\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.

\dfn{Free particle Hamiltonian and Hamilton--Jacobi equation}{
For a free particle of mass $m$ the Hamiltonian is purely kinetic:
\[
\mcH = \frac{p^2}{2m}.
\]
In one dimension, substituting $p = \pdv{\mcS}{x}$ into the Hamilton--Jacobi equation $\mcH + \pdv{\mcS}{t} = 0$ yields
\[
\frac{1}{2m}\left(\pdv{\mcS}{x}\right)^2 + \pdv{\mcS}{t} = 0.
\]
The three-dimensional Hamiltonian is $\mcH = (p_x^2 + p_y^2 + p_z^2)/(2m)$ and the corresponding Hamilton--Jacobi equation is
\[
\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \left(\pdv{\mcS}{y}\right)^2 + \left(\pdv{\mcS}{z}\right)^2\right] + \pdv{\mcS}{t} = 0.
\]}

\thm{Complete integral for a 1D free particle}{
The complete integral of the one-dimensional free-particle Hamilton--Jacobi equation is
\[
\mcS(x,t;E) = \pm\sqrt{2mE}\,x - Et,
\]
where $E > 0$ is the total mechanical energy. Jacobi's theorem $\pdv{\mcS}{E} = \beta$ gives the trajectory
\[
x(t) = \pm\sqrt{\frac{2E}{m}}\,(t+\beta) = v_0 t + x_0,
\]
with constant velocity $v_0 = \pm\sqrt{2E/m}$ and initial position $x_0 = v_0\beta$.}

\pf{Derivation of the 1D and 3D free-particle action}{
Because the free-particle Hamiltonian has no explicit time dependence, $\pdv{\mcH}{t} = 0$ and energy is conserved: $\mcH = E$. Use the time-independent reduction $\mcS(x,t) = W(x) - Et$. Substituting:
\[
\frac{1}{2m}\left(\der{W}{x}\right)^2 - E = 0.
\]
Solve for the derivative:
\[
\der{W}{x} = \pm\sqrt{2mE}.
\]
Integrate with respect to $x$ (absorbing the integration constant into the additive constant of $\mcS$):
\[
W(x) = \pm\sqrt{2mE}\,x.
\]
Reassemble the principal function:
\[
\mcS(x,t) = \pm\sqrt{2mE}\,x - Et.
\]
Jacobi's theorem requires $\pdv{\mcS}{E} = \beta$. Differentiate:
\[
\pdv{\mcS}{E} = \pm\frac{m}{\sqrt{2mE}}\,x - t = \beta.
\]
Solve for $x(t)$:
\[
x = \pm\frac{\sqrt{2mE}}{m}\,(t+\beta) = \pm\sqrt{\frac{2E}{m}}\,(t+\beta).
\]
Setting $v_0 = \pm\sqrt{2E/m}$ and $x_0 = v_0\beta$ gives $x(t) = v_0 t + x_0$.

In three dimensions all Cartesian coordinates are cyclic, so each conjugate momentum is conserved. Setting $\pdv{\mcS}{x} = p_x$, $\pdv{\mcS}{y} = p_y$, $\pdv{\mcS}{z} = p_z$ as constants:
\[
W(x,y,z) = p_x x + p_y y + p_z z,
\]
with energy $E = (p_x^2 + p_y^2 + p_z^2)/(2m)$. The principal function is
\[
\mcS(x,y,z,t) = p_x x + p_y y + p_z z - Et.
\]
Treating $(p_x, p_y, p_z)$ as three independent separation constants, Jacobi's theorem gives
\[
\pdv{\mcS}{p_x} = x - \frac{p_x}{m}\,t = \beta_x,
\qquad
\pdv{\mcS}{p_y} = y - \frac{p_y}{m}\,t = \beta_y,
\qquad
\pdv{\mcS}{p_z} = z - \frac{p_z}{m}\,t = \beta_z.
\]
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.}

\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 $\mathbf{r}(t) = (3.0t,\,4.0t,\,0)\,\mathrm{m}$.
\end{enumerate}}

\sol \textbf{Part (a).} Compute the canonical momenta from the initial velocity:
\[
p_x = m v_{x0} = (2.0)(3.0)\,\mathrm{kg\!\cdot\!m/s} = 6.0\,\mathrm{kg\!\cdot\!m/s},
\]
\[
p_y = m v_{y0} = (2.0)(4.0)\,\mathrm{kg\!\cdot\!m/s} = 8.0\,\mathrm{kg\!\cdot\!m/s},
\qquad
p_z = 0.
\]
The total energy is
\[
E = \frac{p_x^2 + p_y^2 + p_z^2}{2m}
= \frac{(6.0)^2 + (8.0)^2}{2(2.0)}\,\mathrm{J}
= \frac{36 + 64}{4.0}\,\mathrm{J}
= 25\,\mathrm{J}.
\]
Substitute these into the additive ansatz:
\[
\mcS(x,y,z,t) = 6.0\,x + 8.0\,y - 25\,t,
\]
where $x$ and $y$ are in metres, $t$ in seconds, and $\mcS$ in joule-seconds.

\textbf{Part (b).} Apply Jacobi's theorem for each momentum component. For the $x$-coordinate:
\[
\pdv{\mcS}{p_x} = x - \frac{p_x}{m}\,t = \beta_x.
\]
At $t = 0$ the particle is at the origin, so $x(0) = 0$ and $\beta_x = 0$. Therefore,
\[
x(t) = \frac{p_x}{m}\,t = \frac{6.0}{2.0}\,t = 3.0\,t.
\]
For the $y$-coordinate:
\[
\pdv{\mcS}{p_y} = y - \frac{p_y}{m}\,t = \beta_y.
\]
With $y(0) = 0$, we have $\beta_y = 0$ and
\[
y(t) = \frac{p_y}{m}\,t = \frac{8.0}{2.0}\,t = 4.0\,t.
\]
For the $z$-coordinate:
\[
\pdv{\mcS}{p_z} = z - \frac{p_z}{m}\,t = \beta_z.
\]
Since $p_z = 0$ and $z(0) = 0$, we obtain $\beta_z = 0$ and
\[
z(t) = 0.
\]

\textbf{Part (c).} Assembling the three components:
\[
\mathbf{r}(t) = \bigl(x(t),\,y(t),\,z(t)\bigr) = (3.0\,t,\,4.0\,t,\,0).
\]
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).

Therefore,
\[
\mcS(x,y,z,t) = 6.0\,x + 8.0\,y - 25\,t,
\qquad
\mathbf{r}(t) = (3.0\,t,\,4.0\,t,\,0)\,\mathrm{m}.
\]
