion, defined as one-over-$2\pi$ times the phase-space area enclosed by one gyration, is
\[
J = \frac{1}{2\pi}\oint p_x\,\dd x = \frac{1}{\pi}\int_{X_c - R}^{X_c + R} q B_0\sqrt{R^2 - (x - X_c)^2}\,\dd x.
\]
The integral is the area of a semicircle of radius $R$ multiplied by $q B_0$. With the $1/\pi$ factor:
\[
J = \frac{q B_0}{\pi}\cdot\frac{\pi R^2}{2}
= \frac{q B_0 R^2}{2}
= \frac{q B_0}{2}\cdot\frac{2m E_\perp}{q^2 B_0^2}
= \frac{m E_\perp}{q B_0}
= \frac{E_\perp}{\omega_c}.
\]
Geometrically, the full phase-space area $2\pi J/(q B_0) = \pi R^2$ is the area of the real-space circular orbit.

\item Inverting the action--energy relation, the transverse energy as a function of the action is
\[
E_\perp(J) = \omega_c J.
\]
The Hamiltonian expressed in terms of the action variables is $E = \omega_c J + \alpha_z^2/(2m)$, linear in $J$ and quadratic in $\alpha_z$.

\item The Hamilton--Jacobi frequency is $\pdv{E_\perp}{J} = \omega_c$, which already has units of angular frequency (rad/s). It depends only on the charge-to-mass ratio and the field strength, and is independent of the transverse energy $E_\perp$ and the gyroradius $R$. This amplitude independence is the hallmark of uniform circular motion in a constant magnetic field.

\item The angle variable advances linearly in time: $w = \omega_c t + w_0$. The phase of the circular gyration, $\omega_c t + \phi_0$ (all angles in radians), equals $w$ up to a constant phase. The action-angle variables provide a clean canonical description even though the original Cartesian coordinates exhibit coupled oscillatory dynamics.
\end{enumerate}
}

\nt{Comparison with the Lorentz force}{
The Lorentz force law $m\,\ddot{\vec{r}} = q\,\dot{\vec{r}}\times\vec{B}$ with $\vec{B} = B_0\hat{\bm{z}}$ gives the component equations (see also e12-2 for cyclotron motion from Lorentz force):
\[
\ddot{x} = \frac{q B_0}{m}\,\dot{y},
\qquad
\ddot{y} = -\frac{q B_0}{m}\,\dot{x},
\qquad
\ddot{z} = 0.
\]
Introducing $v_x = \dot{x}$ and $v_y = \dot{y}$, these become $\dot{v}_x = \omega_c v_y$ and $\dot{v}_y = -\omega_c v_x$, whose solutions are
\[
v_x = v_\perp\cos(\omega_c t + \phi_0),
\qquad
v_y = -v_\perp\sin(\omega_c t + \phi_0).
\]
Integrating once more gives the same circular trajectory with radius $R = v_\perp/\omega_c = \sqrt{2m E_\perp}/(q B_0)$, and uniform $z$-motion. The Hamilton--Jacobi approach arrives at the identical values of $\omega_c$ and $R$ through a radically different route: solving a first-order nonlinear PDE by separation and quadrature, then differentiating the complete integral. The agreement reaffirms the consistency of the Hamiltonian and Newtonian formulations.}

\qs{Proton cyclotron motion from the HJ complete integral}{
A proton of mass $m = 1.67\times 10^{-27}\,\mathrm{kg}$ and charge $q = e = 1.60\times 10^{-19}\,\mathrm{C}$ moves in a uniform magnetic field $\vec{B} = (1.5\,\mathrm{T})\,\hat{\bm{z}}$. The transverse (perpendicular-to-field) kinetic energy is $E_\perp = 1.0\,\mathrm{keV} = 1.60\times 10^{-16}\,\mathrm{J}$.

\begin{enumerate}[label=(\alph*)]
\item Compute the cyclotron angular frequency $\omega_c = q B_0/m$ and the gyration period $T = 2\pi/\omega_c$.
\item Find the gyroradius $R = \sqrt{2m E_\perp}/(q B_0)$ in meters.
\item Compute the action variable $J = E_\perp/\omega_c$ in SI units and verify by differentiation that $\pdv{E_\perp}{J} = \omega_c$, recovering the cyclotron angular frequency.
\end{enumerate}}

\sol \textbf{Part (a).} The cyclotron angular frequency is
\[
\omega_c = \frac{q B_0}{m}.
\]
Substitute the given values:
\[
q = 1.60\times 10^{-19}\,\mathrm{C},
\qquad
B_0 = 1.5\,\mathrm{T},
\qquad
m = 1.67\times 10^{-27}\,\mathrm{kg}.
\]
Form the ratio:
\[
\omega_c = \frac{(1.60\times 10^{-19})(1.5)}{1.67\times 10^{-27}}\,\mathrm{rad/s}
= \frac{2.40\times 10^{-19}}{1.67\times 10^{-27}}\,\mathrm{rad/s}.
\]
This gives
\[
\omega_c = 1.437\times 10^8\,\mathrm{rad/s}.
\]
Rounding to two significant figures (consistent with the field strength $1.5\,\mathrm{T}$),
\[
\omega_c = 1.4\times 10^8\,\mathrm{rad/s}.
\]
The gyration period is
\[
T = \frac{2\pi}{\omega_c}
= \frac{2\pi}{1.437\times 10^8}\,\mathrm{s}
= 4.37\times 10^{-8}\,\mathrm{s}.
\]
In more convenient units,
\[
T = 4.4\times 10^{-8}\,\mathrm{s} = 44\,\mathrm{ns}.
\]

\textbf{Part (b).} The gyroradius is
\[
R = \frac{\sqrt{2m E_\perp}}{q B_0}.
\]
Evaluate the numerator:
\[
2m E_\perp = 2(1.67\times 10^{-27}\,\mathrm{kg})(1.60\times 10^{-16}\,\mathrm{J})
= 5.34\times 10^{-43}\,\mathrm{kg^2\,m^2/s^2}.
\]
Taking the square root:
\[
\sqrt{2m E_\perp} = \sqrt{5.34\times 10^{-43}}\,\mathrm{kg\,m/s}
= 7.31\times 10^{-22}\,\mathrm{kg\,m/s}.
\]
The denominator is
\[
q B_0 = (1.60\times 10^{-19}\,\mathrm{C})(1.5\,\mathrm{T})
= 2.40\times 10^{-19}\,\mathrm{C\,T}.
\]
Therefore,
\[
R = \frac{7.31\times 10^{-22}}{2.40\times 10^{-19}}\,\mathrm{m}
= 3.05\times 10^{-3}\,\mathrm{m}.
\]
In more convenient units,
\[
R = 3.05\,\mathrm{mm}.
\]

\textbf{Part (c).} The action variable for the transverse cyclotron motion is
\[
J = \frac{E_\perp}{\omega_c}.
\]
Substitute the numerical values:
\[
J = \frac{1.60\times 10^{-16}\,\mathrm{J}}{1.437\times 10^8\,\mathrm{rad/s}}
= 1.11\times 10^{-24}\,\mathrm{J\!\cdot\!s}.
\]
This gives
\[
J = 1.1\times 10^{-24}\,\mathrm{J\!\cdot\!s}.
\]
Now verify the energy--action relation $E_\perp(J) = \omega_c J$. Differentiating with respect to $J$:
\[
\pdv{E_\perp}{J} = \omega_c.
\]
The derivative directly equals the cyclotron angular frequency. For the numerical values,
\[
E_\perp(J) = \omega_c J
= (1.437\times 10^8\,\mathrm{rad/s})(1.11\times 10^{-24}\,\mathrm{J\!\cdot\!s})
= 1.60\times 10^{-16}\,\mathrm{J},
\]
which is exactly the original transverse energy of $1.0\,\mathrm{keV}$. The relation $\pdv{E_\perp}{J} = \omega_c$ holds both algebraically and numerically, confirming that the action-angle formalism reproduces the cyclotron angular frequency exactly.

Therefore,
\[
\omega_c = 1.4\times 10^8\,\mathrm{rad/s},
\qquad
T = 44\,\mathrm{ns},
\qquad
R = 3.1\,\mathrm{mm},
\qquad
J = 1.1\times 10^{-24}\,\mathrm{J\!\cdot\!s}.
\]