Part A · the equation of motion · spine

Where things are: curves, frames, and the frame that lies to you

Differentiation in several variables, the frame a curve carries with it, and the exact price of doing mechanics in a room that is turning.

needs 00~2½ h slides + 5 h problemsproblems: ./mr problems 03

Unit 00 solved $m\ddot x = F$ in one dimension. Three dimensions are not three copies of that: the force at a point depends on where the point is, and "where" is now a vector. So the first half of this unit builds the calculus of maps $\mathbb R^n \to \mathbb R^m$ — the derivative as a linear map, the chain rule as composition, Clairaut's theorem — and the geometry of a single curve in space.

The second half asks a question unit 00 could not even phrase. Newton's law holds in an inertial frame. The room you are in is bolted to a planet that turns once a day. What exactly goes wrong, and by how much? The answer is three extra terms, each of which falls out of differentiating $x_{\mathrm{in}} = R(t)\,x_{\mathrm{rot}}$ twice. Nothing is postulated; the fictitious forces are theorems about $\tfrac{\mathrm d}{\mathrm dt}$.

How to read this deck

One object runs through it: Foucault's pendulum in the Panthéon. Sections 1 and 2 give it a trajectory to be a curve of; section 3 gives the building it hangs in an attitude matrix $R(t)$; sections 4 and 5 turn that into the number $31.79$ hours. Proofs are on the slides where they are one line, and in these notes under Full proof where they are not. Nothing here is asserted without one of the two.

Terms with a dotted underline are glossary entries — click one, or press G for the whole list. N opens these notes, R stacks the deck into one readable page.

03 · overview

One pendulum, five sections

A 67-metre pendulum hangs in the Panthéon. Released, it swings in a vertical plane — and that plane slowly turns, once every $31.79$ hours. Nothing pushes it sideways.

The floor from above: the swing line 0 to 8 hours after release, turning clockwise at $11.33^\circ$ per hour (computed).

the route, in four steps

1. The room's attitude is a rotation $R(t)$, with $x_{\mathrm{in}} = R(t)\,x_{\mathrm{rot}}$. Differentiating $R^{\!\top}R = I$ shows its rate of turning is one vector, $\vec\omega$.

2. Differentiate $x_{\mathrm{in}}$ twice: in the room, $m\vec a = \vec F$ gains three terms. For a moving bob the one that matters is Coriolis, $-2m\,\vec\omega\times\vec v$.

3. Split $\vec\omega$ at latitude $\lambda$ into a vertical part $\omega_{\mathrm E}\sin\lambda$ and a horizontal part. Only the vertical part turns a horizontal swing; the other only presses up or down.

4. In a frame turning back at $\omega_{\mathrm E}\sin\lambda$ the Coriolis term cancels and the plane is fixed; so on the floor it turns once per $2\pi/(\omega_{\mathrm E}\sin\lambda) = 31.79$ h.

The plan. 1 derivatives in $\mathbb R^n$ · 2 curves · 3 frames that turn · 4 the price · 5 the number. Dotted: glossary (G); solid: the running example; N: notes.

The route, slide by slide. Step 1 is section 3: "What a frame is, and what $SO(3)$ is" sets up $x_{\mathrm{in}} = R\,x_{\mathrm{rot}}$, "$\Omega = R^{\!\top}\dot R$ is antisymmetric" proves the antisymmetry, and the hat-map slide turns the antisymmetric matrix into the vector $\vec\omega$. Step 2 is "Differentiate twice" and "The rotating-frame equation" in section 4. Step 3 is "Coriolis: the force that turns cyclones", whose figure draws the split. Step 4 is "Undo the rotation, and Coriolis is gone" in section 5, and the number is "Worked: 31.8 hours at the Panthéon". Sections 1 and 2 supply the calculus all of this uses: the derivative of a map, the product rule, and the fact that a unit vector can only turn.

Why step 3 is the heart of it. A horizontal velocity crossed with a horizontal vector is vertical, so the horizontal part of $\vec\omega$ can only push the bob up or down, never sideways. Crossed with the vertical part it stays horizontal and is turned through a right angle. That is why the answer has $\sin\lambda$ in it, why it vanishes on the equator, and why at the pole it is the whole of $\omega_{\mathrm E}$.

The problem sheet follows the same order: problems 1–3 are the several-variable calculus, 4–7 the curve geometry, 8–9 the frame algebra, 10–13 the consequences with numbers. Three of them — 5, 8 and 13 — are proofs rather than computations.

Foucault hung his first public pendulum in the Panthéon in March 1851, after a trial in the cellar of his house in the rue d'Assas and a second in the Paris Observatory. The point he was making was not subtle and not new — everyone had believed the Earth turned for two centuries — but it was the first demonstration that used no astronomy at all. You can do it indoors, at night, with the shutters closed.

section 1

Differentiating in several variables

The derivative is a linear map. Everything else on this slide deck is that sentence, applied.

03 · derivatives

Partials are not enough

definition

For $f : \mathbb R^n \to \mathbb R$, the partial derivative at $a$ is $$\partial_i f(a) = \lim_{h \to 0} \frac{f(a + h e_i) - f(a)}{h},$$ the ordinary derivative of $f$ along the $i$-th coordinate line. The directional derivative $D_uf(a)$ along a unit vector $u$ replaces $e_i$ by $u$.

both partials exist, $f$ is not even continuous

$$f(x,y) = \frac{xy}{x^2 + y^2}, \qquad f(0,0) = 0.$$ On each axis $f \equiv 0$, so $\partial_1 f(0) = \partial_2 f(0) = 0$. On the line $y = x$, $f \equiv \tfrac12$. So $f$ has no limit at the origin.

$f$ depends only on the angle: in polar coordinates it is $\tfrac12\sin 2\varphi$. Every disc round the origin, however small, contains the values $0$, $\tfrac12$ and $-\tfrac12$. The partials only look along the two dark axes.

Reading the figure. Put $x = r\cos\varphi$, $y = r\sin\varphi$: then $f = r^2\cos\varphi\sin\varphi/r^2 = \tfrac12\sin2\varphi$, with no $r$ left. So $f$ is constant along each line through the origin and the value depends on which line. The partial derivatives ask about two of those lines only, where $f \equiv 0$, and are perfectly happy. Each partial derivative restricts $f$ to one line and asks a single-variable question; that is exactly the weakness.

Why this matters here and not in unit 00. In one variable there is one difference quotient and it either converges or does not. In several variables "differentiable" has to mean something stronger than "differentiable along every line", because a function can be smooth along every line through a point and still be badly behaved on a curve through it. The definition on "The derivative is a linear map" is the right one, and the counterexamples on this slide and the next are the reason it is phrased that way.

These two functions come back as problem 1.

03 · derivatives

Even every direction is not enough

every directional derivative exists, still not differentiable

$$g(x,y) = \frac{x^3}{x^2+y^2}, \qquad g(0,0) = 0 .$$ For a unit vector $u = (u_1,u_2)$, $\big(g(tu) - g(0)\big)/t = u_1^3$ exactly, so $D_ug(0) = u_1^3$ for every $u$.

The partials are $\partial_1 g(0) = 1$, $\partial_2 g(0) = 0$. If $g$ had a derivative, it would be the linear map with those entries, and it would give $D_ug(0) = u_1$. Writing $u = (\cos\varphi, \sin\varphi)$, the truth is $\cos^3\varphi$, not $\cos\varphi$.

And $g$ is continuous at $0$, since $|g| \le |x|$. Continuity plus every directional derivative still does not make a derivative.

The directional derivative of $g$ at the origin, against the direction $\varphi$ of $u$. A linear map would have to give the dashed $\cos\varphi$; the actual $\cos^3\varphi$ agrees with it only on the axes ($\varphi = 0, \pi/2, \pi, \dots$).

03 · derivatives

The derivative is a linear map

definition · the total derivative

$f : U \subseteq \mathbb R^n \to \mathbb R^m$ is differentiable at $a \in U$ if there is a linear map $Df(a) : \mathbb R^n \to \mathbb R^m$ with $$f(a + h) = f(a) + Df(a)\,h + r(h), \qquad \frac{\lVert r(h)\rVert}{\lVert h\rVert} \longrightarrow 0 \ \text{ as } h \to 0 .$$ $Df(a)$ is the linear map that best approximates $f$ near $a$. It is a map, not a number and not a matrix.

uniqueness

There is at most one such $Df(a)$.

Proof. If $\Lambda_1, \Lambda_2$ both work then $(\Lambda_1-\Lambda_2)h = o(\lVert h\rVert)$. Put $h = tu$, $t \to 0$: $(\Lambda_1-\Lambda_2)u = 0$ for every $u$. $\square$

The physics motive: a force field $F : \mathbb R^3 \to \mathbb R^3$ near a point of equilibrium is its derivative, to first order — a $3\times3$ matrix. Every "small oscillations" calculation in this course is that sentence.

For the pendulum that matrix is $-m\omega_0^2 I$ on the horizontal plane; section 5 starts there.

Differentiable $\Rightarrow$ continuous: $\lVert f(a+h)-f(a)\rVert \le \lVert Df(a)\rVert\lVert h\rVert + \lVert r(h)\rVert \to 0$.

Reading the definition. "$r(h)/\lVert h\rVert \to 0$" is the whole content: the error is not merely small, it is small compared with $\lVert h \rVert$, and in every direction at once, uniformly. That last clause is what the counterexamples of the two previous slides fail. The one-variable case is included: $Df(a)$ is multiplication by $f'(a)$.

On the norm. Which norm on $\mathbb R^n$ does not matter — all norms on a finite-dimensional space are equivalent, so the condition $r(h) = o(\lVert h\rVert)$ is the same statement in any of them. The learner has the metric-space topology this needs.

Why "linear map" and not "matrix" is worth insisting on. The matrix depends on the basis; the map does not. This distinction is quietly doing nothing here and will be doing everything in unit 16, where a tensor is defined as a multilinear map and the component transformation law becomes a theorem about change of basis. The habit starts now: define the object, then compute its components.

The operator norm $\lVert \Lambda \rVert = \sup_{\lVert u \rVert = 1}\lVert \Lambda u\rVert$ is finite for a linear map on a finite-dimensional space, which is what the continuity argument on the slide uses.

03 · derivatives

The Jacobian, and when partials suffice

components

If $f$ is differentiable at $a$ then every $\partial_j f_i(a)$ exists, and the matrix of $Df(a)$ in the standard bases is the Jacobian matrix $$\big[Df(a)\big]_{ij} = \partial_j f_i(a).$$

Proof. Put $h = te_j$ in the definition and divide by $t$: $\big(f(a+te_j)-f(a)\big)/t \to Df(a)e_j$, whose $i$-th entry is $\partial_j f_i(a)$. $\square$

converse, with a hypothesis

If every $\partial_j f_i$ exists on a neighbourhood of $a$ and is continuous at $a$, then $f$ is differentiable at $a$. Such an $f$ is called $C^1$.

full proof in the notes (N)

So in practice: compute partials, check they are continuous, and you may use the Jacobian as the derivative. The counterexamples exist to keep the check honest, not to make life hard.

Full proof of the $C^1$ criterion

Take $m = 1$ (do each component separately). Write $h = (h_1,\dots,h_n)$ and walk from $a$ to $a + h$ one coordinate at a time: with $p_k = a + \sum_{j \le k} h_j e_j$, so $p_0 = a$ and $p_n = a + h$, $$f(a+h) - f(a) = \sum_{k=1}^{n}\big(f(p_k) - f(p_{k-1})\big).$$ Each bracket changes one coordinate only, so the single-variable mean value theorem gives $f(p_k) - f(p_{k-1}) = h_k\,\partial_k f(q_k)$ for some $q_k$ on the segment between $p_{k-1}$ and $p_k$; in particular $\lVert q_k - a \rVert \le \lVert h\rVert$. Hence $$f(a+h) - f(a) - \sum_k h_k \partial_k f(a) = \sum_k h_k\big(\partial_k f(q_k) - \partial_k f(a)\big),$$ whose absolute value is at most $\lVert h\rVert \sum_k |\partial_k f(q_k) - \partial_k f(a)|$. By continuity of the partials at $a$, the sum tends to $0$ as $h \to 0$. So the remainder is $o(\lVert h\rVert)$ and $Df(a)h = \sum_k h_k \partial_k f(a)$. $\square$

The hypothesis is not vacuous: $x^2\sin(1/x)$ extended by $0$ is differentiable everywhere with a discontinuous derivative, so "differentiable" is strictly weaker than $C^1$ even in one variable. In several variables there is a further gap below that one — partials existing without any derivative at all — which is what "Partials are not enough" exhibited.

Every map in this course is $C^\infty$ where it is used, so this theorem is the licence to stop worrying. It is invoked silently from here on.

03 · derivatives

The chain rule, once

theorem · the multivariable chain rule

If $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$, then $g \circ f$ is differentiable at $a$ and $$D(g\circ f)(a) \;=\; Dg\big(f(a)\big) \circ Df(a) .$$ The derivative of a composite is the composite of the derivatives. In matrices: multiply the Jacobians, in that order.

full proof in the notes (N)

Written in components it is the familiar page of subscripts, $$\partial_k (g_i \circ f)(a) = \sum_{j} \partial_j g_i\big(f(a)\big)\, \partial_k f_j(a),$$ which is just the $(i,k)$ entry of a matrix product. Learn the line above; derive this one when you need it.

the case used constantly below

$\gamma : \mathbb R \to \mathbb R^n$ a curve, $f : \mathbb R^n \to \mathbb R$: then $$\frac{\mathrm d}{\mathrm dt}\,f(\gamma(t)) = Df(\gamma(t))\,\dot\gamma(t).$$ A $1\times n$ matrix times an $n\times 1$ one: a number.

Full proof of the chain rule

Write $A = Df(a)$, $B = Dg(b)$ with $b = f(a)$. By definition $f(a+h) = b + Ah + r(h)$ with $r(h) = o(\lVert h\rVert)$, and $g(b+k) = g(b) + Bk + s(k)$ with $s(k) = o(\lVert k\rVert)$. Put $k = k(h) = Ah + r(h)$. Then $$g(f(a+h)) = g(b) + B\big(Ah + r(h)\big) + s(k(h)) = g(b) + BAh + \underbrace{B\,r(h) + s(k(h))}_{\text{claim: } o(\lVert h\rVert)} .$$ The first term: $\lVert B r(h)\rVert \le \lVert B\rVert\,\lVert r(h)\rVert = o(\lVert h\rVert)$. The second: $\lVert k(h)\rVert \le (\lVert A\rVert + 1)\lVert h\rVert$ for $h$ small, so $k \to 0$ with $h$, and given $\varepsilon > 0$ we have $\lVert s(k)\rVert \le \varepsilon \lVert k \rVert \le \varepsilon(\lVert A\rVert+1)\lVert h\rVert$ once $\lVert h \rVert$ is small enough. Hence the whole remainder is $o(\lVert h\rVert)$, and by uniqueness $D(g\circ f)(a) = BA$. $\square$

The one place to be careful is that $s$ must be evaluated at $k(h)$, which is not simply $Ah$; the bound $\lVert k(h)\rVert = O(\lVert h \rVert)$ is what makes that harmless.

Where this is used in this unit. Twice, both times invisibly: to differentiate $t \mapsto R(t)x(t)$ (a bilinear map composed with a curve, hence a product rule), and to change to polar coordinates on the problem sheet. It is used visibly from unit 04 onwards, where $\tfrac{\mathrm d}{\mathrm dt}V(\gamma(t)) = \nabla V \cdot \dot\gamma$ is the whole reason potential energy works.

03 · derivatives

Clairaut, and the function that breaks it

Clairaut's theorem

If $\partial_1\partial_2 f$ and $\partial_2\partial_1 f$ exist on a neighbourhood of $a$ and are continuous at $a$, then $$\partial_1\partial_2 f(a) = \partial_2\partial_1 f(a).$$

full proof in the notes (N)

Partial derivatives commute — when the second derivatives are continuous. This is the hypothesis that will later make $\mathrm d^2 = 0$ in unit 34 and kill half the candidate terms in unit 33's curvature identities.

and when they are not

$$f(x,y) = \frac{xy(x^2-y^2)}{x^2+y^2}, \qquad f(0,0) = 0 .$$ Then $\partial_1 f(0,y) = -y$ and $\partial_2 f(x,0) = x$, so $$\partial_2\partial_1 f(0,0) = -1, \qquad \partial_1\partial_2 f(0,0) = +1 .$$

Both mixed partials exist everywhere; neither is continuous at the origin. The theorem is not being evaded, its hypothesis is being failed.

Full proof of Clairaut's theorem

Fix $a = (a_1,a_2)$ and for small $h, k$ define the second difference $$\Delta(h,k) = f(a_1+h,a_2+k) - f(a_1+h,a_2) - f(a_1,a_2+k) + f(a_1,a_2).$$ It is symmetric under swapping the roles of the two coordinates — the same four values with the same signs. Read it one way: with $\varphi(s) = f(s, a_2+k) - f(s, a_2)$ we have $\Delta = \varphi(a_1+h)-\varphi(a_1) = h\,\varphi'(\xi) = h\big(\partial_1 f(\xi, a_2+k) - \partial_1 f(\xi, a_2)\big)$ by the mean value theorem, and a second application in the other variable gives $\Delta = hk\,\partial_2\partial_1 f(\xi,\eta)$ for some $(\xi,\eta)$ in the rectangle. Reading it the other way gives $\Delta = hk\,\partial_1\partial_2 f(\xi',\eta')$ with $(\xi',\eta')$ in the same rectangle. Divide by $hk$ and let $(h,k) \to (0,0)$; both interior points tend to $a$, and by continuity of the two mixed partials at $a$ the limits are $\partial_2\partial_1 f(a)$ and $\partial_1\partial_2 f(a)$. $\square$

The counterexample, computed. For $y \ne 0$, $\partial_1 f(0,y) = \lim_{h\to0} \frac{1}{h}\cdot\frac{hy(h^2-y^2)}{h^2+y^2} = \frac{y(-y^2)}{y^2} = -y$; it is $0$ at $y = 0$ too, so $\partial_1 f(0,y) = -y$ for all $y$ and $\partial_2\partial_1 f(0,0) = -1$. The function is antisymmetric, $f(y,x) = -f(x,y)$, so $\partial_2 f(x,0) = +x$ and $\partial_1\partial_2 f(0,0) = +1$. This is problem 2, where you also check that the mixed partials are genuinely discontinuous at the origin rather than merely unequal there.

The example is Peano's, from 1884; Clairaut had stated the theorem in 1740 without the hypothesis, and Schwarz gave the modern proof. Some books call it Schwarz's theorem for that reason.

section 2

Curves

A trajectory is a map $\mathbb R \to \mathbb R^3$. It carries a frame of its own, and that frame splits acceleration in two.

03 · curves

Vector-valued functions and two product rules

definition

A vector-valued function $\gamma : I \to \mathbb R^3$ on an interval is differentiable iff each component is, and then $\dot\gamma = (\dot\gamma_1,\dot\gamma_2,\dot\gamma_3)$. This is the total derivative of $\gamma$: the linear map $t \mapsto t\,\dot\gamma$.

For a trajectory, $\vec v = \dot\gamma$ is the velocity, $v = \lVert\vec v\rVert$ the speed, $\vec a = \ddot\gamma$ the acceleration.

product rules

For differentiable $u, w : I \to \mathbb R^3$ and scalar $f$, $$(f u)^{\textstyle\cdot} = \dot f u + f\dot u, \qquad (u\cdot w)^{\textstyle\cdot} = \dot u\cdot w + u\cdot\dot w,$$ and the same for any bilinear map in place of $\cdot$ — including the cross product of section 3, where the order must be kept.

Proof. Every bilinear $B$ on finite-dimensional spaces satisfies $B(u(t+h),w(t+h)) - B(u,w) = B(\Delta u, w) + B(u, \Delta w) + B(\Delta u, \Delta w)$; divide by $h$ and let $h \to 0$, the last term going as $h$. $\square$

Why bilinearity is the right level of generality. Writing the dot-product rule out in components proves it, but proves it once per bilinear operation. The argument on the slide proves it for all of them at once — dot, cross, matrix product, the action of a matrix on a vector — and the only thing it needs is that $B$ is bilinear and continuous, which every bilinear map on a finite-dimensional space is. The $B(\Delta u, \Delta w)$ term is $O(h^2)$ because $\lVert B(p,q)\rVert \le C \lVert p\rVert \lVert q\rVert$ for some constant $C$ depending only on $B$.

Order matters for the cross product, since $u \times w = -w\times u$. The rule reads $(u\times w)^{\textstyle\cdot} = \dot u \times w + u \times \dot w$, and swapping either factor's position changes the sign. Nearly every sign error in rotating-frame mechanics is a swapped cross product.

Notation: an overdot always means $\mathrm d/\mathrm dt$ in this course, and $\vec{\ }$ marks a 3-vector. Later units drop both when the objects stop being 3-vectors.

03 · curves

Constant speed means sideways acceleration

lemma

For a differentiable $u : I \to \mathbb R^3$: $\ \lVert u\rVert$ is constant $\iff u \cdot \dot u = 0$ on $I$.

Proof. $\tfrac{\mathrm d}{\mathrm dt}\lVert u \rVert^2 = \tfrac{\mathrm d}{\mathrm dt}(u\cdot u) = 2\,u\cdot\dot u$ by the product rule. A differentiable function on an interval is constant iff its derivative vanishes, and $\lVert u\rVert$ is constant iff $\lVert u\rVert^2$ is. $\square$

Applied to velocity: a particle moves at constant speed $\iff$ its acceleration is perpendicular to its velocity at every instant.

That is the whole content of "circular motion has centripetal acceleration". No circle has been assumed — only $v = \text{const}$.

The pendulum is the opposite case: at the bottom of its swing $\dot v = 0$ and all of $\vec a$ is normal; towards the ends $v^2/\rho$ dies away and what is left is tangential.

Applied to a unit vector: if $\lVert e(t)\rVert = 1$ then $\dot e \perp e$. A unit vector can only turn.

Used three times below: for $\hat T$, for $\hat N$, and for the columns of a rotation matrix — which is where the antisymmetry of $R^{\!\top}\dot R$ comes from.

This one-line lemma is the smallest true statement in the unit and the most used. It is problem 5's first half; the second half asks you to get from it to the two-term decomposition of acceleration without quoting the Frenet formulas.

A dimension check on the applied version. $u \cdot \dot u$ has units of $[u]^2/\mathrm T$. For $u = \vec v$ that is $\mathrm{m^2\,s^{-3}}$ — which is, up to the mass, power. Constant speed means the force does no work, and unit 04 will say exactly that. The lemma is the kinematic shadow of a statement about energy that cannot yet be phrased.

A trap. "$\vec a \perp \vec v$" is an instantaneous statement and does not by itself make the path a circle: a helix has constant speed and acceleration everywhere perpendicular to velocity, and it is not closed, let alone circular. Its curvature is constant too, so even that is not the missing condition: constant curvature forces a circle only for a curve that also stays in a plane, which is what problem 13 proves.

03 · curves

Arc length and the unit tangent

definition

For a $C^1$ curve $\gamma$ on $[a,b]$ the arc length is $$s(t) = \int_a^t \lVert\dot\gamma(\tau)\rVert\,\mathrm d\tau ,$$ so $\dot s = v \ge 0$. If $v > 0$ throughout, $s$ is strictly increasing, hence invertible, and $\tilde\gamma = \gamma\circ t(s)$ is the unit-speed parametrisation: $\lVert\tilde\gamma'(s)\rVert = 1$.

The unit tangent is $\hat T = \vec v / v = \mathrm d\gamma/\mathrm ds$. It depends on the curve and the direction of travel, not on how fast you go.

check

By the one-variable chain rule and the inverse function rule, $\tilde\gamma'(s) = \dot\gamma\,\frac{\mathrm dt}{\mathrm ds} = \dot\gamma/v$, of norm 1. ✓

Inverting $s(t)$ is exactly unit 00's move: there we inverted $t(x)$ to get $x(t)$ after a quadrature. Here we invert $s(t)$ to reparametrise. In both cases monotonicity is what licenses it, and in both cases the inverse rarely has a formula.

Why arc length is the right parameter. A curve is a set of points; a parametrisation is a way of walking along it. Curvature ought to be a property of the set, so it must be defined using a parameter fixed by the set itself. Arc length is that parameter, unique up to the choice of starting point and direction.

That the integral is the length. The definition above is a definition, and it deserves a word of justification: for a partition $a = t_0 < \dots < t_N = b$ the inscribed polygon has length $\sum \lVert \gamma(t_i)-\gamma(t_{i-1})\rVert$, and each term is $\lVert \int_{t_{i-1}}^{t_i}\dot\gamma\rVert$, which for a $C^1$ curve tends to $\int \lVert \dot\gamma \rVert$ as the mesh goes to zero. The supremum over partitions — the honest definition of length — agrees with the integral for $C^1$ curves. Cited: this is the rectifiability theorem, proved in any real analysis text (Rudin, Principles, Thm 6.27); nothing later in this course depends on the proof.

Dimensions. $[s] = \mathrm L$, $[\mathrm d/\mathrm ds] = \mathrm L^{-1}$, $\hat T$ dimensionless. Every formula below can be checked by counting how many $\mathrm d/\mathrm ds$ it carries.

03 · curves

Curvature, and the frame a curve carries

definition

At a point where $\hat T' \ne 0$ (primes are $\mathrm d/\mathrm ds$): $$\kappa = \lVert \hat T'\rVert, \qquad \hat N = \hat T'/\kappa, \qquad \hat B = \hat T \times \hat N .$$ $\kappa$ is the curvature and the radius of curvature is $\rho = 1/\kappa$. $\hat B = \hat T\times\hat N$ is the unit vector perpendicular to both that makes the triple right-handed (the cross product is defined properly in section 3). $(\hat T, \hat N, \hat B)$ is the Frenet–Serret frame: an orthonormal, right-handed basis attached to each point of the curve.

$\hat T' \perp \hat T$ by the lemma, so $\hat N$ really is perpendicular to $\hat T$: the principal normal. $\hat B$ is the binormal, and $\hat T, \hat N$ span the osculating plane — the plane the curve is momentarily bending in.

The osculating circle: same point, same tangent, same curvature as the curve (schematic). $\hat N$ points at its centre, a distance $\rho$ away; $\hat B$ points out of the page.

Why $\kappa = \lVert\hat T'\rVert$ deserves the name. $\hat T$ is a unit vector, so $\hat T'$ measures only how fast the direction of travel turns, per unit of distance travelled. For a circle of radius $\rho$ traversed at unit speed, $\hat T$ goes once round the unit sphere per circuit, of length $2\pi\rho$, so $\lVert \hat T'\rVert = 2\pi/(2\pi\rho) = 1/\rho$. Hence the name "radius of curvature", and hence $\rho = 1/\kappa$ is the radius of the circle that best fits the curve there.

Dimensions. $[\kappa] = \mathrm L^{-1}$, $[\rho] = \mathrm L$. A straight line has $\kappa = 0$ and $\rho = \infty$, and the frame is undefined — which is correct, since a straight line does not single out any perpendicular direction.

The degenerate points. $\hat N$ and $\hat B$ are undefined wherever $\kappa = 0$, and can jump across an isolated zero of $\kappa$. That is a real nuisance in differential geometry and irrelevant here: every trajectory in this unit has $\kappa > 0$ where the frame is used. Unit 32 replaces this entire apparatus with a connection, which has no such defect.

The frame is named for Jean Frédéric Frenet (1847) and Joseph Serret (1851), who found it independently.

03 · curves

The Frenet–Serret formulas

theorem

Along a unit-speed curve with $\kappa > 0$, with $\tau = -\hat B'\cdot\hat N$ the torsion: $$\hat T' = \kappa\,\hat N, \qquad \hat N' = -\kappa\,\hat T + \tau\,\hat B, \qquad \hat B' = -\tau\,\hat N .$$

proof

Each of $\hat T', \hat N', \hat B'$ is expanded in the orthonormal basis $(\hat T,\hat N,\hat B)$, so the whole content is the matrix $M$ with $M_{ij} = (\text{$i$-th vector})' \cdot (\text{$j$-th vector})$. Differentiating $e_i \cdot e_j = \delta_{ij}$ gives $e_i'\cdot e_j + e_i \cdot e_j' = 0$: $M$ is antisymmetric, so only three entries are free. $\hat T' = \kappa\hat N$ is the definition of $\kappa, \hat N$, which fixes $M_{12} = \kappa$ and $M_{13} = \hat T'\cdot\hat B = 0$. Naming $M_{23} = \tau$ leaves nothing else. $\square$

The Frenet–Serret formulas say: the frame rotates, and the rotation is described by two numbers. That antisymmetry argument is the same one that will produce $\vec\omega$ in section 3 — watch for it returning.

Reading the proof again. Three orthonormal vectors evolving in $s$ are three columns of an orthogonal matrix $Q(s)$, and $M = Q^{\!\top}Q'$ is antisymmetric for exactly the reason given. An antisymmetric $3\times3$ matrix has three independent entries; the definition of $\hat N$ forces one of them to vanish and names another $\kappa$, leaving $\tau$. So a space curve has exactly two local invariants, and cited: the fundamental theorem of curves says $\kappa(s) > 0$ and $\tau(s)$ determine the curve up to a rigid motion (do Carmo, Differential Geometry of Curves and Surfaces, §1-5).

Torsion, in words. $\tau$ measures the rate at which the curve leaves its osculating plane. $\tau \equiv 0$ iff the curve is planar: if $\tau = 0$ then $\hat B' = 0$, so $\hat B$ is a fixed vector $\hat B_0$, and $(\gamma \cdot \hat B_0)' = \hat T\cdot\hat B_0 = 0$, so $\gamma$ lies in a fixed plane. Dimensions: $[\tau] = \mathrm L^{-1}$, like $\kappa$.

Sign conventions. Some books define $\tau$ with the opposite sign ($\hat B' = +\tau\hat N$). The convention here — the common one, and do Carmo's — makes a right-handed helix have $\tau > 0$. Check any formula you import against a right-handed helix before believing it.

Notice what is not claimed: nothing about $\kappa$ or $\tau$ requires the curve to be a trajectory, and neither of them knows about time. The next slide puts time back.

03 · curves

Acceleration has exactly two parts

theorem

For any trajectory with $v = \lVert\vec v\rVert > 0$, $$\vec a \;=\; \dot v\,\hat T \;+\; \frac{v^2}{\rho}\,\hat N .$$ Speeding up is tangential; turning is normal. There is no third component: $\vec a \cdot \hat B = 0$ always.

The two pieces are the tangential and normal acceleration.
Proof. $\vec v = v\hat T$, so $\vec a = \dot v \hat T + v\dot{\hat T}$; and $\dot{\hat T} = \hat T' \dot s = \kappa v \hat N$ by the chain rule and Frenet. $\square$

The acceleration at one point (schematic): a part along the curve, $\dot v\,\hat T$, and a part towards the centre of the osculating circle, $(v^2/\rho)\,\hat N$. Nothing sticks out of the page.

Reading the figure. The dark arrow is $\vec a$; the two coloured arrows are its components in the Frenet frame, completed by the dashed parallelogram. Change the speed and only the tangential arrow changes; tighten the bend (smaller $\rho$) at the same speed and only the normal arrow grows. The binormal $\hat B$ points out of the page, and the theorem says the acceleration never has a component there: the curve is always, at that instant, accelerating within its own osculating plane.

The physics. Multiply by $m$: the component of force along the motion changes the speed, the component across it changes the direction, and nothing changes the binormal component because there is nothing there to change. A car's engine and brakes act along $\hat T$; its tyres act along $\hat N$; and the friction budget is shared between them, which is the entire content of the "friction circle" that racing drivers are taught.

03 · curves

Two parts, checked

The decomposition $\vec a = \dot v\,\hat T + (v^2/\rho)\,\hat N$ of the previous slide, put through the course's three standard tests.

dimensions

$[\dot v] = \mathrm{L\,T^{-2}}$ ✓. $[v^2/\rho] = \mathrm{L^2T^{-2}/L} = \mathrm{L\,T^{-2}}$ ✓.

limit: straight line

$\rho \to \infty$ leaves $\vec a = \dot v\hat T$: one-dimensional motion, unit 00. ✓

limit: circle at constant speed

$\dot v = 0$, $\rho = r$: $\vec a = (v^2/r)\hat N$, pointing at the centre. ✓

The centripetal acceleration of school physics is a corollary, not a postulate — and it has arrived without assuming the orbit is a circle.

a number

A car at $30\ \mathrm{m\,s^{-1}}$ on a motorway bend of radius $400\ \mathrm m$, at steady speed: $\dot v = 0$ and $v^2/\rho = 900/400 = 2.25\ \mathrm{m\,s^{-2}} = 0.229\,g$, all of it sideways. Braking in the bend adds a $\dot v\,\hat T$ part, and the tyres must supply both.

03 · curves · example

Worked: a helical ramp

worked example 1

A trolley is hauled up a helical ramp: $\gamma(t) = (3\cos 2t,\ 3\sin 2t,\ 1.5\,t)$ in metres, $t$ in seconds.

1 · velocity and speed. $\dot\gamma = (-6\sin 2t,\ 6\cos 2t,\ 1.5)$, so $v = \sqrt{36 + 2.25} = \sqrt{38.25} = 6.1847\ \mathrm{m\,s^{-1}}$, constant.

2 · unit tangent. $\hat T = \dot\gamma / v = (-0.9701\sin 2t,\ 0.9701\cos 2t,\ 0.2425)$: mostly round, a little up.

3 · acceleration. $\ddot\gamma = (-12\cos 2t,\ -12\sin 2t,\ 0)$, of magnitude $12\ \mathrm{m\,s^{-2}}$, horizontal, pointing at the axis.

Two turns of the ramp (computed from $\gamma$; the far side dashed), with the frame at one point: $\hat T$ along the ramp, $\hat N$ horizontal and pointing at the axis, $\hat B$ tilted back from the vertical by the ramp's slope.

03 · curves · example

Worked: the ramp's curvature and torsion

worked example 1, continued

4 · the split. $\dot v = 0$, so all of $\vec a$ is normal: $\hat N = \ddot\gamma/12 = (-\cos 2t, -\sin 2t, 0)$ and $v^2/\rho = 12$, giving $$\rho = \frac{38.25}{12} = 3.1875\ \mathrm m, \qquad \kappa = 0.31373\ \mathrm{m^{-1}} .$$

5 · torsion. $\hat B = \hat T\times\hat N = (0.2425\sin 2t,\ -0.2425\cos 2t,\ 0.9701)$, and $\hat B' = \dot{\hat B}/v$ gives $\tau = 0.078431\ \mathrm{m^{-1}}$.

Checks. Write the helix as $(a\cos kt,\ a\sin kt,\ ct)$ — here $a = 3$, $k = 2$, $c = 1.5$; the letters $a$, $k$ are chosen so as not to collide with $R(t)$ and $\vec\omega$ of section 3. Then $v^2 = a^2k^2 + c^2$, $\kappa = ak^2/v^2$, $\tau = ck/v^2$, and

$\rho = 3.1875 > a = 3$: the helix is less curved than its shadow circle, as it must be ✓. $\kappa^2+\tau^2 = \big((ak^2)^2 + (ck)^2\big)/(a^2k^2+c^2)^2 = k^2/v^2$ ✓. $\tau \to 0$ as the rise $c \to 0$ ✓.

Step 5 in full. With $\gamma = (a\cos kt, a\sin kt, ct)$ and $v^2 = a^2k^2+c^2$: $\hat B = \hat T \times \hat N$ works out to $(c\sin kt, -c\cos kt, ak)/v$, so $\dot{\hat B} = kc(\cos kt, \sin kt, 0)/v = -kc\,\hat N/v$, and $\hat B' = \dot{\hat B}/v = -(kc/v^2)\hat N$. Since $\hat B' = -\tau\hat N$, $\tau = kc/v^2 = ck/(a^2k^2+c^2)$. With $a = 3$, $k = 2$, $c = 1.5$: $\tau = 3/38.25 = 0.078431\ \mathrm{m^{-1}}$, and likewise $\kappa = ak^2/v^2 = 12/38.25 = 0.31373\ \mathrm{m^{-1}}$.

Why $\rho > a$. Flattening the helix onto the horizontal plane gives a circle of radius $a = 3\ \mathrm m$. Stretching it vertically straightens it, so its radius of curvature must increase; the exact factor is $v^2/(ak)^2 = 38.25/36 = 1.0625$. A helix with a very steep rise is nearly a straight line, with $\rho \to \infty$.

03 · curves

Three dimensions are not three one-dimensional problems

A projectile with drag: $m\vec{\dot v} = m\vec g - b\lVert\vec v\rVert^{\,n-1}\vec v$.

$n = 1$: linear drag decouples

$m\dot v_i = mg_i - b v_i$ for each $i$ separately: three copies of unit 00's $F(v)$ case, each solved by separation of variables. The horizontal motion knows nothing about the vertical.

$n = 2$: quadratic drag does not

$m\dot v_x = -b\sqrt{v_x^2+v_z^2}\,v_x$ contains $v_z$. Falling faster slows the horizontal motion. No separation, no closed form, and unit 02's qualitative methods or unit 21's asymptotics are what is left.

Which case is real? The dimensionless Reynolds number $\mathcal R = \rho_{\text{air}} v d/\mu$ decides, with $d$ the ball's diameter, $\rho_{\text{air}}$ the air's density and $\mu$ its viscosity: $\mathcal R \lesssim 1$ gives linear drag, $\mathcal R \gtrsim 10^3$ quadratic. For a $22\ \mathrm{cm}$ football at $20\ \mathrm{m\,s^{-1}}$, $\mathcal R \approx 3\times 10^5$ — firmly quadratic.

The lesson for this unit: going to three dimensions is not bookkeeping. It is why a geometric language — curves, frames, linear maps — is worth building before the equations get harder.

The terminal velocity limit still works in both cases: set $\vec{\dot v} = 0$ and solve. For $n=1$, $\vec v_\infty = m\vec g/b$; for $n = 2$, $v_\infty = \sqrt{mg/b}$, straight down.

The Reynolds number, computed. With $\rho_{\text{air}} = 1.2\ \mathrm{kg\,m^{-3}}$, $\mu = 1.8\times10^{-5}\ \mathrm{Pa\,s}$, $d = 0.22\ \mathrm m$, $v = 20\ \mathrm{m\,s^{-1}}$: $\mathcal R = 1.2\times20\times0.22/1.8\times10^{-5} = 2.9\times10^5$. Quoted above as $3\times10^5$. A falling mist droplet, $d = 20\ \mu\mathrm m$ at $0.01\ \mathrm{m\,s^{-1}}$, gives $\mathcal R = 0.013$: linear, which is why Stokes' law describes fog and not footballs.

Why $n=1$ separates and $n=2$ does not. The drag force is antiparallel to $\vec v$ in both cases, but only in the linear case is its magnitude a function of one component at a time. Coupling through $\lVert \vec v\rVert$ is the generic situation; the decoupled case is the accident.

This slide is the only place drag appears in this unit. Unit 21 returns to the quadratic projectile as a perturbation problem, and gets the range correction as an asymptotic series in the small parameter $bv_0/(mg)$.

section 3

Frames that turn

One time-dependent orthogonal matrix, differentiated once. Everything after this is bookkeeping.

03 · frames

What a frame is, and what $SO(3)$ is

definition

A frame is an orthonormal, right-handed basis of $\mathbb R^3$ together with an origin. An inertial frame is one in which Newton's second law holds in the form $m\vec a = \vec F$ with $\vec F$ built only from real interactions.

definition

$$SO(3) = \{R \in \mathbb R^{3\times3} : R^{\!\top}R = I,\ \det R = +1\},$$ a group under matrix multiplication: the special orthogonal group. $R^{\!\top}R = I$ says $R$ preserves the inner product; $\det R = +1$ says it preserves orientation.

Two frames sharing an origin are related by exactly one $R \in SO(3)$: if $x_{\mathrm{rot}}$ are a vector's components in one and $x_{\mathrm{in}}$ in the other, then $$\boxed{\;x_{\mathrm{in}} = R\,x_{\mathrm{rot}}\;}$$ and if the frames move relative to each other, $R = R(t)$.

The names anticipate the use: "in" for an inertial frame, "rot" for one that turns with the room.

Why $\det = +1$ and not $-1$. $O(3)$ has two connected components. A frame that moves continuously from another cannot change handedness on the way, so a physically realisable relative motion lives in the identity component, $SO(3)$. The learner's group theory says the rest: $SO(3)$ is a normal subgroup of index 2 in $O(3)$, with quotient $\{\pm1\}$ recording handedness.

Does an inertial frame exist? Strictly, no: the laboratory rotates with the Earth, the Earth orbits the Sun, the Sun orbits the galaxy. What the definition means in practice is "inertial to the accuracy you need", and this unit's business is to compute exactly how large the error is for the first of those. Unit 30 reopens the question and answers it differently: there is no global inertial frame in the presence of gravity, only local ones, and that is the equivalence principle.

The convention. Different books put $R$ or $R^{\!\top}$ in $x_{\mathrm{in}} = R x_{\mathrm{rot}}$. Fix one and never change it; every sign in section 4 depends on this choice. Here: $R$ maps rotating-frame components to inertial-frame components. With the other convention every $\Omega$ below picks up a minus sign and $\vec\omega$ reverses.

03 · frames

One arrow, two lists of numbers

One arrow $u$, drawn once. Dropping perpendiculars onto the axes $e_1, e_2$ gives one list of components; onto the axes $e'_1, e'_2$, turned by $25^\circ$, a different list.

one vector, two lists

$x_{\mathrm{in}}$ and $x_{\mathrm{rot}}$ are the same arrow written in two bases. Nothing physical changes when $R$ acts; what changes is the list of numbers. Keeping that straight is half of this section.

Why $R$ does it. The columns of $R$ are $e'_1, e'_2, e'_3$ written in the basis $e_1, e_2, e_3$. So $R\,x_{\mathrm{rot}} = \sum_i (x_{\mathrm{rot}})_i\,e'_i$ — the arrow rebuilt from its primed components — written in the unprimed basis.

Check, in the figure's plane. Rotating $(2.90,\ 0.78)$ by $25^\circ$: $2.90\cos25^\circ - 0.78\sin25^\circ = 2.30$ and $2.90\sin25^\circ + 0.78\cos25^\circ = 1.93$ ✓.

What to look for in the figure. The arrow does not move; only the axes do. The two dashed constructions land at different places on different axes, and both are correct descriptions of the same $u$. The arrow has length $3$ and makes $40^\circ$ with $e_1$, hence $15^\circ$ with $e'_1$; the two lists are $3(\cos40^\circ, \sin40^\circ)$ and $3(\cos15^\circ, \sin15^\circ)$.

The direction of $R$, once more. A common slip is to think that because the primed axes are turned by $+25^\circ$, the primed components are the unprimed ones turned by $+25^\circ$. It is the other way round: the components turn by $-25^\circ$ (the arrow is at $15^\circ$ from the new axis, not $65^\circ$), and $R$, which turns by $+25^\circ$, takes primed components back to unprimed. Section 4's signs all rest on getting this one right.

03 · frames

$\Omega = R^{\!\top}\dot R$ is antisymmetric

theorem

Let $R : I \to SO(3)$ be differentiable. Then $\Omega(t) = R(t)^{\!\top}\dot R(t)$ is antisymmetric: $\Omega^{\!\top} = -\Omega$.

Proof. Differentiate $R^{\!\top}R = I$ with the product rule: $\dot R^{\!\top}R + R^{\!\top}\dot R = 0$, i.e. $(R^{\!\top}\dot R)^{\!\top} = -R^{\!\top}\dot R$. $\square$

definition

$$\mathfrak{so}(3) = \{A \in \mathbb R^{3\times3} : A^{\!\top} = -A\},$$ the space of antisymmetric matrices, of dimension 3: the diagonal vanishes and the three entries above it are free.

$\dot R$ is not in $SO(3)$ and is not a rotation. But $R^{\!\top}\dot R$ lies in a 3-dimensional space — and three numbers is exactly what an angular velocity vector should be.

This is the lemma on "constant speed" again, in disguise: the columns of $R$ are unit vectors, so each column's derivative is perpendicular to it — the diagonal of $\Omega$ vanishes — and each pair is orthogonal, which pairs up the off-diagonal entries.

Same argument as the Frenet proof, third appearance. The moving-frame trade: orthonormality plus differentiation gives antisymmetry.

The other one. $\tilde\Omega = \dot R R^{\!\top}$ is antisymmetric too, by the same computation applied to $RR^{\!\top} = I$, and $\tilde\Omega = R\Omega R^{\!\top}$. They are the body and space angular velocities, and they differ: $\Omega$ has components in the rotating frame, $\tilde\Omega$ in the inertial one. Problem 8 proves $\tilde\Omega$ corresponds to $R\vec\omega$, which is exactly the statement that $\vec\omega$ is a genuine vector rather than a triple of numbers that happens to have three entries.

Group-theoretic context, for the reader who wants it. $\mathfrak{so}(3)$ is the tangent space to $SO(3)$ at the identity — its Lie algebra — and $\Omega$ is $\dot R$ transported back to the identity by left translation. None of that language is needed here, and it is not used again until unit 31. What is needed is the dimension count: a curve in $SO(3)$ has, at each instant, three degrees of freedom of velocity, no more.

Why $R^{\!\top}\dot R$ and not $\dot R$ alone. $\dot R$ is a general $3\times3$ matrix with nine entries and no structure. Multiplying by $R^{\!\top}$ undoes the rotation already accumulated and leaves only the rate, which is where the structure is.

03 · frames

$\mathfrak{so}(3)$ is $\mathbb R^3$: the hat map and the cross product

definition · the hat map

$$\widehat{\ \cdot\ } : \mathbb R^3 \to \mathfrak{so}(3), \qquad \hat\omega = \begin{pmatrix} 0 & -\omega_3 & \omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0\end{pmatrix},$$ and the cross product is the bilinear map $\ \vec u \times \vec w := \hat u\,w$.

theorem

$\widehat{\ \cdot\ }$ is a linear isomorphism $\mathbb R^3 \to \mathfrak{so}(3)$. Consequently every antisymmetric $3\times3$ matrix is "cross with a unique vector".

Proof. Linear by inspection; injective since $\hat\omega = 0$ forces every $\omega_i = 0$; and $\dim\mathbb R^3 = \dim\mathfrak{so}(3) = 3$. $\square$

Immediate consequences, each one line from the matrix:

$\vec u\times\vec w = -\vec w\times\vec u$, so $\vec u\times\vec u = 0$;
$\vec u \cdot (\vec u \times \vec w) = 0$, since $u^{\!\top}\hat u\,w = -u^{\!\top}\hat u^{\!\top}w = 0$;
$\lVert \vec u\times\vec w\rVert = \lVert\vec u\rVert\lVert\vec w\rVert\sin\theta$;
$\hat e_1 e_2 = e_3$: right-handed.

The further identities — $\vec u\times(\vec w\times \vec z)$ and friends — arrive in unit 08, where they earn their keep. Only the four above are used here.

Definition first, formula second. The cross product is usually introduced by a mnemonic determinant. Here it is defined as the action of the unique antisymmetric matrix attached to a vector, which is why it exists only in three dimensions: $\dim\mathfrak{so}(n) = n(n-1)/2$, and that equals $n$ only when $n = 3$. In every other dimension "the axis of a rotation" is not a vector, and the cross product does not exist. Unit 28 meets the general object — an antisymmetric $(0,2)$-tensor — and unit 34 names it a 2-form.

Full proof of the norm identity

Expand $\lVert \hat u w\rVert^2$ directly: it equals $(u_2w_3-u_3w_2)^2 + (u_3w_1-u_1w_3)^2 + (u_1w_2-u_2w_1)^2$, and multiplying out and adding $(\vec u\cdot\vec w)^2 = (u_1w_1+u_2w_2+u_3w_3)^2$ gives $\lVert\vec u\rVert^2\lVert\vec w\rVert^2$ — every cross term cancels. This is Lagrange's identity, $\lVert\vec u\times\vec w\rVert^2 + (\vec u\cdot\vec w)^2 = \lVert\vec u\rVert^2\lVert\vec w\rVert^2$. With $\vec u\cdot\vec w = \lVert\vec u\rVert\lVert\vec w\rVert\cos\theta$ it gives $\lVert\vec u\times\vec w\rVert = \lVert\vec u\rVert\lVert\vec w\rVert|\sin\theta|$, and $\theta \in [0,\pi]$ makes the modulus unnecessary. $\square$

Dimensions. $\times$ multiplies dimensions like a product: $[\vec\omega\times\vec r] = \mathrm{T^{-1}\,L} = \mathrm{L\,T^{-1}}$, a velocity, as the transport theorem needs.

Handedness. The cross product changes sign under a reflection, which is why it is sometimes called a pseudovector. The honest statement is that $\hat u$ is the real object and the vector $\vec u$ is its image under an isomorphism that depends on an orientation. Unit 28 makes this precise.

03 · frames · example

Worked: reading $\vec\omega$ off $R$

worked example 2

A turntable. A frame turning about $\hat z$ at rate $\alpha$: $$R(t) = \begin{pmatrix}\cos\alpha t & -\sin\alpha t & 0\\ \sin\alpha t & \cos\alpha t & 0 \\ 0&0&1\end{pmatrix}.$$

$\dot R$ is $\alpha$ times the same matrix with $\cos \to -\sin$, $\sin\to\cos$. Then $$\Omega = R^{\!\top}\dot R = \alpha\begin{pmatrix}0&-1&0\\1&0&0\\0&0&0\end{pmatrix} = \widehat{\alpha \hat z},$$ using $\cos^2 + \sin^2 = 1$ in every entry. So $\vec\omega = \alpha\hat z$: magnitude the spin rate, direction the axis, right-hand rule included.

The Earth. Same shape, with $\hat z$ the polar axis and $$\alpha = \omega_{\mathrm E} = \frac{2\pi}{86\,164.09\ \mathrm s} = 7.2921\times10^{-5}\ \mathrm{s^{-1}} .$$

Note the number in the denominator: the sidereal day, $23\ \mathrm h\ 56\ \mathrm{min}\ 4.09\ \mathrm s$, not $86\,400\ \mathrm s$. $R(t)$ measures the ground against the fixed stars, and the Earth's orbit costs one extra turn per year: $366.24$ rotations in $365.24$ days.

Check. $86\,400/86\,164.09 = 1.002738 = 1 + 1/365.25$ ✓. Using the solar day would put the Foucault period out by $0.27\%$ — about $5$ minutes in $32$ hours, which is inside the measurement error but outside the arithmetic's.

The computation in full. Write $c = \cos\alpha t$, $s = \sin\alpha t$. Then $R = \begin{pmatrix} c&-s&0\\ s&c&0\\0&0&1\end{pmatrix}$ and $\dot R = \alpha\begin{pmatrix} -s&-c&0\\ c&-s&0\\0&0&0\end{pmatrix}$, so $R^{\!\top}\dot R = \alpha\begin{pmatrix} c&s&0\\ -s&c&0\\0&0&1\end{pmatrix} \begin{pmatrix} -s&-c&0\\ c&-s&0\\0&0&0\end{pmatrix} = \alpha\begin{pmatrix} -cs+sc & -c^2-s^2 & 0\\ s^2+c^2 & sc - cs & 0 \\ 0&0&0\end{pmatrix} = \alpha\begin{pmatrix} 0&-1&0\\ 1&0&0\\ 0&0&0\end{pmatrix}$. Comparing with the hat matrix, $\omega_3 = \alpha$ and $\omega_1 = \omega_2 = 0$.

Notice what did not happen. $\Omega$ came out constant even though $R$ did not. That is the point of conjugating by $R^{\!\top}$: it strips off the accumulated rotation. For a frame turning about a moving axis, $\Omega$ depends on $t$ and $\vec\omega$ is a genuine curve in $\mathbb R^3$; the Euler term in section 4 is precisely $\dot{\vec\omega} \ne 0$.

Sidereal versus solar, carefully. In one year the Earth makes $N$ rotations against the stars and $N - 1$ noon-to-noon days, because the orbit carries the Sun's direction once round. Hence $T_{\text{sid}} = T_{\text{sol}}\cdot\frac{365.24}{366.24} = 86\,400\times0.997270 = 86\,164.1\ \mathrm s$ ✓. Every rotating-frame number in this unit uses $\omega_{\mathrm E} = 7.2921\times10^{-5}\ \mathrm{s^{-1}}$.

03 · frames

The transport theorem

theorem · the transport theorem

Let $u(t)$ be the rotating-frame components of a vector, $R(t) \in SO(3)$, and $\vec\omega$ the vector with $\hat\omega = R^{\!\top}\dot R$. Then the inertial-frame rate of change, expressed in rotating-frame components, is $$\left(\frac{\mathrm d\vec u}{\mathrm dt}\right)_{\!\text{in}} = \left(\frac{\mathrm d\vec u}{\mathrm dt}\right)_{\!\text{rot}} + \vec\omega\times\vec u .$$

Proof. The inertial components are $Ru$; differentiate: $(Ru)^{\textstyle\cdot} = \dot Ru + R\dot u = R(R^{\!\top}\dot R\,u + \dot u) = R(\dot u + \vec\omega\times\vec u)$. Multiplying back by $R^{\!\top}$ gives the statement. $\square$

dimensions

$[\vec\omega\times\vec u] = [u]/\mathrm T$ ✓, matching both derivatives.

limit $\vec\omega \to 0$

The two derivatives agree: a non-rotating frame tells the truth. ✓

test $\vec u = \vec\omega$

$\vec\omega\times\vec\omega = 0$, so $\dot{\vec\omega}$ is the same in both frames. ✓

What the theorem is really saying. "The rate of change of a vector" is not well defined until you say which frame's basis is being held fixed. The next slide draws the special case that makes it memorable.

It applies to any vector, not just position: velocity, angular momentum, a magnetic field. Applying it to $\vec v$ is the second route through "Differentiate twice" in section 4, and applying it to the angular momentum of a rigid body is how unit 16 gets Euler's equations.

The third check is not a triviality. It says the angular acceleration needs no frame label — a fact used silently in every textbook and proved here in four characters.

Notation warning. The right-hand side mixes two things that look alike: $(\mathrm d\vec u/\mathrm dt)_{\text{rot}}$ means "differentiate the component list $u(t)$", full stop. The subscripts are not decoration, and dropping them is how the theorem gets misremembered as $\dot u = \dot u + \omega\times u$.

03 · frames

A vector painted on a turntable

A turntable turning anticlockwise, seen from above. The arrow $u$ is painted on it, so its components in the turning frame never change; seen from the room, its tip moves sideways at $\vec\omega\times\vec u$. Faint: $u$ a moment later.

Take the transport theorem with $u$ fixed in the rotating frame: $(\mathrm d\vec u/\mathrm dt)_{\text{rot}} = 0$. Then $$\left(\frac{\mathrm d\vec u}{\mathrm dt}\right)_{\!\text{in}} = \vec\omega\times\vec u .$$

For $u$ the position of a point on the turntable this is its velocity: perpendicular to both $\vec\omega$ and $\vec u$, of size $\omega\,\lVert u\rVert$ when $u$ lies in the plane of turning.

the general case is this plus linearity

A vector whose rotating components change has two contributions to its inertial rate: the change you see in the room, $(\mathrm d\vec u/\mathrm dt)_{\text{rot}}$, and the turning of the room's axes underneath it, $\vec\omega\times\vec u$. They add.

section 4

The frame that lies to you

Differentiate $x_{\mathrm{in}} = R(t)x_{\mathrm{rot}}$ twice, and read off what the room has added.

03 · the price

Differentiate twice

Write $x$ for $x_{\mathrm{rot}}$, so $x_{\mathrm{in}} = R(t)x$. Differentiating twice with the product rule: $$\ddot x_{\mathrm{in}} = \ddot R x + 2\dot R\dot x + R\ddot x .$$ Multiply by $R^{\!\top}$ to get inertial acceleration in rotating components: $$R^{\!\top}\ddot x_{\mathrm{in}} = \big(R^{\!\top}\ddot R\big)x + 2\Omega\dot x + \ddot x .$$

the one identity needed

$$R^{\!\top}\ddot R = \dot\Omega + \Omega^2 .$$

Proof. $\dot\Omega = \tfrac{\mathrm d}{\mathrm dt}(R^{\!\top}\dot R) = \dot R^{\!\top}\dot R + R^{\!\top}\ddot R$, and $\dot R^{\!\top}\dot R = (R\Omega)^{\!\top}(R\Omega) = \Omega^{\!\top}\Omega = -\Omega^2$ using $\dot R = R\Omega$ and antisymmetry. $\square$

Substituting, and translating every $\Omega$ into a cross product:

$$R^{\!\top}\ddot x_{\mathrm{in}} = \ddot x + 2\vec\omega\times\dot{\vec x} + \vec\omega\times(\vec\omega\times\vec x) + \dot{\vec\omega}\times\vec x .$$

Newton's law in the inertial frame is $m\,\ddot x_{\mathrm{in}} = F_{\mathrm{in}}$; multiplying by $R^{\!\top}$ turns both sides into rotating components. Rearranged, that is the boxed result on the next slide.

Four terms, and every one of them came from the product rule. Nothing was postulated.

The step that is easy to fumble. $\Omega^2 x = \hat\omega(\hat\omega x) = \vec\omega\times(\vec\omega\times\vec x)$, and $\dot\Omega x = \dot{\hat\omega}x = \widehat{\dot{\vec\omega}}\,x = \dot{\vec\omega}\times\vec x$ because the hat map is linear, so it commutes with $\mathrm d/\mathrm dt$.

A second route, for cross-checking. Apply the transport theorem twice instead. With $\vec v_{\text{in}} = \dot{\vec x} + \vec\omega\times\vec x$ (transport applied to $\vec x$), apply transport again to $\vec v_{\text{in}}$: $$\vec a_{\text{in}} = \frac{\mathrm d}{\mathrm dt}\big(\dot{\vec x}+\vec\omega\times\vec x\big) + \vec\omega\times\big(\dot{\vec x}+\vec\omega\times\vec x\big) = \ddot{\vec x} + \dot{\vec\omega}\times\vec x + 2\vec\omega\times\dot{\vec x} + \vec\omega\times(\vec\omega\times\vec x),$$ the same four terms. The factor 2 in the Coriolis term appears because $\vec\omega\times\dot{\vec x}$ arises once from each application. That is problem 9, and it is worth doing both ways once: the matrix route shows where the terms come from, the transport route shows why there are exactly four.

A frame that also translates. If the origins differ by $d(t)$, add $-m\ddot d$ to the right-hand side of the next slide's equation. Nothing else changes, because translation commutes with the whole calculation. A lift accelerating upwards is that term alone, and it is the seed of unit 30's equivalence principle.

03 · the price

The rotating-frame equation

theorem

In a frame whose attitude relative to an inertial frame is $R(t)$, with $\hat\omega = R^{\!\top}\dot R$, a particle of mass $m$ obeys $$m\,\vec a_{\mathrm{rot}} \;=\; \vec F \;\underbrace{-\;2m\,\vec\omega\times\vec v_{\mathrm{rot}}}_{\text{\small Coriolis}} \;\underbrace{-\;m\,\vec\omega\times(\vec\omega\times\vec r)}_{\text{\small centrifugal}} \;\underbrace{-\;m\,\dot{\vec\omega}\times\vec r}_{\text{\small Euler}} ,$$ with $\vec F$ the real force in rotating-frame components. The three extra terms are the fictitious forces.

dimensions

$m\omega v = \mathrm{M\,T^{-1}\,L\,T^{-1}}$ ✓; $m\omega^2 r$ ✓; $m\dot\omega r$ ✓. All $\mathrm{M\,L\,T^{-2}}$.

limit $\vec\omega\to0$

All three vanish: $m\vec a = \vec F$ returns. ✓

case: at rest, $\vec\omega$ constant

Only centrifugal survives, magnitude $m\omega^2 r_\perp$, pointing away from the axis. ✓

they are not forces

Nothing exerts them; no third-law partner exists; they are proportional to $m$ for the same reason weight is, which is a hint unit 30 will collect. They are the price of writing $\mathrm d/\mathrm dt$ in a basis that is itself moving.

Every symbol, once. $\vec r$ is the particle's position in rotating-frame components; $\vec v_{\mathrm{rot}} = \dot{\vec r}$ and $\vec a_{\mathrm{rot}} = \ddot{\vec r}$ are its component-wise derivatives — what an observer holding a stopwatch and a ruler in the room measures; $\vec F$ is the real force (gravity, tension, normal force), written in the same basis.

Why the names. Coriolis published the $2\vec\omega\times\vec v$ term in 1835 in a paper about waterwheels, not about weather. "Centrifugal" is Huygens' word, from 1659. The Euler term has no agreed name; some books call it the azimuthal or transverse force.

The structure worth remembering. Coriolis depends on velocity but not position; centrifugal on position but not velocity; Euler on position and on $\dot{\vec\omega}$ only. So on a steadily rotating Earth the Euler term is zero, and the two that remain split cleanly by what they depend on.

A number for the third check. On the equator, $r_\perp = R_{\mathrm E} = 6.378\times10^6\ \mathrm m$ and $\omega_{\mathrm E}^2 R_{\mathrm E} = 0.033916\ \mathrm{m\,s^{-2}}$, which is $0.346\%$ of $g$. Small, and not zero: it is the next slide.

03 · the price

Centrifugal: what a plumb line points at

A mass hanging at rest at latitude $\lambda$: $\vec v_{\mathrm{rot}} = 0$ and $\dot{\vec\omega} = 0$, so only the centrifugal force survives. With $R_{\mathrm E}$ the Earth's radius, it has magnitude $m\omega^2 R_{\mathrm E}\cos\lambda$ and points away from the polar axis, not from the centre.

definition

Effective gravity is what a plumb line and a spring balance actually measure: $$\vec g_{\text{eff}} = \vec g_{\text{true}} - \vec\omega\times(\vec\omega\times\vec r).$$ It is not radial and its magnitude is latitude-dependent.

"Vertical" is defined by $\vec g_{\text{eff}}$ — a building is plumb to the local effective vertical, so the deviation is invisible from inside. Which is why the effect shows up in the shape of the planet instead.

A section through the poles, at $\lambda = 45^\circ$. True gravity aims at the centre; the centrifugal term (drawn about 160 times too long) points straight away from the axis; their sum $\vec g_{\text{eff}}$ — the plumb line — aims at a point on the axis south of the centre.

Why $R_{\mathrm E}\cos\lambda$. The distance from the polar axis, not from the centre, is $R_{\mathrm E}\cos\lambda$: the dotted line in the figure. Decompose $-\vec\omega\times(\vec\omega\times\vec r)$ in the local east-north-up frame: its vertical component is $\omega^2R_{\mathrm E}\cos^2\lambda$ and its northward component is $-\omega^2R_{\mathrm E}\sin\lambda\cos\lambda$. At the poles and the equator the horizontal part vanishes, and it peaks at $45^\circ$, where $\sin\lambda\cos\lambda = \tfrac12$.

Reading the figure. The real centrifugal arrow at $45^\circ$ is $0.24\%$ of $g_{\text{true}}$, invisible at any honest scale, so it is drawn about 160 times too long. What the exaggeration preserves is the geometry: the arrow is perpendicular to the axis, so adding it swings $\vec g_{\text{eff}}$ towards the equator, and the extended plumb line misses the centre on the equator's side. At true scale the miss is $1.73\times10^{-3}$ rad, the tilt computed on the next slide.

03 · the price

Effective gravity, in numbers

numbers

At the equator: $\omega_{\mathrm E}^2R_{\mathrm E} = 0.0339\ \mathrm{m\,s^{-2}}$, straight up, so $g$ is reduced by $0.346\%$.

At $45^\circ$: the horizontal component is $\omega_{\mathrm E}^2R_{\mathrm E}\sin\lambda\cos\lambda = 0.01696\ \mathrm{m\,s^{-2}}$, tilting the plumb line by $$\frac{0.01696}{9.81} = 1.73\times10^{-3}\ \text{rad} = 0.099^\circ,$$ about $5.9$ arcminutes, towards the equator. That is problem 11.

Measured $g$ runs from $9.780\ \mathrm{m\,s^{-2}}$ at the equator to $9.832$ at the poles. Rotation explains $0.034$ of the $0.052$ gap; the rest is the equatorial bulge that the rotation itself produced.

limits

At a pole, $\cos\lambda = 0$: no centrifugal term, $\vec g_{\text{eff}} = \vec g_{\text{true}}$ ✓. On the equator it is all vertical, so the plumb line is radial again and only the weight changes ✓.

The bulge. A fluid Earth settles until its surface is perpendicular to $\vec g_{\text{eff}}$ everywhere — otherwise water would flow. That surface is an oblate spheroid, flattening $1/298$, equatorial radius $21\ \mathrm{km}$ larger than polar. The extra mass near the equator then pulls $g$ back up there, which is why rotation accounts for only about two thirds of the measured equator-to-pole difference. Unit 10 computes the bulge's own gravitational field from its quadrupole moment, and gets the famous $J_2$ coefficient.

Recomputed. $\omega_{\mathrm E}^2R_{\mathrm E} = (7.292116\times10^{-5})^2 \times 6.378137\times10^6 = 0.0339157\ \mathrm{m\,s^{-2}}$; as a fraction of $9.81$, $0.34573\%$; the $45^\circ$ tilt is $1.72863\times10^{-3}\ \mathrm{rad} = 0.09904^\circ = 5.94'$.

03 · the price

Coriolis: the force that turns cyclones

In the local frame (east, north, up), $\vec\omega = \omega(0,\cos\lambda,\sin\lambda)$. For a horizontal velocity $\vec v = (v_x, v_y, 0)$, $$-2\vec\omega\times\vec v = 2\omega\sin\lambda\,(v_y,\,-v_x,\,0) \;+\; 2\omega\cos\lambda\,(0,0,v_x).$$ The first part rotates $\vec v$ by $-90^\circ$ in the horizontal plane: in the northern hemisphere, a Coriolis force always to the right of the motion, of magnitude $2m\omega\sin\lambda\,v$. The second is vertical — the Eötvös effect — and is absorbed into $g_{\text{eff}}$.

Only $\omega\sin\lambda$, the vertical component of $\vec\omega$, turns a horizontal plane. Hold onto that: it is the whole of section 5.

$\vec\omega$ is a free vector, so copy it to Paris and split it there. The part along the local vertical, $\omega\sin\lambda$, turns horizontal motion sideways; the part along local north, $\omega\cos\lambda$, lies in the horizontal plane and can only push up or down.

The cross product, done slowly. With $\vec\omega = \omega(0,\cos\lambda,\sin\lambda)$ and $\vec v = (v_x,v_y,0)$: $\vec\omega\times\vec v = \omega(\cos\lambda\cdot 0 - \sin\lambda\cdot v_y,\ \sin\lambda\cdot v_x - 0,\ 0 - \cos\lambda\cdot v_x) = \omega(-v_y\sin\lambda,\ v_x\sin\lambda,\ -v_x\cos\lambda)$. Multiply by $-2$ to get the slide's expression.

Reading the figure. The angle between the polar axis and the local horizontal at latitude $\lambda$ is $\lambda$ itself — the same angle as at the centre, because the local vertical is the radius and the axis meets the equatorial plane at a right angle. So the components are $\omega\cos\lambda$ along north and $\omega\sin\lambda$ along up, which is the $(0, \cos\lambda, \sin\lambda)$ above. A horizontal $\vec v$ crossed with the northward part gives a vertical vector (the Eötvös term); crossed with the upward part it gives a horizontal one at right angles to $\vec v$.

Why "right" is not a convention. Take $\vec v$ due north, $(0,v,0)$, in the northern hemisphere. The horizontal Coriolis term is $2\omega\sin\lambda(v, 0, 0)$: due east, which is to the right of north. The sign flips with $\sin\lambda$, so it is to the left in the southern hemisphere.

03 · the price

Coriolis: why storms turn anticlockwise

a storm, in numbers

A $30\ \mathrm{m\,s^{-1}}$ wind at $45^\circ$ N: Coriolis acceleration $2\omega v\sin\lambda = 3.09\times10^{-3}\ \mathrm{m\,s^{-2}}$. Left alone it would bend the air onto an inertial circle of radius $$\frac{v}{2\omega\sin\lambda} = 291\ \mathrm{km}.$$

Air pushed inward by a low-pressure centre and deflected right instead circulates anticlockwise (northern hemisphere), clockwise in the south, and not at all at the equator, where $\sin\lambda = 0$.

The Rossby number $\mathcal{R}o = v/(2\omega\sin\lambda\,\ell)$, with $\ell$ the size of the flow, compares inertia with Coriolis: $0.29$ for a $1000\ \mathrm{km}$ storm, $\sim 10^4$ for water leaving a bath. Bathtubs are not Coriolis.

Parcels of air released round a low L (computed, with friction; parameters schematic). Each is pushed straight in, is turned to its own right the whole way, and the four together circle L anticlockwise.

What the figure computes. Each parcel obeys $\ddot{\vec r} = -G\hat r - 2\omega_z\hat z\times\dot{\vec r} - \gamma\dot{\vec r}$: a constant pull towards L, the horizontal Coriolis term of the previous slide, and a friction term that lets it spiral in rather than orbit. The constants are chosen to make the picture legible, not to match a real storm. What does not depend on them is the sense: deflection to the right of an inward push is anticlockwise circulation.

Numbers, recomputed. $2\omega v\sin 45^\circ$ with $v = 30$: $3.0937\times10^{-3}\ \mathrm{m\,s^{-2}}$; radius $v/(2\omega\sin45^\circ) = 290.9\ \mathrm{km}$; Rossby number with $\ell = 10^6\ \mathrm m$: $0.291$.

The bathtub. With $v \sim 0.1\ \mathrm{m\,s^{-1}}$ and $\ell \sim 0.1\ \mathrm m$, $\mathcal{R}o \sim 10^4$: Coriolis is four orders of magnitude below the residual swirl left by filling the bath. It has been measured, in 1962, by Shapiro at MIT — in a symmetric tank, filled and then left still for 24 hours. The circulation is real and it is about $10^{-4}$ of the relevant scale.

03 · the price

Euler: only while the frame is spinning up

$-m\dot{\vec\omega}\times\vec r$: the Euler force. Zero whenever $\vec\omega$ is constant, so it plays no part in anything on Earth over a day. It is what throws you backwards when a roundabout starts.

a roundabout

A playground roundabout of radius $2\ \mathrm m$ brought from rest to $1$ revolution per $4\ \mathrm s$ in $8\ \mathrm s$: $\dot\omega = (2\pi/4)/8 = 0.196\ \mathrm{s^{-2}}$, so a child at the rim feels a tangential $\dot\omega r = 0.393\ \mathrm{m\,s^{-2}}$ — and, once up to speed, a centrifugal $\omega^2 r = 4.93\ \mathrm{m\,s^{-2}}$, half a $g$ outwards.

checks

Dimensions: $[\dot\omega r] = \mathrm{T^{-2}L}$, an acceleration ✓. Direction: $-\dot{\vec\omega}\times\vec r$ is tangential and against the sense of spin-up, which is "thrown backwards" ✓. Limit $\dot\omega \to 0$: gone ✓.

$\dot{\vec\omega}$ for the Earth is not exactly zero — precession of the equinoxes, $26\,000$ years — but $|\dot{\vec\omega}|/\omega^2 \sim 10^{-7}$, far below anything this unit computes.

The roundabout numbers. One revolution per $4\ \mathrm s$ is $\omega = 2\pi/4 = 1.5708\ \mathrm{s^{-1}}$. Reaching it in $8\ \mathrm s$ gives $\dot\omega = 0.19635\ \mathrm{s^{-2}}$, so $\dot\omega r = 0.3927\ \mathrm{m\,s^{-2}}$ at $r = 2\ \mathrm m$; and $\omega^2 r = 4.9348\ \mathrm{m\,s^{-2}}$, which is $0.503\,g$. The tangential push lasts only while the roundabout speeds up; the outward one lasts as long as it turns. That matches the experience exactly, and it is the cheapest available test of the boxed equation.

Precession, for the estimate. The Earth's axis traces a cone in $25\,772$ years, so the direction of $\vec\omega$ turns at $2\pi/(25\,772\times3.156\times10^7\ \mathrm s) = 7.72\times10^{-12}\ \mathrm{s^{-1}}$, giving $|\dot{\vec\omega}| \approx \omega_{\mathrm E}\times7.72\times10^{-12} = 5.63\times10^{-16}\ \mathrm{s^{-2}}$. Against $\omega_{\mathrm E}^2 = 5.32\times10^{-9}\ \mathrm{s^{-2}}$ that is a ratio of $1.06\times10^{-7}$: the Euler term on Earth is seven orders of magnitude below the centrifugal one, and nothing in this unit can see it.

03 · the price

Three terms on one roundabout

The roundabout from above, turning anticlockwise and still speeding up. Child A sits on the rim: centrifugal out, Euler backwards. Child B walks outwards: Coriolis to the right of the walk (B also feels the other two, not drawn).

recap of the three

term	depends on	on Earth
Coriolis $-2m\vec\omega\times\vec v$	velocity	weather, Foucault
centrifugal $-m\vec\omega\times(\vec\omega\times\vec r)$	position	$g_{\text{eff}}$, the bulge
Euler $-m\dot{\vec\omega}\times\vec r$	$\dot{\vec\omega}$	nothing

Each arrow in the figure is the cross product read off: with $\vec\omega$ and $\dot{\vec\omega}$ out of the page, "$\vec\omega\times$" turns a horizontal vector anticlockwise by $90^\circ$, and the minus signs turn it back the other way.

Reading the figure. For A, at rest on the roundabout, $\vec v_{\mathrm{rot}} = 0$ and only the position terms act: $-\vec\omega\times(\vec\omega\times\vec r)$ is $\vec r$ turned twice by $+90^\circ$ and negated, which is $+\omega^2\vec r$, straight out; and $-\dot{\vec\omega}\times\vec r$ is $\vec r$ turned by $+90^\circ$ (forwards) and negated, so backwards — clockwise, against the spin-up. For B, walking out with velocity $\vec v$, $-2\vec\omega\times\vec v$ is $\vec v$ turned by $-90^\circ$: to B's right. Walking outwards on an anticlockwise roundabout you are pushed clockwise, which is the same "to the right" as for winds in the northern hemisphere, and for the same reason: the frame turns anticlockwise seen from above.

Why B is pushed back, physically. Each step outwards takes B to a ring that is moving faster round the circle than the ring B came from. Without a sideways push from the ground B lags behind it — which, in the roundabout's frame, is a push against the direction of turning.

03 · the price · example

Worked: falling east

worked example 3

Drop a mass from rest from $h = 100\ \mathrm m$ at the equator. Where does it land?

1 · set up. Local frame east-north-up. At $\lambda = 0$, $\vec\omega = \omega(0,1,0)$. Centrifugal is already inside $g_{\text{eff}}$, and $\dot{\vec\omega} = 0$, so $$\vec a = -g\hat z - 2\vec\omega\times\vec v .$$

2 · iterate once. To zeroth order $\vec v = (0,0,-gt)$. Then $$-2\vec\omega\times\vec v = -2\omega(0,1,0)\times(0,0,-gt) = (2\omega g t,\,0,\,0),$$ eastward — because the ground below is moving east more slowly than the released mass was.

The fall, computed from steps 3–4 (next slide): drift east against height, the horizontal axis in centimetres, so stretched about 3400-fold against the vertical. Dashed: where the mass would land if it simply kept the extra eastward speed $\omega h$ it started with.

Reading the figure. The solid curve is $x = \tfrac13\omega g t^3$ plotted against the height $h - \tfrac12 gt^2$: it leaves the tower vertically (the eastward push starts at zero, because the velocity does) and leans over more and more as the fall speeds up. The dashed curve is the naive estimate of the notes on the next slide, $x = \omega h t$, which drifts at a constant rate from the start and so ends further east. The true answer is two thirds of it.

Why iterating once is legitimate. The exact equation is $\ddot{\vec r} = -g\hat z - 2\vec\omega\times\dot{\vec r}$, a linear system this course cannot yet solve in closed form. But the correction is controlled by the dimensionless $\omega t \approx 3\times10^{-4}$: feeding the first-order eastward velocity back in produces a term smaller by that factor again, of order $10^{-5}\ \mathrm m$. Unit 21 gives this move a name — regular perturbation — and a theory. Here it is honest because the small parameter has been identified and the next term has been bounded.

03 · the price · example

Worked: falling east, integrated

worked example 3, continued

3 · integrate. $\ddot x = 2\omega g t$ with $x(0) = \dot x(0) = 0$ gives $x = \tfrac13\omega g t^3$.

4 · put in numbers. $t = \sqrt{2h/g} = \sqrt{200/9.81} = 4.5152\ \mathrm s$, so $$x = \tfrac13(7.2921\times10^{-5})(9.81)(4.5152)^3 = 0.02195\ \mathrm m,$$ 2.20 cm to the east. At latitude $\lambda$, multiply by $\cos\lambda$.

Checks. Dimensions: $\omega g t^3 = \mathrm{T^{-1}LT^{-2}T^3} = \mathrm L$ ✓. Sign: east, and eastward at every latitude, since $\cos\lambda > 0$ ✓. Limit $\omega\to0$: no deflection ✓. Size: $x/h = 2\times10^{-4}$, so the one-step iteration is justified ✓.

Reich, 1831, down a $158\ \mathrm m$ shaft at Freiberg: $28\ \mathrm{mm}$ measured against $27.5\ \mathrm{mm}$ predicted.

The physical reading, which is exact and not a hand-wave. Released from $100\ \mathrm m$ up, the mass has the eastward speed of a point $100\ \mathrm m$ further from the axis, namely $\omega(R_{\mathrm E}+h)$, while the ground has $\omega R_{\mathrm E}$. Over $t$ seconds the excess $\omega h$ carries it $\omega h t = 7.29\times10^{-5}\times100\times4.515 = 3.29\ \mathrm{cm}$ east — which is larger than the right answer, because the mass also falls inwards and loses some of that lead. The exact bookkeeping is the $\tfrac13$ in $\tfrac13\omega gt^3$ against the naive $\omega h t = \tfrac12\omega gt^3$: the deflection is two thirds of the angular-momentum estimate.

History. Newton proposed the experiment to Hooke in 1679 and got the direction right. It was first measured convincingly by Ferdinand Reich in 1831, down a $158\ \mathrm m$ mineshaft in Freiberg: $106$ drops, mean deflection $28\ \mathrm{mm}$ east, against a predicted $27.5\ \mathrm{mm}$. This is problem 10, at another latitude.

section 5

The Foucault pendulum

The running example, closed: one number, predicted from $R^{\!\top}\dot R$ alone.

03 · foucault

The pendulum's two horizontal equations

A Foucault pendulum is a bob on a wire of length $L$, free to swing in any vertical plane. Displace it horizontally by $\vec r = (x,y)$, with $\lVert\vec r\rVert \ll L$. The wire's tension is $T \approx mg$, and its horizontal component is $-T\vec r/L$: a restoring force $-m\omega_0^2\vec r$ with $\omega_0^2 = g/L$.

Adding the horizontal Coriolis term and writing $\omega_z = \omega\sin\lambda$: $$\ddot{\vec r} = -\omega_0^2\,\vec r \;-\; 2\omega_z\,\hat z\times\dot{\vec r}. $$ In components, $\ \ddot x = -\omega_0^2x + 2\omega_z\dot y$, $\ \ddot y = -\omega_0^2 y - 2\omega_z\dot x$.

$\omega_0 = \sqrt{g/L}$ is forced by dimensions alone (unit 00): $g$ and $L$ admit exactly one frequency. That the motion is sinusoidal is unit 06's business, and is not needed here.

Side view, angle exaggerated. $T$ and its horizontal part form a triangle similar to the one made by $L$ and $r$, so the horizontal part is $-T\,r/L$.

Reading the figure. The wire, the vertical through the pivot and the horizontal displacement $r$ make a right triangle with hypotenuse $L$; the tension and its two components make a similar one with hypotenuse $T$. So the horizontal component is $T\sin\theta = T\,r/L$, pointing back towards the vertical, exactly — the small-angle approximation enters only in $T \approx mg$.

The tension, more carefully. With the wire at angle $\theta$ to the vertical, the vertical balance is $T\cos\theta = mg$ (to lowest order in $\dot\theta^2$) and the horizontal force is $-T\sin\theta\,\hat r = -mg\tan\theta\,\hat r$. For small $\theta$, $\tan\theta \approx \lVert \vec r\rVert/L$, giving $-mg\vec r/L$. The neglected corrections are relative $O(\theta_0^2)$, and $\theta_0 = 6/67 = 0.0896$ gives $\theta_0^2 = 0.8\%$ — far larger than the Coriolis corrections above. The frequency shift they produce is smaller, $\theta_0^2/16 = 0.05\%$ (the standard anharmonic result, derived in unit 21), but it depends on amplitude, and that is why real Foucault pendulums need long wires and small swings: an amplitude-dependent frequency makes an elliptical orbit precess on its own, with no connection to the Earth's rotation at all.

Notation. $\hat z\times\dot{\vec r}$ for a horizontal $\dot{\vec r} = (\dot x,\dot y,0)$ is $(-\dot y,\dot x,0)$: a $+90^\circ$ rotation. The minus sign in front makes the deflection clockwise from above in the northern hemisphere.

03 · foucault

What the horizontal equations leave out

what has been dropped, and why

Centrifugal: constant, absorbed into $g_{\text{eff}}$ and hence into $\omega_0$.

Vertical Coriolis $2\omega\cos\lambda\,\dot x$: changes the tension by a relative $2\omega\cos\lambda\,v/g = 2.3\times10^{-5}$ at the Panthéon's peak speed $2.30\ \mathrm{m\,s^{-1}}$. Ignorable.

Vertical motion of the bob: second order in the amplitude.

What survives is one number, $\omega_z = \omega\sin\lambda$ — the component of $\vec\omega$ along the local vertical. The horizontal part of $\vec\omega$ cannot turn a horizontal plane: that was the split drawn on "Coriolis: the force that turns cyclones".

And the two timescales are far apart: $\omega_z/\omega_0 = 1.435\times10^{-4}$. The plane turns through less than a tenth of a degree per swing.

03 · foucault

Undo the rotation, and Coriolis is gone

theorem

View the motion from a frame $F$ turning at $\vec\Omega_F = -\omega_z\hat z$ relative to the ground. In $F$ the equation is $$\ddot{\vec r}_F = -(\omega_0^2 + \omega_z^2)\,\vec r_F$$ — no Coriolis term. So a swing that starts as a line stays in that line's vertical plane in $F$, and relative to the ground the plane turns at $-\omega_z\hat z$: clockwise from above in the northern hemisphere, at $\omega\sin\lambda$.

The bob's path on the floor for a swing that is a straight line in $F$ (computed; the turning rate exaggerated about 175-fold). Each pass is a line through the centre, and the line turns clockwise: a quarter turn over the ten swings drawn.

Why this argument and not a substitution. The usual route puts $x + \mathrm iy = w\,\mathrm e^{-\mathrm i\omega_z t}$ and reads off the answer. That is unit 06's machinery and it is not available yet. The route above uses only this unit's own transport theorem — applied a second time, to a frame invented purely to kill a term — and it makes the reason transparent: the Coriolis term is the signature of a rotating frame, so it can be removed by rotating back. Unit 06 will do it the other way as an exercise in complex exponentials, and get the same $\omega_0^2 + \omega_z^2$.

Reading the figure. In $F$ the bob goes back and forth along one line, the dashed one. Mapping each instant back to the ground turns that line by $-\omega_z t$, so on the floor the successive passes fan out clockwise. To make the fan visible in ten swings the ratio $\omega_z/\omega_1$ is set to $0.025$; at the Panthéon it is $1.4\times10^{-4}$ and the same quarter turn takes about $1740$ swings, just under eight hours — the overview's figure.

Sign, once more. Seen from above in the northern hemisphere, $\omega_z > 0$ and the plane turns by $-\omega_z t$: clockwise. In Buenos Aires it is anticlockwise. At the equator it does not turn at all, and there is a pendulum in Quito built to show exactly that.

03 · foucault

Undo the rotation: the proof

proof

Apply the transport theorem with the ground frame in the "inertial" role. With $\vec\Omega_F = -\omega_z\hat z$ constant, $\vec a_{\text{ground}} = \vec a_F + 2\vec\Omega_F\times\vec v_F + \vec\Omega_F\times(\vec\Omega_F\times\vec r)$ and $\vec v_{\text{ground}} = \vec v_F + \vec\Omega_F\times\vec r$. Substitute both into $\vec a_{\text{ground}} = -\omega_0^2\vec r - 2\omega_z\hat z\times\vec v_{\text{ground}}$:

$2\vec\Omega_F\times\vec v_F = -2\omega_z\hat z\times\vec v_F$ on the left cancels $-2\omega_z\hat z\times\vec v_F$ on the right. What is left uses $\hat z\times(\hat z\times\vec r) = -\vec r$ for horizontal $\vec r$: the two remaining terms are $-\omega_z^2\vec r_F$ on the left and $-2\omega_z^2\vec r_F$ on the right. Collecting, $\ddot{\vec r}_F = -(\omega_0^2+\omega_z^2)\vec r_F$. $\square$

why "a fixed plane"

Let $\hat n$ be horizontal with $\hat n\cdot\vec r_F(0) = \hat n\cdot\dot{\vec r}_F(0) = 0$, and $f = \hat n\cdot\vec r_F$. Then $\ddot f = -\omega_1^2 f$, writing $\omega_1^2 = \omega_0^2+\omega_z^2$, with $f(0)=\dot f(0) = 0$. Unit 00's trick — multiply by $\dot f$ and integrate — gives $\dot f^2 + \omega_1^2f^2 = 0$, constant and equal to $0$. So $f\equiv 0$. ✓

The correction $\omega_z^2$ is a relative $(\omega_z/\omega_0)^2 = 2.1\times10^{-8}$ on the swing frequency. It is written down because it is free, and then never used again.

The cancellation, written out. Left-hand side after substituting: $\vec a_F - 2\omega_z\hat z\times\vec v_F + \omega_z^2\hat z\times(\hat z\times\vec r)$. Right-hand side: $-\omega_0^2\vec r - 2\omega_z\hat z\times\vec v_F - 2\omega_z\hat z\times(-\omega_z\hat z\times\vec r)$. The middle terms are identical and cancel. With $\hat z\times(\hat z\times\vec r) = -\vec r$, the left keeps $-\omega_z^2\vec r$ and the right keeps $-2\omega_z^2\vec r$, so $\vec a_F = -\omega_0^2\vec r - 2\omega_z^2\vec r + \omega_z^2\vec r$.

The energy trick is unit 00's, exactly. There it turned $m\ddot x = F(x)$ into a first integral by multiplying by $\dot x$. Here the same multiplication proves a uniqueness statement without any theory of ODEs: $\dot f^2 + \omega_1^2 f^2$ is a non-negative constant that starts at zero. Unit 01's Picard–Lindelöf would also do it, and is not a prerequisite of this unit.

03 · foucault · example

Worked: 31.8 hours at the Panthéon

worked example 4 — the running example, closed

1 · the rate. The Panthéon is at $\lambda = 48.8462^\circ$ N, so $$\omega_z = \omega_{\mathrm E}\sin\lambda = 7.29212\times10^{-5}\times0.752946$$ $$= 5.49057\times10^{-5}\ \mathrm{s^{-1}} .$$

2 · the Foucault period. $$T_{\mathrm F} = \frac{2\pi}{\omega_z} = 1.14436\times10^{5}\ \mathrm s = \mathbf{31.79\ h},$$ i.e. $11.33^\circ$ per hour, clockwise seen from above.

3 · what you would see. One swing takes $T_0 = 2\pi/\omega_0 = 16.42\ \mathrm s$, in which the plane turns $\omega_zT_0 = 9.0\times10^{-4}\ \mathrm{rad} = 0.052^\circ$. At the $6\ \mathrm m$ end of the swing that moves the turning point sideways by $5.4\ \mathrm{mm}$ — a couple of centimetres a minute, visible against a scale.

The subscript in $T_{\mathrm F}$ keeps the Foucault period apart from the wire's tension $T$. Checks against every latitude: next slide.

Every ingredient came from $R^{\!\top}\dot R$: no new physics entered after slide "The rotating-frame equation".

Every number, recomputed. $\omega_{\mathrm E} = 2\pi/86\,164.0905 = 7.292116\times10^{-5}\ \mathrm{s^{-1}}$; $\sin(48.8462^\circ) = 0.7529458$; $\omega_z = 5.490568\times10^{-5}\ \mathrm{s^{-1}}$; $T_{\mathrm F} = 114\,435.98\ \mathrm s = 31.7878\ \mathrm h$; $360^\circ/31.7878\ \mathrm h = 11.3251^\circ\,\mathrm h^{-1}$. One swing period $2\pi/\omega_0 = 16.4204\ \mathrm s$, in which the plane turns $9.0157\times10^{-4}\ \mathrm{rad}$; the end of a $6\ \mathrm m$ half-amplitude swing therefore shifts by $6\times9.0157\times10^{-4} = 5.41\ \mathrm{mm}$ between successive arrivals at the same end, one full period apart — about $20\ \mathrm{mm}$ a minute.

Why "about 32 hours" is the right thing to compare with. The figure quoted for Paris in the contemporary accounts is $31\ \mathrm h\ 47\ \mathrm{min}$, and $31.7878\ \mathrm h$ is $31\ \mathrm h\ 47.3\ \mathrm{min}$. The rounder "about 32 hours" is what the popular retellings repeat. A modern pendulum cannot do much better than a few percent, because air drag, wire torsion and the elliptical drift of "What Foucault actually saw" all contaminate the measurement.

The formula is worth memorising in this form: $T_{\mathrm F} = (\text{one sidereal day})/\sin\lambda$. Everything else on this deck exists to prove that the denominator is $\sin\lambda$ and not, say, $\cos\lambda$ or 1.

03 · foucault

The Foucault period at every latitude

$T_{\mathrm F} = T_{\text{sid}}/\sin\lambda$, computed from $15^\circ$ to the pole. It never drops below one sidereal day, and it climbs without bound towards the equator, where $\sin\lambda \to 0$.

checks

$T_{\mathrm F} = T_{\text{sid}}/\sin\lambda$, so $T_{\mathrm F} \ge 23.93\ \mathrm h$ always ✓.

At the pole, $T_{\mathrm F} = 23.93\ \mathrm h$ — the Earth simply turns under a fixed plane ✓.

At the equator, $T_{\mathrm F} = \infty$ ✓.

Foucault reported "about 32 hours" in 1851 ✓.

Dimensions: $[2\pi/(\omega\sin\lambda)] = \mathrm T$ ✓.

03 · foucault

What Foucault actually saw

the pendulum does not swing in a plane

Released from rest relative to the ground, the bob is not at rest in $F$: $\vec v_F(0) = -\vec\Omega_F\times\vec r(0) = \omega_z\hat z\times\vec r(0)$, of size $\omega_z A$. So in $F$ the orbit is a thin ellipse, not a segment, with

$$\frac{\text{minor}}{\text{major}} \approx \frac{\omega_z}{\omega_0} = 1.4\times10^{-4},$$ which at $A = 6\ \mathrm m$ is a minor semi-axis of $0.86\ \mathrm{mm}$. (That the orbit is an ellipse is unit 06's isotropic oscillator; only its aspect ratio is needed here, and it follows from the initial data.)

The bob never passes through the centre: each pass misses it by the minor semi-axis, and each turning point is a sharp cusp, where the bob is momentarily at rest on the floor.

The path on the floor when the bob is released from rest (computed; the turning rate exaggerated about 1000-fold, which fattens the ellipse to match). Cusps at the ends; the dashed circle at the centre is never entered.

The initial velocity, from the transport theorem. $\vec v_{\text{ground}} = \vec v_F + \vec\Omega_F\times\vec r$. Setting $\vec v_{\text{ground}}(0) = 0$ with $\vec\Omega_F = -\omega_z\hat z$ gives $\vec v_F(0) = \omega_z\hat z\times\vec r(0)$, perpendicular to the displacement and of magnitude $\omega_z A = 5.4906\times10^{-5}\times 6 = 3.29\times10^{-4}\ \mathrm{m\,s^{-1}}$. For an oscillator of frequency $\omega_1 \approx \omega_0$, a transverse velocity $u$ at the turning point produces a transverse excursion $u/\omega_0 = 8.6\times10^{-4}\ \mathrm m$ — the $0.86\ \mathrm{mm}$ quoted.

Reading the figure. In $F$ the path is the thin ellipse $x_F = A\cos\omega_1t$, $y_F = (\omega_z A/\omega_1)\sin\omega_1 t$, the isotropic oscillator's orbit for this initial data (that it is an ellipse is quoted from unit 06). Carried back to the floor by the turn $-\omega_z t$, it becomes the rosette. At each end the ellipse's sideways velocity $\omega_z A$ is exactly cancelled by the floor's turning, which is why the bob comes to rest there and the path has a cusp. The ratio $\omega_z/\omega_1$ is drawn as $0.15$; the real $1.4\times10^{-4}$ would make the central circle $0.86\ \mathrm{mm}$ in radius inside a $12\ \mathrm m$ swing.

Honest labelling. The claim that the orbit is exactly an ellipse requires solving $\ddot{\vec r} = -\omega_1^2\vec r$, which needs unit 06. What is proved here is the statement of "Undo the rotation, and Coriolis is gone" — the swing plane is fixed in $F$ when the initial velocity is along the initial displacement — plus the size of the initial transverse velocity, which is exact.

03 · foucault

Burning the thread, and the Charron ring

The bob never passes through the centre. Foucault knew, and launched his pendulum by burning a thread holding it aside — the only way to release it with no sideways velocity at all.

Even then, the amplitude-dependent frequency of a real pendulum — a $0.05\%$ shift here — makes any residual ellipse precess on its own, in a direction set by which way the bob was nudged. Every serious installation fights this: the Paris pendulum uses a "Charron ring", a metal collar that the wire touches at the bottom of each swing and which bleeds off the elliptical component.

The physics of "Undo the rotation" is exact to the order stated. This slide is why a demonstration is harder than a derivation.

the drift that competes

A real pendulum's frequency depends on its amplitude, so an elliptical orbit precesses on its own, at a rate proportional to its area: $\tfrac38\,ab\,\omega_0/L^2$ for semi-axes $a$, $b$ (cited, Airy 1851). A $1\ \mathrm{cm}$ minor axis at the Panthéon gives $3.5\%$ of the Foucault rate; the $0.86\ \mathrm{mm}$ one forced by release, $0.3\%$.

Why the ellipse is a problem and not a curiosity. An anharmonic pendulum's frequency depends on amplitude, so the two perpendicular components of an elliptical orbit run at slightly different rates, and the ellipse's axis rotates at a rate proportional to its area: $\tfrac38\,ab\,\omega_0/L^2$ for semi-axes $a$ and $b$. Cited: that is Airy's result of 1851, and it is the anharmonic perturbation theory of unit 21 applied to two coupled coordinates. At the Panthéon, a $1\ \mathrm{cm}$ minor axis against the $6\ \mathrm m$ major one gives $1.92\times10^{-6}\ \mathrm{s^{-1}}$ — $3.5\%$ of the Foucault rate, already fatal to a careful measurement — and since it grows in proportion to the minor axis it reaches the Foucault rate itself at about $30\ \mathrm{cm}$. The $0.86\ \mathrm{mm}$ ellipse forced by the release above contributes only $0.3\%$. The Charron ring works by interrupting the orbit once per swing near the centre, where the elliptical component is fastest.

03 · summary

What to keep

the four lines worth memorising

$D(g\circ f)(a) = Dg(f(a))\circ Df(a)$

$\vec a = \dot v\hat T + (v^2/\rho)\hat N$

$\Omega = R^{\!\top}\dot R \in \mathfrak{so}(3)$, and $\Omega u = \vec\omega\times\vec u$

$(\mathrm d/\mathrm dt)_{\text{in}} = (\mathrm d/\mathrm dt)_{\text{rot}} + \vec\omega\times$

The last two are one fact. Everything in sections 4 and 5 is that fact applied twice and then counted.

What this unlocks.

04: the derivative of a scalar field is a covector; the gradient is what an inner product makes of it.
08: the cross product, with its identities, conserving angular momentum.
16: $\mathfrak{so}(3)$ again, as the configuration velocity of a rigid body, with $\Omega$ contracted against an inertia tensor.
25–26: the same question one dimension up — what survives a change of frame — where the group is Lorentz rather than $SO(3)$.
30: fictitious forces proportional to $m$, taken seriously.

03 · checkpoint

Done when

checkpoint

Starting only from $x_{\mathrm{in}} = R(t)\,x_{\mathrm{rot}}$ with $R \in SO(3)$, and with nothing else on the page, you can:

derive $m\vec a_{\mathrm{rot}} = \vec F - 2m\vec\omega\times\vec v_{\mathrm{rot}} - m\vec\omega\times(\vec\omega\times\vec r) - m\dot{\vec\omega}\times\vec r$, identifying $\vec\omega$ from $R^{\!\top}\dot R$ and saying why that matrix is antisymmetric;
predict the Foucault period at a latitude you are given, to within an hour.

Falsifiable both ways: a missing factor of 2, a wrong sign, or $\cos\lambda$ in place of $\sin\lambda$ each fails it. Record it with ./mr done 03.

Problems

./mr problems 03 — 13 problems, about 5 hours. Hints and worked solutions fold under each problem on the same page.

04 work and line integrals (needs this unit) · 05 flux and gravity · 16 rigid bodies, where $\mathfrak{so}(3)$ returns · 25 the same question for Maxwell.

Where things are: curves, frames, and the frame that lies to you

One pendulum, five sections

Differentiating in several variables

Partials are not enough

Even every direction is not enough

The derivative is a linear map

The Jacobian, and when partials suffice

The chain rule, once

Clairaut, and the function that breaks it

Curves

Vector-valued functions and two product rules

Constant speed means sideways acceleration

Arc length and the unit tangent

Curvature, and the frame a curve carries

The Frenet–Serret formulas

Acceleration has exactly two parts

Two parts, checked

Worked: a helical ramp

Worked: the ramp's curvature and torsion

Three dimensions are not three one-dimensional problems

Frames that turn

What a frame is, and what $SO(3)$ is

One arrow, two lists of numbers

$\Omega = R^{\!\top}\dot R$ is antisymmetric

$\mathfrak{so}(3)$ is $\mathbb R^3$: the hat map and the cross product

Worked: reading $\vec\omega$ off $R$

The transport theorem

A vector painted on a turntable

The frame that lies to you

Differentiate twice

The rotating-frame equation

Centrifugal: what a plumb line points at

Effective gravity, in numbers

Coriolis: the force that turns cyclones

Coriolis: why storms turn anticlockwise

Euler: only while the frame is spinning up

Three terms on one roundabout

Worked: falling east

Worked: falling east, integrated

The Foucault pendulum

The pendulum's two horizontal equations

What the horizontal equations leave out

Undo the rotation, and Coriolis is gone

Undo the rotation: the proof

Worked: 31.8 hours at the Panthéon

The Foucault period at every latitude

What Foucault actually saw

Burning the thread, and the Charron ring

What to keep

Done when

Problems

Next