Exercises — Hamilton's principle — least action

What the pieces mean: $T=\frac{1}{2}m{\dot{x}}^2$ is the energy of motion (half mass times speed-squared). "Free" means nothing pulls it, so $V=0$. $L=T-V=\frac{1}{2}m{\dot{x}}^2-0=\frac{1}{2}m{\dot{x}}^2.$ The momentum slot: differentiate $L$ treating $\dot{x}$ as the variable, $\frac{\partial L}{\partial \dot{x}}=m\dot{x}(\text{this is the momentum }p).$ The force slot: differentiate $L$ treating $x$ as the variable. Since $x$ does not appear, $\frac{\partial L}{\partial x}=0.$ A coordinate that does not appear in $L$ is called cyclic — its momentum will be conserved. See Noether's Theorem.

Spring potential is $V=\frac{1}{2}k{x}^2$ (energy stored grows as the square of the stretch). Kinetic energy $T=\frac{1}{2}m{\dot{x}}^2$. $L=\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{x}^2.$ $\frac{\partial L}{\partial \dot{x}}=m\dot{x},\frac{\partial L}{\partial x}=-kx.$ The minus sign is the spring pulling back toward $x=0$ — a restoring tendency, not a random sign.

(b) $S={\int }_{t_1}^{t_2}Ldt$ is correct: the action is the Lagrangian accumulated over time. (a) integrates over the wrong variable — action sums a "cost per unit time," so we integrate $dt$, not $dx$. (c) is just a difference of values; it ignores the whole journey in between, which is exactly what "action" is meant to capture.

E–L says $\frac{d}{dt}\frac{\partial L}{\partial \dot{x}}-\frac{\partial L}{\partial x}=0$. From Problem 1.1: $\partial L/\partial \dot{x}=m\dot{x}$ and $\partial L/\partial x=0$. So $\frac{d}{dt}(m\dot{x})-0=0\Rightarrow m\ddot{x}=0\Rightarrow \ddot{x}=0.$ What this means: acceleration is zero, so $\dot{x}=\text{const}$ (constant velocity), and integrating once more $x(t)=x_0+vt$ — a straight line in a position–time plot. This matches Newton's Laws: no force ⇒ no acceleration.

Slots (from Problem 1.2): $\partial L/\partial \dot{x}=m\dot{x}$, $\partial L/\partial x=-kx$. $\frac{d}{dt}(m\dot{x})-(-kx)=0\Rightarrow m\ddot{x}+kx=0.$ Divide by $m$: $\ddot{x}=-\frac{k}{m}x$. Comparing to $\ddot{x}=-{\omega }^{2}x$, $\boxed{\omega =\sqrt{k/m}}.$ The restoring sign $-kx$ becomes the $-{\omega }^{2}x$ that produces oscillation.

Kinetic energy needs the full speed. The bead moves both horizontally and vertically: $v^2={\dot{x}}^2+{\dot{y}}^2={\dot{x}}^2+(2x\dot{x}{)}^2={\dot{x}}^2(1+4x^2).$ So $T=\frac{1}{2}m{\dot{x}}^2(1+4x^2)$. Potential is gravity: $V=mgy=mg{x}^2$. $L=\frac{1}{2}m{\dot{x}}^2(1+4x^2)-mg{x}^2.$ Momentum slot: $\frac{\partial L}{\partial \dot{x}}=m\dot{x}(1+4x^2)$. Time-derivative (product rule — both $\dot{x}$ and $x$ vary): $\frac{d}{dt}[m\dot{x}(1+4x^2)]=m\ddot{x}(1+4x^2)+m\dot{x}\cdot 8x\dot{x}=m\ddot{x}(1+4x^2)+8mx{\dot{x}}^2.$ Force slot: $\frac{\partial L}{\partial x}=\frac{1}{2}m{\dot{x}}^2\cdot 8x-2mgx=4mx{\dot{x}}^2-2mgx.$ E–L (force slot subtracted): $m\ddot{x}(1+4x^2)+8mx{\dot{x}}^2-(4mx{\dot{x}}^2-2mgx)=0.$ $\boxed{\ddot{x}(1+4x^2)+4x{\dot{x}}^2+2gx=0}$ Read it: near $x=0$ the wire is nearly flat, $4x^2\approx 0$, and it reduces to $\ddot{x}\approx -2gx$ — small oscillations, just like a spring. The power of the method: we never resolved the normal force from the wire; choosing $x$ and writing $T,V$ did all the geometry.

Speed: the bob traces a circle of radius $\ell$, so its speed is $\ell \dot{\theta }$ and $T=\frac{1}{2}m(\ell \dot{\theta }{)}^2=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$. Height: measuring height from the pivot, the bob sits at $-\ell \cos \theta$ (lowest when $\theta =0$). So $V=mg(-\ell \cos \theta )=-mg\ell \cos \theta$. $L=T-V=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2+mg\ell \cos \theta .$ Slots: $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }$, and $\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta$. E–L: $\frac{d}{dt}(m{\ell }^2\dot{\theta })-(-mg\ell \sin \theta )=0\Rightarrow m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0$. $\boxed{\ddot{\theta }=-\frac{g}{\ell }\sin \theta }.$ Small angles: $\sin \theta \approx \theta$ (the graph of $\sin$ hugs the line $y=\theta$ near $0$), so $\ddot{\theta }\approx -\frac{g}{\ell }\theta$, giving $\omega =\sqrt{g/\ell }$.

The string tension does no work: it always points along the string, exactly perpendicular to the bob's motion (which is along the circle). Energy methods only ever see forces through the energies $T$ and $V$; a force that stores no energy and does no work simply never enters $L$. The constraint "length is fixed" was already baked in by using the single coordinate $\theta$. This is why generalized coordinates are so powerful — Lagrangian Mechanics sidesteps constraint forces entirely.

Look at the $\varphi$ equation. The $\varphi$ momentum slot is $\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }$. The $\varphi$ force slot is $\frac{\partial L}{\partial \varphi }=0$ because $\varphi$ does not appear in $L$ (only $\dot{\varphi }$ does; $V$ depends on $r$ alone). E–L for $\varphi$: $\frac{d}{dt}(m{r}^2\dot{\varphi })-0=0\Rightarrow m{r}^2\dot{\varphi }=\text{const}.$ That constant is the angular momentum $p_{\varphi }$. Why: $\varphi$ is a cyclic coordinate (absent from $L$), and E–L then says its momentum's time-derivative is zero. This is Noether's Theorem in miniature: the rotational symmetry of $L$ (rotating $\varphi$ leaves $L$ unchanged) is conservation of angular momentum.

Set up E–L. Here the "coordinate" is $y$ and the "time" is $x$, so E–L reads $\frac{d}{dx}\left(\frac{\partial L}{\partial y^{\prime }}\right)-\frac{\partial L}{\partial y}=0.$ Compute each slot for $L=\frac{\sqrt{1+y^{\prime 2}}}{v(y)}$:

Sign convention first. We measure $x,y$ downward (positive = further down). Gravity does positive work as a mass descends, so potential energy must decrease with depth: $V_1=-m_{1}gx$ and $V_2=-m_{2}gy$. That is where each minus sign comes from — it is forced by "PE drops as you fall," not a free choice. Apply the constraint. The rope is inextensible: $x+y=\ell$ (constant), so $y=\ell -x$ and differentiating, $\dot{y}=-\dot{x}$ (as $m_1$ drops, $m_2$ rises at the same rate). Kinetic energy (substitute $\dot{y}=-\dot{x}$, and $(-\dot{x}{)}^2={\dot{x}}^2$): $T=\frac{1}{2}{m}_1{\dot{x}}^2+\frac{1}{2}{m}_2{\dot{y}}^2=\frac{1}{2}{m}_1{\dot{x}}^2+\frac{1}{2}{m}_2{\dot{x}}^2=\frac{1}{2}(m_1+m_2){\dot{x}}^2.$ Potential (substitute $y=\ell -x$): $V=-m_{1}gx-m_{2}g(\ell -x)=-(m_1-m_2)gx\underset{\text{constant}}{\underbrace{-m_{2}g\ell }}.$ Why the constant may be dropped — shown, not asserted. A constant $c$ added to $V$ changes $L$ by $-c$. Its contributions to both slots vanish: $\frac{\partial (-c)}{\partial \dot{x}}=0$ and $\frac{\partial (-c)}{\partial x}=0$ (the derivative of a constant is zero). So it cannot affect the E–L equation at all — we keep it in $L$ but it plays no role. $L=\frac{1}{2}(m_1+m_2){\dot{x}}^2+(m_1-m_2)gx-m_{2}g\ell .$ Slots: $\frac{\partial L}{\partial \dot{x}}=(m_1+m_2)\dot{x}$, $\frac{\partial L}{\partial x}=(m_1-m_2)g$. E–L: $(m_1+m_2)\ddot{x}-(m_1-m_2)g=0$, so $\boxed{\ddot{x}=\frac{(m_1-m_2)}{m_1+m_2}g}.$ Sanity check: if $m_1=m_2$ the acceleration is zero (balanced); if $m_2=0$ it is $+g$ (free fall of $m_1$). This is the classic Atwood-machine result, derived without a single free-body diagram.

The one-line argument: the derivation of E–L only ever forced the first-order change of the action to vanish, ${\frac{dS}{d\epsilon }|}_{\epsilon =0}=0$; but a vanishing first derivative marks a minimum, a maximum, or a saddle — so all the physics guarantees is stationarity, not minimality. A concrete non-minimum case. Consider a mass on a spring (or any oscillator) with period $T_0=2\pi /\omega$. Compare the true path between two points to nearby paths:

• For a short time interval $t_2-t_1<T_0/2$, the true path is a genuine local minimum of the action.
• Once the interval exceeds a conjugate point (here $t_2-t_1=T_0/2$, half a period, where all nearby oscillator paths refocus), the true path becomes a saddle: some neighbouring paths lower the action while others raise it. It still satisfies $\delta S=0$, but it is no longer a minimum. The same happens for light passing through a focus and for orbits past their focal point. Hence the honest name is principle of stationary action.

The extra action is ${\int }_{t_1}^{t_2}\frac{d}{dt}fdt=f(q(t_2),t_2)-f(q(t_1),t_1),$ which depends only on the fixed endpoints. When we vary the path with $\eta (t_1)=\eta (t_2)=0$, this endpoint term does not change at all, so $\delta S^{\prime }=\delta S$. Same variation ⇒ same E–L equation. Why it matters: the Lagrangian is not unique — you may add any total time-derivative ("gauge freedom") without changing physics. This underlies why different textbooks write different-looking $L$'s that agree on dynamics, and it is central to Hamiltonian Mechanics and gauge theory.

Momentum slot: $\frac{\partial L}{\partial \dot{x}}=0$ (no $\dot{x}$ anywhere). Force slot: $\frac{\partial L}{\partial x}=-V^{\prime }(x)$. E–L: $\frac{d}{dt}(0)-(-V^{\prime }(x))=0\Rightarrow V^{\prime }(x)=0.$ What happened: there is no equation for how $x$ evolves in time — E–L only demands $x$ sit at a critical point of $V$ (an algebraic condition, not a differential one). With no inertia, the theory is degenerate: it cannot predict motion, only equilibrium. This is exactly why a genuine kinetic term $T$ (quadratic in $\dot{q}$) is what makes the E–L equation a second-order equation of motion with a well-posed initial-value problem.

$\frac{\partial L}{\partial \dot{x}}=m\dot{x}\Rightarrow \frac{d}{dt}(m\dot{x})=m\ddot{x}$. And $\frac{\partial L}{\partial x}=-V^{\prime }(x)$. E–L: $m\ddot{x}-(-V^{\prime }(x))=0\Rightarrow m\ddot{x}=-V^{\prime }(x)$. Since force is $F=-V^{\prime }(x)$ (force points downhill on the potential), this is $m\ddot{x}=F$, Newton's Laws. The replacement: "$F=ma$ holds at every instant" becomes "the whole path makes $\int Ldt$ stationary." A local, instant-by-instant law is re-expressed as a single global optimisation. Both give the identical trajectory — but the variational form transfers cleanly to fields, relativity, and quantum mechanics, where "force" is hard to define.