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Johannes Kepler

Johannes Kepler, Imperial Mathematician to Rudolf II

A course in two strands: the life of a Württemberg schoolboy who became Imperial Mathematician through war, plague and religious exile, and the physics he built from Tycho Brahe’s observations: elliptical orbits, the area law, the harmonic law, modern optics and the first physical astronomy.

Born 27 December 1571, Weil der Stadt Died 15 November 1630, Regensburg Masterpiece Astronomia nova, 1609 Level advanced undergraduate → graduate
Kepler’s construction for Mars. The auxiliary circle fixes the eccentric anomaly E; the shaded area swept from perihelion is proportional to E − e sin E, the left side of Kepler’s equation. Here e = 0.6.
Lesson 00

How to use this course

Johannes Kepler is remembered for three laws of planetary motion. Those laws were the end of a long argument he had with the sky, recorded in books that show every false start. He was also the founder of modern optics, a pioneer of sphere packing and of the calculus of volumes, and the first astronomer to insist that planetary orbits need physical causes. This course follows that work alongside a life spent in schools, courts and exile during the Counter-Reformation and the Thirty Years’ War.

Prerequisites for Part II
  • Euclidean geometry and trigonometry, including conic sections.
  • Single-variable calculus; Lesson 07 uses Newton’s method and simple series.
  • Some Newtonian mechanics helps for the modern derivations in Lessons 06–08.
  • For Lesson 05, the basics of refraction and thin lenses.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements in modern notation, derivations and exercises with folded solutions. Kepler wrote without algebraic notation or calculus. Where the course uses modern forms, it says so, and it tries to show what Kepler himself could and could not prove.

Reference

Timeline

1571Born 27 December in Weil der Stadt, Württemberg, to Heinrich Kepler, a mercenary soldier, and Katharina Guldenmann.
1577His mother shows him the Great Comet. The family moves to Leonberg.
1584–89Protestant cloister schools at Adelberg (1584) and Maulbronn (1586).
1589Enters the Tübinger Stift; learns Copernican astronomy from Michael Maestlin.
1591Master’s degree, 11 August. Continues in theology.
1594Arrives in Graz on 11 April to teach mathematics at the Protestant school; becomes district mathematician.
1596Mysterium cosmographicum: the planetary spheres nested around the five regular solids.
1597Marries Barbara Müller on 27 April.
1600Meets Tycho Brahe at Benátky on 4 February. Expelled from Graz as a Protestant; moves to Prague in the autumn.
1601Tycho dies on 24 October; Kepler becomes Imperial Mathematician to Rudolf II.
1604Astronomiae pars optica (Ad Vitellionem paralipomena). A new star appears in October.
1606De stella nova.
1609Astronomia nova: elliptical orbits and the area law.
1610Dissertatio cum nuncio sidereo, his response to Galileo’s telescopic discoveries.
1611Dioptrice and Strena (the six-cornered snowflake). His son Friedrich and his wife Barbara die.
1612Rudolf II dies. Kepler moves to Linz as mathematician of Upper Austria.
1613Marries Susanna Reuttinger on 30 October.
1615Nova stereometria doliorum vinariorum. His mother Katharina is accused of witchcraft.
1617–21Epitome astronomiae Copernicanae in three parts.
1618Third law glimpsed on 8 March and confirmed on 15 May.
1619Harmonices mundi libri V.
1620–21Defends his mother in Württemberg; she is released on 4 October 1621.
1624Chilias logarithmorum.
1626Siege of Linz; leaves for Ulm.
1627Tabulae Rudolphinae printed at Ulm.
1628Enters the service of Albrecht von Wallenstein at Sagan in Silesia.
1630Dies 15 November in Regensburg, aged 58.
1631Pierre Gassendi observes the transit of Mercury that Kepler predicted.
1634Somnium published posthumously.
Part I · Formation
Lesson 01

Weil der Stadt and the cloister schools, 1571–1589

Johannes Kepler was born on 27 December 1571 in Weil der Stadt, a small free imperial city near Stuttgart. His grandfather Sebald had been the town’s mayor, but the family was in decline. His father Heinrich served as a mercenary and left home for the last time when Johannes was about five; he is thought to have died in the wars in the Netherlands. His mother Katharina, an innkeeper’s daughter, practised as a herbal healer. The family moved to nearby Leonberg in 1577.

Two memories from these years mattered to him later. In 1577 his mother took him up a hill to see the Great Comet, and in 1580 he was called outside to see the reddened Moon of a lunar eclipse. A childhood bout of smallpox left him with poor eyesight and weakened hands, which made him a calculator more than an observer.

Württemberg’s Lutheran dukes ran a scholarship system to train clergy, and it was Kepler’s way out of poverty. He attended the Latin school at Leonberg, then the cloister schools at Adelberg from 1584 and Maulbronn from 1586. He matriculated at Tübingen in 1587 and passed the baccalaureate examination there on 25 September 1588 while still a Maulbronn pupil. The schools taught Latin, Greek, rhetoric and theology with rigour, and Kepler later wrote harsh, self-mocking horoscopes of his own anxious and quarrelsome adolescence.

Lesson 02

Tübingen and Graz, 1589–1600

In 1589 Kepler entered the Tübinger Stift, the theological seminary of the University of Tübingen. The professor of mathematics and astronomy, Michael Maestlin, taught the geocentric system in lectures but explained Nicolaus Copernicus’s heliocentric system to able students. Kepler became a convinced Copernican, defending the system in student disputations, and argued for it on physical as well as mathematical grounds. He took his master’s degree on 11 August 1591 and began theology.

In 1594, before he finished, the Tübingen faculty recommended him for a vacancy as mathematics teacher at the Protestant seminary school in Graz, in Catholic Styria. He arrived on 11 April 1594. The post included the office of district mathematician, which meant producing annual calendars with astrological forecasts. His first calendar predicted a cold winter and a Turkish incursion, both of which came, and his local reputation rose.

The inspiration of 1595

On 19 July 1595, drawing a figure in class, Kepler noticed that a triangle inscribed between two circles fixed a ratio close to that of the orbits of Saturn and Jupiter. Plane figures did not generalize, but the five regular solids did. His first book, Mysterium cosmographicum, appeared at Tübingen in 1596 with Maestlin’s help; Lesson 04 treats its geometry. He sent copies to leading astronomers, including Tycho Brahe and Galileo.

On 27 April 1597 he married Barbara Müller, a twice-widowed miller’s daughter with a young daughter. Their first two children died in infancy. Religious pressure in Styria grew under Archduke Ferdinand. The Protestant school was closed in 1598, and on 17 July 1600 a decree ordered all Lutherans to convert or leave. Kepler refused to convert and left Graz with his family on 30 September 1600.

Lesson 03

Prague: Tycho and the Emperor, 1600–1612

Tycho Brahe, the Danish nobleman who had built the finest observatory in Europe at Uraniborg, had moved to Bohemia in 1599 as Imperial Mathematician to Rudolf II. His observations of planetary positions were accurate to about two minutes of arc, several times better than anything before. Kepler met him on 4 February 1600 at Benátky castle, north-east of Prague. Relations were tense: Tycho guarded his data, and Kepler needed both the data and a salary. Tycho set him to work on the orbit of Mars, the planet whose motion had defeated Tycho’s assistant Longomontanus.

Tycho died on 24 October 1601. Two days later Kepler was appointed his successor as Imperial Mathematician, and he kept the observations despite disputes with Tycho’s heirs. The Prague years were his most productive. He published his optics in 1604, a book on the new star that appeared in October 1604 (De stella nova, 1606), and in 1609 the Astronomia nova, whose first two laws are the subject of Lessons 06–07.

In 1610 Galileo announced the moons of Jupiter. Kepler replied in support within weeks, before he had looked through a telescope, and then observed the moons himself with a borrowed instrument. In his Narratio of 1611 he introduced the word satellite for them. His Dioptrice (1611) gave the first theory of the telescope.

The year 1611 broke the Prague household. Rudolf was forced to abdicate in favour of his brother Matthias. Kepler’s six-year-old son Friedrich died of smallpox, and his wife Barbara died shortly afterwards. Rudolf died in January 1612, and Kepler left for Linz.

Part II · The physics
Lesson 04

The cosmic mystery: polyhedra and planets

For the general reader. Why are there six planets, and why are they spaced as they are? Kepler’s first answer was that God built the solar system around the five perfect solids of Greek geometry, nested between the planetary spheres. The answer is wrong, but the question was new. Kepler asked for the cause of the planetary distances, and he asked that the cause be checked against measurement.

Goals
  • Prove that there are exactly five convex regular polyhedra.
  • Compute the ratio of inscribed to circumscribed radius for each.
  • Compare Kepler’s model with Copernican distances, and see his first physical idea about the Sun.

The five solids

Theorem (Euclid, Elements XIII)

A convex polyhedron whose faces are congruent regular \(p\)-gons, with \(q\) faces at each vertex, exists exactly for \((p,q)\in\{(3,3),(4,3),(3,4),(5,3),(3,5)\}\): the tetrahedron, cube, octahedron, dodecahedron and icosahedron.

Proof. Each interior angle of a regular \(p\)-gon is \(\pi(1-2/p)\). At a convex vertex the face angles must total less than \(2\pi\), so \(q\pi(1-2/p)<2\pi\), equivalently \((p-2)(q-2)<4\) with \(p,q\ge3\). The solutions are the five pairs listed, and each is realized. ∎

The nesting

Kepler placed the solids, from the inside out, in the order octahedron, icosahedron, dodecahedron, tetrahedron, cube. Each solid is inscribed in the sphere of the planet outside it and circumscribes the sphere of the planet inside it:

\[ \text{Mercury}\ \big|\ \text{octahedron}\ \big|\ \text{Venus}\ \big|\ \text{icosahedron}\ \big|\ \text{Earth}\ \big|\ \text{dodecahedron}\ \big|\ \text{Mars}\ \big|\ \text{tetrahedron}\ \big|\ \text{Jupiter}\ \big|\ \text{cube}\ \big|\ \text{Saturn}. \]

The model therefore predicts that each ratio of neighbouring planetary distances equals \(r/R\), the ratio of inradius to circumradius of the solid between them.

Proposition

The ratios \(r/R\) are: tetrahedron \(\tfrac13\); cube and octahedron \(\tfrac1{\sqrt3}\approx0.577\); dodecahedron and icosahedron \(\sqrt{\tfrac{5+2\sqrt5}{15}}\approx0.795\). Dual solids have equal ratios.

For the cube with vertices \((\pm1,\pm1,\pm1)\), \(R=\sqrt3\) and \(r=1\). Duality exchanges face centres and vertices: the polar of a solid with radii \(r,R\) about the unit sphere has radii \(1/R,1/r\), with the same ratio. For the icosahedron with vertices at cyclic permutations of \((0,\pm1,\pm\varphi)\), \(\varphi=\tfrac{1+\sqrt5}2\), we have \(R^2=1+\varphi^2\), and the inradius is the distance from the origin to the centroid of a face, which gives the stated value.

Copernican distances fit only roughly. Kepler gave each sphere a thickness to contain the planet’s eccentric orbit and made a special adjustment for Mercury, but discrepancies remained. He blamed Copernicus’s numbers, and the desire for better data led him to Tycho. In a second edition of 1621 he kept the model, now subordinated to the harmonies of Lesson 08.

The first physical idea

Chapter 20 of the Mysterium contains the germ of celestial physics. Outer planets move more slowly, and Kepler proposed that a moving power, issuing from the Sun and weakening with distance, drives them. He tested a relation between periods and distances and found it did not fit. The search for the right relation took him another 22 years.

Exercise 4.1Show that the inradius of a regular tetrahedron is one third of its circumradius.

Show solution
Take vertices \((1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)\), so \(R=\sqrt3\). The face through the last three vertices lies in the plane \(x+y+z=-1\), at distance \(1/\sqrt3\) from the origin. So \(r/R=(1/\sqrt3)/\sqrt3=\tfrac13\).

Exercise 4.2Modern mean distances (in AU) are Mars 1.524, Jupiter 5.203, Saturn 9.537. Compare the ratios Mars/Jupiter and Jupiter/Saturn with Kepler’s predictions.

Show solution
Mars/Jupiter \(=0.293\) against the tetrahedron’s \(0.333\); Jupiter/Saturn \(=0.546\) against the cube’s \(0.577\). Both are within about 12 percent. The pattern is suggestive but not exact, which is what Kepler found with Copernicus’s values.

Exercise 4.3Use the vertex condition to find the regular tilings of the plane.

Show solution
Now the angles must sum to exactly \(2\pi\): \((p-2)(q-2)=4\). The solutions are \((3,6),(4,4),(6,3)\): triangles, squares, hexagons. Kepler classified tilings in Book II of Harmonices mundi.
Lesson 05

Optics: light, the eye and the telescope

For the general reader. To measure the sky one must understand the instrument, and the first instrument is the eye. Kepler’s optics of 1604 explained how a pinhole forms an image, showed that the eye focuses an inverted picture on the retina, and stated that light weakens with the square of distance. In 1611 he explained how lenses combine in a telescope and designed the form that astronomers still use.

Goals
  • Understand the ray model of the pinhole camera and the eye.
  • Derive the inverse-square law for illumination.
  • Compute the magnification of the Keplerian telescope.

The pinhole and the eclipse

Observers of solar eclipses through a pinhole saw the Moon looking smaller than it should. Kepler, puzzled by the eclipse of July 1600, reasoned by tracing rays with threads. Each point of the source sends a pencil of rays through the aperture and makes a small copy of the aperture on the screen. The image is the superposition of these copies: the source shape blurred by the aperture shape. An aperture of finite size enlarges a bright image and shrinks a dark intruder such as the Moon’s disc. This settled the eclipse paradox.

The retinal image

Medieval optics placed vision in the crystalline lens. Kepler, using anatomical drawings of Felix Platter, argued that the lens refracts the cone of rays from each object point to a point on the retina, so the eye forms a real, inverted image there, like a camera. How an inverted picture is seen upright he left to the natural philosophers, a clear division between optics and perception.

Inverse-square law (Kepler, 1604)

Light from a point source spreads over spheres centred on the source. The illumination \(I\) at distance \(d\) satisfies \(I(d)\,4\pi d^2=\text{const}\), so \(I\propto d^{-2}\).

Kepler argued that the same quantity of light passes through every sphere about the source, and the area of a sphere grows as the square of its radius. He did not apply the square law to the Sun’s moving force, which he took to weaken as \(1/d\) because it acted only in the plane of the orbits.

The Dioptrice

Kepler did not know the sine law of refraction, found later by Willebrord Snell and René Descartes. He used the small-angle rule that in glass the angle of refraction is about two thirds of the angle of incidence, equivalent to an index of 1.5. With this he worked out the foci of convex and concave lenses, explained real and virtual images, described total internal reflection, and proposed a telescope of two convex lenses. In thin-lens form, with object and image distances \(u,v\) and focal length \(f\),

\[ \frac1u+\frac1v=\frac1f, \qquad m_{\text{angular}}=\frac{f_{\text{obj}}}{f_{\text{eye}}}\ \text{ for a telescope focused at infinity.} \]

The Keplerian telescope gives an inverted image but a wider field than Galileo’s design, and it allows a crosshair or micrometer at the common focus. It became the standard astronomical instrument by the mid-seventeenth century.

Exercise 5.1Mars is at about 1.52 AU from the Sun and Jupiter at 5.20 AU. How much weaker is sunlight at Jupiter than at Mars?

Show solution
\((5.20/1.52)^2=(3.42)^2\approx11.7\). Sunlight at Jupiter is about one twelfth as intense as at Mars.

Exercise 5.2A Keplerian telescope has an objective of focal length 120 cm and an eyepiece of 4 cm. Find the tube length and the magnification. Why is the image inverted?

Show solution
For a distant object the two foci coincide, so the tube is about \(120+4=124\) cm. The magnification is \(120/4=30\). The objective forms a real inverted image at its focus, and the eyepiece acts as a magnifier on it without re-inverting it.

Exercise 5.3Show that Kepler’s rule “refraction angle \(\approx\tfrac23\) of incidence” agrees with Snell’s law \(\sin i=n\sin r\), \(n=1.5\), for small angles. Where does it fail?

Show solution
For small angles \(\sin i\approx i\), \(\sin r\approx r\), so \(r\approx i/n=\tfrac23 i\). The rule fails at large angles; at grazing incidence \(i=90^\circ\), Snell gives \(r=\arcsin(2/3)\approx41.8^\circ\), not \(60^\circ\). Running the rule backwards from glass to air also misses the critical angle.
Lesson 06

The war on Mars and the area law

For the general reader. Since antiquity, planets were assumed to move on circles at uniform speed, viewed from a suitable point. Kepler, working on Tycho’s observations of Mars, found that the best such model still missed by eight minutes of arc, a quarter of the Moon’s width. He refused to accept the error. Instead he looked for a physical rule, and found that a line from the Sun to a planet sweeps out equal areas in equal times.

Goals
  • Understand the equant model and Kepler’s “vicarious hypothesis”.
  • Follow Kepler’s route from a distance law to the area law.
  • Derive the area law from a central force.

The equant and the eight minutes

In Ptolemy’s scheme a planet moves on a circle whose centre \(C\) is offset from the observer, and it moves uniformly in angle as seen from a third point, the equant \(Q\). Ptolemy placed \(C\) midway between the observer and \(Q\), which is called bisecting the eccentricity. Kepler, working with the true Sun and not the mean Sun of Copernicus and Tycho, first freed the ratio and fitted the positions of \(C\) and \(Q\) to four of Tycho’s oppositions of Mars. This vicarious hypothesis reproduced Tycho’s longitudes at opposition to within about two minutes, but it predicted wrong distances. When he forced the physically natural bisected ratio, the errors in longitude rose to about eight minutes elsewhere in the orbit. Since Tycho’s data were good to about two minutes, Kepler concluded that the circle itself had to go.

Earth first

To get distances from the Sun to Mars, Kepler needed Earth’s orbit. He used observations of Mars made one Martian period apart, 687 days, when Mars is at the same point of its orbit: the changing direction to it from Earth then traces Earth’s orbit by triangulation. He found that Earth’s orbit also needs a bisected eccentricity, and that Earth’s speed varies inversely with its distance from the Sun, at least near the apsides.

From the distance law to the area law

Kepler’s physical idea was that the Sun pushes the planets with a force that weakens as \(1/r\), so speed is inversely proportional to distance. Summing distances along the orbit to compute the time was laborious. As a short cut he replaced the sum of distances by the area swept out, and he found the area version predicted positions well. In modern form:

Second law (Kepler, 1609)

The line from the Sun to a planet sweeps out equal areas in equal times: \(\dfrac{dA}{dt}=\tfrac12 r^2\dot\theta\) is constant.

Kepler reached this law first, around 1602, before he had the ellipse. His distance law is correct only at perihelion and aphelion, where the velocity is perpendicular to the radius. The area law is exact, and Isaac Newton showed why.

Newton’s derivation

If the force on the planet points toward the Sun, then the torque about the Sun vanishes and the angular momentum \(\mathbf L=m\,\mathbf r\times\dot{\mathbf r}\) is constant. Its magnitude is \(mr^2\dot\theta=2m\,dA/dt\). So the area law holds for every central force, whatever its dependence on distance. Newton’s Principia proved this geometrically in Book I, Proposition 1.

Exercise 6.1Show from the area law that the speeds at perihelion and aphelion satisfy \(v_p/v_a=r_a/r_p=(1+e)/(1-e)\).

Show solution
At the apsides the velocity is perpendicular to the radius, so \(r^2\dot\theta=rv\). Conservation gives \(r_pv_p=r_av_a\). With \(r_p=a(1-e)\), \(r_a=a(1+e)\), the ratio is \((1+e)/(1-e)\). This is exactly Kepler’s distance law at those two points.

Exercise 6.2Mars has \(e\approx0.093\). By what factor does its orbital speed change between perihelion and aphelion? Earth has \(e\approx0.017\); repeat.

Show solution
Mars: \(1.093/0.907\approx1.21\), a 21 percent change. Earth: \(1.017/0.983\approx1.035\). The large eccentricity of Mars is why it, of all the planets, exposed the failure of circles.

Exercise 6.3To first order in \(e\), the true anomaly on a Kepler orbit is \(\nu\approx M+2e\sin M\). Show that, seen from the empty focus at distance \(2ae\) from the Sun, the planet moves uniformly to first order in \(e\).

Show solution
Moving the viewpoint a distance \(2ae\) along the apse line, away from perihelion, changes the apparent direction of a planet at distance \(r\approx a\) by about \(-(2ae\sin\nu)/a=-2e\sin\nu\). So the angle seen from the empty focus is \(\nu-2e\sin\nu\approx M+2e\sin M-2e\sin M=M+O(e^2)\), which grows uniformly in time. The empty focus is a first-order equant with the centre midway between it and the Sun: Ptolemy’s bisected eccentricity. This is why the equant worked so well, and why only Tycho’s accuracy could expose the second-order error.
Lesson 07

The ellipse and Kepler’s equation

For the general reader. With the area law in hand, Kepler still needed the shape of the orbit. He tried egg-shaped ovals for years. In 1605 he found that Mars moves on an ellipse with the Sun at one focus. Combining the ellipse with the area law gives an equation for where a planet is at a given time. That equation has no solution in closed form, and it has occupied mathematicians ever since.

Goals
  • Understand the numerical clue that led to the ellipse.
  • Derive Kepler’s equation from the auxiliary circle.
  • Solve it by iteration and by Newton’s method.

The clue

Kepler found that the orbit of Mars falls inside the circle on the line of apsides, leaving a crescent whose greatest width is about 0.00429 of the radius. The largest angle between the directions of Mars seen from the Sun and from the orbit’s centre, the “optical equation”, was about \(5^\circ18'\), and \(\sec 5^\circ18'\approx1.00429\). The coincidence led him to a distance formula that, after one more false turn, gave the ellipse. Chapter 58 of the Astronomia nova describes the moment.

First law (Kepler, 1609)

Each planet moves on an ellipse with the Sun at one focus. With semi-major axis \(a\), eccentricity \(e\) and true anomaly \(\nu\) measured from perihelion,

\[ r=\frac{a(1-e^2)}{1+e\cos\nu}. \]

The auxiliary circle

Draw the circle of radius \(a\) about the centre of the ellipse. Through the planet \(P\) draw a perpendicular to the major axis, meeting the circle at \(Q\). The angle \(E\) at the centre from perihelion to \(Q\) is the eccentric anomaly. The ellipse is the circle compressed by \(b/a\), with \(b=a\sqrt{1-e^2}\), so

\[ x=a(\cos E-e),\quad y=b\sin E,\qquad r=a(1-e\cos E),\qquad \tan\frac{\nu}{2}=\sqrt{\frac{1+e}{1-e}}\,\tan\frac{E}{2}, \]

with coordinates measured from the focus.

Kepler’s equation

In the circle, the area swept from the focus between perihelion and \(Q\) is the circular sector at the centre, \(\tfrac12a^2E\), minus the triangle with the centre and the focus, \(\tfrac12a\cdot ae\sin E\). Compressing by \(b/a\) gives the area swept on the ellipse, \(\tfrac12ab(E-e\sin E)\). By the area law this grows uniformly in time, reaching \(\pi ab\) after one period \(T\). Define the mean anomaly \(M=2\pi(t-t_p)/T\).

Kepler’s equation \[ M = E - e\sin E. \]

For \(0\le e<1\) the map \(E\mapsto E-e\sin E\) is strictly increasing, so each \(M\) determines a unique \(E\).

Kepler could not invert the equation and said so, inviting geometers to try. Two methods work well. The fixed-point iteration \(E_{k+1}=M+e\sin E_k\) converges because the map is a contraction with constant \(e\). Newton’s method,

\[ E_{k+1}=E_k-\frac{E_k-e\sin E_k-M}{1-e\cos E_k}, \]

converges quadratically. Joseph-Louis Lagrange gave a power series in \(e\), and Friedrich Bessel expanded the solution in the functions that now bear his name: \(E=M+\sum_{n\ge1}\frac{2}{n}J_n(ne)\sin nM\).

Exercise 7.1Solve \(M=E-e\sin E\) for \(e=0.1\), \(M=1\) rad by three fixed-point iterations starting from \(E_0=M\).

Show solution
\(E_1=1+0.1\sin1=1.08415\). \(E_2=1+0.1\sin1.08415=1.08839\). \(E_3=1+0.1\sin1.08839=1.08859\). The solution is \(E\approx1.0886\); the error falls by about a factor \(e\cos E\approx0.05\) each step.

Exercise 7.2Derive the polar equation \(r=a(1-e^2)/(1+e\cos\nu)\) from the focal definition of the ellipse, \(r+r'=2a\).

Show solution
Put the focus at the origin and the other focus at \((-2ae,0)\). Then \(r'^2=r^2+4a^2e^2+4aer\cos\nu\). Substituting \(r'=2a-r\) gives \(4a^2-4ar=4a^2e^2+4aer\cos\nu\), so \(r(1+e\cos\nu)=a(1-e^2)\).

Exercise 7.3Using \(r=a(1-e\cos E)\), show that the time average of \(r\) over one orbit is \(a(1+e^2/2)\).

Show solution
Time is uniform in \(M\), and \(dM=(1-e\cos E)\,dE\). So \(\langle r\rangle=\frac{1}{2\pi}\int_0^{2\pi}a(1-e\cos E)^2\,dE=a\left(1+\tfrac{e^2}{2}\right)\), since \(\langle\cos E\rangle=0\) and \(\langle\cos^2E\rangle=\tfrac12\).
Lesson 08

The harmony of the world and the third law

For the general reader. Kepler believed the world was built on harmony, in the musical sense. In Harmonices mundi (1619) he looked for musical ratios in geometry, astrology and the planets, and found them in the planets’ speeds. In the same book, almost in passing, he stated the law that links every planet’s period to its distance from the Sun. That law later led Newton to the inverse-square law of gravity.

Goals
  • See Kepler’s classification of polyhedra and tilings in Book II.
  • Understand the planetary harmonies of Book V.
  • State the third law, derive it from Newton’s laws, and apply it.

Book II: congruence of figures

Kepler studied how regular polygons fit together in the plane and in space. He identified the eleven tilings of the plane by regular polygons with all vertices alike, described the thirteen Archimedean solids as a complete family, and introduced two new regular star polyhedra, the small and great stellated dodecahedra. Louis Poinsot later added two more, and the four are now the Kepler–Poinsot polyhedra.

Book V: the harmonies

Kepler compared each planet’s angular speed, as seen from the Sun, at aphelion and perihelion. For Saturn the ratio is close to \(4:5\), a major third; for Jupiter, close to \(5:6\), a minor third; for Earth about \(15:16\), a semitone. Ratios between extreme speeds of neighbouring planets also give consonances. Kepler imagined each planet singing a range of notes and the whole system sounding chords at rare moments. The theory has not survived, but it made him compare periods and distances across all planets.

Third law (Kepler, 1619)

The squares of the periods of the planets are proportional to the cubes of their mean distances from the Sun: \(T^2\propto a^3\), with the same constant for all planets.

Kepler first saw the relation on 8 March 1618, rejected it after a computing error, and confirmed it on 15 May. He stated it as a ratio of periods equal to the ratio of distances raised to the power \(\tfrac32\). In the Epitome he showed it also held for the four moons of Jupiter, using observations by Simon Marius, with its own constant.

Newton’s form

For a circular orbit under gravity, \(GMm/a^2=m(2\pi/T)^2a\), so

\[ T^2=\frac{4\pi^2}{G(M+m)}\,a^3 \]

after including the motion of the Sun about the centre of mass (two-body problem). The same holds for ellipses with \(a\) the semi-major axis. Read backwards, the third law for circles forces the force to fall as \(1/a^2\), which is how Newton, Robert Hooke, Christopher Wren and Edmond Halley each saw the inverse-square law around 1684.

Exercise 8.1Mars’s sidereal period is 1.881 years. Find its semi-major axis in AU.

Show solution
\(a=T^{2/3}=1.881^{2/3}\). Since \(\ln1.881=0.632\), \(a=e^{0.421}\approx1.524\) AU.

Exercise 8.2Io orbits Jupiter in 1.769 days at 421,700 km. Callisto is at 1,882,700 km. Predict Callisto’s period.

Show solution
\(T=1.769\times(1882.7/421.7)^{3/2}=1.769\times4.465^{1.5}\approx1.769\times9.43\approx16.7\) days. The observed value is 16.69 days.

Exercise 8.3Show that if \(T^2\propto a^3\) for circular orbits under a central force \(F(a)\), then \(F\propto a^{-2}\).

Show solution
Circular motion needs \(F=m\omega^2a=4\pi^2ma/T^2\). With \(T^2=ka^3\), \(F=4\pi^2m/(ka^2)\).
Lesson 09

Snowflakes, packings and wine barrels

For the general reader. Two small books show Kepler as a pure mathematician. A New Year’s gift to a patron in 1611 asked why snowflakes have six points and how equal spheres can be packed most tightly. His answer about packing became a conjecture that was proved only in this century. A dispute with a wine merchant in 1613 led him to compute the volumes of curved solids by slicing them into infinitely many thin pieces, a step toward integral calculus.

Goals
  • State the Kepler conjecture and compute the density of the face-centred cubic packing.
  • Use Kepler’s method of infinitesimal slices for areas and volumes.
  • Solve the barrel problem as an early maximization.

The six-cornered snowflake (1611)

In Strena seu de nive sexangula Kepler compared snowflakes with honeycombs and pomegranate seeds. Spheres packed in square layers, each layer sitting in the hollows of the one below, give an arrangement where each sphere touches twelve others. Squeezed together, the spheres become rhombic dodecahedra, as pomegranate seeds do. Kepler asserted that this packing is the tightest possible.

Kepler conjecture (Hales, 1998; formal proof 2014)

No packing of equal balls in \(\mathbb R^3\) has upper density greater than \(\dfrac{\pi}{\sqrt{18}}\approx0.7405\), the density of the face-centred cubic packing.

David Hilbert included the problem in his eighteenth problem in 1900. Thomas Hales, partly with Samuel Ferguson, announced a computer-assisted proof in 1998, published in the Annals of Mathematics in 2005. Doubts about checking the computations led Hales to organize the Flyspeck project, which completed a formal machine-verified proof in 2014.

The new stereometry of wine barrels (1615)

In Linz, buying wine for his second wedding, Kepler saw the merchant measure every barrel by pushing a rod through the bung hole to the far edge of the bottom, whatever the barrel’s shape. Kepler set out to see whether this could be fair. He first recomputed Archimedes’ results by treating a circle as infinitely many thin triangles with apex at the centre, so its area is half the radius times the circumference. He then found volumes of many solids of revolution, which he named after fruit (apple, lemon, olive), by slicing.

Example: the torus

A disc of radius \(r\) revolved about an axis at distance \(R\ge r\) from its centre sweeps a solid of volume \(V=\pi r^2\cdot2\pi R=2\pi^2Rr^2\).

Kepler cut the ring into thin wedges by planes through the axis. Each wedge is thicker on the outside than the inside, but the excess and the deficit cancel about the centre circle, so each acts like a cylinder of length equal to the arc of the centre. This anticipates the centroid theorem of Pappus and Guldin. Kepler also noted that near a maximum a quantity changes very little, an observation that Pierre de Fermat later made into a method.

Exercise 9.1Compute the density of the face-centred cubic packing.

Show solution
The cubic unit cell of side \(s\) contains 4 balls (8 corners × ⅛ + 6 faces × ½). Balls touch along a face diagonal, so \(4\rho=s\sqrt2\), \(\rho=s/(2\sqrt2)\). Density \(=4\cdot\tfrac43\pi\rho^3/s^3=\frac{16\pi}{3}\cdot\frac{1}{16\sqrt2}=\frac{\pi}{3\sqrt2}=\frac{\pi}{\sqrt{18}}\).

Exercise 9.2Model a barrel as a cylinder of radius \(\rho\) and height \(h\). The gauging rod runs from the bung hole, at the middle of the side, to the opposite edge of the base, with length \(d\). For fixed \(d\), which cylinder has the largest volume?

Show solution
\(d^2=(2\rho)^2+(h/2)^2\), so \(\rho^2=\tfrac14(d^2-h^2/4)\) and \(V=\pi\rho^2h=\tfrac{\pi}{4}(d^2h-h^3/4)\). Setting \(dV/dh=0\) gives \(h=2d/\sqrt3\), and then \(2\rho=2d/\sqrt6\). The best ratio is \(h/(2\rho)=\sqrt2\). Kepler found that Austrian barrels were close to the maximizing shape, so small departures changed the volume little and the rod gauge was nearly fair.

Exercise 9.3Use Kepler’s triangle decomposition to explain why a circle of radius \(r\) has area \(\pi r^2\), given that its circumference is \(2\pi r\).

Show solution
Divide the circle into \(n\) thin sectors. Each is nearly a triangle of height \(r\) and base \(2\pi r/n\), area \(\tfrac12 r\cdot2\pi r/n\). Summing \(n\) of them gives \(\pi r^2\), exactly in the limit.
Lesson 10

Logarithms and the Rudolphine Tables

For the general reader. An astronomical theory is useful only if it predicts. Kepler’s duty as Imperial Mathematician was to turn Tycho’s observations into tables from which anyone could compute planetary positions. The tables took him 26 years. Along the way he adopted John Napier’s new logarithms, rebuilt their theory and used them throughout. The tables were far more accurate than any before, and they predicted transits of Mercury and Venus across the Sun.

Goals
  • See why logarithms mattered for astronomical computation.
  • Use the third law and synodic periods in logarithmic form.
  • Understand how the tables predicted the 1631 transit of Mercury.

Logarithms

John Napier published his logarithms in 1614. Kepler saw the book in 1617 and grasped its value in 1618, the year of the third law. Distrusting Napier’s kinematic definition, he gave his own derivation from the theory of proportions, published as Chilias logarithmorum (1624) with a supplement in 1625. The key property, in modern form, is

\[ \log(xy)=\log x+\log y,\qquad \log x^{p}=p\log x, \]

which turns multiplication into addition and roots into division. The third law becomes linear: \(\log T=\tfrac32\log a+\text{const}\).

The tables

The Tabulae Rudolphinae, named for Rudolf II, were finished in 1623 but delayed by war and money. Kepler printed 1,000 copies at his own expense at Ulm in 1627. They contained Tycho’s star catalogue, the planetary theory based on ellipses and the area law, and logarithm tables. They remained the standard for about a century.

Synodic period

If an inner planet has sidereal period \(P\) and Earth has period \(P_\oplus\), the time \(S\) between successive inferior conjunctions satisfies

\[ \frac1S=\frac1P-\frac1{P_\oplus}. \]

A transit occurs when an inferior conjunction happens near a node, where the planet’s orbit crosses the ecliptic. Using his tables, Kepler predicted in 1629 a transit of Mercury on 7 November 1631 and one of Venus a month later. Pierre Gassendi saw Mercury cross the Sun from Paris on 7 November 1631, within about ten arc-minutes of Kepler’s position; older tables had been off by degrees. The Venus transit happened at night in Europe. In 1639 Jeremiah Horrocks, correcting the Rudolphine Tables, predicted and observed a transit of Venus that Kepler had missed.

Exercise 10.1Mercury’s sidereal period is 0.2408 years. Find its synodic period.

Show solution
\(1/S=1/0.2408-1=4.153-1=3.153\), so \(S=0.317\) years, about 116 days.

Exercise 10.2Using \(\log_{10}9.537=0.9794\), compute Saturn’s period in years with logarithms only.

Show solution
\(\log T=1.5\times0.9794=1.4691\), so \(T=10^{1.4691}=10\times10^{0.4691}\approx10\times2.945=29.45\) years.

Exercise 10.3Venus has sidereal period 0.6152 years. Show that 8 Earth years are very nearly 5 Venus synodic periods, and explain why transits of Venus come in pairs eight years apart.

Show solution
\(1/S=1/0.6152-1=0.6255\), so \(S=1.599\) years and \(5S=7.99\) years. After eight years Venus and Earth return almost to the same configuration, so a conjunction near a node recurs once, slightly shifted. The shift then takes it away from the node for over a century.
Part III · Life and legacy
Lesson 11

Linz, the witch trial and the war, 1612–1630

In 1612 Kepler became mathematician to the estates of Upper Austria at Linz, keeping his title of Imperial Mathematician under Matthias and then Ferdinand II. On 30 October 1613 he married Susanna Reuttinger, an orphan raised in a noble household; they had several children, some of whom died young. He chose her methodically after considering eleven candidates, and the marriage proved happy. In Linz the Lutheran pastor barred him from communion because he would not accept every article of the Formula of Concord. Kepler held to his own Lutheranism for life while refusing to condemn Calvinists.

In 1615 his mother Katharina was accused of witchcraft in Leonberg, in the middle of a wave of trials in Württemberg. Kepler took charge of her defence. She was arrested in August 1620 and held for over a year, shown the instruments of torture, and refused to confess. Kepler’s legal writing helped secure her release on 4 October 1621. She died about six months later.

The Linz years produced the Epitome astronomiae Copernicanae (1617–1621), the first textbook of astronomy built on the three laws, and the Harmonices mundi (1619). The first part of the Epitome was placed on the Catholic Index of prohibited books in 1619. In 1626 Linz was besieged during a peasants’ revolt, and the press printing the Rudolphine Tables was damaged. Kepler left for Ulm, finished the tables there, and in 1628 entered the service of the imperial general Albrecht von Wallenstein at Sagan in Silesia, partly in hope of the unpaid imperial salary. He set up a printing press there in 1629.

In October 1630 he set out for Regensburg, where the imperial diet was meeting, to recover money owed him. He fell ill soon after arriving and died there on 15 November 1630. He was buried in a Protestant churchyard outside the city, and the grave was lost when the churchyard was destroyed during the war. He had written his own epitaph: having measured the heavens, he now measured the shadows of the Earth.

DiscussionKepler moved because of religious expulsion (Graz), politics (Prague), war (Linz) and debt (Sagan). Choose one move and describe how it changed what he could work on.

Lesson 12

From Kepler to Newton and beyond

Kepler’s laws were not accepted at once. Many astronomers used his tables without his ellipses, and Galileo never adopted them. The tables’ success, confirmed by the 1631 transit, persuaded the computers first. Jeremiah Horrocks applied elliptical theory to the Moon. Ismaël Boulliau, who first suggested an inverse-square law for a solar force, kept ellipses but replaced the area law by a geometric substitute.

Isaac Newton’s Principia (1687) turned the three laws into consequences of the laws of motion and universal gravitation. The area law follows from any central force, the ellipse from the inverse square, and the third law fixes the constant. Newton also showed where Kepler’s laws fail: the planets disturb each other, and the Sun moves. Kepler’s demand for a physical cause, set out in the subtitle of the Astronomia nova, “a celestial physics”, was the programme that Newton completed.

Legacy

  • Astronomy: the three laws, the first accurate planetary tables, and the methods of orbit determination that run through Gauss to modern celestial mechanics. Kepler’s equation is solved billions of times a day in navigation software.
  • Optics: the retinal image, the inverse-square law of illumination, and the astronomical telescope.
  • Mathematics: the Kepler–Poinsot polyhedra, the classification of Archimedean tilings, infinitesimal volume methods, and the sphere-packing conjecture settled by Hales.
  • Literature: the Somnium, a dream of a journey to the Moon that describes how the Earth would look from there, written around 1609, annotated in 1621 and printed in 1634. It is often called the first work of science fiction.
  • Honours: the NASA Kepler space telescope (2009–2018), which found thousands of exoplanets by transit photometry, and many craters, an asteroid and a supernova remnant (SN 1604) bear his name.

Final projectTake ten positions of Mars from a modern ephemeris spread over two years. Fit first a circle with a bisected equant and then a Kepler ellipse, and report the largest error of each in arc-minutes. Compare with Kepler’s eight minutes.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Johannes Kepler, MacTutor History of Mathematics.
  2. O. Gingerich, Kepler, Johannes, Dictionary of Scientific Biography, vol. 7 (1973).
  3. Johannes Kepler, Wikipedia (with extensive references).
  4. M. Caspar, Kepler, trans. C. D. Hellman, Dover, 1993.
  5. J. Connor, Kepler’s Witch, HarperCollins, 2004.
  6. U. Rublack, The Astronomer and the Witch, Oxford University Press, 2015.

Physics and mathematics

  1. J. Kepler, Mysterium cosmographicum, Tübingen, 1596; trans. A. M. Duncan, The Secret of the Universe, Abaris, 1981.
  2. J. Kepler, Astronomia nova, Heidelberg, 1609; trans. W. H. Donahue, Green Lion Press, 2015.
  3. J. Kepler, Optics: Paralipomena to Witelo and Optical Part of Astronomy, trans. W. H. Donahue, Green Lion Press, 2000.
  4. J. Kepler, The Harmony of the World, trans. E. J. Aiton, A. M. Duncan, J. V. Field, American Philosophical Society, 1997.
  5. J. Kepler, The Six-Cornered Snowflake, trans. C. Hardie, Clarendon Press, 1966.
  6. Epitome Astronomiae Copernicanae, Wikipedia.
  7. C. Wilson, Kepler’s derivation of the elliptical path, Isis 59 (1968).
  8. J. V. Field, Kepler’s Geometrical Cosmology, University of Chicago Press, 1988.
  9. P. Colwell, Solving Kepler’s Equation over Three Centuries, Willmann-Bell, 1993.
  10. T. C. Hales, A proof of the Kepler conjecture, Ann. of Math. 162 (2005).
  11. T. Hales et al., A formal proof of the Kepler conjecture, Forum Math. Pi 5 (2017), arXiv:1501.02155.