Exercises — Principle of least action — Hamilton's principle derivation

WHAT: $T$ is the energy of motion, $V$ the stored energy. Kinetic: $T=\frac{1}{2}m{\dot{x}}^2$. Potential: with height fixed and no spring, $V=0$ (any constant works; a constant never affects Euler–Lagrange since its derivatives are zero). $L=T-V=\frac{1}{2}m{\dot{x}}^2.$ WHY it matters: this is the free particle — the simplest test case. With $L$ in hand every later step is mechanical.

WHAT: one number fixes the bob's position on its circular arc — the swing angle $\theta$ from the downward vertical. So $q=\theta$. WHY this coordinate: the rod is rigid, so the radius never changes; only the angle is free. Using $\theta$ instead of $(x,y)$ means one coordinate instead of two plus a constraint.

Reading the figure below. The dashed vertical line is the reference: it points straight down from the pivot to the lowest point of the arc (the open circle). The solid rod makes angle $\theta$ with that dashed line. The red segment measures the bob's height $h$ above the lowest point. Notice from the geometry: the pivot sits a distance $\ell$ above the lowest point, while the bob sits a distance $\ell \cos \theta$ below the pivot — so the bob is $\ell -\ell \cos \theta$ above the lowest point. That difference is exactly the red segment: $h=\ell -\ell \cos \theta =\ell (1-\cos \theta ).$ Check the extremes on the figure: at $\theta =0$ the rod lies along the dashed line, red segment shrinks to zero ($h=0$); swing out and the red segment grows, so $h$ grows. Good.

$V=mgh=mg\ell (1-\cos \theta ).$

False. The condition is $\delta S=0$ — i.e. the first-order change in the action $S=\int Ldt$ vanishes for every allowed wiggle. That makes the action stationary, not necessarily minimal. It can be a minimum, a saddle point, or (past a conjugate point) a maximum. The honest name is the principle of stationary action.

Step (WHAT/WHY): compute the two pieces the equation needs. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$ (derivative treating $\dot{x}$ as a slot), so $\frac{d}{dt}(m\dot{x})=m\ddot{x}$. $\frac{\partial L}{\partial x}=0$ (no $x$ appears). Euler–Lagrange: $m\ddot{x}-0=0\Rightarrow \ddot{x}=0$. Integrate twice: $\dot{x}=v_0$ (const), $x(t)=x_0+v_{0}t$. Straight line, uniform speed. ✅

Orientation flag: with $y$ positive upward, gravitational potential energy grows with height, so $V=+mgy$ and $L=T-V=\frac{1}{2}m{\dot{y}}^2-mgy$. This sign convention is what makes the answer come out as downward acceleration. $\frac{\partial L}{\partial \dot{y}}=m\dot{y}\Rightarrow \frac{d}{dt}=m\ddot{y}$. $\frac{\partial L}{\partial y}=-mg$. EL: $m\ddot{y}-(-mg)=0\Rightarrow m\ddot{y}=-mg\Rightarrow \ddot{y}=-g$. ✅ Free fall: the minus sign means acceleration points downward (toward decreasing $y$), exactly as gravity should.

The $\dot{x}$ piece (WHY): $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$ — only the $\frac{1}{2}m{\dot{x}}^2$ term contains $\dot{x}$; its slot-derivative is $m\dot{x}$. Then $\frac{d}{dt}(m\dot{x})=m\ddot{x}$. The $x$ piece (WHY): $\frac{\partial L}{\partial x}=-kx$. Reason: differentiate $L$ with respect to $x$ term by term. The kinetic term $\frac{1}{2}m{\dot{x}}^2$ has no bare $x$, so it contributes $0$. The potential term $-\frac{1}{2}k{x}^2$ differentiates by the power rule ($\frac{d}{dx}{x}^2=2x$): $-\frac{1}{2}k\cdot 2x=-kx$. That is the whole $\partial L/\partial x$. EL: $m\ddot{x}-(-kx)=0\Rightarrow m\ddot{x}+kx=0\Rightarrow \ddot{x}=-\frac{k}{m}x=-{\omega }^{2}x$, with $\omega =\sqrt{k/m}$. Numbers: $\omega =\sqrt{8/2}=\sqrt{4}=2\text{rad/s}$.

WHY $T=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$: the bob speed along its arc is $\ell \dot{\theta }$ (arc length $=\ell \theta$, differentiate). Square it, halve, times $m$. $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }\Rightarrow \frac{d}{dt}=m{\ell }^2\ddot{\theta }$. WHY $\partial L/\partial \theta =-mg\ell \sin \theta$: the kinetic term has no bare $\theta$, so it gives $0$. The potential part of $L$ is $-V=-mg\ell (1-\cos \theta )=-mg\ell +mg\ell \cos \theta$. Differentiate with respect to $\theta$: the constant $-mg\ell$ gives $0$, and $\frac{d}{d\theta }\cos \theta =-\sin \theta$, so $mg\ell \cos \theta$ gives $-mg\ell \sin \theta$. EL: $m{\ell }^2\ddot{\theta }-(-mg\ell \sin \theta )=0\Rightarrow m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0\Rightarrow \boxed{\ddot{\theta }=-\frac{g}{\ell }\sin \theta }$. For small $\theta$, $\sin \theta \approx \theta$, giving $\omega =\sqrt{g/\ell }$.

New Lagrangian $L^{\prime }=T-V-C$. In Euler–Lagrange we need $\partial L^{\prime }/\partial q$ and $\partial L^{\prime }/\partial \dot{q}$. $\frac{\partial L^{\prime }}{\partial \dot{q}}=\frac{\partial L}{\partial \dot{q}}$ (a constant has zero derivative). $\frac{\partial L^{\prime }}{\partial q}=\frac{\partial L}{\partial q}$ (same reason). Both pieces are identical, so the equation is identical. WHY it matters: only differences in potential energy have physical meaning — you may put the zero of $V$ wherever is convenient.

The added piece is $g\equiv \frac{df}{dt}=\frac{\partial f}{\partial q}\dot{q}+\frac{\partial f}{\partial t}$. Its contribution to the action is ${\int }_{t_1}^{t_2}\frac{df}{dt}dt=f(q(t_2),t_2)-f(q(t_1),t_1)$ — a pure boundary term. Since the endpoints $q(t_1),q(t_2)$ are fixed, this boundary value is a constant with respect to any path variation. Its variation is zero: $\delta \int gdt=0$. Therefore $\delta S^{\prime }=\delta S$, and the true path — the one making $\delta S=0$ — is unchanged. WHY it matters: the Lagrangian is not unique; this freedom underlies gauge choices and canonical transformations in Hamiltonian Mechanics.

If $q$ is absent then $\frac{\partial L}{\partial q}=0$. Euler–Lagrange becomes $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\right)=0\Rightarrow \frac{\partial L}{\partial \dot{q}}=\text{const}.$ The conserved quantity $p\equiv \frac{\partial L}{\partial \dot{q}}$ is the generalized momentum. WHY it matters: a symmetry (nothing depends on $q$, i.e. shifting $q$ leaves physics unchanged) produces a conservation law. This is the seed of Noether's Theorem.

WHY that $T$: the bead's velocity has a radial part $\dot{r}$ (sliding out) and a tangential part $r\Omega$ (carried around by rotation). Speed${}^2={\dot{r}}^2+(r\Omega {)}^2$. $L=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\Omega }^2$. $\frac{\partial L}{\partial \dot{r}}=m\dot{r}\Rightarrow m\ddot{r}$. $\frac{\partial L}{\partial r}=mr{\Omega }^2$ (differentiate $\frac{1}{2}m{r}^2{\Omega }^2$ by the power rule: $\frac{1}{2}m{\Omega }^2\cdot 2r$). EL: $m\ddot{r}-mr{\Omega }^2=0\Rightarrow \boxed{\ddot{r}={\Omega }^{2}r}$. Interpretation: the bead accelerates outward — the familiar centrifugal push, arising here with no force diagram, purely from the $r^2{\Omega }^2$ term acting like an inverted potential.

Euler–Lagrange applies once per coordinate (each $q_i$ gets its own equation). For $x$: $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$, $\frac{\partial L}{\partial x}=0\Rightarrow m\ddot{x}=0$. For $y$: $\frac{\partial L}{\partial \dot{y}}=m\dot{y}$, $\frac{\partial L}{\partial y}=-mg\Rightarrow m\ddot{y}=-mg$. So $\ddot{x}=0$ (uniform horizontal drift) and $\ddot{y}=-g$ (vertical free fall) — the projectile parabola. WHY it matters: because $x$ is cyclic ($L$ has no bare $x$), $p_x=m\dot{x}$ is conserved — horizontal momentum, exactly as expected.

Reading the figure below. The horizontal black line is the boundary between the two media. Point $A=(0,a)$ sits in the top medium (light speed $v_1$), point $B=(d,-b)$ in the bottom medium (light speed $v_2$). The red bent line is the ray: it travels straight from $A$ down to a crossing point $(x,0)$ on the boundary, then straight on to $B$. The single number we get to choose is $x$, where it crosses. The dashed vertical line is the normal (perpendicular to the boundary) through the crossing point. The ==angle of incidence ${\theta }_1$ is measured between the red incoming ray and that dashed normal; the angle of refraction ${\theta }_2$== between the red outgoing ray and the normal. Those two marked angles are what the algebra will pin down.

Time = (distance/speed) in each medium: $\mathcal{T}(x)=\frac{\sqrt{x^2+a^2}}{v_1}+\frac{\sqrt{(d-x{)}^2+b^2}}{v_2}.$ Stationary means $\frac{d\mathcal{T}}{dx}=0$: $\frac{x}{v_1\sqrt{x^2+a^2}}-\frac{(d-x)}{v_2\sqrt{(d-x{)}^2+b^2}}=0.$ Now read the two right triangles in the figure: in the upper triangle the side opposite ${\theta }_1$ (the horizontal run from the normal across to $A$) is $x$ and the hypotenuse is the red segment $\sqrt{x^2+a^2}$, so $\frac{x}{\sqrt{x^2+a^2}}=\sin {\theta }_1$. Likewise in the lower triangle $\frac{d-x}{\sqrt{(d-x{)}^2+b^2}}=\sin {\theta }_2$. Substituting, $\frac{\sin {\theta }_1}{v_1}=\frac{\sin {\theta }_2}{v_2}.✓$ WHY it matters: the same variational logic that ran mechanics runs optics — one crossing point plays the role of the whole path's wiggle. This kinship is why Feynman Path Integral later unifies both.

Orientation flag: take $y$ positive downward (increasing $y$ = deeper below the start). This is opposite to the upward-positive convention of L2.2 — we choose it here so the drop below the start is simply $+y$ and the speed formula has no stray minus sign. The horizontal endpoint is $x=x_B$. WHAT to minimize: total travel time $\mathcal{T}=\int \frac{ds}{v}$, where $ds$ is a little piece of arc length along the curve and $v$ the bead's speed there. The arc-length piece (WHY): for a graph $y(x)$, a small step $dx$ raises the curve by $dy=y^{\prime }dx$ where $y^{\prime }=dy/dx$. By Pythagoras the slanted step is $ds=\sqrt{d{x}^2+d{y}^2}=\sqrt{1+y^{\prime 2}}dx$. The speed piece (WHY): energy conservation with $y$ measured downward — the bead has dropped a height $y$, converting potential to kinetic: $\frac{1}{2}m{v}^2=mgy\Rightarrow v=\sqrt{2gy}$. Put them together: $\boxed{\mathcal{T}[y]={\int }_0^{x_B}\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{2gy}}dx},F(y,y^{\prime })=\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{2gy}}.$ Here $x$ plays the role that time $t$ played in mechanics, and $y(x)$ plays the role of the path $q(t)$; the integrand $F(y,y^{\prime })$ plays the role of the Lagrangian $L(q,\dot{q})$. WHY this matters: this is precisely a Calculus of Variations problem — the same Euler–Lagrange engine solves it. Since $F$ has no explicit $x$, the Beltrami identity ($F-y^{\prime }\partial F/\partial y^{\prime }=\text{const}$) applies and yields the cycloid. (Solving that ODE is the next step; here we only needed the correct functional and integrand $F$.)

WHAT we test. The true path $x_{\text{true}}(t)$ already satisfies $\delta S=0$ (first variation vanishes). To classify it we examine the second variation ${\delta }^{2}S$ — the leading change in $S$ when we add a small endpoint-vanishing wiggle $\eta (t)$. Writing the perturbed action to second order, $S[x_{\text{true}}+\eta ]=S[x_{\text{true}}]+\underset{=0}{\underbrace{\delta S}}+\frac{1}{2}{\delta }^{2}S+\cdots$ The path is a minimum iff ${\delta }^{2}S>0$ for every allowed $\eta$, and a saddle if some $\eta$ makes ${\delta }^{2}S<0$. WHY the oscillator turns. For $L=\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}m{\omega }^2{x}^2$ one finds ${\delta }^{2}S={\int }_{t_1}^{t_2}(m{\dot{\eta }}^2-m{\omega }^2{\eta }^2)dt.$ Trade the ${\dot{\eta }}^2$ term for $\eta$ by integrating by parts (the boundary dies since $\eta (t_1)=\eta (t_2)=0$): ${\delta }^{2}S=-{\int }_{t_1}^{t_2}m\eta (\ddot{\eta }+{\omega }^2\eta )dt.$ The sign of this is governed by the operator $\ddot{\eta }+{\omega }^2\eta$. A nonzero wiggle $\eta$ that both vanishes at the two endpoints and satisfies $\ddot{\eta }+{\omega }^2\eta =0$ is exactly a sine $\eta \propto \sin (\omega (t-t_1))$ that returns to zero — and the first time it can do that is when $\omega (t_2-t_1)=\pi$, i.e. after half a period $t_2-t_1=\pi /\omega$. That instant is the conjugate point. Reading off the three regimes.

• $t_2-t_1<\pi /\omega$: no allowed wiggle can drive ${\delta }^{2}S$ negative, so ${\delta }^{2}S>0$ for all $\eta$ ⇒ the path is a genuine minimum. WHY: the restoring "spring cost" $m{\omega }^2{\eta }^2$ cannot outweigh the kinetic penalty $m{\dot{\eta }}^2$ over such a short span.
• $t_2-t_1=\pi /\omega$ (the conjugate point): the special sine makes ${\delta }^{2}S=0$ — the second variation goes flat, marking the exact borderline.
• $t_2-t_1>\pi /\omega$: a wiggle now exists that makes ${\delta }^{2}S<0$ ⇒ the stationary path is a saddle, not a minimum. WHY: over a long enough span the potential term wins for that shape, so a competing path lowers the action. WHY it matters. This is the concrete proof behind L1.3: "least action" is a misnomer — past the conjugate point the true path is stationary but not least.

Write $V^{\prime }$ for the derivative of $V$ with respect to its single argument $u=x_1-x_2$. Euler–Lagrange for each particle: $x_1$: $\frac{d}{dt}(m_1{\dot{x}}_1)-\frac{\partial L}{\partial x_1}=0$. Since $\frac{\partial L}{\partial x_1}=-V^{\prime }(x_1-x_2)\cdot \frac{\partial u}{\partial x_1}=-V^{\prime }(x_1-x_2)$, this gives $m_1{\ddot{x}}_1=-V^{\prime }(x_1-x_2)$. $x_2$: $\frac{\partial L}{\partial x_2}=-V^{\prime }(x_1-x_2)\cdot \frac{\partial u}{\partial x_2}=+V^{\prime }(x_1-x_2)$ (the chain rule flips the sign since $x_2$ enters as $-x_2$), giving $m_2{\ddot{x}}_2=+V^{\prime }(x_1-x_2)$. Add the two equations of motion: $\frac{d}{dt}(m_1{\dot{x}}_1+m_2{\dot{x}}_2)=-V^{\prime }(x_1-x_2)+V^{\prime }(x_1-x_2)=0$. So $p=m_1{\dot{x}}_1+m_2{\dot{x}}_2=\text{const}$. ✅ WHY it works: shifting both particles by the same amount ($x_i\to x_i+c$) leaves $x_1-x_2$ — and thus $L$ — unchanged. That translation symmetry is momentum conservation, the Noether's Theorem statement in miniature.

Every problem here is the same move: build $L=T-V$, feed it to $\frac{d}{dt}{\partial }_{\dot{q}}L={\partial }_{q}L$, one equation per coordinate — and read off physics (motion, conservation, optics, minimality).