fix(HJ): A.07 — Surface motivation, cross-ref U7, phase-space ellipse, explain cancellation

concepts/advanced/sho-hj.tex

@@ -1,6 +1,6 @@
 \subsection{Simple Harmonic Oscillator}
 
-This subsection solves the simple harmonic oscillator through the Hamilton--Jacobi equation, obtains the complete integral and trajectory by quadrature, and computes the action-angle variables that confirm isochronous oscillation.
+This subsection solves the simple harmonic oscillator through the Hamilton--Jacobi equation. The quadratic potential turns the Hamilton--Jacobi square-root integral into an elementary trigonometric substitution, making the oscillator one of the few nonlinear Hamilton--Jacobi equations integrable in closed form. This same physical system appears in Unit~7 (oscillations) under Newton\normalsize{}'s law; here we solve it by an entirely different route to obtain the complete integral, trajectory by quadrature, and action--angle variables confirming isochronous oscillation. This integrability makes the harmonic oscillator the prototypical example for testing both the Hamilton--Jacobi method and the action--angle formalism.
 
 \dfn{Hamilton--Jacobi formulation of the simple harmonic oscillator}{
 The Hamiltonian for a one-dimensional simple harmonic oscillator of mass~$m$ and spring constant~$k$ is
@@ -15,9 +15,7 @@ The Hamilton--Jacobi partial differential equation for the principal function~$\
 \[
 \frac{1}{2m}\left(\pdv{\mcS}{x}\right)^2 + \frac{1}{2}m\omega_0^2 x^2 + \pdv{\mcS}{t} = 0.
 \]
-A complete integral $\mcS(x,t;E)$, containing one independent non-additive constant $E$ equal to the total energy, determines the full dynamics by Jacobi's theorem.}
-
-\nt{The simple harmonic oscillator is one of the few nonlinear Hamilton--Jacobi equations that can be integrated in closed form. The quadratic potential turns the square-root integral into an elementary trigonometric substitution, and the resulting complete integral yields the standard sinusoidal trajectory. This integrability makes the harmonic oscillator the prototypical example for testing both the Hamilton--Jacobi method and the action-angle formalism.}
+A complete integral $\mcS(x,t;E)$, containing one independent non-additive constant $E$ equal to the total energy, determines the full dynamics by Jacobi\normalsize{}'s theorem.}
 
 \thm{Complete integral of the SHO Hamilton--Jacobi equation}{
 The complete integral of the Hamilton--Jacobi equation for a simple harmonic oscillator is
@@ -38,6 +36,8 @@ Solve for the spatial derivative:
 \]
 The square root is real for $|x| \le A$, where $A = \sqrt{2E/(m\omega_0^2)}$ is the amplitude. The turning points $x = \pm A$ bound the oscillation and correspond to the points where the kinetic energy vanishes.
 
+The sign carries physical meaning: the upper sign describes forward motion ($p>0$, the mass traveling toward $+A$) and the lower sign describes backward motion ($p<0$, the mass returning toward $-A$). In the closed contour integral $\oint p\,\dd x$ that defines the action variable, the $+$ branch contributes the outward half of the orbit $(-A\to A)$ and the $-$ branch contributes the return half $(A\to -A)$. For finding $W(x;E)$ as a local generating function we choose the positive branch and carry it through the integration; full periodicity is then imposed by the boundary conditions.
+
 Integrate by the trigonometric substitution $x = A\sin\theta$, giving $\dd x = A\cos\theta\,\mathrm{d}\theta$. The radicand becomes
 \[
 2mE - m^2\omega_0^2 A^2\sin^2\theta
@@ -92,8 +92,8 @@ W(x;E) = \frac{E}{\omega_0}\arcsin\!\left(x\sqrt{\frac{m\omega_0^2}{2E}}\right)
 \]
 and the complete integral is $\mcS(x,t;E) = W(x;E) - Et$.}
 
-\cor{Trajectory from Jacobi's theorem}{
-Jacobi's theorem states $\pdv{\mcS}{E} = \beta$, where $\beta$ is a constant fixed by the initial conditions. Differentiate $\mcS = W - Et$ with respect to $E$ at fixed~$x$:
+\cor{Trajectory from Jacobi\normalsize{}'s theorem}{
+Jacobi\normalsize{}'s theorem states $\pdv{\mcS}{E} = \beta$, where $\beta$ is a constant fixed by the initial conditions. Differentiate $\mcS = W - Et$ with respect to $E$ at fixed~$x$:
 \[
 \pdv{\mcS}{E} = \pdv{W}{E} - t.
 \]
@@ -122,7 +122,7 @@ The third term equals
 \frac{xm}{2R} = \frac{xm}{2\sqrt{2mE}\sqrt{1-\chi^2}}
 = \frac{x\sqrt{m}}{2\sqrt{2E}\sqrt{1-\chi^2}},
 \]
-which exactly cancels the second term. This cancellation reflects the fact that the energy dependence of the amplitude and the energy dependence of the integrand conspire to leave only the angular part. Therefore,
+which exactly cancels the second term. This cancellation carries deep physical significance: the energy dependence of the amplitude $A = \sqrt{2E/(m\omega_0^2)}$ and the energy dependence of the integrand $\sqrt{2mE - m^2\omega_0^2 x^2}$ nearly cancel when differentiated with respect to $E$, leaving only the geometric phase $\tfrac{1}{\omega_0}\arcsin(\cdots)$. Because $\pdv{W}{E}$ is independent of the amplitude, the period $T = 2\pi/\omega_0$ is the same for every orbit regardless of how much energy it carries. This is the isochrony of the simple harmonic oscillator---all amplitudes share one period. The algebraic cancellation in $\pdv{W}{E}$ is the Hamilton--Jacobi embodiment of that physical fact. Therefore,
 \[
 \pdv{W}{E} = \frac{1}{\omega_0}\arcsin\!\left(x\sqrt{\frac{m\omega_0^2}{2E}}\right).
 \]
@@ -140,6 +140,24 @@ x(t) = A\sin(\omega_0 t + \phi).
 \]
 The total energy is $E = \tfrac12 m\omega_0^2 A^2 = \tfrac12 k A^2$, and the initial phase $\phi$ is determined by the initial position and velocity through $\sin\phi = x_0/A$ and $\cos\phi = v_0/(\omega_0 A)$. When the oscillator is released from rest at maximum displacement, $\cos\phi = 0$ and $\phi = \pi/2$, giving $x(t) = A\cos(\omega_0 t)$.}
 
+\ex{Phase-space ellipse}{
+The action variable $J = \oint p\,\dd x$ equals the area enclosed by the orbit in the $(x,p)$ phase plane. For the harmonic oscillator the orbit is an ellipse. The orbit equation follows from energy conservation,
+\[
+\frac{p^2}{2m} + \frac{1}{2}m\omega_0^2 x^2 = E,
+\]
+which can be rearranged to standard form:
+\[
+\frac{x^2}{A^2} + \frac{p^2}{p_{\max}^2} = 1,
+\]
+where $A = \sqrt{2E/(m\omega_0^2)}$ is the maximum displacement (semi-axis along the position direction) and $p_{\max} = \sqrt{2mE} = m\omega_0 A$ is the maximum momentum (semi-axis along the momentum direction). The phase-space orbit is an ellipse centered at the origin with these semi-axes, so the enclosed area is
+\[
+J = \pi A\,p_{\max}
+= \pi\sqrt{\frac{2E}{m\omega_0^2}}\cdot\sqrt{2mE}
+= \pi\sqrt{\frac{4E^2}{\omega_0^2}}
+= \frac{2\pi E}{\omega_0}.
+\]
+The action variable is the geometric area in phase space. The linear relation $J \propto E$ reflects the fact that doubling the energy rescales the ellipse uniformly, while the ratio $p_{\max}/A = m\omega_0$ sets the ellipse\normalsize{}'s aspect ratio.}
+
 \mprop{Action-angle variables for the harmonic oscillator}{
 Applying the action-angle formalism to the simple harmonic oscillator gives the following results:
 
@@ -148,7 +166,7 @@ Applying the action-angle formalism to the simple harmonic oscillator gives the
 \[
 J = \oint p\,\dd x = 2\int_{-A}^{A}\sqrt{2mE - m^2\omega_0^2 x^2}\,\dd x.
 \]
-With the substitution $x = A\sin\phi$, the integral reduces to $J = \pi\sqrt{2mE}\cdot A$. Using $A = \sqrt{2E/(m\omega_0^2)}$ this becomes $J = 2\pi E/\omega_0$. Geometrically, $J$ is the area of the elliptical orbit in the $(x,p)$ phase plane.
+With the substitution $x = A\sin\phi$, the integral reduces to $J = \pi\sqrt{2mE}\cdot A$. Using $A = \sqrt{2E/(m\omega_0^2)}$ this becomes $J = 2\pi E/\omega_0$. Geometrically, $J$ equals the area of the elliptical orbit in the $(x,p)$ phase plane, confirming the computation in the example above.
 
 \item Inverting the action relation, the Hamiltonian as a function of the action alone is
 \[
@@ -156,14 +174,14 @@ E(J) = \frac{\omega_0 J}{2\pi}.
 \]
 The Hamiltonian is now linear in $J$, which is the defining feature of an action-angle representation.
 
-\item The HJ frequency is $\hat{\omega} = \pdv{E}{J} = \omega_0/(2\pi)$. The physical angular frequency is $\omega = 2\pi\hat{\omega} = \omega_0$, which is independent of the action $J$. Every oscillation shares the period $T = 2\pi/\omega_0$, regardless of energy or amplitude. This amplitude independence is the isochrony property of the harmonic oscillator.
+\item The frequency conjugate to the action is $\pdv{E}{J} = \omega_0/(2\pi)$, which counts complete cycles per unit time. The physical angular frequency is $2\pi\pdv{E}{J} = \omega_0$, independent of $J$. Every oscillation shares the period $T = 2\pi/\omega_0$, regardless of energy or amplitude. This amplitude independence is the isochrony property of the harmonic oscillator, whose algebraic origin we saw in the cancellation within $\pdv{W}{E}$.
 
-\item The angle variable advances linearly in time: $w = \hat{\omega}t + w_0 = \omega_0 t/(2\pi) + w_0$. The phase of the sinusoidal trajectory, $\omega_0 t + \phi$, equals $2\pi w$ up to a constant, matching the canonical construction.
+\item The angle variable $w\in[0,2\pi)$ tracks the oscillator\normalsize{}'s progress through one complete cycle. At $w=0$ the particle sits at the outward turning point $x=+A$ (maximum displacement, zero velocity). It crosses equilibrium toward $-A$ at $w=\pi/2$, reaches the opposite turning point $x=-A$ at $w=\pi$, and returns through equilibrium at $w=3\pi/2$. At $w=2\pi\equiv 0$ the orbit closes, having completed one period. The angle advances linearly, $w = \omega_0 t + w_0$, advancing exactly $2\pi$ each period $T=2\pi/\omega_0$. The sinusoidal phase $\omega_0 t + \phi$ coincides with $w$ up to a constant offset, confirming that $w$ measures angular position within the cycle.
 \end{enumerate}
 }
 
-\nt{Comparison with Newton's law and energy conservation}{
-Newton's second law for the harmonic oscillator gives $m\ddot{x} + m\omega_0^2 x = 0$, a linear second-order ODE whose solution is $x(t) = A\sin(\omega_0 t + \phi)$. The energy method gives $E = \tfrac12 m\dot{x}^2 + \tfrac12 m\omega_0^2 x^2$ and $\dot{x} = \pm\sqrt{2E/m - \omega_0^2 x^2}$, which integrates to the same sinusoidal motion. The Hamilton--Jacobi approach reaches the identical result through a completely different route: solving a first-order nonlinear PDE by separation, evaluating a quadrature, and applying Jacobi's theorem. The agreement confirms the equivalence of the three formulations -- Newton's, Lagrange's, and Jacobi's -- as different faces of the same underlying mechanics.}
+\nt{Comparison with Newton\normalsize{}'s law and energy conservation}{
+Newton\normalsize{}'s second law for the harmonic oscillator gives $m\ddot{x} + m\omega_0^2 x = 0$, a linear second-order ODE whose solution is $x(t) = A\sin(\omega_0 t + \phi)$. The energy method gives $E = \tfrac12 m\dot{x}^2 + \tfrac12 m\omega_0^2 x^2$ and $\dot{x} = \pm\sqrt{2E/m - \omega_0^2 x^2}$, which integrates to the same sinusoidal motion. The Hamilton--Jacobi approach reaches the identical result through a completely different route: solving a first-order nonlinear PDE by separation, evaluating a quadrature, and applying Jacobi\normalsize{}'s theorem. The agreement confirms the equivalence of the three formulations -- Newton\normalsize{}'s, Lagrange\normalsize{}'s, and Jacobi\normalsize{}'s -- as different faces of the same underlying mechanics.}
 
 \qs{Simple harmonic oscillator from the HJ complete integral}{
 A mass $m = 1.0\,\mathrm{kg}$ is attached to a horizontal spring with spring constant $k = 4.0\,\mathrm{N/m}$. The mass is displaced from equilibrium to $x_0 = 2.0\,\mathrm{m}$ and released from rest, so $v_0 = 0\,\mathrm{m/s}$.