courseshub.world · Lives in Mathematics and Physics

George Gabriel Stokes

Sir George Gabriel Stokes, 1st Baronet

A course in two strands: the life of the rector’s son from County Sligo who held Newton’s chair for 54 years, and the work that gave us the equations of viscous flow, the drag on a falling sphere, fluorescence, the polarization parameters and the theorem every physics student uses.

Born 13 August 1819, Skreen, County Sligo Died 1 February 1903, Cambridge Lucasian Professor 1849–1903 Level advanced undergraduate → graduate
Flow past a sphere, computed from the stream functions. Above the axis: Stokes’s creeping flow (1851), which disturbs the fluid far from the body. Below: ideal flow, which closes up within a radius or two.
Lesson 00

How to use this course

Stokes’s name is attached to more results in mathematical physics than almost anyone of his century: the Navier–Stokes equations, Stokes’s law, Stokes’s theorem, the Stokes shift, the Stokes parameters, Stokes drift and the Stokes phenomenon. He was also, for over thirty years, the man who ran the Royal Society’s scientific business, and the adviser to whom British physicists sent their problems. This course presents the work and the life together.

Prerequisites for Part II
  • Vector calculus: gradient, divergence, curl, line and surface integrals.
  • Ordinary and partial differential equations at the level of separation of variables.
  • Basic fluid mechanics (continuity, pressure) and the wave description of light.
  • For Lessons 09 and 10, real analysis (uniform convergence) and some complex analysis.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. The order is roughly that of his papers: fluids (1842–1851), waves (1847), analysis (1847–1857), light (1849–1852), with the theorem that bears his name placed where it fits the vector calculus.

Reference

Timeline

1819Born 13 August at Skreen, County Sligo, youngest of six children of the Rev. Gabriel Stokes, rector of the parish, and Elizabeth Haughton.
1832Sent to school in Dublin, living with his uncle John Stokes.
1835Enters Bristol College.
1837Enters Pembroke College, Cambridge; coached by William Hopkins.
1841Senior Wrangler and first Smith’s Prizeman; elected Fellow of Pembroke.
1842–43First papers, on the steady motion of incompressible fluids.
1845On the theories of the internal friction of fluids in motion: the Navier–Stokes equations.
1847On the theory of oscillatory waves; paper on the critical values of the sums of periodic series.
1849Elected Lucasian Professor of Mathematics. Papers on diffraction and on the variation of gravity over the Earth.
1850Pendulum memoir read (published 1851), containing Stokes’s law. Computes the zeros of Airy’s integral. Receives from William Thomson the theorem now named after Stokes.
1851Fellow of the Royal Society. Introduces the Stokes parameters for partially polarized light.
1852On the change of refrangibility of light: fluorescence. Rumford Medal.
1854Sets the theorem as a Smith’s Prize question. Becomes Secretary of the Royal Society.
1857Marries Mary Susanna Robinson, 4 July, in Armagh, and gives up his fellowship. Reads his paper on the discontinuity of arbitrary constants (published 1864): the Stokes phenomenon.
1880–1905Mathematical and Physical Papers, five volumes.
1885–90President of the Royal Society.
1887–92Member of Parliament for Cambridge University.
1889Created a baronet.
1893Copley Medal of the Royal Society.
1899Cambridge celebrates his jubilee as Lucasian Professor.
1902Elected Master of Pembroke.
1903Dies 1 February in Cambridge, aged 83. Buried in Mill Road Cemetery.
Part I · Formation
Lesson 01

Skreen, Dublin, Bristol, 1819–1837

George Gabriel Stokes was born on 13 August 1819 at Skreen, a village on the Atlantic coast of County Sligo in the west of Ireland. His father, the Rev. Gabriel Stokes, was the Church of Ireland rector of the parish and a graduate of Trinity College Dublin. His mother, Elizabeth Haughton, was also the daughter of a clergyman. George was the youngest of six children, and his three elder brothers all entered the church. The home was evangelical Protestant, and Stokes kept a serious Christian faith throughout his life.

His father taught him Latin grammar, and he had his first lessons from the parish clerk. He kept a love of the sea; years later he tested his theory of the steepest possible wave by wading, trousers rolled up, into the surf on the north coast of Ireland.

In 1832, aged thirteen, he went to Dublin to attend the Rev. R. H. Wall’s school in Hume Street, living with his uncle John Stokes. His father died while he was there. In 1835 he moved to Bristol College in England, where he was taught by Francis Newman, brother of the future cardinal John Henry Newman. Stokes won prizes in mathematics and prepared for Cambridge.

Lesson 02

Pembroke and the tripos, 1837–1841

Stokes entered Pembroke College, Cambridge, in 1837. Cambridge mathematics meant the Mathematical Tripos, a punishing written examination whose top candidate, the Senior Wrangler, became a national celebrity. Serious candidates hired private coaches. From his second year Stokes was coached by William Hopkins, the most successful coach of the era, whose pupils also included William Thomson (later Lord Kelvin), Peter Guthrie Tait and James Clerk Maxwell. Hopkins steered him toward physical astronomy and physical optics, and more generally toward mathematics applied to nature.

In 1841 Stokes was Senior Wrangler, and he then took first place in the Smith’s Prize examination, the more advanced competition that followed the tripos. Pembroke immediately elected him to a fellowship. A fellowship gave him an income and freedom to research, but it required him to stay unmarried.

The Cambridge school

Stokes belonged to the first generation of Cambridge mathematicians who read the French analysts closely: Joseph Fourier on heat, Augustin-Louis Cauchy on elasticity, Siméon Denis Poisson on everything. The mathematical physics of George Green, published obscurely in Nottingham in 1828, was being rediscovered by Thomson at this time. Stokes and Thomson met as young men and remained close correspondents for more than fifty years. Their letters are one of the main records of how British mathematical physics was built.

Lesson 03

The young fellow, 1841–1849

Stokes chose hydrodynamics as his first subject on Hopkins’s advice. His papers of 1842 and 1843 treated the steady motion of incompressible fluids. He learned that Jean-Marie Duhamel had obtained some of the same results in the context of heat, but published anyway, since the fluid setting was new. In 1845 he produced the great paper on the internal friction of fluids (Lesson 04), and in 1846 a report on recent researches in hydrodynamics for the British Association.

The output of these years is remarkable for its range. In 1845 he wrote on the aberration of starlight and the motion of the ether. In 1847 came water waves (Lesson 06) and a paper on Fourier series with the idea now called non-uniform convergence (Lesson 09). In 1849 he published a dynamical theory of diffraction, using it to argue that the vibrations of light are perpendicular to the plane of polarization. The same year he gave a theorem in geodesy, Stokes’s formula, which determines the shape of the Earth’s geoid from measurements of gravity at its surface.

The Lucasian chair

In October 1849, aged thirty, Stokes was elected Lucasian Professor of Mathematics, the chair once held by Isaac Newton. The Times announced the appointment under his elder brother’s name, which amused him. The chair was poorly paid, and to supplement it he also taught physics at the Government School of Mines in London. He held the Lucasian chair until his death, fifty-four years later.

Part II · The mathematics and physics
Lesson 04

Internal friction: the Navier–Stokes equations

For the general reader. Honey, air and water resist being sheared. Stokes wrote down, from simple physical principles, how that internal friction enters the equations of fluid motion. The resulting Navier–Stokes equations govern everything from blood flow to weather, and whether their solutions in three dimensions always stay smooth is one of the seven Millennium Prize Problems.

Goals
  • Derive the viscous stress from linearity and isotropy.
  • Write the Navier–Stokes equations and the no-slip condition.
  • Solve two exact flows.

History

Claude-Louis Navier wrote equations of this form in 1822 from a model of molecular forces. Cauchy, Poisson and Adhémar Barré de Saint-Venant derived related equations before 1845. Stokes’s 1845 paper derived them from assumptions about the continuum alone, without a molecular model, and applied them to concrete problems. He also argued for the condition that a viscous fluid does not slip at a solid wall, which became standard after comparison with experiment.

The constitutive law

Let \(\mathbf u(\mathbf x,t)\) be the velocity field and \(e_{ij}=\tfrac12(\partial_iu_j+\partial_ju_i)\) the rate-of-strain tensor. Stokes assumed that the stress in a moving fluid differs from the hydrostatic pressure by an amount that (i) depends linearly on the rates of strain, (ii) is the same in all directions (isotropy), and (iii) vanishes for a rigid motion.

Theorem (Stokes, 1845)

Under (i)–(iii), the stress tensor has the form

\[ \sigma_{ij}=-p\,\delta_{ij}+2\mu\,e_{ij}+\lambda\,(\nabla\!\cdot\mathbf u)\,\delta_{ij} \]

for two scalars \(\mu,\lambda\). For an incompressible fluid with constant density \(\rho\) and viscosity \(\mu\), Newton’s second law \(\rho\,D\mathbf u/Dt=\nabla\!\cdot\sigma+\mathbf f\) becomes

\[ \rho\Big(\frac{\partial\mathbf u}{\partial t}+(\mathbf u\cdot\nabla)\mathbf u\Big)=-\nabla p+\mu\nabla^2\mathbf u+\mathbf f,\qquad \nabla\!\cdot\mathbf u=0, \]

with \(\mathbf u=\mathbf 0\) on a fixed solid boundary (no slip).

Derivation sketch. A linear map from symmetric tensors to symmetric tensors that commutes with all rotations must be a combination of the identity and the trace: \(e\mapsto 2\mu e+\lambda(\operatorname{tr}e)I\). This is a short exercise in representation theory, since symmetric tensors split into the trace part and the traceless part, which are irreducible under rotations. Rigid rotations have \(e=0\), so condition (iii) holds automatically. Taking the divergence and using \(\nabla\!\cdot\mathbf u=0\) gives \(\partial_j(2\mu e_{ij})=\mu\nabla^2u_i\). Stokes also proposed that the mean normal stress equal the pressure, \(3\lambda+2\mu=0\). This Stokes hypothesis, of zero bulk viscosity, is still widely used for gases.

The ratio of the inertial to the viscous term, for a flow of speed \(U\) and size \(L\), is the Reynolds number \(Re=\rho UL/\mu\). Stokes identified this combination in 1851. Osborne Reynolds made it famous in 1883.

Exercise 4.1(Pipe flow.) Fluid flows steadily along a circular pipe of radius \(R\) under a pressure gradient \(-dp/dz=G\). Seek \(\mathbf u=w(r)\,\hat{\mathbf z}\). Find \(w\) and the volume flux \(Q\).

Show solution
The convective term vanishes because \(w\) does not depend on \(z\). The equation reduces to \(\mu\frac1r\frac{d}{dr}\big(r\frac{dw}{dr}\big)=-G\). Integrating, \(w=-\frac{G}{4\mu}r^2+A\ln r+B\). Regularity at \(r=0\) gives \(A=0\), and no slip at \(r=R\) gives \(w=\frac{G}{4\mu}(R^2-r^2)\). Then \(Q=\int_0^R 2\pi r\,w\,dr=\frac{\pi GR^4}{8\mu}\), the Hagen–Poiseuille law. The flux scales as \(R^4\).

Exercise 4.2(Stokes’s second problem, 1851.) A plane wall at \(y=0\) oscillates in its own plane with velocity \(U\cos\omega t\). Find the periodic velocity \(u(y,t)\) of the fluid above it and the penetration depth.

Show solution
With \(\mathbf u=u(y,t)\hat{\mathbf x}\), the equation is \(u_t=\nu u_{yy}\), \(\nu=\mu/\rho\). Try \(u=\mathrm{Re}\big(U e^{i\omega t}f(y)\big)\): \(i\omega f=\nu f''\), so \(f=e^{-(1+i)y/\delta}\) with \(\delta=\sqrt{2\nu/\omega}\), choosing the decaying root. Hence \(u=U e^{-y/\delta}\cos(\omega t-y/\delta)\): a damped shear wave. For water (\(\nu\approx10^{-6}\,\mathrm{m^2/s}\)) at \(\omega=1\,\mathrm{s^{-1}}\), \(\delta\approx1.4\) mm. Stokes needed this boundary layer to compute the damping of a pendulum.

Exercise 4.3Show that any rigid motion \(\mathbf u=\mathbf a+\boldsymbol\omega\times\mathbf x\) has \(e_{ij}=0\).

Show solution
\(\partial_iu_j=\partial_i(\epsilon_{jkl}\omega_kx_l)=\epsilon_{jki}\omega_k\), which is antisymmetric in \(i,j\). Its symmetric part \(e_{ij}\) is therefore zero.
Lesson 05

Pendulums and Stokes’s law

For the general reader. A pendulum swinging in air is slowed and its period lengthened by the air it drags along. Precise pendulum measurements of gravity needed a correction for this, and Stokes computed it. Along the way he found the drag on a small sphere moving slowly through a viscous fluid. His formula explains why clouds float, and Robert Millikan used it to measure the charge of the electron.

Goals
  • Write the Stokes (creeping-flow) equations and their range of validity.
  • State the drag on a sphere and split it into pressure and friction.
  • Apply the law to terminal velocity.

Creeping flow

When \(Re\ll1\), drop the inertial terms. The steady equations become linear:

\[ \mu\nabla^2\mathbf u=\nabla p,\qquad\nabla\!\cdot\mathbf u=0 . \]

For a sphere of radius \(a\) held fixed in a stream of speed \(U\) along the axis, Stokes worked with an axisymmetric stream function \(\psi(r,\theta)\) in spherical coordinates. The solution is

\[ \psi=\tfrac12U\sin^2\theta\,\Big(r^2-\frac{3ar}{2}+\frac{a^3}{2r}\Big),\qquad p=p_\infty-\frac{3\mu Ua}{2}\,\frac{\cos\theta}{r^2}. \]
Theorem (Stokes’s law, 1851)

In steady creeping flow past a rigid sphere of radius \(a\) with no slip, the force on the sphere is

\[ \mathbf F=6\pi\mu a\,\mathbf U, \]

of which one third, \(2\pi\mu aU\), comes from pressure and two thirds, \(4\pi\mu aU\), from viscous shear stress.

The far field decays like \(a/r\), as the hero picture shows: a sphere in slow viscous flow disturbs the fluid many radii away. This slow decay causes two famous difficulties. In two dimensions no Stokes solution for flow past a cylinder satisfies both the wall and the far-field conditions (Stokes’s paradox). In three dimensions the neglected inertial terms become comparable to viscous ones at distances of order \(a/Re\), which Carl Wilhelm Oseen repaired in 1910.

Terminal velocity

A sphere of density \(\rho_s\) falling in fluid of density \(\rho_f\) reaches speed \(v\) when drag balances weight minus buoyancy: \(6\pi\mu av=\tfrac43\pi a^3(\rho_s-\rho_f)g\), so

\[ v=\frac{2}{9}\,\frac{(\rho_s-\rho_f)\,g\,a^2}{\mu}. \]

Exercise 5.1An oil droplet of density \(900\ \mathrm{kg/m^3}\) and radius \(1\ \mu\mathrm m\) falls in air (\(\mu=1.8\times10^{-5}\ \mathrm{Pa\,s}\), density negligible). Find its terminal speed and check that \(Re\ll1\).

Show solution
\(v=\frac{2\cdot900\cdot9.81\cdot10^{-12}}{9\cdot1.8\times10^{-5}}\approx1.1\times10^{-4}\ \mathrm{m/s}\), about 0.1 mm/s. With air density \(1.2\ \mathrm{kg/m^3}\), \(Re=\rho_{air}v(2a)/\mu\approx1.2\cdot1.1\times10^{-4}\cdot2\times10^{-6}/1.8\times10^{-5}\approx1.5\times10^{-5}\). Millikan measured such slow speeds to find droplet sizes; at this size a small slip correction to Stokes’s law is also needed.

Exercise 5.2Use dimensional analysis to show that in creeping flow the drag on a sphere must be \(F=C\mu aU\) for a pure number \(C\). Why can \(\rho\) not appear?

Show solution
In the Stokes equations only \(\mu\) appears; \(\rho\) has been dropped with the inertial terms. The available quantities are \(\mu\) (\(\mathrm{kg\,m^{-1}s^{-1}}\)), \(a\) (m) and \(U\) (\(\mathrm{m\,s^{-1}}\)). The only combination with units of force (\(\mathrm{kg\,m\,s^{-2}}\)) is \(\mu aU\). Linearity of the equations confirms the dependence on \(U\) to first power. Stokes’s calculation gives \(C=6\pi\).

Exercise 5.3Check that the stream function above vanishes on the sphere and on the axis, and that it tends to that of a uniform stream as \(r\to\infty\).

Show solution
At \(r=a\): \(a^2-\frac32a^2+\frac12a^2=0\). On the axis \(\sin\theta=0\). As \(r\to\infty\), \(\psi\approx\frac12Ur^2\sin^2\theta=\frac12U\varpi^2\) with \(\varpi=r\sin\theta\) the distance from the axis, which is the stream function of uniform flow \(U\hat{\mathbf z}\). The no-slip condition also requires \(\partial\psi/\partial r=0\) at \(r=a\): \(2a-\frac32a-\frac12a=0\).
Lesson 06

Oscillatory waves and Stokes drift

For the general reader. An ocean wave looks as if water travels with it, but the water mostly goes round in circles. Stokes showed that the circles do not quite close, so each wave carries the water forward a little. He also found that steep waves travel faster than gentle ones, and that the steepest possible wave has a sharp crest with a 120-degree angle.

Goals
  • Derive the linear dispersion relation for deep-water waves.
  • Compute the Stokes drift.
  • State Stokes’s nonlinear results.

Linear theory

For irrotational flow of an inviscid incompressible fluid, \(\mathbf u=\nabla\phi\) with \(\nabla^2\phi=0\). Take deep water occupying \(z<\eta(x,t)\). Linearizing the free-surface conditions at \(z=0\) gives \(\eta_t=\phi_z\) and \(\phi_t+g\eta=0\). A travelling wave \(\eta=a\cos(kx-\omega t)\) has

\[ \phi=\frac{a\omega}{k}\,e^{kz}\sin(kx-\omega t),\qquad \omega^2=gk . \]

Water particles move on circles of radius \(ae^{kz}\), shrinking with depth.

Theorem (Stokes, 1847)

For periodic deep-water waves of small amplitude \(a\) and wavenumber \(k\):

  1. the surface profile is \(\eta=a\cos\theta+\tfrac12ka^2\cos2\theta+O(a^3)\), \(\theta=kx-\omega t\), with crests sharper and troughs flatter than a sine wave;
  2. the phase speed is \(c^2=\dfrac gk\big(1+k^2a^2\big)+O(a^4)\);
  3. fluid particles have a mean forward velocity, the Stokes drift, \(\bar u_S(z)=\omega ka^2e^{2kz}\).

Derivation of the drift

A particle at mean position \((x_0,z_0)\) has displacement \((\xi,\zeta)\) with \(\xi=-ae^{kz_0}\sin\theta\), \(\zeta=ae^{kz_0}\cos\theta\) to first order. Its velocity is the Eulerian velocity at its actual position: \(u(x_0+\xi,z_0+\zeta)\approx u+\xi\,\partial_xu+\zeta\,\partial_zu\). With \(u=a\omega e^{kz}\cos\theta\), the first term averages to zero over a period, and

\[ \langle\xi\,\partial_xu+\zeta\,\partial_zu\rangle=a^2\omega k\,e^{2kz_0}\big\langle\sin^2\theta+\cos^2\theta\big\rangle=\omega ka^2e^{2kz_0}. \]

In 1880 Stokes returned to the problem and argued that the wave of greatest height has a crest angle of \(120^\circ\). The existence of this extreme wave and its crest angle were proved by Charles Amick, Ludwig Fraenkel and John Toland in 1982.

Exercise 6.1Derive \(\omega^2=gk\) by substituting \(\phi=A e^{kz}\sin(kx-\omega t)\) and \(\eta=a\cos(kx-\omega t)\) into the two linear surface conditions.

Show solution
\(\eta_t=\phi_z\) at \(z=0\): \(a\omega\sin\theta=Ak\sin\theta\), so \(A=a\omega/k\). \(\phi_t+g\eta=0\): \(-A\omega\cos\theta+ga\cos\theta=0\), so \(A\omega=ga\). Combining, \(a\omega^2/k=ga\), i.e. \(\omega^2=gk\). The phase speed \(\sqrt{g/k}\) grows with wavelength: long swells outrun short waves.

Exercise 6.2A swell has wavelength 100 m and amplitude 1 m. Find its period, its phase speed, and the Stokes drift at the surface and at 25 m depth.

Show solution
\(k=2\pi/100\approx0.0628\ \mathrm{m^{-1}}\), \(\omega=\sqrt{gk}\approx0.785\ \mathrm{s^{-1}}\), period \(\approx8.0\) s, \(c=\omega/k\approx12.5\ \mathrm{m/s}\). Surface drift \(\omega ka^2\approx0.049\ \mathrm{m/s}\), about 4 km per day. At \(z=-25\) m, the factor \(e^{2kz}=e^{-\pi}\approx0.043\), giving about 2 mm/s.
Lesson 07

Light: reflection, fluorescence, polarization

For the general reader. Stokes made three lasting contributions to optics. He found a neat argument, by running light backwards in time, that links reflection from the two sides of a surface. He explained why some liquids glow blue under ultraviolet light and named the effect fluorescence. And he found four numbers that describe any beam of light, polarized or not, which astronomers still use.

Goals
  • Derive the Stokes relations from reversibility.
  • State Stokes’s rule for fluorescence.
  • Define the Stokes parameters and their additivity.

The Stokes relations (1849)

Let a wave of unit amplitude hit an interface from medium 1, with amplitude reflection and transmission coefficients \(r,t\); from medium 2 the coefficients are \(r',t'\).

Theorem (Stokes relations)

For a lossless interface, \(\ r'=-r\ \) and \(\ tt'=1-r^2\).

Proof. Reverse time. The reflected wave \(r\) and transmitted wave \(t\), sent back, must recombine into the incident wave alone. Sending \(r\) back from side 1 produces \(r\cdot r\) reflected and \(r\cdot t\) transmitted; sending \(t\) back from side 2 produces \(t\cdot r'\) reflected into side 2 and \(t\cdot t'\) transmitted into side 1. The total in side 1 must be 1: \(r^2+tt'=1\). The total going into side 2 must vanish: \(rt+tr'=0\), so \(r'=-r\). Stokes used this to explain the perfectly black centre of Newton’s rings in reflection: the two reflections differ in sign.

Fluorescence (1852)

John Herschel had described in 1845 the blue glow on the surface of a solution of quinine sulphate in sunlight. Stokes dispersed sunlight with a prism and passed the spectrum through the solution. The glow appeared where the invisible ultraviolet entered the liquid. He concluded that the solution absorbs ultraviolet light and re-emits it at visible wavelengths, and he coined the word fluorescence after fluorspar, a mineral that shows the effect. He stated the rule that the emitted light is always of lower refrangibility, that is, longer wavelength, than the light that excites it. The difference is now the Stokes shift. Quantum theory explains it: part of each absorbed photon’s energy is lost to vibrations before emission. Stokes also suggested using fluorescence to make ultraviolet spectra visible.

The Stokes parameters (1851)

A quasi-monochromatic beam along \(z\) has transverse field components \(E_x=a_x e^{i\phi_x}\), \(E_y=a_y e^{i\phi_y}\) with slowly varying amplitudes and phases. With \(\delta=\phi_y-\phi_x\) and \(\langle\cdot\rangle\) a time average, define

\[ I=\langle a_x^2+a_y^2\rangle,\quad Q=\langle a_x^2-a_y^2\rangle,\quad U=\langle2a_xa_y\cos\delta\rangle,\quad V=\langle2a_xa_y\sin\delta\rangle . \]
Theorem (Stokes, 1851)

The parameters of a superposition of mutually incoherent beams are the sums of their parameters. Always \(I^2\ge Q^2+U^2+V^2\), with equality exactly for fully polarized light. The degree of polarization is \(P=\sqrt{Q^2+U^2+V^2}/I\).

Additivity holds because the cross terms between incoherent beams average to zero. The inequality is the Cauchy–Schwarz inequality for the coherency matrix \(\begin{pmatrix}\langle E_xE_x^*\rangle&\langle E_xE_y^*\rangle\\\langle E_yE_x^*\rangle&\langle E_yE_y^*\rangle\end{pmatrix}\), whose determinant is \(\tfrac14(I^2-Q^2-U^2-V^2)\ge0\). Henri Poincaré later represented polarized states as points \((Q,U,V)/I\) on a sphere, and the same mathematics describes a qubit on the Bloch sphere. (The sign convention for \(V\) differs between fields.)

Exercise 7.1Find \((I,Q,U,V)\) for unit-intensity light that is (a) linearly polarized along \(x\), (b) linearly polarized at \(45^\circ\), (c) circularly polarized with \(\delta=\pi/2\), (d) unpolarized.

Show solution
(a) \(a_x=1,a_y=0\): \((1,1,0,0)\). (b) \(a_x=a_y=1/\sqrt2,\delta=0\): \((1,0,1,0)\). (c) \(a_x=a_y=1/\sqrt2,\delta=\pi/2\): \((1,0,0,1)\). (d) \((1,0,0,0)\), e.g. as the incoherent sum of half-intensity beams (a) and its \(y\)-polarized partner \((\tfrac12,-\tfrac12,0,0)\).

Exercise 7.2Equal intensities of \(x\)-polarized light and circularly polarized light, incoherent with each other, are combined. What is the degree of polarization?

Show solution
Add \((\tfrac12,\tfrac12,0,0)\) and \((\tfrac12,0,0,\tfrac12)\): \((1,\tfrac12,0,\tfrac12)\). Then \(P=\sqrt{\tfrac14+\tfrac14}=1/\sqrt2\approx0.71\). The mixture is partially polarized, though each component is fully polarized.

Exercise 7.3Quinine absorbs strongly near 350 nm and fluoresces near 450 nm. Using \(E=hc/\lambda\) with \(hc\approx1240\ \mathrm{eV\,nm}\), how much energy per photon is lost?

Show solution
\(1240/350\approx3.54\) eV and \(1240/450\approx2.76\) eV, so about 0.8 eV per photon goes into molecular vibrations and the solvent. The sign of the difference is Stokes’s rule.
Lesson 08

Stokes’s theorem

For the general reader. The total swirl of a flow over a surface equals the circulation around the surface’s edge. William Thomson stated this in a letter to Stokes in 1850, and Stokes set it as an examination question in 1854. Because of the exam it carries Stokes’s name. It is the curved-surface version of the fundamental theorem of calculus.

Goals
  • State the theorem with its hypotheses and orientation convention.
  • Prove it by reduction to Green’s theorem.
  • Apply it to circulation and irrotational fields.

History

On 2 July 1850 Thomson wrote to Stokes with the result, stated without proof. In February 1854 Stokes, as examiner, asked candidates for the Smith’s Prize to prove it. That year the prize was shared by Edward Routh and James Clerk Maxwell. Maxwell used the theorem in his electromagnetic theory and attributed it to Stokes, and the name stuck. Hermann Hankel published a proof in 1861. The modern form, \(\int_{\partial M}\omega=\int_M d\omega\) for differential forms on manifolds, was developed by Élie Cartan and others in the twentieth century.

Theorem (Kelvin–Stokes)

Let \(S\subset\mathbb R^3\) be a compact, oriented, piecewise smooth surface with unit normal \(\mathbf n\), whose boundary \(\partial S\) consists of finitely many piecewise smooth closed curves, oriented so that \(S\) lies to the left when one walks along \(\partial S\) with head pointing along \(\mathbf n\). If \(\mathbf F\) is continuously differentiable on an open set containing \(S\), then

\[ \oint_{\partial S}\mathbf F\cdot d\mathbf r=\iint_S(\nabla\times\mathbf F)\cdot\mathbf n\,dA . \]

Proof sketch

By cutting \(S\) into pieces, it suffices to treat one smooth patch \(\mathbf r(u,v)\), \((u,v)\in D\), with \(D\) a plane region bounded by a piecewise smooth curve; the integrals over interior cuts cancel because each cut is traversed twice in opposite directions. Pull back the line integral: \(\oint_{\partial S}\mathbf F\cdot d\mathbf r=\oint_{\partial D}(P\,du+Q\,dv)\) with \(P=\mathbf F\cdot\mathbf r_u\), \(Q=\mathbf F\cdot\mathbf r_v\). By Green’s theorem this equals \(\iint_D(Q_u-P_v)\,du\,dv\). The mixed derivatives \(\mathbf r_{uv}\) cancel, and a direct computation gives

\[ Q_u-P_v=(\mathbf r_u\cdot\nabla\mathbf F)\cdot\mathbf r_v-(\mathbf r_v\cdot\nabla\mathbf F)\cdot\mathbf r_u=(\nabla\times\mathbf F)\cdot(\mathbf r_u\times\mathbf r_v), \]

using the vector identity \((\mathbf a\cdot\nabla\mathbf F)\cdot\mathbf b-(\mathbf b\cdot\nabla\mathbf F)\cdot\mathbf a=(\nabla\times\mathbf F)\cdot(\mathbf a\times\mathbf b)\). Since \((\mathbf r_u\times\mathbf r_v)\,du\,dv=\mathbf n\,dA\), the result follows.

Exercise 8.1Verify the theorem for \(\mathbf F=(-y,x,0)\) on the upper unit hemisphere with outward normal.

Show solution
\(\nabla\times\mathbf F=(0,0,2)\). The boundary is the unit circle traversed counterclockwise seen from above: \(\oint(-y\,dx+x\,dy)=\int_0^{2\pi}(\sin^2t+\cos^2t)\,dt=2\pi\). The flux of \((0,0,2)\) through the hemisphere equals its flux through the unit disc (the field is divergence-free and the two surfaces share a boundary), which is \(2\cdot\pi=2\pi\).

Exercise 8.2Let \(\mathbf F=\dfrac{(-y,x,0)}{x^2+y^2}\) on \(\mathbb R^3\) minus the \(z\)-axis. Show \(\nabla\times\mathbf F=\mathbf 0\) but \(\oint\mathbf F\cdot d\mathbf r=2\pi\) around the unit circle. Why does this not contradict the theorem?

Show solution
Direct differentiation gives zero curl away from the axis. Along the unit circle \(\mathbf F\cdot d\mathbf r=dt\), so the integral is \(2\pi\). Any surface bounded by this circle must meet the \(z\)-axis, where \(\mathbf F\) is not defined, so the hypothesis fails. The example shows that zero curl implies a potential only on simply connected domains. In fluid terms, it is the velocity of a line vortex.
Lesson 09

Uniform convergence

For the general reader. Add up infinitely many smooth waves and you can get a function with a jump. Cauchy had claimed that a convergent series of continuous functions is always continuous, which Fourier series contradict. In 1847 Stokes, and independently Philipp Ludwig von Seidel, found what goes wrong: near the jump the series converges “infinitely slowly”. The repair, now called uniform convergence, is a basic idea of analysis.

Goals
  • Define uniform convergence and prove that it preserves continuity.
  • See why a Fourier series converges non-uniformly near a jump.

Stokes’s 1847 paper on the critical values of the sums of periodic series studied Fourier series of discontinuous functions. He noted that near a discontinuity the number of terms needed for a given accuracy grows without bound, and he separated this from ordinary convergence. In modern language:

Definition and theorem

Functions \(f_n:X\to\mathbb R\) converge uniformly to \(f\) if \(\sup_{x\in X}|f_n(x)-f(x)|\to0\). If each \(f_n\) is continuous on a metric space \(X\) and \(f_n\to f\) uniformly, then \(f\) is continuous.

Proof. Given \(\varepsilon>0\) and \(x_0\), choose \(n\) with \(\sup|f_n-f|<\varepsilon/3\), then \(\delta\) with \(|f_n(x)-f_n(x_0)|<\varepsilon/3\) for \(d(x,x_0)<\delta\). Then \(|f(x)-f(x_0)|\le|f(x)-f_n(x)|+|f_n(x)-f_n(x_0)|+|f_n(x_0)-f(x_0)|<\varepsilon\).

Karl Weierstrass made uniform convergence a central tool in his Berlin lectures. Stokes did not pursue the rigorous theory further; his interest was in using series correctly in physics.

Exercise 9.1Show that \(f_n(x)=x^n\) converges on \([0,1]\) but not uniformly. Where is convergence “infinitely slow” in Stokes’s sense?

Show solution
The limit is \(0\) for \(x<1\) and \(1\) at \(x=1\), which is discontinuous, so convergence cannot be uniform. Explicitly \(\sup_{[0,1)}x^n=1\) for every \(n\). To have \(x^n<\tfrac12\) one needs \(n>\ln2/|\ln x|\), which grows without bound as \(x\to1\): convergence becomes infinitely slow at the critical point \(x=1\).

Exercise 9.2The square wave \(f(x)=\operatorname{sign}(x)\) on \((-\pi,\pi)\) has Fourier series \(\frac4\pi\sum_{k\ \mathrm{odd}}\frac{\sin kx}{k}\). Explain without computation why the convergence is not uniform on \((0,\pi)\).

Show solution
Each partial sum \(S_N\) is continuous with \(S_N(0)=0\). So for each \(N\) there are points \(x>0\) close enough to 0 that \(S_N(x)<\tfrac12\), while \(f(x)=1\). Hence \(\sup_{(0,\pi)}|S_N-f|\ge\tfrac12\) for all \(N\). (In fact the partial sums also overshoot by about 9% of the jump, the Gibbs phenomenon.)
Lesson 10

The Airy integral and the Stokes phenomenon

For the general reader. George Biddell Airy had described the bright bands of a rainbow with an integral that was hard to compute. Stokes found a way to approximate it with series that diverge but are very accurate if cut off in time. In doing so he discovered that such approximations switch abruptly from one form to another as you cross certain lines in the complex plane. This switching, the Stokes phenomenon, is now a central topic of asymptotic analysis.

Goals
  • State the asymptotic forms of the Airy function on both sides.
  • Locate its zeros approximately, as Stokes did in 1850.
  • Understand Stokes lines and divergent series.

The Airy function

In 1838 Airy expressed the intensity near a caustic, such as the edge of a rainbow, through the integral now written

\[ \mathrm{Ai}(x)=\frac1\pi\int_0^\infty\cos\Big(\frac{t^3}{3}+xt\Big)\,dt, \]

which solves \(y''=xy\). Airy tabulated it with power series, which converge slowly for large \(|x|\). In 1850 Stokes derived asymptotic expansions. With \(\zeta=\tfrac23|x|^{3/2}\):

Theorem (asymptotics of Ai) \[ \mathrm{Ai}(x)\sim\frac{e^{-\zeta}}{2\sqrt\pi\,x^{1/4}}\quad(x\to+\infty),\qquad \mathrm{Ai}(-x)\sim\frac{\sin(\zeta+\pi/4)}{\sqrt\pi\,x^{1/4}}\quad(x\to+\infty), \]

each the leading term of a divergent asymptotic series in powers of \(\zeta^{-1}\).

The second formula gives the zeros \(-a_n\) of Ai: \(\zeta+\pi/4\approx n\pi\), so \(a_n\approx\big[\tfrac{3\pi}{8}(4n-1)\big]^{2/3}\). Stokes used such formulas to compute the positions of the dark bands of the rainbow far more easily than Airy’s series allowed.

The Stokes phenomenon

The two formulas are puzzling together. On the right the solution is a single decaying exponential; on the left it is a combination of two oscillating exponentials \(e^{\pm i\zeta}\). How can one analytic function have such different asymptotic forms? In his 1857 paper, published in 1864, Stokes explained this. Write the general asymptotic form in the complex plane as \(A\,e^{-\zeta}(\cdots)+B\,e^{+\zeta}(\cdots)\). Where one exponential is maximally dominant over the other, on what are now called Stokes lines, the coefficient of the smaller (subdominant) one can change without any visible effect, because it is hidden below the error of the dominant series. Crossing such a line, the constant \(B\) jumps. By the time one reaches the negative real axis both terms are of equal size, and the switched-on term is visible.

Principle (Stokes, 1857; Berry, 1989)

The constants multiplying exponentially small terms in an asymptotic expansion are discontinuous across Stokes lines. Michael Berry showed in 1989 that with optimal truncation of the dominant series the jump is in fact smooth, following an error function across a narrow region of the Stokes line.

Exercise 10.1Use the formula \(a_n\approx\big[\tfrac{3\pi}{8}(4n-1)\big]^{2/3}\) to estimate the first zero of Ai. The true value is \(-2.33811\). Comment.

Show solution
\(n=1\): \(\big(\tfrac{9\pi}{8}\big)^{2/3}=(3.534)^{2/3}\approx2.320\), so \(-a_1\approx-2.320\), within 1% of \(-2.338\). Even the first zero, far from the asymptotic regime, is well approximated, and the error decreases rapidly with \(n\). Including the next term of the series improves it further.

Exercise 10.2Show by repeated integration by parts that \(I(x)=\int_0^\infty\frac{e^{-t}}{1+xt}\,dt\sim\sum_{n\ge0}(-1)^nn!\,x^n\) as \(x\to0^+\). Why does this series diverge for every \(x>0\), and roughly where should it be truncated?

Show solution
Expanding \(\frac1{1+xt}=\sum_{n=0}^{N-1}(-xt)^n+\frac{(-xt)^N}{1+xt}\) and using \(\int_0^\infty t^ne^{-t}dt=n!\) gives \(I=\sum_{n<N}(-1)^nn!x^n+R_N\) with \(|R_N|\le N!x^N\). The terms \(n!x^n\) eventually grow for any fixed \(x>0\), so the series diverges. The terms are smallest near \(n\approx1/x\), and truncating there gives an error of order \(e^{-1/x}\), exponentially small. Terms of exactly this size are what the Stokes phenomenon switches on and off.
Part III · Life and legacy
Lesson 11

Secretary, president, member of Parliament

On 4 July 1857 Stokes married Mary Susanna Robinson, daughter of Thomas Romney Robinson, the astronomer at Armagh Observatory. During the engagement he sent her a very long letter about his habits and sense of duty, which nearly ended it. Because fellows had to be celibate, he gave up his Pembroke fellowship. A change in the college statutes allowed him to hold a fellowship again from 1862. The couple had five children, two of whom died in infancy. His wife’s journal of their honeymoon records his delight in lightning and in places that looked dangerous.

The Royal Society

From 1854 to 1885 Stokes was one of the two Secretaries of the Royal Society, responsible for the physical sciences. In practice this meant reading, refereeing and editing a large share of the physical papers submitted for publication in Britain. He answered correspondents at length and gave away ideas freely. Around 1854 he suggested to Thomson that the dark Fraunhofer lines in the solar spectrum arise from absorption by atoms in the Sun’s outer layers, the principle that Gustav Kirchhoff published in 1859; Stokes later declined to claim priority. Some historians think the administrative load explains why his own research slowed after the late 1850s. He also turned more to experiment, especially in optics.

From 1885 to 1890 he was President of the Royal Society. Only Newton had previously combined the Lucasian chair, the presidency and a seat in Parliament. Stokes completed that list as Conservative member for Cambridge University from 1887 to 1892, though he rarely spoke in the House. He was created a baronet in 1889.

Science and faith

Stokes wrote and lectured on the relation between science and religion. He was President of the Victoria Institute, a society devoted to that subject, from 1886, and gave the Gifford Lectures on natural theology between 1891 and 1893. His religious views were conservative in most respects, but he rejected the doctrine of eternal punishment, holding that immortality is conditional. He also gave the Burnett Lectures on light at Aberdeen, which were published as a book, the nearest he came to a treatise on optics.

DiscussionStokes spent decades refereeing, editing and advising rather than writing his own papers. How should such work be weighed against original research in judging a scientist’s contribution? Consider what British physics in 1850–1885 might have lacked without him.

Lesson 12

Last years and legacy

In 1899 the University of Cambridge celebrated the jubilee of his election to the Lucasian chair. He received the Copley Medal in 1893, and Prussia’s Pour le Mérite for science in 1879. His collected Mathematical and Physical Papers appeared in five volumes, the first three edited by himself and the last two, completed in 1905, by Joseph Larmor. In 1896 he proposed that the newly discovered X-rays were transverse pulses in the ether rather than ordinary waves of short wavelength. In 1902 he was elected Master of Pembroke. He died in Cambridge on 1 February 1903, aged 83, and was buried in Mill Road Cemetery. A memorial was placed in Westminster Abbey.

Legacy

  • Fluid mechanics: the Navier–Stokes equations, the no-slip condition, Stokes flow and Stokes’s law, which underlie aerodynamics, microfluidics, sedimentation and the biology of swimming microorganisms. The CGS unit of kinematic viscosity, the stokes, bears his name.
  • Waves: Stokes waves, Stokes drift (important in ocean transport of heat, pollutants and plastic), and the Stokes boundary layer.
  • Optics: fluorescence and the Stokes shift, the foundation of fluorescence microscopy; Stokes parameters in astronomy, remote sensing and quantum optics; the Stokes relations.
  • Mathematics: Stokes’s theorem in vector calculus and its generalization on manifolds; the Stokes phenomenon and the modern theory of resurgence; an early form of uniform convergence.
  • Geodesy: Stokes’s formula for the geoid.
  • Honours: Rumford Medal (1852), Copley Medal (1893), baronetcy (1889), craters on the Moon and Mars, the Stokes Society at Pembroke.

Final projectChoose one result from Part II: Stokes’s law, Stokes drift, the Stokes parameters or the Stokes phenomenon. Write a two-page account for a strong undergraduate that states it precisely, derives it, and describes one modern application in detail.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, George Gabriel Stokes, MacTutor History of Mathematics.
  2. Sir George Stokes, 1st Baronet, Wikipedia (with extensive references).
  3. Institute of Mathematics and its Applications, Fluids, fluorescence and a hat full of beetles, Mathematics Today, August 2019.
  4. A. Wood, George Gabriel Stokes 1819–1903: An Irish Mathematical Physicist, Dublin, 2003.
  5. M. McCartney, A. Whitaker and A. Wood (eds.), George Gabriel Stokes: Life, Science and Faith, Oxford University Press, 2019.
  6. D. B. Wilson, Kelvin and Stokes: A Comparative Study in Victorian Physics, Adam Hilger, 1987.
  7. Smith’s Prize, Wikipedia (list of winners and the 1854 question).

Mathematics and physics

  1. G. G. Stokes, Mathematical and Physical Papers, 5 vols., Cambridge University Press, 1880–1905.
  2. G. G. Stokes, On the theories of the internal friction of fluids in motion, and of the equilibrium and motion of elastic solids, Trans. Camb. Phil. Soc. 8 (1845).
  3. G. G. Stokes, On the theory of oscillatory waves, Trans. Camb. Phil. Soc. 8 (1847). See Stokes drift.
  4. G. G. Stokes, On the effect of the internal friction of fluids on the motion of pendulums, Trans. Camb. Phil. Soc. 9 (1851).
  5. G. G. Stokes, On the composition and resolution of streams of polarized light from different sources, Trans. Camb. Phil. Soc. 9 (1852). See Stokes parameters.
  6. G. G. Stokes, On the change of refrangibility of light, Phil. Trans. R. Soc. 142 (1852).
  7. G. G. Stokes, On the discontinuity of arbitrary constants which appear in divergent developments, Trans. Camb. Phil. Soc. 10 (1864).
  8. Kelvin–Stokes theorem, Wikipedia; V. J. Katz, The history of Stokes’ theorem, Mathematics Magazine 52 (1979).
  9. M. V. Berry, Uniform asymptotic smoothing of Stokes’s discontinuities, Proc. R. Soc. Lond. A 422 (1989).
  10. G. K. Batchelor, An Introduction to Fluid Dynamics, Cambridge University Press, 1967.
  11. C. J. Amick, L. E. Fraenkel and J. F. Toland, On the Stokes conjecture for the wave of extreme form, Acta Math. 148 (1982).