Huygens and Bernoulli's brachistochrone - Rijksuniversiteit Groningen

H&B – p.1/27Huygens and Bernoulli’sbrachistochroneHenk BroerJohann Bernoulli Institute for Mathematics and Computer ScienceRijksuniversiteit Groningen

H&B – p.2/27Summaryi. Christiaan Huygens’s isochronous curveii. Evolute and evolventiii. Johann Bernoulli’s surpriseiv. Literaturehttp://www.math.rug.nl/˜broer

H&B – p.3/27Huygens and BernoulliChristiaan Huygens Johann Bernoulli1629–1695 1667–1748

Huygens’s isochronous curveHooke and Newton ⇒equation of motion springNotation:mx ′′ = −kxx ′ (t) = dx(t) ,x ′′ (t) = d2 x(t)dt dt 2Abbreviate ω = √ k/m, get solutionx(t) = R cos(ωt + φ)Period P = 2π/ω independent of amplitude R:isochronyH&B – p.4/27

H&B – p.6/27Commentary1. Potential energy: V (x) = 1 2 ω2 x 2 :harmonic oscillatorx ′′ = −ω 2 x = − dVdx (x)2. Pendulum x ′′ = −ω 2 sinx, ω = √ g/lan-isochronous: period −→ ∞ as x → π3. Gestalt-switch: bead moves along wire profilein vertical plane (gravitation, no friction)pendulum ←→ circleWhat profile provides harmonic oscillations(hence isochrony)?

Parametrization cycloidRoll wheel ( radius ̺) along ceiling −→ξ(ϕ) = ̺(ϕ + sin ϕ),η(ϕ) = ̺(1 − cosϕ)parameter ϕ called rolling angleH&B – p.7/27

H&B – p.8/27Arclength cycloidarclength x = x(ϕ) (using Pythagoras):dx = √ dξ 2 + dη 2 == √ (dξ/dϕ) 2 + (dη/dϕ) 2 dϕ == ̺√2 √ 1 + cos ϕ dϕ = 2̺ cos ϕ 2 dϕ x(ϕ) = 4̺ sin ϕ 2

Cycloidal wire profileVertical heightη(ϕ) = 2̺ sin 2 ϕ 2 = 18̺(x(ϕ))2 potential energy “V = mgη” : V (x) = mg8̺ x2−→ equation of motion beadx ′′ = − g 4̺x :a harmonic oscillator with ω = √ g/(4̺)Conclusion: cycloid isochronous curve(also tautochronous ...)H&B – p.9/27

H&B – p.10/27Evolute and evolventSocietal problem in 17th century:determination of longitude at sealongitude ∼ time (360 = 15 × 24)pendulum clock: period increases with amplitudeTo find adapted, isochronous pendulumIdea: ‘cheeks’ shorten length l,hence shorten period, at larger amplitudeWhat is the shape of these ‘cheeks’?(musea Boerhaave, Hofwijck, Teylers)Related problems- Period as a function amplitude (elliptic integrals)- Geometrical properties of curves and their evolute

H&B – p.13/27Huygens’s solutionK(ϕ) = chord between C(ϕ) en E(ϕ)Do you see it coming ?

The master himself . . .H&B – p.14/27

H&B – p.15/27Evolute and evolventEvolute E contains the centers of curvature of theevolvent C and hence evolves the normal-bundle of Calso think of an optical caustic ...

H&B – p.16/27Detailed formulation1. K(ϕ) is tangent to E and perpendicular to C ;2. |K(ϕ)| is radius of curvature of C in C(ϕ)Comments:- Pendulum attached to cusp of Echord evolves along ‘cheeks’ E=⇒ mass follows C- K⊥C ∼ instantaneous rotation in rolling-wheelconstruction of cycloid;NB: K⊥C ⇒ dynamics of friction-less beadH.W. Broer, Huygens’ isochrone slinger. Euclides 70(4) (1995) 110-117

Check with high school computationsLemma: K(ϕ)//E ′ (ϕ) andK(ϕ)⊥C(ϕ)Proof: On the one handK(ϕ) = C(ϕ) − E(ϕ) =0@2 sin ϕ−2 − 2cos ϕ1A =0101@ 2 sin ϕ 2 cos ϕ 2−2 cos 2 ϕ2A ‖@ sin ϕ 2−cos ϕ 2A;On the other handE ′ (ϕ) =0@ 1 − cos ϕ−sin ϕ1A =0@2 sin 2 ϕ2−2sin ϕ 2 cos ϕ 21A ‖0@ sin ϕ 2−cos ϕ 21Aetc., etc., etc.NB: Check that k(ϕ) = arclength along E from E(ϕ) till extremal:On the one hand k(ϕ) = 2 √ 2 √ 1 + cos ϕ = 4 cos ϕ 2 ,On the other hand the arclength equals 4 sin ϕ+π = 4sin( ϕ 2 2 + π 2 ) = 4 cos ϕ 2 ;“figures !”QEDH&B – p.17/27

H&B – p.18/27Johann Bernoulli’s surpriseThe brachistochrone problem (Groningen 1696):Given: Two points P and Q in vertical plane underinfluence of gravityRequest: Wire profile for bead to slide from P to Q inthe shortest timeOptical metaphore:bead follows light ray through optical medium withsuitably chosen speed op propagation v = v(h)

H&B – p.19/27Two principlesP1x2α jn jjα j+1n j+1j +1NhQSpeed of propagation by Conservation of Energy12 v2 (h) − gh = constant (1)Lightray by Fermat Principle of least time:

H&B – p.20/27DiscretisationHorizontal layers 1, 2,... ,N of equal width and withconstant speed of propagation v j , 1 ≤ j ≤ NLightray follows broken straight line,refraction according to Snell’s lawLimit N → ∞ give cycloid: rolling angle ϕ is doublethe inclination angle with the vertical directionThis assertion needs proof!

H&B – p.21/27Fermat implies Snell (Leibniz 1684)Willebrord Snellius Gottfried Wilhelm Leibniz1580–1626 1646–1716

H&B – p.22/27OpticsPn 1 = 1/v 1α 1RδR ′α 2n 2 = 1/v 2QRefraction between two media:n j = 1/v j ,j = 1, 2 refraction indicesSnell’s Law:n 1 sinα 1 = n 2 sinα 2

H&B – p.23/27Optics ctd.δsinα 1δsinα 2n 1 = 1/v 1Rα 1δα 2R ′n 2 = 1/v 2Proof: Take variation of an arbitrary path P → R → Q to P → R ′ → Q with R ′ − R = δTime difference:δn 1 sin α 1 − δn 2 sin α 2 + O(δ 2 )Optimal choice for R :n 1 sin α 1 = n 2 sin α 2QEDComment: A high school proof exists with rectangles . . .

H&B – p.24/27Deduction cycloidSnell’s law between successive layerssinα jv j= sinα j+1v j+1, j = 1, 2,... ,NAs N → ∞ this gives another conservation law ∗sinαv= C (2)From conservation laws (1) en (2) will obtainparametrized curveα ↦→ (x(α),h(α))∗ Think of translation symmetry in x–direction and Noether’s theorem

H&B – p.25/27A few more high school computationsLemma (abbreviating v ′ = dv/dh)v ′ = g vandcos α = Cv ′ dhdα ,Proof: Differentiate (1) with respect to h and (2) with respect to αQEDThuswhile:dxdαIntegration then givesdhdαlemma== tan αdhdα1Cgv cos α(2)= 12C 2 g(3)= v sin α (2)= 12C 2 (1 − cos2α)gsin2α (3)h(α) = h 0 − 14C 2 g cos2αx(α) = x 0 + 14C 2 (2α − sin2α) :gcycloid with rolling angle ϕ = 2α and radius ̺ = 1/(4C 2 g)QED

H&B – p.26/27Commentary1. Challenge in Acta Eruditorum (twice)Many contemporaries sent in solutionNewton’s was anonymous, however,ex ungue leonem cognavi2. Nowadays excercise in course in Calculus ofqVariations e.g., with Lagrangian L(h, h ′ 1+(h) =′ ) 2−2ghH.H. Goldstine, A History of the Calculus of Variations from the 17th through the 19th Century.Studies in the History of Mathematics and Physical Sciences 5, Springer-Verlag (1980)

H&B – p.27/27Background bibliography1. J.M. Aarts, Meetkunde: facetten van de planimetrie en stereometrie, Epsilon Uitgaven 47(2000)2. V.I. Arnold, Huygens & Barrow, Newton & Hooke. Birkhäuser (1990)3. O. Bottema, Theoretische Mechanica, Epsilon Uitgaven 3 (1985)4. H.W. Broer, Huygens, Bernoulli en enige atmosferische optica. (2011) (in preparation)5. H. Erlichson, Johann Bernoulli’s brachistochrone solution using Fermat’s principle ofleast time. Eur. J. Phys. 20 (1999) 299–3046. J.A. van Maanen, Een Complexe Grootheid, leven en werk van Johann Bernoulli,1667–1748. Epsilon Uitgaven 34 (1995)