courseshub.world · Lives in Mathematics and Physics

Christiaan Huygens

Christiaan Huygens, Heer van Zeelhem

A course in two strands: the life of the Dutch patrician who made the first accurate clock and saw the true shape of Saturn, and the mechanics, probability and wave optics he built with nothing but Greek geometry.

Born 14 April 1629, The Hague Died 8 July 1695, The Hague Pendulum clock patented 1657 Level advanced undergraduate
The cycloidal pendulum of Horologium Oscillatorium (1673). The cord wraps against two cycloidal cheeks; the bob then moves on a congruent cycloid, its involute, and every swing takes the same time.
Lesson 00

How to use this course

Christiaan Huygens stands between Galileo Galilei and Isaac Newton. He found the first moon of Saturn and explained its ring, built the first pendulum clock and gave its complete mathematical theory, wrote the first printed book on probability, found the correct laws of elastic collision, derived the formula for centrifugal force, and proposed the wave theory of light that still carries his name. He did all of this in the language of Euclid and Archimedes, without the calculus that Newton and Gottfried Wilhelm Leibniz were inventing around him.

Prerequisites for Part II
  • Single-variable calculus and simple differential equations, in particular simple harmonic motion.
  • Newtonian mechanics at first-year level: momentum, kinetic energy, moment of inertia.
  • Parametric curves, arc length and curvature, for Lesson 08.
  • Elementary probability (finite sample spaces), for Lesson 05.
  • Reflection and refraction at a plane interface, for Lesson 10.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Huygens argued with similar triangles, proportions and limiting arguments in the style of Archimedes. The course states his results in modern notation and says where his own route differed.

Dates. The Dutch provinces of Holland and Zeeland used the Gregorian calendar in Huygens’s lifetime, so dates in this course are New Style unless marked otherwise. English dates of the same period, as in the Hooke and Newton modules, are usually Old Style, ten days behind.

Reference

Timeline

1629Born 14 April in The Hague, second son of the diplomat and poet Constantijn Huygens and Suzanna van Baerle.
1637His mother dies shortly after the birth of his sister Suzanna.
1644Tutored in mathematics by Jan Jansz Stampioen. René Descartes, a friend of his father, follows his progress.
1645May: enters Leiden University to study law and mathematics; Frans van Schooten becomes his mathematics tutor.
1646–47Correspondence with Marin Mersenne, who sets him problems, among them the hanging chain. March 1647: moves to the Orange College in Breda.
1649Leaves Breda in August; joins a diplomatic mission to Denmark with Henry, Duke of Nassau.
1651Theoremata de quadratura hyperboles, ellipsis et circuli, refuting Grégoire de Saint-Vincent’s quadrature of the circle.
1652–56Works out the laws of elastic collision (published in full only in 1703).
1654De circuli magnitudine inventa: sharp bounds for π.
1655With his brother Constantijn grinds lenses and builds telescopes. 25 March: discovers Titan. First visit to Paris.
1656Explains the ring of Saturn, first in an anagram. December: begins work on a pendulum-regulated clock.
1657Patents the pendulum clock; Salomon Coster builds the first examples. De ratiociniis in ludo aleae appears in van Schooten’s Exercitationes mathematicae.
1658Short tract Horologium describes the clock.
1659Systema Saturnium. Derives the law of centrifugal force and discovers the isochronism of the cycloid.
1660–61Visits Paris, then London, where he meets John Wallis and the founders of the Royal Society.
1663Elected a Fellow of the Royal Society of London.
1666Moves to Paris at the invitation of Jean-Baptiste Colbert as a founding member of the Académie royale des sciences.
1668–69Sends his collision rules to the Royal Society; publishes them in the Journal des sçavans (1669).
1670Serious illness; recovers in Holland and returns to Paris in 1671.
1672Meets Leibniz in Paris and becomes his mathematical mentor.
1673Horologium oscillatorium, dedicated to Louis XIV.
1675Spiral balance spring for watches; priority dispute with Robert Hooke.
1676–78In The Hague after another illness; studies double refraction in Iceland spar. Presents his theory of light to the Académie in 1678.
1681Leaves Paris for the last time.
1687–88His father dies; Huygens makes the country house Hofwijck at Voorburg his home.
1689Third visit to England; meets Newton on 12 June.
1690Traité de la lumière and Discours de la cause de la pesanteur, Leiden.
1691Solution of the catenary problem, with Leibniz and Johann Bernoulli.
1695Dies 8 July in The Hague, aged 66; buried in the Grote Kerk.
1698Cosmotheoros, on life on other planets, published posthumously.
1703Opuscula postuma, including De motu corporum ex percussione and De vi centrifuga.
Part I · Formation
Lesson 01

The Hague, 1629–1645

Christiaan Huygens was born in The Hague on 14 April 1629, the second son of Constantijn Huygens and Suzanna van Baerle. His father was secretary to successive Princes of Orange, a poet in Dutch and Latin, a composer and a friend of the leading minds of Europe. He corresponded with Marin Mersenne and knew René Descartes well. The family was rich, Calvinist and well connected. There were five children: Constantijn the younger (1628), Christiaan, Lodewijk (1631), Philips (1632) and Suzanna (1637). Their mother died in 1637, soon after Suzanna’s birth.

The boys were taught at home until the age of sixteen. The programme was broad and deliberate: Latin, Greek, French and Italian, music, drawing, logic, fencing and riding, and mathematics. Christiaan built mechanical models as a child and showed early skill in geometry. In 1644 his father engaged the mathematician Jan Jansz Stampioen as a tutor. Descartes, who visited the family, took an interest in the boy’s progress, and his Principia philosophiae and Géométrie shaped Huygens’s youth. Later in life Huygens rejected much of Descartes’s physics, but he kept the Cartesian demand that every effect have a mechanical cause.

Lesson 02

Leiden and Breda, 1645–1655

In May 1645 Huygens entered Leiden University to study law and mathematics. On Descartes’s advice his mathematics tutor was now Frans van Schooten, the editor of the Latin edition of the Géométrie. Van Schooten taught him Descartes’s analytic geometry and the methods of François Viète and Pierre de Fermat, and remained his mathematical correspondent for years. In March 1647 Huygens moved to the newly founded Orange College at Breda, where his father was a curator. He left in August 1649 and joined a diplomatic mission of Henry, Duke of Nassau, to Denmark. He hoped to travel on to Stockholm to see Descartes, but the weather prevented it, and Descartes died there in February 1650.

In these years Mersenne began to set him problems by letter. Huygens showed that a chain hanging freely is not a parabola, as Galileo had suggested, and found how a weightless cord must be loaded to hang in a parabola. Mersenne called him a new Archimedes.

Huygens never practised law. His first publication, Theoremata de quadratura hyperboles, ellipsis et circuli (1651), exposed an error in the quadrature of the circle claimed by the Jesuit Grégoire de Saint-Vincent. De circuli magnitudine inventa (1654) gave new and sharper bounds for the ratio of circumference to diameter (Lesson 04). Between 1652 and 1656 he also worked out the rules of elastic collision, discovering that Descartes’s rules were wrong (Lesson 06), but he kept them in manuscript.

Around 1654 Christiaan and his brother Constantijn began to grind and polish their own telescope lenses, and Christiaan devised improved methods for doing so. The craft would give him his first fame.

Lesson 03

Lenses, Saturn and the clock, 1655–1666

On 25 March 1655, with a telescope he and Constantijn had built, Huygens saw a faint star near Saturn that followed the planet. It was a moon, now called Titan, the first satellite found since Galileo’s four moons of Jupiter. He announced it in a short tract, De Saturni luna observatio nova, and hid a second claim in a Latin anagram. When he unscrambled it in Systema Saturnium (1659), it said that Saturn is girdled by a thin flat ring, nowhere touching the planet and inclined to the ecliptic. Galileo had seen Saturn with “ears” that came and went. Huygens explained the changes geometrically: because the ring is tilted, its appearance from Earth changes with the planet’s position in its 29½-year orbit, and twice in each orbit we see it edge on, when it nearly vanishes. Systema Saturnium also described an eyepiece micrometer for measuring small angles.

In 1655 Huygens made his first visit to Paris, where he heard of the correspondence between Blaise Pascal and Fermat on games of chance. Back in Holland he wrote De ratiociniis in ludo aleae, which van Schooten published in 1657 (Lesson 05).

The pendulum clock

Astronomy needed accurate time. Galileo had noticed that a pendulum’s swing takes nearly the same time whatever its amplitude, but nobody had used a pendulum to govern a working clock. Huygens began the project in December 1656. His design let a freely swinging pendulum regulate the verge escapement of a weight-driven clock, and he patented it in the Dutch Republic in 1657. The Hague clockmaker Salomon Coster built the first examples; the oldest surviving one, dated 1657, is in the Museum Boerhaave in Leiden. Good clocks of the day might drift by a quarter of an hour a day; Huygens’s by seconds. He described the design in the short tract Horologium (1658).

A circular pendulum is only approximately isochronous. In 1659 Huygens discovered the curve on which the period does not depend on amplitude at all, the cycloid, and how to make a pendulum bob follow it (Lesson 08). In the same year he derived the law of centrifugal force (Lesson 07). Both results went into the book of 1673.

Longitude and London

A clock that kept time at sea would solve the problem of longitude, and Huygens hoped his would. Sea trials began in the early 1660s with mixed results: the motion of a ship disturbs a pendulum. In 1661 he visited London, met John Wallis and Robert Boyle, saw Boyle’s air pump, and watched the Royal Society take shape. In 1663 he was elected a Fellow. By 1666 he was the most celebrated natural philosopher on the Continent, and Jean-Baptiste Colbert invited him to Paris to help found the new Académie royale des sciences.

Part II · The physics
Lesson 04

Geometry before calculus: π and the hanging chain

For the general reader. Archimedes trapped π between the perimeters of polygons drawn inside and outside a circle. Huygens, at 25, found how to squeeze far more accuracy out of the same polygons by combining them cleverly. Forty years later he helped solve one of the first problems of the new calculus, the shape of a hanging chain, still using geometric methods.

Goals
  • State and prove Huygens’s lower bound for a circular arc.
  • Use it to bound π from a 12-gon better than Archimedes did from a 96-gon.
  • Distinguish the parabola of a loaded cord from the catenary of a free chain.

Chords and arcs

For an arc of a circle let \(c\) be its chord and \(c'\) the chord of half the arc. Archimedes’s inequality \(2c'<s\) has an error of third order in the angle. Huygens proved in De circuli magnitudine inventa a bound that is accurate to fifth order.

Theorem (Huygens, 1654)

For a circular arc of length \(s\) less than a semicircle, with chord \(c\) and half-arc chord \(c'\),

\[ s > 2c' + \tfrac13\,(2c'-c) = \tfrac13(8c'-c). \]

Equivalently, if \(p_n\) and \(p_{2n}\) are the perimeters of regular \(n\)- and \(2n\)-gons inscribed in a circle of circumference \(L\), then \(L>p_{2n}+\tfrac13(p_{2n}-p_n)\).

Modern check. On the unit circle take the arc subtending angle \(2\alpha\): \(s=2\alpha\), \(c=2\sin\alpha\), \(c'=2\sin(\alpha/2)\). Taylor expansion gives

\[ \tfrac13(8c'-c)=\tfrac13\big(16\sin\tfrac\alpha2-2\sin\alpha\big)=2\alpha-\frac{\alpha^5}{240}+O(\alpha^7), \]

so the error is of fifth order and positive. Huygens’s own proof compares areas of inscribed triangles and segments in the manner of Archimedes. The practical meaning is that one step of polygon doubling, plus a one-third correction, does the work of several further doublings. This is an early instance of what is now called Richardson extrapolation: \(p_n=L-a/n^2+O(n^{-4})\), so \(p_{2n}+\tfrac13(p_{2n}-p_n)\) cancels the \(n^{-2}\) term.

The hanging chain

Galileo had suggested that a hanging chain forms a parabola. As a student Huygens showed it does not. A cord carries a parabola when the load is uniform per unit horizontal length, as for the deck of a suspension bridge. A free chain carries a load uniform per unit arc length. In 1690 the Basel mathematician Jacob Bernoulli posed the problem of the chain’s exact shape; in June 1691 the Acta Eruditorum printed solutions by Huygens, Leibniz and Johann Bernoulli. Huygens, who had named the curve catenaria in a letter to Leibniz, found its properties by geometric arguments; in modern form it is \(y=a\cosh(x/a)\).

Proposition (equilibrium of a cord)

Let a flexible cord hang in the plane with horizontal tension component \(H\), constant along the cord. If the vertical load on the part between \(0\) and \(x\) is \(W(x)\), then \(H\,y'(x)=W(x)+\text{const}\). For \(W=wx\) the cord is a parabola; for \(W=\lambda g\,s(x)\), with \(s\) the arc length, it is the catenary \(y=a\cosh(x/a)+b\), \(a=H/(\lambda g)\).

Exercise 4.1Start from the inscribed hexagon and 12-gon in a circle of diameter 1. Use Huygens’s bound to find a lower bound for π, and compare with the inscribed 12-gon alone.

Show solution
With diameter 1, \(p_6=3\) and \(p_{12}=12\sin15^\circ\approx3.105829\). Then \(p_{12}+\tfrac13(p_{12}-p_6)\approx3.105829+0.035276=3.141105\). So \(\pi>3.1411\), whereas the 12-gon alone gives only \(\pi>3.1058\). Archimedes needed a 96-gon to reach \(3\tfrac{10}{71}\approx3.1408\).

Exercise 4.2Show that a cord whose load is uniform per unit horizontal length \(w\) hangs in a parabola, and find the sag \(d\) at the centre for span \(2\ell\) and horizontal tension \(H\).

Show solution
Measure \(x\) from the lowest point. The load between 0 and \(x\) is \(wx\), and the slope there is zero, so \(Hy'=wx\) and \(y=\frac{w}{2H}x^2\). The sag is \(d=y(\ell)=\frac{w\ell^2}{2H}\).

Exercise 4.3Verify that \(y=a\cosh(x/a)\) satisfies \(a\,y'=s(x)\), where \(s\) is arc length from the lowest point, and compare it with the parabola \(y=a+x^2/(2a)\) near \(x=0\).

Show solution
\(y'=\sinh(x/a)\) and \(s=\int_0^x\sqrt{1+\sinh^2(t/a)}\,dt=a\sinh(x/a)\), so \(a y'=s\). Also \(a\cosh(x/a)=a+\frac{x^2}{2a}+\frac{x^4}{24a^3}+\cdots\): the two curves agree to second order, which is why the error is easy to make and hard to see near the bottom.
Lesson 05

Expectation and games of chance

For the general reader. What is a fair price for a gamble? Pascal and Fermat had discussed the question in private letters in 1654. Huygens turned their ideas into a short book, the first printed treatise on probability, built on one principle: a chance of winning something is worth exactly the stake that would buy you into a fair game with the same prospects. Today we call that worth the expected value.

Goals
  • State Huygens’s axiom and derive the formula for expectation.
  • Solve the problem of points by backward induction.
  • Derive the gambler’s ruin formula and solve Huygens’s fifth problem.

The axiom

De ratiociniis in ludo aleae (1657) has fourteen propositions and ends with five problems for the reader. Huygens does not define probability. He defines the value of a chance by a fairness argument: if I have equal chances of getting \(a\) or \(b\), my chance is worth \((a+b)/2\), because with that stake I could set up a fair game with an opponent that pays me \(a\) or \(b\) with equal chance.

Propositions I–III (Huygens, 1657)

If I have \(p\) chances of obtaining \(a\) and \(q\) chances of obtaining \(b\), all chances equally easy, my chance is worth

\[ \frac{p\,a+q\,b}{p+q}. \]

More generally, with equally likely outcomes paying \(a_1,\dots,a_n\), the value is their mean.

In modern terms, the value of a gamble with outcomes \(a_i\) of probability \(\pi_i\) is \(E=\sum_i\pi_ia_i\). Huygens’s word for it was expectatio in van Schooten’s Latin, the origin of the English term.

The problem of points

Two players stake equal sums on a game of several rounds, each round won by either player with probability \(\tfrac12\). The game stops early when one needs \(m\) more rounds and the other \(n\). How should the stake be divided? Huygens solves it by working backwards from positions whose value is known, the method of Pascal. Let \(e(m,n)\) be the first player’s share of a total stake of 1. Then

\[ e(m,n)=\tfrac12\,e(m-1,n)+\tfrac12\,e(m,n-1),\qquad e(0,n)=1,\quad e(m,0)=0. \]

The gambler’s ruin

Huygens’s fifth problem came from Pascal by way of Pierre de Carcavi. Two players A and B throw three dice. When 11 comes up A gains a point from B; when 14 comes up B gains a point from A. Each starts with 12 points, and the first to lose all of them is ruined. What are their chances?

Theorem (gambler’s ruin)

Two players exchange single points; each exchange goes to A with probability \(p\) and to B with probability \(q=1-p\), independently, \(p\ne q\). If A starts with \(a\) points and B with \(b\), the probability that A wins everything is

\[ P_A=\frac{1-r^{a}}{1-r^{a+b}},\qquad r=\frac qp. \]

In particular, for \(a=b=n\), \(P_A:P_B=p^n:q^n\).

Derivation. Let \(u_k\) be A’s chance of winning when A holds \(k\) points. Conditioning on the next exchange, \(u_k=p\,u_{k+1}+q\,u_{k-1}\) with \(u_0=0\), \(u_{a+b}=1\). The characteristic roots are \(1\) and \(r\), so \(u_k=\alpha+\beta r^k\), and the boundary conditions give the formula. For \(a=b=n\), \(P_A=\frac{1-r^n}{(1-r^n)(1+r^n)}=\frac{p^n}{p^n+q^n}\).

Exercise 5.1Solve the fifth problem: with three dice, count the ways to throw 11 and 14, and find the odds that A ruins B.

Show solution
Of the 216 outcomes, 27 give a total of 11 and 15 give 14. Throws that give neither are ignored, so each exchange goes to A with probability \(27/42=9/14\) and to B with probability \(5/14\). With 12 points each, \(P_A:P_B=9^{12}:5^{12}=282\,429\,536\,481:244\,140\,625\), about 1157 to 1. A’s small advantage per throw becomes overwhelming over a long game.

Exercise 5.2A needs one more round and B needs three, with fair rounds. Find A’s share of the stake.

Show solution
B wins only by taking the next three rounds, with probability \(1/8\). So \(e(1,3)=7/8\). The recursion gives the same: \(e(1,1)=\tfrac12\), \(e(1,2)=\tfrac12+\tfrac12\cdot\tfrac12=\tfrac34\), \(e(1,3)=\tfrac12+\tfrac12\cdot\tfrac34=\tfrac78\).

Exercise 5.3How many throws of two dice are needed for an even chance of at least one double six? (Huygens’s Propositions X and XI treat this kind of question.)

Show solution
The chance of no double six in \(n\) throws is \((35/36)^n\). We need \((35/36)^n\le\tfrac12\), so \(n\ge\ln2/\ln(36/35)\approx24.6\). With 24 throws the chance is about 0.491; with 25, about 0.506. The answer is 25.
Lesson 06

The laws of collision

For the general reader. Descartes had given seven rules for what happens when two bodies collide, and most were wrong. Huygens replaced them with correct rules for hard, elastic bodies. His key move was to imagine the collision watched by a man standing on a boat drifting along a canal: the laws must look the same to him as to a man on the bank. This is the principle of relativity, used as a tool for the first time.

Goals
  • State Huygens’s hypotheses and his rules for elastic collision.
  • Derive the general rule from the symmetric case by a change of frame.
  • Identify the conserved quantities: momentum and \(\sum mv^2\).

Hypotheses

Huygens worked on collisions from 1652, sent his results to the Royal Society in 1668 and printed a summary in the Journal des sçavans in 1669; Wallis and Christopher Wren sent the Society results at the same time. The full treatise, De motu corporum ex percussione, appeared only in 1703. Its main hypotheses are:

  1. A body in motion continues in a straight line with constant speed unless something prevents it (inertia, from Galileo and Descartes).
  2. Two equal hard bodies meeting head-on with equal and opposite speeds rebound with the same speeds reversed (symmetry).
  3. The motion of bodies and their relative motions are to be judged relative to a frame that may itself be moving uniformly; the rules hold equally in any such frame (relativity: the boat).
  4. A larger body striking a smaller one at rest gives it some motion and loses some of its own.
Theorem (Huygens’s rules, 1669)

For a head-on collision of two hard (perfectly elastic) bodies of masses \(m_1,m_2\) with velocities \(u_1,u_2\) before and \(v_1,v_2\) after:

  1. the relative velocity is reversed: \(v_2-v_1=-(u_2-u_1)\);
  2. the centre of gravity moves uniformly, unaffected by the collision; equivalently, \(m_1u_1+m_2u_2=m_1v_1+m_2v_2\) with signs;
  3. the sum of each mass times the square of its speed is the same before and after: \(m_1u_1^2+m_2u_2^2=m_1v_1^2+m_2v_2^2\).

Derivation in Huygens’s style

For equal masses, the symmetric hypothesis says that \((+w,-w)\) becomes \((-w,+w)\). Now watch from a boat moving with velocity \(-V\): the velocities \(V+w\) and \(V-w\) become \(V-w\) and \(V+w\). Any pair \((u_1,u_2)\) has this form with \(V=(u_1+u_2)/2\), so equal bodies exchange velocities. For unequal masses Huygens went to the frame of the centre of gravity, \(V=(m_1u_1+m_2u_2)/(m_1+m_2)\), in which the velocities \(w_1,w_2\) satisfy \(m_1w_1+m_2w_2=0\). He argued that in this frame each body simply reverses its velocity: otherwise the common centre of gravity could be raised by the collision, which contradicts the principle that heavy bodies cannot raise their own centre of gravity by their own motions (Lesson 09). Transforming back gives

\[ v_1=2V-u_1,\qquad v_2=2V-u_2,\qquad V=\frac{m_1u_1+m_2u_2}{m_1+m_2}. \]

Rules 1–3 follow at once. Descartes had conserved the scalar sum \(\sum m|u|\), which fails. Huygens saw that the conserved “quantity of motion” must take direction into account, and that \(\sum mv^2\) is also conserved in elastic impact. Leibniz later made \(mv^2\), his vis viva, a central quantity of dynamics, partly on Huygens’s example.

Exercise 6.1A body of mass 3 moving at speed 4 hits a body of mass 1 at rest, head-on and elastically. Find the velocities afterwards and check rules 1–3.

Show solution
\(V=(3\cdot4+0)/4=3\), so \(v_1=6-4=2\) and \(v_2=6-0=6\). Relative velocity: before \(0-4=-4\), after \(6-2=4\). Momentum: \(12=6+6\). \(\sum mv^2\): \(48=12+36\).

Exercise 6.2Descartes’s fourth rule says that a body at rest cannot be moved by a smaller body, whatever its speed; the smaller body rebounds. Use the boat argument to show this contradicts the principle of relativity.

Show solution
Take \(m_1<m_2\), \(u_1=u>0\), \(u_2=0\); Descartes says \(v_1=-u\), \(v_2=0\). View the same event from a boat moving with velocity \(+u\): body 1 is at rest before, and body 2 approaches at \(-u\). After, body 1 has \(-2u\) and body 2 has \(-u\). So in this frame the larger moving body passes through its collision unchanged while launching the smaller one at twice its own speed, and \(\sum m|v|\) rises from \(m_2u\) to \(2m_1u+m_2u\). Descartes’s rules do not give this result when applied to the second description, so they are not frame-independent.

Exercise 6.3Show that rules 1 and 2 together imply rule 3.

Show solution
Rule 2: \(m_1(u_1-v_1)=m_2(v_2-u_2)\). Rule 1: \(u_1+v_1=u_2+v_2\). Multiply: \(m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2)\), which is rule 3.
Lesson 07

Centrifugal force

For the general reader. Whirl a stone on a string and you feel it pull outward. In 1659 Huygens worked out exactly how strong that pull is: it grows with the square of the speed and falls with the radius of the circle. Newton later used the same formula, read inwards as a centripetal force, to connect the motion of the planets with the inverse-square law of gravity.

Goals
  • Derive \(a=v^2/r\) by Huygens’s comparison with free fall.
  • Prove the period of a conical pendulum depends only on its height.
  • Use the result to estimate the flattening of the Earth and to connect Kepler’s third law with an inverse-square force.

The derivation

Huygens imagined a body tied to a wheel turning uniformly. Released, it would fly off along the tangent. Seen by an observer turning with the wheel, it recedes from the circle as a falling body recedes from its starting point: in time \(t\) a body moving along the tangent at speed \(v\) is at distance \(\sqrt{r^2+v^2t^2}-r\approx v^2t^2/(2r)\) from the circle. This distance grows as the square of the time, the signature of a constant acceleration, here \(v^2/r\). The cord must therefore pull the body with a force that bears the same ratio to its weight as \(v^2/r\) to \(g\).

Theorem (Huygens, 1659; published 1673 without proof, 1703 with proof)

A body of weight \(mg\) moving uniformly with speed \(v\) on a circle of radius \(r\) exerts on its constraint a centrifugal force \(F=mv^2/r=m\omega^2r=4\pi^2mr/T^2\). In particular, at equal speeds \(F\propto1/r\); on equal circles \(F\propto v^2\); at equal periods \(F\propto r\).

Huygens stated thirteen propositions on circular motion at the end of Horologium oscillatorium without proofs. Newton studied them closely.

The conical pendulum

A bob on a cord of length \(l\) moving on a horizontal circle of radius \(r\), with the cord at angle \(\theta\) to the vertical, is balanced when the horizontal and vertical components of the tension satisfy \(T\sin\theta=m\omega^2r\) and \(T\cos\theta=mg\). Then \(\omega^2=g/(l\cos\theta)=g/h\), where \(h\) is the height of the cone.

Theorem (Huygens)

All conical pendulums of the same height \(h\) have the same period \(2\pi\sqrt{h/g}\), whatever the radius of the circle.

Huygens used this to design a clock governed by a conical pendulum, which runs smoothly rather than with ticks, and he showed how a bob kept on a paraboloid of revolution would be exactly isochronous for every radius.

Exercise 7.1Compute the ratio \(q=\omega^2R/g\) of centrifugal to gravitational acceleration at the equator, with \(\omega=7.29\times10^{-5}\,\mathrm{s^{-1}}\), \(R=6.378\times10^6\,\mathrm m\), \(g=9.78\,\mathrm{m\,s^{-2}}\). How fast would the Earth have to spin for objects at the equator to float?

Show solution
\(q=(7.29\times10^{-5})^2\cdot6.378\times10^6/9.78\approx3.47\times10^{-3}\approx1/289\). Objects float when \(\omega^2R=g\), i.e. when \(\omega\) is \(\sqrt{289}=17\) times larger: a day of about 1.4 hours.

Exercise 7.2In the Discours de la cause de la pesanteur (1690) Huygens assumed gravity directed to the Earth’s centre with constant strength, and required the sea surface to be perpendicular to the combined pull of gravity and centrifugal force. Show this gives an oblate surface with flattening \(f\approx q/2\). Huygens obtained about 1/578; Newton, with mutually attracting matter, obtained 1/230.

Show solution
The surface is an equipotential of \(\Phi=g r-\tfrac12\omega^2 r^2\sin^2\vartheta\), with \(\vartheta\) the colatitude and gravity of constant strength \(g\) toward the centre. Setting \(\Phi(R_{\rm eq},\pi/2)=\Phi(R_{\rm pole},0)\): \(gR_{\rm eq}-\tfrac12\omega^2R^2=gR_{\rm pole}\), so \(f=(R_{\rm eq}-R_{\rm pole})/R\approx\omega^2R/(2g)=q/2\approx1/578\). (A point-mass inverse-square field gives the same first-order answer.) Newton’s self-gravitating homogeneous fluid gives \(5q/4\approx1/230\); the measured value is about 1/298.

Exercise 7.3Combine \(F=4\pi^2mr/T^2\) with Kepler’s third law \(T^2\propto r^3\) for circular orbits. What law of force follows? (This was the step taken in the 1680s by Hooke, Wren and Edmond Halley.)

Show solution
\(F\propto r/T^2\propto r/r^3=1/r^2\). A central force inversely proportional to the square of the distance.
Lesson 08

The cycloid, the tautochrone and evolutes

For the general reader. A pendulum swinging on a circle keeps almost, but not quite, the same time whatever the size of its swing. Huygens found the curve for which the time is exactly the same: the cycloid, traced by a point on the rim of a rolling wheel. He then invented a theory of curves, evolutes and involutes, to show how to make a hanging bob follow a cycloid, by letting its cord wrap against two cycloidal plates.

Goals
  • Prove that the inverted cycloid is a tautochrone.
  • Define evolute and involute and show the evolute of a cycloid is a congruent cycloid.
  • Compute the amplitude error of a circular pendulum.

The tautochrone

Take the cycloid generated by a circle of radius \(a\), inverted so that its lowest point is at the origin:

\[ x=a(\theta+\sin\theta),\qquad y=a(1-\cos\theta),\qquad -\pi\le\theta\le\pi. \]

Its arc length from the bottom is \(s=\int_0^\theta 2a\cos(t/2)\,dt=4a\sin(\theta/2)\), and the height is \(y=2a\sin^2(\theta/2)=s^2/(8a)\).

Theorem (Huygens, 1659; Horologium oscillatorium, Part II)

A heavy particle sliding without friction on an inverted cycloid reaches the lowest point in time \(\pi\sqrt{a/g}\), whatever its starting point. The motion is exactly simple harmonic in arc length with period \(2\pi\sqrt{4a/g}\).

Proof. Along the curve, \(\ddot s=-g\,dy/ds=-\frac{g}{4a}\,s\). This is simple harmonic motion with \(\omega^2=g/(4a)\), and every solution reaches \(s=0\) a quarter period after it starts from rest: \(\tfrac\pi2\sqrt{4a/g}=\pi\sqrt{a/g}\). Huygens proved it with a long sequence of geometric lemmas comparing descent along the cycloid with descent along a circle and with uniform motion, in the manner of Galileo’s Two New Sciences.

Evolutes and involutes

Unwind a taut thread from a curve \(C\): its end traces an involute of \(C\), and \(C\) is the evolute of the traced curve. Huygens proved the basic facts in Part III of Horologium oscillatorium:

Theorem (Huygens, 1673)
  1. The free part of the thread is normal to the involute, and its length is the radius of curvature of the involute at that point. The evolute is the locus of centres of curvature.
  2. The evolute of a cycloid is a congruent cycloid, shifted by half a period along the base and by the diameter \(2a\) of the generating circle vertically.

So a pendulum of length \(l=4a\) hung from the cusp between two cycloidal “cheeks”, each half an arch of the cycloid generated by radius \(a\), wraps against them and its bob moves exactly on the cycloid of the theorem. Its period is \(2\pi\sqrt{l/g}\) for every amplitude. Part III also contains the rectification of curves: since the free thread has the length of the arc it has unwound, the arc length of the evolute is known. This gave Huygens the arc lengths of the semicubical parabola and of the cycloid, \(8a\) per arch.

In practice the cheeks gave less advantage than Huygens hoped, because friction and the flexing of the cord introduced errors of their own. Clockmakers kept small amplitudes instead. Huygens’s theory of evolutes outlived the clock: it is the theory of curvature in the plane.

Exercise 8.1For a circular pendulum of length \(l\) and amplitude \(\theta_0\), the exact period is \(T=4\sqrt{l/g}\,K(\sin(\theta_0/2))\), where \(K\) is the complete elliptic integral of the first kind. Using \(K(k)=\tfrac\pi2(1+k^2/4+\cdots)\), show \(T\approx2\pi\sqrt{l/g}\,(1+\theta_0^2/16)\), and estimate the daily error of a clock whose amplitude changes from \(4^\circ\) to \(5^\circ\).

Show solution
\(k^2=\sin^2(\theta_0/2)\approx\theta_0^2/4\), so \(T\approx2\pi\sqrt{l/g}(1+\theta_0^2/16)\). The fractional change is \((\theta_1^2-\theta_0^2)/16\) with angles in radians: \((0.0873^2-0.0698^2)/16\approx1.71\times10^{-4}\). Over 86 400 s that is about 15 s a day, the size of error Huygens wanted to eliminate.

Exercise 8.2Show that the cheek curve \(x=a(t-\sin t),\ y=-a(1-\cos t)\), \(-\pi\le t\le\pi\) (cusp at the origin, \(y\) up), has as involute, for a thread of length \(4a\) fixed at the cusp, the cycloid \(x=a(t+\sin t),\ y=-a(3+\cos t)\).

Show solution
The arc length of the cheek from the cusp to parameter \(t>0\) is \(\int_0^t2a\sin(u/2)\,du=4a(1-\cos(t/2))\). The free thread has length \(4a\cos(t/2)\) along the unit tangent \((\sin(t/2),-\cos(t/2))\) at parameter \(t\) (the tangent direction is \((1-\cos t,-\sin t)\) normalized). Its end is at \(x=a(t-\sin t)+4a\cos\tfrac t2\sin\tfrac t2=a(t+\sin t)\) and \(y=-a(1-\cos t)-4a\cos^2\tfrac t2=-a(3+\cos t)\). This is the original cycloid translated, with its lowest point at depth \(4a\) below the cusp.

Exercise 8.3Using the evolute, find the radius of curvature of the cycloid \(x=a(\theta+\sin\theta),\ y=a(1-\cos\theta)\) at the lowest point and at the cusps.

Show solution
The radius of curvature equals the free thread length \(4a\cos(\theta/2)\). At the bottom (\(\theta=0\)) it is \(4a\), equal to the pendulum length, so near the bottom the cycloid matches the circle of radius \(l\). At the cusps (\(\theta=\pm\pi\)) it is 0.
Lesson 09

The compound pendulum

For the general reader. A real pendulum is not a point on a weightless string. It is a rod and a heavy bob, or any rigid body swinging about an axis. Which simple pendulum swings in step with it? Marin Mersenne had asked the question in the 1640s and Descartes and Gilles Personne de Roberval had failed to answer it. Huygens solved it in Part IV of Horologium oscillatorium, and in doing so used, for the first time, the quantity now called the moment of inertia.

Goals
  • Derive the equivalent length \(l=I/(mh)\) from Huygens’s principle on the centre of gravity.
  • Prove the reciprocity of the axis and the centre of oscillation.
  • Use the seconds pendulum as a standard of length.

Huygens’s principle

Huygens took as his first hypothesis a principle he attributed to Evangelista Torricelli: if a system of heavy bodies starts from rest and moves under gravity, its centre of gravity cannot rise above its starting height. He added that if, at any moment, the bodies were separated and each turned upward with its acquired speed, their common centre of gravity would rise exactly to its original height. In modern terms this is conservation of energy for a rigid system, \(\tfrac12\sum m_iv_i^2=g\,M\,\Delta h_G\).

Theorem (Huygens, 1673; Horologium oscillatorium, Part IV)

A rigid body of mass \(M\) oscillating about a horizontal axis at distance \(h\) from its centre of gravity has, for small swings, the same period as a simple pendulum of length

\[ l=\frac{\sum_i m_ir_i^2}{M\,h}=\frac{I_O}{M h}=h+\frac{k^2}{h}, \]

where \(r_i\) are distances from the axis, \(I_O\) is the moment of inertia about the axis and \(k\) the radius of gyration about the parallel axis through the centre of gravity. The point on the line from the axis through the centre of gravity at distance \(l\) is the centre of oscillation.

Derivation. Rotating through angle \(\phi\) from the lowest position raises the centre of gravity by \(h(1-\cos\phi)\). With angular velocity \(\dot\phi\), Huygens’s principle reads \(\tfrac12\dot\phi^2\sum m_ir_i^2+Mgh(1-\cos\phi)=\text{const}\). Differentiating, \(I_O\ddot\phi=-Mgh\sin\phi\), the equation of a simple pendulum of length \(I_O/(Mh)\). The parallel-axis relation \(I_O=M(k^2+h^2)\) gives the last form.

Corollary (reciprocity)

If the body is hung instead from an axis through the centre of oscillation, the old axis becomes the new centre of oscillation and the period is unchanged.

Indeed the new distance is \(h'=l-h=k^2/h\), and \(h'+k^2/h'=k^2/h+h=l\). In 1817 Henry Kater turned this into the reversible pendulum, the standard instrument for measuring \(g\) for a century.

A universal measure

For small swings the period is \(T=2\pi\sqrt{l/g}\). Huygens put it as a proportion: the time of one small swing is to the time of free fall through half the length as the circumference of a circle is to its diameter. He proposed the length of a pendulum beating seconds as a natural unit of length that any nation could reproduce. The proposal was revived in the debates that led to the metre in the 1790s; the metre was defined instead from a meridian of the Earth.

Exercise 9.1A uniform rod of length \(L\) swings about one end. Find the centre of oscillation and the period.

Show solution
\(I_O=ML^2/3\), \(h=L/2\), so \(l=\frac{ML^2/3}{ML/2}=\tfrac23L\) and \(T=2\pi\sqrt{2L/(3g)}\). The rod swings like a simple pendulum two-thirds as long.

Exercise 9.2Show that, for a given body, the period is least when the axis is at distance \(h=k\) from the centre of gravity, and find the minimum equivalent length.

Show solution
\(l(h)=h+k^2/h\) has \(l'(h)=1-k^2/h^2=0\) at \(h=k\), where \(l=2k\). At this point the period is stationary in \(h\), so small errors in the axis position hardly affect it, a property used in precision pendulums.

Exercise 9.3A seconds pendulum takes one second for each swing (period 2 s). Find its length where \(g=9.81\,\mathrm{m\,s^{-2}}\), and verify Huygens’s proportion between one swing and free fall through half the length.

Show solution
\(T=2\pi\sqrt{l/g}=2\) gives \(l=g/\pi^2\approx0.994\) m. One swing takes \(\pi\sqrt{l/g}\); free fall through \(l/2\) takes \(\sqrt{2(l/2)/g}=\sqrt{l/g}\). The ratio is \(\pi\), the ratio of circumference to diameter.
Lesson 10

The wave theory of light

For the general reader. Huygens thought of light as a disturbance travelling through a fine invisible medium, like sound through air. His central idea is a construction: every point that the disturbance reaches becomes a small source of its own, and the new wavefront is the envelope of all these little waves. With it he derived the laws of reflection and refraction and explained the double image seen through a crystal of Iceland spar.

Goals
  • State Huygens’s construction and derive the law of refraction with \(\sin\theta_1/\sin\theta_2=v_1/v_2\).
  • Explain double refraction by spheroidal wavelets.
  • Understand what Fresnel added in 1818.

The construction

Huygens presented his theory to the Académie in 1678 and published it in 1690 as Traité de la lumière. He held that light travels with a finite speed, and reported Ole Rømer’s 1676–77 inference of that speed from the timing of the eclipses of Jupiter’s moon Io.

Huygens’s principle (1678)

Each point of a wavefront at time \(t\) is the source of a secondary wavelet that spreads with the local speed of light in the medium. The wavefront at time \(t+\Delta t\) is the forward envelope of these wavelets. The rays are the lines joining each wavelet’s source to its point of contact with the envelope.

Refraction

A plane wave with speed \(v_1\) meets a plane interface at angle of incidence \(\theta_1\). Let the wavefront touch the interface at \(A\) at time 0, while its other end is still a distance \(v_1\Delta t\) from the interface, reaching it at \(B\) at time \(\Delta t\), so \(AB\sin\theta_1=v_1\Delta t\). By then the wavelet from \(A\) has radius \(v_2\Delta t\) in the second medium, and the refracted front is the tangent from \(B\) to it, so \(AB\sin\theta_2=v_2\Delta t\).

Law of refraction (Huygens’s derivation) \[ \frac{\sin\theta_1}{\sin\theta_2}=\frac{v_1}{v_2}=\frac{n_2}{n_1}. \]

Light bends toward the normal on entering a denser medium because it travels more slowly there. Descartes and later Newton’s corpuscular theory required the opposite. The question was settled for Huygens by Léon Foucault’s measurement of the speed of light in water in 1850. With \(\theta_2=\pi/2\) the construction also gives total internal reflection beyond the critical angle.

Iceland spar

A crystal of calcite splits a ray into two. One, the ordinary ray, obeys the law of refraction. The other, the extraordinary ray, does not: it is refracted even at normal incidence. Huygens explained this by supposing that in the crystal there are two kinds of wavelet, spheres for the ordinary ray and spheroids of revolution about the crystal’s optic axis for the extraordinary ray. The envelope of spheroids is a plane, but the ray to the point of tangency is no longer perpendicular to it. His predictions for many cuts of the crystal agreed with measurement. He also noticed that the two rays from one crystal behave differently when passed through a second crystal turned about the ray, and confessed he could not explain it. The explanation, polarization of a transverse wave, came with Thomas Young and Augustin-Jean Fresnel in the 1810s.

Fresnel’s completion

Huygens’s wavelets had no wavelength and no phase, so his theory could not explain diffraction or interference, and he had no reason why the backward envelope should be ignored. In 1818 Fresnel combined the construction with the principle of interference: the field at a point \(P\) is the superposition of secondary wavelets with their phases. In the Fresnel–Kirchhoff form,

\[ U(P)=\frac{1}{i\lambda}\iint_{\Sigma}U(Q)\,\frac{e^{iks}}{s}\,K(\chi)\,dS,\qquad K(\chi)=\frac{1+\cos\chi}{2}, \]

where \(s\) is the distance from \(Q\) to \(P\) and \(\chi\) the angle between the normal and the direction to \(P\). The obliquity factor vanishes in the backward direction, and Gustav Kirchhoff derived it in 1882 from the wave equation.

Exercise 10.1Use Huygens’s construction to derive the law of reflection.

Show solution
In the same medium the wavelet from \(A\) has radius \(v\Delta t\) when the far end of the front reaches \(B\). So \(AB\sin\theta_i=v\Delta t=AB\sin\theta_r\), giving \(\theta_r=\theta_i\). The two right triangles with hypotenuse \(AB\) are congruent.

Exercise 10.2Find the critical angle for light going from glass (\(n=1.5\)) to air, and explain it with the construction.

Show solution
\(\sin\theta_c=1/1.5\), so \(\theta_c\approx41.8^\circ\). Beyond it, \(AB\sin\theta_1/v_1>AB/v_2\): the front in glass reaches \(B\) later than the wavelet from \(A\) has spread a distance \(AB\) along the interface, so no tangent from \(B\) to the wavelet exists in the air and there is no refracted wave.

Exercise 10.3Suppose, as was argued from Rømer’s observations, that light needs about 22 minutes to cross the diameter of the Earth’s orbit. With the modern value \(1.496\times10^{11}\) m for the Earth–Sun distance, what speed results, and how does it compare with the modern value?

Show solution
\(c\approx2\cdot1.496\times10^{11}/(22\cdot60)\approx2.27\times10^8\,\mathrm{m\,s^{-1}}\), about 24% below \(2.998\times10^8\). The true crossing time is about 16.6 minutes. The order of magnitude was right, which was the point at issue: light is fast but not instantaneous.
Part III · Life and legacy
Lesson 11

Paris and the Académie, 1666–1681

In 1666 Huygens moved to Paris as a founding member of Louis XIV’s Académie royale des sciences, with a royal pension and lodgings in the Bibliothèque du Roi. He was its most eminent member and helped shape its programme of collective experiment and observation. He worked on the vacuum pump, the pendulum clock and the theory of gravity, and in 1669 defended a version of Descartes’s vortex theory of gravity against objections from Roberval and Edme Mariotte.

From 1671 his assistant was the young Denis Papin, with whom he built an engine in which gunpowder exploded in a cylinder drove a piston, an ancestor of the internal combustion engine. In 1672 Gottfried Wilhelm Leibniz arrived in Paris on a diplomatic mission. Huygens directed his reading, set him problems such as the sum of the reciprocals of the triangular numbers, and guided him until 1676, the years in which Leibniz invented his calculus. In the same years Huygens criticized Newton’s new theory of colours, at first without understanding it.

The Dutch War made his position delicate. In 1672 Louis XIV invaded the Dutch Republic, whose new stadtholder, William III of Orange, was the prince his family had served for generations. Huygens stayed in Paris. In 1673 he published Horologium oscillatorium, his masterpiece, with a fulsome dedication to Louis XIV.

In 1675 Huygens devised a watch regulated by a spiral balance spring. He sought a French patent, and the announcement provoked Robert Hooke, who claimed he had invented spring regulation years before; the dispute embittered relations with the Royal Society for a time. Huygens fell seriously ill in 1670, again in 1676, when he spent two years in The Hague and studied Iceland spar, and again in 1679 and 1681. In 1681 he left Paris for the last time. After Colbert’s death in 1683 and with the growing hostility to Protestants in France that culminated in the revocation of the Edict of Nantes in 1685, he did not return.

DiscussionHuygens’s method of publication was slow: the collision rules waited seventeen years for a summary and fifty for the full text, and De vi centrifuga was published only after his death. Compare this with Hooke’s practice of announcing many results quickly. What did each gain and lose?

Lesson 12

Hofwijck, death and legacy, 1681–1695

Back in Holland, Huygens worked on a marine clock for the Dutch East India Company, ground very long-focus lenses with Constantijn for aerial telescopes without tubes, and built a planetarium driven by clockwork whose gearing used continued fractions to approximate the planetary periods. In 1686 he wrote a manual in Dutch, Kort onderwijs, on finding longitude with clocks, and corrected the results of the Company’s trial voyage to the Cape of Good Hope. His father died in 1687, aged 90, and in 1688 Christiaan made the family’s country house, Hofwijck at Voorburg, his home.

In 1689, after William III had become king of England, Huygens made a third visit there. He met Newton on 12 June 1689, and they discussed Iceland spar and motion in resisting media. Huygens admired the Principia but rejected an attraction acting at a distance without mechanical cause. In 1690 he published the Traité de la lumière together with the Discours de la cause de la pesanteur, which proposed a vortex mechanism for gravity and gave his estimate of the flattening of the Earth (Exercise 7.2). He corresponded with Leibniz on the new calculus, which he came to accept late in life, and solved the catenary problem in 1691.

His last book, Cosmotheoros, speculated on life on the planets and the size of the universe. He estimated the distance to Sirius by comparing its brightness with that of the Sun. He died in The Hague on 8 July 1695, aged 66. He never married. He was buried in the Grote Kerk in The Hague, in an unmarked grave like his father’s. Constantijn published Cosmotheoros in 1698, and the Opuscula postuma of 1703 contained the collision treatise and De vi centrifuga.

Legacy

  • Timekeeping: the pendulum clock remained the most accurate timekeeper until quartz clocks in the 1930s. The spiral balance spring still regulates mechanical watches.
  • Mechanics: the relativity principle in collisions, conservation of momentum and of \(\sum mv^2\) in elastic impact, centrifugal force, the moment of inertia and the theory of the physical pendulum. Newton acknowledged his debt in the Principia.
  • Mathematics: evolutes, involutes and curvature; the first printed treatise on probability and the concept of expectation, which Jacob Bernoulli reprinted and annotated in Ars conjectandi (1713).
  • Optics: the Huygens–Fresnel principle, the foundation of diffraction theory, and the two-lens Huygens eyepiece.
  • Astronomy: Titan and the ring of Saturn. The European Space Agency’s Huygens probe, carried by Cassini, landed on Titan on 14 January 2005.
  • Editions: the Dutch Society of Sciences published his complete works, the Œuvres complètes, in 22 volumes between 1888 and 1950.

Final projectChoose one result from Part II. Present Huygens’s geometric argument from the original (the Latin texts have English translations: Blackwell for Horologium oscillatorium, Thompson for the Traité), then the modern derivation, and explain what the calculus made easier and what, if anything, it hid.

Reference

Connections

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

Teachers and advisors

  • Frans van Schooten 1645-1647

Students

Influenced by

Influenced

Colleagues and collaborators

Rivals

Family

  • Constantijn Huygens
  • Constantijn Huygens Jr.

Also mentioned in this life

Fields

On CoursesHub

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Christiaan Huygens, MacTutor History of Mathematics.
  2. Christiaan Huygens, Wikipedia (with extensive references).
  3. C. D. Andriesse, Huygens: The Man Behind the Principle, Cambridge University Press, 2005.
  4. H. J. M. Bos, Huygens, Christiaan, in Dictionary of Scientific Biography, vol. 6, Scribner, 1972.
  5. J. G. Yoder, Unrolling Time: Christiaan Huygens and the Mathematization of Nature, Cambridge University Press, 1988.
  6. H. Aldersey-Williams, Dutch Light: Christiaan Huygens and the Making of Science in Europe, Picador, 2020.

Works and science

  1. C. Huygens, Œuvres complètes, 22 vols., Société hollandaise des sciences, 1888–1950.
  2. C. Huygens, The Pendulum Clock or Geometrical Demonstrations Concerning the Motion of Pendula as Applied to Clocks, transl. R. J. Blackwell, Iowa State University Press, 1986.
  3. C. Huygens, Treatise on Light, transl. S. P. Thompson, Macmillan, 1912. Project Gutenberg.
  4. Horologium Oscillatorium, Wikipedia.
  5. Huygens–Fresnel principle, Wikipedia.
  6. A. Hald, A History of Probability and Statistics and Their Applications before 1750, Wiley, 1990 (chapter on De ratiociniis).
  7. M. Born and E. Wolf, Principles of Optics, Cambridge University Press, 7th ed., 1999 (Huygens–Fresnel and Kirchhoff theory).
  8. R. Dugas, A History of Mechanics, Dover, 1988 (Huygens on collision and the compound pendulum).