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al-Biruni

Abu Rayhan Muhammad ibn Ahmad al-Biruni · أبو ريحان البيروني

A course in two strands: the life of the Khwarazmian scholar who served five dynasties, followed a conquering sultan into India and wrote about 146 works, and his exact sciences: calendars, trigonometry, chords, the size of the Earth, longitudes, astronomy and specific gravity.

Born 4 September 973, Kath, Khwarazm Died c. 1048–1050, Ghazna al-Qanun al-Masʿudi c. 1030 Level undergraduate → advanced undergraduate
The dip method. From a summit of height h the horizon lies an angle α below the horizontal. The sight line is tangent to the Earth, so cos α = R/(R + h). The figure greatly exaggerates h and α.
Lesson 00

How to use this course

Abu Rayhan al-Biruni is often remembered as the author of the finest medieval book about India. That book is one part of an output of about 146 works, roughly two-thirds of them in the exact sciences. He measured the Earth from a hilltop, determined longitudes from eclipse timings, built trigonometry into the foundation of astronomy, and weighed metals and gems with a precision that impressed nineteenth-century chemists. He also argued with Ibn Sina about Aristotle, and he insisted, more firmly than most of his contemporaries, that a claim about nature must answer to observation.

Prerequisites for Part II
  • Trigonometry, including the sine and cosine rules and small-angle approximations.
  • For Lessons 05 and 08, the basics of spherical trigonometry (great circles, spherical triangles); the needed formulas are stated.
  • For Lesson 09, finite differences at the level of a first numerical-analysis course.

The sources for his life are better than for most medieval scientists. His own prefaces and his bibliography of 1036 date many works, and later historians, notably Bayhaqi and Nizami ʿAruzi, add episodes, some of them anecdotal. The lessons say which source supports each claim and mark later tradition as such. All dates are CE; “c.” marks an approximate date. Most of his works are in Arabic, the language of science in his world, though his mother tongue was Khwarezmian, an Eastern Iranian language. This course is a companion to the course on Ibn Sina, his contemporary and sparring partner.

Reference

Timeline

973Born 4 September in the outer district of Kath, capital of Khwarazm (in today’s Karakalpakstan, Uzbekistan). Some sources give 15 September.
c. 990Studies with the mathematician Abu Nasr Mansur ibn ʿIraq, a prince of the ruling Afrighid family. Early solar observations with a meridian ring.
995The Maʾmunids of Gurganj overthrow the Afrighids; al-Biruni leaves Khwarazm, probably for Rayy, where he meets the astronomer al-Khujandi.
997Observes a lunar eclipse at Kath, timed in coordination with Abu’l-Wafaʾ at Baghdad, to find their difference in longitude.
c. 997–999Patronage of the Samanid ruler Mansur II at Bukhara.
c. 998–1003At the court of the Ziyarid amir Qabus at Gurgan.
c. 1000al-Athar al-baqiya (the Chronology of Ancient Nations), dedicated to Qabus. Questions and answers with Ibn Sina.
c. 1004–1017Returns to Khwarazm; adviser at the Maʾmunid court of Gurganj, from 1009 under Abu’l-ʿAbbas Maʾmun.
1017Mahmud of Ghazna conquers Khwarazm; al-Biruni and other scholars are taken to Ghazna.
1018Observations near Kabul.
c. 1020sAccompanies Ghaznavid campaigns into northwest India; learns Sanskrit. Measures the Earth at Nandana in the Punjab.
c. 1021Exhaustive Treatise on Shadows.
1025Tahdid nihayat al-amakin, on determining geographical coordinates.
c. 1027Book on Chords (Istikhraj al-awtar).
c. 1029Kitab al-Tafhim, an introduction to mathematics, astronomy and astrology, in Arabic and Persian versions.
1030Death of Mahmud. India completed; al-Qanun al-Masʿudi dedicated to the new sultan Masʿud.
1036Lists 103 completed works in his bibliography of al-Razi.
c. 1041–1048Kitab al-Jamahir, on precious stones and metals, under the sultan Mawdud.
c. 1048–1050Dies in Ghazna. Uzbek scholars give December 1048; his pharmacology, written when he was over 80, suggests he lived past 1050.
Part I · Formation
Lesson 01

Kath and Khwarazm, 973–995

Al-Biruni was born on 4 September 973 in Khwarazm, the oasis of the lower Amu Darya south of the Aral Sea. His name is usually derived from the Persian birun, “outside”: he came from the outer district of Kath, the capital. He later said little about his family, and nothing reliable is known of his parents. He spent his first twenty-five years or so in Khwarazm.

Khwarazm was then divided between two dynasties: the old Afrighid shahs at Kath and the rising Maʾmunids at Gurganj (Urgench) to the north. The Afrighid family produced al-Biruni’s teacher, Abu Nasr Mansur ibn ʿIraq, a mathematician of real distinction. Abu Nasr worked on spherical trigonometry, and al-Biruni later credited him with proofs of the sine theorem (Lesson 05). Several of Abu Nasr’s treatises are addressed to his pupil.

First observations

As a young man al-Biruni was already observing. Using a meridian ring, he measured the Sun’s noon altitudes at Kath to find the latitude of the town. This was the start of a lifelong preference for measurement: wherever he lived, he set up instruments, and his later books keep records of dated observations.

DiscussionAl-Biruni’s teacher was a member of a ruling family, and his own career depended on princes. How did court patronage shape what kinds of science were possible in tenth-century Central Asia?

Lesson 02

Wandering years, 995–c. 1004

In 995 the Maʾmunid ruler of Gurganj overthrew the Afrighids. Al-Biruni left Khwarazm. He seems to have gone to Rayy, near modern Tehran, where he met the astronomer Abu Mahmud al-Khujandi and discussed the giant mural sextant al-Khujandi had built to measure the obliquity of the ecliptic. Al-Biruni later reported those observations, and their instrumental problems, in his Tahdid.

He was back in Kath for the lunar eclipse of 997. By arrangement, Abu’l-Wafaʾ al-Buzjani observed the same eclipse in Baghdad. Because an eclipse of the Moon is seen at the same physical moment everywhere, the difference between the local times of the two observations gives the difference in longitude. It is one of the earliest documented observations coordinated between distant cities (Lesson 08).

Bukhara and Gurgan

He found patrons in Bukhara under the Samanid ruler Mansur II (997–999), and then at Gurgan, south-east of the Caspian, under the Ziyarid amir Qabus ibn Wushmgir. For Qabus he wrote al-Athar al-baqiya ʿan al-qurun al-khaliya, “the remaining traces of past centuries”, around 1000. It is a comparative study of the calendars, eras and festivals of Persians, Sogdians, Khwarazmians, Jews, Christians, Arabs and others, and the subject of Lesson 04.

The questions to Ibn Sina

Around 1000 al-Biruni sent eighteen questions on natural philosophy to the young Ibn Sina, ten of them aimed at Aristotle’s On the Heavens. Why must celestial motion be circular, and why could the heavens not move in an ellipse? Why is a vacuum impossible? Why could there not be other worlds? Why does ice float, when it is solid water? Ibn Sina answered for the Aristotelians; al-Biruni objected sharply; and Ibn Sina’s pupil Abu Saʿid al-Maʿsumi wrote a final reply. The tone on both sides is impatient. The exchange is our best evidence of al-Biruni’s habit of testing doctrine against experience.

DiscussionAl-Biruni asked whether the heavens might move in an ellipse, about six centuries before Kepler. Was this a prediction, a logical objection, or something else? What would he have needed to turn the question into a theory?

Lesson 03

Gurganj and the conquest, c. 1004–1017

Around 1004 al-Biruni made peace with the Maʾmunids and returned to Khwarazm, now ruled from Gurganj. He served Abu’l-Hasan ʿAli ibn Maʾmun and then his brother Abu’l-ʿAbbas Maʾmun (ruled 1009–1017). According to the historian Bayhaqi, he was a companion and political adviser for seven years under the second ruler. Ibn Sina also served at this court until about 1012. His teacher Abu Nasr Mansur was there too, and the two men continued their trigonometric work.

The court lay in the shadow of Mahmud of Ghazna, the most powerful ruler of the eastern Islamic world. In 1017, after the murder of Abu’l-ʿAbbas Maʾmun in a revolt, Mahmud invaded Khwarazm and annexed it. He carried off its scholars to Ghazna (modern Ghazni, Afghanistan), among them al-Biruni and Abu Nasr. Whether al-Biruni went as a prisoner, a hostage or a reluctant courtier is debated; the early years at Ghazna were hard.

A twelfth-century collection of anecdotes by Nizami ʿAruzi tells that Mahmud once asked al-Biruni to predict by which gate he would leave a pavilion, and then broke through a wall to defeat him, only to find the prediction written down correctly. The story is later tradition and cannot be checked; it shows how al-Biruni’s reputation as an astrologer outlived his own scepticism about astrology.

DiscussionAl-Biruni served five dynasties: Afrighid, Samanid, Ziyarid, Maʾmunid and Ghaznavid. Compare this career with Ibn Sina’s. What did itinerant scholars gain and lose by moving between courts?

Part II · The mathematics
Lesson 04

Calendars and the Chronology

For the general reader. To compare the histories of different peoples, you must convert dates between their calendars. Some count days by the Sun, some by the Moon, some by both, and each starts from a different era. Al-Biruni’s Chronology (c. 1000) gathered the rules for all the calendars he could find and turned the conversion between them into arithmetic.

Goals
  • Model calendars by their mean year and month lengths and their intercalation rules.
  • Convert between eras by counting days modulo 7 and modulo cycle lengths.
  • Work the chessboard sum that al-Biruni computed.

Three kinds of calendar

A solar calendar tracks the year of the seasons (the tropical year, about 365.2422 days). The Julian calendar does this with a year of 365 days and a leap day every fourth year. The old Persian calendar had 12 months of 30 days and 5 extra days, 365 in all, with no regular intercalation, so its new year drifted through the seasons. A lunar calendar, such as the Islamic one, counts 12 synodic months (about 29.5306 days each) and ignores the seasons. A lunisolar calendar, such as the Jewish one, inserts a thirteenth month 7 times in 19 years.

Proposition (the Metonic cycle)

235 synodic months and 19 tropical years differ by about two hours: \(235\times29.53059\approx6939.69\) days and \(19\times365.2422\approx6939.60\) days. Hence a calendar with 12 years of 12 months and 7 years of 13 months in each 19-year cycle stays aligned with both Sun and Moon for centuries.

The Islamic calendar uses a 30-year cycle of 360 months in which 11 years have 355 days and 19 have 354, giving \(10631\) days. The mean month is then \(10631/360=29.530\overline5\) days, very close to the true mean synodic month.

Counting days

To convert a date, count the days from a fixed epoch in each calendar. Al-Biruni tabulated the epochs of the eras: the Seleucid era (“era of Alexander”), the era of Yazdegerd III, the Hijra, and others. The weekday of any day number \(N\) counted from a known epoch follows from \(N\bmod 7\).

The chessboard

In the Chronology, al-Biruni computed the number of grains on a chessboard when one grain is placed on the first square and the number doubles on each square. He gave the result in full.

Proposition

\(\displaystyle\sum_{k=0}^{63}2^k=2^{64}-1=18\,446\,744\,073\,709\,551\,615.\)

One way to compute it, close to al-Biruni’s, is repeated squaring: \(2^{8}=256\), \(2^{16}=65536\), \(2^{32}=4\,294\,967\,296\), and \(2^{64}=(2^{32})^2\).

Exercise 4.1Prove \(\sum_{k=0}^{n-1}2^k=2^n-1\) and compute \(2^{32}\) from \(2^{16}\) by hand-style multiplication.

Show solution
Let \(S=\sum_{k=0}^{n-1}2^k\). Then \(2S-S=2^n-1\). For \(2^{32}=65536^2\): \(65536^2=(65000+536)^2=4\,225\,000\,000+2\cdot65000\cdot536+536^2=4\,225\,000\,000+69\,680\,000+287\,296=4\,294\,967\,296\).

Exercise 4.2A Persian year of exactly 365 days drifts against the tropical year. After how many years has the Persian new year moved through a whole year of seasons?

Show solution
The drift is about \(0.2422\) days per year, so a full cycle takes \(365.2422/0.2422\approx1508\) years. In the Julian calendar the drift is only \(0.0078\) days per year, one day in about 128 years.

Exercise 4.3Compute the mean Islamic year in days from the 30-year cycle, and find after how many lunar years the Islamic calendar has run one full year ahead of a solar calendar.

Show solution
\(10631/30=354.3\overline6\) days. Each lunar year is shorter than the tropical year by \(365.2422-354.3667\approx10.8755\) days. The lunar count gains a full solar year after \(n\) lunar years where \(n\times354.3667\approx(n-1)\times365.2422\), so \(n\approx365.2422/10.8755\approx33.6\). Roughly 34 lunar years correspond to 33 solar years.
Lesson 05

Sines, tangents and shadows

For the general reader. Greek astronomers computed with chords of circles. Indian astronomers replaced the chord by the half-chord, our sine, and astronomers of the Islamic world added the shadow functions, our tangent and cotangent, read off the shadow of a vertical stick. Al-Biruni’s generation turned these into a systematic trigonometry, with clean theorems that replaced Ptolemy’s cumbersome configurations. His Exhaustive Treatise on Shadows (c. 1021) is the fullest medieval account of the shadow functions.

Goals
  • Relate medieval sines and shadows of radius \(R\) to modern functions.
  • Prove the plane sine theorem and state the spherical one.
  • Use the shadow functions to measure heights and the Sun’s altitude.

Medieval functions

Medieval tables used a base circle of radius \(R\), commonly \(R=60\). The sine of an arc \(\theta\) was the half-chord \(\mathrm{Sin}\,\theta=R\sin\theta\). For a gnomon (vertical stick) of length \(g\) and Sun altitude \(h\), the horizontal shadow is \(g\cot h\) and the shadow on a wall perpendicular to the Sun’s azimuth is \(g\tan h\). Gnomons were divided into 12 digits or 60 parts.

Theorem (plane sine theorem)

In any triangle with sides \(a,b,c\) opposite angles \(A,B,C\), \(\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2\rho\), where \(\rho\) is the circumradius.

Proof. In the circumcircle, side \(a\) is a chord subtending the central angle \(2A\), by the inscribed angle theorem. A chord subtending \(2A\) in a circle of radius \(\rho\) has length \(2\rho\sin A\). The same holds for \(b\) and \(c\).

Theorem (spherical sine theorem)

In a spherical triangle with sides (arcs) \(a,b,c\) opposite angles \(A,B,C\), \[\frac{\sin a}{\sin A}=\frac{\sin b}{\sin B}=\frac{\sin c}{\sin C}.\]

Al-Biruni sent proofs of the plane and spherical sine theorems to the geometer al-Sijzi and credited them to Abu Nasr Mansur. Abu’l-Wafaʾ also gave a proof, and priority was disputed at the time. The theorem let astronomers solve spherical triangles directly, replacing Menelaus’s theorem on the complete quadrilateral.

Proof sketch (spherical). Put the triangle on a unit sphere with centre \(O\). Drop the perpendicular from vertex \(C\) to the plane \(OAB\), meeting it at \(P\). Dropping perpendiculars from \(P\) to the lines \(OA\) and \(OB\) and computing \(|CP|\) in two ways gives \(|CP|=\sin b\sin A=\sin a\sin B\).

Exercise 5.1A gnomon of 12 digits casts a horizontal shadow of 7 digits. Find the Sun’s altitude. With a table of radius 60, what is the tabulated Sine of that altitude?

Show solution
\(\cot h=7/12\), so \(\tan h=12/7\approx1.714\) and \(h\approx59.74^\circ\). Then \(\mathrm{Sin}\,h=60\sin h\approx60\times0.8638\approx51.83\), that is \(51;50\) in sexagesimal.

Exercise 5.2Use the spherical sine theorem in a right spherical triangle (\(C=90^\circ\)) to derive the rule \(\sin a=\sin c\sin A\). Apply it to find the Sun’s declination \(\delta\) when its ecliptic longitude is \(\lambda\) and the obliquity is \(\varepsilon\).

Show solution
With \(C=90^\circ\), \(\sin c/\sin C=\sin c\), so \(\sin a=\sin c\sin A\). In the triangle formed by the equinox, the Sun and the foot of the Sun’s hour circle on the equator, the right angle is on the equator, the hypotenuse is \(\lambda\), the angle at the equinox is \(\varepsilon\), and the side opposite is \(\delta\). Hence \(\sin\delta=\sin\lambda\sin\varepsilon\).

Exercise 5.3In a plane triangle, \(A=40^\circ\), \(B=65^\circ\) and \(c=10\). Find \(a\) and \(b\).

Show solution
\(C=75^\circ\), so \(2\rho=10/\sin75^\circ\approx10.353\). Then \(a\approx10.353\times0.6428\approx6.65\) and \(b\approx10.353\times0.9063\approx9.38\).
Lesson 06

Chords, the broken chord and the nonagon

For the general reader. Every sine table needs a starting point, and the hardest is the sine of 1°. Euclid’s constructions give the chords of 36°, 60°, 72° and 90°, and halving and subtracting give all multiples of 3°. Getting to 1° requires trisecting an angle, which ruler and compass cannot do. Al-Biruni’s Book on Chords (c. 1027) gathered many proofs of a theorem he attributed to Archimedes, and his Masʿudic Canon reduced the trisection to a cubic equation and solved it numerically.

Goals
  • Prove the broken chord theorem and derive trigonometric identities from it.
  • Derive the cubic equation for the chord of 40° from the triple-angle formula.
  • Solve it numerically to sexagesimal precision.

The broken chord

Theorem (broken chord; attributed to Archimedes)

Let \(A\) and \(B\) be points on a circle, and let \(C\) be a point on the arc \(AB\) with \(|AC|>|CB|\), so that \(ACB\) is a “broken chord”. Let \(M\) be the midpoint of the arc \(ACB\), and let \(D\) be the foot of the perpendicular from \(M\) to \(AC\). Then \(D\) bisects the broken chord: \(|AD|=|DC|+|CB|\).

Proof. Choose \(E\) on \(AC\) with \(|AE|=|CB|\). Since \(M\) is the midpoint of the arc, \(|MA|=|MB|\). The inscribed angles \(\angle MAC\) and \(\angle MBC\) stand on the same arc \(MC\), so they are equal. Hence triangles \(MAE\) and \(MBC\) are congruent (SAS) and \(|ME|=|MC|\). Triangle \(MEC\) is isosceles, so the foot \(D\) of the perpendicular from \(M\) is the midpoint of \(EC\). Therefore \(|AD|=|AE|+|ED|=|CB|+|DC|\).

Al-Biruni used the theorem to derive trigonometric rules. With chords \(\mathrm{crd}\,\theta=2R\sin(\theta/2)\) on a unit circle and arcs \(AC=2\alpha\) and \(CB=2\beta\), the theorem becomes the sum-to-product rule \(\sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2\) (Exercise 6.1), which is equivalent to the addition formula for sines.

The chord of 40° and the nonagon

The side of a regular nonagon subtends 40°. With the triple-angle formula \(\cos3\theta=4\cos^3\theta-3\cos\theta\) and \(\theta=20^\circ\), \(\cos60^\circ=\tfrac12\), the number \(x=2\cos20^\circ\) satisfies

\[ x^3=3x+1 . \]
Proposition

The equation \(x^3=3x+1\) has three real roots, \(2\cos20^\circ\), \(2\cos140^\circ\) and \(2\cos260^\circ\). None is rational, and none is constructible with ruler and compass. The positive root is \(x=1.8793852\ldots=1;52,45,47,12,43,\ldots\) in sexagesimal.

In the Canon (Book III, chapter 3), al-Biruni set up the cubic \(x^3=1+3x\) in the course of finding the chord of 40° and found its root by an iterative numerical procedure, correct to three sexagesimal places. Non-constructibility, proved by Pierre Wantzel in 1837, explains why such a numerical method was unavoidable.

Exercise 6.1On a unit circle a chord subtending an arc \(2\phi\) has length \(2\sin\phi\). In the broken chord theorem let arc \(AC=2\alpha\) and arc \(CB=2\beta\) with \(\alpha>\beta\). Show that the theorem is equivalent to \(\sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2\), and check it for \(\alpha=50^\circ\), \(\beta=20^\circ\).

Show solution
The theorem says \(|AD|=\tfrac12(|AC|+|CB|)=\sin\alpha+\sin\beta\). Arc \(ACB\) is \(2\alpha+2\beta\), so \(M\) is at arc distance \(\alpha+\beta\) from \(A\) and \(|MA|=2\sin\frac{\alpha+\beta}2\). Arc \(MC\) is \(2\alpha-(\alpha+\beta)=\alpha-\beta\), and the inscribed angle \(\angle MAC\) is half of it. In the right triangle \(MAD\), \(|AD|=|MA|\cos\angle MAC=2\sin\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2\). Numerically: \(\sin50^\circ+\sin20^\circ\approx0.7660+0.3420=1.1081\) and \(2\sin35^\circ\cos15^\circ\approx2\times0.5736\times0.9659=1.1081\). Since this sum-to-product rule is equivalent to the addition formula for sines, the broken chord theorem carries the addition theorem, as al-Biruni observed.

Exercise 6.2Show that \(x=2\cos20^\circ\) satisfies \(x^3=3x+1\), and prove that the equation has no rational root.

Show solution
\(x^3-3x=8\cos^320^\circ-6\cos20^\circ=2(4\cos^320^\circ-3\cos20^\circ)=2\cos60^\circ=1\). A rational root of the monic integer polynomial \(x^3-3x-1\) must be an integer dividing 1, and \(\pm1\) give \(-3\) and \(1\), not 0.

Exercise 6.3Solve \(x^3=3x+1\) by the fixed-point iteration \(x_{k+1}=\sqrt[3]{3x_k+1}\) from \(x_0=2\). How many steps give three sexagesimal places (error below \(60^{-3}\approx4.6\times10^{-6}\))?

Show solution
\(x_1=\sqrt[3]7\approx1.91293\), \(x_2\approx1.88884\), \(x_3\approx1.88206\), \(x_4\approx1.88014\), and so on. The contraction factor is \(g'(x)=(3x+1)^{-2/3}\approx0.28\) near the root, so the error shrinks by about 0.28 each step. Starting from an error of 0.12, one needs about \(\log(4.6\times10^{-6}/0.12)/\log0.28\approx8\) steps; indeed \(x_8\approx1.879390\) has error \(4.9\times10^{-6}\) and \(x_9\) is within the bound. Newton’s method would need three.
Lesson 07

Measuring the Earth

For the general reader. Earlier measurements of the Earth, by Eratosthenes and by the astronomers of the caliph al-Maʾmun, needed long journeys across flat desert to measure a degree of latitude. Al-Biruni found a way to do it from one place. Climb a hill of known height, look at the horizon, and measure how far it lies below the horizontal. He carried this out at Nandana, a fort in the Salt Range of the Punjab, while in Ghaznavid service, and reported it in the Masʿudic Canon.

Goals
  • Derive the dip formula for the Earth’s radius.
  • Measure a hill’s height from two stations by tangents.
  • Quantify the sensitivity of the result to the dip angle and to refraction.

Step 1: the height of the hill

From two points on the plain, a distance \(d\) apart and in line with the summit, measure the angles of elevation \(\theta_1<\theta_2\). If \(x\) is the distance from the nearer point to the foot of the summit, then \(h=x\tan\theta_2=(x+d)\tan\theta_1\), so

\[ h=\frac{d\,\tan\theta_1\tan\theta_2}{\tan\theta_2-\tan\theta_1}. \]

Step 2: the dip

Proposition (the dip method)

Let an observer stand at height \(h\) above a spherical Earth of radius \(R\), and let \(\alpha\) be the dip of the horizon below the horizontal. Then \(\cos\alpha=\dfrac{R}{R+h}\), so

\[ R=\frac{h\cos\alpha}{1-\cos\alpha}\approx\frac{2h}{\alpha^2}\quad(\alpha\text{ in radians}). \]

Proof. The line of sight to the horizon touches the Earth at a point \(T\), so \(OT\perp\) the sight line, where \(O\) is the centre. In the right triangle \(O T P\), with \(P\) the observer, \(|OP|=R+h\) and \(|OT|=R\). The horizontal at \(P\) is perpendicular to \(OP\), so the dip \(\alpha\) equals the angle \(\angle TOP\). Hence \(\cos\alpha=R/(R+h)\). Solve for \(R\). For small \(\alpha\), \(1-\cos\alpha\approx\alpha^2/2\).

His numbers

As reported in the Canon, the height was \(652;3,18\) cubits (about 652.06) and the dip \(0;34^\circ\), that is 34 arcminutes. His stated radius is about 12.8 million cubits (12,803,337 cubits in the usual reading). Converting cubits to kilometres is uncertain, because the length of his cubit is not known exactly; reasonable conversions give values within a few per cent of the modern mean radius of 6371 km. Two cautions: refraction bends the line of sight and makes the dip appear smaller, and the plain at Nandana is not a sea horizon. The close agreement owes something to luck.

Exercise 7.1Recompute \(R\) in cubits from \(h=652.06\) cubits and \(\alpha=34'\), with the exact formula and the small-angle approximation. Then repeat with \(\alpha=33'\) and \(35'\). What do you conclude?

Show solution
\(\alpha=34'=0.0098902\) rad, \(1-\cos\alpha\approx4.8908\times10^{-5}\), so \(R\approx652.06\times0.99995/4.8908\times10^{-5}\approx1.333\times10^{7}\) cubits; the approximation \(2h/\alpha^2\) gives the same to four figures. With \(33'\), \(R\approx1.415\times10^7\); with \(35'\), \(R\approx1.258\times10^7\). One arcminute changes \(R\) by about 6 per cent, since \(R\propto\alpha^{-2}\). Al-Biruni’s stated 12.8 million differs from this recomputation by about 4 per cent; how he evaluated the cosine of so small an angle is part of the explanation, and historians reconstruct his computation differently.

Exercise 7.2From two stations 200 m apart, the summit of a hill is seen at elevations of \(20^\circ\) and \(30^\circ\). Find the height.

Show solution
\(h=\frac{200\tan20^\circ\tan30^\circ}{\tan30^\circ-\tan20^\circ}=\frac{200\times0.36397\times0.57735}{0.57735-0.36397}=\frac{42.028}{0.21338}\approx197\) m.

Exercise 7.3Atmospheric refraction is often modelled by replacing \(R\) with an effective radius \(R/(1-k)\), with \(k\approx0.13\). If the measured dip obeys the formula with the effective radius, by what factor does the naive method overestimate \(R\)?

Show solution
The measured dip satisfies \(\alpha^2\approx2h(1-k)/R\), so the naive estimate \(2h/\alpha^2=R/(1-k)\approx1.15R\). Refraction alone would make the result about 15 per cent too large, which shows how much the final agreement depends on compensating errors.
Lesson 08

Longitudes, the qibla and the Tahdid

For the general reader. Latitude is easy to measure from the Sun or the pole star. Longitude is hard, because it requires knowing the time at two places at once. A lunar eclipse supplies a common moment. In the Tahdid nihayat al-amakin (“determination of the coordinates of places”, 1025), al-Biruni combined eclipse timings, travellers’ distances and spherical trigonometry to compute the longitude of Ghazna. The motive was partly religious: Muslims pray facing Mecca, and the direction, the qibla, is a problem in spherical trigonometry.

Goals
  • Convert eclipse time differences into longitude differences.
  • Compute great-circle distances between places of known coordinates.
  • Derive and apply the formula for the qibla.
Principle

The Earth turns \(360^\circ\) in 24 hours, so a difference of \(\Delta t\) hours in local time for the same instant corresponds to a longitude difference \(\Delta\lambda=15^\circ\times\Delta t\).

Al-Biruni computed the longitude difference between Baghdad and Ghazna as \(24;20^\circ\), within about 18 arcminutes of the modern value.

Spherical distance and direction

Theorem (spherical cosine rule and the qibla)

For places at latitudes \(\varphi_1,\varphi_2\) with longitude difference \(\Delta\lambda\), the central angle \(\sigma\) between them satisfies

\[ \cos\sigma=\sin\varphi_1\sin\varphi_2+\cos\varphi_1\cos\varphi_2\cos\Delta\lambda . \]

The direction \(q\) of the great circle to place 2, measured from north at place 1, satisfies

\[ \tan q=\frac{\sin\Delta\lambda}{\cos\varphi_1\tan\varphi_2-\sin\varphi_1\cos\Delta\lambda}. \]

Both follow from the spherical triangle formed by the pole and the two places, whose sides are the colatitudes \(90^\circ-\varphi_1\), \(90^\circ-\varphi_2\) and the arc \(\sigma\), with angle \(\Delta\lambda\) at the pole. Al-Biruni gave several methods for the qibla, some by projection onto the plane, some by successive applications of the sine theorem.

Exercise 8.1A lunar eclipse begins at 21:12 local time in Baghdad and at 22:48 local time at a city to the east. Find the longitude difference. Compare with Baghdad–Ghazna, \(24;20^\circ\).

Show solution
\(\Delta t=1\) h 36 min \(=1.6\) h, so \(\Delta\lambda=24^\circ\). A difference of \(24;20^\circ\) corresponds to 1 h 37 min 20 s.

Exercise 8.2Take Ghazna at \(\varphi_1=33.55^\circ\) N and Mecca at \(\varphi_2=21.42^\circ\) N, with Mecca \(28.59^\circ\) to the west. Compute the qibla direction at Ghazna.

Show solution
\(\sin\Delta\lambda=0.4785\); \(\cos\varphi_1\tan\varphi_2=0.8334\times0.3923=0.3269\); \(\sin\varphi_1\cos\Delta\lambda=0.5527\times0.8781=0.4853\). So \(\tan q=0.4785/(0.3269-0.4853)=0.4785/(-0.1584)\). The denominator is negative, so \(q\) lies between \(90^\circ\) and \(180^\circ\) from north: \(q\approx108.3^\circ\) towards the west, i.e. about \(18^\circ\) south of due west.

Exercise 8.3Using the same coordinates and \(R=6371\) km, find the great-circle distance from Ghazna to Mecca.

Show solution
\(\cos\sigma=0.5527\times0.3652+0.8334\times0.9309\times0.8781=0.2018+0.6812=0.8830\), so \(\sigma\approx27.99^\circ=0.4885\) rad and the distance is about \(3110\) km.
Lesson 09

The Masʿudic Canon: tables and astronomy

For the general reader. Al-Qanun al-Masʿudi, dedicated around 1030 to the sultan Masʿud, is al-Biruni’s astronomical summa, covering the ground of Ptolemy’s Almagest in eleven books. It is not a copy. It defines \(\pi\) as the ratio of circumference to diameter, builds trigonometric tables with second-order interpolation, shows that the Sun’s apogee moves independently of precession, and gives a star catalogue of 1029 stars.

Goals
  • Derive quadratic (second-difference) interpolation and see why linear interpolation fails for sine tables.
  • Understand the solar apogee and precession as two distinct motions.
  • Place al-Biruni’s discussion of the Earth’s rotation in its context.

Structure

Book III treats trigonometry, with tables of sines and tangents. Book IV covers spherical astronomy, Book V geodesy and mathematical geography with a table of coordinates of about 600 places, Books VI–VII the Sun and Moon, Book VIII eclipses, Book IX the fixed stars, Book X the planets, and Book XI astrology. In the star catalogue he added \(13^\circ\) to Ptolemy’s longitudes to account for precession.

Interpolation

Theorem (quadratic interpolation)

Let \(f\) be tabulated at \(x_{-1},x_0,x_1\) with step \(s\), and let \(\Delta_-=f_0-f_{-1}\), \(\Delta_+=f_1-f_0\). The unique quadratic through the three points gives, for \(x=x_0+ts\),

\[ f(x)\approx f_0+t\,\frac{\Delta_-+\Delta_+}{2}+\frac{t^2}{2}\,(\Delta_+-\Delta_-). \]

If \(f\in C^3\), the error is \(O(s^3)\), compared with \(O(s^2)\) for linear interpolation.

Proof. Write \(p(t)=f_0+bt+ct^2\). The conditions \(p(1)=f_1\) and \(p(-1)=f_{-1}\) give \(b+c=\Delta_+\) and \(-b+c=-\Delta_-\), so \(b=\tfrac12(\Delta_-+\Delta_+)\) and \(c=\tfrac12(\Delta_+-\Delta_-)\). The error bound is the standard Lagrange remainder. Brahmagupta had used a related second-difference rule; al-Biruni knew Brahmagupta’s Khandakhadyaka but developed his own method.

The solar apogee

Ptolemy held that the Sun’s apogee, the direction where the Sun is farthest from Earth, was fixed relative to the equinoxes. Astronomers of the ninth century had already found it in a different place. Al-Biruni concluded that the apogee has a motion of its own, distinct from precession. In modern terms, the apsidal line of the Earth’s orbit advances slowly relative to the stars because of planetary perturbations.

Does the Earth turn?

Al-Biruni knew that Aryabhata and, in his own world, the astronomer Abu Saʿid al-Sijzi had proposed that the Earth rotates daily. He noted that the hypothesis does not change any astronomical computation and judged that the question belonged to natural philosophy rather than mathematics. He seems to have leaned against rotation without claiming a proof; his treatise devoted to the question is lost.

Exercise 9.1Given \(\sin30^\circ=0.5\), \(\sin45^\circ=0.70711\), \(\sin60^\circ=0.86603\), estimate \(\sin50^\circ\) by linear interpolation between 45° and 60° and by the quadratic formula centred at 45°. Compare with \(0.76604\).

Show solution
\(t=1/3\), \(\Delta_-=0.20711\), \(\Delta_+=0.15892\). Linear: \(0.70711+\tfrac13(0.15892)=0.76008\), error \(0.0060\). Quadratic: \(0.70711+\tfrac13(0.18301)+\tfrac1{18}(-0.04819)=0.70711+0.06100-0.00268=0.76543\), error \(0.0006\). The second-order term removes about 90 per cent of the error even with this coarse 15° step.

Exercise 9.2Ptolemy’s catalogue is epoch 137; al-Biruni added \(13^\circ\). Assuming his epoch is about 1030, what precession rate does this imply? Compare with the modern value of about \(50.3''\) per year.

Show solution
Over about 893 years, \(13^\circ=46800''\), so the rate is about \(52.4''\) per year, or about one degree in 69 years. This is within 5 per cent of the modern rate. (Ptolemy’s own longitudes contain a systematic error of about one degree, which complicates the comparison.)
Lesson 10

Specific gravity and the mineral world

For the general reader. How can you tell gold from a cheaper alloy without destroying it? Archimedes’ answer was to compare its weight with the weight of the water it displaces. Al-Biruni built an instrument for doing this precisely, a conical flask with a spout, and measured the specific gravities of metals and precious stones. His results are preserved in his Kitab al-Jamahir on gems and metals and in the Book of the Balance of Wisdom by al-Khazini a century later.

Goals
  • State and prove Archimedes’ principle for a submerged body.
  • Compute specific gravities from displacement and from weighing in water.
  • Analyse a two-metal alloy by its specific gravity.

The instrument

The flask was filled with water to the level of a lateral spout. A sample dropped in displaced exactly its own volume of water, which ran out of the spout into a pan and was weighed. The ratio of the sample’s weight to the weight of displaced water is its specific gravity. Al-Khazini noted a practical difficulty: drops of water clinging inside the spout spoil the reading.

Theorem (Archimedes’ principle)

A body immersed in a fluid at rest experiences an upward force equal to the weight of the fluid it displaces. Consequently, if a body weighs \(W\) in air and \(W'\) when fully immersed in water, its specific gravity is \(\dfrac{W}{W-W'}\).

Proof. Replace the body by fluid of the same shape. That fluid is in equilibrium, so the pressure on its boundary supports exactly its weight. The pressure depends only on depth, not on what is inside, so the same net force acts on the body. The apparent loss of weight \(W-W'\) is therefore the weight of the displaced water, and \(W/(W-W')\) is the ratio of the body’s density to water’s.

Exercise 10.1A sample weighs 96.5 units in air and the water it displaces weighs 5.0 units. Find its specific gravity. Is it plausibly pure gold (modern value 19.3)?

Show solution
\(96.5/5.0=19.3\). Yes: this matches pure gold. Silver (10.5) or copper (8.96) would displace about twice as much water.

Exercise 10.2A crown is an alloy of gold (SG 19.3) and silver (SG 10.5), with measured SG 16.0. Assuming volumes add, find the fraction of gold by weight.

Show solution
For a unit weight with gold fraction \(w\), the volume is \(w/19.3+(1-w)/10.5=1/16.0\). Then \(w(1/19.3-1/10.5)=1/16-1/10.5\), i.e. \(w(-0.04342)=-0.03274\), so \(w\approx0.754\). About three-quarters of the weight is gold.
Part III · Life and legacy
Lesson 11

Ghazna and India, 1017–c. 1050

At Ghazna al-Biruni began again with almost nothing. He recorded observations near Kabul in 1018, improvising instruments, and in the following years he accompanied Mahmud’s campaigns into northwest India. He learned Sanskrit, worked with pandits, and read Indian astronomy, mathematics, philosophy and religion in the original. He translated works in both directions, including Patanjali’s Yoga Sutras into Arabic, and he intended to translate Brahmagupta’s Brahmasphutasiddhanta.

The book on India

His Tahqiq ma li-l-Hind, usually called India, was completed in 1030, shortly after Mahmud’s death. It describes Indian religion, philosophy, literature, geography, calendars, astronomy and astrology, with long accounts of Aryabhata, Brahmagupta and Varahamihira. Al-Biruni states in his preface that he will report what Indians say in their own terms and compare it with Greek ideas, without polemic. Modern scholars often call him the first Indologist. He is also frank: he criticizes Indian astronomers for mixing exact computation with myth, and he records the destruction caused by the Ghaznavid conquests.

The last works

Mahmud’s son Masʿud treated al-Biruni well, and the Masʿudic Canon repays him with a dedication. The Tafhim, an introduction to geometry, arithmetic, astronomy and astrology in question-and-answer form, appeared in Arabic and Persian around 1029. In 1036 he compiled a bibliography of the physician al-Razi and appended a list of his own 103 completed works. Under the sultan Mawdud he wrote the Kitab al-Jamahir on gems and metals. His last book, on pharmacology (Kitab al-Saydana), says he was over eighty and failing in sight and hearing, and that he worked with a helper. Uzbek scholars date his death to December 1048; the Saydana evidence suggests a date after 1050. He died at Ghazna.

Exercise 11.1Indian sine tables used a radius \(R=3438\). Explain this number, and compute the Indian Sine of \(3^\circ45'\), the first entry of Aryabhata’s table.

Show solution
Measuring the circumference in arcminutes gives \(21600\), and \(R=21600/(2\pi)\approx3437.75\approx3438\), so that a small arc and its sine are equal in the same unit. Then \(3438\sin3.75^\circ\approx3438\times0.065403\approx224.9\), tabulated as 225: equal to the arc \(225'\) itself, as expected for a small angle.
Lesson 12

Legacy

Of roughly 146 works, about 22 survive. The losses include his book on the Earth’s rotation, his book on root extraction and most of his astronomical instruments treatises. The survivors were read mainly in the eastern Islamic world. His student tradition is thin, but al-Khazini preserved his specific gravities in the twelfth century, and parts of the Canon passed into later astronomy: an irregularity in its Mercury tables reappears in the Chinese Huihui li of 1384.

Modern recognition began with the German orientalist Eduard Sachau, who edited and translated the Chronology (1879) and India (1888). In the twentieth century, Edward S. Kennedy and his students translated the Shadows (1976) and analysed the Tahdid and the Canon. The millennium of his birth in 1973 was marked by conferences and a Soviet stamp; a town in Karakalpakstan is named Beruniy, and the lunar crater al-Biruni and the asteroid 9936 al-Biruni carry his name.

Assessment

  • Trigonometry: systematic use of sines, tangents and the sine theorems in astronomy; the broken chord as a source of identities; numerical solution of the trisection cubic.
  • Geodesy: the single-station dip method for the Earth’s radius; eclipse longitudes and spherical-trigonometric coordinates.
  • Astronomy: second-order interpolation in tables, the independent motion of the solar apogee, a new star catalogue.
  • Method: repeated measurement, attention to instrumental error, and comparison of sources across languages and cultures.

Exercise 12.1Al-Biruni lists 103 completed works by 1036, and modern estimates give 146 in all. If about 22 survive, what fraction is lost? Discuss what such losses mean for judging a medieval scientist’s originality.

Show solution
\(1-22/146\approx0.85\): about 85 per cent is lost. Claims of priority, for or against, rest on a small and possibly unrepresentative sample; a result absent from the survivors may still have been his, and a result present may have come from a lost source he used.

Final projectChoose one measurement from Part II (the Earth’s radius, a longitude, a qibla or a specific gravity). Write a two-page account for a strong undergraduate that describes how al-Biruni made it, recomputes it with modern data, and identifies the dominant source of error.

Reference

Connections

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

Teachers and advisors

  • Abu Nasr Mansur ibn 'Iraq

Influenced by

Influenced

  • al-Khazini

Colleagues and collaborators

  • Abu Mahmud al-Khujandi
  • Abu Sa'id al-Sijzi
  • Abu'l-Wafa' al-Buzjani 997

Rivals

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Abu Arrayhan Muhammad ibn Ahmad al-Biruni, MacTutor History of Mathematics.
  2. Al-Biruni, Wikipedia.
  3. D. J. Boilot et al., Biruni, Abu Rayhan: Life, and Mathematics and astronomy, Encyclopaedia Iranica.
  4. al-Biruni, Biographical Encyclopedia of Astronomers (via MacTutor).
  5. E. S. Kennedy, Al-Biruni, in Dictionary of Scientific Biography, vol. 2, Scribner, 1970.
  6. S. F. Starr, Lost Enlightenment: Central Asia’s Golden Age from the Arab Conquest to Tamerlane, Princeton, 2013.

Works and mathematics

  1. al-Biruni, The Chronology of Ancient Nations, trans. E. Sachau, London, 1879.
  2. al-Biruni, Alberuni’s India, trans. E. Sachau, London, 1888.
  3. al-Biruni, The Exhaustive Treatise on Shadows, trans. and comm. E. S. Kennedy, 2 vols., Aleppo, 1976.
  4. al-Biruni, The Determination of the Coordinates of Positions for the Correction of Distances between Cities (Tahdid), trans. J. Ali, Beirut, 1967.
  5. J. L. Berggren, Episodes in the Mathematics of Medieval Islam, 2nd ed., Springer, 2016.
  6. G. Van Brummelen, The Mathematics of the Heavens and the Earth: The Early History of Trigonometry, Princeton, 2009.
  7. D. Joyce, al-Biruni’s measurement of the Earth (worked solution), Clark University.
  8. Biruni’s method, Etymological Dictionary of Astronomy and Astrophysics, Paris Observatory.
  9. Al-Biruni’s pyknometer, Chemistry World.