Part A · the equation of motion · spine

The central-force problem, and the two conservation laws that solve it

Six equations, two conserved quantities, and what is left is a one-dimensional problem you already know how to read off a graph — with the Kepler orbit falling out in closed form at the end.

needs 04, 06~2.5 h slides + 5 h problemsproblems: ./mr problems 08

This is the unit where the course's first hard problem is solved completely. Two bodies pulling on each other is twelve first-order equations; by the end of these slides the answer is a formula for the shape of the path and a formula for the time along it, and both have been checked against a real comet.

How to read this deck

Six sections. Section 1 finds the first conserved quantity, angular momentum, and with it the fact that the motion is flat. Section 2 turns two bodies into one. Section 3 uses the second conserved quantity, energy, to collapse the problem to a single graph. Section 4 makes the change of variable that solves it, and section 5 reads Kepler's three laws off the answer. Section 6 is the extra conserved vector that only the inverse-square law has.

Words with a dotted underline are glossary terms; click one, or press G for the index. N opens the notes, where the long computations sit under a Full proof heading. A solid thin underline names a piece of the running example and shows it on hover.

New mathematics here: the cross product's identities (unit 03 defined the product and deferred them), the use of first integrals to lower the order of a system, quadrature in two variables at once, and the Binet substitution $u = 1/r$ — which is worth learning carefully, because unit 37 adds exactly one term to the equation it produces and gets general relativity.

08 · overview

Two conserved quantities, six sections

Two bodies pull on each other along the line joining them, with a potential $V(r)$ of their separation $r$ — for gravity $V = -k/r$ with $k = Gm_1m_2 > 0$. What path does each follow?

reduce · sections 1–3

1. The force lies along $\vec r$, so its torque $\vec r\times\vec F$ is zero and $\vec L = \vec r\times\vec p$ never changes. Since $\vec r\perp\vec L$ always, the orbit is flat, and $r^2\dot\theta$ is fixed.

2. The separation $\vec r = \vec r_1 - \vec r_2$ obeys $\mu\ddot{\vec r} = f(r)\hat r$: one body, of reduced mass $\mu = m_1m_2/(m_1+m_2)$, in a fixed field.

3. In $E = \tfrac12\mu(\dot r^2 + r^2\dot\theta^2) + V$, replace $\dot\theta$ by $L/\mu r^2$: $E = \tfrac12\mu\dot r^2 + V_{\mathrm{eff}}(r)$, with $V_{\mathrm{eff}} = V + L^2/2\mu r^2$. A one-dimensional problem, read off one graph.

solve · sections 4–6

4. Ask for the shape $r(\theta)$, not $r(t)$. With $u = 1/r$ and $' = \mathrm d/\mathrm d\theta$, the chain rule gives $\dot r = -(L/\mu)\,u'$, and for $V = -k/r$ the radial equation becomes $u'' + u = \mu k/L^2$: an oscillator with a constant drive.

5. So $u = (1 + e\cos\theta)/p$ — a conic with the force centre at a focus, $p$ a length and $e$ a pure number. Kepler's three laws are read off it.

6. Differentiate $\vec A = \vec p\times\vec L - \mu k\hat r$: for the inverse square, and only for it, the two terms cancel. A third conserved vector, pointing at perihelion.

The running example. Halley's comet, all the way through: from one perihelion distance and one speed come $h = L/\mu$ and the energy, and from those every other number about the orbit, ending with the period it keeps. The next slide draws it.

Where each step is done. Step 1 is Torque, and the derivative of $\vec r\times\vec p$, Central forces, and the first conservation law and The motion is planar, including when it is not. Step 2 is The problem splits, exactly. Step 3 is Energy, with $\dot\theta$ eliminated, and its graph is Every kind of orbit, read off one graph. Step 4 is The substitution $u = 1/r$ and Inverse square: the right-hand side goes constant; step 5 is Solving it, with unit 06 in one line, then Conics, defined before they are met and section 5's three Kepler slides. Step 6 is $\dot{\vec A} = \vec 0$, by differentiating and The orbit equation, in one line, which gets the conic a second time with no differential equation at all.

The letters. $f(r)$ is the radial component of the force, so $f = -V'$; for gravity $f = -k/r^2$. $\vec p = \mu\dot{\vec r}$ is the momentum and $L$ the (constant) magnitude of $\vec L$. The $p$ in step 5 is a different thing — a length, the semi-latus rectum — and the deck writes the momentum's size as $\lVert\vec p\rVert$ wherever the two could meet.

The order is chosen so that each section removes exactly one difficulty. Sections 1 and 2 reduce the dimension of the problem without solving anything. Section 3 solves it in unit 00's sense — reduced to integrals — which is enough for every qualitative question and for some numbers. Section 4 does better, and produces a formula, but only because the inverse-square law happens to make one particular equation linear.

Two habits to carry out of here. First: a conserved quantity is not a fact to be admired, it is one dimension removed from a system of ODEs, and the arithmetic of that removal is on the slide "Six to two". Second: the right change of variable is worth more than any amount of cleverness with the wrong one — $r(t)$ has no elementary formula and $r(\theta)$ has a two-line one.

Problem sheet: ./mr problems 08, about five hours. P6 and P7 are the two the checkpoint is built from.

08 · overview · the running example

The running example: Halley's comet, to scale

Halley's orbit (thick) and the planets' orbits (dashed), distances to scale. The ellipse is $3.9$ times longer than it is wide; the inset enlarges the inner $1.6$ AU, where perihelion falls inside the orbit of Venus. The real orbit is tilted to the planets' plane and runs the other way round; flattening it changes no distance from the Sun.

what the deck computes, in order

From perihelion, $q$ and $v_q$: the angular momentum per unit mass $h$ and the energy per unit mass $\varepsilon$ (section 1).
From $h$ and $\varepsilon$: both turning points, so aphelion comes out as a prediction (section 3).
The shape: $p$ and $e$ (section 4).
The size $a$, and from it the period, $75.3$ yr (section 5).

Why a comet: the Sun's pull on it is as close to a pure two-body, inverse-square problem as the sky offers, and its orbit is so eccentric that every formula is tested far from the circle.

What to look for in the figure. The Sun sits almost at the right-hand tip of the ellipse, not at its centre: the centre is $17.25$ AU away, between the orbits of Saturn and Uranus. Aphelion, at $35.08$ AU, is beyond Neptune's $30.1$; perihelion, at $0.586$ AU, is inside Venus's $0.723$. The comet covers the inner part of the drawing in a few months and spends decades crawling round the far end — Kepler's second law, which section 5 proves.

How it was drawn. From the deck's own elements, $a = 17.8325$ AU and $e = 0.967137$, parametrised as $\big(a(\cos E - e),\ a\sqrt{1-e^2}\sin E\big)$ with the Sun at the origin; the planets as circles of their mean distances (Jupiter $5.20$, Saturn $9.58$, Uranus $19.2$, Neptune $30.1$ AU), cropped at the frame. The orbit's inclination to the ecliptic is $162^\circ$ — retrograde, tilted $18^\circ$ from the planets' plane — and the drawing lays it flat. Distances from the Sun are unchanged by that, which is all a central-force problem sees.

section 1

Angular momentum, and why the orbit is flat

One vector identity, one derivative, and a three-dimensional problem becomes a two-dimensional one.

08 · angular momentum

The identities unit 03 owed you

Unit 03 defined $\vec u\times\vec w := \hat u\,\vec w$, the action of the antisymmetric matrix the hat map assigns to $\vec u$, and promised the identities here.

the four that get used

For $\vec a,\vec b,\vec c,\vec d\in\mathbb R^3$:

$\vec a\times\vec b = -\,\vec b\times\vec a$, and $\vec a\cdot(\vec a\times\vec b) = 0$;
scalar triple product: $\vec a\cdot(\vec b\times\vec c) = \det[\vec a\,\vec b\,\vec c] = \vec c\cdot(\vec a\times\vec b)$;
BAC–CAB rule: $\vec a\times(\vec b\times\vec c) = \vec b\,(\vec a\cdot\vec c) - \vec c\,(\vec a\cdot\vec b)$;
Lagrange's identity: $(\vec a\times\vec b)\cdot(\vec c\times\vec d) = (\vec a\cdot\vec c)(\vec b\cdot\vec d) - (\vec a\cdot\vec d)(\vec b\cdot\vec c)$.

full proofs in the notes (N)

1, on the slide

Antisymmetry is $\hat a\,\vec b = -\hat b\,\vec a$, checked on basis vectors. Then $\vec a\cdot(\hat a\vec b) = \vec a^{\!\top}\hat a\,\vec b = 0$, because $\vec a^{\!\top}\hat a$ is the transpose of $\hat a^{\!\top}\vec a = -\hat a\vec a = 0$.

a corollary used below

Unit 03's product rule holds for every bilinear operation with the order kept: $\frac{\mathrm d}{\mathrm dt}(\vec u\times\vec w) = \dot{\vec u}\times\vec w + \vec u\times\dot{\vec w}$.

The other corollary — that $\lVert\vec a\times\vec b\rVert$ is an area — gets the next slide.

Full proof of 2, 3 and 4

2. Write $\hat b$ out: its rows are $(0,-b_3,b_2)$, $(b_3,0,-b_1)$, $(-b_2,b_1,0)$, so $\vec b\times\vec c = (b_2c_3-b_3c_2,\ b_3c_1-b_1c_3,\ b_1c_2-b_2c_1)$. Dotting with $\vec a$ gives $a_1(b_2c_3-b_3c_2) - a_2(b_1c_3-b_3c_1) + a_3(b_1c_2-b_2c_1)$, which is the cofactor expansion of $\det[\vec a\,\vec b\,\vec c]$ along its first row. The determinant is unchanged by a cyclic permutation of its columns (two transpositions, two sign changes), which gives the second equality and the two more it implies.

3. Both sides are linear in each of $\vec b$ and $\vec c$, so it is enough to check $\vec b = \vec e_i$, $\vec c = \vec e_j$. If $i = j$ the left side is $\vec a\times\vec 0 = \vec 0$ and the right side is $\vec e_i a_i - \vec e_i a_i = \vec 0$. If $i\ne j$, relabelling cyclically and using antisymmetry reduces to $i = 1, j = 2$: $\vec e_1\times\vec e_2 = \vec e_3$, so the left side is $\vec a\times\vec e_3 = (a_2, -a_1, 0)$, and the right side is $\vec e_1(\vec a\cdot\vec e_2) - \vec e_2(\vec a\cdot\vec e_1) = a_2\vec e_1 - a_1\vec e_2 = (a_2,-a_1,0)$. Equal. $\square$

4. By 2, $(\vec a\times\vec b)\cdot(\vec c\times\vec d) = \vec d\cdot\big((\vec a\times\vec b)\times\vec c\big)$. By 1 and 3, $(\vec a\times\vec b)\times\vec c = -\,\vec c\times(\vec a\times\vec b) = -\vec a(\vec c\cdot\vec b) + \vec b(\vec c\cdot\vec a)$. Dotting with $\vec d$ gives $(\vec a\cdot\vec c)(\vec b\cdot\vec d) - (\vec a\cdot\vec d)(\vec b\cdot\vec c)$. $\square$

On the name. BAC–CAB is a mnemonic for the order on the right-hand side, not a reason. The reason is that $\vec a\times(\vec b\times\vec c)$ is perpendicular to $\vec b\times\vec c$, hence lies in the plane of $\vec b$ and $\vec c$, so it must be some combination $\alpha\vec b + \beta\vec c$; the identity says which.

08 · angular momentum

The cross product measures area

corollary of identity 4

Put $\vec c = \vec a$, $\vec d = \vec b$ in Lagrange's identity: $$\lVert\vec a\times\vec b\rVert^2 = \lVert\vec a\rVert^2\lVert\vec b\rVert^2 - (\vec a\cdot\vec b)^2 .$$ With $\varphi$ the angle between them, $\vec a\cdot\vec b = \lVert\vec a\rVert\lVert\vec b\rVert\cos\varphi$, so

$$\lVert\vec a\times\vec b\rVert = \lVert\vec a\rVert\lVert\vec b\rVert\sin\varphi ,$$

the area of the parallelogram $\vec a$ and $\vec b$ span: base $\lVert\vec a\rVert$ times height $\lVert\vec b\rVert\sin\varphi$.

where it is used

Half of it is the triangle on $\vec a$ and $\vec b$. With $\vec a = \vec r$ and $\vec b = \Delta t\,\dot{\vec r}$, that triangle is the area the radius sweeps in a short time, and $\vec r\times\dot{\vec r} = \vec L/\mu$: Kepler's second law, section 5.

Length and direction of $\vec a\times\vec b$: its length is the shaded area, and it points out of the page — perpendicular to both $\vec a$ and $\vec b$ by identity 1, and with the sign that makes $(\vec a, \vec b, \vec a\times\vec b)$ right-handed.

The corollary, carefully. For nonzero $\vec a,\vec b$ the angle $\varphi\in[0,\pi]$ between them is defined by $\cos\varphi = \vec a\cdot\vec b/\lVert\vec a\rVert\lVert\vec b\rVert$, which Cauchy–Schwarz puts in $[-1,1]$. Then identity 4 gives $\lVert\vec a\times\vec b\rVert^2 = \lVert\vec a\rVert^2\lVert\vec b\rVert^2(1 - \cos^2\varphi)$, and $\sin\varphi \ge 0$ on $[0,\pi]$, so the root is taken with a plus sign. Unit 03 proved this by expanding components; here it drops out of identity 4 in one line. A parallelogram with sides $\lVert\vec a\rVert,\lVert\vec b\rVert$ and included angle $\varphi$ has area $\lVert\vec a\rVert\lVert\vec b\rVert\sin\varphi$ — base times perpendicular height, the dashed segment in the figure.

Why "right-handed" in the caption. $\det[\vec a\,\vec b\,\vec a\times\vec b] = (\vec a\times\vec b)\cdot(\vec a\times\vec b) = \lVert\vec a\times\vec b\rVert^2 \ge 0$ by identity 2, so the three vectors, in that order, have positive orientation whenever $\vec a\times\vec b\ne\vec 0$. In a drawing with $x$ to the right and $y$ up, positive orientation puts the third vector out of the page.

08 · angular momentum

Torque, and the derivative of $\vec r\times\vec p$

definition

For a particle at $\vec r$ with momentum $\vec p = m\dot{\vec r}$, about the origin:

$$\vec L := \vec r\times\vec p,\qquad \vec N := \vec r\times\vec F .$$

Angular momentum $\vec L$ and torque $\vec N$. Both depend on the choice of origin: they are statements about a point, not about the particle alone.

theorem

$\dfrac{\mathrm d\vec L}{\mathrm dt} = \vec N$.

Proof. $\dot{\vec L} = \dot{\vec r}\times\vec p + \vec r\times\dot{\vec p} = \dot{\vec r}\times m\dot{\vec r} + \vec r\times\vec F = \vec 0 + \vec N$, using the product rule and $\vec u\times\vec u = \vec 0$. $\square$

dimensions

$[\vec L] = \mathsf{L}\cdot\mathsf{MLT^{-1}} = \mathsf{ML^2T^{-1}}$, the dimension of action. $[\vec N] = \mathsf{L}\cdot\mathsf{MLT^{-2}} = \mathsf{ML^2T^{-2}}$, an energy — and $\mathsf{ML^2T^{-1}}/\mathsf{T} = \mathsf{ML^2T^{-2}}$ ✓.

what $\vec L$ actually is

Like the curl in unit 04, $\vec L$ is a pseudovector: under a reflection it does not transform as a vector, because the cross product is built from the antisymmetric part of something. Unit 34 identifies it correctly — as a $2$-form, $\vec r\wedge\vec p$ — and the anomaly disappears. Until then, keep to right-handed frames.

$\dot{\vec L} = \vec N$ is Newton's second law with $\vec r\times$ applied to it, and no new physics. What is new is which forces make the right-hand side vanish.

Why $\vec L$ and not $\dot{\vec L}$ is the natural object. Nothing forces us to look at $\vec r\times\vec p$ rather than any other combination. The reason to look is empirical and old: a stone on a string speeds up as the string is shortened, and $r^2\dot\theta$ is what stays fixed while it does. Unit 13 replaces this accident with a theorem — angular momentum is conserved because the Lagrangian is invariant under rotation — and the three components of $\vec L$ are the three conserved quantities of the three-parameter group $SO(3)$ from unit 03.

Origin dependence, concretely. Shift the origin by $\vec s$: $\vec L' = (\vec r - \vec s)\times\vec p = \vec L - \vec s\times\vec p$. So a particle moving in a straight line at constant velocity has $\vec L = \vec 0$ about any point on that line and $\vec L \ne\vec 0$ about any other. Conservation is likewise a statement about a chosen origin; in this unit the origin is always the force centre, and section 2 says what that means when both bodies move.

The name "torque". $\vec N$ is what you feel as twist. The magnitude is $rF\sin\varphi$, which is force times the perpendicular distance from the origin to the line of action — the lever arm. The letter $\vec N$ is Goldstein's; $\vec\tau$ and $\vec M$ are equally common, and $\tau$ is already the proper time of units 26 onwards, so $\vec N$ it is.

08 · angular momentum

Central forces, and the first conservation law

definition

A force field on $\mathbb R^3\setminus\{\vec 0\}$ is central if it points along the radius everywhere:

$$\vec F(\vec r) = f(\vec r)\,\hat r,\qquad \hat r = \vec r/r,\ \ r = \lVert\vec r\rVert .$$

It is isotropic if in addition $f$ depends on $\vec r$ only through $r$, so that $\vec F(\vec r) = f(r)\hat r$. Unless said otherwise, "central" means both from here on.

theorem

For a central force, $\vec L$ is constant in time.

Proof. $\vec N = \vec r\times f\hat r = (f/r)\,(\vec r\times\vec r) = \vec 0$, so $\dot{\vec L} = \vec 0$. $\square$

two different hypotheses

Only the direction condition is needed for $\vec L$ to be conserved: $f$ may depend on the angles, on $\dot r$, on $t$. Isotropy is a separate hypothesis, and it is what makes the field conservative, by unit 04: $f(r)\hat r = -\nabla V$ with $V(r) = -\int^r f$. Energy conservation needs it; angular momentum does not.

check: is $f(r)\hat r$ a gradient?

$\partial r/\partial x_i = x_i/r$, so $\nabla r = \hat r$, and the chain rule (unit 03) gives $\nabla V(r) = V'(r)\,\hat r$. Choosing $V' = -f$ on $r>0$ makes $f(r)\hat r = -\nabla V$ on all of $\mathbb R^3\setminus\{\vec 0\}$. The potential is written down explicitly, so no topology is needed.

The two hypotheses, kept apart. This is the one place in the unit where a definition is doing two jobs and it pays to notice. Direction alone buys planarity and $r^2\dot\theta =$ const. Isotropy buys a potential $V(r)$ and hence $E$. Every result from section 3 on uses both; section 1 uses only the first. A magnetic field is the standard example of the gap in the other direction: it is neither, and neither $\vec L$ nor the naive energy behaves as you would guess.

Where the force laws come from. Newton's law of gravitation, $\vec F = -Gm_1m_2\hat r/r^2$, and Coulomb's law, $\vec F = q_1q_2\hat r/4\pi\varepsilon_0 r^2$, are both central and isotropic and both inverse square. Why the exponent is $-2$ — a field spreading over spheres in three dimensions — is unit 05's business and is not needed here: in this unit the exponent is an input, and the question is what it forces.

When the hypotheses fail slightly. A satellite over an oblate Earth feels a force with a small component towards the equatorial plane. That force is not central, so its torque about the centre is not zero and $\vec L$ turns slowly: the orbital plane precesses. Its radial part is also not exactly inverse square, so the perihelion drifts within the plane. Both drifts are unit 10's subject, and the second is why the Runge–Lenz slides matter: the vector that is exactly conserved here becomes a slowly turning one there.

08 · angular momentum

The motion is planar, including when it is not

theorem

Let $\vec F$ be central and let $\vec L$ be the (constant) angular momentum about the centre.

If $\vec L\ne\vec 0$, the motion stays in the plane through the origin with normal $\vec L$, for all time.
If $\vec L = \vec 0$, the motion stays on a fixed ray from the origin: it is purely radial.

proof

First case: $\vec L\cdot\vec r = (\vec r\times\vec p)\cdot\vec r = 0$ by identity 1, at every time, and $\vec L$ is a fixed nonzero vector. So $\vec r(t)$ lies in $\vec L^{\perp}$, a plane, always. $\square$

Second case: $\vec r\times\dot{\vec r} = \vec 0$ means $\dot{\vec r} = \lambda\vec r$ for a scalar $\lambda(t)$ wherever $\vec r\ne\vec 0$. Then $\dot r = (\vec r\cdot\dot{\vec r})/r = \lambda r$, so $$\frac{\mathrm d}{\mathrm dt}\hat r = \frac{\dot{\vec r}}{r} - \frac{\dot r}{r^2}\vec r = \lambda\hat r - \lambda\hat r = \vec 0 . \qquad\square$$

$\vec L = \vec r\times\vec p$ is perpendicular to both $\vec r$ and $\vec p$. Because $\vec L$ is fixed, the plane through the centre with normal $\vec L$ is fixed, and $\vec r$ can never leave it. Perspective sketch.

Reading the figure. The first case of the proof is the whole picture: $\vec r\cdot\vec L = 0$ at every instant says the position lies in the shaded plane, and the plane does not move because $\vec L$ does not. The momentum lies in the same plane for the same reason, $\vec p\cdot\vec L = 0$, so the body is never even pointed out of it.

Why "for all time" needs saying. The argument shows $\vec r(t)\in\vec L^{\perp}$ at each $t$, which is the whole claim only because $\vec L$ does not move. If the force were not central, $\vec L$ would turn, and $\vec r$ would still be perpendicular to it at each instant while wandering out of any fixed plane. Perpendicularity to a moving vector is no constraint at all in three dimensions.

The second case, in words. $\vec L = \vec 0$ says position and velocity are parallel. The computation shows the unit vector $\hat r$ is then constant, so the motion is along a fixed line — and since the force is central it stays central, so the line is preserved. The proof is valid only while $\vec r\ne\vec 0$; the particle reaching the centre in finite time is a genuine singularity of the equation, where unit 01's existence theorem stops applying because the field is not Lipschitz there, and indeed not defined.

08 · angular momentum

Plane polar coordinates, derived once

Work in the orbital plane. Write $\hat r(\theta) = (\cos\theta,\sin\theta)$ and $\hat\theta(\theta) = (-\sin\theta,\cos\theta)$: an orthonormal pair at every point, turning with the particle. By the chain rule (unit 03),

$$\frac{\mathrm d\hat r}{\mathrm dt} = \dot\theta\,\hat\theta, \qquad \frac{\mathrm d\hat\theta}{\mathrm dt} = -\dot\theta\,\hat r .$$

Then $\vec r = r\hat r$ and two differentiations give

velocity and acceleration $$\vec v = \dot r\,\hat r + r\dot\theta\,\hat\theta,$$ $$\vec a = (\ddot r - r\dot\theta^2)\,\hat r + (r\ddot\theta + 2\dot r\dot\theta)\,\hat\theta .$$

$\hat r$ and $\hat\theta$ are attached to the moving point and turn with it. The velocity splits into $\dot r\,\hat r$, the rate of moving out, and $r\dot\theta\,\hat\theta$, the rate of going round. Inset: as $\theta$ grows by $\Delta\theta$, $\hat r$ changes by about $\Delta\theta\,\hat\theta$ — the first derivative rule.

Full derivation

$\frac{\mathrm d}{\mathrm dt}\hat r(\theta(t)) = \frac{\mathrm d\hat r}{\mathrm d\theta}\dot\theta = (-\sin\theta,\cos\theta)\dot\theta = \dot\theta\hat\theta$, and likewise $\frac{\mathrm d\hat\theta}{\mathrm d\theta} = (-\cos\theta,-\sin\theta) = -\hat r$. Then $\vec v = \frac{\mathrm d}{\mathrm dt}(r\hat r) = \dot r\hat r + r\dot\theta\hat\theta$ and $$\vec a = \ddot r\hat r + \dot r\dot\theta\hat\theta + (\dot r\dot\theta + r\ddot\theta)\hat\theta + r\dot\theta(-\dot\theta\hat r) = (\ddot r - r\dot\theta^2)\hat r + (r\ddot\theta + 2\dot r\dot\theta)\hat\theta .$$

Reading the inset. $\hat r(\theta)$ and $\hat r(\theta+\Delta\theta)$ are two radii of the unit circle; the chord between their tips has length $2\sin(\Delta\theta/2)\approx\Delta\theta$ and points, in the limit, along the tangent — which is $\hat\theta$. Dividing by $\Delta t$ gives $\mathrm d\hat r/\mathrm dt = \dot\theta\,\hat\theta$. The same picture turned through a right angle gives $\mathrm d\hat\theta/\mathrm dt = -\dot\theta\,\hat r$.

The $2\dot r\dot\theta$ term. It is the Coriolis term of unit 03, seen from the inertial frame rather than from the rotating one: $\hat r,\hat\theta$ is a frame turning with angular velocity $\dot\theta\hat z$, and the extra transverse acceleration is what unit 03's transport theorem produces. It is also why $r\ddot\theta + 2\dot r\dot\theta = \frac1r\frac{\mathrm d}{\mathrm dt}(r^2\dot\theta)$, an identity worth checking by hand: it turns the transverse equation of motion straight into $\dot L = rF_\theta$, and for a central force into $\dot L = 0$ again.

08 · angular momentum

What the flat orbit buys, in polar coordinates

what has been bought

Three first integrals — the three components of $\vec L$ — for the price of one line. Two of them (the direction of $\vec L$) fix the plane and cut the six-dimensional phase space of one particle in $\mathbb R^3$ down to four dimensions. The third (the magnitude) is used in section 3.

angular momentum, in these coordinates

$\hat r\times\hat\theta = \hat z$, the normal to the plane, so

$$\vec L = m\,r\hat r\times(\dot r\hat r + r\dot\theta\hat\theta) = m r^2\dot\theta\,\hat z .$$

One scalar, $L = mr^2\dot\theta$, signed: positive anticlockwise. Constant, by the first conservation law. Hence $\dot\theta$ never changes sign, and $\theta$ is a monotone function of $t$ — which section 4 needs.

two checks

Circle at constant rate: $\dot r = \ddot r = \ddot\theta = 0$ gives $\vec a = -r\dot\theta^2\hat r$, the centripetal $v^2/r$ inwards — unit 03's Frenet result, recovered.

Straight radial line: $\dot\theta = 0$ gives $\vec a = \ddot r\hat r$, one-dimensional motion ✓.

the degenerate case is real

$\vec L = \vec 0$ is a body dropped straight at the centre. It is not a curiosity: it is the fall $m\ddot x = -k/x^2$ of unit 00's checkpoint notes, and the third-law slide gives its fall time as the half-period of a squashed ellipse — the same number unit 00 got by quadrature, $C = \pi/2\sqrt2$.

On "six-dimensional phase space". Unit 01's reformulation: one particle in $\mathbb R^3$ has state $(\vec r,\dot{\vec r})\in\mathbb R^6$, and the equation of motion is a first-order system there. Each first integral confines the solution to a level set of one dimension less. Two of $\vec L$'s three components, used as the conditions $\vec r\cdot\hat n = 0$ and $\dot{\vec r}\cdot\hat n = 0$ with $\hat n = \vec L/L$, take $6$ down to $4$.

Only the transverse part counts. In $\vec L = m\,r\hat r\times\vec v$ the radial part of the velocity drops out, because $\hat r\times\hat r = \vec 0$; what is left is $r$ times the transverse speed $r\dot\theta$. In the previous slide's figure, $L/m$ is $r$ times the length of the $r\dot\theta\,\hat\theta$ component — and, by the area corollary, twice the area of the triangle on $\vec r$ and $\vec v$.

Why monotone $\theta$ matters. With $L\ne0$, $\dot\theta = L/mr^2$ has the sign of $L$ and never vanishes, so $t\mapsto\theta(t)$ is strictly monotone and therefore invertible. That is exactly unit 00's move — invert a monotone function to change independent variable — and it is what licenses treating $r$ as a function of $\theta$ in section 4. Without it the orbit could double back and $r(\theta)$ would not be a function.

08 · angular momentum · example

Worked: one comet, two measurements, one constant

Halley's angular momentum

Given. At perihelion the distance is $q = 0.586025\ \mathrm{AU}$ and the speed is $v_q = 54.5698\ \mathrm{km\,s^{-1}}$. One astronomical unit is $1\ \mathrm{AU} = 1.495978707\times10^{11}\ \mathrm{m}$, so $q = 8.76681\times10^{10}\ \mathrm{m}$.

Step 1 — why perihelion is the easy place. $r$ is at a minimum there, so $\dot r = 0$, so $\vec v$ has no radial part and $\vec v\perp\vec r$. Then $\lVert\vec r\times\vec v\rVert = qv_q\sin 90^\circ = qv_q$.

Step 2 — the constant. Divide $\vec L = \mu\vec r\times\dot{\vec r}$ by the reduced mass, which section 2 shows is the comet's own mass to one part in $10^{16}$: $$h := \frac{L}{\mu} = q\,v_q = 4.78403\times10^{15}\ \mathrm{m^2\,s^{-1}} .$$

Step 3 — and the energy, for free. With $GM_\odot = 1.32712440\times10^{20}\ \mathrm{m^3\,s^{-2}}$, $\tfrac12v_q^2 = 1.48893\times10^{9}$ and $GM_\odot/q = 1.51381\times10^{9}$, so $$\varepsilon := \frac{E}{\mu} = \tfrac12 v_q^2 - \frac{GM_\odot}{q} = -2.48740\times10^{7}\ \mathrm{J\,kg^{-1}} .$$

checks

Dimensions. $[h] = \mathsf{L}\cdot\mathsf{LT^{-1}} = \mathsf{L^2T^{-1}}$ ✓ — angular momentum per unit mass. $[\varepsilon] = \mathsf{L^2T^{-2}}$, energy per unit mass ✓.

Sign. $\varepsilon < 0$: the comet is bound, which it had better be, since it comes back. It is bound by only $1.7\%$ of its perihelion kinetic energy — the two big numbers in step 3 nearly cancel, so carry the extra digits.

Limit. Escape from $q$ needs $\sqrt{2GM_\odot/q} = 55.0237\ \mathrm{km\,s^{-1}}$ (unit 04's escape speed). The comet has $54.5698$, which is $99.175\%$ of it.

These two numbers, $h$ and $\varepsilon$, are the whole orbit. Section 4 turns them into a shape and section 5 into a timetable, and nothing else about the comet is ever needed.

Where the inputs come from. $q$ and $v_q$ are not independent measurements in practice: an orbit is fitted to many astrometric positions at once and $q$, $e$, $a$ come out together. They are used here as if measured directly because the point is the arithmetic, not the astrometry. The two values are chosen to reproduce the published elements of the 1986 apparition — $q \approx 0.586$ AU, $e \approx 0.967$, $a \approx 17.8$ AU — and are given to six figures so that the cancellation below does not eat the answer. Every later number on these slides is computed from these two.

Why divide by $\mu$ at all. Halley's mass is known only roughly (estimates run from about $2$ to $3\times10^{14}$ kg), and it cancels out of every orbital quantity. Working with $h = L/\mu$ and $\varepsilon = E/\mu$ throughout means no orbit computation ever needs it. This is a general habit for the gravitational two-body problem, where $k = GMm$ and the combination that appears is always $k/\mu$, and it is the reason the spacecraft literature speaks of the "gravitational parameter" $GM$ and never of $G$ and $M$ separately.

The arithmetic of step 3. $\tfrac12 v_q^2 = \tfrac12(5.45698\times10^4)^2 = 1.488931\times10^9$ and $GM_\odot/q = 1.327124\times10^{20}/8.76681\times10^{10} = 1.513805\times10^9$. The difference is $-2.48740\times10^7$. The two terms agree to within $1.7\%$, so a relative error $\delta$ in either input becomes roughly $60\delta$ in $\varepsilon$: inputs good to six figures leave $\varepsilon$ good to about four, and four-figure inputs would leave it good to about two. This is the standard hazard of a nearly parabolic orbit, and it is why comet elements are quoted as $q$ and $e$ rather than as $a$: $a = -GM/2\varepsilon$ inherits the whole cancellation.

section 2

Two bodies become one

Neither body sits still. A change of coordinates makes that stop mattering.

08 · two bodies

The problem splits, exactly

Two bodies, masses $m_1, m_2$, interacting only with each other by a central force depending on the separation: $\vec F_{12} = f(r)\hat r$ with $\vec r := \vec r_1 - \vec r_2$, and $\vec F_{21} = -\vec F_{12}$ by Newton's third law (unit 00). Set $M := m_1 + m_2$ and

$$\vec R := \frac{m_1\vec r_1 + m_2\vec r_2}{M}, \qquad \mu := \frac{m_1m_2}{m_1+m_2} .$$

theorem

In the coordinates $(\vec R,\vec r)$ the equations of motion become

$$M\ddot{\vec R} = \vec 0, \qquad \mu\,\ddot{\vec r} = f(r)\,\hat r ,$$

and the energy and angular momentum split to match:

$$T = \tfrac12 M\lVert\dot{\vec R}\rVert^2 + \tfrac12\mu\lVert\dot{\vec r}\rVert^2,$$ $$\vec L_{\text{tot}} = M\vec R\times\dot{\vec R} + \mu\,\vec r\times\dot{\vec r} .$$

Masses in the ratio $1:3$, seen from the centre of mass ($\vec R = \vec 0$). Then $\vec r_1 = \tfrac{m_2}{M}\vec r$ and $\vec r_2 = -\tfrac{m_1}{M}\vec r$: the two bodies are always on opposite sides of CM, at distances in the ratio $3:1$, and each traces a scaled copy of the relative orbit. Dashed: the pair at two other instants.

08 · two bodies

Why the centre of mass drops out

proof of the second equation

$\ddot{\vec r} = \ddot{\vec r}_1 - \ddot{\vec r}_2 = \dfrac{\vec F_{12}}{m_1} - \dfrac{\vec F_{21}}{m_2} = \vec F_{12}\Big(\dfrac1{m_1}+\dfrac1{m_2}\Big) = \dfrac{\vec F_{12}}{\mu}$, since $\dfrac1{m_1}+\dfrac1{m_2} = \dfrac{M}{m_1m_2} = \dfrac1\mu$. $\square$

The first is $M\ddot{\vec R} = m_1\ddot{\vec r}_1 + m_2\ddot{\vec r}_2 = \vec F_{12}+\vec F_{21} = \vec 0$: unit 00's momentum conservation, restated.

what this means

The centre of mass is an inertial observer. In its frame $\vec R\equiv\vec 0$ and the whole of the rest of this unit is one particle of mass $\mu$ at position $\vec r$ in a fixed central field — a genuine one-body problem, not an approximation.

Full proof of the splitting of $T$ and $\vec L$

Invert the definitions: $\vec r_1 = \vec R + \frac{m_2}{M}\vec r$ and $\vec r_2 = \vec R - \frac{m_1}{M}\vec r$ (check: the mass-weighted average is $\vec R$, and the difference is $\vec r$). Then $$2T = m_1\Big\lVert\dot{\vec R} + \tfrac{m_2}{M}\dot{\vec r}\Big\rVert^2 + m_2\Big\lVert\dot{\vec R} - \tfrac{m_1}{M}\dot{\vec r}\Big\rVert^2 .$$ Expanding, the cross terms are $2\dot{\vec R}\cdot\dot{\vec r}\big(m_1\frac{m_2}{M} - m_2\frac{m_1}{M}\big) = 0$, and the square terms give $M\lVert\dot{\vec R}\rVert^2 + \frac{m_1m_2^2 + m_2m_1^2}{M^2}\lVert\dot{\vec r}\rVert^2 = M\lVert\dot{\vec R}\rVert^2 + \mu\lVert\dot{\vec r}\rVert^2$, using $m_1m_2^2+m_2m_1^2 = m_1m_2M$. The same substitution in $\vec L_{\text{tot}} = m_1\vec r_1\times\dot{\vec r}_1 + m_2\vec r_2\times\dot{\vec r}_2$ kills the cross terms by the identical cancellation. $\square$

The potential energy needs no splitting: it depends on $\vec r_1 - \vec r_2 = \vec r$ alone, so it is already a function of the relative coordinate. Hence the total energy is $E_{\text{cm}} + \big[\tfrac12\mu\lVert\dot{\vec r}\rVert^2 + V(r)\big]$ with the bracket conserved on its own, and it is the bracket that this unit calls $E$.

What is being used. Newton's third law in its strong form — equal, opposite and along the line joining the bodies. The weak form (equal and opposite) gives $M\ddot{\vec R} = \vec 0$ but not the conservation of $\vec L_{\text{tot}}$; the strong form is needed for the internal torques to cancel. Unit 00 flagged this. Magnetic forces between moving charges violate even the weak form, and the missing momentum is in the field — unit 24.

08 · two bodies

Reduced mass, and how much it actually matters

reduced mass $$\mu = \frac{m_1m_2}{m_1+m_2},\qquad \frac1\mu = \frac1{m_1} + \frac1{m_2} .$$

Half the harmonic mean: $\mu < \min(m_1,m_2)$ always, since $1/\mu$ exceeds both $1/m_1$ and $1/m_2$.

the two limits

$m_2\to\infty$: $\mu\to m_1$ and $\vec R\to\vec r_2$. The heavy body is nailed down and the light one orbits it: the naive picture, recovered as a limit rather than assumed.

$m_1 = m_2 = m$: $\mu = m/2$. Two equal stars each trace an ellipse of half the size of the relative orbit, about their common centre.

how big is the correction?

Gravity is $f(r) = -k/r^2$, i.e. $V(r) = -k/r$, with $k = Gm_1m_2$, so the equation $\mu\ddot{\vec r} = -k\hat r/r^2$ involves the masses only through $k/\mu = GM$ with $M = m_1+m_2$. Section 5 shows the period is $T = 2\pi\sqrt{a^3/GM}$, so using $GM_\odot$ instead of $G(M_\odot + m)$ makes it too long by the factor $(1+m/M_\odot)^{1/2}$.

body	$m/M_\odot$	error in $T$
Jupiter, $T = 11.87$ yr	$9.548\times10^{-4}$	2.07 d
Earth, $T = 1$ yr	$3.003\times10^{-6}$	47 s
Halley, $T = 75.3$ yr	$1.11\times10^{-16}$	0.13 µs

So for the comet, $\mu$ is its own mass and $M$ is the Sun's, to far better than anything is measured. For Jupiter the correction is two days, far larger than the uncertainty in its observed period.

The arithmetic. $(1+x)^{-1/2} - 1 \approx -x/2$ for small $x$. Jupiter: $x = 1/1047.35 = 9.5479\times10^{-4}$, so the fractional shift is $-4.7705\times10^{-4}$, and $4.7705\times10^{-4}\times11.8708\ \mathrm{yr}\times365.25 = 2.068$ days. Earth: $x = 3.0035\times10^{-6}$ gives $1.5017\times10^{-6}\times365.25\times86400 = 47.4$ s. Halley: $x = 2.2\times10^{14}/1.989\times10^{30} = 1.106\times10^{-16}$, and $\tfrac12 x\times75.31\ \mathrm{yr} = 1.3\times10^{-7}$ s.

Why this is not a detail. Kepler's third law as Kepler wrote it says $T^2/a^3$ is the same for every planet. It is not — it is the same only to the accuracy with which the planets' masses are negligible, and Jupiter's departure is the largest in the solar system. Newton's version with $G(M+m)$ is what is actually true, and it is how the masses of binary stars are measured: observe $a$ and $T$, and $M_1 + M_2$ follows with no other input. That is the single most productive equation in observational astronomy.

Reduced mass elsewhere. The same combination appears in the two-body problem of quantum mechanics — hydrogen's lines sit lower than an infinitely heavy nucleus would put them by the fraction $m_e/m_p\approx1/1836$, deuterium's by about half that, so the two isotopes' lines differ by about $2.7\times10^{-4}$, which is how Urey found deuterium in 1931 — and in two free masses joined by a spring of stiffness $\kappa$, whose separation oscillates at $\sqrt{\kappa/\mu}$. It is always the same algebra: $1/\mu = 1/m_1 + 1/m_2$ because accelerations add when forces are equal and opposite.

section 3

The effective potential

Angular momentum is spent to buy a one-dimensional problem. Then one graph answers every qualitative question.

08 · effective potential

Energy, with $\dot\theta$ eliminated

One body of mass $\mu$ in the central isotropic field $f(r)\hat r = -V'(r)\hat r$. Unit 04's work–energy theorem gives the conserved $E = \tfrac12\mu\lVert\vec v\rVert^2 + V(r)$, and the polar velocity formula gives $\lVert\vec v\rVert^2 = \dot r^2 + r^2\dot\theta^2$. Now use $L = \mu r^2\dot\theta$ to remove $\dot\theta$:

$$E = \tfrac12\mu\dot r^2 + \tfrac12\mu r^2\dot\theta^2 + V(r) = \tfrac12\mu\dot r^2 + \frac{L^2}{2\mu r^2} + V(r).$$

the effective potential $$V_{\mathrm{eff}}(r) := V(r) + \frac{L^2}{2\mu r^2},\qquad E = \tfrac12\mu\dot r^2 + V_{\mathrm{eff}}(r).$$

A function of the single variable $r$, with $L$ a fixed parameter: one graph for each value of $L$.

this is unit 00's case 3

$E = \tfrac12\mu\dot r^2 + V_{\mathrm{eff}}(r)$ is exactly the first integral unit 00 got for one-dimensional motion. Everything that unit proved about reading motion off a potential — turning points, bounded and unbounded components, equilibria, separatrices — applies verbatim, with $V_{\mathrm{eff}}$ in place of $V$.

what $V_{\mathrm{eff}}$ is not

It is not the potential energy of any force acting in the plane. The $L^2/2\mu r^2$ term is kinetic energy — the transverse part — wearing a potential's clothes. Two consequences: the total energy is still $E$, not $E$ plus something; and $V_{\mathrm{eff}}$ is valid only along an orbit with that value of $L$. Change $L$ and the graph changes.

Where the trick comes from. $\theta$ appears in the equations only through $\dot\theta$, never on its own: it is what unit 12 will call a cyclic coordinate, and unit 13 will identify as the symmetry that produced $L$ in the first place. Whenever a coordinate is cyclic, its conjugate momentum is conserved and can be used to eliminate the velocity, leaving a system in the remaining coordinates with an extra term in the potential. That is the whole content of the effective potential; the general procedure goes by the name Routh reduction.

Compare unit 03. Unit 03's centrifugal force in a frame turning at rate $\Omega$ has magnitude $\mu\Omega^2 r$, outwards, and at fixed $\Omega$ it is minus the gradient of $-\tfrac12\mu\Omega^2r^2$. Here the frame's rate is the orbit's own, $\Omega = \dot\theta = L/\mu r^2$, so the same force is $L^2/\mu r^3$ — and at fixed $L$ it is minus the gradient of $+L^2/2\mu r^2$. Same force, opposite-signed potentials, because a different quantity is held fixed while differentiating. The safe rule: derive it once from $E$, as here, and never from a force diagram.

Dimensions. $[L^2/\mu r^2] = \mathsf{M^2L^4T^{-2}}/(\mathsf{M}\cdot\mathsf{L^2}) = \mathsf{ML^2T^{-2}}$, an energy ✓.

08 · effective potential

The same radial equation, by two routes

Route 1 — differentiate the energy. $E$ is constant, so

$$0 = \mu\dot r\ddot r + V_{\mathrm{eff}}'(r)\dot r \ \Longrightarrow\ \mu\ddot r = -V_{\mathrm{eff}}'(r)$$

wherever $\dot r\ne0$, and by continuity everywhere.

Route 2 — Newton in polar coordinates. The radial component of $\mu\vec a = f(r)\hat r$ is $\mu(\ddot r - r\dot\theta^2) = f(r)$, so

$$\mu\ddot r = f(r) + \mu r\dot\theta^2 = -V'(r) + \frac{L^2}{\mu r^3} .$$

they agree

$V_{\mathrm{eff}}'(r) = V'(r) - \dfrac{L^2}{\mu r^3}$, so $-V_{\mathrm{eff}}' = -V' + \dfrac{L^2}{\mu r^3}$ ✓ — route 2's right-hand side exactly.

The $L^2/\mu r^3$ term is the centrifugal barrier: an outward push, growing faster than any inverse-square attraction as $r\to0$, and it is the reason a body with $L\ne0$ never reaches the centre.

the barrier can be beaten

"Faster than inverse square" is a statement about $r^{-3}$ against $r^{-2}$. A force $-c/r^3$ gives $V = -c/2r^2$ and $V_{\mathrm{eff}} = (L^2/2\mu - c/2)/r^2$: if $c > L^2/\mu$ the barrier is gone and the particle spirals in. Attractions steeper than inverse cube always win, and P4's stability condition $n > -3$ is the same statement.

Why the two routes are worth doing both. Route 1 is short but it hides where the $L^2/\mu r^3$ came from; route 2 shows it is the $-r\dot\theta^2$ of the polar acceleration, i.e. the centripetal term moved to the other side of the equation. Getting the same answer from the energy and from the force is also the standard check that no factor of two has been dropped in $V_{\mathrm{eff}}$, which is the commonest error in this subject.

"By continuity everywhere". Route 1's division by $\dot r$ is illegitimate exactly at the turning points. But $\mu\ddot r + V_{\mathrm{eff}}'(r)$ is a continuous function of $t$ which vanishes wherever $\dot r\ne0$, so by continuity also on the closure of that set. What is left is the interior of $\{\dot r = 0\}$: an interval on which $r$ is constant, so $\ddot r = 0$ there. Route 1 cannot say that $V_{\mathrm{eff}}' = 0$ on such an interval — any constant $r$ conserves $E$ trivially — but route 2, which is Newton's law itself, does. So the radial equation holds for all $t$, and the energy route alone proves it only away from circular stretches.

The barrier, physically. Squeezing an orbit inwards at fixed $L$ forces $\dot\theta$ up like $r^{-2}$ and the transverse kinetic energy up like $r^{-2}$. There is only so much energy; past some radius, all of it would be transverse, and that radius is the inner turning point. The hydrogen atom's radial equation carries the same barrier term with $L^2 = \hbar^2\ell(\ell+1)$; its ground state has $\ell = 0$, so no barrier at all, and quantum mechanics has to supply a different reason why the electron does not fall in.

08 · effective potential

Every kind of orbit, read off one graph

Gravity's $V = -k/r$ plus the barrier $L^2/2\mu r^2$ (both dashed) gives $V_{\mathrm{eff}}$, in units of $\mu k^2/L^2$, against $r$ in units of $p = L^2/\mu k$. The barrier wins at small $r$, the attraction at large $r$; the minimum is at $r = p$, depth $-\mu k^2/2L^2$. Dots: turning points.

the four cases

$E = V_{\mathrm{eff}}(r_c)$, the minimum: $\dot r\equiv0$, a circle of radius $r_c$. Here $r_c = L^2/\mu k = p$.
$V_{\min} < E < 0$: two turning points, motion bounded in an annulus $r_-\le r\le r_+$.
$E \ge 0$: one turning point, motion unbounded — in comes and out goes, once.
$L\ne0$ always: $V_{\mathrm{eff}}\to+\infty$ as $r\to0$, so $r_->0$ and the body never reaches the centre.

Each case is unit 00's classification of $\{V_{\mathrm{eff}}\le E\}$ into components, applied unchanged. Nothing here is specific to the inverse-square law except the shape of the curve.

The scaling that makes one picture do for all. Put $x = r/p$ with $p = L^2/\mu k$, and divide $V_{\mathrm{eff}}$ by $\mu k^2/L^2 = k/p$. Then $V_{\mathrm{eff}}/(k/p) = -1/x + 1/2x^2$ with no parameters left. Every inverse-square effective potential, for every $k$, $\mu$ and $L$, is this one curve after a rescaling of both axes — which is unit 00's mechanical similarity, in the form that matters here. The minimum sits at $x = 1$, value $-\tfrac12$.

The minimum, computed. $V_{\mathrm{eff}}' = k/r^2 - L^2/\mu r^3 = 0$ at $r_c = L^2/\mu k$, and $V_{\mathrm{eff}}(r_c) = -k/r_c + L^2/2\mu r_c^2 = -k/r_c + k/2r_c = -k/2r_c = -\mu k^2/2L^2$. It is a minimum, not a maximum: $V_{\mathrm{eff}}'' = -2k/r^3 + 3L^2/\mu r^4$, which at $r_c$ is $(-2k + 3k)/r_c^3 = k/r_c^3 > 0$.

The $E\ge0$ boundary is sharp because $V\to0$ at infinity. That normalisation of $V$ is a choice, and it is the one that makes "$E\ge0$ means escape" true. With a different additive constant the criterion would read $E\ge V(\infty)$. Potentials that do not tend to a limit — the harmonic $\tfrac12kr^2$, for instance — have no unbounded orbits at all, whatever $E$, and the graph shows it at once: $V_{\mathrm{eff}}\to\infty$ at both ends.

Marking the turning points. The dots are the roots of $-1/x + 1/2x^2 = E$, i.e. $2Ex^2 + 2x - 1 = 0$. For the bound level drawn, $E = -\tfrac14$: roots $x = 2\mp\sqrt2 = 0.586, 3.414$. For the circular level $E = -\tfrac12$ the two roots coincide at $x = 1$. For $E = 0$ the equation is linear and its one root is $x = \tfrac12$; for the unbound level drawn, $E = 0.18$, the positive root is $x = 0.462$ and the other is negative, so there is one turning point only.

08 · effective potential · example

Worked: Halley on the graph

turning points from $h$ and $\varepsilon$

Idea. Divide everything by $\mu$: with $h = L/\mu$ and $\varepsilon = E/\mu$ and $k/\mu = GM_\odot$, the energy equation is $\varepsilon = \tfrac12\dot r^2 + h^2/2r^2 - GM_\odot/r$. At a turning point $\dot r = 0$; multiply by $2r^2$:

$$2\varepsilon\,r^2 + 2GM_\odot\,r - h^2 = 0 .$$

Roots. With $\varepsilon < 0$ write $|\varepsilon|$ to keep signs visible: $r_\pm = \big(GM_\odot \pm \sqrt{(GM_\odot)^2 - 2|\varepsilon|h^2}\big)/2|\varepsilon|$.

Numbers (SI). $(GM_\odot)^2 = 1.76126\times10^{40}$, $2|\varepsilon|h^2 = 1.13858\times10^{39}$; difference $1.64740\times10^{40}$, root $1.28351\times10^{20}$. The large root: $$r_+ = \frac{(1.32712 + 1.28351)\times10^{20}}{4.97479\times10^{7}} = 5.24773\times10^{12}\ \mathrm m .$$

The small root, without cancellation. The product of the roots is $h^2/2|\varepsilon| = 4.60058\times10^{23}\ \mathrm{m^2}$, so $r_- = 8.76681\times10^{10}\ \mathrm m$. In AU: $r_+ = 35.079$, $r_- = 0.586025$.

checks

The input comes back. $r_-$ reproduces the perihelion $q = 0.586025$ AU that $h$ was built from — it had to, and that it does to six figures says the arithmetic is clean.

$r_+$ is a prediction. Halley's aphelion is $35.08$ AU, beyond Neptune at $30.1$ AU. It was never put in.

Sum rule. $r_-+r_+ = GM_\odot/|\varepsilon| = 5.33540\times10^{12}$ m $= 35.665$ AU, and half of that is $a = 17.832$ AU — the semi-major axis, from the energy alone. Section 5 explains why $a$ depends only on $E$.

Dimensions. $[GM/\varepsilon] = \mathsf{L^3T^{-2}}/\mathsf{L^2T^{-2}} = \mathsf L$ ✓.

Why the quadratic. At a turning point $\dot r = 0$, so $\varepsilon = V_{\mathrm{eff}}(r)/\mu$ exactly. Multiplying through by $2r^2$ clears the denominators and leaves a quadratic — which is special to the inverse-square law and is the first hint that its orbits will be conics. For $V\propto r^{-3}$ the same step gives a cubic, and for a general power it gives no polynomial at all.

Vieta, used twice. For $2\varepsilon r^2 + 2GM r - h^2 = 0$ the sum of the roots is $-GM/\varepsilon$ and the product is $-h^2/2\varepsilon$. So $a = \tfrac12(r_-+r_+) = -GM/2\varepsilon$ — depending on $\varepsilon$ and not at all on $h$ — and $b = \sqrt{r_-r_+}$, the geometric mean of the two extremes. Both facts are re-derived geometrically in section 5; getting them here, from a quadratic, is the cheaper route and a good check on the later one. ($b = \sqrt{r_-r_+}$ holds because $r_\pm = a(1\mp e)$ gives $r_-r_+ = a^2(1-e^2) = b^2$.)

Two measured numbers in, the whole extent of the orbit out, and no differential equation has been solved yet.

Why the small root is got from the product. $r_- = (GM - \sqrt{\cdot})/2|\varepsilon|$ subtracts $1.28351\times10^{20}$ from $1.32712\times10^{20}$: the two agree to within $3.3\%$, so about a digit and a half is lost in the difference, and each further rounding costs more. The large root adds them and loses nothing, and Vieta's product then gives the small one to full accuracy. This is the standard way to solve a quadratic whose roots differ greatly in size, and it is worth making a habit.

Arithmetic, for checking. $h^2 = 2.28869\times10^{31}\ \mathrm{m^4s^{-2}}$; $2|\varepsilon| = 4.97479\times10^{7}$; $2|\varepsilon|h^2 = 1.13858\times10^{39}$; $(GM_\odot)^2 - 2|\varepsilon|h^2 = 1.76126\times10^{40} - 0.11386\times10^{40} = 1.64740\times10^{40}$. Semi-minor axis, as a bonus: $b = \sqrt{r_-r_+} = 6.78276\times10^{11}$ m $= 4.5340$ AU.

08 · effective potential

Quadrature: the timetable and the shape

From $E = \tfrac12\mu\dot r^2 + V_{\mathrm{eff}}(r)$, unit 00's third case: separate and integrate.

solved, in unit 00's sense $$t - t_0 = \pm\int_{r_0}^{r}\frac{\mathrm dr'}{\sqrt{\tfrac{2}{\mu}\big(E - V_{\mathrm{eff}}(r')\big)}}$$ $$\theta - \theta_0 = \pm\int_{r_0}^{r}\frac{L\,\mathrm dr'}{r'^2\sqrt{2\mu\big(E - V_{\mathrm{eff}}(r')\big)}}$$

The first is the timetable, the second the shape. The sign is that of $\dot r$ (with $L>0$) and flips at each turning point.

The second follows from the first by $\dfrac{\mathrm d\theta}{\mathrm dr} = \dfrac{\dot\theta}{\dot r} = \dfrac{L/\mu r^2}{\dot r}$ — the chain rule, legitimate because $\dot r\ne0$ away from the turning points.

solved does not mean elementary

For $V = -k/r$ the second integral is elementary and gives an exact $r(\theta)$ — section 4 gets it by a shortcut. The first integral is elementary too, but inverting it to get $r(t)$ is not: that inversion is Kepler's equation, which has no solution in elementary functions. Unit 09 is that problem, and it is where Bessel functions were born.

08 · effective potential

The apsidal angle, drawn

A bound orbit lives in the annulus $r_-\le r\le r_+$ (dashed circles). From the start at $r_-$ to the next $r_+$ the radius turns through the apsidal angle $\Delta\theta$, here $211.8^\circ$; the orbit then repeats that arc, mirrored. Drawn for the force of P11, $f = -k/r^2 - c/r^3$ (notes). Thin: the next three radial periods.

what the integrals are good for anyway

Definite versions of them are exact answers to real questions with no formula needed in between: the period is $2\int_{r_-}^{r_+}$ of the first, and the apsidal angle — the angle swept from one turning point to the next — is $\int_{r_-}^{r_+}$ of the second. Both are numbers; P4 and P5 compute apsidal angles, P12 a time.

what decides closure

After $n$ radial periods the radius has turned through $2n\,\Delta\theta$. The orbit closes if and only if that is a whole number of turns for some $n$: if and only if $\Delta\theta/\pi$ is rational. Here $\Delta\theta/\pi = 1/0.85 = 20/17$, so it closes after $17$ radial periods and $20$ turns; for an irrational ratio it never closes, and the orbit fills the annulus.

What to look for in the figure. The thick arc is one radial period: out from $r_-$ to $r_+$ and back to $r_-$, sweeping $2\Delta\theta = 423.5^\circ$, so the second closest approach is $63.5^\circ$ further round than the first. Each later loop is the same arc turned by that amount, and each dashed apsidal line is a mirror axis of the whole orbit.

Where the formula in the caption comes from. The Binet equation of section 4 turns $f = -k/r^2 - c/r^3$ into $u'' + (1 - \mu c/L^2)u = \mu k/L^2$, whose solutions are $u = (1 + e\cos\gamma\theta)/\tilde p$ with $\gamma = \sqrt{1 - \mu c/L^2}$ and $\tilde p = \gamma^2 L^2/\mu k$. The radius returns to $r_-$ when $\gamma\theta$ has grown by $2\pi$, so $\Delta\theta = \pi/\gamma$. The figure takes $\gamma = 0.85$ and $e = 0.45$; nothing here depends on having that formula, which is why the slide sits before section 4.

The apsidal angle, and why it decides everything about closure. By the symmetry of the radial motion about a turning point, the orbit is symmetric under reflection in the apsidal line, and the whole path is the arc between $r_-$ and $r_+$ repeated and reflected. In $n$ radial periods the radius turns through $2n\Delta\theta$, so the orbit closes after $n$ of them if and only if $2n\Delta\theta = 2\pi m$ for some whole number $m$ of turns — that is, if and only if $\Delta\theta/\pi = m/n\in\mathbb Q$. For the inverse-square law $\Delta\theta = \pi$ exactly and every orbit closes after one radial period. Bertrand's theorem, on the closure slide at the end of section 6, says how rare that is.

The reflection symmetry, proved. Let $t_-$ be a time with $r(t_-) = r_-$, so $\dot r(t_-) = 0$, and put $\tilde r(t) := r(2t_- - t)$. The radial equation $\mu\ddot r = -V_{\mathrm{eff}}'(r)$ contains only the second derivative, which the substitution leaves unchanged, so $\tilde r$ solves it; and $\tilde r(t_-) = r_-$, $\dot{\tilde r}(t_-) = -\dot r(t_-) = 0$. Same equation, same data at $t_-$: by unit 01's uniqueness theorem $\tilde r = r$. So $r$ is even about $t_-$, and then $\theta(t) - \theta(t_-) = \int_{t_-}^{t}L/\mu r^2$ is odd about $t_-$. The orbit is therefore its own mirror image in the line $\theta = \theta(t_-)$. This is what makes "apsidal angle" a property of the orbit and not of the starting point.

08 · effective potential

Six to two: what a first integral actually buys

state space	dim	what removed it
two bodies, $(\vec r_1,\vec r_2,\dot{\vec r}_1,\dot{\vec r}_2)$	12	—
relative motion, $(\vec r,\dot{\vec r})$	6	$\dot{\vec R}$ and $\vec R - t\dot{\vec R}$: six integrals, section 2
in the orbital plane, $(r,\theta,\dot r,\dot\theta)$	4	direction of $\vec L$: two integrals
$(r,\theta,\dot r)$ with $\dot\theta = L/\mu r^2$	3	magnitude of $\vec L$: one integral
$(r,\dot r)$ — $\theta$ appears nowhere	2	$\theta$ is cyclic
a curve in that plane	1	$E$: one integral

One-dimensional means quadrature: a single separable ODE, integrated and inverted. Then $\theta$ is recovered by one more integration, and $\vec R(t)$ by two.

the accounting rule

Each independent first integral confines the flow to a level set one dimension smaller. Each cyclic coordinate removes a further dimension, because the remaining equations do not mention it. Twelve minus six minus three minus one leaves two, and two minus one leaves one: solved.

counting is not proving

"Enough conserved quantities" is a heuristic here, not a theorem. The theorem is unit 17's Liouville–Arnold: $n$ integrals in involution on a $2n$-dimensional phase space give integrability, and involution is a condition this unit has no language for yet.

Read the table as the plan of the unit. Section 2 did row 2, section 1 rows 3 and 4, section 3 rows 5 and 6. Nothing else is needed, and that is why the central-force problem is solvable while the three-body problem is not: three bodies have an 18-dimensional phase space and the same ten classical integrals (energy, three of momentum, three of angular momentum, three of centre-of-mass drift), which is nowhere near enough. Bruns proved in 1887 that there are no further integrals algebraic in positions and velocities, and Poincaré extended the negative result soon after. Cited; neither proof is in this course.

The two integrations at the end are not free. "Recovered by one more integration" hides the whole of unit 09: the integral for $t(r)$ is elementary but its inverse $r(t)$ is not. In the classical sense of "solved" — reduced to quadrature — the problem is finished at row 6. In the modern sense of "give me the position at a stated time" it is not, and the gap between those two senses is a unit and a half wide.

08 · effective potential

The last two dimensions: the $(r, \dot r)$ plane

The table's last two rows, drawn. Each curve is one level set $\tfrac12\mu\dot r^2 + V_{\mathrm{eff}}(r) = E$ for the inverse square, in the units of the one-graph slide ($r$ in $p$, $\dot r$ in $k/L$). Arrows: the direction of motion — outward above the axis, inward below.

what each curve is

$E < 0$: a closed loop between the turning points $r_\pm$. Going round it once is one radial period, from perihelion out to aphelion and back.
$E \ge 0$: an open curve, in from infinity along the bottom, through one turning point, out along the top.
The dot, $E = V_{\min}$: the circular orbit, a single point.

$\theta$ is not in the picture at all: it is recovered afterwards from $\dot\theta = L/\mu r^2$, one more integration.

What to look for. The curves are the graph of the one-graph slide read sideways: where $V_{\mathrm{eff}}$ is low, $E - V_{\mathrm{eff}}$ is large and the curve is far from the axis; at a turning point $\dot r = 0$ and the curve crosses the axis vertically. The loops are not ellipses — they are squashed towards small $r$, because the body moves fastest near perihelion — but they are closed, and that is all row 6 of the table needs: a one-dimensional set, traversed periodically.

What "independent" means here. The level sets must intersect transversally — the gradients of the integrals must be linearly independent at the point in question. $E$ and $\lVert\vec L\rVert$ are independent except on the circular orbits, where the two conditions coincide; that degeneracy is exactly why a circular orbit is a single point of the $(r,\dot r)$ plane rather than a curve, and why the quadrature integral diverges there.

section 4

The Binet equation

One change of variable, and the hardest case in the course becomes the easiest equation in unit 06.

08 · Binet

The substitution $u = 1/r$

Ask for the shape $r(\theta)$, not the timetable. Since $\dot\theta$ never vanishes when $L\ne0$ (What the flat orbit buys, in polar coordinates), $\theta$ is a legitimate independent variable. Write $u(\theta) := 1/r(\theta)$ and $' := \mathrm d/\mathrm d\theta$. Then $\dot\theta = L u^2/\mu$, and

two derivatives $$\dot r = \frac{\mathrm dr}{\mathrm d\theta}\dot\theta = -\frac{u'}{u^2}\cdot\frac{Lu^2}{\mu} = -\frac{L}{\mu}\,u' ,$$ $$\ddot r = -\frac{L}{\mu}\,u''\,\dot\theta = -\frac{L^2}{\mu^2}\,u^2u'' .$$

The first line is the whole trick: $\mathrm dr/\mathrm d\theta = -u'/u^2$ cancels against the $u^2$ in $\dot\theta$, and $\dot r$ becomes linear in $u'$ with a constant coefficient.

Put that into the radial equation $\mu\ddot r = f(r) + L^2/\mu r^3$:

$$-\frac{L^2}{\mu}u^2u'' = f(1/u) + \frac{L^2}{\mu}u^3 .$$

Divide by $-(L^2/\mu)u^2$:

the Binet equation $$\boxed{\ u'' + u = -\frac{\mu}{L^2u^2}\,f(1/u)\ }$$

An exact second-order ODE for the orbit's shape, for any isotropic central $f$. No approximation has been made.

Named for Jacques Binet. It is the single most useful change of variable in this course.

Dimensions. $[u] = \mathsf{L^{-1}}$, and $\theta$ is a pure number, so $[u''] = \mathsf{L^{-1}}$ ✓. On the right, $[\mu f/L^2u^2] = \mathsf{M}\cdot\mathsf{MLT^{-2}}/(\mathsf{M^2L^4T^{-2}}\cdot\mathsf{L^{-2}}) = \mathsf{M^2LT^{-2}}/\mathsf{M^2L^2T^{-2}} = \mathsf{L^{-1}}$ ✓.

Check on the case already known. Free motion, $f\equiv0$, gives $u'' + u = 0$, so $u = C\cos(\theta-\theta_0)$ and $r\cos(\theta-\theta_0) = 1/C$: a straight line at perpendicular distance $1/C$ from the origin. That a straight line comes out of a sinusoid in $u$ is the clearest possible statement that $u$, not $r$, is the natural variable here.

Why $1/r$ and not something else. Two features conspire. The centrifugal term is $L^2/\mu r^3$, which under $u = 1/r$ becomes $u^3$ — the same power as the $u^2u''$ on the left divided by $u^2$ — so the whole left-hand side collapses to $u''+u$ regardless of the force. And the factor $\dot\theta\propto u^2$ is exactly what is needed to make the first derivative linear. Both are consequences of one fact: $\dot\theta = L/\mu r^2$, with $L$ constant. Any other substitution $r = g(w)$ leaves $\dot r = g'(w)w'\,L/\mu g(w)^2$, which has constant coefficient only when $g'/g^2$ is constant — that is, $g(w) = -1/(cw + d)$, which is $1/r$ up to a scale and a shift of $w$, and the shift would spoil the $u^3$ cancellation.

Where this is going. Unit 37 computes the geodesic equation of the Schwarzschild metric and gets $u'' + u = GM\mu^2/L^2 + 3GMu^2/c^2$ — this equation with one extra term of relative size $3GM/rc^2$. Everything general relativity adds to Newtonian gravity, for a planet, is that term. That is why this slide is worth learning rather than looking up.

08 · Binet

Inverse square: the right-hand side goes constant

Take $f(r) = -k/r^2$ with $k > 0$ — attraction. For gravity $k = Gm_1m_2$; for two opposite charges $k = |q_1q_2|/4\pi\varepsilon_0$. Then $f(1/u) = -ku^2$, and

$$-\frac{\mu}{L^2u^2}\cdot(-ku^2) = \frac{\mu k}{L^2},$$

the orbit equation $$u'' + u = \frac{\mu k}{L^2} =: \frac1p .$$

Unit 06's equation $\ddot x + \omega_0^2 x = f$ with $\omega_0 = 1$, zero damping, and a constant drive. Linear, constant-coefficient, inhomogeneous: the easiest object in that unit.

dimensions

$[k] = \mathsf{MLT^{-2}}\cdot\mathsf{L^2} = \mathsf{ML^3T^{-2}}$, so $[\mu k/L^2] = \mathsf{M}\cdot\mathsf{ML^3T^{-2}}/\mathsf{M^2L^4T^{-2}} = \mathsf{L^{-1}}$ ✓. So $p := L^2/\mu k$ is a length.

this is a coincidence, and an important one

For $f = -cr^n$ the right-hand side is $(\mu c/L^2)u^{-n-2}$, and the orbit equation is linear with constant coefficients only for $n = -2$ (a constant drive) and $n = -3$ (a term in $u$, which shifts the frequency rather than driving it). The inverse square is the one power that gives unit 06's driven oscillator.

check $n = -3$

$f = -c/r^3$ gives $u'' + (1 - \mu c/L^2)u = 0$: solvable, but with no drive there is no bounded orbit — the Cotes spirals (notes). Added to an inverse square, the same term makes the ellipse precess: P11, and unit 37 in miniature.

The "coincidence" is Bertrand's theorem in embryo. Two facts about the inverse-square law keep reappearing: its Binet equation is $u''+u = $ const, and all its bounded orbits close. The second is much harder to prove than the first, and Bertrand's theorem (last section) says only one other force law shares it. The harmonic force $f = -\kappa r$ has Binet right-hand side $\mu\kappa/L^2u^3$, thoroughly nonlinear — yet its orbits are ellipses too, centred on the force centre rather than focused there, which is why $u$ is the wrong variable for it and $x,y$ the right one.

Repulsion, for later. For $f = +\kappa/r^2$ with $\kappa>0$ — like charges — the same computation gives $u'' + u = -\mu\kappa/L^2$, a constant of the other sign. Then $u = -\mu\kappa/L^2 + C\cos\theta$, and $u>0$ forces $C>\mu\kappa/L^2$: only hyperbolae, as it must be, since a repulsive potential has no bound states at all. That is the calculation behind Rutherford's formula on the problem sheet.

The inverse cube alone. $u'' + (1-\mu c/L^2)u = 0$ has no constant term, so its solutions are not conics. With $\gamma^2 = 1 - \mu c/L^2 > 0$, $u = C\cos\gamma\theta$ vanishes at $\gamma\theta = \pm\pi/2$: the particle comes in from infinity and leaves again after turning through $\pi/\gamma > \pi$. With $\mu c > L^2$ the solutions are combinations of $e^{\pm\gamma\theta}$, $\gamma^2 = \mu c/L^2 - 1$, and spiral into the centre or out to infinity. These are Cotes's spirals; the closure slide's notes list every case. None is bounded except the neutral circles of the borderline $\mu c = L^2$, as the effective potential $(L^2/2\mu - c/2)/r^2$ — monotone in $r$ — already says.

Naming $p$. $p = L^2/\mu k$ is the semi-latus rectum, the conics slide's geometric object: the half-width of the conic measured perpendicular to the axis through the focus. Writing the constant as $1/p$ now saves rewriting the solution later.

08 · Binet

Solving it, with unit 06 in one line

$\mathcal D u := u'' + u$ has characteristic polynomial $\lambda^2 + 1$, roots $\pm i$, so $\ker\mathcal D$ is the two-dimensional space spanned by $\cos\theta$ and $\sin\theta$ — by unit 06, and two-dimensional by unit 01's uniqueness theorem, not by counting constants. The constant $u_p \equiv 1/p$ is a particular solution. Hence

$$u(\theta) = \frac1p + C_1\cos\theta + C_2\sin\theta ,$$

or $u = 1/p + C\cos(\theta-\theta_0)$ in amplitude–phase form. The polar axis is ours to choose: rotate it so that $\theta_0 = 0$, and take $C\ge0$ by flipping $\theta_0$ by $\pi$ if need be. Writing $e := Cp \ge 0$,

the orbit $$\frac1r = \frac{1 + e\cos\theta}{p}, \qquad r(\theta) = \frac{p}{1 + e\cos\theta}.$$

two constants, and what they are

A second-order ODE needs two constants and we have used them: $\theta_0$ is fixed by choosing the axis, and $e$ is what is left. So a central inverse-square orbit in a given plane has exactly one shape parameter — with $p$ set by $L$, that is $e$, and The two constants, in terms of $E$ and $L$ identifies it with the energy.

Where the freedom went. The general solution has constants $(C,\theta_0)$; the orbit in space needs, on top of the plane and $L$, only $C$. $\theta_0$ is not physics, it is the choice of where to put $\theta = 0$ — but it does carry information once the axis is fixed by something else, and in section 6 it is the direction of the Runge–Lenz vector. Unit 10's perturbed orbits are exactly orbits whose $\theta_0$ drifts slowly.

The amplitude–phase form. $C_1\cos\theta + C_2\sin\theta = C\cos(\theta-\theta_0)$ with $C = \sqrt{C_1^2+C_2^2}$ and $\tan\theta_0 = C_2/C_1$ — unit 06's identity, proved there from Euler's formula. It is used here purely to make the two constants geometric.

Why no initial conditions were imposed. The two constants of the Binet solution encode the shape, not the position along it. The full initial-value problem needs four numbers in the plane $(r,\theta,\dot r,\dot\theta)$; one is absorbed into $L$ (which appears in $p$), one into the choice of axis, one into $e$, and the last is the time of perihelion passage, which the shape equation cannot see because $t$ has been integrated out. Unit 09 restores it.

08 · Binet

Reading the solution: $u$ is a shifted cosine

$p\,u = 1 + e\cos\theta$: a cosine of amplitude $e$ riding on the constant drive's response $1$. Where it is large the body is close; where it would go negative there is no orbit. For $e = 1.6$ the zeros, at $\theta = \pm128.7^\circ$, are where $r\to\infty$.

$e \ge 0$ and $r > 0$

If $e\ge1$, the denominator $1+e\cos\theta$ vanishes at $\cos\theta = -1/e$; the formula is then valid only on the arc where it is positive, and $r\to\infty$ at the ends. That is not a failure of the solution — it is the orbit going to infinity, and those two angles are the asymptotes.

three readings of one graph

$e < 1$: $u$ stays positive, $r$ stays finite — a closed curve. $e = 1$: $u$ touches zero once, at $\theta = \pm\pi$ — the body just reaches infinity. $e > 1$: $u$ crosses zero — the body comes in along one asymptote and leaves along the other.

What to look for. Every curve has its maximum at $\theta = 0$, which is why the axis was "aimed at perihelion"; its minimum, at $\theta = \pi$, is aphelion when it is positive. The constant $1$ is unit 06's particular solution, the cosine its free oscillation with frequency $1$ — exactly one oscillation of $u$ per turn of $\theta$, which is the analytic reason the orbit closes. A force law whose Binet equation had a frequency other than $1$, as in the apsidal-angle figure, makes $u$ oscillate at a different rate from $\theta$, and successive perihelia drift round: the rosette.

The asymptote angle. $1 + e\cos\theta_\infty = 0$ gives $\cos\theta_\infty = -1/e$; for $e = 1.6$, $\theta_\infty = \arccos(-0.625) = 128.68^\circ$. The shaded strip is the region the formula $r = p/(1+e\cos\theta)$ would map to negative $r$ — not an orbit point behind the focus, but no point at all.

08 · Binet

Conics, defined before they are met

definition (focus–directrix)

Fix a point $F$ (the focus), a line $D$ not through it (the directrix) and a number $e > 0$ (the eccentricity). The conic with these data is

$$\mathcal C = \{\,X : \lVert X - F\rVert = e\cdot\mathrm{dist}(X, D)\,\}.$$

theorem

In polar coordinates centred at $F$ with the axis perpendicular to $D$ and pointing towards it, the part of $\mathcal C$ on the focus's side of $D$ is exactly the set $r = p/(1 + e\cos\theta)$, where $p = e\,d$ and $d = \mathrm{dist}(F,D)$. For $e\le1$ that part is all of $\mathcal C$.

Proof. $D$ is the line $x = d$, and a point $(r,\theta)$ with $x = r\cos\theta < d$ is at distance $d - r\cos\theta$ from it. The condition reads $r = e(d - r\cos\theta)$, i.e. $r(1 + e\cos\theta) = ed = p$. $\square$

The proof, drawn for $e = 0.6$. $X$ is at distance $r$ from the focus and $d - r\cos\theta$ from the directrix; the conic is where the first is $e$ times the second.

Why define conics at all, rather than say "it is an ellipse". House rule: the mathematics carries the definition. "Ellipse" as a shape recognised by eye is not a definition, and the focus–directrix set is, and it is the one that matches the formula the ODE produced without any algebra in between. It also handles all four cases at once, including the circle as a limit, which the sum-of-distances definition does not do gracefully.

Reading the figure. Only two lengths matter, both measured from $X$: to the focus, $r$, and to the directrix along the axis direction, $d - r\cos\theta$ (the dotted drop shows $r\cos\theta$, the part of $d$ already covered). Setting the first equal to $e$ times the second is the whole proof. Moving $X$ round the curve changes both lengths, never their ratio.

08 · Binet

One definition, three names

One focus $F$, one directrix $D$, three values of $e$. Below $1$ the curve is closed and lies wholly on $F$'s side of $D$; at $1$ it just fails to close; above $1$ it has a second branch beyond $D$ (dashed), which the polar formula about $F$ does not describe.

so the three names are one family

$e$	name	bounded?
$0$	circle	yes ($D$ at infinity)
$0<e<1$	ellipse	yes
$e=1$	parabola	no
$e>1$	hyperbola (one branch)	no

$p$ is the semi-latus rectum: $r(\pm\pi/2) = p$, the half-width of the conic measured through the focus, perpendicular to the axis.

The Sun sits at the focus this definition names. An ellipse has a second focus, and nothing is there (notes).

What to look for. All three curves cross the axis between $F$ and $D$ at $p/(1+e) = ed/(1+e)$, which grows with $e$ but never reaches $D$; that crossing is perihelion. The ellipse closes up on the far side of $F$, the parabola runs off to the left, and the hyperbola's near branch bends away even faster, with a second branch on the far side of $D$.

Two foci, one used. An ellipse also has the familiar "sum of distances to two foci is constant" definition; a conic in the present sense has one focus and one directrix. They describe the same curves (cited, with the cone below). The definition here is purely metric — no coordinates, no quadratic equation, no cone — and it names only the focus where the Sun is.

The circle as $e\to0$. Keeping $p = ed$ fixed while $e\to0$ pushes the directrix to infinity, and the condition $\lVert X-F\rVert = e\,\mathrm{dist}(X,D)$ becomes $\lVert X-F\rVert = p$: a circle of radius $p$ about $F$. This is the right way to see why the circle has no directrix.

Why $e\le1$ leaves nothing beyond $D$. A point with $x > d$ is at distance $x - d$ from $D$, and would need $r = e(x-d) \le x - d < x \le r$ — impossible. For $e > 1$ there is a second branch beyond $D$, $r = p/(e\cos\theta - 1)$; it is the path of a repelled particle with the force centre at $F$, which is the case the problem sheet's Rutherford problem needs.

The relation to the cone. Apollonius defined these curves as plane sections of a cone, around 200 BC; the focus–directrix property is Pappus's, five centuries later; Dandelin's spheres (1822) prove the two definitions agree by inscribing a sphere in the cone tangent to the cutting plane, whose point of tangency is the focus. None of that is needed here, and it is cited rather than proved: see Hilbert and Cohn-Vossen, Geometry and the Imagination, chapter I.

Cartesian form, for the record. Squaring $r = e(d - x)$ with $r^2 = x^2+y^2$ gives $(1-e^2)x^2 + y^2 + 2e^2dx - e^2d^2 = 0$, a quadratic with $A = 1-e^2$, $B = 0$, $C = 1$, so $B^2 - 4AC = 4(e^2-1)$ has the sign of $e^2-1$. Unit 18 borrows the three names — hyperbolic, parabolic, elliptic — for second-order PDEs, because the classification there is the sign of the same kind of discriminant.

08 · Binet

The two constants, in terms of $E$ and $L$

$p$ came out of the equation as $L^2/\mu k$. For $e$, evaluate the energy at perihelion $\theta = 0$, where $r = p/(1+e)$ and $\dot r = 0$, so $E = V_{\mathrm{eff}}(r)$:

$$E = -\frac{k(1+e)}{p} + \frac{L^2(1+e)^2}{2\mu p^2} = -\frac{k(1+e)}{p} + \frac{k(1+e)^2}{2p},$$

using $L^2 = \mu k p$. Factor out $k(1+e)/2p$:

$$E = \frac{k(1+e)}{2p}\big[(1+e) - 2\big] = \frac{k(e^2-1)}{2p} = \frac{\mu k^2(e^2-1)}{2L^2}.$$

the orbit, fully determined $$p = \frac{L^2}{\mu k},\qquad e = \sqrt{1 + \frac{2EL^2}{\mu k^2}}\, .$$

checks

Dimensions. $[EL^2/\mu k^2] = \mathsf{ML^2T^{-2}}\cdot\mathsf{M^2L^4T^{-2}}/(\mathsf{M}\cdot\mathsf{M^2L^6T^{-4}}) = 1$ ✓, so $e$ is a pure number.

The circular case. $e = 0$ iff $E = -\mu k^2/2L^2$, which is exactly the minimum of $V_{\mathrm{eff}}$ read off the one-graph slide ✓. And $e$ is real only for $E \ge -\mu k^2/2L^2$: below the minimum there is no orbit, as there should not be.

The classification. $e < 1 \iff E < 0$; $e = 1 \iff E = 0$; $e > 1 \iff E > 0$. The geometric trichotomy of the conics slide and the energetic one of the effective-potential graph are the same trichotomy.

Two conserved numbers in, a complete orbit out. For gravity, divide $E$ and $L$ by $\mu$: then $e^2 = 1 + 2\varepsilon h^2/(GM)^2$ and $p = h^2/GM$ involve the orbiting body's mass not at all.

The same result without choosing perihelion. Energy conservation at a general $\theta$: from $u = (1+e\cos\theta)/p$, $\dot r = -(L/\mu)u' = (L/\mu p)e\sin\theta$ and $r\dot\theta = L u/\mu = (L/\mu p)(1+e\cos\theta)$. Then $$\tfrac12\mu(\dot r^2 + r^2\dot\theta^2) = \frac{L^2}{2\mu p^2}\big[e^2\sin^2\theta + (1+e\cos\theta)^2\big] = \frac{L^2}{2\mu p^2}\big[1 + 2e\cos\theta + e^2\big],$$ and $V = -k/r = -(k/p)(1+e\cos\theta)$. Using $L^2/\mu p^2 = k/p$, the sum is $\frac{k}{2p}(1+2e\cos\theta+e^2) - \frac{k}{p}(1+e\cos\theta) = \frac{k}{2p}(e^2-1)$ — independent of $\theta$, which is simultaneously the value of $E$ and a proof that the solution really does conserve energy. Doing it at perihelion is the same computation with $\theta = 0$ and less algebra.

Why the trichotomy had to come out this way. $e^2 - 1 = 2EL^2/\mu k^2$ and $L^2, \mu, k^2$ are all positive, so $e^2-1$ has the sign of $E$ and nothing else. A one-line statement, but it is the bridge between two entirely different ways of arriving at the same three cases — one from a graph of a function of $r$, one from the geometry of a plane curve — and a physicist should be uncomfortable until such a bridge is built.

$a$ from $E$ alone, foreshadowed. For $e<1$, $a = p/(1-e^2)$ (next section), and substituting both formulas gives $a = -k/2E$: the semi-major axis depends on the energy and not at all on the angular momentum. The Halley-on-the-graph example found the same thing from Vieta's formulas. It is the fact that makes orbital manoeuvring tractable — a burn that changes energy changes the size of the orbit, a burn that changes only direction changes its shape.

08 · Binet · example

Worked: the comet's shape from its two constants

$p$ and $e$ for Halley

Setup. Divide by $\mu$ throughout: $k/\mu = GM_\odot$, so $p = h^2/GM_\odot$ and $e = \sqrt{1 + 2\varepsilon h^2/(GM_\odot)^2}$, with $h = 4.78403\times10^{15}$ and $\varepsilon = -2.48740\times10^{7}$ from the angular-momentum example.

Semi-latus rectum. $h^2 = 2.28869\times10^{31}$, so $$p = \frac{2.28869\times10^{31}}{1.32712\times10^{20}} = 1.72455\times10^{11}\ \mathrm{m} = 1.15279\ \mathrm{AU}.$$

Eccentricity. $2\varepsilon h^2 = -1.13858\times10^{39}$ and $(GM_\odot)^2 = 1.76126\times10^{40}$, so the ratio is $-0.0646457$ and $$e = \sqrt{1 - 0.0646457} = \sqrt{0.935354} = 0.967137 .$$

Semi-major axis, for later: $a = p/(1-e^2) = 1.15279/0.0646457 = 17.8325$ AU.

checks

Against the graph. $r_{\min} = p/(1+e) = 1.15279/1.967137 = 0.586025$ AU and $r_{\max} = p/(1-e) = 1.15279/0.032863 = 35.079$ AU — the two turning points found from the quadratic ✓.

A second route to $e$. At perihelion $p = h^2/GM = q^2v_q^2/GM$ and $q = p/(1+e)$, so $e = qv_q^2/GM_\odot - 1 = 0.967137$ ✓, with no cancellation at all.

Sign and size. $E<0$ gives $e<1$: bound ✓. And $e$ is close to 1, matching the $99.2\%$-of-escape-speed finding ✓.

A circular orbit with Halley's angular momentum would sit at $r = p = 1.1528$ AU, between Earth and Mars. Same $L$, very different $E$.

Where the digits go. $e^2 = 1 - x$ with $x = 2|\varepsilon|h^2/(GM)^2 = 0.0646$ small, so $e$ itself is well conditioned: a relative error $\delta$ in $x$ moves $e$ by only about $0.033\,\delta$. The danger is upstream, in $\varepsilon$, which is the difference of two numbers agreeing to $1.7\%$ and so carries about sixty times the relative error of its inputs. A slip of a power of ten there — $2\varepsilon h^2$ written as $10^{40}$ instead of $10^{39}$ is the classic one — gives $x = 0.646$ and $e = 0.59$, a wildly wrong orbit that still looks like a sensible number. The dimension-free ratio $x$ is the thing to sanity-check: it must be less than 1 for a real $e$, and for a comet that nearly escapes it must be small.

The stable route, for hand work. Get $p = h^2/GM$ from the angular momentum, and $e = qv_q^2/GM - 1$ from the perihelion data directly; or, given $a$, use $e^2 = 1 - p/a$. Each is a ratio of well-conditioned quantities. That is why comet elements are published as $q$ and $e$, never as $\varepsilon$ and $h$.

The circular comparison. Kepler's third law (section 5) gives a circle of radius $1.15279$ AU a period of $1.15279^{3/2} = 1.2377$ yr. Halley, with the same angular momentum, takes $75.3$ yr: all the difference is energy.

08 · Binet

One family, four members

One $p$, hence one $L$; four values of $E$, so four values of $e$. Every curve passes through the point $p$ above the focus; perihelion, at $p/(1+e)$, is on the right of each and moves in as $E$ rises.

same $L$, different $E$

Raising $E$ from the bottom of the well at fixed $L$: circle, ellipses of growing $e$, the parabola at $E = 0$, hyperbolae beyond. Perihelion $p/(1+e)$ shrinks throughout.

08 · Binet

The unbound case: scattering

An attracted body on a hyperbola, $e = 1.5$. Far away it moves on straight lines, the asymptotes (dashed). $b$ is how far the incoming line would miss the centre; $\Theta$ is the angle between the directions of travel before and after — here $83.6^\circ$, from $\sin(\Theta/2) = 1/e$.

the unbound case is scattering

For $e>1$ the asymptotes are at $\cos\theta_\infty = -1/e$, and the deflection angle $\Theta$ between incoming and outgoing directions satisfies $\sin(\Theta/2) = 1/e$, hence

$$\tan\frac{\Theta}{2} = \frac{k}{\mu v_\infty^2 b}$$

with $b$ the impact parameter, $L = \mu v_\infty b$. Rutherford's formula: P9 derives it, for attraction and repulsion alike.

Rutherford used the repulsive case in 1911: alpha particles bouncing back from gold foil at large $\Theta$ need a $b$ far smaller than an atom, and so a nucleus.

What to look for. The body bends towards the centre — the centre is on the inside of the curve — and the closer the incoming line passes (smaller $b$), the larger $\Theta$. The two asymptotes meet at the hyperbola's centre, a distance $ae$ from the focus, not at the force centre. For the figure, $p = 1$ gives $a = p/(e^2-1) = 0.8$ and $b = a\sqrt{e^2-1} = 0.894$, and $\tan(\Theta/2) = 1/\sqrt{e^2-1} = 0.894$ in the same units: in $\tan(\Theta/2) = k/\mu v_\infty^2 b$, $k/\mu v_\infty^2 = a$ and the right-hand side is $a/b$.

The deflection formula, sketched. For a hyperbola, $r\to\infty$ where $1+e\cos\theta = 0$, so the two asymptotic directions are $\pm\theta_\infty$ with $\cos\theta_\infty = -1/e$ and $\theta_\infty\in(\pi/2,\pi)$. The particle sweeps $2\theta_\infty$ in total, and the angle between its initial and final directions of travel is $\Theta = 2\theta_\infty - \pi$. Then $\cos\theta_\infty = \cos(\pi/2 + \Theta/2) = -\sin(\Theta/2)$, so $\sin(\Theta/2) = 1/e$ and $\tan(\Theta/2) = 1/\sqrt{e^2-1}$. Finally $e^2-1 = 2EL^2/\mu k^2$ with $E = \tfrac12\mu v_\infty^2$ and $L = \mu v_\infty b$ gives $e^2-1 = (\mu v_\infty^2 b/k)^2$. Done; the problem sheet (P9) does it in full, for both signs of the force.

Why attraction and repulsion give the same angle. Because the answer depends on $k$ only through $k^2$, by way of $e^2-1$. Rutherford's cross-section therefore cannot distinguish a positive nucleus repelling alpha particles from a negative one attracting them — the deflection angles are identical and only the sense of the swing differs: the attracted particle passes on the near side of the centre, the repelled one on the far side.

section 5

Kepler's three laws, derived

Twenty years of Kepler's work on Tycho Brahe's data, recovered from two conservation laws in three slides.

08 · Kepler

Second law: equal areas in equal times

areal velocity

$\mathrm dA/\mathrm dt$, where $A(t)$ is the area swept by the radius from the centre to the particle.

Kepler's second law

For any central force — isotropic or not —

$$\frac{\mathrm dA}{\mathrm dt} = \tfrac12\lVert\vec r\times\dot{\vec r}\rVert = \frac{L}{2\mu}\ =\ \text{constant} .$$

Computed for $e = 0.6$ from Kepler's equation (unit 09). The two shaded sectors are swept in the same time, and have the same area: short and wide near perihelion, long and thin near aphelion. Dots: positions at sixteen equal time steps.

Why this needs no potential. The proof used only $\vec L = \mu\vec r\times\dot{\vec r}$ constant, which needs only that the force points along $\vec r$. Kepler's second law is therefore the most robust of the three: it survives any change to the radial dependence of gravity, and Newton knew this — it is Proposition I of the Principia, proved before any force law is named, by exactly the triangle argument above in its discrete form.

What to look for in the figure. The dots crowd together at the far end and spread out near the Sun: in equal times the comet covers little arc where it is far and a lot where it is near. The second law says exactly how: $r^2\dot\theta$ is the same everywhere, so the angular rate goes as $1/r^2$. The positions were found by solving Kepler's equation $M = E - e\sin E$, the subject of unit 09; nothing on this slide depends on it except the drawing.

08 · Kepler

The second law, proved: one thin triangle

proof

The area swept between $t$ and $t+\Delta t$ is, to leading order, that of the triangle with vertices $\vec 0$, $\vec r(t)$, $\vec r(t+\Delta t)$, which is $\tfrac12\lVert\vec r(t)\times\vec r(t+\Delta t)\rVert$ by The cross product measures area. Expand: $\vec r(t+\Delta t) = \vec r + \Delta t\,\dot{\vec r} + O(\Delta t^2)$, and $\vec r\times\vec r = \vec 0$, so the area is $\tfrac12\Delta t\lVert\vec r\times\dot{\vec r}\rVert + O(\Delta t^2)$. Divide by $\Delta t$ and let it go to zero. $\square$

check on Halley

$v_q/v_Q = 54.570/0.91164 = 59.86$, and $r_{\max}/r_{\min} = 35.079/0.58602 = 59.86$ ✓ — because $rv$ is the same at both apses, both being places where $\vec v\perp\vec r$.

The proof's picture, exaggerated. In time $\Delta t$ the radius sweeps the shaded triangle plus a sliver between chord and arc. The triangle's area is half the cross product of its two sides; the sliver is too thin to matter in the limit.

The $O(\Delta t^2)$, honestly. "The area swept is that of the triangle to leading order" needs a word. The swept region lies between the triangle and the triangle plus a sliver bounded by the chord and the arc; the sliver's width is $O(\Delta t^2)$ (the curve's deviation from its chord) and its length $O(\Delta t)$, so its area is $O(\Delta t^3)$ for a $C^2$ curve. The stated limit is therefore exact. A fully rigorous treatment defines $A$ as $\int\tfrac12r^2\,\mathrm d\theta$ and differentiates; the triangle version is Newton's, and it needs no integration theory.

The apsis check, in general. At either apsis $\dot r = 0$, so $v = r|\dot\theta|$ and $L = \mu r^2|\dot\theta| = \mu rv$. Hence $r_{\min}v_q = r_{\max}v_Q$, a relation that holds for every central force and is the quickest sanity check on any pair of orbit numbers. For Halley, $0.58602\times54.570 = 31.98$ and $35.079\times0.91164 = 31.98$, in AU·km/s ✓.

08 · Kepler

First law: the ellipse, with its axes named

For $E < 0$ we have $e < 1$, and $r = p/(1+e\cos\theta)$ is bounded, with apses (the turning points)

$$r_{\min} = \frac{p}{1+e},\qquad r_{\max} = \frac{p}{1-e} .$$

the semi-axes $$a = \tfrac12(r_{\min}+r_{\max}) = \frac{p}{1-e^2},$$ $$b = a\sqrt{1-e^2} = \frac{p}{\sqrt{1-e^2}},$$

so $b^2 = ap$ and $r_{\min}r_{\max} = b^2$, and the focus sits a distance $ea$ from the centre.

the fact worth remembering

$E = \dfrac{k(e^2-1)}{2p} = -\dfrac{k}{2a}$: the energy fixes $a$ and nothing else. $L$ then fixes $e$ through $p = a(1-e^2)$.

Drawn with $e = 0.6$. Halley's is $0.967$: an ellipse $3.9$ times longer than it is wide, with the Sun almost at one end.

Full derivation of $b$

$a = \tfrac12\big(\frac{p}{1+e}+\frac{p}{1-e}\big) = \tfrac12 p\frac{(1-e)+(1+e)}{1-e^2} = \frac{p}{1-e^2}$. The centre is the midpoint of the apse line, at distance $a - r_{\min} = \frac{p}{1-e^2}-\frac{p}{1+e} = \frac{p}{1+e}\big(\frac1{1-e}-1\big) = \frac{pe}{(1+e)(1-e)} = ea$ from the focus. For $b$: the point of the curve directly above the centre is at $\theta$ with $r\cos\theta = -ea$, and $r = p - er\cos\theta$ gives $r = p + e^2a = a(1-e^2) + e^2a = a$. Its height above the axis is then $b = \sqrt{r^2 - (ea)^2} = a\sqrt{1-e^2}$. $\square$ Hence also $b^2 = a^2(1-e^2) = a\cdot p$ and $r_{\min}r_{\max} = \frac{p^2}{1-e^2} = ap = b^2$.

Halley's proportions, in numbers. $a = 17.8325$ AU, $b = a\sqrt{1-e^2} = 17.8325\times0.254255 = 4.5340$ AU (the product-of-roots value from the Halley-on-the-graph notes ✓), $ea = 17.2464$ AU. So the ellipse is $2a = 35.7$ AU long and $2b = 9.07$ AU wide, and the Sun is $17.25$ AU from the centre — that is, $0.586$ AU from the near end. The picture of a comet's orbit as a long thin cigar with the Sun near the tip is right.

Kepler got here first, and the hard way. Kepler had Tycho's naked-eye positions of Mars, good to about two arcminutes, and no calculus, no dynamics and no reason to expect an ellipse. It took eight years and the Astronomia Nova (1609) to get laws I and II; law III came a decade later, in Harmonices Mundi (1619). What the last three slides did in fifteen minutes is the inverse problem — from the force to the shape — which is enormously easier than Kepler's, from the data to the shape. Newton's Principia (1687) did the hard direction too.

08 · Kepler

Third law: the constant, not just the power

The areal velocity $L/2\mu$ is constant, so one circuit takes area over rate, $T = \pi ab/(L/2\mu)$. Eliminate $L$: from $p = L^2/\mu k$ and $b^2 = ap$, $L = b\sqrt{\mu k/a}$. So

$$T = \frac{2\pi\mu ab}{L} = 2\pi\sqrt{\frac{\mu a^3}{k}} .$$

Kepler's third law $$T^2 = \frac{4\pi^2\mu}{k}\,a^3 \ \overset{\text{gravity}}{=}\ \frac{4\pi^2}{G(m_1+m_2)}\,a^3 ,$$

since $k = Gm_1m_2$ and $\mu = m_1m_2/M$ give $\mu/k = 1/GM$.

three checks

Dimensions. $[\mu a^3/k] = \mathsf{M}\cdot\mathsf{L^3}/\mathsf{ML^3T^{-2}} = \mathsf{T^2}$ ✓.

Against unit 00. Mechanical similarity gave $T\propto a^{3/2}$ with the constant undetermined; here it is $2\pi/\sqrt{GM}$ ✓.

The degenerate orbit. Let $e\to1$ at fixed $a$: the ellipse collapses to a segment travelled out and back. The fall time from rest at $r_0$ is half the period of an orbit with $a = r_0/2$, i.e. $\pi\sqrt{r_0^3/8GM} = \dfrac{\pi}{2\sqrt2}\sqrt{\dfrac{r_0^3}{GM}}$ — unit 00's checkpoint constant $C = \pi/2\sqrt2$, from an entirely different computation ✓.

Full proof that the area of the ellipse is $\pi ab$

In Cartesian coordinates centred on the ellipse, the curve is $x^2/a^2 + y^2/b^2 = 1$, so the region is $\{|y|\le b\sqrt{1-x^2/a^2}\}$ and its area is $$2\int_{-a}^{a}b\sqrt{1-\frac{x^2}{a^2}}\,\mathrm dx = \frac{2b}{a}\int_{-a}^{a}\sqrt{a^2-x^2}\,\mathrm dx .$$ Substituting $x = a\sin\varphi$, $\mathrm dx = a\cos\varphi\,\mathrm d\varphi$, the integral is $a^2\int_{-\pi/2}^{\pi/2}\cos^2\varphi\,\mathrm d\varphi = a^2\cdot\frac{\pi}{2}$. Hence the area is $\frac{2b}{a}\cdot\frac{\pi a^2}{2} = \pi ab$. $\square$ (Single-variable calculus only; unit 05's change of variables would give it in one line, but is not a prerequisite here.)

The degenerate limit, carefully. Fixing $a$ and letting $e\to1$ sends $r_{\min} = a(1-e)\to0$ and $r_{\max} = a(1+e)\to 2a$, and $L = b\sqrt{\mu k/a}\to0$ since $b = a\sqrt{1-e^2}\to0$. The limit is radial infall from rest at $r_0 = 2a$ (the initial data at aphelion tend to those of the fall, and unit 01's continuous dependence on initial data carries the solutions along over the finite time before the centre is reached), so $a = r_0/2$ and $T = 2\pi\sqrt{(r_0/2)^3/GM} = \frac{2\pi}{2\sqrt2}\sqrt{r_0^3/GM}$, of which the fall is half: $\frac{\pi}{2\sqrt2}\sqrt{r_0^3/GM}$. Unit 00 got the same by substituting $s = r_0\sin^2 u$ in the quadrature integral. Two routes, one number — and the second route never mentions an ellipse.

The stated constant, in solar units. Writing $a$ in AU and $T$ in years, $T^2 = a^3$ exactly when $4\pi^2/GM_\odot$ equals $1\ \mathrm{yr^2/AU^3}$. It does not, quite: $2\pi\sqrt{\mathrm{AU}^3/GM_\odot} = 1.0000189$ Julian years, because the Julian year of $365.25$ days is a definition, and so, since 2012, is the astronomical unit: a fixed number of metres, no longer the radius at which the Sun alone would give a period of exactly one year. The Earth's actual sidereal year differs from both again, partly because the Earth's and Moon's own masses enter (the worked example's notes).

08 · Kepler

Same size, same period

Five orbits with one semi-major axis $a$ and the Sun at a common focus: $e = 0$, $0.5$, $0.8$, $0.95$, and the limit $e\to1$, a straight drop from rest at $2a$. All five take the same time, $T = 2\pi\sqrt{a^3/GM}$ — the drop, there and back.

read off the third law

$T$ depends on $a$ alone. Shape, angular momentum and eccentricity do not enter: the circle and the needle-thin $e = 0.95$ ellipse keep the same time.

why, in one line

$a = -k/2E$, so equal $a$ means equal energy; and the period, area over areal rate, is $\pi ab\cdot2\mu/L$, in which $b\propto L$ cancels.

What to look for. The orbits share a focus, not a centre: as $e$ grows the ellipse slides left and thins, keeping its length $2a$, until at $e = 1$ it is the segment from the Sun to the point $2a$ away. A body on the $e = 0.95$ orbit spends almost all of its period far out near the left end and whips round the Sun in a small fraction of it — but the total is the circle's.

The cancellation, spelled out. From the third-law slide, $T = 2\pi\mu ab/L$ and $L = b\sqrt{\mu k/a}$, so $T = 2\pi\sqrt{\mu a^3/k}$: the $b$ in the area and the $b$ in the angular momentum are the same $b$, and they cancel. A thinner ellipse has less area to sweep and sweeps it proportionally more slowly.

The limit is the fall of unit 00. The dashed segment is the degenerate orbit of the third-law slide's checks: released from rest at $r_0 = 2a$, the body falls to the centre in $T/2$, which is $\frac{\pi}{2\sqrt2}\sqrt{r_0^3/GM}$.

08 · Kepler · example

Worked: the third law on three bodies

periods from semi-major axes

Formula. $T = 2\pi\sqrt{a^3/GM_\odot}$ with $GM_\odot = 1.3271244\times10^{20}\ \mathrm{m^3s^{-2}}$ and $1\ \mathrm{AU} = 1.495978707\times10^{11}$ m. One Julian year is $3.15576\times10^{7}$ s.

Earth, $a = 1$ AU. $a^3 = 3.34793\times10^{33}\ \mathrm{m^3}$; $a^3/GM_\odot = 2.52269\times10^{13}\ \mathrm{s^2}$; square root $5.02264\times10^6$ s; times $2\pi$: $3.15582\times10^{7}$ s $= 1.000019$ yr.

Halley, $a = 17.8325$ AU. $T = 1.000019\times17.8325^{3/2}\ \mathrm{yr} = 1.000019\times75.3038 = 75.305$ yr.

Jupiter, $a = 5.2038$ AU. $5.2038^{3/2} = 11.8708$, so $T = 11.8710$ yr with $M_\odot$ alone. With $G(M_\odot + m_{\mathrm J})$ and $m_{\mathrm J}/M_\odot = 1/1047.35$ it is $11.8710/\sqrt{1+1/1047.35} = 11.8654$ yr, shorter by $2.07$ days.

against the sky

body	computed	observed
Earth, $M_\odot$ only	1.000019 yr	1.000017 yr
Earth, $M_\odot+m_\oplus+m_{\text{Moon}}$	1.000017 yr	1.000017 yr
Halley	75.305 yr	75.3 yr
Jupiter, $M_\odot$ only	11.8710 yr	11.8620 yr
Jupiter, $M_\odot+m_{\mathrm J}$	11.8654 yr	11.8620 yr

For Jupiter the reduced-mass correction removes $2.07$ of the $3.3$ days by which the naive prediction was long. The residual $1.2$ days is the size of the ambiguity in $a$ itself: a planet pulled by Saturn has no single semi-major axis (notes).

What "observed" means in that table. Earth's sidereal year is $365.256363$ days $= 1.0000174$ Julian years. With the Sun alone, $2\pi\sqrt{\mathrm{AU}^3/GM_\odot} = 365.256898$ days; dividing by $\sqrt{1 + (m_\oplus + m_{\text{Moon}})/M_\odot}$, with the ratio $1/328\,900.56$, gives $365.256343$ days — two seconds from the sidereal year. Jupiter's sidereal period is $4332.59$ days $= 11.8620$ Julian years. For Halley the comparison is with the two-body period of the 1986 orbit, about $75.3$ yr, which is the like-for-like number since the computation on the slide is itself a two-body one.

Halley's period is itself only a two-body idealisation: the interval between successive perihelia changes by years from one return to the next, for the same reason as Jupiter's residual.

The honest limit of this agreement. Do not read the Jupiter row as a triumph of the reduced mass over observation, nor the residual as a measured effect of Saturn. Because the other planets perturb Jupiter, its osculating ellipse changes continually and "the" semi-major axis is an average whose value depends on how the average is taken: $5.2038$ AU is one published figure, and JPL's mean element $5.20289$ AU gives, with the same reduced-mass correction, $1.000019\times5.20289^{3/2}/\sqrt{1+1/1047.35} = 11.8623$ yr — within about $0.1$ day of the sidereal period. The spread between the two values of $a$ is the whole residual. What the row does show honestly is the size of the reduced-mass term — $2.07$ days — against the size of the discrepancy it is meant to explain, and that the sign is right. A clean test needs a system where $a$ is measured geometrically, which is what binary-star astrometry provides.

Scaling, to avoid the arithmetic. Once one row is done, every other is a ratio: $T_2/T_1 = (a_2/a_1)^{3/2}$, with no constants at all. That is how the table was actually computed, and it is why Kepler could state the law without knowing $G$, $M_\odot$, or the length of an astronomical unit in metres — none of which were known in 1619.

section 6

The vector that did not have to exist

A third conserved quantity, available only to the inverse-square law, which reads the orbit off in one line.

08 · Runge–Lenz

A conserved vector, and why anyone looked for one

The physics question. Every orbit so far has been described by $E$ and $L$ — two numbers, which fix the size and shape but say nothing about where perihelion points. Yet for the inverse-square law that direction is fixed for all time. Something must be conserving it.

the Laplace–Runge–Lenz vector

For $\vec F = -k\hat r/r^2$, with $\vec p = \mu\dot{\vec r}$:

$$\vec A := \vec p\times\vec L - \mu k\,\hat r .$$

What kind of object it is: a map from states $(\vec r,\vec p)$ to $\mathbb R^3$. $\vec p$ is a vector and $\vec L$ a pseudovector, so $\vec p\times\vec L$ is a genuine vector, as is $\hat r$: unlike $\vec L$, $\vec A$ behaves properly under reflections.

dimension check

$[\vec p\times\vec L] = \mathsf{MLT^{-1}}\cdot\mathsf{ML^2T^{-1}} = \mathsf{M^2L^3T^{-2}}$; $[\mu k] = \mathsf{M}\cdot\mathsf{ML^3T^{-2}} = \mathsf{M^2L^3T^{-2}}$ ✓. So $\vec A/\mu k$ is dimensionless, which is a hint about what it will turn out to be.

three names, one vector, many conventions

Hermann and Bernoulli (1710), Laplace (1799), Hamilton, Gibbs, Runge (1919), Lenz (1924) — rediscovered about six times. Some books use $\vec A/\mu k$ (dimensionless, magnitude $e$), some $\vec A/\mu$, some the opposite sign. Check the convention before comparing a formula with a book.

The name is an accident of citation: Lenz cited Runge's textbook, Pauli cited Lenz, and the eighteenth century was forgotten.

The mathematics first, as promised. $\vec A$ is defined as a map from the phase space $\mathbb R^6$ to $\mathbb R^3$: $(\vec r,\vec p)\mapsto \vec p\times(\vec r\times\vec p) - \mu k\vec r/\lVert\vec r\rVert$. The claim to be proved is that this map is constant along every solution of the equation of motion with the given $k$ — that is, that its three components are first integrals. Nothing about orbits is assumed in the statement, and the next slide's proof is a computation with the chain rule and one vector identity.

Why such a thing is surprising. Section 3's counting said two conserved quantities are enough to reduce the problem to quadrature. A third is not needed, and generically does not exist: a first integral is a function constant on every orbit, and for a typical central force a bounded orbit fills its annulus densely — in phase space, it is dense in the whole two-dimensional set where $E$ and $\vec L$ take their values — so any continuous first integral is constant on that set, a function of $E$ and $\vec L$, and carries no new information. The inverse-square orbits do not fill their annuli — they close — and that is precisely what leaves room for $\vec A$.

Unit 13 will explain it, partly. Noether's theorem attaches a conserved quantity to a symmetry of the Lagrangian; the three components of $\vec L$ come from rotations. $\vec A$ does not come from any symmetry of configuration space, and its symmetry — a rotation in a four-dimensional space mixing position and momentum, making the full group $SO(4)$ for bound orbits — is visible only in phase space. Fock found it in 1935, in the hydrogen atom.

08 · Runge–Lenz

$\dot{\vec A} = \vec 0$, by differentiating

the computation

$\dot{\vec L} = \vec 0$, so $\dot{\vec A} = \dot{\vec p}\times\vec L - \mu k\,\mathrm d\hat r/\mathrm dt$. First term: $\dot{\vec p} = -k\vec r/r^3$ and $\vec L = \mu\,\vec r\times\dot{\vec r}$, so

$$\dot{\vec p}\times\vec L = -\frac{k\mu}{r^3}\,\vec r\times(\vec r\times\dot{\vec r}) .$$

By BAC–CAB, with $\vec r\cdot\dot{\vec r} = r\dot r$ and $\vec r\cdot\vec r = r^2$,

$$\dot{\vec p}\times\vec L = -\frac{k\mu}{r^3}\big[r\dot r\,\vec r - r^2\dot{\vec r}\big] = k\mu\Big[\frac{\dot{\vec r}}{r} - \frac{\dot r}{r^2}\vec r\Big] .$$

Second term: $\dfrac{\mathrm d}{\mathrm dt}\dfrac{\vec r}{r} = \dfrac{\dot{\vec r}}{r} - \dfrac{\dot r}{r^2}\vec r$, the same bracket. Hence $\dot{\vec A} = \mu k\,[\cdot] - \mu k\,[\cdot] = \vec 0$. $\square$

where the inverse square was used

Once: the $r^{-3}$ in $\dot{\vec p}$ against the $r^2$ inside $\vec r\times(\vec r\times\dot{\vec r})$ leaves $r^{-1}$, matching $\mathrm d\hat r/\mathrm dt$. For $\dot{\vec p} = f(r)\hat r$ the first term is $-\mu f(r)r^2\,[\cdot]$, equal to $\mu k[\cdot]$ only if $f(r) = -k/r^2$.

Check the identity $\vec r\cdot\dot{\vec r} = r\dot r$. $r^2 = \vec r\cdot\vec r$; differentiate: $2r\dot r = 2\vec r\cdot\dot{\vec r}$. This is used twice on the slide and is the commonest slip in the computation — it is not $\lVert\dot{\vec r}\rVert r$.

The failure for other forces, quantified. Repeating the computation for a general central $f$ gives $\dot{\vec A} = -\mu\big(f(r)r^2 + k\big)\big[\frac{\dot{\vec r}}{r}-\frac{\dot r}{r^2}\vec r\big] = -\mu(f(r)r^2+k)\,\frac{\mathrm d\hat r}{\mathrm dt}$: the $\dot{\vec p}\times\vec L$ term is $(f/r)\,\mu\,\vec r\times(\vec r\times\dot{\vec r}) = \mu f\,(\dot r\,\vec r - r\dot{\vec r}) = -\mu fr^2[\cdot]$ by BAC–CAB. So $\vec A$ turns at a rate proportional to how far $r^2f$ departs from $-k$. That is the exact statement behind "an almost-inverse-square force gives an almost-fixed perihelion, drifting slowly" — which is what unit 10 computes for the Earth's oblateness and unit 37 for the correction general relativity adds. In both cases the number wanted is the average of $\dot{\vec A}$ over one orbit.

An alternative proof, for the record. Compute $\frac{\mathrm d}{\mathrm dt}(\vec p\times\vec L)$ in polar components directly: with $\vec L = L\hat z$, $\vec p\times\vec L = \mu L(\dot r\hat r + r\dot\theta\hat\theta)\times\hat z = \mu L(r\dot\theta\hat r - \dot r\hat\theta)$, and differentiating with the polar derivative rules of the polar-coordinates slide plus the radial equation gives $\mu k\dot\theta\hat\theta$, which is $\mu k\,\mathrm d\hat r/\mathrm dt$. Same answer, more components, no vector identity — and worth doing once as a check, as the problem sheet's P7 asks.

08 · Runge–Lenz

$\vec A$, drawn round the orbit

At four points of one orbit ($e = 0.6$), $\vec p\times\vec L$ (orange) and $-\mu k\hat r$ (dashed, towards the Sun) drawn head to tail. Both change from point to point; their sum $\vec A$ (green) is the same arrow every time, pointing along the axis towards perihelion. Computed; the vectors share one scale, different from the orbit's.

in the plane

$\vec A\cdot\vec L = (\vec p\times\vec L)\cdot\vec L - \mu k\,\hat r\cdot\vec L = 0$, by identity 1 and $\vec r\perp\vec L$. So $\vec A$ is a fixed vector in the orbital plane; the next slide shows it points at perihelion.

its length, read off the figure

At perihelion $\vec p\times\vec L$ points straight out, with length $\lVert\vec p\rVert L = \mu k(1+e)$, and $\mu k\hat r$ is taken off it: $A = \mu k e$. At aphelion $\vec p\times\vec L$ points out the other way, with length $\mu k(1-e)$, and adding $\mu k$ towards the Sun gives the same $\mu k e$, the same way round.

How the figure was computed. In units with $\mu = k = L = 1$, so that $p = L^2/\mu k = 1$, the orbit is $r = 1/(1 + e\cos\theta)$, and the polar velocity formula with $\dot r = e\sin\theta$, $r\dot\theta = 1 + e\cos\theta$ gives $\vec p\times\vec L = (1 + e\cos\theta)\,\hat r - e\sin\theta\,\hat\theta$ (the notes to the previous slide). Subtracting $\hat r$ leaves $e(\cos\theta\,\hat r - \sin\theta\,\hat\theta)$, which is $e$ times the unit vector along the axis, whatever $\theta$ is. That is the conservation law, checked point by point; the previous slide proved it without coordinates.

Why $\vec p\times\vec L$ is where it is. $\vec L$ points out of the page, so $\vec p\times\vec L$ is the momentum turned a right angle clockwise, scaled by $L$: perpendicular to the velocity, in the plane, on the outside of the orbit. Drawn from one common point instead, its tip runs round a circle of radius $\mu k$ centred on the tip of $\vec A$ as the body goes round — Hamilton's hodograph, turned through a right angle.

08 · Runge–Lenz

The orbit equation, in one line

dot $\vec A$ with $\vec r$

Let $\theta$ be the angle from $\vec A$ to $\vec r$. By the scalar triple product (identity 2), $(\vec p\times\vec L)\cdot\vec r = \vec L\cdot(\vec r\times\vec p) = L^2$, so

$$\vec A\cdot\vec r = L^2 - \mu k r .$$

But also $\vec A\cdot\vec r = A\,r\cos\theta$. Equate and solve for $r$:

$$r = \frac{L^2/\mu k}{1 + (A/\mu k)\cos\theta}. \qquad\square$$

No differential equation was solved: the conic came out of one dot product.

$p = L^2/\mu k$, $\ e = A/\mu k$, and $\vec A$ points at perihelion, since $r$ is least where $\cos\theta = 1$.

consistency

$p$ agrees with the Binet route ✓. Computing $\lVert\vec A\rVert^2$ directly (notes) gives $A^2 = \mu^2k^2 + 2\mu EL^2$, so $A/\mu k = \sqrt{1 + 2EL^2/\mu k^2}$ — the Binet route's $e$ ✓.

For Halley, $A/\mu k = 0.967137$, pointing from the Sun to perihelion.

Full computation of $\lVert\vec A\rVert^2$

$\lVert\vec A\rVert^2 = \lVert\vec p\times\vec L\rVert^2 - 2\mu k\,\hat r\cdot(\vec p\times\vec L) + \mu^2k^2$.

$\vec p\perp\vec L$ (since $\vec L = \vec r\times\vec p$), so $\lVert\vec p\times\vec L\rVert^2 = \lVert\vec p\rVert^2L^2$ by Lagrange's identity with the cross term vanishing;
$\hat r\cdot(\vec p\times\vec L) = \frac1r\vec r\cdot(\vec p\times\vec L) = \frac1r\vec L\cdot(\vec r\times\vec p) = L^2/r$.

So $\lVert\vec A\rVert^2 = \lVert\vec p\rVert^2L^2 - 2\mu kL^2/r + \mu^2k^2 = 2\mu L^2\big(\tfrac{\lVert\vec p\rVert^2}{2\mu} - \tfrac{k}{r}\big) + \mu^2k^2 = 2\mu L^2E + \mu^2k^2$, because $\lVert\vec p\rVert^2/2\mu - k/r$ is exactly the total energy $E$. $\square$ (The momentum is written $\lVert\vec p\rVert$ here, never $p$, because $p$ is the semi-latus rectum — a clash of standard notations worth watching.)

The circular case. If $\vec A = \vec 0$ the angle $\theta$ is undefined, but then $\vec A\cdot\vec r = 0$ gives $r = L^2/\mu k$ directly: a circle, $e = 0$ ✓. For Halley the direction of $\vec A$ is fixed to the accuracy of the two-body problem, and turned slowly by the planets.

What this route does not give. The one-line derivation produces the orbit but says nothing about the timetable, because $\vec A$ is a function of the state and carries no information about $t$. Kepler's equation is still unavoidable, and is unit 09.

The counting of integrals, finished. There are seven conserved scalars here: $E$, three components of $\vec L$, three of $\vec A$. Two relations cut them down — $\vec A\cdot\vec L = 0$ and $A^2 = \mu^2k^2 + 2\mu EL^2$ — leaving five independent. A bounded orbit is a one-dimensional curve in the six-dimensional phase space, so five independent constraints is exactly the number needed to pin it down. That is the sharpest available statement of why inverse-square orbits close: there are just enough conserved quantities to leave the orbit no room to drift.

08 · Runge–Lenz

Which forces close, and why almost none do

Take $\vec F = -c\,r^{n}\hat r$ with $c>0$ — attraction, power $n$. Circular orbits exist at every radius (pick $L$ to suit), and near one, of radius $r_0$, the radial motion is unit 06's harmonic oscillator with $\mu\omega_r^2 = V_{\mathrm{eff}}''(r_0)$. Compare it with the angular rate round the circle, $\omega_\theta = L/\mu r_0^2$.

stability and the apsidal angle $$V_{\mathrm{eff}}''(r_0) = c\,(n+3)\,r_0^{\,n-1},$$ $$\omega_r = \omega_\theta\sqrt{3+n},$$

so circular orbits are stable exactly for $n > -3$, and near one the apsidal angle is $\Delta\theta = \pi/\sqrt{3+n}$: near-circular orbits close only when $\sqrt{3+n}$ is rational.

derived on the problem sheet, P4

Bertrand's theorem — cited

The only central potentials for which every bounded orbit is closed are $V = -k/r$ and $V = \tfrac12\kappa r^2$: that is, $n = -2$ and $n = +1$.

Bertrand 1873. Cited: Goldstein, Classical Mechanics, §3.6 (Appendix A of the 2nd edition); Arnold, Mathematical Methods, §8. For power laws, P5 proves it.

the two survivors, checked

$n = -2$: $\Delta\theta = \pi$, the ellipse with the centre at a focus ✓. $n = +1$: $\Delta\theta = \pi/2$, the ellipse centred on the force ✓. $n = -1$: $\Delta\theta\approx\pi/\sqrt2$ near a circle, and varying with the amplitude — a typical orbit fills an annulus. $n = -3$: $\omega_r = 0$, marginal.

Where the proof stands. For power laws the problem sheet's P5 proves Bertrand's theorem completely except for one lemma, cited there: that the apsidal angle depends continuously on the energy. The general case needs a third-order expansion about the circular orbit — unit 21's kind of argument — and is cited.

Rational is not enough, and that is the content of Bertrand. The condition $\sqrt{3+n}\in\mathbb Q$ makes orbits close only near a circular one, where the harmonic approximation holds. Bertrand's theorem says that for every $n$ except $-2$ and $+1$, orbits far from circular fail to close even when the near-circular ones do — the apsidal angle depends on the amplitude. Proving that is the hard half, and it is the half this unit cites.

Where the formula comes from. With $f = -cr^n$, the circular condition $-f(r_0) = L^2/\mu r_0^3$ gives $L^2 = \mu c\,r_0^{n+3}$. Then $V_{\mathrm{eff}}'' = -f'(r) + 3L^2/\mu r^4$ evaluated at $r_0$ is $cn r_0^{n-1} + 3cr_0^{n-1} = c(n+3)r_0^{n-1}$, and $\omega_\theta^2 = (L/\mu r_0^2)^2 = c r_0^{n-1}/\mu$, so $\omega_r^2/\omega_\theta^2 = 3+n$. The problem sheet does every step.

The two survivors are the two with extra conserved objects. The inverse-square law has $\vec A$; the harmonic force has the Fradkin tensor $T_{ij} = p_ip_j/2\mu + \tfrac12\kappa x_ix_j$, a symmetric $(0,2)$-tensor whose conservation is checked the same way — differentiate and use the equation of motion: $\dot T_{ij} = -\kappa(x_ip_j + p_ix_j)/2\mu + \kappa(p_ix_j + x_ip_j)/2\mu = 0$. Unit 16 supplies the language for calling it a tensor; here it is six conserved numbers, whose trace is the energy. Closure and extra conserved quantities are two views of the same degeneracy, and Bertrand's list is the list of degenerate central forces.

Marginal $n = -3$ in full. There $V_{\mathrm{eff}} = (L^2/2\mu - c/2)/r^2$ is a pure inverse square, so there is no minimum at all unless the bracket vanishes, in which case every radius is neutral. The orbits are the Cotes spirals of the inverse-square slide's $n = -3$ check, solving $u'' + (1-\mu c/L^2)u = 0$; write $\gamma = \sqrt{|1-\mu c/L^2|}$. With $\mu c < L^2$, $u = C\cos\gamma\theta$: in from infinity and out again. With $\mu c > L^2$, $u$ is a combination of $e^{\pm\gamma\theta}$: a spiral into the centre or out to infinity. With $\mu c = L^2$, $u = A + B\theta$: a circle of any radius (every one neutral), or a reciprocal spiral. No bounded orbit except those circles.

08 · Runge–Lenz

Three force laws, one starting point

$\vec F = -r^n\hat r$, each started at $r = 1$ (open circle) moving sideways at $0.7$ of the circular speed, and integrated numerically. The two Bertrand forces close after one turn; $n = -1$ swings through $126^\circ$ from far to near point — close to the near-circular $\pi/\sqrt2 = 127.3^\circ$, not equal to it — and fills an annulus.

Read across. Same start, same speed: only the exponent differs. The force centre (black dot) is a focus of the inverse-square ellipse and the centre of the harmonic one.

Why it matters. Real planets feel $-k/r^2$ plus small extras. The extras make $\Delta\theta$ differ slightly from $\pi$ — the slow rosette of unit 10 and, for Mercury, unit 37.

How the figure was computed. Each orbit solves $\ddot{\vec r} = -r^{n-1}\vec r$ (so $c = \mu = 1$, and the circular speed at $r = 1$ is $1$ for every $n$) by fourth-order Runge–Kutta with step $2\times10^{-4}$, from $\vec r = (1,0)$, $\dot{\vec r} = (0, 0.7)$. Since $0.7$ is below the circular speed, $r = 1$ is the far point of each orbit. The angle from the start to the first near point is $180.0^\circ$ for $n = -2$, $90.0^\circ$ for $n = +1$ and $126.2^\circ$ for $n = -1$; the first two are the slide's $\pi$ and $\pi/2$ to the integrator's accuracy.

What to look for. In the left panel the force centre sits off-centre, at a focus; in the middle one it sits at the centre, and the orbit reaches its near point twice per turn, which is why $\Delta\theta = \pi/2$ there. In the right panel each near point is $2\times126.2^\circ$ round from the last, so successive loops are turned by $252.4^\circ$, or $-107.6^\circ$; after a dozen radial periods the annulus is visibly filling. At this amplitude the $n = -1$ apsidal angle is $1.1^\circ$ below its near-circular value: it depends on the amplitude, which is the hard half of Bertrand's theorem in a single number.

08 · closing

What this unit is the prerequisite for

09	Kepler's equation: inverting $t(r)$, which this unit left as an integral. Bessel functions come out of the series.
10	Real planets are not points: the multipole correction to $-k/r$, and the perihelion drift that $\dot{\vec A}\ne\vec0$ then gives.
12	Lagrangian mechanics: the same reduction with $\theta$ cyclic, done in one line instead of a section.
21	Perturbation theory, with the nearly-closed orbit as the standard example and the secular term as the drift.
37	The classical tests: $u''+u = \mu k/L^2 + 3GMu^2/c^2$. One extra term, and the number 43 arcseconds per century.

the three things to carry forward

A conserved quantity is a dimension removed. Six to two, and then quadrature.
$u = 1/r$. The one substitution in this course worth memorising, because unit 37 reuses it verbatim.
$V_{\mathrm{eff}}$ answers qualitative questions with no solving at all — and the questions it answers are most of the ones asked.

what was assumed and not proved

Bertrand's theorem in general (cited; P5 proves the power-law case). The equivalence of the cone and focus–directrix definitions of a conic (cited, not used). Bruns's theorem on the three-body problem (cited, in the notes). Unit 01's uniqueness theorem is used, not assumed.

The thread to unit 37, spelled out. Schwarzschild's metric gives a timelike geodesic equation which, in the same variable $u = 1/r$ and with $L$ the same conserved quantity of the same rotational symmetry, reads $u'' + u = GM\mu^2/L^2 + 3GMu^2/c^2$. The first term is this unit's; the second is smaller than $u$ by the factor $3GMu/c^2 \approx 7.6\times10^{-8}$ at Mercury's mean distance. Treating it by unit 21's perturbation theory gives a perihelion advance of $6\pi GM/c^2a(1-e^2)$ per orbit, which for Mercury is $43$ arcseconds per century. That number, and the fact that it comes from adding one term to an equation derived on these slides, is the reason this unit is on the spine.

A caution about the LRL vector in perturbed problems. Once $\vec A$ is not exactly conserved, the temptation is to treat it as slowly varying and average $\dot{\vec A}$ over an unperturbed orbit. Differentiating $\vec A$ with the perturbing force included is exact; replacing the result by its average over one unperturbed orbit is not. It is an asymptotic step, correct to first order in the perturbation, whose error needs unit 21's machinery to control.

08 · checkpoint

Done when

checkpoint

From $V(r) = -k/r$ alone you derive $r(\theta) = p/(1+e\cos\theta)$, express $e$ and $p$ in terms of $E$ and $L$, prove Kepler's second and third laws, and verify that the Runge–Lenz vector is conserved by explicit differentiation.

How to fail it, unambiguously

Any of: the Binet equation is quoted rather than derived from $\dot r = -(L/\mu)u'$; the solution of $u''+u = 1/p$ is written down without saying that $\dim\ker\mathcal D = 2$ and why; $e = \sqrt{1+2EL^2/\mu k^2}$ appears without the energy being evaluated somewhere on the orbit; the third law is stated as $T^2\propto a^3$ without the constant $4\pi^2/G(m_1+m_2)$; or $\dot{\vec A} = \vec 0$ is asserted with "one checks that" in place of BAC–CAB and $\vec r\cdot\dot{\vec r} = r\dot r$.

Then

./mr problems 08 — P6 is the full Kepler derivation, P7 the Runge–Lenz differentiation, P4 and P5 the closure question, P9 Rutherford. Each problem's hints and worked solution are folded under it: open them only after an attempt.
./mr done 08 when the checkpoint is met.

Reading, optional: Arnold, Mathematical Methods, §8; Newton, Principia, Book I, Props. I–XVII. Landau & Lifshitz §§13–15 and Goldstein ch. 3 start from a Lagrangian (unit 12): skip those lines.

The falsifiable part is "by explicit differentiation". It is easy to look up $\vec A = \vec p\times\vec L - \mu k\hat r$ and to assert that it is conserved; the checkpoint is the four lines of the "$\dot{\vec A} = \vec 0$" slide, produced without the slide, including the place where the inverse square is used and the identity $\vec r\cdot\dot{\vec r} = r\dot r$. If those four lines are not available, the perturbed versions in units 10 and 37 — where the same computation is done with an extra term and the residual is the answer — will not be either.

Try the checkpoint on a force this deck has not solved: $f(r) = -k/r^2 - c/r^3$, a small extra attraction. The Binet equation becomes $u'' + (1 - \mu c/L^2)u = \mu k/L^2$, still linear, still solvable by unit 06. Read off the apsidal angle and check that it reduces to $\pi$ when $c = 0$. That is the problem sheet's P11, and in miniature it is unit 37's calculation with a different source for the extra term.

When the checkpoint is met: ./mr done 08.