How to use this course
Alexis Clairaut belongs to the generation that turned Isaac Newton’s Principia into working science. Newton had stated the law of gravitation and drawn some of its consequences with geometric arguments that few could extend. Clairaut, with Leonhard Euler and Jean le Rond d’Alembert, recast those arguments in the calculus of Gottfried Wilhelm Leibniz and pushed them to quantitative tests: the shape of the Earth, the motion of the Moon, the orbit of a comet. Each test could have refuted Newton. Each, in Clairaut’s hands, confirmed him.
- Multivariable calculus and first-order ordinary differential equations.
- Newtonian mechanics: central forces, Kepler’s laws, the effective potential.
- For Lesson 06, curves on surfaces; for Lessons 07 and 08, the gravitational potential.
Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Lessons 04 to 06 are pure mathematics from his youth; Lessons 07 to 10 follow his great problems in celestial mechanics and geodesy. Formulas are given in modern notation, which differs from Clairaut’s.
Timeline
| 1713 | Born 13 May in Paris (some sources give 7 May), son of Jean-Baptiste Clairaut, a teacher of mathematics, and Catherine Petit. |
| 1726 | At 13, reads a paper on four new curves to the Paris Academy of Sciences. |
| 1729 | Completes his research on curves of double curvature; proposed for the Academy on 4 September. |
| 1731 | Recherches sur les courbes à double courbure. Elected to the Academy at 18, below the statutory age, with the king’s approval. |
| 1733 | Papers on maxima and minima and on geodesics of surfaces of revolution. |
| 1734 | Studies with Johann Bernoulli in Basel, with Pierre Louis Moreau de Maupertuis. Work on the differential equation that bears his name. |
| 1736–37 | Lapland expedition under Maupertuis, from 20 April 1736 to 20 August 1737. Elected Fellow of the Royal Society, 27 October 1737. |
| 1738 | Becomes a full member (pensionnaire) of the Academy, 12 May. |
| 1739–40 | Memoirs on integrating factors and exact differentials. |
| 1741 | Élémens de géométrie. |
| 1743 | Théorie de la figure de la Terre: hydrostatic equilibrium and Clairaut’s theorem. |
| 1747 | 15 November: tells the Academy that the Moon’s apogee seems to require a correction to the inverse-square law. |
| 1749 | 17 May: announces that a more careful calculation reconciles the apogee with Newton’s law. |
| 1752 | Théorie de la lune, after his prize-winning essay for the St Petersburg Academy. |
| 1754 | Lunar tables. |
| 1756 | French translation of the Principia by Émilie du Châtelet published with his assistance. |
| 1758 | 14 November: predicts that Halley’s comet will pass perihelion in mid-April 1759. |
| 1759 | Perihelion occurs on 13 March, within the month he allowed. |
| 1761–62 | Work on achromatic lenses. |
| 1765 | Dies in Paris on 17 May, aged 52, after a brief illness. |
A prodigy in Paris, 1713–1731
Alexis Claude Clairaut was born in Paris on 13 May 1713. The date 7 May also appears in reference works; 13 May is the date given in the eulogy by Grandjean de Fouchy, the Academy’s secretary. His father, Jean-Baptiste Clairaut, taught mathematics in Paris and was later elected to the Berlin Academy. His mother, Catherine Petit, bore many children, of whom few survived infancy.
His father taught him at home and used Euclid’s Elements as a reading book. By nine Clairaut had worked through Guisnée’s Application de l’algèbre à la géométrie, and at ten he was studying the calculus in the Marquis de l’Hôpital’s Analyse des infiniment petits (1696), the first textbook of the subject. In 1726, aged thirteen, he read to the Academy of Sciences a paper on four new curves. A younger brother was also precocious: he read a paper to the Academy at fourteen and died in 1732 at sixteen.
Curves of double curvature
By 1729 Clairaut had finished a systematic study of curves in three-dimensional space, then called curves of double curvature because their projections onto two coordinate planes are both curved. He was proposed for the Academy on 4 September 1729. The book, Recherches sur les courbes à double courbure, appeared in 1731, and in that year the king confirmed his election as adjoint mécanicien. He was eighteen, below the statutory minimum age, and he remains the youngest person ever elected.
Maupertuis, Basel and the Newtonians, 1731–1736
The Paris Academy of the early 1730s was Cartesian. Most of its members explained planetary motion by René Descartes’s vortices of subtle matter, and they distrusted Newton’s attraction acting at a distance. A small group of younger academicians, led by Pierre Louis Moreau de Maupertuis, set out to bring Newton’s physics to France. Clairaut joined them almost at once.
In 1733 he published on problems of maxima and minima, an early contribution to the calculus of variations, and on the shortest paths on surfaces of revolution (Lesson 06). In 1734 he and Maupertuis went to Basel to work with Johann Bernoulli, the most powerful analyst in Europe and a teacher of Euler. There Clairaut befriended Samuel König, with whom he corresponded for years. In the same period he studied the differential equations now named after him and their singular solutions (Lesson 05).
The shape of the Earth
The great scientific controversy of the decade concerned the shape of the Earth. Newton and Christiaan Huygens had argued from rotation that the Earth is flattened at the poles. The surveys of the Cassini family along the meridian of Paris suggested the opposite: an Earth elongated toward the poles. The question was settled by measuring the length of a degree of latitude at very different latitudes. The Academy sent one expedition to Peru, near the equator, in 1735, and a second to Lapland, near the Arctic Circle, in 1736.
Lapland, 1736–1737
The Lapland expedition left Dunkirk on 20 April 1736 under Maupertuis. Its members included Clairaut, Charles Étienne Louis Camus, Pierre Charles Le Monnier, the abbé Outhier and the Swedish astronomer Anders Celsius. They worked along the Torne River at the head of the Gulf of Bothnia, triangulating across hilltops in summer and measuring a baseline on the frozen river in winter. Astronomical observations of a star at both ends of the arc gave the difference in latitude.
The degree measured in Lapland was markedly longer than the degree near Paris. A longer degree means a flatter meridian, so the Earth is flattened at the poles, as Newton had said. The expedition returned to Paris on 20 August 1737. Voltaire celebrated Maupertuis as the man who had flattened the Earth and the Cassinis. The Royal Society elected Clairaut a Fellow on 27 October 1737, and Louis XV later granted him a pension for his work on the Earth’s figure. In 1738 he became a full member of the Academy.
The measurements raised a deeper question. Given a rotating, self-gravitating body, what shape should it take, and how does its shape relate to gravity at its surface? Clairaut’s first attempts, published in the Philosophical Transactions during the expedition, were incomplete. The full answer was his book of 1743 (Lessons 07 and 08).
DiscussionThe Cassini survey and the Lapland survey disagreed about the sign of a small effect. What sources of error in eighteenth-century geodesy (instrument, refraction, baseline, astronomy) could produce such a disagreement, and which did the Lapland team try hardest to control?
Space curves, mixed partials and exact differentials
For the general reader. Before Clairaut, the calculus was mostly used for curves in a plane. His first book treated curves in space, and his later papers treated functions of several variables. He was among the first to state when an expression like \(M\,dx+N\,dy\) is the differential of a single function, a question that sits at the root of thermodynamics, potential theory and the hydrostatics of Lesson 07.
- Describe a space curve by two projections and compute its tangent and arc length.
- State the equality of mixed partial derivatives and its hypotheses.
- Test exactness and find integrating factors.
Curves of double curvature
Clairaut represented a space curve by two equations \(y=\phi(x)\), \(z=\psi(x)\), its projections onto the \(xy\)- and \(xz\)-planes. The tangent direction is \((1,\phi',\psi')\) and the arc length is \(\int\sqrt{1+\phi'^2+\psi'^2}\,dx\). More generally a curve is the intersection of two surfaces \(F=0\), \(G=0\), and its tangent is orthogonal to both normals \(\nabla F\) and \(\nabla G\).
Let \(f\) be defined on an open set in \(\mathbb R^2\) and suppose \(f_x,f_y,f_{xy}\) exist and \(f_{xy}\) is continuous. Then \(f_{yx}\) exists and \(f_{yx}=f_{xy}\).
Clairaut and Euler used this identity around 1740; the hypotheses were made precise by Hermann Schwarz in 1873. Its main use for Clairaut was the following criterion, which appears in his memoirs of 1739 and 1740 on the integration of first-order equations.
Let \(M,N\) be continuously differentiable on an open rectangle. There is a function \(\Phi\) with \(d\Phi=M\,dx+N\,dy\) if and only if \(\partial M/\partial y=\partial N/\partial x\). Then the equation \(M\,dx+N\,dy=0\) has solutions \(\Phi(x,y)=\text{const}\).
Necessity is the equality of mixed partials. For sufficiency put \(\Phi(x,y)=\int_{x_0}^xM(s,y_0)\,ds+\int_{y_0}^yN(x,t)\,dt\); then \(\Phi_y=N\) and \(\Phi_x=M(x,y_0)+\int_{y_0}^yN_x(x,t)\,dt=M(x,y_0)+\int_{y_0}^yM_y(x,t)\,dt=M(x,y)\). When the criterion fails, one looks for an integrating factor \(\mu\) such that \(\mu M\,dx+\mu N\,dy\) is exact. Clairaut showed such factors exist for first-order equations, a fact anticipated by Nicolas Fatio de Duillier and Johann Bernoulli.
Exercise 4.1Let \(f(x,y)=xy\,\dfrac{x^2-y^2}{x^2+y^2}\) for \((x,y)\ne0\) and \(f(0,0)=0\). Show \(f_{xy}(0,0)\ne f_{yx}(0,0)\). Which hypothesis of the theorem fails?
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Exercise 4.2Show that \((3x^2y+y^3)\,dx+(x^3+3xy^2)\,dy=0\) is exact and solve it. Then find an integrating factor of \(y\,dx-x\,dy=0\) and solve.
Show solution
Clairaut’s equation and singular solutions
For the general reader. Most differential equations have a family of solutions depending on a constant. Clairaut found a simple equation whose solutions are a family of straight lines, and which has one more solution: the curve that all the lines touch. That extra curve cannot be obtained from the family by any choice of constant. It was the first clear example of a singular solution.
- Solve Clairaut’s equation by differentiation.
- Identify the singular solution as an envelope.
- Relate the equation to the Legendre transform.
Let \(f\) be differentiable. The equation \(y=xp+f(p)\), \(p=dy/dx\), has the general solution \(y=cx+f(c)\), a family of lines. If \(f''\ne0\) it also has the singular solution given parametrically by \(x=-f'(p)\), \(y=f(p)-pf'(p)\), which is the envelope of the family.
Derivation
Differentiate \(y=xp+f(p)\) with respect to \(x\): \(p=p+xp'+f'(p)p'\), so
\[ p'\,\big(x+f'(p)\big)=0. \]Either \(p'=0\), so \(p=c\) and \(y=cx+f(c)\); or \(x=-f'(p)\), which with the original equation gives the parametric curve. The envelope of \(F(x,y,c)=y-cx-f(c)=0\) is found by eliminating \(c\) from \(F=0\) and \(\partial F/\partial c=-x-f'(c)=0\), which is the same pair of equations. At each point of the singular curve the solution is not unique: a line of the family and the envelope pass through it with the same slope.
The singular solution is the graph of the Legendre transform. If \(g(x)=\sup_p\big(xp+f(p)\big)\) is attained at an interior point, then \(x=-f'(p)\) there and \(g(x)\) is the singular \(y\). The duality between a curve and its family of tangent lines, implicit in Clairaut’s equation, is the same one used in Hamiltonian mechanics and thermodynamics.
Exercise 5.1Solve \(y=xp-\tfrac14p^2\) completely and sketch the solutions.
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Exercise 5.2Find the curve such that every tangent line cuts off a segment of constant length \(\ell\) between the coordinate axes.
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Geodesics on surfaces of revolution
For the general reader. On a sphere the shortest paths are great circles. On a flattened Earth they are more complicated, and a navigator or surveyor needs to know them. In 1733 Clairaut found a conserved quantity that governs every shortest path on any surface of revolution. It is the geometric form of the conservation of angular momentum.
- State and prove Clairaut’s relation.
- Use it to classify geodesics on a surface of revolution.
Let \(S\) be obtained by rotating a curve about the \(z\)-axis, parametrized by \(\mathbf x(s,\theta)=(r(s)\cos\theta,\ r(s)\sin\theta,\ z(s))\) with \(s\) the arc length of the profile curve. The metric is \(ds^2+r(s)^2\,d\theta^2\).
Along any geodesic on \(S\), parametrized by arc length, let \(\psi\) be the angle between the geodesic and the meridian through the current point, and \(r\) the distance from the axis. Then \(r\sin\psi\) is constant.
Proof
Geodesics are critical points of the energy \(\int\big(\dot s^2+r(s)^2\dot\theta^2\big)\,dt\). The Lagrangian does not depend on \(\theta\), so the conjugate momentum \(r^2\dot\theta\) is conserved (compare Noether’s theorem). For a unit-speed curve the component of the velocity along the parallel is \(r\dot\theta=\sin\psi\), so \(r^2\dot\theta=r\sin\psi\).
Consequences: meridians (\(\psi=0\)) are geodesics; a parallel is a geodesic exactly where \(r'(s)=0\); and a geodesic with Clairaut constant \(c>0\) stays in the region \(r\ge c\), turning back where \(r=c\) with \(\psi=\pi/2\).
Exercise 6.1Use Clairaut’s relation to show that geodesics on the cylinder \(x^2+y^2=1\) are helices, circles and straight lines.
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Exercise 6.2On the hyperboloid of one sheet \(x^2+y^2-z^2=1\), a geodesic starts on the waist circle \(r=1\) making angle \(\psi_0\) with the meridian. Show it never returns to the waist if \(0<\psi_0<\pi/2\) and it moves upward initially. What happens if it starts at height \(z_0>0\) heading downward with Clairaut constant \(c<1\)?
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The figure of the Earth: equilibrium and flattening
For the general reader. A spinning ball of fluid bulges at its equator. How much it bulges depends on how fast it spins and on how its mass is arranged inside. Clairaut put this problem on a sound mathematical footing in 1743, starting from a clean principle of fluid equilibrium and ending with formulas that could be checked against the Lapland and Peru measurements.
- State Clairaut’s condition for fluid equilibrium.
- Relate the length of a degree to the flattening.
- Derive Newton’s flattening for a homogeneous Earth.
Equilibrium
Clairaut imagined thin channels of fluid running through the body. The fluid is in equilibrium only if the forces produce no net flow around any closed channel. In modern terms:
A fluid mass under a force field \(\mathbf F=(P,Q,R)\) per unit mass can be in equilibrium only if \(P\,dx+Q\,dy+R\,dz\) is an exact differential \(dU\), so that \(\mathbf F=\nabla U\). Pressure is then constant on the level surfaces of \(U\), and the free surface is one of them.
This is the exactness criterion of Lesson 04 used in physics. For the rotating Earth, \(U\) is the sum of the gravitational potential and the centrifugal potential \(\tfrac12\omega^2(x^2+y^2)\). Clairaut’s principle introduced the potential, though the name came later, with George Green in 1828.
Degrees of latitude
For an ellipse of equatorial radius \(a\), flattening \(f=(a-b)/a\) and eccentricity \(e^2=f(2-f)\), the radius of curvature of the meridian at geodetic latitude \(\varphi\) is
\[ M(\varphi)=\frac{a(1-e^2)}{(1-e^2\sin^2\varphi)^{3/2}}\approx a\,\big(1-2f+3f\sin^2\varphi\big). \]A degree of latitude has length \(M(\varphi)\,\pi/180\), so it grows from equator to pole in proportion to \(3f\sin^2\varphi\). Comparing degrees at two latitudes gives \(f\).
A slowly rotating, homogeneous, self-gravitating fluid in equilibrium is an oblate spheroid with flattening \(f=\tfrac54m\) to first order, where \(m=\omega^2a/g_e\) is the ratio of centrifugal acceleration to gravity at the equator.
For the Earth \(m\approx1/288\), so \(f\approx1/230\), Newton’s value. Huygens, who assumed all attraction directed toward the centre, found \(f=\tfrac12m\approx1/578\). The true value, about \(1/298\), lies between. Clairaut showed why: a real Earth is denser toward the centre (Lesson 08).
Exercise 7.1Modern values give a degree of latitude of 111.132 km at \(45^\circ\) and 111.512 km at \(66.3^\circ\), near the Lapland arc. Estimate \(f\).
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Exercise 7.2Show that a uniform gravitational field \(\mathbf F=(0,0,-g)\) together with a rotation about the \(z\)-axis gives a paraboloidal free surface. (This is Newton’s bucket, and a check of Clairaut’s principle.)
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Clairaut’s theorem on gravity
For the general reader. A pendulum clock runs slower near the equator, because gravity is weaker there. Jean Richer had noticed this in Cayenne in 1672. Clairaut found a simple formula linking the variation of gravity with latitude to the Earth’s flattening and its rotation. Remarkably, the formula does not depend on how mass is distributed inside, so pendulum measurements at the surface reveal the shape of the planet.
- State Clairaut’s theorem and test it with modern data.
- State Clairaut’s equation for the flattening of internal layers.
- Recover the homogeneous and Huygens limits.
Let the Earth be a rotating fluid in equilibrium whose level surfaces are nearly spherical concentric spheroids. To first order in small quantities, surface gravity at geodetic latitude \(\varphi\) is
\[ g(\varphi)=g_e\Big[1+\Big(\tfrac52m-f\Big)\sin^2\varphi\Big],\qquad m=\frac{\omega^2a}{g_e}, \]where \(f\) is the surface flattening. Equivalently, \(\dfrac{g_p-g_e}{g_e}+f=\tfrac52m\).
Clairaut proved this for a body made of homogeneous spheroidal layers of different densities. George Stokes showed in 1849 that it holds for any internal density distribution, provided the outer surface is a level surface of spheroidal shape.
The internal layers
Let \(\varepsilon(a)\) be the flattening of the level surface of mean radius \(a\), \(\rho(a)\) the density there, and \(\bar\rho(a)\) the mean density inside it. Clairaut derived the second-order equation
\[ \varepsilon''+\frac{6\rho}{a\bar\rho}\,\varepsilon'-\frac{6}{a^2}\Big(1-\frac{\rho}{\bar\rho}\Big)\varepsilon=0, \]with \(\varepsilon\) regular at the centre and the surface condition \(R\,\varepsilon'(R)+2\varepsilon(R)=\tfrac52m\). Since density does not increase outward, \(\rho\le\bar\rho\), and one shows that \(\varepsilon\) increases outward and that \(\tfrac12m\le f\le\tfrac54m\). The two bounds are Huygens’s and Newton’s values. Later, Pierre-Simon Laplace, Adrien-Marie Legendre and Harold Jeffreys used this equation to constrain the density inside the Earth.
Exercise 8.1With \(m=1/288.4\) and \(f=1/298.26\), compute \(\tfrac52m-f\) and compare with the measured value \((g_p-g_e)/g_e=0.005302\).
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Exercise 8.2For a homogeneous Earth (\(\rho=\bar\rho\)), solve Clairaut’s equation and recover \(f=\tfrac54m\). Then use Clairaut’s theorem to find \((g_p-g_e)/g_e\) in this model and in Huygens’s model \(f=\tfrac12m\).
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The Moon’s apogee and the three-body problem
For the general reader. The Moon’s elliptical orbit slowly turns, so that its farthest point from the Earth goes once around the sky in about nine years. The Sun’s pull causes this. But when Clairaut, Euler and d’Alembert each calculated the effect from Newton’s law, all three found only half the observed rate. For eighteen months it looked as if the law of gravitation itself was wrong. Clairaut then found the missing half in terms everyone had neglected.
- Compute the first-order precession of a nearly circular orbit under a perturbing force.
- See why the first approximation gives half the answer.
- Follow the history of the 1747–1749 crisis.
Precession from a perturbing force
For a nearly circular orbit of radius \(r\) under a central attraction \(F(r)\) per unit mass, the angular frequency \(\Omega\) and the frequency \(\kappa\) of small radial oscillations satisfy
\[ \Omega^2=\frac{F(r)}{r},\qquad \kappa^2=F'(r)+\frac{3F(r)}{r}. \]For \(F=\mu/r^2\), \(\kappa=\Omega\): the ellipse closes. If \(\kappa<\Omega\), the apsides advance by \(2\pi(\Omega/\kappa-1)\) per revolution. Averaged over a month, the Sun’s tidal force on the Moon has a mean outward radial component \(\tfrac12n'^2r\), where \(n'\) is the Sun’s mean motion.
With \(m=n'/n\approx0.0748\), the ratio of the Moon’s sidereal month to the sidereal year, the averaged radial force gives an apsidal advance of \(\tfrac34m^2\) revolutions per revolution of the Moon, about \(0.0042\). The observed value is about \(0.0085\).
The crisis and its resolution
On 15 November 1747 Clairaut told the Academy that the discrepancy required adding a small term, varying as \(1/r^4\), to Newton’s inverse-square law. D’Alembert had deposited similar results shortly before, and Euler had written to Clairaut in September supporting the same conclusion. Georges-Louis Leclerc de Buffon objected on metaphysical grounds that a law of nature should be simple.
Clairaut went back to his equations. He had written the orbit in a form \(1/r=\frac1p\big(1-e\cos(c\theta)\big)\) plus perturbation terms, and had dropped products of small quantities that looked negligible. In spring 1748 he found that the transverse component of the solar force, combined with the variation of the orbit, contributes a second-order term almost as large as the first. On 17 May 1749 he announced that the inverse-square law accounted for nearly the whole motion. In modern form the series is
\[ \frac{\dot\varpi}{n}=\frac34m^2+\frac{225}{32}m^3+\frac{4071}{128}m^4+\cdots, \]whose terms are about \(0.0042,\ 0.0029,\ 0.0010\). The series converges slowly; George William Hill and others computed many more terms in the nineteenth century. Euler, now convinced, praised the result warmly, and Clairaut’s essay won the prize of the St Petersburg Academy. He published it as Théorie de la lune (1752) and produced lunar tables in 1754, important for finding longitude at sea.
Exercise 9.1Derive the first-order result. Take \(F(r)=\mu/r^2-kr\) with \(k=\tfrac12n'^2\), compute \(\kappa^2\) and \(\Omega^2\), and show that the apsidal advance per revolution is \(2\pi\cdot\tfrac34m^2\) to leading order.
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Exercise 9.2The sidereal month is 27.32 days and the lunar apsides complete a revolution in about 8.85 years. Compute the observed \(\dot\varpi/n\) and compare with the first three terms of the series.
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The return of Halley’s comet
For the general reader. Edmond Halley had identified the comets of 1531, 1607 and 1682 as one object and predicted its return around 1758. But the comet passes close to Jupiter and Saturn, whose pulls stretch or shrink its orbit. Clairaut, with Joseph Jérôme Lalande and Nicole-Reine Lepaute, spent months computing these effects by hand, and predicted the date of its closest approach to the Sun. The comet arrived within his margin of error. It was the first object other than a planet shown to orbit the Sun, and a public triumph for Newton’s theory.
- Relate a change in orbital energy to a change in period.
- Understand the logic of Clairaut’s computation.
For a Keplerian orbit about a body of gravitational parameter \(\mu\), the specific energy is \(E=-\mu/(2a)\) and the period \(P=2\pi a^{3/2}/\sqrt\mu\). Hence \(\dfrac{\Delta P}{P}=\dfrac32\dfrac{\Delta a}{a}=-\dfrac32\dfrac{\Delta E}{E}\), and a perturbing acceleration \(\mathbf f\) changes the energy at the rate \(\dot E=\mathbf v\cdot\mathbf f\).
The computation
Clairaut did not integrate the full motion. He took the known orbit, estimated how the attraction of Jupiter and Saturn had altered the comet’s orbital elements along each of the two previous returns, and compared the two intervals 1607–1682 and 1682–1759. The integrals over the planets’ positions were evaluated numerically, point by point. Lalande and Lepaute carried out much of the arithmetic. On 14 November 1758 Clairaut announced perihelion for mid-April 1759, with an uncertainty of about a month due to neglected effects and the unknown masses of Saturn and other bodies.
Johann Georg Palitzsch, a farmer and amateur astronomer in Saxony, saw the comet on 25 December 1758. Perihelion came on 13 March 1759, about a month earlier than predicted. Clairaut’s calculation had found that Jupiter and Saturn would delay the return by about 618 days compared with an unperturbed orbit, and the observed return confirmed the bulk of that delay. D’Alembert and others attacked the precision of the prediction, and a public quarrel followed.
Exercise 10.1The comet’s period is about 76 years. What relative change in semi-major axis corresponds to a delay of 618 days?
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Exercise 10.2The comet passed perihelion in late August 1531, late October 1607 and mid-September 1682. Estimate the two intervals and explain why Halley could not predict the 1758–59 return to better than a year or so.
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Textbooks, salons and quarrels
Clairaut was a public figure in Enlightenment Paris. He never married, enjoyed society and moved easily in the salons. He was close to Émilie du Châtelet and Voltaire, and he guided du Châtelet’s French translation of Newton’s Principia, checking her commentary and adding his own results. The translation appeared in part in 1756, after her death in 1749, and remained the standard French version.
He also wrote for beginners. His Élémens de géométrie (1741) taught geometry by leading the reader to rediscover results from practical problems of measuring land, instead of from axioms. A companion Élémens d’algèbre followed in the same spirit. Both went through many editions into the nineteenth century.
His relations with d’Alembert, cordial in the early 1740s, became hostile after the lunar crisis of 1747. Both worked on the same problems, the three-body problem and the figure of the Earth, and each accused the other of errors and borrowings. D’Alembert favoured general analytic methods; Clairaut favoured results that could be compared with observation. The quarrel reached the public after the comet’s return in 1759.
DiscussionClairaut’s Élémens de géométrie begins from the problem of measuring fields, not from definitions and axioms. What are the strengths and weaknesses of this approach for a modern student, and how does it compare with Euclid?
Last years and legacy
In 1756 Clairaut took part in measuring a geodetic baseline at Villejuif, south of Paris. In 1761 and 1762 he worked on achromatic lenses combining two kinds of glass; the work was unfinished at his death. He was a member of the Royal Society of London and the academies of Berlin, St Petersburg, Bologna and Uppsala. He died in Paris on 17 May 1765, aged 52, after a short illness.
Legacy
- Geodesy and geophysics: Clairaut’s theorem and his equation for the internal flattening remain the starting point of the theory of the Earth’s figure; the gravity formula used in geodesy today is a refinement of his.
- Celestial mechanics: his resolution of the lunar apsides was the decisive test of the inverse-square law in the three-body problem, and his comet calculation was the first quantitative prediction of planetary perturbations confirmed by observation. Lagrange and Laplace built on both.
- Mathematics: Clairaut’s equation, Clairaut’s relation for geodesics, the exactness criterion and the potential function are standard. His treatment of space curves opened differential geometry in three dimensions, later developed by Euler, Gaspard Monge and Gauss.
- Commemoration: the lunar crater Clairaut is named after him.
Final projectChoose one of the three great tests of Newton’s law in Part II: the figure of the Earth, the Moon’s apogee or Halley’s comet. Write a two-page account that states the problem, sketches Clairaut’s method, quantifies the agreement with observation, and identifies the approximation that limited his accuracy.
Sources and further reading
Biography
- J. J. O’Connor and E. F. Robertson, Alexis Claude Clairaut, MacTutor History of Mathematics.
- Alexis Clairaut, Wikipedia.
- Académie des sciences, notice on Clairaut (J.-C. Pecker).
- O. Courcelle, Chronologie de la vie de Clairaut (online chronology, 2007 onward).
- J.-P. Grandjean de Fouchy, Éloge de M. Clairaut, Histoire de l’Académie royale des sciences (1765).
- J. Itard, Clairaut, Alexis-Claude, Dictionary of Scientific Biography, vol. 3.
Mathematics and physics
- A. C. Clairaut, Recherches sur les courbes à double courbure, Paris, 1731.
- A. C. Clairaut, Théorie de la figure de la Terre, tirée des principes de l’hydrostatique, Paris, 1743.
- A. C. Clairaut, Théorie de la lune, déduite du seul principe de l’attraction, St Petersburg, 1752.
- A. C. Clairaut, Théorie du mouvement des comètes, Paris, 1760.
- Clairaut’s theorem (gravity) and Halley’s Comet, Wikipedia.
- H. Jeffreys, The Earth, Cambridge University Press, chapter on the figure of the Earth (Clairaut’s equation).
- C. Wilson, The problem of perturbation analytically treated: Euler, Clairaut, d’Alembert, in R. Taton and C. Wilson (eds.), Planetary Astronomy from the Renaissance to the Rise of Astrophysics, Part B, Cambridge, 1995.
- M. C. Gutzwiller, Moon–Earth–Sun: the oldest three-body problem, Reviews of Modern Physics 70 (1998) 589–639.
- M. do Carmo, Differential Geometry of Curves and Surfaces, section on Clairaut’s relation.