diff --git a/concepts/advanced/sho-hj.tex b/concepts/advanced/sho-hj.tex
index fbacc50..6b558ef 100644
--- a/concepts/advanced/sho-hj.tex
+++ b/concepts/advanced/sho-hj.tex
@@ -141,7 +141,7 @@ 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,
+The action variable $J = \frac{1}{2\pi}\oint p\,\dd x$ is $\frac{1}{2\pi}$ times 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,
 \[
 \frac{p^2}{2m} + \frac{1}{2}m\omega_0^2 x^2 = E,
 \]
@@ -149,32 +149,32 @@ 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
+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. The enclosed area is $\pi A\,p_{\max}$. The action variable is $\frac{1}{2\pi}$ times this area:
 \[
-J = \pi A\,p_{\max}
-= \pi\sqrt{\frac{2E}{m\omega_0^2}}\cdot\sqrt{2mE}
-= \pi\sqrt{\frac{4E^2}{\omega_0^2}}
-= \frac{2\pi E}{\omega_0}.
+J = \frac{1}{2\pi}\cdot\pi A\,p_{\max}
+= \frac{1}{2}\sqrt{\frac{2E}{m\omega_0^2}}\cdot\sqrt{2mE}
+= \frac{1}{2}\sqrt{\frac{4E^2}{\omega_0^2}}
+= \frac{E}{\omega_0}.
 \]
-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.}
+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:
 
 \begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
-\item The action variable is the phase-space area enclosed by one complete cycle:
+\item The action variable is $\frac{1}{2\pi}$ times the phase-space area enclosed by one complete cycle:
 \[
-J = \oint p\,\dd x = 2\int_{-A}^{A}\sqrt{2mE - m^2\omega_0^2 x^2}\,\dd x.
+J = \frac{1}{2\pi}\oint p\,\dd x = \frac{1}{\pi}\int_{-A}^{A}\sqrt{2mE - m^2\omega_0^2 x^2}\,\dd x.
 \]
-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, $J$ equals the area of the elliptical orbit in the $(x,p)$ phase plane, confirming the computation in the example above.
+With the substitution $x = A\sin\phi$, the integral $\int_{-A}^{A}\sqrt{2mE - m^2\omega_0^2 x^2}\,\dd x$ evaluates to $\tfrac{\pi}{2}\sqrt{2mE}\cdot A$. With the prefactor $1/\pi$ this gives $J = \tfrac{1}{2}\sqrt{2mE}\cdot A$. Using $A = \sqrt{2E/(m\omega_0^2)}$ we obtain $J = E/\omega_0$. Geometrically, $J$ is $\frac{1}{2\pi}$ times 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
 \[
-E(J) = \frac{\omega_0 J}{2\pi}.
+E(J) = \omega_0 J.
 \]
-The Hamiltonian is now linear in $J$, which is the defining feature of an action-angle representation.
+The Hamiltonian is linear in $J$, which is the defining feature of an action-angle representation.
 
-\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 frequency conjugate to the action is $\pdv{E}{J} = \omega_0$, which is the physical angular frequency itself, 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 $w\in[0,2\pi)$ 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}
@@ -189,7 +189,7 @@ A mass $m = 1.0\,\mathrm{kg}$ is attached to a horizontal spring with spring con
 \begin{enumerate}[label=(\alph*)]
 \item Compute the natural angular frequency $\omega_0 = \sqrt{k/m}$. Write the Hamilton--Jacobi equation for this system, separate the variables to find $\der{W}{x}$, and state the complete integral $\mcS(x,t;E)$ with numerical parameter values.
 \item Use the initial conditions $x(0) = 2.0\,\mathrm{m}$ and $\dot{x}(0) = 0\,\mathrm{m/s}$ to find the total energy $E$ and the amplitude $A = \sqrt{2E/(m\omega_0^2)}$. Write the trajectory $x(t)$ and verify that the maximum speed equals $A\omega_0$.
-\item Compute the action variable $J = 2\pi E/\omega_0$ in SI units and verify numerically that $E(J) = \omega_0 J/(2\pi)$ reproduces the original energy.
+\item Compute the action variable $J = E/\omega_0$ in SI units and verify numerically that $E(J) = \omega_0 J$ reproduces the original energy.
 \end{enumerate}}
 
 \sol \textbf{Part (a).} The natural angular frequency is
@@ -262,24 +262,20 @@ From energy, $v_{\max} = \sqrt{2E/m} = \sqrt{16.0/1.0}\,\mathrm{m/s} = 4.0\,\mat
 
 \textbf{Part (c).} The action variable for the harmonic oscillator is
 \[
-J = \frac{2\pi E}{\omega_0}.
+J = \frac{E}{\omega_0}.
 \]
 Substitute the numerical values:
 \[
-J = \frac{2\pi(8.0\,\mathrm{J})}{2.0\,\mathrm{rad/s}}
-= 8\pi\,\mathrm{J\!\cdot\!s}.
+J = \frac{8.0\,\mathrm{J}}{2.0\,\mathrm{rad/s}}
+= 4.0\,\mathrm{J\!\cdot\!s}.
 \]
-Evaluating numerically:
+Now verify the energy--action relation $E(J) = \omega_0 J$:
 \[
-J = 8\pi\,\mathrm{J\!\cdot\!s} \approx 25\,\mathrm{J\!\cdot\!s}.
-\]
-Now verify the energy--action relation $E(J) = \omega_0 J/(2\pi)$:
-\[
-E(J) = \frac{\omega_0 J}{2\pi}
-= \frac{(2.0\,\mathrm{rad/s})(8\pi\,\mathrm{J\!\cdot\!s})}{2\pi}
+E(J) = \omega_0 J
+= (2.0\,\mathrm{rad/s})(4.0\,\mathrm{J\!\cdot\!s})
 = 8.0\,\mathrm{J}.
 \]
-This returns the original energy exactly, confirming $E(J) = \omega_0 J/(2\pi)$ both algebraically and for the numerical values of this problem.
+This returns the original energy exactly, confirming $E(J) = \omega_0 J$ both algebraically and for the numerical values of this problem.
 
 Therefore,
 \[
@@ -287,5 +283,5 @@ Therefore,
 \qquad
 x(t) = (2.0\,\mathrm{m})\cos\bigl((2.0\,\mathrm{rad/s})\,t\bigr),
 \qquad
-J = 8\pi\,\mathrm{J\!\cdot\!s}.
+J = 4.0\,\mathrm{J\!\cdot\!s}.
 \]
\ No newline at end of file