342 lines
16 KiB
TeX
342 lines
16 KiB
TeX
\subsection{Cyclotron Motion}
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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.
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\dfn{Hamilton--Jacobi formulation of cyclotron motion}{
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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:
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\[
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\nabla\times(0, B_0 x, 0)
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= \left(0,\; 0,\; \pdv{(B_0 x)}{x}\right)
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= B_0\,\hat{\bm{z}}.
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\]
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Set the scalar potential $\varphi = 0$. The Hamiltonian is
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\[
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\mcH = \frac{1}{2m}\Bigl[p_x^2 + \bigl(p_y - q B_0 x\bigr)^2 + p_z^2\Bigr].
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\]
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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
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\[
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\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \bigl(\pdv{\mcS}{y} - q B_0 x\bigr)^2 + \left(\pdv{\mcS}{z}\right)^2\right] + \pdv{\mcS}{t} = 0.
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\]}
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\nt{Why Landau gauge?}{
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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 $\mcS = W_x(x) + \alpha_y y + \alpha_z z - Et$.}
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\nt{Two momenta}{
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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.}
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\thm{Complete integral for cyclotron motion}{
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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
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\[
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\mcS = W_x(x) + \alpha_y y + \alpha_z z - Et,
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\]
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where the $x$-part of the characteristic function is
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\[
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W_x(x) = \frac{E_\perp}{\omega_c}\arcsin\!\left(\frac{x - X_c}{R}\right)
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+ \frac{q B_0}{2}\,(x - X_c)\sqrt{R^2 - (x - X_c)^2}.
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\]
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Here $E_\perp = E - \alpha_z^2/(2m)$ is the energy of motion in the $xy$-plane alone. The complete integral is defined for $|x - X_c| < R$.}
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\pf{Separation of the HJ equation and integration of the x-dependent part}{
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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
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\[
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\frac{1}{2m}\left[\left(\der{W_x}{x}\right)^2 + \bigl(\alpha_y - q B_0 x\bigr)^2 + \alpha_z^2\right] = E.
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\]
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Define the transverse energy $E_\perp = E - \alpha_z^2/(2m)$. The $x$-equation simplifies to
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\[
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\left(\der{W_x}{x}\right)^2 + \bigl(\alpha_y - q B_0 x\bigr)^2 = 2m E_\perp.
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\]
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Solve for the spatial derivative:
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\[
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\der{W_x}{x} = \pm\sqrt{2m E_\perp - \bigl(\alpha_y - q B_0 x\bigr)^2}.
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\]
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The square root is real when $\abs{\alpha_y - q B_0 x} \le \sqrt{2m E_\perp}$. Introduce the guiding-center coordinate
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\[
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X_c = \frac{\alpha_y}{q B_0}.
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\]
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Then $\alpha_y - q B_0 x = -q B_0(x - X_c)$, and the radicand factors as
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\[
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2m E_\perp - q^2 B_0^2(x - X_c)^2
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= q^2 B_0^2\left(\frac{2m E_\perp}{q^2 B_0^2} - (x - X_c)^2\right).
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\]
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Define the gyroradius $R = \sqrt{2m E_\perp}/(q B_0)$. The derivative of $W_x$ becomes
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\[
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\der{W_x}{x} = \pm q B_0\sqrt{R^2 - (x - X_c)^2}.
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\]
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This has the same square-root structure as the simple harmonic oscillator. Integrate by the trigonometric substitution $x - X_c = R\sin\theta$, giving $\dd x = R\cos\theta\,\mathrm{d}\theta$:
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\[
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W_x = \int q B_0\sqrt{R^2 - R^2\sin^2\theta}\cdot R\cos\theta\,\mathrm{d}\theta
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= q B_0 R^2\int \cos^2\theta\,\mathrm{d}\theta.
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\]
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The antiderivative of $\cos^2\theta$ is $\tfrac12(\theta + \sin\theta\cos\theta)$, so
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\[
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W_x = \frac{q B_0 R^2}{2}\bigl(\theta + \sin\theta\cos\theta\bigr).
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\]
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Evaluate the constant prefactor using $R^2 = 2m E_\perp/(q^2 B_0^2)$:
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\[
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\frac{q B_0 R^2}{2}
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= \frac{q B_0}{2}\cdot\frac{2m E_\perp}{q^2 B_0^2}
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= \frac{m E_\perp}{q B_0}
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= \frac{E_\perp}{\omega_c}.
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\]
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Now express the trigonometric factors in terms of $x$:
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\[
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\theta = \arcsin\!\left(\frac{x - X_c}{R}\right),
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\qquad
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\sin\theta = \frac{x - X_c}{R},
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\qquad
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\cos\theta = \frac{\sqrt{R^2 - (x - X_c)^2}}{R}.
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\]
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The product $\sin\theta\cos\theta$ is $(x - X_c)\sqrt{R^2 - (x - X_c)^2}/R^2$. Substituting back,
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\[
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W_x(x) = \frac{E_\perp}{\omega_c}\arcsin\!\left(\frac{x - X_c}{R}\right)
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+ \frac{E_\perp}{\omega_c}\cdot\frac{(x - X_c)\sqrt{R^2 - (x - X_c)^2}}{R^2}.
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\]
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The coefficient of the second term simplifies as
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\[
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\frac{E_\perp}{\omega_c R^2}
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= \frac{E_\perp}{\omega_c}\cdot\frac{q^2 B_0^2}{2m E_\perp}
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= \frac{q^2 B_0^2}{2m\omega_c}
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= \frac{q B_0}{2},
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\]
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since $\omega_c = q B_0/m$. Therefore,
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\[
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W_x(x) = \frac{E_\perp}{\omega_c}\arcsin\!\left(\frac{x - X_c}{R}\right)
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+ \frac{q B_0}{2}\,(x - X_c)\sqrt{R^2 - (x - X_c)^2}.
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\]
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The full complete integral is $\mcS = W_x(x) + \alpha_y y + \alpha_z z - Et$.}
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\cor{Helical trajectory from Jacobi's theorem}{
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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 $x$:
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\[
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\pdv{\mcS}{E_\perp} = \pdv{W_x}{E_\perp} - t.
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\]
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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
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\[
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\pdv{W_x}{E_\perp} = \frac{1}{\omega_c}\arcsin\chi
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+ \chi\cos\chi\cdot\pdv{E_\perp/\omega_c}{E_\perp}\cdot\frac{1}{(E_\perp/\omega_c)}
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+ \frac{q B_0}{2}(x - X_c)\cdot\frac{1}{2U}\pdv{R^2}{E_\perp}.
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\]
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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
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\[
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-\frac{1}{2\omega_c}\frac{\chi}{\sqrt{1-\chi^2}}
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= -\frac{x - X_c}{2\omega_c R\sqrt{1-\chi^2}},
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\]
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\[
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\frac{q B_0}{2}(x - X_c)\cdot\frac{m}{q^2 B_0^2 U}
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= \frac{m}{q B_0}\cdot\frac{x - X_c}{2U}
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= \frac{x - X_c}{2\omega_c R\sqrt{1-\chi^2}},
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\]
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which exactly cancel, as in the harmonic oscillator case. Hence,
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\[
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\pdv{W_x}{E_\perp} = \frac{1}{\omega_c}\arcsin\!\left(\frac{x - X_c}{R}\right).
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\]
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Set $\pdv{\mcS}{E_\perp} = \beta$ (constant):
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\[
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\frac{1}{\omega_c}\arcsin\!\left(\frac{x - X_c}{R}\right) - t = \beta,
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\]
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or equivalently,
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\[
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x(t) = X_c + R\sin\!\bigl(\omega_c(t + \beta)\bigr).
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\]
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Define the initial phase $\phi_0 = \omega_c\beta$, measured in radians.
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Differentiate $\mcS$ with respect to the separation constant $\alpha_y$:
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\[
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\pdv{\mcS}{\alpha_y} = \pdv{W_x}{\alpha_y} + y = \beta_y.
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\]
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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 $v_y = (\alpha_y - q B_0 x)/m$:
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\[
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v_y(t) = \frac{\alpha_y - q B_0 x(t)}{m}
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= \frac{q B_0\bigl(X_c - x(t)\bigr)}{m}
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= -\omega_c R\sin(\omega_c t + \phi_0).
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\]
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Integrating with respect to time,
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\[
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y(t) = Y_c + R\cos(\omega_c t + \phi_0),
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\]
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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:
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\[
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x(t) = X_c + R\sin(\omega_c t + \phi_0),
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\qquad
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y(t) = Y_c + R\cos(\omega_c t + \phi_0),
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\qquad
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z(t) = v_z t + z_0.
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\]
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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 $v_z$.}
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\nt{Hamilton's equations shortcut for $y(t)$}{
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The same $y(t)$ result follows directly from Hamilton's equations without differentiating $\mcS$. Because $y$ is cyclic,
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\[
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\dot{p}_y = -\pdv{\mcH}{y} = 0,
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\]
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so $p_y = \alpha_y$ is constant. Hamilton's velocity equation gives
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\[
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\dot{y} = \pdv{\mcH}{p_y} = \frac{1}{m}\bigl(p_y - q B_0 x\bigr)
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= \frac{\alpha_y - q B_0 x(t)}{m}.
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\]
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Substituting $x(t) = X_c + R\sin(\omega_c t + \phi_0)$ with $X_c = \alpha_y/(q B_0)$:
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\[
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\dot{y} = \frac{q B_0(X_c - X_c - R\sin(\omega_c t + \phi_0))}{m}
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= -\omega_c R\sin(\omega_c t + \phi_0).
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\]
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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.}
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\nt{Guiding-center independence}{
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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.}
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\mprop{Action-angle variables for cyclotron motion}{
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The action-angle formalism applied to cyclotron motion yields the following results:
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\begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
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\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
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\[
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J = \frac{1}{2\pi}\oint p_x\,\dd x = \frac{1}{\pi}\int_{X_c - R}^{X_c + R} q B_0\sqrt{R^2 - (x - X_c)^2}\,\dd x.
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\]
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The integral is the area of a semicircle of radius $R$ multiplied by $q B_0$. With the $1/\pi$ factor:
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\[
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J = \frac{q B_0}{\pi}\cdot\frac{\pi R^2}{2}
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= \frac{q B_0 R^2}{2}
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= \frac{q B_0}{2}\cdot\frac{2m E_\perp}{q^2 B_0^2}
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= \frac{m E_\perp}{q B_0}
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= \frac{E_\perp}{\omega_c}.
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\]
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Geometrically, the full phase-space area $2\pi J/(q B_0) = \pi R^2$ is the area of the real-space circular orbit.
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\item Inverting the action--energy relation, the transverse energy as a function of the action is
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\[
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E_\perp(J) = \omega_c J.
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\]
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The Hamiltonian expressed in terms of the action variables is $E = \omega_c J + \alpha_z^2/(2m)$, linear in $J$ and quadratic in $\alpha_z$.
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\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.
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\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.
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\end{enumerate}
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}
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\nt{Comparison with the Lorentz force}{
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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):
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\[
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\ddot{x} = \frac{q B_0}{m}\,\dot{y},
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\qquad
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\ddot{y} = -\frac{q B_0}{m}\,\dot{x},
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\qquad
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\ddot{z} = 0.
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\]
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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
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\[
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v_x = v_\perp\cos(\omega_c t + \phi_0),
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\qquad
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v_y = -v_\perp\sin(\omega_c t + \phi_0).
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\]
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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.}
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\qs{Proton cyclotron motion from the HJ complete integral}{
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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 $E_\perp = 1.0\,\mathrm{keV} = 1.60\times 10^{-16}\,\mathrm{J}$.
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\begin{enumerate}[label=(\alph*)]
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\item Compute the cyclotron angular frequency $\omega_c = q B_0/m$ and the gyration period $T = 2\pi/\omega_c$.
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\item Find the gyroradius $R = \sqrt{2m E_\perp}/(q B_0)$ in meters.
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\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.
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\end{enumerate}}
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\sol \textbf{Part (a).} The cyclotron angular frequency is
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\[
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\omega_c = \frac{q B_0}{m}.
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\]
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Substitute the given values:
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\[
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q = 1.60\times 10^{-19}\,\mathrm{C},
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\qquad
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B_0 = 1.5\,\mathrm{T},
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\qquad
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m = 1.67\times 10^{-27}\,\mathrm{kg}.
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\]
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Form the ratio:
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\[
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\omega_c = \frac{(1.60\times 10^{-19})(1.5)}{1.67\times 10^{-27}}\,\mathrm{rad/s}
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= \frac{2.40\times 10^{-19}}{1.67\times 10^{-27}}\,\mathrm{rad/s}.
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\]
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This gives
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\[
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\omega_c = 1.437\times 10^8\,\mathrm{rad/s}.
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\]
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Rounding to two significant figures (consistent with the field strength $1.5\,\mathrm{T}$),
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\[
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\omega_c = 1.4\times 10^8\,\mathrm{rad/s}.
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\]
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The gyration period is
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\[
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T = \frac{2\pi}{\omega_c}
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= \frac{2\pi}{1.437\times 10^8}\,\mathrm{s}
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= 4.37\times 10^{-8}\,\mathrm{s}.
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\]
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In more convenient units,
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\[
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T = 4.4\times 10^{-8}\,\mathrm{s} = 44\,\mathrm{ns}.
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\]
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\textbf{Part (b).} The gyroradius is
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\[
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R = \frac{\sqrt{2m E_\perp}}{q B_0}.
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\]
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Evaluate the numerator:
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\[
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2m E_\perp = 2(1.67\times 10^{-27}\,\mathrm{kg})(1.60\times 10^{-16}\,\mathrm{J})
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= 5.34\times 10^{-43}\,\mathrm{kg^2\,m^2/s^2}.
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\]
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Taking the square root:
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\[
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\sqrt{2m E_\perp} = \sqrt{5.34\times 10^{-43}}\,\mathrm{kg\,m/s}
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= 7.31\times 10^{-22}\,\mathrm{kg\,m/s}.
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\]
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The denominator is
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\[
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q B_0 = (1.60\times 10^{-19}\,\mathrm{C})(1.5\,\mathrm{T})
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= 2.40\times 10^{-19}\,\mathrm{C\,T}.
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\]
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Therefore,
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\[
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R = \frac{7.31\times 10^{-22}}{2.40\times 10^{-19}}\,\mathrm{m}
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= 3.05\times 10^{-3}\,\mathrm{m}.
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\]
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In more convenient units,
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\[
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R = 3.05\,\mathrm{mm}.
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\]
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\textbf{Part (c).} The action variable for the transverse cyclotron motion is
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\[
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J = \frac{E_\perp}{\omega_c}.
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\]
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Substitute the numerical values:
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\[
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J = \frac{1.60\times 10^{-16}\,\mathrm{J}}{1.437\times 10^8\,\mathrm{rad/s}}
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= 1.11\times 10^{-24}\,\mathrm{J\!\cdot\!s}.
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\]
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This gives
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\[
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J = 1.1\times 10^{-24}\,\mathrm{J\!\cdot\!s}.
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\]
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Now verify the energy--action relation $E_\perp(J) = \omega_c J$. Differentiating with respect to $J$:
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\[
|
|
\pdv{E_\perp}{J} = \omega_c.
|
|
\]
|
|
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,
|
|
\[
|
|
\omega_c = 1.4\times 10^8\,\mathrm{rad/s},
|
|
\qquad
|
|
T = 44\,\mathrm{ns},
|
|
\qquad
|
|
R = 3.1\,\mathrm{mm},
|
|
\qquad
|
|
J = 1.1\times 10^{-24}\,\mathrm{J\!\cdot\!s}.
|
|
\] |