fix(HJ): A.06 — Add motivation, cross-refs to U1/U3, fill derivation gaps

concepts/advanced/projectile-hj.tex

@@ -1,6 +1,6 @@
 \subsection{Projectile Motion via Hamilton-Jacobi}
 
-This subsection solves the Hamilton--Jacobi equation for projectile motion in a uniform gravitational field, showing that Jacobi's theorem reproduces the standard parabolic kinematics of the AP Physics C curriculum.
+This subsection solves the Hamilton--Jacobi equation for projectile motion in a uniform gravitational field, demonstrating that Jacobi's theorem reproduces the standard parabolic trajectory without ever integrating Newton's second-order differential equations. The kinematics approach (Unit 1, Section m1--5) solves two decoupled ODEs for $x(t)$ and $y(t)$ separately. Here, a single first-order PDE separates into the same two independent problems because the cyclic coordinate $x$ forces the horizontal--vertical split by the structure of the formalism. The energy budget, cross-referencing Unit 3, emerges naturally from the separation constants rather than from the work--energy theorem.
 
 \dfn{Projectile Hamiltonian}{
 A particle of mass $m$ moving in the $xy$-plane under uniform gravity $g$ has the Hamiltonian
@@ -14,6 +14,8 @@ where $y$ is measured upward from ground level and $p_x$, $p_y$ are the canonica
 
 \nt{The coordinate $x$ is cyclic (ignorable) because it does not appear in the Hamiltonian. Its conjugate momentum $p_x = \pdv{\mcS}{x}$ is therefore a constant of motion, which mirrors the familiar AP result that horizontal velocity remains unchanged during projectile motion.}
 
+\nt{Energy budget: the total energy $E$ splits into a horizontal part $E_x = \alpha_x^2/(2m) = \tfrac{1}{2}mv_{0x}^2$, which is constant because $x$ is cyclic, and a vertical part $E_y = \tfrac{1}{2}mv_y^2 + mgy$. The separation constant $E_x$ is the horizontal kinetic energy, carrying no potential contribution. The vertical energy $E_y$ accounts for both the vertical kinetic and gravitational potential energy, so $E_y$ is conserved within the vertical subsystem. The total energy is $E = E_x + E_y = \tfrac{1}{2}m(v_{0x}^2 + v_{0y}^2)$, matching the initial kinetic energy at ground level. This energy partition is equivalent to the mechanical-energy bookkeeping used in Unit 3.}
+
 \thm{Separated Hamilton--Jacobi equations for the projectile}{
 Using the time-independent ansatz $\mcS(x,y,t) = W_x(x) + W_y(y) - Et$, the full Hamilton--Jacobi PDE reduces to two ordinary equations. Because $x$ is cyclic, $\der{W_x}{x} = \alpha_x$ (constant). The remaining vertical equation is
 \[
@@ -55,21 +57,49 @@ Solving for $x(t)$ with $\beta_x = 0$ (launch from the origin):
 \[
 x(t) = \frac{\alpha_x}{m}\,t = v_{0x}\,t.
 \]
-From Jacobi's theorem with respect to $E$, using $\pdv{E_y}{E} = 1$:
+To find $y(t)$, apply Jacobi's theorem with respect to $E$. The principal function depends on $E$ both explicitly in the term $-Et$ and implicitly through $E_y(E) = E - \alpha_x^2/(2m)$. The chain rule gives
 \[
-\pdv{\mcS}{E} = \mp\frac{2\sqrt{2m}}{3mg}\cdot\frac{3}{2}\left(E_y - mgy\right)^{1/2} - t = \beta_E,
+\pdv{\mcS}{E}
+= \pdv{\mcS}{E_y}\,\pdv{E_y}{E} - t.
+\]
+Since $\pdv{E_y}{E} = 1$, substituting the $W_y$ term yields
+\[
+\pdv{\mcS}{E}
+= \mp\frac{2\sqrt{2m}}{3mg}\cdot\frac{3}{2}\left(E_y - mgy\right)^{1/2} - t = \beta_E,
 \]
 which simplifies to
 \[
 \mp\frac{\sqrt{2m}}{mg}\sqrt{E_y - mgy} - t = \beta_E.
 \]
-Solve the squared relation for $y(t)$. At $t = 0$ the particle is at $y = 0$ with vertical speed $v_{0y} = \sqrt{2E_y/m}$. The initial conditions fix $\beta_E$ and yield
+To solve for $y(t)$, choose the sign consistent with an upward launch: $p_y = \der{W_y}{y} > 0$ at $t = 0$ selects the upper square root for $\der{W_y}{y}$, which gives the negative sign in $\mcS$. Rearrange:
+\[
+\frac{\sqrt{2m}}{mg}\sqrt{E_y - mgy} - t = \beta_E.
+\]
+At $t = 0$ with $y = 0$ and $v_{0y} = \sqrt{2E_y/m}$, the integration constant is fixed:
+\[
+\frac{\sqrt{2m}}{mg}\sqrt{E_y} = -\beta_E
+\qquad\Longrightarrow\qquad
+\beta_E = -\frac{\sqrt{2E_y/m}}{g} = -\frac{v_{0y}}{g}.
+\]
+Substitute $\beta_E$ back and isolate the radical:
+\[
+\frac{\sqrt{2m}}{mg}\sqrt{E_y - mgy} = \frac{v_{0y}}{g} - t.
+\]
+Multiply by $mg/\sqrt{2m}$ and square both sides:
+\[
+E_y - mgy = \frac{m}{2}\left(v_{0y} - gt\right)^2.
+\]
+Since $2E_y/m = v_{0y}^2$, divide through by $m$ and expand the right-hand side:
+\[
+\frac{1}{2}v_{0y}^2 - gy = \frac{1}{2}\left(v_{0y}^2 - 2v_{0y}gt + g^2t^2\right).
+\]
+The term $\tfrac{1}{2}v_{0y}^2$ cancels, leaving $gy = v_{0y}gt - \tfrac{1}{2}g^2t^2$, so
 \[
 y(t) = v_{0y}\,t - \frac{1}{2}\,g\,t^2.
 \]
 The two equations combine into the parabolic trajectory $y = (v_{0y}/v_{0x})\,x - \tfrac{g}{2v_{0x}^2}\,x^2$.}
 
-\nt{Comparison to AP kinematics}{The Hamilton--Jacobi equations $x(t) = v_{0x}t$ and $y(t) = v_{0y}t - \tfrac{1}{2}gt^2$ are identical to the standard AP C constant-acceleration kinematics. The separation constant $\alpha_x = mv_{0x}$ is the conserved horizontal momentum, and the transverse energy $E_y = \tfrac{1}{2}mv_{0y}^2$ encodes the initial vertical kinetic energy. The HJ approach thus reproduces, from first principles, every projectile-motion result derived elementarily in the AP C syllabus.}
+\nt{Comparison to AP kinematics}{The Hamilton--Jacobi equations $x(t) = v_{0x}t$ and $y(t) = v_{0y}t - \tfrac{1}{2}gt^2$ are identical to the constant-acceleration kinematics of Unit 1 (Section m1--5, free-fall formulas). The separation constant $\alpha_x = mv_{0x}$ is the conserved horizontal momentum, and the transverse energy $E_y = \tfrac{1}{2}mv_{0y}^2$ encodes the initial vertical kinetic energy. Where kinematics integrates $a_x = 0$ and $a_y = -g$ separately into two decoupled ODEs, the Hamilton--Jacobi approach solves one PDE and lets the cyclic coordinate enforce the exact same horizontal--vertical split. The HJ formalism reproduces, from first principles, every projectile-motion result derived elementarily in the AP C syllabus.}
 
 \qs{Projectile launched from the ground}{
 A projectile of mass $m = 0.50\,\mathrm{kg}$ is launched from the origin with speed $v_0 = 20\,\mathrm{m/s}$ at angle $\theta_0 = 30.0^\circ$ above the horizontal. Use $g = 9.81\,\mathrm{m/s^2}$.