Manifolds and Mechanics (Australian Mathematical Society Lecture Series)

AUSTRALIAN MATHEMATICAL SOCIETY LECTURE SERIES

Editor-in-Chief: Dr. S.A. Morris, Department of Mathematics, La Trobe University, Bundoora, Victoria 3083, Australia Subject Editors: Professor C.J. Thompson, Department of Mathematics, University of Melbourne, Parkville, Victoria 3052, Australia Professor C.C. Heyde, Department of Statistics, University of Melbourne, Parkville, Victoria 3052, Australia Professor J.H. Loxton, Department of Pure Mathematics, University of New South Wales, Kensington, New South Wales 2033, Australia

1

2

Introduction to Linear and Convex Programming, N. CAMERON Manifolds and Mechanics, A. JONES, A. GRAY & R. HUTTON

Australian Mathematical Society Lecture Series. 2

Manifolds and Mechanics Arthur Jones and Alistair Gray Mathematics Department, La Trobe University

Robert Hutton Comalco Ltd

C:J

CAMBRIDGE UNNERSTTY PRESS Cambridge

New York New Rochelle

Melbourne Sydney

CAMBRIDGE UNIVERSITY PRESS Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, Sao Paulo, Delhi

Cambridge University Press The Edinburgh Building, Cambridge CB2 8RU, UK Published in the United States of America by Cambridge University Press, New York www.cambridge.org Information on this title: www.cambridge.org/9780521333757

© Cambridge University Press 1987

This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press. First published 1987 Reprinted 1988 Re-issued in this digitally printed version 2008

A catalogue record for this publication is available from the British Library ISBN 978-0-521-33375-7 hardback ISBN 978-0-521-33650-5 paperback

CONTENTS PROLOGUE 1.

CALCULUS PRELIMINARIES

1.1. 1.2. 1.3. 1.4. 1.5. 2.

Componentwise Calculus Variable-free Elementary Calculus

DIFFERENTIABLE MANIFOLDS

2.1. 2.2. 2. 3. 3.

FrSchet Derivatives The Tangent Functor Partial Differentiation

Charts and Atlases

Definition of a Differentiable Manifold Topologies

SUBMANIFOLDS

3.1. 3.2.

What is a Submanifold? The Implicit Function Theorem

3.3.' A Test of Submanifolds 3.4. Rotations in R 3 4.

DIFFERENTIABILITY

4.1. 4.2. 4. 3.

5.

Local Representatives Maps to or from the Reals Diffeomorphisms

TANGENT SPACES AND MAPS

5.1. Tangent Spaces 5.2. Tangent Maps 5.3. Tangent Spaces via Implicit Functions 6.

TANGENT BUNDLES AS MANIFOLDS

6.1. Charts for TM 6.2. Parallelizability 6.3. Tangent Maps and Smoothness 6.4. Double Tangents 7.

PARTIAL DERIVATIVES

7.1. 7.2. 7. 3.

7.4.

Curves in TQ

Traditional Notation Specialization to T Q Homogeneous Functions

8.

DERIVING LAGRANGE'S EQUATIONS

8.1. Lagrange's Equations for Free-Fall 8.2. Lagrange's Equations for a Single Particle 8.3. Lagrange's Equations for Several Particles 8.4. Motion of a Rigid Body 8.5. Conservation of Energy 9.

FORM OF LAGRANGE'S EQUATIONS 9. 1. Motion on a Paraboloid

9.2. 9. 3.

10.

Quadratic Forms Lagrange's Equations are Second-Order

12.

90 93 95

98 101 104

11.1. Globalizing Theory

110

11.2. Application to Lagrange's Equations 11.3. Back to Newton

113 117

FLOWS 120 125 135

THE SPHERICAL PENDULUM

13.1. Circular Orbits 13.2. Other Orbits, via Charts 14.

83 86 88

LAGRANGIAN VECTORFIELDS

12.1. Flows Generated by Vectorfields 12.2. Flows from Mechanics 12.3. Existence of Lagrangian Flows 13.

81

VECTORFIELDS

10.1. Basic Ideas 10.2. Maximal Integral Curves 10. 3. Second-Order Vectorfields 11,

78

138 141

RIGID BODIES

14.1. Motion of a Lamina 14.2. The Configuration Manifold of a Rigid Body

147 150

14.3. Orthogonality of Reaction Forces

154

REFERENCES

159

INDEX

161

SYMBOL TABLE

165

PROLOGUE

as an attempt to bridge the gap between

These notes began

undergraduate advanced calculus texts (such as Spivak(1965)) and graduate texts on differential topology.

One of the major applications of

differentiable manifolds is to the foundations of mechanics.

Here a huge

gap exists between the classical literature (see for example Goldstein (1980) and the more modern differentiable manifolds approach (see Abraham and Marsden(1978)).

Dieudonne(1972) says "The traditional domain of

differential geometry, namely the study of curves and surfaces in three-

dimensional space, was soon realized to be inadequate, particularly under the influence of mechanics".

Two of the authors of this work have been

involved in the writing of papers on problems in Lagrangian mechanics and they soon realized the need for a simple but modern treatment of the theoretical background to such problems.

Thus our aim is to make some of the basic ideas about manifolds readily available to applied mathematicians and theoretical physicists while at the same time exhibiting applications of an important area of modern mathematics to mathematicians.

Classical texts use vague ideas such as "virtual work" and "infinitesimal displacements" in their derivation of Lagrange's equations.

By contrast the modern texts jump straight to Hamiltonian systems and lose the physical motivation.

In these notes, the derivation of Lagrange's

equations makes direct appeal to geometrical principles.

As far as the

authors are aware, these ideas do not appear elsewhere.

In his original work, Mecanique analytique, Lagrange used no diagrams.

In fact the geometrical ideas necessary for a thorough under-

standing of classical mechanics did not exist at that time.

It was only

with the development of differential geometry and differential topology that geometry once again became an essential part of classical mechanics.

2

PROLOGUE

An advantage of starting with the Lagrangian formulation, as opposed to the Hamiltonian, is that it follows directly from Newton's Laws of Motion and that the mathematical background required is much less formidable. These notes were developed primarily with a view to providing the necessary theory for the study of particle motion as in the papers by Gray et al(1982) and Gray(1983).

As we demonstrate, however, our treatment

of Lagrangian mechanics is equally applicable to the motion of rigid bodies which consist of a finite collection of particles. Our approach can be extended so as to apply to the motion of a rigid body defined by a continuous mass distribution even in nonholonomic cases but this work is beyond the scope of the present volume and will appear separately.

We thank Lindley Scott for the diagrams of page 132, Sid Morris for his encouragement, the referees for their constructive reports and Professor Sneddon for reading and commenting on our preliminary notes. Our debt to Judy Storey is profound and we thank her accordingly.

Without her monumental effort this manuscript would never have

reached its final form.

Arthur Jones Alistair Gray Robert Hutton December, 1986.

CALCULUS PRELIMINARIES

1.

The differential calculus for functions which map one normed vector space into another nonmed vector space is the main prerequisite for the study of manifolds and Lagrangian mechanics in the modern idiom. The mild degree of abstraction involved helps focus attention on the simple geometric ideas underlying the basic concepts and results.

The

insights and techniques which this approach fosters turn out to be very worthwhile in the study of many topics in applied mathematics. The idea of a function as an entity in its own right, independently of any numerical variables which may be used to define it, is fundamental for this approach to calculus.

This idea, although usually

confined to statements of the theory, can easily be used to provide computational tools for particular examples.

This involves the

development of a "variable free" language of calculus in which the functions themselves, as distinct from the values that they take are highlighted.

This enables us to formulate and work with many important

and difficult mathematical ideas in a simple manner.

Some of the

notation does not appear elsewhere but a good account of the basic ideas may be found in Spivak(1965) or in Lang(1964).

1.1. FRECHET DERIVATIVES Here we outline how the idea of a derivative can be formulated for functions which map one normed vector space to another.

The basic idea

is this: a function is differentiable at a point if there is an affine map which approximates the function very closely.

The derivative of the

function at this point is then the linear part of this affine map.

1.1.1. Example.

Let f: R2 -- R2 be given by

f(x,y) = (x2 + y2 + 1, xy + 1)

CALCULUS PRELIMINARIES

1.

4

so that

f(1+h, 2+k) = f(1,2) + (2h+4k, 2h+k) + (h2+k2, hk).

As

(h,k)

approaches (0,0) the quadratic terms become negligible

compared with affine map

and

h

and so

k

is approximated very closely by the

f

with

A: R2 - R2

A(1+h, 2+k) = f(1,2) + (2h+4k, 2h+k). The linear part of

A

with

L: R2 - R2

is the linear map

L(h,k) = (2h+4k, 2h+k).

The precise meaning of "approximates

f

closely"

is clarified in the

following definition by the use of limits. 1.1.2.

spaces and

U

is an open subset of

differentiable at

A: E - F

f: U -+ F

Let

Definition.

where E.

E

and

F

By saying that

are normed vector f

is

a E U we mean that there is a continuous affine map

such that f(a) = A(a)

and

lim

Ilf(x)-A(x)II

x+a 1.1.3.

Definition.

For a map

Frechet derivative at Df(a)

:

E --> F

= 0.

Ox-all f

as in Definition 1.1.1 we define its

to be the unique continuous linear map

a

such that

lim IIf(a+h)-f(a)-Df(a)(h)II= 0. h -> 0

Ilhll

Now let us return to Example 1.1.2. and verify that in that case Df(1,2)(h,k) = (2h+4k, 2h+k).

Notice that the norm of a vector

(x,y) E R2

is given by II(x,y)II =

and so in this case we have Jim

Ilf(1+h,2+k) - f(1,2) - (2h+4k,2h+k)ll

(h,k)->(0,0)

II(h,y)II

= his

II(h2+k2, hk)II

(h,k)-'-(0,0) II(h,y)II

and since the terms in the numerator are quadratic it is easy to show that

FRECHET DERIVATIVES

1.1.

5

this limit is in fact zero.

Notice that the,Jacobian matrix for is just the matrix of the linear map

Df(1,2).

f

at the point

(1,2)

More about this later.

1.1.4. Theorem. (Chain Rule I). Let maps f: U - + F and g: F -+ G be differentiable at points E, F

a E U

and

f(a) E F

and G are nonmed vector spaces and

The composite map

g o f: U -i G

respectively where

is an open subset of E.

U

is then differentiable at

a

and

D(g o f)(a) = Dg(f(a)) o Df(a). A proof of this theorem may be found in Spivak(1965).

A variable-free notation for functions which map normed vector spaces to normed vector spaces (in most applications to mechanics these normed vector spaces are, or are closely related to, Rn) turns out to be very useful for our purposes. 1.1.5.

Definition.

The identity mapping from a normed vector space

to itself is denoted by

for each x E E.

idE(x) = x 1.1.6.

Definition.

E

and defined by

idE

The constant mapping from

vector spaces) which takes the value

c E F

E

to

F

is denoted by

(E, F normed c

and

defined by

c(x) = c

for each x E E.

Usually we will omit the subscript

"E"

from

idE

when the context is

clear.

EXERCISES I.I. 1.

Show that

2.

Show that i6 f .c,6 a eonti.nuoua Li.nean map between nonmed vectors

D idE(a) = idE

bon each

a E E.

4paeea then Df(a) = f. 3.

Suppa4e

o: Rn - Rn . 1)".

word "differentiable" is everywhere replaced by

,9nooth means

Coo.

EXERCISES 4.1. 1.

Let

L, M

and

N

be mant6otd6.

Show that ij

f: L -+ M and

g: M -+ N ate Cr (r > 1) then 40 .c6 gof : L -+ N. 2.

Let m be a .bet and Let N be a mantbotd. Suppoae, 6utthenmone, that there .i,6 a b.i j ectLv a map f : M -+ N. Show how to dej-Lne an ateab bon the 4et m which mahe.6 the map f .to be Cr.

3.

Let M and N

f : M - N .t.6 Cr then 6oh each open bet u c M and each open .6 et V f(U) in N, the ne6t'u.cted map flu : u -+ V i6 a26o Cr, whet the open bet6 are be man i,6otd6.

Show that .%i5

to be negv ded a6 aubmani6otd6. Prove a.t6o the conve.toe. 4.

Let

Nl

and

N2

natuAat projection

be man.L6otd6 and .let

(i = 1,2).

IIi

: Nl x N2 -+ Ni

Show that each

ni

.i,6

be the

Cr.

(The product o6 two man,i.6oLd6 wa6 de6.f.ned in exenai,6e 2.2.6.)

5.

Let M and N be ma,v otd6. Show that i6 f: M -+ N .id continuou.6 in the benbe ob Ve1c.n.itLon 4.1.2, then f .G6 continuoud a6 a map between the topotogLeat apace4 M and N with topotogie,6 given by Veb.ini tLon 2.3.1.

4.2. We may regard

Rn

41

MAPS BETWEEN REAL SPACES

4.2.

MAPS TO OR FROM REAL SPACES as a manifold with its usual differentiable structure

containing the identity map as a chart.

Thus if

M

is any manifold it f: M -> Rn

makes sense now to speak about the differentiability of a map or a map

g: Rn -+ M.

For such maps the definition given in the previous section may be reduced to a somewhat simpler criterion, as follows:

Theorem.

4.2.1.

at a

Let

f: M -} Rn

and let

if and only if for each open set

is an admissible chart

(U, +)

(i)

f(U) c V

(ii)

the map

V

a E M.

in

The map

Rn with

is

f

f(a) E V

Cr there

for M with a E U such that

foo-1

:

O(U) - Rn

is of class

Cr

at

4(a).

The proof of this result is left as one of the exercises,along with the formulation of an analogous result for the differentiability of a map

g

:

Rn - M.

See Figure 4.2.1.

Figure 4.2.1.

4. DIFFERENTIABILITY

42

EXERCISES 4.2. 1.

Let

and Y be open 6et6 in r and

X

x and

Rn

ne.6pectL.veLy, 6o that

can be n.eganded ab manL o1d6.

Y

that a map f

x-Y

a map between man. otd6 £b and oney i.b it .c6 dibbenenti,abte a.6 a map between open 4et6. Show

:

.e.6 d,L46eicentiab.2e as

2.

Prove theorem 4.2.1.

3.

fonmu2ate a theorem anatogou6 to the one in the. text bon a map

g

: x -> M where m .in6 a man,ibo!d and x 4A an open bet in

Rn.

Vhaw a diagxam and phove your .theorem.

4.

Let M be a man.L o!d. Show that a map f : M - Rn

(a)

fi: M -* R

.is and only t6 each o6 .ct6 component 6unction6

Let

(b)

(N

s

be a 4ubmanL6otd ob

a ma .Ljotd)

Show that f S

4.3.

:

.i,6

M and euppo6e

f

£6

Cr Cr.

: M -* N

Cr.

£4

S --+ N

£6 aC6o

Cr.

DIFFEOMORPHISMS

Now that we have a definition of differentiability of functions which map manifolds to manifolds we are in a position to extend the idea of a diffeomorphism to include such functions.

4.3.1. Definition.

A map

f: M --} N, where M and

N

are manifolds is

said to be a diffeomorphism if it is differentiable and has a differentiable inverse.

If

and its inverse are

f

Cr

Cr

it is said to be a

diffeomorphism. 4.3.2.

Definition.

to each other

Two manifolds M and

are said to be diffeomorphic

N

if there exists a diffeomorphism from M

to

N.

We shall regard two manifolds as being "essentiablly the same" if they are diffeomorphic to each other.

This idea is summed up by the following

lemma. 4.3.3.

Lemma.

N and

{(Ui,Oi)

Suppose the manifold M is diffeomorphic to the manifold I

i E I}

is an atlas for

M.

Let

f: M - N

diffeomorphism.

Then

{(f(Ui), 4iof-1)

I

i E I)

is an atlas for

N.

be a

4.3.

Proof.

DIFFEOMORPHISMS

43

Follows from the definition of a chart. The above and the following lemma will be useful when we deal

with the motion of a rigid body.

Lemma.

4.3.4.

Let

Suppose M

m 3 n).

Then

L

:

Rn -- Rm be a linear one-to-one map (where

R.

is a submanifold of

L(M)

is a submanifold of

Rm

which is diffeomorphic to

M.

Proof.

See exercises.

We now mention some celebrated results referred to by Spivak(1970) volume 1, chapter 2 and Stern(1983). There are, up to diffeomorphism, unique differentiable structures on each

Rn

if

n # 4.

differentiable structures on

R"

There are at least three "exotic" and it is speculated that there are

uncountably many.

There are, up to diffeomorphism, unique differentiable structures on

Sn

for

n 5 6.

But on

S7

there are 28 essentially

different differentiable structures, and for

S31

there are over 16

million.

EXERCISES 4.3. 1.

Show that i6 (U, 4) .L,a an adm.iae-ib1!e chant bon a mani6o!d m then 4 -ia a dibbeomon.ph.i,em 6xom u to O(U) a Rn. Compare Exexc.%a a 2.3.3.

2.

The two chants (R, id) and (R, id3) generate two dietLnct di66ehenti.abte atnuctunea bon R as these chants a/re not eompatib.Ce. Show that neventhe e.64 the two neeaCtLng man.i6otda axe di66eomohphi.e.

3.

Prove Lemma 4.3.4.

5. TANGENT SPACES AND MAPS

In the previous chapter the idea of differentiability was generalized so as to apply to maps between manifolds.

In this chapter the

idea of the derivative itself will be generalized to this context. In order to retain the idea of differentiation as a process of linearization it will be convenient to introduce, at each point of a manifold, a vector space known as the tangent space at that point.

These

vector spaces will supply the domains and codomains for the linearized maps. A neat formulation of the chain rule then becomes possible via a generalization of the tangent functor, from Section 1.2, to manifolds. Although all the concepts of this chapter can be formulated for arbitrary manifolds, our discussion will be restricted to submanifolds of Rn.

This will provide an adequate background for later chapters on

mechanics while avoiding unnecessary abstraction.

5.1.

TANGENT SPACES

In elementary geometry one studies tangents to circles and tangent planes to spheres.

The definition of a tangent space will generalize these ideas

to arbitrary curves and surfaces and their higher dimensional analogues.

Let M be a submanifold of we want the tangent space at

from a 5.1.1.

a

Rn

and let

Intuitively,

a E M.

to consist of all the arrows emanating

in directions which are "tangential" to

M

at

as in Figure

a,

The arrows may be regarded as elements of the vector space

introduced in Section 1.2. parametrized

TaRn,

To capture the idea of "tangency" we use

curves in the manifold.

5.1.1. Definition. A parametrized curve in M at a is a map y: I --r M with y(I)

y(O) = a, of the map

where y

said to parametrize

I

is an open interval containing

will be called a curve in this curve.

M

at

a

0.

and

The image y

will be

TANGENT SPACES

5.1.

The parametrized

class

Cr)

if the map

y

45

curve will be called differentiable (or of

is differentiable (or of class

Since M c Rn, the notation in Corollary 1.4.4 to

y

and can be interpreted as follows:

Cr).

is applicable

If

y(t)

is the position of a

t

then

y'(t)

particle moving along the manifold at time

is the velocity

Intuition requires this to be in a direction "tangential"

of the particle.

to the manifold and leads to the following definition,

illustrated in

Figure 5.1.2.

Figure 5.1.1.

5.1.2.

based at by

A tangent vector to

Definition.

of the form a.

A tangent space at

(a, y'(0))

where

y

M

at

a.

a

is an element of

is a smooth parametrized

TaRn

curve in M

For brevity, we shall sometimes denote this tangent vector

[Y]a

0 Y I

0

0 Figure 5.1.2.

A tangent vector.

46

5.1.3.

Definition.

tangent vectors to bundle of by

TANGENT SPACES AND MAPS

5.

TM

M

The tangent space to M at

M at a

a

and will be denoted by

is the set of all TaM.

The tangent

is the union of all its tangent spaces and will be denoted

so that

TM

=

TaM.

U

a E M The definition of vector space

TaRn

TaM

implies that it is a subset of the

from Section 1.2.

The following theorem confirms the

expectation that the tangent space should be "flat", rather than "curved" like a typical manifold. 5.1.4.

Theorem.

The tangent space

is a vector subspace of TaRn

TaM

of the same dimension as the manifold M. Proof.

The proof will use a chart

submanifold property for

M.

(U,

4))

for

Rn

at

a

which has the

Hence,in the notation of Section 3.1,

(Mnu) = Rk X {0} n gU) as illustrated in Figure 5.1.3.

Figure 5.1.3.

We may assume that

m(a) = 0.

A submanifold chart.

We shall check that

TaM

is closed under vector addition and

leave it as an exercise to prove that it is closed under scalar multiplication.

vectors in As

4)

TaM,

So let

with

(a, y'(0)) and y

and

6

(a, d'(0)) be an arbitrary pair of

parametrized

curves in M based at

is a submanifold chart it follows that both

a.

5.1.

TANGENT SPACES

4oy

and

0.6

(4.y)'

and

(4o6)'map into

map into

47

Rk x {0}

and hence also

From Corollary 1.4.4

Rk x {0}.

this implies that

D(¢o6)(0)(1) E Rk x {0}

and

and hence by the chain rule, Theorem 1.1.4, E Rk x {0}.

and

Hence by linearity of

D4,(a)

6'(0)) E Rk x {0}.

The vectors involved are shown in Figure 5.1.4.

Figure 5.1.4.

We may thus define a map

e

from an interval into M by

putting

c(t) = -1(tD4(a)(y'(0) + 6'(0)). Since TaM.

a

is a curve in M at

a,

the vector

(a, e'(0)) is an element of

By the chain rule, however, e'(0) a

6'(0))

= y'(0) + 6'(0) n and so, by the definition of vector addition in TaR,

48

5.

TANGENT SPACE AND MAPS

(a, Y'(0)) + (a, 6'(0)) Thus

TaM

(a, e'(0))

Tam

E

is closed under vector addition. The proof that

TaM

is closed under scalar multiplication is as is the statement that

similar and is left as an exercise, same dimension as

M.

TaM has the

I

EXERCISES 5.1. 1.

Let m be the unit aihaee {(xl,x2): x12+ x22 = 1} in

and .c,6 a panane4'',i.zed

.let a = (al,a2) E M. Show that .i6 y = (Y1.Y2)

R2

curve in m ba6ed at a then Y1(O) Y1'(0) + Y2(0) Y2'(0)

=0

.

Deduce that each vectoh..cn TaM .c,6 onthogona2 to the vector (a,a) T R2 and £Leu6tnate with a sketch. a

in 2.

Show that i4 M d,6 the unit 6pheAe in

Rn

veaton. in TaM £4 onthogona2 to the veaton.

3.

(a,a)

in TaRn

Let m be a 6ubmanL o.2d o6 Rn, let a E M and .let b E Rn Suppo6 a that the distance £nom b .to a point x E M hab a minimum at x = a. Prove that eveicy veaton in TaM .c,6 onthogona2 to the vector.

4.

and a E M then each

(a, a-b) in

and ittu.6thate with a 6ketch.

TaR'1

Complete the proob ob the pant o6 Theorem 5.1.4

TaM

which say-6 that

TaRn by proving the cto4une ob

.c,6 a vector 6ub-6pace of

TaM

under 6cat x mutti.pticatLon. 5.

Let m be a 6ubmanJ6o.2d, and

M n u 4 a 6ubmani6o.2d ob 6.

Rn

u

an open 6ub6et,

and that

TM n Tu

o6

=

Rn. Show that

T(M n u)

Let m be a 6ubmani6otd of Rn and .let a E M. Prove that .i.6 .i.6 a zubman.ibotd chant bon M at a a6 in the proo6 ob Theorem 5.1.4, then TaM

=

{a} x D¢ 1(0)(Rk x {0})

and (ii.) LHS c RHS. used in the proob ob Theorem 5.1.4 wilt come in handy. Do th,i.6 by showing

(.L)

RHS c LHS

Techniques

Use the n.e6u2,t ob Exexa%de 6 to 6how that the tangent space T am ha6 the dimension k (which .« also the dimension ob M), thereby completing the proos ob Theorem 5.1.4.

5.2.

5.2.

TANGENT MAPS

49

TANGENT MAPS

The idea of a "local linear approximation" to a map between manifolds can now be formulated in a precise way with the aid of tangent vectors. this end let

submanifolds of

Rm

and

Rn, and let

To

N are

f: M --> N be differentiable, where M and a E M.

Intuitively, a sufficiently small piece of a curve in a manifold will be indistinguishable from a line segment and hence will The map

correspond to a tangent vector to the manifold.

piece of curve through f(a)

in

N,

a

in M

f

sends a small

to another small piece of curve through

as illustrated on the left of Figure 5.2.1.

Figure 5.2.1.

The tangent map of

f

at

a.

By now imagining the pieces of curve to be so small that we can regard them as tangent vectors, we get the map, shown in the figure as

tangent vectors in

TaM

across to tangent vectors in

is, in a sense, an approximation to

f

near

a

Taf, which sends

Tf(a)M.

This map

and, hopefully, it will

be linear.

By introducing parametrizations 5.2.2

for these curves as in,Figure

and using their derivatives to define the tangent vectors, we are

led to propose the following definition. 5.2.1.

Definition.

The tangent map of

f

at a is the map

Taf : TaM -} Tf(a)N

50

TANGENT SPACES AND MAPS

5.

with

Taf(a,Y1(0)) = (f(a).(foY) 1(0)) Tf([yJa) = [f0Y]

or, equivalently,

for each differentiable parametrized

map of

f

curve

f (a)

y

M

in

at

a.

The tangent

is the map Tf: TM --* TN

given by

TfjTaM = Taf

for each

a E M.

Figure 5.2.2.

It is necessary to check that the proposed definition is unambiguous since two distinct curves

y

and

d

may have the same

We want to be sure that when this occurs the derivatives

derivative at

0.

of the curves

foy

and

fod

will also be equal at

0,

as illustrated in

Figure 5.2.3.

[Y]a

f

.i N

M Figure 5.2.3.

Tangency preserved.

5.2.

5.2.2. d

in

Proposition.

51

For all differentiable parametrized curves

y

and

at a

M

[Y]a = Ib]a Proof.

TANGENT MAPS

..

IfoY]f(a) = Ifod]f(a)

It is sufficient to show that

Y'(0) = 6'(0)

'+

(foy)'(0) = (foe)'(0).

The chain rule, Theorem 1.1.4, cannot be applied directly on the right hand side of this implication because the domain of

(U,

4))

for

Rn

need not be an open set

To overcome this difficulty, we introduce a

in a normed vector space. chart

f

at

a

with the submanifold property for

then write, at least on a neighbourhood of

M and

0 E R,

foy = (fob-1)o4oY Now by Corollary 1.4.4

and the chain rule

(foY)'(0) = D(foY)(0)(1) = D(fob-1)(4(a))oD4(Y(0))oDY(0)(1) =

This, together with a similar expression for

(fo6)'(0), establishes the

desired implication.

Inasmuch as the Freshet derivative of a map at any point of its domain is linear, the above expression for

Theorem.

shows it to be a

y'(0).

linear function of 5.2.3.

(foy)'(0)

The tangent map of

f

at

a

Taf : TaM -* Tf(a)N is linear.

With the aid of the tangent map, the chain rule for maps between manifolds may now be expressed in the following elegant way.

5.2.4. Theorem. The composite

Let g: L -* M and f: M --* N be L, M and N are submanifolds of Euclidean spaces.

(Chain Rule).

differentiable where

fog: L -* N

is then differentiable and T(fog) = TfoTg

If a E M

then we may also write Ta(fog) = Tg(a)foTag.

52

Proof.

TANGENT SPACES AND MAPS

5.

Proving the differentiability of the composite was set as Exercise

4.1.1.

To complete the proof of the theorem, let differentiable parametrized

curve in

L

at

a

differentiable parametrized

curve in M

at

g(a).

y

so that

be a goy

is a

From Definition 5.2.1

of the tangent map T(fog)([Y]a) _ [(fog)oY]f(g(a)) If°(g°Y)]f(g(a))

= Tf([g-ylg(a)) = Tf(Tg([Y]a)) = TfoTg([Y]a).

EXERCISES 5.2. 1.

Show that

2.

CLeanLy Son each = id

TidM = id

W)

TM

Cr (r > 1)

and

os the manL otd M

chant

¢-loo = idu.

Use the chain we. and question

I above to deduce that T$: Tu -> T($(U))

has both a Zest and a

night .Lnvex6e and that T(O 1)

(To) -l =

3.

where

TO(TU)

=

m and N be dubman olds o6 Rm and f is Rn neapectivety, such that f(M) C N. Pnove that A d i enenti.abte then so A A its neatnAatLon f I M : M - N and that Let f: Rm --* Rn and .Let

T(fTM) = TfITM.

5.3.

TANGENT SPACES VIA IMPLICIT FUNCTIONS

We now turn to submanifolds of

Rn+m which have the form

M = {x: f(x) = 0} where

f: Rn+M -- Rm

full rank m

for each

Df(a) has

is a differentiable function for which a E M.

way to find the tangent space:

For such submanifolds there is a very easy just linearize the function

the submanifold near the appropriate point. of the following theorem.

f

defining

This is the intuitive content

IMPLICIT FUNCTIONS

5.3.

Theorem.

5.3.1.

53

a E M

The tangent space to the above submanifold at

is given by

TaM = {(a,h): h E Rn+M with Df(a)(h) = 0}. We will deal only with the case in which the partial derivative

Proof.

a2f(a): Rm - Rm

has full rank m, leaving it as an exercise to derive

the result in the remaining case. Suppose first that curve

y

in

M based at

(a,h) E TaM

so that

h = y'(0)

From the definition of

a.

M,

for some

foy = 0

and

hence by the chain rule

Df (a) (y'(0)) = 0 Thus

.

(a,h) is a member of the set described in the theorem.

Conversely, suppose Df(a)(h1,h2) = 0 where

h = (hl,h2)

curve

y

in

M based at

(a,h) E TaM

that

with

h1 E Rn a

and

such that

(1)

If we now construct a

h2 E Rm.

h = y'(0), then it will follow

so the theorem will be proved.

To this end note that since

32f(a) has full rank, the implicit

function theorem, Theorem 3.2.1, shows that there is a differentiable function

g

such that

f(x) = 0 - x2 = g(x1) for all x = (x1,x2) Y = (Y1'Y2)

in a neighbourhood of

t

We define the curve

by putting Y1(t) = a1 + thl,

for

a.

(2)

sufficiently small, where

follows from (3) that

y(O) = a.

(3)

Y2(t) = g(Y1(t))

a = (al,a2).

Since

a2 = g(a1)

it

From (2) and (3) it follows that

f(Y1(t),YZ(t)) = f(Y1(t), g(Y1(t))) = 0

so by the chain rule

(4)

Df (a) (Y1'(0) , Y2`(0)) = 0. Since, however, hl = y1'(0)

that

32f(a) has full rank, it follows from (1), (4) h2 = Y2(0);

hence

h = y'(0)

as required.

and

54

S.

TANGENT SPACES AND MAPS

EXERCISES 5.3. 1.

We Theonem 5.3.1

to 4how that the tangent apace to the unit epheAe Sn-i = (x E Rn: x.x = 1)

at a point a E Sn 1

.ia

TaSn-' - {(a,h): h E Rn

2.

with

We Theonem 5.3.1 to 4how .that the tangent epace .to

in Exenciae 3.3.2, at the identity matrix I TI 0(n) 3.

0}

0(n),, dej.i.ned

.ce

- { (I,H) : H is a skew-symmetric matrix of order n}

Use. the .cdeaa contained in Exenc.iae 3.3.4 to extend the pnoob ob Theorem 3.3.1 .to the caee in which a2f(a) does not neceaea 2y have Lull rank.

6. TANGENT BUNDLES AS MANIFOLDS

In this chapter we show that the tangent bundle

manifold M is a manifold in its own right.

TM of a

Furthermore it has a

differentiable structure that is induced naturally by the differentiable

structure of M.

These ideas can then be extended to form

which

T(TM)

is the natural setting for second order differential equations which are fundamental in mechanics and other applications.

6.1.

CHARTS FOR TM.

Recall that the tangent bundle

TM

of a submanifold M

of

Rn

is

defined by TM =

u

Tam

a E M This is a subset of

TRn

and hence may be regarded as a subset of

For example, the tangent bundle the tangent spaces to the unit circle. of

R4,

TS1

RnXRn.

is the collection of all

Although it is strictly a subset

it is nevertheless helpful to vizualize it in the plane by means

of tangent lines attached to

Figure 6.1.1.

S1,

as in Figure 6.1.1.

A tangent bundle

TS1

56

TANGENT BUNDLES

6.

It is even easier to vizualize

TS1

if we give each tangent line a

rotation perpendicular to the plane of the circle to form a cylinder, as This latter interpretation is given formal

illustrated in Figure 6.1.1.

justification in Exercise 6.1.1.

lead, via differentiation, to submanifold

how submanifold charts for M charts for

6.1.1.

TM.

Let M be a

Lemma.

dimension

Meanwhile the following results show

If

k.

property for

M,

(U, 0)

C2'(r >. 2) submanifold of Rn of

is a chart for

(TU, TO)

then

Rn

with the submanifold

is a chart for

R2n

such that

TO(Tu n TM) = T(4)(U)) n (Rk x {p} x Rk x {o}).

Proof.

Let

(U,

5.2.2, the map

By Exercise

be as in the hypothesis of the lemma.

4))

TO: TU --* T4(TU)

is one-to-one and onto the set

T4)(TU) = TO(U)) _ 4)(U) x Rn,

which is open in

(TU, TO)

Hence

R2n.

is a chart for

T4)(TU n TM) = T4)(T(U n m)) = T(4)(U n m))

R2n.

Finally,

by Exercise 5.1.5. by Exercise 5.2.2.

= T(4)(U) n (Rk x {o}))

by Definition 3.1.1.

_ (4)(U) n (Rk x {o})) n (Rk x {o} x Rk x {o})

by Definition 1.2.3. = T4)(U) n (Rk x {o} x Rk x {o})

by Definition 1.2.3.

Thus all that remains to get a chart for to the right. P :

R2n --> R2n

TM

is to shift zeros

This can be achieved by applying the permutation

with P(x, w, y, z) _ (x, y, w, z)

for

x,y E Rk and w,z E Rn-k

6.1.2.

Corollary.

property for M

If

then

(U,4))

.

This gives the following corollaries.

is a chart for

(TU, Po(T4)))

submanifold property for

TM.

Rn

with the submanifold

is a chart for

R2n

with the

CHARTS FOR

6.1.

6.1.3. Corollary. If M is a

Cr

TM

R2n.

is a

Cr-l

submanifold of

57

TM

(r > 2) submanifold of Rn then The dimension of TM

is twice the

dimension of M. The following example provides an illustration of Corollary 6.1.2. 6.1.4.

Example.

A chart

(U,(8, r-1)) on

R2

is defined by

U = {(a,b) E R2: a\> 0) e(a,b) = arctan `ab)

(r-1) (a,b)

a2 + b2 - 1

This chart has the submanifold property for a2 + b2 = 1

so

u ft Si,

T(e, r-1) and show that, to within a

We now compute

Let

since, on

(8, r-1)(a,b) _ (arctan (a), 0).

permutation, this is a chart for TS1.

S1

.

(a,b) E U

and

with the submanifold property for

TR2

Then

(h,k) E R2. /b

ids

D8(a,b)(h,k) = D aretanlQ oD(ia)(a,b)(h,k) \

1

b

=

2d1

i/db12

a

+{ a)

+ i2)(h,k)

a

_ -bh + ak

a2 + b2 and

D(r-1)(a,b)(h,k) = D(id/o(id 12+id22)-1)(a,b)(h,k) = Did/(a2+b2)oD(id22+id22)(a,b)(h,k) = ah

a2

+ bk +

b2

Putting this together and identifying

TO, r-1)(a,b)(h,k) =

Restricting this to

with a 2 +b 2-1,

R4

we obtain -bh + ak

ah + bk

a2 + b2

a2 + b2

gives

TS1

T(8, r-1)I

b) a/

TR2

(a,b)(h,k)

TS'

(arctan (b) , 0, -bh + ak

a

a2 + b2

,

0 )

58

TANGENT BUNDLES

6.

ah + bk = 0

since by Exercise 5.1.1.

has the submanifold property for

Theorem.

6.1.5.

M of Rn Proof.

then

If

(U, +)

S1.

Thus (U x R2,T(e, r-1))

as required.

TS1

is a Cr (r > 2)

is a

(TU, Tm)

on

chart for

CI-1

chart for a submanifold TM.

Follows from Corollaries 6.1.2 and 6.1.3.

EXERCISES 6.1. 1.

Show that the map, Snom the tangent bundle o6 the unit c tae to the

cy .indek in

R3, o: TO -+ {(a,b,c) E R3: a2 + b2 = 1}

given by iD((a,b),(h,k)) _ (a,b, -bh + ak)

£6 a

dL

(Hint:

eomoxphi4m. neben to Exampee 6.1.4. and then conattuct char bon

Ts'

and the cytlndeh with nebpect to which the toca2 nepneaentative o 0 2.

£4 the identity mapping).

Let

U - R2\{(0,b)

and

: b > 0}

a

*N(a,b) =

7a (r-1)(a,b) =

- b a2 -+b2 - 1.

(See FLgune 6.1.2.)

N(a,b)

Figure 6.1.4..

Extended. stereographic chart.

(a)

Show that (U, (4PN, r-1)) pnopenty bon S1.

(b)

Use an argument ai ito

(TU, T(,PN, r-1)) bon

£.

59

PARALLELIZABILITY

6.2.

-Lo a chant bon.

R2

with the submani.botd

to that o6 example 6.1.4 to show that

a chant bon.

TR2

with the 4ubmanJ6o2d pn.opWy

Ts'. 6.2. PARALLELIZABILITY

In the previous section we showed that

observation that TO = R' x Rn. that in general 6.2.1. Rn.

and made the

It therefore seems natural to conjecture

This conjecture is false.

TM = M x Rk.

Definitition.

TS1 = S' x R

Let M be

Cr (r > 2)

k-dimensional submanifold of

If there exists a diffeomorphism 0 :

which takes each tangent space

then M

TM -->M x Rk

TaM

by a linear isomorphism to

{a} x Rk

is called parallelizable.

An important example of a manifold that is not parallelizable and which arises in the study of the spherical pendulum is 6.2.2.

Theorem.

S2n

is not paraZlelizable.

S2.

The proof of this theorem

involves the use of the "Hairy Ball Theorem" (see Hirsch(1976)) and for a sketch of the proof itself see Chillingworth(1976). We do, however, have "local" parallelizability as stated in the following lemma. 6.2.3.

Rn

Lemma.

and Zet

Proof.

Let M be a

Cr (r > 2)

k-dimensional submanifold of

(U, 0 be an admissible chart for

M.

Then

TU = U x Rk.

See exercises.

We quote a deep result, mentioned in Chillingworth(1976), whose proof lies outside the scope of this book. 6.2.4.

Theorem.

T Sn = Sn x Rn

only in the cases

n = 0, 1, 3, 7.

EXERCISES 6.2. 1.

Pn.ove Lemma 6.2.3. by quoting the appn.opncate n.e6u.-t bn.om Section 6.1.

60

2.

6.

TANGENT BUNDLES

Suppose M .c.a a Mvb.i.ua 4.tx p (See Figue 6.2.1). 1,6 M panaate izabLe? Gave geome ,i,c %eazonb bon you anbwe&.

Figure 6.2.1.

6.3.

A Mobius Strip

TANGENT MAPS AND SMOOTHNESS

Corollary 6.1.3. enables use to give a tangent bundle the structure of a manifold.

Put simply if

chart for

TM.

(U, )

is a chart for

M

then

(TU, T+)

There are clearly other admissible charts for

is a

TM which

are not obtained in this way but they are not of interest to us since they may not preserve the linearity of the tangent spaces and hence may not preserve the linearity of local representatives of tangent maps. 6.3.1.

submanifold M {(TUi, Tai):

be an atlas for the

Let

{(Ui,4i): i E I}

Rn.

Then the collection of charts for

Definition.

of

is called the natural atlas for

i E I}

TM

Cr (r > 2)

given by

TM.

It is now meaningful to discuss differentiability of maps which have tangent bundles as their domains.

An important example of such a

map is given by the following definition. 6.3.2.

Definition.

Let

M be a

with TM(a,h) = a for each

(a,h) E TaM.

submanifold of

Cr (r > 2)

natural projection on the tangent bundle of

M

is the map

Rn.

The

TM: TM -+ M

6.3.

SMOOTHNESS

61

is of class C.

6.3.3.

Theorem.

Proof.

We check the conditions in Definition 4.1.2.

let

be a

(U, 0)

chart for

TM

and

The natural projection

Cr

chart for

M.

TM

Let

By Theorem 6.1.5,

(a,h) E TM (TU, TO)

and

is a

TM(TU) c U.

The local representative of

TM

relative to these charts is

the map

0°TM°(To)

From Figure 6.3.1, T4),

which is validated by the definition of the tangent map

it is clear that this map just the natural projection

which is

II:

R2n _., Rn,

C°°.

TM

0

T4)

II

Figure 6.3.1.

Local representative of

TM.

Since formation of a tangent map is essentially a process of differentiation, it is not surprising that it may lead to a loss of one degree of differentiability. 6.3.3. Rn

Theorem.

respectively. Cr-1.

of class

Let M and N

be

If

is of class

f

: M -} N

Cr (r > 2)

Cr

submanifoZds of r and then. Tf: TM -+ TN

is

62

6.

Proof.

TANGENT BUNDLES

This again involves straightforward checking of the conditions in

Definition 4.1.2.

EXERCISES 6.3. Let f : M - N be as in Theohem 6.3.3. Show that the Local kepneeewtati.ve ob Tf &ati.6b.i.e6

1.

Tf To T*

(u, 0) .cs a cha. t bon f (u) c v. See f.Lguhe 6.3.2.

whene

=Tf M,

0

(v,

p)

.c.6 a chant bon N and

If

If #

Figure 6.3.2.

Local representative of

If.

We que6tLon I .to 6itt in the detait6 o6 the pnoo 6 o6 Theorem 6.3.3.

2.

6.4.

DOUBLE TANGENTS.

In Section 6.1 we showed that if

M was a manifold then

TM was also a

manifold with a differentiable structure which is induced naturally by that of

M.

T(TM).

It is possible to repeat this process and obtain an atlas for We will denote by

T2M

the manifold

T(TM).

DOUBLE TANGENTS

6.4.

6.4.1.

Lemma.

and suppose

k

is a chart for dimension Proof.

Let M be a

submanifold of Rn

is an admissible chart for

(U, 0)

T2M.

Cr(r > 3)

63

Furthermore

Then

M.

is a submanifold of

T2M

of dimension (T2U, T20) (Rn)"

of

4k.

Follows from Lemma 6.1.1, Corollary 6.1.2. and Corollary 6.1.3.

Various other results follow from those of previous sections for example: T2(fog) = T2foT2g

and

T2f

T2 .T2 =

when each of the above expressions is properly defined.

We will not fill

in the details but the following lemma gives a little insight into the structure of double tangent maps. 6.4.2.

Lemma.

Let

g: Rn - Rm

be twice differentiable.

Then

and

T2g: (Rn)" - (Rm)"

T2g(a,h,k,.e) = (g(a),Dg(a)(h),Dg(a)(k),D2g(a)(h,k) + Dg(a)(.f!)). Proof.

Follows from Definitions

1.1.3

and

1.2.4.

EXERCISES 6.4. 1.

Fitt in the detaitA o4 the pnoob o6 Lemma 6.4.2.

2.

Notice that (a) Suppoae

(b) Show that

TM: TM - M and

TM(a,h) = a.

Find

(a,h,k,L) E T2M.

Tha4

TTM : T2M -+ TM.

TTM(a,h,k,t).

TM.TTM = TM 0 TTM

(c) Figuxe 6.3.1. bhowd the £ocat nepxeaentatLve 06 natunat chant6.

TM with xe6peat to

Find the £ocat xepxeaentative6 ob

with xe6pect to eu i table natunat cha its. (d) Fox what bubbet o6

T2M

.c6

TTM = TTM ?

TTM

and

TTM

7.

PARTIAL DERIVATIVES

The tangent bundle of a differentiable manifold provides a natural setting for the study of particle motion.

Before deriving

Lagrange's equations of motion we establish some technical results on partial differentiation with respect to charts.

The resulting formulas

have a classical Zook but are here given a precise and useful meaning.

7.1. CURVES IN TQ To specify the state of a particle moving on a manifold

at any given

Q

time one specifies both the position and the velocity of the particle.

Hence it is natural to regard the particle as tracing out a curve in rather than in

Q.

TQ

The motion of the particle is thus described by a

parametrized curve

c where

I

is an open interval in

components as

(Cl, c2)

where

:

I ->TQ We may write

R.

cl

TQoc

c

is the projection of :

in terms of its c

onto

Q,

I -+ Q,

which traces out only the position of the particle, as in figure 7.1.1.

TQOC

I

Figure 7.1.1.

The position of a particle.

7.1.

PATHS IN TQ

65

Intuitively

I -+TQ

c : can be thought of as the parametrized attached "arrow" at each point of

curve TQ c

together with an

TQoc(I), as in Figure 7.1.2.

c

I

Figure 7.1.2.

In this interpretation particle on

Q)

The position and velocity.

c(t)

consists of

together with an "arrow"

TQoc(t)

(the position of the

(the velocity of the particle).

The arrow should point in the direction of motion of the particle, that TQoc

is, it should be tangential to the image of

at the point in

question and the length of the arrow should give the speed of the particle.

These requirements may be expressed in terms of the components of

by

c

the condition that

c2(t) = c1'(t) and hence that

c(t) = (cl(t),c1'(t)) This leads to the following definitions. 7.1.1.

Definition.

Let

Q

differentiable parametrized cl'(t) _

be a manifold and curve.

(cl(t), c1'(t)) (

7.1.2.

Definition.

self-consistent

if

c = (cl,c2): I - TQ

We define

= Tc1(t,l))

A parametrized (TQoc)' - c.

curve

c: I -+ TQ

is said to be

a

66

7.

7.1.3.

Example.

t E R.

Here

Let

c

PARTIAL DERIVATIVES

be given by

R --> TR

:

c(t) = (t2, 2t)

for each

(TR°c)(t) = t2

(TRoc)' (t) _ (t2, 2t) and hence (TRoc)' = c

so that the parametrized curve

c

is self-consistent. The effect of

c

c

1

R

TR

0 1

2

3

4

5

6

7

8

-1

Figure T.1.3. some elements of Identifying

R

with

TR

is indicated in Figure 7.1.3 R2

t

R

Figure 7.1.4.

(c.f. Figure 7.1.2).

gives the diagram shown in Figure 7.1.4.

c

on

TRADITIONAL NOTATION

7.2.

67

EXERCISES 7.1.

Show that c : R -+ TR given by c(t) = (sin(t), cos(t)) Ld a aekb-cona.i.atent path in TR and 4fzetch .cta tnajectoicy in TR in and 7.1.4. the flame ma.nnex as FLgwce4 7.1.3

1.

Show that

2.

c

:

R --> TR2

given by

c(t) = ((sin(t), cos(t)), (cos(t), -sin(t))

.c.6 a beC4-cona.i,atent path in

this path in 7.2.

TR2.

Sketch an "a44ow diagram" bon

R2.

TRADITIONAL NOTATION

The traditional notation for partial derivatives is often used in an ambiguous way.

We give a rigorous definition which is valid in the context

of partial differentiation with respect to the component functions of a chart.

To this end let f

:

Q -+ R

Q

be differentiable.

local representative

fo

of

be a manifold of dimension

k

and let

is a chart for

Q

then the

If f

(U, 0)

satisfies f =

as shown in Figure 7.2.1. into

R,

Now

f maps an open set in Euclidean space

hence the Definition 1.4.1

of

is applicable.

R

Rk

Figure 7.2.1.

Local representative of real-valued

f.

68

7.

7.2.1.

Let

Definition.

PARTIAL DERIVATIVES

be a chart for the manifold

(U, 0)

and, in

Q

terms of components, let 0 _ (+11 02,...,+k).

The partial derivative of

f

with respect to 8f

a4i

:

denoted by

4)i,

U --> R

is defined by putting of

where

is the local representative of

f4) = (f°4-1)

f.

Notice that in the above definition all of the components of the chart function

are involved.

4)

This is often emphasised in the literature by

writing af

as

\a /

A good account of the confusion which can arise when this point is overlooked is given in Munroe(1963) Chapter 5.

In the classical approach the called "real-variables". derivative of

f

"coordinate functions"

We can still think of

with respect to

4)i

aji

4i

were

as being the

keeping the other

?

4) s

fixed.

The

traditional rules for calculating with partial derivatives, some of which are contained in the exercises, are also valid in this newer context.

The

following example illustrates this.

Example. Refer to Exercise 2.1.5

7.2.2. R2

was defined with

{(a,b) E R2

:

where the chart (U,(r,6)) for

a > 0} for U.

It is easy to show that

0) (U) = R+ x (- 2' 2) (r,8)-1 = (idl.cosoid2, idl.sin°id2)

f: R2 -* R : (a,b) - a2 + b2 from Definition 7.2.1. we have

Now suppose

so on

U

.

Thus

f = id12 + id22 (=r2 )

= ((id12+id22)°(idl.cosoid2iidl.sinoid2))/lo(r,6)

TRADITIONAL NOTATION

7.2.

69

(id12)/Io(r,e) 21d1°(r,9) = 2r.

= (i$12)/2°(r,e)

and 2f

= 0.

It will be convenient to define the partial derivative of a function

g

which maps into

7.2.3.

Definition.

(U, 0)

and let

g

Let

be a

as a column vector.

Rn

be a manifold of dimension

Q

k

with chart

Cr (r 3 1) map

g: Q- Rn . We define the derivative of g with respect to

as the column vector

¢i

1(g'°4-1)/i°m

l(gn°4_1)/io4j

Notice that if

h: Rn - R

is differentiable then

thought of as the row vector of functions

h/:Rn --f R

(h1 ,h2',....hn').

can be

The

following lemma should be read with this in mind. 7.2.4.

Lemma. (Semi-classical chain rule). Suppose

(U,4))

g: Q --} Rn

and

is a chart for the manifold

hog 30i

=

n

Q

h/`7, g

h: Rn --> R

are differentiable.

If

then

agj

= h /° g

a

a0i

j=1

Proof.

ah'g

_ (h ° g

by Definition 7.2.1.

i

n

= F

(h/j° g ° 4-1(g

j=1 by the chain rule,

Theorem 1.4.5.

.2

70

PARTIAL DERIVATIVES

7.

A further generalization of Definition 7.2.1. which involves the partial derivative of a real valued function with respect to all of a chart's component functions is as follows: 7.2.5.

Suppose

Definition.

f: Q -* R

Q

For each chart

is differentiable.

the partial derivative of f

is a manifold of dimension (U, p)

with respect to

of

of _

3

aiyl

,...,

for

k

and

Q

we define

as the row vector

of ll

a classical looking chain rule.

7.2.6.

Theorem.

f: Q -+ R

manifold, and

Let

(Classical chain rule). be differentiable.

Q

If

be a k-dimensional (U, 4)

and

are

(V, iy)

charts for q with u fl v # , then of

=

of

aa*

ami

aoi Proof.

by Definition 7.2.1

a-Oi = (f° -

J J

j=1

k -

_

(f 4

1)/D°

4-0

by the chain rule for partial derivatives Theorem 1.4.5.

1)

j=1 k

of

by Definition 7.2.1

of

a

a*

aoi

The following definition generalizes Definitions 7.2.1

and

7.2.3.

7.2.7.

Definition.

Let

(U, 0)

for the k-dimensional manifold to

is the

c, denoted by a

a*i

and Q.

k x k

(V, ) where u fl V #

Then the derivative of matrix whose

i-jth

ip

be charts with respect

component is

7.2.

71

TRADITIONAL NOTATION

The above definition will be useful later as will be the following obvious Corollary to Theorem 7.2.6.

a of

7.2.8. Corollary.

The classical notion of a "differential" has a modern analogue as shown by the following theorem.

7.2.9.

If

Theorem.

(U, 4))

is a chart for

Rn and

f

: U -s R

is

differentiable, then

n

Df - ZIl a Proof.

Let

,

DQi

2

a E U.

Df(a) - D(fo4) 1em)(a) = D(fe0-1)(0(a))oD4)(a) =

n I

by Theorem 1.1.4. by Theorem 1.4.3.

i-1

n D4)i) (q)

by Definition 7.2.1.

- 2Il(a i

The following version of the chain rule expresses what is sometimes referred to in more traditional books as the rule for the "total derivative" of a function.

7.2.10 Theorem.

If y : I -+ Q

Let

(U, 4))

be a chart for a k-dimensional manifold

is a differentiable paranetrized curve and

f

: Q - R

differentiable then k

(f. Y) Proof.

(feY)' - (fe4'1 e

' jEl (a,F o Y) (4je Y) '

.

oY).

F (fe4-1/ae40Y

(4)-y)!

by Theorem 1.4.5.

j=1 _

k

j.1 O'j o Y)(4joy)'

by Definition 7.2.1.

is

Q.

72

PARTIAL DERIVATIVES

7.

EXERCISES 7.2. (u,4) wit denote a chant ban a mani.botd

In these exehciaea

and we £et

Q

Show that bon

1.

i,j E {1,2, ...,k}

4i

11

tib

i=j

0

ij

i#j.

4j

Let f:Q --} R and g:Q -- R be di. e e.nti.abte. Pnove each ob the bottowi..ng £ami.Ciaa/c £ookLng uL1ea:

2.

B(f+g) = of

+

Let (x,y)

of

g

(Absume the eonJeaponding hutea bon

3.

+

a(fg) = f

and

(f+g)/Z

(fg)/Z).

denote the identity chant bon R2, that is, .Let

(x,y)(a,b) _ (a,b)

Show that, bon

(a,b) E R2.

bon each

(U,(r,e))

ab in. Exampte 7.2.2,

ar

x

Dr

ax ae _ TX_

-Y

x2+y2

x+y2

De =

x x2+y2

ay

'

y

=

3y

(r, e) Hence white out the mat' x expneae.i.on bon 3a(x,y)

7.3 SPECIALIZATION TO TQ The notation introduced in the previous section is applicable to functions defined on any manifold hence, in particular, it is applicable to functions defined on the tangent bundle

TQ

of a submanifold

Q

of

Rn.

In this

more specialized context additional results on partial differentiation can be obtained.

Q

We assume throughout that

has dimension

k.

The dot notation is used ambiguously in the traditional books on mechanics.

Here we assign a rigorous meaning to one of its two possible

interpretations (and ignore the other in which it is regarded as differentiation with respect to time).

A key role will be played by the

idea of a self-consistent path, introduced in Section 7.1. 7.3.1. f

:

Definition.

TQ -- R

For a differentiable map

by putting, for

[y]a E TQ,

f

:

Q -a R

we define

7.3.

SPECIALIZATION TO

73

TQ

f(ma) _ vr2oTf([Y]a) = (foY)'(0) where

denotes projection onto the second factor.

'rte: R x R -- R

The maps involved in the definition of

are illustrated in

f

Figure 7.3.1.

TR

If

(f (a), (foy)'(0))

R

(feY)'(0)

Figure 7.3.1.

Now let component

i

maps

The

f map

(U,4) be a chart for a manifold U

into

R

Q.

Since each

we may apply the above definition to

obtain a map $i . TU -+ R

such that for

[Y]a E TU

$i([Y]a) - 0ioY)'(0) Hence, by Definition 5.2.1,

we may write

T _ (peTQ, But since, by Definition 6.3.1, bundle

TQ,

of the map

(TU, TO)

is a chart on the tangent

we may partially differentiate with respect to the components ((boTQ, ct).

To avoid repeated use of cumbersome symbolism, we

introduce the following definition. 7.3.2.

Definition.

(TU. 0OTQ, b)

Let

g: TQ - R

be differentiable and let

be a natural chart on ag

TFi

TQ.

: TU - R

For

1 < i < k,

we define

74

PARTIAL DERIVATIVES

7.

to be the map a (c oTQ ) is not applicable here because

Note that the earlier Definition 7.2.1 and

do not have a common domain.

7.3.3.

Theorem. For

as in Definition 7.3.1

f

ZFlla ll

Proof.

For

[y] a

and

as above,

4

.

TQ,si

2

g

E TQ,

f([Yla) = 7T2oTf([Yla)

by Definition 7.3.1.

_

= 7f20T(f4-1),T¢([y]a)

by Theorem 5.2.4.

=

by Definition 5.2.1.

=

k F (fob-1)' i=1 k

by Theorem 1.4.3.

(fob_

F i=1 =

F i=1

7.3.4.

Corollary.

7.3.5. Corollary. 7.3.6.

by Definition 7.2.1.

(a)oi([Yla)

a

i

(Cancellation of dots rule)

U-

3f

ai

4i

TQ c

a

o

= 1iY1

TQ,

i.

Theorem. (Interchange of dot and dash 1).

self-consistent path in

TQ

then, for

f: Q -+ R

foc = (foTQoc)' Proof.

f(c(t)) = 1T2oTf(c(t))

If

c

:

I -+ TQ

is a

differentiable,

.

by Definition 7.3.1.

= ir2oTf (TTQoc(t,1))

by Definition 7.1.1.

= T20T(foTQoc)(t,1)

by the chain rule

= (f0TQoc)'(t).

SPECIALIZATION TO

7.3.

7.3.7.

For

Corollary.

chart for

TQ

75

a

a self-consistent path,

c: I --> TQ

Q,

[not QOC

(fo TQoc) E a=1

Proof. 7.3.8.

See exercises.

ao c = for each self-consistent path Proof. 7.3.9.

c

:

° TQ o

)/

I -> TQ.

See exercises.

Theorem. Let

natural chart for

TQ.

Q

be a submanifold of

If

c

necessarily self-consistent)

:

and

f

:

TQ --

k

j=1

Rn

and let

be a

(TU,

is any differentiable curve (not

I -+ TU

(foc)' = F (a, o Proof.

cid

(Interchange of dot and dash 2).

Corollary.

R

F

is differentiable then

(af

j=1 4j

e

o

c)

This follows as a special case of Theorem 7.2.10, on use of

Definition 7.3.2.

The geometric interpretation of the next theorem is indicated in Figure 7.3.2. for the special case of a two-dimensional submanifold of R3.

The theorem has a traditional counterpart which is usually considered

to be geometrically obvious.

aq2

Figure 7.3.2.

a) aq

(a)

i

76

7.

Theorem.

7.3.10.

identity map on

If Q

Rn

and

PARTIAL DERIVATIVES

is a submanifold of (U, q)

of dimension

R'Z

Q

a chart for

at each

a E Q,

k, X

the

then

(a, q (a)) E TaQ for

1 < i ,

g: TQ -+ R

A map

Definition.

1 if, for all

is said to be homogeneous of degree

t E R

and

(a,u) E TQ

77

g(a, tv) = tr g(a,v). g: TR2 -} R

For example, the map

with

g(x,y, h,k) = cos(x)h2 + 2hk + sin(y)k2 is homogeneous of degree 2.

The following theorem extends the familiar result known as "Euler's relation for homogeneous functions" to the new context. 7.4.2.

degree

Theorem .

r and

(Euler's relation).

If

is homogeneous of

g: TQ --* R

is a natural chart for

(TU,

TQ

then

k

and define

E TQ

f: R -> TQ

by putting

f(t) = g(a,th).

c: R - TQ

Next define a curve

by putting

c(t) _ (a,th). Thus

f = goc

and so we may apply Theorem 7.3.9, to get

f'=E

j=1

But j ° TQ ° c

j-1 4j

(;j ° c)' - ;j ° (a,h).

hence

space TaQ,

j

is a constant function while 4j

`7

is linear on the tangent

Thus

k

f ' = E (L ° c); ° (a,h).

(1)

j=1 Now by homogeneity,

f(t) = trg(a,h)

so that

f'(1) = rg(a,h) -Evaluating (1) at

1

and comparing with (2) shows that

k E

ag

j=1 a¢j at each point

(a,h)

j=rg

in the common domain of these two functions.

(2)

S.

DERIVING LAGRANGE'S EQUATIONS

In this chapter we begin our account of classical mechanics. The physics background required is minimal.

We will assume that the reader

is familiar with the concepts of velocity, acceleration, mass and force at about the level of high school physics.

The results on partial derivatives established in the previous chapter will be used to derive Lagrange's equations for particle motion directly from Newton's laws.

our derivation follows the same general plan

as the traditional one, which may be found in books such as Goldstein(1980), Synge & Griffith(1959) and Whittaker(1952).

The use of manifolds,

however, enables us to give a more geometrical slant to this topic in that we use ideas such as 'orthogonality with respect to a tangent space' in place of 'infinitesimal displacements' and 'virtual work'. To give our calculations a conventional look we will use X = (x,y,z)

for the identity chart on

naturally induced chart on

R3

and

(X V TQ, k)

for the

TR3.

8.1. LAGRANGE'S EQUATIONS FOR FREE-FALL To provide some physical motivation we first derive Lagrange's equations for the very simple mechanical system consisting of a particle falling freely under gravity.

and the acceleration

Regard the force a

F

acting on a particle of mass m

which it produces as vectors.

Newton's laws then

imply that

F=ma provided measurements are made relative to an 'inertial frame of reference'. From the study of motion near the earth's surface it is often useful to suppose that axes fixed relative to the earth provide such a frame of reference; for the study of planetary motion, however, it would be more appropriate to fix the axes relative to the distant stars.

8.1.

FREE-FALL

79

Consider now a particle of mass m moving near the surface of the earth and choose a cartesian coordinate system fixed relative to the earth with the. x

and

y

axes horizontal and the

z

We

axis vertical.

will assume that the particle is acted on by a constant force of magnitude

mg vertically downwards.

See Figure 8.1.1.

z

Figure 8.1.1.

Motion under gravity.

It can be shown that the position and velocity of the particle at any instant determine its subsequent motion uniquely.

natural to represent its state at time

t

by an element

Hence it is c(t) E TR3

with

a self-consistent path, which is defined on some interval

c: I -- TR3

The cartesian coordinates of the particle at time (XOTR3oc)(t)

t

I.

are

and so Newton's laws of motion imply that m(xoTR3oc)"(t) - 0 m(yeTR3oC)"(t) - 0

(1)

m(zoTR3oc)"(t) = -mg.

Dynamically significant quantities associated with the motion of the particle are the kinetic and potential energies, given by the formulas

T - '/rn(xz + yz + z2) V - mgz . We regard

T

and

V

as maps from

TR3

and

R3,

respectively, into

and then define the Lagrangian function L : TR3 --f R

as

R

80

DERIVING LAGRANGE'S EQUATIONS

8.

L - T - V 0 TR3 where

TR3

:

.

is the natural projection.

TR3 --> R3

be written in terms of

L

The equations (1) may

as

aL,

c

ax 1 11

y

GcI

I- ^L 0 c- 0

1,

Ia

0

(2)

Oy

II

(

0

ax

,

cJ

-

aZ

0 c= 0.

The equations (2) are Lagrange's equations for the motion of the free-falling particle. equations of motion (1)

They have an advantage over the Newtonian

in that their form is preserved under change of

coordinates, as will become apparent in the next section.

The gravitational

force in the example above is the simplest example of a conservative field of force, as defined below. Definition.

8.1.1.

F(a)

at each

a E U,

V : U - R

If the force acting on a particle has the form

_13y

an open subset of

we say that the map

ay, Z)(a)) R3,

E

TaR3

for some differentiable function

F : U -> TR3

is a conservative field of

force.

This definition has an obvious generalization to

Rn.

Further

examples of conservative fields of force are given in the exercises.

EXERCISES 8.1. 1.

ue. Lagrange'a equati.on6 (2) Joh the Snee-baP..Ung pantLe2e 6hom the ix Newtonian £o'm (1). You w tt 6.th4t need to apply theonem De

7.3.6 to the Lebt hand a.idea o6 the equations (1). 2.

Show that the gnav.i tati.ona2 6ie d ob £once described in the text .L4 cons ehvativ e.

3.

Show that the £once JLetd defined on R1 by F(a) _ (a, -ka) L6 conservative. (This 1 Leed o4 bo)Lce an,%aes, bon k > o, when Hook e'.6 taw .L4 u6ed to modee the motion ob a pa tEcte attached to an a "tic 6pAing and Lead6 to the 6tudy ob a.impte ha'unonic motion.)

8. 1.

4.

SINGLE PARTICLES

R3\{0} by

Show that the bone b.LeLd deb.Lned on

F(a,b,c) = C(a,b,c), -

81

(a,b,c))

1

(a2+b2+e2)3

(The bone b.Le.ed cov eapond6 to the gnav.i tatona.2 att4act on pfioduced by a paxtLc e. ob unit maba at the oivLgin).

.w conse%vat-Lve.

8.2. LAGRANGE'S EQUATIONS FOR A SINGLE PARTICLE Lagrange's equations will now be derived for the motion of a particle of

mass m which is constrained to move on a submanifold dimension of

Q

being

where

k

to be the identity chart on

1 < k < 3.

Q

of

Again we take

R3,

the

X = (x,y,z)

R3.

It will be assumed, furthermore, that the particle is subject to a conservative field of force together with a reaction force which acts orthogonally to the manifold at each point - as would be the case if there were no friction between the particle and the manifold.

In mathematical

terms this means that the force acting on the particle at each point

a E Q

is a sum

F(a) + R([Yla) = (F°TQ + R)([Yla) E TR3 where

F(a) and

R([Y]a)

have the following forms:

F(a) = (a, for some function

V

:

U -+ R

where R([Y]a)

in the sense of Definition 8.2.1.

Figure 8.2.1.

(1)

(ax , ay , _)(a)) is open,

U

while the reaction force

is orthogonal to

below.

TaQ

(2)

82

8.

8.2.1.

TQ

DERIVING LAGRANGE'S EQUATIONS

Definition. A vector TRn

of

a E Q

at

is orthogonal to the submanifold

(a,v) E TRn

v.h = 0

if the dot product

for every vector

(a,h) E TaQ.

At time its state will be I

and

x,y

Xoc(t) respectively.

Hence Newton's laws imply that

m(XoTQoc)"(t) = 7r2o(FoTQ + R)

- 7r2o(FoTQ + R) where

TQoc(t) and

I -+ TQ, where

:

axes the position and velocity of the particle will be

z

and

XeTQoc(t)

c

Relative to the cartesian coordinate system defined by

is an interval.

the

the position of the particle will be

t

c(t) for some self-consistent path

o

o c(t)

(XOTQ,X)

o c(t)

denotes projection onto the second factor.

7r2: R3 x R3 --> R3

Hence

by Theorem 7.3.6. m(Xoc)'(t) = 7r2o(FoTQ + R) o c(t).

To convert (3)

(3)

to Lagrangian form we first introduce the

kinetic energy

T = I!Vn(k2 + y2 + Z2) =''SmX2 where

denotes

%2

After restricting

introduce the Lagrangian function

L

:

T

TQ -- R

to

as domain, we

TQ

by putting

L = T - VOTQ 8.2.2.

curve

Theorem. c

:

satisfies the equations aagZocl'

-

Proof.

each of

Under the above assumptions the

(Lagrange's equations).

I -+ TQ

for each chart

(4)

Iag2ocJ = 0

(i=1,...,k)

(TU, ((g1,...,gk) o TQ, (4k...... k))) for

Partial differentiation of the kinetic energy qi

and

qi

T

TQ.

with respect to

gives on use of Lemma 7.2.4, the semi-classical

chain rule, aqi

= mX , IX aqi

DT

= mX

aT

34i

aX aqi

(5)

= mX . LX_ o TQ aqi

(6)

SEVERAL PARTICLES

8.3.

83

where the last equation follows from Corollary 7.3.4, the cancellation of dots rule.

Differentiation of (6) along the curve 3T

0

la

C)I=

mi(X0c1.(ax

Ma gi

J

0 TQ o

Cl

+ (x0c)

(ax Ila

11

c

:

I --> TQ

o TQ

o

now gives

c)'}

gd.

gi

= mL

C )r.

F30.

TQ

0

o

CJ

oc taq. J J

+

L

11

by Corollary 7.3.8, the interchange of dot and dash rule. subtraction of (5) composed with

t faT

o

cl ' -

aq i

aT a

c

Hence

gives

o

qi

c

' (ax l

= m(% 0 c)'

3qi

= Tr2o(F oTQ+R)OC

0

C)

TQ 0

lax

o TQ o cl by Newton (3)

aqi

aX

= Tr2oIF

J

Jo TQ o c by (2) and Theorem 7.3.10.

ll

aqi

_ -(aV

LX-

,

by (1)

aq .)

flax

2

aV

o TQ o c

by Theorem 7.2.6.

aqi

Thus

(aT

0

cJ,

IaT

aqi

aqi

o cl = '

v o

TQ0c.

aqi

Lagrange's equations now follow from (4) since a(VOTQ)

a(voTQ)

av

o TQ and aqi

aqi

- 0.

aqi

8.3. LAGRANGE'S EQUATIONS FOR SEVERAL PARTICLES Lagrange's equations will now be set up for a system of of which moves in

R3

subject to certain constraints.

n

particles each

(One obvious

constraint, for example, is that two particles may not simultaneously occupy the same position.)

Here only those constraints will be considered which

place restrictions upon the positions which the particle may occupy - and not those which place constraints upon their velocities, for example.

84

8.

8.3.1.

DERIVING LAGRANGE'S EQUATIONS

Definition. The configuration set of the system of particles will

be defined as the set of all elements a = (a1,bl,cl........... an,bn,cn) E Rsn

such that for

i = 1,...,n

particle may occupy the position

ith

the

in conformity with the constraints.

(ai,bi,ci) E R3,

In this way, instead of considering we consider one particle moving in

n

particles moving in

R3

Only those systems will be

Ran.

considered in which the configuration set can be made into a submanifold Q

of

8.3.2.

Ran .

If the configuration set of a system of

Definition.

is a submanifold

of

Q

then

Ran

manifold and the tangent bundle

TQ

Q

n

particles

will be called the configuration

will be called the velocity phase

space of the system.

Let the identity chart of

Ran

be denoted by

X = (x1,Yl,z1,..... ,xn,yn,zn).

The history of the particles will then be described by a curve such that the position of the

ith

particle at time

is

t

We now generalize

and its velocity is

(xi,yi,zi)°rQ°c(t)

c: I -+ TQ

some of the ideas of Section 8.1. 8.3.3. R3

.The field of force for a system of

Definition.

is the map

F

:

F(a) _ (a,

U e R3n

where the

ith

7f2° F, (a, bl,cl)...... .2° Fn(an,bn,cn))

is open and

particle.

Fi

U --> TR3 is the field of force acting on

:

The field of force

there exists a potential function _(av

Definition.

Yl

Let m1,...,mn

F

is said to be conservative if such that

V : U --} R

av

av

l 8.3.4.

particles in

n

with

U -* R3n

aV

z'

av

T : TR3n --). R

Q e R3n.

given by

n T = / F mi(xi2 + yi2 + z.2). i=1

U

n

be the respective masses of the

particles of a system with configuration manifold energy of the system is the map

av Vi(a))

Yn

x72

71

n

The kinetic

SEVERAL PARTICLES

8.3.

85

The above formula for the kinetic energy can be written more succinctly as a matrix product

T=TMX where M

is the diagonal matrix given in terms of the masses by M = diag(m1,m1,m1,m2,m2,m2,..... ,mn,mn,mn)

and where

XT

is the transpose of the column matrix

X.

We are now able to define the Lagrangian function of the system of particles. Definition.

8.3.5.

and 8.3.3.

to

Restricting the maps

T

and

V

in Definitions 8.3.4.

TQ we define the Lagrangian function L : TQ -+ R

by

putting L = T - V.TQ. 8.3.6.

Theorem. Consider a system of n particles in

R3

which are

constrained to move in such a way that the equivalent single particle moves on a submanifold

Q of On of dimension

k.

Let the field of force

acting on this particle be the sum of a conservative field of force and a reaction which is orthogonal to

TQ

at each point.

Let

L : TQ --> R

be

For each chart

the Lagrangian function of the system.

(TU, ((gj,...,gk)eTQ, (4 ,...,qk)))

for

TQ,

each self-consistent curve

c: I -- TQ

describing the history

of the particle satisfies the equations

aL

aqi Proof.

o

c, -aL a c=0 (1 0,

can then be expressed by the condition that

f (a) - 0 where

f: R6 - R

with f(a) _ (al-a2)2 + (b1-b2)2 + (cl-c2)2 - d2

(1)

We now show that the configuration set is a submanifold of Let f

a E R6

be as above and let

h E R6.

R6.

By (1) the Frechet derivative of

is given by the dot product

Df(a)(h) - 2(al-a2, bl-b2, cl-c2, a2-al, b2-b1, c2-cl)..h Clearly

has full rank

Df(a)

configuration set

f-1(0)

1

(2)

and so Theorem 3.3.1, shows that the

is a eubmanifold of R6, of dimension

5.

The following proposition states a somewhat surprising fact, which is crucial for the valid application of Theorem 8.3.6. 8.4.6.

The reaction force due to the rigid rod connecting

Proposition.

two particles is orthogonal to the configuration manifold at each point. Proof.

Let

denote the configuration manifold

Q

by (1) above, and let

(a,h) E TaQ.

f-1(0), with

f

given

By Theorem 5.3.1,

Df(a)(h) - 0 which by (2), is equivalent to 2(a1-a2, bl-b2, cl-c2, a2-a1, b2-bl, c2-cl)'. h = 0.

(3)

Since the reaction forces on the two particles Act along the rigid rod and are equal in magnitude but opposite in direction, we may write their directions as X(al-a2, bl-b2, cl-c2) respectively.

and

-X(al-a2a bl-b2, cl-c2)

These two reaction forces are combined to give a single

reaction force in

TaR6

R = (a, A(al-a2, b1-b2, cl-c2, a2-al, b2-bl, c2-c1)) and so, by (3), .orthogonal to

R.(a,h) = 0.

Thus the reaction R at the point

a

is

TaQ.

If now we are given a potential function V pair of particles and a submanifold chart

:

R6 --r R

(U,(gl,g2,qa,qa,q5)),

for the for

Q

we may apply Theorem 8.3.6, to obtain Lagrange's equations for the system.

88

8.

DERIVING LAGRANGE'S EQUATIONS

EXERCISE 8.4. Find a eubmani1otd chaht.autitabee ,oh the con6 guuti.on mani.bo.Cd (Hint: b.ix one end ob the Aod and then bind a aubma.n%botd chant Jon S2 c R3.)

8.5. CONSERVATION OF ENERGY Physical systems are subject to the law of conservation of energy.

For

systems of the type discussed in Section 8.3, it will be shown from Newton's laws that the total mechanical energy - kinetic energy plus potential energy - remains constant with the lapse of time.

From a

physical viewpoint, these systems conserve their mechanical energy because they are free from the effects of such forces as friction and airresistance, which dissipate mechanical energy into other forms of energy such as heat.

Returning to the situation described in Section 8.3, we consider a single particle moving on a submanifold

Q

of

R3tt

subject to

a conservative field of force together with a reaction force orthogonal to TQ

at each point.

T : TQ - R

for, and

V : Q -+ R

be a potential function

be the kinetic energy of, the particle.

The total

for the system is defined by putting

E : TQ -+ R

energy function

we let

As before,

E - T+VoTQ . On the basis of Newton's laws, the following theorem can now be proved. 8.5.1.

Theorem.

(Conservation of energy).

If

c: I -+ TQ

is a self-

consistent curve describing the history of the above particle then

E o c

is a constant function. Proof.

The notation being as in) Section 8.3,

by Corollary 7.3.7.

(VeTQoc)' _ (ax o TQ o 11110

-

[Tr 2

-(7r2

0

F o TQ

cJJJl.(Xoc)

o (FoTQ +

where the last step is valid since, by Theorem 7.3.6, (X°TQoc)',

which lies in

at each point.

by Definition 8.3.3.

TQ

Note also that

(Xoc)

equals

and hence is orthogonal to the reaction

R

CONSERVATION OF ENERGY

8.5.

(Toc)' _ (X ° c) 'TM (X ° c) (712°(F°TQ + R)oc)'(Xoc)

Thus

in the interval

(E°c)' = 0

and

I

E°c

89

by Exercise

8.3.4.

by Exercise

8.3.2.

is a constant.

The above proof is a rigorous version of one of the two proofs In the alternative approach Whittaker

given in Whittaker(1952) section 41.

first of all proves a conservation law for a system of Lagrangian equations defined in terms of an arbitrary Lagrangian function specializes the result to the case where

L

L

and then

may be expressed in terms of

In the exercises, the reader is invited

kinetic and potential energies.

to follow through the details of the alternative approach by way of comparison.

EXERCISES 8.5.

In .theze exenci6e6 aubmani.6o!d

Q

o6

c

:

denoteb a chant bon, a k-dimen6.conae

R3n.

Let L : TQ -- R

1.

(U, q)

be any dLb6exentiabee 6unction and 6uppose that

I -- TU i6 a deC6-con6.i,6xent path. aatL66yi,ng Lagnange'a equation6

(.1_oc) aqi

° c0 or 1

i

k.

aqi

7.3.9. and 7.3.6, that the dehivative o6 the

Show, by u6ing Theorem

6unction

k

(E

\i=1 .%6 zero on

I

q.Z

\ - L I° c L aqi

and hence that this 6unction .id a constant.

Let T: TQ -+ R and V : Q --v R be di66enentiab.be 6uncti.onz and auppo.6e that T L6 homogeneous o6 degtee 2 and .Let

2.

L=T - V°TQ (a)

Show, 6)tom Theorem 7.4.2, that

k

E q.aL-L=T+VoT Z aqi

i=1 (b)

Hence prove Theorem 8.5.1.

Q

9. FORM OF LAGRANGE'S EQUATIONS

This chapter begins with a simple example of a mechanical system.

By explicit calculations, the Lagrange equations for this system

can be proved equivalent to a system of second-order differential equations.

For arbitrary mechanical systems of the type studied in Chapter 8, a similar fact can be proved by using standard results about quadratic forms.

In subsequent chapters the theory of differential equations will be

used to show how Lagrange's equations determine the motion of mechanical systems.

9.1. MOTION ON A PARABOLOID This section investigates the explicit form of Lagrange's equations for a

mechanical system consisting of a particle of mass m moving under gravity on a paraboloid in the absence of friction. assumed to be the subset

of

Q

R3

The paraboloid is

consisting of all the points on which

the equation

z - 'I(x2 + y2) holds, where

(x, y, z)

(1)

is the identity map on

R3.

Thus

Q

is a

paraboloid of circular cross-section with its axis vertical, as illustrated in Figure 9.1.1.

A chart for

R3

with the submanifold property for (R3,

Hence

Q

is a submanifold of

Q

is

(x, y, z--/(x2 + y2))). R3

and has

(Q, (x,y)) as a chart.

Incidentally, this chart forms a one-chart atlas for

Q

and so Lagrange's

equations with respect to this chart apply to the whole of the tangent bundle

TQ.

MOTION ON A PARABOLOID

9.1.

Figure 9.1.1. With

x,y

A paraboloid

and

Q.

restricted to

z

91

Q

it follows from (1) and

Theorem 7.3.3. that z = (x°TQ)x + (Y°TQ)Y x,y

Hence, on temporarily writing

in place of

.

x°TQ

and

y°TQ

for

simplicity, we find that the kinetic energy is given by

T =n(x2 + y2 + (xx + YY)2) Since the potential function, moreover, is simply

V - mg z

it follows

that the Lagrangian function is given by L = 'gn ((1+x2)x2 + 2xy xy + (1+,y2)'2) -/mg(x2+y2).

Now let

c: I -* TQ

history of the particle.

(2)

be a self-consistent curve describing the

We assume

c

is of class

C2

and introduce the

notation x = XOTQOC,

for the local representative to the chart

(Q, (x,y)).

Y = Y°TQ°c

(TQ°c)(X,Y)

of the curve

(3)

TQoc with respect

From the interchange of dot and dash rule,

Theorem 7.3.6, it follows that x' = x°c,

Y' = y°c.

An application of Theorem 8.2.2. shows that Lagrange's equations are satisfied by the curve

c

and from (2), (3) and (4) it

(4)

92

FORM OF LAGRANGE'S EQUATIONS

9.

follows that these equations can be written in terms of the local representatives as

(1 + x2)x" + xyy" + x(g + x'2 + yi2) = 0 (5)

+y2)y"+y(g+x12 +y'2) = 0

xY x"+

These equations can be solved to give the acceleration

x"

and

y"

explicitly:

X" _

x[gg + (X1)2 + (y1)2]/[1 + x2 + y2]

(6)

y" _ _y [g + (x,)2 + (y')2]/[1 + X2 + y2].

Thus Lagrange's equations lead to a pair of second-order differential equations for the local representative of the curve

c.

Conversely, the steps can be reversed to show that these equations imply Lagrange's equations for the system.

EXERCISES 9.1. Fox the pnobtem discu66ed in the text concenvvi,ng motion on the

1.

panaboloJd

(a)

Q:

Show that the kinetic energy i.6 given by T + /m(i y)A-TQ {YJ bon a 6uitabte matrix-valued

poA.wtwii.6e .inveue A-1 (b)

:

bunatiol

n A : Q -. R212

with

Q -. R2> 2.

Show that Lagrang e' 6 equation (5) bon .the panabolo-14 can be wn c tten ab: (A'TQ'c)

`y+ (g + x12 +

y12)

(Y] = (0)

and .then deduce the di65enentia2 equation (6) with .the aid o6 A-'. 2.

Show that the 6y6.tem o6 6econd-ondeA di66enentiae equation (6), deachib-i.ng motion on the paicatboto.id, admit6 the 6amity o6 peniodi.c

6otuti.on given by x(t) = a cos(V t),

son each a E R.

y(t) = a sin(V t)

Note that the period .c.s contant a.Cong the 6ami2y.

Use the heEation (3) in the text to heCp you z ketch the set o6 po.int6

9.2.

QUADRATIC FORMS

93

{TQoc(t): t E R} c Q

which .c.6 txaeed out by the pant cte on 3.

Q.

Con6idex a pa&tic a o4 mabe m moving under gxav.c ty on the unit aphexe S2 a R3 in the abaence o4 4/LctLon. Let (u, (e,q)) be the aphen.icaP-pots chant 6o,% S' de6ined in Exexei.6e 2.1.6. (The .6phehi.caC pendulum).

(a) Show that the kinetic and po-tenti,a2 enexgie6 may be expneweed in thiA chant a4 T = /m(62 + sin2oO $2) V = mg coso6. (b) Veti.4y that, with

6 = eQTQoc

and

OOTQoc, Lagrange'4 equations

are expei.city g sin-U) = 0

m(U"-(sinoZ)(cose (m (sin2oU)T')' = 0.

(c) Deduce that Lagrange'a equati,on4 ate equivalent to a ayatem o4

aeeond-oxdex di64e e.ntial equati,on6 in

and .

U

(d) Show that white the pahtLc.Le ataya within the chant domain (sin2oU)T' temainb con6ta.nt. 9.2.

U,

QUADRATIC FORMS

The discussion of Lagrange's equations in the next section will use the basic ideas about quadratic forms which are summarized below.

A fuller

account of this topic is contained in standard linear algebra texts such as Nering(1964).

Our definition of the matrix of a quadratic form differs

from the usual one, however, and is more natural in the context of manifolds.

A quadratic form on a vector space g: V -+ R

V

over

R

is a mapping

whose values are given by g(v) = f(v,v)

for some bilinear map

f: VxV --a R.

If

g

is a quadratic form then there

is only one such bilinear form which is symmetric and we call it the bilinear form generating

g.

A quadratic form g(v) > 0

for all

v 0 0.

g

is said to be positive definite if

It is clear that the restriction of a positive

94

9.

FORM OF LAGRANGE'S EQUATIONS

definite quadratic form to a vector subspace of its domain is also a positive definite quadratic form.

We claim that the map

Example.

9.2.1.

g: R2 -. R

with values given by

g(a,b) - a2 + 6ab + 10b2 is a quadratic form and is positive definite.

To see this it is helpful

to use matrices and write

g(a,b) - (a b) 13

10, (b)

f((a,b), (c,d)) _ (a b)

Hence

g

is a quadratic form.

(1

rcl.

3

3

10)

d

Completing the square shows that

g(a,b) - (a + 3b)2 + b2 > 0 and this vanishes only if

Hence

(a,b) - (0,0).

is positive

g

definite.

..The representation of quadratic forms by matrices will play an

important role in the next section.

The relevant definitions and theorems

will be stated first for the special case of quadratic forms on usual basis for 9.2.2.

Rn

Definition.

will be denoted by

Rn.

The

(el,...,en).

The matrix of a quadratic form

g

on

Rn

has as its

i-jth element

.f(ei,ej) where

f

9.2.3.

for each

is the bilinear function generating

Theorem.

If g

g.

is a quadratic form on

Rn with matrix A

then

v E Rn, regarded as a column matrix, g(v) = vT Av.

Conversely if this relation holds for all

and

A.

v

then

g

is a quadratic form

if symmetric, is the matrix of g. More generally, we now consider quadratic forms on an arbitrary

vector space isomorphism

V

of dimension n

c: V -. Rn,

over

R.

There is then a vector-space

which we may regard as a chart for

V.

LAGRANGE'S EQUATIONS SECOND-ORDER

9.3.

9.2.4.

Theorem.

If

local representative on

Rn.

is a quadratic form on V

g: V -+ R gq,

95

(shown in Figure

If A denotes the matrix of g,

9.2.1.)

then its

is a quadratic form

then the values of g

are given

by g(v) - 0(v)T AO (V)

where we regard

0(v) as a column matrix.

are given by this formula then

g

Conversely if the values of

is a quadratic form on

g

V

g

V.

R

Rn

Figure 9.2.1.

Local representative

of a quadratic form.

The key theorem for our purposes is as follows: 9.2.5.

Theorem.

If a quadratic form

g: V -+ R

is positive definite

then the matrix of each local representative is non-singular.

9.3.

LAGRANGE'S EQUATIONS ARE SECOND-ORDER.

It will now be shown how Lagrange's equations for a system of the type discussed in Section 8.3 can be written as a system of second-order differential equations.

The kinetic energy of the system of particles was

there defined in terms of the identity chart

(XeT, X)

on

TR3n

as

T 1331 XTM X where M - diag(ml, m1, ml, m2, m2, m2,.... ,mn,mn,mn) with each mass mi > 0.

Thus for each a E Ran

positive-definite quadratic form.

the restriction of

T

to

TaR3n

is a

96

FORM OF LAGRANGE'S EQUATIONS

9.

Now suppose that the particle representing the system is constrained to move on a smooth For each

a E Q

k-dimensional submanifold

the restriction of

to

T

TaQ

of

Q

Ran.

is the restriction of

a positive-definite quadratic form to a vector subspace of its domain, hence it is again a positive-definite quadratic form. TQ so that

is a chart for

Suppose furthermore,

is a linear chart on

that

(goTQ, q)

TaQ.

It then follows from Theorem 9.2.4. that the kinetic energy for the

particle moving on

Q

4ITaQ

is given by

T = /qT AOTQ q

A:Q - Rkxk.

for some matrix-valued function

(1)

A simple example

illustrating this is contained in Exercise 9.1.1.

To work out Lagrange's equations explicitly, we will need the

smoothness of the matrix valued function A occurring in (1).

Lemma.

9.3.1.

in

The i-jth entry

of the matrix-valued function A

aij

above is given by

(1)

i

-1 o TQ

=

a2T

ag aq Z

for each Proof.

a

i,j = 1,2,...,k.

Notice that equation (1), written out in full, is

k k

F Z afoTQ

T

R=lm=1

gQ

c1

.

Differentiating this expression gives the required result. 9.3.2.

Corollary.

The matrix-valued function A

faT o c aq

is smooth and

= AoTQoc (goTQoc)" + (AoTQoc)'(goTQoc)'

for each smooth self-consistent path

c: I -+ TQ.

We can now prove the main result of this chapter. 9.3.3.

Theorem.

Lagrange's equations are a system of second-order

differential equations in the local representative Proof.

We may suppose the Lagrangian function

q =

q o TQ o c

L : TQ -+ R

is given by

L - '4TA°TQ q - VOTQ where A

is as in Lemma 9.3.1.

a chart (qoTQ, q)

become

By Theorem 8.3.6. Lagrange's equations in

9.3.

LAGRANGE'S EQUATIONS SECOND-ORDER

aaq (' q A°TQ g)°c I, where

c: I -+ TQ

-

aQ

97

°c-0

is a smooth self-consistent path.

Application of Corollary 9.3.2. then gives (AoTQoc) (goTQoc)" + (AoTQ°c)' (goTQoc)' - aL ° c = 0.

By Theorem 9.2.5. the matrix-valued function

A°TQoc

is

pointwise invertible so the above equation becomes (goTQoc)n = (AoTQ0c)-1 (-(A°TQoc)'(goTQ°c)' + aLoc) as required.

EXERCISES 9.3. 1,

FLU in the detaiA o6 the pnao6 o6 Lemma 9.3.1.

2.

Compare the content6 o6 Section 9.3

Whi ta.kei (1952) .

with. SecUon4 27 and 28 06

i0. VECTORFIELDS

In modern terminology, the right-hand side of a differential equation determines a "vectorfield", which means assigning an "arrow" to each point of the domain, while the solutions of the differential equation are referred to as "integral curves", which are tangential to the arrows.

In the context of manifolds, the integral curves are regarded as maps into the manifold while the "arrows" are elements of its tangent spaces.

The

modern terminology thus invites us to think of differential equations in a very geometrical way.

In this chapter, enough of the theory of vector-

fields will be developed to enable Lagrange's and Newton's equations to be put into proper geometric and analytic perspective.

Among the most basic results in the theory of differential equations are those concerning the existence and uniqueness of solutions and these results translate easily into corresponding results about integral curves.

When studying vectorfields and their integral curves

on a manifold, however, it is important to distinguish between "local" and

"global" properties;

the former refer to what happens in a particular

chart domain whereas the latter refer to what happens on the whole manifold.

Finally, some special properties of vectorfields which arise from second-order differential equations, such as Lagrange's equations, will be studied.

10.1. BASIC IDEAS In this section we introduce the ideas of vectorfields and integral curves and give a vectorfield version of the classical existence and uniqueness theorem for the solutions of differential equations.

A vectorfield may be thought of as a mapping from a manifold

M

to its tangent space

definition is as follows.

TM which preserves the base points.

The formal

10.1.1.

Definition.

99

BASIC IDEAS

10.

Let M be a smooth manifold.

A mapping

Y : M-+ TM with the property

TM a Y - idM

is called a vectorfield on

M

TM

Figure 10.1.1. 10.1.2.

Example.

M.

A vectorfield on

M.

The mapping Y

:

R2 -+ TR2

given by

Y(x,y) _ ((x,y), (-y,x)) is a vectorfield on

R2

since

various arrows to points in

R2

TR2 o Y = idR2.

The assignment of the

is illustrated in Figure 10.1.2.

TR2

I

Figure 10.1.2.

100

VECTORFIELDS

10.

Before reading on, recall (see Definition 7.1.1.) that if

c: I -+ M

is

an interval, M a manifold), then we define

differentiable (I

c'(t) = Tc(t,1) _ (c(t), c'(t)).

The manifold version of a solution to a differential equation is given by the following definition.

Let M be a smooth manifold and suppose Y : M -+ TM

Definition.

10.1.3.

An integral curve of Y

is a vectorfield on

M.

c: I -+ M where

is an interval containing

I

at 0

a E M

is a mapping

such that

c(0) = a

and

c'(t) = Y(c(t)) t E I.

for each

The relationship between a differential equation on

Rk

written as c'(t) = f(c(t))

for some function

f

Rk

:

- Rk

(1)

and its vectorfield counterpart is then

given by including the "base point" as follows: c'(t) = (c(t), c'(t)) _ (c(t), f(c(t))).

Thus the vectorfield for the differential equation (1) is then given by Y : Rk

-+ TRk where Y(x) _ (x, f()) for each x E Rk.

10.1.4.

Example.

Y(x) = (x,3x) Thus

Let

be the vectorfield defined by

Y : R -+ TR

for each x E R. c

:

I -+ R

is an integral curve at

a E R

for

Y

if

and only if c(0) = a

and c'(t) _ (c(t), c'(t)) _ (c(t), 3c(t)).

By elementary differential equation techniques this means that

c: I-)- R

with c(t) = ae3t

Notice that the maximal choice for

for each I

is

t E I.

R.

The key existence-uniqueness theorem which we state below in the language of vectorfields on

Rk will, under certain smoothness assumptions,

be generalized in the next section to arbitrary smooth manifolds.

10.2.

10.1.5. F

:

101

Let U be an open subset of R k and let

Theorem.

U -+ T U

MAXIMAL INTEGRAL CURVES

be a vectorfield on

such that for each

Suppose that there exists a

U.

K > 0

y,z E U

IIg20F(y)-71 2 F(z)II < K By - zll. Then for each

there is an

a E U,

c(0) = a and

c: (-e,E) -+ U

c'(t) = F(c(t)).

See Chillingworth(1976) or any standard analysis text.

Lemma.

10.1.6.

by "F

and a unique

e > 0

t E (-E,E)

such that for each

Proof.

(2)

is

C'

The condition (2) in Theorem 10.1.5. can be replaced on

U".

EXERCISES 10.1. 1.

Show that the veetonbietd Y : R -+ TR given by Y(x) = (x,x2 ) has a unique .integ at cwcve at 1 which .i,6 deb.ined on a maxima-2 open Show neventheteaa that the 4o1 tLon .ia not de4-ined on

.intenvaC.

the whole ob 2.

R.

Show that the veeto,%jieCd

W ass .integiut c: R -

R+

cwtvee

Y :

at

given by

0

R -+ TR

((1+/t)2

,

Y(x) _ (x, x/)

given by

the zexo bunction

0

and the 6uncti.on

t < -2

c(t) = Exp4ain why thin doea not contnadi.et Theonem 10.1.5.

3.

Find the .integiut curve bon the vectong.iekd og Exampke 10.1.2. at the point (a,b) of R2. What i-6 the £angeet open £ntenvat in R on which it may be dejined? 10.2.

MAXIMAL INTEGRAL CURVES

In Section 10.1. we showed that a smooth vectorfield on an open subset of Rk

gave rise to integral curves which were defined on some interval

I.

In this section we show how that theory may be applied to obtain integral curves on the manifold itself.

Finally we show how integral curves may be

patched together in an unambiguous way.

The idea is as follows:

102

10.

VECTORFIELDS

take local representatives of the vectorfield on the manifold obtain integral curves in

on

Rk of the resulting vectorfields

Rk

j

lift these integral curves up to the manifold via the inverses of the chart mappings

4

show that these integral curves may be patched together to give unique maximal integral curves on the manifold.

We begin this process with the manifold version of the existence-uniqueness Theorem 10.1.5.

The relationship between a vectorfield and its local

representative is illustrated in Figure 10.2.1.

Y

TO

TOYo¢-1

Figure 10.2.1. 10.2.1.

Theorem.

and suppose

Local representative Y of a vectorfield Y

Let H be a smooth k-dimensional submanifold of Rn

Y : M -+ TM

is a smooth vectorfield on

M.

Then for each curve

c

Proof.

:

a E M

e > 0

there is an

(-e,e) -+ M of Y at

Suppose

103

MAXIMAL INTEGRAL CURVES

10.2.

and a unique integral

a.

is a chart for M with a E U.

(U,4))

10.1.5. and Lemma 10.1.6. there is an integral curve

Then by Theorem

l;

4)(U)

such that

E'(t) = To o Y e with

(a).

1;(0)

-lob c =

Setting

gives the existence of a suitable integral curve on M.

For uniqueness see the exercises.

We now show how the domain of each integral curve may be extended to yield "maximal" integral curves.

10.2.2. Lemma.

Let Y : M -r TM be a smooth vectorfield on a smooth

k-dimensional submanifold M of Rn

IX --> MIX E J}

:

{cX

and suppose

is the set of all integral curves of Y t E Ix n Iu,

and each

X,u E J

The fact that M

Proof.

c(t) = cX(t) for some 10.2.4.

Definition.

Corollary 10.2.3.

curve for

1.

Y

at

I --} M where

:

Y, M, I

The map a E M.

Rn

and hence is Hausdorff

U IX defined by XEJ is an integral curve for Y at a E M.

c

A E J Let

Then for each pair

See exercises for the details.

The map

Corollary.

a E M.

is a submanifold of

is the key to the proof. 10.2.3.

at

cX(t) = cu(t).

c

:

and

I -- M

c

I =

be as in Lemma 10.2.2. and

is called the maximal integral

U

EXERCISES 10.2. Let M be a amooth submani.botd ob Rn, Y : M -} TM a 4mooth vectonbieed and c : I - M a 6moo.th panamethized curve. Show that

.(,6 an .integut eunve ob Y .i6 and only £b, ban each cheAt bon M, c4) Aa an .i,ntegna2 eu ve ob Y4).

c

2.

(u,4))

Abaume the hypotheb e6 ob Theorem 10.2.1. and 6uppo4e that bon each

I with 0 E i there ane distinct ..ntegnat cunvu I -} M and d : I - M bon Y at a. Show that thi4

Ln.tenua.L

c

:

contnud ct6 Theorem 10.1.5.

104

3.

VECTORFIELDS

10.

Here we prove Lemma 10.2.2.

such that

t

Ads wne the hypoihebes ob the lemma.

be the subset ob the £ntexvaI is n iu eon6J 2Lng ob all

Let I

cX(t) = cu(t).

The aim .t,b to prove

I = IX n Iu.

Thiz £4 done by showing: (a)

I # o

(b)

I

(Easy! Try t = o.)

£4 open in ix n I.

F.ucat suppose s E I and set Now cona.Ldex an £ntegML curve

b = cA(s) = cu(s). d

:

(-e,e) -+ M

at b where d(t) = cX(s+t) bon each

Y

ob

Use Theorem 10.2.1

t E (-e,e).

we obtain

t E (-c,c)

to .chow that bon each

ca(s+t) = cu(s+t) = d(t),

which gives

the requviced re6uP.t.

(c)

I d,6 ceosed in IX n IV.

(Use the baat that m tia Hauodorbb

and the continuity o6

and cu).

cX

is n iu = I.

(d)

10.3.

SECOND-ORDER VECTORFIELDS.

This section is about the special properties of the vectorfields which arise from the study of second-order differential equations, such as Lagrange's equations.

In constructing a vectorfield from a second-order

differential equation, we first apply the usual "reduction of order" procedure.

The ideas of Section 10.1. can then be applied to the resulting

pair of first-order equations.

10.3.1. Example.

Given

f

:

Rk x Rk -+ Rk, consider the second-order

differential equation E" = fo(E.E')

for the unknown function

E

:

I -+ Rk.

To reduce the order put

(1)

n = E'

and thus get the pair of first-order equations n

(2)

n' = fo (,n) These may be combined into the single first-order equation

(E,n)' = Fo (C,n) where

F

:

Rk x Rk -+ R2k with

F(x,y) = (y, f(x,y)).

(3)

10.3.

To obtain the vectorfield "base point"

SECOND-ORDER VECTORFIELDS

Y

105

corresponding to (3) we simply adjoin

corresponding to (2) to the "right-hand side"

(x,y)

the

F(x,y),

to get

as in Section 10.1,

Y(x,y) = ((x,y), F(x,y)) (4)

_ ((x,y), (y, f(x,y)) Thus a vectorfield

has been constructed in a natural way from the

Y

original second-order differential equation (2).

The following remarks concerning the above example are intended to motivate our definition below of a "second-order vectorfield". 10.3.2.

Remarks.

The above vectorfield Y has the rather special

(a)

property that, of the four components occurring on the right-hand side of (4), the second and third components are equal. I -b Rk x Rk

If

(b)

vectorfield

Y

is any integral curve of the

then, in accordance with Section 10.1, it is a solution

of the differential equation (3) and hence also of (2).

This means in

particular that (C. V)

so that

is a self-consistent parametrized curve.

(C,n)

In generalizing these ideas to manifolds, we note that if a vectorfield

Y : M -+ TM

is to have self-consistent integral curves then

the manifold M should itself be the tangent bundle submanifold and

Q

10.3.3. :

TQ

of some

In the following definition, the elements of

Rn.

TQ

are ordered pairs and ordered quadruples, respectively, of

TZQ

R.

elements of

Y

of

Definition.

TQ --r T2Q

(x,y) E TQ

Let

Q

be a smooth submanif old of

is called a second-order vectorfield on

there is

z E Rn

Rn.

A vectorfield

TQ if for each

such that

Y(x,y) _ ((x,y), (y,z)). It is possible to express the idea that vectorfield by using projection maps. 10.3.4.

Lemma.

Y

is a second-order

This depends on the following lemma.

If Q is a submanifold of Rn and

the natural projection then the map TTQ : TZQ --> TQ

TQ : TQ --> Q

is

106

VECTORFIELDS

10.

has its value at ((a,b), (c,d)) E T2Q

given by

TTQ ((a,b),(c,d)) _ (a,c). Proof.

We use Definition 5.2.1.

parametrized curve

y

in

at

TQ

To this end, choose a differentiable (a,b) with

y'(0) _ (c,d).

Now write

y

componentwise as

(y1,y2)

with y1 = TQoy

so that

(TQOY)'(0) = Y1'(0) = C. Hence,

TTQ((a,b),(c,d)) = TtQ((a,b), y'(0)) _ (TQ(a,b), (TQoY)'(0)) (a,c).

The following corollary gives the desired characterization of a second-order vectorfield in terms of projections. 10.3.5.

Corollary.

Y : TQ -- T2Q

Let

be a submanifoZd of R.

Q

A vectorfield

is second-order if and only if TTQ0Y = idTQ .

Proof.

Given a vectorfield

Y : TQ -+ T2Q, let

(a,b) E TQ

and let

Y(a,b) _ ((a,b),(c,d)) E T2Q By Lemma 10.3.4,

the condition (TTQoY)(a,b) = (idTQ)(a,b)

is equivalent to (a,c) = (a,b),

and hence to Y(a,b) = ((a,b),(b,d))

This establishes the desired equivalence of the conditions in the corollary.

The next result shows that second-order vectorfields may be characterized in terms of their integral curves - a possibility which might be guessed from

Example 10.3.1.

10.3.

Theorem.

10.3.6.

SECOND-ORDER VECTORFIELDS

A smooth vectorfield

Y

:

107

TQ -+ T2Q

is second-order

if and only if each of its integral curves is self-consistent. Suppose first that

Proof.

vectorfield.

Let

Y

c: I -+ TQ

is a smooth second-order

TQ --+ T2Q

:

be an integral curve for

Y

so that

c' = Yoc

and hence (TTQ)oc' = (TTQ)oYoc

by Corollary 10.3.5.

= idTQoc = c.

It now follows from Exercise 10.3.3

that

(TQoc)' = c

and hence

c

is

self-consistent.

Conversely, suppose that

Y : TQ -+ T2Q

is a smooth vector-

field with the property that each of its integral curves is selfLet

consistent.

curve

c

:

(a,b) E TQ.

I -+ TQ

of

at

Y

By Theorem 10.2.1, there is an integral Hence

(a,b).

c' = Yoc and so

(TTQ)oc' = (TTQ)oYoc

From Exercise 10.3.3

it now follows that = (TTQ)oYoc

(TQoC)'

and since

c

is self-consistent this implies that c = (TTQ)oYoc.

Evaluating both sides at

gives

0

(a,b)

Thus

TTQoY = idTQ, showing that

= (TTQ)eY(a,b). Y

is second-order by Corollary 10.3.5.

Thus the significant part of the integral curve

c

of a

second-order vectorfield is the first component since the second is simply the derivative of the first.

10.3. 7. Definition.

If

c : I --+ TQ

curve for a second-order vectorfield

is a (self-consistent) integral Y

:

TQ -+ T2Q

then

TQoc: I -+ Q

is called a base integral curve of Y. The following lemma will provide a useful tool for establishing a further criterion for second-order vectorfields, involving local representatives.

108

10.

10.3.8. Lemma. Let c: I -} TQ

curve

(U, 0 for Proof.

Q

VECTORFIELDS

be a smooth submanifold of

A parametrized

Rn.

is self-consistent if and only if, for each chart

Q, the local representative

is self-consistent.

cTO =

See the exercises.

Theorem.

10.3.9.

Let

Q

vectorfield Y : TQ -- T2Q for

(U, Q)

Proof.

Q,

be a smooth submanifold of

is second-order if and only if, for each chart

the local representative Y

Suppose that

Y

by Exercise 10.2.1.

is also self-consistent.

the vectorfield

Put

11:

is second-order.

I -+ TQ

c = TO-loft

be an

so that

By Theorem 10.3.6,

c

c

is an

is

Hence, by Lemma 10.3.8, the local representative

self-consistent. cTO = n

YTO: TU -+ T2U

is second-order and let

integral curve of the vectorfield integral curve of

A smooth

Rn.

YT

It now follows from Theorem 10.3.6. that

is second-order.

The proof in the converse direction is similar.

EXERCISES 10.3. 1.

WA to down the vectoxb.ietd on TR co)t'ce6pond.ing to the bottow.Lng : I --> R: second-oxdeA dibbexent ae. equation bon

"+21;'+1;=0. Find the max ima e babe integna 2 curve o6 this v ec toi Leed at (a,b) E TR. 2.

Let

Q

be a zubmanLbo.ed ob

Show that

Y

Rn

and Zet

Y

: TQ -+ T2Q

£s a second-ondeh vectohb.ieed on

TQ

be any map.

£b and oney ib

TTQoY = idTQ = TTQoY

where TTQ:T2Q -+ TQ and -rQ:TQ -> Q axe the natwu1 pxojecti.onb. 3.

Let Q be a submanifold ob Rn, .let c : I -- TQ be a smooth paxametXLzed cuAve and Let TQ:TQ -. Q be the natty e pxojection. Show bxom DebLn tLon 7.1.1 that (TTQ)oc' = (TQoc)'

4.

Let

Q

be a submcJvL oed ob

Rn with

TQ - Q and

as a chant and Let

4(U) be the p/wjecti.ons. bnom DebLnitLon 5.2.1 that, bon each paxametniz ed curve TQ :

c :

I --; TQ, 4oTQoc

Show

10.3.

5.

Prove Lemma 10.3.8.

6.

Let

Q

the pnecedi.ng exec-L6e. )

Rn and Let Y : TQ -+ T2Q be a amooth Show that Y it second-onden i6 and only £b each og

.W Lntegna.C c.Wwe6 .

(Hint:

be a eubmanii.botd ob

vecxons.Leed. d

109

SECOND-ORDER VECTORFIELDS

I -+ Q

4atA.46Le6

c

:

I --> TQ

d>> = Yod'

can be wnii tten ab .

c = d' where

11. LAGRANGIAN VECTORFIELDS

In this chapter the theory of vectorfields and their integral curves will be related back to the study of the motion of a particle on a submanifold

Q

of R.

Whereas in Section 8.3 we assumed the existence

c: I - TQ satisfying Newton's

of a self-consistent differentiable curve

second law of motion, in this chapter one of the aims is to prove the existence of such curves - thereby verifying that the theory developed in Section 8.3 is not vacuous.

To achieve this aim we shall use the idea that Lagrange's equations, with respect to each chart in an atlas for TQ, define "local" vectorfields which can be patched together to form a "global" vectorfield

on the whole of TQ.

The existence-uniqueness theory of the preceding

chapter can then be used to establish the existence of integral curves for Finally, the integral curves so obtained can be shown

this vectorfield.

to satisfy Newton's equations of motion.

11.1.

GLOBALIZING THEORY

This section sets up some language which is useful for studying how local vectorfields may be patched together to form a global vectorfield on a manifold.

Throughout this section M will denote a

submanifold of 11.1.1. M.

k-dimensional

Rn

Definition.

Let

(U,¢)

and

(V,p)

be two compatible charts for

A pair of local vectorfields Y : O(U) -* TO(U)

and

Z

: (V) - Tip(V)

is said to be patchable with respect to these charts if, on the domain

unv, (To)-1o Y o q

=

(Tip)` o z o

gyp.

11.1.

GLOBALIZING THEORY

111

Y

Figure 11.1.1.

Patchable local vectorfields.

The motivation for this definition is that

Y

and

Z

will

satisfy it whenever there is a global vectorfield on the manifold M of

which Y (U,4)

and

and

Z

are the local representatives with respect to the charts as is illustrated in Figure 11.1.1.

(V,p)

In applications to specific examples it will sometimes be possible to avoid awkward computations by using the following theorem which shifts attention away from the vectorfields to their integral curves. 11.1.2.

Lemma.

Let

(U,c)

and

(V,ip) be two compatible charts for

A pair of local vectorfields Y :

0(U) -

and z : (V) - Tip (v)

M.

112

LAGRANGIAN VECTORFIELDS

11.

is patchable with respect to these charts if the following condition holds:

if n

i --; 4(U fl v) is an integral curve for

:

then

:

where

I - p(U fl v)

is an integral curve for

Let

By the local existence-uniqueness theorem, Theorem 10.1.5, integral curve at

a

Z

1)°n.

= (ipo

Suppose the condition stated in the lemma holds.

Proof.

Y

for

Y,

a E 4(U fl v).

there is an

say

n : I

is an interval containing

Hence

0.

TI= Y°n

(1)

n(0) - a.

(2)

and

The assumed condition in the lemma now implies that

C _ (p ° 4-1) ° n is an integral curve for

Z

(3)

so that

C' = Z o

(4)

Now substitute (3)into(4) and use Definition 7.1.1

and then the chain rule

to get

Hence

T(l»0-1)oY°n = Z o (P°O 1)°n T(po

and so But as

')oY(a) - Z ° (W° 1)(a) by (2)

T 1oY(a) = TV-' ° z o a

by (1)

is an arbitrary element of

o 4-1(a).

4(U fl v)

this implies that, on the

domain w fl v), Tc-1°Y = and hence, on the domain

oz

u fl v,

To-1oYoq =

Tiy-1

oZo

ip.

The next theorem is a trivial consequence of our definition of patchability.

11.1.3.

YX

be a smooth local vectorfield on

and

(UX, ¢.A`)

Suppose that, for

O(UX).

are patchable with respect

There is then a unique smooth

(Uu, ¢ u).

Y on M such that each YX is the local representative of

vectorfield

in the chart

Proof.

Ya and Y u

the vectorfields

A, u E J,

to the charts

Y

be an atlas for M and, for each

Theorem. Let {(UX (k ) : X E J} let

X E J,

each

113

GLOBALIZING LAGRANGE

11.2.

(UX, 0X).

We may define

on each

Y

U X by putting

o YX o 0A

YIUX _

This definition is unambiguous by the definition of patchability.

moreover, is the only possible choice of 11.1.4.

Corollary.

the manifold M is the tangent

If in Theorem 11.1.3

bundle of some other manifold and each local vectorfield

Y X is second-

on M is also second-order.

order then the vectorfield Y Proof.

This,

Y.

This is an immediate consequence of Theorem 10.3.9.

11.2.

APPLICATION TO LAGRANGE'S EQUATIONS

The study of the motion of a particle on a manifold leads, via Lagrange's equations with respect to the charts on the manifold, to local second-order vectorfields.

In this section it will be shown that these vectorfields can

be patched together to give a vectorfield defined on the whole of the tangent bundle of the manifold.

In order to be able to apply the globalizing theory of Section 11.1, we must first check that the local vectorfields obtained from Lagrange's equations are patchable. crucial.

Note that the function

L

For this task, the next lemma is

which occurs in the lemma is not

restricted to being the Lagrangian for any particle motion.

Birkhoff(1927),

page 21, has an alternative albeit more coordinatewise proof. Throughout this section,

Q

will denote a k-dimensional

submanifold of R. 11.2.1.

and

Lemma.

(V, r)

Consider any smooth function

are charts for

and if c

Q

:

L : TQ --> R.

I -} U fl v

If

consistent smooth curve then

raLo eJ `34 J

-aLo c=0 *raLe c) 3q

``ar

(U, q)

is any self-

-al. oc=0 2r

114

LAGRANGIAN VECTORFIELDS

11.

Proof.

A mild extension of Corollary 7.2.8, shows that

-a

aL _ aL 3r aq ar aq

+ LL ar

TQ

Dr 3q

(1)

and also

since

r

aL

aL ar

aq

ar aq

depends only on

(2)

ar

so that

q

We also have by Corollary

= 0.

aq

7.3.4, ar

Dr

C = as a TQ o c

(3)

aq

and hence by Corollary 7.3.8.

(arocl'=arac. J

aq

(4)

aq

Now suppose, in accordance with the hypotheses of the theorem, that r

aLo c

-aLo c=0

(5)

q

aQ

where by (2)

aL

o

c l/ =

aq

(L a c lr ar o c+ LL o c rare cl \ar

/

ar

/

ar

'aq

ar

aq

/

o TQo c +

q

aai o c aro c q

and

by (3)

J

(4)

while by (1) aL aq

aL ai

V

c

ar

oc +

aq

aL

oc -Dra T oc

Dr

aq

Q

Hence (5) becomes

((aL a c) /

aro c

ar

But as

(U,q)

and

l

o

aq

are both charts for

(V,r)

matrix on the right is non-singular and hence

aL a=

r

c

ar

c = 0.

-[Q0 c = 0. Q,

it follows that the

GLOBALIZING LAGRANGE

11.2.

Our next result recasts Lemma 11.2.1

115

into a form which refers

to local representatives of curves rather than to the curves themselves. Motivation for this comes from Chapter 9 where we studied the explicit form of Lagrange's equations for the local representatives of the curves representing the history of the particle. c: I -+ TQ

is a curve and

local representative

c

or

cTq = Tgoc

In addition to the notation of

TQ

then the

is defined by c - (Tq) i°cTq

(goTQoc,

we also used the notation

c,

is a natural chart for

(TU, Tq)

of

cTq

Recall that if

for the local representative

qoc)

(q, q') to give our answers in Chapter 9

a more customary appearance.

Since, however, we now wish to change tack and take the local representatives themselves as our starting point, we shall use letters such as

"c"

and

"n"

In this way we avoid the possibility of

to denote them.

begging the question as to whether we can construct a suitable global curve c

:

I --+ TQ which is independent of charts.

Lemma.

11.2.2.

and

(V,r)

Consider any smooth function Q

be charts for

L : TQ --+ R.

Let

(U,q)

and let

and

n: I -+ (Tq)(u n V)

C: I -+ (Tr)(u n v)

be any self-consistent smooth curves. (8L

If

-

(Tq)-l on)

(31,

then

o (Tr)-ioC-

a'r

ar °

J/

-Tro (Tq)

where

o

(Tq) ion = 0

(1)

(Tr) ioC = 0

(2)

q

aq

ion

.

(3)

Define a parameterized curve

Proof.

c - (Tq)

c: I -- U n V

by putting (4)

ion

From the hypothesis (1) of the lemma 1a L

°c

-Qo c-0

aQ

Since 10.3.8.

n

is self-consistent, however, the same is true of

c, by Lemma

Hence, by Lemma 11.2.1,

I\

ar

°

`)I -ar° c=0.

(5)

116

LAGRANGIAN VECTORFIELDS

11.

But from (4) and the hypothesis (3) in the lemma,

c = (Tr)

1

c

Substituting this in (5) gives the desired conclusion

(2) of the lemma.

The stage has now been set for an application of the globalizing to the local vectorfields defined by Lagrange's

theory of Section 11.1 equations.

The assumption that the function

L : TQ -* R

is the

Lagrangian of a mechanical system will now be used for the first time in this section. 11.2.3.

Let L

Theorem.

moving on the submanifold {(UA, qX):a E J}

If

as described in Section 8.3.

is an atlas for Q

then

there is a smooth local second-order vectorfield

A E J

(a) for each

be the Lagrangian of the particle

T Q -. R

:

Q

YX : TqX(UX) --, T2gA(U)L)

whose integral curves are the solutions

(aL aQ)L

(b)

nX of Lagrange's equations

DL A -1 a a (Tq > o (Tq) n -Q

-1,

J

=0

n

there is a unique smooth second-order vectorfield Y : TQ -- T2Q

for which the local representative in the chart Proof.

By Theorem 9.3.3,

for each

A E J,

(TUX, TqA)

is

YX

.

Lagrange's equations in the

form written above are second-order differential equations for the function

where n X _ (q

q

Thus, as in Section 10.3, they define a smooth

q

local second-order vectorfield By Lemmas 11.2.2 and

Yu

Y X with the stated integral curves. and 11.1.2,

each pair of vectorfields

is locally patchable with respect to the charts

(TUU, Tqu),

so Theorem 11.1.3

vectorfield

Y.

The vectorfield

Y

(TUX, TqX)

YX

and

gives the existence of the desired

will be called

occurring in Theorem 11.2.3

the Lagrangian vectorfield of the mechanical system described in Section 8.3.

11.2.4.

Corollary.

then for each

Y

If Y

a E TQ

is the vectorfield given by Theorem 11.2.3

there is a unique integral curve

at a having maximal domain.

c

:

I ---> TQ

for

This curve is self-consistent and for any

chart

for

(U, q)

it satisfies Lagrange's equations

Q

r

fal,

-aLec=0

oc i

Proof.

117

BACK TO NEWTON

11.3.

q

aq

Left for the exercises.

EXERCISE

11.2.

Prove ConolLany 11.2.4.

11.3.

BACK TO NEWTON.

It will now be shown that the arguments leading from Newton's second law of motion to Lagrange's equations can be reversed. direction is given in Landau

&

A result in this reverse

Liftshitz(1960), page 9, although their

result applies only in the more rudimentary context in which there are no geometric constraints on the motion.

The following lemma will be used in

the proof of our result. 11.3.1.

Let X

Lemma.

be the identity map on

manifold of Rn of dimension for

Q

Rn, let

a E Q.

If

Q

(U,q)

be a subis a chart

at a then the set of vectors

(aq,

is a basis for Proof.

and let

k

For

(a), aq, (a),..., a-k (a))

TaQ.

1 < i < k

define

yi

:

I -+ U

by putting

yi(t) = q 1(q(a) + t ei) so that, as in the proof of Theorem 7.3.10,

yi' (0) = aQ (a) 2

yi

But by the construction of

q u yi = q(a) + idI ei

and hence by the chain rule Tq c Tyi = T(q(a) + idI ei) Tq(a, yi'(0)) _ (q(a), ei).

But since

TgITaQ

is linear and the set of vectors

(e1,e2,...,ek)

is

118

LAGRANGIAN VECTORFIELDS

11.

linearly independent the same is true of the set of vectors which therefore is a basis for the

(yl'(0), y2'(0),...,yk'(0)), k-c'.imensional vector space

TaQ.

Although our theorem is stated and proved only in the context of Section 8.2, where the submanifolds lie in

it can easily be

R3,

extended to the more general situation discussed in Section 8.3.

Consider a particle of mass m moving on a k-dimensional

11.3.2. Theorem. submanifold

Q of

T : TQ -* R

and

V

particle and let

subject to a conservative field of force.

R3

Let

be the kinetic and potential energies of the

Q -- R

L = T-VOTQ

be the Lagrangian.

The reaction force

R. constraining the particle to stay on the

manifold, can then be chosen in just one way at each point of TQ

in order

that the following condition holds:

if c: I -* TQ vectorfield on each

TQ

is an integral curve for the Lagrangian

determined by

m(XoTQoC)n(t) _

lax R

At each point the reaction force Proof.

then Newton's second law holds for

L

t E I:

Let

o TQ +7f2 O R

is orthogonal to

field so that, as in Corollary 11.2.4,

and let (U,q) be a chart for

Q

which shows that for

aqi

J

Let

t EI

The argument

c(t) E U.

oc - m(% c) .

3T

o

aqi

(aX

G T oc I

`aqi

Q

1 5 i 0.

4.

Veni.6y that the map

A : R -+ Diff0(TR) with

A(t)(a,v) _ `k sin(kt) + a cos(kt), v cos(kt) - ak sin(kt))

where

k > 0, -i.6 a blow on TR. Sketch come typA.cat o&".

12.2. FLOWS FROM MECHANICS Some familiar problems from mechanics will be used here to illustrate the idea of the flow of a vectorfield. examples is very straightforward.

The theoretical background for these The vectorfields which arise in these

problems can be proved complete by an application of the results in the next section; the integral curves of these vectorfields then generate flows in accordance with Theorem 12.1.9.

Hence our discussion will concentrate

on the geometrical ideas associated with the flows.

Recall that a flow A : R -+ Diffw(M) vectorfield

Y

for each

(a)

t E R, the time evolution operator for lapse of time t

At for each

(b)

generated by a complete

on a manifold M determines two families of mappings

a E M,

: M -+M : a-+A(t)(a)

the integral curve at

Aa

:

R --+ M .

a

t -+ A(t) (a) .

In addition, there is for each a E M the subset of M consisting of the image of the integral curve through a through

and called the orbit of the flow

a.

In the mechanical problems which follow, the configuration manifolds are 1-dimensional and hence their tangent bundles, which contain the orbits, are 2-dimensional.

This permits direct diagrammatic

representation of the orbits and also, but less directly, of the integral

12.

126

FLOWS Thus in Figure 12.2.1, the

curves and the time evolution operators.

curves with arrows are the orbits of a flow

A.

From a knowledge of the

times at which the particle assumes the various states, it is possible to At

infer what

is doing.

states after time t

initial states

Figure 12.2.1.

12.2.1. Example.

Time evolution operator.

(Motion under gravity in one dimension.)

For a particle

of mass m moving under gravity and with the x-coordinate measuring the height, the Lagrangian

L

is given by

L =11mx2 -mg and hence Lagrange's equation

becomes (xoc)n = -g .

When expressed Y :

R -* TR

as a vectorfield, as in Example 10.3.1, this gives

with Y(x,w) = ((x,w),(w,-g))

for each

(x,w) E TR.

The vectorfield

Y

is shown in Figure 12.2.2.

Elementary differential equations techniques show that the integral curve for

Y

at the point

(a,v) E TR

is given by

c(t) _ (a + vt - '/gt2, v-gt)

and hence the flow

A : R -- Diff°°(TR)

is given by

A(t)(a,v) = (a+vt - 'gt2, v-gt)

12.2.

Figure 12.2.2.

FLOWS FROM MECHANICS

127

Gravitational vectorfield on phase space.

Note that each integral curve is self-consistent and that the orbits consist of parabolas which are tangent to the vectorfield. See Figure 12.2.3.

Figure 12.2.3.

12.2.2.

Example.

Gravitational orbits in phase space.

(The simple pendulum.)

This consists of a particle of

mass m constrained to move under gravity on a circle lying in a vertical plane, in the absence of friction.

Distances in the horizontal and vertical directions will be measured by

x

and

y-coordinates, respectively, and the circle on which

the particle moves with be taken as

S1.

See Figure 12.2.4.

12.

128

FLOWS

x

Figure 12.2.4.

A simple pendulum

Thus the configuration manifold for the particle is phase space is

TS',

S'

and the velocity

which is diffeomorphic to the cylinder

S' x R

in

view of Exercise 6.1.1.

aiT

U2 R

60

Figure 12.2.5.

Charts for

We shall use the two charts which are defined by Figure 12.2.5. submanifold structure of

S1

S1

.

(U,, 61)

and

(U2, 92)

for

S'

These charts are compatible with the

and have the property that

91

on one connected component of

= J02 1 9 2 -2v

Hence

129

FLOWS FROM MECHANICS

12.2.

U1 n U2

on the other .

61= e2

on

Tul n Tut

.

Thus we may define a smooth map, called the angular velocity map, by putting

TS1 -+ R with w =

w :

61 162

on on

TU1 TU2

In addition to its obvious kinematic interpretation, the angular velocity has an important geometrical role as a component of the diffeomorphism, defined in Exercise 6.1.1,

(D :Tsl-->S1xR between the tangent bundle of the circle and the cylinder.

It is set as

an exercise to show that, in fact, this diffeomorphism satisfies (1)

'D - (TS1, W)

The geometrical quantities needed for the study of the simple pendulum have now been defined.

Before giving a formal discussion, however,

we use physical intuition to guess at the possible types of orbit which can appear in the velocity phase space, as in Irwin(1980).

Thus the orbits

shown in Figure 12.2.7, shown in both chart and cylinder representation, can arise in the following ways: (a)

If placed with zero initial velocity at either the lowest or the highest point of the circle, the particle will stay there indefinitely.

These two "equilibrium points" give rise to the

two single-point orbits where the cylinder meets the y-axis. These two orbits are said to be "stable" and "unstable", respectively, for obvious reasons. (b)

If started with zero initial velocity at some point intermediate in height between the two equilibrium points, the particle will oscillate to and fro, rising indefinitely often to the initial height - first on one side of the origin, then on the other.

These motions are represented on the cylinder

by closed orbits which encircle the y-axis.

130

12.

FLOWS

If the particle is started with sufficiently large initial

(c)

velocity it will pass through the highest point of the circle and then continue to make complete circuits. orbits are the closed orbits which

The corresponding

go right around the cylinder.

There are two distinct families of these orbits because the particle may traverse the full circle either clockwise or anticlockwise.

It is left as an exercise to show that there exists an orbit having the unstable equilibrium point orbit as a limit point.

A particle traversing

this orbit approaches arbitrarily close to the equilibrium point without ever actually reaching it.

To complete our discussion of the motion of the simple pendulum, it remains to show how the formal theory from previous chapters leads to the sketches of (a) the vectorfields in Figure 12.2.6. and (b) the orbits in Figure 12.2.7.

As to the vectorfields, we first apply the theory of Section 8.2

to the particle constituting the bob of the pendulum.

the

x

and

In terms of

y-coordinates the kinetic and potential energies are T = /n(x2 + y2) V - mgy.

After restricting these functions to

L : TO -+ R charts

by putting

(U,, 61)

and

(U2, 62)

for

we get the Lagrangian function

TS1

L - T - V°TS1.

We will need to use both of the

S1

so let

i = 1

or

2

and note

that

x = sin°9i

on

Ui

y = -cos°9i

and hence L = &rn62 + mg cos°9i°TS1

Thus, in terms of the local representative curve

c: I -+ TS1

equation

O.

with respect to the chart

on

Ui.

- 9i ° TS1°c (Ui, 9i),

of an integral

Lagrange's

becomes 9i

n+

gsin°Si = 0.

Hence, by the procedure given in Example 10.3.1, we find for

(2)

i - 1

the local vectorfield corresponding to this second-order differential equation is

that

YTe 1

131

FLOWS FROM MECHANICS

12.2.

:

T(-1r,Jr) -} T2R (4,w) i-+ ((q,w),(w,- g sin(k)))

:

This vectorfield is sketched on the left in Figure 12.2.6 in the plane.

The sketch for

YTe

is similar.

and lies flat

Now, by the

z

construction in Section 11.1,

the global vectorfield for the problem,

Y. Ts' -+ T2s',

is related to the above vectorfields by T(Tei)-1oYT6ioTei

Y -

on

Ui

Finally, the vectorfield sketched on the cylinder in Figure 12.2.6. is S' x R -- T(S' x R)

given by

Y, - T o Y o $-1 where

is the diffeomorphism from TS1

to calculate an explicit formula for details here.

to the cylinder.

It is possible

Y4D, although we shall not give the

Intuitively, this vectorfield is obtained by wrapping the

local vectorfields

YTe1

and

around the cylinder, at the same

YTe 2

time allowing their arrows to poke out into

R3

while remaining tangent

to the cylinder.

+w

Figure 12.2.6.

The vectorfield for the simple pendulum.

132

FLOWS

12.

Turning now to the orbits, our aim is to formally justify the claims made earlier on the basis of physical intuition.

In other words

we are going to show how to derive mathematically the sort of orbits shown in Figure 12.2.7.

Pendulum orbits in velocity phase space.

Figure 12.2.7.

Recall that an orbit is a subset of

of the form

TS'

{c(t): t E R}

where

c

:

is an integral curve of the Lagrangian vectorfield of

R -} TS1

the pendulum.

(3)

In Figure 12.2.7

(a) the natural chart

TO,

and

the images of such orbits are shown under (b) the diffeomorphism

(D

onto the

cylinder.

Under the natural chart

Tel

the image of the orbit (3) is the

set

{(e1,S1')(t): t E R} = TR ,

denotes the local representative of

where

the natural chart

Tel.

c

with respect to

A similar set is obtained under

T62.

It is

left as an exercises to check that the image of the orbit (3) under the diffeomorphism TO,

and

T62

(D

is obtained by wrapping the images of the orbit under

around the cylinder.

12.2.

FLOWS FROM MECHANICS

133

To complete our discussion of the orbits we use energy considerations.

Given

e E R, the set of all points in TO at which

the sum of the kinetic and potential energies assume the value out to be a curve in Theorem 8.5.1 c: R -. TS', curves.

and

TO,

TO - called a constant energy curve.

e

turns

Since by

the total energy stays constant along an integral curve it follows that the orbits are subsets of the constant energy

The images of the constant energy curves under the charts

may be obtained by putting /(OZ)2 - g coso6i - e

for

TO,

i = 1,2 and then sketching the graph of

(4)

6i

against

6i.

in fact, the way in which the curves shown in Figure 12.2.7

This is,

were obtained.

Each orbit must lie inside a constant energy curve, although some constant energy curves contain more than one orbit.

Since (4) implies that

e >, -g, the following cases exhaust the possibilities. (a)

Here the constant energy curve is a single

e = -g.

point corresponding to the stable equilibrium point. (b)

-g < e < g.

Here the constant energy curves are closed

curves encircling the stable equilibrium point.

Each

curve is a single orbit. (c)

e = g.

The constant energy curve contains the unstable

equilibrium point.

Removal of this point leaves two

connected components, each of which is a single orbit. (d)

e > g.

The constant energy curves encircle the

cylinder.

Each curve is a single orbit.

These results will be verified in the exercises.

EXERCISES 12.2. 1.

Diz euee the advanxagea ob nepteaent4ng the onbita £on the pendu&um

pnabtem on a cyt i.nden nathet than in the peane v.i,a cha4t.6 (see Inw.in(1980), page 4). 2.

Veniby that the dtbbeomonphi m

Eww..caa 6.1.1

0: TO - S' x R

de4.ined in

can be w&itten in the Lonm (1) in the text.

134

3.

FLOWS

12.

'

Let be a6 in the pnev.iou6 exe'tc a and feet (ui,ei) be the chcvrt bon sl .inxiwduced in the text (i = 1,2). Show that .i6 c: R -. TO .c6 an .i.ntegnat cwwe Jon the Lagnang.ian vector6.ietd o6 the pendulum then Doc(t) _ (sine

6or att

t E R

Ouch that

,

cosoui, ci)(t)

c(t) E Ui.

(Thi,6 veni6.Le6 the ceaim made in the text that the image o6 an orbit under

P

.i6 obtained by wnappLng anaund the cyeinden the.mage6 o6

that a)tb.it under each o6 the chanta 4.

Tel

and

Tee.)

Veni6y 6nom (4) in the text that the contact energy cwcvea have the genenae.4hape ceaimed in the text bon the vani,ou6 nange6 o6 vatue6

o6 the parameter

5.

e.

Let c: R -} Ts' be an .integnae curve o6 the Lagnang.ian vector feed o6 the pendueum. Note that i6 e > g then the anguear.veeocJty, woc, along the £ntegnat curve .c6 bounded away 6nom zero. Deduce

that in this cane the con6tant energy curve .ia a a.ingte orbit. 6.

Let 0 : R --> R be a di66enenti.abte 6unction ouch that (a)

9(t)

(b)

t -oo

lim

.t,6 bounded a6

t-

00

9'(t) exL6t6.

Show 6nom the mean vatu.e theorem that lim

e'(t) - 0.

t->00 7.

figure 12.2.8

cab a and

6how6 a con6tant energy curve bon the penduCum in the

-g < e < g, which cna66 ea the a 1-axi,a at the po.i.nt6 (a,0)

where

0 < a < Tr.

Let

(91,91)

(-a,0)

be the £ocae

nepne6entative o6 an .i.ntegnae curve and 6uppo6e that (91,9,') (t) £Le6 on the conatant energy curve and in the upper ha26 ptane when

t=to. Thi6 exenai.6e .i4 to prove that at home .eaten. time t > to the paint (91,91') (t) pa66ea into the .cower W6-peane. Suppo4e, on the contuty that it 6tay6 in the upper. hae6-pCane bon a L t > to and then derive a contnadicati.on by 6howing, with the aid o6 Exenai,6e 6,

that

LAGRANGIAN FLOWS

12.3.

(a)

(b) (c)

01(t) exd6t6

lim

t+°

135

e1

lim 9i(t) exi.6t6

t-r00

lim 91(t) = 0

-Tr

t+°

-aI

a

Tr

61

lim

(d)

91(t) exi-6t6

t}° el(t) = 0

lim

(e)

t-+ 00 lim

Figure 12.2.8.

911(t) # 0

t + co

8.

Deduce Snom the pn.ev.i.ou6 exenci,6e that in the cadet -g < e < g each conbtant energy curve in the penduLwn p1w& em eon6.i.6.t6 ob a a.ingCe

ohbit. 9.

Suppose that Y: M -> TM AA a amooth veeto/t Lees on the man jo!d m

and Let a E M with Y(a) - 0 (4o that a .(.6 an equtitibAium point o6 the veetonbietd).

Let c: R -} TM be an .LntegnaL eunve o6 the Show that .L6 c(to) # a 6o/t come tQ E R then

veetonb.Leed.

c(t) #a 6ota.UtER. 10.

Prove the dtatement6 made in the text about the oabtt6 ob the

pendulum in the ea6e e = g.

Uae the Miu .t6 o4 Exe&cA.sed 6 and 9.

12.3. EXISTENCE OF LAGRANGIAN FLOWS In this section the major result is a theorem which can be used to establish the completeness of many Lagrangian vectorfields arising from classical problems such as the simple pendulum, motion on a paraboloid (see Section 9.1) and the spherical pendulum (dealt with in Chapter 13).

As noted in Section 12.1, completeness of a vectorfield implies the existence of a flow. 12.3.1.

Theorem.

and suppose

Q

Let

Q

be a smooth k-dimensional submanifold of

is a closed subset of

smooth second-order vectorfield on

TQ

Rn.

Suppose

Y : TQ --+ TZQ

Rn

is a

and that

c: (-6,t) -- TQ is a maximal integral curve for some

K and all

t E

Y

with the property

Then

6 = e = -.

117T. 0c(t)11 0

and all

x, y E U.

U

is

12.3.

Then for each

LAGRANGIAN FLOWS

137

t E (-e,e)

Ilc(t) - d(t)II 5 Ilc(0) - d(0)IleKitl Proof.

See exercises.

EXERCISES 12.3. 1.

Expeai,n why there ex,i.6t6 a Lagnang.i.an b.eow bok the p4ob.eem6 deacxibed

in Exampte 12.2.2 2.

and Section 9.1.

Prove Theorem 12.3.4

FAA4t note that

c'(t) = Foc(t) and ao we

can apply Theorem 1.5.4.(a) to.6 how that

fto

rt J

Now chow that bon

7f2oFoc

t E (O,E)

IIc(t) - d(t)II

K normo(c-d)

Ilc(0) - d(0)II +

J0

where "norm" £a given by norm:

-+ R: x -+ IIxII. Then appl-y Gnanwal'a £nequa.Ci ty, Lemma 1.5.6, to get the %equited ne6uLt. F.ina1y 6hOW how to extend the neautt to a U ob (-c,e). Rk

13. THE SPHERICAL PENDULUM

The spherical pendulum is typical of the sorts of problems which are traditionally studied in Lagrangian mechanics in that it has a In addition it has the

configuration manifold which is 2-dimensional.

very special simplifying property of being "integrable":

in suitable charts

Lagrange's equations "uncouple" leading, in effect, to a pair of A fairly extensive list of integrable problems is

1-dimensional problems.

contained in Whittaker(1952) and a recent addition to this list is given our discussion of the spherical pendulum will

in Gray et al(1982).

illustrate the way in which the preceding theory can be applied to such problems.

13.1. CIRCULAR ORBITS A spherical pendulum consists of a particle of mass m constrained to move under gravity on a sphere, in the absence of friction.

The constraint can

be achieved, for example, by attaching the particle to one end of a light rod while keeping the other end fixed.

The

sphere on which the particle

moves will be taken as S2 - {(a,b,c) E R3: a2 + b2 + CZ - 1}

which is a submanifold of

The kinetic and potential energies are

R3.

given in terms of the identity chart

T

(x, y, z)

on

R3

by

= '-gn(x2 + Y2 + z2)

V -gz. The Lagrangian domain

T82,

L : TS2 -+ R so

L

is then the map

is smooth.

T - VoTS2

restricted to the

13.1.

139

CIRCULAR ORBITS

By Theorem 11.2.3, Lagrange's equations, with respect to the various charts in any atlas for

S2, yield a second-order vectorfield on

TS2 - the Lagrangian vectorfield for the spherical pendulum.

Since

S2

is

compact, moreover, Theorem 12.3.2. shows that this vectorfield is complete.

Hence it generates a flow on

by Theorem 12.1.9.

TS2

Thus there is a unique integral curve of the vectorfield at each point of

TS2

with

R

more, are self-consistent.

The integral of curves, further-

as domain.

Since there is no realistic way to sketch

orbits on the 4-dimensional manifold

TS2, we shall have to be content to

sketch the projections of these orbits on

S2.

These projected orbits are

traced out by the base integral curves.

The existence of the following families of orbits is almost obvious on the basis of physical intuition:

The equilibrium points at the north pole

(a)

and the south pole

S = (0,0,-1).

N = (0,0,1)

If placed at rest

at either of these points, the particle stays there

indefinitely. Orbits along each meridian, obtained by intersecting

(b)

S2

with a vertical plane through the poles.

The

motion of the particle along a meridian is the same as for a simple pendulum.

Orbits around each parallel of latitude below the

(c)

equator, obtained by intersecting

S2

with a plane

perpendicular to the polar axis of the sphere.

A

particle moving in these orbits is called a "conical pendulum".

These orbits are illustrated in Figure 13.1.1.

It is very easy to formally establish the existence of these orbits by using the equivalence we have established between Newton's second law and Lagrange's equations.

To illustrate the method, we prove

the following physically plausible result: 13.1.1.

Proposition.

Let

c: R -+ TS2

be an integral curve for the

Lagrangian vectorfield of the spherical pendulum at the point

where a

is either of the two poles.

For all

t E R

(a,v) E TS2

it then follows that

(a)

if v = 0

then

TS2oe(t) = a

(b)

if v # 0

then

TS2ec(t) lies in the vertical plane

through the poles containing the vector

v E R3.

13.

140

THE SPHERICAL PENDULUM

Figure 13.1.1.

Motion on a meridian and on a parallel of latitude.

We leave the proof of part (a) as an exercise and prove part (b).

Proof.

The theory for the simple pendulum given in Example 12.2.2

be adapted to the meridian M as configuration manifold. an integral curve

d: R - TM

11.3.2, the reaction force R It also lies in the plane of

the integral curve

d

at

for that problem.

(a,v)

may

There is then By Theorem

is orthogonal at each point to the circle

M and hence is orthogonal to

S2.

M.

Thus

satisfies Newton's second law, as formulated in

Section 8.2, for the spherical pendulum problem and hence, by Theorem 8.2.2 and Theorem 11.2.3,

d

is an integral curve for the Lagrangian vectorfield

of the spherical pendulum at the point and so

TS20c

maps into the meridian

(a,v) E TS2.

By uniqueness,

c = d

M.

EXERCISES 13.1. 1.

Prove pa't (a) 05 p/wpoaition 13.1.1.

2.

Let d: R -} Ts2 be an £ntegnat curve Son the Lagnang.ian veatoi 6 etd o6 the apheni.cas pendulum. Show that .is the base .integhat curve TS2°d paabea thkough a pose at home time to then the babe .integral

c.ulwe mapb into a mehi an. (Hint:

Put

c (t) = d (t + t o)

and z how

c

£4 an .i.n tegnae cww e o6

13.2.

141

OTHER ORBITS

ob the veatoii4L Ld at acme point (a,v) ob the 6oht de6cnibed in Pnopo4ition 13.1.1.) 3.

Read about the con cat pendu um in Synge 6 Gni.66.ith(1959), pages 336-337, and then deduce the existence ob cortne6ponding .Lntegaat cutcve6 Got the LaguangA.an vecto' Letd o6 the 6phentc.cJ pendulum. 13.2. OTHER ORBITS, VIA CHARTS

The orbits of the spherical pendulum to be discussed in this section are not as obvious physically as those introduced in the previous section. establish their existence, we shall choose charts for which Lagrange's equations "uncouple". S2

S2

with respect to

Since orbits through the poles of

have already been fully discussed in Section 13.1, moreover, we lose

nothing by choosing charts whose domains exclude these points. To define the desired charts put U1 = S2\{(a,b,c) E S2 : a < 0

and

b = 0}

U2 = S2\{(a,b,c) E S2 : a % 0

and

b = 0}

and let the maps 1, W2, z

These charts are then

be as shown in Figure 13.2.1

1

U1 -+ (-Tf,Tf)

W2

U2 --r (0,2Tf)

z

U1 U U2 --

(Ui, (4i, zjUi))

Figure 13.2.1.

Charts for

To

(-1,1)

for

S2

i = 1,2.

where

142

THE SPHERICAL PENDULUM

13.

Note that the unit sphere in

S2

which is the domain of

U1 U U2,

with both poles removed.

z, consists of

On the other hand,

consists of two open hemispheres and

U1 fl U2

01

on one hemisphere

41+21T

on the other

and hence we may define a smooth map w : T(U1 fl U2) -+ R ¢A,1

on

TU1

`Y2

on

TU2

In this way we get a pair of maps,

and

z

w,

by putting

which have a sort of

global significance throughout the part of the manifold which now concerns us.

The differential equation introduced in the next proposition will play a key role in our discussion of the possible orbits for the As usual, we write

spherical pendulum.

Wi = AioTS2ec

z = zoTS20C,

for the local representatives of an integral curve

c.

13.2.1. Proposition. Let c: R -+ TS2 be an integral curve for the Lagrangian vectorfield of the spherical pendulum satisfying the condition that the corresponding base integral curve never passes through a pole. There are then numbers

e, K E R, depending on the initial value

c(0),

such that (z')2 = 2g((1-z2)((e/g)-z) - K2/2g) - foz, Proof.

say.

(1)

When expressed in terms of the charts the Lagrangian

on the domain

Ui

L

is given

by

L

39n((1-z2);Z

+

z2

) - mgz

1-z2

for

i = 1,2.

Hence by Theorem 8.2.2

the integral curve satisfies the

Lagrangian equations ((1-z2)

')'= 0

(2)

143

OTHER ORBITS

13.2.

z" +z,)2+zz'2+g=0

(3)

(1-z2)2

1-z2

(It is perhaps reassuring to note that our assumptions ensure that can never assume the value

constant K E R

z2

From (2) it follows that there is a

1.)

such that

(1 - z2)$i' = K

(4)

Since l and 2 are restrictions of the smooth map w moreover, that

is the same for both choices of

K

We may assume,

since otherwise (4) would imply that

K # 0

furthermore, that

i.

it follows,

¢i' = 0

and hence the base integral curve would be an orbit along a meridian, contrary to hypothesis.

Now since, by Theorem 8.5.1, the total energy of the system is conserved, there is an

e E R

such that

{ (1-22) (Yi')2 + and, moreover,

e

+ gz = e

zz

is independent of

i.

(5)

and then

Solving (4) for

gives, after some manipulation, the required equation

substituting in (5)

(1). 13.2.2.

Proposition.

The function

f

appearing in the previous proposition

has a graph which looks like one of those in Figure 13.2.2. f

al, a2, a3

must have three zeros

such that

1 < al < a2 < 1 < a3 Both of the cases

al = a2

of the initial value

c(0)

and

In particular,

al < a2

,

actually occur, for suitable choice

for the integral curve.

a2

Figure 13.2.2.

144

THE SPHERICAL PENDULUM

13.

Proof.

Note first that, since K # 0,

f(-1) < 0

and

f(1) < 0.

It is

clear also that lim f(a) =

and

lim f(a) = w .

For motion to be at all possible, we must have must be at least one f

a E (-1,1)

such that

foz

and hence there

0

Hence the graph of

f(a) > 0.

must be of one of the possible types shown. To show that the case

a1 < a2

can be realized by suitable

choice of initial conditions note that from (1)

f (O) = 2e - K2

.

But it is clear from (4) and (5) that we can keep e

K

fixed while making

arbitrarily large by suitably choosing the initial conditions. and hence

f(O) > 0

gives

al < a2.

Our argument that the case from Section 13.1

indirect:

This

can actually occur is

al = a2

there are orbits of the "conical pendulum"

as shown later, these are not the sort of orbits obtained in the

type;

a1 < a2;

case

hence they must belong to the case

al - a2.

The following result completes our discussion of the possible types of orbits for the spherical pendulum. 13.2.3.

Proposition.

Suppose that the numbers

previous proposition satisfy the condition

and

al

a1 < a2.

a2

in the

There is then a

family of orbits which oscillate between the parallels of latitude ("apsidal circles") on which the

z-coordinate assumes the values

al and a2

respectively, as in Figure 13.2.3. Proof.

If

z

foz > 0

on

(a1, a2).

in this range.

lies between Thus

Now suppose

show that

z"(0) # 0

remain at

z = al.

al z

and

a2

then

cannot be zero since

z'

must increase or decrease monotonically

z(0) = a1.

Thus

z'(0) = 0.

We need to

for if this is not the case the particle would

From Lagrange's equations (3) we obtain the expression for z"(0)

Now

subject to our initial conditions a1Kz z"(O) = + (a12 - 1)g a12-1 f(a1) = 0

so from (1) we obtain

.

(6)

13.2.

z a1K2-1

145

OTHER ORBITS

- 2g(a1

)

g and on substituting this into (6) we obtain z"(0) = g(2a1(al - - ) + (a12-1)).

Figure 13.2.3.

(7)

Oscillatory base integral curve.

It is easy to see that the right hand side of (7) is just f'(al)

cannot be zero otherwise

would contradict our assumption that

/f'(a1).

would be a double root of

al

a1 < a2 < 1 < a3.

non-zero so that particle cannot remain at

Thus

f

z"(O)

But

which is

z = al.

Now suppose we have initial conditions given by a1 < z(s) < a2,

z'(s) > 0.

Suppose by way of contradiction that time.

Thus

z(t) < a2

whenever

does not reach

z

t > s.

increases monotonically in this range if

We deduced above that z' > 0

have

z' = / 7

# 0

on

z = a2

( a1,a2).

initially.

in finite z

So here we

146

THE SPHERICAL PENDULUM

13.

Thus

t

t-s

=

foz

a and so

z,

-z (t)

dx 11

z (s)

=t-s

(8)

29(x-a1)(x-a2)(x-a3)

The right hand side of (8) can be made arbitrarily large but the left hand side is bounded above by

a

dx

2

(9)

2g(x-a1)(x-(X2)(x-a3) al

which can be shown to be finite.

The argument which shows

z

Thus

reaches

reaches

z

a1

a2

in finite time.

is the same and the expression (9)

gives the half-period of the motion between

Cl

and

a2.

EXERCISE 13.2. A64ume

-1 < a1 < a2 < 1 < a3

and that

f(a) = 2g/(a-a1)(a-a2)(a-a3) = 2gl (a2-1)(a- 9 ) - 2g).

Uae etementLv y af-gebnx to ehow that a1 + a2 < o and ube thf.6 to thaw

that the mean o6 the two "ap6ida2 eL .c2e6" shown in FLgune 13.2.3 A. below the x-y plane. What doe,6 thi6 teft us about the "conicab pendulum" bah a £nteg .a.C cuxu e5 ?

14. RIGID BODY MOTION

In section 8.4

it was shown that the configuration set for the

motion of a rigid rod was a 5-dimensional submanifold of

R6

and

furthermore that the reaction (or "internal") forces were orthogonal to Thus the problem could be treated as one of a single

this submanifold.

particle moving on a 5-dimensional submanifold of

R6.

In this chapter the motion of a rigid body moving in

R3

is

considered and it is shown that the configuration space can be thought of as

R3 x SO(3).

Thus this problem reduces essentially to the motion of a

single particle on a 6-dimensional configuration manifold.

Furthermore the

reaction forces are shown to be orthogonal to the configuration manifold. The derivation of Lagrange's equations given in Section 8.3. thus applies to rigid body motion when the "external" forces are due to a conservative field of force.

14.1. MOTION OF A LAMINA. In section 8.4. we showed that Lagrange's equations were applicable to the problem of the motion of a rigid rod.

There the "constraint condition"

that the rod's length remained constant was sufficient to define the configuration manifold.

By way of introduction to the problem of the rigid

body we look at the motion on a plane of a "lamina" which we think of as being defined by three "point masses". be a subset of

R6

Its configuration set will clearly

and we will again require that the distances between

the particles remain constant.

These three constraints, however, do not

tell the whole story.

In Figure 14.1.1. we have two configurations which satisfy the same distance constraint conditions but if our lamina cannot "flip" (which is certainly the case because it is constrained to move in a plane) then these configurations cannot both be on the same configuration manifold.

14.

148

RIGID BODY MOTION

,(c1,,2)

(a, a,)

(a1,a2)

d (a,c)

Figure 14.1.1.

Thus there must be another constraint which implies the preservation of the "orientation" of the lamina.

This is obtained by the use of the vector or

cross-product:

(b1-al, b2-a2, 0) x (bl-c1, b2-c2, 0)

which defines a vector normal to the plane in which the lamina moves.

We

will insist on this cross product also remaining constant. The "distance preservation" constraint is f(a1,a2,b1,b2,c1,c2) = 0

where

f(a1,a2,b1,b2,c1,c2) = (u,v,w)

with u = (b1 - a1)2 + (b2 - a2)2 -d2a, b)

v = (cl - a1)2 + (c2 - a2)2 -d2

(a,c)

w = (b1 - a1)2 + (b2 - c2)2 -d2c,b) Thus our configuration set f-1(0)

Q

satisfies

Qe f-1(0)

is a three dimensional submanifold of

R6

.

It turns out that

which consists of the

union of two disjoint open subsets, one of which is

Q.

The details are

left as exercises.

An alternative way of describing the configuration manifold

which uses rotation matrices, is hinted at in the exercises.

Q,

This

alternative approach will be exploited in the next section where a direct description of the configuration manifold of a rigid body via constraints on distance and orientations becomes unweildy.

14.1.

149

MOTION OF A LAMINA

1.

EXERCISES 14.1. Read your /4olut on6 to Fxehaise4 3.3.2. and 3.3.3. and then phove the 4tatement6 made in the text concenni.ng the sets Q and f-1(0).

2.

Coni.i,deA the 6ottowing (distance pnese/w,ing) t4an.56onmati.on o6 a

lamina which cona.c6t6 o6 an anticlockwise notation thiwugh an angle 6 about one o6 the vexti.ce4, which we place at the oAigtn a-, in Figure 14.1.2.

(b1,b2)

Figure 14.1.2.

(a) Find a matAix A a uch that b' j

b2

=A

bjj b2C2

A fCl2j =

0 mathi.x (b) Do these conditions determine the

A

uni,queCy?

(c) Show that det(A) = 1. (d) Does thL6 than.66onmation pneoeAve oaientation? 3.

Now consider the (distance pneseiw.ing) taan46onmation o6 a lamina which con6i.6t6 o6 a ne6leati.on about one o6 .its o.Ldes, ao.6umed to

lie on the 6i ut axis, a6 in Figure 14.1.3.

14.

150

RIGID BODY MOTION

(bl,b2)

(bl,-

Figure 14.1.3.

(a) find the matt ix o6 th . .tican.o $onmation and Zt6 detenmi.nant. (b) Doea thL6 ttan46onmati.on paeaenve oni.entati..on?

14.2. THE CONFIGURATION MANIFOLD OF A RIGID BODY Our mathematical model of a rigid body is a collection of particles in

at least four of which are not coplanar.

R3,

n (>- 4)

They are to

be constrained in such a way that the distances between, and the relative orientation of, the particles remain constant throughout the motion.

We

regard these quantities as being preserved by means of rigid rods, of negligible mass, connecting each pair of particles.

To help us define the configuration set of the rigid body precisely, the following definition is introduced. 14.2.1.

A rigid motion is a map

Definition.

p

:

R3 --> R3

(a)

distances between pairs of points in

(b)

the orientation of triples of vectors in

which preserves

R3,

Algebraically, this definition means that the map

R3. p

is to satisfy the

conditions that

for all

a, b

(a)

Ilp(a)-p(b)II = Ila - bll

(b)

det(p(a),p(b),p(c)) - det(a,b,c)

and

c

in

R3

where

II

II

is the usual norm on

The configuration set can now be defined as follows.

R3.

14.2.

14.2.2. of

The configuration set

Definition.

n - tuples of vectors from

a2....... an

al,

of a rigid body is the set

Q

R3,

{(p(al), p(a2)....... p(an)):

where

151

CONFIGURATION MANIFOLD

p

is a rigid motion}

,

are the respective initial positions in

R3

of the

n particles constituting the rigid body.

We now aim at showing that the configuration set is a submanifold of

Ran.

Intuitively we can think of the configuration set of a

rigid body as consisting of a combination of translations of one of its points together with rotations about axes passing through this point, as illustrated in Figure 14.2.1.

i-

f- rotation --

translation

THIS

THIS

SIDE

SIDE

UP

UP

-

rigid motion

Figure 14.2.1.

A rigid motion.

Thus it would seem that the configuration set may well look like {translations in R3} x {rotations in R3},

which can be represented as the manifold show that this is so. 14.2.3.

Theorem.

there is some

Our task then is to

The key result is as follows:

A map

c E R3

R3 x SO(3).

p: R3 --

R3

and A E SO(3)

is a rigid motion if and only if such that, for all

p(a) - c + Aa.

a E R3,

152

RIGID BODY MOTION

14.

Proof.

It is rather long and technical and appears at the end of this

section.

In the following lemma, the element as a

3 x 3

14.2.4. Lemma.

Let

L

:

R3 x R3x3 _,, (R3)n

L(c, A) _ (c + Aal, C + where R3

n > 4

A E R9

is to be regarded

matrix.

and

with Aa2.... ,c + Aan)

are the position vectors of points in

a', a2,.... an

The map

not all lying in the same plane.

L

is then linear and one-

to-one.

Proof.

This is left as an exercise. By using the above map

L we can easily derive the following

theorem, which is the main result of this chapter. 14.2.5.

Theorem.

The configuration set

of n particles is a submanifold of Ran 6-dimensional manifold Proof.

Q

for a rigid body consisting

diffeomorphic to the

R3 X SO(3).

The idea is that the one-to-one linear map L

R3 x R3X3

:

Ran

gives a diffeomorphism L'

when restricted to the set

:

R3 x SO(3) - Q

R3 x SO(3),

L

Figure 14.2.2.

as illustration in Figure 14.2.2.

14.2.

CONFIGURATION MANIFOLD

153

The remaining details of the proof are left as an exercise.

We now go back and give a proof of Theorem 14.2.3, thereby completing this section.

Proof of Theorem 14.2.3. First, assume

is given by

p : R3 --> R3

p(a) = c + Aa

where that

c E R3 p

vectors.

and

It follows from Lemmas 3.4.3. and 3.4.5.

A E $0(3).

preserves distances between points and orientation of triples of Hence

is a rigid motion.

p

Conversely, assume L = p - p(0).

p

R3 -+ R3

:

It is easily checked that

orientations and that

L

is a rigid motion.

preserves distances and

Thus, for each

L(O) = 0.

Let

a E R3,

IIL(a) II = Hall and since, for all

a,b E R3,

Ila-b112 = I1all2 - 2a.b + 11b 112 we have

IIL(b)112 = Ilall2 - 2a.b + llbll2

and so

Now let

for

i = 1,2,3.

Then for each

and so

{u1,u2,u3}

is an orthonormal basis for

ui = L(ei)

and

dij

Now for each

i = 1,2,3

i,j,

I!u11l = 1

R3.

a,b E R3,

and

L(a).ui =

ai

=bi

L(b).ui and so

(L(a + b) - L(a) -

0

which gives L(a + b) = L(a) + L(b). Similarly

L(Xa) = XL(a)

represented by some

for each

3 x 3

matrix

(ATA).. =

Now since

L

A E R.

Thus

L

is linear and can be

A whose columns are and so

u1,u2,u3.

Thus

ATA = I.

preserves orientations we must have, for three linearly

independent vectors

(al, a2, a3), (bl, b2, b3)

and

(Cl, c2, c3),

154

14.

sgn

RIGID BODY MOTION

b1

C1

b2

A e2

b3

C3

= sgn

al

b1

c1

a2

b2

C2

a3

b3

C3

Now letting

I Z = a2 b2 c2 a3 b3 c3

we have sgn(IAZI) = sgn(IAI) sgn(IZI) = sgn(IZI). Thus

sgn(IAI) = 1

sgn(IZI) # 0

since

A E SO(3).

and so

But

p(a) = p(0) + L(a) = p(0) + Aa which shows that the map

1.

2.

p

has the desired form.

EXERCISES 14.2. Show that -L p1 : R3 -+ R3 and p2 : R3 --+ R3 ane h,Lg.Ld motLonb then so .ins the composite p1op2 : R3 -} R3.

Show that i6 an n-.tupee ob vectonb (b1, b2,...,bn) bnom belongs to the con6igunatLon set o4 a Aig.id body then the conbigunatLon set may be whitten as {(p(bl), p(b2),..., p(bn):

3.

Prove Lemma 14.2.4.

To show L

R3

p is a rigid motion}.

.i.a one-to-one, use the tact that .Lb

R3, say a1, a2, a3, a", are the pos.i tLon vecton6 ob boon non-copeanan po-.nt6 then the di66elcencu al-a2, a2-a3, a3-al {owl. etement6 o{

bonm a tc.neaAty independent set ob vec-tons. 4.

Compete the pnaob ofi Theorem 14.2.5. by using Lemma 4.3.4.

14.3. ORTHOGONALITY OF REACTION FORCES The methods employed here to study the reaction forces within a rigid body are basically the same as those used to study the corresponding problem for the rigid rod in Section 8.4.

Here, however, the computational details

are more complicated and it will be convenient to introduce some extra notation to describe the constraints between the constituting the rigid body.

n

particles

14.3.

REACTION FORCES

155

To this end we write a typical element

a E Ran

in the form

a = (a1, a2,...,a) 1 < i < n,

where,for

is an element of

ai

which we write as

R3

a2 = (al, a2, a3). We think of

ai

furthermore, let

1 < i < j < n,

ith

and

jth

as the point occupied by the

d.,

ith

particle in

R3.

For

denote the distance between the

particles and let Ran -+ R

f.

fij(a) =

(ai-al)2 + (a2-a2)2 +

;jn(n-1)

f: Ran -> R

Finally, let

components are the

with 17)2

- d. 2

(1)

be the map whose respective real-valued

maps

112(n-1)

f 13,.........., fin

f 12,

f23,.......... ,f2n

fnn-1

Since rigid motions preserve distance, it follows that the configuration manifold

of the rigid body satisfies

Q

Q c f-1(0).

(2)

In the following geometrical lemma, notation is used for the components of

h E Ran

which is analogous to that used for

a

above.

14.3.1. Lemma. If (a,h) E TQ then, for 1 < i < j < n, (a2 Proof.

Assume

for each

t

so that

(a,h) E TQ

parametrized curve

y

in

(a2

Q

at

-

h = y'(0)

for some smooth

a, by Definition 5.1.2.

in an interval containing

But by (2),

0,

(f -Y) (t) = 0 and hence by the chain rule Df(a)(h) = Df(Y(0))(Y'(0)) = D(f°Y)(t)(1) = 0.

156

RIGID BODY MOTION

14.

In terms of the components of

this means that,

f

for

1 < i < j < n,

Df..(a)(h) = 0 The explicit form

now gives the result.

(1) of the map fij

As to the reaction forces between the particles constituting the rigid body, observe that the force the

jth

(1 < i