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Jean-Baptiste Biot

Jean-Baptiste Biot

A course in two strands: the long life of a Laplacian physicist who rose in a balloon, chased meteorites and measured the meridian, and the physics he founded, from the magnetic field of a current to the rotation of polarized light by sugar.

Born 21 April 1774, Paris Died 3 February 1862, Paris Biot–Savart law Paris, 1820 Level advanced undergraduate → graduate
Rotatory dispersion in quartz. One plane enters; violet leaves turned far more than red, roughly as 1/λ² (Biot, 1818).
Lesson 00

How to use this course

Jean-Baptiste Biot lived from the last years of Louis XVI to the Second Empire and worked in nearly every branch of physics of his time. Students meet his name twice: in the Biot–Savart law of magnetostatics and in the Biot number of heat transfer. Chemists know him as the founder of polarimetry. Historians know him as the most stubborn defender of the particle theory of light against Augustin-Jean Fresnel. This course follows the man and the physics together.

Prerequisites for Part II
  • Calculus of several variables and ordinary differential equations.
  • Introductory electromagnetism: magnetic field, current, line integrals.
  • Wave optics at the level of polarization and Jones vectors; for Lesson 10, the idea of a crystal’s optic axes (see the Fresnel course, Lesson 09).
  • Newtonian gravitation and the simple pendulum, for Lesson 06.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Biot often found the correct empirical law and the wrong explanation. The lessons therefore state his laws in their modern form and show how later theory, usually wave optics or Maxwell’s equations, derives them. That contrast is the main lesson of his career.

Reference

Timeline

1774Born 21 April in Paris, son of Joseph Biot, an official of the Treasury.
1791Completes classical studies at the Collège Louis-le-Grand.
1792–93Joins the army; serves in the artillery at the battle of Hondschoote, September 1793.
1794January: École des Ponts et Chaussées. Late in the year, the newly founded École Polytechnique.
1797Professor of mathematics at the École Centrale of the Oise in Beauvais. Marries Gabrielle Brisson.
1800Professor of mathematical physics at the Collège de France; associate of the Institut.
1803Investigates the L’Aigle meteorite fall of 26 April. Elected to the mathematics section of the Institut. Son Édouard born.
180424 August: balloon ascent with Joseph Louis Gay-Lussac to about 4000 m. Paper on heat conduction in bars.
1806–08Bureau des Longitudes. With François Arago, extends the French meridian arc to the Balearic Islands.
1809Professor of physical astronomy at the Faculty of Sciences, Paris, a post he holds for over fifty years.
1812First memoir on the rotation of the plane of polarization in quartz.
1814Chevalier of the Légion d’honneur. Memoir on the motion of light molecules (“mobile polarization”).
1815Discovers optical rotation in liquids such as oil of turpentine. Foreign Member of the Royal Society.
1816Traité de physique expérimentale et mathématique, four volumes.
1818Second memoir on rotation: the inverse-square law in wavelength; rotation by turpentine vapour.
182030 October: with Félix Savart, announces the law of the magnetic action of a current.
1822Loses the election for permanent secretary of the Académie to Joseph Fourier.
1830Loses the same post to Arago.
1839Champions William Henry Fox Talbot’s paper photography against Arago’s support for the daguerreotype.
1840Rumford Medal of the Royal Society.
1841Elected to the Académie des Inscriptions et Belles-Lettres.
1848Witnesses and confirms Louis Pasteur’s separation of the two forms of tartrate crystals.
1849Commander of the Légion d’honneur.
1850Death of his son Édouard; Biot completes Édouard’s translation of the Zhouli.
1856Elected to the Académie française.
1862Dies 3 February in Paris, aged 87.
Part I · Formation
Lesson 01

Paris and the army, 1774–1794

Jean-Baptiste Biot was born in Paris on 21 April 1774. His father, Joseph Biot, held a responsible post in the Treasury; the family’s roots were among farmers in Lorraine. The boy received a classical education at the Collège Louis-le-Grand, finishing in 1791. His literary training never left him. Sixty years later it carried him into two academies devoted to letters.

His father wanted a commercial career for him. Biot took private lessons in mathematics from Antoine-René Mauduit of the Collège de France, then was sent to Le Havre as a merchant’s clerk. He found the work dull. In September 1792, as revolutionary France went to war, he enlisted.

He served in the artillery and fought at Hondschoote in September 1793, where French forces broke the British and Hanoverian siege of Dunkirk. Illness followed, and he left the army. Back in Paris, still in uniform, he was arrested as a deserter. A stranger in a passing carriage intervened and obtained his release. Biot never learned who it was.

Lesson 02

Polytechnique and Beauvais, 1794–1800

In January 1794 Biot entered the École des Ponts et Chaussées. Later that year he transferred to the new École Polytechnique, in its first intake. Gaspard Monge, one of its founders, noticed his ability. Biot was also active in student politics. He joined the royalist rising against the Convention in 1795, was captured, and was freed after Monge pleaded for him.

At the school he became a friend of Siméon Denis Poisson. The mathematician Sylvestre-François Lacroix took him up as a patron. Biot graduated in 1797 and, through his fellow student Barnabé Brisson, obtained the chair of mathematics at the École Centrale of the Oise in Beauvais.

The same year he married Brisson’s sister Gabrielle, then sixteen. He taught her mathematics and physics, and she translated Ernst Gottfried Fischer’s German textbook of physics. It appeared in 1806 as Physique mécanique under Biot’s name as translator, as custom then required.

The proofreader of the Mécanique céleste

In late 1799 Biot offered to read the proofs of Pierre-Simon Laplace’s Traité de mécanique céleste. He did the work with care and checked the calculations. Laplace became his protector, and Biot became a member of the circle around Laplace and the chemist Claude-Louis Berthollet that aimed to explain all physics by short-range forces between particles. That programme shaped everything Biot did, including his mistakes.

Lesson 03

Laplace’s protégé, 1800–1809

With Laplace’s support, Biot became professor of mathematical physics at the Collège de France in 1800, aged 26. The same year he was elected an associate of the First Class of the Institut. In 1803, after a memoir on tautochronous curves, he was elected a full member of its mathematics section. His early books included an analysis of the Mécanique céleste (1801), a treatise on curves and surfaces of the second degree (1802), and a Traité élémentaire d’astronomie physique (1805).

He also became a field scientist. In 1803 the Institut sent him to L’Aigle in Normandy to investigate a shower of stones (Lesson 04). In August 1804 he and Gay-Lussac made the first balloon ascent for scientific purposes. In the same year he published a short memoir on heat conduction in metal bars (Lesson 05).

In 1806 he joined the Bureau des Longitudes as an assistant astronomer. In August the Bureau commissioned him and the young François Arago to extend the meridian arc of Jean-Baptiste Delambre and Pierre Méchain from Barcelona to the Balearic Islands. Biot returned to France before the work was finished, while Arago stayed on, and the outbreak of war in Spain in 1808 turned Arago’s journey home into a long ordeal of prison and shipwreck. With Claude-Louis Mathieu, Biot then measured the length of the seconds pendulum at stations along the meridian (Lesson 06). In 1809 he took the chair of physical astronomy at the new Faculty of Sciences in Paris.

Part II · The physics
Lesson 04

Stones from the sky, air from a balloon

For the general reader. Around 1800 most scientists doubted that stones could fall from the sky. Peasants’ stories were dismissed, and the stones were said to come from volcanoes or lightning. On 26 April 1803 thousands of stones fell near L’Aigle in Normandy. Biot’s report settled the question for Europe. A year later he and Gay-Lussac rose in a balloon to test the air and the Earth’s magnetism at altitude.

Goals
  • Understand the structure of Biot’s argument for an extraterrestrial origin.
  • Estimate the terminal speed of a falling stone.
  • Derive the isothermal barometric formula and apply it to the 1804 ascent.

The L’Aigle report

Ernst Chladni had argued in 1794 that fireballs and fallen stones were cosmic bodies. At L’Aigle, early in the afternoon, a fireball exploded and more than 3000 fragments fell. Biot interviewed witnesses village by village and mapped where stones had landed. He used two kinds of evidence. The physical evidence: the stones were all alike, unlike any local rock, and resembled stones reported from other falls. The human evidence: many independent witnesses, of different social standing, described the same event. His report is often taken as the founding document of meteoritics.

Terminal speed

A body of mass \(m\) and cross-section \(A\) falling through air of density \(\rho_a\) with drag coefficient \(C_d\) approaches the speed at which drag balances weight:

\[ v_t=\sqrt{\frac{2mg}{\rho_a C_d A}}. \]

A stone entering the atmosphere at tens of kilometres per second is braked high up, where it glows and fragments. The pieces then fall at a few tens of metres per second, which is why fresh meteorites are not buried deep and can be picked up warm, as some were at L’Aigle.

The balloon ascent of 1804

On 24 August 1804 Biot and Gay-Lussac rose from the garden of the Conservatoire des Arts et Métiers to about 4000 m. They timed the oscillations of a magnetized needle and found no noticeable change in the Earth’s magnetic intensity with altitude, within their accuracy. They also recorded temperature, humidity and pressure. Gay-Lussac went up alone in September and reached about 7000 m.

Isothermal atmosphere

Hydrostatic balance \(dp/dz=-\rho g\) together with the ideal-gas law \(p=\rho RT/M\) at constant \(T\) gives

\[ p(z)=p_0\,e^{-z/H},\qquad H=\frac{RT}{Mg}. \]

Derivation. Substituting \(\rho=pM/(RT)\) gives \(dp/p=-(Mg/RT)\,dz\); integrate from 0 to \(z\). For air, \(M=0.029\ \mathrm{kg\,mol^{-1}}\), and at \(T=288\ \mathrm K\) the scale height is \(H\approx8.4\ \mathrm{km}\). Biot returned to this subject in 1841 with a memoir on the true constitution of the atmosphere.

Exercise 4.1A stony meteorite fragment of mass 1 kg and density \(3500\ \mathrm{kg\,m^{-3}}\) is modelled as a sphere with \(C_d=0.5\). Find its terminal speed in sea-level air (\(\rho_a=1.2\ \mathrm{kg\,m^{-3}}\)).

Show solution
Volume \(V=1/3500=2.86\times10^{-4}\ \mathrm m^3\), radius \(r=(3V/4\pi)^{1/3}\approx0.041\ \mathrm m\), area \(A=\pi r^2\approx5.2\times10^{-3}\ \mathrm m^2\). Then \(v_t=\sqrt{2\cdot1\cdot9.81/(1.2\cdot0.5\cdot5.2\times10^{-3})}\approx79\ \mathrm{m\,s^{-1}}\). Large, but far below cosmic speeds: the atmosphere has removed almost all the original kinetic energy.

Exercise 4.2Using the isothermal model with \(H=8.4\ \mathrm{km}\), what fraction of sea-level pressure did Biot and Gay-Lussac experience at 4000 m? What is the main error of the model?

Show solution
\(p/p_0=e^{-4/8.4}\approx0.62\). The temperature actually falls with height, by roughly \(6.5\ \mathrm K\) per km, so the real atmosphere is colder aloft and the pressure falls slightly faster; the standard atmosphere gives about 0.61 at 4000 m. The balloonists themselves recorded the fall of temperature.
Lesson 05

Heat in a bar

For the general reader. Heat one end of an iron bar and the other end warms, but less and less as the bar gets longer, because the sides lose heat to the air. In 1804 Biot tried to write the equation for this. He got the physics partly right and the mathematics wrong, and his attempt was the starting point for Joseph Fourier’s theory of heat. Engineers still honour him with the Biot number.

Goals
  • Derive the steady temperature profile of a bar losing heat from its surface.
  • See why Biot’s formulation failed and Fourier’s succeeded.
  • Define the Biot number and use it to decide when a body can be treated as isothermal.

Biot’s problem

In his 1804 memoir Biot considered a thin bar heated at one end, losing heat sideways to the air by Newton’s law of cooling. He treated conduction as an action at a distance between slices of the bar, in the Laplacian style. Johann Heinrich Lambert had studied the same set-up earlier and found that, in the steady state, temperatures at equally spaced points fall in geometric progression. Fourier began from Biot’s paper, found that the action-at-a-distance approach led nowhere, and replaced it by a flux proportional to the local temperature gradient.

The fin equation

Let a bar of conductivity \(k\), cross-sectional area \(A\) and perimeter \(P\) lose heat from its surface at rate \(h\,(T-T_a)\) per unit area, where \(T_a\) is the air temperature. In the steady state, \(\theta=T-T_a\) satisfies

\[ \frac{d^2\theta}{dx^2}=\frac{\theta}{\ell^2},\qquad \ell=\sqrt{\frac{kA}{hP}}. \]

For a long bar with \(\theta(0)=\theta_0\), the solution is \(\theta(x)=\theta_0\,e^{-x/\ell}\).

Derivation. Fourier’s law gives the heat flux along the bar as \(-kA\,d\theta/dx\). Balance the heat entering and leaving a slice \([x,x+dx]\) against the loss through its side, \(hP\,\theta\,dx\): \(kA\,\theta''\,dx=hP\,\theta\,dx\). The bounded solution decays exponentially, so equal steps \(\Delta x\) multiply \(\theta\) by the same factor \(e^{-\Delta x/\ell}\). That is Lambert’s geometric progression.

The Biot number

The ratio of the resistance to conduction inside a body to the resistance to convection at its surface is the dimensionless number

\[ \mathrm{Bi}=\frac{hL}{k}, \]

where \(L\) is a characteristic length, usually volume divided by surface area. When \(\mathrm{Bi}\ll1\) (in practice \(\mathrm{Bi}<0.1\)), temperature differences inside the body are small compared with the drop across the surface, and the body cools as a single lump: \(T-T_a\propto e^{-t/\tau}\) with \(\tau=\rho cV/(hS)\). The name honours Biot’s early recognition that internal conduction and surface loss compete.

Exercise 5.1An iron bar of radius 1 cm has \(k=50\ \mathrm{W\,m^{-1}K^{-1}}\) and loses heat to still air with \(h=10\ \mathrm{W\,m^{-2}K^{-1}}\). Find \(\ell\). How far from the hot end does the excess temperature fall to 10 percent of its value at the end?

Show solution
\(A/P=\pi r^2/(2\pi r)=r/2=0.005\ \mathrm m\), so \(\ell=\sqrt{50\cdot0.005/10}=\sqrt{0.025}\approx0.158\ \mathrm m\). The fall to 10 percent needs \(x=\ell\ln10\approx0.36\ \mathrm m\). A bar a metre long behaves as infinitely long, which is why Biot’s and Lambert’s bars showed clean geometric decay.

Exercise 5.2A copper sphere of radius 5 mm (\(k=400\), \(\rho=8900\ \mathrm{kg\,m^{-3}}\), \(c=385\ \mathrm{J\,kg^{-1}K^{-1}}\)) cools in air with \(h=20\ \mathrm{W\,m^{-2}K^{-1}}\). Compute the Biot number with \(L=V/S\), check that lumped cooling is valid, and find the time constant.

Show solution
\(L=r/3\approx1.67\times10^{-3}\ \mathrm m\), so \(\mathrm{Bi}=20\cdot1.67\times10^{-3}/400\approx8\times10^{-5}\ll0.1\). Lumped model valid. \(\tau=\rho cL/h=8900\cdot385\cdot1.67\times10^{-3}/20\approx286\ \mathrm s\), about five minutes.

Exercise 5.3Show that if the bar has finite length \(L\) and an insulated far end, \(\theta(x)=\theta_0\cosh\big((L-x)/\ell\big)/\cosh(L/\ell)\).

Show solution
The general solution is \(\alpha\cosh((L-x)/\ell)+\beta\sinh((L-x)/\ell)\). An insulated end means \(\theta'(L)=0\), which kills the \(\sinh\) term since its derivative at \(x=L\) is \(-\beta/\ell\). Then \(\theta(0)=\theta_0\) fixes \(\alpha=\theta_0/\cosh(L/\ell)\). As \(L\to\infty\) this tends to \(\theta_0e^{-x/\ell}\).
Lesson 06

The meridian and the seconds pendulum

For the general reader. The metre was defined in the 1790s as one ten-millionth of the distance from the pole to the equator along the meridian of Paris. Measuring that arc, and measuring how gravity changes with latitude, also measured the shape of the Earth. From 1806 Biot spent years on both tasks, first with Arago in Spain and the Balearic Islands, then with Mathieu at pendulum stations along the French meridian, from Bordeaux to Dunkirk.

Goals
  • Relate the length of the seconds pendulum to local gravity.
  • State Clairaut’s theorem and use it to estimate the Earth’s flattening.
  • See how arc lengths at different latitudes reveal an oblate Earth.

The seconds pendulum

A seconds pendulum swings from one side to the other in one second, so its full period is \(T=2\ \mathrm s\). For a simple pendulum, \(T=2\pi\sqrt{L/g}\), so

\[ L=\frac{g\,T^2}{4\pi^2}=\frac{g}{\pi^2}\ \ (T=2\ \mathrm s). \]

Biot and Mathieu used a compound pendulum after the method of Jean-Charles de Borda and reduced their results to the equivalent simple pendulum. The Académie awarded them prizes for this work in 1809 and 1812. Biot later carried such observations into Scotland and the Shetland Islands; the results were published with Arago in 1821.

Clairaut’s theorem (1743)

For a rotating, slightly oblate Earth in hydrostatic equilibrium, gravity at the surface at geographic latitude \(\varphi\) is, to first order in small quantities,

\[ g(\varphi)=g_e\big(1+\beta\sin^2\varphi\big),\qquad \beta+f=\tfrac52\,m,\qquad m=\frac{\omega^2a}{g_e}, \]

where \(f\) is the flattening, \(a\) the equatorial radius and \(\omega\) the angular speed of rotation.

Clairaut’s theorem is remarkable because it does not depend on how density varies inside the Earth. Measuring \(\beta\) with pendulums gives \(f\) without any geological assumption. This is why pendulum surveys mattered so much to Alexis Clairaut’s successors, Laplace among them.

Arc lengths

On an ellipsoid with flattening \(f\), the length of one degree of latitude grows toward the pole. To first order,

\[ \Delta s_{1^\circ}\approx\frac{\pi a}{180}\Big(1-f\,\big(2-3\sin^2\varphi\big)\Big). \]

Delambre and Méchain had measured from Dunkirk to Barcelona. The Biot–Arago extension to Formentera pushed the arc further south, so that it straddled 45° more evenly.

Exercise 6.1Take \(g_e=9.780\ \mathrm{m\,s^{-2}}\) and \(\beta=0.0053\). Find the length of the seconds pendulum at the southern end of the extended arc, Formentera (\(\varphi\approx38.7^\circ\)) and at Dunkirk (\(\varphi\approx51.0^\circ\)). What precision did Biot and Mathieu need to see the difference?

Show solution
\(g(38.7^\circ)\approx9.780(1+0.0053\cdot0.391)\approx9.8002\), so \(L\approx0.99297\ \mathrm m\). \(g(51.0^\circ)\approx9.8113\), \(L\approx0.99410\ \mathrm m\). The difference is about 1.1 mm in a metre. They needed precision well below a tenth of a millimetre, about one part in \(10^5\), to map the variation along the arc.

Exercise 6.2With \(\omega=7.29\times10^{-5}\ \mathrm{s^{-1}}\), \(a=6.378\times10^6\ \mathrm m\) and \(g_e=9.780\), compute \(m\). Using \(\beta=0.0053\), estimate the flattening \(f\) from Clairaut’s theorem.

Show solution
\(m=\omega^2a/g_e=(5.31\times10^{-9})(6.378\times10^6)/9.780\approx0.00346\approx1/289\). Then \(f=\tfrac52m-\beta\approx0.00865-0.0053=0.00335\approx1/298\), close to the modern value \(1/298.26\).

Exercise 6.3Use the arc formula with \(a=6378\ \mathrm{km}\) and \(f=1/298\) to compare the length of one degree of latitude at 40° and at 50°.

Show solution
\(\pi a/180\approx111.32\ \mathrm{km}\). At 40°, \(\sin^2\varphi=0.413\), so the bracket is \(1-(2-1.240)/298\approx0.99745\) and \(\Delta s\approx111.04\ \mathrm{km}\). At 50°, \(\sin^2\varphi=0.587\), bracket \(1-(2-1.760)/298\approx0.99919\), \(\Delta s\approx111.23\ \mathrm{km}\). A degree is about 190 m longer at 50°: the Earth is flatter near the poles.
Lesson 07

The Biot–Savart law

For the general reader. In July 1820 Hans Christian Ørsted announced that an electric current deflects a compass needle. Arago repeated the experiment at the Académie on 11 September. Within weeks, André-Marie Ampère in Paris had begun his study of forces between currents, and Biot and the young physician Félix Savart had measured how the force on a magnet falls off with distance from a long wire. Their answer, inversely as the distance, became the first quantitative law of electromagnetism.

Goals
  • Understand Biot and Savart’s measurement by oscillating needles.
  • State the element law, due in its differential form to Laplace, and derive the field of a long wire.
  • Compute fields of loops and bent wires.

The experiment

Biot and Savart hung a small magnetized steel needle from a silk thread near a long vertical wire joined to the poles of a voltaic pile. They cancelled or allowed for the Earth’s horizontal field with a distant magnet, and timed the needle’s small oscillations with a Breguet half-second chronometer at different distances. A needle of magnetic moment \(\mu\) and moment of inertia \(I\) in a horizontal field \(B\) oscillates with period \(T=2\pi\sqrt{I/(\mu B)}\), so \(B\propto1/T^2\). On 30 October 1820 they reported that the force on each pole is perpendicular both to the wire and to the line joining the pole to the wire, and that its intensity varies inversely as that distance.

Biot–Savart law (modern form)

A steady current \(I\) in a thin wire \(C\) produces at the point \(\mathbf r\) the magnetic field

\[ \mathbf B(\mathbf r)=\frac{\mu_0I}{4\pi}\oint_C\frac{d\boldsymbol\ell'\times(\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3}. \]

For a volume current density \(\mathbf J\), replace \(I\,d\boldsymbol\ell'\) by \(\mathbf J(\mathbf r')\,d^3r'\). The field satisfies \(\nabla\cdot\mathbf B=0\) and \(\nabla\times\mathbf B=\mu_0\mathbf J\) when \(\nabla\cdot\mathbf J=0\).

Biot and Savart measured only the total effect of a long wire. Laplace showed that it follows from an element law in which each piece \(d\ell\) contributes a force proportional to \(\sin\theta\,d\ell/r^2\), where \(\theta\) is the angle between the element and the line to the pole. For this reason French texts often call it the Biot–Savart–Laplace law. In later experiments Biot and Savart used wires bent into a V to probe the action of the individual elements.

The long straight wire

Put the wire on the \(z\)-axis and the field point at distance \(s\). An element \(dz\) at height \(z\) contributes \(\frac{\mu_0I}{4\pi}\frac{s\,dz}{(s^2+z^2)^{3/2}}\) in the azimuthal direction. With \(z=s\tan\psi\),

\[ B=\frac{\mu_0I}{4\pi s}\int_{-\pi/2}^{\pi/2}\cos\psi\,d\psi=\frac{\mu_0I}{2\pi s}. \]

This is the \(1/s\) law of 30 October 1820. More generally, a straight segment seen from the field point between angles \(\psi_1\) and \(\psi_2\) (measured from the perpendicular) gives \(B=\frac{\mu_0I}{4\pi s}(\sin\psi_2-\sin\psi_1)\).

Exercise 7.1A wire carries 10 A. Find \(B\) at 5 cm and compare with the Earth’s horizontal field in Paris, about \(2\times10^{-5}\ \mathrm T\). Why did Biot and Savart need to compensate the Earth’s field?

Show solution
\(B=\mu_0I/(2\pi s)=2\times10^{-7}\cdot10/0.05=4\times10^{-5}\ \mathrm T\), only twice the Earth’s horizontal field. With voltaic piles of the time the currents were often smaller, so the wire’s field was comparable to or weaker than the Earth’s. The needle responds to the vector sum, and without compensation the \(1/s\) law would be hidden.

Exercise 7.2With the Earth’s field fully compensated, a needle oscillates with period 1.0 s at 1 cm from the wire. Predict its period at 4 cm.

Show solution
\(B\propto1/T^2\) and \(B\propto1/s\), so \(T\propto\sqrt s\). At 4 cm the period is \(1.0\cdot\sqrt4=2.0\ \mathrm s\). Timing periods gives distance laws through a square, which is why the method needed a good chronometer.

Exercise 7.3Two semi-infinite straight wires meet at a vertex \(O\) and carry current \(I\) in and out. They lie in a plane, each making angle \(\beta\) with a line \(OP\) extended backward through \(O\), so the V opens away from \(P\). Show that the field at \(P\), at distance \(r\) from \(O\), is \(B=\dfrac{\mu_0I}{2\pi r}\tan\dfrac{\beta}{2}\). Check the cases \(\beta=\pi/2\) and \(\beta\to0\).

Show solution
For a semi-infinite wire starting at \(O\) along a unit vector making angle \(\alpha\) with \(\overrightarrow{OP}\), the perpendicular distance is \(d=r\sin\alpha\), and the segment formula gives \(B=\frac{\mu_0I}{4\pi d}(1+\cos\alpha)=\frac{\mu_0I}{4\pi r}\cot\frac{\alpha}{2}\). Here \(\alpha=\pi-\beta\), so each arm gives \(\frac{\mu_0I}{4\pi r}\tan\frac\beta2\), and the two add. For \(\beta=\pi/2\) the arms form one straight wire at distance \(r\): \(\mu_0I/(2\pi r)\), correct. As \(\beta\to0\) both arms point away along the line through \(P\), and the field vanishes.

Exercise 7.4Find the field at the centre of a circular loop of radius \(R\), and on its axis at distance \(z\).

Show solution
Every element is perpendicular to the radius vector from it to the centre, so \(B=\frac{\mu_0I}{4\pi R^2}\cdot2\pi R=\frac{\mu_0I}{2R}\). On the axis, the components perpendicular to the axis cancel and each element contributes \(\frac{\mu_0I\,d\ell}{4\pi(R^2+z^2)}\cdot\frac{R}{\sqrt{R^2+z^2}}\), giving \(B=\frac{\mu_0IR^2}{2(R^2+z^2)^{3/2}}\).
Lesson 08

Chromatic polarization

For the general reader. In 1811 Arago noticed that a thin sheet of mica, placed in polarized light and viewed through a crystal of Iceland spar, shows two images in complementary colours. Biot studied the phenomenon with great precision and explained it in the particle theory: light molecules inside the crystal sheet spin back and forth, a theory he called mobile polarization. Fresnel later explained the same colours by interference. Biot’s formulas survived; his mechanism did not.

Goals
  • Write Biot’s empirical intensity law for a crystal lamina between polarizers.
  • Derive the exact intensity from wave optics and match it to Biot’s form.
  • Explain the complementary colours.

Biot’s law

Let linearly polarized light pass through a thin birefringent lamina whose axis makes angle \(i\) with the plane of polarization, then through an analyser at angle \(\alpha\). Biot first tried to count rays with Malus’s law, treating the two beams inside the lamina as independent. The result predicted half the light at \(\alpha=0,\ i=45^\circ\), where he saw none. He then found that the data obey

\[ F_o=U\cos^2\alpha+A\cos^2(2i-\alpha),\qquad F_e=U\sin^2\alpha+A\sin^2(2i-\alpha), \]

where \(U\) is the fraction of light that keeps its polarization and \(A=1-U\) is the fraction turned through the angle \(2i\), from one side of the axis to the mirror-image side. The fractions depend on colour and on thickness, periodically, like Newton’s rings. In his 1814 memoir Biot attributed this to light molecules whose axes oscillate inside the lamina between the two azimuths 0 and \(2i\).

Wave-optical result

A lamina of birefringence \(\Delta n\) and thickness \(d\) introduces the phase difference \(\delta=2\pi\Delta n\,d/\lambda\). With incident intensity 1, the intensity transmitted by the analyser is

\[ I(\alpha)=\cos^2\!\tfrac\delta2\,\cos^2\alpha+\sin^2\!\tfrac\delta2\,\cos^2(2i-\alpha). \]

So Biot’s law is exact, with \(U=\cos^2(\delta/2)\) and \(A=\sin^2(\delta/2)\).

Derivation. Resolve the incident field \(\hat x\) along the lamina axes \(\mathbf e_1=(\cos i,\sin i)\), \(\mathbf e_2=(-\sin i,\cos i)\): it is \(\cos i\,\mathbf e_1-\sin i\,\mathbf e_2\). The lamina multiplies the second component by \(e^{i\delta}\). Project on the analyser \(\mathbf a=(\cos\alpha,\sin\alpha)\), using \(\mathbf e_1\cdot\mathbf a=\cos(\alpha-i)\), \(\mathbf e_2\cdot\mathbf a=\sin(\alpha-i)\):

\[ \begin{aligned} I&=\big|\cos i\cos(\alpha-i)-\sin i\sin(\alpha-i)\,e^{i\delta}\big|^2\\ &=\cos^2\alpha+\sin2i\,\sin(2\alpha-2i)\,\sin^2\!\tfrac\delta2. \end{aligned} \]

The identity \(\cos^2(2i-\alpha)-\cos^2\alpha=-\sin2i\,\sin(2i-2\alpha)\) turns this into the boxed form. A lamina with \(\delta=\pi\), a half-wave plate, turns all the light through \(2i\): this is Biot’s “affected” light in pure form.

Exercise 8.1Show that \(F_o+F_e=U+A\) for every \(\alpha\) and \(i\). Explain why the two images seen through Iceland spar have complementary colours, and why they are white when \(i=0\).

Show solution
\(\cos^2+\sin^2=1\) in each term. For white light, \(U\) and \(A\) depend on \(\lambda\) through \(\delta\), but \(F_o(\lambda)+F_e(\lambda)\) is independent of \(\lambda\), so the two colours add to white: they are complementary. When \(i=0\), \(\cos^2(2i-\alpha)=\cos^2\alpha\), so \(F_o=(U+A)\cos^2\alpha\) has no colour dependence: both images are white, as Arago observed at four positions of the lamina.

Exercise 8.2Biot’s first ray-counting model gave \(F_o=U\cos^2\alpha+A\cos^2i\cos^2(\alpha-i)+A\sin^2i\sin^2(\alpha-i)\). Show that this is what wave optics gives if the two beams in the lamina are added incoherently.

Show solution
Expanding \(|\cos i\cos(\alpha-i)-\sin i\sin(\alpha-i)e^{i\delta}|^2\) gives \(\cos^2i\cos^2(\alpha-i)+\sin^2i\sin^2(\alpha-i)-2\cos i\sin i\cos(\alpha-i)\sin(\alpha-i)\cos\delta\). Averaging over random \(\delta\) removes the last term and leaves the incoherent sum. Biot’s failure at \(\alpha=0,\ i=45^\circ\) was a failure to include interference, which Fresnel and Arago made the basis of their laws of polarized light.

Exercise 8.3Gypsum has \(\Delta n\approx0.009\). What thickness gives a half-wave plate for green light of 550 nm? Why do thick plates look white between crossed polarizers?

Show solution
\(\delta=\pi\) requires \(d=\lambda/(2\Delta n)\approx550\times10^{-9}/0.018\approx31\ \mu\mathrm m\). For thick plates \(\delta\) changes by many multiples of \(2\pi\) across the visible spectrum, so \(\sin^2(\delta/2)\) oscillates rapidly with \(\lambda\) and averages to \(\tfrac12\): the eye sees a pale, nearly white mixture. Biot’s colours occur only for thin laminae, in the same way as Newton’s rings.
Lesson 09

Optical rotation and polarimetry

For the general reader. Some substances twist the plane of polarized light. Arago saw the effect in quartz in 1811. Biot made it a science: he found its laws, found it in liquids such as turpentine and sugar solutions, and showed that it persists in vapour. A property that survives when the molecules fly apart must belong to each molecule. This led directly to Pasteur’s discovery of molecular handedness and to the polarimeter, still used to measure sugar.

Goals
  • State Biot’s laws of rotation.
  • Use specific rotation to measure concentration.
  • Connect Biot’s laws to Fresnel’s circular birefringence.
Biot’s laws of rotatory polarization (1812–1818)
  1. The angle of rotation is proportional to the thickness traversed.
  2. For quartz cut perpendicular to its axis, some crystals rotate to the right and others to the left by equal amounts.
  3. The rotation varies with colour, approximately as the inverse square of the wavelength: \(\rho\propto1/\lambda^2\).
  4. In a solution of an active substance in an inactive solvent, the rotation is proportional to the concentration, and rotations of several active substances add.

The first memoir (1812) recorded that the rotation in quartz increases from red to violet. In 1815 Biot found rotation in liquids, including oil of turpentine and solutions of sugar and camphor, and in 1818 he announced the inverse-square law and observed rotation in turpentine vapour. He measured liquids by balancing their rotation against quartz plates of known thickness. John Herschel linked the sense of rotation of quartz to the handedness of its crystal faces in 1820. The inverse-square law is only an approximation; Biot himself found deviations in 1836.

Specific rotation

Laws 1 and 4 combine into the definition used in every chemistry laboratory:

\[ \alpha=[\alpha]^{T}_{\lambda}\;l\;c, \]

with \(\alpha\) the observed rotation in degrees, \(l\) the path length in decimetres, \(c\) the concentration in grams per millilitre, and \([\alpha]\) the specific rotation at temperature \(T\) and wavelength \(\lambda\), usually the sodium D line. For sucrose, \([\alpha]_D^{20}\approx+66.5^\circ\).

The wave explanation

Fresnel showed in 1822 that an optically active medium has different refractive indices \(n_L,n_R\) for the two circular polarizations. Linear light is their sum, and after thickness \(d\) its plane is turned by \(\rho=\pi(n_L-n_R)d/\lambda\) (see the Fresnel course, Lesson 07). Law 1 is immediate. Law 3 follows when \(n_L-n_R\) varies slowly with \(\lambda\) compared with \(1/\lambda\); a fuller dispersion theory, due to Paul Drude, gives \(\rho=\sum_k A_k/(\lambda^2-\lambda_k^2)\), which reduces to Biot’s law far from the absorption wavelengths \(\lambda_k\).

Exercise 9.1Quartz rotates sodium light (589 nm) by \(21.7^\circ\) per millimetre. Use Biot’s inverse-square law to predict the rotation for the mercury blue line at 436 nm. The measured value is about \(41.5^\circ\,\mathrm{mm^{-1}}\). Comment.

Show solution
\(21.7\times(589/436)^2\approx21.7\times1.825\approx39.6^\circ\,\mathrm{mm^{-1}}\). The measured value is about 5 percent higher. The rotation rises faster than \(1/\lambda^2\) toward the ultraviolet, as the Drude form predicts when \(\lambda\) approaches an absorption \(\lambda_k\): \(1/(\lambda^2-\lambda_k^2)\) exceeds \(1/\lambda^2\).

Exercise 9.2A sucrose solution in a 2 dm tube rotates sodium light by \(+13.3^\circ\). Find the concentration.

Show solution
\(c=\alpha/([\alpha]l)=13.3/(66.5\times2)=0.100\ \mathrm{g\,mL^{-1}}\), that is 10 g per 100 mL. This non-destructive measurement made the polarimeter, or saccharimeter, an instrument of the sugar trade and of customs offices.

Exercise 9.3Acid hydrolysis converts sucrose into an equimolar mixture of glucose (\([\alpha]_D\approx+52.7^\circ\)) and fructose (\([\alpha]_D\approx-92^\circ\)), with total mass increasing by the factor \(360/342\). Using Biot’s additivity law, find the specific rotation of the mixture per gram of original sucrose, and explain the name “invert sugar”.

Show solution
Per gram of mixture, \([\alpha]=\tfrac12(52.7-92)\approx-19.7^\circ\). Per gram of original sucrose, multiply by \(360/342\approx1.053\): about \(-20.7^\circ\). The sign of the rotation changes from \(+66.5^\circ\) to about \(-21^\circ\): the rotation is inverted, hence the name. Following the rotation in time lets one measure the reaction rate, as Ludwig Wilhelmy did in 1850.
Lesson 10

Biaxial crystals: the sine and dihedral laws

For the general reader. David Brewster found that many crystals, such as topaz and mica, have two optic axes instead of one. Biot found two simple rules for them: one for how the speeds of the two waves differ, the other for how they are polarized. He stated both as generalizations of the one-axis case. When Fresnel built his theory of double refraction in 1821, he showed that it reproduced Biot’s rules exactly. They became the first test the new theory passed.

Goals
  • State Biot’s sine law and dihedral law.
  • Derive the sine law from Fresnel’s normal velocities.
  • Check the uniaxial limit.

Use the notation of the Fresnel course: principal speeds \(a>b>c\), and two optic axes (directions of equal wave speed) in the plane of \(a\) and \(c\). For a wave normal \(\mathbf s\), let \(\theta_1,\theta_2\) be its angles with the two optic axes.

Normal velocities in a biaxial crystal

The two phase speeds \(v_1\ge v_2\) for wave normal \(\mathbf s\) satisfy

\[ v_{1,2}^2=\tfrac12\Big[(a^2+c^2)+(a^2-c^2)\cos(\theta_1\mp\theta_2)\Big]. \]

Biot’s sine law. Consequently \(v_1^2-v_2^2=(a^2-c^2)\sin\theta_1\sin\theta_2\).

Biot’s dihedral law. The planes of polarization of the two waves bisect the two dihedral angles formed by the planes through \(\mathbf s\) and each optic axis.

Derivation of the sine law. Subtract the two expressions: \(v_1^2-v_2^2=\tfrac12(a^2-c^2)\big[\cos(\theta_1-\theta_2)-\cos(\theta_1+\theta_2)\big]=(a^2-c^2)\sin\theta_1\sin\theta_2\). The speeds agree exactly when \(\mathbf s\) lies along an optic axis, as they must. The formula for \(v_{1,2}^2\) itself is obtained from Fresnel’s equation of wave normals by a rotation of coordinates; Exercise 10.2 checks it along the principal axes. Note that \(\sin\theta\) is unchanged by \(\theta\mapsto\pi-\theta\), so it does not matter which end of each optic axis is chosen.

For a uniaxial crystal Biot’s rules reduce to what Huygens and Malus knew: the difference of squared speeds varies as \(\sin^2\theta\), and the ordinary wave is polarized in the principal plane containing the axis.

Exercise 10.1Let \(b\to a\), so that the two optic axes merge along \(z\) and \(\theta_1=\theta_2=\theta\). Show that \(v_1^2-v_2^2=(a^2-c^2)\sin^2\theta\), and identify the two waves with the ordinary and extraordinary waves of a uniaxial crystal (Fresnel course, Exercise 9.1).

Show solution
With \(\theta_1=\theta_2\) the boxed formula gives \(v_1^2=a^2\) and \(v_2^2=\tfrac12[(a^2+c^2)+(a^2-c^2)\cos2\theta]=a^2\cos^2\theta+c^2\sin^2\theta\). So one wave travels at constant speed \(a\) (the ordinary wave, with \(v_o=a\)), the other has \(v^2=v_o^2\cos^2\theta+v_e^2\sin^2\theta\) with \(v_e=c\), the extraordinary wave. The difference is \((a^2-c^2)\sin^2\theta\).

Exercise 10.2The optic axes are at \(\pm\psi\) from the \(z\)-axis in the \(xz\)-plane, with \(\cos^2\psi=(b^2-c^2)/(a^2-c^2)\). Check the formula for \(v_{1,2}^2\) for \(\mathbf s\) along \(z\) (the speeds should be \(a\) and \(b\)) and along \(x\) (speeds \(b\) and \(c\)).

Show solution
Along \(z\): \(\theta_1=\theta_2=\psi\). Then \(\cos0=1\) gives \(a^2\), and \(\cos2\psi=2\cos^2\psi-1\) gives \(\tfrac12[(a^2+c^2)+2(b^2-c^2)-(a^2-c^2)]=b^2\). Along \(x\): \(\cos\theta_1=\sin\psi\), \(\cos\theta_2=-\sin\psi\), so \(\theta_2=\pi-\theta_1\). Then \(\theta_1+\theta_2=\pi\) gives \(\tfrac12[(a^2+c^2)-(a^2-c^2)]=c^2\), and \(\cos(\theta_1-\theta_2)=-\cos2\theta_1=1-2\sin^2\psi=2\cos^2\psi-1\) gives \(b^2\). Both correct.

Exercise 10.3For a wave normal in the plane of the optic axes, show that the dihedral law puts one polarization plane in that plane and the other perpendicular to it.

Show solution
If \(\mathbf s\) lies in the plane of the axes, the planes through \(\mathbf s\) and each axis both coincide with that plane. The dihedral angle between them is 0 (or \(\pi\)), and its bisectors are the plane itself and the plane through \(\mathbf s\) perpendicular to it. So the two waves are polarized in and across the plane of the optic axes, as Fresnel’s analysis of the \(xz\)-plane confirms.
Part III · Life and legacy
Lesson 11

Rivalries, photography, Pasteur

By 1816 Biot was among the best known physicists in Europe. His four-volume Traité de physique expérimentale et mathématique and the shorter Précis élémentaire de physique expérimentale taught a generation. He was elected a Foreign Member of the Royal Society in 1815. Yet the following decade turned against him.

His collaboration with Arago soured into a lasting enmity, fed by disputes over priority in polarization. On the Académie’s commission for the 1819 prize on diffraction, Biot sided with the particle theory, and Fresnel’s wave theory won anyway. Biot never accepted it. He twice sought the post of permanent secretary of the Académie and lost both times, to Fourier in 1822 and to Arago in 1830. Charles-Augustin Sainte-Beuve, who admired him, judged that he had every intellectual gift except originality. Others found him difficult; the mathematician Olinthus Gregory, who met him in Shetland, left a hostile portrait.

Photography, 1839

When photography was announced in 1839, Arago championed Louis Daguerre’s silvered plates. Biot preferred William Henry Fox Talbot’s sensitized paper and corresponded with him often. He thought that light carried a “chemical radiation” distinct from what the eye sees, and that photography should serve science first.

Recognition and Pasteur

The Royal Society awarded him the Rumford Medal in 1840 for his work on the polarization of light in chemical solutions. He continued to measure rotations through the 1830s and 1840s, including a long study of tartaric acid, and argued that rotation was a window onto molecular structure. In 1848 the young Louis Pasteur found that crystals of the inactive racemic tartrate came in two mirror-image forms, which rotated light in opposite senses when separated and dissolved. Biot was sceptical. He made Pasteur repeat the separation with materials Biot supplied, prepared the solutions himself and looked through the polarimeter. The left-handed solution turned the light strongly to the left. Pasteur recalled that the old man took his arm in emotion. From then on Biot was one of Pasteur’s strongest supporters.

DiscussionBiot found exact empirical laws of chromatic polarization and of biaxial crystals, then defended a mechanism that was wrong. Was his commitment to the particle theory a handicap, or did it drive the precision of his measurements?

Lesson 12

The scholar of old astronomies; death and legacy

From the 1820s Biot wrote more and more on the history of astronomy. He studied Egyptian astronomy (1823) and the zodiac of Dendera (1844), and published a long biography of Newton (1822). His son Édouard Constant Biot, born in 1803, trained as an engineer and became a sinologist; father and son worked together on ancient Chinese astronomy. Édouard died in 1850, and Biot completed his translation of the Zhouli (Rites of Zhou), checking the technical chapter on crafts with the help of Stanislas Julien and visits to workshops. His last book, on Indian and Chinese astronomy, appeared in 1862.

He was elected to the Académie des Inscriptions et Belles-Lettres in 1841 and to the Académie française in 1856. He was made a commander of the Légion d’honneur in 1849. He continued to teach physical astronomy at the Faculty of Sciences for more than fifty years. Biot died in Paris on 3 February 1862, aged 87.

Legacy

  • Electromagnetism: the Biot–Savart law is the standard way to compute magnetic fields of steady currents; the CGS unit of current, the biot, is named for him.
  • Heat transfer: the Biot number governs every lumped-capacitance and fin calculation.
  • Optics and chemistry: the laws of optical rotation, specific rotation and the polarimeter, which led through Pasteur to stereochemistry.
  • Earth and space: the L’Aigle report founded meteoritics; the balloon ascent opened scientific study of the upper air; the meridian and pendulum work refined the figure of the Earth.
  • Names: the mineral biotite, the lunar crater Biot and the rue Biot in Paris commemorate him.

Final projectChoose one law in Part II that Biot found empirically. Write a two-page account that states it as Biot did, derives it from modern theory, and identifies exactly which part of his physical explanation failed.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Jean-Baptiste Biot, MacTutor History of Mathematics.
  2. Jean-Baptiste Biot, Wikipedia.
  3. Jean-Baptiste Biot, Encyclopaedia Britannica.
  4. L. Pasteur, On the asymmetry of naturally occurring organic compounds (1860 lecture, English translation).
  5. M. Crosland, The Society of Arcueil: A View of French Science at the Time of Napoleon I, Heinemann, 1967.

Physics

  1. J.-B. Biot, F. Savart, Note sur le magnétisme de la pile de Volta, Annales de chimie et de physique 15 (1820). Text and commentary, e-magnetica.
  2. Biot–Savart law, Wikipedia.
  3. J.-B. Biot, Recherches expérimentales et mathématiques sur les mouvements des molécules de la lumière autour de leur centre de gravité, Paris, 1814.
  4. J.-B. Biot, Traité de physique expérimentale et mathématique, 4 vols., Paris, 1816.
  5. T. M. Lowry, Optical rotatory dispersion, Nature 117 (1926), on Biot’s laws and their limits.
  6. J. Z. Buchwald, The Rise of the Wave Theory of Light, University of Chicago Press, 1989.
  7. J. Bain, Arago and Biot, lecture notes on the history of light.
  8. L’Aigle meteorite, Wikipedia.
  9. Optical rotation, Wikipedia.
  10. A. Fresnel, Letter to Arago on biaxial birefringence (1821), Wikisource translation, on Biot’s laws.
  11. T. N. Narasimhan, Thermal conductivity through the 19th century, arXiv:1005.2119.
  12. M. Born and E. Wolf, Principles of Optics, Cambridge University Press, chapter on crystal optics.