How to use this course
George Biddell Airy is remembered in two ways. Historians of astronomy know the organizer who turned the Royal Observatory at Greenwich into a scientific factory, fixed the prime meridian of the world, and was blamed for the British failure to find Neptune. Physicists, engineers and geologists know a list of eponyms: the Airy disk, the Airy function, Airy waves, Airy isostasy, the Airy stress function and the Airy points. This course connects the two. Behind each eponym is a specific problem that Airy met as an astronomer, a surveyor or a government adviser, and solved with the mathematics of his day.
- Calculus, ordinary differential equations and a little linear algebra.
- Wave optics: interference and Fraunhofer diffraction at the level of a first course.
- Newtonian gravitation and the two-body problem for Lessons 07 and 09.
- Potential flow and elementary elasticity are introduced as needed in Lessons 08 and 10.
Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. The lessons follow the order of Airy’s work: physiological optics (1825), diffraction (1835), caustics (1838), planetary theory (1828–1832), tides (1841), geodesy (1854–1855) and elasticity (1845–1863).
Timeline
| 1801 | Born 27 July at Alnwick, Northumberland, son of William Airy, an excise officer, and Ann Biddell. The family moves to Hereford in 1802. |
| 1814–19 | Colchester Grammar School; long stays with his uncle Arthur Biddell at Playford, near Ipswich. |
| 1819 | Enters Trinity College, Cambridge, as a sizar. |
| 1823 | Senior Wrangler and first Smith’s Prize. |
| 1824 | Fellow of Trinity. Meets Richarda Smith. |
| 1825 | On a peculiar defect in the eye: astigmatism and its correction by a cylindrical lens. |
| 1826 | Lucasian Professor of Mathematics. Mathematical Tracts. |
| 1828 | Plumian Professor of Astronomy and director of the Cambridge Observatory. |
| 1830 | Marries Richarda Smith. Figure of the Earth: the Airy ellipsoid. |
| 1831 | Copley Medal of the Royal Society. |
| 1832 | Publishes the long inequality of Earth and Venus. Report on astronomy to the British Association. |
| 1833 | Gold Medal of the Royal Astronomical Society (again in 1846). |
| 1835 | Diffraction by a circular aperture: the Airy disk. June: appointed Astronomer Royal, succeeding John Pond. |
| 1838 | The intensity of light near a caustic: the rainbow integral. Experiments on compass correction in the iron ship Rainbow (published 1839). |
| 1841 | “Tides and Waves” in the Encyclopaedia Metropolitana. |
| 1845 | Royal Medal. Airy points. Member of the Railway Gauge Commission. October–November: correspondence with John Couch Adams. |
| 1846 | July: asks James Challis to search for the trans-Uranian planet. 23 September: Johann Galle finds Neptune at Berlin from Urbain Le Verrier’s prediction. |
| 1850–51 | Airy transit circle installed at Greenwich and first used: the Greenwich meridian. |
| 1854 | Pendulum experiments in Harton colliery, South Shields: mean density of the Earth 6.566. |
| 1855 | Isostasy: mountains supported by light roots. |
| 1862–63 | On the strains in the interior of beams: the Airy stress function. |
| 1871–73 | President of the Royal Society. Knighted (KCB) in 1872. |
| 1874 | Organizes British expeditions to observe the transit of Venus. |
| 1875 | Death of Richarda Airy. Freedom of the City of London. |
| 1881 | Retires on 15 August. |
| 1884 | The International Meridian Conference in Washington adopts the meridian of his transit circle. |
| 1892 | Dies 2 January at Greenwich, aged 90; buried at Playford. |
Alnwick, Colchester, Playford, 1801–1819
George Biddell Airy was born at Alnwick in Northumberland on 27 July 1801. His father, William Airy, came from a Lincolnshire farming family, had educated himself, and worked as an officer of the excise, the tax service that moved its men around the country. His mother, Ann, was the daughter of a Suffolk farmer named Biddell. In 1802 the family moved to Hereford, and later to Essex.
Airy went to a small private school in Colchester, where he came top, and from 1814 to 1819 to Colchester Grammar School. He was a prodigious memorizer and a systematic worker, and he built his own instruments and kept careful notes from an early age.
Playford
From 1812 the boy spent summers, and later about half of each year, with his mother’s brother Arthur Biddell, who farmed at Playford near Ipswich. Biddell had a good library and scientific friends, and Airy later counted him as the strongest influence of his youth. Biddell’s friends included the abolitionist Thomas Clarkson of Playford Hall, who took an interest in the boy’s progress. Biddell helped pay for his studies at Cambridge. Airy kept a house at Playford for the rest of his life, and he is buried there.
Trinity and the Lucasian chair, 1819–1828
Airy entered Trinity College, Cambridge, in 1819 as a sizar, a poor student who received reduced fees in exchange for menial duties. He arrived already knowing more mathematics than most graduates. The Cambridge of 1819 was being converted from Newton’s fluxions to the continental calculus by George Peacock, Charles Babbage and John Herschel, and Peacock helped and advised him. Airy became a scholar of Trinity in 1822, and in January 1823 he graduated as Senior Wrangler, first in the Mathematical Tripos, and won the first Smith’s Prize. He was elected a fellow of Trinity in 1824.
In 1824, on a walking tour in Derbyshire, he met Richarda Smith, daughter of a clergyman, and proposed within days. Her father refused until Airy had an adequate income. They married in 1830.
Lucasian Professor
In 1826 Airy was elected Lucasian Professor of Mathematics, the chair once held by Isaac Newton. The stipend was small. He revived lectures on experimental philosophy, mechanics, optics and the undulatory theory of light, with demonstrations, and they drew large audiences. His Mathematical Tracts on Physical Astronomy (1826) became a standard Cambridge text and, in later editions, presented the wave theory of light of Thomas Young and Augustin Fresnel to English students. His paper on the defect in his own left eye (Lesson 04) dates from these years, and he joined the Board of Longitude in 1826.
The Cambridge Observatory, 1828–1835
In February 1828 Airy moved to the Plumian Professorship of Astronomy, which came with the direction of the new Cambridge University Observatory. He found it short of instruments and staff. He set rules that defined his whole career: observations were to be reduced promptly by fixed procedures and published in full each year. The Cambridge volumes were models of their kind.
In the same years he did his most ambitious mathematical work. He found an inequality in the Earth’s motion caused by Venus (Lesson 07), received the Copley Medal of the Royal Society in 1831, gave the British Association its report on the progress of astronomy in 1832, and wrote the 1835 paper on diffraction by a circular lens (Lesson 05). His 1830 computation of the dimensions of the Earth gave the Airy ellipsoid, which the Ordnance Survey of Great Britain still uses as the reference surface of its national grid.
When John Pond resigned as Astronomer Royal in 1835, the government offered the post to Airy. He accepted at the age of 33 and moved to Greenwich, where he would remain for 46 years.
Astigmatism and the cylindrical lens
For the general reader. Airy noticed as a young man that his left eye saw a point of light as a short line, and that the line changed direction depending on how far away the light was. He worked out that the eye focused more strongly in one direction than in the perpendicular one, and he had a lens made with a cylindrical surface to correct it. His 1825 paper was the first description and correction of astigmatism, and millions of spectacle prescriptions use his idea.
- Describe an astigmatic lens by its power in each meridian.
- Represent thin lenses by symmetric power matrices and add them.
- Write a prescription in sphere–cylinder–axis form.
Power in a meridian
A thin lens bends a ray that crosses it at height \(\mathbf r=(x,y)\) from the axis by an angle proportional to \(\mathbf r\). For a spherical lens of power \(P\) (in dioptres, the inverse of the focal length in metres), the deviation is \(-P\mathbf r\), the same in every direction. A thin cylindrical lens of power \(C\) acts only across its axis: rays in the plane of the axis pass undeviated, rays in the perpendicular plane are focused. If the cylinder axis makes angle \(\alpha\) with the \(x\)-axis, let \(\mathbf n=(-\sin\alpha,\cos\alpha)\) be the unit vector across the axis.
To first order every thin lens deviates rays by \(-F\mathbf r\) with \(F\) a real symmetric \(2\times2\) matrix, its power matrix. For a sphere \(F=PI\); for a cylinder \(F=C\,\mathbf n\mathbf n^{\mathsf T}\). Thin lenses in contact add their power matrices. The power in the meridian of direction \(\mathbf u=(\cos\theta,\sin\theta)\) is \(\mathbf u^{\mathsf T}F\mathbf u\), which for the cylinder equals \(C\sin^2(\theta-\alpha)\). The eigenvalues of \(F\) are the two principal powers, attained in perpendicular meridians.
Why symmetric. A thin lens acts on the wavefront by a phase \(\tfrac{k}{2}\,\mathbf r^{\mathsf T}F\mathbf r\), a quadratic form; the deviation is minus its gradient divided by \(k\). Any quadratic form can be written with a symmetric matrix. By the spectral theorem every thin lens is therefore a sphere plus a cylinder: \(F=S\,I+C\,\mathbf n\mathbf n^{\mathsf T}\), the sphere–cylinder–axis form of an optician’s prescription.
Airy’s eye
An astigmatic eye is a spherical eye plus an unwanted cylinder. A point source is imaged not to a point but to two perpendicular focal lines, one for each principal meridian, with the most compact blur, the circle of least confusion, between them. Airy measured the far point of his left eye in its two principal meridians, found them at different distances, and in 1825 had a lens made, with one spherical and one cylindrical surface, whose power matrix cancelled the difference. He reported that it gave him nearly normal vision.
Exercise 4.1Two cylinders, of power \(C_1\) with axis along \(x\) and \(C_2\) with axis along \(y\), are placed in contact. Find the combined power matrix and write it as a sphere plus a single cylinder. When is the combination a pure sphere?
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Exercise 4.2An eye sees sharply out to 0.50 m in the horizontal meridian and only to 0.25 m in the vertical meridian. Find a correcting lens in sphere–cylinder–axis form, ignoring the distance between lens and eye.
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Exercise 4.3Show that a cylinder of power \(C\) at axis \(\alpha\) has power matrix \(\frac C2\begin{pmatrix}1-\cos2\alpha&-\sin2\alpha\\-\sin2\alpha&1+\cos2\alpha\end{pmatrix}\). Deduce that cylinders at different axes do not combine by simply adding powers, but behave like vectors at double the angle.
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The Airy disk
For the general reader. Even a perfect telescope cannot show a star as a point. Light spreads as it passes through the round lens, so the star appears as a small bright disk surrounded by faint rings. Airy calculated the size of this disk in 1835. It sets the sharpest detail any telescope, microscope or camera of a given size can resolve.
- Derive the Fraunhofer pattern of a circular aperture.
- Find the radius of the first dark ring, \(1.22\lambda/D\).
- Compute the energy in the central disk and apply the Rayleigh criterion.
The diffraction integral
Light of wavelength \(\lambda\), \(k=2\pi/\lambda\), falls normally on a circular aperture of radius \(a\) and diameter \(D=2a\). In the Fraunhofer (far-field or focal-plane) limit, the amplitude in direction \(\theta\) is proportional to the Fourier transform of the aperture:
\[ A(\theta)\propto\int_0^{a}\!\!\int_0^{2\pi}e^{-ik\rho\sin\theta\cos\phi}\,\rho\,d\phi\,d\rho=2\pi\int_0^a J_0(k\rho\sin\theta)\,\rho\,d\rho, \]using \(\frac1{2\pi}\int_0^{2\pi}e^{-iz\cos\phi}d\phi=J_0(z)\). With \(v=ka\sin\theta\) and the identity \(\frac{d}{dt}\big(tJ_1(t)\big)=tJ_0(t)\), the integral is \(2\pi a^2J_1(v)/v\).
The intensity in the focal plane of a perfect lens with circular aperture of diameter \(D\) is
\[ I(\theta)=I_0\left(\frac{2J_1(v)}{v}\right)^2,\qquad v=\frac{\pi D}{\lambda}\sin\theta. \]The first zero is at \(v=3.8317\), that is \(\sin\theta=1.22\,\lambda/D\). The fraction of the total power inside the first dark ring is \(1-J_0^2(3.8317)\approx0.838\).
Airy did not have Bessel’s notation in this form; he expanded the integral in power series and computed the rings numerically. The encircled-energy formula, \(E(v)=1-J_0^2(v)-J_1^2(v)\), is due to Lord Rayleigh. In 1879 Rayleigh proposed that two stars of equal brightness are just resolved when the central maximum of one falls on the first dark ring of the other, an angular separation of \(1.22\lambda/D\).
Exercise 5.1Find the Rayleigh resolution limit, in arcseconds, of a telescope of aperture 2.4 m at \(\lambda=500\) nm, and of the human eye with a 3 mm pupil at the same wavelength.
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Exercise 5.2Show that the peak intensity \(I_0\) grows as \(D^4\) while the total power through the aperture grows only as \(D^2\). Explain the difference.
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Exercise 5.3Show that \(I(\theta)\to I_0\) as \(\theta\to0\) by expanding \(J_1\), and find the fractional drop in intensity at \(v=1\).
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The rainbow integral
For the general reader. Ray optics says that a rainbow is infinitely bright along its edge, which is impossible. Airy treated the light as a wave and showed in 1838 that near the bright edge the light forms a series of bands, brightest just inside the geometric edge and fading smoothly outside it. The faint extra bands inside the main bow, called supernumerary bows, are part of the same pattern. The function he introduced, now called the Airy function, appears wherever waves meet a turning point, from quantum mechanics to the bending of light by gravity.
- Find the rainbow angle from the minimum-deviation condition.
- Derive the Airy integral from a cubic wavefront.
- Know the differential equation and asymptotic forms of Ai.
The geometric rainbow
A ray entering a spherical drop of refractive index \(n\) at incidence angle \(i\), refracted to angle \(r\) (\(\sin i=n\sin r\)), reflected once inside and refracted out, is deviated by \(\Delta(i)=\pi+2i-4r\). René Descartes found that \(\Delta\) has a minimum. Rays near the minimum pile up, so the light is concentrated at that angle: a caustic. In ray optics the intensity is proportional to \(1/|d\Delta/di|\), which is infinite at the minimum.
The cubic wavefront
Airy observed that the outgoing wavefront near the rainbow ray has an inflection point: to lowest order its shape is cubic, \(z=c\,s^3\) in suitable coordinates. Summing the contributions of all points of such a wavefront in a direction at a small angle from the rainbow ray gives an integral of the form \(\int_{-\infty}^{\infty}\cos\big(\tfrac\pi2(w^3-mw)\big)dw\), where \(m\) is proportional to the angle. This is Airy’s rainbow integral. In modern notation, after rescaling,
converges for real \(x\) (conditionally) and solves Airy’s equation \(y''=xy\). It is the solution that tends to 0 as \(x\to+\infty\), with \(\mathrm{Ai}(0)=3^{-2/3}/\Gamma(\tfrac23)\approx0.3550\) and \(\mathrm{Ai}'(0)=-3^{-1/3}/\Gamma(\tfrac13)\approx-0.2588\). With \(\zeta=\tfrac23|x|^{3/2}\), as \(|x|\to\infty\):
\[ \mathrm{Ai}(x)\sim\frac{e^{-\zeta}}{2\sqrt\pi\,x^{1/4}}\ \ (x\to+\infty),\qquad \mathrm{Ai}(x)\sim\frac{\sin(\zeta+\pi/4)}{\sqrt\pi\,|x|^{1/4}}\ \ (x\to-\infty). \]Derivation of the equation. Formally, \(\mathrm{Ai}''(x)-x\,\mathrm{Ai}(x)=-\frac1\pi\int_0^\infty(t^2+x)\cos(\tfrac{t^3}3+xt)\,dt=-\frac1\pi\big[\sin(\tfrac{t^3}3+xt)\big]_0^\infty\). The bracket vanishes at \(t=0\) and oscillates without limit at infinity; giving the integral a convergence factor, or rotating the contour into the complex plane where the integrand decays, makes the boundary term zero. The intensity of the rainbow near the caustic is proportional to \(\mathrm{Ai}^2\) of a variable proportional to the angle from the geometric rainbow: oscillating bands on the lit side (the primary bow and its supernumeraries) and exponential decay on the dark side.
Airy evaluated his integral by power series, which converge slowly for large arguments, and computed only the first few maxima. In 1841 William Hallowes Miller measured up to 30 bands produced by a thin cylinder of water and confirmed Airy’s theory. In 1850 George Gabriel Stokes derived the asymptotic forms above and so computed many more bands easily; in doing so he discovered the Stokes phenomenon. The notation \(\mathrm{Ai}\) is due to Harold Jeffreys.
Exercise 6.1Show that \(\Delta(i)=\pi+2i-4r\) is stationary when \(\cos^2i=(n^2-1)/3\). For water, \(n=4/3\), find \(i\), the deviation, and the angle of the rainbow measured from the antisolar point.
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Exercise 6.2Use the oscillatory asymptotic form to show that the zeros of \(\mathrm{Ai}\) are approximately \(-a_n\) with \(a_n\approx\big[\tfrac{3\pi}8(4n-1)\big]^{2/3}\). Compare with the exact first zero \(-2.3381\).
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Exercise 6.3Show that the integral \(\int_{-\infty}^\infty e^{i(t^3/3+xt)}\,dt\) equals \(2\pi\,\mathrm{Ai}(x)\), and use stationary phase to find which values of \(x\) have real stationary points. Interpret this in terms of rays.
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The long inequality of Earth and Venus
For the general reader. Thirteen years of the Earth last almost exactly as long as eight years of Venus. Because of this near coincidence, the small tugs of Venus on the Earth do not average out quickly. They build up into a slow swing in the Earth’s position that takes about 240 years to complete. Airy found this effect at the end of the 1820s through a calculation of great length, published in 1832, and with it removed a puzzling error in the tables of the Sun.
- Understand near-commensurability and small divisors in perturbation theory.
- Use the d’Alembert rules to find the order of a term in the eccentricities.
- Compute the period of a long inequality.
Small divisors
Let planets with mean motions \(n\) and \(n'\) have mean longitudes \(\lambda=nt+\varepsilon\), \(\lambda'=n't+\varepsilon'\). The perturbing function, the part of the potential energy due to the other planet, is a Fourier series in the angles \(\lambda,\lambda'\) and the longitudes of perihelion and node. A term \(A\cos(p\lambda'-q\lambda+\dots)\) perturbs the semimajor axis and hence the mean motion. Integrating twice to obtain the mean longitude produces the factor \(1/(pn'-qn)^2\).
If \(pn'-qn\) is small compared with \(n\), the corresponding term in the mean longitude of each planet has period \(2\pi/|pn'-qn|\) and amplitude of order
\[ \delta\lambda\sim\frac{m'}{M_\odot}\,\frac{n^2}{(pn'-qn)^2}\;C\,e^{|p-q|}, \]where \(e\) stands for the eccentricities and inclinations and \(C\) is a coefficient of order one. The power \(|p-q|\) follows from the d’Alembert rules: the coefficient of a term whose longitude multipliers sum to \(p-q\) is at least of that order in eccentricities and inclinations.
The classic case is the great inequality of Jupiter and Saturn, with \(5n_S\approx2n_J\), explained by Pierre-Simon Laplace in 1785. For the Earth and Venus, \(13n_E\approx8n_V\). The term is of fifth order in the small eccentricities, so its coefficient is tiny, but the divisor is very small too. Airy computed it, an enormous labour of algebra and arithmetic, and published it in the Philosophical Transactions in 1832. The resulting oscillation, of a few arcseconds in the Sun’s apparent longitude, removed a discrepancy between observed solar positions and the tables. Laplace had considered such terms negligible; Airy showed that fifth-order terms can matter when the divisor is small enough.
Exercise 7.1Using sidereal periods \(T_E=365.256\) d and \(T_V=224.701\) d, compute the period of the term with argument \(13\lambda_E-8\lambda_V\).
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Exercise 7.2Repeat for Jupiter and Saturn (\(T_J=11.862\) yr, \(T_S=29.457\) yr) with argument \(2\lambda_J-5\lambda_S\). What is the order in eccentricities? Why is this inequality much larger than Airy’s?
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Tides and waves
For the general reader. Why do long waves travel faster in deep water, and why do short ripples behave differently from ocean swell? In a long article of 1841, Airy gave the theory of small water waves of any length in water of any depth. It is still called Airy wave theory and is the starting point of every course on ocean waves.
- Derive the linear dispersion relation \(\omega^2=gk\tanh kh\).
- Find the deep- and shallow-water limits.
- Use Airy’s canal theory to see why tides can be inverted.
Linear waves
Consider irrotational flow of an inviscid incompressible fluid of depth \(h\), with velocity potential \(\phi(x,z,t)\) and free surface \(z=\eta(x,t)\). For small amplitude, linearize the boundary conditions about \(z=0\):
\[ \nabla^2\phi=0\ \ (-h<z<0),\qquad \partial_z\phi=0\ \ (z=-h),\qquad \partial_t\eta=\partial_z\phi,\quad \partial_t\phi+g\eta=0\ \ (z=0). \]Seek \(\eta=a\cos(kx-\omega t)\) and \(\phi=B\cosh k(z+h)\sin(kx-\omega t)\), which satisfies Laplace’s equation and the bottom condition.
Small-amplitude progressive waves of wavenumber \(k\) on water of depth \(h\) have frequency
\[ \omega^2=gk\tanh kh,\qquad c=\frac\omega k=\sqrt{\frac gk\tanh kh}. \]For \(kh\gg1\) (deep water) \(c\approx\sqrt{g/k}\); for \(kh\ll1\) (shallow water, long waves) \(c\approx\sqrt{gh}\), independent of wavelength.
Derivation. The two surface conditions give \(a\omega=Bk\sinh kh\) and \(B\omega\cosh kh=ga\). Eliminating \(a/B\) gives \(\omega^2=gk\tanh kh\). Airy also analysed long waves of finite amplitude in shallow water and showed that their crests travel faster than their troughs, so the front of a long wave steepens, the mechanism of tidal bores.
The canal theory of tides
Isaac Newton’s equilibrium theory predicts high water under the Moon. Airy studied the dynamic response of water in a canal around the equator, forced by the Moon’s tidal pull moving westward at the speed of the Earth’s rotation relative to the Moon. A forced wave in a channel follows the forcing in phase if the free long-wave speed \(\sqrt{gh}\) exceeds the forcing speed, and is inverted if it is smaller. The ocean is too shallow for the first case, so the canal theory predicts low water under the Moon, an example of how dynamics can reverse the naive answer. Airy applied the theory of waves in channels to tides in rivers and estuaries, and his 1845 Royal Medal was for work on the tides of Ireland.
Exercise 8.1Find the speed of a tsunami (wavelength hundreds of kilometres) in an ocean 4000 m deep, and of a deep-water swell of wavelength 100 m. Check that the approximations are justified.
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Exercise 8.2The lunar tidal bulge moves around the equator once per lunar day, 24.84 hours. Find the depth of an equatorial canal in which free long waves would keep pace with it (Earth radius 6378 km).
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Exercise 8.3Show that the group velocity \(c_g=d\omega/dk\) equals \(c/2\) in deep water and \(c\) in shallow water.
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Weighing the Earth and isostasy
For the general reader. Airy weighed the Earth by timing a pendulum at the top and bottom of a coal mine. He then argued that mountains float. A mountain range, he suggested, is like an iceberg: it is supported by a deep root of light rock pushing down into denser material below. Both ideas came from the same question: how does gravity change as you move through the Earth’s outer layers?
- Derive gravity at depth in a layered Earth and its use to find the mean density.
- State the Airy and Pratt hypotheses of isostasy.
- Compute the depth of a mountain root.
Gravity in a mine
Model the Earth as a sphere of radius \(R\) and mean density \(\bar\rho\), with an outer shell of density \(\rho_s\). At depth \(d\ll R\), the shell above exerts no net force (Newton’s shell theorem), and the matter below is the whole Earth minus the shell of thickness \(d\).
Gravity increases with depth whenever \(\rho_s<\tfrac23\bar\rho\). A pendulum clock’s rate changes by the fraction \(\tfrac12\,\delta g/g\).
Derivation. \(g(d)=G(M-4\pi R^2d\,\rho_s)/(R-d)^2\approx g(0)\big(1-\frac{3d\rho_s}{R\bar\rho}\big)\big(1+\frac{2d}R\big)\), using \(M=\tfrac43\pi R^3\bar\rho\). Measuring \(\delta g/g\) and the density of the rock \(\rho_s\) then gives \(\bar\rho\).
Airy first tried this in 1826 and 1828 at the Dolcoath copper mine in Cornwall; both attempts failed through accidents. In 1854 he succeeded at Harton colliery near South Shields, comparing pendulums at the surface and at the bottom of the shaft by electric telegraph signals. He found \(\bar\rho\approx6.566\ \mathrm{g\,cm^{-3}}\), well above the modern value of 5.51, mainly because the density of the overlying rock was hard to estimate.
Isostasy (1855)
The Great Trigonometrical Survey of India had found a discrepancy between astronomical and geodetic latitudes at stations south of the Himalaya. John Henry Pratt calculated in 1855 that the attraction of the mountains should have deflected the plumb line about three times more than observed. Airy replied the same year that the mountains could not be supported by a rigid crust. He proposed that the crust floats on a denser, yielding layer, and that a mountain range is held up by a root of light crust projecting downward, whose mass deficit offsets the mountain’s excess. Pratt proposed instead that the rock under mountains is less dense. Both are models of what Clarence Dutton later named isostasy.
For crust of density \(\rho_c\) floating on mantle of density \(\rho_m\), a topographic load of height \(h\) is compensated by a root of depth
\[ r=\frac{\rho_c}{\rho_m-\rho_c}\,h, \]so that the mass of every vertical column above a deep compensation level is the same.
Exercise 9.1With \(\rho_c=2.75\) and \(\rho_m=3.30\ \mathrm{g\,cm^{-3}}\), find the root under a plateau 5 km high, and the crustal thickness there if normal crust is 35 km thick.
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Exercise 9.2An ocean of depth \(w\) (water density \(\rho_w=1.03\)) sits on thinned crust. Show that the crust is thinner by \(w(\rho_c-\rho_w)/(\rho_m-\rho_c)\), and evaluate for \(w=4\) km.
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Exercise 9.3Take \(d=380\) m, \(R=6371\) km, \(\rho_s=2.5\) and \(\bar\rho=5.5\). By how many seconds a day does a pendulum clock at the bottom gain on one at the top?
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Bars, beams and the stress function
For the general reader. A metal bar that defines the length of a yard sags under its own weight, and the sag changes the distance between marks on it. Airy calculated where to support such a bar so that its ends stay parallel. Later, thinking about the stresses inside the iron girders of Victorian bridges, he invented a single function from which all the stresses in a flat elastic body can be computed. Engineers still use it.
- Derive the Airy points of a uniformly loaded bar.
- Show that the Airy stress function solves the equilibrium equations.
- Use polynomial stress functions for beam problems.
The Airy points (1845)
As a member of the commission for restoring the imperial standards of length after the fire at the Houses of Parliament in 1834, Airy considered a uniform bar of length \(L\) resting on two symmetric supports. By Euler–Bernoulli beam theory the bending moment \(M(x)\) satisfies \(EI\,y''=M\), so the slope is \(\theta(x)=\frac1{EI}\int M\,dx\).
A uniform bar of length \(L\) on two symmetric supports has its end faces exactly vertical (zero end slope) when the supports are \(L/\sqrt3\approx0.577L\) apart, that is \(\frac{3-\sqrt3}{6}L\approx0.2113L\) from each end.
Proof. Put the origin at one end and the support at \(x=a\). With load \(w\) per unit length, \(M=-wx^2/2\) for \(x<a\) and \(M=-wx^2/2+\tfrac{wL}2(x-a)\) for \(a<x<L/2\). By symmetry the slope at the centre is zero, so the end slope vanishes when \(\int_0^{L/2}M\,dx=0\): \(-\frac{wL^3}{48}+\frac{wL}{4}\big(\frac L2-a\big)^2=0\), giving \(\frac L2-a=\frac L{2\sqrt3}\). Standards of length defined by marks on the end faces are supported at the Airy points. A bar with marks on its neutral surface is better supported at the Bessel points, about \(0.559L\) apart, which minimize the change in length between the marks.
The stress function (1862)
In a paper read to the Royal Society in 1862 and published in 1863, On the strains in the interior of beams, Airy introduced a potential for plane stress. In two dimensions without body force, equilibrium requires \(\partial_x\sigma_{xx}+\partial_y\sigma_{xy}=0\) and \(\partial_x\sigma_{xy}+\partial_y\sigma_{yy}=0\).
On a simply connected plane region, a smooth symmetric stress field satisfies the equilibrium equations without body force if and only if there is a function \(\Phi\) with
\[ \sigma_{xx}=\frac{\partial^2\Phi}{\partial y^2},\qquad \sigma_{yy}=\frac{\partial^2\Phi}{\partial x^2},\qquad \sigma_{xy}=-\frac{\partial^2\Phi}{\partial x\,\partial y}. \]For a linear isotropic elastic material, compatibility of the strains is equivalent to \(\nabla^4\Phi=0\).
Proof sketch. The first equilibrium equation says the field \((\sigma_{xx},\sigma_{xy})\) is divergence-free, so on a simply connected region it is the rotated gradient of a function \(\psi\): \(\sigma_{xx}=\partial_y\psi\), \(\sigma_{xy}=-\partial_x\psi\). Similarly \(\sigma_{xy}=\partial_y\chi\), \(\sigma_{yy}=-\partial_x\chi\). Equating the two expressions for \(\sigma_{xy}\), \(\partial_x\psi+\partial_y\chi=0\), so \((\psi,-\chi)\) is again a gradient: \(\psi=\partial_y\Phi\), \(\chi=-\partial_x\Phi\). Airy did not impose compatibility, so his method did not single out the true elastic solution. James Clerk Maxwell added the biharmonic condition in 1868. The Airy stress function, with complex-variable methods developed by Kolosov and Muskhelishvili, became the standard tool of plane elasticity.
Exercise 10.1Show that \(\Phi=\frac{M}{6I}y^3\) (per unit thickness, \(I\) the second moment of area) gives the stress of pure bending, and that it is biharmonic.
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Exercise 10.2For which constants is \(\Phi=Ax^4+Bx^2y^2+Cy^4\) biharmonic? Check that every \(\Phi\) gives stresses satisfying equilibrium.
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Exercise 10.3Where should a 1 m standard bar be supported to make its end faces vertical? How far apart would the supports be for a 36-inch yard bar?
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Astronomer Royal, 1835–1881
Airy ran Greenwich as a disciplined organization. He hired young computers, often boys, to carry out routine reductions by fixed rules, and he reduced and published the backlog of Greenwich observations of the planets and the Moon from 1750 onward. He added a magnetic and meteorological department in 1838, and designed new instruments: an altazimuth (1847), the reflex zenith tube, and above all the transit circle installed in 1850 and first used in 1851. From 1852 Greenwich time was sent along the telegraph lines of the railways, which helped make it the standard time of Britain. The historian Olin Eggen summed him up: not a great scientist, but one who made great science possible.
He was also the government’s scientific adviser on almost everything. His experiments of 1838 in the iron ship Rainbow showed how to correct a ship’s compass for the magnetism of an iron hull with fixed magnets and masses of soft iron. He served on the Railway Gauge Commission of 1845–46, which favoured the narrow gauge, against Isambard Kingdom Brunel’s broad gauge. He clashed with Charles Babbage, whose calculating engine he judged useless to the government.
Neptune
In 1845 John Couch Adams, a young Cambridge fellow, computed the position of an unseen planet disturbing Uranus. In October he called twice at Greenwich without seeing Airy. In November Airy wrote to ask whether the theory also explained the errors in Uranus’s distance from the Sun; Adams did not reply. In June 1846 Urbain Le Verrier published a prediction close to Adams’s, and in July Airy asked James Challis to search with the Cambridge telescope. Challis recorded the planet in August without recognizing it. On 23 September 1846 Johann Galle at Berlin found Neptune within a degree of Le Verrier’s position. In the English controversy that followed, Airy was blamed for inaction. Adams’s papers, missing from the Greenwich archives for decades and recovered in 1999, suggest that his prediction was less definite than the English account had claimed.
DiscussionAiry believed an observatory should make long, uniform series of measurements, not chase discoveries. Using the Neptune affair, discuss the strengths and costs of that philosophy. Was it reasonable to ask Adams about the radius vector before searching?
Last years and legacy, 1871–1892
Airy was President of the Royal Society from 1871 to 1873 and was knighted (KCB) in 1872, having earlier declined the honour three times because of the cost. He organized the British expeditions to observe the transit of Venus in 1874. His wife Richarda died in 1875; of their nine children, three died young. Their son Wilfrid became a civil engineer, and Hubert a physician who described the visual aura of migraine. In 1875 Airy became the first scientist to receive the freedom of the City of London. He advised on wind loading for the Tay and Forth bridges and gave evidence at the inquiry into the Tay Bridge disaster of 1879.
He retired on 15 August 1881, aged 80, and lived with two unmarried daughters at the White House near Greenwich Park. He continued to work on lunar theory, without success. He died at Greenwich on 2 January 1892, aged 90, and was buried at St Mary’s church, Playford. Wilfrid edited his Autobiography, published in 1896.
Legacy
- The prime meridian: in 1884 the International Meridian Conference in Washington chose the meridian through his transit circle as longitude zero. On Mars, the crater Airy-0 defines the zero meridian.
- Optics: the Airy disk sets the resolution of every circular aperture; the Airy function describes caustics, turning points in quantum mechanics and much else.
- Earth science: Airy isostasy, Airy wave theory and the Airy ellipsoid of the Ordnance Survey.
- Engineering: the Airy points of length standards and the Airy stress function of plane elasticity.
- Honours: Copley Medal (1831), Gold Medals of the Royal Astronomical Society (1833, 1846), Lalande Prize (1834), Royal Medal (1845), Albert Medal (1876); president of the Royal Astronomical Society four times and of the British Association in 1851.
Final projectChoose one eponym from Part II. Write a two-page account for a strong undergraduate that explains the practical problem Airy faced, derives the result, works one example, and traces one modern application.
Sources and further reading
Biography
- J. J. O’Connor and E. F. Robertson, George Biddell Airy, MacTutor History of Mathematics.
- George Biddell Airy, Wikipedia.
- W. Airy (ed.), Autobiography of Sir George Biddell Airy, Cambridge University Press, 1896 (Project Gutenberg).
- Airy, George Biddell, Biographical Encyclopedia of Astronomers, via MacTutor.
- Orwell Astronomical Society, Airy and Playford.
- Discovery of Neptune, Wikipedia; W. Sheehan, N. Kollerstrom and C. B. Waff, The case of the pilfered planet, Scientific American, December 2004.
Physics
- G. B. Airy, On a peculiar defect in the eye, and a mode of correcting it, Trans. Camb. Phil. Soc. 2 (1827; read 1825).
- G. B. Airy, On an inequality of long period in the motions of the Earth and Venus, Phil. Trans. R. Soc. 122 (1832).
- G. B. Airy, On the diffraction of an object-glass with circular aperture, Trans. Camb. Phil. Soc. 5 (1835) 283–291.
- G. B. Airy, On the intensity of light in the neighbourhood of a caustic, Trans. Camb. Phil. Soc. 6 (1838) 379–402.
- G. B. Airy, Tides and waves, Encyclopaedia Metropolitana, 1841.
- G. B. Airy, On the computation of the effect of the attraction of mountain-masses, Phil. Trans. R. Soc. 145 (1855).
- G. B. Airy, On the strains in the interior of beams, Phil. Trans. R. Soc. 153 (1863).
- O. Vallée and M. Soares, Airy Functions and Applications to Physics, Imperial College Press, 2004.
- M. Born and E. Wolf, Principles of Optics, Cambridge University Press (diffraction by a circular aperture).
- H. M. Nussenzveig, The theory of the rainbow, Scientific American 236 (1977).
- Airy function, Airy disk and Isostasy, Wikipedia.