Worked examples — Principle of least action — Hamilton's principle derivation

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The scenario matrix

Before solving anything, let us name every case class this topic can throw at you. Each row is a "flavour" of problem; the last column tells you which worked example below covers it.

#	Case class	What makes it distinct	Covered by
A	No potential ($V=0$)	$\partial L/\partial q=0$ ⟹ a conserved momentum	Ex 1
B	Constant force ($V$ linear in $q$)	$\partial L/\partial q=\text{const}$ ⟹ constant acceleration	Ex 2
C	Restoring force ($V$ quadratic)	oscillation, sign of stiffness matters	Ex 3
D	Sign flip / degenerate stiffness ($k<0$ or $k=0$)	runaway vs. free motion — limiting behaviour	Ex 4
E	Curvilinear / non-Cartesian coordinate ($q=\theta$)	kinetic energy is not $\frac{1}{2}m{\dot{q}}^2$; new geometry	Ex 5
F	Cyclic (ignorable) coordinate	$L$ doesn't contain $q$ ⟹ Noether's Theorem conserved quantity	Ex 6
G	Explicit time dependence ($L$ has $t$ in it)	energy not conserved; forced system	Ex 7
H	Real-world word problem	translate words → $T$, $V$, $L$	Ex 8
I	Exam twist: free endpoint	boundary term does not die ⟹ natural boundary condition	Ex 9

We will now walk A → I in order. Every numeric answer at the bottom of the page is machine-checked.

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Ex 1 — Case A: free particle (no potential)

Step 1. Compute the two partial derivatives. $\frac{\partial L}{\partial x}=0,\frac{\partial L}{\partial \dot{x}}=m\dot{x}.$ Why this step? Euler–Lagrange needs exactly these two ingredients. Here $L$ contains no $x$, only $\dot{x}$ — that is what "no potential" means.

Step 2. Plug into $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)-\frac{\partial L}{\partial x}=0$: $\frac{d}{dt}(m\dot{x})-0=0\Rightarrow m\ddot{x}=0\Rightarrow \ddot{x}=0.$ Why this step? A zero $\partial L/\partial x$ means the momentum $m\dot{x}$ has zero time-derivative — momentum is conserved. This is your first taste of Noether's Theorem: no $x$ in $L$ ⟹ $x$-momentum conserved.

Step 3. Integrate: $\dot{x}=\text{const}=v$. With $x(0)=0$, $x(3)=6$: $v=\frac{6-0}{3}=2\text{m/s}.$ Why this step? Constant velocity is the only motion with $\ddot{x}=0$; fixed endpoints pin the constant.

Step 4. Momentum $p=m\dot{x}=2\times 2=4\text{kg⋅m/s}$.

Verify: Units: $\text{kg}\cdot \text{m/s}$ ✅. Sanity: no force ⟹ straight uniform motion. This constant-speed path makes $\int \frac{1}{2}m{\dot{x}}^{2}dt$ stationary (here genuinely a minimum for short times — any wobble raises the average of ${\dot{x}}^2$), exactly as Calculus of Variations promises. Answer $p=4$.

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Ex 2 — Case B: falling body (constant force)

Step 1. $\frac{\partial L}{\partial \dot{y}}=m\dot{y}$, so $\frac{d}{dt}(m\dot{y})=m\ddot{y}$. Why this step? This is the "momentum rate" term of Euler–Lagrange.

Step 2. $\frac{\partial L}{\partial y}=-mg$. Why this step? Differentiate the potential debt $-mgy$; the linear potential gives a constant slope — that is the signature of a constant force.

Step 3. EL: $m\ddot{y}-(-mg)=0\Rightarrow m\ddot{y}=-mg\Rightarrow \ddot{y}=-g=-10{\text{m/s}}^2$. Why this step? Mass cancels — heavy and light stones fall alike, and the minus says "downward".

Step 4. From rest at origin, $y(t)=-\frac{1}{2}g{t}^2$. At $t=2$: $y=-\frac{1}{2}(10)(4)=-20\text{m}$.

Verify: Units of $\frac{1}{2}g{t}^2$: $({\text{m/s}}^2)({\text{s}}^2)=\text{m}$ ✅. Sign negative = fell downward ✅. Answer $y(2)=-20\text{m}$, recovering Newton's Second Law $m\ddot{y}=-mg$.

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Ex 3 — Case C: simple harmonic oscillator (restoring force)

Step 1. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}\Rightarrow \frac{d}{dt}(m\dot{x})=m\ddot{x}$. Why this step? Momentum-rate term again.

Step 2. $\frac{\partial L}{\partial x}=-kx$. Why this step? A quadratic potential differentiates to a linear restoring force — the hallmark of oscillation. The minus pulls back toward $x=0$.

Step 3. EL: $m\ddot{x}+kx=0\Rightarrow \ddot{x}=-\frac{k}{m}x=-{\omega }^{2}x$ with $\omega =\sqrt{k/m}$. Why this step? Matching to $\ddot{x}=-{\omega }^{2}x$ reads off the frequency by inspection.

Step 4. $\omega =\sqrt{8/0.5}=\sqrt{16}=4\text{rad/s}$; period $T_{\text{per}}=\frac{2\pi }{\omega }=\frac{2\pi }{4}=\frac{\pi }{2}\approx 1.571\text{s}$.

Verify: Units of $\sqrt{k/m}$: $\sqrt{(\text{N/m})/\text{kg}}=\sqrt{1/{\text{s}}^2}=\text{1/s}$ ✅. Stiffer $k$ ⟹ larger $\omega$ ⟹ faster ✅. Answers $\omega =4$, $T_{\text{per}}=\pi /2$.

Figure s01 — the three stiffness regimes. This plot draws the potential $V(x)=\frac{1}{2}k{x}^2$ for $k>0$ (lavender, a valley: motion oscillates), $k=0$ (mint, dead flat: free particle), and $k<0$ (coral, an upside-down hill: motion runs away). Its pedagogical job is to make the sign of $k$ visible as the shape of the landscape — a stable minimum versus an unstable maximum — which is precisely the distinction Ex 4 exploits algebraically.

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Ex 4 — Case D: degenerate & sign-flipped stiffness (limiting behaviour)

Step 1. The EL equation from Ex 3 holds for any $k$: $m\ddot{x}+kx=0$. Why this step? Reuse the machinery; only the number $k$ changes.

Step 2 — case (a), $k=0$. Then $m\ddot{x}=0\Rightarrow \ddot{x}=0$: the oscillator degenerates into the free particle of Ex 1. Why this step? This is the limiting value $k\to 0$ — the SHM answer $\omega =\sqrt{k/m}\to 0$ means "infinitely long period", i.e. no oscillation at all.

Step 3 — case (b), $k<0$. Write $k=-|k|$: $\ddot{x}=-\frac{k}{m}x=+\frac{|k|}{m}x={\gamma }^{2}x$ with $\gamma =\sqrt{|k|/m}$. Why this step? A positive coefficient on $x$ (from the flipped sign) turns sines into exponentials: $x(t)=A{e}^{\gamma t}+B{e}^{-\gamma t}$ — the solution runs away. This is why $\delta S=0$ gives stationary, not minimum: on the hilltop the action is a saddle/maximum, matching the parent's first "mistake" callout.

Step 4. Numeric growth rate for (b): $\gamma =\sqrt{2/0.5}=\sqrt{4}=2{\text{s}}^{-1}$.

Verify: (a) $\omega =\sqrt{0/m}=0$ ✅ degenerate. (b) $\gamma =2$, so displacement grows like $e^{2t}$ — unstable ✅. This edge case is exactly where "least action" is a lie and "stationary action" is the truth.

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Ex 5 — Case E: simple pendulum (curvilinear coordinate)

Step 1 — kinetic energy in the angle. The bob moves on a circle of radius $\ell$; its speed is $v=\ell \dot{\theta }$. So $T=\frac{1}{2}m{v}^2=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2.$ Why this step? This is the whole point of a curvilinear coordinate: distance travelled per unit angle is $\ell$, so the "effective mass" for $\theta$ is $m{\ell }^2$ — a moment of inertia. Look at the figure: the arc length is $\ell \theta$.

Step 2 — potential energy. Height above the lowest point is $\ell (1-\cos \theta )$, so $V=mg\ell (1-\cos \theta )$. Why this step? Raising the bob costs potential debt; the geometry of the circle sets the height.

Step 3 — assemble and apply EL. $L=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2-mg\ell (1-\cos \theta )$.

• $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }\Rightarrow \frac{d}{dt}=m{\ell }^2\ddot{\theta }$.
• $\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta$.

EL: $m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0\Rightarrow \ddot{\theta }=-\frac{g}{\ell }\sin \theta .$ Why this step? Same six-step machine; only the symbols changed because the geometry did.

Step 4 — small angle limit. For small $\theta$, $\sin \theta \approx \theta$, so $\ddot{\theta }=-\frac{g}{\ell }\theta =-{\omega }^2\theta$, $\omega =\sqrt{g/\ell }$. $\omega =\sqrt{10/2.5}=\sqrt{4}=2\text{rad/s}.$

Verify: Units $\sqrt{({\text{m/s}}^2)/\text{m}}=\text{1/s}$ ✅. Structure matches Ex 3's SHM with $g/\ell$ playing the role of $k/m$ ✅. Answer $\omega =2\text{rad/s}$.

Figure s02 — pendulum geometry. This diagram earns every symbol used in Ex 5: the lavender rod of length $\ell$, the coral bob $m$, the mint angle $\theta$ measured from the dashed vertical, the butter-coloured arc whose length is $\ell \theta$ (so speed $v=\ell \dot{\theta }$), and the vertical drop marking the height $\ell (1-\cos \theta )$. Its pedagogical role is to show why $T=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$ and $V=mg\ell (1-\cos \theta )$ before we ever write them — the geometry is the derivation.

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Ex 6 — Case F: cyclic coordinate & a conserved quantity

Step 1. Test the $\varphi$-Euler–Lagrange equation: $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\varphi }}\right)-\frac{\partial L}{\partial \varphi }=0$. Why this step? We must build the two ingredients for the $\varphi$ coordinate specifically.

Step 2. $\frac{\partial L}{\partial \varphi }=0$ (cyclic!) and $\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }$. Why this step? A cyclic (ignorable) coordinate is one absent from $L$; its $\partial L/\partial q$ vanishes.

Step 3. EL then reads $\frac{d}{dt}(m{r}^2\dot{\varphi })=0\Rightarrow p_{\varphi }=m{r}^2\dot{\varphi }=\text{const}.$ Why this step? Zero force-slot ⟹ the conjugate momentum is conserved. $p_{\varphi }$ is angular momentum — this is Noether's Theorem made concrete: rotational symmetry ↔ conserved angular momentum.

Step 4. $p_{\varphi }=m{r}^2\dot{\varphi }=1\times 2^2\times 3=12{\text{kg⋅m}}^2/\text{s}.$

Verify: Units $\text{kg}\cdot {\text{m}}^2/\text{s}$ = angular momentum ✅. Answer $p_{\varphi }=12$. (The link to Hamiltonian Mechanics: $p_{\varphi }$ is exactly the momentum you'd carry into a Legendre transform.)

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Ex 7 — Case G: explicitly time-dependent Lagrangian (forced system)

Step 1. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}\Rightarrow \frac{d}{dt}(m\dot{x})=m\ddot{x}$. Why this step? Momentum-rate term.

Step 2. $\frac{\partial L}{\partial x}=F_0\cos (\Omega t)$. Why this step? Differentiating $F_{0}x\cos (\Omega t)$ w.r.t. $x$ treats $t$ as frozen — this exposes the time-varying driving force.

Step 3. EL: $m\ddot{x}-F_0\cos (\Omega t)=0\Rightarrow m\ddot{x}=F_0\cos (\Omega t).$ Why this step? This is Newton with an oscillating external force. Because $L$ depends on $t$ explicitly, the total energy is not conserved — energy is pumped in and out by the drive.

Step 4. At $t=\frac{\pi }{6}$: the drive phase is $\Omega t=2\cdot \frac{\pi }{6}=\frac{\pi }{3}$, and $\cos \frac{\pi }{3}=\frac{1}{2}$, so $\ddot{x}\left(\frac{\pi }{6}\right)=\frac{F_0}{m}\cos \left(\frac{\pi }{3}\right)=\frac{4}{1}\times \frac{1}{2}=2{\text{m/s}}^2.$ Why this step? Divide the EL equation by $m$ and evaluate the cosine at the given instant — the acceleration simply tracks the driving force.

Verify: Units $\text{N/kg}={\text{m/s}}^2$ ✅. Answer $\ddot{x}\left(\frac{\pi }{6}\right)=2{\text{m/s}}^2$. Contrast Ex 6 (symmetry ⟹ conservation) with this (broken time-symmetry ⟹ no energy conservation).

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Ex 8 — Case H: real-world word problem (Atwood machine)

Step 1 — one coordinate for the whole system. The string ties them: if $m_1$ drops $x$, $m_2$ rises $x$, and both move at speed $\dot{x}$. Kinetic energy: $T=\frac{1}{2}(m_1+m_2){\dot{x}}^2.$ Why this step? The constraint collapses two objects into one degree of freedom $x$ — the elegance of Lagrangian bookkeeping.

Step 2 — potential energy. Take down as positive $x$ and put the zero of height at the pulley. Then $m_1$ has descended $x$ (height $-x$, PE $=-m_{1}gx$) while $m_2$ has risen $x$ (height $+x$, PE $=+m_{2}gx$). Total: $V=-m_{1}gx+m_{2}gx=(m_2-m_1)gx.$ Why this step? Height changes are equal and opposite; the potential debt therefore depends only on the difference of the weights. This is the piece missing from a naive "just write $T$" attempt — no $V$, no forces.

Step 3 — assemble & apply EL. $L=T-V=\frac{1}{2}(m_1+m_2){\dot{x}}^2-(m_2-m_1)gx.$

• $\frac{\partial L}{\partial \dot{x}}=(m_1+m_2)\dot{x}\Rightarrow \frac{d}{dt}=(m_1+m_2)\ddot{x}.$ Why? momentum-rate term for the combined mass.
• $\frac{\partial L}{\partial x}=-(m_2-m_1)g=(m_1-m_2)g.$ Why? derivative of the linear potential — the net driving weight.

EL: $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right)-\frac{\partial L}{\partial x}=0\Rightarrow (m_1+m_2)\ddot{x}-(m_1-m_2)g=0.$

Step 4 — solve for the acceleration. $\ddot{x}=\frac{m_1-m_2}{m_1+m_2}g=\frac{3-1}{3+1}\times 10=\frac{2}{4}\times 10=5{\text{m/s}}^2.$

Verify: Since $m_1>m_2$, $\ddot{x}>0$: the heavier side descends ✅. Magnitude $5<g=10$ because both masses must be accelerated together ✅. Units ${\text{m/s}}^2$ ✅. Answer $\ddot{x}=5{\text{m/s}}^2$, matching the standard Atwood result from Newton's Second Law.

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Ex 9 — Case I: exam twist — a free endpoint

Step 1 — vary the path, keeping the boundary term. Following the parent's integration by parts but not discarding the endpoints: $\frac{dS}{d\epsilon }{|}_0={\left[\frac{\partial L}{\partial \dot{x}}\eta \right]}_{t_1}^{t_2}+{\int }_{t_1}^{t_2}\left[\frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}\right]\eta dt=0.$ Why this step? Same machine as the parent — but now we may not throw the boundary term away, because $\eta (t_2)$ is no longer forced to zero.

Step 2 — split the two demands. The interior integral must vanish for arbitrary $\eta$ (giving the usual EL equation), and the surviving boundary piece must vanish on its own. At the fixed start $\eta (t_1)=0$ kills that end; at the free end $\eta (t_2)$ is arbitrary, so its coefficient must be zero: ${\frac{\partial L}{\partial \dot{x}}|}_{t_2}=0\Rightarrow m\dot{x}(t_2)=0.$ Why this step? Whenever a factor multiplies an arbitrary quantity in a sum forced to zero, that factor must itself vanish. This is the natural boundary condition the parent's last "mistake" callout warned about.

Step 3 — interpret. $m\dot{x}(t_2)=0$ says the bead arrives at the free end with zero momentum — it coasts to a halt. Physically, with nothing pinning the finish, the laziest path spends the least $\int \frac{1}{2}m{\dot{x}}^{2}dt$ by simply not moving: $\dot{x}\equiv 0$. Why this step? The interior EL equation still gives $\ddot{x}=0$ (constant velocity); the free-end condition then pins that constant to zero velocity.

Step 4 — numeric check of the resulting motion. With $x(t_1)=0$ and $\dot{x}=0$ everywhere, $x(t)=0$ for all $t$, and the momentum at the free end is $p(t_2)=m\dot{x}(t_2)=0$.

Verify: The natural boundary condition gives $p(t_2)=0$. For $x(0)=0$ the whole path is $x\equiv 0$, momentum $0$ ✅. Contrast with Ex 1, where a fixed finish forced $p=4\ne 0$: same $L$, different boundary data, different physics. Answer: $p(t_2)=0$.

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Recall Which case gives a

conserved quantity, and why? A coordinate missing from $L$ (cyclic) ::: its conjugate momentum $\partial L/\partial \dot{q}$ is conserved — this is Noether's Theorem (Ex 6, and Ex 1's momentum).

Recall Why is "least action" a misnomer, shown by which example?

Ex 4(b): with negative stiffness the action is a saddle/maximum, so only stationary $\delta S=0$ is correct, not minimum. ::: correct.

Recall When does the boundary term survive integration by parts?

When an endpoint is free (Ex 9) ::: giving the natural boundary condition $\partial L/\partial \dot{q}=0$ there.

See also: Fermat's Principle (the optics twin of stationary action) and the Feynman Path Integral (where all paths contribute, and the stationary one dominates).