Huygens and Bernoulli’sbrachistochrone

Henk Broer

Johann Bernoulli Institute for Mathematics and Computer Science

RijksuniversiteitGroningen

H&B – p.1/27

Summary

i. Christiaan Huygens’s isochronous curve

ii. Evolute and evolvent

iii. Johann Bernoulli’s surprise

iv. Literature

http://www.math.rug.nl/˜broer

H&B – p.2/27

Huygens and Bernoulli

Christiaan Huygens Johann Bernoulli1629–1695 1667–1748

H&B – p.3/27

Huygens’s isochronous curveHooke and Newton⇒equation of motionspring

mx′′ = −kx

Notation:

x′(t) =dx(t)

dt, x′′(t) =

d2x(t)

dt2

Abbreviateω =√

k/m, get solution

x(t) = R cos(ωt + φ)

PeriodP = 2π/ω independent of amplitudeR:isochrony

H&B – p.4/27

Spring

Spring modelled as harmonic oscillator

H&B – p.5/27

Commentary

1. Potential energy:V (x) = 1

2ω2x2 :

x′′ = −ω2x = −dV

dx(x)

harmonic oscillator

2. Pendulumx′′ = −ω2 sin x, ω =√

g/ℓan-isochronous: period−→∞ asx→ π

3. Gestalt-switch: bead moves along wire profilein vertical plane (gravitation, no friction)pendulum←→ circle

What profile provides harmonic oscillations(hence isochrony)?

H&B – p.6/27

Parametrization cycloid

Roll wheel (radius ̺) along ceiling−→

ξ(ϕ) = ̺(ϕ + sin ϕ), η(ϕ) = ̺(1− cos ϕ)

parameterϕ called rolling angleH&B – p.7/27

Arclength cycloid

arclengthx = x(ϕ) (using Pythagoras):

dx =√

dξ2 + dη2 =

=√

(dξ/dϕ)2 + (dη/dϕ)2 dϕ =

= ̺√

2√

1 + cos ϕ dϕ = 2̺ cosϕ

2dϕ

x(ϕ) = 4̺ sin ϕ2

H&B – p.8/27

Cycloidal wire profileVertical height

η(ϕ) = 2̺ sin2ϕ

2=

1

8̺(x(ϕ))2

potential energy“V = mgη” : V (x) = mg8̺

x2

−→ equation of motion bead

x′′ = − g

4̺x :

a harmonic oscillator withω =√

g/(4̺)

Conclusion: cycloid isochronous curve(also tautochronous. . .)

H&B – p.9/27

Evolute and evolventSocietal problem in 17th century:determination of longitude at sea

longitude∼ time (360 = 15× 24)pendulum clock: period increases with amplitude

To find adapted,isochronous pendulumIdea: ‘cheeks’ shorten lengthℓ,hence shorten period, at larger amplitude

What 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 evoluteH&B – p.10/27

Pendulum clock adapted by ‘cheeks’

Clock made by Solomon Coster by order ofChristiaan Huygens

J.M. Aarts and H.W. Broer, Schoolmeetkunde in het Horologium Oscillatorium van Christiaan

Huygens. InLiber Amicorum voor Agnes Verweij.(2011) (to appear)

J.G. Yoder,Unrolling Time, Christiaan Huygens and the mathematisation of nature.Cambridge

University Press (1988)

H&B – p.11/27

Two cycloidsHow to implement isochronous curvein pendulum set-up?

For simplicity taking̺ = 1, considercycloidsC = C(ϕ) andE = E(ϕ):

ξC(ϕ) = ϕ + sin ϕ, ηC(ϕ) = 1− cos ϕ

ξE(ϕ) = ϕ− sin ϕ, ηE(ϕ) = 3 + cos ϕ,

|ϕ| ≤ π

E andC congruentE is the evoluteof C andC the evolventof E

H&B – p.12/27

Huygens’s solution

K(ϕ) = chord betweenC(ϕ) enE(ϕ)

Do you see it coming ?

H&B – p.13/27

The master himself. . .

H&B – p.14/27

Evolute and evolvent

EvoluteE contains the centers of curvature of theevolventC and hence evolves the normal-bundle ofC

also think of an optical caustic. . .

H&B – p.15/27

Detailed formulation

1. K(ϕ) is tangent toE and perpendicular toC ;

2. |K(ϕ)| is radius of curvature ofC in C(ϕ)

Comments:

- Pendulum attached to cusp ofEchord evolves along ‘cheeks’E=⇒ mass followsC

- K⊥C ∼ instantaneous rotation in rolling-wheelconstruction of cycloid;

NB: K⊥C ⇒ dynamics of friction-less bead

H.W. Broer, Huygens’ isochrone slinger.Euclides70(4) (1995) 110-117

H&B – p.16/27

Check with high school computationsLemma:K(ϕ)//E′(ϕ) and K(ϕ)⊥C(ϕ)

Proof: On the one hand

K(ϕ) = C(ϕ) − E(ϕ) =

0

@

2 sin ϕ

−2 − 2 cos ϕ

1

A =

0

@

2 sin ϕ

2cos ϕ

2

−2 cos2 ϕ2

1

A ‖

0

@

sin ϕ

2

− cos ϕ2

1

A ;

On the other hand

E′(ϕ) =

0

@

1 − cos ϕ

− sin ϕ

1

A =

0

@

2 sin2 ϕ

2

−2 sin ϕ

2cos ϕ

2

1

A ‖

0

@

sin ϕ

2

− cos ϕ

2

1

A

etc., etc., etc. QED

NB: Check thatk(ϕ) = arclength alongE from E(ϕ) till extremal:

On the one handk(ϕ) = 2√

2√

1 + cos ϕ = 4 cos ϕ

2,

On the other hand the arclength equals4 sin ϕ+π

2= 4 sin( ϕ

2+ π

2) = 4 cos ϕ

2;

“figures !”H&B – p.17/27

Johann Bernoulli’s surpriseThe brachistochroneproblem (Groningen 1696):

Given: Two pointsP andQ in vertical plane underinfluence of gravity

Request: Wire profile for bead to slide fromP to Q inthe shortest time

Optical metaphore:bead follows light ray through optical medium withsuitably chosen speed op propagationv = v(h)

H&B – p.18/27

Two principlesx

h

1

2

j

j + 1

N

P

Q

nj

nj+1αj+1

αj

Speed of propagation byConservation of Energy

12v2(h)− gh = constant (1)

Lightray by Fermat Principle of least time:

H&B – p.19/27

DiscretisationHorizontal layers1, 2, . . . , N of equal width and withconstant speed of propagationvj, 1 ≤ j ≤ N

Lightray follows broken straight line,refraction according to Snell’s law

Limit N →∞ give cycloid: rolling angleϕ is doublethe inclination angle with the vertical direction

This assertion needs proof!

H&B – p.20/27

Fermat implies Snell (Leibniz 1684)

Willebrord Snellius Gottfried Wilhelm Leibniz1580–1626 1646–1716

H&B – p.21/27

Optics

P

Q

R

R′δ

n1 = 1/v1

n2 = 1/v2

α2

α1

Refraction between two media:nj = 1/vj, j = 1, 2 refraction indices

Snell’s Law:

n1 sin α1 = n2 sin α2

H&B – p.22/27

Optics ctd.

R

R′δ

n1 = 1/v1

n2 = 1/v2

α2

α1

δ sinα1

δ sinα2

Proof: Take variation of an arbitrary pathP → R → Q to P → R′ → Q with R′ − R = δ

Time difference:

δn1 sin α1 − δn2 sin α2 + O(δ2)

Optimal choice forR :

n1 sin α1 = n2 sin α2

QED

Comment: A high school proof exists with rectangles. . .

H&B – p.23/27

Deduction cycloidSnell’s law between successive layers

sin αj

vj

=sin αj+1

vj+1

, j = 1, 2, . . . , N

As N →∞ this gives anotherconservation law∗

sin α

v= C (2)

From conservation laws (1) en (2) will obtainparametrized curve

α 7→ (x(α), h(α))

∗Think of translation symmetry inx–direction andNoether’s theoremH&B – p.24/27

A few more high school computationsLemma (abbreviatingv′ = dv/dh)

v′ =g

vand cos α = Cv′

dh

dα,

Proof: Differentiate (1) with respect toh and (2) with respect toα QED

Thusdh

dα

lemma=

1

Cgv cos α

(2)=

1

2C2gsin 2α (3)

while:dx

dα= tan α

dh

dα

(3)= v sin α

(2)=

1

2C2g(1 − cos 2α)

Integration then gives

h(α) = h0 − 1

4C2gcos 2α

x(α) = x0 +1

4C2g(2α − sin 2α) :

cycloid with rolling angle ϕ = 2α and radius ̺ = 1/(4C2g) QED

H&B – p.25/27

Commentary

1. Challenge inActa Eruditorum (twice)Many contemporaries sent in solutionNewton’s was anonymous, however,ex ungue leonem cognavi

2. Nowadays excercise in course inCalculus ofVariations e.g., with LagrangianL(h, h′) =

q

1+(h′)2

−2gh

H.H. Goldstine,A History of the Calculus of Variations from the 17th throughthe 19th Century.

Studies in the History of Mathematics and Physical Sciences5, Springer-Verlag (1980)

H&B – p.26/27

Background bibliography1. J.M. Aarts,Meetkunde: facetten van de planimetrie en stereometrie, Epsilon Uitgaven47

(2000)

2. V.I. Arnold,Huygens & Barrow, Newton & Hooke.Birkhäuser (1990)

3. O. Bottema,Theoretische Mechanica, Epsilon Uitgaven3 (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 of

least time.Eur. J. Phys.20 (1999) 299–304

6. J.A. van Maanen,Een Complexe Grootheid, leven en werk van Johann Bernoulli,

1667–1748.Epsilon Uitgaven34 (1995)

H&B – p.27/27