From 2c04812ff6f1c2fc9c60ee56b898aa73365e3715 Mon Sep 17 00:00:00 2001 From: Krishna Ayyalasomayajula Date: Sat, 2 May 2026 13:07:39 -0500 Subject: [PATCH] =?UTF-8?q?fix(HJ):=20A.11=20=E2=80=94=20Radian=20conventi?= =?UTF-8?q?on=20fix=20for=20action=20variable=20J?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- concepts/advanced/cyclotron-hj.tex | 49 +++++++++++++++--------------- 1 file changed, 25 insertions(+), 24 deletions(-) diff --git a/concepts/advanced/cyclotron-hj.tex b/concepts/advanced/cyclotron-hj.tex index fb2b432..2fbd5d4 100644 --- a/concepts/advanced/cyclotron-hj.tex +++ b/concepts/advanced/cyclotron-hj.tex @@ -188,28 +188,29 @@ The guiding center $X_c = \alpha_y/(q B_0)$ depends only on the conserved canoni 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: +\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 \[ -J = \oint p_x\,\dd x = 2\int_{X_c - R}^{X_c + R} q B_0\sqrt{R^2 - (x - X_c)^2}\,\dd x. +J = \frac{1}{2\pi}\oint p_x\,\dd x = \frac{1}{\pi}\int_{X_c - R}^{X_c + R} q B_0\sqrt{R^2 - (x - X_c)^2}\,\dd x. \] -The integral is twice the area of a semicircle of radius $R$ (multiplied by $q B_0$), so +The integral is the area of a semicircle of radius $R$ multiplied by $q B_0$. With the $1/\pi$ factor: \[ -J = q B_0 \cdot \pi R^2 -= q B_0 \cdot \pi\cdot\frac{2m E_\perp}{q^2 B_0^2} -= \frac{2\pi m E_\perp}{q B_0} -= \frac{2\pi E_\perp}{\omega_c}. +J = \frac{q B_0}{\pi}\cdot\frac{\pi R^2}{2} += \frac{q B_0 R^2}{2} += \frac{q B_0}{2}\cdot\frac{2m E_\perp}{q^2 B_0^2} += \frac{m E_\perp}{q B_0} += \frac{E_\perp}{\omega_c}. \] -Geometrically, $J/(q B_0)$ is the area of the circular orbit in the $xy$-plane. +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 \[ -E_\perp(J) = \frac{\omega_c J}{2\pi}. +E_\perp(J) = \omega_c J. \] -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 $\alpha_z$. +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 $\hat{\omega} = \pdv{E_\perp}{J} = \omega_c/(2\pi)$. 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 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: $w = \hat{\omega}t + w_0 = (\omega_c/2\pi)t + w_0$. 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. +\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} } @@ -236,7 +237,7 @@ A proton of mass $m = 1.67\times 10^{-27}\,\mathrm{kg}$ and charge $q = e = 1.60 \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. +\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 @@ -306,28 +307,28 @@ R = 3.05\,\mathrm{mm}. \textbf{Part (c).} The action variable for the transverse cyclotron motion is \[ -J = \frac{2\pi E_\perp}{\omega_c}. +J = \frac{E_\perp}{\omega_c}. \] Substitute the numerical values: \[ -J = \frac{2\pi(1.60\times 10^{-16}\,\mathrm{J})}{1.437\times 10^8\,\mathrm{rad/s}} -= \frac{1.005\times 10^{-15}}{1.437\times 10^8}\,\mathrm{J\!\cdot\!s}. +J = \frac{1.60\times 10^{-16}\,\mathrm{J}}{1.437\times 10^8\,\mathrm{rad/s}} += 1.11\times 10^{-24}\,\mathrm{J\!\cdot\!s}. \] This gives \[ -J = 7.0\times 10^{-24}\,\mathrm{J\!\cdot\!s}. +J = 1.1\times 10^{-24}\,\mathrm{J\!\cdot\!s}. \] -Now verify the energy--action relation $E_\perp(J) = \omega_c J/(2\pi)$. Differentiating with respect to $J$: +Now verify the energy--action relation $E_\perp(J) = \omega_c J$. Differentiating with respect to $J$: \[ -\pdv{E_\perp}{J} = \frac{\omega_c}{2\pi}. +\pdv{E_\perp}{J} = \omega_c. \] -The physical angular frequency is recovered as $\omega = 2\pi(\pdv{E_\perp}{J}) = \omega_c$. For the numerical values, +The derivative directly equals the cyclotron angular frequency. For the numerical values, \[ -E_\perp(J) = \frac{\omega_c J}{2\pi} -= \frac{(1.437\times 10^8\,\mathrm{rad/s})(7.0\times 10^{-24}\,\mathrm{J\!\cdot\!s})}{2\pi} +E_\perp(J) = \omega_c J += (1.437\times 10^8\,\mathrm{rad/s})(1.11\times 10^{-24}\,\mathrm{J\!\cdot\!s}) = 1.60\times 10^{-16}\,\mathrm{J}, \] -which is exactly the original transverse energy of $1.0\,\mathrm{keV}$. The relation $\pdv{E_\perp}{J} = \omega_c/(2\pi)$ holds both algebraically and numerically, confirming that the action-angle formalism reproduces the cyclotron frequency exactly. +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, \[ @@ -337,5 +338,5 @@ T = 44\,\mathrm{ns}, \qquad R = 3.1\,\mathrm{mm}, \qquad -J = 7.0\times 10^{-24}\,\mathrm{J\!\cdot\!s}. +J = 1.1\times 10^{-24}\,\mathrm{J\!\cdot\!s}. \] \ No newline at end of file