\subsection{Free Particle in 1D and 3D}

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:
\[
\mcH = \frac{p^2}{2m}.
\]
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 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 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:
\[
p_x = m v_{x0} = (2.0)(3.0)\,\mathrm{kg\!\cdot\!m/s} = 6.0\,\mathrm{kg\!\cdot\!m/s},
\]
\[
p_y = m v_{y0} = (2.0)(4.0)\,\mathrm{kg\!\cdot\!m/s} = 8.0\,\mathrm{kg\!\cdot\!m/s},
\qquad
p_z = 0.
\]
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).} The position vector at $t = 2.5\,\mathrm{s}$ is
\[
\mathbf{r}(2.5\,\mathrm{s}) = \bigl(3.0(2.5),\,4.0(2.5),\,0\bigr)\,\mathrm{m}
= (7.5,\,10.0,\,0)\,\mathrm{m}.
\]
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,
\[
\mcS(x,y,z,t) = 6.0\,x + 8.0\,y - 25\,t,
\qquad
\mathbf{r}(t) = (3.0\,t,\,4.0\,t,\,0)\,\mathrm{m}.
\]