diff --git a/concepts/advanced/free-particle-hj.tex b/concepts/advanced/free-particle-hj.tex
index 9783d95..2818c11 100644
--- a/concepts/advanced/free-particle-hj.tex
+++ b/concepts/advanced/free-particle-hj.tex
@@ -1,6 +1,6 @@
 \subsection{Free Particle in 1D and 3D}
 
-This subsection solves the Hamilton--Jacobi equation for a free particle in one and three dimensions, demonstrating that Jacobi's theorem reproduces the familiar result of uniform straight-line motion.
+The free particle is the simplest test of the Hamilton--Jacobi formalism, showing the full machinery in the least cluttered setting. With no potential to complicate the Hamiltonian, every step of the method -- separation of variables, identification of the complete integral, application of Jacobi\normalsize{}'s theorem -- can be seen clearly. Here we solve the equation in both one and three dimensions and recover the familiar result of uniform straight-line motion.
 
 \dfn{Free particle Hamiltonian and Hamilton--Jacobi equation}{
 For a free particle of mass $m$ the Hamiltonian is purely kinetic:
@@ -72,14 +72,22 @@ Treating $(p_x, p_y, p_z)$ as three independent separation constants, Jacobi's t
 \]
 Each coordinate evolves linearly with time, confirming uniform straight-line motion in three dimensions.}
 
-\nt{Connection to Newton's second law}{Each coordinate equation $q_i(t) = (p_i/m)t + \beta_i$ integrates a constant velocity $\dot{q}_i = p_i/m$. The acceleration vanishes, $\ddot{q}_i = 0$, which is precisely the result of Newton's second law for zero applied force. The Hamilton--Jacobi formalism therefore reproduces the familiar kinematic result of uniform motion along a straight line.}
+\nt{Connection to earlier material}{Free-particle motion was already solved in Unit 1 (Kinematics), where constant velocity and linear position functions $x(t) = v_0 t + x_0$ were obtained directly from Newton\normalsize{}'s second law for zero net force. Unit 4 (Momentum and Impulse) used conservation of linear momentum to analyze collisions of free particles.
+
+Here we recover all of those results without writing a differential equation: each coordinate evolves as $q_i(t) = (p_i/m)t + \beta_i$, carrying a conserved momentum $p_i$ and zero acceleration $\ddot{q}_i = 0$. The Hamilton--Jacobi formalism reproduces familiar kinematics from the geometry of the action, treating the entire trajectory as a consequence of the action principle rather than Newton\normalsize{}'s second law.}
+
+\nt{Geometric picture of the action}{In three dimensions the principal function $\mcS(x,y,z,t) = p_x x + p_y y + p_z z - Et$ is a linear function, so its level sets are planes in configuration space. The surfaces $\mcS = \text{const}$ are parallel planes $p_x x + p_y y + p_z z = \text{const} + Et$, propagating along the direction of $\mathbf{p}$ with speed $|\mathbf{p}|/m$.
+
+This is the wavefront picture from the optics analogy discussed in A.01. The action $\mcS$ plays the role of an optical phase, and curves orthogonal to the wavefronts are the ray paths. Those ray paths coincide with the particle trajectories. For a free particle there is no potential to refract the rays, so the wavefronts remain perfectly planar and propagate without distortion. The constant velocity of the wavefronts reflects the constant speed of the particle itself.}
+
+As stated in Jacobi\normalsize{}'s theorem (A.01), setting $\pdv{\mcS}{\alpha_i} = \beta_i$ for each separation constant $\alpha_i$ gives the equations of motion.
 
 \qs{Free particle in three dimensions}{A free particle of mass $m = 2.0\,\mathrm{kg}$ passes through the origin at $t = 0$ with initial velocity $(3.0,\,4.0,\,0)\,\mathrm{m/s}$.
 
 \begin{enumerate}[label=(\alph*)]
 \item Write Hamilton's principal function $\mcS(x,y,z,t)$ using the additive separation ansatz $\mcS = p_x x + p_y y + p_z z - Et$, substituting numerical values for the momenta and energy.
 \item From Jacobi's theorem, $\pdv{\mcS}{p_x} = \beta_x$, $\pdv{\mcS}{p_y} = \beta_y$, $\pdv{\mcS}{p_z} = \beta_z$, find $x(t)$, $y(t)$, and $z(t)$ using the given initial conditions.
-\item Verify that the trajectory matches $\mathbf{r}(t) = (3.0t,\,4.0t,\,0)\,\mathrm{m}$.
+\item At $t = 2.5\,\mathrm{s}$, compute the position vector $\mathbf{r}(t)$, the speed $|\mathbf{v}|$, and the kinetic energy $K$. Verify that $K$ equals the total energy $E$ found in Part (a).
 \end{enumerate}}
 
 \sol \textbf{Part (a).} Compute the canonical momenta from the initial velocity:
@@ -129,11 +137,20 @@ Since $p_z = 0$ and $z(0) = 0$, we obtain $\beta_z = 0$ and
 z(t) = 0.
 \]
 
-\textbf{Part (c).} Assembling the three components:
+\textbf{Part (c).} The position vector at $t = 2.5\,\mathrm{s}$ is
 \[
-\mathbf{r}(t) = \bigl(x(t),\,y(t),\,z(t)\bigr) = (3.0\,t,\,4.0\,t,\,0).
+\mathbf{r}(2.5\,\mathrm{s}) = \bigl(3.0(2.5),\,4.0(2.5),\,0\bigr)\,\mathrm{m}
+= (7.5,\,10.0,\,0)\,\mathrm{m}.
 \]
-This matches the expected trajectory $\mathbf{r}(t) = (3.0t,\,4.0t,\,0)\,\mathrm{m}$, confirming the consistency of the Hamilton--Jacobi formalism for the free particle. The speed is $|\mathbf{v}| = \sqrt{3.0^2 + 4.0^2} = 5.0\,\mathrm{m/s}$, and the kinetic energy $E = \tfrac{1}{2}(2.0)(5.0)^2 = 25\,\mathrm{J}$ matches the energy from part (a).
+The velocity is constant, $\mathbf{v} = (3.0,\,4.0,\,0)\,\mathrm{m/s}$, so the speed is
+\[
+|\mathbf{v}| = \sqrt{3.0^2 + 4.0^2} = 5.0\,\mathrm{m/s}.
+\]
+The kinetic energy at this instant is
+\[
+K = \tfrac{1}{2}\,m\,|\mathbf{v}|^2 = \tfrac{1}{2}(2.0)(5.0)^2\,\mathrm{J} = 25\,\mathrm{J}.
+\]
+This equals the total energy $E = 25\,\mathrm{J}$ from Part (a), confirming energy conservation for the free particle.
 
 Therefore,
 \[