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Edmond Halley

Edmond Halley, Savilian Professor of Geometry, Astronomer Royal

A course in two strands: the life of a soap-maker’s son who catalogued the southern sky at twenty, steered Newton’s Principia into print and commanded a navy ship, and the science he built from Newton’s laws: periodic comets, the transit method for the Sun’s distance, life tables, charts of winds and magnetism, and a cubically convergent root-finder.

Born 8 November 1656 (N.S.), Haggerston, London Died 25 January 1742 (N.S.), Greenwich Prediction the comet’s return in 1758 Level advanced undergraduate
The orbit of Halley’s comet to scale (a ≈ 17.8 AU, e ≈ 0.967), projected on the ecliptic with circles for Earth, Jupiter, Saturn, Uranus and Neptune. Dots mark equal four-year intervals from Kepler’s equation: the comet spends most of its 76 years far from the Sun.
Lesson 00

How to use this course

Edmond Halley’s name is attached to a comet, but his range was far wider. He was an observer, a calculator, a geophysicist, a demographer, a classical scholar, a ship’s captain and an editor. He was the first person to make a successful prediction from Isaac Newton’s gravitation, and without him the Principia might never have been written. This course follows his science alongside the life of a sociable, energetic man at the centre of the early Royal Society.

Prerequisites for Part II
  • Single-variable calculus, including logarithms and simple integrals.
  • Newtonian mechanics: circular motion and the law of gravitation.
  • Kepler’s laws; the course on Kepler covers them in full.
  • For Lesson 07, elementary probability; for Lesson 09, Taylor series.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements in modern notation, derivations and exercises with folded solutions. England used the Julian calendar in Halley’s lifetime, ten days behind the Gregorian calendar until 1700 and eleven days after. The course gives Gregorian dates for birth and death and the dates used in England for other events.

Reference

Timeline

1656Born 8 November (29 October O.S.) at Haggerston, near London, son of Edmond Halley, a prosperous soap-maker.
1671Captain of St Paul’s School.
1673Enters The Queen’s College, Oxford.
1676Leaves Oxford without a degree and sails for St Helena to map the southern sky.
1677Observes a complete transit of Mercury from St Helena, 7 November.
1678Returns in May. Elected Fellow of the Royal Society on 30 November; MA by royal mandate on 3 December. Catalogus stellarum australium (dated 1679) lists 341 southern stars.
1679Sent to Danzig to examine the instruments of Johannes Hevelius.
1680–81Tour of France and Italy; observes the great comet of 1680 and works with Giovanni Domenico Cassini in Paris.
1682Marries Mary Tooke. Observes the comet later named after him.
1684January: discusses the inverse-square law with Christopher Wren and Robert Hooke. August: visits Newton in Cambridge.
1686Salaried officer of the Royal Society. Paper on the trade winds and monsoons with the first meteorological chart; paper on barometric heights.
1687The Principia is published at Halley’s expense.
1691Diving bell trials in the Thames. Fails to obtain the Savilian chair of astronomy.
1692Proposes a hollow, layered Earth to explain the drift of magnetic variation.
1693Life table for Breslau and the valuation of annuities.
1694Method for the roots of equations.
1696–98Deputy comptroller of the mint at Chester.
1698–1701Commands HMS Paramour on two Atlantic voyages and a survey of Channel tides. Chart of magnetic variation with isogonic lines (1701).
1703–04Savilian Professor of Geometry at Oxford.
1705Astronomiae cometicae synopsis: identifies the comets of 1531, 1607 and 1682 and predicts a return in 1758.
1710Edition of Apollonius’s Conics, with a reconstruction of the lost Book VIII.
1715Map predicting the path of the total solar eclipse of 22 April (O.S.) over England.
1716Method of using transits of Venus to measure the distance of the Sun.
1718Publishes the proper motion of stars.
1720Astronomer Royal at Greenwich, succeeding John Flamsteed.
1742Dies at Greenwich, 25 January (14 January O.S.), aged 85.
1758–59The comet is sighted on 25 December 1758 and passes perihelion on 13 March 1759.
Part I · Formation
Lesson 01

London and St Paul’s, 1656–1673

Edmond Halley was born on 8 November 1656 (29 October in the English calendar) at Haggerston, then a village north-east of the City of London. His father, also Edmond, came from a Derbyshire family and had grown rich making soap in London. The Great Fire of 1666 destroyed much of the family’s property, but the business survived and the father could still afford the best education for his son. His mother Anne died in 1672.

Halley was taught first at home and then at St Paul’s School, where he was elected captain of the school in 1671. He was already an observer. Before leaving school he had measured the variation of the compass needle in London, a quantity that would occupy him for the rest of his life, and he had begun to collect astronomical instruments.

Lesson 02

Oxford and St Helena, 1673–1678

In July 1673 Halley entered The Queen’s College, Oxford, arriving with a set of instruments his father had bought, including a telescope of 24 feet. He wrote to John Flamsteed, the first Astronomer Royal, and from 1675 assisted him with observations at the new Royal Observatory in Greenwich. His first paper, in 1676, gave a direct geometric method for finding the aphelia and eccentricities of planetary orbits. That year he also observed an occultation of Mars by the Moon.

Flamsteed was cataloguing the northern sky. Halley saw that the southern sky, invisible from Europe, had no accurate catalogue at all. In November 1676 he left Oxford without a degree and sailed on an East India Company ship for St Helena, the southernmost British possession, with his father’s money and a letter from Charles II. The weather on the island was cloudy, and he stayed about a year and a half. He catalogued the positions of 341 stars with a sextant and on 7 November 1677 observed a complete transit of Mercury across the Sun. Lesson 06 explains why that transit mattered.

He returned in May 1678. His Catalogus stellarum australium, the first catalogue of southern stars made with a telescopic sight, made his name. Flamsteed called him the southern Tycho. On the King’s instruction Oxford granted him the degree of Master of Arts on 3 December 1678 without the usual residence and examinations.

Lesson 03

The young Fellow, 1678–1684

Halley was elected a Fellow of the Royal Society on 30 November 1678, aged 22. In 1679 the Society sent him to Danzig to settle a dispute between Robert Hooke and Johannes Hevelius. Hooke claimed that Hevelius’s naked-eye sights could not match telescopic ones. Halley observed with Hevelius and reported that his measurements were remarkably accurate.

In 1680 and 1681 he toured France and Italy with his school friend Robert Nelson. Near Calais he saw the great comet of 1680, and in Paris he worked with Giovanni Domenico Cassini at the new Observatory on its orbit. He spent much of 1681 in Italy. In 1682 he married Mary Tooke and settled in Islington, where he set up an observatory. That August he observed a bright comet, the one that would carry his name.

In early 1684 his father disappeared and was found dead about five weeks later. The same months brought the conversation, recounted in Lesson 04, that led to Newton’s Principia.

Part II · The physics
Lesson 04

From Kepler’s third law to the Principia

For the general reader. By 1684 several people in London suspected that the Sun pulls the planets with a force falling off as the square of distance. Halley had shown it from Kepler’s third law for circular orbits. Nobody could prove the harder converse, that such a force produces elliptical orbits. Halley took the question to Newton, who said he had already solved it. Halley then persuaded him to write the solution out, and the result was the Principia.

Goals
  • Derive the inverse-square law from Kepler’s third law for circles.
  • Understand why the ellipse problem was hard and how Newton answered it.
  • See Halley’s role as editor and publisher.

The circular argument

Christiaan Huygens had published in 1673 that a body moving uniformly on a circle of radius \(r\) with speed \(v\) needs a central force proportional to \(v^2/r\). Combine this with Kepler’s third law.

Proposition (Halley, Wren, Hooke, Newton, independently)

If bodies move uniformly on concentric circles with periods satisfying \(T^2=ka^3\), then the central accelerations are \(\dfrac{4\pi^2}{k}\cdot\dfrac{1}{a^2}\).

Proof. \(v=2\pi a/T\), so the acceleration is \(v^2/a=4\pi^2a/T^2=4\pi^2/(ka^2)\). ∎

The converse and the coffee-house

In January 1684 Halley, Christopher Wren and Robert Hooke discussed the question after a meeting of the Royal Society. Hooke claimed he could derive all celestial motions from an inverse-square law. Wren offered a book worth forty shillings to whoever could produce a proof within two months, and none came. In August 1684 Halley visited Newton in Cambridge and asked what curve a planet would follow under an inverse-square attraction. Newton answered at once that it would be an ellipse, but could not find his calculation. He promised to redo it and in November sent Halley a nine-page tract, De motu corporum in gyrum.

Theorem (Newton, 1684–87)

Under an attractive central force of magnitude \(GMm/r^2\), a body moves on a conic section with the centre of force at a focus: \(r=\dfrac{\ell}{1+e\cos\nu}\) with \(\ell=\dfrac{L^2}{GMm^2}\). The orbit is an ellipse, parabola or hyperbola according as the energy \(\tfrac12mv^2-GMm/r\) is negative, zero or positive.

In modern form, with \(u=1/r\) and \(L=mr^2\dot\nu\) constant (the area law), the radial equation becomes Binet’s equation \(u''+u=GMm^2/L^2\), whose solutions are \(u=(1+e\cos\nu)/\ell\).

Halley as publisher

Halley reported the tract to the Royal Society and urged Newton to expand it. Over the next two and a half years he handled the correspondence, corrected the proofs, smoothed a priority quarrel with Hooke, and wrote a Latin ode to Newton printed at the front. The Society had agreed to publish the book but had no money, having spent heavily on a History of Fishes. Halley paid for the printing himself. The Principia appeared in July 1687.

Exercise 4.1The Moon is about 60 Earth radii away and orbits in 27.32 days. Taking Earth’s radius as 6371 km, compute its centripetal acceleration and compare with \(g/60^2\).

Show solution
\(a=4\pi^2r/T^2\) with \(r=3.823\times10^8\) m and \(T=2.360\times10^6\) s gives \(a\approx2.71\times10^{-3}\) m/s². And \(9.81/3600=2.72\times10^{-3}\) m/s². This is Newton’s “Moon test” of the inverse-square law.

Exercise 4.2Solve \(u''+u=c\) and show the orbit \(r=1/u\) is a conic with focus at the origin.

Show solution
The general solution is \(u=c+A\cos(\nu-\nu_0)\). Choosing \(\nu_0=0\) and writing \(A=ce\), \(r=\dfrac{1/c}{1+e\cos\nu}\): the focus–directrix form of a conic with focus at the origin and semi-latus rectum \(\ell=1/c\).
Lesson 05

Comets: parabolas and a periodic return

For the general reader. Before Newton, comets were thought to travel in straight lines or to be one-time visitors. Newton showed that a comet near the Sun moves almost on a parabola. Halley computed such orbits for every well-observed comet in history and noticed that three of them, seen 75 or 76 years apart, had nearly the same path. He concluded they were one object on a long ellipse, and predicted its return in 1758. The comet came back after his death, and it has carried his name since.

Goals
  • Derive Barker’s equation for parabolic motion.
  • Relate period, distance and speed for Halley’s comet.
  • Understand why the return was delayed by planetary perturbations.

The synopsis of 1705

Halley’s Astronomiae cometicae synopsis, printed at Oxford in 1705 and in English the same year, tabulated parabolic orbits for 24 comets observed between 1337 and 1698. Each orbit is fixed by five elements: the node, inclination, longitude of perihelion, perihelion distance \(q\) and time of perihelion. The comets of 1531, 1607 and 1682 had very similar elements, all with retrograde motion. Halley argued that their differences in period, slightly more and slightly less than 76 years, came from the attraction of Jupiter and Saturn. He predicted a return about the end of 1758 and asked posterity to remember that an Englishman had foreseen it.

Barker’s equation (parabolic orbit)

For a parabolic orbit with perihelion distance \(q\) about a mass \(M\), with \(D=\tan(\nu/2)\),

\[ t-t_p=\sqrt{\frac{2q^3}{GM}}\left(D+\frac{D^3}{3}\right), \qquad r=q\,(1+D^2). \]

Derivation. For a parabola \(e=1\) and \(\ell=2q\), so \(r=2q/(1+\cos\nu)=q\sec^2(\nu/2)\). The area law gives \(r^2\,d\nu=h\,dt\) with \(h=\sqrt{GM\ell}=\sqrt{2GMq}\). With \(d\nu=2\,dD/(1+D^2)\), we get \(r^2d\nu=2q^2(1+D^2)\,dD\). Integrating from perihelion gives the formula. The cubic in \(D\) has exactly one real root, so the inverse problem is easy, unlike Kepler’s equation.

The ellipse behind the parabola

Near perihelion a very eccentric ellipse is indistinguishable from the parabola with the same \(q\). Halley’s identification therefore needed only the parabolic elements plus a period. With period \(T\) in years and distances in AU, Kepler’s third law gives \(a=T^{2/3}\), and the energy equation gives the speed at any distance:

\[ v^2=GM\left(\frac2r-\frac1a\right)\qquad\text{(vis-viva)}. \]

The comet was sighted on 25 December 1758 by Johann Georg Palitzsch, a farmer and amateur astronomer near Dresden, and passed perihelion on 13 March 1759. Alexis Clairaut, Joseph-Jérôme Lalande and Nicole-Reine Lepaute had computed the perturbations by Jupiter and Saturn and predicted perihelion within a month of 13 April 1759. The comet’s return was the first confirmed prediction of a periodic comet and a public triumph for Newton’s gravitation.

Exercise 5.1Take Halley’s comet to have period 75.3 years and perihelion distance 0.59 AU. Find \(a\), \(e\) and the aphelion distance.

Show solution
\(a=75.3^{2/3}\approx17.8\) AU. \(e=1-q/a=1-0.59/17.8\approx0.967\). Aphelion \(=a(1+e)\approx35.0\) AU, beyond Neptune.

Exercise 5.2Earth’s mean orbital speed is 29.8 km/s. Find the speed of Halley’s comet at perihelion (\(q=0.587\) AU) and compare with the parabolic speed there.

Show solution
In units where Earth’s circular speed at 1 AU is 29.8 km/s, \(v=29.8\sqrt{2/0.587-1/17.8}\approx29.8\times1.829\approx54.5\) km/s. The parabolic speed is \(29.8\sqrt{2/0.587}\approx55.0\) km/s. The difference is under 1 percent, which is why parabolic elements served Halley so well.

Exercise 5.3A comet on a parabolic orbit has \(q=1\) AU. How many days does it take to go from perihelion to \(\nu=90^\circ\)?

Show solution
In AU and years \(GM=4\pi^2\), so \(\sqrt{2q^3/GM}=\sqrt2/(2\pi)\approx0.2251\) yr. With \(D=1\), \(D+D^3/3=4/3\). The time is \(0.3001\) yr \(\approx109.6\) days.
Lesson 06

Transits and the distance of the Sun

For the general reader. Kepler’s laws give the relative sizes of planetary orbits but not the scale. One measured distance fixes everything. Halley saw at St Helena that the timing of a planet’s passage across the Sun’s face could supply that distance. Venus, closer to us than Mercury, was the better choice. In 1716 he urged astronomers to observe the transits of 1761 and 1769 from widely separated places. The international expeditions that followed, including James Cook’s voyage to Tahiti, measured the Sun’s distance to within a few percent.

Goals
  • Define the solar parallax and relate it to the astronomical unit.
  • Understand Halley’s method of durations.
  • Compute the parallax displacement of Venus on the solar disc.

Parallax

The solar parallax \(\pi_\odot\) is the angle subtended by Earth’s equatorial radius \(R_\oplus\) at the distance \(A\) of the Sun: \(\sin\pi_\odot=R_\oplus/A\). The modern value is \(8.794''\). In Halley’s time estimates ranged widely, and he himself suspected a value near \(12.5''\).

Halley’s method of durations (1716)

Observers at two places of known separation time the whole transit of Venus, from first to last contact. Parallax shifts Venus’s path across the solar disc by different amounts for the two observers, so the chords, and the durations, differ. The difference in durations, together with the known geometry of the disc, gives the relative displacement of the chords and hence the solar parallax.

Only clocks are needed, not absolute positions, which suited the instruments of the time. Absolute time is not needed either: each observer measures an interval locally. The parallax of Venus against the Sun is enhanced because Venus is much nearer. At inferior conjunction Venus is at about \(1-0.723=0.277\) AU, so the relative displacement of Venus on the disc for two observers separated by a baseline \(B\) perpendicular to the line of sight is

\[ \Delta\theta=\frac{B}{A}\left(\frac{1}{0.277}-1\right)\approx2.61\,\frac{B}{A}. \]

The chord displacement is a small fraction of the solar radius, so the durations differ by minutes out of several hours. In practice the “black drop” effect blurred the moment of contact by tens of seconds and limited the 1761 and 1769 results. Johann Franz Encke’s reduction gave about \(8.6''\), and later analysis of the eighteenth-century data gave values near the modern one.

Exercise 6.1Use \(\pi_\odot=8.794''\) and \(R_\oplus=6378\) km to find the astronomical unit.

Show solution
\(8.794''=4.263\times10^{-5}\) rad, so \(A=6378/4.263\times10^{-5}\approx1.496\times10^8\) km.

Exercise 6.2Two observers are separated by 10,000 km perpendicular to the line of sight. How far apart, in arc-seconds, do they see the centre of Venus on the solar disc during a transit?

Show solution
\(B/A=10^4/1.496\times10^8=6.68\times10^{-5}\) rad \(=13.8''\). Multiplying by 2.61 gives about \(36''\), about 4 percent of the Sun’s radius of roughly \(960''\).

Exercise 6.3A chord of the solar disc of radius \(\rho\) is at distance \(d\) from the centre and is crossed at a constant rate. Show that a small shift \(\delta d\) changes the duration \(\tau\) by \(\delta\tau/\tau\approx-d\,\delta d/(\rho^2-d^2)\). Why did Halley prefer transits whose path lies well away from the centre?

Show solution
The chord length is \(2\sqrt{\rho^2-d^2}\), proportional to \(\tau\). Differentiating, \(\delta\tau/\tau=-d\,\delta d/(\rho^2-d^2)\). The sensitivity vanishes for a central transit and grows for an off-centre chord, so off-centre paths turn a small parallax shift into a measurable change in duration.
Lesson 07

Life tables and annuities

For the general reader. How much should a government charge for a pension that pays a fixed sum each year until the buyer dies? The answer depends on how long people of each age live. In 1693 Halley used records of births and deaths from the city of Breslau to build one of the first life tables and showed how to price such an annuity. The English government was then selling annuities at the same price regardless of age.

Goals
  • Understand the stationary-population assumption behind Halley’s table.
  • Compute survival probabilities from a life table.
  • Value a life annuity by discounting expected payments.

The Breslau data

Caspar Neumann, a pastor in Breslau (now Wrocław), had compiled the city’s records of births and deaths by age, and they reached the Royal Society through Gottfried Wilhelm Leibniz’s circle. London’s bills of mortality, used earlier by John Graunt, did not record ages at death and were distorted by migration. Breslau was a city with few migrants, and births and deaths were nearly balanced. Halley assumed a stationary population: the number born each year is constant, and so is the age pattern of deaths. Then the number of people alive at age \(x\) in the city equals the number of a single birth cohort who survive to age \(x\). Breslau averaged about 1,238 births a year.

Definitions

Let \(l_x\) be the number alive at exact age \(x\) out of a fixed cohort. The probability that a person aged \(x\) survives \(t\) more years is \({}_tp_x=l_{x+t}/l_x\). With interest rate \(i\) and discount factor \(v=1/(1+i)\), the present value of an annuity paying 1 at the end of each year while the holder lives is

\[ a_x=\sum_{t\ge1} v^t\,\frac{l_{x+t}}{l_x}. \]

Each payment is made only if the holder is alive, so its expected present value is \(v^t\,{}_tp_x\), and the annuity is the sum. Halley computed such values at 6 percent interest for each age. He also answered questions such as the odds that a person of a given age dies within a year, and the age to which a person has even odds of living. His table and method were extended by Abraham de Moivre and founded actuarial science.

Exercise 7.1A life table has \(l_{40}=100\), \(l_{41}=90\), \(l_{42}=75\), \(l_{43}=50\), \(l_{44}=0\). Find the value at 5 percent interest of an annuity of 1 per year for a person aged 40.

Show solution
\(a_{40}=\frac{0.90}{1.05}+\frac{0.75}{1.05^2}+\frac{0.50}{1.05^3}=0.857+0.680+0.432=1.969\).

Exercise 7.2In the same table, what is the probability that a person aged 41 dies before 43? At what age does a 40-year-old have even odds of still being alive?

Show solution
\(1-l_{43}/l_{41}=1-50/90=4/9\). Half of 100 survive to 43, so the even-odds age is 43.

Exercise 7.3Show that in a stationary population with \(B\) births per year, the total population is \(B\,e_0\), where \(e_0=\frac{1}{l_0}\int_0^\infty l_x\,dx\) is the life expectancy at birth.

Show solution
The people aged between \(x\) and \(x+dx\) today were born between \(x+dx\) and \(x\) years ago, about \(B\,dx\) of them, of whom a fraction \(l_x/l_0\) survive. Integrating over \(x\) gives \(B\int_0^\infty(l_x/l_0)\,dx=Be_0\). Halley used this kind of reasoning to estimate Breslau’s population from its births.
Lesson 08

Winds, pressure and the magnetic Earth

For the general reader. Halley was a geophysicist before the word existed. He drew the first map of the world’s winds and explained the trade winds by the Sun’s heating of the tropics. He worked out how air pressure falls with height. He tested a diving bell in the Thames. On a navy ship he crossed the Atlantic to map the compass’s deviation from true north, and published the first chart with lines of equal magnetic variation. To explain the slow drift of those lines, he proposed that the Earth has an inner core turning inside an outer shell.

Goals
  • Derive the exponential fall of pressure with height.
  • Understand Halley’s explanation of the trade winds and its later correction.
  • Read an isogonic chart and Halley’s model of magnetic drift.

The trade winds (1686)

Halley’s 1686 paper gathered sailors’ reports into a chart of the prevailing winds over the oceans. He argued that the Sun heats the air most strongly near the equator, where the heated air rises and cooler air flows in from north and south. To explain why the trades blow from the east, he appealed to the daily westward passage of the Sun. George Hadley showed in 1735 that the easterly component comes from Earth’s rotation: air moving toward the equator keeps less eastward speed than the ground beneath it.

Pressure and height (1686)

Halley combined Boyle’s law, density proportional to pressure, with the weight of the air column, and found that heights in arithmetic progression correspond to pressures in geometric progression. He used the hyperbola area, then the standard form of the logarithm, to compute them.

Barometric formula (isothermal atmosphere)

If \(dp/dz=-\rho g\) and \(\rho=p/(gH)\) for a constant scale height \(H\), then \(p(z)=p_0\,e^{-z/H}\), where \(H=p_0/(\rho_0g)\approx8.4\) km at sea level and \(0^\circ\)C conditions.

The magnetic Earth

In 1683 Halley proposed that the Earth has four magnetic poles. In 1692 he gave a physical model: a solid outer shell and an inner sphere separated by a fluid, each magnetized and turning at slightly different rates, so the pattern of variation drifts slowly westward. On his Paramour voyages of 1698–1700 he measured variation across the Atlantic, and his 1701 chart joined points of equal variation by curves, now called isogonic lines. Contour lines on maps of all kinds descend from this idea. The modern picture of a fluid outer core with a solid inner core keeps the westward drift that Halley identified.

Exercise 8.1At what height is the pressure half its sea-level value, with \(H=8.4\) km? Halley estimated the height of the atmosphere from such reasoning. Why does the formula have no top?

Show solution
\(z=H\ln2\approx5.8\) km. The exponential never reaches zero, so the atmosphere has no sharp edge; a finite “height” only appears if one assumes constant density, which gives \(H\) itself.

Exercise 8.2Halley’s diving bell was used at a depth of about 60 feet (18.3 m) of water. Treating the air as isothermal, to what fraction of its surface volume is the air in an open bell compressed? Why did he send down barrels of fresh air?

Show solution
The pressure is \(1+\rho gh/p_0\approx1+(1000)(9.81)(18.3)/101325\approx2.8\) atmospheres in fresh water, so the air shrinks to about 0.36 of its volume and water rises in the bell. Weighted barrels with a hose at the bottom forced fresh air in, which pushed the water out and replaced the air the divers had breathed.

Exercise 8.3The ground at the equator moves east at 465 m/s, and at latitude 30° at \(465\cos30^\circ\approx403\) m/s. Air starting at rest relative to the ground at 30° moves to the equator conserving its angular momentum about Earth’s axis. Find its velocity relative to the ground there, and explain Hadley’s correction to Halley.

Show solution
Angular momentum per unit mass is \(uR\cos\varphi\). Starting with \(u=403\) m/s at \(R\cos30^\circ\), at the equator it has \(u=403\cos30^\circ\approx349\) m/s, about 116 m/s slower than the ground: a strong easterly wind. Friction reduces this to the observed trade winds of a few metres per second, but the direction comes from rotation, as Hadley saw.
Lesson 09

Halley’s method and Apollonius

For the general reader. Halley was a strong mathematician in the classical style. In 1694 he published a method for solving equations that roughly triples the number of correct digits at each step. Later, as professor of geometry at Oxford, he learned Arabic to recover the lost books of Apollonius of Perga on conic sections, and published a standard edition of the Conics.

Goals
  • State Halley’s method and prove its cubic convergence.
  • Relate it to Newton’s method.
  • See what Apollonius’s Conics contain and what Halley restored.

Roots of equations (1694)

Halley’s paper treated polynomial equations, using the first two terms of the expansion of \(f(x+h)\) to correct an approximate root. In modern form, for any twice differentiable \(f\):

Halley’s method \[ x_{n+1}=x_n-\frac{2f(x_n)f'(x_n)}{2f'(x_n)^2-f(x_n)f''(x_n)}. \]

If \(f\) is three times continuously differentiable near a simple root \(\alpha\) (\(f'(\alpha)\ne0\)), then for \(x_0\) close enough to \(\alpha\) the iterates converge to \(\alpha\) with \(|x_{n+1}-\alpha|\le C|x_n-\alpha|^3\).

Proof sketch. Let \(g=f/\sqrt{|f'|}\). Then \(g\) has the same simple root, and a short computation gives \(g'/g=f'/f-f''/(2f')\). Newton’s step for \(g\), \(x-g/g'\), equals Halley’s step for \(f\). Moreover \(g''(\alpha)=0\): writing \(f=f'(\alpha)\varepsilon+\tfrac12f''(\alpha)\varepsilon^2+O(\varepsilon^3)\) with \(\varepsilon=x-\alpha\), the quadratic term of \(g\) cancels. Newton’s error constant is \(g''/(2g')\), so the quadratic error term vanishes and convergence is cubic. ∎

Each step costs one more derivative than Newton’s, so the gain is real only when \(f''\) is cheap. Halley’s method is the second member of Householder’s family of methods, of which Newton’s is the first.

Apollonius

Apollonius of Perga wrote eight books on conic sections around 200 BCE. Books I–IV survive in Greek, V–VII only in Arabic translation, and Book VIII is lost. In 1706 Halley published a Latin translation of Apollonius’s De sectionis rationis from the Arabic, and in 1710 a Greek and Latin edition of the Conics, with Books V–VII from the Arabic and his own reconstruction of Book VIII from hints in Pappus. The edition remained standard for two centuries.

Exercise 9.1Apply one step of Halley’s method to \(f(x)=x^2-2\) from \(x_0=1.5\), and compare with one step of Newton’s method.

Show solution
\(f=0.25\), \(f'=3\), \(f''=2\). Halley: \(1.5-\frac{2(0.25)(3)}{2(9)-0.25(2)}=1.5-\frac{1.5}{17.5}=1.414286\), error \(7\times10^{-5}\). Newton: \(1.5-0.25/3=1.416667\), error \(2.5\times10^{-3}\).

Exercise 9.2Show that for \(f(x)=x^2-c\), Halley’s method becomes \(x_{n+1}=x_n\dfrac{x_n^2+3c}{3x_n^2+c}\).

Show solution
\(x-\frac{2(x^2-c)(2x)}{2(4x^2)-2(x^2-c)}=x-\frac{2x(x^2-c)}{3x^2+c}=\frac{x(3x^2+c)-2x(x^2-c)}{3x^2+c}=x\frac{x^2+3c}{3x^2+c}\).

Exercise 9.3Use Halley’s method to solve Kepler’s equation \(E-0.967\sin E=M\) for \(M=0.1\), starting from \(E_0=1\). Do one step.

Show solution
\(f=1-0.967\sin1-0.1=1-0.8137-0.1=0.0863\); \(f'=1-0.967\cos1=0.4775\); \(f''=0.967\sin1=0.8137\). The step is \(\frac{2(0.0863)(0.4775)}{2(0.2280)-0.0863(0.8137)}=\frac{0.0824}{0.3858}=0.2137\), so \(E_1\approx0.786\). The exact root is \(E\approx0.780\). Newton’s step from the same start gives \(0.819\), with an error about six times larger. High eccentricity, as for Halley’s comet, is where the extra derivative pays.
Lesson 10

The Moon, eclipses and moving stars

For the general reader. Halley spent his last decades on the Moon, because an accurate lunar theory would let sailors find longitude at sea. He noticed that ancient eclipse records did not fit a Moon moving at a constant average rate, the first hint that the Earth’s rotation is slowing. He predicted the path of the 1715 eclipse across England within about 20 miles. Comparing his star positions with those in Ptolemy’s catalogue, he found that some bright stars had moved, and so ended the idea of fixed stars.

Goals
  • Understand the saros cycle and why eclipses repeat.
  • See the secular acceleration of the Moon and its modern explanation.
  • Compute stellar proper motion over historical time.

The saros

An eclipse needs a new or full Moon near a node of the Moon’s orbit. Halley gave the name saros to the ancient cycle after which these conditions repeat.

The saros cycle

223 synodic months \(=6585.32\) days, 242 draconic months \(=6585.36\) days, and 239 anomalistic months \(=6585.54\) days. After about 18 years 11 days the Sun, Moon and node return to nearly the same configuration, with the Moon at nearly the same distance, and the eclipse repeats with similar character.

As Astronomer Royal from 1720, already over sixty, Halley set out to observe the Moon through a full saros, so that errors in lunar tables could be corrected cycle by cycle. He completed it and used the results in a method of finding longitude.

The secular acceleration

In 1695 Halley suggested, from ancient eclipse records, that the Moon’s mean motion was slowly increasing. Pierre-Simon Laplace later attributed part of the effect to the slow change of Earth’s orbital eccentricity, but John Couch Adams showed this explains only about half. The rest is apparent: tidal friction slows Earth’s rotation, so days lengthen, and a clock based on Earth’s rotation runs slow against the Moon. Tidal torque also pushes the Moon outward, today by about 3.8 cm per year.

Proper motion (1718)

Halley compared his positions of bright stars with those in Ptolemy’s Almagest, after allowing for precession. Sirius and Arcturus, among others, had shifted in latitude by more than half a degree, far beyond the errors of the old observations. The stars therefore move relative to each other. This was the first evidence that the stars lie at different distances with their own motions.

Exercise 10.1The synodic month is 29.530589 days. Compute 223 synodic months, and explain why each eclipse in a saros series is seen about 120° of longitude west of the last.

Show solution
\(223\times29.530589=6585.32\) days. The extra 0.32 day is about a third of a rotation, so the Earth has turned about \(0.32\times360^\circ\approx116^\circ\), roughly \(120^\circ\), less far at the time of the next eclipse, which is seen correspondingly further west.

Exercise 10.2Arcturus has a proper motion of about \(2.28''\) per year. How far has it moved between Hipparchus (about 130 BCE, whose positions Ptolemy’s catalogue largely reflects) and 1718?

Show solution
About 1,850 years, so \(2.28\times1850\approx4200''\approx70'\), more than a degree in total. Part of this lies in latitude, which is not affected by precession, so Halley could separate it from the precessional shift.

Exercise 10.3Show that 223 synodic months and 19 eclipse years (346.62 days each) agree to within a day. Why is the eclipse year shorter than the calendar year?

Show solution
\(19\times346.62=6585.78\) days, against 6585.32. The eclipse year is the time for the Sun to return to the same node of the Moon’s orbit. The nodes regress westward along the ecliptic once in about 18.6 years, so the Sun meets the node again about 19 days before completing a full year.
Part III · Life and legacy
Lesson 11

Captain, professor, Astronomer Royal, 1686–1742

In 1686 Halley gave up his fellowship to take a salaried post at the Royal Society that included editing the Philosophical Transactions, which he ran until 1693. He was a busy contributor himself, on optics, tides, evaporation and the salinity of the sea, which he proposed as a measure of the age of the Earth. In 1691 he tested his diving bell in the Thames, and a company was formed to exploit it for salvage. The same year he applied for the Savilian chair of astronomy at Oxford. Flamsteed opposed him, doubts were raised about his religious orthodoxy, and David Gregory was chosen.

In 1696 Newton, now Warden of the Mint, made him deputy comptroller of the mint at Chester during the recoinage, a post abolished in 1698. In August 1698 he received a commission to command the navy pink Paramour, an unusual appointment for a civilian. The first voyage turned back after trouble with his first lieutenant. The second, from September 1699 to September 1700, reached the ice of the South Atlantic and produced the magnetic chart of Lesson 08. A third cruise in 1701 surveyed tides in the English Channel. In 1702 and 1703 Queen Anne sent him on diplomatic missions to Vienna and the Adriatic to advise on harbour fortifications.

On the death of John Wallis he was elected Savilian Professor of Geometry at Oxford in late 1703, and he gave his inaugural lecture in 1704. The Oxford years produced the comet synopsis and the Apollonius edition. In 1712 he edited Flamsteed’s observations for publication against Flamsteed’s wishes, a bitter episode in which Newton backed Halley. Oxford made him a Doctor of Civil Law in 1710.

In 1720 he succeeded Flamsteed as Astronomer Royal. Flamsteed’s heirs had removed the instruments, and Halley re-equipped the observatory, adding a transit instrument and a mural quadrant. Already 64, he began and completed his eighteen-year programme of lunar observations. His wife Mary died in 1736. He died at Greenwich on 25 January 1742 (14 January 1741/2 in the English calendar), aged 85, and was buried in the churchyard of St Margaret’s, Lee.

DiscussionHalley was a fixer: editor, publisher, diplomat and organizer of other people’s work. Compare his role in the Principia with his role in the publication of Flamsteed’s catalogue. What made one a service to science and the other a quarrel?

Lesson 12

The return of the comet and Halley’s legacy

Halley did not live to see his comet return. When it was sighted on Christmas night 1758 and passed perihelion in March 1759, it confirmed that comets are members of the solar system bound by gravitation, and it made Newton’s theory convincing to the wider public. Earlier records have since been matched to the comet back to 240 BCE, including the comet shown in the Bayeux Tapestry in 1066. In 1986 the Giotto spacecraft of the European Space Agency flew past its nucleus, and the comet will return to perihelion on 28 July 2061.

Legacy

  • Celestial mechanics: the first prediction from Newton’s gravitation, the first periodic comet, and the publication of the Principia itself.
  • Astronomy: the first telescopic southern catalogue, proper motion of the stars, the transit method that gave the first good value of the astronomical unit, and the secular acceleration of the Moon.
  • Geophysics and cartography: the first wind chart, the barometric law, the isogonic chart and the contour line, and the idea of a layered Earth.
  • Probability and statistics: the life table and the actuarial valuation of annuities.
  • Mathematics: Halley’s method for roots and the standard edition of Apollonius.
  • Honours: comet 1P/Halley, lunar and Martian craters, and the British Antarctic Survey’s Halley Research Station bear his name.

Final projectTake orbital elements for the apparitions of Halley’s comet in 1531, 1607 and 1682 from a modern catalogue. Compare their perihelion distances, inclinations and periods, and explain in one page why Halley concluded they were one comet and why the periods differ.

Reference

Connections

People linked to Edmond Halley across this collection. A name without a link has no page here yet. See Edmond Halley in the network of all 337 lives →

Influenced by

Influenced

Colleagues and collaborators

Rivals

Friends

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Edmond Halley, MacTutor History of Mathematics.
  2. Edmond Halley, Wikipedia (with extensive references).
  3. C. A. Ronan, Halley, Edmond, Dictionary of Scientific Biography.
  4. A. Cook, Edmond Halley: Charting the Heavens and the Seas, Clarendon Press, 1998.
  5. E. F. MacPike (ed.), Correspondence and Papers of Edmond Halley, Clarendon Press, 1932.

Science and mathematics

  1. E. Halley, Astronomiae cometicae synopsis, Oxford, 1705; description.
  2. Halley’s Comet, Wikipedia.
  3. E. Halley, An estimate of the degrees of the mortality of mankind, Phil. Trans. 17 (1693); the life table as a dataset (HistData).
  4. E. Halley, An historical account of the trade winds, and monsoons, Phil. Trans. 16 (1686).
  5. E. Halley, Methodus nova accurata & facilis inveniendi radices æquationum, Phil. Trans. 18 (1694).
  6. E. Halley, Methodus singularis qua Solis parallaxis … ope Veneris intra Solem conspiciendæ, Phil. Trans. 29 (1716).
  7. E. Halley, Considerations on the change of the latitudes of some of the principal fixt stars, Phil. Trans. 30 (1718).
  8. Solar eclipse of May 3, 1715, Wikipedia.
  9. “Calculating the return of Halley’s comet”, History Today.
  10. T. R. Scavo, B. Thoo, On the geometry of Halley’s method, Amer. Math. Monthly 102 (1995).
  11. E. Maor, Venus in Transit, Princeton University Press, 2000.