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Advanced undergraduate textbook on classical mechanics by Stephen T. Thornton and the late Jerry B. Marion. The front matter shown covers the preface, course suitability, teaching aids and acknowledgments. The preface describes vector methods, nonlinear oscillations and chaos, calculus of variations, and rigid bodies. It is a published book kept in Phil's downloaded physics books, not his own writing.

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CLASSICA DYNAMIC Tl-IOlVlSC)I\I BROOKS/COLEOFPARTICLES AND SYSTEMS FIFTH EDITION Stephen T.Thornton Proﬁzssor ofPhysics, University ofVirginia Jerry B.Marion LateProfessor ofPhysics, University ofMaryland Australia Canada 'Mexico 'Singapore vSpain United Kingdom 'United Slates THOIVISCJN BROOKS] COLE Acquisitions Editor: Chris Hall Assistant Editor: Alyssa White Editorial Assistant: Seth Dobrin Technology Project Manager: Sam Subity Marketing Manager: Kelley M<:Allister Marketing Assistant: Sandra Perin Project Manager, Editorial Production: Karen Haga Print/ Media Buyer: Kris Waller COPYRIGHT ©2004 Brooks/ Cole, adivision ofThomson Learning, Inc.Thomson LearningTM isatrademark used herein under license. ALL RIGHTS RESERVED. Nopartofthiswork covered bythecopyright hereon mayberepro- duced orused inanyform orbyanymeans-— graphic, electronic, ormechanical, including butnotlimited tophotocopying, recording, taping, Web distribution, information net- works, orinformation storage andretrieval sys- tems--without thewritten permission ofthe publisher. Printed intheUnited States ofAmerica 12345670706050403 Formore information about ourproducts, contact usat: Thomson Learning Academic Resource Center 1-800423-0563 Forpermission tousematerial from thistext, contact usby:Phone: 1-800-730-2214 Fax: 1-800-730-2215 Web: http://www.Lhomsonn'ghts.com Library ofCongress Control Number: 20031 05243 ISBN O-534-40896-6Permissions Editor:_]oohee Lee Production Service andCompositor: Nesbitt Graphics, Inc. Copy Editor:_]ulie M.DeSilva Illustrator: Rolin Graphics, Inc. Cover Designer: Ross Calron Text Printer: Maple-Vail Book Mfg. Group Cover Printer: Lehigh Press Brooks/C0le—-Thomson Learning 10Davis Drive Belmont, CA94002 USA Asia Thomson Learning 5Shenton Way #01-O1 UIC Building Singapore 068808 Australia /New Zealand Thomson Learning 102Dodds Street Southbank, Victoria 3006 Australia Canada Nelson 1120 Birchmount Road Toronto, Ontario MIK 5G4 Canada Eur0pe/ Middle East/Africa Thomson Learning High Holbom House 50/51Bedford Row London WCIR 4LR United Kingdom Latin America Thomson Leaming Seneca, 53 Colonia Polanco 11560 Mexico D.F. Mexico Spain/Portugal Paraninfo Calle/Magallanes, 25 28015 Madrid, Spain To DrKathryn C.Thornton Astronaut and I/Vzfe Asshesoars and walks through space, May herlzﬁebesafeandfulﬁlling, And letourchildren’s minds beopen forallthatlzﬁzhastooﬂei: Preface Oftheﬁveeditions ofthistext, thisisthethird edition thatIhave prepared. In doing so,Ihave attempted toadhere tothe1ate_]erry Marion’s original purpose of producing amodem andreasonably complete account oftheclassical mechanics ofparticles, systems ofparticles, andrigid bodies forphysics students atthead- vanced undergraduate level. The purpose ofthebook continues tobethreefold: 1.Topresent amodern treatment ofclassical mechanical systems insuch away that thetransition tothequantum theory ofphysics canbemade with the least possible difﬁculty. 2.Toacquaint thestudent with new mathematical techniques wherever possi- ble,andtogive him/ hersufficient practice insolving problems sothat the student may become reasonably proﬁcient intheir use. 3.Toimpart tothestudent, atthecrucial period inthestudent’s career be- tween “introductory” and“advanced” physics, some degree ofsophistication inhandling both theformalism ofthetheory andtheoperational technique ofproblem solving. After aﬁrm foundation invector methods ispresented inChapter 1,further mathematical methods aredeveloped inthetextbook astheoccasion demands. Itisadvisable forstudents tocontinue studying advanced mathematics insepa- ratecourses. Mathematical rigor must belearned andappreciated bystudents of physics, butwhere thecontinuity ofthephysics might bedisturbed byinsisting oncomplete generality and mathematical rigor, thephysics hasbeen given precedence. Changes fortheFifth Edition The comments andsuggestions ofmany users ofClassical Dynamics have been in- corporated into thisﬁfth edition. Without thefeedback ofthemany instructors v Vi PREFACE who have used thistext, itwould notbepossible toproduce atextbook ofsignif- icant value tothephysics community. After theextensive revision forthefourth edition, thechanges inthisedition have been relatively minor. Only afewre- arrangements ofmaterial have been made. Butseveral examples, especially nu- merical ones, andmany end-of-chapter problems have been added. Users have notwanted extensive changes inthetopics covered, butmore examples forstu- dents andawider range ofproblems arealways requested. Astrong effort continues tobemade tocorrect theproblem solutions avail- able intheInstructor and Student Solutions Manuals. Ithank themany users who sent comments concerning various problem solutions, and many oftheir names arelisted below. Answers toeven-numbered problems have again been in- cluded attheendofthebook, and theselected references andgeneral biblio- graphy have been updated. Course Suitability The book issuitable foreither aone-semester ortwo-semester upper level (jun- iororsenior) undergraduate course inclassical mechanics taken after anintro- ductory calculus-based physics course. AttheUniversity ofVirginia weteach a one-semester course based mostly ontheﬁrst 12chapters with several omissions ofcertain sections according totheInstructor’s wishes. Sections that canbe omitted without losing continuity aredenoted asoptional, buttheinstructor can alsochoose toskip other sections (orentire chapters) asdesired. Forexample, Chapter 4(Nonlinear Oscillations andChaos) might beskipped initsentirety foraone-semester course. Some instructors choose nottocover thecalculus of variations material inChapter 6.Other instructors may want tobegin with Chapter 2,skip themathematical introduction ofChapter 1,andintroduce the mathematics asneeded. This technique ofdealing with themathematics intro- duction isperfectly acceptable, andthecommunity isdivided onthisissue with a slight preference forthemethod used here. The textbook isalsosuitable fora fullacademic year course with anemphasis onmathematical and numerical methods asdesired bytheinstructor. The textbook isappropriate forthose who choose toteach inthetraditional manner without computer calculations. However, more and more instructors andstudents areboth familiar andadept with numerical calculations, andmuch canbelearned bydoing calculations where parameters canbevaried andreal- world conditions likefriction and airresistance canbeincluded. Idecided be- fore the4thedition toleave thechoice ofmethod totheinstructor and/ orstu- dent tochoose thecomputer techniques tobeused. That decision hasbeen conﬁrmed, because there aremany excellent software programs (including Mathematica, Maple, andMathcad tomention three) available touse. Inaddi- tion, some Instructors have students write computer programs, which isanim- portant skill toobtain. PREFACE Vii Special Feature The author haskept onepopular feature ofjerry Marion’s original book: thead- dition ofhistorical footnotes spread throughout. Several users have indicated how valuable these historical comments have been. The history ofphysics has been almost eliminated from present-day curricula, andasaresult, thestudent is frequently unaware ofthebackground ofaparticular topic. These footnotes are intended towhet theappetite andtoencourage thestudent toinquire into the history ofhisﬁeld. Teaching Aids Teaching aids to accompany the textbook are available online at http:ﬂinfo.brookscole.com/thomton. The Instructor’s Manual (ISBN 0-534- 40898-2) contains solutions toalltheend-of-chapter problems inaddition to Transparency Masters ofselected keyﬁgures from thetext. This password- protected resource iseasily printable in.pdf format. Toreceive your password, justgototheabove website andregister; ausemame andpassword willbesent to you once the information you have provided isveriﬁed. The veriﬁcation procedure ensures thatyouareaninstructor teaching thiscourse. Ifyouarenot able todownload theInstructor’s Manual ﬁlesandwould likeaprinted copy sent toyou, please contact your local sales representative. Ifyou donotknow who your sales representative is,please visit www.brookscole.com, and click onthe Find your Rep tab,which islocated atthetopofthewebpage. Please donotdis- tribute theInstructor’s Manual tostudents, orpost thesolutions ontheInternet. Students arenotpermitted toaccess theInstructor’s Manual. Student Solutions Manual AStudent Solutions Manual byStephen T.Thomton, which contains solutions to 25% oftheproblems, isavailable forsaletothestudents. Instructors areencour- aged toorder theStudent Solutions Manual fortheir students topurchase atthe school bookstore. Topackage theStudent Solutions Manual with thetext, use ISBN 0-534-08378-1, ortoorder theStudent Solutions Manual separately useISBN 0-534-40897-4. Students canalso purchase themanual online atthepublisher’s website www.brookscole.com /physics. Acknowledgments Iwould liketograciously thank those individuals who Wrote mewith suggestions onthetextorproblems, who returned questionnaires, orwho reviewed parts of the4thedition. They include ... William L.Alford, Auburn University Philip Baldwin, University ofAkron Robert P.Bauman, University of Alabama, Birmingham Michael E.Browne, University ofIdaho Melvin G.Calkin, Dalhousie University F.Edward Cecil, Colorado School ofMines Arnold Dahm, Case Western Reserve University George Dixon, Oklahoma State University John Dykla, Loyola University of Chicago Thomas A.Ferguson, Carnegie Mellon University Shun-fu Gao, University ofMinnesota, Morris Reinhard Graetzer, Pennsylvania State University Thomas M.Helliwell, Harvey Mudd College Stephen Houk, College oftheSequoias Joseph Klarmann, Washington University atSt.Louis Kaye D.Lathrop, Stanford University Robert R.Marchini, Memphis State UniversityPREFACE Robert B.Muir, University ofNorth Carolina, Greensboro Richard P.Olenick, University ofTexas, Dallas Tao Pang, University ofNevada, Las Vegas Peter Parker, YaleUniversity Peter Rolnick, Northeast Missouri State University Albert T.Rosenberger, University of Alabama, Huntsville Wm. E.Slater, University ofCalifornia, LosAngeles Herschel Snodgrass, Lewis andClark College J.C.Sprott, University ofWisconsin, Madison Paul Stevenson, RiceUniversity Larry Tankersley, United States Naval Academy Joseph S.Tenn, Sonoma State University Dan deVries, University ofColorado The present 5thedition would nothave been possible without theassistance ofmany people who made suggestions fortextchanges, sent meproblem solu- tion comments, answered aquestionnaire, orreviewed chapters. Isincerely ap- preciate their help andgratefully acknowledge them: Jonathan Bagger,_]ohns Hopkins University Arlette Baljon, SanDiego State University Roger Bland, SanFrancisco State University John Bloom, Biola University Theodore Burkhardt, Temple University Kelvin Chu, University ofVermont Douglas Cline, University ofRochester Bret Crawford, Gettysburg College Alfonso Diaz-Jimenez, Universidad Militar Nueva Granad, ColombiaAvijit Gangopadhyay, University of Massachusetts, Dartmouth Tim Gfroerer, Davidson College Kevin Haglin, Saint Cloud State University Dennis C.Henry, Gustavus Adolphus College John Hermanson, Montana State University YueHu,Wellesley College Pawa Kahol, Wichita State University Robert S.Knox, University ofRochester Michael Kruger, University ofMissouri Whee KyMa, Groningen University PREFACE ix Steve Mellema, Gustavus Adolphus Keith Riles, University ofMichigan College Lyle Roelofs, Haverford College Adrian Melott, University ofKansas Sally Seidel, University ofNew Mexico William A.Mendoza, jacksonville Mark Semon, Bates College University Phil Spickler, Bridgewater College Colin Momingstar, Carnegie Mellon Larry Tankersley, United States Naval University Academy Martin M.Ossowski, Naval Research LiYou, Georgia '1ech Laboratory Iwould especially liketothank Theodore Burkhardt ofTemple University who graciously allowed metouseseveral ofhisproblems (and provided solutions) forthenewend-of~chapter problems. Thehelp ofPatrick]. Papin, SanDiego State University, andLyle Roelofs, Haverford College, inchecking theaccuracy ofthe manuscript isgratefully acknowledged. Inaddition Iwould liketoacknowledge theassistance ofTran ngoc Khanh who helped considerably with theproblem so- lutions fortheﬁfth edition aswellasWarren Griﬂith andBrian Giambattista who didasimilar service forthefourth andthird editions, respectively. The guidance and help oftheBrooks/ Cole Publishing professional staff is greatly appreciated. These persons include Alyssa White, Assistant Editor; Chris Hall, Acquisitions Editor; Karen Haga, Project Manager; Kelley McAllister, Marketing Manager; Stacey Puwiance, Advertising Project Manager; Samuel Subity, Technology Project Manager; Seth Dobrin, Editorial Assistant, andMaria McColligan andstaffatNesbitt Graphics, Inc.fortheir production help. Iwould appreciate receiving suggestions ornotices oferrors inanyofthese materials. Icanbecontacted byelectronic mail [email protected]. Stephen T.Thornton Charlottesville, Virginia Contents Matrices, Vectors, andVector Calculus 1 1.1 Introduction 1 1.2 Concept ofaScalar 2 1.3 Coordinate Transformations 3 1.4 Properties ofRotation Matrices 6 1.5 Matrix Operations 9 1.6 Further Deﬁnitions 12 1.7 Geometrical Signiﬁcance ofTransformation Matrices 14 1.8 Deﬁnitions ofaScalar andaVector inTerms of Transformation Properties 20 1.9 Elementary Scalar andVector Operations 20 1.10 Scalar Product ofTwo Vectors 21 1.11 Unit Vectors 23 1.12 Vector Product ofTwoVectors 25 1.13 Differentiation ofaVector with Respect toaScalar 29 1.14 Examples ofDerivatives—Velocity andAcceleration 30 1.15 Angular Velocity 34 1.16 Gradient Operator 37 1.17 Integration ofVectors 40 Problems 43 Newtonian Mechanics—Single Particle 48 2.1 Introduction 48 2.2 Newton’s Laws 49 2.3 Frames ofReference 53 2.4 TheEquation ofMotion foraParticle 55 xi CONTENTS 2.5 Conservation Theorems 76 2.6 Energy 82 2.7 Limitations ofNewtonian Mechanics 88 Problems 90 Oscillations 99 3.1 Introduction 99 3.2 Simple Harmonic Oscillator 100 3.3 Harmonic Oscillations inTwo Dimensions 104 3.4 Phase Diagrams 106 3.5 Damped Oscillations 108 3.6 Sinusoidal Driving Forces 117 3.7 Physical Systems 123 3.8 Principle ofSuperposition—Fourier Series 126 3.9 The Response ofLinear Oscillators toImpulsive Forcing Functions (Optional) 129 Problems 138 Nonlinear Oscillations andChaos 144 4.1 Introduction 144 4.2 Nonlinear Oscillations 146 4.3 Phase Diagrams forNonlinear Systems 150 4.4 Plane Pendulum 155 4.5 jumps, Hysteresis, andPhase Lags 160 4.6 Chaos inaPendulum 163 4.7 Mapping 169 4.8 Chaos Identiﬁcation 174 Problems 178 Gravitation 182 5.1 Introduction 182 5.2 Gravitational Potential 184 5.3 Lines ofForce andEquipotential Surfaces 194 5.4 When IsthePotential Concept Useful? 195 5.5 Ocean Tides 198 Problems 204 Some Methods intheCalculus ofVariations 207 6.1 Introduction 207 6.2 Statement oftheProblem 207 6.3 Euler’s Equation 210 CONTENTS 6.4 6.5 6.6 6.7--- The “Second Form” oftheEuler Equation 216 Functions with Several Dependent Variables 218 Euler Equations When Auxiliary Conditions AreImposed 219 The 8Notation 224 Problems 226 7 Hamilton’s Principle—Lagrangian and Hamiltonian Dynamics 228 7.1 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.11 7.12 7.13Introduction 228 Hamilton’s Principle 229 Generalized Coordinates 233 Lagrange ’sEquations ofMotion in Generalized Coordinates 237 Lagrange’s Equations with Undetermined Multipliers 248 Equivalence ofLagrange’s andNewton’s Equations 254 Essence ofLagrangian Dynamics 257 ATheorem Concerning theKinetic Energy 258 Conservation Theorems Revisited 260 Canonical Equations ofM0tion—Hamiltonian Dynamics 265 Some Comments Regarding Dynamical Variables and Variational Calculations inPhysics 272 Phase Space andLiouville’s Theorem (Optional) 274 Virial Theorem (Optional) 277 Problems 280 8 Central-Force Motion 287 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 8.9 8.10Introduction 287 Reduced Mass 287 Conservation Theorems—-First Integrals oftheMotion 289 Equations ofMotion 291 Orbits inaCentral Field 295 Centrifugal Energy andtheEffective Potential 296 Planetary Motion—Kepler’s Problem 300 Orbital Dynamics 305 Apsidal Angles andPrecession (Optional) 312 Stability ofCircular Orbits (Optional) 316 Problems 323 9 Dynamics ofaSystem ofParticles 328 9.1 9.2 9.3Introduction 328 Center ofMass 329 Linear Momentum oftheSystem 331 XIV 10 ll 129.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11CONTENTS Angular Momentum oftheSystem 336 Energy oftheSystem 339 Elastic Collisions ofTwo Particles 345 Kinematics ofElastic Collisions 352 Inelastic Collisions 358 Scattering Cross Sections 363 Rutherford Scattering Formula 369 Rocket Motion 371 Problems 378 Motion inaNonintertial Reference Frame 387 10.1 10.2 10.3 10.4Introduction 387 Rotating Coordinate Systems 388 Centrifugal andCoriolis Forces 391 Motion Relative totheEarth 395 Problems 408 Dynamics ofRigid Bodies 411 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9Introduction 411 Simple Planar Motion 412 Inertia Tensor 415 Angular Momentum 419 Principal Axes ofInertia 424 Moments ofInertia forDifferent Body Coordinate Systems 428 Further Properties oftheInertia Tensor 433 Eulerian Angles 440 Euler’s Equations foraRigid Body 444 11.10 Force-Free Motion ofaSymmetric Top 448 11.11 Motion ofaSymmetric Topwith One Point Fixed 454 11.12 Stability ofRigid-Body Rotations 460 Problems 463 Coupled Oscillations 468 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8Introduction 468 Two Coupled Harmonic Oscillators 469 Weak Coupling 473 General Problem ofCoupled Oscillations 475 Orthogonality oftheEigenvectors (Optional) 481 Normal Coordinates 483 Molecular Vibrations 490 Three Linearly Coupled Plane Pendula—an Example of Degeneracy 495 CONTENTS 12.9 The Loaded String 498 Problems 507 13 Continuous Systems; Waves 512 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 13.9 14 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8 14.9Introduction 512 Continuous String asaLimiting Case ofthe Loaded String 513 Energy ofaVibrating String 516 Wave Equation 520 Forced andDamped Motion 522 General Solutions oftheWave Equation 524 Separation oftheWave Equation 527 Phase Velocity, Dispersion, andAttenuation 533 Group Velocity andWave Packets 538 Problems 542 Special Theory ofRelativity 546 Introduction 546 Galilean Invariance 547 Lorentz Transformation 548 Experimental Veriﬁcation oftheSpecial Theory 555 Relativistic Doppler Effect 558 _ Twin Paradox 561 Relativistic Momentum 562 Energy 566 Spacetime andFour-Vectors 569 14.10 Lagrangian Function inSpecial Relativity 578 14.11 AppendicesRelativistic Kinematics 579 Problems 583 A Taylor’s Theorem 589 Problems 593 B Elliptic Integrals 594 B.1 B.2 B.3Elliptic Integrals oftheFirst Kind 594 Elliptic Integrals oftheSecond Kind 595 Elliptic Integrals oftheThird Kind 595 Problems 598 XVI C D E F G HCONTENTS Ordinary Differential Equations ofSecond Order 599 C.1 Linear Homogeneous Equations 599 C.2 Linear Inhomogeneous Equations 603 Problems 606 Useful Formulas 608 D.1 Binomial Expansion 608 D.2 Trigonometric Relations 609 D.3 Trigonometric Series 610 D.4 Exponential andLogarithmic Series 610 D.5 Complex Quantities 611 D.6 Hyperbolic Functions 611 Problems 612 Useful Integrals 613 E.1 Algebraic Functions 613 E.2 Trigonometric Functions 614 E.3 Gamma Functions 615 Differential Relations inDifferent Coordinate Systems 617 F.1 Rectangular Coordinates 617 F.2 Cylindrical Coordinates 617 F.3 Spherical Coordinates 619 A“Proof” oftheRelation Exf,=gig 621I-L Numerical Solution forExample 2.7 623 Selected References 626 Bibliography 628 Answers toEven-Numbered Problems 633 Index 643 CHAPTER Matrices, Vectors, and Vector Calculus 1.1 Introduction Physical phenomena canbediscussed concisely andelegantly through theuseof vector methods.* Inapplying physical “laws” toparticular situations, theresults must beindependent ofwhether wechoose arectangular orbipolar cylindrical coordinate system. The results must alsobeindependent oftheexact choice of origin forthecoordinates. The useofvectors gives usthisindependence. A given physical lawwillstillbecorrectly represented nomatter which coordinate system wedecide ismost convenient todescribe aparticular problem. Also, the useofvector notation provides anextremely compact method ofexpressing even themost complicated results. 'Inelementary treatments ofvectors, thediscussion may start with thestate- ment that “avector isaquantity that canberepresented asadirected lineseg- ment.” Tobesure, thistype ofdevelopment willyield correct results, and itis even beneﬁcial toimpart acertain feeling forthephysical nature ofavector. We assume that thereader isfamiliar with thistype ofdevelopment, butweforego theapproach here because wewish toemphasize therelationship that avector bears toacoordinate transfonnation. Therefore, weintroduce matrices andma- trixnotation todescribe notonly thetransformation butthevector aswell. We alsointroduce atype ofnotation thatisreadily adapted totheuseoftensors, al- though wedonotencounter these objects until thenormal course ofevents re- quires their use(seeChapter 11). *]0siah Willard Gibbs (1839-1903) deserves much ofthe credit fordeveloping vector analysis around 1880-1882. Much ofthepresent-day vector notation wasoriginated byOliver Heaviside (1850-1925), anEnglish electrical engineer, anddates from about 1893. 1 2 1/MATRICES, VECTORS, AND VECTOR CALCULUS Wedonotattempt acomplete exposition ofvector methods; instead, we consider only those topics necessary forastudy ofmechanical systems. Thus in thischapter, wetreat thefundamentals ofmatrix andvector algebra andvector calculus. 1.2 Concept ofaScalar Consider thearray ofparticles shown inFigure 1-la. Each particle ofthearray is labeled according toitsmass, say,ingrams. The coordinate axes areshown so thatwecanspecify aparticular particle byapair ofnumbers (x,y).The mass M oftheparticle at(x,y)canbeexpressed asM(x, y);thus themass oftheparticle atx=2,y=3canbewritten asM(x=2,y=3)=4.Now consider theaxes ro- tated anddisplaced inthemanner shown inFigure 1-lb. The 4gmass isnow lo- cated atx’=4,y’=3.5;thatis,themass isspeciﬁed byM (x'=4,y’=3.5) =4. And, ingeneral, MWJ’)=MWN’) (1.1) because themass ofanyparticle isnotaffected byachange inthecoordinate axes. Quantities thatareinvariant under coordinate tran.q"ormation—those thatobey anequation ofthistype—are termed scalars. Although wecandescribe themass ofaparticle (orthetemperature, orthe speed, etc.) relative toanycoordinate system bythesame number, some physical properties associated with theparticle (such asthedirection ofmotion ofthe particle orthedirection ofaforce thatmay actontheparticle) cannot bespeci- ﬁedinsuch asimple manner. The description ofthese more complicated quan- tities requires theuseofvectors. ]ust asascalar isdeﬁned asaquantity that re- mains invariant under acoordinate transformation, avector mayalsobedeﬁned interms oftransformation properties. Webegin byconsidering how thecoordi- nates ofapoint change when thecoordinate system rotates around itsorigin. J.,.,.it,.+;=,':3, (K) (b) FIGURE 1-1 Anarray ofparticles intwodifferent coordinate systems. 1.3 COORDINATE TRANSFORMATIONS 3 1.3 Coordinate Transformations Consider apoint Pwith coordinates (x1,x2,x3)with respect toacertain coordi- nate system.* Next consider adifferent coordinate system, onethatcanbegen- erated from theoriginal system byasimple rotation; letthecoordinates ofthe point Pwith respect tothenew coordinate system be(xi,xé,xé).The situation is illustrated foratwo-dimensional case inFigure 1-2. The new coordinate xiisthesum oftheprojection ofx1onto thexi-axis (thelineE)plus theprojection ofx2onto thexf-axis (thelineH;+R);thatis, xi=x1cosl9 +x2sin9 11'=x1cos9+x2cos<§- —0) (1.2a) The coordinate xéisthesum ofsimilar projections: xé=W—Fe,butthe linedeisalsoequal tothelineOf.Therefore xé=—x1sin6+xgcos 9 =x1cos(g +9)+x2cos6 (l.2b) Letusintroduce thefollowing notation: wewrite theangle between the x{-axis andthex1-axis as(xi,x1),andingeneral, theangle between thexf-axis andthexj-axis isdenoted by(xf,xj).Furthermore, wedeﬁne asetofnumbers Agby /\,<jEcos(x{, xj) (1.3) X2-3.XlS xé-axis R.\\\\\\$1II \I\|\1\1‘|\\|\| __;)I‘ta,.\xi x1-axis \ NX\M‘ ¢\\\\\ ®Q / I / I ¢________w1‘) O ’x 1 I, 1/ / / I / f FIGURE l-2 Theposition ofapoint Pcan berepresented intwocoordinate systems, onerotated from theother. *Welabel axes asxl,x2,x3instead ofx,y,ztosimplify thenotation when summations areperformed. Forthemoment, thediscussion islimited toCartesian (orrectangular) coordinate systems. 4 1/MATRICES, VECTORS, AND VECTOR CALCULUS Therefore, forFigure 1-2,wehave /111=cos(x{, x1)=cos6 A12=cos(xi, x2)=cos(% -0)=sin9 A21=cos(x§, x1)=cos(% +6)=—sin 9 /122=cos(x2, x2)=cos6 The equations oftransformation (Equation 1.2)now become xi=x1cos(x{, x1)+x2cos(x{, x2) =Anxi +/\12x2 xé=x1cos(x§, x1)+x2cos(x§, x2) =421X1 '4''\22x2 Thus, ingeneral, forthree dimensions wehave xi=411x1 +412x2 +413%‘ xi=421-xi +422952 +/\2sxs i xi=451-xi '1'432942 +453965) or,insummation notation, 3 x,7=2/\,--x», i=1,2,3 J-=11] The inverse transformation is x1=xicos(x{, x1)+xécos(x§, x1)+x§cos(x§, =/\11xi +/\21xi +431-‘xi or,ingeneral, 3 x-=Ex--xi i=1,2,3 IJ,-=1J’1’xl(1.4) (1.5a) (1.5b) (1.6) (1.7) (1.8) The quantity AUiscalled thedirection cosine ofthex}-axis relative tothe xj--axis. Itisconvenient toarrange the/\,-jinto asquare array called amatrix. The boldface symbol Itdenotes thetotality oftheindividual elements Agwhen arranged asfollows: 411 412 /\1s A=421 422 425 (1-9) A31 A32 A33 Once weﬁnd thedirection cosines relating thetwosetsofcoordinate axes, Equations 1.7and 1.8give thegeneral rules forspecifying thecoordinates ofa point ineither system. When Aisdeﬁned thiswayandwhen itspeciﬁes thetransformation proper- tiesofthecoordinates ofapoint, itiscalled atransformation matrix orarota- tionmatrix. 1.3COORDINATE TRANSFORMATIONS 5 Apoint Pisrepresented inthe(x1,x2,x3)system byP(2, 1,3).Inanother coor- dinate system, thesame point isrepresented asP(x{, xé,x§)where x2hasbeen rotated toward x3around thex1-axis byanangle of30°(Figure 1-3). Find the rotation matrix anddetermine P(x{, xé,x_€,). FIGURE 1-3 Example 1.1.Apoint Pisrepresented intwocoordinate-systems, onexx3’ 3 ~14 30°P9 , x2 *1 xi’K\\\\\ rotated from theother by30°. Solution. The direction cosines A,-jcanbedetermined from Figure 1-3using thedeﬁnition ofEquation 1.3. /\11= M2 /\1s A21 A22 A23 /\s1 /\s2 /\ss¢0S(xi. xl)= ¢<>S(xi, x2)= ¢OS(xi, xs)= ws(xé.xi) ¢<>S(~é. *2) cos(x§, x3)= ¢0$(x§, X1)= ¢<>s(x§, X2)= ¢0S(xé. X3)= A:cos(0°) =1 cos(90°) =0 cos(90°) =0 cos(90°) =0 cos(30°) =0.866 cos(90° -30°) =cos(60°) =0.5 c0s(90°) =0 cos(90° +30°) = cos(30°) =0.866 1 0 0 00.866 0.5 0——0.5 0.866-0.5 and using Equation 1.7,P(x{, xé,xé)is x{= x§= x§=/\11-xi +/\12x2 +Aisxs 021941 '1'/\22x2 +/\2s~’¢s /\s1x1 +A529‘? +/\ssxs=x1 =2 0.866x2 +0.5963 =2.37 —0.5x2 +0.866x3 =2.10 6 1/MATRICES, VECTORS, ANDVECTOR CALCULUS Notice thattherotation operator preserves thelength oftheposition vector. r= \/xi{+x§+x§= \/x{2+x§2+x§2=3.74 1.4 Properties ofRotation Matrices* Tobegin thediscussion ofrotation matrices, wemust recall twotrigonometric results. Consider, asinFigure 1-4a, alinesegment extending inacertain direc- tion inspace. Wechoose anorigin forourcoordinate system that liesatsome point ontheline. The line then makes certain deﬁnite angles with each ofthe coordinate axes; welettheangles made with thex1-,xi,-,x3-axes bea,B,y.The quantities ofinterest arethecosines ofthese angles; cosa,cosB,cos'y.These quantities arecalled thedirection cosines oftheline. The ﬁrstresult weneed is theidentity (seeProblem 1-2) coszoz +cos2B +cos2'y =1 (1.10) Second, ifwehave twolines with direction cosines cosoz,cosB,cos‘yandcosa’, cosB’,cos‘y’,then thecosine oftheangle 0between these lines (seeFigure 1-4b) isgiven (seeProblem 1-2)by cos0=cosa cos01'+cosBcosB’ +cos')/cos 'y’ (1.11) With asetofaxes x1,x2,x3,letusnow perform anarbitrary rotation about some axis through theorigin. Inthenew position, welabel theaxes xi,xé,xg. x3 (I1BY) x3,’’ (mm) I:__"'_____'____\\IS‘Q5.‘Q‘<_ Y 9 G / ____ /I x2 IIIIIIIIIIIIIIIIIIIIK DC1 xl (a) (b) FIGURE 1-4 (a)Alinesegment isdeﬁned byangles (oz,/3,'y)from thecoordinate axes. (b)Another linesegment isadded thatisdeﬁned byangles (0z’,B’,'y’). *Much ofSections 1.4—l.l3 deals with matrix methods andtransformation properties andwillnotbe needed bythereader until Chapter ll.Hence thereader may skip these sections until then ifde- sired. Those relations absolutely needed—-scalar andvector products, forexample——-should already befamiliar from introductory courses. 1.4 PROPERTIES OFROTATION MATRICES 7 The coordinate rotation may bespeciﬁed bygiving thecosines ofalltheangles between thevarious axes, inother words, bytheAil-. Notallofthenine quantities Ag»areindependent; infact, sixrelations exist among theA,-J-,soonly three areindependent. Weﬁnd these sixrelations by using thetrigonometric results stated inEquations 1.10 and1.11. First, thex{-axis maybeconsidered alone tobealineinthe(x1,x2,x3)coor- dinate system; thedirection cosines ofthislineare(A11, A12,A15). Similarly, thedi- rection cosines ofthexé-axis inthe(x1,x2,x3)system aregiven by(A21, A22,A23). Because theangle between thexi-axis and thexé-axis is11'/2, wehave, from Equation 1.11, AHAQI +A12A22 +ABA23 =cos9=cos('n'/2) =0 or* 21,,-1,,-= 0 And, ingeneral, §2t,,»,\,,- =0,wek (1.12a) Equation 1.12a gives three (one foreach value of2'ork)ofthesixrelations among theA,-J. Because thesum ofthesquares ofthedirection cosines ofalineequals unity (Equation 1.10), wehave forthex{-axis inthe(x1,x2,x3)system, ' 011+ AI2+/\is=1 or Z,\%—E,\ ,1-1 .'-.1'1"- ]1J11 and, ingeneral, 21/1,,-1t,,,.= 1,i=k (1.12b) which aretheremaining three relations among theAg-. Wemay combine theresults given byEquations 1.12a and1.12b as 211,1,-,_2t,, =5,, (1.13) where 55,,istheKronecker delta symboll 0,ifiakk5.= 1.1"'i1,ifi=k (4) Thevalidity ofEquation 1.13 depends onthecoordinate axes ineach ofthe systems being mutually perpendicular. Such systems aresaid tobeorthogonal, *AIlsummations here areunderstood torunfrom lto3. llntroduced byLeopold Kronecker (1823-1891). 8 1/MATRICES, VECTORS, ANDVECTOR CALCULUS andEquation 1.13 istheorthogonality condition. The transformation matrix A specifying therotation ofanyorthogonal coordinate system must then obey Equation 1.13. Ifwewere toconsider thex,--axes aslines inthexfcoordinate system and perform acalculation analogous toourpreceding calculations, wewould ﬁnd therelation E/\,-J-21,, =5,-,, (1.15) The twoorthogonality relations wehave derived (Equations 1.13 and 1.15) appear tobedifferent. (Note: InEquation 1.13 thesummation isover thesecond indices oftheA,-J-,whereas inEquation 1.15 thesummation isover theﬁrst in- dices.) Thus, itseems that wehave anoverdetermined system: twelve equations innine unknowns.* Such isnotthecase, however, because Equations 1.13 and 1.15 arenotactually different. Infact, thevalidity ofeither ofthese equations implies thevalidity oftheother. This isclear onphysical grounds (because the transformations between thetwo coordinate systems ineither direction are equivalent), andweomit aformal proof. Weregard either Equation 1.13 or1.15 asproviding theorthogonality relations foroursystems ofcoordinates. Inthepreceding discussion regarding thetransformation ofcoordinates and theproperties ofrotation matrices, weconsidered thepoint Ptobeﬁxed andallowed thecoordinate axes toberotated. This interpretation isnotunique; wecould equally well have maintained theaxes ﬁxed andallowed thepoint to rotate (always keeping constant thedistance totheorigin). Ineither event, the transformation matrix isthesame. Forexample, consider thetwocases illustrated inFigures 1-5aandb.InFigure 1-5a, theaxes x1andx2arereference axes, and thex{-and xé-axes have been obtained byarotation through anangle 6. x2 x2 xé\ \ \\ .P P \ ,'\ / \ / \ I I \ \ ’ Ix’ / P\\9 ,/ l / ,/ I K. /’ I /’// 2\ / 9 / \ z’ / 1’ \ /I / /’\ / / I I I\ /I / 1 \ I/’ 9 1’ \/ x /,1 ' xiII1 (E) (b) FIGURE 1-5 (a)The coordinate axes x1,x2 arerotated byangle 0,butthepoint P remains ﬁxed. (b)Inthiscase, thecoordinates ofpoint Pare rotated toanew point P’,butnotthecoordinate system. *Recall that each oftheorthogonality relations represents sixequations. 1.5MATRIX OPERATIONS 9 Therefore, thecoordinates ofthepoint Pwith respect totherotated axes may befound (seeEquations 1.2aand1.2b) from xi=x1cos6 +x2sin9 xé=—x1sin0 +x2cos6} (1.16) However, iftheaxes areﬁxed andthepoint Pisallowed torotate (asinFigure 1-5b) through anangle 6about theorigin (but intheopposite sense from that oftherotated axes), then thecoordinates ofP’areexactly those given by Equation 1.16. Therefore, wemay elect tosayeither thatthetransformation acts onthepoint giving anew state ofthepoint expressed with respect toaﬁxed co- ordinate system (Figure 1-5b) orthat thetransformation actsontheframe ofref- erence (the coordinate system), asinFigure 1-5a. Mathematically, theinterpreta- tions areentirely equivalent. 1.5 Matrix Operations* The matrix Agiven inEquation 1.9hasequal numbers ofrows andcolumns and istherefore called asquare matrix. Amatrix need notbesquare. Infact, theco- ordinates ofapoint may bewritten asacolumn matrix x1 X= x2 (l.l7a) xs X=(x1x2x3) (1.17b)orasarowmatrix Wemust now establish rules tomultiply twomatrices. These rules must be consistent with Equations 1.7and 1.8when wechoose toexpress thexiandthe x§inmatrix form. Letustake acolumn matrix forthecoordinates; then wehave thefollowing equivalent expressions: x;=11,,xj (l.18a) x’=Ax (1.l8b) xi /\11 A12 /\1s x1 Xé = A21 A22 A23 X2 0 xii /I31 /I32 /\ss xs xi=)\11x1 +x12x2 +Msxs xé =/\21X1 +A22x2 +/\.23x3 xi=)Is1x1 +xs2x2 '1'Assxs *The theory ofmatrices wasﬁrst extensively developed byA.Cayley in1855, butmany ofthese ideas were thework ofSirWilliam Rowan Hamilton (1805-1865), who haddiscussed “linear vector opera- tors” in1852. Theterm matrix wasﬁrstused by_]._]. Sylvester in1850. 10 1/MATRICES, VECTORS, ANDVECTOR CALCULUS Equations l.18a—d completely specify theoperation ofmatrix multiplication foramatrix ofthree rows and three columns operating onamatrix ofthree rows and one column. (Tobeconsistent with standard matrix convention we choose Xand X’tobecolumn matrices; multiplication ofthetype shown in Equation 1.18c isnotdeﬁned ifXandX’arerowmatrices.)* Wemust now ex- tend ourdeﬁnition ofmultiplication toinclude matrices with arbitrary numbers ofrows andcolumns. The multiplication ofamatrix Aandamatrix Bisdeﬁned only ifthenum- berofcolumns ofAisequal tothenumber ofmwsofB.(The number ofrows of Aand thenumber ofcolumns ofBareeach arbitrary.) Therefore, inanalogy with Equation 1.18a, theproduct ABisgiven by C=AB 1.190,,={AB},=21.1,,3,, () Asanexample, letthetwomatrices AandBbe 3-2 2A= (4 -3 5) abc B=(ti e gh1' Wemultiply thetwomatrices by b03-22aAB=(4 _3 5)(d e (1.20) gh1' The product ofthetwomatrices, C,is C=AB=3a2d+2g3b28+2h362f+2]) (L21) 4a—3d+5g 4b—3e+5h 40- 3f+ 5] Toobtain theC,-jelement intheithrowandjthcolumn, weﬁrstsetthetwo matrices adjacent aswedidinEquation 1.20 intheorder Aandthen B.Wethen multiply theindividual elements intheithrowofA,one byone from leftto right, times thecorresponding elements inthejthcolumn ofB,one byone from toptobottom. Weaddallthese products, andthesum istheC,-7element. Now itiseasier toseewhyamatrix Awith mrows and ncolumns must bemulti- plied times another matrix Bwith nrows andanynumber ofcolumns, sayp.The result isamatrix Cofmrows and[Jcolumns. *Although whenever weoperate onXwith theAmatrix thecoordinate matrix Xmust beexpressed asacolumn matrix, wemayalsowrite Xasarowmatrix (xl,x2,x5),forother applications. 1.5MATRIX OPERATIONS 11 Find theproduct ABofthetwomatrices listed below: 2 1 3 A=-2 2 4 -1 -3 -4 -1 -2 B= 1 2 3 4 Solution. Wefollow theexample ofEquations 1.20 and1.21 tomultiply thetwo matrices together. 2 1 3 -1 -2 AB=-2 2 4 1 2 -1 -3 -4 3 4 —2+1+9 -4+2+12 810 AB= 2+2+12 4+4+16 =16 24 1-2-12 2-6-16 -14 -20 Theresultofmultiplyinga3 ><3matrix timesa3 ><2matr1X isa3><2matrix. Itshould beevident from Equation 1.19 that matrix multiplication isnot commutative. Thus, ifAandBareboth square matrices, then thesums ;A,,,B,,,and23,,/1,, areboth deﬁned, but,ingeneral, they willnotbeequal. Show thatthemultiplication ofthematrices AandBinthisexample isnon- commutative. Solution. IfAandBarethematrices 21 -1 2A= B= (-13)»(4-2) 22 AB_(1s -8)then 12 1/MATRICES, VECTORS, ANDVECTOR CALCULUS but -4 5BA=(.._.) AB2*BAthus 1.6 Further Definitions Atransposed matrix isamatrix derived from anoriginal matrix byinterchange ofrows andcolumns. Wedenote thetranspose ofamatrix AbyA‘.According to thedeﬁnition, wehave (A‘)‘ =A (1.23)Evidently, Equation 1.8may therefore bewritten asanyofthefollowing equivalent expres- sions: x,»=2}/\,,x; (1.24a)I x,=Ztfjx; (1.246)I x=A‘x’ (1.24c) x1 A11 A21 A51 xi .762 = A12 A22 A32 xé x5 A15 A25 A55 xi Theidentity matrix isthatmatrix which, when multiplied byanother matrix, leaves thelatter unaffected. Thus 1A=A, B1=B (1.25) 1-(2.§)(:;)=<:;>=ALetusconsider theorthogonal rotation matrix Aforthecase oftwodimensions: A=(A1.16.)A21 A22thatis, 1.6FURTHERDEFINITIONS 13 M,=(A1.A12)(A1.A2.) A21 A22 A12 A22Then _ Aii'1'Ai2 A11A21 '1'A12A22) _A21A11 '1'A22A12 A21'1'A22 Using theorthogonality relation (Equation 1.13), weﬁnd Aii+Ai2=A21'1'A22=1 A21A11 +A22A12 =A11A21 +A12A22 =0 sothatforthespecial case oftheorthogonal rotation matrix Awehave* 10AA‘— (0 1)—1 (1.26) The inverse ofamatrix isdeﬁned asthat matrix which, when multiplied by theoriginal matrix, produces theidentity matrix. The inverse ofthematrix Ais denoted byA'1: AA"1=1 (1.27) Bycomparing Equations 1.26 and1.27, weﬁnd fororthogonal matrices ' (1.28) Therefore, thetranspose andtheinverse oftherotation matrix Aareidentical. Infact, thetranspose ofanyorthogonal matrix isequal toitsinverse. Tosummarize some oftherules ofmatrix algebra: 1.Matrix multiplication isnotcommutative ingeneral: AB9*BA (1.29a) The special case ofthemultiplication ofamatrix anditsinverse iscommu- tative: AA“ =A“A=1 (1.29b) The identity matrix always commutes: 1A=A1=A (1.29c) 2.Matrix multiplication isassociative: [AB]C =A[BC] (1.30) 3.Matrix addition isperformed byadding corresponding elements ofthetwo matrices. The components ofCfrom theaddition C=A+Bare Cij=Aij+B,-j (1.31) Addition isdeﬁned only ifAandBhave thesame dimensions. *This result isnotvalid formatrices ingeneral. Itistrueonly fororthogonal matrices. 14 1/MATRICES, VECTORS, ANDVECTOR CALCULUS 1.7 Geometrical Significance ofTransformation Matrices Consider coordinate axes rotated counterclockwise* through anangle of90° about thex3-axis, asinFigure 1-6.Insuch arotation, xi=x2,xi=-xi, xi,=x3. The only nonvanishing cosines are C0$(xi, x2)=1=A12 COS(xé, X1) =_1 =A21 c0s(x§, X3) = 1=A33 sotheAmatrix forthiscase is 010 A1: —1 0 0 001 Next consider thecounterclockwise rotation through 90°about thexi-axis, asinFigure 1-7.Wehave xi=xi,xi,=x3,xi,=-x2, andthetransformation ma- trixis 100 1t,=0 01 0-10 Toﬁnd thetransformation matrix forthecombined transformation forrota- tion about thex3-axis, followed byrotation about thenew xi-axis (see Figure 1-8), wehave x’=Aix (1.32a) and x"=A2x’ (1.32b) OI‘ x”=A2/\ix (1.33a) x'i 1 0 0 O 1 0 X1 0 1 0 X1 X2 xg =0 01 -1 00 x2=001x2=x3 xii 0-1 0 001 x3 100 x3 xi (1.33b) *We determine thesense oftherotation bylooking along thepositive portion oftheaxisofrotation attheplane being rotated. This deﬁnition isthen consistent with the“right-hand rule,” inwhich the positive direction ofadvance ofaright-hand screw when turned inthesame sense. 1.7 GEOMETRICAL SIGNIFICANCE OFTRANSFORMATION MATRICES 15 xx xé *2 imam xi _ A1 xl about x3-axis FIGURE 1-6 Coordinate system xi,x2,x3isrotated 90°counter-clockwise (ccw) about thex3-axis. This isconsistent with theright-hand ruleof rotation. *3 xé 5 Ag _,_,,_i> xé 90°rotation about xi-axis I x1 xi FIGURE 1-7 Coordinate system xi,x2,x3isrotated 90°ccwabout thexi-axis. xs xi xi xéA x 90°rotation 90°rotation I’ 1 about x5-axis ' about xi’-axis x5 FIGURE 1-8 Coordinate system xi,x2,x3isrotated 90°ccwabout thex3-axis followed bya90°rotation about theintermediate xi-axis. 16 1/MATRICES, VECTORS, ANDVECTOR CALCULUS Therefore, thetworotations already described may berepresented byasingle transformation matrix: 0 1 0 1 0 0 andtheﬁnal orientation isspeciﬁed byxi’=x2,xg=x3,xii=xi.Note that the order inwhich thetransfonnation matrices operate onXisimportant because themultiplication isnotcommutative. Intheother order, A4: A1A2 010 1 00 = -1 0O 0 01 001 0-1 0 0 01 =-1 00¢A3 (1.35) 0-1 0 and anentirely different orientation results. Figure 1-9illustrates thedifferent ﬁnal orientations ofaparallelepiped thatundergoes rotations corresponding to tworotation matrices AA,ABwhen successive rotations aremade indifferent order. The upper portion oftheﬁgure represents thematrix product ABAA,and thelower portion represents theproduct AAAB. AA AB— Q 90°rotation 90°rotation about x3-axis about x2-axis x5 x2 xi AB AAO} O} about x2-axis about x5-axis FIGURE l-9 Aparallelepiped undergoes twosuccessive rotations indifferent order. The results aredifferent. 1.7 GEOMETRICAL SIGNIFICANCE OFTRANSFORMATION MATRICES 17 Next, consider thecoordinate rotation pictured inFigure 1-10 (which isthe same asthat inFigure 1-2). The elements ofthetransformation matrix intwo dimensions aregiven bythefollowing cosines: cos(x{, x1)=cos6=A11 c0s(xi, x2)=cos(% —9)=sin9=A12 cos(x§, x1)=cos(% +0)=—sin 6=A21 cos(x§, x2)=cos6=A22 Therefore, thematrix is cos6sin6A5= _ ) (l.36a)—sin I9cosI9 Ifthisrotation were athree-dimensional rotation with x’=x,wewould s s have thefollowing additional cosines: cos(x{, x3)= cos(x§, x3) cos(x§, x5) cos(x§, x1)= cos(x§, x2)=©®'—'©©Ms /\ Ass A31 /\3223 andthethree-dimensional transformation matrix is cos6 sin90 A5= —sin6 cos6 0 (l.36b) 0 0 1 x3=xé la X1 I xQ *1 FIGURE 1-10 Coordinate system x1,x2,x3isrotated anangle 6ccwabout thex3-axis. 18 1/MATRICES, VECTORS, ANDVECTOR CALCULUS *3 1 x,~ ., .¢§@~ lT ',2tI1(Inversion) , . X>.-\ ,5‘ iii 1 ;:x_\..._._.....iA‘ --._ -42»-. x; FIGURE 1-11 Anobject undergoes aninversion, which isareﬂection about theorigin ofalltheaxes. Asaﬁnal example, consider thetransfonnation thatresults inthereﬂection through theorig-in ofalltheaxes, asinFigure 1-11. Such atransformation is called aninversion. Insuch acase, xi=—x1, xé=—x2, xé=—x3, and -1 00 A6=0-1 0 (1.37) 00~1 Inthepreceding examples, wedeﬁned thetransformation matrix A3tobe theresult oftwosuccessive rotations, each ofwhich wasanorthogonal transfor- mation: A3=AQAI. Wecanprove that thesuccessive application oforthogonal transformations always results inanorthogonal transformation. WeWrite =§Aij'xj> X’):=2!/~kt' Combining these expressions, weobtain xi= (;!1~kiAy) xj =gll-‘Al kjxj Thus, weaccomplish thetransformation from xitox’,-'byoperating onxiwith the (pk) matrix. The combined transformation willthen beshown tobeorthogonal if(;.tA)‘ =(pA)“1. The transpose ofaproduct matrix istheproduct ofthe transposed matrices taken inreverse order (see Problem 1-4); that is,(AB)’ = B‘A‘.Therefore um’=Mu’ <1-38> 1.7 GEOMETRICAL SIGNIFICANCE OFTRANSFORMATION MATRICES 19 But, because Aandpareorthogonal, A‘=A'1and pt=pf]. Multiplying the above equation by,uAfrom theright, weobtain (/M)‘M =/\’M‘rM\ =MIA =wt =1 =(uA)“/M Hence UM’) =(/M)" (1-39) andthe;.¢Amatrix isorthogonal. The detenninants ofalltherotation matrices inthepreceding examples can becalculated according tothestandard rule fortheevaluation ofdeterminants ofsecondor third order: A A |A|=11 12=A11A22 _A12A21 (1-40) A21 A22 A11 A12 A13 |A|=A21 A22 A23 A31 A32 A33 _ A22 A23_ A21 A23 A21 A22—A A +A 1.41 HA32 A33 12A31 A33 13A31 A32 ( ) where thethird-order determinant hasbeen expanded inminors oftheﬁrst row. Therefore, weﬁnd, fortherotation matrices used inthissection, |)(1| =|A2| = =|A5| =1 but |A6|=_1 Thus, allthose transformations resulting from rotations starting from theoriginal set ofaxeshave determinants equal to+1.Butaninversion cannot begenerated by anyseries ofrotations, andthedeterminant ofaninversion matrix isequal to-1. Orthogonal transfomiations, thedeterminant ofwhose matrices is+1, are called proper rotations; those with determinants equal to—larecalled im- proper rotations. Allorthogonal matrices must have adeterminant equal toeither +1or—l.Here, weconﬁne ourattention totheeffect ofproper rotations anddo notconcern ourselves with thespecial properties ofvectors manifest inimproper rotations. 20 1/MATRICES, VECTORS, ANDVECTOR CALCULUS Showthat|A2| =land |A6| =—l. Solution. 100 01 |)t|=0 01=+1 =0—(—1)=12 -100-10 -1 00 _1 O |)t6|= 0-1 0=—1} 01}=—1(1—0)=—1 001 1.8Defmitions ofaScalar andaVector inTerms ofTransformation Properties Consider acoordinate transformation ofthetype x;=2%-.~xj (1.42) with I ;2t,~,)t,.j =5,, (1.43) If,under such atransformation, aquantity qbisunaffected, then gbiscalled a scalar (orscalar invariant). Ifasetofquantities (A1,A2,A3)istransformed from thex,system tothex{ system byatransformation matrix Awith theresult A;=Ex,-,A, (1.44) then thequantities A,transform asthecoordinates ofapoint (i.e., according to Equation 1.42), andthequantityA =(A1,A2,A3)istermed avector. 1.9Elementary Scalar andVector Operations Inthefollowing, Aand Barevectors (with components Aiand B,-)and cb,1/1,and §arescalars. Addition A,+B,=B,+A,- Commutative law (1.45) A5+(Bi+Q»)=(A,+B,-)+CiAssociative law (1.46) 1.10 SCALAR PRODUCT OFTWOVECTORS 21 ¢>+1/1=1/1+qb Commutative law (1.47) qb+(1/1+§)=(qb+1/1)+§Associative law (1.48) Multiplication byascalar§ §A=Bisavector (1.49) ﬁqb=1/1isascalar (1.50) Equation 1.49 canbeproved asfollows: B5=;A=1‘5'1 =;A11"§A1" =5221,2,-1,.=§/1; (1.51)J and§Atransforms asavector. Similarly, ﬁqbtransforms asascalar. 1.10 Scalar Product ofTwo Vectors The multiplication oftwovectors AandBtoform thescalar product isdeﬁned tobe A-B=Z,-1,12, _(1.52) where thedotbetween AandBdenotes scalar multiplication; thisoperation is sometimes called thedotproduct. ThevectorA hascomponents A1,A2,A3,andthemagnitude (orlength) ofA isgiven by _ IAI=+\/A? +/13+A§EA (1.53) where themagnitude isindicated by|A|or,ifthere isnopossibility ofconfu- sion, simply byA.Dividing both sides ofEquation 1.52 byAB,wehave A-B A-B-——=E—’—’ 1.4AB tAB (5) A1/A isthecosine oftheangle ozbetween thevector Aand thex1-axis (see Figure 1-12). Ingeneral, A,-/A and B,‘/B arethedirection cosines A;-Aand A?of thevectors AandB: %=;A{‘AP (1.55) The sum 2,-A§‘A‘,B isjust thecosine oftheangle between AandB(seeEquation 1.11); cos(A,B)=2/\;‘A,B or t IA-B=ABcos(A, B)I (1.56) 22 1/MATRICES, VECTORS, AND VECTOR CALCULUS *3 \\\\ \\\A \ 3 \ "“""é.5“IS‘ \\a \ \ A \I I lIIIIIIA/I// l/1/‘_____\\\\\ *1 FIGURE 1-12 AvectOrA isshown incoordinate system x1,x2,x2with itsvector components A1,A2,andA3.Thevector Aisoriented atanangle oz with thex]-axis. That theproduct A-Bisindeed ascalar may beshown asfollows. AandB transform asvectors: A;=Z2t,,A,-, B;=§k§,\,-,,B,, (1.57)1 Therefore theproduct A’-B’becomes = 212.-1,-) A,-,,Bk) Rearranging thesummations, wecanwrite 1-11.131: §(E)t,-J,-1t2)A,B,,A’-B’ =2A{B§ y Butaccording totheorthogonality condition, theterm inparentheses isjust5,-,,. Thus, A’ 1B’ = =%4@ =A-B (1.53) Because thevalue oftheproduct isunaltered bythecoordinate transformation, theproduct must beascalar. Notice that thedistance from theorigin tothepoint (x1,x2,x2)deﬁned by thevector A,called theposition vector, isgiven by |A|=\/A-A= \/x¥+x§+x§= \/ ___-_.-- .~C-v-_. 40 *3 <21.22.23>A'E.- (-1.2.=2) B A *2 xi FIGURE 1-13 The vectorA istheposition vector ofpoint (x1,x2,x3),and vector Bis theposition vector ofpoint (E1,x2,2,)ThevectorA —Bisthe position vector from (E1,E2,E2)to(x1,x2,x2). Similarly, thedistance from thepoint (x1,x2,x3)toanother point (21,22,E3)de- ﬁned bythevector Bis \/Zn.-—r=.~>2= \/(A-B)-<A—B> =|A—B| That is,wecandeﬁne thevector connecting anypoint with anyother point as thedifference oftheposition vectors that deﬁne theindividual points, asin Figure 1-13. The distance between thepoints isthen themagnitude ofthedif- ference vector. And because thismagnitude isthesquare root ofascalar prod- uct,itisinvariant toacoordinate transformation. This isanimportant factand canbesummarized bythestatement that orthogonal transformations aredistance- preserving transformations. Also, theangle between twovectors ispreserved under anorthogonal transformation. These tworesults areessential ifwearetosuc- cessfully apply transformation theory tophysical situations. The scalar product obeys thecommutative anddistributive laws: A-B=EA,-B, =EB,-A, =B-A (1.59) A-(B+<1)=Z-4.~<B+ c).=EA.-(B. +C) =Z(A,B,.+ A,C,)=(A-B) +(A-C) (1.50) 1.11 Unit Vectors Sometimes wewant todescribe avector interms ofthecomponents along the three coordinate axes together with aconvenient speciﬁcation ofthese axes. For thispurpose, weintroduce unitvectors, which arevectors having alength equal totheunit oflength used along theparticular coordinate axes. Forexample, the unit vector along theradial direction described bythevector RiseR=R/(|R|). 24 1/MATRICES, VECTORS, AND VECTOR CALCULUS There areseveral variants ofthesymbols forunit vectors; examples ofthemost common setsare(i,j,k),(e1,e2,e3),(e,,ea,e¢),and (;~,0,41). The following ways ofexpressing thevector Aareequivalent: A=(A1,A2,A3) or A=e1A1 +e2A2 +e3A3 =ze,-A, (1.61) Or A=A1i +A2]+Agk Although theunit vectors (i,j,k)and (2,6,(fa)aresomewhat easier touse,we tend touseunitvectors such as(cl,e2,e3),because oftheease ofsummation no- tation. Weobtain thecomponents ofthevector Abyprojection onto theaxes: Al‘ = e1"A Wehave seen (Equation 1.56) that thescalar product oftwovectors hasa magnitude equal totheproduct oftheindividual magnitudes multiplied bythe cosine oftheangle between thevectors: A-B=ABcos(A, B) (1.63) Ifanytwounitvectors areorthogonal, wehave Two position vectors areexpressed inCartesian coordinates asA=i+2j—2k andB=4i+2j—3k.Find themagnitude ofthevector from point Atopoint B,theangle 0between AandB,andthecomponent ofBinthedirection ofA. Solution. Thevector from point Atopoint BisB—A(seeFigure l-13). B—A=4i+2j—3k—(i+2j—2k)=3i—k |B—Ah=V9+1=w55 From Equation 1.56 6_A.B_ (i+2j—2k)-(4i+2j—3k) cs— — °AB \/§\/23445¢mo=—i—i—=o%7m»@§ e=3m The component ofBinthedirection ofAisBcos9and, from Equation 1.56, A-B 14Bcos6= =—=4.67 1.12VECTOR PRODUCT OFTWOVECTORS 25 1.12Vector Product ofTwo Vectors Wenext consider another method ofcombining twovectors—the vector prod- uct(sometimes called thecross product). Inmost respects, thevector product of twovectors behaves likeavector, andweshall treat itassuch.* The vector prod- uctofAandBisdenoted byabold cross X, C=AXB (1.65) where Cisthevector resulting from thisoperation. The components ofCare deﬁned bytherelation C,E8,-,A,B, (1.65)J, where thesymbol s,-J-,,isthepermutation symbol or(Levi-Civita density) andhas thefollowing properties: 0, ifanyindex isequal toanyother index s,-J2=+1, ifi,j,kform anevenpermutation of1,2,3 (1.67) -1, ifi,j,kform anoddpermutation of1,2,3 Aneven permutation hasaneven number ofexchanges ofposition oftwosym- bols. Cyclic permutations (forexample, 123 —>231 —>312) arealways even. Thus 3122=3313:8211=0.em- 8123=8231=8312=+1 9132=3213=9321=-1 Using thepreceding notation, thecomponents ofCcanbeexplicitly evaluated. Fortheﬁrst subscript equal to1,theonly nonvanishing t-:22are.9123ands132-- thatis,forj,k=2,3ineither order. Therefore C1=.2k81jkA]'Bk =8123/1233 +813241332 =A2B3 —A3B2 (1.68a) Similarly, C2=A3B1 —AIBS (1.68b) C3=AIB2 —A2B1 (1.68c) Consider now theexpansion ofthequantity [ABsin(A,B)]2=(ABsin(9)2: A2B2sin2l9 =A2B2 —A2B2cos29 =(Z./-1?)(Z B?)—(EA.-B.-)2 I(A233 _A332)2 +(A331 _A133)2 +(A132 _A2B1)2 (1-59) *The product actually produces anaxial vector, buttheterm vector product isused tobeconsistent with popular usage. 26 1/MATRICES, VECTORS, ANDVECTOR CALCULUS C B \\\-__‘ "~ _\ 9 -’z, r”g,4, FIGURE l-14 The magnitude ofthevector Cdetermined byC=AXBhasa magnitude given bythearea oftheparallelogram ABsin0,where 0 istheangle between thevectors AandB. where thelastequality requires some algebra. Identifying thecomponents ofC inthelastexpression, wecanwrite (ABsino)2=C?+cg+cg=|C2|=C2 (1.70) Ifwetake thepositive square root ofboth sides ofthisequation, C=ABsin6 (1.71) This equation states thatifC=AXB,themagnitude ofCisequal totheprod- uctofthemagnitudes ofAandBmultiplied bythesine oftheangle between them. Geometrically, ABsin0isthearea oftheparallelogram deﬁned bythe vectors AandBandtheangle between them, asinFigure 1-14. EX.-\MPl.F. 1.6 Show byusing Equations 1.52 and1.66 that A-(B><D)=D-(A><B) (1.72) Solution. Using Equation 1.66, wehave (B><D),=§5,,,.t2,-D, Using Equation 1.52, wehave A-(B><D)=25,,-,A,12,-D, (1.73)1.1.»! Similarly, fortheright-hand sideofEquation 1.72, wehave 0-(A><B)=§.-;,-I-,0,-A,-B, From thedeﬁnition (Equation 1.67) of8,)-,,,wecaninterchange twoadjacent in- dices ofB,-J2,which changes thesign. 0-(A><B)=—ej,-,,D,-A]-B,,1y =gcsj-,,,-A,-B,,1), (1.74) Because theindices i,j,karedummy andcanberenamed, theright-hand sides ofEquations 1.73 and 1.74 areidentical, and Equation 1.72 isproved. Equation 1.12VECTOR PRODUCT orTWOVECTORS 27 1.72 canalsobewritten asA-(BXD)=(AXB)-D,indicating thatthescalar andvector products canbeinterchanged aslong asthevectors stayintheorder A,B,D.Notice that, ifwe letB=A,wehave A-(AXD)=D-(A><A)=0 showing thatA XDmust beperpendicular toA. AXB(i.e., C)isperpendicular totheplane deﬁned byAandBbecause A-(AXB)=0andB-(AXB)=0.Because aplane area canberepresented byavector normal totheplane and ofmagnitude equal tothearea, Cisevi- dently such avector. The positive direction ofCischosen tobethedirection of advance ofaright-hand screw when rotated from AtoB. The deﬁnition ofthevector product isnow complete; components, magni- tude, andgeometrical interpretation have been given. Wemay therefore reason- ably expect that Cisindeed avector. The ultimate test, however, istoexamine thetransformation properties ofC,and Cdoes, infact, transform asavector under aproper rotation. Weshould note thefollowing properties ofthevector product that result from thedeﬁnitions: (5) A><B=—BxA (1.75) but,ingeneral, (b) AX(BXC)#(AXB)XC (1.76) Another important result (seeProblem 1-22) is AX(BXC)=(A-C)B —(A-B)C (1.77) E" PLE1.7 _ I _ i I_ Find theproduct of(AXB)-(CXD). Solution. (AX13).": §9g'kAj3t (CXD)t=5'8ilmClDm The scalar product isthen computed according toEquation 1.52: (A><B)-(<1><D)=E,(§~3,-1. AJ'Bk)(§8ilmClDm) Rearranging thesummations, wehave (AXB)‘(CXD)=5 Bjkiszmiy AjBkClDm kj, where theindices ofthes’shave been permuted (twice each sothatnosign change occurs) toplace inthethird position theindex over which thesum is LO 1/1V1!\1l\1\JLD, VLLJ1\Jl\D, 1‘\lYLJ VB.LJl\Jl\k.4.l"\1_1\J\J1_|\JD carried out.Wecannow useanimportant property ofthesq),(seeProblem 1-22); Wetherefore have (A><B)-(C><1))=§(3,,3,,,, -5,,,5,,)A,B,c,1),, Lmsijkslmk =5715;)» _532521 (1-73) Carrying outthesummations overjandk,theKronecker deltas reduce theex- pression to (A><B)-(0><D)=E<A.B..CD.. -A..B)C)D..) This equation canberearranged toobtain (A><B)-(C><1))=(E)-B0,) (§;B,,,1),,,) —(gag) (2)-t,,,1),,,) Because each term inparentheses ontheright-hand sideisjustascalar product, wehave, ﬁnally, <A><B)-(C><B)=(A-c)(B-B) -(B-<:)<A-B) The orthogonality oftheunit vectors e,-requires thevector product tobe e2Xej=cki,j,kincyclic order (1.79a) Wecannow usethepermutation symbol toexpress thisresult as Thevector product C=AXB,forexample, cannow beexpressed as C = gksy-kei/1j.Bk Bydirect expansion andcomparison with Equation l.80a, wecanverify ade- terminantal expression forthevector product: C1 C2 C3 C=AXB=A1 A2 A3 (1.80b) 313233 Westate thefollowing identities without proof: A-(B><C)=B-(CXA)=C-(A><B)EABC AX(BXC) (AXB)-(CXD) (AXB)X(C><D)(A-C)B —(A-B)C A-[Bx(C><D)] A-[(B-D)C— (B-C)D] (A-C)(B-D) —(A-D)(B-C) [(AxB)-D]C —[(AxB)-C]D (ABD)C —(ABC)D =(ACD)B —(BCD)A(1.31) (1.32) (1.33) (1.34) 1.13 DIFFERENTIATION OFAVECTOR YVITH RESPECT TOASCALAR 29 1.13 Differentiation ofaVector with Respect toaScalar Ifascalar function ¢>=¢>(s) isdifferentiated with respect tothescalar variable s, then, because neither part ofthederivative can change under acoordinate transformation, thederivative itself cannot change and must therefore bea scalar; thatis,inthexiandxicoordinate systems, qb=c,b'ands=s’,sodqb=dqﬁ’ andds= ds'.Hence d_¢>_2t'_ d_¢'ds_48''ds Similarly, wecanformally deﬁne thedifferentiation ofavector Awith re- spect toascalar s.The components ofAtransform according to A;=Z,\,-,-A], (1.35)1 Therefore, ondifferentiation, weobtain (because the)t.,-jareindependent ofs’) 4,4! d dA- ds' ds'j UJ jUnis’ Because sands’areidentical, wehave 4/1; dA,-’ dA~ ds ds J ds Thus thequantities dAj/ds transform asdothecomponents ofavector and hence arethecomponents ofavector, which wecanwrite asdA/ds. Wecangiveageometrical interpretation tothevector dA/dsasfollows. First, fordA/ds toexist, Amust beacontinuous function ofthevariable stA=A(s). Suppose thisfunction isrepresented bythecontinuous curve FinFigure 1-15; at thepoint P,thevariable hasthevalue s,and atQithasthevalue s+As.The de- rivative ofAwith respect tosisthen given instandard fashion by LA =lim LA =llm A________(s +As) _.A__(s) (1.863) d3 As—>0 A5' As—>0 A5' *2 F($) Q AA , A(s+As) P AU) ,_, , xl FIGURE 1-15 Thevector A(s) traces outthefunction F(s) asthevariable schanges. 30 1/MATRICES, VECTORS, AND VECTOR CALCULUS The derivatives ofvector sums andproducts obey therules ofordinaiy vec- torcalculus. Forexample, l(A+B‘)=‘ill+dl (186b‘>ds ' ds ds '' a dBdA—A-B‘ =A 1.36‘ds( " dsds (°" d dB dA—(A xB)=A><—+—><B (1.86d)ds ds ds d dA d¢—A‘=—+—A 1.86 ‘ds(¢ "¢ds ds ( e) andsimilarly fortotal differentials andforpartial derivatives. 1.14 Examples ofDerivatives— Velocity andAcceleration Ofparticular importance inthedevelopment ofthedynamics ofpoint particles (and ofsystems ofparticles) istherepresentation ofthemotion ofthese parti- clesbyvectors. Forsuch anapproach, werequire vectors torepresent theposi- tion, velocity, andacceleration ofagiven particle. Itiscustomary tospecify the position ofaparticle with respect toacertain reference frame byavector r,which isingeneral afunction oftime: r=r(t). The velocity vector vandtheacceleration vector aaredeﬁned according to vE3=t (1.87) a=@—5'~l~2—-- (188—at at? r ') where asingle dotabove asymbol denotes theﬁrst time derivative, andtwodots denote thesecond time derivative. Inrectangular coordinates, theexpressions forr,v,andaare r=xlel +x2e? +xgeg, = x,e,- PositionI .2. dxi .v=1-=Aate,=ZEe,- Velocity (1.89)2 1 d2x~a=v=i‘= e,-= ﬁe, AccelerationI Z Calculating these quantities inrectangular coordinates isstraightforward because theunit vectors e,-areconstant intime. Innonrectangular coordinate systems, however, theunit vectors attheposition oftheparticle asitmoves inspace are 1.14 EXAMPLES OFDERIVATIVES--VELOCITY AND ACCELERATION 31 notnecessarily constant intime, andthecomponents ofthetime derivatives ofr arenolonger simple relations, asinEquation 1.89. Wedonotdiscuss general cun/ilinear coordinate systems here, butplane polar coordinates, spherical coordi- nates, andcylindrical coordinates areofsufﬁcient importance towarrant adiscus- sion ofvelocity andacceleration inthese coordinate systems.* Toexpress vand ainplane polar coordinates, consider thesituation in Figure 1-16. Apoint moves along thecurve s(t) and inthetime interval t2—ll=dtmoves from P“)toP(2>. The unit vectors, e,and ea,which areor- thogonal, change from eil)toelmandfrom egl)toe§2).The change ine,is 69>-an=de, (1.90) which isavector normal toe,(and, therefore, inthedirection ofea).Similarly, thechange ine,,is e52’-<15"=deg (1.91) which isavector normal toea.Wecanthen write dc,=d0e@ (1.92) and deg=—d6e, (1.93) where theminus sign enters thesecond relation because degisdirected opposite toe,(seeFigure 1-16). . s(t) Pa) dr ds rat?‘-_ P11) 1'2(t) 1'1(t)cg) def (1) er de°<2)\ , 9e9 FIGURE 1-16 Anobject traces outthecurve s(t)over time. The unitvectors e,and66 andtheir differentials areshown fortwoposition vectors r1andr2. *Refer totheﬁgures inAppendix Fforthegeometry ofthese coordinate systems. 32 1/MATRICES, VECTORS, AND VECTOR CALCULUS Equations 1.92and1.93areperhaps easier toseebyreferring toFigure 1-16. Inthiscase, dc,subtends anangle d6with unitsides, soithasamagnitude ofd9. Italsopoints inthedirection ofeg,sowehave dc,=d0eg. Similarly, degsubtends anangle d6with unit sides, soitalsohasamagnitude ofd6,butfrom Figure 1-16 weseethat degpoints inthedirection of—e,sowehave deg=—d6e,. Dividing each sideofEquations 1.92 and1.93bydt,wehave e,=ée, (1.94) ég=-90, (1.95) Ifweexpress vas _5.1:_ig)Vatate’ =re,+ré, (1.96) wehave immediately, using Equation 1.94, sothat thevelocity isresolved into aradial component rand anangular (or transverse) component r6. Asecond differentiation yields theacceleration: d_ .a=Z!}(re, +r6eg) =iie,+re,+réeg +r§eg +Tééa =(r-¢é=’)¢,+ (T6+2ré)e,, » (1.99) sothat theacceleration isresolved into aradial component (F—r92) and an angular (ortransverse) component (rlil+2&9). The expressions fords,ds2,112,andvinthethree most important coordi- nate systems (see also Appendix F)are Rectangular coordinates (x,y,z) ds=dxlel +dx2e2 +dxgeg ds2=dxif+dx§+dx§ v2= + + iv=iclel +icgeg +icgeg(1.99) Spherical coordinates (r,9,¢>) ds=dre,+rd9eg +rsin 6dqbeg ds2=dr2+r2d62 +r2sin26 d¢>2 v2=#2+r292 +r2sin20ql>2 v=ie,+rfieg +rsin 0dieg,(1.100) 1.14EXAMPLES orDERIVATIVES--VELOCITY ANDACCELERATION 33 (The expressions forplane polar coordinates result from Equation 1.100 byset- ting 41¢=0.) Cylindrical coordinates (r,qb,z) ds=dre,+rdqbeg, +dze, ds2=dr2+r2d¢2 +dz2 v2=12+r2¢)2 +22 v=re,+rrpeg +ﬁe, Find thecomponents oftheacceleration vector aincylindrical coordinates.(1.101) Solution. Thevelocity components incylindrical coordinates were given in Equation 1.101. The acceleration isdetermined bytaking thetime derivative ofv. d d .a=——v=——re +r e+'e dt dt( r ¢d> Zz) =re,+ it-=,+ rqleg+niieg+rqlbi-=g,+ '2e,+ ze, Weneed toﬁnd thetime derivative oftheunit vectors e,,eg,and e,.The cylindrical coordinate system isshown inFigure 1-17, andinterms _ofthe(x,y,z) components, theunit vectors e,,cg,ande,are e,=(cos qb,sinqb,0) eg,=(—sin ¢>,cos¢>,O) e,=(0,0, 1) Z ez 9A.99} \,.....~.--H» an-an!, J’11111111' I JC FIGURE 1-17 Thecylindrical coordinate system (r,qb,z)areshown with respect tothe Cartesian system (x,y,z). 34 1/MATRICES, VECTORS, ANDVECTOR CALCULUS The time derivatives oftheunitvectors arefound bytaking thederivatives of thecomponents. 9.=<—sin<1».01cos¢.0)=-9-.ég,=(~qb cos¢,-¢> sincb,0)=—¢>e, é,=0 Wesubstitute theunitvector time derivatives into theabove expression fora. a=re,+iqieg, +rqﬁeg, +rciieg, —rqB2e, +Ze, =(r-rql>2)e,+(T15;+2t¢)e_, +2e, 1.15Angular Velocity Apoint oraparticle moving arbitrarily inspace may always beconsidered, ata given instant, tobemoving inaplane, circular path about acertain axis; that is, thepath aparticle describes during aninﬁnitesimal time interval 5tmay berep- resented asaninﬁnitesimal arcofacircle. The linepassing through thecenter ofthecircle and perpendicular totheinstantaneous direction ofmotion is called theinstantaneous axis ofrotation. Astheparticle moves inthecircular path, therateofchange oftheangular position iscalled theangular velocity: as.=—=0 1.100’at (2) Consider aparticle that moves instantaneously inacircle ofradius Rabout anaxisperpendicular totheplane ofmotion, asinFigure 1-18. Lettheposition vector roftheparticle bedrawn from anorigin located atanarbitrary point O ontheaxis ofrotation. The time rate ofchange oftheposition vector isthe linear velocity vector oftheparticle, 1»=v.Formotion inacircle ofradius R,the instantaneous magnitude ofthelinear velocity isgiven by dv=R2‘; =Rm (1.103) The direction ofthelinear velocity visperpendicular torandintheplane ofthe circle. Itwould bevery convenient ifwecould devise avector representation of theangular velocity (say, to)sothat allthequantities ofinterest inthemotion oftheparticle could bedescribed onacommon basis. Wecandeﬁne adirection fortheangular velocity inthefollowing manner. Iftheparticle moves instanta- neously inaplane, thenormal tothat plane deﬁnes aprecise direction in space-—-or, rather—two directions. Wemaychoose aspositive thatdirection correspon- ding tothedirection ofadvance ofaright-hand screw when tumed inthesame sense astherotation oftheparticle (seeFigure 1-18). Wecanalsowrite themag- nitude ofthelinear velocity bynoting thatR=rsinoz. Thus v=rwsinoz (1.104) 1.15ANGULAR VELOCITY 35 (D V 0:r O FIGURE 1-18 Aparticle moving ccwabout anaxis according totheright-hand rule hasanangular velocity to=vXrabout that axis. Having deﬁned adirection and amagnitude fortheangular velocity, wenote thatifwewrite <1-1"5>then both ofthese deﬁnitions aresatisﬁed, andwehave thedesired vector rep- resentation oftheangular velocity. Weshould note atthispoint animportant distinction between ﬁnite andin- ﬁnitesimal rotations. Aninf'mitesi1nal rotation canberepresented byavector (actually, anaxial vector), butafinite rotation cannot. The impossibility ofde- scribing aﬁnite rotation byavector results from thefactthat such rotations do notcommute (seetheexample ofFigure 1-9), andtherefore, ingeneral, differ- entresults willbeobtained depending ontheorder inwhich therotations are made. Toillustrate thisstatement, consider thesuccessive application oftwoﬁ- niterotations described bytherotation matrices A1andA2.Letusassociate the vectors AandBinaone-to-one manner with these rotations. Thevector sumisC= A+B,which isequivalent tothematrix A3=A2A1. Butbecause vector addition iscommutative, wealso have C=B+A,with A4=A1A2. Butweknow that matrix operations arenotcommutative, sothat ingeneral A3¢A4.Hence, the vector Cisnotunique, and therefore wecannot associate avector with aﬁnite rotation. Inﬂnitesimal rotations donotsuffer from thisdefect ofnoncommutation. We aretherefore ledtoexpect thataninﬁnitesimal rotation canberepresented bya vector. Although thisexpectation is,infact, fulﬁlled, theultimate testofthevec- tornature ofaquantity iscontained initstransformation properties. Wegive only aqualitative argument here. Refer toFigure 1-19. Iftheposition vector ofapoint changes from rtor+ 5r,thegeometrical situation iscorrectly represented ifwewrite 51*=50Xr (1.106) 36 1/MATRICES, VECTORS, ANDVECTOR CALCULUS 50 / r r+51' FIGURE 1-19 Theposition vector rchanges tor+5rbyaninﬁnitesimal rotation angle 50. where 50isaquantity whose magnitude isequal totheinﬁnitesimal rotation angle and that hasadirection along theinstantaneous axis ofrotation. The mere fact that Equation 1.106 correctly describes thesituation illustrated in Figure 1-19isnotsufﬁcient toestablish that50isavector. (Wereiterate thatthe true testmust bebased onthetransformation properties of50.) Butifweshow thattwoinﬁnitesimal rotation “vectors”—501 and502—-actually commute, thesole objection torepresenting aﬁnite rotation byavector willhave been removed. Letusconsider thatarotation 501takes rinto r+5r1,where 5r1=501Xr. Ifthisisfollowed byasecond rotation 502around adifferent axis, theinitial po- sition vector forthisrotation isr+5r1.Thus 5r2=502X(r+5r1) andtheﬁnal position vector for501followed by502is r+5r12=r+ [501><r+502X (r+5r1)] Neglecting second-order inﬁnitesimals, then, 5r12=601><r+502><r (1.107) Similarly, if502isfollowed by501,wehave r+5r21=r+[502><r+501>< (r+5r2)] or 5r21 = 502 X1‘+ 601 Xr Rotation vectors 5r12and5r21areequal, sotherotation “vectors” 501and502do commute. Ittherefore seems reasonable that 50inEquation 1.106 isindeed a vector. 1.16GRADIENT OPERATOR 37 Itisthefactthat50isavector thatallows angular velocity toberepresented byavector, because angular velocity istheratio ofaninﬁnitesimal rotation angle toaninﬁnitesimal time: 50m=—5t Therefore, dividing Equation 1.106 by5t,wehave 5r 50—=—Xr6:5: or,inpassing tothelimit, 5t—>O, v=toxr asbefore. 1.16 Gradient Operator Wenow turn tothemost important member ofaclass called vector differential operators—-the gradient operator. Consider ascalar ¢>that isanexplicit function ofthecoordinates xiand, moreover, isacontinuous, single-valued function ofthese coordinates through- outacertain region ofspace. Under acoordinate transformation thatcarries the x,into thexf,¢'(x1, x2,xg)=<;b(x1, x2,x3),andbythechain rule ofdifferentia- tion, wecanwrite ' 6x- 2=ZBib—’, (1.109)8x1 j(ix)5x1 The case issimilar for6gb'/8.102 and8gb’/8x;.'1, soingeneral wehave 395'_23¢ 3’?—— —— 1.110{ix} 1'Bxjﬁxj ( ) The inverse coordinate transformation is lg=§kL\,,x,; (1.111) Differentiating, ﬁx) a , ax;(Tc; =‘Td(%/1),]-Xk) =g/\k]~((TQ) (1.112) Buttheterm inthelastparentheses isjust5,1,,so 8x]- =;A,,,»5,-1, =Ag (1.113) 38 1/MATRICES, VECTORS, ANDVECTOR CALCULUS Substituting Equation 1.113 into Equation 1.110, weobtain 0' 03}=2/\,-)2 (1.114) Because itfollows thecorrect transformation equation ofavector (Equation 1.44), thefunction 19¢)/Bx) isthejthcomponent ofavector termed thegradient ofthefunction Note that even though qbisascalar, thegradient of¢isavector. The gradient of¢iswritten either asgrad ¢>orasV¢>(“de1” t,b). Because thefunction qbisanarbitrary scalar function, itisconvenient tode- ﬁnethedifferential operator described inthepreceding interms ofthegradient operator: (grad), =V,= (1.115) Wecanexpress thecomplete vector gradient operator as grad =V=_»— Gradient (1.116) The gradient operator can (a)operate directly onascalar function, asin V¢>;(b)beused inascalar product with avector function, asinV-A(the diver- gence (div) ofA);or(c)beused inavector product with avector function, asin VXA(the curlofA).Wepresent thegrad, divergence, andcurl: 8 gradqb =Vqb=20,5 (1.117a) 9A.divA=V-A=2a—' (l.117b)xi 6Acl.1r1A=v><A=Z8,-,.,,—"e, (1.117¢)ﬂak (ix) Toseeaphysical interpretation ofthegradient ofascalar function, consider thethree-dimensional andtopographical maps ofFigure 1-20. The closed loops ofpart brepresent lines ofconstant height. Letqbdenote theheight atanypoint ¢=¢(xls x2: x3)- Then -14>=Ziﬁdx. =E_<v¢>.-as 1ﬂxi i The components ofthedisplacement vector dsaretheincremental displace- ments inthedirection ofthethree orthogonal axes: ds=(dx1, dx2,dxs) (1.118) Idqb=(Vcb) -dsi (1.119)Therefore 1.16GRADIENT OPERATOR 39 (=1) 3’ I 1 ds 0. ~ 0.102 . 1 O I I I x (b) FIGURE 1-20 (a)Athree-dimensional contour map canberepresented by(b)a topographical map oflines ¢representing constant height. The gradient V4’)represents thedirection perpendicular totheconstant r,blines. Letdsbedirected tangentially along oneoftheisolatitude lines (i.e., along aline forwhich qb=const.), asindicated inFigure 1-20. Because cb=const. for thiscase, d¢=0.But,because neither Vqbnordsisingeneral zero, theymust there- fore beperpendicular toeach other. Thus Vqbisnormal totheline (orinthree dimensions, tothesurface) forwhich qb=const. 40 1/MATRICES, VECTORS, ANDVECTOR CALCULUS The maximum value ofdq5results when Vqband dsareinthesame direc- tion; then, (a¢),,,,,= |v¢>|a.<, forvqbllas _@ |v¢|_(dag) (1.120)IIIHXOf Therefore, Vqbisinthedirection ofthegreatest change in¢. Wecansummarize these results asfollows: 1.Thevector V¢is,atanypoint, normal tothelines orsurfaces forwhich cb= const. 2.Thevector Vqbhasthedirection ofthemaximum change incl). 3.Because anydirection inspace canbespeciﬁed interms oftheunit vector n inthatdirection, therateofchange ofqbinthedirection ofn(the directional derivative of¢>)canbefound from n-VgbE8(1)/8n. The successive operation ofthegradient operator produces 86 62 x»~ x~ This important product operator, called theLap1acian,* isalsowritten 2 V2=(% (1.122) When theLaplacian operates onascalar, wehave, forexample, val= (1.123) 1.17 Integration ofVectors Thevector resulting from thevolume integration ofavector function A=A(x,-) throughout avolume Visgiven byl JAdv =(IA1dv, J’A2dv, [A3dv> (1.124) v V V v *After Pierre Simon Laplace (1749-1827); the notation V2isascribed toSirWilliam Rowan Hamilton. lThe symbol f1,actually represents atriple integral over acertain volume V.Similarly, thesymbol fs stands foradouble integral over acertain surface S. 1.17INTEGRATION orVECTORS 41 19.4%‘ ii, ‘A110:1 -i{§.;§?1é1'§1(s*§'5I 5 {iii 1 >liftRE‘555: iii/1faigilxy wt to *ﬁlm jg ,7Si“)§Hg;’§(I '1;ii1“*9:~s=-1,10'§§21_‘;13§§5 ,‘;‘5;e;:;_, 1:)-~-ai la.»..1.-.»§:..<<...)-.. 1:1 ,-aﬁwc 191-W2:M5 373.: “Qsiiffli‘ii“ﬁn. 1 11..1,‘I Y.-£@1¥Lt= z ...'iIIﬁ2l*i... .1.2,-2‘? .:,§;'__,..~-:.,,,_..::;;g g ‘7‘M FIGURE 1-21 The differential dais anelement ofarea ofthesurface. Itsdirection is normal tothesurface. Thus, weintegrate thevector Athroughout Vsimply byperforming three sepa- rate, ordinary integrations. Theintegral over asurface Softheprojection ofavector function A=A(x,-) onto thenormal tothatsurface isdeﬁned tobe J'A-da S where dais anelement ofarea ofthesurface (Figure 1-21). Wewrite daasavec- torquantity because wemay attribute toitnotonly amagnitude dabutalsoadi- rection corresponding tothenormal tothesurface atthepoint inquestion. If theunit normal vector isn,then da=nda (1.125) Thus, thecomponents ofdaaretheprojections oftheelement ofarea onthe three mutually perpendicular planes deﬁned bytherectangular axes: da1 ==dx2dxg, etc. (1.126) Therefore, wehave [A-da =[A-nda (1.127)S S Of [A-da= (EA,-aa, (1.128)S S1 Equation 1.127 states that theintegral ofAover thesurface Sistheintegral of thenormal component ofAover thissurface. The normal toasurface may betaken tolieineither oftwopossible direc- tions (“up” or“doWn”); thus thesign ofnisambiguous. Ifthesurface isclosed, we adopt theconvention thattheoutward normal ispositive. The lineintegral ofavector function A=A(x,-) along agiven path extend- ingfrom thepoint Btothepoint Cisgiven bytheintegral ofthecomponent of 42 1/MATRICES, VECTORS, ANDVECTOR CALCULUS C ds A Q A ds P B FIGURE 1-22 Theelement dsisanelement oflength along thegiven path from Bto C.Itsdirection isalong thepath atagiven point. Aalong thepath 1A-ds=1ZA,-dx, (1.129)BC BC1 The quantity dsisanelement oflength along thegiven path (Figure 1-22). The direction ofdsistaken tobepositive along thedirection thepath istraversed. In Figure 1-22 atpoint P,theangle between dsandAislessthan rr/2, soA-dsis positive atthispoint. Atpoint Q,theangle isgreater than 1r/2,andthecontri- bution totheintegral atthispoint isnegative. Itisoften useful torelate certain surface integrals toeither volume integrals (Gauss’s theorem) orline integrals (Stokes’s theorem). Consider Figure l-23, which shows aclosed volume Venclosed bythesurface S.LetthevectorA andits ﬁrst derivatives becontinuous throughout thevolume. Gauss’s theorem states thatthesurface integral ofAover theclosed surface Sisequal tothevolume in- tegral ofthedivergence ofA(V-A)throughout thevolume Venclosed bythe surface S.Wewrite thismathematically as J'A-da =iv-Aav (1.180)S V Gauss stheorem issometimes also called thedivergence theorem. The theorem is particularly useful indealing with themechanics ofcontinuous media. SeeFigure l-24 forthephysical description needed forStokes’s theorem, which applies toanopen surface Sandthecontour path Cthat deﬁnes thesur- face. The curlofthevector A(VXA)must exist andbeintegrable over theen- tiresurface S.Stokes’s theorem states that theline integral ofthevector A around thecontour path Cisequal tothesurface integral ofthecurl ofAover thesurface deﬁned byC.Wewrite itmathematically as [A-ds =1(v><A)-da (1.191)C S where thelineintegral isaround theclosed contour path C.Stokes’s theorem is particularly useful inreducing certain surface integrals (two dimensional) to,it PROBLEMS 43 _:.'a- :§§:“ﬂN-1.;-.2-4-. 20'E>'12.'3-=I I.1 If0*1' J, :~~:11. 42“““""T‘ 1.92-11112.;'{~¢'.1§ :.=-.;=*§?=.(,»= I 'i'=1;;,, gm _ Surface S ‘if Volume V FIGURE 1-23 The differential daisanelement ofarea onasurface Sthat surrounds aclosed volume V .1...‘-..-zazjaaaifi-teal‘ eel=-‘=~>. ~"====-.. A '5I?.1‘“:EEE"’EE£- ==..~:-"Q-;E‘$ if"..1'.1‘ ~ ‘‘=11. Q.A Skiﬂfggsei’111"if.;=-t)2=;.=...===; l*:=-r_~€Z¢i=“¥é( 1-. 1 Surface S ‘~=2E:5522::/siiiiisiliiil *=1-;(=.-. _’:.I>s. .:-2= - 1.15.,‘_:li€E2"’1!;iI§::.. '1';22:“.'.5:'H-‘"=1-»‘I .21":-;== ,,,».-.<,,...,..:,,,.,,,_ .......». .. -=¥i§1=i211;: 2:‘;:= 1,..4,.....,!..‘ ,E;'1¥= I=L.1. I. =i;i-__?2:2=1E£.;1Z T,...\4T F ézil/7,1A I*1;IsI32&<;3*) Ei.1‘. I'~‘1.1.1. FIGURE 1-24 Acontour path Cdeﬁnes anopen surface S.Alineintegral around the path Candasurface integral over thesurface Sisrequired forStokeS’s theorem. ishoped, asimpler line integral (one dimensional). Both Gauss’s andStokes’s theorems have wide application invector calculus. Inaddition tomechanics, they arealsouseful inelectromagnetic applications andinpotential theory. PROBLEMS 1-1. Find thetransformation matrix thatrotates theaxisxgofarectangular coordinate system 45°toward x1around thex2-axis. 1-2. Prove Equations 1.10and1.11from trigonometric considerations. 1-3. Find thetransformation matrix that rotates arectangular coordinate system through anangle of120° about anaxismaking equal angles with theoriginal three coordinate axes. 1-4. Show (a)(AB)‘ =B‘A’ (b)(AB)‘1 =B"1A“ 1-5. Show bydirect expansion that |A2=1.For simplicity, take Atobeatwo- dimensional orthogonal transformation matrix. 44 1-6. 1-7. 1-8. 1-9. 1-10 1-11 1-12 1-13.1/MATRICES, VECTORS, AND VECTOR CALCULUS Show thatEquation 1.15 canbeobtained byusing therequirement thatthetrans- formation leaves unchanged thelength ofalinesegment. Consider aunit cube with onecorner attheorigin andthree adjacent sides lying along thethree axes ofarectangular coordinate system. Find thevectors describ- ingthediagonals ofthecube. What istheangle between anypairofdiagonals? LetAbeavector from theorigin toapoint Pﬁxed inspace. Letrbeavector from theorigin toavariable point Q(x1, x2,x3).Show that A-r =A2 istheequation ofaplane perpendicular toAandpassing through thepoint P. Forthetwovectors A=i+2j—k, B=—2i+3j+k ﬁnd (a)A—Band |A—B| (b)component ofB alongA (c)angle between AandB (d)A><B (e)(A—B)X(A+B) Aparticle moves inaplane elliptical orbit described bytheposition vector r=Zbsin wti +bcos wtj (a)Find v,a,andtheparticle speed. (b)What istheangle between vandaattime t=11"/2w? Show thatthetriple scalar product (AXB)-Ccanbewritten as A1 A2 As <A>-c= B1B2B3ClQC3 Show alsothattheproduct isunaffected byaninterchange ofthescalar andvector product operations orbyachange intheorder ofA,B,C,aslong asthey arein cyclic order; thatis, (AXB)-C=A-(BXC)=B'(CXA)=(CXA)-B, etc. Wemaytherefore usethenotation ABC todenote thetriple scalar product. Finally, give ageometric interpretation ofABC bycomputing thevolume oftheparal- lelepiped deﬁned bythethree vectors A,B,C. Leta,b,cbethree constant vectors drawn from theorigin tothepoints A,B,C. What isthedistance from theorigin totheplane deﬁned bythepoints A,B,C? What isthearea ofthetriangle ABC? Xisanunknown vector satisfying thefollowing relations involving theknown vec- torsAandBandthescalar cf), AXX=B, A-X=¢. Express Xinterms ofA,B,qb,andthemagnitude ofA. PROBLEMS 45 1-14. Consider thefollowing matrices: 1-15. 1-16. 1-17 1-18 1-19. 1-20 1-21 1-2212-1 210 21 A= 031,B=0-12,c=43 201 113 10 Find thefollowing (a)|AB| (b)AC (c)ABC (d)AB —BtAt Find thevalues ofozneeded tomake thefollowing transformation orthogonal. 10 0 0a—a 0a as What surface isrepresented byr-a=const. thatisdescribed ifaisavector ofcon- stant magnitude anddirection from theorigin andristheposition vector tothe point P(x1, x2,x3)onthesurface? Obtain thecosine lawofplane trigonometry byinterpreting theproduct (A—B)' (A—B)andtheexpansion oftheproduct. Obtain thesine lawofplane trigonometry byinterpreting theproduct AXBand thealternate representation (A~'B)XB. Derive thefollowing expressions byusing vector algebra: (a)cos(01—B)=cosa cosB +sina sinB (b)sin(a—B)=sinacosB —cosasinB Show that izjsﬁk :0 jgksgk 8;]-k :2651 (C) 51891 83:71 :6 Show (seealsoProblem 1-11) that ABC :UEkE€"kA;BjCk Evaluate thesum 21$,-1,, elm),(which contains 3terms) byconsidering theresult for allpossible combinations ofi,j,l,m;thatis, (a)i=j (b)i=l (c)i=m (d)j=l (e)j=m (f)l=m (g)i¢lorm (h)j¢lorm Show that 29¢)";-‘ilm =5a5jm _51111311 andthen usethisresult toprove AX (BX C)=(A-C)B— (A-B)C 46 1-23. 1-24. 1-25 1-26 1-27 1-28. 1-29 1-30 1-31 1-32 1-331/MATRICES, VECTORS, AND VECTOR CALCULUS Usethesijknotation andderive theidentity (AXB)X(CXD)=(ABD)C —(ABC)D LetAbeanarbitrary vector, andletebeaunitvector insome ﬁxed direction. Show that A=e(A-e) +eX (AXe) What isthegeometrical signiﬁcance ofeach ofthetwoterms oftheexpansion? Find thecomponents oftheacceleration vector ainspherical coordinates. Aparticle moves with v=const. along thecurve r==k(1+cos9)(acardioid). Find 1’-e, =a-e,,|a|, and9. Ifrandi"=vareboth explicit functions oftime, show that d 2 2gt[r X(vXr)]=ra+(r-v)v -(v+r-a)r Show that v<1n|rl> =%T Find theangle between thesurfaces deﬁned byr2=9andx+y+22=1atthe point (2,—2,1). Show thatV(¢\//) =qbV1/1 +1,!/Vqb. Show that (a)Vr"=nr(”‘2)r (b)Vf(r) = (c)V2(ln r)=% Show that J’(2ar-r +2bi'- i‘)dt =arg+bi?+const. where risthevector from theorigin tothepoint (xl,x2,x3).Thequantities rand if arethemagnitudes ofthevectors randi',respectively, andaandbareconstants. Show that i'ri" rJ’<— -2)dt =-+C rr r where Cisaconstant vector. PROBLEMS 47 1-34. Evaluate theintegral 1-35 1-36 1-37 I-38 I-39 1-40. 1-41[AxAd: Show thatthevolume common totheintersecting cylinders deﬁned byx2+)1?=a2 andx2+z2=a2isV=16a3/3. Find thevalue oftheintegral fSA-da,where A=xi—yj+zkand Sistheclosed surface deﬁned bythecylinder 02=x2+y2.The topandbottom ofthecylinder areatz=dandO,respectively. Find thevalue oftheintegral fSA- da,where A=(x2+yg+z2)(xi+yj+zk)and thesurface Sisdeﬁned bythesphere R2=xi+y2+z2.Dotheintegral directly and also byusing Gauss’s theorem. Find thevalue oftheintegral fs(VXA)'daifthevectorA =yi+zj+xkandSis thesurface deﬁned bytheparaboloid z=1—x2—3:2,where z20. Aplane passes through thethree points (x,y,z)=(1,O,O),(O,2,0),(O,O,3). (a)Find aunit vector perpendicular totheplane. (b)Find thedistance from the point (1,1,1)totheclosest point ofthePlane andthecoordinates oftheclosest point. The height ofahillinmeters isgiven byz=2xy-3x2—4y?—18x+28y+12, where xisthedistance eastandyisthedistance north oftheorigin. (a)Where is thetopofthehilland how high isit?(b)How steep isthehillatx==y=1,that is, what istheangle between avector perpendicular tothehillandthezaxis? (c)In which compass direction istheslope atx=y=1steepest? Forwhat values ofaarethevectors A=2ai—2j+akandB=ai+2aj+2k perpendicular? CHAPTER W Newtonian Mechanics— Single Particle 2.1Introduction The science ofmechanics seeks toprovide aprecise and consistent descrip- tion ofthedynamics ofparticles andsystems ofparticles, that is,asetofphys- icallaws mathematically describing themotions ofbodies and aggregates of bodies. Forthis, weneed certain fundamental concepts such asdistance and time. The combination oftheconcepts ofdistance and time allows usto deﬁne thevelocity and acceleration ofaparticle. The third fundamental concept, mass, requires some elaboration, which wegive when wediscuss Newton’s laws. Physical laws must bebased onexperimental fact. Wecannot expect apri- orithat thegravitational attraction between twobodies must vary exactly as theinverse square ofthedistance between them. Butexperiment indicates that thisisso.Once asetofexperimental data hasbeen correlated andapos- tulate hasbeen formulated regarding thephenomena towhich thedata refer, then various implications canbeworked out. Ifthese implications areallveri- ﬁed byexperiment, wemay believe that thepostulate isgenerally true. The postulate then assumes thestatus ofaphysical law. Ifsome experiments dis- agree with thepredictions ofthelaw,thetheory must bemodiﬁed tobecon- sistent with thefacts. Newton provided uswith thefundamental laws ofmechanics. Westate these lawshere inmodem terms, discuss their meaning, andthen derive theimplications 48 2.2NEWTON’S LAWS 49 ofthelaws invarious situations.* Butthelogical structure ofthescience ofme- chanics isnotstraightforward. Ourlineofreasoning ininterpreting Newton’s laws isnottheonly onepossible? Wedonotpursue inanydetail thephilosophy ofme- chanics butrather giveonly sufficient elaboration ofNewton’s laws toallow usto continue with thediscussion ofclassical dynamics. Wedevote ourattention inthis chapter tothemotion ofasingle particle, leaving systems ofparticles tobedis- cussed inChapters 9and11-13. 2.2 Newton’s Laws Webegin bysimply stating inconventional form Newton’s laws ofmechanicsi: I.Abodyremains atrestorinuniform motion unless acted upon byaforce. II.Abodyacted upon byaforce moves insuch amanner thatthetimerateofchange of momentum equals theforce. IH. Iftwobodies exert forces oneach other; these forces areequal inmagnitude andoppa siteindirection.~. These laws aresofamiliar thatwesometimes tend tolosesight oftheir true signiﬁcance (orlack ofit)asphysical laws. The First Law, forexample, ismean- ingless without theconcept of“force,” aword Newton used inallthree laws. In fact, standing alone, theFirst Law conveys aprecise meaning only forzeroforce; thatis,abody remaining atrestorinuniform (i.e., unaccelerated, rectilinear) motion issubject tonoforce whatsoever. Abody moving inthismanner is termed afree body (orfree particle). The question oftheframe ofreference with respect towhich the“uniform motion” istobemeasured isdiscussed inthe following section. Inpointing out thelack ofcontent inNewton’s First Law, SirArthur Eddington§ observed, somewhat facetiously, that allthelawactually saysisthat “every particle continues initsstate ofrestoruniform motion inastraight line *Truesdell (Tr68) points outthat Leonhard Euler (1707-1783) clariﬁed and developed the Newtonian concepts. Euler “put most ofmechanics into itsmodern form” and“made mechanics simple andeasy” (p.106). ’rErnst Mach (1838-1916) expressed hisview inhisfamous book ﬁrstpublished in1883; E.Mach, Die Mechanic inihrerEntwicklung histmcisch-kritisch dargestellt [The science ofmechanics] (Prague, 1883). Atranslation ofalater edition isavailable (Ma60). Interesting discussions arealso given by R.B.Lindsay andH.Margeneau (Li36) andN.Feather (Fe59). IEnunciated in1687 bySirIsaac Newton (1642-1727) inhisPhilosophiae naturalis principia mathemat- ica[Mathematical principles ofnatural philosophy, normally called Principia] (London, 1687). Previously. Galileo (1564-1642) generalized theresults ofhisown mathematical experiments with statements equivalent toNewton’s First andSecond Laws. ButGalileo wasunable tocomplete thedescription of dynamics because hedidnotappreciate thesigniﬁcance ofwhat would become Newton’s Third Law—and therefore lacked aprecise meaning offorce. §SirArthur Eddington (Ed30, p.124). 50 2/NEWTONIAN MECHANICS-SINGLE PARTICLE except insofar asitdoesn’t.” This ishardly fairtoNewton, who meant something very deﬁnite byhisstatement. Butitdoes emphasize that theFirst Law byitself provides uswith only aqualitative notion regarding “force.” The Second Law provides anexplicit statement: Force isrelated tothetime rate ofchange ofmomentum. Newton appropriately deﬁned momentum (al- though heused theterm quantity ofmotion) tobetheproduct ofmass andveloc- ity,such that pEmv (2.1) Therefore, Newton’s Second Law canbeexpressed as dp dF—E—dt(mv) (2.2) The deﬁnition offorce becomes complete andprecise only when “mass” isde- ﬁned. Thus theFirst andSecond Laws arenotreally “laws” intheusual sense; rather, they may beconsidered deﬁnitions. Because length, time, and mass are concepts normally already understood, weuseNewton’s First andSecond Laws astheoperational deﬁnition offorce. Newton’s Third Law, however, isindeed a law.Itisastatement concerning therealphysical world andcontains allofthe physics inNewton’s laws ofmotion.* Wemust hasten toadd, however, that theThird Law isnotageneral lawof nature. The lawdoes apply when theforce exerted byone (point) object onan- other (point) object isdirected along theline connecting theobjects. Such forces arecalled central forces; theThird Law applies whether acentral force is attractive orrepulsive. Gravitational andelectrostatic forces arecentral forces, soNewton’s laws can beused inproblems involving these types offorces. Sometimes, elastic forces (which areactually macroscopic manifestations ofmi- croscopic electrostatic forces) arecentral. Forexample, twopoint objects con- nected byastraight spring orelastic string aresubject toforces that obey the Third Law. Any force thatdepends onthevelocities oftheinteracting bodies is noncentral, and theThird Law may notapply. Velocity-dependent forces are characteristic ofinteractions that propagate with ﬁnite velocity. Thus theforce between moving electric charges does notobey theThird Law, because theforce propagates with thevelocity oflight. Even thegravitational force between mov- ingbodies isvelocity dependent, buttheeffect issmall and difficult todetect. The only observable effect istheprecession oftheperihelia oftheinner planets (see Section 8.9). Wewillreturn toadiscussion ofNewton’s Third Law in Chapter 9. Todemonstrate thesigniﬁcance ofNewton’s Third Law, letusparaphrase it inthefollowing way,which incorporates theappropriate deﬁnition ofmass: *The reasoning presented here, viz.,thattheFirst andSecond Laws areactually deﬁnitions andthat theThird Law contains thephysics, isnottheonly possible interpretation. Lindsay andMargenau (LL36), forexample, present theﬁrst twoLaws asphysical laws and then derive theThird Law asa consequence. 2.2NEWTON’S LAWS 51 III’. Q‘twobodies constitute anideal, isolated system, thentheaccelerations ofthese bodies arealways inopposite directions, andtheratio ofthemagnitudes oftheaccelerations isconstant. Thisconstant ratio istheinverse ratio ofthemasses ofthebodies. With thisstatement, wecan 'vearactical deﬁnition ofmass andtherefore ive 8‘ P __ 8 precise meaning totheequations summarizing Newtonian dynamics. Fortwo isolated bodies, 1and2,theThird Lawstates that F1=—F2 (2.3) Using thedeﬁnition offorce asgiven bytheSecond Law, wehave dP1 dP2i=—i .4at at (23) d d and, because acceleration isthetime derivative ofvelocity, m1(31) =m2(_a2) (2-4c)or,with constant masses, Hence, m2 G1 ml a2 _ (2.5) where thenegative sign indicates only that thetwoacceleration vectors areop- positely directed. Mass istaken tobeapositive quantity. Wecanalways select, say,mlastheunitmass. Then, bycomparing theratio ofaccelerations when mlisallowed tointeract with anyother body, wecande- termine themass oftheother body. Tomeasure theaccelerations, wemust have appropriate clocks andmeasuring rods; also, wemust choose asuitable coordi- nate system orreference frame. The question ofa“suitable reference frame” is discussed inthenext section. One ofthemore common methods ofdetermining themass ofanobject is byweighing~—for example, bycomparing itsweight tothat ofastandard by means ofabeam balance. This procedure makes useofthefactthatinagravita- tional ﬁeld theweight ofabody isjustthegravitational force acting onthebody; thatis,Newton’s equation F=mabecomes W=mg,where gistheacceleration due togravity. The validity ofusing thisprocedure rests onafundamental as- sumption: thatthemass mappearing inNewton’s equation anddeﬁned accord- ingtoStatement III’isequal tothemass mthatappears inthegravitational force equation. These twomasses arecalled theinertial mass andgravitational mass, respectively. The deﬁnitions may bestated asfollows: Inertial Mass: That moss determining theacceleration ofabodyunder theaction ofa given force. Gravitational Mass: That mass determining thegravitational forces between abody andother bodies. 52 2/NEWTONIAN MECHANICS—SINGLE PARTICLE Galileo wastheﬁrsttotesttheequivalence ofinertial andgravitational mass inhis(perhaps apocryphal) experiment with falling weights attheTower ofPisa. Newton also considered theproblem and measured theperiods ofpendula of equal lengths butwith bobs ofdifferent materials. Neither Newton norGalileo found anydifference, but themethods were quite crude.* In1890 Eotvosl de- vised aningenious method totesttheequivalence ofinertial and gravitational masses. Using twoobjects made ofdifferent materials, hecompared theeffect of theEarth’s gravitational force (i.e., theweight) with theeffect oftheinertial force caused bytheEarth’s rotation. The experiment involved anullmethod using asensitive torsion balance andwastherefore highly accurate. More recent experiments (notably those ofDickei), using essentially thesame method, have improved theaccuracy, andweknow now thatinertial andgravitational mass are identical towithin afewparts in1012. This result isconsiderably important inthe general theory ofrelativity.§ The assertion oftheexact equality ofinertial and gravitational mass istenned theprinciple ofequivalence. Newton’s Third Law isstated interms oftwobodies that constitute aniso- lated system. Itisimpossible toachieve such anideal condition; every body inthe universe interacts with every other body, although theforce ofinteraction maybe fartooweak tobeofanypractical importance ifgreat distances areinvolved. Newton avoided thequestion ofhow todisentangle thedesired effects from all theextraneous effects. Butthispractical difﬁculty only emphasizes theenormity ofNewton’s assertion made intheThird Law. Itisatribute tothedepth ofhis perception andphysical insight that theconclusion, based onlimited observa- tions, hassuccessfully borne thetestofexperiment for300years. Only within the 20th century didmeasurements ofsuﬁicient detail reveal certain discrepancies with thepredictions ofNewtonian theory. The pursuit ofthese details ledtothe development ofrelativity theory andquantum mechanics." Another interpretation ofNewton’s Third Law isbased ontheconcept of momentum. Rearranging Equation 2.4agives ti Zt(P1 +P2)=0 or pl+p2=constant (2.6) The statement that momentum isconserved intheisolated interaction oftwo particles isaspecial case ofthemore general conservation oflinear momen- tum. Physicists cherish general conservation laws, and theconservation oflin- earmomentum isbelieved always tobeobeyed. Later weshall modify ourdeﬁ- *InNewton’s experiment, hecould have detected adifference ofonly onepartin103. ’yRoland von Eiitvos (1848-1919), aHungarian baron; hisresearch ingravitational problems ledto thedevelopment ofagravimeter, which wasused ingeological studies. IP.G.Roll, R.Krotkov, andR.H.Dicke, Ann. Phys. (N.Y.) 26,442(1964). SeealsoBraginsky and Pavov, Sov.Phys.-]ETP 34,463(1972). §See, forexample, thediscussions byP.G.Bergmann (Be46) and_].Weber (We61). Weber’s book alsoprovides ananalysis oftheEotvos experiment. ||See alsoSection 2.8. 2.3FRAMES orREFERENCE 53 nition ofmomentum from Equation 2.1forhigh velocities approaching the speed oflight. 2.3 Frames ofReference Newton realized that, forthelaws ofmotion tohave meaning, themotion of bodies must bemeasured relative tosome reference frame. Areference frame is called aninertial frame ifNewton ’slaws areindeed valid inthatframe; thatis,if abody subject tonoexternal force moves inastraight linewith constant velocity (orremains atrest), then thecoordinate system establishing thisfactisaniner- tialreference frame. This isaclear-cut operational deﬁnition and one that also follows from thegeneral theory ofrelativity. IfNewton’s laws arevalid inonereference frame, then they arealsovalid in any reference frame inunifonn motion (i.e., not accelerated) with respect to theﬁrstsystem.* This isaresult ofthefactthattheequation F=mi‘involves the second time derivative ofr:Achange ofcoordinates involving aconstant velocity does notinﬂuence theequation. This result iscalled Galilean invariance orthe principle ofNewtonian relativity. Relativity theory hasshown usthat theconcepts ofabsolute restand anab solute inertial reference frame aremeaningless. Therefore, even though wecon- ventionally adopt areference frame described with respect tothe“ﬁxed” stars— and, indeed, insuch aframe theNewtonian equations arevalid toahigh degree ofaccuracy—such aframe is,infact, not anabsolute inertial frame. Wemay, however, consider the“ﬁxed” stars todeﬁne areference frame that approxi- mates an“absolute” inertial frame toanextent quite sufﬁcient forourpresent purposes. Although theﬁxed-star reference frame isaconveniently deﬁnable system and onesuitable formany purposes, wemust emphasize that thefundamental deﬁnition ofaninertial frame makes nomention ofstars, ﬁxed orotherwise. Ifa body subject tonoforce moves with constant velocity inacertain coordinate sys- tem, that system is,bydeﬁnition, aninertial frame. Because precisely describing themotion ofarealphysical object intherealphysical world isnormally difﬁ- cult, weusually resort toidealizations and approximations ofvarying degree; that is,weordinarily neglect thelesser forces onabody ifthese forces donotsig- niﬁcantly affect thebody’s motion. Ifwewish todescribe themotion of,say,afree particle andifwechoose for thispurpose some coordinate system inaninertial frame, then werequire that the(vector) equation ofmotion oftheparticle beindependent oftheposition of theorigin ofthecoordinate system and independent ofitsorientation inspace. Wefurther require that time behomogeneous; that is,afree particle moving with acertain constant velocity inthecoordinate system during acertain time *InChapter 10,wediscuss themodiﬁcation ofNewton’s equations thatrnust bemade ifitisdesired todescribe themotion ofabody with respect toanoninertial frame ofreference, thatis,aframe that isaccelerated with respect toaninertial frame. 54 2/NEWTONIAN MECHANICS-—SINGLE PARTICLE C\ \ \ \ \ \ \ \ \ VP \\\ A 3 \\ \ 3 \\ V6 \\ B‘ ‘ 2 1 1 FIGURE 2-1 Wechoose todescribe thepath ofafreeparticle moving along thepath ACinarectangular coordinate system whose origin moves inacircle. Such asystem isnotaninertial reference frame. interval must not,during alater time interval, befound tomove with adifferent velocity. Wecanillustrate theimportance ofthese properties bythefollowing exam- ple.Consider, asinFigure 2-1,afreeparticle moving along acertain path AC.To describe theparticle’s motion, letuschoose arectangular coordinate system whose origin moves inacircle, asshown. Forsimplicity, welettheorientation of theaxes beﬁxed inspace. The particle moves with avelocity vi,relative toanin- ertial reference frame. Ifthecoordinate system moves with alinear velocity v, when atthepoint B,andifv,=vp,then toanobserver inthemoving coordinate system theparticle (atA)willappear tobeatrest.Atsome later time, however, when theparticle isatCand thecoordinate system isatD,theparticle willap- pear toaccelerate with respect totheobserver. Wemust, therefore, conclude that therotating coordinate system does notqualify asaninertial reference frame. These observations arenotsufficient todecide whether time ishomoge- neous. Toreach such aconclusion, repeated measurements must bemade in identical situations atvarious times; identical results would indicate thehomo- geneity oftime. Newton’s equations donotdescribe themotion ofbodies innoninertial sys- tems. Wecandevise amethod todescribe themotion ofaparticle byarotating coordinate system, but,asweshall seeinChapter 10,theresulting equations con- tainseveral terms thatdonotappear inthesimple Newtonian equation F=ma. Forthemoment, then, werestrict ourattention toinertial reference frames to describe thedynamics ofparticles. 2.4 THE EQUATION OFMOTION FORA PARTICLE 2.4The Equation ofMotion foraParticle Newton’s equation F=dp/dt canbeexpressed alternatively as dF=£(mv) =mi =mi‘ (2.7) ifweassume that themass mdoes notvary with time. This isasecond-order dif- ferential equation that may beintegrated toﬁnd r=r(t)ifthefunction Fis known. Specifying theinitial values ofrand i"=vthen allows ustoevaluate the twoarbitrary constants ofintegration. Wethen determine themotion ofaparti- clebytheforce function Fandtheinitial values ofposition randvelocity v. The force Fmay beafunction ofanycombination ofposition, velocity, and time andisgenerally denoted asF(r,v,t).Foragiven dynamic system, wenor- mally want toknow randvasafunction oftime. Solving Equation 2.7willhelp usdothisbysolving forii.Applying Equation 2.7tophysical situations isanim- portant part ofmechanics. Inthischapter, weexamine several examples inwhich theforce function is known. Webegin bylooking atsimple force functions (either constant orde- pendent ononly oneofr,v,and t)inonly onespatial dimension asarefresher ofearlier physics courses. Itisimportant toform good habits inproblem solving. Here aresome useful problem-solving techniques. g>o:>w»-.Make asketch oftheproblem, indicating forces, velocities, andsoforth. .Write down thegiven quantities. ' .Write down useful equations andwhat istobedetemiined. Strategy andtheprinciples ofphysics must beused tomanipulate theequa- tions toﬁnd thequantity sought. Algebraic manipulations aswell asdiffer- entiation orintegration isusually required. Sometimes numerical calcula- tions using acomputer aretheeasiest, ifnottheonly, method ofsolution. 5.Finally, putintheactual values fortheassumed variable names todetermine thequantity sought. Letusﬁrstconsider theproblem ofablock sliding onaninclined plane. Let theangle oftheinclined plane be6and themass oftheblock be100g.The sketch oftheproblem isshown inFigure 2—2a. J’ N N “ f\ Fé.cos6 ' 1/ /I9F // Ex g l\ Fe Igsin0‘ ‘ 9 9 (11) (b) FIGURE 2-2 Examples 2.1and2.2. 56 2/NEWTONIAN MECHANICS—SINGLE PARTICLE EXAMPLE 2.l Ifablock slides without friction down aﬁxed, inclined plane with 6=30°,what istheblock’s acceleration? Solution. Twoforces actontheblock (seeFigure 2-2a): thegravitational force Fg andtheplane’s normal force Npushing upward ontheblock (nofriction inthis example). The block isconstrained tobeontheplane, and theonly direction the block canmove isthex-direction, upanddown theplane. Wetake the+x-direc- tiontobedown theplane. The total force Fm,isconstant; Equation 2.7becomes Fm=Fg+N andbecause Fmisthenetresultant force acting ontheblock, Fnet :mi: or Fg+N=mi‘ (2.8) This vector must beapplied intwodirections: xand y(perpendicular tox). The component offorce inthey-direction iszero, because noacceleration oc- curs inthisdirection. The force Fgisdivided vectorially into itsx-andy-compo- nents (dashed lines inFigure 2—2a). Equation 2.8becomes y-direction —I'l, cos6+N= O (2.9) x-direction I1,sin6=mil (2.10) with therequired result ii=Ilrsinti =L-——mgsin6 =gsintim m .._ . ,,_8'_ 2x—gsin(3O )—E—4.9m/s (2.11) Therefore theacceleration oftheblock isaconstant. Wecanﬁnd thevelocity oftheblock after itmoves from restadistance x0 down theplane bymultiplying Equation 2.11 by25candintegrating 2ic55 =2icgsin6 d2 _ dx_' = 6_dt(x )2gsin dt j0d(:E2) =2gsinBjOdx 0 0 2.4 THE EQUATION OFMOTION FOR APARTICLE 57 Att=0,both x=ii=0,and, att=tﬁnal, x=x0,andthevelocity ic=v0. 113=2gsin6x0 v0=\/2gsin6x0 EXAMPLE 2.2 Ifthecoefficient ofstatic friction between theblock andplane intheprevious example isus=0.4,atwhat angle 6willtheblock start sliding ifitisinitially at rest? Solution. Weneed anewsketch toindicate theadditional frictional force f(see Figure 2-2b). The static frictional force hastheapproximate maximum value fmax=/J-.N (2-12) andEquation 2.7becomes, incomponent form, y-direction —Ii;,cos 6+N= 0 (2.13) x-direction —f,+ Fgsin6 =mii (2.14) The static frictional force jﬂwillbesome value f,Sfmxrequired tokeep 56=0 —that is,tokeep theblock atrest. However, astheangle 6oftheplane in- creases, eventually thestatic frictional force willbeunable tokeep theblock at rest. Atthatangle 6',jﬂbecomes f,(6=6’)=fnax=/.t,N= /.t,1'1,cos 6 and mil=Igsin 6—f,,,,,,, mil=Fgsin 6—uslil, cos6 (2.15) 56=g(sin 6—/1.,cos6) just before theblock starts toslide, theacceleration 56=0,so sin6 —/.t,cos6 =0 tan6=/is=0.4 6=tan‘1(0.4) =22° After theblock intheprevious example begins toslide, thecoefficient ofki- netic (sliding) friction becomes ;.t,,=0.3.Find theacceleration fortheangle 6=30°. 58 2/NEWTONIAN MECHANICS-—SINGLE PARTICLE Solution. Similarly toExample 2.2,thekinetic friction becomes (approxi- mately) ﬁ,=/.t,,N= /.t,,I'l, cos6 (2.16) and mié=Fgsin 6—ﬁ,=mg(sin6—/.t,,cos6) (2.17) 56=g(sin6—/.t,,cos6)=0.24g (2.18) Generally, theforce ofstatic friction (fm,,,, =/.t,N)isgreater than that of kinetic friction ()1=/.1,N).This canbeobserved inasimple experiment. Ifwe lower theangle 6below 16.7°, weﬁnd that 56<0,andtheblock eventually stops. Ifweraise theblock back upabove 6=16.7°,weﬁnd thattheblock does notstart sliding again until 6222°(Example 2.2). The static friction deter- mines when itstarts moving again. There isnotadiscontinuous acceleration as theblock starts moving, because ofthedifference between ].Lsanduh.Forsmall speeds, thecoefﬁcient offriction changes rather quickly from ti,to/.L,,. The subject offriction isstillaninteresting andimportant area ofresearch. There arestillsurprises. Forexample, even though wecalculate theabsolute value ofthefrictional force asf=/.tN,research hasshown thatthefrictional force isdirectly proportional, nottotheload, buttothemicroscopic area of contact between thetwoobjects (asopposed totheapparent contact area). We use/.tNasanapproximation because, asNincreases, sodoes theactual contact area onamicroscopic level. Forhundreds ofyears before the1940s, itwasac- cepted thattheload—and notthearea-—was directly responsible. Wealsobe- lieve thatthestatic frictional force islarger than thatofkinetic friction because thebonding ofatoms between thetwoobjects does nothave asmuch time to develop inkinetic motion. Effects ofRetarding Forces Weshould emphasize thattheforce FinEquation 2.7isnotnecessarily constant, andindeed, itmay consist ofseveral distinct parts, asseen intheprevious exam- ples. Forexample, ifaparticle falls inaconstant gravitational ﬁeld, thegravita- tional force isFg=mg,where gistheacceleration ofgravity. If,inaddition, a retarding force F,exists that issome function oftheinstantaneous speed, then thetotal force is F=Fg+F, (2.19)=mg+F,,(v) Itisfrequently sufﬁcient toconsider that F,(v) issimply proportional tosome power ofthespeed. Ingeneral, realretarding forces aremore complicated, but thepower-law approximation isuseful inmany instances inwhich thespeed does notvary greatly. Even more tothepoint, ifE.ocv”,then theequation of motion canusually beintegrated directly, whereas, ifthetrue velocity depend- ence were used, numerical integration would probably benecessary. With the 2.4 THE EQUATION OFMOTION FORA PARTICLE 59 power-law approximation, wecanthen write vF=mg—mkv”; (2.20) where kisapositive constant that speciﬁes thestrength oftheretarding force and where v/visaunit vector inthedirection ofv.Experimentally, weﬁnd that, forarelatively small object moving inair,nE1forvelocities lessthan about 24m/s(~80ft/s).Forhigher velocities butbelow thevelocity ofsound (~330m/sor1,100 ft/s),theretarding force isapproximately proportional to thesquare ofthevelocity.* Forsimplicity, thev2dependence isusually taken forspeeds uptothespeed ofsound. The effect ofairresistance isimportant foraping-pong ballsmashed toan opponent, ahigh-ﬂying softball hitdeep totheoutﬁeld, agolfer’s chip shot, and amortar shell lofted against anenemy. Extensive tabulations have been made formilitary ballistics ofprojectiles ofvarious sorts forthevelocity asafunction of ﬂight time. There areseveral forces onanactual projectile inﬂight. The airre- sistance force iscalled thedrag Wandisopposite totheprojectile’s velocity as shown inFigure 2-3a. The velocity visnormally notalong thesymmetry axisof theshell. The component offorce acting perpendicular tothedrag iscalled the liftLa.There may alsobevarious other forces duetotheprojectile ’sspin andos- cillation, andacalculation ofaprojectile’s ballistic trajectory isquite complex. The Prandtl expression fortheairresistancel is W=%cwpAv2 (2.21) where cwisthedimensionless drag coefﬁcient, pistheairdensity, vistheveloc- ity,and Aisthecross-sectional area oftheobject (projectile) measured perpen- dicularly tothevelocity. InFigure 2-3b, weplotsome typical values forcw,andin Figures 2-3canddwedisplay thecalculated airresistance Wusing Equation 2.21 foraprojectile diameter of10cmand using thevalues ofcwshown. The airre- sistance increases dramatically near thespeed ofsound (Mach number M= speed/speed ofsound). Below speeds ofabout 400m/s itisevident that an equation ofatleast second degree isnecessary todescribe theresistive force. For higher speeds, theretarding force varies approximately linearly with speed. Several examples ofthemotion ofaparticle subjected tovarious forces are given below. These examples areparticularly good tobegin computer calcula- tions using anyoftheavailable commercial math programs andspreadsheets or forthestudents towrite their own programs. The computer results, especially theplots, canoften becompared with theanalytical results presented here. Some oftheﬁgures shown inthissection were produced using acomputer, and *The motion ofaparticle inamedium inwhich there isaresisting force proportional tothespeed ortothesquare ofthespeed (ortoalinear combination ofthetwo) wasexamined byNewton inhis Principia (1687). The extension toanypower ofthespeed wasmade byjohann Bernoulli in1711. The term Stokes’ lawofresistance issometimes applied toaresisting force proportional tothespeed; Newton’s lawofresistance isaretarding force proportional tothesquare ofthespeed. ’tSee thearticle byE.Melchior andM.Reuschel inHandbook onWeaponry (Rh82, p.137). 60 2/NEWTONIAN MECHANICS—SINGLE PARTICLE 0.5 € \\“F1“‘N<2: Dragcoefﬁcientcw0.4 0.3 0.2 0.1 0.1 0.2 0.5 1 2 510 Mach numberM (a) (b) 600 . l. 000 --— -- 400 000 -- - -300 20° '51000 - - - 100 . 0T100200 3002400 500 0 300 210005 1500 2000 Velocity (m/s) Velocity (m/s) (C) (<1) FIGURE 2-3 (a)Aerodynamic forces acting onprojectile. Wisthedrag (airresistive force) andisopposite thevelocity oftheprojectile v.Notice thatvmay beatanangle orfrom thesymmetry axisofprojectile. The component offorce acting perpendicular tothedrag iscalled theliftLa.Thepoint Disthecenter ofpressure. Finally, thegravitational force Fgactsdown. Ifthecenter ofpressure isnotattheprojectile’s center ofmass, there is alsoatorque about thecenter ofmass. (b)The drag coefﬁcient cw, from theRheinmetall resistance law(R1182), isplotted versus theMach number M.Notice thelarge change near thespeed ofsound where M==1.(c)Theairresistive force W(drag) isshown asafunction of velocity foraprojectile diameter of10cm.Notice theinﬂection near thespeed ofsound. (d)Same as(c)forhigher velocities.(N)U18 (N)Q9 Airresistiveforce Airresstiveforceso several end-of-chapter problems aremeant todevelop thestudent’s computer experience ifsodesired bytheinstructor orstudent. EX.-\l\"l PLE 2.4 Asthesimplest example oftheresisted motion ofaparticle, ﬁnd thedisplace- ment andvelocity ofhorizontal motion inamedium inwhich theretarding force isproportional tothevelocity. Solution. Asketch oftheproblem isshown inFigure 2-4.The Newtonian equa- tionF=maprovides uswith theequation ofmotion: 2.4 THE EQUATION OFMOTION FOR APARTICLE 61 Ix . V0 —+ <-— Resisting force F=kmv FIGURE 2-4 Example 2.4. x-direction d ma=mi=—kmv (2.22) where krnv isthemagnitude oftheresisting force (k=constant). Wearenot implying bythisform thattheretarding force depends onthemass m;thisform simply makes themath easier. Then dv_=_k d iv it (2.23) lnv= —kt+ C1 The integration constant inEquation 2.23 canbeevaluated ifweprescribe the initial condition v(t=0)Ev0.The C1=lnv0,and v=v0e_k‘ - (2.24) Wecanintegrate thisequation toobtain thedisplacement xasafunction of time: v=g=v0e'k‘ x=v0je'k‘dt =—lZe_k‘ +C2 (2.25a) The initial condition x(t=0)E0implies C2=v0/k. Therefore x=lid—e"“) (2.25b) This result shows that xasymptotically approaches thevalue v0/kast—>oo. Wecanalsoobtain thevelocity asafunction ofdisplacement bywriting £i2_ﬁ’ﬁ_d”ldx dtdx dtv sothat vd—1)=ﬁ)= —kvdx dt or Q1:_,,dx 62 2/NEWTONIAN MECI—1ANICS—SINGLE PARTICLE from which weﬁnd, byusing thesame initial conditions, v=vo—kx (2.26) Therefore, thevelocity decreases linearly with displacement. 7 4 1’ 1 _. ' 1 ' _ r nnl I _ r EXAMPLE 2.5 -I -I Find thedisplacement andvelocity ofaparticle undergoing vertical motion ina medium having aretarding force proportional tothevelocity. Solution. Letusconsider thattheparticle isfalling downward with aninitial velocity v0from aheight hinaconstant gravitational ﬁeld (Figure 2-5). The equation ofmotion is z-direction dvF= mE= —mg— kmv (2.27) where —kmv represents apositive upward force since wetake zand v=itobe positive upward, andthemotion isdownward—that is,v<0,sothat —kmv >0. From Equation 2.27, wehave dvkn+g——dt (2.28) Integrating Equation 2.28 andsetting v(t=0)Ev0,wehave (noting thatv0<0) 1-kln(kv+ g)=—t+ C kv+g=e—kt+kc dz gkv+g_ v=5=—;+—°T-e '1‘ (2.29) hA‘IU jGravitational force =mg TResisting force =kmvz FIGURE 2-5 Example 2.5. 2.4THEEQUATION orMOTION FORAPARTICLE 63 ‘U /lvoi >lvtl Tbrnnnalspeed,v, Speedlvsl<lvil \ g/k-------------- --g---------------------- --1 -7 i i \v°:0 0.//1 I I I |_,0Tnne FIGURE 2-6 Results forExample 2.5indicating thedownward speeds forvarious initial speeds v0asthey approach theterminal velocity. Integrating once more andevaluating theconstant bysetting z(t=0)Eh,we ﬁnd gt kv+g _ z=h—Z+—0ké—(1—e M) (2.30) Equation 2.29 shows thatasthetime becomes very long, thevelocity ap- proaches thelimiting value —g/k;thisiscalled theterminal velocity, v,. Equation 2.27 yields thesame result, because theforce willvanish—and hence nofurther acceleration willoccur—when v=-—g/k.Iftheinitial velocity ex- ceeds theterminal velocity inmagnitude, then thebody immediately begins to slow down andvapproaches theterminal speed from theopposite direction. Figure 2-6illustrates these results forthedownward speeds (positive values). EXAMPLE 2.6 -I - - - _ _ Next, wetreat projectile motion intwodimensions, ﬁrst without considering air resistance. Letthemuzzle velocity oftheprojectile bev0andtheangle ofeleva- tionbe6(Figure 2-7). Calculate theprojectile’s displacement, velocity, andrange. Solution. Using F=mg,theforce components become x-direction 0=mii (2.3la) y-direction —mg=my (2.3lb) 64 2/NEWTONIAN MECHANICS-—SINGLE PARTICLE _____.____ 1-‘O’, "*~_ 4’ __, /” Z _\ v I \I (), \\ \ \ // yj '-:- \\\\ \0 \_‘\ \ - ------_-.-...-_:f2.>l'/.1 -*"’r'\\-——>x FIGURE 2-7 Example 2.6. Neglect theheight ofthegun, andassume x=y=0att=0.Then 55=0 ii=v0cos6 x=votcos6 (2.32) and 00 J’I"g y=-—gt+ v0sin6 __g-t2 I y=T +votsin6 (2.33) The speed andtotal displacement asfunctions oftime arefound tobe v=\/22+)2=(1)3+g2t2—20.,gtsino)1/2 (2.34) and g2t2 . 1/2 r= Vx2+y?=v§t2 +—-4- —U0g'l3S1I1 6 (2.35) Wecanﬁnd therange bydetermining thevalue ofxwhen theprojectile falls back toground, thatis,when y=0. t y=t<—2g— +v0sin6)=O (2.36) One value ofy=0occurs fort=0andtheother onefort=T T ?g+v0sin6=0 T=;—2”°:1‘6 (2.37) 2.4 THE EQUATION OFMOTION FOR APARTICLE 65 The range Risfound from 211%.x(t=T)=range :?sin6cos6 (2.38) 2 R=range ==%sin26 (2.39) Notice thatthemaximum range occurs for6=45°. Letususesome actual numbers inthese calculations. The Germans used a long-range gunnamed BigBertha inWorld War Itobombard Paris. Itsmuzzle velocity was1,450 m/s.Find itspredicted range, maximum projectile height, andprojectile time ofﬂight if6=55°.Wehave v0==1450 m/s and6=55°,so therange (from Equation 2.39) becomes (1450 m/s)2 _R=i—-— 110° =0k 9.8m/S2 [s1n( )] 22m BigBertha’s actual range was120km.The difference isaresult ofthereal effect ofairresistance. Toﬁnd themaximum predicted height, weneed tocalculated yforthe time T/2where Tistheprojectile time ofﬂight: (2)(1450 m/s) (sin55°)r=~ - =49.8111/S2 225 _T__gT2 T/0T .y,,,,,,,<t—2)— 8+2sin6 :—(9.8 m/s)(242 s)2+(1450 m/s)(242 s)ggs_i_n(55°) 8 2 =72km EXAMPLE 2.7 Next, weaddtheeffect ofairresistance tothemotion oftheprojectile inthe previous example. Calculate thedecrease inrange under theassumption that theforce caused byairresistance isdirectly proportional totheprojectile’s velocity. Solution. The initial conditions arethesame asintheprevious example. x(t=0) =0=y(t=0) 2Z(t=0)=v0cos6EU (2.40) y(t= 0)=v0sin6E V However, theequations ofmotion, Equation 2.31, become mii=—kmk (2.41) =—kmy —mg (2.42) 66 2/NEWTONIAN MECHANICS-SINGLE PARTICIE Equation 2.41 isexactly thatused inExample 2.4.The solution istherefore Ux=Z(1—W") (2.43) Similarly, Equation 2.42 isthesame astheequation ofthemotion inExample 2.5.Wecanusethesolution found inthat example byletting h=0.(The fact thatweconsidered theparticle tobeprojected downward inExample 2.5isofno consequence. The sign ofthe initial velocity automatically takes this into ac- count.) Therefore kV i=—%t+—,l—g<1 —ck‘) <2-44> The trajectory isshown inFigure 2-8forseveral values oftheretarding force constant kforagiven projectile ﬂight. The range R’,which istherange including airresistance, canbefound as previously bycalculating thetime Trequired fortheentire trajectory andthen substituting thisvalue into Equation 2.43 forx.The time Tisfound asprevi- ously byﬁnding t=Twhen y=0.From Equation 2.44, weﬁnd :r=K2 (1—e_kT) (2.45)gk This isatranscendental equation, and therefore wecannot obtain ananalytic expression forT.Nonetheless, westillhave powerful methods tousetosolve N 1.5— (104IT1)1.0— k=0(Parabolic motion) ight0.005 C H 0.01 Verticah0.5- 0.02 0.04 R 0.08 0 . I I I I I I I x 1 2 3 4 -0.3- I Horizontal distance (104m) FIGURE 2-8 The calculated trajectories ofaparticle inairresistance (Fm =—kmv) forvarious values ofk(inunits ofs_1). The calculations were performed forvalues of6=60°andv0=600m/s. Thevalues ofy(Equation 2.44) areplotted versus x(Equation 2.43). 2.4 THE EQUATION OFMOTION FORA PARTICLE 67 such problems. Wepresent twoofthem here: (1)aperturbation method toﬁnd an approximate solution, and (2)anumerical method, which cannormally beasac- curate asdesired. Wewillcompare theresults. Perturbation Method Tousetheperturbation method, weﬁnd anexpansion pa- rameter orcoupling constant thatisnormally small. Inthepresent case, thisparam- eteristheretarding force constant k,because wehave already solved thepresent problem with k=0,andnow wewould liketoturn ontheretarding force, but letkbesmall. Wetherefore expand theexponential term ofEquation 2.45 (see Equation D.34 ofAppendix D)inapower series with theintention ofkeeping only thelowest terms ofk”,where kisourexpansion parameter. kV+ g 1 1 T= it kT— -k2T2 +-k3T3 — (2.46) gk 2 6 IfWekeep only terms intheexpansion through kg,thisequation canbere- arranged toyield 2V 1T=——/g— +-kT2 (2.47)1+kV/g 3 Wenow have theexpansion parameter kinthedenominator oftheﬁrstterm on theright-hand sideofthisequation. Weneed toexpand thisterm inapower se- ries(Taylor series, seeEquation D.8ofAppendix D): - 1-———=1—kv +kV 2--~ .48 1+W/g </g)</g) <2> where wehave kept only terms through k2,because weonly have terms through kinEquation 2.47. Ifweinsert thisexpansion ofEquation 2.48 into theﬁrst term ontheright-hand sideofEquation 2.47 andkeep only theterms inktoﬁrst order, wehave T=%/+ -if-2/j)k +0(k2) (2.49) where wechoose toneglect O(k2) ,theterms oforder k2and higher. Inthelimit k—>0(noairresistance), Equation 2.49 gives usthesame result asintheprevi- ousexample: 2 '6T(k=0)=T0=%/= Therefore, ifkissmall (but nonvanishing), theﬂight time willbeapproximately equal toT0.Ifwethen usethisapproximate value forT=T0intheright-hand sideofEquation 2.49, wehave V kVTE2—(l —M) (2.50) g 3g which isthedesired approximate expression fortheﬂight time. 68 2/NEWTONIAN MECHANICS-—SINGLE PARTICLE Next, wewrite theequation forx(Equation 2.43) inexpanded form: U 1 1x=Z(kt—-ék2t2+-ék5t3— (2.51) Because x(t=T)ER’,wehave approximately fortherange 1R’EU(T —EkT2) (2.52) where again wekeep terms only through theﬁrstorder ofk.Wecannow evalu- atethisexpression byusing thevalue ofTfrom Equation 2.50. Ifweretain only terms linear ink,weﬁnd 4kVREE/(1 F) (2.55g 3g The quantity 2UV/gcannow bewritten (using Equations 2.40) as UV22 22?=%s1no¢oso =-1é9sin20 =R (2.54) which willberecognized astherange Roftheprojectile when airresistance is neglected. Therefore 12'ER(1-43/) (2.55)3g Over what range ofvalues forkwould weexpect ourperturbation method tobe correct? Ifwelook attheexpansion inEquation 2.48, weseethat theexpansion willnotconverge unless kV/g <1ork<g/I/I and infact, wewould likek<< g/V= g/(v0 sin6). Numerical Method Equation 2.45 canbesolved numerically using acomputer byavariety ofmethods. Wesetupaloop tosolve theequation forTfor many values ofkupto0.08 s“1: Ti(k,-). These values ofTiand k,-areinserted into Equation 2.43 toﬁnd therange R,7,which isdisplayed inFigure 2-9.The range drops rapidly forincreased airresistance, just asonewould expect, butitdoes notdisplay thelinear dependence suggested bytheperturbation method solu- tionofEquation 2.55. Fortheprojectile motion described inFigures 2-8and 2-9,thelinear ap- proximation isinaccurate forkvalues aslowas0.01 s_1andincorrectly shows therange iszero forallvalues ofklarger than 0.014 s'1.This disagreement with theperturbation method isnotsurprising because thelinear result fortherange R’wasdependent onk<<g/(vo sin6)=0.02 s'1,which ishardly true foreven k=0.01 s‘1.The agreement should beadequate fork=0.005 s'1.The results shown inFigure 2-8indicate thatforvalues ofk>0.005 s_1,thedrag canhardly beconsidered aperturbation. Infact, fork>0.01 s'1thedrag becomes the dominant factor intheprojectile motion. 2.4 THE EQUATION OFMOTION FOR APARTICLE 69 I I I I I3 _ \\I \ ‘ \\\\ (104m)I\2'“\ _\\ Range\ I \\ \ \ II"_ \\ 1 ‘\XApproximation ,.-NuII1eriCal \\ . \ L \\ \I _J I L00.02 0.04 0.06 0.08 0.1 Retarding force constant, k(s'1) FIGURE 2-9 The range values calculated approximately andnumerically forthe projectile data given inFigure 2-8areplotted asafunction ofthe retarding force constant k. The previous example indicates how complicated therealworld canbe.Inthat example, westillhadtomake assumptions thatwere n0nphysical—in assuming, forexample, thattheretarding force isalways linearly proportional totheveloc- ity.Even ournumerical calculation isnotaccurate, because Figure 2-3shows us that abetter assumption would betoinclude a112retarding term aswell. Adding such aterm would notbedifﬁcult with thenumerical calculation, and weshall doasimilar calculation inthenext example. Wehave included theau- thor’s Mathcad ﬁlethatproduced Figures 2-8and2-9inAppendix Hforthose students who might want toreproduce thecalculation. Weemphasize thatthere aremany ways toperform numerical calculations with computers, andthestu- dent willprobably want tobecome proﬁcient with several. EXAMPLE 2.8 Usethedata shown inFigure 2-3tocalculate thetrajectory foranactual pro- jectile. Assume amuzzle velocity of600m/s,gun elevation of45°,andapro- jectile mass of30kg.Plot theheight yversus thehorizontal distance xand plot y,5c,andjversus time both with andwithout airresistance. Include only theairresistance andgravity, andignore other possible forces such asthe lift. Solution. First, wemake atable ofretarding force versus velocity byreading Figure 2-3.Read theforce every 50m/sforFigure 2-3candevery 100m/sfor Figure 2-3d. Wecanthen useastraight lineinterpolation between thetabular 70 2/NEWTONIAN MECHANICS—SINGLE PARTICLE values. Weusethecoordinate system shown inFigure 2-7.The equations of motion become Fat=—— (2.56) 5.315..5;=——g (2.57) where Fxand1*;aretheretarding forces. Assume gisconstant. Fxwillalways bea positive number, butF;>0fortheprojectile going up,andlg<0forthepro- jectile coming back down. Let6betheprojectile’s elevation angle from thehor- izontal atanyinstant. v=\/922+9'12 (2.58) tan6= (2.59) 1';=Fcos 6 (2.60) F,=Fsino (2.61) Wecancalculate 1'}and 1*;atanyinstant byknowing icand Over asmall time interval, thenext oiand)3canbecalculated. "t ic= 55dt+v0cos6 (2.62) .0 ”t )3= jidt+v0sin6 (2.63) 40 "t x= 5cdt (2.64) .0 N.‘ y=0)3dt (2.65) Wewrote ashort computer program tocontain ourtable fortheretarding forces and toperform thecalculations forvi,3'1,x,and yasafunction oftime. We must perform theintegrals bysummations over small time intervals, because theforces aretime dependent. Figure 2-10shows theresults. Notice thelarge difference thattheairresistance makes. InFigure 2-10a, thehorizontal distance (range) thattheprojectile travels isabout 16kmcom- pared toalmost 3'7kmwith noairresistance. Our calculation ignored thefact thattheairdensity depends onthealtitude. Ifwetake account ofthedecrease intheairdensity with altitude, weobtain thethird curve with arange of18km shown inFigure 2-10a. Ifwealso included thelift,therange would bestill greater. Notice thatthechange invelocities inFigures 2-10c and2-10d mirror theairresistive force ofFigure 2-3.The speeds decrease rapidly until thespeed reaches thespeed ofsound, andthen therateofchange ofthespeeds levels off somewhat. 2.4 THE EQUATION OFMOTION FOR APARTICLE 71 10 N0air 10 resistance Height(km)A0)oo \”Height(km)8 Noair With air 5 resistance '::\\,/resistance g,.-.._~ ‘‘andaltitude 4 1’ X\ I \ ,.dependence ,' ~ 2Withat.‘.‘.<>fairdwsiw 2 Withair/"\\resistance \‘\\ resistance \\ I\_)_I I I I__J LII. .J 0 10 20 30 40 0 20 40 60 80 Horizontal distance (km) Time (sec) (a) (b) Honztavelocity(m/s)»—-toonAowooooo<:><:>0III *III Verticalcity(m/s)tooo400 \ 00 X \ LNoairres'stance . .1 Noairresistance V60~_ —' 0'_ ~‘~ ‘-1 / ~___" ‘Ty“ s I“ 00 /I ‘~ _2 1 Q“ _ With airresistance with air ~_ OI1 400 resistance I ,I "J J 0 20 40 60 80 0 20 40 60' 80 Time (sec) Time (sec) (¢) (<1) FIGURE 2-10 Theresults ofExample 2.8.Thesolid lines aretheresults ifnoairresis- tance isincluded, whereas thedashed lines include theresults ofadding theairresistive force. In(a)wealsoinclude theeﬁect oftheairdensity dependence, which becomes smaller astheprojectile rises higher. This concludes oursubsection ontheeffects ofretarding forces. Much more could bedone toinclude realistic effects, butthemethod isclear. Normally, one ef- fectisadded atatime, andtheresults areanalyzed before another effect isadded. Other Examples ofDynamics Weconclude thissection with twoadditional standard examples ofdynamical particle-like behavior. EXAMPLE 2.9 Atwood’s machine consists ofasmooth pulley with twomasses suspended from alight string ateach end (Figure 2-11).Find theacceleration ofthemasses and thetension ofthestring (a)when thepulley center isatrest and (b)when the pulley isdescending inanelevator with constant acceleration a. 72 2/NEWTONIAN MF.CHANICS—SINGLE PARTICLE Fixed _ . I II I I x1 J/Elevator xi I xé’ T X2 "I2I“ 'at ' (P1) (b) FIGURE 2-11 Example 2.9;Atwood’s machine.3* S>6 <1.3?8)—I r"““"“““"“"|+:II|I'-I|I|I|III1I|I|I|I|I|I|:<I| L_____________—————— Solution. Weneglect themass ofthestring andassume thatthepulley is smooth—that is,nofriction onthestring. The tension Tmust bethesame throughout thestring. The equations ofmotion become, foreach mass, for case (a), WL1551 :mlg —T m255 =m2g— T (2.67) Notice again theadvantage oftheforce concept: Weneed only identify the forces acting oneach mass. The tension Tisthesame inboth equations. If thestring isinextensible, then 562=-—3&1,and Equations 2.66 and 2.67 may be combined "Z1551 :m1g"" (m2g —7212562) =mig“ (m2g'I' "Z2550 Rearranging, ,,_g(m1_m2) _ __x1——-"—i——xm1+m22 (2-68) Ifml>m2,then 561>0,and 5&2<0.The tension canbeobtained from Equations 2.68 and2.66: T: "I18_"I1551 T:m_m"ﬁlm-"121 18' lgml+m2 T: 2m1m2g ml+m2(2.69) 2.4THEEQUATION orMOTION FORAPARTICLE 73 Forcase (b),inwhich thepulley isinanelevator, thecoordinate system with origins atthepulley center isnolonger aninertial system. Weneed anin- ertial system with theorigin atthetopoftheelevator shaft (Figure 2-11b). The equations ofmotion intheinertial system (x§’=x{+x1,xg=2;+x2)are m1$5I1I= + 551) Z m1g— T "@552 ="I2(5(2 'I'5(2)="I2g_ T SO m15c'1= m1g— T— m1551= m1(g— a)—T .. ... (2-70)m2x2 =m2g— T— m2x2 =m2(g— a)—T where 5&1’==5E§=a.Wehave 562=—561,sowesolve for561asbefore byeliminat- ingT: ("I1_"I2)551=-552=(g—a)m (2-71) and T= Q ml'I'"I2 Notice thattheresults fortheacceleration andtension arejustasiftheacceler- ation ofgravity were reduced bytheamount oftheelevator acceleration a. The change foranascending elevator should beobvious. Inourlastexample inthislengthy review oftheequations ofmotion foraparti- cle,letusexamine particle motion inanelectromagnetic ﬁeld. Consider a charged particle entering aregion ofuniform magnetic ﬁeld B—for example, theearth’s ﬁeld—as shown inFigure 2-12. Determine itssubsequent motion. Solution. Choose aCartesian coordinate system with itsy-axis parallel tothe magnetic ﬁeld. Ifqisthecharge ontheparticle, vitsvelocity, aitsacceleration, andBtheearth’s magnetic ﬁeld, then v=ici+ +ik a=iii+ +'z'k B=Boj The magnetic force F=qvXB=ma,so m(5ii+yj+zk)=q(5¢i+yj+zk)xBoj:qB0(5¢k —zi) 74 2/NEWTONIAN MECHANICS~—SINGLE PARTICLE Subsequent xvo / particle motion Sui BZ IV x FIGURE 2-12 Example 2.10; amoving particle enters aregion ofmagnetic ﬁeld. Equating likevector components gives mii=-qB0i =0 (2.73) mi=qBO5c Integrating thesecond ofthese equations, =0,yields I=5’0 where )0isaconstant andistheinitial value of)3.Integrating asecond time gives 2=W+yo where yoisalsoaconstant. Tointegrate theﬁrstandlastequations ofEquation 2.73, leta=qB0/m,so that sa=-'...°“} (2.14) Z=01.76 These coupled, simultaneous differential equations canbeeasily uncoupled by differentiating oneandsubstituting itinto theother, giving =aﬁé=—-a22 =-—a1i =—a2ic sothat '2': —-0:22 Z__a29_c} (2.75) 2.4 THE EQUATION OFMOTION FOR APARTICLE 75 Both ofthese differential equations have thesame form ofsolution. Using the technique ofExample C.2ofAppendix C,wehave x=Acosat+ Bsinat+ xo z=A’cosat+B'sinat+z0 where A,A’,B,B’,xo,andzoareconstants ofintegration thataredetermined by theparticle ’sinitial position andvelocity andbytheequations ofmotion, Equation 2.74. These solutions canberewritten (x—x0)=Acos at+Bsin at (2-JI0)=55¢ (2-76) (z—zo)=A’cosat+B’sinat The x-andz-coordinates areconnected byEquation 2.74, sosubstituting Equations 2.76 into theﬁrstequation ofEquation 2.74 gives —a2A cosat—a2Bsinat=~a(—aA’ sinat+aB'cosat) (2.77) Because Equation 2.77 isvalid forallt,inparticular t=0andt=11'/201, Equation 2.77 yields —a2A =—a2B’ sothat AIB’ and —a2B =a2A' gives B=—A’ Wenow have (x— xo)=Acos at+Bsinat (y—yo)=jot (2.78) (z—zo)=—Bcosat+Asin at Ifatt=0,i=20andat=0,then from Equation 2.78, differentiating andset- ting t=0gives aB=0 and CYA 1:-'£0 76 2/NEWTONIAN MECHANICS—SINGLE PARTICLE SO 5-0(x—x0)=Z;cosat (JI—yo)Ijot io.(z—zo)=Z;sinat <2)<2)x—x0IZ cos TqB0 m (J’_yo):LI’o( (2-79) 50"‘ .‘I301(2-20) =Z s1nZ 930 I” These aretheparametric equations ofacircular helix ofradius iom/qB0. Thus, thefaster theparticle enters theﬁeld orthegreater itsmass, thelarger the radius ofthehelix. And thegreater thecharge ontheparticle orthestronger themagnetic ﬁeld, thetighter thehelix. Notice alsohow thecharged particle is captured bythemagnetic ﬁeld—just drifting along theﬁeld direction. Inthis example, theparticle hadnoinitial component ofitsvelocity along thex-axis, buteven ifithaditwould notdrift along thisaxis (seeProblem 2-31). Finally, notice thatthemagnetic force ontheparticle always actsperpendicular toits velocity andhence cannot speed itup.Equation 2.79 veriﬁes thisfact. Theearth ’smagnetic ﬁeld isnotassimple astheuniform ﬁeld ofthisexam- ple.Nevertheless, thisexample gives some insight intooneofthemechanisms by which theearth’s magnetic ﬁeld traps low-energy cosmic raysandthesolar wind to create theVanAllen belts.nI— I7 uni lull mull! mFinally, 2.5 Conservation Theorems Wenow turn toadetailed discussion oftheNewtonian mechanics ofasingle particle andderive theimportant theorems regarding conserved quantities. We must emphasize that wearenotproving theconservation ofthevarious quanti- ties. Wearemerely deriving theconsequences ofNewton’s laws ofdynamics. These implications must beputtothetestofexperiment, andtheir veriﬁcation then supplies ameasure ofconﬁrmation oftheoriginal dynamical laws. Thefact that these conservation theorems have indeed been found tobevalid inmany instances furnishes animportant part oftheproof forthecorrectness of Newton’s laws, atleast inclassical physics. The ﬁrstoftheconservation theorems concerns thelinear momentum ofa particle. Iftheparticle isfree, that is,iftheparticle encounters noforce, then Equation 2.2becomes simply =0.Therefore, pisavector constant intime, andtheﬁrstconservation theorem becomes 2.5CONSERVATION THEOREMS 77 I. Thetotal linear momentum pofaparticle isconserved when thetotalforce onitis zero. Note that thisresult isderived from avector equation, pI0,andtherefore applies foreach component ofthelinear momentum. Tostate theresult in other terms, weletsbesome constant vector such thatF-sI0,independent of time. Then p-sIF-sI0 or,integrating with respect totime, p'sIconstant (2.80) which states that thecomponent oflinear momentum inadirection inwhich theforce vanishes isconstant intime. The angular momentum Lofaparticle with respect toanorigin from which theposition vector rismeasured isdeﬁned tobe The torque ormoment offorce Nwith respect tothesame origin isdeﬁned tobe -<2-82>where ristheposition vector from theorigin tothepoint where theforce Fis applied. Because FImyfortheparticle, thetorque becomes NIr><m\'rIr><p)=<r>+<r> but i'XpIi'XmvIm(1"><i")I0 <22Ifnotorques actonaparticle (i.e., ifNI0),then I0and Lisavector con- stant intime. The second important conservation theorem isSO II. Theangular momentum ofaparticle subject tonotorque isconserved. Weremind thestudent thatajudicious choice oftheorigin ofacoordinate system willoften allow aproblem tobesolved much more easily than apoor choice. Forexample, thetorque willbezero incoordinate systems centered 78 2/NEWTONIAN MECHANICS~—SINGLE PARTICLE along theresultant line offorce. The angular momentum willbeconserved in thiscase. Ifwork isdone onaparticle byaforce Fintransforming theparticle from Condition 1toCondition 2,then thiswork isdeﬁned tobe 2 W12IIF-dr (2.84)1 IfFisthenetresultant force acting ontheparticle, dvdr dvF I 1 O‘L i W 0dr mdtdtdt mdtvdt d d 1I*2"-zzi-t(v-v)dt IgZt(v2)dt Id(§mv2) (2.85) The integrand inEquation 2.84 isthus anexact differential, and thework done bythetotal force Facting onaparticle isequal toitschange inkinetic energy: 121I/V12 :('é’!!lU2)l = —U?) :T2_T1 where TI%mv2 isthekinetic energy oftheparticle. IfT1>T2then W12<0, andtheparticle hasdone work with aresulting decrease inkinetic energy. Itis important torealize that theforce Fleading toEquation 2.85 isthetotal (i.e., netresultant) force ontheparticle. Letusnow examine theintegral appearing inEquation 2.84 from adiffer- entstandpoint. Inmany physical problems, theforce Fhastheproperty thatthe work required tomove aparticle from one position toanother without any change inkinetic energy depends only ontheoriginal and ﬁnal positions and notontheexact path taken bytheparticle. Forexample, assume thework done tomove theparticle from point 1inFigure 2-13 topoint 2isindependent ofthe actual paths a,b,orctaken. This property isexhibited, forexample, byacon- stant gravitational force ﬁeld. Thus, ifaparticle ofmass misraised through a height h(byanypath), then anamount ofwork mghhasbeen done ontheparti- cle,andtheparticle candoanequal amount ofwork inreturning toitsoriginal position. This capacity todowork iscalled thepotential energy oftheparticle. Wemaydeﬁne thepotential energy ofaparticle interms ofthework (done bytheforce F)required totransport theparticle from apoint 1toapoint 2 (with nonetchange inkinetic energy): 2 [F-drI U1—[@ (2.87) 1 The work done inmoving theparticle isthus simply thedifference inthepoten- tialenergy Uatthetwopoints. Forexample, ifweliftasuitcase from position 1 ontheground toposition 2inacartrunk, weastheexternal agent aredoing 2.5 CONSERVATION THEOREMS 79 '1 2 b 1 C Origin FIGURE 2-13 Forsome forces (identiﬁed later asconservative), thework done bythe force tomove aparticle from oneposition 1toanother position 2is independent ofthepath (a,b,orc). work against theforce ofgravity. Lettheforce FinEquation 2.87 bethegravita- tional force, andinraising thesuitcase, F-drbecomes negative. The result of theintegration inEquation 2.87 isthat U1—U2isnegative, sothatthepotential energy atposition 2inthecar’s trunk isgreater than that atposition 1onthe ground. The change inpotential energy Lg—U1isthenegative ofthework done bythegravitational force, ascanbeseen bymultiplying both sides of Equation 2.87 by—1.Astheexternal agent, wedopositive work (against gravity) toraise thepotential energy ofthesuitcase. ' Equation 2.87 canbereproduced* ifwewrite Fasthegradient ofthescalar function U: IF=—g'rad U:—VUI (2.88) Then 2 2 2 IF-drI—I(VU)'drI"(dUIU1-Ué (2.89) 1 1 1 Inmost systems ofinterest, thepotential energy isafunction ofposition and, possibly, time: UIU(r) orU=U(r,t).Wedonotconsider cases inwhich thepotential energy isafunction ofthevelocity.I Itisimportant torealize thatthepotential energy isdeﬁned only towithin an additive constant; that is,theforce deﬁned by—VUisnodifferent from that de- ﬁned by—V(U+constant). Potential energy therefore hasnoabsolute meaning; only differences ofpotential energy arephysically meaningful (asinEquation 2.87). *The necessary andsufficient condition thatpermits avector function toberepresented bythegra- dient ofascalar function isthatthecurlofthevector function vanishes identically. I-Velocity-dependent potentials areneccessary incertain situations, e.g., inelectromagnetism (the so-called Liénard—Wiechert potentials). 80 2/NEWTONIAN MECHANICS-SINGLE PARTICLE Ifwechoose acertain inertial frame ofreference todescribe amechanical process, thelaws ofmotion arethesame asinanyother reference frame inuni- form motion relative totheoriginal frame. Thevelocity ofaparticle isingeneral different depending onwhich inertial reference frame wechose asthebasis for describing themotion. Wetherefore ﬁnd thatitisimpossible toascribe anab- solute kinetic energy toaparticle inmuch thesame waythat itisimpossible to assign anyabsolute meaning topotential energy. Both ofthese limitations are theresult ofthefactthat selecting anorigin ofthecoordinate system used to describe physical processes isalways arbitrary. The nineteenth-century Scottish physicistjames Clerk Maxwell (1831-1879) summarized thesituation asfollows.* Wemust, therefore, regard theenergy ofamaterial system asaquantity of which wemay ascertain theincrease ordiminution asthesystem passes from onedeﬁnite condition toanother. The absolute value oftheenergy inthestan- dard condition isunknown tous,anditwould beofnovalue tousifwedid know it,asallphenomena depend onthevariations ofenergy andnotonitsab- solute value. Next, wedeﬁne thetotal energy ofaparticle tobethesum ofthekinetic andpotential energies: <2-90> mHw—=— — 4mm+m QM Toevaluate thetime derivatives appearing ontheright-hand side ofthisequa- tion, weﬁrstnote thatEquation 2.85 canbewritten asThe total time derivative ofEis 1F-drId(-5 mv2) IdT (2.92) Dividing through bydt, dT dr——IF-—IF-' .93 m m I Q) Wehave also dU 6Udx,- BU_:§__+_ E6U_ GU Z ixi + i I6x, 6t I(VU) -i-+%] (2.94) C.Maxwell, Matter andMotion (Cambridge, 1877), p.91. 2.5CONSERVATION THEOREMS 81 Substituting Equations 2.93 and2.94 into 2.91, weﬁnd dE 6U_: _- V _- _dt Fr+( U)r+at UI(F+VU) -i-+L6t _Q’p—at (2.95) because theterm F+VUvanishes inview ofthedeﬁnition ofthepotential en- ergy (Equation 2.88) ifthetotal force istheconservative force FI—VU IfUisnotanexplicit function ofthetime (i.e., if6U/8t I0;recall thatwedo notconsider velocity-dependent potentials), theforce ﬁeld represented byFis conservative. Under these conditions, wehave thethird important conservation theorem: III. Thetotalenergy Eofaparticle inaconservative forcefield isaconstant intime. Itmust bereiterated thatwehave notproved theconservation laws oflinear momentum, angular momentum, andenergy. Wehave only derived various con- sequences ofNewton’s laws; that is,ifthese laws arevalid inacertain situation, then momentum and energy willbeconserved. Butwehave become soenam- ored with these conservation theorems thatwehave elevated them tothe-status oflaws andwehave come toinsist that they bevalid inanyphysical theory, even those thatapply tosituations inwhich Newtonian mechanics isnotvalid, as,for example, intheinteraction ofmoving charges orinquantum-mechanical sys- tems. Wedonotactually have conservation laws insuch situations, butrather conservation postulates thatweforce onthetheory. Forexample, ifwehave two isolated moving electric charges, theelectromagnetic forces between them are notconservative. Wetherefore endow theelectromagnetic ﬁeld with acertain amount ofenergy sothatenergy conservation willbevalid. This procedure issat- isfactory only iftheconsequences donotcontradict anyexperimental fact, and thisisindeed thecase formoving charges. Wetherefore extend theusual con- cept ofenergy toinclude “electromagnetic energy” tosatisfy ourpreconceived notion that energy must beconserved. This may seem anarbitrary and drastic step totake, butnothing, itissaid, succeeds asdoes success, andthese conserva- tion “laws” have been themost successful setofprinciples inphysics. The refusal torelinquish energy and momentum conservation led Wolfgang Pauli (1900-1958) topostulate in1930 theexistence oftheneutrino toaccount for the“missing” energy andmomentum inradioactive Bdecay. This postulate al- lowed Enrico Fermi (1901-1954) toconstruct asuccessful theory ofBdecay in 1934, butdirect observation oftheneutrino wasnotmade until 1953 when Reines and Cowan performed their famous experiment.* Byadhering tothe conviction that energy and momentum must beconserved, anew elementary *C.L.Cowan, F.Reines, F.B.Harrison, H.W.Kruse, andA.D.McGuire, Science 124,103(1956). 82 2/NEWTONIAN MECHANICS-—SINGLE PARTICLE particle wasdiscovered, one that isofgreat importance inmodern theories of nuclear andparticle physics. This discovery isonly oneofthemany advances in theunderstanding oftheproperties ofmatter that have resulted directly from theapplication oftheconservation laws. Weshall apply these conservation theorems toseveral physical situations in theremainder ofthisbook, among them Rutherford scattering and planetary motion. Asimple example here indicates theusefulness oftheconservation theorems. EX.»~\MPLE 2.11 ITTlT Amouse ofmass mjumps ontheoutside edge ofafreely turning ceiling fanof rotational inertia Iandradius R.Bywhat ratio does theangular velocity change? Solution. Angular momentum must beconserved during theprocess. Weare using theconcept ofrotational inertia leamed inelementary physics torelate angular momentum Ltoangular velocity w:L=Iw.The initial angular momen- tum L0=Iwomust beequal totheangular momentum L(fanplus mouse) after themouse jumps on.The velocity oftheoutside edge isvIwR. L=Iw+mvR= %(1+ mR2) L=L0=Iw0 U U0 ~1+ R2=1»R( m) R l)__ I U0 and no I (U0 I+MR2 2.6Energy The concept ofenergy wasnotnearly aspopular inNewton’s time asitistoday. Later weshall study twonewformulations ofdynamics, different from Newton’s, based onenergy—-the Lagrangian andHamiltonian methods. Early inthenineteenth century, itbecame clear that heat wasanother form ofenergy andnotaform ofﬂuid (called “caloric”) thatﬂowed between hotand cold bodies. Count Rumford* isgenerally given credit forrealizing that the *Benjamin Thompson (1753-1814) wasborn inMassachusetts andemigrated toEurope in1776 asa loyalist refugee. Among theactivities ofhisdistinguished military and, later, scientific career, hesu- pervised theboring ofcannons ashead oftheBavarian wardepartment. 2.6ENERGY 83 great amount ofheat generated during theboring ofacannon wascaused by friction andnotthecaloric. Iffrictional energy isjust heat energy, interchange- able with mechanical energy, then atotal conservation ofenergy canoccur. Throughout thenineteenth century, scientists performed experiments on theconservation ofenergy, resulting intheprominence given energy today. Hermann vonHelmholtz (1821-1894) fomiulated thegeneral lawofconserva- tion ofenergy in1847. Hebased hisconclusion largely onthecalorimetric ex- periments ofJames Prescottjoule (1818-1889) begun in1840. Consider apoint particle under theinﬂuence ofaconservative force with potential U.The conservation ofenergy (actually, mechanical energy, tobepre- ciseinthiscase) isreﬂected inEquation 2.90. 1E=T+U= gmv2 +U(x) (2.96) where weconsider only theone-dimensional case. Wecanrewrite Equation 2.96 as v(t)=%:=i1/%[E —U(x)] (2.97) andbyintegrating X idx 1-: =ii (2.98) 0L,\/i[EU(x)] where x=x0att—to.Wehave formally solved theone-dimensional case in Equation 2.98; that is,wehave found x(t). Allthat remains istoinsert thepo- tential U(x) into Equation 2.98 and integrate, using computer techniques if necessary. Weshall study later insome detail thepotentials U=ékxi’ forhar- monic oscillations and U=—k/xforthegravitational force. Wecanlearn agood deal about themotion ofaparticle simply byexamin- ingaplot ofanexample ofU(x) asshown inFigure 2-14. First, notice that, be- cause §m1/2 —T20,E2U(x) foranyrealphysical motion. WeseeinFigure 2-14 that themotion isbounded forenergies E1andE2.ForE1,themotion isperiodic between theturning points xaandx,,.Similarly, forE2themotion isperiodic, but there aretwopossible regions: x,SxSxdand x,SxSxf.The particle cannot “jump” from one“pocket” totheother; once inapocket, itmust remain there forever ifitsenergy remains atE2.The motion foraparticle with energy E0has only onevalue, x=x0.The particle isatrestwith T=O[E0=U(x0)]. The motion foraparticle with energy E2,issimple: The particle comes in from inﬁnity, stops andturns atx—xg,andreturns toinﬁnity—much likeaten- nisballbouncing against apractice wall. Fortheenergy E4,themotion isun- bounded andtheparticle maybeatanyposition. Itsspeed willchange because it depends onthedifference between E4and U(x). Ifitismoving totheright, it willspeed upandslow down butcontinue toinﬁnity. 84 2/NEWTONIAN MECI-IANICS—SINGLE PARTICLE U(x) E4 -- E3___- _ E2?"""" “ MElL____ I 02*-""-"T-"F"XIIH—--_-_-_—_-_—_-i-_-_-_-1'"-s<fll1 5"---— =?$““‘-“““i§'“““““““ ~R-___“-_—-_" t§--"--""-RE0‘~~~~~~ - - H, FIGURE 2-14 Potential energy U(x) curve with various energies Eindicated. For certain energies, forexample E1andE2,themotion isbounded. The motion ofaparticle ofenergy E1issimilar tothat ofamass attheend ofaspring. The potential intheregion xa<x<x,,canbeapproximated by U(x) =:1;k(xPx0)2.Aparticle with energy barely above E0willoscillate about the point x=x0.Werefer tosuch apoint asanequilibrium point, because ifthepar- ticle isplaced atx=xoitremains there. Equilibrium may bestable, unstable, or neutral. The equilibrium just discussed isstable because iftheparticle were placed oneither side ofx=x0itwould eventually return there. Wecan usea hemispherical mixing bowl with asteel ballasanexample. With thebowl right side up,theballcanrollaround inside thebowl; butitwilleventually settle to thebottom-in other words, there isastable equilibrium. Ifweturn thebowl upside down andplace theballprecisely outside atx=qco,theballremains there inequilibrium. Ifweplace theballoneither side ofxIx0ontherounded sur- face, itrolls off;wecallthisunstable equilibrium. Neutral equilibrium would apply when theballrolls onaﬂat,smooth, horizontal surface. Ingeneral, wecanexpress thepotential U(x) inaTaylor series about a certain equilibrium point. Formathematical simplicity, letusassume that the equilibrium point isatx=Orather than x=x0(ifnot, wecanalways redeﬁne thecoordinate system tomake itso).Then wehave dU x2d2U x3d3UU(x) :U0+x(dx)0 +2!(dx2 )0+E3!(‘beg )0+ (2.99) The zero subscript indicates that thequantity istobeevaluated atx=O.The po- tential energy U0atx=0issimply aconstant thatwecandeﬁne tobezero without anylossofgenerality. IfxIOisanequilibrium point, then 2.6ENERGY 85 dU=0Equilibrium point (2.100) 0 andEquation 2.99 becomes 2d2U 3 U(x)=-3!-(ﬁx +9;-!(ﬁ)0 + (2.101) Near theequilibrium point x=0,thevalue ofxissmall, and each term in Equation 2.101 isconsiderably smaller than theprevious one. Therefore, we keep only theﬁrst term inEquation 2.101: 2 2 U(x)=-’;—(%])0 (2.102) Wecandetermine whether theequilibrium atx—0isstable orunstable by examining (d2U/dx2)0. Ifx=0isastable equilibrium, U(x) must begreater (more positive) oneither side ofx=0.Because x2isalways positive, thecon- ditions fortheequilibrium are d2U ...E >0Stable equilibrium _ 0 (2.103) d2U ...Z <0Unstable equilibrium dx2 0 If(d2U/dx2)0 iszero, higher-order terms must beexamined (see Problems 2-45 and2-46). Consider thesystem ofpulleys, masses, andstring shown inFigure 2-15. Alight string oflength bisattached atpoint A,passes over apulley atpoint Blocated a distance 2daway, andﬁnally attaches tomass ml.Another pulley with mass m2 attached passes over thestring, pulling itdown between Aand B.Calculate the distance x1when thesystem isinequilibrium, anddetermine whether theequi- librium isstable orunstable. The pulleys aremassless. Solution. Wecansolve thisexample byeither using forces (i.e., when 561=0= :21)orenergy. Wechoose theenergy method, because inequilibrium theki- netic energy iszero andweneed todeal only with thepotential energy when Equation 2.100 applies. WeletU=0along thelineAB. U= —m1gx1 —m2g(x2 +c) (2.104) 86 2/NEWTONLAN MECHANICS~—SINGLE PARTICLE B__________ __2_'1_________________ __A b_x1 x2 b—x1 2 2 xi C FIGURE 2-15 Example 2.12.' Weassume thatthepulley holding mass m2issmall, sowecanneglect the pulley radius. The distance cinFigure 2-15 isconstant. x2=\/[(b—x1)2/4] —d2 U: _m1gx1_ m2gV [(5—xi)2/4] _d2_"Z285 Bysetting dU/dxl =0,wecandetermine theequilibrium position (x1)0 Ix0: (av) _ m2g(b—xo)~— ——m1g +c * —I0 dxi0 4\/[(0 —x0)2/4] —.12 4m,\/[(0 —22,)?/4] —d2=m2(b—x0) (b—x0)2(-im% —mg)=10m%d2 x0=b——i-7'-'-“ll (2.105)‘\/ 2_ 24777.1 m2 Notice thatarealsolution exists only when 4m? > Under what circumstances willthemass m2pull themass mluptothepul- leyB(i.e., x1I0)?WecanuseEquation 2.103 todetermine whether theequi- librium isstable orunstable: aw: 2—m2g + m2g(b— mi dxi4{[(b—x1)2/4] —a2}”2 16{[(b —x1)2/4] —a2}3’2 Now insert x1IJ60. d2U g(4mi —mi)?” M),_T4m§d The condition fortheequilibrium (real motion) previously wasfor4m? >m2, sotheequilibrium, when itexists, willbestable, because (d2U/dx2)0 >0. 2.6ENERGY 87 EXAMPLE 2.13 T Consider theone-dimensional potential —Wd2(x2 +d2)U(x) I7-? (2.106) Sketch thepotential anddiscuss themotion atvarious values ofx.Isthemotion bounded orunbounded? Where aretheequilibrium values? Arethey stable or unstable? Find theturning points forEI—W/8.The value ofWisapositive constant. Solution. Rewrite thepotential as U<-<2+1)Z(y)=7”)=ﬁf where y=E (2.107) First, ﬁnd theequilibrium points, which willhelp guide usinsketching the potential. dZ I231 4y3(J’2 +1)Z +— :0 dy y4+8 (y4+8)2 This isreduced to y(y4 +23:2 I8)=0 y(y2+4)(y2 I2)=0 yi=2.0 so .7001 =0 x02=\/2.1 (2.103) X05 : There arethree equilibrium points. Wesketch U(x)/ Wversus x/dinFigure 2-16. The equilibrium isstable atx02andx03butunstable atx01.The motion is bounded forallenergies E<0.Wecandetermine turning points foranyen- ergy Ebysetting EIU(x). W IW(y2+1)E=-5IU(y) =W (2.109) )4+s:8)?+8 2*‘=822 2=-;2\/2.0 (2.110) Theturning points forE I—W/8 arexI-2\/2d and+2\/2d, aswellasxI0— which istheunstable equilibrium point. 88 2/NEWTONIAN MECHANICS-—SINGLE PARTICLE U(x)/W LI I I I -10 -5 0 5 10x/d -0.10- -0.20 p _0.25T FIGURE 2-16 Example 2.13. Sketch ofU(x)/W 2.7Limitations ofNewtonian Mechanics Inthischapter, wehave introduced such concepts asposition, time, momentum, andenergy. Wehave implied that these areallmeasurable quantities and that they canbespeciﬁed with anydesired accuracy, depending only onthedegree ofsophistication ofourmeasuring instruments. Indeed, thisimplication appears tobeveriﬁed byourexperience with allmacroscopic objects. Atanygiven in- stant oftime, forexample, wecanmeasure with great precision theposition of, say,aplanet initsorbit about thesun.Aseries ofsuch measurements allows usto determine (also with great precision) theplanet’s velocity atanygiven position. When weattempt tomake precise measurements onmicroscopic objects, however, weﬁnd afundamental limitation intheaccuracy oftheresults. For example, wecanconceivably measure theposition ofanelectron byscattering a light photon from theelectron. The wave character ofthephoton precludes an exact measurement, and wecan determine the position ofthe electron only within some uncertainty Axrelated totheextent (i.e., thewavelength) ofthe photon. Bythevery actofmeasurement, however, wehave induced achange in thestate oftheelectron, because thescattering ofthephoton imparts momen- tumtotheelectron. This momentum isuncertain byanamount Ap.Theproduct AxApisameasure oftheprecision with which wecansimultaneously determine theelectron’s position andmomentum; Ax—>0,Ap—>0implies ameasurement with allimaginable precision. Itwasshown bytheGerman physicist Werner Heisenberg (1901-1976) in1927 that thisproduct must always belarger than a certain minimum value.* Wecannot, then, simultaneously specify both theposition *This result alsoapplies tothemeasurement ofenergy ataparticular time, inwhich case theprod- uctoftheuncertainties isAEAt(which hasthesame dimensions asAxAp). 2.7LIMITATIONS orNEWTONIAN MECHANICS 89 andmomentum oftheelectron with inﬁnite precision, forifAx—>0, then we must have Ap—>ooforHeisenberg’s uncertainty principle tobesatisﬁed. The minimum value ofAxApisoftheorder of10'5"] -s.This isextremely small bymacroscopic standards, soforlaboratory-scale objects there isnopracti- caldifﬁculty inperforming simultaneous measurements ofposition and momen- tum. Newton’s laws cantherefore beapplied asifposition andmomentum were precisely deﬁnable. Butbecause oftheuncertainty principle, Newtonian mechan- icscannot beapplied tomicroscopic systems. Toovercome these fundamental difﬁculties intheNewtonian system, anew method ofdealing with microscopic phenomena wasdeveloped, beginning in1926. The work ofErwin Schrodinger (1887-1961), Heisenberg, Max Born (1872-1970), Paul Dirac (1902-1984), and others subsequently placed thisnew discipline onaﬁrm foundation. Newtonian mechanics, then, isperfectly adequate fordescribing large-scale phenomena. But weneed thenew mechanics (quantum mechanics) toanalyze processes inthe atomic domain. Asthesizeofthesystem increases, quantum mechanics goes over into thelimiting fomi ofNewtonian mechanics. Inaddition tothefundamental limitations ofNewtonian mechanics asap- plied tomicroscopic objects, there isanother inherent difﬁculty inthe Newtonian scheme—one that rests ontheconcept oftime. IntheNewtonian view, time isabsolute, thatis,itissupposed thatitisalways possible todetermine unambiguously whether twoevents have occurred simultaneously orwhether onehaspreceded theother. Todecide onthetime sequence ofevents, thetwo observers ofthe events must beininstantaneous communication,- either through some system ofsignals orbyestablishing twoexactly synchronous clocks atthepoints ofobservation. Butthesetting oftwoclocks into exact synchronism requires theknowledge ofthetime oftransit ofasignal inonedirection from one observer totheother. (Wecould accomplish thisifwealready hadtwosynchro- nous clocks, butthisisacircular argument.) When weactually measure signal velocities, however, wealways obtain anaverage velocity forpropagation inoppo- sitedirections. And todevise anexperiment tomeasure thevelocity inonly one direction inevitably leads totheintroduction ofsome new assumption that we cannot verify before theexperiment. Weknow that instantaneous communication bysignaling isimpossible: Interactions between material bodies propagate with ﬁnite velocity, and aninter- action ofsome sortmust occur forasignal tobetransmitted. The maximum ve- locity with which anysignal canbepropagated isthat oflight infree space: cE3X108m/s.* The difﬁculties inestablishing atime scale between separate points lead us tobelieve that time is,after all,notabsolute and that space and time aresome- how intimately related. The solution tothedilemma wasfound during thepe- riod 1904-1905 byHendrik Lorenz (1853-1928), Henri Poincaré (1854-1912), and Albert Einstein (1879-1955) and isembodied inthespecial theory ofrela- tivity (seeChapter 14). *The speed oflight hasnow been deﬁned tobe299,792,458.0 m/s tomake comparisons ofother measurements more standard. The meter isnow deﬁned asthedistance traveled bylight inavac- uum during atime interval ofI/299,792,458 ofa second. 90 2/NEWTONIAN MECHANICS-—SINGLE PARTICLE Newtonian mechanics istherefore subject tofundamental limitations when small distances orhigh velocities areencountered. Diﬁiculties with Newtonian me- chanics may alsooccur when massive objects orenormous distances areinvolved. Apractical limitation alsooccurs when thenumber ofbodies constituting thesys- tem islarge. InChapter 8,weseethat wecannot obtain ageneral solution in closed form forthemotion ofasystem ofmore than twointeracting bodies even fortherelatively simple caseofgravitational interaction. Tocalculate themotion in athree-body system, wemust resort toanumerical approximation procedure. Although such amethod isinprinciple capable ofanydesired accuracy, thelabor involved isconsiderable. The motion ineven more complex systems (forexam- ple, thesystem composed ofallthemajor objects inthesolar system) can like- wise becomputed, buttheprocedure rapidly becomes toounwieldy tobeof much useforanylarger system. Tocalculate themotion oftheindividual mole- cules in,say,acubic centimeter ofgascontaining I1019 molecules isclearly out ofthequestion. Asuccessful method ofcalculating theaverage properties ofsuch systems wasdeveloped inthelatter part ofthenineteenth century byBoltzmann, Maxwell, Gibbs, Liouville, and others. These procedures allowed thedynamics ofsystems tobecalculated from probability theory, andastatistical mechanics was evolved. Some comments regarding theformulation ofstatistical concepts in mechanics arefound inSection 7.13. PROBLEMS 2-1. Suppose that theforce acting onaparticle isfactorable into one ofthefollowing forms: (8)F06, 1)=f(X.-)g(i) (b)F(5¢.-» I)==f(5¢I)g(l) (<1)F(%.-» 561;)=f(XI)g(5¢i) Forwhich cases aretheequations ofmotion integrable? 2-2. Aparticle ofmass misconstrained tomove onthesurface ofasphere ofradius R byanapplied force F(6, 05).Write theequation ofmotion. 2-3. Ifaprojectile isﬁred from theorigin ofthecoordinate system with aninitial veloc- ityvoandinadirection making anangle awith thehorizontal, calculate thetime required fortheprojectile tocross alinepassing through theorigin andmaking an angle ﬁ<awith thehorizontal. 24. Aclown isjuggling four balls simultaneously. Students useavideo tape todeter- mine thatittakes theclown 0.9stocycle each ballthrough hishands (including catching, transferring, andthrowing) andtobeready tocatch thenext ball. I/Vhat istheminimum vertical speed theclown must throw upeach ball? 2-5. Ajetﬁghter pilot knows heisable towithstand anacceleration of9gbefore black- ingout.The pilot points hisplane vertically down while traveling atMach 3speed and intends topull upinacircular maneuver before crashing into theground. (a)Where does themaximum acceleration occur inthemaneuver? (b)I/Vhat isthe minimum radius thepilot cantake? PROBLEMS 91 2-6. 2-7. 2-8. 2-9. 2-10. 2-11 2-12. 2-13.Intheblizzard of’88,arancher wasforced todrop haybales from anairplane to feed hercattle. The plane ﬂew horizontally at160km/hr anddropped thebales from aheight of80mabove theﬂatrange. (a)Shewanted thebales ofhaytoland 30mbehind thecattle soastonothitthem. Where should shepush thebales out oftheairplane? (b)Tonothitthecattle, what isthelargest time error shecould make while pushing thebales outoftheairplane? Ignore airresistance. Include airresistance forthebales ofhayintheprevious problem. Abale ofhay hasamass ofabout 30kgandanaverage area ofabout 0.2m2.Lettheresistance be proportional tothesquare ofthespeed andletcwI0.8.Plot thetrajectories with a computer ifthehaybales land 30mbehind thecattle forboth including airresis- tance andnot. Ifthebales ofhaywere released atthesame time inthetwocases, what isthedistance between landing positions ofthebales? Aprojectile isﬁred with avelocity v0such thatitpasses through twopoints both a distance habove thehorizontal. Show that ifthegun isadjusted formaximum range, theseparation ofthepoints is 11-%\/03 -420 Consider aprojectile ﬁred vertically inaconstant gravitational field. Forthesame initial velocities, compare thetimes required fortheprojectile toreach itsmaxi- mum height (a)forzero resisting force, (b)foraresisting force proportional tothe instantaneous velocity oftheprojectile. _ Repeat Example 2.4byperforming acalculation using acomputer tosolve Equation 2.22. Usethefollowing values: mI1kg,-00I10m/s, x0I0,andkI 0.1s'1. Make plots ofvversus t,xversus t,and '0versus x.Compare with theresults ofExample 2.4toseeifyour results arereasonable. Consider aparticle ofmass mwhose motion starts from restinaconstant gravita- tional ﬁeld. Ifaresisting force proportional tothesquare ofthevelocity (i.e., kmvg) isencountered, show thatthedistance stheparticle fallsinaccelerating from v0to v1isgiven by 1 g— kvg s(v0—>-01) =2—kln l_ kt)? Aparticle isprojected vertically upward inaconstant gravitational ﬁeld with an initial speed v0.Show thatifthere isaretarding force proportional tothesquare oftheinstantaneous speed, thespeed oftheparticle when itreturns totheinitial position is U01), \/vg+v% where v,istheterminal speed. Aparticle moves inamedium under theinﬂuence ofaretarding force equal to mk(v3 +a2v), where kand aareconstants. Show thatforanyvalue oftheinitial 92 2-14. 2-15 2-16 2-17 2-18 2-19. 2-20 2-21 2-22 *See62/NEWTONIAN MECHANICS-—SINGLE PARTICLE speed theparticle willnever move adistance greater than 11'/2ka andthattheparti- clecomes torestonly fort——>oo. Aprojectile isﬁred with initial speed v0atanelevation angle ofaupahillofslope B01>I5)- (a)How farupthehillwilltheprojectile land? (b)Atwhat angle awilltherange beamaximum? (c)I/Vhat isthemaximum range? Aparticle ofmass mslides down aninclined plane under theinﬂuence ofgravity. If themotion isresisted byaforce fIkmv2, show that thetime required tomove a distance dafter starting from restis cosh 'l(e"" ) r=—-———- \/kgsin0 where 9istheangle ofinclination oftheplane. Aparticle isprojected with aninitial velocity v0upaslope thatmakes anangle a with thehorizontal. Assume frictionless motion andﬁnd thetime required forthe particle toreturn toitsstarting position. Find thetime forv0I2.4m/sandaI26°. Astrong softball player smacks theballataheight of0.7mabove home plate. The ballleaves theplayer’s batatanelevation angle of35°andtravels toward afence 2 mhigh and60maway incenter ﬁeld. What must theinitial speed ofthesoftball be toclear thecenter ﬁeld fence? Ignore airresistance. Include airresistance proportional tothesquare oftheball’s speed intheprevious problem. Letthedrag coefﬁcient becwI0.5,thesoftball radius be5cmandthe mass be200g.(a)Find theinitial speed ofthesoftball needed now toclear the fence. (b)Forthisspeed, ﬁnd theinitial elevation angle thatallows theballtomost easily clear thefence. Byhow much does theballnow vertically clear thefence? Ifaprojectile moves such thatitsdistance from thepoint ofprojection isalways in- creasing, ﬁnd themaximum angle above thehorizontal with which theparticle could have been projected. (Assume noairresistance.) Agunﬁres aprojectile ofmass 10kgofthetype towhich thecurves ofFigure 2-3 apply. The muzzle velocity is140m/s.Through what angle must thebarrel beele- vated tohitatarget onthesame horizontal plane asthegun and 1000 maway? Compare theresults with those forthecase ofnoretardation. Show directly thatthetime rateofchange oftheangular momentum about theori- ginforaprojectile ﬁred from theorigin (constant g)isequal tothemoment of force (ortorque) about theorigin. Themotion ofacharged particle inanelectromagnetic ﬁeld canbeobtained from theLorentz equation* fortheforce onaparticle insuch aﬁeld. Iftheelectric ﬁeld vector isEandthemagnetic ﬁeld vector isB,theforce onaparticle ofmass mthat ,forexample, Heald and Marion, Classical Electromagnetic Radiation (95, Section 1.7). PROBLEMS 93 2-23. 2-24.carries acharge qandhasavelocityv isgiven by FIqE+qvXB where weassume thatv<<c(speed oflight). (a)Ifthere isnoelectric ﬁeld andiftheparticle enters themagnetic ﬁeld inadi- rection perpendicular tothelines ofmagnetic ﬂux, show thatthetrajectory isa circle with radius THU ‘U rqB co, where co,IqB/m isthecyclotron frequency. (b)Choose thez-axis tolieinthedirection ofBandlettheplane containing Eand Bbetheyz—plane. Thus B=Bk, E=Eyj+E,k Show that thezcomponent ofthemotion isgiven by __ 6 qEZ 2 Z(l) ——Z0‘I’Zol ‘I’"-"l 2m where 1(0) I20and 2(0) I20 (c)Continue thecalculation andobtain expressions for5c(t)andj1(t).Show thatthe timeaverages ofthese velocity components are . E7 . (X)="E,(J1)=0 (Show thatthemotion isperiodic andthen average over onecomplete period.) (d)Integrate thevelocity equations found in(c)andshow (with theinitial condi- tions x(O) I—A/01,, a'c(0) IE,/B,y(0) I0,37(0) IA)that A E Ax(t)IIcosco,t+-1-;t,y(t)I20-sinco,t These aretheparametric equations ofatrochoid. Sketch theprojection ofthe trajectory onthexy—-plane forthecases (i)A>|Ey/BI, (ii)A<IE,/BI, and (iii)A=IE,/Bl.(Thelastcaseyieldsacycloid.) Aparticle ofmass mI1kgissubjected toaone-dimensional force F(t)Ikte"", where kI1N/sandaI0.5s'1.Iftheparticle isinitially atrest, calculate andplot with theaidofacomputer theposition, speed, andacceleration oftheparticle asa function oftime. Askier weighing 90kgstarts from rest down ahillinclined at17°. Heskis 100 m down thehillandthen coasts for70malong level snow until hestops. Find thecoef- ﬁcient ofkinetic friction between theskisandthesnow. 1/Vhat velocity does theskier have atthebottom ofthehill? 94 2/NEWTONIAN MECHANICS~—SINGLE PARTICLE 2-25. Ablock ofmass mI1.62 kgslides down africtionless incline (Figure 2-A). The 2-26 2-27 2-28. 2-29 2-30block isreleased aheight hI3.91 mabove thebottom oftheloop. (a)What istheforce oftheinclined track ontheblock atthebottom (point A)? (b)What istheforce ofthetrack ontheblock atpoint B? (c)Atwhat speed does theblock leave thetrack? (d)How faraway from point Adoes theblock land onlevel ground? (e)Sketch thepotential energy U(x) oftheblock. Indicate thetotal energy onthe sketch. h \\m\\\\ ———————/“;/U!//0/U:./s, )—->x A FIGURE 2-A Problem 2-25. Achild slides ablock ofmass 2kgalong aslick kitchen ﬂoor. Iftheinitial speed is4 m/s andtheblock hitsaspring with spring constant 6N/m, what isthemaximum compression ofthespring? I/Vhat istheresult iftheblock slides across 2mofa rough ﬂoor thathasukI0.2? Arope having atotal mass of0.4kgand total length 4mhas0.6moftherope hanging vertically down offawork bench. How much work must bedone toplace alltherope onthebench? Asuperball ofmass Mandamarble ofmass maredropped from aheight hwith the marble just ontopofthesuperball. Asuperball hasacoefficient ofrestitution of nearly 1(i.e., itscollision isessentially elastic). Ignore thesizes ofthesuperball and marble. The superball collides with theﬂoor, rebounds, and smacks themarble, which moves back up.How high does themarble goifallthemotion isvertical? How high does thesuperball go? Anautomobile driver traveling down an8%grade slams onhisbrakes andskids 30 mbefore hitting aparked car.Alawyer hires anexpert who measures thecoefﬁ- cient ofkinetic friction between thetires androad tobeuhI0.45. Isthelawyer correct toaccuse thedriver ofexceeding the25-MPH speed limit? Explain. Astudent drops awater-ﬁlled balloon from theroof ofthetallest building intown trying tohither roommate ontheground (who istooquick). The ﬁrst student ducks back buthears thewater splash 4.021 safter dropping theballoon. Ifthe speed ofsound is331m/s,ﬁndtheheight ofthebuilding, neglecting airresistance. PROBLEMS 95 2-31 2-32. 2-33. 2-34. 2-35. 2-36.InExample 2.10, theinitial velocity oftheincoming charged particle hadnocom- ponent along thex-axis. Show that, even ifithad anxcomponent, thesubsequent motion oftheparticle would bethesame-—that only theradius ofthehelix would bealtered. Two blocks ofunequal mass areconnected byastring over asmooth pulley (Figure 2-B). Ifthecoefficient ofkinetic friction ispk,what angle 6oftheincline allows the masses tomove ataconstant speed? MF - FIGURE 2—B Problem 2-32. Perform acomputer calculation foranobject moving vertically inairunder gravity andexperiencing aretarding force proportional tothesquare oftheobject’s speed (seeEquation 2.21). Usevariables mformass andrfor theobject’s radius.'All the objects aredropped from restfrom thetopofa100-m-tall building. Useavalue of cwI0.5andmake computer plots ofheight y,speed v,andacceleration aversus t forthefollowing conditions and answer thequestions: (a)Abaseball ofmI0.145 kgandrI0.0366 m. (b)Aping-pong ball ofmI0.0024 kgand rI0.019 m. (c)Araindrop ofrI0.003 m. (d)Doalltheobjects reach their terminal speeds? Discuss thevalues ofthetermi- nalvelocities andexplain their differences. (e)I/Vhy canabaseball bethrown farther than aping-pong balleven though the baseball issomuch more massive? (f)Discuss thetemiinal speeds ofbigandsmall raindrops. What aretheterminal speeds ofraindrops having radii 0.002 mand0.004 m? Aparticle isreleased from rest(yI0)andfalls under theinﬂuence ofgravity and airresistance. Find therelationship between vand thedistance offalling ywhen theairresistance isequal to(a)avand(b)Bv2. Perform thenumerical calculations ofExample 2.7forthevalues given inFigure 2-8.Plot both Figures 2-8and 2-9.Donotduplicate thesolution inAppendix H; compose your own solution. Agunislocated onabluff ofheight hoverlooking ariver valley. Ifthemuzzle ve- locity isv0,ﬁnd theexpression fortherange asafunction oftheelevation angle of thegun. Solve numerically forthemaximum range outinto thevalley foragiven h and U0. 96 2-37 2-38. 2-39 2-40. 2-41. 2-42.2/NEWTONIAN MECHANICS-—SINGLE PARTICLE Aparticle ofmass mhasspeed vIas/ac where xisitsdisplacement. Find theforce F(x) responsible. The speed ofaparticle ofmass mvaries with thedistance xasv(x) Iax”. Assume v(xI0)I0attI0.(a)Find theforce F(x) responsible. (b)Determine x(t)and (¢)1'l(l)- Aboat with initial speed v0islaunched onalake. Theboat isslowed bythewater by aforce FI-ae-3". (a)Find anexpression forthespeed v(t). (b)Find thetime and (c)distance fortheboat tostop. Aparticle moves inatwo-dimensional orbit deﬁned by x(t) IA(2at Isinat) y(t) IA(1 —-cosat) (a)Find thetangential acceleration atandnormal acceleration anasafunction of time where thetangential andnormal components aretaken with respect tothe velocity. (b)Determine atwhat times intheorbit anhasamaximum. Atrain moves along thetracks ataconstant speed u.Awoman onthetrain throws aballofmass mstraight ahead with aspeed vwith respect toherself. (a)What isthe kinetic energy gain oftheballasmeasured byaperson onthetrain? (b)byaper- sonstanding bytherailroad track? (c)How much work isdone bythewoman throwing heballand(d)bythetrain? Asolid cube ofuniform density and sides ofbisinequilibrium ontopofacylinder ofradius R(Figure 2-C). Theplanes offour sides ofthecube areparallel totheaxis ofthecylinder. The contact between cube andsphere isperfectly rough. Under what conditions istheequilibrium stable ornotstable? b FIGURE 2-C Problem 2-42. 2-43. Aparticle isunder theinfluence ofaforce FI—-kx +kx?’/012, where kandaare constants andkispositive. Determine U(x) anddiscuss themotion. What happens when EI(1/4)ka2? 2-44. Solve Example 2.12 byusing forces rather than energy. How canyoudetermine whether thesystem equilibrium isstable orunstable? PROBLEMS 97 2-45 2-46 2-47 2-48 2-49 2-50 2-51 2-52 2-53 2-54.Describe how todetermine whether anequilibrium isstable orunstable when (d2U/dx2)0 I0. Write thecriteria fordetermining whether anequilibrium isstable orunstable when allderivatives upthrough order n,(d"U/dx") 0IO. Consider aparticle moving intheregion x>0under theinﬂuence ofthepotential ma=%G+§ where U0I1]andozI2m.Plot thepotential, ﬁnd theequilibrium points, and determine whether they aremaxima orminima. Two gravitationally bound stars with equal masses m,separated byadistance d,re- volve about their center ofmass incircular orbits. Show that theperiod 1'ispropor- tional tod3/2 (Kepler’s Third Law) and ﬁnd theproportionality constant. Two gravitationally bound stars with unequal masses mland m2,separated byadis- tance d,revolve about their center ofmass incircular orbits. Show thattheperiod r isproportional tod3/2 (Kepler’s Third Law) and ﬁnd theproportionality constant. According tospecial relativity, aparticle ofrestmass m0accelerated inonedimen- sion byaforce Fobeys theequation ofmotion dp/dt IFHere lbIm0v/ (1— v2/c2) 1/2istherelativistic momentum, which reduces tom0vforv2/c2 <<1.(a)For thecase ofconstant Fand initial conditions x(0) I0Iv(0), ﬁnd x(t)and v(t). (b)Sketch your result forv(t). (c)Suppose thatF/m0 I10m/s? (IgonEarth). How much time isrequired fortheparticle toreach halfthespeed oflight andof 99% thespeed oflight? Letusmake the(unrealistic) assumption thataboat ofmass mgliding with initial velocity v0inwater isslowed byaviscous retarding force ofmagnitude bvg,where b isaconstant. (a)Find andsketch v(t). How long does ittake theboat toreach a speed ofv0/1000? (b)Find x(t). How fardoes theboat travel inthistime? LetmI 200kg,v0I2m/s, andbI0.2Nm‘2s2. Aparticle ofmass mmoving inone dimension haspotential energy U(x) I U0[2(x/(1)2 -(x/a)4], where U0and aarepositive constants. (a)Find theforce F(x), which actsontheparticle. (b)Sketch U(x). Find thepositions ofstable and unstable equilibrium. (c)I/Vhat istheangular frequency coofoscillations about the point ofstable equilibrium? (d)What istheminimum speed theparticle must have attheorigin toescape toinﬁnity? (e)AttI0theparticle isattheorigin anditsve- locity ispositive andequal inmagnitude totheescape speed ofpart (d).Find x(t) and sketch theresult. I/Vhich ofthefollowing forces areconservative? Ifconservative, ﬁnd thepotential energy U(r). (a)F,Iayz+bx+c,F,Iaxz+bz,F,Iaxy+by.(b)F,I -ze”‘, F,Ilnz,F,Ie”‘+y/z.(c)FIera/r(a, b,care constants). Apotato ofmass 0.5kgmoves under Earth’s gravity with anairresistive force of Ikmv. (a)Find theterminal velocity ifthepotato isreleased from restand kI 0.01 s‘1.(b)Find themaximum height ofthepotato ifithasthesame value ofk, 98 2-55.2/NEWTONIAN MECHANICS-—SINGLE PARTICLE butitisinitially shot directly upward with astudent-made potato gunwith aninitial velocity of120m/s. Apumpkin ofmass 5kgshot outofastudent-made cannon under airpressure at anelevation angle of45°fellatadistance of142mfrom thecannon. The students used light beams and photocells tomeasure theinitial velocity of54m/s. Iftheair resistive force wasFI-kmv, what wasthevalue ofk? CHAPTER Oscillations 3.1 Introduction Webegin byconsidering the oscillatory motion ofaparticle constrained to move inonedimension. Weassume that aposition ofstable equilibrium exists fortheparticle, andwedesignate thispoint astheorigin (seeSection 2.6). Ifthe particle isdisplaced from theorigin (ineither direction), acertain force tends torestore theparticle toitsoriginal position. Anexample isanatom inalong molecular chain. The restoring force is,ingeneral, some complicated function ofthedisplacement and perhaps oftheparticle’s velocity oreven ofsome higher time derivative oftheposition coordinate. Weconsider here only cases inwhich therestoring force Fisafunction only ofthedisplacement: FIF(x). Weassume thatthefunction F(x) thatdescribes therestoring force possesses continuous derivatives ofallorders sothat thefunction canbeexpanded ina Taylor series: dF 1 d2F 1 d5FF(x) IF0+x(dx)0 +2!x2(dx2 )0+3!x5(dx3)0 + (3.1) where F0isthevalue ofF(x) attheorigin (xI0),and (d”I*7dx")0 isthevalue of thenthderivative attheorigin. Because theorigin isdeﬁned tobetheequilib- rium point, F0must vanish, because otherwise theparticle would move away from theequilibrium point andnotreturn. If,then, weconﬁne ourattention todis- placements oftheparticle that aresufﬁciently small, wecannormally neglect all terms involving x2andhigher powers ofx.Wehave, therefore, theapproximate relation 99 100 3/OSCILLATIONS where wehave substituted kI—"(dI*7dx)0. Because therestoring force isalways directed toward theequilibrium position (the origin), thederivative (dF/dx)0 is negative, andtherefore kisapositive constant. Only theﬁrstpower ofthedisplace- ment occurs inF(x),sotherestoring force inthisapproximation isalinearforce. Physical systems described interms ofEquation 3.2obey Hooke’s Law.* One oftheclasses ofphysical processes thatcanbetreated byapplying Hooke’s Lawis thatinvolving elastic deformations. Aslong asthedisplacements aresmall andthe elastic limits arenotexceeded, alinear restoring force canbeused forproblems ofstretched springs, elastic springs, bending beams, andthelike. Butwemust em- phasize thatsuch calculations areonly approximate, because essentially every real restoring force innature ismore complicated than thesimple Hooke’s Lawforce. Linear forces areonlyuseful approximations, andtheir validity islimited tocases in which theamplitudes oftheoscillations aresmall (butseeProblem 3-8). Damped oscillations, usually resulting from friction, arealmost always the type ofoscillations thatoccur innature. Welearn inthischapter how todesign an efficiently damped system. This damping oftheoscillations may becounteracted ifsome mechanism supplies thesystem with energy from anextemal source ata rate equal tothat absorbed bythedamping medium. Motions ofthistype are called driven (orforced) oscillations. Normally sinusoidal, they have important applications inmechanical vibrations aswell asinelectrical systems. The extensive discussion oflinear oscillatory systems iswarranted bythe great importance ofoscillatory phenomena inmany areas ofphysics and engi- neering. Itisfrequently permissible tousethelinear approximation intheanaly- sisofsuch systems. The usefulness ofthese analyses isdueinlarge measure to thefact that wecanusually useanalytical methods. When welook more carefully atphysical systems, weﬁnd thatalarge number ofthem arenonlinear ingeneral. Wewilldiscuss nonlinear systems inChapter 4. 3.2 Simple Harmonic Oscillator The equation ofmotion forthesimple harmonic oscillator may beobtained by substituting theHooke’s Law force into theNewtonian equation FIma.Thus -kx Imié (3.3) Ifwe deﬁne (03Ik/m (3.4) *Robert Hooke (1635-1703). The equivalent ofthisforce lawwasoriginally announced byHooke in 1676 intheform ofaLatin cryptogram: CEIIINOSSSTTUV. Hooke later provided atranslation: ut temio sicvis[thestretch isproportional totheforce].Equation 3.3becomes 5.2SIMPLE HARMONIC OSCILLATOR 101 According totheresults ofAppendix C,thesolution ofthisequation canbe expressed ineither oftheforms x(t) IAsin(w0t -5) (3.6a) x(t) IAcos(0o0t -—qﬁ) (3.6b) where thephases* 5andoidiffer by11/2.(Analteration ofthephase angle corre- sponds toachange oftheinstant thatwedesignate tI0,theorigin ofthetime scale.) Equations 3.6a andbexhibit thewell-known sinusoidal behavior ofthe displacement ofthesimple harmonic oscillator. Wecanobtain therelationship between thetotal energy oftheoscillator andtheamplitude ofitsmotion asfollows. Using Equation 3.6a forx(t), weﬁnd forthekinetic energy, [Q1-—I NJ"-‘l\9r—‘TIImx2IImw§A2 cos?-’(w0t —5) IIkA2cos2(w0t I5) (3.7) The potential energy may beobtained bycalculating thework required to displace theparticle adistance x.The incremental amount ofwork dWnecessary tomove theparticle byanamount dxagainst therestoring force Fis dWI —-Fdx Ikxdx ,(3.8) Integrating from 0toxandsetting thework done ontheparticle equal tothe potential energy, wehave 1UI-kx2 (3.9)2 Then 1UI-gkA2 sin2(w0t -—5) (3.10) Combining theexpressions forTand Utoﬁnd thetotal energy E,wehave 1EIT+ UI-gkA2[cos2(o)0t —5)+sin2(w0t -—5)] 1E=r+ UI—ékA2 (3.11) sothatthetotal energy isproportional tothesquare oftheamplitude; thisisagen- eralresult forlinear systems. Notice alsothatEisindependent ofthetime; thatis, *The symbol 5isoften used torepresent phase angle, and itsvalue iseither assigned ordetermined Within thecontext ofanapplication. Becareful when using equations within thischapter because 5 inoneapplication maynotbethesame asthe5inanother. Itmight beprudent toassign subscripts, forexample, 5land 52,when using different equations. 102 3/OSCILLATIONS energy isconserved. (Energy conservation isguaranteed, because wehave been considering asystem without frictional losses orother external forces.) Theperiod 1'0ofthemotion isdeﬁned tobethetime interval between succes- siverepetitions oftheparticle’s position anddirection ofmotion. Such aninter- valoccurs when theargument ofthesine inEquation 3.6a increases by211: o)0'r0 I211 (3.12) or 1'0I21'r\/Z}: (3.13) From thisexpression, aswell asfrom Equation 3.6,itshould beclear that010rep- resents theangular frequency ofthemotion, which isrelated tothefrequency I/0 by* k010I211110 I\/gt (3.14) 11kV0_25_5&1 (3.15) Note that theperiod ofthesimple harmonic oscillator isindependent ofthe amplitude (ortotal energy); asystem exhibiting this property issaid tobe isochronous. For many problems, ofwhich thesimple pendulum isthebest example, theequation ofmotion results in6+010sin6I0,where 6isthedisplacement angle from equilibrium, and(00I\/g/6,where 6isthelength ofthependu- lum arm. Wecanmake this differential equation describe simple harmonic motion byinvoking thesmall oscillation assumption. Iftheoscillations about theequilibrium aresmall, weexpand sin6and cos6inpower series (see Appendix A)andkeep only thelowest terms ofimportance. This often means sin6I6and cos6I1I62/2, where 6ismeasured inradians. Ifweusethe small oscillation approximation forthesimple pendulum, theequation ofmo- tion above becomes 6+0106I0,anequation that does represent simple har- monic motion. Weshall often invoke thisassumption throughout thistextand initsproblems. *Henceforth weshall denote angular frequencies byo)(units: radians perunit time) andfrequencies byv(units: vibrations perunit time orHertz, Hz). Sometimes towillbereferred toasa“frequency” forbrevity, although “angular frequency” istobeunderstood. 3.2 SIMPLE HARMONIC OSCILLATOR 103 EXAMPLE3.l ___ TT‘ T‘——— Tn '- Find theangular velocity andperiod ofoscillation ofasolid sphere ofmass m andradius Rabout apoint onitssurface. SeeFigure 3-1. Solution. Lettherotational inertia ofthesphere beIabout thepivot point. Inele- mentary physics weleam thatthevalue oftherotational inertia about anaxis through thesphere’s center is2/5mR2. Ifweusetheparallel-axis theorem, therota- tional inertia about thepivot point onthesurface is2/5mR2 +mR2 =7/5mR2. Theequilibrium position ofthesphere occurs when thecenter ofmass (center of sphere) ishanging directly below thepivot point. Thegravitational force F=mg pulls thesphere back towards theequilibrium position asthesphere swings back andforth with angle 6.Thetorque onthesphere isN=Ia,where a==5isthean- gular acceleration. Thetorque isalsoN=RXF,with N=RFsin6=Rmgsin 6. Forsmall oscillations, wehave N=Rmg6. Wemust have I5="-Rmg 6forthe equation ofmotion inthiscase, because as6increases, iiisnegative. Weneed to solve theequation ofmotion for6. nR0+~%b=0 This equation issimilar toEquation 3.5andhassolutions fortheangular fre- quency andperiod from Equations 3.14 and3.15, 0,:\/Rmg: Rmg:\/5g1 ZmR2 7R 5 and 7_,mR2 I 5 7R1'I211,/i: 21'r\{ i-I 21?,/—Rmg Rmg 5g Pivot /l “~11/' FIGURE 3-1 Example 3.1.The physical pendulum (sphere)., f 104 3/OSCILLATIONS Note that themass mdoes notenter. Only thedistance Rtothecenter of mass determines theoscillation frequency. 3.3 Harmonic Oscillations inTwo Dimensions Wenext consider themotion ofaparticle thatisallowed twodegrees offreedom. Wetake therestoring force tobeproportional tothedistance oftheparticle from aforce center located attheorigin and tobedirected toward theorigin: F=—kr (3.16) which canberesolved inpolar coordinates into thecomponents Fx= —krcos6 =-kx (3.17) F,=——krs1n6 =-—ky The equations ofmotion are 56+wgx=03.185‘+@132»=0} () where, asbefore, (11%=k/m.The solutions are x(t) ZAcos(w0t ——a) ya)=Bc<>s<w@¢ —#3)} (319) Thus, themotion isoneofsimple harmonic oscillation ineach ofthetwodirec- tions, both oscillations having thesame frequency butpossibly differing inam- plitude andinphase. Wecanobtain theequation forthepath oftheparticle by eliminating thetime tbetween thetwoequations (Equation 3.19). First wewrite y(t)=Bcos[w0t —a+(a"-B)] =Bcos(a)0t —a)cos(a —B)-—Bsin(a)0t -—a)sin(a —-B) (3.20) Deﬁning 5EaWBandnoting thatcos(w0t ——oz)=x/A, wehave B 2 y=Zxcos5— B,/1—(§5)sin5 Ay~Bxcos5=—-B\/ A2wx2sin5 (3.21) Onsquaring, thisbecomesOI‘ A2y2 *2ABxy cos5+B2x2 cos25 =AQB2 sin25 *—B2002 sin25 sothat B2x2 ~2ABxy cos5+A2y2 =A2B2 sin25 (3.22) 3.3 HARMONIC OSCILLATIONS INTWO DIMENSIONS 105 If5issetequal toi1r/2,thisequation reduces totheeasily recognized equation foranellipse: x2 y2F+E=1, 5=in/2 Iftheamplitudes areequal, A=B,andif5=i11'/ 2,wehave thespecial case of circular motion: x2+312=A2, forA =Band 5=in/2 (3.24) Another special case results ifthephase 5vanishes; then wehave B2x2 "2ABxy +A2322 =0, 5=0 Factoring, (Bx—Ay)2 =0 which istheequation ofastraight line: y=-Ex, 5=0 (3.25) Similarly, thephase 5=taryields thestraight lineofopposite slope: By=——Ax, 5=:t1'r (3.26) The curves ofFigure 3-2illustrate Equation 3.22 forthecase A=B;5fl90° or270° yields acircle, and5==180° or360°(0°) yields astraight line. Allother values of5yield ellipses. Inthegeneral case oftwo-dimensional oscillations, theangular frequencies forthemotions inthex-and)1-directions need notbeequal, sothat Equation 3.19 becomes x(t) =Acos(wxt —a) y(t) IBcos(wyt —B) 5=90° 5=120° 5=150° 5=180° 5=210° 5=240° 5=270° 5=300° 5=330° 5=360° FIGURE 3-2 Two-dimensional harmonic oscillation motion forvarious phase angles 5=a—B.} (3.27) 106 3/OSCILLATIONS y 2B x .-————2A—?—-1 FIGURE 3-3 Closed two-dimensional oscillatory motion (called Lissajous curves) occurs under certain conditions forthexand ycoordinates. The path ofthemotion isnolonger anellipse butaLissajous curve.* Such a curve willbeclosed ifthemotion repeats itself atregular intervals oftime. This willbepossible only iftheangular frequencies 0),,andmyarecommensumble, that is,if0),,/my isarational fraction. Such acaseisshown inFigure 3-3,inwhich w,=20),, (also oz=B).Iftheratio oftheangular frequencies isnotarational fraction, the curve willbeopen; thatis,themoving particle willnever pass twice through the same point with thesame velocity. Insuch acase, after asufﬁciently long time haselapsed, thecurve willpass arbitrarily close toanygiven point lying within therectangle 2AX2Bandwilltherefore “ﬁll” therectangle? The two-dimensional oscillator isanexample ofasystem inwhich aninﬁni- tesimal change canresult inaqualitatively different type ofmotion. The motion willbealong aclosed path ifthetwoangular frequencies arecommensurable. Butiftheangular frequency ratio deviates from arational fraction byeven anin- ﬁnitesimal amount, then thepath willnolonger beclosed anditwill“ﬁll” the rectangle. Forthepath tobeclosed, theangular frequency ratio must beknown to bearational fraction with inﬁnite precision. Iftheangular frequencies forthemotions inthex-andy-directions aredif- ferent, theshape oftheresulting Lissajous curve strongly depends onthephase difference 5Ea—B.Figure 3-4shows theresults forthecase my=20),, for phase differences of0,11/3,and1-r/2. 3.4 Phase Diagrams The state ofmotion ofaone-dimensional oscillator, such asthat discussed in Section 3.2,willbecompletely speciﬁed asafunction oftime iftwoquantities *The French physicistjules Lissajous (1822-1880) demonstrated thisin1857 and isgenerally given credit, although Nathaniel Bowditch seems tohave reported in1815 twomutually orthogonal oscil- lations displaying thesame motion (Cr8l). 1Aproof isgiven, forexample, byHaag (Ha62, p.36). 3.4 PHASE DIAGRAMS 107 J’ _.i\‘I1III FIGURE 3-4 Lissajous curves depend strongly onthephase differences oftheangle 5. aregiven atone instant oftime, that is,theinitial conditions x(t0) and a2(t0). (Two quantities areneeded because thedifferential equation forthemotion isof second order.) Wemayconsider thequantities x(t)anda'c(t)tobethecoordinates of apoint inatwo-dimensional space, called phase space. (Intwo dimensions, the phase space isaphase plane. Butforageneral oscillator with ndegrees offreedom, thephase space isa2n-dimensional space.) Asthetime varies, thepoint P(x,5:) describing thestate oftheoscillating particle willmove along acertain phase path in thephase plane. Fordifferent initial conditions oftheoscillator, themotion willbe described bydifferent phase paths. Anygiven path represents thecomplete time his- tory oftheoscillator foracertain setofinitial conditions. The totality ofallpossible phase paths constitutes thephase portrait orthephase diagram oftheoscillator.* According totheresults ofthepreceding section, wehave, forthesimple har- monic oscillator, x(t) =Asin(co0t —5) (3.28a) a2(t) =Arno cos(w0t ~5) (3.28b) Ifweeliminate tfrom these equations, weﬁnd fortheequation ofthepath x2 :22—+i =1 (3.29A2A2w§ ) Thisequation represents afamily ofellipses,l several ofwhich areshown inFigure 3-5. Weknow that thetotal energy Eoftheoscillator isék/12 (Equation 3.11), and be- cause mg=k/m, Equation 3.29 canbewritten as x2 022i +—~— =1 (3.30) 2E/k 2E/m Each phase path, then, corresponds toadeﬁnite total energy oftheoscillator. This result isexpected because thesystem isconservative (i.e., E=const.). Notwophase paths oftheoscillator can cross. Ifthey could cross, thiswould imply that foragiven setofinitial conditions x(t0), :Z(t0) (i.e., thecoordinates ofthe *These considerations arenotrestricted tooscillating particles oroscillating systems. The concept of phase space isapplied extensively invarious ﬁelds ofphysics, particularly instatistical mechanics. "l‘The ordinate ofthephase plane issometimes chosen tobe9?;/wo instead of:2;thephase paths are then circles. 108 3/OSCILLATIONS x X FIGURE 3-5 Phase diagram forasimple harmonic oscillator foravariety oftotal energies E. crossing point), themotion could proceed along different phase paths. But thisis impossible because thesolution ofthedifferential equation isunique. Ifthecoordinate axes ofthephase plane arechosen asinFigure 3-5, the motion oftherepresentative point P(x, :2)willalways beinaclockwise direction, because forx>0thevelocity 5:isalways‘ decreasing andforx<0thevelocity is always increasing. Toobtain Equations 3.28 forx(t)and a'c(t), wemust integrate Equation 3.5,a second-order differential equation: d2 gig‘+wgx=0 (3.31) Wecanobtain theequation forthephase path, however, byasimpler procedure, because Equation 3.31 canbereplaced bythepairofequations dx _ dxE=x, Z;=——w§x (3.32) Ifwedivide thesecond ofthese equations bytheﬁrst, weobtain 4E:=-wgi‘ (ass) This isaﬁrst-order dilferential equation forat=a2(x), thesolution towhich isjust Equation 3.29. Forthesimple harmonic oscillator, there isnodifﬁculty inobtaining thegeneral solution forthemotion bysolving thesecond-order equation. Butinmore complicated situations, itissometimes considerably easier todirectly ﬁnd theequation ofthephase path at=:i(x)without proceeding through thecalculation ofx(t). 3.5 Damped Oscillations The motion represented bythesimple harmonic oscillator istermed afree oscilla- tion; once setinto oscillation, themotion would never cease. This oversimpliﬁes theactual physical case, inwhich dissipative orfrictional forces would eventually damp themotion tothepoint that theoscillations would nolonger occur. Wecan analyze themotion insuch acase byincorporating into thedifferential equation a 3.5 DAMPED OSCILLATIONS 109 term representing thedamping force. Itdoes notseem reasonable thatthedamping force should, ingeneral, depend onthedisplacement, butitcould beafunction ofthevelocity orperhaps ofsome higher time derivative ofthedisplacement. Itis frequently assumed that thedamping force isalinear function ofthevelocity,* Fd=av.Weconsider here only one-dimensional damped oscillations sothatwe canrepresent thedamping term by-—boZ. The parameter bmust bepositive inorder that theforce indeed beresisting. (Aforce —-bk with b<Owould acttoincrease the speed instead ofdecreasing itasanyresisting force must.) Thus, ifaparticle of mass mmoves under thecombined inﬂuence ofalinear restoring force -—kx and a resisting force -—b5c, thedifferential equation describing themotion is mﬁé+biz+kx=0 (3.34) which wecanwrite as Here BEb/2m isthedamping parameter andmo=\/k/misthecharacteristic angular frequency intheabsence ofdamping. The roots oftheauxiliary equation are(cf.Equation C.8, Appendix C) T1Z"*3'1'VB2_W5 fie (3.36) T2Z*3" B2*H15 The general solution ofEquation 3.35 istherefore xv)=@“”[A1@XP(\/B2 —wit)+A2eXp(_\/B2 —@501i(3-37) There arethree general cases ofinterest: Underdamping: mg>B2 Critical damping: mg?)===B2 Overdamping: mg<B2 Themotion ofthethree Cases isshown schematically inFigure 3-6forspeciﬁc initial conditions. Weshall seethat only thecase ofunderdamping results inoscillatory motion. These three cases arediscussed separately. Underdamped Motion Forthecase ofunderdamped motion, itisconvenient todeﬁne 1 (0%Emg4B2 (3.38) *See Section 2.4foradiscussion ofthedependence ofresisting forces onvelocity. 110 3/OSCILLATIONS X Underdamping,B2 <mg \{~‘~__ Critical damping,B2 =(03 \ ~~.\ '22 \ __~_ Overdamp1ng,B >(00 \ _______/ \ -\ ~--.~-_______ \ §-_—___ _ "__ ———— I FIGURE 3-5 Damped oscillator motion forthree cases ofdamping. where ml>0;then theexponents inthebrackets ofEquation 3.37 areimaginary, andthesolution becomes x(t)=e_f”[Ale‘°’1‘ +A2e_“"1‘1 (3.39) Equation 3.39 canberewritten as* x(t) =Ae_B"cos(mlt ~5) (3.40) Wecallthequantity mltheangularﬁequency ofthedamped oscillator. Strictly speaking, wecannot deﬁne afrequency when damping ispresent, because the motion isnotperiodic—that is,theoscillator never passes twice through agiven point with thesame velocity. However, because ml=211'/(2Tl), where Tlisthe time between adjacent zero x-axis crossings, theangular frequency mlhasmeaning foragiven time period. Note that2Tlwould bethe“period” inthiscase, notTl. Forsimplicity, werefer tomlasthe“angular frequency” ofthedamped oscillator, andwenote that thisquantity islessthan thefrequency oftheoscillator intheab- sence ofdamping (i.e., ml<mo). Ifthedamping issmall, then ‘"1: V‘"5*B2E‘1)0 sotheterm angular frequency may beused. Butthemeaning isnotprecise unless B=0. The maximum amplitude ofthemotion ofthedamped oscillator decreases with time because ofthefactor exp(—Bt), where B>O,andtheenvelope ofthe displacement versus time curve isgiven by xﬁn=iAe'B‘ (3.41) This envelope andthedisplacement curve areshown inFigure 3-7forthecase 5=0. Thesinusoidal curve forundamped motion (B=0)isalsoshown inthisﬁgure. A close comparison ofthetwocurves indicates that thefrequency forthedamped case isless(i.e., that theperiod islonger) than that fortheundamped case. *See Exercise D-6,Appendix D. 3.5 DAMPED OSCILLATIONS ll1 I-'\ ,'\ \ I \ :0 I \\\ Ae—Bl II \\ﬁ I’ \\ \\ I \ \ g I \uI1 \~ \ I ~-____'l___ l " I I" _.."_..;.- I1-I I =0.2(00" k §. 4 ll’-/1e"5tAmplitude "¥€‘@§“0" J.- ’P P”\______‘§” // I \ r \\'l \_{I FIGURE 3-7 The underdamped motion (solid line) isanoscillatory motion (short dashes) thatdecreases within theexponential envelope (long dashes). Theratio oftheamplitudes oftheoscillation attwosuccessive maxima is Ae_BT BIejgﬁ Ie1 (3.42) where theﬁrstofanypair ofmaxima occurs attITand where 'rlI211/ml. The quantity exp(B1"l) iscalled thedecrement ofthemotion; thelogarithm ofexp(B1'l)-— thatis,B1-l--is known asthelogarithmic decrement ofthemotion. Unlike thesimple harmonic oscillator discussed previously, theenergy ofthe damped oscillator isnotconstant intime; rather, energy iscontinually given upto thedamping medium anddissipated asheat (or,perhaps, asradiation inthefonn ofﬂuid waves). The rate ofenergy lossisproportional tothesquare ofthevelocity (seeProblem 3-11), sothedecrease ofenergy does nottake place uniformly. The lossratewillbeamaximum when theparticle attains itsmaximum velocity near (but notexactly at)theequilibrium position, and itwill instantaneously vanish when theparticle isatmaximum amplitude andhaszero velocity. Figure 3-8shows thetotal energy andtherateofenergy lossforthedamped oscillator. EXAMPLE _ I Construct ageneral phase diagram analytically forthedamped oscillator. Then, using acomputer, make aplot forxand:2versus tand aphase diagram forthe following values: AI1cm, moI1rad/s, BI0.2s_1, and 5ITr/2 rad. Solution. First, wewrite theexpressions forthedisplacement andthevelocity: x(t) IAe_B‘cos(mlt —-5) s(t)I—Ae'3‘[B cos(mlt —5)+mlsin(mlt —5)] These equations canbecoverted into amore easily recognized form byintroducing achange ofvariables according tothefollowing linear transformations: uImlx, w=Bx+ :2 Then uImlAe'B‘ cos(mlt —5) wI—mlAe_B‘ sin(mlt —5) 112 3/OSCILLATIONS E \ 0 1 l_ I I t—> 0 Edt FIGURE 3-8 The total energy andrateofenergy lossforthedamped oscillator. w l ¢ P 11. FIGURE 3-9 Example 3.2. Ifwerepresent uand win polar coordinates (Figure 3-9), then pI\/u2+w2, ¢Imlt Thus pZ a)lAe“(B/wl)¢ which istheequation ofalogarithmic spiral. Because thetransformation from x, 5ctou,wislinear, thephase path hasbasically thesame shape intheu-wplane (Figure 3-10a) and 5c-xplane (Figure 3-10b). They both show aspiral phase path oftheunderdamped oscillator. The continually decreasing magnitude oftheradius vector forarepresentative point inthephase plane always indicates damped motion oftheoscillator. 3.5 DAMPED OSCILLATIONS 113 w 5c(m/s) 1 0.51 \ ‘ 0.5 0 1u () x(m —0.5 -1- -0.5 1 _l . -0. FIGURE 3-105 0 0.5 1 -0.5 0 0.5 1 (a) (b) x,5c 1 Position x Q/\ sition(m)eed(m/s)Speed 5: g_/ O1 4’, E”~_~'‘\I I I\8 ,- ~_ ’ T'*" 54 Po3P \ \I’ *1 _1 1. I | p. ll 0 5 10 15 20 25 Time (s) (C) Results forExample 3.2.The phase path (a)ofthew,ucoordinates and(b)ofthe:21,xcoordinates, and(c)anumerical calculation of position andspeed versus time. Thespiral path ischaracteristic ofthe underdamped oscillator. The actual calculation using numbers canbedone byvarious means with a computer. Wechose touseone ofthecommercially available numerical pro- grams thathasgood graphics output. Wechose thevalues AI1,BI0.2,kI1, mI1,and5I1'r/2intheappropriate units toproduce Figure 3-10. Forthe particular value of5chosen, theamplitude hasxI0attI0,but5chasalarge positive value, which causes xtorisetoamaximum value ofabout 0.7mat2s (Figure 3-10c). Theweak damping parameter Ballows thesystem tooscillate about zero several times (Figure 3-10c) before thesystem ﬁnally spirals down to zero. The syst lessthan 10“3emcrosses thexI0lineeleven times before xdecreases ﬁnally to ofitsmaximum amplitude. The phase diagram ofFigure 3-10b displays theactual path. 114 3/OSCILLATIONS Critically Damped Motion Ifthedamping force issufﬁciently large (i.e., ifB2>m2l), thesystem isprevented from undergoing oscillatory motion. Ifzero initial velocity occurs, thedisplace- ment decreases monotonically from itsinitial value totheequilibrium position (xI0). The case ofcritical damping occurs when B2isjust equal tom§.The roots ofthe auxiliary equation arethen equal, and thefunction xmust bewritten as(cf., Equation C.11, Appendix C) x(t)I(A+Bt)e“5‘ (3.43) This displacement curve forcritical damping isshown inFigure 3-6forthecase in which theinitial velocity iszero. For agiven setofinitial conditions, acritically damped oscillator willapproach equilibrium atarate more rapid than that foreither anoverdamped oranunderdamped oscillator. This isimportant indesigning certain practical oscillatory systems (e.g., galvanometers) when thesystem must return to equilibrium asrapidly aspossible. Apneumatic-tube screen-door closure system isa good example ofadevice thatshould becritically damped. Iftheclosure were under- damped, thedoor would slam shut asother doors with springs always seem todo,If itwere overdamped, itmight take anunreasonably long time toclose. Overdamped Motion Ifthedamping parameter Biseven larger than mo,then overdamping results. Because B2>m2,theexponents inthebrackets ofEquation 3.37 become real quantities: x(t) Ie”B'[Ale“'2‘ +A2e"”2t] (3.44) where m2I\/B2Img (3.45) Note thatm2does notrepresent anangular frequency, because themotion isnot periodic. The displacement asymptotically approaches theequilibrium position (Figure 3-6). Overdamping results inadecrease oftheamplitude tozero that may have some strange behavior asshown inthephase space diagram ofFigure 3-11. Notice thatforallthephase paths oftheinitial positions shown, theasymptotic paths at longer times arealong thedashed curve 5cI—(B—(U2)x.Only aspecial case (see Problem 3-22) hasaphase path along theother dashed curve. Depending onthe initial values oftheposition andthevelocity, achange insign ofboth xandicmay occur; forexample, seethephase path labeled IIIinFigure 3-11. Figure 3-12 dis- plays xand 5casafunction oftime forthethree phase paths labeled I,II,and IIIin Figure 3-11. Allthree cases have initial positive displacements, x(0) Ix0>0.Each ofthethree phase paths hasinteresting behavior depending ontheinitial value, a2(O) I220,ofthevelocity: I.:20>0,sothat x(t)reaches amaximum atsome t>0before approaching zero. Thevelocity :2decreases, becomes negative, andthen approaches zero. 3.5DAMPED OSCILLATIONS 115 it :5=-(B+w2) x\_ \\ Dots represent \ initial values \\ \ \ \ \ \ \ \ \\\ I .2:-(ﬁ—~G)2)X&‘\__ \\\\ . \ 7 x \\\\ \ \ \ \ \ \ \\\\ l \ \\ III FIGURE 3-11The phase paths foroverdamped motion areshown forseveral initial values of(x,5c).Weexamine more closely thepaths labeled I,II,andIH. II.5:0<O,with x(t)and:2(t)monotonically approaching zero. HI. :20<0,butbelow thecurve isII(B+m2)x, sothat x(t)goes negative before approaching zero, and aZ(t)goes positive before approaching zero. The motion inthiscasecould beconsidered oscillatory. The initial points lying between thetwodashed curves inFigure 3-11 seem to have phase paths decreasing monotonically tozero, whereas those lying outside those twolines donot. Critical damping hasphase paths similar totheoverdamp- ingcurves shown inFigure 3-11 (seeProblem 3-21), rather than thespiral paths of Figure 3-10b. EXAMPLE 3.3 L -T 1 2 . -T2 - I Consider apendulum oflength t’and abob ofmass matitsend (Figure 3-13) moving through oilwith 6decreasing. Themassive bobundergoes small oscilla- tions, buttheoilretards thebob’s motion with aresistive force proportional to thespeed with FmI2m\/J (£25). The bobisinitially pulled back attI0with BI aandflI0.Find theangular displacement 0andvelocity Basafunction oftime. Sketch thephase diagram ifVg/l’I10s'1andozI1O'2 rad. Solution. Gravity produces therestoring force, and thecomponent pulling the bobback toequilibrium ismgsin6.Newton’s Second Lawbecomes Force Im(£’fl) IRestoring force +Resistive force met;=-mgsintl —2m\/g/me) (3.46) 116 X QPosition3/OSCILLATIONS Velocity x Case I 550>0 l Lt 0 |W 71. ,_l Ll_ Case II 3'60<0 | 1.)3 0Case III 3'60<0 _ i_ —,-"L T1II1C FIGURE 3-12 The position andvelocity asfunctions oftime forthethree phase paths labeled I,II,and IIIshown inFigure 3-11. Check thattheforce direction iscorrect, depending onthesigns of6andB.For small oscillations sin6I9,and Equation 3.46 becomes 5+2\/g/to +go=0 (3.47) Comparing thisequation with Equation 3.35 reveals thatmgIg/l’,andB2Ig/t’. Therefore, mgIB2andthependulum iscritically damped. After being initially pulled back and released, thependulum accelerates and then decelerates as9 goes tozero. The pendulum moves only inonedirection asitreturns toits equilibrium position. 3.6SINUSOIDAL DRIVING FORCES 117 9(rad/S) 0.05 - I 0 0.005 0.01 2222) I l Q‘.l - ..’{‘2:.Ai :2‘1:2 5'1jg: _ FIGURE 3-13 Example 3.3.The FIGURE 3-14 Phase diagram forExample 3.3. bob ismoving with decreasing 6. The solution ofEquation 3.47 isEquation 3.43. Wecandetermine theval- uesofAandBbysubstituting Equation 3.43 into Equation 3.47 using theinitial conditions. 6(t)=(A+B¢)e-B’ o(¢=0)=aIA (3.43) o(t)=Be-B’—B(A+B¢)e-B’ 0(t=0) =0=B-BA B=BA=Ba (3.43) 0(1)=a(1+\/Q¢)e-\/Wt (3.49) 6(1)=$16-Wt (3.30) Ifwecalculate 6(t)andtl(t)forseveral values oftime uptoabout 0.5s,wecan sketch thephase diagram ofFigure 3-14. Notice that Figure 3-14 isconsistent with thetypical paths shown inFigure 3-11. Theangular velocity isalways negative after thebobstarts until itreturns toequilibrium. Thebobspeeds upquickly andthen slows down. 3.6 Sinusoidal Driving Forces The simplest case ofdriven oscillation isthat inwhich anexternal driving force varying harmonically with time isapplied totheoscillator. The total force onthe particle isthen FI -kx Ibaé+F0cosmt (3.51) 118 3/OSCILLATIONS where weconsider alinear restoring force andaviscous damping force inaddition tothedriving force. Theequation ofmotion becomes mi?+bi‘:+kxIF0cosmt (3.52) or,using ourprevious notation, 55+2B:2+mgxIAcosmtl (3.53) where AIF0/m andwhere mistheangular frequency ofthedriving force. The solution ofEquation 3.53 consists oftwoparts, acomplementary function x,(t), which isthesolution ofEquation 3.53 with theright-hand side setequal tozero, andaparticular solution xp(t), which reproduces theright-hand side. Thecomple- mentary solution isthesame asthatgiven inEquation 3.37 (seeAppendix C): xc(t) Ie'»3‘[Alexp( VB2 Imgt)+A2exp(I VB2 Imgt)] (3.54) Fortheparticular solution, wetry xp(t) IDcos(mt I5) (3.55) Substituting xp(t) inEquation 3.53andexpanding cos(mt I5)andsin(mtI5),we obtain {AID[( Im2)cos5 +2mB sin5]}cosmt I Im2)sin 5I2mB cos5]}sinmtIO (3.56)HIQ2%3.Because sinmtandcosmtare linearly independent functions, thisequation canbe satisﬁed ingeneral only ifthecoefﬁcient ofeach term vanishes identically. From thesinmtterm, wehave 2)3tan5 I (3.57) 0 sowecanwrite . 2013s1n5 II I I2 22 22V(m0Im) +4mB (3.53) 8 mgIm2 cos I I V(mgIm2)2 +4m2B2 And from thecoefﬁcient ofthecosmtterm, wehave 1)=-- A _(mgIm2)cos 5+2mB sin5 A I I 3.59 V(mgIm2)2 +4m2B2 ( ) 3.6 SINUSOIDAL DRIVING FORCES 119 Thus, theparticular integral is A Xp(l) = COS((.0l '"5) (3.60) with 2B3=mn_1( ) | (3.61) The quantity 5represents thephase difference between thedriving force and theresultant motion; arealdelay occurs between theaction ofthedriving force and theresponse ofthesystem. Foraﬁxed mo,asmincreases from 0,thephase increases from 5I0atmI0to5ITl’/2 atmIml,and to77'asm—>oo.The varia- tionof5with misshown later inFigure 3-16. The general solution is x0)=x.<0+no (3.32) But x,,(t) here represents transient effects (i.e., effects that dieout), and theterms contained inthissolution damp outwith time because ofthefactor exp(IBt). The term xp(t) represents thesteady-state effects andcontains alltheinformation fort large compared with 1/B.Thus, _ x(t>>1/B) Ixp(t) The steady-state solution isimportant inmany applications and problems (see Section 3.7). The details ofthemotion during theperiod before thetransient effects have disappeared (i.e., tS1/B) strongly depend ontheoscillator’s conditions atthe time thatthedriving force isﬁrstapplied andalsoontherelative magnitudes of thedriving frequency mand thedamping frequency VmgIB2inthecase ofun- derdamped, undriven oscillations. This canbeshown bynumerically calculating xp(t) ,x,(t), and thesum x(t)(see Equation 3.62) fordifferent values ofBand mas wehave done forFigure 3-15. The student may proﬁt from solving Problems 3-24 (underdamped) and3-25 (critically damped) where such aprocedure issuggested. Figure 3-15 illustrates thetransient motion ofanunderdamped oscillator when driving frequencies lessthan and greater than mlIVmg IB2areapplied. If m<ml(Figure 3-15a), thetransient response oftheoscillator greatly distorts the sinusoidal shape oftheforcing function during thetime interval immediately after theapplication ofthedriving force, whereas ifm>ml(Figure 3-15b), theeffect isamodulation oftheforcing function with little distortion ofthehigh- frequency sinusoidal oscillations. Thesteady-state solution (xp)iswidely studied inmany applications andprob- lems (seeSection 3.7). The transient effects (xc), although perhaps notasimpor- tant overall, must beunderstood and accounted forinmany cases, especially in certain types ofelectrical circuits. 120 3/OSCILLATIONS X;.(l) 1 _,(0xp ,1LJ2l\ t __.--_1 (4I,_‘— 1 (I) )6=0.15 1ll 2" m=ml/7 >00) 5=0.30 ll co=5ml x(t) x(t) ll 1 ll (B) (b) FIGURE 3-15 Examples ofsinusoidal driven oscillatory motion with damping. The steady-state solution xp,transient solution x,,and sum xareshown in (a)fordriving frequency mgreater than thedamping frequency ml(m >ml)and in(b)form <ml. Resonance Phenomena Toﬁnd theangular frequency ml;atwhich theamplitude D(Equation 3.59) isa maximum (i.e., theamplitude resonance frequency), weset dDm —_; 0 dw m=mll Performing thedifferentiation, weﬁnd mllIVmg I2B2 (3.63) Thus, theresonance frequency mRislowered asthedamping coefﬁcient Bisin- creased. Noresonance occurs ifB>mo/2, forthen ml;isimaginary and Dde- creases monotonically with increasing m. Wemaynowcompare theoscillation frequencies forthevarious cases wehave considered: 1.Free oscillations, nodamping (Equation 3.4): kmgIE 2.Free oscillations, damping (Equation 3.38): wi=wt?IB2 3.6 SINUSOIDAL DRIVING FORCES 121 3.Driven oscillations, damping (Equation 3.63): wi=w3"2W andwenote thatmo>ml>mll. Wecustomarily describe thedegree ofdamping inanoscillating system in terms ofthe“quality factor” Qofthesystem: QE2 (3.64)23 Iflittle damping occurs, then Qisvery large and theshape oftheresonance curve approaches thatforanundamped oscillator. Buttheresonance canbecompletely destroyed ifthedamping islarge and Qisvery small. Figure 3-16 shows thereso- nance andphase curves forseveral different values ofQ.These curves indicate the lowering oftheresonance frequency with adecrease inQ(i.e., with anincrease of thedamping coefﬁcient B).The effect isnotlarge, however; thefrequency shift is lessthan 3%even forQassmall as3andisabout 18% forQI1. Foralightly damped oscillator, wecanshow (seeProblem 3-19) that “)0Q‘I’AI (3.65)(1) where Amrepresents thefrequency interval between thepoints ontheamplitude resonance curve thatare1/\/2 I0.707 ofthemaximum amplitude. Q ; ii‘.-/Q“Q=13 5 ______________ __ _ j Jr /I '_______ ___,’- I Q: ® Ir’ "______ _ | Q:,7 QZ13 // ”’,’_- _ g E Z I, _ Q=3 2- Q:1Q 0 I Q=3 Q=1 3Q=7 2 1QZO .1_t_Q:w_ rm-___ 1 1 I I 400 If -2 2) 1 _““u “("0 2) (3) (b) FIGURE 3-16 (a)The amplitude Disdisplayed asafunction ofthedriving frequency mforvarious values ofthequality factor Also shown is(b)thephase angle 5,which isthephase angle between thedriving force and the resultant motion._.;__ 122 3/OSCILLATIONS The values ofQfound inrealphysical situations vary greatly. Inrather ordi- nary mechanical systems (e.g., loudspeakers), thevalues may beintherange from afewto100orso.Quartz crystal oscillators ortuning forks may have Qsof104. Highly tuned electrical circuits, including resonant cavities, may have values of104 to105. Wemay also deﬁne Qsforsome atomic systems. According totheclassical picture, theoscillation ofelectrons within atoms leads tooptical radiation. The sharpness ofspectral lines islimited bythedamping due totheloss ofenergy byradiation (radiation damping). The minimum width ofalinecanbecalculated classically andis*AwE2><10'8w. The Qofsuch anoscillator istherefore ap- proximately 5><107. Resonances with thelargest known Qsoccur intheradia- tionfrom gaslasers. Measurements with such devices have yielded Qsofapproxi- mately 1014. Equation 3.63 gives thefrequency foramplitude resonance. Wenow calculate thefrequency forkinetic energy resonance—that is,thevalue ofwforwhich Tisa maximum. The kinetic energy isgiven byT=émkg, and computing icfrom Equation 3.60, wehave —A22= w sin(wt -—5) (3.66) \/(0)3-(1)2)? +4w2B2 sothatthekinetic energy becomes mA2 wgT: -aw "2 -5 am2 (0)3—-w2)2 +4w2B2 Sm(wt ) ( ) Toobtain avalue ofTindependent ofthetime, wecompute theaverage ofTover one complete period ofoscillation: m/12 co? , _<T>» 2-(mg-w2)2 (DB(s1n2(wt 5)> (3.68)__ +422 Theaverage value ofthesquare ofthesinefunction taken over oneperiod isl 211'/w1(sing(wt—-6))=1} sin2(wt ~5)dt =- (3.69)277' 0 2 Therefore, mA2 (1)2 (T)I4'(wg __w2)2 _|_40,232 (3-70) Thevalue ofwfor(T)amaximum islabeled wEandisobtained from d<T)dw —0 nJ=wEman *See Marion andHeald (M2180). TThe reader should prove theimportant result that theaverage over acomplete period ofsinzwtor cos2 wtisequal t0éz(sin2wt) =(cos2wt) = 3.7PHYSICAL SYSTEMS 123 Differentiating Equation 3.70 andequating theresult tozero, weﬁnd (DE = (U0 sothekinetic energy resonance occurs atthenatural frequency ofthesystem for undamped oscillations. Weseetherefore thattheamplitude resonance occurs atafrequency \/mg—232, whereas thekinetic energy resonance occurs atmo.Because thepotential energy is proportional tothesquare oftheamplitude, thepotential energy resonance must also occur at\/mg-2B2. That thekinetic and potential energies resonate atdif- ferent frequencies isaresult ofthefactthatthedamped oscillator isnotaconser- vative system. Energy iscontinually exchanged with thedriving mechanism, and energy isbeing transferred tothedamping medium. 3.7 Physical Systems Westated intheintroduction tothischapter that linear oscillations apply tomore systems thanjustthesmall oscillations ofthemass—spring andthesimple pendulum. The same mathematical formulation applies toawhole host ofphysical systems. Mechanical systems include thetorsion pendulum, vibrating string ormembrane, andelastic vibrations ofbars orplates. These systems may have overtones, and each overtone canbetreated much thesame aswedidintheprevious discussion. Wecanapply ourmechanical system analog toacoustic systems. Inthiscase, theairmolecules vibrate. Wecanhave resonances thatdepend ontheproperties and dimensions ofthemedium. Several factors cause thedamping, including fric- tionandsound-wave radiation. Thedriving force canbeatuning fork orvibrating string, among many sources ofsound. Atomic systems canalso berepresented classically aslinear oscillators. When light (consisting ofelectromagnetic radiation ofhigh frequency) fallsonmatter, it causes theatoms andmolecules tovibrate. \/Vhen light having oneoftheresonant frequencies oftheatomic ormolecular system falls onthematerial, electromag- netic energy isabsorbed, causing theatoms ormolecules tooscillate with large amplitude. Large electromagnetic ﬁelds ofthesame frequency areproduced by theoscillating electric charges. Wave mechanics (orquantum mechanics) useslin- earoscillator theory toexplain many ofthephenomena associated with light ab- sorption, dispersion, and radiation. Even todescribe nuclei, linear oscillator theory isused. One ofthemodes of excitation ofnuclei iscollective excitation. Neutrons andprotons vibrate invari- ouscollective motions. Resonances occur, and damping exists. The classical me- chanical analog isvery useful indescribing themotion. Electrical circuits are,however, themost noted examples ofnonmechanical oscillations. Indeed, because ofitsgreat practical importance, theelectrical example hasbeen sothoroughly investigated thatthesituation isfrequently reversed, and mechanical vibrations areanalyzed interms ofthe“equivalent electrical circuit.” Wedevote twoexamples toelectrical circuits. 124 3/OSCILLATIONS l*IXAl\"l PLE 3.4 Find theequivalent electrical circuit forthehanging mass—spring shown inFigure 3-17a and determine thetime dependence ofthecharge qinthesystem. Solution. Letusﬁrstconsider theanalogous quantities inmechanical andelectri- calsystems. Theforce F(=mgin themechanical case) isanalogous totheemf5. The damping parameter bhastheelectrical analog resistance R,which isnot present inthisCase. Thedisplacement xhastheelectrical analog charge q.We show other quantities inTable 3-1.Ifweexamine Figure 3-17a, wehave 1/k—>C,m—>L,F—> 8,x—>q,and k—> I.Without theweight ofthemass, the equilibrium position would beatx=O;theaddition ofthegravitational force extends thespring byanamount h=mg/k and displaces theequilibrium position tox==h.Theequation ofmotion becomes m5Z+k(xPh)=0 (3.73) or mi+kx=kh with solution x(t)==h+Acos wot (3.74) where wehave chosen theinitial conditions x(t=0)=h+Aand:'c(t=0)=0. Wedraw theequivalent electrical circuit inFigure 3-17b. Kirchoff’ sequation around thecircuit becomes 411 ql—-1==— . Ldt+Ci at5C (375) h L fill|Ct my: F=mg (a) (b) FIGURE 3-17 Example 3.4(a)hanging mass-spring system; (b)equivalent electrical circuit. 3.7PHYSICAL SYSTEMS 125 TABLE 3-1 Analogous Mechanical andElectri tities calQuan Mechanical Electrical Displacement Charge Velocity =I Current Mass Inductance Damping resistance Resistance Mechanical compliance Capacitance Amplitude ofimpressed force Amplitude ofimpressed emf >§§w-3x-x <‘v:Q>;[-e-a where qlrepresents thecharge that must beapplied toCtoproduce avoltage 5. IfweuseI=tj,wehave ..qqLq+E=E‘ (3.76) Ifq=qoand I=0att=O,thesolution is q(t)=ql+(qo~—ql)costoot (3.77) which istheexact electrical analog ofEquation 3.74. Consider theseries RLC circuit shown inFigure 3-18 driven byanalternating emf ofvalue E0sinwt.Find thecurrent, thevoltage VLacross theinductor, and theangular frequency watwhich I/)4isamaximum. Solution. Thevoltage across each ofthecircuit elements inFigure 3-18 are dI V:L—=L"L dt q V LI Ldq L' R‘ *dt“ ‘I 9'Vc Z E sothevoltage drops around thecircuit become .. .‘I_ .Lq+ Rq+ E— E0sinwt L R E0sincot C FIGURE 3-18 Example 3.5.RLC circuit with analternating emf. 126 3/OSCILLATIONS Weidentify thisequation assimilar toEquation 3.53, which wehave already solved. Inaddition totherelationships inTable 3-1,wealso have B=b/2m —>R/2L, too=\/k/m—>1/\/EC, andA=F0/m—>E0/L. Thesolution forthecharge qis given bytranscribing Equation 3.60, andtheequation forthecurrent Iisgiven by transcribing Equation 3.66, which allows ustowrite __E0 1=~—————--——— sin(wt —5) wC_- where 5canbefound bytranscribing Equation 3.61. Thevoltage across theinductor isfound from thetime derivative ofthecurrent. dl _Q)LEO VL=L ==* — —cos(wt —~6)dl 1 2 2_|_ m _MlR <0‘: (UL) =V(to) cos(tutP5) Toﬁnd thedriving frequency 0)max,which makes VLamaximum, wemust take the derivative ofVLwith respect towand settheresult equal tozero. Weonly need to consider theamplitude V(w) and notthetime dependence. LEO <12? ‘_'21: + d‘/(ml _ C toC do) T 1 23/QR2+(—— —wL)wC Wehave skipped afewintermediate steps toarrive atthisresult. Wedetermine thevalue wmax sought bysetting theterm inparentheses inthenumerator equal tozero. Bydoing soand solving forwmx gives ____l__._ LC-—-2 which istheresult weneed. Note thedifference between thisfrequency and those given bythenatural frequency, too=1/\/LC,andthecharge resonance frequency (given bytranscribing Equation 3.63), toR=V1/LC —2R2/L2.im _ I-1 L 7 i l bx Z 7 i Ii l I-1 l_ -I Ii I 3.8 Principle ofSuperposition--Fourier Series Theoscillations wehave been discussing obey adifferential equation oftheform 0:2 d _Z1?+.1-it+bx(t)—Acoswt (3.78) 3.8 PRINCIPLE OFSUPERPOSITION—FOURIER SERIES 127 The quantity inparentheses ontheleft-hand side isalinear operator, which we may represent byL.Ifwegeneralize thetime-dependent forcing function onthe right-hand side, wecanwrite theequation ofmotion as |.x(t) =F(t) (3.79) Animportant property oflinear operators isthatthey obey theprinciple ofsuper- position. This property results from thefactthatlinear operators aredistributive, that is, |.(x1 +x2)=|..(x1) +L(x2) (3.80) Therefore, ifwehave twosolutions, x1(t) andx2(t), fortwodifferent forcing func- tions, F1(t)andF2(t), Lxl=F1(t), Lxg=F2(t) (3.81) wecanadd these equations (multiplied byarbitrary constants 011andoz2)and obtain l-(011941 +0129(2) :0l1F1(t) +012F2(t) (3-82) Wecanextend thisargument toasetofsolutions x,,(t), each ofwhich isappropri- ateforagiven F,,(t): N N . L2110z,,x,,(t)) =§1a,,F,,(t) (3.83) This equation isjust Equation 3.79 ifweidentify thelinear combinations as N xv)=,§1a..x..<¢> Fa)=§1a.r..<¢> Ifeach oftheindividual functions F,,(t)hasasimple harmonic dependence on time, such ascoscont, weknow that thecorresponding solution x,,(t) isgiven by Equation 3.60. Thus, ifF(t)hastheform F(t)=201,,cos(w,,t —¢,,) (3.85) thesteady-state solution is —1E Q“ -—8 386 x(t) —m n COS(wnt ¢n 11) (' ) where 8,,=tan-1 (3.87) Wecanwrite down similar solutions where F(t) isrepresented byaseries of terms, sin(wnt —¢,,). Wetherefore arrive attheimportant conclusion that ifsome arbitrary forcing function F(t)canbeexpressed asaseries (ﬁnite orinﬁnite) of 128 3/OSCILLATIONS harmonic terms, thecomplete solution canalsobewritten asasimilar series of harmonic terms. This isanextremely useful result, because, according toFourier’s theorem, anyarbitrary periodic function (subject tocertain conditions that are notveryrestrictive) canberepresented byaseries ofharmonic terms. Thus, inthe usual physical case inwhich F(t)isperiodic with period 1'3217/co, F(t+1')=F(t) (3.88) wethen have OO 1F(t) =500 +;1(a,, cosnwt +bnsinnwt) (3.89) where 1' an= F(t')cos ntot'dt' 2°, (3.90) b,,=—(F(t')sin nwt'dt’ T0 or,because F(t)hasaperiod 1',wecanreplace theintegral limits 0and1-bythe limits -%1' =~17/to and +%1' =+'rr/w: w +11’/w an= F(t')cos ntot'dt’ w <3-91>b,,=—J F(t')sin nwt'dt' Tr -7|’/w Before wediscuss theresponse ofdamped systems toarbitrary forcing func- tions (inthefollowing section), wegiveanexample oftheFourier representation ofperiodic functions. l~§XAl\"l PLE 3.6 _ - _ _ _ - Asawtooth driving force function isshown inFigure 3-19. Find thecoefﬁcients anandb,,,andexpress F(t)asaFourier series. Solution. Inthiscase, F(t)isanoddfunction, F(- t)=-F(t) ,and isexpressed by AF(t) ==A-i =&t,-1'/2 <t<1'/2 (3.92)1' 217' F(¢) A/2 If —/I/2 |<—1—>l FIGURE 3-19 Example 3.6.Asawtooth driving force function. 3.9 THE RESPONSE OFLINEAR OSCILLATORS 129 Because F(t)isodd, thecoefﬁcients anallvanish identically. The b,,aregiven by w2A +7’/°’bn=——§ t'sin mot’ dt' / 277 -1rto _“of/1 t'cos mot’ +sinmot’ Hr/"J 2112 nw nit»? —1r/w w2A 211' A='§;§'$'(-'1)"+1=;;("1)"+1 (3.93) where theterm (—1)"*1 takes account ofthefactthat +1, nodd-—cos nrr= (3.94)W1, neven Therefore wehave =— snw——s1nw —1nw——--- . F(t) A't1'2t+1s'3t (395)Tr1 2 3 Figure 3-20 shows theresults fortwoterms, ﬁveterms, andeight terms of thisexpansion. The convergence toward thesawtooth function isnone too rapid. Weshould note twofeatures oftheexpansion. Atthepoints ofdiscontinu- ity(tIii-/2) theseries yields themean value (zero), andintheregion imme- diately adjacent tothepoints ofdiscontinuity, theexpansion “overshoots” the original function. This latter effect, known astheGibbs phenomenon,* occurs inallorders ofapproximation. The Gibbs overshoot amounts toabout 9%on each side ofanydiscontinuity, even inthelimit ofaninﬁnite series. In_ —.._ n._ I1. l_ ii i Q 3.9 The Response ofLinear Oscillators toImpulsive Forcing Functions (Optional) Intheprevious discussions, wehave mainly considered steady-state oscillations. For many types ofphysical problems (particularly those involving oscillating electrical circuits), thetransient effects arequite important. Indeed, thetran- sient solution may beofdominating interest insuch cases. Inthissection, wein- vestigate thetransient behavior ofalinear oscillator subjected toadriving force that actsdiscontinuously. Ofcourse, a“discontinuous” force isanidealization, because italways takes aﬁnite time toapply aforce. Butiftheapplication time is small compared with thenatural period oftheoscillator, theresult oftheideal caseisaclose approximation totheactual physical situation. *]osiah Willard Gibbs (1839-1903) discovered thiseffect empirically in1898. Adetailed discussion is given, forexample, byDavis (D2163, pp.113-118). The amount ofovershoot isactually 8.9490 --*%. 130 3/OSCILLATIONS \\\\\\\\\\\ _____________\ ___7__--—I%__-___\I I I 2terms 5temrsI I I I I I I I I I I ‘________\\\\\\\\ ‘_______\\\\ -.1;-u-—-I?-___-_-I 8terms I I T___"__\I FIGURE 3-20 Results ofExample 3.6.Fourier series representation ofsawtooth driving force function. The differential equation describing themotion ofadamped oscillator is .. . F0)x-1" -1-(1)396 Z7 (3.96) Thegeneral solution iscomposed ofthecomplementary andparticular solutions: x(t)=x,(t) +x‘,,(t) (3.97) Wecanwrite thecomplementary solution as x,(t) =e_B'(A1 coswlt+A2sinwlt) (3.98) where wlE\/.33 -B2 (3.99) Theparticular solution x[,(t) depends onthenature oftheforcing function F(t). Two types ofidealized discontinuous forcing functions areofconsiderable in- terest. These arethestep function (orHeaviside function) andtheimpulse func- tion, shown inFigures 3-21a andb,respectively. The stepfunction Hisgiven by t<z00,H(t0) —-{(1, t>to (3.100) 3.9THERESPONSE orLINEAR OSCILLATORS 131 F(l) F(l)m Tn HU0) IU0-I1)a a t t to to I1 (H) (b) FIGURE 3-21 (a)Step function; (b)impulse function. where aisaconstant with thedimensions ofacceleration and where theargu- ment toindicates thatthetime ofapplication oftheforce ist=to. The impulse function Iisapositive step function applied att=to,followed byanegative step function applied atsome later time t1.Thus I00,ti)=H00) "H01) 0,z<to I(t0. $1):ll» to<t< '71 (3101) 0,¢>:1 Although wewrite theHeaviside and impulse functions asH(to)and I(to,t1)for simplicity, these functions depend onthetime tand aremore properly written asH(t; to)and I(t;to,t1). Response toaStep Function Forstepfunctions, thedifferential equation thatdescribes themotion fort>tois 52+2[-3:2 +wgx Ia, t> to (3.102) Weconsider theinitial conditions tobex(t0) ==0and 5c(t0) =0.The particular solution isjust aconstant, and examination ofEquation 3.102 shows that itmust bea/tug. Thus, thegeneral solution fort>tois x(t)=e'B<“‘@)[A1 COS(1)1(i -to)+A2sinw1(t—— ¢.,)]+é(3.103)0 Applying theinitial conditions yields A,=-1,, A2=—ﬂ; (3.104)(U0 (t)l(O0 Therefore, fort>to,wehave a 36-80-10) x(t) =—2 1-e_B(“‘°)cos w1(t—- to)-ti sinw1(t- to) (3.105)(U0 (1)1 andx(t)='-0fort<to. 132 3/OSCILLATIONS X0) -“ ,,~," ,-_\N ,/ ______‘\-_\ ,—’/ ,,_-'___,_, -,,\,)0=0.2010 )3=02(Z/CD3 I\\ Ia‘/ ’,\\ \ \ ..~—, ../..g-- _-- __,--”\ ,-\_- 2t__ r‘~_7-."_ 1,\- Pb 0 __ FIGURE 3-22 Response function solution tothestepforce function. If,forsimplicity, wetake t0=0,thesolution canbeexpressed as H0 '3‘ x(t)=—Q 1We“B‘cos w1tW EL sinwlt (3.106)010 (01 This response function isshown inFigure 3-22forthecaseB=0.2to0. Itshould be clear that theultimate condition oftheoscillator (i.e., thesteady-state condi- tion) isSimply adisplacement byanamount a/(1)0. Ifnodamping occurs, B=0andwl=to0.Then, fort0=0,wehave _Hm) _ 07(7)-7n —cosco0t], BW0 (3.107)0 The oscillation isthus sinusoidal with amplitude extremes x=0andx=2a/tog (seeFigure 3-22). Response toanImpulse Function Ifweconsider theimpulse function asthedifference between twostep functions searated byatime t1Wt0=7',then, because thesystem islinear, thegeneral so P lution fort>t1isgiven bythesuperposition ofthesolutions (Equation 3.105) forthetwostepfunctions taken individually: 0 _v 30-B<t—v.> _ x(t)=—2|i1 WeB”‘°)cos w1(tW t0)WWi sintu1(t Wt0) a e_B(t_ t0_T) ' W—2|:1 We'B("‘"-T) costo1(t Wt0W1-)WL sinw1(t Wt0W7-)(U0 (1)1 aAg'B('7‘ Io) =T eta’cosco1(t Wt0W7')Wcosw1(t Wt0) 0 BB“ . B.+Z sinto1(t Wt0W7')W—sinto1(t Wt0), t> t1 (3.108(1)1 (1)1) 3.9THERESPONSE orLINEAR OSCILLATORS 133 The totalresponse (i.e., Equations 3.105 and3.108) toanimpulse function of duration 7-=5><277/ml applied att=t0isshown inFigure 3-23 forB=0.20:0. Ifweallow theduration 7-oftheimpulse function toapproach zero, there- sponse function willbecome vanishingly small. Butifweallow a—>ooas1-—>0so that theproduct a7-isconstant, then theresponse willbeﬁnite. This particular limiting case isconsiderably important, because itapproximates theapplication ofadriving force that isa“spike” att=t0(i.e., 7'<<277'/w1).* Wewant toex- pand Equation 3.108 byletting 7-—>0, butwith b=a1-=constant. LetA=tWt0 andB=t,then useEquations D.11 andD.12 (from Appendix D)toobtain ag_B(3* I0) x(t)=TT{eB’[cos w1(t Wt0)cos7011-+sinw1(t Wt0)sin0117-] Bet- Wcosw1(t Wt0)+?1—[sin w1(t Wt0)cos0117-Wcosw1(t Wt0)sin0111-] B.—JSln(U1(t"_ to) , to 1 x(t) ﬂ= _1___ _____ __ __ __ 5."‘*AW-) .5‘-I----_-_ t 1=5(27:/col) FIGURE 3-23 Response function solution totheimpulse force function. *A“spike” ofthistype isusually termed adelta function andiswritten 6(t—t0).Thedelta function hastheproperty that 5(1) =0fort=#0and 5(0) =0°,but +00 I8(tWt0)dt =1 “O0 This istherefore notaproper function inthemathematical sense, butitcanbedeﬁned asthelimit ofawell-behaved and highly local function (such asaGaussian function) asthewidth parameter approaches zero. SeealsoMarion andHeald (Ma80, Section 1.11). 134 3/OSCILLATIONS Because 7-issmall, wecanexpand e5",cos0117-, andsin0117-using Equations D.34, D.29, and D.28, keeping only theﬁrst two terms ineach. After multiplying out alltheterms containing 7-,wekeep only thelowest-order term of7-. ae-B(l_ tn) . 21' 8(7)=-—.,~—s1nw,(t—t0) 00,1+L,¢>to(U0 (U| Using Equation 3.99 for010and 7-=b/agives us,ﬁnally, b x(t) =;)W1e'B("t“) sin0)1(t Wt0), t>t0 (3.110) This response function isshown inFigure 3-24 forthecase B=0.2030. Notice that, astbecomes large, theoscillator returns toitsoriginal position of equilibrium. The factthattheresponse ofalinear oscillator toanimpulsive driving force canberepresented inthesimple manner ofEquation 3.110 leads toapowerful technique fordealing with general forcing functions, which was developed by Green.* Green’s method isbased onrepresenting anarbitrary forcing function asaseries ofimpulses, shown schematically inFigure 3-25. Ifthedriven system is linear, theprinciple ofsuperposition isvalid, and wecanexpress theinhomoge- neous part ofthedifferential equation asthesum ofindividual forcing functions F,,(t)/m, which inGreen’s method areimpulse functions: EX] 1:1” TX) . at+2357+030= 5;}?= l,,(z) (3.111) 3(1) I - \_/WW . -1 '0 FIGURE 3-24 Response function solution toaspike (ordelta functiori) force function. *(§t-urge (Z1-t-en (1793-1841), aself-educated English llrzrtllt-rrurticiztrr. 3.9 THE RESPONSE OFLINEAR OSCILLATORS 135 Em Fn(t)/m F(i)/m l l l /4 “Ct tn tn+1 FIGURE 3-25 Anarbitrary force function canberepresented asaseries ofimpulses, amethod known asGreen’s methods. where Ina) :I(tm tn-I-1) a'n(tn)a tn< t< t+1 = ” 3.11 {On <2)Otherwise The interval oftime over which Inacts istn+1 Wtn==7-,and 7-<<27-r/0:1. The so- lution forthenthimpulse is,according toEquation 3.110, Tl tn . xn(t) =%”_e'B("t")s1n 0)1(t Win), t>tn+7- (3.113) 1 and thesolution foralltheimpulses uptoand including theNth impulse is N@000)?‘ 1_.x(t) ==-“_2°oTe B“t")s1n0)1(t““ in), tN< t< tN+1 (3.114) “W” 1 Ifweallow theinterval 7-toapproach zero and write tnast’,then thesum be- comes anintegral: Ia(t') , 0(x = WWWe'B("‘ )sin w1(t Wt’)dt' (3.115) —<><>‘"1 136 3/OSCILLATIONS Wedeﬁne l .me‘/3("')sin 0)1(tW t'), t2t'mall G(t,t’)E0 t<t, (3.116) Then, because ma(t’) =F(t') (3.117) wehave x(t)= F(t')G(t, t’)dt’ (3.118) The function G(t,t’)isknown astheGreen’s function forthelinear oscillator equation (Equation 3.96). The solution expressed byEquation 3.118 isvalid only foranoscillator initially atrestinitsequilibrium position, because thesolu- tionweused forasingle impulse (Equation 3.110) wasobtained forust such an initial condition. For other initial conditions, thegeneral solution may beob- tained inananalogous manner. Green’s method isgenerally useful forsolving linear, inhomogeneous differ- ential equations. The main advantage ofthemethod lies inthefact that the Green’s function G(t,t’),which isthesolution oftheequation foraninﬁnitesi- malelement oftheinhomogeneous part, already contains theinitial c0nditi0ns—so thegeneral solution, expressed bytheintegral ofF(t') G(t,t’),automatically also contains theinitial conditions. EXAMPLE 3.7 Find x(t)foranexponentially decaying forcing function beginning att=0and having thefollowing form fort>0: F(x)=F0e-Y‘, t>0 (3.119) Solution. The solution forx(t)according toGreen’s method is F0 t ’ '- I Ix(t) =-W— e_*"e_B(‘_‘ )s1n to1(t Wt)dt (3.120) mwl 0 Making achange ofvariable toz=011(t Wt’),weﬁnd F0 0_ .x(t) =Win e"'e[(l’*5)/"’1]‘s1nz dzml] F/m __ 7WB. = [e ""We 3‘(cos 011tW T1 sin011t):| (3.121) 3.9 THE RESPONSE OFLINEAR OSCILLATORS 137 x(t)F0/m_2 2(75)+@1 ﬂzonmo y=0.30J0 t (y_ﬁ)2+ co? ﬂ:0.2w0 y=0.2c00 t F0/m xv) (2/_ﬂ)2+ (012 ﬂ:0_3m0 y=0.lC00 l I 1 I 4|‘. 1 I I t FIGURE 3-26 Response function forExample 3.7. This response function isillustrated inFigure 3-26 forthree different combi- nations ofthedamping parameters Band -y.When -yislarge compared With B,and ifboth aresmall compared with 010,then theresponse approaches that fora“spike”; compare Figure 3-24 with theupper curve inFigure 3-26. When ‘yissmall compared with B,theresponse approaches theshape oftheforcing function itself—that is,aninitial increase followed byanexponential decay. The lower curve inFigure 3-26 shows adecaying amplitude onwhich issu- perimposed aresidual oscillation. When Band-yareequal, Equation 3.121 becomes F x(t) =It-1W2 e”B"(l Wcoswlt), B=y (3.122) 1 Thus, theresponse isoscillatory with a“period” equal to27.-/011 butwith an exponentially decaying amplitude, asshown inthemiddle curve ofFigure 3-26. Aresponse ofthetype given byEquation 3.121 could result, forexample, if aquiescent butintrinsically oscillatory electronic circuit were suddenly driven bythedecaying voltage onacapacitor. 3/OSCILI ATIONS PROBLEMS Asimple harmonic oscillator consists ofa100-g mass attached toaspring whose force constant is104dyne/cm. The mass isdisplaced 3cmand released from rest. Calculate (a)thenatural frequency 1/0andtheperiod 7'0,(b)thetotal energy, and (c)themaximum speed. Allow themotion inthepreceding problem totake place inaresisting medium. After oscillating for10s,themaximum amplitude decreases tohalf theinitial value. Calculate (a)thedamping parameter B,(b)thefrequency 1/1(compare with theundamped frequency v0),and(c)thedecrement ofthemotion. The oscillator ofProblem 3-1issetinto motion bygiving itaninitial velocity of 1cm/s atitsequilibrium position. Calculate (a)themaximum displacement and (b)themaximum potential energy. Consider asimple harmonic oscillator. Calculate thetimeaverages ofthekinetic and potential energies over one cycle, and show that these quantities areequal. I/Vhy isthisareasonable result? Next calculate thespace averages ofthekinetic and potential energies. Discuss theresults. Obtain anexpression forthefraction ofacomplete period that asimple harmonic oscillator spends Within asmall interval Axataposition x.Sketch curves ofthis function versus xforseveral different amplitudes. Discuss thephysical signiﬁcance oftheresults. Comment ontheareas under thevarious curves. Two masses ml=100gand m2=200gslide freely inahorizontal frictionless track and areconnected byaspring whose force constant isk=0.5N/m. Find thefre- quency ofoscillatory motion forthissystem. Abody ofuniform cross-sectional area A=1cm? and ofmass density p=0.8 g/cm?’ ﬂoats inaliquid ofdensity p0=1g/cm3 andatequilibrium displaces avol- ume V= 0.8cm3. Show that theperiod ofsmall oscillations about theequilibrium position isgiven by 7'=277'\/Wg./1 where gisthegravitational ﬁeld strength. Determine thevalue of7-. Apendulum issuspended from thecusp ofaCyc1oid* cutinarigid support (Figure 3-A). Thepath described bythependulum bobiscycloidal andisgiven by x=a(¢Wsin¢), y=a(cos¢W1) where thelength ofthependulum isl=4a,andwhere <12istheangle ofrotation ofthecircle generating thecycloid. Show thattheoscillations areexactly isochro- nous with afrequency 010=\/g/l,independent oftheamplitude. *The reader unfamiliar with theproperties ofcycloids should consult atext onanalytic geometry. PROBLEMS 139 3-9. 3-10. 3-11. 3-12. 3-13 3-14 3-15 3-16_______?________._,tnn gla L ‘#7 FIGURE 3-A Problem 3-8. Aparticle ofmass misatrestattheendofaspring (force constant =k)hanging from aﬁxed support. Att=0,aconstant downward force Fisapplied tothemass and acts foratime t0.Show that, after theforce isremoved, thedisplacement ofthe mass from itsequilibrium position (x=x0,where xisdown) is F xWx0=E[cos0)0(t Wt0)Wcosw0t] where (00=k/m. Iftheamplitude ofadamped oscillator decreases to1/eofitsinitial value after nperiods, show that the frequency ofthe oscillator must beapproximately [1W(87'r2n2)"1] times thefrequency ofthecorresponding undamped oscillator. Derive theexpressions fortheenergy and energy-loss curves shown inFigure 3-8 forthedamped oscillator. Foralightly damped oscillator, calculate theaverage rate atwhich thedamped oscillator loses energy (i.e., compute atime average over one cycle). Asimple pendulum consists ofamass msuspended from aﬁxed point byaweight- less, extensionless rod oflength l.Obtain theequation ofmotion and, inthe approximation thatsin6E6,show that thenatural frequency is010=\/g7l, where g isthegravitational ﬁeld strength. Discuss themotion intheeyent that themotion takes place inaviscous medium with retarding force 2m\/gfl 6. Show that Equation 3.43 isindeed thesolution forcritical damping byassuming a solution oftheform x(t) =y(t)exp(WBt) and determining thefunction y(t). Express thedisplacement x(t)andthevelocity a2(t)fortheoverdamped oscillator in terms ofhyperbolic functions. Reproduce Figures 3-10b and cforthesame values given inExample 3.2,but instead letB=0.1s-1and5=7-rrad.How many times does thesystem cross thex= 0line before theamplitude ﬁnally falls below 1072 ofitsmaximum value? Which plot, borc,ismore useful fordetermining thisnumber? Explain. Discuss themotion ofaparticle described byEquation 3.34 intheevent that b<0 (i.e., thedamping resistance isnegative). 140 3-17. 3-18. 3-19. 3-20 3-21 3-22. 3-23 3-24. 3-25.3/OSCILLATIONS Foradamped, driven oscillator, show thattheaverage kinetic energy isthesame at afrequency ofagiven number ofoctaves* above thekinetic energy resonance asat E1frequency ofthesame number ofoctaves below resonance. Show that, ifadriven oscillator isonly lightly damped anddriven near resonance, theQof thesystem isapproximately QE277XQlinergy lossduring oneperiod)Touﬂ energy Foralightly damped oscillator, show thatQE010/A0) (Equation 3.65). Plotavelocity resonance curve foradriven, damped oscillator with Q=6,andshow thatthefullwidth ofthecurve between thepoints corresponding tokm“/\/2 isap- proximately equal to0:0/6. Useacomputer toproduce aphase space diagram similar toFigure 3-11 forthe case ofcritical damping. Show analytically that theequation oftheline that the phase paths approach asymptotically is92:=WBx. Show thephase paths foratleast three initial positions above andbelow theline. Lettheinitial position and speed ofanoverdamped, nondriven oscillator bex0and v0,respectively. (a)Show thatthevalues oftheamplitudes A1andA2inEquation 3.44have thevalues A1= E-gﬁiﬁ andA2 =W whereB1= BW0)2andB2 =B+012. B2_B1 B2“B1 (b)Show that when A1=0,the phase paths ofFigure 311 must bealong the dashed curve given byair=WB2x, otherwise theasymptotic paths arealong the other dashed curve given by0?:=WB1x. Hint: Note that B2>B1and ﬁnd the asymptotic paths when t—>oo. Tobetter understand underdamped motion, useacomputer toplot x(t)ofEquation 3.40 (with A=1m)anditstwocomponents [e_F3‘ and cos(mlt W5)]andcompar- isons (with B=0)onthesame plot asinFigure 3-6.Let0:0=1rad/ sand make sep- arate plots forB2/020 =0.1,0.5,and 0.9andfor5(inradians) =0,77'/2, and 77'.Have only onevalue of5andBoneach plot (i.e., nine plots). Discuss theresults. ForB=0.2s71, produce computer plots likethose shown inFigure 3-15 forasinu- soidal driven, damped oscillator where xn(t), x,(t), andthesum x(t)areshown. Let k=1kg/s2 and m=1kg.D0 thisforvalues ofa)/0)1of1/9,1/3,1.1, 3,and 6.For thex,(t) solution (Equation 3.40), letthephase angle 5=0and theamplitude A=W1m.For thexn(t) solution (Equation 3.60), letA=1m/s2 butcalculate 5. What doyou observe about therelative amplitudes ofthetwo solutions as0)in- creases? Why does thisoccur? For0)/0:1 =6,letA=20m/s2 forxn(t) and produce theplotagain. Forvalues ofB==1s71,k=1kg/s2,and m=1kg,produce computer plots likethose shown inFigure 3-15 forasinusoidal driven, damped oscillator where xn(t), x,(t), *Anoctave isafrequency interval inwhich thehighest frequency isjusttwice thelowest frequency. PROBLEMS 141 3-26 3-27. 3-28. 3-29. 3-30. 3-31.andthesum x(t)areshown. D0thisforvalues ofa)/col, of1/9, 1/3, 1.1,3,and6.For thecritically damped xE(t) solution ofEquation 3.43, letA=—1mand B=1m/s. Forthexl,(t) solution ofEquation 3.60, letA=1m/s2 andcalculate 5.What doyou observe about therelative amplitudes ofthetwosolutions ascoincreases? Why does thisoccur? Forw/wo =6,letA=20m/s2 forxl,(t) and produce theplot again. Figure 3-Billustrates amass mldriven byasinusoidal force whose frequency isw. The mass mlisattached toarigid support byaspring offorce constant kand slides onasecond mass mg.Thefrictional force between mlandm2isrepresented bythe damping parameter bl,and thefrictional force between m2and thesupport isrep- resented bybg.Construct theelectrical analog ofthissystem and calculate the impedance. “’ b »-\ l k FIGURE 3-B Problem 3-26. Show that theFourier series ofEquation 3.89 canbeexpressed as 1 O0 F(t)=5all+Z1c,lcos(nwt —<15") Relate thecoefficients c,ltotheanandbnofEquation 3.90. Obtain theFourier expansion ofthefunction _-1, ~11‘/w<t<O F(t)_{+1, 0<t<11'/cu intheinterval -11"/to <t<11'/w. Take cu=1rad/s. Intheperiodical interval, cal- culate andplot thesums oftheﬁrst twoterms, theﬁrst three terms, andtheﬁrst four terms todemonstrate theconvergence oftheseries. Obtain theFourier series representing thefunction _O, ~2rr/w <t<O Fa) _{sinwg 0<t< 211/to Obtain theFourier representation oftheoutput ofafull-wave rectiﬁer. Plottheﬁrst three terms oftheexpansion and compare with theexact function. Adamped linear oscillator, originally atrestinitsequilibrium position, issubjected toaforcing function given by O t<0 t 7 @= a><(t/1'), 0<t<1'ma, t>1' Find theresponse function. Allow 1'—>Oand show that thesolution becomes that forastep function. 142 3-32 3-33 3-34 3-35. 3-36 3-37 3-38 3-39. 3-40 3-413/OSCILLATIONS Obtain theresponse ofalinear oscillator toastepfunction andtoanimpulse func- tion (inthelimit ‘T-—>O)foroverdamping. Sketch theresponse functions. Calculate themaximum values oftheamplitudes oftheresponse functions shown inFigures 3-22 and3-24. Obtain numerical values forB=0.2100 when a=2m/s2, mo=1rad/s, and to=0. Consider anundamped linear oscillator with anatural frequency mo=0.5rad/s and thestep function a=1m/s2. Calculate and sketch theresponse function for animpulse forcing function acting foratime 1'=211'/mo. Give aphysical interpre- tation oftheresults. Obtain theresponse ofalinear oscillator totheforcing function O, t< O F0) .7= aslnwt, O<t<77/0) 0, t> 11/to Derive anexpression forthedisplacement ofalinear oscillator analogous to Equation 3.110 butfortheinitial conditions x(t0) =x0and a2(to)=9&0. Derive theGreen’s method solution fortheresponse caused byanarbitrary forcing function. Consider thefunction toconsist ofaseries ofstep functions--—that is,start from Equation 3.105 rather than from Equation 3.110. Use Green’s method toobtain theresponse ofadamped oscillator toaforcing function oftheform 0 t<0F(t) ={ F0e"‘Y‘ sinwt t>0 Consider theperiodic function Fm_sinwt, 0<t<11'/w 0, Tr/w <t<211'/w which represents thepositive portions ofasine function. (Such afunction repre- sents, forexample, theoutput ofahalf-wave rectifying circuit.) Find theFourier representation andplotthesum oftheﬁrstfour terms. Anautomobile with amass of1000 kg,including passengers, settles 1.0cmcloser to theroad forevery additional 100kgofpassengers. Itisdriven with aconstant hori- zontal component ofspeed 20krn/hover awashboard road with sinusoidal bumps. The amplitude and wavelength ofthesine curve are5.0cmand 20cm,respectively. The distance between thefront andback wheels is2.4m.Find theamplitude of oscillation oftheautomobile, assuming itmoves vertically asanundamped driven harmonic oscillator. Neglect themass ofthewheels andsprings andassume that thewheels arealways incontact with theroad. (a)Use thegeneral solutions x(t)tothedifferential equation d2x/dt2 +2Bdx/dt + wgx=Oforunderdamped, critically damped, andoverdamped motion andchoose theconstants ofintegration tosatisfy theinitial conditions x=x0andv=110=0at t=0.(b)Useacomputer toplot theresults forx(t)/x0 asafunction ofwotinthe PROBLEMS 143 3-42 3-43. 3-44 3-45.three cases B=(1/2)w0, ,81w0,andB=20:0. Show allthree curves onasingle plot. Anundamped driven harmonic oscillator satisﬁes theequation ofmotion m(d2a<;/dt2+ w02x) =F(t).The driving force F(t) =P0sin(wt) isswitched onatt=0.(a)Find x(t) fort>0fortheinitial conditions x=0and v=0att=0.(b)Find x(t)forw=w0 bytaking thelimit to—>w0 inyour result forpart (a).Sketch your result forx(t). Hint: Inpart (a)look foraparticular solution ofthedifferential equation ofthe form x=Asin(wt) and determine A.Add thesolution ofthehomogeneous equa- tion tothistoobtain thegeneral solution oftheinhomogeneous equation. Apoint mass mslides without friction onahorizontal table atone end ofa massless spring ofnatural length aandspring constant kasshown inFigure 3-C.Thespring isattached tothetable soitcanrotate freely without friction. The netforce onthe mass isthecentral force F(r) =—k(r —-a).(a)Find andsketch both thepotential energy U(r) and theeffective potential Uel.f(1"). (b)What angular velocity woisre- quired foracircular orbit with radius T0?(C)Derive thefrequency ofsmall oscillations toabout thecircular orbit with radius r0.Express your answers for(b)and(c)interms ofk,m,r0,and a. /%m~+ FIGURE 3-C Problem 3-43. Consider adamped harmonic oscillator. After four cycles theamplitude oftheos- cillator hasdropped to1/eofitsinitial value. Find theratio ofthefrequency ofthe damped oscillator toitsnatural frequency. Agrandfather clock hasapendulum length of0.7mand mass bob of0.4kg.A mass of2kgfalls0.8minseven days tokeep theamplitude (from equilibrium) of thependulum oscillation steady at0.03 rad. What istheQof thesystem? _____-_C?L”¥PTER Nonlinear Oscillations and Chaos 4.1Introduction The discussion ofoscillators inChapter 3was limited tolinear systems. When pressed todivulge greater detail, however, nature insists ofbeing nonlinear; ex- amples aretheﬂapping ofaﬂag inthewind, thedripping ofaleaky water faucet, andtheoscillations ofadouble pendulum. The techniques learned thus farforlinear systems may notbeuseful fornonlinear systems, butalarge num- beroftechniques have been developed fornonlinear systems, some ofwhich we address inthischapter. Weusenumerical techniques tosolve some ofthenon- linear equations inthischapter. The equation ofmotion forthedamped anddriven oscillator ofChapter 3 moving inonly one dimension canbewritten as msa+f(l2) +g(x)=h(t) (4.1) Iff(5c) org(x) contains powers ofisorx,respectively, higher than linear, then the physical system isnonlinear. Complete solutions are not always available for Equation 4.1,and sometimes special treatment isneeded tosolve such equa- tions. For example, wecan learn much about aphysical system byconsidering the deviation oftheforces from linearity and byexamining phase diagrams. Such asystem isthesimple plane pendulum, asystem that islinear only when small oscillations areassumed. Inthebeginning ofthenineteenth century, thefamous French mathemati- cian Pierre Simon deLaplace espoused theview thatifweknew theposition and velocities ofalltheparticles intheuniverse, then wewould know thefuture for alltime. This isthe deterministic view ofnature. Inrecent years, researchers in 144 4.1INTRODUCTION 145 many disciplines have come torealize that knowing thelaws ofnature isnot enough. Much ofnature seems tobechaotic. Inthis case, werefer todetermin- istic chaos, asopposed torandomness, tobethemotion ofasystem whose timeevo- lution hasasensitive dependence oninitial conditions. The deterministic develop- ment refers totheway asystem develops from one moment tothenext, where thepresent system depends ontheonejust past inawell-determined way through physical laws. Wearenotreferring toarandom process inwhich the present system hasnocausal connection totheprevious one (e.g., theﬂipping ofacoin). Measurements made onthestate ofasystem atagiven time may notallow us topredict thefuture situation even moderately farahead, despite thefactthat the governing equations areknown exactly. Deterministic chaos isalways associated with anonlinear system; nonlinearity isanecessary condition forchaos butnota sufﬁcient one. Chaos occurs when asystem depends inasensitive way onitspre- vious state. Even atinyeffect, such asabutterﬂy ﬂying nearby, may beenough to vary theconditions such that thefuture isentirely different than what itmight have been, notjustatinybitdifferent. Theadvent ofcomputers hasallowed chaos tobestudied because wenow have thecapability ofperforming calculations of thetime evolution oftheproperties ofasystem thatincludes these tinyvariations inthe initial conditions. Chaotic systems can only besolved numerically, and there arenosimple, general ways topredict when asystem willexhibit chaos. Chaotic phenomena have been uncovered inpractically allareas ofscience and eng-ineering—in irregular heartbeats; themotion ofplanets inoursolar sys- tem; water dripping from atap;electrical circuits; weather patterns; epidemics; changing populations ofinsects, birds, andanimals; andthemotion ofelectrons inatoms. The listgoes onand on.Henri Poincaré* isgenerally given credit for ﬁrst recognizing theexistence ofchaos during hisinvestigation ofcelestial me- chanics attheendofthenineteenth century. Hecame totherealization thatthe motion ofapparently simple systems, such astheplanets inoursolar system, can beextremely complicated. Although various investigators also eventually came tounderstand theexistence ofchaos, tremendous breakthroughs didnothap- pen until the1970s, when computers were readily available tocalculate thelong- time histories required todocument thebehavior. The study ofchaos hasbecome widespread, andwewillonly beable tolook atthe rudimentary aspects ofthe phenomena. Specialized textbooksi onthe subject have become abundant forthose desiring further study. Forexample, space does notpermit ustodiscuss thefascinating area offractals, thecompli- cated patterns that arise from chaotic processes. *Henri Poincare (1854-1912) was amathematician who could also beconsidered aphysicist and philosopher. Hiscareer spanned theerawhen classical mechanics wasatitsheight, soon tobeover- taken byrelativity and quantum mechanics. Hesearched forprecise mathematical formulas that would allow him tounderstand thedynamic stability ofsystems. lParticularly useful books arebyBaker and Gollub (Ba96), Moon (M092), Hilborn (Hi00), and Strogatz (St94). 146 4/NONLINEAR OSCILLATIONS AND CHAOS 4.2 Nonlinear Oscillations Consider apotential energy oftheparabolic form 1 U(x) =5kxg (4.2) Then thecorresponding force is F(x) =—kx (4.3) This isjust thecase ofsimple harmonic motion discussed inSection 3.2.Now, suppose aparticle moves inapotential well, which issome arbitrary function of distance (asinFigure 4-1). Then, inthevicinity oftheminimum ofthewell, we usually approximate thepotential with aparabola. Therefore, iftheenergy of theparticle isonly slightly greater than Umln, only small amplitudes arepossible and themotion isapproximately simple harmonic. Iftheenergy isappreciably greater than Umln, sothat theamplitude ofthemotion cannot beconsidered small, then itmay nolonger besufﬁciently accurate tomake theapproximation U(x) '-=ékxg andwemust deal with anonlinear force. Inmany physical situations, thedeviation oftheforce from linearity issym- metric about theequilibrium position (which wetake tobeatxI0).Insuch cases, themagnitude oftheforce exerted onaparticle isthesame at—xasatx; thedirection oftheforce isopposite inthetwocases. Therefore, inasymmetric situation, theﬁrst correction toalinear force must beaterm proportional tox3; hence, F(x) E—kx +ex?’ (4.4) where sisusually asmall quantity. The potential corresponding tosuch aforce is 1 1 U(x) =Ekxg —Zex4 (4.5) U(x) . \ I ' l‘ Parabolic -"T, »\_ I \ I \ I \ I\ / om,“t____________ __ y ' ' ' ' X FIGURE 4-1 Arbitrary potential U(x) indicating aparabolic region where simple harmonic motion isapplicable. 4.2NONLINEAR OSCILLATIONS 147 F(x) F(x) \\ ‘ \ \ \Linear Iv /‘x \ \ . \ Linear - - X X s>0 e<0 (Soft) \ (Hard) \ l \ \ \ \ \ \ \ \ U06) U00 Parabolic \ . / I I\ \ I\ I \ I\ I \ I\ I \ I I Parabolic _ x tix 0 0 FIGURE 4-2 Force F(x)andpotential U(x)forasoftandhard system when anx3 term isadded totheforce. Depending onthesign ofthequantity s,theforce may either begreater orless than thelinear approximation. Ifs>0,then theforce isless than thelinear term alone and thesystem issaid tobesoft; ifs<O,then theforce isgreater and thesystem ishard. Figure 4-2shows theform oftheforce andthepotential fora softand ahard system. EXAMPLE 4.1 _- -I ___ _ _- -_ _ Consider aparticle ofmass msuspended between twoidentical springs (Figure 4-3). Show that thesystem isnonlinear. Find thesteady-state solution foradriving force F0coswt. Solution. Ifboth springs areintheir unextended conditions (i.e., there isno tension, and therefore nopotential energy, ineither spring) when theparticle isinitsequilibrium position—and ifweneglect gravitational forces—then when theparticle isdisplaced from equilibrium (Figure 4-3b), each spring exerts a force —k(s —l)ontheparticle (kistheforce constant ofeach spring). The net (horizontal) force ontheparticle is F= —2k(s —l)sinI9 (4.6) Now, s= \/l2+x2 148 4/NONLINEAR OSCILLATIONS AND CHAOS l s m x m z 6‘ (a)Equilibrium position (b)Extended position FIGURE 4-3 Example 4.1.Adouble spring system in(a)equilibrium and (b)extended positions. so _ x xs1nB=—=i—?S \/[2_|_x2 Hence, _ Qkx A _A _ 1F——i—F_+_x.5- (V12 +x21)—2kx(1 (4.7) Ifweconsider x/ltobeasmall quantity and expand theradical, weﬁnd ~~<:>l1~i(€i+i Ifweneglect allterms except theleading term, wehave, approximately, ' F(x)E—(k/l2)x5 (4.3) Therefore, even iftheamplitude ofthemotion issufﬁciently restricted sothat x/lisasmall quantity, theforce isstillproportional tox3.The system isthere- fore intrinsically nonlinear. However, ifithadbeen necessary tostretch each spring adistance dtoattach ittothemass when attheequilibrium position, then wewould ﬁnd fortheforce (see Problem 4-1): F(x) E-2(kd/l)x —[k(l— d)/l5]x3 (4.9) and alinear term isintroduced. For oscillations with small amplitude, themo- tionisapproximately simple harmonic. From Equation 4.9weidentify 8'=—k(l— d)/Z3<0 4.2 NONLINEAR OSCILLATIONS 149 Thus thesystem ishard. Ifwe have adriving force F0coswt,theequation ofmotion forthe stretched spring (force ofEquation 4.9)becomes kd kldmii==-—2Tx —LEAK’ +F0coswt (4.10) Let ' 2kd Ft-:=5, a=*, and G=—0 (4.11)m ml m then 55=-*ax +ex?’+Gcos wt (4.12) Equation 4.12 isadifﬁcult differential equation tosolve. Wecanﬁnd theimpor- tant characteristics ofthesolution byamethod ofsuccessive approximations (perturbation technique). First, tryasolution xl==Acoswt,and insert xlinto theright-hand side ofEquation 4.12, which becomes 562=—~aA coswt+r:A3cos?’ wt+Gcoswt (4.13) where thesolution ofEquation 4.13 isx=x2.This equation canbesolved for x2using theidentity cos?’ wt==écoswt+1cos3tot4 4 Using thisequation inEquation 4.13 gives ~- 35 13x2=—-aA—Z614 —Gcoswt+;sA cos3wt (4.14) Integrating twice (with integration constants setequal tozero) gives 1 3 3A3 x2=E(aA —*Z3143 —-G)cos cot—g-6; cos3wt (4.15) This isalready acomplicated solution. Under what conditions fors,a,andxis x2asuitable solution? Numerical techniques with acomputer canquickly yield aperturbative solution quite accurately. Wehave found that theamplitude depends onthedriving frequency, butnoresonance occurs atthenatural frequency ofthesystem. Further discussion ofsolution methods forEquation 4.12 would take ustoo faraﬁeld ofourpresent discussion. The result isthatforsome values ofthe driving frequency co,three different amplitudes may occur with “jumps” be- tween theamplitudes. The amplitude may have adifferent value foragiven to depending onwhether toisincreasing ordecreasing (hysteresis). Wepresent a simple case ofthiseffect inSection 4.5. 150 4/NONLINEAR OSCILLATIONS ANDCHAOS F(x) Linear ‘ \\\/ \\\ l \ U(x) \ I \ I ‘x ‘ Q/Parabolic \ I \ I l \ I \\ \ I \ I l \ ,\ \ \ \ \ ‘ \ (Soft) (Hard) i‘ HZ 00TE Ex FIGURE 4-4 Example ofasymmetric forces andpotentials. Inreal physical situations, weareoften concerned With symmetric forces andpotentials. Butsome cases have asymmetric forms. Forexample, F(x) =-kx +/\x2 (4.16) Thepotential forwhich is 1 1 U(x) ==5kx2—gAx?’ (4.17) This case isillustrated inFigure 4-4forA<0;thesystem ishard forx>0and softforx<O. 4.3 Phase Diagrams forNonlinear Systems The construction ofaphase diagram foranonlinear system may beaccom- plished byusing Equation 2.97: .9iZ(X)<><\/.1:—U(x) (4.18) When U(x) isknown, itisrelatively easy tomake aphase diagram fora2(x). Computers, with their ever-improving graphics capability, make this aparticu- larly easy task. However, inmany cases itisdifﬁcult toobtain U(x), andwemust resort toapproximation procedures toeventually produce thephase diagram. Ontheother hand, itisrelatively easy toobtain aqualitative picture ofthephase diagram forthemotion ofaparticle inanarbitrary potential. Forexample, con- sider theasymmetric potential shown inFigure 4-5a, which represents asystem that issoftforx<Oand hard forx>O.Ifnodamping occurs, then because isis proportional to\/E—U(x), thephase diagram must beoftheform shown in Figure 4-5b. Three ofthe oval phase paths aredrawn, corresponding tothe 4.3PHASE DIAGRAMS FORNONLINEAR SYSTEMS 151 \i/Is-I amj .‘U3|'|'I'|'I||I X (a) E1 $1 .5“X (b) FIGURE 4-5 (a)Asymmetric potential and (b)phase diagram forbounded motion. three values ofthetotal energy indicated bythedotted lines inthepotential di- agram. Foratotal energy only slightly greater than that oftheminimum ofthe potential, theovalphase paths approach ellipses. Ifthesystem isdamped, then theoscillating particle will“spiral down thepotential well” and eventually come torest attheequilibrium position, x=O.The equilibrium point atx=Ointhis case iscalled anattractor. Anattractor isasetofpoints (oronepoint) inphase space toward which asystem is“attracted” when damping ispresent. For thecase shown inFigure 4-5, ifthetotal energy Eoftheparticle isless than theheight towhich thepotential rises oneither side ofx=0,then thepar- ticle is“trapped” inthepotential well (cf., theregion xl,<x<xl,inFigure 2-14). The point x==0isaposition ofstable equilibrium, because (d2U(x)/dx2)0 >O (seeEquation 2.103), andasmall disturbance results inlocally bounded motion. Inthevicinity ofthemaximum ofapotential, aqualitatively different type of motion occurs (Figure 4-6). Here thepoint x==0isone ofunstable equilibrium, 152 4/NONLINEAR OSCILLATIONS AND CHAOS -—-—_--E, ————————E0 U __i_E2 -x (=1) 5: l ..>>>>j({,l<I(b) FIGURE 4-6 (a)Inverted asymmetric potential and(b)phase diagram for unbounded motion. because ifaparticle isatrestatthispoint, then aslight disturbance willresult inlo- cally unbounded motion.* Similarly, (d2U(x)/dx2)0 <0gives unstable equilibrium. Ifthepotential inFigure 4-6a were parabolic—if U(x) =—%kx2—then the phase paths corresponding totheenergy E0would bestraight lines and those corresponding totheenergies Eland E2would behyperbolas. This isthere- fore thelimit towhich thephase paths ofFigure 4-6would approach if thenonlinear term intheexpression fortheforce were made todecrease in magnitude. Byreferring tothephase paths forthepotentials shown inFigures 4-5and 4-6,wecanrapidly construct aphase diagram foranyarbitrary potential (such as that inFigure 2-14) . \ *The deﬁnition ofinstability must bestated interms oflocally unbounded motion, forifthere are other maxima ofthepotential greater than theone shown atx=0,themotion willbebounded by these other potential barriers. 4.3PHASE DIAGRAMS FORNONLINEAR SYSTEMS 153 I3 I I I I /» ~\, \1 \ /' /--" __ ‘*--\ ‘-\ Limitcycle/1 /4 /,_'-:___:::___\ \ , 4 4 414,- ___ -_ \ III /A 4 " \§ \ \1 , 1,-,- 1-, -.4 2 , I’,’,_, _ \___ 4 ’ Q \ \/1 /I /T I’, I \ I’//\ /I 1/ \\\\,/ I \\\ / \\:/ \\,, \\\ \ 1___ ,' ///,;’ \\\ \\ __ 1I 1 I 1 I I I K X Ii inIII \\ //I,I I‘ \\\ /' I I\ \\\\ I1II 1I _1— \ \\\Q\ 1/’, /—I\ \\\ ,/1 I\\\~, ,1,,;I /\ \\\\\ -’ I\ \\\\\\ /,’/ ,', /\\ \\\\Q\\ ’ / ’/ \ \ /\ \‘\\\\ // /\\\ / I\ \ \ / , I \\ \\ -, I, I,,,2_. \ \ \‘\;~_‘_ -1’- , _— \ \ \ -“\Z~ 1. ,*:’,’ __ _-- 4 \ \ \ \ \_ _ ___. » /\ \ \\\___--__- __,» ' ,/ / \\ ‘ ____- w 1 /\\ \ — © ,_»"“‘~___\_»—-~\ a”,._-—-._T-_T“\\\r”\\ 1.-——-“—_\\-;::==-=-=-‘-1-.‘x\‘\\ .»\\\\\\\\ \\\\ /\\\ //\\/11\///¢//4//1;I;~I’\\\,,I\\\/,,’\.. \\\\\’’(\\\\\1'\\// \\\\\\4/,’//\4//1//..-'/ \‘HgJigF,’I/1 \_—-*____-’_/’//,\1-1’-1*,’/’/\-_“--"T-",*I-_::::,-,,,-______,-"T1’_‘I *~_-__-”TT._§<_ \ ~_ ,_-—' - / \ \_ __.____. 1, ,\ \ - ,\ \~_ ___’ /\ ___- - r\ - -3 | J“"""" "'| I -3 -2 -1 O 1 2 3 ' FIGURE 4-7 Phase diagram forthesolution ofthevanderPolEquation 4.20. The damping term is/.4.=0.05, and thesolution very slowly approaches the limit cycle at2.Positive and negative damping occur, respectively, for lxlvalues outside andinside thelimit cycle at2.The solid anddashed lines have initial (x,ab)values of(1.0, 0)and(3.0, 0),respectively. Animportant type ofnonlinear equation wasextensively studied byvander Polinhisinvestigation ofnonlinear oscillations invacuum tube circuits ofearly radios.* This equation hastheform 55+/_t(x2 —a2)ic +wgx =O (4.19) where II.isasmall, positive parameter. Asystem described byvanderPol’s equa- tion hasthefollowing interesting property. Iftheamplitude exceeds thecrit- icalvalue |a|,then thecoefﬁcient ofScispositive and thesystem isdamped. But if|x|<|a|,then negative damping occurs; that is,theamplitude ofthemotion increases. Itfollows thatthere must besome amplitude forwhich themotion nei- ther increases nordecreases with time. Such acurve inthephase plane iscalled thelimit cyclel (Figure 4-7) and isthe attractor forthis system. Phase paths *B.vanderPol,Phil. Mag. 2,978(1926). Extensive treatments ofvanderPol’s equation may be found, forexample, inMinorsky (Mi47) orinAndronow andChaikin (An49); brief discussions are given byLindsay (Li51, pp.64-66) and byPipes (Pi46, pp.606-610). TThe term wasintroduced byPoincare and isoften called thePoincaré limit cycle. 154 4/NONLINEAR OSCILLATIONS AND CHAOS outside thelimit cycle spiral inward, andthose inside thelimit cycle spiral outward. Inasmuch asthelimit cycle deﬁnes locally bounded motion, wemay refer tothe situation itrepresents asstable. Asystem described byvanderPol’s equation isself-limiting; that is,once set into motion under conditions that lead toanincreasing amplitude, theampli- tude isautomatically prevented from growing without bound. The system has thisproperty whether theinitial amplitude isgreater orsmaller than thecritical (limiting) amplitude x0. Now letusturn tothenumerical calculation ofvanderPol’s Equation 4.19. Inorder tomake thecalculation simpler andtobeable toexamine thesystem’s motion, weleta=1and (1)0==1with appropriate units. Equation 4.19 becomes 55+;.I.(x2 ~—1)ic +x=O (4.20) Inourcase, weused Mathcad tosolve thisdifferential equation. Weuseavalue ofit=0.05, which willgiveasmall damping term. Itwilltake some time forthe solution toreach thelimit cycle. Weshow thecalculation fortwoinitial values of x(x0 =1.0and 3.0) inFigure 4-7;inboth cases, welettheinitial value ofat==0. Note that inthiscase thelimit cycle isacircle ofradius 2.Inboth cases, when theinitial values areboth inside and outside thelimit cycle, thesolution spirals toward thelimit cycle. Ifwesetx0=2(with 5:0=O),themotion remains atthe limit cycle. The solution ofthecircle inthiscase isaresult ofourspecial values foraand w0above. Ifweusealarge damping term, lu.=0.5, thesolution reaches thelimit cycle much more quickly, and thelimit cycle isdistorted asshown in _4_.._;<.’/Limit cycle 2— Il - I /1 I I 0- - -- -— -—--x\ I I /\ ,,’ \‘ II» \ ,-’\ /r \ / \ I’ \ / _2 ,_ \\ I’ ._ \ 4"’\ A__ _¢ I_____i__ _,l_. _*_.___I_ _I, __ -3 -2 -1 O 1 2 3 FIGURE 4-8 Similar calculation toFigure 4-7forthesolution ofthevanderPol Equation 4.20. Inthiscase thedamping parameter /.t=0.5.Note that thesolution reaches thelimit cycle (now skewed) much more quickly. 4.4PLANE PENDULUM 155 Figure 4-8.Forasmall value of,I.L(0.05) thexand itterms aresinusoidal with time, butforhigher values oflu(0.5) thesinusoidal shapes become skewed (see Problem 4-26). ThevanderPoloscillator isanice system forstudying nonlinear behavior andwillbefurther examined intheproblems. 4.4 Plane Pendulum The solutions ofcertain types ofnonlinear oscillation problems can beex- pressed inclosed form byelliptic integrals.* Anexample ofthistype istheplane pendulum. Consider aparticle ofmass mconstrained byaweightless, extension- less rod tomove inavertical circle ofradius l(Figure 4-9). The gravitational force actsdownward, butthecomponent ofthisforce inﬂuencing themotion is perpendicular tothesupport rod. This force component, shoum inFigure 4-10, is simply F(6) =—*mg sin6.The plane pendulum isanonlinear system with asym- metric restoring force. Itisonly forsmall angular deviations that alinear ap- proximation maybeused. Weobtain theequation ofmotion fortheplane pendulum byequating the torque about thesupport axistotheproduct oftheangular acceleration andthe rotational inertia about thesame axis: 16=IF or,because I=ml2and F: —-mgsin 6, cogE€ (4.22)where I I 4\\\\ /\rIt\ _-_,isQ SI U=0 FIGURE 4-9 The plane pendulum where themass misnotrequired tooscillate in small angles. The angle 9>0isinthecounterclockwise direction so that00<0. ‘SeeAppendix Bforalistofsome elliptic integrals. 156 4/NONLINEAR OSCILLATIONS ANDCHAOS Linear 3 FOX] [DQ111011 W .am\ \ \ \ \ \ \ \ ” e-1: \ \ \ \ \ U(6) ;|___________Q-1: 0 FIGURE 4-10 The component oftheforce, F(6), anditsassociated potential that actsontheplane pendulum. Notice thattheforce isnonlinear. Iftheamplitude ofthemotion issmall, wemay approximate sin6 EI9,and the equation ofmotion becomes identical with thatforthesimple harmonic oscillator: §+%o=0 Inthisapproximation, theperiod isgiven bythefamiliar expression 1'E2Tr —-VU g Ifwewish toobtain thegeneral result fortheperiod intheevent that the amplitude isﬁnite, wemay begin with Equation 4.21. Butbecause thesystem is conservative, wecanusethefactthat T+ U= E: constant toobtain asolution byconsidering theenergy ofthesystem rather than bysolv- ingtheequation ofmotion. Ifwetake thezero ofpotential energy tobethelowest point onthecircular path described bythependulum bob (i.e., 6=O;seeFigure 4-10), thekinetic and potential energies canbeexpressed as 1 .T=élwg ==gmlg 62 4.23 U: mgl(l —cos6) ( ) 4.4PLANE PENDULUM 157 Ifwelet6=60atthehighest point ofthemotion, then T(6I60)=0 U(9 =90)==E== mgl(1 —cos60) Using thetrigonometric identity cos6=1-2sin2(6/2) wehave E=2mgl sin2(60/2) (4.24) and U= 2mgl sin2(I9/2) (4.25) Expressing thekinetic energy asthedifference between thetotal energy andthe potential energy, wehave T=E—U, émlgtig =2mgl [sin2(60/2) —sin2(B/2)] or 0=2\/%[sin2(90/2) -*sin2(6/2)]1/2 (4.26) from which at-%\/5[sin2(60/2) -sin2(o/2)]'1/we This equation may beintegrated toobtain theperiod 1-.Because themotion is symmetric, theintegral over 0from 6=0to9=60yields 1-/4; hence 90 T=2(El[sin2(I90/2) -sin2(6/2) 1-1/2d6 (4.27)0 That thisisactually anelliptic integral ofthefirstkind* maybeseen more clearly by making thesubstitutions '9/2 .z= , k=s1n(90/2) Then d cos(I9/2) d6 \/1—-k2z2d6Z I: ' I: 2Sin(60/2) 2k *Refer toEquation B.2, Appendix B. 158 4/NONLINEAR OSCILLATIONS AND CHAOS from which Z 1 T=4\/él [(1-z2)(1—- 1%)]-1/241 (4.2s)0 Numerical values forintegrals ofthistype canbefound invarious tables. Foroscillatory motion toresult, |60|<11',or,equivalently, sin(60/2) =k, where —*l<k<+1.Forthiscase, wecanevaluate theintegral inEquation 4.28 byexpanding (1—-k2z2)'1/2 inapower series: k22 k4 4 (1-11212)-1/2 =1+ Y2+-3%+ Then, theexpression fortheperiod becomes =4\Fl1-—--dz 1+ﬁ+L1Z4+T gto—Z2)" 28 4 :4 — --|— -|—o-—|-3k ¢E..Tr+ III 2 882 \/7 k2we=21'r —1+—+—+---g 464 If|k|islarge (i.e., near 1),then weneed many terms toproduce areasonably accurate result. Butforsmall k,theexpansion converges rapidly. And because k=sin(60/2), then kE(60/2) —(60/48); the result, correct tothe fourth order,is =rr — — i . ~2 J? 1+102+ 1104 (429)T g 16°3072" Therefore, although theplane pendulum isnotisochronous, itisvery nearly so forsmall amplitudes ofoscillation.* Wemayconstruct thephase diagram fortheplane pendulum inFigure 4-11 because Equation 4.26 provides thenecessary relationship =6(6). The param- eter60speciﬁes thetotal energy through Equation 4.24. If6and60aresmall an- gles, then Equation 4.26 canbewritten as00%-.Tilw=\w=‘t-.l\Di—l=1 2 (\/gt) +02E93 (4.30) Ifthecoordinates ofthephase plane are6and 6/\/gw, then thephase paths near 6=0areapproximately circles. This result is expected, because forsmall 60,themotion isapproximately simple harmonic. For-rr<6<1randE<2mg! EE0,thesituation isequivalent toaparticle bound inthepotential well U(6) =mgl(l —cos6)(seeFigure 4-10). The phase *This wasdiscovered byGalileo inthecathedral atPisain1581. The expression fortheperiod of small oscillations wasgiven byChristiaan Huygens (1629-1695) in1673. Finite oscillations were ﬁrst treated byEuler in1736. 4.4PLANE PENDULUM 159 é I _‘*—‘_ _/_ E=E0P21 ___\/____£1 -7:/1 ” i9 A _l\ I Il Stable equilibrium Unstable equilibrium FIGURE 4-11 The phase diagram fortheplane pendulum. Note thestable and unstable equilibrium points and theregions ofbounded and unbounded motion. paths aretherefore closed curves forthisregion andaregiven byEquation 4.26. Because thepotential isperiodic in6,exactly thesame phase paths exist forthe regions 7T<6<317, -317 <6<-Tr, and soforth. The points 6= , -2'rr, 0,211', along the6-axis arepositions ofstable equilibrium andaretheat- tractors when theundriven pendulum isdamped. For values ofthe total energy exceeding E0,the motion isnolonger oscillatory—although itisstillperiodic. This situation corresponds tothepen- dulum executing complete revolutions about itssupport axis. Normally the phase space diagram isplotted foronly onecomplete cycle ora“unit cell,” in this case over theinterval -rr <6<7T.Wedenote this region inFigure 4-11 between thedashed lines atangles -Trand 7T.One can follow aphase path by noting that motion that exits ontheleftofthecellre-enters ontheright and viceversa. Ifthetotal energy equals E0,then Equation 4.24 shows that 60=in‘. Inthis case, Equation 4.26 reduces to 6=i2\/gr)cos(6/2) (4.31) 160 4/NONLINEAR OSCILLATIONS ANDCHAOS sothephase paths forE=E0arejust cosine functions (see theheavy curves in Figure 4~l1). There aretwobranches, depending onthedirection ofmotion. The phase paths forE=E0donotactually represent possible continuous motions ofthependulum. Ifthependulum were atrestat,say,6=7T(which isa point ontheE==E0phase paths), then anysmall disturbance would cause the motion tofollow closely butnotexactly onone ofthephase paths that diverges from 6=rr,because thetotal energy would beE=E0+6,where 5isasmall but nonzero quantity. Ifthemotion were along one oftheE=E0phase paths, the pendulum would reach oneofthepoints I9=mrwith exactly zero velocity, but only after aninﬁnite time! (This may beveriﬁed byevaluating Equation 4.27 for 90==Tr;theresult is1'—->00.) Aphase path separating locally bounded motion from locally unbounded motion (such asthepath forE=E0inFigure 4-ll) iscalled aseparatrix. Asep- aratrix always passes through apoint ofunstable equilibrium. Themotion inthe vicinity ofsuch aseparatrix isextremely sensitive toinitial conditions because points oneither side oftheseparatrix have very different trajectories. 4.5 Jumps, Hysteresis, and Phase Lags InExample 4.1weconsidered aparticle ofmass msuspended between two springs. Weshowed that thesystem was nonlinear and mentioned thephenom- enaofjumps inamplitude andhysteresis effects. Now, wewant toexamine such phenomena more carefully. Wefollow closely thedescription byjanssen and col- leagues* who developed asimple method toinvestigate such effects. Consider aharmonic oscillator subjected toanexternal force F(t)== F0coswtand aresistive viscous force "T02, where risaconstant. The equation of motion foraparticle ofmass mconnected toaspring with force constant kis m5E=—r:I¢ -—kx+F0coswt (4.32) Asolution toEquation 4.32 is W)=/1(0)) COS[wt"¢(w)] (4-33) where F /‘(ml :[(k__mw2)20_|_ (M0211/2 (434) and tan[¢(w)] =(T_T—‘;’nw—% (4.35) The reader canverify that Equation 4.33 isaparticular solution bysubstitution into Equation 4.32. *H._]._]ansser1, etal.,Am.]. Phys, 51,655 (1983). 4.5JUMPS, HYSTERESIS, ANDPHASE LAGS 161 A(w) ¢(w) Z’, _, / it 11 / 1 ll ‘Q "2’/V ' '>co co(U0 CO1 CO2 C00 CO1 CO2 FIGURE 4-12 The amplitude A(o)) and phase angle ¢(o)) asafunction ofthe angular frequency co.Notice the‘jumps” atwland (02depending onthedirection ofchange ofo). Ifthespring constant kdepends onxask(x), then wehave anonlinear oscil- lator. Anoften used dependence is k(x) =(1+Bx2)k0 (4.36) and theresulting equation ofmotion inEquation 4.32 isknown astheDuﬂing equation. Ithasbeen widely studied through perturbation techniques with solu- tions similar toEquation 4.33 butwith complicated results forA(w) and¢(w) as shown inFigure 4-12. Astoincreases, A(¢o) increases toitspeak until itreaches w=m2,where theamplitude suddenly decreases byalarge factor.’ Astode- creases from large values, theamplitude slowly increases until to=(01,where the amplitude suddenly approximately doubles. These arethe“jumps” referred to earlier. The amplitude between coland(1)2depends onwhether wisincreasing ordecreasing (hysteresis effect). Similarly strange phenomena occur forthe phase ¢(w) inFigure 4-12. The physical explanation ofFigure 4-12 isnotvery transparent, soweconsider asimpler dependence ofkasshown inFigure 4-13. F(x)==~kx xSa =-k'x~c x2a(4.37) -F(x) kl Pv Q,_____>36 FIGURE 4-13 Asimpler dependence ofF(x)onthespring constant kthan thatin Equation 4.36. 162 4/NONLINEAR OSCILLATIONS ANDCHAOS A(co) k k’ a------ ---- ----- || ,0, CO0 CDO’ FIGURE 4-14 The values ofA(w) forthetwovalues ofkshown inFigure 4-12. The Dufﬁng equation represents asituation with many values ofa,because k(x) continuously varies inEquation 4.36. Our example ofananharmonic oscillator allows simpler mathematics. Figure 4-14 shows theharmonic response curves A(w) forkand k’(with k<k’).Forvery large values ofa(a-—><><>),wehave alinear oscillator with force constant k(because x<a,see Figure 4-13) and aresonance frequency w0=(k/m)1/2. Forvery small values ofa(a->0),theforce constant isk’and m0=(k’/m)1/2. Wewant toconsider intermediate values ofa,where both kandk’areeffec- tive. Weconsider thesituation inwhich aismuch smaller than themaximum amplitude ofA(w). Ifwestart atsmall values ofco,oursystem hassmall vibrations that follow theamplitude curve fork.The amplitude moves upthetailofthe A(to) curve forkasshown inFigure 4-15. However, when thevibration amplitude A(w) islarger than thecritical am- plitude a,theforce constant k’iseffective. Forthese larger amplitudes, thesys- tem follows A'(¢o) forforce constant k’.This isrepresented bythesolid bold line from BtoCinFigure 4-15. AFB) G 4 k "\k/ BTCI ‘\ A 'E\D 1g _J J ,0, 601 (02 FIGURE 4-15 Thebold lines andarrows help follow thepath aswincreases anddecreases.\\\. 4.6CHAOS INAPENDULUM 163 ¢(w) ¢(w)A A 7E— __ 7Ir- __.- ,_ 1 Ac/> kkl " I / / i"""' .. ,0, 1L .0,(00 ‘"6 ")1402 (Q) (b) FIGURE 4-16 The phase angle ¢(w) forkand k'isshown in(a),and thesystem’s path isshown in(b). Between Aand B,asthefrequency increases, thesystem follows thesimpli- ﬁed amplitude rise shown bythedashed line inFigure 4-15. Continuing toin- crease thedriving frequency toatC,weagain reach thecritical amplitude aat point D.Iftoisonly slightly increased, thesystem must follow A(¢o) fork,andthe amplitude suddenlyjumps down from A’(w) atpoint DtoA(o)) atpoint Fat to=(02.Astocontinues increasing above (1)2,thesystem follows theA(w) curve. Now letusseewhat happens ifwedecrease tofrom large values. The system follows A(w) until to=col,where A(o)) :a.Ifwisbarely decreased, theampli- tude increases above a,andthesystem must follow A’(w). Therefore theampli- tude jumps from EtoG.Astocontinues decreasing, itfollows asimilar path as before. Ahysteresis effect occurs because thesystem behaves differently depending onwhether toisincreasing ordecreasing. Two amplitude jumps occur, one forcu increasing andoneforcudecreasing. The system’s paths areABGCDF (o)increas- ing) and FEGBA (todecreasing). Similar phenomena occur forthephase lagq5(w). Figure 4-16a shows the phase curves ¢(w) and¢'(w) forthelinear harmonic oscillators. Using thesame arguments asapplied toA(w), wedepict thesystem ’spaths inFigure 4-16b bythe bold lines and thearrows. The reader isreferred tothearticle byjanssen etal.for anexperiment suitably demonstrating these phenomena. 4.6 Chaos inaPendulum Wewill usethedamped and driven pendulum tointroduce several chaos con- cepts. The simple motion ofapendulum iswell understood after hundreds of years ofstudy, butitschaotic motion hasbeen extensively studied only inthe past few years. Among the motions ofpendula that have been found tobe chaotic areapendulum with aforced oscillating support asshown inFigure 164 4/NONLINEAR OSCILLATIONS ANDCHAOS I/lcos mt (a)Forced pivot (b)Double pendulum ‘ii; 3% '3““M (c)Coupled pendulums (d)Magnetic pendulum FIGURE 4-17 Examples ofpendulums thathave chaotic motion. 4-17a, thedouble pendulum (Figure 4-17b), coupled pendulums (Figure 4-17c), andapendulum oscillating between magnets (Figure 4-17d). The damped and driven pendulum thatwewillconsider isdriven around itspivot point, andthe geometry isdisplayed inFigure 4-18. Forced K_\ motion .1‘Q: ._f,»;£¥l m FIGURE 4-18 Adamped pendulum isdriven about itspivot point. 4.6CHAOS INAPENDULUM 165 The torque around thepivot point canbewritten as die .. . _N= IF ==I6=—-b6 -—mgl’ sin6+Ndcoswdt (4.38) where Iisthemoment ofinertia, bisthedamping coefﬁcient, and Ndisthedriv- ingtorque ofangular frequency wd.Ifwedivide byI=ml’2,wehave ‘ --___ b - g _ Nd 6--‘$9-zs1n6+ﬁcoswdt (4.39) Wewilleventually want todeal with thisequation with acomputer, anditwillbe much easier inthat case touse dimensionless parameters. Let usdivide Equation 4.39 byw02=g/L’ and deﬁne thedimensionless time t’=t/t0with t0=1/w0 and thedimensionless driving frequency w=(dd/(1)0. The new dimen- sionless variables andparameters are x=6 oscillating variable (4.40a) b c=-T-nﬁgg damping coefﬁcient (4.40b) NN F2424mg20,02 mg‘, driving force strength (4.400) t’=5-=\/5t dimensionless time (4.40d) 0 ll’ w=gg=Jim‘, driving angular frequency (4.40e) 0 Note that _dx d6dt d61 XZHZEHTEJO ___d2x__d26(dt)2_d26 1_6 xpdﬂ dz?dz’Wdt2w02 (1)02 Using these variables andparameters, Equation 4.39 becomes 55==~eo'c~sinx +Fcoswt’ (4.41) Equation 4.41 isanonlinear equation oftheform ﬁrst presented in Equation 4.1.Wewillusenumerical methods tosolve this equation forx,given theparameters c,F,andw.The techniques mentioned inChapter 3areused to solve thisequation, depending ontheaccuracy desired and computer speed available, and commercial software programs areavailable. Weusetheprogram Chaos Demonstrations bySprott andRowlands (Sp92). Equation 4.41, asecond-order differential equation, canbereduced totwo ﬁrst-order equations bymaking thesubstitution dx)1==E (4-42) 166 4/NONLINEAR OSCILLATIONS ANDCHAOS Equation 4.41 becomes aﬁrst-order differential equation dy .E: -—cy— s1nx+Fcosz (4.43) where wehave also made thesubstitution z=cot’. Equations 4.42 and 4.43 are theﬁrst-order differential equations. Wepresent theresults ofnumerical methods solutions inFigure 4-19. We leave theparameters cand tosetat0.05 and 0.7, respectively, and vary only the driving strength Fin steps of0.1from 0.4to1.0.The results arethatthemotion isperiodic forFvalues of0.4,0.5,0.8,and 0.9butischaotic for0.6,0.7,and 1.0. These results indicate thebeautiful and surprising results obtained from nonlin- eardynamics. The leftside ofFigure 4-19 displays y==dx/dt' (angular velocity) versus time long after theinitial motion (i.e., transient effects have died out). The value ofF =0.4shows simple harmonic motion, buttheresults for0.5, 0.8, and0.9,although periodic, arehardly simple. Wecanlearn more byexamining thephase space plots, shown inthemiddle column ofFigure 4-19 (note that wepresent only aunit cell ofthephase dia- gram from —1'rto1-r).Asexpected, theresult forF=0.4shows theresults seen previously inChapter 3(Figure 3-5). The phase plot forF: 0.5shows one long cycle that includes twocomplete revolutions and twooscillations. The entire al- lowed area inthephase plane isaccessed chaotically forF=0.6and0.7,butfor F"-10.8, themotion becomes periodic again with one complete revolution and anoscillation. The result forF=0.9isinteresting, because there appears tobe twodifferent revolutions inonecycle, each similar totheoneforF-"=0.8.This result iscalled period doubling (i.e., theperiod forF=0.9istwice theperiod for F=0.8). After close inspection, thiseffect canalsobeobserved from thedx/dt' versus time plot, shown ontheleftcolumn ofFigure 4-19. Poincaré Section Henry Poincaré invented atechnique tosimplify therepresentations ofphase space diagrams, which canbecome quite complicated. Itisequivalent totaking a strobdscopic view of_the phase space diagram. Athree-dimensional phase dia- gram plots y(=5c=0)versus x(==9)versus z(=wt’). The leftcolumn ofFigure 4-19 isaprojection ofthisplot onto ay-zplane, showing points thatcorrespond tovarious values ofphase angle x.Themiddle column ofFigure 4-19isaprojection onto ay-xplane, showing points belonging tovarious values ofz.InFigure 4-20we show thethree-dimensional phase space diagram intersected byasetofy-x planes, perpendicular tothez-axis, atequal zintervals. APoincaré section plotis thesequence ofpoints formed bytheintersections ofthephase path with these parallel planes inphase space, projected onto one ofthe planes. The phase path pierces theplanes asafunction ofangular speed (y=ti),time (z=wt’), and phase angle (x=6).The points ontheintersections arelabeled as A1,A2,A3,etc.This setofpoints A,forms apattern when projected onto one of theplanes (Figure 4-20b) thatsometimes willbearecognizable curve, butsome- times willappear irregular. Forsimple harmonic motion, such asF=0.4in CHAOS INAPENDULUM Poincaré section F=0.4 F=0.5Phase-space plot I IIW ‘ CD 0 0 F:0.6 0 rveocity,y=dx/dtAF=0.7 <3. -—< -—a AnguaF=0.8 0 F=0.90 0 0 . . ti)»0 A ’\ 0'/I 2 '[‘\. §t>Y<<.(CL §*“( .0 1/ 1 0 * 0 r 0 IF=1.0 01 0"" *10"" r I - 0Or 'Ev114))0 4/ ,_ 0 FIGURE 4-190101: 201: -rt 0 rt—rr 0 rt Time t’ Angle x Angle x The damped and driven pendulum forvarious values ofthedriving force strength. The angular velocity versus time isshown ontheleft, and phase diagrams areinthecenter. Poincaré sections areshown on theright. Note that motion ischaotic forthedriving force Fvalues of 0.6,0.7,and 1.0.167 168 4/NONLINEAR OSCILLATIONS ANDCHAOS .1I(Ij) ‘I/Phase - - hPoincare Pat sections \1-X, 7 l .==ea'ae:;1:i%:éz; As II 2:1iii-';%¢l:'§)§§ x (2 iiii|:aa5;Z>!.=es£ -===_-at )1 Interval 0I10 0 I1 A A A1 3 2‘,.('.'.-»'.'.~.-.\-,y,;‘:,‘.1 ____-,--___‘. . / —————— F- x z(=(0t’) (3) (b) FIGURE 4-20 (a)Poincaré plot, athree-dimensional phase diagram, showing three Poincaré sections and thephase path. The sections areprojections along the_y—xplane. (b)The points A,arethephase path intersections with thesection plots. They areplotted here onthey-xplane tohelp visualize themotion inphase space. Figure 4-19, allthepoints projected arethesame (orinasmooth curve, de- pending onthezspacing ofthey-xplanes). Poincaré realized that thesimple curves represent motion with possibly analytic solutions, butthemany compli- cated, apparently irregular, curves represent chaos. The Poincaré section curve effectively reduces anN-dimensional diagram to(N—1)-dimensions forgraph- icalpurposes andoften helps visualize themotion inphase space. Forthecase ofthedamped anddriven pendulum, theregularity ofthedy- namical motion isduetotheforcing period, andacomplete description ofthe dynamical motion depends onthree parameters. Wecantake those parameters tobex(angle 6),y=dx/dt' (angular frequency), and z=wt’(phase ofthedriv- ingforce). Acomplete description ofthemotion inphase space would require three-dimensional phase diagrams rather than displaying just twoparameters asin Figure 4-19. Allthevalues ofzareincluded inthemiddle column ofFigure 4-19, sowechoose totake thestroboscopic sections ofthemotion forjust thevalues ofzI21211" (n=0,1,2,...),which isatafrequency equal tothatofthedriving force. Weshow thePoincaré section forthependulum intheright column of Figure 4-19 forthesame systems displayed intheleftand middle columns. For thesimple motion ofF=0.4, thesystem always comes back tothesame position of(x,y)after zgoes through 2'rr.Therefore, weexpect thePoincaré section to show only one point, and that iswhat weﬁnd inthetopﬁgure oftheright column ofFigure 4-19. The motion forF=0.8also shows only one point, butF=0.5and 4.7MAPPING 169 0.9show three and twopoints, respectively, because ofthemore complex mo- tion. The number ofpoints nonthePoincaré section here shows that thenew period T=T0n/m,where T0=21r/co istheperiod ofthedriven force and mis aninteger (m=2fortheF=0.5plot and m=1fortheF=0.9plot). The chaotic motions forF=0.6, 0.7, and 1.0display thecomplicated variation of points expected forchaotic motion with aperiod T—>oo.The Poincaré sections arealsorichinstructure forchaotic motion. Onthree occasions thus far(Figures 4-5,4-7,and 4-11), wehave pointed out attractors, asetofpoints (orapoint) onwhich themotion converges fordissipa- tivesystems. The regions traversed inphase space arestrictly bounded when there isanattractor. Inchaotic motion, nearby trajectories inphase space are continually diverging from oneanother butmust eventually return totheattrac- tor.Because theattractors inthese chaotic motions, called strange orchaotic at- tractors, arenecessarily bounded inphase space, theattractors must fold back into thenearby regions ofphase space. Strange attractors create intricate pat- terns, because thefolding andstretching ofthetrajectories must occur such that notrajectory inphase space intersects, which isruled outbythedeterministic dyamical motion. The Poincaré sections ofFigure 4-19 reveal thefolded, layered structure oftheattractors. Chaotic attractors arefractals, butspace does notper- mitfurther discussion ofthisextremely interesting phenomenon. 4.7 Mapping Ifweusentodenote thetime sequence ofasystem and xtodenote aphysical observable ofthesystem, wecandescribe theprogression ofanonlinear system ataparticular moment byinvestigating how the(n+1)th state (oriterate) de- pends onthenthstate. Anexample ofsuch asimple, nonlinear behavior is x,,+1 =(2xn +3)2. This relationship, xmtl =f(x,,), iscalled mapping and is often used todescribe theprogression ofthesystem. The Poincaré section plots previously discussed areexamples oftwo-dimensional maps. Aphysical example appropriate formapping might bethetemperature ofthespace shuttle orbiter tiles while theshuttle descends through theatmosphere. After theorbiter has been ontheground forsome time, thetemperature TH] isthesame asTn,but thiswasnottrue while theshuttle plummeted through theatmosphere from its earth orbit. Modeling thetiletemperatures correctly with amathematical model isdifﬁcult, and linear assumptions areoften ﬁrst assumed insuch calculations with nonlinear terms added tomake more realistic calculations. Wecanwrite adifference equation using f(a,xn)where xnisrestricted toareal number intheinterval (0,1)between 0and 1,and aisamodel-dependent parameter. x,,+1 =f(a, xn) (4.44) The function f(a, x,,)generates thevalue ofx,,+1 from xn,and thecollection ofpoints generated issaid tobeamap ofthefunction itself. The equations, which areoften nonlinear, areamenable tonumerical solution byiteration, 170 4/NONLINEAR OSCILLATIONS AND CHAOS starting with x1.Wewillrestrict ourselves here toone-dimensional maps, but two-dimensional (and higher order) equations arepossible. Mapping canbest beunderstood bylooking atanexample. Letusconsider the“logistic” equation, asimple one-dimensional equation given by f(a, x)=ax(1 —x) (4.45) sothattheiterative equation becomes x,,+1 =ax,,(1 —x,,) (4.46) Wefollow thediscussion ofBessoir andWolf (Be91) who usethelogistic equa- tion forabiological application example ofstudying thepopulation growth of ﬁsh inapond, where thepond isWell isolated from external effects such as weather. The iterations, ornvalues, represent theannual ﬁshpopulation, where x1isthenumber ofﬁshinthepond atthebeginning oftheﬁrst year ofthe study. Ifx1issmall, theﬁsh population may grow rapidly intheearly years be- cause ofavailable resources, butoverpopulation may eventually deplete the number ofﬁsh.Thepopulation x,,isscaled sothatitsvalue ﬁtsintheinterval (0,1) between 0and 1.The factor atisamodel-dependent parameter representing av- erage effects ofenvironmental factors (e.g., ﬁshermen, ﬂoods, drought, preda- tors) thatmay affect theﬁsh. The factor amaybevaried asdesired inthestudy, butexperience shows that ashould belimited inthisexample totheinterval (0,4) toprevent theﬁshpopulation from becoming negative orinﬁnite. The results ofthelogistic equation aremost easily observed bygraphical means inamap called thelogistic map. The iteration x,,+1 isplotted versus x,,in Figure 4-21a foravalue ofa=2.0.Starting with aninitial value x1onthehori- zontal (xn)axis, wemove upuntil weintersect with thecurve x,,,t1 =2x,,(1 —x,,), and then wemove totheleftwhere weﬁnd x2onthevertical axis (x,,+1). We then start with thisvalue ofx2onthehorizontal axisandrepeat theprocess to ﬁnd x3onthevertical axis. Ifwedothisforafewiterations, weconverge onthe value x=0.5,andtheﬁshpopulation stabilizes athalfitsmaximum. Wearrive at thisresult independent ofourinitial value ofx]aslong asitisnot0or1. Aneasier waytofollow theprocess istoaddthe45°line, x,,,.1 =x,,,tothe same graph. Then after initially intersecting thecurve from x1,onemoves hori- zontally tointersect with the45°linetoﬁnd x2andthen moves upvertically to ﬁnd thenext iterative value ofx3.This process cangoonand reach thesame re- sultasinFigure 4-21a. Weshow theprocess inFigure 4—21b toindicate thatthis method iseasier tousethan theonewithout the45°line. Inpractice, wewant tostudy thebehavior ofthesystem when themodel pa- rameter ctisvaried. Inthepresent case, forvalues ofalessthan 3.0,stable pop- ulations willresult (Figure 4-22a). The solutions follow asquare spiral path to thecentral, ﬁnal value. Forvalues ofajust above 3.0, more than one solution fortheﬁshpopulation occurs (Figure 4—22b). The solutions follow apath simi- lartothesquare spiral, which converges tothetwopoints atwhich thesquare intersects the“iteration line,” rather than toasingle point. Such achange in thenumber ofsolutions toanequation, when aparameter such asaisvaried, is called abifurcation. 4.7MAPPING 171 1l_ xn+l xn+l =2x1: (1"xn) X4 . 1 *5 *2 J I. 0 xl *2 xa"4 1 x71. (a) 1l_ xn+l I 1 I . 0 xn 1 (b) FIGURE 4-21 Techniques forproducing amap ofthelogistics equation. Weobtain amore general view oftheglobal picture byplotting abifurcation diagram, which consists ofx,,,determined after many iterations toavoid initial ef- fects, plotted asafunction ofthemodel parameter oz.Many new interesting ef- fects emerge indicating regions and windows ofstability aswell asthose of chaotic dynamics. Weshow thebifurcation diagram inFigure 4-23 forthelogis- ticequation over therange ofavalues from 2.8to4.0. Forthevalue ofoz=2.9 shown inFigure 4—22a, weobserve that after afewiterations, astable conﬁgura- tion forx=0.655 results. AnNcycleisanorbit that returns toitsoriginal posi- tion after Niterations, that is,xN+,- =x,-.The period fora=2.9isthen aone cycle. Fora=3.1(Figure 4-22b), thevalue ofxoscillates between 0.558 and 172 4/NONLINEAR OSCILLATIONS AND CHAOS 1|_ or=2.9 xn+1 0 4,, 1 ta) 1|_ oz=3.1 xn+1 0 xn 1 (bi FIGURE 4-22 Logistic equation map foravalues of2.9and 3.1,indicating stable populations in(a)andmultiple possible solutions fora>3.0in(b). 0.765 (two cycle) after afewiterations evolve. The bifurcation occuring at3.0is called apitchfork bifurcation because oftheobvious shape ofthediagram caused bythesplitting. Ata=3.1,theperiod doubling effect hasx,,+2 =x,.,.Ata= 3.45, thetwo-cycle bifurcation evolves into afour cycle, andthebifurcation and period doubling continues uptoaninﬁnite number ofcycles near a=3.57. Chaos occurs formany oftheavalues between 3.57 and4.0,butthere arestill windows ofperiodic motion, with anespecially Wide window around 3.84. Are- allyinteresting behavior occurs forat=3.82831 (Problem 4-11). Anapparent periodic cycle of3years seems tooccur forseveral periods, butthen itsuddenly violently changes forafewyears, andthen returns again tothe3-year cycle. This intermittent behavior could certainly prove devastating toabiological study oper- ating over several years thatsuddenly turns chaotic without apparent reason. 4.7 MAPPING 1.0‘- F X00.5 " L ___‘ -., \ D‘.2-31-.a-'*':r7I_v*" -441“~"‘=='".‘.r't~.-'.-. 5"-:3-:1’ SE3‘ I-4!.‘Q‘-'01‘ poi} E‘? 4‘,4155"4 » r"'-Alli‘ . . 3F _.,."'ii¥’»' 4.».2.»-1» r I u ...415"--fli-'4' 1. '..J‘ . '-."3.'-.1“V -vi _--I-,#i\j \*--1,; 1,:-.=.»» =~-rrefh ;,_..\.;._.g_,_\ _~ .iii.‘‘iii}‘ "0-.-''-'.- '" .'- ,.;41--:42-=. 13;...1' 1.. 1* .lLlgi..'=.:».'.»~.»~.‘**:;,;J?‘-’\nu1 '=%i‘"-‘";7?’i.';:v:.F7”(474,4-173 n-I‘;\/z F“ ., P .\ﬁg}?,. 0.0 2.8 3.0 3.2 3.4 3.6 38 4.0 FIGURE 4-23 Bifurcation diagram forthelogistic equation map EXAMPLE 4.2 J- LetAan =an—a,,_1 bethewidth between successive period doubling bifurca- tions ofthelogistic map that wehave been discussing. Forexample, from Figure 4-23, welet011=3.0where theﬁrst bifurcation occurs and 012= 3.449490 where thenext oneoccurs. Let5,,bedeﬁned astheratio A5,,=—°i (4.47)Aan+l andlet8,,—>5asn—>oo.Find 5,,fortheﬁrstfewbifurcations andthelimit 5. Solution. Although wecould program thisnumerical calculation with acom- puter, wewilluseone ofthecommercially available software programs (Be91)to work thisexample. Wemake atable oftheanvalues using thecomputer pro- gram, ﬁnd Aan, andthen determine afewvalues ofan. n an Aa 5.. 01>-I-uamtd3.0 3.449490 3.544090 3.564407 3.568759 <><- 3.56994560.449490 0.094600 0.020317 0.0043524.7515 4.6562 4.6684 4.6692 As01,,approaches thelimit 3.5699456, thenumber ofperiod doublings approaches inﬁnity, andtheratio 5",called Feigenbaumiv number, approaches 4.669202. This result wasﬁrstfound byMitchell Feigenbaum inthe1970s, and hefound that thelimit 6wasauniversal property oftheperiod doubling route 174 4/NONLINEAR OSCILLATIONS AND CHAOS tochaos when thefunction f(a, x)hasaquadratic maximum. Itisaremarkable factthatthisuniversality isnotconﬁned toone-dimensional mappings; itisalso true fortwo-dimensional maps and hasbeen conﬁrmed forseveral cases. Feigenbaum claims tohave found thisresult using aprogrammable hand calcu- lator. The calculation obviously hastobecarried tomany signiﬁcant ﬁgures to establish itsaccuracy, and such acalculation wasnotpossible before such calcu- lators (orcomputers) were available. 4.8 Chaos Identification Inourdriven anddamped pendulum, wefound thatchaotic motion occurs for some values oftheparameters, butnotforothers. What arethecharacteristics of chaos andhow canweidentify them? Chaos does notrepresent periodic motion, and itslimiting motion willnotbeperiodic. Chaos cangenerally bedescribed as having asensitive dependence oninitial conditions. Wecandemonstrate thisef- fectbythefollowing example. EXAMPLE 4.3 I I - _ _ - _ Consider thenonlinear relation x,,+1 =f(a, x,,)=orx,,(1 —x,,2). Leta=2.5 andmake twonumerical calculations with initial x1values of0.700000000 and 0.700000001. Plottheresults andﬁnd theiteration nwhere thesolutions have clearly diverged. Solution. The iterative equation that weareconsidering is x,,+1=ctx,,(1—x02) (4.48) Weperform ashort numeric calculation and plot theresults ofiterations for thetwoinitial values onthesame graph. The result isshown inFigure 4-24 where there isnoobserved difference forx,,+1 until nreaches atleast 30.By n=39,thedifference inthetworesults ismarked, despite theoriginal values differing byonly 1part in108. Ifthecomputations aremade without error, and thedifference between it- erated values doubled ontheaverage foreach iteration, then there willbean exponential increase such as 2n=enln2 where nisthenumber ofiterations undergone. Fortheiterates tobeseparated bytheorder ofunity (the sizeoftheattractor), wewillhave 2410-8~1 4.8CHAOS IDENTIFICATION 175 1 I I I I I E3/III/II- 0.2— -—0700000000 ‘T I ----I0700000001 ' Iii -% 0 0 10 20 30 40 50 Iteration, n FIGURE 4-24 Example 4.3.The n+1iterative state isplotted versus thenumber of iterations and shows twoeventual results forslightly different initial conditions ofx1. which gives n=27.That is,after 27iterations, thedifference between thetwo iterates reaches thefullrange ofxn.Tohave theresults differ byunity forn= 40iterations, wewould have toknow theinitial values with aprecision of1part in1012!_ I III t 7 it *1 ma I The previous example indicates thesensitive dependence oninitial condi- tions that ischaracteristic ofchaos. The tworesults canstillbedetermined in thiscase, butitisrare toknow theinitial values toaprecision of10-8. Ifweadd another factor of10totheprecision ofx1,wegain only four interative steps of agreement inthecalculation. Wemust accept thereality that increasing thepre- cision oftheinitial conditions only gains usalittle intheaccuracy oftheulti- mate measurement. This exponential growth ofaninitial error willultimately prevent usfrom predicting theoutcome ofameasurement. The effect ofsensitive dependence oninitial conditions hasbeen called the “butterﬂy” effect. Abutterﬂy moving slowly through theairmay cause anex- tremely small effect onthe airﬂow that will prevent usfrom predicting the weather patterns next week. Background noise orthermal effects willusually adduncertainties larger than theones wehave discussed here, andwecannot distinguish these effects from measurement errors. Precise predictive power of many steps isjustnotpossible. Lyapunov Exponents One method toquantify the sensitive dependence oninitial conditions for chaotic behavior uses theLyapunov characteristic exponent. Itisnamed after the Russian mathematician A.M.Lyapunov (1857-1918). There are asmany 176 4/NONLINEAR OSCILLATIONS AND CHAOS Lyapunov exponents foraparticular system asthere arevariables. Wewilllimit ourselves atﬁrst toconsidering only one variable and therefore one exponent. Consider asystem with twoinitial states differing byasmall amount; wecallthe initial states x0and x0+s.Wewant toinvestigate theeventual values ofx,, after niterations from thetwo initial values. The Lyapunov exponent )trepre- sents thecoefﬁcient oftheaverage exponential growth perunit time between thetwostates. After niterations, thedifference dnbetween thetwox,,values is approximately d,,=sew‘ (4.49) From thisequation, wecanseethat if)tisnegative, thetwoorbits willeventually converge, butifpositive, thenearby trajectories diverge andchaos results. Letuslook ataone-dimensional map described byx,,+1 =f(x,,). The initial difference between thestates isd0=s,and after one iteration, thedifference d1 1S 4/ d1=f(x0+ s)—j(x0) =85)-C where thelastresult ontheright side occurs because sisvery small. After niter- ations, thedifference d,,between thetwoinitially nearby states isgiven by d,,=f"(x +s)—f"(x0) =se"* (4.50) where wehave indicated thenthiterate ofthemap f(x)bythesuperscript n.If wedivide by8andtake thelogarithm ofboth sides, wehave ln(jQ~K(x +82- fn(x0)> =ln(e"") =n)t andbecause sisvery small, wehave for)t, A=—1n(f——-L 8)f(9%))=-In—-f(X) (4.51)n 8 n dxxo Thevalue off"(x0) isobtained byiterating thefunction f(x0)ntimes. f”(X0) =f(f( (f(X@)) )) Weusethederivative chain ruleofthenthiterate toobtain d/"<0 :5; Q‘dx ,0 dx clxxnz, dxxn—1 r xi) Wetake thelimit asn—>ooandﬁnally obtain 1n—l d i A=lim-Z15-1% (4.52)x n—>oo TI,i=0 4.8CHAOS IDENTIFICATION 177 1_lIIl|III|11ll__ L .- 0.)) A .- _1.. __ _2- ..- 731 ..I1II-I-1IJWIILIIII2.8 3 3.2 3.4 3.6 3.8 4 (1 FIGURE 4-25 Lyapunov exponent asafunction ofaforthelogistic equation map. Avalue ofA >0indicates chaos. Weplot theLyapunov exponent asafunction ofainFigure 4-25 forthelo- gistic map. Wenote theagreement ofthesign ofAwith thediscussion ofchaotic behavior inSection 4.6.The value ofAiszero when bifurcation occurs, because |df/dxl =1,and thesolution becomes unstable (see Problem 4-16). Asuper- stable point occurs where df(x)/ dx=0,and this implies that A=-00. From Figure 4-25 asAgoes above 0,weseethere arewindows where Areturns toA<0 andperiodic orbits occur amid thechaotic behavior. The relatively wide window justabove 3.8isapparent. Remember that forndimensional maps, there willbenLyapunov expo- nents. Only oneofthem need bepositive forchaos tooccur. Fordissipative sys- tems, thephase space volume willdecrease astime passes. This means thesum oftheLyapunov exponents willbenegative. The calculation ofLyapunov exponents forthedamped anddriven pendu- lum isdifﬁcult, because one hastodeal with thesolutions ofdifferential equa- tions rather than maps such asthose ofthelogistic equation. Nevertheless, these calculations have been done, andweshow inFigure 4-26 theLyapunov exponents, three ofthem because ofthethree dimensions (calculated using Baker’s program [Ba90]). The parameters arethesame asthose discussed in Section 4.6:c=0.05, 0)=0.7,andF=0.4(periodic) andF=0.6(chaotic). For both cases, wemust make atleast several hundred iterations tomake sure tran- sient effects have died out. Note that one oftheLyapunov exponents iszero, because itdoes notcontribute totheexpansion orcontraction ofthephase space volume. For the case ofF=0.4, none ofthe Lyapunov exponents is greater than zero after 350iterations, butfortheF=0.6driven case, oneofthe exponents isstillwellabove zero. The motion ischaotic forFI0.6,aswefound earlier inFigure 4-19. However, because themotion described inFigure 4-26 is damped, thesum ofthethree Lyapunov exponents isnegative forboth cases, as itshould be. 178 0.4 0.2 A0 -0.2 -0.4 -0.6 -0.8 0.4 0.2 A0 -0.2 -0.4 -0.6 -0.84/NONLINEAR OSCILLATIONS AND CHAOS Iy- L \_ I-I F I I I II I VI’ T F ‘ “I \ ' /1 - F=0.4 I --I 4 --L *_ I_ 1 I I l. - I- 0 100 200 300 400 I_ I- F II '1 I I I l I I ---- --4 I F=0.6 I I I I I I l I 0 100 200 300 400 Number ofdrive cycles FIGURE 4-26 The three Lyapunov exponents forthedamped and driven pendulum. The values ofAarethose approached ast—>oo (large number ofcycles). PROBLEMS 4-1. Refer toExample 4.1.Ifeach ofthesprings must bestretched adistance dtoattach theparticle attheequilibrium position (i.e., initsequilibrium position, theparticle issubject totwoequal and oppositely directed forces ofmagnitude kd), then show that thepotential inwhich theparticle moves isapproximately 4-2. 4-3. 4-4.U(x) E(kd/l)x2 +[k(l— d)/4l3]x4 Construct aphase diagram forthepotential inFigure 4-1. Construct aphase diagram forthepotential U(x) =—(A/3) x3. Lord Rayleigh used theequation ae—(a—1»z2)s.+ w§x= 0 inhisdiscussion ofnonlinear effects inacoustic phenomena.* Show that differenti- ating thisequation with respect totime and making thesubstitution y=y0\/3b/aai W.S.Rayleigh, Phil. Mag. 15(April 1883); seealso R2194, Section 68a. PROBLEMS 179 4-5. 4-6. 4-7. 4-8. 4-9. 4-10. 4-llresults invanderPol’s equation: 5‘-fgtyt -y2)i+ w3y= 00 Solve byasuccessive approximation procedure, and obtain aresult accurate tofour signiﬁcant ﬁgures: (a)x+x2+1=tanx, 0$xS'rr/2 (b)x(x+ 3)=10sin x, x>0 (c)1+x+cosx=e", x>0 (Itmay beprofitable tomake acrude graph tochoose areasonable ﬁrst approximation.) Derive theexpression forthephase paths oftheplane pendulum ifthetotal energy isE>2mgl. Note thatthisisjustthecase ofaparticle moving inaperiodic poten- tialU(6) =mgl(1 —cos0). Consider thefree motion ofa plane pendulum whose amplitude isnotsmall. Show that thehorizontal component ofthemotion may berepresented bytheapproximate expression (components through thethird order areincluded) 2 5E+w§(1+%)x—ex3=0 where tog=g/land s=3g/2Z3, with lequal tothelength ofthesuspension. Amass mmoves inone dimension and issubject toaconstant force +F0 when x<0and toaconstant force —F0 when x>0.Describe themotion byconstructing aphase diagram. Calculate theperiod ofthemotion interms ofm,F0,and theam- plitude A(disregard damping). Investigate themotion ofanundamped particle subject toaforce oftheform _k , <F(x): x Ixl a —(k+5)x+5a, >a where kand 5arepositive constants. The parameters F=0.7and c=0.05 areﬁxed forEquation 4.43 describing the driven, damped pendulum. Determine which ofthevalues forto(0.1, 0.2,0.3, ..., 1.5)produce chaotic motion. Produce aphase plot forw=0.3.Dothisproblem numerically. Areally interesting situation occurs forthelogistic equation, Equation 4.46, when a=3.82831 and xl=0.51. Show that athree cycle occurs with theapproximate x values 0.16, 0.52, and 0.96 fortheﬁrst 80cycles before thebehavior apparently turns chaotic. Find forwhat iteration thenext apparently periodic cycle occurs and forhow many cycles itstays periodic. 180 4-12 4-13. 4-14 4-15. 4-16 4-17 4-18 4-19 4-20 4-21.4/NONLINEAR OSCILLATIONS AND CHAOS Letthevalue ofainthelogistic equation, Equation 4.46, beequal to0.9.Make a map like that inFigure 4-21 when x1=0.4.Make theplot forthree other values of x1forwhich0 <x1<1. Perform thenumerical calculation done inExample 4.3andshow thatthetwocal- culations clearly diverge byn=39.Next, letthesecond initial value agree towithin another factor of10(i.e., 0.700 000 000 1),and conﬁrm thestatement inthetext that only four more iterations aregained intheagreement between thetwoinitial values. Use thefunction described inExample 4.3, x,,+1 =ax,,(1 —x,,2) where oz=2.5. Consider two starting values ofx1that aresimilar, 0.900 000 0and 0.900 000 1. Make aplot ofx,,versus nforthetwo starting values and determine thelowest value ofnforwhich thetwovalues diverge bymore than 30%. Use direct numerical calculation toshow that themap f(x) =ctsin1rxalso leads to theFeigenbaum constant, where xand aarelimited totheinterval (0,1). The curve x,,+1 =f(x,,) intersects thecurve x,,+1 =x,,atx0.The expansion ofx,,+1 about x0is x,,+1 —x0=B(x,, —x0)where B=(df/dx) atx=x0. (a)Describe thegeometrical sequence thatthesuccessive values ofx,,+1 —x0form. (b)Show that theintersection isstable when <1and unstable when >1. The tentmap isrepresented bythefollowing iterations: x,,+1= 2ax,, for0 <x<1/2 x,,+1= 2a(1— x,,) for1/2<x<1 where 0<a<1.Make amap upto20iterations fora=0.4and 0.7with x1=0.2. Does itappear that either ofthemaps represent chaotic behavior? Plot thebifurcation diagram forthetentmap oftheprevious problem. Discuss the results forthevarious regions. Show analytically that theLyapunov exponent forthetentmaps isA=ln(2a). This indicates that chaotic behavior occurs fora>1/2. Consider theHenon map described by x,,+1= y,,+1— axg, yn+l :bxn Leta=1.4and b=0.3,and useacomputer toplot theﬁrst 10,000 points (x,,,y,,) starting from theinitial values x0=0,y0=0.Choose theplot region as—1.5 <x< 1.5and -0.45 <y<0.45. Make aplot oftheHenon map, this time starting from theinitial values x0=0.63135448,y0 =018940634. Compare theshape ofthis plot with that ob- tained intheprevious problem. Istheshape ofthecurves independent oftheini- tialconditions? PROBLEMS 131 4-22 4-23. 4-24. 4-25. 4-26.Acircuit with anonlinear inductor canbemodeled bytheﬁrst-order differential equations Q‘_aty Q=—k —x5+B stat 5' °° Chaotic oscillations forthissituation have been extensively studied. Use acom- puter toconstruct thePoincaré section plot forthecase k=0.1and 9.8SBS13.4. Describe themap. The motion ofabouncing ball, onsuccessive bounces, when theﬂoor oscillates sinusoidally canbedescribed bytheChirikov map: pn+1= P1» _Ksinqn ¢I..+1= ll”+pn+1 where —1rSp517and -17$qS1r.Construct two-dimensional maps forK= 0.8, 3.2,and6.4bystarting with random values ofpandqanditerating them. Useperi- odic boundary conditions, which means that iftheiterated values ofporqexceed 1r,avalue of21rissubtracted andwhenever they arelessthan —1r,avalue of21ris added. Examine themaps after thousands ofiterations and discuss thedifferences. Assume that x(t)=bcos(w0t) +u(t)isasolution ofthevanderPolEquation 4.19. Assume thatthedamping parameter ,u.issmall andkeep terms inu(t)toﬁrstorder in].L.Show that b=2aand u(t) =—(,u.a3/4w0) sin(3w0t) isasolution. Produce a phase diagram ofkversus xand produce plots ofx(t) and aZ(t)forvalues ofa=1, w0=1, and ].L=0.05. Use numerical calculations toﬁnd asolution forthevan der Pol oscillator of Equation 4.19. Let x0and m0equal 1forsimplicity. Plot thephase diagram, x(t), and 5c(t)forthefollowing conditions: (a),u.=0.07, x0=1.0,5:0=0att=0;(b)/.t= 0.07, x0=3.0,£0=0att=0.Discuss themotion; does themotion appear toap- proach alimit cycle? Repeat theprevious problem with p.=0.5.Discuss also theappearance ofthelimit cycle, x(t), and40). W _ CHAPTER L} Gravitation 5.1Introduction By1666, Newton hadformulated and numerically checked thegravitation law heeventually published inhisbook Principia in1687. Newton waited almost 20 years topublish hisresults because hecould notjustify hismethod ofnumerical calculation inwhich heconsidered Earth and theMoon aspoint masses. With mathematics formulated oncalculus (which Newton later invented), wehave a much easier time proving theproblem Newton found sodifﬁcult intheseven- teenth century. Newton ’slawofuniversal gravitation states that each mass particle attracts every other particle intheuniverse with aforce thatvaries directly astheproduct ofthetwo masses andinversely asthesquare ofthedistance between them. Inmathematical form, wewrite thelawas MF=-G-T’-‘?e, (5.1)T where atadistance rfrom aparticle ofmass Masecond particle ofmass mexpe- riences anattractive force (see Figure 5-1).The unit vector e,points from Mtom, andtheminus sign ensures thattheforce isattractive—that is,that misattracted toward M. Alaboratory veriﬁcation ofthelawand adetermination ofthevalue ofGwas made in1798 bytheEnglish physicist Henry Cavendish (1731-1810). Cavendish’s experiment, described inmany elementary physics texts, used atorsion balance with twosmall spheres ﬁxed attheends ofalight rod. The twospheres were at- tracted totwoother large spheres that could beplaced oneither side ofthe smaller spheres. The ofﬁcial value forGis6.673 i0.010 ><10*] N-m2/kg2. Interestingly, although Gisperhaps theoldest known ofthefundamental constants, 182 5.1 INTRODUCTION 183 Fm ,,,' r er ,-''' 01' M FIGURE 5-1 Particle mfeels anattractive gravitational force toward M. weknow itwith lessprecision than weknow most ofthemodern fundamental constants such ase,c,andh.Considerable research isongoing today toimprove theprecision ofG. Intheform ofEquation 5.1,thelawstrictly applies only topoint particles. If one orboth oftheparticles isreplaced byabody with acertain extension, we must make anadditional hypothesis before wecancalculate theforce. Wemust assume that thegravitational force ﬁeld isalinear field. Inother words, weas- sume that itispossible tocalculate thenetgravitational force onaparticle due tomany other particles bysimply taking thevector sum ofalltheindividual forces. Forabody consisting ofacontinuous distribution ofmatter, thesum be- comes anintegral (Figure 5-2): F=—Gmj B£?dv' (5.2)V T where p(r') isthemass density anddv'istheelement ofvolume attheposition deﬁned bythevector r’from the(arbitrary) origin tothepoint within themass distribution. Ifboth thebody ofmass Mandthebody ofmass rnhave ﬁnite extension, a second integration over thevolume ofmwillbenecessary tocompute thetotal gravitational force. ‘I77 1' -'~p ~I ,=‘=aI2%%*;i5 ' .I‘E5.~;;i:5.E1.},';.éza'§:' . *¥€;;‘;§I'' .lag, ‘.;'-'_'='~£~.I 1MI T'fi1.2;; er FIGURE 5-2 Toﬁnd thegravitational force between apoint mass mand acontinuous distribution ofmatter, weintegrate themass density over thevolume. 184 5/GRAVITATION The gravitational field vector gisthevector representing theforce perunit mass exerted onaparticle intheﬁeld ofabody ofmass M.Thus F M g=E= —GFe, Or 0').,2=—cjv9-IF-e-at (5.4) Note thatthedirection ofe,varies with r'(inFigure 5-2). The quantity ghasthedimensions offorce perunit mass, also equal toaccelera- tion.Infact, near thesurface oftheearth, themagnitude ofgisjustthequantity thatwecallthegravitational acceleration constant. Measuremennwith asimple pendulum (orsome more sophisticated variation) issufﬁcient toshow that lg]is approximately 9.80 m/s? (or9.80 N/kg) atthesurface oftheearth. 5.2 Gravitational Potential The gravitational ﬁeld vector gvaries as1/r2andtherefore satisﬁes therequire- ment* that permits gtoberepresented asthegradient ofascalar function. Hence, wecanwrite where <15iscalled thegravitational potential andhasdimensions of(force perunit mass) X(distance), orenergy perunitmass. Because ghasonly aradial variation, thepotential (Pcanhave atmost avari- ation with r.Therefore, using Equation 5.3forg,wehave d<D Mv<1>=—-=—dre’ Gr2er M The possible constant ofintegration hasbeen suppressed, because thepotential isundetermined towithin anadditive constant; thatis,only differences inpoten- tialaremeaningful, notparticular values. Weusually remove theambiguity in thevalue ofthepotential byarbitrarily requiring that <15—>0 asr—>oo; then Equation 5.6correctly gives thepotential forthiscondition.Integrating, weobtain *Thatis,vxgE0. 5.2GRAVITATIONAL POTENTIAL 185 The potential duetoacontinuous distribution ofmatter is ¢=—ci—p('1)dv' (5.7)V T Similarly, ifthemass isdistributed only over athin shell (i.e., asurface distri- bution), then <1>=-cl5‘dd’ (5.8)ST where p,isthesurface density ofmass (orareal mass density). Finally, ifthere isalinesource with linear mass density pl,then <1>=-0}3’ds' (5.9)1"T The physical signiﬁcance ofthegravitational potential function becomes clear ifweconsider thework perunit mass dW' thatmust bedone byanoutside agent onabody inagravitational ﬁeld todisplace thebody adistance dr.Inthis case, work isequal tothescalar product oftheforce and thedisplacement. Thus, forthework done onthebody perunit mass, wehave dW' =—g-dr =(V<D) -dr =2Qd __dd) (5.10). xi_ 1dxi because <15isafunction only ofthecoordinates ofthepoint atwhich itismeas- ured: (P=<P(x1, x2,x3)=<P(x,-). Therefore theamount ofwork per unit mass thatmust bedone onabody tomove itfrom oneposition toanother inagravi- tational ﬁeld isequal tothedifference inpotential atthetwopoints. Iftheﬁnal position isfarther from thesource ofmass Mthan theinitial posi- tion, work hasbeen done ontheunit mass. The positions ofthetwopoints arearbi- trary, and wemay take one ofthem tobeatinﬁnity. Ifwedeﬁne thepotential to bezero atinﬁnity, wemay interpret (Patanypoint tobethework perunit mass required tobring thebody from inﬁnity tothat point. The potential energy is equal tothemass ofthebody multiplied bythepotential <15.IfUisthepotential energy, then U= m<P (5.11) andtheforce onabody isgiven bythenegative ofthegradient ofthepotential energy ofthat body, F=—VU (5.12) which isjust theexpression wehave previously used (Equation 2.88). Wenote thatboth thepotential andthepotential energy increase when work isdone onthebody. (The potential, according toourdeﬁnition, isalways nega- tiveand only approaches itsmaximum value, that is,zero, asrtends toinﬁnity.) 186 5/GRAVITATION Acertain potential energy exists whenever abody isplaced inthegravita- tional ﬁeld ofasource mass. This potential energy resides intheﬂeld,* butitis customary under these circumstances tospeak ofthepotential energy “ofthe body.” Weshall continue thispractice here. Wemay also consider thesource mass itself tohave anintrinsic potential energy. This potential energy isequal to thegravitational energy released when thebody wasformed or,conversely, is equal totheenergy thatmust besupplied (i.e., thework that must bedone) to disperse themass over thesphere atinﬁnity. Forexample, when interstellar gas condenses toform astar, thegravitational energy released goes largely into the initial heating ofthestar.Asthetemperature increases, energy isradiated away aselectromagnetic radiation. Inalltheproblems wetreat, thestructure ofthe bodies isconsidered toremain unchanged during theprocess wearestudying. Thus, there isnochange intheintrinsic potential energy, and itmay beneg- lected forthepurposes ofwhatever calculation wearemaking. EXAMPLE 5.1 What isthegravitational potential both inside andoutside aspherical shell of inner radius band outer radius a? Solution. One oftheimportant problems ofgravitational theory concerns the calculation ofthegravitational force duetoahomogeneous sphere. This prob- lem isaspecial case ofthemore general calculation forahomogeneous spheri- calshell. Asolution totheproblem oftheshell canbeobtained bydirectly com- puting theforce onanarbitrary object ofunit mass brought into theﬁeld (see Problem 5-6), butitiseasier tousethepotential method. Weconsider theshell shown inFigure 5-3andcalculate thepotential at point Padistance Rfrom thecenter oftheshell. Because theproblem hassym- metry about theline connecting thecenter ofthesphere and theﬁeld point P, theazimuthal angle qbisnotshown inFigure 5-3andwecanimmediately inte- grate over d¢intheexpression forthepotential. Thus, (P=—Gi Bﬂdv’V T G 77 ‘ 6 =—21TpGLr'2dr’ L3511-40 (5.13) where wehave assumed ahomogeneous mass distribution fortheshell, p(r') =p.According tothelawofcosines, r2=r'2+R2—2r'R cos6 (5.14) Because Risaconstant, foragiven r'wemay differentiate thisequation and obtain 2rdr=2r'R sin6d6 *See, however, theremarks attheend ofSection 9.5regarding theenergy inaﬁeld. 5.2GRAVITATIONAL POTENTIAL 187 ""‘iW5w' *sgﬁ.5...Q\€1'$"AMY “ﬁgﬁa¢*~‘?r.5&...,_5-=5m‘Q wM K‘ Vii,1-+5-g»1 d l if T ===;m=~ =1=%J,-kw-‘L,-.3.-.-it .-.£,.;1~..J’-V , FIGURE 5-3 Thegeometry forﬁnding thegravitational potential atpoint Pdue toa spherical shell ofmass. OI’ 591-650=i (5.15)T rR Substituting thisexpression into Equation 5.13, wehave 2 G“ "W <P=— r'dr'i dr (5.16) R 5 Tmin The limits ontheintegral over drdepend onthelocation ofpoint P.IfPisout- sidetheshell, then R+r’2 G“ <P(R >a)=— r'dr'[ dr R b R-1' 4 Ll =._;2§J,2M,R 5 __éW@3_.-3R(a b) (5.17) Butthemass Moftheshell is 4M=§rrp(a3 —I23) (5.18) sothepotential is |<1>(R> .5)=—%/1| (5.19) 188 5/GRAVITATION Iftheﬁeld point liesinside theshell, then 2 G a r'+R <P(R <b)=— r’dr'i dr R b r'—R Ll =—4rrpGJ r’dr’ 5 =—2'rrpG(a2 —b2) (5.20) Thepotential istherefore constant andindependent ofposition inside theshell. Finally, ifwewish tocalculate thepotential forpoints within theshell, we need only replace thelower limit ofintegration intheexpression for(R <b) bythevariable R,replace theupper limit ofintegration intheexpression for <P(R >a)byR,and add theresults. Weﬁnd 4 G <1><b<R<a)=——;'f7<R3 —11*)—2¢rp@<a2 —R2) 2bs R2 =-41____ . 'rrpG(2 3R 6) (521) WeseethatifR—> a,then Equation 5.21 yields thesame result asEquation 5.19 forthesame limit. Similarly, Equations 5.21 and5.20 produce thesame result forthelimit R—> b.The potential istherefore continuous. Ifthepotential were notcontinuous atsome point, thegradient ofthepotential—and hence, the force——would beinﬁnite atthatpoint. Because inﬁnite forces donotrepresent physical reality, weconclude that realistic potential functions must always be continuous. Note thatwetreated themass shell ashomogeneous. Inorder toperform calculations forasolid, massive body like aplanet that hasaspherically symmet- ricmass distribution, wecould addupanumber ofshells or,ifwechoose, we could allow thedensity tochange asafunction ofradius. The results ofExample 5.1arevery important. Equation 5.19 states thatthe potential atanypoint outside ofaspherically symmetric distribution ofmatter (shell orsolid, because solids arecomposed ofmany shells) isindependent of thesizeofthedistribution. Therefore, tocalculate theexternal potential (orthe force), weconsider allthemass tobeconcentrated atthecenter. Equation 5.20 indicates that thepotential isconstant (and theforce zero) anywhere inside a spherically symmetric mass shell. And ﬁnally, atpoints within themass shell, the potential given byEquation 5.21 isconsistent with both oftheprevious results. The magnitude oftheﬁeld vector gmaybecomputed from g=—d<P/dR for each ofthethree regions. The results are gR<m=0 4rrpG b3d)=T EE_R mi 5R>”=”E? 5.2 GRAVITATIONAL POTENTIAL 189 ..-==="i§ “iffijzz... i (D=const,8‘)H -m .9_.@ no--------------——- Q-—————-———-——-———————————————'9'R R1 QP7l -<1> 0.1. 2;“i—g R2 g: FIGURE 5-4 The results ofExample 5.1indicating thegravitational potential and magnitude oftheﬁeld Vector g(actually —g) asafunction ofradial distance. Weseethat notonly thepotential butalso theﬁeld vector (and hence, the force) arecontinuous. The derivative oftheﬁeld vector, however, isnotcontinu- ousacross theouter andinner surfaces oftheshell. Allthese results forthepotential and theﬁeld vector canbesummarized as inFigure 5-4. EXAMPLE 5.2 Astronomical measurements indicate that theorbital speed ofmasses inmany spiral galaxies rotating about their centers isapproximately constant asafunc- tionofdistance from thecenter ofthegalaxy (like ourown Milky Way andour nearest neighbor Andromeda) asshown inFigure 5-5.Show that thisexperi- mental result isinconsistent with thegalaxy having itsmass concentrated near thecenter ofthegalaxy andcanbeexplained ifthemass ofthegalaxy increases with distance R. Solution. Wecanﬁnd theexpected orbital speed vduetothegalaxy mass M that iswithin theradius R.Inthiscase, however, thedistance Rmay behundreds oflight years. Weonly assume themass distribution isspherically symmetric. The gravitational force inthiscase isequal tothecentripetal force duetothe 190 Orbitalspeed(km/s)O9OQ IQCC >-ICCA ._\I/ ‘\I \ / ‘\ I ‘\/ \'~_ I \_ / 1 I, ‘[1? I7-5/GRAVITATION /"> i J 7, l__ _ _Ii i L M s 0 20 40 50 so 100Radius from galactic center (thousands oflight years) FIGURE 5-5 Example 5.2.The solid line represents data fortheorbital speed ofmass asafunction ofdistance from thecenter oftheAndromeda galaxy. The dashed line represents the1/\/R behavior expected from theKeplerian result ofNewton’s laws. mass mhaving orbital speed v: Wesolve thisequation for-u:GMm _mug r2 R /GMv= —-R Ifthiswere thecase, wewould expect theorbital speed todecrease as1/\/R as shown bythedashed lineinFigure 5-5,whereas what isfound experimentally is thatvisconstant asafunction ofR.This canonly happen intheprevious equa- tionifthemass Mofthegalaxy itself isalinear function ofR,M(R)ocR. Astrophysicists conclude from thisresult thatformany galaxies there must be matter other than thatobserved, andthatthisunobserved matter, often called “dark matter,” must account formore than 90percent oftheknown mass inthe universe. This area ofresearch isattheforefront ofastro hsicstoda . PY Y EX"-\l\1PLE5.3 F. rIIF TI74 Consider athin uniform circular ring ofradius aandmass M.Amass mis placed intheplane ofthering. Find aposition ofequilibrium and determine whether itisstable. Solution. From symmetry, wemight believe thatthemass mplaced inthecen- terofthering (Figure 5-6)should beinequilibrium because itisuniformly sur- rounded bymass. Putmass matadistance r’from thecenter ofthering, and place thex-axis along thisdirection. 5.2 GRAVITATIONAL POTENTIAL 191 z IFIGURE 5-6 Example 5.3.The geometry ofthepoint mass mand ring ofmass M The potential isgiven byEquation 5.7where p=M/2"n'a: dM Gdo=—c— =-Edq'> (5.25)b b where bisthedistance between dMand m,anddM=pad¢. Letrandr’bethe position vectors todMandm,respectively. b=|r—r'l=|acos</>e1+asin¢e2—r’e1| I =|(acos¢> —r')e1+ asin¢e2| =[(acosqﬁ —r')2+a2sin2¢]1/2 I 2 I 1/2 =(a2+r’2—2ar’cos <11)‘/2 =a':1 + —2%cos¢>:| (5.24) Integrating Equation 5.23 gives I 211' <P(r) =—GJT =—paGL T 211' d¢ =—pGJ 5,2 , -1/2 (5.25)0 r 2r[1+(—) ——cos¢:|a a The integral inEquation 5.25 isdifﬁcult, soletusconsider positions close to theequilibrium point, r’=0.Ifr’<<a,wecanexpand thedenominator in Equation 5.25. I2 ' -1/2 1 I2 I [1+ —2%cos¢:| =1-5&2) —2%cos¢:| 3 I2 2' 2 .,[(;)-7»-.....].I I2 =1+%cos¢ + (3cos2¢ -1) + (5.26) 192 5/GRAVITATION Equation 5.25 becomes 11' I 1 12 <D(r') =—pGJ'2 {I+Lacos¢ +E (3cos2</> —1)+---}d¢ (5.27) 0 which iseasily integrated with theresult MG 1’2<D(r') =—-—~':1 +—(L) + (5.28)a 4a The potential energy U(r’) isfrom Equation 5.11, simply U(r’) =m<D(r') =—-—1Zl%;|i1 +g(E)? + (5.29) The position ofequilibrium isfound (from Equation 2.100) by dU(r') __mMG1 L’dr,-0_ a2Q,+ (5.50) sor’=0isanequilibrium point. WeuseEquation 2.103 todetermine thestability: d2U(r’) mMG —JTT§'"=-"'2?+ <0 (5.31) sotheequilibrium point isunstable. This lastresult isnotobvious, because wemight beledtobelieve thatasmall displacement from r’=0might stillbereturned tor’=0bythegravitational forces from allthemass inthering surrounding it.mi M’ Tm m m um — Poisson’s Equation Itisuseful tocompare these properties ofgravitational ﬁelds With some ofthefa- miliar results from electrostatics that were determined intheformulation of Maxwell’s equations. Consider anarbitrary surface asinFigure 5-7with amass m placed somewhere inside. Similar toelectric ﬂux, let'sﬁnd thegravitational ﬂux (Pmemanating from mass mthrough thearbitrary surface S. (Pm=Jn-g da (5.32) s where theintegral isover thesurface Sand theunit vector nisnormal tothe surface atthedifferential area da.Ifwesubstitute gfrom Equation 5.3for 5.2GRAVITATIONAL POTENTIAL 193 Surface S FIGURE 5-7 Anarbitrary surface with amass mplaced inside. The unitvector nis normal tothesurface atthedifferential area da. thegravitational ﬁeld vector forabody ofmass m,wehave forthescalar product n-g, cos6n-g= —GmT where 6istheangle between nandg.Wesubstitute thisinto Equation 5.32 and obtain (Pm=—Gmi lgadaS T The integral isover thesolid angle ofthearbitrary surface andhasthevalue 41r steradians, which gives forthemass ﬂux <P,,,= In~gda= —41rGm (5.33) s Note thatitisimmaterial where themass islocated inside thesurface S.Wecan generalize thisresult formany masses miinside thesurface Sbysumming over themasses. in-gda= —41rG2m, (5.34)S 1 Ifwechange toacontinuous mass distribution within surface S,wehave in-g da=—41rGi pdv (5.35) 5 V where theintegral ontheright-hand side isover thevolume Venclosed byS,pis themass density, and dvisthedifferential volume. WeuseGauss’s divergence theorem torewrite this result. Gauss’s divergence theorem, Equation 1.130 where da=nda,is in-gda= [V-gdv (5.36) S v 194 5/GRAVITATION Ifwesettheright-hand sides ofEquations 5.35 and5.36 equal, wehave I(—41rG)pdv =IV-g dv V V andbecause thesurface S,anditsvolume V,iscompletely arbitrary, thetwointe- grands must beequal. V~g=—41rGp (5.37) This result issimilar tothedifferential form ofGauss’s lawforelectric ﬁeld, V-E=p/s, where pinthiscase isthecharge density. Weinsert g=—VQ5from Equation 5.5into theleft-hand side ofEquation 5.37 andobtain V-g=—V-V<P=—V245. Equation 5.37 becomes V2<15 =41rGp (5.38) which isknown asPoisson ’sequation and isuseful inanumber ofpotential theory applications. I/Vhen theright-hand side ofEquation 5.38 iszero, theresult VQQ5 =0isaneven better known equation called Laplace’s equation. Poisson’s equation isuseful indeveloping Green’s functions, whereas weoften encounter Laplace’s equation when dealing with various coordinate systems. 5.3 Lines ofForce andEquipotential Surfaces Letusconsider amass thatgives risetoagravitational ﬁeld thatcanbedescribed byaﬁeld vector g.Letusdraw alineoutward from thesurface ofthemass such that thedirection oftheline atevery point isthesame asthedirection ofgat thatpoint. This linewillextend from thesurface ofthemass toinﬁnity. Such a lineiscalled alineofforce. Bydrawing similar lines from every small increment ofsurface area ofthe mass, wecanindicate thedirection oftheforce ﬁeld atanyarbitrary point in space. The lines offorce forasingle point mass areallstraight lines extending from themass toinﬁnity. Deﬁned inthisway, thelines offorce arerelated only tothedirection oftheforce ﬁeld atanypoint. Wemay consider, however, thatthe density ofsuch lines——that is,thenumber oflines passing through aunit area ori- ented perpendicular tothelines——is proportional tothemagnitude oftheforce atthatarea. The lines-of-force picture isthus aconvenient waytovisualize both themagnitude andthedirection (i.e., thevector property) oftheﬁeld. The potential function isdeﬁned atevery point inspace (except attheposi- tion ofapoint mass). Therefore, theequation <15=<P(x], x2,x5)=constant (5.39) deﬁnes asurface onwhich thepotential isconstant. Such asurface iscalled an equipotential surface. The ﬁeld vector gisequal tothegradient of(P,sogcan 5.4 WHEN ISTHE POTENTIAL CONCEPT USEFUL? 195 P Tl T2 FIGURE 5-8 The equipotential surfaces due totwopoint masses M. have nocomponent along anequipotential surface. Ittherefore follows that every lineofforce must benormal toevery equipotential surface. Thus, theﬁeld does nowork onabody moving along anequipotential surface. Because thepo- tential function issingle valued, notwoequipotential surfaces canintersect or touch. The surfaces ofequal potential that surround asingle, isolated point mass (oranyspherically symmetric mass) areallspheres. Consider twopoint masses Mthatareseparated byacertain distance. Ifr1isthedistance from one mass tosome point inspace and ifr2isthedistance from theother mass tothe same point, then l l <15=—GM(-—+—)=constant (5.40) T1 T2 deﬁnes theequipotential surfaces. Several ofthese surfaces areshown inFigure 5-8forthistwo-particle system. Inthree dimensions, thesurfaces aregenerated byrotating thisdiagram around thelineconnecting thetwomasses. 5.4 When IsthePotential Concept Useful? The useofpotentials todescribe theeffects of“action-at-a-distance” forces isan extremely important and powerful technique. We should not, however, lose sight ofthefactthattheultimate justiﬁcation forusing apotential istoprovide a 196 5/GRAVITATION convenient means ofcalculating theforce onabody (ortheenergy forthebody intheﬁeld)——for itistheforce (and energy) and notthepotential that isthephys- ically meaningful quantity. Thus, insome problems, itmaybeeasier tocalculate theforce directly, rather than computing apotential andthen taking thegradi- ent.The advantage ofusing thepotential method isthat thepotential isascalar quantity*: Weneed notdeal with theadded complication ofsorting outthe components ofavector until thegradient operation isperformed. Indirect cal- culations oftheforce, thecomponents must becarried through theentire com- putation. Some skill, then, isnecessary inchoosing theparticular approach to use. Forexample, ifaproblem hasaparticular symmetry that, from physical considerations, allows ustodetermine that theforce has acertain direction, then thechoice ofthatdirection asoneofthecoordinate directions reduces the vector calculation toasimple scalar calculation. Insuch acase, thedirect calcu- lation oftheforce may besufﬁciently straightforward toobviate thenecessity of using thepotential method. Every problem requiring aforce must beexamined todiscover theeasiest method ofcomputation. EXAMPLE 5.4 Consider athin uniform disk ofmass Mand radius a.Find theforce onamass mlocated along theaxisofthedisk. Solution. Wesolve thisproblem byusing both thepotential and direct force approaches. Consider Figure 5.9.The differential potential d<Patadistance zis FIGURE 5-9 Example 5.4.Weusethegeometry shown here toﬁnd thegravitational force onapoint mass mdue toathin uniform disk ofmass M.liq. 2 *Weshall seeinChapter 7another example ofascalar function from which vector results maybeob- tained. This isthe function, which, toemphasize thesimilarity, issometimes (mostly in older treatments) called thekinetic potential. 5.4 WHEN ISTHE POTENTLAL CONCEPT USEFUL? 197 given by d<15=—o5i‘—4 (5.41) The differential mass dMisathin ring ofWidth dx,because wehave azimuthal symmetry. dM=pdA =p211'x dx (5.42) xdx xdx61¢ ——2WpGT —_2WpG a 2xd<P(z) =—"n'pGL @ =—21'rpG(x2 +z2)1/2 =—21rpG[(a2 +z2)1/2 —z] (5.43) Weﬁnd theforce from F=—VU= —mV<D (5.44) From symmetry, wehave only aforce inthezdirection, F=—m§ip@ =+27rmpG -9- -1 '(5.45)Z 62 (a2+z2)1/2 Inoursecond method, wecompute theforce directly using Equation 4.2: dM’dF=—Gm? e, (5.46) where dM’ refers tothemass ofasmall differential area more like asquare than athin ring. Thevectors complicate matters. How cansymmetry help? Forevery small dM’ononesideofthethin ring ofwidth dx,another dM’exists onthe other sidethatexactly cancels thehorizontal component ofdFonm.Similarly, allhorizontal components cancel, andweneed only consider thevertical com- ponent ofdFalong z. 6d1\4’dE,= cos6\dF| =—mG£(-)3?r and, because cos6=z/r, dM’til‘; =—mGL? T Now weintegrate over themass dM’ =p21rx dxaround thering andobtain __ZEL“ .1r,- mop T, 198 5/GRAVITATION and “ 2xdx1.;=—1rmpGz J0ﬁg+362),,/2 _2 a =—1rmpGz|%z2 +X2),/2:|L Z =2WmpG|: —l:| (5.47) which isidentical toEquation 5.45. Notice that thevalue ofF,isnegative, indi- cating thattheforce isdownward inFigure 5-9andattractive. 5.5 Ocean Tides The ocean tides have long been ofinterest tohumans. Galileo tried unsuccess- fully toexplain ocean tides butcould notaccount forthetiming oftheapproxi- mately twohigh tides each day. Newton ﬁnally gave anadequate explanation. The tides arecaused bythegravitational attraction oftheocean toboth the Moon and theSun, butthere areseveral complicating factors. The calculation iscomplicated bythefactthat thesurface ofEarth isnotan inertial system. Earth and Moon rotate about their center ofmass (and move about theSun), sowemay regard thewater nearest theMoon asbeing pulled away from Earth, and Earth asbeing pulled away from thewater farthest from theMoon. However, Earth rotates while theMoon rotates about Earth. Let’s ﬁrst consider only theeffect oftheMoon, adding theeffect oftheSunlater. Wewill assume asimple model whereby Earth’s surface iscompletely covered with water, and weshall add theeffect ofEarth’s rotation atanappropriate time. We setupaninertial frame ofreference x'y’z' asshown inFigure 5.10a. WeletMm bethemass oftheMoon, rtheradius ofacircular Earth, and Dthedistance from thecenter oftheMoon tothecenter ofEarth. Weconsider theeffect of both theMoon’s andEarth’s gravitational attraction onasmall mass mplaced on thesurface ofEarth. Asdisplayed inFigure 5-10a, theposition vector ofthemass infrom theMoon isR,from thecenter ofEarth isr,and from ourinertial system rl,,.The position vector from theinertial system tothecenter ofEarth isrg.As measured from theinertial system, theforce onm,due totheearth and the Moon, is __, GmME GmM,,,mr,,,= —Te,— 7%-é—eR (5.48) Similarly, theforce onthecenter ofmass ofEarth caused bytheMoon is ..,METE: _ eD 5.5 OCEAN TIDES 199 IZ yl xi rt.rfg m 4".’ Y Mm R G5- -DMoon Earth (3) Y 6 CR 8 @—>,,, t 4Q a,4-. Polar _> Moon FT axis FT it era (b) FIGURE 5-10 (a)Geometry toﬁnd ocean tides onEarth due totheMoon. (b)Polar view with thepolar axis along thez-axis. Wewant toﬁnd theacceleration Fasmeasured inthenoninertial system placed atthecenter ofEarth. Therefore, wewant 5;: 5:,_F2: mr'f,,_MEi"'g ”' m ME __GME _GM”, +GM", _ T2er R2eR D2cl’ GM e e=—-7-,—Ee, —GM",<3’;-51;) (5.50) The ﬁrst part isdue toEarth, and thesecond part istheacceleration from the tidal force, which isresponsible forproducing theocean tides. Itisduetothe difference between theMoon’s gravitational pull atthecenter ofEarth and on Earth’s surface. 200 5/GRAVITATION Wenext ﬁnd theeffect ofthetidal force atvarious points onEarth as noted inFigure 5-10b. Weshow apolar view ofEarth with thepolar axis along thez-axis. The tidal force FTon themass monEarth’s surface is _ E5_E2FT-—GmM,,, (R,D2) (5.51) where wehave used only thesecond part ofEquation 5.50. Welook ﬁrst atpoint a,thefarthest point onEarth from theMoon. Both unit vectors eRand enare pointing inthesame direction away from theMoon along thex-axis. Because R >D,the second term inEquation 5.51 predominates, and the tidal force is along the+x-axis asshown inFigure 5-10b. Forpoint b,R<Dand thetidal force hasapproximately thesame magnitude asatpoint abecause r/D<<1,but isalong the~x-axis. The magnitude ofthetidal force along thex-axis, FTx,is 1 1 1 1F;-x =—GmMm<F — =TGmMm( — GmM,,,( 1 )=__-i- _i-i;_1 D2 T21 _ (+0) Weexpand theﬁrst term inbrackets using the(1+x)-2expansion inEquation D.9. GmM,,, r r2 2GmM,,,r FTx=—7 1-25-I-35 —"'—l =+T (5.52) where wehave kept only thelargest nonzero term intheexpansion, because r/D= 0.02. Forpoint c,theunit vector eR(Figure 5-10b) isnotquite exactly along eD, butthex—axiscomponents approximately cancel, because R=Dandthex-com- ponents ofeRandeDaresimilar. There willbeasmall component ofeRalong they-axis. Weapproximate they-component ofeRby(r/D)j, andthetidal force atpoint c,callitFTy,isalong they-axis andhasthemagnitude 1r_ GmM,,,r FT),=—GmMm ——T (5.53) Note thatthisforce isalong the—y-axis toward thecenter ofEarth atpoint c.We ﬁnd similarly atpoint Dthesame magnitude, butthecomponent ofeRwillbe along the—y-axis, sotheforce itself, with thesign ofEquation 5.53, willbealong the+y-axis toward thecenter ofEarth. Weindicate thetidal forces atpoints a,b, c,anddonFigure 5-lla. 5.5OCEAN TIDES 201 Tidal force M Moon (a) ”’;’ ’f’—UU U Moon \ I T-._ _,’ (b) FIGURE 5-11 (a)The tidal forces areshown atvarious places onEarth’s surface including thepoints a,b,c,anddofFigure 5-10. (b)Anexaggerated view ofEarth’s ocean tides. Wedetermine theforce atanarbitrary point ebynoting that thex-andy- components ofthetidal force canbefound bysubstituting xandyforrinFT, and F7-y,respectively, inEquations 5.52 and 5.53. F:2GmM,,,x Tx D3 F:_GmMm y Ty D5 202 5/GRAVITATION Then atanarbitrary point such ase,weletx=rcos 6andy=rsin6,sowehave 2GmMm1‘ cos6F7-x =T (5.543) GmM,,,r sin6FT), ='—T (5.54b) Equations 5.54a andbgive thetidal force around Earth forallangles 6.Note that they give thecorrect result atpoints a,b,c,and d. Figure 5-1lagives arepresentation ofthetidal forces. Foroursimple model, these forces lead tothewater along they-axis being more shallow than along thex-axis. Weshow anexaggerated result inFigure 5-1lb.AsEarth makes arev- olution about itsown axisevery 24hours, wewillobserve twohigh tides aday. Aquick calculation shows thattheSun’s gravitational attraction isabout 175 times stronger than theMoon’s onEarth’s surface, sowewould expect tidal forces from theSun aswell. The tidal force calculation issimilar totheonewe have just performed fortheMoon. The result (Problem 5-18) isthat thetidal force due tothe Sun is0.46 that ofthe Moon, asizable effect. Despite the stronger attraction duetotheSun, thegravitational force gradient over thesur- face ofEarth ismuch smaller, because ofthemuch larger distance totheSun. EXAMPLE 5.5 5 - T Calculate themaximum height change intheocean tides caused bytheMoon. Solution. Wecontinue touseoursimple model oftheocean surrounding Earth. Newton proposed asolution tothiscalculation byimagining thattwo wells bedug, onealong thedirection ofhigh tide (our x-axis) andonealong thedirection oflowtide (our y-axis). Ifthetidal height change wewant tode- termine ish,then thedifference inpotential energy ofmass mduetothe height difference ismgh. Let’s calculate thedifference inwork ifwemove the mass mfrom point cinFigure 5-12 tothecenter ofEarth and then topoint a. This work Wdone bygravity must equal thepotential energy change mgh. The work Wis O r+52 W= JFndy +JF;-xdxr+51 0 where weusethetidal forces FT,andFT,ofEquations 5.54. The small distances 81and 52aretoaccount forthesmall variations from aspherical Earth, but these values aresosmall they canbehenceforth neglected. The value forW becomes _GmM,,, 0 'W-— —-D? T(—y)dy +02xdx GmM,,, .2 3GmM,,,r2=-- ~+.2=~=-—D3(2)2.3 5.5 OCEAN TIDES 203 J‘ C X (Z Earth FIGURE 5-12 Example 5.5.Wecalculate thework done tomove apoint mass mfrom point ctothecenter ofEarth andthen topoint a. Because thiswork isequal tomgh, wehave 3GmMmr2 nigh=Mi2D3 1.-3GM"’T2 555)_2gD3 (l Note thatthemass mcancels, andthevalue ofhdoes notdepend onm.Nor does itdepend onthesubstance, sototheextent Earth isplastic, similar tidal effects should be(and are) observed forthesurface land. Ifweinsert the known values oftheconstants into Equation 5.55, weﬁnd 3(6.67 X10'“m3/kg-s2)(7.350 Xl022kg) (6.37 ><l06m)2 1.=-4 - -=0.542(9.s0 m/s2)(3.84 ><l08m)3 m The highest tides (called spring tides) occur when Earth, theMoon, andthe Sun arelined up(new moon and fullmoon), and thesmallest tides (called neap tides) occur fortheﬁrst and third quarters oftheMoon when theSun and Moon areatright angles toeach other, partially cancelling their effects. The maximum tide, which occurs every 2weeks, should bel.46h =0.83 mforthe spring tides. Anobserver who hasspent much time near theocean hasnoticed that typi- caloceanshore tides aregreater than those calculated inExample 5.5. Several other effects come into play. Earth isnotcovered completely with water, andthe continents play asigniﬁcant role, especially the shelfs and narrow estuaries. Local effects can bedramatic, leading totidal changes ofseveral meters. The tides inmidocean, however, aresimilar towhat wehave calculated. Resonances canaffect thenatural oscillation ofthebodies ofwater and cause tidal changes. 204 5/GRAVITATION Tidal distortion (highly exaggerated) I I I I -eMoon 4- Moon's orbit Earth FIGURE 5-13 Some effects cause thehigh tides tonotbeexactly along theEarth-Moon axis. Tidal friction between water andEarth leads toasigniﬁcant amount ofenergy lossonEarth. Earth isnotrigid, anditisalsodistorted bytidal forces. Inaddition totheeffects just discussed, remember that asEarth rotates, the Moon isalsoorbiting Earth. This leads totheresult thatthere arenotquite ex- actly twohigh tides perday, because they occur once every 12hand 26min (Problem 5-19). The plane ofthemoon’s orbit about Earth isalso notperpendi- cular toEarth’s rotation axis. This causes one high tideeach daytobeslightly higher than theother. The tidal friction between water and land mentioned pre- viously also results inEarth “dragging” theocean with itasEarth rotates. This causes thehigh tides tobenotquite along theEarth-Moon axis, butrather sev- eraldegrees apart asshown inFigure 5-13. PROBLEMS 5-1. Sketch theequipotential surfaces and thelines offorce fortwopoint masses sepa- rated byacertain distance. Next, consider one ofthemasses tohave aﬁctitious negative mass —M. Sketch theequipotential surfaces and lines offorce forthis case. Towhat kind ofphysical situation does this setofequipotentials and ﬁeld lines apply? (Note that thelines offorce have direction; indicate thiswith appropri- atearrows.) 5-2. Iftheﬁeld vector isindependent oftheradial distance within asphere, ﬁnd the function describing thedensity p=p(r) ofthesphere. PROBLEMS 205 5-3. 5-4. 5-5. 5-6. 5-7. 5-8. 5-9. 5-10. 5-11 5-12. 5-I3.Assuming thatairresistance isunimportant, calculate theminimum velocity apar- ticle must have atthesurface ofEarth toescape from Earth’s gravitational ﬁeld. Obtain anumerical value fortheresult. (This velocity iscalled theescape velocity.) Aparticle atrest isattracted toward acenter offorce according totherelation FI -mk2/xi. Show that thetime required fortheparticle toreach theforce center from adistance disd2/k. - Aparticle falls toEarth starting from restatagreat height (many times Earth’s radius). Neglect airresistance and show that theparticle requires approximately T91 ofthetotal time offalltotraverse theﬁrsthalfofthedistance. Compute directly thegravitational force onaunit mass atapoint exterior toaho- mogeneous sphere ofmatter. Calculate thegravitational potential due toathin rodoflength land mass Mata distance Rfrom thecenter oftherodand inadirection perpendicular totherod. Calculate thegravitational ﬁeld vector due toahomogeneous cylinder atexterior points ontheaxis ofthecylinder. Perform thecalculation (a)bycomputing the force directly and (b)bycomputing thepotential ﬁrst. Calculate thepotential due toathin circular ring ofradius aand mass Mforpoints lying intheplane ofthering andexterior toit.The result canbeexpressed asan elli ticinte ral.* Assume that thedistance from thecenter oftherin totheﬁeldP 8 8 oint islarecom ared with theradius oftherin .Exand theexression fortheP 8 P 8 P P potential and ﬁnd theﬁrst correction tenn. Find thepotential atoff—axis points due toathin circular ring ofradius aand mass M.LetRbethedistance from thecenter ofthering totheﬁeld point, and let6be theangle between theline connecting thecenter ofthering with theﬁeld point and theaxis ofthering. Assume R>> asothat terms oforder (a/R)?’ and higher may beneglected. Consider amassive body ofarbitrary shape and aspherical surface that isexterior toand does notcontain thebody. Show that theaverage value ofthepotential due tothebody taken over thespherical surface isequal tothevalue ofthepotential at thecenter ofthesphere. Intheprevious problem, letthemassive body beinside thespherical surface. Now show that theaverage value ofthepotential over thesurface ofthesphere isequal tothevalue ofthepotential thatwould exist onthesurface ofthesphere ifallthe mass ofthebody were concentrated atthecenter ofthesphere. Aplanet ofdensity pl(spherical core, radius R1)with athick spherical cloud of dust (density p2,radius R2)isdiscovered. “That istheforce onaparticle ofmass m placed within thedust cloud? *See Appendix Bforalistofsome elliptic integrals. 206 5-14 5-15 5-16 5-17 5-18. 5-19. 5-20 5-215/GRAVITATION Show thatthegravitational self-energy (energy ofassembly piecewise from inﬁnity) ofauniform sphere ofmass Mand radius Ris 5011/12U=-—-—5R Aparticle isdropped into ahole drilled straight through the center ofEarth. Neglecting rotational effects, show that theparticle’s motion issimple harmonic if you assume Earth hasuniform density. Show that theperiod oftheoscillation is about 84min. Auniformly solid sphere ofmass Mand radius Risﬁxed adistance habove athin inﬁnite sheet ofmass density p,(mass/area). With what force does thesphere at- tract thesheet? Newton’s model ofthetidal height, using thetwowater wells dug tothecenter of Earth, used thefact that thepressure atthebottom ofthetwowells should bethe same. Assume water isincompressible and ﬁnd the tidal height difference h, Equation 5.55, due totheMoon using thismodel. (Hint: f§'“"pg,dy =f§‘"‘“pg,,dx; h=xmx —ymax, where xm,,,, +ym =2Re,,,,h, and Rcmh isEarth’s median radius.) Show thattheratio ofmaximum tidal heights duetotheMoon andSunisgiven by Mm RE,3 M,D and that thisvalue is2.2.RE,isthedistance between theSun and Earth, and M,is theSun’s mass. The orbital revolution oftheMoon about Earth takes about 27.3 days and isinthe same direction asEarth’s rotation (24h).Usethisinformation toshow that high tides occur everywhere onEarth every 12hand 26min. Athin disk ofmass Mand radius Rliesinthe(x,y)plane with thez-axis passing through thecenter ofthedisk. Calculate thegravitational potential (z) andthe gravitational ﬁeld g(z) =—V(z) =—l§d¢'(z)/dz onthez.-axis. Apoint mass mislocated adistance Dfrom thenearest end ofathin rodofmass M and length Lalong theaxis oftherod. Find thegravitational force exerted onthe point mass bytherod. Z I gCHAPTER \) Some Methods inthe Calculus ofVariations 6.1Introduction Many problems inNewtonian mechanics aremore easily analyzed bymeans of alternative statements ofthelaws, including Lagrange’s equation andHamilton’s princip1e.* Asaprelude tothese techniques, weconsider inthischapter some general principles ofthetechniques ofthecalculus ofvariations. Emphasis willbeplaced onthose aspects ofthetheory ofvariations that have adirect bearing onclassical systems, omitting some existence proofs. Our primary interest here isindetermining thepath that gives extremum solutions, forexample, theshortest distance (ortime) between twopoints. Awell-known example oftheuseofthetheory ofvariations isFermat’s principle: Light travels bythepath that takes theleast amount oftime (see Problem 6-7). 6.2 Statement oftheProblem The basic problem ofthecalculus ofvariations istodetermine thefunction y(x) such that theintegral ]= J2f{y(x), y'(x); x}dx (6.1) *The development ofthecalculus ofvariations wasbegun byNewton (1686) andwasextended by _]ohann andjakob Bernoulli (1696) andbyEuler (1744). Adrien Legendre (1786),]oseph Lagrange (1788), Hamilton (1833), andJacobi (1837) allmade important contributions. Thenames ofPeter Dirichlet (1805-1859) and Karl Weierstrass (1815-1879) areparticularly associated with theestab- lishment ofarigorous mathematical foundation forthesubject. 207 208 6/SOME METHODS INTHECALCULUS OFVARIATIONS 9 y(x)+w1(x) Varied path Extremum path, y(x) 3.._--___>—lR------___.B9_. ffi W x FIGURE 6-1 The function y(x)isthepath that makes thefunctional ]anextremum. Theneighboring functions y(x) +a"r](x) vanish attheendpoints and may beclose toy(x),butarenottheextremum. isanextremum (i.e., either amaximum oraminimum). InEquation 6.1, y'(x) Edy/dx, andthesemicolon infseparates theindependent variable xfrom thedependent variable y(x) anditsderivative y'(x). The functional* ]depends onthefunction y(x), andthelimits ofintegration areﬁxed.l Thefunction y(x)is then tobevaried until a”nextreme value of]isfound. Bythiswemean thatifa function y=y(x)gives theintegral ]aminimum value, then anynez'ghbm"ingfunc- tion, nomatter how close toy(x), must make ]increase. The deﬁnition ofa neighboring function may bemade asfollows. Wegiveallpossible functions ya parametric representation yIy(a, x)such that, foraI0,y=y(0,x)Iy(x)is thefunction thatyields anextremum for Wecanthen write )’(¢Y,X)=y(0,X)+¢Y"'7(X) (5-2) where 1](x) issome function ofxthat hasacontinuous ﬁrst derivative and that vanishes atx1and x2,because thevaried function y(a, x)must beidentical with y(x) attheendpoints ofthepath: n(x1) ="r](x2) =0.The situation isdepicted schematically inFigure 6-1. Iffunctions ofthetype given byEquation 6.2areconsidered, theintegral ] becomes afunctional oftheparameter a: ](a) =r2f{y(0z, x),y'(a, x);x}dx (6.3) *The quantity ]isageneralization ofafunction called afunctional, actually anintegral functional in thiscase. Tltisnotnecessary thatthelimits ofintegration beconsidered ﬁxed. Ifthey areallowed tovary, theprob- lemincreases toﬁnding notonly y(x) butalso xlandx2such that] isanextremum. 6.2 STATEMENT OFTHE PROBLEM 209 The condition thattheintegral have astationary value (i.e., thatanextremum re- sults) isthat ]beindependent oforinﬁrst order along thepath giving theex- tremum (aI0),or,equivalently, that 6] _ -6;“=0—0 (6.4) forallfunctions 17(x). This isonly anecessary condition; itisnotsufﬁcient. EXAMPLE 6.1 -I - - - IT Consider thefunction fI(dy/dx)2, where y(x) Ix.Add toy(x)thefunction 1)(x) Isinx,andﬁnd](a) between thelimits ofxIOandxI21r.Show that thestationary value of](a) occurs foraI0. Solution. Wemay construct neighboring varied paths byadding toy(x), )’(X)=X (5-5) thesinusoidal variation 0:sinx, y(a,x)Ix+asinx (6.6) These paths areillustrated inFigure 6-2fororIOand fortwodifferent nonvan- ishing values ofoz.Clearly, thefunction "r)(x) Isinxobeys theendpoint condi- tions, thatis,17(0) I0I"r](21r)_ Todetermine f(y,y’;x)weﬁrstdetermine d,2% 1+acosx (6.7)x J’ y(0l,x)=+asin x r‘_"""_"""""""-""-X . I ls. I 0 7: 2n: FIGURE 6-2 Example 6.1.The various paths y(a, x)Ix+ozsinx.The extremum path occurs foraI0. 210 6/SOME METHODS INTHE CALCULUS OFVARIATIONS then _dy(a,x) 2 22 f— T I1+2acosx+a cos x (6.8) Equation 6.3now becomes 211' ](a) Ii(1+2acosx+0:2cos? x)dx (6.9) 0 I27r+(1217 (6.10) Thus weseethevalue of](a) isalways greater than ](0),nomatter what value (positive ornegative) wechoose for01.The condition ofEquation 6.4isalso satisﬁed. 6.3 Euler’s Equation Todetermine theresult ofthecondition expressed byEquation 6.4,weperform theindicated differentiation inEquation 6.3: 6] 3*1 .—I— ’;d .1160,aaLf{M X}X (6) Because thelimits ofintegration areﬁxed, thedifferential operation affects only theintegrand. Hence, 6 *1’66 66' l=l(ll+—f,l)dx (6.12)6a x,6y6a 6y6a From Equation 6.2,wehave 6y 6y’ dn —I ;—I— 6.1360: nu) 6a dx ( ) Equation 6.12 becomes a] Maf afen)—I — +—— d 6.14 aa (ayn<»<> ay,dxx <) The second term intheintegrand canbeintegrated byparts: JudvIuv—ivdu (6.15) *2afdn er*2"2daf—— dI— — —— 6.16 ix,6y’dx x6y’n(x) dx(6y' T'(x)dx ( ) 6.3EULER’S EQUATION 211 The integrated term vanishes because ’T](X1) I"r)(x2) I0.Therefore, Equation 6.12 becomes 6 *26 6 i: [55/I(x) —y€;(§)n(x)]dx _*2allii _Ll(6y dxay,)"r)(x)dx (6.17) The integral inEquation 6.17 now appears tobeindependent ofa.Butthe functions yand y’with respect towhich thederivatives offaretaken arestill functions ofa.Because (6]/6a)|a:0 must vanish fortheextremum value and be- cause n(x) isanarbitrary function (subject totheconditions already stated), the integrand inEquation 6.17 must itself vanish foraI0: 6f d6f ———— I0 El’ ' .1 ay dxay, uersequation (68) where now yand y’aretheoriginal functions, independent ofa.This result is known asEuler’s equation,* which isanecessary condition for]tohave anex- tremum value. EXAl\1PLE 6.2 Wecanusethecalculus ofvariations tosolve aclassic problem inthehistory of physics: thebrachistochronel Consider aparticle moving inaconstant force ﬁeld starting atrestfrom some point (x1,yl)tosome lower point (x2,3:2).Find the path thatallows theparticle toaccomplish thetransit intheleast possible time. Solution. The coordinate system may bechosen sothat thepoint (xl,yl)isat theorigin. Further, lettheforce ﬁeld bedirected along thepositive x-axis as inFigure 6-3.Because theforce ontheparticle isconstant—and ifweignore thepossibility offriction—the ﬁeld isconservative, andthetotal energy ofthe particle isT+UIconst. Ifwemeasure thepotential from thepoint xIO [i_e., U(x I0)IO],then, because theparticle starts from rest, T+UI0. The kinetic energy isTI%mv2, andthepotential energy isUI—Fx I—mgx, where gistheacceleration imparted bytheforce. Thus -u=\/2gx (6.19) Thetime required fortheparticle tomake thetransit from theorigin to(x2,yg)is (12J2)ds (dx2 +dy2)l/2 tI I Z (11191) (2gx)]/2 lUl12 1+y'2 1/2 Ii:0W dx (6.20) *Derived ﬁrst byEuler in1744. When applied tomechanical systems, thisisknown astheEuler- Lagrange equation. {First solved by_]ohann Bernoulli (1667-1748) in1696. 212 6/SOME METHODS INTHE CALCULUS OFVARIATIONS (xvF1) 7 T 1’ 1.W2»N2) l X FIGURE 6-3 Example 6.2.The brachistochrone problem istoﬁnd thepath ofaparticle moving from (x1,y1) to(X2, yg)that occurs intheleast possible time. Theforce ﬁeld acting ontheparticle isF,which isdown andconstant. The time oftransit isthequantity forwhich aminimum isdesired. Because the constant (2g)"1/2does notaffect theﬁnal equation, thefunction fmaybeiden- tiﬁed as 1+y'2 1/2 fI 7-— (6.21) And, because 6f/6y IO,theEuler equation (Equation 6.18) becomes d6_l:0dx6y' or 6 iiIconstant I(2a)"1/2 as where aisanew constant. Performing thedifferentiation 6f/6y’ onEquation 6.21 andsquaring the result, wehave 9'2 _L x(1+y'2) 2a This may beputintheform(6.22) _ xdx y—(Qax —x2)1/2 Wenow make thefollowing change ofvariable:(6.23) xIa(1—cos6) dxIasin6d6 (6.24) The integral inEquation 6.23 then becomes yIia(1— cos6)d6 6.3EULER’S EQUATION 213 (X1,311) Ira 277-'11’V r \\\\\\-Q‘~\\ \\\E‘TN\_-----_DI l “\\ A y \ I‘\ 1\ \ I \ \ I \ I 1 \- I \ i / \ F / \ / / \ ’ P(x,y) \ ,’ /(x2»J’2) ‘ 2a’ \“' I Cycloid X FIGURE 6-4 Example 6.2.The solution ofthebrachistochrone problem isacycloid. and yIa(6—sin6)+constant (6.25) The parametric equations foracycl0id* passing through theorigin are xIa(1—cos6) yIa(6—sin6)} (6.26) which isjust thesolution found, with theconstant ofintegration setequal to' zero toconform with therequirement that (0,0)isthestarting point ofthe motion. The path isthen asshown inFigure 6-4,andtheconstant amust be adjusted toallow thecycloid topass through thespeciﬁed point (x2,yg). Solving theproblem ofthebrachistochrone does indeed yield apath theparti- cletraverses inaminimum time. Buttheprocedures ofvariational calculus are designed only toproduce anextremum—either aminimum oramaximum. It isalmost always thecase indynamics thatwedesire (and ﬁnd) aminimum for theproblem. Consider thesurface generated byrevolving alineconnecting twoﬁxed points (x1,yl)and (x2,yg)about anaxis coplanar with thetwopoints. Find theequa- tionofthelineconnecting thepoints such thatthesurface area generated by therevolution (i.e., thearea ofthesurface ofrevolution) isaminimum. Solution. Weassume thatthecurve passing through (x1,y1) and (x2,y2)isre- volved about they-axis, coplanar with thetwopoints. Tocalculate thetotal area ofthesurface ofrevolution, weﬁrst ﬁnd thearea dAofastrip. Refer toFigure 6-5. *Acycloid isacurve traced byapoint onacircle rolling onaplane along alineintheplane. Seethe dashed sphere rolling along x=0inFigure 6-4. 214 6/SOME METHODS INTHE CALCULUS OFVARIATIONS III\\\<¢$=(M+4%)"? (X141) M. M 4 I__ I, I I xdA FIGURE 6-5 Example 6.3.The geometry oftheproblem and area dAareindicated to minimize thesurface ofrevolution around they-axis. dAI21Tx dsI21rx(dx2 +dy2)‘/2 (6.27) AI21rj x(1+ y'2)1/2 dx (6.28) where y’Idy/dx.Toﬁnd theextremum value welet f=xu+y"*>1/2 (6.29) andinsert into Equation 6.18: 6IZ0 as 6f icy’ 6),! (1+yI2)l/2 therefore, AXi’ _0dx (1+y'2)1/2 W (630) I Iconstant Ia .(1+yI2)1/2 From Equation 6.30, wedetermine ,_ a 3’_(x2_a2)1/2 (631) d,= (6.32) 6.3EULER’S EQUATION 215 The solution ofthisintegration is y=666$I1-1(5) +6 (6.63) where aandbareconstants ofintegration determined byrequiring thecurve to pass through thepoints (x1,y1) and(xg,y2). Equation 6.33 canalsobewritten as _ iIb x—acosh T (6.34) which ismore easily recognized astheequation ofacatenary, thecurve ofaﬂex- iblecord hanging freely between twopoints ofsupport. Choose twopoints located at(x1,y1) and (x2,y2)joined byacurve y(x). We want toﬁnd y(x)such thatifwerevolve thecurve around thex-axis, thesurface area oftherevolution isaminimum, This isthe“soap ﬁlm” problem, because a soap ﬁlmsuspended between twowire circular rings takes thisshape (Figure 6-6). Wewant tominimize theintegral ofthearea dAI21ry dswhere dsI \/1+ y'2dx and y’Idy/dx. A=27Tj )1V1 -l"y'2dX (6.35) Weﬁndtheextremum bysetting fIy\/1+y’2andinserting intoEquation 6.18. The derivatives weneed are if: \/1+y'2 5)’ Li9)’, \/1+y'2 J’ W2»Y2) (*1,)1) /1 II I’llIy ii 1 ‘\\\\ \\ \ ‘/R * z \(is: (dx2 +dy2)1/2__._L.____ ;g,;v_,,,H.E;X FIGURE 6-6 The “soap ﬁlm” problem inwhich wewant tominimize thesurface area of revolution around thex-axis. 216 6/SOME METHODS INTHECALCULUS OFVARIATIONS Equation 6.18 becomes 1+'2Il-L 6.6\/J’dxTy, <6) Equation 6.36 does notappear tobeasimple equation tosolve fory(x). Let’s stop and think about whether there might beaneasier method ofsolution. You may have noticed that this problem isjust like Example 6.3, butinthat case we were minimizing asurface ofrevolution about they-axis rather than around the ac-axis. The solution tothesoap ﬁlm problem should beidentical toEquation 6.34 ifweinterchange xand y.Buthow didweend upwith such acomplicated equation asEquation 6.36? Weblindly chose xastheindependent variable and decided toﬁnd thefunction y(x). Infact, ingeneral, wecanchoose theinde- pendent variable tobeanything wewant: x,6,t,oreven y.Ifwechoose yasthe independent variable, wewould need tointerchange xandyinmany ofthepre- vious equations that leduptoEuler’s equation (Equation 6.18). Itmight beeas- ierinthebeginning tojust interchange thevariables that westarted with (i.e., callthehorizontal axisyinFigure 6-6andlettheindependent variable bex).(In aright-handed coordinate system, theso-direction would bedown, butthatpres- ents nodifﬁculty inthis case because ofsymmetry.) Nomatter what wedo,the solution ofourpresent problem would justparallel Example 6.3.Unfortunately, itisnotalways possible tolook ahead tomake thebest choice ofindependent variable. Sometimes wejust have toproceed bytrial and error. 6.4 The “Second Form” oftheEuler Equation Asecond equation may bederived from Euler’s equation thatisconvenient for functions thatdonotexplicitly depend onx:6f/6xI0.Weﬁrstnote thatforany function f(y,y’;x)thederivative isasum ofterms df d 6fdy 6fdy' 6f _=_ ,'; I—_ _I+—dx dxf{y yX} 6ydx+6y’dx 6x 6 6 6 Iy'l+ y”—Lfj +If (6.37) 6y 6y 6x Also d /af //af —y—. =J’—.+>’——.dx 6y 6y dx6y or,substituting from Equation 6.37 fory"(6f/6y’), d 6f df 6f 6f d6f ' I — —' ’ 6.38dx(y 6y’) dx 6x y6y+ydx6y' ( ) 6.4 THE “SECOND FORM” OFTHE EULER EQUATION 217 The lasttwoterms inEquation 6.38 maybewritten as (12_2‘ydx6y’ 6y which vanishes inview oftheEuler equation (Equation 6.18). Therefore, ill _/if_6x dx y6y') -0 (639) Wecanusethisso-called “second form” oftheEuler equation incases inwhich f does notdepend explicitly onx,and6f/6xI0.Then, ,"’f_ af_ f—yI,—constant for— —0 (6.40)6y 6x EXAIVIPLE 6.4 Ageodesic isalinethatrepresents theshortest path between anytwopoints when thepath isrestricted toaparticular surface. Find thegeodesic ona sphere. Solution. The element oflength onthesurface ofasphere ofradius pisgiven ‘ (see Equation F.15 with drI0)by dsIp(d62 +sin2 6d¢2)1/2 (6.41) The distance sbetween points 1and 2istherefore 2 sIpj] +sin? 6:|l/2d</J (6.42) and, ifsistobeaminimum, fisidentiﬁed as fI (6'2 +sin2 6)]/2 (6.43) where 6'Id6/dd). Because 6f/6¢ I0,wemay usethesecond form ofthe Euler equation (Equation 6.40), which yields 6 (6'2 +sin? 6)]/2 —6'-a0,(6'2 +sin? 6)]/2 Iconstant Ia (6.44) Differentiating andmultiplying through by]§wehave sin26Ia(6'2 +sin?6)‘/2 (6.45) This may besolved fordd)/d6 I6'71, with theresult 61¢_ acsc26 646 d6_(1—a2csc26)1/2 (') 218 6/SOME METHODS INTHE CALCULUS OFVARIATIONS Solving forqb,weobtain cot6 4,=sin-1(—E—) +6. (6.47) where aistheconstant ofintegration andB2I(1—a2)/a2. Rewriting Equation 6.47 produces cot6 IBsin(¢—a) (6.48) Tointerpret thisresult, weconvert theequation torectangular coordinates by multiplying through bypsin6toobtain, onexpanding sin(¢ —a), (Bcosa)psin6sinqb—(Bsina)psin6cosqbIpcos6 (6.49) Because aandBareconstants, wemaywrite them as BcosaIA, BsinaIB (6.50) Then Equation 6.49 becomes A(psin6sin(fa)—B(psin6cos¢)I(pcos6) (6.51) The quantities intheparentheses arejusttheexpressions fory,x,andz,respec- tively, inspherical coordinates (seeFigure F-3,Appendix F);therefore Equation 6.51 maybewritten as Ay—BxIz (6.52) which istheequation ofaplane passing through thecenter ofthesphere. Hence thegeodesic onasphere isthepath thattheplane forms attheintersec- tionwith thesurface ofthesphere—a great circle. Note thatthegreat circle isthe maximum aswellastheminimum “straight-line” distance between twopoints onthesurface ofasphere. 6.5 Functions with Several Dependent Variables The Euler equation derived inthepreceding section isthesolution ofthevaria- tional problem inwhich itwasdesired toﬁnd thesingle function y(x) such that theintegral ofthefunctional fwasanextremum. The case more commonly en- countered inmechanics isthat inwhich fisafunctional ofseveral dependent variables: f=f{).<»<).I'.<»<).)2<»<>. yaw). .X} <6-53) orsimply _fIf{y,(x), y’,-(x); x}, iI1,2,, n (6.54) Inanalogy with Equation 6.2,wewrite 3’-"(OhX)=)1.-(0,X)+M7.-(X) (5-55) 6.6EULER’S EQUATIONS WHEN AUXILIARY CONDITIONS AREIMPOSED 219 The development proceeds analogously (cf.Equation 6.17), resulting in 6 "2 6 6 i= —Ia;f,i)17,(x)dx (6.56) Because theindividual variations—the 17,-(x)—are allindependent, thevanishing ofEquation 6.56 when evaluated ataI0requires theseparate vanishing ofeach expression inthebrackets: 6f d6f ‘-6 _ ___‘? I 0! II 1: a"' s 1 6% dxayg Z 2 n (657) 6.6 Euler’s Equations When Auxiliary Conditions AreImposed Suppose wewant toﬁnd, forexample, theshortest path between twopoints ona surface. Then, inaddition totheconditions already discussed, there isthecon- dition that thepath must satisfy theequation ofthesurface, say,g{y,-; x}I0. Such anequation wasimplicit inthesolution ofExample 6.4forthegeodesic on asphere where thecondition was g= —p2=0 (6.56) thatis, rIpIconstant (6.59) Butinthegeneral case, wemust make explicit useoftheauxiliary equation or equations. These equations arealsocalled equations ofconstraint. Consider the case inwhich f=f{i.»)2;X}=fl)’-3"-1-1'; X} (6-60) The equation corresponding toEquation 6.17 forthecase oftwovariables is i[_ “Qaf_daf 6y 6f_d6f 5; 6a—ix,i:(6y dx6y’)6a +(62 dx6z’)6a:|dx (6.61) Butnow there alsoexists anequation ofconstraint oftheform g{y.-;X}Igly-1; X}I0 (5-52) and thevariations 6y/6a and 6z/6a arenolonger independent, sotheexpres- sions inparentheses inEquation 6.61 donotseparately vanish ataI0. Differentiating gfrom Equation 6.62, wehave 6g6y6g61=—— —— d= 6.6dg(6)1601 6626(1)60 (3) 220 6/SOME METHODS INTHE CALCULUS OFVARIATIONS where noterm inxappears since 6x/6aI0.Now y(¢X.X)Iy(X)+m71(X) z(a,x)Iz(x)+a"r)2(x)} (6.64) Therefore, bydetermining 6y/6a and 62/6a from Equation 6.64 and inserting into theterm inparentheses ofEquation 6.63, which, ingeneral, must bezero, weobtain §mo=—§mw 6%) Equation 6.61 becomes g= -jigi)-).<x) +-;dxg)~rt<»<)j dx Factoring "r]1(x) outofthesquare brackets andwriting Equation 6.65 as "'12(x) ag/6)’ ""I1(x) 33‘/51 wehave a_66;<1a _a_(1a anyM—ix,ii<6y dx6j/6 (6z dx6z') (6g/6z):|6l (6)66 (6.66) This latter equation now contains thesingle arbitrary function r]1(x), which is notinanywayrestricted byEquation 6.64, and onrequiring thecondition of Equation 6.4,theexpression inthebrackets must vanish. Thus wehave (21ii’) :(2112') (66,,6ydx6y' 6y 6zdx6z' at ' The left-hand side ofthisequation involves only derivatives offand gwith re- spect toyand y’,and theright-hand side involves only derivatives with respect to zandz’.Because yandzareboth functions ofx,thetwosides ofEquation 6.67 maybesetequal toafunction ofx,which wewrite as—)t(x): 2:i2.....2-.6 d6' 6 6;dx; 62 (6.68) ————+)t(x)—I062 dx6z' 6z The complete solution totheproblem now depends onﬁnding three functions: y(x), z(x), and)t(x). Butthere arethreerelations thatmay beused: thetwoequa- tions (Equation 6.68) and theequation ofconstraint (Equation 6.62). Thus, there isasufﬁcient number ofrelations toallow acomplete solution. Note that here )t(x) isconsidered tobeundetermined *andisobtained asapart ofthesolu- tion. Thefunction )t(x) isknown asaLagrange undetermined multiplier. *The function )t(x)wasintroduced inLagrange’s Mécanique analytique (Paris, 1788). 6.6 EULER’S EQUATIONS I/VI-IEN AUXILIARY CONDITIONS ARE IMPOSED 221 Forthegeneral case ofseveral dependent variables and several auxiliary conditions, wehave thefollowing setofequations: 6 d6 6»I-—l: +Z1,-(x)§ =0 (6.69)33% dxayt 1 3).: gj-{y,-; x}I0 (6.70) IfiI1,2, ,m,andjI1,2, ...,n,Equation 6.69 represents mequations in m+nunknowns, butthere arealso thenequations ofconstraint (Equation 6.70). Thus, there arem+nequations inm+nunknowns, and thesystem is soluble. Equation 6.70 isequivalent tothesetofndifferential equations l—ll—lNJNJa- -=25.1),-= 0,{', ’m (6.71)'52:" J: in Inproblems inmechanics, theconstraint equations arefrequently differential equations rather than algebraic equations. Therefore, equations such asEquation 6.71 aresometimes more useful than theequations represented byEquation 6.70. (See Section 7.5foranampliﬁcation ofthispoint.) Consider adisk rolling without slipping onaninclined plane (Figure 6-7). 6 Determine theequation ofconstraint interms ofthe“coOrdinates”* yand 6. Solution. The relation between thecoordinates (which arenotindependent) is yIR6 (6.72) where Ristheradius ofthedisk. Hence theequation ofconstraint is g(y, 6)Iy—R6I0 (6.73) Ir rIIt 1 1‘1 V /I' r / (X FIGURE 6-7 Example 6.5.Adisk rolls down aninclined plane without Slipping. *These areactually thegeneralized coordinates discussed inSection 7.3;seealsoExample 7.9. 222 6/SOME METHODS INTHECALCULUS OFVARIATIONS and ag__ ag_ay-1, 69—R (6.74) arethequantities associated with )1,thesingle undetermined multiplier forthis case.nu ___|_ i 13$ 7 _——I@n_ _ The constraint equation canalso appear inanintegral form. Consider the isoperimetric problem that isstated asﬁnding thecurve yIy(x) forwhich the functional 1» ][y] Ij_fb1,y'; x}dx (6.75) hasanextremum, andthecurve y(x)satisﬁes boundary conditions y(a) IAand y(b) IBaswell asthesecond functional b K[y] Ijg{y,y’; xjdx (6.76) thathasaﬁxed value forthelength ofthecurve (6).This second functional rep- resents anintegral constraint. Similarly towhat wehave done previously,* there willbeaconstant Asuch thaty(x) istheextremal solution ofthefunctional b j(f+ )tg)dx. (6.77) The curve y(x) then willsatisfy thedifferential equation 6fdaf asdag—~ +)t —I I0 6.786y dx6y' (6)) dx6_y') ( ) subject totheconstraints y(a) IA,y(b) IB,andK[y] I6.Wewillwork anex- ample forthisso-called Dido Probleml EXAMPLE 6.6 6 One version oftheDido Problem istoﬁnd thecurve y(x) oflength 6bounded bythex-axisonthebottom thatpasses through thepoints (—a, 0)and (a,0) andencloses thelargest area. Thevalue oftheendpoints aisdetermined bythe problem. *For aproof, seeGe63, p.43. TThe isoperimetric problem wasmade famous byVirgil’s poem Aeneid, which described Queen Dido ofCarthage, whoin900B.C.wasgiven byalocal king asmuch land asshecould enclose with anox’s hide. Inorder tomaximize herclaim, shehadthehide cutintothinstrips andtiedthem endtoend. Sheapparently knew enough mathematics toknow thatforaperimeter ofagiven length, themaxi- mum area enclosed isacircle. 6.6 EULER’S EQUATIONS WHEN AUXILIARY CONDITIONS ARE IMPOSED 223 T dx /ﬁx) d€ y __ l x "-11 £1 FIGURE 6-8 Example 6.6.Wewant toﬁnd thecurve y(x)thatmaximizes thearea above theyI0line consistent with aﬁxed perimeter length. The curve must gothrough acI—aanda.The differential area dAIydx,andthe differential length along thecurve isdf. Solution. Wecanusetheequations justdeveloped tosolve thisproblem. We show inFigure 6-8thatthedifferential area dAIydx.Wewant tomaximize the area, soWewant toﬁnd theextremum solution forEquation 6.75, which becomes G ]I jydx (6.79) The constraint equations are y(x):y(—a) IO,y(a) I0and KI jd€ I6. (6.80) The differential length along thecurve dtI(dx2 +dy2)1/2 I(1+j/2)]/2 dx where y’Idy/dx. The constraint functional becomes ll KI L[1+j/2]]/2dx I6. (6.81) Wenow have y(x) Iyand g(x) I\/1+y'2,and weusethese functions in Equation 6.78. 6 6 6 6 ' 6) at 6) 6)(1-t1) Equation 6.78 becomes d y _11dxL1+ygWj 0 wsm Wemanipulate Equation 6.82 toﬁnd d y’ _I +y!2)]/2:| _A 224 6/SOME METHODS INTHE CALCULUS OFVARIATIONS Weintegrate over xtoﬁnd )ty' (i) where C1isanintegration constant. This canberearranged tobe _ i(x—C1)dx dy—II-Q‘-__ A2I(XIC1)6 This equation isintegrated toﬁnd )1: I \/A6 — (x— C])2 +C2 where C2isanother integration constant. Wecanrewrite thisastheequation of acircle ofradius )t. (X—CO2+(YIC2)2I/\2 (5-35) The maximum area isasemicircle bounded bytheyI0line. The semicircle must gothrough (x,y)points of(—a, 0)and (a,0),which means thecircle must becentered attheorigin, sothat C1I0IC2,and theradius IaI)t. The perimeter ofthetophalfofthesemicircle iswhat wecalled 6,andthe perimeter length ofahalf circle is"Ira.Therefore, wehave '1I'£lI(Z,and aI6/7r. 6.7 The 5Notation Inanalyses that usethecalculus ofVariations, wecustomarily useashorthand notation torepresent thevariation. Thus, Equation 6.17, which canbewritten as a 2-a,1aalint =j(l——l)l do:dx (6.86)6a ,.,6y dx6y’ 6a X2a a5]: l—I 5)dx (6.87)may beexpressed as ix,<6y dx6y) where at_d E 6a6 6]6)) (6.88) —da I5y6a The condition ofextremum then becomes 6]:6)f{y,y’;x)dx=0 (6.89) 6.7THE5NOTATION 225 y Varied path (x2’)9) Actual path (x1»Y1) —— x FIGURE 6-9 The varied path isavirtual displacement 5yfrom theactual path consistent with alltheforces and constraints. Taking thevariation symbol 5inside theintegral (because, byhypothesis, the limits ofintegration arenotaffected bythevariation) ,wehave 5]:i25fdx *2a a=T(lay+—J€5y')dx (6.90)x.By 6y But dy d 5'=5— =—5 6.91 9 (M) dx(3*) () so *2af af45= —5 ——5 6.9 J Qy+aydx >’)dx (2) Integrating thesecond term byparts asbefore, weﬁnd _x’allii’ 5]—Ll(6)) away’) 5ydx (6.93) Because thevariation 5yisarbitrary, theextremum condition 5]=Orequires the integrand tovanish, thereby yielding theEuler equation (Equation 6.18). Although the5notation isfrequently used, itisimportant torealize thatitis only ashorthand expression ofthemore precise differential quantities. Thevaried path represented by5ycanbethought ofphysically asavirtual displacement from theactual path consistent with alltheforces andconstraints (seeFigure 6-9). This variation 5yisdistinguished from anactual differential displacement dybythe condition thatdt=O—thatis,thattime isﬁxed. Thevaried path 5y,infact, need noteven correspond toapossible path ofmotion. The variation must vanish atthe endpoints. 226 6/SOME METHODS INTHE CALCULUS OFVARIATIONS PROBLEMS 6-1 6_ 6-3 6-4 U 6-5. 6-6. 6-7. 6-8. 6-9.2.Consider theline connecting (x1,yl)=(0,0)and (x2,y2)=(1,1).Show explicitly that thefunction y(x) =xproduces aminimum path length byusing thevaried function y(a, x)=x+asin1r(1 —x).Use theﬁrst fewterms intheexpansion of theresulting elliptic integral toshow theequivalent ofEquation 6.4. Show that theshortest distance between twopoints onaplane isastraight line. Show that theshortest distance between twopoints in(three-dimensional) space is astraight line. Show thatthegeodesic onthesurface ofaright circular cylinder isasegment ofa helix. Consider thesurface generated byrevolving aline connecting two ﬁxed points (x1,y1) and (x2,312)about anaxiscoplanar with thetwopoints. Find theequation oftheline connecting thepoints such that thesurface area generated bytherevo- lution (i.e., thearea ofthesurface ofrevolution) isaminimum. Obtain thesolu- tion byusing Equation 6.39. Reexamine theproblem ofthebrachistochrone (Example 6.2) and show that the time required foraparticle tomove (frictionlessly) totheminimum point ofthecy- cloid is17\/a/g, independent ofthestarting point. Consider light passing from one medium with index ofrefraction n1into another medium with index ofrefraction n2(Figure 6-A). UseFermat’s principle tomini- mize time, andderive thelawofrefraction: n1sin61=",2sin62. 5° :l—| 'Fl>’!'-E"2<2.> .5“ FIGURE 6-A Problem 6-'7. Find thedimensions oftheparallelepiped ofmaximum volume circumscribed by (a)asphere ofradius R;(b)anellipsoid with semiaxes a,b,c. Find anexpression involving thefunction d>(x1, x2,x3)that hasaminimum average value ofthesquare ofitsgradient within acertain volume Vof space. PROBLEMS 227 6-10 6-11 6-12 6-13. 6-14. 6-15 6-16. 6-17 6-18.Find theratio oftheradius Rtotheheight Hofaright-circular cylinder ofﬁxed volume Vthat minimizes thesurface area A. Adisk ofradius Rrolls without slipping inside theparabola y=ax? Find theequa- tion ofconstraint. Express thecondition thatallows thedisk torollsothatitcon- tacts theparabola atone and only one point, independent ofitsposition. Repeat Example 6.4, ﬁnding theshortest path between anytwopoints onthesur- face ofasphere, butusethemethod oftheEuler equations with anauxiliary con- dition imposed. Repeat Example 6.6butdonotusetheconstraint thatthey=0lineisthebottom artofthearea. Show that the lane curve ofaiven len th,which encloses amax- P P g 8 imum area, isacircle. Find theshortest path between the(x,y,z)points (O,-*1, O)and (O,1,O)onthe conical surface z=1-\/x2+312.What isthelength ofthepath? Note: thisisthe shortest mountain path around avolcano. (a)Find thecurve y(x) that passes through theendpoints (O,O)and (1,1)and min- imizes thefunctional I[y] =f6[(dy/six)? —-y2]dx. (b)VVhat istheminimum value oftheintegral? (c)Evaluate I[y] forastraight line y=xbetween thepoints (O,O) and (1,1). (a)What curve onthesurface z=x3/Qjoining thepoints (x,y,z)=(O,O,O)and (1,1,1)hastheshortest arclength? (b)Use acomputer toproduce aplot showing thesurface and theshortest curve onasingle plot. The corners ofa rectangle lieontheellipse (x/a) 2+(y/b) 2=1.(a)‘Where should thecorners belocated inorder tomaximize thearea oftherectangle? (b)What fraction ofthearea oftheellipse iscovered bytherectangle with maximum area? Aparticle ofmass misconstrained tomove under gravity with nofriction onthe surface xy=z.What isthetrajectory oftheparticle ifitstarts from restat(x,y,z)= (1,-1, -1) with thez-axis vertical? I CPLRPTER Hamilton ’sPrincijile--— Lagiangian and Hamiltonian Dynamics 7.1Introduction Experience hasshown that aparticle’s motion inaninertial reference frame is correctly described bytheNewtonian equation F=p.Iftheparticle isnotre- quired tomove insome complicated manner andifrectangular coordinates are used todescribe themotion, then usually theequations ofmotion arerelatively simple. Butifeither ofthese restrictions isremoved, theequations canbecome quite complex and difﬁcult tomanipulate. Forexample, ifaparticle iscon- strained tomove onthesurface ofasphere, theequations ofmotion result from theprojection oftheNewtonian vector equation onto that surface. The repre- sentation oftheacceleration vector inspherical coordinates isaformidable expression, asthereader who hasworked Problem 1-25 canreadily testify. Moreover, ifaparticle isconstrained tomove onagiven surface, certain forces must exist (called forces ofconstraint) thatmaintain theparticle incon- tactwith thespeciﬁed surface. Foraparticle moving onasmooth horizontal sur- face, theforce ofconstraint issimply F,=—mg.But, iftheparticle is,say,abead sliding down acurved wire, theforce ofconstraint canbequite complicated. Indeed, inparticular situations itmaybedifficult oreven impossible toobtain ex- plicit expressions fortheforces ofconstraint. Butinsolving aproblem byusing theNewtonian procedure, wemust know alltheforces, because thequantity F thatappears inthefundamental equation isthetotalforce acting onabody. Tocircumvent some ofthepractical difficulties that arise inattempts to apply Newton’s equations toparticular problems, alternate procedures may be 228 7.2HAMILTON’S PRINCIPLE 229 developed. Allsuch approaches areinessence aposteriori, because weknow before- hand thataresult equivalent totheNewtonian equations must beobtained. Thus, toeffect asimplification weneed notformulate anewtheory ofmechanics—the Newtonian theory isquite correct—but only devise analternate method ofdeal- ingwith complicated problems inageneral manner. Such amethod isCon- tained inHamilton’s Principle, andtheequations ofmotion resulting from the application ofthisprinciple arecalled Lagrange’s equations. IfLagrange’s equations aretoconstitute aproper description ofthedynam- icsofparticles, they must beequivalent toNewton’s equations. Ontheother hand, Hamilton’s Principle canbeapplied toawide range ofphysical phenom- ena (particularly those involving ﬁelds) notusually associated with Newton’s equations. Tobesure, each oftheresults thatcanbeobtained from Hamilton’s Principle wasﬁrst obtained, aswere Newton’s equations, bythecorrelation of experimental facts. Hamilton’s Principle hasnotprovided uswith anynewphysical theories, butithasallowed asatisfying uniﬁcation ofmany individual theories by asingle basic postulate. This isnotanidleexercise inhindsight, because itisthe goal ofphysical theory notonly togiveprecise mathematical formulation toob- served phenomena butalsotodescribe these effects with aneconomy offunda- mental postulates andinthemost unified manner possible. Indeed, Hamilton’s Principle isoneofthemost elegant andfar-reaching principles ofphysical theory. Inview ofitswide range ofapplicability (even though thisisanafter-the-fact discovery), itisnotunreasonable toassert that Hamilton’s Principle ismore “fundamental” than Newton’s equations. Therefore, weproceed byﬁrstpostulat- ingHamilton’s Principle; wethen obtain Lagrange’s equations and show that these areequivalent toNewton’s equations. Because wehave already discussed (inChapters 2,3,and4)dissipative phe- nomena atsome length, wehenceforth conﬁne ourattention toconservative systems. Consequently, wedonotdiscuss themore general setofLagrange’s equations, which take into account theeffects ofnonconservative forces. The reader isreferred totheliterature forthese details.* 7.2Hamilton’s Principle Minimal principles inphysics have along andinteresting history. The search for such principles ispredicated onthenotion thatnature always minimizes certain important quantities when aphysical process takes place. The ﬁrst such mini- mum principles were developed intheﬁeld ofoptics. Hero ofAlexandria, inthe second century B.C.,found thatthelawgoverning thereﬂection oflight could be obtained byasserting thatalight ray,traveling from onepoint toanother byare- ﬂection from aplane mirror, always takes theshortest possible path. Asimple geometric construction verifies thatthisminimum principle does indeed lead to *See, forexample, Goldstein (G080, Chapter 2)or,foracomprehensive discussion, Whittaker (Wh37, Chapter 8). 230 7/HAMILTON’S PRINCIPLE—LAGRANGIAN ANDHAMILTONIAN DYNAMICS theequality oftheangles ofincidence and reﬂection foralight rayreﬂected from aplane mirror. Her0’s principle oftheshortest path cannot, however, yield a correct lawforrefraction. In1657, Fermat reformulated theprinciple bypostulat- ingthat alight rayalways travels from onepoint toanother inamedium bya path that requires theleast time.* Fermat’s principle ofleast timeleads immedi- ately, notonly tothecorrect lawofreﬂection, butalsotoSnell’s lawofrefraction (seeProblem 6-7).l Minimum principles continued tobesought, andinthelatter part ofthesev- enteenth century thebeginnings ofthecalculus ofvariations were developed by Newton, Leibniz, andtheBernoullis when such problems asthebrachistochrone (seeExample 6.2)andtheshape ofahanging chain (acatenary) were solved. Theﬁrstapplication ofageneral minimum principle inmechanics wasmade in1747 byMaupertuis, who asserted thatdynamical motion takes place with min- imum action? Maupertuis’s principle ofleast action wasbased ontheological grounds (action isminimized through the“wisdom ofGod”), andhisconcept of “action” wasrather vague. (Recall thataction isaquantity with thedimensions of length Xmomentum orenergy Xtime.) Only later wasaﬁrm mathematic foundation oftheprinciple given byLagrange (1760). Although itisauseful form from which tomake thetransition from classical mechanics tooptics and toquantum me- chanics, theprinciple ofleast action islessgeneral than Hamilton’s Principle and, indeed, canbederived from it.Weforego adetailed discussion here.§ In1828, Gauss developed amethod oftreating mechanics byhisprinciple of least constraint; amodiﬁcation waslater made byHertz and embodied inhis principle ofleast curvature. These principles" areclosely related toHamilton’s Principle andaddnothing tothecontent ofHamilton’s more general formula- tion; their mention only emphasizes thecontinual concern with minimal princi- plesinphysics. Intwopapers published in1834 and 1835, Hamilton‘ announced thedy- namical principle onwhich itispossible tobase allofmechanics and, indeed, most ofclassical physics. Hamilton’s Principle may bestated asfollows“: Ofallthepossible paths along which adynamical system maymove from one point toanother within aspecified time interval (consistent with anycon- straints), theactual pathfollowed isthatwhich minimizes thetimeintegral ofthe difference between thekinetic andpotential energies. *Pierre deFermat (1601-1665), aFrench lawyer, linguist, andamateur mathematician. ﬂn1661, Fermat correctly deduced thelawofrefraction, which hadbeen discovered experimentally inabout 1621 byWillebrord Snell (1591-1626), aDutch mathematical prodigy. IPierre-Louise-Moreau deMaupertuis (1698-1759), French mathematician and astronomer. The ﬁrstusetowhich Maupertuis puttheprinciple ofleast action wastorestate Fermat’s derivation of thelawofrefraction (1744). §See, forexample, Goldstein (G080, pp.365-371) orSommerfeld (S050, pp.204-209). |lSee, forexample, Lindsay and Margenau (Li36, pp. 112-120) orSommerfeld (S050, pp. 210-214). 1Sir William Rowan Hamilton (1805-1865), Irish mathematician andastronomer, andlater, Irish Astronomer Royal. **The general meaning of“thepath ofasystem” ismade clear inSection 7.3. 7.2HAMILTON’S PRINCIPLE 231 Interms ofthecalculus ofvariations, Hamilton’s Principle becomes 5[t2(T— U)dt=0 (7.1) where thesymbol 5isashorthand notation todescribe thevariation discussed in Sections 6.3and 6.7.This variational statement oftheprinciple requires only that theintegral ofT——Ubeanextremum, notnecessarily aminimum. Butinal- most allimportant applications indynamics, theminimum condition occurs. The kinetic energy ofaparticle expressed inﬁxed, rectangular coordinates isafunction only ofthe913,-,andiftheparticle moves inaconservative force ﬁeld, thepotential energy isafunction only ofthex,-: T:T1951), U: U(xi) Ifwedeﬁne thedifference ofthese quantities tobe LET—U= L(x,-, :2,-) (7.2) 52 6iL(x,-,i2,)dt=0 (7.3)tl The function Lappearing inthisexpression maybeidentiﬁed with thefunction fofthevariational integral (seeSection 6.5),then Equation 7.1becomes 8rim.-<x>. y£(x);xldx ifwemake thetransformations x-—>t )’i(-7‘) "2xi(t) >»:<x>->ii.-<0 ffy.-(x),)’i(x);x}->Lo.-.it.-) The Euler-Lagrange equations (Equation 6.57) corresponding toEquation 7.3 aretherefore BL d5L Q""Z: =O,i=1,2,3 Lagrange equations ofmotion (7.4) These aretheLagrange equations ofmotion fortheparticle, andthequantity L iscalled theLagrange fl.l1'lCl210Il orLagrangian fortheparticle. 232 7/HAMILTON’S PRINCIPLE-—LAGRANGIAN AND HAMILTONIAN DYNAMICS Bywayofexample, letusobtain theLagrange equation ofmotion forthe one-dimensional harmonic oscillator. With theusual expressions forthekinetic andpotential energies, wehave 1 1L=T—U=§mF—§M2 BL—=-kx(ix BL _,=mx5x d€ﬁ u— ,=mxdt5x Substituting these results into Equation 7.4leads to mii+kx=O which isidentical with theequation ofmotion obtained using Newtonian mechanics. The Lagrangian procedure seems needlessly complicated ifitcanonly du- plicate thesimple results ofNewtonian theory. However, letuscontinue illustrat- ingthemethod byconsidering theplane pendulum (see Section 4.4). Using Equation 4.23 forTand U,wehave, fortheLagrangian function 1 . L==§ml262 —-mgl(_1 -ecos6) Wenow treat 6as itwere arectangular coordinate and apply theoperations speci- ﬁedinEquation 7.4;weobtain 5Lg=—mgl sin6 6L . .=ml26 66 d8L -—( Iml26dim 5+€mm=0 which again isidentical with theNewtonian result (Equation 4.21). This isa remarkable result; ithasbeen obtained bycalculating thekinetic andpotential energies interms of6rather than xand then applying asetofoperations de- signed forusewith rectangular rather than angular coordinates. Wearetherefore ledtosuspect thattheLagrange equations aremore general anduseful than the fOrIn ofEquation 7.4would indicate. Wepursue thismatter inSection 7.4. Another important characteristic ofthemethod used inthetwopreceding simple examples isthatnowhere inthecalculations didthere enter anystatement 7.3 GENERALIZED COORDINATES 233 regarding force. The equations ofmotion were obtained only byspecifying certain properties associated with theparticle (the kinetic and potential energies), and without thenecessity ofexplicitly taking into account thefactthat there wasan external agency acting ontheparticle (the force). Therefore, insofar asenergy can bedeﬁned independently ofNewtonian concepts, Hamilton’s Principle allows us tocalculate theequations ofmotion ofabody completely without recourse to Newtonian theory. Weshall return tothisimportant point inSections 7.5and7.7. 7.3Generalized Coordinates Wenow seek totake advantage oftheﬂexibility inspecifying coordinates that thetwo examples ofthepreceding section have suggested isinherent in Lagrange’s equations. Weconsider ageneral mechanical system consisting ofacollection ofndis- crete point particles, some ofwhich may beconnected toform rigid bodies. We discuss such systems ofparticles inChapter 9andrigid bodies inChapter 11.To specify thestate ofsuch asystem atagiven time, itisnecessary tousenradius vectors. Because each radius vector consists ofthree numbers (e.g., therectan- gular coordinates), 3nquantities must bespeciﬁed todescribe thepositions of alltheparticles. Ifthere exist equations ofconstraint that relate some ofthese coordinates toothers (aswould bethecase, forexample, ifsome oftheparticles‘ were connected toform rigid bodies orifthemotion were constrained tolie along some path oronsome surface), then notallthe3ncoordinates areinde- pendent. Infact, ifthere aremequations ofconstraint, then 3n~mcoordinates areindependent, andthesystem issaidtopossess 3n—mdegrees offreedom. Itisimportant tonote thatifs=3n—-mcoordinates arerequired inagiven case, weneed notchoose srectangular coordinates oreven scurvilinear coordi- nates (e.g., spherical, cylindrical). Wecanchoose anysindependent parameters, aslong asthey completely specify thestate ofthesystem. These squantities need noteven have thedimensions oflength. Depending ontheproblem athand, it may prove more convenient tochoose some oftheparameters with dimensions ofenergy, some with dimensions of(length)2, some that aredimensionless, andso forth. InExample 6.5,wedescribed adisk rolling down aninclined plane in terms ofonecoordinate thatwasalength andonethatwasanangle. Wegivethe name generalized coordinates toanysetofquantities that completely speciﬁes thestate ofasystem. The generalized coordinates arecustomarily written as ql,qg,...,orsimply astheqj.Asetofindependent generalized coordinates whose number equals thenumber sofdegrees offreedom ofthesystem andnot restricted bytheconstraints iscalled aproper setofgeneralized coordinates. In certain instances, itmay beadvantageous tousegeneralized coordinates whose number exceeds thenumber ofdegrees offreedom and toexplicitly take into account theconstraint relations through theuseoftheLagrange undetermined multipliers. Such would bethecase, forexample, ifwedesired tocalculate the forces ofconstraint (seeExample 7.9). 234 7/HAMILTON’S PRINCIPLE—LAGRANGIAN ANDHAMILTONIAN DYNAMICS The choice ofasetofgeneralized coordinates todescribe asystem isnot unique; there areingeneral many setsofquantities (infact, aninfinite number!) thatcompletely specify thestate ofagiven system. Forexample, intheproblem ofthedisk rolling down theinclined plane, wemight choose ascoordinates the height ofthecenter ofmass ofthedisk above some reference level andthedis- tance through which some point ontherimhastraveled since thestart ofthe motion. The ultimate testofthe“suitability” ofaparticular setofgeneralized coordinates iswhether theresulting equations ofmotion aresufﬁciently simple toallow astraightforward interpretation. Unfortunately, wecanstate nogeneral rules forselecting the“most suitable” setofgeneralized coordinates foragiven problem. Acertain skillmust bedeveloped through experience, andwepresent many examples inthischapter. Inaddition tothegeneralized coordinates, wemay deﬁne asetofquantities consisting ofthetime derivatives of zjl,()2,...,orsimply Inanalogy with the nomenclature forrectangular coordinates, wecall thegeneralized velocities. Ifweallow forthepossibility thattheequations connecting x,,',-andqjexplic- itlycontain thetime, then thesetoftransformation equations isgiven by* b—4D—4 pageusa= ...,nxag.=xw-(ql, q2,... ,qs,t), {i =xay,-(qj, t), j==1,2,, s (7.5) Ingeneral, therectangular components ofthevelocities depend onthegeneral- ized coordinates, thegeneralized velocities, andthetime: 5%,;:*a,i(qj> 47]",17) (7-6) Wemay alsowrite theinverse transformations as qj“q,*(x..,.-. t) (7-7) ti;=qi,‘(X..,.-. 99.1,.»1‘) (7-3) Also, there arem=3n—-sequations ofconstraint oftheform }§,(xa,,~, t)=O,k=1,2, ,m (7.9) Find asuitable setofgeneralized coordinates forapoint particle moving onthe surface ofahemisphere ofradius Rwhose center isattheorigin. Solution. Because themotion always takes place onthesurface, wehave x2+y2+z2-R2=O, z2O (7.10) Letuschoose asourgeneralized coordinates thecosines oftheangles between thex-,y-,andz-axes andthelineconnecting theparticle with theorigin. *Inthischapter, weattempt tosimplify thenotation byreserving thesubscript itodesignate rectan- gular axes; therefore, wealways have i=1,2,3. 7.3 GENERALIZED COORDINATES 235 Therefore, x y z 91: E, ‘I2=E» ‘Is=E (7.11) Butthesum ofthesquares ofthedirection cosines ofalineequals unity. Hence, qi+115+11%=1 (7-12) This setoflbdoes notconstitute aproper setofgeneralized coordinates, because wecanwrite qgasafunction ofqlandq2: ‘Is=\/1~qi—qi (7-13) Wemay, however, choose qlKx/Randqg=y/Rasproper generalized coordi- nates, andthese quantities, together with theequation ofconstraint (Equation 7.13) .1=\/R2-x2-)2 (7.14) aresufﬁcient touniquely specify theposition oftheparticle. This should bean obvious result, because only twocoordinates (e.g., latitude andlongitude) are necessary tospecify apoint onthesurface ofasphere. Buttheexample illus- trates thefactthattheequations ofconstraint canalways beused toreduce a trialsetofcoordinates toaproper setofgeneralized coordinates. EXAMPLE 7.2 ___ _- Usethe(x,y)coordinate system ofFigure 7-1toﬁnd thekinetic energy T,po- tential energy U,andtheLagrangian Lforasimple pendulum (length 6,mass bob m)moving inthex,yplane. Determine thetransformation equations from the(x,y)rectangular system tothecoordinate 6.Find theequation ofmotion. Solution. Wehave already examined thisgeneral problem inSections 4.4and 7.1.When using theLagrangian method, itisoften useful tobegin with J’ Q:N §x FIGURE 7-1 Example 7.2.Asimple pendulum oflength L’and bob ofmass m. 236 7/HAMILTON’S PRINCIPLE-LAGRANGIAN AND HAMILTONIAN DYNAMICS rectangular coordinates andtransform tothemost obvious system with the simplest generalized coordinates. Inthiscase, thekinetic andpotential energies andtheLagrangian become 1 1 T='2'mX2+§my2 U=mo 1. 1.L‘-= T-U=gmx2+§my2—-mgy Inspection ofFigure 7-1reveals thatthemotion canbebetter described by using 6and6.Let’s transform xandyinto thecoordinate 6andthen ﬁnd Lin terms of6. xI6sin6 y=—-6cos6 Wenow ﬁnd foriiand i'c=€6cos6 y=€6 sin6 L=g(€262cos26 +€262sin2 6)+mglicos6 =$6262 +mgtfcos6 The only generalized coordinate inthecase ofthependulum istheangle 6, andwehave expressed theLagrangian interms of6byfollowing asimple procedure ofﬁnding Linterms ofxandy,ﬁnding thetransformation equations, andthen inserting them into theexpression forL.Ifwedoaswedidinthe previous section andtreat 6asifitwerearectangular coordinate, wecanﬁnd the equation ofmotion asfollows: 6L (Q"-=—-mgtl sin6 BL . .==m€26 66 ear $5— .=m dt60 Weinsert these relations into Equation 7.4toﬁnd thesame equation ofmotion asfound previously. é+%mw=0 7.4 LAGRANGE’S EQUATIONS OFMOTION INGENERALIZED COORDINATES 237 The state ofasystem consisting ofnparticles and subject tomconstraints that connect some ofthe3nrectangular coordinates iscompletely speciﬁed by s=3n-mgeneralized coordinates. Wemay therefore represent thestate of such asystem byapoint inans-dimensional space called configuration space. Each dimension ofthisspace corresponds tooneoftheqjcoordinates. Wemay represent thetime history ofasystem byacurve inconﬁguration space, each point specifying theconﬁguration ofthesystem ataparticular instant. Through each such point passes aninﬁnity ofcurves representing possible motions of thesystem; each curve corresponds toaparticular setofinitial conditions. We may therefore speak ofthe“path” ofasystem asit“moves” through conﬁguration space. Butwemust becareful nottoconfuse thisterminology with thatapplied to themotion ofaparticle along apath inordinary three-dimensional space. Weshould alsonote that adynamical path inaconﬁguration space consis- ting ofproper generalized coordinates isautomatically consistent with thecon- straints onthesystem, because thecoordinates arechosen tocorrespond only to realizable motions ofthesystem. 7.4 Lagrange’s Equations ofMotion inGeneralized Coordinates Inview ofthedeﬁnitions inthepreceding sections, wemaynow restate Ham_ilton’s Principle asfollows: Ofallthepossible paths along which adynamical system maymove from one point toanother inconﬁguration space within aspecified timeinterval, theac- tualpath followed isthatwhich minimizes thetimeintegral oftheLagrangian function forthesystem. Tosetupthevariational form ofHamilton’s Principle ingeneralized coordi- nates, wemay take advantage ofanimportant property oftheLagrangian we have notsofaremphasized. The Lagrangian forasystem isdeﬁned tobethedif- ference between thekinetic andpotential energies. Butenergy isascalar quantity andsotheLagrangian isascalar function. Hence theLagrangian must beinvari- antwithrespect tocoordinate transformations. However, certain transformations that change theLagrangian but leave theequations ofmotion unchanged areallowed. For example, equations ofmotion are unchanged ifLisreplaced by L+d/dt[f(q,-, t)]forafunction f(q,-, t)with continuous second partial deriva- tives. Aslong aswedeﬁne theLagrangian tobethedifference between theki- netic andpotential energies, wemay usedifferent generalized coordinates. (The Lagrangian is,however, indeﬁnite toanadditive constant inthepotential energy U)Itistherefore immaterial whether weexpress theLagrangian interms ofxm, andaka’,or(5and L=r(»e,,,,.)-U(x,,y,-) =T(q,,1;,-,1)-U(q5~,t) (7.15) 238 7/HAMILTON’S PRINCIPLE-—LAGRANGIAN AND HAMILTONIAN DYNAMICS thatis, L=I-(411, 412’ ’qt?61,62> »‘ls;t) =L(q]~, rjj,t) (7.16) Thus, Hamilton’s Principle becomes $2 5fL(@, 6],-,t)dt==0 Hamilton’s Principle (7.17) tl Ifwerefer tothedeﬁnitions ofthequantities inSection 6.5and make the identiﬁcations x-—>t y.-(x)—>q,-(t) rE(x)—>(Lit) fir.-.91.‘;-r}—>L(¢1,» ii»'1) then theEuler equations (Equation 6.57)corresponding tothevariational prob- lemstated inEquation 7.17 become 6L ‘ML 0 '12 (718) i’_Ti:: s L: 1 )"')s ~ aq,diaq, 7 These aretheEuler-Lagrange equations ofmotion forthesystem (usually called simply Lagrange’s equations*). There aresofthese equations, andtogether with themequations ofconstraint and theinitial conditions that areimposed, they completely describe themotion ofthesystem.* Itisimportant torealize that thevalidity ofLagrange’s equations requires thefollowing twoconditions: 1.The forces acting onthesystem (apart from anyforces ofconstraint) must bederivable from apotential (orseveral potentials). 2.The equations ofconstraint must berelations that connect thecoordinates of theparticles andmay befunctions ofthetime—that is,wemust have con- straint relations oftheform given byEquation 7.9. Iftheconstraints canbeexpressed asincondition 2,they aretermed holonomic constraints. Iftheequations donotexplicitly contain thetime, theconstraints aresaid tobefixed orscleronomic; moving constraints arerheonomic. *First derived foramechanical system (although not,ofcourse, byusing Hamilton’s Principle) by Lagrange andpresented inhisfamous treatise Mécanique analytique in1788. Inthismonumental work, which encompasses allphases ofmechanics (statics, dynamics, hydrostatics, andhydrodynam- ics), Lagrange placed thesubject onaﬁrm anduniﬁed mathematical foundation. The treatise is mathematical rather than physical; Lagrange wasquite proud ofthefactthattheentire work con- tains notasingle diagram. 1“Because there aressecond-order differential equations, 2sinitial conditions must besupplied to determine themotion uniquely. 7.4 LAGRANGE’S EQUATIONS OFMOTION INGENERALIZED COORDINATES 239 Here weconsider only themotion ofsystems subject toconservative forces. Such forces canalways bederived from potential functions, sothatcondition 1is satisﬁed. This isnotanecessary restriction oneither Hamilton’s Principle or Lagrange ’sequations; thetheory canreadily beextended toinclude nonconser- vative forces. Similarly, wecanformulate Hamilton’s Principle toinclude certain types ofnonholonomic constraints, butthetreatment here isconﬁned toholo- nomic systems. Wereturn tononholonomic constraints inSection 7.5. Wenowwant towork several examples using Lagrange’s equations. Experience isthebest waytodetermine asetofgeneralized coordinates, realize thecon- straints, and setuptheLagrangian. Once thisisdone, theremainder ofthe problem isforthemost part mathematical. Consider thecase ofprojectile motion under gravity intwodimensions aswas discussed inExample 2.6.Find theequations ofmotion inboth Cartesian and polar coordinates. Solution. WeuseFigure 2-7todescribe thesystem. InCartesian coordinates, we usex(horizontal) andy(vertical). Inpolar coordinates weuser(inradial direc- tion) and6(elevation angle from horizontal). First, inCartesian coordinates we have 1. 1. .T= -gmx2 +Emy2 (7.19) U=me) whereU=0aty=0. 1_2 1_2L==T-U=gmx +:z-my -emgy (7.20) Weﬁnd theequations ofmotion byusing Equation 7.18: XI algae,6x dt572 e0-—'=0 dtmx 55=0 (7.21) )1! ateat ____ ‘:0 5y dt6y d—-mg —-;t(my) =O 5;=-—g (7.22) 240 7/HAMILTON’S PR1NCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS Byusing theinitial conditions, Equations 7.21 and7.22 canbeintegrated to determine theappropriate equations ofmotion. Inpolar coordinates, wehave 1 1 .T= -2-mi"? +~2-m(r6)2 U=mgrsin6 where U:Ofor6=O. 1 1 . L=T—U==g-mi? +§mr262 —mgrsin 6 (7.23) r: <E__d6L__0 6r dtiir mr62 rmgsin6 -£(mi) =O 762-gsin6 -'7=0 (7.24) 6: L %_l‘9 ao drab=0 d . —-mgr cos6 -E:(mr26) ==0 —-grcos6—2ri6 ~r26=0 (7.25) Theequations ofmotion expressed byEquations 7.21and7.22areclearly simpler than those ofEquations 7.24 and7.25. Weshould choose Cartesian co- ordinates asthegeneralized coordinates tosolve thisproblem. The keyin recognizing thiswasthatthepotential energy ofthesystem only depended onthe coordinate. Inpolar coordinates, thepotential energy depended onboth rand6. 9’ EXAM PLE 7.4 Aparticle ofmass misconstrained tomove ontheinside surface ofasmooth cone ofhalf-angle oz(seeFigure 7-2). The particle issubject toagravitational force. Determine asetofgeneralized coordinates anddetermine thecon- straints. Find Lagrange’s equations ofmotion, Equation 7.18. Solution. Lettheaxisofthecone correspond tothez-axis andlettheapex of thecone belocated attheorigin. Since theproblem possesses cylindrical sym- metry, wechoose r,6,andzasthegeneralized coordinates. Wehave, however, theequation ofconstraint z=rcot a (7.26) 7.4 LAGRANGE’S EQUATIONS OFMOTION INGENERALIZED COORDINATES 241 ,...;1i¢.:.=i%W*1“= 9ﬁg. .“‘~.*11:15;ta»'7 :7.7' ~ ’ e ‘i 1. ’ /e FIGURE 7-2 Example 7.4.Asmooth cone ofhalf-angle a.Wechoose r,6,and zasthe generalized coordinates. sothere areonly twodegrees offreedom forthesystem, andtherefore only two proper generalized coordinates. WemayuseEquation 7.26 toeliminate either the coordinate zorr;wechoose todotheformer. Then thesquare ofthevelocity is 112=72+#62+22 =172+ r262+ i2cot20z - =i2csc2a +r262 (7.27) The potential energy (ifwechoose U=Oatz=O)is U=mgz=mgrcotoi sotheLangrangian is 1 . L=gm(i2cscga +r262) —mgr cota (7.28) Wenote ﬁrstthatLdoes notexplicitly contain 6.Therefore 6L/66 ==0,and theLagrange equation forthecoordinate 6is i6L dt66=O Hence 6L . 69=mr26 =constant (7.29) Butmr26 =mr2w isjusttheangular momentum about thez-axis. Therefore, Equation 7.29 expresses theconservation ofangular momentum about theaxis ofsymmetry ofthesystem. The Lagrange equation forris L dLa———a, =0 (7.30)6r dt6r 242 7/HAMILTON’S PRINCIPLE-—LAGRANGIAN AND HAMILTONIAN DYNAMICS Calculating thederivatives, weﬁnd r'—-r62sin2a +gsina cosa =0 (7.31) which istheequation ofmotion forthecoordinate r. Weshall return tothisexample inSection 8.10 andexamine themotion in more detail. The point ofsupport ofasimple pendulum oflength bmoves onamassless rim ofradius arotating with constant angular velocity w.Obtain theexpression for theCartesian components ofthevelocity andacceleration ofthemass m. Obtain alsotheangular acceleration fortheangle 6shown inFigure 7-3. Solution. Wechoose theorigin ofourcoordinate system tobeatthecenter of therotating rim. The Cartesian components ofmass mbecome x=acoswt+bsin6} (7.32)y=asin wt- bcos6 Thevelocities are -=1-+b66 at awsinwt '‘cos (7.33) y=awcoswt+ b6s1n6 J’ -1°’/ >x ________&____cr- 3 FIGURE 7-3 Example 7.5.Asimple pendulum isattached toarotating rim. 7.4 LAGRANGE’S EQUATIONS OFMOTION INGENERALIZED COORDINATES 243 Taking thetime derivative once again gives theacceleration: 52=—-aw? coswt+b(6cos6—-62sin 6) y= -aw2 sinwt+b(6sin6+62¢0s 6) Itshould nowbeclear thatthesingle generalized coordinate is6.Thekinetic and potential energies are T=%m(:22 +5)?) v=mo where U==0aty==0.The Lagrangian is L=T—-U=i;'[a%ﬁ +(>262+2b6aw sin(0-011)] —-mg(a sinwt—-bcos6) (7.34) The derivatives fortheLagrange equation ofmotion for6are 6i =mb26 +mbaw(6 -w)cos(6 —-wt)dt66 6L .E=mb6aw cos(6 —-wt)—-mgbsin6 which results intheequation ofmotion (after solving for6) II 2 0=95-“cos(6-wt)--ﬁrsin0 (7.35) Notice thatthisresult reduces tothewell-known equation ofmotion forasim- plependulum ifw =O. Find thefrequency ofsmall oscillations ofasimple pendulum placed inarail- road carthathasaconstant acceleration ainthex-direction. Solution. Aschematic diagram isshown inFigure 7-4aforthependulum of length 6,mass m,anddisplacement angle 6.Wechoose afixed cartesian coordi- nate system with x=Oand5c=v0att=0.Theposition andvelocity ofmbecome 1 x=v0t+ Eat? +6sin6 y=—6cos6 ii=v0+at+66cos6 y=66sin6 244 7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS i- 8 I F (Q) i 8 I P (1)) FIGURE 7-4 Example 7.6.(a)Asimple pendulum swings inanaccelerating railroad car.(b)The angle 6,istheequilibrium angle duetothecar’s acceleration aandacceleration ofgravity g. The kinetic andpotential energies are 1T=-gm(i?2 +)2) U= —mg6 cos6 andtheLagrangian is 1 . 1 .L=T—U=§m(v0 +at+66cos6)?+§m(66 sin6)?+mg6cos 6 The angle 6istheonly generalized coordinate, andafter taking thederiva- tives forLagrange’s equations andsuitable collection ofterms, theequation of motion becomes (Problem 7-2) ..__g‘_ _it 6— 6sin6,,cos6 (7.36) Wedetermine theequilibrium angle 6=6,bysetting 6=0, O=gsin 6,+acos6, (7.37) The equilibrium angle 6,,shown inFigure 7-4b, isobtained by ei ii I tan6—a (7ss)s Because theoscillations aresmall andareabout theequilibrium angle, let 6=6,+7;,where 7)isasmall angle. 6=7']=—%sin(6, +7))—%cos(6, +17) (7.39) 7.4LAGRANGE’S EQUATIONS orMOTION INGENERALIZED COORDINATES 245 Weexpand thesine andcosine terms andusethesmall angle approximation forsin1)andcos1;,keeping only theﬁrstterms intheTaylor series expansions. .. g. . 4 . .7)=—E(s1n6, cos1)+cos6,sin1))—E(cos 6,cos1)—sin6,sin1;) _g. 61 .——Z(S1I'1 6,+77cos6,)—Z(cos 6,—1|sin6,) 1 . .=—z[(g sin6,+acos6,)+r;(gcos6,—asin 6,)] The ﬁrstterm inthebrackets iszero because ofEquation 7.37, which leaves 17')=—z(gcos6,—asin 6,)7) (7.40) WeuseEquation 7.38 todetermine sin6,andcos6,andafter alittle manipula- tion (Problem 7-2), Equation 7.40 becomes ".. ~/we=——m 7.41 17 ,"'1 () Because thisequation now represents simple harmonic motion, thefrequency toisdetermined tobe 2Va2+g2 to=‘T (7.42) This result seems plausible, because to-—>\/g/6fora=0when therailroad car isatrest. 'i_—_ 7 “-Abead slides along asmooth wire bent intheshape ofaparabola z=cr2 (Figure 7-5). The bead rotates inacircle ofradius Rwhen thewire isrotating about itsvertical symmetry axiswith angular velocity co.Find thevalue ofc. Solution. Because theproblem hascylindrical symmetry, wechoose r,6,andzas thegeneralized coordinates. The kinetic energy ofthebead is T—971’'2+'2 7 —2[ +2 (r6)] (.43) Ifwechoose U=0atz=O,thepotential energy term is U=mgz (7.44) Butr,z,and6arenotindependent. The equation ofconstraint fortheparabola is z=cr2 (7.45) i=2cir (7.46) 246 7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS z 74s (2\\____g r__________ 6 FIGURE 7-5 Example 7.7.Abead slides along asmooth wire thatrotates about thez-axis. Wealsohave anexplicit time dependence oftheangular rotation 6=wt 6=(.1 (7.47) Wecannow construct theLagrangian asbeing dependent only onr,because there isnodirect 6dependence. L=T“ U _ =g(i2 +4c2r2i2 +r2w2) —mgcr2 (7.48) The problem stated that thebead moved inacircle ofradius R.The reader might betempted atthispoint toletr=R=const. andi=0.Itwould bea mistake todothisnow intheLagrangian. First, weshould ﬁnd theequation ofmotion forthevariable randthen letr=Rasacondition oftheparticular motion. This determines theparticular value ofcneeded forr=R. 67=g(21'»+ 862727) datE5=l;'(2r+1662772 +seer) 6L—=m(4c2rr2 +rw2—2gcr)6r Lagrange’s equation ofmotion becomes ii(1+4c2r2) +i2(4c2r) +r(2gc —(112) =0 (7.49) which isacomplicated result. If,however, thebead rotates with r=R=constant, then i=F=0,andEquation 7.49 becomes R(2gc—(.12)=0 7.4 LAGRANGE’S EQUATIONS OFMOTION INGENERALIZED COORDINATES 247 and 0,2 c=——— (7.50) 2g istheresult wewanted.i *1 F.X.~\MPl.E 7.8 Consider thedouble pulley system shown inFigure 7-6.Usethecoordinates in- dicated, anddetermine theequations ofmotion. Solution. Consider thepulleys tobemassless, andletllandL,bethelengths of rope hanging freely from each ofthetwopulleys. The distances xand yare measured from thecenter ofthetwopulleys. ml: U1=2 77121 d Pulley 1 X ll--76 l2—y ’ M FIGURE 7-6 Example 7.8.The double pulley system. 248 7/HAMILTON’S PR1NCIPLE—LAGRANG1AN AND HAMILTONIAN DYNAMICS 17131 d ..v3=E:(l1— x+l2—y) =—.x—y (7.53) 121212T: 51777101 + 51712112 '1' 57"/3'Ug 1. 1 .. 1 ..=§m1x2 +gm,(y —x)2+§m3(—x —y)2 (7.54) Letthepotential energy U=0atx=0. U: U1+ U2 + U3 =—m1gx —m2g(l1— x+y)—m3g(l1— x+Z2—y) (7.55) Because Tand Uhave been determined, theequations ofmotion canbeob- tained using Equation 7.18. The results are "1155 +m2(55 _)3)+ms(55 +ll)=(ml_m2_ma)g (7-55) —m2<r—1')+met+1)=on.—mag <7-57> Equations 7.56 and7.57 canbesolved for55and . Examples 7.2-7.8 indicate theease andusefulness ofusing Lagrange’s equa- tions. Ithasbeen said, probably unfairly, thatLagrangian techniques aresimply recipes tofollow. The argument isthatwelosetrack ofthe“physics” bytheir use. Lagrangian methods, onthecontrary, areextremely powerful and allow usto solve problems thatotherwise would lead tosevere complications using Newtonian methods. Simple problems canperhaps besolved justaseasily using Newtonian methods, buttheLagrangian techniques canbeused toattack awide range of complex physical situations (including those occurring inquantum mechanics*). 7.5Lagrange’s Equations with Undetermined Multipliers Constraints that canbeexpressed asalgebraic relations among thecoordinates areholonomic constraints. Ifasystem issubject only tosuch constraints, wecan always ﬁnd aproper setofgeneralized coordinates interms ofwhich theequa- tions ofmotion arefreefrom explicit reference totheconstraints. Any constraints that must beexpressed interms ofthevelocities oftheparti- clesinthesystem areoftheform .f(xa,i7 '7.ca,ia t):O *See Feynman andHibbs (Fe65). 7.5 LAGRANGE’S EQUATIONS WITH UNDETERMINED MULTIPLIERS 249 andconstitute nonholonomic constraints unless theequations canbeintegrated toyield relations. among thecoordinates.* Consider aconstraint relation oftheform ZA,;¢,- +B=0,z=1,2,3 (7.59) Ingeneral, thisequation isnonintegrable, andtherefore theconstraint isnon- holonomic. ButifA,-andBhave theforms 8 8 A)=B=f=f<x,,=1) (7.60) then Equation 7.59 maybewritten as afdxi af_ 2097) dz+at_O "'61) Butthisisjust d_fZ0dt which canbeintegrated toyield f(xi, t)—constant =O (7.62) sotheconstraint isactually holonomic. From thepreceding discussion, weconclude that constraints expressible in differential form as 912 512Ejaqjdqj+atat0 (7.62) areequivalent tothose having theform ofEquation 7.9. Iftheconstraint relations foraproblem aregiven indifferential form rather than asalgebraic expressions, wecanincorporate them directly into Lagrange ’s equations byusing theLagrange undetermined multipliers (see Section 6.6) without ﬁrstperforming theintegrations; thatis,forconstraints expressible asin Equation 6.71, 61$,_j=l,2,...,sZdqj-0 {k1,2, (7.64)Jiiqj = ...,m theLagrange equations (Equation 6.69) are 6 aLdL 6];———_+Zi(¢)—=0 (7.65)6%dtiirb k"6% Infact, because thevariation process involved inHamilton’s Principle holds the time constant attheendpoints, wecould addtoEquation 7.64 aterm (Gﬁ,/8t)dt *Such constraints aresometimes called “semiholonomic.” 250 7/HAMILTON’S PRINCIPLE-—LAGRANGIAN AND HAMILTONIAN DYNAMICS without affecting theequations ofmotion. Thus constraints expressed byEquation 7.63alsolead totheLagrange equations given inEquation 7.65. The great advantage oftheLagrangian formulation ofmechanics isthatthe explicit inclusion oftheforces ofconstraint isnotnecessary; thatis,theempha- sisisplaced onthedynamics ofthesystem rather than thecalculation ofthe forces acting oneach component ofthesystem. Incertain instances, however, it might bedesirable toknow theforces ofconstraint. Forexample, from anengi- neering standpoint, itwould beuseful toknow theconstraint forces fordesign purposes. Itistherefore worth pointing outthat inLagrange’s equations ex- pressed asinEquation 7.65,theundetermined multipliers 1\,,(t) areclosely re- lated totheforces ofconstraint.* The generalized forces ofconstraint Qjare given by 6 @=§M§ aw)‘I1 EX1\l\-‘l PLE 7.9 Letusconsider again thecase ofthediskrolling down aninclined plane (see Example 6.5andFigure 6-7). Find theequations ofmotion, theforce ofcon- straint, andtheangular acceleration. Solution. The kinetic energy maybeseparated into translational androtational termsl 1 1.T=—M*+—m22y2 1 1 .=—M*+-MW022y4 where Misthemass ofthedisk andRistheradius; I=%MR2 isthemoment of inertia ofthediskabout acentral axis. The potential energy is U=Mg(l —y)sina (7.67) where listhelength oftheinclined surface oftheplane andwhere thediskis assumed tohave zero potential energy atthebottom oftheplane. The Lagrangian istherefore L=T— U 1, 1 . p=-éMy2 +ZLMR262 +Mg(y —l)s1na (7.68) *See, forexample, Goldstein (G080, p.47).Explicit calculations oftheforces ofconstraint insome speciﬁc problems arecarried outbyBecker (Be54, Chapters lland 13)andbySymon (Sy7l, p.372ff). ‘fWe anticipate here awell-known result from rigid-body dynamics discussed inChapter ll. 7.5 LAGRANGE’S EQUATIONS WITH UNDETERMINED MULTIPLIERS 251 The equation ofconstraint is f(y.6)=y-R6=0 (7.69) The system hasonly onedegree offreedom ifweinsist thattherolling takes place without slipping. Wemay therefore choose either yor6astheproper co- ordinate anduseEquation 7.69 toeliminate theother. Alternatively, wemay continue toconsider bothyand6asgeneralized coordinates andusethe method ofundetermined multipliers. The Lagrange equations inthiscase are L L 6@__15,,1:O6y dtdy 8y atd6L 6/ 0'70)————.+,\—=0ae41:60 as Performing thedifferentiations, weobtain, fortheequations ofmotion, Mgsin a— +A=O (7.7la) 1 ..—5MR26 —/\R=O (7.7lb) Also, from theconstraint equation, wehave y=R6 (7.72) These equations (Equations 7.71 and7.72) constitute asoluble system forthe three unknowns y,6,A.Differentiating theequation ofconstraint (Equation 7.72), weobtain ..= 6R (7.73) Combining Equations 7.7lb and7.73, weﬁnd 1A=— (7.74) andthen using thisexpression inEquation 7.7lathere results with A:_Mgsina 3 (7.76) sothatEquation 7.7lbyields ..2' 6=gsina 3R Thus, wehave three equations forthequantities ,,andAthatcanbeimme- diately integrated.(7.77) 252 7/HAMILTON’S PRINCIPLE-—LAGRANGIAN AND HAMILTONIAN DYNAMICS Wenote thatifthediskwere toslide without friction down theplane, we would have =gsin a.Therefore, therolling constraint reduces theaccelera- tionto§ofthevalue offrictionless sliding. The magnitude oftheforce offric- tionproducing theconstraint isjust/\—that is,(Mg/3) sina. The generalized forces ofconstraint, Equation 7.66, are Of Mgsinoz Qy 6y 3 8f MgRsina=,\_=_,\ =m_ Q9 68 R 3 Note that Q,and Q0areaforce andatorque, respectively, andthey arethegen- eralized forces ofconstraint required tokeep thediskrolling down theplane without slipping. _ Note thatwemay eliminate from theLagrangian bysubstituting 6=j1/R from theequation ofconstraint: L=%Mj>2+ Mg(y— l)sina (7.78) The Lagrangian isthen expressed interms ofonly oneproper coordinate, and thesingle equation ofmotion isimmediately obtained from Equation 7.18: 3Mgsina —§Mji =0 (7.79) which isthesame asEquation 7.75. Although thisprocedure issimpler, itcan- notbeused toobtain theforce ofconstraint. Aparticle ofmass mstarts atrestontopofasmooth ﬁxed hemisphere ofradius a.Find theforce ofconstraint, anddetermine theangle atwhich theparticle leaves thehemisphere. Solution. SeeFigure 7-7.Because weareconsidering thepossibility oftheparti- cleleaving thehemisphere, wechoose thegeneralized coordinates tobe7and 6.The constraint equation is f(r,6) =r— a=0 (7.80) The Lagrangian isdetermined from thekinetic andpotential energies: T=glue’+#61’) U=mgrcos6 L=T—U L=gm+.162)—mgrCOS6 (7.81) 7.5 LAGRANGE’S EQUATIONS WITH UNDETERMINED MULTIPLIERS ///1 Q1 /T_ T“~_e /‘___ / -_‘_/253 FIGURE 7-7 Example 7.10.Aparticle ofmass mmoves onthesurface ofaﬁxed smooth hemisphere. where thepotential energy iszero atthebottom ofthehemisphere. The Lagrange equations, Equation 7.65, are Performing thedifferentiations onEquation 7.80 givesL L 8a——ia_ +)1l=O 87" dt8r 87" 895-i"’?+1_f=680 dt89 80 8f 8f —=l, —=O8r 89 Equations 7.82 and7.83 become 6662- mgcos6 —617+11=0 mgr sin6—711.728. —2mr1'"8 =O Next, weapply theconstraint r=atothese equations ofmotion: r=a, i=O=i* Equations 7.85 and7.86 then become ma82— mgcos6+)t=O From Equation 7.88, wehave Wecanintegrate Equation 7.mga sin6-— =O fl=gsin6G 89todetermine 82. d.1616116.16 .(j= Weintegrate Equation 7.89,= = =6 dtdt dt d6dt d6 (6.16=21516616(7.82) (7.83) (7.84) (7.65) (7.86) (7.87) (7.88) (7.s9) (7.90) (7.91) 254 7/HAMILTON’S PR1NCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS which results in 62-~—=—-5cos6+5 (7.92)2 a a where theintegration constant isg/a,because 6=Oatt=Owhen 6=0. Substituting 62from Equation 7.92 into Equation 7.87 gives, after solving forA, A=mg(3 cos6—2) (7.93) which istheforce ofconstraint. The particle falls offthehemisphere atangle 60 when A=0. A=O=mg(3 cos60—2) (7.94) _,260=cos 5 (7.95) Asaquick check, notice thattheconstraint force isA=mgat6=0when the particle isperched ontopofthehemisphere. The usefulness ofthemethod ofundetermined multipliers istwofold: 1.The Lagrange multipliers areclosely related totheforces ofconstraint that areoften needed. . 2.When aproper setofgeneralized coordinates isnotdesired ortoodifﬁcult toobtain, themethod may beused toincrease thenumber ofgeneralized coordinates byincluding constraint relations between thecoordinates. 7.6 Equivalence ofLagrange’s andNewton’s Equations Aswehave emphasized from theoutset, theLagrangian andNewtonian formu- lations ofmechanics areequivalent: The viewpoint isdifferent, butthecontent isthesame. Wenow explicitly demonstrate thisequivalence byshowing thatthe twosetsofequations ofmotion areinfactthesame. InEquation 7.18, letuschoose thegeneralized coordinates tobetherectan- gular coordinates. Lagrange’s equations (forasingle particle) then become 8L d8L‘——— =0, '=1, ,3 7.96 8x, 2 2 ( ) OI‘ 8(Tf U)_d8(T— U)IO 8x,- dt 85¢, 7.6EQUIVALENCE orLAGRANGE’S ANDNEWTON’S EQUATIONS 255 Butinrectangular coordinates andforaconservative system, wehave T=TUE,-) and U= U(x,-), so 8T 8U —=Oand ,=O836, 8x,- Lagrange’s equations therefore become _¥_/-2” 7.97 8x, ( ) Wealsohave (foraconservative system) 8U 8.76,- and 66T1631 d— =Z 2 — '2 =— '_ = . 1116.6. .1166,(F12mx’ 111(mx‘) '2’ soEquation 7.97yields theNewtonian equations, asrequired: F1=.51 (7-93) Thus, theLagrangian andNewtonian equations areidentical ifthegeneralized coordinates aretherectangular coordinates. Now letusderive Lagrange’s equations ofmotion using Newtonian con- cepts. Consider only asingle particle forsimplicity. Weneed totransform from thex,--coordinates tothegeneralized coordinates qj.From Equation 7.5,wehave x,=x,-(qj, t) (7.99) '22”‘'+ax‘ (7100) X,» Z "'_ ' W 6 18%-q] 8t and 22‘—% (7101aq.Bq. ') Ageneralized momentum pfassociated with (5iseasily determined by 8T 11,-8% (7.102) Forexample, foraparticle moving inplane polar coordinates, T=(72+T262) m/2, wehave p,=mi"forcoordinate randpa=mr26 forcoordinate 6.Obviously p,isa linear momentum andpgisanangular momentum, soourgeneralized momen- tumdeﬁnition seems consistent with Newtonian concepts. 256 7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS Wecandetermine ageneralized force byconsidering thevirtual work 8W done byavaried path 8x,asdescribed inSection 6.7. 8. 6W=212,61, =ZF,i‘6q, (7.105)1 51 8% EQ,-sq, (7.104) sothatthegeneralized force Q,associated with qjis 896,- Q,-F,5-(Z (7.105) just aswork isalways energy, soistheproduct ofQq.Ifqislength, Qisforce; ifq isanangle, Qistorque. Foraconservative system, Qjisderivable from thepo- tential energy: au-=—— (7.106) Q,6% Now weareready toobtain Lagrange’s equations: 8T 8 1 2 P‘: -=J8q]- 8q}- 12 6'- xj 1 aq] .ax;,6,=1.116% (7.107) where weuseEquation 7.101 forthelaststep. Taking thetime derivative of Equation 7.107 gives .2(..@»<.+ .66.) 7108-= mx,-— mx,_— _121 aqj 616% ( ) Expanding thelastterm gives d8-xi 2 a2xi ' 8296, __ = M qk+___ dt8q]- h8q,,8q]- 8qj-8t andEquation 7.108 becomes - 51 521 521,6,=Zmse,-3+Zms¢,—i- 4,+Zm;¢,—i (7.109)1 811,- ah 8q,,8(b 1 8%-8t The ﬁrst term ontheright side ofEquation 7.109 isjust Q]-(F, =m5E,- and Equation 7.105). The sum oftheother twoterms is8T/8%: 5 51 31=Zms.,—(Ei .),+i) (7.110)1 8q]- k8q,, 8t where wehave used T=Z,1/2 andEquation 7.100.6T 2 . "— = mx, _—" 8q] i 8q] 7.7ESSENCE orLAGRANGIAN DYNAMICS 257 Equation 7.109 cannow bewritten as . 8T-=.— 7.111 or,using Equations 7.102 and7.106, d8T 8T 8U-—<,)——=Qj=—— (7.112)dt8%» 8qJ.,- 8q]- Because Udoes notdepend onthegeneralized velocities tjj,Equation 7.112 can bewritten 8T— U 8T— U d‘:( _U—( )= 0 (7.113)dt 8q]- 8% d— —2"]:=0 (7.114)dt8q]- 8q]- which areLagrange’s equations ofmotion.andusing L=T—U, 7.7 Essence ofLagrangian Dynamics Inthepreceding sections, wemade several general and important statements concerning theLagrange formulation ofmechanics. Before proceeding further, weshould summarize these points toemphasize thedifferences between the Lagrange andNewtonian viewpoints. Historically, theLagrange equations ofmotion expressed ingeneralized co- ordinates were derived before thestatement ofHamilton’s Principle.* We elected todeduce Lagrange’s equations bypostulating Hamilton’s Principle be- cause thisisthemost straightforward approach andisalso theformal method forunifying classical dynamics. First, wemust reiterate thatLagrangian dynamics does notconstitute anew theory inanysense oftheword. The results ofaLagrangian analysis ora Newtonian analysis must bethesame foranygiven mechanical system. The only difference isthemethod used toobtain these results. 1/Vhereas theNewtonian approach emphasizes anoutside agency acting ona body (the force), theLagrangian method deals only with quantities associated with thebody (the kinetic and potential energies). Infact, nowhere inthe Lagrangian formulation does theconcept offorce enter. This isaparticularly im- portant property—and foravariety ofreasons. First, because energy isascalar quantity, theLagrangian function forasystem isinvariant tocoordinate transfor- mations. Indeed, such transformations arenotrestricted tobebetween various *Lagrange’s equations, 1788; Hamilton’s Principle, 1854. 258 7/HAMILTON’S PRINCIPLE—LAGRANGIAN ANDHAMILTONIAN DYNAMICS orthogonal coordinate systems inordinary space; they mayalsobetransformations between ordinary coordinates andgeneralized coordinates. Thus, itispossible to pass from ordinary space (inwhich theequations ofmotion may bequite com- plicated) toaconﬁguration space thatcanbechosen toyield maximum simpliﬁ- cation foraparticular problem. Weareaccustomed tothinking ofmechanical systems interms ofvector quantities such asforce, velocity, angular momentum, andtorque. ButintheLagrangian formulation, theequations ofmotion areob- tained entirely interms ofscalar operations inconﬁguration space. Another important aspect oftheforce-versus-energy viewpoint isthatincer- tainsituations itmay noteven bepossible tostate explicitly alltheforces acting onabody (asissometimes thecase forforces ofconstraint), whereas itisstill possible togive expressions forthekinetic andpotential energies. Itisjust this fact that makes Hamilton’s Principle useful forquantum-mechanical systems where wenormally know theenergies butnottheforces. The differential statement ofmechanics contained inNewton’s equations or theintegral statement embodied inHamilton’s Principle (and theresulting Lagrangian equations) have been shown tobeentirely equivalent. Hence, nodis- tinction exists between these viewpoints, which arebased onthedescription of physical cyfects. Butfrom aphilosophical standpoint, wecanmake adistinction. In theNewtonian formulation, acertain force onabody produces adeﬁnite motion—that is,wealways associate adeﬁnite effect with acertain cause. According toHamilton’s Principle, however, themotion ofabody results from theattempt ofnature toachieve acertain purpose, namely, tominimize thetime integral ofthedifference between thekinetic andpotential energies. The opera- tional solving ofproblems inmechanics does notdepend onadopting one or theother ofthese views. Buthistorically such considerations have had apro- found inﬂuence onthe development ofdynamics (as, for example, in Maupertuis’s principle, mentioned inSection 7.2). The interested reader isre- ferred toMargenau’s excellent book foradiscussion ofthese matters.* 7.8 ATheorem Concerning theKinetic Energy Ifthekinetic energy isexpressed inﬁxed, rectangular coordinates, theresult isa homogeneous quadratic function of5c,,,,: ,.,1its:64-..§_ T=— 3,, (7.115) Wenow wish toconsider inmore detail thedependence ofTon thegeneralized coordinates andvelocities. Formany particles, Equations 7.99 and7.100 become =x..,.<e.0.1=1.2.--as (7.116) 5axa i axe: i 11,,=Z—’q,+—’ (7.117)J=18qj- 8t *Margenau (Ma77, Chapter 19). 7.8 ATHEOREM CONCERNING THE KINETIC ENERGY 259 Evaluating thesquare of22%,,weobtain _2 28%,,-8x,,,, __ 28xa,,8x,,,,- _ 8x,,,,» 2 xay,=p qjqk+2j qj+ (7.118) M8q]-8q,, ,18qj 8t 8t andthekinetic energy becomes 1 aaiaai atliaai 1 aaig T=2z-...,:i.,,,.2z,.,iL,..22_..,(L) .._...)fl‘M2 8qj 81],, (15] 8q]- 8t 1112 8t Thus, wehave thegeneral result T=Z.1.q-q+Z6-4+6 (7.120) MJkik JJ] Aparticularly important case occurs when thesystem isscleronomic, sothatthe time does notappear explicitly intheequations oftransformation (Equation 7.116); then thepartial time derivatives vanish: 8x,,,,» =0, =0, c=08t Therefore, under these conditions, thekinetic energy isahomogeneous quadratic function ofthegeneralized velocities: T=6.61. (7.121) Next, wedifferentiate Equation 7.121 with respect to6,: 8T _ _ 661:Ta”‘q"+;a"’2 Multiplying thisequation by6,andsumming over l,wehave .3T .. ..2(I16&1=%alkqhql + a,-lq,-ql Inthiscase, alltheindices aredummies, soboth terms ontheright-hand side areidentical: ar ,_Z6155 =212,‘a,,,q,-q,=2T (7.122),. This important result isaspecial case ofEuler’s theorem, which states thatiff(y,) is ahomogeneous function oftheykthatisofdegree n,then af_Ey,,a—yk-nf (7.125) 260 7/HAMILTON’S PRINCIPLE—LAGRANGIAN ANDHAMILTONIAN DYNAMICS 7.9 Conservation Theorems Revisited Conservation ofEnergy Wesawinourprevious arguments* that timeishomogeneous within aninertial reference frame. Therefore, theLagrangian that describes aclosed system (i.e., a system notinteracting with anything outside thesystem) cannot depend explic- itlyontime,1 thatis, 8L—=0 7.148t (2) sothat thetotal derivative oftheLagrangian becomes dL aL aL—=Z—q,-+2, _2,1, (7.125)dt J8q]- J8%- where theusual term, 8L/8t, does notnow appear. ButLagrange’s equations are %_i21_¥ _ (7.126)8q]~ dt8qj Using Equation 7.126 tosubstitute for8L/8qj inEquation 7.125, wehave é_-2 L aL a__ dt—2qjd16'- +26? qjJ q] J'6 OI‘ dL 2d8L_-- _'. =0 dz 1d1(2’a¢j,) sothat .1 aL—L—E'»=0 7.17dt( 1'q]8(j]) (2) The quantity intheparentheses istherefore constant intime; denote thiscon- stant by—H: 8LL—Z1;_=—H= constant (7.128)18q]- Ifthepotential energy Udoes notdepend explicitly onthevelocities 56,),orthe time t,then U=U(x,,’,). The relations connecting therectangular coordinates and thegeneralized coordinates areoftheform x,,,,~=xm,-(qj-) orqj=q]-(x,,,,-), ___.-M...-iM.. *See Section 2.3. 1‘The Lagrangian islikewise independent ofthetime ifthesystem exists inauniform force ﬁeld. 7.9 CONSERVATION THEOREMS REVISITED 261 where weexclude thepossibility ofanexplicit time dependence inthetransfor- mation equations. Therefore, U=U(¢b), and6U/(iqj =0.Thus aL_a(T— U)_g 6% 6% aqj Equation 7.128 canthen bewritten as ,ar(T—U)—qj%q= —H (7.129) and, using Equation 7.122, wehave (T— U)—2T= —H or T+U=E=H= constant (7.130) The total energy Eisaconstant ofthemotion forthiscase. The function H,called theHamiltonian ofthesystem, may bedeﬁned asin Equation 7.128 (butseeSection 7.10). Itisimportant tonote thattheHamiltonian Hisequal tothetotal energy Eonly ifthefollowing conditions aremet: 1.The equations ofthetransformation connecting therectangular andgen- eralized coordinates (Equation 7.116) must beindependent ofthetime, thus ensuring that thekinetic energy isahomogeneous quadratic function ofthe 2.The potential energy must bevelocity independent, thus allowing theelimi- nation oftheterms aU/aqj from theequation forH(Equation 7.129). The questions “Does H=Eforthesystem?” and“Isenergy conserved forthesys- tem?,” then, pertain totwodifferent aspects oftheproblem, and each question must beexamined separately. Wemay, forexample, have cases inwhich the Hamiltonian does notequal thetotal energy, butnevertheless, theenergy iscon- served. Thus, consider aconservative system, andletthedescription bemade in terms ofgeneralized coordinates inmotion with respect toﬁxed, rectangular axes. The transformation equations then contain thetime, and thekinetic en- ergy isnotahomogeneous quadratic function ofthegeneralized velocities. The choice ofamathematically convenient setofgeneralized coordinates cannot alter thephysical factthatenergy isconserved. Butinthemoving coordinate sys- tem, theHamiltonian isnolonger equal tothetotal energy. Conservation ofLinear Momentum Because space ishomogeneous inaninertial reference frame, theLagrangian of aclosed system isunaffected byatranslation oftheentire system inspace. Consider aninﬁnitesimal translation ofevery radius vector rasuch thatra—>ra +8r;this amounts totranslating theentire system by81'.For simplicity, letusexamine a system consisting ofonly asingle particle (byincluding asummation over ozwe could consider ann-particle system inanentirely equivalent manner), andletus 262 7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS write theLagrangian interms ofrectangular coordinates L=L(x,, iv,-).The change inLcaused bytheinﬁnitesimal displacement Br=Z;8x,-e, is 8L 8L8L=2_5—8x,~+ Ea,ax,-=0 (7.131)x- x- Weconsider only avaried displacement, sothat the8x,arenotexplicit orimplicit functions ofthetime. Thus, dx, d6..Z 8——— =i .E .x, dt dt5x, O (7132) Therefore, 5Lbecomes BLat=2-ax,=0 (7.133)1ax‘: Because each ofthe8x,isanindependent displacement, 8Lvanishes identically only ifeach ofthepartial derivatives ofLvanishes: Lit=O (7.134)(ix, Then, according toLagrange’s equations, i6L dt6a2,-=O“ (7.135) and _=constant (7.136)6x,- OI‘ 6T— U 6T 1__(____) =__=_‘.3_(_ 6x, 6x, 6x,2 J . =mi,=pi=constant (7.137) Thus, thehomogeneity ofspace implies thatthelinear momentum pofaclosed system isconstant intime. This result may alsobeinterpreted according tothefollowing statement: If theLagrangian ofasystem (not necessarily closed) isinvariant with respect to translation inacertain direction, then thelinear momentum ofthesystem in thatdirection isconstant intime. Conservation ofAngular Momentum Westated inSection 2.3thatonecharacteristic ofaninertial reference frame isthat space isisotropic thatis,thatthemechanical properties ofaclosed system areun- affected bytheorientation ofthesystem. Inparticular, theLagrangian ofaclosed system does notchange ifthesystem isrotated through aninﬁnitesimal angle.* *We limit therotation toaninﬁnitesimal angle because wewish tobeable torepresent therotation byavector; seeSection 1.15. 7.9 CONSERVATION THEOREMS REVISITED 263 A 189 ‘iv FIGURE 7-8 Asystem isrotated byaninﬁnitesimal angle 56. Ifasystem isrotated about acertain axisbyaninﬁnitesimal angle 86(see Figure 7-8), theradius vector rtoagiven point changes tor+8r,where (see Equation 1.106) 8r=as><r (7.133) The velocity vectors also change onrotation ofthesystem, and because the transformation equation forallvectors isthesame, wehave at=so><i~ (7.139) Weconsider only asingle particle andexpress theLagrangian inrectangular coordinates. The change inLcaused bytheinﬁnitesimal rotation is at=295ax,+Eafax,=0 (7.140)1(ix, idxi Equations 7.136 and7.137 show thattherectangular components ofthemo- mentum vector aregiven by p,-= (7.141)6*,- Lagrange ’sequations may then beexpressed by .8L Hence, Equation 7.140 becomes at=Zii,-ax,+Zp,-Bk,=0 (7.143) or 1')-8r+p-8i'=0 (7.144) 264 7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS Using Equations 7.138 and7.139, thisequation may bewritten as }')*(59 Xr) +p-(80 Xi‘) =0 (7.145) Wemay permute incyclic order thefactors ofatriple scalar product without al- tering thevalue. Thus, m-uxpytm-uxp)=0 OI‘ w¢uxpy+&xm]=0 (zmm The terms inthebrackets arejustthefactors thatresult from thedifferentiation with respect totime ofrXp: rXp)=O (zmn Because 86isarbitrary, wemust have dZt(r Xp)=O (7.148) so rXp=constant (7.149) ButrXp=L;theangular momentum oftheparticle inaclosed system isthere- fore constant intime. Animportant corollary ofthistheorem isthefollowing. Consider asystem in anexternal force ﬁeld. Iftheﬁeld possesses anaxis ofsymmetry, then the Lagrangian ofthesystem isinvariant with respect torotations about thesymme- tryaxis. Hence, theangular momentum ofthesystem about theaxisofsymmetry isconstant intime. This isexactly thecase discussed inExample 7.4;thevertical direction wasanaxis ofsymmetry ofthesystem, and theangular momentum about thataxiswasconserved. The importance oftheconnection between symrrwtvy properties andtheinvari- anceofphysical quantities canhardly beoveremphasized. The association goes be- yond momentum conservation—indeed beyond classical systems-—and ﬁnds wide application inmodern theories ofﬁeld phenomena andelementary particles. Wehave derived theconservation theorems foraclosed system simply by considering theproperties ofaninertial reference frame. The results, summa- rized inTable 7-1,aregenerally credited toEmmy Noether.* There arethen seven constants (orintegrals) ofthemotion foraclosed sys- tem: total energy, linear momentum (three components), andangular momen- tum (three components). These and only these seven integrals have theprop- erty that they areadditive fortheparticles composing thesystem; they possess thisproperty whether ornotthere isaninteraction among theparticles. *Emmy Noether (1882-1935), one oftheﬁrst female German mathematical physicists, endured poor treatment byGerman mathematicians early inhercareer. Sheistheoriginator ofNoether’s Theorem, which proves arelationship between symmetries and conservation principles. 7.10 CAN ONICAL EQUATIONS OFMOTION—HAMILTONIAN DYNAMICS 265 TABLE 7-1 Characteristic ofinertial frame Property ofLagrangian Conserved quantity Time homogeneous Not explicit function oftime Total energy Space homogeneous Invariant totranslation Linear momentum Space isotropic Invariant torotation Angular momentum 7.10 Canonical Equations ofMotion—Hamiltonian Dynamics Intheprevious section, wefound thatifthepotential energy ofasystem isveloc- ityindependent, then thelinear momentum components inrectangular coordi- nates aregiven by aLpi-8&1 (7.150) Byanalogy, weextend thisresult tothecase inwhich theLagrangian isexpressed ingeneralized coordinates anddeﬁne thegeneralized momenta* according to -E_A (7.151) (Unfortunately, thecustomary notations forordinary momentum andgeneral- ized momentum arethesame, even though thetwoquantities may bequite dif- ferent.) The Lagrange equations ofmotion arethen expressed by '—% 7152 P)6% (-) Using thedeﬁnition ofthegeneralized momenta, Equation 7.128 forthe Hamiltonian maybewritten as H=gpjéj —L (7.153) The Lagrangian isconsidered tobeafunction ofthegeneralized coordinates, thegeneralized velocities, andpossibly thetime. The dependence ofLonthe time may arise either iftheconstraints aretime dependent orifthetransforma- tionequations connecting therectangular andgeneralized coordinates explicitly contain thetime. (Recall thatwedonotconsider time-dependent potentials.) We may solve Equation 7.151 forthegeneralized velocities andexpress them as aj=Q]-(qk, pk,t) (7.154) *The terms generalized coordinates, generalized velocities, andgeneralized momenta were introduced in 1867 bySirWilliam Thomson (later, Lord Kelvin) and P.G.Tait intheir famous treatise Natural Philosophy. 266 7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS Thus, inEquation 7.153, wemay make achange ofvariables from the(qj,4]]-,t) settothe(qj,pj,t)set*andexpress theHamiltonian as H(qk> Pk»t)=.Pﬂlj_L(qk> ‘lie5) (7-155) This equation iswritten inamanner thatstresses thefactthat theHamiltonian is always considered asafunction ofthe(qk,pk,t)set,whereas theLagrangian isafunction ofthe (qk,ak,t)set: ‘H: H(qk> Pk»5)» L=L(‘1k>‘lh»t) 1 (7-156) The total differential ofHistherefore H H HdH=E<L dqk+6-711),)+6-at (7.157)k dqk dpk According toEquation 7.155, wecanalsowrite . ._Qé _Q1;._%dH_2k(qk dpk+Pkdqk aqdqk at?dqt) atdi (7-158)k k Using Equations 7.151 and7.152 tosubstitute for6L/6q,, and6L/élrjk, thesecond andfourth terms intheparentheses inEquation 7.158 cancel, andthere remains dH=§(¢2.dp. —rm.)—%at (7.159) Ifweidentify thecoefﬁcientsl ofdqk,dpk,and dtbetween Equations 7.157 and 7.159, weﬁnd ,_6H qk—éfph (7.160) .6H Hamilton’s equations ofmotion -1)"I(Ta (7.161) and 6L 6H—at—at (7.162) Furthermore, using Equations 7.160 and 7.161 inEquation 7.157, theterm in theparentheses vanishes, anditfollows that dHaH-=— 7.163dt 6t ( ) % *This change ofvariables issimilar tothatfrequently encountered inthemiodynamics andfallsin thegeneral class oftheso-called Legendre transformations (used ﬁrstbyEuler andperhaps even by Leibniz). Ageneral discussion ofLegendre transformations with emphasis ontheir importance in mechanics isgiven byLanczos (La49, Chapter 6). 1-The assumptions implicitly contained inthisprocedure areexamined inthefollowing section. 7.10CANONICAL EQUATIONS orMOTION—HAMILTONIAN DYNAMICS 267 Equations 7.160 and 7.161 areHamilton’s equations ofmotion.* Because of their symmetric appearance, they arealsoknown asthecanonical equations ofmo- tion. Thedescription ofmotion bythese equations istermed Hamiltonian dynamics. Equation 7.163 expresses thefactthat ifHdoes notexplicitly contain the time, then theHamiltonian isaconserved quantity. Wehave seen previously (Section 7.9) thattheHamiltonian equals thetotal energy T+Uifthepotential energy isvelocity independent andthetransformation equations between x,,,,andqj donotexplicitly contain thetime. Under these conditions, andif6kV6t =0,then H=E=constant. There are2scanonical equations andthey replace thesLagrange equations. (Recall thats=3n—misthenumber ofdegrees offreedom ofthesystem.) But thecanonical equations arefirst-order differential equations, whereas theLagrange equations areofsecond ordeni Tousethecanonical equations insolving aproblem, wemust ﬁrstconstruct theHamiltonian asafunction ofthegeneralized coordi- nates andmomenta. Itmay bepossible insome instances todothisdirectly. In more complicated cases, itmay benecessary ﬁrst tosetuptheLagrangian and then tocalculate thegeneralized momenta according toEquation 7.151. The equations ofmotion arethen given bythecanonical equations. EXAMPLE 7.1l UsetheHamiltonian method toﬁnd theequations ofmotion ofaparticle of mass mconstrained tomove onthesurface ofacylinder deﬁned by . x2+yg=R2.The particle issubject toaforce directed toward theorigin and proportional tothedistance oftheparticle from theorigin: F=—kr. Solution. The situation isillustrated inFigure 7-9.The potential corresponding totheforce Fis 1 1 U=§kr2 =§k(x2 +312 +z2) 1 =§k(R2 +22) (7.164) Wecanwrite thesquare ofthevelocity incylindrical coordinates (seeEquation 1.101) as 62=R2+R262+£2 (7.165) Butinthiscase, Risaconstant, sothekinetic energy is T=%m(R1’é2 +62) (7.166) *This setofequations wasﬁrstobtained byLagrange in1809, andPoisson alsoderived similar equa- tions inthesame year. Butneither recognized theequations asabasic setofequations ofmotion; thispoint wasﬁrst realized byCauchy in1831. Hamilton ﬁrst derived theequations in1834 from a fundamental variational principle andmade them thebasis forafar-reaching theory ofdynamics. Thus thedesignation “Hamilton’s” equations isfully deserved. 1‘This isnotaspecial result; anysetofssecond-order equations canalways bereplaced byasetof2s ﬁrst-order equations. 268 7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS Z _-;;;:"a=-rrf ~. iiillgiii,-t§. jg‘.-==’ - , 1'‘.. it 'T‘l1é""*§:.~ ,'l§1:'.-5.5 " ~35?):== '1. 3i 22¢ r%’9"1'~-6-. .- N-H"9 ~rm“ ‘8I I:7.:»:,~,(' ‘~'=s¢~z::=$5 191;“ it,4.=f's“=,;"~ii _ ».=1“:Z;._:§ “X (1 ~~: ~':i")';¢r.I ii‘5’-=-=.§2I:--4-- : -l:5*.L _Il _A VatX ‘X‘llt k'Z§$_‘_.s ..:i;§évi§'1ii § ' .31’,-ii" FIGURE 7-9 Example 7.11. Aparticle isconstrained tomove onthesurface ofacylinder. Wemay now write theLagrangian as L=T— U=%m(R292 +22)—%k(R2 +z2) (7.167) The generalized coordinates are6andz,andthegeneralized momenta are a=%=mmé umm pz= =mi (7.169) Because thesystem isconservative andbecause theequations oftransformation between rectangular andcylindrical coordinates donotexplicitly involve the time, theHamiltonian Hisjustthetotal energy expressed interms ofthevari- ables 6,pa,z,andpz.But6does notoccur explicitly, so H(Z,p@,pz) =T-1' U 1»?11%’1 =2%? +27” '1'5kZ2 where theconstant term ék.R2hasbeen suppressed. The equations ofmotion aretherefore found from thecanonical equations: . 8H . 8H E--M--a mum 7.10 CAN ONICAL EQUATIONS OFMOTION—I-IAMILTONIAN DYNAMICS 269 .6H p6=5=71-1% (7.173)5 z="1'-I=E (7.174)anm Equations 7.173 and1.174 justduplicate Equations 7.168 and7.169. Equations 7.168 and7.171 give pk=mR2(i =constant (7.175) The angular momentum about thez-axis isthus aconstant ofthemotion. This result isensured, because thez-axis isthesymmetry axisoftheproblem. Combining Equations 7.169 and7.172, weﬁnd "z"+wgz=0 (7.176) where (0%Ek/m (7.177) The motion inthezdirection istherefore simple harmonic. The equations ofmotion forthepreceding problem canalsobefound by theLagrangian method using thefunction Ldeﬁned byEquation 7.167. Inthis case, theLagrange equations ofmotion areeasier toobtain than arethecanoni- calequations. Infact, itisquite often true that theLagrangian method leads more readily totheequations ofmotion thatdoes theHamiltonian method. But because wehave greater freedom inchoosing thevariable intheHamiltonian formulation ofaproblem (the qkandthepkareindependent, whereas theqkand theakarenot), weoften gain acertain practical advantage byusing theHamiltonian method. Forexample, incelestial mechanics—particularly intheevent that the motions aresubject toperturbations caused bytheinﬂuence ofother bodies-—it proves convenient toformulate theproblem interms ofHamiltonian dynamics. Generally speaking, however, thegreat power oftheHamiltonian approach to dynamics does notmanifest itself insimplifying thesolutions tomechanics prob- lems; rather, itprovides abase wecanextend toother ﬁelds. The generalized coordinate qkandthegeneralized momentum pkarecanon- ically conjugate quantities. According toEquations 7.160 and 7.161, ifqkdoes notappear intheHamiltonian, then pk=0,and theconjugate momentum pkis aconstant ofthemotion. Coordinates notappearing explicitly intheexpres- sions forTand Uaresaid tobecyclic. Acoordinate cyclic inHisalsocyclic inL. But, even ifqkdoes notappear inL,thegeneralized velocity qkrelated tothisco- ordinate isingeneral stillpresent. Thus L:L(q1! "'1qk—1s qk+1s "-J1.’ q‘l ML, t) andweaccomplish noreduction inthenumber ofdegrees offreedom ofthesys- tem, even though onecoordinate iscyclic; there arestillssecond-order equations 270 7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS tobesolved. However, inthecanonical formulation, ifqkiscyclic, pkisconstant, pk=ak,and H: H(ql: >qk—1> qk+1> #1., Pb »Pk—1> alv pk+1! >P;> t) Thus, there are2s—2ﬁrst-order equations tobesolved, andtheproblem has,in fact,been reduced incomplexity; there areineffect only s—1degrees offreedom remaining. Thecoordinate qkiscompletely separated, anditisignorableas farasthe remainder oftheproblem isconcerned. Wecalculate theconstant akbyapplying theinitial conditions, andtheequation ofmotion forthecyclic coordinate is -_if= qk—aak—wk (7.178) which canbeimmediately integrated toyield The solution foracyclic coordinate istherefore trivial toreduce toquadrature. Consequently, thecanonical formulation ofHamilton isparticularly well suited fordealing with problems inwhich one ormore ofthecoordinates arecyclic. The simplest possible solution toaproblem would result iftheproblem could beformulated insuch awaythat allthecoordinates were cyclic. Then, each co- ordinate would bedescribed inatrivial manner asinEquation 7.179. Itis,in fact, possible toﬁnd transformations thatrender allthecoordinates cyclic,* and these procedures lead naturally toaformulation ofdynamics particularly useful inconstructing modern theories ofmatter. The general discussion ofthese top- ics,however, isbeyond thescope ofthisbook.l EXAMPLE 7112 UsetheHamiltonian method toﬁnd theequations ofmotion foraspherical pendulum ofmass inandlength b(seeFigure 7-10). Solution. The generalized coordinates are6and.The kinetic energy is 1 . 1 _ .T= 5mb262 +gmbg S1112 6<;b2 The only force acting onthependulum (other than atthepoint ofsupport) is gravity, andwedeﬁne thepotential zero tobeatthependulum’s point of attachment. U=—mgbcos6 *Transformations ofthistypewere derived byCarl Gustavjacob Jacobi (l804—l85l ).]acobi’s investi- gations greatly extended theusefulness ofHamilton’s methods, andthese developments areknown asHamiltmi-jacobi theory. tSee, forexample, Goldstein (G080, Chapter 10). 7.10CANONICAL EQUATIONS orMOTION—HAMILTONIAN DYNAMICS 271 U=0 Q Q‘ LUQ ¢ m CDI FIGURE 7-10 Example 7.12. Aspherical pendulum with generalized coordinates 6and The generalized momenta arethen 6L -pa= =mb26 (7.180) 6L .pd,= =mb2sin?6 (7.181) 6 Wecansolve Equations 7.180 and7.181 for6and interms ofpkandp¢. Wedetermine theHamiltonian from Equation 7.155 orfrom H= ' T+U(because theconditions forEquation 7.130 apply). H=T+U 1 pg 1mb2sing6p§, =—62 +- -~~—662m (mb2)2 2(mb2 sini’6)? mg COS _P3 Pi —2mb2 +2mb2 sin?6mgb COS6 The equations ofmotion are - OH Po6:--=— dpg mb2 -aH Pd»¢=_=_._i Bpd, mbgsin?6 _ 5H p§cos6 _ pa: -66: mbg sin36 _mgbsme .__6_I"_I: 116- 64) 0 Because <j>iscyclic, themomentum pd,about thesymmetry axisisconstant. 272 7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 7.11Some Comments Regarding Dynamical Variables andVariational Calculations inPhysics Weoriginally obtained Lagrange’s equations ofmotion bystating Hamilton’s Principle asavariational integral and then using theresults ofthepreceding chapter onthecalculus ofvariations. Because themethod andtheapplication were thereby separated, itisperhaps worthwhile torestate theargument inan orderly butabbreviated way. Hamilton’s Principle isexpressed by $2 slL(¢h,.),,t)dt=0 (7.182) '31 Applying thevariational procedure speciﬁed inSection 6.7,wehave —q-.q-= FGLB +6L8’) dt0tl J J Next, weassert that the8q]-andthe861-arenotindependent, sothevariation op- eration andthetime differentiation canbeinterchanged: .dd<18qj—5(E —2256],» (7.183) The varied integral becomes (after theintegration byparts inwhich the5%are setequal tozero attheendpoints) l.(621, d6%)'2 tQ5— Sq]-dt= 0 (7.184) The requirement that the8q_,-beindependent variations leads immediately to Lagrange’s equations. InHamilton’s Principle, expressed bythevariational integral inEquation 7.182, theLagrangian isafunction ofthegeneralized coordinates andthegen- eralized velocities. But only theq}areconsidered asindependent variables; the generalized velocities aresimply thetime derivatives oftheqj.When theintegral isreduced totheform given byEquation 7.184, westate that the8q_,~areinde- pendent variations; thus theintegrand must vanish identically, andLagrange’s equations result. Wemay therefore pose thisquestion: Because thedynamical motion ofthesystem iscompletely determined bytheinitial conditions, what is themeaning ofthevariations 8%-PPerhaps asufﬁcient answer isthat thevari- ables aretobeconsidered geometrically feasible within thelimits ofthegiven constraints—-although they arenotdynamically possible; that is,when using a variational procedure toobtain Lagrange’s equations, itisconvenient toignore temporarily thefactthatwearedealing with aphysical system whose motion is completely determined andsubject tonovariation andtoconsider instead only acertain abstract mathematical problem. Indeed, thisisthespirit inwhich any variational calculation relating toaphysical process must becarried out. In adopting such aviewpoint, wemust notbeoverly concerned with thefactthat 7.11 SOME COMMENTS REGARDING DYNAMICAL VARLABLES 273 thevariational procedure may becontrary tocertain known physical properties ofthesystem. (For example, energy isgenerally notconserved inpassing from thetrue path tothevaried path.) Avariational calculation simply tests various possible solutions toaproblem andprescribes amethod forselecting thecorrect solution. The canonical equations ofmotion canalsobeobtained directly from avari- ational calculation based ontheso-called modiﬁed Hamilton’s Principle. The Lagrangian function canbeexpressed as(seeEquation 7.153): L=1%—H(q,,p,-,t) (7.135) andthestatement ofHamilton sPrinciple contained inEquation 7.182 canbe modiﬁed toread Apl-(L—Hdt=0 (7.186) Carrying outthevariation inthestandard manner, weobtain (2 Z(-31+ '_3~—‘-its-—‘-l-I—“l3-)di=0 7.137 L,at in aqji aka <> IntheHamiltonian formulation, theqjand thepjareconsidered tobeinde- pendent. The rjjareagain notindependent oftheqj,soEquation 7.183 canbe used toexpress theﬁrst term inEquation 7.187 as ' £2 ' £2 d L;P15‘I1d‘ =§r>,-;,5<t 1” Integrating byparts, theintegrated term vanishes, andwehave 1, t, It;pj6¢jjdt =—J $15]-Sqj dt (7.188) 1 51 Equation 7.187 then becomes £2 -_ELI _-ill _ LE{(q,- apj)3,6, (,6,+8%)3.1,}at-0 (7.139) If8%andBpjrepresent independent variations, theterms intheparentheses must separately vanish and Hamilton’s canonical equations result. Inthepreceding section, weobtained thecanonical equations bywriting two different expressions for the total differential ofthe Hamiltonian (Equations 7.157 and7.159) andthen equating thecoefﬁcients ofdqjand dpj. Such aprocedure isvalid iftheqjand thepjareindependent variables. Therefore, both intheprevious derivation andinthepreceding variational cal- culation, weobtained thecanonical equations byexploring theindependent na- ture ofthegeneralized coordinates andthegeneralized momenta. The coordinates and momenta arenotactually “independent” intheulti- mate sense oftheword. Forifthetime dependence ofeach ofthecoordinates is 274 7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS known, qj=qj(t), theproblem iscompletely solved. The generalized velocities canbecalculated from in=5;,46> andthegeneralized momenta are 6.= L 3'.’ P.6%(‘I191’) The essential point isthat, whereas theqjandthe arerelated byasimple time derivative independent ofthemanner inwhich thesystem behaves, theconnection be- tween theqjandthepjaretheequations ofmotion themselves. Finding therelations that connect theabandthepj(and thereby eliminating theassumed independ- ence ofthese quantities) istherefore tantamount tosolving theproblem. 7.12 Phase Space and Liouville’s Theorem (Optional) Wepointed outpreviously thatthegeneralized coordinates qjcanbeused tode- ﬁne ans-dimensional conﬁguration space with every point representing acertain state ofthesystem. Similarly, thegeneralized momenta deﬁne ans-dimensional momentum space with every point representing acertain condition ofmotion of thesystem. Agiven point inconﬁguration space speciﬁes only theposition of each oftheparticles inthesystem; nothing canbeinferred regarding themo- tion oftheparticles. The reverse istrue formomentum space. InChapter 3,we found itproﬁtable torepresent geometrically thedynamics ofsimple oscillatory systems byphase diagrams. Ifweusethisconcept with more complicated dynam- icalsystems, then a2s—dimensional space consisting oftheqjandthepjallows us torepresent both thepositions andthemomenta ofallparticles. This general- ization iscalled Hamiltonian phase space or,simply, phase space.* EXAMPLE 7.13 ___ - _ -_ - - Construct thephase diagram fortheparticle inExample 7.11. Solution. The particle hastwodegrees offreedom (6,z),sothephase space for thisexample isactually four dimensional: 6,pa,z,pk.Butpgisconstant and therefore may besuppressed. Inthezdirection, themotion isSimple harmonic, andsotheprojection onto thez-p,plane ofthephase path foranytotal energy Hisjustanellipse. Because =constant, thephase path must represent motion increasing uniformly with 6.Thus, thephase path onanysurface H=constant isauniform elliptic spiral (Figure 7-11). *We previously plotted inthephase diagrams theposition versus aquantity proportional totheve- locity. InHamiltonian phase space, thislatter quantity becomes thegeneralized momentum. 7.12 PHASE SPACE AND LIOUVILLE’S THEOREM (OPTIONAL) 275 P. , Surface H=const. ll /I / /| I I III, I | |U1 1 I | |1 I I I ,'L"'1""'- 1----——— 1————— -1- —— 6I / I // 1 I‘ I4’ ,1 I/ I I I / r ~ FIGURE 7-11 Example 7.13. The phase path fortheparticle inExample 7.11. If,atagiven time, theposition and momenta ofalltheparticles inasys- temareknown, then with these quantities asinitial conditions, thesubsequent motion ofthesystem iscompletely determined; that is,starting from apoint q]-(0), pi-(0) inphase space, therepresentative point describing thesystem moves along aunique phase path. Inprinciple, thisprocedure canalways be followed andasolution obtained. Butifthenumber ofdegrees offreedom of thesystem islarge, thesetofequations ofmotion may betoocomplicated to solve inareasonable time. Moreover, forcomplex systems, such asaquantity ofgas,itisapractical impossibility todetermine theinitial conditions foreach constituent molecule. Because wecannot identify anyparticular point inphase space asrepresenting theactual conditions atanygiven time, wemust devise some alternative approach tostudy thedynamics ofsuch systems. Wetherefore arrive atthepoint ofdeparture ofstatistical mechanics. The Hamiltonian for- mulation ofdynamics isideal forthestatistical study ofcomplex systems. We demonstrate thisinpart bynow proving atheorem that isfundamental for such investigations. Foralarge collection ofparticles—say, gasmolecules-—we areunable to identify theparticular point inphase space correctly representing thesystem. Butwemay ﬁllthephase space with acollection ofpoints, each representing a possible condition ofthesystem; that is,weimagine alarge number ofsystems (each consistent with theknown constraints), anyofwhich could conceivably betheactual system. Because weareunable todiscuss thedetails oftheparti- cles’ motion intheactual system, wesubstitute adiscussion ofanensemble of equivalent systems. Each representative point inphase space corresponds toa single system oftheensemble, andthemotion ofaparticular point represents theindependent motion ofthat system. Thus, notwoofthephase paths may ever intersect. Wemay consider therepresentative points tobesufﬁciently numerous that wecandeﬁne adensity inphase space p.The volume elements ofthephase space deﬁning thedensity must besufﬁciently large tocontain alarge number ofrep- resentative points, butthey must also besufﬁciently small sothat thedensity 276 7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS Pr Iiti>- dpk i» dqk (‘IrvPi) I_ Pi ‘Ir FIGURE 7-12 Anelement ofarea dA=dqkdpkintheqk—pk plane inphase space. varies continuously. The number Nofsystems whose representative points lie within avolume doofphase space is N= pd-o (7.190) where dv=dqkdqk dq,dp1 dpg dp, (7.191) Asbefore, sisthenumber ofdegrees offreedom ofeach system intheensemble. Consider anelement ofarea intheqk~pk plane inphase space (Figure 7-12), The number ofrepresentative points moving across theleft-hand edge into the area perunit time is %d='d PdtPk pqiPi andthenumber moving across thelower edge into thearea perunit time is dp , PFldqk=PMdqk sothatthetotal number ofrepresentative points moving intothearea dqkdpkper unit time is P(‘lkdPk +bide.) (7-192) ByaTaylor series expansion, thenumber ofrepresentative points moving outof thearea perunit time is(approximately) [Pit +(9i(P‘lt)dqk]dPi +lippi +“§”(Pf5k) dhldqi (7193)qt 51614 Thus, thetotal increase indensity indqkdpkperunit time isthedifference be- tween Equations 7.192 and7.193: 6 6 6 . £2dqkdpk=_ligxpik) +5l;;(PPk):|d‘lk dﬁi (7-194) 7.13VIRIAL THEOREM (OPTIONAL) 277 After dividing bydqkdpkandsumming thisexpression over allpossible values of k,weﬁnd an (P <99].5P @113_- -'+ -+- + —0 (7.195 ZaqkqtPaqkamﬁrpap )32-|' 3" -54“QJ From Hamilton’s equations (Equations 7.160 and7.161), wehave (ifthesecond partial derivatives ofHarecontinuous) 6' 8' -Q?+-131‘=0 (7.196)3% apt soEquation 7.195 becomes d§'9+Z(§~3@‘+59-‘lg =0 (7.197)atkaqkdt 61),,at Butthisisjustthetotal time derivative ofp,soweconclude that 1°._ This important result, known asLiouville’s t.heorem,* states that thedensity of representative points inphase space corresponding tothemotion ofasystem of particles remains constant during themotion. Itmust beemphasized that we have been able toestablish theinvariance ofthedensity ponly because theprob- lem was formulated inphase space; anequivalent theorem forconﬁguration space does notexist. Thus, wemust useHamiltonian dynamics (rather than Lagrangian dynamics) todiscuss ensembles instatistical mechanics. Liouville’s theorem isimportant notonly foraggregates ofmicroscopic par- ticles, asinthestatistical mechanics ofgaseous systems andthefocusing proper- tiesofcharged-particle accelerators, butalsoincertain macroscopic systems. For example, instellar dynamics, theproblem isinverted andbystudying thedistri- bution function pofstars inthegalaxy, thepotential Uofthegalactic gravita- tional ﬁeld may beinferred. 7.13 Virial Theorem (Optional) Another important result ofastatistical nature isworthy ofmention. Consider a collection ofparticles whose position vectors raand momenta paareboth bounded (i.e., remain ﬁnite forallvalues ofthetime). Deﬁne aquantity sEZpa-r, (7.199) *Published in1838 byjoseph Liouville (1809-1882). 278 7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS The time derivative ofSis ii=20>.-r..+r'>..-r...) <7-200) Ifwecalculate theaverage value ofdS/dtover atime interval 7,weﬁnd 1 -so<%j>=%L%§d¢=%—(—l (7.201) Ifthesystem’s motion isperiodic—and if7issome integer multiple ofthe period-—then S(t)=S(0), and(5')vanishes. Buteven ifthesystem does notex- hibit anyperiodicity, then-—because Sisbyhypothesis abounded function-—we canmake (5')assmall asdesired byallowing thetime 7'tobecome sufﬁciently long. Therefore, thetime average oftheright-hand sideofEquation 7.201 canal- ways bemade tovanish (oratleast toapproach zero). Thus, inthislimit, wehave (21%,-r,)= -(élpa-rd) (7.202) Ontheleft-hand side ofthisequation, pa-iy,istwice thekinetic energy. Onthe right-hand side, paisjusttheforce Faontheathparticle. Hence, (2;Ta)=-F,-13,) (7.203) The sum over Taisthetotal kinetic energy Tof thesystem, sowehave thegen- eralresult (T)=—éFa-1'“) (7.204) The right-hand side ofthisequation wascalled byClausius* thevirial ofthesys- tem, andthevirial theorem states that theaverage kinetic energy ofasystem ofparticles isequal toitsvirial EXAMPLE 7.14 _ I“ J - Consider anideal gascontaining Natoms inacontainer ofvolume V,pressure P,andabsolute temperature T1(not tobeconfused with thekinetic energy T). Usethevirial theorem toderive theequation ofstate foraperfect gas. Solution. According totheequipartition theorem, theaverage kinetic energy ofeach atom intheideal gasis3/2kT1,where kistheBoltzmann constant. The total average kinetic energy becomes (T)=gNkT1 (7.205) *Rudolph julius Emmanuel Clausius (1822-1888), aGerman physicist andoneofthefounders of thermodynamics. 7.12VIRIAL THEOREM (OPTIONAL) 279 The right-hand side ofthevirial theorem (Equation 7.204) contains the forces Fa.Foranideal perfect gas,noforce ofinteraction occurs between atoms. The only force isrepresented bytheforce ofconstraint ofthewalls. The atoms bounce elastically oﬁ’thewalls, which areexerting apressure onthe atoms. Because thepressure isforce perunit area, weﬁnd theinstantaneous dif- ferential force over adifferential area tobe dFa =—nPdA (7.206) where nisaunitvector normal tothesurface dAandpointing outward. The right-hand sideofthevirial theorem becomes 1 P—-2—(§Fa-r,,> —EJII-I'dA (7.207) Weusethedivergence theorem torelate thesurface integral toavolume integral. in-rdA= [V-rdV= 3JdV= 3V (7.208) Thevirial theorem result is 3 3PV—NkT =—-2 2 _ NkT =PV (7.209) which istheideal gaslaw. Iftheforces Facanbederived from potentials U0),Equation 7.204 may be rewritten as (T)=5r,-vua) (7.210) Ofparticular interest isthecase oftwoparticles thatinteract according toacen- tralpower-law force: Focr".Then, thepotential isoftheform U=kr"+1 (7.211) Therefore avr~VU=E=k(n+1)r"+1=(n +1)U (7.212) andthevirial theorem becomes (T)=5-"~‘;—1<u) (7.212) 280 7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS Iftheparticles have agravitational interaction, then n=—2,and 1 (T)=-§(U). n=-2 This relation isuseful incalculating, forexample, theenergetics inplanetary motion. PROBLEMS 7-1. Adisk rolls without slipping acr0ss ahorizontal plane. The plane ofthedisk re- mains vertical, butitisfreetorotate about avertical axis. What generalized coordi- nates may beused todescribe themotion? Write adifferential equation describing therolling constraint. Isthisequation integrable? Justify your answer byaphysical argument. Istheconstraint holonomic? 7-2. Work out Example 7.6 showing allthe steps, inparticular those leading to Equations 7.36 and 7.41. Explain why thesign oftheacceleration acannot affect thefrequency 0).Give anargument why thesigns ofa2and g2inthesolution of(02 inEquation 7.42 arethesame. 7-3. Asphere ofradius pisconstrained torollwithout slipping onthelower half ofthe inner surface ofahollow cylinder ofinside radius R.Determine theLagrangian function, theequation ofconstraint, and Lagrange’s equations ofmotion. Find the frequency ofsmall oscillations. 7-4. Aparticle moves inaplane under theinﬂuence ofaforce f=—Ar"_l directed to- ward theorigin; Aand LY(>0)areconstants. Choose appropriate generalized co- ordinates, and letthepotential energy bezero attheorigin. Find theLagrangian equations ofmotion. Istheangular momentum about theorigin conserved? Isthe total energy conserved? 7-5. Consider avertical plane inaconstant gravitational ﬁeld. Lettheorigin ofacoor- dinate system belocated atsome point inthisplane. Aparticle ofmass mmoves in thevertical plane under theinﬂuence ofgravity and under theinﬂuence ofanad- ditional force f=—Ar"_1 directed toward theorigin (risthedistance from the origin; Aanda[sﬁ0or1]areconstants). Choose appropriate generalized coordi- nates, and ﬁnd theLagrangian equations ofmotion. Istheangular momentum about theorigin conserved? Explain. 7-6. Ahoop ofmass mand radius Rrolls without slipping down aninclined plane of mass M,which makes anangle 01with thehorizontal. Find theLagrange equations and theintegrals ofthemotion iftheplane canslide without friction along ahori- zontal surface. 7-7. Adouble pendulum consists oftwosimple pendula, with one pendulum suspended from thebob oftheother. Ifthetwopendula have equal lengths and have bobs of equal mass and ifboth pendula areconﬁned tomove inthesame plane, ﬁnd Lagrange’s equations ofmotion forthesystem. Donotassume small angles. PROBLEMS 281 7-8. 7-9. 7-10 7-ll 7-12 7-13 7-14. 7-15 7-16. 7-17Consider aregion ofspace divided byaplane. The potential energy ofaparticle in region 1isU1and inregion 2itisLg.Ifaparticle ofmass mand with speed v1inre- gion 1passes from region 1toregion 2such that itspath inregion 1makes an angle 01with thenormal totheplane ofseparation and anangle 62with thenormal when inregion 2,show that sin01 U1—[51/2 __Z(1.D) sin02 T1 where T1= What istheoptical analog ofthisproblem? Adisk ofmass Mandradius Rrolls without slipping down aplane inclined from thehorizontal byanangle ct.The disk hasashort weightless axle ofnegligible ra- dius. From thisaxis issuspended asimple pendulum oflength l<Rand whose bob hasamass 7n.Consider that themotion ofthependulum takes place intheplane of thedisk, and ﬁnd Lagrange’s equations forthesystem. Two blocks, each ofmass M,areconnected byanextensionless, uniform string of length l.One block isplaced onasmooth horizontal surface, andtheother block hangs over theside, thestring passing over africtionless pulley. Describe themo- tion ofthesystem (a)when themass ofthestring isnegligible and (b)when the string hasamass m. Aparticle ofmass misconstrained tomove onacircle ofradius R.Thecircle rotates inspace about onepoint onthecircle, which isﬁxed. The rotation takes place in theplane ofthecircle andwith constant angular speed w.Intheabsence ofagravi- tational force, show that theparticle’s motion about one end ofadiameter passing through thepivot point and thecenter ofthecircle isthesame asthat ofaplane pendulum inauniform gravitational ﬁeld. Explain why thisisareasonable result. Aparticle ofmass mrests onasmooth plane. The plane israised toaninclination angle 9ataconstant rate a(0I0att=O),causing theparticle tomove down the lane. Determine themotion ofthe article.P P Asimple pendulum oflength bandbobwith mass misattached toamassless sup- port moving horizontally with constant acceleration a.Determine (a)theequations ofmotion and (b)theperiod forsmall oscillations. Asimple pendulum oflength bandbobwith mass misattached toamassless sup- port moving vertically upward with constant acceleration a.Determine (a)the equations ofmotion and (b)theperiod forsmall oscillations. Apendulum consists ofamass msuspended byamassless spring with unextended length band spring constant k.Find Lagrange’s equations ofmotion. The point ofsupport ofasimple pendulum ofmass mandlength bisdriven hori- zontally byx=asinwt.Find thependulum’s equation ofmotion. Aparticle ofmass mcanslide freely along awire ABwhose perpendicular distance totheorigin Oish(see Figure 7-A, page 282). The line OCrotates about theorigin 282 7-18.7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS J’ A C 1/ m//h/ //\9 me 0 Bx FIGURE 7-A Problem 7-17. ataconstant angular velocity =w.The position oftheparticle can bedescribed interms oftheangle 6and thedistance qtothepoint C.Iftheparticle issubject to agravitational force, and iftheinitial conditions are 0(0) =0,q(0) =0,(}(0) =0 show thatthetime dependence ofthecoordinate qis q(t)=2%;(coshwt —coswt) Sketch thisresult. Compute theHamiltonian forthesystem, andcompare with the total energy. Isthetotal energy conserved? - Apendulum isconstructed byattaching amass mtoanextensionless string of length l.The upper end ofthestring isconnected totheuppermost point onaver- tical disk ofradius R(R<l/71') asinFigure 7-B. Obtain thependulum’s equation ofmotion, and ﬁnd thefrequency ofsmall oscillations. Find theline about which theangular motion extends equally ineither direction (i.e., 91=02). 8 8 8 8 8 8 8 8 8 8 8 8 /~~_o----- .' 91 \ 1 s1 s\ 1 s\ '1\ s‘ \ 1\ I’ 62 ‘s\, 8 O ‘~‘ D \ \ ,’\\ ,/ \ / \\ /L 1 \\ 4’~__ _- FIGURE 7-B Problem 7-18. PROBLEMS 283 7-19. Two masses mland 7712(ml9*1'12)areconnected byarigid rodoflength dandof negligible mass. Anextensionless suing oflength llisattached tomland con- nected toaﬁxed point ofsupport P.Similarly, astring oflength Q(l1 =ﬁL2)con- nects mgand P.Obtain theequation describing themotion intheplane ofml,"Z2, andP,andﬁnd thefrequency ofsmall oscillations around theequilibrium position. 7-20. Acircular hoop issuspended inahorizontal plane bythree strings, each oflength l,which areattached symmetrically tothehoop and areconnected toﬁxed points lying inaplane above thehoop. Atequilibrium, each string isvertical. Show that thefrequency ofsmall rotational oscillations about thevertical through thecenter ofthehoop isthesame asthat forasimple pendulum oflength l. 7-21. Aparticle isconstrained tomove (without friction) onacircular wire rotating with constant angular speed toabout avertical diameter. Find theequilibrium position oftheparticle, and calculate thefrequency ofsmall oscillations around this posi- tion. Find and interpret physically acritical angular velocity w=co,that divides the particle’s motion into twodistinct types. Construct phase diagrams forthetwocases w<wcandw >0),. 7-22. Aparticle ofmass mmoves inone dimension under theinﬂuence ofaforce k Fx,t=—e_(‘/T) <)x. where kand1'arepositive constants. Compute theLagrangian andHamiltonian functions. Compare theHamiltonian and thetotal energy, and discuss theconser- vation ofenergy forthesystem. 7-23. Consider aparticle ofmass mmoving freely inaconservative force ﬁeld whose po- tential function isU.Find theHamiltonian function, andshow thatthecanonical equations ofmotion reduce toNewton’s equations. (Use rectangular coordinates.) 7-24. Consider asimple plane pendulum consisting ofamass mattached toastring of length Z.After thependulum issetinto motion, thelength ofthestring isshort- ened ataconstant rate dl—=—a=constantdt The suspension point remains ﬁxed. Compute theLagrangian and Hamiltonian functions. Compare theHamiltonian and thetotal energy, and discuss theconser- vation ofenergy forthesystem. 7-25. Aparticle ofmass mmoves under theinﬂuence ofgravity along thehelix z=k6,r= constant, where kisaconstant and zisvertical. Obtain theHamiltonian equations ofmotion. 7-26. Determine theHamiltonian andHamilton’s equations ofmotion for(a)asimple pendulum and (b)asimple Atwood machine (single pulley). 7-27. Amassless spring oflength bandspring constant kconnects twoparticles ofmasses mlandmg.Thesystem rests onasmooth table andmayoscillate androtate. 284 7-28. 7-29. 7-30. 7-31 7-32.7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS (a)Determine Lagrange ’sequations ofmotion. (b)What arethegeneralized momenta associated with anycyclic coordinates? (c)Determine Hamilton’s equations ofmotion. Aparticle ofmass misattracted toaforce center with theforce ofmagnitude k/r2. Useplane polar coordinates andﬁnd Hamilton’s equations ofmotion. Consider thependulum described inProblem 7-15. The pendulum’s point ofsup- port rises vertically with constant acceleration a. (a)Use theLagrangian method toﬁnd theequations ofmotion. (b)Determine theHamiltonian and Hamilton’s equations ofmotion. (c)What istheperiod ofsmall oscillations? Consider any two continuous functions ofthe generalized coordinates and mo- menta g(<Ii»11:.) andh(q,,,pk).ThePoisson brackets aredeﬁned by ﬁgah ﬁgahlg./1152 ————~"aqn'31’: 31%‘391 Verify thefollowing properties ofthePoisson brackets: dg as . -(a)5=[g.H1 +5 (b)q,~=[q.-.H].p,»=[p,<.H] ('3)1111,11,-1 :0,lqr:q,~]:O id)[(11-11)] =5y where HistheHamiltonian. IfthePoisson bracket oftwoquantities vanishes, the quantities aresaid tocommute. IfthePoisson bracket oftwoquantities equals unity, thequantities aresaid tobecanonically conjugate. (e)Show that anyquantity that does notdepend explicitly onthetime and that commutes with theHamiltonian is aconstant ofthemotion ofthesystem. Poisson-bracket formalism isofconsider- able importance inquantum mechanics. Aspherical pendulum consists ofabob ofmass mattached toaweightless, exten- sionless rodoflength l.The endoftherodopposite thebobpivots freely (inalldi- rections) about some ﬁxed point. SetuptheHamiltonian function inspherical co- ordinates. (Ifp¢=O,the result isthe same asthat forthe plane pendulum.) Combine theterm that depends onp4,with theordinary potential energy term to deﬁne aseffective potential V(0,11¢).Sketch Vasafunction of6forseveral values of p¢,including 12¢=0.Discuss thefeatures ofthemotion, pointing outthediffer- ences between pd,=Oand pd,=#O.Discuss thelimiting case oftheconical pendu- lum (6=constant) with reference totheV-6diagram. Aparticle moves inaspherically symmetric force ﬁeld with potential energy given byU(r) =_k/T. Calculate theHamiltonian function inspherical coordinates, and obtain thecanonical equations ofmotion. Sketch thepath that arepresentative point forthesystem would follow onasurface HIconstant inphase space. Begin byshowing that themotion must lieinaplane sothat thephase space isfour di- mensional (r,6,p,,pg,butonly theﬁrstthree arenontrivial). Calculate theprojec- tion ofthephase path onthe1'-p,plane, then take into account thevariation with 0. PROBLEMS 285 7-33. Determine theHamiltonian and Hamilton’s equations ofmotion forthedouble Atwood machine ofExample 7.8. 7-34. Aparticle ofmass mslides down asmooth circular wedge ofmass Masshown in Figure 7-C.Thewedge rests onasmooth horizontal table. Find (a)theequation of motion ofmandMand(b)thereaction ofthewedge onm. mR M -.»~:-~:.~'.~ ..5.;,;_.:~;--, ,~.".;>g~::;;':*§',;*:; 33 FIGURE 7-C Problem 7-34. 7-35. Four particles aredirected upward inauniform gravitational ﬁeld with thefollow- 7-36 7-37 7-38. 7-39.inginitial conditions: (1)Z(0)=Z0; M0) =P0 (2)1(0)=Z0+A10; ﬂ.(0) =P0 (3)1(0)=Z0; P=(0) IPo+AP0 (4)z(0)=Z0+A10; 112(0) =Po+AP0 Show bydirect calculation that therepresentative points corresponding tothese particles always deﬁne anarea inphase space equal toA20Apo. Sketch thephase paths, and show forseveral times t>Otheshape oftheregion whose area remains constant. Discuss the implications ofLiouville’s theorem onthe focusing ofbeams of charged particles byconsidering thefollowing simple case. Anelectron beam of circular cross section (radius R0)isdirected along thez-axis. The density ofelec- trons across thebeam isconstant, butthemomentum components transverse to thebeam (p,andpy)aredistributed uniformly over acircle ofradius poinmomen- tum space. Ifsome focusing system reduces thebeam radius from R0toR1,ﬁnd the resulting distribution ofthetransverse momentum components. VVhat isthephysi- calmeaning ofthis result? (Consider theangular divergence ofthebeam.) Usethemethod ofLagrange undetermined multipliers toﬁnd thetensions inboth strings ofthedouble Atwood machine ofExample 7.8. The potential forananharmonic oscillator isU=kx2/2 +bx4/4 where kandbare constants. Find Hamilton’s equations ofmotion. Anextremely limber rope ofuniform mass density, mass mand total length blieson atable with alength zhanging over theedge ofthetable. Only gravity acts onthe rope. Find Lagrange’s equation ofmotion. 286 7/HAMILTON’S PRINCIPLE—LAGRANGIAN AND HAMILTONIAN DYNAMICS 7-40. Adouble pendulum isattached toacartofmass 2mthatmoves without friction on ahorizontal surface. SeeFigure 7-D.Each pendulum haslength bandmass bobm. Find theequations ofmotion. b m b m i U0 FIGURE 7-D Problem 7-40. 7-41. Apendulum oflength band mass bob misoscillating atsmall angles when the length ofthependulum string isshortened atavelocity of01(db/dt =—a). Find ~Lagrange’s equations ofmotion. CPLELPTER Central-Force Motion 8.1Introduction The motion ofasystem consisting oftwobodies affected byaforce directed along thelineconnecting thecenters ofthetwobodies (i.e., acentral force) isan extremely important physical problem—one wecan solve completely. Th_e im- portance ofsuch aproblem liesinlarge measure intwoquite different realms of physics: themotion ofcelestial bodies—planets, moons, comets, double stars, and thelike—and certain two-body nuclear interactions, such asthescattering ofaparticles bynuclei. Intheprequantum-mechanics days, physicists also de- scribed thehydrogen atom interms ofaclassical two-body central force. Although such adescription isstilluseful inaqualitative sense, thequantum- theoretical approach must beused foradetailed description. Inaddition to some general considerations regarding motion incentral-force ﬁelds, wediscuss inthisandthefollowing chapter several oftheproblems oftwobodies encoun- tered incelestial mechanics andinnuclear andparticle physics. 8.2 Reduced Mass Describing asystem consisting oftwoparticles requires thespeciﬁcation ofsix quantities; forexample, thethree components ofeach ofthetwovectors r1and r2fortheparticles.* Alternatively, wemay choose thethree components ofthe center-of-mass vector Randthethree components ofrEr1—r2(seeFigure 8-1a). Here, werestrict ourattention tosystems without frictional losses and forwhich *The orientation oftheparticles isassumed tobeunimportant; thatis,they arespherically symmet- ric(orarepoint particles) . 287 288 s/CENTRAL-FORCE MOTION ml ml I1 CM ‘lCM l R50 T2 1'2 m2 "*2 (8) (b) FIGURE 8-1 Two methods todescribe theposition oftwoparticles. (a)From an arbitrary coordinate system origin, and(b)from thecenter ofmass. Theposition vectors arerland1'2,thecenter-of-mass vector isR,and therelative vector r=rl—1'2. thepotential energy isafunction only ofr=|rl—r2The Lagrangian forsuch asystem may bewritten as 1 1L=§m.|r.|*‘+§m..Ir2|2- 00> <8-1) Because translational motion ofthesystem asawhole isuninteresting from thestandpoint oftheparticle orbits with respect toone another, wemay choose theorigin forthecoordinate system tobetheparticles’ center ofmass—that is, RE0(seeFigure 8-1b). Then (seeSection 9.2) mlrl +7!l21'2 20 This equation, combined with r=rl—r2,yields rT‘ m2 rl__i___ ml+ "12 (8.3) ,___Jﬁ_,2.. 7111+ m2 Substituting Equation 8.3into theexpression fortheLagrangian gives L=%uM2—Um (ab where ].Listhereduced mass, _ "$1702 M=mm (8-5)ml+mg Wehave therefore formally reduced theproblem ofthemotion oftwobod- iestoanequivalent one-body problem inwhich wemust determine only themotion ofa“particle” ofmass /.tinthecentral ﬁeld described bythepotential function 8.3 CONSERVATION THEOREMS—FIRST INTEGRALS OFTHE MOTION 289 U('r). Once weobtain thesolution forr(t)byapplying theLagrange equations to Equation 8.4,wecanﬁnd theindividual motions oftheparticles, rl(t)and r2(t) , byusing Equation 8.3.This latter step isnotnecessary ifonly theorbits relative tooneanother arerequired. 8.3 Conservation Theorems— First Integrals oftheMotion Thesystem wewish todiscuss consists ofaparticle ofmass ;.(.moving inacentral- force ﬁeld described bythepotential function U(r). Because thepotential en- ergy depends only onthedistance oftheparticle from theforce center and not ontheorientation, thesystem possesses spherical symmetry; thatis,thesystem’s rotation about anyﬁxed axisthrough thecenter offorce cannot affect theequa- tions ofmotion. Wehave already shown (see Section 7.9) that under such condi- tions theangular momentum ofthesystem isconserved: L=rXpIconstant . (8.6) From thisrelation, itshould beclear that both theradius vector andthelinear momentum vector oftheparticle liealways inaplane normal totheangular mo- mentum vector L,which isﬁxed inspace (see Figure 8-2). Therefore, wehave only atwo-dimensional problem, andtheLagrangian may then beconveniently expressed inplane polar coordinates: L-$002+7202)-00> (8.7) Because theLagrangian iscyclic in6,theangular momentum conjugate to thecoordinate 6isconserved: .at ear.=**—'=0:_""'- 8.8 1°“a0 dt66 () L P FIGURE 8-2 Themotion ofaparticle ofmass p.moving inacentral-force ﬁeld is described bytheposition vector r,linear momentum p,andconstant angular momentum L. 290 s/CENTRAL-FORCE MOTION A.<b~xQ.Cb‘('71) l'(l2) r(t) FIGURE 8-3 Thepath ofaparticle isdescribed byr(t).The radius vector sweeps out anarea dA=%r2d6 inatime interval dt. or 6L .pl,E—.-=/.tr26 =constant (8.9) 66 The system’s symmetry hastherefore pennitted ustointegrate immediately oneoftheequations ofmotion. The quantity pl,isaﬁrst integral ofthemotion, and wedenote itsconstant value bythesymbol l: IIE1.1.126 iconstant] (8.10) Note thatlcan benegative aswellaspositive. That lisconstant hasasimple geometric interpretation. Referring toFigure 8-3,weseethat indescribing the path r(t), theradius vector sweeps outanarea %r2d6 inatime interval dt: 1dA=§r2d6 (8.11) Ondividing bythetime interval, theareal velocity isshown tobe dA 1d6 1-_____ 2____ 29 dt 27at 27 lI5,;=constant (8.12) Thus, theareal velocity isconstant intime. This result wasobtained empirically byKepler forplanetary motion, and itisknown asKepler’s Second Law.* Itis important tonote that theconservation oftheareal velocity isnotlimited toan inverse-square-law force (the case forplanetary motion) butisageneral result forcentral-force motion. Because wehave eliminated from consideration theuninteresting uniform motion ofthesystem’s center ofmass, theconservation oflinear momentum adds nothing new tothedescription ofthemotion. The conservation ofenergy isthus theonly remaining ﬁrstintegral oftheproblem. The conservation ofthe Q *Published by_]ohannes Kepler (1571-1630) in1609 after anexhaustive study ofthecompilations made byTycho Brahe (1546-1601) ofthepositions oftheplanet Mars. Kepler’s First Lawdeals with theshape ofplanetary orbits (seeSection 8.7). s.4EQUATIONS orMOTION 291 total energy Eisautomatically ensured because wehave limited thediscussion to nondissipative systems. Thus, T+U=E=constant (8.13) and 1 . E=5/.t(i"2 +r262) +U(r) or E=1,172+1L2+U(r) (8.14)2 2/.1.r2 8.4 Equations ofMotion W'hen U(r) isspeciﬁed, Equation 8.14 completely describes thesystem, andthe integration ofthisequation gives thegeneral solution oftheproblem interms of theparameters Eandl.Solving Equation 8.14 forit,wehave ,dr /2 l2T-‘=5: i ;(E** U)**”72 (8.i5) This equation canbesolved fordtandintegrated toyield thesolution t=t(r). Aninversion ofthisresult then gives theequation ofmotion inthestandard form r=r(t).Atpresent, however, weareinterested intheequation ofthepath interms ofrand 6.Wecanwrite d6at 6=———d =—d 8.16‘MatdrT7T () Into this relation, wecan substitute =l//J.r2 (Equation 8.10) and theexpres- sion forrfrom Equation 8.15. Integrating, wehave 0(7)=l (8.17) 1l2,4L(E —U— Furthermore, because lisconstant intime, cannot change sign andthere- fore 6(t) must increase ordecrease monotonically with time. Although wehave reduced theproblem totheformal evaluation ofaninte- gral, theactual solution canbeobtained only forcertain speciﬁc forms ofthe force law. Ifthe force isproportional tosome power ofthe radial distance, F(r) o<r",then thesolution canbeexpressed interms ofelliptic integrals for certain integer andfractional values ofn.Only forn=1,-2,and -3aretheso- lutions expressible interms ofcircular functions (sines and cosines).* The case *See, forexample, Goldstein (G080, pp.88-90). 292 8/CENTRALFORCE MOTION n=1isjustthatoftheharmonic oscillator (seeChapter 3),andthecase n=-2 istheimportant inverse-square-law force treated inSections 8.6and 8.7. These twocases, n=1,--2,areofprime importance inphysical situations. Details of some other cases ofinterest willbefound intheproblems attheend ofthis chapter. Wehave therefore solved theproblem inaformal waybycombining the equations that express theconservation ofenergy andangular momentum into asingle result, which gives theequation oftheorbit 6=6(r). Wecanalso attack theproblem using Lagrange’s equation forthecoordinate r: 6L d6L___ :0 6r dtiii" Using Equation 8.7forL,weﬁnd __ . 6U/.t(r-—r62) =——5 =F(r) (8.18) Equation 8.18 canbecastinaform more suitable forcertain types ofcalcu- lations bymaking asimple change ofvariable: _W1u=_ r First, wecompute du 1dr 1drdt 17 d6 r2d6 r2dtd6 r26 Butfrom Equation 8.10, =l//.1.r2, so £13__&d6 1 Next, wewrite d62 d6 l d6dt l ll’) and with thesame substitution for6,wehaver din #2.. Ta='1-it Therefore, solving forrandr62interms ofu,weﬁnd __ l22d2ur=——u ——M2 0362 r62jféui’ /-4'(8.19) 3.4EQUATIONS orMOTION 293 Substituting Equation 8.19 into Equation 8.18, weobtain thetransformed equation ofmotion: d2 /11F93;+u=-FEF(1/u) (8.20) which wemay alsowrite as at11 /1.12W +;——"' This form oftheequation ofmotion isparticularly useful ifwewish toﬁnd the force lawthatgives aparticular known orbit r=r(6). MT _'Find theforce lawforacentral-force ﬁeld that allows aparticle tomove ina logarithmic spiral orbit given byr=ke"“9, where kandozareconstants. Solution. WeuseEquation 8.21 todetermine theforce lawF(r). First, we determine 1(1)Z1 Zd6r d6 k k d21 oz2e_°“’ 012 (.)-T--From Equation 8.21, wenow determine F(r). __[2 2 1 F(r)=—(5+-)pm? r r __12 F(r) =——3(a2 +1) (8.22)].LT Thus, theforce lawisanattractive inverse cube. ' 'Determine r(t)and 6(t)fortheproblem inExample 8.1. Solution. From Equation 8.10, weﬁnd . l l6=—=i— 8.3[M2 ,_Lk2e2a6 (2) Rearranging Equation 8.23 gives 6220.10 =id:/.tk2 294 8/CENTRALFORCE MOTION andintegrating gives 20:6 L :K +C’ 2a uh? where C'isanintegration constant. Multiplying by2aandletting C=2aC' gives @220=gilt+c (8.24);.tk2 Wesolve for6(t)bytaking thenatural logarithm ofEquation 8.24: 1 2a:lt 6 =—lZ C 8. (t) 2an(,.Lk2 +) (25) Wecansimilarly solve forr(t)byexamining Equations 8.23 and 8.24: r2 20 2alt—-; 0' :—+ C 1.2”are r(t) =[g-E! t+k2C]l/2 (8.26) The integration constant Cand angular momentum lneeded forEquations 8.25 and 8.26 aredetermined from theinitial conditions. F.XAMPl.E 8.3 What isthetotal energy oftheorbit oftheprevious twoexamples? Solution. The energy isfound from Equation 8.14. Inparticular, weneed i‘ and U(r). +l2U(r) =-Fdr= T012 +I)r_3dr 2 2 U0)=-l—(3‘-5-1-)% (8.27) where wehave letU(<><>) =0. Werewrite Equation 8.10 todetermine i: . d6 d6dT 16:——:—-izi dt drdt /.1.r2 '--iii-i— o16i—glr-d6m2—ake [M2-par (8.28) 8.5ORBITS INACENTRAL FIELD 295 Substituting Equations 8.27 and8.28 into Equation 8.14 gives E'1(ill)? +[2mag+1)—2” T 2;.tr2 2/.tr2 E=0 (8.29) The total energy oftheorbit iszero ifU(r=0°)=0. 8.5 Orbits inaCentral Field The radial velocity ofaparticle moving inacentral ﬁeld isgiven byEquation 8.15. This equation indicates that ivanishes attheroots oftheradical, thatis,at points forwhich 2 E—U(r) —L =0 (8.30)2/.LT2 The vanishing ofrimplies that aturning point inthemotion hasbeen reached (see Section 2.6). Ingeneral, Equation 8.30 possesses tworoots: rm,and rmln. The motion oftheparticle istherefore conﬁned totheannular region speciﬁed byrm,‘ 2r2rmln. Certain combinations ofthepotential function U(r) and, the parameters Eandlproduce only asingle root forEquation 8.30. Insuch acase, 7=0forallvalues ofthetime; hence, r=constant, and theorbit iscircular. Ifthemotion ofaparticle inthepotential U(r) isperiodic, then theorbit is closed; that is,after aﬁnite number ofexcursions between theradial limits rmll, and rm“, themotion exactly repeats itself. Butiftheorbit does notclose onitself after aﬁnite number ofoscillations, theorbit issaid tobeopen (Figure 8-4). From Equation 8.17, wecancompute thechange intheangle 6thatresults from onecomplete transit ofrfrom rmll,torm,andback tormln.Because themotion is ’¢ ___ 4 4\ / N / \\ /I \/\ II \ I\ l \ 1 I I1 \ I \ / \ I I I / \ / \ I \ /' xx I’ \ /I\\ f\__ ”/ FIGURE 8-4 Anorbit thatdoes notclose onitself after aﬁnite number ofoscillations issaidtobeopen. 296 8/CENTRAL-FORCE MOTION symmetric intime, thisangular change istwice thatwhich would result from the passage from rmllltorm“; thus A0-=2l%“-——-—£Zfi55————— (881)rmln [2 \l2/-L(E-TU-T The path isclosed only ifA6isarational fraction of271'—that is,ifA6=-277'- (a/b),where aand bareintegers. Under these conditions, after bperiods the radius vector oftheparticle willhave made acomplete revolutions and willhave returned toitsoriginal position. Wecanshow (see Problem 8-35) that ifthepo- tential varies with some integer power oftheradial distance, U(r) ocr"+1, then a closed noncircular path canresult only* ifnI-2or+1.The case n=-2cor- responds toaninverse-square-law force—for example, thegravitational orelec- trostatic force. The n=+1case corresponds totheharmonic oscillator poten- tial. For the two-dimensional case discussed inSection 3.4, wefound that a closed path forthemotion resulted iftheratio oftheangular frequencies for thexandymotions were rational. 8.6 Centrifugal Energy and theEffective Potential Inthepreceding expressions for7,A6,andsoforth, acommon term istheradical [2 JE-U————2r2/4 The lastterm intheradical hasthedimensions ofenergy and, according to Equation 8.10, canalso bewritten as F1. i:_ r292 2a# 2” Ifweinterpret thisquantity asa“potential energy,” l2 U6E2,172 (8.32) then the“force” thatmust beassociated with U,is F___£i]‘-_i.._ Q2 (833) C 6r /.1.r3 I'M ' *Certain fractional values ofnalsolead toclosed orbits, butingeneral these cases areuninteresting from aphysical standpoint. 8.6 CENTRIFUGAL ENERGY AND THE EFFECTIVE POTENTIAL 297 This quantity istraditionally called thecentrifugal force,* although itisnota force intheordinary sense oftheword.1 Weshall, however, continue tousethis unfortunate terminology, because itiscustomary andconvenient. Weseethattheterm Z2/2/.tr2 canbeinterpreted asthecentrifugal potential en- ergyoftheparticle and, assuch, canbeincluded with U(r) inaneﬂfective potential energy deﬁned by 2 v(t)Ev(t)+2,% (8.84) V(r) istherefore aﬁctitious potential that combines thereal potential function U(r) with theenergy term associated with theangular motion about thecenter offorce. For the case ofinverse-square-law central-force motion, theforce is given by kF(r) =-E (8.35) from which U(r) =-IF(r)dr=-2 (8.36) The effective potential function forgravitational attraction istherefore __ k l2 I/(T) —--T7’+ﬁ This effective potential and itscomponents areshown inFigure 8-5.The value of thepotential isarbitrarily taken tobezero atr=0°.(This isimplicit inEquation 8.36, where weomitted theconstant ofintegration.) Wemay now draw conclusions similar tothose inSection 2.6onthemotion ofaparticle inanarbitrary potential well. Ifweplot thetotal energy Eofthepar- ticle onadiagram similar toFigure 8-5,wemay identify three regions ofinterest (seeFigure 8-6). Ifthetotal energy ispositive orzero (e.g., El20),then themo- tion isunbounded; theparticle moves toward theforce center (located atr=0) from inﬁnitely faraway until it“strikes” thepotential barrier attheturning point r1rlandisreﬂected back toward inﬁnitely large r.Note that theheight ofthe constant total energy lineabove V(r)atanyr,such asr5inFigure 8-6,isequal to éuri. Thus theradial velocity 1'"vanishes and changes sign attheturning point (orpoints). *The expression ismore readily recognized intheform F,=mrcog. The ﬁrst real appreciation ofcen- trifugal force wasbyHuygens, who made adetailed examination inhisstudy oftheconical pendu- lum in1659. 'lSee Section 10.3foramore critical discussion ofcentrifugal force. 298 Energy~i=‘°'1» V(rF_-—-—%#1 \ \ \ \ \ \ \ \ \292 ‘\ \ >‘~~-8/CENTRAL-FORCE MOTION V(o°)E0>T ____-_-__-_--I- ¢""v‘v aﬂu aI \s ~1>a~\\ ’__1 11 I FIGURE 8-5 Theeffective potential forgravitational attraction V(r)iscomposed of therealpotential —k/rterm andthecentrifugal potential energy Z2/2p. T2. Energy 0¢_____l/(T) ----------------------------------------------------- --E1 <3‘ii”? T2 T4 T5 Y>-I _______: l__ _, >7- —'----------------------- --E2 ''''''''''''''_'''''_''''''''___'"E3 FIGURE 8-6 Wecantellmuch about motion bylooking atthetotal energy Eona potential energy plot. Forexample, forenergy E1theparticle’s motion isunbounded. Forenergy E2theparticle isbounded with 12SrSr4. Forenergy E3themotion has1"=13and iscircular. 8.6 CENTRIFUGAL ENERGYAND THE EFFECTIVE POTENTIAL 299 wt IIII 40- — F=25h 30 — (MeV)201‘; VTolalIQC G§\‘ 10T 120+288i Nucleus nucleus potential 0 I I I I _10 I".__I I I 6 8 10 12 14 Distance between nuclei centers (1045 m) FIGURE 8-7 The total potential (coulomb, nuclear, andcentrifugal) forscattering 283i nuclei from 12Cforvarious angular momentum lvalues asafunction of distance between nuclei. Forl=20hashallow pocket exists where the twonuclei maybebound together forashort time. Forl=25hthe nuclei arenotbound together. Ifthetotal energy isnegative* and liesbetween zero and theminimum value ofV(r), asdoes E2,then themotion isbounded, with 1'25rS1'4.The val- uesr2and r4aretheturning points, ortheapsidal distances, oftheorbit. IfE equals theminimum value oftheeffective potential energy (see E3inFigure 8-6), then theradius oftheparticle’s path islimited tothesingle value T3,and then r=Oforallvalues ofthetime; hence themotion iscircular. Values ofEless than Vmin ==—(/J.k2/2Z2) donotresult inphysically real mo- tion; forsuch cases 1'2<Oand thevelocity isimaginary. The methods discussed inthissection areoften used inpresent-day research ingeneral ﬁelds, especially atomic, molecular, and nuclear physics. Forexample, Figure 8-7shows effective total nucleus-nucleus potentials forthescattering of 28Siand12C.The total potential includes thecoulomb, nuclear, andthecentrifu- galcontributions. The potential forl=Oﬁindicates thepotential with nocen- trifugal tenn. Forarelative angular momentum value ofI=20ﬁ, a“pocket” ex- istswhere thetwoscattering nuclei may bebound together (even ifonly fora short time). ForlI25?», thecentrifugal “barrier” dominates, and thenuclei can- notform abound state atall. *Note thatnegative values ofthetotal energy arise only because ofthearbitrary choice ofV(r)=0at 1=Q0. 300 8/CENTRALFORCE MOTION 8.7 Planetary Motion—Kepler’s Problem The equation forthepath ofaparticle moving under theinﬂuence ofacentral force whose magnitude isinversely proportional tothesquare ofthedistance be- tween theparticle and theforce center canbeobtained (see Equation 8.17) from __I (l/r2) dr6('r) -——*~"-*"~——i +constant (8.38) k Z2,/2,.(E+T M) The integral canbeevaluated ifthevariable ischanged touEl/'r(see Problem 8-2). Ifwedeﬁne theorigin of6sothat theminimum value of1"isat6=O,we ﬁnd l21-——1/.tkr cost) =———; (8.39)I 21+2El /M2 Letusnow deﬁne thefollowing constants: [2 aE—/.tk(8.40) _I 2El2-‘J= 1+—— /.tk2 Equation 8.39 canthus bewritten as This istheequation ofaconic section with onefocus attheorig1'n.* The quan- titysiscalled theeccentricity, and 2aistermed thelatus rectum oftheorbit. Conic sections areformed bytheintersection ofaplane andacone. Aconic sec- tion isformed bythelociofpoints (formed inaplane), where theratio ofthe distance from aﬁxed point (the focus) toaﬁxed line (called thedirectrix) isa constant. The directrix fortheparabola isshown inFigure 8-8bythevertical dashed line, drawn sothat 1*/r’==1. Theminimum value forrinEquation 8.41 occurs when 6=0,orwhen cos6 isamaximum. Thus thechoice oftheintegration constant inEquation 8.38 cor- responds tomeasuring 6from rmin, which position iscalled thepericenter; rmax corresponds totheapocenter. The general term forturning points isapsides. The corresponding terms formotion about theSun areperihelion and aphelion, and formotion about Earth, perigee and apogee. *]ohann Bemoulli (1667-1748) appears tohave been theﬁrsttoprove that allpossible orbits ofa body moving inapotential proportional to1/rareconic sections (1710). 8.7PLANETARY MOTION—KEPLER’S PROBLEM 301 Hyperbola, £>1 Parabola, 5=1 Directrix forparabola ﬁx ..I__._______Ellipse, 0<e<1 IT, . I - Is-0 II Focus FIGURE 8-8 TheOrbits ofthevarious conic sections areshown together with their eccentricities s. Various values oftheeccentricity (and hence oftheenergy E)classify theor- bitsaccording todifferent conic sections (seeFigure 8-8): s>1, E>O Hyperbola s="-1, E=O Parabola O<s<1, Vmin <E<0Ellipse eZO, E=Vmin Circle For planetary motion, the orbits are ellipses with major and minor axes (equal to2aand2b,respectively) given by a1—s2_2|E| ') oz l b=i=i— (8.43) \/1—~s2 \/2/.t|E| Thus, themajor axis depends only ontheenergy oftheparticle, whereas the minor axisisafunction ofboth ﬁrstintegrals ofthemotion, Eandl.Thegeometry ofelliptic orbits interms oftheparameters oz,s,a,and bisshown inFigure 8-9;P 302 s/CENTRALFORCE MOTION hi (1 Ib ___1;,S2~>|_I <—q8-->| IP’ FIGURE 8-9 Thegeometry ofelliptic orbits isshown intenns ofparameters a,s,a, and b.Pand P’arethefoci. and P’arethefoci. From thisdiagram, weseethat theapsidal distances (rm, and rm“asmeasured from thefocitotheorbit) aregiven by 0! Tmin=a(1*8)i (8.44)G ¢m,,,=a(1+.-3)-=1?"‘S Toﬁnd theperiod forelliptic motion, werewrite Equation 8.12 fortheareal velocity as 2at=3atl Because theentire area Aoftheellipse isswept outinone complete period 1', 1' 2 A Idr=iIaA0 lo 1'=QTMA (8.45) The area ofanellipse isgiven byA=1rab, and using aand bfrom Equations 8.42 and8.43, weﬁnd 'r= 1rab= --2/.¢ 2/IW k l l l 2|E| \/2,,|E| :1rk\/%- |E|“”’2 (8.46) Wealso note from Equations 8.42 and 8.43 that thesemiminor axis* can be written as b=\/aa (8.47) *The quantities aandbarecalled semimajar andsemiminar axes, respectively.___ 3.7PLANETARY MOTION—KEPLER’S PROBLEM 303 Therefore, because a=l2//.Lk, theperiod 1'canalsobeexpressed as 4 2 T2 3%{a3 This result, that thesquare oftheperiod isproportional tothecube ofthe semimajor axisoftheelliptic orbit, isknown asKepler’s Third Law.* Note that this result isconcerned with theequivalent one-body problem, soaccount must betaken ofthefactthat itisthereduced mass 1.4.that occurs inEquation 8.48. Kepler actually concluded that thesquares oftheperiods oftheplanets were proportional tothecubes ofthemajor axes oftheir orbits—with thesame pro- portionality constant forallplanets. Inthissense, thestatement isonly approxi- mately correct, because thereduced mass isdifferent foreach planet. Inparticu- lar,because thegravitational force isgiven by Gmlmg k F<'>=rT"="2 weidentify k1Gmlmg. The expression forthesquare oftheperiod therefore becomes 4'rr2a3 4"rr2a32=-i ZZ << s.49. Tcm,+m2) om,’ m‘mi I) andKepler’s statement iscorrect only ifthemass mlofaplanet canbeneglected with respect tothemass m2oftheSun. (But note, forexample, that themass of Jupiter isabout 1/1000 ofthemass oftheSun, sothedeparture from theap- proximate lawisnotdifﬁcult toobserve inthiscase.) Kepler’s laws cannow besummarized: I. Planets move inelliptical orbits about theSun with theSun atonefocus. II. Theareaperunit timeswept outbyaradius vector from theSun toaplanet is constant. III. Thesquare ofaplanet’s period isproportional tothecube ofthemajor axisofthe planet’s orbit. SeeTable 8-1forsome properties oftheprincipal objects inthesolar system. *Published byKepler in1619. Kepler’s Second Lawisstated inSection 8.3.TheFirst Law (1609) dic- tates thattheplanets move inelliptical orbits with theSunatonefocus. Kepler’s work preceded by almost 80years Newton’s enunciation ofhisgeneral laws ofmotion. Indeed, Newton's conclusions were based toagreat extent onKepler’s pioneering studies (and onthose ofGalileo andHuygens). 304 s/CENTRAL-FORCE MOTION TABLE 8-1 Some Properties ofthePrincipal Objects intheSolar System Semirnajor axisoforbit Mass (inunits of Name (inastronomical units“) Period (yr) Eccentricity Earth’s mass”) Sun — Mercury 0.3871 Venus 0.7233 Earth 1.0000 Eros (asteroid) 1.4583 Mars 1.5237 Ceres (asteroid) c ]upiter 5.2028 Saturn 9.5388 Uranus 19.1910.2408 0.6152 1.0000 1.7610 1.8809 4.6035 c 29.456 84.07 Neptune 30.061 104.81 Pluto 39.529 248.53 Halley (comet) 18 760.2056 0.0068 0.0167 0.2229 0.0934 0.0789 0.0483 0.0560 0.0461 0.0100 0.2484 0.967332,830 0.0552 0.814 1.000 2><10-9 (P) 0.1074 1/8000 (P) 317.89 C 14.56 17.15 0.002 ,_,10~10 “One astronomical unit (A.U.) isthelength ofthesemimajor axisofEarth’s orbit. One A.U. ¥1.495 X1011mE 93><1011miles. "Earth’s mass isapproximately 5.976 X102‘kg. ’SeeProblem 8-19. EXAMPLE 8.4 Halley’s comet, which passed around thesunearly in1986, moves inahighly el- liptical orbit with aneccentricity of0.967 and aperiod of76years. Calculate its minimum andmaximum distances from theSun. Solution. Equation 8.49 relates theperiod ofmotion with thesemimajor axes. Because m(Halley’s comet) <<msun, _ /GmSunT2 1/3 a—\4"rr2 1/3‘ Nm2 365day 24hr3600 s26.67 10-11 "1" 1.99 10311 l< 76 —""i —'i i- iX kg?l( X gliYryr day hr =I I I4'rr2 W C P a=2.68 ><1012m Using Equation 8.44, wecandetermine rm,“and rm“. rm,“=2.68 ><1012m(1 —0.967) =8.8><101°m rm,=2.68 ><1012m(1 +0.967) =5.27 ><1012m This orbit takes thecomet inside thepath ofVenus, almost toMercury’s orbit, and outpast even theorbit ofNeptune and sometimes even tothemoderately eccentric orbit ofPluto. Edmond Halley isgenerally given thecredit forbringing asORBITAL DYNAMICS 305 Newton’s work ongravitational andcentral forces totheattention oftheworld. After observing thecomet personally in1682, Halley became interested. Partly asaresult ofabetbetween Christopher Wren andRobert Hooke, Halley asked Newton in1684 what paths theplanets must follow iftheSun pulled them with aforce inversely proportional tothesquare oftheir distances. Totheastonish- ment ofHalley, Newton replied, “Why, inellipses, ofcourse.” Newton had worked itout20years previously buthad notpublished theresult. With painstaking effort, Halley wasable in1705 topredict thenext occurrence ofthe comet, now bearing hisname, tobein1758. 8.8 Orbital Dynamics The useofcentral-force motion isnowhere more useful, important, and inter- esting than inspace dynamics. Although space dynamics isactually quite com- plex because ofthegravitational attraction ofaspacecraft tovarious bodies and theorbital motion involved, weexamine tworather simple aspects: aproposed trip toMars and ﬂybys past comets and planets. Orbits arechanged bysingle ormultiple thrusts oftherocket engines. The simplest maneuver isasingle thrust applied intheorbital plane that does not change thedirection oftheangular momentum butdoes change theeccentric- ityand energy simultaneously. The most economical method ofinterplanetary transfer consists ofmoving from one circular heliocentric (Sun-oriented mo- tion) orbit toanother inthesame plane. Earth andMars represent such asystem reasonably well, and aHohmann transfer (Figure 8-10) represents thepath of minimum total energy expenditure.* Two engine burns arerequired: (1)the ﬁrst burn injects thespacecraft from thecircular Earth orbit toanelliptical transfer orbit that intersects Mars’ orbit; (2)thesecond burn transfers thespace- craft from theelliptical orbit into Mars’ orbit. Wecancalculate thevelocity changes needed foraHohmann transfer by calculating thevelocity ofaspacecraft moving intheorbit ofEarth around the Sun (r1inFigure 8-10) and thevelocity needed to“kick” itinto anelliptical transfer orbit that canreach Mars’ orbit. Weareconsidering only thegravita- tional attraction oftheSunandnotthatofEarth andMars. Forcircles andellipses wehave, from Equation 8.42, kE=——2a Foracircular path around theSun, thisbecomes =——=—m-0 —— . E '112'1 (850)271 2 1 T1 i_u *See Kaplan (Ka76, Chapter 3)fortheproof. Walter Hohmann, aGerman pioneer inspace travel research, proposed in1925 themost energy-efﬁcient method oftransferring between elliptical (planetary) orbits inthesame plane using only twovelocity changes. 8/CENTRALFORCE MOTION Mars atarrival "2 O ¢'I /I I I I‘Earth at arrival \ \\\ T1 *3 "1 Earth atdeparture'—ﬂI—‘-____ n"u v’*~vWN: QMars at departure nsfer foraround tripbetween Earth andMars. It FIGURE 8-10 The Hohmann tra represents theminimum energy expenditure. -T+UWesolve Equation 8.50 for'01: where wehave E— . k'01—, (8.51) Wedenote thesemimajor axisofthetransfer ellipse bya): 2a; =T1‘I’T2 Ifwecalculate theenergy attheperihelion forthetransfer ellipse, wehave —k 1 kE,=i =§mv?1 —I1 (8.52) T1 "I" T2 'stheperihelion transfer speed. The direction of-0,1isalong v1in iveswhere -0,11 Figure 8-10. Solving Equation 8.52 for11,1g 1),,=,/Z1£(l) (8.53)+r m'I’1 T1 2 The speed transfer A-01needed isjust A111 = UH — U1 Similarly, forthetransfer from theellipse tothecircular orbit ofradius r2, (8.55)wehave A112=1/2—1/:2 8.8ORBITAL DYNAMICS 307 Ik ‘U2= Kg (8.56) ‘Ur?:\I%l (E) + 2 'Ut2 Z 1I% L mrﬂ T1 + T2 The direction of-0,2isalong V2inFigure 8-10. The total speed increment canbe determined byadding thespeed changes, Av=Av,+A112. The total time required tomake thetransfer T,isahalf-period ofthetrans- ferorbit. From Equation 8.48, wehavewhere and (8.57) T T,=5’ T,="rr\[L;&a§‘/2 (8.58) Calculate thetime needed foraspacecraft tomake aHohmann transfer from Earth toMars and theheliocentric transfer speed required assuming both planets areincoplanar orbits. Solution. Weneed toinsert theappropriate constants inEquation 8.58. m m 1 k: GmMSun :“GMSun 1 (6.67 X10-11 II13/S2 -kg)(1.99 )<10311 kg) =7.53 ><10_21s2/m3 (8.59) Because k/moccurs sooften insolar system calculations, wewrite itaswell. k 2082 7”=1.33 ><10m/s >—lI\Q>—4at=_(TEarth—Sun *1’TMars —Sun) =2(1.50 ><1011m +2.28 ><1011 m) =1.89><1011m T,="rr(7.53 ><10“21s2/m3)1/2(1.89 ><1011m)3/2 =2.24 ><107s =259days (8.60) 308 8/CENTRALFORCE MOTION The heliocentric speed needed forthetransfer isgiven inEquation 8.53. _2(1.33 ><l02°m3/s2) (2.28 ><l011m) 1/2 "*1'(1.50X1011111) (8.78><l011m) =3.27 ><104m/s =32.7 km/s Wecancompare -onwith theorbital speed ofEarth (Equation 8.51). 1.33 ><102°m3/s2 1/2 I/1=ii"? =29-8 km/51.50 ><1011m Fortransfers totheouter planets, thespacecraft should belaunched inthe direction ofEarth’s orbit inorder togain Earth’s orbital velocity. Totransfer to theinner planets (e.g., toVenus), thespacecraft should belaunched opposite Earth’s motion. Ineach case, itistherelative velocity Avlthatisimportant tothe spacecraft (i.e., relative toEarth). Although theHohmann transfer path represents theleast energy expendi- ture, itdoes notrepresent theshortest time. Foraround tripfrom Earth to Mars, thespacecraft would have toremain onMars for460days until Earth and Mars were positioned correctly forthereturn trip (seeFigure 8-11a). The total trip (259 +460 +259 =978 days =2.7yr)would probably betoolong. Other schemes either usemore fuel togain speed (Figure 8-11b) orusetheslingshot effect ofﬂybys. Such aﬂyby mission pastVenus (seeFigure 8-11c) could bedone inlessthan 2years with only afewweeks near (oron) Mars. Several spacecraft inrecent years have escaped Earth’s gravitational attrac- tion toexplore oursolar system. Such interplanetary transfer canbedivided into three segments: (1)theescape from Earth, (2)aheliocentric transfer tothe area ofinterest, and (3)anencounter with another body—so far,either aplanet oracomet. The spacecraft fuelrequired forsuch missions canbeenormous, but aclever trick hasbeen designed to“steal” energy from other solar system bodies. Because themass ofaspacecraft issomuch smaller than theplanets (ortheir moons), theenergy lossoftheheavenly body isnegligible. Weexamine asimple version ofthis ﬂyby orslingshot effect that utilizes gravity assist. Aspacecraft coming from inﬁnity approaches abody (labeled B), interacts with B,andrecedes. The path isahyperbola (Figure 8-12). The initial and ﬁnal velocities, with respect toB,aredenoted by-0,7and -0},respectively. The neteffect onthespacecraft isadeﬂection angle of8with respect toB. Ifweexamine thesystem insome inertial frame inwhich themotion ofBoc- curs, thevelocities ofthespacecraft canbequite different because ofthemotion of B.The initial velocity viisshown inFigure 8-13a, andboth -0,andofareshown in Figure 8-13b. Notice that thespacecraft hasincreased itsspeed aswell as changed itsdirection. Anincrease invelocity occurs when thespacecraft passes behind B’sdirection ofmotion. Similarly, adecrease invelocity occurs when the spacecraft passes infront ofB’smotion. During the1970s, scientists atthe_]etPropulsion Laboratory oftheNational Aeronautics and Space Administration (NASA) realized that thefour largest planets ofoursolar system would beinafortuitous position toallow aspacecraft 8.8ORBITAL DYNAMICS 309 1.Earth departure 2.Mars arrival g1---- —~ 3.Mars departure I,-,7’ __“~‘ 3 4_Eartharrival ,-'''_7" \ /1 ~. »:~---- ~I \ 2 ‘x \\2 I’ \\ \\\ I’ \ Q \ \ \\ 1’1\ II II 01¢ \\\ ', (K) (b)Z4i—| O5O G) W1’ _____/v \ 6:4\!—l 1010 1.Earth dearture /'7“? P , _ 2.Mars arrival I ,’ \\ 3.Mars departure 4.Venus passage 5.Earth arrival 4-—-_ ’__\O \ \ *1/T’ I /' I Ir I/I/IIJ O01NJ O0\\-."Q (C) FIGURE 8-ll Round trips from Earth toMars. (a)Theminimum energy mission (Hohmann transfer) requires along stopover onMars before returning toEarth. (b)Ashorter mission toMars requires more fuelandacloser orbit totheSun. (c)Thefuelrequired fortheshorter mission of(b)can befurther improved ifVenus ispositioned foragravity assist during ﬂyby. toﬂypast them andmany oftheir 32known moons inasingle, relatively short “Grand Tour” mission using thegravity-assist method justdiscussed. This oppor- tunity oftheplanets’ alignment would notoccur again for175years. Because of budget constraints, there wasnottime todevelop thenew technology needed, andamission tolastonly 4years tovisitjustjupiter and Saturn wasapproved and planned. Nospecial equipment wasputonboard thetwin Voyager space- crafts foranencounter with Uranus and Neptune. Voyagers 1and 2were launched in1977 forvisits tojupiter in1979 andSatum in1980 (Voyager 1)and 1981 (Voyager 2).Because ofthesuccess ofthese visits toJupiter and Saturn, funding waslater approved toextend Voyager 2’smission toinclude Uranus and Neptune. The Voyagers arenow ontheir wayoutofoursolar system. Thepath ofVoyager 2isshown inFigure 8-14. Theslingshot effect ofgravity al- lowed thepath ofVoyager 2toberedirected, forexample, toward Uranus asit passed Satum bythemethod shown inFigure 8-12. The gravitational attraction from Saturn wasused topullthespacecraft offitsstraight path andredirect itata different angle. The effect oftheorbital motion ofSaturn allows anincrease inthe 310 s/CENTRALFORCE MOTION 1»; / 5 ‘ > Direction of inertial motion ofB \ I "1" 1 _ ‘tP1\Spacecraft 5?’»,131 FIGURE 8-12 Aspacecraft ﬂiesbyalarge body B(like aplanet) andgains speed when itﬂiesbehind B’sdirection ofmotion. Similarly, thespacecraft loses speed when itpasses infront ofB’sdirection ofmotion. The direction ofthespacecraft alsochanges. "1' Vii "B (a) i(b) FIGURE 8-13 Thevectors v§andv}aretheinitial andﬁnal velocities ofthespacecraft with respect toB.Thevectors viandvfarethevelocities inaninertial frame. (a)v,»=vB+vg.(b)vf= VB+v}. spacecraft’s speed. Itwasonly byusing thisgravity-assist technique thatthespectac- ular mission ofVoyager Zwas made possible inonly abrief 12-year period. Voyager 2 passed Uranus in1986 andNeptune in1989 before proceeding into interstellar space inoneofthemost successful space missions everundertaken. Most planetary missions now take advantage ofgravitational assists; forexample, theGalileo satel- 3.9APSIDAL ANGLES ANDPRECESSION (OPTIONAL) 311 1979 _.-—— Saturn Earth Z Umnus Neptune ___->_ ‘ i , 1986 1989 Voyager 2 FIGURE 8-14 Voyager 2was launched in1977 andpassed byjupiter, Saturn, Uranus, andNeptune. Gravitational assists were used inthemission. <:—;==§‘,- <" *..‘_ Comet Giacobini-Zinnert*1I»'¢'é"~*'~ Spacecraft ,----\_Moon,/ /I \ Moon orbit," /2% \\ \ \ ICESpacecraft I/ \ previous orbitN I _ \\Earth""- Ia\ _ _ \Q0‘ ‘ \ 1I\’d;>J". ;‘ \\ P '\=‘»'?'1 1 *---"I/'r,-Qga V SpacecraftToSun<4----_-- FIGURE 8-15 TheNASA spacecraft initially called [SEE-3 wasreprogrammed tobethe International Cometary Explorer andwassentonaspectacular three-year journey utilizing gravity assists onitswaybytheComet Giacobini-Zinner. lite,which photographed thespectacular collisions oftheShoemaker-Levy comet with_]upiter in1994 and reachedjupiter in1995, waslaunched in1989 butwent by Earth twice (1990 and1992) aswellasVenus (1990) togain speed andredirection. Aspectacular display ofﬂybys occurred intheyears 1982-1985 byaspace- craft initially called theInternational Sun-Earth Explorer 3(ISEE-3). Launched in1978, itsmission wastomonitor thesolar wind between theSun andEarth. For 4years, thespacecraft circled intheecliptical plane about 2million miles from Earth. In1982—because theUnited States had decided nottoparticipate inajoint European, japanese, and Soviet spacecraft investigation ofHalley’s comet in1986-—NASA decided toreprogram the ISEE-3, renamed itthe International Cometary Explorer (ICE), and sent itthrough theGiacobini-Zinner comet inSeptember 1985, some 6months before theﬂybys ofother spacecraft 312 s/CENTRAL-FORCE MOTION with Halley’s comet. The subsequent three-year journey ofICEwasspectacular (Figure 8-15). The path ofICEincluded twoclose trips toEarth and fiveﬂybys of themoon along itsbillion-mile triptothecomet. During oneﬂyby, thesatellite came within 75miles ofthelunar surface. The entire path could beplanned precisely because theforce lawisvery well known. The eventual interaction with thecomet, some 44million miles from Earth, included a20-minute tripthrough thecomet—about 5,000 miles behind thecomet’s nucleus. 8.9 Apsidal Angles andPrecession (Optional) Ifaparticle executes bounded, noncircular motion inacentral-force ﬁeld, then the radial distance from theforce center totheparticle must always beintherange rum 2r2rmin; that is,rmust bebounded bytheapsidal distances. Figure 8-5indi- cates thatonly twoapsidal distances exist forbounded, noncircular motion. But inexecuting onecomplete revolution in9,theparticle maynotretum toitsorig- inal position (see Figure 8-4). The angular separation between twosuccessive val- uesofr=rumdepends ontheexact nature oftheforce. The angle between any twoconsecutive apsides iscalled theapsidal angle, and because aclosed orbit must besymmetric about anyapsis, itfollows that allapsidal angles forsuch motion must beequal. The apsidal angle forelliptical motion, forexample, isjust 1r. Iftheorbit isnotclosed, theparticle reaches theapsidal distances atdifferent points ineach revolution; theapsidal angle isnotthen arational fraction of2"rr, asisrequired foraclosed orbit. Iftheorbit isalmost closed, theapsides precess, or rotate slowly intheplane ofthemotion. This effect isexactly analogous totheslow rotation oftheelliptical motion ofatwo-dimensional harmonic oscillator whose natural frequencies forthexand ymotions arealmost equal (seeSection 3.3). Because aninverse-square-law force requires that allelliptical orbits beex- actly closed, theapsides must stay ﬁxed inspace foralltime. Iftheapsides move with time, however slowly, thisindicates thattheforce lawunder which thebody moves does notvary exactly astheinverse square ofthedistance. This important factwasrealized byNewton, who pointed outthat anyadvance orregression ofa planet’s perihelion would require theradial dependence oftheforce lawtobe slightly different from 1/r2. Thus, Newton argued, theobservation ofthetime dependence oftheperihelia oftheplanets would beasensitive testofthevalid- ityoftheform oftheuniversal gravitation law. Inpoint offact, forplanetary motion within thesolar system, one expects that, because oftheperturbations introduced bytheexistence ofalltheother planets, theforce experienced byanyplanet does notvary exactly as1/r2, ifris measured from theSun. This effect issmall, however, andonly slight variations ofplanetary perihelia have been observed. The perihelion ofMercury, forexample, which shows thelargest effect, advances only about 574” ofarcpercentury.* Detailed calculations oftheinﬂuence oftheother planets onthemotion of *This precession isinaddition tothegeneral precession oftheequinox with respect tothe“ﬁxed” stars, which amounts to5025.645” i0.050” percentury. 8.9APSIDAL ANGLES ANDPRECESSION (OPTIONAL) 313 Mercury predict thattherateofadvance oftheperihelion should beapproximately 531” percentury. The uncertainties inthis calculation areconsiderably lessthan thedifference of43”between observation andcalculation,* andforaconsiderable time, thisdiscrepancy wastheoutstanding unresolved difﬁculty intheNewtonian theory. Wenowknow thatthemodiﬁcation introduced intotheequation ofmotion ofaplanet bythegeneral theory ofrelativity almost exactly accounts forthediffer- ence of43”. This result isone ofthemajor triumphs ofrelativity theory. Wenext indicate thewaytheadvance oftheperihelion canbecalculated from themodiﬁed equation ofmotion. Toperform thiscalculation, itisconven- ient tousetheequation ofmotion intheform ofEquation 8.20. Ifweusethe universal gravitational lawforF(r),wecanwrite dzu m1WP ‘Ll: -F;,F(1/U) Gm2M=? 8.61 Z, <> where weconsider themotion ofabody ofmass minthegravitational ﬁeld ofa body ofmass M.The quantity uistherefore thereciprocal ofthedistance be- tween mandM. The modiﬁcation ofthegravitational force lawrequired bythegeneral the- ory ofrelativity introduces into the force asmall component that varies as 1/r4( =u‘*).Thus, wehave dgu HGm2M 3GM 2 'ﬁ+u—T+Tu (8.62) where cisthevelocity ofpropagation ofthegravitational interaction and isiden- tiﬁed with thevelocity oflight.l Tosimplify thenotation, wedeﬁne _1_=Gm2M oz [2 3GM (ass) 5ETC *In1845, theFrench astronomer Urbain _]ean joseph LeVerrier (1811-1877) ﬁrstcalled attention to theirregularity inthemotion ofMercury. Similar studies byLeVerrier andbytheEnglish astronomer john Couch Adams ofirregularities inthemotion ofUranus ledtothediscovery oftheplanet Neptune in1846. Aninteresting account ofthisepisode isgiven byTurner (TuO4, Chapter 2).Wemust note, in thisregard, thatperturbations maybeeither periodic orsecular (i.e., everincreasing with time). Laplace showed in1773 (published, 1776) thatanyperturbation ofaplanet’s mean motion thatiscaused by theattraction ofanother planet must beperiodic, although theperiod maybeextremely long. This is thecase forMercury; theprecession of531” percentury isperiodic, buttheperiod issolong that the change from century tocentury issmall compared with theresidual effect of43”. TOne halfoftherelativistic term results from effects understandable interms ofspecial relativity, viz., time dilation (1/3) and therelativistic momentum effect (1/6);thevelocity isgreatest atperi- helion andleast ataphelion (seeChapter 14).The other halfoftheterm arises from general rela- tivistic effects andisassociated with theﬁnite propagation time ofgravitational interactions. Thus, theagreement between theory andexperiment conﬁrms theprediction thatthegravitational propa- gation velocity isthesame asthatforlight. 314 8/CENTRAL-FORCE MOTION andwecanwrite Equation 8.62 as dgu 1 2ﬁ+u==a+8u (8.64) This isanonlinear equation, andweuseasuccessive approximation procedure toobtain asolution. Wechoose theﬁrst solution tobethesolution ofEquation 8.64 inthecase thattheterm 81¢?isneglected*: 1 ulZZ(1+acos 6) (8.65) This isthefamiliar result forthepure inverse-square-law force (seeEquation 8.41). Note that orishere thesame asthat deﬁned inEquation 8.40 except that /,4.has been replaced bym.Ifwesubstitute thisexpression into theright-hand side of Equation 8.64, weﬁnd dzu 1 8ﬁ+u=a+;[1+28COS9+t-:2C0S29] 1 5 £2=—+—2 1+2scos6+—(1+ cos26) (8.66)oz oz 2 where cos? 6hasbeen expanded interms ofcos26.The ﬁrst trial function ul, when substituted into theleft-hand side ofEquation 8.64, reproduces only the ﬁrstterm ontheright-hand side: 1/oz. Wecantherefore construct asecond trial function byadding toulaterm that reproduces theremainder oftheright-hand side (inEquation 8.66). Wecanverify thatsuch aparticular integral is 8 s2 _ £2u,,=¥[<1+E—)+ t-:9s1n9-€COS 26] (8.67) The second trial function istherefore 112=ul+u1, Ifwestop theapproximation procedure atthispoint, wehave uEug=ul+up =[l(1+ scos 0)+Z26sin9](1 G 8 .92 8.92+ + -‘W COS where wehave regrouped theterms inuland u1,. *Weeliminate thenecessity ofintroducing anarbitrary phase intotheargument ofthecosine term bychoosing tomeasure 0from theposition ofperihelion; i.e.,ulisamaximum (and hence rlisa minimum) atH=0. 8.9APSIDAL ANGLES ANDPRECESSION (OPTIONAL) 315 Consider theterms inthesecond setofbrackets inEquation 8.68: theﬁrst ofthese isjust aconstant, and thesecond isonly asmall and periodic distur- bance ofthenormal Keplerian motion. Therefore, onalong time scale neither ofthese terms contributes, ontheaverage, toanychange inthepositions ofthe apsides. Butintheﬁrst setofbrackets, theterm proportional to9produces sec- ular andtherefore observable effects. Letusconsider theﬁrstsetofbrackets: 1 86 _ umula, =-1+scos6+—6S11'l6 (8.69)a or Next, wecanexpand thequantity 8 8 _ _81+ecos 9--*6 =1+s cos9cos—6+s1n6s1n—~6oz or a 8s _E1+scos0+;0s1n6 (8.70) where wehave used thefactthat8issmall toapproximate 8 _8 8cos-651, sin-BE-9or oz or Hence, wecanwrite usecula, as 1 8 'usecular E;[1 +scos(6—56)] (8.71) Wehave chosen tomeasure 0from theposition ofperihelion att=0. Successive appearances atperihelion result when theargument ofthecosine term inusemlar increases to21r,41r,...,andsoforth. Butanincrease oftheargu- ment by2'rrrequires that 80--0=21ra or _2" Z § 6—~1__(8/a)—21'r(1+a) Therefore, theeffect oftherelativistic term intheforce lawistodisplace the perihelion ineach revolution byanamount aAE2i (8.72a)oz thatis,theapsides rotate slowly inspace. Ifwerefer tothedeﬁnitions ofozand8 (Equations 8.63), weﬁnd 2 AE6-” (8.72b) 316 8/CENTRAL-FORCE MOTION TABLE 8-2 Precessional Rate forthePerihelia ofSome Planets S Precessional rate(seconds ofarc/century) Planet Calculated Observed Mercury 43.03 1'0.03 43.11 i"0.45 Venus 8.63 8.41'24.8 Earth 3.84 5.01'1.2 Mars 1.35 — jupiter 0.06 — From Equations 8.40 and 8.42, wecanwrite l2=].Lk(l(1 —t-:2):then, because k= GmM anditEm,wehave 67rGMA=; (8.726) Weseetherefore thattheeffect isenhanced ifthesemimajor axisaissmall and iftheeccentricity islarge. Mercury, which isthe planet nearest thesun and which hasthemost eccentric orbit ofanyplanet (except Pluto), provides the most sensitive testofthetheory.* The calculated value oftheprecessional rate forMercury is43.03” i0.03” ofarcpercentury. The observed value (corrected fortheinﬂuence oftheother planets) is43.11" i0.45",i sotheprediction of relativity theory isconﬁrmed instriking fashion. The precessional rates forsome oftheplanets aregiven inTable 8-2. 8.10 Stability ofCircular Orbits (Optional) InSection 8.6,wepointed outthat theorbit iscircular ifthetotal energy equals theminimum value oftheeffective potential energy, E=Vmin. More generally, however, acircular orbit isallowed foranyattractive potential, because theattrac- tiveforce can always bemade tojust balance thecentrifugal force bytheproper choice ofradial velocity. Although circular orbits aretherefore always possible in acentral, attractive force ﬁeld, such orbits arenotnecessarily stable. Acircular orbit atr==pexists if1‘,=p=0forallt;thisispossible if(6V/61") |,=p =0.Butonly iftheeffective potential hasatrueminimum does stability result. Allother equilib- rium circular orbits areunstable. Letusconsider anattractive central force with theform Fa)=-7’: (8.73) *Altematively, wecansaythattherelativistic advance oftheperihelion isamaximum forMercury because theorbital velocity isgreatest forMercury and therelativistic parameter -u/clargest. TR.L.Duncombe, Astr0n.]. 61,174(l956); seealsoG.M.Clemence, Rev.Mod. Phys. 19,361(1947). 8.10 STABILITY OFCIRCULAR ORBITS (OPTIONAL) 317 The potential forsuch aforce is k 1U(r)=—n_1TWU (8.74) andtheeffective potential function is k l2 V(r) =-* T(n_1) +% (8.75) The conditions foraminimum ofV(r)andhence forastable circular orbit with aradius pare av agv _ Z d M > _ at,=,,0an aﬂ,=,, 0 (876) Applying these criteria totheeffective potential ofEquation 8.75, wehave 6 k l2 6r.=,, Pup OI‘ p I2 ( and _ a2v nk 812a— =——+—4>0Y2.=,. Pm" I-LP so nk 3l2 -g +I>0 (8.78) Substituting p(”_5)from Equation 8.77 into Equation 8.78, wehave [2 (3-72);>0 (8.79) The condition thatastable circular orbit exists isthus n<3. Next, weapply amore general procedure andinquire about thefrequency ofoscillation about acircular orbit inageneral force ﬁeld. Wewrite theforce as 6U F(Y)='“I-L80) Zr5 (8-30) Equation 8.18 cannow bewritten as r-192=—-g(r) (8.81) Substituting for from Equation 8.10, 2 t‘-Iﬁ=-go) (8.82) 318 8/CENTRAL-FORCE MOTION Wenow consider theparticle tobeinitially inacircular orbit with radius pand apply aperturbation oftheform r—>p +x,where xissmall. Because pIcon- stant, wealso have ii—> Thus .. I2 "" ZW*8 <8-*3’Butbyhypothesis (x/p) <<1,sowecanexpand thequantity: [1+(X/P)l_3 =1r3(X/P) + (3-34) Wealso assume thatg(r) =g(p+x)canbeexpanded inaTaylor series about thepoint r=p: g(r>+X)=8(0)+Xg'(P) + (8-85) where ._dgg(p)—dTr=p Ifweneglect allterms inx2and higher powers, then thesubstitution of Equations 8.84 and8.85 into Equation 8.83yields 2 se-[ﬁn—so/p>1 E-[gm+xg'1 <8-86> Recall thatweassumed theparticle tobeinitially inacircular orbit with r=p. Under such acondition, noradial motion occurs—that is,r|.=,, =0.Then, also, i’|,=p ==0.Therefore, evaluating Equation 8.82 atr=p,wehave [2 )=-i 8.87) 8(1) #2’), ( Substituting thisrelation into Equation 8.86, wehave, approximately, 55“8(P)[1— 3(x/P)l E“[g(P) +x8’(P)l or 55+ +g'(p):|x E0 (8.88) Ifwedeﬁne 3 wtE$+g' <8-89> then Equation 8.88 becomes thefamiliar equation fortheundamped harmonic oscillator: 55+w§x =0 (8.90) The solution tothisequation is x(t)=Ae"“"”">‘ +Be-W (8.91) 8.10STABILITY orCIRCULAR ORBITS (OPTIONAL) 319 If0,3<0,sothat cooisimaginary, then thesecond term becomes Bexp(|w0| t), which clearly increases without limit astime increases. The condition foroscilla- tion istherefore co?)>0,or y +g'(p)>0 (8.92a) Because g(p) >0(see Equation 8.87), wecandivide through byg(p) and write thisinequality as 3-él@+->0 (8.92b)8(9) P or,because g(r) and F(r)arerelated byaconstant multiplicative factor, stability results if F'(p) 3-—-~ +—>0 8.93 ( ) Wenow compare thecondition ontheforce lawimposed byEquation 8.93 with thatpreviously obtained forapower-law force: F(r)=-2 (8.94) Equation 8.93 becomes —(n+1) ﬁ£fT+§>0—kP P or 1 (3—~n) >0 (8.95) andweareledtothesame condition asbefore—that is,n<3.(Wemust note, however, thatthecase n=3needs further examination; seeProblem 8-22.) .Investigate thestability ofcircular orbits inaforce ﬁeld described bythe potential function‘ -kU(r) =-7-e‘('/“l (8.96) where k>0anda>0. Solution. This potential iscalled thescreened Coulomb potential (when k=“—Ze2/4m;0, where Zistheatomic number andeistheelectron charge) 320 8/CENTRAL-FORCE MOTION because itfallsoffwith distance more rapidly than 1/randhence approximates theelectrostatic potential oftheatomic nucleus inthevicinity ofthenucleus by taking intoaccount thepartial “cancellation” or“screening” ofthenuclear charge bytheatomic electrons. The force isfound from 6U 1 1F ::—---—- ::-— -—— -— "(T/1“)(r) 67 k(M+T2)e and 6F 1 2 2_=k J _ _ -(1/0) at (a2r+ av"?+rsle The condition forstability (seeEquation 8.93) is Fl (P)>0 () “PITTherefore 1 1 '"k m-|--5 ap p a2+ap——p2>01 Pkin*ale* 3+ >0 which simpliﬁes to Wemaywrite thisas a23+9»1>0P P Stability thus results forallqEa/pthatexceed thevalue satisfying theequation f+q-1=0 The positive (and therefore theonly physically meaningful) solution is 1q=;y€-1)Eo@ If,then, theangular momentum andenergy allow acircular orbit atr=p,the motion isstable if 920%p or psrwa ow) The stability condition fororbits inascreened potential isillustrated graph- ically inFigure 8-16, which shows thepotential V(r)forvarious values ofp/a. The force constant kisthesame forallthecurves, butl2/2/.t hasbeen adjusted 8.10 STABILITY OFCIRCULAR ORBITS (OPTIONAL) I/(1') Curve p/a.l2/2'“ 130.09 1.98-v(, 2 0.40 1.821/0 09 0.61 1."/svo >P 1.50)/03 j g 0.92 6 5 1.62-EV0 F6 8.40 0.59v0 5 4-}-3 \/_1 T 1 FIGURE 8-16 Example 8.7.Potentials 1-4produce astable, circular orbit forvalues of p/aS1.62. tomaintain theminimum ofthepotential atthesame value oftheradius asais changed. Forp/a<1.62, atrue minimum exists forthepotential, indicating that thecircular orbit isstable with respect tosmall oscillations. Forp/a>1.62, there isnominimum, socircular orbits cannot exist. Forp/a=1.62, thepoten- tialhaszero slope attheposition thatacircular orbit would occupy. The orbit is unstable atthisposition, because 02%iszero inEquation 8.90 and thedisplace- ment xincreases linearly with time. Aninteresting feature ofthispotential function isthat under certain condi- tions there canexist bound orbits forwhich thetotal energy ispositive (see, for example, curve 4inFigure 8-16).I I I — ‘I I | | EXAMPLE 8.7 L‘F Determine whether aparticle moving ontheinside surface ofacone under the inﬂuence ofgravity (see Example 7.4) canhave astable circular orbit. 322 8/CENTRAL-FORCE MOTION Solution. InExample 7.4,wefound thattheangular momentum about the z-axis wasaconstant ofthemotion: l==mr20 =constant Wealso found theequation ofmotion forthecoordinate r: it‘—r02sin2a +gsinozcosa=0 (8.98) Iftheinitial conditions areappropriately selected, theparticle canmove in acircular orbit about thevertical axiswith theplane oftheorbit ataconstant height zoabove thehorizontal plane passing through theapex ofthecone. Although thisproblem does notinvolve acentral force, certain aspects ofthe motion arethesame asforthecentral-force case. Thus wemay discuss, forex- ample, thestability ofcircular orbits fortheparticle. Todothis, weperform a perturbation calculation. First, weassume thatacircular orbit exists forr==p.Then, weapply the perturbation r—>p+x.The quantity r02inEquation 8.98 canbeexpressed as _ [2 l2 1-62 : 7-0-4- :: 4 "L274 m2T3 l2 [2 x-3 :--- -32.-4-1 -—m2(p+x) mgpgﬁ +p) l2 x 5% 1"?’-mp P where wehave retained only theﬁrst term intheexpansion, because x/pisby hypothesis asmall quantity. Then, because ii==0,Equation 8.98 becomes, approximately, I2sin?oz xE—W(1 -3;)+gsinozcosoz==0 55+ ——T-x—--7-+gsinacosa-r-"0 (8.99)or (312sin2oz) l2sina "lip mP Ifwe evaluate Equation 8.98 atr=p,then F=0,and wehave gsinozcosoz=p02sin2oz 12_2==——'ﬁS1nO! mp Inview ofthisresult, thelasttwoterms inEquation 8.99cancel, andthere remains 2'2 at+ =0 (8.100)P PROBLEMS 323 The solution tothisequation isjustaharmonic oscillation with afrequency to, where \/51co=Lmp2sinoz (8.101) Thus, thecircular orbit isstable. PROBLEMS 8-1 8-2 8-3 8-4 8-5 8-6 8-7. 8-8Insection 8.2,weshowed thatthemotion oftwobodies interacting only with each other bycentral forces could bereduced toanequivalent one-body problem. Show byexplicit calculation that such areduction isalso possible forbodies moving inan external uniform gravitational ﬁeld. Perform theintegration ofEquation 8.38 toobtain Equation 8.39. Aparticle moves inacircular orbit inaforce ﬁeld given by F(r)=—k/r2 Show that, ifksuddenly decreases tohalf itsoriginal value, theparticlefs orbit be- comes parabolic. Perform anexplicit calculation ofthetime average (i.e., theaverage over one com- plete period) ofthepotential energy foraparticle moving inanelliptical orbit ina central inverse-square-law force ﬁeld. Express theresult interms oftheforce constant oftheﬁeld and thesemimajor axis oftheellipse. Perform asimilar calculation forthe kinetic energy. Compare theresults andthereby verify thevirial theorem forthiscase. Two particles moving under theinﬂuence oftheir mutual gravitational force de- scribe circular orbits about one another with aperiod 1'.Ifthey are suddenly stopped intheir orbits andallowed togravitate toward each other, show thatthey willcollide afteratime7/4\/§. Two gravitating masses mland m2(m1 +mg=NI)areseparated byadistance r0and released from rest. Show that when theseparation isr(<r0),thespeeds are 2G1 1 2G1 1 "1="'2nrzg’ "1="*1ET7,, Show thattheareal velocity isconstant foraparticle moving under theinﬂuence of anattractive force given byF(r) =—-kr. Calculate thetime averages ofthekinetic and potential energies and compare with theresults ofthevirial theorem. Investigate themotion ofaparticle repelled byaforce center according tothelaw F(r) =kr.Show thattheorbit canonly behyperbolic. 324 8-9. 8-10 8'1lo 8-12. 8-13. 8-14 8-15 8.16. 8-178/CENTRAL-FORCE MOTION Acommunications satellite isinacircular orbit around Earth atradius Randveloc- ity~u.Arocket accidentally ﬁres quite suddenly, giving therocket anoutward radial velocity vinaddition toitsoriginal velocity. (a)Calculate theratio ofthenewenergy andangular momentum totheold. (b)Describe thesubsequent motion ofthesatellite andplot T(r), V(r), U(r), and E(r) after therocket ﬁres. Assume Earth’s orbit tobecircular andthat theSun’s mass suddenly decreases by half. What orbit does Earth then have? Will Earth escape thesolar system? Aparticle moves under theinﬂuence ofacentral force given byF(r) =—k/r". If theparticle’s orbit iscircular and passes through theforce center, show that n=5. Consider acomet moving inaparabolic orbit intheplane ofEarth’s orbit. Ifthe distance ofclosest approach ofthecomet totheSun is[3rE, where rEistheradius of Earth’s (assumed) circular orbit and where B<1,show that thetime thecomet spends within theorbit ofEarth isgiven by \/2(1 —B)-(1+2B)/317 Xlyear Ifthecomet approaches theSuntothedistance oftheperihelion ofMercury, how many days isitwithin Earth’s orbit? Discuss themotion ofaparticle inacentral inverse-square-law force ﬁeld forasu- perimposed force whose magnitude isinversely proportional tothecube ofthedis- tance from theparticle totheforce center; thatis, k )1 F(r)=—F—F k,)t>0 Show that the motion isdescribed byaprecessing ellipse. Consider the cases )1<l2/,u., A=l2/,u., and)t >l2/,u.. Find theforce lawforacentral-force ﬁeld thatallows aparticle tomove inaspiral orbit given byr=k62,where kisaconstant. Aparticle ofunit mass moves from inﬁnity along astraight line that, ifcontinued, would allow ittopass adistance b\/2 from apoint P.Iftheparticle isattracted to- ward Pwith aforce varying ask/r5, andiftheangular momentum about thepoint Pis\/E/b, show that thetrajectory isgiven by r=bcoth(0/\/2) Aparticle executes elliptical (but almost circular) motion about aforce center. At some point intheorbit atangential impulse isapplied totheparticle, changing the velocity from vto-u+8v.Show thattheresulting relative change inthemajor and minor axes oftheorbit istwice therelative change inthevelocity andthattheaxes areincreased if8v>0. Aparticle moves inanelliptical orbit inaninverse-square-law central-force ﬁeld. If theratio ofthemaximum angular velocity totheminimum angular velocity ofthe PROBLEMS 325 particle initsorbit isn,then show thattheeccentricity oftheorbit is szx/it-1 \/0+1 8-18. UseKepler’s results (i.e., hisﬁrst andsecond laws) toshow that thegravitational force must becentral andthattheradial dependence must be1/r2. Thus, perform aninductive derivation ofthegravitational force law. 8-19. Calculate themissing entries denoted bycinTable 8-1. 8-20. Foraparticle moving inanelliptical orbit with semimajor axisaandeccentricity s, show that ((a/r)4 cos0) =s/(1 —.<-:2)?’/2 where theangular brackets denote atime average over one complete period. 8-21. Consider thefamily oforbits inacentral potential forwhich thetotal energy isa constant. Show thatifastable circular orbit exists, theangular momentum associ- ated with thisorbit islarger than thatforanyother orbit ofthefamily. 8-22. Discuss themotion ofaparticle moving inanattractive central-force ﬁeld de- scribed byF(r) =—k/132* Sketch some oftheorbits fordifferent values ofthetotal energy. Can acircular orbit bestable insuch aforce ﬁeld? 8-23. AnEarth satellite moves inanelliptical orbit with aperiod 7,eccentricity s,and semimajor axis a.Show that themaximum radial velocity ofthesatellite is 2'rras/(r V1—s2). 8-24. AnEarth satellite hasaperigee of300kmandanapogee of3,500 kmabove Earth’s surface. How faristhesatellite above Earth when (a)ithas rotated 90°around Earth from perigee and(b)ithasmoved halfway from perigee toapogee? 8-25. AnEarth satellite hasaspeed of28,070 km/hr when itisatitsperigee of220km above Earth’s surface. Find theapogee distance, itsspeed atapogee, anditsperiod ofrevolution. 8-26. Show thatthemost efficient waytochange theenergy ofanelliptical orbit forasin- gleshort engine thrust isbyﬁring the rocket along the direction oftravel at perigee. 8-27. Aspacecraft inanorbit about Earth hasthespeed of10,160 m/s ataperigee of 6,680 kmfrom Earth’s center. \'Vhat speed does thespacecraft have atapogee of 42,200 km? 8-28. VVhat istheminimum escape velocity ofaspacecraft from themoon? *This particular force lawwasextensively investigated byRoger Cotes (1682-1716), andtheorbits areknown asCotes’ spirals. 326 8-29. 8-30. 8-31. 8-32. 8-33. 8-34 8-35. 8-36. 8-378/CENTRAL-FORCE MOTION The minimum and maximum velocities ofamoon rotating around Uranus are vmm =v—v0and um,‘ =v+v0.Find theeccentricity interms ofvand v0. Aspacecraft isplaced inorbit 200kmabove Earth inacircular orbit. Calculate the minimum escape speed from Earth. Sketch theescape trajectory, showing Earth andthecircular orbit. What isthespacecraft’s trajectory with respect toEarth? Consider aforce lawoftheform kk’ FmZ'5_F Show that ifp2k>k’,then aparticle canmove inastable circular orbit atr=p. Consider aforce lawoftheform F(r) =—(k/r2) exp(—r/a). Investigate thestability ofcircular orbits inthisforce ﬁeld. Consider aparticle ofmass mconstrained tomove onthesurface ofaparaboloid whose equation (incylindrical coordinates) isr2=4az.Iftheparticle issubject toa gravitational force, show that thefrequency ofsmall oscillations about acircular orbit with radius p=V4az0 is w=,/-“La+z0 Consider theproblem oftheparticle moving onthesurface ofacone, asdiscussed inExamples 7.4and 8.7.Show that theeﬁective potential iS [2 V(r)=% +mgrcota (Note thathere ristheradial distance incylindrical coordinates, notspherical co- ordinates; seeFigure 7-2.) Show that theturning points ofthemotion canbefound from thesolution ofacubic equation inr.Show further thatonly twooftheroots arephysically meaningful, sothatthemotion isconﬁned toliewithin twohorizon- talplanes thatcutthecone. Analmost circular orbit (i.e., s<<1)canbeconsidered tobeacircular orbit to which asmall perturbation hasbeen applied. Then, thefrequency oftheradial mo- tion isgiven byEquation 8.89. Consider acase inwhich theforce law is F(r)=—k/1"(where nisaninteger), andshow thattheapsidal angle is1r/V3~n. Thus, show that aclosed orbit generally results only fortheharmonic oscillator force andtheinverse-square-law force (ifvalues ofnequal toorsmaller than -6 areexcluded). Aparticle moves inanalmost circular orbit inaforce ﬁeld described by Hr) =—(k/12)exp(—r/a). Show that theapsides advance byanamount approxi- mately equal to11'p/aineach revolution, where pistheradius ofthecircular orbit andwhere p<<a. Acommunication satellite isinacircular orbit around Earth atadistance above Earth equal toEarth’s radius. Find theminimum velocity Avrequired todouble the height ofthesatellite andputitinanother circular orbit. PROBLEMS 327 8-38. Calculate theminimum Avrequired toplace asatellite already inEarth’s heliocen- tricorbit (assumed circular) into theorbit ofVenus (also assumed circular and coplanar with Earth). Consider only thegravitational attraction oftheSun. What time ofﬂight would such atriptake? 8-39. Assuming arocket engine canbeﬁred only once from alowEarth orbit, does a Mars ﬂyby oraVenus ﬂyby require alarger Av?Explain. 8-40. Aspacecraft isbeing designed todispose ofnuclear waste either bycarrying itout ofthesolar system orcrashing into theSun. Assume that noplanetary ﬂybys are permitted and that thrusts occur only intheorbital plane. \'Vhich mission requires theleast energy? Explain. 8-41. Aspacecraft isparked inacircular orbit 200kmabove Earth’s surface. Wewant to useaHohmann transfer tosend thespacecraft totheMoon's orbit. \IVhat arethe total Avand thetransfer time required? 8-42. Aspacecraft ofmass 10,000 kgisparked inacircular orbit 200 kmabove Earth’s surface. What istheminimum energy required (neglect thefuelmass burned) to place thesatellite inasynchronous orbit (i.e., 1'=24hr)? 8-43. Asatellite ismoving incircular orbit ofradius Rabout Earth. Bywhat fraction must itsvelocity vbeincreased forthesatellite tobeinanelliptical orbit with rm,“=R andrm“I2R? 8-44. TheYukawa potential adds anexponential term tothelong-range Coulomb poten- tial,which greatly shortens therange oftheCoulomb potential. Ithasgreat useful- ness inatomic and nuclear calculations. V1" kI/(T) =ﬂeﬁ/M» =1--err/a T T Find aparticle’s trajectory inabound orbit oftheYukawa potential toﬁrstorder in 1/a. 8-45. Aparticle ofmass mmoves inacentral force ﬁeld thathasaconstant magnitude E), butalways points toward theorigin. (a)Find theangular velocity m4,required for theparticle tomove inacircular orbit ofradius r0.(b)Find thefrequency w,of small radial oscillations about thecircular orbit. Both answers should beinterms of F0,m,andT0. 8-46. Two double stars ofthesame mass asthesunrotate about their common center of mass. Their separation is4light years. What istheir period ofrevolution? 8-47. Two double stars, one having mass 1.0M5,", and theother 3.0Mm, rotate about their common center ofmass. Their separation is6light years. \'Vhat istheir period ofrevolution? L CHAPTER g Dynamics ofaSystem ofParticles 9.1Introduction Thus far,wehave treated our dynamical problems primarily interms ofsingle particles. Even though wehave considered extended objects such asprojectiles andplanets, wehave been able totreat them assingle particles. Generally, we have nothadtodeal with theinternal interactions between themany particles thatmake uptheextended body. Later, when wetreat thedynamics ofrigid bodies, wemust describe rota- tional aswell astranslational motion. Weneed toprepare thetechniques that willallow ustodothis. Weﬁrst extend our discussion todescribe thesystem ofnparticles. These particles mayform aloose aggregate—such asapileofrocks oravolume ofgas molecules—or form arigid body inwhich theconstituent particles arere- strained from moving relative toone another. Wedevote thelatter part ofthe chapter toastudy oftheinteraction oftwoparticles (n=2).Forthethree-body problem (n=3),thesolutions become formidable. Perturbation techniques often areused, although great progress hasbeen made through theuseofnu- merical methods with high-speed computers. Finally, weshall examine rocket motion. Newton’s Third Law plays aprominent role inthedynamics ofasystem of particles because oftheinternal forces between theparticles inthesystem. We need tomake twoassumptions concerning theinternal forces: 1.The forces exerted bytwoparticles aandBoneach other areequal inmag- nitude and opposite indirection. Let fa’;represent theforce ontheath 328 9.2CENTER orMASS 329 ao——>- -------qioptag tﬁa FIGURE 9-1 Example ofthestrong form ofNewton’s Third Law, where theequal and opposite forces between twoparticles must liealong astraight line joining thetwoparticles. The force isattractive, asinthemolecular attraction inasolid. particle due totheBth particle. The so-called “weak” form ofNewton’s Third Lawis fag :"'fBa 2.The forces exerted bytwoparticles orandBoneach other, inaddition to being equal and opposite, must lieonthestraight linejoining thetwoparti- cles. This more restrictive form ofNewton’s Third Law, often called the -“strong” form, isdisplayed inFigure 9-1. Wemust becareful toremember when each form ofNewton’s Third Law ap- plies. Werecall from Section 2.2that theThird Law isnotalways valid formov- ingcharged particles; electromagnetic forces arevelocity dependent. Forexample, magnetic forces, those forces exerted onamoving charge qinamagnetic ﬁeld B(F=qvXB),obey theweak form, butnotthestrong form, oftheThird Law. 9.2 Center ofMass Consider asystem composed ofnparticles, with each particle’s mass described byma,where aisanindex from oz=1toa=n.The total mass ofthesystem is denoted byM, M-§m. <9-2)where thesummation over oz(asinallsummations carried outover Greek in- dices) runs from ozI-"1toct=n.Such asystem isdisplayed inFigure 9-2. Ifthevector connecting theorigin with theathparticle isra,then thevector deﬁning theposition ofthesystem’s center ofmass is 1 R= mar“ (9.3) 330 9/DYNAMICS OFASYSTEM orPARTICLES .FIGURE 9-2 Theposition vectors toparticles 1,2,and3inthebody areindicated, along with thecenter ofmass position vector R. Foracontinuous distribution ofmass, thesummation isreplaced byanintegral, 1R=IIjrdm (9.4) The location ofthecenter ofmass ofabody isuniquely deﬁned, buttheposition vector Rdepends onthecoordinate system chosen. Iftheorigin inFigure 9-2 were chosen elsewhere, thevector Rwould bedifferent. EXAMPLE 9.1 T T -_ ——. I. ___ 1-T Find thecenter ofmass ofasolid hemisphere ofconstant density. Solution. Letthedensity bep,thehemispherical mass beM,andtheradius bea. M P:W“2—as3'rr Wewant tochoose theorigin ofourcoordinate system carefully (Figure 9-3) tomake theproblem assimple aspossible. The position coordinates ofRare (X,Y,Z).From symmetry, X==O,Z=O.This should beobvious from Equation 9.4, 1 (1 X=Xjjwx dm 1 G Z=Mjwz dm 9.3LINEAR MOMENTUM OFTHESYSTEM 331 y 9.102_)2 "litZ Z (a) (b) FIGURE 9-3 Example 9.1.(a)Wechoose athin slice dyofasolid hemisphere of constant density toﬁnd thecenter ofmass position value Y. (b)The area oftheslice dyiscircular. because weareintegrating over anoddpower ofavariable with symmetric limits. For Y,however, thelimits areasymmetric. . 1'1 Y==— d Min”"’ Construct dmsoitisplaced ataconstant value ofy.Acircular slice perpendicu- lartothey-axis sufﬁces (seeFigure 9-3). dm=pdV= p1r(a2 -y2)dy 1 G Y: I/ILp"rry(a2 -y2)dy Y__1'rpa4 __@ 4M 8 The position ofthecenter ofmass is(O,3a/8,0). 9.3 Linear Momentum oftheSystem Ifacertain group ofparticles constitutes asystem, then theresultant force acting onaparticle within thesystem (say, theathparticle) isingeneral composed of twoparts. One part istheresultant ofallforces whose origin liesoutside ofthe system; thisiscalled theexternal force, Fff). The other part istheresultant of theforces arising from theinteraction ofalloftheother n—-1particles with the athparticle; thisiscalled theinternal force, fa.Force faisgiven bythevector sum ofalltheindividual forces fag, fa=gr“, (9.5) 332 9/DYNAMICS OFASYSTEM OFPARTICLES where fa);represents theforce ontheathparticle duetotheBthparticle. The total force acting ontheathparticle istherefore Fa=Fff)+fa (9.6) Also, according totheweak statement ofNewton’s Third Law, wehave fag=~—fB,,, (9.1) Newton’s Second Law fortheathparticle canbewritten as 16>.=ma.=Fr+f. <9-1) or d2 ;l;<m.r.> -F2?+§f.,. <9-8) Summing thisexpression over oz,wehave 2 iimara=EF$;>+EEL, (9.9) ‘#2“ “ 12.2”where theterms ozIBdonotenter inthesecond sum ontheright-hand side, because fadEO.The summation ontheleft-hand side just yields MR (see Equation 9.3), andthesecond time derivative is The ﬁrstterm onthetight- hand side isthesum ofalltheexternal forces and canbewritten as _ EFg;>EF (9.10) The second term ontheright-hand side inEquation 9.9canbeexpressed* as E a, afaB ———: + 01¢B which vanishesl according toEquation 9.1.Thus, wehave theﬁrst important result MR=F (9.11) *This equation canbeveriﬁed byexplicitly calculating both sides forasingle case (e.g., n==3). 1'The lastsummation symbol means “sum over allaandBsubject totherestrictions a<B.”Note thatwecanprove thevanishing of F§f-Br1¢B byappealing tothefollowing argument. Because thesummations arecarried outover both ozandB, these indices aredummies; inparticular, wemay interchange aandBwithout affecting thesum. Using themore compact notation, wehave Er=Za,B¢cr up B,|1¢BfBa But,byhypothesis, fa),=—fB,,, so Zr=-Ea,B¢a “B a,B$|::fnB andifaquantity isequal toitsnegative, itmust vanish identically. 9.3LINEAR MOMENTUM OFTHESYSTEM 333 which wecanexpress asfollows: I. Thecenter ofmass ofasystem moves asitwereasingle particle ofmass equal tothe total mass ofthesystem, acted onbythetotal external force, andindependent of thenature oftheinternal forces (aslong astheyfollow fag=——fﬂa, theweak form of Newton’s Third Law). The total linear momentum ofthesystem is , d d .P=Z%%=EE%%=wMm=MR mm) and P:MR=F am) Thus, thetotal linear momentum ofthesystem isconserved ifthere isnoexter- nalforce. From Equations 9.12 and9.13, wenote oursecond andthird impor- tant results: II. Thelinear momentum ofthesystem isthesame as asingle particle ofmass Mwere located attheposition ofthecenter ofmass and moving inthemanner thecenter of mass moves. III. Thetotallinear momentum forasystem freeofexternal forces isconstant andequal to thelinear momentum ofthecenter ofmass (the lawofconservation oflinear mo- mentum forasystem). Allmeasurements must bemade inaninertial reference system. Anexam- pleofthelinear momentum ofasystem isgiven bytheexplosion ofanartillery shell above ground. Because theexplosion isaninternal effect, theonly external force affecting thecenter ofmass velocity isduetogravity. The center ofmass of theartillery shell fragments immediately after theexplosion must continue with thevelocity oftheshell just before theexplosion. EXAMPLE 9.2 — — _- Achain ofuniform linear mass density p,length b,andmass M(p=M/b) hangs asshown inFigure 9-4.Attime t=0,theends AandBareadjacent, butendB isreleased. Find thetension inthechain atpoint Aafter endBhasfallen adis- tance xby(a)assuming freefalland (b)byusing energy conservation. Solution. (a)Inthecase offree fall, let’s assume theonly forces acting onthe system attime tare thetension Tacting vertically upward atpoint Aandthe gravitational force Mgpulling thechain down. The center ofmass momentum reacts tothese forces such that _ i~A@—T mm) The right side ofthechain, with mass p(b—x)/2,ismoving atthespeed ic,and theleftside ofthechain isnotmoving. The total momentum ofthesystem is 334 9/DYNAMICS orASYSTEM orPARTICLES AB A Tj .BCM CM t=0 time t>0 (a) (b) FIGURE 9-4 Example 9.2. (a)Achain ofuniform linear mass density hangs atpoints Aand Bbefore Bisreleased attime t=O.(b)Attime tthe end Bhas fallen adistance x. b~.. P=p(?’“))z P=gt-se +v(t-x)] (9.15)therefore and Forfreefall,wehave x=gt2/2,sothat ai:=gt= \/2gx 56=g and .p __ P== §(gb—- 3gx) —-Mg—- T and ﬁnally, Mg 3xT- +1) (9.16) (b)Calkin andMarch (Am. Phys. 57,154[1989]) have found that chains actmuch like aperfectly ﬂexible, inextensible rope that conserves energy when itfalls, with nodissipative mechanisms. Wetreat thechain asone- dimensional motion, ignoring thesmall horizontal motion. Letthepotential energy Ubemeasured relative totheﬁxed end ofthechain, sothat theinitial potential energy U(t=O)=U0=-pgbz/4. Acareful geometric construction shows that thepotential energy after thechain hasdropped adistance xis 9.3LINEAR MOMENTUM orTHESYSTEM 335 1 U= -;pg(b2 +2bx~" x2) The kinetic energy (where weuseKinstead ofTtoavoid confusion with ten- sion) isdetermined from thespeed ii:oftheright side ofthechain, sothat P .K=—b—- 2 4t»<=>»<= Because energy isconserved, wemust have K+U=U0. P .1 1—b—- 2-— b2 b—-x2=—-— b2 4( vow 4pg( +2X ) 40g Wesolve for£2toobtain 2b—2,22M (9,7)b-x Toﬁnd thetension from Equations 9.14 and 9.15, weneed todetermine We take thederivative ofEquation 9.17 andﬁnd 55: +g(2bx -x2) g2(1>—X)2 ' Wenow insert £2and iifrom thetwoprevious equations into Equation 9.15 to determine Pandinsert thisvalue ofPinto Equation 9.14. After collection of terms andsolving forT,weobtain T=M5---1—--(2122 +2bx—3362) (9.1s)4b(b-X) Note thedifference between thetworesults, Equations 9.16 and9.18, forthe free falland energy conserving methods. Itshould berather easy byexperimen- tation todetermine which iscorrect, because thelatter result hasthetension rising dramatically (T—) oo)attheendwhen x—>b.Experiments byCalkin and March conﬁrm that thetension does increase rapidly attheend toamaximum ofabout 25times thechain’s weight, andtheobservations asafunction ofx agree wellwith thecalculations. Real chains cannot have aninﬁnite tension. Forthefree fallcase, thetension inthechain isdiscontinuous oneither sideofthebottom bend; thetension isT1=p5c2/2ontheﬁxed sideand T2=0 onthefreeside. Fortheenergy conserving case, thetension T2onthefreeside isnotzero, and thistension helps gravity pull thechain down. The result isthat thechain fallsabout 15% faster than calculated forthefreefallcase. For energy-conserving chains, thetension iscontinuous: T1=T2=pa2:2/4. We examine further properties ofthefalling chain intheproblems. 336 9/DYNAMICS OFASYSTEM orPARTICLES 9.4 Angular Momentum oftheSystem Itisoften more convenient todescribe asystem byaposition vector with respect tothecenter ofmass. The position vector raintheinertial reference system (see Figure 9-5)becomes ra=R+r; (9.19) where r;istheposition vector oftheparticle ozwith respect tothecenter of mass. The angular momentum oftheath particle about theorigin isgiven by Equation 2.81: La=raXpa (9.20) Summing thisexpression over oz,and using Equation 9.19, wehave L=EL,=gin,><pa)=Eu,><mag) =Z@+mx%@+m =§m,,,[(r;, ><r,;)+(rg,><R)+(R><r,;)+(R><R)](9.21) The middle twoterms canbewritten as <Emar,',) XR+RXdit<2m,,r;) which vanishes because Emaré, =2m,,(r,, —-R)=Emara —-REmaC! (X G (X Zmar; =MR4MRE0 (9.22) R ra FIGURE 9-5 Wecanalsodescribe asystem byposition vectors 1",;with respect tothe center ofmass. 9.4 ANGULAR MOMENTUM OFTHE SYSTEM 337 This indicates that Eamar; speciﬁes theposition ofthecenter ofmass inthe center-of-mass coordinate system andistherefore anull vector. Thus, Equation 9.21 becomes L=MR><R+§1r;><p;=R><P+Er,;><p; (9.29) Our fourth important result is IV. Thetotal angular momentum about anorigin isthesumoftheangular momentum ofthecenter ofmass about thatorigin andtheangular momentum ofthesystem about theposition ofthecenter ofmass. The time derivative oftheangular momentum oftheozthparticle is,from Equation 2.83, L,=r,><pa (9.24) and, using Equations 9.7and9.8,wehave La:1-0,x(Fae) +élfaﬂ) (9.25) Summing thisexpression over oz,wehave L=ELa=E(r,,><F2»)+Q;(1,,><£22) (9.26) The lastterm may bewritten as a,%a(ra XfazB) :a;ﬁ[(ra X£043) +(rB XfBa)] The vector connecting theathand Bthparticles (see Figure 9-6) isdeﬁned tobe ragEra——r2 (9.27) and then, because fa);=-fﬁa, wehave a,;2a(r.,, xfaﬁ)Za;p(1‘a ""1'2)XfaB =a§B(r,,,, ><r,,,,) (9.29) Now wewant tolimit thediscussion tocentral internal forces and apply the “strong” version ofNewton’s Third Law. Hence, fa);isalong thesame direction asiraﬁ and ray;Xfa),E0 (9.29) and L=-E[r,,,><Fg;>] (9.30) The right-hand sideofthisexpressioii isjustthesum ofalltheextemal torques: L=EN,<;> =N") (9.31) 338 9/DYNAMICS OFASYSTEM OFPARTICLES (1 fag I3 ra 1 rﬁ FIGURE 9-5 The vector from theBthparticle totheathparticle inthesystem is represented byrag. This leads toournext important result: V.Ifthenetresultant external torques about agiven axisvanish, thenthetotalangular momentum ofthesystem about thataxisremains constant intime. Note alsothatthetenn Er,><ta, (9.32) isthetorque onthe01thparticle due toalltheinternal forces—that is,itisthe internal torque. Because thesum ofthisquantity over alltheparticles ctvanishes (seeEquation 9.28), a;;:a(r0, ><faﬁ)=a;ﬁ(r,,2 ><fag)=0 (9.33) thetotal internal torque must vanish, which wecanstate as VI. Thetotal internal torque must vanish iftheinternal forces arecentral—that is,if fa),=—fB,,,, andtheangular momentum ofanisolated system cannot bealtered without theapplication ofexternalforces. EXAMPLE 9.3 I-I Alight string oflength ahasbobs ofmass mlandm2(m2 >ml)onitsends. The end with mlisheld and m2iswhirled vigorously byhand above thehead ina counterclockwise direction (looking down from above) andthen released. 9.5ENERGY orTHESYSTEM 339 7/12 b /__2CMV2 _——-—_____ __-__——___—_- a "l1 V0 =.w=.,»=. =52‘?=' i ..-v. FIGURE 9-7 Example 9.3.Alight string with masses mland m2atitsends iswhirled around byhand above thehead andreleased. Describe thesubsequent motion, and ﬁnd thetension inthestring after release. Solution. The system isshown inFigure 9-7.The center ofmass isadistance b=[ml/(ml+m2)]a from mass m2.After being released, theonly forces onthe system arethegravitational forces onmlandm2.Assume thatv0istheinitial ve- locity ofthecenter ofmass CM. The CMwillcontinue inaparabolic path under theinﬂuence ofgravity asifallthemass (ml+m2)were concentratedat theCM. Butwhen released, mass m2isrotating around mlrapidly. Because no external torque exists, thesystem willcontinue torotate. Butnow both mland m2rotate about theCM, andtheangular momentum isconserved. Ifmass m2is traveling with thelinear velocity v2when released, then wemust have v2== [similarly, vl=(a-b)(i]. The tension inthestring is,however, due tothecentrifugal reaction ofthemasses rotating, which is,inthiscase, . m.<1>é>2 .Centrifugal force =-T =Tension '2_ . mla . mlm2a0Tension ==m2b02 =m2—i~— 62=—~—“~— ml+m2 ml+m2 9.5 Energy oftheSystem 2 The ﬁnal conservation theorem, that ofenergy, may bederived forasystem of particles asfollows. Consider thework done onthesystem inmoving itfrom a Conﬁguration 1,inwhich allthecoordinates rl,arespeciﬁed, toaConﬁguration 2,inwhich thecoordinates rahave some different speciﬁcation. (Note that the individual particles mayjust berearranged insuch aprocess, and that, forexam- ple,theposition ofthecenter ofmass could remain stationary.) Inanalogy with 340 9/DYNAMICS OFASYSTEM OFPARTICLES Equation 2.84, wewrite W12=ErF,,-er, (9.34)<11 where Faisthenetresultant force acting onparticle ct.Using aprocedure simi- lartothat used toobtain Equation 2.86, wehave 1W122 d(§mav2) =T2-T, (9.35) where 1T=ET,=Egmavi (9.36) Using therelation (see Equation 9.19) rd,=E;+R (9.37) wehave 2,,-2,,=v2=(2,;+R).(2;+R) =(2;-2;) +2(P;,-R) +(R-R) =9;?+2(r;,-R) +V2 where v’Ei"andwhere Visthevelocity ofthecenter ofmass. Then 1T=2—m,,,v2012 1 1 .dE 4+2 mV2+R 2mr' (9.33) 1 a 5 mava a5 G .It a G CY But, byaprevious argument, 2.,mar; =O,andthelastterm vanishes. Thus, 1 1T=E5mav,§2+EMV2 (9.39) which canbestated: VII. Thetotalkinetic energy ofthesystem isequal tothesumofthekinetic energy ofapar- ticleofmass Mmoving withthevelocity ofthecenter ofmass andthekinetic energy of motion oftheindividual particles relative tothecenter ofmass. 9.5ENERGY OFTHESYSTEM 341 Next, thetotal force inEquation 9.34 canbeseparated asinEquation 9.6: VVl2=2j2Ff,f)-dr,+ 2ff -dr, (9.40) <11 a,B=#o1 1043 Iftheforces Ff,”andf,llareconservative, then they arederivable from potential functions, andwecanwrite F$;>=—V,U,}_ 9.41 fa1B :__VaI]aB ( ) where U,andU02;arethepotential functions butwhich donotnecessarily have thesame form. The notation V,means that thegradient operation isperformed with respect tothecoordinates oftheathparticle. The ﬁrstterm inEquation 9.40 becomes 2jQF§;’)-dr.,, =—Ej2 (v,U.,)-er.CY 1 Cl’ 1 =-2111, (9.42) The second term* inEquation 9.40 is algaf t,,,-dr,=Elf(f,,,-dr,+t,,,-drp) =Efr,-(dr,—dr,,)=Eﬁjqr, -dr, (9.43)C¥<B 1 B (1< 1 B B where, following thedeﬁnition inEquation 9.27, dr,ll =dr,—drll. Because U04;isafunction only ofthedistance between maand mll,itthere- fore depends onsixquantities—that is,thethree coordinates ofma(the x,_,-) and thethree coordinates ofml;(the xlll,-). The total derivative ofU043istherefore the sum ofsixpartial derivatives andisgiven by _ av, av,da,,=E(—2dx,, +43222,) (9.44).‘6&3. ’1 ax‘); ,. where thexlllareheld constant intheﬁrstterm andthex,,,areheld constant in thesecond. Thus, dt7,,,=(v,t7,,,) -er,+(v,,r7,,,) -er, (9.45) *Note that, unlike theterm Em3,4,,f,l;thatappears inEquation 9.9,theterm f-d isnotantisymmetric inaandBandtherefore does not,ingeneral, vanish. 342 9/DYNAMICS OFASYSTEM OFPARTICLES Now v,E',,, =—r,,, (9.43) butU04; =UB0,,so v,,U,,, =V252, =-£2,=t,,, (9.47) Therefore, dt7,,,=-r,,,-(dr,-dr,,) ==—-fall -drall (9.48) Using thisresult inEquation 9.43, wehave 2 a,;af£,,.dr, =—a;3jTdU,B =_a§<)BU..,,1 (9.49) Combining Equations 9.42 and9.49 toevaluate Wl2inEquation 9.40, weﬁnd 2 _2 W,2=-2U,—Es11,, (9.50)0' 1 1a< Weobtained this equation assuming that both the external and internal forces were derivable from potentials. Insuch acase, thetotal potential energy (both internal andexternal) forthesystem canbewritten as U=EU,+E5,2 (9.51)or 0z<B Then, W12=—U|%=U,-U2 (9.52) Combining thisresult with Equation 9.35, wehave T2“T1=U1“U2 or T1+(]1=T2+U2 (9.53) which expresses theconservation ofenergy forthesystem. This result isvalid for asystem inwhich alltheforces arederivable from potentials thatdonotdepend explicitly onthetime; wesaythatsuch asystem isconservative.sothat VIII. Thetotal energy foraconservative system isconstant. 95ENERGYoFTHEsnuEM 343 InEquation 9.51, theterm gt;U22 represents theinternal potential energy ofthesystem. Ifthesystem isarigid body with theconstituent particles restrained tomaintain their relative positions, then, inanyprocess involving thebody, theinternal potential energy remains constant. Insuch acase, theinternal potential energy canbeignored when computing the total potential energy ofthesystem. This amounts simply toredeﬁning theposition ofzero potential energy, butthisposition isarbitrarily chosen anyway; thatis,itis only thedifference inpotential energy that isphysically signiﬁcant. The absolute value ofthepotential energy isanarbitrary quantity. EXAMPLE 9.4 T Aprojectile ofmass Mexplodes while inﬂight into three fragments (Figure 9-8). One mass (ml=M/2) travels intheoriginal direction oftheprojectile, mass m2 (=M/6)travels intheopposite direction, andmass m3(=M/3)comes torest. The energy Ereleased intheexplosion isequal toﬁvetimes theprojectile’s ki- netic energy atexplosion. What arethevelocities? Solution. Letthevelocity oftheprojectile ofmass Mbev.The three fragments have thefollowing masses andvelocities: ml= vl=klv Forward direction, kl>0 m2=— =-k2v Opposite direction, k2>0 t5 m3=— =0 Atrest ¢$ ,---_-,_‘a’ "~>_ a’ u4 ~ a’ Ts _" _~ | ." A42‘ ///,/’ \ Before ,1’ / T‘ explosion x /\\v\ ’I I I ’ 1I$Z& I - _'~_ After¢’ ¢”TT’ ~_~ VI I _ 1’ 1" TX 1’ ,’T T §,,1’ m2 Q 3I I’ v / explosion / Q.2’ m I2 1 Q I I I I I I FIGURE 9-8 Example 9.4.Aprojectile ofmass Mexplodes inﬂight into three fragments ofmasses ml,m2,andm2. 344 9/DYNAMICS OFASYSTEM OFPARTICLES The conservation oflinear momentum andenergy give M MMv=5klv—€k2v (9.54) .1. 2_i_L/I 2 2E+ 2Mv —22(klv) +26(k2v) (9.55) From Equation 9.54, k2=3kl—6,which wecaninsert into Equation 9.55: 1 1 M112 Mv25_M 2 __ 22M M _ 2(2 v)+2Mv 4k2+12(3kl 6) which reduces tok2"—3klIO,giving theresults kl=0and kl=3.ForklI0, thevalue ofk2=“-6,which isinconsistent with k2>O.Forkl=3,thevalue of k2=3.Thevelocities become vl=3v v2="*3v v3=() EXAl\"lPl.E 9.5 Arope ofuniform linear density pandmass miswrapped onecomplete turn around ahollow cylinder ofmass Mand radius RThe cylinder rotates freely about itsaxisastherope unwraps (Figure 9-9). The rope ends areatx=0 (one ﬁxed, oneloose) when point Pisat6=0,andthesystem isslightly dis- placed from equilibrium atrest. Find theangular velocity asafunction of angular displacement 9ofthecylinder. P \l?to ration ~\\\\\\\\\\‘ “@\\\\\\\\‘\‘ : \Q‘. .@. \‘“ Q QI 9 I 9 .RS1Il¢ -_'£. 0 i '0 RI I '0 '0 0,. ‘\\\\\\\\\\ a\\\\\\\1~‘\\\\\\\\\‘\\l"9'0 I‘\\\\‘\\‘\‘\\“\\\“‘—-11-R.RX‘\\v~'\\\\“‘\\\ E’'fI I\‘\\‘\“‘\‘\27\\\\\§\\\*‘/4.4Q.X‘S- () (b) FIGURE 9-9 Example 9.5.(a)Arope iswound around acylinder. Both ends areat x=0when 6=0.(b)Work isdone toplace section dxback upnext to thecylinder. 9.6 ELASTIC COLLISIONS OFTWO PARTICLES 345 Solution. Gravity hasdone work onthesystem tounwind therope. Consider a section dxoftherope located adistance xfrom where itunwinds. The mass of thissection ispdx.Ifwewere toperform work byreaching upandwrapping thisloose end oftherope against thecylinder, how farupwould thesection dx actually travel? The distance xwould beonthecircumference ofthecylinder (seeFigure 9-9),anddxwould beRsin(x/R) below xK0.The total distance thesection dxwould move upis Distance dxmoves Ix""Rsin Work done =(pdx)gI:x -Rsin (9.56) The total work done bygravity inunwrapping therope through anangle 6is, therefore, R6 “ll W== WR'— dx Lpgiix sin(R 2 W: pgR2(6; +cos0—-1) (9.57) The work done bygravity must equal thekinetic energy gained bytheropeand thecylinder. 1.1 .T=§m(RB)2 +§M(RB)2 (9.58) Because W= Tand p=m/(2"rrR), R62 1 . mi —+cost? —1:|= —-(m +M)R262 21'r 2 2 and . mg(62+ 2cos0— 2) “Q= ? <9-59> 9.6 Elastic Collisions ofTwo Particles Forthenext fewsections, weapply theconservation laws totheinteraction oftwo particles. When twoparticles interact, themotion ofoneparticle relative tothe other isgoverned bytheforce lawthat describes theinteraction. This interaction may result from actual contact, asinthecollision oftwobilliard balls, orthein- teraction may take place through theintermediary ofaforce ﬁeld. Forexample, afreeobject (i.e., onenotbound inasolar orbit) may “scatter” from thesunbya 346 9/DYNAMICS OFASYSTEM OFPARTICLES gravitational interaction, orana-particle may bescattered bytheelectric ﬁeld of anatomic nucleus. Wedemonstrated intheprevious chapter that once theforce lawisknown, thetwo-body problem can becompletely solved. But even ifthe force ofinteraction between twoparticles isnotknown, agreat deal canstillbe learned about therelative motion byusing only theresults oftheconservation of momentum and energy. Thus, iftheinitial state ofthesystem isknown (i.e., ifthe velocity vector ofeach oftheparticles isspeciﬁed), theconservation laws allow us toobtain information regarding thevelocity vectors intheﬁnal state.* Onthebasis oftheconservation theorems alone, itisnotpossible topredict, forexample, theangle between theinitial andﬁnal velocity vectors ofoneofthe particles; knowledge oftheforce lawisrequired forsuch details. Inthissection and thenext, wederive those relationships that require only theconservation of momentum andenergy. Then, weexamine thefeatures ofthecollision process, which demand thattheforce lawbespeciﬁed. Welimit ourdiscussion primarily toelastic collisions, because theessential features oftwo-particle kinematics are adequately demonstrated byelastic collisions. The results obtained under the assumption only ofmomentum and energy conservation arevalid (inthenon- relativistic velocity region) even for quantum mechanical systems, because these conservation theorems areapplicable toquantum aswell astoclassical systems. Wedemonstrated onseveral occasions that thedescription ofmany physical processes isconsiderably simpliﬁed ifone chooses coordinate systems atrest with respect tothesystem’s center ofmass. Intheproblem wenow discuss—the elastic collision oftwo particles—the usual situation (and theone towhich we conﬁne ourattention) isoneinwhich thecollision isbetween amoving particle andaparticle atresti Although itisindeed simpler todescribe theeffects ofthe collision inacoordinate system inwhich thecenter ofmass is‘atrest, theactual measurements aremade inthelaboratory coordinate system inwhich theob- server isatrest. Inthis system, one oftheparticles isnormally moving, and the struck particle isnormally atrest. Wehere refer tothese twocoordinate systems simply astheCMandtheLAB systems. Wewish totake advantage ofthesimpliﬁcations that result bydescribing an elastic collision intheCMsystem. Itistherefore necessary toderive theequa- tions connecting theCMandLAB systems. *The “initial state" ofthesystem isthecondition oftheparticles when they arenotyetsufﬁciently close tointeract appreciably; the“ﬁnal state” isthecondition after theinteraction hastaken place. For acontact interaction, these conditions areobvious. Butiftheinteraction takes place byaforce ﬁeld, then therateofdecrease oftheforce with distance must betaken intoaccount inspecifying theinitial andﬁnal states. TAcollision iselastic ifnochange intheintemal energy oftheparticles results; thus, theconserva- tion ofenergy may beapplied without regard totheinternal energy. Notice that heat may begener- atedwhen twomechanical bodies collide inelastically. Heat isjustamanifestation oftheagitation of abody’s constituent particles andmay therefore beconsidered apart oftheinternal energy. The lawsgoverning theelastic collision oftwobodies were ﬁrstinvestigated by]ohn Wallis (1668), Wren (1668), andHuygens (1669). 9.6 ELASTIC COLLISIONS OFTWO PARTICLES 347 Weusethefollowing notation: ml=Mass ofthe{movmg} particlem2= struck Ingeneral, primed quantities refer totheCMsystem: ul=Initial 1_f _thLAB t eo min e ssevl=Final Vcltyo I ym velocity ofmlintheCMsystemul=Initial vl=Final andsimilarly foru2,v2,ué,andvé(but I12=0): T= LABT2=Total initial kinetic energy in{CM }system T= LAB Tli=Final kinetic energy ofmlin{CM }system andsimilarly forT2and Té, q>=r~=“%<=velocity ofthecenter ofmass intheLAB system =angle through which mlisdeﬂected intheLAB system =angle through which mgisdeﬂected intheLAB system =angle through which mlandml,aredeﬂected intheCMsystem Figure 9-10 illustrates thegeometry ofanelastic collision* inboth theLAB andCMsystems. Theﬁnal state intheLAB andCMsystems forthescattered par- ticle mlmaybeconveniently summarized bythediagrams inFigure 9-11. Wecan interpret these diagrams inthefollowing manner. Tothevelocity VoftheCM, wecanaddtheﬁnal CMvelocity vlofthescattered particle. Depending onthe angle 9atwhich thescattering takes place, thepossible vectors vllieonthecir- cleofradius vlwhose center isattheterminus ofthevector V.The LAB velocity vlandLAB scattering angle titarethen obtained byconnecting thepoint ofori- ginofVwith theterminus ofvl. IfV< vl,only onepossible relationship exists between V,vl,vland 6(see Figure 9-lla). ButifV>vl,then forevery setV,vl,there exists twopossible scattering angles andlaboratory velocities: vl,l,,0,,andvlf,Bf(seeFigure 9-11b), where thedesignations bandfstand forbackward andforward. This situation re- sults from thefactthatiftheﬁnal CMvelocity vlisinsufficient toovercome the velocity Vofthecenter ofmass, then, even ifmlisscattered into thebackward direction intheCMsystem (6>*rr/2), theparticle willappear ataforward angle *We assume throughout thatthescattering isaxially symmetric sothatnoazimuthal angle need be introduced. However, axial symmetry isnotalways found inscattering problems; thisisparticularly true incertain quantum mechanical systems. 348 9/DYNAMICS orASYSTEM orPARTICLES Laboratory System Center-of-Mass System ml ul 'm2 ml ui ué 7”? --VI} u2 =0 (a)Initial condition (b)Initial condition ml "1I "1 0 9 "2 =.__ --------------- ---"——-¢1|:9 vé ¢=n—6 "*2 (c)Final condition (d)Final condition FIGURE 9-10 Geometry andnotations ofanelastic collision intheLAB andCMsystems. (a)Initial condition with u2=0intheLAB system, (b)initial condition intheCMsystem, (c)ﬁnal condition intheLAB system, and(d)ﬁnal condition intheCMsystem. Note carefully thescattering angles. 9-1/I‘P "lb atitV<‘(Ii V>vi (H) (b) FIGURE 9-ll Theﬁnal state ofmass mlfortheelastic collision oftwoparticles forthe case (a)V<vlforwhich there isonetrajectory and(b)V>v{for which there aretwopossible trajectories (bstands forbackward andf forforward). intheLAB system (lb<11'/2). Thus, forV>vl,thevelocity vlintheLAB system isadouble-valued function ofvl.Inanexperiment, weusually measure 4,0,not thevelocity vector vl,sothat asingle value of1/1cancorrespond totwodifferent values of0.Note, however, that aspeciﬁcation ofthevectors Vand vlalways leads toaunique combination vl,6;butaspeciﬁcation ofVandonly thedirec- tionofvl(i.e., 1/1)allows thepossibility oftwoﬁnal vectors, vl,,andvl,],ifV>vl. 9.6 ELASTIC COLLISIONS OFTWO PARTICLES 349 Having given aqualitative description ofthescattering process, wenow ob- tain some oftheequations relating thevarious quantities. According tothedeﬁnition ofthecenter ofmass (Equation 9.3), wehave mlrl +m2r2 =MR (9.60) Differentiating with respect tothetime, weﬁnd mlul +m2u2 =MV (9.61) Butu2=0andM=ml+m2;thecenter ofmass must therefore bemoving (in theLAB system) toward ml,with avelocity T/#1111V=mi (9.62)ml+m2 Bythesame reasoning, because m2isinitially atrest, theinitial CMspeed ofm2 must just equal V: ,_ mlulu2—V=mi (9.63)ml+m2 Note, however, that themotion andthevelocities areopposite indirection and thatvectorially u§=—V. The great advantage ofusing theCMcoordinate system isbecause thetotal linear momentum insuch asystem iszero, sothatbefore thecollision theparti- clesmove directly toward each other andafter thecollision they move inexactly opposite directions. Ifthecollision iselastic, aswehave speciﬁed, then themasses donotchange, andtheconservation oflinear momentum andkinetic energy is sufﬁcient toprovide thattheCMspeeds before andafter collision areequal: ul=vl, ué=vé (9.64) Term ulistherelative speed ofthetwoparticles ineither theCMortheLAB system, ul=ul+ué.Wetherefore have, fortheﬁnal CMspeeds, _U,_ "hut2_-ii ml+m2(9.65a) ._ ._"W11'Ul—ul—112—Y5; (9.65b) I Wehave (seeFigure 9-1la) vlsin0=vlsinqb (9.66a) and vlcos6+V=vlcost/1 (9.66b) Dividing Equation 9.66a byEquation 9.66b, vlsin6 sin6 rand’ :vlcos6+V: dos6+(V/vl) (9.67) 350 9/DYNAMICS orASYSTEM orPARTICLES According toEquations 9.62 and9.65b, V/-ul isgiven by if: mlul/(ml mg) = ml U1 m2"1/(mi 'm2) m2 Thus, theratio ml/m2 governs whether Figure 9-lla orFigure 9-llb describes thescattering process: Figure 9-11a: V<vl, ml<mg Figure 9-11b: V>vl, ml>m2 Ifwecombine Equations 9.67 and9.68 andwrite tant//=-—-§39—6—— (9.69)cos6 +(ml/m2) weseethat ifml<<m2,theLAB and CMscattering angles areapproximately equal; thatis,theparticle m2isbutlittle affected bythecollision with mlandacts essentially asaﬁxed scattering center. Thus 1/1E6, ml<<m2 (9.70) However, ifml=m2,then n=m= n-tall’ sin6 ta0 cos0+1 2 and theLAB scattering angle isone half theCMscattering angle. Because the maximum value of9is180°, Equation 9.71 indicates thatforml=777/2,there can benoscattering intheLAB system atangles greater than 90°. Letusnow refer toFigure 9-10c andconstruct adiagram fortherecoil parti- clem2similar toFigure 9-lla. The situation isillustrated inFigure 9-12, from which weﬁndsothat '02sinQ’=v§sin6 (9.72a) -02cos§=V—vécos6 (9.72b) Dividing Equation 9.72a byEquation 9.72b, wehave 5 vésin9 sin9ta = = n V—v§cos9(V/v§) —cos6 But, according toEquations 9.63 and9.65a, Vand véareequal. Therefore, tan§'= =cot3 (9.73) 9.6ELASTIC COLLISIONS OFTwoPARTICLES 351 / I V I///\6 C 4w@ ¢=x-6 vé FIGURE 9-12 Theﬁnal state ofrecoil mass m2intheelastic collision oftwoparticles. which wemaywrite as tan§= tan(%—g) 2§=rr—6=¢ (9.74)Thus, Forparticles with equal mass, ml=m2,wehave 6=21/1.Combining thisresult inEquation 9.74, wehave g+ti!= ml=m2 (9.75) Hence, thescattering ofparticles ofequal mass always produces aﬁnal state in which thevelocity vectors oftheparticles areatright angles ifoneoftheparti- clesisinitially atrest(seeFigure 9-13) .* 7111 ml ml mg lp ml mg ll’~ . it Z —*" C m2 7722 FIGURE 9-13 Fortheelastic scattering oftwoparticles ofequal mass (ml=m2)with oneofthem initially atrestintheLAB system, theﬁnal velocities (trajectories) ofthetwomasses areatright angles toeach other. Two such possibilities areshown. *This result isvalid only inthenonrelativistic limit; seeEquation 14.131 fortherelativistic expres- siongoverning thiscase. 352 9/DYNAMICS orASYSTEM orPARTICLES "1 ‘lg ____ __ V FIGURE 9-14 Example 9.6.Thecase ofFigure 9-11b isshown for4,/1m,,,,. I/Vhat isthemaximum angle thatitcanattain forthecase V>vl?What islllmax for T/L1 771,2 and ml =7/L2? Solution. Forthecase ofl//max, Figure 9-11b becomes asshown inFigure 9-14. The angle between vlandvlis90°forittobeamaximum. sin¢,,,,,,=% (9.76) According toEquation 9.68, thisisjust from which .11=sin-1E (9.77) Forml>>m2,1//max =0(noscattering), andforml="Z2,tbmax =90°. Generally, forml>m2,noscattering ofmlbackward of90°canoccur. 9.7 Kinematics ofElastic Collisions Relationships involving theenergies oftheparticles may beobtained asfollows. First, wehave simply To=gmlul’ (9.78) and, intheCMsystem, Ti)=%(m1ui2 +m2"§2) which, onusing Equations 9.65a and9.65b, becomes 1 777/17/Z2 mg T'=-? 2=—T 9.79 02m1+m2"I ml+m2 0 ( ) 9.7KINEMATICS orELAsTIC COLLISIONS 353 This result shows that theinitial kinetic energy intheCMsystem T6isalways a fraction my(ml+m2)<1oftheinitial LAB energy. Fortheﬁnal CMenergies, weﬁnd I 1 I 1 m 2 1112 2 Tl =—mlUl2 =-777/l $ U7? = W To 2 2 ml +7112 ml +m2 and ,1 , 1 ml 2_ m1m2 T2=§’""2”2"=5% “i‘aw?" “"8” Toobtain Tlinterms ofT0,wewrite l 2 T1 2m1v1 vi9.8 To émluf ui (2) Referring toFigure 9-11a andusing thecosine law,wecanwrite vl2=1112+ V2—2-ulVcos¢ or §5 §G)-pg>-(IO§§_>-mgM<1IQv==-u,+21:,cos¢ (9.99)I I From theprevious deﬁnitions, wehave -0' m V m—l=L2 and ——=ml (9.84)ul ml +7712 ul ml +177/2 The squares ofthese quantities give thedesired expressions fortheﬁrst two terms ontheright-hand side ofEquation 9.83. Toevaluate thethird term, we write, using Equation 9.66a. 5..<vlV _ Isin027% COS(D—2vl COS(U (9.85) The quantity of-u'lV/ul canbeobtained from theproduct oftheequations in Equation 9.84, andusing Equation 9.69, wehave sin6cosIP sin9 ml+i=-——= cos9+—s1n(l1 tant/1 m2 sothat UlV 2ml7712 ‘ml 27% cost/1= (C0s H+ (9.86) 354 9/DYNAMICS OFASYSTEM OFPARTICLES Substituting Equations 9.84 and9.86 into Equation 9.83, weobtain T1 m2 2 ml 2 2m1m2 mli = 9 — 9 -|-9-i COS 0-|-_~_ To ml + "12 ml + "1/2 (ml + "1/2) m2 which simpliﬁes to 5=1—l-“£0 -cos0) (987a)To (ml+m2)2 Similarly, wecanalso obtain theratio Tl/Tointerms oftheLAB scattering angle lb: T m2 / 2 2 ‘ft=Tlﬁ [COS (Di _Sin211]] (9.87b) ml m2 ml where theplus (+)sign fortheradical istobetaken unless ml>m2-—in which case theresult isdouble-valued, andEquation 9.77 speciﬁes themaximum value allowed for¢. The LAB energy oftherecoil particle m2canbecalculated from T2 T1 4m1m2 i= 1—';b= COS2§, {E17/2 (9.88) 2 Ifml=m2,wehave thesimple relation 5=cos2 1,11 m=m (989a) T # l 2 ° 0 with therestriction noted inthediscussion following Equation 9.71 that 1/1S90°. Also, g = I2 Zsin 1/1, ml "12 (9.89b) To Several further relationships are ._ m1T1 .sin4,"——é S1I11/1 (9.90)m2 T2 tan1/1= (9.91)(ml/m2) —cos2§ sin¢ tan1/1=mi? (9.92) (ml/m2) _C054) Asanexample ofapplying thekinematic relations wehave derived, consider thefollowing situation. Suppose thatwehave abeam ofprojectiles, allwith mass mlandenergy T0.Wedirect thisbeam toward atarget consisting ofagroup of particles whose masses m2may notallbethesame. Some oftheincident parti- clesinteract with thetarget particles andarescattered. The incident particles all move inthesame direction inabeam ofsmall cross-sectional area, andweassume 9.7KINEMATICS orELASTIC COLLISIONS 355 10—----------------- -- ”‘2/ml : tp=90° Intensity—> m--—-----—---———-P_>4- .°_ w-______§°___P -----------wiw-.__--------.-.‘Sl“‘_O5--------------- -- 0 I 0 0.2 1/To Energy—> FIGURE 9-15 Results ofparticles ofmass mlandenergy T0being scattered from particles ofvarious masses m2atangle tﬂ=90°.Bottom: Histogram ofnumber ofparticles detected within anenergy range AT. Top: Curve giving scattered energy Tlinterms ofToasafunction of themass ratio m2/ml. that thetarget particles arelocalized inspace sothat thescattered particles emerge from asmall region. Ifweposition adetector at,say,90°totheincident beam andwith thisdetector measure theenergies ofthescattered particles, we candisplay theresults asinthelower portion ofFigure 9-15. This graph isahis- togram thatplots thenumber ofparticles detected within arange ofenergy AT attheenergy T.This particle histogram shows thatthree energy groups were ob- served intheparticles detected at1/1=90°. The upper portion oftheﬁgure shows acurve giving thescattered energy Tlinterms ofToasafunction ofthe mass ratio m2/ml (Equation 9.87b). The curve canbeused todetermine the mass m2oftheparticle from which oneoftheincident particles wasscattered to fallinto oneofthethree energy groups. Thus, theenergy group with TlE0.8T0 results from thescattering bytarget particles with mass m2=10ml, and the other twogroups result from target masses 5mland2ml. Themeasurement oftheenergies ofscattered particles istherefore amethod ofqualitative analysis ofthetarget material. Indeed, thismethod isuseful inprac- ticewhen theincident beam consists ofparticles (protons, say) that have been given high velocities inanaccelerator ofsome sort. Ifthedetector iscapable of precise energy measurements, themethod yields accurate information regarding thecomposition ofthetarget. Quantitative analysis canalsobemade from thein- tensities ofthegroups ifthecross sections areknown (see thefollowing section). Applying thistechnique hasbeen useful indetermining thecomposition ofair pollution. 356 9/DYNAMICS orASYSTEM OFPARTICLES Inahead~on elastic collision oftwoparticles with masses mland m2,theinitial velocities areuland u2=aul(a >O).Iftheinitial kinetic energies ofthetwo particles areequal intheLAB system, ﬁnd theconditions onul/ua and ml/mg so that mlwillbeatrestintheLAB system after thecollision. SeeFigure 9-16. Solution. Because theinitial kinetic energies areequal, wehave 1 1 1—m112=—m 2=—a2m 112 21122112 2 21 or mlE=a2 (9.93) Ifmlisatrest after thecollision, theconservation ofenergy requires gmlui +gmgug =%mQv§ or mlul =%m2v§ (9.94) The conservation oflinear momentum states that mlul +mgug =(ml+am2)ul =m2v2 (9.95) Substituting v2from Equation 9.95 into Equation 9.94 gives 1 ml+CY?"/2 22__ ____i 2mlul —m2 "1 Or _1 ml 2 ml-51712E+a (9.96) "1 w "2 1 m2 Before collision "*2 After collision FIGURE 9-16 Example 9.7.Velocities areindicated fortwoparticles ofdifferent masses inahead-on elastic collision before andafter thecollision. 9.7 KINEMATICS OFELASTIC COLLISIONS 357 Substituting ml/m2=a2from Equation 9.93 gives 2a2=(a2+a)2 with theresult a=\/2—1=0.414 012=0.172 sothat 3=<12=0.172m2 and 3=<1=0.414"1 Because cc>0,both particles aretraveling inthesame direction; thecollision is shown inFigure 9-16. Particles ofmass mlelastically scatter from particles ofmass ml,atrest. (a)At what LAB angle should amagnetic spectrometer besettodetect particles that loseone-third oftheir momentum? (b)Over what range ml/m2 isthispossible? (c)Calculate thescattering angle forml/m2 =1. Solution. Wehave 2 2mlvl =gmlul and vl=-5ul Using Equations 9.82 and9.87a, wehave T1 vi 22 2m1m2—— =i = — = —i—— — _7 T0 ui 1 (ml+m2)2(1 COS9) (99) This equation canbesolved forcos6,yielding cos6=l— =l—y (9.98)187/£17712 where 5(m +m)2 y=——1—g;z-ii (9.99) 358 9/DYNAMICS orASYSTEM orPARTICLES Butweneed th,which canbeobtained from Equation 9.69. an = = .t¢ sinB 2y-3? (9100) cos9 +ml/m2 1—y+ml/mg where wehave used Equation 9.98 forcos6andfound sin6=\/2y- Because tan1/1must bearealnumber, only values forml/m2 where 2-y20 arepossible. Therefore, 5+ 22—ﬁ--"32->0 (9.101)187nlm2 T which canbereduced to ml 2 ml —5 — +26—) —520 m2 m2 -5062 +26x —52O (9.102)OI‘ where x=ml/m2.The solutions forxwhen Equation 9.102 isequal tozero are x=1/5,5.Substitution veriﬁes that 1 1 ml -s—s55 mg satisﬁes Equation 9.101, butvalues ofml/m2outside thisrange donot. Substituting ml/ml,=1intoEquation 9.99 gives 53+12_5(m1+ m2)2 _ m2 y l8mlm2 18ml/mg =5(1+1)2_@ 1s 9 andsubstituting foryinto Equation 9.100 gives 1/1=48°. 9.8 Inelastic Collisions When twoparticles interact, many results arepossible, depending ontheforces involved. Intheprevious two sections, wewere restricted toelastic collisions. But, ingeneral, multiparticles may beproduced iflarge changes ofenergy are involved. Forexample, when aproton collides with some nuclei, energy may be 9.8INELASTIC CoLLisioNs 359 “i "2 ..O F» O I xInitial ml "'2 V ~-W -- {X3 P Collision ml"12 "1 "2 _ O>O---I .Fmalml "*2 FIGURE 9-17 Direct head-on collision between twobodies indicating theinitial conditions, thecollision, andtheresulting situation. released. Inaddition, theproton may beabsorbed, and thecollision may pro- duce aneutron oralpha particle instead. Allthese possibilities arehandled with thesame methods: conservation ofenergy andlinear momentum. Wecontinue torestrict ourconsiderations tothesame particles intheﬁnal system aswere considered intheinitial system. Ingeneral, theconservation ofenergy is 1 1 1 1 Q+-gmlul +§m2u§ =gmlvl +-gmgvg (9.103) where Qis called theQ-value and represents theenergy loss orgain inthe collision. Q=O:Elastic collision, kinetic energy isconserved Q>0:Exoergic collision, kinetic energy isgained Q<0:Endoergic collision, kinetic energy islost Aninelastic collision isanexample ofanendoergic collision. The kinetic energy may beconverted tomass-energy, as,forexample, inanuclear collision. Orit maybelostasheat energy, as,forexample, byfrictional forces inacollision. The collisions ofallmacroscopic bodies areendoergic (inelastic) tosome degree. Two sillyputty balls with equal masses andspeeds striking head-on may come to acomplete stop, atotally inelastic collision. Even twobilliard balls colliding do notcompletely conserve kinetic energy; some small fraction oftheinitial kinetic energy isconverted toheat. Ameasure oftheinelasticity oftwobodies colliding may beconsidered by referring toadirect head-on collision (seeFigure 9-17) inwhich norotations are involved (translational kinetic energy only). Newton found experimentally that theratio oftherelative initial velocities totherelative ﬁnal velocities wasnearly constant foranytwobodies. This ratio, called thecoefﬁcient ofrestitution (1-:),is deﬁned by l _ 8=W?“ (9.104)lu2_"il This issometimes called Newton’s rule. Foraperfectly elastic collision, s=1; andforatotally inelastic collision, s=0.Values ofshave thelimits 0and 1. 360 9/DYNAMICS OFASYSTEM OFPARTICLES . ul l , b &&& &it“_“_ Iii‘! __ _ ———— b W9_._._....._gg.5 m2 m2§_ FIGURE 9-18 Anoblique collision between twobodies. Forsmooth surfaces, the velocity components along theplane ofcontact bb’arehardly changed bythecollision. Wemust becareful when applying Equation 9.104 tooblique collisions, be- cause Newton’s rule applies only tothevelocity components along thenormal (aa') totheplane ofcontact (bb') between thetwobodies, asshown inFigure 9-18. Forsmooth surfaces, thevelocity components along theplane ofcontact are hardly changed bythecollision. EXAMPLE 9.9 if n_ _- Foranelastic head-on collision described inSections 9.6and9.7,show thats=1. The mass mgisinitially atrest. Solution. Because theﬁnal velocities arealong thesame direction asul,we state theconservation oflinear momentum andenergy as mlul =mlvl +m2U2 1 1 1 gmlul =5117,11)? +5'”!/2U% Wesolve Equation 9.105 for-02andsubstitute into theequation fors mlul _'l7'll'Ul Will 8=U2 'Ul = 77'!/2 = ml _ml 'Ul _ Ul "1 "1 m2 m2"i "1 Wecanﬁnd theratio vl/ul from Equation 9.106 after substituting forv2from Equation 9.105: 1 1 1 mlul _ mlvl 2 -émlul =gmlvf +-2-m2 2mi 2- 2_ 2 2_mlul —mlvl +m(ul+vl 2ulvl) 2 9.8INELASTIC CoLLISIoNS 361 Dividing bymlul’ andletting x=vl/ul gives 1=x2+ﬁ(1+x2—2x)m2 1+5 .x2—%x+ 3-1 =0m2 m2 m2 Using thequadratic equation tosolve forx,weﬁndCollecting terms, x=1 and 3-1m2x=—i mi—+ 1 m2 The solution x=1istrivial (vl=ul,v2=0),sowesubstitute theother solution for;xinto Equation 9.107: m m m_1__1_l _1_l ml M21712 171,2 s=m— m +m 2 —‘+1 —‘+1771.2 m2 2 2ml ml ml ml ml 2+ — 2+ — +1. 111,2 m2 111,2 m2 m2 s= W =1 mi—+1 m2 During acollision (elastic orinelastic), theforces involved may actover a very short period oftime andarecalled impulsive forces. Ahammer striking a nailand twobilliard balls colliding areexamples ofimpulsive forces. Newton’s Second Law isstillvalid throughout thetime period Atofthecollision: dF=Zt(mv) (9.108) After multiplying bydtand integrating, wehave *2 J’FdtEP (9.109) ti where At=t2—tl.Equation 9.109 deﬁnes theterm impulse P.The impulse may bemeasured experimentally bythechange ofmomentum. Anideal impulse 362 9/DYNAMICS OFASYSTEM OFPARTICLES represented bynodisplacement during thecollision would becaused byaninﬁ- niteforce acting during aninﬁnitesimal time. EXAMPLE 9.10 ml Consider arope ofmass perunit length pandlength asuspended justabove a table asshown inFigure 9-19. Iftherope isreleased from restatthetop, ﬁnd theforce onthetable when alength xoftherope hasdropped tothetable. Solution. Wehave agravitational force ofmg=pxgbecause therope lieson thetable, butweneed toconsider theimpulsive force aswell. _Q’F-dt (9.110) During thetime interval dt,themass ofrope equal top(vdt)drops tothetable. The change inmomentum imparted tothetable is dp=(pvdt)v =pv2dt and g)— 2-F 9111 dt_p-U _ impulse (' ) The velocity visrelated toxattime tby112=2gx, because each part ofthere- maining rope isunder constant acceleration g. Fimpulse :P7/2 = The total force isthesum ofthegravitational andimpulsive forces: F: Fg+Fimpulse : which isequivalent totheweight ofalength 3xoftherope. ________ \ a ‘I FIGURE 9-19 Example 9.10. Arope oflength aisreleased while suspended justabove atable. Wewant toﬁnd theforce Fonthetable after therope has dropped adistance x. 9.9SCATTERING CROSS SECTIONS 363 9.9 Scattering Cross Sections Inthepreceding sections, wederived various relationships connecting theinitial state ofamoving particle with theﬁnal states oftheoriginal particle andastruck particle. Only kinematic relationships were involved; that is,noattempt was made topredict ascattering angle oraﬁnal velocity—only equations connecting these quantities were obtained. Wenow look more closely atthecollision process and investigate thescattering intheevent that theparticles interact with aspeciﬁed force ﬁeld. Consider thesituation depicted inFigure 9-20, which il- lustrates such acollision intheLAB coordinate system when arepulsive force ex- istsbetween mland m2.The particle mlapproaches thevicinity ofm2insuch a waythat ifthere were noforce acting between theparticles, mlwould pass m2 with adistance ofclosest approach b.Thequantity biscalled theimpact parameter. Ifthevelocity ofmlisul,then theimpact parameter bdetermines theangular momentum lofparticle mlabout m2: Z: mlulb Wemay express ulinterms oftheincident energy Tobyusing Equation 9.78: l=b\/2mlT0 (9.115) Evidently, foragiven energy T0,theangular momentum andhence thescatter- ingangle 6(orqb)isuniquely speciﬁed bytheimpact parameter biftheforce law isknown. Inthescattering ofatomic ornuclear particles, wecanneither choose normeasure directly theimpact parameter. Wearetherefore reduced, insuch situations, tospeaking interms oftheprobability forscattering atvarious angles 9. Wenow consider thedistribution ofscattering angles that result from colli- sions with various impact parameters. Toaccomplish this, letusassume that we have anarrow beam ofparticles, each having mass mlandenergy T0.Wedirect thisbeam toward asmall region ofspace containing acollection ofparticles, each ofwhich hasmass mgandisatrest(intheLAB system). Wedeﬁne theintensity (orflux density) Ioftheincident particles asthenumber ofparticles passing in unit time through aunit area normal tothedirection ofthebeam. Ifweassume that theforce lawbetween mland mgfalls offwith distance sufﬁciently rapidly, I / / / I I / / I r 121 -— ,’//‘ll!’ 9 I»..V. ._,. ..,.. H,._._ ._.._.... E E "'2 FIGURE 9-20 Particle mlapproaches m2,initially atrest, intheLAB system, andthe repulsive force between particles results inscattering. Had ml continued inastraight line, itsclosest distance ofapproach tom2 would have been b,theimpact parameter. 364 9/DYNAMICS OFASYSTEM orPARTICLES then after anencounter, themotion ofascattered particle asymptotically ap- proaches astraight linewith awell-deﬁned angle 9between theinitial andﬁnal directions ofmotion. Wenow deﬁne adifferential scattering cross section 0(9) intheCMsystem forthescattering into anelement ofsolid angle dQ'atapartic- ularCMangle 9: Number ofinteractions pertarget particle that) lead toscattering intodQ’at theangle9 9= 9.116O-() Number ofincident particles perunit area ( ) IfdNisthenumber ofparticles scattered into dﬂ.’perunit time, then theproba- bility ofscattering into dQ'foraunit area oftheincident beam is 0(9)dQ' =%Y (9.117a) Wesometimes write, alternatively, d0 1dN9=— =—i 9.1l7b <r()da, Ida, ( ) Thefactthat0(9) hasthedimensions ofareapersteradian gives risetotheterm cross section. Ifthescattering hasaxial symmetry (asforcentral forces), wecan immediately perform theintegration over theazimuthal angle toobtain 217,and then theelement ofsolid angle dﬂ’isgiven by dQ'=211'sin9d9 (9.118) Ifwereturn, forthemoment, totheequivalent one-body problem discussed inthepreceding chapter, wecan consider thescattering ofaparticle ofmass 1.1. byaforce center. Forsuch acase, Figure 9-21 shows thatthenumber ofparticles with impact parameters within arange dbatadistance bmust correspond tothe number ofparticles scattered into theangular range d9atanangle 9.Therefore, 1-21Tbdb =—I-0(9) -2*rrsin9d9 (9.119) where db/d9 isnegative, because weassume that theforce lawissuch that the amount ofangular deﬂection decreases (monotonically) with increasing impact parameter. Hence, s1n9 d9 Wecanobtain therelationship between theimpact parameter bandthescatter- ingangle 9byusing Figure 9-22. Inthepreceding chapter, wefound (inEquation 8.31) thatthechange inangle foraparticle ofmass /J.moving inacentral-force ﬁeld wasgiven by .6l0(9) = (9.120) =Tm“~-e(I-/T2)” (9.121)\/21.1112 -0-(12/21111)] 9.9SCATTERING CROSS SEC'I‘IONS 365 I/ / 1/ / db , '19/ I I b I . 1.1 ./ft” -Scattering center dA=2:1:bdb FIGURE 9-21 Theequivalent one-body problem hasmass ,u.scattered byaforce center intheCMsystem. Particles within arange dbaround impact parameter bscatter into theangular range d9attheangle 9. The motion ofaparticle inacentral-force ﬁeld issymmetric about thepoint ofclosest approach totheforce center (see point AinFigure 9-22). The angles 0: andBaretherefore equal and, infact, areequal toG.Thus, 9=11'—26 (9.122) Forthecase that rm,‘=oo,theangle 6isgiven by 6=loo (b/'2)” (9.123)"MnV1—(52/T2) —(U/T6) where usehasbeen made oftheone-body equivalent ofEquation 9.115: l=b\/211.T6 where, asinEquation 9.79, Tl’)=é/1.11%’. Wehave also used E=Tl’,because the total energy Emust equal thekinetic energy Tl’,atr=oowhere U= 0.The value ofrmlnisaroot oftheradical inthedenominator inEquations 9.121 or 9.123—that is,rmlnisaturning point ofthemotion andcorresponds tothedis- tance ofclosest approach oftheparticle totheforce center. Thus, Equations 9.122 and 9.123 give thedependence ofthescattering angle 9ontheimpact parameter b.Once weknow b=b(9) foragiven potential U(r) and agiven value ofT5,wecan calculate thedifferential scattering cross section from Equation 9.120. This procedure leads tothescattering cross section intheCM system, because wehave been considering m2asaﬁxed force center. If m2>>ml,thecross section soobtained isvery close totheLAB system cross A 1; 7 C1 B 6 lb <~>Scattering center FIGURE 9-22 The geometry ofparticle scattering inacentral-force ﬁeld. Point Ais thedistance ofclosest approach. 366 9/DYNAMICS orASYSTEM orPARTICLES section; butifmlcannot beconsidered negligible compared with mg,theproper transformation ofsolid angles must bemade. Wenow obtain thegeneral relations. Because thetotal number ofparticles scattered into aunit solid angle must bethesame intheLAB System asintheCMsystem, wehave 0(9)dQ' =0(lb)a1Q 0(9) -2*rrsin9d9 =0(lb) -20sinlbdlb (9.124) where 9andlbrepresent thesame scattering angle butmeasured intheCMor LAB system, respectively, andwhere dQ'and dﬂrepresent thesame element of solid angle butmeasured intheCMorLAB system, respectively. Therefore, 0(9) and0(lb) arethedifferential cross sections forthescattering intheCMandLAB systems, respectively. Thus, sin9d9= 9- — 9.15 aw) <r<>Slnl//dw <2) The derivative d9/d lbcanbeevaluated byﬁrstreferring toFigure 9-11a andwrit- ing, from thesine law(and using Equations 9.63 and 9.65b) , sin(9 —lb) ml =$2Ex (9.126) Wesetthedifferential dx=0andﬁnd a adx=0=5—l‘idlb+£d9 which gives, after taking thepartial derivatives andcollecting terms, d9=Sin(9 —lb)coslb +1 dlb cos(9 —lb)sinlb Expanding sin(9—lb)andsimplifying, wehave d9= sin(9 g dlb cos(9 —lb)sin lb andso sin29= 9---i——— 9.1O-(‘I’) O-()cos(9 —11/)sin2 lb (27) Multiplying both sides ofEquation 9.126 bycoslbandthen adding cos(9 —lb)to both sides, wehave sin(9 —lb)coslb -w-" +cos(9 —lb)=xcoslb +cos(9 —lb)sinlb Expanding sin(9—lb)andcos(9 —lb)ontheleft-hand side, weobtain sin9_ =xcoslb+ cos(9—lb)sinlb 9.9SCATTERING CROSS SECTIONS 367 Substituting thisresult into Equation 9.127, 6_ 2 O"(l/I)=9(9).[X°°S‘£’O:(B°‘iS(¢) ‘M,(x<1) (9.128) And from Equation 9.126, wehave cos(9 —lb)=\/1—362811121/I Hence, [xcoslb +\/1- x2sin2lb]2 0(lb) =0(9) -t (9.129) \/1—x2S1I12lb Equation 9.126 canbeused towrite I9=sin*1(x sinlb)+lbl (9.130) Equations 9.129 and9.130 therefore specify thecross section entirely interms of theangle lb.*Forthegeneral case (i.e., foranarbitrary value ofx),theevalua- tion of0(lb) iscomplicated. Tables exist, however, sotheparticular cases canbe computed with relative ease.1 The transformation represented byEquations 9.129 and 9.130 assumes a simple form fortwocases. Forx=ml/m2 =1,wehave from Equation 9.71, 9= 21b,andEquation 9.129 becomes 0(lb) =0(9)|l;,:2,l,-4cos lb, ml=m2 (9.131) andforml<<m2,xE0,and9élb,sothat O'(1l/) EO'(9)l9=¢, m1<< "12 (9-132) EXAMPLE 9.1l Consider molecules ofradius Rlmoving toward theright with identical veloci- tiesscattering from dust particles ofradius R2thatareatrest. Consider both as hard spheres andﬁnd thedifferential andtotal cross sections. Solution. The dust particles areatrest, andwewillsolve thisscattering prob- lemintheLAB system. Consider thegeometry ofscattering inFigure 9-23. The particles with impact parameter bwill bescattered atangle lb.Similarly to Equation 9.119, incident particles entering within arange ofimpact parameters dbscatter into anangular range dlb,andwehave 2"n'bdb=—0(lb) -211'sinlbdlb (9.133) *TheSe equations apply notonly forelastic collisions butalsoforinelastic collisions (inwhich thein- ternal potential energy ofoneorboth oftheparticles isaltered asaresult oftheinteraction) ifthe parameter xiswritten asV/z/’l instead ofml/m2 (seeEquation 9.68). Note thatthepreceding equa- tions refer only totheusual case x<l. tSee, forexample, thetables byMarion etal.(Ma59). 368 9/DYNAMICS OFASYSTEM OFPARTICLES \ \ Ir‘ T T \___, R1 b\ \ \ B B ‘II’ \\ lb ___1_________IQ,—’II,’l.»"'I /(\l/I\ laI’|\"Q1/I\I/\II\I11$\1II\ IIIII \ \ \ \ FIGURE 9-23 Amolecule ofradius Rlapproaches adust particle ofradius R2from theleftandscatters atangle l,//. Inorder toﬁnd thedifferential cross section 0(lb) ,wehave toﬁnd therelation- ship between theimpact parameter bandscattering angle lb.Weseefrom Figure 9-23 that b=(Rl+R2)cosa,soweneed toﬁrstﬁnd therelationship be- tween angles aandlb. Look closely atFigure 9-23 toseethat2,8+lb=11',oz=lb+B—11-/2, and ' a=¢+(g-1;»)-g=% (9.134) Before using Equation 9.133, weneed toﬁnd thedifferential db.Wehave b=(Rl+R2)cos0z =(Rl+R2)cos% and db=—————(R‘J;R2)S1n£;‘dlII. Wenow insert theterms intoEquation 9.133 toﬁnd —2?w(Rl +R2)2 coslgsingdlb=—0(lb) -21Tsinlb dlb Ifwe usetheidentity, sinlb=2sin(lb/2)cos(lb/2) ,weﬁnally ﬁnd 0(1)=%1(R1+R92 (9-135) Weﬁrstnote thatthedifferential scattering cross section isisotropic, thescatter- ingisthesame inevery direction. This issomewhat surprising, because differ- ential scattering cross sections normally have anangular dependence. 9.10 RUTHERFORD SCATTERING FORMULA 369 Wewilldiscuss thetotal cross section inthenext section, butbrieﬂy itis proportional totheprobability that anyscattering takes place. Inorder todeter- mine thetotal cross section, wemust integrate Equation 9.135 over allpossible lb.Note thatwehave already done soover theazimuthal angle todetermine Equation 9.118. Wehave 0-,=l-Zgdo =l¢(¢l)do <=+sl.(:tI—4=—(R,+R2)221r sinl/ldlb 11' "_ vr 2 W=5(R,+R2)2Ls1n lbdlb=-5(Rl+R2)coslb U’) =7T(Rl ‘l'R2)2 This isprecisely what wewould expect forthetotal cross section forscattering oftwohard spheres. The maximum area occurs when themolecule anddust particle have justaglancing blow, angle a=0.The impact parameter willbe simply b=Rl+R2,andthearea is'rrb2. 9.10 Rutherford Scattering Formula* One ofthemost important problems that makes useoftheformulas developed inthepreceding section isthescattering ofcharged particles inaCoulomb or electrostatic ﬁeld. The potential forthiscase is U(r) =é (9.136) where k=qlqg/4"rrt-:0, with qland(12theamounts ofcharge thatthetwoparticles carry (kmay beeither positive ornegative, depending onwhether thecharges areofthesame oropposite sign; k>0corresponds toarepulsive force andk<0 toanattractive force). Equation 9.123 then becomes ' bd 6= (9.137)n.....\/r —(k/T6)r— b2 which canbeintegrated toobtain (seetheintegration ofEquation 8.38): cosG=mi (9.138) \/1+(K/b)2 *E.Rutherford, Phil. Mag. 21,669 (1911). 370 9/DYNAMICS OFASYSTEM OFPARTICLES where kKEi2T6 Equation 9.138 canberewritten as b2=K2C2II126 ButEquation 9.122 states that9=1r/2 —9/2, so b=Kcot(9/2) (9.139) Thus, E__51;.d9 2sin2(9/2) Equation 9.120 thus becomes K2 cot(9/2) U(6)=2sin9sin2(9/2) Now, sin9=2sin(9/2)cos(9/2) Hence, K2 1 0(9)T1'sin4(9/2) or 0(9) = (9.140)(4T6)2 s1n4(9/2) which istheRutherford scattering formula* and demonstrates thedependence oftheCMscattering cross section ontheinverse fourth power ofsin(9/2). Note that0(9) isindependent ofthesign ofk,sothattheform ofthescattering distri- bution isthesame foranattractive force asforarepulsive one. Itisalsorather re- markable thatthequantum-mechanical treatment ofCoulomb scattering leads to exactly thesame result asdoes theclassical derivation? This isindeed afortunate circumstance because, ifitwere otherwise, thedisagreement atthisearly stage be- tween classical theory andexperiment might have seriously delayed theprogress ofnuclear physics. Forthecase ml=m2,Equation 9.79 states that T6=$76,sothat k2 0(0)=H5Sinlw/2), ,=m2 (9.141) *This form ofthescattering lawwasveriﬁed fortheinteraction ofaparticles andheavy nuclei bythe experiments ofH.Geiger and E.Marsden, Phil Mag. 25,605 (1913). 1-N.Bohr showed thattheidentity oftheresults isaconsequence ofthe1/r2nature oftheforce; it cannot beexpected foranyother typeofforce law. 9.11ROCKET MOTION 371 Or,from Equation 9.131, k2coslb=E?E, ml=771,2 (9.142) Allthepreceding discussion applies tothecalculation ofdifferential scattering cross sections. Ifitisdesired toknow theprobability thatanyinteraction whatsoever willtakeplace, itisnecessary tointegrate 0(9) [or0(lb)] over allpossible scattering angles. Theresulting quantity iscalled thetotal scattering cross section (0,)andis equal totheeffective area ofthetarget particle forproducing ascattering event: 71' <1,=L0(9)dQ’ =21'rl00(9)sin 9d9 (9.143) where theintegration over 9runs from 0torr.The totalcross section isthesame intheLAB asintheCMsystem. Ifwewish toexpress thetotal cross section in terms ofanintegration over theLAB quantities, 0,=J'O'(l/f)d.Q then ifml<m2,lbalso runs from 0to11'.Ifml2mg,lbruns only uptolbmm, (given byEquation 9.77), andwehave ¢mi\x 0,=2'rrl 0(lb) sinlbdlb (9.144) 0 Ifweattempt tocalculate 0,forthecase ofRutherford scattering, weﬁnd thattheresult isinﬁnite.This occurs because theCoulomb potential, which varies as1/r,falls offsoslowly that, astheimpact parameter bbecomes indeﬁnitely large, thedecrease inscattering angle istooslow toprevent theintegral from di- verging. Wehave, however, pointed outinExample 8.6thattheCoulomb ﬁeld of areal atomic nucleus isscreened bythesurrounding electrons sothat thepoten- tialisetfectively cutoffatlarge distances. The evaluation ofthescattering cross section forascreened Coulomb potential according totheclassical theory is quite complicated and isnotdiscussed here; thequantum-mechanical treatment isactually easier forthiscase. 9.11Rocket Motion The motion ofasimple rocket isaninteresting application ofelementary Newtonian dynamics and might have been covered inChapter 2.However, we want toinclude more complicated rockets with exhaust masses and multiple rocket stages, sowedeferred thediscussion tothischapter onsystems ofparticles. The twocases weexamine arerocket motion infree space and thevertical as- cent ofrockets under gravity. The ﬁrst case requires anapplication ofthecon- servation oflinear momentum. The second case requires amore complicated application ofNewton’s Second Law. 372 9/DYNAMICS orASYSTEM orPARTICLES Free space F=O Vi dm'L Q m U Inertial reference system FIGURE 9-24 Arocket moves infreespace atvelocity v.Inthetime interval dt,amass dm'isejected from therocket engine with velocity uwith respect tothe rocket ship. Rocket Motion inFree Space Weassume here that therocket (space ship) moves under theinfluence ofno external forces. Wechoose aclosed system inwhich Newton’s Second Law can beapplied. Inouter space, themotion ofthespace ship must depend entirely onitsown energy. Itmoves bythereaction ofejecting mass athigh velocities. To conserve linear momentum, thespace ship willhave tomove intheopposite di- rection. The diagram ofthespace ship motion isshown inFigure 9-24. Atsome time t,theinstantaneous total mass ofthespace ship inm,andtheinstantaneous speed ofthespace ship isvwith respect toaninertial reference system. Weas- sume that allmotion isinthexdirection and eliminate thevector notation. During atime interval dt,apositive mass dm'isejected from therocket engine with aspeed —uwith respect tothespace ship. Immediately after themass dm'is ejected, thespace ship’s mass andspeed arem—dm'andv+dv,respectively. Initial momentum =mv (attime t) (9.145) Final momentum =(m—dm')(v +dv)+dm'(v —u)(attime t+dt) space shiplessdm' rocket exhaust dm' (9.146) Notice that thespeed oftheejected mass dm'with respect tothereference sys- tem isv—u.The conservation oflinear momentum requires that Equations 9.145 and9.146 beequal. There arenoexternal forces (IPCX, =0). pinitial =pﬁnal p(t)=p(t+dt) mv=(m—dm’)(-u +dv)+dm'(v —u) (9.147) m-u= m-0+ mdv— vdm' —dm'd-0+ vdm' —udm' mdv =udm' d I do=Ml (9.148)m 9.11ROCKET MOTION 373 where wehave neglected theproduct oftwodifferentials dm'dv. Wehave consid- ered dm'tobeapositive mass ejected from thespace ship. The change inmass ofthespace ship itself isdm,where dm=—dm' (9.149) and ddv=—U-1-” (9.150)m because dmmust benegative. Letm6andv6betheinitial mass andspeed ofthe space ship, respectively, andintegrate Equation 9.150 toitsﬁnal values mandv. ‘U Md ldv= -16‘, -LL "0 mom 9-~06=911.6%) (9.151) v=v6+uln (9.152) The exhaust velocity uisassumed constant. Thus, tomaximize thespace ship’s speed, weneed tomaximize theexhaust velocity uandtheratio m6/m. Because theterminal speed islimited bytheratio m6/m,engineers have con- structed multistage rockets. The minimum mass (less fuel) ofthespace ship is limited bystructural material. However, ifthefuel container itself isjettisoned after itsfuelhasbeen burned, themass oftheremaining space ship iseven less. The space ship cancontain twoormore fuel containers, each ofwhich canbe jettisoned. Forexample, let m6=Initial total mass ofspace ship ml=ma‘l'mb ma=Mass ofﬁrst-stage payload ml,=Mass ofﬁrst-stage fuelcontainers, etc. vl=Terminal speed ofﬁrststage of“burnout” after allfuelisburned vl=-06+uln<%) (9.153) ml Atbumout, theterminal speed vloftheﬁrststage isreached, andthemass mbis released into space. Next, thesecond-stage rocket ignites with thesame exhaust 374 9/DYNAMICS OFASYSTEM orPARTICLES velocity, andwehave ma=Initial total mass ofspace ship second stage "Z2=m,+ml, m,=Mass ofsecond-stage payload md=Mass ofsecond-stage fuelcontainer, etc. vl=Initial speed ofsecond stage -02=Terminal speed ofsecond stage atburnout ‘U2=5,+uln (9.154)2 mo ma v2=v6+uln i (9.155) mil"/2 The product (m6m,,/mlmg) can bemade much larger than just mo/ml. Multistage rockets aremore commonly used inascent under gravity than infree space. Wehave seen that thespace ship ispropelled asaresult oftheconservation oflinear momentum. Butengineers andscientists liketorefer totheforce term46 asrocket thrust.” Ifwemultiply Equation 9.150 bymanddivide bydtwehave d dmi’=-ll-3’ (9.156)dt dt Since theleftsideofthisequation “appears” asma(force) ,theright sideiscalled thrust: dThrust E-9-;-Z3 (9.157) Because dm/dt isnegative, thethrust isactually positive. Vertical Ascent Under Gravity The actual motion ofarocket attempting toleave Earth’s gravitational ﬁeld is quite complicated. Foranalytical purposes, webegin bymaking several assump- tions. The rocket willhave only vertical motion, with nohorizontal component. Weneglect airresistance andassume that theacceleration ofgravity isconstant with height. Wealsoassume that theburn rate ofthefuel isconstant. Allthese factors that areneglected canreasonably beincluded with anumerical analysis bycomputer. Wecanusetheresults oftheprevious case ofrocket motion infree space, butwenolonger have Ex,=0.The geometry isshown inFigure 9-25. Weagain have dm'aspositive, with dm=—dm'. The external force Ifmis d 1'15 = 9.11ROCKET MOTION 375 9 V ls dm'° ul FIGURE 9-25 Arocket invertical ascent under Earth’s gravity. Mass dm'isejected from therocket engine with velocity uwith respect totherocket ship.Earth OI‘ 1~;,,,dl=d(mv)=dp=p(t+dt)—p(t) (9.158) over asmall differential time. For thespace ship system, wefound theinitial and ﬁnal momenta in Equations 9.l45—9.150. Wenow usetheresults leading uptoEquation 9.150 to obtain p(t+dt)—p(t) =mdv +udm (9.159) Infree space, EX,=0,butinascent, Em=—mg. Combining Equations 9.158 and9.159 gives E,X,dt =—mgdt =mdv+udm —mg =ml)+um (9.160) Because thefuelburn rateisconstant, let dmm=-3;=—a, a>0 (9.161) andEquation 9.160 becomes adv—<—g+ ;tu)dt This equation, however, hasthree unknowns (v,m,t),soweuseEquation 9.161 toeliminate time, giving dv=(5—3)dm. (9.162)a 171. 376 9/DYNAMICS OFASYSTEM OFPARTICLES Assume theinitial andﬁnal values ofthevelocity tobe0and vrespectively and ofthemass m6andmrespectively, sothat Idv=I —E)dm0 mo Q! Tn g m0=—;(m6 —m)+u1n(;°) (9.163) Wecanintegrate Equation 9.161 toﬁnd thetime: m t ldm =—aldt mo 0 m6—m=at (9.164) Equation 9.163 becomes 9=—g't+064%) (9.165) Wecould continue with Equation 9.163 and integrate once more todeter- mine theheight oftherocket, butweleave that toExample 9.13 and theprob- lems. Such integrations aretedious, andtheproblem ismore easily handled by computer methods. Even atburnout, therocket willcontinue rising because it stillhasanupward velocity. Eventually, with thepreceding assumptions, thegrav- itational force willstop therocket (because weassumed aconstant gnotde- creasing with height). Aninteresting situation occurs iftheexhaust velocity uisnot sufﬁciently great tomake vinEquation 9.165 positive. Inthiscase, therocket would remain ontheground. This situation occurs because ofthelimits ofintegration weas- sumed leading toEquation 9.163. Wewould need toburn offsufﬁcient fuelbe- fore therocket thrust would liftitofftheground (seeProblem 9-59). Ofcourse, rockets arenotdesigned thisway; they aremade toliftoffastherockets reach a fullburn rate. Consider theﬁrststage ofaSatum Vrocket used fortheApollo moon program. The initial mass is2.8X106kg,andthemass oftheﬁrst-stage fuelis2.1X106 kg.Assume amean thrust of37X106N.The exhaust velocity is2600 m/s. Calculate theﬁnal speed oftheﬁrststage atburnout. Using theresult of Problem 9-57 (Equation 9.166), also calculate thevertical height atburnout. Solution. From thethrust (Equation 9.157), wecandetermine thefuelburn rate: dm thrust 37X106N= = =-1.4 1041<at -0 -2600 m/s 2X g/S 9.11ROCKET MOTION 377 The ﬁnal rocket mass is(2.8 X106kg—2.1X105kg)or0.7><105kg.Wecan determine therocket speed atburnout (vb)using Equation 9.163. 9.8m/s2(2.1 ><106kg) 2.8><106kg vb__ 1.42T>< 1O4kg/s +(2600 m/s) In0.7X106kg vl,=2.16 X103m/s The time toburnout tb,from Equation 9.164, is m6—m 2.1X106kgl,= = =148$C! 1.42 X1O4kg/s orabout 2.5min. Weusetheresult ofProblem 9-57 toobtain theheight atburnout yl: 1 y,=ul,,--églg-%‘1n(%) (9.166) 1 yb=(2600 m/s) (148 s)—“(9.8 I11/S2) -(148 s)2 2 (0.7X106kg) -(2600m/s)l (2.8X106kg) 1.42X1041<g/S n0.7X1O6kg yb=9.98 X104m=100km The actual height isonl about two-thirds ofthisvalue. Y Asounding rocket leaves Earth ’ssurface under gravity, typically inavertical direction andretums toEarth. Theexhaust velocity isii,andtheconstant fuel burn rateisa.Lettheinitial mass bem6andthemass atfuelburnout bemf. Calculate thealtitude andspeed oftherocket atfuelburnout interms of u,oz,ml,m6,andg. Solution. Wedetermine thetime Tatburnout from Equation 9.164, T=(m6—mf)/oz. Weintegrate over thevelocity, Equation 9.165, toﬁnd Hbo, theheight atfuelburn out. T T m_ _ 0H6,—Lvdt— Lli—g't+ uln<;)]dt WeuseEquation 9.161, dt=—(dm)/ct, forthelastintegral and integrate over dm. T my H)",=—gl dt+El ln<ﬂ)dm 0 aml) m0 378 9/DYNAMICS orASYSTEM orPARTICLES Weintegrate thelastterm using thedeﬁnite integral, flnxdx=xlnx—x,to obtain after collecting terms, g(m(> _Inf)? u mfHbu =———* '1' '1'mo — Ifweinsert thenumbers from thelastexample, weﬁnd thesame answer forthe burnout height. The speed atburnout canbedetermined directly from Equation 9.165. ——:r+ 1@ 'UbO— g un mf =-giw +016(E) (9.168)01 mf PROBLEMS 9-1. Find thecenter ofmass ofahemispherical shell ofconstant density andinner ra- dius rlandouter radius 1'2. 9-2. Find thecenter ofmass ofauniformly solid cone ofbase diameter 2aandheight h. 9-3. Find thecenter ofmass ofauniformly solid cone ofbase diameter 2aandheight h andasolid hemisphere ofradius awhere thetwobases aretouching. 9-4. Find thecenter ofmass ofauniform wire thatsubtends anarc9iftheradius ofthe circular arcisa,asshown inFigure 9-A. I /I/ 0/’ //I /// Q, 2 <\\ H x\ _ \2\ \ \ \ \ \ \ FIGURE 9-A Problem 9-4. 9-5. Thecenter ofgravity ofasystem ofparticles isthepoint about which external grav- itational forces exert nonettorque. Forauniform gravitational force, show that thecenter ofgravity isidentical tothecenter ofmass forthesystem ofparticles. 9-6. Consider twoparticles ofequal mass m.The forces ontheparticles areFl=0and F2=F6i.Iftheparticles areinitially atrestattheorigin, what istheposition, veloc- ity,andacceleration ofthecenter ofmass? PROBLEMS 379 9-7. 9-8. 9-9. 9-10. 9-11 9-12Amodel ofthewater molecule H20 isshown inFigure 9-B.Where isthecenter of mass? y/‘H I I a,’/ / I’ D .1 52/ (X o.\52° \ \\G\\ \ \ l \ 0H FIGURE 9-B Problem 9-7. Where isthecenter ofmass oftheisosceles right triangle ofuniform areal density shown inFigure 9-C? 3' G Cl J6 FIGURE 9-C Problem 9-8. Aprojectile isﬁred atanangle of45°with initial kinetic energy E6.Atthetopofits trajectory, theprojectile explodes with additional energy E6into twofragments. One fragment ofmass mltravels straight down. 1/Vhat isthevelocity (magnitude and direction) ofthesecond fragment ofmass ml,and thevelocity oftheﬁrst? What istheratio ofml/mg when mlisamaximum? Acannon inafortoverlooking theocean ﬁres ashell ofmass Matanelevation angle 9and muzzle velocity v6.Atthehighest point, theshell explodes into two fragments (masses ml+m2=M), with anadditional energy E,traveling inthe original horizontal direction. Find thedistance separating thetwofragments when they land intheocean. Forsimplicity, assume thecannon isatsealevel. Verify thatthesecond term ontheright-hand sideofEquation 9.9indeed vanishes forthecase n=3. Astronaut Stumblebum wanders toofaraway from thespace shuttle orbiter while repairing abroken communications satellite. Stumblebum realizes thattheorbiter ismoving away from himat3m/s. Stumblebum andhismaneuvering unit have a mass of100kg,including apressurized tank ofmass 10kg.The tank includes only 2kgofgasthatisused topropel him inspace. The gasescapes with aconstant ve- locity of100m/s. 380 9-13 9-14 9-15 9-16 9-17. 9-18. 9-19.9/DYNAMICS OFASYSTEM OFPARTICLES (a)Will Stumblebum runoutofgasbefore hereaches theorbiter? (b)With what velocity willStumblebum have tothrow theempty tank away toreach theorbiter? Even though thetotal force onasystem ofparticles (Equation 9.9) iszero, thenet torque maynotbezero. Show thatthenettorque hasthesame value inanycoordi- nate system. Consider asystem ofparticles interacting bymagnetic forces. AreEquations 9.11 and9.31 valid? Explain. Asmooth rope isplaced above ahole inatable (Figure 9-D). One endoftherope fallsthrough thehole att=0,pulling steadily ontheremainder oftherope. Find thevelocity andacceleration oftherope asafunction ofthedistance totheendof therope x.Ignore allfriction. The total length oftherope isL. I3NM .3 !<’:v’“".:|.,;%5‘£ﬂI::r, _I " I I ... KI ea’ !zta; _5 it_. .-\ FIGURE 9-D Problem 9-15. Fortheenergy-conserving case ofthefalling chain inExample 9.2,show that the tension oneither sideofthebottom bend isequal andhasthevalue p222/4. Integrate Equation 9.17 inExample 9.2numerically and make aplot of thespeed versus thetime using dimensionless parameters, kw vs.t/‘V2b/g where V2b/gisthefree falltime, tl,"fall.Find thetime ittakes forthefree endto reach thebottom. Deﬁne natural units by1-Et\/g/2b,a Ex/2b and integrate dr/da from a=s(some small number greater than 0)tooz=1/2.One can’t in- tegrate numerically from a=0because ofasingularity ind'r/da. The expression dr/da is Q1_l1-20: da 20z(1 -—oz) Useacomputer tomake aplot ofthetension versus time forthefalling chain in Example 9.2. Use dimensionless parameters (T/Mg) versus t/tl,“ ﬁll,where tlmﬁll=V2b/g. Stop theplotbefore T/Mgbecomes greater than 50. Achain such astheoneinExample 9.2(with thesame parameters) oflength band mass pbissuspended from oneendatapoint thatisaheight babove atable sothat PROBLEMS 381 9-20. 9-21. 9-22. 9-23. 9-24. 9-25.thefreeendbarely touches thetabletop. Attime t=0,theﬁxed endofthechain is released. Find theforce that thetabletop exerts onthechain after theoriginal ﬁxed endhasfallen adistance x. Auniform rope oftotal length 2ahangs inequilibrium over asmooth nail. Avery small impulse causes therope toslowly rolloffthenail. Find thevelocity ofthe rope asitjust clears thenail. Assume therope isprevented from lifting offthenail andisinfreefall. Aﬂexible rope oflength 1.0mslides from africtionless table topasshown inFigure 9-E.Therope isinitially released from restwith 30cmhanging over theedge ofthe table. Find thetime atwhich theleftendoftherope reaches theedge ofthetable. ......“-\\.....-\_._,\‘,., 3 .. , ll. 1ll= It ; 12-‘ 11 l FIGURE 9-E Problem 9-21. Adeuteron (nucleus ofdeuterium atom consisting ofaproton andaneutron) with speed 14.9 km/s collides elastically with aneutron atrest. Use theapproximation thatthedeuteron istwice themass oftheneutron. (a)Ifthedeuteron isscattered through aLAB angle lb=10°,what aretheﬁnal speeds ofthedeuteron andneu- tron? (b)\/Vhat istheLAB scattering angle oftheneutron? (c)What isthemaxi- mum possible scattering angle ofthedeuteron? Aparticle ofmass mlandvelocity ulcollides with aparticle ofmass m2atrest. The twoparticles stick together. Vlrhat fraction oftheoriginal kinetic energy islostin thecollision? Aparticle ofmass mattheendofalight string wraps itself about aﬁxed vertical cylinder ofradius a(Figure 9-F). Allthemotion isinthehorizontal plane (disre- gard gravity). The angular velocity ofthecord is(06when thedistance from thepar- ticle tothepoint ofcontact ofthestring andcylinder isb.Find theangular velocity andtension inthestring after thecord hasturned through anadditional angle 9. &t>0 a) \ Motion 9> Om t=0 5 .60 FIGURE 9-F Problem 9-24. Slow-moving neutrons have amuch larger absorption rate in255U than fast neu- trons produced by235U' ﬁssion inanuclear reactor. Forthat reason, reactors con- sistofmoderators toslow down neutrons byelastic collisions. What elements are besttobeused asmoderators? Explain. 382 9-26 9-27 9-28 9-29 9-30 9-31. 9-32. 9-33.9/DYNAMICS OFASYSTEM OFPARTICLES Theforce ofattraction between twoparticles isgiven by ft.=lilo.—r.)-10.—mlU0 where kisaconstant, 116isaconstant velocity, and1"E|r2—rl|.Calculate theinternal torque forthesystem; whydoes thisquantity notvanish? Isthesystem conservative? Derive Equation 9.90. Aparticle ofmass mlelastically collides with aparticle ofmass mlatrest. What is themaximum fraction ofkinetic energy lossforml?Describe thereaction. Derive Equation 9.91. Atennis player strikes anincoming tennis ballofmass 60gasshown inFigure 9-G. The incoming tennis ball velocity is11,-=8m/s,and theoutgoing velocity is vf=16m/s. (a)What impulse wasgiven tothetennis ball? (b)Ifthecollision time was0.01 s,what wastheaverage force exerted bythetennis racket? \\ vf \ \-15°\ .-1 \\ -2l 0 \\ 45 =>.1;<’/l V1 .l=_l_ =35’.l.~. ..-lg.33. FIGURE 9-G Problem 9-30. Derive Equation 9.92. Aparticle ofmass mandvelocity ulmakes ahead-on collision with another particle ofmass 2matrest. Ifthecoefﬁcient ofrestitution issuch tomake thelossoftotal ki- netic energy amaximum, what arethevelocities vlandv2after thecollision? Show that 71/T6canbeexpressed intemis ofm2/mlEozandcoslbEyas T1 -2 2 2 \/2W?=(1+a) (2)+0.-1+2) 5+y2—1)0 Plot Tl/T6 asafunction oflbfora=1,2,4,and12.These plots correspond tothe energies ofprotons orneutrons after scattering from hydrogen (oz=1),deuterium (a=2),helium (oz=4),andcarbon (oz=12),orofalpha particles scattered from helium (oz=1),oxygen (a=4),and soforth. PROBLEMS 383 9-34. 9-35. 9-36 9-37 9-38. 9-39 9-40.Abilliard ballofinitial velocity ulcollides with another billiard ball(same mass) initially atrest.Theﬁrstballmoves offatlb=45°.Foranelastic collision, what aretheveloci- tiesofboth balls after thecollision? Atwhat LAB angle does thesecond ballemerge? Aparticle ofmass mlwithinitial laboratory velocity ulcollides withaparticle ofmass ml atrestintheLAB system. Theparticle mlisscattered through aLAB angle lbandhasa ﬁnal velocity vl,where vl=vl(lb).Find thesurface such thatthetime oftravel ofthe scattered particle from thepoint ofcollision tothesurface isindependent ofthescat- tering angle. Consider thecases (a)ml=ml,(b)ml==2ml,and(c)m2=oo.Suggest anapplication ofthisresult interms ofadetector fornuclear particles. Inanelastic collision oftwoparticles with masses mland m2,theinitial velocities areulandu2=ozul. Iftheinitial kinetic energies ofthetwoparticles areequal, ﬁnd theconditions onul/ul and ml/m2 such that mlisatrestafter thecollision. Examine both cases forthesign ofa. VVhen abullet ﬁres inagun, theexplosion subsides quickly. Suppose theforce on thebullet isF=(360 —107t2s'2) Nuntil theforce becomes zero (and remains zero). The mass ofthebullet is3g. (a)1/Vhat impulse actsonthebullet? (b)W'hat isthemuzzle velocity ofthegun? Show that 5=mi 2.S2 To (mi+m2) where SEcoslb +-L09 —W) ml m2 Aparticle ofmass mstrikes asmooth wallatanangle 9from thenormal. The coef- ﬁcient ofrestitution iss.Find thevelocity and therebound angle oftheparticle after leaving thewall. Aparticle ofmass mlandvelocity ulstrikes head-on aparticle ofmass m2atrest. The coefﬁcient ofrestitution iss.Particle mlistiedtoapoint adistance aaway as shown inFigure 9-H. Find thevelocity (magnitude and direction) ofmland m2 after thecollision. ll Oma&{// // ///// H1 ml FIGURE 9-H Problem 9-40. 384 9-41. 9-42. 9-43. 9-44 9-45. 9-46 9-47. 9-48. 9-49 9-50. 9-519/DYNAMICS OFASYSTEM OFPARTICLES Arubber ballisdropped from restonto alinoleum ﬂoor adistance hlaway. The rubber ballbounces uptoaheight hq.What isthecoefﬁcient ofrestitution? VVhat fraction oftheoriginal kinetic energy islostinterms of2? Asteel ballofvelocity 5m/sstrikes asmooth, heavy steel plate atanangle of30° from thenormal. Ifthecoefﬁcient ofrestitution is0.8,atwhat angle andvelocity does thesteel ballbounce offtheplate? Aproton (mass m)ofkinetic energy Tocollides with ahelium nucleus (mass 4m)at rest. Find therecoil angle ofthehelium if4/1=45°andtheinelastic collision has Q=—nm. Auniformly dense rope oflength bandmass density ,u.iscoiled onasmooth table. One endislifted byhand with aconstant velocity v0.Find theforce oftherope held bythehand when therope isadistance aabove thetable (b>a). Show thattheequivalent ofEquation 9.129 expressed interms of6rather than tpis 1+xcos6 0-(6) =0-(I/I) l(1+2xcos0+x2)5/2 Calculate thedifferential cross section 0(0) and thetotal cross section 0',forthe elastic scattering ofaparticle from animpenetrable sphere; thepotential isgiven by - 0 r>aU = ’(T) {oo, r<a Show that theRutherford scattering cross Section (for thecase ml=m2)canbeex- pressed interms oftherecoil angle as k2 1 T§cos5 §<TtAB(§) = Consider thecaseofRutherford scattering intheevent thatml>>m2.Obtain anap- proximate expression forthedifferential cross section intheLAB coordinate system. Consider thecase ofRutherford scattering intheevent that m2>>ml.Obtain an expression ofthedifferential cross section intheCMsystem thatiscorrect toﬁrst order inthequantity m1/mg.Compare thisresult with Equation 9.140. Aﬁxed force center scatters aparticle ofmass maccording totheforce law F(r) =k/F’. Iftheinitial velocity oftheparticle isuo,show thatthedifferential scat- tering cross section is 0(6) = k1r2(1r —6) mu§t92(2'n' —6)?sin0 Itisfound experimentally that intheelastic scattering ofneutrons byprotons (mug mp)atrelatively lowenergies, theenergy distribution oftherecoiling pro- tons intheLAB system isconstant uptoamaximum energy, which istheenergy of theincident neutrons. \/Vhat istheangular distribution ofthescattering intheCM system? PROBLEMS 385 9-52 9-53 9-54 9-55. 9-56 9-57 9-58 9-59 9-60Show that theenergy distribution ofparticles recoiling from anelastic collision is always directly proportional tothedifferential scattering cross section intheCM system. The most energetic a-particles available toErnest Rutherford andhiscolleagues forthefamous Rutherford scattering experiment were 7.7MeV. Forthescatter- ingof7.7MeV oz-particles from 258U (initially atrest) atascattering angle inthe labof90°(allcalculations areintheLAB system unless otherwise noted), ﬁnd thefollowing: (a)therecoil scattering angle of258U. (b)thescattering angles ofthea-particle and238UintheCMsystem. (c)thekinetic energies ofthescattered oz-particle and238U. (d)theimpact parameter b. (e)thedistance ofclosest approach rmin. (f)thedifferential cross section at90°. (g)theratio oftheprobabilities ofscattering at90°tothatof5°. Arocket starts from restinfreespace byemitting mass. Atwhat fraction oftheini- tialmass isthemomentum amaximum? Anextremely well-constructed rocket hasamass ratio (mo/m)of10.Anewfuelis developed thathasanexhaust velocity ashigh as4500 m/s.Thefuelburns atacon- stant ratefor300s.Calculate themaximum velocity ofthissingle-stage rocket, as- suming constant acceleration ofgravity. Iftheescape velocity ofaparticle from the earth is11.3km/s,canasimilar single-stage rocket with thesame mass ratio andex- haust velocity beconstructed thatcanreach themoon? Awater droplet falling intheatmosphere issphelical. Assume that asthedroplet passes through acloud, itacquires mass atarate equal tokAwhere kisacon- stant(>0) andAitscross-sectional area. Consider adroplet ofinitial radius 1'0that enters acloud with avelocity v0.Assume noresistive force andshow (a)thatthera- dius increases linearly with thetime, and (b)that if10isnegligibly small then the speed increases linearly with thetime within thecloud. Arocket inouter space inanegligible gravitational ﬁeld starts from restandaccel- erates uniformly atauntil itsﬁnal speed isv.The initial mass oftherocket ismo. How much work does therocket’s engine do? Consider asingle-stage rocket taking offfrom Earth. Show that theheight ofthe rocket atburnout isgiven byEquation 9.166. How much farther inheight willthe rocket goafter burnout? Arocket hasaninitial mass ofm.andafuelburn rateofa(Equation 9.161). VVhat is theminimum exhaust velocity thatwillallow therocket toliftoffimmediately after ﬁring? Arocket hasaninitial mass of7><104kgandonﬁring burns itsfuelatarateof250 kg/s.The exhaust velocity is2500 m/s.Iftherocket hasavertical ascent from rest- ingontheearth, howlong after therocket engines ﬁrewilltherocket liftoff?What iswrong with thedesign ofthisrocket? 386 9-61 9-62 9-63. 9-64 9-65. 9-66. 9-67.9/DYNAMICS OFASYSTEM OFPARTICLES Consider amultistage rocket ofnstages, each withexhaust speed u.Each stage of therocket hasthesame mass ratio atburnout (k=m,-/mf). Show that theﬁnal speed ofthenthstage isnulnk. Toperform arescue, alunar landing craft needs tohoverjust above thesurface of themoon, which hasagravitational acceleration ofg/6.The exhaust velocity is 2000 m/s,butfuel amounting toonly 20percent ofthetotal mass may beused. How long canthelanding craft hover? Anew projectile launcher isdeveloped intheyear 2023 that canlaunch a104kg spherical probe with aninitial speed of6000 m/s.Fortesting purposes, objects are launched vertically. (a)Neglect airresistance andassume that theacceleration ofgravity isconstant. Determine how high thelaunched object canreach above thesurface ofEarth. (b)Iftheobject hasaradius of20cmandtheairresistance isproportional tothe square oftheobject’s speed with cw=0.2,determine themaximum height reached. Assume thedensity ofairisconstant. (c)Now alsoinclude thefactthattheacceleration ofgravity decreases astheobject soars above Earth. Find theheight reached. (d)Now addtheeffects ofthedecrease inairdensity with altitude tothecalcula- tion. Wecanvery roughly represent theairdensity byl0g10(p) =—0.05h +0.11 where pistheairdensity inkg/m5 and histhealtitude above Earth inkm. Determine how high theobject now goes. Anewsingle-stage rocket isdeveloped intheyear 2023, having agasexhaust veloc- ityof4000 m/s.The total mass oftherocket is105kg,with 90% ofitsmass being fuel. The fuel burns quickly in100sataconstant rate. Fortesting purposes, the rocket islaunched vertically atrestfrom Earth’s surface. Answer parts (a)through (d)oftheprevious problem. Inatypical model rocket (Estes Alpha III)theEstes C6solid rocket engine provides atotal impulse of8.5N-s.Assume thetotal rocket mass atlaunch is54gandthatit hasarocket engine ofmass 20gthatbums evenly for1.5s.The rocket diameter is 24mm. Assume aconstant burn rateofthepropellent mass (11g),arocket exhaust speed 800m/s,vertical ascent, anddrag coefﬁcient cw=0.75. Detemiine (a)Thespeed andaltitude atengine burnout, (b)Maximum height and time itoccurs, (c)Maximum acceleration, (d)Total ﬂight time, and (e)Speed atground impact. Produce aplot ofaltitude andspeed versus time. Forsimplicity, because thepro- pellent mass isonly 20% ofthetotal mass, assume aconstant mass during rocket burning. Fortheprevious problem, take into account thechange ofrocket mass with time andomit theeffect ofgravity. (a)Find therocket’s speed atburn out. (b)How far hastherocket traveled atthatmoment? Complete thederivation fortheburnout height H,,,,inExample 9.13. Usethenum- bers fortheSatum Vrocket inExample 9.12 anduseEquations 9.167 and9.168 to determine theheight andspeed atburnout. g .........1UMotion inaNoninertial Reference Frame 10.1Introduction The advantage ofchoosing aninertial reference frame todescribe dynamic processes wasmade evident inthediscussions inChapters 2and7.Itisalways possi- bletoexpress theequations ofmotion forasystem inaninertial frame. Butthere aretypes ofproblems forwhich these equations would beextremely complex, andit becomes easier totreat themotion ofthesystem inanoninertial frame ofreference. Todescribe, forexample, themotion ofaparticle onornear thesurface of Earth, itistempting todosobychoosing acoordinate system ﬁxed with respect toEarth. Weknow, however, that Earth undergoes acomplicated motion, com- pounded ofmany different rotations (and hence accelerations) with respect to aninertial reference frame identiﬁed with the“ﬁxed” stars. Earth’s coordinate system is,therefore, anoninertial frame ofreference; and, although thesolutions tomany problems canbeobtained tothedesired degree ofaccuracy byignoring thisdistinction, many important effects result from thenoninertial nature ofthe Earth coordinate system. Infact, wehave already studied noninertial systems when westudied ocean tides (Section 5.5). Tidal forces due toEarth-Moon and Sun-Earth orbits are observed onEarth’s surface, which isanoninertial system. Space does notallow usfurther study inthischapter ofthisinteresting subject, butquite reasonable accounts canbefound elsewhere.* *See, forexample, Knudsen andHjorth (Kn00, Chapter 6)andM.S.Tiersten andH.Soodak, Am.]. Phys. 68,129(2000). 387 388 10/MOTION INANONINERTIAL REFERENCE FRAME xé xg P r r’ x2 R xé xl xiFIGURE 10-1 The x,-'arecoordinates intheﬁxed system, andxiarecoordinates in therotating system. Thevector Rlocates theorigin oftherotating system intheﬁxed system. Inanalyzing themotion ofrigid bodies inthefollowing chapter, wealsoﬁnd itconvenient tousenoninertial reference frames and therefore make useof much ofthedevelopment presented here. 10.2 Rotating Coordinate Systems Letusconsider twosetsofcoordinate axes. Letonesetbethe“ﬁxed” orinertial axes, andlettheother beanarbitrary setthat may beinmotion with respect to theinertial system. Wedesignate these axes asthe“ﬁxed” and“rotating” axes, re- spectively. Weusex,5ascoordinates intheﬁxed system and xiascoordinates in therotating system. Ifwechoose some point P,asinFigure 10-1, wehave r’=R+r (10.1) where r’istheradius vector ofPin theﬁxed system andristheradius vector of Pintherotating system. The vector Rlocates theorigin oftherotating system in theﬁxed system. Wemay always represent anarbitraiy inﬁnitesimal displacement byapure rotation about some axiscalled theinstantaneous axis ofrotation. Forexample, theinstantaneous motion ofadisk rolling down aninclined plane canbede- scribed asarotation about thepoint ofcontact between thedisk andtheplane. Therefore, ifthex,-system undergoes aninﬁnitesimal rotation 50,correspon- ding tosome arbitrary inﬁnitesimal displacement, themotion ofP(which, for themoment, weconsider tobeatrestinthex,system) canbedescribed interms ofEquation 1.106 as (dloﬁxed = X1' where thedesignation “ﬁxed” isexplicitly included toindicate that thequantity drismeasured inthexf,orﬁxed, coordinate system. Dividing thisequation bydt, thetime interval during which theinﬁnitesimal rotation takes place, weobtain thetime ofratechange ofrasmeasured intheﬁxed coordinate system: dr d0 — =— 10.3(dtjﬁxed dtxr ( ) 10.2ROTATING COORDINATE SYSTEMS 389 or,because theangular velocity oftherotation is d0E—— 10.4 wdt () Wehave d=0.)Xr (forPﬁxed in.~x,-system) (10.5) ﬁxed This same result wasdetermined inSection 1.15. Ifweallow thepoint Ptohave avelocity (dr/dt),oming with respect tothex,- system, thisvelocity must beadded to0.)Xrtoobtain thetime rateofchange of £15 —55 +0.)><r (106)dtﬁxed dtrotating . l<LX.~\Ml-’l.l<Il0.l _——. — -_ - -- Consider avector 1-=xlel +x2e? +xsesintherotating system. Lettheﬁxed androtating systems have thesame origin. Find i-'intheﬁxed system bydirect differentiation iftheangular velocity oftherotating system isoointheﬁxed system.rintheﬁxed system: Solution. Webegin bytaking thetime derivative directly dr d (dt)ﬁxed Tdb Xiei) =§(t,e,. +x,-e,) (10.7) Theﬁrstterm issimply i‘,intherotating system, butwhat aretheéi? _ drl': " r dtrotating d(E5) =i-,+zx,-é,» (10.8)tﬁxed Look atFigure 10-2andexamine which components ofco,tend torotate e1. Weseethat(02tends torotate e1toward the—e5direction andthatmstends to rotate e1toward the+e2direction. Wetherefore have deTil ':(1)562 T"(1)283 390 10/MOTION INANONINERTIAL REFERENCE FRAME xs ms es 91 x82 2 ‘"2 ‘"1 xl FIGURE 10-2 Theangular velocity components to,rotate thesystem around thee,-axis, sothat, forexample, (D3tends torotate eltoward the+e2direction. Similarly, wehave d%=-w,e, +w,e,, (10.9b) de73 :“(1)281 ‘_'(1)182 Ineach case, thedirection ofthetime derivative oftheunit vector must beper- pendicular totheunit vector inorder nottochange itsmagnitude. Equations 10.9a—c canbewritten éi=toXel. (10.10) andEquation 10.8 becomes d(-5) =i‘,+21.0 Xx,e, diﬁxed =i-,+0.»Xr (10.11) which isthesame result asEquation 10.6. Although wechoose thedisplacement vector 1-forthederivation ofEquation 10.6, thevalidity ofthisexpression isnotlimited tothevector r.Infact, foranar- bitrary vector Q,wehave Q _@ (dt )ﬁxed ___ (dt )rotating +0’ XQ Equation 10.12 isanimportant result. Wenote, forexample, that theangular acceleration ti)isthesame inboth theﬁxed androtating systems: d d .(.33) = +.,,><noEno (10.13)dt ﬁxed dt rotating because 0.)X0.)vanishes and6:designates thecommon value inthetwosystems. 10.3 CENTRIFUGAL AND CORIOLIS FORCES 391 Equation 10.12 maynow beused toobtain theexpressions forthevelocity of thepoint Pasmeasured intheﬁxed coordinate system. From Equation 10.1, we have dtﬁxed difixed dtfixed sothat ’ R d(di) =(L) +(-5) +00><r (10.15)dt ﬁxed dt ﬁxed dt rotating Ifwedeﬁne __dr’vj-E rf= (Elm d (10.16a) .anvERf;(zlxed (10.16b) .1v,E1,5 (10.16c)rotating wemaywrite vf= V+v,+0.)Xr (10.17) where vf=Velocity relative totheﬁxed axes V=Linear velocity ofthemoving origin v,=Velocity relative totherotating axes 00=Angular velocity oftherotating axes 0.)Xr=Velocity duetotherotation ofthemoving axes 10.3 Centrifugal and Coriolis Forces Wehave seen thatNewton’s equation F=maisvalid only inaninertial frame of reference. The expression fortheforce onaparticle can therefore beobtained from <0F=ma=—- (10.18) m f dtﬁxed where thedifferentiation must becarried outwith respect totheﬁxed system. Differentiating Equation 10.17, wehave (@) =(ﬂ) +(d—"') +a.><r+e,><(5i5) (1019)dbﬁXed dtﬁxed dtﬁxed dtﬁxed . 392 10/MOTION INANONINERTIAL REFERENCE FRAME Wedenote theﬁrstterm byRf: .. JV RfE (10.20) ﬁxed The second term canbeevaluated bysubstituting v,forQinEquation 10.12: “"""' = “‘""' 0.) V,(dv <dv +X dt ﬁxed dt rotating =a,+0.)Xv, (10.21) where a,istheacceleration intherotating coordinate system. The lastterm in Equation 10.19 canbeobtained directly from Equation 10.6: wxfil *wX(£l£) +wX(o.)Xr) dtﬁxed dtrotating =wxv,+wx (ooxr) (10.22) Combining Equations 10.18-10.22, weobtain F=maf= mRf+ ma,+ mtbXr+mmX(0.)X1-)+2mo.) Xv,(10.23) Toanobserver intherotating coordinate system, however, theeffective force onaparticle isgiven by* F...Ema, (10.24) =F—mRf—m6.)Xr—-m00X(o0Xr)—2mwXv, (10.25) The ﬁrstterm, F,isthesum oftheforces acting ontheparticle asmeasured in theﬁxed inertial system. The second (-—mRf) andthird (-—m(b Xr)terms result because ofthetranslational and angular acceleration, respectively, ofthemov- ingcoordinate system relative totheﬁxed system. The quantity —-mo.) X(0.)Xr)istheusual centrzfugal force term andreduces tomw2r forthecase inwhich toisnormal totheradius vector. Note that the minus sign implies thatthecentrifugal force isdirected outward from thecenter ofrotation (Figure 10-3). The lastterm inEquation 10.25 isatotally new quantity thatarises from the motion oftheparticle intherotating coordinate system. This term iscalled the Coriolis force. Note that theCoriolis force does indeed arise from themotion of theparticle, because theforce isproportional to-0,andhence vanishes ifthere isnomotion. Because wehave used (onseveral occasions) theterm cenmfugalforce and have now introduced theCoriolis force, wemust now inquire about thephysical meaning ofthese quantities. Itisimportant torealize that thecentrifugal and Coriolis forces arenotforces intheusual sense oftheword; they have been *This result waspublished byG.G.Coriolis in1835. Thetheory ofthecomposition ofaccelerations wasanoutgrowth ofCoriolis’s study ofwater wheels. 10.3CENTRIFUGAL ANDCORIOLIS FORCES 393 (1)XI‘ -tox(toxr) l r (D FIGURE 10-3 Diagram indicating thatthevector -0)X((1.)Xr)points outward, away from theaxisofrotation along co.The term —mw X(0.)Xr) istheusual centrifugal force. introduced inanartiﬁcial manner asaresult ofourarbitrary requirement that webeable towrite anequation resembling Newton’s equation thatisatthesame time valid inanoninertial reference frame; thatis,theequation F:maf isvalid only inaninertial frame. If,inarotating reference frame, weWish to write (letRfand (I)bezero forsimplicity) Fm=ma, then wecanexpress such anequation interms ofthereal force mafas Fcff=maf+(noninertial terms) where the“noninertial terms” areidentiﬁed asthecentrifugal and Coriolis “forces.” Thus, forexample, ifabody rotates about aﬁxed force center, theonly realforce onthebody istheforce ofattraction toward theforce center (and gives risetothecmtdpetal acceleration). Anobserver moving with therotating body, however, measures thiscentral force andalsonotes thatthebody does not falltoward theforce center. Toreconcile thisresult with therequirement that thenetforce onthebody vanish, theobserver must postulate anadditional force—the centrifugal force. Butthe“requirement” isartiﬁcial; itarises solely from anattempt toextend theform ofNewton’s equation toanoninertial sys- tem, and thiscanbedone only byintroducing aﬁctitious “correction force.” The same comments apply fortheCoriolis force; this“force” arises when anat- tempt ismade todescribe motion relative totherotating body. Despite their artiﬁciality, theconcepts ofcentrifugal andCoriolis forces are useful. Todescribe themotion ofaparticle relative toabody rotating with re- spect toaninertial reference frame isacomplicated matter. Buttheproblem canbemade relatively easy bythesimple expedient ofintroducing the“nonin- ertial forces,” which then allows theuseofanequation ofmotion resembling Newton’s equation. 394 10/MOTION INANONINERTIAL REFERENCE FRAME EXAMPLF. 10.2 T _ F Astudent isperforming measurements with ahockey puck onalarge merry- go-round with asmooth (frictionless) horizontal, ﬂatsurface. The merry-go- round hasaconstant angular velocity 0.)androtates counterclockwise asseen from above. (a)Find theeffective force onthehockey puck after itisgiven a push. (b)Plot thepath forvarious initial directions andvelocities ofthepuck as observed bytheperson onthemerry-go-round thatpushes thepuck. Solution. The ﬁrstthree terms forFcﬂ»inEquation 10.25 arezero, sotheeffec- tiveforce asobserved bytheperson onthemerry-go-round is Feff=—m00 X(00Xr)—2mm Xv, (10.26) Wehave taken thefrictional force tobezero. Remember thatv,isthevelocity asmeasured bytheobserver ontherotating surface. The effective accelera- tion is F.acff=if=-00X(00Xr)—20.)Xv, (10.27) Thevelocity andposition aregiven byintegration, intum, oftheacceleration. Vcff Z J'aeffdt ref,=[V65dt ,(10.28b) Weputtheorigin ofourrotating coordinate system atthecenter ofthe merry-go-round. Wewillneed theinitial positions andvelocities ofthepuck to plot themotion. Forthisexample, welettheradius ofthemerry-go-round be Rand thevelocities beinunits ofwR.The initial position ofthepuck willalways beatan(x,y)position of(—O.5R, 0). Weperform anumerical calculation todetermine themotion andshow theresults forseveral directions andvalues oftheinitial velocity inFigure 10-4. Forpurposes ofcalculation, weletw=1rad/ sandR=1m,sotheunits of-00 (initial speed) and T(time forpuck toslide offthesurface) shown inFigure 10-4 areinm/sands,respectively. Forparts (a)-(d), theinitial velocity isin the+y-direction, andtheinitial speed decreases ineach succeeding view. In (a),thepuck slides offquickly. For(b)and (d),thepuck slides offatsimilar positions, butnote thedifferences ininitial speeds aswell asthetime ittakes thepuck toreach theedge. Foraspeed intermediate between these two speeds, asseen in(c),thepuck may make several paths around themerry-go- round; atsome speed, thepuck must stayon.The lasttwoviews show theini- tialvelocity atanangle of45°tothex-axis. In(e), thepuck loops around its path along thewaytoexiting themerry-go-round, andin(f),itchanges direc- tion rather abruptly. The real challenge istoperform such experiments tocompare theactual paths intheﬁxed androtating coordinate systems with thecomputer calculations. 10.4 MOTION RELATIVE TOTHEEARTH 395 vo= 1.5 vo=0.8 v0=0.45 + + + T=0.80 T=2.9 T=17.3 a c <) y (b) <> —i7x vo=0.328 vo=0.47 1/O:0,283 + "+ + T=5.0 T=3.83 T=3.3 (<1) (1?) (0 FIGURE 10-4 The motion ofthehockey puck ofExample 10.2asobserved inthe rotating system forvarious initial directions andvelocities 00atthe times Tnoted. The angular velocity co(1rad/s)isoutofthepage. Ineach ofthecases above, thepuck willmove inastraight lineintheﬁxed sys- tem, because there isnofriction orexternal force intheplane.Q Ti 1 *_ I tn -I 10.4 Motion Relative totheEarth The motion ofEarth with respect toaninertial reference frame isdominated by Earth’s rotation about itsown axis. The effects oftheother motions (e.g., the revolution about theSunandthemotion ofthesolar system with respect tothe local galaxy) aresmall bycomparison. Ifweplace theﬁxed inertial frame x'y’z' atthecenter ofEarth and themoving reference frame xyzonthesurface of Earth, wecandescribe themotion ofamoving object close tothesurface of Earth asshown inFigure 10-5. Wethen apply Equation 10.25 tothedynamical motion. Wedenote theforces asmeasured intheﬁxed inertial system as F=S+mgo, where Srepresents thesum oftheexternal forces (e.g., impulse, electromagnetic, friction) other than gravitation, and mgorepresents thegravi- tational attraction toEarth. Inthiscase, gorepresents Earth’s gravitational ﬁeld vector (Equation 5.3), go * ER where MEisthemass ofEarth, Ristheradius ofEarth, andtheunit vector eRis aunit vector along thedirection ofRinFigure 10-5. Weareassuming Earth is 396 10/MOTION INANONINERTIAL REFERENCE FRAME Z: --._ z . R. ~=i'&'§;;7&:2:-' .‘z&'"*s*-‘LR. .1. ~.;--~.;\..;...;.-,,, :::IY;‘@2.<I:£%:):@sa,;. *I;e;;es, .1.==s:¢.we:('s)gI=»):>";Ia:)..;ai ‘ qéiﬁ-’l/< I..X-I;;j»gr~,;~§;.;- *~»'11»), -=t¢~1..t». 1.. ‘it*5" zL" Eli ‘iii?.1‘: la,‘,5».;;;.",::..".j iti£.:=Y1f*e<@~.%l§> Iit4 .»=\ is.=1Wmzfgz _,,,=" in”Hv/.2: MUJ >022;.2?. .-')1./2%%§Y@f‘~ii'Z““;;".i@i}"¢.'1. . .;¢&1#;‘@?i.i>§li§u.1§i;‘§§i" 1,~,-3 »-4/datawig»: -.1.22.1.2‘/~\t§3%.’.»§m)§I 2):-Qt-gs iii?Li ¥?‘”¥a!%¢iI;i)- M’/1‘*~/i‘)%>I I-1’ *9‘3*‘if"*5! y-1 KI1E vi“Q.#W"W' 3”‘*1’ rsﬁa it-M%-E5.Mr-‘W1it?”.2:-tr"amEm 2;.,..I:¢.2it.§Z§§5;§;<3§i>'.. ;a1<%i.:%. iii1z!)~‘iI@tWI=%ii,' Ev ;)=.;/.(..§-mg,» 1 "14‘.-at§~‘1;g!;tpg‘ig§e;@1';,,3;: ﬁggi wag;'ii;*¢§§¥§§}<§;1'rEZ"‘'- ‘=..i.:%%:1§: ‘:1..-Ie?.i...§;%'.ft'.§¢ Am’1%ﬂee»*-“-'~~~.~.'‘1:¢‘::¢‘1'»4~“‘wi'.;“¢ W”~->0 xi FIGURE 10-5 Inorder tostudy themotion ofanobject near Earth’s surface, we place aﬁxed inertial frame x'y'z’ atthecenter ofEarth andthe moving frame xyzonEarth’s surface. spherical andisotropic and thatRoriginates from thecenter ofEarth. The ac- celeration ofgravity varies over thesurface ofEarth due toEarth’s oblateness, density nonuniformities, and altitude. Wechoose, atthepresent time, notto addthiscomplexity tomotion relative toEarth, butwehave previously pointed outthat effects such asthese canbeconsidered indue course byperforming computer calculations. The effective force Fcffasmeasured inthemoving system placed onthesur- face ofEarth becomes, from Equation 10.25, r,,,=s+mgo—mRf— md.)><r—mt.»><(00><r)—2mm><v,(10.30) WeletEarth’s angular velocity 00bealong theinertial system’s z'~direction (e;). The value ofwis7.3><10-5rad/ s,which isarelatively slow rotation, butitis 365times greater than therotation frequency ofEarth about theSun. Thevalue oftoispractically constant intime, andtheterm doXrwillbeneglected. According toEquation 10.12, wehave forthethird term above, R,=00><(00><R) (10.21) Equation 10.30 now becomes Fe);=S+mgo—mo) X[00X(r+R)]T2mm Xv, (10.32) The second andthird terms (divided bym)arewhat weexperience (and meas- ure) onthesurface ofEarth astheeffective g,andwewillhenceforth denote it asg.Itsvalue is g=g0—o.>X [cox (r+R)] (10.33) 10.4 MOTION RELATIVE TOTHE EARTH 397 The second term ofEquation 10.33 isthecentrifugal force. Because wearelim- iting ourpresent consideration tomotion near thesurface ofEarth, wehave r<< R,andthe0.)X(0.)XR)term totally dominates thecentrifugal force. For situations faraway from thesurface ofEarth, wewould have toconsider both thevariation ofgwith altitude aswell asthe00X(00Xr)term. The centrifugal force isresponsible fortheoblateness ofEarth. Earth isnotreally asolid sphe- roid; itismore like astrongly viscous liquid with asolid crust. Because of Earth’s rotation, Earth hasdeformed sothat itsequatorial radius is21.4 km greater than itspolar radius, and the acceleration ofgravity is0.052 m/s2 greater atthepoles than attheequator. The surface ofcalm ocean water isper- pendicular tog,notgoand ontheaverage, theplane ofEarth’s surface isalso perpendicular t0g. Werewrite Equation 10.32 insimpler terms as Fff=S+mg—2mo.)Xv (10.34) Itisthis equation that wewill usetodiscuss themotion ofobjects close tothe surface ofEarth. Butﬁrst, let’s return totheeffective gofEquation 10.33. The period ofa pendulum determines themagnitude ofg,andthedirection ofaplumb bobin equilibrium determines thedirection ofg.The value ofw2R is0.034 m/s2,and thisisasigniﬁcant enough amount (0.35%) ofthemagnitude ofgtobeconsid- ered. Wedetermined thedirection ofthecentrifugal term 00X[00X(r+R)] inFigure 10-3 (where therisourr’ofFigure 10-5). The direction ofthecen- trifugal term (—00 X[OJX(r+R)] isoutward from theaxis oftherotating Earth. The direction ofaplumb bobwillinclude thecentrifugal term. Because ofthisfact, thedirection ofgatagiven point isingeneral slightly different from thetrue vertical (deﬁned asthedirection ofthelineconnecting thepoint with thecenter ofEarth; seeProblem 10-12). The situation isrepresented schemati- cally (with considerable exaggeration) inFigure 10-6. ("I -t0><(t0xR) 804 8 I I FIGURE 10-6 Near Earth’s surface theterms go(Earth’s gravitational ﬁeld vector) and—o.)X(0.)XR)(main centrifugal term) make uptheeffective g (other smaller temis have been neglected). The effect ofthecentrifu- galterm ongisexaggerated here. 398 10/MOTION INANONINERTIAL REFERENCE FRAME t0,e, vr _m’e’ XV’ Deﬂected path FIGURE 10-7 IntheNorthern Hemisphere, aparticle projected inahorizontal plane willbedirected toward theright oftheparticle’s motion. Inthe Southern Hemisphere, thedirection willbetotheleft. Coriolis Force Effects Theangular velocity vector 00,which represents Earth’s rotation about itsaxis, is directed inanortherly direction. Therefore, intheNorthern Hemisphere, 00 hasacomponent w,directed outward along thelocal vertical. Ifaparticle ispro- jected inahorizontal plane (inthelocal coordinate system atthesurface of Earth) with avelocity v,,then theCoriolis force —2mo.) Xv,hasacomponent in theplane ofmagnitude 2mw,v, directed toward theright oftheparticle’s motion (seeFigure 10-7), andadeﬂection from theoriginal direction ofmotion results.* Because themagnitude ofthehorizontal component oftheCoriolis force is proportional tothevertical component of00,theportion oftheCoriolis force producing deﬂections depends onthelatitude, being amaximum attheNorth Pole and zero attheequator. IntheSouthern Hemisphere, thecomponent to,is directed inward along thelocal vertical, andhence alldeﬂections areintheop- posite sense from those intheNorthern Hemisphere.I Perhaps themost noticeable effect oftheCoriolis force isthat ontheair masses. Asairﬂows from high-pressure regions tolowpressure, theCoriolis force deﬂects theairtoward theright intheNorthern Hemisphere, producing cyclonic motion (Figure 10-8). The airrotates with high pressure ontheright andlowpressure ontheleft.The high pressure prevents theCoriolis force from deﬂecting theairmasses farther totheright, resulting inacounterclockwise ﬂow ofair.Inthetemperate regions, theairﬂow does nottend tobealong the pressure gradients, butrather along thepressure isobars due totheCoriolis force and theassociated centrifugal force oftherotation. *Poisson discussed thedeviation ofprojectile motion in1837. 'l‘During thenaval engagement near theFalkland Islands early inWorld War I,theBritish gunners were surprised toseetheir accurately aimed salvos falling 100yards totheleftoftheGerman ships. The designers ofthesighting mechanisms were wellaware oftheCoriolis deﬂection andhadcare- fully taken thisintoaccount, butthey apparently were under theimpression thatallseabattles took place near 50°N latitude andnever near 50°S latitude. TheBritish shots, therefore, fellatadistance from thetargets equal totwicetheCoriolis deflection. 10.4MOTION RELATIVE TOTHEEARTH 399 N High /“‘\Wm Low \_/ S FIGURE 10-8 The Coriolis force deﬂects airintheNorthern Hemisphere totheright producing cyclonic motion. Near theequatorial regions, thesun heating Earth’s surface causes hotsur- faceairtorise. IntheNorthern Hemisphere, thisresults incooler airmoving in asoutherly direction toward theequator. The Coriolis force deﬂects thismoving airtotheright, resulting inthetrade winds, which provide abreeze toward the southwest intheNorthern Hemisphere andtoward thenorthwest intheSouthern Hemisphere. Note thatthisparticular effect does notoccur attheequator because ofthedirections oftoandtheair’s surface v. The actual motion ofairmasses ismuch more complicated than thesimple picture described here, butthequalitative features ofcyclonic motion and the trade winds arecorrectly given byconsidering theeffects oftheCoriolis force. The motion ofwater inwhirlpools is(atleast inprinciple) asimilar situation, butinac- tuality, other factors (various perturbations and residual angular momentum) dominate theCoriolis force, andwhirlpools arefound with both directions ofﬂow. Even under laboratory conditions, itisextremely difﬁcult toisolate theCoriolis ef- fect. (Reports ofwater inﬂush toilets and bathtubs circulating inopposite direc- tions ascruise ships cross theequator aremost likely highly exaggerated.) I~:xAMPI.E10.:) II-1 IIIITTTT Find thehorizontal deﬂection from theplumb line caused bytheCoriolis force acting onaparticle falling freely inEarth’s gravitational ﬁeld from aheight h above Earth ’ssurface. Solution. WeuseEquation 10.34 with theapplied forces S=0.Ifweset Feff=ma,,wecansolve fortheacceleration oftheparticle intherotating coor- dinate system ﬁxed onEarth. a,=g—200Xv,. Theacceleration duetogravity gistheeffective oneandisalong theplumb line. Wechoose az-axis directed vertically outward (along —g)from thesurface of 400 10/MoTIoN INANONINERTIAL REFERENCE FRAME (.0 N ez /g .~-o~""'- __ ~ » \ ‘,1III _._.__.T_I O2 ;I‘IIIII __i_t_n»RI ,*2'-"" T ‘~"“~1‘mﬁjgi,=§';;&‘~;.t .1.»‘H7"..;§¢ Ii;‘~';g;;1 W15’);.:i’°.. 1- o or . ._._ .Evar~;¥;t;;§',;§E;;s,/»',»s1;:...:r>: er-.W’ W29* wgz))2;2'+13%..s:~..;1(=1 =2‘‘ifit *ii’ id_.._.- S FIGURE 10-9 The coordinate system onEarth’s surface forﬁnding thehorizontal deﬂection ofafalling particle from theplumb line caused bythe Coriolis force. Thevector exisinthesoutherly direction, andeyisin theeasterly direction. Earth. With thisdeﬁnition ofe,,wecomplete theconstruction ofaright-hand co- ordinate system byspecifying thate,,beinasoutherly andeyinaneasterly direc- tion, asinFigure 10-9. Wemake theapproximation thatthedistance offallis sufﬁciently small that gremains constant during theprocess. Because wehave chosen theorigin Ooftherotating coordinate system to lieintheNorthern Hemisphere, wehave wx=-tocosA toy=0 to,=tosin)t Although theCoriolis force produces small velocity components intheey and exdirections, wecancertainly neglect 22and compared with 5;,thevertical velocity. Then, approximately, DlE0 E0 E—-gt where weobtain 2byconsidering afallfrom rest. Therefore, wehaveN.Q- ex ey e, wXv,E -—wcos)t 0wsin/\ 0 0 ——gt E——(wgt cos)t)ej 10.4 MOTION RELATIVE TOTHE EARTH 401 The components ofgare g.=0 gy=0 gr:'8 sotheequations forthecomponents ofa,(neglecting terms* inm2;see Problem 10-13) become (at).=55E0 (a,)y =jiE2wgt cosA (a.).=EE-g Thus, theeffect oftheCoriolis force istoproduce anacceleration intheey, oreasterly, direction. Integrating twice, wehave -.., 1 3 y(t) =gwgt cosA where y=0and3')=0att=0.The integration ofiyields thefamiliar result for thedistance offall, 1z(t)2z(0)—§g't2 andthetime offallfrom aheight h=z(0)isgiven by tE\/2h/g Hence theresult fortheeastward deﬂection dofaparticle dropped from restat aheight handatanorthem latitude Aisi 1 (8/23 dEgm COSA'? (10.35) Anobject dropped from aheight of100matlatitude 45°isdeﬂected approxi- mately 1.55 cm(neglecting theeffects ofairresistance). *According toM.S.Tiersten andH.Soodak, Am.j.Phys. 68,129(2000), thesoutherly deﬂection ison theorder ofamillion times smaller than theeasterly deﬂection foradrop ofabout 100m,andthere is nocredible evidence thatthesoutherly deﬂection hasbeen correctly measured, despite many attempts. TThe eastward deﬂection waspredicted byNewton (1679), andseveral experiments (notably those ofRobert Hooke) appeared toconfirm theresults. The most careful measurements were probably those ofF.Reich (1831; published 1833), who dropped pellets down amine shaft 188mdeep and observed amean deﬂection of28mm. This issmaller than thevalue calculated from Equation 10.35, thedecrease being duetoairresistance effects. Inalltheexperiments, asmall southerly component ofthedeﬂection wasobserved—and remained unaccounted foruntil Coriolis’s theorem wasappre- ciated (see Problems 10-13 and 10-14). 402 10/MOTION INANONINERTIAL REFERENCE FRAME EXAMPLE 10.4 -_ — _— _ — — ___ - Todemonstrate thepower oftheCoriolis method forobtaining theequations ofmotion inanoninertial reference frame, rework thelastexample butuse only theformalism previously developed—the theory ofcentral-force motion. Solution. Ifwerelease aparticle ofsmall mass from atower ofheight habove Earth’s surface, thepath theparticle describes isaconic section—an ellipse with sE1and with one focus very close toEarth’s center. IfRisEarth’s radius andAthe(northern) latitude, then atthemoment ofrelease, theparticle hasa horizontal velocity intheeastward direction: vhm =nocosA=(R+h)wcosA andtheangular momentum about thepolar axisis l=mr-ohm =m(R +h)2co cosA (10.36) The equation ofthepath is* 9;=1-—scos0 (10.37) ifwemeasure 0from theinitial position oftheparticle (see Figure 10-10). Att=0, wehave a__..__ =1_ R+h 8 soEquation 10.37 canbewritten as (1-s)(R+ h) ,:_______ 10.31Escos9 ( 8) From Equation 8.12 fortheareal velocity, wecanwrite 1d9 1_,2__2_ 2 dt 2m Thus, thetime trequired todescribe anangle 0is 6 ¢=@j Faelo Substituting into thisexpression thevalue oflfrom Equation 10.36 andrfrom Equation 10.38, weﬁnd 1 6 __ 2 ¢= j(18)d6 (10.39)wcosA 01—scos0 *Notice thatthere isachange ofsignbetween Equation 10.37 andEquation 8.41 duetothediffer- entorigins for0inthetwocases. 10.4MoTIoN RELATIVE ToTHEEARTH 403 h .;I.",'~. e2.=.;;i1; @ .1EF 5! 3 .-Eat»---**.;.| kn 3, 1 _ 1 3'14l§§§i‘§Y“'..5.; Y3.) "1“v T V‘\ 1"l"}¥i‘§i1‘~“:§~¥<i1i"1'_ l :;..»====_€\=;-.a=l.-5:‘ ,=.;,I- ....~.. .1’:.~I .~:.~..-Ir:' reJ‘5;,““".‘~'-I12.I,-;=;[7'-3- “M?tit?.';""i1i‘ ‘-‘———-*"”*.f3l1, om,~;‘W-val‘.'..~5.5.5:=;;1’,~,.._-’t’€“.:@»E:EE‘-)..:;"1»:"..*%t1#~:[email protected]=: -*5""-- ".,.,..1»-.::".-%;;»*, 2£.@?;r‘%§..'§ii=i§L. ‘‘I;lI =-:\.;;;<;-;;= 11%; , --;:= ;'=.;;=r.§*i=:=1'55“:--e.:=.--1!-';ii.rag2~,,===9jfIZEgﬁft Eel‘ .2: ’ FIGURE 10-10 Therather complicated geometry fordescribing themotion ofa falling particle inanoninertial system using central-force motion. Ifwelet6=Bowhen theparticle hasreached Earth’s surface (r=R),then Equation 10.38 becomes R_ 1-rs R+h 1-scosBo or,inverting, 1+_Ii=1-t-:cos0o R 1 2: Z1-—t~:[1-—2sin2(6o/2)]g 1——s 0=1+iisin2—0 (10.40)1-—s 2 from which wehave h 28 .90_Z___ 2_R 1__8s1n2 Because thepath described bytheparticle isalmost vertical, little change oc- curs intheangle 0between theposition ofrelease andthepoint atwhich the particle reaches thesurface ofEarth; Boistherefore small andsin(6o/ 2)canbe approximated byitsargument: h t-:95 —EL 10.41R 2(1—-s) ( ) 404 10/MOTION INANONINERTIAL REFERENCE FRAME Ifweexpand theintegrand inEquation 10.39 bythesame method used to obtain Equation 10.40, weﬁnd W1" d0 t-wCOSA0{I+[28/(1 -s)]sin2(6/2)}2 andbecause 0issmall, wehave tE1F d6 cocosA 0[1+.962/2(1 —s)]2 Substituting fors/2(1 -—5:)from Equation 10.41 and writing t(B=Bo)=Tfor thetotal time offall,weobtain TE(1,,jg“O d6 cocosA 0[1+(h62/R6§)j2 E?-"Tr 1""'—"-02 d0 wcosA() R95 1 2/.=—-- I-— 0wcosA< 312)° 9E—-——tOSA EwTcosA1+%°E1-2h/3R I 3RI-427* /5I\9E" \__/ Solving for9o,weﬁnd During thetime offallT,Earth turns through anangle a>T,sothepoint on Earth directly beneath theinitial position oftheparticle moves toward theeast byanamount RwTcos A.During thesame time, theparticle isdeﬂected toward theeastbyanamount R6o. Thus, theneteasterly deviation dis d=R6o-—RwTcos A =-3;hwT cosA andusing TEV2h/g asinthepreceding example, wehave, ﬁnally, 1 8h5dE-tocosA-—" 3 g which isidentical with theresult obtained reviousl E ‘on .. p y(quati 1035) The effect oftheCoriolis force onthemotion ofapendulum produces apreces- sion, orrotation with time oftheplane ofoscillation. Describe themotion of thissystem, called aFoucault pendulum.* *Devised in1851 bytheFrench physicist jean-Bemard-Leon Foucault, pronounced FE-c5 (1819-1868). 10.4 MOTION RELATIVE TOTHE EARTH 405 Solution. Todescribe thiseffect, letusselect asetofcoordinate axes with ori- ginattheequilibrium point ofthependulum and z-axis along thelocal vertical. Weareinterested only intherotation oftheplane ofoscillation—that is,we wish toconsider themotion ofthependulum bobinthex—yplane (the hori- zontal plane). Wetherefore limit themotion tooscillations ofsmall amplitude, with thehorizontal excursions small compared with thelength ofthependu- lum. Under thiscondition, 2issmall compared with )2and5»andcanbeneg- lected. The equation ofmotion is T a,=g+7”-—20>Xv, (10.42) where T/mistheacceleration produced bytheforce oftension Tinthependu- lum suspension (Figure 10-11). Wetherefore have, approximately, =\\é~,;<Tx= _T. T,=-T-— (10.43) T,ET Asbefore, g.=0 gy=0 gz=“g and wx=-ea)cosA wy=0 w,=wsinA z Suspension point \atgreat height l _ ,_ ._ y /ll T ”Tx=_T. % x T7:_TH3_ mg FIGURE 10-11 Geometry fortheFoucault pendulum. Theacceleration gvector is along the—z-direction, andthetension Tisseparated into x-,y-, andz-components. 406 10/MOTION INANONINERTIAL REFERENCE FRAME with (W).=it (V-))=i (W).=iE0 Therefore, ex ey e, 00XVE—wcosA 0tosinA it jr 0 sothat (0:XV,),, E—ywsinA (toXv,)y ExwsinA (10.44) (00Xv,), E—jzwcosA Thus, theequations ofinterest are §'~i§’~l=-\~£@-X(a,),, =55E---— +2jrtosinA (10.45) (a,)y E E———— —2xwsinA Forsmall displacements, TEmg.Deﬁning 012ET/mlEg/l,andwriting to,= wsinA,wehave 52+a2x E2w,j1 10.46 53+o.'2_y E—2<.»,s¢I ( ) Wenote thattheequation for55contains aterm injtand thattheequation forjicontains aterm inx.Such equations arecalled coupled equations. Asolu- tionforthispair ofcoupled equations canbeeffected byadding theﬁrstofthe above equations toitimes thesecond: (55+2'32)+a2(x +iy)E—2w,(i5c— jw)=—2iw,(5¢+ ijw) Ifwe write qEx+iy wethen have g'+2iw,rj +a2qE0 This equation isidentical with theequation thatdescribes damped oscillations (Equation 3.35), except thathere theterm corresponding tothedamping factor 10.4 MOTION RELATIVE ToTHEEARTH 407 ispurely imaginary. The solution (seeEquation 3.37) is q(t) Eexp[—z'w,t][A exp( V-103 —012t)+Bexp( —\/—wf —a2t)](10.47) IfEarth were notrotating, sothatw,=0,then theequation forqwould become ¢']"+a2q' E0, w,=0 from which itisseen thatacorresponds totheoscillation frequency ofthepen- dulum. This frequency isclearly much greater than theangular frequency of Earth’s rotation. Therefore, a>>(0,,andtheequation forq(t)becomes q(t)Ee"""-'(Ae“"“+ Be“°“) (10.48) Wecaninterpret thisequation more easily ifwenote thattheequation for q’hasthesolution q’(t)=x’(t)+iy'(t) =Ae“"' +Be""°" Thus, q(t)=q'(t)-@""”Z’ or 4(1)+0(1)=[(x'(r) +i)»'(t)]-e“"‘"1' =(x'+iy')(cos w,t—isin w,t) =(x'cosw,t+y’sin w,t) +z'(—x'sin wzt+y’cos w,t) Equating realandimaginary parts, x(t)=x’cosw,_t+y’sinw,t y(t)=—x'sinw,t+y’cosant} Wecanwrite these equations inmatrix form as (x(t)) :(cosw,t sinw,t)(xI (t)) (10.49) y(t) -sinw,t cosw,t y(t) from which (x,y)may beobtained from (x',y’)bytheapplication ofarotation matrix ofthefamiliar form cos0sin9A= 10.50(—sin9cos0) ( ) Thus, theangle ofrotation is9=w,t,andtheplane ofoscillation ofthependu- lumtherefore rotates with afrequency w,=wsinA.The observation ofthisro- tation gives aclear demonstration oftherotation ofEarth.* *Vincenzo Viviani (1622-1703), apupil ofGalileo, hadnoticed asearly asabout 1650 thatapendu- lum undergoes aslow rotation, butthere isnoevidence that hecorrectly interpreted thephenome- non. Foucault’s invention ofthegyroscope intheyear following thedemonstration ofhispendulum provided even more striking visual proof ofEarth’s rotation. 408 10/MOTION INANONINERTIAL REFERENCE FRAME PROBLEMS 10-1. Calculate thecentrifugal acceleration, duetoEarth’s rotation, onaparticle onthe surface ofEarth attheequator. Compare thisresult with thegravitational accelera- tion. Compute alsothecentrifugal acceleration duetothemotion ofEarth about theSunandjustify theremark made inthetextthatthisacceleration may beneg- lected compared with theacceleration caused byaxial rotation. 10-2. Anautomobile drag racer drives acarwith acceleration aandinstantaneous veloc- ity'u.The tires (ofradius ro)arenotslipping. Find which point onthetirehasthe greatest acceleration relative totheground. VVhat isthisacceleration? 10-3. InExample 10.2, assume thatthecoefﬁcient ofstatic friction between thehockey puck andahorizontal rough surface (onthemerry-go-round) is/1,.How faraway from thecenter ofthemerry-go-round canthehockey puck beplaced without sliding? 10-4. InExample 10.2, forwhat initial velocity anddirection intherotating system will thehockey puck appear tobesubsequently motionless intheﬁxed system? W'hat willbethemotion intherotating system? Lettheinitial position bethesame asin Example 10.2. Youmaychoose todoanumerical calculation. 10-5. Perform anumerical calculation using theparameters inExample 10.2 andFigure 10-4e, butﬁnd theinitial velocity forwhich thepath ofmotion passes back over the initial position intherotating system. Atwhat time does thepuck exitthemerry-go- round? 10-6. Abucket ofwater issetspinning about itssymmetry axis. Determine theshape of thewater inthebucket. 10-7. Determine how much greater thegravitational ﬁeld strength gisatthepole than at theequator. Assume aspherical Earth. Iftheactual measured difference is Ag=52mm/s2, explain thedifference. How might youcalculate thisdifference between themeasured result andyour calculation? 10-8. Ifaparticle isprojected vertically upward toaheight habove apoint onEarth’s sur- face atanorthern latitude A,show that itstrikes theground atapoint %tocosA- \/8h?’/g tothewest. (Neglect airresistance, and consider only small vertical heights.) 10-9. Ifaprojectile isﬁred dueeastfrom apoint onthesurface ofEarth atanorthern latitude Awith avelocity ofmagnitude Voandatanangle ofinclination tothehor- izontal ofa,show thatthelateral deﬂection when theprojectile strikes Earth is 41/ti’ . .d=j-ws1nA- smga cosa g where toistherotation frequency ofEarth. PROBLEMS 409 10-10. 10-ll 10-12. 10-13. 10-14. 10-15Inthepreceding problem, iftherange oftheprojectile isR6forthecase co=0, show thatthechange ofrange duetotherotation ofEarth is \/2R6?’ tocosA(cot ‘/20:—1tan3/2:1)g 3 Obtain anexpression fortheangular deviation ofaparticle projected from the North Pole inapath thatliesclose toEarth. Isthedeviation signiﬁcant foramis- silethatmakes a4,800-km ﬂight in10minutes? What isthe“miss distance” ifthe missile isaimed directly atthetarget? Isthemiss distance greater fora19,300-km ﬂight atthesame velocity?AR’= Show thatthesmall angular deviation sofaplumb linefrom thetruevertical (i.e., toward thecenter ofEarth) atapoint onEarth’s surface atalatitude Ais Rw2sinAcosA 8:?go—Ru) cosA where Ristheradius ofEarth. What isthevalue (inseconds ofarc) ofthemaxi- mum deviation? Note thattheentire denominator intheanswer isactually theef- fective g,andgodenotes thepure gravitational component. Refer toExample 10.3concerning thedeﬂection from theplumb lineofaparticle falling inEarth’s gravitational ﬁeld. Take gtobedeﬁned atground level anduse thezeroth order result forthetime-of-fall, T=\/2h/g. Perform acalculation in second approximation (i.e., retain terms in(02)andcalculate thesoutherly deﬂec- tion. There arethree components toconsider: (a)Coriolis force tosecond order (C1), (b)variation ofcentrifugal force with height (C2), and(c)variation ofgravi- tational force with height (C3). Show thateach ofthese components gives aresult equal to h2 _ C,-E ax?sinAcosA with C1=2/3,C2=5/6,andC3=5/2.The total southerly deﬂection istherefore (4h2w2 sinAcosA)/g. Refer toExample 10.3andtheprevious problem, butdrop theparticle atEarth’s sur- face down amineshaft toadepth h.Show thatinthiscase there isnosoutherly de- ﬂection duetothevariation ofgravity andthatthetotal southerly deﬂection isonly §h2w22gsinAcosA Consider aparticle moving inapotential U(r). Rewrite theLagrangian interms of acoordinate system inuniform rotation with respect toaninertial frame. Calculate theHamiltonian anddetermine whether H=E.IsHaconstant ofthe motion? IfEisnotaconstant ofmotion, why isn’t it?The expression forthe Hamiltonian thus obtained isthestandard formula 1/2mv2+Uplus anaddi- tional term. Show that theextra term isthecentrifugal potential energy. Use the Lagrangian youobtained toreproduce theequations ofmotion given inEquation 10.25 (without thesecond andthird terms). 410 10-16 10-17 10-18 10-19. 10-20 10-21 10-22.10/MOTION INANONINERTIAL REFERENCE FRAME Consider Problem 9-63 butinclude theeffects oftheCoriolis force ontheprobe. The probe islaunched atalatitude of45°straight up.Determine thehorizontal deﬂection intheprobe atitsmaximum height foreach part ofProblem 9-63. Approximate Lake Superior byacircle ofradius 162kmatalatitude of47°. Assume thewater isatrestwith respect toEarth andﬁnd thedepth thatthecenter isdepressed with respect totheshore duetothecentrifugal force. ABritish warship ﬁres aprojectile due south near theFalkland Islands during World War Iatlatitude 50°S. Iftheshells areﬁred at37°elevation with aspeed of 800m/s, byhow much dotheshells miss their target and inwhat direction? Ignore airresistance. Find theCoriolis force onanautomobile ofmass 1300 kgdriving north near Fairbanks, Alaska (latitude 65°N) ataspeed of100km/h. Calculate theeffective gravitational ﬁeld vector gatEarth’s surface atthepoles and theequator. Take account ofthediﬂerence intheequatorial (6378 km) and polar (6357 km) radius aswell asthecentrifugal force. How well does the result agree with thedifference calculated with theresult g=9.780356[1 + 0.0052885 sin2A—0.0000059 sin2(2A)]m/s2 where Aisthelatitude? Water being diverted during aﬂood inHelsinki, Finland (latitude 60°N) ﬂows along adiversion channel ofwidth 47minthesouth direction ataspeed of3.4m/s.On which sideisthewater thehighest (from thestandpoint ofnoninertial systems) and byhow much? Shot towers were popular intheeighteenth and nineteenth centuries fordrop- ping melted lead down talltowers toform spheres forbullets. The lead solidiﬁed while falling andoften landed inwater tocool thelead bullets. Many such shot towers were built inNew York State. Assume ashot tower wasconstmcted atlati- tude 42°N, andthelead felladistance of27m.Inwhat direction andhowfardid thelead bullets land from thedirect vertical? llDynamics ofRigid Bodies 11.1 Introduction Wedeﬁne arigid body asacollection ofparticles whose relative distances are constrained toremain absolutely ﬁxed. Such bodies donotexist innature, be- cause theultimate component particles composing every body (the atoms) are always undergoing some relative motion likevibrations. This motion, however, is microscopic, and ittherefore usually may beignored when describing the macroscopic motion ofthebody. However, macroscopic displacement within the body (such aselastic deformations) cantake place. Formany bodies ofinterest, wecansafely neglect thechanges insizeandshape caused bysuch deformations andobtain equations ofmotion valid toahigh degree ofaccuracy. Weusehere theidealized concept ofarigid body asacollection ofdiscrete particles orasacontinuous distribution ofmatter interchangeably. The only change isthereplacement ofsummations over particles byintegrations over mass density distributions. The equations ofmotion areequally valid foreither viewpoint. Wehave studied rigid bodies inintroductory physics andhave already seen examples inthisbook ofhoops andcylinders rolling down inclined planes. We alsoknow how toﬁnd thecenter ofmass ofvarious rigid objects (Section 9.2). Such problems can'be handled with concepts already presented including rota- tional inertia, angular velocity and momentum, and torque. Wecanusethese techniques tosolve many problems, such assome simple examples ofplanar mo- tion inSection 11.2. When weallow complete three-dimensional motion, the mathematical complexity considerably escalates. The classic example, ofcourse, isofthecat,who invariably lands onitsfeetafter being dropped (under acare- fully controlled experimental situation) with itsfeetinitially pointing upwards. 411 412 11/DYNAMICS orRIGID BODIES Wehave learned todescribe themotion ofabody bythesum oftwoinde- pendent motions—a linear translation ofsome point ofthebody plus arotation about thatpoint.* Ifthepoint ischosen tobethecenter ofmass ofthebody, then such aseparation ofthemotion into twoparts allows theuseofthedevelopment inChapter 9,which indicates that theangular momentum (see Equation 9.39) canbeseparated into portions relating tothemotion ofthecenter ofmass and tothemotion around thecenter ofmass. Itisthegeneral rotation thatincreases thecomplexity. Wewillﬁnd ituseful tohave twocoordinate systems, one aninertial coordinate system (ﬁxed) and theother acoordinate system ﬁxed with respect tothebody. Sixquantities must bespeciﬁed todenote theposition ofthebody. Wenormally usethree coordi- nates todescribe theposition ofthecenter ofmass (which canoften conve- niently bemade tocoincide with theorigin ofthebody coordinate system) and three independent angles that give theorientation ofthebody coordinate sys- temwith respect totheﬁxed (orinertial) system.* The three independent angles arenormally taken tobetheEulerian angles, described inSection 11.8. Unfortunately, themathematical level increases inthischapter. Wewillﬁnd itprudent tointroduce tensor andmatrix algebra inorder todescribe thecom- plete motion ofsimple looking dynamical systems likerotating tops (either free orinagravitational ﬁeld), dumbbells, gyroscopes, ﬂywheels, and automobile wheels out-of-balance. Wewillusethedumbbell, because ofitssimplicity, asour system ofinterest asweintroduce theneeded mathematics. 11.2 Simple Planar Motion Wehave already solved theproblem ofadisk rolling down aninclined plane (see Examples 6.5and7.9,andFigure 6-7). Several end-of-chapter problems in Chapter 7concerned simple rigid bodies. Wediscussed center ofmass inSection 9.2,angular momentum ofasystem ofparticles inSection 9.4,andtheenergy of thesystem inSection 9.5.Werestrict ourselves inthissection tothemotion ofa rigid body inaplane and present examples asareview ofourintroductory physics. Astring attached totheceiling iswrapped around ahomogenous cylinder of mass Mandradius R(seeFigure 11-1). Attime t=0,thecylinder isdropped from restandrotates asthestring unwinds. Find thetension Tinthestring, the linear andangular accelerations ofthecylinder, andtheangular velocity about thecylinder’s center. *Chasles’ Theorem, which iseven more general than thisstatement (itsaysthatthelineoftranslation andtheaxisofrotation canbemade tocoincide), wasproven bytheFrench mathematician Michel Chasles (1793-1880) in1830. The proof isgiven, e.g., byE.T.Whittaker (Wh37, p.4). Tlnthischapter, weusethedesignation bodysystem inplace oftheterm rotating system used inthepre- ceding chapter. Theterm ﬁxed system willberetained. Tl 5,Fs FIGURE 11-1 Example 11-1. Astring attached totheceiling iswrapped around a cylinder. The cylinder isreleased from rest. Solution. The center ofmass moves duetothesum oftheforces, which areall inthevertical direction. Weletypoint downward. My=Fg— T=Mg— T, (11.1) where thecenter ofmass acceleration isji,andwehave used Fg=Mg.The ro- tation about thecylinder’s center ofmass atOisduetothetension T. ‘T=RT=15 (11.2) where 1'isthetorque about O,andIistherotational inertia ofthecylinder (MR2/2). Welety=Oand9=0att=0when thecylinder isreleased. Then we have y=R0,5»=V=R0,and ji=R0.Wecancombine these relations with Equations 11.1 and11.2 todetermine theacceleration. --_-2_J_5_ _MR2i_ _iygMgMRg2MR2g2 which gives ji=2g/3 fortheacceleration, andtheangular acceleration, a=9'=ji/R=2g/3R. The tension Tisthen found from Equation 11.2 tobe 15MR2}? M2gT=—=i~=——=M 3 11.3R2122 2s 5/ () The angular velocity isor=0=V/R. Weintegrate jitoobtain V=y=2gt/3 andw=2gt/3R. Aphysical orcompound pendulum isarigid body thatoscillates duetoitsown weight about ahorizontal axisthatdoes notpass through thecenter ofmass of thebody (Figure 11-2). Forsmall oscillations, ﬁnd thefrequency andperiod of oscillation ifthemass ofthebody isMandtheradius ofgyration isk. 414 11/DYNAMICS OFRIGID BODIES ,I // _______?-_______Qt,1QI I I I /P‘ \CMXI I I I I / I Fa FIGURE 11-2 Example 11-2. The physical orcompound pendulum. The body rotates about anaxispassing through O.The body rotates dueto thegravitational force acting atthecenter ofmass. Solution. WeusetheLagrangian method tosolve thisexample, although we could justaseasily solve forthetorque toﬁnd theequation ofmotion. The ro- tation axispasses through thepoint Oofthebody. The radius ofgyration isde- ﬁned such thattherotational inertia Iabout thegiven axisofrotation (Ointhis case) isgiven byI=Mk2. The kinetic energy ofrotation andthepotential energy is 1. T=—I92 2 92 U= —MgLc0s0= —MgL 1-5 where wehave deﬁned thezero ofthepotential energy tobeatpoint 0and have used thesmall angle approximation forcos9.Weﬁnd theLagrangian function andtake theappropriate derivatives toform theLagrange equation of motion. The generalized coordinate isclearly 0. 1. WL=T—U=?W+A@L1—5 %——MMw s L .6.=I6 89 d6L ..—.=I0 dt69 11.3 INERTIA TENSOR 415 The Lagrange equation ofmotion is ..ML6+—;L 9=O Wehave seen thisequation several times, andtheangular frequency isgiven by to?=MgL/I. From thisweﬁnd thefrequency vandperiod T, co 1lMgL 1)MgL 1/gLVi Z Z Z 211' 211' I 211' Mk2 211' k2 1 2 T--=2».../iV gL Now thatwehave brieﬂy reviewed ourprevious study ofrigid body motion, let’sproceed tothemore general cases. Forthiswewillneed theinertia tensor. 11.3 Inertia Tensor Wenow direct ourattention toarigid body composed ofnparticles ofmasses ma, a=1,2, Ifthebody rotates with aninstantaneous angular velocity to about some point ﬁxed with respect tothebody coordinate system and ifthis point moves with aninstantaneous linear velocity Vwith respect totheﬁxed co- ordinate system, then theinstantaneous velocity oftheathparticle intheﬁxed system canbeobtained byusing Equation 10.17. Butwearenow considering a rigid body, so dv,= EO dt rotating where thesubscript ﬂdenoting theﬁxed coordinate system, hasbeen deleted from thevelocity va,itnow being understood that allvelocities aremeasured in theﬁxed system. Allvelocities with respect totherotating orbody system now vanish because thebody isrigid. Because thekinetic energy oftheathparticle isgiven byTherefore, 1Ta=gmv2aa wehave, forthetotal kinetic energy, 1T=§Em,,(v +or><r,,)2 416 11/DYNAMICS orRIGID BODIES Expanding thesquared term, weﬁnd 1 1T=§EmJ%+Emyw»x%+§§M4wxqy arm This isageneral expression forthekinetic energy andisvalid foranychoice of theorigin from which thevectors r,,aremeasured. Butifwemake theorigin of thebody coordinate system coincide with thecenter ofmass oftheobject, acon- siderable simpliﬁcation results. First, wenote that inthesecond term onthe right-hand side ofthisequation neither Vnor0)ischaracteristic oftheathpar- ticle, andtherefore, these quantities may betaken outside thesummation: 2m,,,V-or Xra=V-or X<2mara) Butnow theterm 2%q=MR isthecenter-of-mass vector (see Equation 9.3), which vanishes inthebody sys- tembecause thevectors raaremeasured from thecenter ofmass. The kinetic energy canthen bewritten as T=Tlrans + Trot where 1 1 Tlrans 2§;maV2 = 1 T...=5§m..<<» ><-..>2 (11.61)) Tirans and 7}.»designate thetranslational androtational kinetic energies, respec- tively. Thus, thekinetic energy separates into twoindependent parts. The rotational kinetic energy term canbeevaluated bynoting that (A><B)2= (AXB)-(AXB) =AQB2 —(A-B)? Therefore, 1nm=§Em$ﬁ&-osnﬁ] urn Wenow express Tmtbyusing thecomponents or,andrd),ofthevectors orand ra.Wealso note that ra=(xaj, xa,2,xag) inthebody system, sowecanwrite TIL,‘ = xa’,-. TIILIS, =ism-it-rm-2>(a ~(Z--X--)(%-I-1)] 11.3INERTIA TENSOR 417 Now, wecanwrite w,-=2]‘(1)]-8,]-, sothat 1 not =5; gjma xgnk) _wiwj‘xa,i xa,j:| 1 =ggwiw]-;ma<5,j~; xi’,—xay,-xmj) (11.9) Ifwedeﬁne theijthelement ofthesum over atobeI,]», 19.Egm,,(5,»j-gxij, —xa,,-xaj) | (11.10) then wehave 12 T...=5,].I.,-w.»w,- <11-11> This equation initsmost restricted form becomes 1Tm,=51¢»? (11.12) where Iisthe(scalar) rotational inertia (moment ofinertia) about theaxisofro- tation. This equation willberecognized asthefamiliar expression fortherota- tional kinetic energy given inelementary treatments. The nine terms 1,;constitute theelements ofaquantity wedesignated by{I}. Inform, {I}issimilar toa3X3matrix. Itistheproportionality factor between therotational kinetic energy and theangular velocity and hasthedimensions (mass) X(length)? Because {I}relates twoquite different physical quantities, weexpect that itisamember ofasomewhat higher class offunctions than has heretofore been encountered. Indeed, {I}isatensor andisknown astheinertia tensor.* Note, however, that Tm,canbecalculated without regard toanyofthe special properties oftensors, byusing Equation 11.9, which completely speciﬁes thenecessary operations. The elements of{I}canbeobtained directly from Equation 11.10. Wewrite theelements ina3X3array forclarity: ;ma(x?1,2 +x¢21,3) _%maxa,l xa,2 _%maxa,1xa,3 = _%maxa,2xa,1 ?ma(xZ,l +x?r,3) _%maxa,2xa,3 _%maxa,3xa,1 _%maxa,3xa,2 %ma(xg,1 +363,2) *The true testofatensor liesinitsbehavior under acoordinate transformation (seeSection 11.7). 418 11/DYNAMICS orRIGID BODIES Equation 11.10 isacompact waytowrite theinertia tensor components, but Equation 11.13a isanimposing equation. Byusing components (xa,ya,za)in- stead of(x,,,1,x,,,2, x,,,3) and letting rf=xi’+yﬁ+Z3,Equation l1.13a canbe written as f %ma(T¢2r _xi) _%maxaya _%maxaZa {I}=<—§m..y..x.. §m..(r?. —yi)—§m..y..z.. <11-1%) —2mazaxa _Emazaya i which islessimposing andmore recognizable. Wecontinue, however, with the x,,,,-notation because ofitsutility. The diagonal elements, I11,I22,and I33,arecalled themoments ofinertia about thex1-,x2-,and x3-axes, respectively, andthenegatives oftheoff-diagonal elements I12,I13,andsoforth, aretermed theproducts ofinertia.* Itshould be clear thattheinertia tensor issymmetric; thatis, 1,,=I, (11.14) and, therefore, thatthere areonly sixindependent elements in{I}.Furthermore, theinertia tensor iscomposed ofadditive elements; theinertia tensor forabody canbeconsidered tobethesum ofthetensors forthevarious portions ofthe body. Therefore, ifweconsider abody asacontinuous distribution of-matter with mass density p=p(r) ,then 11].=jvpo) (5,-,.;x§ — do (11.15) where do=dx1dx2dx2 istheelement ofvolume attheposition deﬁned bythe vector r,andwhere Visthevolume ofthebody. 7 7T777 Vii 1 Calculate theinertia tensor ofahomogeneous cube ofdensity p,mass M,and sideoflength b.Letonecorner beattheorigin, andletthree adjacent edges lie along thecoordinate axes (Figure 11-3). (For thischoice ofthecoordinate axes, itshould beobvious thattheorigin does notlieatthecenter ofmass; we return tothispoint later.) Solution. According toEquation 11.15, wehave b b b In=pjdx3jdx2(x2 + dxlO 0 0 2 2=-b5=—Mb2sp 3 *Introduced byHuygens in1673; Euler coined thename. 11.4ANGULAR MOMENTUM 419 *3 .s@;z1ii1-?1;»: 51:. T iii. iij:‘TY, HIiI i'X*1»; ';,_s b I -/' 5if%::il%..*i§@ t‘Ii L I’ b xr FIGURE 11-3 Example 11-3. Ahomogeneous cube ofsides bwith theorigin at onecorner. I I I I12=—pjx1dx1jx2dx2jdx3 0 0 0 1 1 =——b5=——Mb24" 4 Itshould beeasy toseethatallthediagonal elements areequal and, fur- thermore, thatalltheoff-diagonal elements areequal. Ifwedeﬁne BEMb2, we have 2 I11: I22=I:-as=55 1 1122113253: -23 The moment-of-inertia tensor then becomes ).|>)—l,.|>)—\OON>‘CbRWI-I>r—'UDI\'>H>I—‘“QR‘Q<.>OI\'>).I>>—1,.;>)-I“G:——-B {|}= -- — ‘*5 Weshall continue theinvestigation ofthemoment-of-inertia tensor forthe cube inlater sections. 11.4 Angular Momentum With respect tosome point Oﬁxed inthebody coordinate system, theangular momentum ofthebody is L=Er,><pa (11.16) 420 11/DYNAMICS orRIGID BODIES The most convenient choice fortheposition ofthepoint Odepends onthepar- ticular problem. Only twochoices areimportant: (a)ifone ormore points of thebody areﬁxed (intheﬁxed coordinate system), Oischosen tocoincide with onesuch point (asinthecase oftherotating top,Section 11.11); (b)ifnopoint ofthebody isﬁxed, Oischosen tobethecenter ofmass. Relative tothebody coordinate system, thelinear momentum pais pa=mava =mac.) Xra Hence, theangular momentum ofthebody is L=grmara X(coXra) (11.17) Thevector identity AX(BXA)=A2B —A(A-B) canbeused toexpress L: L=§m,,[I§Is -r,,(1-0,-12)] | (11.18) The same technique weused towrite 11",,“intensor form cannow beapplied here. Buttheangular momentum isavector, sofortheithcomponent, wewrite Li:Ema wizxgk H“xai2xa'w'a k ’ 'J "7J =;I,j§m,,(o,,;x3,,, -x,,,x,,,,-) (11.19) The summation over acanberecognized (see Equation 11.10) astheijthele- ment oftheinertia tensor. Therefore, L=-"{I}-or (11.20b) Thus, theinertia tensor relates asumover thecomponents oftheangular veloc- ityvector totheithcomponent oftheangular momentum vector. This may at ﬁrst seem asomewhat unexpected result; for,ifweconsider arigid body for which theinertia tensor hasnonvanishing off-diagonal elements, then even if0.) isdirected along, say,thexl-direction, or=(col,0,0),theangular momentum vector ingeneral hasnonvanishing components inallthree directions: L= (L1,L2,L3);that is,theangular momentum vector does notingeneral have the same direction astheangular velocity vector. (Itshould beemphasized that this statement depends onI,-jab0foriabj;wereturn tothispoint inthenext section.)or,intensor notation, 11.4 ANGULAR MOMENTUM 421 or "1 ‘1'1 L 1 O 1'2 ,,,V2 Rotation axis FIGURE 11-4 Adumbbell connected bymasses mlandm2attheends ofitsshaft. Note that0)isnotalong theshaft, andthatnoandLarenotcollinear. Asanexample oftoand Lnotbeing collinear, consider therotating dumb- bellinFigure 11-4. (Weconsider theshaft connecting maandm2tobeweightless andextensionless.) The relation connecting ra,va,andoris va=0.)Xra andtherelation connecting r,,,va,andLis L= Emara Xva(1 Itshould beclear thatorisdirected along theaxisofrotation andthatLisper- pendicular tothelineconnecting mlandm2. Wenote, forthisexample, thattheangular-momentum vector Ldoes notre- main constant intime butrotates with anangular velocity toinsuch a_waythatit traces outacone whose axis istheaxis ofrotation, Therefore L=#0.But Equation 9.31 states that i.=N (11.21) where Nistheexternal torque applied tothebody. Thus, tokeep thedumbbell rotating asinFigure 11-4, wemust constantly apply atorque. Wecanobtain another result from Equation 11.20a bymultiplying L,»by%co,- andsumming over i: 1 15 £261’;-Z41‘ Z 2 Trot 422 11/DYNAMICS orRIGID BODIES where thesecond equality isjustEquation 11.11. Thus, Equations 11.20b and 11.22b illustrate twoimportant properties oftensors. The product ofatensor andavector yields avector, asin L={I}-0.) andtheproduct ofatensor andtwovectors yields ascalar, asin 1ZI"m,=%(.o-L=é-or-{I}-or Weshall not, however, have occasion tousetensor equations inthisform. We useonly thesummation (orintegral) expressions asinEquations 11.11, 11.15, and11.20a. Consider thependulum shown inFigure 11-5 composed ofarigid rodoflength bwith amass mlatitsend. Another mass (m2) isplaced halfway down therod. Find thefrequency ofsmall oscillations ifthependulum swings inaplane. _ Solution. Weusethemethods ofthischapter toanalyze thesystem. Letthe ﬁxed andbody systems have their origin atthependulum pivot point. Letelbe along therod, e2beintheplane, ande2beoutoftheplane (Figure 11-5). The angular velocity is 0.)=(I)3e3 ==deg (11.23) WeuseEquation 11.10 toﬁnd theinertia tensor. Allthemass isalong el,with xlll=bandx2,l=b/2.Allother components ofxmlequal zero. II)"=m1(5I)"xi,1 ‘x1,ix1,j) +m2(8i;’x2,1 ‘X2,t*2,j) (11-24) The inertia tensor, Equation 11.13a, becomes '52 Oowb Outof § plane .§ L g2 ml \-FIGURE 11-5 Example 11.4. Arigid rodrotating asapendulum hasamass mlatits endandanother mass m2halfway. 11.4ANGULAR MOMENTUM 423 0 0 0 b2 Omlb2 +m2— 0 {I}= 4 (11.25) b2 0 0 m1b2 + "Z21 Wedetermine theangular momentum from Equation 11.20a: L1 7'0 L2Z0 (11.26) b2 - L3 Z133(1):; : (7n1b2 + The only external force isgravity, which causes atorque Nonthesystem. Because L=N,wehave b2..(mlb2+11121502, =Zr,><F, (11.27) Because thegravitational force isdown, g=gcos9el -gsin0e2 Thus, rlXFl=belX(cos Bel—sin6e2)mlg= —~mlgb sinGel, b br2XF2==éel X(cos Bel-sinOe2) m2g =—~m2gg sinGel, Equation 11.27 becomes b2(ml + =—bgsin9(ml+ (11.28) andthefrequency ofsmall oscillations is +miml Z 2E 2=___. _(00 m2 b ml + T Wecancheck Equation 11.29 bynoting that(0%==g/bforml>>m2and toil-'=2g/b form2>>mlasitshould. This example could have justaseasily been solved byﬁnding thekinetic en- ergy from Equation 11.22a andusing Lagrange’s equations ofmotion. We 424 11/DYNAMICS OFRIGID BODIES would then have 1 1 Yiot=E0):-1L3 =5wilss 12 b2.2 bU= *mlgb cos0—~m2g§ cos9 (11.31) Where U=0attheorigin. The equation ofmotion (Equation 11.28) follows di- rectly from astraightforward application oftheLagrangian technique. 11.5 Principal Axes ofInertia* Itshould beclear thataconsiderable simpliﬁcation intheexpressions forTand Lwould result iftheinertia tensor consisted only ofdiagonal elements. Ifwe could write Ill=Il5,-j (11.32) then theinertia tensor would be 1,00 {I}=01,0 (11.32) 001, L,=21,5,-jwj =go, (11.24)JWewould then have and 1 1 1;...=521.-6.-,<».»<»,~ =521.-wt <11-25> I] l Thus, thecondition that{I}have only diagonal elements provides quite sim- pleexpressions fortheangular momentum and therotational kinetic energy. Wenow determine theconditions under which Equation 11.32 becomes thede- scription oftheinertia tensor. This involves ﬁnding asetofbody axes forwhich theproducts ofinertia (i.e., theoff-diagonal elements of{I})vanish. Wecall such axes theprincipal axes ofinertia. Ifabody rotates around aprincipal axis, both theangular velocity and the angular momentum are, according toEquation 11.34, directed along this axis. *Discovered byEuler in1750. 11.5 PRINCIPAL AXES orINERTIA 425 Then, ifIistherotational inertia (moment ofinertia) about thisaxis, wecanwrite L=Im (11.36) Equating thecomponents ofLinEquations 11.20a and11.36, wehave L1 2 [(1)1 Z Illﬂll + I12(.U2 + Ilgwg L2 =[(1)2 :121(1)] +IQQIUQ +I2g(l)g LsZIws:Isrwr +Is2w2 +Issws Or,collecting terms, weobtain (I11“DWI +I12¢"2 +Irsws 20 I21(U1 +(I22 "'I)(1)2 +I23(1)3 =20 Isrwr +I:-s2w2 +(I:-rs“I)‘*’s T"0 The condition that these equations have anontrivial solution isthat thede- terminant ofthecoefﬁcients vanish: (I11‘*I) I12 I12 I2] (I22 "‘ I23 =0 I31 I32 (I33“I) The expansion ofthisdeterminant leads tothesecular orcharacteristic equa- tion* forI,which isacubic. Each ofthethree roots corresponds toamoment of inertia about oneoftheprincipal axes. These values, Il,I2,andI3,arecalled the principal moments ofinertia. Ifthebody rotates about theaxiscorresponding to theprincipal moment Il,then Equation 11.36 becomes L=Ilw—-that is,both 0) andLaredirected along thisaxis. The direction oforwith respect tothebody coordinate system isthen thesame asthedirection oftheprincipal axiscorre- sponding toIl.Therefore, wecandetermine thedirection ofthisprincipal axis bysubstituting IlforIinEquation 11.38 and determining theratios ofthecom- ponents oftheangular-velocity vector: wl:w2:w2. Wethereby determine thedirec- tion cosines oftheaxisabout which themoment ofinertia isIl.The directions corresponding toI2andI2canbefound inasimilar fashion. That theprincipal axes determined inthis manner are indeed real and orthogonal isproved in Section 11.7; these results also follow from themore general considerations given inSection 12.4. The factthat thediagonalization procedure just described yields only the ratios ofthecomponents oforisnohandicap, because theratios completely de- termine thedirection ofeach oftheprincipal axes, anditisonly thedirections ofthese axes that arerequired. Indeed, wewould notexpect themagnitudes of theco,tobedetermined, because theactual rate ofthebody’s angular motion cannot bespeciﬁed bythegeometry alone. Wearefree toimpress onthebody anymagnitude oftheangular velocity wewish. *Socalled because asimilar equation describes secular perturbations incelestial mechanics. The mathematical terminology isthecharacteristic polynomial 426 11/DYNAMICS OFRIGID BODIES Formost oftheproblems encountered inrigid-body dynamics, thebodies areofsome regular shape, sowecandetermine theprincipal axes merely byex- amining thesymmetry ofthebody. Forexample, anybody thatisasolid ofrevo- lution (e.g., acylindrical rod) hasoneprincipal axisthatliesalong thesymmetry axis (e.g., thecenter lineofthecylindrical rod), andtheother twoaxes areina plane perpendicular tothesymmetry axis. Itshould beobvious thatbecause the body issymmetrical, thechoice oftheangular placement ofthese other twoaxes isarbitrary. Ifthemoment ofinertia along thesymmetry axisisIl,then I2=I3 forasolid ofrevolution--that is,thesecular equation hasadouble root. Ifabody hasIl=I2=I3,itistermed aspherical top; ifIl=I2abI3,itis termed asymmetric top; iftheprincipal moments ofinertia arealldistinct, itis termed anasymmetric top. Ifabody hasIl=0,Q=I3,as,forexample, two point masses connected byaweightless shaft, oradiatomic molecule, itiscalled arotor. Find theprincipal moments ofinertia andtheprincipal axes forthecube in Example 11.3. Solution. InExample 11.3, wefound that themoment-of-inertia tensor fora cube (with origin atonecorner) hadnonzero off-diagonal elements. Evidently, thecoordinate axes chosen forthatcalculation were notprincipal axes. If,for. example, thecube rotates about thex3-axis, then or=w3e3 and theangular momentum vector L(seeEquation 11.37) hasthecomponents 1 L1: “zﬁws 1 L2=":lB¢°s 2 Ls=5.3"’:-I Thus, 1 1 2LZ Z61“ Zep '1'5153) which isnotinthesame direction asor. Toﬁnd theprincipal moments ofinertia, wemust solve thesecular equation 2 1 1 §B—1 ‘Z5 ‘EB M2 1 I-Pi-‘r-IR)-—‘‘Qpk)--IEUQNJ‘Q "~1—Bgo-I -Z5 =0 (mam 11.5PRINCIPAL AXESorINERTIA 427 The value ofadeterminant isnotaffected byadding (orsubtracting) anyrow (orcolumn) from anyother row(orcolumn). Equation 11.40 canbesolved more easily ifwesubtract theﬁrstrowfrom thesecond: 22-1 -is -is —~%B+I %B-I 0 =0 -22 -22 §B—1 Wecanfactor (%B —-I)from thesecond row: NJ ,.|>)--I*-‘gar-II-—| 5B-I —"B -ZB (EB-1) -1 0=0 12 1 2 ‘EB ""'B §B*1 Expanding, wehave 11 2 21212 _ (123Iiiigﬁ I)554B(3B Iii“) which canbefactored toobtain <II~><I-I->sI->-Thus, wehave thefollowing roots, which give theprincipal moments ofinertia: 1 11 11I=~,I=—-,I=— 16B212B312B The diagonalized moment-of-inertia tensor becomes 1EB 0 0 11{I}— 0 EB 0 (11.41) 11O 0 EB Because twooftheroots areidentical, I2=I3,theprincipal axisassociated with Ilmust beanaxisofsymmetry. 428 11/DYNAMICS orRIGID 11001125 Toﬁnd thedirection oftheprincipal axisassociated with I1,wesubstitute forIinEquation 11.38 thevalue I=I1=%B: 2 1 1 (3B6B) B ,.|>»--ll--‘gmch»-*'**"E N>;/ 1-1“‘*’ W11‘ w21_:1"Bws1 =0 1 2 1 ‘ZBw11+ 53“ ¢°21_:1B¢°s1=0 ““'Bw11 "ZBw21+ (55 “gB)‘°s1 =0 where thesecond subscript 1onthewisigniﬁes thatweareconsidering the principal axisassociated with I1.Dividing theﬁrsttwoofthese equations byB/4, wehave 2¢°11 ““4°21_‘"31Z0 (11.42) ‘("11 +2°°21 -wsl=0 Subtracting thesecond ofthese equations from theﬁrst, weﬁnd wn=(1:21. Using thisresult ineither oftheEquations 11.42, weobtain wn=(021=(1)31, andthedesired ratios are (1)11I(U21I(031 = Therefore, when thecube rotates about anaxisthathasassociated with it themoment ofinertia I1=éB=%Mb2, theprojections of0:onthethree coor- ‘ dinate axes areallequal. Hence, thisprincipal axiscorresponds tothediagonal ofthecube. Because themoments I2andI3areequal, theorientation oftheprincipal axes associated with these moments isarbitrary; they need only lieinaplane normal tothediagonal ofthecube. 11.6 Moments ofInertia forDifferent Body Coordinate Systems Forthekinetic energy tobeseparable into translational androtational portions (seeEquation 11.6), itis,ingeneral, necessary tochoose abody coordinate sys- tem whose origin isthecenter ofmass ofthebody. Forcertain geometrical shapes, itmay notalways beconvenient tocompute theelements oftheinertia tensor using such acoordinate system. Wetherefore consider some other setof coordinate axes X,-,alsoﬁxed with respect tothebody andhaving thesame ori- entation asthexi-axes butwith anorigin Qthat does notcorrespond with the origin O(located atthecenter ofmass ofthebody coordinate system). Origin Q may belocated either within oroutside thebody under consideration. The elements oftheinertia tensor relative totheX,-—axes canbewritten as J1]=gma(8ij; X511 _Xr,1"Xz,;') (11-43) 11.6 MOMENTS OFINERTIA FOR DIFFERENT BODY COORDINATE SYSTEMS X3 *3 O 1' x2 I a R *1 Q X2 X1 FIGURE 11-6 The coordinate axes X,»areﬁxed inthebody andhave thesame orientation asthex,~-axes, butitsorigin Qisnotlocated atthe origin O(atcenter ofmass ofbody system) . Ifthevector connecting Qwith Oisa,then thegeneral vector R(Figure 11-6) canbewritten as R= a+r (11.44) with components X“ 2 (Ii + X,» Using Equation 11.45, thetensor element jgbecomes ,/13'=21% (5¢j;(xa,k +at)?‘(Km; +ai)(-‘$11,; +411)) 2?7na(8§j;'x§,k T'xa,ixa,j) +§m,,(8,j ;(2x,,,,,,a,, +ai)—(a,~x,,,]- +ajxa,,- +a,-aj-)) (11.46) Identifying theﬁrstsummation asIi]-,wehave, onregrouping, jg=I,-j+§m,,<5,-j 241?, —aiaj) +§m,,(25,j gxmkak -a,-x,,_j -ajxai) (11.47) Buteach term inthelastsummation involves asum oftheform 2m,,,x,,,,,I1 Weknow, however, thatbecause Oislocated atthecenter ofmass, Emara =O(Z 430 11/DYNAMICS orRIGID BODIES or,forthekthcomponent, ;m,,x,,,,k =0 Therefore, allsuch terms inEquation 11.47 vanish andwehave 1,3.=1,,+gm,(5,,;a%, —a,-4,) (11.48) But 2111,, =M and gai, Ea2 Solving forI,-J-,wehave theresult IIt]=]ij —M(a25,~]- -a,-aj) ‘ (11.49) which allows thecalculation oftheelements Iijofthedesired inertia tensor (with origin atthecenter ofmass) once those with respect totheX,--axes areknown. The second term ontheright-hand side ofEquation 11.49 istheinertia tensor referred totheorigin Qforapoint mass M. Equation 11.49 isthegeneral form ofSteiner’s parallel-axis theorem,* the simpliﬁed form ofwhich isgiven inelementary treatments. Consider, forexam- ple,Figure 11-7. Element I11is I11:]11 ‘"M[(a1+ 4%+“@511 "ail :]11 '_M(ag +4%) X3 xs O *2 IIII1/QIIIII11$‘IlNII1»:,_IIII\--——-—------\TsX."' /' I / I I / /(ll X1 FIGURE 11-7 The elements I,-7inthex,--axes arerelated tothose (jg)intheX,’--axes byEquation 11.49. Thevector aconnects theorigin Qwith theorigin O. *_]ac0b Steiner (1796-1863). 11.6 MOMENTS OFINERTIA FOR DIFFERENT BODY COORDINATE SYSTEMS 431 which states thatthedifference between theelements isequal tothemass ofthe body multiplied bythesquare ofthedistance between theparallel axes (inthis case, between thex1-andX1-axes). EXAMPLE 11.11 _1 Find theinertia tensor ofthecube ofExample 11.3 inacoordinate system with origin atthecenter ofmass. Solution. InExample 11.3, with theorigin atthecorner ofthecube, wefound theinertia tensor tobe USN)Q" 1.51-IB" 1->1-I_M2 ___M2 __“Mb2 1-I>*-'1-P1-' 1-B1-—*OOI\'J DEN)»-I>>-'{J}=—-Mb? -Mb? --Mb? (11.50) ——Mb2 ———Mb2 —Mb2 Wemay now useEquation 11.49 toobtain theinertia tensor {I}referred toaco- ordinate system with origin atthecenter ofmass. Inkeeping with thenotation ofthissection, wecallthenewaxes x,with origin Oandcalltheprevious axes X,- with origin Qatonecorner ofthecube (Figure 11-8). The center ofmass ofthecube isatthepoint (b/2,b/2,b/2)intheX,»coor- dinate system, andthecomponents ofthevector atherefore are a1==a2=a3=b/2 From Equation 11.50, wehave 2 J11=12 :]ss =§Mb2 1 2]12=]1s=}2s= *j1'Mb X3 xg M _><U‘\\ e~gg-1-—-—-X‘\<2__ D‘£<-5‘I ___ I ,» xl ’ FIGURE 11-8 Example 11.6. The X,--axes have their origin Qatonecorner ofacube of sides b.Thesystem xihasitsorigin Oatthecube’s center ofmass. 432 11/DYNAMICS OFRIGID BODIES And applying Equation 11.49, weﬁnd I11’-“J11” M(a2 ‘ 2:111 -—M/3Q|\3|\3+@ §:~2:B 2 1 1=—Mb2 -~Mb2 =~Mb23 2 6 and I12=]12 “M("a1a2) 1 1==——Mb2 +—Mb2= 04 4 Altogether, wehave 1 I11: I22ZIsa2gMb2 I12=11:1=I23=0 The inertia tensor istherefore diagonal: 1'Mb?’ 0 6 O 1{I}= 0 6Mb2 0 (11.51) 0 0 -1-Mb?6 Ifwefactor outthecommon term %Mb2 from thisexpression, wecanwrite 1{I}‘———EMbg{1} (11.52) 100 {115 010 (11.53) 001 Thus, weﬁnd that, forthechoice oftheorigin atthecenter ofmass ofthe cube, theprincipal axes areperpendicular tothefaces ofthecube. Because, from aphysical standpoint, nothing distinguishes anyoneofthese axes from an- other, theprincipal moments ofinertia areallequal forthiscase. Wenote fur- ther that, aslong aswemaintain theorigin atthecenter ofmass, then theinertia tensor isthesame foranyorientation ofthecoordinate axes andthese axes are equally valid principal axes.*I — I 1 1 | | I I __1 I 1 1 I I I 1where {1}istheunittensor: *Inthisregard, thecube issimilar toasphere asfarastheinertia tensor isconcerned (i.e., foranori- ginatthecenter ofmass, thestructure oftheinertia tensor elements isnotsufﬁciently detailed to discriminate between acube andasphere). 11.7 FURTHER PROPERTIES OFTHE INERTIA TENSOR 433 11.7 Further Properties oftheInertia Tensor Before attacking theproblems ofrigid-body dynamics byobtaining thegeneral equations ofmotion, weshould consider thefundamental importance ofsome oftheoperations wehave been discussing. Letusbegin byexamining theprop- erties oftheinertia tensor under coordinate transformations.* Wehave already obtained thefundamental relation connecting theinertia tensor and theangular momentum and angular velocity vectors (Equation 11.20), which wecanwrite as L,=§1,,,w, (11.5-15) Because thisisavector equation, inacoordinate system rotated with respect to thesystem forwhich Equation l1.54a applies, wemust have anentirely analo- gous relation, L;=21;-J-cu} (115415)J where theprimed quantities allrefer totherotated system. Both Land0.)obey thestandard transformation equation forvectors (Equation 1.8): x-2)t‘x' —EA--x'~i'"]- =7'J'—j J11 Wecantherefore write L,=§1,,,,,L;, (11.55a) and 51,=E11,-,5); (ll.55b) Ifwesubstitute Equations 11.55a andbinto Equation 11.54a, weobtain §21,,,,L;,, =21,,E11,“);. (11.56) Next, wemultiply both sides ofthisequation by)1,-1,and sum over kt €,<§1,,,21,,,,) 1;,=§<%/1-,,21,,1,,,) 1.1; (11.57) The term inparentheses ontheleft-hand sideisjust8,-m,soperforming thesum- mation over mweobtain L;=;(%11,,,21j,1,,,)w; (11.58) Forthisequation tobeidentical with Equation 11.54b, wemust have 1;,=§21,,,1,-,1,,, (11.59)P This istherefore therule that theinertia tensor must obey under acoordinate transformation. Equation 11.59 is,infact, thegeneral rulespecifying themanner *We conﬁne ourattention torectangular coordinate systems s0thatwemay ignore some ofthe more complicated properties oftensors thatmanifest themselves ingeneral curvilinear coordinates. 434 ll/DYNAMICS OFRIGID BODIES inwhich anysecond-rank tensor must transform. Foratensor {T}ofarbitrary rank, thestatement is* T;......-,,,E,___1..1.,1.11..-~- ir.,»..__ <11.B0> Note thatwecanwrite Equation 11.59 as 11¢%1,,,1,,,1g- (11.61) Although matrices and tensors aredistinct types ofmathematical objects, the manipulation oftensors isinmany respects thesame asformatrices. Thus, Equation 11.61 canbeexpressed asamatrix equation: I’==MA‘ (11.62) where weunderstand Itobethematrix consisting oftheelements ofthetensor {I}.Because weareconsidering only orthogonal transformation matrices, the transpose ofAisequal toitsinverse, sowecanexpress Equation 11.62 as 1'=11>.-1 (11.66) Atransformation ofthisgeneral type iscalled asimilarity transformation (I’is similar toI). l . Prove theassertion stated inExample 11.6 thattheinertia tensor foracube (with origin atthecenter ofmass) isindependent oftheorientation ofthe axes. Solution. The change intheinertia tensor under arotation ofthecoordinate axes canbecomputed bymaking asimilarity transformation. Thus, iftherota- tionisdescribed bythematrix A,wehave I’=MA“ (11.64) Butthematrix I,which isderived from theelements ofthetensor {I}(Equation 11.52 ofExample 11.4), isjusttheidentity matrix 1multiplied byaconstant: 1 1001 |=g1\/162 010=51/1621 (11.65) 001 *Note thatatensor oftheﬁrstrank transforms as T;=21.11". Such atensor isinfactavector. Atensor ofzero rank implies that T’=T,orthatsuch atensor isa scalar. Theproperties ofquantities thattransform inthismanner were ﬁrstdiscussed byC.Niven in 1874. Theapplication oftheterm tensor tosuch quantities canbetraced to].Willard Gibbs. 11.7 FURTHER PROPERTIES OFTHE INERTIA TENSOR 435 Therefore, theoperations speciﬁed inEquation 11.64 aretrivial: 1 1 1I’=5Mb2)t1)t'1 =5Mb2)¢A"1 =EMb21 =I (11.66) Thus, thetransformed inertia tensor isidentical totheoriginal tensor, inde- pendent ofthedetails oftherotation. Letusnext determine what condition must besatisﬁed ifwetake anarbi- trary inertia tensor and perform acoordinate rotation insuch away that the transformed inertia tensor isdiagonal. Such anoperation implies thatthequan- tityIinEquation 11.59 must satisfy (seeEquation 11.32) therelation lg»=I,-5,7 (11.67) Thus, 1,-6,,=%21,,,1,.,1,,, (11.68) Ifwemultiply both sides ofthisequation by21,",andsum over i,weobtain Z1.1...B., =§(Z1..1..)1,.1.. <11-B9) The term inparentheses isjust8,,,,,,sothesummation over iontheleft-hand side oftheequation andthesummation over kontheright-hand sideyield 151,,=E1,-,1,,,, (11.70) Now theleft-hand side ofthisequation canbewritten as 1,1,,=21,21,-,6,,,, (11.71) soEquation 11.70 becomes E1571,-,8,,,, =E1,-,1,,, (11.72a) or §(I..1* Ij8ml)Ajl I0 (11-72b) This isasetofsimultaneous linear algebraic equations; foreach value ofjthere arethree such equations, oneforeach ofthethree possible values ofm.Fora nontrivial solution toexist, thedeterminant ofthecoefﬁcients must vanish, so theprincipal moments ofinertia, I1,I2,andI3,areobtained asroots ofthesecu- lardeterminant forI: Ilml HIamli :U This equation isjustEquation 11.39; itisacubic equation thatyields theprincipal moments ofinertia. 436 11/DYNAMICS orR1011) BODIES Thus, foranyinertia tensor, theelements ofwhich arecomputed foragiven origin, itispossible toperform arotation ofthecoordinate axes about that ori- gininsuch awaythat theinertia tensor becomes diagonal. The new coordinate axes arethen theprincipal axes ofthebody, andthenew moments aretheprin- cipal moments ofinertia. Thus, foranybody andforanychoice oforigin, there always exists asetofprincipal axes. Forthecube ofExample 11.3, diagonalize theinertia tensor byrotating theco- ordinate axes. Solution. Wechoose theorigin tolieatonecorner andperform therotation insuch amanner thatthex1-axis isrotated totheoriginal diagonal ofthecube. Such arotation canconveniently bemade intwosteps: ﬁrst, werotate through anangle of45°about thex3-axis; second, werotate through anangle of cos‘1(\/2) about thexé-axis. The ﬁrstrotation matrix is '_ — 0 A1=_1 i 0 (11.74) 0 1 andthesecond rotation matrix is B0-3 \/5 A2=( 0 1 0) (11.75) 1 2__\730\/g The complete rotation matrix iso§$d$~§7§W _§~§“ ,-s-s-§m‘°§“3.. .8-$7siat1%..)1=11,11,= --— ——— =-—- ‘‘ ' 0(11.76) \/6 W The matrix form ofthetransformed inertia tensor (seeEquation 11.62) is I’=MA‘ (11.77) 11.7 FURTHER PROPERTIES OFTHEINERTIA TENSOR 437 or,factoring BoutofI, 1 2_1_r _1._,_ 1 1 1 3 |I=g _ T T “T 2 “‘"' _ __,_ 3 -'— —"'— -'" —" r 1 0 1_..1_1.\/B _1_1ﬁ6 122 122 1 6097% Z3,$$7$E"°$'°' §m%|_§am%1O..13° M’)I-I§*'-‘I-I31‘-' 11>»-Os11> U9I\DH;h_|h>)_|Z""'€—$&'&1 . —~— - E\/3 ..1_1_\/3=* 6122 122 ---~\/B-0Eva 1 1 1 11 /163 0 0 =-0gr; 0 (11.78) 11.0 0 — \ 12B Equation 11.78 isjust thematrix form oftheinertia tensor found bythe diagonalization procedure using the secular determinant (Equation 11.41 of Example 11.5). Wehave demonstrated twogeneral procedures todiagonalize theinertia tensor. Wepreviously pointed outthatthese methods arenotlimited totheiner- tiatensor butaregenerally valid. Either procedure canbeVery complicated. For example, ifwewish tousetherotation procedure inthemost general case, we must ﬁrst construct amatrix that describes anarbitrary rotation. This entails three separate rotations, one about each ofthecoordinate axes. This rotation matrix must then beapplied tothetensor inasimilarity transformation. The off- diagonal elements oftheresulting matrix* must then beexamined andvalues of therotation angles determined sothat these off-diagonal elements vanish. The actual useofsuch aprocedure cantaxthelimits ofhuman patience, butinsome simple situations, thismethod ofdiagonalization canbeused with proﬁt. This is particularly true ifthegeometry oftheproblem indicates thatonly asimple rota- tionabout oneofthecoordinate axes isnecessary; therotation angle canthen be evaluated without difﬁculty (see, forexample, Problems 11-16, 11-18, and11-19). *Alargresheet ofpaper should beused! 438 11/DYNAMICS orR1011) BODIES Inpractice, there aresystematic procedures forﬁnding principal moments andprincipal axes ofanyinertia tensor. Standard computer programs andhand- calculator methods areavailable toﬁnd thenroots ofannth-order polynomial andtodiagonalize amatrix. When theprincipal moments areknown, theprin- cipal axes areeasily found. The example ofthecube illustrates theimportant point thattheelements of theinertia tensor, thevalues oftheprincipal moments ofinertia, andtheorien- tation oftheprincipal axes forarigid body alldepend onthechoice oforigin forthesystem. Recall, however, that forthekinetic energy tobeseparable into translational and rotational portions, theorigin ofthebody coordinate system must, ingeneral, betaken tocoincide with thecenter ofmass ofthebody. However, foranychoice oftheorigin foranybody, there always exists anorienta- tion oftheaxes that diagonalizes theinertia tensor. Hence, these axes become principal axes forthatparticular origin. Next, weseek toprove that theprincipal axes actually form anorthogonal set.Letusassume thatwehave solved thesecular equation andhave determined theprincipal moments ofinertia, allofwhich aredistinct. Weknow thatforeach principal moment there exists acorresponding principal axiswith theproperty that, iftheangular velocity vector coliesalong thisaxis, then theangular mo- menturn vector Lissimilarly oriented; thatis,toeach there corresponds anan- gular velocity mjwith components ml]-,(02,,603]‘.(Weusethesubscript onthevec- toro)and thesecond subscript onthecomponents of0)todesignate the principal moment with which weareconcerned.) Forthemthprincipal" mo- ment, wehave L,-,,,=I,,,w,-,,, (11.79) Interms oftheelements ofthemoment-of-inertia tensor, wealsohave 1,,=§1r,,,w,,,, (11.80) Combining these tworelations, wehave §1,,w,,, =1,16,, (ll.81a) Similarly, wecanwrite forthenthprincipal moment: E11,,(6,,=1,,¢.1,,,, (11.811>) Ifwemultiply Equation 11.81a bywinandsum over iand then multiply Equation 11.81b bywk”,andsum over k,wehave Eklikwkmwin 2 wimwin ’ (11.82) Eklkiwinwkm Zglnwknwkm1 The left-hand sides ofthese equations areidentical, because theinertia tensor is symmetrical (I,-k==I,,,»).Therefore, onsubtracting thesecond equation from the 11.7 FURTHER PROPERTIES OFTHE INERTIA TENSOR 439 ﬁrst, wehave Imzwimwin —Ingwkm wkn :0 Because iand kareboth dummy indices, wecanreplace them byl,say,and obtain (1..P1..>§<»,..w,. =0 <11-84> Byhypothesis, theprincipal moments aredistinct, sothat I",9*I,,.Therefore, Equation 11.84 canbesatisﬁed only if 20),,w,,,=0 (11.85) Butthissummation isjust thedeﬁnition ofthescalar product ofthevectors mm and0),,Hence, o:,,,-0),,==0 (11.86) Because theprincipal moments I,,,andInwere picked arbitrarily from thesetof three moments, weconclude that each pair ofprincipal axes isperpendicular; thethree principal axes therefore constitute anorthogonal set. Ifadouble root ofthesecular equation exists, sothattheprincipal moments areI1,Q=I3,then thepreceding analysis shows thattheangular velocity vectors satisfy therelations o:1J_o:2, wlloag _ butthat nothing may besaid regarding theangle between 0:2and 0:3.Butthe factthatI2=I3implies that thebody possesses anaxisofsymmetry. Therefore, 0:1liesalong thesymmetry axis, and 0:2and 0:3arerequired only tolieinthe plane perpendicular to0:1.Consequently, there isnolossofgenerality ifwealso choose (x)2J_(|.)3. Thus, theprincipal axes forarigid body with anaxisofsymme- trycanalsobechosen tobeanorthogonal set. Wehave previously shown that theprincipal moments ofinertia areob- tained astheroots ofthesecular equation—a cubic equation. Mathematically, at least one oftheroots ofacubic equation must bereal, butthere may betwo imaginary roots. Ifthediagonalization procedures fortheinertia tensor areto bephysically meaningful, wemust always obtain only realvalues fortheprincipal moments. Wecanshow inthefollowing waythatthisisageneral result. First, we assume theroots tobecomplex anduseaprocedure similar tothatused inthe preceding proof. Butnow wemust alsoallow thequantities w,,,,,tobecome com- plex. There isnomathematical reason why wecannot dothis, andwearenot concerned with anyphysical interpretation ofthese quantities. Wetherefore write Equation 1l.81a asbefore, butwetake thecomplex conjugate ofEquation 1l.81b: %Iikwkm :Imwim §1;:;<»;:.=Izwz. “"8” Next, wemulti ltheﬁrst ofthese euations bw’?andsum over iand multi-PY q Ym plythesecond bywk",andsum over k.The inertia tensor issymmetrical, andits 440 11/DYNAMICS OFRIGID BODIES elements areallreal, sothat I,-,,=If,-.Therefore, subtracting thesecond of these equations from theﬁrst, weﬁnd (1,,-1;;)§w,,,w;;, =0 (11.88) Forthecase m=n,wehave (1,,-1;,)§w,,,w;;, =0 (11.89) The sum isjustthedeﬁnition ofthescalar product ofmmand01);; mm-win =|o:,,,|2 20 (11.90) Therefore, because thesquared magnitude ofmmisingeneral positive, it must betrue thatI,,,—I;forEquation 11.89 tobesatisﬁed. Ifaquantity andits complex conjugate areequal, then theimaginary parts must vanish identically. Thus, theprincipal moments ofinertia areallreal. Because {I}isreal, thevec- torsox,"must alsobereal. Ifm95ninEquation 11.88 andifIm#5In,then theequation canbesatisﬁed only if0),,-can=0;thatis,these vectors areorthogonal, asbefore. Inalltheproofs carried outinthissection, wehave referred totheinertia tensor. Butexamining these proofs reveals thattheonly properties oftheinertia tensor that have actually been used arethefacts that thetensor issymmetrical and that theelements arereal. Wemay therefore conclude that anyreal, sym- metric tensor* hasthefollowing properties: 1.Diagonalization may beaccomplished byanappropriate rotation ofaxes, thatis,asimilarity transformation. 2.The eigenvaluesi areobtained asroots ofthesecular determinant andare real. 3.The eigenvectorsl arerealandorthogonal. 11.8 Eulerian Angles The transformation from onecoordinate system toanother canberepresented byamatrix equation oftheform X=AX' Ifweidentify theﬁxed system with X’andthebody system with X,then therota- tionmatrix Acompletely describes therelative orientation ofthetwosystems. The rotation matrix Itcontains three independent angles. There aremany possible *Tobemore precise, werequire only thattheelements ofthetensor obey therelation I“,=If-;thus weallow thepossibility ofcomplex quantities. Tensors (and matrices) with thisproperty aresaidto beHermitean. TThe terms eigenvalues andeigenvectmrs arethegeneric names ofthequantities, which, inthecase of theinertia tensor, aretheprincipal moments andtheprincipal axes, respectively. Weshall encounter these terms again inthediscussion ofsmall oscillations inChapter 12. 11.8EULERIAN ANGLES 441 M e [<13 0 Ty;x2 'ti1V .._.’ \! ffta) 6'” "I Line ofnodes *1=x1 (6) (b) FIGURE 11-9 TheEulerian angles areused torotate from thex,Ysystem tothex, system. (a)First rotation iscounterclockwise through anangle qb about thexg-axis. (b)Second rotation iscounterclockwise through anangle 6about thex,”-axis. (c)Third rotation iscounterclockwise through anangle 111about thex§”—axis. choices forthese angles; weﬁnd itconvenient tousetheEulerian ang1es* ¢>,0, and11!. The Eulerian angles aregenerated inthefollowing series ofrotations, which takes thex}system into thex,system.* 1.The ﬁrst rotation iscounterclockwise through anangle ¢about thexg-axis (Figure 11-9a) totransform thex;into thex§’.Because therotation takes place inthexi-x5 plane, thetransformation matrix is cos<1; sin¢0 A4,==rsin d)cos¢0 (11.91) 0 0 1 and x”=)t¢x’ (11.92) 2.The second rotation iscounterclockwise through anangle 9about thex’1'- axis (Figure 11-9b) totransform thex’,-’into thex’1".Because therotation is now inthexg-xg’ plane, thetransformation matrix is 1 0 0 I16=O cos9 sin0 (11.93) 0*sin0cos6 and x'”==Aax” (11.94) *The rotation scheme ofEuler wasﬁrstpublished in1776. 1-The designations oftheEuler angles andeven themanner inwhich theyaregenerated arenotuni- versally agreed upon. Therefore, some care must betaken incomparing anyresults from different sources. Thenotation used here isthatmost commonly found inmodern texts. 442 11/DYNAMICS OFRIGID BODIES 3.The third rotation iscounterclockwise through anangle allabout thex’5.’,'-axis (Figure 11-9c) totransform thex’;-’into thexi.The transformation matrix is cos11/ sin11/0 AwI—sin1/1cos1/10 (11.95) 0 0 1 and X=2.,,x'" (11.96) The line common totheplanes containing thex1-and sq-axes and thex{- and x§-axes iscalled theline ofnodes. The complete transformation from the system tothex,-system isgiven by X :: Awxllf =: Awhoxlf ==)t,,,A0)t¢x’ (11.97) andtherotation matrix )1is A=)t,),A0)t¢ (11.98) The components ofthismatrix are A11: cost!/cos¢~ cos(9 S1I1¢SiI11// ‘ )t21== -—sin|,U cos¢ —-cos6 sinqb cos¢r A31=sin0sin¢ A12=cost]; sin¢ +cos6 cos¢ sin|,!/ A22=rsin ((1sin<15+cos0cos¢cos11/i> (11.99) A32==-sin 0cosda A13==sinillsin9 A23==cos11:sin6 A33=cos9 J (The components Agareoffset inthepreceding equation toassist inthevisuali- zation ofthecomplete Amatrix.) Because wecanassociate avector with aninﬁnitesimal rotation, wecanasso- ciate thetime derivatives ofthese rotation angles with thecomponents ofthean- gular velocity vector 0.).Thus, (U9=0 (11.100) cu,/,==|,l/ The rigid-body equations ofmotion aremost conveniently expressed inthe body coordinate system (i.e., thex,-system), and therefore wemust express the components of0.)inthissystem. Wenote thatinFigure 11-9theangular velocities 11.8EULERIAN ANGLES 443 cl), and aredirected along thefollowing axes: (inalong thexg-(ﬁxed) axis 9along thelineofnodes 11;along thex3-(body) axis Thecomponents ofthese angular velocities along thebody coordinate axes are <13,=<13sin(9Sin1// <1?=<1.)sin19cos1/1 (11.10la) ¢3=4:cos9 91=9cos1,11 92=-9sin1/1 (11.1011>) ég =:0 121=01,112=0 (11.101c) 1/'3Z1/’ Collecting theindividual components of1.0,wehave, ﬁnally, w1==(£1"l-é1'l-J11:(f3S1Il9SiI1I,U+éCOSl/I 1»,=¢2+152+112=4351110 costb-— 9sin1,l1 '(11.102) w3=¢3+03+1ﬂ3=¢cos9+1b These relations willbeofuselater inexpressing thecomponents oftheangular momentum inthebody coordinate system. EXAMPLE 11.9 In I Using theEulerian angles, ﬁnd thetransformation thatmoves theoriginal x’1- axistothexg-xg plane halfway between x§andxgandmoves xéperpendicular tothex;'_.-xg plane (Figure 11-10). Solution. The keytotransformations using Eulerian angles isthesecond rota- tion about thelineofnodes, because thissingle rotation must move xgtox3. From thestatement oftheproblem, x3must beinthex§-xgplane, rotated 45° from xg.The ﬁrstrotation must move xitox’{tohave thecorrect position to rotate .->1;==x'5.’;tox'§'=x3. Inthiscase, xg=xgisrotated 6=45°about theoriginal xi=x'1'-axis so that <11==Oand A,»=1 (11.103) 1 0 0 A,=01/\/5 1/\/§ (11.104) 0-1/\/5 1/\/5 444 11/DYNAMICS OFRIGID BODIES IX I3'5 I ‘ I\\ I \ II $2 \\ I /I \ I I’ \ ' I\ I I \ / ,'\ ' 1\ II 1/ \ \ I’ /I\ I \ I,’\ I1 \ II\1,’ , -*2 xi FIGURE ll-10 Example 11.9. WeuseEulerian angles torotate thexisystem into thex,-system. The lastrotation, 11/=90°,moves xi=x’i=x'i’toxitotheposition desired in theoriginal mg-x3 plane. 010 Ad,==-1 00 (11.105) 001 The transformation matrix AisA=A,),A@A¢ ==A,),A0: 0101 0 0 )1=-10001/\/§ 1/\/5 0010F1/\/§ 1/\/5 01/\/5 1/\/5 11=-1 0 0 111.106) 0-1/\/5 1/\/5 Direction comparison between thex,--andxi--axes shows thatArepresents asin- glerotation describing thetransformation. 11.9 Euler’s Equations foraRigid Body Letusﬁrstconsider theforce-free motion ofarigid body. Insuch acase, thepo- tential energy Uvanishes andtheLagrangian Lbecomes identical with therota- tional kinetic energy T.*Ifwechoose thex,--axes tocorrespond totheprincipal *Because themotion isforce free, thetranslational kinetic energy isunimportant forourpurposes here. (Wecanalways transform toacoordinate system inwhich thecenter ofmass ofthebody isat rest.) 11.9 EULER’S EQUATIONS FORARIGID BODY 445 axes ofthebody, then from Equation 11.35 wehave 1T=521,1»? (11.10?) Ifwe choose theEulerian angles asthegeneralized coordinates, then Lagrange’s equation forthecoordinate tilis ar.1ar—-—=0 (11.10s)awdt61/I which canbeexpressed as §H‘a&_i§£‘<"°1idw,-at/I 53(1),-alﬂ=0 (11.109) Ifwedifferentiate thecomponents ofto(Equation 11.102) with respect to11/and th,wehave "& <91" 6(1)? -. _ - J==-—¢s1n9s1n1,!/-— 6cos1,lr== -—wi (11.110) éaZ031/1==qisin9cos1,b-—9sin1/1== m2 and awl :81? ::0 53¢’3*” (11.111) =1(93 31!’ From Equation 11.107, wealsohave 6T—=I,-mi (11.1l2)81:0, Equation 11.109 therefore becomes dI1(l)1Cl)2 +I2(1)2(*_(D1) '_Ztlgwg ::0 or (I1 '_ I2)(1)1(U2 '_ Igélg =2 0 446 11/DYNAMICS OFRIGID BODIES Because thedesignation ofanyparticular principal axisasthex3-axis isen- tirely arbitrary, Equation 11.113 canbepermuted toobtain relations for(biand (1)2: (I2TI3)¢"2‘1-'3 T11°31 Z0 (I3-—Ii)w3wi -—I202 =0 (11.114) (I1T§)w1‘1’2 TIs‘3’s T0 Equations 11.114 arecalled Euler’s equations forforce-free motion.* Itmust be noted that, although Equation 11.113 for0'13isindeed theLagrange equation for thecoordinate 1,0,theEuler equations forobiand12:2arenottheLagrange equa- tions for0and¢>. Toobtain Euler’s equations formotion inaforce ﬁeld, wemay start with the fundamental relation (seeEquation 2.83) forthetorque N: dL<7) ~N (11.115)tﬁxed where thedesignation “ﬁxed” hasbeen explicitly appended toLbecause thisre- lation isderived from Newton’s equation andistherefore valid only inaniner- tialframe ofreference. From Equation 10.12 wehave d dI +0)XL (11.116) dt ﬁxed dt body OI’ d (i) +1,,xL=N (11.11?)dt body The component ofthisequation along thex3—axis (note thatthisisabodyaxis) is L,+wiL2-w2Li=N3 (11.11s) Butbecause wehave chosen thex,--axes tocoincide with theprincipal axes of thebody, wehave, from Equation 11.34, L,-==Iiw, sothat 13033 -—(Ii-—I2)wiw2 =N3 (11.119) Bypermuting thesubscripts, wecanwrite allthree components ofN: I191 T(I2TIs)¢"2“’s TN1 12452 T(IsTI1)wsw1 ZM (11-120) I55’:-1 T(I1TI2)w1w2 TNa *Leonard Euler. 1758. 11.9 EULER’S EQUATIONS FORARIGID BODY 447 Using thepermutation symbol, wecanwrite, ingeneral (1,-Ij)w-w- ~(1,121,-N,)8,,,=0 (11.121) Equations 11.120 and11.121 arethedesired Euler equations forthemotion ofa rigid body inaforce ﬁeld. The motion ofarigid body depends onthestructure ofthebody only through thethree numbers Ii,I2,andI3—that is,theprincipal moments ofiner- tia.Thus, anytwobodies with thesame principal moments move inexactly the same manner, regardless ofthefactthat they may have quite different shapes. (However, effects such asfrictional retardation may depend ontheshape ofa body.) The simplest geometrical shape that abody having three given principal moments may possess isahomogeneous ellipsoid. The motion ofanyrigid body cantherefore berepresented bythemotion oftheequivalent ellips0id.* The treatment ofrigid-body dynamics from this point ofview was originated by Poinsot in1834. The Poinsot construction issometimes useful fordepicting the motion ofarigid body geometrically.i EXAMPLE11.10 -___------——- Consider thedumbbell ofSection 11.4. Find theangular momentum ofthesys- temandthetorque required tomaintain themotion shown inFigures 11-4 and 11-11. A1"1/1;. ml a I‘L 1 82 0 1'2 "12 "2 FIGURE 11-11 Example 11.10. The dumbbell with masses miandm2attheends of itsshaft hasitsangular momentum Lperpendicular totheshaft and Lrotates around to.The shaft maintains anangle awith to. (ii, *The momental ellipsoid wasintroduced bytheFrench mathematician Baron Augustin Louis Cauchy (1789-1857) in1827. TSee, forexample, Goldstein (G080, p.205). 448 11/DYNAMICS OFRIGID BODIES Solution. Let|ri|=|r2|==b.Letthebody ﬁxed coordinate system have itsori- ginatOandthesymmetry axisx3bealong theweightless shaft toward mi. L=Emmi], xv, (11.122) Because Lisperpendicular totheshaft andLrotates around 0)astheshaft ro- tates, lete2bealong L: L=:L262 Ifaistheangle between 0:andtheshaft, thecomponents Of(11) are (U1 :0 (1)2==tosina (11.124) (03==wcosa The principal axes arexi,xi,andxii,andtheprincipal moments ofinertia are, from Equation 11.13a, I1=(ml+m2)b2 I2=(mi+m2)b2 (11.125) Combining Equations 11.124 and 11.125 L1 =: Illﬂl :: 0 L2=12102 ==(mi+m2)b2w sina (1l.l26) L3 :Igwg :0 which agrees with Equation 11.123. Using Euler’s equations (Equation 11.120) and11;==0,thetorque compo- nents are Ni=-—(mi +m2)b2w2 sinacosa N2=0 (11.127) N3 :0 The torque required tomaintain themotion ifd1==0isdirected along the xi-axis. 7 _7 11.10 Force-Free Motion ofaSymmetric Top Ifweconsider asymmetric top, that is,arigid body with Ii=I2viI3,then the force-free Euler equations (Equation 11.114) become (I1TI3)w2w3 T11511 Z0 (I3 T_ I1)(l)3(l)1 TT IICUQ Z 0 I3(b3 ==0 11.10 FORCE-FREE MOTION OFASYMMETRIC TOP 449 where Iihasbeen substituted forI2.Because forforce-free motion thecenter of mass ofthebody iseither atrestorinuniform motion with respect totheﬁxed orinertial frame ofreference, wecan, without lossofgenerality, specify thatthe body’s center ofmass isatrestandlocated attheorigin oftheﬁxed coordinate system. Weconsider thecase inwhich theangular velocity vector 0.)does notlie along aprincipal axisofthebody, otherwise, themotion istrivial. The ﬁrst result forthemotion follows from thethird part ofEquations 11.128, 6:3=0,or w3(t) =const. (11.129) The ﬁrsttwoparts ofEquation 11.128 canbewritten as .__ IaTI1(D1*— TT Twg (1)2 11.130 _I1 1>I (U2 : (3 I1 (O3)(U1 Because theterms intheparentheses areidentical andcomposed ofconstants, wemay deﬁne 1—10E%co3 (11.1s1)1 sothat (1)1 + 1Q(U2 = 0 11.1 These arecoupled equations offamiliar form, andwecaneffect asolution by multiplying thesecond equation byiand adding totheﬁrst: (obi+£612) -—i.Q(wi +iw2) ==0 (1l.l33) Ifwe deﬁne 17Emi+'i(1)2 (ll.l34) then 1*)-—LOT] =0 (11.135) with solution* 17(t) ==Aeim (11.1.36) Thus, wi+iwi,==Acos.Qt+ z'Asin.(2t (11.137) *Ingeneral, theconstant coefﬁcient iscomplex, soweshould properly write Aexp(i5). Forsimplic- ity,hOwever, wesetthephase 5equal tozero; thiscanalways bedone bychoosing anappropriate in- stant tocallt=0. 450 11/DYNAMICS orRIGID BODIES X5 ‘>-im, FIGURE 11-12 Theangular velocity toofaforce freesymmetric topprecesses with constant angular velocityfl about thesymmetric xii-axis ofthebody. Thus totraces outacone around thebody symmetric axis. andtherefore wi(t) ==Acos Qt 012(1)=Asinm) (11138) Because 103=constant, wenote thatthemagnitude of0:isalsoconstant: |o.)|=w=\/mi +103+ wit,’=\/A2 +(0%=constant (11.139) Equations 11.138 aretheparametric equations ofacircle, sotheprojection of thevector 0.)(which isofconstant magnitude) onto thexi-x2 plane describes a circle with time (Figure 11-12). The x3-axis isthesymmetry axisofthebody, soweﬁnd that theangular ve- locity vector 0.)revolves orprecesses about thebody xii-axis with aconstant angular frequency Q.Thus, toanobserver inthebody coordinate system, (11)traces outa cone around thebody symmetry axis, called thebody cone. Because weareconsidering force-free motion, theangular-momentum vec- torLisstationary intheﬁxed coordinate system andisconstant intime. Anad- ditional constant ofthemotion fortheforce-free case isthekinetic energy, orin particular, because thebody’s center ofmass isﬁxed, therotational kinetic energy isconstant: Tm,=gm-L=constant (11.140) Butwehave L=constant, sowmust move such that itsprojection onthesta- tionary angular-mornentum vector isconstant. Thus, toprecesses around and makes aconstant angle With thevector L.Insuch acase, L,0),and thex3-(body) axis (i.e., theunit vector e3)alllieinaplane. Wecanshow thisbyproving that L-(coXe3)=0.First, 0.)Xeii==wiei -—wiei. Ifwetake thescalar product of 11.10 FORCE-FREE MOTION orASYMMETRIC TOP 451 / X3 m L SPace 3,» cone x2 =-.1 I,2; E;-~ '='-:=az=.Tflifiii .-i .11,...~¢i_§;;~ ~'1,,2§= 3*-7,, *1I‘ .= W;.1,ti:1:.~’;=.=<11=1=.:i==.--_1.115;“ 1555:‘: Body cone X1 xi FIGURE 11-13 Welettheangular momentum Lbealong theﬁxed xi-axis. The angular velocity totraces outthebody cone asitprecesses about the x5-axis inthebody system, andittraces outthespace cone asit precesses around thexii-axis inthespace-ﬁxed system. Wecan imagine thebody cone rolling around thespace cone. thisresult with L,wehave L-(coXe3)==Iiwiwi -—Iiwiwi =0,because Ii=I2 forthesymmetric top.Therefore, ifwedesignate thexii-axis intheﬁxed coordi- nate system tocoincide with L,then toanobserver intheﬁxed system, wtraces outacone around theﬁxed xii-axis, called thespace cone. The situation isthen described (Figure 11-13) byonecone rolling onanother, such that 0:precesses around thexii-axis inthebody system and around thexii-axis (orL)inthe space-ﬁxed system. i The rate atwhich 0:precesses around thebody symmetry axisisgiven by Equation 11.131: I:-ITIIQ "T I1 (1)3 IfIiEI3,then (2becomes very small compared with (U3.Earth isslightly ﬂat- tened near thepoles,* soitsshape canbeapproximated byanoblate spheroid with IiEI3,butwith I3>Ii.IfEarth isconsidered tobearigid body, then the moments IiandI3aresuch thatQEmg/300. Because theperiod ofEarth’s ro- tation is211'/w =1day,andbecause mi,Ecu,theperiod predicted forthepreces- sion oftheaxisofrotation is1/Q E300days. The observed precession hasan irregular period about 50percent greater than thatpredicted onthebasis ofthis simple theory; thedeviation isascribed tothefacts that (1)Earth isnotarigid body and(2)theshape isnotexactly thatofanoblate spheroid, butrather hasa higher-order deformation and actually resembles aﬂattened pear. Earth’s equatorial “bulge” together with thefactthat Earth’s rotational axis isinclined atanangle ofapproximately 23.5° totheplane ofEarth’s orbit around thesun (the plane oftheecliptic) produces agravitational torque (caused byboth theSun and theMoon), which produces aslow precession of MM *The ﬂattening atthepoles wasshown byNewton tobecaused byEarth’s rotation; theresulting pre- cessional motion wasﬁrstcalculated byEuler. 452 11/DYNAMICS OFRIGID BODIES Earth’s axis. The period ofthisprecessional motion isapproximately 26,000 years. Thus, indifferent epochs, different stars become the“pole star.”* EX.-\l\I-1I’I.l*‘. ll.ll Show thatthemotion depicted inFigure 11-13 actually refers tothemotion ofa prolate object such asanelongated rod(Ii>I3),whereas foraﬂatdisk (I3>Ii) thespace cone would beinside thebody cone rather than outside. Solution. IfLisalong x3,then theEuler angle 9(between thex3-andx3-axes) istheangle between Landthex3-axis. Atagiven instant, wealign e2tobein theplane deﬁned byL,ox,ande3.Then, atthissame instant, Li=0 L2=1LSiI1 9 (1l.l41) L3 Z LCOS 6 Letabetheangle between 0.)andthex3-axis. Then, atthissame instant, we have (U1 =:0 w2‘-=tosina (1l.142) w3=wcosa Wecanalsodetermine thecomponents ofLfrom Equation 11.34: L1 ZIlwl 2:O L2=Iiw2 =Iiwsina (11.143) L3==I3w3 -‘I13wcosa Wecanobtain theratio L2/L3 from Equations 11.141 and11.143, —=ta =—ta . L2 0I1 11144 L3 n I3na ( ) sowehave Prolate spheroid Ii>I3, 0>a (11.145a) Oblate spheroid I3>Ii, a>0 (11.145b) The twocases areshown inFigure 11-14. From Equation 11.131, wedeter- mine thatQandw3have thesame sign ifI3>Iibuthave opposite signs if *This precession oftheequinoxes wasapparently discovered bytheBabylonian astronomer Cidenas inabout 343B.C. 11.10 FORCE-FREE MOTION OFASYMMETRIC TOP 453 x3 xii L L *3 “'"' m m ....._; »..iz:;:= . ‘522?’”',3 1WI" 1 *3 ‘cg‘ts._asSpace Space ‘iii ﬁxed ﬁxed 3Q cone cone I Bd ’Cg“: .0. Body cone‘III Prolate, Ii>I3 Oblate, I3>Ii Q,033have opposite signs. Q,033have same sign. (K) (b) FIGURE ll-14 Example 11.11. (a)VVhen thebody isprolate (Ii>I3),wehave the situation here andinFigure 11-13. (b)When thebody isoblate (I3>Ii),theinside ofthebody cone rotates around theoutside of thespace cone. The space cone isatrestineither case.A‘I Ii>I3.Thus, thesense ofprecession isopposite forthetwocases. This factand Equation 11.145 canbereconciled only ifthespace cone isoutside thebody cone fortheprolate case butinside thebody cone fortheoblate case. The an- gular velocity 0)deﬁnes both cones asitrotates about L(space cone) andthe symmetry axisx3(body cone). The lineofcontact between thespace andbody cones istheinstantaneous axisofrotation (along 0.)).Atanyinstant, thisaxisis atrest, sothatthebody cone rolls around thespace cone without slipping. In both cases, thespace cone isﬁxed, because Lisconstant. With what angular velocity does thesymmetry axis (x3)and0:rotate about the ﬁxed angular momentum L? Solution. Because e3,co,andLareinthesame plane, e3andtoprecess about Lwith thesame angular velocity. InSection 11.8welearned that istheangu- larvelocity along thex3-axis. Ifweusethesame instant oftime considered in theprevious example (when e2wasintheplane ofe3,co,andL),then theEuler angle 1/;=0,andfrom Equation 11.102 (02= sin0 and '—“)2 11146¢—sin6 (l) Substituting forw2from Equation 11.142, wehave .tosina(D= (ll.l47) 454 11/DYNAMICS OFRIGID BODIES Wecanrewrite bysubstituting sinafrom Equation 11.143 and sin9from Equation 11.141: '_£211_2 ¢_ _ wliwL2I1 (11.14s) 11.11 Motion ofaSymmetric Top with One Point Fixed Consider asymmetric topwith tipheld ﬁxed* rotating inagravitational ﬁeld. In ourprevious development, wehave been able toseparate thekinetic energy into translational androtational parts bytaking thebody’s center ofmass tobetheori- ginoftherotating orbody coordinate system. Alternatively, ifwecanchoose the origins oftheﬁxed andthebody coordinate systems tocoincide, then thetrans- lational kinetic energy vanishes, because V=R=0.Such achoice isquite con- venient fordiscussing thetop,because thestationary tipmay then betaken asthe origin forboth coordinate systems. Figure 11-15 shows theEuler angles forthis situation. The x3-(ﬁxed) axis corresponds tothevertical, and wechoose the x3-(body) axistobethesymmetry axisofthetop.The distance from thefixed tip tothecenter ofmass ish,andthemass ofthetopisM. Because wehave asymmetric top, theprincipal moments ofinertia about thexi-and x2-axes areequal: Ii=I2.Weassume I3=15Ii.The kinetic energy is Rto wt‘. 0/TT' \Mg I \/’zV\ 1 \ 1’, ¢ \¢ f’/I \\\ xl xi \Line ofnodes FIGURE 11-15 Asymmetric topwith itsbottom tipﬁxed rotates inagravitational ﬁeld. TheEuler angles relate thexi-(ﬁxed) axes with thex,~-(body) axes. Theangle 1/;represents therotation around thex3symmetry axis. *This problem wasﬁrstsolved indetail byLagrange inMécanique analytiquxz. 11.11 MOTION OFASYMMETRIC TOPWITH ONEPOINT FIXED 455 then given by 1 1 1T=5211,16? =5Ii(to2+(1)2)+5131.13 (11.149) According toEquation 11.102, wehave (112=(dosin9 sin1,l1 +9cos1[1)2 =2sin29sin2([1+2&9sin9 sin1,9cos111+92cos21p (112=(dosin9 cos1,11—9sin1p)2 =<32sin29cos21,0—29sin9sinitcos1p+92 sin2it sothat (oi+602=Q32sin20+192 (11.150a) and mg=(<13cos0+(1'))? (11.150b) Therefore, 1 ., . 1 . .T=gIi(¢2 sin29 +92)+gl3(</> cos9+1/1)2 (11,151) Because thepotential energy isMgh cos9,theLagrangian becomes 1 . . 1 . . L=gIi(¢2sin29 +92)+gI3(¢ cos9+1/1)2—Mgh cos9(11.152) The Lagrangian iscyclic inboth the¢-and1,0-coordinates. The momenta conju- gate tothese coordinates aretherefore constants ofthemotion: BL . . pi,=64.)=(Iisin29 +I3cos2 9)¢+I31/1cos9=constant (11.153) 6L . .pi,=W.’=l3(¢ +¢cos9)=constant (11.154) Because thecyclic coordinates areangles, theconjugate momenta areangu- larmomenta—the angular momenta along theaxes forwhich </>and1/1arethero- tation angles, thatiis, thex3-(orvertical) axis and thex3-(orbody symmetry) axis, respectively. Wenote that thisresult isensured bytheconstruction shown inFigure 11-15, because thegravitational torque isdirected along theline of nodes. Hence, thetorque canhave nocomponent along either thex3-orthe x3-axis, both ofwhich areperpendicular tothelineofnodes. Thus, theangular momenta along these axes areconstants ofthemotion. 456 11/DYNAMICS orRIGID BODIES Equations 11.153 and 11.154 canbesolved for and interms of9.From Equation 11.154, wecanwrite 4)=p‘if5"Idis—6- (11.155)3 andsubstituting thisresult into Equation 11.153, weﬁnd (l1sin20 +I3cos29)¢i+(pl/I—13¢.)cos9)cos 0=12¢ or . (I1sin?9)q§+pl],cos0=10¢ sothat J)=12¢—p,y,cos9 11.155 I1sin20 ( ) Using thisexpression for inEquation 11.155, wehave ,12¢ (pq,—pd,cos9)cos9 =- 11.157‘J’1, 11811120 ( ) Byhypothesis, thesystem weareconsidering isconservative; wetherefore have thefurther property thatthetotal energy isaconstant ofthemotion: 1 . . 1E=5I1(¢>2 sin?9+62)+513¢»; +Mgh cos9=constant (11.158) Using theexpression for(03(e.g., seeEquation 11.102), wenote that Equation 11.154 canbewritten as 12¢=I3w3 =constant (1l.159a) OI‘ 2 I3w§ =p7:=constant (11.l59b) Therefore, notonly isEaconstant ofthemotion, butsoisE—%I3w§; weletthis quantity beE'1 1 - . E’EE—élgwg =511((1)2sin29 +62)+Mgh cos9=constant (11.160) Substituting into thisequation theexpression for (Equation 11.156), wehave I1, (p¢— pcos0)2 1;=51,02+—i%—— +Mghcos0 (11.1s1) which wecanwrite as 1.E’=51162 +V(0) (H.162) 11.11 MOTION OFASYMMETRIC TOP WITH ONE POINT FIXED 457 where V(9) isan“effective potential” given by (p—pcos6)? 1/(0)E +Mghcos0 (11162)1 Equation 11.162 canbesolved toyield t(0): d9t(9)=]i——ir—i—— (l1.l64) V(2/I1) [E—V(9)1 This integral can(formally, atleast) beinverted toobtain 6(t), which, inturn, canbesubstituted into Equations 11.156 and 11.157 toyield qb(t) and 1/1(t)._ Because theEuler angles 9,¢,tbcompletely specify theorientation ofthetop, theresults for0(t), ¢(t), and¢r(t) constitute acomplete solution fortheprob- lem. Itshould beclear thatsuch aprocedure iscomplicated andnotvery illumi- nating. Butwecanobtain some qualitative features ofthemotion byexamining thepreceding equations inamanner analogous tothatused fortreating themo- tion ofaparticle inacentral-force ﬁeld (seeSection 8.6). Figure 11-16 shows theform oftheeffective potential V(9) intherange 05 9S11',which clearly isthephysically limited region forB.This energy diagram indicates thatforanygeneral values ofE’(e.g., thevalue represented byEi)the motion islimited bytwoextreme values of9-—that is,91and92,which correspond totheturning points ofthecentral-force problem andareroots ofthedenomi- nator inEquation 11.164. Thus weﬁnd thattheinclination oftherotating topis, 1 Ei “1-3.?TV(6) .* y Eé Q$______._I—l <?1____-®1_____.M>4 9_. FIGURE ll-I6 The effective potential V(6) fortherotating topofFigure 11-15 is plotted Versus theangle 6.Wecanstudy theangular limits ofthe inclination ofthetopbyknowing themodiﬁed energy E'. 458 11/DYNAMICS OFRIGID BODIES ingeneral, conﬁned totheregion 91S9S92.Forthecase thatE’=E5=Vmin, 9islimited tothesingle value 90,andthemotion isasteady precession ataﬁxed angle ofinclination. Such motion issimilar totheoccurrence ofcircular orbits inthecentral-force problem. The value of90canbeobtained bysetting thederivative ofV(9) equal to zero. Thus, 51/ —cos90(1),, —11¢cos90)2+pd,sin?90(1),, —10,’,cos90) _ — =—~ _3 —Mgh sin90=0899:90 I1sin 90 (11.165) Ifwedeﬁne BE12¢—pd,cos90 (l1.166) then Equation 11.165 becomes (cos 90)B2 —(ply,sin?90)B +(Mghll sin490)=0 (ll.l67) This isaquadratic in,8andcanbesolved with theresult sin?9B=Li(, 1/1 (1,168,2cos 90 pi Because Bmust bearealquantity, theradicand inEquation 11.168 must bepos- itive. If90<11'/2, wehave pg,24MghI1 cos00 (11169) Butfrom Equation 11.159a, 11¢=Igwg; thus, 2(1)32T\/Mghll cos90 (1l.170) 3 Wetherefore conclude that asteady precession canoccur attheﬁxed angle of inclination 90only iftheangular velocity ofspin islarger than thelimiting value given byEquation 11.170. From Equation 11.156, wenote thatwecanwrite (for9=90) 430= (11.171)ISIII2 9 1 0 Wetherefore have twopossible values oftheprecessional angular velocity </30, oneforeach ofthevalues ofBgiven byEquation 11.168: <13,“1)—>Fastprecession and (fink) —>Slow precession If(1)3(orply)islarge (afastt0P), then thesecond term intheradicand of Equation 11.168 issmall, andwemay expand theradical. Retaining only theﬁrst 11.11 MOTION OFASYMMETRIC TOP WITH ONE POINT FIXED 459 nonvanishing term ineach case, weﬁnd . I3w3 ¢0(+) =I1cos90 . Mgh ¢0( )Zi—_Isws(11.172) Itistheslower ofthetwopossible precessional angular velocities, qim_),that is usually observed. The preceding results apply if90<1r/2; butif*90>1r/2, theradicand in Equation 11.168 isalways positive and there isnolimiting condition on(03. Because theradical isgreater than unity insuch acase, thevalues ofciaoforfast andslow precession have opposite signs; thatis,for90>11'/2, thefastprecession isinthesame direction asthatfor90<11'/2, buttheslow precession takes place intheopposite sense. Forthegeneral case, inwhich 91<9<92,Equation 11.156 indicates that may ormay notchange sign as9varies between itslimits—depending onthe values of11¢andpd,Ifdoes notchange sign, thetopprecesses monotonically around thexff,-axis (seeFigure 11-15), andthex3-(orsymmetry) axisoscillates between 9=91and 9=92.This phenomenon iscalled nutation; thepath de- scribed bytheprojection ofthebody symmetry axisonaunit sphere intheﬁxed system isshown inFigure 11-17a. Ifdoes change signbetween thelimiting values of9,theprecessional angu- larvelocity must have opposite signs at9=91and9=92.Thus, thenutational- (a) (b) (<1) FIGURE ll-17 The rotating topalsonutates between thelimit angles 91and92.In (a) does notchange sign. In(b) does change sign, andwesee looping motion. In(c)theinitial conditions include9 = =0;this isthenormal cusp-like motion when wespin atopandrelease it. *If90>'11"/2, theﬁxed tipofthetopisataposition above thecenter ofmass. Such motion ispossible, forexample, with agyroscopic topwhose tipisactually aballandrests inacupthatisfixed atop a pedestal. 460 ll/DYNAMICS OFRIGID BODIES precessional motion produces thelooping motion ofthesymmetry axisdepicted inFigure 11-17b. Finally, ifthevalues of10¢and12,)aresuch that (p¢—[2,/1cos9)|1,=91 =0 (1l.173) then (£l9=01= 0.él..=.,=0 <11-1v4> Figure 11-17c shows theresulting cusplike motion. Itisjust thiscase that corre- sponds totheusual method ofstarting atop. First, thetopisspun around its axis, then itisgiven acertain initial tiltandreleased. Thus, initial conditions are 9=91and9=0= Because theﬁrst motion ofthetopistobegin tofallin thegravitational ﬁeld, theconditions areexactly those ofFigure 11-17c, andthe cusplike motion ensues. Figures 11-17a and 11-17b correspond tothemotion in theevent that there isaninitial angular velocity either inthedirection ofor opposite tothedirection ofprecession. 11.12 Stability ofRigid-Body Rotations Wenow consider arigid body undergoing force-free rotation around oneofits principal axes andinquire whether such motion isstable. “Stability” here means, asbefore (seeSection 8.10), thatifasmall perturbation isapplied tothesystem, themotion willeither return toitsformer mode orwillperform small oscilla- tions about it. Wechoose forourdiscussion ageneral rigid body forwhich alltheprincipal moments ofinertia aredistinct, and welabel them such that I3>I2>I1.Welet thebody axes coincide with theprincipal axes, andwestart with thebody rotat- ingaround thex1-axis-~that is,around theprincipal axisassociated with themo- ment ofinertia I1.Then, (I) =(0161 Ifweapply asmall perturbation, theangular velocity vector assumes theform 0)=w1e1 +)te2+1u.e3 (l1.176) where )1and ,u.aresmall quantities and correspond totheparameters used pre- viously inother perturbation expansions. (Aandp.aresuﬁiciently small sothat their product canbeneglected compared with allother quantities ofinterest to thediscussion.) The Euler equations (seeEquation 11.114) become (12—1,)/\,1-110,1=0 (1,—11),“), —1211=0 (11.17?) (A—Q)/W1 -éﬂ=0 11.12 STABILITY OFRIGID-BODY ROTATIONS 461 Because A1u.*-=10,theﬁrst ofthese equations requires rb1=0,orm1=constant. Solving theother twoequations forAand)1,weﬁnd .1—121= w1)1u. (11.17s)2 . 1112p.=-i—w1 A (11.179) Is where theterms inparentheses areboth constants. These arecoupled equa- tions, butthey cannot besolved bythemethod used inSection 11.10, because theconstants inthetwoequations aredifferent. Thesolution canbeobtained by ﬁrstdifferentiating theequation forA: an_ I3 _ I1 _ A——i (01p. (l1.l80) 12 The expression forﬁtcannow besubstituted inthisequation: ..(1—1)(1 -1)A+( w§ 11=0 (11.1s1)23 The solution tothisequation is /1(1)=/-1@*"~‘+ B@"*"»‘ (11.1s2) Q1,E2,1,/ (11.1ss)23 andwhere thesubscripts 1andAindicate thatweareconsidering thesolution forAwhen therotation isaround thex1-axis. Byhypothesis, I1<I3andI1<I2,so01,1isreal. The solution forA(t)there- fore represents oscillatory motion with afrequency (2111. Wecansimilarly investi- gate ;.t(t), with theresult that(21,,=(211E(21.Thus, thesmall perturbations in- troduced byforcing small x2-and x2-components on0:donotincrease with time but oscillate around theequilibrium values A"———Oand p.=O.Conse- quently, therotation around thex1-axis isstable. Ifweconsider rotations around thex2»and x21-axes, wecanobtain expres- sions forQ2andQ3from Equation 11.183 bypermutation: I—II—Q1=121,/ (l1.184a)23 .02=0,2,/———~—-O2_Jig?_I5) (11.184b) (22=.23,/-———-O_ FI‘) (11.1s4¢)where 462 11/DYNAMICS OFRIGID BODIES Butbecause I1<I2<I3,wehave Q1,Q3real, .Q2imaginary Thus, when therotation takes place around either thex1-orx3-axes, thepertur- bation produces oscillatory motion andtherotation isstable. When therotation takes place around x2,however, thefactthat [22isimaginary results intheper- turbation increasing with time without limit; such motion isunstable. Because wehave assumed acompletely arbitrary rigid body forthisdiscus- sion, weconclude that rotation around theprincipal axiscorresponding toei- ther thegreatest orsmallest moment ofinertia isstable andthatrotation around theprincipal axis corresponding totheintermediate moment isunstable. We candemonstrate thiseffect with, say,abook (kept closed bytape orarubber band). Ifwetossthebook into theairwith anangular velocity around oneofthe principal axes, themotion isunstable forrotation around theintermediate axis andstable fortheother twoaxes. Iftwoofthemoments ofinertia areequal (I1=I2,say), then thecoefﬁcient ofAinEquation 11.179 vanishes, and wehave ,4].=0or,u.(t) =constant. Equation 11.178 forAcantherefore beintegrated toyield 21(1)=c+12¢ (11.1ss) andtheperturbation increases linearly with thetime; themotion around thex1- axisistherefore unstable. Weﬁnd asimilar result formotion around thex2-axis. Stability exists only forthex3-axis, independent ofwhether I3isgreater orless than I1=I2. Agood example ofthestability ofrotating objects isseen bythesatellites putinto space bythespace shuttle orbiter. When thesatellites areejected from thepayload bay, they arenormally spinning inastable conﬁguration. InMay 1992, when theastronauts attempted tograb inspace theIntelsat satellite (which originally hadfailed togointo itsdesigned orbit) toattach arocket thatwould insert itinto geosynchronous orbit, thespinning satellite wasslowed down and stopped before theastronaut attempted toattach agrappling ﬁxture tobring it intothepayload bay.After each futile attempt, when thegrappling ﬁxture failed, thesatellite tumbled even more. After spending twounsuccessful days trying to attach thegrappling ﬁxture, theastronauts hadtoabort their attempts because oftheincreased tumbling. Ground controllers required afewhours torestabi- lizethesatellite using jetthrusters. The satellite wasleftinastable conﬁguration ofspinning slowly about itscyclindrical syinmetry axis(aprincipal axis) until the next recovery attempt. Finally, onthethird day, three astronauts went outside theorbiter, grabbed theslightly rotating satellite, stopped it,andputitinto the payload baywhere therocket skirt wasattache d.The Intelsat satellite wasﬁnally successfully placed into orbit intime tobroadcast the1992 Barcelona Olympic summer games. PROBLEMS 463 PROBLEMS ll-l. 11-2. ll-3. ll-4. 11-50 ll-6. ll-7. ll-8. ll-9. ll-10.Calculate themoments ofinertia I1,I2,andI2forahomogeneous sphere ofradius Randmass M.(Choose theorigin atthecenter ofthesphere.) Calculate themoments ofinertia I1,I2,andI3forahomogeneous cone ofmass Mwhose height ishandwhose base hasaradius R.Choose thex3-axis along the axis ofsymmetry ofthecone. Choose theorigin attheapex ofthecone, and calculate theelements oftheinertia tensor. Then make atransformation such that thecenter ofmass ofthecone becomes theorigin, andﬁnd theprincipal moments ofinertia. Calculate themoments ofinertia I1,I2,andI2forahomogeneous ellipsoid ofmass Mwith axes’ lengths 2a>2b>2c. Consider athin rodoflength landmass mpivoted about oneend. Calculate the moment ofinertia. Find thepoint atwhich, ifallthemass were concentrated, the moment ofinertia about thepivot axiswould bethesame astherealmoment of inertia. The distance from thispoint tothepivot iscalled theradius ofgyration. (a)Find theheight atwhich abilliard ballshould bestruck sothatitwillrollwith noinitial slipping. (b)Calculate theoptimum height oftherailofabilliard table. Onwhat basis isthecalculation predicated? Two spheres areofthesame diameter and same mass, butoneissolid and the other isahollow shell. Describe indetail anondestructive experiment todeter- mine which issolid andwhich ishollow. Ahomogeneous diskofradius Randmass Mrolls without slipping onahorizontal surface andisattracted toapoint adistance dbelow theplane. Iftheforce ofat- traction isproportional tothedistance from thedisk’s center ofmass totheforce center, ﬁnd thefrequency ofoscillations around theposition ofequilibrium. Adoor isconstructed ofathin homogeneous slab ofmaterial: ithasawidth of1 m.Ifthedoor isopened through 90°,itisfound thatonrelease itcloses itself in2 s.Assume that thehinges arefrictionless, andshow that thelineofhinges must make anangle ofapproximately 3°with thevertical. Ahomogeneous slab ofthickness aisplaced atop aﬁxed cylinder ofradius R whose axisishorizontal. Show that thecondition forstable equilibrium ofthe slab, assuming noslipping, isR>a/2.VVhat isthefrequency ofsmall oscillations? Sketch thepotential energy Uasafunction oftheangular displacement 9.Show thatthere isaminimum at9=0forR>a/2butnotforR<a/2. Asolid sphere ofmass Mandradius Rrotates freely inspace with anangular ve- locity toabout aﬁxed diameter. Aparticle ofmass m,initially atonepole, moves with aconstant velocity valong agreat circle ofthesphere. Show that, when thepar- ticle hasreached theother pole, therotation ofthesphere willhave been retarded 464 11-ll 11-12 ll-13. ll-14 ll-15 ll-16ll/DYNAMICS OFRIGID BODIES I2Ma—wT<1 2M+5m) where Tisthetotal time required fortheparticle tomove from onepole tothe other.byanangle Ahomogeneous cube, each edge ofwhich hasalength l,isinitially inaposition of unstable equilibrium with oneedge incontact with ahorizontal plane. The cube isthen given asmall displacement andallowed tofall.Show that theangular ve- locity ofthecube when oneface strikes theplane isgiven by w2=A€(VE—1) where A=3/2iftheedge cannot slide ontheplane andwhere A=12/5ifslid- ingcanoccur without friction. Show that none oftheprincipal moments ofinertia canexceed thesum ofthe other two. Athree-particle system consists ofmasses m,~andcoordinates (x1,x2,x11)asfollows: m1=3m, (b,0,b) m2=4m, (b,b,—b) ms=2m,(-b>1&9) Find theinertia tensor, principal axes, and principal moments ofinertia. Determine theprincipal axes and principal moments ofinertia ofauniformly solid hemisphere ofradius bandmass mabout itscenter ofmass. Ifaphysical pendulum hasthesame period ofoscillation when pivoted about ei- ther oftwopoints ofunequal distances from thecenter ofmass, show that the length ofthesimple pendulum with thesame period isequal tothesum ofsepa- rations ofthepivot points from thecenter ofmass. Such aphysical pendulum, called Kater’s reversible pendulum, atonetime provided themost accurate way (toabout 1partin105)tomeasure theacceleration ofgravity.* Discuss theadvan- tages ofKater’s pendulum over asimple pendulum forsuch apurpose. Consider thefollowing inertia tensor: . 1 1 ' —A+B —A—B 0 2( )2( ) _1 1{H-—m-m %A+& 02 2 O O C *First used in1818 byCaptain Henry Kater (1777-1835), butthemethod wasapparently suggested somewhat earlier byBohnenberger. The theory ofKater’s pendulum was treated indetail by Friedrich Wilhelm Bessel (1784-1846) in1826. PROBLEMS 465 ll-17. ll-18. ll-19 ll-20. ll-21Perform arotation ofthecoordinate system byanangle 9about thex3-axis. Evaluate thetransformed tensor elements, andshow thatthechoice 9=11/4ren- ders theinertia tensor diagonal with elements A,B,andC. Consider athin homogeneous plate thatliesinthex1-x2 plane. Show thatthein- ertia tensor takes theform A —C O {I}=—C B O 0 OA+B If,intheprevious problem, thecoordinate axes arerotated through anangle 9 about thex3-axis, show thatthenewinertia tensor is A’ —'C’ O {I}=—C' B’ O 0 O A’+B’ where A’=Acos29 —Csin29 +Bsin29 B’=Asin29 +Csin29 +Bcos29 1 .C’=Ccos 29—g(B— A)s1n 29 andhence show thatthex1-andx2-axes become principal axes iftheangle ofrota- tionis 1 C9=gtan“ Consider aplane homogeneous plate ofdensity pbounded bythelogarithmic spi- ral1"=ke""andtheradii 9=0and9=11.Obtain theinertia tensor fortheorigin atr=0iftheplate liesinthex1-x2 plane. Perform arotation ofthecoordinate axes toobtain theprincipal moments ofinertia, andusetheresults oftheprevi- ousproblem toShow that they are If=pk4P(Q_— R), I2=pk4P(Q+ R), I2:=If+I2 where e4"“—1 1+4a2P=——€, =-—-, R=\/1+4?16(1+4a2) Q 211 a Auniform rodoflength bstands vertically upright onarough floor andthen tips over. \/Vhat istherod’s angular velocity when ithitstheﬂoor? The proof represented byEquations 11.54-11.61 isexpressed entirely inthesum- mation convention. Rewrite thisproof inmatrix notation. 466 11-22 ll-23 11-24 ll-25 ll-26 11-27 1l-28. 1l-29ll/DYNAMICS OFRIGID BODIES Thetrace ofatensor isdeﬁned asthesum ofthediagonal elements: tI'{|} E2111 Show, byperforming asimilarity transformation, that thetrace isaninvariant quantity. Inother words, show that tr{|} =tr{|’} where {I}isthetensor inonecoordinate system and{|}'isthetensor inacoordi- nate system rotated with respect totheﬁrstsystem. Verify thisresult forthediffer- entforms oftheinertia tensor foracube given inseveral examples inthetext. Show bythemethod used intheprevious problem thatthedeterminant oftheele- ments ofatensor isaninvariant quantity under asimilarity transformation. Verify thisresult alsoforthecase ofthecube. Find thefrequency ofsmall oscillations forathin homogeneous plate ifthemo- tion takes place intheplane oftheplate and iftheplate hastheshape ofanequi- lateral triangle andissuspended (a)from themidpoint ofonesideand(b)from oneapex. Consider athin diskcomposed oftwohomogeneous halves connected along adi- ameter ofthedisk. Ifonehalfhasdensity pandtheother hasdensity 2p,ﬁnd the expression fortheLagrangian when thedisk rolls without slipping along ahori- zontal surface. (The rotation takes place intheplane ofthedisk.) Obtain thecomponents oftheangular velocity vector to(seeEquation 11.102) di- rectly from thetransformation matrix A(Equation 11.99). Asymmetric body moves without theinﬂuence offorces ortorques. Letx3bethe symmetry axisofthebody andLbealong x2',.Theangle between toandx2,isa.Let toandLinitially beintheX2"-7C3 plane. What istheangular velocity ofthesymmetry axisabout Linterms ofI1,I3,to,anda? Show from Figure 11-9c thatthecomponents oftoalong theﬁxed (x,-')axes are co;=9cos¢ +1lrsin9sinqb w2=9sin¢— 1Lsin9cos¢ w§=1l(cos9+q'5 Investigate themotion ofthesymmetric topdiscussed inSection 11.11 forthe case inwhich theaxisofrotation isvertical (i.e., thex§-and x3-axes coincide). Show thatthemotion iseither stable orunstable depending onwhether thequan- tity4I1Mhg/I§w%, islessthan orgreater than unity. Sketch theeffective potential V(9) forthetwocases, andpoint outthefeatures ofthese curves that determine whether themotion isstable. Ifthetopissetspinning inthestable conﬁguration, what istheeffect asfriction gradually reduces thevalue of(U3?(This isthecase of the“sleeping top.”) PROBLEMS 467 ll-30 ll-31 ll-32 11-33. ll-34.Refer tothediscussion ofthesymmetric topinSection 11.11. Investigate theequa- tion fortheturning points ofthenutational motion bysetting 9=OinEquation 11.162. Show that theresulting equation isacubic incos9and hastworeal roots andoneimaginary root for9. Consider athin homogeneous plate with principal momenta ofinertia I1along theprincipal axis x1 I2>I1along theprincipal axisx2 I3=I1+I2along theprincipal axisx2, Lettheorigins ofthex,and x;systems coincide and belocated atthecenter of mass Ooftheplate. Attime t=0,theplate issetrotating inaforce-free manner with anangular velocity Qabout anaxisinclined atanangle afrom theplane of theplate andperpendicular tothex2-axis. IfI1/I2 Ecos201,show that attime t theangular velocity about thex2-axis is w2(t) =Qcos atanh(.O tsina) Solve Example 11.2 forthecase when thephysical pendulum does notundergo small oscillations. The pendulum isreleased from restat67°attime t=0.Find theangular velocity when thependulum angle isat1°.The mass ofthependulum is340g,thedistance Lis13cm,andtheradius ofgyration kis17cm. Doaliterature search and explain how acatcanalways land onitsfeetwhen dropped from aposition atrestwith itsfeetpointing upward. Estimate themini- mum height acatneeds tofallinorder toexecute such amaneuver. Consider asymmetrical rigid body rotating freely about itscenter ofmass. Afric- tional torque (I\§=—bm) actstoslow down therotation. Find thecomponent of theangular velocity along thesymmetry axisasafunction oftime. CHAPTER _LA|( Coupled Oscillations 12.1Introduction InChapter 3,weexamined themotion ofanoscillator subjected toanexternal driving force. The discussion waslimited tothecase inwhich thedriving force is periodic; thatis,thedriver isitself aharmonic oscillator. Weconsidered theaction ofthedriver ontheoscillator, butwedidnotinclude thefeedback effect oftheos- cillator onthedriver. Inmany instances, ignoring thiseffect isunimportant, butif two(ormany) oscillators areconnected insuch awaythat energy canbetrans- ferred back andforth between (oramong) them, thesituation becomes themore complicated case ofcoupled oscillations.* Motion ofthistype canbequite com- plex (the motion may noteven beperiodic), butwecanalways describe themo- tion ofanyoscillatory system interms ofnormal coordinates, which have the property thateach oscillates with asingle, well-deﬁned frequency; thatis,thenor- malcoordinates areconstructed insuch awaythat nocoupling occurs among them, even though there iscoupling among theordinary (rectangular) coordi- nates describing thepositions ofparticles. Initial conditions canalways bepre- scribed forthesystem sothatinthesubsequent motion only onenormal coordi- nate varies with time. Inthiscircumstance, wesaythatoneofthenormal modes ofthesystem hasbeen excited. Ifthesystem hasndegrees offreedom (e.g., n-cou- pled one-dimensional oscillators orn/3-coupled three-dimensional oscillators), there areingeneral nnormal modes, some ofwhich may beidentical. The gen- eralmotion ofthesystem isacomplicated superposition ofallthenormal modes ofoscillation, butwecanalways ﬁnd initial conditions such thatanygiven oneof thenormal modes isindependently excited. Identifying each ofasystem’s normal *The general theory oftheoscillatory motion ofasystem ofparticles with aﬁnite number ofdegrees offreedom wasformulated byLagrange during theperiod 1762-1765, butthepioneering work had been done in1753 byDaniel Bernoulli (1700-1782). 468 12.2TWOcouruzn HARMONIG OSCILLATORS 469 modes allows ustoconstruct arevealing picture ofthemotion, even though the system’s general motion isacomplicated combination ofallthenormal modes. Itisrelatively easy todemonstrate some ofthecoupled oscillator phenom- enadescribed inthischapter. Forexample, twopendula coupled byaspring be- tween their mass bobs, twopendula hung from arope, andmasses connected by springs canallbeexperimentally examined intheclassroom. Similarly, thetri- atomic molecule discussed here isareasonable description ofCO2. Similar mod- elscanapproximate other molecules. Inthefollowing chapter, weshall continue thedevelopment begun here anddiscuss themotion ofvibrating strings. This example bynomeans exhausts theusefulness ofthenorrnal-mode approach tothedescription ofoscillatory sys- tems; indeed, applications canbefound inmany areas ofmathematical physics, such asthemicroscopic motions incrystalline solids and theoscillations ofthe electromagnetic ﬁeld. ' 12.2 Two Coupled Harmonic Oscillators Aphysical example ofacoupled system isasolid inwhich theatoms interact by elastic forces between each other andoscillate about their equilibrium positions. Springs between theatoms represent theelastic forces. Amolecule composed of afewsuch interacting atoms would beaneven simpler model. Webegin bycon- sidering asimilar system ofcoupled motion inonedimension: twomasses con- nected byaspring toeach other andbysprings toﬁxed positions (Figure 12-1). Wereturn tothisexample throughout thechapter aswedescribe various in- stances ofcoupled motion. Weleteach oftheoscillator springs have aforce constant* K:theforce constant ofthecoupling spring isK12.Werestrict themotion totheline connecting the masses, sothesystem hasonly twodegrees offreedom, represented bythecoordi- nates x1andx2.Each coordinate ismeasured from theposition ofequilibrium. m1==M m2-M |c1=|r K12 K‘2==K *1 *2 FIGURE 12-l Two masses areconnected byaspring toeach other andbysprings to ﬁxed positions. This isasystem ofcoupled motion inone dimension. *Henceforth, wedenote force constants byKrather than byItas heretofore. The symbol kisre- served for(beginning inChapter 13)anentirely different context. 470 12/COUPLED OSCILLATIONS Ifm1and m2aredisplaced from their equilibrium position byamounts x1 andx2,respectively, theforce onm1is—Kx1—K12(x1 —x2),andtheforce onm2is —Kx2 —K12(x2 —x1).Therefore theequations ofmotion are M591 '1'(K+K12)x1_K12x2:0} M£2 + (K + K12)x2 _ K12.?C1 Z 0 - Because weexpect themotion tobeoscillatory, weattempt asolution ofthe form x1(t)=B1e‘*”‘ :B2€iwt where thefrequency wistobedetermined andwhere theamplitudes B1andB2 maybecomplex.* These trialsolutions arecomplex functions. Thus, intheﬁnal step ofthesolution, therealparts ofx1(t) and x2(t) willbetaken, because the realpart isallthat isphysically signiﬁcant. Weusethismethod ofsolution be- cause ofitsgreat efﬁciency, andweuseitagain later, leaving outmost ofthede- tails. Substituting these expressions forthedisplacements into theequations of motion, weﬁnd(12.2) —Mw2B1e""‘ +(K +K12)B1€iwt _K12B2€iwt :O . . . 12.3 _M(U2B2elwt + (K + K12)B26””t _ K12B16“”° : ( ) Collecting terms andcanceling thecommon exponential factor, weobtain (K + K12 _ M(1)2)B1 _ KIQBQ I O 1.4 _ K12B1 + (K + K12 — M(U2)B2 Z (2 ) Foranontrivial solution toexist forthispair ofsimultaneous equations, thede- terminant ofthecoefﬁcients ofB1andB2must vanish: K+K12—Ma)? —K122=0 (12.5)—K12 K+K12—Mw The expansion ofthissecular determinant yields (K+K12—M102)? —K12=O (12.6) Hence, K+K12—Mw2= iK12 -1- i o=./—-—--K K;K” (12.7) *Because acomplex amplitude hasamagnitude andaphase, wehave thetwoarbitrary constants nec- essary inthesolution ofasecond-order differential equation; that is,wecould equally wellwrite 21(1)=IBIexp[i(wt—5)]or1(1)=|B|cos(wt—5),asinEquation 3.6b.Later (seeEquation 12.9), weshall ﬁnd itmore convenient tousetwodistinct realamplitudes and thetime-varying factors exp(iwt) andexp( —itut). These various forms ofsolution areallentirely equivalent.Solving forco,weobtain 12.2 TWO COUPLED HARMONIC OSCILLATORS 471 Wetherefore have twocharacteristic frequencies (oreigenfrequencies) forthe system: /K+2K /K (1)1= TE, (02= M (12.8) Thus, thegeneral solution totheproblem is x1(t) :Bil-leizu1t_|_ Bl-le-im1¢_|_ Bf-2eiw21_|_ Bﬁe—im2t x2(t) =B“§1e“"1‘ +BQe_“"1‘ +B§e“"2‘ +B§e_“"*‘ '¢ where wehave explicitly written both positive andnegative frequencies, because theradicals inEquations 12.7 and12.8 cancarry either sign. InEquation 12.9, theamplitudes arenotallindependent, aswemay verify bysubstituting wlandw2into Equation 12.4. Weﬁnd for w=col: B11=—B21 for U}:(U2: B12 :B22 The only subscripts necessary ontheBsarethose indicating theparticular eigen- frequency (i.e., thesecond subscripts). Wecantherefore write thegeneral solution as x1(t)=B1+ei"’1‘ +Bfe_“"1‘ +B§*e"‘“2‘ +B2‘e_""2’ x2(t) :_Bi§-eiw1i_ Bl—e—iw1t_|_ B;-eiwgt _|_ e-2'01?! Thus, wehave fourarbitrary constants inthegeneral solution—just asweexpect- because wehave twoequations ofmotion that areofsecond order. Wementioned earlier thatwecould always deﬁne asetofcoordinates that have asimple time dependence and that correspond totheexcitation ofthe various oscillation modes ofthesystem. Letusexamine thepair ofcoordinates deﬁned by TI1Ex1_x2 °'I2Ex1+x2} (12.11) or N);-iI\'>r—\xiI“(TF2 +171) (12.12) *2=-(112-m) Substituting these expressions forxlandx2into Equation 12.1, weﬁnd M(ﬁ1 +1'52)+(K+2K12)TI1 +K772ZO} (1213) M(ﬁ1 _112)+(K+2K12)TI1 _K772:0 which canbesolved (byadding andsubtracting) toyield M" =0 TI1+(K 2K12)"71 } (1214) M172 +K112=O 12/COUPLED OSCILLATIONS 0)=C01 w= <1-1- ib <1 4; Antisymmetrical mode Symmetrical mode (out ofphase) (inphase) FIGURE 12-2 The twocharacteristic frequencies areindicated schematically. One is theantisymmetrical mode (masses areoutofphase) and theother is thesymmetrical mode (masses areinphase). The coordinates 171and172arenow uncoupled andaretherefore independent. The solutions are (t) :C+ ia:1t+ C1- —ia:1t 21¢)—c1+eew+052-W ("'15) 2 _ 2 where thefrequencies wland (U2aregiven byEquations 12.8. Thus, 171and172 arethenormal coordinates oftheproblem. Inalater section, weestablish agen- eralmethod forobtaining thenormal coordinates. Ifweimpose thespecial initial conditions x1(O) I—x2(O) and a%1(O) = —nZ2(0), weﬁnd 172(0) =Oand1’;;,(0) =O,which leads toC‘;=C2‘=O;thatis, 172(t) EOforallvalues oft.Thus, theparticles oscillate always outofphase and with frequency (01;thisistheantisymmetrical mode ofoscillation. However, ifwe begin with x1(O) =x2(0) andaE1(O) =a%2(O), weﬁnd 171(t) E0,andtheparticles oscillate inphase andwith frequency (U2;thisisthesymmetrical mode ofoscilla- tion. These results areillustrated schematically inFigure 12-2. The general mo- tion ofthesystem isalinear combination ofthesymmetrical andantisymmetri- calmodes. The factthat theantisymmetrical mode hasthehigher frequency andthe symmetrical mode hasthelower frequency isactually ageneral result. Ina complex system oflinearly coupled oscillators, themode possessing thehigh- estdegree ofsymmetry hasthelowest frequency. Ifthesymmetry isdestroyed, then thesprings must “work harder” intheantisymmetrical modes, and the frequency israised. Notice that ifwewere tohold "Z2ﬁxed and allow mltooscillate, thefre- quency would be\/(K+K12)/M. Wewould obtain thesame result forthefre- quency ofoscillation ofm2ifmlwere held ﬁxed. The oscillators areidentical andintheabsence ofcoupling have thesame oscillation frequency. The effect ofcoupling istoseparate thecommon frequency, with one characteristic fre- quency becoming larger and one becoming smaller than thefrequency for uncoupled motion. Ifwedenote bymothefrequency foruncoupled motion, then wl>coo>(U2,and wemay schematically indicate theeffect ofthecou- pling asinFigure 12—3a. The solution forthecharacteristic frequencies inthe problem ofthree coupled identical masses isillustrated inFigure 12-3b. Again, 12.3 WEAK COUPLING 473 .___..._ Q) ’ /I, 1 1/, cooiq wo—-——<—\— ——i (02=wo \ \\._.i(g \2 \_______ wg n=2 n=3 (H) (b) FIGURE 12-3 (a)Coupling separates thecommon frequency fortwoidentical masses, with one characteristic frequency being higher and one being lower than thefrequency tooforuncoupled motion. (b)Forthree coupled identical masses, one characteristic frequency issmaller than moand one islarger. Forn(number ofoscillators) odd, one characteristic frequency isequal tocoo.The separations areonly schematic. wehave asplitting ofthecharacteristic frequencies, with one greater and one smaller than wo.This isageneral result: Foraneven number nofidentical near- estneighbor coupled oscillators, n/2characteristic frequencies aregreater than wo,and n/2characteristic frequencies aresmaller than mo.Ifnisodd, onechar- acteristic frequency isequal tomo,and theremaining n—1characteristic fre- quencies aresymmetrically distributed above andbelow wo.The reader familiar with thephenomenon oftheZeeman effect inatomic spectra willappreciate the similarity with thisresult: Ineach case, there isasymmetrical splitting ofthefre- quency caused bytheintroduction ofaninteraction (inonecase bytheapplica- tionofamagnetic ﬁeld andintheother bythecoupling ofparticles through the intermediary ofthesprings). 12.3 Weak Coupling Some ofthemore interesting cases ofcoupled oscillations occur when thecou- pling isweak--that is,when theforce constant ofthecoupling spring issmall com- pared with that oftheoscillator springs: K12<<K.According toEquations 12.8, thefrequencies colandw2are K+2K f 12 IK(01= T, (02= M (12.16) Ifthecoupling isweak, wemay expand theexpression forcol: w1=\/ﬁ\/l+$=\/ﬁ\/1+4a where _K12a=—<<1 (12.17)2K 474 12/COUPLED OSCILLATIONS The frequency mlnow reduces to w,E,/id +2.2) (12.18) The natural frequency ofeither oscillator, when theother isheld ﬁxed, is K+K (U0=,/~—1\/I-‘-2 E\[4(1+2) (12.19) \/kg/IE wO(1 —s) (12.20) Therefore, thetwocharacteristic frequencies aregiven approximately by (01%,/%1(1+28), w,=,/i Ewo(1 —@)(1+28) Ew0(1 —8) Eto0(1 +s)OI‘ (12.21) Wecannow examine thewayaweakly coupled system behaves. Ifwedis- place Oscillator 1adistance Dandrelease itfrom rest, theinitial conditions for thesystem are x1(0) =D,12(0) =0,21(0) =0,952(0) =0 (12-22) Ifwesubstitute these initial conditions into Equation 12.10 forx1(t) and x2(t), weﬁnd theamplitudes tobe DBf=B;=B;=B;=Z (12.23) Then, x1(t) becomes N>bH>Ux1(t) :_[(euu1Z +e—iw1¢) + (eim2t_|_ e—im2t)] =—(cos w1t+ coswgt) — (U1 +(U2 (U1 _(U2 —Dcos —Tt cos T12 (12.24) But, according toEquation 12.21, + _ ﬁgi Imo;% Zwe (1225) Theref0re,* x1(t)=(Dcosscoot) coswot (12-26a) mm *Note thatinthisfortuitous case, xlandx2were always real, sotherealpart didnothave tobeex- pressly taken inthefinal stepasoutlined after Equation 12.2. 12.4 GENERAL PROBLEM OFCOUPLED OSCILLATIONS 475 ‘A_-_ I1 - _ ,x1(t) ~_ I,, _ ,\ I \ I \\ I 1 I \ I’ \\ \ [—>’ \ I’ \ I _ 1 _I’ \\_, _/ 2 ,, __ "T T*~_ 4’; ‘\‘J \\\ _x(t) 12 I; \ I \ J \, 1\ \ I\ I_\4-/ —>- , i\ I~. z\ I~.\'1~ I~ 4 ___ ’ ,4 FIGURE 12-4 Thesolutions forx1(t) andx2(t) have ahigh frequency component (too) thatoscillates inside aslowly varying component (stoo). Energy is transferred back and forth between thetwooscillators. Similarly, xi,(t)=(DsinewOt)sin wot -(l2.26b) Because sissmall, thequantities Dcosscoot andDsinscoot vary slowly with time. Therefore, xl(t)and x2(t)areessentially sinusoidal functions with slowly varying amplitudes. Although only x1isinitially different from zero, astime in- creases theamplitude ofx1decreases slowly with time, andtheamplitude ofx2in- creases slowly from zero. Hence, energy istransferred from theﬁrstoscillator to thesecond. When t=11'/2ew0, then Dcosswot =0,andalltheenergy hasbeen transferred. Astime increases further, energy istransferred back totheﬁrstoscil- lator. This isthefamiliar phenomenon ofbeats and isillustrated inFigure 12-4. (Inthecase illustrated, s=0.08.) 12.4 General Problem ofCoupled Oscillations Inthepreceding sections, wefound that theeffect ofcoupling inasimple sys- tem with twodegrees offreedom produced twocharacteristic frequencies and twomodes ofoscillation. Wenow turn ourattention tothegeneral problem of coupled oscillations. Letusconsider aconservative system described interms of asetofgeneralized coordinates q,,andthetime t.Ifthesystem hasndegrees of freedom, then k=1,2, ,n.Wespecify that aconﬁguration ofstable equilib- rium exists forthesystem and that atequilibrium thegeneralized coordinates have values qko.Insuch aconﬁguration, Lagrange’s equations aresatisﬁed by qkzqkfls 01,20, qkzos k=ls2s"-an 476 12/COUPLED OSCILLATIONS Every nonzero term oftheform (d/dt) (61/6q,,) must contain atleast either ohorﬁk, soallsuch terms vanish atequilibrium. From Lagrange’s equation, wetherefore have 6L 6T 6U— =— —— I0 (12.27) 5(1):05%05(1)» where thesubscript 0designates thatthequantity isevaluated atequilibrium. Weassume that theequations connecting thegeneralized coordinates and therectangular coordinates donotexplicitly contain thetime; thatis,wehave xa,i :xa,i(qj) or :qj(xa,i) The kinetic energy isthus ahomogeneous quadratic function ofthegeneralized velocities (seeEquation 7.121): 1 .. Therefore, ingeneral, 6T(T =0, k=1,2,...,n qk0 (12.29) andhence, from Equation 12.27, wehave 6U — =0, k= 1,2,,..,n (12.30) aqk 0 Wemay further specify that thegeneralized coordinates q),bemeasured from theequilibrium positions; that is,wechoose qko=0.(Ifweoriginally had chosen asetofcoordinates qj,such that q},09*0,wecould always effect asimple linear transformation oftheform qk=qj,+01,,such that qko=0,) The expansion ofthepotential energy inaTaylor series about theequilib- rium conﬁguration yields 6U 1 62U U(q1.q2.-.-,q..) —U0+2&2. +-§j,E,?qkO2q. + <12-31) The second term intheexpansion vanishes inview ofEquation 12.30, and—with- outlossofgenerality—we may choose tomeasure Uinsuch awaythat U0E0. Then, ifwerestrict themotion ofthegeneralized coordinates tobesmall, wemay neglect allterms intheexpansion containing products oftheq,,ofdegree higher than second. This isequivalent torestricting ourattention tosimple harmonic oscillations, inwhich case only terms quadratic inthecoordinates appear. Thus, 1U=5§Aj,,qjq,, (12.32) where wedeﬁne 62U A-E"-m (12.33) 1"6qj6q,, 0 12.4 GENERAL PROBLEM OFCOUPLED OSCILLATIONS 477 Because theorder ofdifferentiation isimmaterial (ifUhascontinuous second partial derivatives), thequantity A1,,issymmetrical; thatis,Aﬂ,IA,g-. Wehave specified thatthemotion ofthesystem istotake place inthevicin- ityoftheequilibrium conﬁguration, andwehave shown (Equation 12.30) that Umust have aminimum value when thesystem isinthisconﬁguration. Because wehave chosen U=0atequilibrium, wemust have, ingeneral, UZ0.Itshould beclear thatwemust alsohave T20.* Equations 12.28 and12.32 areofasimilar form: 1 ..T:5ﬂzkmjkqjqk (12.34)1 U:5§Ajkqj‘11 The quantities A]-,,arejust numbers (see Equation 12.33); butthemi-,,may be functions ofthecoordinates (seeEquation 7.119): 6aria ai Wecanexpand themy),about theequilibrium position with theresult Gm), m,2(q1.22..2.)=m,2(qm)+23$112+ .(12.35)0 Wewish toretain only theﬁrstnonvanishing term inthisexpansion; but, unlike theexpansion ofthepotential energy (Equation 12.31), wecannot choose the constant term ml-,,(q,0) tobezero, sothisleading term becomes theconstant value ofmjkinthisapproximation. This isthesame order ofapproximation asthat used forU,because thenext higher order term inTwould involve thecubic quantity qjqkq, and thenext higher order term inUwould contain qjqkql. Inthe small oscillation approximation, Tshould betreated similarly toU,andjustlike Uisnormally expanded toorder q2,oneneeds toexpand Ttoorder Q2,andnahis evaluated atequilibrium. Thus, inEquation 12.34, thernjkandtheA]-,,arenXn arrays ofnumbers specifying thewaythemotions ofthevarious coordinates are coupled. Forexample, ifmm#50forWes,then thekinetic energy contains aterm proportional toq,i],,andacoupling exists between therthandsthcoordinate. If, however, mﬂ,isdiagonal, sothat)“ mjh9*0forj=kbutvanishes otherwise, then the kinetic energy isoftheform T= *That is,both UandTare positive deﬁnite quantities, inthatthey arealways positive unless thecoor- dinates (inthecaseofU)orthevelocities (inthecaseofT)arezero, inwhich casetheyvanish. Hfadiagonal element ofml-,,(say, m,,)vanishes, then theproblem canbereduced tooneofn—1 degrees offreedom. 478 12/COUPLED OSCILLATIONS where m,,,hasbeen abbreviated tom,.Thus, thekinetic energy isasimple sum of thekinetic energies associated with thevarious coordinates. Asweseebelow, if, inaddition, A]-,,isdiagonal sothat Uisalsoasimple sum ofindividual potential energies, then each coordinate behaves inanuncomplicated manner, undergo- ingoscillations with asingle, well-deﬁned frequency. The problem istherefore toﬁnd acoordinate transformation that simultaneously diagonalizes both mjk and A]-,,and thereby renders thesystem describable inthesimplest possible terms. Such coordinates arethenormal coordinates. The equations ofmotion ofthesystem with kinetic and potential energies given byEquation 12.34 areobtained from Lagrange’s equation Q_d6L_0 dqk dqk Butbecause Tisafunction only ofthegeneralized velocities and Uisafunction only ofthegeneralized coordinates, Lagrange’s equation forthekthcoordinate becomes 1’+1"P"=0 (12.2.6)dqk dtdqk From Equations 12.34, weevaluate thederivatives: 6U T=§A1*q('1" (12.37) I;”‘1h‘l1 6Q), The equations ofmotion then become §1(Aj,,qj +mj,z;,)=0 (12.33) This isasetofnsecond-order linear homogeneous differential equations with constant coefﬁcients. Because wearedealing with anoscillatory system, weex- pect asolution oftheform q]-(t)=a,-e"(""' 8) (12.39) where theajarereal amplitudes andwhere thephase 8hasbeen included to give thetwoarbitrary constants (ajand8)required bythesecond-order nature ofeach ofthedifferential equations.* (Only therealpart oftheright-hand side istobeconsidered.) The frequency wandthephase 5aretobedetermined by theequations ofmotion. Iftoisareal quantity, then thesolution (Equation 12.39) represents oscillatory motion. That toisindeed realmay beseen bythe following physical argument. Suppose thattocontains animaginary part iw,(in which w,-isreal). This produces terms oftheform e"’(‘ande“"*‘intheexpression *This isentirely equivalent toourprevious procedure ofwriting x(t)=Bexp(iwt) (seeEquations 12.2) with Ballowed tobecomplex. InEquations 12.9, weexhibited therequisite arbitrary constants asreal amplitudes byusing exp(z'tot) and exp(— iwt) rather than byincorporating aphase factor asin Equation 12.39. 12.4 GENERAL PROBLEM OFCOUPLED OSCILLATIONS 479 ofqj.Thus, when thetotal energy ofthesystem iscomputed, T+Ucontains fac- torsthat increase ordecrease monotonically with thetime. Butthisviolates the assumption that wearedealing with aconservative system: therefore, thefre- quency tomust bearealquantity. With asolution oftheform given byEquation 12.39, theequations ofmotion become ;(Aj,,—w2mj,,)a]- =0 (12.40) where thecommon factor exp[i(wt —8)]hasbeen canceled. This isasetofn linear, homogeneous, algebraic equations that theajmust satisfy. Foranontrivial solution toexist, thedeterminant ofthecoefﬁcients must vanish: IA]-k—w21n]~k| =O (12.41) Tobemore explicit, thisisannXndeterminant oftheform A11—¢°2m11 A12_w2m12 A13—w2m1:-s A12_°)2m12 A22_°)2m22 A25_¢°2m2s :0 (1242) A13—w2m13 A22 —w2m2s A33_¢°2ms3 where thesymmetry oftheA1,,and‘H21-khasbeen explicitly included. The equation represented bythisdeterminant iscalled thecharacteristic equation orsecular equation ofthesystem andisanequation ofdegree nin0:2. There are,ingeneral, nroots wemaylabel 41$.Thew,arecalled thecharacteristic frequencies oreigenfrequencies ofthesystem. (Insome situations, twoormore ofthew,canbeequal; thisisthephenomenon ofdegeneracy andisdiscussed later.) Just asintheprocedure fordetermining thedirections oftheprincipal axes forarigid body, each oftheroots ofthecharacteristic equation may besub- stituted intoEquation 12.40 todetermine theratios a1:a2:a3: ---ranforeach value ofw,.Because there arenvalues ofw,,,wecanconstruct nsetsofratios oftheaj. Each ofthesetsdeﬁnes thecomponents ofthen-dimensional vector a,,called an eigenvector ofthesystem. Thus a,,istheeigenvector associated with theeigenfre- quency (0,.Wedesignate byaj,the component ofthertheigenvector. Because theprinciple ofsuperposition applies forthedifferential equation (Equation 12.38), wemust write thegeneral solution forqjasalinear combina- tion ofthesolutions foreach ofthenvalues ofr: q]‘(t)=2:1]-,e"(“’*"5') (12.43) Because itisonly therealpart ofq]~(t) that isphysically meaningful, weactually have* q/,():)=Regaj-,e"(""“5') =aj,cos(W—5,) (12.44) *Notice here, unlike theexample ofweak coupling described inSection 12.3 (Equation 12.26), the realpart ofq]-(t)hastobeexplicitly taken sothattheq]-(t)inEquation 12.44 isnotthesame asthe q]-(t)inEquation 12.43. Buthere andelsewhere, because oftheir close relationship, weusethesame symbol (e.g., qj(t)) forconvenience. 480 12/COUPLED OSCILLATIONS The motion ofthecoordinate qjistherefore compounded ofmotions with each ofthenvalues ofthefrequencies (0,.The qjevidently arenotthenormal coordi- nates thatsimplify theproblem. Wecontinue thesearch fornormal coordinates inSection 12.6. Find thecharacteristic frequencies forthecase ofthetwomasses connected by springs ofSection 12.2bymeans ofthegeneral formalism justdeveloped. Solution. The situation isthat shown inFigure 12-1. The potential energy of thesystem is I\9l—‘I\Dl—‘NJ I_ 1 _212 U K961+2K12(.')C2 x1)+2Kxg (12.45) 1 I“(K+K12)xi +§(K+K12)x2 —K12x1x2 The term proportional tox1x2isthefactor thatexpresses thecoupling inthesys- tem. Calculating theA]-,,,weﬁnd 62UA=— =+ 11 ax?0KK12 82UA12 I £5-‘I2 0Z —K12 Z A21 62UA=—~ I+ 22 axg0KK12 The kinetic energy ofthesystem is 1 l2 1 O2 T=5Mxl +5Mxg (12.47) According toEquation 12.28, 1 Identifying terms between these twoexpressions forT,weﬁnd "111: "122 :M} m12Zm21= 0(12.49) Thus, thesecular determinant (Equation 12.42) becomes "+K”_M“’2 _"” 2=0 (12.50)—K12 K+K12 _MO) 12.5 ORTHOGONALITY OFTHEEIGENVECTORS (OPTIONAL) 481 This isexactly Equation 12.5, sothesolutions arethesame (seeEquations 12.7 and12.8) asbefore: (0: /K+K12iK1;, M The eigenfrequencies are K+ 2K12 K ‘"1=_;vz—~ ‘"22M The results ofthetwoprocedures areidentical. 12.5 Orthogonality oftheEigenvectors (Optiona1)* Wenow wish toshow thattheeigenvectors a,form anorthonormal set.Rewriting Equation 12.40 forthesthroot ofthesecular equation, wehave 0)?;mjkak, IEhAjkak, (12.51) Next, wewrite acomparable equation fortherthroot bysubstituting rfor sand interchanging jandk: _ (02 mjkaj, =%lA]-kaj, (12.52) where wehave used thesymmetry ofthemi),andAjk.Wenow multiply Equation 12.51 byaj,andsum overjandalsomultiply Equation 12.52 byah,andsum over k: mggmjkajraks =jzA}kaﬂaks (1253)to?gm]-,,a]~,ak, =]zAj,,aj,a,“ ' The right-hand sides ofEquations 12.53 arenow equal, sosubtracting theﬁrstof these equations from thesecond, wehave <0»?-@552,(4.11,-.11..=0 (12.54) Wenow examine thetwopossibilities r=sand r¢s.Forrabs,theterm (co?—(of)is,ingeneral, different from zero. (The case ofdegeneracy, ormulti- pleroots, isdiscussed later.) Therefore thesum must vanish identically: J2’)?/mj-,,a]~,a,,, =O,r¢s (12.55) *Section 12.5may beomitted without losing physical understanding. This highly mathematical sec- tion isincluded forcompleteness. The method used here isageneralization ofthesteps used in Section 11.6fortheinertia tensor. 482 12/COUPLED OSCILLATIONS Forthecase r=s,theterm (0)3—co?)vanishes andthesum isindeterminate. The sum, however, cannot vanish identically. Toshow this, wewrite thekinetic energy forthesystem andsubstitute theexpressions forQ,andQ),from Equation 12.44: 1 .. 1 . .=§§nrr% w,a]-, s1n(to,t —5,)] wsak, s1n(w,t —5% 1=5w,co, sin(w,t —5,)sin(w,t —5,) m]-ha,-,a,,, Thus, forr=s,thekinetic energy becomes 1T=E4»;sin2(w,t —5,);mjka]-,a,,, (12.56) Wenote ﬁrstthat <02sin2(w,t —5,)2O Wealsoknow that Tispositive andcanbecome zero only ifallthevelocities van- ishidentically. Therefore, gimjka]-,a,,, 2O Thus, thesum is,ingeneral, positive and canvanish only inthetrivial instance thatthesystem isnotinmotion—that is,thatthevelocities vanish identically and T=0 Wepreviously remarked thatonly theratios oftheaj,aredetermined when thew,aresubstituted into Equation 12.40. Wenow remove thisindeterminacy byimposing anadditional condition onthea]-,.Werequire that .2m‘ka'ﬂk¢ =1 (12-57) J’,JJ The aj,arethen saidtobenormalized Combining Equations 12.55 and12.57, we maywrite Because aj,isthejthcomponent ofthertheigenvector, werepresent a,by a,= a]-,ej (12.59)J Thevectors a,deﬁned inthiswayconstitute anorthonormal set;thatis,they are orthogonal according totheresult given byEquation 12.55, and they have been normalized bysetting thesum inEquation 12.57 equal tounity. Allthepreceding discussion bears astriking resemblance totheprocedure given inChapter 11fordetermining theprincipal moments ofinertia and the principal axes forarigid body. Indeed, theproblems aremathematically identi- cal,except that wearenow dealing with asystem with ndegrees offreedom. 12.6 NORMAL COORDINATES 483 The quantities mi),andA1,,areactually tensor elements, because mandAaretwo- dimensional arrays thatrelate different physical quantities,* andassuch, wewrite them as{I11} and{A}. The secular equation fordetermining theeigenfrequen- ciesisthesame asthat forobtaining theprincipal moments ofinertia, andthe eigenvectors a,correspond totheprincipal axes. Indeed, theproof oftheorthog- onality oftheeigenvectors ismerely ageneralization oftheproof given inSection 11.6oftheorthogonality oftheprincipal axes. Although wehave made aphysical argument regarding thereality oftheeigenfrequencies, wecould carry outa mathematical proof using thesame procedure used toshow that theprincipal moments ofinertia arereal. 12.6 Normal Coordinates Aswehave seen (Equation 12.43), thegeneral solution forthemotion ofthecoor- dinate qjmust beasum over terms, each ofwhich depends onanindividual eigen- frequency. Intheprevious section, weshowed that thevectors a,areorthogonal (Equation 12.55) and, asamatter ofconvenience, weeven normalized their com- ponents a,-,(Equation 12.57) toarrive atEquation 12.58; thatis,wehave removed allambiguity inthesolution fortheqj,soitisnolonger possible tospecify anarbi- trary displacement foraparticle. Because such arestriction isnotphysically mean- ingful, wemust introduce aconstant scale factor a,(which depends ontheinitial conditions oftheproblem) toaccount forthelossofgenerality introduced bythe arbitrary normalization. Thus, q,(i)=241,11,-.,¢('<'~1*‘-¢*'> (12.60) Tosimplify thenotation, wewrite to=;B.a,~.@""’*‘ (12.61)where thequantities ,8,arenew scale factorsi (now complex) that incorporate thephases of5,. Wenow deﬁne aquantity 17,, q,~<(>=Za,~.n.<(> (12.63) The 17,,bydeﬁnition, arequantities that undergo oscillation atonly one fre- quency. They may beconsidered asnew coordinates, called normal coordinates, forthesystem. The 17,satisfy equations oftheform +(vim=0 (12.64)sothat *See thediscussion inSection 11,7 concerning themathematical deﬁnition ofatensor. 1-There isacertain advantage innormalizing theaj,tounity andintroducing thescale factors a,and B,rather than leaving thenormalization unspeciﬁed. The (1,,arethen independent oftheinitial con- ditions, andasimple orthonormality equation results. 484 12/COUPLED OSCILLATIONS There arenindependent such equations, sotheequations ofmotion expressed innormal coordinates become completely separable. EXAl\~"IPI.E 12.2 _- Derive Equation 12.64 directly byusing Lagrange’s equations ofmotion. Solution. Wenote that from Equation 12.63 qj=gay‘/ilr and from Equation 12.34 wehave, forthekinetic energy, 1 .. ='5 “hit aj/fir) aksils) 1 ..=E1% Tlrns The sum intheparentheses isjust 5,,,according totheorthonorrnality condition (Equation 12.58). Therefore, 1 1 - T=§Emmm=§E4? u2w> : Similarly, from Equations 12.34 wehave forthepotential energy, 1 U=2*‘(*%= 1=5(].2kA,-/.<1_,-.a2.)11n1. The ﬁrstequation inEquation 12.53 is giA,-ka,-,ak, =wf m]~,,a,-,a,,, =(0515,, sothepotential energy becomes 1 1 v=§§w%nm=§Zw%$ uzw) 9 Using Equations 12.65 and12.66, theLagrangian is 1E-2 22L=5T(11.-wm.) (12-67) 12.6 NORMAL COORDINATES 485 andLagrange’s equations are 6L d6L____._. =0 611. dt611. or +win.=0 asfound inEquation 12.64. Thus, when theconﬁguration ofasystem isexpressed innormal coordi- nates, both thepotential andkinetic energies become simultaneously diagonal. Because itistheoff-diagonal elements of{m} and{A}thatgiverisetothecou- pling oftheparticles’ motions, itshould beevident thatachoice ofcoordinates thatrenders these tensors diagonal uncouples thecoordinates andmakes the problem completely separable into theindependent motions ofthenormal coordinates, each with itsparticular normal frequency.* The foregoing hasbeen amathematical description ofthemethods used to determine thecharacteristic frequencies co,andtodescribe thecoordinates 1],of thenormal mode motion. The actual application ofthemethod canbesumma- rized byseveral statements: 1.Choose generalized coordinates andﬁnd Tand Uinthenormal Lagrangian method. This corresponds tousing Equations 12.34. - 2.Represent A,-,,and mi,astensors innXnarrays, anduseEquation 12.42 to determine thenvalues ofeigenfrequencies w,. 3.Foreach value ofw,,determine theratios a1,:a2, :a3,: :a,,,bysubstituting into Equation 12.40: ;(.-1,, —wZm,,,)(1,,=0 (12.68) 4.Ifneeded, detennine thescale factors ,8,(Equation 12.60) from theinitial conditions. 5.Determine thenormal coordinates 17,byappropriate linear combinations of theqjcoordinates that display oscillations atthesingle eigenfrequency (1),. The description ofmotion forthissingle normal coordinate 17,iscalled a normal mode. The general motion (Equation 12.63) ofthesystem isacom- plicated superposition ofthenormal modes. Wenow apply these steps inseveral examples. Determine theeigenfrequencies, eigenvectors, andnormal coordinates ofthe mass-spring example inSection 12.2 byusing theprocedure justdescribed. Assume K12=K. *The German mathematician Karl Weierstrass (1815-1897) showed in1858 that themotion ofa dynamical system canalways beexpressed interms ofnormal coordinates. 486 12/COUPLED OSCILLATIONS Solution. The eigenfrequencies were determined inExample 12.1, where we found Tand U(step 1).Wecanﬁnd thecomponents forA,-,,directly from Equation 12.46 orbyinspection from Equation 12.45, making sure A],issym- metrical. +K —K A:K 12 12 . {}1_K12 K'1'K12 (1269) The array m,,,caneasily bedetermined from Equation 12.47: {I11}= (12.10) WeuseEquation 12.42 todetermine theeigenfrequencies (0,. K+K12—Mw2 —K12 _ _K12 K'1' K12 _ Mwg which isidentical toEquation 12.50 with theresults ofEquation 12.8for(01and(02. WeuseEquation 12.68 todetermine theeigenvector components a,-,.We have twoequations foreach value ofr,butbecause wecandetermine only the ratios a1,/a2,, one equation foreach rissufﬁcient. Forr=1,k=1,wehave (A11 _wim11)a11 '1'(A21 _‘"im21)a21 =0 (12-71) or,inserting thevalues forA11,A21,(0%,andmu,andusing thesimpliﬁcation that K12 zK, 2K—MM a11—Ka21=0 with theresult an = —a21 Forr=2,k=1, wehave (QK _ 11:4 M)a/12 _ Kagg = 0 with theresult an=a22 (12.73) The general motion (Equation 12.63) becomes K1=011771 '1'012172 K2=a21°'l1 '1'@2172} Using Equations 12.72 and 12.73, thisbecomes(12.74) K1=a11"l1 '1'(122712} (12.75) K2=_a11"11 '1'(122712 12.6NORMAL COORDINATES 487 TABLE 12-1 Normal Mode Motions Normal Particle mode Eigenfrequency oscillation Particle velocities /3K1 0)] = M Out ofphase Equal butopposite 2 (02=,1% Inphase Equal Adding x1and x2gives 1172=i(x1 +x2) (12.76) Subtracting x2from x1gives 1 '71I_*(x1 _K2) (12-77)2a22 24111 The normal coordinate 172canbedetermined byﬁnding theconditions when theother normal coordinate 171remains equal tozero. From Equation 12.77, 171=0when x1=x2.Thus, fornormal mode 2(172), thetwomasses oscil- lateinphase (thesymmetrical mode). The distance between theparticles isal- ways thesame, andthey oscillate asifthespring connecting them were arigid, weightless rod. Similarly, wecanﬁnd theconditions forthenormal coordinate 171bydeter- mining when 172=0(x;) =—x1). Innormal mode 1(171), theparticles oscillate outofphase (the antisymmetrical mode). This analysis (summarized inTable 12-1) conﬁrms ourprevious results (Section 12.2) ,andtheparticle motion isasshown inFigure 12-2. Such mo- tions foratoms inmolecules arecommon. Remember thatweletK=K12dur- ingthisexample. (U1: 31Wemay determine thecomponents oftheeigenvectors (Equation 12.59), (U2: 32 byusing Equations 12.72 and12.73. 31: a2:=@1161 '1'@2162 =@1261 +(12262 ¢l11(e1 _62) a22(e1 +e2)(12.78) (12.79) Although normally notrequired, wemay determine thevalues ofanandax, from theorthonormality condition ofEquation 12.58 with theresult 1 (111: _a21= —“—* \/2M 1 012=(122=WI(12.80) 488 12/COUPLED OSCILLATIONS Inthisexample, itwasnotnecessary todetermine thescale factors B,norto write down thecomplete solution, because theinitial conditions were notgiven. l‘lX.»\l\'1l’l.E I2.4 Determine theeigenfrequencies anddescribe thenormal mode motion fortwo pendula ofequal lengths bandequal masses mconnected byaspring offorce constant Kasshown inFigure 12-5. The spring isunstretched intheequilibrium position. Solution. Wechoose 01and62(Figure 12-5) asthegeneralized coordinates. The potential energy ischosen tobezero intheequilibrium position. The kinetic andpotential energies ofthesystem are,forsmall angles, 1. 1.:r=§m(b61)2 +~§m(b62)2 (12.81) 1U= mgb(1 —cos61)+mgb(1 —cos62)+§K(b sin61—bsin62)? (12.82) Using thesmall oscillation assumption sin6~(9andcos6~1—62/2, wecan write 7,2 U=$70; +0;)+K?(6,—0,)? (12.33) The components of{A}and{I11} are mb2 0 {I11} —{O mm} (12-34) b2 —bi’ 1»-(2.:.2...) The determinant needed toﬁnd theeigenfrequencies wis mgb+Kb?—w2mb2 —Kl72 _ —Kb2 mgb +K02—to2mb2 (12186) which gives thecharacteristic equation b2(mg+ Kb—w2mb)2 —(Kb2)2 =0 (mg+ Kb—w2mb)2 =(Kb)2 or mg+Kb—w2mb =i'Kb (12.87) 12.6NORMAL COORDINATES 489 __ 2l 6___->—4 o~ Vi.9 Q- m m FIGURE 12-5 Example 12.4. Two pendula ofequal lengths having equal masses are connected byaspring. Taking theplus sign, w=wl, mg+ Kb—aﬁmb =Kb (8%=§ (12.88) Taking theminus sign inEquation 12.87, w=(O2, mg+ Kb— w§mb= —Kb mg=§+2% (12.89) Putting thevalues ofwland(U2into Equation 12.40 gives, fork=1, (mgb +Kb2—w§mb2)a1, —Kb2(l2,. =0 (12.90) Ifr=1,then 2g2 2_mgb+Kb —bmb d11—Kb 1121-0 and all=421 (12-91) Ifr=2,then + Kb2 _ %mb2 _ 2%mb2)a12 _ Kb2a22 = 0 and Wewrite thecoordinates 61and02interms ofthenormal coordinates by 91=a11"71 +412772} (1293) 92=a21°'I1 +a22"72 490 12/COUPLED OSCILLATIONS UYTYTYTSpring not <~——-—----+ compressed or Spring isextended ¢Xl¢11d¢d andthen compressed ($Ymm@Tl'i¢) (antisymmetric) Normal mode 1 Normal mode 2 FIGURE 12-6 Example 12.4. The twonormal mode motions areshown. Using Equations 12.91 and 12.92, Equations 12.93 become 91=(111771 _922772(12.94) 92=(111171 +022172} The normal modes areeasily determined, byadding andsubtracting 01and02, tobe 1 TI1="551:(91+92) 1 (12.95) 172=T99 2_91)4122 Because normal coordinate 171occurs when 172=O,then 62=61fornormal mode 1(symmetrical). Similarly, normal coordinate 172occurs when 171=0 (61=—62), andnormal mode 2isantisymmetrical. The normal mode motions areshown inFigure 12-6. Notice thatformode 1,thespring isneither com- pressed norextended. The twopendula merely oscillate inunison with their natural frequencies (col=coo=\/g/b). These motions canbeeasily demon- strated inthelaboratory orclassroom. The higher frequency ofnormal mode 2 iseasily displayed forastiffspring. 12.7 Molecular Vibrations Wementioned previously thatmolecular vibrations aregood examples oftheappli- cations ofthesmall oscillations discussed inthischapter. Amolecule containing natoms generally has3ndegrees offreedom. Three ofthese degrees offreedom areneeded todescribe the translational motion, and, generally, three are needed todescribe rotations. Thus, there are3n—6vibrational degrees offree- dom. Formolecules with collinear atoms, only twopossible rotational degrees of 12.7MOLECULAR VIBRATIONS 491 freedom exist, because rotation about theaxis through theatoms isinsigniﬁ- cant. Inthiscase, there are3n—5degrees offreedom forvibrations. Wewant toconsider here only thevibrations occurring inaplane. Weelimi- nate thetranslational and rotational degrees offreedom byappropriate trans- formations and choice ofcoordinate systems. Formotion inaplane, there are 2ndegrees offreedom. Because twoaretranslational andoneisrotational, gen- erally 2n—3normal vibrations occur intheplane [leaving (312—6)—(2n—3) =n—3degrees offreedom forvibrations oftheatoms outoftheplane]. Linear molecules mayhave both longitudinal andtransverse vibrations. The longitudinal vibrations occur along thelineoftheatoms. Fornatoms, there are ndegrees offreedom along theline, butoneofthem corresponds totranslation. Thus, there aren—1possible vibrations inthelongitudinal direction forn atoms inalinear molecule. Ifatotal of3n—5vibrational degrees offreedom exist foralinear molecule, there must be(3n—5)—(n—1)=2n—4transverse vibrations causing theatoms tovibrate perpendicular totheline ofatoms. But from symmetry, anytwomutually perpendicular directions sufﬁce—so there are really only halfthenumber oftransverse frequencies, orn-2. EXAMPLE 12.5 if I - Determine theeigenfrequencies anddescribe thenormal mode motion ofa symmetrical linear triatomic molecule (Figure 12-7) similar toCO2. The central atom (carbon) hasmass M,andthesymmetrical atoms (oxygen) have masses m. Both longitudinal andtransverse vibrations arepossible. Solution. Forthree atoms, thepreceding analysis indicates thatwehave two longitudinal andonetransverse vibrational degrees offreedom ifweeliminate thetranslational androtational degrees offreedom. ..M ... <9aaaa"Q"""""9 |—~~11»-2l—~~-2(a)Linear triatomic molecule (b)Longitudinal d€SCI‘ip[iO11< .__b .._>< I__b .7» O-—> <—--( ) O—>Mode1 __A______yEt_Q“_a _____ __ 9.. Q4-—OMode2 1110 oi»(c)Longitudinal normal modes (d)Transverse normal mode FIGURE 12-7 Example 12.5. (a)Alinear triatomic molecule (forexample, CO2 with central mass Mandsymmetrical masses m.(b)The elastic forces between atoms arerepresented bysprings; theatomic displacements from equilibrium arex1,x2,and x5.(c)The twolongitudinal normal modes. (d)The transverse nonnal mode. 492 12/COUPLED OSCILLATIONS Wecansolve thelongitudinal andtransverse motions separately, because they areindependent. InFigure 12-7b, werepresent theatomic displacements from equilibrium byxl,xi),x3.The elastic forces between atoms arerepresented bysprings offorce constant K1.Wehave three longitudinal variables butonly two degrees offreedom. Wemust eliminate thetranslational possibility byrequiring thecenter ofmass tobeconstant during thevibrations. This issatisﬁed if Therefore, wecaneliminate thevariable x2: 2,=—ﬁ(x, +2,) (12.97) The kinetic energy becomes 1_ 1_ 1_ T= + + 1 12=émi?+Emg+5%(23+2%+22,21) (12.98) Having the£3221coupling term inthekinetic energy (called “dynamic cou- pling”) canbeinconvenient when solving Equation 12.42 fortheeigenfrequen- cies. Weuseatransformation toeliminate thedynamic coupling. Let q‘ixi+*1} (l2.99a)‘I2_xs_x1 Then 1 xs=§(‘I1 '1'92) x=— — 1( )12q‘‘I2 (12991))and Tn x2=‘E41 andthekinetic energy (Equation 12.98) becomes m_ (mM+ 2m2) _T: Zqg+—T qg (12-100) The potential energy is 1 1U=§K1(X2 —x1)2+§K1(x3 —x2)? (l2.101) 12.7 MOLECULAR VIBRATIONS 493 andwith thetransfonnations, Equations 12.99, thepotential energy becomes (after considerable reduction) 2m+M221 2U= W Klql +;K1q2 (12.102) The eigenfrequencies aredetermined byinspection, using Equation 12.42 12m+M2 mM +2m2 5T "1-‘"2ﬁn" °=0 (12.103) O Q_w2ll 2 2 tobe w,_<2m+M)1 mM K1(12.104) K12__mi 2m Because thetensor formed bythecoefﬁcients ofEquation 12.40 isalready diag- onal, thevariables qlandq2represent thenormal coordinates (unnormalized). = +Q1_(711711 1112712} (12105) (12—021711 '1'022712 But G12 = 0 and (Z21 = 0 111=a117I1 (I2=(122712 Asusual, wedetermine themotion ofonenormal mode when theother iszero. Thedescriptions ofthelongitudinal nonnal mode motion aregiven inTable 12-2. Normal mode 1hastheendatoms insymmetrical motion, butthecentral atom (from Equation 12.97) moves opposite tox1andx3.Normal mode 2hasthe endatoms vibrating antisymmetrically, butthecentral atom isatrest. This mo- tion isdisplayed inFigure 12-7c. TABLE 12-2 Longitudinal Normal Mode Motions Mode Eigenfrequencies Variable Motion 1 ,/ K1 q1=x3+x1 x3=x1(q2=O) 2mN2 Z —“-15.761 K1 2 Z (12:75:-1-771 xs=—x1(q1=0) x2=0 494 12/COUPLED OSCILLATIONS Because wehave eliminated rotations inoursystem, thetransverse vibra- tions must beasshown inFigure 12-7d, with theendatoms moving inphase (yl=yg)opposite tothatofy2.Anequation similar toEquation 12.97 relates y2 toylandygtokeep thecenter ofmass constant. "101+)3)+M02)=0 (12-106) m )2=-501 +)3) (12-107) Werepresent thesingle degree offreedom forthetransverse vibration bythe angle arepresenting thebending ofthelineofatoms. Weassume aissmall. _(J11-72) '1'()l3_)’2)a——i—-—I;i— The kinetic energy forthetransverse mode is 1 T=émoi’+13)+51‘/11% Because yl=ygandusing Equation 12.107, aand Tbecome a=%(2m+M) (12.108)bM :r=gm+2m))% M122 _:r= a2 (12.109) The potential energy represents thebinding ofthelineofatoms. Weassume therestoring force tobeproportional tothetotal deviation from astraight line (ba), sothepotential energy is 1U=§K2(ba)2 (12.110) Equations 12.109 and12.110 aresimilar tothose forthemass-spring, with the vibrational frequency determined tobe 2(M +2m)w§=———iK2 (12.11l)mM The transverse normal mode isrepresented by )’1=J73 (12-112) m T2I_M()’1 '1'ya) (12-113) asalready discussed andshown inFigure 12-7d. 12.8 THREE LINEARLY COUPLED PLANE PENDULA 495 The CO2 molecule isanexample ofthesymmetrical linear molecule just discussed. Electromagnetic radiation resulting from theﬁrstandthird normal modes isobserved, because theelectrical center ofthemolecule deviates from thecenter ofmass (m:O';M:C”). Butnoradiation emanates from normal mode 2,because theelectrical center iscoincident with thecenter ofmass and thus thesystem hasnodipole moment.* 12.8 Three Linearly Coupled Plane Pendula— anExample ofDegeneracy EXAMPLE 12.6 F Consider three identical pendula suspended from aslightly yielding support. Because thesupport isnotrigid, acoupling occurs between thependula, and energy canbetransferred from onependulum totheother. Find theeigenfre- quencies andeigenvectors anddescribe thenormal mode motion. Figure 12-8 shows thegeometry oftheproblem. Solution. Tosimplify thenotation, weadopt asystem ofunits (sometimes called natural units) inwhich alllengths aremeasured inunits ofthelength of thependula l,allmasses inunits ofthependula masses M,andaccelerations in units ofg.Therefore, inourequations thevalues ofthequantities M,l,andg arenumerically equal tounity. Ifthecoupling between each pair ofthepen- dula isthesame, wehave = 1'l'ég'l'é§) '“l I\Ql_‘1\D1—‘/%$1IQ (12.114) U=-(0%+0‘;+9;—260,0, —280,0, —26020,) s> :_sz E.s>rm i- \ q=z @=z @=z m1=M m2= m5= l FIGURE 12-8 Example 12.6. Three identical pendula aresuspended from aslightly yielding support that allows energy tobetransferred between pendula. Such anexperiment iseasy tosetupand demonstrate. *For aninteresting discussion ofpolyatomic molecules, seeD.M.Dennison, Rev.Mod. Phys. 3,280 (1931). 496 12/COUPLED OSCILLATIONS Thus, thetensor {I11} isdiagonal, 100 {m}={0 10 (12.115) 001 1-s -s {A} =-8 1-s (12.116) —e -s 1but{A}hastheform The secular determinant is 1-(02 -s -e —e 1-(02 -e =0 (l2.117) -e —e 1-(02 Expanding, wehave (1—w2)3 -263—3e2(1 —(1)2)=0 which canbefactored to ((02-1-e)2(co2-1+2s)=0 andhence theroots are w]= 1+s (1)2= e (12.118) isNotice thatwehave adouble root: wl=(1)2=\/1+s.The normal modes corre- sponding tothese frequencies aretherefore degenerate-—that is,these twomodes areindistinguishable. Wenow evaluate thequantities of-,,beginning with ajg.Again wenote that, because theequations ofmotion determine only theratios, weneed consider only twoofthethree available equations; thethird equation isautomatically sat- isﬁed. Using theequation ;(Aj/1 _wgmjk) (1)9=O weﬁnd 280513 _80123 —B0333 =0 } (12.119)—ea13 +2.91123 -9:133 =0 Equations 12.119 yield 1113=1123:1133 (12.120) andfrom thenormalization condition wehave 2 2 2-¢119'l'a23'l'1132_1 12.8 THREE LINEARLY COUPLED PLANE PENDULA 497 OI‘ 1 1119=1129=1193=“V? (12-121) Thus, weﬁnd thatforr=3there isnoproblem inevaluating thecompo- nents oftheeigenvector a3.(This isageneral rule: There isnoindeﬁniteness in evaluating theeigenvector components foranondegenerate mode.) Because allthecomponents ofa2areequal, thiscorresponds tothemode inwhich all three pendula oscillate inphase. Letusnow attempt toevaluate theaﬂanda]-2.From thesixpossible equa- tions ofmotion (three values ofjandtwovalues ofr),weobtain only twodiffer- entrelations: 8<a11 '1'G21 +G31) =0 s(a12 +a22+a32) =0 *(l2.l23) The orthogonality equation is %'.mj,aj,a,,, =0, r9*s but,because mj-,,=8]-2,thisbecomes 221,11), =0,r211 (12124) which leads toonly onenew equation: 11111112 '1'11211122 '1'11311192 =O *(12-125) (The other twopossible equations areidentical with Equations 12.122 and 12.123 above.) Finally, thenormalization conditions yield a%1+(121+agl=1 *(12.l26) (122+(122+a§2=1 *(12.127) Thus, wehave atotal ofonlyﬁve(starred,*) equations forthesixunknowns aﬂanda]-2.This indeterminacy intheeigenvectors corresponding toadouble root isexactly thesame asthatencountered inconstructing theprincipal axes forarigid body with anaxisofsymmetry; thetwoequivalent principal axes may beplaced inanydirection aslong asthesetofthree axes isorthogonal. There- fore, weareatliberty toarbitrarily specify theeigenvectors a1anda2,aslong as theorthogonality andnormalizing relations aresatisﬁed. Forasimple system such aswearediscussing, itshould notbedifﬁcult toconstruct these vectors, so wedonotgiveanygeneral rules here. Ifwearbitrarily choose a31=0,theindeterminacy isremoved. Wethen ﬁnd 1 1a=——(1,—1,0), a=——(1,1,-2) (l2.128) 1\/2 2\/5 from which wecanverify thatthestarred relations areallsatisﬁed. 498 12/COUPLED OSCILLATIONS Recall thatthenondegenerate mode corresponds tothein-phase oscilla- tionofallthree pendula: a3=?;E(1,1,1) (12129) Wenow seethat thedegenerate modes each correspond toout-of-phase oscil- lation. Forexample, a2inEquation 12.128 represents twopendula oscillating together with acertain amplitude, whereas thethird isoutofphase andhas twice theamplitude. Similarly, a1inEquation 12.128 represents onependu- lum stationary andtheother twoinout-of-phase oscillation. The eigenvectors a1and a2already given areonly one setofaninﬁnity ofsets satisfying thecon- ditions oftheproblem. Butallsuch eigenvectors represent some sortofout- of-phase oscillation. (Further details ofthisexample areexamined in Problems 12-19 and 12-20.) 12.9 The Loaded String* Wenow consider amore complex system consisting ofanelastic string (ora spring) onwhich anumber ofidentical particles areplaced atregular intervals. The ends ofthestring areconstrained toremain stationary. Letthemass ofeach ofthenparticles bem,andletthespacing between particles atequilibrium bed. Thus, thelength ofthestring isL=(n+1)d. The equilibrium situation'is shown inFigure 12-9. Wewish totreat thecase ofsmall transverse oscillations oftheparticles about their equilibrium positions. First, weconsider thevertical displacements ofthemasses numbered j—1,j,andj+1(Figure 12-10). Ifthevertical dis- placements qj-_1, qj,andq]~+1aresmall, then thetension 1'inthestring isapprox- imately constant andequal toitsvalue atequilibrium. Forsmall displacements, thestring section between anypair ofparticles makes only small angles with the equilibrium line. Approximating thesines ofthese angles bythetangents, the expression fortheforce that tends torestore thejthparticle toitsequilibrium position is 1;=—§<q.—1,»)—201,—q,~.1> <12-130) The force is,according toNewton’s law, equal tomijj; Equation 12.130 can therefore bewritten as gj=mld(q,~_, —2q,+q]'+l) (12.1s1) *The ﬁrstattack ontheproblem oftheloaded string (orone-dimensional lattice) wasbyNewton (in thePrincipia, 1687). The work wascontinued byJohann Bemoulli andhissonDaniel, starting in 1727 and culminating inthelatter’s formulation oftheprinciple ofsuperposition in1753. Itisfrom thispoint thatthetheoretical treatment ofthephysics ofsystems (asdistinct from particles) begins. 12.9 THE LOADED STRING 499 ', (n+1)d 0 d 2d (j-1)d jd (j+1)d (n—1)d nd=L m m m m m m M ------- --oi-o-io-------- I 2 j—l j+l n—l 11, FIGURE 12-9 Aschematic oftheloaded string. Inequilibrium, identical masses are spaced equidistantly. The ends ofthestring areﬁxed. J’m j+l 1'1 .m q] m ‘I;-1 d d ‘I141 Equilibrium line FIGURE 12-10 Vertical displacements (q,--1, qj,andq]-+1) ofmasses ontheloaded string. which istheequation ofmotion forthejthparticle. The system iscoupled, because theforce onthejthparticle depends onthepositions ofthe(j-1)th and (j+1)th particles; thisistherefore anexample ofnearest neighbor interaction, inwhich thecoupling isonly totheadjacent particles. Itisnotnecessary thatthe interaction beconﬁned tonearest neighbors. Iftheforce between pairs ofparti- cleswere electrostatic, forexample, then each particle would becoupled toallthe other particles. Theproblem canthen become quite complicated. Buteven ifthe force iselectrostatic, the1/r2 dependence ondistance frequently permits usto neglect interactions atdistances greater than oneinterparticle spacing, sothatthe simple expression fortheforce given inEquation 12.130 isapproximately correct. Wehave considered only themotion perpendicular tothelineofthestring: transverse oscillations. Itiseasy toshow thatexactly thesame form fortheequa- tions ofmotion results ifweconsider longitudinal vibrations—that is,motions along thelineofthestring. Inthiscase, thefactor 1/disreplaced byK,theforce constant ofthestring (seeProblem 12-24). Although weused Newton's equation toobtain theequations ofmotion (Equation 12.131), wemay equally well usetheLagrangian method. The poten- tialenergy arises from thework done tostretch then+1string segments*: n+1 ‘T U:2-01];(qj_1-q]‘)2 (12.132) where qoandq,,+1areidentically zero, because these positions correspond tothe ﬁxed ends ofthestring. Wenote thatEquation 12.132 yields anexpression forthe force onthejthparticle thatisthesame astheprevious result (Equation 12.130): *We consider thepotential energy tobeonly theelastic energy inthestring; thatis,wedonotcon- sider theindividual masses tohave anygravitational (oranyother) potential energy. 500 12/COUPLED OSCILLATIONS _high all __ 2 __ 21? aqj 2daq]_ l(‘Ij-1 qj)'l'(qj qj+1) 1 =§,<q,~-.—Qq,+q,~..1> (12.1%) The kinetic energy isgiven bythesum ofthekinetic energies ofthenindividual particles: 1"_T—gm;/lqf (12124) Because (1,... E0,wemay extend thesum inEquation 12.134 toj=n+1so thattherange ofjisthesame asthatintheexpression forthepotential energy. Then, theLagrangian becomes 111-1-1 _ 1'L== -:i(qj_1 -qj)2:| (l2.135) Itshould beobvious that theequation ofmotion forthejthparticle must arise from only those terms intheLagrangian containing qjorQ]-.Ifweexpand thesum inL,weﬁnd 1_ 11' 11' L= 'l' "5:10];-1 "qj)2"55011‘ —‘1j+1)2 " (12-136) where wehave written only those terms that contain either qjor1]]-.Applying Lagrange’s equation forthecoordinate qj,wehave II T ___E(q]~_1 "' + q]-+1) Z O Thus, theresult isthesame asthatobtained byusing theNewtonian method. Tosolve theequations ofmotion, wesubstitute, asusual, q]-(t) =aje"”‘ (l2.l38) where ajcanbecomplex. Substituting this expression forq]-(t) into Equation 12.137, weﬁnd 1' 1'-—;!a]-_1 +(25 —mw2)aJ- —Eajﬂ ==0 (12.139) where j=1,2,...,n, butbecause theends ofthestring arefixed, wemust have a0:an+1 :O- Equation 12.139 represents alinear difference equation that canbesolved fortheeigenfrequencies to,bysetting thedeterminant ofthecoefﬁcients equal tozero. Wetherefore have thefollowing secular determinant: 12.9 THE LOADED STRING 501 T___ 0 O O .. d -coo_‘>-___ A __Z O O .. d d 1' 1'—— A ~- 0 ---=0 1.4 d d (210) 1' 1'_..~ A ___. .. d d 0 0 -0 where wehave used AE2E-moi’ (12.141) This secular determinant isaspecial case ofthegeneral determinant (Equation 12.42) that results ifthetensor misdiagonal and thetensor Ainvolves acou- pling only between adjacent particles. Thus, Equation 12.140 consists only ofdi- agonal elements plus elements once-removed from thediagonal. Forthecase n=1(i.e., asingle mass suspended between twoidentical springs), wehave A=0,or .1=,/31md Wemayadapt thisresult tothecase oflongitudinal motion byreplacing 1'/dbyK; wethen obtain thefamiliar expression, 2Kw==,/-m Forthecase n==2,andwith 1'/dreplaced byK,wehave A2=K2,or 2Ki1<w==,('——-—m which arethesame frequencies asthose found inSection 12.2 fortwocoupled masses (Equation 12.8). The secular equation should berelatively easy tosolve directly forsmall val- uesofn,butthesolution becomes quite complicated forlarge n.Insuch cases, it issimpler tousethefollowing method. Wetryasolution oftheform aj=aei<j~/-6) (12442) where aisrealThe useofthisdevice isjustiﬁed ifwecanﬁnd aquantity 7anda phase 5such that theconditions oftheproblem areallsatisﬁed. Substituting aj- inthisform into Equation 12.139 andcanceling thephase factor, weﬁnd -§e"'Y+<2§— mw2)—ie"°’=0 502 12/COUPLED OSCILLATIONS Solving for(02,weobtain (02=it-1-—l6i(e1Y +e"“')m m 2=l(1—cos7) (12.14.2)md 41.2?=—sin —-md 2 Because weknow thatthesecular determinant isoforder nandtherefore yields exactly nvalues for(02,wecanwrite w,=2./l S1111’, T:1,2,...,n (12.144)md 2 Wenow evaluate thequantity 7,andthephase 8,byapplying theboundary condition thattheends ofthestring remain ﬁxed. Thus, wehave ajr=aTei(j‘Y’!‘—81‘) (12145) or,because itisonly therealpart thatisphysically meaningful, aj,=a,cos(j'y, -5,) (12.146) The boundary condition is (1013a(n+1)1 E0 (12-147) ForEquation 12.146 toyield a]-,='-0forj=0,itshould beclear that5,must be 11'/2 (orsome oddinteger multiple thereof). Hence, - _ 1r aj,=a,cos]'y,-5 ==a,sinjy, (12.148) Forj= n+1,we have a(,,+1), I0=a,sin(n +1)'y, Therefore, (n+1)'y,=s'n', s=1,2,... or s11'=—- =1,71 n+1’ S 2 Butthere arejust ndistinct values of7,because Equation 12.144 requires ndis- tinct values ofw,.Therefore, theindex sruns from 1to11.Because there isaone- to-one correspondence between thevalues ofsandthevalues ofr,wecansimply replace sinthislastexpression bytheindex r: T7!‘=——-——— =1 1.149 yr n+11 T 221 an (2 ) The aj,(Equation 12.148) then becomes _ _T71‘ J,=a,s1n(]---n +1)l (12.150) 12.9 THE LOADED STRING 503 The general solution forqj(seeEquation 12.61) is 1.- = P ' ‘___TL-_ im,,l;B,a, s1n<] n+1)e =B,sin(j Z’-1-_:-'T)ei<-¢ (12151) where wehave written B,EB§a,.Furthermore, forthefrequency wehave m 2(n+11'_ r17= ——s1n ' (l2.l52) Wenote thatthisexpression yields thesame results found forthecase oftwo coupled oscillators (Equations 12.8) when weinsert n==2,r=1,2andreplace 1'/dbyK( =K12). Notice alsothatifeither r=0orr=n+1issubstituted intoEquation 12.150, then alltheamplitude factors aj,vanish identically. These values ofrtherefore refer tonullmodes. Moreover, ifrtakes onthevalues n+2,11+3,..., 2n+1,then theaj,arethesame (except foratrivial signchange andinreverse order) asforr= 1,2,...,n;also, r=2n+2yields thenext nullmode. Weconclude, therefore, that there areindeed only ndistinct modes andthatincreasing rbeyond nmerely du- plicates themodes forsmaller n.(Asimilar argument applies forr<0.)These conclusions areillustrated inFigure 12-11 forthecase n=3.The distinct modes arespeciﬁed byr=1,2,3;r=4isanullmode. The displacement patterns aredu- plicated forr=7,6,5,8,butwith achange insign. InFigure 12-11, thedashed curves merely represent thesinusoidal behavior oftheamplitude factors aj,for various values ofr;theonly physically meaningful features ofthese curves arethe values atthepositions occupied bytheparticles (j=1,2,3).The“high frequency” ofthesinecurves forr=5,6,7,8isthus notatallrelated tothefrequency ofthe particles’ motions; these latter frequencies arethesame asforr=1,2,3,4. The normal coordinates ofthesystem (Equation 12.62) are 771(1)EBrew" (12-153) sothat q,-(1)=211,sin(j;%) (12.154) This equation forqjissimilar totheprevious expression (Equation 12.63) ex- cept thatthequantities a,-,arenow replaced bysin[j(r1r)/(n +1)]. Because B,maybecomplex, wewrite fortherealpart ofqj, real: q,(t) =2sin(j —%)(p, cosw,t-—1/,sin(0,1) (12.155)1 n 504 ~r 1\r \ I \ / \ I, .. !\\ ' I12/COUPLED OSCIL \\ I1, \ I\ , \ I \\ I \ =1 1'=5I \ I \ I ‘\ I \ \_'I \, I’\\ I‘\ I I\I\ =2\ I \ I\ I\ /\ I /"\ ’\ I '‘II_.r" /\ "‘ \ I\ , \ \ \ ' T\ I \ II \I I 3___,"r ...»*'—_-___-1 ,- ,_~___‘O r ’,,'\ '\I »*" f\\~__,- "T f*~1‘: I I r=7 I\ I '\ '\I \ I II\ 1 ' 111 \ ' \ FIGUREr=4 ‘ I, 1 I’ ‘ I 1 I1 I 1 I1I 1I \I \J 12-11 Thenormal mode-_ motion fohA 3are d''rtecase ofn— istlnct mod,-»" _.» ‘Q-. ,...~-_ ,_.\‘Q ,_ r\I \, \_,' I ~_‘r \ ___-'— .--'I \I]\ I ,-v" ___.’~_A ~_____...— __.-p _-— —3masses. Only 1"=1,2, es,because 1*=4isanull areduplicates of1mode and - ,2,3,and4 ' dash dI ~8 , r—7,6,5,8 ,respectively, with achange insign. The ecurves represent thesinusoidal behavior oftheamplitude factors air,andarenotphysically meaningf lufeatures ofthemotion.LATIONS 12.9THELOADED STRING 505 where [3,=p.,+iv,1 (l2.l56) The initial value ofq]-(t)canbeobtained from Equation 12.155: q,-(0)=;.I.,sin(j5_€—-i) (12.157) 1'77 q,(0)=_§.,,,»,s1n(j 71-1?) (12.15s) Ifwemultiply Equation 12.157 bysin[j(s1r)/(n +1)]andsum overj,weﬁnd __s1r __T77 _,511'guy]-(0) S11'1(]7_|:“i) I%,u.,sin(] sin(y (12.159) Arelationship intheform ofatrigonometric identity isavailable forthesine terms: n r11' s11" n+1J;Sin(j;1-gt?) Si1'1(j;L-_":"I) -'1-T 51$) T,S=1,2, ,1’),(12.160) sothatEquation 12.159 becomes __s'n' n+1 __n+1 2I-4': or 2 $11",1,=;z—_1—l]2t],-(0) sin(j;Z—_.-T-T) (l2.l61a) Asimilar procedure for11,yields 2_ 2 . ..W1»,4in+1)?q,-(0)S11'1(]n+1) (12.161b) Thus, wehave evaluated allthenecessary quantities, andthedescription ofthe vibrations ofaloaded string istherefore complete. Weshould note thefollowing point regarding thenormalization procedures used here. First, inEquation 12.57 wearbitrarily nonnalized theaj,tounity. Thus, thea,-,arerequired tobeindependent oftheinitial conditions imposed onthesys- tem. The scale factors 01,andB,then allowed themagnitude oftheoscillations to bevaried bytheselection oftheinitial conditions. Next, intheproblem ofthe loaded string, wefound thatinstead ofthequantities a,-,,there arose thesinefunc- tions sin[j(r1r)/ (n+1)],andthese functions possess anormalization property (Equation 12.160) thatisspeciﬁed bytrigonometric identities. Therefore, inthis case itisnotpossible arbitrarily toimpose anormalization condition; weareauto- matically presented with thecondition. Butthisisnorestriction; itmeans only that thescale factors ,8,forthiscase have aslightly different form. Thus, there are 506 12/COUPLED OSCILLATIONS certain constants thatoccur inthetwoproblems that, forconvenience, aresepa- rated indifferent ways inthetwocases. Consider aloaded string consisting ofthree particles regularly spaced onthe string. Att=0thecenter particle (only) isdisplaced adistance aandreleased from rest. Describe thesubsequent motion. Solution. The initial conditions are q2(0) =a,111(0) Iq3(0) =0 111(0) =12(0) =124(0) =0 Because theinitial velocities arezero, thev,vanish. TheII,aregiven by (Equation 12.161a):} (l2.l62) 2 __T’IT MT? n+ 1J2qj(0)S1n(]n +1) =—asin— . 1'T”) (12162)2 2 because only theterm j=2contributes tothesum. Thus, 1 1 I11:5% I-1220, I-1'3:*5“ (12-164) Thequantities sin[j(r11')/ (11+1)]thatappear intheexpression forqj-(t) (Equation 12.155) are r=1 2 3 J'= 1__ 1l/22 »s-s(12.l65) 2 0 —-1 3—-—~ *1 —\/22 The displacements ofthethree particles therefore are \/2q1(t) =—4—a(cos wltecoswgt) 1q2(t) =§a(cos w1t+ coswgt) (12.166) \/2q3(t) =—4—a(cos colt-cosw3t) =q1(t) PROBLEMS 507 where thecharacteristic frequencies aregiven byEquation 12.152: 5),:2./it854%), T:1,2,2 (12.157) Notice thatbecause themiddle particle wasinitially displaced, novibration mode occurs inwhich thisparticle isatrest; thatis,mode 2with frequency (02 (seeFigure 12-11) isabsent. PROBLEMS C 12-1. Reconsider theproblem oftwocoupled oscillators discussed inSection 12.2 inthe event that thethree springs allhave different force constants. Find thetwocharac- teristic frequencies, and compare themagnitudes with thenatural frequencies of thetwooscillators intheabsence ofcoupling. 12-2. Continue Problem 12-1, and investigate thecase ofweak coupling: K12<<I<1,K2. Show that thephenomenon ofbeats occurs butthat theenergy-transfer process is incomplete. 12-3. Two identical harmonic oscillators (With masses Mand natural frequencies too)are coupled such thatbyadding tothesystem amass mcommon toboth oscillators the equations ofmotion become 12,+(m/1v1)sa, +.531,=0 552+(In/l\/I)551 +5,32,=0 Solve thispair ofcoupled equations, and obtain thefrequencies ofthenormal modes ofthesystem. 12-4. Refer totheproblem ofthetwocoupled oscillators discussed inSection 12.2. Show thatthetotal energy ofthesystem isconstant. (Calculate thekinetic energy ofeach of theparticles andthepotential energy stored ineach ofthethree springs, andsum theresults.) Notice thatthekinetic andpotential energy terms thathave K12asacoef- ﬁcient depend onC,andto,butnotonC2orm2.\/Vhy issuch aresult tobeexpected? 12-5. Find thenormal coordinates fortheproblem discussed inSection 12.2 and in Example 12.1 ifthetwomasses aredifferent, ml95m2.You may again assume all theKareequal. 12-6. Twoidentical harmonic oscillators areplaced such thatthetwomasses slide against oneanother, asinFigure 12-A. Thefrictional force provides acoupling ofthemo- tions proportional totheinstantaneous relative velocity. Discuss thecoupled oscil- lations ofthesystem. 508 l2-7 12-8. 12-9 12-10 12-1112/COUPLED OSCILLATIONS OOOO0O iaawaaﬁR asaga.» K is000000 gi as ass*a“ls,. ,..t.,x~sr' Vl’. ..., FIGURE 12-A Problem 12-6. Aparticle ofmass misattached toarigid support byaspring with force constant K. Atequilibrium, thespring hangs vertically downward. Tothismass-spring combina- tion isattached anidentical oscillator, thespring ofthelatter being connected to themass oftheformer. Calculate thecharacteristic frequencies forone-dimensional vertical oscillations, and compare with thefrequencies when one ortheother ofthe particles isheld ﬁxed while theother oscillates. Describe thenormal modes of motion forthesystem. Asimple pendulum consists ofabob ofmass msuspended byaninextensible (and massless) string oflength LFrom thebobofthispendulum issuspended a second, identical pendulum. Consider the case ofsmall oscillations (sothat sin6E‘6),and calculate thecharacteristic frequencies. Describe also thenormal modes ofthesystem (refer toProblem 7-7). The motion ofa pair ofcoupled oscillators may bedescribed byusing amethod similar tothat used inconstructing aphase diagram forasingle oscillator (Section 3.4). For coupled oscillators, thetwo positions x1(t)and x2(t) may be represented byapoint (the system point) inthetwo-dimensional conﬁguration space xl-x2. Astincreases, thelocus ofallsuch points deﬁnes acertain curve. The locioftheprojection ofthesystem points onto thex1-and x2-axes repre- sent themotions ofmland 1"/2,respectively. Inthegeneral case, xl(t)and x2(t) arecomplicated functions, andsothecurve isalsocomplicated. Butitisalways possible torotate thexl-x2 axes toanew setxf-xé insuch awaythat thepro- jection ofthesystem point onto each ofthenew axes issimple harmonic. The projected motions along thenew axes take place with the characteristic fre- quencies andcorrespond tothenormal modes ofthesystem. The new axes are called normal axes. Find thenormal axes fortheproblem discussed inSection 12.2 and verify thepreceding statements regarding themotion relative tothis coordinate system. Consider twoidentical, coupled oscillators (asinFigure 12-1). Leteach oftheos- cillators bedamped, andleteach have thesame damping parameter B.Aforce E, coswtisapplied toml.Write down thepair ofcoupled differential equations that describe the motion. Obtain the solution byexpressing the differential equations interms ofthenomial coordinates given byEquation 12.11 andby comparing these equations with Equation 3.53. Show that thenormal coordi- nates 111and172exhibit resonance peaks atthecharacteristic frequencies wland m2,respectively. Consider theelectrical circuit inFigure 12-B. Use thedevelopments inSection 12.2 toﬁnd thecharacteristic frequencies interms ofthecapacitance C,inductance L, PROBLEMS 509 andmutual inductance MTheKirchhoff circuit equations are O’;$Q'$L.i1+ +Mi2=0 Li2+—+Mf,=0 ‘I1 ‘I2 C C QL LQ 1, M 1, FIGURE 12-B Problem 12-11. 12-12. Show that theequations inProblem 12-11 canbeputinto thesame form as Equation 12.1bysolving thesecond equation above forT2andsubstituting there- sult into theﬁrst equation. Similarly, substitute for inthesecond equation. The characteristic frequencies may then bewritten down immediately inanalogy with Equation 12.8. 12-13. Find thecharacteristic frequencies ofthecoupled circuits ofFigure 12-C. C1 C2 Ll L12 L2 FIGURE 12-C Problem 12-13. 12-14. Discuss thenormal modes ofthesystem shown inFigure 12-D. L1 L2 Cl C12i C2T FIGURE 12-D Problem 12-14. 12-15. InFigure 12-C, replace L12byaresistor andanalyze theoscillations. 510 12-16 12-17 12-18. 12-19 12-20. 12-21. 12-22.12/COUPLED OSCILLATIONS Athinhoop ofradius Rand mass Moscillates initsown plane hanging from asin- gleﬁxed point. Attached tothehoop isasmall mass Mconstrained tomove (ina frictionless manner) along thehoop. Consider only small oscillations, andshow thattheeigenfrequencies are g \/2g ‘"1=‘/§\/ii '"2=?\E Find thetwosets ofinitial conditions that allow thesystem tooscillate initsnor- malmodes. Describe thephysical situation foreach mode. Find theeigenfrequencies and describe thenormal modes forasystem such as theone discussed inSection 12.2 butwith three equal masses mand four springs (allwith equal force constants) with thesystem ﬁxed attheends. Amass Mmoves horizontally along asmooth rail. Apendulum ishung from M with aweightless rodand mass matitsend. Find theeigenfrequencies and de- scribe thenormal modes. Intheproblem ofthethree coupled pendula, consider thethree coupling con- stants asdistinct, sothat thepotential energy may bewritten as 1 U: + + _28126162 _28136165 _23236265) with 612,613,823alldifferent. Show that nodegeneracy occurs insuch asystem. Show also that degeneracy canoccur Only if812=315=823. Construct thepossible eigenvectors forthedegenerate modes inthecase ofthe three coupled pendula byrequiring an=2a21. Interpret thissituation physically. Three oscillators ofequal mass marecoupled such that thepotential energy of thesystem isgiven by 1 U= 5lK1(xi +xi.)+K2942 +K5(x1x2 +x2xs)] where K3=\/2K1K2. Find theeigenfrequencies bysolving thesecular equation. What isthephysical interpretation ofthezero-frequency mode? Consider athin homogeneous plate ofmass Mthatliesinthexl-x2 plane with its center attheorigin. Letthelength oftheplate be2A(inthex2-direction) and let thewidth be2B(inthex1-direction). Theplate issuspended from aﬁxed support byfour springs ofequal force constant Katthefour corners oftheplate. Theplate is free tooscillate butwith theconstraint that itscenter must remain onthex3-axis. Thus, wehave three degrees offreedom: (1)Vertical motion, with thecenter ofthe plate moving along thex5-axis; (2)atipping motion lengthwise, with thex1-axis serving asanaxisofrotation (choose anangle 9todescribe thismotion); and(3)a tipping motion sidewise, with thex2-axis sewing asanaxisofrotation (choose an angle c/>todescribe thismotion). Assume only small oscillations andshow thatthe secular equation hasadouble root, and hence that thesystem isdegenerate. Discuss thenormal modes ofthesystem. (Inevaluating thea]~,,forthedegenerate PROBLEMS 511 12-23 12-24. 12-25 12-26 12-27 12-28.modes, arbitrarily setoneoftheallequal tozero toremove theindeterminacy.) Show thatthedegeneracy canberemoved byadding totheplate athinbarofmass mandlength 2Asituated (atequilibrium) along thex2-axis. Find theneweigenfre- quencies forthesystem. Evaluate thetotal energy associated with anormal mode, andshow thatitiscon- stant intime. Show thisexplicitly forthecase ofExample 12.3. Show thattheequations ofmotion forlongitudinal vibrations ofaloaded string are ofexactly thesame form astheequations fortransverse motion (Equation 12.131), except that thefactor 1'/d must bereplaced byK,theforce constant of thestring. Rework theproblem inExample 12.7, assuming that allthree particles aredis- placed adistance aand released from rest. Consider three identical pendula instead ofthetwoshown inFigure 12-5with a spring ofconstant 0.20 N/m between thecenter pendulum and each oftheside ones. The mass bobs are250g,andthependula lengths are47cm.Find thenor- malfrequencies. Consider thecase ofadouble pendulum shown inFigure 12-E where thetoppen- dulum haslength Llandthebottom length isL2,andsimilarly, thebobmasses are mlandm2.The motion isonly intheplane. Find anddescribe thenormal modes andcoordinates. Assume small osillations. - 5°F ml _gb____ .5L2 FIGURE 12-E Problem 12-27. Find thenormal modes forthecoupled pendulums inFigure 12-5 when thepen- dulum onthelefthasmass bob ml=300gand theright hasmass bob mg=500g. Thelength ofboth pendula is40cm,andthespring constant is0.020 N/m.When theleftpendulum isinitially pulled back toBl=—7° and released from restwhen 02=02=0,what isthemaximum angle that62reaches? Usethesmall angle ap- proximation. CHAPTER _l_pl Continuous Systems; Waves 13.1 Introduction Wehave sofarbeen considering particles, systems ofparticles, orrigid bodies. Now, wewant toconsider bodies (gases, liquids, orsolids) thatarenotrigid, that is,bodies whose particles move (however slightly) with respect toone another. The general study ofsuch bodies isquite complex. However, oneaspect ofthese continuous bodies isvery important throughout physics—the ability totransmit wave motion. Adisturbance ononepart ofthebody canbetransmitted bywave propagation throughout thebody. The simplest example ofsuch phenomena isavibrating string stretched under uniform tension between twoﬁxed supports. Asusual, thesimple example repre- sents many oftheimportant results needed tounderstand other physical examples, such asstretched membranes andwaves insolids. Waves maybe either transverse or longitudinal. Anexample ofalongitudinal wave isthevibration ofmolecules along thedirection ofpropagation ofawave moving inasolid rod. Longitudinal waves occur inﬂuids andsolids andareofgreat importance inacoustics. \/Vhereas both transverse and longitudinal waves may occur insolids, only longitudinal waves occur inside ﬂuids, inwhich shearing forces arenotpossible. Wehave already considered (Chapter 12)both kinds ofvibrations forasystem of particles. Adetailed study ofthetransverse vibrating string isimportant forsev- eral reasons. Astudy ofaone-dimensional model ofsuch string vibrations allows amathematical solution with results that areapplicable tomore complex two- und three-dimensional problems. The modes ofoscillation aresimilar. lnpartic- ular, theapplication ofboundary conditions (ﬁxed ends), which areofextreme importance inmany areas ofphysics, iseasiest inone-dime.nsional problems. 512 13.2 CONTINUOUS STRING ASALIMITING CASE orTHELOADED STRING 513 Boundary conditions play aroleintheuseofpartial differential equations simi- lartotherole initial conditions play inordinary differential equations using Newtonian orLagrangian techniques. Inthischapter, weextend thediscussion ofthevibrations ofaloaded string presented inChapter 12byexamining theconsequences ofallowing thenum- berofparticles onthestring tobecome inﬁnite (while maintaining aconstant linear mass density). Inthisway, wepass tothecase ofacontinuous string. All theresults ofinterest forsuch astring canbeobtained bythislimiting process- including thederivation oftheimportant wave equation, oneofthetruly funda- mental equations ofmathematical physics. The solutions ofthewave equation areingeneral subject tolimitations im- posed bycertain physical restrictions peculiar toagiven problem. These limita- tions frequently take theform ofconditions onthesolution thatmust bemetat theextremes oftheintervals ofspace andtime that areofphysical interest. We must therefore deal with aboundary-value problem involving apartial differen- tialequation. Indeed, such adescription characterizes essentially thewhole of what wecallmathematical physics. Weconﬁne ourselves here tothesolution toaone-dimensional wave equa- tion. Such waves candescribe atwo-dimensional wave intwodimensions andcan describe, forexample, themotion ofavibrating string. The compression (or sound) waves thatmay betransmitted through anelastic medium, such asagas, can also beapproximated asone-dimensional waves ifthemedium islarge enough that theedge effects areunimportant. Insuch acase, thecondition of themedium isapproximately thesame atevery point onaplane, andtheprop- erties ofthewave motion arethen functions only ofthedistance along aline normal totheplane. Such awave inanextended medium, called aplane wave, is mathematically identical totheone-dimensional waves treated here. 13.2 Continuous String asaLimiting Case oftheLoaded String Inthepreceding chapter, weconsidered asetofequally spaced point masses suspended byastring. Wenowwish toallow thenumber ofmasses tobecome in- ﬁnite sothat wehave acontinuous string. Todothis, wemust require that as n—>oowesimultaneously letthemass ofeach particle andthedistance between each particle approach zero (m—>0,d—>0)insuch amanner thattheratio m/d remains constant. Wenote that m/dEpisjust thelinear mass density ofthe string. Thus, wehave n—>oo, d—>0, suchthat(n +1)d=L (13.1) m—> 0, d—>O,such thatg =p=constant 514 13/CONTINUOUS SYSTEMS; WAVES From Equation 12.154, wehave to=211.0)sin(j (13-2) Wecannow write T7T jd x'—— = me“ = — 13.3]n+1 T77-(n+1)d "TL ( ) where jd=xnow speciﬁes thedistance along thecontinuous string. Thus, qj-(t) becomes acontinuous function ofthevariables xandt: q(x,1:)=211(1) 8114?‘) (13.4) Of T71’X q(x,1)=23,613‘ S114?) (13.3) Inthecase ofaloaded string containing nparticles, there arendegrees of freedom ofmotion andtherefore nnormal modes and ncharacteristic frequen- cies. Thus, inEquation 12.154 (orEquation 13.2) thesum isover therange r=1 tor=n.Butnow thenumber ofparticles isinﬁnite, sothere isaninﬁnite setof normal modes andthesum inEquations 13.4 and13.5 runs from r=1tor=00. There are,then, inﬁnitely many constants (the realandimaginary parts oftheB,) that must beevaluated tocompletely specify themotion ofthecontinuous string. This isexactly thesituation encountered inrepresenting some function asaFourier series—the inﬁnitely many constants arespeciﬁed bycertain inte- grals involving theoriginal function (seeEquations 3.91). Wemayview thesitua- tioninanother way: There areinﬁnitely many arbitrary constants inthesolution oftheequation ofmotion, butthere arealsoinﬁnitely many initial conditions available fortheir evaluation, namely, thecontinuous functions q(x, 0)and rj(x,0).The realandimaginary parts ofthe,8,canthus beobtained interms of theinitial conditions byaprocedure analogous tothat used inSection 12.9. Using B,=11,.+i1/,,wehave from Equation 13.5, q(x,0)=21¢,s1n(%;x) (13.63) q(x,0)=-EW»,314%‘) (l3.6b) Next, wemultiply each ofthese equations bysin(s1rx/ L)andintegrate from x=0 tox=L.Wecanmake useofthetrigonometric relation L Lsin(1Jc)sin(%c) dx=gs, (13.7) 13.2 CONTINUOUS STRING ASALIMITING GASE OFTHELOADED STRING 515 from which weobtain 2L . ll?=ZLq(x,0)s1n(%c)dx (13.8a) 2L. .1»,=-ﬁloqor, 0)s1n(?)dx (13.31)) The characteristic frequency w,may alsobeobtained asthelimiting value of theresult fortheloaded string. From Equation 12.152, wehave Io,=2,/it sin[ ] (13.9) _2I-mi co, d\/; s1n( 2L) (13.10) When d—>O,wecanapproximate thesine term byitsargument, with theresult T7T T =—— - 13.11 w.L,/; <> .EXAMPLEl3.l ' C2ITTi T Find thedisplacement q(x,t)fora“plucked string,” where onepoint ofthe string isdisplaced (such thatthestring assumes atriangular shape) andthen released from rest. Consider thecase shown inFigure 13-1, inwhich thecenter ofthestring isdisplaced adistance h.which canbewritten as Solution. The initial conditions are glléx, 0SxSL/2 q(x,0) =2h (13.12) f(L—x), L/2SxSL a(x,0) =0 r—L/2—+ /1 __________ _ti____________ L FIGURE 13-1 Example 13.1. Astring is“plucked” bypulling thecenter ofthestring adistance hfrom equilibrium sothatthestring hasatriangular shape. Thestring isreleased from restinthisposition. 516 13/CONTINUOUS SYSTEMS; WAVES Because thestring isreleased from rest, allthe1»,vanish. The ii,aregiven by 4hL/2 _r'n'x 4hL _Mrx 11.,=FL xS1n('T) dx+FL/2(L —x)s1n(T) dx Integrating, _8h _T77‘ '3-W81“?sothat 0, reven = 8h FLT ﬁ(—1)i('_1), TOClCl r'11" Therefore, 8h 1 3q(x,t)=§[sin(%) coscolt—gsin(—%c) coswgt+ (13.13) where thew,areproportional torand aregiven byEquation 13.11. From Equation 13.13, weseethatthefundamental mode (with frequency wl) andalltheoddharmonics (with frequencies (03,(05,etc.) areexcited butthat none oftheevenharmonics areinvolved inthemotion. Because theinitial dis- placement wassymmetrical, thesubsequent motion must alsobesymmetrical, sonone oftheeven modes (forwhich thecenter position ofthestring isa- node) areexcited. Ingeneral, ifthestring isplucked atsome arbitrary point, none oftheharmonics with nodes atthatpoint willbeexcited. Asweprove inthenext section, theenergy ineach oftheexcited modes isproportional tothesquare ofthecoefﬁcient ofthecorresponding term in Equation 13.13. Thus, theenergy ratios forthefundamental, third harmonic, ﬁfth harmonic, andsoonare12%:F13: .Therefore theenergy inthesystem (ortheintensity oftheemitted sounci) isdominated bythefundamental. The third harmonic is19dB*down from thefundamental andtheﬁfth harmonic is down by28dB. 13.3 Energy ofaVibrating String Because wehave made theassumption thatfrictional forces arenotpresent, the total energy ofavibrating string must remain constant. Wenow show thisexplic- itly;moreover, weshow that theenergy ofthestring isexpressed simply asthe *The decibel (dB) isaunit ofrelative sound intensity (oracoustic power). The intensity ratio ofa sound with intensity Itoasound with intensity Illisgiven by10log(I/Ill) dB.Thus, forthefunda- mental (lll)andthird harmonic (I),wehave 10log(1/81) =—19.1 dBor“19dBdown” inintensity. Aratio of3dBcorresponds approximately toafactor oftwoinrelative intensity. 13.3ENERGY orAVIBRATING STRING 517 sum ofcontributions from each ofthenormal modes. According toEquation 13.4, thedisplacement ofthestring isgiven by q(x,t)=217,(t)sin(I—Eic) (13.14) where thenormal coordinates are 17.0) =Brew" (13-15) Asalways, theB,arecomplex quantities and thephysically meaningful normal coordinates areobtained bytaking therealpartofEquation 13.15. The kinetic energy ofthestring isobtained bycalculating thekinetic energy foranelement ofthestring, %(pdx) (12,and then integrating over thelength. Thus, 1L61152T=— —d 13.16 2pi)la»: x ( ) or,using Equation 13.14, L 2 T=éplo 1'),sin('T%)] dx (13.17) Thesquare oftheseries canbeexpressed asadouble sum, thistechnique ensur- ingthatallcross terms areproperly included: . 1 L T=5p17,1110 sin sin dx (13.18) The integral isnow thesame asthatinEquation 13.7, so r,rLT=%Z11.11.§.. L=pZE»7§ (13.19) Intheevaluation ofthekinetic energy, wemust becareful totake theproduct ofrealquantities. Wemust therefore compute thesquare oftherealpart ofT1,: d 2 (Re1=)T)2=(Re-gt[(11, +i11,)(cos wlt+isinw,t)]) =(—w,;,t, sinro,t—w,v,cosw,t)2 The kinetic energy ofthestring istherefore LT=PIEw$(,1,s1n 0),;+1»,cosw,t)2 (13.20) The potential energy ofthestring canbecalculated easily bywriting down theexpression fortheloaded string and then passing tothelimit ofacontinuous 518 13/CONTINUOUS SYSTEMS; WAVES string. (Recall thatweconsider thepotential energy tobeonly theelastic energy inthestring.) Fortheloaded string, _1T 2 v-2d§<q,--1 11,-) Multiplying anddividing byd, 1 q;‘—1"qj2 U=— E?—— d27-J'( d ) Inpassing tothelimit, d—>O,theterm inparentheses becomes just thepartial derivative ofq(x,t)with respect tox,and thesum (including thefactor d)be- comes anintegral: 1L61’U=—'rj (2) dx (13.21)206x Using Equation 13.14, wehave 6q T7T r'n'x —=E— — 13. ax TLn,cos L (22) sothat _ L 2 U=%¢L %n,¢os('—1”‘)j dx (13.23) Again, thesquared term can bewritten asadouble sum, and because the trigonometric relation (Equation 13.7) applies forcosines aswell assines, we have U-Z2-Tllsl Fcos it cos£72 d2-.sLL"’"‘ 0 L Lx 'r r'n's'n' L=_2___ ._527,8 LL7lr7ls2 T5 'r r2172 L=_E ._22TL,211. L=%Ewin? (13.24) where Equation 13.11hasbeen used inthelastlinetoexpress theresult interms of<33.Evaluating thesquare oftherealpart of17,,wehave, ﬁnally, LU=%2w§(/.1, COS(3,;-1»,sinw,t)2 (13.25) 13.3 ENERGY OFAVIBRATING STRING 519 The total energy isnow obtained byadding Equations 13.20 and 13.25, in which thecross terms cancel andthesquared terms addtounity: E=T+U =%2wne+o (mas 01' E=%EwZ|e,|2 (132611) The total energy istherefore constant intime and, furthermore, isgiven bya sum ofcontributions from each ofthenormal modes. The kinetic and potential energies each vary with time, soitissometimes useful tocalculate thetime-averaged kinetic andpotential energies—that is,the averages over onecomplete period ofthefundamental vibration r=1: L (T)=%;wE((/1., sinco,t+1/,cosw,t)2) (13.27) where theslanted brackets denote anaverage over thetime interval 211'/wl. The averages ofsin2wlt orcos2wltover thisinterval areequal to Similarly, theav- erages ofsin2to,t and cos2w,t forr22arealso %,because theperiod ofthe fundamental vibration isalways some integer times theperiod ofahigher har- monic vibration. The averages ofthecross terms, cosw,tsinw,t, allvanish. Therefore, <r>=%2<»%(1e+ V2) L =%EwZle,|2 (13.28) Forthetime-averaged potential energy, wehave asimilar result: PL .(U)=I§Tlw§((/1, cosw,t—1/,sinw,t)2) L =$2-we +2?) L =%§MmP (mm Wetherefore have theimportant result that theaverage kinetic energy ofavibrating string isequal totheaverage potential energy.* (T)=(U) (13.30) *This result alsofollows from thevirial theorem. 520 13/CONTINUOUS SYSTEMS; WAVES Notice alsothesimpliﬁcation thatresults from theuseofnormal coordinates: Both (T)and(U)aresimple sums ofcontributions from each ofthenormal modes. 13.4 Wave Equation Our procedure thus farhasbeen todescribe themotion ofacontinuous string asthelimiting case oftheloaded string forwhich wehave acomplete solution; wehave notyetwritten down thefundamental equation ofmotion forthecon- tinuous case. Wemay accomplish thisbyreturning totheloaded string and again using thelimit technique—but now ontheequation ofmotion rather than onthesolution. Equation 12.131 canbeexpressed as m_,_1'q,1'—1 _qj_Zqj_qj+1 dqi_d( dDd( dD (msl) Asdapproaches zero, wehave qﬁ'11-+1 >gtx)—q(x+d)>_6q d d axx+d/2 which isthederivative atx+d/2. Fortheother term inEquation 13.31, wehave ‘11-1_‘I1)'19‘_d)-q(x) )5‘! Vd 7 d —69¢ x_d/2 which isthederivative atx—d/2. The limiting value oftheright-hand side of Equation 13.31 istherefore 6q 6q _ axx+1:/2 axx—(1/2 62q 62¢]11111 '1' {E33 = T—'* = TL 4->0 d 6x2x 6x2 Also inthelimit, m/dbecomes p,sotheequation ofmotion is "—agq 1332 Pq —Taxg ( ' ) 62¢!P32‘! This isthewave equation inonedimension. InSection 13.6, weshall discuss the solutions tothisequation. Wenowwant toshow thatEquation 13.33 canalsobeeasily obtained bycon- sidering theforces onacontinuous string. Only transverse waves areconsidered. Aportion ofthestring ﬁxed atboth ends, asdiscussed sofarinthischapter, is shown inFigure 13-2.OI‘ 12.4WAVE EQUATION 521 61 Z+dq___t___________________ _l___,4/KA x-————————---:1.>< _.----I-__-A.~e x+dx FIGURE 13-2 Aportion, length ds,ofastring ﬁxed atboth ends isshown. The displacement from equilibrium isqontheleftandq+dqonthe right. The tensions 1'ateach endareequal inmagnitude, butnot indirection. Only transverse waves areconsidered. Weassume that thestring hasaconstant mass density p(mass/ length). We consider alength dsofthestring described bys(x,t).The tensions 1-oneach end ofthestring areequal inmagnitude butnotindirection. This imbalance leads toaforce andthus anacceleration ofthesystem. Weassume that thedis- placement q(perpendicular tox)issmall. The mass dmofthelength ofstring dsispds.The horizontal components ofthetension areapproximately equal andopposite, soweneglect themovement ofthestring inthex-direction. The force intheq~direction is 62¢] AF= pdsa? (13.34) where AFrepresents thedifference intension atxandx+dx.Weusepartial de- rivatives todescribe theacceleration, 62q/6t2, because wearenotconsidering the x-dependence ofthedisplacement q(x,t). The force canbefound from thedifference inthey-components ofthe tension. (AF)y= —rsin61 +'rSin92 =—'rtan01+'rtan02 6 8 =-1"—q +1' x ax x+dx =rﬁdx (13.35) where weletsin6=tan6because theangles 6aresmall forsmall displace» ments. Wenow setEquations 13.34 and13.35 equal, letting ds~"*~'dx: 62q 62q Tgcgdx =pdxﬁ aiqP52‘!—=-~ 13.366x2 1'6t2 ( ) 522 13/CONTINUOUS SYSTEMS; WAVES Equation 13.36 isidentical toEquation 13.33 butdoes notprovide theuseful in- formation obtained earlier from thenormal coordinate method. 13.5 Forced andDamped Motion Wecaneasily determine Lagrange’s equations ofmotion forthevibrating string byusing thekinetic energy from Equation 13.19 andthepotential energy from Equation 13.24: L= T~ U Pb.Pb=;§w~;§ww b ="Z201? —wﬁni) <13-31> where thelength ofthestring hasbeen setequal tobtoavoid confusion between theLs.The ease ofthenonnal coordinate description isapparent. The equa- tions ofmotion follow from Equation 13.37: a§,+1.1317,: 0 (13.33) Next, weaddaforce perunitlength F(x,t)acting along thestring. Wealsoadd adamping force proportional tothevelocity. Thewave equation (Equation 13.33) now becomes ‘f3+D‘-31-¢ﬁ=F(x1) (1339)p611’ at 6x2 ’ ' where each term represents aforce perunit length, andDisthedamping (resis- tive) term. Equation 13.39 issolved using normal coordinates. Aswedidin Section 13.2, weuseasolution q(x,1)=Z1741) sin (13.40) Substitution ofEquation 13.40 into Equation 13.39 gives Lagrange’s equations ofmotion—similar toEquation 13.38 butwith thedamping andforced terms added: O0 2 2 §1Up~;§, +01),+———’Z;Tn,)s1n(1;'5)] =F(x,1) (13.41) The sum over risagain from 1to00because weareconsidering acontinuous string. The solution ofEquation 13.41 parallels that ofSection 13.2 (which we donotrepeat here indetail) bycomparing realandimaginary components. We multiply each side ofEquation 13.41 bysin(s'n'x/ b)and integrate over dxfrom O tob(remember that b=L=length ofstring). Using Equation 13.7, wehave‘i °° 22 b 11 §1(p'ij,+ D1‘),+———’Z;T11,)55,,=lord,1)s1n(5?)dx (13.42)‘I 13.5 FORCED ANDDAMPED MOTION 523 which becomes D s2'n'2'r 2I’ s1rx"-'—- =— ‘—d 1. 17,+pr), +pbg 07, pbLF(x, t)s1n( 1)) x (343) Wenow letf,(t) betheFourier coefﬁcient oftheFourier expansion ofF(x,t), which isontheright side ofEquation 13.43: 1- f,(¢)=LF(x,1)s1n(%3‘) dx (13.44) Innormal coordinate terms, Equation 13.43 simply becomes D s21r2'r 2“,+—',+i— ,=—,(t 13.45 np11 Pb,11pbf) ( ) Itisnow apparent thatf,(t) isthecomponent ofF(x,t)effective indriving the normal coordinate s. Reconsider Example 13.1. Asinusoidal driving force ofangular frequency w drives thestring atx=b/2.Find thedisplacement. Solution. The driving force perunit length is F(x,t)=F0coswt, x=b/2 ....1} M»The driving Fourier coefﬁcient becomes f,(1)=F0coswtsing5 (13.47) Notice thatf,(t) =0foreven values ofs.Only theodd terms aredriven. Ifweinclude asmall damping term, Equation 13.45 becomes D s21r21' 2 s11‘ﬁs+F1’),+W 17,=HJFO coswtsin5- (13.48) With thedamping term effective, weneed notdetermine acomplementary so~ lution, which willbedamped out.Weneed only ﬁnd aparticular (steady-state) solution, aswasdone inSection 3.6.Equation 13.48 may becompared with Equation 3.53, where D_=2p B s2'n'2'rW =(1)3 (13.49) 2Fsin(s'rr/2)0Z____i_=A pb 524 13/CONTINUOUS SYSTEMS; WAVES The solution (seeEquation 3.60) forn(t)becomes "s(t) :2F0sin(s1r/2) cos(wt —5) (18.50) 22 D2 Pb\/(“z'TTJ —(02)+F-(02 where Dw 6=tan_1 7;;;_“‘“— (13.51) .(_.».1)pb andthedisplacement ofq(x,t)is 2F0sin%cos(wt —5)sin(Z%§) q(x,1)=Z A (13.52)1 2 pb\/(%l —-(1)2)+%-£02 P P where wehave neglected thepart ofthesolution thatisdamped out. Equation 13.52 represents many ofthefeatures discussed previously. Depending onthe driving frequency, only afewofthenormal coordinates may dominate because oftheresonance effects inherent inthedenominator. Ifthedamping term is negligible, thedominant normal coordinate terms are r2=#2 (13.53)1rr andbecause ofthesin(r'n'/ 2)term ofEquation 13.52, only oddvalues ofrare effective. 13.6 General Solutions oftheWave Equation Theone-dimensional wave equation foravibrating string (seeEquation 13.33) is* 6211' p6‘"I/F — =0 (13.54) where pisthelinear mass density ofthestring, 'risthetension, and ‘I’iscalled thewave function. The dimensions ofpare[ML'1] andthedimensions ofrare those ofaforce, namely, [MLT -2]. The dimensions ofp/"r aretherefore [T2L'2] that is,thedimensions ofthereciprocal ofasquared velocity. Ifwe write \/'r/p=v,thewave equation becomes 62‘? 16211/ *Weusethenotation ‘I’=1I’(x, t)todenote atime-dependentwave function and1/1=1/1(x) todenote a ﬁmindepmdent wave function. 13.6 GENERAL SOLUTIONS OFTHEWAVE EQUATION 525 One ofourtasks istogiveaphysical interpretation ofthevelocity "0;itisnotsuf- ﬁcient tosaythat visthe“velocity ofpropagation” ofthewave. Toshow that Equation 13.55 does indeed represent ageneral wave motion, weintroduce twonewvariables, §E x+ vtT’Ex_vt (13.56) Evaluating thederivatives of‘I’=‘I/(x, t),which appear inEquation 13.55, we have 61F 611/6f 611/61‘) 611/ 61?=—*— =~ ~ 13.76x 6f6x+ 6176x 6§+617 (5) Then, 6211/: ear/_ 6611/+611») 6x2 6x6x 6x6f 81) =a(aw+611/)6; +6(aw+611/>617 6f6§ 61)6x 61761f 61)6x 6211/ 6211/ 6211/_662+2666”+an, (13.53) Similarly, weﬁnd 1611’ 611/ 6‘? Z3? —55 —E (13.59) and 1922-12 1211-122114202612 v61v61 v616§ 61) 6211/ 621.? 6211/=-—— —— 4 13.6:2264611+W (6°) Butaccording toEquation 13.55, theright-hand sides ofEquations 13.58 and 13.60 must beequal. This canbetrue only if 6211/i E0 (13.61) 66<11! The most general expression for‘I’that cansatisfy thisequation isasum oftwo terms, oneofwhich depends only onEandtheother only on17;nomore com- plicated function of§and1]permits Equation 13.61 tobevalid. Thus, 11/=f(§)+g(1)) (l3.62a) or,substituting for§and17, ‘F=f(x +vt)+g(x—vt) (l3.62b) 526 13/CONTINUOUS SYSTEMS; WAVES where fandgarearbitrary functions ofthevariables x+vtand x—vt,respec- tively, which arenotnecessarily ofaperiodic nature, although they may be. Astime increases, thevalue ofxmust also increase inorder tomaintain a constant value forx—vt.The function gtherefore retains itsoriginal form as time increases ifweshift ourviewpoint along thex-direction (inapositive sense) with aspeed v.Thus, thefunction gmust represent adisturbance thatmoves to theright (i.e., tolarger values ofx)with aspeed v,whereas frepresents theprop- agation ofadisturbance totheleft.Wetherefore conclude that Equation 13.55 does indeed describe wave motion and, ingeneral, atraveling (orpropagating) wave. Letusnow attempt tointerpret Equation 13.62b interms ofthemotion ofa stretched string. Attime t=0,thedisplacement ofthestring isdescribed by q(x,0) =f(x)+g(x) Ifwetakeidentical triangular forms forf(x)andg(x), theshape ofthestring att= Oisasshown atthetopofFigure 13-3. Astime increases, thedisturbance repre- sented byf(x+vt)propagates totheleft,whereas thedisturbance represented byg(x—vt)propagates totheright. This propagation oftheindividual distur- bances totheleftandright isillustrated inthelower part ofFigure 13-3. Consider next theleft-going disturbance alone. Ifweterminate thestring (atx=O)byattaching ittoarigid support, weﬁnd thephenomenon ofreﬂec- tion. Because thesupport isrigid, wemust have f(vt) E0forallvalues oftime. This condition cannot bemet bythefunction falone (unless ittrivially van.- ishes). Wecansatisfy thecondition atx=Oifweconsider, inaddition tof(x+vt), (a) /\L /1 \\ i I \z \/ \ (b) <— /xv,’ \ —-> (c) _:"/\/K; (d) _"/\__/KFIGURE 13-3 The propagation ofastring isshown asafunction oftime from (a)to (d).Attime t=0thestring isdescribed asf(x) +g(x) asshown in(a). Astime progresses, thedisturbance f(x+vi)propagates totheleft andg(x—vt)propagates totheright. 13.7SEPARATION orTHEwAvE EQUATION 527 (H) ._______\\ ,__.__ \\\//i (b) l ___________ _.\\\\ ll ‘4>J (<1) i //\\\ (<1) ‘ G /\\ 1 / \\ ___.’ \\ _______ FIGURE 13-4 Consider only theleftmoving disturbance ofFigure 13-3. Theendof thestring isﬁxed, andthewave reﬂects, because f(vt)=0always at theend. Wecanvisualize themotion asifanimaginary disturbance (dashed line) wasmoving from thelefttotheright astime proceeds from (a)to(d). animaginary disturbance, —-f(—x +vt),which approaches theboundary point from theleft, asinFigure 13-4. The disturbance f(x+vt)continues topropa- gate totheleft,even into theimaginary section ofthestring (x<0),while the disturbance —f(—-x +vt)propagates across theboundary and along thereal string. The neteffect isthat theoriginal disturbance isreﬂected atthesupport andthereafter propagates totheright. Ifthestring isterminated byrigid supports atx=0andalso atx=L,the disturbance propagates periodically back andforth with aperiod 2L/v. 13.7 Separation oftheWave Equation Ifwerequire ageneral solution ofthewave equation thatisharmonic (asforthe Vibrating string Or,forthat matter, foralarge number ofproblems ofphysical in- terest), wecanwrite \.P'(x, t)=I/1(x) e“"‘ (13.63) sothattheone-dimensional wave equation (Equation 13.55) becomes 62¢ (02 — —— =0 13.646x2+v22] ( ) where t/1isnow afunction ofxonly. 528 13/CONTINUOUS SYSTEMS; wAvEs The general wave motion ofasystem isnotrestricted toasingle frequency w.Forasystem with ndegrees offreedom, there arenpossible characteristic fre- quencies, andforacontinuous string there isaninﬁnite setoffrequencies.* If wedesignate therthfrequency by0),,thewave function corresponding tothis frequency is !F,(x, t)=¢,(x) em" (13.65) The complete wave function isasuperposition (recall thatwearedealing with a linear system) ofalltheparticular wave functions (ormodes). Thus ‘I'(x,1)=§rhI',(11, 1)=t/1,(x)e"“"‘ (13.66) InEquation 13.63, weassumed that thewave function wasperiodic intime. Butnow weseethatthisassumption entails norealrestriction atall(apart from theusual assumptions regarding thecontinuity ofthefunctions andtheconver- gence oftheseries), because thesummation inEquation 13.66 actually gives a Fourier representation ofthewave function andistherefore themost general expression forthetrue wave function? Wenow wish toshow thatEquation 13.65 results naturally from apowerful method that canoften beused toobtain solutions topartial differential equa- tions—the method ofseparation ofvariables. First, weexpress thesolution as ‘I’(x.l)E1/10¢)'X(¢) (13-67) thatis,weassume thatthevariables areseparable andtherefore thatthecomplete wave function canbeexpressed astheproduct oftwofunctions, oneofwhich is aspatial function only, and one ofwhich isatemporal function only. Itisnot guaranteed thatwewillalways ﬁnd such functions, butmany ofthepartial differ- ential equations encountered inphysical problems areseparable inatleast one coordinate system; some (such asthose involving theLaplacian operator) are separable inmany coordinate systems. Inshort, thejustiﬁcation ofthemethod ofseparation ofvariables, asisthecasewith many assumptions inphysics, isinits success inproducing mathematically acceptable solutions toaproblem that eventually arefound toproperly describe thephysical situation, i.e.,are“experi- mentally veriﬁable.” Substituting ‘I’=I/1Xinto Equation 13.55, wehave if‘/i_ifl_0Xdx2 "02dt2 mi? *Aninﬁnite setoffrequencies would exist foratruly continuous string, butbecause arealstring is composed fundamentally ofatoms, there does exist anupper limit forw(seeSection 13.8). TEuler proved in1748 thatthewave equation foracontinuous string issatisﬁed byanarbitrary func- tion ofxivt,and Daniel Bernoulli showed in1753 that themotion ofastring isasuperposition of itscharacteristic frequencies. These tworesults, taken together, indicated thatanarbitrary ﬁinction could bedescribed byasuperposition oftrigonometric functions. This Euler could notbelieve, and sohe(aswellasLagrange) rejected Bernoullis superposition principle. TheFrench mathematician Alexis Claude Clairaut (1713-1765) gave aproof inanobscure paper in1754 that theresults of Euler andBernoulli were actually consistent, butitwasnotuntil Fourier gave hisfamous proof in 1807 thatthequestion wassettled. 13.7 SEPARATION OFTHEWAVE EQUATION 529 OT 21121/1 111211%E =-iF (13.63) But, inView ofthedeﬁnitions of¢(x) and x(t), theleft-hand side ofEquation 13.68 isafunction ofxalone, whereas theright-hand side isafunction oft alone. This situation ispossible only ifeach part oftheequation isequal tothe same constant. Tobeconsistent with ourprevious notation, wechoose this con- stant tobe—w2. Thus, wehave 222+‘"21/1—0 (1369)dx2 v2 'a and %+w2,\/ =0 (13.69b) These equations areofafamiliar form, andweknow thatthesolutions are ¢(x) =Ae"(“’/">" +Be“'(“’/"l" (13.70a) ,\/(t) =Ce“"‘ +De“"" (l3.70b) where theconstants A,B,C,Daredetermined bytheboundary conditions. We maywrite thesolution ‘I/(x, t)inashorthand manner as ‘I/(x, t)=¢(x),\/(t) ~exp[i'i(w/-0) x]exp [iiwt] ' ~exp[i-z'(w/'0) (x1'vt)] (13.71) This notation means that thewave function ‘I’varies asalinear combination ofthe terms exp[z'(w/'0) (x+vt)] exp[z'(w/'0) (x—vt)] exp[— i(w/v) (x+vt)] exp[—i(w/v)(x —1/t)] The separation constant forEquation 13.68 waschosen tobe—w2. There is nothing inthemathematics oftheproblem toindicate that there isaunique value ofw;hence, there must exist aset*ofequally acceptable frequencies w,.To each such frequency, there corresponds awave function: ‘F,(x, t)~exp[i“i(w,/v) (x1*'ut)] The general solution istherefore notonly alinear combination oftheharmonic terms butalsoasum over allpossible frequencies: ‘I’(x, t)~ET:aﬂl’, ~Zia, exp[i‘i(w,,/v) (x1'vt)] (13.72) *Atthisstage ofthedevelopment, thesetisinfactinﬁnite, because nofrequencies have yetbeen eliminated byboundary conditions. 530 13/CONTINUOUS SYSTEMS; WAVES The general solution ofthewave equation therefore leads toavery compli- cated wave function. There are, infact, aninﬁnite number ofarbitrary constants a,.This isageneral result forpartial differential equations; butthisinﬁnity of constants must satisfy thephysical requirements oftheproblem (the boundary conditions), and therefore they canbeevaluated inthesame manner that the coefﬁcients ofaninﬁnite Fourier expansion canbeevaluated. Formuch ofourdiscussion, itissufﬁcient toconsider only oneofthefour possible combinations expressed byEquation 13.71; that is,weselect awave propagating inaparticular direction andwith aparticular phase. Then, wecan write, forexample, ‘1’1(x, I)'”°XP[—i(w./v) (X—1/5)] This istherthFourier component ofthewave function, and thegeneral solu- tion isasummation over allsuch components. The functional form ofeach component is,however, thesame, andsothey canbediscussed separately. Thus, weshall usually write, forsimplicity, ‘I’(x, t)~exp[—z'(w/v)(x —vt)] (13.73) The general solution must beobtained byasummation over allfrequencies that areallowed bytheparticular physical situation. Itiscustomary towrite thedifferential equation for1//(x) as d2 7::+1.21/1=0 (13.74) which isthetime-dependent form oftheone-dimensional wave equation, also called theHehnholtz equation,* andwhere 2 112E9’; (13.73)U The quantity k,called thepropagation constant orthewave number (i.e., pro- portional tothenumber ofwavelengths perunit length), hasdimensions [L'1]. The wavelength /\isthedistance required forone complete vibration ofthe wave, ’\=_1_J=2'7T'U V (1) andthus therelationi between kand)1is 1=2l)1 *Hermann von Helmholtz (1821-1894) used this form ofthewave equation inhistreatment of acoustic waves in1859. tMore properly thewave number should bedeﬁned ask=1/Arather than 271/)1,because 1/Aisthe number ofwavelengths perunitdistance. However, k=27r//\ ismore commonly used intheoretical physics, andwefollow that usage here. 13.7 SEPARATION OFTHEwAvE EQUATION 531 Wecantherefore write, ingeneral, (P-Ax, t),_,e:11..(.¢61) or,forthesimpliﬁed wave function, !P(x, t)~e_”‘(""' "2=e"(“"_ 2”) (13.76) Ifwesuperimpose twotraveling waves ofthetype given byEquation 13.76 andifthese waves areofequal magnitude (amplitude) butmoving inopposite directions, then ‘P=‘IQ.+!P_=Ae_”‘("+"‘) +Ae_”‘("_"‘) (13.77) or III=A11-"'==1(@('~' +e_2“") =2Ae"""‘ coswt therealpart ofwhich is El’=2Acoskxcoswt (13.78) Such awave nolonger hastheproperty that itpropagates; thewave form does notmove forward with time. There are,infact, certain positions atwhich there isnomotion. These positions, thenodes, result from thecomplete cancellation ofone wave bytheother. The nodes ofthewave function given byEquation 13.78 occur atx=(2n+1)7T/2k, where nisaninteger. Because there areﬁxed positions inwaves ofthistype, they arecalled standing waves. Solutions tothe problem ofthevibrating string areofthisform (but with aphase factor attached totheterm kxsuch that thecosine istransformed into asinefunction satisfying theboundary conditions). EXAMPLE 13.. Consider astring consisting oftwodensities, p1inregion 1where x<0andp2 inregion 2where x>0.Acontinuous wave train isincident from theleft(i.e., from negative values ofx).I/Vhat aretheratios ofthesquare oftheamplitude magnitudes forthereﬂected andtransmitted waves totheincident wave?Q3 Solution. Thewave willbeboth reﬂected andtransmitted atx=0where the mass density discontinuity occurs. Therefore, inregion 1wehave thesuperposi- tionoftheincident andreﬂected waves, andinregion 2wehave only thetrans- mitted wave. Iftheincident wave isAe"(‘““ 21”‘),then wehave forthewaves ‘P1(x,t)and ‘F2(x,t)inregions 1and 2,respectively (see Equation 13.77) 11'/1(x, t)=q‘/inc +qfreﬁ =Aei(wt—k1x) +Bei(mt+k1x) . 1. W261)==c@1<~»*-12> i ‘372’ 532 I3/CONTINUOUS SYSTEMS; WAVES InEquation 13.79, wehave explicitly taken into account thefactthatthe waves inboth regions have thesame frequency. Butbecause thewave velocity onastring isgiven by -u=-\/? P wehave v16*1/2,andtherefore kl6*k2.Wealsohave 11=-"3=w\/E (13.30)v 7' so,interms ofthewave number oftheincident wave, P11,=11,,/F2 (13.31) The amplitude Aoftheincident wave (seeEquation 13.79) isgiven andis real. Wemust then obtain theamplitudes Band Cofthereﬂected andtransmit- tedwaves tocomplete thesolution oftheproblem. There areasyetnorestric- tions onBand C,andthey may becomplex quantities. The physical requirements ontheproblem may bestated interms of theboundary conditions. These are, simply, that thetotal wave function EV=‘F1+W2anditsderivative must becontinuous across theboundary. The continuity ofEVresults from thefactthatthestring iscontinuous. The condi- tion onthederivative prevents theoccurrence ofa“kink” inthestring, forif 8!?/6x01 6*6!?/6x0_, then 6211’/6x2 isinﬁnite atx=0;butthewave equation re- lates 62'?‘/6x2 and62!?/6t2; andiftheformer isinﬁnite, thisimplies aninﬁnite acceleration, which isnotallowed bythephysical situation. Wehave, therefore, forallvalues ofthetime t, '1'1|,.=0 ='p2lx=0 (l3.82a) i‘1_’_1 _L5 13.8 b ax x=0 ax x=0 ( 2) From Equations 13.79 and13.82a, wehave A+B=C (l3.83a) andfrom Equations 13.79 and13.82b weobtain —k1A +k1B=—k2C (l3.83b) The solution ofthispair ofequations yields B=BEA (l3.84a)kl+k2 and c—in (13346)1,+11, ' 13.3PHAsE VELOCITY, DISPERSION, ANDATTENUATION 533 Thewave numbers klandk2areboth real, soamplitudes Band Carelikewise real. Furthermore, kl,k2,and Aareallpositive, soCisalways positive. Thus, the transmitted wave isalways inphase with theincident wave. Similarly, ifkl>kg, then theincident andreﬂected waves areinphase, butthey areoutofphase for k2>kl,that is,forp2>pl. The reﬂection coefﬁcient Risdeﬁned astheratio ofthesquared magni- tudes oftheamplitudes ofthereﬂected andincident waves: |B|2 kl_k22RE——= i- 13.35|A|2 k1+ k2 ( ) Becausethe energy content ofawave isproportional tothesquare oftheampli- tude ofthewave function, Rrepresents theratio ofthereﬂected energy tothe incident energy. The quantity |B|2represents theintensity ofthereﬂected wave. Noenergy canbestored inthejunction ofthetwostrings, sotheincident energy must beequal tothesum ofthereﬂected andtransmitted energies; that is,R+T=1.Thus, T—1R—4k‘k2 (1336)(k1+kz)2 ' OI‘ kC2 T=_.| |21.|A|Inthestudy ofthereﬂection andtransmission ofelectromagnetic waves, we ﬁnd quite similar expressions forRand T.(13.87) 13.8 Phase Velocity, Dispersion, andAttenuation Wehave seen inEquations 13.71 that thegeneral solution tothewave equation produces, even intheone-dimensional case, acomplicated system ofexponen- tialfactors. Forthepurposes offurther discussion, werestrict ourattention to theparticular combination 1 This equation describes thepropagation totheright (larger x)ofawave possess- ingawell-deﬁned angular frequency (U.Certain physical situations can bequite adequately approximated byawave function ofthis type—for example, the propagation ofamonochromatic light wave inspace orthepropagation ofasi- nusoidal wave onalong (strictly, inﬁnitely long) string. Iftheargument oftheexponential inEquation 13.88 remains constant, then thewave function ‘I'(x, t)alsoremains constant. The argument oftheex- ponential iscalled thephase _¢ofthewave, 41E61-1111 (13.39) 534 13/CONTINUOUS SYSTEMS; wAvEs Ifwemove ourviewpoint along thex-axis atavelocity such that thephase at every point isthesame, wealways seeastationary wave ofthesame shape. The velocity Vwith which wemust move, called thephase velocity ofthewave, corre- sponds tothevelocity with which thewave form propagates. Toensure d)=con- stant, weset dd)=0 (13.90) or todt=kdx from which V—22—-(2— (13916111U ') sothatthephase velocity inthiscase isjust thequantity originally introduced as thevelocity. Itispossible tospeak ofaphase velocity only when thewave func- tion hasthesame form throughout itslength. This condition isnecessary sowe can measure thewavelength bytaking thedistance between anytwo successive wave crests (orbetween anytwosuccessive corresponding points onthewave). If thewave form were tochange asafunction oftime orofdistance along the wave, these measurements would notalways yield thesame results. The wave- length isnotafunction oftime orspace (i.e., thatwispure) only ifthewave train isofinﬁnite length. Ifthewave train isofﬁnite length, there must beaspectrum offrequencies present inthewave, each with itsown phase velocity. Wewilloften assign asingle frequency andphase velocity toawave ofﬁnite length asacon- venient approximation. Letusreturn totheexample oftheloaded string andexamine theproper- tiesofthephase velocity inthat case. Wehave previously found (Equation 12.152) that thefrequency fortherthmode oftheloaded string when termi- nated atboth ends isgiven by l7'_ r7T0),—2ElS1I1|iF(n +1)] (13.92) where thenotation isthesame asinChapter 12.Recall thatwetake only positive values forthefrequencies. I/Vhen r=1,there isanode ateach end, andnone between; hence, thelength ofthestring isone-half ofawavelength. Similarly, when r=2,then L=Aand, ingeneral, )1,=2L/r. Therefore, r71’ r71'd r7Td 7rd k,d ’2(1-1+ 1)'2a(11+ 1)_21'1,'2 (1292) .1=2 lsinkii (1394 .,/d 2 -> m Because thisexpression nolonger contains norL,itapplies equally well toater- minated orinﬁnite loaded string.and 13.3 PHAsE VELOCITY, I)IsPERsION. ANDATTENUATION 535 Tostudy thepropagation ofawave intheloaded string, weinitiate adistur- bance byforcing one oftheparticles, say,thezeroth one, tomove according to q0(t) =Ae“"‘ (13.95) Ifthe string contains many particles,* then any angular frequency less than 2V7"/md isanallowed frequency (actually aneigenfrequency), satisfying Equation 13.95. After thetransient effects have subsided and thesteady-state conditions areattained, thephase velocity ofthewave isgiven byl’ _9 »1_d|sin(kd/2)| _v-k-\/;—-Gd/2 -V(k) (13.96) Thus thephase velocity isafunction ofthewave number; thatis,Visfrequency- dependent. When V=V(k) foragiven medium, that medium issaid tobedis- persive, andthewave exhibits dispersion. The best-known example ofthisphe- nomenon isthesimple optical prism. The index ofrefraction oftheprism depends onthewavelength oftheincident light (i.e., theprism isadispersive medium foroptical light); onpassing through theprism, thelight isseparated into aspectrum ofwavelengths (i.e., thelight wave isdispersed). Foralongitudinal wave propagating down along, slender rod, most ofthe energy isassociated with thedirection ofthelongitudinal wave propagation. There is,however, asmall amount ofenergy dissipated inatransverse wave mov- ingatright angles. This lateral disturbance causes thephase velocity ofthelon- gitudinal wave tobedecreased, andtheeffect depends onwavelength. Forlarge wavelengths, theeffect issmall; forshort wavelengths, especially those approach- ingtheradius oftherod, thevelocity dispersion ispronounced. From Equation 13.96, weseethat, asthewavelength becomes very long (A-9ooork—>0),thephase velocity approaches theconstant value V(/\—~>oo) =\/Tag (13.97) Otherwise, V=V(k), andthewave isdispersive. Wenote that thephase velocity forthecontinuous string (seeEquation 13.55) is =1»=(E (13.98) and because m/d fortheloaded string corresponds topforthecontinuous string, thephase velocities forthetwocases areequal inthelong-wavelength limit (but onlyinthislimit). This isareasonable result because asAbecomes *Strictly, weneed aninﬁnite number ofparticles forthistype ofanalysis, butwemay approach the ideal conditions asclosely asdesired byincreasing theﬁnite number ofparticles. TinEquation 13.92 thevalues ofrare required tobesn(seeEquation 12.144), soweautomatically have w.Z0because sin[r7r/2(n +1)]20for05rsn.Wenolonger have such arestriction onkd, sosin(kd/ 2)canbecome negative. Wecontinue toconsider only positive frequencies byalways tak- ingonly themagnitude ofsin(kd/2). This result was obtained byBaden-Powell in1841, but William Thomson (Lord Kelvin) (1824-1907) realized thefullsigniﬁcance only in1881. 536 13/CONTINUOUS SYSTEMS; WAVES large compared with d,theproperties ofthewave arelesssensitive tothespacing between particles, andinthelimit, dmay vanish without affecting thephase ve- locity. InEquation 13.94, therestriction onris15r5n.Then, because k,='r11'/L, weseethatthevalue ofk,thatmaximizes w,inEquation 13.94 is km=rr/d (13.99) The corresponding frequency, from Equation 13.96, is2\/'1'/md. VVhat isthere» sultofforcing thestring tovibrate atafrequency greater than 2V1'/md? Forthis purpose, weallow ktobecome complex andinvestigate theconsequences: kEK—iB, K,B>0 (13.100) The expression forw(Equation 13.94) then becomes w=2, sinligk —z'B)] _2/LSing@'B_d_QS,n#*_d) md 2COS 2 COS 2 2 d d=2,/i1(s1n“§‘l¢osh-I-35 —icosgisinhgi) (13.101) Ifthefrequency istobeareal quantity, theimaginary part ofthis expression must vanish. Thus, wemayhave either cos(Kd/ 2)=0orsinh(Bd/ 2)=0.Butthe latter choice requires B=0,contrary totherequirement that kbecomplex. We therefore have cos? =0 (l3.l02) Forthiscase, wemust alsohave sin? =1 (13.103) The expression fortheangular frequency becomes _ I'1' Bdw——2mdcosh 2 (l3.104) Thus, wehave theresult that, forws2\/"r/md, thewave number kisrealand the relation between wand kisgiven byEquation 13.94; whereas, for w>2\/"r/md, kiscomplex with therealpart Kﬁxed byEquation 13.102 atthe value K=11'/d andwith theimaginary part Bgiven byEquation 13.104. The situ- ation isshown inFigure 13-5. What isthephysical signiﬁcance ofacomplex wave number? Our original wave function wasoftheform q‘/':Aei(wt— kx) 13.8PHASE VELOCITY} DISPERSION, ANDATTENUATION 537 7;‘ K_.________ __,...r d B”,- Q—7i AD\\\ —\\\ |-kt: — an FIGURE 13-5 Inthecase oftheloaded string thewave number kisrealforangular frequencies w52V'1'/md. Forw>2V'1'/md, thewave number kis complex (K—iB)with therealpartdenoted byK(displayed asﬁxed at 1r/d) andtheimaginary part denoted byB(dashed line). but,ifk=K—iﬁ,then 1?canbewritten as ‘I’=Ae"3"e‘(‘"“"") (13.105) andthefactor exp(—Bx) represents adamping, orattenuation, ofthewave with increasing distance x.Wetherefore conclude that thewave ispropagated with- outattenuation forwS2V1'/md (this region iscalled thepassing band offre- quencies), and that attenuation setsinatwt=2V1'/md (called thecritical or cut0ﬂfrequemy*) andincreases with increasing frequency. 1 The physical signiﬁcance oftherealandimaginary parts ofkisnow appar- ent:Bistheattenuation coefﬁcientl (and exists only ifw>wc),whereas Kisthe wave number inthesense thatthephase velocity V’isgiven by I_E2=L V K Rek (13.106) rather than byV=w/k. Ifkisreal, these expressions forVand V’areidentical. This example emphasizes thefact that thefundamental deﬁnition ofthe phase velocity isbased ontherequirement oftheconstancy ofthephase andnot ontheratio w/k. Thus, ingeneral, thephase velocity Vand theso-called wave ve- locity varedistinct quantities. Wenote alsothatiftoisrealandifthewave num- berkiscomplex, then thewave velocity vmust alsobecomplex sothattheprod» uctkvyields arealquantity forthefrequency through therelation to=kv.On theother hand, thephase velocity, which arises from therequirement that 4)= constant, isnecessarily always arealquantity. Inthepreceding discussion, weconsidered thesystem tobeconservative andargued thatthisrequires wtobearealquantity? Wefound thatifwexceeds *The occurrence ofacutoff frequency wasdiscovered byLord Kelvin in1881. TThe reason forwriting k=K—i/3rather than k=K+iBinEquation 13.100 isnow clear; ifﬁ>0 forthelatter choice, then theamplitude ofthewave increases without limit rather than decreasing toward zero. ISee thediscussion inSection 12.4intheparagraph following Equation 12.39. 538 13/CONTINUOUS SYSTEMS; WAVES thecritical frequency wc,attenuation results and thewave number becomes complex. Ifwerelax thecondition thatthesystem isconservative, thefrequency may then becomplex and thewave number real. Insuch acase, thewave is damped intimerather than inspace (seeProblem 13-13). Spatial attenuation (w real, kcomplex) isofparticular signiﬁcance fortraveling waves, whereas tempo- ralattenuation (tucomplex, kreal) isimportant forstanding waves. Although attenuation occurs intheloaded string ifw>wc,thesystem isstill conservative and noenergy islost. This seemingly anomalous situation results because theforce applied totheparticle intheattempt toinitiate atraveling wave is(after thesteady-state condition ofanattenuated wave issetup)exactly 90°outofphase with thevelocity oftheparticle, sothatthepower transferred, P= F-v,iszero. Inthistreatment oftheloaded string, wehave tacitly assumed anideal situa- tion; that is,thesystem wasassumed tobelossless. Asaresult, Wefound that there wasattenuation forw>w,butnone forw<wt.However, every realsys- temissubject toloss, soinfactthere issome attenuation even forw<wt. 13.9 Group Velocity andWave Packets Itwasdemonstrated inSection 3.9thatthesuperposition ofvarious solutions of alinear differential equation isstill asolution totheequation. Indeed, we formed thegeneral solution totheproblem ofsmall oscillations (see Equation 12.43) bysumming alltheparticular solutions. Letusassume, therefore, thatwe have twoalmost equal solutions tothewave equation represented bythewave functions ‘I/1and ‘P2,each ofwhich hasthesame amplitude, 11; J=Ai(wt-kx)T12:ta_AZ,"(.()t—Kx)} (13407) 2i_ butwhose frequencies andwave numbers differ byonly small amounts: Q=w+AwK:HM} (13.10s) Forming thesolution thatconsists ofthesum ‘I’1and ‘P2,wehave 11/(at) =qr,+11»,=A[€Xp(iwt)(-":Xp(—ikx) +exp{i(w +Aw)t} exp{—i(k +Ak)x}] IAiwlitw%>¢}w»{~¢(k+%)x}l~i@XP{~i(e@l;~M)} +<e~—;e>}1 12.9GROUP VELOCITY ANDWAVE PACKETS 539 /T _ \ Ir _ T \ ’ _ ‘\\ / / \ I I \ , \ \ / \ I / \ \ / \ \\ / \ ’ \ \/ \ / \\ , \ I’ \ r \ / \ \ \ / \ / \ /I /\ / \\ 1 , \ /\1 \ \ , \ \ fz ~_— \_’ ~, FIGURE 13-6 When twowave functions having frequencies very close together are summed, thephenomenon ofbeats (slowly varying amplitude) is observed. The second bracket isjust twice thecosine oftheargument oftheexponential, and therealpart oftheﬁrst bracket isalso acosine. Thus, therealpart ofthe wave function is ‘I’(x, t)=2Acos|: :|cos|:(w +%)t —(k+ (13.109) This expression issimilar tothatobtained intheproblem oftheweakly coupled oscillators (seeSection 12.3), inwhich wefound aslowly varying amplitude, cor- responding totheterm 2AC0S|:(Aw)t; (At)? which modulates thewave function. The primary oscillation takes place atafre- quency w+(Aw/2), which, according toourassumption thatAwissmall, differs negligibly from w.The varying amplitude gives risetobeats (Figure 13-6). The velocity U(called thegroup velocity*) with which themodulations (or groups ofwaves) propagate isgiven bytherequirement that thephase ofthe amplitude term beconstant. Thus, dx Aw=—-=— 1. U dt Ah (3110) Inanondispersive medium Aw/Ak =I/Isothegroup and phase velocities are identica1.* Ifdispersion ispresent, however, Uand Vare distinct. Thus far,wehave considered only thesuperposition oftwowaves. Ifwewish tosuperpose asystem ofnwaves, wemust write !P'(x,t) =§n1A,¢xp[t(w,¢ —k,x)] (l3.1l1a) *The concept ofgroup velocity isduetoHamilton, 1839; thedistinction between phase andgroup velocity wasmade clear byLord Rayleigh (Theory ofSound, 1stedition, 1877; seeR2194). TThis identity isshown explicitly inEquation 13.117. 540 13/CONTINUOUS SYSTEMS; WAVES where A,represents theamplitudes oftheindividual waves. Intheevent that n becomes very large (strictly, inﬁnite), thefrequencies arecontinuously distrib- uted, andwemay replace thesummation byanintegration, obtaining* +00 11/(at) =IA(k)e‘(‘°“*")dk (13.111b) where thefactor A(k) represents thedistribution amplitudes ofthecomponent waves with different frequencies, that is,thespectral distribution ofthewaves. The most interesting cases occur when A(k) hasasigniﬁcant value only inthe neighborhood ofaparticular wave number (say, k0)and becomes vanishingly small forkoutside asmall range, denoted byk0iAk.Insuch acase, thewave function canbewritten as k0+Ak !F(x, t)=I A(k)e"(°"‘*")dk (l3.l12) ko—Ak Afunction ofthistype iscalled awave packetf The concept ofgroup velocity canbeapplied only tothose cases thatcanberepresented byawave packet, that is,towave functions containing asmall range (orband) offrequencies. Forthecase ofthewave packet represented byEquation 13.112, thecon- tributing frequencies arerestricted tothose lying near w(k0). Wecantherefore expand w(k) about k=k0: w(k) =w(k0) + '(k—k0)+ (13.113a) k=k0 which wecanabbreviate as cu=wo+wf,(k —k0)+ (l3.ll3b) The argument oftheexponential inthewave packet integral becomes, approxi- mately, wt—kx=(wot —kox) +w{,(k —k0)t—(k—k0)x where wehave added andsubtracted theterm kox.Thus, wt—kx=(wot —kox) +(k—k0)(w6t —x) (13.114) andEquation 13.112 becomes 0+A 'F(x, t)=F kA(k)exp[i(k —k0)(w6t —x)]exp[i(w0t —k0x)]dk (l3.115) k0—Ak *We have previously made thetacit assumption thatk20.However, kisdeﬁned byk2=m2/v2 (see Equation 13.75), sothere isnomathematical reason whywemaynotalsohave k<0.Wemaythere- fore extend theregion ofintegration toinclude —w<k<0without mathematical difﬁculty. This procedure allows theidentiﬁcation oftheintegral representation of‘I'(x, t)asaFourier integral. TThe term wave packet isduetoErwin Schrodinger. 13.9 GROUP VELOCITYAND WAVE PACKETS 541 The wave packet, expressed inthisfashion, may beinterpreted asfollows. The quantity A<k>exp1i<k —k.,><<»:.¢—x>1 constitutes aneffective amplitude that, because ofthesmall quantity (k—k0)in theexponential, varies slowly with time and describes themotion ofthewave packet (orenvelope ofagroup ofwaves) inthesame manner thattheterm 2ACOS[(Aw)¢; (Akpcjl describes thepropagation ofthepacket formed from twosuperposed waves. The requirement ofconstant phase fortheamplitude term leads to do)='=— 13.116U°’°(dk).=., (’ forthegroup velocity. Asstated earlier, only ifthemedium isdispersive does U differ from thephase velocity VToshow thisexplicitly, wewrite Equation 13.116 as _1__QU dw0 where thesubscript zero means “evaluated atk=k0or,equivalently, atw=coo.” Because k=(v/v, _1__1(9) Iv0-(wdv/dw)0 U dw v0 -0% Thus, "0U= i" (13.117) I-A-<e>U0 dw 0 Ifthemedium isnondispersive, "0=V=constant, sodv/dw =0(seeEquation 13.91); hence U= 110=V. The remaining quantity inEquation 13.115, exp[i(w0t —-k0x)], varies rap- idlywith time; andifthiswere theonly factor inEV,itwould describe aninﬁnite wave train oscillating atfrequency (00andtraveling with phase velocity V=mo/k0. Weshould note that aninﬁnite train ofwaves ofagiven frequency cannot transmit asignal orcarry information from onepoint toanother. Such transmis- sion can beaccomplished only bystarting and stopping the wave train and thereby impressing asignal onthewave—in other words, byforming awave 542 13/CONTINUOUS SYSTEMS; WAVES packet. Asaconsequence ofthisfact, itisthegroup velocity, notthephase ve- locity, thatcorresponds tothevelocity atwhich asignal may betransmitted.* PROBLEMS I T7 TW 13-1. Discuss themotion ofacontinuous string when theinitial conditions are q'1(x,0) =0,q(x,0) =Asin(3¢rx/L). Resolve thesolution intonormal modes. 13-2. Rework theproblem inExample 13.1 intheevent that theplucked point isadis- tance L/3from oneend. Comment onthenature oftheallowed modes. 13-3. Refer toExample 13.1. Show byanumerical calculation that theinitial displace- ment ofthestring iswell represented bytheﬁrst three terms oftheseries in Equation 13.13. Sketch theshape ofthestring atintervals oftime of-£5ofaperiod. 13-4. Discuss the motion ofastring when the initial conditions are q(x,0) = 4x(L —x)/L2, rj(x,0) =0.Find thecharacteristic frequencies and calculate the amplitude ofthenthmode. 13-5. Astring with noinitial displacement issetinto motion bybeing struck over a length 2sabout itscenter. This center section isgiven aninitial velocity vo. Describe thesubsequent motion. 13-6. Astring issetinto motion bybeing stmck atapoint L/4from oneendbyatrian- gular hammer. Theinitial velocity isgreatest atx=L/4anddecreases linearly to zero atx=0and X=L/2. The region L/2 SxSLisinitially undisturbed. Determine thesubsequent motion ofthestring. Why arethefourth, eighth, and related harmonics absent? How many decibels down from thefundamental are thesecond andthird harmonics? 13-7. Astring ispulled aside adistance hatapoint 3L/7from oneend. Atapoint 3L/7 from theother end, thestring ispulled aside adistance hintheopposite direc- tion. Discuss thevibrations interms ofnormal modes. 13-8. Compare, byplotting agraph, thecharacteristic frequencies w,asafunction ofthe mode number rfor aloaded string consisting of3,5,and 10particles andfora continuous string with thesame values of1'andm/d=p.Comment ontheresults. *The group velocity corresponds tothesignal velocity only innondispersive media (inwhich case thephase, group, andsignal velocities areallequal) andinmedia ofnormal dispersion (inwhich case thephase velocity exceeds thegroup andsignal velocities). Inmedia with anomalous disper- sion, thegroup velocity mayexceed thesignal velocity (and, infact, mayeven become negative orin- ﬁnite). Weneed only note here that amedium inwhich thewave number kiscomplex exhibits at- tenuation, andthedispersion issaidtobeanomalous. Ifkisreal, there isnoattenuation, andthe dispersion isnormal What iscalled anomalous dispersion (due toahistorical misconception) is,in fact, normal (i.e., frequent), andso-called normal dispersion isanomalous (i.e., rare). Dispersive ef- fects arequite important inoptical andelectromagnetic phenomena. Detailed analyses oftheinterrelationship among phase, group, andsignal velocities were made byArnold Sommerfeld andbyLéon Brillouin in1914. Translations ofthese papers aregiven inthe book byBrillouin (Br60) . PROBLEMS 543 13-9. 13-10. 13-11 13-12. 13-13. 13-14. 13-15 13-16. 13-17.InExample 13.2, thecomplementary solution (transient part) wasomitted. If transient effects areincluded, what aretheappropriate conditions forover- damped, critically damped, and underdamped motion? Find thedisplacement q(x,t)that results when underdamped motion isincluded inExample 13.2 (as- sume that themotion isunderdamped forallnormal modes). Consider thestring ofExample 13.1. Show thatifthestring isdriven atanarbi- trary point, none ofthenormal modes with nodes atthedriving point willbe excited. I/Vhen aparticular driving force isapplied toastring, itisobserved thatthestring vibration ispurely ofthenthharmonic. Find thedriving force. Determine thecomplementary solution forExample 13.2. Consider thesimpliﬁed wave function 1I’(x, t)=Ae‘(°"""") Assume that0)andvarecomplex quantities andthatkisreal: to=a+iB v=u+iw Show thatthewave isdamped intime. Usethefactthat k2=w2/v2 toobtain ex- pressions foraandBinterms ofuandw.Find thephase velocity forthiscase. Consider anelectrical transmission line that hasauniform inductance perunit length Landauniform capacitance perunit length C.Show that analternating current Iinsuch alineobeys thewave equation 621 621——-LC—— =06x2 6t2 sothatthewave velocity isv=l/\/LC. Consider thesuperposition oftwoinﬁnitely long wave trains with almost thesame frequencies butwith different amplitudes. Show that thephenomenon ofbeats occurs butthatthewaves never beat tozero amplitude. Consider awave g(x—vt)propagating inthe+x-direction with velocity v.Arigid wallisplaced atx=xo.Describe themotion ofthewave forx<xo. Treat theproblem ofwave propagation along astring loaded with particles oftwo different masses, m’andm”,which alternate inplacement; thatis, m’, forjevenm. Z J m", for odd Show thatthew—kcurve hastwobranches inthiscase, andshow thatthere isat- tenuation forfrequencies between thebranches aswell asforfrequencies above theupper branch. 544 13-18. 13-19 13-20. 13-21. 13-22.13/CONTINUOUS SYSTEMS; WAVES Sketch thephase velocity V(k) andthegroup velocity U(k) forthepropagation of waves along aloaded string intherange ofwave numbers 0SkS11'/d. Show that U(1r/d) =0,whereas V(1r/d) does notvanish. What istheinterpretation ofthisre- sultinterms ofthebehavior ofthewaves? Consider aninﬁnitely long continuous string with linear mass density plforx<0 andforx>L,butdensity p2>plfor0<x<L.Ifawave train oscillating with an angular frequency toisincident from theleftonthehigh-density section ofthe string, ﬁnd thereﬂected andtransmitted intensities forthevarious portions ofthe string. Find avalue ofLthat allows amaximum transmission through thehigh- density section. Discuss brieﬂy therelationship ofthisproblem totheapplication ofnonreﬂective coatings tooptical lenses. Consider aninﬁnitely long continuous string with tension 1'.Amass Misattached tothestring atx=0.Ifawave train with velocity 00/kisincident from theleft, show that reﬂection andtransmission occur atx=0andthat thecoefﬁcients R and Tare given by R= sin2O,T= cos20 where n 2kr Consider carefully theboundary condition onthederivatives ofthewave functions atx=O.1/Vhat arethephase changes forthereﬂected andtransmitted waves? Consider awave packet inwhich theamplitude distribution isgiven by 1,It—kol<AhAk=()l0, otherwise Show that thewave function is 'W(x, t)= ei(wot—k0x) (Dot '_X Sketch theshape ofthewave packet (choose t=0forsimplicity). Consider awave packet with aGaussian amplitude distribution A(k) =Bexp[—~cr(k —ko)2] where 2/\/disequal tothe1/eWidth* ofthepacket. Using thisfunction forA(k), show that 1I’(x,0) =BI+°°exp[—(r(k —k0)2]exp(—ikx)dk =B\/2 eXp(—x2/40')exp(—ik0x) *Atthepoints k=koi1/\/;, theamplitude distribution is1/eofitsmaximum value A(kO). Thus 2/\/; isthewidth ofthecurve atthe1/eheight. PROBLEMS 545 Sketch theshape ofthiswave packet. Next, expand w(k) inaTaylor series, retain theﬁrsttwoterms, andintegrate thewave packet equation toobtain thegeneral result ‘I'(x, t)=BE exp[—(w Qt—x)2/4tr]exp[i(w0t —kox)] Finally, take oneadditional term intheTaylor series expression ofw(k) andshow that 0'isnow replaced byacomplex quantity. Find theexpression forthe1/e width ofthepacket asafunction oftime forthiscase andshow that thepacket moves with thesame group velocity asbefore butspreads inwidth asitmoves. Illustrate thisresult with asketch. emit.141:‘Special Theory ofRelativity 14.1 Introduction InSection 2.7,itwaspointed outthat theNewtonian idea ofthecomplete sepa- rability ofspace and time and theconcept oftheabsoluteness oftime break down when they aresubjected tocritical analysis. The ﬁnal overthrow ofthe Newtonian system astheultimate description ofdynamics wastheresult ofsev- eralcrucial experiments, culminating with thework ofMichelson andMorley in 1881—1887. The results ofthese experiments indicated that thespeed oflight is independent ofanyrelative uniform motion between source andobserver. This fact, coupled with theﬁnite speed oflight, required afundamental reorganiza- tion ofthestructure ofdynamics. This wasprovided during theperiod 1904- 1905 byH.Poincaré, H.ALorentz, andA.Einstein,* who formulated thethe- oryofrelativity inorder toprovide aconsistent description oftheexperimental facts. The basis ofrelativity theory iscontained intwopostulates: *Although Albert Einstein (1879-1955) isusually accorded thecredit fortheformulation ofrelativ- itytheory (see, however, Wh53, Chapter 2),thebasic formalism hadbeen discovered byPoincaré and Lorentz by1904. Einstein wasunaware ofsome ofthisprevious work atthetime (1905) ofthepubli- cation ofhisﬁrst paper onrelativity. (Einstein’s friends often remarked that “heread little, but thought much.”) Theimportant contribution ofEinstein tospecial relativity theory wasthereplace- ment ofthemany adhocassumptions made byLorentz andothers with buttwobasic postulates from which alltheresults could bederived. [The question ofprecedence inrelativity theory isdiscussed byG.Holton, Am.].Phys. 28,627(1960); seealsoArn63.] Inaddition, Einstein later provided the fundamental contribution totheformulation ofthegeneral theory ofrelativity in1916. Hisﬁrstpub- lication onatopic ofimportance ingeneral relativity—-speculations ontheinﬂuence ofgravity on light—was in1907. Itisinteresting tonote that Einstein’s 1921 Nobel Prize wasawarded, notforcon- tributions torelativity theory, butforhiswork onthephotoelectric effect. 546 14.2GALILEAN INVARIANCE 547 I. Thelawsofphysical phenomena arethesame inallinertial reference frames (that is, only therelative motion ofinertial frames canbemeasured; theconcept ofmotion rela- tiveto“absolute rest” ismeaningless). II. Thevelocity oflight (infreespace )isauniversal constant, independent ofanyrela- tivemotion ofthesource andtheobserver Using these postulates asafoundation, Einstein wasable toconstruct a beautiful, logically precise theory. Awide variety ofphenomena that take place athigh velocity andcannot beinterpreted intheNewtonian scheme areaccu- rately described byrelativity theory. Postulate I,which Einstein called theprinciple ofrelativity, isthefundamental basis forthetheory ofrelativity. Postulate II,thelawofpropagation oflight, follows from Postulate Iifweaccept, asEinstein did,thatMaxwell’s equations arefunda- mental lawsofphysics. Maxwell’s equations predict thespeed oflight invacuum to bec,andEinstein believed thistobethecaseinallinertial reference frames. Wedonotattempt here togivetheexperimental background forthetheory of relativity; such information canbefound inessentially every textbook onmodern physics andinmany others concerned with electrodynamics.* Rather, wesimply ac- cept ascorrect theabove twopostulates andwork outsome oftheir consequences forthearea ofmechanics.l Thediscussion here islimited tothecaseofspecial rela- tivity, inwhich weconsider only inertial reference frames, thatis,frames thatarein uniform motion with respect tooneanother. Themore general treatment ofaccel- erated reference frames isthesubject ofthegeneral theory ofrelativity. 14.2 Galilean Invariance InNewtonian mechanics, theconcepts ofspace andtime arecompletely separa- ble;furthermore, time isassumed tobeanabsolute quantity susceptible ofpre- cisedeﬁnition independent ofthereference frame. These assumptions lead to theinvariance ofthelawsofmechanics under coordinate transformations ofthe following type. Consider twoinertial reference frames Kand K’,which move along their x1-and xi-axes with auniform relative velocity v(Figure 14-1). The transformation ofthecoordinates ofapoint from one system totheother is clearly oftheform xi=x1—vt xé=x2 (14.la) xé=xi Also, wehave t’=1: (14.1b) *Aparticularly good discussion oftheexperimental necessity forrelativity theory canbefound in Panofsky andPhillips (Pa62, Chapter 15). 1Relativistic effects inelectrodynamics arediscussed inHeald andMarion (He95, Chapter 14). 548 14/SPECIAL THEORY OFRELATIVITY I *2 *2K! K U O; i xi 0- —--W1 xi xs FIGURE 14-1 Twoinertial reference frames KandK’move along their x1—and xi-axes with auniform relative velocity v. Equations 14.1 deﬁne aGalilean transformation. Furthermore, theelement of length inthetwosystems isthesame andisgiven by es?=Edit}1 =Eaxf =are (14.2) The factthat Newton’s laws areinvariant with respect toGalilean transforma- tions istermed theprinciple ofNewtonian relativity orGalilean invariance. Newton’s equations ofmotion inthetwosystems are 1'}="551" =mat;=F; (14.3) The form ofthelawofmotion isthen invariant toaGalilean transformation. The individual terms arenotinvariant, however, butthey transform according to thesame scheme andaresaidtobecovariant. Wecaneasily show that theGalilean transformation isinconsistent with Postulate II.Consider alight pulse emanating from aﬂashbulb positioned in frame K’.The velocity transformation isfound from Equation 14.la, where we consider thelight pulse only along x1: aif=ail—v (14.4) Insystem K’,thevelocity ismeasured asici=c;Equation 14.4 therefore indi- cates thespeed ofthelight pulse tobeicl=c+v,clearly inviolation of Postulate II. 14.3 Lorentz Transformation The principle ofGalilean invariance predicts thatthevelocity oflight isdifferent intwoinertial reference frames thatareinrelative motion. This result isincon- tradiction tothesecond postulate ofrelativity. Therefore, anew transformation law that renders physical laws relativistically covariant must befound. Such a transformation lawistheLorentz transformation. The original use ofthe 14.3LORENTZ TRANSFORMATION 549 Lorentz transformation preceded thedevelopment ofEinsteinian relativity the- ory,* butitalsofollows from thebasic postulates ofrelativity; wederive itonthis basis inthefollowing discussion. Ifalight pulse from aﬂashbulb isemitted from thecommon origin ofthe systems KandK’(seeFigure 14-1) when they arecoincident, then according to Postulate II,thewavefronts observed inthetwosystems must bedescribedl by ‘lM-_ —c2t2=0 5 (14.5) Ex‘? —c2t'2 =0 1-11 Wecanalready seethatEquations 14.5, which areconsistent with thetwopostu- lates ofthetheory ofrelativity, cannot bereconciled with theGalilean transfor- mations ofEquations 14.1. The Galilean transformation allows aspherical light wavefront inonesystem butrequires thecenter ofthespherical Wavefront inthe second system tomove atvelocity vwith respect totheﬁrstsystem. The interpre- tation ofEquations 14.5, according toPostulate II,isthateach observer believes that hisspherical wavefront hasitscenter ﬁxed athisown coordinate origin as thewavefront expands. Wearefaced with aquandary. Wemust abandon either thetworelativity postulates ortheGalilean transformation. Much experimental evidence, includ- ingtheMichelson-Morley experiment and theaberration ofstarlight, requires thetwopostulates. However, thebelief intheGalilean transformation isen- trenched inourminds byoureveryday experience. The Galilean transformation hadproduced satisfactory results, including those ofthepreceding chapters of thisbook, forcenturies. Einstein’s great contribution wastorealize that the Galilean transformation wasapproximately correct, butthatweneeded toreexam- ineourconcepts ofspace andtime. Notice thatwedonotassume t=t’inEquations 14.5. Each system, KandK’, hasitsownclocks, andweassume thataclock maybelocated atanypoint inspace. These clocks areallidentical, runthesame way,andaresynchronized. Because the ﬂashbulb goes offwhen theorigins arecoincident andthesystems move only inthe x1-direction with respect toeach other, bydirect observation wehave xi=x214.6 xiIxi} ( ) Attime t=t’=0,when theﬂashbulb goes off,themotion oftheorigin O’ofK’ ismeasured inKtobe x1—vt=0 (14.7) *The transformation wasoriginally postulated byHendrik Anton Lorentz (1853-1928) in1904 toex- plain certain electromagnetic phenomena, buttheformulas hadbeen setupasearly as1900 by_]._]. Larmor. Thecomplete generality ofthetransformation wasnotrealized until Einstein derived theresult. W.Voigt wasactually theﬁrsttousetheequations inadiscussion ofoscillatory phenomena in1887. 1'See Appendix G. 550 14/SPECLAL THEORY orRELATIVITY andinsystem K’,themotion ofO’is xi=0 (14.8) Attime t=t’=0wehave xi=x1—vt,butweknow that Equation 14.1a isin- correct. Letusassume thenext simplest transformation, namely, xi='y(x1 —vt) (14.9) where 'yissome constant thatmaydepend onvandsome constants, butnotonthe coordinates x1,xi,t,ort’.Equation 14.9 isalinear equation andassures usthat each event inKcorresponds tooneandonly oneevent inK’.This additional as- sumption inourderivation willbevindicated ifwecanproduce atransformation thatisconsistent with alltheexperimental results. Notice that'ymust normally be veryclose to1tobeconsistent with theclassical results discussed inearlier chapters. Wecanusethepreceding arguments todescribe themotion oftheorigin O ofsystem Kinboth Kand K’toalso determine x1=')/(xi +vt’) (14.10) where weonly have tochange therelative velocities ofthetwosystems. Postulate Idemands that thelaws ofphysics bethesame inboth reference systems such that ‘y=')/'.Bysubstituting xifrom Equation 14.9 into Equation 14.10, wecansolve theremaining equation fort’: t'='yt+—x—l(1 —72) (1411)rv ' Postulate IIdemands that thespeed oflight bemeasured tobethesame in both systems. Therefore, inboth systems wehave similar equations fortheposi- tionoftheﬂashbulb light pulse: x‘,at (14.12)x1=ct Algebraic manipulation ofEquations 14.9-14.12 gives (seeProblem 14-1) 1y=———-——-—— (14.13) V1—v2/c2 The complete transformation equations cannow bewritten as x1—vt xl V1—v2/c2 x x?2 (14.14)x5 =X5 Uxlti i I . '5=LV1—v2/c2 These equations areknown astheLorentz (orLorentz-Einstein) transfor- mation inhonor oftheDutch physicist H.A.Lorentz, who ﬁrstshowed that the 14.3 LORENTZ TRANSFORMATION 551 equations arenecessary sothatthelaws ofelectromagnetism have thesame form inallinertial reference frames. Einstein showed that these equations arere- quired forallthelaws ofphysics. The inverse transformation caneasily beobtained byreplacing vby—vand exchanging primed andunprimed quantities inEquations 14.14. xf+vt' x1=i V1—v2/c2 e=%(14.15)e=m ,vxft+C2 t= 1 V1—v2/c2 Asrequired, these equations reduce totheGalilean equations (Equations 14.1) when v—>0(orwhen c—>oo). Inelectrodynamics, theﬁelds propagate with thespeed oflight, soGalilean transformations arenever allowed. Indeed, thefactthattheelectrodynamic ﬁeld equations (Maxwell’s equations) arenotcovariant toGalilean transformations wasamain factor intherealization oftheneed foranew theory. Itseems rather extraordinary that Maxwell’s equations, which areacomplete setofequations fortheelectromagnetic ﬁeld andarecovariant toLorentz transformations, were de- duced from experiment long before theadvent ofrelativity theory. ' The velocities measured ineach ofthesystems aredenoted byu. _dx, M‘ dz dxl7-; u‘—dt’(mm) Using Equations 14.14, wedetermine Idx{ dxl—vdt u1=ir=__W'__Tdt v C 'l.l1_'U 1.;=1——u; (l4.17a) C2 Similarly, wedetermine u§==---9?--- (14.17b)U11) ((1"rlI=_iiU3 ulv Y1'F 552 14/SPECIAL THEORY orRELATIVITY Now wecandetermine whether Postulate IIissatisﬁed directly. Anobserver in system Kmeasures thespeed ofthelight pulse from theﬂashbulb tobeul=cin thex1-direction. From Equation 14.17a, anobserver inK’measures , C-11 C-1) "1=i=@"_— =6 U C U1__ C asrequired byPostulate II,independent oftherelative system speed v. Determine therelativistic length contraction* using theLorentz transformation. Solution. Consider arodoflength llying along thex,-axis ofaninertial frame K.Anobserver insystem K’moving with uniform speed valong thex1-axis (as inFigure 14-1) measures thelength oftherodintheobserver’s own coordinate system bydetermining atagiven instant oftimet’thedifference inthecoordi- nates oftheends oftherod, x{(2) —x{(l). According tothetransformation equations (Equations 14.14), [x1(2) —x1(1)] r1/[t(2) —1(1)]x’ —x’1= - 14.18 where x1(2) -x1(l) =l.Note thattimes t(2)andt(l)arethetimes intheK system atwhich theobservations aremade; they donotcorrespond tothein- stants inK’atwhich theobserver measures therod. Infact, because t’(2) =t’(1), Equations 14.14 give »<2>—41>=141(2)—4(1)]; The length l’asmeasured intheK’system istherefore 1'=xi(2) -xi(1) Equation 14.18 now becomes length contraction l'=l\/1—v2/c2 (14.19) and, toastationary observer inK,objects inK’alsoappear contracted. Thus, to anobserver inmotion relative toanobject, thedimensions ofobjects arecon- tracted bafactor \/1—2inthedirection ofmotion, inwhich BEv/c. Y I3 Aninteresting consequence oftheFitzGerald-Lorentz contraction oflength wasreported in1959 byJames Terrelll Consider acube ofside lmoving with uniform velocity vwith respect toanobserver some distance away. Figure 14—2a *The contraction oflength inthedirection ofmotion wasproposed byG.F.FitzGerald (1851-1901) in1892 asapossible explanation oftheMichelson-Morley ether-drift experiment. This hypothesis wasadopted almost immediately byLorentz, who proceeded toapply itinhistheory ofelectrody- namics. 1*].Terrell, Phys. Rev.116, 1041 (1959). 14.2LORENTZ TRANSFORMATION 553 C! E 1) c:"" D»\ \ \ \ \ \ V\ I M} \ \\\ B, \\ I A B A’ toLN1 -52 (a) (b) Observer FIGURE 14-2 (a)Anobserver faraway seesacube ofsides latrestinsystem K (b)Terrell pointed outthatsurprisingly, thesame cube appears to berotated ifitismoving totheright with velocity vrelative to system K shows theprojection ofthecube ontheplane containing thevelocity vector v and theobserver. The cube moves with itsside ABperpendicular totheob- server’s lineofsight. Wewish todetermine what theobserver “sees”; thatis,ata given instant oftime intheobserver’s rest frame, weWish todetermine therela- tiveorientation ofthecorners A,B,C,andD.The traditional view (which went unquestioned formore than 50years!) wasthattheonly effect isaforeshortening ofthesides ABand CDsuch that theobserver sees adistorted tube ofheight l butoflength lV1 —B2.Terrell pointed outthat thisinterpretation overlooks certain facts: Forlight from corners AandDtoreach theobserver atthesame instant, thelight from D,which must travel adistance lfarther than thatfrom A, must have been emitted when corner Dwas atposition E.Thelength DEisequal to (l/c)v=lB.Therefore, theobserver sees notonly face AB,which isperpendicu- lartothelineofsight, butalsoface AD,which isparallel tothelineofsiht.Also, thelength oftheside ABisforeshortened inthenormal waytolV1—B2.The netresult (Figure 14-2b) corresponds exactly totheview theobserver would have ifthecube were rotated through anangle sin"1B. Therefore, thecube is not distorted; itundergoes anapparent rotation. Similarly, thecustomary state- ment* thatamoving sphere appears asanellipsoid isincorrect; itappears stillas asphere.l Computers canbeused toshow extremely interesting results ofthe typex wehave been discussing (Figure 14-3). *See, forexample,]oos andFreeman (I050, p.242). TAn interesting discussion ofapparent rotations athigh velocity isgiven byV.F.Weisskopf, Phys. Today 13,no.9,24(1960), reprinted inAm63. ISee also aninteresting website attheAustralian National University, §egrlg[ind§x,htg)1, Antony C.Searle (2003) andanarticle byM.C.Chang, F.Lai,andW.C.Chen, ACM Transactions onGraphics 15,No.4,265(1996). FIGURE 14-3 Anarray ofrectangular barsisseen from above atrestintheﬁgure on theleft. Intheright ﬁgure thebars aremoving totheright with v= 0.9c. Thebars appear tocontract androtate. Quoted from P.-K. Hsiung andR.H.P. Dunn, seeScience News 137, 232(1990). EXAMPLE 14.2 UsetheLorentz transformation todetermine thetime dilation effect. Solution. Consider aclock ﬁxed atacertain position (x1)intheKsystem that produces signal indications with theinterval A¢=¢(2) ~z(1) According totheLorentz transformation (Equations 14.14), anobserver inthe moving system K’measures atime interval At’(onthesame clock) of Ar’=t’(2)~z'(1) 1[ML L[K1,L_ c c \/1-v2/c2 Because x1(2) =x1(1) and because theclock isﬁxed intheKsystem, Wehave At,=5(2) "5(1) V1—-v2/02 A:At’ = (14.20) _' U C Thus, toanobserver inmotion relative totheclock, thetime intervals appear tobelengthened. This istheorigin ofthephrase “moving clocks runmore slowly.” Because themeasured time interval onthemoving clock islengthened, theclock actually ticks slower. Notice that theclock isﬁxed intheKsystem, x1(1) =x1(2), butnotintheK’system, xi(1) HEx’1(2). 14.4 EXPERIMENTAL VERIFICATION OFTHE SPECIAL THEORY 555 The argument intheprevious example canbereversed andtheclock ﬁxed intheK’system. The same result occurs; moving clocks run slower. The effect is called time dilation. Itisimportant tonote that thephysical system isunimpor- tant. The same effect occurs foratuning fork, anhourglass, aquartz crystal, and aheartbeat. The problem isoneofsimultaneity. Events simultaneous inonesys- tem may notbesimultaneous inanother one moving with respect totheﬁrst. The same clock may beviewed from ndifferent reference frames andfound to berunning atndifferent rates, simultaneously. Space andtime areintricately in- terwoven. Weshall return tothispoint later. The time measured onaclock ﬁxed inasystem present attwoevents is called theproper time and given thesymbol 1'.Forexample, At=A1"when a clock ﬁxed insystem Kispresent forboth events, x,-(1) andx,-(2). Equation 14.20 becomes At’='yAr (14.21) Notice that theproper time isalways theminimum measurable time difference between twoevents. Moving observers always measure alonger time period. 14.4 Experimental Verification oftheSpecial Theory The special theory ofrelativity explains thedifﬁculties existing before 1900 with optics andelectromagnetism. Forexample, theproblems with stellar aberration and theMichelson-Morley experiment aresolved byassuming noether butre- quiring theLorentz transformation. Butwhat about thenew startling predictions ofthespecial theory—length contraction andtime dilation? These topics areaddressed every dayintheaccel- erator laboratories ofnuclear and particle physics, where particles areacceler- ated tospeeds close tothatoflight, andrelativity must beconsidered. Other ex- periments canbeperformed with natural phenomena. Weexamine twoofthese. Muon Decay When cosmic raysenter theearth’s outer atmosphere, they interact with parti- clesand create cosmic showers. Many oftheparticles inthese showers are1r- mesons, which decay toother particles called muons. Muons arealso unstable and decay according totheradioactive decay law, N=N0exp(~0.693 t/t1/2), where N0andNarethenumber ofmuons attime tI0and t,respectively, and t1/2isthehalf-life. However, enough muons reach theearth’s surface thatwecan detect them easily. Letusassume thatwemount adetector ontopofa2,000-m mountain and count thenumber ofmuons traveling ataspeed near -u=0.980. Over agiven pe- riod oftime, wecount 103muons. The half-life ofmuons isknown tobe 1.52 X10‘5 sintheir own restframe (system K’).Wemove ourdetector tosea level andmeasure thenumber ofmuons (having v=0.980) detected during an equal period oftime. What doweexpect? 556 14/SPECIAL THEORY OFRELATIVITY Determined classically, muons traveling ataspeed of0.980 cover the2,000 m in6.8X10‘6 s,and45muons should survive theﬂight from 2,000 mtosealevel according totheradioactive decay law. Butexperimental measurement indicates that542muons survive, afactor of12more. This phenomenon must betreated relativistically. The decaying muons are moving atahigh speed relative totheexperimenters ﬁxed ontheearth. We therefore observe themuons’ clock toberunning slower. Inthemuons’ rest frame, thetime period ofthemuons’ ﬂight isnotAt=6.8X10_6 sbutrather At/‘y. Forv=0.98c, y=5,sowemeasure theﬂight time onaclock atrestin themuons’ system tobe1.36 Xl0‘6 s.The radioactive decay lawpredicts that 538muons survive, much closer toourmeasurement and within theexperi- mental uncertainties. Anexperiment similar tothishasveriﬁed thetime dila- tion prediction.* Examine themuon decay justdiscussed from theperspective ofanobserver moving with themuon. Solution. The half-life ofthemuon according toitsown clock is1.52 X10-6 s. Butanobserver moving with themuon would notmeasure thedistance from thetopofthemountain tosealevel tobe2,000 m.According tothatobserver, thedistance would beonly 400m.Ataspeed of0.98c,ittakes themuon only 1.36 X10‘6 stotravel the400m.Anobserver inthemuon system would pre- dict538muons tosurvive, inagreement with anobserver ontheearth. Muon decay isanexcellent example ofanatural phenomenon that canbe described intwosystems moving with respect toeach other. One observer sees time dilated andtheother observer sees length contracted. Each, however, pre- dicts aresult inagreement with experiment. Atomic Clock Time Measurements Aneven more direct conﬁrmation ofspecial relativity was reported bytwo American physicists, C.Hafele and Richard E.Keating, in1972.* They used four extremely accurate cesium atomic clocks. Two clocks were ﬂown onregu- larly scheduled commercial jetairplanes around theworld, one eastward and onewestward; theother tworeference clocks stayed ﬁxed ontheearth atthe U.S. Naval Observatory. Awell-deﬁned, hyperﬁne transition intheground state ofthe135Cs atom hasafrequency of9,192,631,770 Hzandcanbeused asanac- curate measurement ofatime period. *The experiment wasreported byB.Rossi andD.B.Hall inthePhys. Rem, 59,223(1941). Aﬁlmen- titled “Time Dilation-—An Experiment with p.-Mesons” byD.H.Frisch and].H.Smith isavailable from theEducation Development Center; Newton, Mass. SeealsoD.H.Frisch and]. H.Smith, Am. ].Phys., 31,342(1963). TSee_]. C.Hafele andRichard E.Keating, Science, 177, 166-170 (1972). 14.4 EXPERIMENTAL VERIFICATION OFTHE SPECIAL THEORY 557 The time measured onthetwomoving clocks wascompared with thatofthe tworeference clocks. The eastward triplasted 65.4 hours with 41.2 ﬂight hours. The westward trip, aweek later, took 80.3 hours with 48.6 ﬂight hours. The pre- dictions arecomplicated bytherapid rotation oftheearth andbyagravitational effect from thegeneral theory ofrelativity. Wecangain some insight totheexpected effect byneglecting thecorrec- tions and calculating thetime difference asiftheearth were notrotating. The circumference oftheearth isabout 4X107m,andatypical jetairplane speed is almost 300m/s.Aclock ﬁxed ontheground measures aﬂight time T0of 7 To=1 @ ==1.33 X105s(~ 37hr) (14.22) Because themoving clock runs more slowly, theobserver ontheearth would say thatthemoving clock measures only T=7},\/1~B2.The time difference is AT= To—T=T0(1—— \/1~32) 1 (14.23) "7_B2To2 where only theﬁrst and second terms ofthepower series expansion for \/1—B2arekept because B2issosmall. 1 300 2AT= (1.33 X105s) - 3X108m/s (14.24) =6.65 X10-85 =66.5 ns This time difference isgreater than theuncertainty ofthemeasurement. Notice that inthiscase, theclock leftontheearth actually measures more time insec- onds than themoving clock. This seems atvariance with ourearlier comments (see Equation 14.21 and discussion). But the time period referred toin Equation 14.21 isthetime between twoticks, inthiscase, atransition in1?’3Cs, which wemeasure inseconds. Itiseasy toremember that moving clocks run more slowly, sothatinseconds themeasured time difference involves fewer ticks and, according tothedeﬁnition ofasecond, fewer seconds. The actual predictions andobservations forthetime difference are Travel Predicted Observed Eastward -40 i23ns ~59 I10ns Westward 275i21ns 273i7ns Again, thespecial theory ofrelativity isveriﬁed within theexperimental uncer- tainties. Anegative sign indicates that thetime onthemoving clock islessthan theearth reference clock. The moving clocks losttime (ran slower) during the eastward tripandgained time (ranfaster) during thewestward trip.This difference iscaused bytherotation oftheearth, indicating that theﬂying clocks actually ticked faster orslower than thereference clocks ontheearth. The overall posi- 558 14/SPECIAL THEORY orRELATIVITY tivetime difference isaresult ofthegravitational potential effect (which wedo notdiscuss here). Wehave only brieﬂy described twoofthemany experiments that have veri- ﬁedthespecial theory ofrelativity. There arenoknown experimental measure- ments thatareinconsistent with thespecial theory ofrelativity. Einstein’s work in thisregard hassofarwithstood thetestoftime. 14.5 Relativistic Doppler Effect The Doppler effect insound isrepresented byanincreased pitch ofsound asa source approaches areceiver andadecrease ofpitch asthesource recedes. The change infrequency ofthesound depends onwhether thesource orreceiver is moving. This effect seems toviolate Postulate Iofthetheory ofrelativity until we realize thatthere isaspecial frame forsound waves because there isamedium (e.g., airorwater) inwhich thewaves travel. Inthecase oflight, however, there isnosuch medium. Only relative motion ofsource andreceiver ismeaningful in thiscontext, andweshould therefore expect some differences intherelativistic Doppler effect forlight from thenormal Doppler effect ofsound. Consider asource oflight (e.g., astar) andareceiver approaching onean- other with relative speed v(Figure 14-4a). First, consider thereceiver ﬁxed in system Kand thelight source insystem K’moving toward thereceiver with speed v.During time Atasmeasured bythereceiver, thesource emits nwaves. During thattime At,thetotal distance between thefront andrear ofthewaves is length ofwave train =cAt-vAt (14.25) The Wavelength isthen cAt~-uAt)1=T‘ (14.26) andthefrequency is c cn VT/\TcAt~vAt (14.27) According tothesource, itemits nwaves offrequency 1/0during theproper time At’: n=1/OAt’ (14.28) This proper time At’measured onaclock inthesource system isrelated tothe time Atmeasured onaclock ﬁxed insystem Kofthereceiver by AtAt’=— 14.29 7 ( ) The clock moving with thesource measures theproper time, because itispres- entatboth thebeginning andendofthewaves. 14.5 RELATIVISTIC DOPPLER EFFECT 559 System K’ -1System K bi(a)Source andreceiver approaching qi v "\/L, (b)Source and receiver receding FIGURE 14-4 (a)Anobserver insystem Kseeslight coming from asource ﬁxed in system K’.System K’ismoving toward theobserver with speed v.The frequency ofthelight isobserved inKtobeincreased over thevalue observed inK’.(b)VVhen system K’ismoving away from theobserver, thefrequency ofthelight decreases (thewavelength increases). This isthesource oftheterm redshzfted. Substituting Equation 14.29 into Equation 14.28, which inturn issubsti- tuted forninEquation 14.27, gives 1 V0 V“-:?"—"’ii (1—v/c) ‘Y =l—~'1"1F/C2», (14.30)1—v/c which canbewritten as \/1+B . .11=ii 110source andreceiver approaching (14.31) \/1*B Itisleftforthereader (Problem 14-14) toshow thatEquation 14.31 isalsovalid when thesource isﬁxed andthereceiver approaches itwith speed v. Next, weconsider thecase inwhich thesource and receiver recede from each other with velocity v(Figure 14-4b). The derivation issimilar totheone 560 14/SPECIAL THEORY OFRELATIVITY just presented—with one small exception. InEquation 14.25, thedistance be- tween thebeginning andendofthewaves becomes length ofwave train =cAt+vAt (14.32) This change insign ispropagated through Equations 14.30 and14.31, giving \/1-112/c2v-"=—————"-1/01+v/c \/1*Bv=———--11 source andreceiver receding (14.33) v1+p° Equations 14.31 and 14.33 canbecombined into oneequation, v1+pv=\/——l_-— 110relativistic Doppler effect (14.34) 1-B ifweagree tousea+sign forB(+v/ c)when thesource and receiver areap- proaching each other anda—-signforBwhen they arereceding. The relativistic Doppler effect isimportant inastronomy. Equation 14.34 in- dicates that, ifthesource isreceding athigh speed from anobserver, then a lower frequency (orlonger wavelength) isobserved forcertain spectral lines or characteristic frequencies. This istheorigin oftheterm redshift; thewavelengths ofvisible light areshifted toward longer wavelengths (red) ifthesource isreced- ingfrom us.Astronomical observations indicate that theuniverse isexpanding. The farther away astaris,thefaster itappears tobemoving away (orthegreater itsredshift). These data areconsistent with the“big bang” origin oftheuni- verse, which isestimated tohave occurred some 13billion years ago. '7_"_"'During aspaceﬂight toadistant star,anastronaut andhertwin brother onthe earth send radio signals toeach other atannual intervals. What isthefrequency oftheradio signals each twin receives from theother during theﬂight tothe stariftheastronaut ismoving atv=0.8c? \/Vhat isthefrequency during there- turn ﬂight atthesame speed? Solution. WeuseEquation 14.34 todetermine thefrequency ofradio signals thateach receives from theother. The frequency 1/0=1signal/ year. Ontheleg ofthetripaway from theearth, B==——0.8 andEquation 14.34 gives VV1-as=_i_iiV v1+us° V0 3 The radio signals arereceived once every 3years. 14.6 TWIN PARADOX 561 Onthereturn trip, however, B=+0.8 andEquation 14.34 gives 11=3110,so theradio signals arereceived every 4months. Inthisway, thetwin ontheearth canmonitor theprogress ofhisastronaut twin. i 1 mm mm I mm _m 1 iL i lL 1 I l L j 14.6 Twin Paradox Consider twins who choose different career paths. Mary becomes anastronaut, andFrank decides tobeastockbroker. Atage30,Mary leaves onamission toa planet inanearby star’s system. Mary willhave totravel atahigh speed toreach theplanet andreturn. According toFrank, Mary’s biological clock willtickmore slowly during hertrip, soshewillagemore slowly. Heexpects Mary tolook and appear younger than hedoes when shereturns. According toMary, however, Frank willappear tobemoving rapidly with respect tohersystem, andshethinks Frank willbeyounger when shereturns. This istheparadox. Which twin, ifei- ther, isyounger when _M_ary (themoving twin) returns totheearth where Frank (the ﬁxed twin) hasremained? Because thetwoexpectations aresocontradic- tory, doesn’t Nature have awaytoprove they willbethesame age? This paradox hasexisted almost since Einstein ﬁrst published hisspecial theory ofrelativity. Variations oftheargument have been presented many times. The correct answer isthat Mary, theastronaut, willreturn younger than hertwin brother, Frank, who remains busy onWall Street. The correct analysis isasfollows. According toFrank, Mary’s spaceship blasts offand quickly reaches acoasting speed ofv=0.8c, travels adistance of8ly(ly=a light year, thedistance light travels in1year) totheplanet, andquickly decel- erates forashort visit totheplanet. The acceleration and deceleration times arenegligible compared with thetotal travel time of10years totheplanet. The return tripalsotakes 10years, soonMary’s return toEarth, Frank willbe 30+10+10=50years old. Frank calculates that Ma ’sclock isticking slower and that each legofthetriptakes only 10V1~0.82=6years. Mary therefore isonly 30+6+6I42years oldwhen shereturns. Frank’s clock is (almost) inaninertial system. When Mary performs thetime measurements onherclock, they may bein- valid according tothespecial theory because hersystem isnotinaninertial frame ofreference moving ataconstant speed with respect totheearth. Sheac- celerates and decelerates atboth theearth and theplanet, and tomake valid time measurements tocompare with Frank’s clock, shemust account forthisac- celeration anddeceleration. The instantaneous rateofMary’s clock isstillgiven byEquation 14.20, because theinstantaneous rateisdetermined bytheinstanta- neous speed v.*Thus, there isnoparadox ifweobey thetwopostulates ofthe *See theclock hypothesis ofW.Rindler (Ri82, p.31). 562 14/SPECIAL THEORY OFRELATIVITY special theory. ‘Itisalso clear which twin isintheinertial frame ofreference. Mary willactually feeltheforces ofacceleration anddeceleration. Frank feels no such forces. VVhen Mary returns home, hertwin brother hasinvested her20 years ofsalary, making herarich woman attheyoung ageof42.Shewaspaid a 20-year salary forajobthattook heronly 12years! Mary andFrank send radio signals toeach other at1-year intervals after she leaves Earth. Analyze thetimes ofreceipt oftheradio messages. Solution. InExample 14.4, wecalculated thatsuch radio signals arereceived every 3years onthetripoutandevery %year onthetripback. First, weexamine thesignals Mary receives from Frank. During the6-year triptotheplanet, Mary receives only tworadio messages, butonthe6-year return trip, shereceives eighteen signals, soshecorrectly concludes that hertwin brother Frank has aged 20years andisnow 50years old. InFrank’s system, Mary’s triptotheplanet takes 10years. Bythetime Mary reaches theplanet, Frank receives 10/3 signals (i.e., three signals plus one-third ofthetime tothenext one). However, Frank continues toreceive asignal every 3years forthe8years ittakes thelastsignal Mary sends when shereaches the planet totravel toFrank. Thus, Frank receives signals every 3years for8more years (total of18years) foratotal ofsixradio signals from theperiod oftravel to theplanet. Frank hasnowayofknowing thatMary hasstopped andturned around until theradio message, which takes 8years, isreceived. Oftheremain- ing2years ofMary’s journey according toFrank (20—18=2),Frank receives signals every fl-gyear, orsixmore signals. Frank correctly determines thatMary hasaged 6+6=12years during herjourney because hereceives atotal of12 signals. Thus, both twins agree about their own ages andabout each other’s. Mary is 42andFrank is50years old. 14.7 Relativistic Momentum Newton’s Second Law, F=dp/dt,iscovariant under aGalilean transformation. Therefore, Wedonotexpect ittokeep itsform under aLorentz transformation. Wecanforesee difﬁculties with Newton’s laws andtheconservation laws unless wemake some necessary changes. According toNewton’s Second Law, forexam- ple, anacceleration athigh speeds might cause aparticle ’svelocity toexceed c, animpossible condition according tothespecial theory ofrelativity. Webegin byexamining theconservation oflinear momentum inaforce- free (noexternal forces) collision. There arenoaccelerations. Observer Aat restinsystem Kholds aballofmass m,asdoes observer Binsystem K’moving to 14.7RELATIVISTIC MOMENTUM 563 I V I x2 x2 i x2 x2 m B m B v4-1 A m A m 1<' xf 1<' xf .-‘K in K1 K 2 xl (a)Collision according tosystem K (b)Collision according tosystem K’ FIGURE 14-5 Observer A,atrestinﬁxed system K,throws aballstraight upinsystem K.Observer B,atrestinsystem K’,which ismoving totheright with velocity v,throws aballstraight down sothatthetwoballs collide. (a)Thecollision according toobserverA insystem K.(b)Thecollision according toobserver Binsystem K’.Each observer measures thespeed ofhisorherballtobeuo.Weexamine thelinear momentum oftheball. theright with relative speed vwith respect tosystem K,asinFigure 14-1. The twoobservers throw their (identical) balls along their respective x2-axes, which results inaperfectly elastic collision. The collision, according toobservers inthe twosystems, isshown inFigure 14-5. Each observer measures thespeed ofhisor herballtobeuo. Weﬁrst examine theconservation ofmomentum according tosystem K. Thevelocity oftheballthrown byobserver Ahascomponents =0“’“ } (14.35)"A2 The momentum ofballAisinthe.762-(I1I‘CC[lOI1Z PA2 =muo The collision isperfectly elastic, sotheball returns down with speed uo.The change inmomentum observed insystem Kis 411,,=—2mu0 (14.37) Does Equation 14.37 alsorepresent thechange inmomentum oftheballthrown byobserver Binthemoving system K’?Weusetheinverse velocity transforma- tion ofEquations 14.17 (i.e., weinterchange primes and unprimes and let v—>—v)todetermine "B1: "U1.38 um=—u0\/1 —v2/c2} (4 ) 564 14/SPECIAL THEORY OFREIATIVITY where ufn=0and uf,2=—u0. The momentum ofballBanditschange inmo- mentum during thecollision become pm=—mu0 V1—v2/c2 (14.39) A1232 =+2mu0 \/1—v2/c2 (14.40) Equations 14.37 and 14.40 donotaddtozero: Linear momentum isnotconserved according tothespecial theory weusetheconventions formomentum ofclassical physics. Rather than abandoning thelawofconservation ofmomentum, welook fora solution thatallows ustoretain both itandNewton’s Second Law. AswedidfortheLorentz transformation, weassume thesimplest possible change. Weassume that theclassical form ofmomentum muismultiplied bya constant thatmaydepend onspeed k(u): p=k(u) mu (14.41) InExample 14.6, weshow thatthevalue k(u)=ii (14.42) \/1—u2/c2 allows ustoretain theconservation oflinear momentum. Notice thattheform of Equation 14.42 isthesame asthat found fortheLorentz transformation. Infact, theconstant k(u) isgiven thesame label: 'y.However, this‘ycontains thespeed oftheparticle u,whereas theLorentz transformation contains therelative speed vbetween thetwoinertial reference frames. This distinction must bekept in mind; itoften causes confusion. Wecanmake aplausible calculation fortherelativistic momentum ifweuse theproper time 1'(seeEquation 14.21) rather than thenormal time t.Inthiscase. dx dxdt=—= —— 14.43P mdr mdtdr ( ) dx 1= ——i—-—— (14.44) mdtV1—u2/c2 p=——-2-ri— =ymu relativistic momentum (14.45) V1—u2/c2 where weretain u=dx/dt asused classically. Although allobservers donot agree astodx/dt, they doagree astodx/do-, where theproper time drismeas- ured bythemoving object itself. The relation dt/dr isobtained from Equation 14.21, where thespeed uhasbeen used in'ytorepresent thespeed ofarefer- ence frame ﬁxed intheobject thatismoving with respect toaﬁxed frame. Equation 14.45 isournew deﬁnition ofmomentum, called relativistic mo- mentum. Notice that itreduces totheclassical result forsmall values ofu/c.It wasfashionable inpastyears tocallthemass inEquation 14.45 therestmass mo andtocalltheterm m=-3-2% (old-fashioned notation) (14.46)V —U C 14.7 RELATIVISTIC MOMENTUM 565 therelativistic mass. The term restmass resulted from Equation 14.46 when u=0, andtheclassical form ofmomentum wasthus retained: p=mu.Scientists spoke ofthemass increasing athigh speeds. Weprefer tokeep theconcept ofmass as aninvariant, intrinsic property ofanobject. The useofthetwoterms relativistic andrestmass isnow considered old-fashioned, although theterms arestillsome- times used. Wealways refertothemass m,which isthesame astherestmass. The useof relativistic mass often leads tomistakes when using classical expressions. Show thatlinear momentum isconserved inthex2-direction forthecollision shown inFigure 14-5ifrelativistic momentum isused. Solution. Wecanmodify theclassical expressions formomentum already ob- tained forthetwoballs. The momentum forballAbecomes (from Equation 14.36) pm=—mi (14.47)\/1—ug/c2 and -2Ap,,,=-i (14.48)V1—ug/c2 Before modifying Equation 14.39 forthemomentum ofballB,wemust ﬁrst ﬁnd thespeed ofballBasmeasured insystem K.WeuseEquation 14.38 tode- terrnine "B=V"ii+"in =Vv2+'u%(1 —v2/c2) (14.49) The momentum [132isfound bymodifying Equation 14.39: P32='_"m0'Y V1_"U2/C2 where L_;_ 7\/1-‘W PM= \/711% Using ul,from Equation 14.49 gives PM= -—mu0 V1—v2/c2 g \/<1—at/@2><1 —v2/C2)_ —mu0 —\/1T7./C5 Ap=+2722“ (14.52) B2 If_Us/C2(14.50) (14.51) 566 14/SPECIAL THEORY orREIATIVITY Equations 14.48 and14.52 addtozero, asrequired fortheconservation oflin- ear I1'lOI1'lCIlI1lIl'1. 14.8 Energy With anew deﬁnition oflinear momentum (Equation 14.45) inhand, weturn ourattention toenergy and force. Wekeep ourformer deﬁnition (Equation 2.86) ofkinetic energy asbeing thework done onaparticle. The work done is deﬁned inEquation 2.84 tobe 2 W12=LF-dr=E—T1 (14.53) Equation 2.2forNewton ’sSecond Law ismodiﬁed toaccount forthenew deﬁ- nition oflinear momentum: dp d F=—=— 14.54 dtdtwmu) (> Ifwestart from rest, T1=0,andthevelocity uisinitially along thedirection of theforce. dW= T= IE0)/mu) 'udt (14.55) ll =miud('yu) (14.56) 0 Equation 14.56 isintegrated byparts toobtain “ dT=,,m,2_.,,,( Li0 2\/1- u/c2 ='ymu2 +mc2V1 —u2/c2 ='ymu2 +mc2V1—u2/c2 —mc2 (14.57) With algebraic manipulation, Equation 14.57 becomes relativistic kinetic energy (14.58) Equation 14.58 seems toresemble innowayourformer result forkinetic en- ergy, T=%mu2. However, Equation 14.58 must reduce to%mu2 forsmall values ofvelocity. 14.8ENERGY 567 Show thatEquation 14.58 reduces totheclassical result forsmall speeds, u<<c. Solution. The ﬁrstterm ofEquation 14.58 canbeexpanded inapower series: T= mc2(1 —u2/c2)_1/2 -mc2 12 =mc2 1+-12 + —mc2 (14.59)2c where allterms ofpower (u/c)4 orgreater areneglected because u<<c. 1T=mc2+§mu2 —mc2 1=§mu2 (14.60) which istheclassical result. Itisimportant tonote that neither %mu2 nor%'ymu2 gives thecorrect rela- tivistic value forthekinetic energy. The term mc2inEquation 14.58 iscalled therestenergy andisdenoted byE0. restenergy (14.61) Equation 14.58 isrewritten 'ymc2 =T+ mc2 Thus, E=T+E0 (14.62) EEymc2 =T+E0 total energy (14.63)where The total energy, E=ymc2, isdeﬁned asthesum ofkinetic energy andtherest energy. Equations 14.58-14.63 aretheorigin ofEinstein’s famous relativistic re- sultoftheequivalence ofmass andenergy (energy =mc2). These equations are consistent with thisinterpretation. Note thatwhen abody isnotinmotion (u= 0=T),Equation 14.63 indicates thatthetotal energy isequal totherestenergy. Ifmass issimply another form ofenergy, then wemust combine theclassical conservation laws ofmass andenergy into oneconservation lawofmass-energy represented byEquation 14.63. This lawiseasily demonstrated intheatomic nu- cleus, where themass ofconstituent particles isconverted totheenergy that binds theindividual particles together. 568 14/SPECIAL THEORY orRELATIVITY Usetheatomic masses oftheparticles involved tocalculate thebinding energy ofadeuteron. Solution. Adeuteron iscomposed ofaneutron andaproton. Weuseatomic masses, because theelectron masses cancel. mass ofneutron =1.008665 u mass ofproton (1H) =1.007825 umi-ii sum =2.016490 u mass ofdeuteron (2H) =2.014102 u difference =0.002388 u This difference inmass-energy isequal tothebinding energy holding theneu- tron andproton together asadeuteron. The mass units areatomic mass units (u),which canbeconverted tokilograms ifnecessary. However, theconversion ofmass toenergy isfacilitated bythewell-known relation between mass and energy: 1uc2=931.5 MeV (14.64) The binding energy ofthedeuteron istherefore Mv0.002388 6.2><931.5% =2.22MeV11C Nuclear experiments oftheform y+2H—>n+pindicate thatgamma raysof energy justgreater than 2.22 MeV arerequired tobreak thedeuteron apart into aneutron andaproton. Conversely, when aneutron andproton join atrestto form adeuteron, 2.22 MeV ofenergy isreleased intheform ofkinetic energy ofthedeuteron andgamma ray. Because physicists believe that momentum isamore fundamental concept than kinetic energy (forexample, there isnogeneral lawofconservation ofki- netic energy) ,wewould likearelation formass-energy thatincludes momentum rather than kinetic energy. Webegin with Equation 14.45 formomentum: P=rm“ P262 =y2m2u2C2 2 ='y2m2c4 (14.65) Itiseasy toshow that u2 1—=1-— 14.66 C.7. <> 14.9SPACETIME ANDFOUR-VECTORS 569 soEquation 14.65 becomes 1262=y2m2c4 1__ =.y2m2c4 _m2C4 =E2—E3 Equation 14.67 isavery useful kinematic relationship. Itrelates thetotal energy ofaparticle toitsmomentum andrestenergy. Notice thataphoton hasnomass, sothatEquation 14.67 gives E=pc photon (14.68) There isnosuch thing asaphoton atrest. 14.9 Spacetime andFour-Vectors InSection 14.3 (Equation 14.5), wenoticed thatthequantities R-m-t2r2=0 $5._[*43i"MQ_ -c2t’2 =0 areinvariant because thespeed oflight isthesame inallinertial systems inrela- tivemotion. Consider twoevents separated byspace andtime. Insystem K, Ax,=x,-(event 2)-x,(event 1) At=t(event 2)—t(event 1) The interval As2isinvariant inallinertial systems inrelative motion (see Problem 14-34): 3 A52=Z(Ax,-)2 —MR (14.69)Q. s As2=As'2 =]g(Ax;)2 —c2At'2 (14.70) Equation 14.69 canbewritten asadifferential equation: ds2=dx2+ax;+dx§—c2dt2 (14.71) Consider thesystem K’,where theparticle isinstantaneously atrest. Because dx{=dxé=dxl,=0inthiscase, dt'=d1",theproper time interval discussed 570 14/SPECIAL THEORY orRELATIVITY ct ¢Futureif B @‘Y?,‘}5 ’§ii,i;aI1L, 2'A. 2.-‘,_-E" 2' 11'}_5llxvi..511 ‘1'3f>7I 51:2Past Light cone FIGURE 14-6 Thevariable ctisplotted versus xwith theorigin being thepresent. The heavy solid lines indicate thepastandfuture paths oflight andform a lightcone. Totheright andleftofthese lines isconsidered “elsewhere,” because wecannot reach thisregion from thepresent. Thepath from A toBrepresents aworldline, apath thatwecantake traveling atspeeds less than orequal tolight. above (Equation 14.21). Equation 14.70 becomes —c2d1'2 =dx12+dxg+dx§—c2dt2 (14.72) Using theLorentz transformation, Equation 14.72 gives asimilar result to Equation 14.21: dtd=— 14.73 1'7 ( ) Theproper time 1'is,along with thelength quantity As2, another Lorentz invari- antquantity. Auseful concept inspecial relativity isthat ofthelight cone. The invariant length As2suggests adding ctasafourth dimension tothethree space dimen- sions x1,x2,andx3.InFigure 14-6, weplot ctversus oneoftheEuclidean space coordinates. The origin of(x,ct)isthepresent (0,0).The solid lines represent thepaths taken inthepast and inthefuture bylight. Aparticle traveling the path from AtoBissaidtobemoving along itsworldline. Fortime t<0,thepar- ticle hasbeen inthelower cone, thepast. Similarly, for‘t>0theparticle will move intheupper cone, thefuture. Itisnotpossible forustoknow about events outside thelight cone; thisregion, called “elsewhere,” requires~v> c. There aretwopossibilities concerning thevalue ofAs2. IfAs2>0,thetwo events have aspacelike interval. One canalways ﬁnd aninertial frame traveling with v<csuch thatthetwoevents occur atdifferent space coordinates butatthe 14.9SPACETIME ANDFOUR-VECTORS 571 same time. When As2<0,thetwoevents aresaid tohave atimelike interval. One canalways ﬁnd asuitable inertial frame inwhich theevents occur atthe same point inspace butatdifferent times. Inthecase As2=0,thetwoevents areseparated byalight ray. Only events separated byatimelike interval can becausally connected. The present event inthelight cone canbecausally related only toevents in thepast region ofthelight cone. Events with aspacelike interval cannot be causally connected. Space and time, although distinct, arenonetheless intri- cately related. The previous discussion ofspace and time suggests using ctasafourth di- mensional parameter. Wecontinue thislineofthought bydeﬁning x4Eictand xiEict’.The useoftheimaginary number i(V-—1) does notindicate thatthis component isimaginary. The imaginary number simply allows ustorepresent therelations inconcise, mathematical form. The restofthissection could justas well becarried out without the use ofi(e.g., x4=ct), but the mathematics would bemore cumbersome. The useful results areinterms ofreal, physical quantities. Using x4=ictand xi=ict’, wecanwrite Equations 14.5 as* _l‘4*~1:2»=0,,=(14.74) =Q ‘ ‘F_M"*><1.; From these equations, itisclear thatthetwosums must beproportional, andbe- cause themotion issymmetrical between thesystems, theproportionality con- stant isunity.l Thus, gxi =gist’? (14.75) This relation isanalogous tothethree-dimensional, distance-preserving, ortho- gonal rotations wehave studied previously (see Section 1.4) and indicates that theLorentz transformation corresponds toarotation inafour-dimensional space (called world space orMinkowski spacei). The Lorentz transformations arethen orthogonal transformations inMinkowski space: x,',=2:11,“, x,, (14.76) *Inaccordance with standard convention, weuseGreek indices (usually p.orv)toindicate summa- tions thatrunfrom 1to4;inrelativity theory, Latin indices areusually reserved forsummations that runfrom 1to3. 1-A“proof” isgiven inAppendix G. $Herman Minkowski (1864-1909) made important contributions tothemathematical theory ofrela- tivity andintroduced ictasafourth component. 572 14/SPECLAL THEORY orRELATIVITY where the}l,,,,aretheelements oftheLorentz transformation matrix. From Equations 14.14, thetransformation Ais -QOOQ‘< Q©i—‘© ©>—'©©‘iOiB)' A= 0 (14.77) —z Aquantity iscalled afour-vector ifitconsists offour components, each of which transforms according totherelation* 14;,=2110.4, (14.78) where the11,0,deﬁne aLorentz transformation. Such afour-vectorl is X=(x1,x2,x3,ict) (14.79a) 0...... where thenotation ofthelastline means that theﬁrst three (space) compo- nents ofXdeﬁne theordinary three-dimensional position vector xandthat the fourth component isict.Similarly, thedifferential ofXisafour-vector: dX=(dx, icdt) (14.80) InMinkowski space, thefour-dimensional element oflength isinvariant. Its magnitude isunaffected byaLorentz transformation, and such aquantity is called afour-scalar orworld scalar. Equation 14.71 canbewritten as ds=\/E425, (14.81)I-"or andEquation 14.72 as at=ivEar; =ids (14.82) The proper time drisinvariant because itissimply i/ctimes theelement of length ds.The ratio ofthefour-vector dXtotheinvariant dristherefore alsoa four-vector, called thefour-vector velocity V: \/—g— 5'3’ 1483_d7'_ d7"wd~r (‘l The components oftheordinary velocity uare .,.=K‘1dt *We donotdistinguish here between covariant andcontravariant vector components; see,forexam- ple,Bergmann (Be46, Chapter 5). TFour-vectors aredenoted exclusively byopenface capital letters. 14.9 SPACETIME AND FOUR-VECTORS 573 so,using Equations 14.71 and14.82, drcanbeexpressed as 2d7'= dt,/1—— _dx’ OI‘ at=an/1 —62 (14.84)<1,»-1-Mgal aswefound inEquation 14.73. The four-vector velocity cantherefore bewritten as V=i,1%B? (ll,it‘) (14.85) where urepresents thethree space components ofordinary velocity, ul,u2,U3. (Remember that theparticle’s velocity isnow denoted byutodistinguish itfrom themoving frame velocity v.)Thefour-vector momentum isnow simply themass times four-vector velocity,* because mass isinvariant: P=mv (14.86) = , 2p4)1 where p,Ei (14.88)V1—B2 The ﬁrstthree components ofthefour-vector momentum Parethecompo- nents oftherelativistic momentum (Equation 14.45): P,=p,=ymuj, J"=1,2,3 (14.89) Using Equation 14.63, thefourth component ofthemomentum isrelated tothe total energy E: E [24=ymc =Z (14.90) The four-vector momentum cantherefore bewritten as P=<p,i7E) (14.91) where pstands forthethree space components ofmomentum. Thus, inrelativ- itytheory, momentum andenergy arelinked inamanner similar tothatwhich joins theconcepts ofspace andtime. Ifwe apply theLorentz transformation matrix *Afour-vector multiplied byafour-scalar isalso afour-vector. 574 14/SPECIAL THEORY OFRELATIVITY (Equation 14.77) tothemomentum P,weﬁnd I_pl—(v/c2)E P1— /‘F_B2 g;Z (14.92) E,:E_11121 \/1-,6? Using themethods ofthissection, derive Equation 14.67. Solution. Ifweplace theorigin ofthemoving system K’ﬁxed ontheparticle, wehave u=v.The square ofthefour-vector velocity (Equation 14.85) is invariant: U2LLC2 \/2=Evg=1—;-FL; =—C2 (14.93) Hence, thesquare ofthefour-vector momentum isalsoinvariant: P2=E123=m2\/2=-4.282 (14.94)P- From Equation 14.91, wealsohave, using p-p=p2=p2+pg+p§, E2 P2=p2—F (14.95) Combining thelasttwoequations gives Equation 14.67. E2=p2c2 +m2c4 =p2c2 +E2, Ifwedeﬁne anangle qbsuch thatB=sind>, therelativistic relations between velocity, momentum, andenergy canbeobtained bytrigonometric relations in- volving theso-called “relativistic triangle” (Figure 14-7). .Derive thevelocity addition rule. Solution. Suppose thatthere arethree inertial reference frames, K,K’,andK”, which areincollinear motion along their respective x1-axes. Letthevelocity of K’relative toKbev1andletthevelocity ofK”relative toK’bez/2.The speed of K"relative toKcannot bev1+v2,because itmust bepossible topropagate asig- nalbetween anytwoinertial frames, andifboth v1andv2aregreater than c/2 (but lessthan c),then vl+112>c.Therefore, therulefortheaddition ofveloci- tiesinrelativity must bedifferent from thatinGalilean theory. The relativistic 14.9SPACETIME ANDFOUR-VECTORS 575 l 77162 \<sin'1%/ E Ei = mcmc2“< "Q ¢=sin'1ﬁ 1-1l__ !l FIGURE 14-7 The relativistic triangle allows ustoﬁnd relations between velocity, momentum, andenergy byusing trigonometric relations. velocity addition rule canbeobtained byconsidering theLorentz transforma- tionmatrix connecting KandK".The individual transformation matrices are it-1.7.) Y 00$ co»-oo»-oo0 AK'—>K = 0 \‘"iB171 71/ O ©©l—‘© ©l—'©©GO/'72 (B272) 0 AK”—>K' = V"25272 ‘Y2/ The transformation from K"toKisjust theproduct ofthese twotransforma- tions: GOP-‘C ©P—‘©©'Y1'Y2(1 +B152) i')’1'Y2(B1 +B2) 0 0 AK”—>K =AK”—>K"\K'—>K Z 0 0 -271’)/2(/'31 +132) ')’1'Y2(1 +B152) Sothat theelements ofthismatrix correspond tothose ofthenormal Lorentz matrix (Equation 14.77), wemust identify Band~yfortheK"—>K transforrna- tionas 7=m'2(1 +B182) B7='Y1'}’2(B1 +B2)} (14.96) from which weobtain 6=ii (14.97)1+B152 576 14/SPECLAL THEORY orRELATIVITY Ifwemultiply thislastexpression byc,wehave theusual form ofthevelocity (speed) addition rule: 2‘+U2 (1498) v=ii . 1+(7/1222/(32) Itfollows thatifv<candv< then v<calso. 1 2Q Even though signal velocities cannever exceed c,there areother types ofve- locity thatcanbegreater than c.Forexample, thephase velocity ofalight wave in amedium forwhich theindex ofrefraction islessthan unity isgreater than c, butthephase velocity does notcorrespond tothesignal velocity insuch a medium; thesignal velocity isindeed lessthan c.Orconsider anelectron gun thatemits abeam ofelectrons. Ifthegunisrotated, then theelectron beam de- scribes acertain path onascreen placed atsome appropriate distance. Ifthean- gular velocity ofthegun and thedistance tothescreen aresufficiently large, then thevelocity ofthespot traveling across thescreen canbeanyvelocity, arbi- trarily large. Thus, thewriting speed ofanoscilloscope canexceed c,butagain the writing speed does notcorrespond tothesignal velocity; that is,information cannot betransmitted from onepoint onthescreen toanother bymeans ofthe electron beam. Insuch adevice, asignal canbetransmitted only from thegun tothescreen, andthistransmission takes place atthevelocity oftheelectrons in thebeam (i.e., <c). Derive therelativistic Doppler effect iftheangle between thelight source and direction ofrelative motion oftheobserver is9(Figure 14-8). Solution. This example caneasily besolved using themomentum-energy four- vector bytreating thelight asaphoton with total energy E=hv.The light source isatrestinsystem Kandemits asingle frequency 1/0. E=hi/0 (14.99) E hvp=2=-69 (i4.100) The observer moving totheright insystem K’measures theenergy E’forapho- tonoffrequency 11’.From Equation 14.92, wehave E’='y(hv0 —vpl) (14.101) vhi/0 hi/'='y(hv0 —-7 cos(9) (14.102) where pl=pcos9.Equation 14.102 reduces to 11'=')/1/0(1 —Bcos9) (14.103) 14.9SPACETIME ANDFOUR-VECTORS 577 22 K’ v 1 Observer xi x2 K Light (bl 1/ 0 “R x1 (4) . FIGURE 14-8 Alight source ﬁxed insystem Kemits light atasingle frequency 1/0.An observer insystem K’,moving totheright atvelocity vwith respect toK,measures thelight frequency tobe1/’. which isequivalent toEquation 14.34, depending onthevalue ofB.Foranearly time, theobserver isfartotheleftofthesource, andastheobserver ap- proaches thesource (0=77'), v1+pv’=1/—i— observer approaching source (14.104) 0r?_B asinEquation 14.31. Atamuch later time, theobserver isreceding (6=0)and \/1-B11'=11ni observer receding from source (14.105) °v1+6 asinEquation 14.33. 1/Vhen theobserver justpasses thesource (0='rr/2), 11'=% observer passing source (14.106) \/1B Wecanalsotreat thecase where theobserver isatrestandthesource is moving (see Problem 14-18). ‘Westillobtain Equations 14.104—14.106 because, according totheprinciple ofrelativity, itisnotpossible todistinguish between themotion oftheobserver andthemotion ofthesource. 578 14/SPECIAL THEORY OFRELATIVITY 14.10 Lagrangian Function inSpecial Relativity Lagrangian andHamiltonian dynamics (discussed inChapter 7)must beadjusted inlight ofthenew concepts presented here. Wecanextend theLagrangian for- malism into therealm ofspecial relativity inthefollowing way. Forasingle (non- relativistic) particle moving inavelocity-independent potential, therectangular momentum components (see Equation 7.150) may bewritten as 8Lpi= (14.107) According toEquation 14.87, therelativistic expression fortheordinary (i.e., space) momentum component is p——-1 (14108) 7 /fl_B2 ' Wenow require that therelativistic Lagrangian, when differentiated with respect tou,-asinEquation 14.107, yield themomentum components given byEquation 14.108: at .--=-1 (i4.109)Bu.\/1-62 This requirement involves only thevelocity oftheparticle, soweexpect that the velocity-independent part oftherelativistic Lagrangian isunchanged from the nonrelativistic case. Thevelocity-dependent part, however, may nolonger beequal tothekinetic energy. Wetherefore write L=T*—U (14.110) where U=U(x,-) and T*=T*(u,-). The function T*must satisfy therelation g—i (14111)8u,- \/1_B2 ' Itcanbeeasily veriﬁed that asuitable expression forT*(apart from apossible constant ofintegration that canbesuppressed) is T*=—mc2V1 —B2 (14.112) Hence, therelativistic Lagrangian canbewritten as 1L=—mc2\/1 -62-Ul (i4.ii3) andtheequations ofmotion areobtained inthestandard wayfrom Lagrange’s equations. ' Notice thattheLagrangian isnotgiven byT—U,because therelativistic ex- pression forthekinetic energy (Equation 14.58) is =i —mc . T W2 2 (14114)\/1-62 14.11 RELATIVISTIC KINEMATICS 579 The Hamiltonian (seeEquation 7.153) canbecalculated from =2L2C22+L2+ U‘V7725 '7 where wehave used Equations 14.108 and14.113 andchanged V1—B2to')/'1. Thus, 22 2 1 H=Lg+%+U=—2(p2c2+m2c4)+U ymc ymc E2 =fa,+U =E+U= T+U+E0 (14.115) The relativistic Hamiltonian isequal tothetotal energy deﬁned inSection 14.8 plus thepotential energy. Itdiffers from thetotal energy used previously in Chapter 7bynow including therestenergy. 14.11 Relativistic Kinematics Intheevent thatthevelocities inacollision process arenotnegligible with respect tothevelocity oflight, itbecomes necessary touserelativistic kinematics. Inthedis- cussion inChapter 9,wetook advantage oftheproperties ofthecenter-of-mass co- ordinate system inderiving many ofthekinematic relations. Because mass anden- ergy areinterrelated inrelativity theory, itnolonger ismeaningful tospeak ofa “center-of-mass” system; inrelativistic kinematics, oneusesa“center-of-momentum” coordinate system instead. Such asystem possesses thesame essential property asthe previously used center-of-mass system—the total linear momentum inthesystem is zero. Therefore, ifaparticle ofmass mlcollides elastically with aparticle ofmass 711.), then inthecenter-of-momentum system wehave pi=pé (14.116) Using Equation 14.87 ,thespace components ofthemomentum four-vector can bewritten as m1u{'y{ =m2u§'y§ (14.117) where, asbefore, 'yE1/V1—B2andBEu/c. Inacollision problem, itisconvenient toassociate thelaboratory coordi- nate system with theinertial system Kand thecenter-of-momentum system with K’(see Figure 14-9). Asimple Lorentz transformation then connects the twosystems. Toderive therelativistic kinematic expressions, theprocedure is 580 14/SPECIAL THEORY OFRELATIVITY System K System K' Laboratory System Center-of-Momentum System "*1 "1 '2? ml ui uémi’ } i — =0 V ‘12 (a)Initial condition (b)Initial condition ml "1I "1 6 9 v2 ———————--——--—--——-—————— vé "I2 (c)Final condition (d)Final condition FIGURE 14-9 Theelastic collision schematic ofFigure 9-10 isredisplayed with systems KandK’indicated. toobtain thecenter-of-momentum relations and then perform aLorentz transformation back tothelaboratory system. Wechoose thecoordinate axes sothat mlmoves along thex-axis inKwith speed M1.Because m2isinitially at restinK,u2=0.InK’,m2moves with speed uéandsoK’moves with respect to Kalsowith speed uéandinthesame direction astheinitial motion ofml. Using thefact that By=\/'y2—1,wehave I_ I I_ I I P1—m1"1')'1 —"$155171 =m1¢\/vii-1=m2¢\/752-1 =pg (14.11s) which expresses the equality ofthemomenta inthe center-of-momentum system. According toEquation 14.92, thetransfonnation ofthemomentum pl (from Kto K’)is I 'I.L 1»;=(1>1—7?E1)v§ 04.119) Wealsohave Pi=m1"1')’1 E1="$15271} (I4.l20) 14.11 RELATIVISTIC KINEMATICS 581 soEquation 14.118 canbeused toobtain "L16V'Yi2_1=(m1¢B1')'1 _i3§m1C'Y1)'Y§ =m1¢(vé\/vi —1-v1\/"152 -1) =m2c\/'y§2 —1 (14.121) These equations canbesolved foryiandyéintenns of'y1: ml 71+E 'y{= *ml ml2 (l4.l22a) 1+2')/{Z + 2 2 '71+E,_ rml3/2—1+2T2+E2 (14.122b) ‘Y1ml ml Next, wewrite theequations ofthetransformation ofthemomentum components from K’back toKafter thescattering. Wenow have both x-andy-components: ,"5,,pl,x = P1,): +?E1 y? =(m1@Bi'Yi C059 +7711555’)/i)')’§ ="#167/i7'é(Bi COS9+B5) (14-1233) (Note that, because thetransformation isfrom K’toK,aplus sign occurs before thesecond term, incontrast toEquation 14.119.) Also, p1_,=m1cB{'y{ sin0 (14.123b) The tangent ofthelaboratory scattering angle ti:isgiven bypl’),/pm; therefore, dividing Equation 14.123b byEquation 14.123a, weobtain ta¢ 1 sinB n=——-i-~*— YéC089 +(B5/Bi) Using Equation 14.117 toexpress gym, theresult is ta.111 Sine (14124) n = r r 1 ' 72C059 +(m1'Y1/m2'Y2) Fortherecoil particle, wehave I I I P2,»=P2,);+?E2 72 =(—m2¢Bévé¢0s9 +m»z¢Bévé>vé =m2cB§'y§2(1 —cos9) (l4.l25a) 582 14/SPECIAL THEORY orRELATIVITY where aminus sign occurs intheﬁrst term because p§_,,isdirected opposite to P1.»A150, pm=—m2cB§'y§ sin0 (l4.l25b) Asbefore, thetangent ofthelaboratory recoil angle §isgiven bypl’,/pm: 1 '0tang=-—,-i‘3-- (14426)'y21—cos6 The overall minus sign indicates that ifmlisscattered toward positive values of mp,then m2recoils inthenegative §-direction. Acase ofspecial interest isthat inwhich ml=mg.From Equations 14.122, weﬁnd ,_,_1+71 __Y1 — Y2 — T, ml — "Z2 The tangents ofthescattering angles become 2 sin0=,/ 14.1 8tanw 1+yl 1+cos0 ( 2) i9 tan =— 7 'i—:S-P;-)'s-6 (14.1.29) 1 The product istherefore tant/1tan§=—i, ml=I"/2 (14.1.30)1+'yl (The minus sign isofnoessential importance; itonly indicates thatalland4”are measured inopposite directions.) Wepreviously found thatinthenonrelativistic limit there wasalways aright angle between theﬁnal velocity vectors inthescattering ofparticles ofequal mass. Indeed, inthelimit yl—>1,Equations 14.128 and14.129 become equal to Equations 9.69 and 9.73, respectively, and so1/1+{=1r/2. Equation 14.130, however, shows thatintherelativistic case ab+Z<11'/2; thus, theincluded angle inthescattering isalways smaller than inthenonrelativistic limit. Forequal scat- tering andrecoil angles (¢=4"),Equation 14.130 becomes 2 1/2 W11/1= E , m1=m2 andtheincluded angle between thedirections ofthescattered andrecoil parti- cles is ¢>=¢+Z=2¢ 2 1/2 =2tan_1 i , ml=m2 (I4.I3l)1+ 71 PROBLEMS 583 90°- m1=”‘260°- Included - Scattering Angle, ¢ ~ 20°- ()0 I J _.I 1. I1 .I 1 5 10 15 20 Y1_’ FIGURE 14-10 Theincluded scattering angle, qb=l/I+4’,isshown asafunction of therelativistic parameter ylforml=m2.Fornonrelativistic scattering ("yl=1),thisangle isalways 90°. Figure 14-10 shows 4:asafunction ofyluptoyl=20.Atyl=10,theincluded angle isapproximately 46°.This value ofylcorresponds toaninitial velocity that is99.5% ofthevelocity oflight. According toEquation 14.58, thekinetic energy isgiven byTl=mlc2(yl —1);therefore, aproton with yl=10would Ihave a kinetic energy ofapproximately 8.4GeV, whereas anelectron with thesame ve- locity would have TlE4.6MeV.* Byusing thetransformation properties ofthefourth component ofthemo- mentum four-vector (i.e., thetotal energy) ,itispossible toobtain therelativistic analogs ofalltheenergy equations wehave previously derived inthenonrela- tivistic limit. PROBLEMS 14-1. Prove Equation 14.13 byusing Equations l4.9—l 4.12. 14-2. Show that the transformation equations connecting the K’and Ksystems (Equations 14.14) canbeexpressed as xl=xlcosh a—ctsinha I__ I_ x2_x2» xa_xs xl.t’=tcosha ——sinhac where tanh ac=v/c.Show thattheLorentz transformation corresponds toarota- tionthrough anangle iainfour-dimensional space. *These units ofenergy aredeﬁned inProblem 14-39: 1GeV =103MeV =109eV=1.602 X lO'3erg =1.602 XlO'1°_]. 584 14-3. 14-4. 14-5. 14-6. 14-7. 14-8. 14-9. 14-10. 14-11 14-12 14-13.14/SPECIAL THEORY OFRELATIVITY Show thattheequation 1621!’ V2‘?-GT=0 c6t isinvariant under aLorentz transformation butnotunder aGalilean transforma- tion. (This isthewave equation that describes thepropagation oflight waves in freespace.) Show thattheexpression fortheFitzGerald-Lorentz contraction (Equation 14.19) canalsobeobtained iftheobserver intheK’system measures thetime necessary fortherodtopass aﬁxed point inthatsystem andthen multiplies theresult byv. What istheapparent shape ofacube moving with auniform velocity directly toward oraway from anobserver? Consider twoevents thattakeplace atdifferent points intheKsystem atthesame in- stant t.Ifthese twopoints areseparated byadistance Ax,show thatintheK’system theevents arenotsimultaneous butareseparated byatime interval At’=-v'yAx/c2. Two clocks located attheorigins oftheKand K’systems (which have arelative speed v)aresynchronized when theorigins coincide. After atime t,anobserver at theorigin oftheKsystem observes theK’clock bymeans ofatelescope. 1/Vhat does theK’clock read? Inhis1905 paper (seethetranslation inL023), Einstein states: “Weconclude that abalance-clock attheequator must gomore slowly, byavery small amount, than a precisely similar clock situated atoneofthepoles under otherwise identical con- ditions.” Neglect thefactthat theequator clock does notundergo uniform mo- tionandshow thatafter acentury theclocks willdiffer byapproximately 0.0038 s. Consider arelativistic rocket whose velocity with respect toacertain inertial frame isvandwhose exhaust gases areemitted with aconstant velocity Vwith respect to therocket. Show thattheequation ofmotion is dv dm—+V-— —2=mdt dt(1 B) 0 where m=m(t) isthemass oftherocket initsrestframe andB=v/c. Show byalgebraic methods thatEquations 14.15 follow from Equations 14.14. Astick oflength lisfixed atanangle 0from itsxl-axis initsown restsystem K. What isthelength andorientation ofthestick asmeasured byanobserver moving along xlwith speed v? Aracer attempting tobreak theland speed record rockets bytwomarkers spaced 100mapart ontheground inatime of0.4itsasmeasured byanobserver onthe ground. How farapart dothetwomarkers appear totheracer? What elapsed time does theracer measure? What speeds dotheracer andground observer measure? Amuon ismoving with speed v=0.9990 vertically down through theatmosphere. Ifitshalf-life initsown restframe is1.5].LS,what isitshalf-life asmeasured byan observer onEarth? PROBLEMS 585 14-14. Show thatEquation 14.31 isvalid when areceiver approaches aﬁxed light source with speed v. 14-15. Astarisknown tobemoving away from Earth ataspeed of4><10‘m/s. This speed isdetermined bymeasuring theshift oftheHaline()1=656.3 nm). Byhow much andinwhat direction istheshift ofthewavelength oftheHaline? 14-16. Aphoton isemitted atanangle 6'byastar(system K’)andthen received atan angle 6onEarth (system K).The angles aremeasured from alinebetween the starandEarth. Thestarisreceding atspeed vwith respect toEarth. Find therela- tionbetween 0and6';thiseffect iscalled theaberration oflight. 14-17. Thewavelength ofaspectral linemeasured tobeAonEarth isfound toincrease by50% onafardistant galaxy. VVhat isthespeed ofthegalaxy relative toEarth? 14-18. Solve Example 14.11 forthecase oftheobserver atrestandthesource moving. Show thattheresults arethesame asthose given inExample 14.11. 14-19. Equation 14.34 indicates thatared(blue) shift occurs when asource andobserver arereceding (approaching) with respect tooneanother inpurely radial motion (i.e., B=I3,).Show that, ifthere isalso arelative tangential speed B1,Equation 14.34 becomes 7\o_v__\/__1-B?-B? AI/0 1-is, . andthatthecondition foralways having aredshift (i.e., noblue shift), A>A0or V< I/0,1S* Bi>2I3r(1- I3.) 14-20. Anastronaut travels tothenearest starsystem, 4light years away, andreturns at speed 0.30. How much hastheastronaut aged relative tothose people remaining onEarth? 14-21. Theexpression fortheordinary force is F_it_ﬂ_dt\/1 _B2 Take utobeinthexl-direction andcompute theComponents oftheforce. Show that F1=mlih, F2=mcua» F3=mails where mlandm,are,respectively, thelongitudinal mass andthetransverse mass: _ m _ m ml_(1_ B2)3/2, ml_y/1_ B2 *See_]._]. Dykla, Am. Phys. 47,381(1979). 586 14-22. 14-23 14-24 14-25. 14-26. 14-27 14-28 14-29. 14-30. 14-31 14-32 14-33.14/SPECIAL THEORY OFRELATIVITY The average rate atwhich solar radiant energy reaches Earth isapproximately 1.4><105W/m2. Assume that allthisenergy results from theconversion ofmass toenergy. Calculate therate atwhich thesolar mass isbeing lost. Ifthisrate is maintained, calculate theremaining lifetime oftheSun. (Pertinent numerical data canbefound inTable 8-1.) Show that themomentum and thekinetic energy ofaparticle arerelated by 11202 =2Tmc2 +T2. VVhat istheminimum proton energy needed inanaccelerator toproduce an- tiprotons I2bythereaction P+P—>l>+P+(P+T>) Themass ofaproton andantiproton ismp. Aparticle ofmass m,kinetic energy T,andcharge qismoving perpendicular toa magnetic ﬁeld Basinacyclotron. Find therelation fortheradius roftheparti- cle’s path interms ofm,T,q,andB. Show thatanisolated photon cannot beconverted intoanelectron-positron pair, y—>e_+e*.(The conservation lawsallow thistohappen only near another object.) Electrons and ositrons collide from oosite directions head-on with eualener- P PP q giesinastorage ring toproduce protons bythereaction e_+e*—>p+I» ' The restenergy ofaproton andantiproton is938MeV. VVhat istheminimum ki- netic energy foreach particle toproduce thisreaction? Calculate therange ofspeeds foraparticle ofmass minwhich theclassical rela- tionforkinetic energy, %mv2, iswithin onepercent ofthecorrect relativistic value. Find thevalues foranelectron andaproton. The 2-mile long Stanford Linear Accelerator accelerates electrons to50GeV (50X109eV). VVhat isthespeed oftheelectrons attheend? Afree neutron isunstable anddecays into aproton andanelectron. How much energy other than therestenergies oftheproton andelectron isavailable ifaneu- tron atrestdecays? (This isanexample ofnuclear beta decay. Another particle, called aneutrino—actually anantineutrino 5isalsoproduced.) Aneutral pion 1r°moving atspeed v=0.980 decays inﬂight into twophotons. If thetwophotons emerge oneach side ofthepion’s direction with equal angles 0, ﬁnd theangle 6andenergies ofthephotons. The restenergy of1r°is135MeV. Innuclear andparticle physics, momentum isusually quoted inMeV/c tofacili- tatecalculations. Calculate thekinetic energy ofanelectron andproton ifeach hasamomentum of1000 MeV/c. Aneutron (mn=939.6 MeV/c2) atrestdecays intoaproton (mp=938.3 MeV/02), anelectron (me=0.5MeV/02), andanantineutrino (ml~0).Thethree particles PROBLEMS 587 14-34. 14-35 14-36 14-37.emerge atsymmetrical angles inaplane, 120° apart. Find themomentum andki- netic energy ofeach particle. Show that As2isinvariant inallinertial systems moving atrelative velocities to each other. Aspacecraft passes Saturn with aspeed of0.90 relative toSaturn. Asecond spacecraft isobserved topass theﬁrst one (going inthesame direction) atrel- ative speed of0.20. What isthespeed ofthesecond spacecraft relative to Saturn? Wedeﬁne thefour-vector force IF(called theMinkowski force) bydifferentiating thefour-vector momentum with respect toproper time. d|P’IF=—d1" Show thatthefour-vector force transformation is F1’="Y(F1+iﬂli) Fé=1% Fé=1% F4=Y(1‘Ti-iBF1) Consider aone-dimensional, relativistic harmonic oscillator forwhich the Lagrangian is L=m¢2(1 —\/1—52)—gm Obtain theLagrange equation ofmotion andshow that itcanbeintegrated to yield 1 E= mc2+§ka2 where aisthemaximum excursion from equilibrium oftheoscillating particle. Show thattheperiod Xza 1'=4] dt x==0 canbeexpressed as 21/214.22 2 ,=_q _L<2_i,,,,,K60 \/1‘i'K2COS2¢ Expand theintegrand inpowers ofKE(a/2) Vk/mc2 and show that, toﬁrst order inK, ___ 1+3ka21'= ———-T0 16mc2 where 1'0isthenonrelativistic period forsmall oscillations, 2'rrVm/k. 588 14/SPECIAL THEORY OFRELATIVITY 14-38. Show thattherelativistic form ofNewton’s Second Lawbecomes 14-39 14-40. 14-41 14-42.du -3/2 F= —1-— MFlAcommon unit ofenergy used inatomic andnuclear physics istheelectron volt (eV), theenergy acquired byanelectron infalling through apotential difference ofonevolt: 1MeV =106eV=1.602 X10"” Inthese units, themass ofanelec- tron ismec2 =0.511 MeV andthatofaproton ismpc2 =938MeV. Calculate the kinetic energy andthequantities Band'yforanelectron andforaproton each having amomentum of100MeV/e. Show thattheelectron is“relativistic” whereas theproton is“nonrelativistic.” Consider aninertial frame Kthatcontains anumber ofparticles with masses ma, ordinary momentum components pm]-,andtotal energies Ea.The center-of-mass system ofsuch agroup ofparticles isdeﬁned tobethatsystem inwhich thenetor- dinary momentum iszero. Show that thevelocity components ofthecenter-of- mass system with respect toKaregiven by vi_§1’~-1‘c 2:5“ Show thattherelativistic expression forthekinetic energy ofaparticle scattered through anangle 1/1byatarget particle ofequal mass is T1_ 2cos21l: It‘(v1+1)—<v1—1>cos2-11 The expression evidently reduces toEquation 9.89a inthenonrelativistic limit -yl—>1.Sketch T1(:,l/) forneutron-proton scattering forincident neutron energies of100MeV, 1GeV, and10GeV. The energy ofalight quantum (orphoton) isexpressed byE=hv,where his Planck’s constant and visthefrequency ofthephoton. The momentum ofthe photon ishv/e. Show that, ifthephoton scatters from afreeelectron (ofmass me), thescattered photon hasanenergy -1EE’=E|;1 +——?(1 —cos0):l me where 6istheangle through which thephoton scatters. Show alsothat theelec- tron acquires akinetic energy T_gE? 1—cos0 02 Eme 1+i(1—cos0)mec2 “Better istheendofathing than thebeginning thereof”—-Ecclesiastes APPENDIX Ta)»l01"’s Theorem Atheorem ofconsiderable importance inmathematical physics isTaylor’s theo- rem,* which relates totheexpansion ofanarbitrary function inapower series. Inmany instances, itisnecessary tousethistheorem tosimplify aproblem toa tractable form. Consider afunction f(x) with continuous derivatives ofallorders within a certain interval oftheindependent variable x.Ifthisimenial includes .1},5 xSx0+h,wemaywrite x0+h IE I f'(-"7)dx =f(-Y0 +h)—f(x0) (A-1)*0 where f'(x) isthederivative off(x) with respect tox.Ifwemake thechange of variable x= x(,+ h—t (A.2) wehave 1- I=ff'(x0 +h—t)dt (A.3) 0 Integrating byparts I=tf'(x0 +h-1) +tf”(x(, +/1-‘mu =hf'(x(,) +Ftf"(x(, +h—t)dt (A.4) 0 "‘First published in1715 bytheEnglish mathematician Brook Taylor (I685-I731). 589 590 A/TAYLOR’S THEOREM Integrating thesecond term byparts, weﬁnd 2II 1=hf’(x(,) +i2f"(x(,) +It—f'”(x0 +h-Mt (A.5)2! 02! Continuing thisprocess, wegenerate aninﬁnite series forI.From thedeﬁnition ofI,wethen have f(x0 ‘l’h)=f(x0) +hfI(x0) +gf"(x0) '1' (A6) This istheTaylor series expansion* ofthefunction f(x0 +h).Amore common form oftheseries results ifwesetx0=0andh=x[i.e., thefunctionf(x) isex- panded about theorigin]: f(x) +xf1(0) +§j-11(0) +gf///(0) + +-’3§Tnf(n)(0) + where /<"><0>-%/<x> (A-8)x0 X = Equation A.7isusually called theMaclaurin’s seriesl forthefunction f(x). Theseries expansions given inEquations A.6andA.7possess twoimportant properties. Under very general conditions, they may bediﬁerentiated orinte- grated term byterm, andtheresulting series converge tothederivative orinte- graloftheoriginal function. 1‘:X:'\l\11)1.1‘: A.l - Find theTaylor series expansion ofe". Solution. Because thederivative ofexp(x) ofanyorder isjustexp(x),theexpo- nential series is - 2 3 ex;-1+x+£+£+... (A9) 2! 3! This result isofconsiderable importance andwillbeused often. *The remainder term ofaseries that isterminated after aﬁnite number ofterms isdiscussed, for example, byKaplan (Ka84). 1'Discovered by_]ames Stirling in1717 andpublished byColin Maclaurin in1742. 592 A/TAYLOR'S THEOREM Taylor's series canbeused torestructure afunction aswellastoapproximate it. Forsome applications, such arestructuring maybemore useful towork with. We may, forexample, want toexpand thepolynomialf(x) =4+6x+3x2+2x3+x" about x=2rather than x=O. Solution. First, wecompute thevarious derivatives andevaluate them atx=2: f(2) =60 f'(2) =(6+6x+6x2+4x3)|,,=2 =74 f"(2) =(6+12x+l2x2)|,,=2 =78 f"'(2) =(12+24x)l,=2 =60 fl"(2) =24 f"<2>=0 Using Equation A.6with h=(x—2) f(x) =60+74(x —2)+39(x —2)2+10(x —2)3+(x—2)‘ (A.l5) There areagreat many important integrals arising inphysics thatcannot bein- tegrated inclosed form, that is,interms ofelementary functions (polynomials, exponentials, logarithms, trigonometric functions, andtheir inverses). Integrals with integrands e"", %, xtanx, sinxi’, 1/lnx, (sinx)/x, or 1/\/I-14" areafewsuch examples. Nevertheless, thevalues oftheintegrals orgood ap- proximations oftheir values areneeded. ATaylor series expansion ofallorpart oftheintegrand followed byaterm-by-term integration oftheresulting series produces ananswer asprecise asiswished. Asanexample, solve thefollowing integral: X e‘I—dt (A.l6)1t Solution. Using Equation A.9, 1’131+z+—+—+ dz~81 2:3: 5£?dt= ll t (A.l7) A/TAYLOR'S THEOREM 591 Find theTaylor series expansion ofsinx. Solution. Toexpand f(x) =sinx,weneed f(x) f'(X)fl! Z fill Z Therefore,sinx, f(0) =0 cosx, f'(O) =l —sin x, f"(O) =0 —cosx, f"'(O) =—-l xii x5 sinx =x—§+§— | (A,1()) Similarly, ‘Z 4_ x xcosx—l—;+;— (A.ll) l~'.X.»\.'\ll’l.li .-\.I§ UsetheTaylor series expansion of(1+t)"tointegrate I‘dz ()l+l Solution. Aseries expansion canoften beproﬁtably used intheevaluation ofa deﬁnite integral. (This isparticularly trueforthose cases inwhich theindeﬁnite integral cannot befound inclosed form.) X X J——t= LU—12+z3— ---)dt, lzl<101+ lntegrating term byterm, weﬁnd lo1"l" Because“dt 2“__j=x_%+%_.H (AH) d l IclIl(l +x)—1-T,‘ (A.13) Wealsohave theresult tZ *3 mu+o=x-%+%-~- (Am PROBLEMS 593 Xdt X It Xtﬂ _L7+ Ldz+ L§dz+ L§!dt+ =lnx—(x—1)+l(x2— l)+—1-(x"—1)+ (A.18)4 18 PROBLEMS A-l. Show bydivision andbydirect expansion inaTaylor series that 1 1-——=1+x+x2+x5+ +x"+—-x Forwhat range ofxistheseries valid? A-2. Expand cosx about thepoint x=17/4. A-3. Useaseries expansion toshow that l x__ _X J5-—‘—ax= 21145....0 X A-4. UseaTaylor series toexpand sin” x.Verify theresult byexpanding theintegral in therelation __, I‘dzsm x= i—-o\/1—2’ A-5. Evaluate tothree decimal places: r Jexp(—x’/2)dx0 Compare theresult with thatdetermined from tables oftheprobability integral. A-6. Show thatiff(x) =(1+x)"(with |x|<1)isexpanded inaTaylor series, there- sultisthesame asabinomial expansion. APPENDIX Elliptic Integrals There isalarge andimportant class ofintegrals called elliptic integrals thatcan- notbeevaluated inclosed form interms ofelementary functions. Elliptic inte- grals occur inmany physical situations; forexample, seetheexact solution tothe plane pendulum inSection 4.4.Any integral oftheform I(asin9+bcos0+0):‘/2 d0, or JR(x,\/yldx (B.1) where Risarational function, y=ax‘+bxs+ax?+dx+e,with distinct linear factors anda,b,c,d,andeconstants (with notboth a,bzero) isanelliptic inte- gral. Itiscustomary, however, totransform allelliptic integrals intooneormore ofthree standard forms. These standard forms have been much studied andtab- ulated. Several handbooks areavailable with tables ofvalues forthem* B.l Elliptic Integrals oftheFirst Kind F(k,d>) =l¢——-19-—, r2<1 (B.2a)0 2\/l—k2sin 9 orifz=sin0 — " dzF(k, x)=Ii-—~——i, k2<1 (B.2b)0\/(1— z2)(1— 1812) *One ofthebestofthese isAbramowitz andStegun (Ab65). Seealsoextensive numerical tables in Adams andHippisley (Ad22) andshort tables inDwight (Dw6l). 594 B/ELLIPTIC INTEGRALS 595 B.2 Elliptic Integrals oftheSecond Kind E(k,d>)=EV1 —k2sin20d6, k2<1 (B.3a) orifz=sin0 — "1—k2zE(k,x)= i—-dz, k2<1 (B.3b)01—22 B.3 Elliptic Integrals oftheThird Kind 4’ d0 “‘""""” 'l<)(1+ Wm ‘M’ orifz=sin0 ﬁ(n,rt,x)=ix dz (B.4b)<>(1+m=’)\/(1 -z2)(l-1&2) These standard forms obey thefollowing identities, which areoften helpful: Fck.¢)=F(k.-tr)—F(k,1r—¢)} (B5)Eck.¢)=Err.Tr)—Eu.w—¢) ' and F(k,m1r+¢)=mF(k,1r) +F(k,¢)} (B6) E(k,mvr+¢)=mE(k. Tr)+E(k.¢) ' where misaninteger. Iftables arenothandy orif4:orxisneeded asavariable, thestandard inte- grals maybeapproximated byexpanding theintegrand inaninﬁnite series and integrating term byterm. Forexample, consider 45 E(k,¢)= J\/1- k2sin20d00 Using thebinomial theorem ontheintegrand , . 1 , 1 _(I—k2srn20)'” =1—§k2s1n20 —gk‘s1n“0 — 5S6 B/ELLIPTIC INTEGRALS so lb 12-2 14-4E(k,¢)= 01-5): s1n0—§k srn0—"~ 1-3-5---<21)-3) n_2"2_4_6___(2n) r2Sln0 jlao _we.5-5..6»;,1re4*2 2»-3)=-— -ode---~-———-——12~4’2lo5”‘ <2") ¢ XLsin2"0d0- (B.7) Similarly, thebinomial theorem canbeused toexpand (1—k2sin?0)"/2 to yield 1¢ 3"’F(k,¢)=¢-+-§k2‘[ sin20d0+§k“J sin“0d0+ 0 0 —-——-—— 12sin?"0d0+ (11.8)+I-3-5-"(2n— 1)"It 2-4-6---(2n) 0 ¢ Puttheintegral I2V1—k2sin20 d0intostandard form. ¢r b Solution. Recall from calculus thatforanyintegral Jf(x)dxitispossible to write “ b c b lf(x)dx= Jf(x)dx+ Jf(x)dx so 4': 0 4'1 J\/1- ksin20d0 =I\/1- k2sin20d0 -l-J VI—k2sin20d0 451 ¢r 0 Butthere isanother property ofintegrals: b a jj(x)dx =—Lf(x)dx so 4': 4': ¢r J\/l—k2sin20d0=J \/l—k2sin20d0—J \/l—k2sin20d0¢, 0 0 B/ELLIPTIC INTEGRALS 597 OI‘ 4': IV1—k2sin2 0d0=E(k,¢2)—E(k,(bl) (B.9) ¢r The terms ontheright canbelooked upinahandbook. Transform theelliptic integral ¢ I-—-—-ig——-— where n2>I0 2 2\/1—nsin0' intoastandard form. Solution. Toreduce thisintegral tostandard form, theradical must betrans- formed toV1—Ir’sin’0,with k2<1.Todothis, consider thetransformation nsin0=sinB.Differentiating, wehave ncos0:10=cosBdB so _cosBdBd6-—--ncos0 Using theidentity sin?0+cos?0=1leads to cos0= V1—sin20=,/1_<§lll£)2 n Also, cosB=V1—sin?B,and \/l—n2sin?0=\/1—sin?B.Hence thein- tegral becomes Id: _Jsin"(n sin45) -\/1_singﬁ dﬁ 0\/1—n2sin20 0 n/1_(sinB)2( /1_singﬁ Tl sin'l(n sin -1 ¢) dB -tl.y——*—,—""1—(;)sin"’ B 598 B/ELLIPTIC INTEGRALS SO ¢ d 1srn“(n srn¢) d 1--—2——— =-1 —--—--5-——— (11.10)<>V1—n2sin20 "0 1 1_2 I—;l;)s1n B 2where 1/n <1.Theintegral ontheright isnow instandard form. liX.r\Y\ll'l.li Bil Transform theelliptic integral 4'.10 Vcos20 Lintoastandard form. Solution. Let11. =sin0;then d11.=cos0d6.Because cos2 0+sin20=1, cos0=Vl—sin20=VI—11.2,sod0=d11/\/1 —11.2.Byanother trigono- metric identity, cos20=1—2sin?0=1—211.2.Thus Vcos29=VI—2112, and jib jsllltll dp 0Vcos20— 0VI—11.2Vl —211.2 Letz= \/211.;thendz= \/2d11.,so \/2sin dz =— (11.11)Fd0 1I .11 0Vcos 20 \/20 \/(1- z2)(] -$9) Theintegral ontheright isinstandard form. PROBLEMS B-I. Evaluate thefollowing integrals using asetoftables. (a)F(0.27, 1r/3) (b)E(0.27, 1r/3) (c)F(0.27, 71r/4) (d)E(0.27, 71r/4) B-2. Reduce tostandard fonn: 1r/6 3/4 __(19 25 4 4 b Jig 11..(alL\/1—4511120 (lI-1/-1 1"12 B-3. Find thebinomial expansion of(1—k2sin?0)“/2 andthen derive Equation B.8. APPENDIX Ordinary Differential Equations of Second Order* C.1 Linear Homogeneous Equations Byfar,themost important typeofordinary differential equation encountered in problems inmathematical physics isthesecond-order linear equation with con- stant coeﬁicients. Equations ofthistype have theform diy dy _E +GI‘ +by— (C-18) or,denoting derivatives byprimes, y"+ay'+by=f(x) (C.1b) Aparticularly important class ofsuch equations arethose forwhich f(x)=0. These equations (called homogeneous equations) areimportant notonly in themselves butalsoasreduced equations inthesolution ofthemore general type ofequation (Equation C.l). Weconsider thelinear homogeneous second-order equation with constant coefﬁcients ﬁrstf yr! +ayl +by: O ‘Astandard treatise ondifferential equations isthatofInce (In27). Alisting ofmany types ofequa- tions andtheir solutions isgiven byMurphy (Mu60). Amodem viewpoint iscontained inthebook byHochstadt (H064). 1'The ﬁrstpublished solution ofanequation ofthistypewasbyEuler in1743, butthesolution appears tohave been known toDaniel andjohann Bemoulli in1739. 599 600 C/ORDINARY DIFFERENTIAL EQUATIONS OFSECOND ORDER These equations have thefollowing important properties: a.Ify,(x)isasolution ofEquation C.2,then cly,(x)isalsoasolution. b.Ifyl(x)andy2(x) aresolutions, then y|(x) +y2(x) isalsoasolution (principle ofsuperposition). c.Ify,(x)andy2(x) arelinearly independent solutions, then thegeneral solution totheequation isgiven byc|y|(x) +c2y2(x). (The general solution always contains twoarbitrary constants.) The functions y,(x) andy2(x) arelinearly independent ifand only ifthe equation /\yr(x) +/1y2(x) E0 (C-3) issatisﬁed only by)1=11.=0.IfEquation C.3canbesatisﬁed with )1and11.dif- ferent from zero, then y,(x)andy2(x) aresaidtobelinearly dependent. The general condition (i.e., thenecessary andsufﬁcient condition) that a setoffunctions yl,yg,ya,...belinearly dependent isthattheWronskian determi- nant ofthese functions vanish identically: yr 12 >2 '1.. Jli 1% 15 -11. W=1’; 1'1» 1'5 '111=0 (C-4) y(n—l) ygn—1) ygn—l) y(ln—l) where y("listhenthderivative ofywith respect tox. The properties (a)and (b)above canbeveriﬁed bydirection substitution, but(c)isonly asserted here toyield thegeneral solution. These properties apply onlytothehomogeneous equation (Equation C.2) andnottothegeneral equa- tion (Equation C.l). Equations ofthetype C.2arereducible through thesubstitution y=e"‘ (C.5) Now y’=re", y"=r2e"‘ (C.6) Using these expressions fory'andy"inEquation C.2,weﬁndanalgebraic equa- tioncalled theauxiliary equation: r2+ar+b=0 (C.7) Thesolution ofthisquadratic inris r=—gi%\/a2—4b (cs) Weﬁrstassume that thetworoots, denoted byr1and r2,arenotidentical and write thesolution as y=e""+6"“ (C.9) C/ORDINARY DIFFERENTIAL EQUATIONS OFSECOND ORDER 601 Because theWronskian determinant ofexp(r1x) andexp(r2x) does notvanish, these functions arelinearly independent. Thus, thegeneral solution is ly=618"’+¢2@""» T14‘T2 (C-10) Ifithappens that rl=1'2=r,then itcanbeveriﬁed bydirect substitution thatxexp(rx) isalsoasolution, andbecause exp(rx) andxexp(rx) arelinearly independent, thegeneral solution foridentical roots isgiven by l7=61¢"+¢~zx@"‘» T1=T2ET (C-11) Solve theequation y"—2y’—3y=O (C.l2) Solution. The auxiliary equation is r2-2r—3=(r—3)(r+1)=O (C.13) Theroots are 1|=3,1'2=—1 (C.14) Thegeneral solution istherefore y=ales‘ +c2e"‘ (C.15) Solve theequation y"+4y’+4y=0 (C.l6) Solution. Theauxiliary equation is r2+4r+4=(r+2)2=0 (C.l7) Theroots areequal, arer=-2.The general solution istherefore y=c|e‘2" +c2xe'2" (C.18) Iftheroots 1',and12oftheauxiliary equation areimaginary, thesolutions given byclexp(r1x) and02exp(r2x) arestillcorrect. 502 C/ORDINARY DIFFERENTIAL EQUATIONS OFSECOND ORDER Togivethesolutions entirely interms ofrealquantities, weusetheEuler re- lations toexpress theexponentials. Then, e""=e°”‘e‘5" =e°"‘(cos Bx+isinBx) e""=e°”‘e_'5" =e°"‘(cos Bx—isinBx)} (C.l9) andthegeneral solution is y=Clem: +c2er,x =e“"[(c, +C2)cosBx+i(c1—02)sinBx] (C.20) Now clandc2arearbitrary, butthese constants may becomplex. However, not allfour elements canbeindependent (because there would befour arbitrary constants rather than two). The number ofindependent elements canbere- duced totherequired twobymaking cland c2complex conjugates. Then the combinations AEcl+02and BEi(c1—02)become apair ofarbitrary, real constants. Using these quantities inthesolution, wehave y=e°"‘(A cosBx+BsinBx) (C.21) Equation C.21 maybeputintoaform thatissometimes more convenient by multiplying anddividing by11= : y=11e°”‘[(A/11.) cosBx+(B/11) sinBx] (C.22) Next, wedeﬁne anangle 8(seeFigure C-1) such that sin8=A/11, cos8=B/11., tan8=A/B (C.23) Then, thesolution becomes y=11e""( sin8cosBx+cos8sinBx) =11e°”‘ sin(Bx +8) Depending ontheexact deﬁnition ofthephase 8,wemay write thesolution alternatively as y=11.e‘“ sin(Bx +8) (C.24a) y=11e“" cos(Bx +8) (C_24b) X 11=\lA2+B2 A 5 B J? FIGURE C-1 C/ORDINARY DIFFERENTIAL EQUATIONS OFSECOND ORDER 603 Solve theequation y"+2y’+4y=0 (C.25) Solution. Theauxiliary equation is r2+2r+4=O (C.26) with -_\/4-11'= = -1:t\/5 (0.27) Hence, 11=-1,1;=\/5 (c.2s) andthegeneral solution is y=1-*(¢,cos\/51+C2sin\/51) (0.29) or =11e"‘ sin[(\/Bx +8)] (C.30) 9 Summarizing, then, there arethree possible types ofgeneral solutions to homogeneous second-order linear differential equations, asindicated in Table C-1. ABLE C-l T Roots oftheauxiliary equations General solution Real, unequal (r,#=r-1) c,e"" +r2e'*" Real, equal (r,=r-2Ir) ele"+c2xe"‘ Imaginary (ctiiB) e""‘(c| cosBx+c2sinBx) or 110'“ sin(Bx +8) C.2 Linear Inhomogeneous Equations Tosolve thegeneral (i.e., inhomogeneous) second-order linear differential equation, consider thefollowing. Lety=ubethegeneral solution of y"+ay'+by=O (C.3l) andlety=vbe anysolution of y"+ay'+by=f(x) (C.32) 604 C/ORDINARY DIFFERENTIAL EQUATIONS OFSECOND ORDER Then, y=u+visasolution ofEquation C32, because y"+ay'+by=(u"+au'+bu)+(v"+av’+bv) =0+f(x) Because ucontains thetwoarbitrary constants clandC2,thecombinations u+v satisﬁes alltherequirements ofthegeneral solution toEquation C.32. Thefunc- tionuisthecomplementary ftmction andvistheparticular integral oftheequa- tion. Because ageneral method ofﬁnding uhasbeen given above, itonly re- mains toﬁnd, byinspection orbytrial, some function vthatsatisﬁes v”+av’+bv=f(x) (C33) Solve theequation y"+5y’+6y=xi’+2x (C.34) Solution. Theauxiliary equation is r2+5r+6=(r+3)(r+2)=0 (C.35) rl=-3, r2=-2 (C.36) sothecomplementary function is u=c,e‘3" +c2e'2" (C.37) Because theright-hand sideoftheoriginal equation isasecond-degree polyno- mial, weguess aparticular integral oftheform v=Ax? +Bx+C (C.38) Then, v’=2Ax+B (C.39) v"=2A (CAO) Substituting intothedifferential equation, wehave 2A+5(2Ax +B)+6(Ax2 +Bx+C)=x2+2x (C.4l) or (6/l)x2 +(10A +6B)x +(2A+5B+6C)=x2+2x (CA2) Equation coefﬁcients oflikepowers ofx: 6/1=1 10,1+613=2 ((2.43) 2/1+513+6c=0 C/ORDINARY DIFFERENTIAL EQUATIONS OFSECOND ORDER 605 Solving, 1 1 11A=-1 =——, =—i ' 6B 18 C 108 (C44) Hence, 1 1 11 "_6’?+is* 108 18x2 +6x—ll— 108 (CA5) The general solution istherefore I82+ 6—11 y=u+v=c1e‘3" +c2e‘2" + — (CA6) The type ofsolution illustrated inthisexample iscalled themethod of undetermined coefficients. Solve theequation y”+4y=3xcosx (C.47) Solution. Theauxiliary equation is 12+4=(r+2z)(r—2i)=O (CA8) with roots r*=<>+g;} ...... r2=a—2B=0— so oz=0,B=2 (C.50) andthecomplementary function is u=e“"(q cosBx+e2sinBx) =clcos2x+c2sin2x (C.5l) Toﬁndaparticular integral, wenote thatfrom xcosxanditsderivatives itispos- sible togenerate only terms involving thefollowing functions: xcosx,xsinx,cosx,sinx 606 c/ORDINARY DIFFERENTIAL EQUATIONS orSECOND ORDER Therefore, because these functions arelinearly independent, thetrialparticular integral is v=Axcosx+Bxsinx+Ccosx+Dsinx (C.52) v’=A(cos x—xsinx)+B(sin x+xcosx) —Csinx+Dcosx (C.53) v"=—A(2 sinx+xcosx)+B(2cosx—xsinx) —Ccosx—Dsinx (C.54) Substituting intotheoriginal differential equation, (SD—2./l)sin x+(2B+3C)cos x+3(A—l)xcosx+(3B)x sinx=0(C.55) Thecoefﬁcient ofeach term must vanish (because ofthelinear independence oftheterms): SD=2A, 2B=—3C, A=1, SB=0 (C.56) from which A=l, B=0, C=0, D=§ (C.57) Thegeneral solution istherefore 2y=c1sin2x+e2cos2x+xcosx+§sinx (C.58) Iftheright-hand side, f(x),ofthegeneral equation (Equation C.1orC32) is such thatf(x) anditsﬁrsttwoderivatives (only second-order equations arebeing considered) contain only linearly independent functions, then alinear combina- tionofthese functions constitutes thetrialparticular integral. Intheevent thatthe trialfunction contains aterm thatalready appears inthecomplementary func- tion, usetheterm multiplied byx;ifthiscombination alsoappears inthecomple- mentary function, usetheterm multiplied byx2.Nohigher powers areneeded be- cause only second-order equations arebeing considered andonly exp(rx) orx exp(rx) occur assolutions tothereduced equation; (x2)exp(rx) never occurs. PROBLEMS C-I. Solve thefollowing homogeneous second-order equations: (a)y"+2y'—3y=0 (b)y"+y=0 (c)y"—2y'+2y=O (d)y"—2y'+5y=O PROBLEMS 607 C-2. Solve thefollowing inhomogeneous equations bythemethod ofundetermined co- efﬁcients: (a)y"+2y’—8y=16x (b)y"—2y’+y=2e?" (c)y"+y=sinx (d)y"—2y’+y=3xe" (e)y"—4y’+5y=e2”+4sinx C-3. UseaTaylor series expansion toobtain thesolution of yn+y2:__ x2 thatobeys theconditions y(0) =1andy'(0) =0.(Differentiate theequation suc- cessively toobtain thederivatives thatoccur intheTaylor series.) 1\I0?El€I)I)( Useful Formulas * D.1 Binomial Expansion n(n— 1) n(n—.1)(n—2)(1+x)"=1+nx+TH++E .+_ .+_(n)xr+ ...’ Ix‘ <1 r n_ _ n(n—1) _n(n—l)(n—2)(1—x)-1 nx+ T! x2 —-i3%———x3 + +(_1)r<:)xr+ ..., |x| <1 n n!where thebinomial coefficient is <1") (n—r)lr! Some particularly useful cases oftheabove are 1 1 11: I/2=1i_ __2:_3_... (2 2*ax16* 1 1 5 (1iJ6)‘/3 =1i'§x—§x2 ig-1-x3 -‘ "'Anextensive listmaybefound, forexample, inDwight (Dwﬁl). 608(D.l) (D.2) (D.3) (D.4) (D.5) D/USEFUL FORMULAS I 3 (Iix)"/2 =1 T-§x+§x2 1 (1ix)“'/3 =1 T-§x+ Ex? (11- (Ii (1: Forconvergence ofalltheabove series, wemust have |x|<1.x)-‘=1 1x+x2 1 x)‘2=l 12x+Sx2 x)‘3=l 1Sx+ 6x2 D.2 Trigonometric Relations1._5__x3+ . 16 1159+-81 x3+ ;4x3+ -T1Ox3 + sin(A 1B)=sinAcosBicosAsinB cos(A 1B)=cosAcosB1sinAsinB sin2A=2sinAcosA= cos2A=2cos2A—12tanA _2A Isin;=§(1— cosA) A 1 cos2§ =§(l+cosA) 1sin2A=§(l—cos2A) 1 sinsA=;(SsinA—sinSA) 1 sin‘A=§(S—4cos2A+cos4A) 1 cos? A=E111+cos2A) 1 cos3A =z(Scos A+ 1cos“A =§(S+4cos 2A+cos4A)cosSA)609 (D.6 (D.7 (D.8 (D.9 (D.l0 (D.ll (D.l2 (D.13 (D.14 (D.l5 (D.16 (D.17 (D.l8 (D.19 (D.20 (D.2l (D.22 610 1)/USEFUL FORMULAS tan/1+ tnB ““(A+B)= A 1-cosAtan2§=i_ 1+cosA ell’ _ e_lX _m8|: +8-1:: COS X=i 2sinx e”‘=cosx+isinx D.3 Trigonometric Series _ x3 x5 x7s1nx= x——+———+3! 5! 7! x2 x4 xe_. i + ..- 2! 4! 6!cosx=1— tanx=x+£3+—2—x5+--- |xI<1r/23 15 ' sin"x=x+is+—3—x5+--- |xI<16w ’|m*n<wm Cos_.x=z_x_£’_lx5_ |*|<12 6 40 ' 0<cos"x<1r 3 5 7 tan"x=x—%+%—%+---, |x|<1 D.4 Exponential andLogarithmic Series OM8.=';."’=x2 x3 e‘=1+x+—+—+ =I9. 93 3 2 3 4 hﬂ+@=x—%+%—%+"y|ﬂ<1,x=l 1n[\/ (x2/a2) +1+(x/a)] =sinh"x/a =—ln[\/(x2/a2) +1" (X/4)](D.23) (D24) (D25) (D.26‘D A (D27) (D.28) (D.29) (D.30) (D.3l) (D32) (D.33) (D.34) (D.35) (D.36) (D.37) n/USEFUL FORMULAS D.5 Complex Quantities Cartesian form: z=x+iy,complex conjugate z‘=x—iy,i=\/-1 Polar form: z=|z|e'9 z*=|z|e"" zz*=|z|2=x2+y2 1Real partofz: Rez=§(z+z*)=x 1 Imaginary partofz: Imz=——é(z —z*)=y Euler’s formula: e"'=cos6+isin0 D.6 Hyperbolic Functions _ e"—e“sinh x=i2 _e"+e“ 2 e2“—1tanh =—-i x22*+1cosh x sinix=isinh x cosix=cosh x sinh ix=isinx cosh ix=cosx sinh“'x= tanh"<—L— \/x2 +1 =ln(x+ \/x2+ 1) >O, =cosh"(\/x2 +1), {< 0 cosh—'x=i' tanh“'<-%), x> 1 =iln(x+ \/x2—l), x>1x>O x<0611 (D38) (D39) (DAO) (D.41) (D.42) (D.43) (D.44) (DA5) (D.46) (D.47) (D.48) (DA9) (D.50) (D.51‘b 1 (D52) (D.53) (D54) (D.55) (D.56) 512 D/USEFUL FORMULAS cosh"x =i'sinh"(\/x2— 1), x>1 d;ysinhy= coshy d5cosh y=sinh y sinh(x, +x2)=sinh x,cosh x2+cosh xlsinh x2 cosh(x, +x2)=cosh x,cosh x2+sinh x,sinh x2 cosh2x —sinh2x =1(D.57) (D.58) (D.59) (D.60) (D.6l) (D.62) PROBLEMS D-l. lsitpossible toascribe ameaning totheinequality z,<z2?Explain. Does thein- equality |z,|<|z2|have adifferent meaning? D-2. Solve thefollowing equations: (a)z2+2z+2=0 (b)2z2+z+2=O D-3. Express thefollowing inpolar form: (8)l|=i (b)12=_1 (¢)z,=1+i\/5 (<1)zi=1+2i (e)Find theproduct zlzg (f)Find theproduct 2,13 (g)Find theproduct 1314 D-4. Express (:2—1)"/2 inpolar form. D-5. Ifthefunction w=sin": isdeﬁned astheinverse ofz=sinw,then usetheEuler relation forsinwtoﬁndanequation forexp(iw). Solve thisequation andobtain theresult w= sin"z= —iln(iz+ \/1- .22) D-6. Show that y=Ae"‘+Be"" canbewn'tten as y=Ccos(x —5) where AandBarecomplex butwhere Cand5arereal. D-7. Show that (a)sinh(x, +x2)=sinh x,cosh x2+cosh x,sinh x2 (b)cosh(x, +x2)=cosh x,cosh x2+sinh x,sinh x2 APPENDIX Useful Integrals * E.1Algebraic Functions dx 1_x _x 1r pIﬁ2=-'?lLﬂl'l 1(3), Itan 1(2) <-5 (15.1) xdx 1 ‘Jm =§ln(a2 +x2) (15.2) (E.3) g‘:-I/'\§ /xv“io-I dx =1I x2 x(a2 +x2) 202 n<a2 +x2 I dx 1 ax— a2x2 b2_ =Qabln (E.4a) =—-—coth" — a2x2 >I12 (E.4b) _ 1 -1 22 2-—Emm —» ax<b mu) dx 2——~——=—V b asl\/;,T;1»“+" " lV§%i=mu+ve+$) mmG *This listisconﬁned tothose (nontrivial) integrals thatarise inthetextand intheproblems. Extremely useful compilations are,forexample, Pierce andFoster (Pi57) andDwight (Dw6l). 613 614 x2dx x/i a2__x - a2—x2+—s1n' 1I——L=iln(2WVax2+bx+c+2m+ \/ax2+bx+c \/Z1 1s_nh_,( 2ax+ b){a>0::i—l a, \/; \/4ac—b2 4‘1C>b2 <1 2ax+b “0E/usnrut INTEGRALS E.7 2 2 a () b), a>0 (E.8a) (E.8b) =—isin"%— b2>40¢ '_a b2—4 l|2ax+b|< \/b2—4ac (E.8c) d 1 b dx J€—ic-——,;--=—\/ax2+bx+c——l——-"———? (E.9) \/ax2+bx+c a 2“ \/ax2+bx+c Jm=_;Si,,h_1(1;ﬂ1__)x\/ax2+bx+c \/2 |x|\/4ac—b2, 4>0ac>b2 (E.l0a) 1 <0=——sin-'<—ibx +2“ {C (E.10b)V-0 |x|\/b2—4ac b2> 4”‘ 1 =—7ln<L:/;Vax2+bx+c+2;c+b), c>0 c (E.l0c) b 4-1)?lx/w¢2+bx+ cdx=%1xT:"Vax2+bx+c+ ‘“ E.2 Trigonometric Functions Isin2xdx= 5-lsin 2x2 4 1 Jcos2xdx=g+ Zsin2x atan(x/2) +bI dx 8” \/ax2+bx+c (E.ll) (E.12) (E.13) = ta‘ ———-—-——-, a2>b2 (12.14)dx 2 la+bsinx \/a2_b? nll: \/a2_b2 :| E/USEFUL INTEGRALS dx 2 2 ta_l(a—b)tan(x/2) H2>b2 la+b¢<>sx \/a2—b2 nl \/(12-02 y dx _ bsinx _ a I dx (a+bcosx)2 (b2—a2)(a+bcosx) b2—a2 a+bcosx Jtan xdx=—lnIcos x| Itanh xdx=lncosh x eax ax ' d =_-_i_._ ' _.fasinx xa2+1(as1nx cosx) I‘ eax _ _ 2 e“"sin2xdx= 3? as1n2x— 2s1nx cosx+-a+4 H Fe_‘""dx= \/1r/a E.3 Gamma Functions O0 F(n) =Lx""e"‘dx 1 =L[ln(1/x)]""dx F(n) =(n—1)!, forn=positive integer nF(n) 1\ r<§1 I‘(1‘>=1I 1\r(1;1 r<1%) =0.919=F(n+l) =\/; =0.906615 (E.15) (E.16) (E.l7a) (E.17b) (E.18a) (E.18b) (E.18c) (E.19a) (E.19b‘> A (E.19c) (E.20) (E.2l) (E22) (E23) (E.24) E/USEFUL INTEGRALS I‘(2)=1 (12.25) Idx \/F F"<1) —— =—-—~—— (12.26)lm/_J nI.(1+1)n2 1‘(n+1)1"<%> x"‘(1—x2)"dx =—i—i— (E.27a) 21_'<n + ~<=1'ﬁn+1 1r/2 (2> Icos"xdx=——i———, n> -1 (E.27b) 0 71 1 l‘(2+ ) APPENDIX Differential Relations inDifferent Caordinate F.1Rectangular Coordinates UgradU= VU= 2e,(;lx- 6AdivA= V-A= E—'Iax, 6AcurlA =VXA=2s,],,i"e,|1]'h ax] Wt/= v-vu=Zax? F.2 Cylindrical Coordinates Refer toFigures F-1andF-2. x1= rcos¢, x2=rsinrb, x5=z _*2r= \/x{"+x§, ¢= tan‘I, z= 1Systems (F-1) (E2) (F-3) (F.4) (F.5) (F.6) 617 618 F/DlFFERENTlAL RELATIQNS INDIFFERENT CQORDINATE SYSTEMS Z=X3 I¢¢ ;*2 *1 Cylindncal coordinates. dv=rdrd¢dz d r¢ dr r dz N /' ¢rd / drPlane polar coordinates: da=rdrd¢ FIGURE F-2 ds2=dﬁ+#d¢’ +dz? (F.7) -dv=rdrdd)dz (F.8) __iii 1% Q/1 grad!/1—V¢—e,ar+ed,Ta¢+e,az (F.9) , 1a 16/is a/1,d1vA —7_5_(r/1,) +7_64>+az (F.l0) lA_ la/1‘ ‘E41 + ill’ (211 _+_ la/1’ F11 °“r _e’ ra¢> 61 ed’6281" e‘rar('A“’) ra¢ (') _1aat/1 1a2-p a2¢V2,’,_raT(r ar)+T26¢2 +612 (E12) F/DIFFERENTIAL RELATIONS INDIFFERENT COORDINATE SYSTEMS F.3 Spherical Coordinates Refer toFigures F-3andF-4 x1=rsin0c0s¢, x2=rsin6sind>, x3=rcos6 xs _x2r=\/x‘f’+x§+x§, 6=cos'1—, ¢=tan‘—T xl ¢%=mL+ﬂw2+ r2sin?6d¢2 dv=T2sin6drd6 dd) *1 rsin6*2 cf’ €¢ \ ¢l‘\“‘i I X2 FIGURE F-3 Spherical coordinates: dv=r2sin0drd0d¢ d ¢dr da=12sin6d6d¢ r de rsinedq) 6rd6 ¢ 11¢ FIGURE F-4619 (F.13) (F.14) (F.15) (F.l6) F/DIFFERENTIAL RELATIONS INDIFFERENT COORDINATE SYSTEMS 6 gradl/J =V1]!=e,'£_" +e9%% +e¢,'T—s%l-6% (E17) 1a 1a 1 BA.»(l'A==—— 2/l *i— ' mi ""— F.l W r2<')r(r ')+rsin060(A0sm6)+rsin6+ 69¢ (8) curlA =e,;-gil-n—é I:%(/14, sin9)— 1 6A, _ 8 16 6A,+693:6 '55‘ -'Sln65.(TA¢) +845; -'E (E19) 1a 1 1a2 W=ml”ill+we§l(s‘"”3%)* $5<1’-2°’ APPENDIX A“Proof” oftheRelation = Consider thetwoinertial systems KandK’that aremoving relative toonean- other with aspeed v.Attheinstant when thetwoorigins coincide (t=0,t’=0), letalight pulse beemitted from thecommon origin. The equations that de- scn'be thepropagation ofthewave fronts arerequired, bythesecond Einstein postulate, tobeofthesame form inthetwosystems: ~t~/1~31»-tie=Z1935 S2=0,inK ((1.111)P Ex?-er?=Esq;ES'2=0,inK' (G.lb) These equations state that thevanishing ofthefour-dimensional interval be- tween twoevents inoneinertial reference frame implies thevanishing ofthein- terval between thesame twoevents inanyother inertial reference frame. Butwe need more than this;wemust show, infact, thats2=s'2ingeneral. Ifwerequire thatthemotion ofaparticle observed tobelinearin thesystem Kalsobelinear inthesystem K’,then theequations oftransformation thatcon- nect thexﬁandthexi,must themselves belinear. Insuch acase, thequadratic forms s2ands'2canbeconnected by,atmost, aproportionality factor: s'2=KS2 (G.2a) Thefactor Kcould conceivably depend onthecoordinates, thetime, andtherela- tivespeed ofthetwosystems. Aspointed outinSection 2.3,thespace andtime as- sociated with aninertial reference frame arehomogeneous, sotherelation between 621 622 G/A"PROOF" orTHERELATION Ex;=21,71I4 s2ands’2cannot bedifferent atdifferent points inspace noratdifferent instants oftime. Therefore, thefactor Kcannot depend oneither thecoordinates orthe time. Adependence onvisstillallowed, however, buttheisotropy ofspace forbids a dependence onthedirection ofv.Wehave therefore reduced thepossible depend- ence ofs'2ons?toafactor that involves atmost themagnitude ofthespeed v;that is,wehave s'2=K(U)S2 (G.2b) Ifwemake thetransformation from K’back toK,wehave theresult s2=K(-"U)S'2 where —voccurs because thevelocity ofKrelative toK'isthenegative oftheve- locity ofK'relative toK.Butwehave already argued that thefactor Kcande- pend only onthemagnitude ofv.Wetherefore have thetwoequations s’2=K(U)S2 s2=K(v)s'2} (G.3) Combining these equations, weconclude that K2=1,orK(v) =:1.The value ofK(v) must notbeadiscontinuous function ofv;thatis,ifwe change vatsome rate, Kcannot suddenlyjump from +1to-1.Inthelimit ofzero velocity, the systems Kand K’become identical, sothat K(v=0)=+1.Hence, K=+1 (G.4) forallvalues ofthevelocity, andwehave, ﬁnally, 52=s'2 (G.5) This important result states thatthefour-dimensional interval between two events isthesame inallinertial reference frames. APPENDIX Numerical Solution forExample 2.7 Inthisappendix, weshow theMathCad solution thatproduced Figures 2-8and2-9 forExample 2.7.This program waswritten forMathCad forWindows, version 4.0. g==9.8 acceleration ofgravity th==60- initialangle180 vo1=600 u1=vo-cos(th) initial velocity v==vo-sin(th) initial horizontal velocity i==1..6 initial vertical velocity kl;= 000 table ofdrag coefficients .0o 9999980000moo-Ana»-OE t==0,1..130 range oftime values x(t,K)== (1—exp(—K- t)) calculate horizontal position 623 624 H/NUMERICAL SOLUTION FOR EXAM K'v t +gy(t,K) ==_g._ +__. 1 K (K)2 (—exp( —K-t))calculate vertical [Now ploty(t,k])versus x(t,kJ) toproduce Figure 2-8.] 1.5-104 )’(t>k|), y(tvk2)a Z91"}1' .Y$E'l‘;Q'. ! y(t'k6) I05000—_I, .,,-o---o-_ 1-104— .-.-'-'7'-0,.‘l r"-Es\ \\ '¢ -"p ¢"T---1-0 --"-—_——-_F‘$¢v"T' ¢—'F‘us.-"' .-..-'',4-u ks -5000.-- -n4,-Iuoo=k2 ".ks ._l . Ik1 -n-000PLE27 position 0 1-10* 2-104 x(t,k|), x(t,k )x(t,k21 3): x(trk-1): X“-11(5): X(t1k6) FIGURE 2-8 Now setupanequation tosolve Equation 2.45 forTfor an I ingforceyvaueoftheretard- constant k. . - exp(—k-T)k-v+g f(k,T) ==root Vi (1—gk j1,2..s1 K1==—0.001 + Tro==100 Tr]==f(KJ,TrJ_1) Tr,==106.0740.001 +0.0000000l),T:| (2.45) Setuprange ofvalues tocal- culate; 80values. This willallow ustocalculate over arange ofkvalues from 0to0.08. The time value fork=0is 106s.This isaguess toget thecalculation started. Wenowdetermine thesolu- tion forthetime Tforall thevalues ofk.Solve Equation 2.45. This isthevalue ofT forkl. Wedonotbother tocalcu- latealltheothers here.3-104 4-10 1-1/NUMERICAL SOLUTION FOREXAMPLE 2.7 625 Now wewant tocalculate therange Rforallthevalues ofT(asafunction ofk) thatwehave justfound. Todothis, weneed tosolve Equation 2.43 foreach of thevalues ofk and t=Tthat wehavejust found. x==100 This istheguess fortheﬁrst value ofx.The actual value ofthe essdoes notmatter. f(k,T)==root[X—2-(1—exp(—k'T)), x:| This iguthe Equation 2.43 k thatweneed tosolve toﬁnd therange R. Rj==f(Kj,TrJ-) Now calculate therange R forallthevalues. R1=3.182 -10‘ Wejust listtheﬁrstvalue and plot theremainder. This is therange fornoairresis- tance, thatisk=0. Now let’scalculate andplottherange determined from theapproximate calcu- lation. Calculate Equation 2.55. 1 4' 'V R.2= R- —P] 1 3.g [Now plotRjandRpjversus Kjtoproduce Figure 2-9.] Plotapproximate andnumerical solutions. Figure 2-9. I I I I R L2-104 \‘ Rpl ‘\"' \ \ ‘\ 0 II I I I I 0 0.02 0.04 0.00 0.08 0.1 KI FIGURE 2-9 Selected References The following texts areparticularly recommended asgeneral sources ofcollat- eralreading material. A.General Theoretical Physics Blass (B162), Theoretical Physics. Lindsay andMargenau (L136), Foundations ofPhysics. Wangsness (Wa63), Introduction toTheoretical Physics. B.Elementary Mechanics Baierlein (Ba83) ,Newtonian Dynamics. Barger andOlsson (Ba73), Classical Mechanics. Davis (Da86) ,Classical Mechanics. Fowles andCassiday (F099), Analytical Mechanics. French (Fr7l), Newtonian Mechanics. Knudsen andHjorth (Kn00) ,Elements ofNewtonian Mechanics. McCall (Mc0l), Classical Mechanics. Rossberg (R083), Analytical Mechanics. C.Intermediate Mechanics Arya (A198) ,Introduction toClassical Dynamics. Becker (Be54), Introduction toTheoretical Mechanics. Lindsay (L161), Physical Mechanics. Scheck(Sc99) ,Mechanics. Slater andFrank (S147), Mechanics. Symon (Sy71), Mechanics. D.Advanced Mechanics Baruh (Ba99), Analytical Dynamics. Goldstein (G080), Classical Mechanics. 626 SELECTED REFERENCES Landau andLifshitz (La76), Mechanics. McCuskey (Mc59), AnIntroduction toAdvanced Dynamics. E.Mathematical Methods Abramowitz andStegun (Ab65) ,Handbook ofMathematical Functions. Arﬂ<en (Ar85), Mathematical Methods forPhysicists. Byron andFuller (By69), Mathematics ofClassical andQuantum Physics. Churchill (Ch78), Fourier Series andBoundary Value Problems. Davis (D2161), Introduction toVector Analysis. Dennery andKrzywicki (De67), Mathematics forPhysicists. Dwight (Dw61), Tables ofIntegrals andOther Mathematical Data. Kaplan (Ka84), Advanced Calculus. Mathews andWalker (Ma70), Mathematical Methods ofPhysics. Pipes andHarvill (Pi70), Applied Mathematics forEngineers andPhysicists. F.Special Relativity Einstein (Eiﬁl), Relativity. French (Fr68) ,Special Relativity. Resnick (Re72) ,Basic Concepts inRelativity andEarly Quantum Theory. Rindler (R182), Introduction toSpecial Relativity. I Taylor andWheeler (Ta66) ,Spacetime Physics. G.Chaos Baker and Gollub (Ba90) ,Chaotic Dynamics. Bessoir andWolf (Be9l), Chaos Simulations. Hilborn (Hi94), Chaos andNonlinear Dynamics. Moon (M092), Chaotic andFractal Dynamics. Rasband (R2190), Chaotic Dynamics ofNonlinear Systems. Rollins (R090), Chaotic Dynamics Workbench. Sprott and Rowlands (Sp92), Chaos Demonstrations. Strogatz (St94), Nonlinear Dynamics andChaos. H.Numerical Methods Dejong (De91), Introduction toComputational Physics. johnson andReiss (I082), Numerical Analysis. Press, Teukolsky, Vetterling, andFlannery (Pr92), Numerical Recipes. Ab65 Ad22 Am63 An49 Ar85 Ar98 Ba73 Ba83 Ba96 Ba99 Be46 Be54 Be91 B162 Br6O Br68 By69 628Bibliography M.Abramowitz andI.Stegun, Handbook ofMathematical Functions. Dover, NewYork, 1965. E.P.Adams andR.L.Hippisley, Smithsonian Mathematical Formulae and Tables ofElliptical Functions. Smithsonian Institution, Washington, D.C., 1922. American Association ofPhysics Teachers, Special Relativity Theory, Selected Reprints. American Institute ofPhysics, New York, 1963. A.A.Andronow and C.E.Chaikin, Theory ofOscillations, transl. ofthe 1937 Russian ed.Princeton University Press, Princeton, New jersey, 1949. - G.Arﬂdn, Mathematical Methods forPhysicists, 3rded.Academic Press, Orlando, Florida, 1985. A.P.Arya, Introduction toClassical Mechanics 2nd ed. Prentice-Hall, Englewood Cliffs, Newjersey, 1998. V.Barger andM.Olsson, Classical Mechanics. McGraw-Hill, NewYork, 1973. R.Baierlein, Newtonian Dynamics. McGraw-Hill, New York, 1983. G.L.Baker and]. P.Gollub, Chaotic Dynamics, 2nded.Cambridge, New York, 1996. H.Baruh, Analytical Dynamics, WCB/McGraw-Hill, Boston, 1999. P.G.Bergman, Introduction totheTheory ofRelativity, Prentice-Hall, Englewood Cliffs, Newjersey, 1946 (reprinted byDover, New York, 1976). R.A.Becker, Introduction toTheoretical Mechanics. McGraw-Hill, New York, 1954. T.Bessoir andA.Wolf, Chaos Simulations. American Institute ofPhysics, College Park, 1991. Available from Physics Academic Software, Box8202, North Carolina State University, Raleigh, North Carolina 27695-8202. G.A.Blass, Theoretical Physics. Appleton-Century-Crofts, New York, 1962. L.Brillouin, Wave Propagation andGroup Velocity. Academic Press, New York, 1960. T.C.Bradbury, Theoretical Mechanics. Wiley, New York, 1968 (reprinted byKrieger, Melbourne, Florida, 1981). F.Byron and R.Fuller, Mathematics ofClassical and Quantum Physics. Addison-Wesley, Reading, Massachusetts, 1969. — BIBLIOGRAPHY 629 Ch78 C053 C060 Cr81 Da61 Da63 Da86 De62 De67 De82 De91 Dw61 Ed3O E161 Fe59 Fe65 F099 Fr68 Fr71 Ge63 (1080 Ha62 He95 lﬁOO lIo64 In27 ]o50R.V.Churchill, Fourier Series andBoundary Value Problems, 3rded.McGraw- Hill, New York, 1978. R.Courant and D.Hilbert, Methods ofMathematical Physics, Vol. 1.Wiley (Interscience), New York, 1953. H.C.Corben andP.Stehle, Classical Mechanics, 2nded.Wiley, New York, 1960. A.D.Crowell, Am.]. Phys. 49,452(1981). H.F.Davis, Introduction toVector Analysis. Allyn 8cBacon, Boston, 1961. H.F.Davis, Fourier Series andOrthogonal Functions. Allyn 8cBacon, Boston, 1963. A.D.Davis, Classical Mechanics. Academic Press, Orlando, Florida, 1986. J.W.Dettman, Mathematical Methods inPhysics andEngineering McGraw- Hill, NewYork, 1962. P.Dennery and A.Krzywicki, Mathematics forPhysicists. Harper 8cRow, New York, 1967. E.A.Desloge, Classical Mechanics, Vols. 1and2.Wiley, New York, 1982. M.L.Dejong, Introduction toComputational Physics. Addison-Wesley, Reading, Massachusetts, 1991. H.B.Dwight, Tables ofIntegrals and Other Mathematical Data, 4thed. Macmillan, New York, 1961. SirA.S.Eddington, TheNature ofthePhysical World. Macmillan, New York, 1930. A.Einstein, Relativity, 15th ed.Crown, New York, 1961. y N.Feather, ThePhysics ofMass, Length, andTime. Edinburgh University Press, Edinburgh, 1959. R.P.Feynman andA.R.Hibbs, Quantum Mechanics andPath Integrals. McGraw-Hill, New York, 1965. G.R.Fowles and G.L.Cassiday, Analytical Mechanics, 6thed.Harcourt, Philadelphia, 1999. A.P.French, Special Relativity. W.W.Norton, New York, 1968. A.P.French, Newtonian Mechanics. W.W.Norton, New York, 1971. I.M.Gelfand and S.V.Fomin, Calculus ofVariations, Prentice—Hall, Englewood Cliffs, Newjersey, 1963. H.Goldstein, Classical Mechanics, 2nd ed.Addison-Wesley, Reading, Massachusetts, 1980. ].Haag, Oscillatory Motions. Wadsworth, Belmont, California, 1962. M.Heald and].B.Marion, Classical Electromagnetic Radiation, 3rded. Saunders, New York, 1995. R.C.Hilbom, Chaos andNonlinear Dynamics, 2nd ed.Oxford, New York, 2000. H.Hochstadt, Diﬂerential Equations——A Modern Approach. Holt, New York, 1964. E.L.Ince, Ordinary Differential Equations. Longmans, Green, New York, 1927 (reprinted byDover, New York, 1944). G.]oos andI.M.Freeman, Theoretical Physics, 2nded.Hafner, New York, 1950. 630 ]o82 K2176 Ka84 Kn00 La49 La76 Li36 Li51 Li61 L023 Ma59 Ma60 Ma65 Ma70 Ma77 Mc59 Mc01 Mi47 M053 Mo53a M058 M092BIBLIOGRAPHY LeeW.johnson andR.Dean Reiss, Numerical Analysis. Addison-Wesley, Reading, Massachusetts, 1982. M.Kaplan, Modern Spacecraft Dynamics and Control. Wiley, New York, 1976. W.Kaplan, Advanced Calculus, 3rd ed. Addison-Wesley, Reading, Massachusetts, 1984. J.M.Knudsen andP.G.Hjorth, Elements ofNewtonian Mechanics, 3rded. Springer-Verlag, Berlin, 2000. C.Lanczos, TheVariational Principles ofMechanics. University ofToronto Press, Toronto, 1949. L.D.Landau and E.M.Lifshitz, Mechanics, 3rded.Pergammon, New York, 1976. R.B.Lindsay andH.Margenau, Foundations ofPhysics. Wiley, New York, 1936 (reprinted byDover, New York, 1957; reprinted byOxBow, Woodbridge, Connecticut, 1981). R.B.Lindsay, Concepts andMethods ofTheoretical Physics. Van Nostrand, Princeton, Newjersey, 1951. R.B.Lindsay, Physical Mechanics, 3rded.Van Nostrand, Princeton, New jersey, 1961. H.A.Lorentz, A.Einstein, H.Minowski, and H.Weyl, ThePrinciple of Relativity, original papers. Translated in1923; reprinted byDover, New York, 1952. ].B.Marion, T.I.Arnette, and H.C.Owens, “Tables for the Transformation Between theLaboratory andCenter—of-mass Coordinate Systems andfortheCalculation oftheEnergies ofReaction Products.” Oak Ridge National Lab. Rept. ORNL-2574, 1959. E.Mach, TheScience ofMechanics, 6thAm. ed.Open Court, LaSalle, Illinois, 1960 (original German edition published 1883). ].B.Marion, Principles ofVector Analysis. Academic Press, New York, 1965. ].Matthews and R.Walker, Mathematical Methods ofPhysics, 2nd ed. Benjamin, New York, 1970. H.Margenau, TheNature ofPhysical Reality, 2nd ed.McGraw-Hill, NewYork, 1977. S.W.McCuskey, AnIntroduction toAdvanced Dynamics. Addison-Wesley, Reading, Massachusetts, 1959. M.W.McCall, Classical Mechanics._]ohn Wiley, Chichester, 2001. N.Minorsky, Introduction toNon-linear Mechanics. Edwards, Ann Arbor, Michigan, 1947. P.M.Morse and H.Feshbach, Methods ofTheoretical Physics, 2vols. McGraw-Hill, New York, 1953. P.Morrison, “ASurvey ofNuclear Reactions,” inExperimental Nuclear Physics (E.Segré, ed.), Vol.II.Wiley, New York, 1953. F.R.Moulton, AnIntroduction toCelestial Mechanics, 2nd ed.Macmillan, New York, 1958. Frances Moon, Chaotic andFractal Dynamics. Wiley, New York, 1992. BIBLIOGRAPHY 631 Mu60 G.M.Murphy, Ordinary Diﬂerent Equations andTheir Solutions. VanNostrand, Ob83 Pa62 Pi57 Pi70 Pr92 Ra90 R2194 Re72 Rh82 Ri82 R083 R090 Sc99 Se58 S147 S050 Sp92 St94 Sy71 Ta66 Tr68 Tu04 Wa63 We61Princeton, Newjersey, 1960. ].E.Oberg, Mission toMars. New American, New York, 1983. W.K.H.Panofsky andM.Phillips, Classical Electricity andMagnetism, 2nd ed.Addison-Wesley, Reading, Massachusetts, 1962. B.O.Pierce and R.M.Foster, AShort Table ofIntegrals, 4thed.Ginn, Boston, 1957. L.Pipes and L.Harvill, Applied Mathematics forEngineers andPhysicists. McGraw-Hill, New York, 1970. W.H.Press, S.A.Teukolsky, W.T.Vetterling, andB.P.Flannery, Numerical Recipes, 2nded.Cambridge, New York, 1992. S.N.Rasband, Chaotic Dynamics ofNonlinear Systems. Wiley, New York, 1990. ].W.S.Rayleigh, The Theory ofSound, 2nd ed., 2vols. Macmillan, London, 1894 (reprinted byDover, New York, 1945). R.Resnick, Basic Concepts inRelativity andEarly Quantum Theory. Wiley, New York, 1972. Rheinmetall GmbH, Handbook onWeaponry. Dusseldorf, 1982. W.Rindler, Introduction toSpecial Relativity. Clarendon, Oxford, 1982. K.Rossberg, Analytical Mechanics. Wiley, New York, 1983. R.W.Rollins, Chaotic Dynamics Workbench. American Institute ofPhysics, College Park, 1990. Available from Physics Academic Software, Box 8202, North Carolina State University, Raleigh, North Carolina 27695- 8202. F.Scheck, Mechanics, 3rded.Springer-Verlag, Berlin, 1999. F.W.Sears, Mechanics, Wave Motion, andHeat. Addison-Wesley, Reading, Massachusetts, 1958. ].C.Slater andN.H.Frank, Mechanics. McGraw-Hill, New York, 1947. A.Sommerfeld, Mechanics. Academic Press, New York, 1950. ].C.Sprott andG.Rowlands, Chaos Demonstrations. American Institute of Physics, College Park, 1992. Available from Physics Academic Software, Box 8202, North Carolina State University, Raleigh, North Carolina 27695-8202. S.H.Strogatz, Nonlinear Dynamics andChaos. Perseus Books, 1994. K.R.Symon, Mechanics, 3rded.Addison-Wesley, Reading, Massachusetts, 1971. E.F.Taylor and]. A.Wheeler, Spacetime Physics. Freeman, SanFrancisco, 1966. C.Truesdell, Essays intheHistory ofMechanics. Springer-Verlag, New York, 1968. H.H.Turner, Astronomical Discovery, Arnold, London, 1904 (reprinted bytheUniversity ofCalifornia Press, Berkeley, California, 1963). R.K.Wangsness, Introduction toTheoretical Physics. Wiley, New York, 1963. J.Weber, General Relativity andGravitational Waves. Wiley (Interscience), New York, 1961. 632 Wh37 Wh53BIBLIOGRAPHY E.T.Whittaker, ATreatise ontheAnalytical Dynamics ofParticles andRigid Bodies, 4thed.Cambridge University Press, London andNew York, 1937 (reprinted byDover, New York, 1944). E.T.VVhittaker, AHistory oftheTheories ofAether andElectricity; Vol.II:The Modem Theories. Nelson, London, 1953 (reprinted byHarper andBros., New York, 1960). Answers toEven- Numbered Problems Chapter 1 10.(a) v=2bw cosarti—boosinwtj (b)90° a= —or2r |v|=bw[3cos2wt+ 11% 12.h= ~la'bx°|hxb+bx¢+¢xd 1\9r—1l\9r—-/-\U‘r—¢ A=— —=\)X(¢—b)|=§|(a—<1)><(b—a)| =d®-b)X@—¢H 9 7 ‘*5 -5 0 —3 —4 1.4. (3) -104 (1)) 13 9 (C) 3 -5 ((1) 3 0 6 5 2 25 14 4 -'6 0 3v2 3v2 2 - v6. . =——i~ =— r =——i- 2ae’ 4k’|al 4k\l1+cos6’6 @ 34.[(AXA)dt=(AXA)+C,where Cisaconstant vector 36. rrcgd 38.—1r 40.(a)x=—2m,y= 3m,zm,,,, =72m; (c)SE 633 634 ANSWERS TOEVEN-NUMBERED PROBLEMS Chapter 2 2.E,=mR(l§—ql>2sin6cos6) F,=mR(2é qbc0s6 +<2;sin6) 4.13.2m-s"1 6.(a)210mbehind (b)canbenomore than 0.68 slate _2v§cos asin(a —,3) 1r B _ vi, 14'(a)d_ gcos2B (b)4+2(C)dm'“ _g(1+sinﬁ) 21/0 gsina 1s.(a)35.2m's“1 (b)407°;1.1m 20.17.4°16. 22.(c)s(t) =C1cosw,t+C2sinwct+% y(t)=—C1 sinw,t+C2cos(uct 24.;.t,,=0.18; vB=15.6m/s 26.2.3m;1.1m 28.hmarble =h(?%%)2; hsuperban =h(%_T3§)2 where a=m/M 30.71m _1i,u.,,V3+4);} 32. sin90 + 2 34.(a)y=——v 1 —— (b)y=—%1n(1-%) Us.5‘... /_\I§’_§_‘+Q_Ug:3 /7I-1 ...0%?5-‘..r§_‘/||36.R= —cos6 sin6 + sin26 +—~) 38.(a)F(x) =—mna2x“(2”*1) 1 (b)x(t) =[(n+1)at]"+1 11¢)F(t)=—mna2[(n +1)at]"(2”+1)/("+1) 2Aa2 sinat Aa2|2 cosat—1| 1')oi?" a..=ii——~— 40. (3 I \/5—4cosat \/5—4cosat V 0 (b)—— where n=integera 42.Stable ifR>b/2;unstable ifRSb/2 ANSWERS TOEVEN-NUMBERED PROBLEMS 635 I24_= 3/2 ___81' rrd mG _e/ﬁre _Fr 50.(a)x(t)~F‘ m3+-6? mo), v(t)— F2? M?) +T c (c)t(v=c/2) =0.55 yr;t(v=0.99c) =6.67 yr. _ 4U0x X2 ___ _ 2U0 52-(a)F(x)—_'aT 1-}, (C)¢°— E! ((1) vmin_ Y ()(t) a[exp(t \/8U0/ma?) —1] ex = A A — [exp(t V8U0/ma?) +1] 54.(a)v=g/k=1000 m/s, (b)height =%+iiln =680m Chapter3 __ 10 __2.(a)6.9X102s1(b)27(1 —2.40 X105)s1(c)1.0445 mA2w§._ 1._ m./120034. <T> = =T-U=§T=—-GT" 6.2.74 rad-s“1 12.0=—€sin 9 14.x(t)=(cosh Bt—sinh Bt)[(A1 +A2)cosh co2t+(A1—A2)sinh w2t] aE(t)=(cosh Bt—sinh Bt)[(A1or2 —A1B) (cosh cogt+sinh (ﬂgt) —(A23 +A2Q)2) (cosh (U2t—sinh w2t)] R,[R,(R, +R1)+@213] +i[R,aL, +((oL1—1/wC)((R1 +R2)?+(021.326.(R,+R2)?+ML; 4. 4. 4.28.F(t) =—sin t+—sin 3t+—sin 5t+11' 31r 511' 2 4 430.F(t) —Tr—gwcos 2cot— 157Tcos 4wt H0 '3’32.(a)x(t)=—~Q 1—e“~5’cosh (Llgt—& sinh (U2t(U?) (U2 b (b)x(t) =Jeh!“sinh orgt; t>0 2 636 ANSWERS TOEVEN-NUMBERED PROBLEMS 0 t<0 34.x(t)=4[1—cos(0.5t)] m 0<t<411' 0 t>411' ' b36.x(t)=e‘5("’")[x0 cosw1(t ~to)+(E+Big+—)sinco1(t —t0):|; t>to(O1 (U1 (O1 x(t)=F0 to ' m[(B- r)2+ (w+w1)2][(B— r)2+ (w-@02138 ><l:e"°"[2('y -B)coswt+([3—'y]2+or?—or?) sincolt +e'”’[2(B -r)¢0s w1t+ ([6-112+<02-wi)1 40.Amplitude =-0.16 mm, minus sign indicates spring iscompressed 42(a)x(t)=F 1 (cosinwt—wsinw t)(b)x(t)=1%' mwo ((00+cu)((00—or) 0 0l 6m '‘"0\/6422 +1 Chapter 4 - /26.6= ;lE[E— mgl(1 —cos6)]1/2 8.1'=4,{L772/1 F0 10.Only 0.6and0.7arechaotic 14.n=30 22.Transitions atB1=9.8—9.9, B2=11.6 —11.7, and B3=13.3 —13.4. Behavior: (i)one period perthree drive cycles when B<B1,(ii)chaotic when B1<B<B2,(iii)mixed chaotic/one period perdrive cycle (depend- ingoninitial conditions) when B2<B<B2,,and (iv)oneperiod perdrive cycle when B>B5 Chapter 5 C 6gb.=——— h C=-—= . 2pQWGT were GT const GM 6.g=—?e,, ANSWERS TOEVEN-NUMBERED PROBLEMS s.g,=—21TGp(\/a2 +(1,,-1)?-\/a2+zg+1) GM la? 310.R2-— 1--— 1——' 204”) Rl2114 2”“ 16.I§=21'rp,GM 2GM 2GM 22+ 2—z 20. <l)(z) =—?“‘(\/Z2 +R2—Z),g(Z) =—k?,—( iiChapter 6 2 2 2 28.(a)a1=b1=c1=%R (b)a1=a7_é;b1=b—\7-g;c1=c—\/—:; 110.R= —H2 14.length =2\/58111-L2\/E 8 93/216. y(x) = -1- —1:’and z=x3/2 18.x=—y=\/—z where x>0,y<0,z<0.Parabolic line. Chapter 7 .. - d -4.mr-mr02 +Ara” =0;Et(mr26) =0;yes;yes 6.2mS+ m§c0sa— mgsina =0 (m+M)§+ mScosa=0 10.(a)y(t)=—ft? (b)y(t)=ijﬂ -cosh 'yt) 12.r(t)=rocosh at+55-2(sin at—sinh at)oz ..a+g_ 1; 14.(a)6+—is1n6=0 (b)211' —ib a+g 16.ii+€sino—i;w2sinwi¢<>so=0 gsin60 rr1s.=,/~——; 0=-°’ 1-Re, °2 3828638 ANSVVERS TOEVEN-NUMBERED PROBLEMS I\9l—‘l\9I—*réar-NJ 22.L=—mx ——e_‘/T; H= 235+ﬁe"/Tx 2m x 24.L=—m(a2 +1262)+ 115 1H=———2-z0 2ml2 27760 mg COSmglcos6 P2 -1’ . .26.(a)H=27‘}— mglcos6; o=;Z-:2; p,,=—mgls1n9 Pf H: — — — 2('”b1 + m2 + I/(12) mlgxin p= m+m +laZ x 1 2 ag ﬁx:g(7n1_m2)2g(l—x) .. P5 -..p,=mr;p,=$§—T2 ,,=mr26;p,,=0 _1 113 Pi k. k113 232.H—2m(p,3+;"§+ 2.26 V1), 34. £‘_%’_P———+--+———L -rsin r2 mr?’ mr3sin26’ _ p§,cot6 _ O ‘b0_mr2sin26’p¢ .._ . ..5Zsin6+gc0s6 m(a)52=aR(6 sin6 +62cos6);6=—T; whereaﬁ 4 mM 3M (b)A_ g(sin6 —asin56 "+m -2sin60) (M+ m)(1— asin?6)2 dp 6Hi-ii:ii:ik x3 at0;,mdt Bx (x+b) 40.0_d2x d261 d262 d6,2_ d622_—4225+ b2Fcos01+Fcos62 —b2-65 s1n61+ E s1n62 d26 at 2—2gsino=2b—1 xd61 dig+2;; cos61+bﬁcoswl —62) +1;3622' (0o dtS1Il1 2) _ d262 d2x d26gsin62=b~—— +dt, gt;cos62+b———1 cos61—62)dt2 ( ___S_ bd012'(6 0)atml 2 ANSWERS TOEVEN-NUMBERED PROBLEMS 639 Chapter 8 k 4. =——; <T> =1ll 2a 10.Parabola; yes 12.76days l26k 1 14. HT) = + 22.No 24.(a)1590 km (b)1900 km 28.2380 m/s 30.Av=3.23 km/s; parabola 32.Stable ifr<a 38.Av=5275 m/s(opposite todirection ofmotion); 146days 40.Carrying thewaste outofthesolar system requires lessenergy than crashing itinto thesun 42.2.57><1011]2 2 / 44.g=l+scos6, wherea=£-’s= 1+2-€——E—ﬁT Mk 14,152 a If0<s<1,theorbit isellipsoid. Ifs=0,theorbit iscircular 46.T=9><1O7yr. Chapter 9 32.Ontheaxis; Ehfrom vertex 64.7c=%lsin§; y=0 F F F6.rem=Z:ln¢2i; vcm=gilnti; am,:§:—ni (I s.x=0;§=——-3\/5 E10.Esin0\/2 m‘+\/mg)g ml + m2 m2 ml 12.(a)yes (b)11m/s 14.No 20.\/Q; 22.(a)twosetsofsolutionsg vn=5.18km/s, vd=14.44 km/s and-u,,=19.79 km/s, vd=5.12 km/s. (b)74.8° and5.2°. (c)30° 640 ANSWERS TOEVEN-NUMBERED PROBLEM to24_w=-—O?—; T= mbwow 1——6b k 26-N=—’<r1—r2)><<i1—*2)‘"0 4m1m2 (ml+m2)2 30.(a)(—0.09i +1.27j)N-s (b)(—9i +127j)N28. "1 34.‘U1='02=$5;0=45° :<ss.ﬁ=s:2\/5; 5=—(1:\/é) with{+a OT11»; U1 —:a>0 u1(m1 S1112a—sm2)40.vi='**i—-———, 2 ; along ulmlsina+mg (8+1)m1u1 sina _ -02=——i———_ 2 ; straight upmlsina+m2 42.4.3m/s,36°from normal Hi44.;.1.ag<1 +Zg) a2 46.0(0) =Z;0,=1ra2 W2mgT0 48-(TLAB(‘//) E ‘ * 54.e71 v§ Q60.25s 62.273s 64.(a)3700 km (b)890km (c)950km (d)8900 km 66.(a)131m/s. (b)108m58. ANSWERS TOEVEN-NUMBERED PROBLEMS 641 Chapter 10 2.Thelocation isgiven bytan6=2?,where 6istheangle between theradiusv _ I vi‘andthehorizontal; |af|=a+ a2+—2T0 4.vo=0.5wR,inydirection; acircle . wt’6.pZI'3.l)OlO1d(Z =Zgr2+const.) 12.0.0018 rad=6min 16.(a)77km (b)8.9km (c)10km (d)160km(alltothewest) 18.260mtotheleft 20.g(poles) =9.832 m/s2, g(equator) =9.780 m/s2 22.2.26 mmtotheright Chapter 11 3 3 2- I1 = I2 = + I3 = I I 3 1 I I1: I2= + I3=I3 1 €4.I=—ml2; a=—- 3 \/5 83 2 14. I1=I2="@0402; I3=EMIJ2 /3g20. b g 12g24.(),/\/§— (b),/%—a a \/ga5 32.53.7 rad/s 34.(ox=wxoexp(—bt/Ix) Chapter 12 10.m5E1+125:1+(K+K12)x1 —K12X2 =E,coswt ‘H1532 +b5C2 +(K +K12)x2 _K12x1 =0 642 ANSWERS TOEVEN-NUMBERED PROBLEM 116.60=—E¢>0,Mode 1;60=(I20,Mode 2 18.co1=0;0.2=,/%)(M+ m) L ii 3 i 4 ii L 20' = 1 i—T): =(ii — 1 al \/14 a2\/‘E 22.w1=2 f/I;w2=2 #’;w5=2 ﬁ 26.4.57, 4.64, 4.81 rad/s 28.Hum =0.96 rad(but atthisangle, small angle approximation isnotcom pletely valid, sothisisarough estimate)/5gmr-P9-H-> 9.»M6";/ Chapter 13 4.wn=M\/E LP The amplitude ofthenthmode isgiven bypm=0, neven 32Ta TlOdd n1-r 6.The second harmonic isdown 4.4dB;thethird, 13.3 dB 212.n,(t) =e'D‘/2" A1exp 2-_ﬁt +Azexp —23-ﬁt 4/>2 Pb 402 Pb 20-¢B,_¢A,=t3T1T1(¢0t 9) ¢/12_¢A,=-9 Chapter 14 12.55.3 m;0.22 /ts;2.5><108m/s; 2.5><108m/s cos6’—B 1—Bcos6' 20.The astronaut ages 25.4 years; those onEarth age26.7 years. 22.4.4><109kg/s; 1.4><1015years 24.7mpc2, including therestmass oftheproton (kinetic energy is61/21,02) 28.-uS0.1150 30.0.8MeV 32.Tcmon =999.5MeV Tpm,=439MeV16. cos6= Note: Page numbers followed bynindicate foot- notes. A Acceleration, 30-34 centripetal, 393 force and, 56-58 Acceleration vector, 30-31 Acoustic systems, oscillation in,123 Action, 230 Adams,]ohn Couch, 3l3n Airmasses, motion of,398-399 Airresistance, 59,65-71 Algebraic functions, 613-614 Amplitude resonance frequency, 120-122, 123 Angular frequency ofdamped oscillations, 109, 110 ofharmonic oscillations, 102 Angular momentum conservation of,77-78, 262-265, 289-290 ofrigid body, 419-424, 454-455 insystem ofparticles, 336-339 Angular velocity, 34-37 inertia tensor and, 420 Anomalous dispersion, 542n Aphelion, 300 Apocenter, 300 Apogee, 300 Apsidal angle, 311-312 Apsidal distance, 299, 311 Apsides, 295, 299, 300, 311-312 Areal velocity, 290 Asymmetric forces/ potentials, 150 Asymmetric top,426 Atomic clock, 556-558 Atoms, oscillation of,123 Attenuated wave, 537, 538 Attractor, 151, 153, 169 chaotic, 169 strange, 169 Atwood’s machine, 71-73 Auxiliary equation, 600-603 Axes ofinertia, 424-428INDEX Axial vector, 25n Axis ofrotation, instantaneous, 34 B Baden-Powell, G.,535 Beats, 475, 539 Bernoulli Daniel, 468n, 599n _]akob, 207n Johann, 207n, 211n, 300n, 599n Bessel, F.W.,464n Bdecay, 81-82 Bifurcation, 170 pitchfork, 172 ' Bifurcation diagram, 171 Binomial expansion formulas, 608-609 Body cone, 450 Boltzmann, L.,90 Boundary-value problem, 513 Bowditch, Nathaniel, 106n Brachistochrone problem, 211-213 Brahe, Tyco, 290n Brillouin, Léon, 542n Butterﬂy effect, 145, 175 C Calculus ofvariations, 207-225, 272-274 with auxiliary conditions, 219-224 basic problem in,207-210 brachistochrone problem and, 211-213 constraint equations in,219-222 Dido Problem and, 222-224 Euler’s equation in,210-211, 219-224 second form of,216-218 extremum solutions and, 207-210 forgeodesic onsphere, 217-218 Hamilton’s principle in,229-233 8notation in,224-226 withseveral dependent variables, 218-219 soap ﬁlm problem and, 215-216 Canonical conjugates, 269 Canonical equations ofmotion, 265-274 Catenary, 215 643 Cauchy, August Louis, 447n Cavendish, Henry, 182 Cayley, A.,9n Center ofmass, insystem ofparticles, 329-331, 333, 411-412 position vectors for,336-337 Center-of-mass coordinate system, 346-353 elastic collisions in,346-358 laboratory coordinate system and, 346-347 Center-of-momentum system, 579-583 Central-force motion, 287-323 apsides and, 295, 299, 300, 311 areal velocity and, 290 aspidal angle and, 311-312 centrifugal energy and, 296-299 conservation theorems for,289-290 effective potential and, 296-299 elliptic, 301-303 equation ofmotion for,291-295 equivalent one-body problem for,288-289 ﬁrstintegral of,290 Kepler’s Laws and, 290, 303 Lagrangian for,289 innoninertial reference frame, 393, 402-404 orbital, 295-296, 300-323. SeealsoOrbit(s) reduced mass and, 287-289 inspace dynamics, 305-311 Central forces, 50 Centrifugal force, 296-299, 391-395 onEarth, 397 Centiipetal acceleration, 393 Ceres (asteroid), 304 Chaos, 144-178. SeeaLsoNonlinear oscilla- tions/ system butterﬂy effect and, 145, 175 deterministic, 145 identiﬁcation of,174-178 initial conditions and, 145, 175 inpendulum, 163-169 Chaotic attractor, 169 Characteristic equation forcoupled oscillations, 479 formoment ofinertia, 425 Characteristic frequencies, ofcoupled oscilla- tions, 471-473, 474, 479, 483-490 Characteristic polynomial, 425n Chasles’ theorem, 412n Cidenas, 452n Circular orbit, 301 stability of,316-323 Clausius, R.]. E.,278n Coeﬁicient ofrestitution, 359 Collisions. SeealsoScattering elastic, 345-358 incenter-of-mass coordinate system, 346-358 conservation theorems for,346-352 geometry of,347-348 kinematics of,352-358, 359 inlaboratory coordinate system, 347-348 velocity vectors for,346,351 endoergic, 359 exoergic, 359impact parameter for,363 impulsive forces in,361-362 inelastic, 358-362 oblique, 360 Q-value of,359 relativistic, 579-583 Rutherford scattering formula for,369-371 scattering angles for,363-369 Column matrix, 9 Comet Giacobini-Zinner, 311 Halley’s, 304-305, 311 orbit of,304-305 Shoemaker-Levy, 311 Complementary function, 118, 604 Complex quantities, 611 Compound pendulum, 413-415 Cone body, 450 space, 451 Conﬁguration space, 237, 274 Conservation theorems, 260-266 forangular momentum, 77-78, 262-264, 289-290 forcentral-force motion, 289-291 forcollisions, 346-352 forenergy, 78-81, 260-261, 290, 339-345 inLagrangian mechanics, 260-266 forlinear momentum, 52,76-77, 261-262, 265, 290-291, 331-339 forrocket infree space, 372-374 special relativity and, 562-566 formass-energy, 567 aspostulates vs.laws, 81 forsystem ofparticles, 289-291, 331-352 Conservative force, 81 Conservative system, 342 Constraint equations, 219-222 Constraints, 228 computation of,450-452 holonomic, 238-248 nonholonomic, 248-250 rheonomic, 238 scleronomic, 238 semiholonomic, 249n undetermined multipliers and, 250 Continuous string, 513-516, 528n Continuous systems, 513-542 Coordinates, rotating, 53-54, 388-391 Coordinate systems, 3-6 center-of-mass, 346-353 cyclic, 269-270 cylindrical, 31-34, 617-618 diﬁerential relations in,617-620 generalized, 221n, 233-248, 274 ininertial reference frame, 53-54 forinertia tensor, 428-432 laboratory, 346-358 inLagrangian mechanics, 257-258 moments ofinertia in,428-432 innoninertial reference frame, 387-407 normal, 468, 471-472, 478, 485-490 orthogonal, 7-8 plane polar, 31-33 rectangular, 3-9,617 rotating, 53-54, 388-391 spherical, 31-33, 619-620 transformation of,3-20. SeealsoTrans- formation(s); Transformation matrix Coriolis force, 392-395, 398-407 Cosine, direction, 4,6 Cotes, Roger, 325n Cotes’ spirals, 325n Coulomb scattering, 369-371 Coupled equations, 406 Coupled oscillations, 468-507 antisymmetrical, 472 beats and, 475 characteristic frequencies (eigenfrequencies) for,471-473, 474, 479, 483-490 damped, 522-524 degeneracy of,379,495-498 eigenvectors for,379, 479-483, 485-490 forced, 522-524 general problem of,475-481 harmonic, 469-481 molecular vibrations as,490-495 nearest neighbor interaction of,499 normal coordinates for,468, 471-472, 478, 485-490 symmetrical, 472 ofthree linearly coupled plane pendula, 495-498 ofvibrating suing, 498-507, 513-538. Seealso Vibrating string wave equation for,520-542 weakly coupled, 473-475 Coupled pendula, 164, 495-498 Cowan, C.L.,81 Critical damping, 114 Critical frequency, 537 Cross product, 25-28 Curl, ofvector, 38,43,79n Cutoff frequency, 537 Cyclic coordinates, 269-270 Cycloid, 213n Cylindrical coordinates, 31-34, 617-618 D Damped oscillations, 100, 108-117. Seealso Oscillations amplitude of,110, 111 amplitude resonance frequency of,120-123 angular frequency of,109, 110 coupled, 522-524 critically damped, 114 damping force and, 109 damping parameter for,109 decrement ofmotion for,111 inelectrical circuits, 123-126 equation ofmotion for,109, 144 kinetic energy resonance of,122-123 logarithmic decrement ofmotion for,111 overdamped, 114potential energy resonance of,123 resonance phenomena and, 120-123 sinusoidal driving forces and, 117-123 superposition principle and, 126-128 total energy of,111 underdamping and, 109-113 Damped wave, 537, 538 Damping negative, 153 radiation, 122 Damping parameter, 109 Dark matter, 190 Decibel, 516n Decrement ofmotion, 111 Deﬁnitions, vs.physical laws, 50 Degeneracy, 379, 495-498 Delta function, 133n Determinism, 144-145 Deterministic chaos, 145 Deuteron, 381 binding energy of,568 Dido Problem, 222-224 Diﬁerence equation, 169 Diﬁerential equations ﬁrst-order, 267 partial, separation ofvariables for,528 second-order, 267 Diﬁerential scattering cross section, 364 Dirac, Paul, 89 Direction cosine, 4,6 Dirichlet, Peter, 207n ' Discontinuous driving forces, 129-137 Dispersion, 535 anomalous, 542n normal, 542n Divergence, ofvector, 38 Divergence theorem, 42-43 Doppler effect, relativistic, 558-561, 576-577 Dotproduct, 21 Double pendulum, 164 Drag, 59,65-71 Driven oscillations, 100 coupled, 522-524 discontinuous, 129-137 sawtooth, 128-129 sinusoidal, 117-123 Dufﬁng equation, 161-162 E Earth Coriolis force on,398-401 data for,304 gravitational force on,395-397 motion relative to,395-407 asnoninertial reference frame, 387 orbit around, 300. SeeaLsoOrbit(s) precession of,316, 451-452, 451n shape of,451 Eccentricity, orbital, 300 Eddington, Arthur, 49 Effective potential, 296-299 ofrigid body, 457 Eigenfrequencies, forcoupled oscillations, 471-473, 474, 479, 483, 485-490 Eigenvalues, 440n Eigenvectors, 440n forcoupled oscillations, 379, 479-483, 485-490 orthogonal, 481-483 orthonormal, 482 Einstein, Albert, 89,546n, 549n, 551 Elastic collisions, 345-358. SeealsoCollisions, elastic Elastic deformations, restoring forces for,100 Elastic forces, 50 Electrical circuits, oscillations in,123-126 Electromagnetic ﬁeld, particle motion in, 73-76, 81 Electrostatic scattering, 369-371 Ellipsoid equivalent, 447 momental, 447n Elliptical integrals, 594-598 Elliptical orbit, 301-305 Endoergic collisions, 359 Energy, 82-87 centrifugal, 296-299 conservation of,78-87, 260-261, 265, 290, 339-345 ofelastic collisions, 352-358 gravitational, 186 heat as,82-83 kinetic. SeeKinetic energy mass and, 567-569 potential. SeePotential energy rest, 567 special relativity and, 566-569 ofsystem ofparticles, 339-345 total. SeeTotal energy ofvibrating string, 516-520 Eotvés, Roland von, 52 Equation (s) auxiliary, 600-603 characteristic (secular) forcoupled oscillations, 479 formoment ofinertia, 425 constraint, 219-222 coupled, 406 difference, 169 Dufﬁng, 161-162 Euler-Lagrange, 211n, 238 Euler’s. SeeEuler’s equations ﬁrst-order differential, 267 Helmholtz, 530 Lagrange’s, 229, 231-258 Laplace’s, 194 linear homogeneous, 599-603 inhomogeneous, 603-606 linear difference, 500 logistic, map of,170-172 Lorentz, 92 Maxwell’s, 547, 551 partial differential, separation ofvariables for, 528Poisson’s, 193-194 second-order differential, 267 Van derPol, 153-155 wave, 520-542 Equation(s) ofmotion canonical, 265-273 forcoupled oscillations, 478-479 fordamped oscillations, 109, 144 Hamiltonian principle and, 232-233 Hamilton’s, 265-273 forharmonic oscillations, 100-101, 232-233 Lagrange’s, 229, 231-258, 267n, 269 fornoninertial reference frame, 393, 402-404, 405 fornonlinear oscillations, 149 fororbit, 291-295, 313 forparticle, 55-76 forplane pendulum, 155-156, 232 fortwo-body systems, 291-295 forvibrating string, 500-501, 522 Equilibrium stable, 151 unstable, 151-152 Equilibrium points, 84-85 Equinox, precession of,312n, 451-452, 452n Equipotential surface, 194-195 Equivalence principle, 52 Equivalent electric circuits, 123-126 Equivalent ellipsoid, 447 Eros (asteroid), 304 Euler,Leonhard, 4911,20711,41811,42411,441'n,451n, 599n Eulerian angles, 412 forrigid body, 440-444 Euler-Lagrange equation, 211n, 238. Seealso Lagrange’s equations Euler’s equations, 210-211, 219-224 with auxiliary conditions, 219-224 forforce ﬁeld, 446 forforce-free motion, 446, 448-450 forrigid body, 444-448 second form of,216-218 with several dependent variables, 218-219 Euler’s theorem, 259 Exoergic collisions, 359 Exponential series, 610 External force, insystem ofparticles, 331-332 Extemal potential energy, 342 External torque, 337-338 Extremum solutions, 207-210. SeealsoCalculus ofvariations F Feigenbaum's number, 173-174 Fermat, Pierre de,230n Fermat's principle, 207, 230 Fermi, Enrico, 81 Field vector, 188-189 Finite rotation, 34 First-order differential equations, 267 FitzGerald, G.F.,552n FitzGerald-Lorentz length contraction, 552-553 Fixed-star reference frame, 53 Flux, gravitational, 192-194 Flux density, 363 Flybys, 308-311 Force, 55-76 acceleration and, 56-58 asymmetric, 150 central, 50 centrifugal, 296-299 conservative, 81 ofconstraint. SeeConstraints Coriolis, 392-395, 398-407 deﬁnition of,50 discontinuous driving, 129-137 elastic, 50 external, 331-332 frictional, 57-58 gravitational, 56,58,183-184, 194-198 impulsive, 361 internal, 331-332 inLagrangian mechanics, 257-258 lineof,194 moment of,77 inNewton’s First Law, 49-50 inNewton’s Second Law, 49,50 inNewton’s Third Law, 49,50-52 nonlinear, 146-147 vs.potential, 195-196 restoring, 99-100 retarding, 58-71 insystem ofparticles, 331-332 tidal, 199-204 total, 78 velocity-dependent, 50,59 zero, 49 Forced oscillations, 100 coupled, 522-524 discontinuous, 129-137 sawtooth, 128-129 sinusoidal, 117-123 Forced-pivot pendulum, 163-164 Foucault,]. L.,404n, 407n Foucault pendulum, 404-407 Fourier series, 127-129, 134, 498n Four-scalar, 572 Four-vector, 572-574 Frame ofreference ﬁxed-star, 53 inertial inLagrangian mechanics, 260-264 inNewtonian mechanics, 53 noninertial, 387-407 Free body (particle), 49 Free oscillations, 108 Frequency characteristic, ofcoupled oscillations, 471-473, 474, 479, 483-490 cutoff (critical), 537 ofdamped oscillations, 109, 110, 120-123 ofelectrical oscillations, 124-126 ofharmonic oscillations, 102Friction sliding (kinetic), 57-58 static, 57,58 tidal, 204 Functional, 208n G Galaxy, orbital speed in,188-189 Galilean invariance, 53,547-548 Galilean transformation, 547-548, 549 Galileo, 49n, 52,158n, 198, 407n Galileo (satellite), 310-311 Gamma functions, 615-616 Gaussian function, 133n Gauss’ theorem, 42-43 Generalized coordinates, 221n, 233-248, 274 deﬁnition of,233 Lagrange’s equations in,237-248 proper, 233 suitability of,234 Generalized momenta, 265 Generalized velocities, 234 General relativity, 546n, 547 Geodesic, 217-218 Giacobini-Zinner comet, 311 Gibbs,]. W.,1n,90,434n Gibbs phenomenon, 129 Grad, 38 Gradient, 38 Gradient operator, 37-40 Gravitation, 182-204 innoninertial reference frame, 395-397 ocean tides and, 198-204 rocket invertical ascent and, 374-378 Gravitational acceleration constant, 184 Gravitational energy, 186 Gravitational ﬁeld vector, 183-184 Gravitational flux, 192-194 Gravitational force, 56,58 computation of,183-184, 195-198 direction of,194-195 magnitude of,194-195 Gravitational mass, 51-52 Gravitational potential, 184-198, 297 continuous, 188 equipotential surface and, 194-195 Laplace’s equation and, 194 lines offorce and, 194 orbital speed and, 189-190 Poisson’s equation and, 192-194 asscalar quantity, 196 ofspherical shell, 186-188 ofthin ring, 190-192, 196-198 Gravitational potential energy, 186, 192-193 Great circle, 218 Green, George, 134n Green’s function, 136, 194 Group velocity, 539-542 H Hafele,_]. C.,556 Halley’s comet, 304-305, 311 Hamilton, William Rowan, 9n,40n, 207n, 230n, 539n Hamiltonian dynamics, 267-277 Hamiltonian function, 261 Lagrangian function and, 237, 261, 265-274 relativistic, 579 Hamilton-Jacobi theory, 270n Hamilton’s equations ofmotion, 265-274 Hamilton’s principle, 229-233, 272-274 Lagrange’s equations and, 257-258 modiﬁed, 273 Newtonian mechanics and, 257-258 variational, 237-239 Hard system, 146-147, 149, 150 Harmonic oscillations, 100-137. Seealso Oscillations inacoustic systems, 123 amplitude of,101, 102 amplitude resonance frequency of,120-122 angular frequency of,102 inatomic systems, 123 coupled, 469-481. SeealsoCoupled oscilla- tions damped, 100, 108-117. SeealsoDamped oscil- lations driven (forced), 100 coupled, 522-524 discontinuous, 129-137 sawtooth, 128-129 sinusoidal, 117-123 inelectrical circuits, 123-126 equation ofmotion for,100-101 free, 108 impulsive forcing functions and, 129-137 inisochronous system, 102 kinetic energy resonance of,122-123 Lagrange equation ofmotion for,232 linear, 99-137 inmechanical systems, 123 one-dimensional, 100-104 period ofmotion in,102, 110 phase diagram for,106-108 potential energy resonance of,123 representative point for,108 resonance phenomena and, 120-123 restoring forces for,99-100 small oscillations assumption for,102 superposition principle and, 126-128 total energy of,101 twodimensional, 104-106 Heat, asenergy, 82-83 Heaviside function, 1n,130-132 Heisenberg, Max Born, 89 Heisenberg’s uncertainty principle, 89 Helmholtz, Hennann von, 83,530n Helmholtz equation, 530 Hermitean tensor, 440n Hero ofAlexandria, 229 Histogram, 355 Hohmann, Walter, 305n Hohmann transfer, 305-308 Holonomic constraints, 238-248Homogenous equations, linear, 599-603 Homogenous function, Euler’s theorem of,259 Hooke, Robert, 305, 401n Hooke’s law,100 Huygens, Christiaan, 158n, 297n, 346n, 418n Hyperbolic functions, 611-612 Hyperbolic orbit, 301 Hysteresis, 163 I Identity matrix, 12-13 Impact parameter, 363 Improper rotations, 19 Impulse, 361-362 Impulse function, 130-137 Impulsive forces, 361 Inelastic collisions, 358-362 Inertial mass, 51-52 Inertial moment. SeeMoment ofinertia Inertial products, 418 Inertial reference frame four-dimensional interval between twoevents in,621-622 inLagrangian mechanics, 260-264 inNewtonian mechanics, 53 Inertia tensor, 415-440 angular momentum and, 419-424 angular velocity and, 420 under coordinate transformations, 433-435 diagonalization of,435-437 indifferent coordinate systems, 428-432 elements of,417-418 ofﬁrst rank, 434n asmatrix, 434 moment-of-inertia, 418, 425-428, 439-440, 447 principal axes of,424-432, 438-439 principal moments of,418, 425-432, 439-440, 447 product-of-inertia, 418 symmetric, 440 transformation of,433-440 asvector, 434n Inﬁnitesimal rotation, 34-37 Inhomogeneous equations, linear, 603-606 Instability, unbounded motion and, 151-152 Instantaneous axisofrotation, 34,388 Integral(s), 613-616 elliptical, 594-598 line, 41-42 particular, 604 Intelsat satellite, 462 Intensity, ofscattered particles, 363-364 Intemal force, insystem ofparticles, 331-332 Intemal potential energy, 342-343 Intemal torque, 338 International Cometary Explmvr, 311 International Sun-Earth Explorer 3,311 Interplanetary transfer, 308 Inversion, 13,18 Inversion matrix, 13,18,19 Isochronous system, 102 JJacobi, C.G.S.,207n, 270n Joule,James Prescott, 83 Jumps, 161-163 Jupiter data for,304 precession of,316 travel to,309, 311 K Kater, Henry, 464n Kater’s reversible pendulum, 464 Keating, Richard, 556 Kelvin, Lord, 265n, 535n Kepler,Johannes, 290n Kepler’s Laws, 290, 303 Kinetic energy, 78-81, 258-259, 278, 340-341. SeealsoEnergy ofcenter-of-mass system, 352-353 ofelastic collisions, 352-358 ofrigid body, 415-417, 438 rotational, 415-417, 438 ofscleronomic systems, 259 special relativity and, 566-569 ofsystem ofparticles, 340-341, 352-358 time-averaged, 519 translational, 415-417, 438 ofvibrating string, 517 virial and, 278 Kinetic energy resonance, 122-123 Kinetic friction, 57-58 Kinetic potential, 196n Kronecker delta symbol, 7 L Laboratory coordinate system, 346-358 Lactus rectum, 300 Lagrange,Joseph, 207n, 230, 238n, 267n, 454n, 468n Lagrange’s equations ofmotion, 229, 231-258, 267n, 269. SeealsoEquation(s) ofmotion ingeneralized coordinates, 237-248 Hamilton’s equations and, 237, 261, 265-274 Hamilton's principle and, 257-258, 272-273 holonomic constraints in,238-248 Newton’s equations and, 254-258 nonholonomic constraints in,248-250 rheonomic constraints in,238 scleronomic constraints in,238 with undetermined multipliers, 248, 249, 250 utility of,248, 250 variational, 272-273 Lagrange undetermined multiplier, 221 Lagrangian function, 196n, 231 Hamiltonian and, 237, 261, 265-271 invariance of,260-264 relativistic, 578-579 asscalar function, 237, 258 Lagrangian mechanics conservation theorems in,260-266 energy vs.force in,257-258Newtonian mechanics and, 254-258 scalar operations in,258 Laplace, Pierre Simon de,40n, 144 Laplace’s equation, 194 Laplacian operator, 40 Larmor,J.J., 549 Law(s) altemative statements of.SeeCalculus ofvaria tions conservation, vs.postulates, 81 Hooke’s, 100 Kepler’s, 290, 303 Newton’s First, 48-49 Newton’s Second, 49 Newton’s Third, 49-53 physical, 1,48,50 ofrefraction, 230 ofresistance, 59t ofuniversal gravitation, 182-184 Least action principle, 230,258 Least constraint principle, 230 Least curvature principle, 230 Legendre, Adrien, 207n Legendre transfornrations, 266 LeVerrier, UrbainJ.J., 3l3n Levi-Civita density, 25 Lift, 59 Light oscillation and, 123 speed of,89n Light cone, 570 Limit cycle, 153-154 Linear difference equation, 500 Linear equations homogeneous, 599-603 inhomogeneous, 603-606 Linearly dependent functions, 600 Linearly independent functions, 600 Linear momentum conservation of,52,261-262, 265, 290-291 forrocket infree space, 372-374 insystem ofparticles, 331-339 special relativity and, 562-566 Linear operator, 127 Linear oscillations, 99-137. Seealso Oscillations Linear velocity, 30-34 direction of,35 magnitude of,35 Line integral, 41-42 Line offorce, 194 Line ofnodes, 442 Liouville,J., 90 Liouville’s theorem, 277 Lissajous curve, 106 Loaded string problem, 498-507. Seealso Vibrating string Logarithmic decrement ofmotion, 111 Logarithmic series, 610 Logistic map, 170-172 Longitudinal vibrations, 491-495 Longitudinal wave, 512 Lorentz, Hendrik A.,89,546n, 549n, 550-551, 552n Lorentz equation, 92 Lorentz transformation, 548-555 Lyapunov exponents, 175-178 M Mach, Ernest, 49n Maclaurin, Colin, 590n Maclaurin’s series, 590n Magnetic ﬁeld, particle motion in,73-76, 81 Magnetic pendulum, 164 Mapping, 169-174 logistic, 170-172 Margenau, H.,258 Mars data for,304 precession of,316 travel to,308 Mass center of position vectors for,336-337 insystem ofparticles, 329-331, 333, 411-412 gravitational, 51-52 inertial, 51-52 inNewton’s Third Law, 50-52 reduced, 287-289, 303 relativistic, 565 rest, 564 unit, 51 Mass-energy equivalence kinetic energy and, 567 momentum and, 568-569 Mathematical physics, 513 Matrix, 4 addition of,13 column, 9 geometrical signiﬁcance of,14-20 identity, 12-13 inversion, 13,18,19 multiplication of,9-12 orthogonality of,8,18-19 properties of,6-8 rotation (transformation), 4-20 rotation of,14-20 row, 9 square, 9 tensors and, 434 transposed, 12,18-19 Matrix operations, 9-12 Matrix theory, development of,9n Maupertuis, P.L.M.de,230, 258 Maupertuis’s principle ofleast action, 230, 258 Maxwell,James Clerk, 80,90 Maxwell’s equations, 547, 551 Mécaniqw: analytiqwe (Lagrange), 238n Mechanical quantities, analogous electric quan- tities and, 125, 125t Mechanics Lagrangian, 231-258 Newtonian, 48-90. SeealsoNewtonian me- chanicsquantum, 89 statistical, 90 Mercury, 304 precession of,312-313, 316 Method ofundetermined coefﬁcients, 605 Michelson-Morley experiment, 546, 549, 552n Minimal principles, 229-233 Minkowski, Herrrian, 571 Minkowski space, 571 Modiﬁed Hamilton’s principle, 273 Molecular vibrations, 123, 490-495 Momental ellipsoid, 447n Moment offorce, 77 Moment ofinertia, 418, 425-428 indifferent coordinate systems, 428-432 principal, 425-432, 438-439, 447 secular (characteristic) equation for,425 Moment-of-inertia tensor, 418, 425-428, 439-440 Momentum angular. SeeAngular momentum conservation of,52 deﬁnition of,50 four-vector, 572-574 generalized, 265 linear. SeeLinear momentum mass-energy and, 568-569 position and, 88-89 relativistic, 562-566 Momentum space, 274 Moon, tides and, 198-204 Motion incomplex systems, 90 decrement of,111 equations of.SeeEquation (s)ofmotion logarithmic decrement of,111 ofparticle, 55-76 inAtwood’s machine, 71-73 conservation theorems for,76-82 inelectromagnetic ﬁeld, 73-76, 81 energy and, 82-87 resistive forces on,58-71 unbounded, 152 Muon decay, 555-556 N Natural Philosophy (Thomson 8cTait), 265n Neap tides, 203 Neighboring function, 208 Neptune data for,304 travel to,309, 310 Newton, Isaac, 49-50, 52,59,182, 198, 207n, 305, 499n Newtonian mechanics, 48-90 conservation theorems in,76-82 energy in,82-87 equation ofmotion forparticle in,55-76 First Law of,48-49 frames ofreference for,53-54 Lagrangian mechanics and, 254-258 lawofuniversal gravitation in,182-184 limitations of,88-90 quantum mechanics and, 89 Second Law of,49 system sizeand, 89-90 Third Law of,49-53 strong form of,329 weak form of,328-329 time in,89 Newtonian relativity, 548 principle of,53 Newton’s rule, 359-360 Niven, C.,434n Nodes lineof,442 ofwave function, 531 Nonholonomic constraints, 248-250 Noninertial reference frame, 387-407 centrifugal force in,391-395, 397 Coriolis force in,392-395, 398-401 Earth as,387, 395-407 motion relative toEarth in,395-407 rotating coordinates in,388-391 tides as,387 Nonlinear oscillations/ system amplitude of,149 attractor for,151, 153 chaotic nature of,145. SeealsoChaos deterministic chaos and, 145 equation ofmotion for,149 hard, 146-147, 149, 150 hysteresis in,163 jumps in,161-163 limit cycle for,153-154 Lyapunov exponents for,175-178 mapping and, 169-174 phase diagram for,150-155 phase lagsin,163 plane pendulum as,155-160 progression of,169-174 self—limiting, 154 separatrix in,160 soft, 146-147, 150 vanderPolequation for,153-155 Nonsyrnmetrical coupled oscillations, 472 Normal coordinates forcoupled oscillations, 471-472, 478, 485-490 deﬁnition of,468 Normal dispersion, 542n 8notation, 224-226 Nuclei, collective excitation of,123 Numerical method, forretarding forces, 68-69 Nutation, ofrigid body, 459-460 O Oblique collisions, 360 Ocean tides, 198-204 neap, 203 asnoninertial system, 387 spring, 203 Orbit(s) apocenter of,300aspidal angle of,311-312 aspidal distance of,299, 311-312 incentral ﬁeld, 295-296 circular, 301 stability of,316-323 closed, 295, 311 ofcomet, 304-305 conic sections of,300-301 Cotes’ spiral, 325n eccentric, 300 elliptical, 301-305 equation ofmotion for,291-295, 313 hyperbolic, 301 latus rectum of,300 major/ minor axes of,301-302, 304 open, 295, 311-312 parabolic, 301 pericenter of,300 period of,302-303 planetary, 301-305 precessional ratefor,312-316 ofrocket, 305-311 turning points (apsides) of,295, 299, 300, 311-312 Orbital dynamics, 305-311 Orbital speed, 188-189 Orthogonal coordinate systems, 7-8 Orthogonal eigenvectors, 481-483 Orthogonality, ofrotation matrix, 8,18-19 Orthogonality condition, 8 Orthogonal transformations, 8,18-19 angle-preserving property of,23 distance-preserving property of,23 geometrical representation of,14-20 Orthonormal eigenvectors, 482 Oscillations, 99-178 inacoustic systems, 123 amplitude resonance frequency of,120-122 inatomic systems, 123 coupled, 468-506. SeealsoCoupled oscilla- tions damped, 100, 108-117. SeealsoDamped oscil lations driven (forced), 100 coupled, 522-524 discontinuous, 129-137 sawtooth, 128-129 sinusoidal, 117-123 inelectrical circuits, 123-126 free, 108 harmonic, 100-137. SeealsoHamionic oscilla tions impulsive forcing functions and, 129-137 kinetic energy resonance of,122-123 linear, 99-137 longitudinal, 491-495 inmechanical systems, 123 molecular, 123, 490-495 nonlinear, 144-178 nonsymmetrical, 472 ocean tides and, 203 potential energy resonance of,123 Oscillations (continued) resonance phenomena and, 120-123 restoring forces for,99-100 insteady-state systems, 100-129 superposition principle and, 126-128 symmetrical, 472 transverse, 491-495 Overdamping, 114-115 P Parabolic orbit, 301 Partial differential equations, separation ofvari- ables for,528 Particle systems. SeeSystem ofparticles Particular integral, 604 Particular solution, 118 Pauli, Wolfgang, 81 Pendulum(a) chaotic motion of,163-169 compound, 413-415 coupled, 164, 495-498 double, 164 forced-pivot, 163-164 Foucault, 404-407 Kater’s reversible, 464 Lyapunov exponents for,177-178 magnetic, 164 asnonlinear system, 155-160 phase diagram for,158-160, 168-169 physical, 413-415 plane, 155-160 equation ofmotion for,155-156, 232 Pericenter, 300 Perigee, 300 Perihelion, 300 precession of,312-316 Period, orbital, 302-303 Period doubling, 166 Periodic functions, Fourier’s theorem and, 127-128 Permutation symbol, 25 Perturbation method fornonlinear forces, 149 forretarding forces, 67-68 Phase (¢),533 Phase angle, 101n Phase diagram, 107 forharmonic oscillations, 106-108 fornonlinear oscillations, 150-155 forplane pendulum, 158-160, 168-169 Poincaré sections in,166-169 Phase lags, 163 Phase plane, 107 Phase space, 107 particle density in,274-277 Phase velocity, 534-537, 576 Physical laws, 1,48,50 vs.deﬁnitions, 50 Physical pendulum, 413-415 Physical systems, oscillation in,123-125 Physics, mathematical, 513 Pitchfork bifurcation, 172Plane oftheelliptic, 451 Plane pendula, 155-160. SeealsoPendulum(a) equation ofmotion for,155-156, 232 three linearly coupled, 495-498 Plane polar coordinates, 31-33 Planets data for,304 motion of,300-305, 312-316. SeealsoCentral- force motion; Orbit(s) reduced mass of,303 travel to,308-311 Plane wave, 513 Pluto, 304, 316 Poincaré, Henri, 89,145n, 166, 546n Poincaré section, 166-168 Poinsot construction, 447 Poisson, S.D.,193, 267n, 398n Poisson brackets, 284 Poisson’s equation, 193-194, 267n Polar coordinates, 31-34 Poles, precession of,451-452, 451n Polynomial, characteristic, 425n Position momentum and, 88-89 inNewtonian mechanics, 48-53, 88 Position vector, 22,30 Positive deﬁnite quantities, 477n Potential asymmetric, 150 effective, 296-299 ofrigid body, 457 gravitational, 184-198, 297. Seealso Gravitational potential kinetic, 196n screened Coulomb, 319-320 Potential energy, 78-80, 185-186. SeealsoEnergy ofthebody, 186 centrifugal, 296-299 extemal, 342 gravitational, 186, 192-193 intemal, 342-343 ofsystem ofparticles, 341-345 time-averaged, 519 total, 342 ofvibrating string, 517-518 Potential energy resonance, 123 Precession Coriolis force and, 404-407 deﬁnition of,312 ofequinox, 312n, 451-452, 452n ofplanets, 312-316 ofpoles, 451-452, 451n ofrigid body, 450-453, 458-460 Principal axes ofinertia, 424-432, 438-439 Principal moments ofinertia, 418, 425-432, 439-440, 447 Principia (Newton), 49n, 182, 498n Principle (s) ofequivalence, 52 Fermat’s, 207, 230 Hamilton’s, 229-233, 237-239, 257-258, 272-274 Heisenberg’s uncertainty, 89 ofleast action, 230,258 ofleast constraint, 230 ofleast curvature, 230 minimal, 229-233 ofNewtonian relativity, 53,548 ofrelativity, 547 ofsuperposition, 127-129, 134, 471, 498n Probability theory, 90 Problem-solving techniques, 55 Product-of-inertia tensor, 418 Products ofinertia, 418 Propagating wave, 526 Propagation constant, 530 Proper rotations, 19 Proper time, 555 Pulleys, inAtwood’s machine, 71-73 QQualitative analysis, 355 Quantitative analysis, 355 Quantum mechanics, 89 Q-value, 359 R Radiation damping, 122 Radius ofgyration, 463 Rayleigh, Lord, 539n Rectangular coordinates, 3-9, 617 Red shift, 560 Reduced mass, 287-289, 303 Reference frame ﬁxed-star, 53 inertial four-dimensional interval between two events in,621-622 in mechanics, 260-264 inNewtonian mechanics, 53 noninertial reference, 387-407 Reﬂection coefﬁcient, 533 Refraction, Snell’s lawof,230 Reich, F.,401n Reines, F.,81 Relativistic collisions, 579-583 Relativistic Doppler effect, 558-561, 576-577 Relativistic Hamiltonian, 579 Relativistic kinematics, 579-583 Relativistic Lagrangian, 578-579 Relativistic length contraction, 552-553 Relativistic mass, 565 Relativistic momentum, 562-566 Relativistic triangle, 574 Relativity, 546 general, 546n, 547 mass-energy equivalence in,567 Newtonian, 548 principle of,547 special, 89,546-583 theory of,546-547 Representative point, 108 Resonance amplitude, 120-122, 123kinetic energy, 122-123 potential energy, 123 Rest energy, 567 Rest mass, 564 Restoring forces, 99-100 Retarding forces, 58-71 numerical method for,68-69 perturbation method for,67-68 Rheonomic constraints, 238 Right-hand nile, 14n Rigid body, 411-462. SeealsoSystem ofparticles angular momentum of,419-424, 454-455 asymmetric topas,426 center ofmass of,339-341, 411-412 deﬁnition of,411 effective potential of,457 equations ofmotion for,442-443 equivalent ellipsoid for,447 Eulerian angles for,440-444 force-free motion of,448-454 inertia tensor of,415-440. SeealsoInertia tensor kinetic energy of,415-417, 438 nutation of,459-460 Poinsot construction for,447 precession of,450-453, 458-460 principal axes ofinertia for,424-432, 438-439 principal moments ofinertia for,418, 425-432, 439-440, 447 , rotational stability of,460-462 rotor as,426 simple planar motion of,412 spherical topas,426 symmetric topas,426 inuniform force ﬁeld, 454-460 Rockets. SeealsoSpace travel infree space, 371-374 orbital dynamics and, 305-311 vertical ascent under gravity of,374-378 Rotating coordinates, 53-54, 388-391 Rotation direction of,14 ﬁnite, 34 improper, 19 inﬁnitesimal, 34-37 instantaneous axisof,34 proper, 19 Rotational kinetic energy, 438 ofrigid body, 415-417, 438 Rotation matrix. SeeTransformation matrix Rotation vectors, 35-36 Rotor, 426 Row matrix, 9 Rumford, Count, 82-83 Rutherford scattering formula, 369-371 S Satellites, rotational stability of,462 Satum data for,304 travel to,309-310 Sawtooth driving forces, 128-129 Scalar deﬁnition of,2,20 world, 572 Scalar function, gradient of,37-40 Scalar product, 21-23 vector product and, 26-27 Scattering, 345. SeealsoCollisions axially symmetric, 347n electrostatic, 369-371 ﬂuxdensity (intensity) in,363-364 inforce ﬁeld, 345-346 histogram for,355 Rutherford, 369-371 Scattering angle, 350, 363-369 Scattering cross section, 363-369 differential, 364 isotropic, 368 total, 369, 371 Schrodinger, Erwin, 89,540n Scleronomic constraints, 238 Scleronomic systems, kinetic energy in,259 Screened Coulomb potential, 319-320 Second-order differential equations, 267 Secular equation forcoupled oscillations, 479 formoment ofinertia, 425 Self-limiting system, 154 Semiholonomic constraints, 249n Semimajor axis, oforbit, 304 Separation ofvariables, 528 Separatrix, 160 Shoemaker-Levy comet, 311 Signal velocity, 541-542, 576 Similarity transformation, 434 Sinusoidal driving forces, 117-123 Sliding friction, 57-58 Small oscillations assumption, 102 Snell, Willebrord, 230n Snell’s lawofrefraction, 230 Soap ﬁlm problem, 215-216 Softsystem, 146-147, 150 Solar system objects in,data for,304 orbital motion in.SeeOrbit(s) Sommerfeld, Arnold, 542n Space homogeneity of,261-262, 265 isotropic, 53,262, 265 momentum, 274 phase, 107, 274-277 world (Minkowski), 571 Space cone, 451 Spacetime, 569-579 Space travel. SeealsoRockets central-force motion in,305-311 Hohmann transfer in,305-308 interplanetary, 308-311 Special relativity, 89,546-583covariance in,548-555 Doppler effect and, 558-561, 576-577 energy and, 566-569 experimental veriﬁcation of,555-558 four-vector and, 572-574 Galilean invariance and, 547-548 Hamiltonian in,579 kinematics in,579-583 Lagrangian in,578-579 length contraction and, 552-553 light cone and, 570-571 mass and, 565 momentum and, 562-566 muon decay and, 555-556 spacelike interval and, 570-571 spacetime and, 569-579 time dilation and, 554-555, 556-558 timelike interval and, 571 twin paradox and, 561-562 velocity addition rule for,574-576 worldline and, 570 world space and, 571 Speed, orbital, 188-189 Speed oflight, 89n Spherical coordinates, 31-33, 619-620 Spherical symmetry, 289 Spherical top,426 Spiral galaxy, orbital speed in,188-189 Spring tides, 203 Square matrix, 9 Stable equilibrium, 151 Standing waves, 531 Static friction, 57,58 Statistical mechanics, 90 Steady—state solution, 119 Steiner’s parallel-axis theorem, 430 Step function, 130-132 Stirling,James, 590n Stoke’s lawofresistance, 59 Stokes’s theorem, 42-43 Strange attractor, 169 Sun mass of,304 orbit around, 300. SeealsoOrbit(s) tides and, 202 Superposition principle, 127-129, 134, 471, 498n Sylvester, J.,9n Symmetrical coupled oscillations, 472 Symmetric tensor, 440 Symmetric top,426 force-free motion of,448-454 inuniform force ﬁeld, 454-460 System ofparticles angular momentum of,336-339 center ofmass in,329-331, 333, 339-341 411-412 position vectors for,336-337 central-force motion in,287-323 center-of-momentum system in,579-583 collisions in,345-362. SeealsoCollisions collisions and. 579-583 conservation theorems for.289-290 conservative, 342 energy of,339-345 equations ofmotion for,291-295 equivalent one-body problem for, 288-289 external force in,331-332 external torque in,337-338 ﬁnal state of,346n initial state of,346, 346n internal force in,331-332 internal torque in,338 linear momentum of,331-335 qualitative analysis of,355 quantitative analysis of,355 rigid body as,411. SeealsoRigid body T Tait, P.G.,265n Taylor, Brook, 589n Taylor's theorem, 589 Tensor Hermitean, 440n inertia, 415-440 matrices and, 434 symmetric, 440 unit, 432 asvector, 434n Terminal velocity, 63 Theorem(s) Chasles,' 412n conservation, 260-266, 289-291 divergence, 42-43 Euler’s, 259 Fourier’s, 127-128 Gauss,’ 42-43 Liouville’s, 277 Steiner’s parallel axis, 430 Stokes’s, 42-43 Taylor's, 589 virial, 277-278 Theory ofrelativity. SeeRelativity Thomson, William, 265n, 535n Tides, 198-204 neap, 203 asnoninertial system, 387 spring, 203 Time absolute, 89 homogeneity of,53-54, 260, 265 inNewtonian mechanics, 89 proper, 555 inspecial theory ofrelativity, 89 Time-dependent wave function, 524n Time dilation, 554-558 twin paradox and, 561-562 Time-independent wave function, 524n Timelike interval, 571 Top. SeealsoRigid body asymmetric, 426 spherical, 426 symmetric, 426force-free motion of,448-454 inuniform ﬁeld, 454-460 Torque, 77 extemal, 337-338 internal, 338 Total energy, 80.SeealsoEnergy special relativity and, 567-569 ofsystem ofparticles, 339-345 ofvibrating string, 519 Total force, 78 Trade winds, 399 Transformation(s) Eulerian angles in,412, 440-444 Galilean, 547-548, 549 ofinertia tensor, 433-440 Legendre, 266 Lorentz, 548-555 orthogonal. SeeOrthogonal transformations similarity, 434 Transformation matrix, 4-20. SeealsoMatrix addition of,13 column, 9 deﬁnition of,4 geometrical signiﬁcance of,14-20 identity, 12-13 inverse of,13,18 multiplication of,9-12 orthogonality of,8,18-19 properties of,6-8 rotation of,14-20 row, 9 square, 9 transposed, 12,18-19 transpose of,18-19 Transient effects, 119, 129 Translational kinetic energy, 438 ofrigid body, 415-417, 438 Transposed matrix, 12,18-19 Transverse vibrations, 491-495 Transverse wave, 512 Traveling wave, 526 Trigonometric functions, 614-615 Trigonometric relations, 609-610 Trigonometric series, 610 Turning points (apsides), 295, 299, 300, 311-312 Twin paradox, 561-562 Two-body system, 287-323, 346. SeealsoCentral- force motion; Orbit(s); System ofparticles collisions in,345-362 conservation theorems for,289-290, 331-352 equations ofmotion for,291-295 equivalent one-body problem for,288-289 spherical symmetry in,289 U Unbounded motion, 152 Uncertainty principle, 89 Underdamping, 109-113 Unit mass, 51 Unit tensor, 432 656 Index Unit vector, 23-24 Venus Unstable equilibrium, unbounded motion and, data for,304 151-152 Uranus data for,304 travel to,309, 310 V VanderPolequation, 153-155 Variational calculus. SeeCalculus ofvariations equation ofmotion for,500-501 522 Vector(s), 1-2, 23-43 acceleration, 30-31 addition of,20-21 axial, 25n components of,20-21, 29 curlof,38,43,79n deﬁnition of,2,20 differentiation of,29-30 direction cosines of,21 divergence of,38 ﬁeld, 188-189 gradient operator of,38-40 gravitational ﬁeld, 183-184 integration of,40-43 lineintegral of,41-42 magnitude of,21 multiplication of,21-23 scalar product of,21-23precession of,316 travel to,311 Vibrating suing, 498-507, 513-538 Seealso Coupled oscillations continuous, 513-516, 528n damped, 522-524 energy of,516-520 phase velocity of,534-537 plucked, 515-516 Vibrations. SeealsoOscillations longitudinal, 491-495 molecular, 123, 490-495 transverse, 491-495 Virial theorem, 277-278 Viviani, Vincenzo, 407n Voigt, W.,549 Voyager spacecraft, 309-310 W Wallis,John, 346n Wave damped (attenuated), 537, 538 dispersion of,535 longitudinal, 512 phase(¢) of,533 vector product of,25-28, 30-34 phase velocity of,534 position, 22,30 rotation, 35-36 tensor as,434nplane, 513 standing, 531 transverse, 512 transformation properties of,20-21, 36 traveling (propagating), 526 unit, 23-24 velocity, 30-33Wave equation, 513, 520-522 boundary conditions for,532 Vector differential operators, 37-40 general solutions of,524-527 Vector product, 25-28 derivatives of,30-34 scalar product and, 27 Vector sums, 20-21 derivatives of,30-34 Velocity, 30-34 angular, 34-37 inertia tensor and, 420 areal, 290 force and, 50,59 four-vector, 573 generalized, 234 group, 539-542 linear, 30-34, 35 phase, 534, 576 inquantum mechanics, 89 retarding forces and, 58-71 signal, 541-542, 576 temrinal, 63 wave, 537 Velocity addition rule, 574-576 Velocity vector, 30-33one-dimensional, 513 separation of,527-533 wave function in,524 Wave form, 534 Wave function, 524 nodes of,531 time-dependent, 524n time-independent, 524n Wave number, 530 complex, 536 Wave packet, 540-541 Wave velocity, 537 Weakly coupled oscillations, 473-475 Weierstrass, Karl, 207n, 485n Work, 78-79. SeealsoEnergy Worldline, 570 World scalar, 572 World space, 571 Wren, Christopher, 305, 346n Wronskian determinant, 600 Z Zero force, 49