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Ole Rømer

Ole Christensen Rømer

A course in two strands: the life of a Danish astronomer who went from Tycho Brahe’s manuscripts to the Paris Observatory and back to run the police of Copenhagen, and the science of timing, from the eclipses of Io that revealed the finite speed of light to the transit instrument and the thermometer.

Born 25 September 1644, Aarhus Died 19 September 1710, Copenhagen Light time presented in Paris, 21 November 1676 Level advanced undergraduate
Io emerges from Jupiter’s shadow at fixed intervals. Seen from the far side of the Earth’s orbit the signal has farther to travel, and the emergences appear late.
Lesson 00

How to use this course

Ole Rømer is remembered for one result: light takes time to travel, and the eclipses of Jupiter’s moon Io show how much. That result came from a working astronomer whose daily business was precise timing. Rømer built instruments, gearing and calculating machines, standardized weights and measures for a kingdom, reformed its calendar and ran the police of its capital. This course treats the speed of light in detail and places it among the other problems of measurement he solved.

Prerequisites for Part II
  • Circular orbits, angular velocity and Kepler’s third law.
  • Plane geometry and parametric curves; a little calculus.
  • For Lesson 08, continued fractions are introduced from scratch.
  • For Lesson 09, the celestial sphere: right ascension, declination, sidereal time.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. The course on Giovanni Domenico Cassini covers the same events in Paris from Cassini’s side; the two are written to agree.

Reference

Timeline

1644Born 25 September in Aarhus, son of the merchant and skipper Christen Pedersen, who had taken the name Rømer, and Anna Olufsdatter Storm.
1662Leaves Aarhus Cathedral School and enrols at the University of Copenhagen.
1664Becomes assistant to Rasmus Bartholin in editing Tycho Brahe’s observations, and lives in his house.
1671With Jean Picard and Bartholin on Hven, at the site of Tycho’s Uraniborg; observes eclipses of Io to fix its longitude relative to Paris.
1672Goes to Paris with Picard; works at the new Observatory under Cassini. Soon a member of the Académie royale des sciences.
1674Proposes the epicycloid as the ideal profile for gear teeth.
c. 1675Designs planetary machines, a Jovilabe and an eclipse machine; teaches mathematics to the Dauphin.
167622 August: Cassini announces a new inequality in the eclipses of Jupiter’s satellites. 21 November: Rømer presents the finite speed of light to the Academy. 7 December: report in the Journal des sçavans.
1677English translation of the report in the Philosophical Transactions.
1679Visits England; meets John Flamsteed and Edmond Halley.
1681Recalled to Copenhagen; professor of astronomy. Marries Anne Marie Bartholin.
16831 May: first national system of weights and measures for Denmark–Norway, prepared by Rømer as royal mathematician.
c. 1689–91Builds a transit instrument at his house in Store Kannikestræde.
1700Denmark–Norway adopts the reformed (Gregorian) calendar: 18 February is followed by 1 March.
c. 1702Calibrates thermometers on a scale with two fixed points.
1704Builds the Observatorium Tusculanum at Vridsløsemagle, with the first meridian circle.
1705Becomes police director and mayor of Copenhagen.
170620–23 October: the Triduum, three nights of meridian observations.
1708Daniel Gabriel Fahrenheit visits him in Copenhagen.
1710Dies 19 September in Copenhagen, six days before his 66th birthday.
1728The Copenhagen fire destroys most of his observations and instruments. James Bradley explains stellar aberration, confirming the finite speed of light.
1735Peder Horrebow describes Rømer’s instruments and methods in Basis astronomiae.
Part I · Formation
Lesson 01

Aarhus and Copenhagen, 1644–1671

Ole Christensen Rømer was born in Aarhus on 25 September 1644. His father, Christen Pedersen, came from the island of Rømø and took the surname Rømer when he settled in Aarhus in 1642. He was a merchant and skipper who owned navigational instruments and books. His mother, Anna Olufsdatter Storm, was the daughter of an Aarhus alderman. The father died in 1663.

Rømer left Aarhus Cathedral School in 1662 and went to the University of Copenhagen, then the only university in Denmark–Norway. There the decisive influence was Rasmus Bartholin, whose discovery of double refraction in crystals of Iceland spar was published while Rømer lived in his house. In 1655 King Frederik III had bought Tycho Brahe’s original observation logs from Johannes Kepler’s son, and in 1664 he ordered Bartholin to publish them. Bartholin took the student Rømer, after two years of study, as his assistant and lodged him in his house.

The editing gave Rømer an education no lecture course could provide. Tycho’s logs were the most accurate naked-eye observations ever made, recorded with notes on instruments, corrections and errors. The edition was abandoned when Frederik III died in 1670 and was printed only in the twentieth century, but Rømer had learned at first hand what precision astronomy demanded.

Lesson 02

Hven with Picard, 1671

In 1671 the French Academy of Sciences sent Jean Picard to Denmark. Picard had just measured an arc of the meridian near Paris, and he needed the exact position of Tycho’s observatory of Uraniborg on the island of Hven to compare Tycho’s observations with those made in Paris. Uraniborg had been built around 1580 and torn down by 1601; Picard found the site overgrown and partly used as a dump. Bartholin and Rømer went with him to locate the foundations and set up instruments.

The key measurement was longitude. Picard and Rømer observed eclipses of Jupiter’s innermost satellite, Io, at Hven, while Cassini observed the same events in Paris. The difference of local times gives the difference of longitude (Lesson 04). Over several months they recorded a long series of eclipses; Wikipedia’s account gives the figure of about 140. One of the Hven–Paris comparisons gave a time difference of 42 minutes 10 seconds, within about 45 seconds of the modern value.

Picard was impressed by the young Dane and obtained King Christian V’s permission to take him to Paris. Rømer travelled there in 1672, taking Tycho’s observations with them.

Lesson 03

Paris, 1672–1681

The Paris Observatory had opened in 1671 under Giovanni Domenico Cassini. Rømer lived and worked there for nearly nine years and was soon admitted to the Académie royale des sciences. He observed Jupiter’s satellites as part of Cassini’s programme to perfect their tables, and he built instruments, including an improved micrometer.

He also served the court. He taught mathematics to the Dauphin, the heir of Louis XIV, and for teaching designed mechanical models: a planetary machine showing the motions of the six known planets around the Sun, a Jovilabe for Jupiter’s satellites and an eclipse machine (Lesson 08). The models were built in four copies with the help of a Paris clockmaker. One set stayed in Paris, one went to Christian V in Copenhagen and is now at Rosenborg Castle, and two were sent as gifts to the King of Siam and the Emperor of China. Rømer also worked on the hydraulics and levelling of the water supply for the royal fountains at Versailles. In 1674 he proposed the epicycloid as the correct profile for gear teeth (Lesson 07).

The work that made him famous began as a correction to the tables of Io. On 22 August 1676 Cassini announced to the Academy that his eclipse tables would include a new inequality, and mentioned that it could be explained by a finite speed of light. Rømer developed that explanation. He predicted that an emergence of Io in November 1676 would be seen about ten minutes later than the old tables gave, and on 21 November he presented his theory to the Academy: light takes about 22 minutes to cross the diameter of the Earth’s orbit. An anonymous summary appeared in the Journal des sçavans on 7 December 1676, and an English version in the Philosophical Transactions in 1677. Rømer never published a full account himself.

Cassini then turned against the idea. The other satellites did not show the effect clearly, and Io had other irregularities. Christiaan Huygens, by contrast, corresponded with Rømer and in 1677 wrote to the minister Colbert in support. In 1679 Rømer visited England, where he met John Flamsteed, the Astronomer Royal, who began to apply the light-time correction to his own tables of the satellites, and Edmond Halley. In 1681 Christian V recalled him to Copenhagen.

Part II · The physics
Lesson 04

A clock in the sky: longitude from Io

For the general reader. Latitude can be read from the height of the pole star; longitude is a difference of clock times. To find how far east of Paris you stand, you need to know what time it is in Paris at the moment when it is noon where you are. Galileo proposed using the moons of Jupiter as a clock visible from everywhere at once. When a moon enters Jupiter’s shadow, every observer on Earth sees it disappear at the same instant, whatever their local time.

Goals
  • Compute longitude differences from eclipse timings.
  • Find the synodic period of Io and predict eclipses from a table.
  • Estimate how timing errors become position errors.
Proposition (longitude from a simultaneous event)

If an event is seen at the same absolute instant from two places, at local mean times \(T_A\) and \(T_B\) (in hours), then the longitude of \(A\) east of \(B\) is

\[ \lambda_A-\lambda_B=15^\circ\times(T_A-T_B). \]

The Earth turns \(360^\circ\) in 24 hours of mean solar time, \(15^\circ\) per hour, and local time is fixed by the Sun’s position relative to the local meridian. An eclipse of a Jovian satellite is, to a very good approximation, seen simultaneously from all points of the Earth. The light time across the Earth is about 0.04 s, negligible here (though the light time across the Earth’s orbit is the subject of Lesson 05).

Periods

The eclipses of Io recur with the synodic period, the time for Io to return to the same position relative to the Sun–Jupiter line, because Jupiter’s shadow points away from the Sun. If Io’s sidereal period is \(P_I\) and Jupiter’s is \(P_J\),

\[ \frac{1}{S}=\frac{1}{P_I}-\frac{1}{P_J}. \]

With \(P_I=1.769138\) d and \(P_J=4332.59\) d, \(S\approx1.76986\) d, or 42 h 28.6 min. Near the opposition of Jupiter, eclipses are seen as immersions (disappearances) on one side of the planet before opposition and as emergences after it; near conjunction Jupiter is lost in the Sun’s glare and nothing is seen for weeks.

Exercise 4.1An emergence of Io is seen at Uraniborg at 10 h 45 min 30 s local time. Cassini in Paris sees the same emergence at 10 h 03 min 20 s. Find the longitude of Uraniborg east of Paris.

Show solution
\(T_A-T_B=42\) min 10 s \(=0.70278\) h, so \(\Delta\lambda=15\times0.70278=10.542^\circ=10^\circ32'30''\). This is the figure obtained in 1671; the modern value, 41 min 26 s, gives \(10^\circ21'30''\).

Exercise 4.2A navigator times an eclipse with an error of 30 s. How large is the resulting error in longitude, in degrees and in kilometres at latitude \(50^\circ\)?

Show solution
30 s is \(1/120\) h, so \(15/120=0.125^\circ\). A degree of longitude at latitude \(\phi\) is about \(111.3\cos\phi\) km, which is 71.5 km at \(50^\circ\). The error is about 9 km. On land this was excellent; at sea, the real obstacle was holding a long telescope steady on a moving deck.

Exercise 4.3Verify the synodic period of Io quoted above, and find how many eclipses occur in 30 days.

Show solution
\(1/S=1/1.769138-1/4332.59=0.565247-0.000231=0.565016\) per day, so \(S=1.76986\) d \(=42.477\) h. In 30 days there are \(30/1.76986\approx16.95\), so 16 or 17 eclipses, depending on the starting point.
Lesson 05

The delay of light, 1676

For the general reader. If light takes time to travel, an event at Jupiter is seen late, and the delay depends on how far away Jupiter is. As the Earth moves along its orbit away from Jupiter, each eclipse of Io is seen a little later than the one before it should have been; as the Earth approaches, a little earlier. Over months, the small differences add up to many minutes. Rømer saw that this single cause explained the drift of Io’s eclipses, and that the drift measures the time light needs to cross the Earth’s orbit.

Goals
  • Derive the light-time inequality in the eclipses of a satellite.
  • Reproduce the size of Rømer’s November 1676 prediction.
  • Understand why the effect was clearest for Io.
Proposition (light-time inequality)

Let \(\Delta(t)\) be the Earth–Jupiter distance and \(c\) the speed of light. An eclipse occurring at Jupiter at time \(t\) is seen at \(t+\Delta(t)/c\). If eclipses occur at intervals \(S\), the observed intervals are

\[ S_{\text{obs}}\approx S\Big(1+\frac{\dot\Delta}{c}\Big), \]

and the observed time of the eclipse numbered \(k\) after a reference eclipse at \(t_0\) is

\[ t_k^{\text{obs}}=t_0+kS+\frac{\Delta(t_0+kS)-\Delta(t_0)}{c}+\frac{\Delta(t_0)}{c}. \]

The total range of the effect, from conjunction to opposition of Jupiter, is \((\Delta_{\max}-\Delta_{\min})/c=2a_\oplus/c\) for circular orbits.

Derivation. Differentiate \(t+\Delta(t)/c\) with respect to \(t\): consecutive events separated by \(S\) at the source arrive separated by \(S(1+\dot\Delta/c)\) to first order. Summing gives the second formula. For circular coplanar orbits, \(\Delta\) ranges from \(a_J-a_\oplus\) at opposition to \(a_J+a_\oplus\) at conjunction.

Rømer’s argument

The observable effect is not \(2a_\oplus/c\) directly, since eclipses cannot be seen near conjunction. What one sees is that a period measured over many orbits while the Earth recedes from Jupiter is longer than one measured while it approaches, and that the accumulated difference grows in step with the change of distance. Cassini’s tables were reset at the opposition of 8 July 1676. By November the Earth had moved far around its orbit and the distance to Jupiter had grown by more than an astronomical unit, so emergences should be seen roughly ten minutes later than tables built without the correction. The November observations agreed. Rømer combined his long series of observations, which by his own count exceeded seventy by the end of 1677, into the estimate of about 22 minutes for the diameter of the orbit.

Why Io

Io has the shortest period and the most frequent, sharpest eclipses. The light-time delay is the same for every satellite, but Cassini’s objection had force: the outer satellites have longer periods, so fewer eclipses, and their orbits have perturbations, mostly from their mutual resonances, which Rømer could not separate from the light-time effect. The full theory of the satellites, with mutual perturbations, had to wait for Pierre-Simon Laplace at the end of the eighteenth century.

Exercise 5.1Treat the orbits as circles with \(a_\oplus=1\) AU, \(a_J=5.20\) AU and light time 499 s per AU. Opposition was on 8 July 1676. The Earth gains on Jupiter at \(0.902^\circ\) per day. Find the Earth–Jupiter distance 30 days and 124 days after opposition (7 August and 9 November), and the difference in light time.

Show solution
The angle between the planets seen from the Sun is \(\theta=0.902^\circ\times\) days. Then \(\Delta=\sqrt{1+a_J^2-2a_J\cos\theta}\). At 30 days, \(\theta=27.1^\circ\), \(\Delta=4.34\) AU, 36.1 min. At 124 days, \(\theta=111.9^\circ\), \(\Delta=5.65\) AU, 47.0 min. The difference is about 10.9 minutes, the size of the delay Rømer predicted.

Exercise 5.2Find the maximum value of \(\dot\Delta\) for the circular model, and the largest lengthening of one synodic period of Io.

Show solution
With \(\theta=\omega t\), \(\dot\Delta=a_\oplus a_J\omega\sin\theta/\Delta\). Numerically the maximum, near \(\theta\approx79^\circ\), is about 27 km/s, a little less than the Earth’s orbital speed of 29.8 km/s because part of the Earth’s velocity is perpendicular to the line of sight. Then \(S\dot\Delta/c=152\,900\ \text{s}\times27.3/299\,792\approx14\) s per period.

Exercise 5.3One account of Rømer’s 1672 data finds a period of 42 h 28 min 31¼ s from four orbits near opposition and 42 h 29 min 3 s from thirty orbits. What error accumulates over thirty orbits if the first value is used for the whole interval? What does this illustrate?

Show solution
The difference is 31.75 s per orbit; over 30 orbits it amounts to \(30\times31.75\approx950\) s, about 16 minutes. Small differences in the measured period, from orbits at different phases of the Earth’s motion, accumulate to errors comparable to the whole light-time effect. That is why the inequality had to be explained, not merely averaged away.
Lesson 06

From light time to the speed of light

For the general reader. Rømer measured a time, not a speed. To turn “22 minutes to cross the Earth’s orbit” into kilometres per second, one needs the size of the orbit, which in 1676 was known only roughly. Huygens made the conversion in 1690. Fifty years after Rømer, James Bradley found a second, independent effect of the finite speed of light: the stars appear shifted in the direction of the Earth’s motion. The two methods agreed, and the finite speed of light was no longer in doubt.

Goals
  • Convert light times to speeds with historical and modern distances.
  • Derive the aberration of starlight and Bradley’s light time.
  • See why aberration settled the question.

Huygens’s conversion

In his Traité de la lumière (1690), Huygens took Rømer’s 22 minutes and a diameter of the Earth’s orbit of about 22 000 Earth diameters. Light then travels about 1000 Earth diameters per minute, or about \(16\tfrac23\) Earth diameters per second. In modern units this is about 212 000 km/s. The error comes from both inputs: the true light time across the orbit is 16.6 minutes, and the true diameter of the orbit is about 23 500 Earth diameters. In 1704 Isaac Newton gave seven or eight minutes for light to come from the Sun in his Opticks, much closer to the truth.

Aberration

Proposition (aberration of starlight, first order)

An observer moving with speed \(v\ll c\) sees a star whose true direction makes angle \(\theta\) with the velocity displaced towards the direction of motion by

\[ \delta\theta\approx\frac{v}{c}\sin\theta. \]

Over a year a star near the pole of the ecliptic describes a circle of angular radius \(\kappa=v_\oplus/c\approx20.5''\), the constant of aberration.

The proof is the rain analogy made exact. In the observer’s frame the light’s velocity is the vector difference of its velocity in the Sun’s frame and the observer’s velocity \(\mathbf v\). Resolving along and across \(\mathbf v\), the apparent direction \(\theta'\) satisfies \(\tan\theta'=\sin\theta/(\cos\theta+v/c)\), so \(\theta-\theta'\approx(v/c)\sin\theta\) to first order. Bradley, working with Samuel Molyneux from 1725 on the star γ Draconis, found an annual displacement he explained in 1728 in this way, and published in 1729. His constant of about \(20.2''\) implied that light takes 8 minutes 12 seconds to reach us from the Sun, in good agreement with eclipse timings.

Aberration confirmed Rømer independently of any theory of Jupiter’s satellites. It also showed that the Earth moves, which was the observational proof of Copernican motion that Rømer himself had sought through stellar parallax (Lesson 09).

Exercise 6.1Using Rømer’s 22 minutes and the modern astronomical unit, \(1.496\times10^8\) km, find the implied speed of light. Repeat with the modern light time of 16.6 minutes.

Show solution
\(2.992\times10^8/1320\approx2.27\times10^5\) km/s, about 24% low. With \(16.6\times60=996\) s, \(c\approx3.00\times10^5\) km/s.

Exercise 6.2Check Huygens’s \(16\tfrac23\) Earth diameters per second from his inputs, and convert to km/s with the modern Earth diameter of 12 742 km.

Show solution
22 000 diameters in 22 minutes is 1000 per minute, or \(1000/60=16.67\) per second. Times 12 742 km gives \(2.12\times10^5\) km/s.

Exercise 6.3Compute the constant of aberration from \(v_\oplus=29.78\) km/s. From Bradley’s \(20.2''\), find the light time from the Sun, taking the Earth’s orbital speed as \(2\pi\) AU per year.

Show solution
\(\kappa=29.78/299\,792=9.93\times10^{-5}\) rad \(=20.5''\). With \(\kappa=20.2''=9.79\times10^{-5}\) rad, \(c=v/\kappa\) and the light time for 1 AU is \(\tau=1\ \text{AU}/c=\kappa\,(1\ \text{yr})/(2\pi)=9.79\times10^{-5}\times3.156\times10^{7}/6.283\approx492\) s \(=8\) min 12 s.
Lesson 07

Epicycloidal gear teeth

For the general reader. Two gears with badly shaped teeth turn jerkily: the driven wheel speeds up and slows down as each pair of teeth slides past. Clocks and planetary models need smooth, steady motion. In 1674, according to standard accounts, Rømer proposed that teeth be cut to the shape of an epicycloid, the curve traced by a point on a circle rolling around another circle. With such teeth the ratio of speeds stays exactly constant.

Goals
  • State and prove the fundamental law of gearing.
  • Show that epicycloid and hypocycloid profiles generated by the same circle satisfy it.
  • Compute with epicycloids.

The law of gearing

Theorem (fundamental law of gearing)

Two wheels turn about fixed centres \(O_1,O_2\) with angular velocities \(\omega_1,\omega_2\), their tooth profiles touching at a point \(C\). Let the common normal to the profiles at \(C\) meet the line \(O_1O_2\) at \(P\). Then

\[ \frac{\omega_1}{\omega_2}=\frac{O_2P}{O_1P}. \]

The velocity ratio is constant if and only if \(P\) is fixed throughout the engagement. \(P\) is then the pitch point, and the circles about \(O_1,O_2\) through \(P\) roll on each other without slipping.

Proof. The profiles stay in contact, so the velocities of the contact point as a point of each body have equal components along the common normal \(\mathbf n\): \(\omega_1\,d_1=\omega_2\,d_2\), where \(d_i\) is the distance from \(O_i\) to the normal line. The triangles formed by \(O_1,O_2\), the normal and \(P\) are similar, so \(d_1/d_2=O_1P/O_2P\).

Cycloidal profiles

Theorem (cycloidal gearing)

Let a generating circle roll, without slipping, simultaneously on the outside of pitch circle 1 and on the inside of pitch circle 2, touching both at the pitch point \(P\). A point on the generating circle traces an epicycloid attached to wheel 1 and a hypocycloid attached to wheel 2. These two profiles are conjugate: used as tooth flanks, they transmit a constant velocity ratio.

Idea. At each instant the generating circle rotates about its point of contact with the pitch circles, which is \(P\). The tracing point therefore moves perpendicular to the line from \(P\) to itself, so the normal to each traced curve passes through \(P\). Both profiles share the tracing point and its normal, so they touch there with a common normal through the fixed point \(P\), and the law of gearing applies.

Cycloidal teeth dominated clockmaking for centuries. Engineering gears later moved to the involute profile, which tolerates errors in the distance between centres, but watch and clock trains still often use cycloidal forms.

Exercise 7.1A circle of radius \(r\) rolls outside a fixed circle of radius \(R\). Show that a point of the rolling circle traces \(x=(R+r)\cos t-r\cos\big(\tfrac{R+r}{r}t\big)\), \(y=(R+r)\sin t-r\sin\big(\tfrac{R+r}{r}t\big)\). For \(R=kr\) with \(k\) an integer, how many cusps does the curve have?

Show solution
The centre moves on a circle of radius \(R+r\) at angle \(t\). Rolling without slipping means the arc \(Rt\) on the fixed circle equals the arc \(r\phi\) on the moving one, and the moving circle’s own turning angle relative to the fixed frame is \(t+\phi=\tfrac{R+r}{r}t\). The tracing point is at \(-r\) times the unit vector at that angle from the centre, giving the formula. The point touches the fixed circle each time \(\phi\) is a multiple of \(2\pi\), that is, \(k\) times per revolution: \(k\) cusps.

Exercise 7.2Show that the length of one arch of the epicycloid is \(8r(R+r)/R\). Check the case \(R=r\) (the cardioid).

Show solution
Differentiating, the speed is \(|\dot{\mathbf x}|=(R+r)\sqrt{2-2\cos(Rt/r)}=2(R+r)|\sin(Rt/2r)|\). One arch is \(0\le t\le2\pi r/R\). The integral is \(2(R+r)\cdot\frac{2r}{R}\int_0^{\pi}\sin u\,du=\frac{8r(R+r)}{R}\). For \(R=r\) this is \(16r\), the length of a cardioid.

Exercise 7.3Two meshing wheels have 60 and 12 teeth. The small wheel is driven at 1 turn per minute. At what rate does the large wheel turn, and where does the pitch point lie if the centres are 72 mm apart?

Show solution
Pitch radii are proportional to tooth counts, \(r_1:r_2=60:12\) with \(r_1+r_2=72\) mm, so \(r_1=60\) mm, \(r_2=12\) mm. The large wheel turns at \(12/60=1/5\) turn per minute. \(P\) lies on the line of centres 12 mm from the small wheel’s centre.
Lesson 08

Planetary machines and gear ratios

For the general reader. A planetarium driven by a crank must make Saturn go round once while the Earth goes round about 29½ times. Gears with whole numbers of teeth can only produce ratios of whole numbers, and the true ratios are not simple fractions. Every maker of astronomical machines, from Rømer in Paris onwards, had to choose tooth counts that approximate the real periods well enough with gears small enough to cut.

Goals
  • Compute the gear train ratios for planetary motions.
  • Use continued fractions to find best approximations with small tooth counts.
  • Model the resonance of Jupiter’s inner satellites in a Jovilabe.

Best approximations

Every real \(x>0\) has a continued fraction \(x=a_0+\cfrac{1}{a_1+\cfrac{1}{a_2+\cdots}}\), written \([a_0;a_1,a_2,\dots]\). Its truncations \(p_n/q_n\) are the convergents, computed by \(p_n=a_np_{n-1}+p_{n-2}\), \(q_n=a_nq_{n-1}+q_{n-2}\).

Theorem (best approximation)

Each convergent satisfies \(\big|x-p_n/q_n\big|<1/(q_nq_{n+1})\). If \(p/q\) is any fraction with \(0<q\le q_n\) and \(p/q\ne p_n/q_n\), then \(|qx-p|>|q_nx-p_n|\). In this sense no fraction with smaller denominator does better.

The first statement follows from the identity \(p_{n+1}q_n-p_nq_{n+1}=(-1)^n\) and the fact that \(x\) lies between consecutive convergents. The second is Lagrange’s theorem on best approximations. For a gear train, \(q\) and \(p\) are tooth counts or products of tooth counts, so the convergents tell the instrument maker which small gears are worth using.

The Jovilabe

The sidereal periods of Io, Europa and Ganymede are 1.769, 3.551 and 7.155 days, very nearly in ratio \(1:2:4\). Laplace later showed that the mean motions satisfy \(n_I-3n_E+2n_G=0\) exactly, so the three moons can never be in line on the same side of Jupiter. A Jovilabe geared in the simple ratios \(1:2:4\) reproduces the satellites’ configurations well for some days; any real instrument must be reset from observations.

Exercise 8.1Jupiter’s sidereal period is 11.8626 years. Find the first five convergents of its continued fraction, and choose a pair of wheels with fewer than 100 teeth each to drive Jupiter from the Earth’s yearly wheel.

Show solution
\(11.8626=[11;1,6,3,1,\dots]\). Convergents: \(11,\ 12,\ 83/7,\ 261/22,\ 344/29\). The fraction \(83/7=11.857\) is good to 0.05%; using wheels of 83 and 7 teeth (driving 7 → 83) gives Jupiter one revolution per 11.857 years. With more room, \(344/29=11.8621\) is better.

Exercise 8.2Saturn’s period is 29.457 years. A maker proposes wheels of 206 and 7 teeth. How good is the ratio \(206/7\)? Compare it with the convergents of the continued fraction.

Show solution
\(206/7=29.4286\), an error of 0.1%. The continued fraction is \(29.457=[29;2,5,\dots]\) with convergents \(29,\ 59/2=29.5,\ 324/11=29.4545\). The fraction \(206/7\) lies between 29 and 29.5 but is not itself a convergent; \(324/11\) is much better. Real makers also weighed the practical sizes of the wheels.

Exercise 8.3Using the periods 1.769138, 3.551181 and 7.154553 days, check numerically that \(n_I-3n_E+2n_G\approx0\), where \(n=360^\circ/P\).

Show solution
\(n_I=203.489^\circ\)/d, \(n_E=101.375^\circ\)/d, \(n_G=50.318^\circ\)/d. Then \(203.489-304.124+100.635=0.000^\circ\)/d, zero to the accuracy of the periods.
Lesson 09

The transit instrument and meridian circle

For the general reader. As the Earth turns, every star crosses the local meridian, the north–south line through the zenith, once a day. If a telescope can swing only in the plane of the meridian, then the clock time at which a star crosses the crosshair gives its east–west coordinate, and the angle of the telescope gives its north–south coordinate. Rømer built such a transit instrument around 1690 and added a precisely divided circle in 1704. This design became the standard tool of positional astronomy for two hundred years.

Goals
  • Relate transit times to right ascension and meridian altitudes to declination.
  • Find the latitude from circumpolar stars.
  • Quantify the effect of collimation error.
Proposition (meridian astronomy)

A star of right ascension \(\alpha\) and declination \(\delta\), observed at latitude \(\phi\) (northern hemisphere, star south of the zenith), crosses the meridian when local sidereal time equals \(\alpha\), at altitude

\[ h=90^\circ-\phi+\delta. \]

A circumpolar star passing north of the zenith (\(\delta>\phi\)) crosses the meridian twice, above and below the pole, at altitudes \(\phi\pm(90^\circ-\delta)\) measured from the north horizon. Their mean is \(\phi\), independent of \(\delta\).

Sidereal time is defined as the hour angle of the vernal equinox, and a star’s hour angle is \(\mathrm{LST}-\alpha\). At transit the hour angle is zero. For the altitudes, the pole stands at altitude \(\phi\) above the north horizon, and a star circles it at angular distance \(90^\circ-\delta\). Refraction raises both culminations slightly, by different amounts, and must be corrected.

Rømer’s instrument turned on a horizontal east–west axis supported by two piers, so the telescope could only sweep the meridian. Compared with the large quadrants and sextants of Tycho, it reduced the measurement of one coordinate to timing with a good pendulum clock, which, since Huygens, could keep time to a few seconds a day. Rømer hoped to detect the parallax of the stars and so prove the Earth’s motion. The Triduum, three nights of observations on 20–23 October 1706, gave positions of 88 stars from about 250 observations made at the Tusculanum.

Exercise 9.1At Copenhagen (\(\phi=55^\circ41'\)) a star transits at altitude \(52^\circ19'\) south of the zenith. Find its declination. If local sidereal time at transit is \(6^{\mathrm h}45^{\mathrm m}\), what is its right ascension?

Show solution
\(\delta=h+\phi-90^\circ=52^\circ19'+55^\circ41'-90^\circ=18^\circ00'\). Right ascension equals the sidereal time at transit, \(\alpha=6^{\mathrm h}45^{\mathrm m}\).

Exercise 9.2A circumpolar star is seen at altitudes \(82^\circ10'\) and \(29^\circ12'\) above the north horizon at its two culminations (after correcting for refraction). Find the latitude and the star’s declination.

Show solution
\(\phi=(82^\circ10'+29^\circ12')/2=55^\circ41'\). The polar distance is \((82^\circ10'-29^\circ12')/2=26^\circ29'\), so \(\delta=63^\circ31'\).

Exercise 9.3The line of sight makes a small angle \(c\) with the plane perpendicular to the axis (collimation error). Show that a star of declination \(\delta\) is seen to transit late or early by \(c\sec\delta/15\) seconds of time, with \(c\) in arcseconds. How can the error be eliminated?

Show solution
The instrument sweeps a small circle parallel to the meridian, offset by \(c\). A star at declination \(\delta\) moves along its diurnal circle at \(15\cos\delta\) arcseconds per second of time, so it needs \(c/(15\cos\delta)\) seconds to cover the offset. Reversing the axis on its piers changes the sign of \(c\); the mean of transits observed in the two positions is free of it. Near the pole \(\sec\delta\) is large, so circumpolar stars reveal the error most clearly.
Lesson 10

Two fixed points: the Rømer thermometer

For the general reader. Seventeenth-century thermometers had arbitrary marks, so readings from two instruments could not be compared. Around 1702 Rømer made alcohol thermometers calibrated at two reproducible temperatures, the freezing and the boiling of water, with the degrees in between spaced equally. Daniel Gabriel Fahrenheit visited him in 1708, learned the method, and built on it the scale that still bears Fahrenheit’s name.

Goals
  • Understand calibration by two fixed points and linear interpolation.
  • Convert between the Rømer, Celsius and Fahrenheit scales.
  • Follow the steps from Rømer’s scale to Fahrenheit’s.
Definition (two-point linear scale)

A thermometer with reading \(L\) (for example, the length of a liquid column) is calibrated by assigning values \(\Theta_1,\Theta_2\) to two fixed points where it reads \(L_1,L_2\). Its temperature is then defined by

\[ \Theta=\Theta_1+(\Theta_2-\Theta_1)\frac{L-L_1}{L_2-L_1}. \]

On Rømer’s scale, water freezes at \(7.5\) and boils at \(60\). Any two linear scales are related by an affine map; in particular \(T_{\mathrm{C}}=\tfrac{40}{21}(T_{\text{Rø}}-7.5)\).

The scale is only as good as the assumption that the liquid expands uniformly. Different liquids expand nonlinearly in different ways, so alcohol and mercury thermometers calibrated at the same two points disagree slightly in between. The problem was resolved only with the thermodynamic temperature of the nineteenth century. Rømer’s choice of 7.5 for freezing has no documented explanation; one proposal is that his zero was first set at the freezing point of a salt solution. His notes on the thermometer survive in his notebook, the Adversaria, published in 1910.

Fahrenheit, according to his own later correspondence with Herman Boerhaave, adopted the scale, raised the freezing point from 7.5 to 8 and the mark near body heat from 22.5 to 24, and then divided each degree into four. Freezing became 32 and body heat 96.

Exercise 10.1Convert normal body temperature, 37 °C, to the Rømer scale. Convert 22.5 °Rø to Celsius.

Show solution
\(T_{\text{Rø}}=\tfrac{21}{40}\times37+7.5=26.9\). Conversely \(\tfrac{40}{21}(22.5-7.5)=28.6\) °C, well below body temperature, so whatever 22.5 originally marked, it was not the modern 37 °C.

Exercise 10.2Show that one Rømer degree equals \(24/7\) Fahrenheit degrees, using the two fixed points of water. Then check the conversion \(T_{\mathrm F}=\tfrac{24}{7}(T_{\text{Rø}}-7.5)+32\) at the boiling point.

Show solution
Between freezing and boiling there are \(60-7.5=52.5\) Rømer degrees and \(212-32=180\) Fahrenheit degrees, so one Rømer degree is \(180/52.5=24/7\) °F. At 60 °Rø, \(\tfrac{24}{7}\times52.5+32=180+32=212\).
Part III · Life and legacy
Lesson 11

Copenhagen, 1681–1710

Rømer returned to Copenhagen in 1681 as professor of astronomy at the university; from 1686 he directed the observatory on top of the Round Tower. The same year he married Anne Marie Bartholin, daughter of his teacher. Over the next three decades he held a remarkable range of offices, and much of his time went to the state.

As royal mathematician he prepared the first national system of weights and measures for Denmark–Norway, introduced on 1 May 1683 and refined in 1698. He defined the Danish mile as 24 000 Danish feet, about 7.5 km. His survey of the main roads led to the first Danish milestones, some of which still stand. He helped establish schools of navigation. In 1700, after discussing the reform with Gottfried Wilhelm Leibniz, he persuaded the king to adopt the reformed calendar: in Denmark–Norway, 18 February 1700 was followed directly by 1 March.

In 1705 he became police director and mayor of Copenhagen. He set building regulations, improved water supply, sewers and fire equipment, introduced oil lamps for street lighting and organized measures on poverty and begging. He also served as a judge in the Supreme Court, as university librarian and twice as rector of the university, and sat on many royal commissions.

He did not give up astronomy. Disturbed by the bells of the church beneath the Round Tower, he built a transit instrument at his house in Store Kannikestræde around 1690, and in 1704 a country observatory, the Tusculanum, at Vridsløsemagle west of the city, on Bartholin family land. Its meridian circle produced the Triduum of October 1706. His assistant Peder Horrebow later published an account of his instruments. Rømer died in Copenhagen on 19 September 1710, six days before his 66th birthday, and was buried in the Church of Our Lady, Copenhagen’s cathedral.

Exercise 11.1Under the Julian calendar every fourth year is a leap year; under the Gregorian, century years are leap years only if divisible by 400. By 1700 the Julian date was 10 days behind the Gregorian. Explain why Denmark–Norway dropped 11 dates, from 19 to 29 February, in 1700.

Show solution
1700 is a leap year in the Julian calendar but not in the Gregorian. Before Julian 29 February 1700 the difference was 10 days; that extra Julian day raised it to 11. Passing from Julian 18 February directly to Gregorian 1 March skips the 11 Julian dates 19–29 February.
Lesson 12

Fire, survival and legacy

On 20 October 1728 a fire broke out in Copenhagen and burned for three days. It destroyed much of the old city, including university buildings, and with them most of Rømer’s observations and instruments. Three things survived. The Triduum survived, of which Rømer had five copies made. His working notebook, the Adversaria, survived and was published in 1910. And Peder Horrebow, who had worked with him, described his instruments and methods in Basis astronomiae (1735), which spread his designs across Europe. A folio listing his eclipse observations of 1668–1677 was found in 1913.

  • The speed of light. Rømer’s light time was accepted by Huygens, Flamsteed, Halley and Newton, resisted in Paris, and confirmed beyond doubt by Bradley’s aberration of 1728. Pehr Wargentin’s ephemerides of Jupiter’s satellites (1746) included the light-time correction, and Jean-Baptiste Delambre in 1809 derived 8 minutes 12 seconds for the light time from the Sun from a long series of eclipses. The first terrestrial measurement, by Hippolyte Fizeau in 1849, came 173 years after Rømer’s announcement.
  • Instruments. The transit instrument and the meridian circle were the backbone of positional astronomy until photography and, much later, astrometric satellites.
  • Measurement and the state. His weights and measures, milestones and calendar reform belong to the same programme as his astronomy: reproducible standards, carefully defined.
  • Temperature. Through Fahrenheit, his two-point calibration became standard practice.
  • Honours. The Ole Rømer Medal of the Danish research councils, the Ole Rømer Observatory of Aarhus University, the Kroppedal Museum beside the excavated Tusculanum (rediscovered in 1978), the lunar crater Römer, and streets in Aarhus and Copenhagen bear his name.

DiscussionCassini first mentioned a finite speed of light and then rejected it, while Rømer developed and defended it. Using Lessons 03 and 05 and the course on Cassini, argue for and against the view that Cassini’s caution was justified by the data available in 1676.

Final projectChoose one topic from Part II. Write a two-page account for a strong undergraduate that explains the measurement problem, derives the key formula, works a numerical example with seventeenth-century data, and identifies the main source of error.

Reference

Sources and further reading

Biography

  1. Ole Rømer, Wikipedia.
  2. Niels Bohr Institute, Ole Rømer: education, Copenhagen, Machines and Observatorium Tusculanum.
  3. Ole Rømer, Den Store Danske (lex.dk).
  4. Ole Rømer and the speed of light, Optics and Photonics News 20 (2009).

Physics and astronomy

  1. [O. Rømer], Démonstration touchant le mouvement de la lumière, Journal des sçavans, 7 December 1676; English translation in Philosophical Transactions 12 (1677).
  2. Rømer’s determination of the speed of light, Wikipedia (detailed chronology and numbers).
  3. L. Bobis and J. Lequeux, Cassini, Rømer and the velocity of light, Journal of Astronomical History and Heritage 11 (2008).
  4. I. B. Cohen, Roemer and the first determination of the velocity of light (1676), Isis 31 (1940).
  5. C. Huygens, Traité de la lumière, Leiden, 1690.
  6. J. Bradley, An account of a new discovered motion of the fix’d stars, Philosophical Transactions 35 (1729); Aberration (astronomy), Wikipedia.
  7. Rømer scale, Wikipedia.
  8. P. Horrebow, Basis astronomiae, Copenhagen, 1735.
  9. R. Willis, Principles of Mechanism, 1841 (cycloidal and involute gearing).
  10. A. Ya. Khinchin, Continued Fractions, Dover (best approximations).