Nonholonomic dynamics, optics and the least time.

Anthony M. Bloch

with A. Rojo

• Least Action

• Optical Mechanical Analogy

• Hamiltonization

• Nonlinear Constraints

1 Introduction

Well known that there is an analogy between optics and me-chanics that inspired much of the classical theory of mechanicsand indeed extended to the theory of quantum mechanics.

Here we develop the optical mechanical analogy for a proto-typical non-holonomic mechanical system (a system with non-integrable velocity constraints): a knife-edge moving on theplane subject to a potential force.

Nonholonomic systems are not Hamiltonian or indeed varia-tional so this analogy is quite subtle. There is an interest goingback to the work of Chaplygin in finding a time transformationthat “Hamiltonizes” a (reduced) nonholonomic system (see alsothe work of Federov and Jovanovic, Borisov and Mamaev andFernandez, Mestdag and Bloch. See also work with Zenkov.

We show that our analysis provides a somewhat differentapproach to this idea. A key in all our analysis is to notethat the time variable is changed and so while trajectoriesare mapped to trajectories the dynamics along the trajectorieschange. Also important is the role played by gyroscopic forcesand the gyroscopic-like terms in the nonholonomic equations.

A key difference in our analysis is that our time change isdependent on the trajectory. We normally choose zero energy,as in the classical analysis of the knife edge on the plane, butour treatment can be extended to arbitrary energy without lossof generality. In this sense our analysis is closer the principle ofleast action than to the Lagrange D’Alembert principle, whichis of course the standard approach to nonholonomic systemsand is fundamental also to the Chaplygin Hamiltonization.

We explore various potentials for the nonholonomic systemthat gives rise to classical dynamic orbits in the plane and de-rive the associated index of refraction for the correspondingoptical system.

2 Nonholonomic Systems

The general equations of motion for a nonholonomic systemmay be formulated as follows. Let Q, a smooth manifold, bethe configuration space of the system. Let {ωa} be a set of m in-dependent one-forms whose vanishing describes the constraintson the system; that is, the constraints on system velocities aredefined by the m conditions ωa · v = 0, a = 1, . . . ,m. Using thefact that these m one-forms are independent one can chooselocal coordinates such that the one-forms ωa have the form

ωa(q) = dsa + Aaα(r, s)drα, a = 1, . . . ,m, (2.1)

where q = (r, s) ∈ Rn−m × Rm.

With this choice, the constraints on virtual displacements(variations) δq = (δr, δs) are given by the conditions

δsa + Aaαδr

α = 0. (2.2)

Now the Lagrange-D’Alembert principle gives the equations

−δL =

(d

dt

∂L

∂qi− ∂L

∂qi

)δqi = 0, (2.3)

for all variations δq such that δq that satisfy the constraints.Substituting (2.2) into (2.3) and using the fact that δr is ar-

bitrary gives(d

dt

∂L

∂rα− ∂L

∂rα

)= Aa

α

(d

dt

∂L

∂sa− ∂L

∂sa

), α = 1, . . . , n−m. (2.4)

The equations (2.4) combined with the constraint equations

sa = −Aaαr

α, a = 1, . . . ,m, (2.5)

give a complete description of the equations of motion of the sys-tem. Notice that they consist of n−m second-order equationsand m first-order equations.

We now define the “constrained” Lagrangian by substitutingthe constraints (2.5) into the Lagrangian:

Lc(rα, sa, rα) = L(rα, sa, rα,−Aa

α(r, s)rα).

The equations of motion (2.4) can be written in terms of theconstrained Lagrangian in the following way, as a direct coor-dinate calculation shows:

d

dt

∂Lc∂rα− ∂Lc∂rα

+ Aaα

∂Lc∂sa

= −∂L∂sb

Bbαβr

β, (2.6)

where

Bbαβ =

(∂Ab

α

∂rβ−∂Ab

β

∂rα+ Aa

α

∂Abβ

∂sa− Aa

β

∂Abα

∂sa

). (2.7)

Now one can show that the system is holonomic if and onlyif the the coefficients (2.7) vanish. More generally the systemis Lagrangian if the right hand side of (2.6) vanishes. One canview the goal of Hamiltonization as finding a change in the timevariable such that this occurs.

Examples:

θ

x

z

y

(x, y)

A

ξ

η

Ca

O

Figure 2.1: The Chaplygin sleigh is a rigid body moving on two sliding posts and one knife edge.

θ

x

z

y

(x, y)φ

d1 d

2

Figure 2.2: The geometry for the roller racer.

Figure 2.3: The rattleback.

Knife Edge on Inclined Plane:

m = mass

g

ϕ

(x, y)

xy

α

J = moment of inertia

Figure 2.4: Motion of a knife edge on an inclined plane.

The knife edge Lagrangian is taken to be

L =1

2m(x2 + y2

)+

1

2Jϕ2 + mgx sinα (2.8)

with the constraintx sinϕ = y cosϕ . (2.9)

The equations of motion:

mx = λ sinϕ + mg sinα ,

my = −λ cosϕ ,

Jϕ = 0 .

We assume the initial data x(0) = x(0) = y(0) = y(0) = ϕ(0) = 0and ϕ(0) = ω. The energy:

E =1

2m(x2 + y2

)+

1

2Jϕ2 −mgx sinα

and is preserved along the flow. Since it is preserved, it equalsits initial value

E(0) =1

2Jω2 .

Hence, we have1

2

x2

cos2ϕ−mgx sinα = 0 .

Solving, we obtain

x =g

2ω2sinα sin2 ωt

and, using the constraint,

y =g

2ω2sinα

(ωt− 1

2sin 2ωt

).

Hence the point of contact of the knife edge undergoes a cycloidmotion along the plane, but does not slide down the plane.

Different from the vakonomic (variational) sleigh (Kozlov,Arnold...).

3 The knife edge constraint

We now consider our prototypical example, the knife edge. Wedevelop first the geometry of the trajectories of a knife edgemoving on the plane co-ordinatized by (x, y) with blade angle θwith the y-axis.

Call φ(s) the tangent angle to a the trajectory [and s the arclength, not to be confused with the variable s in the pair (r, s)of the previous section] of a particle moving in two dimensions

x(s) =

(∫ s

0

ds sinφ(s),

∫ s

0

ds cosφ(s)

), (3.1)

and(x, y) = s (sinφ(s), cosφ(s)) . (3.2)

From which follows the relation

x cosφ− y sinφ = 0, (3.3)

valid for any unconstrained curve. Imposing a knife edge con-straint amounts to imposing the equality between the tangentangle to the curve, and the knife edge angle θ, which in princi-ple is independent of φ.

In other words, ifθ = φ (3.4)

then we have the knife edge in the usual form:

x cos θ − y sin θ = 0. (3.5)

4 Spatial dependence of the trajectory’s curvature

Once the constraint is imposed we can analyze the propertiesof the center of mass motion parametrizing the curve in termsof the arc length s and the tangent θ(s) to the curve:

θ = θ(s). (4.1)

We restrict to the case of “free” knife edge motion, meaningthat the knife edge variable θ is not subject to a θ dependentpotential. Now, for a knife edge constraint we have

θ = ω . (4.2)

and with ω a constant we obtain:

ω =dθ(s)

ds

ds

dt=

1

ρ(x)

p(x)

m, (4.3)

where ρ is the radius of curvature of the curve and p the mo-

mentum of the particle (we are considering fixed energy sincethe system conserves energy). So, a knife edge trajectory sat-isfies a simple relation between the radius of curvature and themomentum:

ρ(x) =1

ωmp(x) (4.4)

5 Knife Edge Dynamics

Using the constraint in the form x = y tan θ the reduced La-grangian (with the constraint substituted) and mass equal tounity becomes

Lc =1

2(y2 sec2 θ + θ2)− V (5.1)

In this case, in the notation above the variable s is equalto x and the variables (y, θ) are the r-variables. Assume forsimplicity that V is independent of x.

In the constraint equation (2.5) we thus have A11 = − tan θ while

A12 = 0.Thus

B112 = −B1

21 =∂A1

1

∂θ− ∂A1

2

∂y= sec2 θ . (5.2)

Then the dynamic equations of motion for y and θ follow from(2.6) and are given respectively by

y sec2 θ + 2y sec2 θ tan θθ = x sec2 θθ − ∂V

∂y= yθ sec2 θ tan θ − ∂V

∂y

θ − yθ tan θ sec2 θ = −x sec2 θθ − ∂V

∂θ= −yθ sec2 θ tan θ − ∂V

∂θwhere we used the constraints.

Hence we obtain

y = −yθ tan θ − cos2 θ∂V

∂y(5.3)

θ = − ∂V

∂θ. (5.4)

This, together with the constraints, defines the dynamics.As an example, consider V = 0, and θ = ω, where the above

equations, together with the constraint imply:

y = −ωxx = +ωy, (5.5)

which corresponds to the knife edge moving in a circular orbitand rotating at angular velocity ω

6 Optical mechanical analogy for the knife edge

The classical optical mechanical analogy stems from the iso-morphsim between trajectories of a particle of mass m, movingat constant energy E in a potential V (x) (the momentum beingp(x) =

√2m(E − V (x)), and that of a light ray that propagates,

at constant frequency, in a medium of index of refraction n(x).In each case, if xi and xf are the initial and final points, thetrajectories are the extrema of their corresponding action func-tionals:

So =∫ xfxinds (geometric optics)

Sm =∫ xfxipds (mechanics).

(6.1)

The analogy results from the equivalence of two conservationlaws: conservation of momentum in the direction parallel to thesurfaces of constant potential (Newton’s second law for parti-cles) and conservation of wave vector (or “slowness”) in thedirection parallel to the surfaces of constant index of refraction(Snel’s law for light rays).

The analogy implies that the physical trajectories between xiand xf can be either computed for a light ray or for a particle,provided one has the equivalence

p(x) =√

2m(E − V (x) = n(x). (6.2)

Notice that p and n have different units, but this is irrele-vant in determining the geometry of the trajectories since therespective units amount to multiplicative constants in their ac-tions.

The optical mechanical analogy elevated its status with theadvent of quantum mechanics, and the early search of a wavemechanics for particles. The natural question is: if geomet-ric optics is the small wave length limit of wave optics, whatplays the role of a wave length λ for particles, in such a waythat Newtonian mechanics is recovered in the limit of smallλ? The optical mechanical analogy provides the natural corre-spondence:

p(x) ∝ n(x) ∝ 1

λ(x). (6.3)

Since p and λ have different units there must be a constantof proportionality between them: p = h/λ, the celebrated DeBroglie’s relation, with the proportionality constant (Planck’suniversal constant) determined experimentally.

Now we explore an extension of the optical mechanical anal-ogy to a nonholonomic system. The special interest of thisproblem results from the fact that the nonholonomic trajec-tories are not determined by a Least Action Principle, so theanalogy in principle does not apply in the usual sense.

Consider the trajectory of a light ray propagating in an ar-bitrary two–dimensional index of refraction n(x), and choosen such that the local curvature of the ray κ = 1/ρ is preciselythat of the trajectory of the knife edge, as given by Eq. (4.4).This step leads us, following Hamilton’s program of the opticalmechanical analogy, to an equivalent Hamilton-Jacobi equationfor a non-holonomic system.

The problem of the curvature of a light ray in an arbitraryindex of refraction was treated by Born and Wolf in their classic“Principles of Optics”. Here we re-derive the same result usinga slightly different approach for completeness. We discretizethe problem into lines of constant n, as in Figure (6.1).

Figure 6.1: Discretization of the trajectory of a light ray in a spatially dependent index of refraction n

Snel’s law for a ray refracting on one of this lines is

n(s) sinα(s) = n(s + ds) sin (α + dθ) , (6.4)

where α(s) is the angle the light ray makes with the normalto the surface of constant n, s is the arc length and dθ is thechange of the angle of the tangent to the curve [See Figure(6.1)]. Notice that in this general case, whereas θ is the anglethat the tangent makes with a fixed direction in space, α is theangle that the tangent makes with the gradient of n.

Now expand the right hand side of Equation (6.4) to obtain

dθ

ds= −n

′(s)

n(s)tanα(s). (6.5)

Since α is the angle of the tangent to the ray with the normalto the surface,

dn(s)

ds

1

cosα(s)= |∇n|, (6.6)

and from this equation we obtain the general expression for thecurvature of the light ray

dθ

ds≡ κ(s) = −|∇n|

n(s)sinα(s), (6.7)

which leads us to the optical mechanical analogy for the knifeedge, relating the translational momentum p (a quantity deter-mined by the local potential at constant energy), and the indexof refraction:

ωm

p(s)= −|∇n|

n(s)sinα(s). (6.8)

This equation is the main result, relating geometric opticswith nonholonomic mechanics. Notice the difference with theusual optical mechanical analogy, for which p = n.

In order to apply the optical mechanical analogy we need tofind, explicitly, n given p, and this we were able to do for caseswhere there is a constant of motion that relates sinα(s) withposition.

7 Examples

Example: n(x, y) = n(y) and the brachistochroneHere we consider the case where the index of refraction varies

in one of the spatial directions only, as is the case for modelsof mirages and in the brachistochrone, one of the paradigmaticvariational problems. For this case |∇n| sinα = (dn/dy) sin θ (weput sinα = sin θ since the normal to n has a constant directionin space). Also, Snel’s Law in this case gives n sin θ = C, with Ca constant, and we can integrate (6.8) to obtain

n(y) ∝(∫

dy

p(y)

)−1. (7.1)

As a particular example consider the case of the knife edgefalling on an inclined plane (with potential proportional to y),for which (at zero translational energy) p(y) = a

√y, with a a

constant. Substituting in (7.1) we obtain

n(y) =c√y, (7.2)

and this corresponds to the classical mapping proposed first byBernoulli between the brachistochrone, a minimization of timeproblem, and the motion of a light ray. The point pertinent toour present treatment is that the brachistochrone trajectory isa cycloid–that is, the motion of a light ray moving in an indexof refraction such as that of Eq. (7.2) is a cycloid–and so is themotion of a knife edge in an inclined plane.

Example: Constant nConsider the case p = constant. For a knife edge the motion

corresponds to circular motion. According to (7.1) this wouldcorrespond to an index of refraction

n(y) ∝ 1

y. (7.3)

The trajectory of a light ray with an index of refraction withthis functional dependence is given by Snell’s law

n(y) sin θ = C,

sin θ = Cy, (7.4)

which of course is the equation of a circle.

Example: Rotational Symmetry, n(x, y) = n(r)In the case of a central potential we make use of the Formula

of Bouguer –the conservation of angular momentum for lightrays:

rn(r) sinα(r) = L (Constant), (7.5)

and the curvature of a light ray in a central potential, substi-tuting (7.5) in (6.8) is

κ(s) =dθ(s)

ds= −1

r

dn

dr

L

n2(7.6)

Using Equation (4.4) we can invert (7.6) to obtain

1

n(r)∝∫r dr

p(r)(7.7)

Example: Logarithmic spiralIn order to exemplify our treatment for central potentials we

proceed in reverse. We consider a few curves for which thecurvature as a function of the radius can be computed easily,and derive the potential for which a knife edge will describethe curve in question. Consider the curve of the logarithmicspiral

r(θ) = aebθ (7.8)

In order to find the corresponding p(r) for the knife edge wecompute the curvature κ. In cylindrical coordinates the curva-ture is given by the following relation:

κ =|r2 + 2r′2 − rr′′|

(r2 + r′2)3/2, (7.9)

which, for the logarithmic spiral gives

κ =1√2r, R =

√2r. (7.10)

Using (4.4),

p(r) =√

2ωmr, V (r) = −mω2r2, (7.11)

a repulsive quadratic potential. In other words, given ω as aconstant of motion, the knife edge “falls” in a repulsive har-monic potential following a logarithmic spiral. Now let’s seehow this connects with the corresponding optical problem.

According to our mapping (7.7), the corresponding index ofrefraction is, in this case,

n(r) =C

r, (7.12)

with C a constant.

Using this dependence in the Formula of Bouguer, Eq. (7.5),we obtain

sinα(r) = LC, (7.13)

which means that the trajectory of a light ray in an index ofrefraction n = C/r forms a constant angle of incidence with thesurfaces of constant n. This is a well known property of thelogarithmic spiral: the tangent to the curve forms a constantangle with the radius vector.

We illustrate this with a simple calculation:

r = aebθ(t)ur. (7.14)

The tangent vector to the spiral is therefore

t =r

|r|=bur + uθ√

1 + b2. (7.15)

which means that, indeed, for this curve the tangent forms aconstant angle with ur,

sinα =1√

1 + b2, (7.16)

the classic property of the logarithmic spiral We see that theparameter b is related to the “angular momentum” L of theray. For L = 0, b→∞ we get a straight line. for b = 0 we obtaina circle, which is also an orbit for the knife edge.

It is instructive to consider this dynamics from the standardnonholonomic point of view, enforcing the constraint by La-grange multipliers.

Consider the knife edge in a repulsive potential

V (r) = −1

2mω2

0r2 (7.17)

The non-holonomic equations of motion are

x = ω20x− 2

λ

mcos θ(t)

y = ω20y + 2

λ

msin θ(t), (7.18)

where time dependent Lagrange multipliers are introduced inthe equations of motion to enforce the constraint.

We want to check that the parametric equations for a loga-rithmic spiral, given by

x(t) = r0ebt cosαt

y(t) = r0ebt sinαt, (7.19)

are indeed solutions of the non-holonomic equations. The ini-tial conditions for the spiral (for zero translation energy of theparticle)

1

2mx2 =

1

2mr20(α

2 + b2)

=1

2mω2

0r20, (7.20)

which implies that the parameters of the spiral satisfy:

α2 + b2 = ω20. (7.21)

Also, the tangent vector to the spiral is

t =(b cosαt− α sinαt, b sinαt + α cosαt)√

b2 + α2

= (cos(αt + δ), sin(αt + δ)) (7.22)

withcos δ = b/

√b2 + α2 = b/ω0.

Since the tangent angle is rotating at a constant rate, we have

α = ω

with α the initial angular velocity of the knife edge–a constantof motion–and where δ is the initial angle that the knife edgemakes with the x axis.

Now we compute the second derivative of the spiral equation(Eq. (7.19))

x = (b2 − ω2)x− 2bωr0ebt cosωt

= (b2 + ω2)x− 2ωr0ebt (b cosαt− ω sinωt)

= ω20x− 2r0e

btω√ω2 + b2 cos(ωt + δ)

= ω20x−

λ

mcos θ(t), (7.23)

and, similarly

y = ω20y +

λ

msin θ(t), (7.24)

with

θ(t) = ωt + δ

λ = 2m2r0ebtω√ω2 + b2 . (7.25)

In summary, we have verified that, if a knife edge in a re-pulsive potential starts (at zero translational kinetic energy)with initial conditions r = r0, ω for the angular frequency of theknife, and δ for the initial orientation of the velocity (and theedge), the corresponding motion is a logarithmic spiral givenby

x(t) = r0eω0t cos δ cosωt

y(t) = r0eω0t cos δ sinωt (7.26)

7.0.1 Lemniscate of Bernoulli

This curve is described in polar coordinates as

r2(φ) = 2a2 cos 2φ, (7.27)

and has the interesting property that the radius of curvaturevaries as the inverse of the polar radius:

R =2a2

3r. (7.28)

According to the basic mapping of Eq. (4.4), the knife edgeproblem has this orbit for p ∝ 1/r, or for a potential

V (r) ∝ − 1

r2, (7.29)

(an attractive potential that decreases as the inverse squarepower) and zero total energy.

Figure 7.1: Lemniscate of Bernoulli. Notice that Eq. (7.27) describes a curve with two “lobes” that in principle could be oriented in any direction inthe plane. Since the direction of rotation is continuous, the motion of the knive edge will describe four lobes, corresponding to the two lemniscatesshown in the figure. If the knife edge is rotating clockwise, it will transition from lobe to lobe counterclockwise.

The corresponding index of refraction for this problem is

n ∝ 1

r3(7.30)

Finally, we verify that indeed the lemniscate motion corre-sponds to a holonomic potential V ∝ n2, which in this case isV ∝ −1/r6.

In general, for a particle moving in a central potential we have

dr

dθ=

√2mEr4

L2− r2 − 2mr4V (r)

L2

(7.31)

For V = −αr−6 and E = 0

dr

dθ=√β − r4/r

(7.32)

with β = 2mα/L2, and which is obeyed, as expected by,

r2(θ) = β sin 2θ. (7.33)

We have so far presented the optical mechanical analogy andillustrated its applicability in several specific examples. Uponmapping the problem to one of geometric optics (where theLeast Action Principle applies) we are in fact converting theproblem into a Lagrangian or Hamiltonian problem. This ispossible only because we are concerned with the geometry ofthe trajectories and not with their specific dynamics. In thefollowing sections we describe a related approach by Chaplygin,who “Hamiltonizes” the problem through a reparametrizationof time.

8 Chaplygin Analysis

The simplest setting is when the constraint functions Aaα and

L are independent of s, in which case the last term on the lefthand side of equation (2.6) vanishes as do the last two termsof (2.7).

For the classical case of Chaplygin Hamiltonization we assumethat there are only two base variables r1 and r2 and that theAaα depend only on these variables.The constraints take the form

sa = −Aa1r

1 − Aa2r

2, a = 1 . . .m . (8.1)

In this case we can compute that the equations (2.6) become

d

dt

∂Lc∂r1− ∂Lc∂r1

= r2S (8.2)

d

dt

∂Lc∂r2− ∂Lc∂r2

= −r1S (8.3)

where

S = −∂L∂sb

(∂Ab

1

∂r2− ∂Ab

2

∂r1

). (8.4)

Our goal is to make these equations Lagrangian.To this end change to the new time variable

dτ = N(q)dt . (8.5)

Denote the derivative with respect to new time variable asprimed, i.e. qi = N(q)q

′i. Also, denote Lc in terms of this timevariable by Lc.

Then we have

∂Lc∂rα

=1

N

∂Lc∂r′α

(8.6)

∂Lc∂rα

=∂Lc∂rα− 1

N

∂N

∂rα

2∑α=1

r′α Lc∂r′α

(8.7)

Then a computation shows that the equations (8.3) become

d

dτ

∂Lc∂r′1− ∂Lc∂r1

= r′2R (8.8)

d

dτ

∂Lc∂r′2− ∂Lc∂r2

= −r′1R , (8.9)

where

R = NS − 1

N

(∂N

∂r2∂Lc∂r′1− ∂N

∂r1∂Lc∂r′2

)(8.10)

Hence if we can choose N such that R is zero we have reducedthe equations to Lagrangian and hence Hamiltonian form. Fur-ther generalizations are possible.

9 Chaplygin Hamiltonization for the knife edge

Now let us return to the knife edge.The reduced Lagrangian was written in (5.1) (and may be

generalized in this Chaplygin setting to include any potentialwhich does not depend on x).

As above the classical Chaplygin Hamiltonization proceeds byintroducing a time change of the form dτ = N(q)dt which makesthe reduced dynamics Hamiltonian or Lagrangian.

Here we can show that N = cos θ satisfies the Hamiltonizationcondition, i.e. sets R, equation (8.10), to zero. Thus the deriva-tive with respect to τ of a variable q, which we will denote q′ isrelated to that with respect to t by

q = q′ cos θ, . (9.1)

Setting the potential equal to zero for convenience and usingthe constraint in the form x = y tan θ the reduced kinetic en-ergy (with the constraint substituted) and mass equal to unitybecomes

T =1

2(y2 sec2 θ + θ2) (9.2)

while T in the τ-time becomes

T =1

2(y′

2+ θ′

2cos2 θ) . (9.3)

The Lagrangian equations in the τ-time are thus

y′′ = 0 (9.4)

θ′′ = (θ′)2 tan θ . (9.5)

Now to see that these are the correct nonholonomic equationsreplace the τ derivatives by the derivatives with respect to t andwe obtain

y = −yθ tan θ (9.6)

θ = 0 . (9.7)

which are the nonholonomic equations which one can supple-ment with the constraint giving the dynamics in x. It is thenpossible to introduce any potential function which depends ony and θ.

In order to make the connection between the Chaplygin treat-ment and ours, let τ be the time parameter of our “Hamil-tonized” problem (using the optical analogy) and t that of orig-inal nonholonomic knife edge

ds

dτ≡ pHm

= an, (9.8)

with a a constant, and n the index of refraction of the opticalmechanical problem, and pH is the momentum of the Hamil-tonized, nonholonomic particle. For an index of refraction thatdepends on one coordinate n = n(y) we can relate n to θ usingSnel’s law n cos θ = const, where θ is the angle with respect tothe x axis:

ds

dτ=

A

cos θ, (9.9)

with A a constant. For the true dynamics, on the other hand,we have:

ds

dt=p

m. (9.10)

The time parameters are therefore related through

dτ = Cpdt cos θ, (9.11)

and the parametrizations coincide when p is constant (no po-tential).

Magnetic Analogy:Present here another mapping of the trajectories of constant

E to a holonomic system, and show that the trajectories areequivalent to those of a spatially dependent magnetic field withno external potential.

We can map the knife edge trajectories to those of a particlein a magnetic field as follows:

A particle of unit mass and unit charge, moving in a spatiallydependent magnetic field B(x) , and in the absence of a poten-tial, has a local radius of curvature ρ = v0/B, with v0 a constantof the motion. This means that the trajectory of a knife edgeof translational kinetic energy E and moving in a potential V (x)is equivalent to that of a particle of velocity v0 in a magneticfield given by

B(x) =v0mω

p(x)(9.12)

9.1 Example: Knife edge falling in an inclined plane

In this case V (x) = −α2y/2 and we treat the case of E = 0, forwhich the solution is known to be a cycloid. So we have

B(x) =v0mω

α√y≡ b√y

(9.13)

The equations of motion are

x = − b√yy (9.14)

y =b√yx (9.15)

From (9.14) we have

x + b√y = C = 0, (9.16)

where we chose the initial velocity in the x direction equalto zero. Substituting the above relation in (9.15) we obtainy = −b2, dy/dt = b

√ymax − y, with ymax = u20/2b2. With these we

obtaindy

dx= −

√ymax

y− 1, (9.17)

the equation of the cycloid or radius ymax/2. We stress that thisanalogy allows us to get the trajectories but not the dynamicsof the particle. The true velocity of a particle in a spatiallyvarying magnetic field is not necessarily related to the true dy-namics of the knife edge. But the trajectories are the same. Weremark that the “magnetic” terms here are different from thecoefficients Bb

αβ that arise in the general nonholonomic equa-tions (2.7).

Conclusion: We have shown that there is an extension of theclassical optical-mechanical analogy to the nonholonomic set-ting that enable one to write a Hamiltonian for the trajectoriesof a knife edge system on the plane subject to a potential. Fordifferent potentials one obtains various classic orbital dynamicmotions in the plane. The accompanying table summarizes thelink between the potential of the nonholonomic system, therelated index of refraction of the optical problem and the po-tential of the classical holonomic mechanical system that willyields the same orbit.

Table 1: Summary and examples of the non holonomic mechanical analogy for central potentials and for V (x, y) = V (y)Non-hol. potential Index of refraction Effective hol. potential orbit

V (r) n(r) = c(∫

rdr/V 1/2)−1

V (r) = −d(∫

rdr/V 1/2)−2

-kr2 c/r −d/r2 Logarithmic spiral−k/r2 c/r3 −d/r6 Lemniscate of Bernoulli

V (y) c(∫

dy/V 1/2)−1 −d

(∫dy/V 1/2

)−2

C c/y −d/y2 Circleky c/

√y −d/y Cycloid

Nonlinear Constraints: simple example:Consider an N dimensional vector V = (x1, · · · , xN) and an N

dimensional force F = (f1, · · · , fN). The constraint is imposedby a “time dependent viscosity” η(t).

For the velocity dependent constraint

G(v) = 0

v = F− η(t)∇Gand

v = F− ∇G · F(∇G)2

∇G

guarantees that the constraint is satisfied dG/dt = 0.Constant velocity constraint:

G = v2 ≡ v20, (9.18)

v = F− F · vv20

v

=F(v · v)− (F · v)v

v20

=v × (F× v)

v20. (9.19)

Using the constancy of the speed we have t = v/v0, and

v =dv

dsv0

=dt

dsv20, (9.20)

which, combined with (9.19) gives

dt

ds= t×

(F

v20× t

). (9.21)

Compare with

dt

ds= t×

(∇ ln(n)× t

). (9.22)

Given (9.19) and (9.22) we have the equivalence

F = −v20∇ ln(n), (9.23)

In other words, for the constant velocity constraint, the opti-cal mechanical analogy is expressed in the equation

U(x)

v20= lnn(x) . (9.24)

Example–constant gravity

F = gj

vy = g − gv2yv20

(9.25)

vx = −gvyvxv20

(9.26)

Since the speed is constant, we write

v = v0(sin θ, cos θ) (9.27)

and rewrite (9.26) as

vx = −g sin θ cos θ. (9.28)

Also,

vx = v0d sin θ

dy

dy

dt

= v20d sin θ

dycos θ,

which, combined with (9.28) gives

d sin θ

dy= − g

v20sin θ (9.29)

or

sin θ = Ce−αy,

with α = g/v20. Now, using Snell’s law

n(y) sin θ = Const

we get in fact that

n(y) ∝ eαy.

In general, using (9.29)

nd(1/n)

dy=

1

v20

dV

dy, (9.30)

or

−d ln(n)

dy=

1

v20

dV

dy, (9.31)

and

lnn(y) = −V (y)

v20+ Constant (9.32)