Newtonian Dynamics 0070030162

Table of contents :
Front Cover
Contents
Preface
1. A Reveiw of Some Basics
1.1 Qualitative Reasoning
1.2 Vectors and the Scalar Product
1.3 Tossing Rocks
Tossing up the Hill
1.4 A Consequence of Newton’s Second Law: Energy
F and grad U
Cartesian Representation
1.5 Gravitational Potential Energy
1.6 Potential Energy and Stability
1.7 Some Implications of Invariance
Translational Invariance
Rotational Invariance
Invariance under Time Translation
1.8 Newton’s Laws: Structure and Meaning
1.9 Reviewing the Review
Worked Problems
Problems
2. The Harmonic Oscillator
2.1 Damped Harmonic Oscillator
Damped Oscillations
Energy
2.2 Phase Space: An Introduction
2.3 Harmonic Oscillation in Two Dimensions
2.4 Sinusoidally Driven Oscillator
The Lortentz Atom
2.5 Quality Factor Q
2.6 A Guide to the Major Ideas
Worked Problems
Problems
3. Nonlinear Oscillators
3.1 A Nonlinear Oscillator
Small Oscillations
Phase Space Again
Instability
Summary of Principles
3.2 Amplitude Jumps and Hysteresis
Duffing’s Equation without Damping
3.3 Van der Pol’s Equation: The Limit Cycle
3.4 More Uses for the Averaging Method
Newton’s Law of Damping
3.5 Series Expansions
Secular Terms
A Note About Appearances
Synopsis
3.6 About the Methods
Worked Problems
Problems
4. Lagrangian Formulation
4.1 Fermat’s Principle
4.2 Calculus of Variations
The Economical Lampshade
4.3 Newton II as an Extremal Principle
Newton II in Cylindrical Coordinates
4.4 Lagrangians and Constraints
Constraints
4.5 Another Instance of Constrained Motion
Equilibrium
Interlude on Newton II
Stability
4.6 Conversion to First-Order Equations: Hamilton’s Equations
Examples
Examples Continued
The Example Again
4.7 Liouville’s Theorem*
4.8 The Lagrangian and Quantum Mechanics
4.9 A Sense of Perspective
Worked Problem
Problems
5. Two-Body Problem
5.1 Reduction to Motion in a Plane
Plane Polar Coordinates
5.2 Effective Potential Energy
Gravitational Attraction
Spring Force
5.3 Orbit Shape
5.4 Orbits Around the Spherical Sun
The Typical Bound Orbit
Classifying the Objects
5.5 The Oblate Sun
5.6 Stability of Circular Orbits
How a Pertubation Evolves in Time
Stability Criteria
5.7 The Orbit in Time
5.8 Compendium on Central-Force Motion
Worked Problems
Problems
6. Rotating Frames of Reference
6.1 Vectors in a Rotating Frame of Reference
Uniform Rotation
General Rotation
Illustration
The Primary Derivation Resumed
6.2 Physics on a Rotating Table
6.3 The Rotating Earth
The Plumb Bob
Motion near the Earth’s Surface
The Wind
6.4 Foucault Pendulum
6.5 The Figure of the Earth
Geopotential
6.6 A Perspectie on Rotating Frames
Worked Problem
Problems
7. Extended Bodies in Rotation
7.1 Equations for Location and Orientation
Equation for Location
Equation for Orientation
More about Torque
7.2 Simple Precession
The Dust Grain: An Example
7.3 How L is Related to ω
Familiar Ground
7.4 A Novel Pendulum
Fixed Axis of Rotation
7.5 L is not Necessarily Parallell to ω
7.6 Diagonal Form for the Inertia Tensor
7.7 Euler’s Equations for a Rigid Body
7.8 Axisymmetric and Torque-Free
Evolution of ω
Orientation of the Disk in Space
ω as Seen from the Inertial Frame
Summary
The Dust Grain Again
7.9 Chandler Wobble
7.10 An Interlude on Kinetic Energy
7.11 The Symmetric, Supported Top
Conserved Quantities
How the Top’s Orientation Changes
Extracting the Implications
7.12 Precession of the Equinoxes
7.13 Survey of the Critical Notions
Worked Problems
Problems
8. Cross Sections
8.1 Scattering Effectiveness: The Idea Behind the Cross Section
8.2 A Capture Cross Section
8.3 A Differential Cross Section
8.4 Rutherford Scattering
8.5 Major Ideas
Worked Problems
Problems
Appendices
A. Expansions, Identities, and Miscellany
A.1 Expansions
A.2 Trigonometric Identities
A.3 Miscellany
B. Vector Product
Components and Orthogonal Unit Vectors
Some Identities
A x B as an Axial Vector
C. The Averaging Method
D. The Craft of the Physicist
Index
Back Cover

Citation preview

-

EM

NEWTONIAN DYNAMICS

Ralph Baierlein Charlotte A yres Prof essor of Phys ics Wes leya n Uni, •ersity

McGraw-Hill Book Company New York St . Loui s San Francisco Au ckla nd Bogota Hamburg Johanne sb urg London Madrid Mexic o Montrea l New De lhi Pa nama Paris Siio Paulo Singapore Sydney Tokyo Toronto

This book was se t in Times Roman by A Grap hic Meth od Inc. The ed itors were John J . Co rrigan and J ames S. Amar; the production supervi sor was Leroy A. Young . The drawing s were done by Allyn-Ma so n. Inc. The cov er was de signed by Robin Hesse l. Hallida y Lithograph Co rporation was printer and binder .

NEWTONIAN DYNAMICS

Co pyright © 1983 by McG raw-Hill , Inc. All right s reserved . Print ed in the United Sta tes of America . Exce pt as permitted under the United States Co pyright Act of 1976. no part of thi s publicatio n may be reprodu ce d or distributed in any form or by any mea ns, or stored in a data base or retrieval sys tem. without the prior written permis sion of th e publi sher. 234567890

HALHAL

89876543

ISBN 0-07-003016-2 Library of Co ngress Ca taloging in Publica tion Data Baierlein , Ralph. Newtonian d ynamics. Includes index. I. Dynamics. I. Title . QA845 .B33 1983 53 1'. I I ISBN 0-07-00301 6-2

82- 17944

To MY FAMILY Je an , Eric, and Jeffrey

CONTENTS

Preface A Note to the Problem Solver Chapter 1 1.1 1.2 1.3 1.4 1.5 1.6 I. 7 1.8 1.9

Chapter 2 2.1 2.2 2.3 2.4 2.5 2.6

Chapter 3 3. I 3.2 3.3 3.4 3.5 3.6

Chapter 4 4. I 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9

VII IX

A Review of Some Basic Qualitative Rea oning Vectors and the calar Product To ing Rock A on equence of ewton' Second Law : Energy Gravitational Potential Energy Potential Energy and tability ome Impli cation of Inv ariance ewton' Law : Structure and Meaning Reviewing the Review

20 23 27 30 34

The Harmonic Oscillator

45

Damped Harmonic O cillator Pha e Space: An Introduction Harmonic Oscillation in Two Dimen sions inu oidally Driven O cillator Quality Factor Q A Guide to the Major Ideas

45 54 56 61 63

Non linear O cillator

71

A onlinear O cillator Amplitude Jump and Hy teresi van der Pol' Equation: The Limit Cycle More U e for the Averaging Method erie s Expan ion About the Method s

71

97 104

Lagrangian Formulation

116

Fermat' Principle alculu of Variation ewton I I a an Extremal Principle Lagrangian and on traints Another In tance of Con trained Motion onver ion to First-Order Equations: Hamilton 's Equations Liouville' Theorem The Lagrangian and Quantum Mechanic A en e of Per pective

116

I 7 8 15

51

81 88 93

I 18 123 128 I 31 136

140 144

146 V

vi

CONTENTS

Chapter 5 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8

Chapter 6 6.1 6.2 6.3 6.4 6.5 6.6

Chapter 7 7. 1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7 .1 1 7 . 12 7 . 13

Chapter 8 8.1 8.2 8.3 8.4 8.5

A B D

Two-Body Problem

156

Reduction to Motion in a Plane Effective Potential Energy Orbit Shape Orbit s around the Spherical Sun The Oblate Sun Stability of Circular Orb its The Orbit in Time Compendium on Central-Force Motion

156 161 164 168 175 181 187 189

Rotating Frame of Reference

202

Vector in a Rotating Frame of Reference Physics o n a Rotating Table The Rot ati ng Earth Foucault Pendulum The Figure of the Earth A Per pective on Rotating Frames

202 209 210 217 223 228

Extended Bodie in Rotation

235

Equations for Location and Orientation Simple Precession How L Is Related tow A Novel Pendulum L l s Not Neces arily Parallel tow Diagonal Form for the Inerti a Ten or Euler's Equation s for a Rigid Body Axisym metric and Torque-Free Chandler Wobble An Interlu de on Kinetic Energy The Symmetric, Supported Top Prece ion of the Equinoxe urve y of Critical Notion

23 240 244 247 249 251 254 25 260 264 265

Cross Sections

286

Scatterin g Effectiven es : The Idea Behind the Cro s Section A Capture ros Section A Differenti al Cr o s Section Rutherford Scattering Majo r ldea s

286 289 291 295 300

Appendixes

30

Expansion , ldentitie , and Mi cellany Vector Product The Averaging Method The Craft of the Physici t

308 311 317 321

Index

324

272 275

PREFACE

A book on clas ical mechanics for junior and senior is more than ju t that. The topics cho en and the method u ed reveal the author's attitude toward the subject. When I look back at what l empha ized in class and then in writing the book , I find a tre s on analytical methods , on techniques that are useful throughout phy ics and engineering - thing uch as expan ions , dimensional analysi , sketche (rat her than just equations ), and tability analysi . These are the tock in trade of an a trophysici t, say, and of anyone who i working on the back of an envelope . I have written with the aim to free more of class time for doing problems , for di c u sion , and for glo s. To this end , the text develop the theory with enough word to upply motivation and context. A difficult piece of theory is immediately followed by an illu tration of how the principle work in practice. Thus one can ask student to read sections before the topic arise in class. While thi may be uncommon , the benefit are great : both a richer use of class time and a maturing ability to under tand cience directly from the printed page. Some les central theoretical developments have been left to the homework problems for two reason . Fir t, a text can become overburdened with informatio n. Then readers have an unnece arily difficult time as imilating the more ba ic part of the subject. Second , nothing sinks in like a problem tackled and olved. Those homework problem s that extend the theor y in any notable way have been flagged in the margin with a superscript T , like this: T Equations that are e pecially notable , either as tarting points or as results, are starre d in the margin : Something should be said about the sections that are essential for continuity. Within a given chapter , the progression and prerequisites are I hope , clear. It is the continuity from one chapter to another that merits an outline . Chapter

*

vii

Viii PREFACE

I, as a whole , is essential. Then come sections 2.1, 2.2, and 3. 1. These sections introduce the harmonic oscillator, phase space, and motion near an equilibrium point. They provide the basis for stabi lity analysis and "s mall oscil lations" theory , and thus they lay the foundation for frequent applications in the succeeding chapters. No subsequent material depends in an esse ntial fashion on Chapter 4. Portions of Chapter 5 are needed in the last chapter, on cross sections, specifically Sections 5.1 through 5.4. Section 6.1 is a prerequi site for Chapter 7. Even after so me picking and choosing , there is enough to occupy a class productively for a se me ster . An equation with the struct ure x = - w/x + J(x, i), where J(x, x) incorporates all the complicated terms, ar ises surprisingly often in mechanics and elsewhere in science. When / has only a small influence on the motion during any single time interval of order 2TT/w0 , a general method for solving the equation can be given. The method, called here the averaging method , is derived in Appendix C and is used in three contexts: nonlinear osci llators in Chapter 3, perturbed planetary motion in Chapter 5, and the Foucault pendulum in Chapter 6. (Ho mework problems illustrate its usefulness still elsewhere , for example, in lase r physics.) One can pick up the method the first time one needs it. Every instru ctor uses personal experience as a guide in selecting some topics from a text and omitting others. Still, it may be useful to ask, If I were strapped for time, what would I omit? Formulating a reply is painful. Nonetheless, let me specify massive omissions (so that there may be space to reinstate a few topic s). I would omit Chapter 3 on nonlinear oscillators, except for Section 3.1; Sections 4.6 and 4.7 on Hamilton 's equatio ns and Liouville's theorem; Section 5.6 on the stabil ity of circular orbits; Sections 6.4 and 6.5 on the Foucault pendulum and the figure of the earth; Section 7.11 on the symmetric, supported top; and Sectio ns 8.3 and 8.4 on differential cross sec tion s. My thank s for help and advice go to many people : to my classe at Wesleyan; to Janet Morgan , George Zepko , a nd D av id Todd in the Wesleyan Computer Center; to my colleagues Robert Behringer , Richard Lindquist, Ronald Ruby, Peter Scott , and Alan Spero . Appe ndi x Dis an adapted version ofan article that Ron Ruby and I wrote; without our conversations, the appendix might never have been thought of. Peter Scott allowed me to include some of his homework problems, for which I tender my special thanks . Alan Spero offered to teach from a preliminary edition. His experience was a rich source of comments; I have incorporated many of Alan' s suggestions and express here my appreciation. A solutions manual for instructors is avai lable from the publisher upon request. Often readers hesitate to write the author. Please do not hesitate . I will be grateful for corrections (ty pographica l or otherwise), for suggestio ns, and for comments .

Ra/pl, Baierlei11

A NOTE TO THE PROBLEM SOLVER

There are two basic approaches to solving homework problems. You can read the chapter from epigraph through the end of the assigned sections, look at the worked problems , and then tackle the homework problems. Or you can start with the homework problems and flip through the chapter for relevant equations or sections. Thi s book is designed for the first approach . A colleague taught from a preliminary edition; here are so me of his comments, made at the semester's end. This is a book for reading - each chapter should be read in its entiret y and ideally in one sitting. The student s who realize that will benefit the most. ... In the preface. tell your reader s that they must read eac h chapter throu gh comp letely first. Then they can go back and read sections. and parts of sec tions, when doing the problems. Mayb e even advi se them to make a flowchart or compendium of ideas as they read .

An entire chapter at one sitting is severe- I usually assign half a chapter at a time- but the general drift makes exce llent advice: read first.

ix

CHAPTER

ONE A REVIEW OF SOME BASICS

1.1 Qualitative rea oning

1.2 Vector and the calar product ing rock 1.4 A con equence of Newton ' second law: Energy 1.5 Gravitational potential energy 1.6 Potential energy and stability 1.7 ome implication of invariance 1.8 Newton ' law : tructure and meaning 1.9 Reviewing the review

1.3 To

If we pass in review the period in which the development of dynamics fell - a period inaugurated by Galileo, continued by Huygens, and brought to a close by ewton - its main result will be found to be the perception that bodies mutually determine in each other acce ler ations dependent on definite spatial and material circum lances and that there are ma se . Ernst Mach The Science of Mechanics

1.1 QUALITATIVE REASONING uppo e you drop a small rubber ball from the highe st building accessible to you. At what rate doe it fall and how doe s the distance fallen depend on time? To an wer the e que tion we need to know the force that act on the ball. As sketched in figure 1.1-1, they are gravity and air resi tance. Here of course, m denote the ball' mas and g denotes the force per unit ma s exerted gravita-

2

A REVIEW

OF SOME BA SICS

F air re,slstance

F-.,a v11y

= mg Figure 1.1-1 The forces on the ball.

tionally by the earth. The fo rces go into Newt on's second law, which her e asserts d dt (mv)

= mg -

F air resistance·

( I. 1-1)

The dow nward direction ha s been taken as positive. We need to assess the force due to air res ista nce. Look at the is ue from the point of view of momentum gain and loss. The moving ba ll set s air into motion, impart momentum to the air it disturb s. At what ra te does the ba ll transfer mo mentum to the ambie nt air ? In a time tl.t. the ball swee ps out a cylindrica l volume in space equ a l to 7TR 2 v tl.t, as ske tched in figure 1. 1-2, and o the ball affect s an amount of air apR 2 v tl.t. (H ere R denote the ball's radiu s, a nd Pair proximately eq ual to Pair7T denotes the density of air .) To the mass of air, the ball give s a velocity of ord er v, a nd so R 2 v2 tl.t. Mome ntum given to air in tl.1 = Pair7T

T he ga in in momentum by the a ir co mes at the cost of momentum lost by th e ball. Thu s the temporal rate of mome ntum tran sfer goe s into Newton 's seco nd law as F air resl~tance: ( 1.1-2) wher e # denote s a dimen s ionless co nstant of order I .

T t

At

J_ Figure 1.1-2 The volume of space swept out by the ball.

I. I

QUALITAT IVE REASO

I G

3

Equatio n ( I . 1-2) is our basic wo rking differenti al equation . Rat her th an try to solve it immediately, however, we should u e it first to infer the qua litative behav ior of the motion . A dropped ball starts from re t and he nce moves lowly at fir t. Because the a ir re si ta nce term is quadrati c in the ve loci ty v, it is in significant at sma ll v; the speed mu st build up approx imately as v = g t. But as v grows, so doe s th e air re sistance. The net force , give n by the right -hand side of equation ( l. 1-2), decrea ses tow a rd zero; thu s the rate of grow th of v decre ases. There is a characteri tic spee d for which the net force would be preci se ly zero: _ (

v, =

mg

)112.

( 1.1-3 )

R2

#P a1r7T

As v approaches v1, the net force approache s zero, impl ying that v will ne ver exce ed (or eve n reach ) the " terminal velocit y" v1• Th e evo lution of v is depicted in figure 1.1-3 . On ce we kn ow how v depend s on tim e, at lea st qu alit ativel y, we can com pute the distanc e fallen. In the beginning , w hile v = gt holds, the di stance w ill increase approximately as ½gt 2, a qu adrati c increase with time. The sit uatio n is quite different when v get s clo se to its te rmin al va lue v1 and hen ce changes little with time. Then , with v = v1, di stance will inc rea se on ly line arl y with time , though at a high rate . Th e full run of beha vior is indi ca ted in figure I . 1-4. Now let us go back and solve e qu ation (1.1-2) a nalytica lly . On ce characte ristic qua nt itie s have emerge d , it is wise to displa y them prominently ; so we should rew rite equ ation ( 1.1-2) in terms of v1 , the terminal ve loci ty:

d mg 2 m v)=dt V/ (v I -v

-(

t,

2 ).

( 1.1-4)

---------=~----------

Figure 1.1-3 Th e evo lution of v. as pred icted by qualitative reaso ning. Th e das hed, interpolating section ca n be inferred from the co ntinu ous decrease in slope, predicted by eq uation ( 1.1-2), as v grows toward v,.

4

A REVIEW

OF

OME BA I

= L' I

constant

1

/ /

/

/

/ /

/

/ 1 (on

= ½gt"

a co mpre ssed cale)

Figure 1.1-4 The di tance fallen. a inferred from qualitative reasoning.

Only velocity , time, and constants appear: no function of position per se. Hence we can try to solve immediately for If we transfer all dependence on to the left-hand side , we have

v(I).

dv

v,2- v2

=_K_ dt

(t.1-5)

vi2 ·

To integrate the left-hand ide, we can expand in partial fraction ,

----,--- = I /(2v1) + I /(2v1) v,2-

V2

V1

+V

Vt -

,

V

and then integrate each term to get a logarithm. Upon combining the log arithms, we arrive at

(v,+v)= -gt

-2I In -Vi V1 -

V

Vt

+ constant.

2

Because t = 0 is intended to imply v = 0 (and because In I = 0), the constant i actually zero. Finally, we can olve for v by transferring 2v1 to the right -hand ide, taking anti-logarithms, and rearranging terms : e2nt1v1 _ V

=

e2nt/u ,

I

+ I v,.

( 1. 1-6)

We can check this result. When t ~ vtfg, the numerator and denominator are dominated by the exponential terms , and o v = v,. To deduce the behavior

I.)

at early times when t

t Jorce): No force acts on a body if it moves with constant velocity in an inertial frame . l s there no way to break the circ le? Perhap we can defin e a special ca e of no net force without reference to an inertial frame. The prescription might be • Albert Einstein and Leopold lnfeld, The Evo/11tio11 of Phys ics (Si mon and Schuster. New York, 1961).

32

A REVIEW

01-' SOME B SIC

this: Get far away from all nearby bodie an d locate your elf ymmetrically relative to the large- ca le univer e. Our experience is that phy ical forces dimini h with di tance; that plus ymmetry may uffice to e tabli h a 'no net force" situation. Then we may turn Newton I into a definition of an inertial frame . If we can establi h one frame as an inerti al frame , we are all et. We can test any other frame by asking, Doe it move with con tant veto ity relative to thee tabli hed inertial frame? If yes, then it, too , i an inertial frame . Ifno. then it is not. Whether this trategy i logically soun d is debatable. Certainly it is not practical. What, indeed, doe the practical person ay? Structural engineers say that the earth is an inerti al frame, a far as the y are concerned. Meteorologist ay no, the trade winds and a hurricane are understandable only if we recognize that the earth rotate relative to an ideal inertial frame. A frame fixed with respect to the nearby star will, however, erve a an inertial frame for meteorologi t . But radio astronomers mapping the Milk y Way by the radio emi ion of hydrogen or carbon monoxide find that the galaxy rotate relative to an ideal inertial frame. For them, the local tar do not suffice as an inertial frame, but a grid based on the nearby galaxie would be sufficiently inerti al. Here we have variou approximations to an ideal inertial frame, indeed a hierarchy of them. The practical per on merely pick a frame of suitable quality and continue working. Whether the ideal inertial frame even exist i a question that the practical perso n can defer. We defer the question a lso. T he issue have been rai ed. The notion of an inertial frame is a topic more for di cu sio n than for definitive expo ition and that is true for all three of Newto n laws. Within the community of pby sici ts and philo ophers. there is no unani mit y on how the foundation hould be conce ived. Now we go on to Newton' seco nd law. We uppo e we ha ve at lea t a practical approximation t an inertial frame. Then Newton 11 de cribe the effect of a physical force when it is observed in an inertial frame. If we want to, we may regard Newton 11 a defining the force F. The u efulnes of Newt on II then lie in the empirical property that force can be con tructed out of po ition and velocitie , with no need for higher derivative ; and after the force have been constructed in one context, they may be u ed in another. We have, however, a natura l sense of a force as a push or a pull. We pu h a wing or pull a stake out of the grou nd and know that we are exerting a force. The import of Ne wton I I i clearer, then. The physical force determin e d(mv)/dt, rather than some other kinematic quantity , such a c/3(mv4v)/d t 3 , which i ab tractly conce ivab le. Newton ' third law was a n exce llent a pproxim ation for the phy ic of hi day . When objects a re sharp ly acce lerated or move at high speed, however the approximation break down. An electron hot clo e to an atomic nucleus will

1.8 N EWT O 's

LA W S:

TR UC T U RE A

D M EA

I

G

33

Fi cm 2:

/

• move q 2 here

I I I I

----

. q,

f.-,~

I

q, • ___

..,

Figure 1.8-1 A violation of Newton 111 (for a time of o rder t ic).

be abruptly accelerated and will emit electromagnetic radiation, a photon , say. Momentum i con erved and the photon itself carrie off some momentum. The um of the electron and nuclear momenta mu t change, but that could not occur if Newton III were literally true. The field - electromagnetic and gravit ational - may ab orb and contribute momentum , and thereby they invalidate Newton I II , which consider only the masse a carrier of momentum. Figure 1.8- I hows another in tance in which Newton 111breaks down. The two po itive charge q 1 and q2 have long been at rest and exert electric force s on each other that are equal but opposite. Now move q2 quickly to the new location. Since the electric field produced by q 1 points radially away from qi, the new force on q2 differ s from the old. The charge q., however , cannot 'know" anything about q2 ' motion until a time t' /c has elapsed , because change in the electric field propagate with a finite speed , the peed of light c. For an interval t'/c the force on q 1 " due to q 2 " remains unchanged, but the force on q2 " due to q," ha been changing. Obviously Newton 111fails here . The quotation marks suggest why: We really have three entities-the two charges and the electric field. Whenever field are changing , the amount of momentum they carry may change , and Newton III may be violated . The effect is vital ; without it, we would have no electromagnetic communication and could not even see. Nonethele , Newton 111is an excellent approximation in mo t terre strial and a trophy ical context where one is at all inclined to invoke it. The que tion raised in thi ection beg to be followed up . Here are several route s. Surely it i proper to tart with the author of the three law : Sir I aac Newton , Th e Math emati ca l Prin cipl es of Natural Philosoph y , as translated by Motte from the Latin original. (The original , with the title Phil osophia e Na turali s Prin c ipia Math em atica , was published in 1687, while Samuel Pepy wa president of the Royal ociety and only after Newton had been prodded into publi hing by Edmund Halley. ) Critique s of Newtonian mechanics that are now classic were written by Ern t Mach , Th e S cfr nce of M ec hani cs (Open

34

A REV I EW OF

OME BASI

Cou,1 Publi shing , La Salle, 111., 1960). and by Henri Poincare Science and H ypothesis (Dover New York 1952). The law of inertia, another name for Newton I, is the focus of a paper by G. J. Whitrow, Bri1 . J . Phil. ci .. 1, 92 ( 1950). Whitrow provide a good se n e of how many different view and interpretation a ingle law can engender. Some reas urance may be found in the paper "La w of C la ical Motion : What's F ? What' m ? What ' a?' by Rob ert Wein tock, Am.J. Ph s., 29,698 (1961). Finally. Mary Hes e provide an annotated bibliogr ap hy in " Reso urce Letter on Philo ophical Foundation of Cla ical Mechanic ," Am. J . Phys., 32, I (1964).

1.9 REVIEWING THE REVIEW A review a long and com plex a this chapter deserve a review it elf . Here i a recapitulation of the major point . There are two geo me tricall y defined ways to multiply vector : the calar and the vector product. With the ca lar product we can decompo e a ector equation into a et of equat ion in the component ; we take the ca lar produ ct with unit vector uch as y,and i . In general , we can determine the amount of a vector (A, ay) alo ng a pec ific dir ec tion (B, ay) by forming a calar produ ct: A · B = IAI co (A, B). The vector product figure prominently in the definiti on of angular momentum, r X mv, and, in general, whenever the notion of r tation ari e . Vector s provide a natural language for writing Newton 's second law :

x,

d

J/mv)

= F.

T he derivative of a vector with respect to time mean " vector at , t:,.1 min u vector at I , all divided by t:,.1, with the limit t:,.1- 0 then taken ." (Bec au ew e take the difference f two vector , which it elf i a vector, the re ult of the lim it proces i al o a vector.) Sometime we tay at thi relatively ab tract le el. a when writ ing v = dr/dt . At other time . we expres the vector in terms of un it vector and component , ap ply the definition to th at exp re ion. and emer ge with derivatives of the component and of the unit vector (which need not be temporally con Lant). Taking the calar product of N ewton II with an infinite imal displacemen t a nd then summing along the object' path leads to the idea s of kinetic energy and work. T hu . it fo llows from N ew ton 11that the c ha nge in kinetic energy i equal to the work done by the force actually ac ting on the ma . Often the work can be computed as a change in a potenti al energy . Th en we arrive at energy con ervation: The um of the kinetic and potential energie rema in constant. Beyond that , we can compute the force from the gradient f U: F = - grad U. By definition , grad U is a vector formed by finding the dire ction in which U change the fa te t. The vector grad U ha that direction. and it magnitude i that maximum patia l rate of change .

WORKED

PROBL: MS

35

The pot en tial energy may be co mput ed in tw o way . They are not I gically distinct; the econd method fo llow fro m the fir t, but i often pr efer red beca use it i ea ier to implement. I. We may co mput e U from the defin ition , a - I t ime

a line integral of the force . take n from a fiducial point to th e locatio n where we want U. T hi i di played in eq uation l .4-4). 2. We may plit the force Finto cont ributi on for w hic h th e integral i ea y or known. compute the correspo nd ing co ntributi ons to the total pote nti al energy. and then add t ho e contr ibuti on . The additi o n is ju t the add iti on of ca lars, not vector , and o i often ea ily done. (We u ed this method in ection 1.6.) We tou hed on the idea of a table eq uilib rium . (A sy te matic tudy co me later. in ect ion 3.1, and in tances recur throughout th e boo k.) The que tio n of tability i u uaJly und erstood to a k, l there a re toring force for a ll po ible mall di placement ? The an wer we found. i ye if-a nd on ly if- the pote ntial energy U ha a minimum at th e point in q ue tion. The con ervation law are intimately co nnect ed with invaria nce propertie . If the poten tia l e ne rgy doe no t change num er ical val ue und er so me mentally conceived motion-a tran lat ion o r a rotatio n - t hen the corre ponding "mo mentum" com ponen t i con erved : a compone nt of linear momentum for a trans lation , a component of a ngul a r momentum fo r a rot atio n. Last. but not lea t, come Newton ' three law . They can be stated s uccinctly , but a thoughtful inspecti .on raises que tion . H ow do we avoid circ ularity in the definit ion ? How do we elect an inertial frame of reference ? When will ewton third law be va lid or at lea t an ade quat e ap p roximation? And other question ari e. o r ome of the e que tions, sectio n 1.8 provide an an wer or a partial an wer. Other que tion remain as foo d for th ought or ubjects for debate . The c hapter a l o illu tra ted ome problem-solving te c hniqu es and commented on th em (a in the final paragrap h of ectio n 1.1 a nd 1.3) . The topi c of how be t to tackle a problem is a wide one , but there are some technique to bear in mind or to turn to whe n in de s pair. A number of those tec hnique s are collected in appendix D. We h ave not met a ll of them yet but now is a good time for a fir t reading of that appendix.

WORKED PROBLEMS WPI-1 The force Fi pecified to at i fy the co ndition s under which a potential energy exist . l n co mpone nt form, th e force is F

with

= F.cx + Fyy

F x= Fo(x

+ by)

( I)

36

A RE IEW OF

OME BASI

F11= bF 0

and

+ by),

where F0 and b are con tants . Problem. Determine the potential ene rgy function U( r ). Then ketch c ntour of con taat U . According to equation (1.4-4), we need to evaluate an integral of the form U( r )

=-

J

(.r,u)

(o.ol

F(r ') · dr '

+ U(O).

(2)

Becau e the integral is pecified to be independent of the path we take be twee n the origin and the poin t (x, y), we ca n simplify our calculation by ch o ing a path such that dr runs parallel to the carte ian coordinate axe . Figure WP 1- 1a how a good c hoice for the two leg that compri e the tot al path . On the fir t leg we have dr ' = i dx' ,

and o .r,O)

-

~

U (0), the mass may pass through the spatial origin , and so the trajectory in phase space extends into both half planes , x > 0 and x < 0. The kinetic energy is largest when the mass passe s through the potential minim a; hence the trajectory bulges out to ward large !xiat those times.

78

NONLINEAR

OSC IL LA T O RS

The behavior when E = U (0) is subtle . Suppose we place the mas s at rest at the positive location x 0 where the U curve intersects the E = U (0) curve. That is the positive turning point. Beca use -(d U/dx)x. is negative , there is an inward force , and the mas s starts inward , picking up speed as it goes (at le ast for a while). How long does it take the mass to reach the origin? Equ ati on (3.1-12) was derived to answer this kind of question. Taking the potenti al energy U(x)from equation (3. 1-2) and remembering that E = U ( O) here , we can write

-

- +

k t'o - t' f'

X

2[1_4 /'2 ( t'o t' o -

t')

X

2]•

Since {t' 0 - t') > 0 in the present context , we may write Time to travel ) ( from x to x= 0 =0

[

mt' 2k(t' 0 -t')

] 112 J o

dx

xo

[i- 4f'2(f'ot' x2- t') ] 0

X

112 •

We need to examine the integrand near x = 0.The square root approaches l , but the factor° 1/x blows up badly; indeed , the integral diverges logarithmicall y. Formally, that means an infinite travel time . A better way to put it is this : the mass continues to approac h the origin indefinitely. At an y finite time , the mass is still some dista nce away , but creeping inward. We ca n find the reas on· for this beh avior if we examine the force that slows the particle . The force. whos e magnitude is given by the slope of the potenti al energy curve , gets progress ive ly weaker as the particle approaches x = 0. The or igin in phase space cor res pond s to x = 0 and x = 0, that is, to no mo tion. Because the retarding force gets progressively weaker , the stopping time can be indefinitely long, and the mass never reac hes the origin. The situation at a turnin g point suc h as x 0 is quite different. There the slope dU /dx is nonzero ; there is a nonvani shing force ; and the turning-around process is accomplished in a finite time . Although the integran d will diverge as U (x) approaches E, the integral will co nverge to a finite time . But back to the neighborhood of the origin (where the mass is still creepin g in). The indefinitely prolonged appro ac h saves us from an inconsistency. In section 2.2 we note d that a trajectory in pha se space co uld be computed fr om a pair of coupled first-order equations . Their structure was di

F (x , i)

dx

.

dt=

dt=x,

m

(3. l- l 3a)

(3. l - 13b )

where F (x, i) denote s the force as a function of position and, possibly, ofveloci-

3.1

A NONLINEAR

OSCILLATOR

79

ty. These equations enable us to ex tend the trajector y a step forward in time: x(t

+ ~t ) = x(t) + _!__ F[x(t)

x( t

+ ~t ) = x(t) + i(t) At .

m

, x (t ) ] M,

(3. l-14a ) (3. l- 14b)

The step forward is uniquel y determined by the locat ion in phase space at time t. The same would be true of a step backward, if we took ~t to be negative. The implication of this uniqueness is that distinct trajectories can never intersect each other or merge. The proposed "imposs ibilities" are sketched in figure 3.1-4. The proof is by contradiction. If two trajectorie s were to inters ect or merge, then at that location in pha se space there would be two different directions for a forward step or two directions for a backward step or both. Equa tions (3. l-14a and b) guarantee uniqueness, however , and so the proposed intersections or mergers cannot occur . The uniqueness does not prevent two trajectories from coming closer and closer to each other. What is required is merely that they never meet in finite time; they may approach each other asymptotically, as time goes to infinity , which is "almost meeting, but not literally." That is the situation near the origin in the phase space of figure 3.1-3. The point at the origin, x = 0 and x = 0, is a trajectory all by itself. If the mass is placed at x = 0 with zero velocity, it will remain there forever, ideally at least, for that is a location of (unstab le) equilibrium. The lobes on either side are two other, distinct trajectories . If the mass is started at a point on either lobe , eventually the mass will move close to the origin, but it will never get there literally. The three trajectorie s-two lobes and the central point - remain distinct non-intersecting trajectorie s for all finite time s.

---------If-----------~

No!

X

Figure 3.1-4 Inter sectio ns and

mergers such as these cannot occu r in phase space.

80

NONLINEAR

OSCILLATORS

Instability We just noted , in passing , thatthe potential hill at x = 0 (when t < t 0) is a location of unstable eq uilibrium . What doe that imply analytically? lfwe insert the potential energy (3.1-2) into Newton 11 and retain just the leading term in the force, we have d 2x __ k 2 m dt2 -

t-to {'

X (3. l -15)

If the displacement x is positive, so is the force; it is anti-restoring, tending to increase the displacement. Because equation (3.1-15) is linear and homogeneous in x, with constant coefficients, the solutions are exponentials. The exponents will be ± at where a. =

r

+[2kl~; ti

2

(3.1-16)

•

To meet initial conditions, we can combine the two distinct solutions as x( t )

= x(0)

e111 + e--111+ x(0)

2

~1-

e-'.

2a.

(3.1-17)

If the mass starts at rest but with x(0) ,6- 0 , the displacement grows exponentially. (We can afford to ignore the decreasing exponential relative to the growing exponential.) Similarly, if the mass is initially at the origin but x(0) ,6- 0, it will depart exponentially. Generic initial conditions-a random choice of x( 0) and x(0)- lead, sooner or later, to rapid and accelerated departure. The sole exception is the special trajectory of the previous subsection , where we co uld head the mass inward and have it spend infinite time creeping toward the center. In real life, achievi ng that trajectory is a practical impossibilit y. A random choice of initial co nditions is almost certain to see the ma ss make a speedy departure.

Summary of Principles In this sectio n we have worked our way from example to general principle and back again, indeed, more than once. Such a sequence is natural when principle s are developed as the need arises, but now a summary of the principle s is in order. The first item is sma ll oscillations about an equilibrium. We expand the potential energy (or the force itself) in a Taylor series about the equilibrium location . Then we convert Newton II into an equation for the displacement from equilibrium. If there is a linear restoring term, we have harmonic oscillation s. If the linear term is abse nt , there may still be a restoring term farther out in the

3.2

AMPLITUDE

J U MPS AND H YSTERES IS

81

Tay lor ser ies , but the equation will be inher ent ly nonline ar, and so a quick answer may not be forthcoming. Finally , if there is a line ar a nti-r es tori ng term. then the equilibrium is unstable. and the displacement will (typica lly) grow exponentia lly in time. The travel time between two spe cific points can be ca lcul ated as the int egral of dx/i. This proposition is useful if all the force can be derived from a potentia l energy, so that i can be expressed in terms of U(x) a nd£. To map out trajectories in phase space, we ca n co nv ert Newton 11 to a pair of first-order equations. The evol ution of x and x then follows by stepwise integration , one step t:.rafter another. When the entire force is derivable from a potential energy U, a graph of U versus x provides a powerful aid to rea soning abo ut the motion. We ca n read off equilibriu m loca tion s and can determine by inspection whether each is stab le or unstable. After drawing total energy lines and recalling that ½mJ2 = £ - U, we can see even more : we can discern the qualitative behavior of the act ua l motion for various initial cond ition s as well as the qualitative shape of the tr ajecto ry in ph ase space. Trajectories in phase space can never intersect eac h ot her or merge. They may, however. approach eac h other a ymptotica lly. as time goes to infinity, or they may have betn close in the past. [Our proof depended on the force being a function of x and i only, so that its va lue wa determined by the location in phase space. If the force depends o n time expl ic itly, F(x , x, t). then new possibilities are opened. Problem 3-10 explores them. ]

3.2 AMPLITUDE JUMPS AND HYSTERESIS A hac ksaw blade cla mped at one end makes a fine vibrating osc illator. If we ce ment a ceramic magnet to the free end , we can drive the osci lla tor sinusoidally by running an altern ating current through a coil placed near the ceramic magnet. The apparatus is sketc hed in figure 3.2-1. For sma ll amp litudes of osci llation, the restori ng force is proportion al to the displacement , a linear restori ng force. At amplitudes o f a ce ntimeter or so, the stee l ceases to respon d linearly. We can figure out the genera l form of the first co rrection to linear restori ng. The linear force has a potential energy ½kx2 that is quadratic and sy mmetric between +x and -x . The co rre ct ion s to the potential energy also should be symmetric, and so a term in x 4 is the mo st likely ca ndidate. ( It would be the next nonzero term in the Taylor ex pa nsion of a symmetric potent ial.) To obtain the force , we need to differentiate the potenti al energy; the term in x 4 will produ ce a correct ion force proportional to x3. Thus we are led to an equation of the form

mx.. = - kx

k xa + x.z

-yx. + F 0 cos wt.

The length X. is a distance suc h th at , when

µ-I= >..,the

(3.2- 1) magnitude of the

82

NON I !NEAR OSCILI A"I ORS

H

Figure 3.2-1 The saw blade osc illator . The frequen cy w of the driving force can be adjusted at the voltage source.

nonline a r force equa ls the magnitude of the linear re storing force (or would equal it , if no higher terms were needed). The sign for the correction term is a guess; if it turn s out to be wrong , we can just go back through the analysis, 2 2 replacing 'A. by - 'A. • The damping term i line ar in the velocity: quadr atic might be more realistic , but we have little need for the damping term , and so the issue is not significa nt. As it tand s, equation (3 .2-1) is a clas ic nonline ar equation , D11ffi11 g's equation , named after Georg Duffing , a German engineer who tudied it exten ively aro und 1915.

Duffing's Equation without Damping For the moment , let us drop the damping term and write ..

X

+ Wo 2X -

2

Wo

x3 \ • "-

Fo

=-

m

cos wt '

(3.2-2)

where w 0 = (k / 111)112, as before . As long as the max imum value of lxl is small relative to .>t,as we will supp ose. the nonlinear term remains a small correction to the linea r re storing term, and we can be guided - so mewhat - by our experience with the driven harmoni c oscillator . That oscillator , when driven at w, has a steady state that is periodic with the period of the driving force. For equation (3.2 -2), can we find an a na logou s periodic so lution , one whose major amplitud e (at least ) comes in a term th at oscillate at the drivin g frequency w? A trial form for the olution would start as X (t )

= C t COS WI +

· ··,

(3.2-3)

where c , is a consta nt to be determined. Difficulty a rises when we insert the

3.2

AMPLIT U DE JUMPS

AND HYST ERESIS

83

trial form into the nonline ar term: it ge ner ates cos 3 wt . and th at cube amounts to a linear combination of cos wt a nd cos 3wt. A way to see this is through the identity e ia

cos a =

+ e - ia 2

whose cube yields

= ¼(cos

+ 3 cos

3a

(3.2-4)

a).

Since we will generate cos 3wt via the differenti a l equation . let us include such a term in the tri al solu tion : x(r )

= c 1 cos

+ ···

wt + c 3 cos 3wt

(3.2-5)

We hope to find lc:Jc ,I ~ I , or at least we hope to be ab le to choose parameter values so that the strong inequ al ity holds . Now we can substitute the augmented tri al form into equation (3 .2-2) and collect all term s in each frequency: [

(w

0

2 -

w 2)c _ 1_w 2 .S:.( 1 + c3 I 4 0 >,_2 C' 1

+ [... Then the nonlinear term in equ ati on (3.2- 1) should be negligible relative to the linear re storing force , and the sys tem should reduce to the driven , damped harm onic osc illa tor. Figure 3.2-6 co nfiirm s this. As F 0 is decrease d, the amplitude curve evo lves from trip le-va lued to "single-valued but tipped ove r" and then to the almost-symmet ric shape of the harm onic osci llator . The hys ter es is and amplitud e jumps of Duffing's equation ca n be demonstrated in ma ny ways , far more ways than just the saw blade osci llator. Several are described in four articles in the Ameri can J ournal of Physics: D . P . Stockard , T. L. John son , and F. W . Sears , 35,961 ( 1967); J . N. Fox and J . J . Arlotta, 36 , 326 ( 1968); J . A. Wa rden, 38, 773 ( 1070); and T . W . Arn old and W. Cas e , 50, 220 ( 1982).

88

NONLINEAR

OSCILLATORS

lc,I

w,,i

wt

Figure 3.2-6 When the driving force F, is dec reased in magnitude , the shape of the amplitude-versus-w • curve redu ces continuou sly to the sing le-valued, almost-symmetric shape of the (damped ) harmonic oscillator.

3.3 VAN DER POL'S EQUATION: THE LIMIT CYCLE An old-fashioned electric circuit with vacuum tubes and iron-core inductor s led Balthasar van der Pol (around 1920) to the eq uation that now bear s his name :

x = - w/x -

µ,

(x 2

-

t/)x,

(3.3-1)

whereµ, is a positive constant. It is a nonlinear equation whose structure arises surpr isingly often ; for example, the structure reappeared in the 1960s in a theory of the laser. We will ta lk about van der Pol's equation as it might arise in a mechanical context. If lxlis greater than t 0 , which is the characteristic length in the problem, the velocity-dependent term describes some kind of damping. If , however, lxl is les s than t' 0 , then the velocity-dependent force is in the directi on of the moti on, not opposed , and so energy is fed into the motion . This is so metime s called anti-damping. Let's see what qualitative reasoning in phase space has to suggest abo ut the motion. Suppose we start the system point at point a in figure 3.3- 1. Ifµ, were zero , the trajectory would be an e llip se with lxl < t' 0 always, and so in the actual situation, energy will go into the moti on: the trajectory will spira l outward , at least until some portion crosses the line s IxI= t' 0 • (The first crossing comes where I.xi is small and so ha s little effect; the spira l shou ld cont inue outward for some substantial distance .) Now switc h to point b. Ifµ, were equa l to zero , the sy stem point would move from bin a large elliptical trajector y. Most of such a trajectory would lie in the damping region , a nd so- on the actual trajectory -a lot of energy is ex-

3.3

VAN DER POL'S EQUA TI ON: T HE LIMIT CYCLE

89

tracted from the system . To be sure, energy is fed in when lx l < f 0 and lx l is large there, but if bis very large relative to t' 0 , then damping will sure ly dominate . The tr ajecto ry shou ld sp iral inw ard . From point a, the traj ectory would move outward. From b, it would move inwar d . Co uld they conve rge on a co mmon trajector y that would be maintained inde finitely ? And what shape wou ld it have ? Let us se e . Y a n der Pol' s equation is a n instance of the general form

i

= - w/x + f(x,

i).

where f(x, i) incorporates al l the complicated term s. App e ndix C outlines a metho d, the averag ing method , for so lvin g such equations. The form of solution is X

=

(1, ( f ) COS [ Wol

+ \0( f) ] ,

where a is a slow ly var ying amplitud e and \0 is a time-dependent Equa tion s for a and \0 are deri ve d in appe nd ix C. In the co ntext of van der Pol 's equation , f

=-

µ.(x 2

-

t'/)x.

phase ang le.

(3. 3-2)

i

damping

dampi ng

anti- dampin g -{

..

+(.,

Figure 3.3·1 Tw o trajectorie s for van der Pol's equation . Damp ing occ urs outside the vertical dashed line s; anti-damping , between them .

90

NONLINEAR

OSC ILLATORS

Then equation (C.8a) implie s 1

Ci=--

(- µ.(,.2 =

2k t m

f

u 12

t-

(0

'

(3.4 -3)

fo .

(3 .4-4)

t'o

The fre qu e ncy w 0 is the small-oscillations frequ ency ca lculate d earlier. The para meter >-.is the characteristic length of the nonline ar te rm ; when the co ndition lxl ~ >-.ceas es to hold, the nonline ar term beco mes sig nificant. If we co mpare equ ation (3.4-2) with equ ation (3.2 -2), we see that th e stru ctur e is Duff ing' s equ ation without damping or dri ving. T he averag ing method provide s us with equations for the a and ..

(3.5-1)

where (3.5-2) 2

( d U / dx

2

)

>..=- 2 (d3U/dxa)

I > 0.

(3.5-3)

o

Th e parameter >..is the characteristic length of the nonlinear part of the force. As long as the displacement is small relative to >..,one may investigate the motion with the averaging method. But one so on finds that both and cj,are zero . There is no effect that accumulates from one nominal period , of duration 21r/Wo, to another. (At least that is so at the level of approximation considered here .) Nonetheless , the asymmetry in U(x) must have some effect. We will look for a seri es expansion in the dimensionless parameter

a

where xmaxis the right-hand maximum displ ace ment (if the left i different ). Thi s means (3.5-4) an expansion that should converge well if e is sma ll, for example , if e = 0.1. In fact , let us be satisfied with just the first two terms. We need to substitute them into the basic equation, which we can write as ..

x

+ w/ x -

e-

w 02 -

Xmax

x2

=O

'

to display the dependence on e. The substitution generates

(3.5-5)

3.5

E Ri ES EX PA

10

S

99

Now we collect term s with like powers of e: (x• •0

2

+ w02x 0 ) + (x•• 1 + w0 2x 1 -

Wo

-

x 02 )

Xmax

E

+ · (e2 ) = 0.

(3.5-6)

The symbol · (e2 ) denote s the terms of order e2 and higher. Those terms a re to be ignored ,just as we ignored the cor re sponding term e2x2 in the expansio n. We may look at equation (3.5-6) as a problem in mathemati cs , where we are free to change>..and hen ce e. The funct ions x 0 and x 1 are independent of e. To ma inta in a zero va lue on the left if we change e, the full coefficient of each power of e must be zero. Thi generates th e equation we need :

x0 + w/x 0 = 0, ••

Xi +

wl

2

Wo Xi ---Xo

2 _

-

(3.5-7) (3.5- 8)

0.

Xmax

We ca n solve them equenti ally. A so lution to eq uation (3. 5-7) is x0

=A

co w 0 I.

(3.5-9)

We ca n subs titut e this into equation (3.5-8) to get 2

i 1 =- w0 2x 1 + xwo

A 2

max

cos2 w0 t

Expressing the squ are of the cosi ne in term s of cos 2w0 t a nd a co nsta nt make s it eas ier to gue ss a form for the so luti o n. Ind eed , a goo d gue ss for x 1 wou ld be a term in cos 2w0 t plu s a co n tant. It wor ks, and we find X 1 =-

A 2

max

Now we ca n go bac k to the write x

A 2

2x

ba sic expansion,

= A cos w0 t + E ( -2--A2

A 2

Xmax

=A

A 2

CO

w0t

(3.5- 10)

- - -co 2w0 t . 6Xmax

+ 2X -

)

- - - - cos 2w 0 t , 6Xmax

A2

6

equa ti on (3.5-4) , a nd

>..CO 2w 0 t .

(3.5 - 1 I a) (3.5- ll b)

The cons ta nt term gives an offse t to the right of the pot e ntial minimum . A look back at figure 3.5 - 1 make that understandable: the steep repul sive rise on th e left ca n readil y pu sh the ave rage po sition (x) to the right of th e minimum. The osc illato ry motio n is still pe riodi c with period 21r/w0 • A harmonic , 2w0 , has appeared , however , and more would a rise if we ca rried th e expa nsion to higher powe rs of e.

100 NONLINEAR

OSCILLATORS

[In fairness , one must acknowledge that the averaging meth od would have produced the offset if we had pursued that route . The ave rage of f alone is nonzero : (/) = w 0 2 (x 2 )/'A.> 0. According the the last par agraph of appendix C, the nonzero average would require a con stant term in the trial form for x, namel y (/)/wl, which here would be (x 2 )/A.=(i, 2/(2>...),in agreement with equation (3.5-1 l b). ] Sometimes we need to meet specific initial conditions, x(O) and x( O). If the form in equation (3.5- 1 lb) will not suffice, we can go back and replace w 0 t everywhere with w0 t + 'Po, where 'Po is a constant phase term . Then the tw o free constant s, A and 'Po, will always enable us to meet the initial condition s.

Secular Terms When a series expansion works, it provides more detail than the averagin g method . Let us apply the series method to the first context in sect ion 3.4, the symmetric potential with an x 4 term. (That will also offer another comp arison with the averaging method. ) The differential equation is

..

x=-w

2

o x-w

2

x3

o -·>..2

(3.5-12)

The characteristic length >..enters squared, and we might as well use it that way in the dimensionles s expansion par ameter , writing X

e =-l!!!!L..

2

>.. 2

If we expand x(t) as X

= Xo(t) + EX1 (t ) + .. ·,

(3.5-13)

substitute into equation (3.5- 12), with I/>..there expres sed in terms of e, collect term s of like powers of e, and set the full coefficients to zero , then we find the two equations we need : 2

(3.5-14) (3.5-15) A solution to equation (3.5-14) is x0

= A cos

w0 t,

(3.5- 16)

and so equation (3.5-15) becomes

=-

w0

2 x1 --

w

2A3

° 2 4Xmax

(cos 3Wot + 3 cos w 0 t) ,

(3.5-17)

3.5

SER I ES EXPANS IONS

101

after we use identity (3.2-4). For a so lution Xi, we can guess that we will need a term in cos 3w0t, but to match the term in cos w0 t, we need something special: a term in t sin w 0 t . Ind eed, a solution is 3

X1

A x2( I = 4Xma 8

3 w0 t sm . w0 t ) . cos 3w0 t - 2

(3 .5-18)

[This is, moreover , the only possible solution, except for a n additive term , a solution to the part of equation (3.5-17) that is homogeneou in x i ; that additive term would have the form cos (w 0 t + const). The pecial term in the sol ution arises, mathematically , because the inhomogeneous term 3 cos w 0 t in equation (3.5-17) oscillates at the nat ural frequen cy w 0 of the homogeneou s portion .] The term in w 0 t sin w 0 t grows with time . It threatens to destroy the utility of the expansio n in e; no matter how small e is, the product ew0 t will grow to unity a nd beyond. Moreover, the term is not periodic. lt ha a long -term tr e nd , which leads to its name , a secular term. Yet motion in a potential well, in the absence of damping, must be periodic. Wh at has gone wrong? We get a clue by looking at the solution provided by the averaging method. Sectio n 3.4 gave us X

=

(i (0) COS

[wot+ ..

and y=-.

X

Xmax

In terms of these quant ities, equatio n (3.5-1) takes the neat form Since IYI :s; I by construction, we immediately see £ as mea suring the relat ive significance of the nonlinear term. After olving the problem in terms of the variab le y and the parameter £, we can revert to x and >... This two-step procedure improves appea rance s and may even make the mathem at ics ea ier to follow. There is a cost : introducing yet another variable , here y = x/ X max· For that rea so n, the sec tion was written in terms of x.

Synopsis A series expansion will work with an equation of the general form

x= -

w/x

+ f(x,

x)

provided the (unknown) exact so luti on is truly periodic in time. The basic computational need is to get equations for the successive terms in the expansion. Here , in outline form , are the steps to take (as inferred from the preceding examp les). 1. Co nstruct a dimensionless expansion parameter £ that characterizes the size of the nonlinear term f relative to the linear term wlx. Often we can take € to be the ratio fl(w 0 2x) , evaluated at x = Xma ,:

_ fat

€ -

2

Xmax

Wo Xmax

•

104

NONLINEAR

OS C II I.ATOR

2. Rewrite the differential equation so that equation may take the form ••

X

=-

2

w0 X

E

appear

+ EWo 2X ma x f

at

explicitly. Thus the

J Xm ax

3. Express the solution as a series in e: X

= Xo(I)

+ a.(t)

+ E2X2(/) + • · ·.

4. Insert the series into the differential equation, and collect terms of like order in e: e0, e 1, e2, . ... 5. Equate to zero each set of terms of like order . 6. Solve the equations sequentially, tarting with the equation of zeroth order in E. If secular term s arise or if we anticipate that the true period differ from 27T/w0 by a significant amount , then we must adjust the fundamental frequenc y in the series. Here are the additional tep .

7. Expand the true angular frequency win power of e: w 2 = w 02

+ Ea1

+ e2a2+ · · ·.

8. Replace wa2by w2 - ea 1 - • • • in the differential equation , a part of step 2. (This insert s w into the mathematic and en ure that the fundamental frequency in the series will be w, not w0 .) 9. In step 6, choose the a,'s to preclude ecular term .

3.6 ABOUT THE METHODS Three methods for solving nonlinear equation have appeared in this chapter : the averaging method, series ex pan ion, and the method of judicious guessing. The third was not formally announced, but it is a fair de cription of how we solved Duffing's equation in section 3.2 . The met hod often consi t of guessing some trial functional form and then adjusting con tant to get equality, at all times, between the two sides of the equation. Sometime it work exceedingly well , but one cannot count on it. The first and second methods are ystematic. If ome mode t conditions hold, they are pretty sure to work. What are the conditions? The averaging method arose in the context

X = - Wo2X + f (x , x ) . As long as the function f ha only a mall influence on the motion during any single time interval of order 27T/w0 , we may average as we did. (More specifically, the changes in a and 'P must be small during each interval of length 21r/wo, so that a and 'P may be legitimate ly taken a con tant in the

WORKED

PROBLEMS

105

averaging operation. Thus the changes must satisfy the following strong inequalities: ti a~ a and tiip ~ 21r.) There is no reason why J could not depend explicitly on time , too: f(x, x, t). The averaging method is powerful and quite general, but it sacrifices some detail; things that happen on a time scale shorter than 21r/w0 are averaged out and lost from sight. The series method can be expected to work well if the motion is truly periodic. To be sure , we may need to adjust the fundamental frequency , but , once recognized, that is no great difficulty. A damped system, however , cannot be handled well. The long-term decay of the oscillation cannot be represented well-for long times-by a sum of ostensib ly periodic term s. Always , of course , the convergence of the series is a real question. typically difficult to answer. And even if the series co nverges , it may do so slowly. If the first few terms do not give an approximat ion of the desired accuracy, the series is of little use. The limit cycle of van der Pol's equation provides an example. The motion is certainly periodic . When the dimensionless parameter has the value µ,c'0 2/w0 = 0.1, a small value , we would expect a series expansion to describe the limit cycle nicely. Indeed, the trajectory is close to the ellipse that we would obtain ifµ, were zero. If we carry the series expansion through second order in the small parameter , the calculated trajectory does depart from the ellipse , heading toward the shape of the limit cycle as determined by digitial computer. At a typical specified value of x, the values of x agree within a few tenths of a percent. That is about what we would expect ; the neglected terms in the series commence with terms of order (µ,c'// w0 )3, which are fra ctionally of order 10- 3 , or a tenth ofa percent. Often the first correction term in a series does remarkably well , and the first and second, taken together , satisfy all practical needs. The first section of this chapter developed small-oscillations theory , a general way to describe motion near an equilibrium point. What were the objectives of the other sections? One was to develop method s of approximation; we have just reviewed them. The other objective was to show a few of the surprises that nonlinear systems generate. Amplitude jumps , hystere sis, limit cycles-t hese are marvelous modes of behavior , and so different from what a harmonic oscillator can show. Theories of turbulen ce and of biochemical selforganization build on these properties of nonlinear systems - and on others that are even more curious.

WORKED PROBLEMS WP3-1 A small mass m starts from rest at a great distance from the su n. At

some later time it passes the earth 's orbital distance re and finally hits the so lar surface (of radius R0 ). Problem. How long is the time interval between those two events?

We can adapt the analysis that led to equation (3.1- 12), writing

106

NON LI NEA R O C ILLATORS

. .

Time interv al = -

JR.

' EB dr

(,n /2) 112 [ E _ V (r ) ] 1/2·

( I)

The minu s sign applie s be cause dr < 0 as rdecre ase from re to R0 . The potential energy is V (r) =-G M0 m/r, whe re M 0 denote s the sun 's mass. Sin ce V = 0 "at a grea t distance from the sun, " a ma s that tart from re st th ere ha E = 0. Thu s equation ( I ) becomes ,-112

R.

Time interval = -

J

r Elldr

=I3

( 2G M ) 112

0

312312 (2GM 0 ) 112 ( r e R . ). j

The ratio R 0 /rEll is about 5 x 10- 3, and so th e term in R . is negligible. Numerical eva luation , usi ng data from appendix A, yields an interv a l of about 2 weeks. WP3-2 For a mass m co nstrain ed to one- dimensional

moti on. the potential

energ y function is

where>.. and V 0 are pos iti ve co nstant s with the unit s of length and energy , respectively. Problem . (a) Dete rmine the point s at whic h the mass could be placed at re st and would remain at rest , i.e., the equilibrium position s. (b) Co mpute the frequency of small o cillat ions abo ut the point (s) of table eq uilibrium . A goo d first step is to work out th e shape of the potentia l energy curv e. The c urv e has a maximum or minimum where its derivative is zero: dV dx

=

[-3(~) + 24 ~>..- 45]Vo = 0 >.. >.. ' 2

with so lutions x = 3>..and x = 5>...To see whether we have a maximum or minimum , we can eva luate the seco nd de rivati ve: 2

d V dx2

I X• 3>.

=+

6V 0

.

>,_2 '

Thu x = 3>..is a minimum , while x = 5>..is a loca l max imum . Since x = 0 implies V = 0, th e cur ve mu st have a shape like that sketch ed in figure WP 3-2. An equilibrium pos ition is charac te rized by a ze ro va lue of the force F. Since F (x) = -dV/ dx, che point s x = 3>..and x = 5>..are equilibrium point s (a nd the only one s).

PROBL E MS

107

X

Figure WP3-2 A qu alitative re nditi o n of the po tenti a l ene rgy c urve.

The pote ntial minimu m at x = 3>..produ ces a stable equilibri um . A ppe a l to equatio n (3. 1-5) gives us m -d22 (x - 3>..)

dt

= - -d2V

dx2

=-

I

(x - 3>..)

x • 3>.

6i o (x-

3>..).

The displacemen t from equilibrium x - 3>..is su bject to a re storing force 2 ) (x - 3>.. -( 6U0/>.. ). Th e co rr es pondin g angular frequ e ncy of osc illatio n is [6Uo/(m>,.2)] 112 .

PROBLEMS 3-1 The pote nt ial e ne rgy of a mass m is

V = ( 17 + cos h

f) V

0,

where V 0 and >,. are pos itive co nsta nts. Sket ch V and the n co mput e the freque ncy of sma11osc i11a tions about the equilibrium loca tion. 3.2 An osc illator is made by a11o wing a negatively charged bead (charge = q) to slide on a thin insulating rod . The rod is hori zont al, and a pos itive charge of magnitude Q is loca ted a fixed distan ce d from the rod , as sketched in figure

108 NONLINEAR

OSC ILLATORS

~x--i

--~tr--~1------•------rod q,111

_J_t d

Figure PJ.2

P3 -2. Of course , q is attracted toward Q, but it is co nstrained to move on the rod and so oscillates about x = 0. Ignore friction. Re ca ll that Coulomb's law for point charges gives an inverse-s qu are force law . A note abo ut electrical un its appears at the end of appendix A . (a) What is the potential energy V (x)? (b) Wh at is the restoring force F ( x ) ? (c) Sketch a rough gra ph of V (x) vers us x and of F (x). (d) What is the frequency of sma ll osci llations about x = 0? (e) Now imagine a microscopic versio n of thi s situ ation . If q = I electr on 0 charge and Q = lql, if d = 2 x I meter s and if the m ass were m = I elect ron mass , what would be the oscillation frequency (numeric ally) and where in the electromagnetic spectru m wou ld yo u loo k for radi atio n? 3-3 Calculate the period of small osci llat ion s about the equilibrium locati on when the potenti al energy is V = V 0 tan 2 (x/'11.), wher e V 0 and '11.are po sitive constants . [For this potential energy one ca n determine the period exactl y: the 2 integral J dx/igives the period as 21r[m'11. / (2V 0 )] 112 ( I + E/V 0 ) - 112, where Eis the energy of the ma ss m. How does yo ur re sult co mpare with this exp re sio n?] 3-4 Refer to WP3-2 and the potentia l ene rgy ske tched there. Wh at are the qu alitatively distinct total e nergies? In a phase space with axes x and x, sketch each of the qualitatively distin ct trajectories. Wh at is spec ia l about the traject ories having£ = U(5'11.)= - 50U 0 ?

o-•

3-5 Suppose a mass m restricted

to one-di men sio nal motion has a poten tial

ene rgy U (x )

=

u

(x4 - >,._2x2) >,._.°'

wh ere '11. is a fixed le ngth and U O is a positive co nsta nt. (a) S ketc h the potential ene rgy. De termine the poi nt s of stable equilibri um . (b) De rive the frequency of sma ll osc illation s abo ut the stable equi librium point s. (c) Suppose th e particle has energy £ = - U 0/8. Sketc h the trajectory (or possible trajectories) in phase space. Do the sa me for E = + UofB.

PROBLEMS

109

(d) Suppose mass mis released from rest at x = 10- 4 X..Compute (in some approximation) the motion x (t ) during the interval O :!S; t :!S; 5 (mX.2/ U 0 ) 112• What

is the location at the end of that interval? 3-6 The potential energy is U (x ) = - C .x3,let us suppose , where C is a positive constant. Sketch U(x ), the associated force F (x) , and one each of the qualitatively distinct trajectories in phase space . What is special about the trajectories having E = O? 3-7 The force F (x), acting on a mass m in one-dimensional motion , is specified to be discontinuous: F (x ) = + F 0 when x < 0, but F (x ) = -F 0 when x > 0. Here F 0 is a positive con stant , and so the force i alway s inward . (We may take F = 0 at x = 0.) Draw, with some care , the typical trajectory in phase space. [You may find it helpful first to draw graphs of F (x ) and U(x).] What is the period of the motion , expressed in terms of F 0 , m , and Xmax? 3-8 The context is section 3.1, with t = l.5 t 0 • Use the integral expression (3.1-12) to calculate the period of oscillation , correct through the term of order Xma.2 /t/ . How does the associated frequency compare with those calculated by the averaging method or series expansion (in sections 3.4 and 3.5)? Note that we may write

.2.) [I+ 2 (Xm9(' a/ +r) ]

_ k ( U (Xmax) - U ( X ) - 3 Xmax2 - .-\-

0

2

•

When a root is taken , the quantity in square brackets may be expanded by the binomial theorem. Useful integrals appear in appendix A. 3-9 A positive charge (of mass m and charge q) is free to slide along a rigid plastic rod. Also present are two fixed positive charges, as indicated in figure P3-9. (a) Sketch the potential energy of q that arises from interaction with the right-hand Q alone . A qualitative rendition will suffice. Next , do the same for the left-hand Q, and then indicate what the total potential energy might look like. (b) What inequality must the ratio tj ft satisfy if the location x=O is to be a point of stabl e equilibrium? .J.

Q __

t

I

r.1.

I

•

q,m

J__l ____ Q

~

t ,~

Figure P3-9

110

NONLINEAR

OSC ILLATORS

(c) Assume that your inequality in (h) is satisfied , and compute the frequency of small oscillations about x = 0. (d) Separately from (c), assume that yo ur inequality in (b) is satisfied . and then describe in words how you would determine the maximum amplitude of the oscillatory motion. TJ-10 The context is phase space , as discussed in section 3.1, a nd a force that depends on time explicitly: F(x , i, t). For example, suppose F = -k(t ) x. th at is. a spring constant can be changed as the experimenter wishes. (a) Suppose the mass has been osci llating for several periods with /.. = l..1. Then when x = 0 but x > 0, the experimenter suddenly changes k to k

= k2 = 4k1.

How does the trajectory continue? Describe (and sketch) it for the next fev. periods. (b) Next , at an instant when both x and x are positive , the experimenter changes from k 2 to k = 1..3 = 0. How doe the trajectory continue into the future? (c) In the context of this problem, does a trajectory ever interse ct itsel f? l s the continuation of a trajectory uniquely determined by the current location (x and .r) in phase space ? By the location plus time t? T3-ll Grc1vita tional co llapse . Imagine a sp herica lly symmetri c, hom ogene ou cloud of dust grains , all the grains initially at rest and separated from o ne another by distances large relative to a grain diameter. The cloud 's initiaJ mass density and radius are p 0 and R0 . respectively. On the assumption th at collisions and pressure may be ignored, calculate the time for gravitation al collapse to zero radius. (The assumption will not be valid for the last moment s of an actual collapse . Can you reason why you would still get a collap e tim e that is correct in order of magnitude?) The co llap e time so ca lculated i commonly used to estimate the star formatio n time in gaseou clouds (or at leas t to provide a time scale against which to co mpare the effects of oth er pro ce ses, such as magnetic interaction s). TJ-12 The context is section 3.2. If the driving force F 0 is small enough , th e cubic term in Ouffing's equation is negligible. How small is "small enough "? (The limitation should be couched in terms of F 0 , >.,m, w 0 , and T only .) You might try to construct a self-consisten t argument , commencing with some expressions in section 2.4. 3-13 In equation (3.2-1 ), the nonlinear term "sof ten s" the spring ; you can ee that best if the full spring force is written as -kx( I - x 2/>.2 ). What would hap2 2 pen to the solution in section 3.2 if we repla ced .>. by - .>. , so that the nonlinearity " hardens" the spring? TJ-14 The saw blade oscillator of sec tion 3.2 could be driven with two ele ctromagnets simultaneously, one fed by current from a sou rce oscillating at frequency Wi, the other, at frequency w 2 • What frequencies would you expe ct the blade 's response to show?

PROBLEMS

111

TJ-15 Suppose the drivin g frequ e ncy in sec tion 3.2 is fixed a t w = 0.9w0 , but the magnitude F 0 of the driving force is c hanged slowly over a la rge range. Would you expect any di sco ntinu ous chan ge in the saw blade's oscillation amplitude? 3-16 Suppose a n osc illat ing mass slide s on a surface whose roughne ss varies with posi tion . In particular , suppo se the total force is-kx - -y(x2/t' 2 )x, where t' is a fixed length and 'Yhas th e dimen sions of th e usual damping coe fficient. Use the averag ing method to analyze the motion of this nonlin ear osc illator. Ca n you inco rpor ate the initial con dition s x(0) = O and i(O) = v0 ? 3-17 Supposeµ, were repla ce d in va n der Pol's equation by - µ,' , whereµ, ' > 0. What would the new traj ec torie s be like ? 3-18 An equation doe s not need to be nonlinear for the ave raging meth od to be a usefu l appro ac h. Co nsider , for example , a moder ately damped harmon ic oscillator .. x=-w

with w0T

=

2

0

x--,

X

T

tl ( 10) a nd the init ial co ndition s x( 0)

= x0 ,

x(0)

= 0.

Trea t the da mping term as the "co mplicated function f(x, x)" in the ave raging method , a nd so lve approximately for x( t ). Co mpare yo ur so lution with the exact so lution. If w 0T = IO pre c isely , how close to the exact so lution is your approxim ation at time t* = 3 (21r/w0 ), th at is, afte r three peri ods , more or less? ls the failure of the approxim ation to gener ate a phase or frequen cy shift important ? Wh at abou t time t ** = 300 (21r/w0 ), for both questions? T3-19 Co nsider a nonline ar oscillator for which the entire force is deri vab le from a potential energy : mx= -dU(x)/dx. Further , supp ose the motion is inve stigated with the ave raging me thod . Ca n yo u co nstru ct an argument showing why (j, = 0 will hold ? 3-20 An acoustical problem that Lord Ra yleigh tackled in 1883 led to a different ial equati on of the following form :

i

= - w/x + (a -

bi 2 ).x,

r

where co nstant s a and bare pos itive . As sume that the term s in .xand have a small effect on the motion during any single interv al 21r/w0 • C hoose so me approx imation schem e and solve for x(t). Lord Rayleigh 's paper a ppeared in the Philosop hica l Ma gaz ine 15, 229 (1883). The anti-damping term +ax would produce unbounded motion if not compensated - at large I.xi- by so me damping te rm . Rayleigh propo sed to "fo rm an idea of the state of thin gs which then arises by adding to [his] equation a term proportional to a highe r power of the velocity." He adopted the term -b.x3 • 3-21 This problem is a co ntinuation of problem 3-20 on Ra yleigh' s nonlinear

112

NONLINEAR

OSC ILL ATORS

equation. Use a digital compute r or a programmable hand caJculator to compute numerically three trajectories in phase space. Cont inue your numeri cal integration until the asymptotic behavior is evident. The ultimate output should be in graph ical form. Dimensi onal analysis suggests that Rayleigh 's equation has two characte ristic time s, I/a and 2n/w 0 • Be sure to investigate both a small and a large value for their ratio 2na /w0 • There is also a characteris tic length, [a /( bw/)] 112• The qualitative behavior of a traject ory may depend on how x( O) compare s with that length . Choose your three trajec tories to be qualitatively distinct. Is your approximate analytical solution (in problem 3-20) supported [provided the param eters and initial values sat isfy relations such as 2na /w 0 ~ 1 and 2 x(0)/[a/(bw/)]li = I] ? 3-22 More about limit cycles. Suppose Newton 's second law for a nonli near oscillator has the form

x = -w/x + µ..xr (r2 - 4r + 3), = (x 2 + x2/w0 2 ) 112• Try polar coordinates

whereµ. > 0 and r in a phase plane (constructed with coo rdinate s x and .x/w0) as a route to determining the limit cycles for this oscillator. (Specifically, look at r.) For each limit cycle that you discover , determin e whether nearby trajectories spiral toward the limit cycle or away from it. Also, dete rmine how long it takes the system point to tra vel around each limit cycle. T3-23 In discu ssing the harmonic osc illator in chapter 2, we worked hard to incorporate initial conditio ns. Typica lly, the solution to a dynamical equati on depend s on initial condit ions in a significant and detailed fashion. D oes that "typical " situation hold if the equation has a limit cycle? Should we make a distinction between the short-term and the long-term behavior of the so luti on? Can you illustrate your respo nse by refere nce to figure 3.3-3? To pr oblem 3-22? 3-24 A pendulum of length t and bob mass m osci llate s with moderate amplitude. The maximum angular excursion from the downward vertical is 8max, specified to be less than or equa l to 0.5 radian. Determine the frequency of oscillation as a function of 8max, correct through effects of order 0max2 • Note: The energy expression for the pendulum may be taken as

½m(t'0) 2 + mg[t' - t' co s 8] = £. Differenti ating the expression will generate an equation in 0.Trigonometi c expansion s are provided in appendix A. Comparisons of theory and experime nt are provided by M . K . Smith , Am. J . Phy s., 32, 632 ( 1964) and by L. P. Ful cher and B. F. Davis,Am. J . Ph ys., 44, 51 ( 1976). An intriguing harmonic analysis is given by R. Simon and R . P. Riesz, Am. J. Phys., 47, 898 ( 1979) and 48,582 (1980).

PR O BLE MS

113

3-25 Extend our series solution of equation (3.5-1) to second order in e. Do you find that you need to adjust the frequency of the fundamental ? 3-26 Th e relativisti c harm onic osc illat or. For a particle of rest mass m0 subject to a linear restoring force , the rela tivistic generalization of Newton 's second law might be

d[( I -

dt

m0 x

] k x2/c.2)112 = - x.

Suppose the initial conditions are x (0) = x1" and .i (0) = 0 and that (-h) , where w0 = (k/m0 ) 112 • (a) l s there an energy conservation law , perhaps of the form 2

( 1 _ mx2c/c 2 ) 1, 2 0

+ U (x ) _-

w 0x 1n/ c

=

constant ?.

(b) Determine the subsequent motion of the mass , using some approxi mation scheme that you justify. Your result must go beyond a purely newtonian approximation; in different words , the speed of light c mJst remain in your final result in a significa nt way . 3-27 When the electric field of a laser oscillate s in only a single mode , the amplitude E ( t ) satisfie s an equation of the form

d 2£

I dE dt 2 = - w/ E - -:;:dt +

-

( g - g£2 )

dE

dt'

in some good approx imation. Here w 0 is the mode 's natural frequency , the term in -r represe nt s losses (including the laser output ), the g term describes gain (from atomic emission) , and the g term describes saturation of suc h gain (as the supply of excited atom s is exhau sted). (a) If g were zero , how would E vary with time? [Assume £ (0) =I=0 and g

> 1/-r.]

(b) How doe s E evolve when g > O? No te: If you examine the structur e of the equation s in (a) and (b} and compare them with structures that you have met, you can arrive at an swers with very little algebraic effort. 3-28 A mass is acted on by a linear restoring force and a spe cial velocity dependent force , so that Newton II reads as follows :

mx = - kx - -yxi

with -y > 0 .

The second term produce s damping when x > 0 but feed s in energy when x < 0. T he initial conditions are x(0) = x,n and i (O) = 0. Moreover, the parameter s and initial conditions have numeric al values such th at

(mk)

-y-

12 1

--x,/ _ -0 kx 1n

(- I )

20 ·

114

NONLINEAR

OSC I LLATORS

De ter mine the subsequent motion of the particle , correct through effects of fir t order in the dimensionless combination cited above. (And there is a nonze ro first-order effect.) 3-29 A mass m co nstra ined to one-dimensional motion is acted o n by tw o force s. The first force is purel y attr ac tive (tow ard the origin) and h a a magni tude inver sely proportional to the distance. The seco nd force arises from an ideal spring , anchored at the origin, whose relaxed length i t' 0 • Thu the force has the analytic form

both ha nd k are positive co nstants. Only motion with x > 0 needs to be con idered. T he parameter s b, k , and t' 0 have numeri ca l values such th at bf ( kt'/)= 0 (-:rn-);that is, th e comb ination is of order (a) Determine the point (or points) where the mass 111could be pla ced at rest and would remain at rest, i.e., the equilibrium position(s). (b) Compute the frequency of sma ll oscillations about the point (s) of stable equi librium. (c) Suppose m is released from rest. Co nsider different locat ion s for the release , and draw a trajectory in phase space for each (kind of) loc ation th at yields a qualitatively distinct trajectory. Take ome pains to dr aw traject orie that are true to the qualita tive features of the motion . (d) Would the physica l behavior of the mass be qualitativel y different if b/( kt/) = 10 were the numerical value of the combination ? Wh y? An d in which respect s?

n-

p(r)

= Po for O ::

Figur e P3-30

r :0: R .,

p

(r)

= {Ofor O:: r
0 for all time but z oscillates between ±z max ¥- O? Describe the motion in words . \ 4-15 The hamilt onian and the energy. Often the kinetic energy is a quadrati c ~tion of the generalized velocity q: KE = f(q)it 2 , wheref (.{tilde ) is some function of the generalized coordinate q. Often, too, the potenti a l energy doe s not depend on q: U = U(q , t ). Can you show that under these circumstances the hamiltonian is equal to the energy ? Can you cite an example ? T/4-16 The hamilt onia n as a cons tant of the motion . In a context with one genera 1zed coordinate and the associated generalized momentum , the hamiltoni an can depend on time in three ways: H = H[q (t), p (t), t]. Two ways are implicit - through the temporal evolution of q and p. The third is explicit - through an explicit appearance of/ in H. (For example , the change with time of an external electric field would introduce an explicit time dependence .) Work out the total time derivative of H , and then use Hamilton's equations to eliminate q and p. Under which circumstances is the hamiltonian a "constant of the motion," that is, a function whose value remains constant by virtue of the equations of motion? Can you cite examples? 4-17 Apropos section 4.7. Start with one spatial dimen sion. Suppose the particles in a bunch are free except for a damping force F

= _!!!_ q = _J_ p. 'T

'T

How does the density f evolve as you ride along? Can you solve explicitly for f in your neighborhood as a function of time? How would f evolve if there were three spatial dimensions and the for ce were

F = - kr --

I 'T

p?

T,4-18 J;ou p/ed pendula. Two simple pendula, each of length t and mass m, are Scrpported from a ceiling at a horizontal separation t 0 • Th e ma sses are con nected bY.,a spring of relaxed length equal to t 0 , as shown in figure P4- I 8. (a) Construct a lagrangian for this system of two coupled pendula. (b) Anticipate small oscillations about the equilibrium: expand the trig onometric functions in the lagrangian , retaining only terms linear or quadratic in 01 and 02 • (c) From the ensuing equations of motion , construct equations for 0 1 + 02 (which is proportional to the displacement of the center of mass from its equilibrium location) and for 0 1 - 02 (which describes the relative separation). What are the corresponding angular frequencies?

PROBLEMS

I"'-...

- i ..--

155

........ .-~ 1

Figure P4-18

(d) Wh at would the genera l so lution for 81 as a function of time look like? What frequencies would appear? Under which circumst a nces would on ly a single frequency appear , and what would the corres ponding motio n of m 1 (and of m2) be like? (The mode s of motion that occ ur with a single frequency are called the normal modes of th e system.) (e) Ca n you co nfirm your theoretica l resu lts by experiment ?

CHAPTER

FIVE TWO-BODY PROBLEM

S.1 S.2 S.3 S.4

S.S S.6 S.7 S.8

Reduction to motion in a plane Effective potential energy Orbit shape Orbits around the spherica l sun The oblate sun Stability of circular orbits The orbit in time Compendium on central-force motion

Now man y things lead me to believe that the come t of the yea r 1531 , ob served by Apian, is the same as that which, in the year 1607, was des cribed by K epler and Longom ontanus , and which 1 saw and observed myself, at its return in 1682 .... 1 may , therefor e, with confidence predict its return in the year 1758. If this prediction be fulfilled, there is no reason to doubt that the other comers will return. Edmund Halley Philosophical Transactions, Mar ch 1705

S.1 REDUCTION TO MOTION IN A PLANE Newton 's analysis of planetar y motion , Bohr's mode l of the atom, and the launching of an artificial earth satellite share certain features. Each was a big step in the development of our civilization. Each is an instance of the two-b od} problem: what are the motions of the two inter acting bodies ? The questio n con -

156

5.1

RED UCT I ON TO MOTIO N I N A PLANE

157

tinues to arise in research , and so-for reasons both historical and anticipatory - this chapter is devoted to the two-body problem . Two bodies of mass m1 and m 2 interact with one another and move about in space. For each we can write Newton II : (5. 1-la) (5.1-lb)

With these equation s as the basis , what can we say about the motion? Typically we are intere sted in the relative motion. So let us introduce the relative separation r and the center of mass R:

R= m 1r 1 + m2r 2 _ m 1 + m2

These vectors are illustrated in figure 5.5- 1. Adding equation s (5.1-1 a and b) gives an equation for R: (5. 1-2)

Divid ing each of those equations by the corresponding mass and subtracting gives an equatio n for r :

dr

-=dt

1 I F - -F m 1 1 m2

2

(5. 1-3) •

m,

point fixed in some inertial frame

Figure 5.1-1 The relative separatio n and center-ofmass vecto rs. In this illustration , the mass ratio is m,/m, = ½,and so R is close tom ,.

158

TWO-BODY

PROBLEM

IfF 2 equals -Fi, that is , if the two forces satisfy Newton III , then the right hand side of equation (5 .1-2) is zero. The equation becomes dR./dt = 0, whi ch implies that the center of mas s moves with constant velocity . We can summ arize this conclusion with the statement

R = constant vector. -F

Furthermore, when F2 equals - F 1, we can replace F2 in equation (5.1-3) by That equation simplifies to

1•

dr = (-1-+- 1 - ) F,= dt

m,

m2

m,

+ m2 F ,

m,m 2

where

the so-called reduced ma ss . Note that µ,

< lesser of m 1 and m2

because we can factor the expression for µ,

µ,

as

greater mass )

= (lesser mass) ( 1esse r + grea t er .

The inequality makes sense . There is a force acting at each end of r, with both forces acting to expand r or to contact r (or to swing it arou nd). So, of course , we would expect r to change more strongly than if just one force (and one mass) were present. We will specify F2 = - F 1 henceforth. Moreover , we will take R = 0, which we can do without loss of generality by choosing an inertial fr ame th at moves with the center of mass. T he essential equation is then d

.

(5.1-4)*

dt µ,r = F, .

If F 1 a: f, that is, if F 1 lies along the line joining the two masses, then the vector product ofr with equ ation (5. 1-4) yields

J 1

[ r X ( µ,i-) ]

= r X F1 =

0,

whence r x (µ,r)

= constant vector =

L.

This is angular momentum conservation, and it has several implication s. The first is that rand i- lie in a.fixed plane perpendicular to L. Figure 5.1-2 help s us see this. Whereas the tip ofr might have wandered about in a three-dimen sional

5. 1

RE D UC TION

TO MOTION

I N A PLAN E

159

L

Figure S.1-2 Co nservat ion of angular mo me ntum implies that r and i- lie, at all times, in a fixe d plane perpendicular to L. Th e proo f is by co ntr adiction. If r began to dep art from the plane , r would

have a compon ent out of the plane. Th en r x i- would not be along the fixed vector L. which wo uld be a contrad iction.

fashion, it is, in fact, restricted to moving in a fixed plane , motion th at is inherently two-dimensional. When would we no r have F 1 a: f? Suppose m2 were a small dumbbell , a s in figure 5.1-3. Each ball of the dumbbell pulls gravitationally on mi, but the unequal distance s imply unequal force magnitudes. The sum of the two forces is F 1 , but that sum is not antiparallel to f. We could rotate the dumbbell to generate a "mass doughnut" and get qualitatively the same effect. And then we could supplement the doughnut with a huge mass at its center. The net effect would be equivalent to the earth with an

m,

!m,

!m,

Figure S.1-3 A situation where F, would not be along

r.

160 TWO - BODY PROBLEM

eq uatori al bulge. For an earth satellite, consequently , we should not expect the orbit to remain in a fixed plane . Let us specify (5.1-5 ) and

IF1 I=

functio n of r only.

(5.1-6 )

The first st ipulation ensures co nservation of angular momen tum. The two , take n together, guarantee that a potential energy U (r ) exists. We can solve F1 =

r

aV ( r ) ar

forU (r): U (r)

=-J'

dr ' F1 (r ') ·

r ' + U (rA) .

'A

Thus , given stipu lations (5. 1-5) and (5.1-6) , equation (5.1-4) implies two conserved quantitie s: A vector: A scalar:

= L, ½µ, r . r + u (r ) = £ . r

X

(µ,r)

(5. 1-7) * (5. 1-8) *

A force that meets co nditi o ns (5.1-5) and (5. 1-6) is ca lled a ce ntral f orce. (In the most common usage , both condition s must be met if a force is to be ca lled a ce ntral force. We could be satisfied with ju st the first condition: any force directed along the line joining the ce nter s of the two interacting objects would be called a ce ntral force , even if the magnitud e were not a function of r only. We will, howeve r, insist on both co nditions .)

Plane Polar Coordinates To rea p the benefits of equation (5. 1-7)- that the relative mot ion is restricted to a plane fixed in space - we should go to pola r coo rdinates in th at plane, as indicated in figure 5. 1-4 . Althou gh in real life both m 1 and m2 move , in the mathematical prob lem- the problem set by equation (5. 1-4)- we may treat one end of r as fixed. Mo st of the kinematic analysis has been don e alread y, in sectio n 4 .3. Equation (4 .3-7) implies

r = ,=r + riJIJ.

(5. 1-9)

(Note that we must distinguish between i- and ;_ The former is the derivative of a vector ; the latter, the derivati ve of a scalar.) T he angular momentum is then L = µ,r X (i f +

=

µ,r 2

0f

X

8.

r08)

5.2

EFFECTIVE

POTENTIAL

ENERGY

161

y

,.

Figure 5.1-4 Polar coo rdin ates in the plane of the motion .

X

The vector product f x {Jis a fixed unit vector perpendicul ar to the plane of the motion. (In figure 5. 1-4, the product f x {Jalways point s out of the plane of the paper. ) The constancy of the vector L implies the constancy of µ,, 2 0,which we abbreviate as L

= µ,r 20.

(5.1-10)*

To cast the energy expression into polar form , we need only squa re the right-hand side of equation (5.1-9) and insert it: E

= ½µ,{,:2 + r 2(J2)+ U (r ).

As it stands, the energy expression depend s on ; , r, and iJ.The last two variables, however , are connected by angular momentum co nserva tion. We may use L to eliminate 0 in terms of r and L: E

= ½µ,;2 + 2L\ + U(r). µ,r

(5.1-1 I)*

The implications of this form are central to the chapter .

S.2 EFFECTIVE POTENTIAL ENERGY The energy expression (5. 1-1 l ) contains only r and r as variables. We may regard µ,?/2 as the kinetic energy for the radial motio n of a part icle of mass µ,. The n it is natural to regard the other terms as an effec tive potential energy for the radial part of the motion : L2 Uerr (r; L ) = - 2 + U(r ). (5.2- 1)* . 2 µ,r Thus we may write, suggestively,

½µ,r2 + Uerr(r; L) = E.

(5.2-2)

We can be st see the implication s of these two equatio ns ifwe look at examples.

162 TWO - BODY PROBLEM

Grav itational Attrac tion If ma sses m 1 and m 2 a ttr ac t each other gravitationally sy mmetric objects , then U (r )

=_

and do so as spheri cally

Gm 1m2

r

and U err(r ,· L )

= .J:::... 2µr 2 + (-

Gm,.1m 2 )

.

Which term in Uerr dominate s? We should ex a mine the ratio L2/( 2µ.r2 )

IUI

I

ex.--;: ·

The ratio goes to zero as r - oo, and so U domin ate s at large r . Wh en r - 0 , however , the ratio diverges , and so L2/( 2µ.,r 2 ) mu st dominate at sma ll r . Thi s be havior is di splaye d in figure 5.2-1. For the radia l moti o n , ther e is an effective potential we ll, as sketc hed in figure 5.2-2. If E < 0, th e ra dial mo tion is co nstr ained to lie between the two radia l di stances at which Uerr( r ; L ) equals E. At the se turnin g point s, I' = 0 instantane ou sly, as the radial motion re ve rse dire ct ion. Why do th ese two bound s to the motion arise? The outer turning point is easy to und ersta nd : The attr ac tive fo rce pr event s the relative se paration from growing indefinitel y. The inner turning point is subtle . Re ca ll th at a n inw ard force i ne eded even for circular motion. [ We saw th at explicitly in eq ua tion (4 .3- 15).] W hen L # 0, it be co me s diffic ult- at small r- to shrink r furth e r beca use most of the

t: C:

::,

>.

=---r .. - ----= ========== ..~~-iC:

Figure 5.2-1 Th e effec tive potential en ergy U,,ff( r ; L) when the force is grav itational attra ction .

5.2

.fC

U,,rr(r ;

EFFECT I VE POTENTIA i E ERGY

163

/.)

:::, >,

f!l ...

51---+--- -- -----------

-

r

I: " 0

Figure 5.2-2 The effective potential we ll fo r rad ial motion .

attractive force is used just to turn the velocity r in the dire ction needed for almost circ ular motion , to keep the massµ, from flying off along the instantaneous tangent to the orbit. The behavior at sma ll r remind s us sharpl y th at we ar e really de aling with motion in two dimen sions. The co nse rvat ion of a ngular momentum exe rt s a powerf ul influence on the radial motio n. Beca use th e dependen ce on L is crucia l, that para meter is displaye d explicit ly in the sy mb ol U err(r; L ) . Th e effective pote ntial ener gy for the radial motion will prove to be a powerful idea , bu t it will also produ ce puzz les unle ss we remember the ess ential two-d imensional qual ity of the orbital motion.

Spring Force The force holding together the two atom s of a diatomic mole cule may be repre sented by a spring whose relaxed length t' 0 is th e nor ma l se para tion of the ato ms. Thi s leads us to in vest igate a potenti al e nergy of the form U (r )

k

=2

(r - t' o)2,

where k is th e "s pring co nstant. " For thi s sys tem the effec tive potential energy is Uerr ( r; L )

=

L2 µ,r 2 2

k

+2

(r - t' 0 ) 2.

Without que stion, L2/( 2µ,r2 ) dominate s at small r , and U( r ) dominates at la rge r. Thu s the cur ve for U err co me s down from infinit y at each end of the range 0 s r s x . If we were sketching , rather th an plotting numeri ca lly, how shou ld we jo in th ose two pie ces of the full curve? A single potential well see m s cor-

164

T W O - BO D Y PRO BL EM

Figure 5.2-3 A graphical olution of equ ation (5.2-3). Th e left-hand side plots as a straig ht line; the nghthand side decrea ses monotonically.

- kl .

rect , but we should che c k. We need to find the place - or place s- wher e Uerr has a zero slope : dU err = _ _!:!__ + k (r _ t' ) = O. dr O

µr

Thus we need the solutions of

u

k (r - t' o)=-· µr 3

(5.2-3)

Multiplying by r1 wo uld give a quartic equation , not a pleasant prospect . We ca n learn what we ne ed by a method that often work s for compli cated algebraic

I.'/ 2µ.r•\

\ \ \

\

\

U, rr

\ \

\ \

\

\

\

\

\

\

\

\

\.

"

'' '

---r

Figure 5.2-4 Th e effec tive poten tial energy assoc iated with a spr ing of relaxed length t •·

5.3

ORBIT SHAPE

16S

equations: Plot both sides as a function of the dependent variable, and look for the points of intersection. This is done in figure 5.2-3. In the range Os rs oo, there is always one - but only one - intersection. Thus U errhas a zero slope at one and only one location, which lies at r > t 0 • The completed graph of Uerr is shown in figure 5.2-4. As long as the mass µ has on ly finite energy, it will be trapped between a nonzero inner radius and a finite outer radius. The radia l motion will consist of oscillation between those bounding radii. While that oscillation is occurring, however , the vector r will be moving azimuthally, too. The motion is inher ently two-dimensional.

5.3 ORBIT SHAPE The comments just made prompt us to look for the shape of the orbital motion. We will ignore the time dependence of the motion if we can extract the shape without it. And indeed we can , by combining the two conservation laws. Conservation of angular momentum gives us (5.3-l) and we can solve for ,: from energy conservation: dr { 2 dt = ,;:[E

- Ue11(r; L )]

}1 12

(5.3-2)

.

The ratio of these two equations gives us d()

dr =

L/( µ.r2 ) [ (2/µ ) (£ - Ue11) ]' '2'

(5.3-3)

which we may integrate , at least in principle , as 0 + const

=

I

[L/( µ.r2) ] dr _ Uerr(r; L ) ]} 112

(5.3-4)

{(2/µ.)[£

This relation is quite general , but only for a few special forms of U (r) can the integral be evaluated in terms of tabulated functions. Numerical integration will always work, of course , but it gives results only for the chosen values of the parameters, such as L and £. So that we can do the integral analytically , let us use the spring force of section 5.2 but set t 0 = 0, so that (5.3-5) Equation (5.3-4) becomes L 0 + canst = ( 2µ ) 112

I

( 1/r ) dr

[£ _ L2/ (Zµ.r2) _ kr2/i]

112·

*

166 TWO-BODY

PROBLEM

The variable r appears under the square root sign in terms that differ by r2or r4. We may make this quartic expression look like merely a quadratic form by the substitution w = -,

r2

with an accompanying

change in differential:

½w- 312 dw .

dr = d ( w- 112) = The integral becomes 8

L + con st = - -2I --(2µ.) 112

J[£ -

ww - 312dw wL2/ kl

-==-=--

---:--::-:--:7""""-:------:--:-:-::--;,-:;--;-:::-• (2µ.) (2w) ] 112

The net factor in the numerator , w- 112 , can be transferred to the square root as wand will clear it of the inverse power of w. The integrand is then just I over the square root of a quadratic expression . A table of integraJs provides the integral as an inverse sine function:

8 + con st

I

.

= 2 arc sm

£ - L2w/µ. 2 _ kL2/ µ.) 112

(£

We can multiply by 2, take the sine of both sides , and solve for w: kL2)1 7"I _- µ.£{ L2 I - ( I - µ.£ 2

12

.

sm[2 ( 8

+ const

}

)] .

(5.3-6 )

The basic orbit shape is determined by the values of L and £. They do th at, of course , within a framework set by the specific potential energy function U ( r) . What determines the value of " const "? The constant is associated with how we orient our polar coordinates relative to the actual orbit. For defini teness, specify that the direction 8 = 0 should line up with an extreme value of r, either a minimum or a maximum. That stipulation implies

:0[right-hand

side of equation (5.3-6)]

I

= 0,

9= 0

which reduces to cos (2 const)

= 0,

a solution of which is const = rr/4. Then sin[2 (8

+ const ) ] = sin = cos

(28+ 2~) 28

by the identity (A.2-3) for the sine of a sum. When we choose this orientation for the direction 8 = 0, the expression for the orbit shape is

[ ( kL2)1 12 cos -r2I = -µ.£ L2 I - I - -µ.£-2

28 ] .

(5.3- 7)

5.3

ORBIT SHAPE

Figure 5.3-1 The shape of the orbit , as computed from equation (5.3-7). when kL 2 /(µE

2

)

167

= 0.6 .

The radius r goes through a full cycle when 8 goes merely through an angle 7T. Figure 5.3-1 displays the orbit , computed from equation (5.3-7) by solving for r as a function of 8. To the eye , the shape is quite like an ellipse. We cannot , however , trust the eye to distinguish an ellipse from a curve of similar shape but different structure in detail. For example , a graph of x2 y• -+-= a2 b4

1

looks more or less elliptical , but the graph is not a genuine ellipse. The orbit in figure 5 .3-1 , however , is a bona fide ellipse; a proof comes readily from our analysis in section 2.3. The present potential energy , given in equation (5.3-5), implies a radially inward linear restoring force. That is precisely the same as the force in equation (2.3-1 ). We found that the shape of the periodic motion was an ellipse , with a circle and a straight line as special cases. Q.E .O. The orbit given by equation (5.3-7) becomes a circle when the coefficient of cos 28 is zero , that is, when E = (kl2 /µ.) H2. A square root then gives the radius as 114

112

Radius of circular orbit=(~;)

=

(t;)

•

Let us compare these statements with predictions based on the effective potential energy. If we set t 0 = 0 in the latter half of section 5.2, we find, via equation (5.2-3) , that the minimum in V eeroccurs at r = [L2/( kµ.) ] 114 • If we want circular motion , we need to keep ;-= 0 for all time. That means we need to reduce the energy £ to its lowest possible value , so that the inner and outer turning points of the radial oscillation coalesce into a single radial location. From the general equation (5.2-2), namely,

½µr 2 + Uerr(r; L ) = £,

168

TWO-BODY

PROBLEM

we extract the requirement on E as

E = Uerr (r; L )

I

r that gives minimum

This energy will guarantee motion at the radius that gives the minimum is U namely, r = [L2/ (kµ.)] 114 • The agreeme nt with the det ailed olution i rea uring. Two comments are in order before we leave this section. The radial com ponent of velocity ; may be either positive or negative. Thus the square root on the right -hand side of equation (5.3-2) may be negativ e as well as positive. The distinction has no substantial effect on the route we took to the orbit olution in equation (5.3- 7) , but we should keep the plus-or-minus po sibility in the back of our minds. The second comment concerns the periodicity of the motion . Whene ver there are both outer and inner turning points, the radial coordinate is periodi c in time . (An analysis with Uerr tells us that.) The variation of r with 0-the orbi t shape, independent of time - is also periodic , but the period in 0 may not be 2r or 21r/(integer). The linear restori ng force gives a period 1r, and so the orbit in two dimensions begins to retrace itself whenever 0 ha gone through ar. An )' period of the form 21r/(integer) will give a shape that retraces itself after 2rr. Tha t behavior, however , is not typical of attractive force laws. Such retracin g happens to be true for the two force laws mo st commonly studied-linear restoring and 1/r - but it is not generic. In ection 5.5 we will find an orbit - indeed , a very realistic orbit - that does not ret race itself.

S.4 ORBITS AROUND THE SPHERICAL SUN This section is both an end in itself and a prelude. Taking the sun to be perfectl y spherica l and calculating orbits will give us a goo d approximation to the actu al orbits of planets and comets. Moreover, we will establish a framework for sect ion 5.5, in which the a sumption of sp heri c ity is relaxed. If we let m2 denote the sun' s ma s and m 1 , the mass of the object in orbit , then their mutual potential energy is U = - G 11111112/r. We can construc t Uerr(r ; L) , and we co uld ca lc ulate an orbit shape via equation (5.3-3), namely , by integrating the equation d8

L/ (µ., 2 )

cf; = {(2 /µ.}[£ - Uerr(r ; L)

]}t 12

(5.4-1)

While that route would go well here , it would be diffic ult to follow when we drop the spheri city as umption and must cope with a mor e co mplicated potential energy. In stea d of integrating equation (5.4- 1), we look for a differential equa -

5.4

O RBIT S A RO UN D T H E SP H ERI CAL SUN

169

tion in which later we can use the techniques we learned when studying nonlinear oscillators . The reciprocal of equation (5 .4-1) gives us an equation in dr/dO. We ca n take Oas the independent variable. In the present situation, the r dependence in Uerr actua lly appears as powe rs of 1/r; that- together with experience -s uggests a change in dependent variable to u =-

1

(5.4-2)

·

r

Then du I dr dO =-~ do'

and the reciproc al of equation (5 .4- 1) becomes du µ, dO = - L

(2) ,;

12 '

(E - Uerr)112_

(5.4-3)

To get rid of the square root -a nd to get a second-order equation of the kind we studied in chapter 3-we ca n differentiate again : 2

d u d0 2

(2) ½(£ _ U )_ µ, err 12 '

= + J!:... L

=

2

µ,r L2

112

dUerr ~ dr dO

(-_!!__+ dU ) µ,r dr 3

so that (5. 4-4) *

I

The step to the second line above follows when we use equation (5 .4-1) to express . dr/dO. The outcome , equation (5.4-4), is quite general and app lies in any context in which the potential energy is a functi on of r only . Substituting our potential energy into equation (5.4-4) yields

=-

u

1 + -, a

(5 .4-5)

L2 µ,G m 1m 2

(5.4-6)

where a=---

170

TWO - BODY PRO BLEM

is a convenient abbreviation. In stru ct ure, th e differential equation re emble s that for a harmoni c osc illator : it has a second derivative term a nd a linear resto ring term. To make th e resemblance perfect , we ca n combi ne the con ta nt term with u to form a new dep e ndent va riable . Upon lettin g -

I

LI =

II -

(5.4-7 )

~'

we get (5.4-8) N ow we ca n write a so lution immediately :

ii = A co s

(8

+ cp0 ),

which implie s

-rI = -aI + A

co s (8

+ rp0 ).

(5.4-9 )

Although a is determined by the masse s and L2, we do not know imm ediatel y what value the co nsta nt s A and 'Po shou ld have. That i be cau e we los t the energ y E in the process of going from d8/dr to d21i/ d8 2 • We can determine A by returning to equation (5 .4-3). Th at equ at ion and ou r soluti o n give u two equi valent wa ys to e xpress du / d8 :

r-Asin (8+ cp d8 = 2) -L -;;, [E du

0)

µ, (

1/2

Ifw e eva luate the right-hand s ide when 8 + rp0 expression s, we get

-A ----

µ, (

L

2)

µ,

i 12 (

Ucrr(r ; L )]

112 •

= rr/2 a nd equate

L2 £ ---+---'2

im2)

Gm ~ a

2µ,a

the equi valent

i12

·

Thi s gives us A , which can be written neatly a s A = ~,

a

where e denote s the ecc entricit y (di scu sect sho rtl y):

r

2

e=

[I +

Gm1 ,! /( 2a )

•

(5.4-10 )

Fin a lly, equation (5 .4-9 ) beco mes

rI = ~1 [ I + e cos ( 8 + rp

0} ] •

(5.4 - l t)*

5.4

ORBIT S AROUN D TH E SPH E RICA L SUN

171

T he constant 'Po is ass oci ated merely with the orientati on of the line () = 0 rela tive to the actual orbit. The essential dependence of the orbit is on E (thro ugh its presence in th e expression for e) and on L (through the param eter a).

The Typical Bound Orbit Fig ure 5. 2-2 showed us th at a bound orbit necessaril y has £ < 0. Th en e is less than 1. T o lea rn what the sha pe of the typical bound orbit is like, we ca n tabu late r for several values of the argument 0 + 'Po in equ ation (5.4-11 ), as is done in tab le 5.4- 1. Int erpo lation between those values and many interm ediate va lues ge nerates a continuou s c urve. Fore = 0.6, such a curve is shown in figur e 5.4 -1. Th e shape appe ars to be elliptical ; indeed , it is the ju stly fa mo us kep le rian ellipse . For a proof, we convert the orbit equation from polar form (r and 8) to a ca rte sian expre ss ion in x and y. W e write

x = rco s (0 + -.of the point where a radia l line along e 3 emerges from the earth' s surface.

212 ROTATIN G FRAMES OF REFEREN CE

string tension must supply to balance g. Either way , equation (6.3-3) tells us that the local , empirical g is given by the sum of the second and third terms: g=

F grav1ty -

w X (w X r ) .

(6.3-4)

111

The vectors are sketched in figure 6.3-2. If the earth were a perfect sphere, then F grav1ty/ m would be purely radial and equal to - ( G m e /r 2 )e3 • but even then g would not be purely radial. The contribution to g from the earth 's rotation has its maximum magnitude at the equator, where it is

lw X

(w

X

r ) Imax=

w 2R e

= 3.4

x 10- 2 newton s/ kilogram or meters /seco nd 2 •

Since the average value of g is 9.8 in these units, the maximum effect is about 0.3 percent. The earth 's aspherical shape also alters gas we proceed from pole to equator.

Motion near the Earth's Surface Now consider objects in motion near the earth 's su rface. If air resistance is negligible and if there are no forces of propulsion - the rocket engine is off, the ball has left the hand-then the equation of motion follows from equations (6.3-2) and (6.3-4):

a = g - 2w Xv.

(6.3-5)

For motion within Connecticut , say, the direction and magnitude of g are essentially constant. Then we can integrate equ a tion (6.3-5), term by term, once with respect to time: v

= v 0 + gt -

2w

X

(r - r0 ),

(6.3-6)

w

_I_ m

Figure 6.3-2 A plumb bob at rest in the

rotating

frame .

The

magnitude of

- w x ( w x r ) is w' times the perpendic-

ular distan ce from the rotation axis to the mass. The direction i perpendicular to the rotation axis and radially outward. [The e propertie s always hold for the vector - w x ( w X r).)

6.3

rH E RO r AT IN G EA RTH

213

where v0 and r0 are the initial velocity and position, respectively. That is, however , the only easy , exact integration. To integrate again, we would need to know r - r 0 as a function of time , which is precise ly what we are trying to compute . The recalcitr ant term in equation (6.3-6) is already of order w , however , and so we may proceed by suc cessive approximations. First we integrate equatio n (6.3-6) to zeroth order in w to get r - r0 = v0 t + !gt2 + · ( w ), where th e symbol · (w) ha s been written bec ause the term s that we cannot eva lua te are of order w. Second , we substitute this lowest-order expression for r - r 0 back into equation (6.3-6): v = v0

+ gt -

2w

X [ v0 t

+ ½gt2 +

0 ( w) ].

N ow we can integrate again: r - r0 = v0 t

+ Jgt

2

-

2w X (½v0 t2

+ kgt + · (w 3

)

2

) .

(6.3-7)

Here we have r - r 0 correct to first order in w. l f we need greater accur acy , we may substitute the current result into equation (6 .3-6) and integrate again. lt may seem too easy to be true, but the process is va lid . Not unconditionally , of course. The expansion cannot literally be in pow er s of w bec ause w is not dimensionles s. If we compare the two term s in v0 in e qu ation (6.3-7), we see that the second term is of order wt time s the first te rm. Th e sa me is true for the two term s in g. Thu s the expansion is in powers of the dimen sionless product wt. If the time of flight is 10 minute s or les s, the n w t ~ 4.4 x I0- 2 • The neglected term in equation (6.3-7) , which is of ord e r (w t ) 2 • repre sents a fraction a l error of order I0- 3 , at most. Anyway , so lving by succe ssive approximation s work s as long as wt is less tha n unity. Let 's be content with motion calculated correctly through order wt ; then we may use equation (6.3-7) as it stands. Two exa mple s will illustrate its impli ca tion s. Fall from rest If we drop an object from rest , so that v 0

= 0, the displ acement

as a function of time is

r - r0

= ½gt2 - ½w X gt 3 + · · · .

The ea rth 's rotation induces a deflection in the direction - w x g. If we look at figure 6.3-2 , we can work out that the deflection is eastward. What is the size of the deflection for a rea sonable falling distance , 30 meters , ay? We need the flight time Ip as a function of di stance fallen h. Knowing that in zeroth order suffice s:

Ji = ½gt/+ a nd so

tp

=

(2h/g)

112•

X

gt/ ),

= g( ~ 3 ) , we can use figure 6.3-1 g I = wg I wX e3 I = wg cos >,...

Sinceg

Iw

(wlp X

to deduce that

214 ROTATI NG F RAMES OF REFEREN CE

Now the deflection follows as Deflection eastward =

t

3 wg

(2'7)3 g 12 cos >...

(6.3-8)

For h=30 meters and >..= 41°45 ' 58" N, the latitude of the Old State House in Hartford , Connecticut, whence cos >..= 0.746 , the eastward deflection is 2.68 x I0 - 3 meter, that is, 2.68 millimeters. If the fall occurs at the equator, the eastward deflection is not difficult to understand. Let us look at the motion from an inertial frame centered on the earth's center. At the instant of release , the object has an angu lar momentum L = m (R e+ h ) 2 w because it sta rted at a distance R e + hand was rotating with the earth's angular velocity w. Since the gravitational force is purely radial, the angular momentum is conserved in time ; that is, mr 2 0 = canst. As the mass falls and r decreases, the angular velocity must incre ase: fJ > w. This increase carries the object eastward relative to the line from the earth 's center through the starting location (which rotates at the rate w). Lateral deflection of a projectile If the object has an initial velocity v0 , then equation (6.3- 7) gives us two terms of order w. The second one, proportional to - w x g, gives an eastward displacement. The fir t, proportional to - w x v0 . is perpendicular to both w and v0 and so changes direction if v0 is changed. We should examine two instances. If a snowball is thrown from the north pole , then v0 points south and upward , and so -w X v0 point s westward (relative to the initial velocity). The term w x g is zero at the pole . Thus the snowball ' s path is deflected westward. From the viewpoint of the inertial frame , that makes sense : the earth rotates eastward under the snowball , and so, of course , the snowball lands somewhere to the west of the initial line of sight. If a projectile is launched eastward at some middle northern latitude , say, then - w x v0 produces a deflection southward (and influences the vertical motion). To compute the southward deflection , we can take the scalar product of equation (6.3-7) with e 1, the unit vector in the southerly direction: e1 · (r - r0 ) = 0

+ (=0)

- e1

•

( w X v0 ) / 2 + 0.

(6.3-9)

The first zero arises because v0 points eastward and upw ard ; the second, because the component of gin a southerly direction is of order w 2 • The last zero arises because the term in - w x g points eastward and so may influence the total distance traveled , but does not "deflect" a projectile launched eastward. The triple product is unchanged under a cyclic permutation (as proved in appendix B), and so we can tidy up equation (6.3- 9) by writing e1 · (r - r 0 )

=-

v0

•

(e, x w ) t 2

= - vo · (- w sin >..e2 ) t2

=

(e2 · v 0 ) w sin>.. 12•

(6.3-10)

6.3

THE ROTATING

EARTH

215

w

Figure 6.3-3 The angle between e, and -w is also >..

Figure 6.3-3 help s us see how the sin X.factor arises . We need the time of flight to zeroth order only. If the projectile is launched atan angle a with respect to the horizontal , then t p = (2v 0 sin a.)/g, a result we derived in section 1.3. Inserting thi s into equation (6.3-10) and noting that e2 · v0 = v0 cos a. gives Southerly deflection

= 4;

03

sin X.cos a. sin2 a..

(6.3-11)

A projectile fired eastward at the equator should-by symmetry - experience no lateral deflection. The limit of equation (6.3-11) as the lat itude X. goes to zero is indeed zero, a reassuring check on the calculation. But we had set out to calculate the deflection in middle northern latitudes. During World War I, the Germany army shelled Paris with its Big Bertha cannon from a distance of 76 miles. The shell left the cannon at a speed of 1700 meters/second , about 5 times the speed of sound. Thus the product wv03 / g 2 was 3.7 kilometers. The latitude of Paris is 49°, and the cannon was fired at a large elevatio n, a= 55°. For the product of the trigonometric functions in equation (6.3-1 I), these values give 0 .29, and so the entire right-hand side is 4 kilometers, in round figures. To be sure, the cannon was not fired due east, and air resistance may not be ignored. Still, under the conditions in the field, the deflection was about a mile. The ballistics experts had neatly included the effect in their calculations. Since the times of Archimedes and Leonardo da Vinci , physics has been brought to bear on war. Sometimes it ha s been on the side of justice , at other times not. The legacy is a mixture of pride and shame. But probably we should not look at the history as a legacy of the profession. Rather, it is the history of individuals who happen to share an intellectual pursuit. As with all large groups of individuals, some members are admirable and others are not.

216 ROTATING

FRAMES

OF REFE RENCE

The Wind If the earth were not rotating, the wind wou ld blow directly from high pressure to low . As things really are, the pre ssu re gradient does push the air toward low pressure , but the earth's rotatio n deflects the air to the right (in the northern hemi sphere). The mathematic al expression of this is the Coriolis term in equation (6.3-2), the term in - w X v. The deflection builds up a circu lating pattern, as sketched in figure 6.3-4. If we begin with just the pressure gradient and the Coriolis effect, the steady- state motion emerges as circular. The Co riolis effect turns the wind perpendicular to the pressure gradient; the two effects ca nce l, except for a residual inward force to keep the air in circu lar motion around the low-pressure region. In reality, there is also friction between the wind and the earth's surface . This dissipates the kinetic energy of the wind; to maint ain a steady speed, the wind must head somewhat toward low pressure, picking up energy from the pre ssure gradient. The net result - a balance among pressure gradie nt, friction , and Cor iolis effect - is the pattern sketc hed in figure 6 .3-4. In a sense , the Co riolis effect delays the motion of air from high to low pre ssure. The delay enab les the processes that produce the pressure difference in the first place - primarily heating at the earth's surface and condensation-to build up a substantial pressure difference. The result can be a storm of great violence o r at least major (and ofte n welco me) changes in the weather. Very close to the equator , the delay ca nnot occur. Th ere the rotation vector w is almost parallel to the earth's surface , and so a vector product with a surface wind produces a Coriolis effect th at is predominantly vertical (either upward or downw ard). No delay means no chance for pre ssure difference s to grow large , and hence little change in the weather ca n be initiated . The equat orial belt where so little occurs was well known to the sailors of bygone days: the Doldrum s.

Figure 6.3-4 How the earth' s rotation build\

a circulating wind pattern. The light lines a re contou rs of con stant pre ssure ; the heav> lines denote the actual wind pattern. Th e da shed lines indicate how the wind would blow if the re were no Coriol is effect.

6.4

FOUCAULT

PENDULUM

217

We think of hurricanes a nd typhoons as tropical storms. Yes, but they are not from the very equatorial tropics. These storms always or iginate outside a n equatorial belt of some ± 5° in latitude. The Corio lis effect is essential to th eir formation. For a moment, let us go back to the beginni ng of this subsection. The most ideal wind pattern of all would be this: the Coriol is effect swings the wind arou nd perpendicular to the pressure gradient and then balances the gradient with no remainder, so that the motion is in a straig ht line (on the rotating earth). The meteorologist s call such an ideal wind the geos trophic wind , a euphonious name and an apt one, for the Greek roots mean "turned by the earth." At altitudes of 700 meters and more, the friction with the earth's surface ceases to have a major effect. If the wind flows with a radius of curvature that is large- not near the eye of a hurricane -t he wind is nearly geostrophic. Rather than blowing from high pressure to low. a steady wind moves at right angles, following the contours of constant pressure. Analysis in detail relies on Newton II when it is couched in the language of the rotating frame, the contribution made by Gaspard Corio lis in 1835. A qualitative understanding goes back much further, to 1700 or so. George Hadley , a London lawyer, recognized the influence of the eart h's rotation. All portions of the su1face are carried around at the same angular rate. Those farthest from the rotation axis have the highest speed (relative to the inertial frame provided by the stars). Consider now , with Hadley , a wind blowing southward in the northern hemisphere. It is moving toward the equato r , the region where the linear velocity of the earth's surface is largest. The earth will move eastward under the wind. From the viewpoint of a sa iling ship on the earth , the wind is deflected westward-and becomes part of the trad e winds .

6.4 FOUCAULT PENDULUM For an introduction , we should return to the rotating table in section 6.2. I magine a stand attached to the table, with an arm that projects over the table 's center. From the arm hangs a pendulum , its point of support directly over the rotation axis. This is sketched in figure 6.4-J. First the table is set into uniform motion, and then the hitherto quiescent pendulum bob is given a sharp horizontal push. What is the bob 's motion ? From the viewpoint of the inertial frame , the bob merely swings in a fixed plane. After all, the pendulum' s point of support (located on the rotation axis) is not moving, and so the pendulum is "unaware" of any rotation. For observers on the table , the perceived motion is quite different. A Co riolis effect accelerates the bob to the right (as they look along the path) , a nd so the trajectory is a sequence of loops , as shown in figure 6.4-2 . (The inertial observer might say to them , " You see that deflection because the table is moving under the swinging bob .") When the pendulum's oscillation frequency w 0 is

218

ROT A TI NG FRAMES

OF REFE RENCE

Figure 6.4-1 Pendulum on the rotating tab le.

high relative to the table 's rotation frequency w, the pendulum makes many swings while the tab le turns 1°, say. The loop s are very narr ow , and their orientation changes on ly slowl y. To the ob server on the table , the motion appears as a slow prec essio n of the osci llation plane. Otherwise , the pendulum acts perfectly norm ally. The rotating table is a fine model for the vicinity of the eart h's north pole. From the table , we have a prediction of how a pendulum at the north pole would beh ave: The plane of oscillation would drift we stward (as seen by us) while the earth moved eastward under it (as see n from an inertial frame). Aw ay from the pole , the analysis is more difficult and presents a challenge to the method s th at we have developed. The forces on the pendulum bob are gravity and the tension Tin the string . Equations (6.3 -2) and (6.3-4) give us the

Figure 6.4-2 Look ing downward in the rotatin g fra me . The tab le. when vie~ ed fro m the inertia l frame , rotate counterc loc kw ise.

6.4

FOUCAULT

PE D U L U M

219

eq uation of motion for an observer on the earth: T a = g +-

111

-

(6.4-1)

2w x v.

The ten sion is directed a long the st ring, away from the bob . Figur e 6.4-3 shows why we may write

T

=+ rs-

f

r T

'

whe re r 8 is the fixed upper terminu s of the string a nd f is the length of the pendulum . With the unit vectors of equation (6.3- 1), we can des cribe the bob 's location by

The Co riolis term require s

= e. (- w3y) + e2(W3X-

w.z)

+ e3(W1)

1

) •

The angul ar velocity w lies in the (e 1 , e3 ) plane, a nd so w2 = e2 • w = 0. To determine the precession of the pendulum's oscillation plane , we need to find the (x, y) component s of motion . Taking the scalar product of equation (6.4- 1) with e 1 gives

upper terminu s of strin g

w

Figure 6.4-3 The tension is along the vector difference r, - r. which ha lengt h I'. the pendu lum' length.

220

ROT A TIN G FRAMES OF REFEREN CE

r"- r T x.. = ( = O) + e 1 • -·---+ t m

.. 2w3.Y

Because g is almost purely along---1!3 , having only a portion of order w 2 along e1 , the term e 1 • g is of order w 2 and may be ignored here . The support vector r, is along e3 , and so e 1 • r, = 0 , but e 1 • r = x. The upshot is ..

x

x =---+ t

T

2wy3

m

.

(6.4-2 )

'

correct through first order in w. The scalar product with e2 gives ..

r, - r T = 0 + e2 ' -t/11

_Y

V __ T = _...,__

f'

111

2w3 X.

-

2 ( W3X.

-

.)

W 1Z

(6.4-3 ) '

z

x,

because the vertic al motion is sma ll relative to a horizontal component like provided that the amplitude of swing is small relative to the length t . Equations (6.4-2) and (6.4-3) are coupled equations in x and y- plus tension T. If the pendulum bob were at rest , we would have Tim = g ; this was part of our definition of g in section 6.3. When the pendulum swings with sm all amp litud e , the tens ion need vary only a little from the static value. To an adequate approximation , we may use T/m = g in equations (6.4-2) and (6.4 -3). They become the pair

x = - wlx + 2w3)', Y= - w0 2y - 2w3x,

(6.4-4 a) (6.4-4 b)

where wl = g/t' . Now to solve these equations. If the terms in w 3 were absent, the soluti on would be oscillation a long a straight line . We start the pendulum so that such motion will ensue. We could write such a solution as

+ rp0 ),

(6.4- 5a)

A u cos (w 0 t + 'Po),

(6.4-5 b)

x

= A,,, cos

y

=

( w0 t

so that (6.4-6 ) Figure 6.4-4 shows how the position in the plane would oscillate a long a fixed line. In equations (6.4-4a and b), the ratio of the terms that depend on w 3 to th e linear restoring terms is of order

6.4

FOU C A U LT PE DULUM

221

Figure 6.4-4 1f the earth did not rot ate, equation (6.4-6 ) wou ld describe the motion . The bo b's po ition along the diagonal line is propor ti onal to cos (w 0t + / * > 0.

With this much information, we can go back to equation (7 .8- I) and assert that (7.8-2a ) at all times and th a t (7.8-2b) when the right-hand side is evaluated at t = 0. Since / 33 "F I *' we find that Li s not parallel to w(O). The situation is geometrically similar to what we studied in sec tion 7.5, where we found L not parallel tow . A difference in direction s need not surprise us. To see where L points, we first write equation (7.8-2b) as

L = / *[ w,(O)e,(0) + wa(O)ea(O)]+ (/ 33 - l .) w3 (0)ei0). Because / 33 > I * ' we haveL asa linear combination ofw ( O) ande 3 { 0) with positive coefficients. Thu s the sum that gives L will lie between w (O) and ea(O), as shown in figure 7.8-1. This completes the prelimin arie s; we can go on to investigate the disk 's motion.

Evolution of w What can we learn about w from Eu ler's equations for a rigid body ? Takin g the first equation in the set (7. 7-5) and bearing in mind our speci al form for f u, we get

7.8

AX I SYMMET RI C A

1.wi - (I. - I 33) w2w3 =

D TORQUE -F REE

257

o,

which we can rearrange as (7.8-3) The seco nd of Euler's equations gives a similar re sult: • W2

=

+ / 33- / •

/*

W3W1.

(7.8- 4)

La t, the third equation gives / 33w3 - (I* - / *) w 1w2 = 0 ,

and so

w3 = 0. We find that w3 retains its original value: W3

= W3(0).

The co nstanc y of w 3 enables us to write equations (7.8-3) and (7.8 -4) as

w,= -D.w 2, w2 = + D.w.,

(7.8-5a) (7.8-5b)

D. = / 33- / *

where

I*

W3.

(7.8-6)

As defined here , D. is a constant with the dimension s ofan angular velocity . To solve the pair of coupled equations, we differentiate the first and then use the seco nd to get

In this equation we can recognize a familiar form: the harm onic oscillator, once again. The so lution that meets our initial conditions is w1 Then

w2

= wi(0) cos D.t.

follows algebraically from equation (7.8-5a ): w2

=-*

w = w (0) 1

1

sin D.t.

The su mmary form is w

= w 1(0) [cos D.t e 1 + sin D.t e2 ] + w3(0)e3 •

(7.8-7)*

The vecto r in square brackets is a unit vec tor ; it rotates with angular frequency .nrela tive to axes e 1 and e2 • The implication is this : w prece sses about the symmetry axis e3 with angular frequency D. as viewed fr om the body frame. In section 7 .9 we will see how this prece ssion applies to the earth's rotation.

258

EXT EN D ED BODI ES I N ROTA TI ON

Orientation of the Disk in Space A que stion of more intuitive intere st is this : How does the orientation of the di sk change (relative to the inertial frame )? The question amounts to asking, How does the orientation of the symmetry axis e3 change with time? After rec a lling two fact s, we can make rapid progress to the a nswer. Fir st, we do have one spatially fixed vector in the problem, the angular momentum L. Any motion of the symmetry axis ultimately mu st be motion relative to L. Second , the general equation (6. 1-7) applies to e3 : (7.8-8) Our strategy is to combine these two observation s. We need the c onnection between L and w. Let us go back to equation (7.8-2a ), add / * w3 e3 to get I *w, and compensate in the last term :

+ (/33-

L= /*w

/ *) w 3e3.

(7.8-9) j

We can solve thi s equ ation for w :

L w = T;

-

(/ 33- l *) 1* w3e3.

(7.8-10)

When we insert this expression for w into equation (7.8-8) and remember that e3 X e3 = 0, we get (7.8- 11)* This equation has preci sely the structure we studied in section 7.2 and described verbally in equation (7.2-3 ). The general results of that section give us our answer: The symmetry axis e3 preces ses about L with angular frequency Lil *. Surprisingly, there is a commonplace example of this precession: a poor ly thrown U.S. football. The ball spins , but the ends wobble around in the air. Th e symmetry axis is precessing around the fixed direction of the angular momentum.

eu as Seen from the Inertial Frame Our analysis with Euler's equ ations gave us the motion of was viewed from the body frame . How does the evolution of w appear when it is viewed from the inertial frame? The answer comes readily . In equation (7.8-10) we have w expressed as a linear combination of L and e3 • Thus w lies in the plane that L and e3 instant aneously define. As e3 precesse s around the spatially fixed vector L, so must w , and it must precess with precisely the same angular frequency , L i l *. Becau se

7.8

AXISYMMEl

RI C AND

r O RQ UE·F RE E

259

the linear combination of L and e3 in equation (7 .8-10 ) is formed with constant coefficients, w has a constant magnitude and maintains a con tant angle of inclination relative to L.

Summary This paragraph and figure 7 .8-2 summarize what we derived in this section. The wLe3 plane preces ses around L with angular frequency L i l • . Thus w traces the cone centered on L (ca lled the space cone). Simultaneously, w precesses about e3 with respect to the body axes; the angular frequency is n = (/33 - / *) w3/J*. Thus w traces the cone centered on e3 and fixed in the body. (It is ca lled the body cone.) The vector w traces the two cones simult aneou sly. The points of the body that lie (instantaneously) along the direction of ware at rest (relative to the center of mass) because a rotation does not move points on the very ax is of rotation. Thus the body cone "rolls without slipping" on the space cone.

The Dust Grain Again When we studied a spinnin g dust grain in section 7 .2, we took w ex L but noted that we shou ld later examine the proposition. The disk of this section tells us that we should not expect the proportionality to be literally true. Nonetheless ,

L

Figure 7.8-2 The body cone (centered on e3 ) roll s wi thout slipping on th e space cone (centered on L). The line of contact gives w as a function of time . Because the body cone is fixed in the di sk, the movement of that cone through space shows how the di sk's orien tation changes.

260

EXTENDED

BODIES

I N ROTATION

the proportionality hold s in an average sense that is amply sufficient to justify our work . Here is why. The magnetic torque on the grain is tiny. The characteristic time for it to change L is thousands of years. Over an hour, say , L is essentially constant. The angu lar velocity w is of order I 04 radian s/seco nd, with a period of 21r/ I 0 4 = J 0- 3 second. Any prece ss ion of w relative to L will have a similar characteristic time sca le. (We may approximate the shape of a dust grain by a thick disk. Then our work ear lier in this section tells us that w pre cesses with ang ular frequency L i l *" For a cylinder who se height is about equal to its diameter , L is approximately I *w, in order of magnitude , and so the prece ssion frequency is approximately w.) In an hour 's time , the prece ssion of w relative to L will give an ave rage w that lies along Land is proportional to Lin magnitude. Using the average in a torque expression that has a time scale of I 0,000 years is sufficient , a nd so the ste p from equation (7.2-4) to equation (7.2 -5) follows as an amply justified approximation.

7.9 CHANDLER WOBBLE Let us ignore for now the small torque s that the sun and moon exert on the earth. Then the results of section 7.8 apply. To see them in some detail , we need to estimate the earth's inertia ten sor. We may approximate the oblate earth by a sphere plu s an equatorial ring of mass. The summation form in equation (7.3-4) implies that the earth's inerti a tensor will be composed additivel y of (/ u) sphere and (/ u)rtng· We have worked them out in equations (7 .6-1) and (7.3-7), respectively. Their sum gives us an approximation to the earth's inerti a tensor : 0

I+

if

0

if = ~

where

M ,;ng M sphere ·

4

Acco rding to equations (7.8-6 ) and (7.8-7), the eart h's angular velocit y w will pre cess about the symmetry axis e3 at the rate n H

=

/ 33- / * I*

= 2. M ,1ng 4 M

W3

sphere

if = I+R

w.

W3

(7.9-1)

The approximation in the second line follow s bec ause if is mu ch less than I and because w is always very close to the symmetry axis, so that w 3e3 is almo st the whole of w. Thi s precession is sketched in figure 7 .9-1. In section 6.5 we

7.9 e,

C H ANDLE R WO BB LE

261

w

\ \.

'-

'-

---

Figur e 7.9- 1 H ow w move s as een from the body frame. that is. from the viewp0i nt of us on eart h . T he size of the preces sion circ le is muc h exagge rated.

adopted a value M, 1ng/M e = 2. 17 x 10- 3, base d on sa tellite tr acking da ta. lf we u e thi va lue in equ ation (7 .9-1 ), we infer

fl = :n\HW, which impl ie a prece sional perio d of 368 days. We may not take thi s as mo re than a n es tim ate beca use th e "s phe rical pa rt" of th e ea rth is not homogene ous (as in our a pp rox imation) nor is th e bulge co ncentr ated in an equ atorial rin g of negligib le c ross sec tion. N oneth e le s, we oug ht to hav e a good orde r-of-ma gnitude es tim ate , a nd ind ee d we do. O bse rvations th at a re independent of the motion imme diately before us giv e / 33- / * _ I / 33 - 306"

Thi ratio and th e ratio of mo ment s in equ ation (7 .9- 1) ca n differ only by tenth of a percen t, a nd o a period of esse ntially 306 days is predi cted . T he observe d per iod, howeve r, is markedl y different. Th e w ax is move relative to the surface of th e ear th with a pe riod of so me 427 days. Thi s mu c h longe r period was di cove red in 189 1 by S. C. C handl er, an assoc iate of th e Harvard Co llege Ob se rva tory. C handler prese nt ed his di covery, meti culously sub ta ntiated, in a se ries of papers in the As tronom ical J ournal. Eve n as th e pape r were appeari ng- litera lly be tween in ta llment s 3 and 4- Sim on N ewcomb pro pose d an exp lanation for th e longer per iod . In the A stronom ical J ournal of Dece mbe r 23 , I 89 I , N ewco mb wro te: Mr . C hand ler 's remark ab le disc ov er y. th at the apparen t variations in terre strial la titudes may be acco unted for by suppo s ing a revoluti on of the axis of rota tion of the earth around that of

262

EXTENDED

BODIES

IN ROTATION

figure, in a period of 427 days, is in such disaccord with the received theory of the eanh' s rotation that. at first. I was disposed to doubt its possibility. But I am now able to point out a ,·era ca11sa which affo rd s a comp lete exp lanation of this period .

Newcomb recapitulates the kind of calculation we have made and cites the prediction: a period of 306 days . Then he continues, as follows: The question now arises whether Mr . C hand ler 's res ult can be reconciled with dynamic theor y. I answer th a t it can . because the theor y which assigns 306 days as the time of revolution is ba sed on th e hypothesi s that the earth is an a bsol utely rigid body . But , as a mailer of fact, the fluidity of the ocean plays an important pan in the phenomenon. as does also the ela sticity of the earth .

New comb goes o n to say that when the rotation axis dep arts from the symmetry axis e3 of the earth , it carries some of the bulge with it. The so lid earth stret ches, and the ocea ns flow a little. The shift in shape is equivalent to reducing the differen ce / 33 - / * and hence lengthens the period . The motion of w relative to the earth's surface carries a curiously informal na me, the Chandler 111obb /e. More preci sely, the motion with the 427-day period carries that name ; there are variations with other periods , too. For example , there is an annua l variation, believed to be caused by a shift in air mass, from ove r the oceans in summer to over Siberia in winter. From his stellar observations, C handler inferred "a revolution of the earth 's [ rotation axis w] in a period of 427 day s, from west to ea st, with a radius of thirty feet, measured at the earth's surface." The modem value for the period is a trifle lo nger, perhaps 2 percent longer , with an uncertainty of simila r size . The radius va rie s, by a factor of 2, in an irregul ar manner. The motion is damped on a tim e sca le of perhaps 25 year s, although that estimate is thorou ghly uncertain . Equally unce rtain is the energ y sourc e that rejuvenate s the motion. Earthquakes , as the source, have been proposed by some geophysicists and dismissed by others. Although Newcomb point ed out the essential element - an earth that flexes and flows-a great share of the puzzle remain s unsolved . Although the geophysical problem may remain a puzzle , we can understand what Newc om b mea nt when he spoke of "t he apparent variations in terre strial latitude s." Understanding that will also help us under stand the precessional motions as a whole. Chandler's observations implied that w and e3 are separated by some 10 meter s where the y poke through the arctic ice. The corresponding separation between w and L is much less , as we calculate here . An efficient way to ca lculate the angle betwe en two nearly parallel vectors is to form their vector product. That produ ct is proportional to the sine of the angle between the two vectors, and for a small angle , the sine is approximately equal to the angle itself. Using equation (7 .8-9), we get w x L=

(/ 33

-

/

*) w 3 w x e3

7.9

because w x w = 0. If we divide by the magnitudes absolute va lues, we get

Iw.

X

t i=

1 33

L/w

1

3

*1·w

X

w

C H ANDLER

WOBBLE

263

and L and then form

e 3 I.

Beca use Li s almost alonge 3 , the ratio L/w3 is approximately (/ 33 w3 ) /w3 • Thu s on the right-hand side we have the rati o (I 33 - / *) / / 33 , which we know to be about Tihf.Therefore the angle between w and Li s smaller, by a factor of about "!f3lr, than the angle between w and e 3 • On the earth's surface wand L are separated by only 3 cen timeter s. We should summari ze the orientational features of the three vectors as noted by an observer in the inertial frame : I. Vector L is fixed in space. 2. Vector w precesses about L in a circle of radius 3 centimeters and so is almos t fixed in space. 3. Vector e3 prece sses about Lin a circle of radiu s 10 meters. Co nsider now the great circle that passes through Cambridge, Massachusetts (where C handler worked) , and through the point where e3 emerges from the ea rth 's surface. As w moves around e3 (as see n on earth ), it crosses the great circle twice . For those two instants , figure 7.9 -2 shows the relative

L

L

(a)

(b)

Figure 7.9-2 (a} The relative orie ntatio ns of L, w, e3 , and the local vertical in Cambridge , Massac hu sett s, when w lies between e3 and Cambridge on the grea t circle. (b) As in (a}, but after w ha s precessed aro und e3 by 180°, in some½ (427) days. (To make the di agram s clear, the angle s between w and L and be twee n e3 and L are great ly exagge rated . The ang le between e3 and the vertical in Ca mbridge rem ains constant and is faithfully drawn .)

264

EXTENDED

BODIES

I N ROTATION

orientations of L, w , e3 , and the vertical at Cambridge. Between the time of one crossing and the other, the earth tips bodily. We may take the direction to the north star to be along the fixed direction L (a good ap proxim ation here). The tipping cha nge s the angle between the local vertical an d the north star. If that angle is taken to define the co latitude , then the colatitude changes. Becau se the latitude of a city is 90° minus the colatitude, we-have here an explanation of why Newcomb spoke of " the apparent variations in terrestrial latitude s."

7.10 AN INTERLUDE ON KINETIC ENERGY The kinetic energy of a rotating body figures prominentl y in the theory of molecular energy leve ls. It is part of the energy conservation law that we will use to analyze a spinning top. Thus an interlude to work out a general expression for such kinetic energy is time we ll spent. We need to start with a typical mass mo.in the rigid body and with its velocity as see n from an inertial reference frame: v,a = R + (t)

X

ro,.

The goal is to express the kinetic energy in terms of the two natural velocities - the velocity R of the body frame's origin and the angular velocity of rotation w-toge ther with constants that describe the mass a nd its distribution throughout the body. Summing the con tributions from all the mass points , we may write the kinetic energy as KE =

L ½mo,V/o.

• Vi a

Ot

= L ½ma(R + w

X

r a) · (R + w

X

ra).

(7. I 0-1)

0.

When we expand the sca lar product , we get cross terms of the form

L ma R · (w X ro,) = R · w X 0,

(Lmara) · 0.

Such terms are numerically zero ( I) if we choose the center of mass as the origin of the body frame, so that ~m ar a= 0, or (2) if the body ha s a point spatially fixed in the inertial frame and we place the origin of the body frame there, so that R = 0. These conditions appeared in section 7. I; we agreed to impose one or the other thenceforth because they suffice for our purposes (and for most applications). Thu s the cross terms are zero for us, and we may drop them. A few steps simplify the term with the vector products:

7.1 1 T H E

a

SYMMET RI C. SUPPO RT E D TOP

265

a

=½ w·L

= ½w · UuwJei ) = ! W1f uWJ.

(7.10-2)

A triple product is uncha nged by cycl ic permut at ion of the vectors (a proved in appe ndi x B). If, on the left-hand side , we rega rd w x ra, w , and r a as three vec tors in a triple product , then we may permute cyc lica lly to get an w ou tside. We ca n recog nize the sum as the angula r momentum L, expre ssed in the form displayed in equation (7 .3-3), and so the succin ct result ½w · L follows. But we can a lso expand that result , by using (7 .3-6), and thereby a rrive at the last line . Returning now to equation (7. I0-1 ), we may write

KE =½

(I ma) IRl + ½wiluwJ. 2

(7. 10-3) *

a

Ifwe c hoose the center of mass as the origin of the body frame, we find that the kinetic energy decompo ses neatly into the energy of the center-of-mass motion plus the energy of rotati on about the center of mass. If we exploit a spati ally fixed point and opera te under condition 2, then R = 0, and the kinetic energ y is, of course , purely rotational e ner gy. Section 7 . 11 provide s such an instan ce , and an application under co ndition I is made in WP7-2 .

7.11 THE SYMMETRIC, SUPPORTED TOP Figure 7. I 1- 1 shows a spinnin g top. One point is spa tially fixed ; the tip rests in a slight depre ssion at the apex of the support . We should take the point of support as the origin of th e body frame , so that the co ntac t force will drop out of the torq ue exp ression . The net torque co mes solely from the grav ita tion al forces:

L ra X Fa = L r a X (mag ) a

a

We ma y express the center of mass as (7 . 11- 1) a

whe re R cM denote s the distance from the support point to the center of the ma ss,

266

EXTENDED

BODIES

I N ROTAT ION

i

point of suppo n , fixed in

inertial frame. Figure 7. 11-1 A spinning top .

which mu st lie a long th e sy mmetr y ax is e3 . Bea ring in mind that g = -g i, we ca n write th e ang ula r moment um equa tio n (7 . 1- 12) in the prese nt co ntext as dL

dt = -

M RcMg

e3 x z.

(7.11-2 )

W e ca n so lve for th e top's motio n fro m th is dynamica l eq uat ion. We will find it eas ier , howeve r, to use the co nserve d qua ntities.

Conserved Quantities In sect ion 1.7 we deve lope d the co n se qu ence of ro tational invar ia nce. If an objec t's po tentia l ene rgy d oes not change w hen th e o bject is rota ted about some ax is, then th e co m po nent of angu la r mome ntu m a long that axis is co nserved in tim e. We can ro tate th e top abo ut th e z ax is (at fixed a ngle 0, acco rdi ng to the ske tch) witho ut c hanging the grav ita tiona l pote nt ial energy. Th at rotational inva riance implies

z · L = Lz =

co nstant.

(7 .11 -3)

Th at is no t a ll, h owever. T he top is sy mme tri c abo ut th e ax is e3 , a nd so we may ce rt ainly ro tate th e to p about e 3 w ith out cha nging the poten tial e nergy. Th e impli ca tion is e3

•

L = L 3 = co nstant.

(7. 11-4)

Alth ough the orient atio n of e3 may cha nge wit h time (re lati ve to the inerti al

7. 11 T H E

frame). the angu lar momentum

SYMM ET RI C . SUPPORTED

TOP

267

mainta ins a co mp onent of co n tant size a lo ng

e3. Both implication from rotational invarian ce can be co nfirmed with equation (7. 11-2). They cou ld , of cou r e, be der ived from it , but the inva riance argument i quicker, at leas t for L3 • There i one more co n erved quantity : energy. Let us note the kineti c part fir t. Becau e the origin of the body frame is fixed in the ine rtial frame , eq ua tio n (7. 10-3) reduce to ½w1f uw1 , the kinetic energy of rotation . The grav itatio na l potential energy is Mg tim e the hei ght of the ce nt e r of mass . which i ( R 0 1e 3 ) • i . We may ignore th e potenti al ener gy of th e forces that hold the top toget her becau e that energy does not cha nge. Th e co nt ac t force act on the point of th e body that is spa ti ally fixed, and so it cannot cha nge the energy . In ummary, we need include o nly th e grav itational energy. Energy conserva tio n takes the form (7. 11-5) We now have the three dynamical quantities that are conse rved during the top's motion .

How the Top's Orientation Changes How does the ymmetry axis e3 move? That amou nt s to as king how th e angle 0 (betwee n e3 a nd z) c hange s and how e3 precesses about i (at some instanta ne ous 8). Let us recall th at w de sc rib es how th e o rient atio n changes with tim e a nd thus mu t co ntain the information we need . We can usefully deco mpo e w into three orthogo na l piece , a sket ched in figur e 7. 11-2. There i a portion alo ng the sy mmetry axis , ju t w3e 3 . There is a portion perpendicular to the ie 3 plane , M itten w.di .1. And there i a portion lying in that plane , denoted w1u 1. Thus

i

e.,

Figure 7. l 1-2 The decomposit ion of w int o thr ee natur al orthogon al pieces. T he portion w.1. ii.1.points int o the paper, perpendi cular to the ie 3 plane. T he sy mbols ii.1. and ii denote unit vect ors.

268

EXTENDED

BODIES

I N ROTATION

= W J. ll .L +

W i ll

I

+

W3e 3 .

The first line gives w in terms of ca rte sia n unit vectors; the seco nd , in term s of the unit vectors u.L, fin, and e3 . The derivative dO/dt is given by

{)=

(7.11-6)

W J.

because that part of the angular ve locity will sw ing e3 toward £ or away from it. To begin with, can we extract 0 from the conservation law s? The top has the same symmetry as the di sk we studied in sec tion 7.8, and so its inertia tensor mu st have the stru ct ure displayed in equation (7.6-2). Energy conservation appears as

½I*(w12 + w22 ) + ½/33 w/ + M g RcM cos 0 =

E,

while the angular momentum takes the form

L = / * (w 1e1 + w2e 2) + /33 w3e 3 . The conservation

of e3

•

L becomes

and so

L3

W3

(7 .11-7)

=T;;·

Thus w 3 is constant in time , and we may repl ace it in the energy expression with the ratio L 3 / / 33 . The energy co nt ains al so the sum w 12 + w/ , which we may rewrite as (7 . 11-8) The first term on the right is 02,the quantity that we are looking for , and the seco nd term we ca n get from Lz, as follows. Figure 7.11-2 shows the essential vectors. The co nse rvation off· L takes the form f •L

= /* f

• ( W J. u.L + Wjjll ll)

+ /33W 3 f

• e3 = Lz.

The sketch tells us that

and

z · u1 =

sin 0.

Solving for w 1, we get

_ Lz - L3 cos 0 / * sin 0

w11-

(7.11-9)

7.11

TIIESYMMETRIC.

UPPORTED lOP

269

Now we may u e equation s (7. 11-6) through (7. 11-9) in the energy expression to arrive at

1/ [0+( *

2

L., ~ Linc~s

or]+;~+M gR cMcos 0 = £.

(7.11- 10)

We can olve this equation for d0/dt as a function of 0 and then integrate to find 0 as a function of time. We come back to this equation shortly; let us go on for now. The next que stion i this: How doe s e 3 pre cess about i (at some insta nta neous 0)? Figure 7. 11-2 is aga in a help : the portion eu, 1u will swing e3 a round in the prece s ional se n e. (And that portion of w i the entire sto ry: The part of w a long e3 doe s not move e3 itself, and the third orthogo nal part of w is what gives 0.)Thu s so me of what we need is ready for u in equation (7. 11-9). In a time flt , the tip of e3 will swing a round i by a distance wJtlt . Really the part of e3 perpendi cular to i doe s the prece ing ; its magnitude is in 0. The angle through which it wing is w flt / in 0. (A top view would be simi lar to figure 7 .2- 1.) Dividing the angle by tlt , we get

. Prece ss1o na l a ngular frequen cy

0 = L " - L.3 cos 2

(7. 11-11)

1 .s m 0

*

Extracting the Implications Now we have the nece ssa r y equations ; we can co mmence extracting their implication s. We may construe equation (7 . 11-10) as giving kineti c and effective potential energies for a " particle " of " mass" / * and "coordinate location " ():

*

where

· (L., - L3 cos 0) 2 Uerr(0;L.,,La )= 2J .s in20

La2

+ 2133+ MgR cMcos0.

(7 . 11-12)

This tactic will help us to think about a complicated equation . Even so , U erris difficult to graph for general L ., and L 3 • For a start, consider the ituation when L., = 0. We can achieve that experimentally by starting the top with e3 perpendicular to z and with the top spinning purely about e3 , there being no other motion initially . (Some forms of support will permit such a starting ang le.) The func tion Uerr redu ces to Uerr(8; 0, L3 )

=

L 3 2 cos 2 0 ~ 21* S tn 0

2

L3 + 21 + Mg RcMcos

0.

(7. 11-13)

33

The three contributions a nd their sum are plotted in figure 7 . 11-3. Beca u~e sin 0 goes to zero as () approac hes O or 1r, the effective potential energy rises steeply toward infinity a t the ends of the angular range .

270 EXTENDED BODIES IN ROTATION

Figure 7.11-3 The heav y line show s equation (7.11-13).

u .,, 0, then the factor multiplying cos 8 is negative. As 8 drops from rr/2 toward Bmax, cos 8 goes negative and precession commences. The preces sion will be in the positive sense around t. The precession rate has a fixed sign (when not equal to zero) but a

7. 11 THE

SYMMET RI C, SUPPORTED TOP

271

variable magnitude (as () oscillates). The combined effect of precession and nutation is s ketched on the left in figure 7. I 1-4. Our expression for the precession rate came from the i · L conservation law. That law can give us a qualitative picture of why precess ion must occur. As the top drops away from () = TT/2, the spin about its symmetry axis gives some angular momentum along z, really, along - t. But t · L is zero to start with. Hence the top must move bodily around t to keep z · L = 0 at all times . Th at bodily motion is the precession. Why does the top not fall all the way to () = TT?The interplay of angular momentum and energy conservation prevents that from happening. Gravity pulls down on the top; when released , the top commences to fall over. Gravita-

1T

9= -

2

-

Figure 7.11-4 The motion of the sy mmetr y axis. Shown is the curve tra ce d by the tip of e3 on the

unit s phere. (a) The initial co nditi o ns are 9(0) = 1r/2, 8(0) = 0, a nd no motio n other than spin about e3 • (b) Now 8( 0) < 0, but still IJ(O) = rr/2. and the re is no az imuthal motion initially .

272

EXTENDED

BODI ES IN ROTATIO N

tional potential energy is converted to kinetic energy , but the constancy of£ , L shunts much of that energy into preces sio nal motion. The situation is then similar (only) to a rubber stopper swung more or less horizontaJly at the end of a string (as in figure P4-6) ; the support force plays the role of the string tension. Now imagine starting the top a bit differently. We will have 0(0) = 1r/2,as before , but we give the top a push upward , so that 0(0) < 0. As before , there is no azimuthal or precessional motion at t = 0, and so Lz = 0 still holds. Because iJ(O) ~ 0, the total energy is larger than before. Figure 7.11-3 tells us that 8 will now oscillate between Omin < 7T/2and a new Omax, somewhat larger than before. Beca use O now crosses 7T/2 in its oscillations , the cosine in equation (7. 11- 14) will change sign regularly . The precession will be in the positive sense when 0 > 1T/2, as before , a nd in the negative se nse when O < 1r/ 2. The en suing motion of the symme try axis is sketched on the right in figure 7.11-4. What we have described so far is the ideal situation. When we try the experiment , we find the nutation and the prece ss ion, but the nutation often damps out after a dozen oscillations or so. That is due to friction at the support point. Nutation appears as a transient while the top settles toward uniform precession. But friction at the support also nibbles away La, the angular momentum about the symmetry axis . The term in U err (0; 0, L a) with the l/sin 2 8 factor is proportional to L3 2 a nd will diminish . That term is crucial in setting the size of Omax· As Ladecreases , Omax increases. Sooner or later the real support will fail in the role so easily filled by the ideal support: keep ing the tip of the top spatially fixed. And so the top slide s off the support-but only then "falls down." If L, is not zero, the graph of Uerr versus Ois qualit at ivel y the same as in figure 7 .11-3. The first term in equation (7.11-12) sends Uerr to infinity at each end of the angu lar range , and the curve ha s only a single minimum . (That minimum may now lie at O < 7T/2.) To avoid difficulties with the support , we may want to relea se the top with O small, 30°, say. If we do that with 0(0) = 0 and no azimuthal motion initially, the top will nutate and precess as sketched in the left portion of figure 7.11-4, except that the turning points in O will be at smaller values. The situation when L, = L 3 is spec ial: U err no longer goes to infinity as 0 goes to ze ro , the vertical orientation. The implic ation s are explored in problem 7-20.

7.12 PRECESSION OF THE EQUINOXES The earth's bulge is tilted relative to the earth 's orbital plane , as sketched in figure 7 . 12-1. The sun pulls more strongl y on the nea rer portion of the bulge than on the more distant part an d thus exerts a net torque on the earth . Certainly the earth is not an object " with one point spatially fixed in an inertial frame ," and so we shou ld take torques relative to the earth ' s center of mas s. Doing that in the sketch indicate s a net torque perpendi c ular to the plane of the page and

7.12 PRE C ESSION OF T H E EQUINOXES

273

Figure 7.12-1 The sun exert s a torque on the earth's bulge. (T he unit vector ri denotes the perpen dicular to the orbital plane.)

toward us . Such a torque will cause L to prece s around the perpendicular to the orbital plane. Sketche s for other location of the earth in its orbit will show torques that induce prece ssion in the sa me ense . (There are two except ions : on the first day of spring and of fall, L is perpendicular to the sun-earth line, and ymmetry sets the torque to zero.) We ca n readil y estima te the torque and the precession rate, at least in o rder of magnitude. Th e solar gravitatio nal force per unit mass , as it acts on the bulge, is

=

GM 0 (r ± R ) 2

=

GM2 0 r

(t+ 2 .!i.) r ·

where ra nd R are the earth 's orbital and surface radii , respectively. The far and near parts of the bulge differ in dista nce by roughly ±R . Previously, we regarde d the bulge as a ring with mass M, 1ng· Continui ng with that description , we ca n e ti mate the net solar torque as " R times the solar force on the near part of the ring" minu s "R time s the solar fo rce on the far part of the ring." Thi difference leads to the estimate

GM0 R Net solar torque = R-- 2 -- M, ing; r r factors of 2 and the like are ignored. In figure 7 . 12- 1, the torque is perpendicular tot and ton , the perpendicular to the o rbit al plane ; in sho rt , the torque is proportional to - n x L. If we av erage the torque over the orbit (that is, over I year) , the average direction must be similarl y expressi ble in terms oft and fl (because r has been averaged) . Thu we must be ab le to write

Average so Iar torque

=

G M~r1nv.R ,-

2

..._ X 11

r. L

(7 . 12- 1)

This form is particularly useful when we ask abo ut the preces ion rate . We

274

EXTENDED

BODI E S I N ROTA TI ON

have here the torque structure tha t we studied in sec tion 7 .2. If we compare equation (7 . 12-1) with equ ation (7 .2-2) and reca ll the co nclusion s of that section , we can write immediatel y that

=

Prece ss ional angula r frequency

1ngR G M0 Af;_

2

1

For the earth's angular momentum (relative to its ce nter of mass) , we can make the estimate L = M eR 2w, ba sed on a spheri ca l approxim ation to the shape and with all the sub tletie s of the C handler wobble ignored. Inserting this. we get Prece ss .ion frequency

=

GM0M r tngR 2 r1MeR 2w

(7. 12-2)

We can simplif y thi s. Since the earth ' s orbit is essentially circ ular, N ewton II implie s [ via equation (5.8- 7)] that .,,

1n e fW orb1t

z-

GM 0Me ,-2

Thus we can e liminate several factors on th e right in equation (7 . 12-2) in term s of the earth 's orb ita l angular frequency Worbit: Prece ss ion frequenc y

=

M-

EB

=

w

b'

Mrmg --2!:...!.!.Worbit W

2 X I Q5 Wo rbit·

The estimate of M rtng/Me .comes from equation (6.5-9 ). The preces sion peri od, as es timated so far , is of order 2 x I 0 5 yea rs. Th e moon also exerts a torque on the earth's bulge. Th e average of that torqu e is about twice the solar average. The full a verage is then 3 time s as large as the torque we used , which reduces the period by a factor of 3. The numerical factor s that we implicitly set to I in the order-of-magnitud e calculation work out to a simila r redu c tion. The earth's angular mom entu m is indeed observed to prece ss with a period of 26 ,000 years. (The di sco very itself goes back to Hipp a rchu s, in the seco nd century a .c.) Current ly the north star coin cides in dir ec tion with L, but the pre cessi on is slowl y carrying L away from Pola ris. Because the tilt (shown in figure 7.12-1) is as large as 23½ the earth' s axis will move - in several thousand years -fa r from the north star. Indeed , in 13,000 years, the directi on on th e sky will have shifted by more than the size of the co nstellation that form s th e Big Dipper. A s L creeps around, the two locations in the earth 's orbit where L is pe rpendicular to r creep also. Th ey are the location s of equal day time and night time , the equinoxes. The loca tion of th e equinoxes precesses as L prece ss es 0

,

7.13

URVE )

or

THE CRl fl CAL

OTIO

and provide the histor ical name for the entire phenomenon : the prece the eq uinoxes.

s 275 ion of

7.13 SURVEY OF THE CRITICAL NOTIONS To de cribe a rotati ng extended body , we need eq ua tion for locat ion and orientatio n . Thi chapter focu e on the latter. Compu ting the orientatio n a it evolve in time i comp lica ted becau se everal vectors play es entia l roles: I . The orthogo nal unit vector e1 tell u the instantaneou orie nt atio n. 2. The a ngular velocity w de cribes how the orienta ti on change wit h time . 3. The a ngu lar momentum L is the dynamical quantity that change in repon e to torque .

In chapter 6 we learned how the rotation alters the body 's orientat ion :

de1 dt= w x e1• The vectors L and w are not neces arily parallel. The hape and mas distribution of the body determine how much angu lar momentum along the direction e1 i produced by a rotation rate w1 along the direction e1. The essential information i ummarized in the moment -of-inertia tensor: fu =

L

111a ( Bur/

-

X1aX Ja ).

a

The relation between L and w i expressed by equation (7 .3-6):

L = f uwi! ,. Equation (7. 1-12) ay that the time rate of c hange of the angu lar momentum equa l th e net torque : dL

dt = L r

0

X

Fa.

a

All orientation equations in the chapter follow from equatio ns (7.1- J2) a nd (7 .3-6). Equation for the time derivatives of components w; are worked out in section 7 .7. Euler's equations are quite generally app licable, but we can often do better w ith L and co nse rvation laws . For applicat ions th at you are likely to meet af ter a cour e in mech anic , th e equatio n for imple prece ion and its olutio n are probably the most sig nificant. Equat ion (7.2-2) displayed th e structure of th e equation: dL = ;J c, X dt

L,

276

EXTENDED

where the se:

BODIE S I N ROT ATION

c is a constant

unit vector and J is so me fun ction. The consequence s are

I. The vector L ha s co nsta nt magnitude . 2. It pre cesses about c at a constant angle of inclination and with an angular frequency given by ff. Sometimes a torque will take the form ff c x L only after an approxim ation has been made . Co mputing an average often suffices . Second in importance is likel y to be an expre ssio n for the kinetic ener gy of rotation and translation. We worked out such an expression in sect ion 7.10 . When no point in the body is spatially fixed, the natural origin for the body frame is the ce nter of mass. The kinetic energy splits neatl y into two contribution s: energ y of center-of-mass

motion ;

energy of rotational motion about center of ma ss. There are two kinem atic quantitie s, the velocities R and w . With each is assoc iated some cha racterization of the body 's mass: the total mass I.ma with R and the mome nt of inertia tensor/ u with w . The way in which the total mass is distributed throughout the body is crucial for the rotational energy.

WORKED PROBLEMS WP7-1 The star s in a n oblate elliptical galaxy provide a mas s di stributi on that

we can approximate by a uniform sphere of mas s plu s an equatorial ring. For a du st grain in orbit within the galaxy, the potential energy is the su m of two contributi ons. Acco rdin g to figure 1.5- 2, the sphere contributes

Usphere

= ½kr2 -

con st

for so me appropriate po sitiv e co nst a nt k (me ant to look like a spr ing constant beca use the assoc iated grav itationa l force is a linear restoring force) . For the equatorial ring , we ca n turn to equation ( 1.6-6) and abbreviate it s contributi on as

u ring=

-

e ( r2 -

3z2 )

-

const ',

where e is a pos itiv e constant proportion a l to the ring 's mass. The du st grain is , let us suppose , in almost circular orbit about the galactic center. Problem. How does the orientation of the grain 's orbit change with tim e (ove r the long run )? An order-of-magnit ude re spon se will suffice .

WORKED

/

/

PROBLEMS

277

/

/

Figure WP7-t The orbit relative to the galactic equatorial plane .

i

An orientation question calls for the angular momentum L. If a torque tips L, the orbital plane will change, too. The du t grain experiences a force

F = -grad

U = -!ff

+ e (2r -

6zt).

The radial portion produce no torque (relative to the galactic center). To see what the portion - 6ezi does, let us look at figure WP7- I. That force pushes down where the orbit rises above the galactic equatorial plane ; it pu shes up when the orbit is below. We can infer an average torque directed along the unit vector (L x t)/(L sin 0). What is the size of the average torque, in order of magnitude ? When z ha s its maximum value. namely r sin 0, the magnitude of r x Fi s 6er sin 0 (r cos 0) because Ir x t i = r cos 0. The ave rage torque i a ignificant fraction of this product. For the angular momentum equation, we may write

dL -- 6r., • 0 co 0 LL X. t , \Jr 2 sin dt

Sin

0

whic h is valid as an average over one orbital period. The structure is akin to the "s imple precession" form we studied in sectio n 7.2. [A tran sposition of Landi in the vector product , which costs only a minus sign, put s the present equation in the form of equation (7 .2-2). ] The angular momentum will preces about i with a precession frequency given, in order of magnitude , by (6er 2 cos 0)/ L. Because the orbital plane is perpendi cular to L, the orbital plane will precess about t also. WP7-2 A molecule consisting of a carbon atom and a sulfur atom rotates abo ut an axis perpendicular to the line connecting the two atoms, as sketched in figure WP7-2. The center of mass is at rest , and the square of the angular momentum is L2 = 2.22 x 10- 5s joule · meter 2 · kilogram. The carbon-sulfur

278

EXTENDED

BODIES

I N ROTA TION

w

Figure WP7-2 The rotating ca rbon-sulfur molec ule.

separation is t' = 1.54 x 10- 10 meter . To adequate accuracy , the masses are m e= 12mP and m8 = 32mp , where m p is the proton 's mass. Problem. What is the kinetic energy of rotation ? To answer the question , we can use equation (7.10-3) , provided we have the connection between L (whose square is given) and w. A calculation of the moment-of -inertia tensor can follow the pattern in section 7.4 . If e 1 points from the center of mass toward the carbon atom along the symmetry axis , then

and Using these expressions in equation (7 .3-4) leads to

~

~) ,

0

I

where I

=

m e m s t72_ m e+ m s

(We can recognize the combination of masse s as the reduced mass.) The problem spec ifies that w is perpendicular to the symmetry axis e1 ,

and so w 1 = 0. Thus, by equatio n (7 .3-6), we have L = f uw1e1

= I (w2 e2 + w3 e3 ) = lw . Equation (7. I0-3) now implies

WORKED

KE = ½I(w/ + w/ )

PROBLEMS

279

= ½/w2

IU

=1- ,. Numerical evaluation gives / = 3.46 x 10- 46 kilogram · meter 2 and 23 KE = 3.21 x 10- joule. The problem pecified L2 becau se that quantity would determine the allowed rotational energie in a quantum mechanical treatment of the carbonsulfur molecule . WP7-3 A disk of con tant densit y p has a radius R and a height h. Problem. Compute the moment-of -inertia tensor . The origin of the body axes is to be placed at the center of ma s, and the unit vector e3 is to lie along the disk's symmetry axis. Qualitatively , a disk can be thin , like a pancake, or elongated , like a cigar. What value of the ratio h/R separates these two regimes? The moment-of-inertia tensor for the disk has the form displayed in equation (7 .6-2). We need compute only /33 and I * " We may decompose the disk into annuli and add their contributions. Because/ u is defined by a su m [in equation (7 .3-4)], such a superposition principle holds. The " summation " is best done by integration in cylindrical coordinates. We need to distinguish the radius variable in cylindrical coordinates from the position magnitude r. Let us write r cyt

= YX 2 + )'2 •

The element / 33 follows as

Because the elements I II and / 22 must be equal , we can use the same trick that we used for a ring in section 7.3. We work out the sum:

= 21TP[ 4R _ I

4

R h + 22

2

2(h)a]

3 2 2

- 2

h ) M dt sk R 2 ( I +} I R2 ·

Therefore _ I I * - 4 M d tsk R

2

2 (

1 h ) I + 3 R2

•

280

EXTENDED

BODIES IN ROTATION

Now that we have expressions for / 33 and/* in terms of h and R, we can compare their magnitudes. The ratio I */133 is I*

/

33

2

= 2I ( I + 3I Rh 2 ) •

This ratio is less than I if hIR < V'3= I. 73. A kinematic distinction between a thin disk and an elongated disk can be extracted from equation (7.8-9). If / 33 is greater than' *· then L lies between w a nd e3 . If / 33 is less than/ *• then L lies outside. Thus the ratio I */133 = I forms the boundary between the two regimes. Our calculation says that the boundar y is equivalent to the ratio h/R = \1'3.

PROBLEMS 7-1 The nucleus of the helium isotope ?He acts as a tiny spinning bar magnet. In an externally imposed magnetic field B, the nucleus experiences a torque

-yL x B, where L is the nuclear angular momentum and the constant -y is - 2.04 x 108 per tesla · second. In a typical laboratory field of0.6 tesla , how long does it take for the nucleus to pre cess 90°? 7-2 If the polar ice caps were to melt , the level of the seas would rise, the increase in height being /:,.h= 60 meters. Estimate the change that would ensue in the length of the day. (Describe your simplifying assumptions as you make them.) 7-3 A satellite circles the earth, moving close to its surface . The orbit is

inclined at 45° to the earth's equatorial plane. The earth 's equatorial bulge will exert a torque on the satellite. Use a sketch to help you to determine the direction of the average torque (i.e., the torque averaged over one round trip ). Next, use dimensional analysis to estimate the magnitude of the average torque . At what rate, expressed symbolically , does the satellite's angular momentum precess ? If L precesses through a full cycle in 7 weeks , how massive is the ear th' s equatorial bulge (relative to me), in order of magnitude ? 7-4 How would the results in section 7.4 change if the rod had a nonnegligible mass m, 00 ? That mas s is distributed uniformly along length t. 7-5 A piece of sheet metal , in the shape of an equilateral triangle , is pivoted at one vertex and sw ings in the plane of the triangle. Why must the frequen cy of small oscillations be some numerical multiple of (g/t) 112, where t is the length of a side? Determine the numerical factor . 7-6 A disk is suspended by a stiff wire , aligned along the disk 's symmetry axis. The wire , when twisted, produces a restoring torque proportional to the angular displacement (J from the equilibrium orientation: torque = - k' (J, where k' is a torsional constant. The disk has a radius of 0. 1 meter and a mass of 0.2 kilogram. The measured oscillation period is 6 seconds. (a) Use the data to calculate the torsional constant k'.

PROBLEMS

281

(b) Now a U.S. football is suspended vertically below the di k, so that the two objects rotate as a single rigid body. The oscillation period increa e to 9.82 econds. What is the U.S. football's moment of inerti a for rotation about its sym metry ax is? (c) When the U.S. football is suspen ded horizontall y, the period increa e again, to 11.70 seconds. What is / 33 / / • for the football? 7-7 To teach circular motion in an introductor y laborator y, one often swi ngs a rubber stopper around at the end of a string . The st ring trace a co nica l surface . Suppose one replaces the tring with a rigid rod (but a mas sless one), pivo ted at one end, as sketched in figure P7-7 . The rotating syste m consi ts of the massless rod (of length t') and the concentrated mas m . At the instant shown. the veloci ty of the ma ss is into the paper . (a) Describe or calculate the angular momentum L of the system and the gravitationa l torque on it. (b) Calcu late the rotational angular velocity as a function of 8, t', and g by s({lrting with th e dy nami ca l connection between angular momentum and torque. Be sure to include careful , faithful sketc hes. (c) C heck you r result in part (b) by using a more elementary method to calcu late the angular velocity. 7-8 Return to the shell and four masses of ection 7 .5. Although w is constant in time, the angular momentum L is not (because e3 move s through pace). Calcu late the torque that is being ap plied externally to maintain the hell and masses in uniform rotation. For which angles of inclination 8 need no torque be applied? Explain with sketc he s. T7-9 Transforming the inertia tensor. Suppose we change from one set of body axes, the set e1, to another set , the set e;. The origins are specified to coincide, and so the change in axes is to a set rotated relative to the original set. How is the new moment-of -inertia tensor related to the old one? For mas s m,., the new po sition coordinate x 1a is given by

Ca n you use this co nnection to prove that f u= (e; . ek) (ej . e,)

,,., ?

pivot

The instanta ne ou5 ve loc ity is into the paper .

m

Figure P7-7

282

EXTENDED

BOD I ES I

ROTAT ION

If you are familiar with matrice , can you cast this tran sfo rmation law into matrix multiplication form ? For any given se t of body axes fixed in a rigid body , the moment-of-inert ia tensor is an array of con tant . The array undergoes no dynamical change. If new axes are chose n, however, then the array change s (but the new values are dynamically constant, too). T7-10 Parall el-axis theor em. Suppose we displ ace the origin of the body axes from the ce nter of mas s to a location a di stance t away. The or ientation of the axes remains unaltered. Can you show that the new moment-of-inertia tensor / 0 is given by f u = M (f'2 ou - Cf1)

+ U ulc M?

Here M denote s the body's total ma ss, and Uu)0 1 i the inertia ten or ca lcu lated with the ce nter of mas as origin. The connection between inertia tensors is called the parallel-axis theo rem. When a body has ymmetries. o that(/ ulc Mis ea y to calculate, but the required origin is at an awkward location, such as on the periphery of the body, the theorem i handy. 7-11 A U.S. footba ll ha s the symmetry of a disk but with / 33 = 0.6/ * < I.,. We ma y take / * = 3 x I0- 3 kilogram · meter 2. Suppo e that when the ball ha been poorly thrown, the ymmetry axi e3 de scribes a cone in pace with a half-a ngle of 10°. The axis make s six circuits of the cone per second, as seen from the ground. (a) Calculate the angular momentum L, the component w 3 of the angular velocity along the sy mmetry axis, and the magnitude w. (b) Suppose an insect were riding along on the ball. At what rate would the insect see w precess about the ymmetr y axis? Would the prece sion be in the sa me sense as the rotation w itself? (c) Sketch the space and body cone , a nalogous to tho e in figure 7.8-2. Why is there a qualitative difference? 7-12 The average rotational kinetic energy of a dust grain i ½kT , where /.. is Boltzmann's co n tant. 1.38 x 10- 23 jou le/kelvin , and T i the temperature of th e interstellar du t c loud. A value of T = 80 kelvins would be typical. Dust gra ins have irregular shapes; let us merely say the grain is about 5 x 1Q- 7 meter across. The den ity of grains is similar to that of ice or silicate rock; take a density of 2 x 103 kilogram per cubic meter . With this information , estimate the rotational frequency of a grain (to within a factor of 3 or so). 7-13 Suppose the novel pendulum of section 7.4 and its support were inverted , o that the rod stood above th e pivot. For a pict ure , ju t turn figure 7 .4- 1by 180 . The vertical orie nt at.io n i unstable. Y ou balance the rod as best you can and then let it go. Ca lcu late the angu lar velocity iJas a function of orientation . What is its va lue w hen th e rod hit s the urface , havin g tipped over 90°? (No te that yo u need not so lve a complica ted dynamical eq uation for O as a function of time .) Cou ld you ge nera lize your method to an inverted pendulum with a more comp licated distribution of mas s?

PROBLEM S

283

T7-14 A di k suc h as the one in sect ion 7.8 rotates abou t its center of mas ,

which is at re t. Derive an expression for the kinetic energy in term of L2 and L3 • the angular momentum along the the symmetry axi , together with th e two const ant I * and / 33 th at characterize the moment-of -inertia tensor . (The structure carrie s over directly to the quantum mechanical treatment of a molecule or nucle us with the ame ymmetry .) 7-15 Use equa tion (7. I 1-2) to confirm conservation laws (7. I 1-3) and (7 . I 1-4). Bea r in mind that vector e3 move as seen from the inertial frame . T 7-16 Th e f as t top. If the top in ection 7.11 pin very fast around its symmetry ax is. then L = Lae3 may be a good approximation. (This amount to ignoring- here - the contributio n of nutation and precession to L.) Given the ap pro xim ation. we ca n write the torque in term of L rat her then e3 • What do you predict for the motion of a " fast top " in this approximation ? If the top i about the size of your fi t and you sp in it o that w 3 / (27T) = 30 revolutions/second, e · timate the prece ion frequency . T7-17 Th e top more formall y. Thi s problem outlines more formal route s to equation (7.11-6) and (7.11-11). (a) To derive an expression for iJin terms of w, star t with cos 0 = i · e3 . Differentiate both sides with respect to time, and then use equatio n (6. 1-7) on the right. Can you ~olve for 8 as 8 = w · ( ome unit vector) ? (b) We can compute the top's pre cessional angular frequency by determin ing the precessiona l motion of the unit vector tl .1, equa l to i x e3/sin 0. Calcu late dii.J./dt, u ing equa tion (6. I-7). That step will introduce w ; write it in terms of uw, ul.. a nd e3 • With the aid of figure 7.11-2 and equa tion (7. 11-6), reduce your equ ation to dtl .1 WI -d = - .t

m0

,

ZX

tl .1.

How doe s thi re sult confirm equation (7. I 1-11)? T 7-18 The context i ection 7. I I. Suppose we started the top with 8(0) = 0 but with some azimuthal motion , by giving the top a gentle sideways push . How would the symmetry axis move ? Only a qualitative discu sion and a ketch are a ked for. 7-19 I it po ible to tart the top (of section 7 . 11) o that it prece e s only ? (T his means " no nutation " or 0 = constant.) Can you describe the condition on L3 and Lz that mu t be met ? (There is no need to work them out in detail.) T7-20 Th e s/eepillR top . Suppo e the top in ection 7 . 11 is initially vertica l, so that Lz = L 3 and 8(0) = 0. Exami ne Uerr near 0 = 0, calcu lating its first and second derivative at 0 = 0 a well as its value. How fa t must the top be spinning if the vertical orie ntat ion i to be stable (against perturbation by a puff of wind that produces a tiny 8 # 0)? When spi nnin g sufficient ly fast , the top remains - seemingly motionles -in the vert ical orient a tion; it "sleeps" there . 7-21 A dumbbell of length D. with a mass m at each end, i suspended

from a

284 EX f EN DED BO DI ES IN ROTA rION poinl of suppon

Figure P7-21

point by tw o strings , eac h of length t. (See figure P7-2 I .) The dumbbell is in gentl e osc illation in th e plane of th e ske tch. Th e string s remain tau t throu ghout the mot ion. Moreover , their ma ss a nd that of the rod conn ec ting the two end masses are neg ligible re lative to 111. Co mpute th e frequen cy of the mall oscillations, expre ss ing yo ur final answe r in term s of no paramete rs ot her than m, g, t, and D. D oes yo ur re sult redu ce to a familiar expression if yo u let D go to zero ?

lJI is slightly cigar-s haped and spins ra pidl y abo ut it s sy mm etr y ax is. Supp ose the ato mic electro ns produce an electric field which, in th e vo lume occ upied by the nuc leus , may be expressed a 7-22 Th e nuc le us of th e iodi ne isotope

E (x, y . z)

1

= C[-(x x + yy) + 2zi] = C[-(xx + yy + zi) + 3zi],

where C is a pos itive co nsta nt. (The field has cy lindr ical - bu t not phericalsy mmetr y beca use th e distributio n of a tomic electron s ha the fir t sym metry but not the seco nd .) F o r purpo ses of est imation , we may rega rd th e nucleus a a spher e (of c ha rge Q. a nd radiu s R ) ce nt e red on the origi n plu s two thin caps (of c ha rge Q c eac h) wher e the a ngular momentum ax is emerges from the nucleus. Th a t ax is is inclined (initi ally, at leas t) a t a n angle 0 relat ive to the z axi . Ca n yo u use sy mmetr y to show th at the net e lec tric force on the nucl eus is zero ? Next , es tim ate th e magnitud e an d direction of the net torqu e. What i the resultant nucl ea r mot io n like? Fin a lly, for whic h quantiti es wo uld yo u need numeri ca l va lues in order to estim ate num e rica lly the natur a l frequ ency in the problem ? 7-23 Suppo se a torque is applied to a disk whose height-to -radiu s ra tio is 3. Wou ld the di sk' s re pon se to the torque be differe nt from th at of a phere (having the ame radiu s and a mass 1.25 tim es la rger)? Wh y or why not? T7-24 Int erm ed iat e-ax is theorem. Co nsider an objec t with no rotatio nal sym-

PROBLEM

285

metry, such a a tennis racket. When principal axes are used for the body axes , the moment-of -inerti a ten or will be diagonal , but all three diagonal element u ua lly differ from one another in numeri ca l value . (a) Suppose the object is set spinning with w(O) precisely aligned along one princi pal ax is, which we may call the e3 axi , for definitenes s. No torque s act. Will the object continue to rotate purely about that bod y axis ? (b) Suppose w(O) is almost aligned withe 3 • Co mponent w 1(0) and wiO) are no nzero but small relative to wa(O). Is suc h rot ation stable ? That is, will w con tinue to lie close to e3 ? You will need to exa mine three situations: / 33 is the large t , mallest , or intermediate moment of inertia . Eu ler 's equations provide a route to the answer . You can te t your prediction b y to ss ing a tennis racket or a book held closed with an e lastic band . T 7-25 Angular mom entum relative to th e origin of th e inertial fram e. In sec tions 7. 1 a nd 7 . 3 we computed torque s a nd the angular momentum by forming vector products with the displacement r O from the origin of the body frame. Wha t form do the equations take if you use the displacement R + rO from the origin of the inertia l frame , under the specification that the origin of the body frame is placed at the center of mas s? After you have eliminated those terms that are identica lly zero , can yo u interpret the remaining term s? 7-26 In a classical picture , an electron moves aro und a mass ive nucleu s with orbita l angular momentum Lorbtt· As an object with intrin ic angular momentum, the electron has angular momentum Lspln relative to its center of mass. When the spinning charge move s through the spherically ymmetri c electric field produced by the nucl eus , torque s arise, so that Lspln obeys the equation dLspln -"' L L dt - '-=- orbitX spin• For a ne arly circular orbit , we may take e to be a con tant. (a) On general principle s, what equation would yo u expect Lorbltto sa tisfy ? (b) For general initial conditions , determine the evolution ofL sptn and Lorbtt as far as you can . 7-27 A dumbbell- haped sa tellite circle s th e Earth at a dista nce r 0 • In general , w hat kind of tumbling or oscillatory motion should we expect? (Motion about the sa tellite ' s center of ma ss is me ant he re.) Be quantitative where yo u can be . (You are welcome to simplif y by specifying that aJI motion is confined to the orbital plane .)

CHAPTER

EIGHT CROSS SECTIONS

8.1 8.2 8.3 8.4 8.5

Scattering effectiveness: The idea behind the cross sec tion A capture cross sect ion A differenti al cross sect ion Rutherford scattering Major ideas

I had observed the scatt ering of alpha particles, and Dr. Geiger in my laboratory had examin ed it in detail. H e found , in thin pieces of metal, that the scattering was usually sma ll, of the order of one degree. One day Geiger came to m e and said, "Don't yo u think that yo ung Marsd en, whom I am trainin g in radioactive methods, ought to begin a small resea rch?" Now I had thought that, too, so I said, " Why not let him see if any alpha particles can be scatter ed through a large angl e?" Lord Rutherford Background to Modern Science

8.1 SCATTERING EFFECTIVENESS: THE IDEA BEHIND THE CROSS SECTION A young hot star produces a fierce stellar wind , a stream of particle s moving radially outward into space. (Particles in the stellar surface absorb light, thus acqu iring momentum from the photons , and recoil outward, forming the wind.) Suppose a dust grai n finds itself in thi s wind. Some of the hydrogen atoms in the 286

8.1

SCATTER ING EFFEC TIVENESS : T H E IDEA BEHIND

D1rec11on of wind

I

----o I

0 ]II,

0 0

287

o-------·

I

0

0

fH E CROSS SECTION

]II,

]II,

o---'1]11,~ --

---

o---'1]11,~ - -

Figure 8.1-1 A dust grain in a wind of hydrogen atoms. The grai n will scaue r some atoms (the ben t dashed line depicts the traje ctor y of one suc h atom) but leave other atoms unaffected . as indicated b} the undeviated dashed line.

wind will bounce off the grain ; ot her s, thoug h nearby, will miss the grain. The situation is sket ched in figure 8.1- 1. How many scatterings per seco nd hould we expect? Fir t we need to determine which initial tr aje c torie s lead to a co llision. Let R denot e the grain's radius, a nd let R Hdenote the radiu s of a hydroge n atom. If an initial trajectory lies within a distance R + RH of the ce nterline , as sketc hed in figure 8.1-2 , the atom will sca tter off the grain . An ato m must be in a specific volume if it is to strike the grain within the next seco nd . In time tlt , an atom moves a distance v M . Thus, with tlt = I second, the a tom must be so mewhere in the cy lindri ca l volume 1r( R + R H)2 v to the left of the grain. If the number of hydroge n ato ms per unit volume is n 11, we can write

---

·

---------r~r--------~1 Figure 8.1-2 An initial tr ajec tory within the perpe ndicular dista nce R lead to a coll isio n.

+ R 11 of the

ce nterline will

288 C ROSS SECT ION S

Number of scatterings per second =

1111v1r(R

+ Ru )2 .

(8.1-1)

If the wind become s more inten se, through an increas e in nu or v or both, the scattering rate goes up . The grain is not more effective at scattering than before ; it just gets more chances to scatter. To define an honest measure of scattering eff ec tiven ess, we need to divide the sca ttering rate by some measure of the scattering opportunities. A good choice is the produ ct ntt v, called the incident flux. It is the number of pr ojectile atoms that flow across unit area (oriented perpendi cular to the wind's direction) per second. Thus . . number of sca ttering s per second Mea sure of sca ttenng effectiveness = · ·ct t fl mc1 en ux (8. 1-2)

The ratio th at provides the effect iveness meas ure has the unit s of an area and is called the cross sec tion for sca ttering . Our c ro ss sec tion depend s on the sca tterer (through R) and on the incident partic le (through R u). It is not a property of either se parately. Thinking of the cross section as an area that th e sca tterer pre sents to the projectile wind is a good idea. Any par ticle whose incident trajectory would intersect the area will be scattered. We can recover -o r com pute - the scattering rate from the product of the cross sec tion and the incident flux. Remember, however , that the scattering cros s section is not the sa me thing as the geometrical cross section of the target. The distinction is evident here: 1r(R + Rttl 2 versus 1rR2 • Later in thi s chapter we study the capture of incident ion s by a charged dust grain. We study the deflection of alp ha particles by a heavy nucle us. These are two more insta nces of a perv asiv e context: a "ta rget" object does something to some "projecti le" particle s in an inciden t beam. The "somet hing" can be scatteri ng or capture or molecul e forma tion or . ... Ca ll it, for short, an event of type X. We can always define a measure for the effectiveness of the proces s by a ratio: Effectivene ss meas ure _ number of eve nts of type X per seco nd for process X incident flux

*

The ratio is then ca lled the cross sec tion for events of type X. A co mmon symbol would be CTx. To se e the notat ion in use , we can write equation (8. 1-2) as CT scattering=1r( R

+ RH)2.

What good is a cros s sect ion ? There are at leas t three facets to the answer. Fir st , a measured cross sect ion conveys inform at io n about the inter action of projectile and targe t and about their individual properties . In our exampl e, we could ex tract R + Rtt, the sum of the rad ii, from a meas urement of CT scatterina· Second , a theory about the behavior of the proj ec tile and target particle s can be tested by compar ing the predicted a nd measured cross sec tio ns. Thi s strategy

8.2

A CA P I U RE C RO S

1:.Cl ION

289

has been a major theme in phy ic since the time of Ruth erford during the first deca de s of thi s century. Third , wit h a known cross section we can compu te the event rate to be expected for any specified incident flux.

8.2 A CAPTURE CROSS SECTION An inter tella r du st grain is often charged, u ua lly by an exce s of electrons; so let us endow our grai n with a n electric charge Z (-e) . where Z is ome positive intege r and -e is the charge on an electron. ( Plain e denotes the proton' s charge.) The tellar wind contains a variet y of pa rticle s, not ju t hydrogen atoms . Let us focu on ome po sitive ions: cha rge Z 1e , ma s 111, a nd incident peed v0 • If uch an ion trike s the surface of our grain , it will- let us suppose - be held there : it has been captured . What is the cro s section for such ca pture ? Figure 8.2-1 ske tche s the situation when the ion is still far from the grain and shows also the trajectory that will be followed . The attractive electric for ce betwee n ion and grain bend s the initial trajec tory toward the grain. The distance b is the decisive quantity . As sket ched, b is small enough that capture occurs. If b were made much larger , however , the tr ajectory would bend in but then wing around behind the grain , and the ion would depart to infinity : no ca pture . Thi s pos ibility is indi ca ted by the equence of short dashes in the figure . The cruci al qu e tion for us is this: How sma ll a value must b have if the io n is to strike the grain? To answer the question , we ca n use an effective potenti al energy to tell us the minimum distance between the ion and the du st grai n on a trajectory that is pecified by band the other pa rameter s in the problem . Equation (5.8- 4) gives the struc ture of such an energy as

----------

---

r _________ :'·\ Vu

---------~

-.

-

\ I

I I I

; Figure 8.2- 1 Tw o tr ajec torie s of the ion rela tive to the grain .

290 CROSS SECT IONS

Uerr(r; L ) =-

L2 2mr 2

+ U (r) ,

where r denotes the ion 's position relative to the center of the dust grain. Becau se the grain is so much more massi ve th a n th e ion, we may regard it as perm anently a t rest , and the reduced mas s become s merel y m, the ion's mass. To evaluate the angul ar momentum L , we can use figure 8.2-2. Reasoning from the length s a nd a ngles dr aw n there , we have

L

=

lmr0 X v0 1=

mv 0 r0 sin f3 (8.2-1)

beca use f3= {3' a nd b = r0 sin /3'. The e lectri ca l attrac tion va ries as I /r2, and so the cor respo nding Cou lomb pot en tia l energy will va ry as 1/r: U

=

K (- Ze ) (Z;e). r

Here K is th e co nsta nt in Cou lomb' s law, dependent for its numeri ca l value on the unit s used for charge; the mo st common choices are listed in appendix A. The total ene rgy £ can be eva luated mo st easily in the initial co nfiguration where the distance is so larg e that U = 0. Thus

E=½mv/. The run of Uerr with rand the tot a l e nergy line a re sket ched in figure 8.2-3. The minimum separat io n along the trajectory occurs where th e c ur ve and line intersect. If we write L in term s of b, we have, at the minimum ,

(mv 0 b) 2mrmln

2

+ U(r

)

=

£.

min

The value of b for which rm in equals R , the grain 's rad ius, i the cri tica l value . Ca lling it be, we can so lve for it from

(mvobc) 2m R 2

2

+

U ( ) _ 1_ 2 R - :imvo '

which yield

b/=

R 2 [1 - U( R ) I (½niv/)] .

v.,

Figure 8.2-2 The initial configuration. Beca use th e centerline is draw n para llel to v0• angles /3 and {3' a re equal.

8.3 /\

Dll · F E Rl' I'< I IA I C ROS

FC

no

29 1

Figure 8.2-3 A graph of the effec tive potential energy for th e ion.

All ion with b > b,. mi the grain; tho e with b s hr strike it and are ca ptured . In effect. the grain hold s up a circ ula r area 1rbr 2 to the distant wind and captu re a ll ion w ho e initial tr ajec tor y wou ld take them through it. Thu s we ca n write the capture cross sec tion as CTcap tur e

= 1rb/ =

2

1rR2[ I

+ ( KZZ R '-e ) /

(imvoz) ] .

The cros section depend s, we find, on the energy of the ions. If that energy i muc h larger tha n the pot enti al ener gy (in ab olute value) at the grain 's surface , then CTrap tur e i approximately 1iR 2 , the geometric cross sec tion . We can unde rsta nd th at. The traje c tor y for an ion of ve ry lar ge energy will be a lm ost a straight line , and o the ion mu st be headed for a dire c t colli ion . independent of the electri ca l attra ction , if it i to be ca ptured. If, however , the ion moves lowly w hen far away , then th e e lec tri ca l attraction can pull it in to a ca pture eve n if b is very large. H e nce th e cros ection rise as the energy decrease . The quantit y b ha s bee n a valuable tool in our ana lysi and will continue in that role throughout the c ha pter. It s definition is be st extracted from figure 8.2-2: the perpendi cula r di ta nce from the initi al trajectory to a parallel line running through the target 's center. It s co mmon name , the impa ct parameter , ari e becau e b pec ifies where a strai f?ht -li11e trajectory would pier ce a plane at the target. a pla ne oriented perpendicular to the centerline .

8.3 A DIFFERENTIAL CROSS SECTION Let u return no w to the neutral du st gra in and hydrogen atom . A glance at figure 8. 1- 1 is enou gh to sugge st that atoms are cattered in variou s direction s. Some ato ms are deflected only a little ; others go off at angle ne ar 90° ; till

292 CROSS SECT IONS

v,.

b

Figure 8.3-1 Relating the scatter ing angle to th e impa c l parameter .

others are sca ttered almost dead backward. The angle of scattering depend s on the impact parameter. Let us derive the exact relationship. Figure 8.3-1 displays the essential geometry. The angle 0 is the scattering ang le, the angle through which the int eract ion deflects the initial trajector y. The di stance between centers at the instant of collision is R + R 11• Provided the atom and grain act as hard , smooth spheres, the area of contact acts as a locall y flat surface. During imp act , the component of ve locity alo ng the co nne cting line is reduced to zero and then is reversed. The perpendicular component is not changed. Figure 8.3-2 shows how the reversa l transmutes the incident velocit y v0 to the final velocity. From the figure we can also see why the angle labeled t0 there and in figure 8.3-1 has that value. The angle between Vr;n ai and v0 is 0. Thus the angle between v .1. and v 0 is !0. That angle, plus the a ngle between v0 and the connecting line, equals 90°. But the latter angle, plus the angle labeled ½0,also equals 90". Thus the label indeed denotes th e true value of the angle. Now we can expre ss the scatteri ng angle in terms of the impact paramete r. The triangle with a ng le t0 in figure 8 .3-1 lets us write

b= (R + R 11) cost0.

(8.3-1 )

Moreover, we can see how b must cha nge to give scattering through an infinite sima lly larger a ngle 0 + d0. Differentiating equation (8.3- 1), we find

db= (R

+ R 11) (-t

sin ½0)d0.

(8.3-2)

The change db is negat ive. implying a sma ller impact parameter. If we look at figure 8.3-1 and mentally reduce b, we see that , inde ed. th e scattering angle will increase.

8.3

A DIFFEREN

fl AL C ROSS SECT ION 293

V

rever~al

Figure 8.3-2 The reversal of 1he velocit y component that lies along the con necting lin e. The vect o r v . denotes the component of velocity perpendicular to the connecting lin e.

Figure 8.3-3 is a useful prelude to the next step . It shows a sphere of radius R l!:>R ce ntered on the du st grain. Tw o circles a re dr aw n by rotat ing the radiu s vec tor , first at polar angle() a nd then at() + d0. Th at generates a band. Finally a sec tion of the band i marked off by an incre ment di{) in the azimuthal ang le: arcs at "' and '{)+ di(). The patch thu s defined is a n infinitesima l rectangle with side CRd() a nd cH sin () di(). It de fines a ce rtain set o f direc tions: tho se str aight lines that ema nate from th e origin and pass through the patch. To cha racterize th at bund le of dire ctions, we speci fy angles () a nd '{) togeth er with the ratio (patc h a re a)/