fix(HJ): A.08 — Radian convention fix for action variables
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@@ -203,32 +203,32 @@ The eccentricity $\varepsilon = \sqrt{1 + 2EL^2/(\mu k^2)}$ determines the shape
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\nt{Action--angle variables and Kepler\normalsize{}'s third law}{
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\nt{Action--angle variables and Kepler\normalsize{}'s third law}{
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The three independent action variables for the Kepler problem are computed by integrating the appropriate momenta over one complete cycle of each coordinate. For the azimuthal coordinate, $J_\phi = \oint p_\phi\,\mathrm{d}\phi = 2\pi L_z$. For the polar coordinate, $J_\theta = \oint p_\theta\,\mathrm{d}\theta = 2\pi(L - |L_z|)$. For the radial coordinate, the integral $J_r = \oint p_r\,\dd r$ requires careful evaluation between the two radial turning points for bound orbits ($E < 0$). The result is $J_r = 2\pi(-L + k\sqrt{\mu/(2|E|)})$.
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The three independent action variables for the Kepler problem are computed by integrating the appropriate momenta over one complete cycle of each coordinate. For the azimuthal coordinate, $J_\phi = \frac{1}{2\pi}\oint p_\phi\,\mathrm{d}\phi = L_z$. For the polar coordinate, $J_\theta = \frac{1}{2\pi}\oint p_\theta\,\mathrm{d}\theta = L - |L_z|$. For the radial coordinate, the integral $J_r = \frac{1}{2\pi}\oint p_r\,\dd r$ requires careful evaluation between the two radial turning points for bound orbits ($E < 0$). The result is $J_r = -L + k\sqrt{\mu/(2|E|)}$.
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Each action variable carries a distinct physical meaning. The azimuthal action $J_\phi = 2\pi L_z$ measures the conserved rotation about the vertical axis: it is the circulation of angular momentum along the $\phi$ direction and sets the rate of azimuthal precession. The polar action $J_\theta = 2\pi(L - |L_z|)$ measures the inclination of the orbital plane: when the orbit is equatorial ($L = |L_z|$) we have $J_\theta = 0$, and larger values of $L - |L_z|$ correspond to a more tilted orbit with greater polar oscillation between $\theta_{\min}$ and $\pi - \theta_{\min}$. The radial action $J_r = 2\pi(-L + k\sqrt{\mu/(2|E|)})$ measures the extent of the radial excursion between periapsis and apoapsis: for a circular orbit $J_r = 0$ (no radial oscillation), and for highly eccentric orbits $J_r$ grows as the particle swings farther from the center.
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Each action variable carries a distinct physical meaning. The azimuthal action $J_\phi = L_z$ measures the conserved rotation about the vertical axis: it is the circulation of angular momentum along the $\phi$ direction and sets the rate of azimuthal precession. The polar action $J_\theta = L - |L_z|$ measures the inclination of the orbital plane: when the orbit is equatorial ($L = |L_z|$) we have $J_\theta = 0$, and larger values of $L - |L_z|$ correspond to a more tilted orbit with greater polar oscillation between $\theta_{\min}$ and $\pi - \theta_{\min}$. The radial action $J_r = -L + k\sqrt{\mu/(2|E|)}$ measures the extent of the radial excursion between periapsis and apoapsis: for a circular orbit $J_r = 0$ (no radial oscillation), and for highly eccentric orbits $J_r$ grows as the particle swings farther from the center.
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Adding all three actions eliminates the angular--momentum dependence:
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Adding all three actions eliminates the angular--momentum dependence:
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\[
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\[
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J_{\mathrm{tot}} = J_r + J_\theta + J_\phi
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J_{\mathrm{tot}} = J_r + J_\theta + J_\phi
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= 2\pi k\sqrt{\frac{\mu}{2|E|}}.
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= k\sqrt{\frac{\mu}{2|E|}}.
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\]
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\]
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Inverting this expression gives $|E| = 2\pi^2\mu k^2/J_{\mathrm{tot}}^2$, which expresses the energy in terms of the single total action $J_{\mathrm{tot}}$. Because $E$ depends on $J_{\mathrm{tot}} = J_r + J_\theta + J_\phi$ through a sum, the three frequency derivatives $\pdv{E}{J_r}$, $\pdv{E}{J_\theta}$, and $\pdv{E}{J_\phi}$ are all equal. Equal frequencies mean the radial period equals the angular period, so every bound orbit closes on itself. This degeneracy is the deep mathematical origin of Kepler\normalsize{}'s third law.}
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Inverting this expression gives $|E| = \mu k^2/(2J_{\mathrm{tot}}^2)$, which expresses the energy in terms of the single total action $J_{\mathrm{tot}}$. Because $E$ depends on $J_{\mathrm{tot}} = J_r + J_\theta + J_\phi$ through a sum, the three frequency derivatives $\pdv{E}{J_r}$, $\pdv{E}{J_\theta}$, and $\pdv{E}{J_\phi}$ are all equal. Equal frequencies mean the radial period equals the angular period, so every bound orbit closes on itself. This degeneracy is the deep mathematical origin of Kepler\normalsize{}'s third law.}
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\ex{Action--angle derivation of $E(J_{\mathrm{tot}})$ and the degenerate frequencies}{
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\ex{Action--angle derivation of $E(J_{\mathrm{tot}})$ and the degenerate frequencies}{
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We evaluate the three action variables for the Kepler problem explicitly.
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We evaluate the three action variables for the Kepler problem explicitly.
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\textbf{Azimuthal action.} The momentum conjugate to $\phi$ is $p_\phi = L_z$, a constant. Integrating over one full revolution:
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\textbf{Azimuthal action.} The momentum conjugate to $\phi$ is $p_\phi = L_z$, a constant. Integrating over one full revolution:
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\[
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\[
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J_\phi = \oint p_\phi\,\mathrm{d}\phi = \int_{0}^{2\pi} L_z\,\mathrm{d}\phi = 2\pi L_z.
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J_\phi = \frac{1}{2\pi}\oint p_\phi\,\mathrm{d}\phi = \frac{1}{2\pi}\int_{0}^{2\pi} L_z\,\mathrm{d}\phi = L_z.
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\]
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\]
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\textbf{Polar action.} The polar momentum is $p_\theta = \sqrt{L^2 - L_z^2/\sin^2\theta}$. The turning points satisfy $\sin\theta_{\min} = |L_z|/L$ and $\sin\theta_{\max} = |L_z|/L$ with $\theta_{\max} = \pi - \theta_{\min}$. The integral over one oscillation is
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\textbf{Polar action.} The polar momentum is $p_\theta = \sqrt{L^2 - L_z^2/\sin^2\theta}$. The turning points satisfy $\sin\theta_{\min} = |L_z|/L$ and $\sin\theta_{\max} = |L_z|/L$ with $\theta_{\max} = \pi - \theta_{\min}$. The integral over one oscillation is
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\[
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\[
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J_\theta = 2\int_{\theta_{\min}}^{\pi-\theta_{\min}}\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}\;\mathrm{d}\theta.
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J_\theta = \frac{1}{2\pi}\,2\int_{\theta_{\min}}^{\pi-\theta_{\min}}\sqrt{L^2 - \frac{L_z^2}{\sin^2\theta}}\;\mathrm{d}\theta.
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\]
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\]
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The substitution $u = \cos\theta$ converts the integrand to $\sqrt{L^2 - L_z^2/(1-u^2)}$, and the integral evaluates to
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The substitution $u = \cos\theta$ converts the integrand to $\sqrt{L^2 - L_z^2/(1-u^2)}$, and the integral evaluates to
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\[
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\[
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J_\theta = 2\pi\bigl(L - |L_z|\bigr).
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J_\theta = L - |L_z|.
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\]
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\]
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\textbf{Radial action.} The radial momentum is $p_r = \pm\sqrt{2\mu E + 2\mu k/r - L^2/r^2}$. For bound orbits ($E < 0$) write $|E| = -E$. The turning points are the roots of $2\mu|E|r^2 - 2\mu kr - L^2 = 0$, which are
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\textbf{Radial action.} The radial momentum is $p_r = \pm\sqrt{2\mu E + 2\mu k/r - L^2/r^2}$. For bound orbits ($E < 0$) write $|E| = -E$. The turning points are the roots of $2\mu|E|r^2 - 2\mu kr - L^2 = 0$, which are
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@@ -237,41 +237,41 @@ r_{\pm} = \frac{\mu k \pm L\sqrt{\mu^2 k^2 - 2\mu|E|L^2}}{2\mu|E|}.
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\]
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\]
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The radial action integral is
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The radial action integral is
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\[
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\[
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J_r = 2\int_{r_-}^{r_+}\sqrt{2\mu|E| + \frac{2\mu k}{r} - \frac{L^2}{r^2}}\;\frac{\dd r}{r^2/r^2}.
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J_r = \frac{1}{2\pi}\,2\int_{r_-}^{r_+}\sqrt{2\mu|E| + \frac{2\mu k}{r} - \frac{L^2}{r^2}}\;\frac{\dd r}{r^2/r^2}.
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\]
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\]
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The standard contour--integration or elliptic--integral evaluation gives
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The standard contour--integration or elliptic--integral evaluation gives
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\[
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\[
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J_r = 2\pi\!\left(-L + k\sqrt{\frac{\mu}{2|E|}}\right).
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J_r = -L + k\sqrt{\frac{\mu}{2|E|}}.
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\]
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\]
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\textbf{Total action and energy.} Adding the three actions:
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\textbf{Total action and energy.} Adding the three actions:
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\[
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\[
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J_{\mathrm{tot}} = J_r + J_\theta + J_\phi
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J_{\mathrm{tot}} = J_r + J_\theta + J_\phi
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= 2\pi\!\left(-L + k\sqrt{\frac{\mu}{2|E|}}\right) + 2\pi(L - |L_z|) + 2\pi L_z
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= \left(-L + k\sqrt{\frac{\mu}{2|E|}}\right) + \left(L - |L_z|\right) + L_z
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= 2\pi k\sqrt{\frac{\mu}{2|E|}}.
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= k\sqrt{\frac{\mu}{2|E|}}.
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\]
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\]
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The angular--momentum terms $L$ and $|L_z|$ cancel exactly. Invert the total--action relation to obtain the energy:
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The angular--momentum terms $L$ and $|L_z|$ cancel exactly. Invert the total--action relation to obtain the energy:
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\[
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\[
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\sqrt{\frac{\mu}{2|E|}} = \frac{J_{\mathrm{tot}}}{2\pi k},
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\sqrt{\frac{\mu}{2|E|}} = \frac{J_{\mathrm{tot}}}{k},
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\qquad
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\qquad
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\frac{2|E|}{\mu} = \frac{4\pi^2 k^2}{J_{\mathrm{tot}}^2},
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\frac{2|E|}{\mu} = \frac{k^2}{J_{\mathrm{tot}}^2},
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\]
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\]
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\[
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\[
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E(J_{\mathrm{tot}}) = -\frac{2\pi^2\mu k^2}{J_{\mathrm{tot}}^2}.
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E(J_{\mathrm{tot}}) = -\frac{\mu k^2}{2J_{\mathrm{tot}}^2}.
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\]
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\]
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\textbf{Degenerate frequencies.} The three action--angle frequencies are
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\textbf{Degenerate frequencies.} The three action--angle frequencies are
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\[
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\[
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\omega_r = \pdv{E}{J_r} = \pdv{E}{J_{\mathrm{tot}}}\pdv{J_{\mathrm{tot}}}{J_r}
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\omega_r = \pdv{E}{J_r} = \pdv{E}{J_{\mathrm{tot}}}\pdv{J_{\mathrm{tot}}}{J_r}
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= \frac{4\pi^2\mu k^2}{J_{\mathrm{tot}}^3}\cdot 1,
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= \frac{\mu k^2}{J_{\mathrm{tot}}^3}\cdot 1,
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\]
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\]
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\[
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\[
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\omega_\theta = \pdv{E}{J_\theta} = \pdv{E}{J_{\mathrm{tot}}}\pdv{J_{\mathrm{tot}}}{J_\theta}
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\omega_\theta = \pdv{E}{J_\theta} = \pdv{E}{J_{\mathrm{tot}}}\pdv{J_{\mathrm{tot}}}{J_\theta}
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= \frac{4\pi^2\mu k^2}{J_{\mathrm{tot}}^3}\cdot 1,
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= \frac{\mu k^2}{J_{\mathrm{tot}}^3}\cdot 1,
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\]
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\]
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\[
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\[
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\omega_\phi = \pdv{E}{J_\phi} = \pdv{E}{J_{\mathrm{tot}}}\pdv{J_{\mathrm{tot}}}{J_\phi}
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\omega_\phi = \pdv{E}{J_\phi} = \pdv{E}{J_{\mathrm{tot}}}\pdv{J_{\mathrm{tot}}}{J_\phi}
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= \frac{4\pi^2\mu k^2}{J_{\mathrm{tot}}^3}\cdot 1.
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= \frac{\mu k^2}{J_{\mathrm{tot}}^3}\cdot 1.
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\]
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\]
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All three frequencies are equal: $\omega_r = \omega_\theta = \omega_\phi$. The period of radial oscillation equals the period of angular advance. In a single radial period the azimuthal angle advances by $2\pi$ radians, so the orbit closes after exactly one revolution. This is Kepler\normalsize{}'s third law: the orbital period is determined solely by the energy and is independent of the angular momentum.}
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All three frequencies are equal: $\omega_r = \omega_\theta = \omega_\phi$. The period of radial oscillation equals the period of angular advance. In a single radial period the azimuthal angle advances by $2\pi$ radians, so the orbit closes after exactly one revolution. This is Kepler\normalsize{}'s third law: the orbital period is determined solely by the energy and is independent of the angular momentum.}
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