Part B · the action · spine

Lagrangian mechanics: coordinates stop mattering

Write one scalar, T − V, in whatever coordinates the constraints suggest, and the equations of motion follow — correct in every coordinate system at once, with the constraint forces never named.

needs 11, 08~2 h slides + 5 h problemsproblems: ./mr problems 12

Nothing in this unit says anything Newton did not. Every equation of motion derived here could be derived from $m\ddot{\vec x} = \vec F$ with enough patience, and the first theorem of the unit proves exactly that. The value is elsewhere, in three things the new language makes obvious.

Constraints disappear. A bead on a wire, a rod of fixed length, a hoop turned by a motor: choose coordinates in which the constraint is automatically satisfied and the forces that enforce it never enter the calculation.
Coordinates stop mattering. The Euler–Lagrange equations have the same form in every coordinate system, including time-dependent ones. Newton's equations do not: $m\ddot r = F_r$ is false in polar coordinates. This is the idea that ends the course as general covariance.
Structure becomes visible. A coordinate absent from $L$ hands you a conserved quantity for free, and a time-independent $L$ hands you another. Unit 13 turns this into one theorem.

How to read this deck

One system runs through it: a bead on a hoop spun about its vertical diameter. Section 1 builds the principle and writes the bead's Lagrangian. Section 2 proves that coordinates do not matter and uses it to get the Coriolis force for free. Section 3 reads the equations: momenta, the energy function, and the bifurcation where the bottom of the hoop goes unstable. Section 4 brings the constraint forces back when they are wanted. Section 5 asks how unique $L$ is, and writes down the Lagrangian of a charge in a magnetic field. Section 6 linearises, and finishes with the double pendulum's normal modes.

Proofs are on the slides where they are one line, and in these notes under Full proof where they are not. Results this unit does not prove are marked cited, with where the proof is. Terms with a dotted underline are glossary entries — click one, or press G. N opens these notes; R stacks the deck into one readable page.

12 · overview

One bead, one hoop, three lines

the running example

A bead ($m = 0.12$ kg) slides freely on a hoop of radius $R = 0.35$ m. A motor spins the hoop about its vertical diameter at a fixed $\Omega$. Where can the bead rest, and is it stable there?

By forces: two unknown reaction components, an acceleration in a rotating plane, a page of algebra. By this unit: one kinetic energy, one potential, three lines, and the answer:

$$\begin{aligned}\Omega^2 \lt g/R &:\ \text{stable only at the bottom};\\ \Omega^2 \gt g/R &:\ \text{stable at } \cos\theta_* = g/R\Omega^2.\end{aligned}$$

the programme

$L = T - V$ in any coordinates $q$ that fit the constraints. The motion makes $\int L\,\mathrm dt$ stationary, so it obeys the Euler–Lagrange equations $\ \dfrac{\mathrm d}{\mathrm dt}\dfrac{\partial L}{\partial\dot q_a} = \dfrac{\partial L}{\partial q_a}$ — unit 11's equation with $x$ renamed $t$.

how the answer is reached

1. The constraints leave one coordinate, $\theta$; write $L = T - V$ in it.

2. No explicit $t$ in $L$, so $h = \tfrac12 mR^2\dot\theta^2 + V_{\mathrm{eff}}(\theta)$ is conserved.

3. So the bead moves in a one-dimensional well, $V_{\mathrm{eff}}(\theta)$.

4. $V_{\mathrm{eff}}''(0) \propto g/R - \Omega^2$: the bottom is a minimum until $\Omega^2 = g/R$; after, the minima sit at $\cos\theta_*$ above.

Getting around. Dotted underline: glossary (G). Solid, like the hoop: running example, on hover. N: notes.

The route, slide by slide. Step 1 is Generalised coordinates and constraints (the two constraints and the one coordinate) and Worked: the bead's Lagrangian in three lines. Step 2 is The energy function, which also explains why $h$ is not $T + V$ here. Step 3 is The hoop's effective potential, where the three wells are plotted. Step 4 is the theorem on the same slide, with the bifurcation drawn in The pitchfork, in unit 02's normal form and the motions on either side in the two phase-portrait slides. Hover over $V_{\mathrm{eff}}$ on any slide to see it again.

What is being claimed. Unit 04's problem P10 found the same equilibria by a different route: go to the rotating frame, write the centrifugal potential, and minimise subject to the hoop constraint with a multiplier. That worked for statics. For the dynamics — how the bead moves, whether the equilibria are stable, what the oscillation frequencies are — the rotating-frame route needs the Coriolis force as well, and the inertial-frame route needs the hoop's reaction, which has a component out of the plane of the hoop. This unit does the dynamics without either.

The numbers. $\Omega_c = \sqrt{g/R} = \sqrt{9.81/0.35} = 5.294\ \mathrm{s^{-1}}$, about 50.6 revolutions per minute. At $\Omega = 6\ \mathrm{s^{-1}}$, $\cos\theta_* = 9.81/(0.35 \times 36) = 0.77857$, so $\theta_* = 38.87^\circ$, as in unit 04. Both are recomputed where they are derived.

Notation. $q = (q_1, \dots, q_n)$ is a tuple of coordinates, $\dot q_a$ its time derivatives, and $\partial L/\partial\dot q_a$ means: differentiate $L(q, \dot q, t)$ with respect to its $(n+a)$-th argument, holding the others fixed. $L$ is a function on $\mathbb R^n \times \mathbb R^n \times \mathbb R$; only after the partial derivative is taken is a curve $q(t)$ substituted. This is exactly unit 11's $F_{y'}$, and the same care is needed: $\dot q$ is an independent slot, not "the derivative of $q$", until a curve is inserted. Angular momentum was $L$ in unit 08; here $L$ is the Lagrangian, and angular momentum will be written $p_\theta$.

section 1

The principle

Constraints, the forces nobody wants to compute, and the one assumption that removes them.

12 · the principle

Generalised coordinates and constraints

definitions

$N$ particles have positions $x = (\vec x_1, \dots, \vec x_N) \in \mathbb R^{3N}$. A holonomic constraint is an equation $G(x, t) = 0$ restricting positions. It is scleronomic if $G$ has no explicit $t$, rheonomic if it does.

With $k$ independent constraints ($\mathrm DG$ of rank $k$), the allowed positions at each $t$ form, locally, the image of an injective map $q \mapsto x(q, t)$ from an open set of $\mathbb R^n$, $n = 3N - k$, with $\partial x/\partial q$ of rank $n$. The $q_a$ are generalised coordinates; $n$ is the number of degrees of freedom.

The running example, schematic. The wire holds the bead at distance $R$ from the centre; the motor holds the hoop's plane at azimuth $\varphi = \Omega t$ from a fixed direction. Of the bead's three coordinates, only $\theta$ is left free.

Reading the figure. In spherical coordinates $(r, \theta, \varphi)$ about the hoop's centre, with $\theta$ measured from straight down, the bead could be anywhere in $\mathbb R^3$. The wire is one equation, $r - R = 0$, which does not mention time. The motor is a second, $\varphi - \Omega t = 0$, which does. The dashed horizontal circle is where the motor would carry a bead held at fixed $\theta$: the constraint set at each instant is the hoop, and it sweeps round the sphere of radius $R$ as $t$ runs. The next slide writes $x(\theta, t)$ for it.

Why "locally, the image of a map". This is unit 04's implicit function theorem. If $G : \mathbb R^{3N} \times \mathbb R \to \mathbb R^k$ is $C^2$ and $\mathrm D_xG$ has rank $k$ at a point of $G = 0$, then near that point one can solve for $k$ of the coordinates in terms of the other $3N - k$ and $t$; those remaining coordinates (or any smooth invertible function of them) serve as $q$. The rank condition on $\partial x/\partial q$ is the statement that no direction of $q$ is wasted — it is what will make the kinetic energy a positive definite quadratic form in $\dot q$ in section 6.

Terminology. Holonomic (Hertz, 1894) is from the Greek for "whole law": the constraint is an equation between the coordinates themselves. Scleronomic and rheonomic ("rigid law", "flowing law") are Boltzmann's.

12 · the principle

On the hoop: two constraints, one coordinate

on the hoop

One particle, so $\mathbb R^3$. In spherical coordinates about the centre, two constraints: $r = R$ (scleronomic) and azimuth $\varphi = \Omega t$ (rheonomic: the motor). $n = 3 - 2 = 1$, and

$$x(\theta, t) = R\begin{pmatrix}\sin\theta\cos\Omega t\\ \sin\theta\sin\Omega t\\ -\cos\theta\end{pmatrix}.$$

$\partial x/\partial\theta = R(\cos\theta\cos\Omega t,\ \cos\theta\sin\Omega t,\ \sin\theta)$ has length $R \ne 0$: rank 1, as the definition asks.

The set of allowed positions, as a space in its own right, is the configuration space. For the hoop it is a circle: $\theta$ and $\theta + 2\pi$ are the same bead.

At each instant the allowed set is the hoop drawn on the previous slide, and it moves: the set itself depends on $t$, which is what "rheonomic" means. The coordinate $\theta$ does not see the motion; $x(\theta, t)$ carries it.

The rank check. $\partial x/\partial\theta$ is $R$ times a unit vector (its squared length is $\cos^2\theta(\cos^2 + \sin^2) + \sin^2\theta = 1$), so it never vanishes: no direction of $\theta$ is wasted anywhere on the hoop. It is the tangent to the hoop at the bead, and the next slide's virtual displacements are its multiples.

"Locally" is not pedantry. The circle has no single coordinate that is injective and continuous on all of it: $\theta \in (-\pi, \pi]$ jumps at $\pi$. Two overlapping charts are needed, and the slide on configuration space as a manifold says why this costs nothing.

A rheonomic constraint is still holonomic. The motor's constraint is a condition on position that changes in time; it restricts where the bead may be at each instant. What it is not is a constraint on velocities alone — those are section 4's non-holonomic constraints, and they are genuinely different.

12 · the principle

Virtual displacements and d'Alembert's principle

definition

At a configuration $x(q, t)$, a virtual displacement is a tangent vector to the constraint set with time frozen:

$$\delta\vec x_k = \sum_{a=1}^n \frac{\partial\vec x_k}{\partial q_a}\,\delta q_a,\qquad \delta q \in \mathbb R^n \text{ arbitrary}.$$

The constraints are ideal if the forces $\vec R_k$ enforcing them do no work on any virtual displacement: $\sum_k \vec R_k\cdot\delta\vec x_k = 0$.

d'Alembert's principle

Newton gives $m_k\ddot{\vec x}_k = \vec F_k + \vec R_k$, with $\vec F_k$ the applied forces. If the constraints are ideal, then for every virtual displacement

$$\sum_{k=1}^N \big(\vec F_k - m_k\ddot{\vec x}_k\big)\cdot\delta\vec x_k = 0.$$

Proof. The bracket is $-\vec R_k$. ∎

The unknown forces are gone, at the price of one physical assumption.

What "ideal" means physically, case by case. A smooth wire or surface: the reaction is normal to it, and a virtual displacement is tangent, so the product vanishes. A rigid rod between two particles: the forces on the two ends are equal, opposite and along the rod, and a virtual displacement preserves the rod's length to first order, so $\vec R\cdot(\delta\vec x_1 - \delta\vec x_2) = 0$. Rolling without slipping: the contact point is instantaneously at rest, so the static friction force at it does no work on an allowed displacement. Sliding friction is the standard failure: it is tangent to the surface, and does work on every tangential displacement.

History. D'Alembert's Traité de dynamique (1743) treated $-m\ddot{\vec x}$ as an "inertial force" and reduced dynamics to statics, where Johann Bernoulli's principle of virtual work (1715, in a letter to Varignon) already said that a system is in equilibrium when the applied forces do no work on any virtual displacement. Lagrange's Mécanique analytique (1788) built all of mechanics on the combination, and famously contains no figures.

Status of the principle. Given Newton's second law, d'Alembert's principle is equivalent to the ideality of the constraints; it is not a new law. What makes it useful is that ideality is easy to judge physically and the reaction forces are hard to compute.

12 · the principle

Frozen time: real work, no virtual work

On the hoop the reaction has two parts: $R_r$ along the radius (the wire) and $R_\varphi$ perpendicular to the hoop's plane (the motor turning it).

A virtual displacement $\delta x$ is taken at frozen $t$: it runs along the hoop as it stands, so both parts of the reaction are perpendicular to it. No virtual work: the constraints are ideal.

The actual displacement $\mathrm dx$ in time $\mathrm dt$ also carries the bead round with the hoop, a part $R\sin\theta\,\Omega\,\mathrm dt$ along $R_\varphi$. So $R_\varphi\cdot\mathrm dx \ne 0$: the motor does real work.

ideal is not energy-conserving

Ideal constraints remove the reactions from the equations of motion. They need not conserve energy, and on the hoop they do not (section 3).

The hoop at $t$ and, rotated by an exaggerated angle, at $t + \mathrm dt$. The virtual $\delta x$ lies along the hoop at $t$; the actual $\mathrm dx$ adds the dotted part carried round with the hoop, out of its plane. $R_\varphi$ is perpendicular to the plane, so it is perpendicular to $\delta x$ but not to $\mathrm dx$. Schematic.

Why freeze time. On the hoop, the motor's constraint $\varphi = \Omega t$ is enforced by a force perpendicular to the plane of the hoop. That force does real work on the bead — the motor feeds it energy whenever the bead moves away from the axis — but the virtual displacements, taken at fixed $t$, lie in the plane of the hoop, and the force is perpendicular to them. So the constraint is ideal though it is not energy-conserving. This is exactly why the energy of the bead will turn out not to be conserved while something else is (section 3).

In formulas. From $x(\theta, t)$ of On the hoop: two constraints, one coordinate, $\mathrm dx = \frac{\partial x}{\partial\theta}\,\mathrm d\theta + \frac{\partial x}{\partial t}\,\mathrm dt$ and $\delta x = \frac{\partial x}{\partial\theta}\,\delta\theta$. The first term is along the hoop; the second, $\partial x/\partial t = R\Omega\sin\theta\,(-\sin\Omega t, \cos\Omega t, 0)$, is the velocity of the point of the hoop under the bead, horizontal and perpendicular to the hoop's plane. Freezing time drops exactly that term. Worked: the motor's torque finds $R_\varphi = 2mR\Omega\dot\theta\cos\theta$, and its power $R_\varphi R\Omega\sin\theta$ is what changes the bead's energy on the energy-function slide.

12 · the principle

From d'Alembert to Lagrange's equations

theorem · Lagrange's equations

Holonomic ideal constraints, applied forces $\vec F_k = -\nabla_k V(x, t)$. Put $T = \sum_k \tfrac12 m_k\lvert\dot{\vec x}_k\rvert^2$ written in $(q, \dot q, t)$, and $L = T - V$. Then along the motion, for each $a = 1, \dots, n$,

$$\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot q_a} - \frac{\partial L}{\partial q_a} = 0 .$$

two identities and d'Alembert — full proof in the notes (N)

the two identities

From $\dot{\vec x}_k = \sum_a \dfrac{\partial\vec x_k}{\partial q_a}\dot q_a + \dfrac{\partial\vec x_k}{\partial t}$:

$$\frac{\partial\dot{\vec x}_k}{\partial\dot q_a} = \frac{\partial\vec x_k}{\partial q_a},\qquad \frac{\mathrm d}{\mathrm dt}\frac{\partial\vec x_k}{\partial q_a} = \frac{\partial\dot{\vec x}_k}{\partial q_a}.$$

For forces without a potential, the right-hand side is the generalised force $Q_a = \sum_k\vec F_k\cdot\partial\vec x_k/\partial q_a$, and $\frac{\mathrm d}{\mathrm dt}\frac{\partial T}{\partial\dot q_a} - \frac{\partial T}{\partial q_a} = Q_a$.

Dimensions: $[\partial L/\partial q_a] = \mathrm{J}/[q_a]$, and so is $[\tfrac{\mathrm d}{\mathrm dt}\partial L/\partial\dot q_a] = \mathrm J\,\mathrm s/([q_a]\,\mathrm s)$ ✓. For an angle, both are torques.

Full proof

Everything is evaluated along the motion $q(t)$, and $x(q, t)$ is $C^2$. The first identity: $\dot{\vec x}_k$ is affine in $\dot q$ with coefficient $\partial\vec x_k/\partial q_a$ on $\dot q_a$. The second: by the chain rule (unit 03), $\frac{\mathrm d}{\mathrm dt}\frac{\partial\vec x_k}{\partial q_a} = \sum_b\frac{\partial^2\vec x_k}{\partial q_b\,\partial q_a}\dot q_b + \frac{\partial^2\vec x_k}{\partial t\,\partial q_a}$, while differentiating the formula for $\dot{\vec x}_k$ in $q_a$ gives $\sum_b\frac{\partial^2\vec x_k}{\partial q_a\,\partial q_b}\dot q_b + \frac{\partial^2\vec x_k}{\partial q_a\,\partial t}$; Clairaut's theorem (unit 03) makes them equal.

Now insert $\delta\vec x_k = \sum_a(\partial\vec x_k/\partial q_a)\,\delta q_a$ into d'Alembert's principle. Since $\delta q$ is arbitrary, the coefficient of each $\delta q_a$ vanishes: $\sum_k m_k\ddot{\vec x}_k\cdot\frac{\partial\vec x_k}{\partial q_a} = \sum_k\vec F_k\cdot\frac{\partial\vec x_k}{\partial q_a} = Q_a$. The left side, by the product rule and then the two identities, is

$$\frac{\mathrm d}{\mathrm dt}\Big(\sum_k m_k\dot{\vec x}_k\cdot\frac{\partial\vec x_k}{\partial q_a}\Big) - \sum_k m_k\dot{\vec x}_k\cdot\frac{\mathrm d}{\mathrm dt}\frac{\partial\vec x_k}{\partial q_a} = \frac{\mathrm d}{\mathrm dt}\Big(\sum_k m_k\dot{\vec x}_k\cdot\frac{\partial\dot{\vec x}_k}{\partial\dot q_a}\Big) - \sum_k m_k\dot{\vec x}_k\cdot\frac{\partial\dot{\vec x}_k}{\partial q_a} = \frac{\mathrm d}{\mathrm dt}\frac{\partial T}{\partial\dot q_a} - \frac{\partial T}{\partial q_a}.$$

If $\vec F_k = -\nabla_kV$, then $Q_a = -\partial V/\partial q_a$ by the chain rule, and since $V$ does not depend on $\dot q$, $\partial V/\partial\dot q_a = 0$, so $V$ can be moved inside both derivatives: that is $L = T - V$. ∎

The converse. Lagrange's equations give $n$ equations; together with the constraints (which the parametrisation satisfies automatically) they determine the motion, by unit 01's Picard–Lindelöf, provided they can be solved for $\ddot q$ — which section 6 shows they can, because $\partial^2T/\partial\dot q\,\partial\dot q$ is positive definite. Knowing the motion, $\vec R_k = m_k\ddot{\vec x}_k - \vec F_k$ recovers the reactions, and they satisfy the ideality condition because Lagrange's equations are that condition written in the basis $\partial\vec x_k/\partial q_a$ of virtual displacements. So the two descriptions are equivalent.

Why "$T - V$" and not "$T + V$". The minus sign is forced by the derivation: $Q_a$ enters on the right with a plus sign and equals $-\partial V/\partial q_a$. There is no deeper reason at this stage. Unit 27 finds one: $L = -mc^2\sqrt{1 - v^2/c^2}$ for a free relativistic particle, whose expansion $-mc^2 + \tfrac12 mv^2 + \dots$ makes the kinetic term appear with the sign it has here.

12 · the principle

Hamilton's principle

definition

The Lagrangian is $L(q, \dot q, t) = T - V$. The action of a path $q : [t_0, t_1] \to \mathbb R^n$ is the functional

$$S[q] = \int_{t_0}^{t_1} L\big(q(t), \dot q(t), t\big)\,\mathrm dt .$$

Hamilton's principle

For holonomic ideal constraints and potential forces, a $C^2$ path is a motion of the system if and only if it is a stationary point of $S$ among $C^2$ paths with the same endpoints $q(t_0)$, $q(t_1)$.

Proof. Unit 11's theorem for several dependent variables: stationary $\iff$ the Euler–Lagrange equations of $L$. The previous slide: those are Lagrange's equations, which hold $\iff$ the path is a motion. ∎

An oscillator ($m = k = 1$) from $x = 0$ at $t_0 = 0$ to $x = 1$ at $t_1 = 1.2$: the motion, and the motion plus $\pm0.6\sin(\pi t/t_1)$. Same ends, different paths, and the motion has the smallest action.

What the principle buys: a scalar, one per system, in any coordinates. The equations are its derivative. The next section proves that "any coordinates" costs nothing.

The figure's numbers. The motion is $x = \sin t/\sin 1.2$. On it, $S = \int_0^{1.2}\tfrac12(\dot x^2 - x^2)\,\mathrm dt = \tfrac12\big[x\dot x\big]_0^{1.2} = \tfrac12\cot 1.2 = 0.194$, integrating by parts and using $\ddot x = -x$. For $x + \varepsilon\eta$ with $\eta = \sin(\pi t/1.2)$, the term linear in $\varepsilon$ is the first variation, zero on the motion, and the quadratic term is $\varepsilon^2\cdot\tfrac12\int_0^{1.2}(\dot\eta^2 - \eta^2)\,\mathrm dt = \varepsilon^2\cdot 0.3\big((\pi/1.2)^2 - 1\big) = 1.756\,\varepsilon^2$. At $\varepsilon = \pm 0.6$: $0.194 + 0.632 = 0.827$, the same on both sides, as a stationary point requires. Here $t_1 = 1.2 \lt \pi$, so by the next slide the motion is a true minimum.

The logical status. Here Hamilton's principle is a theorem, derived from Newton plus ideal constraints. From unit 25 onwards it is used the other way round, as the starting point: the Maxwell field (unit 29), the relativistic particle (unit 27) and the metric itself (unit 35) are given actions, and the equations of motion are whatever those actions' Euler–Lagrange equations say. That reversal is only safe because this unit shows that, where Newton applies, the two agree.

Hamilton. William Rowan Hamilton stated the principle in this form in 1834–35, as part of the optical–mechanical analogy that unit 15 develops. The word "action" is older and meant different integrals to Maupertuis, Euler and Lagrange.

12 · the principle

Stationary, not least: the oscillator's conjugate point

stationary, not least

For the oscillator $L = \tfrac12 m\dot x^2 - \tfrac12 kx^2$, the second variation is $\int(\tfrac12 m\dot\eta^2 - \tfrac12 k\eta^2)\,\mathrm dt$ — unit 11's $\int(y'^2 - y^2)$ rescaled. Its conjugate point is at $t_1 - t_0 = \pi/\omega$: the true motion minimises $S$ only over less than half a period.

The picture is the reason. Every motion leaving $x = 0$ at $t_0$ is $x = (v_0/\omega)\sin\omega(t - t_0)$, and all of them are back at $x = 0$ together at $\omega(t - t_0) = \pi$, whatever $v_0$.

At that time the endpoint no longer picks out one motion — a whole family joins the same ends — and beyond it a varied path has smaller action than the motion.

$x = (v_0/\omega)\sin\omega(t - t_0)$ for six starting velocities. They spread, then refocus at $\omega(t - t_0) = \pi$: the conjugate point, half a period on.

The conjugate point, computed. The accessory equation of unit 11 for $F = \tfrac12 m\dot x^2 - \tfrac12 kx^2$ is $m\ddot h + kh = 0$ with $h(t_0) = 0$, so $h = \sin\omega(t - t_0)$ with $\omega = \sqrt{k/m}$, whose next zero is at $t_0 + \pi/\omega$. By Jacobi's theorem (unit 11) the motion is a local minimum of $S$ if $t_1 - t_0 \lt \pi/\omega$, and for $t_1 - t_0 \gt \pi/\omega$ the variation $\eta = \sin\big(\pi(t - t_0)/(t_1 - t_0)\big)$ gives second variation $\tfrac14(t_1 - t_0)\big(m\pi^2/(t_1-t_0)^2 - k\big) \lt 0$, so it is not. Half a period is the time for all paths from one point to refocus, which is what a conjugate point is: the curves in the figure are the solutions $h$ of the accessory equation, scaled, and they all vanish where $h$ does. "Least action" is Maupertuis' phrase (1744) and is wrong as a general statement; "principle of stationary action" is the honest name.

Why the family matters. At exactly $t_1 - t_0 = \pi/\omega$ with $x(t_1) = 0$, every curve in the figure is a motion with the same ends, and all have the same action (zero, by the integration by parts of the previous slide's notes). The second variation in the direction $\eta = \sin\omega(t - t_0)$ is then exactly zero: the borderline between minimum and saddle.

12 · the principle

Newton, in Cartesian coordinates

theorem

For free particles in Cartesian coordinates, $L = \sum_k\tfrac12 m_k\lvert\dot{\vec x}_k\rvert^2 - V(x, t)$, the Euler–Lagrange equations are Newton's equations $m_k\ddot{\vec x}_k = -\nabla_kV$, and conversely.

Proof. $\partial L/\partial\dot{\vec x}_k = m_k\dot{\vec x}_k$, whose time derivative is $m_k\ddot{\vec x}_k$; and $\partial L/\partial\vec x_k = -\nabla_kV$. The Euler–Lagrange equation for $\vec x_k$ is their difference set to zero, which is Newton's law for particle $k$. ∎

check

Dimensions: $m\ddot{\vec x}$ and $\nabla V$ are both newtons ✓. Special case: $V = 0$ gives straight lines at constant speed, Newton's first law, and for one particle $S = \tfrac12 m\lvert\Delta\vec x\rvert^2/\Delta t$ on the straight path.

what $L$ is not

$L$ is not unique. $cL$ ($c \ne 0$) and $L + \text{const}$ give the same equations, and section 5 finds a much larger freedom. Nor is $L$ an energy: it has units of energy, but $T - V$ is not conserved.

This is the base case on which everything else rests: in the one coordinate system where Newton's law is known to hold, the Euler–Lagrange equations say the same thing. The next section's invariance theorem transports it to every other coordinate system, and to every constrained system through the parametrisation $x(q, t)$ — which is a second proof of Lagrange's equations, and the cleanest one.

The free-particle action on the straight path. With $\vec x(t) = \vec x_0 + (t - t_0)\Delta\vec x/\Delta t$, the integrand is the constant $\tfrac12 m\lvert\Delta\vec x\rvert^2/\Delta t^2$, so $S = \tfrac12 m\lvert\Delta\vec x\rvert^2/\Delta t$, of dimension $\mathrm{J\,s}$ ✓. Any other path with the same ends has larger $S$: writing $\vec x = \vec x_{\text{line}} + \vec\eta$ with $\vec\eta$ vanishing at both ends, the cross term $\int m\dot{\vec x}_{\text{line}}\cdot\dot{\vec\eta}\,\mathrm dt$ is a constant vector dotted into $\int\dot{\vec\eta} = \vec 0$, so $S = S_{\text{line}} + \tfrac12 m\int\lvert\dot{\vec\eta}\rvert^2\,\mathrm dt$. With no potential there is no conjugate point, and the minimum is global.

Why $cL$ is harmless. Both sides of the Euler–Lagrange equation are linear in $L$. Adding a constant changes neither side. The second fact generalises to adding a total time derivative $\mathrm df/\mathrm dt$, which is section 5's gauge freedom and is far from trivial.

12 · the principle · example

Worked: the bead's Lagrangian in three lines

the hoop, from scratch

1. Velocity. Differentiate $x(\theta, t) = R(\sin\theta\cos\Omega t,\ \sin\theta\sin\Omega t,\ -\cos\theta)$:

$$\dot{\vec x} = R\begin{pmatrix}\dot\theta\cos\theta\cos\Omega t - \Omega\sin\theta\sin\Omega t\\ \dot\theta\cos\theta\sin\Omega t + \Omega\sin\theta\cos\Omega t\\ \dot\theta\sin\theta\end{pmatrix}.$$

Square and add; the cross terms $\pm 2\dot\theta\Omega\cos\theta\sin\theta\sin\Omega t\cos\Omega t$ cancel, and $\cos^2 + \sin^2 = 1$ twice:

$$\lvert\dot{\vec x}\rvert^2 = R^2\big(\dot\theta^2 + \Omega^2\sin^2\theta\big).$$

2. Lagrangian. $V = mgz = -mgR\cos\theta$, so

$$L = \tfrac12 mR^2\big(\dot\theta^2 + \Omega^2\sin^2\theta\big) + mgR\cos\theta .$$

3. the equation

$\dfrac{\partial L}{\partial\dot\theta} = mR^2\dot\theta$, $\quad\dfrac{\partial L}{\partial\theta} = mR^2\Omega^2\sin\theta\cos\theta - mgR\sin\theta$. Equate the time derivative of the first to the second and divide by $mR^2$:

$$\boxed{\ \ddot\theta = \sin\theta\,\Big(\Omega^2\cos\theta - \frac gR\Big)\ }$$

Checks. Dimensions: $\mathrm{s^{-2}}$ on both sides ✓. $\Omega = 0$: $\ddot\theta = -(g/R)\sin\theta$, the pendulum of length $R$ ✓. $g = 0$: equilibria at $0$, $\pm\pi/2$ and $\pi$ only, the bead flung to the equator ✓.

The page it replaces. In the inertial frame the bead's acceleration has three components in the local basis $(\hat r, \hat\theta, \hat\varphi)$ of spherical coordinates. With $r = R$ and $\dot\varphi = \Omega$ they are $a_r = -R\dot\theta^2 - R\Omega^2\sin^2\theta$, $a_\theta = R\ddot\theta - R\Omega^2\sin\theta\cos\theta$ and $a_\varphi = 2R\Omega\dot\theta\cos\theta$. Deriving these means differentiating the rotating unit vectors twice (unit 03's transport theorem, twice over). Then the forces: gravity has components along $\hat r$ and $\hat\theta$; the hoop's reaction has an unknown component along $\hat r$ (normal to the hoop, in its plane) and an unknown component along $\hat\varphi$ (perpendicular to the plane, the motor pushing). Only the $\hat\theta$ equation is free of unknowns, $mR\ddot\theta - mR\Omega^2\sin\theta\cos\theta = -mg\sin\theta$, and it is the boxed equation. Getting there by forces requires knowing in advance which component to look at; the Lagrangian route cannot look anywhere else.

Where the $\hat\varphi$ reaction went. It is perpendicular to every virtual displacement, so d'Alembert's principle discards it. It is not zero: $F_\varphi = ma_\varphi = 2mR\Omega\dot\theta\cos\theta$. Section 4 recovers it with a multiplier and identifies it as the Coriolis force the hoop must supply.

The $g = 0$ check in detail. With $g = 0$ the equation is $\ddot\theta = \Omega^2\sin\theta\cos\theta = \tfrac12\Omega^2\sin2\theta$, zero at $\theta \in \{0, \pi/2, \pi\}$ in $[0, \pi]$. Linearising at $\pi/2$ gives $\ddot\delta = -\Omega^2\delta$: the equator is stable, as a bead flung outward by the centrifugal effect should be.

section 2

Coordinates stop mattering

The theorem the whole formalism exists for, its proof twice, and a fictitious force obtained for nothing.

12 · invariance

The theorem the formalism exists for

theorem · form invariance

Let $q = \phi(Q, t)$ be a $C^2$ change of coordinates with $\partial\phi/\partial Q$ invertible at every point (a point transformation, allowed to depend on time). Define $\tilde L(Q, \dot Q, t) = L\big(\phi(Q,t),\ \tfrac{\partial\phi}{\partial Q}\dot Q + \tfrac{\partial\phi}{\partial t},\ t\big)$. Then $q(t)$ satisfies the Euler–Lagrange equations of $L$ if and only if $Q(t)$ satisfies those of $\tilde L$.

proof · without computing

$\tilde L$ is $L$ evaluated on the same path, so $\tilde S[Q] = S[q]$. The map $Q \mapsto \phi(Q, \cdot)$ is a bijection between paths, with fixed endpoints going to fixed endpoints. So $Q$ is stationary for $\tilde S$ iff $q$ is stationary for $S$, and by unit 11 stationarity is the Euler–Lagrange equations. ∎

"Write $L$ in any coordinates, then apply the same rule" is what the course later calls general covariance. Newton's law does not have it, as Newton's form is not invariant shows after the component proof.

What the slick proof uses. Three facts. (1) $\tilde L(Q, \dot Q, t) = L(q, \dot q, t)$ whenever $q(t) = \phi(Q(t), t)$ — this is the definition of $\tilde L$, and the chain rule for $\dot q$. (2) The correspondence of paths is a bijection that preserves "same endpoints": invertibility of $\phi(\cdot, t)$ at each fixed $t$ is needed for that, locally guaranteed by unit 04's inverse function theorem from the invertibility of $\partial\phi/\partial Q$. (3) Variations correspond: if $Q_\varepsilon = Q + \varepsilon H$ then $q_\varepsilon = \phi(Q_\varepsilon, t)$ is a one-parameter family through $q$ whose derivative at $\varepsilon = 0$ is $\eta = (\partial\phi/\partial Q)H$, and every admissible $\eta$ arises this way from $H = (\partial\phi/\partial Q)^{-1}\eta$. The first variation of $S$ at $q$ is therefore zero in every direction iff that of $\tilde S$ at $Q$ is.

There is one honest gap in the "without computing" version: unit 11 defined the first variation along straight lines $q + \varepsilon\eta$ in the space of paths, and $\phi(Q + \varepsilon H, t)$ is a curved family. The derivative at $\varepsilon = 0$ of $S$ along a curved family with velocity $\eta$ equals the derivative along the straight line with the same $\eta$ — that is the chain rule in the space of paths, and to avoid proving it there, the next slide gives the proof in components, which needs only unit 03's chain rule in finite dimensions.

12 · invariance

The proof in components, and how the equations transform

lemma

Write $\mathcal E_i = \frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot q_i} - \frac{\partial L}{\partial q_i}$ and $\tilde{\mathcal E}_a$ likewise for $\tilde L$. Along any $C^2$ path,

$$\tilde{\mathcal E}_a = \sum_{i=1}^n \mathcal E_i\,\frac{\partial q_i}{\partial Q_a}\qquad\text{i.e.}\qquad \tilde{\mathcal E} = J^{\top}\mathcal E,\quad J = \frac{\partial\phi}{\partial Q}.$$

proof

The same two identities as before, now for $\phi$: $\partial\dot q_i/\partial\dot Q_a = \partial q_i/\partial Q_a$ and $\partial\dot q_i/\partial Q_a = \frac{\mathrm d}{\mathrm dt}\,\partial q_i/\partial Q_a$. The chain rule gives $\partial\tilde L/\partial\dot Q_a = \sum_i\frac{\partial L}{\partial\dot q_i}\frac{\partial q_i}{\partial Q_a}$; differentiate that in $t$ and subtract $\partial\tilde L/\partial Q_a$. The notes do it line by line. ∎

$J$ is invertible, so $\mathcal E = 0 \iff \tilde{\mathcal E} = 0$: the theorem, again.

The rule $\tilde{\mathcal E} = J^{\top}\mathcal E$ is how the components of a derivative transform — unit 04's covector, not a vector. Unit 16 makes that a theorem about dual spaces.

Full proof

Fix a $C^2$ path $Q(t)$ and put $q(t) = \phi(Q(t), t)$, so $\dot q_i = \sum_a\frac{\partial q_i}{\partial Q_a}\dot Q_a + \frac{\partial q_i}{\partial t}$. As functions of $(Q, \dot Q, t)$: $\dot q_i$ is affine in $\dot Q$, so $\partial\dot q_i/\partial\dot Q_a = \partial q_i/\partial Q_a$; and $\partial\dot q_i/\partial Q_a = \sum_b\frac{\partial^2q_i}{\partial Q_a\partial Q_b}\dot Q_b + \frac{\partial^2q_i}{\partial Q_a\partial t} = \frac{\mathrm d}{\mathrm dt}\frac{\partial q_i}{\partial Q_a}$ by Clairaut. The chain rule applied to $\tilde L(Q, \dot Q, t) = L(q(Q,t), \dot q(Q, \dot Q, t), t)$ gives

$$\frac{\partial\tilde L}{\partial\dot Q_a} = \sum_i\frac{\partial L}{\partial\dot q_i}\frac{\partial q_i}{\partial Q_a},\qquad \frac{\partial\tilde L}{\partial Q_a} = \sum_i\Big(\frac{\partial L}{\partial q_i}\frac{\partial q_i}{\partial Q_a} + \frac{\partial L}{\partial\dot q_i}\frac{\mathrm d}{\mathrm dt}\frac{\partial q_i}{\partial Q_a}\Big).$$

Differentiate the first in $t$ by the product rule: $\frac{\mathrm d}{\mathrm dt}\frac{\partial\tilde L}{\partial\dot Q_a} = \sum_i\Big(\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot q_i}\Big)\frac{\partial q_i}{\partial Q_a} + \sum_i\frac{\partial L}{\partial\dot q_i}\frac{\mathrm d}{\mathrm dt}\frac{\partial q_i}{\partial Q_a}$. Subtract the second: the terms in $\frac{\mathrm d}{\mathrm dt}\frac{\partial q_i}{\partial Q_a}$ cancel, leaving $\tilde{\mathcal E}_a = \sum_i\mathcal E_i\,\partial q_i/\partial Q_a$. ∎

Where invertibility was used. Only in the last step, $\tilde{\mathcal E} = 0 \Rightarrow \mathcal E = 0$. The identity itself holds for any $C^2$ map $\phi$, invertible or not. That is exactly why Lagrange's equations for a constrained system follow from Newton's: the parametrisation $x(q, t)$ of the constraint set is a map from $n$ coordinates into $3N$, not invertible, and the identity says the Euler–Lagrange expressions of $L(q, \dot q, t)$ are the Newtonian residuals $m\ddot{\vec x}_k + \nabla_kV$ paired with the basis $\partial\vec x_k/\partial q_a$ of virtual displacements. Setting those pairings to zero is d'Alembert's principle. So this lemma is the previous section's theorem in a single line.

The covector remark, and why it matters later. A vector's components transform with $J^{-1}$ (if $\dot q = J\dot Q$ then $\dot Q = J^{-1}\dot q$); these transform with $J^{\top}$. The difference is the difference between $V$ and its dual $V^*$, and the course does not define either transformation law as the definition of anything: unit 16 defines tensors as multilinear maps and proves both laws as theorems about change of basis.

12 · invariance

Newton's form is not invariant

Newton does not have this. A free particle in polar coordinates does not obey $m\ddot r = 0$; it obeys $\ddot r - r\dot\theta^2 = 0$. "Mass times the second derivative of the coordinate equals the force" is true only in inertial Cartesian coordinates.

The figure is the free particle passing the origin at distance $b$ with speed $v$: $r(t) = \sqrt{b^2 + v^2t^2}$. It curves upward though nothing pushes, $\ddot r = b^2v^2/r^3 \gt 0$.

The Lagrangian route gets this without thought: $\tilde L = \tfrac12 m(\dot r^2 + r^2\dot\theta^2)$, and $\partial\tilde L/\partial r = mr\dot\theta^2$ is the term "$m\ddot r = F_r$" misses.

Left: no force, so a straight line traversed in equal steps. Right: the polar coordinate $r$ along it. Its graph bends, so $\ddot r \ne 0$ with $F_r = 0$: Newton's law is not true "in polar coordinates".

The polar check. For a free particle $L = \tfrac12 m(\dot x^2 + \dot y^2)$, and with $x = r\cos\theta$, $y = r\sin\theta$ one gets $\tilde L = \tfrac12 m(\dot r^2 + r^2\dot\theta^2)$. Its equations: $m\ddot r = mr\dot\theta^2$ and $\frac{\mathrm d}{\mathrm dt}(mr^2\dot\theta) = 0$. The first is the centripetal term that "$m\ddot r = F_r$" would miss. Problem P3 checks the component transformation law on exactly this example.

The figure's curve, checked. With the line $x = b$, $y = vt$: $r^2 = b^2 + v^2t^2$, so $r\dot r = v^2t$ and, differentiating again, $\dot r^2 + r\ddot r = v^2$, giving $r\ddot r = v^2 - v^4t^2/r^2 = v^2b^2/r^2$. Meanwhile $\tan\theta = vt/b$ gives $r^2\dot\theta = bv$ (the constant angular momentum per unit mass), so $r\dot\theta^2 = b^2v^2/r^3$, which is $\ddot r$: the radial equation holds ✓. For large $\lvert t\rvert$ the curve approaches the dashed lines $r = v\lvert t\rvert$, and the bending is concentrated near closest approach, where $\dot\theta$ is largest.

12 · invariance · example

Worked: Coriolis for free

a time-dependent change of coordinates

A frame turns at constant $\vec\omega$. Unit 03's transport theorem: the inertial velocity, in rotating components, is $\dot{\vec x} + \vec\omega\times\vec x$. Rotation preserves length, so

$$L = \tfrac12 m\lvert\dot{\vec x} + \vec\omega\times\vec x\rvert^2 - V(\vec x).$$

Expand: $L = \tfrac12 m\lvert\dot{\vec x}\rvert^2 + m\,\dot{\vec x}\cdot(\vec\omega\times\vec x) + \tfrac12 m\lvert\vec\omega\times\vec x\rvert^2 - V$.

$\partial L/\partial\dot{\vec x} = m\dot{\vec x} + m\,\vec\omega\times\vec x$, with time derivative $m\ddot{\vec x} + m\,\vec\omega\times\dot{\vec x}$.

$\partial L/\partial\vec x = m\,\dot{\vec x}\times\vec\omega - m\,\vec\omega\times(\vec\omega\times\vec x) - \nabla V$. Subtract:

$$m\ddot{\vec x} = -\nabla V - 2m\,\vec\omega\times\dot{\vec x} - m\,\vec\omega\times(\vec\omega\times\vec x).$$

Coriolis and centrifugal, exactly as unit 03 found them — here from one scalar and no differentiation of rotating unit vectors.

back on the hoop

In the frame of the hoop, $\vec x = R(\sin\theta, 0, -\cos\theta)$ and $\vec\omega\times\vec x = \Omega R\sin\theta\,\hat y$. The cross term $\dot{\vec x}\cdot(\vec\omega\times\vec x)$ is $0$ — $\dot{\vec x}$ has no $\hat y$ part — and $\lvert\vec\omega\times\vec x\rvert^2 = \Omega^2R^2\sin^2\theta$: the same $L$, as the theorem promised.

The two gradients. By the cyclic property of the scalar triple product (unit 08), $\dot{\vec x}\cdot(\vec\omega\times\vec x) = \vec x\cdot(\dot{\vec x}\times\vec\omega)$, which is linear in $\vec x$ with gradient $\dot{\vec x}\times\vec\omega = -\vec\omega\times\dot{\vec x}$. And $\lvert\vec\omega\times\vec x\rvert^2 = \omega^2\lvert\vec x\rvert^2 - (\vec\omega\cdot\vec x)^2$ (Lagrange's identity, unit 08), with gradient $2\omega^2\vec x - 2(\vec\omega\cdot\vec x)\vec\omega = -2\,\vec\omega\times(\vec\omega\times\vec x)$ by the BAC–CAB rule. So $\partial L/\partial\vec x = -m\,\vec\omega\times\dot{\vec x} - m\,\vec\omega\times(\vec\omega\times\vec x) - \nabla V$, and the Euler–Lagrange equation $m\ddot{\vec x} + m\,\vec\omega\times\dot{\vec x} = -m\,\vec\omega\times\dot{\vec x} - m\,\vec\omega\times(\vec\omega\times\vec x) - \nabla V$ rearranges to the slide's result. The factor 2 in the Coriolis term is one $\vec\omega\times\dot{\vec x}$ from each side.

Why this counts as a use of the theorem. The map from rotating to inertial components, $\vec x_{\text{in}} = R(t)\vec x$, is a point transformation depending on time. The inertial $L$ is $\tfrac12 m\lvert\dot{\vec x}_{\text{in}}\rvert^2 - V$, and $\tilde L$ is that same function rewritten in rotating coordinates — which is what the slide wrote. The theorem guarantees its Euler–Lagrange equations describe the true motion, with no fictitious forces put in by hand: they come out.

With a changing $\vec\omega$. If $\vec\omega$ depends on $t$, then $\frac{\mathrm d}{\mathrm dt}(m\,\vec\omega\times\vec x)$ has an extra term $m\,\dot{\vec\omega}\times\vec x$, and the equation acquires unit 03's Euler force $-m\,\dot{\vec\omega}\times\vec x$. Nothing else changes.

Look at the cross term again. $m\,\dot{\vec x}\cdot(\vec\omega\times\vec x)$ is linear in the velocity. Section 5 meets a term of exactly this form, $q\,\dot{\vec x}\cdot\vec A$, for a particle of charge $q$ in a magnetic field with vector potential $\vec A$, and the comparison says the Coriolis force is a magnetic force with $q\vec A = m\,\vec\omega\times\vec x$. Unit 04's problem P13 noted the substitution $qB = 2m\omega$; the curl of $\vec\omega\times\vec x$ is $2\vec\omega$, and that is where the 2 comes from.

12 · invariance

Configuration space is a manifold

The theorem holds chart by chart. Where two coordinate systems overlap, their Euler–Lagrange equations describe the same motions. So "the equations of motion" is a statement about the space of configurations, not about any chart of it.

system	configuration space	one chart misses
bead on hoop	circle $S^1$	one point
double pendulum	torus $S^1\times S^1$	two circles
spherical pendulum	sphere $S^2$	at least a point
rigid body	rotations $\times$ positions, $SO(3)\times\mathbb R^3$	any three angles fail somewhere

informal, on purpose

"A space covered by overlapping charts, each a bijection with an open subset of $\mathbb R^n$, with smooth changes of chart" is a manifold. It is defined properly in unit 31. Here it is enough to know that no single chart need cover everything, and that nothing in the equations depends on which chart is used.

The spherical pendulum in $(\theta, \varphi)$ breaks at the bottom, where $\varphi$ is undefined. Problem P5 treats it a second way, in Cartesian coordinates with a multiplier, where nothing breaks.

The rigid body's row. A rigid body's configuration is where its centre of mass is (a point of $\mathbb R^3$) and how it is turned (a rotation, that is, a $3\times3$ matrix $A$ with $A^{\top}A = I$ and $\det A = 1$; the set of these is called $SO(3)$). Three angles suffice locally — unit 16 uses Euler's — but every choice of three angles has orientations where it breaks, as latitude and longitude break at the poles.

What unit 31 adds. Tangent vectors to the configuration space are the possible velocities $\dot q$, so the Lagrangian is really a function on the tangent bundle, and the lemma of The proof in components is the statement that the Euler–Lagrange expression is a covector field along the motion. Unit 31 opens with exactly this: the double pendulum's torus and the rigid body's $SO(3) \times \mathbb R^3$, and why Euler angles have singularities for reasons that are about $SO(3)$ and not about carelessness.

12 · invariance

Two charts for a circle, a square for a torus

The hoop. $\theta \in (-\pi, \pi)$ covers every position but the top; $\theta \in (0, 2\pi)$ every one but the bottom. Where both apply they differ by $0$ or $2\pi$, so their Euler–Lagrange equations agree there.

No single chart will do. A chart lays the space continuously and one-to-one onto an open interval, and a closed loop cannot be laid flat without cutting it somewhere (the notes prove it).

The double pendulum. Draw $(\theta_1, \theta_2)$ in the square $[-\pi, \pi]^2$ and glue opposite edges: leaving through the right edge is re-entering through the left. That is the torus; the chart $(-\pi, \pi)^2$ misses the two glued edges, two circles.

Left: two charts on the hoop's circle; each misses one point (open dot), and together they cover it. Right: the double pendulum's torus as a square with opposite edges glued, matching arrows to matching arrows.

Why a circle needs two charts. A chart must be a homeomorphism from an open subset of the space onto an open subset of $\mathbb R$. If one chart covered the whole circle $S^1$, then $S^1$ would be homeomorphic to an open subset of $\mathbb R$; but $S^1$ is compact and connected, so its image would be a compact connected subset of $\mathbb R$, a closed interval $[a, b]$, which is not open unless empty. Contradiction (the metric-space topology you came in with is all this uses). Two charts suffice: $\theta \in (-\pi, \pi)$ misses the top, $\theta \in (0, 2\pi)$ misses the bottom, and on the overlap the change of chart is $\theta \mapsto \theta$ or $\theta \mapsto \theta + 2\pi$, smooth with derivative 1.

The torus. The double pendulum's configuration is a pair of angles, each defined modulo $2\pi$: the product of two circles. The same compactness argument shows it is not homeomorphic to an open subset of $\mathbb R^2$. The sphere likewise. For the sphere the obstruction is visible in the standard chart: at $\theta = 0$ every $\varphi$ gives the same point, so the map $(\theta, \varphi) \mapsto$ point is not injective there and its derivative drops rank — a coordinate singularity, which is a property of the chart and not of the pendulum.

Reading the square. The gluing is what the arrows say: a point on the left edge, $\theta_1 = -\pi$, is the same configuration as the point at the same height on the right edge, $\theta_1 = \pi$ (the upper rod pointing straight up either way); likewise top and bottom for $\theta_2$. Glue left to right and the square becomes a tube; glue the tube's two end circles and it is the surface of a doughnut.

section 3

Reading the equations

What falls out before any equation is solved: conserved momenta, a conserved energy function, and the hoop's bifurcation.

12 · reading

Generalised momenta and cyclic coordinates

definitions

The generalised momentum conjugate to $q_a$ is $p_a = \partial L/\partial\dot q_a$. A coordinate that does not appear in $L$ (only its velocity does) is cyclic.

theorem

If $q_a$ is cyclic, $p_a$ is constant along every motion.

Proof. The Euler–Lagrange equation reads $\dot p_a = \partial L/\partial q_a = 0$. ∎

$p$ is not always $m\dot q$

For an angle, $p$ is an angular momentum. In the rotating frame, $\vec p = m\dot{\vec x} + m\,\vec\omega\times\vec x$, the inertial momentum in rotating components. For a particle of charge $q$ — a constant, not a coordinate — in a magnetic field described by a vector potential $\vec A$ (both in section 5), $\vec p = m\dot{\vec x} + q\vec A$. Being cyclic also depends on the chart: in Cartesian coordinates the Kepler problem has none.

On the hoop, $p_\theta = mR^2\dot\theta$ and $\theta$ is not cyclic: gravity and the spin both see it.

Why "cyclic". The standard examples are angles — the azimuth of a central-force problem, the rotation angle of a symmetric top — and a coordinate going round a circle without $L$ noticing is a cycle. The alternative name "ignorable" says what it is for: the equation for $p_a$ integrates at once, and the other equations can then be solved with $p_a$ as a fixed parameter. Routh's procedure (1877) formalises that reduction.

What is and is not coordinate-free here. "$\partial L/\partial q_a = 0$" depends on the whole chart, not on $q_a$ alone, because $\partial/\partial q_a$ means "vary $q_a$ holding the other coordinates fixed". Unit 13 replaces "a cyclic coordinate" by the invariant idea underneath it: a one-parameter group of transformations of configuration space that leaves $L$ unchanged. Every cyclic coordinate gives one (translate that coordinate), and Noether's theorem gives the conserved $p_a$ for any such group, whether or not a chart exists in which it is a translation of one coordinate.

A hoop that is not driven. If the motor is switched off and the hoop, of moment of inertia $I$ about the axis, spins freely, then $\varphi$ is no longer constrained, the system has two degrees of freedom, and $L = \tfrac12 I\dot\varphi^2 + \tfrac12 mR^2(\dot\theta^2 + \dot\varphi^2\sin^2\theta) + mgR\cos\theta$. Now $\varphi$ is cyclic and $p_\varphi = (I + mR^2\sin^2\theta)\dot\varphi$ is conserved: the total angular momentum about the axis. As the bead climbs, the hoop slows. The driven hoop has no such law, because the motor supplies whatever torque it takes (section 4).

12 · reading · example

Worked: Kepler, with no force drawn

unit 08 in five lines

1. Relative coordinate in the orbital plane, reduced mass $\mu$, polar $(r, \theta)$ (the free-particle $\tilde L$ of Newton's form is not invariant):

$$L = \tfrac12\mu\big(\dot r^2 + r^2\dot\theta^2\big) - V(r).$$

2. $\theta$ is cyclic, so $p_\theta = \mu r^2\dot\theta$ is constant: angular momentum, unit 08's $L$, and Kepler's second law.

3. The $r$ equation: $\mu\ddot r - \mu r\dot\theta^2 + V'(r) = 0$.

4. Eliminate $\dot\theta = p_\theta/\mu r^2$:

$$\mu\ddot r = -\frac{\mathrm d}{\mathrm dr}\Big(V(r) + \frac{p_\theta^2}{2\mu r^2}\Big) = -V_{\mathrm{eff}}'(r).$$

5. A one-dimensional problem in unit 08's effective potential.

check

$[p_\theta^2/\mu r^2] = (\mathrm{kg\,m^2\,s^{-1}})^2/(\mathrm{kg\,m^2}) = \mathrm J$ ✓. Circular orbit, $\ddot r = 0$: $\mu r\dot\theta^2 = V'(r)$, centripetal force equals attraction ✓.

Step 2 is where $L$ saved work: $\theta$ is absent from $L$, so its momentum is conserved before any equation is written. The next slide is the one way to spoil that.

Step 1 in detail. Unit 08 reduced the two-body problem to one body of reduced mass $\mu = m_1m_2/(m_1 + m_2)$ at the relative position, and showed the motion is planar because the angular momentum vector is constant. In the plane, Cartesian $L = \tfrac12\mu(\dot x^2 + \dot y^2) - V(\sqrt{x^2+y^2})$, and the point transformation to polar coordinates gives $\dot x^2 + \dot y^2 = \dot r^2 + r^2\dot\theta^2$: from $\dot x = \dot r\cos\theta - r\dot\theta\sin\theta$, $\dot y = \dot r\sin\theta + r\dot\theta\cos\theta$, square and add, and the cross terms $\mp 2r\dot r\dot\theta\sin\theta\cos\theta$ cancel.

Step 3 in detail. $\partial L/\partial\dot r = \mu\dot r$ and $\partial L/\partial r = \mu r\dot\theta^2 - V'(r)$. Unit 08 obtained the same equation by differentiating the rotating unit vector $\hat r$ twice to get the radial acceleration $\ddot r - r\dot\theta^2$. Here the $-r\dot\theta^2$ comes from the $r$-dependence of the kinetic energy, and no unit vector was differentiated.

12 · reading

The trap: eliminating a cyclic velocity in $L$

the trap

Substituting $\dot\theta = p_\theta/\mu r^2$ into $L$ and then varying gives $V - p_\theta^2/2\mu r^2$: the centrifugal term with the wrong sign. Eliminate a cyclic velocity in the equations, not in $L$.

The figure shows what the sign does for gravity, $V = -k/r$. The right effective potential has a barrier at small $r$ and a well; its minimum is the circular orbit at $r_c = p_\theta^2/\mu k$.

The wrong one has neither: every orbit with $\dot r \lt 0$ falls to the centre. An answer this unphysical is the symptom to recognise.

Gravity in units of $r_c$ and $k/r_c$. Solid: $V + p_\theta^2/2\mu r^2$, minimum $-\tfrac12$ at $r = r_c$. Dashed: the trap's $V - p_\theta^2/2\mu r^2$, with no minimum at all. Grey: $V$ alone.

Full proof of the trap

Define $\hat L(r, \dot r) = L$ with $\dot\theta$ replaced by $p_\theta/\mu r^2$: $\hat L = \tfrac12\mu\dot r^2 + p_\theta^2/(2\mu r^2) - V(r)$. Its Euler–Lagrange equation is $\mu\ddot r = -p_\theta^2/(\mu r^3) - V'(r)$, whereas the true one is $\mu\ddot r = +p_\theta^2/(\mu r^3) - V'(r)$. The error is that $\partial L/\partial r$ at fixed $\dot\theta$ is not $\partial\hat L/\partial r$ at fixed $p_\theta$: they differ by $\frac{\partial L}{\partial\dot\theta}\frac{\partial\dot\theta}{\partial r}\big|_{p_\theta} = p_\theta\cdot(-2p_\theta/\mu r^3)$, which is exactly twice the centrifugal term, so it flips its sign. The correct reduced function is Routh's $\mathcal R = L - p_\theta\dot\theta$: with $\dot\theta = p_\theta/\mu r^2$ it is $\tfrac12\mu\dot r^2 + \frac{p_\theta^2}{2\mu r^2} - V - \frac{p_\theta^2}{\mu r^2} = \tfrac12\mu\dot r^2 - V_{\mathrm{eff}}(r)$, whose Euler–Lagrange equation is the right one. Unit 14's Legendre transform is the systematic version of this bookkeeping.

The figure's numbers. With $x = r/r_c$ and energies in units of $k/r_c$, the right function is $-1/x + 1/(2x^2)$: derivative $1/x^2 - 1/x^3$, zero at $x = 1$, value $-\tfrac12$ — the circular orbit's $E = -k/2r_c$ of unit 08. The wrong one is $-1/x - 1/(2x^2)$, increasing for all $x \gt 0$ with no critical point.

12 · reading

The energy function

theorem · the energy function

Put $h(q, \dot q, t) = \sum_a\dot q_a\,\dfrac{\partial L}{\partial\dot q_a} - L$. Along any motion, $\dfrac{\mathrm dh}{\mathrm dt} = -\dfrac{\partial L}{\partial t}$. So if $L$ has no explicit $t$, $h$ is conserved.

Proof. $\dot h = \sum_a(\ddot q_a p_a + \dot q_a\dot p_a) - \big(\partial_tL + \sum_a(\partial_{q_a}L\,\dot q_a + p_a\ddot q_a)\big) = -\partial_tL + \sum_a\dot q_a(\dot p_a - \partial_{q_a}L)$, and the bracket vanishes. ∎

This is unit 11's Beltrami identity with $n$ variables. Unit 13 explains it as time-translation symmetry; unit 14 calls $h$ the Hamiltonian.

on the hoop

$L$ has no explicit $t$ — though the constraint does. So

$$h = \tfrac12 mR^2\dot\theta^2 \underbrace{- \tfrac12 mR^2\Omega^2\sin^2\theta - mgR\cos\theta}_{V_{\mathrm{eff}}(\theta)}$$

is conserved. It is not the energy $E = T + V$, which has $+\tfrac12 mR^2\Omega^2\sin^2\theta$: $E - h = mR^2\Omega^2\sin^2\theta$, and $E$ changes whenever the bead moves, because the motor does work on it.

When $h = T + V$. If the constraints are scleronomic, $x = x(q)$ with no $t$, then $\dot{\vec x}_k = \sum_a(\partial\vec x_k/\partial q_a)\dot q_a$ is linear in $\dot q$ and $T$ is a homogeneous quadratic in $\dot q$. Euler's theorem on homogeneous functions, $\sum_a\dot q_a\,\partial T/\partial\dot q_a = 2T$ (differentiate $T(\lambda\dot q) = \lambda^2T(\dot q)$ in $\lambda$ at $\lambda = 1$), then gives $h = 2T - (T - V) = T + V$. With a rheonomic constraint $T$ acquires terms of degree 1 and 0 in $\dot q$ — on the hoop, the degree-0 term $\tfrac12 mR^2\Omega^2\sin^2\theta$ — and $h \ne T + V$. Unit 13 makes this distinction precise, and it matters again in unit 36, where the conserved "energy" of a geodesic in a static spacetime is an $h$, not a $T + V$.

The motor's power, checked. The motor's force on the bead is $F_\varphi = 2mR\Omega\dot\theta\cos\theta$ (notes of the bead's worked Lagrangian), applied at distance $R\sin\theta$ from the axis, so its torque is $2mR^2\Omega\dot\theta\sin\theta\cos\theta$ and its power is torque times $\Omega$: $P = 2mR^2\Omega^2\dot\theta\sin\theta\cos\theta = \frac{\mathrm d}{\mathrm dt}\big(mR^2\Omega^2\sin^2\theta\big)$. Since $h$ is constant, $\frac{\mathrm d}{\mathrm dt}(E - h) = \dot E = P$: the energy changes by exactly the motor's work, as it must. Section 4 obtains the same torque as a Lagrange multiplier.

The name. "Energy function" and the letter $h$ follow Goldstein. It is Jacobi's integral in celestial mechanics, where the same situation — a rotating frame in which the Lagrangian is time-independent — gives the conserved Jacobi constant of the restricted three-body problem.

12 · reading

The hoop's effective potential

Since $h = \tfrac12 mR^2\dot\theta^2 + V_{\mathrm{eff}}(\theta)$ is conserved, the bead moves exactly like a particle in the effective potential $V_{\mathrm{eff}}$. Equilibria are its critical points:

$$V_{\mathrm{eff}}'(\theta) = mR^2\sin\theta\,\big(\omega_0^2 - \Omega^2\cos\theta\big),$$

with $\omega_0^2 = g/R$, so $\omega_0 = \Omega_c = 5.294\ \mathrm{s^{-1}}$.

theorem · the equilibria

$\theta = 0$ and $\theta = \pi$ always. If $\Omega \gt \omega_0$, also $\pm\theta_*$ with $\cos\theta_* = \omega_0^2/\Omega^2$. Stability, from $V_{\mathrm{eff}}''$:

bottom: $mR^2(\omega_0^2 - \Omega^2)$, stable iff $\Omega \lt \omega_0$; top: $-mR^2(\omega_0^2 + \Omega^2)$, never; tilted: $mR^2\Omega^2\sin^2\theta_* \gt 0$, always.

$V_{\mathrm{eff}}(\theta) - V_{\mathrm{eff}}(0)$ for $m = 0.12$ kg, $R = 0.35$ m. Slow spin: one well, at the bottom. At $\Omega_c$ the bottom goes flat. Fast spin: the bottom is a hump, with a well on each side, 13.0 mJ deep at $\pm 38.87^\circ$.

Why a minimum of $V_{\mathrm{eff}}$ is stable. This is unit 02's Lyapunov argument with $h$ as the Lyapunov function. At a strict local minimum $\theta_0$, the function $h - V_{\mathrm{eff}}(\theta_0)$ is positive near $(\theta_0, 0)$ except at that point, and it is constant along motions, so a motion starting on a small level set stays on it: Lyapunov stable. It is not asymptotically stable — nothing dissipates — and the linearisation is a centre, which is exactly the case where Hartman–Grobman is silent and the conserved function has to do the work. At a strict maximum the linearisation has a positive eigenvalue and Hartman–Grobman does apply: unstable.

The second derivatives. Differentiating $V_{\mathrm{eff}}' = mR^2(\omega_0^2\sin\theta - \Omega^2\sin\theta\cos\theta)$ gives $V_{\mathrm{eff}}'' = mR^2(\omega_0^2\cos\theta - \Omega^2\cos2\theta)$. At $0$: $mR^2(\omega_0^2 - \Omega^2)$. At $\pi$: $mR^2(-\omega_0^2 - \Omega^2)$. At $\theta_*$, using $\omega_0^2 = \Omega^2\cos\theta_*$: $mR^2\Omega^2(\cos^2\theta_* - \cos2\theta_*) = mR^2\Omega^2\sin^2\theta_*$.

Reading the figure. The curves are computed from $V_{\mathrm{eff}} - V_{\mathrm{eff}}(0) = mR^2\big(\omega_0^2(1 - \cos\theta) - \tfrac12\Omega^2\sin^2\theta\big)$, in millijoules ($mR^2 = 0.0147\ \mathrm{kg\,m^2}$). Near the bottom it is $\tfrac12 mR^2(\omega_0^2 - \Omega^2)\theta^2 + O(\theta^4)$, so the curvature there changes sign exactly at $\Omega = \omega_0$; at $\Omega_c$ itself the bottom is flat to fourth order, which is why the period diverges there (next slide). At $\Omega = 6$ the wells are $mR^2\big(\omega_0^2(1 - \cos\theta_*) - \tfrac12\Omega^2\sin^2\theta_*\big) = -12.97$ mJ deep — $0.88\ \mathrm{s^{-2}}$ times $mR^2$, shallow compared with the climb to the equator.

12 · reading

The equilibria, checked and timed

checks

Dimensions: $\omega_0^2/\Omega^2$ is a pure number ✓. $\Omega \downarrow \omega_0$: $\theta_* \to 0$, the tilted pair is born at the bottom ✓. $\Omega \to \infty$: $\theta_* \to 90^\circ$, the $g = 0$ equator ✓.

$\Omega$ ($\mathrm{s^{-1}}$)	$\theta_*$	period at the minimum
0	—	1.187 s (bottom)
4.0	—	1.812 s (bottom)
$\omega_0 = 5.294$	$0^\circ$	$\infty$
6.0	$38.87^\circ$	1.669 s (tilted)
8.0	$64.03^\circ$	0.874 s (tilted)

Small-oscillation period about the stable equilibrium, $2\pi/\sqrt{\omega_0^2 - \Omega^2}$ at the bottom and $2\pi/\Omega\sin\theta_*$ at the tilted pair. It diverges at $\Omega_c$ from both sides; the dots are the table's rows.

The table. Small-oscillation angular frequency is $\sqrt{V_{\mathrm{eff}}''/mR^2}$: at the bottom $\sqrt{\omega_0^2 - \Omega^2}$, at $\theta_*$ it is $\Omega\sin\theta_* = \sqrt{\Omega^4 - \omega_0^4}/\Omega$. With $\omega_0^2 = 28.0286\ \mathrm{s^{-2}}$: $\Omega = 0$ gives $2\pi/5.2942 = 1.187$ s; $\Omega = 4$ gives $2\pi/\sqrt{12.0286} = 2\pi/3.4682 = 1.812$ s; $\Omega = 6$ gives $\cos\theta_* = 0.77857$, $\Omega\sin\theta_* = 3.7653$, period $1.669$ s; $\Omega = 8$ gives $\cos\theta_* = 0.43795$, $\theta_* = 64.03^\circ$, $\Omega\sin\theta_* = 7.1920$, period $0.874$ s. The period diverges as $\Omega \to \omega_0$ from either side — the restoring force at the minimum weakens to nothing, and the system responds ever more slowly. That is the signature of a bifurcation, and it is measurable.

12 · reading

The pitchfork, in unit 02's normal form

Expand the equation of motion about the bottom, with $\mu = \Omega^2 - \omega_0^2$. Using $\sin\theta = \theta - \theta^3/6 + \dots$ and $\cos\theta = 1 - \theta^2/2 + \dots$,

$$\ddot\theta = \mu\,\theta - \Big(\frac{\Omega^2}2 + \frac\mu6\Big)\theta^3 + O(\theta^5).$$

The right side is unit 02's supercritical pitchfork bifurcation normal form $\mu x - x^3$ after rescaling $\theta$: one equilibrium for $\mu \lt 0$, three for $\mu \gt 0$, the outer two carrying the stability the middle one lost. It is a pitchfork because the hoop is symmetric under $\theta \mapsto -\theta$.

check against the exact answer

Balancing $\mu\theta$ against the leading cubic, $\tfrac12\omega_0^2\theta^3$, predicts $\theta_* \approx \sqrt{2\mu}/\omega_0$. At $\Omega = 5.5\ \mathrm{s^{-1}}$: $22.81^\circ$, against the exact $22.09^\circ$. The error is $O(\mu)$, as a normal form promises.

$\theta_*$ against $\Omega$ from $0$ to $10\ \mathrm{s^{-1}}$, exact: solid stable, dashed unstable. The thin dashed pair is the normal form's $\pm\sqrt{2\mu}/\omega_0$: it leaves $\Omega_c$ with the exact branches and pulls away as $\mu$ grows. The dots are the check at $\Omega = 5.5$.

The expansion. The right-hand side is $\sin\theta\,(\Omega^2\cos\theta - \omega_0^2)$. With $\Omega^2\cos\theta - \omega_0^2 = \mu - \tfrac12\Omega^2\theta^2 + O(\theta^4)$ and $\sin\theta = \theta - \tfrac16\theta^3 + O(\theta^5)$, the product is $\mu\theta - \tfrac12\Omega^2\theta^3 - \tfrac16\mu\theta^3 + O(\theta^5)$. At $\mu = 0$ the cubic coefficient is $\tfrac12\omega_0^2 \gt 0$: supercritical. Rescaling $\theta = x\sqrt{2}/\omega_0$ turns $\mu\theta - \tfrac12\omega_0^2\theta^3$ into $(\sqrt2/\omega_0)(\mu x - x^3)$, unit 02's normal form up to a positive constant factor.

What the normal form does and does not say here. Unit 02's normal form is for a first-order system $\dot x = f(x, \mu)$. Here the system is second order, $\ddot\theta = f(\theta, \mu)$, and conservative. The set of equilibria and which of them are minima of $V_{\mathrm{eff}}$ follows the pitchfork exactly, because both are decided by $f$ and $f' = -V_{\mathrm{eff}}''/mR^2$. What differs is the local picture: stable equilibria are centres, not nodes, and unstable ones are saddles. Add a little friction and the stable ones become spirals and the unit 02 picture returns entirely.

The numbers. At $\Omega = 5.5$: $\mu = 30.25 - 28.0286 = 2.2214\ \mathrm{s^{-2}}$, $\sqrt{2\mu}/\omega_0 = 2.1078/5.2942 = 0.39813$ rad $= 22.81^\circ$; exact $\arccos(28.0286/30.25) = \arccos 0.92656 = 22.09^\circ$. Keeping the next coefficient, $\theta_*^2 \approx \mu/(\tfrac12\Omega^2 + \tfrac16\mu)$ gives $21.69^\circ$ — overshooting on the other side, since the neglected $O(\theta^5)$ term is now comparable. Problem P10 does this carefully. The diagram's curves are the exact $\pm\arccos(\omega_0^2/\Omega^2)$ from $\Omega_c = 5.294$ to $10\ \mathrm{s^{-1}}$ (where $\theta_* = 73.7^\circ$), on a vertical scale where $\pm 90^\circ$ is the edge.

12 · reading

Phase portraits on each side

Because $h$ is conserved, each motion stays on one level set of $h$ in the $(\theta, \dot\theta)$ plane: the phase portrait is a contour map, drawn without solving anything.

$\Omega = 4\ \mathrm{s^{-1}}$, below $\Omega_c$. One centre at the bottom, period 1.812 s for small swings. Closed curves round it are swings to and fro.

The separatrix through the top saddle (only its ends near $\theta = \pm\pi$ are on this scale) divides swinging from going over the top, the open curves at the edges.

Level sets of $h$ at $\Omega = 4\ \mathrm{s^{-1}}$, $\theta \in [-\pi, \pi]$, $\dot\theta \in [-6, 6]\ \mathrm{s^{-1}}$. Green dot: a centre. Red dots: the top saddle, at $\pm\pi$, which are one point. Heavy curves: the separatrix.

How the figure was made. Each curve is a level set $\tfrac12\dot\theta^2 + U(\theta) = c$ with $U = V_{\mathrm{eff}}/mR^2 = -\tfrac12\Omega^2\sin^2\theta - \omega_0^2\cos\theta$, drawn as $\dot\theta = \pm\sqrt{2(c - U(\theta))}$. No differential equation was integrated: the conserved $h$ turns the phase portrait into a contour plot, which is the practical reason to find conserved quantities. Unit 02 drew the pendulum's portrait the same way.

The levels. $\Omega = 4$: $c = -27, -24, -18, -10, 0, 12\ \mathrm{s^{-2}}$ in the trunk colour, and the separatrix $c = U(\pi) = \omega_0^2 = 28.03\ \mathrm{s^{-2}}$ heavy, which reaches $\dot\theta = \sqrt{2 \cdot 56.06} = 10.6\ \mathrm{s^{-1}}$ at the bottom — off the plotted range, so only its ends near $\pm\pi$ show. The bottom of the well is $U(0) = -28.03\ \mathrm{s^{-2}}$, so the innermost curve, $c = -27$, is a swing of about $\pm 23^\circ$ at nearly the small-oscillation period.

Topology. The plane of the picture is really a cylinder: $\theta = -\pi$ and $\theta = \pi$ are the same point, the top of the hoop. Unit 02 met this for the pendulum. An open curve leaving the right edge re-enters on the left: the bead going round and round the hoop.

12 · reading

Phase portrait above the critical spin

$\Omega = 6\ \mathrm{s^{-1}}$. Two centres at $\pm 38.87^\circ$, period 1.669 s. The bottom is now a saddle.

Its separatrix is a figure of eight around both wells: inside a lobe the bead rocks on one side; outside, it swings through the bottom and across.

The wells are shallow: the lobes reach only $\dot\theta = \pm 1.33\ \mathrm{s^{-1}}$, so a gentle push sends the bead across the bottom from one side to the other.

Level sets of $h$ at $\Omega = 6\ \mathrm{s^{-1}}$, same axes as the previous slide. Green dots: centres. Red dots: saddles, at the bottom and the top. Heavy curves: the figure-of-eight separatrix and, at the edges, the one through the top.

The levels. $\Omega = 6$: two levels inside each lobe, $U(\theta_*) + 0.3$ and $U(\theta_*) + 0.65\ \mathrm{s^{-2}}$; then $c = -27, -24, -18, -8, 4$ outside the eight; heavy, the separatrices $c = U(0) = -\omega_0^2 = -28.03$ (the eight) and $c = U(\pi) = +28.03\ \mathrm{s^{-2}}$ (off scale except near $\pm\pi$). The wells have depth $U(0) - U(\theta_*) = -28.03 - (-28.91) = 0.88\ \mathrm{s^{-2}}$, so the lobes of the figure of eight reach only $\dot\theta = \pm\sqrt{2 \times 0.88} = \pm 1.33\ \mathrm{s^{-1}}$.

Counting equilibria. The count obeys unit 02's index theorem on the cylinder: centres have index $+1$, saddles $-1$, and the totals are $1 - 1 = 0$ below $\Omega_c$ and $2 - 2 = 0$ here. The bifurcation cannot change the total, and it does not: the bottom goes from $+1$ to $-1$ while two new $+1$'s are born.

section 4

Constraint forces, when you want them

Keep a redundant coordinate, add a multiplier, and the force appears as its value. Then the constraints that do not fit.

12 · constraint forces

Multipliers bring the forces back

theorem · constraint force by multiplier

Describe the system by more coordinates than it needs, $q \in \mathbb R^{n+k}$, with holonomic ideal constraints $G_j(q, t) = 0$, $j = 1..k$, and $\mathrm D_qG$ of rank $k$. Then there are functions $\lambda_j(t)$ with

$$\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot q_a} - \frac{\partial L}{\partial q_a} = \sum_{j=1}^k\lambda_j\,\frac{\partial G_j}{\partial q_a},$$

and the right-hand side is the generalised constraint force: $\sum_j\lambda_j\partial G_j/\partial q_a = \sum_{i=1}^N\vec R_i\cdot\partial\vec x_i/\partial q_a$.

Proof. Unit 11's pointwise-constraint theorem, with several constraints: the implicit function theorem reduces to $n$ free coordinates, and the left side must annihilate every direction tangent to $G = 0$, so it lies in the span of the $\nabla_qG_j$. The notes give the linear algebra.

Use. $n + k$ equations plus $k$ constraints, for $n + k$ coordinates and $k$ multipliers. Keep only the coordinates whose forces you want: to get the hoop's normal force, keep $r$ and impose $r - R = 0$.

Full proof

Let $\mathcal E_a$ be the left-hand side, computed along the motion. By the component lemma of section 2 applied to the (non-invertible) map from free coordinates into $q$, the Euler–Lagrange equations of the reduced system say exactly that $\sum_a\mathcal E_a\,\delta q_a = 0$ for every $\delta q$ tangent to the constraint set at fixed $t$, that is, for every $\delta q \in \ker\mathrm D_qG$. (Directly: this is d'Alembert's principle written in the coordinates $q$, with $\mathcal E$ the generalised residual.) In linear algebra, a linear functional vanishing on $\ker A$ is in the row space of $A$: the annihilator of $\ker A$ is $\operatorname{im}A^{\top}$. With $A = \mathrm D_qG$, $k \times (n + k)$ of rank $k$, that gives unique $\lambda_j$ with $\mathcal E = \sum_j\lambda_j\nabla_qG_j$. Uniqueness comes from the rows being independent, and continuity in $t$ from solving the linear system $A A^{\top}\lambda = A\,\mathcal E$, whose matrix is invertible.

Why the right side is the constraint force. In Cartesian coordinates $\mathcal E_k = m_k\ddot{\vec x}_k + \nabla_kV = \vec R_k$ by Newton's law, so the multiplier term is the reaction. In other coordinates both sides transform by $J^{\top}$ (section 2's lemma), and so does the force into a generalised force. Sign: with this convention a positive $\lambda$ pushes in the direction of increasing $G$.

The same theorem as unit 04. For statics, $\dot q = 0$, it reduces to $\nabla V = \sum\lambda_j\nabla G_j$ up to sign: unit 04's Lagrange multipliers, whose multiplier in problem P10 was the hoop's force. This slide is that problem, in motion.

12 · constraint forces · example

Worked: what the hoop pushes with

the in-plane reaction

Keep $r$; release it from the constraint. With $\varphi = \Omega t$ still imposed,

$$\begin{aligned}L ={}& \tfrac12 m\big(\dot r^2 + r^2\dot\theta^2 + r^2\Omega^2\sin^2\theta\big)\\ &+ mgr\cos\theta,\end{aligned}$$

with constraint $G = r - R = 0$.

The $r$ equation: $m\ddot r - mr\dot\theta^2 - mr\Omega^2\sin^2\theta - mg\cos\theta = \lambda$. On the hoop $r = R$, $\ddot r = 0$:

$$\lambda = -mR\big(\dot\theta^2 + \Omega^2\sin^2\theta\big) - mg\cos\theta,$$

the outward radial reaction ($\Omega = 0$, at rest at the bottom: $-mg$, the hoop holding the bead up ✓). At rest at $\theta_*$, with $g\cos\theta_* = R\Omega^2\cos^2\theta_*$: $\lambda = -mR\Omega^2$.

$\lvert\lambda\rvert = 0.12 \times 0.35 \times 36 = 1.512$ N, inward — unit 04's P10 answer ✓.

The bead at rest at $\theta_* = 38.87^\circ$, $\Omega = 6\ \mathrm{s^{-1}}$, forces to scale. Gravity plus the hoop's push along the radius add to a horizontal $mR\Omega^2\sin\theta_*$ toward the axis: exactly the centripetal force for the bead's circle round it.

The kinetic energy with $r$ free. Differentiate $\vec x = r(\sin\theta\cos\varphi,\ \sin\theta\sin\varphi,\ -\cos\theta)$ in $t$, letting $r$, $\theta$ and $\varphi$ all move: $\dot{\vec x} = \dot r\,\hat r + r\dot\theta\,\hat\theta + r\sin\theta\,\dot\varphi\,\hat\varphi$, where $\hat r = (\sin\theta\cos\varphi, \sin\theta\sin\varphi, -\cos\theta)$, $\hat\theta = (\cos\theta\cos\varphi, \cos\theta\sin\varphi, \sin\theta)$ and $\hat\varphi = (-\sin\varphi, \cos\varphi, 0)$ are of unit length and mutually orthogonal, as three dot products confirm. Hence $\lvert\dot{\vec x}\rvert^2 = \dot r^2 + r^2\dot\theta^2 + r^2\sin^2\theta\,\dot\varphi^2$, and with $\dot\varphi = \Omega$ that is the slide's $L$. $\partial L/\partial\dot r = m\dot r$ and $\partial L/\partial r = mr\dot\theta^2 + mr\Omega^2\sin^2\theta + mg\cos\theta$, which gives the $r$ equation.

Reading $\lambda$. The three terms are, in order: the centripetal force needed for motion along the hoop, the part of the centripetal force for the circle around the axis that is directed along $\hat r$, and the component of gravity along $\hat r$ that the hoop must cancel. At $\theta_*$ at rest: $\lambda = -mR\Omega^2\sin^2\theta_* - mg\cos\theta_*$, and $g = R\Omega^2\cos\theta_*$ gives $-mR\Omega^2(\sin^2\theta_* + \cos^2\theta_*) = -mR\Omega^2$. Unit 04 checked the magnitude 1.512 N in the inertial frame, from the horizontal $0.94886$ N and vertical $1.1772$ N components.

When the multiplier changes sign. A bead threaded on a hoop is held on both sides, so any sign of $\lambda$ is allowed. A particle resting on the outside of a surface is held only one way, and it leaves the surface when the multiplier passes through zero. Problem P8 finds where a particle leaves a sphere this way.

12 · constraint forces · example

Worked: the motor's torque

the motor's torque

Now keep $\varphi$ too, with $G_2 = \varphi - \Omega t$ and $\partial G_2/\partial\varphi = 1$. The $\varphi$ equation, on the constraint:

$$\begin{aligned}\lambda_2 &= \frac{\mathrm d}{\mathrm dt}\big(mR^2\sin^2\theta\,\dot\varphi\big)\\ &= 2mR^2\Omega\,\dot\theta\sin\theta\cos\theta .\end{aligned}$$

A torque about the axis: zero at rest, and exactly the Coriolis term the hoop must supply when the bead moves. It is the $F_\varphi$ discarded when $L$ was built.

Check: $[\lambda_2] = \mathrm{kg\,m^2\,s^{-2}} = \mathrm{N\,m}$ ✓, a torque.

Seen from above, schematic. A bead sliding outward along the turning hoop must be pushed sideways to keep up with it — farther out, the hoop moves faster. That push, perpendicular to the hoop's plane, is $\lambda_2$ divided by the lever arm $R\sin\theta$.

The torque. With $\varphi$ free, the $\varphi$-part of $L$ is $\tfrac12 mr^2\sin^2\theta\,\dot\varphi^2$, and $\varphi$ is cyclic, so $\frac{\mathrm d}{\mathrm dt}(mr^2\sin^2\theta\,\dot\varphi) = \lambda_2$. On the constraints, $r = R$ and $\dot\varphi = \Omega$, the time derivative is $2mR^2\Omega\sin\theta\cos\theta\,\dot\theta$. Dividing by the lever arm $R\sin\theta$ gives the force $2mR\Omega\dot\theta\cos\theta$ — the $ma_\varphi$ of the worked Lagrangian's notes ✓. Without the motor this torque would be zero and $p_\varphi$ conserved: the free-hoop case of the cyclic-coordinate slide.

Reading the figure. The bead's distance from the axis is $\rho = R\sin\theta$, so it moves outward at $\dot\rho = R\dot\theta\cos\theta$. Its sideways velocity, carried by the hoop, is $\Omega\rho$, and keeping up as $\rho$ grows needs a sideways acceleration $\Omega\dot\rho$ — and another $\Omega\dot\rho$ because the outward velocity itself turns with the hoop. Total $2\Omega\dot\rho = 2R\Omega\dot\theta\cos\theta$, times $m$: the Coriolis force of Worked: Coriolis for free, supplied here by the hoop and paid for by the motor.

12 · constraint forces

Non-holonomic constraints, honestly

definition

A constraint $\sum_a A_a(q)\,\dot q_a = 0$ on velocities is non-holonomic if it is not equivalent to any constraint $f(q) = 0$ on positions.

the knife edge

A blade at $(x, y)$ with heading $\varphi$ cannot move sideways: $\dot x\sin\varphi - \dot y\cos\varphi = 0$. Yet it can reach every $(x, y, \varphi)$: turn, roll, turn. So no $f$ with $\nabla f \ne 0$ is constant along allowed motions.

Left: the blade may roll along its heading, never sideways. Right: it gets sideways anyway — turn on the spot, roll, turn back — so the constraint restricts velocities without restricting where the blade can be.

12 · constraint forces

Lagrange–d'Alembert, and what does not work

the Lagrange–d'Alembert principle

Virtual displacements obey the constraint, $\sum_aA_{ja}\,\delta q_a = 0$. Ideal forces do no work on them. The row-space argument then gives

$$\frac{\mathrm d}{\mathrm dt}\frac{\partial L}{\partial\dot q_a} - \frac{\partial L}{\partial q_a} = \sum_j\lambda_jA_{ja}(q).$$

what does not work

Making $S$ stationary among curves obeying the constraint (vakonomic mechanics) gives different equations — here, a knife edge whose heading swings like a pendulum. Experiment sides with d'Alembert (cited: Lewis & Murray, 1995). Problem P9.

Solved by the principle, the knife edge runs at constant speed $v$ and constant turning rate $\omega$: a circle of radius $v/\omega$, with the multiplier $\lambda = -mv\omega$ as the sideways grip. The notes do it; problem P9 in full.

The knife edge solved. $L = \tfrac12 m(\dot x^2 + \dot y^2) + \tfrac12 I\dot\varphi^2$ and $A = (\sin\varphi, -\cos\varphi, 0)$. Lagrange–d'Alembert: $m\ddot x = \lambda\sin\varphi$, $m\ddot y = -\lambda\cos\varphi$, $I\ddot\varphi = 0$. So $\dot\varphi = \omega$ is constant, and the forward speed $v = \dot x\cos\varphi + \dot y\sin\varphi$ has $\dot v = (\lambda/m)(\sin\varphi\cos\varphi - \cos\varphi\sin\varphi) + \omega(-\dot x\sin\varphi + \dot y\cos\varphi) = 0$, the last bracket being the constraint. Constant speed, constant turning rate: a circle of radius $v/\omega$. Projecting sideways, $m(\ddot x\sin\varphi - \ddot y\cos\varphi) = \lambda$, and the differentiated constraint gives $\ddot x\sin\varphi - \ddot y\cos\varphi = -\omega v$, so the multiplier $\lambda = -mv\omega$ is the sideways grip force providing the centripetal acceleration, as it should be. Problem P9 does this in full.

The warning, and its source. The vakonomic equations come from making $\int(L + \sum_j\mu_j(t)\sum_aA_{ja}\dot q_a)\,\mathrm dt$ stationary with free multiplier functions $\mu_j$ — the tempting generalisation of unit 11's pointwise constraint. For holonomic constraints the two routes agree. For non-holonomic ones they do not, and the vakonomic system even has an extra free initial datum $\mu(0)$ that no physical initial state determines. The standard references: A. D. Lewis and R. M. Murray, "Variational principles for constrained systems: theory and experiment", Int. J. Non-Linear Mechanics 30 (1995) 793–815, which compares both predictions with a physical experiment; A. M. Bloch, Nonholonomic Mechanics and Control (2003), chapters 1 and 5. The claim "experiment sides with d'Alembert" is cited from Lewis and Murray, not proved here. The honest summary: non-holonomic mechanics is not a variational principle in Hamilton's sense, and many published treatments get this wrong.

section 5

What $L$ is allowed to be

A total derivative changes nothing that can be measured, and a potential may depend on velocity.

12 · freedom in L

Gauge freedom

theorem · gauge freedom

For any $C^2$ function $f(q, t)$, the Lagrangians $L$ and $L' = L + \dfrac{\mathrm df}{\mathrm dt} = L + \sum_a\dfrac{\partial f}{\partial q_a}\dot q_a + \dfrac{\partial f}{\partial t}$ have the same Euler–Lagrange equations.

Proof. $S'[q] = S[q] + f(q(t_1), t_1) - f(q(t_0), t_0)$, and the added term is the same for every path with those endpoints. ∎

but not the same momenta

$p'_a = p_a + \partial f/\partial q_a$ and $h' = h - \partial f/\partial t$. The canonical momentum is not gauge-invariant, so it is not by itself a measurable quantity.

a Galilean boost is a gauge transformation

A free particle seen from a frame moving at $\vec u$: $\vec x = \vec x' + \vec ut$, so

$$\begin{aligned}\tfrac12 m\lvert\dot{\vec x}'+\vec u\rvert^2 &= \tfrac12 m\lvert\dot{\vec x}'\rvert^2 + \frac{\mathrm df}{\mathrm dt},\\ f &= m\,\vec u\cdot\vec x' + \tfrac12 m\lvert\vec u\rvert^2t.\end{aligned}$$

Same equations in both frames — Galilean relativity, in one line. The momenta differ by $m\vec u$, as they must.

On the hoop, $L$ and $L + c\,\dot\theta$ give identical motion; $p_\theta$ shifts by $c$ while $mR^2\dot\theta$ does not.

The proof by computation. For $F = \mathrm df/\mathrm dt = \sum_b f_{q_b}\dot q_b + f_t$: $\partial F/\partial\dot q_a = f_{q_a}$, whose time derivative is $\sum_b f_{q_aq_b}\dot q_b + f_{q_at}$; and $\partial F/\partial q_a = \sum_b f_{q_bq_a}\dot q_b + f_{tq_a}$. Equal by Clairaut, so the Euler–Lagrange expression of $\mathrm df/\mathrm dt$ vanishes identically — for every path, not only for motions. The converse (cited: Olver, Applications of Lie Groups to Differential Equations, 2nd ed., §4.1, on null Lagrangians) is that a Lagrangian whose Euler–Lagrange expression vanishes identically is a total derivative, locally.

Which quantities are physical. The motion $q(t)$ is. The velocity, the kinetic energy, and here $m\dot{\vec x}$, are. The canonical momentum $p$ and the energy function $h$ depend on the choice of $f$. This does not make them useless: conservation of $p_a$ is a gauge-invariant statement when $\partial f/\partial q_a$ is itself conserved, and unit 13's quasi-invariant case of Noether's theorem is exactly the bookkeeping for when a symmetry changes $L$ by a total derivative. The Galilean boost is that case: its conserved charge is $m\vec x - \vec pt$, the centre-of-mass theorem.

The expansion. $\tfrac12 m\lvert\dot{\vec x}' + \vec u\rvert^2 = \tfrac12 m\lvert\dot{\vec x}'\rvert^2 + m\vec u\cdot\dot{\vec x}' + \tfrac12 m\lvert\vec u\rvert^2$, and with $\vec u$ constant, $m\vec u\cdot\dot{\vec x}' = \frac{\mathrm d}{\mathrm dt}(m\vec u\cdot\vec x')$ and $\tfrac12 m\lvert\vec u\rvert^2 = \frac{\mathrm d}{\mathrm dt}(\tfrac12 m\lvert\vec u\rvert^2t)$. Problem P6 does the same for a uniform gravitational field, and identifies which conserved quantity moves.

Where the word comes from. Weyl's 1918 "Eichinvarianz" — invariance under a change of gauge, meaning scale. The name stuck to the electromagnetic freedom of the next slide, and from there spread to any redundancy of description.

12 · freedom in L

Velocity-dependent potentials

definition

A generalised potential is a function $U(q, \dot q, t)$ with $Q_a = \dfrac{\mathrm d}{\mathrm dt}\dfrac{\partial U}{\partial\dot q_a} - \dfrac{\partial U}{\partial q_a}$. Then $L = T - U$ gives the motion, by the d'Alembert slide.

theorem · a charge in a field

A particle of mass $m$ and charge $q$ (a constant here, not a coordinate) in fields given by potentials $\phi$, $\vec A$: $\vec E = -\nabla\phi - \partial_t\vec A$, $\vec B = \nabla\times\vec A$. Then

$$L = \tfrac12 m\lvert\dot{\vec x}\rvert^2 - q\phi(\vec x, t) + q\,\dot{\vec x}\cdot\vec A(\vec x, t)$$

gives $m\ddot{\vec x} = q(\vec E + \dot{\vec x}\times\vec B)$, the Lorentz force.

BAC–CAB in components — full proof in the notes (N)

cited, not proved here

That the force on a charge is $q(\vec E + \vec v\times\vec B)$, and that every field can be written through potentials $\phi$, $\vec A$, are physics and a theorem of unit 24. Take them as given for now.

two things this explains

$\vec p = m\dot{\vec x} + q\vec A$, and the kinetic momentum $m\dot{\vec x}$ is the gauge-invariant part. Changing potentials by $\vec A \to \vec A + \nabla\chi$, $\phi \to \phi - \partial_t\chi$ adds $q\,\mathrm d\chi/\mathrm dt$ to $L$: the previous slide's freedom, with $f = q\chi$.

Full proof

Write components $x_i$, $v_i = \dot x_i$, and sum over repeated indices explicitly. $\partial L/\partial v_i = mv_i + qA_i$. Along the path, $\frac{\mathrm d}{\mathrm dt}A_i(\vec x(t), t) = \partial_tA_i + \sum_jv_j\partial_jA_i$. And $\partial L/\partial x_i = -q\partial_i\phi + q\sum_jv_j\partial_iA_j$. The Euler–Lagrange equation:

$$m\dot v_i = q\big(-\partial_i\phi - \partial_tA_i\big) + q\sum_jv_j\big(\partial_iA_j - \partial_jA_i\big).$$

The first bracket is $E_i$. For the second, BAC–CAB (unit 08) with $\vec v$ constant in space gives $\big(\vec v\times(\nabla\times\vec A)\big)_i = \sum_jv_j\partial_iA_j - \sum_jv_j\partial_jA_i$, so the second term is $q(\vec v\times\vec B)_i$. ∎

Dimensions. $q\phi$ is an energy, so $[\phi] = \mathrm{J\,C^{-1}} = \mathrm V$; $q\vec v\cdot\vec A$ is an energy, so $[q\vec A] = \mathrm{J\,s\,m^{-1}} = \mathrm{kg\,m\,s^{-1}}$, a momentum, as $\vec p = m\vec v + q\vec A$ requires ✓. $[\vec B] = [\vec A]/\mathrm m = \mathrm{kg\,s^{-1}\,C^{-1}} = \mathrm T$ ✓.

The gauge check. Under $\vec A \to \vec A + \nabla\chi$, $\phi \to \phi - \partial_t\chi$, the fields $\vec E$ and $\vec B$ are unchanged (curl of a gradient is zero, unit 04; and the two $\partial_t\nabla\chi$ terms cancel by Clairaut), and $L$ changes by $q(\partial_t\chi + \vec v\cdot\nabla\chi) = q\,\mathrm d\chi/\mathrm dt$. So electromagnetic gauge freedom is a special case of Lagrangian gauge freedom. Problem P7 does a uniform magnetic field in two gauges and watches the cyclic coordinate move.

Magnetism and rotation. Compare with the rotating frame's $m\dot{\vec x}\cdot(\vec\omega\times\vec x)$: it is $q\dot{\vec x}\cdot\vec A$ with $q\vec A = m\vec\omega\times\vec x$, whose curl is $2m\vec\omega$. So the Coriolis force is a magnetic force with $q\vec B = 2m\vec\omega$ — Larmor's theorem, and the reason unit 04's P13 found the same substitution.

Why it is only written down now. Unit 24 has Maxwell's equations, the potentials and gauge transformations; unit 28 shows that this $L$ is the non-relativistic limit of $-mc^2\sqrt{1 - v^2/c^2} + qA_\mu\dot x^\mu$, which is where the velocity-dependence stops looking odd.

section 6

Small oscillations

Expand $L$ to second order about an equilibrium, and a matrix problem comes out.

12 · small oscillations

Linearising a Lagrangian

theorem

Let the constraints be scleronomic, so $T = \tfrac12\dot q^{\top}M(q)\,\dot q$, and let $q_0$ be a critical point of $V(q)$. With $q = q_0 + \eta$, the Euler–Lagrange equations truncated at first order in $(\eta, \dot\eta)$ are those of

$$L_2 = \tfrac12\dot\eta^{\top}M\dot\eta - \tfrac12\eta^{\top}K\eta,$$

with $M = M(q_0)$ and $K = \mathrm{Hess}\,V(q_0)$, namely $M\ddot\eta + K\eta = 0$.

$M$ is the mass matrix, symmetric and positive definite; $K$ is the stiffness matrix, symmetric.

proof

Terms of $L$ of degree three or more in $(\eta, \dot\eta)$ contribute degree two or more to the equations. $V(q_0)$ is a constant and the linear term of $V$ vanishes at a critical point. In $T$, $M(q_0 + \eta) = M(q_0) + O(\eta)$, and the $O(\eta)$ part is cubic. What remains is $L_2$. ∎

the hoop, as one degree of freedom

$T$ has a $\dot\theta$-free term, but it moves into $V_{\mathrm{eff}}$ and the theorem applies: $M = mR^2$, $K = V_{\mathrm{eff}}''$, $\omega^2 = K/M = \omega_0^2 - \Omega^2$ at the bottom ✓.

Why $M$ is positive definite. $T = \tfrac12\sum_km_k\lvert\sum_a(\partial\vec x_k/\partial q_a)\dot q_a\rvert^2 \ge 0$, with equality only if $\sum_a(\partial\vec x_k/\partial q_a)\dot q_a = \vec 0$ for all $k$, which by the rank condition on generalised coordinates (the generalised-coordinates slide) forces $\dot q = 0$. Symmetric by construction. Positive definiteness is also what lets the Euler–Lagrange equations be solved for $\ddot q$ in general: they read $M(q)\ddot q + (\text{terms without } \ddot q) = 0$.

The hoop's extra term. With the rheonomic constraint, $T = \tfrac12 mR^2\dot\theta^2 + \tfrac12 mR^2\Omega^2\sin^2\theta$. The second term has no velocity in it, so for the Euler–Lagrange equations it behaves exactly like $-(\text{a potential})$; moving it into $V_{\mathrm{eff}}$ changes nothing. What would break the theorem is a term linear in $\dot q$, like the rotating frame's $m\dot{\vec x}\cdot(\vec\omega\times\vec x)$: it contributes a first-order term $G\dot\eta$ with $G$ antisymmetric, and the equations become $M\ddot\eta + G\dot\eta + K\eta = 0$. That is the gyroscopic case (unit 16's tops, and the Lagrange points of the restricted three-body problem, which are stable though they are maxima of the effective potential).

Stability and linearisation. If $K$ is positive definite, $q_0$ is a strict minimum of $V$ and $h = T + V$ is a Lyapunov function as on the effective potential slide: the equilibrium is stable, and this does not depend on linearisation. If $K$ has a negative eigenvalue, linearisation gives a growing solution, and Hartman–Grobman (unit 02) makes the equilibrium unstable.

12 · small oscillations

Modes: $\det(K - \omega^2M) = 0$

Try $\eta = v\cos(\omega t + \alpha)$ with $v \in \mathbb R^n$ fixed. Then $M\ddot\eta + K\eta = (K - \omega^2M)v\cos(\omega t + \alpha)$, so $v \ne 0$ works exactly when

$$(K - \omega^2M)\,v = 0,\qquad \det(K - \omega^2M) = 0 .$$

Each root $\omega_j^2$ is a normal frequency squared, each $v_j$ a normal mode.

theorem

$M$ symmetric positive definite, $K$ symmetric. Then the $n$ roots are real, positive if $K$ is, and there are $n$ modes with $v_i^{\top}Mv_j = 0$ for $i \ne j$. Every solution is a sum of modes.

proof · the two short parts

Real: for complex $v$, $\bar v^{\top}Kv = \omega^2\,\bar v^{\top}Mv$, and both quadratic forms are real with $\bar v^{\top}Mv \gt 0$. Positive: take $v$ real, then $\omega^2 = v^{\top}Kv/v^{\top}Mv$. $M$-orthogonal: $v_i^{\top}Kv_j = \omega_j^2v_i^{\top}Mv_j$ and, by symmetry, $= \omega_i^2v_i^{\top}Mv_j$; subtract, for $\omega_i \ne \omega_j$. ∎

completeness by the spectral theorem — notes (N)

Unit 07, if you have met it, develops this at length. Nothing here depends on it.

Full proof of completeness

$M$ is symmetric positive definite, so by the spectral theorem $M = P\,\mathrm{diag}(\mu_i)P^{\top}$ with $P$ orthogonal and $\mu_i \gt 0$; put $M^{1/2} = P\,\mathrm{diag}(\sqrt{\mu_i})P^{\top}$, symmetric, positive definite, invertible, squaring to $M$. Substitute $\eta = M^{-1/2}\xi$ and multiply $M\ddot\eta + K\eta = 0$ on the left by $M^{-1/2}$: $\ddot\xi + \tilde K\xi = 0$ with $\tilde K = M^{-1/2}KM^{-1/2}$, symmetric. The spectral theorem again gives an orthonormal eigenbasis $w_j$ of $\tilde K$, eigenvalues $\omega_j^2$. In that basis the system is $n$ uncoupled equations $\ddot\xi_j + \omega_j^2\xi_j = 0$, each solved by unit 06 (cosines, or $\cosh$ and $\sinh$ when $\omega_j^2 \lt 0$, or $a + bt$ when $\omega_j^2 = 0$). By Picard–Lindelöf the solution space has dimension $2n$ and these span it. Back-substituting, $v_j = M^{-1/2}w_j$ satisfies $Kv_j = \omega_j^2Mv_j$ and $v_i^{\top}Mv_j = w_i^{\top}w_j = \delta_{ij}$. ∎

Why "diagonalise $K$" is wrong. The eigenvectors of $K$ alone, or of $M^{-1}K$ treated as if it were symmetric, are not the modes in general: $M^{-1}K$ is not symmetric, and its eigenvectors are orthogonal in the $M$ inner product $\langle u, v\rangle_M = u^{\top}Mv$, not the Euclidean one. The double pendulum's modes, two slides on, show it: $(1, \sqrt2)\cdot(1, -\sqrt2) = -1 \ne 0$, while the $M$-product is zero.

Dimensions. For $\det(K - \omega^2M)$ to make sense, $[K] = [M]\,\mathrm s^{-2}$, entry by entry. For angles, $M$ is in $\mathrm{kg\,m^2}$ and $K$ in $\mathrm{N\,m} = \mathrm{kg\,m^2\,s^{-2}}$ ✓.

12 · small oscillations · example

Worked: the double pendulum

two bobs, two rods, angles from the vertical

1. Positions. $\vec x_1 = l_1(\sin\theta_1, -\cos\theta_1)$, $\ \vec x_2 = \vec x_1 + l_2(\sin\theta_2, -\cos\theta_2)$.

2. Lagrangian. $\lvert\dot{\vec x}_2\rvert^2 = l_1^2\dot\theta_1^2 + l_2^2\dot\theta_2^2 + 2l_1l_2\dot\theta_1\dot\theta_2\cos(\theta_1 - \theta_2)$, so

$$\begin{aligned}L ={}& \tfrac12(m_1{+}m_2)l_1^2\dot\theta_1^2 + \tfrac12 m_2l_2^2\dot\theta_2^2\\ &+ m_2l_1l_2\dot\theta_1\dot\theta_2\cos(\theta_1{-}\theta_2)\\ &+ (m_1{+}m_2)gl_1\cos\theta_1 + m_2gl_2\cos\theta_2 .\end{aligned}$$

3. Linearise at $\theta = 0$. $\cos(\theta_1 - \theta_2) \to 1$ in $T$; $\cos\theta \to 1 - \theta^2/2$ in $V$:

$$M = \begin{pmatrix}(m_1{+}m_2)l_1^2 & m_2l_1l_2\\ m_2l_1l_2 & m_2l_2^2\end{pmatrix},$$ $$K = \begin{pmatrix}(m_1{+}m_2)gl_1 & 0\\ 0 & m_2gl_2\end{pmatrix}.$$

Both angles from the vertical, each measured at its own rod's upper end. The rods' fixed lengths are built into the coordinates, so their tensions never enter.

Step 2 in detail. $\dot{\vec x}_1 = l_1\dot\theta_1(\cos\theta_1, \sin\theta_1)$ and $\dot{\vec x}_2 = \dot{\vec x}_1 + l_2\dot\theta_2(\cos\theta_2, \sin\theta_2)$. The cross term in $\lvert\dot{\vec x}_2\rvert^2$ is $2l_1l_2\dot\theta_1\dot\theta_2(\cos\theta_1\cos\theta_2 + \sin\theta_1\sin\theta_2) = 2l_1l_2\dot\theta_1\dot\theta_2\cos(\theta_1 - \theta_2)$. The heights are $-l_1\cos\theta_1$ and $-l_1\cos\theta_1 - l_2\cos\theta_2$, so $V = -(m_1 + m_2)gl_1\cos\theta_1 - m_2gl_2\cos\theta_2$. No tension in either rod was named, and the rods' constraints were never written as equations: choosing the two angles as coordinates satisfied them.

Step 3. $M(\theta)$ has entries $(m_1 + m_2)l_1^2$, $m_2l_2^2$ and $m_2l_1l_2\cos(\theta_1 - \theta_2)$; at $\theta = 0$ the last is $m_2l_1l_2$. The Hessian of $V$ at $0$ is diagonal, because $V$ is a sum of a function of $\theta_1$ and a function of $\theta_2$, with $\partial^2V/\partial\theta_1^2 = (m_1 + m_2)gl_1$ and $\partial^2V/\partial\theta_2^2 = m_2gl_2$.

12 · small oscillations · example

The double pendulum's two modes

4. equal masses and lengths

$M = ml^2\begin{pmatrix}2&1\\1&1\end{pmatrix}$, $K = mgl\begin{pmatrix}2&0\\0&1\end{pmatrix}$. With $\omega^2 = \lambda g/l$:

$$\det\begin{pmatrix}2-2\lambda & -\lambda\\ -\lambda & 1-\lambda\end{pmatrix} = \lambda^2 - 4\lambda + 2 = 0,$$

so $\lambda = 2 \mp \sqrt2$. From the first row, $v_2/v_1 = (2 - 2\lambda)/\lambda = \pm\sqrt2$:

slow, in phase: $(1, \sqrt2)$, $\ \omega^2 = (2 - \sqrt2)\,g/l$;
fast, in antiphase: $(1, -\sqrt2)$, $\ \omega^2 = (2 + \sqrt2)\,g/l$.

$l = 1$ m: periods 2.621 s and 1.086 s, ratio $1 + \sqrt2$.

Checks: $v_1^{\top}Mv_2 \propto 2 + 1\cdot(-\sqrt2) + \sqrt2 \cdot 1 + \sqrt2 \cdot (-\sqrt2) = 0$ ✓. Single pendulum, 2.006 s, lies between ✓.

Each mode at both ends of its swing, amplitude exaggerated. Slow: the rods swing together, the lower angle $\sqrt2$ times the upper. Fast: the lower rod swings against the upper, again $\sqrt2$ times as far.

A limit, for general masses. With $l_1 = l_2 = l$ and $\beta = m_2/(m_1 + m_2)$, dividing by $(m_1 + m_2)l^2$ and $(m_1 + m_2)gl$, the determinant is $(1 - \lambda)^2\beta - \beta^2\lambda^2$, zero when $1 - \lambda = \pm\sqrt\beta\,\lambda$, that is $\lambda = 1/(1 \pm \sqrt\beta)$. Equal masses: $\beta = \tfrac12$, $\lambda = 1/(1 \pm 0.70711) = 0.58579$ and $3.41421$, which are $2 \mp \sqrt2$ ✓. Light lower bob, $\beta \to 0$: both $\lambda \to 1$, two pendulums of length $l$ that barely interact ✓. Problem P11 uses this formula.

The numbers. $\omega_1 = \sqrt{0.58579 \times 9.81} = 2.3972\ \mathrm{s^{-1}}$, period $2.6211$ s; $\omega_2 = \sqrt{3.41421 \times 9.81} = 5.7874\ \mathrm{s^{-1}}$, period $1.0857$ s; ratio $\sqrt{(2 + \sqrt2)/(2 - \sqrt2)} = 1 + \sqrt2 = 2.4142$. A single pendulum of length 1 m has $2\pi/\sqrt{9.81} = 2.006$ s. The Euclidean dot product of the modes is $1 - 2 = -1$: not orthogonal, as the notes to the modes slide warned.

What linearisation does not see. At large amplitude the double pendulum is the standard example of deterministic chaos: nearby initial conditions separate exponentially (cited: Shinbrot, Grebogi, Wisdom and Yorke, "Chaos in a double pendulum", Am. J. Phys. 60 (1992) 491–499, who measure it). The normal modes describe the motion only while every angle stays small; the configuration space is the torus of the manifold slide, and the modes live on a small patch of it.

Reading the figure. In a mode every coordinate moves as the same cosine, so the whole shape scales up and down in time: the faint drawing is the same shape half a period later. The bobs are drawn with $\theta_1 = \pm0.22$ rad, $\theta_2 = \pm0.31$ rad, beyond where "small" is accurate, so that the shapes are visible. A general small motion is a sum of the two drawings with independent amplitudes and phases, and because the periods are in the irrational ratio $1 + \sqrt2$, it never exactly repeats.

12 · closing

The hoop, closed

the running example, start to finish

1. Constraints $r = R$, $\varphi = \Omega t$ eliminated by the choice of $\theta$; $L$ in three lines; the reaction never named.

2. In the hoop's rotating frame, the same $L$; the Coriolis term drops out of the $\theta$ equation because the hoop absorbs it.

3. $h = \tfrac12 mR^2\dot\theta^2 + V_{\mathrm{eff}}$ conserved, though the energy is not; a pitchfork at $\Omega_c = 5.294\ \mathrm{s^{-1}}$; at $\Omega = 6$, $\theta_* = 38.87^\circ$, period $1.669$ s.

4. Multipliers returned the reaction, $1.512$ N at rest at $\theta_*$, and the motor's Coriolis torque.

5. Linearised, one mode, $\omega^2 = \omega_0^2 - \Omega^2$ below $\Omega_c$.

What this unit hands on

13 — cyclic coordinates and the energy function become one theorem: Noether's.

14 — $p = \partial L/\partial\dot q$ becomes a coordinate, and $h$ the Hamiltonian.

16 — $\tilde{\mathcal E} = J^{\top}\mathcal E$ is a covector's law, proved from a definition.

24, 28 — the charge's Lagrangian, justified and made relativistic.

31, 35 — configuration spaces as manifolds; then general covariance for spacetime itself.

Every number above was computed, not quoted: $\Omega_c = \sqrt{9.81/0.35} = 5.2942\ \mathrm{s^{-1}}$; $\theta_* = \arccos(28.0286/36) = 38.870^\circ$; the tilted period $2\pi/(6\sin\theta_*) = 1.6687$ s; the reaction $mR\Omega^2 = 1.512$ N.

What was assumed and what was proved. Proved: Lagrange's equations from d'Alembert's principle; their equivalence with Hamilton's principle and with Newton's equations in Cartesian coordinates; form invariance under time-dependent point transformations, twice; conservation of cyclic momenta and of $h$; the multiplier theorem; gauge freedom; the Lorentz-force Lagrangian given the Lorentz force; the normal-mode theorem. Assumed as physics: that constraints are ideal, which is a statement about the hardware. Cited: the Lorentz force law and the potentials (unit 24); the experimental verdict on non-holonomic constraints (Lewis and Murray); chaos in the double pendulum (Shinbrot et al.); the converse of gauge freedom (Olver).

Reading. Goldstein, Poole and Safko, Classical Mechanics (3rd ed.), chapter 1 (§§1.3–1.6) and chapter 2 (§§2.1–2.3, 2.6–2.7): the standard account; §1.5's treatment of velocity-dependent potentials takes the Lorentz force as given, which is what this deck did. Its §2.4 on non-holonomic constraints is the kind of treatment the warning on this deck's non-holonomic slide is about, and has been criticised on exactly that point (Flannery, "The enigma of nonholonomic constraints", Am. J. Phys. 73 (2005) 265–272); read it after Lewis and Murray, if at all. José and Saletan, Classical Dynamics: A Contemporary Approach, chapter 2, is closer to this deck's view of configuration space. Landau and Lifshitz, Mechanics, §§1–5 and §23, is the shortest route and the source of the Galilean-boost argument; its §§2–4 assume the principle rather than deriving it.

12 · checkpoint

Done when

checkpoint

You can, on paper and without the deck, take one system — the spherical pendulum, say, or a particle on a cone — and

write its Lagrangian in two genuinely different coordinate systems (angles in one; Cartesian with a multiplier, or a rotating frame, in the other);
derive both sets of Euler–Lagrange equations and show, by an explicit substitution, that they describe the same motion; and
state the theorem that guaranteed it in advance: form invariance of the Euler–Lagrange equations under any $C^2$ point transformation $q = \phi(Q, t)$ with invertible $\partial\phi/\partial Q$, with $\tilde{\mathcal E} = J^{\top}\mathcal E$.

Falsifiable: two sets of equations that are not shown to agree, or agree only because one was derived from the other, fail it. Record it with ./mr done 12.

Problems

./mr problems 12 — 12 problems, about 5 hours. Hints and worked solutions fold under each problem on the same page.

13 Noether · 14 Hamilton · 15 Hamilton–Jacobi · 16 rigid bodies and the first real tensor.

Try it on the spherical pendulum cold, then compare with problem P5, which is exactly this checkpoint with numbers. The step that usually fails is the last bullet's "explicit substitution": it is not enough that both sets of equations look reasonable. Substitute $x = l\sin\theta\cos\varphi$, $y = l\sin\theta\sin\varphi$, $z = -l\cos\theta$ into the Cartesian equations, eliminate the multiplier, and recover the angular equations — or show that the angular equations are the Cartesian ones paired with $\partial\vec x/\partial\theta$ and $\partial\vec x/\partial\varphi$.

If it stalls at the multiplier, go back to the hoop's worked reaction. If it stalls at the theorem, rewrite the component proof from memory: two identities, the chain rule, one cancellation.

Lagrangian mechanics: coordinates stop mattering

One bead, one hoop, three lines

The principle

Generalised coordinates and constraints

On the hoop: two constraints, one coordinate

Virtual displacements and d'Alembert's principle

Frozen time: real work, no virtual work

From d'Alembert to Lagrange's equations

Hamilton's principle

Stationary, not least: the oscillator's conjugate point

Newton, in Cartesian coordinates

Worked: the bead's Lagrangian in three lines

Coordinates stop mattering

The theorem the formalism exists for

The proof in components, and how the equations transform

Newton's form is not invariant

Worked: Coriolis for free

Configuration space is a manifold

Two charts for a circle, a square for a torus

Reading the equations

Generalised momenta and cyclic coordinates

Worked: Kepler, with no force drawn

The trap: eliminating a cyclic velocity in $L$

The energy function

The hoop's effective potential

The equilibria, checked and timed

The pitchfork, in unit 02's normal form

Phase portraits on each side

Phase portrait above the critical spin

Constraint forces, when you want them

Multipliers bring the forces back

Worked: what the hoop pushes with

Worked: the motor's torque

Non-holonomic constraints, honestly

Lagrange–d'Alembert, and what does not work

What $L$ is allowed to be

Gauge freedom

Velocity-dependent potentials

Small oscillations

Linearising a Lagrangian

Modes: $\det(K - \omega^2M) = 0$

Worked: the double pendulum

The double pendulum's two modes

The hoop, closed

What this unit hands on

Done when

Problems

Next