Worked examples — Hamilton's equations of motion

Before the examples, one reminder of notation so nothing is used unearned:

The scenario matrix

Every worked example below is tagged with the cell it covers. The matrix names the kinds of situation Hamilton's machine must survive.

Example 1 — Baseline: falling mass under gravity [C1] [C8]

1. Momentum. $p=\frac{\partial L}{\partial \dot{x}}=m\dot{x}$. Why this step? $p$ is the variable that will replace $\dot{x}$; we also need it to invert $\dot{x}=p/m$.
2. Hamiltonian. $H=p\dot{x}-L=p\cdot \frac{p}{m}-(\frac{1}{2}m\frac{p^2}{m^2}-mgx)=\frac{p^2}{2m}+mgx$. Why this step? Substituting $\dot{x}=p/m$ purges every $\dot{x}$; what remains is a function of $(x,p)$ only. Here $V=mgx$ (gravity's stored energy) and it reads as $T+V$.
3. Canonical equations. $\dot{x}=\frac{\partial H}{\partial p}=\frac{p}{m}$, and $\dot{p}=-\frac{\partial H}{\partial x}=-mg$. Why this step? These are the machine's two rules; the second says the "push" decreases at a steady rate.
4. Recover acceleration. $\ddot{x}=\frac{\dot{p}}{m}=\frac{-mg}{m}=-g=-9.8{\text{m/s}}^2$. Why this step? Differentiating $\dot{x}=p/m$ and inserting $\dot{p}$ turns two first-order equations back into one Newton equation.

Example 2 — Sign flip: the repulsive (inverted) spring [C2]

1. Read off the equations. $\dot{x}=\frac{\partial H}{\partial p}=p$, $\dot{p}=-\frac{\partial H}{\partial x}=-(-4x)=+4x$. Why this step? We must apply the minus rule literally — two minus signs (one from the rule, one from $V<0$) make a plus.
2. Case $x>0$. $\dot{p}=+4x>0$: momentum grows more positive, pushing the bead further right. Why this step? We check the sign region-by-region because a sign error would only reveal itself in one region.
3. Case $x<0$. $\dot{p}=+4x<0$: momentum grows more negative, pushing the bead further left. Why this step? Both regions push outward — the origin is unstable, exactly what an inverted potential should do.
4. Case $x=0$, $p=0$. Both $\dot{x}=0$ and $\dot{p}=0$: a fixed point, but an unstable one (any nudge grows). Why this step? The degenerate equilibrium must be classified; here $\ddot{x}=\dot{p}=4x$ gives exponential growth $x\sim e^{2t}$, not oscillation.

The figure below draws this potential as a hilltop. Look at the amber arrows: they sit on the curve at four sample points and always point away from the centre — leftward when $x<0$, rightward when $x>0$. That is the geometric meaning of $\dot{p}=+4x$ having the same sign as $x$. The amber dot at the top marks the unstable fixed point: perfectly balanced, but the tiniest slip sends the bead sliding down either side.

Example 3 — Degenerate case: the free particle [C3]

1. Hamiltonian. $p=m\dot{x}$, so $H=\frac{p^2}{2m}$ (pure kinetic, since $V=0$ means no stored energy anywhere). Why this step? Setting $V=0$ is the cleanest test that the machine handles the "nothing happens" case.
2. Equations. $\dot{x}=\frac{p}{m}$, $\dot{p}=-\frac{\partial H}{\partial x}=0$. Why this step? $H$ has no $x$ in it, so ${\partial }_{x}H=0$ — momentum is exactly conserved.
3. Solve. $p=\text{const}=p_0$, hence $\dot{x}=p_0/m$ constant, hence $x(t)=x_0+\frac{p_0}{m}t$. Why this step? Constant velocity, straight line — this is Newton's first law falling out of the formalism.
4. Phase-space picture. The point $(x,p)$ moves along a horizontal line $p=p_0$, drifting right at speed $p_0/m$. Why this step? We picture the trajectory to see the degenerate limit of the closed SHO ellipse: as the restoring force $\to 0$ the ellipse opens into a flat line (we return to this exact limit in Example 7).

Example 4 — Angular coordinate: the pendulum [C4]

1. Angular momentum. $p_{\theta }=\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }$. Why this step? The conjugate of an angle is not linear momentum — it is angular momentum (units ${\text{kg⋅m}}^2/\text{s}$). Invert: $\dot{\theta }=p_{\theta }/(m{\ell }^2)$.
2. Hamiltonian. $H=p_{\theta }\dot{\theta }-L=\frac{p_{\theta }^2}{2m{\ell }^2}-mg\ell \cos \theta$. Why this step? The potential energy is $V=-mg\ell \cos \theta$ (lowest at $\theta =0$). Substituting $\dot{\theta }$ leaves $H$ in $(\theta ,p_{\theta })$ only.
3. Equations. $\dot{\theta }=\frac{\partial H}{\partial p_{\theta }}=\frac{p_{\theta }}{m{\ell }^2}$, ${\dot{p}}_{\theta }=-\frac{\partial H}{\partial \theta }=-mg\ell \sin \theta$. Why this step? The second, combined with the first, gives $m{\ell }^2\ddot{\theta }=-mg\ell \sin \theta$, the exact nonlinear pendulum.
4. Small-angle reduction. For $|\theta |\ll 1$, $\sin \theta \approx \theta$, so $\ddot{\theta }\approx -\frac{g}{\ell }\theta$ — SHO with $\omega =\sqrt{g/\ell }$. Why this step? We check the nonlinear pendulum degrades to the harmonic oscillator we already trust, tying this back to the parent's Example B. (True limiting behaviour of the SHO itself — $\omega \to 0$ and large energy — is dissected separately in Example 7.)

Example 5 — The trap: charge in a magnetic field, $p\ne m\dot{q}$ [C5]

1. Momentum — the surprise. $p=\frac{\partial L}{\partial \dot{x}}=m\dot{x}+a$. Why this step? The velocity term $a\dot{x}$ contributes $a$. So $p\ne m\dot{x}$; the "canonical" momentum differs from the "kinetic" momentum $m\dot{x}$ by $a$.
2. Invert correctly. $\dot{x}=\frac{p-a}{m}$. Why this step? We must solve $p=\partial L/\partial \dot{x}$ for the actual $L$, not reuse $p/m$. This is the whole point of the cell.
3. Hamiltonian. $H=p\dot{x}-L=p\cdot \frac{p-a}{m}-\left(\frac{1}{2}m\right(\frac{p-a}{m}{)}^2+a\cdot \frac{p-a}{m})$. Simplify: $H=\frac{(p-a{)}^2}{2m}$. Why this step? Substituting the correct $\dot{x}$ and expanding, the linear pieces cancel, leaving the kinetic energy written with the shifted momentum $p-a$ — exactly the structure of $\frac{(p-qA{)}^2}{2m}$ in electromagnetism.
4. Equations. $\dot{x}=\frac{\partial H}{\partial p}=\frac{p-a}{m}$ ✓ (matches step 2), and $\dot{p}=-\frac{\partial H}{\partial x}=0$ (no $x$ in $H$). Why this step? $p$ is conserved, but note $m\dot{x}=p-a$ is what actually moves — the physical velocity, not the canonical momentum, is what you'd measure.

Example 6 — $H$ conserved but $H\ne E$: bead on a rotating wire [C6]

1. Momentum. $p=\frac{\partial L}{\partial \dot{r}}=m\dot{r}$. Why this step? Only $\dot{r}$ appears in the velocity part, so $p$ is the ordinary radial momentum; invert $\dot{r}=p/m$.
2. Hamiltonian. $H=p\dot{r}-L=\frac{p^2}{m}-(\frac{1}{2}m\frac{p^2}{m^2}+\frac{1}{2}m{\Omega }^2{r}^2)=\frac{p^2}{2m}-\frac{1}{2}m{\Omega }^2{r}^2$. Why this step? Notice the minus on the ${\Omega }^2{r}^2$ term — the Legendre transform flipped its sign. This is why $H$ is not the sum $T+V$ you might expect.
3. Is $H$ conserved? $H$ has no explicit $t$ (the motor speed $\Omega$ is constant), so $dH/dt={\partial }_{t}H=0$. Conserved. Why this step? Conservation depends only on absence of explicit time, which holds here.
4. Is $H=E$? The true kinetic energy is $T=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{\Omega }^2{r}^2$ (both radial and rotational parts), and there is no potential so $E=T=\frac{p^2}{2m}+\frac{1}{2}m{\Omega }^2{r}^2$. But $H=\frac{p^2}{2m}-\frac{1}{2}m{\Omega }^2{r}^2$, so $H\ne E$: they differ by $H-E=-m{\Omega }^2{r}^2$. Why this step? We must directly compare the two quantities. They differ by the full rotational term $m{\Omega }^2{r}^2$. So $H$ is a conserved quantity that is not the energy — the flagship of cell C6. Whenever $r\ne 0$ the difference is nonzero; only at the axis $r=0$ do they momentarily agree.

Example 7 — Limiting behaviour of the SHO orbit [C7]

1. Ellipse equation. Here $H=E$ genuinely holds (quadratic kinetic energy, time-independent — contrast Example 6). Set $H=E$: $\frac{p^2}{2m}+\frac{1}{2}m{\omega }^2{x}^2=E$. Why this step? $H$ is conserved (parent §3), so the orbit lies on a level curve of $H$; and since $H=E$ here, that curve is a curve of constant physical energy — an ellipse.
2. Semi-axes. In $x$: set $p=0\Rightarrow x_{\max }=\sqrt{\frac{2E}{m{\omega }^2}}$. In $p$: set $x=0\Rightarrow p_{\max }=\sqrt{2mE}$. Why this step? The axis intercepts are found by asking "where does the ellipse cross each axis?" — on the $x$-axis the momentum is momentarily zero (turning point, all energy is potential), on the $p$-axis the position is zero (all energy is kinetic). Those two crossing points are the semi-axis lengths.
3. Enclosed area. Area of ellipse $=\pi x_{\max }{p}_{\max }=\pi \sqrt{\frac{2E}{m{\omega }^2}}\sqrt{2mE}=\frac{2\pi E}{\omega }$. Why this step? This confirms the parent's $\oint pdx=2\pi E/\omega$ — the action, a hint of quantization.
4. Limit $\omega \to 0$ (weak spring). $x_{\max }=\sqrt{2E/(m{\omega }^2)}\to \infty$ while $p_{\max }=\sqrt{2mE}$ stays fixed; the area $2\pi E/\omega \to \infty$. The ellipse stretches horizontally until it becomes the free-particle line $p=p_0$ of Example 3. Why this step? Removing the restoring force must recover the free particle; checking this ties C7 back to C3 and confirms the formula degrades sensibly.
5. Limit $E\to \infty$ (large energy). Both semi-axes grow as $\sqrt{E}$, so the ellipse keeps the same shape (aspect ratio $p_{\max }/x_{\max }=m\omega$ is energy-independent) but blows up in size; area grows linearly, $\propto E$. Why this step? Large-energy behaviour must be bounded/self-similar for the model to make sense — no shape distortion appears, so the SHO is well-behaved at all energies.
6. Limit $E\to 0$ (collapse). Both axes $\to 0$; the ellipse shrinks to the fixed point at the origin — the bead permanently at rest. Why this step? Checking the vanishing limit confirms the degenerate rest state, closing the loop with the unstable/stable fixed-point discussion of Example 2.

Example 8 — Real-world word problem: rocket sled on a spring bumper [C1] [C8]

1. Translate words to $L$. Kinetic $T=\frac{1}{2}m{\dot{x}}^2$, elastic potential $V=\frac{1}{2}k{x}^2$, so $L=T-V=\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{x}^2$. Why this step? The word problem must first become a Lagrangian; that is the only entry point to the machine.
2. Momentum & Hamiltonian. $p=m\dot{x}\Rightarrow \dot{x}=p/m$; $H=\frac{p^2}{2m}+\frac{1}{2}k{x}^2$. Why this step? Building $H$ in $(x,p)$ lets us read the equations directly; here $H=T+V=E$ (Example 1-type, scleronomic).
3. Equations & frequency. $\dot{x}=p/m$, $\dot{p}=-kx$, so $m\ddot{x}=-kx$, i.e. $\ddot{x}=-\frac{k}{m}x$, giving $\omega =\sqrt{k/m}$. Why this step? Eliminating $p$ recovers Newton and identifies the angular frequency.

Example 9 — Exam twist: reverse the Legendre transform [C9]

1. Get $\dot{q}$ from $H$. $\dot{q}=\frac{\partial H}{\partial p}=\frac{p}{m}\Rightarrow p=m\dot{q}$. Why this step? The inverse Legendre transform uses the same pairing $\dot{q}={\partial }_{p}H$; we solve it for $p$ in terms of $\dot{q}$.
2. Apply $L=p\dot{q}-H$. $L=(m\dot{q})\dot{q}-(\frac{(m\dot{q}{)}^2}{2m}+V(q))=m{\dot{q}}^2-\frac{1}{2}m{\dot{q}}^2-V(q)$. Why this step? The Legendre transform is its own inverse (up to sign convention): $L=\sum p\dot{q}-H$ with $p$ re-expressed via $\dot{q}$.
3. Simplify. $L=\frac{1}{2}m{\dot{q}}^2-V(q)$. Why this step? We recover exactly the Lagrangian we'd expect — kinetic $T=\frac{1}{2}m{\dot{q}}^2$ minus potential $V(q)$, i.e. $L=T-V$ — confirming the round trip.

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Related: Legendre Transform · Phase Space and Liouville's Theorem · Lagrangian Mechanics · Poisson Brackets · Canonical Transformations