Part A · the equation of motion · spine

Newton's second law is an ODE — and dimensions tell you the answer's shape before you solve it

Three forces can be integrated in closed form, and one page of linear algebra fixes the form of the answer to almost anything else.

needs nothing~2 h slides + 5 h problemsproblems: ./mr problems 00

This is the first unit of the course and it assumes no physics at all. What it does assume is that you are comfortable with proofs, with linear algebra over $\mathbb{R}$, and with single-variable real analysis: limits, the mean value theorem, Taylor's theorem with remainder, uniform convergence, and improper integrals with the comparison test. Everything else — vector calculus, ODE theory, complex analysis, variational calculus, tensors, differential geometry — arrives later, pulled in by a problem that stalls without it.

How to read this deck

One example runs through every section: the plane pendulum. It is done twice, and the second time is not the important one. Dimensional analysis will settle the whole structure of its period in two lines; the calculus is then left to determine a single function of a single dimensionless variable, and nothing else. That division of labour is the habit this unit is trying to install.

Words with a dotted underline are glossary terms: click one for its definition, G for the whole index. A solid underline, like the one on "the plane pendulum" above, names a piece of the running example and shows it on hover. N opens the notes, where the longer proofs are. Proofs that fit on a line are on the slide.

Nothing here is stated without proof or without a label saying where the proof is. There are exactly two imported results in this unit, both flagged in place: the value of the Sedov constant for a blast wave, and the standard properties of the complete elliptic integral, which is named but not developed.

00 · overview

Two ways to answer one question

A bob on a rod, released from rest at an angle $\theta_0$. How long does it take to swing back?

route 1 · change the units, no calculus

1. The period can depend only on $l$, $g$, $m$ and $\theta_0$.

2. Pick units to suit this pendulum: the bob's mass as the unit of mass, the rod as the unit of length, and the unit of time that makes $g = 1$ — it is $\sqrt{l/g}$ seconds. Now $m = l = g = 1$.

3. What is left is $\theta_0$, an angle, which no change of units touches. So in these units the period is some number $f(\theta_0)$; back in seconds,

$T = \sqrt{l/g}\ f(\theta_0)$

The mass is gone; a rod twice as long is $\sqrt2$ times slower. Section 2.

route 2 · solve the differential equation

1. Newton II along the arc gives $\ddot\theta = -(g/l)\sin\theta$.

2. Multiply by $\dot\theta$ and integrate once (energy is conserved): $\dot\theta^2 = \tfrac{2g}{l}(\cos\theta - \cos\theta_0)$.

3. Time is $\int\mathrm d\theta/|\dot\theta|$, and a full swing is four quarter-swings from $\theta_0$ to $0$: $T = 4\sqrt{\tfrac{l}{2g}}\int_0^{\theta_0}\!\frac{\mathrm d\theta}{\sqrt{\cos\theta - \cos\theta_0}}$.

4. Substituting $\sin\tfrac\theta2 = \sin\tfrac{\theta_0}2\sin\varphi$ turns it into a tabulated integral $K$:

$T = 4\sqrt{l/g}\,K(\sin\tfrac{\theta_0}{2})$

Sections 3–5. All of that bought one function: $f = 4K(\sin(\cdot/2))$.

The plan. 1 what the three laws assert · 2 dimensions, and the theorem behind them · 3 the three forces that integrate · 4 reading the motion off the potential · 5 the pendulum, exactly, and the amplitude at which the schoolbook formula is 1 per cent wrong.

Why step 2 of route 1 is allowed. A change of units multiplies every mass by one positive number, every length by another and every time by a third, and those three numbers can be chosen freely. Taking the kilogram to be $m$ sets $m = 1$; taking the metre to be $l$ sets $l = 1$, and $g$ becomes $9.81/l$ per second squared; a time unit of $\tau$ seconds then multiplies $g$ by $\tau^2$, so $\tau = \sqrt{l/g}$ makes $g = 1$. Three free numbers, three quantities set to $1$ — this works because $m$, $l$, $g$ have independent dimensions, and section 2 states that condition precisely (a matrix of rank 3). The step that cannot be proved is the assumption that the law of the pendulum does not care which units you use; section 2 names it and says where it fails.

Route 2, in brief. Each step is a slide later on: step 1 is The pendulum, from the laws (the rod's tension is eliminated by taking the component along the arc), step 2 is case 3 of section 3, step 3 is The period integral in section 4, step 4 is The half-angle substitution and the elliptic integral in section 5, which also defines $K$. Hover over the boxed formula to see $K$ written out now.

The two routes are not competitors. Route 1 costs nothing and is almost never wrong; route 2 is expensive and is available for perhaps a dozen force laws in the whole of mechanics. The professional habit is to run route 1 first, always, and to use its answer as a check on route 2. In this course you will be asked for the dimensional form before you are allowed to integrate anything.

A warning about the word "period". A pendulum released from rest at $\theta_0$ swings to $-\theta_0$ and back; $T$ is the time for the whole round trip, four times the time from $\theta_0$ to $0$. Keep the factor of four in view: it is the commonest slip in the problems.

Numbers in this deck are all computed, not quoted. Where a pendulum is given numbers they are $l = 1.000$ m and $g = 9.81\ \mathrm{m\,s^{-2}}$, for which $2\pi\sqrt{l/g} = 2.0061$ s.

section 1

What the three laws assert

One of them is a definition, one is a scheme rather than an equation, and one is false as usually stated.

00 · the laws

The first law is a definition, not a special case

Newton's first law

There exist frames of reference — inertial frames — in which a body subject to no force moves with constant velocity.

Read as "$F = 0 \Rightarrow a = 0$" it is the case $F = 0$ of the second law and says nothing. Read as above it is an existence claim about frames, and it is what makes the second law falsifiable: without it, any observed acceleration could be explained by declaring a force.

the circularity, honestly

"No force" and "inertial frame" are defined in terms of each other. The escape is empirical, not logical: a frame far from all matter works, and the same frames then work for every experiment. General relativity (unit 30) removes the circle by making free fall, not uniform motion, the primitive.

Given one inertial frame, every frame moving uniformly relative to it is inertial too. If $\vec r' = \vec r - \vec u\,t$ with $\vec u$ constant, then differentiating twice componentwise,

$$\ddot{\vec r}{\,}' = \ddot{\vec r},$$

so accelerations agree. That family of changes of frame is the Galilean group, and unit 25 shows that Maxwell's equations do not survive it — which is where relativity starts.

check

A rotating frame is not inertial: a free body appears to accelerate. Unit 03 computes what appears in its place.

Newton's own statement (Principia, 1687, Law I) is "Every body perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed thereon." Newton had absolute space and did not need the existence claim; the modern reading is due to Lange (1885) and Thomson, who defined an inertial frame as one in which three mutually non-collinear free particles move in straight lines.

Why this matters for the rest of the course. Everything in Parts A to C is written in an inertial frame, and every time the frame is changed you must ask whether the new one is inertial. Unit 03 does rotating frames and derives the terms that appear (centrifugal, Coriolis) rather than asserting them; unit 30 shows that in a gravitational field no global inertial frame exists, which is the equivalence principle and the reason spacetime has to curve.

On "differentiating componentwise". A motion is a map $\vec r : I \to \mathbb{R}^3$ from a time interval. Its derivative is defined by the limit of difference quotients in $\mathbb{R}^3$, which — because the norm is equivalent to the max of the coordinates — exists exactly when each of the three coordinate functions is differentiable, and then equals the triple of their derivatives. That is all the multivariable calculus this unit uses, and it is a theorem of single-variable analysis, not of unit 03's.

00 · the laws

The second law is a scheme

Newton's second law

In an inertial frame, a body of mass $m$ under a total force $\vec F$ obeys $\;\vec F = m\,\ddot{\vec r}$. Equivalently $\vec F = \dot{\vec p}$ with momentum $\vec p = m\dot{\vec r}$.

As written this is not one equation but a template. It becomes an equation only once someone supplies $\vec F$ as a function of the things the body can depend on — position, velocity, time:

$$m\,\ddot{\vec r} = \vec F(\vec r, \dot{\vec r}, t).$$

That is a second-order ordinary differential equation, and the whole of Part A is one long answer to "which one, and how do you solve it?".

dimension check

$[m\ddot r] = \mathsf{M}\,\mathsf{L}\,\mathsf{T}^{-2}$, so force has dimension $\mathsf{MLT^{-2}}$ — the newton is $\mathrm{kg\,m\,s^{-2}}$ by definition, not by measurement. Force is not an independent base dimension; that choice is made here and never revisited.

what is not claimed

That a solution exists, or is unique. Both are true under a Lipschitz hypothesis on $\vec F$ and are proved in unit 01. Nothing in this unit may lean on uniqueness; where it looks as if it does, the notes say so.

Is $F = ma$ a definition of force or a law of nature? Neither on its own. Taken with a stock of independently specified force laws — gravity, springs, drag, the Lorentz force — the package is falsifiable and has been falsified (at high speed, unit 27; at small scale, not in this course). Taken alone, it defines force in terms of acceleration and is empty. Mach's analysis is the standard treatment; Eddington's joke is that the second law says "force equals mass times acceleration, where force is what makes mass accelerate".

Mass, briefly. Two notions are in play. Inertial mass is the $m$ here, the body's resistance to acceleration. Gravitational mass is the coupling constant in the law of gravitation. That they are equal is an experimental fact, tested to about one part in $10^{13}$ by torsion balances and by lunar laser ranging, and to about one part in $10^{15}$ by the MICROSCOPE satellite (final result, 2022); it is the empirical content of the equivalence principle. Unit 30 builds general relativity on it. In this unit the coincidence shows up as the mass dropping out of every free-fall answer.

The momentum form is the better one. $\vec F = \dot{\vec p}$ survives to variable-mass problems and to relativity, where $\vec p = \gamma m \vec v$ and $\vec F = m\vec a$ is simply false (unit 27). Prefer it.

00 · the laws

Force laws are input, not theorem

A force law is an extra, independent physical hypothesis. Mechanics does not derive it; mechanics takes it and integrates it. The ones this unit uses, all in one dimension:

law	form	type
uniform gravity near the ground	$-mg$	constant
Hooke's law	$-kx$	$F(x)$
linear (Stokes) drag	$-b v$	$F(v)$
quadratic (Newton) drag	$-c\,v\|v\|$	$F(v)$
a driving force	$F_0\cos\omega t$	$F(t)$

Each is an approximation with a stated range of validity, not an identity.

which drag, and when

Linear drag is the low-Reynolds-number regime — small, slow, viscous. Quadratic drag is the high-Reynolds regime — large, fast, inertial. A human in air is firmly quadratic; a $1\ \mathrm{\mu m}$ droplet in oil is firmly linear. Using the wrong one is not a small error, it is a different power law.

why $v|v|$ and not $v^2$

Drag opposes motion. $-cv^2$ would push a body moving in the $-x$ direction further in that direction. Writing $-c\,v|v|$ keeps the sign right for both directions and agrees with $-cv^2$ for $v > 0$.

Where the two drag laws come from. Both are dimensional statements about the force on a body of size $a$ moving at speed $v$ through a fluid of density $\rho$ and viscosity $\mu$. With $\mu$ dominant the only combination with the dimensions of force is $\mu a v$, giving Stokes' $6\pi\mu a v$ for a sphere; with $\rho$ dominant it is $\rho a^2 v^2$, giving $\tfrac12 C_D \rho A v^2$. Which dominates is decided by the Reynolds number $\mathrm{Re} = \rho a v/\mu$, itself a dimensionless group in the sense of section 2. That is dimensional analysis doing the whole job, and it is worth returning to this note after the Buckingham slide.

Hooke's law is the first two terms of a Taylor expansion of any smooth force about a stable equilibrium, which is why it is ubiquitous rather than fundamental. Unit 06 makes that precise and unit 07 does the many-variable version.

Gravity as $-mg$ is the leading term of the inverse-square law near the surface. With $R_\oplus = 6371$ km, the fractional change in $g$ over a 100 m fall is $2\times 100/R_\oplus = 3.1 \times 10^{-5}$, so treating $g$ as constant is good to four significant figures over a building. Unit 08 does the exact problem.

00 · the laws

The third law, and what it does not say

Newton's third law

The force exerted by body 1 on body 2 is equal and opposite to that exerted by 2 on 1: $\vec F_{12} = -\vec F_{21}$.

consequence: conservation of momentum

For an isolated pair, $\dot{\vec p}_1 + \dot{\vec p}_2 = \vec F_{12} + \vec F_{21} = \vec 0$, so $\vec p_1 + \vec p_2$ is constant.

Proof: that is the whole proof — add the two second laws and use the third. It extends to $N$ bodies because the internal forces cancel in pairs.

it is false for fields

Two charges in relative motion exert forces on each other that are not equal and opposite: the field carries momentum and takes time to get there. Unit 24 exhibits the failure and unit 29 repairs it, by deriving momentum conservation from a symmetry instead of from the third law.

The lesson to carry forward: momentum conservation is the durable statement and the third law is a special case of it, not the other way round.

dimension check

$[\,p\,] = \mathsf{MLT^{-1}}$ and $[\,F\,] = \mathsf{MLT^{-2}}$, so $\vec F = \dot{\vec p}$ balances.

Unit 13 proves Noether's theorem: momentum conservation follows from the invariance of the laws under translations in space. That is the version that survives into field theory and relativity.

Strong and weak forms. The weak third law is the statement above. The strong third law adds that the mutual forces lie along the line joining the bodies; that is what is needed for angular momentum conservation, and it holds for gravity and electrostatics but not for the magnetic force between current elements. Unit 08 needs the strong form and says so.

The concrete failure. Charge 1 at the origin moving along $x$, charge 2 on the $y$ axis moving along $y$. The magnetic force on each from the other is computed in unit 24; they are perpendicular, not antiparallel, so their sum is not zero. Total momentum is still conserved once the momentum stored in the electromagnetic field is counted, and that quantity falls out of Noether's theorem in unit 29 rather than being postulated.

For this unit the third law is used exactly once — to say that the pendulum's rod and bob exert equal and opposite forces, which is why the tension can be treated as a single unknown scalar. It is eliminated on the next slide and never appears again.

00 · the laws · example

The pendulum, from the laws

worked example 1 · eliminating a constraint force

Put the pivot at the origin, $y$ upwards. The rod is rigid, so the bob's position is $\vec r(t) = \big(l\sin\theta(t),\, -l\cos\theta(t)\big)$ for a single unknown function $\theta$. Differentiate componentwise, twice, with the product and chain rules:

$\dot x = l\cos\theta\ \dot\theta$, $\quad\ddot x = l\cos\theta\,\ddot\theta - l\sin\theta\,\dot\theta^{\,2}$,

$\dot y = l\sin\theta\ \dot\theta$, $\quad\ddot y = l\sin\theta\,\ddot\theta + l\cos\theta\,\dot\theta^{\,2}$.

Two forces act: gravity $(0, -mg)$, and the rod's pull along its own length towards the pivot, $\;-T(\sin\theta, -\cos\theta)$, with $T(t)$ unknown.

The rod's pull is along the rod; the direction the bob can actually move, $(\cos\theta, \sin\theta)$, is across it. The next slide uses exactly that right angle.

The bob at angle $\theta$. Gravity and the unknown tension $T$ act on it; the tangent direction is at right angles to the rod, so $T$ has no component along it, while gravity's component along it is $-mg\sin\theta$.

Why these coordinates. The rod fixes the distance from the pivot, so of the two Cartesian coordinates only one combination is free; $\theta$ is that combination. Writing $\vec r$ in terms of $\theta$ builds the constraint in, and differentiating it twice gives the acceleration of a body that is forced to stay on the circle — the $\dot\theta^2$ terms are the centripetal part, pointing to the pivot, and the $\ddot\theta$ terms the tangential part.

Reading the figure. The unit vector $(\cos\theta,\sin\theta)$ is $\mathrm d\vec r/\mathrm d\theta$ divided by $l$: the direction the bob moves when $\theta$ increases. It is perpendicular to $(\sin\theta,-\cos\theta)$, the direction from pivot to bob, because their dot product is $\cos\theta\sin\theta - \sin\theta\cos\theta = 0$. Gravity $(0,-mg)$ dotted with it gives $-mg\sin\theta$, the arrow drawn in the second colour: it always points back towards the bottom.

What has been assumed. That the rod is rigid and massless, that the bob is a point, that the pivot is frictionless, and that the motion stays in one vertical plane. Each is a modelling choice. Unit 16 does the pendulum with a distributed mass (the compound pendulum) and finds the same equation with $l$ replaced by an effective length $I/(m d)$.

00 · the laws · example

The pendulum: the tension drops out

worked example 1, continued

Newton's second law, componentwise, with the forces of the previous slide:

$m\big(l\cos\theta\,\ddot\theta - l\sin\theta\,\dot\theta^{\,2}\big) = -T\sin\theta$,

$m\big(l\sin\theta\,\ddot\theta + l\cos\theta\,\dot\theta^{\,2}\big) = -mg + T\cos\theta$.

Take $\cos\theta$ times the first plus $\sin\theta$ times the second. On the right, $-T\sin\theta\cos\theta + T\sin\theta\cos\theta = 0$: the tension cancels. On the left $\dot\theta^{\,2}$ cancels too and $\cos^2+\sin^2 = 1$, leaving $ml\,\ddot\theta = -mg\sin\theta$:

$$\boxed{\ \ddot\theta = -\tfrac{g}{l}\sin\theta\ }$$

Mass gone, tension gone, one second-order ODE in one unknown: the running example. The notes recover $T$.

checks

Dimensions: $[g/l] = \mathsf{T^{-2}}$ and $\theta$ is dimensionless, so both sides are $\mathsf{T^{-2}}$. Limit: for small $|\theta|$, $\sin\theta \approx \theta$ gives $\ddot\theta = -(g/l)\theta$, solved by $\theta_0\cos(\sqrt{g/l}\,t)$ with period $2\pi\sqrt{l/g}$ — the schoolbook answer, obtained rather than asserted.

Why the cancellation was inevitable. The combination taken is $\cos\theta$ times the $x$ equation plus $\sin\theta$ times the $y$ equation, which is the component of $\vec F = m\vec a$ along the unit vector $(\cos\theta, \sin\theta)$ — the direction tangent to the circle. The rod's force is perpendicular to that direction by construction, so it has no tangential component and must drop out. A constraint force does no work and enters no tangential equation: that observation, promoted to a principle, is d'Alembert's, and unit 12 turns it into Lagrangian mechanics, where this entire slide becomes one line.

Check the second combination. $\sin\theta\times$(first) gives $m l \sin\theta\cos\theta\,\ddot\theta - ml\sin^2\theta\,\dot\theta^2 = -T\sin^2\theta$; $\cos\theta\times$(second) gives $ml\sin\theta\cos\theta\,\ddot\theta + ml\cos^2\theta\,\dot\theta^2 = -mg\cos\theta + T\cos^2\theta$. Subtract the second from the first: $-ml\dot\theta^2 = mg\cos\theta - T$, so $T = ml\dot\theta^2 + mg\cos\theta$. At the bottom of a swing released from $\theta_0$, energy (section 3) gives $l\dot\theta^2 = 2g(1 - \cos\theta_0)$, so $T = mg(3 - 2\cos\theta_0)$: at $\theta_0 = \pi/2$ the rod carries three times the bob's weight at the bottom. A rod released from horizontal and a string released from horizontal behave identically; released from above horizontal they do not, because a string cannot push.

section 2

Dimensions, before any calculus

A change of units is a group action; dimensional analysis is the study of its orbits. The theorem is a rank argument.

00 · dimensions

Units as a group action

Fix $d$ base dimensions; for mechanics $d = 3$ and they are mass $\mathsf M$, length $\mathsf L$, time $\mathsf T$. Every quantity's dimension is a product of their powers — $[g] = \mathsf L\,\mathsf T^{-2}$ — and we record the exponents as a column vector, $(0, 1, -2)$ for $g$. Write $D_{iq}$ for the exponent of base dimension $i$ in $q$; the columns $D_{\cdot q} \in \mathbb{Q}^d$ of all the quantities side by side are the matrix $D$ of the next slide.

rescaling

Change the base units so that a measurement in base dimension $i$ is multiplied by $\lambda_i$ — shrink the metre by $100$ and $\lambda_{\mathsf L} = 100$. For $\lambda = (\lambda_1,\dots,\lambda_d) \in (\mathbb{R}_{>0})^d$ this replaces the number $q$ by

$$q \;\longmapsto\; \Big(\textstyle\prod_i \lambda_i^{\,D_{iq}}\Big)\, q .$$

This is an action of the group $(\mathbb{R}_{>0})^d$ on measured values, and $\lambda \mapsto \log\lambda$ turns it into the additive action of $\mathbb{R}^d$.

the hypothesis, stated as a hypothesis

dimensional homogeneity: a physical law relating $q_1,\dots,q_n$ holds in one consistent system of units if and only if it holds in all of them. Formally, if $f(q_1,\dots,q_n) = 0$ then $f(q_1',\dots,q_n') = 0$ for every $\lambda$, where $q_j'$ is $q_j$ rescaled as on the left. Shorthand: $f(\lambda^D q) = 0$.

This is not a theorem. It is an assumption about what counts as a physical law: that nature does not know which metre stick you own. Everything in this section is a consequence of it.

where it can fail

If a law secretly involves a dimensional constant you left out of the list — $c$, $\hbar$, $G$, a material property — the assumption is violated for your list, not for nature's. Every failed dimensional argument is a missing variable.

Why $\mathbb{Q}^d$ and not $\mathbb{R}^d$. Nothing forbids irrational exponents formally, but every quantity that occurs in practice has rational exponents, and rationality is what lets the kernel of the dimension matrix be spanned by integer vectors after clearing denominators — which is what makes the dimensionless groups monomials with a chance of being recognisable.

Which dimensions are "base" is a convention. SI takes seven (mass, length, time, current, temperature, amount, luminous intensity). Mechanics needs three. Gaussian units make charge derived; natural units set $c = \hbar = 1$ and collapse length and time into one dimension. The theorem below does not care: it only cares about the rank of the resulting matrix. What does change is the count of dimensionless groups, and here is the trap: choosing fewer base dimensions gives a smaller rank and therefore more groups, which is a weaker conclusion. Take as many independent base dimensions as you honestly can.

Angles. An angle in radians is a ratio of two lengths and is dimensionless, which is why $\theta_0$ appears as a group in its own right on the next slides. That is also why $\sin\theta$ makes sense and $\sin(3\ \mathrm{metres})$ does not: the argument of a transcendental function must be dimensionless, because the function is defined by a power series and the terms would otherwise have different dimensions. Use this as a free error-check on every formula you write.

00 · dimensions

The dimension matrix and its kernel

List the relevant quantities $q_1,\dots,q_n$ and build the dimension matrix $D \in \mathbb{Q}^{d\times n}$, one column per quantity, one row per base dimension.

lemma · the kernel is the answer

The monomial $\prod_j q_j^{a_j}$ is dimensionless if and only if $Da = 0$.

Proof. Its dimension vector is $\sum_j a_j D_{\cdot j} = Da$, and a quantity is dimensionless exactly when its dimension vector is zero. $\square$

So the dimensionless monomials are $\ker D$, a subspace of dimension $n - r$ with $r = \operatorname{rank} D$. Picking a basis of $\ker D$ picks a set of dimensionless groups $\pi_1,\dots,\pi_{n-r}$, and every dimensionless monomial is a product of powers of them.

the pendulum's matrix

Quantities $T, l, g, m, \theta_0$:

$$D = \begin{array}{c|ccccc} & T & l & g & m & \theta_0\\\hline \mathsf M & 0&0&0&1&0\\ \mathsf L & 0&1&1&0&0\\ \mathsf T & 1&0&-2&0&0\end{array}$$

Columns $m, l, T$ are the standard basis vectors of $\mathbb{Q}^3$, so $r = 3$ and $\dim\ker D = 5 - 3 = 2$.

Solving $Da = 0$: the $\mathsf M$ row forces $a_m = 0$, the $\mathsf L$ row gives $a_l = -a_g$, the $\mathsf T$ row gives $a_T = 2a_g$. Free parameters $a_g$ and $a_{\theta_0}$:

$$\pi_1 = \frac{T^2 g}{l}\ \ (a_g = 1), \qquad \pi_2 = \theta_0 .$$

The basis is not unique, and that is fine. $\pi_1' = \pi_1\pi_2^2$ is an equally good first group. What is canonical is the subspace $\ker D$, not the basis; every choice gives the same physics, and a good choice is one whose groups have names (Reynolds number, Mach number, fine structure constant). When you compare two textbooks' groups, check whether one set is a multiplicative basis change of the other before concluding that someone is wrong.

Why "list the relevant quantities" is the hard step. The mathematics below is mechanical; the physics is entirely in the list. Put in one quantity too many and you get a spurious extra group and a weaker conclusion. Leave one out and you get a false statement. There is no algorithm for the list — it is where judgement lives, and it is what the problems drill.

Checking rank by hand. Row-reduce $D$, or spot that some subset of columns is already triangular. Here $m$ is the only quantity carrying $\mathsf M$, so its column is independent of the rest; after deleting the $\mathsf M$ row and the $m$ column, $l$ is the only remaining quantity carrying $\mathsf L$ alone, and so on. A quantity with a dimension nothing else has always contributes to the rank, and it always drops out of the answer — which is exactly why the pendulum's period cannot depend on $m$.

00 · dimensions

The Buckingham pi theorem

Buckingham pi theorem

Let $f(q_1,\dots,q_n) = 0$ be a dimensionally homogeneous relation, $D$ its dimension matrix, $r = \operatorname{rank} D$. Then there are $n - r$ independent dimensionless groups $\pi_1,\dots,\pi_{n-r}$, and a function $F$ of $n-r$ variables with

$$f(q_1,\dots,q_n) = 0 \iff F(\pi_1,\dots,\pi_{n-r}) = 0 .$$

Proof, in one sentence. Reorder so that columns $1..r$ of $D$ are independent; because $D_{[1..r]}^{\!\top}$ has rank $r$ it is surjective onto $\mathbb{R}^r$, so units can be chosen making $q_1 = \dots = q_r = 1$; homogeneity says the relation is unchanged, and in those units every remaining $q_j$ is its dimensionless group. $\square$

full proof in the notes (N)

two checks

Count. $r \le \min(d, n)$, so there is at least one group whenever $n > d$; and $n - r = 0$ means the relation is vacuous — you have not put in enough quantities.

Special case. $n - r = 1$ is the powerful case: a single group, so $\pi_1 = \text{const}$, and the law is determined up to one pure number. The blast wave below is of this kind.

Full proof

Write the action additively: with $\mu_i = \log\lambda_i$ and $Q_j = \log q_j$ (assume all $q_j > 0$), rescaling sends $Q \mapsto Q + D^{\!\top}\mu$.

Step 1. Reorder the quantities so that the first $r$ columns of $D$ are linearly independent; call that $d\times r$ block $B$. Since $\operatorname{rank}B = r$, the map $\mathbb{R}^d \to \mathbb{R}^r$, $\mu \mapsto B^{\!\top}\mu$, is surjective (its matrix has rank $r$ and $r$ rows). Hence there is $\mu^\star$ with $B^{\!\top}\mu^\star = -(Q_1,\dots,Q_r)$, that is, a choice of units in which $q_1 = \dots = q_r = 1$.

Step 2. For $j > r$, the column $D_{\cdot j}$ lies in the span of $B$'s columns — the first $r$ columns span the column space, since the rank is $r$. So there is a unique $c^{(j)} \in \mathbb{R}^r$ with $D_{\cdot j} = B\,c^{(j)}$. Put

$$\pi_j \;=\; q_j \prod_{i \le r} q_i^{-c^{(j)}_i}, \qquad j = r+1,\dots,n .$$

Its exponent vector is $e_j - \sum_i c^{(j)}_i e_i$, and $D$ applied to it is $D_{\cdot j} - Bc^{(j)} = 0$: each $\pi_j$ is dimensionless. These $n - r$ exponent vectors are linearly independent, because the $j$-th has a $1$ in coordinate $j$ and zeros in the other coordinates $> r$; and $\dim\ker D = n - r$ by rank–nullity. So they are a basis of $\ker D$, and every dimensionless monomial is a product of powers of the $\pi_j$.

Step 3. In the units of step 1 the values are $(1,\dots,1,\pi_{r+1},\dots,\pi_n)$ — the $j$-th becomes $q_j\prod q_i^{-c_i^{(j)}}$, which is exactly $\pi_j$. Define $F(\pi_{r+1},\dots,\pi_n) := f(1,\dots,1,\pi_{r+1},\dots,\pi_n)$. Homogeneity gives $f(q) = 0 \iff f(\lambda^{\star D}q) = 0 \iff F(\pi) = 0$. $\square$

History. The counting argument is Vaschy's (1892) and Buckingham's (1914); Riabouchinsky and Bridgman sharpened the statement. Bridgman's Dimensional Analysis (1922) is still the clearest book-length treatment, and Barenblatt's Scaling (2003) is the modern one, including the cases where the naive argument fails because a group tends to zero or infinity and the limit is not uniform — "incomplete similarity", which is where anomalous exponents come from.

The positivity assumption. Taking logs needs $q_j > 0$. For quantities that can change sign, apply the theorem to $|q_j|$ and carry the signs separately; nothing essential changes, since a sign is itself dimensionless.

00 · dimensions · example

The pendulum, dimensionally

worked example 2 · two lines, no calculus

1. The list. What can the period depend on? The rod's length $l$, gravity $g$, the bob's mass $m$, the release angle $\theta_0$. Not the rod's material (it is rigid and massless by assumption), not the time of day.

2. The matrix. $n = 5$, $d = 3$, $r = 3$, so $n - r = 2$ groups, computed two slides ago: $\pi_1 = T^2g/l$ and $\pi_2 = \theta_0$.

3. Buckingham. The relation is $F(T^2g/l,\ \theta_0) = 0$. Solving for the first argument — legitimate whenever $T$ is determined by the rest, which it is —

$$\frac{T^2g}{l} = f(\theta_0)^2 \quad\Longrightarrow\quad \boxed{\ T = \sqrt{l/g}\ f(\theta_0)\ }$$

checks

Dimensions: $[\sqrt{l/g}] = (\mathsf L/\mathsf{LT^{-2}})^{1/2} = \mathsf T$. ✓ Limit: the small-angle solution on "The pendulum, from the laws" gives $T \to 2\pi\sqrt{l/g}$, so $f(0) = 2\pi$ — a value the dimensional argument could never have produced, and a value the calculus must reproduce. Number: $l = 1.000$ m, $g = 9.81\ \mathrm{m\,s^{-2}}$ gives $\sqrt{l/g} = 0.31928$ s and $2\pi\sqrt{l/g} = 2.0061$ s.

00 · dimensions · example

What the dimensional form already says

Everything except the single function $f$ is now settled, and three separate physical facts have been proved without touching the equation of motion:

No mass dependence. $m$ is the only quantity carrying $\mathsf M$, so its exponent must vanish. A lead bob and a cork bob keep the same time.
Length scaling. $T \propto \sqrt{l}$ at fixed amplitude: quadruple the rod, double the period.
Amplitude enters only through $\theta_0$ — a dimensionless angle, so no combination of $l$ and $g$ can trade against it.

What is not settled: the function $f$. Sections 3–5 compute it, and that is all they compute.

Both released at the same $\theta_0$. Quadruple the rod and the period doubles; make the bob eight times heavier and nothing changes. Neither fact needed the equation of motion.

A pendulum on the Moon. $g_{\mathrm{Moon}} = 1.62\ \mathrm{m\,s^{-2}}$, so the same pendulum runs slower by $\sqrt{9.81/1.62} = 2.461$: a seconds pendulum on Earth becomes a 4.92-second pendulum there. No calculus was used, and the answer is exact within the model.

Historically this is how $g$ was measured. Kater's reversible pendulum (1817) turns $T = 2\pi\sqrt{l/g}$ into a determination of $g$ to five significant figures, because periods and lengths are both easy to measure precisely. The 1-per-cent amplitude correction of section 5 is exactly the correction Kater had to make — and the reason the definition of the metre as the length of a seconds pendulum, proposed in 1668 and again in 1790, was abandoned: $g$ varies with latitude, so the metre would have too.

00 · dimensions

Mechanical similarity: scaling the equation, not the units

mechanical similarity

Suppose $m\ddot x = F(x)$ with $F$ homogeneous of degree $k$, that is $F(\alpha x) = \alpha^k F(x)$ for all $\alpha > 0$. If $x(t)$ is a solution, so is

$$\tilde x(t) := \alpha\, x(t/\beta) \quad\text{provided}\quad \beta^2 = \alpha^{\,1-k}.$$

Proof. $m\ddot{\tilde x}(t) = (m\alpha/\beta^2)\,\ddot x(t/\beta) = (\alpha/\beta^2)F(x(t/\beta))$, while $F(\tilde x(t)) = \alpha^k F(x(t/\beta))$. These agree for all $t$ exactly when $\alpha/\beta^2 = \alpha^k$. $\square$

So geometrically similar motions of size ratio $\alpha$ have time ratio $\alpha^{(1-k)/2}$: $\;T \propto a^{(1-k)/2}$ for amplitude $a$.

$F$	$k$	$T \propto$	name
$-\kappa x$ (spring)	1	$a^0$	isochronous
const (free fall)	0	$a^{1/2}$	Galileo
$-x/\|x\|^3$ (gravity)	$-2$	$a^{3/2}$	Kepler III

check, with a number

Kepler: Mars' semi-major axis is $1.5237$ AU, so its year should be $1.5237^{3/2} = 1.8808$ of ours. The measured sidereal period is $1.8808$ yr. Two lines of algebra, no orbit solved — unit 08 does that.

The inverse-square case, componentwise. With $m\ddot x_i = -GMm\,x_i/|x|^3$ for $i = 1,2,3$, put $\tilde x(t) = \alpha x(t/\beta)$. Then $\ddot{\tilde x}_i = (\alpha/\beta^2)\ddot x_i(t/\beta)$ and the right-hand side becomes $-GM\alpha x_i/(\alpha^3|x|^3)$, so the requirement is $\alpha/\beta^2 = \alpha^{-2}$, that is $\beta^2 = \alpha^3$ — Kepler's third law, $T^2 \propto a^3$, with no vector calculus and no orbit. Note it also proves that every orbit of a given semi-major axis has the same period, whatever its eccentricity, which is the non-obvious half of Kepler III.

How this relates to Buckingham. Buckingham scales the units and leaves the physics alone; similarity scales the solution and leaves the units alone. They are two readings of the same invariance, and either can be derived from the other for a given problem. Similarity is sharper when you have the equation, because it uses the actual form of $F$; Buckingham is available when you do not.

Landau and Lifshitz open Mechanics (§10) with this argument and get Kepler III, the isochrony of the harmonic oscillator, and the virial theorem out of it in two pages. It is also the seed of the renormalisation group: a scaling transformation that maps solutions to solutions, with the exponent as the invariant.

The pendulum is not covered, because $\sin\theta$ is not homogeneous of any degree. That is precisely why the period depends on amplitude, and why section 5 has work to do.

00 · dimensions · example

Taylor's blast wave: a classified number from a photograph

worked example 3 · one group, one photograph

The list. Energy $E$ released at a point in still air of density $\rho$ at $t = 0$ drives a shock of radius $R(t)$. Early on the ambient pressure and viscosity are negligible against the shock's own: the list is $R, E, \rho, t$, and nothing else.

The matrix. $R: (0,1,0)$, $E: (1,2,-2)$, $\rho: (1,-3,0)$, $t: (0,0,1)$ in $(\mathsf M,\mathsf L,\mathsf T)$; the last three columns have determinant $-5$, so $r = 3$ and $n - r = 1$: a single group. $Da = 0$ gives $a_E + a_\rho = 0$, $a_t = 2a_E$, $a_R = -5a_E$, so with $a_E = 1$,

$$\pi = \frac{E t^2}{\rho R^5} = \text{const} \ \Longrightarrow\ \boxed{\ R(t) = C\left(\frac{E t^2}{\rho}\right)^{1/5}}$$

The story. G. I. Taylor did this in 1947 from photographs published in Life magazine, which carried a scale bar and frame times because nobody thought a radius and a clock could be secret. The yield was classified. He published the method in 1950 (Proc. Roy. Soc. A 201, 159 and 175) and, by several accounts, irritated the authorities considerably.

The two-fifths law in five words: the shock decelerates because it sweeps up mass. $R \propto t^{2/5}$ means $\dot R \propto t^{-3/5}$, so the front slows without limit — and the solution must fail once $\dot R$ approaches the ambient sound speed, because then the ambient pressure, which was dropped from the list, stops being negligible. That is the standard signal that a dimensional argument has a domain of validity: it holds while the neglected group is small.

00 · dimensions · example

Taylor's blast wave: the test and the yield

The test, on Taylor's table of frame times and radii from the declassified Trinity films. A log–log fit over the five late frames (15–62 ms, radii 106.5–185 m) gives the exponent 0.391 against the predicted $2/5$. The law is right.

The yield. Invert with $\rho = 1.25\ \mathrm{kg\,m^{-3}}$ and $C = 1.033$ (cited: the Sedov self-similar solution for $\gamma = 1.4$; dimensions cannot supply it). At $t = 25$ ms, $R = 130$ m,

$E = \rho R^5/(C^5t^2) = 6.31\times10^{13}\ \mathrm{J} = \mathbf{15.1}$ kt, and the five late frames give 14.1–15.5 kt.

Five frames, 15–62 ms. On log–log axes a power law is a straight line whose slope is the exponent: the points sit on slope $2/5$ (fitted: $0.391$).

checks, and the honest gap

Dimensions: $[Et^2/\rho] = (\mathsf{ML^2T^{-2}})(\mathsf{T^2})/(\mathsf{ML^{-3}}) = \mathsf L^5$. ✓ Limit: $R \to 0$ as $t \to 0$. ✓ Against reality: the declassified yield is 21 kt, so this is 28 per cent low — but $E \propto R^5$, so reading the fireball edge at 139 m rather than 130 m, a 7 per cent difference on a 1945 photograph, gives exactly 21 kt.

Why the estimate is low, in order of importance. (1) The radius is read from a photograph of a luminous, irregular fireball whose edge is not the shock front; a few per cent there is 20-odd per cent in $E$. (2) Trinity was fired at 1470 m altitude, where $\rho \approx 1.0$, not $1.25\ \mathrm{kg\,m^{-3}}$ — that lowers the estimate further, and the two errors do not cancel. (3) Some of the energy goes into radiation and into the ground rather than into the blast wave, so the $E$ this measures is the blast energy, not the total yield. What dimensional analysis delivered is the exponent, and it delivered it exactly; the prefactor was always going to be a measurement problem.

Numbers. Frame by frame, with $\rho = 1.25$ and $C = 1.033$: 15.5 kt at 15 ms, 15.1 at 25 ms, 14.1 at 34 ms, 14.8 at 53 ms, 14.3 at 62 ms. With $C = 1$, which is what you would assume knowing only that $C$ is a pure number of order one, the same frames give 16.6–18.2 kt.

Why log–log. $R = A t^{p}$ becomes $\log R = \log A + p\log t$, a straight line of slope $p$. The dimensional argument predicts the slope and says nothing about the intercept; the intercept carries $E$, and that is the part that needs $C$ and the density.

00 · dimensions

What dimensional analysis cannot do

1 · it never gives a pure number

$f(\theta_0)$, the $C$ in the blast wave, the $2\pi$ in the small-angle period: all invisible to it. When there is one group, you get the law up to one constant. When there are two, you get an unknown function of one variable — much less.

2 · the answer is only as good as the list

Omit a relevant quantity and the conclusion is false. Include an irrelevant one and the conclusion is weaker than it needed to be, with a spurious group that the true law does not depend on.

3 · a small group is not always a negligible one

"$\pi \ll 1$, so drop it" assumes the limit $\pi \to 0$ exists and is non-zero. When it does not, the exponent of $\pi$ is not $0$ and not predicted by any counting argument. Barenblatt calls that incomplete similarity; unit 21 meets it as a singular perturbation.

what it always does

It checks. Every equation you write has both sides of the same dimension or it is wrong, and that test costs seconds and catches most algebra slips. Use it on every line of every problem in this course.

An instructive failure. Ask for the period of the pendulum but forget $g$. Then the quantities are $T, l, m, \theta_0$ with rank 3 and $4 - 3 = 1$ group, namely $\theta_0$ — and the "conclusion" is $F(\theta_0) = 0$, which fixes $\theta_0$ and says nothing about $T$. The machinery returned a vacuous statement rather than a wrong one. That is the usual symptom of a missing variable, and it is why you should always look at what the groups are, not just how many there are.

An instructive over-inclusion. Add the bob's radius $a$ to the pendulum list. Now $n = 6$, $r = 3$, and there are three groups: $T^2g/l$, $\theta_0$ and $a/l$. The conclusion becomes $T = \sqrt{l/g}\,f(\theta_0, a/l)$, which is true but weaker. It is also honest: a real bob does have a size, and unit 16 shows the correction is $O((a/l)^2)$.

The 1D-versus-3D point. Dimensional analysis knows nothing about the dimension of space, and it cannot tell you that the gravitational potential goes as $1/r$ in three dimensions and $\log r$ in two. Unit 23 derives that dependence rather than asserting it, and the mechanism is geometry, not dimensions.

section 3

Three forces you can integrate

A second-order ODE reduces to two integrations in exactly three cases. Everything solvable in this course begins here.

00 · quadrature

Newton II as a second-order ODE

Restrict to one dimension for the rest of the unit: $m\ddot x = F$. A general force depends on all three of $x, \dot x, t$, and then nothing can be done in closed form. But if it depends on only one of them, the equation reduces to quadrature: the answer is an integral of a known function, whether or not that integral has a name.

	force	the move	what you integrate
1	$F(t)$	none needed	twice, in $t$
2	$F(v)$	treat $v$ as the unknown	separate, then invert
3	$F(x)$	multiply by $\dot x$	energy, then separate

quadrature

A problem is reduced to quadrature when its solution has been expressed in terms of integrals of known functions and their inverses. That is the classical notion of "solved", and it is weaker than "has a formula": $\int e^{-x^2}\mathrm dx$ counts.

why exactly two integrations

A second-order ODE needs two constants. Each integration supplies one, fixed by $x(0)$ and $\dot x(0)$. Any method that produces fewer has lost a solution; any that produces more has invented one.

The pendulum, $\ddot\theta = -(g/l)\sin\theta$, is case 3.

Why the three cases and no others. Each move removes one variable from the problem. In case 1 the right-hand side already involves neither of the unknowns, so it is two direct integrations. In case 2 the unknown $x$ itself does not appear, so $v$ satisfies a first-order equation. In case 3 the trick of multiplying by $\dot x$ makes both sides exact derivatives, which produces a first integral; the independent variable $t$ does not appear, which is what makes that possible. A force $F(x, v)$ has neither property in general, and a force $F(x, t)$ has neither. Unit 02 says what can still be learned in those cases without a formula.

"Reduced to quadrature" as a historical standard. For eighteenth- and nineteenth-century mechanics this was the definition of solving a problem, and the great theorems — Liouville's on integrable systems (unit 17), Jacobi's on the Hamilton–Jacobi equation (unit 15) — are theorems about when a reduction to quadrature exists. Unit 17 also shows how rare it is: the generic system is not integrable, and the honest twentieth-century answer to "solve it" is unit 02's qualitative theory.

All three cases are exercised in the problem sheet, each with its own characteristic trap.

00 · quadrature · case 1

Case 1: $F(t)$, integrate twice

$m\ddot x = F(t)$ with $F$ continuous. Integrate from $0$ to $t$:

$$\dot x(t) = v_0 + \frac1m\int_0^t F(s)\,\mathrm{d}s,$$ $$x(t) = x_0 + v_0 t + \frac1m\int_0^t\!\!\int_0^{s} F(u)\,\mathrm{d}u\,\mathrm{d}s .$$

Both steps are the fundamental theorem of calculus, and both constants are fixed by the data. Case 1 is always solvable; the only difficulty is whether the integral has a name.

the single-integral form

Swapping the order over the triangle $0 \le u \le s \le t$,

$$x(t) = x_0 + v_0t + \frac1m\int_0^t (t-u)F(u)\,\mathrm{d}u,$$

the first appearance of a Green's function: the kernel $(t-u)/m$ is unit 24's retarded propagator in one dimension.

the trap in case 1

A force whose average is zero does not give motion whose average is zero. Both constants of integration survive, and the first integration leaves a term $v_0 t$ that no amount of oscillation cancels.

If $F$ is periodic with mean zero then $\dot x$ is periodic about $v_0$, and $x$ drifts at rate $v_0$ plus a bounded oscillation about a shifted centre. "It is pushed back and forth, so it returns to where it started" is false; the next slide computes by how much.

dimension check

$\big[\tfrac1m\int(t-u)F\,\mathrm du\big] = \mathsf M^{-1}\,\mathsf T\,\mathsf{MLT^{-2}}\,\mathsf T = \mathsf L$. ✓

The swap of integration order needs no Fubini — which is just as well, because multiple integrals do not arrive until unit 05. One line of single-variable integration by parts does it. Put $P(s) = \int_0^s F(u)\,\mathrm du$, so that $P' = F$ and $P(0) = 0$; then

$$\int_0^t\!\!\int_0^s F(u)\,\mathrm du\,\mathrm ds = \int_0^t P(s)\,\mathrm ds = \big[sP(s)\big]_0^t - \int_0^t sF(s)\,\mathrm ds = \int_0^t (t-s)F(s)\,\mathrm ds,$$

using $tP(t) = \int_0^t tF(s)\,\mathrm ds$ in the last step. Seen from unit 05 this is Fubini on the triangle $\{0 \le u \le s \le t\}$; seen from here it is the product rule, and nothing in this unit is borrowed.

Why this is the easiest case and also the least interesting. The force does not respond to where the body is or how fast it is going, so there is no feedback, and every phenomenon that makes mechanics interesting — equilibria, oscillation, resonance, instability, chaos — requires feedback. Case 1 is the trivial fibre of the theory. It becomes interesting only when combined with cases 2 or 3, which is unit 06's forced oscillator: $m\ddot x + b\dot x + kx = F(t)$, where the same Green's function reappears with an oscillatory kernel.

Continuity of $F$ is more than needed. Riemann or Lebesgue integrability suffices for the formula, and the resulting $x$ is then $C^1$ with $\ddot x = F/m$ almost everywhere. Impulsive forces — $F$ a delta — need unit 20's distributions to be handled honestly, and the formula above continues to hold with $x$ merely continuous and $\dot x$ jumping by the impulse divided by $m$.

00 · quadrature · example

A periodic push, and where the body ends up

worked example 4 · $F(t) = F_0\cos\omega t$, from rest at the origin

$m = 0.500$ kg, $F_0 = 2.00$ N, $\omega = 4.00\ \mathrm{s^{-1}}$, $x(0) = 0$, $\dot x(0) = 0$.

First integration. $\displaystyle \dot x(t) = 0 + \frac{1}{m}\int_0^t F_0\cos\omega s\,\mathrm ds = \frac{F_0}{m\omega}\sin\omega t .$

Second integration. $\displaystyle x(t) = 0 + \frac{F_0}{m\omega}\int_0^t \sin\omega s\,\mathrm ds = \frac{F_0}{m\omega^2}\big(1 - \cos\omega t\big).$

Numbers: $F_0/(m\omega^2) = 2.00/(0.500\times16.0) = 0.250$ m. The body oscillates between $x = 0$ and $x = 0.500$ m with period $2\pi/\omega = 1.571$ s, and its mean position is $0.250$ m — not the origin.

checks

Dimensions: $[F_0/(m\omega^2)] = \mathsf{MLT^{-2}}/(\mathsf M\,\mathsf{T^{-2}}) = \mathsf L$. ✓ Limit $\omega \to 0$: $1 - \cos\omega t \to \omega^2t^2/2$, so $x \to F_0t^2/(2m)$ — constant force, uniform acceleration, Galileo. ✓ The $\omega^{-2}$ divergence as $\omega\to0$ is therefore not a singularity; it is cancelled by the bracket.

The $\omega\to0$ check done carefully. $1 - \cos u = u^2/2 - u^4/24 + O(u^6)$, so $\frac{F_0}{m\omega^2}(1-\cos\omega t) = \frac{F_0t^2}{2m} - \frac{F_0\omega^2t^4}{24m} + O(\omega^4)$. The limit is uniform on compact $t$-intervals, so the free-fall answer is recovered along with the first correction. This kind of check — take a parameter to a limit where you already know the answer — is worth doing on every result in this course, and unit 21 makes it into a method.

Dimensionally, in advance. The list is $x, F_0, m, \omega, t$ with $r = 3$, so $5 - 3 = 2$ groups: $x m\omega^2/F_0$ and $\omega t$. Buckingham therefore predicts $x = \frac{F_0}{m\omega^2}\,h(\omega t)$ before any integration, and all the calculus did was find $h(u) = 1 - \cos u$. The habit again: the shape is free, the function costs work.

00 · quadrature · example

The periodic push, plotted

$x(t) = 0.250\,(1 - \cos 4t)$ m. The force averages to zero over each period; the position does not. The body starts at the bottom of its oscillation, so it lives entirely on one side of where it started.

both traps, in one example

(i) Writing $x = -\frac{F_0}{m\omega^2}\cos\omega t$ solves the ODE but violates $x(0) = 0$. The constant is not optional.

(ii) The force averages to zero over a period, yet the body never returns to the origin except at $t = 2\pi n/\omega$, and its time-average displacement is $+0.250$ m. Zero mean force means zero mean acceleration, and nothing more.

Why the mean displacement is exactly the amplitude. Because the body starts at rest at the extreme of its oscillation rather than at the centre. Start it at rest at $x(0) = -F_0/(m\omega^2)$ instead and the solution is $-\frac{F_0}{m\omega^2}\cos\omega t$, centred on the origin. No choice of $\dot x(0)$ can do the same: a nonzero $v_0$ adds the term $v_0t$ and the motion is not bounded at all, so $v_0 = 0$ is forced and the centre is set by $x(0)$ alone. The mean position is a property of the initial data, not of the force.

What to look for in the plot. The curve touches $x = 0$ only at multiples of the period, and with zero slope each time: those are the instants the body is back at rest at its start. Shift the curve down by $0.250$ m and it is the textbook oscillation centred on the origin, which is the solution for a different initial position, not the solution of this problem.

00 · quadrature · case 2

Case 2: $F(v)$, separate then invert

$m\dot v = F(v)$: the position never appears, so this is a first-order equation for $v$. On any interval where $F(v) \neq 0$, separation of variables gives

$$t = m\int_{v_0}^{v}\frac{\mathrm{d}w}{F(w)} \;=:\; G(v),$$

an implicit solution: it gives $t$ as a function of $v$, the wrong way round. Then $x$ comes from one more integration, $x = x_0 + \int_0^t v$.

when the inversion is legitimate

$G$ is $C^1$ with $G'(v) = m/F(v) \ne 0$ on the interval, so $G$ is strictly monotone there, hence injective with a $C^1$ inverse (inverse function theorem in one variable). So $v = G^{-1}(t)$ exists and is differentiable — on that interval.

the trap in case 2

Everything stops at a zero of $F$. If $F(v_\infty) = 0$ then $v \equiv v_\infty$ is a solution, the integral $\int^{v_\infty}\mathrm dw/F(w)$ typically diverges, and $v_\infty$ is reached only as $t \to \infty$.

Dividing by $F(v)$ silently assumed $F(v) \ne 0$. Check the zeros first and treat them separately.

terminal velocity

A zero of the total force, $F(v_\infty) = 0$, approached but not attained. For gravity plus drag it is where the drag balances the weight.

A useful variant when you want $v(x)$ rather than $v(t)$: the chain rule gives $\dot v = \frac{\mathrm dv}{\mathrm dx}\dot x = v\,\frac{\mathrm dv}{\mathrm dx}$, so $m v\,\mathrm dv = F(v)\,\mathrm dx$, separable again.

The inverse function theorem in one variable, since it is doing real work here. If $G : I \to \mathbb{R}$ is $C^1$ on an interval with $G' > 0$ throughout, then $G$ is strictly increasing (mean value theorem), hence a bijection onto $G(I)$, which is an interval; the inverse is continuous (a continuous strictly monotone bijection between intervals has continuous inverse) and differentiable with $(G^{-1})'(t) = 1/G'(G^{-1}(t))$, by the usual difference-quotient argument. All of that is single-variable analysis. Unit 03 proves the $n$-dimensional version, which is a genuinely harder theorem.

Uniqueness is unit 01's, not ours. The argument above shows the separated formula defines a solution. That it is the only one through the given initial data needs Picard–Lindelöf, and the hypothesis that fails at a zero of $F$ is exactly the one that can fail elsewhere: $m\dot v = 3v^{2/3}$ with $v(0) = 0$ has both $v \equiv 0$ and $v = (t/m)^3$, because $v^{2/3}$ is not Lipschitz at $0$. Separation of variables happily produces the second and never mentions the first. Unit 01 is about exactly this.

Why $v\,\mathrm dv/\mathrm dx$ is worth remembering. It converts any $F(v)$ or $F(x)$ problem into one with $x$ as the independent variable, which is what you want when the question is "how far", not "how long". It is also the mechanism behind case 3: multiplying by $\dot x$ is the same substitution wearing a different hat.

00 · quadrature · example

Falling with drag, both laws

worked example 5 · $m\dot v = mg - \text{drag}$, dropped from rest, $v$ measured downward

Linear drag, $m\dot v = mg - bv$. Zero of the force: $v_\infty = mg/b$. Separate, with $\tau := m/b$:

$\displaystyle t = \int_0^v \frac{m\,\mathrm dw}{mg - bw} = -\tau\ln\!\Big(1 - \frac{v}{v_\infty}\Big)$

Invert (legitimate for $0 \le v < v_\infty$, where the integrand is finite and positive):

$$\begin{aligned} v(t) &= v_\infty\big(1 - e^{-t/\tau}\big)\\ x(t) &= v_\infty\big(t - \tau(1 - e^{-t/\tau})\big)\end{aligned}$$

A $2.00$ g bead with $b = 1.00\times10^{-2}\ \mathrm{kg\,s^{-1}}$ has $v_\infty = 1.962\ \mathrm{m\,s^{-1}}$, $\tau = 0.200$ s, and reaches $99$ per cent of $v_\infty$ at $t = \tau\ln 100 = 0.921$ s.

Quadratic drag, $m\dot v = mg - cv^2$ for $v > 0$. Zero: $v_\infty = \sqrt{mg/c}$. Separate:

$\displaystyle t = \int_0^v\frac{m\,\mathrm dw}{mg - cw^2} = \frac{v_\infty}{g}\,\operatorname{artanh}\frac{v}{v_\infty}$

$$\begin{aligned} v(t) &= v_\infty\tanh\frac{gt}{v_\infty}\\ x(t) &= \frac{v_\infty^2}{g}\ln\cosh\frac{gt}{v_\infty}\end{aligned}$$

A skydiver, $m = 80.0$ kg, $v_\infty = 55.0\ \mathrm{m\,s^{-1}}$, so $c = 0.259\ \mathrm{kg\,m^{-1}}$: $99$ per cent of $v_\infty$ at $t = (v_\infty/g)\operatorname{artanh}0.99 = 14.84$ s, having fallen $604$ m. Without drag she would be doing $146\ \mathrm{m\,s^{-1}}$.

The second integral, linear case. $\int_0^t v_\infty(1 - e^{-s/\tau})\mathrm ds = v_\infty[s + \tau e^{-s/\tau}]_0^t = v_\infty(t + \tau e^{-t/\tau} - \tau)$, which is the formula given. For $t \gg \tau$ it is $v_\infty(t - \tau)$: the body ends up moving at $v_\infty$ but lagging a fixed distance $v_\infty\tau$ behind where it would be had it always moved at $v_\infty$.

The second integral, quadratic case. $\int_0^t v_\infty\tanh(gs/v_\infty)\mathrm ds = \frac{v_\infty^2}{g}\big[\ln\cosh(gs/v_\infty)\big]_0^t$, since $\frac{\mathrm d}{\mathrm du}\ln\cosh u = \tanh u$ and $\ln\cosh 0 = 0$. For large $t$ this is $\frac{v_\infty^2}{g}(gt/v_\infty - \ln 2) = v_\infty t - \frac{v_\infty^2\ln2}{g}$, the same structure: asymptotically uniform motion with a constant lag, here $214$ m.

The partial fractions. $\frac{m}{mg - cw^2} = \frac{1}{g}\cdot\frac{1}{1 - (w/v_\infty)^2}$, and $\int_0^v \frac{\mathrm dw}{1-(w/v_\infty)^2} = v_\infty\operatorname{artanh}(v/v_\infty)$ for $|v| < v_\infty$. The artanh blows up at $v = v_\infty$: that is the divergent integral promised on the previous slide, and it is the analytic statement that terminal velocity is never attained in finite time.

Numbers, all computed. Linear: $v_\infty = mg/b = 0.00200\times9.81/0.0100 = 1.962\ \mathrm{m\,s^{-1}}$, $\tau = m/b = 0.200$ s, $\tau\ln100 = 0.9210$ s. Quadratic: $c = mg/v_\infty^2 = 80.0\times9.81/55.0^2 = 0.2594\ \mathrm{kg\,m^{-1}}$; $\operatorname{artanh}0.99 = 2.6467$, times $v_\infty/g = 5.6065$ s gives $14.838$ s; $x = (55.0^2/9.81)\ln\cosh(2.6467) = 604$ m; free fall would give $gt = 145.6\ \mathrm{m\,s^{-1}}$. A real skydiver in a stable spread reaches about $55\ \mathrm{m\,s^{-1}}$, so the model is not a toy.

00 · quadrature · example

Two drag laws, one picture

Measure speed in units of $v_\infty$ and time in units of $v_\infty/g$. For linear drag $v_\infty/g = m/b = \tau$, so with $s := gt/v_\infty$ the two solutions are

$$\begin{aligned} \text{linear:}\quad \frac{v}{v_\infty} &= 1 - e^{-s},\\ \text{quadratic:}\quad \frac{v}{v_\infty} &= \tanh s,\end{aligned}$$

the same for every bead and every skydiver. Both start along the free-fall line $v = gt$ and both bend over towards $v_\infty$ without reaching it.

checks

Dimensions: $[\tau] = \mathsf T$ and $[\sqrt{mg/c}] = (\mathsf{MLT^{-2}}/\mathsf{ML^{-1}})^{1/2} = \mathsf{LT^{-1}}$. ✓ Limit $t \ll \tau$: $1 - e^{-t/\tau} \approx t/\tau$ and $\tanh u \approx u$ both give $v \approx gt$, free fall. ✓ Limit $t \to \infty$: both tend to $v_\infty$, never reaching it. ✓

Quadratic drag is gentler at low speed — it grows like $v^2$, not $v$ — so the body keeps up with free fall longer, then closes on $v_\infty$ faster: $1-\tanh s \approx 2e^{-2s}$ against $e^{-s}$.

Why one picture serves both examples. The bead and the skydiver have nothing in common numerically — $v_\infty = 1.962$ against $55.0\ \mathrm{m\,s^{-1}}$ — but each problem has exactly one speed scale, $v_\infty$, and one time scale, $v_\infty/g$. In those units the parameters vanish and only the form of the drag law is left. That is the dimensional habit of section 2 used on a solution rather than on a list: each drag law has two dimensionless groups, $v/v_\infty$ and $gt/v_\infty$, so it has exactly one curve.

Reading the 99 per cent times off the plot. $1 - e^{-s} = 0.99$ at $s = \ln 100 = 4.61$; $\tanh s = 0.99$ at $s = \operatorname{artanh}0.99 = 2.65$. Multiply by $v_\infty/g$ — $0.200$ s for the bead, $5.607$ s for the skydiver — to recover $0.921$ s and $14.84$ s.

00 · quadrature · case 3

Case 3: $F(x)$ — where energy comes from

$m\ddot x = F(x)$. The trick is to multiply by $\dot x$, because then both sides become exact derivatives:

$$m\ddot x\,\dot x = \frac{\mathrm d}{\mathrm dt}\Big(\tfrac12 m\dot x^2\Big), \qquad F(x)\,\dot x = \frac{\mathrm d}{\mathrm dt}\Big(\int^{x(t)}\!\! F\Big)$$

by the chain rule, in both cases. Define the potential energy $V$ by $V'(x) = -F(x)$, that is $V(x) = -\int_{x_{\text{ref}}}^x F(s)\,\mathrm ds$. Then

conservation of energy, in one dimension

$\displaystyle \frac{\mathrm d}{\mathrm dt}\Big(\tfrac12 m\dot x^2 + V(x)\Big) = m\ddot x\dot x + V'(x)\dot x = \big(m\ddot x - F(x)\big)\dot x = 0,$

so $E := \tfrac12 m\dot x^2 + V(x)$ is constant along every solution. That is the entire proof.

first integral

A function of $(x,\dot x, t)$ that is constant along every solution. Each one found lowers the order of the problem by one — here from second order to first, which is what makes the remaining step a quadrature.

Solve for the velocity and separate again:

$$\dot x = \pm\sqrt{\tfrac2m\big(E - V(x)\big)},$$ $$t - t_0 = \pm\int_{x_0}^{x}\frac{\mathrm ds}{\sqrt{\tfrac2m(E - V(s))}} .$$

Two constants, as required: $E$ from the first integration, $t_0$ from the second. The sign is chosen by the direction of travel and flips at every turning point.

Why "multiply by $\dot x$" is not a trick. It is the search for an integrating factor, and the reason $\dot x$ works is that the equation has no explicit $t$ in it: the system is invariant under time translation, and Noether's theorem (unit 13) says that this symmetry has a conserved quantity attached to it, which is the energy. Every conservation law in this course will eventually be re-derived that way. For now, notice only that case 1 has the opposite property — $F(t)$ is not time-translation invariant — and correspondingly has no energy integral.

The reference point. $V$ is determined only up to an additive constant, because only $V'$ appears. Changing $x_{\text{ref}}$ changes $V$ and $E$ by the same constant and changes nothing physical. Fix it once per problem and say what you fixed; most of the sign errors in this subject come from not doing so.

Examples of $V$. Uniform gravity $F = -mg$ gives $V = mgx$. Hooke $F = -kx$ gives $V = \tfrac12kx^2$. The pendulum, with $x \to \theta$, $m \to ml^2$ and generalised force $-mgl\sin\theta$, gives $V(\theta) = mgl(1 - \cos\theta)$, taking the bottom as the reference — that is the running example's potential, and section 4 reads the entire motion off its graph.

Dimension and limit checks on $E$. $[\tfrac12 mv^2] = \mathsf M(\mathsf{LT^{-1}})^2 = \mathsf{ML^2T^{-2}}$, and $[V] = [\int F\,\mathrm dx] = \mathsf{MLT^{-2}}\cdot\mathsf L$, the same. ✓ Limit $F \equiv 0$: $V$ constant, so $\dot x$ constant — Newton's first law recovered. ✓

00 · quadrature

Two warnings about energy

1 · this is a one-dimensional theorem

In three dimensions the step "$\int F\,\mathrm dx$ depends only on the endpoint" is false in general. Whether a force field has a potential is a question about line integrals and closed loops, and it is settled in unit 04. Nothing here licenses $\vec F = -\nabla V$; that symbol has not been defined and will not be until unit 04.

2 · it fails as soon as $F$ depends on $v$

With drag, $\frac{\mathrm d}{\mathrm dt}(\tfrac12m\dot x^2 + V) = F_{\text{drag}}\dot x \le 0$. Energy decreases. Nothing is conserved, and the quadrature of case 3 is unavailable — which is why the falling-with-drag problem had to be done by case 2 instead.

what survives, and where

In one dimension, every continuous $F(x)$ has a potential: define $V(x) = -\int_{x_0}^x F$. That is the fundamental theorem of calculus and it has no three-dimensional analogue.

So the whole of section 4 — turning points, bounded motion, period integrals — is available for any position-dependent force in one dimension, and is genuinely special to one dimension. Unit 08 recovers it in three dimensions for central forces, by finding enough conserved quantities to reduce the problem back to one.

Dimension check on the drag statement: $[F\dot x] = \mathsf{MLT^{-2}\cdot LT^{-1}} = \mathsf{ML^2T^{-3}}$, a power — energy per time. ✓

The three-dimensional obstruction, in advance. Take $F_1 = -y$, $F_2 = x$ in the plane. Carrying a body once anticlockwise round the unit circle, this force is always tangent to the motion and does positive work every time; no function $V$ can have the property that the work equals $V(\text{start}) - V(\text{end})$, because the start and end coincide while the work is $2\pi \neq 0$. Unit 04 turns this observation into the theorem that a force field is conservative if and only if its work around every closed loop vanishes, and finds a local test (vanishing curl) that is sufficient on simply connected regions and not otherwise.

Where the energy goes. "Energy decreases" means mechanical energy decreases; it becomes heat in the fluid, and the total is conserved once thermodynamics is included. This course does not do thermodynamics, so in it drag is simply a non-conservative force and energy is simply not conserved. Say that, rather than gesturing at heat.

A useful special case that does survive: $F(x, v) = F_{\text{cons}}(x) + F_{\text{drag}}(v)$ still gives $\dot E = F_{\text{drag}}\dot x$, so $E$ is a Lyapunov function — non-increasing rather than constant. Unit 02 builds a whole qualitative theory on exactly that observation, and it is how the damped oscillator is shown to settle without solving it.

section 4

Reading the motion off the potential

Where the body can go, whether it comes back, and how long it takes — from the graph of $V$, before any integral is evaluated.

00 · potentials

Turning points, wells and barriers

$\tfrac12m\dot x^2 = E - V(x) \ge 0$, so the motion is confined to $\{x : V(x) \le E\}$. Everything follows from that one inequality.

A turning point is a point with $V(a) = E$: the speed is zero there.
An equilibrium is a point with $V'(x_\star) = 0$, so $F = 0$ and $x \equiv x_\star$ is a solution. It is stable if $V$ has a strict local minimum there and unstable if a strict local maximum.
Motion in a component of $\{V \le E\}$ bounded on both sides by turning points is bounded motion, and periodic unless an endpoint is itself an equilibrium; a component unbounded on one side gives unbounded motion — the body escapes.

Which component you are in is decided by the initial position, not by $E$ alone: the same energy can give a bounded orbit in one well and an unbounded one elsewhere.

Four turning points at this $E$. A body in well A is trapped there for ever; one released to the right of the last turning point escapes. Raise $E$ above the barrier and the two wells merge into one.

Why a strict local minimum is stable. Start at $x_\star + \varepsilon$ with small energy $E$ slightly above $V(x_\star)$. The set $\{V \le E\}$ has a component containing $x_\star$ that shrinks to $\{x_\star\}$ as $E \downarrow V(x_\star)$, by continuity of $V$ and the strictness of the minimum. So the motion stays in a small neighbourhood for all time — which is precisely Lyapunov stability, with $E$ as the Lyapunov function. Unit 02 proves the general theorem (Lagrange–Dirichlet) and shows that the converse needs care.

Why a strict local maximum is unstable is less obvious than it looks, and the honest statement needs the machinery of unit 02. The heuristic: arbitrarily close to the top there are points from which the body accelerates away, and it never returns because it would have to arrive at the top with zero speed, which takes infinite time (next slide but one).

The barrier is impassable, classically. A body in well A with the $E$ drawn can never reach well B, however long you wait, because $\dot x^2 < 0$ is not available. Quantum mechanics removes that conclusion (tunnelling), and the fact that this course's mechanics forbids it absolutely is worth noticing as a place where the classical theory is not merely approximate but qualitatively different.

00 · potentials

The period integral

period of a bounded motion

Let the motion at energy $E$ run between consecutive turning points $a < b$ with $V < E$ on $(a,b)$, $V(a) = V(b) = E$, and both endpoints genuine: $V'(a) \ne 0 \ne V'(b)$. Then the motion is periodic with period

$$T(E) \;=\; \sqrt{2m}\int_a^b \frac{\mathrm dx}{\sqrt{E - V(x)}} .$$

Proof. On $(a,b)$ the body moves right with $\dot x = \sqrt{\tfrac2m(E-V)}$, so the time from $a$ to $b$ is $\int_a^b \mathrm dx/\dot x$, which is half the expression above. At $b$ the velocity is zero and the force $-V'(b) < 0$ pushes it back; the return leg is the mirror image and takes the same time, because the integrand depends on $x$ only. Hence the total is $T(E)$, and the motion repeats. $\square$

The integral is improper at both ends, where the integrand blows up. That it converges is the content of the next slide, and it is exactly what "genuine" buys: drop that hypothesis and the pendulum's separatrix is a counterexample — $V < E$ strictly between $-\pi$ and $\pi$, yet the motion never returns.

checks

Dimensions: $\sqrt{\mathsf M}\cdot\mathsf L/\sqrt{\mathsf{ML^2T^{-2}}} = \mathsf T$. ✓

Harmonic oscillator, $V = \tfrac12kx^2$, turning points $\pm a$ with $E = \tfrac12ka^2$:

$T = \sqrt{2m}\int_{-a}^{a}\frac{\mathrm dx}{\sqrt{\tfrac12 k(a^2-x^2)}} = 2\sqrt{\tfrac{m}{k}}\big[\arcsin\tfrac xa\big]_{-a}^{a} = 2\pi\sqrt{m/k}$,

independent of $a$ — the isochrony that mechanical similarity predicted from homogeneity of degree 1, now confirmed by the integral. ✓

The harmonic computation in full. $E - V = \tfrac12k(a^2 - x^2)$, so $T = \sqrt{2m}\int_{-a}^a \big(\tfrac k2\big)^{-1/2}(a^2-x^2)^{-1/2}\mathrm dx = 2\sqrt{m/k}\int_{-a}^a\frac{\mathrm dx}{\sqrt{a^2-x^2}}$. The substitution $x = a\sin u$ turns the integral into $\int_{-\pi/2}^{\pi/2}\mathrm du = \pi$, giving $T = 2\pi\sqrt{m/k}$. Note the substitution also shows why the improper integral converges: near $x = a$ the integrand behaves like $(a-x)^{-1/2}$, which is integrable.

"The return leg takes the same time" deserves a word. Formally, if $x(t)$ solves $m\ddot x = F(x)$ then so does $x(-t)$, since the equation has no first derivative in it; the motion is reversible. The outward and return legs are related by that symmetry, so they take equal times. This is the first appearance of time-reversal symmetry, and it is exactly what drag destroys.

Periodicity needs a moment's care. Having returned to $a$ with zero velocity, the body is in the same state $(x,\dot x) = (a, 0)$ it started in, so — granted uniqueness of solutions, which is unit 01's theorem — the subsequent motion repeats the previous one. Without uniqueness one could imagine the body doing something else the second time round. This is the second of the two places in the unit where a later result is borrowed — the first was reading $F(\pi_1,\pi_2) = 0$ as $\pi_1 = f(\pi_2)^2$ on The pendulum, dimensionally — and both are borrowed knowingly.

$T(E)$ is a function worth studying. Unit 14 shows it is $\mathrm dI/\mathrm dE$ where $I$ is the area enclosed in phase space — the action variable — which makes the period computable without solving anything, and is the beginning of the theory of integrable systems.

00 · potentials

When is the time finite?

convergence at a turning point

Let $V$ be $C^2$ near $a$ with $V(a) = E$ and $V < E$ just to the right. The time to travel from a nearby point to $a$, $\ \displaystyle\int_{a}^{a+\delta}\frac{\mathrm dx}{\sqrt{\tfrac2m(E-V(x))}}$, is

finite if $V'(a) \neq 0$ — a genuine turning point; the body arrives, stops, returns;
infinite if $V'(a) = 0$ and $V''(a) < 0$ — a barrier top; the body takes for ever.

Proof. Taylor with remainder about $a$. If $V'(a) = -c$ with $c > 0$ then $E - V(x) = c(x-a) + O((x-a)^2)$, so the integrand is $\asymp (x-a)^{-1/2}$, and $\int_0^\delta u^{-1/2}\mathrm du = 2\sqrt\delta$ converges. If $V'(a) = 0$ and $V''(a) = -2\kappa < 0$ then $E - V(x) = \kappa(x-a)^2 + O((x-a)^3)$, so the integrand is $\asymp (x-a)^{-1}$, and $\int_0^\delta u^{-1}\mathrm du$ diverges. Comparison test in both directions. $\square$

the consequence, and a number

A motion that ends at an unstable equilibrium is not periodic: it is a separatrix, taking infinite time to arrive. For the pendulum that is the swing that just reaches the top.

The divergence is logarithmic, so it is slow: with $l = 1.000$ m the period is $2.006$ s at $\theta_0 \to 0$, $7.826$ s at $\theta_0 = 179^\circ$, and $13.71$ s at $\theta_0 = 179.99^\circ$. Each extra factor of ten in $\pi - \theta_0$ costs only about $2.9$ s.

Full proof, with the constants

Case $V'(a) \ne 0$. Since $V < E$ to the right of $a$ and $V(a) = E$, necessarily $V'(a) \le 0$; assume $V'(a) = -c < 0$. Taylor's theorem with Lagrange remainder gives $E - V(x) = c(x-a) - \tfrac12V''(\xi)(x-a)^2$ for some $\xi$ between $a$ and $x$. With $M = \sup_{[a,a+\delta]}|V''|$ and $\delta$ small enough that $M\delta \le c$, we get $\tfrac12 c(x-a) \le E - V(x) \le \tfrac32 c(x-a)$. So the integrand lies between $\sqrt{m/(3c)}\,(x-a)^{-1/2}$ and $\sqrt{m/c}\,(x-a)^{-1/2}$, and $\int_a^{a+\delta}(x-a)^{-1/2}\mathrm dx = 2\sqrt\delta$. Finite, by comparison.

Case $V'(a) = 0$, $V''(a) = -2\kappa < 0$. Taylor to second order with remainder: $E - V(x) = \kappa(x-a)^2 + R(x)$ with $|R(x)| \le M'|x-a|^3$ near $a$ (using $V \in C^3$; for $V \in C^2$ use the Peano form, $E - V(x) = \kappa(x-a)^2(1 + o(1))$, which suffices). Then for small $\delta$, $E - V(x) \le 2\kappa(x-a)^2$, so the integrand is at least $\sqrt{m/(4\kappa)}\,(x-a)^{-1}$, and $\int_a^{a+\delta}(x-a)^{-1}\mathrm dx = \infty$. Infinite, by comparison. $\square$

The intermediate cases. If $V'(a) = V''(a) = 0$ but $V'''(a) \ne 0$, the integrand is $\asymp (x-a)^{-3/2}$ and the time is again infinite; in general the time is finite exactly when $E - V$ vanishes to order $< 2$ at $a$. The threshold is order 2, and the threshold is exactly the condition $F(a) = 0$, that is, $a$ being an equilibrium. The clean statement: the time to reach a point is infinite if and only if that point is an equilibrium.

The logarithm, explicitly. For the pendulum, the asymptotic $K(k) \to \ln(4/\sqrt{1-k^2})$ as $k \to 1$ (cited: a standard property of the complete elliptic integral, DLMF 19.12.1; not proved here) gives, with $k = \sin(\theta_0/2)$ and $\sqrt{1-k^2} = \cos(\theta_0/2)$, $\;T \approx 4\sqrt{l/g}\,\ln\!\big(4/\cos(\theta_0/2)\big)$ as $\theta_0 \to \pi$. At $\theta_0 = 179^\circ$ this predicts $7.826$ s and the exact value is $7.826$ s; at $179.99^\circ$ it predicts $13.707$ s against an exact $13.707$ s. Each factor of ten in $\cos(\theta_0/2)$ adds $4\sqrt{l/g}\ln 10 = 2.941$ s.

00 · potentials · example

The pendulum's potential, read off

$V(\theta) = mgl(1-\cos\theta)$, and the bob released from rest at $\theta_0$ has $E = mgl(1 - \cos\theta_0)$. Everything about the motion is now visible without solving anything:

$\theta = 0$ is a strict minimum: a stable equilibrium. $V''(0) = mgl > 0$.
$\theta = \pm\pi$ are strict maxima, value $2mgl$: unstable equilibria.
$E < 2mgl$: turning points at $\pm\theta_0$ with $V'(\theta_0) = mgl\sin\theta_0 \ne 0$. Bounded, periodic — swinging.
$E > 2mgl$: no turning point, $\dot\theta$ never vanishes. Unbounded in $\theta$ — whirling over the top, forever in the same direction.
$E = 2mgl$ exactly: the separatrix. Infinite time, by the previous slide.

Two qualitatively different motions and a critical energy dividing them, all read off a graph. This is unit 02's subject, arriving early.

One period of $V$. Below the separatrix the level line closes on itself; above it, it runs off to the right for ever.

Deriving $V$. The equation is $ml^2\ddot\theta = -mgl\sin\theta$, so with $\theta$ as the coordinate the "mass" is $ml^2$ and the "force" is $-mgl\sin\theta$; then $V(\theta) = -\int_0^\theta(-mgl\sin s)\,\mathrm ds = mgl(1-\cos\theta)$, taking $V(0) = 0$. Geometrically this is $mg h$ with $h = l(1-\cos\theta)$ the height of the bob above the lowest point, which is the check that the abstraction did not go wrong.

The whirling case is only unbounded in $\theta$, not in space: the bob goes round and round. Whether you call that periodic depends on whether you regard the configuration space as $\mathbb R$ or as the circle $\mathbb R/2\pi\mathbb Z$. On the circle both motions are periodic and the separatrix is the boundary between two families of closed curves. That is the first hint that configuration spaces want to be manifolds rather than copies of $\mathbb R^n$, which is unit 31.

The critical energy as a number. With $m = 0.100$ kg and $l = 1.000$ m, $2mgl = 1.962$ J, and the speed needed at the bottom to just reach the top is $v = \sqrt{4gl} = 6.264\ \mathrm{m\,s^{-1}}$ — from $\tfrac12mv^2 = 2mgl$. Any less and it swings; any more and it whirls; exactly that, and it approaches the top asymptotically and never arrives.

Why "just barely" is physically unobservable. The separatrix is a single energy out of a continuum, and any real system has friction and measurement error. What is observable is the divergence of the period near it, which is the slow logarithm of the previous slide — and which is why a pendulum clock's escapement keeps the amplitude small and fixed.

section 5

The pendulum, exactly

Case 3 carried to the end: a first integral, a substitution, an elliptic integral, a series, and one number.

00 · the pendulum

The first integral, and the period as an integral

Multiply $\ddot\theta = -(g/l)\sin\theta$ by $\dot\theta$ and integrate, exactly as in case 3:

$$\tfrac12\dot\theta^2 - \tfrac{g}{l}\cos\theta = \text{const}.$$

Released from rest at $\theta_0$, the constant is $-(g/l)\cos\theta_0$, so

$$\dot\theta^{\,2} = \frac{2g}{l}\big(\cos\theta - \cos\theta_0\big).$$

The quarter period is the time from $\theta_0$ to $0$:

$$T = 4\sqrt{\frac{l}{2g}}\int_0^{\theta_0}\frac{\mathrm d\theta}{\sqrt{\cos\theta - \cos\theta_0}} .$$

two checks before going on

Dimensions. $\sqrt{l/g}$ is a time and the integral is dimensionless (an angle over the square root of a pure number). ✓

Convergence. At $\theta = \theta_0$ the integrand blows up, but $V'(\theta_0) = mgl\sin\theta_0 \ne 0$ for $0 < \theta_0 < \pi$, so the previous slide's theorem gives a finite integral. At $\theta_0 = \pi$ it diverges, and the pendulum takes for ever. ✓

This is already the answer, in the classical sense: reduced to quadrature. What follows is cosmetics — but the cosmetics turn the integral into a tabulated function and then into a usable series.

Where the factor 4 comes from. The motion is $\theta_0 \to 0 \to -\theta_0 \to 0 \to \theta_0$. The four legs take equal times: the first and second by the reflection symmetry $\theta \to -\theta$ of the equation (since $\sin$ is odd), the return legs by time-reversal symmetry as on the period-integral slide. So $T$ is four times the time from $\theta_0$ to $0$.

Consistency with the general formula. The period integral slide gives $T = \sqrt{2m}\int_a^b\mathrm dx/\sqrt{E - V}$. Here the coordinate is $\theta$, the effective mass is $ml^2$, $V = mgl(1-\cos\theta)$, $E = mgl(1-\cos\theta_0)$, and $a, b = \mp\theta_0$: $T = \sqrt{2ml^2}\int_{-\theta_0}^{\theta_0}\frac{\mathrm d\theta}{\sqrt{mgl(\cos\theta - \cos\theta_0)}} = \sqrt{\tfrac{2l}{g}}\int_{-\theta_0}^{\theta_0}\frac{\mathrm d\theta}{\sqrt{\cos\theta-\cos\theta_0}}$, and the integrand is even, so that is $2\sqrt{\tfrac{2l}{g}}\int_{0}^{\theta_0} = 4\sqrt{\tfrac{l}{2g}}\int_0^{\theta_0}$ — the formula on the slide. Every $m$ cancels, as the dimensional argument promised two sections ago.

The sign of the square root. Going from $\theta_0$ down to $0$, $\dot\theta < 0$, so $\dot\theta = -\sqrt{(2g/l)(\cos\theta - \cos\theta_0)}$ and $\mathrm dt = -\mathrm d\theta/|\dot\theta|$; integrating from $\theta_0$ down to $0$ flips the limits and the sign back, which is why the formula has no minus in it. Getting this wrong produces a negative period, which is at least a visible error.

00 · the pendulum

The half-angle substitution and the elliptic integral

Step 1. $1 - \cos\theta = 2\sin^2(\theta/2)$, so $\cos\theta - \cos\theta_0 = 2\big(\sin^2\tfrac{\theta_0}2 - \sin^2\tfrac\theta2\big)$. Write $k := \sin(\theta_0/2) \in (0,1)$. Then

$$T = 2\sqrt{\tfrac lg}\int_0^{\theta_0}\frac{\mathrm d\theta}{\sqrt{k^2 - \sin^2(\theta/2)}} .$$

Step 2. Substitute $\sin(\theta/2) = k\sin\varphi$, which maps $\theta \in [0,\theta_0]$ bijectively onto $\varphi \in [0,\pi/2]$. Differentiating, $\tfrac12\cos(\theta/2)\,\mathrm d\theta = k\cos\varphi\,\mathrm d\varphi$, and $\cos(\theta/2) = \sqrt{1 - k^2\sin^2\varphi}$, while $\sqrt{k^2 - \sin^2(\theta/2)} = k\cos\varphi$. The $k\cos\varphi$ cancels:

$$\frac{\mathrm d\theta}{\sqrt{k^2 - \sin^2(\theta/2)}} = \frac{2\,\mathrm d\varphi}{\sqrt{1 - k^2\sin^2\varphi}} .$$

the exact period

$\displaystyle T = 4\sqrt{\frac lg}\,K(k), \qquad k = \sin\frac{\theta_0}{2},$

where $\displaystyle K(k) = \int_0^{\pi/2}\frac{\mathrm d\varphi}{\sqrt{1-k^2\sin^2\varphi}}$ is the complete elliptic integral of the first kind.

$K$ is not an elementary function — that is a theorem of differential Galois theory, cited, not proved here. It is a tabulated special function, like $\sin$ or $\log$; naming it is the answer, not an evasion.

check

$k \to 0$: the integrand $\to 1$, so $K \to \pi/2$ and $T \to 2\pi\sqrt{l/g}$. The small-angle formula is the $k = 0$ value, and $f(\theta_0) = 4K(\sin(\theta_0/2))$ with $f(0) = 2\pi$, as "The pendulum, dimensionally" demanded. ✓

Why that substitution and not another. The integrand has square-root branch points where $\sin(\theta/2) = \pm k$, i.e. at the turning points. The substitution $\sin(\theta/2) = k\sin\varphi$ is designed to put those at $\varphi = \pm\pi/2$ where $\cos\varphi$ vanishes to exactly the order that cancels the singularity — this is the standard device for an integral with two branch points, and it is the same move that rationalises $\int\mathrm dx/\sqrt{a^2-x^2}$ with $x = a\sin u$. The general theory says that an integral of a rational function of $x$ and $\sqrt{Q(x)}$ with $Q$ of degree 3 or 4 is an elliptic integral, reducible to Legendre's three normal forms; degree 2 is the elementary case.

Checking the bijection. On $[0,\theta_0]$, $\sin(\theta/2)$ increases from $0$ to $k$ (since $\theta_0 < \pi$ means $\theta/2 < \pi/2$), and $\sin\varphi$ increases from $0$ to $1$ on $[0,\pi/2]$. Both are strictly increasing, so the correspondence is a $C^1$ bijection with $C^1$ inverse, and the change of variables is legitimate — including at the endpoint, where both sides are improper and the equality of improper integrals follows by taking limits of the proper ones.

Conventions. Some references write $K(m)$ with $m = k^2$, and most numerical tables of $K$ are tabulated that way. Confusing $K(k)$ with $K(k^2)$ is the single commonest error in this corner of the subject. Here $k = \sin(\theta_0/2)$ is the modulus, and $\theta_0$ itself is sometimes called the modular angle.

History. These integrals were studied because of exactly this problem and the arc length of an ellipse, by Fagnano, Euler and Legendre; Abel and Jacobi then inverted them, which produced elliptic functions, doubly periodic functions, and eventually the whole theory of Riemann surfaces. A pendulum is where a large part of nineteenth-century complex analysis came from.

00 · the pendulum

Expanding $K$: the amplitude correction

Step 1: binomial series. For $|u| < 1$, $(1-u)^{-1/2} = \sum_{n\ge0}\binom{2n}{n}\frac{u^n}{4^n}$, convergent, and uniformly so on $|u| \le k^2 < 1$. Put $u = k^2\sin^2\varphi$ and integrate term by term — legitimate by uniform convergence:

$$\begin{aligned} K(k) &= \sum_{n\ge0}\binom{2n}{n}\frac{k^{2n}}{4^n} \int_0^{\pi/2}\!\!\sin^{2n}\varphi\,\mathrm d\varphi\\ &= \frac\pi2\sum_{n\ge0}\left[\binom{2n}{n}4^{-n}\right]^2 k^{2n},\end{aligned}$$

using Wallis' formula $\int_0^{\pi/2}\sin^{2n} = \frac\pi2\binom{2n}{n}4^{-n}$. So $\;\frac{2}{\pi}K = 1 + \frac{k^2}{4} + \frac{9k^4}{64} + \cdots$.

Step 2: re-expand in $\theta_0$. Since $k^2 = \frac{\theta_0^2}{4} - \frac{\theta_0^4}{48}+\cdots$,

$$\frac{T}{2\pi\sqrt{l/g}} = 1 + \frac{\theta_0^2}{16} + \frac{11\,\theta_0^4}{3072} + \frac{173\,\theta_0^6}{737280} + \cdots$$

checks

$\theta_0 \to 0$ gives $1$. ✓ Every term is positive, so a real pendulum is always slower than the schoolbook formula — as it must be, since $|\sin\theta| \le |\theta|$ makes the restoring force weaker than the linear one at every angle. ✓

At $\theta_0 = 30^\circ = 0.5236$ rad: two terms give $1.017135$, three give $1.0174039$, and the exact ratio is $1.0174088$. The series is doing its job.

the $\theta_0^4$ coefficient

$-\frac1{192} + \frac{9}{1024} = \frac{-1024 + 1728}{196608} = \frac{11}{3072}$. Both steps contribute: two series, both carried to the same order.

Wallis' formula, since it is used. $I_n := \int_0^{\pi/2}\sin^{n}\varphi\,\mathrm d\varphi$ satisfies $I_n = \frac{n-1}{n}I_{n-2}$ by integration by parts, with $I_0 = \pi/2$. Iterating on even indices, $I_{2n} = \frac{(2n-1)!!}{(2n)!!}\cdot\frac\pi2 = \frac{(2n)!}{4^n(n!)^2}\cdot\frac\pi2 = \frac\pi2\binom{2n}{n}4^{-n}$.

Justifying term-by-term integration. On $|u| \le k^2 < 1$ the binomial series has terms bounded by $\binom{2n}{n}4^{-n}k^{2n} \le k^{2n}$ (since $\binom{2n}{n} \le 4^n$), so by the Weierstrass M-test it converges uniformly in $\varphi$; a uniformly convergent series of continuous functions on a compact interval may be integrated term by term. The restriction $k < 1$ is exactly $\theta_0 < \pi$, and the series diverges at the separatrix, where the period is infinite — the analysis and the physics fail together, which is a good sign.

The $\theta_0^4$ coefficient, done slowly. $\frac{2}\pi K = 1 + \frac{k^2}4 + \frac{9k^4}{64} + O(k^6)$. With $k^2 = \frac{\theta_0^2}{4} - \frac{\theta_0^4}{48} + O(\theta_0^6)$ and $k^4 = \frac{\theta_0^4}{16} + O(\theta_0^6)$: the $\frac{k^2}{4}$ term contributes $\frac{\theta_0^2}{16} - \frac{\theta_0^4}{192}$, and $\frac{9k^4}{64}$ contributes $\frac{9\theta_0^4}{1024}$. Summing the $\theta_0^4$ parts over the common denominator $196608 = 192\times1024$ gives $(-1024 + 1728)/196608 = 704/196608 = 11/3072$.

Is the series convergent or asymptotic? Convergent, with radius of convergence $k = 1$ in the $k$ variable — this is one of the few series in this course that is genuinely convergent. Unit 21 is about the far more common situation where it is not, and where the series is still the right answer.

00 · the pendulum

The number: where the schoolbook formula breaks

Question. At what amplitude is $T = 2\pi\sqrt{l/g}$ wrong by 1 per cent?

Estimate, from the leading term. $\theta_0^2/16 = 0.01$ gives $\theta_0 = 0.400$ rad $= 22.92^\circ$.

Exactly, solving $\frac2\pi K(\sin(\theta_0/2)) = 1.01$ numerically: $\theta_0 = 0.39818$ rad $= \mathbf{22.81^\circ}$. The quartic term moves the answer by a tenth of a degree, which is the honest measure of how good the leading term is.

$\theta_0$	$T$ (s), $l=1.000$ m	error	clock loses
$2^\circ$	2.00622	0.0076 %	6.6 s/day
$5^\circ$	2.00702	0.048 %	41 s/day
$10^\circ$	2.00989	0.19 %	2.7 min/day
$23^\circ$	2.02646	1.02 %	14.6 min/day
$90^\circ$	2.36784	18.0 %	—

why this number mattered

A pendulum clock whose amplitude drifts from $5^\circ$ to $10^\circ$ loses about two extra minutes a day. Huygens understood this in 1673 and solved it by making the bob swing on a cycloid rather than a circle, which is exactly isochronous — the tautochrone, the same curve that unit 11 derives as the brachistochrone.

The practical fix is cheaper: keep the amplitude small and constant. That is what an escapement is for.

All five rows computed from $T = 4\sqrt{l/g}\,K(\sin(\theta_0/2))$ with $l = 1.000$ m, $g = 9.81\ \mathrm{m\,s^{-2}}$; "clock loses" is the error times 86 400 s.

The one per cent is a statement about $f$, not about the pendulum. $T/2\pi\sqrt{l/g}$ is $f(\theta_0)/2\pi$, a dimensionless function of a dimensionless variable — exactly the object "The pendulum, dimensionally" said dimensional analysis leaves undetermined. Everything on this slide is a property of that one function, and nothing on it could have been guessed from dimensions.

How the exact root was found. Newton's method on $h(\theta) = \frac2\pi K(\sin(\theta/2)) - 1.01$, started from the leading-term estimate $0.4$, converges in three iterations to $\theta_0 = 0.3981768$ rad $= 22.8138^\circ$. For comparison the 0.1-per-cent amplitude is $7.24^\circ$ and the 0.01-per-cent amplitude is $2.29^\circ$: the error scales as $\theta_0^2$, so each factor of ten in accuracy costs a factor of $\sqrt{10}$ in amplitude.

The seconds pendulum. A clock ticking once per second has $T = 2$ s, so $l = gT^2/(4\pi^2) = g/\pi^2 = 0.9940$ m at $g = 9.81$. This is why the metre was very nearly defined as the length of a seconds pendulum, and why it was not: $g$ ranges from about $9.780$ at the equator to $9.832$ at the poles, a 0.5 per cent spread, which would have made the metre a function of latitude.

Huygens' cycloidal pendulum worked in principle and failed in practice: the flexible suspension needed to make the bob follow a cycloid introduced more error through friction and stiffness than the circular-arc correction it removed. The lesson is a general one about corrections that cost more than they save, and it is the reason the escapement approach won.

00 · the pendulum

The period across every amplitude

$f(\theta_0)/2\pi = \tfrac2\pi K(\sin\tfrac{\theta_0}2)$ from $0$ to $180^\circ$, with the first correction $1 + \theta_0^2/16$ beside it.

Flat at first. The curve leaves $1$ with zero slope, which is why the schoolbook formula survives to $23^\circ$.
The series holds to about $90^\circ$ — $1.154$ against the exact $1.180$ — and then falls hopelessly behind.
It diverges at $180^\circ$, the separatrix: the bob released upside down takes for ever, as When is the time finite? proved — but only logarithmically, so the curve stays modest until very close.

Exact (solid) against the two-term series (dashed). The marked point is the 1-per-cent amplitude, $22.81^\circ$.

00 · the pendulum

What the calculus bought

two lines of linear algebra gave

$T = \sqrt{l/g}\;f(\theta_0)$

no dependence on $m$;
$T \propto \sqrt l$ exactly;
amplitude enters only as an angle;
$g$ enters only as $g^{-1/2}$.

five slides of calculus gave

$f(\theta_0) = 4K(\sin(\theta_0/2))$

$f(0) = 2\pi$;
$f$ increasing, with $f(\theta_0) \to \infty$ as $\theta_0 \to \pi$;
$f = 2\pi(1 + \theta_0^2/16 + \cdots)$;
the $1$-per-cent amplitude, $22.81^\circ$.

the division of labour, and the habit

Every quantity in the problem was pinned down by counting; one function of one variable was left, and all the analysis went into it. That ratio is typical, not lucky. From here on, in this course: write the dimensional form of the answer before you integrate anything. It costs two lines, it tells you which limits to check, and it catches the algebra error you are about to make.

When the ratio is less favourable. With three or more groups, Buckingham leaves an unknown function of two or more variables, and that is close to knowing nothing — a function of two variables cannot be tabulated on a page or fitted from a handful of experiments. The art of experimental design is largely the art of arranging for $n - r$ to be small: hold some groups fixed, or find a regime in which one of them is irrelevant. Barenblatt's book is about exactly when you may do the latter.

Where this unit leaves you. You can now solve every one-dimensional problem whose force depends on one variable, and you can write down the shape of the answer to problems you cannot solve. What you cannot yet do: say anything about $m\ddot x = F(x,\dot x)$ — no first integral, no quadrature, and this is the generic case. Unit 01 proves that a solution exists and is unique anyway, and unit 02 shows how much can be learned about it without a formula. That pair is where mechanics stops being a catalogue of tricks.

00 · checkpoint

Done when

checkpoint

Given any force of the form $F(t)$, $F(v)$ or $F(x)$:

you classify which of the three it is before starting;
you write the dimensional form of the answer and name the dimensionless group the calculus is left to determine;
you reduce to quadrature and produce the explicit solution or the period integral, with that case's trap handled — the constants in case 1, the zero of $F$ in case 2, the root's sign and the improper endpoint in case 3.

Failing it is unambiguous: a wrong classification, a dimensional form that does not check, or a quadrature that loses a constant.

Reading

Landau & Lifshitz, Mechanics, §10 and §§11–12.
Barenblatt, Scaling, ch. 1.
Taylor, Proc. Roy. Soc. A 201 (1950) 159, 175.

01 determinism: why a solution exists and is unique, which this unit assumed twice and said so · 03 motion in space · then 02, equations with no formula.

Try the checkpoint on a force this deck has not touched: $m\ddot x = -k/x^2$ for $x > 0$, a body falling towards a centre from rest at $x = a$. Classification: case 3. Dimensional form: the quantities are $t, a, m, k$ with $r = 3$ and $n - r = 1$, so $t = C\sqrt{ma^3/k}$ with $C$ a pure number — the fall time scales as $a^{3/2}$, which is mechanical similarity with $k = -2$ again. Quadrature: $V = -k/x$, $E = -k/a$, and $t = \sqrt{m/2}\int_x^a \mathrm ds/\sqrt{k/s - k/a}$, which the substitution $s = a\sin^2 u$ evaluates to give $C = \pi/(2\sqrt2)$. Then check: the answer is the half-period of a degenerate Kepler orbit, and unit 08 gets the same number from Kepler's third law.

The problem sheet is 11 problems and about five hours. Attempt each one before opening its hints, and open the solutions only afterwards — they are written to be read after a real attempt, and they are the only assessment this course has.

In the reading: Landau & Lifshitz's §§1–5 are the Lagrangian formalism, which is unit 12. Start at §10 and come back to the opening sections after that unit, not before.

When the checkpoint is met: ./mr done 00.

Newton's second law is an ODE — and dimensions tell you the answer's shape before you solve it

Two ways to answer one question

What the three laws assert

The first law is a definition, not a special case

The second law is a scheme

Force laws are input, not theorem

The third law, and what it does not say

The pendulum, from the laws

The pendulum: the tension drops out

Dimensions, before any calculus

Units as a group action

The dimension matrix and its kernel

The Buckingham pi theorem

The pendulum, dimensionally

What the dimensional form already says

Mechanical similarity: scaling the equation, not the units

Taylor's blast wave: a classified number from a photograph

Taylor's blast wave: the test and the yield

What dimensional analysis cannot do

Three forces you can integrate

Newton II as a second-order ODE

Case 1: $F(t)$, integrate twice

A periodic push, and where the body ends up

The periodic push, plotted

Case 2: $F(v)$, separate then invert

Falling with drag, both laws

Two drag laws, one picture

Case 3: $F(x)$ — where energy comes from

Two warnings about energy

Reading the motion off the potential

Turning points, wells and barriers

The period integral

When is the time finite?

The pendulum's potential, read off

The pendulum, exactly

The first integral, and the period as an integral

The half-angle substitution and the elliptic integral

Expanding $K$: the amplitude correction

The number: where the schoolbook formula breaks

The period across every amplitude

What the calculus bought

Done when

Reading

Next