fix(HJ): A.11 — Verify ∇×A, explain gauge choice, canonical vs kinetic momentum, cross-ref e12-2

concepts/advanced/cyclotron-hj.tex

@@ -3,7 +3,13 @@
 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
+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
 \[
 \mcH = \frac{1}{2m}\Bigl[p_x^2 + \bigl(p_y - q B_0 x\bigr)^2 + p_z^2\Bigr].
 \]
@@ -12,7 +18,11 @@ In this gauge, the coordinate $x$ appears explicitly in $\mcH$ while $y$ and $z$
 \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.}
+\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
@@ -128,7 +138,7 @@ or equivalently,
 \[
 x(t) = X_c + R\sin\!\bigl(\omega_c(t + \beta)\bigr).
 \]
-Define the initial phase $\phi_0 = \omega_c\beta$.
+Define the initial phase $\phi_0 = \omega_c\beta$, measured in radians.
 
 Differentiate $\mcS$ with respect to the separation constant $\alpha_y$:
 \[
@@ -154,6 +164,26 @@ z(t) = v_z t + z_0.
 \]
 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:
 
@@ -179,12 +209,12 @@ The Hamiltonian expressed in terms of the action variables is $E = \omega_c J/(2
 
 \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.
+\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, $\omega_c t + \phi_0$ (all angles in radians), 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
+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):
 \[
 \ddot{x} = \frac{q B_0}{m}\,\dot{y},
 \qquad
@@ -308,4 +338,4 @@ T = 44\,\mathrm{ns},
 R = 3.1\,\mathrm{mm},
 \qquad
 J = 7.0\times 10^{-24}\,\mathrm{J\!\cdot\!s}.
-\]
+\]