Worked examples — Hamilton's principle — least action

Before we start, one reminder of the recipe you will use every single time:

Here $\dot{q}$ means "the rate of change of $q$ with time" and $\partial$ (curly d) means "differentiate treating the other slots as frozen constants". That is the only notation we need.

The scenario matrix

Every problem in this topic falls into one of these cells. The examples below each carry a tag like [Cell A] so you can see the whole grid gets covered.

We also touch every sign of the force slot: positive (Ex 2), negative (Ex 1, 4), zero (Ex 3, 5, 6).

Example 1 — Bowl-shaped potential (spring) · [Cell A]

Forecast: guess before reading — the mass should oscillate; will the frequency go up or down if we stiffen the spring (larger $k$)?

1. Momentum slot. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$. Why this step? Only the $\frac{1}{2}m{\dot{x}}^2$ term contains $\dot{x}$; differentiating gives $m\dot{x}$ — ordinary momentum.
2. Force slot. $\frac{\partial L}{\partial x}=-kx$. Why this step? Only $-\frac{1}{2}k{x}^2$ contains $x$; its slope is $-kx$. The minus sign means the force pulls back toward $x=0$ — a bowl (positive curvature $V^{\prime \prime }=+k>0$).
3. Assemble E–L. $\frac{d}{dt}(m\dot{x})-(-kx)=m\ddot{x}+kx=0$. Why this step? This is the recipe's line 4 verbatim.
4. Read the physics. $\ddot{x}=-\frac{k}{m}x=-{\omega }^{2}x$ with $\omega =\sqrt{k/m}$. Solution $x(t)=A\cos (\omega t+\varphi )$.

Verify: $\omega =\sqrt{8/2}=\sqrt{4}=2\text{rad/s}$. Units: $\sqrt{(\text{N/m})/\text{kg}}=\sqrt{({\text{kg⋅m/s}}^2/\text{m})/\text{kg}}=\sqrt{1/{\text{s}}^2}={\text{s}}^{-1}$ ✓. Stiffer spring ⇒ larger $k$ ⇒ higher $\omega$ ✓ (matches forecast).

Example 2 — Hill-shaped potential (unstable) · [Cell B]

Forecast: the sign of $V$ flipped. Does the mass still oscillate, or run away?

1. Read off $V$ from $L=T-V$. We are given $L=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}k{x}^2$. Match against the recipe's $L=T-V$: the kinetic part is $T=\frac{1}{2}m{\dot{x}}^2$, so what remains must equal $-V$. That is, $-V=+\frac{1}{2}k{x}^2$, hence $V=-\frac{1}{2}k{x}^2$. Why this step? The potential is not whatever term follows the minus sign in $L$ — it is minus the non-kinetic part. Since $L$ has $+\frac{1}{2}k{x}^2$, algebra forces $V=-\frac{1}{2}k{x}^2$: curvature $V^{\prime \prime }=-k<0$, a hilltop.
2. Momentum slot. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$. Why this step? Recipe line 3, first half — never skip it. Only $\frac{1}{2}m{\dot{x}}^2$ contains $\dot{x}$; its derivative is the ordinary momentum $m\dot{x}$ (identical to Ex 1, since the kinetic energy is unchanged).
3. Force slot. $\frac{\partial L}{\partial x}=+kx$. Why this step? Only $+\frac{1}{2}k{x}^2$ contains $x$; its slope is $+kx$. The plus sign means the force pushes away from the origin — consistent with the hilltop found in step 1.
4. Assemble E–L. $\frac{d}{dt}(m\dot{x})-(+kx)=m\ddot{x}-kx=0\Rightarrow \ddot{x}=+\frac{k}{m}x$. Why this step? Positive coefficient means acceleration points outward — the opposite sign to Ex 1.
5. Read the physics. $\ddot{x}={\omega }^{2}x$ has solutions $x=A{e}^{\lambda t}+B{e}^{-\lambda t}$ with $\lambda =\sqrt{k/m}$ — exponential growth, not oscillation. Why $e^{\lambda t}$ and not $\cos$? We need a function whose second derivative equals plus a constant times itself. $\cos$ gives minus; the exponential $e^{\lambda t}$ gives ${\lambda }^2{e}^{\lambda t}$ — same sign. That is precisely the tool that answers "second derivative = $+$ itself".

Verify: $\lambda =\sqrt{4/1}=2{\text{s}}^{-1}$. Check $x=e^{2t}\Rightarrow \ddot{x}=4e^{2t}=4x$ and ${\omega }^{2}x=4x$ ✓. Runaway confirmed — this is an unstable equilibrium; least action still holds, but the extremum is a saddle, not a minimum.

Example 3 — Constant force (linear potential) · [Cell C]

Forecast: a constant force means constant acceleration. Which classic formula do you expect to fall out?

1. Momentum slot. $\frac{\partial L}{\partial \dot{y}}=m\dot{y}$. Why this step? Only the kinetic term $\frac{1}{2}m{\dot{y}}^2$ depends on $\dot{y}$; its derivative is the ordinary momentum $m\dot{y}$.
2. Force slot. $\frac{\partial L}{\partial y}=-mg$ (constant — the force slot does not depend on $y$; here it is a nonzero constant, our "constant-force" case). Why this step? Only $-mgy$ contains $y$; its slope is the constant $-mg$, the weight pulling downward.
3. Assemble. $\frac{d}{dt}(m\dot{y})-(-mg)=m\ddot{y}+mg=0\Rightarrow \ddot{y}=-g$. Why this step? A linear potential $mgy$ has constant slope, so the acceleration is a constant.
4. Integrate twice. $\dot{y}=-gt$, $y=y_0-\frac{1}{2}g{t}^2$. Why this step? Antiderivatives of a constant give the familiar kinematic equations — E–L reproduces them exactly.

Verify: set $y=0$: $20=\frac{1}{2}(10)t^2=5t^2\Rightarrow t^2=4\Rightarrow t=2\text{s}$. Units: $\sqrt{\text{m}/({\text{m/s}}^2)}=\sqrt{{\text{s}}^2}=\text{s}$ ✓.

Example 4 — Angular coordinate: the pendulum · [Cell D]

Forecast: with no forces to resolve, can a single angle carry all the geometry?

1. Build $T,V$ from geometry. Speed of the bob $=\ell \dot{\theta }$ (arc length rate), so $T=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$. Height above the lowest point $=\ell (1-\cos \theta )$, so $V=mg\ell (1-\cos \theta )$. Drop the constant: $L=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2+mg\ell \cos \theta$. Why this step? Look at the figure: the amber dotted arc shows the bob's path; its length rate is the speed. The cyan vertical shows the height. No tension, no force components needed.
2. Momentum slot. $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }$ (this is angular momentum). Why this step? Only the kinetic term $\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$ contains $\dot{\theta }$; differentiating gives $m{\ell }^2\dot{\theta }$, the angular momentum.
3. Force slot. $\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta$ (negative for $0<\theta <180°$: pulls back down — a restoring torque). Why this step? Only $mg\ell \cos \theta$ contains $\theta$; the derivative of $\cos \theta$ is $-\sin \theta$, giving $-mg\ell \sin \theta$.
4. Assemble. $m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0\Rightarrow \ddot{\theta }=-\frac{g}{\ell }\sin \theta$. Why this step? This is the recipe's line 4; dividing by $m{\ell }^2$ isolates the angular acceleration.

Verify: at $\theta =30°=\pi /6$, $\sin 30°=0.5$, so $\ddot{\theta }=-9.8/1\times 0.5=-4.9{\text{rad/s}}^2$. The minus sign says it accelerates back toward $\theta =0$ ✓. Units ${\text{m/s}}^2÷\text{m}={\text{s}}^{-2}$, correct for angular acceleration ✓.

Example 5 — Cyclic coordinate: conserved momentum · [Cell E]

Forecast: what happens to E–L when a coordinate itself never appears in $L$ (only its rate does)?

1. Spot the missing coordinate. $L$ contains $\dot{\varphi }$ but not $\varphi$. We call such a coordinate cyclic (or ignorable). Why this step? If $\varphi$ is absent, then $\frac{\partial L}{\partial \varphi }=0$ — the force slot for $\varphi$ is exactly zero.
2. Write E–L for $\varphi$. $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\varphi }}\right)-0=0\Rightarrow \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\varphi }}\right)=0$. Why this step? Zero force slot means the momentum slot is constant in time — that is a conservation law, the seed of Noether's Theorem.
3. Identify the constant. $\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }=p_{\varphi }=$ const. This is angular momentum. Why this step? Only the $\frac{1}{2}m{r}^2{\dot{\varphi }}^2$ term contains $\dot{\varphi }$; its derivative is $m{r}^2\dot{\varphi }$, which step 2 just proved is constant.

Verify: $p_{\varphi }=m{r}^2\dot{\varphi }=3\times 2^2\times 5=3\times 4\times 5=60{\text{kg⋅m}}^2/\text{s}$. Units: ${\text{kg⋅m}}^2/\text{s}$ = angular momentum ✓. (Contrast Ex 1–4 where the force slot was nonzero, so momentum changed.)

Example 6 — Degenerate case: the free particle · [Cell F]

Forecast: with $V=0$ everything is "kinetic". Which single path minimizes the total?

1. Force slot is zero. $\frac{\partial L}{\partial x}=0$ (no $x$ in $L$ — the degenerate case). Why this step? $L$ depends only on $\dot{x}$, never on $x$ itself, so its $x$-derivative vanishes.
2. E–L. $\frac{d}{dt}(m\dot{x})=0\Rightarrow \dot{x}=$ const. Constant velocity ⇒ straight line $x(t)=2t$ (slope $6/3=2$). Why this step? Zero force slot ⇒ momentum constant ⇒ uniform speed. The figure's straight cyan line is the true path.
3. Action of the true path. $\dot{x}=2$, so $S_{\text{true}}={\int }_0^3\frac{1}{2}m(2{)}^{2}dt=\frac{1}{2}m\cdot 4\cdot 3=6m$. Why this step? With constant $\dot{x}=2$ the integrand is constant, so the integral is just value $\times$ length.
4. Action of a wiggly rival. Try $x_w(t)=2t+\sin \left(\frac{\pi t}{3}\right)$ (same endpoints: $\sin 0=\sin \pi =0$). Then ${\dot{x}}_w=2+\frac{\pi }{3}\cos \left(\frac{\pi t}{3}\right)$ and $S_w={\int }_0^3\frac{1}{2}m{\dot{x}}_w^{2}dt$. Why this step? The cross-term $\int 2\cdot \frac{\pi }{3}\cos dt$ integrates to zero over a half-period, leaving extra positive $\int (\frac{\pi }{3}\cos {)}^2$ — pure added cost.
5. Why is this a genuine minimum (not a saddle)? Write any rival as $x=\bar{x}+\eta$ with $\eta (0)=\eta (3)=0$. The action difference is exactly $S-S_{\text{true}}=\frac{1}{2}m{\int }_0^3{\dot{\eta }}^{2}dt$. Why this step? The linear term in $\eta$ vanishes (that is E–L holding), and the second variation is $\frac{1}{2}m\int {\dot{\eta }}^{2}dt\ge 0$ — a sum of squares, so it can never be negative. Hence the true path is a strict minimum here, for any interval length, not just "short" ones. Free particles are the one case where the minimum is guaranteed globally; the "short interval" caveat (conjugate points) only bites when the potential can bend nearby paths back together, which a flat $V=0$ never does.

Verify: $S_w=6m+\frac{1}{2}m{\int }_0^3(\frac{\pi }{3}\cos \frac{\pi t}{3}{)}^{2}dt=6m+\frac{1}{2}m\cdot \frac{{\pi }^2}{9}\cdot \frac{3}{2}=6m+\frac{m{\pi }^2}{12}\approx 6m+0.822m>6m$. The straight line wins ✓ — and step 5 proves it is a true minimum, since $\frac{1}{2}m\int {\dot{\eta }}^2\ge 0$ always.

Example 7 — Limiting behaviour: small-angle pendulum · [Cell G]

Forecast: for tiny swings, does the nonlinear pendulum become the linear oscillator of Example 1?

1. Approximate $\sin \theta$. For small $\theta$, $\sin \theta \approx \theta$ (the arc and its chord coincide). Why this step? $\sin \theta =\theta -\frac{{\theta }^3}{6}+\dots$; as $\theta \to 0$ the leading term $\theta$ dominates. This is the limit that turns a hard equation into an easy one.
2. Linearize. $\ddot{\theta }\approx -\frac{g}{\ell }\theta$. This is exactly $\ddot{\theta }=-{\omega }^2\theta$ with $\omega =\sqrt{g/\ell }$ — the same shape as Ex 1's spring. Why this step? We recognize the SHM template; its solution is a cosine of angular frequency $\omega$.
3. Period. $T_{\text{period}}=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{\ell }{g}}$. Why this step? For $\cos (\omega t)$ one full cycle takes time $2\pi /\omega$ — the definition of period.

Verify: $T=2\pi \sqrt{1/9.8}=2\pi \times 0.3194\approx 2.007\text{s}$. Units $\sqrt{\text{m}/({\text{m/s}}^2)}=\text{s}$ ✓. As amplitude → 0 the nonlinear system limits to the linear one ✓.

Example 8 — Word problem: bead on a parabolic wire · [Cell H]

Forecast: translate the sentence into $T$ and $V$. Which single number ($x$) pins the bead down?

1. Constraint into coordinates. On the wire, $y=\frac{1}{2}a{x}^2$, so differentiating in time $\dot{y}=ax\dot{x}$. The bead's speed${}^2={\dot{x}}^2+{\dot{y}}^2={\dot{x}}^2+a^2{x}^2{\dot{x}}^2={\dot{x}}^2(1+a^2{x}^2)$. Why this step? The bead is forced to stay on the curve, so $y$ is not free — it is a function of $x$. One coordinate suffices, exactly as the figure's single pinned bead suggests.
2. Energies. $T=\frac{1}{2}m{\dot{x}}^2(1+a^2{x}^2)$ and $V=mgy=\frac{1}{2}mga{x}^2$. So $L=T-V=\frac{1}{2}m{\dot{x}}^2(1+a^2{x}^2)-\frac{1}{2}mga{x}^2$. Why this step? Kinetic energy is $\frac{1}{2}m(\text{speed}{)}^2$ using step 1; potential is weight times height $y=\frac{1}{2}a{x}^2$.
3. Momentum slot. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}(1+a^2{x}^2)$. Why this step? Differentiate the kinetic term in $\dot{x}$, treating $x$ as frozen — the factor $(1+a^2{x}^2)$ rides along.
4. Force slot. $\frac{\partial L}{\partial x}=m{\dot{x}}^2{a}^{2}x-mgax$. Why this step? $x$ appears in both the kinetic term (through $1+a^2{x}^2$) and the potential; differentiate each. The first piece comes from $\frac{1}{2}m{\dot{x}}^2\cdot a^2\cdot 2x$, the second from $-\frac{1}{2}mga\cdot 2x$.
5. Assemble E–L. $\frac{d}{dt}[m\dot{x}(1+a^2{x}^2)]-(m{\dot{x}}^2{a}^{2}x-mgax)=0$. Expanding the time-derivative: $m\ddot{x}(1+a^2{x}^2)+2m{a}^{2}x{\dot{x}}^2-m{a}^{2}x{\dot{x}}^2+mgax=0$, i.e. $m\ddot{x}(1+a^2{x}^2)+m{a}^{2}x{\dot{x}}^2+mgax=0$. Why this step? Recipe line 4. The product rule on the momentum slot produces the ${\dot{x}}^2$ terms; one of them cancels against the kinetic force-slot term, leaving the clean exact equation.
6. Small oscillations near $x=0$. For small $x$: $a^2{x}^2\to 0$ and $x{\dot{x}}^2\to 0$ (higher order), leaving $m\ddot{x}+mgax=0\Rightarrow \ddot{x}=-gax$. Why this step? Small swings drop the higher-order-in-$x$ terms — the same limiting move as Ex 7 — giving SHM with $\omega =\sqrt{ga}$.

Verify: $\omega =\sqrt{ga}=\sqrt{9.8\times 2}=\sqrt{19.6}\approx 4.427\text{rad/s}$. Units $\sqrt{({\text{m/s}}^2)(1/\text{m})}=\sqrt{{\text{s}}^{-2}}={\text{s}}^{-1}$ ✓ (here $a$ has units $1/\text{m}$).

Example 9 — Exam twist: two coupled coordinates · [Cell I]

Forecast: with two coordinates, how many E–L equations do we get, and how do they talk to each other?

1. E–L for $x_1$. Momentum slot $\frac{\partial L}{\partial {\dot{x}}_1}=m{\dot{x}}_1$; force slot $\frac{\partial L}{\partial x_1}=+k(x_2-x_1)$. So $m{\ddot{x}}_1=k(x_2-x_1)$. Why this step? One E–L equation per coordinate — the parent's "one per generalized coordinate" rule. The chain rule on $-\frac{1}{2}k(x_2-x_1{)}^2$, differentiating in $x_1$, gives $-k(x_2-x_1)\cdot (-1)=+k(x_2-x_1)$.
2. E–L for $x_2$. Momentum slot $\frac{\partial L}{\partial {\dot{x}}_2}=m{\dot{x}}_2$; force slot $\frac{\partial L}{\partial x_2}=-k(x_2-x_1)$. So $m{\ddot{x}}_2=-k(x_2-x_1)$. Why this step? Same potential, but differentiating in $x_2$ gives $-k(x_2-x_1)\cdot (+1)=-k(x_2-x_1)$ — the inner sign flips.
3. Decouple into the relative coordinate. Let $u=x_2-x_1$ (the stretch). Subtract eq 1 from eq 2: $m({\ddot{x}}_2-{\ddot{x}}_1)=-k(x_2-x_1)-k(x_2-x_1)=-2ku$, so $m\ddot{u}=-2ku\Rightarrow \ddot{u}=-\frac{2k}{m}u$. Why this step? Subtracting cancels the shared centre-of-mass motion, exposing a clean SHM in the relative coordinate — frequency $\omega =\sqrt{2k/m}$.
4. Centre-of-mass check. Add eq 1 and eq 2: $m({\ddot{x}}_1+{\ddot{x}}_2)=+k(x_2-x_1)-k(x_2-x_1)=0$: the centre of mass drifts at constant velocity (a conservation echoing the cyclic case Ex 5). Why this step? Adding cancels the equal-and-opposite spring terms, leaving no net force on the pair.

Verify: $\omega =\sqrt{2\times 6/1}=\sqrt{12}\approx 3.464\text{rad/s}$. Two coordinates → two equations ✓; the factor $2$ comes from both masses sharing the one spring ✓. Units $\sqrt{(\text{N/m})/\text{kg}}={\text{s}}^{-1}$ ✓.

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