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Mary Somerville

Mary Somerville, born Mary Fairfax

A course in two strands: the life of the Scottish admiral’s daughter who taught herself mathematics in secret and became the most widely read scientific writer of her age, and the physics she explained, from Laplace’s celestial mechanics and the tides to the unseen planet beyond Uranus.

Born 26 December 1780, Jedburgh Died 29 November 1872, Naples Mechanism of the Heavens 1831 Level advanced undergraduate
Uranus and Neptune, 1782–1846, at eight-year steps from their mean motions. Before the conjunction of about 1822 Neptune pulled Uranus ahead; afterwards it held it back. In 1842 Somerville wrote that such irregularities might reveal an unseen planet.
Lesson 00

How to use this course

Mary Somerville made no great discovery, and she did not claim one. Her achievement was of another kind. She mastered the hardest mathematical physics of her time, Pierre-Simon Laplace’s Traité de mécanique céleste, without a teacher or a university, and explained it in English to a generation of readers. Her four books ran to many editions. One of them gave the occasion for the word “scientist”, and another pointed to the possibility of a planet beyond Uranus a few years before Neptune was found.

Part II follows the content of her books, especially The Mechanism of the Heavens (1831) and On the Connexion of the Physical Sciences (1834). It teaches the physics she explained: the gravitational potential and Laplace’s equation, the shape of the Earth, the tides, the long-term stability of planetary orbits, the perturbations of Uranus, the invisible parts of the solar spectrum and the physics of the atmosphere.

Prerequisites for Part II
  • Vector calculus: gradient, divergence and the Laplacian.
  • Newtonian gravitation and Kepler’s laws.
  • Basic ordinary differential equations and Legendre polynomials (introduced in Lesson 04).
  • For Lesson 10, the ideal gas law.

Each technical lesson opens with an italic paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Companion modules cover her friend Caroline Herschel, who was elected to the Royal Astronomical Society with her, and William Herschel, whose discovery of infrared light figures in Lesson 09.

Reference

Timeline

1780Born 26 December at the manse in Jedburgh, daughter of William George Fairfax, a naval officer, and Margaret Charters. Grows up in Burntisland, Fife.
c. 1790A year at a boarding school in Musselburgh, her only formal schooling of any length.
c. 1794Hears the painter Alexander Nasmyth say that Euclid underlies perspective and astronomy; begins to study geometry and algebra in secret.
1804Marries Samuel Greig, an officer in the Russian navy, and moves to London. Sons born 1805 and 1806.
1807Greig dies; she returns to Scotland and studies mathematics openly.
1811Silver medal for the solution of a problem in the Mathematical Repository.
1812Marries her cousin William Somerville, an army doctor.
1816Moves to London; William joins the Army Medical Board.
1817Visits Paris and meets Laplace, Arago, Biot and Poisson.
1826Her first paper, on the magnetizing power of violet light, is published by the Royal Society.
1827Henry Brougham asks her to render Laplace’s Mécanique céleste into English.
1831The Mechanism of the Heavens.
1834On the Connexion of the Physical Sciences; William Whewell’s review proposes the word “scientist”. Honorary member of the Royal Irish Academy.
1835Honorary member of the Royal Astronomical Society, with Caroline Herschel. Civil list pension of £200 a year, raised to £300 in 1837.
1838Leaves for Italy, where she spends the rest of her life.
1842Sixth edition of the Connexion suggests that Uranus may be perturbed by an unseen planet. Neptune is found in 1846.
1848Physical Geography.
1860Death of William Somerville in Florence.
1869On Molecular and Microscopic Science. Patron’s Gold Medal of the Royal Geographical Society.
1872Dies 29 November in Naples, aged 91.
1873Personal Recollections, edited by her daughter Martha.
1879Somerville Hall, later Somerville College, founded in Oxford and named after her.
Part I · Formation
Lesson 01

Burntisland, 1780–1797

Mary Fairfax was born on 26 December 1780 at the manse in Jedburgh, the home of her mother’s sister and her husband, the Reverend Thomas Somerville. Her father, William George Fairfax, was a naval officer often at sea; he was knighted after the battle of Camperdown in 1797 and rose to vice-admiral. Her mother, Margaret Charters, brought up the children in Burntisland, a small port on the Fife shore of the Firth of Forth. Mary ran free on the beach, collecting shells and watching the birds and the tides.

When her father returned from a voyage and found that, at about ten, she could barely read, he sent her to a boarding school in Musselburgh for a year. She hated it and learned little beyond reading. At home again she read everything she could find, taught herself enough Latin to read Caesar, and in Jedburgh studied with her uncle Thomas Somerville, who encouraged her.

Euclid by accident

At about thirteen she came upon algebraic symbols in a puzzle page of a ladies’ fashion magazine, and wanted to know what they meant. Soon after, in Edinburgh, she attended the painting classes of Alexander Nasmyth. She heard him tell another pupil that Euclid’s Elements were the foundation of perspective, of astronomy and of all mechanical science. She persuaded her younger brother’s tutor to buy her a copy of Euclid and an algebra book. Her family disapproved, and the servants were told to take away her candles. She worked through the first six books of Euclid at night, from memory, in the dark.

Lesson 02

Two marriages and a library, 1797–1816

In 1804 Mary Fairfax married her cousin Samuel Greig, a captain in the Russian navy posted to London, and went to live there. Greig had a low opinion of learned women and gave her no encouragement. They had two sons, born in 1805 and 1806. Greig died in 1807, and she returned to Scotland as a young widow with an independent income and, for the first time, freedom to study as she pleased.

In Edinburgh she found mentors. John Playfair, professor of natural philosophy at the university, encouraged her. William Wallace, then a master at the Royal Military College and later professor of mathematics at Edinburgh, corresponded with her and recommended a course of reading: the calculus and mechanics of the French school, culminating in Laplace’s Mécanique céleste. She bought the books and worked through them alone. She solved problems posed in the Mathematical Repository, a journal edited by Thomas Leybourn, and in 1811 won a silver medal for one of her solutions.

In 1812 she married another cousin, William Somerville, son of the uncle who had taught her Latin. He was an army surgeon and an inspector of hospitals, and, unlike Greig, he was proud of her mathematics and helped her work. They had four children: Margaret (1813–1823), Thomas (1814–1815), Martha (1815–1879) and Mary Charlotte (1817–1875). Of her six children, three survived to adulthood: Woronzow Greig, Martha and Mary.

Lesson 03

London, Paris and Laplace, 1816–1831

In 1816 William Somerville joined the Army Medical Board and the family moved to London. They soon belonged to the city’s scientific society. Their friends included William and John Herschel, George Airy, George Peacock and Charles Babbage, whose calculating engines Mary visited. In 1817 they toured the Continent. In Paris, with introductions from Jean-Baptiste Biot and François Arago, she met Laplace, Siméon Denis Poisson and Louis Poinsot, and she kept up a correspondence with several of them afterwards. In 1824 William became physician to the Royal Hospital at Chelsea, and they lived there.

The first paper and the commission

In 1825 she made experiments to test whether violet sunlight could magnetize a steel needle, as some Italian physicists claimed. She reported a positive result, and in 1826 the Royal Society published her paper, presented by her husband. The effect was later shown to be spurious (Lesson 09).

In March 1827 Henry Brougham, founder of the Society for the Diffusion of Useful Knowledge, wrote to William asking whether Mary would prepare an English account of the Mécanique céleste for the society’s cheap educational series. She agreed, on condition that the manuscript be burned if it failed. She worked between household duties and visitors. The book outgrew the series. On John Herschel’s advice it was published separately by John Murray in 1831 as The Mechanism of the Heavens. It did not simply translate Laplace. It filled in the steps he had omitted, often with the comment that it was easy to see, and added diagrams and a long preliminary dissertation for the general reader, published separately in 1832. The book made her famous. David Brewster called her “certainly the most extraordinary woman in Europe.”

Part II · The physics
Lesson 04

The attraction of bodies and Laplace’s equation

For the general reader. Newton showed that a planet attracts as if all its mass were at its centre, provided it is a perfect sphere. Real planets are flattened, lumpy and spinning. Laplace’s great tool for such bodies was a single equation satisfied by the gravitational potential in empty space. Somerville’s Mechanism opened Laplace’s theory to English readers through this equation and the special functions, now called spherical harmonics, that solve it.

Goals
  • Show that the potential outside the masses satisfies Laplace’s equation.
  • Prove the shell theorem.
  • Expand the potential of an axisymmetric planet in Legendre polynomials.
Theorem (Laplace’s and Poisson’s equations)

For a mass density \(\rho\) occupying a bounded region, the potential \(V(\mathbf x)=-G\int\frac{\rho(\mathbf y)}{|\mathbf x-\mathbf y|}\,d^3y\) satisfies \(\nabla^2V=4\pi G\rho\). In particular \(\nabla^2V=0\) outside the masses.

Sketch. A direct computation in spherical coordinates shows \(\nabla^2(1/r)=\frac1{r^2}\frac{d}{dr}\big(r^2\cdot(-1/r^2)\big)=0\) for \(r\ne0\), so each term of the integral is harmonic away from its source. Outside the masses one may differentiate under the integral. Inside, the flux of \(\nabla(1/r)\) through any small sphere about the origin is \(-4\pi\), which gives \(\nabla^2(1/r)=-4\pi\delta\) and hence Poisson’s equation. Laplace stated only the outside equation; Poisson added the right-hand side in 1813. \(\square\)

Theorem (shell theorem)

A spherically symmetric body of mass \(M\) attracts an external point mass as if all its mass were concentrated at the centre: \(V=-GM/r\) for \(r\) greater than the body’s radius.

Proof via Laplace’s equation. Outside the body, \(V\) depends only on \(r\) and is harmonic, so \(\frac{1}{r^2}(r^2V')'=0\), giving \(V=A+B/r\). As \(r\to\infty\), \(V\to0\), so \(A=0\). By Gauss’s theorem, the flux of \(\nabla V\) through a sphere enclosing the body is \(4\pi GM\), so \(B=-GM\). \(\square\)

Legendre expansion

For a body symmetric about the \(z\)-axis, the general solution of Laplace’s equation outside it, decaying at infinity, is a series in Legendre polynomials \(P_n\):

\[ V(r,\theta)=-\frac{GM}{r}\Big[1-\sum_{n=2}^{\infty}J_n\Big(\frac{R}{r}\Big)^{n}P_n(\cos\theta)\Big], \qquad P_2(x)=\tfrac12(3x^2-1). \]

Here \(R\) is the equatorial radius, \(\theta\) the colatitude, and the \(n=1\) term vanishes when the origin is at the centre of mass. The dimensionless coefficient \(J_2\) measures the equatorial bulge. For the Earth \(J_2\approx1.083\times10^{-3}\).

Exercise 4.1Show that \(r^2P_2(\cos\theta)=z^2-\tfrac12(x^2+y^2)\) and \(r^{-3}P_2(\cos\theta)\) are both harmonic.

Show solution
\(r^2P_2=\tfrac12(3z^2-r^2)=z^2-\tfrac12(x^2+y^2)\), whose Laplacian is \(2-\tfrac12\cdot2-\tfrac12\cdot2=0\). For the second, in spherical coordinates a function \(r^\ell Y(\theta)\) is harmonic iff \(Y\) satisfies Legendre’s equation with eigenvalue \(\ell(\ell+1)\), and \(r^{-\ell-1}Y\) gives the same eigenvalue since \((-\ell-1)(-\ell)=\ell(\ell+1)\). With \(\ell=2\) both are harmonic.

Exercise 4.2Show that inside a uniform spherical shell the gravitational force vanishes.

Show solution
Inside the cavity \(V\) depends only on \(r\) and is harmonic, so \(V=A+B/r\). It must be finite at \(r=0\), so \(B=0\) and \(V\) is constant. The force \(-\nabla V\) is zero.

Exercise 4.3Using the \(J_2\) term, find the ratio of the radial gravitational acceleration \(\partial V/\partial r\) at the pole to that at the equator on a sphere \(r=R\), to first order in \(J_2\). Ignore rotation.

Show solution
\(V=-\frac{GM}{r}+\frac{GMJ_2R^2}{r^3}P_2\), so \(\partial_rV=\frac{GM}{r^2}\big(1-3J_2(R/r)^2P_2\big)\), directed inward. At \(r=R\): pole \(P_2=1\) gives \(1-3J_2\); equator \(P_2=-\tfrac12\) gives \(1+\tfrac32J_2\). The ratio is \(1-\tfrac92J_2\approx0.9951\) for the Earth: on a sphere of the same radius, the bulge makes radial gravity weaker at the pole. On the real flattened Earth the pole is closer to the centre, and with rotation the polar gravity ends up larger (Lesson 05).
Lesson 05

The figure of the Earth

For the general reader. A spinning fluid planet bulges at the equator. Newton predicted the Earth’s flattening, and French expeditions to Lapland and Peru in the 1730s confirmed it. Alexis Clairaut then found a formula that links the flattening to the way gravity changes from equator to pole, so that pendulum measurements could weigh in on the shape of the planet. Laplace extended the theory, and Somerville’s Mechanism devoted a book to it.

Goals
  • Define the flattening and the rotational parameter \(m\).
  • State and use Clairaut’s theorem.
  • Relate the flattening to \(J_2\).

Let the Earth have equatorial radius \(a\), polar radius \(c\), and flattening \(f=(a-c)/a\). Let \(\omega\) be its angular velocity and \(g_e\) gravity at the equator. The ratio of centrifugal acceleration to gravity at the equator is

\[ m=\frac{\omega^2a}{g_e}\approx\frac{1}{288}. \]
Theorem (Clairaut, 1743)

For a rotating body in hydrostatic equilibrium, made of nearly spherical layers of constant density, to first order in small quantities, the effective gravity at latitude \(\varphi\) is \(g(\varphi)=g_e(1+\beta\sin^2\varphi)\) with

\[ f+\beta=\frac52\,m. \]

Moreover \(J_2=\tfrac13(2f-m)\).

Sketch. Outside the body the effective potential is \(U=-\frac{GM}{r}+\frac{GMJ_2a^2}{r^3}P_2(\sin\varphi)-\frac12\omega^2r^2\cos^2\varphi\), and to first order \(m=\omega^2a^3/GM\). The surface is an equipotential. Write \(r=a(1-f\sin^2\varphi)\) and require \(U\) to be the same at equator and pole: this gives \(f=\tfrac32J_2+\tfrac12m\), i.e. \(J_2=\tfrac13(2f-m)\). The magnitude of gravity is \(\partial U/\partial r\). At the equator, \(g_e=\frac{GM}{a^2}\big(1+\tfrac32J_2-m\big)\). At the pole, \(r=a(1-f)\) and \(g_p=\frac{GM}{a^2}\big(1+2f-3J_2\big)\). Hence

\[ \beta=\frac{g_p-g_e}{g_e}\approx2f-\tfrac92J_2+m=2f-\big(3f-\tfrac32m\big)+m=\tfrac52m-f. \]

Only the external field enters, so the result does not depend on how density varies with depth, which is what makes it powerful. \(\square\)

For a uniform incompressible fluid, Newton found \(f=\tfrac54m\approx1/230\). The real Earth is more centrally condensed, and \(f\approx1/298\).

Exercise 5.1Compute \(m\) from \(\omega=2\pi/86{,}164\ \mathrm{s}^{-1}\), \(a=6{,}378\) km and \(g_e=9.780\ \mathrm{m\,s^{-2}}\). Given polar gravity \(g_p=9.832\ \mathrm{m\,s^{-2}}\), use Clairaut’s theorem to find \(f\).

Show solution
\(\omega=7.292\times10^{-5}\), so \(m=\omega^2a/g_e=3.468\times10^{-3}\approx1/288\). \(\beta=(9.832-9.780)/9.780=5.32\times10^{-3}\). Then \(f=\tfrac52m-\beta=8.67\times10^{-3}-5.32\times10^{-3}=3.35\times10^{-3}\approx1/298\). Pendulums alone give the shape of the Earth.

Exercise 5.2Use \(f=1/298.3\) and \(m=1/288.4\) to predict \(J_2\), and compare with the satellite value \(1.083\times10^{-3}\). Then find Newton’s flattening for a homogeneous Earth.

Show solution
\(J_2=\tfrac13(2\times3.352-3.467)\times10^{-3}=1.079\times10^{-3}\), within half a per cent of the measured value. Newton: \(f=\tfrac54m=1/231\), too large. The discrepancy shows that density increases towards the centre.
Lesson 06

The tides

For the general reader. The Moon pulls harder on the near side of the Earth than on the centre, and harder on the centre than on the far side. The result is a stretching force that raises two bulges of water, one facing the Moon and one opposite, and the Earth turns beneath them. The Sun does the same, less strongly. Newton explained this, and Laplace turned it into a dynamical theory of the oceans. Somerville explained both in the Mechanism and the Connexion, and she knew the tides of the Forth from childhood.

Goals
  • Derive the tidal potential.
  • Compute the equilibrium tide of the Moon and the Sun.
  • Explain spring and neap tides.
Proposition (tidal potential)

A body of mass \(M\) at distance \(d\) from the Earth’s centre produces, at a point at distance \(r\ll d\) from the centre and angle \(\psi\) from the sub-body point, the tidal potential (after removing the uniform acceleration of the Earth)

\[ W(r,\psi)=-\frac{GMr^2}{d^3}P_2(\cos\psi)+O\Big(\frac{r^3}{d^4}\Big). \]

Proof. Expand \(-GM/|\mathbf d-\mathbf r|=-\frac{GM}{d}\sum_n(r/d)^nP_n(\cos\psi)\). The \(n=0\) term is constant. The \(n=1\) term, \(-GMr\cos\psi/d^2\), gives a uniform force equal to the one that accelerates the whole Earth, and so has no effect relative to the Earth. The \(n=2\) term remains. \(\square\)

Equilibrium tide

If the ocean surface is an equipotential of gravity plus \(W\), on a rigid Earth of radius \(R\) and mass \(M_E\), its height above the mean is

\[ h(\psi)=\frac{M}{M_E}\Big(\frac{R}{d}\Big)^3R\,P_2(\cos\psi), \]

with range from high water (\(\psi=0\)) to low water (\(\psi=90^\circ\)) equal to \(\tfrac32\frac{M}{M_E}(R/d)^3R\).

The surface rises by \(h\) where \(g h=-W\), and \(g=GM_E/R^2\). The real tides differ greatly from this equilibrium value, because the oceans have a natural response time comparable to the forcing and are broken up by continents. Laplace’s tidal equations of 1775–76 treat the ocean as a thin rotating fluid layer forced by \(W\). They are the starting point of modern tidal theory.

Exercise 6.1Compute the equilibrium tidal range for the Moon (\(M/M_E=0.0123\), \(d=384{,}400\) km) and for the Sun (\(M/M_E=333{,}000\), \(d=1.496\times10^8\) km), with \(R=6{,}371\) km.

Show solution
Moon: \((R/d)^3=4.55\times10^{-6}\), so \(h_0=0.0123\times4.55\times10^{-6}\times6.371\times10^6\ \mathrm m=0.357\) m and the range is \(0.54\) m. Sun: \((R/d)^3=7.73\times10^{-14}\), \(h_0=0.164\) m, range \(0.25\) m. The Sun’s tide is 0.46 of the Moon’s, although its pull on the Earth is about 180 times greater: tides depend on the gradient of the force, which falls as \(1/d^3\).

Exercise 6.2Find the ratio of spring to neap tidal ranges in the equilibrium theory. Why do spring tides follow new and full Moon?

Show solution
At new and full Moon the two bulges align and the ranges add: \(0.54+0.25=0.79\) m. At the quarters the Sun’s high water falls at the Moon’s low water: \(0.54-0.25=0.29\) m. The ratio is about 2.7. Both new and full Moon give alignment because the tidal bulge is symmetric: \(P_2(\cos\psi)\) is unchanged when \(\psi\to\psi+180^\circ\).
Lesson 07

Jupiter, Saturn and the stability of the system

For the general reader. Comparing ancient and modern observations, astronomers had found that Jupiter seemed to be speeding up and Saturn slowing down. If this went on, Jupiter would fall towards the Sun and Saturn would drift away. Newton had suspected that the solar system might need divine correction. Laplace showed in 1785 that the effect is a long oscillation of about 900 years, caused by the near ratio of 5 to 2 between their periods. The result was a showpiece of the Mécanique céleste, and Somerville explained it with care.

Goals
  • State Laplace’s theorem on the invariance of the mean distances.
  • Compute the period of the great inequality from a near-resonance.
  • Understand small divisors and the exchange of energy between planets.
Theorem (Laplace, 1773; Lagrange)

In the planetary problem, to first order in the planetary masses, the semi-major axes have no secular (steadily growing) terms: they undergo only periodic oscillations.

Sketch. By Lagrange’s planetary equations, \(\dot a=\frac{2}{na}\frac{\partial\mathcal R}{\partial\lambda}\), where \(\mathcal R\) is the disturbing function and \(\lambda\) the mean longitude. Expand \(\mathcal R\) as a Fourier series in the mean longitudes, \(\mathcal R=\sum C_{j,k}\cos(j\lambda+k\lambda'+\ldots)\). The terms independent of \(\lambda\) and \(\lambda'\) (the secular part) have zero derivative with respect to \(\lambda\). So every term in \(\dot a\) oscillates with a frequency \(jn+kn'\ne0\). \(\square\)

Small divisors

Integrating a term \(\cos\big((jn+kn')t\big)\) brings in the factor \(1/(jn+kn')\) in \(a\), and in the longitude, which is obtained by integrating \(n\), the factor \(1/(jn+kn')^2\). When \(jn+kn'\) is nearly zero, a small coefficient can produce a large, slow inequality. For Jupiter and Saturn, \(2n_J-5n_S\) is small.

The great inequality

The argument \(2\lambda_J-5\lambda_S\) has period

\[ T=\Big|\frac{2}{P_J}-\frac{5}{P_S}\Big|^{-1}, \]

and the corresponding terms in the longitudes of Jupiter and Saturn have opposite signs.

Because the interaction energy is periodic and small, the sum of the two orbital energies, \(-\frac{GM_\odot m_J}{2a_J}-\frac{GM_\odot m_S}{2a_S}\), is nearly constant. When Jupiter’s orbit shrinks (and its mean motion grows), Saturn’s must expand (and its mean motion fall).

Exercise 7.1With \(P_J=11.862\) yr and \(P_S=29.457\) yr, compute the period of the great inequality and the amplification factor \(\big(n_J/(2n_J-5n_S)\big)^2\).

Show solution
\(2/11.862-5/29.457=0.16861-0.16974=-0.00113\) per year, so \(T\approx880\) years. The factor is \((0.0843/0.00113)^2\approx74^2\approx5{,}500\). A term in the disturbing function thousands of times smaller than the main ones produces displacements of tens of arcminutes.

Exercise 7.2Use conservation of total orbital energy, and \(\Delta n/n=-\tfrac32\Delta a/a\), to show that the amplitudes of the great inequality in the longitudes satisfy \(|\Delta\lambda_S/\Delta\lambda_J|\approx(n_S/n_J)(m_J/m_S)(a_S/a_J)\). Evaluate it with \(m_J/m_S=3.34\) and \(a_S/a_J=1.833\), and compare with Laplace’s amplitudes of roughly \(20'\) for Jupiter and \(48'\) for Saturn.

Show solution
Energy: \(m_J\Delta a_J/a_J^2=-m_S\Delta a_S/a_S^2\). Then \(\Delta n_S/\Delta n_J=\frac{n_S\Delta a_S/a_S}{n_J\Delta a_J/a_J}\), and substituting \(\Delta a_S/\Delta a_J=-(m_J/m_S)(a_S/a_J)^2\) gives \(-(n_S/n_J)(m_J/m_S)(a_S/a_J)\). The longitude amplitudes have the same ratio, since both oscillate with the same period. Numerically \(0.403\times3.34\times1.833\approx2.5\), compared with \(48/20=2.4\).
Lesson 08

Uranus and the unseen planet

For the general reader. After William Herschel found Uranus in 1781, astronomers computed its orbit, allowing for the pulls of Jupiter and Saturn. By the 1830s the planet was clearly straying from the tables. In the sixth edition of the Connexion (1842), Somerville wrote that the discrepancies might one day reveal the existence and even the mass and orbit of a planet beyond the reach of sight. In 1845–46 John Couch Adams and Urbain Le Verrier independently computed where such a planet should be, and on 23 September 1846 Johann Galle found Neptune within about a degree of Le Verrier’s position.

Goals
  • Estimate the size of Neptune’s pull on Uranus.
  • Compute the synodic period of the two planets.
  • See how the Titius–Bode rule helped and misled the predictions.
Synodic period

Two planets with periods \(P_1<P_2\) return to the same relative configuration after

\[ S=\Big(\frac1{P_1}-\frac1{P_2}\Big)^{-1}. \]

The relative angle grows at \(2\pi(1/P_1-1/P_2)\) per unit time, which gives the formula. For Uranus and Neptune it is long, so the interval 1781–1846 covered only one conjunction, around 1822. Before it, the faster Uranus was catching up with Neptune, which lay ahead and pulled it forward along its orbit; after it, Neptune lay behind and held it back. This is the pattern in the residuals that Adams and Le Verrier explained, shown in the figure at the top of this page.

Perturbing acceleration

At conjunction, the ratio of Neptune’s pull on Uranus to the Sun’s is \(\dfrac{m_N}{M_\odot}\Big(\dfrac{a_U}{a_N-a_U}\Big)^2\). The displacement it produces over a time \(t\), starting from rest relative to the predicted path, is roughly \(\tfrac12at^2\).

Neither Adams nor Le Verrier knew Neptune’s distance. Both started from the Titius–Bode rule, \(a_n=0.4+0.3\times2^n\) AU, which had fitted Uranus well. For the next planet it gave 38.8 AU, while Neptune’s true distance is 30.1 AU. Their predicted orbits were too large, but they placed the planet correctly in the 1840s, because the direction of the pull near conjunction depends mostly on where the planet is, and less on how far.

Exercise 8.1With \(P_U=84.0\) and \(P_N=164.8\) years, find the synodic period. With \(a_U=19.2\) AU, \(a_N=30.1\) AU and \(m_N/M_\odot=5.15\times10^{-5}\), find the ratio of Neptune’s pull to the Sun’s at conjunction.

Show solution
\(S=(1/84.0-1/164.8)^{-1}\approx171\) years. Ratio: \(5.15\times10^{-5}\times(19.2/10.9)^2\approx1.6\times10^{-4}\). The pull is tiny but persistent near conjunction.

Exercise 8.2Estimate the displacement of Uranus produced in 20 years by a constant acceleration equal to Neptune’s pull at conjunction, and convert it to an angle seen from 18 AU. Take \(GM_\odot=1.327\times10^{20}\ \mathrm{m^3\,s^{-2}}\).

Show solution
\(a=GM_\odot m_N/M_\odot/(10.9\ \text{AU})^2=1.327\times10^{20}\times5.15\times10^{-5}/(1.63\times10^{12})^2\approx2.6\times10^{-9}\ \mathrm{m\,s^{-2}}\). With \(t=6.3\times10^8\) s, \(\tfrac12at^2\approx5\times10^8\) m. Seen from 18 AU this is \(5\times10^8/2.7\times10^{12}\) rad \(\approx40''\). Residuals of this order, which grew to over an arcminute by the 1840s, were far larger than the observational errors.

Exercise 8.3What period does Kepler’s third law give for a planet at the Titius–Bode distance of 38.8 AU, and for Neptune’s real 30.1 AU?

Show solution
\(38.8^{3/2}\approx242\) years and \(30.1^{3/2}\approx165\) years. The predicted planet would move about two-thirds as fast as the real one, so predictions made from it drift away from the true position within a few decades of the fit.
Lesson 09

Rays beyond the visible

For the general reader. In 1800 William Herschel found invisible heat rays beyond the red end of the Sun’s spectrum, and in 1801 Johann Wilhelm Ritter found invisible rays beyond the violet that darken silver salts. Light, heat and chemical action seemed to be different agents, all coming from the Sun. Somerville’s first experiment asked whether violet light could also magnetize steel. She thought it could; later workers could not reproduce her result. A real link between light and magnetism was found by Michael Faraday in 1845. This lesson explains why the ends of the spectrum act so differently.

Goals
  • Use Wien’s law to locate the peak of the Sun’s spectrum.
  • Compute photon energies and explain why violet and ultraviolet light drive chemistry.
  • State Faraday’s magneto-optical effect.

Her experiment

Following claims by Domenico Morichini in Rome, she exposed half of a steel sewing needle to violet light from a prism while covering the other half, and then tested it for magnetism with a compass. The needle appeared magnetized, and her paper of 1826 reported the result. Within a few years other experimenters, among them Peter Riess and Ludwig Moser, repeated such experiments with careful controls and found no effect. Magnetism picked up from the surroundings, or from handling, probably explains the positive results. She later studied the action of different parts of the spectrum on paper treated with plant juices, a more fruitful question that connects with early photography.

Wien’s displacement law and photon energy

A black body at temperature \(T\) emits per unit wavelength most strongly at \(\lambda_{\max}=b/T\), with \(b=2.898\times10^{-3}\) m K. A photon of wavelength \(\lambda\) has energy \(E=hc/\lambda\), and \(hc=1240\) eV nm.

Derivation of Wien’s law. Planck’s law per unit wavelength is \(B_\lambda\propto\lambda^{-5}(e^{x}-1)^{-1}\) with \(x=hc/\lambda kT\). Setting \(dB_\lambda/d\lambda=0\) gives \(5(1-e^{-x})=x\), whose positive root is \(x\approx4.965\). So \(\lambda_{\max}T=hc/(4.965k)\). \(\square\)

Faraday effect (1845)

The plane of polarization of light travelling a distance \(L\) through a transparent medium along a magnetic field \(B\) rotates by \(\beta=\mathcal V BL\), where \(\mathcal V\) is the Verdet constant of the medium.

Exercise 9.1Take the Sun’s effective temperature as 5,772 K. Find \(\lambda_{\max}\). Find the photon energies at 400 nm, 700 nm and 1,000 nm.

Show solution
\(\lambda_{\max}=2.898\times10^{-3}/5772=502\) nm, in the blue-green. Photon energies: \(1240/400=3.1\) eV, \(1240/700=1.8\) eV, \(1240/1000=1.24\) eV. Typical chemical bonds and the reduction of silver halides need a few eV per event, which violet and ultraviolet photons can supply and infrared photons cannot. Herschel’s infrared rays heat; Ritter’s ultraviolet rays drive chemistry.

Exercise 9.2Verify that \(x=4.965\) satisfies \(5(1-e^{-x})=x\), and use it with \(h=6.626\times10^{-34}\) J s, \(c=2.998\times10^8\) m/s and \(k=1.381\times10^{-23}\) J/K to compute \(b\).

Show solution
\(e^{-4.965}=0.00698\), so \(5(1-0.00698)=4.965\). Then \(b=hc/(4.965k)=1.986\times10^{-25}/(6.857\times10^{-23})=2.897\times10^{-3}\) m K.

Exercise 9.3Dense flint glass has a Verdet constant of about \(30\) rad T\(^{-1}\) m\(^{-1}\) for visible light. Find the rotation in a 5 cm block in a field of 0.5 T. Would an effect of this kind magnetize a needle?

Show solution
\(\beta=30\times0.5\times0.05=0.75\) rad, about \(43^\circ\). The Faraday effect is magnetism acting on light, the reverse of what Morichini and Somerville sought. Light can magnetize matter only through very weak effects, far below what a compass needle could show.
Lesson 10

Physical geography: the atmosphere

For the general reader. Physical Geography (1848) described the Earth as a single system of land, sea, air and life. It was the first English textbook on the subject and was used in schools and universities for decades. Two simple physical laws govern much of what it says about the air: pressure falls by a constant factor for each fixed step upward, and rising air cools as it expands. Together they explain why mountains are cold and why snow lies on the peaks of the tropics.

Goals
  • Derive the barometric formula and the scale height.
  • Derive the dry adiabatic lapse rate.
  • Apply both to mountains and to the measurement of heights by barometer.
Proposition (isothermal atmosphere)

An ideal gas of molar mass \(\mu\) at constant temperature \(T\) in hydrostatic equilibrium has pressure \(p(z)=p_0e^{-z/H}\), with scale height \(H=RT/(\mu g)\).

Proof. Hydrostatic equilibrium gives \(dp/dz=-\rho g\), and the ideal gas law gives \(\rho=p\mu/(RT)\). So \(dp/p=-dz/H\). \(\square\)

Proposition (dry adiabatic lapse rate)

A parcel of dry air rising without exchanging heat cools at the rate

\[ -\frac{dT}{dz}=\Gamma_d=\frac{g}{c_p}, \]

where \(c_p\) is the specific heat at constant pressure per unit mass.

Proof. For an adiabatic change of an ideal gas, the first law gives \(c_p\,dT=dp/\rho\). With \(dp=-\rho g\,dz\) from hydrostatic balance, \(c_p\,dT=-g\,dz\). \(\square\)

Moist air cools more slowly, because condensing water releases latent heat. The observed average fall in the lower atmosphere is about 6.5 K per kilometre. In the nineteenth century barometers were the standard way to measure the heights of mountains, using a version of the first proposition with corrections for temperature.

Exercise 10.1Compute \(H\) for air (\(\mu=0.029\) kg/mol) at 288 K, and estimate the pressure at the summit of Mont Blanc, 4.81 km, from a sea-level pressure of 1,013 hPa.

Show solution
\(H=8.314\times288/(0.029\times9.81)\approx8.4\) km. \(p=1013\,e^{-4.81/8.4}\approx572\) hPa. Observed summit pressures are somewhat lower, because the real air is colder than 288 K aloft and its scale height is smaller.

Exercise 10.2Find \(\Gamma_d\) with \(c_p=1005\) J kg\(^{-1}\) K\(^{-1}\). A dry wind rises over a 3 km range from 20°C at the coast and descends to the lee side at sea level. What are its temperatures at the crest and on arrival, if it is dry throughout? What changes if it loses its moisture as rain while rising, cooling on the windward side at 6 K/km?

Show solution
\(\Gamma_d=9.81/1005\approx9.8\) K/km. Dry throughout: \(20-29=-9\)°C at the crest, and back to \(20\)°C on arrival. With rain: \(20-18=2\)°C at the crest, then dry descent warms it by 29 K to about 31°C. The lee side is warmer and drier: this is the föhn effect, part of the physical geography of the Alps that she described.
Part III · Life and legacy
Lesson 11

The Connexion and fame, 1831–1838

In 1834 Somerville published On the Connexion of the Physical Sciences. It had no equations. It presented astronomy, light, heat, electricity, magnetism and sound as aspects of a few general principles, and argued that the sciences were growing together. The book went through ten editions, and she revised it for each, bringing in the latest work on electromagnetism, light and astronomy.

William Whewell reviewed it in the Quarterly Review in 1834. Discussing the lack of a general name for those who studied nature, he reported a suggestion of “scientist”, by analogy with “artist”. It is the first known use of the word in English. Whewell used the occasion of Somerville’s book to argue that the sciences needed people who saw them whole.

Recognition

Honours followed quickly. In 1834 she was elected an honorary member of the Royal Irish Academy and of the Société de Physique et d’Histoire Naturelle of Geneva. In 1835 the Royal Astronomical Society elected her and Caroline Herschel honorary members, the first women it admitted. In 1835 the government of Robert Peel granted her a civil list pension of £200 a year, raised to £300 in 1837 under Lord Melbourne. Francis Chantrey sculpted a marble bust of her for the Royal Society. Women could not become Fellows of the Royal Society.

In these years she was a friend and mentor to Ada, daughter of Lord Byron, later Ada Lovelace. She helped Ada with mathematics, and the two shared an interest in Charles Babbage’s engines.

DiscussionSomerville thought of herself as an expositor, not a discoverer. Her books shaped how a generation understood physical science and gave the occasion for a new word for its practitioners. What is the place of synthesis and exposition in science, and why is it so often valued below discovery?

Lesson 12

Italy, last books, legacy, 1838–1872

In 1838, for the sake of William’s health and their finances, the Somervilles went to Italy for the winter. They never settled in Britain again, and Mary returned only for two brief visits. They lived in Rome, Florence, Turin, La Spezia and Naples. She went on working. The sixth edition of the Connexion, in 1842, contained her remark on the irregularities of Uranus (Lesson 08); after Neptune’s discovery, a later edition recorded that the prediction had been fulfilled. MacTutor records that her passage prompted John Couch Adams’s investigation, though Adams had already noted the problem in a memorandum of 1841.

Physical Geography appeared in 1848 and became her most successful book, used as a textbook for half a century. William Somerville died in Florence in 1860, and her son Woronzow Greig in 1865. In 1869, at the age of 88, she published On Molecular and Microscopic Science, a two-volume survey of chemistry, light and the microscopic world. The same year the Royal Geographical Society awarded her its Patron’s Gold Medal. She supported the education of women and signed John Stuart Mill’s petition to Parliament for women’s suffrage, at the head of the list.

She died in her sleep in Naples on 29 November 1872, aged 91, and was buried in the English cemetery there. Her daughter Martha edited her Personal Recollections, published in 1873.

Legacy

  • Books: The Mechanism of the Heavens (1831), which brought Laplace’s celestial mechanics into English; On the Connexion of the Physical Sciences (1834, ten editions); Physical Geography (1848); On Molecular and Microscopic Science (1869).
  • Firsts: with Caroline Herschel, the first women elected to the Royal Astronomical Society (1835); the occasion of the first recorded use of the word “scientist”.
  • Names: Somerville Hall, founded in Oxford in 1879, later Somerville College; the lunar crater Somerville; her portrait on the Royal Bank of Scotland £10 note issued in 2017.

Final projectChoose one topic from Part II. Write a two-page account for a strong undergraduate in Somerville’s manner: begin with a page that needs no equations, then give the mathematics, working one example in full, and explain how the topic connects to at least one other branch of physics.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Mary Fairfax Somerville, MacTutor History of Mathematics.
  2. Somerville, Mary, Oxford Dictionary of National Biography article reproduced by MacTutor.
  3. Mary Somerville, Wikipedia (with extensive references).
  4. M. Somerville, Personal Recollections, from Early Life to Old Age, of Mary Somerville, ed. M. Somerville, John Murray, 1873.
  5. K. Neeley, Mary Somerville: Science, Illumination, and the Female Mind, Cambridge University Press, 2001.
  6. E. C. Patterson, Mary Somerville and the Cultivation of Science, 1815–1840, Martinus Nijhoff, 1983.

Physics

  1. M. Somerville, Mechanism of the Heavens, John Murray, 1831.
  2. M. Somerville, On the Connexion of the Physical Sciences, John Murray, 1834; 6th ed. 1842.
  3. M. Somerville, On the magnetizing power of the more refrangible solar rays, Royal Society, 1826.
  4. M. Somerville, Physical Geography, John Murray, 1848; On Molecular and Microscopic Science, John Murray, 1869.
  5. W. Whewell, review of On the Connexion of the Physical Sciences, Quarterly Review 51 (1834).
  6. C. D. Murray and S. F. Dermott, Solar System Dynamics, Cambridge University Press, 1999 (disturbing function, resonances, the great inequality).
  7. W. M. Kaula, An Introduction to Planetary Physics, Wiley, 1968 (figure of the Earth, Clairaut’s theorem).
  8. D. E. Cartwright, Tides: A Scientific History, Cambridge University Press, 1999.