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James Bradley

The Reverend James Bradley, Savilian Professor of Astronomy and third Astronomer Royal

A course in two strands: the life of a Gloucestershire clergyman who became the most exact observer of his century, and the astronomy he made from a few arcseconds of unexplained motion: the aberration of light, the speed of light, the nutation of the Earth’s axis, and the modern reduction of star positions.

Born 1692 or 1693, Sherborne, Gloucestershire Died 13 July 1762, Chalford, Gloucestershire Aberration announced January 1729 Level advanced undergraduate
γ Draconis through the year. Solid: aberration, amplitude κ sin β ≈ 19.8″, extremes in March and September. Dashed: the parallax pattern Bradley was looking for (scaled to the same size), extremes in June and December.
Lesson 00

How to use this course

James Bradley set out to measure the distance of a star and failed. The failure gave astronomy two discoveries. The aberration of light was the first direct observational proof that the Earth moves round the Sun, and it gave the first accurate value for the speed of light. Nutation showed that the Earth’s axis nods under the pull of the Moon, exactly as Newtonian gravitation required. Together they explained almost every residual in the star positions of the time, and they made possible the precise catalogues of the next century.

Prerequisites for Part II
  • Vectors, spherical coordinates and small-angle approximations.
  • Basic celestial coordinates: right ascension, declination, ecliptic longitude and latitude.
  • Newtonian mechanics: circular orbits, torque, the simple pendulum.
  • For part of Lesson 06, the Lorentz transformation of special relativity.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Angles are in arcseconds (″) unless stated; one radian is \(206\,265''\). The course pairs naturally with the modules on John Flamsteed, who first saw the effect without understanding it, and on Friedrich Wilhelm Bessel, who turned Bradley’s Greenwich observations into the foundation of modern astrometry.

Reference

Timeline

1692/93Born at Sherborne, Gloucestershire, third son of William Bradley and Jane Pound. The parish register is lost; March 1693 is the traditional date.
1711Enters Balliol College, Oxford, on 15 March, after Westwood’s Grammar School at Northleach.
1714BA. Around this time begins observing with his uncle James Pound at Wanstead, Essex.
1717MA. Survives smallpox.
17186 November: elected Fellow of the Royal Society on Halley’s recommendation.
1719Ordained deacon (24 May) and priest; vicar of Bridstow.
1721Savilian Professor of Astronomy at Oxford; resigns his church livings.
1722Measures the diameter of Venus with a 212-foot aerial telescope.
1724November: death of James Pound.
1725November–December: with Samuel Molyneux at Kew, begins observing γ Draconis with a zenith sector by George Graham.
1726Longitudes of Lisbon and New York from eclipses of Jupiter’s satellites.
172719 August: begins observing with his own Graham sector at Wanstead.
1728Death of Molyneux. About September, Bradley finds the explanation: aberration of light.
1729January: letter to Halley on aberration read to the Royal Society. Begins lecturing on experimental philosophy at the Ashmolean Museum.
1734Paper on pendulum experiments in London and Jamaica.
1742Succeeds Halley as Astronomer Royal at Greenwich.
174425 June: marries Susannah Peach of Chalford.
1748Nutation announced in a letter to the Earl of Macclesfield. Copley Medal. Foreign associate of the Paris Academy.
1750Greenwich re-equipped with instruments by John Bird, including an 8-foot mural quadrant.
1752Crown pension of £250 a year.
1757Death of his wife.
1761Organizes British observations of the transit of Venus.
1762Dies 13 July at Chalford; buried at Minchinhampton.
1798–1805Greenwich observations of 1750–1762 published in two volumes by the Clarendon Press.
1818Bessel’s Fundamenta Astronomiae, built on Bradley’s observations.
Part I · Formation
Lesson 01

Sherborne, Oxford and Wanstead

James Bradley was born at Sherborne, a village near Cheltenham in Gloucestershire, the third son of William Bradley and Jane Pound. No baptismal entry survives in the parish register, so even the year is uncertain; the traditional date of March 1693 has no documentary support, and a bishop’s transcript points to the autumn of 1692. He went to Westwood’s Grammar School at Northleach and entered Balliol College, Oxford, on 15 March 1711 as a commoner, a student without a scholarship. He took the BA in 1714 and the MA in 1717.

The decisive person in his education was not at Oxford. His mother’s brother, the Reverend James Pound, was rector of Wanstead in Essex and one of the best observers in England. Pound had worked with Edmond Halley and supplied observations to Isaac Newton. From about 1715 Bradley spent long periods at the Wanstead rectory, learning to use telescopes, clocks and micrometers, and Pound helped him with money. In 1719 uncle and nephew observed Mars together to estimate the solar parallax.

Recognition

Halley noticed the young man early. He published two of Bradley’s observations of the aurora in 1718, and on his recommendation Bradley was elected a Fellow of the Royal Society on 6 November 1718, before he had any post or income of his own. Bradley caught smallpox in 1717 and recovered.

Lesson 02

Clergyman and Savilian professor, 1719–1725

For an Oxford graduate without family money, the Church was the obvious career. Bradley was ordained deacon on 24 May 1719 and priest that summer, and became vicar of Bridstow, near Ross-on-Wye. His friend Samuel Molyneux also found him a small sinecure in Wales. He spent as much time as he could at Wanstead.

In 1721 he was elected Savilian Professor of Astronomy at Oxford, the chair founded in 1619 by Sir Henry Savile and once held by Christopher Wren. The professorship was incompatible with his livings, and he resigned them. He gave his inaugural lecture on 26 April 1722. He never returned to parish work, and later declined the living of Greenwich.

Early work

  • Planetary diameters. In 1722 he measured the diameter of Venus with an aerial telescope of 212 feet focal length, a lens on a pole with no tube.
  • Comets. He computed orbits of comets in the manner of Newton and Halley and published on them from 1723 onward.
  • Longitude. In 1726 he used eclipses of Jupiter’s satellites to find the longitudes of Lisbon and New York relative to London (Lesson 10).

These were careful, competent pieces of work. Nothing yet suggested that Bradley would make discoveries of the first rank.

Lesson 03

Kew, 1725–1729: the search for parallax

If the Earth orbits the Sun, a nearby star should shift a little over the year as we view it from opposite sides of the orbit. This annual parallax was the missing proof of the Copernican system. Robert Hooke had claimed in 1669 to see it in the star γ Draconis, which passes almost through the zenith of London, and Flamsteed had claimed it for the Pole Star. The claims were not convincing.

Samuel Molyneux, a wealthy amateur, decided to repeat Hooke’s experiment properly. In November 1725 he and Bradley fixed a zenith sector made by George Graham, about 24 feet long, to a chimney stack of Molyneux’s house on Kew Green. A zenith sector is a long telescope that swings only a few degrees from the vertical in the meridian, with a plumb line and a fine micrometer screw; pointing straight up avoids most of the problems of refraction and flexure. They began observing γ Draconis in December.

Within days the star was moving south. Parallax predicted that in December the star’s declination should be near its extreme and nearly stationary. By March it was about 20″ south of its December place; it then turned north, reached its northernmost point in September, and returned. The motion was real, regular and large, but a quarter of a year out of phase with parallax.

Wanstead

To test whether the effect applied to all stars, Bradley ordered a shorter Graham sector, about 12 feet long, with a wider field, and set it up at Wanstead on 19 August 1727. Many stars showed similar annual motions, with sizes that depended on their position in a way that pointed to the Earth’s motion. Molyneux died in 1728.

Around September 1728 Bradley found the explanation: the observed direction of a star depends on the velocity of the observer, because light travels at a finite speed. The story that he saw it while watching a wind vane on a boat on the Thames is probably apocryphal; his own account does not mention it. His letter to Halley was read to the Royal Society in January 1729. The Bologna astronomer Eustachio Manfredi had observed similar shifts independently in the 1720s and gave the effect its name, aberration, but did not connect it with the Earth’s orbital motion.

DiscussionBradley’s sector was designed to detect a parallax of about one arcsecond. Why would a zenith instrument be better suited to this than a quadrant pointed at a star near the horizon? List three sources of error it avoids.

Show solution
Atmospheric refraction vanishes at the zenith and is very insensitive to temperature and pressure near it. A vertical tube does not sag sideways under its own weight. Only small zenith distances are measured, so the divided arc and screw cover a tiny range and can be made very fine, and the plumb line gives the reference direction directly.
Part II · The astronomy
Lesson 04

The geometry of aberration

For the general reader. Walk through vertical rain and you must tilt your umbrella forward. Light from a star behaves the same way: because the Earth moves at about 30 km/s and light at about 300,000 km/s, a telescope must be tilted very slightly forward to catch it. As the Earth’s velocity turns round during the year, every star traces a tiny ellipse about 41″ across.

Goals
  • Derive the classical aberration formula from velocity addition.
  • Obtain the annual aberration of a star in ecliptic coordinates.
  • Show that each star traces an ellipse with axes \(2\kappa\) and \(2\kappa\sin\beta\).

Velocity addition

Let \(\hat s\) be the true direction to a star and \(\mathbf v\) the observer’s velocity. In the rest frame of the Sun, light arrives with velocity \(-c\hat s\). In the observer’s frame (Galilean addition) it arrives with velocity \(-c\hat s-\mathbf v\), so the telescope must point along \(c\hat s+\mathbf v\).

Proposition (classical aberration)

Let \(\theta\) be the angle between \(\mathbf v\) and \(\hat s\). The apparent direction lies in the plane of \(\mathbf v\) and \(\hat s\), displaced toward \(\mathbf v\) by an angle \(\Delta\theta\) with

\[ \tan(\theta-\Delta\theta)=\frac{\sin\theta}{\cos\theta+v/c},\qquad \Delta\theta=\frac{v}{c}\sin\theta+O\big((v/c)^2\big). \]

Proof. Resolve \(c\hat s+\mathbf v\) along and perpendicular to \(\mathbf v\): the components are \(c\cos\theta+v\) and \(c\sin\theta\), which gives the tangent formula. Expanding to first order in \(v/c\), the displacement vector is \(\Delta\hat s=(\mathbf v-(\mathbf v\cdot\hat s)\hat s)/c\), whose length is \((v/c)\sin\theta\).

Annual aberration

Take the Earth’s orbit as circular, with speed \(v_\oplus\), and let \(\lambda_\odot\) be the Sun’s ecliptic longitude as seen from the Earth. The Earth then moves toward longitude \(\lambda_\odot-90^\circ\) on the ecliptic, the apex. The constant of aberration is \(\kappa=v_\oplus/c\approx20.50''\).

Theorem (annual aberration in ecliptic coordinates)

For a star at ecliptic longitude \(\lambda\) and latitude \(\beta\), to first order in \(\kappa\),

\[ \Delta\lambda\cos\beta=-\kappa\cos(\lambda_\odot-\lambda),\qquad \Delta\beta=-\kappa\sin\beta\,\sin(\lambda_\odot-\lambda). \]

As \(\lambda_\odot\) runs through a year, the star traces an ellipse centred on its true place, with semi-major axis \(\kappa\) parallel to the ecliptic and semi-minor axis \(\kappa\sin\beta\).

Derivation. In ecliptic Cartesian coordinates the unit velocity is \(\hat v=(\sin\lambda_\odot,-\cos\lambda_\odot,0)\). The local unit vectors at the star are \(\mathbf e_\lambda=(-\sin\lambda,\cos\lambda,0)\) and \(\mathbf e_\beta=(-\sin\beta\cos\lambda,-\sin\beta\sin\lambda,\cos\beta)\). The displacement components are \(\kappa\,\hat v\cdot\mathbf e_\lambda\) and \(\kappa\,\hat v\cdot\mathbf e_\beta\):

\[ \begin{aligned} \hat v\cdot\mathbf e_\lambda&=-\sin\lambda_\odot\sin\lambda-\cos\lambda_\odot\cos\lambda=-\cos(\lambda_\odot-\lambda),\\ \hat v\cdot\mathbf e_\beta&=-\sin\beta\,(\sin\lambda_\odot\cos\lambda-\cos\lambda_\odot\sin\lambda)=-\sin\beta\,\sin(\lambda_\odot-\lambda). \end{aligned} \]

Writing \(x=\Delta\lambda\cos\beta\) and \(y=\Delta\beta\), we get \(x^2/\kappa^2+y^2/(\kappa\sin\beta)^2=1\). A star at the ecliptic pole moves on a circle of radius \(\kappa\); a star on the ecliptic oscillates on a line segment of length \(2\kappa\).

Exercise 4.1γ Draconis has ecliptic latitude \(\beta\approx75^\circ\). Find the axes of its aberration ellipse, taking \(\kappa=20.50''\).

Show solution
Semi-major axis \(20.50''\) parallel to the ecliptic; semi-minor axis \(20.50''\sin75^\circ\approx19.80''\). The ellipse is nearly a circle, about \(41''\) across.

Exercise 4.2The Earth’s rotation gives an observer on the equator a speed of 465 m/s toward the east. Find the maximum diurnal aberration.

Show solution
\(v/c=465/(2.998\times10^8)=1.551\times10^{-6}\) rad \(=1.551\times10^{-6}\times206\,265''\approx0.32''\). It is greatest for a star on the meridian at the equator and shrinks as \(\cos\varphi\) with latitude \(\varphi\). Bradley’s instruments could not resolve it.

Exercise 4.3Show that the second-order term in the classical formula changes \(\Delta\theta\) by at most about \(0.001''\). Was Bradley right to ignore it?

Show solution
With \(\beta=v/c\approx10^{-4}\), expanding \(\tan(\theta-\Delta\theta)=\sin\theta/(\cos\theta+\beta)\) gives \(\Delta\theta=\beta\sin\theta-\tfrac12\beta^2\sin2\theta+O(\beta^3)\). The second term is at most \(\tfrac12\beta^2\approx5\times10^{-9}\) rad \(\approx0.001''\), far below anything measurable before the twentieth century.
Lesson 05

Reading γ Draconis

For the general reader. Parallax and aberration both make a star wander in a small yearly ellipse. What gave Bradley the answer was timing. Parallax pushes the star toward the Sun; aberration pushes it toward the direction the Earth is moving. These directions are always a right angle apart, so the two motions run three months out of step.

Goals
  • Compute the parallactic displacement in ecliptic coordinates.
  • Predict the declination curve of γ Draconis under each hypothesis.
  • Turn the absence of parallax into a lower bound on stellar distances.
Proposition (annual parallax)

For a star at distance \(D\) with parallax \(\varpi=a_\oplus/D\) (radians, \(a_\oplus\) = 1 AU), on a circular orbit,

\[ \Delta\lambda\cos\beta=\varpi\sin(\lambda_\odot-\lambda),\qquad \Delta\beta=-\varpi\sin\beta\cos(\lambda_\odot-\lambda). \]

The displacement points toward the Sun, while the aberration displacement points toward the apex, \(90^\circ\) further back along the ecliptic.

Derivation. The Earth is at \(-a_\oplus\hat u\), where \(\hat u=(\cos\lambda_\odot,\sin\lambda_\odot,0)\) points from Earth to Sun. The apparent direction is that of \(D\hat s+a_\oplus\hat u\), so to first order the displacement is \(\varpi(\hat u-(\hat u\cdot\hat s)\hat s)\). Dot with \(\mathbf e_\lambda\) and \(\mathbf e_\beta\) as in Lesson 04.

The zenith star

γ Draconis lies at right ascension about \(17^{\mathrm h}56^{\mathrm m}\), declination \(+51^\circ29'\), very close to the solstitial colure. Its ecliptic longitude is about \(270^\circ\) and its latitude about \(75^\circ\), and on that meridian an increase in \(\beta\) is an increase in declination. So \(\Delta\delta\approx\Delta\beta\) and

\[ \Delta\delta_{\text{ab}}\approx-\kappa\sin\beta\cos\lambda_\odot,\qquad \Delta\delta_{\text{par}}\approx+\varpi\sin\beta\sin\lambda_\odot. \]

Aberration puts the star furthest south at the March equinox (\(\lambda_\odot=0^\circ\)) and furthest north at the September equinox, exactly as observed. Parallax would put the extremes at the solstices.

Exercise 5.1Using the formula above, find the total annual range of declination of γ Draconis due to aberration, and the dates when the star moves fastest in declination.

Show solution
Range \(2\kappa\sin\beta\approx2\times20.5''\times0.966\approx39.6''\), the “about 40″” of the Kew observations. The rate \(d\Delta\delta/d\lambda_\odot=\kappa\sin\beta\sin\lambda_\odot\) is extreme at \(\lambda_\odot=90^\circ\) and \(270^\circ\), around 21 June (moving north fastest) and 21 December (moving south fastest). That is why the star was already drifting south in December 1725, when parallax predicted it would be stationary.

Exercise 5.2Bradley concluded that a parallax of \(1''\) for γ Draconis would not have escaped him. What lower bound on its distance does this give, in AU and in light years? Its modern parallax is about \(0.021''\).

Show solution
\(D=a_\oplus/\varpi>206\,265\) AU, which is 1 parsec or about 3.26 light years. The modern parallax corresponds to about \(1/0.021\approx48\) pc, roughly 150 light years. Bradley’s bound was correct, and it showed that the stars were at least hundreds of thousands of times further than the Sun.

Exercise 5.3Suppose a star shows both parallax \(\varpi\) and aberration \(\kappa\). Show that its declination curve (for a star on the colure as above) is a single sinusoid of amplitude \(\sin\beta\sqrt{\kappa^2+\varpi^2}\), shifted in phase by \(\arctan(\varpi/\kappa)\). How would Bradley have detected a parallax of \(1''\)?

Show solution
\(-\kappa\cos\lambda_\odot+\varpi\sin\lambda_\odot=-\sqrt{\kappa^2+\varpi^2}\cos(\lambda_\odot+\phi)\) with \(\tan\phi=\varpi/\kappa\). For \(\varpi=1''\), \(\phi\approx2.8^\circ\), a shift of about 3 days in the dates of the extremes, and the amplitude grows by only \(0.02''\). The clean test is the declination at the solstices, where aberration contributes nothing and parallax contributes \(\pm\varpi\sin\beta\). Bradley found no such residual.
Lesson 06

The speed of light and the physics of aberration

For the general reader. The size of the aberration ellipse is the ratio of two speeds, the Earth’s and light’s. Since the Earth’s speed follows from the size of its orbit and the length of the year, Bradley could compute how long light takes to come from the Sun. His answer, 8 minutes 12 seconds, was within a few per cent of the truth. Aberration later became one of the clues that led to special relativity.

Goals
  • Derive the light time across 1 AU from \(\kappa\) without knowing the size of the AU.
  • Compare the classical and relativistic aberration formulas.
  • See why aberration troubled the ether theories of the nineteenth century.
Proposition (light time from aberration)

For a circular orbit of radius \(a_\oplus\) and sidereal period \(P\), \(v_\oplus=2\pi a_\oplus/P\), so the light time across one astronomical unit is

\[ \tau_A=\frac{a_\oplus}{c}=\frac{\kappa\,P}{2\pi}. \]

The remarkable feature is that \(a_\oplus\) cancels. In 1729 the distance to the Sun was known only within a factor of perhaps two, yet the light time could be measured to about one per cent. Ole Rømer had inferred a finite speed of light in 1676 from delays in the eclipses of Jupiter’s satellite Io, and Newton in the Opticks estimated seven or eight minutes for light from the Sun. Bradley’s value came from a completely independent effect and settled the question.

Theorem (relativistic aberration)

Let \(\theta\) be the angle between the source direction and the observer’s velocity \(\mathbf v\) in the Sun’s frame, and \(\theta'\) the same angle in the observer’s frame. With \(\beta=v/c\),

\[ \cos\theta'=\frac{\cos\theta+\beta}{1+\beta\cos\theta},\qquad \tan\theta'=\frac{\sin\theta}{\gamma(\cos\theta+\beta)},\quad \gamma=(1-\beta^2)^{-1/2}. \]

Derivation. A photon arriving from direction \(\hat s\) has momentum along \(-\hat s\). Its four-momentum is \(k=(\omega/c)(1,-\hat s)\). Boosting with velocity \(\mathbf v\) along the \(x\)-axis gives \(\omega'=\gamma\omega(1+\beta\cos\theta)\) and \(-k'_x=\gamma(\omega/c)(\cos\theta+\beta)\), so \(\cos\theta'=(\cos\theta+\beta)/(1+\beta\cos\theta)\). The perpendicular component is unchanged, which gives the tangent form. To first order in \(\beta\) it agrees with Bradley’s formula.

Aberration and the ether

In a wave theory of light, aberration is natural only if the ether is not dragged by the Earth. Yet experiments on the Earth’s surface showed no effect of the Earth’s motion. In 1810 François Arago found that refraction of starlight in a prism did not depend on the Earth’s motion, and in 1871 George Biddell Airy filled a telescope with water: if light were simply carried along a tube at speed \(c/n\), the telescope would need a larger tilt, but the measured aberration was unchanged. Augustin-Jean Fresnel’s partial-drag coefficient \(1-1/n^2\) explained both results, and special relativity later derived it from velocity addition.

Exercise 6.1Bradley’s constant was about \(20.2''\). Compute the light time across 1 AU and the ratio \(c/v_\oplus\). Use \(P=365.26\) days.

Show solution
\(\kappa=20.2/206\,265=9.79\times10^{-5}\). \(P/2\pi=365.26\times86\,400/2\pi=5.023\times10^6\) s. So \(\tau_A=9.79\times10^{-5}\times5.023\times10^6\approx492\) s \(=8^{\mathrm m}12^{\mathrm s}\), Bradley’s value. The ratio \(c/v_\oplus=1/\kappa\approx10\,200\).

Exercise 6.2Repeat with the modern \(\kappa=20.4955''\), then use \(a_\oplus=1.496\times10^{8}\) km to find \(c\). How large was Bradley’s error?

Show solution
\(\tau_A=(20.4955/206\,265)\times5.023\times10^6\approx499.1\) s \(=8^{\mathrm m}19^{\mathrm s}\). Then \(c=1.496\times10^8/499.1\approx2.998\times10^5\) km/s. Bradley’s time was about 7 s, or 1.4 per cent, too short.

Exercise 6.3Airy’s water-filled telescope, naive version: in a stationary ether, starlight from a star at right angles to \(\mathbf v\) travels down the water-filled tube at speed \(c/n\), while the tube moves sideways at \(v\). Show that the tube would need a tilt of about \(n\kappa\), and evaluate it for water (\(n=1.33\)).

Show solution
While the light descends the tube at \(c/n\), the tube moves sideways at \(v\). For the light to stay on the axis the tube must be tilted by \(v/(c/n)=n\kappa\approx1.33\times20.5''\approx27.3''\). Airy found the ordinary \(20.5''\). The naive picture fails because moving water partly carries light with it; Fresnel’s drag coefficient \(1-1/n^2\), combined with refraction at the moving objective, restores the tilt to exactly \(\kappa\) to first order in \(v/c\).
Lesson 07

Nutation

For the general reader. Once aberration was removed, small yearly discrepancies remained in Bradley’s star positions, and they grew in one direction for years before reversing. He suspected the Moon. The Moon’s orbit is tilted to the ecliptic, and that tilt swings round every 18.6 years, so the Moon’s pull on the Earth’s bulging equator changes slowly. Bradley watched for twenty years, a whole cycle, before he published. The Earth’s axis nods by about 9″ each way.

Goals
  • Understand precession and nutation as responses of a spinning oblate Earth to lunisolar torques.
  • State the leading nutation terms and the ellipse traced by the pole.
  • Compute the effect of nutation on right ascension and declination.

Precession and the lunar node

The Sun and Moon pull on the Earth’s equatorial bulge and try to turn the equator into the plane of their orbits. A spinning body responds by precessing: the celestial pole moves round the pole of the ecliptic in about 25,800 years, carrying the equinox westward by about \(50.3''\) per year. The lunar part of the torque depends on the Moon’s declination range, which is set by the longitude \(\Omega\) of the ascending node of the Moon’s orbit on the ecliptic. The node regresses with a period of 18.61 years. The torque, and with it the precession rate, therefore oscillates with that period.

Leading nutation terms

The true pole differs from the mean (precessing) pole by a nutation in longitude \(\Delta\psi\) and a nutation in obliquity \(\Delta\varepsilon\). The dominant terms are

\[ \Delta\psi\approx-17.20''\sin\Omega,\qquad \Delta\varepsilon\approx+9.20''\cos\Omega. \]

On the sky the true pole moves round the mean pole on an ellipse with semi-axes \(9.20''\) (toward the ecliptic pole) and \(17.20''\sin\varepsilon\approx6.86''\) (perpendicular), once every 18.6 years.

Bradley’s twenty-year series gave an ellipse of about \(18''\) by \(16''\); the modern figures for the full axes are \(18.4''\) by \(13.7''\). Within a year of his announcement in 1748, Jean le Rond d’Alembert derived precession and nutation from Newton’s law of gravitation, and Leonhard Euler took up the problem soon after. The amplitude of nutation also measures the mass of the Moon, which was otherwise hard to obtain.

Nutation in equatorial coordinates

To first order, for a star at \((\alpha,\delta)\),

\[ \begin{aligned} \Delta\alpha&=(\cos\varepsilon+\sin\varepsilon\sin\alpha\tan\delta)\,\Delta\psi-\cos\alpha\tan\delta\,\Delta\varepsilon,\\ \Delta\delta&=\sin\varepsilon\cos\alpha\,\Delta\psi+\sin\alpha\,\Delta\varepsilon. \end{aligned} \]

The structure is the same as the precession formulas of the Flamsteed module, with \(\Delta\psi\) in place of the accumulated precession in longitude, plus the terms from the change of obliquity.

Exercise 7.1For γ Draconis (\(\alpha\approx269^\circ\), \(\delta\approx51.5^\circ\)), show that the nutation in declination is almost exactly \(-\Delta\varepsilon\). How large is its swing, and how does it compare with the annual precession in declination, \(20.04''\cos\alpha\)?

Show solution
\(\sin\alpha\approx-1\), \(\cos\alpha\approx-0.017\), so \(\Delta\delta\approx-0.017\times0.398\,\Delta\psi-\Delta\varepsilon\approx-\Delta\varepsilon\) (the first term is under \(0.12''\)). The declination swings by \(\pm9.2''\) over 18.6 years. Annual precession in declination is only \(20.04''\times(-0.017)\approx-0.35''\) per year. A star on the solstitial colure is the ideal probe of nutation in obliquity, which is one reason Bradley could see it so clearly.

Exercise 7.2A star at \(\alpha\approx90^\circ\) near the pole shows nutation in declination of the opposite sign to γ Draconis. Explain, and say why this pattern rules out an annual or instrumental cause.

Show solution
At \(\alpha=90^\circ\), \(\Delta\delta\approx+\Delta\varepsilon\); at \(\alpha=270^\circ\), \(\Delta\delta\approx-\Delta\varepsilon\). When the pole tilts toward one side of the sky, stars on that side gain declination and stars on the opposite side lose it. An error of the instrument’s zero or of the plumb line would shift all stars the same way, and an annual effect would repeat every year. Opposite drifts over many years point to a motion of the axis itself.

Exercise 7.3Why did Bradley need to observe for a full node cycle before publishing? What could he have confused with nutation in a five-year series?

Show solution
Over a fraction of a cycle a sinusoid is indistinguishable from a polynomial drift, which could be absorbed into an error in the adopted precession constant or in the stars’ proper motions. Only a full period shows that the drift reverses and returns, and fixes both the period (18.6 years, matching the lunar node) and the amplitude.
Lesson 08

Pendulums and the figure of the Earth

For the general reader. Newton predicted that the spinning Earth is flattened at the poles, so that gravity is weaker at the equator. A pendulum clock is a gravity meter: where gravity is weaker it swings more slowly. Bradley compared a clock in London with the same clock in Jamaica and found the expected loss of about two minutes a day.

Goals
  • Relate a pendulum clock’s daily rate to local gravity.
  • Use Clairaut’s theorem to predict the variation of gravity with latitude.
  • Check Bradley’s London–Jamaica result against modern gravity.
Proposition (pendulum rate)

A simple pendulum of length \(\ell\) has small-amplitude period \(T=2\pi\sqrt{\ell/g}\). A clock regulated by it gains or loses at a fractional rate \(\Delta T/T=-\tfrac12\,\Delta g/g\). A seconds pendulum, which beats once per second (\(T=2\) s), has length \(\ell=g/\pi^2\).

Theorem (Clairaut, 1743)

For a rotating fluid body in hydrostatic equilibrium, slightly flattened with flattening \(f\), surface gravity at geographic latitude \(\varphi\) is, to first order,

\[ g(\varphi)=g_e\big(1+(\tfrac52 m-f)\sin^2\varphi\big),\qquad m=\frac{\omega^2a}{g_e}\approx\frac1{289}. \]

With \(f\approx1/298\), the coefficient is \(\tfrac52 m-f\approx0.0053\). Alexis Clairaut published the theorem in 1743, after the French expeditions to Lapland and Peru had made the figure of the Earth a central question of the day. Bradley’s 1734 paper gave tables of pendulum lengths for every five degrees of latitude, computed on Newton’s theory, and reported the London seconds pendulum as 39.126 inches.

Exercise 8.1From the London seconds pendulum of 39.126 inches, compute \(g\) in London.

Show solution
\(\ell=39.126\times0.0254=0.99380\) m, so \(g=\pi^2\ell=9.8696\times0.99380\approx9.808\) m/s². The modern value in London is about 9.812 m/s²; the small difference comes mainly from treating a real clock pendulum as a simple one.

Exercise 8.2The clock lost 1 minute 58 seconds a day in Jamaica (latitude about \(18^\circ\)) compared with London (\(51.5^\circ\)). Find \(\Delta g/g\) and compare with Clairaut’s formula.

Show solution
Fractional loss \(118/86\,400=1.366\times10^{-3}\), so \(\Delta g/g=2\times1.366\times10^{-3}=2.73\times10^{-3}\). Clairaut: \(0.0053(\sin^2 51.5^\circ-\sin^2 18^\circ)=0.0053(0.6125-0.0955)=2.74\times10^{-3}\). The agreement is excellent, though temperature effects on the pendulum rod make it partly fortunate.
Lesson 09

Meridian astronomy at Greenwich

For the general reader. At Greenwich Bradley measured each star’s position as it crossed the meridian: the time of crossing gives right ascension, and the angle above the horizon gives declination. To reach accuracies of a second of arc he had to correct every measurement for the bending of light by the atmosphere, for the observatory’s exact latitude, for precession, nutation and aberration. The resulting system of corrections is still the backbone of positional astronomy.

Goals
  • Find latitude from upper and lower transits of a circumpolar star.
  • Apply the standard refraction formula and see why refraction matters for latitude.
  • Understand the chain from observed to mean place.

Latitude

Proposition

Let \(\varphi\) be the latitude and measure altitudes from the northern horizon. A circumpolar star of declination \(\delta\) crosses the meridian above the pole at altitude \(h_u=\varphi+(90^\circ-\delta)\) and below the pole at \(h_l=\varphi-(90^\circ-\delta)\). Hence

\[ \varphi=\tfrac12(h_u+h_l),\qquad \delta=90^\circ-\tfrac12(h_u-h_l), \]

independent of any catalogue, provided both altitudes are corrected for refraction.

If the upper transit passes south of the zenith, \(h_u\) is replaced by \(180^\circ\) minus the altitude measured from the south. Bradley’s latitude for Greenwich, \(51^\circ28'38.5''\), is within about \(1.3''\) of the modern value.

Refraction

Refraction near the zenith

For a plane-parallel atmosphere, Snell’s law gives \(n_0\sin z_{\text{obs}}=\sin z_{\text{true}}\) regardless of the layering, so the refraction \(R=z_{\text{true}}-z_{\text{obs}}\) satisfies \(R\approx(n_0-1)\tan z\). At standard temperature and pressure \(n_0-1\approx2.9\times10^{-4}\), giving

\[ R\approx58''\tan z \]

for zenith distances up to about \(70^\circ\). The curvature of the Earth adds a small negative term in \(\tan^3 z\).

Bradley built refraction tables from his own observations with the Bird quadrant in the early 1750s, including the dependence on temperature and pressure. Bessel later refined them.

From observation to mean place

An observed position must be corrected, in order, for instrumental errors, refraction, diurnal aberration, annual aberration (Lesson 04), nutation (Lesson 07) and precession, to give the mean place at a standard epoch. Bradley’s Greenwich observations of 1750–1762 were the first large body of data for which every one of these effects was known.

Exercise 9.1A circumpolar star is observed at Greenwich at refraction-corrected altitudes \(h_u=89^\circ20'00''\) (north of the zenith, above the pole) and \(h_l=13^\circ37'17''\). Find the latitude and the star’s declination.

Show solution
\(\varphi=\tfrac12(89^\circ20'00''+13^\circ37'17'')=\tfrac12(102^\circ57'17'')=51^\circ28'38.5''\). \(\delta=90^\circ-\tfrac12(89^\circ20'00''-13^\circ37'17'')=90^\circ-\tfrac12(75^\circ42'43'')=90^\circ-37^\circ51'21.5''=52^\circ8'38.5''\).

Exercise 9.2In Exercise 9.1, suppose the observer forgot refraction. Estimate the error in latitude, using \(R=58''\tan z\).

Show solution
At \(h_u\approx89.3^\circ\), \(z\approx0.7^\circ\), \(R\approx0.7''\). At \(h_l\approx13.6^\circ\), \(z\approx76.4^\circ\), \(\tan z\approx4.1\), \(R\approx240''\) (the simple formula is rough this low; the true value is close to \(4'\)). Refraction raises both stars, so the uncorrected latitude is too large by about \(\tfrac12(0.7''+240'')\approx2'\). This is why the latitude and the refraction table had to be determined together.
Lesson 10

Jupiter’s satellites and longitude

For the general reader. An eclipse of one of Jupiter’s moons happens at the same instant for everyone who can see it. If two observers note the local time of the same eclipse, the difference in their clocks gives the difference in their longitudes. Bradley used this method in 1726 for Lisbon and New York, and later studied how the moons pull on one another.

Goals
  • Convert a local-time difference into a longitude difference.
  • Verify the resonance among Jupiter’s inner satellites.
Principle

Local solar (or sidereal) time advances by \(1^{\mathrm h}\) for each \(15^\circ\) of longitude eastward. If an instantaneous event is recorded at local times \(t_1\) and \(t_2\) at two places, their longitudes differ by \(\Delta\lambda=15^\circ\times(t_1-t_2)/1^{\mathrm h}\), the place with the later local time lying east.

The method worked well on land but not at sea, where a long telescope cannot be held steady on a moving deck. The sea problem was solved after Bradley’s death by marine chronometers and by the lunar-distance method, whose tables Bradley helped test (Lesson 11).

The satellites’ mutual perturbations

In the 1730s Bradley recognized that irregularities in the motions of the satellites were linked to one another by their mutual attraction. The explanation was completed by Pierre-Simon Laplace in 1788, who showed that the mean motions \(n_1,n_2,n_3\) of Io, Europa and Ganymede are locked in a resonance.

Laplace resonance \[ n_1-3n_2+2n_3=0,\qquad \lambda_1-3\lambda_2+2\lambda_3=180^\circ, \]

where \(\lambda_i\) are the mean longitudes. As a consequence, the three inner satellites can never all be in conjunction.

Exercise 10.1An eclipse of Io is seen at 11:42:10 local time at Greenwich and at 6:46:04 local time in New York. Find New York’s longitude.

Show solution
The difference is \(4^{\mathrm h}56^{\mathrm m}06^{\mathrm s}=4.935^{\mathrm h}\), so \(\Delta\lambda=15^\circ\times4.935=74.03^\circ\), that is \(74^\circ01.5'\) west of Greenwich.

Exercise 10.2The sidereal periods of Io, Europa and Ganymede are 1.769138, 3.551181 and 7.154553 days. Check the Laplace relation numerically.

Show solution
In degrees per day: \(n_1=203.4889\), \(n_2=101.3747\), \(n_3=50.3176\). Then \(n_1-3n_2+2n_3=203.4889-304.1241+100.6352=0.0000\), zero within the rounding of the periods.
Part III · Life and legacy
Lesson 11

Astronomer Royal, 1742–1762

From 1729 Bradley also held the Oxford readership in experimental philosophy. Over some thirty years he gave 79 courses of lectures at the Ashmolean Museum, with audiences that averaged more than fifty. In 1732 he moved into a house in New College Lane, and the Wanstead sector stayed in Essex.

When Halley died in January 1742, Bradley succeeded him as Astronomer Royal, the third holder of the post. The salary was £100 a year. Greenwich was poorly equipped, and Bradley’s reputation won him £1,000 from the Admiralty to re-equip it. By 1750 the observatory had a new transit instrument and an 8-foot mural quadrant by John Bird, the finest instruments of their kind. Between 1750 and 1762 Bradley and his assistants made at least 60,000 observations. In 1752 the Crown granted him a pension of £250 a year.

On 25 June 1744 he married Susannah Peach of Chalford, Gloucestershire. They had one daughter. His wife died in 1757.

Navigation and the transit of Venus

The Board of Longitude asked Bradley to test the lunar tables of the Göttingen astronomer Tobias Mayer. He compared them with more than 230 Greenwich observations and reported in 1756 that they could give a ship’s longitude to about half a degree, subject to trials at sea. This work led, after his death, to the Nautical Almanac, first published for 1767. For the transit of Venus in 1761 he was too frail to travel. He arranged for Charles Mason to go to the East Indies and Nevil Maskelyne to St Helena; war delayed Mason and clouds spoiled Maskelyne’s view.

Death

Ill health forced him to stop observing in 1761. He retired to his wife’s family home at Chalford and died there on 13 July 1762 of a chronic abdominal inflammation. He was buried in the churchyard of Holy Trinity, Minchinhampton, near his mother and his wife. Nathaniel Bliss succeeded him as Astronomer Royal.

DiscussionBradley published very little: two great papers and a handful of shorter ones in twenty years at Greenwich. What does his career suggest about the value of long, unpublished series of observations, and what were the risks?

Lesson 12

The afterlife of the observations

Bradley’s Greenwich notebooks passed to his daughter and her relatives, who refused to give them to the Board of Longitude. After a long dispute they went to the University of Oxford on condition that they be published, and the Clarendon Press issued them in two folio volumes in 1798 and 1805. The first volume included aberration tables for sixteen stars and a catalogue of more than 380 bright stars.

The real value of the observations was released by Friedrich Wilhelm Bessel. Starting in 1807, Bessel reduced Bradley’s observations of 1750–1762 for every effect Bradley had identified, together with refraction and instrumental errors, and in 1818 published Fundamenta Astronomiae, positions of 3,222 stars for 1755. Because Bradley’s positions were so good, comparison with later catalogues gave reliable proper motions; it was such a comparison that singled out 61 Cygni, the first star whose parallax was measured, in 1838. Arthur Auwers published a fresh reduction in three volumes between 1882 and 1903.

Legacy

  • Copernicanism: aberration was the first direct observational evidence of the Earth’s orbital motion.
  • Light: the first accurate value of the speed of light, and an effect that every later theory of light, including special relativity, had to explain.
  • Gravitation: nutation confirmed the Newtonian theory of the Earth’s response to lunar and solar torques.
  • Astrometry: the separation of aberration, nutation and precession, the reduction to mean places, and a standard of observing accuracy near one arcsecond.
  • Honours: Fellow of the Royal Society (1718); Copley Medal (1748); member of the Berlin Academy (1746), foreign associate of the Paris Academy (1748), member of the St Petersburg (1754) and Bologna (1757) academies. The historian Jean-Baptiste Delambre ranked him after Hipparchus and Kepler. A lunar mountain, Mons Bradley, bears his name, and a plaque on the sundial at Kew marks where the aberration observations began.

Final projectWrite a two-page account, for a strong undergraduate, of how Bradley distinguished aberration from parallax. Plot the declination of γ Draconis over a year under both hypotheses, explain the quarter-year phase difference from the geometry, and derive the light time from the amplitude.

Reference

Connections

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

Students

  • Nevil Maskelyne

Influenced by

Influenced

Colleagues and collaborators

Family

  • James Pound 1715-1724

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, James Bradley, MacTutor History of Mathematics.
  2. James Bradley, Wikipedia.
  3. Dictionary of Scientific Biography, Bradley, James.
  4. Dictionary of National Biography, Bradley, James.
  5. Linda Hall Library, James Bradley, Scientist of the Day.
  6. S. P. Rigaud (ed.), Miscellaneous Works and Correspondence of the Rev. James Bradley, Oxford, 1832.

Science

  1. J. Bradley, A letter … giving an account of a new discovered motion of the fix’d stars, Phil. Trans. 35 (1729).
  2. J. Bradley, A letter … concerning an apparent motion observed in some of the fixed stars, Phil. Trans. 45 (1748).
  3. J. Bradley, Astronomical Observations made at the Royal Observatory at Greenwich, 1750–1762, 2 vols., Oxford, 1798–1805.
  4. F. W. Bessel, Fundamenta Astronomiae pro anno 1755, Königsberg, 1818.
  5. Aberration (astronomy) and Astronomical nutation, Wikipedia.
  6. Perth Observatory, Further refining the measurement of the speed of light.
  7. W. M. Smart, Textbook on Spherical Astronomy, 6th ed., Cambridge, 1977, chapters on aberration, precession and nutation.
  8. P. K. Seidelmann (ed.), Explanatory Supplement to the Astronomical Almanac, University Science Books, 1992.