courseshub.world · Lives in Mathematics and Physics

Léon Foucault

Jean Bernard Léon Foucault

A course in two strands: the life of a medical-school dropout and science journalist who became the most inventive experimental physicist of Second Empire Paris, and the experiments that showed the Earth turning, timed light in water and in air, and gave astronomers the modern reflecting telescope.

Born 18 September 1819, Paris Died 11 February 1868, Paris Pendulum Panthéon, 1851 Level advanced undergraduate
The Panthéon pendulum seen from above. At the latitude of Paris the plane of swing turns clockwise by about 11.3° each hour, a full turn in about 31.8 hours. Lines show the plane at hourly intervals.
Lesson 00

How to use this course

Léon Foucault had no degree in science until he was 33, held no professorship, and earned his living for years as a newspaper columnist. He was also the finest experimental physicist in France in his generation. His pendulum made the rotation of the Earth visible to anyone who stood under it. His rotating mirror showed that light is slower in water than in air and then measured its speed to better than one percent. His gyroscope, his discovery of the heating of conductors by eddy currents, and his method for making and testing silvered glass telescope mirrors are still part of everyday technology.

Prerequisites for Part II
  • Newtonian mechanics, including rotating frames and the Coriolis force (reviewed in Lesson 06).
  • Rigid-body rotation and angular momentum for Lesson 07.
  • Faraday’s law of induction for Lesson 08.
  • Geometrical optics of mirrors for Lesson 09.
  • For Lesson 04, the idea of blackbody radiation is helpful.

Each technical lesson opens with a paragraph for the general reader, then gives precise statements, derivations and exercises with folded solutions. Foucault was a close friend, partner and later rival of Hippolyte Fizeau; the Fizeau course tells the shared story from the other side.

Reference

Timeline

1819Born 18 September in Paris, son of the publisher Jean Léon Fortuné Foucault.
1829Death of his father at Nantes. Léon and his mother settle in Paris, at the corner of the rue de Vaugirard and the rue d’Assas.
1839Enters medical school in Paris and soon leaves. September: attends Louis Daguerre’s public demonstrations of the daguerreotype, with Fizeau.
c. 1840–43Assistant to Alfred Donné in his course on microscopic anatomy.
1845With Donné, an atlas of microscopy illustrated from daguerreotypes. Succeeds Donné as science writer for the Journal des débats. With Fizeau, takes the first detailed photograph of the Sun.
1847With Fizeau, shows that radiant heat produces interference fringes.
1849Finds that the electric arc both emits and absorbs light at the position of the solar D lines.
185027 April: rotating-mirror experiment shows light travels more slowly in water than in air, about seven weeks before Fizeau and Breguet.
1851January: pendulum in his cellar. 3 February: demonstration at the Paris Observatory. From late March: the 67 m pendulum in the Panthéon. Stands unsuccessfully for the Academy.
1852September: the gyroscope, which he names, gives a second proof of the Earth’s rotation.
1853Doctorate from the Paris Faculty of Sciences with a thesis on the speeds of light in air and water.
1855Copley Medal of the Royal Society. Appointed physicist at the Imperial Observatory under Urbain Le Verrier. September: eddy-current heating of a rotating copper disc.
1857February: reflecting telescope with a silvered glass mirror. The Foucault polarizing prism.
1858–59Knife-edge test and memoir on silvered-glass telescopes.
186018 July: observes and photographs the total solar eclipse in Spain with Le Verrier.
1862Measures the speed of light in the laboratory: 298 000 ± 500 km/s. 80 cm silvered-glass reflector completed. Member of the Bureau des longitudes; officer of the Légion d’honneur.
1864Foreign Member of the Royal Society.
1865Elected to the Académie des sciences, mechanics section. Papers on the regulation of Watt’s centrifugal governor.
1867Onset of a rapidly progressing paralysis.
1868Dies 11 February at his home in the rue d’Assas, aged 48. Buried in Montmartre Cemetery.
Part I · Formation
Lesson 01

A publisher’s son, 1819–1839

Jean Bernard Léon Foucault was born in Paris on 18 September 1819, five days before Hippolyte Fizeau. His father, Jean Léon Fortuné Foucault, was a publisher. Poor health led the father to retire with his family to Nantes, where he died in 1829, when Léon was nine. Léon and his mother returned to Paris and lived at the corner of the rue de Vaugirard and the rue d’Assas, near the Luxembourg gardens. The house remained his home, and he died there in 1868.

He was a delicate child, educated partly at home. At the Collège Stanislas his teachers thought him lazy, and his mother engaged private tutors to bring him through his studies. His gift was in his hands: he was a patient and skilful builder of small mechanisms. At Stanislas he met Fizeau, who became a close friend.

Like Fizeau, he began in medicine. He entered the Paris medical school in 1839, but he could not bear the sight of blood and suffering, and he left after his first experience of hospital work.

Lesson 02

Microscopes and daguerreotypes, 1839–1845

Foucault found his way into science through Alfred Donné, a physician who taught a course in microscopic anatomy. Donné engaged Foucault as his laboratory assistant for about three years. Foucault prepared the specimens and the projections for the lectures, and the two used the new daguerreotype process to photograph what the microscope showed. In 1845 they published an atlas of microscopy illustrated with engravings made from these daguerreotypes.

The daguerreotype had been made public in August 1839, and in September Foucault and Fizeau attended Louis Daguerre’s demonstrations in Paris. Projection microscopy and photography both need a powerful, steady light, and Foucault began to work on the carbon arc, in which an electric current crosses the gap between two carbon rods. The rods burn away and the gap widens, so the light flickers and dies unless the gap is regulated. Foucault built a regulator that adjusted the gap automatically, the first of many self-regulating mechanisms he designed.

In 1845 Donné handed over to Foucault his position as scientific writer for the Journal des débats, one of the most influential Paris newspapers. Foucault reported on the sessions of the Academy of Sciences for years. His reviews were clear and often sharp, and some academicians whose votes he would later need resented them.

Lesson 03

Fizeau, Arago and the arc, 1845–1850

At the suggestion of François Arago, director of the Paris Observatory, Foucault and Fizeau photographed the image of the Sun formed by a telescope. Their daguerreotype of 1845 shows sunspots and the darkening of the solar disc towards its edge. They compared the brightness of sunlight with the light of the carbon arc and the limelight, and in 1847 they showed that the invisible radiant heat beyond the red end of the spectrum produces interference fringes like visible light. They also studied chromatic polarization and interference over long path differences.

In 1849 Foucault examined the spectrum of the carbon arc and found a bright double line at the position of the dark D lines in sunlight. When he passed sunlight through the arc, the D lines in the sunlight became darker. The arc therefore absorbed the same light it emitted. He published the observation, but it attracted little notice. Ten years later Gustav Kirchhoff, working independently, made the same observation (Lesson 04) the starting point of spectrum analysis and of the law relating emission and absorption.

By 1850 the partnership with Fizeau had become a rivalry. Both wanted to perform the experiment Arago had proposed in 1838: to compare the speed of light in air and in water. A sealed letter that Foucault deposited with the Academy on 27 May 1850 shows that the two had worked together that spring on whether moving air drags light, and found no effect. But on the air–water question they raced, and Foucault won (Lesson 05).

Part II · The physics
Lesson 04

Arc light and the D lines, 1849

For the general reader. Sunlight spread into a spectrum is crossed by dark lines. In 1849 Foucault found that the light of an electric arc contains a bright double line at exactly the place of the dark D lines of the Sun, and that the arc, placed in the path of sunlight, made those dark lines darker. The same vapour that emits a colour also absorbs it. Ten years later Gustav Kirchhoff turned the same observation into the law that founded spectrum analysis and, with it, the chemistry of the stars.

Goals
  • State Kirchhoff’s law and see how it explains Foucault’s observation.
  • Solve the transfer equation for a uniform slab and find when a line appears bright or dark.
  • Understand why the carbon arc needed a regulator.

The observation

Joseph von Fraunhofer had mapped the dark lines of the solar spectrum and noted that a lamp flame showed a bright double line at the position of his dark D line. Foucault, studying the arc for projection and photography, saw the bright D lines in its spectrum. He then sent sunlight through the arc and into the spectroscope. The solar D lines did not fill in; they became darker. When he sent light from the incandescent carbon tips, which give a continuous spectrum, through the arc, dark D lines appeared. He concluded that the arc is a medium that emits the D rays and absorbs them when they come from elsewhere.

Kirchhoff’s law (1859)

In thermal equilibrium at temperature \(T\), the ratio of the emission coefficient \(j_\nu\) to the absorption coefficient \(\alpha_\nu\) of any material is a universal function of frequency and temperature, the Planck function:

\[ \frac{j_\nu}{\alpha_\nu}=B_\nu(T)=\frac{2h\nu^3}{c^2}\,\frac{1}{e^{h\nu/kT}-1}. \]

Kirchhoff’s argument was thermodynamic: if some body absorbed less than it emitted at a given frequency, it could be used to drive heat from a cooler to a hotter body. The explicit form of \(B_\nu\) came only with Planck in 1900. The law explains why a strong absorber at a frequency is also a strong emitter there.

Bright or dark?

Proposition (uniform slab)

Light of intensity \(I_0\) enters a uniform slab of gas at temperature \(T\) and optical depth \(\tau_\nu\) at frequency \(\nu\). The emerging intensity is

\[ I_\nu=I_0\,e^{-\tau_\nu}+B_\nu(T)\,\big(1-e^{-\tau_\nu}\big). \]

At a line, where \(\tau_\nu\) is large, the output approaches \(B_\nu(T)\). The line is dark against the background if \(I_0>B_\nu(T)\), bright if \(I_0<B_\nu(T)\).

Proof. The transfer equation along the ray is \(dI_\nu/ds=-\alpha_\nu I_\nu+j_\nu\). With \(d\tau=\alpha_\nu ds\) and Kirchhoff’s law, \(dI_\nu/d\tau=-(I_\nu-B_\nu)\), whose solution is the formula. In Foucault’s experiment the background was sunlight or the white-hot carbon, brighter at the D lines than the arc vapour’s \(B_\nu\), so the lines appeared dark. Without a background, \(I_0=0\) and the lines appeared bright. The solar Fraunhofer lines arise in the same way, from cooler gas above hotter layers.

The arc regulator

An arc between carbon rods burns the rods away. As the gap lengthens, the current falls and the arc goes out. Foucault’s regulator used the arc current itself, through an electromagnet, to release a clockwork that moved the rods together whenever the current fell, a feedback mechanism.

Exercise 4.1A sodium-laden slab has optical depth \(\tau=3\) at the centre of a D line and nearly 0 away from it. The background continuum has intensity \(I_0=2B\), where \(B\) is the slab’s Planck intensity at the line. Find the emerging intensity at line centre and the line’s depth relative to the continuum.

Show solution
\(I=2Be^{-3}+B(1-e^{-3})=B(1+e^{-3})\approx1.05B\). The continuum emerges unchanged at \(2B\), so the line centre is about 52% of the continuum: a strong dark line.

Exercise 4.2Using the Wien approximation \(B_\nu\propto e^{-h\nu/kT}\), compare \(B_\nu\) at 589 nm for carbon tips at 3800 K and arc vapour at 3000 K. Which is brighter, and so are the lines bright or dark when the tips are seen through the vapour?

Show solution
\(h\nu/k=hc/(\lambda k)=0.014388/5.89\times10^{-7}\approx24\,430\) K. The ratio is \(\exp\big(24\,430(1/3000-1/3800)\big)=\exp(1.71)\approx5.6\) in favour of the tips. The background is brighter, so the D lines appear dark, as Foucault saw. (Temperatures are illustrative.)
Lesson 05

Light in air and water, 1850

For the general reader. Is light a stream of particles or a wave? Both theories explained why a ray bends on entering water, but in opposite ways: Newton’s particles speed up in water, Huygens’s and Fresnel’s waves slow down. Arago proposed in 1838 to race light through air and water using a rapidly rotating mirror, an idea of Charles Wheatstone’s. Foucault made it work. In April 1850 he showed that light is slower in water.

Goals
  • Derive the deflection produced by a rotating mirror, with Foucault’s concave return mirror.
  • Compare the particle and wave predictions quantitatively.
  • Understand why the method needs no knowledge of \(c\).

Foucault’s arrangement

Light from an illuminated slit passes through a lens and falls on a plane mirror \(R\) that rotates rapidly about an axis in its own plane. The lens would form an image of the slit at a point on a concave mirror \(M\) whose centre of curvature lies on the axis of \(R\). Whatever the angle of \(R\), within the range for which the light strikes \(M\), the light returns along its own path to \(R\), and through the lens back to the slit. If \(R\) has turned during the round trip, the returning beam is deflected and the image of the slit is displaced. A glass plate at 45° reflects part of the returning light to a micrometer eyepiece.

Proposition (rotating mirror)

Let \(R\) turn at angular velocity \(\omega\), and let \(M\) be at distance \(h\). Light returns after \(\tau=2h/v\), where \(v\) is its speed on the path. The returning beam is deflected by \(2\omega\tau=4\omega h/v\), and the image of the slit, a distance \(\ell\) from \(R\) along the optical path, is displaced by

\[ \delta = \frac{4\omega h\ell}{v}. \]

A mirror turned through \(\alpha\) deflects a reflected beam through \(2\alpha\), and \(R\) turns through \(\omega\tau\) during the round trip. Foucault used two concave mirrors at equal distances from \(R\) and placed a 3 m tube of water in front of one of them. The rotating mirror, driven by a small steam turbine adapted from a siren of Charles Cagniard de la Tour, turned at 600 to 800 revolutions per second. Mirrors of tin amalgam broke up above about 200 revolutions per second, so he used silvered ones. The two return images, one from each path, appeared side by side. The image from the water path was displaced more. Each displacement was only a few tenths of a millimetre.

Proposition (the two predictions)

If a length \(L\) of a path of total one-way length \(h\) is filled with water of index \(n\), the round-trip time becomes \(\tau=\frac{2}{c}\big(h-L+nL\big)\) in the wave theory and \(\tau=\frac2c\big(h-L+L/n\big)\) in the particle theory. The wave theory predicts the larger displacement for the water path.

Foucault announced the result on 27 April 1850 and presented it to the Academy shortly after. Fizeau and Louis Breguet confirmed it on 17 June. Foucault concluded, more modestly than Arago had hoped, that the emission theory was incompatible with the facts. In 1853 he made the experiment the subject of his doctoral thesis at the Sorbonne. The experiment was qualitative: it compared two displacements and needed neither \(c\) nor the exact rotation rate. Pierre Duhem later used it as his example of why no single experiment can be strictly crucial between two theories, since it refutes only the emission theory together with its auxiliary assumptions.

Exercise 5.1Take \(h=4.0\) m, \(\ell=4.0\) m, 800 revolutions per second, and a 3.0 m water tube with \(n=1.333\). Find the displacement of each image and their difference according to each theory.

Show solution
\(\omega=2\pi\times800=5027\) rad/s. Air path: \(\tau=8.0/3.0\times10^8=2.67\times10^{-8}\) s, so \(\delta=2\omega\tau\ell=2\times5027\times2.67\times10^{-8}\times4.0\approx1.07\times10^{-3}\) m. Water path, wave theory: one-way optical length \(1.0+3.0\times1.333=5.0\) m, so \(\delta\) scales by \(5.0/4.0\) to about 1.34 mm. Particle theory: \(1.0+3.0/1.333=3.25\) m, so about 0.87 mm. The water image lies about 0.27 mm beyond the air image in the wave theory and about 0.20 mm short of it in the particle theory. The sign of the difference decides between them.

Exercise 5.2Why does the concave mirror \(M\) have to be centred on the axis of \(R\)? What would happen with a plane return mirror at the same distance?

Show solution
As \(R\) turns, the beam sweeps across \(M\). A concave mirror centred on \(R\) reflects every ray from \(R\) straight back to \(R\), so the returning light retraces its path for a range of mirror angles and a steady image forms. A plane mirror returns the light only at the single instant when the beam strikes it at normal incidence, giving a much fainter flash, unless a lens images the slit onto it as in Michelson’s later arrangement.

Exercise 5.3Show that the ratio of the displacements for the water and air paths, in the wave theory, does not depend on \(\omega\) or \(c\). Could Foucault have measured \(n\) this way?

Show solution
\(\delta_{\text{water}}/\delta_{\text{air}}=(h-L+nL)/h\), independent of \(\omega\) and \(c\). In principle yes, but the displacements were fractions of a millimetre read to perhaps 0.05 mm, so the ratio was known only roughly. The experiment was decisive about the sign, not precise about the size.
Lesson 06

The pendulum, 1851

For the general reader. A heavy pendulum, hung so that it can swing in any direction, keeps swinging in the same plane relative to the stars. The Earth turns beneath it. At the North Pole the floor would turn once a day under a fixed plane of swing. In Paris the effect is slower, a full turn in about 32 hours, but within minutes visitors under the dome of the Panthéon could see the pendulum’s line drifting clockwise. For the first time the rotation of the Earth was shown without looking at the sky.

Goals
  • Write the equations of motion of a pendulum in the rotating Earth frame.
  • Prove Foucault’s sine law exactly for the linearized equations.
  • Understand the practical difficulties and the role of the suspension.

History

In January 1851, in the cellar of his house, Foucault hung a 5 kg brass bob on a steel wire 2 m long and saw the plane of swing turn. Arago allowed him to repeat the experiment with an 11 m wire in the meridian room of the Paris Observatory, and on 3 February 1851 Arago presented Foucault’s note to the Academy. A few weeks later, with the support of the Prince-President Louis-Napoléon Bonaparte, a 28 kg bob was hung from the dome of the Panthéon on a wire 67 m long. From late March the public was admitted. A stylus beneath the bob cut a mark in a ring of sand at each swing.

Foucault stated that the plane turns at a rate proportional to the sine of the latitude. The mathematician Jacques Binet gave a derivation to the Academy on 17 February, and Giovanni Plana followed in Turin. Two centuries earlier, Vincenzo Viviani, a pupil of Galileo, had noticed that a pendulum’s swing drifts, but he did not connect it with the Earth’s rotation.

Equations of motion

Use local coordinates with \(x\) east, \(y\) north and \(z\) up, at latitude \(\phi\). The Earth rotates with angular velocity \(\boldsymbol\Omega\) of magnitude \(\Omega=7.292\times10^{-5}\ \text{rad/s}\), whose vertical component is \(\Omega_z=\Omega\sin\phi\). For small oscillations of a pendulum of length \(\ell\), with \(\omega_0^2=g/\ell\), the Coriolis acceleration \(-2\boldsymbol\Omega\times\mathbf v\) gives, neglecting vertical motion and terms of order \(\Omega^2\) in the centrifugal force,

\[ \ddot x=-\omega_0^2x+2\Omega_z\dot y,\qquad \ddot y=-\omega_0^2y-2\Omega_z\dot x. \]
Theorem (sine law)

Let \(z=x+iy\). The solutions of the linearized equations are exactly

\[ z(t)=e^{-i\Omega_z t}\,w(t),\qquad \ddot w+(\omega_0^2+\Omega_z^2)\,w=0. \]

Thus the motion is that of an ordinary isotropic pendulum seen from a frame turning clockwise (viewed from above, in the northern hemisphere) at angular velocity \(\Omega\sin\phi\). The plane of swing completes a turn in \(T=T_{\text{sid}}/\sin\phi\), where \(T_{\text{sid}}\approx23.93\) h is the sidereal day.

Proof. The two equations combine into \(\ddot z+2i\Omega_z\dot z+\omega_0^2z=0\). Substitute \(z=e^{-i\Omega_zt}w\): then \(\dot z=e^{-i\Omega_zt}(\dot w-i\Omega_zw)\) and \(\ddot z=e^{-i\Omega_zt}(\ddot w-2i\Omega_z\dot w-\Omega_z^2w)\). Adding, the terms in \(\dot w\) cancel and \(\ddot w+(\omega_0^2+\Omega_z^2)w=0\). If the bob is released from rest in the rotating frame, \(w\) moves on a very thin ellipse, essentially a line, and that line turns as \(e^{-i\Omega_zt}\).

At the poles the plane turns once per sidereal day; at the equator it does not turn at all. In Paris, at latitude 48°51′, it turns about 11.3° per hour. In practice a pendulum develops a slight elliptical motion from imperfect release or asymmetric suspension, and an elliptical orbit precesses on its own at a rate proportional to its area. Long wires, heavy bobs and careful release keep this spurious precession small; Heike Kamerlingh Onnes analysed it in his doctoral thesis of 1879. Modern museum pendulums often use a Charron ring or an electromagnetic drive to suppress it.

Exercise 6.1Find the period of the Panthéon pendulum (\(\ell=67\) m) and the time for its plane to turn once at latitude \(48.85^\circ\). By how much does the plane turn during one complete swing?

Show solution
\(T_0=2\pi\sqrt{67/9.81}\approx16.4\) s. \(T=23.93/\sin48.85^\circ\approx31.8\) h, so the rate is \(360^\circ/31.8\ \text{h}\approx11.3^\circ\) per hour. In 16.4 s the plane turns \(11.3\times16.4/3600\approx0.052^\circ\). With a 3 m amplitude, the end of the swing shifts by about \(3\times0.052\times\pi/180\approx2.7\) mm per swing, easily visible in the sand.

Exercise 6.2A pendulum is set swinging in Sydney (latitude \(33.9^\circ\) S). Which way does the plane turn, and how long does a full turn take?

Show solution
In the southern hemisphere \(\sin\phi<0\), so the rotation is anticlockwise viewed from above. The time is \(23.93/\sin33.9^\circ\approx42.9\) h.

Exercise 6.3Explain the sine law without equations: at the pole the floor turns under a fixed plane. Why, at latitude \(\phi\), does only the vertical component \(\Omega\sin\phi\) matter?

Show solution
Decompose the Earth’s rotation into a component about the local vertical, \(\Omega\sin\phi\), and one about the horizontal north axis, \(\Omega\cos\phi\). The vertical component turns the floor under the pendulum just as at the pole. The horizontal component tilts the local vertical, carrying the pivot and the direction of gravity together; it exerts no torque that turns the swing plane about the vertical, so to first order it does not rotate the plane relative to the floor.
Lesson 07

The gyroscope, 1852

For the general reader. The pendulum shows only the component of the Earth’s rotation about the vertical, and slowly. Foucault looked for a more direct demonstration. A rapidly spinning, well-balanced wheel, mounted in gimbals so that no force can twist it, keeps its axis pointing in a fixed direction in space. Relative to the turning Earth the axis slowly appears to move. Foucault called the device a gyroscope, from Greek words for “rotation” and “to see”.

Goals
  • Use conservation of angular momentum to explain the free gyroscope.
  • Derive the precession of a gyroscope under a torque.
  • Derive the oscillation of a gyrocompass about the meridian.

Foucault presented the gyroscope in September 1852. A heavy, carefully balanced rotor was spun up to high speed by a gear train, then placed in a gimbal mounting pivoted at its centre of mass. Through a microscope he watched the axis move relative to a scale fixed to the laboratory. The motion matched the Earth’s rotation and lasted only a few minutes before friction slowed the wheel, but it required no assumption about pendulum suspensions.

Proposition (free and forced gyroscope)

(a) A body mounted at its centre of mass feels no gravitational torque, so its angular momentum \(\mathbf L\) is constant in an inertial frame. If it spins fast about a symmetry axis, \(\mathbf L\approx I\omega_s\hat{\mathbf n}\) and the axis \(\hat{\mathbf n}\) is fixed relative to the stars.

(b) If a torque \(\boldsymbol\tau\) perpendicular to \(\mathbf L\) acts, then to leading order in \(1/\omega_s\) the axis precesses with angular velocity \(\boldsymbol\Omega_p\) satisfying \(\boldsymbol\tau=\boldsymbol\Omega_p\times\mathbf L\). For a gyroscope with its centre of mass a distance \(r\) from the pivot and its axis horizontal, \(\Omega_p=mgr/(I\omega_s)\).

Part (b) follows from \(d\mathbf L/dt=\boldsymbol\tau\): a torque perpendicular to \(\mathbf L\) changes its direction without changing its size, at rate \(\tau/L\).

Pointing north

Foucault also described in 1852 what happens if the axis is constrained to stay horizontal. The Earth’s rotation then forces the axis towards the meridian, about which it oscillates. This is the principle of the gyrocompass, which Hermann Anschütz-Kaempfe made into a practical ship’s instrument in the first decade of the twentieth century.

Proposition (gyrocompass)

Let the spin axis be constrained to the horizontal plane at latitude \(\phi\), with angle \(\alpha\) east of north, spin angular momentum \(L=I\omega_s\), and moment of inertia \(J\) about the vertical. Then, neglecting friction,

\[ J\ddot\alpha=-L\,\Omega\cos\phi\,\sin\alpha, \]

so the axis oscillates about north with small-amplitude period \(2\pi\sqrt{J/(L\Omega\cos\phi)}\).

Sketch. In the inertial frame the constraint must keep turning the horizontal plane, which rotates about the north horizontal axis at rate \(\Omega\cos\phi\). The component of that rotation that changes the direction of \(\mathbf L\) is \(\Omega\cos\phi\sin\alpha\), and the constraint supplies a torque of size \(L\Omega\cos\phi\sin\alpha\) about the horizontal, whose reaction acts about the vertical and drives \(\alpha\) towards 0.

Exercise 7.1Foucault’s rotor might have had \(I=2\times10^{-3}\ \text{kg m}^2\) and spun at 150 revolutions per second. If the centre of mass is displaced by 0.1 mm from the pivot and the mass is 2.7 kg, find the precession rate. Compare it with the apparent motion produced by the Earth’s rotation, at most \(15^\circ\) per hour.

Show solution
\(L=2\times10^{-3}\times2\pi\times150\approx1.9\ \text{kg m}^2/\text{s}\). \(\tau=2.7\times9.81\times10^{-4}\approx2.6\times10^{-3}\) N m. \(\Omega_p=\tau/L\approx1.4\times10^{-3}\) rad/s \(\approx4.8^\circ\) per minute, about 290° per hour. An imbalance of a tenth of a millimetre would swamp the effect, which is why Foucault balanced the rotor with extreme care.

Exercise 7.2A gyrocompass rotor has \(L=50\ \text{kg m}^2/\text{s}\) and \(J=0.05\ \text{kg m}^2\). Find its period of oscillation about north at latitude \(45^\circ\). Why do practical gyrocompasses add damping?

Show solution
\(L\Omega\cos\phi=50\times7.29\times10^{-5}\times0.707\approx2.58\times10^{-3}\) N m. Period \(=2\pi\sqrt{0.05/2.58\times10^{-3}}\approx27.7\) s. Without damping the compass would swing about north indefinitely; damping makes it settle on the meridian.
Lesson 08

Eddy currents, 1855

For the general reader. Move a lump of metal through a magnetic field and electric currents swirl inside it. They oppose the motion and heat the metal. Foucault showed this in 1855 with a copper disc spun between the poles of an electromagnet: when the magnet was switched on, the disc became hard to turn and grew hot. In France these eddy currents are still called courants de Foucault. They brake trains and roller coasters, heat induction cookers, and waste energy in transformers.

Goals
  • Derive the eddy-current density in a moving conductor.
  • Estimate the braking torque and power for Foucault’s disc.
  • Understand why transformer cores are laminated.

Arago had found in 1824 that a spinning copper disc drags a magnetic needle suspended above it, and Michael Faraday explained the effect by induction in 1831. Foucault reversed the arrangement in September 1855. He spun a copper disc by hand through a gear train between the poles of an electromagnet. With the magnet on, much more effort was needed to keep it turning, and the work done reappeared as heat: the disc grew noticeably warm within minutes.

Proposition (eddy-current braking, simple model)

A thin conducting sheet of conductivity \(\sigma\) and thickness \(d\) moves with velocity \(\mathbf v\) through a perpendicular uniform field \(B\) confined to a region of area \(A\), small compared with the sheet. Neglecting the electric field of accumulated charge and the field of the eddy currents themselves, the current density in the field region is \(\mathbf J=\sigma\,\mathbf v\times\mathbf B\), the dissipated power is

\[ P=\sigma v^2B^2Ad, \]

and the drag force is \(F=P/v=\sigma vB^2Ad\), proportional to the speed.

In the frame of the sheet the field moves, and Faraday’s law gives an electric field \(\mathbf E'=\mathbf v\times\mathbf B\) in the field region. Ohm’s law gives \(\mathbf J=\sigma\mathbf E'\), and the power is \(\int J^2/\sigma\,dV\). The currents close through the field-free part of the sheet, and charges at the edges of the field region reduce the current, typically by a factor of order 2, which the simple model ignores. The force \(\mathbf J\times\mathbf B\) opposes the motion, as Lenz’s law requires. At high speed the eddy currents’ own field partly expels the applied field, and the drag falls off.

Laminations

In a conductor exposed to an alternating field \(B_0\sin\omega t\), eddy currents waste power. For a lamination of thickness \(d\) with the field parallel to its faces, the loss per unit volume at low frequency is

\[ p=\frac{\sigma\,\omega^2B_0^2d^2}{24}, \]

averaged over a cycle. The loss falls as \(d^2\), which is why transformer and motor cores are built from thin insulated sheets.

Exercise 8.1A copper disc (\(\sigma=5.9\times10^7\ \text{S/m}\), \(d=5\) mm) turns at 10 revolutions per second between poles of area \(4\ \text{cm}\times4\) cm centred 8 cm from the axis, with \(B=0.5\) T. Estimate the braking torque and power with the simple model.

Show solution
\(v=2\pi\times10\times0.08\approx5.0\) m/s. \(P=\sigma v^2B^2Ad=5.9\times10^7\times25\times0.25\times1.6\times10^{-3}\times5\times10^{-3}\approx3\,000\) W. Torque \(=P/\omega=3\,000/62.8\approx47\) N m. The real values are smaller by a factor of order 2 because of the edge charges, but still far beyond what a hand crank can sustain for long, consistent with Foucault’s observation that the disc became very hard to turn and heated quickly.

Exercise 8.2Derive the lamination formula: take a slab \(-d/2<x<d/2\) with uniform field \(B(t)\hat{\mathbf z}\) and find the induced electric field from Faraday’s law, assuming \(E\) is along \(y\) and depends only on \(x\).

Show solution
Faraday: \(\partial E_y/\partial x=-\dot B\), and by symmetry \(E_y(0)=0\), so \(E_y=-x\dot B\). Power density \(\sigma E_y^2=\sigma x^2\dot B^2\). Averaging over the thickness, \(\langle x^2\rangle=d^2/12\). With \(B=B_0\sin\omega t\), \(\langle\dot B^2\rangle=\omega^2B_0^2/2\). So \(p=\sigma\omega^2B_0^2d^2/24\).

Exercise 8.3Using the lamination formula, compare the eddy-current loss at 50 Hz and 1.5 T in a solid iron block of thickness 2 cm with that in laminations 0.35 mm thick (\(\sigma=2\times10^6\ \text{S/m}\) for silicon steel).

Show solution
For the laminations, \(p=2\times10^6\times(2\pi\times50)^2\times2.25\times(3.5\times10^{-4})^2/24\approx2.3\times10^3\ \text{W/m}^3\), about 0.3 W per kg of steel. For a 2 cm block the formula gives \((20/0.35)^2\approx3\,300\) times more; in reality the skin depth, about a millimetre in iron at 50 Hz, limits it, but the loss would still be enormous.
Lesson 09

Silvered glass and the knife edge, 1857–1862

For the general reader. Before Foucault, large telescope mirrors were cast in speculum metal, a brittle alloy of copper and tin that reflected only about two thirds of the light and tarnished quickly. Foucault made mirrors of glass, which can be ground and polished accurately, and coated them with a thin chemical layer of silver that can be renewed. Then he invented a simple test, using only a light, a razor edge and the naked eye, that reveals errors in a mirror’s shape of a fraction of a wavelength. Every large telescope until the 1930s, and amateur telescope-making to this day, rests on these two ideas.

Goals
  • Show why a telescope mirror must be a paraboloid and how much it differs from a sphere.
  • Derive the knife-edge test and its zonal readings.
  • Follow Foucault’s telescopes from 10 cm to 80 cm.

Sphere and paraboloid

Proposition

A concave paraboloid \(z=r^2/(2R)\) brings all rays parallel to its axis to a focus at \(z=R/2\). The sphere with the same vertex radius of curvature \(R\) has \(z=r^2/(2R)+r^4/(8R^3)+O(r^6/R^5)\). To make a sphere into a paraboloid one removes glass of depth about \(r^4/(8R^3)\) at radius \(r\), relative to the centre.

The paraboloid focuses parallel light by the focus–directrix property. The sphere expansion is \(R-\sqrt{R^2-r^2}=\frac{r^2}{2R}+\frac{r^4}{8R^3}+\cdots\). For a 40 cm mirror with \(R=5\) m, the edge correction is \(0.2^4/(8\times125)=1.6\ \mu\text{m}\), about three wavelengths of light. The figure must be correct to a fraction of this.

The knife-edge test

Place a point source near the centre of curvature of the mirror, and bring a straight edge into the returning cone of light just at its focus. To an eye behind the edge, a perfect sphere with the source at its centre darkens uniformly all at once. If the edge is slightly inside or outside focus, a shadow sweeps across the mirror from one side or the other. A surface error that tilts the local normal sends that zone’s light past the edge or onto it, so the mirror appears as a relief map lit from the side, with hills and hollows of a few nanometres made visible.

Proposition (zonal readings for a paraboloid)

For a paraboloid of vertex radius \(R\), light from the zone at radius \(r\) crosses the axis behind the centre of curvature of the vertex by

\[ \Delta=\frac{r^2}{2R}\ \text{ if the source moves with the knife edge},\qquad \Delta=\frac{r^2}{R}\ \text{ if the source is fixed at the centre of curvature}. \]

Proof. At radius \(r\) the normal to \(z=r^2/(2R)\) has direction \((-r/R,1)\), so it meets the axis at \(z=r^2/(2R)+R=R+r^2/(2R)\). A source at that point lies on the normal and is returned to itself, which gives the first case. With the source fixed at \(z=R\), the source sits a distance \(r^2/(2R)\) in front of the normal’s crossing point. By the law of reflection, for small angles the reflected ray crosses the axis at the mirror-image point on the other side, a further \(r^2/(2R)\) beyond, so at \(R+r^2/R\) to leading order. The optician measures the knife position at which each zone darkens and compares with these values.

Foucault presented a 10 cm silvered-glass reflector on 16 February 1857, unaware that Carl August von Steinheil had made a similar proposal in Munich the year before. The silver was deposited chemically from solution. He published the knife-edge test in 1858 and a full memoir on silvered-glass telescopes in 1859. A 40 cm reflector was mounted at the Observatory in 1859, and in 1862 he completed a mirror of 80 cm aperture (78 cm clear) and 4.5 m focal length, then the largest of its kind. It was installed at the Marseille Observatory, where Édouard Stephan later used it to test Fizeau’s method of stellar interferometry and to discover galaxies. Foucault also invented the polarizing prism that bears his name (1857) and, in 1866, a solar eyepiece with a semi-transparent silver film on the objective.

Exercise 9.1An 80 cm mirror has focal length 4.5 m. Find \(R\), the depth of glass to be removed at the edge in parabolizing it, and the knife-edge reading for the edge zone with a moving source.

Show solution
\(R=2f=9.0\) m, edge radius \(r=0.40\) m. Correction \(r^4/(8R^3)=0.0256/5832\approx4.4\ \mu\text{m}\). Zonal reading \(r^2/(2R)=0.16/18\approx8.9\) mm, easily measured with a ruler.

Exercise 9.2Speculum metal reflects about 65% of visible light and fresh silver about 95%. A Newtonian telescope uses two reflections. Compare the light reaching the eyepiece.

Show solution
Speculum: \(0.65^2\approx0.42\). Silver: \(0.95^2\approx0.90\). The silvered telescope delivers more than twice the light for the same aperture, and the glass blank is lighter and easier to figure.
Lesson 10

The speed of light and the size of the solar system, 1862

For the general reader. In 1862 Foucault turned his rotating mirror into an absolute measuring instrument. With the light path folded into the space of a room, he measured the speed of light as 298 000 km/s, nearly 4% less than the value then accepted. The result mattered to astronomers. Combined with the aberration of starlight, it gave the speed of the Earth in its orbit and hence the distance to the Sun, and it agreed with Le Verrier’s calculations from the motions of the planets.

Goals
  • Derive \(c\) from the rotating-mirror displacement.
  • Use the aberration constant to turn \(c\) into the astronomical unit.
  • Understand the precision Foucault needed and achieved.

The measurement

The 1850 apparatus compared two paths. To measure \(c\), Foucault needed an exact rotation rate and a known path. He folded the path with five concave mirrors to a total one-way length of about 20 m, so that the return image was displaced by less than a millimetre. The rotation rate was fixed by comparing the mirror with a toothed disc driven by a clock. He presented the result to the Academy on 22 September 1862 and described the apparatus on 24 November.

Proposition

With the notation of Lesson 05, if the image is displaced by \(\delta\) when the mirror turns at \(n\) revolutions per second, then

\[ c=\frac{8\pi n h\ell}{\delta}. \]

Foucault’s result was \(c=298\,000\pm500\) km/s.

From c to the Sun

James Bradley discovered in 1728 that stars appear displaced towards the direction of the Earth’s motion by an angle up to \(\kappa\approx v_\oplus/c\), the constant of aberration. The value then accepted, \(\kappa=20.445''\), gives \(v_\oplus\) once \(c\) is known. The Earth’s orbital speed and the length of the year give the radius of the orbit, and the radius of the Earth then gives the solar parallax \(\pi_\odot\), the angle the Earth’s equatorial radius subtends at the Sun.

Proposition (solar parallax from c) \[ a=\frac{\kappa\,c\,T_{\text{yr}}}{2\pi},\qquad \pi_\odot=\frac{R_\oplus}{a}=\frac{2\pi R_\oplus}{\kappa\,c\,T_{\text{yr}}}. \]

With Foucault’s \(c\) this gives \(\pi_\odot=8.86''\), against the 8.57″ derived from the transits of Venus of 1761 and 1769 in Johann Encke’s reduction. Le Verrier had argued from planetary theory that the parallax was larger than Encke’s value; Foucault’s result supported him and reduced the accepted distance to the Sun by about 3%. The modern value is \(\pi_\odot=8.794''\).

Exercise 10.1Reproduce the 8.86″ value from \(c=298\,000\) km/s, \(\kappa=20.445''\), a sidereal year of 365.256 days and \(R_\oplus=6378\) km. What astronomical unit does it imply?

Show solution
\(\kappa=20.445/206\,265=9.912\times10^{-5}\) rad, so \(v_\oplus=29.54\) km/s. \(a=v_\oplus T/(2\pi)=29.54\times3.1558\times10^{7}/6.2832\approx1.484\times10^{8}\) km. \(\pi_\odot=6378/1.484\times10^8\) rad \(=8.87''\), matching Foucault’s 8.86″ to rounding. The modern astronomical unit is \(1.496\times10^8\) km.

Exercise 10.2Take a one-way path \(h=20\) m and a rotation rate of 400 revolutions per second. Find the angle through which the returning beam is deflected, and the displacement of an image 1.0 m from the rotating mirror. To measure \(c\) to 0.2%, how accurately must this displacement be read?

Show solution
\(\tau=40/2.98\times10^8=1.34\times10^{-7}\) s and \(\omega=2513\) rad/s, so the deflection is \(2\omega\tau\approx6.7\times10^{-4}\) rad and the displacement at 1.0 m is about 0.67 mm. A precision of 0.2% requires reading it to about 1.3 µm, at the limit of a micrometer eyepiece. Michelson’s main improvement was to make this displacement more than a hundred times larger (see the Michelson course, Lesson 04).
Part III · Life and legacy
Lesson 11

The Observatory and the Academy, 1855–1868

The pendulum made Foucault famous, but it did not bring him a seat in the Academy. In March 1851 he stood for election, as Fizeau did, and failed; his reviews in the Journal des débats and his lack of formal qualifications counted against him. His doctorate of 1853 answered the second objection. Honours came from abroad and from the state. The Royal Society awarded him the Copley Medal for 1855 for the pendulum and gyroscope. Napoleon III, who had followed the pendulum demonstrations, had a post created for him: physicist attached to the Imperial Observatory, where Urbain Le Verrier had become director in 1854. The post, held from 1855, gave him a workshop and a modest salary.

At the Observatory he built the silvered-glass telescopes and the regulators that drove the instruments at constant speed. He went with Le Verrier to Spain to observe and photograph the total solar eclipse of 18 July 1860. In 1862 he became a member of the Bureau des longitudes and an officer of the Légion d’honneur. The Royal Society elected him a Foreign Member on 9 June 1864. He was also elected to the Leopoldina. In 1865, after earlier defeats, he was elected to the mechanics section of the Académie des sciences, in the seat of Émile Clapeyron. That year he published studies of the centrifugal governor of James Watt and of the regulation of electric light.

In 1867 he began to lose sensation in his hands, and paralysis followed quickly. The illness is usually described as a rapid form of multiple sclerosis; some writers suggest that a lifetime of work with mercury contributed. Near the end of his life he returned to the Catholic faith of his childhood. He died at his home in the rue d’Assas on 11 February 1868, aged 48, and was buried in Montmartre Cemetery.

DiscussionFoucault and Fizeau, born five days apart and trained together, ended as rivals. Using both courses, compare their careers: their means, their posts, their styles of experiment and the recognition they received. What did competition do for nineteenth-century optics?

Lesson 12

Legacy

Foucault was an instrument-maker of genius who always aimed his instruments at a question of principle. His work divides between demonstrations that made physics visible and techniques that made it precise.

  • The Earth’s rotation. Foucault pendulums swing in hundreds of science museums and universities. The original Panthéon bob is in the Musée des Arts et Métiers in Paris, and a pendulum again hangs in the Panthéon.
  • Gyroscopes. Gyrocompasses guided ships and submarines through the twentieth century, and gyroscopes, mechanical, optical and micro-machined, stabilize aircraft, spacecraft and phones. Gravity Probe B used spinning quartz spheres as gyroscopes to measure relativistic precession.
  • The speed of light. The rotating mirror was the basis of Michelson’s measurements from 1879 to 1926. Foucault’s 1862 value gave the first accurate laboratory determination of the scale of the solar system.
  • Telescopes. Silvered (and later aluminized) glass mirrors made possible the great reflectors of the twentieth century. The knife-edge test is still taught to every amateur mirror-maker, and its descendants are used to test large optics.
  • Electromagnetism. Eddy currents brake trains, heat induction furnaces and cookers, and detect cracks in metal; their losses shaped the design of every transformer.

His name is one of the 72 inscribed on the Eiffel Tower. A lunar crater, the asteroid 5668 Foucault and a street in the 16th arrondissement of Paris bear his name.

Exercise 12.1A MEMS gyroscope in a phone measures angular velocity through the Coriolis force on a vibrating mass. If a proof mass vibrates with velocity amplitude 0.1 m/s and the phone rotates at 1 rad/s, what is the Coriolis acceleration? Compare it with the Earth’s rotation rate.

Show solution
\(a=2\Omega v=2\times1\times0.1=0.2\ \text{m/s}^2\). For the Earth’s rotation, \(\Omega=7.3\times10^{-5}\) rad/s, the acceleration is \(1.5\times10^{-5}\ \text{m/s}^2\), far below what consumer devices resolve. This is the same Coriolis term that turns the pendulum’s plane.

Final projectChoose one experiment from Part II. Write a two-page account for a strong undergraduate that describes Foucault’s apparatus, derives the effect, estimates the size of the signal he had to detect, and explains a modern technology that uses the same principle.

Reference

Sources and further reading

Biography

  1. J. J. O’Connor and E. F. Robertson, Léon Foucault, MacTutor History of Mathematics.
  2. Royal Society, Obituary notice of Léon Foucault, Proceedings of the Royal Society (1868).
  3. Léon Foucault and Léon Foucault (French), Wikipedia.
  4. W. Tobin, The Life and Science of Léon Foucault: The Man Who Proved the Earth Rotates, Cambridge University Press, 2003.
  5. J. Lequeux, Le Verrier: Magnificent and Detestable Astronomer, Springer, 2013 (for the Observatory years).
  6. A. D. Aczel, Pendulum: Léon Foucault and the Triumph of Science, Atria, 2003.

Physics

  1. L. Foucault, Sur les vitesses relatives de la lumière dans l’air et dans l’eau, doctoral thesis, Paris, 1853.
  2. L. Foucault, Démonstration physique du mouvement de rotation de la Terre au moyen du pendule, Comptes rendus 32 (1851).
  3. L. Foucault, Sur les phénomènes d’orientation des corps tournants entraînés par un axe fixe à la surface de la Terre, Comptes rendus 35 (1852).
  4. L. Foucault, Mémoire sur la construction des télescopes en verre argenté, Annales de l’Observatoire impérial de Paris 5 (1859).
  5. L. Foucault, Détermination expérimentale de la vitesse de la lumière, Comptes rendus 55 (1862).
  6. Foucault’s measurements of the speed of light, Foucault pendulum and Foucault knife-edge test, Wikipedia.
  7. Historical synthesis supporting Duhem’s demonstration of the impossibility of the experimentum crucis, arXiv:2310.17982.
  8. J. Texereau, How to Make a Telescope, Willmann-Bell, 2nd ed. 1984 (the knife-edge test in practice).