courseshub.world · Lives in Mathematics and Physics

Johann Heinrich Lambert

Johann Heinrich Lambert · Jean-Henri Lambert

A course in two strands: the life of the tailor’s son from Mulhouse who taught himself science by night and became one of the most productive members of the Berlin Academy, and the mathematics he left in half a dozen fields: the irrationality of π, the measurement of light, the mapping of the sphere, the theory of parallels, and the orbits of comets.

Born 26 August 1728, Mulhouse Died 25 September 1777, Berlin π irrational memoir published 1768 Level advanced undergraduate
The whole sphere in Lambert’s azimuthal equal-area projection (equatorial aspect, 1772). Each ellipse is the image of the same small circle on the globe: shapes are distorted toward the rim, but every ellipse has the same area.
Lesson 00

How to use this course

Johann Heinrich Lambert left school at twelve and never attended a university. By his death at 49 he had written more than 150 works on mathematics, physics, astronomy, cartography and philosophy. Several of his results are now standard in their fields under his name: Lambert’s cosine law in photometry, the Lambert projections in cartography, Lambert’s theorem in orbital mechanics, and the Lambert W function in analysis. He gave the first proof that π is irrational, and his unpublished study of parallel lines came closer to hyperbolic geometry than anyone before Gauss.

Prerequisites for Part II
  • Calculus in one and several variables, including surface integrals and polar coordinates.
  • Power series and the basic properties of continued fractions (convergents) for Lesson 04.
  • Spherical coordinates for Lesson 08, and Kepler’s laws for Lesson 10.
  • For Lesson 09, the first lessons of the courses on Saccheri and Lobachevsky are helpful.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, proofs or derivations, and exercises with folded solutions. The lessons can be read in any order. They follow, roughly, the order of the questions Lambert took up: series (1758), light (1760), orbits and the cosmos (1761), π (1761–68), parallels (1766) and maps (1772).

Reference

Timeline

1728Born 26 August in Mulhouse, then a free city allied to the Swiss Confederacy, son of the tailor Lukas Lambert and Elisabeth Schmerber.
c. 1740Leaves school at about twelve to work in his father’s shop; studies alone in the evenings.
c. 1743Clerk at the ironworks at Seppois, south of Mulhouse.
c. 1745Secretary to Johann Rudolf Iselin, editor of the Basler Zeitung, in Basel.
1748Tutor in the household of Count Peter von Salis at Chur.
1755First paper, on heat, in Acta Helvetica.
1756–58Grand tour with two pupils: Göttingen, Utrecht, Paris, Marseille, Nice, Turin, Milan.
1758First book, on the passage of light through media (The Hague). Observationes variae in mathesin puram: the trinomial equation.
1759Die freye Perspektive, on perspective drawing.
1760Photometria, Augsburg. Euler recommends him for a professorship at St Petersburg.
1761Cosmologische Briefe; Insigniores orbitae cometarum proprietates.
1759–62Involved in founding the Bavarian Academy of Sciences in Munich; leaves after disputes.
1764Neues Organon, Leipzig. Arrives in Berlin at Euler’s invitation.
176510 January: member of the Berlin Academy. Begins corresponding with Immanuel Kant.
1766Writes Theorie der Parallellinien.
1768The memoir proving that π is irrational appears in the Berlin Academy’s Mémoires.
1770Appointed Oberbaurat (senior councillor for public works).
1771Anlage zur Architectonic.
1772Anmerkungen und Zusätze zur Entwerfung der Land- und Himmelscharten: seven map projections. The colour pyramid.
1777Dies in Berlin on 25 September, aged 49.
1786Theorie der Parallellinien published by Johann Bernoulli and Carl Friedrich Hindenburg.
Part I · Formation
Lesson 01

Mulhouse and Basel, 1728–1748

Johann Heinrich Lambert was born on 26 August 1728 in Mulhouse, in southern Alsace. Some sources give 28 August. Mulhouse was then a small Protestant republic allied to the Swiss Confederacy; it joined France only in 1798. The Lamberts were Calvinists whose ancestors had left Lorraine for Mulhouse as religious refugees in 1635. His father, Lukas Lambert, was a tailor. Johann Heinrich was one of five sons, and the family had little money.

He attended the town school until about the age of twelve, then left to help in his father’s workshop. He kept studying on his own in the evenings, from whatever books he could obtain, and he never had any further formal education. At about fifteen he became a clerk at the ironworks at Seppois, south of the town, and took private pupils. At seventeen he moved to Basel as secretary to Johann Rudolf Iselin, a scholar who edited the Basler Zeitung. The post gave him time to read and access to a learned city.

DiscussionLambert is one of the few major eighteenth-century scientists with no university education at all. What resources were available to a self-taught student in Basel in the 1740s, and what would he have missed by not attending a university?

Lesson 02

Chur and the grand tour, 1748–1759

In 1748, at twenty, Lambert became tutor in the household of Count Peter von Salis at Chur, in the Grisons. He stayed about eight years. The family had a large library, and Lambert used it to study mathematics, astronomy, physics and philosophy systematically. He built his own astronomical instruments and kept regular meteorological records. He was elected to the Literary Society of Chur and to the Swiss Scientific Society in Basel, whose Acta Helvetica printed his first paper, on heat, in 1755.

The grand tour

In 1756 he set out with two of his pupils on a tour of the learned cities of Europe. At Göttingen he met the mathematician Abraham Gotthelf Kästner and the astronomer Tobias Mayer, and was elected to the Göttingen Society of Sciences. When French troops occupied Göttingen in 1757, he took the pupils to Utrecht for about two years. His first book, on the passage of light through various media, was printed at The Hague in 1758. In Paris he met Jean le Rond d’Alembert; the party returned through Marseille, Nice, Turin and Milan.

Also in 1758 he published the paper on series that contains his trinomial equation (Lesson 05). He was thirty, and had a growing European reputation but no position.

Lesson 03

Augsburg, Munich and the road to Berlin, 1759–1764

After leaving the Salis household, Lambert settled for a time in Augsburg, where he found publishers. In quick succession there appeared Die freye Perspektive (1759) on perspective drawing, Photometria (1760) on the measurement of light, and in 1761 both Cosmologische Briefe, letters on the structure of the universe, and Insigniores orbitae cometarum proprietates on the orbits of comets. These are the subjects of Lessons 07 and 10.

In 1760 Leonhard Euler recommended him for a professorship of astronomy at the St Petersburg Academy, but the post was not filled. In the same period Lambert was drawn into the founding of the Bavarian Academy of Sciences in Munich, which offered him a salary; he quarrelled with colleagues there and left in 1762. He then helped survey the border between the Duchy of Milan and the Grisons.

Berlin

The Berlin Academy, on Euler’s proposal, had elected him in 1761, but King Frederick II did not confirm the election. In 1764 Lambert went to Berlin to press his case. The King found his dress and manner strange and at first refused him. In the same year Lambert’s major philosophical work, the Neues Organon, appeared in Leipzig. On 10 January 1765 he was admitted as a member of the Academy’s physical class, with a salary. Frederick came to value him highly.

Part II · The mathematics
Lesson 04

The irrationality of π

For the general reader. The ratio of a circle’s circumference to its diameter, π, is about 3.14159. For two thousand years nobody knew whether it could be written exactly as a fraction. In a memoir presented to the Berlin Academy, Lambert proved that it cannot. His proof expands the tangent function as an infinite continued fraction and shows that such a fraction cannot have a rational value.

Goals
  • State Lambert’s continued fraction for \(\tan x\).
  • Prove an irrationality criterion for continued fractions.
  • Deduce that \(\tan x\) is irrational for rational \(x\ne0\), and hence that \(\pi\) is irrational.

The memoir, Mémoire sur quelques propriétés remarquables des quantités transcendantes circulaires et logarithmiques, is dated 1761 and was printed in the Berlin Academy’s volume of 1768. Euler had believed π irrational but had not proved it.

Theorem (Lambert’s continued fraction)

For every real \(x\) with \(\cos x\ne0\),

\[ \tan x=\cfrac{x}{1-\cfrac{x^2}{3-\cfrac{x^2}{5-\cfrac{x^2}{7-\cdots}}}} \]

and the convergents converge to \(\tan x\).

Where the fraction comes from

For \(k\ge-1\) define the entire functions

\[ t_k(x)=\sum_{n\ge0}\frac{(-1)^n\,x^{2n}}{2^n\,n!\,(2k+2n+1)!!},\qquad (2m+1)!!=1\cdot3\cdot5\cdots(2m+1),\quad (-1)!!=1. \]

Since \(2^n n!\,(2n-1)!!=(2n)!\) and \(2^n n!\,(2n+1)!!=(2n+1)!\), we have \(t_{-1}(x)=\cos x\) and \(t_0(x)=\frac{\sin x}{x}\). Comparing coefficients of \(x^{2n}\) gives the three-term recurrence

\[ t_{k-1}(x)=(2k+1)\,t_k(x)-x^2\,t_{k+1}(x)\qquad(k\ge0). \]

Dividing by \(t_k\) gives \(\frac{t_{k-1}}{t_k}=(2k+1)-\frac{x^2}{t_k/t_{k+1}}\). Iterating from \(k=0\) gives \(\frac{\cos x}{\sin x/x}=1-\cfrac{x^2}{3-\cfrac{x^2}{5-\cdots}}\), which is the theorem. Convergence follows because \(t_{k+1}/t_k\to0\) as \(k\to\infty\) (Exercise 4.3). Lambert reached the expansion by a long direct calculation, dividing series by series; the recurrence is the modern way to organize it.

The irrationality criterion

Lemma (after Lambert and Legendre)

Let \(a_k,b_k\) be integers with \(a_k\ne0\) and \(|b_k|\ge|a_k|+2\) for all \(k\ge1\). Then the continued fraction \(\xi_1=\cfrac{a_1}{b_1+\cfrac{a_2}{b_2+\cdots}}\) converges, and its value is irrational.

Proof. Convergence under \(|b_k|\ge|a_k|+1\) is Pringsheim’s criterion, which also gives \(|\xi_k|\le1\) for every tail \(\xi_k=\cfrac{a_k}{b_k+\cfrac{a_{k+1}}{b_{k+1}+\cdots}}\). Then \(|\xi_k|=\frac{|a_k|}{|b_k+\xi_{k+1}|}\le\frac{|a_k|}{|b_k|-1}\le\frac{|a_k|}{|a_k|+1}<1\), and \(\xi_k\ne0\) since \(a_k\ne0\). Suppose \(\xi_1=u_1/u_0\) with integers \(u_0,u_1\). From \(\xi_k=a_k/(b_k+\xi_{k+1})\) we get \(\xi_{k+1}=a_k/\xi_k-b_k\). Define integers by \(u_{k+1}=a_k u_{k-1}-b_k u_k\). Induction shows \(\xi_k=u_k/u_{k-1}\) for all \(k\). Since \(0<|\xi_k|<1\), we get \(|u_0|>|u_1|>|u_2|>\cdots>0\), an infinite strictly decreasing sequence of positive integers. Contradiction. ∎

Theorem (Lambert, 1768)

If \(x\) is a nonzero rational number, then \(\tan x\) is irrational. Consequently \(\pi\) is irrational.

Proof. Let \(x=p/q\). Multiplying numerators and denominators through by \(q\),

\[ \tan\frac pq=\cfrac{p}{q-\cfrac{p^2}{3q-\cfrac{p^2}{5q-\cdots}}}. \]

Choose \(K\) with \((2K+1)q\ge p^2+2\). The tail from level \(K\) on has \(a_k=-p^2\) and \(b_k=(2k+1)q\), so by the lemma it is irrational. The value \(\tan(p/q)\) is obtained from that tail by finitely many maps \(t\mapsto a/(b+t)\) with rational \(a\ne0\) and \(b\), each of which sends irrationals to irrationals. So \(\tan(p/q)\) is irrational. If \(\pi\) were rational, so would be \(\pi/4\), but \(\tan(\pi/4)=1\). ∎

The same memoir treats the hyperbolic tangent, whose continued fraction has plus signs, and shows in the same way that \(e^x\) is irrational for rational \(x\ne0\). Lambert conjectured that \(e\) and \(\pi\) are transcendental. Adrien-Marie Legendre proved in 1794 that \(\pi^2\) is irrational, Charles Hermite proved \(e\) transcendental in 1873, and Ferdinand von Lindemann proved \(\pi\) transcendental in 1882.

Exercise 4.1Compute the first four convergents of Lambert’s continued fraction for \(\tan\frac12\) as exact fractions, and compare with \(\tan\frac12=0.5463024898\ldots\).

Show solution
With \(p=1,q=2\): \(\tan\frac12=\cfrac{1}{2-\cfrac{1}{6-\cfrac{1}{10-\cdots}}}\). The convergents are \(\frac12\), \(\frac{1}{2-1/6}=\frac{6}{11}\), \(\frac{1}{2-\frac{1}{6-1/10}}=\frac{59}{108}\), and \(\frac{820}{1501}\). In decimals: \(0.5,\ 0.54545,\ 0.546296,\ 0.5463024650\). The fourth convergent is correct to 7 decimal places.

Exercise 4.2Show that the hypothesis \(|b_k|\ge|a_k|+2\) cannot be weakened to \(|b_k|\ge|a_k|+1\) for every \(k\): find a continued fraction with \(a_k=-1\), \(b_k=2\) that has a rational value.

Show solution
Let \(\xi=\cfrac{-1}{2-\cfrac{1}{2-\cdots}}\) and \(\eta=\cfrac{1}{2-\cfrac{1}{2-\cdots}}\). The tail satisfies \(\eta=1/(2-\eta)\), so \(\eta^2-2\eta+1=0\) and \(\eta=1\); the convergents \(\frac12,\frac23,\frac34,\dots\) confirm it. Then \(\xi=-\eta=-1\), which is rational. The descent argument fails because \(|\xi_k|=1\), not \(<1\).

Exercise 4.3(a) Verify the recurrence for the coefficient of \(x^2\). (b) Show that \((2k+1)!!\,t_k(x)\to1\) uniformly for \(|x|\le X\) as \(k\to\infty\), and deduce \(t_{k+1}/t_k\to0\). (c) Show that if \(r\ne0\) is rational then \(\arctan r\) is irrational.

Show solution
(a) The coefficient of \(x^2\) in \(t_{k-1}\) is \(-\frac{1}{2(2k+1)!!}\). On the right, \((2k+1)\) times the \(x^2\) coefficient of \(t_k\) is \(-\frac{2k+1}{2(2k+3)!!}\), and \(-x^2t_{k+1}\) contributes \(-\frac{1}{(2k+3)!!}\). The sum is \(-\frac{2k+3}{2(2k+3)!!}=-\frac{1}{2(2k+1)!!}\). (b) \((2k+1)!!\,t_k(x)=1+\sum_{n\ge1}\frac{(-1)^nx^{2n}}{2^nn!\,(2k+3)\cdots(2k+2n+1)}\), and the sum is at most \(\frac{e^{X^2/2}}{2k+3}\) in absolute value, which tends to 0. Hence \(\frac{t_{k+1}}{t_k}\approx\frac{(2k+1)!!}{(2k+3)!!}=\frac1{2k+3}\to0\). (c) If \(x=\arctan r\) were rational, it would be nonzero, and \(\tan x=r\) would be irrational by the theorem below. The theorem says nothing about \(\arctan r/\pi\): for \(r=1\) it equals \(\frac14\).
Lesson 05

The trinomial equation and the W function

For the general reader. Many equations, like \(x=q+x^3\), have no solution in radicals that is easy to use. Lambert found an infinite series that solves every equation of this shape at once. Euler turned his series into the function now called Lambert’s W.

Goals
  • State Lambert’s series for the trinomial equation and check it.
  • See how Euler’s 1783 paper leads to the W function.

The trinomial equation (1758)

In Observationes variae in mathesin puram (1758) Lambert considered the equation \(x=q+x^m\) and found the solution as a power series in \(q\). In modern notation, for the root that tends to 0 with \(q\):

Theorem (Lambert’s series)

For integers \(m\ge2\), \(n\ge1\), and \(|q|\) small enough, the root \(x\) of \(x=q+x^m\) with \(x\to0\) as \(q\to0\) satisfies

\[ x^n=\sum_{k\ge0}\frac{n}{n+mk}\binom{n+mk}{k}\,q^{\,n+(m-1)k}. \]

The modern proof is Joseph-Louis Lagrange’s inversion formula (1770): if \(x=q+\phi(x)\) then \(H(x)=H(q)+\sum_{k\ge1}\frac{1}{k!}\frac{d^{k-1}}{dq^{k-1}}\big[\phi(q)^kH'(q)\big]\). With \(\phi(x)=x^m\) and \(H(x)=x^n\) the coefficients reduce to those above. Lagrange’s formula contains Lambert’s series as a special case.

From Lambert to W

In 1783 Euler rewrote Lambert’s equation in the symmetric form \(x^\alpha-x^\beta=(\alpha-\beta)v\,x^{\alpha+\beta}\) and, letting \(\beta\) tend to \(\alpha\), arrived at the equation \(\ln x=v\,x^{\alpha}\). Its solution is the function now written \(W\), defined by

\[ W(z)\,e^{W(z)}=z,\qquad W_0(z)=\sum_{n\ge1}\frac{(-n)^{n-1}}{n!}z^n\quad\big(|z|<1/e\big). \]

The name “Lambert W function” and its standard notation were fixed by Robert Corless, Gaston Gonnet, David Hare, David Jeffrey and Donald Knuth in 1996. It appears in combinatorics (counting rooted trees), delay differential equations, and quantum mechanics.

Exercise 5.1For \(m=3\), \(n=1\), \(q=0.1\), sum the first three terms of Lambert’s series and compare with the root of \(x=0.1+x^3\) found by iteration.

Show solution
The terms are \(k=0\): \(q=0.1\); \(k=1\): \(\frac14\binom41q^3=q^3=0.001\); \(k=2\): \(\frac17\binom72q^5=3q^5=0.00003\). The sum is \(0.10103\). Iterating \(x\mapsto0.1+x^3\) from \(x=0.1\) gives \(0.101,\ 0.1010303,\dots\to0.1010313\). The next term, \(k=3\): \(\frac1{10}\binom{10}{3}q^7=12q^7=1.2\times10^{-6}\), brings the series to \(0.1010312\).

Exercise 5.2Show that \(y=W(z)\) solves \(y e^y=z\) by checking the first three terms of the series, and find the radius of convergence of the series by the ratio test.

Show solution
The series begins \(W=z-z^2+\tfrac32z^3-\cdots\). Then \(e^W=1+(z-z^2)+\tfrac12z^2+O(z^3)=1+z-\tfrac12z^2+O(z^3)\) and \(We^W=(z-z^2+\tfrac32z^3)(1+z-\tfrac12z^2)+O(z^4)=z+(1-1)z^2+(-\tfrac12-1+\tfrac32)z^3+O(z^4)=z+O(z^4)\). The ratio of consecutive coefficients in absolute value is \(\frac{(n+1)^n}{(n+1)!}\cdot\frac{n!}{n^{n-1}}=\big(1+\tfrac1n\big)^{n-1}\to e\), so the radius is \(1/e\), where the branch point \(z=-1/e\) lies.
Lesson 06

Hyperbolic functions

For the general reader. Sine and cosine describe a point moving round a circle. Hyperbolic sine and cosine do the same for a point moving along a hyperbola, and they turn up in hanging chains, in relativity and in non-Euclidean geometry. Lambert gave them their first systematic treatment, with formulas and tables, and linked them to ordinary trigonometry by a single angle.

Goals
  • Derive the basic identities of hyperbolic functions.
  • See the parameter \(u\) as an area, as for the circle.
  • Use the Gudermannian angle that connects circular and hyperbolic functions.

Vincenzo Riccati had introduced hyperbolic functions in the 1750s. Lambert developed them systematically, with their addition formulas and tables, in his Observations trigonométriques (Berlin, 1770), giving credit to Riccati and to Daviet de Foncenex. With \(\cosh u=\frac{e^u+e^{-u}}2\), \(\sinh u=\frac{e^u-e^{-u}}2\):

\[ \cosh^2u-\sinh^2u=1,\qquad \sinh(u+v)=\sinh u\cosh v+\cosh u\sinh v,\qquad \cosh(u+v)=\cosh u\cosh v+\sinh u\sinh v. \]

Lambert related the circular and hyperbolic functions through the angle \(\varphi\) defined by \(\tan\varphi=\sinh u\), now called the Gudermannian \(\varphi=\operatorname{gd}u\). It satisfies \(\sin\varphi=\tanh u\), \(\cos\varphi=1/\cosh u\), and \(u=\ln\tan(\tfrac\pi4+\tfrac\varphi2)\). The last formula is the vertical coordinate of the Mercator projection (Lesson 08).

Proposition (hyperbolic sector)

The point \((\cosh u,\sinh u)\) lies on the hyperbola \(x^2-y^2=1\), and the region bounded by the hyperbola and the segments from the origin to \((1,0)\) and to \((\cosh u,\sinh u)\) has area \(u/2\). On the unit circle, the sector with angle \(\theta\) likewise has area \(\theta/2\).

Exercise 6.1Prove the proposition on the hyperbolic sector.

Show solution
The first claim is \(\cosh^2u-\sinh^2u=1\), which follows from \((e^u+e^{-u})^2-(e^u-e^{-u})^2=4\). For the area use the sector formula \(\frac12\int(x\,dy-y\,dx)\) along the arc from \(t=0\) to \(t=u\), the two radial segments contributing nothing: \(\frac12\int_0^u(\cosh^2t-\sinh^2t)\,dt=\frac u2\).

Exercise 6.2Prove that if \(\tan\varphi=\sinh u\) with \(|\varphi|<\pi/2\), then \(u=\ln\tan\big(\frac\pi4+\frac\varphi2\big)\).

Show solution
Let \(t=\tan(\frac\pi4+\frac\varphi2)\). Then \(\frac{t-t^{-1}}2=\frac{\tan^2(\frac\pi4+\frac\varphi2)-1}{2\tan(\frac\pi4+\frac\varphi2)}=-\cot\big(\tfrac\pi2+\varphi\big)=\tan\varphi\), using \(\cot2\theta=\frac{1-\tan^2\theta}{2\tan\theta}\) with \(2\theta=\frac\pi2+\varphi\). So \(\sinh(\ln t)=\tan\varphi=\sinh u\), and since \(\sinh\) is injective, \(u=\ln t\).
Lesson 07

Photometria: the measurement of light

For the general reader. How bright is a surface lit at a slant, and how bright does a matte wall look from different angles? Lambert’s Photometria (1760) answered these questions with a few simple laws and so made the measurement of light a part of mathematical physics. Computer graphics, astronomy and the design of lighting still use his laws.

Goals
  • State Lambert’s laws of illumination and diffuse emission.
  • Derive the relation \(M=\pi L\) for a Lambertian surface.
  • Use the exponential law of absorption.

Lambert worked in a period with no light detectors except the eye. His method was to compare two surfaces side by side and adjust one source until they looked equally bright. From such comparisons and from the assumption that light travels in straight lines, he obtained the laws below. He also introduced the word albedo for the fraction of incident light that a surface reflects diffusely.

Laws of illumination (Lambert, 1760)

(i) A point source of intensity \(I\) at distance \(r\) produces on a surface element whose normal makes angle \(\theta\) with the direction to the source the illuminance

\[ E=\frac{I\cos\theta}{r^2}. \]

(ii) A perfectly diffuse (Lambertian) surface has the same radiance \(L\) in every direction. The intensity it sends from area \(dA\) into direction \(\theta\) from the normal is \(dI=L\cos\theta\,dA\), and its total emitted flux per unit area is \(M=\pi L\).

Derivation of \(M=\pi L\)

The flux per unit area into a solid angle \(d\Omega\) at angle \(\theta\) is \(L\cos\theta\,d\Omega\). In spherical coordinates \(d\Omega=\sin\theta\,d\theta\,d\phi\), so

\[ M=\int_0^{2\pi}\!\!\int_0^{\pi/2}L\cos\theta\sin\theta\,d\theta\,d\phi=2\pi L\Big[\tfrac12\sin^2\theta\Big]_0^{\pi/2}=\pi L. \]

A Lambertian surface looks equally bright from every direction, because the factor \(\cos\theta\) in its emitted intensity is exactly compensated by the foreshortening of its apparent area. The full Moon is close to a uniformly bright disc partly for this kind of reason, though lunar soil is not perfectly Lambertian.

The law of absorption

Bouguer–Lambert law

Light of intensity \(I_0\) traversing a thickness \(x\) of a homogeneous absorbing medium emerges with intensity \(I=I_0e^{-\kappa x}\), where \(\kappa\) is the absorption coefficient.

The law follows from the assumption that each thin layer \(dx\) removes the same fraction \(\kappa\,dx\) of the light reaching it: \(dI=-\kappa I\,dx\). Pierre Bouguer had stated it in 1729, and Lambert cited him. August Beer added in 1852 the proportionality of \(\kappa\) to concentration, and the combined statement is now called the Beer–Lambert law.

Exercise 7.1A lamp of intensity \(I\) hangs at height \(h\) above a table. At what height is the illuminance greatest at a point of the table at horizontal distance \(d\) from the foot of the lamp?

Show solution
Here \(r^2=h^2+d^2\) and \(\cos\theta=h/r\), so \(E(h)=\frac{Ih}{(h^2+d^2)^{3/2}}\). Then \(E'(h)=I\frac{(h^2+d^2)-3h^2}{(h^2+d^2)^{5/2}}\), which vanishes when \(d^2=2h^2\). The best height is \(h=d/\sqrt2\).

Exercise 7.2A uniform Lambertian disc of radius \(a\) and radiance \(L\) faces a small surface element on its axis at distance \(z\). Show that the illuminance there is \(E=\pi L\frac{a^2}{a^2+z^2}\), and check the limits \(z\to0\) and \(z\to\infty\).

Show solution
A ring of radius \(\rho\), width \(d\rho\) is at distance \(r=\sqrt{\rho^2+z^2}\) and is seen at angle \(\theta\) with \(\cos\theta=z/r\) both from the element and from the disc normal. Its contribution is \(dE=\frac{L\cos\theta\cdot2\pi\rho\,d\rho\cdot\cos\theta}{r^2}=2\pi Lz^2\frac{\rho\,d\rho}{(\rho^2+z^2)^2}\). Integrating from 0 to \(a\): \(E=\pi Lz^2\big(\frac1{z^2}-\frac1{a^2+z^2}\big)=\pi L\frac{a^2}{a^2+z^2}\). As \(z\to0\), \(E\to\pi L=M\). As \(z\to\infty\), \(E\approx\frac{\pi a^2L}{z^2}\), the inverse-square law for a source of intensity \(\pi a^2L\).

Exercise 7.3A glass plate 5 mm thick transmits 80% of incident light (ignore reflection at the surfaces). What fraction do three such plates transmit? What thickness transmits half?

Show solution
Thicknesses add and transmissions multiply: \(0.8^3=0.512\). From \(e^{-5\kappa}=0.8\), \(\kappa=\frac{\ln1.25}{5}\approx0.0446\) per mm. Half is transmitted at \(x=\frac{\ln2}{\kappa}\approx15.5\) mm.
Lesson 08

Map projections

For the general reader. A globe cannot be flattened without distortion. A map can keep the shapes of small regions or their areas, but not both. In a single essay of 1772 Lambert posed the problem of mapping in general terms, proved that no map has both properties, and introduced seven new projections. Three of them are still among the most widely used in cartography.

Goals
  • Describe Lambert’s cylindrical equal-area, azimuthal equal-area and conformal conic projections.
  • Prove the equal-area and conformal properties.
  • Understand why no map of the sphere is both.

The essay Anmerkungen und Zusätze zur Entwerfung der Land- und Himmelscharten appeared in the third volume of Lambert’s Beiträge zum Gebrauche der Mathematik und deren Anwendung (1772). He did not name his projections. Today they are known as the Lambert conformal conic, the transverse Mercator, the Lambert azimuthal equal-area, the Lagrange projection, the Lambert cylindrical equal-area, the transverse cylindrical equal-area, and the Lambert conic equal-area.

Local distortion

Write a point of the sphere of radius \(R\) by latitude \(\varphi\) and longitude \(\lambda\). The metric is \(ds^2=R^2(d\varphi^2+\cos^2\varphi\,d\lambda^2)\). For a map whose image curves of latitude and longitude meet at right angles, let \(h\) be the scale factor along meridians and \(k\) the scale factor along parallels. The map is conformal when \(h=k\) everywhere and equal-area when \(hk=1\) everywhere.

Three Lambert projections

Cylindrical equal-area: \(x=R\lambda,\ y=R\sin\varphi\).

Azimuthal equal-area (centre at a point \(P\)): a point at angular distance \(c\) from \(P\) in azimuth \(\alpha\) is mapped to polar coordinates \((\rho,\alpha)\) with \(\rho=2R\sin\frac c2\).

Conformal conic (cone constant \(0<n<1\)): polar coordinates \(\theta=n\lambda\) and \(\rho=R\,F\,\big[\tan\big(\tfrac\pi4+\tfrac\varphi2\big)\big]^{-n}\), with a constant \(F>0\) chosen to fix the scale.

For the cylindrical map, \(h=\frac{1}{R}\frac{dy}{d\varphi}=\cos\varphi\) and \(k=\frac{R\,d\lambda}{R\cos\varphi\,d\lambda}=\frac1{\cos\varphi}\), so \(hk=1\). This is Archimedes’ theorem that the sphere and its circumscribed cylinder have equal lateral areas between any two parallel planes. For the azimuthal map, the area of the cap of angular radius \(c\) is \(2\pi R^2(1-\cos c)=4\pi R^2\sin^2\frac c2=\pi\rho^2\), and rotational symmetry then gives equal area for every region. The whole sphere maps onto a disc of radius \(2R\), as in the figure at the top of this page.

Theorem (no perfect map)

No map from an open region of the sphere to the plane is both conformal and equal-area.

Proof. If \(h=k\) and \(hk=1\), then \(h=k=1\): the map preserves all lengths and is a local isometry. By Gauss’s Theorema Egregium (1827), isometries preserve Gaussian curvature. The sphere has curvature \(1/R^2\) and the plane has curvature 0. ∎ Lambert reached the incompatibility by analysing the differential conditions directly. Gauss’s theorem explains it, and also shows that no map at all can preserve lengths.

Exercise 8.1Show that the conformal conic is conformal: compute \(h\) and \(k\) and show they are equal.

Show solution
Along a parallel, the arc \(R\cos\varphi\,d\lambda\) maps to an arc of the circle of radius \(\rho\) with angle \(n\,d\lambda\), so \(k=\frac{n\rho}{R\cos\varphi}\). Along a meridian, \(h=\frac{1}{R}\big|\frac{d\rho}{d\varphi}\big|\). Since \(\frac{d}{d\varphi}\ln\tan(\frac\pi4+\frac\varphi2)=\frac1{\cos\varphi}\), we have \(\frac{d\rho}{d\varphi}=-\frac{n\rho}{\cos\varphi}\), so \(h=\frac{n\rho}{R\cos\varphi}=k\).

Exercise 8.2On the cylindrical equal-area map of the whole Earth, what fraction of the map’s height lies above latitude \(60^\circ\) N? What fraction of the Earth’s surface lies there?

Show solution
The map runs from \(y=-R\) to \(y=R\), and latitude \(60^\circ\) is at \(y=R\sin60^\circ\approx0.866R\). So the fraction of height is \(\frac{1-0.866}{2}\approx6.7\%\). Since the map is equal-area and its width is constant, the same fraction, \(\frac{1-\sin60^\circ}{2}\approx6.7\%\), of the Earth’s surface lies north of \(60^\circ\).

Exercise 8.3In the azimuthal equal-area projection, find the scale factors \(h\) (radial) and \(k\) (tangential) at angular distance \(c\) from the centre. Where is the distortion of shape largest?

Show solution
Radially, \(h=\frac1R\frac{d\rho}{dc}=\cos\frac c2\). Tangentially, a small circle of radius \(R\sin c\) maps to a circle of radius \(\rho\), so \(k=\frac{2R\sin(c/2)}{R\sin c}=\frac{1}{\cos(c/2)}\). Then \(hk=1\). The ratio \(k/h=1/\cos^2\frac c2\) grows without bound as \(c\to\pi\): the antipode of the centre is spread over the whole boundary circle. The ellipses in the hero figure show this.
Lesson 09

The theory of parallels

For the general reader. Like Saccheri thirty years before, Lambert tried to prove Euclid’s parallel postulate by assuming it false and looking for a contradiction. He found none. Instead he found that if the postulate fails, the angles of a triangle determine its area, and he observed that the formulas look like those of a sphere with an imaginary radius. This is exactly the geometry Lobachevsky and Bolyai published sixty years later.

Goals
  • Set up Lambert’s quadrilateral and three hypotheses.
  • Prove that area is proportional to angular defect.
  • Understand the imaginary sphere and check it with hyperbolic trigonometry.

Lambert wrote Theorie der Parallellinien in 1766 and did not publish it. Johann Bernoulli and Carl Friedrich Hindenburg printed it in the Leipziger Magazin für reine und angewandte Mathematik in 1786, nine years after his death. Its three parts review earlier attempts (§§1–11), develop propositions that do not use the postulate (§§12–26), and examine three hypotheses (§§27–88). Lambert knew Georg Simon Klügel’s 1763 dissertation, which examined about thirty attempted proofs, Saccheri’s among them.

The quadrilateral

Lambert’s figure has three right angles. Such a quadrilateral is half of a Saccheri quadrilateral, cut along the line joining the midpoints of base and summit. The three hypotheses concern its fourth angle: right, obtuse or acute.

  • Right angle (§§40–51): implies the parallel postulate.
  • Obtuse angle (§§52–64): Lambert shows that the perpendiculars from one side of the quadrilateral to the opposite line shrink steadily and would reach zero, so two lines would meet twice. He rejects the hypothesis, as Saccheri did, because Euclid’s lines meet at most once.
  • Acute angle (§§65–88): no contradiction appears.
Results under the acute hypothesis (Lambert, 1766)

(a) Every triangle has angle sum less than \(\pi\) (§§73–74). (b) The defect \(\pi-(A+B+C)\) of a triangle, and more generally the defect of a polygon, is proportional to its area (§§81–82). (c) There is an absolute unit of length: a length can be determined by an angle, for example as the side of the equilateral triangle with a given angle (§80). (d) Lines with a common perpendicular diverge on both sides of it (§70).

Why area is proportional to defect

For a polygon \(P\) with \(n\) sides define \(\delta(P)=(n-2)\pi-\Sigma(P)\), where \(\Sigma\) is the angle sum. When a polygon is cut into two along a segment joining two points of its boundary, the defects add (Exercise 9.1). So \(\delta\) is a positive additive function on polygons, unchanged by congruence, exactly like area. Under mild continuity assumptions, any two such functions are proportional: \(\text{area}=k^2\,\delta\) for a constant \(k\). Lambert argued along these lines. Since the defect of a triangle is less than \(\pi\), it follows that no triangle has area as large as \(\pi k^2\), however long its sides.

The imaginary sphere

On a sphere of radius \(R\), a triangle has area \(R^2(A+B+C-\pi)\), the formula of Albert Girard. Replace \(R\) by \(ik\): the area becomes \(-k^2(A+B+C-\pi)=k^2(\pi-A-B-C)\), Lambert’s formula. He remarked (§§81–82) that one could almost conclude that the third hypothesis holds on an imaginary sphere. Lobachevsky made the same substitution systematic in his Imaginary Geometry of 1835, and Eugenio Beltrami’s pseudosphere of 1868 gave the observation a real meaning.

Modern form: Lambert quadrilaterals

In the hyperbolic plane of curvature \(-1/k^2\), a quadrilateral with right angles at three vertices, with sides \(a,b\) meeting at the right angle opposite the fourth vertex, has fourth angle \(\varphi\) given by

\[ \cos\varphi=\sinh\frac ak\,\sinh\frac bk. \]

Exercise 9.1A triangle \(ABC\) is cut by a segment from \(A\) to a point \(D\) of \(BC\). Show that the defects add: \(\delta(ABC)=\delta(ABD)+\delta(ADC)\). Extend to a convex quadrilateral cut by a diagonal.

Show solution
The angles at \(D\) sum to \(\pi\), and the two angles at \(A\) sum to \(\angle A\). So \(\delta(ABD)+\delta(ADC)=2\pi-(\angle A+\angle B+\angle C+\pi)=\pi-(\angle A+\angle B+\angle C)\). For a quadrilateral \(ABCD\) cut by \(AC\): \(\delta(ABC)+\delta(ACD)=2\pi-\Sigma(ABCD)\), which is the defect of a quadrilateral \((4-2)\pi-\Sigma\).

Exercise 9.2Take \(k=1\) and a Lambert quadrilateral with \(a=b=\frac12\). Compute the fourth angle and the area. Check that the angle is acute and tends to \(\pi/2\) as \(a,b\to0\).

Show solution
\(\sinh\frac12\approx0.5211\), so \(\cos\varphi=0.2715\) and \(\varphi\approx1.296\) rad \(\approx74.2^\circ\). The area is the defect, \(2\pi-(3\cdot\frac\pi2+\varphi)=\frac\pi2-\varphi\approx0.275\). For small sides, \(\cos\varphi\approx ab\to0\), so \(\varphi\to\pi/2\), and the area \(\approx ab\), as for a Euclidean rectangle.

Exercise 9.3Use the hyperbolic law of cosines for angles, \(\cos C=-\cos A\cos B+\sin A\sin B\cosh(c/k)\), to find the side \(a\) of an equilateral triangle with angle \(\alpha<\pi/3\). This is Lambert’s “absolute unit”. Evaluate it for \(\alpha=\pi/4\).

Show solution
With all angles \(\alpha\): \(\cos\alpha=-\cos^2\alpha+\sin^2\alpha\cosh(a/k)\), so \(\cosh(a/k)=\frac{\cos\alpha(1+\cos\alpha)}{\sin^2\alpha}=\frac{\cos\alpha}{1-\cos\alpha}\). For \(\alpha=\pi/4\): \(\cosh(a/k)=\frac{\sqrt2/2}{1-\sqrt2/2}=\frac{1}{\sqrt2-1}=1+\sqrt2\), so \(a=k\operatorname{arcosh}(1+\sqrt2)\approx1.529\,k\). The angle alone fixes the length.
Lesson 10

Comets, orbits and the Milky Way

For the general reader. To find the path of a newly seen comet, astronomers had to work back from a few observed positions to an orbit around the Sun. Lambert discovered a surprising fact: the time a body takes to travel between two points of its orbit depends on only three lengths. In the same year he published a picture of the universe as a vast system of star systems, each like the Milky Way.

Goals
  • State Lambert’s theorem on the time of flight.
  • Recover Euler’s parabolic case as a limit.
  • Summarize the cosmology of the Cosmologische Briefe.

Lambert’s interest in comets began with the great comet of 1744. In Insigniores orbitae cometarum proprietates (Augsburg, 1761) he developed geometric methods for comet orbits, including the theorem below. Euler had found the case of a parabolic orbit earlier, and Lagrange later gave an analytic proof of the general statement.

Theorem (Lambert, 1761)

A body moves on an ellipse with semi-major axis \(a\) about a centre of attraction with gravitational parameter \(\mu=GM\). Let \(r_1,r_2\) be its distances from the centre at two points and \(c\) the length of the chord joining them. Then the time of flight depends only on \(r_1+r_2\), \(c\) and \(a\). In the basic case (transfer angle less than \(\pi\), with the arc not passing around the empty focus),

\[ \sqrt{\mu}\,t=a^{3/2}\big[(\alpha-\sin\alpha)-(\beta-\sin\beta)\big],\qquad \sin\frac\alpha2=\sqrt{\frac{r_1+r_2+c}{4a}},\quad \sin\frac\beta2=\sqrt{\frac{r_1+r_2-c}{4a}}. \]

Lambert established the first statement by geometric arguments. One modern way to see it: the time is unchanged when the orbit is deformed keeping \(a\), \(r_1+r_2\) and \(c\) fixed, and the orbit can be deformed into a degenerate ellipse, a straight segment through the centre, on which the time is easy to compute. Today the inverse problem, finding the orbit that joins two given positions in a given time, is called Lambert’s problem. Its solution is used daily to plan spacecraft transfers and rendezvous.

The Cosmologische Briefe (1761)

In a series of letters Lambert described the Milky Way as a flattened system of stars around the Sun, with its plane serving as an ecliptic for the stars. Each star was a sun, presumably with planets. Stars move in groups, and the Milky Way itself might be one member of a system of higher order, and so on, though he held the whole to be finite. He developed these ideas independently of Thomas Wright (1750) and Immanuel Kant (1755), whose work he learned of only later. Unlike Kant he did not consistently regard the nebulae as other Milky Ways.

Exercise 10.1Check the formula for a circular orbit of radius \(r=1\) with \(\mu=1\), between two points separated by a quarter turn.

Show solution
The period is \(2\pi\), so the time for a quarter turn is \(\pi/2\). Here \(a=1\), \(r_1+r_2=2\), \(c=\sqrt2\). Then \(\sin\frac\alpha2=\sqrt{\frac{2+\sqrt2}{4}}=\cos\frac\pi8\), so \(\alpha=\frac{3\pi}4\), and \(\sin\frac\beta2=\sqrt{\frac{2-\sqrt2}4}=\sin\frac\pi8\), so \(\beta=\frac\pi4\). The formula gives \(\frac{3\pi}4-\frac\pi4-\big(\sin\frac{3\pi}4-\sin\frac\pi4\big)=\frac\pi2\).

Exercise 10.2Let \(a\to\infty\) to obtain Euler’s equation for a parabolic orbit: \(6\sqrt\mu\,t=(r_1+r_2+c)^{3/2}-(r_1+r_2-c)^{3/2}\).

Show solution
For large \(a\), \(\alpha\approx2\sqrt{\frac{r_1+r_2+c}{4a}}\) is small and \(\alpha-\sin\alpha\approx\frac{\alpha^3}6=\frac{(r_1+r_2+c)^{3/2}}{6a^{3/2}}\). Similarly for \(\beta\). Multiplying by \(a^{3/2}\), the factors of \(a\) cancel: \(\sqrt\mu\,t\to\frac16\big[(r_1+r_2+c)^{3/2}-(r_1+r_2-c)^{3/2}\big]\).

DiscussionLambert’s cosmology was hierarchical and finite and Kant’s was infinite. What observations available in 1761 could have decided between them, and what would have been needed?

Part III · Life and legacy
Lesson 11

The Berlin Academy, 1765–1777

Lambert spent the last twelve years of his life in Berlin as a member of the Academy’s physical class. He was the only member who regularly presented papers to other classes as well, and his output was enormous: more than 150 works in those twelve years. The four volumes of his Beiträge zum Gebrauche der Mathematik und deren Anwendung (1765–1772) collected work on applied mathematics, including the map projections of Lesson 08 and, in 1772, a mathematical treatment of mortality statistics. His Observations trigonométriques (1770) systematized hyperbolic functions.

His relations with Euler, who had brought him to Berlin, became strained over the Academy’s finances, which depended on its monopoly on selling calendars. Euler left for St Petersburg in 1766. When Lagrange arrived later that year as director of the mathematics class, Lambert became one of his colleagues. In 1770 Frederick II appointed him Oberbaurat, a senior councillor for public works, which raised his income substantially.

Philosophy and other work

The Neues Organon (1764) treats logic, the theory of truth, signs, and appearances; it contains one of the first uses of the word “phenomenology”. From 1765 Lambert corresponded with Immanuel Kant, who valued his judgement and intended to dedicate the Critique of Pure Reason to him; Lambert died before it appeared. The Anlage zur Architectonic (1771) tried to build philosophy deductively in the manner of Euclid. Lambert also invented a practical hygrometer, wrote on hygrometry and on heat (his Pyrometrie appeared posthumously in 1779), argued for a worldwide network of weather stations, and in 1772, with the court painter Benjamin Calau, published a colour pyramid arranging mixtures of three pigments with white.

DiscussionLambert worked across mathematics, physics, astronomy, cartography and philosophy. Pick two of his works from different fields and identify a method or idea they share.

Lesson 12

Death and legacy

Lambert fell ill in 1775 and declined medical treatment. He died in Berlin on 25 September 1777, aged 49, of a lung disease; the standard German biography attributes it to tuberculosis. He never married. His friend Johann Bernoulli, the astronomer and grandson of Johann I Bernoulli, edited his correspondence and some unpublished papers, including, with Hindenburg, the Theorie der Parallellinien in 1786.

Legacy

  • Number theory: the first proof that \(\pi\) is irrational, the starting point for the theory of irrationality and transcendence developed by Legendre, Hermite and Lindemann.
  • Geometry: the Lambert quadrilateral and the area–defect relation; his remark on the imaginary sphere anticipated the hyperbolic geometry of Lobachevsky and Bolyai.
  • Cartography: the Lambert conformal conic is standard for aeronautical charts and for mapping regions of large east–west extent; the azimuthal equal-area projection is used for continental and polar maps; the transverse Mercator underlies the Universal Transverse Mercator grid.
  • Physics and astronomy: Lambert’s cosine law, Lambertian reflectance, albedo, and the Beer–Lambert law; Lambert’s theorem and Lambert’s problem in orbital mechanics. The lambert was a unit of luminance. The asteroid 187 Lamberta and a lunar crater bear his name.
  • Analysis: the trinomial equation, the Lambert W function, and the systematic use of hyperbolic functions.

Final projectWrite a two-page account for a strong undergraduate of Lambert’s proof that \(\pi\) is irrational: derive the continued fraction for \(\tan x\) from the recurrence for \(s_k\), prove the lemma, and compute convergents of \(\tan1\) until they agree with \(\tan1=1.5574077\ldots\) to six decimal places.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Johann Heinrich Lambert, MacTutor History of Mathematics.
  2. Johann Heinrich Lambert, Wikipedia.
  3. Lambert, Johann Heinrich, Neue Deutsche Biographie and Allgemeine Deutsche Biographie, Deutsche Biographie.
  4. H. K. Strick, Johann Heinrich Lambert, MacTutor (translated from Spektrum der Wissenschaft).
  5. H. Frommert, Johann Heinrich Lambert, SEDS (on his astronomy and cosmology).

Mathematics and science

  1. J. H. Lambert, Mémoire sur quelques propriétés remarquables des quantités transcendantes circulaires et logarithmiques, Histoire de l’Académie Royale des Sciences et des Belles-Lettres de Berlin (1761), printed 1768. Reprinted in L. Berggren, J. Borwein and P. Borwein, Pi: A Source Book, Springer.
  2. Proof that π is irrational, Wikipedia (Lambert’s, Niven’s and Laczkovich’s proofs).
  3. M. Laczkovich, On Lambert’s proof of the irrationality of π, Amer. Math. Monthly 104 (1997) 439–443.
  4. J. H. Lambert, Photometria, sive de mensura et gradibus luminis, colorum et umbrae, Augsburg, 1760. English translation by D. L. DiLaura, Illuminating Engineering Society, 2001.
  5. J. H. Lambert, Notes and Comments on the Composition of Terrestrial and Celestial Maps (1772), translated by W. R. Tobler, University of Michigan, 1972.
  6. J. P. Snyder, Flattening the Earth: Two Thousand Years of Map Projections, University of Chicago Press, 1993.
  7. A. Papadopoulos and G. Théret, Hyperbolic geometry in the work of Johann Heinrich Lambert, arXiv:1503.01865.
  8. A. Papadopoulos and G. Théret, La théorie des lignes parallèles de Johann Heinrich Lambert, Blanchard, Paris, 2014 (critical edition with French translation).
  9. P. Stäckel and F. Engel, Die Theorie der Parallellinien von Euklid bis auf Gauss, Teubner, Leipzig, 1895 (includes Lambert’s memoir).
  10. R. M. Corless, G. H. Gonnet, D. E. G. Hare, D. J. Jeffrey and D. E. Knuth, On the Lambert W function, Adv. Comput. Math. 5 (1996) 329–359.
  11. Lambert’s problem, Wikipedia.