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fix(HJ): A.11 — Radian convention fix for action variable J

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Krishna Ayyalasomayajula
2026-05-02 13:07:39 -05:00
parent 0f244f6874
commit 2c04812ff6
1 changed files with 25 additions and 24 deletions
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concepts/advanced/cyclotron-hj.tex
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@@ -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,
\[
@@ -337,5 +338,5 @@ T = 44\,\mathrm{ns},
\qquad
R = 3.1\,\mathrm{mm},
\qquad
J = 7.0\times 10^{-24}\,\mathrm{J\!\cdot\!s}.
J = 1.1\times 10^{-24}\,\mathrm{J\!\cdot\!s}.
\]