N

VECTOR ANALYSIS

The author will be grateful to any readerwho will call his attention to an error in thisbook, or who will suggest new problems,changes in the text, or additions thereto.Careful consideration will be given to suchsuggestions and criticisms, as it is the author'sdesire to make his book as nearly perfect aspossible. Communications of this natureshould be addressed to Dr. J. G. Coffin, 199Elizabeth Ave., Hempstead, N. Y.

VECTOR ANALYSISAN INTRODUCTION

TO

VECTOR-METHODSAND THEIR VARIOUS APPLICATIONS

TO

PHYSICS AND MATHEMATICS

BY

JOSEPH GEORGE COFFIN, B.S., PR. D.MASS. INST. TECH. '98 AND CLARK UNIVERSITY '03)

EX-ASSOCIATE PROFESSOR OF PHYSICS AT THE COLLEGE OF THECITY OF NEW YORK

CONSULTING AND DEVELOPMENT ENGINEER WITH THEGENERAL BAKING COMPANY

SECOND EDITION

NEW YORK

JOHN WILEY & SONS, INC.LONDON : CHAPMAN & HALL, LIMITED

COPYRIGHT, 1909, 1911,

BY

J. G. COFFIN

Printed in U. S. A.

%tanbope pmlsP. M. GILSON COMPANY

BOSTON. U.s.A. 4_26

PREFACE

EVER since the development of Quaternion analysis, bySir William Rowan Hamilton, and of the "Ausdehnungs-lehre," by Grassmann, there has been a growing feeling thatthe older and more.. common processes of analysis were insome way artificial and complex.

This fact exists, for it is such, because these newermethods and ideas apply more naturally, more simply andmore directly to many of the conceptions of geometry,mechanics and mathematical physics, than those longaccepted.

Why then have these admitted advantages not led to a moreuniversal adoption of these methods? The answer seems to bethat the required change of ideas, of manner of thought andof notation, was too radical. It is well known that changesevolve slowly, and although to many, evolution is far too slowa process, the only way to proceed is to aid to the best ofone's ability in bringing about the desired result.

One who has studied and labored over the applicationsof mathematical analysis to physical and geometrical prob-lems, naturally has reluctance to discard the old familiarlooking formulae and start anew in an unknown and radicallydifferent language.

However great the skill and ingenuity shown by thepioneer in solving problems by Quaternions, there wasalways left the thought to the unbiased student that a lackof parallelism existed between the old and the new methodsof treatment. Such a lack undoubtedly does exist, but it isonly during the last few years that a method has been

Ti PREFACE.

evolved which avoids this fatal defect. It is chiefly throughthe labors of Gibbs and Heaviside that an analysis hasbeen perfected which not only does away with the unnec-essary complexity and artificiality of other analyses butoffers a strictly natural and therefore as direct and simple asubstitute as possible. and, at the same time in no wise is atvariance, but runs parallel to them.

This new, yet old method is Vector Analysis; it com-bines within itself most of the advantages of both Quater-nions and of Cartesian Analysis.

The adoption of Vector Analysis is urged on the groundsof naturalness, simplicity and directness; with it the truemeaning of processes and results is brought out as clearlyas possible, and desirable abbreviation is obtained.

It is admitted, that to a straight and clear thinker, almostany notation or mathematical method suffices, and to sucha one, changes in notation or method may appear hardlyworth while. He has already attained one of the resultswhich, perforce, follow the intelligent assimilation of avector method of thinking. To him there is left but theattainment of a simple notation which is the logical accom-paniment of clear thought. A few examples of vector con-centration are to be found in the exercises of the last chapterof this book. But the sole use of vector notation, withoutthe insight and clear conceptions which should obtain at thesame time, is without any value whatsoever, vitiates thevector point of view, and is contrary to the spirit of it.

It is almost unnecessary to state that the mind of thephysicist ought to be of the visual type so well exemplified inthe mind of Faraday. He should see the lines of forceemerging from the magnet; see that they are continuouswithin the metal; follow them, in his mind's eye, asthey are displaced by various causes; he should have somesort of a visual conception of the manner in which theelectro-magnetic waves are traveling through the ether

PREFACE.

around him; to him the divergence and the divergencetheorem should have a simple meaning.

To a mind other than this, the study of mathematicalphysics must be merely a series of analytical transforma-tions without the vitality of their visual significance. Topurely analytical minds, as distinct from the visual or intu-itive type, the methods of Vector Analysis reduce to littlemore than an analytical shorthand. To the intuitive mind,however, they are illuminative and simplifying, allowingthe mind to grasp and the hand to write the essential factsand transformations, unembarrassed with the generally unde-sirable complexity of Cartesian symbolism. It is impossibleto study and to apply Vector Analysis to problems and notto have one's ideas and thought made clearer and better bythe labor involved.

There are very good reasons for all these advantages. InNature we are confronted with quantities called scalarswhich have size or magnitude only, and also with otherquantities called vectors which have direction as well asmagnitude. In order to manipulate vector quantities bythe older methods, they were decomposed into three com-ponents along three arbitrary axes and the operations madeupon these components. Is it not evident that the bringingin of three arbitrary axes is an artificial process, and thatthe decomposition of the vector into components along theseaxes is also artificial, unnatural even? Why not go directlyto the vector itself and manipulate it without axes andwithout components? To do this, is possible, and in thefollowing pages, an attempt is made to show how it may bedone.

There is still another ground for urging more extendedstudy of Vector Analysis than now obtains. So many physi-cists of renown have been converted to its methods and usethat to ignore their leadership is an impossibility. Whensuch men as Lorentz, Foppl, Heaviside, Bucherer, Gibbs,

PREFACE.

Abraham, Bjerknes, Sommerfeld, Cohn and many others areconverted to its use, it is high time that the student famil-iarize himself at least with vector notation, even if not tobecome an expert in its use.

No one can deny the vast improvement that has takenplace, in recent years, in our conceptions of physical pro-cesses; and few will deny that a large part of this improve-ment has been due to the ideas introduced with the adventof vector methods of thought.

That Lagrange reduced all of mechanics to a purely ana-lytical basis without, as he boasts, necessitating diagrams,is certainly a wonderful accomplishment. Yet how muchclearer and more elegant if the equations become alive withmeaning, if to the algebraic transformations a mental pic-ture of what is taking place is obtained!

Maxwell gave a splendid reference in favor of the newmethods when he said, in speaking about the motion ofthe top, "Poinsot has brought the subject under the powerof a more searching analysis than that of the calculus, inwhich ideas take the place of symbols and intelligible propo-sitions supersede equations."

Vector Analysis has the advantages of Lagrange's ana-lytical method as well as those of the idealogical methodof Poinsot.

The writer does not, in any way, urge the rejection ofanything of value in any method whatsoever. It is notwell nor is it intended that the methods of Vector Analysisshould be essentially different from those to which thestudent is supposed to be accustomed. In fact, it has beenthe aim throughout this book to evolve an analysis to whichall the knowledge of the reader can be immediately applied,and to so expound this analysis, that Cartesian equations maybe immediately written in,vector notation and conversely.

There is still another important advantage, which shouldnot be overlooked, that is, vector notation just as vector

PREFACE. ix

thought, is entirely independent of any choice of axes, orplanes of reference, and yet the transformation of the vectorequations into other systems, requiring these axes or planesis always extremely easy. To prove that a natural invari-ant is invariant to a change of axes, has always appearedto the writer an extremely foolish operation and a wasteof time. This is not saying, that in a mathematical theoryof invariants such a property of an algebraic expression isnot instructive or interesting. But to say, for example, thatthe properties of the lines of force which cut a set of equipo-tential surfaces at right angles, (i.e., the lines F = - V V)may be dependent upon the particular set of axes used toinvestigate them, is a waste of time to say the least. Howcan a truth vary with the language used in expressing it?

No attempt at mathematical rigor is made. Such refine-ments serve only to conceal the simplicity of fact, which it isthe aim of these pages to elucidate. The appearance ofextended proofs, the writer considers to be entirely out ofplace in a book of this kind. On the other hand, no one ismore in favor of mathematical rigor than he; the point issimply to eliminate discussions whose presence would leadthe attention astray from the main ideas oo the argument.In any case, whenever a demonstration does not satisfy thefastidious, the results may be found more rigorously, if notmore clearly, established in works devoted to mathemat-ically rigorous demonstrations.

The student will find with a little study that he may easilytake down lectures, given in Cartesian notation, directlyinto vector notation. Serious trial will convince him that timeis gained and what is still more important, that equationswill be, must be, understood if this is done. It is by pre-cisely such a process that the writer familiarized himselfwith the subject.

The notation adopted is that of Prof. Willard Gibbs, oneof the too few great American physicists and mathema-

PREFACE.

ticians. - The reasons leading to this choice are fully setforth in the Appendix.

The first part of the book is devoted to a concise treat-ment of the fundamental principles of the subject, theremaining chapters, to the application of the analysis to thebeginnings of mathematical physics, including geometry,mechanics, magnetism, electricity, heat and hydrodynamics.It was found necessary to omit many beautiful applicationsin elasticity, electron theory and other parts of physics inorder to keep the size of the volume within bounds.

The student who takes up the later chapters, is supposed tobe familiar, to a certain extent, with the subjects therein.contained, and these chapters are intended to show thebeginner how to translate and demonstrate the theoremsinto the new calculus. The writer therefore makes this hisapology, for a certain necessary lack of logical sequence in thetreatment of the various subjects.

The treatment of alternating currents and allied subjectshas been omitted, because in practically every modern bookon the subject the notation of the special vector methodemployed, is fully explained in some part of it.

It is hoped that but few errors still remain in the text.The author, alone, corrected the proof, but numerousequations and special difficulties met with in printing ina new notation, rendered the corrections very difficult andlaborious.

The' copy has been read by Prof. Saurel, professor ofmathematics in the College of the City of New York, andthe author wishes to acknowledge here his indebtedness forthe kindness as well as for many valuable suggestions.

A detailed list of works on Quaternions is rendered unnec-essary by Professor Macfarlane's " Bibliography " publishedby the "Association for the Promotion of the Study of Qua-ternions and Allied Mathematics," Dublin, 1904, but a list ofworks which have been especially consulted is appended to

PREFACE. xi

the preface, and the writer here acknowledges his obligationsto all of them. If this book succeeds in making plain theauthor's particular point of view; in simplifying ideas, or incausing simple ideas to seem clearer than before, he will feelamply repaid for any pains taken in producing what was tohim a labor of love.

J. G. COFFIN.

NEw Yo x, April 9, 1909.

PREFACE TO SECOND EDITION

IN this new edition a number of small errors which arepeculiarly difficult of discovery in a work involving somany different kinds of type have been corrected. Thesincere thanks of the writer are due to the large number ofcorrespondents who have greatly helped him in this revision.

The author is glad to be able to state that to his knowledgebut one theoretical error has been discovered up to the presenttime.

Certain portions have been rewritten and fourteen pagesof notes have been added to the appendix.

In particular a short digression on different varieties ofvectors; certain additional definitions of differential geometrywith reference to curves in space which seemed interestingand useful; the demonstration of Frenet's valuable formulaefor space curves; an interesting example of vector reasoningas applied to the solution of the differential equation ofmotion of an electron in a magnetic field; two new proofsof Stokes' Theorem not found as far as we know in anytreatise of vector analysis; an additional proof of Gauss'sTheorem; and proofs of two theorems in integration analo-gous to the Divergence Theorem.

zii PREFACE.

Both the publisher and the writer are delighted with thereception accorded this little book in this country andabroad.

The writer is of the opinion that a great many results ofmathematical physics are elementary and easily understoodby the student if explained in the right way, and the studentthereby finds himself in a position to go right ahead in themore difficult extensions, when he comes to them. Thisbook was written with that end in view. It is practicallyan elementary course in mathematical physics.

He also hopes that not only will this volume help thestudent to an acquisition of the fundamentals of VectorAnalysis, but that also, and not least, it will awaken in him,adesire for further study in that most beautiful and extensiveof all branches of study, - Mathematical Physics.

He believes that in this country there is a wealth of mate-rial for the making of brilliant investigators in this line, ifthey are encouraged to approach the higher branches with-out the fear that it is beyond their capabilities.

He therefore makes a plea for the encouragement of stu-dents having ability in this direction so that soon it can nolonger be said that we are not up to the standard of the in-vestigators of the Old World. True, they had a long startand we have been handicapped, but we hope in the courseof a few years to be abreast of them.

J. G. COFFIN.NEw YORx, June, 1911.

CONTENTS.

CHAPTER I.

ULEMENTARY OPERATIONS OF VECTOR ANALYSIS.LaT. PAGE

1. Definitions - Vector - Scalar .......................... 1

2. Graphical Representation of a Vector ... .............. 1

3. Equality of Vectors - Negative Vector - Unit Vector -Reciprocal Vector .................................... 2

4. Composition of Vectors - Addition and Subtraction - VectorSum as an Integration ................................ 4

5. Scalar and Vector Fields - Point-Function - Definition ofLame - Continuity of Scalar and Vector Functions ...... 6

6. Decomposition of Vectors ............................... 87. The Unit Vectors I j k ................................. 9

8. Vector Equations - Equations of Straight Line and Plane. . 11

9. Condition that Three Vectors Terminate in a Straight Line -Examples ............................................. 13

10. Equation of a Plane .................................... 1611. Plane Passing through Ends of Three Given Vectors........ 16

12. Condition that Four Vectors Terminate in a Plane......'.... 1813. To Divide a Line in a Given Ratio - Centroid .......... 1814. Relations Independent of the Origin - General Condition... 21

EXERCISES AND PROBLEMS ............................. 22

CHAPTER II.

SCALAR AND VECTOR PRODUCTS of Two VECTORS.

15. Scalar or Dot Product - Laws of the Scalar Product ...... 2816. Line-Integral of a Vector ............................... 31

17. Surface-Integral of a Vector ............................. 3218. Vector or Cross Product - Definition .................... 3419 Distributive Law of Vector Products - Physical Proof ..... 3520. Cartesian Expansion of the Vector Product ................ 38

Eiii

xiv CONTENTS.

ART. PAGE

21. Applications to Mechanics - Moment .................... 3922. Motion of a Rigid Body ................................. 4123. Composition of Angular Velocities ........................ 41

EXERCISES AND PROBLEMS ............................. 43

CHAPTER III.

VECTOR AND SCALAR PRODUCTS OF THREE VECTORS.

24. Possible Combinations of Three Vectors ................... 4825. Triple Scalar Product V =26. Condition that Three Vectors lie in a Plane - Manipula-

48

tion of Scalar Magnitudes of Vectors .................. 50

27. Triple Vector Product q - ax(bxc) - Expansion and Proof. 51

28. Demonstration by Cartesian Expansion .................. 53

29. Third Proof ........................................... 54

30. Products of More than Three Vectors ..................... 55

31. Reciprocal System of Vectors ............................32. Plane Normal to a and Passing through End of b - Plane

through Ends of Three Given Vectors - Vector Perpen-

57

dicular from Origin to a Plane ......................... 58

33. Line through End of b Parallel to a ...................... 60

34. Circle and Sphere ......................................

34a. Resolution of System of Forces Acting on a Rigid Body -61

Central Axis - Minimum Couple ...................... 63

EXERCISES AND PROBLEMS .............................

CHAPTER IV.

DIFFERENTIATION OF VECTORS.

35. Two Ways in which a Vector may Vary - Differentiation with

66

Respect to Scalar Variables ........................... 70

36. Differentiation of Scalar and Vector Products ............. 72

37. Applications to Geometry - Tangent and Normal .........38. Curvature - Osculating Plane - Tortuosity-Geodetic Lines

73

on a Surface ....................................... 76

CONTENTS. xv

ART. PAL39. Equations of Surfaces - Curvilinear Coordinates - Ortho-

gonal System ........................................ 79

40. Applications to Kinematics of a Particle - Hodographs -Equations of Hodographs ............................. 80

41. Integration with Respect to a Scalar Variable - Orbit of aPlanet - Harmonic Motion - Ellipse .................. 83

42. Hodograph and Orbit under Newtonian Forces ............ 87

43. Partial Differentiation - Origin of the Operator V ......... 90

EXERCISES AND PROBLEMS ............. ............ . 91

CHAPTER V.THE DIFFERENTIAL OPERATORS.

V - iax +ia +k .

y44. Scalar and Vector Fields ................................ 94

45. Scalar and Vector Functions of Position - Mathematicaland Physical Discontinuities .......................... 95

46. Potential - Level or Equipotential Surfaces - Relationbetween Force and Potential .......................... 98

47. V applied to a Scalar Function - Gradient - Independenceof Axes - Fourier's Law ............................. 102

48. V applied to Scalar Functions - Effect of V on ScalarProduct ............................................ 104

49. The Operator S , V, or Directional Derivative - Total Deriva-tive ................................................. 106

50. Directional Derivative of a Vector - V applied to a VectorPoint-Function ...................................... 107

51. Divergence - The Operator V ......................... 109

52. The Divergence Theorem - Examples - Equation of Flowof Heat ............................................. 112

53. Equation of Continuity - Solenoidal Distribution of a Vector 116

54. Curl- The Operator Vx - Example of Curl .............. 117

55. Motion of Rotation without Curl - Irrotational Motion...... 119

56. V, Vx applied to Various Functions - Proofs of Formulae 120

57. Expansion Analogous to Taylor's Theorem ................ 124

58. Stokes' Theorem ....................................... 124

59. Condition for Vanishing of the Curl - Conservative Systemof Forces ........................................... 127

RVl CONTENTS.

ART. PAGE

60. Condition for a Perfect Differential ....................... 129

61. Expression for Taylor's Theorem - The Operator e,v( ). 131

62. Euler's Theorem on Homogeneous Functions .............. 131

63. Operators Involving V Twice - Possible Combinations -The Operator V2 - V -V ............................. 133

64. Differentiation of rm by V ............................. 135EXERCISES AND PROBLEMS ............................. 136

65.

CHAPTER VI.APPLICATIONS TO ELECTRICAL THEORY.

Gauss's Theorem - Solid Angle - Gauss's Theorem for thePlane - Second Proof ................................ 138

66. The Potential Function - Poisson's and Laplace's Equations- Harmonic Function ............................... 143

67. Green's Theorems ...................................... 148

68. Green's Formulae - Green's Function .................... 148

69. Solution of Poisson's Equation - The Integrating Operator(Pot = f

( )dv ................................ 152

70.

J JOD r

Vector-Potential ....................................... 153

71. Separation of a Vector-Function into Solenoidal and LamellarComponents - Other Systems of Units ................. 154

72. Energy in Terms of Potential... ........................ 156

73. Energy in Terms of Field Intensity ...................... 157

74. Surface and Volume Density in Terms of Polarization ...... 159

75. Electro-Magnetic Field - Maxwell's Equations ............ 16076. Equation of Propagation of Electro-Magnetic Waves....... 163

77. Poynting's Theorem - Radiant Vector ......:............ 164

78. Magnetic Field due to a Current ......................... 165

79. Mechanical Force on an Element of Current ............... 167

80. Theorem on Line Integral of the Normal Component of aVector Function ..................................... 168

81. Electric Field at any Point due to a Current .............. 170

82. Mutual Energy of Circuits - Inductance - Neumann'sIntegral ............................................ 171

83. Vector-Potential of a Current - Mutual Energy of Systemsof Conductors - Integration Theorem .................. 173

84. Mutual and Self-Energies of Two Circuits ................. 175

EXERCISES AND PROBLEMS ........................ 176

CONTENTS. xvu

CHAPTER VII.

APPLICATIONS TO DYNAMICS, MECHANICS AND HYDRODYNAMICS.

ART. PAGL

85. Equations of Motion of a Rigid Body-D'Alembert's Equa-tion - Equations of Translation - Motion of Center ofMass .............................................. 178

86. Equations of Rotation - Kinetic Energy of Rotation -Moment of Inertia.................................. 180

87. Linear Vector-Function -Instantaneous Axis............ 182

88. Motion of Rotation under No Forces-Poinsot Ellipsoid-Moments and Products of Inertia - Coordinates of aLinear Vector-Function -Principal Moments of Inertia-Principal Axes ..................................... 184

89. Geometrical Representation of the Motion - InvariablePlane - Invariable Axis ............................

90. Polhode and Herpolhode Curves - Permanent Axes - 191

Equations of Polhode and Herpolhode ................. 192

91. Moving Axes and Relative Motion - Theorem of Coriolis. 194

92. Transformation of Equations of Motion-Centrifugal Couple- Gyroscope ....................................... 198

93. Euler's Equations of Motion ...........................94. Analytical Solution of Euler's Equations under No Impressed

199

Forces............................................. 200

95. Hamilton's Principle - Lagrangian Function.............96. Extension of Vector to More than Three Dimensions -

202

Definitions ........................................

97. Lagrange's Generalized Equations of Motion - The Oper-204

ator VL = 0 Contains the Whole of Mechanics ....... 205

98. Hydrodynamics - Fundamental Equations - Equation ofContinuity - Euler's Equations of Motion of a Fluid ... 207

99. Transformations of the Equations of Motion .............. 211

100. Steady Motion - Practical Application ................. 212

101. Vortex Motion - Non-creatable in a Frictionless System -Helmholtz's Equations ............................. 212

102. Circulation - Definition . ..............................103. Velocity-Potential - Circulation Invariable in a Friction-

214

less Fluid .......................................... 216

EXERCISES AND PROBLEMS ............................ 217

xviii CONTENTS.

APPENDIX.

NOTATION AND FORMULA.PACK

Various Notations in Use .................................. 221

Hamilton ........... ............... .. .......... .. ... 221

Heaviside ............................................... 221

Grassmann................................................ 221

Gibbs ............ .. .................... ...... .... 222

Comparison of Formulae in Different Notations................ 222Notation of this Book .................. . . . . .... . . .. . . . . 224

FORMULA.

Resume of the Principal Formulae of Vector Analysis.......... 229Vectors... ................... .. ...... 229Vector and Scalar Products - Products of Two Vectors ........ 230Products of Three Vectors ..... .... ..... . ... . .. ..... . 231Differentiation of Vectors ................ . ..... . . ... 233The Operator V, del ...................................... 233Linear Vector Function .................................. 237Note on Different Varieties of Vectors ....................... 240Definitions of the Normal, Normal Plane, Principal Normal, Bi-

normal and Rectifying Plane for a Space Curve............ 242Frenet's Formula; for a Space Curve ......................... 244Motion of an Electron in a Uniform Magnetic Field ........... 245Two Proofs of Stokes' Theorem ............................... 249Proof of Gauss's Theorem ..................................... 251Other Integration Theorems .................................. 252Index ....................................................... 255

BIBLIOGRAPHY

Works Specially Consulted in the Preparation of this Book.

APPELL. Traitr de MecaniqueRationelle.

BJERKNES. Vorlesungen fiber Hy-drodynamischen Fernkrafte.

BUCHERER. Elemente der Vek-tor-Analysis.

BURNSIDE and PANTON. Theoryof Equations.

CLIFFORD. Elements of Dynamic.DRUDE. Theory of Optics.EMTAGE. Introduction to the

Mathematical Theory of Elec-tricity and Magnetism.

FEHR. Methode Vectorielle deGrassmann.

FISCHER. Vektordifferentiationand Vektorintegration.

FOPPL. Maxwell'sche Theorieder Elektricitat.

Vorlesungen fiber Tech-nische Mechanik.

GANS. Einfuhrung in die Vek-toranalysis.

GIBBS. Collected Papers.HEAVISIDE. Electrical Papers.

Electro-magnetie Theory.HENRICI and TURNER. Vectors

and Rotors with Applications.IBBETSON. Mathematical Theory

of Elasticity.JAUMANN. Bewegungslehre.JoLY. Manual of Quaternions.

KELLAND and TAIT. Introduc-tion to Quaternions.

KIRCHHOFF. Vorlesungen fiberMathematisehe Physik.

LAGRANGE. ME canique Analy-tique.

LOVE. Theory of Elasticity.MAXWELL. Electricity and Mag-

netism.McAULAY. Utility of Quaternions

in Physics.Octonions.

MINCHIN. Treatise on Statics.PIERCE, B. O. Elements of the

Theory of the Newtonian Po-tential Function.

PoINSOT. Theorie Nouvelle de IsRotation des Corps.

ROUTH. Rigid Dynamics.STEINMETZ. Alternating Current

Phenomena.TAIT. Dynamics:

Quaternions.WALTON. Problems in Mechan-

ics.WEBSTER. The D mics of a

Particle and of Rigid, Elasticand Fluid Bodies.

The Theory of Electricityand Magnetism.

WILLIAMSON and TARLETON. Ele-mentary Treatise on Dynamics.

WILSON, GIBBS-. Vector An-alysis.

xix

SUGGESTIONS FOR WRITING VECTOR ANALYSISON THE BOARD

A number of inquiries have come in asking how to writevectors on the blackboard. It seems that the bold-facedtype or Clarendon is perfectly satisfactory as far as print isconcerned, but it is impracticable to produce such a differ-ence in chalk-written symbols. To a great extent these sametroubles also occur in manuscript.

There are several methods of differentiating vectors frompurely scalar symbols which have proved satisfactory.

The notation given in the text is entirely practicable anddefinite. That is, if a, b or r denote vectors in any discussion,let ao, bo or ro denote their magnitudes and a1, b1 or r1 denotetheir directions or unit vectors along a, b or r respectively.

Thus a = anal.

There is here a slight chance of ambiguity in the equation

a = a1i + a2 j + ask

where the i-component of a might be confounded with theunit vector along a. The writer does not consider this aserious objection. Like Tait we say that anybody findingdifficulty with this small matter has begun the study ofvectors too soon!

Another method is to place a line or dash over the vector-symbol. So that if a denotes a vector, then a is its magnitude,al is its direction and a1 is its i-component.

Still another method to which the writer is very partial,having been brought up on Hamilton's notation, is to reservethe Greek alphabet for vectors.

So that if a is any vector, ao or a is its magnitude, and alis its direction, while a1, a2, as are its i-, j-, k-components.

xxi

xxii SUGGESTIONS

After reading the book notices and reviews we are still ofthe same opinion as to the essential superiority of Gibb'snotation over others, notwithstanding the criticisms of itwhich we expected. We have never claimed that aXb was asymmetrical function, but we do claim that thenotation of botha-b and aXb is symmetrical. The minus sign in aXb = - bxa,does not make the notation unsymmetrical.

Almost simultaneously with this text was issued a vectoranalysis by the Italian mathematicians Burali-Forti andMarcolongo.

These gentlemen have invented still another notationwhich is similar to ours but which employs the X (large cross)for a scalar product and an inverted V (A) for a vectorproduct. With the symmetry of their notation we are infavor, but why introduce any more notations when thereare already so many to pick from?

This question of notation, which has nothing to do withthe spirit of the method, is for each individual to solve forhimself. We have employed what we believe to be the sim-plest and best and we have presented at length our argu-ments in favor of it.

J. G. COFFIN, 1911

VECTOR ANALYSIS

CHAPTER I.

ELEMENTARY OPERATIONS OF VECTOR ANALYSIS.

Definitions.

1. A Vector is a directed segment of a straight line on whichare distinguished an initial and a terminal point. A vectorthus has a magnitude and a direction. Any quantity whichcan be represented by such a segment may be called a vectorquantity. The importance of this generalized conception iseasily understood when it is considered that motion or dis-placement, velocity, acceleration, force, electric current,magnetic flux, lines of force, stresses and strains due to anycause, flow of heat and of fluids, all involve two parts, i.e.,magnitude and direction. All such quantities are vectorquantities.

A Scalar on the other hand is any quantity which althoughhaving magnitude does not involve direction. For example,mass, density, temperature, energy, quantity of heat, electriccharge, potential, ocean depths, rainfall, numerical statisticssuch as birth rates, mortality or population, are all scalarquantities.

A scalar, then, reduced to its simplest terms is merely anumber and as such obeys all the laws of ordinary alge-braic analysis. A vector, however, involving direction inaddition to its numerical magnitude has an analysis pecul-iar to itself, the laws of which are to be derived.

2. Graphical Representation of a Vector. Any vectorquantity may be represented graphically by an arrow.

1

2 VECTOR ANALYSIS.

The tail of the arrow, 0, is called the origin; the head, A,is called the end or terminus.

Symbolically a vector may be denoted by two letters,the first one indicating the origin, the second one theterminus.

A small' arrow is often placed over these letters to indi-cate more exactly that the quantity considered is a vector.Thus, OA denotes the vector beginning at 0, ending at A,and pointing in the direction from 0 to A. This notationwhile useful is at times cumbersome. Hence more usuallya vector will be denoted by a single letter, which involvingmore than a mere scalar is printed differently to distinguishit from purely scalar quantities, i.e., in Bold-faced Type.

Thus the vector a* means the going of the distance OAin the direction 0 to A from any point in space as origin.

3. Equality of Vectors. All lines having the same lengthor magnitude and the same sense are equal vectors whatevertheir origin may be. Thus in Fig. 1, OA and O'A' are equalvectors.

Negative Vector. The vector 0"A" having the samelength and direction as a but the opposite sense, is defined

* The terms Step, Stroke, or Directed Magnitude are sometimes usedas synonyms of Vector.

VECTOR ANALYSIS. 8

as the negative of a and is written - a. Evidently also OAis the negative of O"A".

Unit Vector. The directional part of any vector a maybe concisely represented by a vector having the same senseand direction as a but of unit length. Such a vector iscalled a unit vector and will be denoted by adding thesuffix 1 to the symbol representing the vector. Thus a, isa vector having the same direction as a, but of unitlength.

The length of a vector is termed its magnitude, size, orits absolute value. Sometimes, also, the term tensor isused. The magnitude of a vector a will be written a,using the same letter as that which denotes the vector butprinted in italic type. It will be sometimes convenient alsoto denote the magnitude of a by adding the subscript 0 to athus:

ao - a.

The vector a then may be considered as one, a times aslong as a, and hence we may write:

a = aal or = anal. (1)

Any vector then may be represented by the product of itsunit vector into its magnitude as in (1).

The expression m a denotes a vector m times as long as a,having the same direction but m times its magnitude. Themultiplier - 1 from what has been said about negativevectors, reverses a vector.

Parallel vectors whatever their magnitude are said to becollinear.

Reciprocal Vector. The vector parallel to a but whoselength is the reciprocal of the length of a is said to be thereciprocal of a.

So that if a = ca-at

1=a-' =a,. 2)a a

4 VECTOR ANALYSIS.

Composition of Vectors.*

4. Addition and Subtraction. To obtain graphically thesum of the two vectors a and b, draw b starting from the

FIG. 2.

end of a; the line joining theorigin of a with the end of bis the sum in question. In otherwords, it is the diagonal of theparallelogram of which the twovectors a and b are the sides.

Evidently the sum (a + b) isthe same as (b + a). If thereare more than two vectors to beadded, the sum of the first twomay be taken and the third added

to it as above, then to the resultant add the next one andso on. A moment's consideration of Fig. 3 will show that

if we draw the vectors one after the other in a chain, eachnew one from the end of the last one drawn, the line joining

See Appendix, p. 240, Note on Different Varieties of Vectors.

VECTOR ANALYSIS. 5

the origin of the first one to the terminus of the last one isthe vector sum of them all.

A consideration of Fig. 3 will also show that the order inwhich they are taken is immaterial. The same construc-tion then is used to find the sum of any number of vectorsas is used in finding the resultant of the forces which wouldbe represented by these vectors. Hence the importance ofvectors in mechanics.

To subtract two vectors, add to the first the second onereversed. The extension of these rules to both positiveand negative is obvious.

A

Fia. 4.

Vector Sum as an Integration. Any curve may be consid-ered to be built up of an infinite number of infinitely short.vectors, their directions being at every point along thetangent to the curve.

The sum of such a series of vectors differs in no wayfrom the sum of a finite number of finite vectors. If d arepresents any one of these small vectors, then by adding

them all the resultant AB is obtained. The operation of

6 VECTOR ANALYSIS.

adding this infinite number of infinitesimal vectors may berepresented by an integration sign thus:

AB =fB

If the curve is a closed one, whether a plane curve or not,

then A and B coincide and AB = 0 or fda around a closed

path is zero.

Scalar and Vector Fields and their Addition.

5. Point-Function. Definition of Lame. If for everyposition of a point in a region of space a quantity has oneor more definite values assigned to it, it is said to be a func-tion of the point, or more concisely, a point-function. Wemay have both scalar and vector point-functions.

As an Example of a Scalar Point-Function, consider thepotential at any point due to any distribution of matterM, and let its value be V,. Now consider the potential atthe same point P due to any other distribution of matterM2 and let its value be V2. Then the potential at P dueto both masses together is simply V, + V 2. This value isfound by adding together the two scalars V, and V2.

* Perhaps the following example of scalar field will be clearer tosome minds. Consider a point P and let it be illuminated by a sourceof light M,. Evidently every point in the vicinity of the source is illu-minated to a greater or lesser extent according to its distance from thesource. The illumination or intensity of light at all points of the spaceconsidered may be represented by a scalar point-function. Let nowanother source of light M, be brought into the space under considera-tion. This source produces a certain 'intensity of illumination atevery point of the space, of course including the point P. The totalamount of illumination now received at the point P is the scalar sumof the amounts it receives from each individual source. This is trueof every other point in the field. So that in general in order to find theillumination at any point due to separate sources, one simply adds,the

VECTOR ANALYSIS. 7

Practical Definition of Continuity of a Scalar Point Func-tion. If, as we go from any point in space to any nearadjacent point, the magnitude of the scalar point-function

Fia. 5.

suffers no abrupt change, the function is said to be con-tinuous.

As an Example of a Vector Point-Function consider theforce of attraction at any point P due to the attraction ofthe mass Mt. This force is evidently a vector quantity, asit has a definite magnitude and a definite direction, so thatits representation requires the use of a vector at P; let F,be this vector. Similarly let F2 be the vector representingthe force at P due to the matter M2. Then the force at Pdue to the combined action of M, and M. is the vector sumof F1 and F2 and must be obtained by the laws of vector addi-tion; i.e., the parallelogram law. If we go from the point,P to another point Q in space the magnitudes and direc-

values of the separate intensities at the point due to the separatesources respectively. This constitutes an addition of scalar fields.The fields are here scalar fields because we are considering only theamounts of the illumination received at any point.

S VECTOR ANALYSIS.

tions of these forces F, and F. at Q, and hence, in general,their sum, F, + F2, at Q undergo changes.

Fic. 6.

Practical Definition of the Continuity of a Vector PointFunction. If, as we go from any point in space to any nearadjacent point, the direction as well as the magnitude ofthe vector point-function suffers no abrupt change, thefunction is said to be continuous.

6. Decomposition of Vectors Into Components. From § 4it is evident that any vector q may be considered as the

e sum of any number of compo-

Fra. 7.

nent vectors, which when joinedend to end, as in vector addi-tion, the first one begins at theorigin of q, and the last one endsat the terminus of q. Thus:

q=a+b+c+d+e.These vectors need not lie inone plane. Vectors all of which

lie in or parallel to the same plane are said to be coplanar.In particular it is often convenient to decompose a vector

VECTOR ANALYSIS. 9

into two or three components at right angles to each other;two in case all the vectors under consideration are coplanar;three, when they are not coplanar.

7. The Three Unit Vectors i j k. Consider the right-handedCartesian system of axes. The three unit vectors along thex y z axes are called i j k respectively. It is evident thatany vector r is equivalent to a certain vector OA along OX,

plus a vector AB along OZ, plus a vector BC along OY.

FIG. 8.

In other words, if x y z denote the magnitudes of thesevectors respectively, we may write for any vector r whosecomponents are x, y, z,

r=xi+yj+zk. (4)

10 VECTOR ANALYSIS.

X i, y j, and z k are the three projections of r along the threeaxes respectively. If a, (9, r be the direction angles of anyvector parallel to OC, then evidently

x = r cos a,y = r cos p,z = r cos r.

(5)

This decomposition of a vector into two or three rectan-gular components is of the utmost importance and is theconnecting link between the two or three dimensional Car-tesian and Vector Analyses, respectively.

If two vectors are given,

a=a,i+a2j+a3k,b=b,i+b2j+b3k,

their sum is evidently

(a+b)=(a,+b,)i+(a2+b2)j+(a3+b3)k (6)

This may be extended to any number of vectors andshows that the components of the sum are equal to thesums of the components, so that

Za=ia,+iZa2+kZa3. (7)

This theorem is of use in the composition of forces. It ispossible to resolve any vector r into three components par-allel to any three non-coplanar vectors; and such a resolu-tion is easily seen to be unique. Practically, in order tofind the rectangular components of a vector, equations (5)are employed, so that

r = r (i cos a + j cos R + k cos r). (8).

If we divide through by the magnitude of r there remains

r =r1=icosa+jcosP +kcosr,r (9)

VECTOR ANALYSIS. 11

so that the rectangular components of a unit vector arealways its direction cosines.

By inspection of Fig. 8 it is evident thatr2 = x2 + y2 + za.

- 11

Fio. 9.

Fio. 10.

Vector Equations..8. Equations of the Straight Line and Plane. Let r be a

variable vector, with origin at 0, and s a variable scalar;it is then evident on inspection (Fig. 9) that

r = s a (10)

is the equation of a straight line passing through the originand parallel to a. It is also easily seen (Fig. 10) that

r=b+sa (11)

12 VECTOR ANALYSIS.

is the equation of the straight line through the terminus ofb and parallel to a. By means.of equation (11) the equa-tion of a line passing through the ends of any two givenvectors a and b may easily be derived.

FIG. 11.

The vector AB is (b - a), hence by equation (11) theline through the terminus of a parallel to (b - a) is

r = a + t (b - a),where t is a scalar variable. These equations may be putinto the easily remembered forms

r=tb..-I-(1-t)a,and by analogy

r = sa + (1 - s) b. (12)

It is evident that if the directions of the coordinate axesbe taken along. a and b, then the magnitudes of a and b

VECTOR ANALYSIS. 13

respectively are the intercepts the line makes with theseaxes, the corresponding Cartesian equation being

a + b1...

All problems in line geometry are now readily solvable.If all the lines of the problem lie in one plane, two, and onlytwo, arbitrary non-parallel vectors are chosen and all othersexpressed in terms of them. For a problem in three dimen-sions all the lines are expressed in terms of three, and onlythree, arbitrary non-coplanar vectors.

9. Condition that Three Vectors should Terminate in theSame Straight Line. Putting equation (12) in the form

s a + (1 - s) b- r = 0

it is seen that in the linear relation connecting three vectorswhich end in the same straight line the sum of the coeffi-cients is equal to zero. Or in other words, if

xa+ yb+zc=0 (13)

and x+y+z=0,the three vectors a, b, and c necessarily end in the samestraight line, and are said to be termino-collinear.

Example. As a simple example of the general method ofprocedure, let us prove that the diagonals of a parallelo-gram meet in a point which bisects them both. Take theorigin at the corner 0, and write down the equation of thediagonals OC and AB in terms of a and b, the vectors OAand AB. Notice that the origin may be chosen arbitrarilyand hence, may be taken so as to simplify the equations.Very often, however, it is better not to place the originat any special or definite point, so that more symmetry isproduced in the resulting equations.

14 VECTOR ANALYSIS.

The equation of OC isr = s (a + b), by equation (10)

and that of AB is I

r = t a + (1 - t) b, by equation (12)where s and t are variable scalars. For intersection, bothequations must be satisfied by the same value of r; hence,equating,

s (a + b) =ta+(1 - t) b. (14)

This vector equation is actually equivalent to two scalarequations and suffices to determine s and t, for the vector rto the point of intersection is uniquely determined in termsof the vectors a and b, so that the scalar coefficients of thesevectors on both sides of equation (14) must be respectivelyequal. The coefficients of a give

s=tandofb s= (1 -t).This makes s = t = , and the vector to the point of inter-section is then, by substituting this value for sin r = s (a + b),

OD= (a+b)=SOC.This principle is applicable to any kind of line problem

in two or three dimensions. The method of equating the

VECTOR, ANALYSIS. 15

coefficients of the same vector on both sides -of an equationis analogous to tfie conditions for equality of two compleximaginary expressions; that is, if

s+it=s'+it',then s = s' and t = t'.

Example. As an example of the symmetrical method toprove that the medians of a triangle meet in a single pointwhich trisects each of them. Choose any point not in theplane of the triangle for origin, and define the triangle bythe three vectors a, b, and c from the origin to its verticesAl B, and C. We choose the origin out of the plane ofthe triangles so that we may use three independent vectorsinstead of but two, as would be necessary if the origin weretaken in the same plane.

Then OA' = (b + c),OB' = (c+a),OC'= (a + b),

so that the equation of

AA' is r=xa+(1-x)3(b+c), (a)BB' is r = y b + (1 - y) (c + a), (b)

CC' is r = z e + (1 -. z) (a + b). (c)

Equate the coefficients in (a) and (b) for intersection,

of a, x= (1-y),of b,of c,

(1 - x) = y,(1 - x) = 1 (1 - y),

so that x = y = and the vector to their point of intersec-tion is

OD =r=a+b+c3

This is evidently the point of intersection of the third linewith either of the first two, by symmetry. It is also the

16 V L+'U1'UR ANAbYSIS.

B

Fia. 13.

mean point of A, B, and C, as explained below. It is the

point of trisection, because adding to a, I of AA' we obtainthe same result, thus:

OA'= (b+c),AA'= [-a+I(b+c)],

and ODa++[-a+I(b+c)]a+b+cl

3

Fia. 14.

By choosing the origin at one of the vertices the sym-metry is lost but a gain in directness and shortness is made.In problems involving algebraic coefficients instead of nu-merical ones the symmetrical method is generally preferable.

VECTOR ANALYSIS. 17

10. Equation of a Plane. The vector to any point in thelane determined by the vectors a and b and passing through

the origin is evidentlyr=sa+tb, (14)

where s and t are two independent scalar variables. If theorigin be removed to the origin of a vector c, through theterminus of which the plane parallel to a and b passes,then the vector to any point P in the plane is now given by

r=c+sa+tb. (15)

11. To find the equation of a plane passing through the endsof the three non-coplanar vectors a, b,and c, notice that the vec-tors (a-c) and (b-c) evidently lie in the plane. By employ-ing the previous equation (15), the equation may be written

r=c+s(a -c) +t(b -c),

Or . FIG}. 15.

which may be put into easily remembered form, analogousto equation (12)

r = sa + tb + (1 - s - t) c. (16)It is evident that if the directions of the coordinate axes

be taken along a, b, and c, then the intercepts made by the

18 VECTOR ANALYSIS.

FIG. 16.

plane with these axes are the lengths of a, b, and c respec-tively, the corresponding Cartesian equation being

a + b + c = 1.

12. Condition that Four Vectors Terminate in the SamePlane. Rearranging equation (16),

s a + t b + (1 - s - t) c - r =0,it is seen that whenever there is a linear relation betweenany four vectors they terminate in one and the same planeif the sum of the coefficients is zero. Or in other words, if

xa + y b + zc + wd = 0and x+,y+z+w=0; (17)

a, b, c, and d terminate in the same plane and are said to betermino-coplanar.

13. To Divide a Line in a Given Ratio. Centroid. To findthe value of a vector which divides the distance betweentwo points A and B in a given ratio, m to n say, it is simply

VECTOR ANALYSIS. 19

necessary to express the vector r in the form, evident oninspection,

r=a + m (b -a) = nnb(18)

m nm

+

Fio. 17.

It is a well-known result in mechanics that the center ofgravity of two masses m, and m2 divides the line joiningthem. inversely as these masses, so that by (18)

r2 = m,a, + m,aa"h + M2

M1

ML mm

Ma

a, r2r3

a,

a,

is the vector to their center of mass or their centroid. Ifnow there is a third point a, with mass rn3 added to the

20 VECTOR ANALYSIS.

system, the new centroid will be that of the two masses r2with mass (ml + m2) and a3 with mass m3i or again by (18),

r3 =(ml+m2) r2+m3a3= meal+m2a2+m3a3 = Ema (19)

3 MI +m2+m3 ml+mz.+m3 imThe generalization is immediate. If M = E m denotes thetotal mass of the system of particles and r the vector totheir center of mass,

Mr = E ma. (20)

If the masses form a continuous body, the formulabecomes

fffpfffpadv

(21)fffpdv

where p is the density and dv is the element of volume.The integrations are taken throughout the volume.

If xi,+yj+zkand an = x,J + yn.j + zk,

formula (20) breaks up into the three well-known ones for thethree coordinates of the center of mass,

nMx = Mx,,,

ivy=myo, (22)

Mz = mz,,.

Similarly (21) gives three of the form

x= fffxp d xdydz

t 3e c

'( )

fffpdxdydz

VECTOR ANALYSIS. 21

14. Relations Independent of the Origin. That the centerof gravity and therefore all the formulae just derived areindependent of the origin may be shown by the followingreasoning.

Taking the origin at 0, the vector to R, the center ofgravity of the two masses m and n is, by (18),

Now change the origin to 0', the new vectors to themasses being a' and b' and the vector to the first originfrom the new one being c.

Fia. 19.

The vector to the center of gravity from 0' is now given by

r, = m a' + n b' .m+n

But since a' = a + C and b' = b + c, this equation may bewritten

=m(c+a)+n(c+b)ma+nb+c,m+n m+n

22 VECTOR ANALYSIS.

which says that the new center of gravity is the same asbefore, as r' = r + c.

It will be noticed on writing (24) 'in the form(m + n) r - ma - n b = 0

that the algebraic sum of the scalar coefficients is zero.This leads to the

. General Condition for a Relation Independent of theOrigin. The necessary and sufficient condition that a linearvector equation represent a relation independent of theorigin is that the sum of the scalar coefficients of the equa-tion be equal to zero. Let the equation be

mla1 + mZa2 + ... = 0. (25)

Change the origin from 0 to 0' by adding a constantvector 1, the distance from 0 to 0', to each of the vectors,a1, a2, a3, etc.; the equation then becomes

m,(a,+1) +m2(a2+1) +... =0or m1a1 + mZa2 + ... + 1(m1 + m2 + ... ) = 0.

If this is to be independent of the origin, i.e., the same as(25), the coefficient of I must vanish, or

m1+m2+...=0.

EXERCISES AND PROBLEMS.

1. Prove that the vectors±a±b±c

when drawn from a common origin terminate at the vertices of aparallelopiped.

2. A person traveling eastward at a rate of 3 miles an hour findsthat the wind seems to blow directly from the north; on doublinghis speed it appears to come from the northeast. Find the vectorwind velocity.

3. A ship whose head is pointing due south is steaming across acurrent running due west; at the end of two hours it is found thatthe ship has gone 36 miles in the direction 15° west of south. Findthe velocities of the ship and current, graphically and analytically.

VECTOR ANALYSIS. 28.

4. A weight W hangs by a string and is pushed aside by a hori-zontal force until the string makes an angle of 45° with the vertical.Find the horizontal force and the tension of the string.

5. A vector r is the resultant of two vectors a and b which makeangles of 30° and 45° with it on opposite sides. How large are thelatter vectors?

6. A car is running at 14 miles an hour and a man jumps fromit with a velocity of 8 feet per second in a direction making an angleof 30° with the direction of the car's motion. What is his velocityrelative to the ground? .

7. Verify, by drawing, the truth of the laws of association andcommutation, taking a number of vectors, a, b, c, d, etc., to scale,and show that the resultant is independent of the order of additionor subtraction.

8. Given the vector

r=a,1+ajderive the vector of same length perpendicular to it through theorigin.

Derive the vector perpendicular to the one you find. Comparewith the original one.

9. Find the relative motion of two particles moving with thesame speed v, one of which describes a circle of radius a while theother moves along a diameter.

10. Two particles move with speeds v and 2 v respectively inopposite directions, in the circumference of a circle. In whatpositions is their relative velocity greatest and least, and whatvalues has it at those positions?

11. Draw the vectors

a= 61- 4j+10kb=-61+ 4j-10kc= 41- 6j-10kd= lOj+ 4k

Find their sum graphically and analytically.12. The equation

(r-a).=(r-b)0represents the plane bisecting at right angles the line AB.

24 VECTOR ANALYSIS.

13. Find the equation of the locus of a point equidistant fromtwo fixed planes.

14. The line which joins one vertex of a parallelogram to themiddle point of an opposite side trisects the diagonal.

15. To find a line which passes through a given point and cutstwo given lines in space.

16. Ifxa+yb=0

and x+y=0show that a and b are equal in magnitude and directioq. Or whatis the same thing, that measured from the same origin, a and b endat the same point.

17. Ifxa+yb+zc=0

and x+y+z=0show that a, b, and c terminate in the same straight line; they arethen said to be termino-collinear.

18. Ifxa+yb+zc+wd=0

and x+y+z+w=0show that a, b, c, and d terminate in the same plane; they are thensaid to be termino-coplanar.

19. A triangle may be constructed whose sides are equal andparallel to the medians of any given triangle.

20. Given a quadrilateral in space. Find the middle point ofthe line which joins the middle points of the diagonals. Find themiddle point of the line joining the middle points of two oppositesides. Show that these two points are the same and coincide withthe center of gravity of a system of equal masses placed at thevertices of the quadrilateral.

21. Discuss the conditions imposed upon three, four, or fivevectors if they satisfy two equations, the sum of the coefficients ineach of which is zero.

22. Take a number of points at random on a sheet of paper,assigning arbitrary masses to them. Verify by drawing that theircenter of mass is independent of the origin chosen in finding it.

VECTOR ANALYSIS. 25

23. If a system of masses, each mass concentrated at a point, bedivided into a number of partial systems, and each of these bereplaced by its resultant mass, then the new system has the samecenter of mass as the original one.

24. A cardboard square is bent along a diagonal until the twoparts are at right angles. Find the position of the center of gravity.

25. Forces acting at a point 0 are represented by OA, OB, OC,.. ; ON. Show that if they are in equilibrium 0 is the centroid

of the points A, B, C, ... , N.26. The middle points of the ' lines which join the points of

bisection of the opposite sides of a quadrilateral coincide whetherthe four sides be in the same plane or not.

27. The bisectors of the angles of a triangle meet in a point whichtrisects each of them.

Employ unit vectors along two of the sides as independentvectors. The bisectors are then a, + b etc.

28. If two forces acting at a point 0 are represented by thevectors n a and b their resultant is represented in magnitude and

direction by the vector (n + 1) OG, the point G being taken on ABso that BG = nAG.

This allows the resultant of two forces to be drawn knowing oneand part of another.029. If two forces are equal to n.OA and m.OB, the resultant

passes through the point G determined so that BG =

m

n and is

equal to (m + n) OG in magnitude.30. Forces F,, F2, ... , F,,, acting in a plane at 0 are in equi-

librium. Any transversal cuts their lines of action in points L,, L2,

L,,; and a length 0L1 is positive when in the same direction as

OF. Prove thatF

OL=O.

31. Show that the resultant of any number of concurrent forces,F F2, F2, ... may be found thus: measure off any lengths l l,,I ... from the point of meeting along them respectively; place at

the ends of these lines particles of masses proportional to F ,F2

I. I.

26 VECTOR ANALYSIS.

1, , ... ; let G be the center of gravity of these particles; then OG

is, the line of action of the resultant of the given forces and its mag-nitude is

OGXEF.l

32. A particle placed at 0 is acted upon by forces represented inmagnitudes and directions by the lines OA,, OA,f . . . OA,,, whichjoin 0 to any fixed points A A,, . . . A,,,; where must 0 be placedso that the magnitude of the resultant force may be constant?

Ans. If r represent the magnitude of the resultant, 0 may be

placed anywhere on a sphere of radius r described around the cen-n

troid of the fixed points as center.33. ABCD is a quadrilateral of which A and C are opposite

vertices. Two forces acting at A are represented by the sides ABand AD; two at C by CB and CD. Prove that the resultant isrepresented in magnitude and direction by four times the linejoining the middle points of the diagonals of the quadrilateral.(34. Show that the resultant of the three vector diagonals of a

parallelopiped meeting at a point 0 is represented by twice thediagonal of the parallelopiped drawn from the same point.

35. If through any point within a parallelogram, parallels bedrawn to the sides, the corresponding diagonals of the two newparallelograms thus formed and of the original one meet in a point.

36. The middle points P, Q, R of the diagonals of any completequadrilateral ABCDEF are collinear.

37. Any point 0 is joined to the vertices of a parallelogram;show that the sum of the vectors to the vertices is four times thevector to the intersection of the diagonals.

What conclusion do you derive from this fact?38. ABCDEFA.is a regular hexagon. Show that the resultant

of the forces represented by AB, 2 AC, 3 AD, 4 AE, 5 AF is repre-sented by a vector of magnitude 351 AB, and find its direction.

39. ABCDEFA is a regular hexagon. Find the resultant of theforces represented by the lines AB, AC, AD, AE, AF.

40,, 0 is any point in the plane of a triangle ABC, and D, E, Fare the middle points of the sides. Show that the system of forcesOA, OB, OC is equivalent to the system OD, OE, OF.

VECTOR ANALYSIS. 27

41. ABC is a triangle with a right angle at A ; AD is the per-

pendicular on BC. Prove that the resultant of forces AB acting

along AB and AC acting along AC is AD acting along AD.

42. P P2, ... P. are points which divide the circumference ofa circle into n equal parts. If a particle G lying on the circum-ference be acted upon by forces represented by GP,, GP27 . . . GPn,show that the magnitude of the resultant is constant wherever G istaken on that circumference.

It is n X OG, 0 being the center of the circle.043. If 0 be the center of the circumscribed circle of a triangleABC, and L the intersection of the perpendiculars from the verticeson the sides, prove that the resultant of forces represented by LA,LB, LC will be represented in magnitude and direction by 2 LO.Q44. D is a point in the plane of the triangle ABC, and I is the

center of its inscribed circle. Show that the resultant of the vectors

aAD, bBD, cCD is (a + b + c) ID, where a, b, c are the lengths ofthe sides of the triangle.046. The chords APB and CPD of a circle intersect at right

angles. Show that the resultant of PA, PB, PC, and PD is repre-sented by twice the vector P0, where 0 is the center of the circle.

46. Prove that the mean center of a tetrahedron is (a) the inter-section of bisectors of opposite edges; (b) the intersection of linesjoining the vertices to the mean points of the opposite faces. Showthat the former lines bisect one another, and that the latter quad-risect one another.

47. A, B, and C being three given points in a plane show thatany point in this plane can be made their centroid by giving suit-able weights to these points.

48. Show that the medians of a triangle intersect in a pointwhich is the mean center of the vertices A, B, C with weights1, 1, 1; that the altitudes intersect in a point which is the cen-troid of the vertices with weights, tan A, tan B, tan C, respectively;that the bisectors intersect in a point which is the centroid of thevertices with weights equal to the lengths of the opposite sides.-

CHAPTER II.

SCALAR AND VECTOR PRODUCTS OF TWO VECTORS.

The Scalar or Dot Product.

15. The Scalar Product of two vectors a and b, denotedby Sab, ab or (ab) by various writers, is a scalar de-fined by the equation

a b cos (ab) = b.a. (26)

FIG. 20.

This equation shows that the scalar product may be looked'upon as the product of the length of one of the two vec-tors multiplied by the projection of the other upon it, or

OAXOD=OBXOC.Evidently, if the two vectors a and b are perpendicular

to each other cos (ab) = 0 and their scalar product is zero.The condition, then, of perpendicularity of two finite vectorsis that their scalar product be zero.

Or, if a 1 b. (27)28

VECTOR ANALYSIS. 29

If a and b are parallel vectors, cos (ab) = 1 anda

b = a,a2.

The scalar product of a vector into itself is often writtenas the square of the vector, thus,

In general, to obtain the magnitude of a vectorial expressionit is only necessary to square it, and the result is the squareof its 4bsolute value or magnitude.

The Scalar Product Obeys the Ordinary Laws of Multipli-cation. Consider the two vectors c and d as well as theirsum (c + d). Consider also their projections upon anyother vector b.

FIG. 21.

The projection of c on b is OE, the projection of d onb is EF, the projection of (c + d)' on b is OF; hence

c.b + (c + b.(c + d). (28)

This result is easily extended to the scalar product of thesums of any number of vectors.

The application of these results to the unit vectors i, j ,and k is of great importance, giving immediately

j2=k2= 1,i.j = j.i = jk = kj = ki = ik.= 0. (29)

30 VECTOR ANALYSIS.

If the two vectors a and b be given in terms of theircoordinates,

a=a,i+a2j+a3kand

b=b,i+bZj+b3k,then, by (28) and (29),

(a,i + a2j + bZj + b3k)= alb, + a2b2 + a3b3. (30)

If a, and b, are unit vectors, their projections on the threeaxes are equal to their direction cosines; and since in thiscase a,.b, = cos (a,b,), then, by (30),

a,.bl = cos (a,b,) = cos (a,i) cos (b,i) + cos (a,j) cos (b,j)+ cos (a,k) cos (b,k),

the familiar formula of Cartesian geometry for the anglebetween two lines in terms of their direction cosines.

b

Fia. 22.

The well-known and useful formula giving directly themagnitude of the resultant of any two vectors in terms oftheir magr itudes and the angle between them, may bederived in the following manner. In the triangle ABC

c=a+'b.

VECTOR ANALYSIS. 31"

Squaring to find its magnitude,

c2 = a b or

c2 = a2 2 a b cos (c) + b2,

where qS is the supplement to the angle between a and b.

16. Line-Integral of a Vector. The scalar product playsa very important role in mechanics and physics. For ex-ample, the work done by a force F in the displacement dris by definition

F dr cos (F dr) =

If the force is known in direction and magnitude for every

Fia. 23.

point of its path, the workB done in overcoming the forces

from A to B may be found byevaluating the integral

I.a

(31)

This is called the line-integralof the vector F along the curveAB. The term "line-integralof a vector along a curve" thusdenotes the integral of thetangential component along it.unless expressly stated other-wise.

If q denote the vector

velocity at any point of a fluid, the integral

C =fqdr

32 VECTOR, ANALYSIS.

over any path in the fluid is called the circulation alongthat path. If e denote the electric force at any point inspace, the integral

E = ftaken along any path gives the electro-motive force alongthat path. This kind of an integral is thus of great- impor-tance in all branches of physics.

17. Surface-Integral of a Vector. As another example,imagine a surface S drawn in any vector field; for example,in a moving fluid. Let q be the vector velocity, determinate.

FIG. 24.

at every point in the region considered. The lines of flow ofthe fluid are therefore known and may be drawn. Theamount of liquid which passes outward through the ele-

VECTOR ANALYSIS. 33

ment dS in unit time at any point on the surface is theoutward normal component of q multiplied by the area dS,of

q cos (nq) dS = dS

where n is the unit outward drawn normal to dS. Thetotal outward flux through the surface is, then, the surfaceintegral.

Total Flux = f fs dS (32)

taken over the surface in question. It may easily be seenthat in this example the vector q may be any physicalvector such as electric force, magnetic force, gravitationalforce, or flux of heat, and others.

The term surface-integral of a vector over any surfacewill in the following denote the integral of the outwardnormal component over the surface, unless otherwise ex-pressly stated in the context.

The surface integral (32) expresses a very simple fact.If, for instance, we know the motion of every part of afluid, it should be possible, at least theoretically, to findout how much of the fluid leaves or enters a given regionby considering how much passes through every part of thebounding surface of the region and adding the resultstogether. To find the amount passing through any ele-ment of the surface we must evidently consider only thenormal component of the current of fluid. The tangentialcomponent of the current does not pass through the surface.The integral is the mathematical expression of this concep-tion and represents the total outward flux through thesurface S. Of course if the flow is inwards the result willbe negative, and if as much flows outwards through oneportion of the surface as there flows inwards elsewhere theresult will be zero.

34 VECTOR ANALYSIS.

i'he Vector or Cross Product.

18. The vector product of two vectors a and b is a vector,written axb (in distinction from the dot product), alsoV ab or [ab] by different authors, and is defined by theequation

axb = e a b sin (ab) = - bxa, (33)

where a is a vector, normal to the plane of a and b and sodirected that as you turn the first named vector a into thesecond one b, e points in the direction that a right-handedscrew (cork-screw) would progress if turned in this same

Fic. 25.

manner. In other words, axb is a vector perpendicular toboth a and b and whose magnitude may be represented bythe area of the parallelogram of which a and b are the adja-cent sides. The sense of this vector is purely conventionalbut is taken to conform with the more usual system of axes,i.e., the right-handed one.

According to this convention if the factor b came firstinstead of a, in the product, the only difference would bein the reversal of the sense of e, so that

axb = - bxa.

VECTOR ANALYSIS. 85

It is in this change of sign, when the order of the factors ischanged, that the vector product differs from the productof ordinary algebraic or scalar quantities. It is thereforenecessary when manipulating vector products to preservethe order of the factors unchanged or, at every change oforder, to introduce a minus sign as a factor.

In particular if a and b be finite vectors andaXb = 0, then a 11 b

as the sine of their included angle must be zero. This, then,is the condition for parallelism of the two vectors a and b.Since any vector is parallel to itself,

axa = 0. (34)

Remembering that the unit vectors i, j, and k are mutuallyperpendicular, it follows immediately from the definitionthat jxk=ikxj,

kxi = j = - i-k, (35)ixj = k = - JXi.

Notice the cyclical order of the factors in the above equa-tions.

We have also, by (34),jxj = kxk 0.

19. Distributive Law for Vector Products. It is obviousfrom the definition of aXb that

aXb = a'-b, (36)

where a' is the component of a 1 to b. Because in Fig. (26),as a' and b are in the same plane as a and b, a is the same asbefore, and a0' = ao sin 0. We may also say that the vectorproduct of b with the component of a parallel to b is zero.So that in any vector product we may, if we wish, replaceone of the vectors by its normal component to the other, andvice versa, without changing the value of the product.

36 VECTOR ANALYSIS.

Keeping this in mind, we may prove that the distributivelaw holds for vector prgducts, or, in symbols, that

(a + b)xc = axc + bxc. (37)

where a and b are any two vectors.Let c be drawn (Fig. 27) 1 tothe plane of the paper at 0 andtowards the reader. Let a' andb' be the components of a andb 1 to c and hence lying in theplane of the paper, The vectors

FIG. 26. FIG. 27.

a'xc and b'xc will also lie in the plane of the paper perpen-dicular to a' and b' respectively.

= o

b'csin_b'( =2

Since_ A'B'

(a' c)

n a' = m,

x 0 a'csin2

the triangles OAB and OA'B' are similar, hence OB m,

=

and ON is L to OB. Consequently

OB' = (a' + b')xc = OA' + A'B' = a'xc + b'xc.We may now replace a' and b' by a and b according to (36)above, so that

(a + b)xc = axc + bxc.* See equation (1) for notation.

VECTOR ANALYSIS. 87

If c itself be considered to be made up of two vectorse and f, then by the same reasoning

ax (e + f) = axe + axfand bx (e + f) = bxe + bxf,

so that(a + b)x(e+ f) = axe + axf + bxe + bxf (38)

and so on for any number of vectors.

Physical Proof of the Distributive Law. - It is interest-ing to prove the distributive law for vector products bymeans of the following hydrostatic theorem. It is well

Fia. 28.

known that any closed polyhedral surface immersed in afluid is in equilibrium under the normal hydrostatic pres-sures exerted upon its faces by the liquid. These pressuresproduce forces normal to the faces of the polyhedron whichare proportional to their areas and may therefore be repre-sented by vectors perpendicular to them, the length ofeach one being proportional to the total pressure on theface to which it is perpendicular. The condition for equi-librium is then that the sum of these vectors be zero, i.e.,that they have no resultant. This result is seen to be also

-38 VECTOR ANALYSIS.

true for any curved surface by considering it as the limitingcase of a polyhedron with an infinite number of infinitelysmall, plane facets. Let a, b and - (a + b) be the threesides of a triangle, taken in order. Form the prism ofwhich this triangle and any third vector c is the slant heightor edge. The areas of the lateral faces of this prism arerespectively, viewing them from the'outside,

axc bxc and - (a + b)-c;the areas of the end faces are similarly

I axb and - j axb.Now by the preceding hydrostatic theorem the vector sumof the faces of any closed surface is zero, hence

axc+bxc-(a+b)xc +I axb- axb=0,giving again (a + b)xc = axc + bxc.

This proof, which is given purely for its physical interest,amounts to saying that the vector area of any closed surfaceis equal to zero. The relation holds, however nearly par-allel to the plane of a and b, c may be. It may also beshown to hold when c lies in the plane of a and b. Con-versely, assuming that the distributive law holds, thehydrostatic theorem employed in the above proof followsimmediately.*

20. Cartesian Expansion for the Vector Product. It isoften convenient to express a vector product in terms of thecomponents of its vectors.

Let a = ali + a2j + a3k,b = b1i + b2j + b3k;

then axb = (ali + a2j + a3k)x(bli + b2j + b,k)which by the extension of (38) becomesaxb = (azbs- asb2) i + (a3b1 - alb3) j+ (a1b2 - a2b1) k. (39)

* Still another proof of the distributive law may be found in Foppl :Einfiihrung in die Maxwell'sche Theorie der Elektricitit, pp. 16 and 17.

VECTOR ANALYSIS. 39

This expression may be conveniently condensed into thedeterminant

axb =i. j kai a2 asbi b2 b3

(40)

This is a useful mnemonic form for the vector product. Aspreviously stated, if the vector product is zero the vectors,if finite, are parallel. This condition in terms of theirprojections on the axes is given by noticing that in (39) thethree coefficients of i, j, and k must separately vanish, oragain from the determinant form by noticing that tworows must be proportional, or that

ai a2 as

a well known result.If al and bi are unit vectors alxbl is the sine of their

included angle B; the quantities a,, a2; a3, and bi, b2, b3,being then their direction cosines respectively. Squaringformula (39) there results

sine d = (a2b3 - a3b2)2 + (a3b1 - albs)-+ (aib2 - a2bi)2.

If we express the distributive law in the determinantform we obtain the following addition theorem in determi-nants of the third order.

i j k i j k l j k-ai a2 a3 = ai a2 a3

+al a2 a3

(bi + cl) (b2 + c2) (b3 + c3) bi b2 b3 ci C2 c3

21. Application to Mechanics. Moment. The moment ofa force F about a point 0 is defined as the product of theforce into its perpendicular distance from the point 0, or insymbols, by

FXOA=FXrsin6. (41)

* Conversely, assuming the addition theorem for determinants, thedistributive law of vector products follows immediately.

40 VECTOR ANALYSIS.

This moment, in the figure, is right-handed about a vectorperpendicular to the paper and pointing directly up, sothat a vector of magnitude F X Od in this direction wouldrepresent the moment of F about 0 in a very convenientmanner. According to this convention the vector

M= rxF (42)

represents in magnitude and direction the moment of Fabout 0, where r is the vector to the point of applicationof the force. If the force F is the resultant of a numberof forces Fl, F, . . . acting at the same point of application,then by (38)

rxF = rx(FI+ F2+ ...) = rxF,+ r.F2+ .. .or, the moment of the

JJ`/J

F sin e x r=rsin 6 xF

M=rxF=

Qo ;

FIG. 29.

i j kxx Y Z

resultant of any number of forcesabout a point is equal to the sumof the separate moments. Thistheorem also shows that momentsobey the parallelogram law. Con-versely, assuming the truth of thistheorem of moments, the distribu-tive law for vector products is anecessary consequence.

If F have components X Y Z,r components x y z, and M com-ponents M. M,, MZ, the momentof F about the origin may beimmediately written down by (40)

=i(yZ-zY)+j(zX-xZ)+k(xY-yX)

So that M2 = (y7, - zY),My = (zX - xZ), (43)M2 = (xY - yX ).

VECTOR ANALYSIS. 41

22. Motion of a Rigid Body. Consider the motion ofrotation of a rigid body about an axis, with a constantangular velocity w. A velocity of rotation being of neces-sity about some axis, it is convenient to represent this kindof motion by a vector whose magnitude is proportional tothe angular velocity and whose direction coincides with theaxis of rotation. Its direction and that of the correspond-ing rotation may be simply representedby the symbol in Fig. 30.

Notice that this is also the relationexisting between direction of currentand corresponding magnetic field.

Choose an origin on the axis ofrotation Fig. 31 and consider a pointP anywhere in the body, to find thevelocity of the point P. Let P be

Fio. 30.

determined by the radius vector r drawn to it from theorigin. The velocity q of P is at right angles to w andto r, its magnitude being given by the expression

q=c,wXrsin8,

as is easily seen in the figure. In other words q is repre-sented not only in magnitude but in direction as well, by

q = (ox r. (44)

23. Composition of Angular Velocities. Since angularvelocities may be represented by vectors let us see whetherthey compound according to the parallelogram law. Toprove this definitely, let the body have several angularvelocities w co, w3 ... about axes passing through theorigin. Then , the linear velocities of P separately due tothese are

q, = w1X r,q2=o2xr,q3 = w3" r,

42 VECTOR ANALYSIS.

and hence the velocity of P due to them all acting simulta-neously is

q=ql+q2+q3+... = W,xr+WZxr+W3xr+---

=(W1+WZ+(a3+...)xr,

or the resultant velocity q of P is the same as if the bodyrotated with an angular velocity w about an axis through 0given in magnitude and direction by

0) =0),+WZ+W3+...=FW.%

FIG. 31.

If the body have in addition to its angular velocity avelocity of. translation q, the resultant velocity q of thepoint P is simply

q = q, + Wxr. (45)

VECTOR ANALYSIS 43

In the case that qt is 1 to w, there must be a line of pointswhich are instantaneously at rest. This line is determinedby the condition q = 0 or

rxw = qt, (46)

which is a straight line parallel to w. Change the origin to apoint 0' on this line, the expression for the velocity reduces tothe form q = wxr,,

where r' is the vector from 0' to any point in the body. If qtis not 1 to co, decompose it into two components qt' and qi'such that qt = qt' + q,".

Let qt' be 11 to co, and qt" J_ to w, we may then proceed asbefore with qt". It is thus seen that the most general motionpossible of a rigid body is that of a rotation about a certainaxis and a velocity of translation along it; in other words, ascrew motion.

If co and qt are variable this holds true at any instant,although the direction and pitch of the screw motion may berapidly and of course continuously changing. The axis ofrotation about which a rigid body is rotating at any giveninstant is called the instantaneous axis of rotation. If thebody has one point fixed the velocity qt is zero and the instan-taneous axis, of rotation always passes through this fixedpoint. The equation of the instantaneous axis is thengiven by the condition that

rcco = qt".

EXERCISES AND PROBLEMS.1. Show that the two vectors

a=91+ j-6kand b=41-6j +5kare at right angles to each other.

2. The coordinates of two points are (3, 1, 2) and (2, - 2, 4) ;find the cosine of the angle between the vectors joining these pointsto the origin.

44 VECTOR ANALYSIS.

3. Write out in the form

a=a,i+a,j+a,kseveral pairs of mutually perpendicular vectors.

4. Write out in the form

a=a,i+a,j+a,kthe expressions for several unit vectors.

5. Find a vector in the ij-plane which has the same length as thevector

a=41-2j+3 k.Find a vector in the jk-plane having the same length, and the samej-projection as a.

6. Let a and b be two unit vectors lying in the i j-plane. Let abe the angle that a makes with i, and Q the angle b makes with i;then

a=icosa+jsina,b=icosj+jsini9.

Form the dot and cross products and show that the additiontheorems for the cosine and sine follow from their interpretation.

7. Leta=a,i+a,j+a,k

and b = b,i + b,j + b3kbe the unit vectors to the pointsA and B. Find the distance betweenA and B and its direction cosines in terms of a a a, and b,, b,, b,.

8. Three vectors of lengths a, 2 a, 3 a meet in a point and aredirected along the diagonals of the three faces of a cube, meeting atthe point. Determine the magnitude of their resultant. Findthe resultant in the form

r=xi+yj+zkand from'this calculate its magnitude.

9. The sum of the squares of the diagonals of a parallelogram isequal to the sum of the squares of the sides.

10. Parallelograms upon the same base and between sameparallels are equal in area.

11. The squares of the sides of any quadrilateral exceed thesquares of the diagonals by four times the square of the line whichjoins the middle points of the diagonals.

VECTOR ANALYSIS. 45

12. Under what conditions will the resultant of a system ofvectors of magnitudes 7, 24, and 25 be equal to zero?

13. Three vectors of lengths a, a, and a V2 meet in a point andare mutually at right angles. Determine the magnitude of theresultant and the angles between its direction and that of eachcomponent.

14. ABC is a triangle, and P any point in BC. If PQ representthe resultant of the forces represented by AP, PB, BC, show thatthe locus of Q is a straight line parallel to BC.

15. The angle in a semicircle is a right angle.Take equation of circle

r and interpret.16. If two circles intersect, the line joining the centers is per-

pendicular to the line joining the points of intersection.17. 0 is a fixed point, AB a given straight line. A point Q is

taken in the line OP drawn to a point P in AB, such that

OP OQ = k2 (a const.).

To find the locus of Q.Application to problems in Inversion.18. If any line pass through the centroid of a number of points,

the sum of the perpendiculars on this line from the different points,measured in the same direction, is zero.

Application to method of Least Squares.19. Write out the vector product of the two vectors

a 61+0.3j- 5kand b=0.11-4.2j+2.5kand show by calculation that the resulting vector is perpendicularto each of the constituent vectors of the product.

20. Find the area of the triangle determined by the two vectors

a= 31+4jand b= -51+7j.

46 VECTOR ANALYSIS.

21. Find the area of the parallelogram determined by the vectors

a= i+2j+3kand b= -31-2j+ k.

22. Express the relations between the sides and opposite anglesof a triangle.

In any triangle of vector sides a, b, c,

a=b - c,take the vector product of a with this and interpret.

23. By means of the equation of § 20 find the sine of the anglebetween the two vectors

a=31+ j+2kand b=21-2j+4k.

24. Show that the equation of a Tine perpendicular to the twovectors b and c is

r = a + xfixc.

25. Find the perpendicular from the origin on the line

ax (r - b) = 0.

26. Derive an expression for the area of a square of which

r=all+a2jis the semi-diagonal.

27. If the middle point of one of the non-parallel sides of atrapezoid be joined to the extremities of the opposite side, a triangleis obtained whose area is one-half of that of the trapezoid.

28. Find the relations between two right-handed systems ofthree mutually perpendicular unit vectors. See Gibbs-Wilson,p. 104.

29. Given c = a + b.

Expand he right-hand side of each of the equations

(a +(a + b)-(a + b),

and give the geometric interpretation of the result.

VECTOR ANALYSIS. 47

30. Given r = x a + y b + z cwhere a b c are three non-coplanar vectors. Expand the right-hand side of the equation

(xa +yb +yb+zc)and give the geometric interpretation of the result.

31. Show that the work done by a force during a displacementis equal to the sum of the quantities of work performed by itscomponents during the displacement.

32. A fluid is flowing across a plane surface with a uniformvelocity which is represented in magnitude and direction by thevector q. If n is the unit normal to the plane, show that thevolume of the fluid that passes through the unit area of the planein unit time is q n.

33. Show that a system of forces represented in magnitude,direction, and position by the successive sides of a plane polygonis equivalent to a couple whose moment is equal to twice thearea of the polygon.

34. If 0 be any point whatever, either in the plane of the tri-angle ABC or out of that plane, the squares of the sides of the tri-angle fall short of three times the squares of the distances of theangular points from 0, by the square of three times the distanceof the mean point from 0.

35. The sum of the squares of the distances of any point 0 fromthe angular points of the triangle exceeds the sum of the squaresof its distances from the middle points of the sides by the sum ofthe squares of half the sides.

36. Show that(a - b)-(a + b) = 2 axb.

and give its geometric interpretation,37. Show that

(a - b) = a2 - b2

and interpret.

CHAPTER III.

VECTOR AND SCALAR PRODUCTS INVOLVINGTHREE VECTORS.

24. From the three vectors a, b, and c the following com-binations may be derived :

1. a (a vector) 4. a (bxc) (not defined)2. a (bXc) (a scalar) 5. a (absurd) (47)3. ax(bxc) (a vector) 6. (absurd).

Of these six expressions, 5 and 6 are meaningless andabsurd, because they are the scalar product and vectorproduct, respectively, of a vector (a) and a scalar andsuch products require a vector on each side of the dot or cross.As to 4, since no definition of the product of two vectors with-out a dot or a cross has been made, it is as yet meaningless.In this book we shall not consider such products. Weshall consider in detail the three remaining triple products.The first one of these, a is simply the vector a multipliedby the scalar quantity and is a vector in the same direc-tion as a, but be cos (bc) times longer. This triple product,then, offers no new difficulties, and means

a (bc)

25. The Triple Product V = a (bxc) is a scalar and rep-resents the volume of a parallelopiped of which the threeconterminous edges are a, b, and c. This is easily seen tobe the case, as bxc is the area of the base represented bya vector OS L to this base; the scalar product of a and thevector OS will be this area multiplied by the projection ofthe slant height a along it, or, in other words, the volume.As evidently this volume, V, may be obtained by forming the

48

VECTOR ANALYSIS. 49

vector products of any two of the three vectors a, b, and c(thus giving the area of one of the faces) and forming thescalar product of this vector-area with the remaining thirdvector, it follows that

V =

If the vectors (cxa), (axb), and (bxc) are taken so that theyform an acute angle with b, c, and a, respectively, then thevolume is to be considered positive, the cosine term in the

FIG. 32.

scalar product being positive. Otherwise the volume is to beconsidered negative. Of course the inversion of the factorsin the vector products should change the sign, by (33), sothat we have

V = b(cxa) = (cxa)b = - b(axc) _ - (axc)b= c (axb) = (axb) c = - c (bxa) _ - (bxa) c (48)

= a.(bxc) = (bxc)a = - a.(cxb) = - (c-b)-a.

By a consideration of these equalities the following lawsmay be seen to hold:

1. The sign of the scalar triple product is unchanged aslong as the cyclical order of the factors is unchanged.

50 VECTOR ANALYSIS.

2. For every change of cyclical order a minus sign isintroduced.

3.. The dot and the cross may be interchanged ad libitum.The equalities (48) are called by Heaviside the Parallelo-

piped Law.The product may be written in terms of the com-

ponents of its vectors along any three rectangular Cartesianaxes as

a (bxc) = a, (b2c3 - bsc2) + a2 (bac, - b,c3) +a3(b,c2 - b2ci)

= a, b2 b3

C2 C3

+ a2 b3 b,

CS C1

+ a9 b, b2 l

C, C2

a, a2 a3I (49)

b,

b2 bsC, C2 C$

This is the familiar determinant expression for the volumeof a parallelopiped with one corner at the origin.

The parallelopiped principle, then, expresses the fact that aslong as the cyclical order of the rows is unchanged the deter-minant is also unchanged, but that every interchange ofcyclical order introduces a minus sign as a factor. To thestudent familiar with determinants this is a well knownproperty. Conversely, assuming this property of a deter-minant as proven, the equations (48) immediately follow.

The twelve expressions (48) are often written in one, as[abc], a special symbol of abbreviation taken from Grass-mann.

26. Condition that Three Vectors Lie in One Plane.Should the three finite non-parallel vectors a, b, and c liein one plane the volume of the parallelopiped they deter-mine is zero. Hence the condition that the three non-parallel vectors should lie in a plane is that

[abc] = 0. (50)

VECTOR ANALYSIS. 51

In the expression [abc], if any two of the vectors are parallelthe volume of the parallelopiped is again evidently zero.Hence, in general, [aab] = 0. (51)

To look at it in another way, we may put, by (48)(a-a)-b,

and as axa = 0, then a (axb) = 0, so that in a triple scalarproduct if any two of the vectors are parallel their triplescalar product is zero.

In the determinant (49) above, this corresponds to havingany two rows proportional to each other, the result being,as is well known, identically zero.

The parenthesis in an expression such as a (bxc) is inreality unnecessary, as its only other interpretationis without meaning, being the vector product of a scalar(a.b) and a vector c. The parentheses are introduced,however, when by so doing the interpretation is made easier.

Scalar magnitudes of the vectors, it is important toremember, which occur in any kind of scalar or vectorproducts may be placed in any part of the expression asfactors.

For example,

abc (52)= etc., etc.

27. The Triple Product q = ax(b"c) is a vector. In thisexpression the parenthesis, or some separating symbol, isnecessary, as ax(bxc) 34 (axb)xc. The sign of this productchanges every time the order of the factors a and (bxc) ischanged in ax(bxc), or whenever the order of the factors band c is changed in (bxc). The vector product being alwaysperpendicular to both of its components, q is perpendicularto a as well as to bxc, hence

q (bxc) = 0. (53)

52 VECTQR`+ANALYSIS.

Equation (53) shows that q lies in the same plane as b andc, either by (50) or by seeing that it is perpendicular to a linewhich is itself perpendicular to b and c. It is important thatthis result be clearly visualized. The habit of visualizationshould be cultivated, as it is of great importance to the studentwhatever kind of analysis he be using, but particularly so inthis. To a purely analytical mind vector analysis offers butf ew advantages.

As q lies in the plane of b and c it is possible to express q inthe form

q=xb - yc,where x and y are scalar multipliers. Let us try to determinethe quantities x and y. Since q is perpendicular to a,

a.q =and, therefore,

x:y=a.c:a.b orx=na.c,

where n is a scalar factor of proportionality. So thatq = ax(bxc) = n (54)

We shall now prove that 'n is independent of the magni-tudes and inclinations of the vectors a, b, and c. It is inde-pendent of their magnitudes because they may be taken outby (52) as scalar coefficients and eliminated from the equa-tion. Since we are dealing with the mutual relations betweenany three vectors, we may choose one of them arbitrarily.Let that one be a. Let us now replace one of the remainingvectors, c, for instance, by the sum of two other vectors dand e. Then

ax(bx(d + e)) = n [b a.(d + e) - (d + e)or

ax(bxd)+ ax(bxe) = n [b e) - (d+e)or finally,n'[b ad-d ab]+n"[b ae-e ab]=n[b a.(d+e)-(d+e)ab].

VECTOR ANALYSIS. 53

If d and e have been chosen so that b, d, and e are not co-planar, then we may equate coefficients of the vectors b, d, ande on both sides. Thus

n a. (d + e).n' nn" n

which necessitates thatn=n=n.

The coefficient n in equation (54) is thus independent ofa, b, and c, and is, therefore, a numerical constant. To findits value we are now at liberty to consider a special case. Leta, b, c be unit vectors. Let a = c and let b be perpendicu-lar to c. This is the equivalent to writing a = k, bc = k. We then have for equation (54),

kx(jxk) = n [j k(k.j)].but jxk = i and kxi = j,

1 and k.j = 0:therefore the equation reduces to

j = n(j); hence n = 1.

We have thus proved the very important relationax(bxc) = (55)

which should be memorized.

28. Demonstration by Cartesian Expansion. A demon-stration of this equation may also be obtained by expandingin terms of the Cartesian' components of the vectors.This method is a very useful one when no other demonstra-tion readily offers itself, but generally (not in this case)has the disadvantage of being long and cumbersome. Nobetter examples of the concentration of the vector notationmay be found than by carrying through a number of suchtransformations. On account of the importance of the equa-

54 VECTOR ANALYSIS.

tion (55), and also to give an example of the expansionmethod in general, its demonstration by this method will becarried out.

As the components of (bxc) are

we may write, by (40),

ax(bxc) =

ia1

b2b3

c2c3

b2b3

CZC3,

b3b1

C3C1and

b1b2

CiC2, see (49)

i ka a2 3

= i [a2 (b1C2 - b2C1)-a3 (b3C1 - b1C3)]b3b11

Ibib2l + [a3 (b2c3 - b3c2) - a1(b1c2 - b2c1)]C3C1 CIC2 + k [a1 (b9c1 - b1c3) - a2(b2C9b c2)]3

The terms may now be rearranged into

+ j b2 (a1c1 + a2C2 + a3c3)

+ k b3 a2C2 + a3c3)

- i c1 (albs + a2b2 + a3b3)

j c2 (a1b1 + a2b2 + a3b3)

- k c3 (a1b1 + a2b2 + a3b3)

The new underlined terms have been added and subtracted.The first three lines are

(i b1 + j b2 + k b3)

the last three are

- (i c1 + j c2 + k c3)

hence

ax (bxc) = b c

29. Third Proof. That n = 1 in (54) may also be provedas follows: Consider first the triple vector product in whichtwo of the vectors are the same,

bx(bxc) = n (b -c

ax(bxc) = i b1 (a1c1 + a2C2 + a3c3)

55VECTOR ANALYSIS.

Taking the scalar product of this and c, or, in shorter language,applying c dot (.) to it, we obtain

bac21.

But' by an interchange of the dot and the cross ana onechange of cyclical order the left-hand side becomes

(c- b) (bxc) = - (bxc) (bxc) (bxc)2. (56)

We know, however, that

b2c2, (56a)

as by definitionb2c2 cos2 (bc) + b2c2 sine (bc) = b2c2

is equivalent to it; hence the right hand of the first equationis nothing more than-n (bxc)2, and comparing with (56) wesee that n must be unity. The theorem is thus true, whentwo of the vectors are the same. Consider now the generalcase,

ax(bxc) = n (b c (57)

Apply b dot to it, obtainingn (b.b a.b)

= n b

which as we have just proved may be written= na.[- bx(bxc)).

But on the left-hand side we have

by an interchange of dot and cross and one of cyclical order.Comparing the last two equations we see that n = 1 ingeneral.

The parenthesis in ax(bxc) is necessary, for (axb)xc is quitedifferent from the first expression, as one may readily see byexpanding-the two, or by the reasoning of § 27.

30. Products of More than Three Vectors. In practicalapplications to physics more complicated products than thoseof three vectors seldom arise. Whenever they do, they may

56 VECTOR ANALYSIS.

be reduced by successive applications of the preceding prin-ciples. In any case they represent extremely complicatedCartesian expansions. As an example of such a reduction,consider the scalar expression containing four vectors,(axb) (cxd)

Interchange the first cross and dot and expand the vectortriple product, which will give,

a.(c d

bc_ ac bc

(58)ad bd

This formula will be used in the deduction of Stokes' Theo-rem in § (58).

Again consider the quadruple vector product (axb)x(cxd),which may be expanded by (55),

(axb)x(cxd)= c (axb)d - d (axb)cor into = b (cxd )a - a (c-d)-b, (59)

taking in the first case (axb), c, and d as the three vectors ofa triple product, and in the second case a, b, and (cxd). Bysubtracting these two equal expressions from each other wehave

a b(cxd)- b a(cxd)+ c d(axb)- d c(axb)-0, (60)

an important relation holding between any four vectors.Putting d = r, this equation may be written

r [abc] _ [rbc]a + [rca] b + [rab]c,

so thatr = [rbc] a -{- [rca b + [rab c (6-1)

or

,[abc] [abc] [abc]

cb + r 62),r r.

[abc]a

+r.

[abc] ' [abc] (

VECTOR ANALYSIS. 57

an important and useful formula which gives the coefficientsnecessary to express r in terms of any three arbitrary vectorsnot lying in the same plane. This expansion is under theseconditions always possible as explained in § (7).

31. Reciprocal System of Vectors. The three vectors

b (63)[abc[abc]' and [abc]'

perpendicular respectively to the planes of b and c, c and a,and a and b, occur frequently in important relations and aresaid to be the system reciprocal to a, b, and c. They havepeculiar and interesting properties which the student willfind fully demonstrated in another work.*

It will be noticed that only two kinds of products of vectorshave been defined, i.e., the scalar product and the vectorproduct. One should carefully remember as well that thescalar product and the vector product have been defined interms of two simple vectors, but that instead of simple vec-tors any expression which is itself a vector may be used inplace of the simple vectors to form these products. If this iscarefully kept in mind it will make clear that in vector analy-sis certain combinations of symbols are meaningless.

For example, axb being a vector, it may be used in con-junction with another simple vector e, or another vectorproduct cxd, to form new scalar products and vector prod-ucts, such as

and

(axb)xe and (axb)x(cxd),or even [(a-b)-(c-d)]-[(e-b)-(g-h)], etc.,

which are all legitimate expressions. But, on the otherhand, neither nor nor may beused again to form either scalar or vector products becausethey are merely scalars.

* Gibbs-Wilson, Vector Analysis, pp. 82-92.

58 VECTOR ANALYSIS.

Equations of Plane, Line, and Sphere.

32. The Plane Perpendicular to a and Passing through theTerminus of b. Let r be the radius vector to any point in it.The projection of r upon a is evidently constant and equalto the projection of b upon a as long as the terminus of r isin the plane; this condition is expressed by the equation

constant = (64)

or a. (r - b) = 0,

which is, therefore, the equation of the plane. It also statesthat (r - b) is perpendicular to a and hence parallel to the

0Fia. 33.

plane which is an evident truth and could be used to derivethe equation of the plane. If the origin is in the plane, b = 0and 0 (65)is the equation, which is otherwise evident, as r is then alwaysperpendicular to a. If the equation of a plane is desired, theplane being parallel to two given vectors c and d and passingthrough the terminus of b, simply remember that cxd is avector perpendicular to the plane, and putting cxd in placeof a above, its equation is

b) = 0. (66)

VECTOR ANALYSIS. 59

It the equation of a plane passing through the ends of threegiven vectors a, b, and c is desired, remember that the vec-tors (r - a), (a - b), and (b - c) lie in the same plane andexpress this fact by (50), giving

(r - b)x(b - c) = 0,or expanding

bxc + cxa) = a.(axb + bxc + cxa)or (67)

+ (r - a) = 0, where 4 _ (axb + bxc + cxa).

Comparing this last equation with (64), we see that 4) is avector perpendicular to the plane.

To find the vector-perpendicular p from the origin to theplane. Referring to the plane in Fig. 33, the equation ofwhich is

b) = 0,

60 VECTOR ANALYSIS.

let r become perpendicular to the plane and hence somemultiple of a, say x a, then

or x a2 = a

a a a=a a aa

( a)

This result is also evident on inspection. For is theprojection of b on a multiplied by the magnitude of a; hence,to obtain the value of the projection we must divide by themagnitude of a, so that directly

a

In equation. (67) the perpendicular p is thenp = 4-1 +.a. (68b)

33. The equation of a straight line through the end of bparallel to a is, since a and (r - b) are always parallel,

ax (r - b) = 0. (69)

This is compatible with theequation derived previously,

r = b + x a,

for applying ax to it, we obtain69 in the form

axr = axb.

Again, the equation of a lineperpendicular to c and d andpassing through the end of bis, because cxd is parallel to ain the above equation,

FIG. 35.

(cxd)x(r - b) = 0. (70)

VECTOR ANALYSIS 61

34. Equations of the Circle and the Sphere. In a circle orsphere, with origin at the center, the length of the vector tothe surface from the origin is constant and equal to theradius. Hence

r a, (not r=a)or r2 = a2, (71)

is the equation of the circle or of the sphere according astwo or three dimensional space is considered.

FIG. 36.

If the origin is removed to 0 at a distance - c from 0, then(71) becomes

(r - c)2 = a2,or r2 - 2 a2 - c2 = const. (72)

If c = a, that is, if the origin is on the circumference of thecircle, the equation reduces to

r2 - 2 0. (73)

This, in polar coordinates, is nothing more thanr=2acos8,

62 VECTOR ANALYSIS.

where 0 is the angle between r and any predetermined radius,from which 0 is measured.

In rectangular coordinates (73) considered as the equationof a sphere is written immediately

x2 + y2 + z2 = 2 (xal + ya2 + za3),

Fra. 37.

where a,, a2, a3 are the projections of any chosen radius alongthe three axes. If the plane of yz be taken perpendicular tothis radius, a2 and a3 are zero, so that

x2 + y2 + z2 = 2 aix.

If we drop the z-coordinate the equation reduces to that of acircle tangent at the origin to the y axis and with its centeron the x axis.

The equation with origin at center may be put in the form

r2 - a2 = 0,

or

which says, see Fig. 38, that the two lines AD and DB arealways at right angles, a familiar result. Such illustrationsmay be multiplied indefinitely and show the ease with which

VECTOR ANALYSIS. 63

equations may be written down to fit almost any condi-tions. When translated into their Cartesian equivalentsthey give familiar forms.

Such books as Tait, "Quaternions," Kelland and Tait,"Introduction to Quaternions," may be consulted with ad-vantage at this point by the student. In these books thewhole treatment of the line, plane, circle, sphere, and conicsections,with few exceptions, is one of vector analysis pure and

Fia. 38.

simple. The occurrence of a quaternion is a very rare event.The only difference to be noted particularly in reading theseworks is that the scalar product has the opposite sign to ours,and that

is written S aband axb is written V ab.

34a. Resolution of a System of Forces Acting on a RigidBody. Consider any point 0 as origin. This point may beanywhere, even outside of the body. The system of forces

64 VECTOR ANALYSIS.

F1, F2, F3, etc., acting on the body is equivalent to a singleforce R at 0, where

R =ZFand a couple whose strength is

C = YaxF,

where a1, a2 ... are the vectors to the forces F1, F2 .

from 0.To prove this, consider one of the forces F acting at P; we

may introduce the zero system + F - F at 0 without alter-

Fm. 39.

ing in any way the effect of F on the body. By combining- F with the F at P we get a couple of strength axF,leaving a force F acting at 0. We may do the same for eachof the forces F1, F21 etc., so that finally we have

(a) all the forces F1, . . . F2 acting at 0and (b) an equal number of couples a1xF1, a2xF2 ..

VECTOR ANALYSIS. 65

Combine all the forces into a resultant force, R =yFand all the couples in a resultant couple C = axF, whichproves the theorem. Such a resolution may be made for anypoint whatever.

Central Axis. Couple a Minimum for Points on this Axis.In general the axis of the couple C derived above is notparallel to the resultant force R.

Let us inquire if there are any points for which an analo-gous resolution will give a couple whose axis is parallel to R.Notice that whatever point 0' is chosen R is the same vector.

Let O' be such a point and let 00' = r, and for O' the couplewould be, since 0'P = a - r,

Z (a - r) -F.

Then the condition that this couple be parallel to R becomes

Z(a - r)'F = x R = ,axF - rxXF,

so thatx R = C - rxR.

To find x, multiply by R-1, because R-1 is parallel to R,and the triple scalar product term vanishes.

HenceX =

rxR=C -

a linear equation for r in terms of R and C (given vectors bylast theorem). There is then a line of points for which thiskind of resolution is possible. This line is called the CentralAxis of the System, which is then reduced to a force R and acouple about R of a certain magnitude = xR = R

(a)

Considering the equation of the central axisrxR = (R-1xC)xR,

66 VECTOR ANALYSIS.

it is seen that one of the values of r is (R-'xC), and sincethis last vector is evidently perpendicular to R, it must be theline ON; so that the equation may also be written

r = R-1 x C + yR = ON + yR,

where R-1 -C is the normal vector from 0 to the centralaxis.

0Fio. 39A.

By (a) we see that the couple about the central axis is thecomponent of the couple at any other point along R, andhence is always less than for any other point; so that it is aminimum and equal to

along R.

It is, however, the same, i.e., constant for all points on thecentral axis.

EXERCISES AND PROBLEMS,

L Prove the following formulae:

axIbx(cxd)[ _ [acd] b - cxd= axc - axd

[axb cxd exf] = [abd] [cef] - [abc] [def]= [abe] [fed] - [abf] [ecd]= [cda] [bef] - [cdb) [aef].

VECTOR ANALYSIS. 67

2. Prove[axb bxc cxa] _ [abc]'

and interpret the result by determinants.3. Show that

4. Show that

p[pqr] (axb) = q.a q.b q

r

ax(bxc) + bx(cxa) + cx(axb) = 0.

5. Show that

[axp bxq cxr] + [axq bxr cxp] + [axr bxp cxq] = 0.

6. Prove that(a.c)(b.d) - (a.d)(b.c)

and that(axb)x(cxd) = b [aed] - a [bed]

= c [abd] - d [abc].

7. Deduce the fundamental formulae of spherical trigonometryfrom the equations

(axb)(cxd) = ac bd - ad bc,

(axb)x(cxd) = [acd] b - [bed] a

= [abd] c - [abc] d.

Make the vectors unit vectors and take an origin at center of sphereof unit radius, thus making all the vectors terminate upon the sur-face of the sphere.

8. Show that the components of b parallel and perpendicular toa are respectively

b' = a-ab = ai aibaaax (ax b)and b" _ - = - alx(a1xb).aa

9. The second vector a may be omitted from ax (a + b). Mayit be omitted in a7'-(a + b) or in ax(a + b)-'?

68 VECTOR ANALYSIS.

10. The perpendiculars from the vertices of a triangle to thesides opposite meet in a point.

11. Find the point of intersection of a line and a plane and dis-cuss the result.

12. The perpendicular bisectors of the sides of a triangle meetin a point.

13. Find an expression for the common perpendicular to twolines not lying in the same plane.

14. Determine the vector-perpendicular drawn from the originto the plane determined by the three points a, b, c.

15. Find the equation of a plane passing through a given point cand parallel to each of two given straight lines b' and W.

Ans. (r - c)-b'-b" = 0.16. Find the length of the common perpendicular to each of two

given straight lines parallel to b, and bz and passing through al andaz respectively, and show that it is

d = y (bixbz),

y = (a, -where(b,xb2)2

17. Find the equation of the line of intersection of two planes.18. Deduce the Cartesian equation for the volume of the tetra-

hedron whose vertices-area, b, c, d,

where a = a,1 + az j + a3k, etc.

19. Deduce the Cartesian equation for the area of the trianglewhose vertices are

a = a,i + azj + a3k,b=b,i+bzj+b3k,c= c, l+ c z j+ c3k.

20. Find by translating into Cartesian notation that the volumeof a pyramid, of which the vertex is a given point (xyz) and thebase a triangle formed by joining three given points aoo, obo, o'ocin the rectangular coordinate axes, is

Va b c

VECTOR ANALYSIS. 69

21. Show how to determine the directions of two vectors of givenmagnitude so that their resultant shall be of given magnitude anddirection. When is this impossible?

22. The moment of the force AB about the line CD is six timesthe volume of the tetrahedron ABCD divided by the number ofunits of length in CD.

Find vector moment at C, and then the component of this alongCD.

23. The laws of refraction of light from a medium of index a intoone of index p' are comprised in the relation

pnxa = fa'nxa',

where n, a, and a' are unit vectors along the normal, the incidentand refracted rays respectively.

24. Write out the equations of problems 1 to 8 inclusive inCartesian notation.

25. If r = a + b x is the equation of a straight line, b being aunit vector, prove that the line through the origin perpendicularto it is

and that its length isr = y (a - b)

a2 -26. The equation of the plane through the origin perpendicular

to the vector a may be written in either of the forms0 0 (r -f- a)o = (r-a)o.

27. Let r = b + x d be the equation of a straight line. Forwhat values of x does it meet the sphere r2 = c2? By the theoremon the coefficients of an equation as related to the roots, derive thetheorems for the product and sum of the intercepts respectively.

28. Derive the equation of the sphere (or circle) from the equa-tion

(r - c)o = a.This equation states that the end of r must remain at a constantdistance a from the end of c, hence, etc. . . .

29. Prove that if the sum and difference of two forces are atright angles the two forces are equal.

30. Prove that if the lengths of the sum and difference respec-tively of two forces are equal, the two forces are at right angles.

%

CHAPTER IV.

DIFFERENTIATION OF VECTORS.

35. Two Ways in which a Vector may Vary. If a smallvector d a be added to the vector a, the result in general willbe a new vector differing by a small amount from a not onlyin length but also in direction.

If the small vector which is added be perpendicular to a,then the length of a will remain unchanged.

If the small vector which is added be parallel to a, then itsdirection will remain unchanged.

These three cases are shown in Fig. 40.Differentiation with Respect to Scalar Variables. Let the

vector a (t) be a function of a scalar variable t. In other70

VECTOR ANALYSIS. 71

words, let its length and direction be known and determinateas soon as a value of t is given.

Let OA, be the value of a (t) when t = t1, and let OA_ bethe new value of a (t) when I = t2. Then the change in a(t)due to the change in t of t2 - t1 is

A1A2=a(t2)-a(t,).

Dividing this equation by t2 - t1 inorder to find the rate of change andmaking t2 - t1 infinitely small, wedefine

da = lim a (t2)-aa (t1) (74)dt t2 - t1 + 0 t2 - t1

If t2 - t1 = h, where h is some smallscalar, this may be written in the morefamiliar form,

da = lim a (t + h) - a (t) (75)dt h+ 0 h

Evidently the rate change of thevector a with respect to t is made upvectorially of the three rates of changeof its componentsA along i, d 2 along j, and da3 along k,dt dt dt

Fio. 41.

or da =i + j + k. (76)dt dt dt dt

Precisely the same reasoning holds for the rate of change

with respect to t of the vector representingda

writtendt

(La ) ord

a

Similarly for the higher derivatives we may then writedna = final i + dna2 j + dna3 k. ' (77)dtn dtn dtn dtn

72 VECTOR ANALYSIS.

The vectors i j k are to be considered constant in length(being unit vectors) and in direction (being along fixed axes)in all of these differentiations'

If d be denoted by p, then

daat pa = p (all + a2j +a3k) = pa,i + pa2j + pa3k

anddndna =pna=pn(a,i+a2j+a3k) =pna,i+pna2j+pna3k.n

(78)

It will be noticed that the operator p - dt acts like ascalar multiplier.

36. Differentiation of Scalar and Vector Products. Thedifferential of for instance, is defined just as in scalarcalculus as

db)- a-b = dExpanding and neglecting small quantities of the second

order there remainsd a-b + a-b

=

hence, dividing through by dt,

d (a.b) = da.b+ a db,dt dt dt

or p p a + pb.a.In a similar manner

d (axb) = dtxb + ax db,

(79)

or p (axb) = p axb + axp b. (80)

The differentiations then take place very much as they doin scalar calculus, but with this important difference, that the

VECTOR ANALYSIS. 73

order of the factors must remain unchanged in all expressionswhere a change in order of the vectors, as previously ex-plained, would not be allowable. For example, in (79) theorder is immaterial, while in (80) it is essential.

The formulae

dt [a (bxc) ] = da (bxc) + a. (db xcl+ a. (bx dc)

or (81)

and dt [ax (bxc) ] =da x (b x c) + ax (A xc) + ax (bx dt

or p [ax(bxc) ] = p ax(bxc) + ax

(p\baxc)

+ ax(bxp c)

are in the same way easily seen to be true results.It is instructive to notice' the manner in which the opera-

tor p operates in turn on each one of the factors.If a vector is to remain constant in length, then

a const.or const.;hence 0,

or da is perpendicular to a, as is geometrically evident byconsidering

a2 = const.as the equation of a sphere or circle.

37. Applications to Geometry. We shall obtain someinteresting and useful results, as well as a clearer insight intothe calculus of vectors, by the following applications togeometry.

Let a variable vector r be drawn from a fixed origin O. Weshall assume that the terminus of r can be located as soon asa value of t, an independent scalar variable, is given. By aslight extension of mathematical nomenclature and symbolswe shall express this result by writing

r = f (t), (82)

74 VECTOR ANALYSIS.

reading it as: r equals a vector function of t. To indicate thevector character of the function, f is printed in bold-facedtype.

As t varies continuously, the terminus of r describes somecurve or curves in space, depending upon whether r is a single-valued or multi-valued function of t.

Fia. 42.

We assume in the following that the function f is a con-tinuous and single-valued function of the independent scalarvariable t.

Let t = s be the distance along the curve

r = f (s)

from any point P on the curve. The increment dr is evi-

dently a vector along the curve, and of length ds, hence ds

is a unit vector tangent to the curve at the point under con-sideration, M, when M' has approached indefinitely near to

M. For convenience we shall write

ds

= t, where t is a unit

vector along the tangent to the curve, or as we call it the unittangent,

VECTOR ANALYSIS.

and

So that

drt=dxi+ j+dzkds ds ds ds

= tli + t2j + t3k.

ti - !LX' t2 - ds' t3- ds

75

are the direction cosines of the tangent.Tangent and Normal. The equation of the tangent line

at r is then the equation of a line through the terminus of r

and parallel to t =

ds

, and by (69) is written

(r-g)xds 0, (83)

where g is the variable vector to this line from o.Expanding this vector product by means of (40) we

obtain the familiar Cartesian equations

it j kt(x-Sl) (z-5s)

dx ady dz

ds ds ds

0

or making t:i: three components along i, j, and k equal tozero.

xx-e,=y-tz=z-e3. (84)dx dy dzds ds ds

The plane normal to the curve at r is, by (64),

(r - g).ds 0, (85)

or expanding in its Cartesian form,

(x-E,)ds +(y-E2) d +(z-E3)ds=0. (86)

76 VECTOR ANALYSIS.

38. Curvature.* Consider three adjacent points on thecurve, M,, i122i and M3; the unit tangents through M, and M2and through M2 and M3 differ only in direction, hence thevector added to the first one to obtain the second one isat right angles to both and therefore measures the angle

Fza. 43.

through which it is turned in going from M, to M2. By

definition the curvature is defined to be the magnitude of ds

It is convenient to call the vector dt = d?r the vector-ds ds2

k

curvature, c, as it has the same magnitude as At and, being'

normal to the tangent, points towards the center of curva-ture. The vector-curvature, being perpendicular to twoconsecutive tangents, lies in their plane. The radius ofcurvature p. has a length inversely proportional to the mag-nitude of the curvature, but points in the same direction,and hence may be written

p c1 - (87)- (ft)'.* See Appendix, p. 242, for other definitions.

VECTOR ANALYSIS. 77

Osculating Plane. The plane containing two consecu-tive tangents is called the osculating plane. If g be thevector to any point in this plane from 0, since the threevectors t, ds and (r - g) lie in it, its equation may be

written down at once by (50) as

(r - ) txsl = 0,

or r - ) drxd2r = 0.( (ds ds2

(88)

This by (49) may be written in the familiar form of adeterminant

dx d2x

ds ds2

dy dds ds2

dz d2z

ds ds2

=0.

(89)

78 VECTOR ANALYSIS.

Tortuosity. A twisted curve in space twists in two dis-tinct ways. Any small portion of the curve lies in itsosculating plane at that point, and this small portion ofthe curve has a curvature as described above. As we goalong the curve, however, the osculating plane turns througha certain angle; the limit of the ratio of the angle turnedthrough by the osculating plane to the arc traversed to pro-duce that change is called the tortuosity.

Hence if n be a unit normal to the osculating plane,

T=dnds

where ds is the magnitude of the arc.Geodetic Lines on a Surface. The differential equation

to a geodetic line on a surface may be obtained in the fol-lowing simple manner from the definition:

A geodetic line is a curve on a surface, the osculatingplane of the curve being everywhere normal to the surface.

It is the curve a stretched string would lie along if thesurface were a perfectly smooth one, the reaction of the sur-face to the pressure of the string being everywhere along thenormal to the surface where it is in contact with the string.

Let t be the unit tangent to the geodetic, let n be theunit normal to the curve, and let m = nxt be a unit vectorlying therefore in the surface normal to n and to t.

The osculating plane is determined by t and dt which liein it by definition. If the curve is a geodetic, the normalto this plane txdt lies in the surface, and is hence perpen-dicular to n.

Expressing this fact,n (txdt) = 0.

Since t lies along dr (§ 37) and dt lies along d2r (87), thisequation becomes

n (drxdzr) = 0, (89a)

which is the differential equation to the geodetic.

VECTOR ANALYSIS. 79

If the surface is of the form

V (r) = const.,

then VV is a vector normal to it. See (106). But OV has directioncosines proportional to

aV aV aVax , ay , az '

and

n is along i 8z + l a yv"-+ k 'V

Hence (89a) becomes, by (49),

aV aV aVax ay az

dx dy dz

d2x d2y d2z

= 0.

39. Equations of Surfaces. Curvilinear Coordinates. Theequation

r = f (u, v), (90)

where u and v are two independent scalar variables, repre-sents a surface.

If a particular value u, be given to u while v is unre-stricted,

r=f(u1,v)

being of the form (82), is some curve lying wholly on thesurface. If a particular value v, be given to v while u isunrestricted,

r = f (u, v1)

is some other curve lying wholly on the surface. Thesetwo curves intersect at the point or points r determined bythe equation

rl = f (u,, v}),

80 VECTOR ANALYSIS.

We may then determine a point on the surface by givingparticular values u, and v17 say, to u and v. This point willbe found at the intersection of the two curves

r=f (ul,v) andr=f(u,v1).Curvilinear coordinates is the name given to these vari-

ables u and v, such a series of curves divides the surfaceup into a network of curvilinear quadrilaterals, the anglesof which may have any value. In the particular case thatthese curves cut each other always at right angles they

Fro. 45.

are said to form an Orthogonal System of curves. Whenthe two systems of curves divide the surface up intoinfinitesimal square elements they are said to form anIsothermal System. Such systems are of the greatest impor-tance in mathematical physics. The student should consulton this subject an excellent book by Fehr, " Applicationsde la Methode Vectorielle A la Geometrie Infinitesimale "(Carre et Naud, Paris, 1899). His notation is differentfrom ours, and is fully explained in his introduction, buthis methods are quite similar.

40. Applications to the Kinematics of a Particle. Letthe independent variable t now denote the time; then dt = v

is the vector velocity along the curve r = f (t). Notice that

VECTOR ANALYSIS. 81

dr is no longer a unit vector, as here dt 34 ds, the element

of arc, but the direction is still along the tangent. If t isthe unit tangent to the curve, then

v=vt,where v, the magnitude of v, is called the " speed."

The acceleration

a =

dt= d (vt) = dt t + v At, , (91)

dt is the increase of speed along the curve. Speed is here

used to denote the velocity irrespective of its direction.And by (87)

dt=dt ds=v,dt ds dt

2

(92)p

or the acceleration of a particle on a curve may be resolvedinto two Components at right angles to each other, one

d increasing the linear speed along the curve, the other

one v2c, or P, where p is the vector radius of curvature and

is directed towards the center of curvature, merely changingthe direction of the motion.

Hodograph. The . hodograph is a curve obtained in anygiven case of motion of a particle, by laying off from anarbitrary origin vectors equal to the velocities of the par-ticle for all points of the path. The locus of the extremitiesof these vectors is the hodograph.

When a particle describes a curve, there is then a pointrelated to it simultaneously describing the hodograph. This

82 VECTOR ANALYSIS.

conception was introduced by Hamilton, and is an efficientaid to the study of curvilinear motion.

Evidently the hodograph itself may have a hodograph,and this perhaps another, depending upon the complexityof the motion, and so on.

The hodograph of a particle at rest is a point at the arbi-trary origin.

The hodograph of uniform motion in a straight line is apoint at the end of a vector of length equal to the velocity.

FIG. 46.

The hodograph of uniformly accelerated motion in astraight line is another straight line parallel to the first,described with uniform speed by the hodograph variable.

The hodograph of uniform motion in a circle is anothercircle, since the speed is constant, of radius equal to thespeed. The vector velocity in the circle is always perpen-dicular to the radius vector in the original path. Evidentlythe points P and P move around their origins with thesame angular velocities. The velocity of P in general isthe rate of change of v, and hence is the accelerationof P.

VECTOR ANALYSIS. 83

Since the two circles must be described in the sametimes,

2 irr 2 7rv

v ahence

a familiar result.Equation of the Hodograph. If

r = f (t)be the equation of the path described by a particle, con-taining not merely the form of the path but the law of itsdescription as well, then

dtf' (t)

is the equation of the hodograph and the law of its descrip-tion. Again

d 2 r = f (t)dt2

is the hodograph of the hodograph, and so on.

41. Integration with Respect to Scalar Variables. (Recon-sult paragraphs 4 and 16.)

The inverse of differentiation offers merely the difficul-ties of scalar integration. The constants of integration,however, are constant vectors. As a simple example con-sider the motion of a particle under constant acceleration,under gravity for instance. The differential equation of themotion may be written

d2r =dt2

a,

where a is a constant vector. Integrating once,

dr =at+vo,dt

84 VECTOR ANALYSIS.

where vo is a constant vector, as it is a vector equation and

vo is determined by the value of dt when t = 0. Integrat-

ing again we obtain

r = I a t2+vot+so, (93)

where aghin41% is a second constant vector and whose valueis 'than of r . when t = 0. Equation (93) gives the value ofr at any 4tiii t. ' The equation says that starting from the

; iQiit 5.S (i.e. -at S), r, the vector to any point of the path, may

FIG. 46A.

be found by adding to so the vector sum of the two motionsvot and I a t2. The terminus of r evidently describes aparabola passing through so, because the coordinates ofany point on the curve referred to the oblique axes parallelto Vol and a are proportional to the first power and thesecond power of the same quantity t, respectively.

Orbit of a Planet. Central Acceleration. As anotherexample, consider the motion of a particle under a centralacceleration; that is, one always directed towards or awayfrom a fixed point, the exact law of the force of attractionas a function of the distance being left indeterminate. The

VECTOR ANALYSIS. 85

planetary motions are of this description. In this case thedifferential equation of the motion is

z

dtz= rl f (r). (94)

As the acceleration is always along and therefore parallelto the radius vector, the product

dt2 xr = 0.

This may be written d I dt xr) = 0,

Fia. 47.

for, carrying out the differentiations, we obtain from thislast

z

dt2xr

+ dt x dt0,

of which the second term on the left is zero, because anyvector product containing parallel vectors is zero. Hence,integrating,

dt

xr = const. vector = c, (95)

where c is a vector perpendicular to the plane of r anddt

86 VECTOR ANALYSIS.

But dt is the vector velocity along the tangent and r is

the radius vector, so that by § 18 the above equation istwice the area swept out by the radius vector in unit time.We obtain the result, then, that under any central accelera-tion the rate of description of areas is a constant, and theorbit lies in a plane perpendicular to a constant vector c.

Harmonic Motion. Equation of Ellipse. As another ex-ample to integrate

z

dt2 + m2r = 0, (96)

which is the equation of a central acceleration proportionalto the distance from the center. Such motions take placewherever Hooke's law is followed.

We know that the two solutions of the scalar equation

d2 + m2r = 0,

are r = a cos mt and r = b sin mt,

and that the complete solution is the sum of these two. Ifwe replace the arbitrary constants a and b by the arbitraryconstant vectors a and b, obtaining

r = a cos mt + b sin mt, (97)

it is easily seen, by differentiation, that this equation is thecomplete solution of the vector differential equation

d +mzr=0.By an extension of this process, which is easily seen to hold,

we may then state the rule for the solution of linear differ-ential equations to any order with constant coefficients: Findthe solutions, assuming the vector variable to be a scalar variable,multiply these each by an arbitrary vector and add. The resultwill be the complete vector solution.

VECTOR ANALYSIS. 87

These arbitrary vectors are to be determined from theinitial or final conditions of the problem exactly in the samemanner as we do with scalar equations. Equation (97) isthe composition .of two simple harmonic motions along direc-tions determined by a and b, and is easily seen to represent in

general an ellipse inscribable in the parallelogram whosesides are 2 a and 2 b respectively.

r = a cos mt + b sin mt

is therefore one form of the equation of an ellipse, if m is real.

42. Hodograph and Orbit under Newtonian Forces. Asanother example in vector differentiation and integrationconsider the case of motion under a force directed along theradius vector and inversely proportional to the square ofthe distance in magnitude. This is the ordinary planetarymotion. We. are to solve the differential equation

d2r mr, .dt2 r2

(a)

Multiplying by IN, we have at once

rxdr=0,and hence

rx dt = c, (b)

* This equation states that the acceleration is directed outwards,i.e. the forces are repulsive. For attractive forces change m to -m.

88 VECTOR ANALYSIS.

where c is a constant vector. This last equation states thatthe rate of description of areas is constant, as above.

Lemma in Differentiation. Consider now the identity

r rrthen dr = rdr, + r,dr,

hence multiplying by r,x,

rxdr = r r,xdr,,

so that rlxdr, = rxdr?a y

a result we might have written down immediately by similartriangles.

Multiply this last result by r,x, obtaining from the left-hand side of equation

r,x(r,xdr,) = r, dr, drsince r, and dr, are perpendicular and r1.r, = 1, so that

r,x (rxdr)dr,rz

rlx (rx dt )dr,and dr,dt rz

Now the parenthesis on the right is a constant vector forcentral forces by (b), so that, by (a),

d2xc=mrlxc=-md- l,

and integrating,

drdt

xc=d-mr (c)

where d is a constant vector perpendicular to c, as is seen bymultiplying by c

VECTOR ANALYSIS 89

Multiplying equation (c) by x we obtain

1. dr _ d_r 1 dr. 1 1 1c(dtxc

This becomes, since C- = 1, and because by (b) dt is

normal to c or to J ,

dt = c-' x d - me-1 x rl = x (d - mrl) , . (d)

the equation to the hodograph.Since c and d are constant vectors, c-'xd is a constant

vector normal to c, and hence lies in the plane of the orbit;m c-'xr, is a vector constant in length, so that the extremity

of

dt

lies in a circle drawn around the point c-'xd as center

and in the plane of the orbit. This length, since c and r, areperpendicular, is m .

C

To obtain the equation to the path multiply the equationof the hodograph by rx,

rxLrdt

= c = rx(c-'xd)- mrx(c-1 xr,).

Expanding the two triple vector products and remembering(b),

Multiplying by c.,

So that

c2= m r.

C2

C2

mr= dcosa - m1 -d cosa

m

90 VECTOR ANALYSIS.

the polar equation of a conic; the angle a being measuredfrom the line d.

Comparing withlr= 1-ecosa

the general polar equation of a conic where the focus is theorigin, where'l is the semi latus rectum, and where e is theeccentricity, we find that the path of the orbit is a conic of

eccentricity d ; that d is along the major axis and that them

magnitude of the major axis is mc2

m2- d2

43. Partial Differentiation. When any vector is a func-tion of more than one scalar variable it can be differentiatedpartially with respect to each one, the remaining variablesbeing considered constant during the differentiations. Suchpartial differential coefficients are written just as in ordinary

scalar calculus as as ,a s

, etc., where x, z, ... are the inde-

pendent scalar variables. The total change in a due to simul-taneous changes in the variables dx, dy, dz, . . . is written

da = ax dx + aa dy + az dz. (98)

Symbolically we may write this as

da = (ax dx+ ay dy+dz

dz) a, (99)

where the expression in parentheses is to be considered asa differential operator to be applied to a, as in (78).

Origin of the Operator Del (7). The operator (99) hasthe form, of a scalar product, the two constituent vectorsof which are

iax+ a +kazl and (idx+jdy+kdz)=dr,y* See Appendix, p. 245, for Path Described by an Electron in a Uni-

form Magnetic Field

VECTOR ANALYSIS. 91

If the single symbol V (read del) be used to denote thefirst expression

v.(iax

+i ay +k.-', (100)//

the equation may be writtenda = (V.dr) a. . (101)

We are thus led naturally to the consideration of theproperties of this symbolic vector V.

PROBLEMS AND EXERCISES.1. If r = const.If rxdr = 0, show that r,= const.If 0, show that rxdr has a fixed direction and that r

is always parallel to a fixed plane.2. Show that

drr

a particle moving in a plane curve, in the plane of ii,

obtain the components of dt along and perpendicular to the

radius vector.They are

dr r, and r d I r,.dt dt

A unit line 1 to r, is kxr,, where k is normal to the plane.4. Obtain similarly by differentiation of

dr __ dr r, + d9x r)

dt dt ` dt

the accelerations along r, and perpendicular to r,.They are

d2r _ r dB z r, and r 9 +2 dr d 1rl.dt2 (dt C de dt

dt

92 VECTOR ANALYSIS.

6. If r, 0, 6 be a system of polar coordinates in space, where r isthe distance of a point from the origin, 0 the meridional angle, and0 the polar angle, obtain the expressions for the components of thevelocity along the radius vector, the meridian, and a parallel oflatitude.

6. Find the accelerations along the same directions in the problemabove.

Express them in Cartesian form.7. The curve

r = a cost+bsintrepresents an ellipse of which a and b are conjugate radii. Thevector

r'= dr -asint+bcostdt

= acos(2 +t'\+ bsin (2 +t)

is the radius conjugate to r,\and parallel to the tangent at r.8. The parallelogram determined by the conjugate radii of an

ellipse is constant in area.[r-r' = const.].

9. An elliptical helix is represented byr = a cos t + b sin t + et.

10. Show that the tangent line and the osculating plane of anycurve r = f(s) may be respectively written in the forms,

p=r+xr',p = r + xr' + yr",

dfand r" andwhere r'- ds

x and y are variable scalars.11. Find the tortuosity, T, of any curve where T is defined as the

rate change of the. normal n to the osculating plane with respect tothe are ds.

dr d'-r d3r

To = do ds ds2 ds3° (ds ° d2

ds2ds2Express this in Cartesian notation.

12. Find the curvature of a circular helix. Find the tortuosityof a circular helix.

VECTOR ANALYSIS. 98

13. The equationr=x4(t) + a,

where a is a constant vector, represents a cone standing on thecurve r = (t) with its vertex at the extremity of a.

14. The equationr=4(t) + xa,

where a is a constant vector, represents a cylinder standing on thecurve r = 4(t) and having its generators parallel to a.

15. Prove that the acceleration of a particle moving in a circlewith uniform speed is given by

d'r v2

dt2 r16. Write out the equations of the hodograph for uniformly

accelerated motion in a straight line.17. Find the hodograph to the motion

d"`rmgr, (a)

dt2

where the acceleration varies as the distance from the origin.

The solution of (a) being that of (t ± m2) r = 0, that is

r = A cos mt + B sin mt,

and r = Aemt + Be-mt.Interpret these equations and those of the resulting hodographs.18. Show that if the hodograph be a circle, and the acceleration

be directed to a fixed point, the orbit must be a conic section, whichis limited to being a circle if the acceleration follow any other lawthan the inverse square.

19. In the hodograph corresponding to acceleration f(r) directedtowards a fixed center, the curvature is inversely as r2f(r).

20. Show directly without analysis that1 xdr = 1 xdr, =r,xdr1,r r1

and hence that- r1x rxdr

dr, t dt

dt r2

CHAPTER V.

THE DIFFERENTIAL OPERATORS.

The Vector Operator V (read del). This sign is some-times called " nabla " (Heaviside) and also " atled," whichis " delta " (p) reversed. The term " del " is, however,well worthy of adoption, as it is short, easy to pronounceand conflicts with no other terminology. As V is the mostimportant differential operator in mathematical physics itsproperties will be studied in detail.

Definition. V is defined by the equation

v= i x+ j a+ k a . (102)y

We have already come across the scalar differential oper-ator p on page 72. The paragraphs concerning p shouldbe consulted at this point. As by its definition v is madeup of three symbolic components along the three axesi j k, the symbolic magnitudes of them being 5x, a , and

azT,

respectively, it may be looked upon as a symbolic vectoritself. This view of V as a vector, is important and of greathelp in the comprehension of what follows. The employmentof V in the treatment of the physical properties of space isof the most frequent occurrence. It is, therefore, extremelydesirable to have a geometric or visual representation ofsuch physical properties in space, or fields, as they arecalled, and of the effect of operators upon them.

44. Scalar and Vector Fields. Reconsult § 5 at this point.Definition. If to every point in a region, finite or not,

there corresponds a definite value of some physical property,the region so defined is called a field. Should this property

94

VECTOR ANALYSIS. 95

be a scalar one the field is called a scalar field. As examplesof such may be mentioned the temperature at any giveninstant, at all points of a body; or the density at all points;or the potential at all points, due to electrical, magnetic, orgravitational matter-respectively.

On the other hand, if the property is a vector one it issaid to be a vector field. As examples of these are thevelocity at all points of a fluid; the electrical, magnetic, orgravitational intensity (of force) at all points of a region dueto electrical, magnetic, or gravitational matter, respectively.

45. Scalar and Vector Functions of Position. Assumeany arbitrary origin 0 and from it draw a variable radiusvector r. This vector r may extend to and determine anypoint in space. By the term " value of r " is meant theit position of the terminus of r." Now, if to every value ofr there corresponds a definite scalar quantity V, V is said tobe a scalar point-function of r and is written

V =f (r). (103)

If to every value of r there corresponds a definite vectorquantity F, F is said to be a vector point-function of r andis written

F = f (r). (104)

V= f (r) and F = f (r) are thus the functional representa-tions of scalar and vector fields respectively.

Mathematical and Physical Discontinuities. The functionsmet with in physics are almost always continuous andsingle-valued except perhaps at isolated points, lines or sur-faces finite in number. If not, they can be made so byvarious devices, such as by inserting diaphragms to preventpassing into a region by two or more different paths, etc.

The functions dealt with, in what follows, are supposed tobe of this description.

The most common kinds of discontinuities that occur inmathematics are those in which, either the value itself of

96 VECTOR ANALYSIS.

a function suffers an abrupt change, or where the rateof change of the function abruptly takes on a new valueas the independent variable is continuously increased ordiminished.

Graphically these mean a break in the curve, or a suddenchange in the direction of the curve representing thefunction, as in Fig. 48 (a), at P and at Q. In nature suchdiscontinuities do not take place. For example, the tem-

(a)

FIG. 48.

perature cannot have one value on one side of a surface andanother value on the other side, where the two sides of thesurface are infinitely near to each other. In reality thereis a continuous but very rapid change in the temperaturefrom its value on one side to its value on the other as wepass through the surface. Besides, infinitely thin surfacesdo not exist except in our imagination.

If the temperature gradient in a body has one definitevalue and seems to change abruptly to another value quitedifferent from the first, we know that in reality there is avery rapid but finite rate of change of the gradient at theplace in question. This absence of discontinuity in anynatural function is indicated in Fig. 48 (b), which shows thecontinuous function, which to all intents and purposes

VECTOR ANALYSIS. 91

replaces the discontinuous function (a). It is for thisreason that the usual attention will not be paid to the con-sideration of discontinuities in what follows. All naturalfunctions being in reality continuous, such considerationis physically superfluous. We may state the same ideaexplained above, by saying that on sufficient magnificationof finite amount, all curves representing natural phenomenawill be found to be continuous.

(b)FIG. 48.

We do not, however, wish to convey the idea that thestudy of discontinuities is unimportant, as on the contrarythe mathematical results derived from their study are ofthe greatest importance, and teach us how to attack prob-lems involving sudden natural changes, or as we might callthem, " apparent discontinuities " in physical functions.In fact the methods generally employed are to assumethem to be actual mathematical discontinuities and treatthem as such. The point we wish to make is that in thegeneral analytical expression of natural phenomena it isunnecessary to complicate the formulw by the separateconsideration of discontinuities, but to let the studenttreat them by the recognized mathematical methods when-ever it is convenient or necessary to do so.

98 VECTOR ANALYSIS.

46. The Potential. For the sake of definiteness we shallconsider the potential due to electrical matter. The wholeargument applies almost identically to magnetic or gravita-tional potential; to the distribution of temperature in a body,or to the velocity-potential in moving fluids, etc., etc.

Definition. The potential at any point in space due to adistribution of electrical matter may be defined as the workdone on a unit positive quantity of electricity as it is broughtby any path from infinity to that point. As like charges re-pel each other, it will require positive work to be done on theunit charge to bring it in the neighborhood of any positivedistribution of electricity, and hence the potential aroundsuch a distribution will be positive, increasing as the points aretaken nearer and nearer to it. It is evident, also, that theforces acting on the unit charge are repelling forces and thatthey act in the direction opposite to the increase of potential.

For instance in the electric field due to the charge + q ona small sphere, the unit positive charge at P is repelled bya force acting radially outward, of amount calculated byCoulomb's Law

F = 1 r1f

where r is the distance from P to the center of the sphere.The force F evidently becomes greater as P approaches thesphere, and work has to be done upon the unit charge in orderto make it do so.

Level or Equipotential Surfaces. Let all the points havingthe same potential be found, or, in other words, find all thosepoints which require the same amount of work to be ex-pended upon the unit positive charge to bring it from infinityup to them. If C be this amount of work and V be thepotential function, then the equation to the locus defined bythese points will be

V(r)=C,where r is measured from some arbitrary origin.

VECTOR ANALYSIS. 99

Find similarly all points which require a small amountmore of work C + dC; the equation to this locus will be

V(r)=C+dC.In the special case of the sphere, Fig. 48A, these points will

lie on spheres concentric with the charged sphere.

FIG. 48A.

These equations define surfaces which are called Level orEquipotential Surfaces of the function V(r). Let many suchsurfaces be constructed and let the quantities of work em-ployed in reaching the successive ones differ by equal amounts.

100 VECTOR ANALYSIS.

It requires no work to carry the unit charge from one pointto another having the same potential, for by definition itrequires the same amount of work to bring the unit chargefrom infinity to either of these points by any path, and wemay choose the path leading to the second point to pass

Fia. 49. Showing Lines of Force and Equipotential Surfaces Arounda Charged Conductor.

through the first point. Hence the work done in going fromthe first point to the second point must be zero.

Relation between Force and Potential. Consider in par-ticular two adjacent level surfaces, the difference in poten-tial between them being dV. This means that it requires

VECTOR ANALYSIS. 101

d V units of work to carry the unit charge from one of thesurfaces to the other in any manner.

Since the amount of work is constant in going from onelevel surface to the next one, the greatest forces will beencountered in going by the shortest path from one to theother, so that at any given point P, the maximum force isalong the common perpendicular to the two surfaces. Com-

paring the forces at different pointsP1 or PZ these maximurnforces F,. or F2 will be found greater the nearer the surfacesare together. Hence the forces in the field are normal to thelevel surfaces and are inversely proportional to their distanceapart. But V increases most rapidly along the normal to alevel surface and its rate of increase is greatest where the sur-faces lie closest together. We are therefore led to expect arelation between the force at a point and the rate of increaseof V at the same point. All this is concisely represented by

102 VECTOR ANALYSIS.

writing for the work done in going from one surface to thenext,

dV =

where do is the normal distance between the two surfaces, sothat

Fdo

(--n). (105)

where n is the unit normal pointing in the direction of increas-ing potential. This important equation states that the forceat any point is normal to the level surface passing through thatpoint; opposite to the direction of fastest increase in V, andequal in magnitude to this fastest rate of increase.

Thus a knowledge of the potential everywhere gives aknowledge of the forces everywhere not only in magnitude butin direction as well.

A scalar point-function, as it does not involve direction, isclearly simpler of representation on a diagram than a vectorone. The potential function is very useful for this reason, asa complete knowledge of its value everywhere immediatelygives us a complete knowledge of the forces everywhere.Thus the comparatively simple scalar function intrinsicallycontains all that we wish to know about the comparativelymore complicated vector function. This property alone issufficient to justify its invention and use.

47. V Applied to a Scalar Point-Function. Gradient orSlope of a Scalar Point-Function.

Definition. The vector, perpendicular to the level surfaceat any point, equal in magnitude to the fastest rate of increaseof V, and pointing in the direction of this fastest increase,is called the gradient or the slope of V at that point andis written

grad V or slope V,

preferably the first.

VECTOR ANALYSIS. 103

Grad V Independent of Choice of Axes. The force F actingon the unit charge is, by the above definition in connectionwith § 46, evidently equal to - grad V, but as F is entirelyindependent of any choice of axes, so is - grad V independ-ent of them.

It remains to show that the operator V applied to V gives

the grad as defined above. The work done on a unit chargeas it is carried from M to M' is, by § 16, where MM' = dr,

grad

V as a function of x y z,

dV = aV dx + aV dy + aVdz,ax ay az

so that

grad V dr = (i ax + i 'V + k aY). (i dx + j dy + k dz)y /

= vV dr.

104 VECTOR ANALYSIS.

As this equation is true whatever path dr is taken betweenthe two surfaces,

grad V - VV. (106)

Thus the application of the operator v to a scalar pointfunction is a vector which gives its rate of most rapidincrease in magnitude and direction. The significance andimportance of this operator is now easily understood.

The vector vV is often called after Lame'* the first differ-ential parameter of V.

Fourier's Law. If instead of potential we consider tem-perature, the level surfaces are then isothermal surfaces,and the v of the temperature function gives the rate ofthe most rapid increase of the temperature in magnitudeand direction. As the flow of heat takes place in the direc-tion of most rapid decrease, q, the intensity of flow, is given

by q = - kve,where k is a characteristic of the medium at the point inquestion and called its conductivity, and 0 is the tempera-ture at any point. This is called Fourier's Law for theflow of heat.

48. Illustrations of the Application of V to Scalar Func-tions of Position. By means of ordinary partial differentia-tion on the functions

r = (x2 + y2 + z2) i

andn

rn= (x2 + y2 +z2)1

the following important results are obtained:

+ k az) (x2 + y2 + z2)jVr = (i ax + J ay

xi+yl+zk=r(x2+y2+z2)i r

* G. Lame. Lecons sur lea coordonneea curvilignes et leurs diversesapplications. Paris, 1859.

VECTOR ANALYSIS. 105

and

(x2 + y2 + z2)i

= nrn-'Or = nrn-1 r1 = nrn-2 r. (108)

So that V differentiates the function rn similarly to thescalar differentiation of un by d

dun

dx

ax

= nun-1 dudx

A shorter method for obtaining this formula will be givenin § 49.The particular cases

Vr = r1 and V I = - 1 (109)

are important results.. Application of p to a Scalar Product. It is frequentlynecessary to apply V to a scalar product such as a.b.Remembering the laws for the ordinary differentiation ofscalar products and the definition of V it is easily shownthat

V (a.b)

where the subscript affixed to the p shows which vector isconsidered variable, i.e., the one it acts upon.

The notationV(a.b)b (110)

is also sometimes used, where the subscript in the second termshows which vector is considered constant during the differ-entiation. Va( ), then, from this point of view differen.

Orn=(i a + ay +k a)(x2+y2+e)3

n=n(x2+y2+z2)2-1 (xi + yj + zk)

= n (x2 + y2 -f- z2)na l x i+ y J+ z k

106 VECTOR ANALYSIS.

tiates partially, the vector a alone being considered variablein the parenthesis. In particular, if the vector to, say, ofthe scalar product is a constant vector, we obtain theimportant result

v Vr(wix + wly t wsz)=iwl+jw2+kws=w.

49. The Scalar Operator or Directional Derivative.Since VV is a vector, its scalar product with any other vectors1 may be taken, and by definition this would give the com-ponent of the magnitude of VV in the direction of that vec-tor. This is ordinarily written

a v _ av av avas -s1 ax +82ay + 83 az'

where s s2, ss are the components of the unit vector s17or what is the same thing, the direction cosines of the direc-tion s1.

This expression may be looked upon as the operation of

s1 v = s1 az + s2 ay + s3 az = d(112)

upon V, which is the familiar directional derivative of V inthe direction s1 or, otherwise,

f V), (113)

so that the directional derivative in any direction is the com-ponent of the magnitude of the gradient in that direction.The directional derivative then applied to the potential func-tion gives the component of the force in the direction in whichit is taken.

VECTOR ANALYSIS. 107

Total Derivative of a Function. If s, be replaced by thesmall vector dr, the operator applied to a scalar pointfunction gives the total derivative of the function because

V = (a dx + a dy+ az dzl V =dV. (114)y / _Sy

This is the same thing as the directional derivative along(dr), multiplied by dr, or

dr.VV = dr (115)

We may now obtain equation (108) for the differentiationof rn by means of the identity (114)

dr (

dr = nrn-1

so that the factor of dr, nrn-1 r, is the Vrn

and

as before.

Vrn = nrn-1 r, = nrn 2r

50. The Scalar Operator Applied to a Vector. Theoperator may be also applied to a vector point-functionF, giving as a result a new vector function, thus:

F = (i F, + j F2 + kF3)= i j s1.vF2 + k

s a1 ax 2 ay 3 az /J

+j (s, a +s2+ S30FI1 (116)

y

+k (s,as3

+s2aaF

+s3aOZ

.

y

This is the directional derivative of the vector function F inthe direction s,. It is also the vector whose components arethe directional derivatives of the components of F.

108 VECTOR ANALYSIS.

The parentheses may be omitted in s1.V(F) and the ex-pression written simply but it does not mean that

because VF has no significance in thisanalysis. s,-VF may then be interpreted as nothing elsebut (s, 0) F.

We may prove that the component along any constantvector a of the directional derivative along b, of a vectorfunction F, is the directional derivative along b of thecomponent of F along a; that is,

al.(b,.VFF) =And even without restriction to unit vectors that

(117)

This follows directly because V differentiating F alone,a may be placed after the V. Also because (116)

a.(b.VF) = b.VF1 + a.jF, + F2 + F3

= a, F, + a3F3)=

Applying to r, the radius vector, gives

((o. )ri aax +``2

a+[d3

a az)(ix+jy+kz)

ax az/ azay

=iW,+j102+kw3=w.This expression should not be confounded with where

the r is the magnitude of r. Since Vr = r, the value of thislast expression would be

w with equation (111), we see that

r = Vr co. (118)

VECTOR ANALYSIS. 109

The Operation V on a Vector Point Function. Any vectorpoint-function F may be resolved into three componentsalong i j and k so that

F=iF1+jF2+kF3.F1, F2, and F3 are scalar functions of x y z, or of r. Con-sidering V as a vector, the product vF can have no meaningunless the definition of the product of two vectors a and b beextended so that the product ab (without dot or cross) shallhave a meaning.* But the scalar product of V and a vectorF and the vector product of V and a vector F may be foundby rules already given. The two expressions and vxFare of such importance that special names have been given tothem.

51. Divergence. The operator v.( ) or div ( ) [readdel dot ( ) or divergence of ( ) ] when applied to thefunction F gives a scalar which in Cartesian notation is

V.F_(`ax+j am

a1+aF2+aa =div F. (119)

In order to obtain a physical interpretation of this quantityconsider any vector field, the field of force due to an electri-cal distribution for instance. The convention usually adoptedis that from every unit positive charge there originate 4 nlines of force and into every negative unit charge there end4 7r lines of force. The exact number of lines of force thatissue from unit charge which convention has adopted, is ofabsolutely no consequence in this argument, and the newsystem which assumes the unit charge to give rise to but one

* This has not been done in this book, although Professor Gibbs hasachieved beautiful results in his researches using this extended defini-tion. The product ab he calls a dyad. See Gibbs-Wilson, VectoiAnalysis, Chapter V.

110 VECTOR ANALYSIS.

line may be adopted if desirable. If an element of volumebe considered, for example a small parallelopiped with itssides dx dy dz parallel to the axes of x y and z respectively,the amount of electrical matter within it may evidently bemeasured by finding the excess of the lines which come outof it, over those which go into it. For every unit of positive

Y

d

FIG. 52.

electricity there would emanate 47r lines outward, and forevery negative unit 47r lines would enter into it. Consider-ing these lines to cancel each other when going in oppositedirections it is easily seen that the algebraic sum of thecharges within the box may be found in amount and signby an examination of the lines which leave or enter the box.Hence the lines which diverge from the element will be a

VECTOR ANALYSIS. 111

measure of the positive charge within it. If the charge isnegative, lines will end inside of the box, and therefore willconverge into it.

To obtain an analytical expression for this quantity resolveF, or the flux of force, as it is called, into its three componentsparallel to i j and k. The flux into the face parallel to theyz-plane nearest the origin is Fl dy dz, the flux out of theopposite face is

(F1+ /1dx) dy dz,

so that the amount which comes out in excess of that whichgoes in, as far as the x-component of F is concerned, is

(F1+ adxl dydz-F,dydz =ax dxdydz.

Similarly, for the other two components, which are obtainedin the same manner,

"F2 dxdydz,ay

a dxdyd

so that the total amount of the flux F which diverges fromthe box dx dy dz is

(aF, + a + )dxdydz.

Dividing by dx d y dz, the element of volume, to obtain theamount of flux which would come out of a unit volume underthe same conditions, there remains precisely

div F = v-F = MI + 2F2 + (120)ax ay az

Strictly then the term divergence means the number of lineswhich diverge per unit volume.

112 VECTOR ANALYSIS.

If the operator v be applied to the vector function repre-senting the flux of heat or the velocity of a fluid, it will giveby exactly the same reasoning the rate at which heat is issu-ing from a point per unit volume or the rate at which thefluid is originating at a point per unit volume. In the case ofheat, if the divergence exists and is positive, there must be atthe point in question a source of heat, heat actually created,or"else at the point where the heat is leaving the temperaturemust be diminishing.

In the case of fluids, if the divergence exists and is positive,there must be either a source of fluid, fluid actually created,or else the density of the body at the point must be diminish-ing. If the divergence is negative, the opposite conditionshold in both the above examples. For instance, if the diver-gence of heat is negative, or in other words, if it converges,there must be a sink of heat, heat actually destroyed, anni-hilated, or else the temperature at the place must be rising,etc. In the case of electricity, the existence of a positivedivergence proves the existence of positive electrical matterat the point. The negative of divergence is sometimes calledconvergence. It is better, however, to retain but one of theterms and use the negative sign to indicate convergence.

52. The Divergence Theorem.* This important theoremhas a significance almost axiomatic when considered in thelight of the foregoing. Consider any closed surface S, Fig. 24,lying in a vector field q, the velocity, say, in a moving incom-pressible fluid. It is evident that the excess of fluid whichcomes out, over that which goes in, may be measured in twodistinct ways: first, by finding the total outward normal fluxover the surface, or second, by going throughout the interiorand taking the algebraic sum of the sources and sinks or diver-

* For a rigid mathematical proof see A. G. Webster, Electricityand Magnetism, pp. 60-62; Dynamics, pp. 340-342; also R. Gans,Einfahrung in die Vektoranalysis, pp. 29-33. See Appendix, p. 252,for other theorems analogous to the Divergence Theorem. Also P.Appell, Trait6 de M6canique rationnelle, tome III, p. 2.

VECTOR ANALYSIS. 113

gences for every infinitesimal volume element contained withinthe surface. In symbols this is most conveniently expressedas

f fn.q dS =tfffv.q dv, (121)

where n is the outward-drawn unit normal, dS the element ofsurface, dv the element of volume.

In words this reads: In a vector field the surface integralof the outward flux (i.e., normal component of flux, see § 17)over any closed surface S is equal to the volume integral ofthe divergence taken throughout the volume enclosed by S.This is the divergence theorem.

From a mathematical point of view this demonstrationmay not be considered rigorous, but the ideas that this inter-pretation gives should be clearly understood by every student.In Cartesian dress this theorem becomes

j,f8[i cos (nx) + q2 cos (ny) + q3 cos (nz)]dS

dx dy dz. (122)= fff (ax + a + E Oz)

The idea of divergence is evidently independent of anychoice of axes since none are required for its conception.Considered as the result of operating by V. it is invariant tochange of axes because V has been shown to be invariant.Its invariant character may also be directly proved, as usual,by a transformation of axes, but this is a long and unnecessaryprocess.

Examples. In particular

V.r=(i a+ j a +k a ).(ix+j y+kz)J //

= axa+5Z00 1=3. (123)

y

"114 VECTOR ANALYSIS.

Let us apply this result, using r for q in tfie divergencetheorem. We obtain immediately

ffn.rcis = 3fJfdv=3Xvol.

Or in Other words, three times the volume included by anyclosed surface is obtained by multiplying every element ofsurface by the perpendicular from the origin to its plane andadding the results.

In a sphere, for instance, taking the origin at the center, r isperpendicular to every element of surface and is of constantlength, therefore r,

3 X vol sphere = rfJ dS = 4 mr3,

a true result.To obtain By (123) and ;128),

dr (by (112))

Hence

=rV.r1+1.2r

Equation of the Flow of Heat. As another example of theuse of the Divergence Theorem (121) consider the generallaws of thermal flow. Consider a volume of matter throughwhich heat is flowing, and consider a surface S drawn any-where in this space. Let q be the flux of heat, or in otherwords, the amount of heat which crosses unit area drawnnormally to the lines of flow per unit time; q is also calledthe heat current-density.

The amount of heat which escapes through the surfacein any time is furnished at the expense of the materialinside that surface which must then be cooling oil at acertain rate.

By Fourier's Law, § 47, the heat flows in the direction ofgreatest decrease in temperature, 0, and with an intensity

VECTOR ANALYSIS. 115

proportional to a property of the material through which itis flowing, called its heat or thermal conductivity k. So that,

q = - kvO.

The coefficient k may vary from point to point of themedium, and may be also a function of the temperature..In most practical applications it is assumed to be constant.

If there are no sources nor sinks of heat within the sur-face any elementary volume dv is cooling at some rate- a©

at. The amount of heat which leaves this elementary

volume in unit time must then be, if p be its density and cits specific heat,

- a0 c p dv.

For the whole volume, S, the heat lost, which must be equalto that passing through the surface, is

ff1:01- at cpdv = ffn.qds.

By the Divergence Theorem the surface integral is

fff01 vq dv,

and since q = - kVO, vq =So that

Since this equation holds whatever surface be considered,the integrands are equal everywhere and

80at cp = 7-k vO.

116 VECTOR ANALYSIS.

If k be assumed constant this becomes

a©_ k V2©

at cp

= a2V20, where a2 = k (107)cp

This is the general differential equation for the flow ofheat in a body.

If the steady state is reached, that is, if the temperaturesare everywhere constant (this does not mean the sameeverywhere), it becomes

V2©=0,

independently of the values of k, p, and c; that is, thedistribution of temperature follows the same law as thedistribution of potential according to Laplace's Equation(157). So that, what is true about the potential, is underanalogous conditions true of temperature, and the twosubjects, temperature distribution and potential, becomeidentical in mathematical treatment.

53. Equation of Continuity. Considering again a movingliquid, if there are no sources nor sinks of the fluid in theregion considered, then the equation

vq=divq=0 (124)

expresses the condition that the fluid does not concentratetowards nor expand from any point, as this is the onlyremaining way by which more liquid can leave any smallclosed surface than can enter it, or conversely. In otherwords, it means incompressibility. This equation is calledthe equation of continuity. It is of great importance inelectricity, as according to the theory of Maxwell the electricdisplacement behaves like an incompressible fluid. If thedivergence does exist it means that at the point considered

VECTOR ANALYSIS. 117

there must be a source of lines of force or what is the samething, electricity.

Solenoidal Distribution of a Vector. Should the divergenceof a vector function be zero everywhere, then always as muchvector flux enters any volume element as leaves it, or inother words, the lines of vector flux cannot end nor begin infree space. They must then form closed curves or end atinfinity. Such a vector distribution is called solenoidal. Forexample, the motion of any incompressible fluid such aswater, gives a velocity distribution which is solenoidal.

54. Curl. The Operator vx applied to F or curl F (readdel cross F or curl of F), also sometimes written in Germanbooks, rot F (read rotation of F), is a new vector derived fromF. Like 0 it is invariant to choice of axes and has an im-portant significance in physics. It may be defined by theequation

OXF - curlF -

i j ka a aax ay az

F1 FZ F3

=iaFa _j)+j(ay az J ` az ax

+ k(a - -af). (125)y

The new vector, curl F, has components

CaF3-Pl, ( OF, _al, (aFj- aFll,ay az

//az ax // ax ay JJ

along the three axes. When applied to a vector functionVx gives a result independent of the axes because V itself isindependent of them. We may say in general that all com-binations of V with vector or scalar point-functions giveresults independent of any choice of axes. By a directtransformation from one set of axes to another we may provethe invariant property of these operators and thus eliminate

118 VECTOR ANALYSIS.

any lingering doubt in the mind of the skeptic. To a physi.cist, however, to say that the operations of V upon any func-tions are dependent upon a choice of axes, is like saying thatthe physical properties of any medium depend upon the lan-guage in which you express them. For instance, we haveshown (§ 47) that V V, where V is the potential say, gives riseto a vector showing the direction in which V changes mostrapidly and its magnitude.

What have axes to do with such a result? It is truewhatever kind of coordinates are used, however placed,, oreven if none are used at all.

We are here dealing with the properties themselves, and.not. with any particular method of representing them. It isin this respect that the analysis of vectors is extremely useful,as by its intelligent study clear conceptions must necessarilybe obtained.

Example of Curl. In order to give an idea of the meaningof the Vx or curl of a vector function, consider the generalmotion of a rigid body. We have seen (§ 22) that the motionmay be resolved into a velocity of translation qo of the originchosen arbitrarily and an angular velocity of rotation coabout a line passing through this origin. The velocity q ofany point r is then given by

q=q0+wxr,

where qo and co are the same for all points in the body at anygiven instant. Taking the curl of this equation, or, in otherwords, applying or operating with V-, we obtain

Vxq = Vxg0 + Vx(wxr).

Since qo is a constant throughout the body Vxgo = 0 and

Vxq = Vx(wxr).

In this product V differentiates r alone because w is a con-stant throughout the body at any time. In order to find the

VECTOR ANALYSIS. 119

value of this expression expand the triple vector product by(55), considering V as an ordinary vector,

Ox(wxr)=

As V cannot act upon co, we interpret the last term as (o)-V,-)r,

but 3 by (123) and r = co by (118),

Vxq=3w-w=2 w,and w = I Vxq = 4 curl q. (126)

See also equation 131.We see then that when a rigid system is in motion the

operator Vx applied to its velocity-function gives twice Itsangular velocity in magnitude and direction. We maythen write q = q0 + } curl qxr.

Consider now a very small portion of a fluid such that theportion may be considered to move as a rigid body for theinstant; it is fairly evident that the curl of the velocity therewould give similarly twice its angular velocity of rotation.The curl or Vx is an operator such that when applied to anyvelocity-function it gives twice the angular velocity of rota-tion at any point in direction and magnitude.

55. Motion of Rotation which has No Curl. IrrotationalMotion. A clearer idea of curl may perhaps be given by aconsideration of the following two possible motions of a fluidabout an axis. Considering Fig. 53, if the infinitesimalportions of the fluid, indicated by short straight lines, movefrom position 1 to position 2, as indicated, then evidentlyevery elementary portion of the fluid has rotated by thesame amount and the operator Vx would give this rotationmultiplied by 2. On the contrary, if the infinitesimal ele-ments in moving about the axis 0 do not rotate but remainfacing one way as in B, then the curl of such a motion would

120 VECTOR ANALYSIS.

be zero.* Superficially, however, to the eye the two motionshere described would look the same. If we assume that themolecules of iron are free to rotate we may realize these twomotions. If a piece of iron were rotated in a strong magneticfield, the molecules constantly pointing in the fixed directionof magnetic induction, we should obtain a motion such as B,

2 12

B A

Fia. 53.

while if rotated in a non-magnetic field the motion would besimilar to A.

Any motion which has a curl is said to be rotational or vor-tical; if it has no curl it is called irrotational or non-vortical.Any motion represented by a function whose curl is zero isone in which the infinitesimal elements do not rotate, andconversely.

56. V, V. and Vx Applied to Various Functions. It is fre-quently necessary in many cases to apply the operatorsformed with V to combinations of scalar and vector func-tions. The following rules will be found useful for reference.

* A rigid body whose elementary parts move with it as in A, Fig. 53,might be called atomically-rigid; a rigid body whose elementary partsmove as in B would then be non-atomically-rigid. A calculation wasmade to see whether the difference in the moment of inertia of thesetwo kinds of motion could be observed in the case of iron. But mole-cules are so small that the calculated difference in moment of inertiacould not be observed by the most sensitive laboratory methods.

VECTOR ANALYSIS. 121

Let u and v be scalar point-functions; u and v vector point-functions.Then

V (u+v)=Vu+Vv,v(u + v) = vu + V.v,V?(u + v) = Vxu + V.V.

V (uv) = vvu + UVV

(127)

V. (u v) = uV. (U v) = vuxv + u vXv.

(128)

ux(vxv) + vx(vxu)= V (u.v) + vu

(129)

V- (uxv) =uu vv u.V v. (131)

The convention here used, is that the operator v applies tothe nearest term when there are no parentheses or else thevariables on which it operates are indicated by subscripts toit. So that, for example,

Vuv means (vu) v and not V(uv),

and means (Vu) v. In this last case it could meannothing else, as in V(u.v), being a scalar product of ascalar and a vector can have no meaning.

In the above fundamental formulae subscripts have beenused to render ambiguity impossible.

Methods of Proof of the Formulae. These formulae may allbe verified by expanding the quantities in terms of their com-ponents along i j k, differentiating, and rearranging.

If we remember that the symbolic components of v,a , a , a obey with the components of any other vector

ax ay azor of any other V all the laws of common algebra, we shouldexpect V to obey the same laws as any other vector in com-bination with vectors or other V's. With this in mind the

122 VECTOR ANALYSIS.

majority of the formulT on page 121 can be written down atonce without relying upon the demonstration outlined above.

It is evident at once from the definition of V that

V(u+v)=Vu+Vv.Take the formula for instance

V (u v) =

V is supposed to operate upon both u and v. All thepossible combinations of V and a dot with u and v are formedwhich can have a meaning, letting V act once upon eachvariable. In the above example, as u is a scalar, V can acton it only as Vu. The dot, which is as yet unemployed, isused in forming the scalar product of Vu, a vector, with v,another vector. As yet v has not differentiated v. In thesecond term the only way it can act on v is by forming ascalar product, giving V v, which multiplied by u is u,7-v,the correct result.

As another example, consider the expansion for Vx(uxv).We expand this triple vector product as usual, considering Vto be an ordinary vector,

Vx(uxv) =u (Vuyv)- v

where be it remembered that v is to differentiate both u and vin each of the terms. This is here specifically indicated bythe use of subscripts. From u can be formed u(Vyv)and only; so that

u u

similarly from v (Vuyu), v and v can bederived, so that

v v v.

The single subscripts as here used are not necessary accord-ing to the convention explained above. We then have

Vx(uxv) = uVV + vVu - VVu - uVv, (132)

VECTOR ANALYSIS. 123

where the V on the left is to operate on both u and v, whileon the right it operates only on the vector following it. Thiskind of notation is exactly similar to

ddx

(uv) _ (dx u) v + u

C

x V

_dx v+udx

Consider the expression vx(vuxu) in which v is to act upon ualone. Expanding,

vx(vuxu) = vuvu (u.v) - (v'vu) U.

SimilarlyU.7111-V) = Ov (u.v)- (u.V) V.

Adding the two equations, we combine vu (u.v) + Vv (u.v)into v(u.v) by definition, hence equation (129).

The notationv(u.v) = vu (u.v) + vv (u.v)

is strictly analogous to

d (u.v) = du (u.v) + dv (u.v),

which corresponds to partial differentiation, and is true forthe same reasons.

We may write also

V- (U v)= V'-(U v)+ V'-(U v),

or Vx(uxv) = Vu- (uxv) + vvx (uxv), etc.

The process outlined above will always lead to correctresults. It is something more than a help to the memory.A general rigid mathematical proof of its validity has beengiven.*

* See to this effect Joly. Manual of Quaternions, p. 75.

124 VECTOR ANALYSIS.

57. Expansion Analogous to Taylor's Theorem. Expand-ing as a triple vector product and assuming that V acts on valone, we have

andor

ux(Vxv) = vv v

vvvv ux(Vxv). (133)

If u = r a unit vector, and v = q, then (133) becomes

V. (r1.q) - r,x curl q, (134)

and states that the directional derivative of a vector func-tion q in the direction r, is equal to the derivative of the pro-jection of q in that direction plus the vector product of thecurl of q into that direction.

Multiplying the directional derivative by dro, we obtain thedifference in q due to a displacement dr in the direction r,;this gives then, if qr is the value of q at the end of r and qr+dris the value of q at the end of r + dr,

dq = qr+dr - qr = V. drx(vgxq)

So thatqr+dr = qr + V. (v"q)xdr, (135)

or q (r + dr) = q (r) + vq (r)]+[vxq (r)]-dr.

This equation is analogous to the expansion of a functionby means of Taylor's theorem.

58. Theorem Due to Stokes. The line-integral of a vectorfunction F around any closed contour is equal to the surfaceintegral of the curl of that function over any surface of whichthe contour is a hounding edge.

In symbols this is

fF.dr = fia l F dS. (136)p

VECTOR ANALYSIS. 125

This important theorem may be demonstrated in a numberof different ways. The following is a demonstration givenby Helmholtz depending upon the variation of a line-integral.

The principle of commutativity of 8 with d and J is all

that is needed to assume here. Consider the line-integralJ of the vector point-function F (r) along the path ACB.

J = fAF.dr (path ACB).

The possibility of computing this integral in generaldepends entirely on the path ACB, and with this under-

FIG. 54.

standing it is perfectly definite. It is now required to findthe variation in this integral when the path ACB is variedinto an adjacent one AEB infinitesimally close to ACBbut differing from it in an arbitrary manner. The two paths,however, are to begin at A and end at B, two fixed points

126 VECTOR ANALYSIS.

on the contour. Taking the variation of the integral weobtain

8J = SI,B fB

8 B 8 f B F.Sdr.e A A A A

This becomes, by an interchange of d and 8 in the lastintegral and with an integration,y rts,

81 =B

+f B(8 dA AAJ

The integrated term is zero, for since the limits A and B arefixed, there can be no variation 8 r at these points. Remem-bering also (§ 50) that

8 F = and d F = dr.VF F,

8J =fBF - 8r (dr VF) F

fB5r. d using (117)

where v acts on F alone.By (58) this may be written as

8J =fBOr

more directly by (129),

hence

8rx(V F),

But by (117) the left side is 8r-VFHence

(8r.F) =

Referring to the figure it is seen that 8rxdr is the vectorarea of the infinitesimal surface formed by or and dr, so

VECTOR ANALYSIS. ' 127

that calling n the unit normal to the elementary area dS, wemay write arxdr = n dS, so that

M =JB(F)dsn

Another infinitesimal transformation is now made to a newcurve AGB having the same fixed ends A and B and so on,until the movable path has swept over the surface includedbetween the limiting curves I and II. Let now the sum ofall the variations in J be added together. The result will beequal to the difference between J, and J2, the values of J forthe extreme paths; hence

J2 _J, = lim 4 8J =f fn (VxF) d&a r o

But -J, is the line-integral from A to B along BCA, andtherefore J2 _J, is the value of the line-integral around thecontour ADBCA. So that

.fF.clr = f fa F) dS. (136)p

This is Stokes' Theorem.*

59. Condition for the Vanishing of the Curl of a Vector-Function. In the above demonstration, if the value of theintegral J is the same whatever path is taken between A andB, and if this is true wherever the points A and B are taken,it follows that J, always equals J2i so that for any surface S

fi:F dS = 0 and hence vxF = curl F = 0 (137)

must be true everywhere. In this case the value of I dependsonly upon the position of the ends of the path and in no wise

* See also for other demonstrations of Stokes' Theorem,l3ueherer, Elemente der Vektor-Analysis, pp. 42-44.Gibbs-Wilson, Vector Analysis, pp. 188-190.Gans: Einfiihrung in die Vektoranalysis, pp. 35-39.aee Appendix, p. 249, for two other proofs.

128 VECTOR ANALYSIS.

upon the shape of it. Conversely, if the curl F = 0 in aregion, then the line-integral of F between any two pointsA and B in the region is independent of the path chosenbetween them. In this case if one end A of a curve is fixed,

the value of the integralfB is simply a function of itsA

upper limit B whatever the path from A to B may be. Let0 denote this scalar function. The integrand mustthen be a perfect differential and hence of the form do; whichby (114) is the same as

4 = d r-pq,

so that for all values of dr,

dr-F

and hence F = v o. (138)

Or in other words, if F has no curl, it is the rate of fastestincrease Vo or grad 0 of a scalar function

OB = JAB F-dr + OA,

where 95A is a constant.The scalar function 0 thus determined is called the poten-

tial of F. As we have seen before, a scalar function dividesspace up into shells or laminae by means of its level surfaces.The vector-function F = Vq! derived from such a functionis for this reason said to be a lamellar vector (Maxwell). Thecurl of a lamellar vector is then always zero,

or curl (pt) = 0. (139)

Conservative System of Forces. By Stokes' Theorem (136)we see that if the line-integral of i.e., the work aroundany closed path, always vanishes, then the forces have nocurl; and also that in this case the forces in the field are

VECTOR ANALYSIS. 129

derivable from a potential functionq5. Such asystem of forcesis a Conservative System, and we may define such a system :

When the forces acting on a system of bodies are of such anature that the algebraic total of the work done in performingany series of displacements which bring the system back toits original configuration is nil, the system of forces is said tobe conservative. The condition, then, for a system of forcesF to be conservative is

curl F - VxF=O.

60. Condition for a Perfect Differential. If the totaldifferential do of a scalar point-function 5(r) = C be taken,we obtain, by (§ 49), the equation

This is of the form 0,

which in general is not a perfect differential. In Cartesiannotation this equation becomes, if f1, f2, f3 are the compo-nents of f along the three axes,

f,dx + f2dy + f3dz = 0,

and our problem is to find the condition for integrability of it.Assume that, by multiplying this differential equation bysome scalar factor It, it may be made a perfect differential, orthat for every value of dr

f -It is called an integrating factor. In this case, then,

/If=vc.In order to eliminate 0 take the curl of this equation, becausewe know that the curl of a lamellar vector (139) is zero,

Vx(,t f) = 0,

130 VECTOR ANALYSIS.

which may be expanded as

V- (4 f) = fcpxf + (VP) -f = 0.

Applying f- to eliminate the second term, because 0in general, there remains

f) = 0.

So that finally we find that the condition that the equationdefined by 0 should be integrable is that f and its curl,Vxf, shall be at right angles or that the curl vanish. To putthis discussion in a more familiar form, we have proved thatthe condition of integrability of

is that 0 (140)

or X 8Z _ aY + I, aX _ aZ + Z aY _ aX _ 0(ay az) (8z ax) \ax 8y) '

a well-known result.If the

curl F = 0,

this equation is evidently satisfied.If the curl F is not zero, the equation says that it must beeverywhere perpendicular to F.

For example, t T_

yz dx + zx dy + xy dz

has no curl and is therefore derivable from a single primitive,

xyz = const.

On the other hand,

ay2z2dx + bz2x2dy + cx2y2dz

has a curl, but this curl,

2x2(cy- bz) i + 2 j2(az- cx) j + 2z2(bx- ay)k,

VECTOR ANALYSIS. 131

and the vector F,ay2z2i + bz2x2j + cx2y2k,

are at right angles, and this equation is also derivable from asingle primitive, i.e.,

a+b+const.x y z

61. Taylor's Theorem. The Operator e'*V( ). Taylor'sTheorem is often written concisely as

f(xyz)y /

z

+la)f(xyz)+2! hax+kay /s

+3 (+ k a + 1a3

(xyz) + .. .y

If the components of r be x, y, z, and those of e be h, k, 1, itmay still further be condensed into

Remembering the expansion for ex,

ex=1+ X +2!+X +...,1! 3!

we may write the last equation symbolically in the stillshorter form,

f (r + e) = e``f (r), (141)

so that the symbolic differential operator e`'° acting on anyfunction f (r) gives its value when r becomes r + e.

62. Euler's Theorem on Homogeneous Functions. Wemay employ this equation to demonstrate a useful theoremdue to Euler and known by his name. A function 95 of

132 VECTOR ANALYSIS.

degree n in a variable r is said to be homogeneous when roccurs the same number of times in every term of it. It isone such that

0(a r) = ano(r), (142)

where a is any constant. Apply Taylor's Theorem to thehomogeneous function 4(r) and let e = g r, where g is a smallscalar multiplier. Then

e9 r.°o(r) _ o(r + g r) = 0[r(1 + g)] = (1 + g)no(r),

and hence

0(r) + g r Vc (r) + 2 (r.p)20 (r) _+ ..... .

+ ng + n(n - 1)1.2

gz -{- ... o(r).C

l

Subtracting 0(r) from both sides and dividing through byg there remains

r.VV(r) + 2 (r) + ... = [n + n(1.21) g+ ..] (r).

This equation being true for an infinite number of values ofg, we may equate the coefficients of the same powers of g onboth sides of the equation, giving

no (r) =n(n - 1)c(r) = (143)

n(n - 1) (n - 2)o (r) =

The first of equations (143) is known as Euler's Theoremon homogeneous functions, which in terms of x y z becomes

nci(x y z) = x a- + y a- + z -- . (144)

The remaining equations are extensions of the theorem,involving derivatives of higher orders.

VECTOR ANALYSIS. 188

Operators Involving V Twice.

63. Possible Expressions Containing V Twice. Given ascalar point-function V and a vector function F, the follow-ing six combinations, involving V twice, are possible ones:

1° V2V2° VXVV3° F4° V(V.F)5° V. (VXF)6° Vx (V- F)

(div grad V, a scalar)

(curl grad V = 0)

(V2F, a vector).

(grad div F, a vector).

(div curl F = 0).

(curl curl F = curls F, a vector.)

Two of these expressions vanish identically,

VXVV-0, curl (grad V) =0

because any vector product containing two like vectors iszero, and

V.VxF _ 0, div (curl F) =0

because any triple scalar-product with two like vectors is alsozero. These two results may be proved, if not sufficientlyevident, by expanding according to the ordinary rules.

The 6th may be expanded into

V.(VxF) = F (V.V) (145)

so that the 6th 4th - 3d.

Equation (145) is sometimes written as

curl (curl F) = curl2 F = grad div F - V2F. (146)

This last important equation would be, as it is written,rather difficult to remember, but the advantage of retainingthe notation in dels is made evident by the previous equationwhich may always be written out according to the rules forthe expansion of a triple vector-product. As another exam-ple of- the advantage of the symbolic notation it is perfectly

134 VECTOR ANALYSIS.

easy to remember whether it is curl grad V or grad curl V,which is identically zero. For when they are written interms of V, i.e., VxVV and V(VXV), one is evidently zero andthe other can have no meaning at all. In other words, thedel-notation when interpreted according to the ordinary rulesof vector products, leads us to correct results independentlyof any physical or other considerations.

The Operator V2 or V.V. (Read del square of . ..)Operating on

VV - iax + iayV+kaz

with V., or in other words taking the div of the grad of V,.weobtain,

0.0V=V2V= ax + 8y2 + az(X2

+aye+az2)V,(147)

the well known Laplacian' operator, which when equated tozero is satisfied by the potential function in free space.

It is evident on inspection that

V.(VV) = (V.V) V= V2V.

Since the curl of grad V, or in symbols VxV V, is zero iden-tically there can be no ambiguity whatever when V is twiceapplied to a scalar-function. When V2 is applied to a vectorfunction, it means that it is applied to the three scalar-func-tion components of the vector function, and hence offers nonew difficulty.If V2F = 0

then

V2F,=OV2F2 = 0V2F3=0.

that is the three components of F satisfy Laplace's equation.

VECTOR ANALYSIS. 136

Since VxV V = 0 = curl grad V,

it follows that the vector VV is a lamellar vector, by (§ 59).

Since 0 = div. curl F,

it follows that the curl of any vector is a solenoidal vector,by (§ 53).

64. Differentiation of the Scalar Function r- by V. Weproved equation (108), that

Vrm = mrm-1 r1 = mr" 1-2 r.

Take the divergence of this vector.

V.V rm= V2rm= m rm-2 V.r + rm 2

= m 3 rm-2+ r(m - 2)rm_3

,because by (112),

3 and r r dr

so that V2rm= rm-2 3 m + (m - 2) ml,and finally V2rm= m(m + 1)rm-2. (148)

The two value§ of m which will satisfy the differential equa-tion

V2rm = 0,

are easily seen to be m = 0 and m = - 1, so that the scalar

function satisfies Laplace's equation, orr

V21=0.r

This may also be shown, of course, by direct differentiation

of the function 1r

136 VECTOR ANALYSIS.

EXERCISES AND PROBLEMS.

1. Prove that is an operator independent of choice of axes,by actually carrying out the transformation to a new set of axes.If the coordinates of the new set be denoted by primes, it shouldbe found that

ax ay aZ aX' aY' aZ'ax + ay + az ax' + ay' + az'

where X,Y,Z, are the components of a vector function F, and whereX',Y',Z' are its components referred to the new axes.

2. Prove directly by a change of axes that

VxF or Curl F

is invariant to that change.3. Prove that

a,Vr = r1,Vr2= 2 r,

where a is a constant vector. These follow from the relation

d( )4. Prove that

5. Verify that .

where

6. Prove that

3,

2 v. (axr) = 0.

=(xa2

e

02 02

aye + az2) F,

F='iX+jY+kZ.

V = a.(OV)

and state the resulting theorem. Apply this to a simple problemin potential.

VECTOR ANALYSIS. 137

?. Show that(1) = _a.V - ,r r3 r2

br

b are constant vectors.8. Show by direct expansion that

Vx0V=0and vvxv- 0.

9. Find the resultant attraction at the origin of the masses 12,16, and 20 units respectively concentrated at the ends of the vectorsa=31+4j, b= -51+12j, and c=81-6j.

10. From the expression for the attraction at a point P due to amass M, its density p at any point being a ppint-function of r,

dFJl Jv01 prr2 v

deduce the ordinary Cartesian expressions for the componentattractions along the axes.

1L What theorems do equations (111), (117), (118) express?Write them out in Cartesian notation.12. Explain paragraphs 46 and 47, considering instead of

potential(a) Temperature distribution in a body.(b) Velocity distribution in a fluid.

CHAPTER VI.

APPLICATIONS TO ELECTRICAL THEORY.

65. Gauss's Theorem. Solid Angle. Consider a point 0,and any small area in space dS. Join every point in theboundary of the area dS to 0, thus forming a small cone. Wedefine as the solid angle subtended by the area dS at 0, the

value of 9 , where d Z is the area cut out, by the cone, on

any sphere of radius r, described about 0 as center. This isnumerically equal to the area dw, cut out by the same cone on

FIG. 55. O

the sphere of unit radius, described about 0. The dimen-sions of solid angle are evidently zero. Since the total areaof a unit sphere is equal to 4 7r, this is also the solid angle sub-tended by the whole of space or by any surface which com-pletely surrounds the point 0. d Z evidently subtends thesame solid angle at 0 as dS. -Calling n the unit normal to dS,its sense being taken in any conventional manner previouslyagreed upon (outward from the surface in the following),

dE = f dS cos (rn),138

VECTOR ANALYSIS. 139

according as it makes an acute or an obtuse angle with r,respectively.

Now because dw and dl; are parallel sections of the samecone we may write

d E = rzdw,

so that the solid angle

dw= LZ(149)

If a point 0 is chosen inside any closed surface, any smallcone with vertex at 0 will cut out. through the-- surfacealways once more than it cuts into it, but if the point be out-side the surface it will cut in as many times as it cuts out. Ifn be chosen positive when drawn outwards, then the anglebetween n and r will be acute wherever the cone cuts out,obtuse wherever it cuts in. Thus an elementary cone whenits vertex is inside a closed surface contributes an element ofsolid angle, +dw. As for example, in the figure the solidangles + dw and - dw, due to 1 and 2, annul each other, leav-ing + dw due-to 3. When the vertex is outside, the resultingsolid angle is zero; as for example, the solid angles -dw and+ dw, due to 4 and 5, completely annul each other. So alsodo the elements at 6 and 7 and 8 and 9 in pairs. If we inte-grate or sum up all the solid angles due to all the elementaryareas dS over the whole surface S, we shall obtain 47r for thesum, if the point 0 is inside and zero if the point is outside.

Expressing these results in symbols we have

ff dw = ffcos(rn) dS =f f ! dS = 4 7r

0 inside of S (150)=ffcos(rn) dS =J e dS= 0

0 outside of S.These results, which are purely mathematical, are known asGauss's Theorem.

140 VECTOR ANALYSIS.

Gauss's Theorem for the Plane. In a plane we may obtainan analogous theorem, i.e., the plane angle subtended by aclosed contour in a plane at a point 0 is 2 it or 0 according asthe point is inside or outside of the closed contour. In thefigure (56) consider a point A connected to 0 by a radiusvector which starts from B and moves once around the con-tour until it reaches B again. Evidently whatever the shape

FIG. 56.

of the contour may be, the radius vector r has made but onerevolution about 0 and therefore covered an are equal to 2 iton the unit circle about 0. In the case that the contour iscompletely outside of 0 the same reasoning shows that theradius vector when it reaches B has not rotated around 0 at all.

In a plane the angle do subtended by an are dr at a point0 is, using a notation similar to (159),

do = ± dr cos (rn) = tr r

VECTOR ANALYSIS. 141

and the integrals are

fd = fcos () dr = f ndr = 2ir or =r r

'J J Jaccording as the point 0 is inside or outside of the contour.

We may apply analogous reasoning to Gauss's Theoremin space. Think of the solid angle subtended by an exten-sible sheet which gradually is extended completely in anymanner around any point 0, or which is made to form aclosed surface of any shape completely outside of 0, and thetwo results of equation (150) will become visually self-evident.

The importance of this theorem in physics is the factthat the surface-integral of the normal-component of thevery important vector-function varying inversely as thesquare of the distance r from a point 0 is in symbols

ff cos n ) dS = J f'114

r2

and Gauss's Theorem reduces these integrals.Second Proof of Gauss's Theorem. We may obtain another

proof of Gauss's Theorem from the following physical con-siderations. The field of force around a point at which isconcentrated a unit of matter, electrical say, is by Coulomb'sLaw

(151)

This means that the vector F is directed radially outwards,but that its magnitude varies inversely as r2. Consider twoclosed surfaces S, and S2, the first surrounding the point 0and the other lying completely outside of 0. About 0 drawtwo spheres, one of which is completely outside of both of thetwo surfaces S, and 82, and the other of a small enough radiusso that it does not touch either SL or Sz. Since the magni-

142 VECTOR ANALYSIS.

tude of F falls off exactly as fast as the areas of the spheresincrease with increasing radius, the surface integral of thisvector over each of the spheres is the same. Or, in other

FIG. 57.

words, the flux that gets through one reaches the other.Evidently the same amount of flux must have passed throughthe surface St, so that we may write for the flux

fn-rtdS= 1 J 1, R,2 Jsphere R1z

of rad. R,

12 4 7f R12 = 4 7r.1

VECTOR ANALYSIS. 143

As for the surface S2, since no flux is gained or lost in goingfrom sphere R, to sphere R2i whatever flux went into S2 mustalso have come out of it again, and therefore

ffs.

dS = 0.r2

Hence the theorem.*

66. The Potential. Poisson's and Laplace's Equations.From the definition of the potential in § 46 as the work doneon a unit positive charge in bringing it from infinity up tothe point at which the potential is desired, we may show thatthe scalar function 'n is the potential function correspond-

sing to the inverse square, or Coulomb's (in gravitation, New-

ton's) law of force Fa M I.

Consider a quantity of positive matter, m, at 0.The potential at any point P, or in other words the work

done on a unit positive charge in bringing it from infinity to

FIG. 58.

P, is equal to the line-integral of the force function F from Pto 0o along the path PQ traversed by the unit charge.

VP =fF.dr =- see §16r

which may be written, because v 1 = - Z by (109),r r

Vp = f p dr.V(m).

(152)

* See Appendix, p. 251, for another proof of Gauss's Theorem.

144 VECTOR ANALYSIS.

This last expression is the total derivative of m', so that ther

integral is independent of the path and depends only uponthe limits, that is, upon the starting point and ending point,giving

m / rP mVP =ap P

Should there be other masses present, the potential function

FIG. 59.

due to them all is the sum of the separate functions due toeach, or

V ="+Mz+ ... =j m4' (153)rpI rPZ rpe

where rP8 is the distance from the point at which the potentialis to be found to the mass m,.. If the masses instead of beingat discrete points form a continuous distribution, the sum-mation becomes a volume integral; dm, the element of mass,

VECTOR ANALYSIS. 145

becomes p dv, where p is the volume density of the matterunder consideration. We have then,

V=JJJ. dm_ffvol. (154)mass r r

The integral thus defined which is to be taken over thevolume occupied by the masses may be shown to be finite,continuous, uniform, as well as its first derivatives. Itvanishes itself at infinity to the first order, and its firstderivatives to the second order.

A system of forces for which the line-integral between anytwo points is independent of the path is called a Conserva-tive System.

If we multiply Gauss's Integrals by m, a mass concentratedat the point 0, we shall obtain

fm 47rm. (155)

This integral states that the outward normal component offlux of force (according to the inverse square or Newtonianlaw) through a closed surface surrounding 0 is 4 7r times theamount of matter within; any matter lying outside of thesurface contributing nothing. As every element of mass mcontributes 4 arm we have the proposition that the outwardflux of force through any closed surface due to any distribu-tion is 4ir times the total amount of matter within the sur-face. It is in this form that Gauss's Integral is usuallygiven, but evidently, from what precedes, it is a geometricaltheorem rather than an electrical or gravitational one.

The force at any point due to any distribution of matteris - grad V or - VV, by § 47, where

V= ffrfris the potential function due to the distribution. The signco denotes that the integral is to be taken over the whole

146 VECTOR ANALYSIS.

of space. This is equivalent to integrating over-the matteralone, as wherever there is no matter p = 0 and the integralcontributes nothing. So that we may write (155) as

- fJs n.VV dS = 4 7r f f f p dv,

fff p dv being the total quantity of matter within S.

By means of the divergence theorem (121) the surface inte-gral above may be transformed into a volume integral takenthroughout the volume enclosed by S.

- ff dS = - fff V.V V dv = 4 7r i rf p dv..l ./ s

As this equality holds whatever surface S is taken, it followsthat the integrands are everywhere equal and

V.VV=V2V=-4irp. (156)

This is Poisson's Equation.In free space where o = 0 this becomes

V2V= 0. (157)

which is Laplace's Equation.We may interpret these equations as follows: Every

quantity of matter emits lines of force, 4 7r lines per unit quan-tity. This numeric 4 it is purely conventional and appearsbecause the intensity at unit distance from unit charge isdefined as unity ; and since unit intensity corresponds toone line per unit area, there must be 4 it lines emitted in orderto have one for each of the 4 7r units of area in the surface ofthe unit sphere. So then, if a surface of volume dv be drawnaround a point where the density is p, the lines passingthrough the surface are equal to 4 7r times the quantity ofmatter p dv within it, or

div F (due to volume dv) = 4 irp dv,

and per unit volumediv F = 4 irp.

VECTOR ANALYSIS. 147

Now if F have a potential, that is, if F can be representedas the grad or V of some scalar function W, then puttingF=VW,

This equation is true for the potential W due to attractingmatter.

In the case of repelling forces, since the force is oppositeto the direction of increase in the scalar function, we mayplace W = - V, and we may write as before

V2V =- 47rp.

In free space where there is no density p the equation becomes

div F = 0,

which says that the lines of force are solenoidally distributed,that is, the flux takes place in unbroken continuous paths,and hence cannot end nor begin at any point of space devoidof matter.

The reason for the term (div) divergence is evident fromthe foregoing.

Harmonic Function. A function which in a region issingle-valued, continuous, and satisfies Laplace's equation issaid to be harmonic in that region.

A Spherical Harmonic of degree n is any homogeneous(143) harmonic point-function of space. That is, if V sat-isfies the equations'

VZV=0

and r VV = n V,

it is a spherical harmonic of degree n. The study and use ofsuch functions is of great importance in all branches of math-ematical physics.

148 VECTOR ANALYSIS.

67. Green's Theorems. Two theorems due to Green ofvery important application in theoretical physics followimmediately by an application of the divergence theorem,

ff dS= fff01 v.Wdv,

to the functionW = UVV, wheereU and V are two scalarpoint-functions which with their derivatives are uniform andcontinuous in the space considered. Applying V. to W,

V .W = V. (UVV) = UV2V +

Substituting in the equation above

ffn.UV V dS = fffuv2 V dv + f f f vU-vV dv.J(158 )

Similarly, by symmetry, putting forW = VV U, we have

ffn.vvuds= fffvv2udv + fffvu.vvdv.Subtracting these two equations there remains

ffn.(uvv - VVU) dS = f f f (UV2V -VV2U) dv.

(159)

The surface integrals are to be taken over the surfaces bound-ing the region under consideration, and the volume integralsthroughout the volumes enclosed by these surfaces. Equa-tions (158) and (159) are called Green's Theorem in its firstand second forms respectively.

68. Green's Formulae. Apply Green's Theorem in itssecond form to two functions U and V. Let U be the func-tion U = 1 , and let V be the potential due to any distribu-

tion of matter. The region to be considered is the spacelying between the infinite sphere, S., any surfaces S which

VECTOR ANALYSIS. 149

surround the distribution, and the infinitesimal sphere ofradius e, surrounding 0, the. point from which r is measured.

The equation

ffn.(uvv - VV U) dS = fff(Uv2v_ VV2 U) dv(160)

becomes, since V2U = V2 1 = 0, (148)r

ff n. 1vv- vv itdS= f f f(1V2V)dv.\r r/ \r

Fia. 60.

The surface integral is to be taken over the bounding sur-faces S to the region. The infinite sphere contributes nothing,

as at infinity 1 V V and VV 1 become zero to the thirdr , 1 r

order, both containing r8 in the denominator. For the small1

150 VECTOR ANALYSIS.

sphere about 0 the first part of the integral may be trans-formed,

ffn. VVdS=

f ff ffn.vv dw,

where dcw is the solid angle subtended by an element of thesmall sphere at 0, and, where e, its small radius, is constantduring integration. As e becomes smaller and smaller andbecause n.VV, the normal force on the surface, is finite, theintegral vanishes in the limit.

Considering the second part of the surface integral overthe small sphere, we may write

- f ffiw = - V0 X 47r,J J \ r)1 dS = solid angle due to dS. Asbecause + r dS = r2

the radius of the small sphere diminishes V approaches V0,its value at 0.So that finally,

V° 4 7rJ J J rvdv+ 47rJ

.lOV - VV r ) dS,

Region Surfaces (161)

the surface integral being taken over the bounding surfacesS and the volume integral over the region bounded by them,shaded in the figure.

If V2 V is equal to zero in the region considered, the poten-tial function at any point 0 is

V°=4 f VVr)dS, (162)

which shows that it is completely determined everywhere ifthe values of the potential V and of its normal derivative n.VVare known over the bounding surfaces S. If the matter

VECTOR. ANALYSIS. 151

producing this potential and distributed in any mannerwithin S be taken out and replaced by a surface densityof matter a on S of amount

a = 4 n (VV - rVOr) , (163)

the potential at O will be exactly the same as before, becauseby substituting this value for a in

V0= ff8fwe obtain equation (162). We shall call this distributionan Equivalent Layer. But in general this will not neces-sarily make the surface an equipotential surface.

If the point 0 is inside the surface S, a similar deductiongives the formula k

V° 4 7rfffYdv

r 7r+ 4 f fn(r VV- VV r 1 dS,

/ (164)

where n is to be drawn as the external normal to the regionin which 0 lies. With this convention the two formulae due toGreen (161) and (164) are identical in form.

Green's Function. Adding together Green's equation (160),which may be written

0 =f f f (Uv2V-VV2U)dv- f f

and (1.61), which hold under the same conditions, we obtain

47rV0 ffJ' U- rl V2V dv- f f fVv2Udv

- f rrrU - r)11 VV -VO (U - 1 ]-n dS. (165).l J K r/ ` r/

152 VECTOR ANALYSIS.

This equation is of especial importance in the theoriesof light and electricity. The quantity (U - 1) which ap-

rpears in the integral is sometimes known as Green's Function.

69. Solution of Poisson's Equation. Equation (161) statesthat if the quantity V2V is known throughout a regionbounded by any surface S, and if the quantities V and VVare known at all points of the surface, then V is completelydetermined within the surface. Allow the surface S torecede to infinity so that we are now considering the poten-tial in the whole of space. Then in the equation the surfaceintegral contributes nothing, as all the quantities multipliedby dS approach zero to a sufficiently high order. Thereremains, then, in the limit only.

V° 41 f f f V dv. (166)

Now, by Poisson's Equation, V satisfies the relationV2V=- 47rp.

So that, by (166), if the value of p, the density, be given atevery point in space, V is determined by the integral

V = f f f (167)

which is therefore a solution of Poisson's Equation.The Integrating Operator Pot. The operation of finding

the potential due to a distribution whose density is definedeverywhere by the scalar function p(r) plays such an impor-tant role in mathematical physics that Prof. Gibbs has givento this operation a special name and defines

Pot p =- f f p dv (168)fJ x r(read potential of p).

The sign oo indicates that the limits may be taken over thewhole of space, as wherever there is no matter the integral

VECTOR ANALYSIS. 153

contributes nothing. We shall call the operation indicatedby the above equation " the potential of p," even when pdoes not represent a volume density.

Considering again equation (166)

V=-4-fJ fvrV dv,

and using the notation (168), we see that we may write

V =- 41-r pot V2V. (169)

So that the application of

- 41-Pot ( ),

to a function nullifies the effect of V2 on that same function,or, in other words,

- 417rpot ( ) is the inverse operator to V2 ( ).

70. Vector-Potential. In the same way that the potentialdue to the scalar function p is formed, that is,

V=potp fffedv,we may define the potential of the vector function

P=pit+P2j+p3k,where p P2 and p3i are given scalar functions of r, as

V=potp= fffPdv

fff 1dv+j fff f f f °8dv.

(read vector-potential of p)

154 VECTOR ANALYSIS.

The vector function V so defined is called the vector-potential of p. Its three components evidently satisfy therelations satisfied by the scalar potential, so that we have

V2V = - 4 « p. (170)

In strict analogy with the solution of Poisson's Equationfor a scalar potential we have then for the solution of a vec-tor-potential

V _ 4 J J J rV dv. (171)

71. Separation of a Vector Point-Function W, which hasa Vector-Potential, into Solenoidal or Rotational and Lamellaror Irrotational Components. This means that the vectorfunction W is to be separated into two parts, one of whichhas no divergence and the other no curl. - We then assume

W=X+Y, (172)

where 0 and VXY = 0.

Consider the scalar function 95 and the vector function V,related to X and Y, respectively, in the following manner.

X = vxV,

ThenW=vxV - vc.

If it is possible to determine V and 0 the problem is solved.To do this take the divergence of W, giving

VW =- 1720.

So that the solution for q is, by-(166),

4 f fJ r dv 4 L fffV rV dv (173)

= 4 W7r Pot

(V ).

VECTOR ANALYSIS. 155

Similarly, taking the curl of W,VxW = V.(VxV)= V(V.V) - V2 V.

Now since V is as yet undetermined, we may assume thatits divergence is zero or that 0, hence

VxW = - V2V.

So that VxW is 4,r times the vector function of which V is thevector potential, and by (171)

V Iff TrV d,, 4Lfff W dv47r r=

4pot (VxW ), which determines V. (174)

Finally, sinceW=VxV - Vof

W4nvJJJ7rVdv+4 VxJJ.I Vx-dv, (175)7r r

the decomposition is thus accomplished. This decomposi-tion is sometimes known as Helmholtz's Theorem.*

Other Systems of Units. The factor 47r which occurs inmany of these equations is due to the definition of unit quan-tity of matter. In virtue of this definition it is necessary toassume that every unit of matter emits 4 7r lines of force, sothat, for example, the number of lines cutting through anyclosed surface around any amount of matter will be 4 7r timesas many as there are units of matter inside. Of late it is thefashion to eliminate this " eruption of pr's" as Heaviside hasit. This may be done in various ways, one of which is toredefine the unit quantity in such a manner that it emitsbut one line of force, in which case the equations

F = 1 r, div F = 4 prp and V2 V = - 4 prp,7-1

become respectively

F = 4 1r2 r, div F = p and V2 V = - p. (176)

* Wiss. Abh. Band I, p. 101. For a mnemonic if not another proof

of this theorem, multiply equation (145) by 47rr

r and integrate, remem-

bering equation (171).

156 VECTOR ANALYSIS.

Such a choice of units eliminates 47r in a number of for-mulae but introduces it in others. It is nevertheless the mostconvenient assumption in the modern theory, where theenergy is located in the space between the acting matter, andin which action at a distance no longer holds first place. Thepotential at a distance r from a mass m, for example, becomes4 m instead of M. The operation of forming the potential,

or pot V, would in this system consist in forming the integral,

pot p = .f f pdv

P

4 7rr(177)

and the theorem of Helmholtz would become in this notation

W = - V pot (V.W) + Vxpot (VXW), (178)

72. Energy of a System in Terms of Potential. Con-sider two particles of matter acting according to the inverse

r,,

FIG. 60A.

square law, mp and mq, respectively, separated by a distancerpq. In order to bring the mass mp from infinity to its positionan amount of work (§ 66)

m4Wpq= r mp,

must be expended on the mass mp if the masses repel, or bymp if they attract. For definiteness assume the matter to berepelling.

VECTOR ANALYSIS. 157

The expression above may be written in two ways,

Wpq= Vgmp or Vpmq

because Vq = ms' and Vp = M2'rpq rpq

Similarly if we have any two systems of particles, theenergy obtainable by allowing the two systems to disperseto an infinite distance apart is

WpqEpEqmpmg,

rnq

where the summation signs extend to every pair of points,one point from each of the systems. If we consider thetwo systems as one, a factor I must be introduced, as in thesummation every term would appear twice, so that

Wpq=

2

ZpEqmPmq (179)

rq

represents the mutual potential energy of a single systemof particles. If the system forms a continuous distributionthe summation (179) becomes

W=1 ff fffdmdm'iJ'ffvdm ,

where V is the potential function due to the total distribution.

73. Energy of a Distribution in Terms of Field Intensity.If the distribution consists of a surface and a volume distribu-tion of surface density a and volume density p, the aboveintegral takes the form

W= 2f fV a dS + 2 f f f V p dv. (180)

The integrals are taken over the surfaces and throughoutthe volumes of the matter under consideration respectively.

158 VECTOR ANALYSIS.

At a surface distribution there is a discontinuity or changein the normal component of the force due to the surface dis-tribution, given by the well-known expression

Fn = 4 irv =

By drawing a surface completely surrounding the surface dis-tributions we may apply Green's Theorem to the whole ofspace outside of these surfaces. Remembering that from thelast equation

v = 41 and that p = - 4V2 V,

the two integrals which are now

W87r f ff vV2Vdv

become by Green's Theorem (158)

W (181)

If the medium is any other than vacuo, the element in theintegral, FZ dv must be multiplied by a factor a characteristicof the medium. The energy of the distribution is in this case

W=817, fff eF2dv. (182)

This may also be written as

f f f .F dv, (183)W= I f f J. e dv =87r8 74 6

wheree F F'S (184)

is a vector called the Induction.

VECTOR ANALYSIS. 159

74. Expressions for Surface and Volume Densities of a Dis-tribution in Terms of the Intensity of Polarization. Startingagain (180) with the energy of a surface and volume distribu-tion of densities a and p, respectively,

W=2 ffvis +2 fffVpdv.'Let us assume that this energy may be written also as

W=-1ffd v=2 fffJ.vv dv,where I is called the intensity of polarization, H is the fieldstrength, and V is the potential corresponding to H. Thensince V.(VI) =the integral may be transformed into

W=2 fffv.(vI)dv-2 fffvv.idv.Transforming the first integral by the divergence theorem andcomparing this with the expression for the energy in terms ofa and p,

W fJv(n.I) dS - 2 fffvv.i dv,

we see that the polarization I produces a surface densitya = (185)

and a volume density

Conversely, assuming a distribution to consist of a surfacedensity a and a volume density p, it is easy to show by rea-soning backwards that there is a quantity I related to a and pby the equations

a=n-I and p=-17.1such that the energy of the distribution may be representedby the integral throughout the volume,

W= -2fffI.H dv. (186)

160 VECTOR ANALYSIS.

Equations of the Electro-Magnetic Field.

75. Maxwell's Equations. By experiment Faraday showedthat when the magnetic flux through a linear circuit is variedthere is induced in the circuit an electro-motive force. Ifthe circuit is a closed one this induced electro-motive forceproduces a current in it. He also showed that this electro-motive force is equal to the negative rate of change of themagnetic flux. The positive direction of rotation in a cir-cuit is connected with the positive direction of flux throughit, according to the adjoining diagram which symbolizes theso-called cork-screw rule. If the arrow shows direction ofincrease of magnetic flux, the arrow-head in circuit showsthe direction opposite to the induced current. The figure asdrawn shows the direction of the magnetic flux due to thecurrent in the circuit. By Lenz's law such a magnetic fluxwould induce a current opposite to this; hence the negative

Fia. 61.

sign in equation 187 below. Since elec-tricity tends to flow from places of highto places of low potential we may con-sider this electro-motive force as some-thing in the nature of an electrostaticfield which is induced in the space bythe varying flux. That is, the electro-motive force is induced in the spaceeven when unoccupied by a conductor.

In a conductor this electro-motive force produces a currentand in a non-conductor tends to produce a current. Toobtain the total electro-motive force around any circuit, weevaluate the line-integral of this electrostatic field F alongthat circuit. Then the induced e.m.f. may be written

fF.dr = fLD (V-F)-n dS,J

by Stokes' Theorem. But Faraday's experiment shows that

VECTOR ANALYSIS 161

this is equal to the negative of the rate change of the mag-netic induction 3C through this circuit, so that

ff dS =_dt J

JCac.n dS = ff-C.nt

dS.ap

As this is true whatever circuit is considered and whatevercap is taken as long as it is bounded by the circuit we maywrite

vxF axat

(187)

By experiment Ampere proved that a current I is equiva-lent to any magnetic shell of a certain strength which isbounded by the current. He also showed that a current maybe measured by means of the magnetic field that it produces,and quantitatively that 4 7r times the current in any sectionof a conductor is equal to the line integral of the magneticforce H taken once around any path linked positively withthe conductor. If q be the current density, then, symboli-cally,

f 47r f f q.n dS,J J JcapJwhere the surface integral is taken over any cap to the sur-face. Transforming the first integral by Stokes' Theorem,we have

f f dS = 4 ;r f, f dS,J .l J land as this equation is true whatever portion of space is con-sidered and whatever path is taken around that portion ofspace, we may write

VxH = 47r q. (188)

In order to explain the effect of an electro-motive forceupon dielectric non-conductors Maxwell assumed that instead

162 VECTOR ANALYSIS.

of a current q there is produced a so-called displacement-motion or current q' of electricity, which on the release of theinducing electro-motive force springs back and takes up itsoriginal position. He assumes that this current-displacementproduces the same magnetic effect as would be produced by acurrent of density

1 arq4 7r at

where ff = e F

(189)

and e is a constant of the medium called electric inductivityat every point of the field. We therefore, in considering adielectric, introduce this displacement current density insteadof q, giving

VxH = a (190)

This assumption has been completely verified by the experi-ments of Rowland in America.* If the dielectric is also con-ducting we retain the term in q, giving

VxH = 4 it q + a.

atThe term

q+ e47r at(191)

is called the total current and being equal to a curl is sole-noidal, and has no divergence, i.e.

V. q+ 1057 _o.

C 47r at)

This current therefore moves in closed circuits or paths. Itis because of this equation that electricity is said to act likean incompressible fluid. (See § 53.) According to the Elec-tron Theory its compressibility is at most one part in a million.

* Since repeated also by Cremieu and Pender in France.

VECTOR ANALYSIS. 163

The complete system of equations for media at rest holdingin an insulating dielectric are therefore

L5 _ VxH)

at

ax= VxF, (192)at

in combination withT=eF and X=pH.

e and u are the electric and magnetic inductivities respec-tively, and are defined by these equations. They are con-stants for a given homogeneous isotropic medium, that is,for non-crystalline media.*

76. Equation of Propagation of Electro-Magnetic Waves.Let us assume that there are no permanent magnets in thespace considered, or, in other words, that there is no intrinsicmagnetization, symbolically this is expressed by writing

vH = div H = 0.

Take the curl of the first of these equations (192),

Vx LT = a VxF = v- ('7-H) V2H + V(V.H).at at

Changing from F to F and from H to 3C and rememberingthat vH = 0,

epat

VxF = - V23e.

Differentiating with respect to the time, we obtainz

elt a2(VxF)_- V2!=V2(VXF)at at

and also (193)2

eltate

3e = V23e.

* In order to investigate the form these equations take for crystal-line media it is necessary to employ the linear vector-function.

164 VECTOR ANALYSIS.

In a similar manner we could show that F, T, H and X andtheir curls also satisfy the equation

aZ = a2V24P. (194)eat

This is the differential equation for wave motion, one of thefundamental, partial differential equations of mathematicalphysics. It can be shown that the velocity of propagationis equal to a, and therefore for an electro-magnetic pulse

equal to 1 This turns out to be identical with the'Vey

velocity of light in vacuo, as it should be if the etheris the common medium for the propagation of electricalwaves as well as of light. We now believe in fact, thatlight waves and electrical waves are identical.

77. Poynting's Theorem. Radiant Vector. The energyof the electric field is given by (183) as

We 1 filEF.Fdv,877

and of the magnetic field, similarly, by

Wm=8LffJ.According to Joule the energy due to a current of density q inthe electric field F is

Wj= fffq.Fdv.e

Let us find the variation of the sum of these three with thetime, assuming e and a not to vary; then

ae

aFz = 2 E =2F a137= 2 F (VxH - 4 it q),

at at at ata X.H =/I a H2=2 it aH.H = 2 H.aX = - 2 H. (VxF).*at at at at

* Heaviside introduces a fictitious magnetic current density qm inorder to produce a symmetry in the equations. The last parenthesiswould then read (pxF - 4 n qm).

VECTOR ANALYSIS. 165

So that the rate change of energy or the activity is

8 g,c f f f (F.VxH - dv.

By means of (130) and the divergence theorem this may bewritten

aW 47rfff V.(FxH)dv=4 f fOr, in other words, the rate loss of energy per unit volumemay be accounted for, by supposing a flux of energy throughthe bounding surface in unit time per unit surface of amount

R =FXH47r

(195)

where R, the energy flux, or radiant-vector, is by its form,a vector product, perpendicular both to F and to H.

78. Magnetic Field Due to a Current. It was provedexperimentally by Ampere that the magnetic scalar potentialCl at a point due to a current I whose circuit subtends at thepoint a solid angle w is proportional to the product of the cur-rent and the solid angle. If we so choose our unit of currentthat we may write,

11=Iw

we thus define a unit called the electro-magnetic unit of cur-rent. The magnetic intensity of the field is given, similarlyto F =-VV (106), by

H = - V11 = - IVw.

Its component in any direction h, is by the definition of direc-tional derivative (§ 49)

I1.hl= =-I

166 VECTOR ANALYSIS.

To find the variation, 8w, in the solid angle at a point dueto a small displacement 8h of the point, notice that it will bethe same that will take place, supposing the circuit to movea distance 8h in the opposite direction or - 8h, the pointremaining fixed. This motion will cause every element dr ofthe circuit to describe a small area dr x 8h whose component

FIG. 61A.

as a vector along r divided by r2 is the element of solid angled(dw) at the point, due to it. The total change in solid angleOw, due to the motion of the whole circuit, will be the integralof this expression around the contour, thus,

d8w=rl. rx8h=8h

7a 7a

whose integral around the circuit, after dividing by 8h, is

8hhlf drx0 r,

0

VECTOR ANALYSIS. 167

so that I

Lh

LO) =I h1. fVr xdr. (196)

0

Since this is true for any direction h we may write that theelement of magnetic intensity dH, at a point, due to theelement dr of the circuit, is

dH = 1 (ollxdr=-I r'xdr\ r r2 (197)

This expression is determined to any function of r prPs which,when integrated around the circuit, vanishes. So that tothis extent it is arbitrary. The equation shows that the forcedue to an element is perpendicular to the element and to theradius vector. The radius r being drawn from the point tothe circuit, and the current being positive in the direction ofdr, the order of the factors is taken so as to give the rightdirection to H. This is the familiar expression for the mag-netic intensity at 0 due to dr.

dH = I dr sin 0 1 to element and to r.

79. Mechanical Force on an Element of Circuit. Themagnetic intensity dH, above, is the force with which a unitpositive pole placed at the point would be acted upon in thefield due to the current I in dr. By the principle of equalaction and reaction the element of circuit would be actedupon by this amount but in the opposite direction. Theforce on the element dr due to unit pole at the origin is there-fore

dF = I drx !11 = I drx. (198)r

So that the force on an element of current is proportional tothe current strength I, to the length of the element dr, andto the strength of the field at the element 4. The factor of

168 VECTOR ANALYSIS.

proportionality is the sine of the angle between dr and + andthe force is at right angles to their plane.

Hence the force on the elementary current I'dr' in a fielddue to another elementary current I"dr" at 0 whose field atthe element dr' is by (197) and (198)

dr"Xr,rz

where the order of the vector product is reversed to take intoaccount the change of direction in r, or

d2F= .(I'dr')X(I"dr"Xr,) = I,I"dr'x(dr"xr,).r` r`

(199)

We may resolve this expression immediately into compo-nents along the radius vector r and along the element dr byan expansion of the triple vector product, giving

d2F = III dr'dr" [r, cos dr" cos (dr'r) . (200)

The X component

isLL

d2F,, = I,I"r2

dr'dr" cos (rx) cos (dr'dr")

- cos (dr"x) cos (dr'r)1, etc.

These are well-known results.80. Theorem on the Line Integral of the Normal Compo-

nent of a Vector Around a Closed Circuit. By means ofStokes' Theorem the following useful transformation analo-gous to it may be proved :

fqxdr = fp[(V-) n - V dS. (201)J

E)

q is any vector function of r, and n is the unit normal to theelement of surface dS. In this case the line integral of thenormal component of the vector q along a closed circuit istaken, instead of the more usual, tangential component.

VECTOR ANALYSIS. 169

The result is a vector one, and it is shown to be expressibleas the surface integral of a certain other vector quantityrelated to q, taken over any cap which is bounded by the

FIG. 62.

circuit. Let H be this line integral, and form its scalarproduct with the arbitrary constant vector c,

c. gxdr = f ffl J'J J J

This last expression being in the form of a tangential lineintegral may be transformed by Stokes' Theorem (136) into

ff [n V'c(cxq)] dS,ap

where the V differentiates q alone. Now

Vq (cxq) = c (V.q) - (c.V) q,so that

ff [c n (V.q)- q] dS.ap

170 VECTOR ANALYSIS.

But (117)n q.

The differentiation refers to q alone, as in the above integralthe properties of the region are independent of the surfaceconsidered, and we may substitute (q.n) for the secondterm in the integral, hence,

c f f [nVq.q - dS,JJcapand finally, since c is an arbitrary vector,

H = f f gxdr = f f f [nV.q - dS, (202)J J can`J

which sign to take, depends upon the direction of integra-tion around the contour. This theorem is originally due toTait and to McAulay, who gave it in a much more generalform, including Stokes' and other theorems, as special cases.

81. Expression for the Field at any Point In Space Due toa Current. We may use this theorem to transform the inte-gral, giving the magnetic force H at any point in space dueto a current of electricity in a closed circuit.

If the magnetic potential is 0, I the current in the circuit,and w the solid angle subtended by the circuit at the point,we have, by definition,

H=-Vfl=- IVw =- IV ff=t dSap V.

the surface integral being taken over any surface with thecircuit for bounding edge, but not passing through the point.

Now V f f n'r, dS = - V ff dScap ap r

I dS.S.p ` r/

Employing the above theorem (202) and remembering that

1 =V2 1 = 0, by (148),r r

VECTOR ANALYSIS. 171

H=1 f f fv1xdrcap \ r/ 0 r

I rdrxv 1, (203)J r0

which result is in agreement with (197). See also (202).We may thus write

dH=Idrxvl+$,rin Cartesian

dHx= L I dy, (z - z1)- dz, (y - y,)#, etc.

where x1, y z1, are the coordinates of a point in the circuit,x, y, z those of the point P, so that the small magnetic fielddH due to an element dr is determined to a function 4) pressuch that when integrated around a closed circuit the resultvanishes. So to a certain extent this resolution of the fieldis artificial, and may or may not be the correct one, but inany case this as well as any other possible resolution willgive the correct value for H above when integrated arounda closed circuit ; we have but very scanty knowledge of thefields due to unclosed circuits.

82. Mutual Energy of Two Circuits. Inductance. Neu-mann's integral. Consider two circuits carrying currents I'and P. The mechanical force on one of them due to thefield of the other is the integral of equation (199) taken oncearound each circuit. Since

yr=-,we may write

f rF = I'I"l J dr'x(v rxdr")J, 0= I'I" f v 1 dr" (dr'.v 1) .f r

172 VECTOR, ANALYSIS.

Let one circuit be now displaced in any arbitrary manner,so that any point on it is moved a distance W. The workdone in this displacement is

I'1" ff S r).

FIG. 63.

Integrating the second term by parts and rememberingthat d 8r = 8 dr,

f(or'.dr") 1 dr/ r .l rthe integrated portion vanishes for a closed circuit, hence

III" r f)r/

'fl f2 rAs the assumed motion of the circuit is arbitrary we may

then find the force in any direction by finding the change in

I'I"f, (204)

VECTOR ANALYSIS. 173

This integral due to Neumann represents the mutual energyof the two circuits. When the currents I' and I" are eachunity, the integral gives the mutual-inductance of the cir-cuits. When taken twice around a single circuit it gives theself-Inductance of the circuit. It is sometimes called theElectro-Dynamic Potential, as by its variation we obtainthe electro-dynamic forces.

Vector Potential Due to a Current.

83. Mutual Energy of Two Systems of Conductors. If themagnetic force due to a current be denoted by H, a solenoidalvector, i.e., V.H = 0, we may write, by means of the theoremof Helmholtz (175),

H 4 LOX f f

f VrH dv.

But (188)VxH = 4 it q,

where q is the current density, so that substituting, we obtain

H = OxfJ 9 dv = VxQ,,rwhere

Q =J '. 4J.0 gr dv.

(205)

Q is called the potential due to the current distribution q, orthe vector-potential belonging to the magnetic force H. Theword potential is used because it is formed in a manneranalogous to the potential due to a scalar distribution ofmatter p. Notice also that the force vector H is obtainedfrom the vector Q in a manner analogous to the way theforce vector F is obtained from the scalar V, where

V=f f f Ir dv and F= VV.,hence the name vector-potential.

174 VECTOR ANALYSIS.

We may now transform the magnetic energy in terms ofthe field H,

Wm8 7r

III H2 dv 8 fffH H dv,

into

Wm8 7r

fffH.vxQdv. (206)

Integration Theorem. In general (130) we have, whereH and Q are any two vectors,

V.(HXQ) =

the minus sign belonging to the term in which the cyclicalorder has been changed. Integrating over all space andusing the divergence theorem (121), S being the boundingsurface,

fff V.(HxQ) dv =f f. (Q.VxH - dv

ffn.(HxQ) dS.

Substituting (206) in this equation and remembering thatthe surface integral vanishes at infinity, because there themagnetic force vanishes, and also that V -H = 4 7r q thereremains

Wm 1 ffJ Q.OXHdv87,

fffQ.qdv.

Now replacing Q by its value, we have finally

Wm 2 f f f fff9 dv dv'. (207)

This sextuple integral covers the whole of space twice.

VECTOR ANALYSIS. 175

Let the only portions of space having any current densitybe two closed circuits. The wire forming the circuits mayhave a small but finite cross-section.

Place q dv - I'dr' q'dv' = I"dr",

where I' and I" are called the currents in the two circuits,respectively. The integrals then reduce to a double lineintegral each integral to be taken once around each of thecircuits, so that

Wm= Lffdr!.dr"(208)

This expression is really identical with Neumann's Integral(204). The factor I is due to the fact that by the conven-tion in (207) the integrals cover each of the circuits twiceand hence would give twice the value of (204).

84. Mutual and Self-Energies of Two Circuits. Each inte-gral being taken over both circuits, (208) may be broken upinto four parts,

Wm =2z J

(' dr' dr" + I,IZffdr' dr"1 1

+ (' +12 ? J:J:dr;drF.Z

l

The second and third parts are evidently equal, so that wemay write for their sum

1112f1 2J r

where here, each integral is taken around its correspondingcircuit once.

176 VECTOR ANALYSIS.

If we call the integrals

ffdr'.dr",L3 = La ffdrF.drlP,r r

M12 =xx r 7

we may write for the magnetic energy of the field due to bothcurrents

W,. L1I32 + M121112 + I L2I22. (209)

The integrals L1 or L2 and M12 are called the self-inductancesand mutual-inductance of the circuits respectively.

EXERCISES AND PROBLEMS.

1. If the line integral of the forces in any field around a closedcontour is zero for any such contour, the forces in the field form aconservative system.

2. Show that the surface integral of a scalar point-function Vtaken over any closed surface is equal to the volume integral of itsgrad (VV) taken throughout the volume of that surface; that is,

f1s",

3. Show that the line integral of a scalar point-function, V,around a closed contour is equal to the surface integral of thevector product of the normal by its gradient taken over any capto the contour; that is, prove

J,V dr = r r, n-VV dS.Jcap

04. Using the divergence theorem, let q = r1, and prove that the

potential of a body may be represented by the surface integral

V= 2 Ja p dS - 2 JfJri.vdv.tea'

VECTOR ANALYSIS. 177

5. If Poisson's equation holds,

i.e., V2V=4irp=divF and if Vp=0show that the potential of a body in the last example becomes

V8lirff dS = 81-f f div F dS,

so that if the force at every point of S be known it is possible tocompute the potential.

6. By drawing a small cylindrical box enclosing a portion dSof a surface charged with a surface density of electricity a, andmaking the cylindrical sides everywhere parallel to the lines offorce, show that there is a change in the normal component of theflux of moment 4 ;ra.

7. The curl of the curl of a solenoidal vector such that the threefunctions which give the strengths of its components parallel to1, j and k satisfy Laplace's Equation, vanishes.

8. If the lines of a vector, F, are all parallel to a plane and thevector has the same value at all points in any line perpendicularto the plane, the vector is perpendicular to its curl,

i.e., 0.

9. Compare the results of the last problem with those of § 60.Can you devise any other functions, the lines of which are every-where perpendicular to its curl?

10. If the lines of a vector are circles parallel to the lj-plane withcenters on the k axis, and if the intensity of the vector is a functionf(r) of the distance from the k axis, a vector everywhere parallel to

the k axis, of intensity F(r), where f(r) dF is a vector poten-dr

tial-function of the original vector. Is the original vector solenoidal?

CHAPTER VII.

APPLICATIONS TO DYNAMICS, MECHANICS, ANDHYDRODYNAMICS.

Equ ions of Motion of a Rigid Body.

$6..,E(4uatiaps for Translation. D'Alembert's Principle,up which Lagrange founded the whole subject of analytical

"mechanics, may be written

(m d? - F).8r = 0, (210)dt'

where 8r is any possible arbitrary or virtual displacementcompatible with the constraints imposed upon the system,and where the E sums for all the particles.

In order to deduce the equations of motion of translationassume the virtual displacement to be the same for all pointsof the system, as this is the definition of pure translatorymotion. It then follows, since we may now take the 8r fromunder E sign, that

8r (m !f - Fl= 0,dtz

and since 8r is arbitrary, that

(m_F)=O(211)which are the ordinary equations:

y l x

xX0.dt

Y,(Mq!y- Y)= 0.dt2

See papers by Ziwet and Field in American Mathematical Month-ly, 1914, pp. 105-113 and by Rees same journal 1923, pp. 290-296 forinteresting vectorial treatments of kinematics and motion of rigid bodies.

178

VECTOR ANALYSIS. 179

Motion of Center of Mass. Let `r be the vector to thecenter of mass or centroid of the system; then, by the defi-nition of this point for which (20)

rIm=Zmr,we have by differentiation

&r d2rd2%m=dt2'

so that finally equation (211) may be written

9Zm=F, (212)

or, in words, the motion of translation of the centroid of asystem of bodies moves precisely as if all the forces of thesystem were applied to the total mass concentrated at thatpoint. This reduces the problem of the translatory motionof the system to that of the motion of a single point. Aninteresting example of this property is seen in the case of themotion of a shell which explodes while describing its pathin space. As the resultant of the actions and the reactionswhich are produced when the shell explodes is zero, the pathof the center of mass of the fragments is the identical parabolathe center of mass of the shell would have described hadit not exploded. In other words, the path of the centerof mass remains unchanged by the explosion. The center ofmass of a thrown stick describes a smooth parabola, as itwhirls through the air.

The kinetic energy of translation of the body is evidentlygiven by

T2 m (dt

dr\22

Mq2, (213)

where q is the velocity of the center of mass of the system,where M = Em is its total mass and because all points of thesystem have the same velocity.

180 VECTOR ANALYSIS.

86. Equations for Rotation. To deduce the equations ofmotion for rotation, let Sw be an elementary rotation, then

8r, = 8w-r,,

where 8r, is an arbitrary possible infinitesimal motion dueto the rotation of any particle m, of the system about someaxis, w. Substituting this in d'Alembert's equation, weobtain

2 m d? 8wxr = 2

and with obvious transformations

Em w rXdtr = E 8w rxF.

We shall now assume that the particles of the systemrotate about the same axis, so that 8w shall be the same forall the particles and may be divided out, and rememberingthat (34)

drxdr-0,dt dt

we obtain Emrx°2r= d 2mrxdr= rxF (214)de dt 4 dt

for the equation of motion of rotation of a system about anaxis. The motion of a rigid system is of course a specialcase of this. This equation expands into the familiar Car-tesian ones,

dtZr(ydt - z dt)= 2 (yZ -zY),

dt 2 m (z dt - x dt) _ (zX - xZ), (215)

T, Em(xd- !)= E(xY-yX),

about the three rectangular axes, by the ordinary rules.

Defining M - E rcF (216)

VECTOR ANALYSIS. 181

as the moment of the applied forces about the axis of rota-tion, co, and

H-_Z mrxdt (217)

as the moment of momentum about the same axis, the aboveequation (214) may be written

dtH=M, (218)

or even I=i = M,

or, in words, the rate of increase of angular momentum tabout the axis of rotation of a system is equal to themoment of the impressed forces about that same axis.

Kinetic Energy of Rotation. Moment of Inertia. * Thekinetic energy of rotation of the system rotating with angularvelocity co is (44)

T Em q' Em ((*-r)'.

If the system moves as a rigid body all of the w's are thesame, so that

T = j w2Em ((*,xr)2. (219)

But (o),-r)2 is the perpendicular squared from the point r tothe axis of rotation co (A P in Fig. 31), so that the expression

Em (o),-r)2

means that every elementary mass is to be multiplied bythe square of its distance from the axis of rotation and thattheir sum is to be taken. This quantity is called the Mo-ment of Inertia* of the system about the axis cal; it evidentlyvaries with the direction of co. We may then define themoment of inertia, I, about an axis co, by the equation

I. - Em (o),-r)2 = Alk.2, (220)

where kW, also defined by the above equation, is called theRadius of Gyration about the axis co. The radius of gyra-tion is, therefore, the distance from the axis of rotation at

* Or, better, Rotational Mass.t Same as Moment of Momentum.

182 VECTOR ANALYSIS.

which, if the total mass M of the system were placed, itsmoment of inertia would remain unchanged.

The total kinetic energy of a rigid system moving in anymanner may then be written

T= I M q2 + 11.62. (221)

M, the mass of the body, is an absolute constant,* but I., as

FIG. 64.

stated above, varies with the directionof the axis about which the systemrotates, and hence the treatment ofrotation is essentially more complicatedthan that of pure translation.

We shall treat of the motion ofrotation more in detail, not only for itsintrinsic interest, but also because itintroduces naturally the Linear Vector-Function and some of its elementaryproperties.

87. Linear Vector-Function. Instan-taneous Axis. Consider a rigid bodyof mass M rotating in any mannerabout a fixed point. This precludesany translatory motion of the bodywhich is now one of pure rotation atany instant about some axis necessarilypassing through this fixed point. This

* In the Electron Theory of Matter the inertia of a particle, at leastin part, is accounted for by the electrical charge which we know theparticle carries. The resistance of an electrical charge to accelerationis not constant but is a function of the velocity, and theoretically becomesinfinite when its velocity approaches that of light. The apparent inertiaof such a particle is therefore not constant. But for any velocities withwhich we are likely to deal in mechanical systems these variations ininertia are inappreciable. The ordinary equations of mechanics are thenfirst approximations only, but for ordinary velocities, up to 10,000 km.per sec. say, are extremely close to the truth.

VECTOR ANALYSIS. 183

axis, which may vary continuously in direction, is called theInstantaneous Axis of rotation. Let the angular velocityabout this axis at any instant be represented by a vectorof length w = f(t) in the direction of it and in the conven-tional sense, i.e., that of the motion of progression and direc-tion of rotation of a corkscrew.

As the velocity of any point r is

q dtdr _

-wxr,

we may write for the moment of momentum H

H =4 mr`dt

=J m rx((*xr) =y m(w r2 - r (222)

thus H is a vector-function linear in co, 4w say.This particular function 4w has a number of important

properties which are evident upon inspection. If r and ware any two vectors, the following equations hold:

(a)

a = const. 4a'r = a+ r, (b)

d(4T)$dT, (C)

and 0 = O44 T. (d)

(223)

In particular when a linear vector-function has the prop-erty represented by (223) (d) it is said to be Self-Conjugate.

Form the scalar product of H and w. .

w = Em 0. (rc ((oxr) )

= Em (o)-r) (wxr) = Em (wxr)2

= w2I., = 2 T by (221). (224)

184 VECTOR ANALYSIS.

Now, since +co is linear in w, the scalar product (o.+0 is aquadratic scalar function of co, and hence represents whenequated to a constant, a quadric surface.

88. Motion under No Forces. Invariable Plane. Assumingno applied forces,

dt H=0,

so that H = const. vector,

or, in words, under no applied forces the moment of momen-tum of a rigid system remains constant in magnitude anddirection. It remains perpendicular to the plane, calledInvariable Plane, whose equation is

const.

Also, since the energy (kinetic) of the system is conserved, itfollows that the moment of inertia I. about any direction wis inversely proportional to the square of the radius vector inthe quadric const., because

2 T = w2I.so that

w2(225)

This equation also says that with a given amount ofenergy the body rotates the faster the smaller I. is; i.e.,

1woe Iw

Poinsot Ellipsoid. Since evidently no finite body has aninfinite or a zero moment of inertia about any axis, thequadric surface const. must be one the radius vector ofwhich has a finite minimum and maximum value; that is, itmust be an ellipsoid. This ellipsoid is called the Momentalor Poinsot Ellipsoid. Let us consider this surface more indetail and incidentally show its expansion in Cartesian form.

I,=2T=const.

VECTOR ANALYSIS. 1$5

If H1, H2i H3 are the components of H about the three axesi, j, and k, then (222)

H-+w = I:m(w r2 - r Hli + H2j + H3k

= {Emw1(x2+y2+z2) - Emx(W1x+w2y+w3z)} i+ {Emw2(x2+y2+z2)- Emy(w1x+w2y+w3z)} j+ {Emw3(x2+y2+z2) - Emz(w1x+w2y+w3z)} k

{W1Em(y2+z2) -w2Emxy -w3Emxz} i+ {-w1Emyx +w2Em(z2+x2)-w3Emyz} j (226)

+ {-w1Emxz -w2Em yz +W3Em(x2+y2)} k.

Moments and Products of Inertia. Coordinates of a Self-conjugate Linear Vector-Function. The scalar coefficientswhich occur in the above expansion and are reprinted below,assuming for definiteness A > B > C,

A Em(y2+z2), D Emyz,B = Em(z2+x2), E = Emzx, (227)

C = Em(x2+y2), F = Emxy,

are called the Moments of Inertia about the axes x, y, and z,and Products of Inertia with respect to the planes of yz, zx,and xy, respectively. The quantities A, B, and C are essen-tially positive, but D, E, and F may be of either sign. Theymay be found in any particular case by integration, forinstance, since

and

dm = p dx dy dz,

A -fff(?+pdxdYdzBody

D-fffyz pdxdydz,Body

(228)

where p is the density at any point of the body.

186 VECTOR ANALYSIS.

Knowing these six coefficients (A, B, C, D, E, F), thefunction 4w is completely determined, and for this reasonthey are sometimes called the coordinates of the self-conju-gate linear vector-function.

Consider now the ellipsoid (224)

w+) = 2 T = const.w12A - w1 w2 F - w1w3E

-40201F + w22 B - w2w3D (229)- w3w1E - ws o2D+ w3 C,

which may also be written

w.40) = w12 A + w22 B + wag C - 2 w2w3 D - 2 w3w1 E - 2w1w2 F.

It may be easily seen that in order that a .4T, it isnecessary that the coefficients D, E, F should occur in pairs asabove. If however they do not occur in pairs, there willbe nine coefficients in (229) all different. In this case thefunction is not said to be self-conjugate; but it is still alinear vector-function. In this last case T40 is not equalto Q4T. If we write o .4'T, c)' is said to be the con-jugate of 4, and 4) the conjugate of c)'.

Principal Moments of Inertia. Principal Axes. The f unc-tion as is seen by its expansion, is homogeneous in w,and the ellipsoid it represents when equated to a positive con-stant, 2 T, is referred to an origin at its center. We maynow refer the ellipsoid to its three principal axes, its equationthen becoming, as is well known,

qAw12 + Bw22 + C(O32 = const. (230)

The axes have lengths proportional to 11 , and_ C

The coefficients A, B, and C, in the above equation calledthe principal moments of inertia, are the moments of inertiaof the body about these three principal axes and in generaldiffer from the values they had before, but they are defined in

VECTOR ANALYSIS. 187

the same manner with respect to the new axes. The prod-ucts of inertia have all vanished. There are thus threedirections in any rigid body for which the products of inertiawhen referred to them vanish.

Referred to these axes, since0).4 w = Aw12 + Bw22 + Cw32,

4 w then becomes 4 (o = Aw1i + BwJ + ?w3k.

and H= Aw1i + Bw2 j + Cw3k. (231)

So that the components of H are

= Aw1fH.Hb = BW2,

H=

Looking upon 4 ( ) as an operator, we see from equation(231) that when it is applied to any vector w

w = w1i + w2j + w3k,

thus 4 w = Aw1i + Bw2j + Cw3k,

it multiplies the components of w by the quantities A, B, andC, respectively. Applying 4 again to 4 (w) we should obtainfrom

40 =Aw1i+Bw2j+Cw3k,

44w- +20) = AZw1i + BZw2j + C2w3k, (232)

and so on. Defining 4-1 as that operator which when appliedto 4 annuls its effect, 4-1 must then evidently divide the com-ponents of any vector by and C, respectively, so that

+-'4x *= w = 4-' { Aw1i + Bw2j + Cw3k}

= AAw1i +BBw2j+1 V..k

=w ii +w2j + w3k,

so that 4'1w =' i + 2 j + `-'3 k.A B 0

(233)

188 VECTOR ANALYSIS.

Applying 4-' to this we obtain

-'AZ

i + B j + k.

and so on.Lemma. We shall now show that fw is perpendicular from

the origin to the tangent plane at w, and that its magnitudeis inversely proportional to the distance from the origin tothis tangent plane, or, in other words, that ($w)-' * is theperpendicular vector from the origin to the tangent plane, atw of the quadric

Consider the quadrico o= 1.

)4w = const.Differentiate this function, considering w as a variable

dw4w 0,

using c and d of equations (223).

FIG. 65.

Hence 4w is perpendicular to dw. But do) is a small vector inthe surface at the extremity of w and therefore lies in thetangent plane, so that +w is perpendicular to this plane. Ifa is the running coordinate of the plane its equation is (64)

const.,* Do not confound the reciprocal operator, and [+( )1-i,

the reciprocal of +.

VECTOR ANALYSIS. 18$

and by (68a) the perpendicular vector from the origin to thisplane is

const.P = w = const. (+w)-'.

(234)If the constant is unity, p = (4(o)-1.

Physically this means, referring to (231), that the angularmomentum H = +w is normal to the tangent plane at w andinversely proportional to the length of the perpendicular fromthe origin on it.

Axes of a Central Quadric. The principal axes of acentral quadric may be defined as those directions forwhich the magnitude of the radius vector is a maximum ora minimum. That is,

w2

is to be a max. or min. subject to the condition thatconst.

Multiplying the first by an arbitrary multiplier and add-ing, the condition is obtained by writing the derivative of

A(o) = const.to zero; this is

Aw) + co.(4dw- Ado)) = 0.

This becomes, using (223) (c), and (d),A(o) = 0

and since this must be true, independently as to how covaries, i.e., it is to be a true maximum or minimum, itfollows that 4w = Aw

is the condition required.This is already an interesting result, for it states (234)

that the radius-vector being parallel to +w, is thereforefor those directions, perpendicular to the surface. We mighthave started with this condition as a definition of principalaxes.

190 VECTOR ANALYSIS.

This last equation is by (226) and (227) equivalent to(A-A)w, -

- F w, + (BF w2 -

- A) w2 -Ewa=0,D wa = 0, (a)

-Ew, - Dw2 +(C-A)w3=0.The condition that these equations shall be compatible

for values of w1, w2, w3, other than zero, is that the deter-minant of the coefficients shall vanish, i.e.,

A-A -F -E-F B--A -D 0. (b)

-E -D ('--This is a cubic in 2, and may be shown to always have

three real roots. Each of these three values for A insertedinto equations (a) will allow for their solution, obtainingfrom each of them values for w1, w2 and w3 and hence a

Cl) =W I + w2 J + w3 kdirection for each 2.

There are then always at least three principal axes to acentral quadric surface.

The Principal Axes Intersect Normally. Let A 4, A3 bethe roots of the determinantal cubic (b), and w,, w2, w3 thecorresponding axes.

Then wl = Awland +w2 = 4(02-

Multiplying the first by and the second by (o,- andsubstracting, there results, using (223) (d),

(Al - 22) 0,

which means that if A, is not equal to X12, then the twoprincipal axes w, and W2 are perpendicular to each other;similarly if A 74 A3 and 23 =/ 21we could show that the three principal axes are mutuallyperpendicular to each other.

If two roots of the cubic are equal, the position of thecorresponding axes becomes indeterminate, and it may beshown that all radii perpendicular to the direction given by

VECTOR ANALYSIS. 191

the third root are principal axes of the same length. Thesurface is then one of revolution about the determinateaxis. If all three roots are equal, the surface is a sphere,and any axis is a principal axis. It may also be shownthat the three roots of the cubic are equal to the squares ofthe reciprocals of the lengths of the semi-axes, the Cartesianequation then being

AIW12 + A24 + A3W32 = const.Comparing with (230) we see that the roots of the deter-minantal cubic are proportional to the principal momentsof inertia,

A B C

89. Geometrical Representation of the Motion. InvariablePlane. If no impressed forces act upon the rotating bodythe equation of motion (218) becomes

dH=O,dt

the solution of which isH = w = const. vector, (235)

hence H or 4w is a vector constant in magnitude and directionthroughout the motion, so that the tangent plane of theellipsoid to which it is always perpendicular must remainfixed in space and is for this reason called the InvariablePlane. The point where this plane is touched by the ellip-soid is on the extremity of the instantaneous axis or pole, sothat the ellipsoid is always rolling without sliding on thisplane. In other words, having constructed the ellipsoid ofinertia, and having determined the position of the invariabletangent plane in space, the motion of the body is the same asif it were rigidly attached to this ellipsoid which is rollingwithout sliding on the invariable plane.

This geometrical condition, in addition to the fact thatthe angular velocity of rotation is proportional to the radiusvector to the point of contact of the ellipsoid and plane,completely determines the motion.

192 VECTOR ANALYSIS.

It is easy to see, since the radius vector always passesthrough a fixed point, the origin, that it must describe a conein space, the vector H being its axis fixed in space. Forthis reason, H, is called the Invariable Line.

It also describes a cone in the ellipsoid. The vector Hdescribes a cone in the ellipsoid because the ellipsoid movesrelatively to it. This description of the motion is due toPoinsot.

FIG. 66.

90. Polhode and Herpolhode Curves. If the paths de-scribed by the point of contact of the ellipsoid and inva-riable plane be determined on them, for instance by placingcarbon paper between them as they roll on each other, twocurves are obtained: one on the invariable plane, called theHerpolhode (sinuous path),* and one on the surface of theellipsoid, called the Polhode (path of the pole). These curvesare the directing curves of the cones described by the radiusvector w in space and in the ellipsoid, called respectivelyHerpolhode Cone and Polhode Cone.

Permanent Ages. It is easy to see that in three cases H andw coincide in direction, i.e., when co is perpendicular to thetangent plane; in this case when both H and w coincide in

* The above name is a misconception, because as a matter of fact theHerpolhode can be proved to have no point of inflection, and hence isAot " SZUUQUS,"

VECTOR ANALYSIS. 193

direction along some one of the three principal axes of themomental ellipsoid these curves reduce to points and theellipsoid rotates without rolling, permanently about theseaxes. There are then at least three directions at every pointin a body about which if the body be set rotating it willcontinue to do so forever. Further consideration shows thattwo of these permanent axes are stable and one unstable, thislast being the mean axis. The most stable is the least axis.

Equations of Polhode and Herpolhode Curves. The inter-section of the cones described in the ellipsoid by theinstantaneous axis with its surface will determine the pol-hode curves. In the quadric

w must always satisfy the condition that the distance, p, tothe tangent plane is constant, or that

2

(4-1)) = p2 = const.,

or (4w)2 = p2 = (a quadric),

where we define +2( ) _ 4 (4 ( )) etc., see (232),so that the two equations

const. = kand

(236)1

Pmust be simultaneously satisfied. Combining them, weobtain from

PZo)-+o) = p2 and k ps

by subtraction

or finally

(0.+0) = 0,

0)4 (k+ - p21 to = 0I (237)

194 VECTOR ANALYSIS.

a homogeneous equation of the second degree, and hence acone with vertex at the origin. Its equation in Cartesiancoordinates may be immediately written down by (231) and(232) as

wl2 CAzk - P) + w22 (jj2k - ) + w32 (02k - = 0,

or \(,fk/ (238)

A - -\)w12+B1 Bk - ) 2+V (k - )w32= 0.9 p2 P;

The intersections of this `cone for different values of p with theellipsoid k,give the polhode curves, which are, therefore, twisted curvesof the fourth degree, lying on the momental or Poinsotellipsoid.

Since the herpolhode is traced out by the points of contactof an ellipsoid rotating on its center with an invariable tan-gent plane, these curves must lie between two concentriccircles on the plane, their centers being at the intersection ofthe invariable line H with that plane, and touching themalternately.

Moving Axes and Relative Motion.

91. Theorem of Coriolis. It is often convenient in dynam-ics to use axes which themselves move in space and to whichthe motions of the body under consideration are referred.

In order to determine at any time the position of the mov-ing axes, one method is to refer them to axes which remain atrest throughout the motion. According to this device thefixed axes are left behind by the moving ones. However, itis found to be more advantageous to refer the moving axes atall times to fixed axes instantaneously coinciding with them.

No generality is lost by referring the motion of a body tomoving axes which simply turn about a fixed point in space,as any motion of translation of the moving axes with refer-ence to fixed ones may be compensated for by giving to

VECTOR ANALYSIS. 195

every point of the body considered a motion equal and oppo-site to that of these moving axes. This condition, then, doesnot limit the generality of the choice of moving axes.

Consider any vector OP= r drawn from a fixed origin 0and for definiteness let it be the vector to a point P in amoving body. We shall now consider the motion of the

I

FIG. 67.

point P in two ways. Refer it to two different spaces ini-tially coincident, one revolving about the axis 01 with anangular velocity w, the other remaining at rest (or fixed).

Let PR be the motion seen in a time dt by an observerremaining in the fixed space, or, in other words, the absolute

motion in space. Denote this vector PR as drf8, the sub-script denoting its reference to fixed space. Consider nowthe point P as remaining at rest with reference to the mov-ing space; it will therefore move relatively to fixed spacewith the velocity of the moving space alone and will describethe path

PQ = wdt xr = wxr dt,

as wdt is the angle described by NP in time dt.

196 VECTOR ANALYSIS.

But as the particle P by its own motion actually reachesthe point R, the vector QR must represent the path of theparticle as seen by an observer moving with the moving space;in other words, QR = dr, the subscript denoting its refer-ence to moving space. From the figure

dr18= drfl8 + wxr dt.

Dividing through by dt we obtain the very important equa-tion

(239)

Letting OP = r represent any directed quantity such asforce, velocity, moment of a couple, or angular momentum,etc., equation (239) shows how to refer them to a movingspace.

The vector r always represents the vector at the beginningof the motion and referred to either space, as initially they areboth coincident. If r represent a displacement and q thevelocity of the point P,

v8 . (dt )+ wxr.

The acceleration of a body whose motion is known rela-tively to moving space, and the motion of moving spaceknown relatively to fixed space, may be obtained by a secondapplication of this equation.

Replace r by q18 thus,

at8 (Ls)= + wxgjB

\d at(wxr) +.. (!)

d t , ,,8 + Ox((Oxr)

(d.r)9na

+2x() + dw xr + wx(wxr). (240)

VECTOR ANALYSIS. 197

This equation will readily expand to the familiar onesreferred to Cartesian axes, as ordinarily given, for example:

(ai8)(X)M+ 2(o'2dt-o'$dt)+(z d -y d )-x(w32 w32)

+w1w2Z+w1w2y. (241)

The last three terms may be written out as above byremembering that

wx((axr) = r w2.

If the point P is attached to the moving space,

Id2r) = 0, and also 2ms

wxdr) = 0,dt2 dt ms

so that the remaining expressions are the accelerations pro-duced by the motion of the moving space itself.

Hence the so-called

Acceleration of moving space = dwx r+wx((oxr). (242)

If the angular velocity of moving space is constant,

dw = 0 and thereforedt

xr = 0.

The remaining term wx((oxr) is the acceleration produced onthe body by its individual rotation about the axis 01.Since wxr is perpendicular to w, wx ((*xr) is perpendicular bothto wxr and to co and directed normally towards the axis co.This is the ordinary Centripetal Acceleration. Because in thevector product wxr, NP instead of r may be used (see § 36)without changing its value, and as co and NP are at rightangles,

(*.NP = 0,

this centripetal acceleration may then be written

wx(wxr) = -w2LY `.

198 VECTOR ANALYSIS.

The term 2 wx (t) is called the Compound Centripetal\ /.Acceleration of Coriolis.

We may then consider the moving axes to be at rest if tothe actual forces applied to the body fictitious ones be addedcapable of producing accelerations equal and opposite to theacceleration of moving space and to the compound centripetalacceleration. This is the theorem of Coriolis.

92. Transformation of the Equation of Motion. Centri-fugal Couple. Let us utilize equation (239) to refer themotion of a rigid body to a space moving with it, or, in otherwords, to axes in the body. The equation of motion of arigid body about a point referred to fixed space is (218)

dH = M.dt

If there are no impressed forces, M = 0 anddH=0dt

states that H, the moment of momentum, remains constantin magnitude and direction in the fixed space, howeverpeculiar the motion of the body may seem to be.

Employing now the equation of Coriolis, we substitute for

(dH)18 its equivalent for moving space, obtaining as the equa-

tion of motion of a rigid body about a fixed point referred toa space moving with the body

(c!i)+ wxH = M. (243)

If there are no applied forces, M = 0, and (243) becomesCH

-Co.t), = H . (244)

Hxw is called the Centrifugal Couple and, as is evident byits form, is perpendicular to both co and H. The above equa-

VECTOR ANALYSIS. 199

tion then states that the rate of change of the angular momen-tum H in the body is equal to the centrifugal couple, Hxw. Ifthe change in H is always normal to itself, then H is neverincreased or decreased in length but only changes in direc-tion at a rate proportional to H -to. Thus H describes a conein the body, although it remains fixed in fixed space.

If H and a) ever become parallel then Hxw vanishes, and inthe body too we have

(dH = 0,

or, in other.words, the body must continue to rotate foreverabout the Invariable Line H in fixed space; it is then aninvariable line in the body also. We have seen that thereare at least three such directions, called permanent axes, orprincipal axes, for which the above condition is fulfilled. Asymmetrical body supported at its center of mass and rotat-ing about its axis of symmetry will give this kind of motion.

Gyroscope. The property that a rotating body possessesof rotating permanently about a principal axis was utilizedby Foucault in the gyroscope. When a symmetrical top israpidly spinning in gimbals, it keeps its axis pointing in thesame direction (invariable line) in space, so that if the top iscarried around by the earth's motion, the axis of the topremaining fixed in fixed space will describe a cone with refer-ence to the earth (or moving space). By observations onsuch an instrument not only can the rotation of the earthbe proved but the latitude of the locality at which the experi-ment is performed may be determined.

93. Euler's Equations. If cvl, w27 w$ be the three compo-nents of w along the principal axes of a rigid body at a point,and if A, B, U be the principal moments of inertia aboutthose same axes, we may write (231)

H=1W11+Bw2j+Cw-,k_ ew.

200 VECTOR ANALYSIS.

There are no products of inertia entering into this equation,as for the principal axes they vanish (§ 89).

Substituting this value for H in (243) there result thethree equations

4d+ (Z9-B)W2W3=M1,

B d + (A - V)w3W1 = M2, (245)

+(B-A)w1W2=M3.dt

N.

These are the dynamical equations of Euler for the motion ofa rigid body about a fixed point, referred to axes moving withthe body. Of course

dH+wxN=M

is the corresponding vector equation.

94. Analytical Solution of Euler's Equations for Motionunder No Impressed Forces. For convenience we rewritethe following:

w = W1i + W2j + 403k. (a)

Then

H = Aw1i + BW2j + G'w3k = +w.

+-I to = W1 i + -`'2 j + W3 k.

(b)

(c)A B C

Alsod ( ) = 0 d (). (d)

v+w = 0).00 % (e)

Euler's equation for this case is

da = H-0) or d-d = 0wxw. (246)

VECTOR ANALYSIS. 201

Since the right-hand side is perpendicular to 4)w and to w,let us take the scalar products of the equation with 4w andwith to respectively. Multiplying with w. we obtain

(OAL(O = 0.

which we must integrate.By differentiating const. with respect to the time thus,

w.dt+w + 0,

which becomes by (e) and (d)

4w + w'dt 4)w = 20.dt

4w = 0

we see that the integral of the equation is

2 T, - (247)

where T is an arbitrary constant. Those having read thepreceding pages will recognize in (247) the equation of thePoinsot Ellipsoid, and in T the kinetic energy of the body.

Multiplying the. original equation with +co we have imme-diately

w = 0,

and by (e) and (d)

whose integral isI 0,

(+0))2 = H2, (248)

where H2 is an arbitrary constant. We recognize here theconstancy of H in magnitude only in the body, hence anychange in H must be perpendicular to it. See § 92.

In order to obtain a third integral, multiply (246) byr-lw, the 4-' and the 4) annulling each other.

4_lw.ddt = w.Lo) _ 4-'w.40Xw. (249)

202 VECTOR ANALYSIS.

From these three equations (247), (248), (249) it is possible tofind w for all time and hence all about the motion. Supposeit is desired to expand these three integrals into Cartesianform; we have immediately for (247) and (248)

o)+o = 2 T = T4w12 + Bw22 + ZS'w32 (250)

and (+w)2 = H2 = a2w12+ B2ct122 + j7 crly2. (251)

The third is more complicated, but easy,

dt -+

_\al+Bj

+ kl.i j yk

.w1 Hw2 Cw3

w1 w2 w3

B+A-B www. (252)

A C 1 2 3

By solving equations (250) and (251) combined with

W2 = W12 + W22 + W32,

for wl, WZ and (03, an_%/ dsubstituting in (252), we find

W at = (X1 - W2) (X2 - W2) (X3 - W2),

where the X's are functions of A, H, C, T and H, an equa-tion for w in terms of these constants whose general solutioninvolves elliptic functions.*

95. Hamilton's Principle. Starting again with d'Alem-bert's equation,

`

// 2

Z(m dt2 - F). r = 0, (253)

the Sr's being any variations consistent with the constraintsimposed upon the system, or, what is the same thing, satisfy

* See article by Professor Greenhill, in fourteenth volume of theQuarterly Journal, pp. 182 and 265, 1876.

VECTOR ANALYSIS. 208

certain equations of condition, we may transform it in thefollowing manner:

dzr' d dr'8r) - dr d8rdt28r

__

Q tdt dt ' dtZ

_ d (dt'8r) - 82

(dt) .

Treat each term of (253) in this way; d'Alembert's equationthen becomes

) +gEm(r) = 82 4m(dtz

= 8T + Iwhere T is the kinetic energy due to the velocities of themasses of the system. As the first term is an exact deriva-tive, let us integrate with respect to the time from t = tl tot =tz,

y (ST+dt / t

If the positions of the system are given at the times tl and tthen the 8r's are zero for those times, and the left-hand termsvanish, leaving

ft,t2(3T + (254)

This equation is true whatever the system of forces is thatacts; if, however, the system is a conservative one, the workdone (by definition) in going from any point to any otherpoint against these forces is independent of the path chosen,and is therefore a function solely of the initial and final pointsof the path. In this case, then, there must be a scalar point-function, (- W say), such that knowing its value everywherewe can calculate the work done in going from any point toany other by any path, simply by knowing the values of - W,for those points. (Consult § 59.) In other words, the work

204 VECTOR ANALYSIS.

in going from position 1 to position 2 is the differ-ence in value of the function - W at positions 1 and 2, or

E, 8r = -(W, - W2) W.

In the case of conservative forces, then, 2; may bereplaced by -8W and (254) becomes

af(T - W) dt = 0. (255)

a,

This is Hamilton's Principle.Lagranglan Function. The function T - W is called the

Lagrangian Function, and is often written L, so that Hamil-ton's integral becomes

8ft2L dt = 0. (256)

96. Extension of the Conception of Vector to More thanThree Dimensions. Certain processes occur in mathematicalphysics in which more than three independent variables areconcerned. In such cases as this the vector notation is stillapplicable to the manipulation of these quantities. Ifq1, q2, q3 . . . be these independent quantities, we conceive ofa vector q,

q = g1i1 + g2i2 + g3i3 +

where i,, i2f i3 . . . are independent unit vectors. By anextension of the idea of a vector, we are to consider q as avector existing in more than three dimensions, as many asthere are q's. An example will show that we are not goingvery far beyond the manipulation of ordinary vectors.

Definitions. Consider the generalized vector in n-dimen-sions,

q = g1i1 + g2i2 + ... +gnin. (257)

In analogy to V a i+ a j+ a kax ay az

write V, = a i, + a i2 + ... + a in. (258)aq, aq2 aqn

VECTOR ANALYSIS. 205

If we write q,' for , we may have also

q' = q1'i, + g2'i2 + . . + gn'in (259)

and define also V,,'

age aq2 agnin. (260)

Let us apply this notation to the following transformation.97. Lagrange's General Equations of Motion. By means

of Hamilton's Integral we may deduce Lagrange's Equations.Let the position r8 of any point s of the dynamical (or analo-gous) system be expressible in terms of the independentparameters, q1, q2, . . . qn, so that

r8 = 48 (q, q2 q3 ... qn). (261)

This means that to every point

q = q1i, + q2i2 + + gnin

in a space of n-dimensions there corresponds a definite valuefor all of the parameters of the dynamical system and hencea definite and determinable configuration of every particlein the system. If it is possible to find what functions ofthe time the q8's are, subject to the dynamical equationsof condition, included in Hamilton's principle, the positionof every particle for all instants will be known. Hence thesimple motion of one point in n-dimensions includes theproblem of the motion of a system of points in three orless dimensions (or for that matter in more than three).

Differentiating (261),

+ = d 8- 03 q1' + a4 q2'+ ... = q'.On4B. (262)

Thus every 4' is expressible as a linear function of the q"s.

206 VECTOR ANALYSIS.

The kinetic energy function T becomes

(263)T=2 m(d122

every term of the sum being linear and homogeneous in q'.The square is a homogeneous quadratic function of the q"s;often written

T = I Q11q'12 + I Q22q'22 + ... + Q12gi'g2' +... (264)

Performing the variation with respect to q' and q indicatedby Hamilton's integral (255), and using L = T - W, we havesince

But because

8L = 'L,fi2 'L) dt = 0.

8q'8 dtq dt

8%

the second term becomes

J't2 [(8q).Vn L] dt',

which may be integrated by parts, with respect to the time,into

It, -ft2aclq V'L't

of which the first term vanishes, as q is fixed at the limits andhence suffers no variation there, leaving altogether

dt Vn'L) dt = 0.

But as 8q is arbitrary, it follows that

at VnL - VnL = 0. (265)

VECTOR ANALYSIS. 207

This is a vector in n-dimensional space, whose componentsmust all be zero, for example:

d a(T-W)\ a(T-W)dt C aql' / aq,d /a(T-W)1-a(T-W)=0, etc.a ` aq2' f aq2

(266)

These are Lagrange's Generalized Equations of Motion.If then for any system the functions T and W are known,

and it is possible to express them in terms of n-independentparameters q1, q2, ... qn, these n-equations (266) make itpossible to determine the values of these parameters for alltime; thus the path of the point

q =gill+gxi2+...gnin

is determined, in n-dimensional space.Defining the operator v ( ),

V ( ) = Cat vn., - vn) ( ),

all dynamics is included under the Remarkable Formula

VL=0. (267)

Hydrodynamics.

98. Fundamental Equations. We shall now derive thefundamental equations of hydrodynamics for a frictionlessfluid and some of their most important consequences bymeans of the previous principles. The directness of attackand absence of artificiality is especially noticeable in thisapplication of the vector method.

Let p represent the density of the fluid; we shall assumethat it is a function of the pressure p alone, so that

P = f(p). (268)

Equation of Continuity. Let q (u v w) be the velocity of thefluid and F(X YZ) be the force per unit mass acting on the

208 VECTOR ANALYSIS.

fluid. Consider a fixed surface S in the fluid. By fixed ismeant that the imagined surface retains its position in spaceirrespective of the motion of the fluid itself. The rate ofincrease of matter in it is measured by the surface integralof the flux of the fluid though the surface taken along theinward drawn normal. No fluid is supposed to be creatednor destroyed inside of the surface. As we use generally theoutward drawn normal n in our formulae, - n will representthe inward drawn normal, so that we may then write usingthe divergence theorem

as

-J Js p q.n dS = -ffJ V.pq dv

UfLpdv=because in = fffPdv.

As this relation holds whatever surface is taken we mayequate the integrands to each other,

at + V pq = 0. (269)

In Cartesian this is

atp+axpu+a pv+azpw=0.y

This equation is called the equation of continuity. It statesthat matter is neither created nor destroyed at any point inthe fluid.

It is convenient here to employ a special notation to beused when we follow the fluid in its motion as distinct fromconsidering the fluid as it passes by a fixed region in space.For example, the rate change of density of a definite portion ofthe fluid as it is followed in its motion, symbolized by the

VECTOR ANALYSIS. 209

special notation

DE

, is equal to the rate of change of its den-

sity

atobserved as it goes by a fixed point in space plus the

rate of change due to its velocity q, so that

DP =Dt a + (270)

In fact whatever point-function is placed in the parenthesis

Dt( ) at(This corresponds in the Cartesian notation to

Dtat+axu+aay + azw

p+ap dx+ap dy+ap dzat ax dt ay dt az dt

We may write the equation of continuity in a slightly differ-ent form by (128) and using (270),

at + V.(pq) = at + pV.q + q.Vp

so that 2° + pV.q = 0.

In either form, if the fluid is incompressible, p does not varyeither with time or with position, and hence

vq = 0 = div q, (272)

which shows that q is then a solenoidal vector and its streamlines form closed curves or end at infinity, just as a solenoidaldistribution of electrical force acts. In fact the two theoriesof Hydrodynamics of incompressible fluids and of Electricityare identical.

Consider now a small surface always containing the samefluid of volume v. This surface may be distorted as itmoves with the liquid, but it is supposed to be always

210 VECTOR ANALYSIS.

made up of the same small portions of the fluid with whichit started. For such a surface, evidently the mass isconstant, or

Dm=0=P=P Dv+v Do,

Dt Dt Dt Dt

PV -2so that Dt = - Dt = V.q by (271). (273)

V P

We may then interpret V.q as the fractional decrease ofdensity per unit of time, or as the rate of increase of volumeper unit volume, or as the time rate of dilatation, a divergence.Equation (272) follows also from (273) if p is constant.

Euler's Equations of Motion of a Fluid. Consider theforces acting upon a definite mass of the fluid enclosed inthe surface S.

Let F per unit mass or p F per unit volume be the externalforce function, and let p be the pressure function. actingnormally over the enclosing surface and along the inwardlydrawn normal. By Newton's law the rate of increase ofmomentum (Zp q dv) of the fluid is equal to the appliedforces F acting directly on the mass of the fluid and to theforces (Zp dS) resulting from the pressures acting on thesurrounding surface, or

pgdv=J pFdv+ f fpndS,Vol

or fffDt (p q dv) =fff(pF - Vp) dv.

D Dq D(p dv)Now Dt (p q dv) = Dt

p dv -{- q Dtand the last term vanishes as the mass remains constantthroughout the motion, so that the integral becomes

fff( p) dU f f f(F- Vp) dv. (274)

* See note to §52, p./252, for transformation of last term by diver-gence theorem.

VECTOR ANALYSIS. 211

As this is true for any volume whatever, the integrandsare equal and we have, using (270),

P Dq = p aq + p pF - Vp. (275)

This is Euler's equation of motion which, in connectionwith (268), (269) and (270), forms the basis of theoreticalhydrodynamics.

99. Transformation of the Equation of Motion. If wedivide (275) by p and employ the identity (129)

V1(g1.q) - gx(V1xgi),the subscripts indicating precisely on what the V acts, or

a Vq2 - gx(Vxq),we may transform this equation into

at - qx(Vxq) = F - Op - 2Vg2. (276)

If the externally applied forces have a potential, V, forinstance forces due to gravity or any other conservativesystem of forces, then F = -7V.

If the pressure p at any point depends only upon thedensity p, we may define a quantity P such that

p=VP,orP= fddp,P P

so that our equation becomes

a9 - qx curl q = -V (V + P+ 2q2/

(277)

= V U,

where U = - (V + P + 4 q2).Referring back to equation (126) where it was shown that

curl q = 2 w,

where w is the vorticity or angular velocity of rotation of thefluid at the point considered, (277) becomes

a -2gxw=-O(V+P+2q2).

212 VECTOR ANALYSIS.

100. Steady Motion. Definition. Steady motion is one inwhich F, q, p, and p are independent of the time. If such isthe case and co = 0, that is, if the motion is non-vortical,

-VU=V(V+P+ jq2)=0,or integrating V + P + J q2 = const.

If p is constant,

P = 2,p

and if there are no applied forces,

F=Oand hence V = const.,

so that2 +

2= const. (278)

p

In other words, where the pressure is great the velocitymust be small, and where the velocity is great the pressure issmall. For example, in a constricted pipe the pressure isleast at the constriction where the velocity of the incom-pressible fluid necessarily is the greatest. Air pumps andwater meters are constructed on this principle.

101. Vortex Motion. Theorem of Helmholtz. Take thecurl, or apply Vx to

aq+2wxq=-0(V+P+22)'

giving

curl 8 + 2 vx(wxq) = 0, as curl grad - 0

or at (curl q) + 2 (wV.q + 0.

Remembering that

curl q = 2 w; that V.co = 3 V.Vx (q) = 0,

VECTOR ANALYSIS. 213

as w is a solenoidal vector, i.e., co = I curl q, and that (270)

aw + Dw,

we have Dw+

This transforms intoD o _ co._vq, (279)D pt

because identically, using (273)1 Dv'q o

p Dt'Dw Do) w Do

°D+ wv w

Dt q Dt p Dt - DI p)Hence, also, differentiating (279) again

Dt' p) lp)). vq + p ' Dt V q, (280)

so that if w ever vanishes (279) and (280) likewise vanish,and similarly all the successive derivatives may be shown tovanish. Hence if w is ever zero it will always remain zero byTaylor's theorem, because all of its derivatives vanish at acertain instant. This theorem due to Helmholtz says that ifno vorticity exists in any incompressible, frictionless fluid atany time it is impossible to produce any by means of a con-servative system of forces, and the motion will remain forevernon-vortical.

* If the ether be considered to be a frictionless medium, then avortex once set up in it would be indestructible; and conversely, if novortices existed, it would be impossible to create any. It is conceivable,however, that some "Cataclysm "might have rendered the ether tempo-rarily viscous to some extent. By this we mean that it is conceivable,for example, that under extraordinary conditions say of temperature theether might acquire unusuvv1 properties, in which case, if it becamefrictionless again after vortical motion had been produced in it while inthis state, such vortical motion would persist forever. This speculationis of interest in connection with the vortex-atom theory of matter.

214 VECTOR ANALYSIS.

102. Circulation. The circulation along any path in afluid is defined as the line-integral of the velocity along thatpath. If OAB denote the circulation along the path AB, bydefinition

sAB =A

B(281)

If the path is a closed one, we may express the circulationaround it as a surface integral over any cap bounded by it,by means of Stokes' Theorem, for

fq.dr=ff xq dS = 2 f f dS,ap J cap

where2w=Vxq by(126).

Roughly speaking this equation says, if there is a pre-ponderance of motion of a liquid in one direction or the other

A

FIG. 68.

around any closed path, that the liquid inside of the closedpath must be rotating.

Consider a tube made up of the lines of the vector w, andconsider a portion of it bounded by two caps S, and S.

VECTOR ANALYSIS. 215

Apply the divergence theorem to this closed surface S,, S3and sides. Remembering that w is solenoidal, we have

ffn.o dS = fffot SV.0 dv = 0.

As the sides contribute nothing to the surface integralthere must be as much flux of co inward at S, as there is out-ward at SZ, or the flux is constant throughout the tube. If

FIG. 69.

this tube be chosen very small it is called a vortex filament,and if the section of such a filament be denoted by s theabove result. expresses the fact that

s

n is the normal to the cross-section. This product iscalled the strength of the filament. It shows that if w isfinite, or, in other words, if there does exist any vorticity, scannot vanish, hence a filament cannot end anywhere in thefluid. Such filaments must then either form closed curves orend in the surface of the liquid or at infinity. All vortices,then, form closed curves in the fluid or else end in the surface.This also follows from the fact that 0, that is, w is asolenoidal vector.

216 VECTOR ANALYSIS.

103. Velocity-Potential. If w is zero everywhere, the cir-culation around any closed curve is zero, hence the circula-tion from any point A to any other point B is independentof the path. In this case is a perfect differential; thatis, it is of the form

dr.q = do =so that q = Oct. (283)

The velocity q is thus derivable from the function 95 in thesame way (except for sign), that'the force is derivable fromthe ordinary potential. Accordingly 0 is called the Velocity-Potential, and is a scalar point-function of the space occupiedby the fluid. All the results of the theory of potential aretherefore directly applicable to the function q5.

Production of a Vortex Impossible in a Frictionless Fluid.Let us find the time-rate of variation of the circulationalong any path, assuming a velocity-potential to exist. Thispath is made up of certain elements of the fluid which areto be followed in their motion, however distorted the pathmay become. Differentiating (281),

Do= DDt

JB q.dr= fB . dr + f q dr.

Since the velocity-potential exists,

curl q =2w = 0,

and the equation of motion (275) becomes, if W = (- V + Pl

t = VW, (284)

so that PA dr = dr.VW = dW,Dt

also

and hence

qDt dr = q.d Dr = q.dq = df 2

=f B d (W + 2) = CW + 2218 (285)ADt

VECTOR ANALYSIS. 217

If the path is a closed one,

{w+cT=o,

as W and are scalar point-functions of position and have

identical values at the limits, so that finally

1286)

Equation (286) then states that the circulation around anyclosed curve, formed of a chain of particles of the fluid, cannotchange as these particles are carried about by the liquid. Aswe have assumed the circulation to be zero at the beginningit remains so forever, or, in other words, it is impossible tocreate vorticity in frictionless fluid by means of a conserva-tive system of forces. Also, as it is impossible to conceivehow any system of forces could act on a frictionless fluidin a non-conservative manner, it follows that it is impos-sible to create vorticity in any manner in a frictionlessmedium.

It was from these peculiar properties of vortices in africtionless fluid, discovered by Helmholtz, that Lord Kelvinwas led to his Vortex Atom Theory of Matter.

PROBLEMS AND EXERCISES

1. Show that the center of gravity of a system of particles, andhence of any body, continues to move uniformly in a straight linewhen no impressed forces act upon the system.

Find the equation of the path.

2. Show that the total momentum of a system of particles, andhence of any body, remains constant as long as there are no appliedforces.

218 VECTOR ANALYSIS.

3. Prove that a system of forces acting along and representedby the sides of a plane polygon taken in order is equivalent to acouple whose moment is represented by twice the area of the poly-gon. Extend this to forces acting along a closed plane curve.

4. By means of the theorem (202)

fqxdr=ff[nv.q - dSapJ

show that if forces equal in magnitude act everywhere along thetangents to a plane contour, that the moment of these forces aboutany point is measured by twice the area of the contour.

5. If a rigid body has a velocity of translation qt and anangular velocity of rotation co, the velocity q at any instant ofa point r in the body may be represented,by

q = qt + wxr. see § 22.

Show that if qt and w are constants the path of any point in thebody is a circular helix described with uniform velocity, and findits equation.

6. Show that two equal rotations in opposite directions abouttwo parallel axes produce a motion perpendicular to the plane ofthe two axes.

7. The motion of a point in a plane being given, refer it to(a) fixed rectangular vectors in the plane;(b) rectangular vectors in the plane, revolving uniformly about afixed point.

Translate into Cartesian in both cases.8. Prove that the central axis of two forces F1 and F2 intersects

the shortest distance between their lines of action and divides it inthe ratio

F2 (F2 + F, cos 0) : F, (F1 + F2 cos 0),

B being the angle between their directions. Also prove that themoment of the principal couple is

cF,F2 in 021 + F22 + 2 F1F., cos 0

9. Show that

ff nxF dS = f f f VXF dv.

What conclusion in Hydrodynamics does this theorem lead to?

VECTOR ANALYSIS. 219

10. Express the following equation in vector notation:

`' J

raw avIL ay az

cos (nx) + au _ awcos

az ax

ax 8cos (nz) dS.T I+ U

(ny)

Ans. ff(vxq).n dS.

11. Express the following equations in vector notation:

au au 1 aat X Pav

+ 2(u w __ aU 1

at ay pay '

at+2(v$-urn) =au-1a

P

Let q have components u, v and w, and w have componentsrJ, ( and $.

Ans. + 2wxq = VU - 1 Op.N

12. Express in vector notation the following equations whichoccur in the theory of Elasticity :

02U

= ( + a+ /82u + 82u a2UI

ate ax axz ayz azz

azv as azv azv azvatz = (A + p)

ay+ p (axz + ay + 1,

z z z 2w

atw =(x + p)

as+ ; (az + ay + )82

where u, v, w are the components of a vector q, where a is ascalar variable, and where p, A, and p are constants.

L2Ans. P atq= (A + p)Vo + pVq.

220 VECTOR ANALYSIS.

13. Express the following equations in vector notation :

au au au au v as aZU a2u a2u

at+u ax

+aaywaz 3ax v(axe+ aye+azz

av av av av v asat + u ax- waz - 3 ay y

X-pa

ra2v a2v a2vv

1 ay2 az2)

=Y-1 1,Pay

aw aw aw aw v as a2w a2w a2wll

at+u ax +vay +waz

3az -v(ax2+ay2

+az2

zpawhere u, v, and w are components of the vector q, where X, Y,and Z are components of the vector F, where p and a are scalarvariables, and where v is a scalar constant.

Ans. L + 3 0 or - vV'q = F - 1 Op.P

APPENDIX.

NOTATION AND FORMULIE.

The Various Notations.

Whenever new quantities are introduced it is well to haveas simple and as convenient a notation as possible. Thenotation devised by the late Professor Willard Gibbs seemedto us, after much thought on the matter, to be the simplestand most symmetrical of any of the existing kinds.

Hamilton, the inventor of quaternions, used the lettersS and V for the scalar and vector products respectively ofthe vectors that followed them; thus

S ab and V ab

represented respectively the scalar and vector products ofthe vectors a and b.

The letter T, standing for tensor, represented the magni-tude of the vector following it. This notation has manyadvantages, but after deliberation it was discarded.

Oliver Heaviside, the English electrician, used a notationsimilar to Hamilton's, but rendered it unsymmetrical bydiscarding the S for the scalar product while retaining theV for a vector product. This seemed to us to be a stepbackward, although. he was followed in its use by Foppland Bucherer in Germany and by others.

The disciples of Grassmann, who had devised a notationof his own, adapted it to the analysis of vectors, and at thepresent time the resulting notation has a number of adhe-rents in Germany and elsewhere. Our main objection to it

221

222 VECTOR ANALYSIS.

is that it uses different kinds of parentheses to distinguishthe two products, thus preventing the use of these paren-theses for other purposes. It is also quite cumbersome andtakes much longer to write than any of the other systems,besides there being a liability of error due to the fact that allparentheses necessarily look somewhat alike.

Gibbs, on the other hand, puts the distinguishing productmark between the two vectors instead of in front or aroundthem.

This is essentially a symmetrical notation, and to ourmind and to many others the best. The two symbols usedto indicate the " variety " of product are the dot andthe cross (x ). In order to avoid any confusion with theordinary dot and cross used for ordinary products, and anecessity in any analysis, we have ventured to use a specialdot and a special cross. That is, the dot is above thewriting line and the cross is a small one and when usedis placed in the same position as the dot. Thus

and axb

are the scalar and vector products of the vectors a and brespectively.

They are easy to write, easily distinguished and con-nected with the idea of a product. They do not interferewith parentheses, neither do they render the use of an ordi-nary dot (.) or a cross (X) undesirable nor ambiguous inother parts of the work. They are symmetrically placed.

Comparison of Notations.

A Few Examples of Formulae in the four systems ofnotation will render the foregoing clear to the student. Weshall give Hamilton's notation the benefit of our bold-facedtype and avoid the wholesale use of Greek letters whichwere employed by him to represent vectors.

VECTOR ANALYSIS. 223

The formulae are in the order:1. Gibbs' Notation.2. Hamilton's Notation.3. Heaviside's Notation.4. Gans' Notation. (Grassmannian.)

1. a or aa.2. Ta.3. a.4. lal.

1. ab cos-(ab).

2. Sab = Sba = - Ta Tb cos (ab).3. ab = ba = ab cos (ab).4. (ab) = (ba) = lallbl cos (ab).1. axb = - bxa = e ab sin (ab).2. Vab = -Vba = E Ta Tb sin (ab).3. Vab = -Vba = e ab sin (ab).4. [ab] - [ba] = e lallbl sin (ab).

1. c) =

2. Sa(b + c) = Sab + Sac. SaVbc = SbVca.3. a(b + c) = ab + ac. aVbc = bVca.4. (a, b + c) = (ab) + (ac). (a[bc]) = (b[ca]).

1. ax (bxc) = b c2. VaVbc = cSab bSac.3. VaVbc = b ac - c ab.4. [a[bc]] = b(ac) - c(ab).

1. r = r-bxc a + r.cxa b + r.axb c.[abc] [abc] [abc]

2. r = SrVbc a+ SbVca b + SrVabC.

SaVbc SaVbc SaVbcrVbc a

+rVca b

+rVab

c3. r =

aVbc aVbc aVbc

4. r =(r[bc) a + (r ca b + (r ab)

c.

(a[bc]) (a [bc]) (a[bc])

These examples are sufficient to show the characteristicsof the various notations.

224 VECTOR ANALYSIS.

Notation of this Book.

It seems to be the consensus of opinion that vectors arebest represented by single letters printed in some sort ofbold-faced type. If this is not done, in order to distinguisha vector from a scalar we are obliged to employ Greek orother special alphabets and thus deprive ourselves of theconvenience of using that alphabet if desired, and also weare prevented from using any other letter as a vector.

The Magnitude of a vector is represented by the sameletter as the vector itself but in ordinary or italic type.

A Unit Vector parallel to any vector is represented bythat vector with the subscript unity. Thus the vector

a

has a magnitude a

and a direction al,

so that we may write a = a al.

Sometimes the subscript zero to a vector, or particularlyto a vector-expression, will mean that its magnitude alone isexpressed. Thus

(axb)o

denotes the magnitude of axb.

In order to connect the Analysis of Vectors with Carte-sian Analysis it is necessary to relate the vector a to itsthree. components along the three Cartesian axes. We shalldenote these components by adding the subscripts 1, 2, and3 to the italic letter a.

As mutually perpendicular axes are by far the most im-portant of any, three unit vectors

i, j, and k

have been universally adopted to represent- their directionsin space.

VECTOR ANALYSIS. 225

The vector a then is made up ofa vector along i, of length al,a vector along j, of length a,,

and a vector along k, of length aso that

a = a a1 = a,i + a2j + a3k.

There can be no confusion between the letters a1 and al,for obvious reasons.

A Voluntary Exception to this convention is in the case ofthe radius vector r to any point from the origin. Its compo-nents will be denoted by x, y, and z instead of r r2i and r3 inorder to approach more closely to the usual Cartesian char-acter of the work when translated into that notation. Forthis reason we write

r=xi+yj+zk,and for similar reasons when desirable,

F= X i +Yj +Zk,q= ui+ vj+wk,w= Ei+ fl +Ck.

The Unit Tangent along and the Unit Normal to a curveor surface are denoted by the unit vectors t and n. Sincethe components of a unit vector are its direction cosines in

t = t1i + t2j + t3kand n = n1i + n2j + n3k,

t1, t2, and t3 and n17 n2, and 9i3 are the direction cosines of thetangent and normal respectively.

The Scalar or Dot Product of two vectors is representedby placing a dot between them thus:

ab cos (ab),

where cos (ab) is the notation used for the cosine of theangle included by the positive directions of a and b.

226 VECTOR ANALYSIS.

The Vector or Cross Product of two vectors is representedby placing a special cross between them thus :

a-b = - bxa = cab sin (ab).

The unit vector a is a vector perpendicular to the twovectors in the product and is taken in such a " sense " thatas you turn a into b, a points in the direction a cork-screwwould advance if so rotated.

Then (axb), = e

and (axb)o = ab sin (ab).

A Scalar Point-Function is represented by writing thefunctional symbol in italic. Thus

V = f(r)

means that V is a scalar point-function of the radius vec-tor r. That is, for every value of r, V has a determinatemagnitude. This is equivalent to one Cartesian equation.

A Vector Point-Function is represented by writing thefunctional symbol in bold-faced type. Thus

F = f(r)

means that F is a vector point-function of the radius vec-tor r. That is, for every value of r, F has a determinatemagnitude and direction. This is equivalent to three Car-tesian equations.

A Linear Vector Function, in particular, is represented bythe special symbols

w, $ or X.

Velocity and Angular Velocity are represented by the sym-bols respectively

q and w.

The Scalar Potential Function is represented by the italics

Vori'2.

VECTOR ANALYSIS. 227

The Vector Potential is represented by the bold-faced

V.

Electric or Magnetic Intensities or Forces in General arerepresented by

ForH.

The Differential Vector Operator 'V (read del) is equivalentto

V()=Ciax+jdy+ka)(Besides obeying the laws obeyed by ordinary vectors, it

is a differentiating operator, and the same care should betaken in its use and interpretation as should be taken withother differentiating operators.

As Professor Joly* puts it:" Of course some little care is necessary when V is ex-

pressed in the general form, but it is precisely of the samekind as the care required to distinguish between

(x22

(X2 a (X2a = x, a a +x2a xz

1 ax ax ax ax ax ax Cax )a 2 a a Y)and x' Cax =

x'ax Cax)

Del sometimes differentiates partially. Generally a sub-script attached to it, indicates the variable which it differ-entiates. Thus Va a-b

means that in the scalar product the vector a above is tobe considered variable.

Sometimes the same process may be indicated by writingas a subscript to the expression the quantity which is toremain constant during the differentiation. Thus

is the same as

Charles Jasper Joly, Manual of Quaternions, Macmillan & Co.,1905, Art. 57, p. 75.

228 VECTOR ANALYSIS.

Which notation is preferable will depend on whether allbut one of the vectors are to be considered variable orwhether all but one are constant, respectively.

Sometimes also the notation

V'(a'.a +

is useful, the variables having the same accent as the V,being alone considered variable during the differentiation.

Gradient (grad), or vector of greatest slope of a scalarfunction V, is a vector normal to the level surfaces of thefunction V and is given by the operation of V upon thefunction thus:

VV = gradient or slope of V = grad V.

The Divergence of a vector function F is given by form-ing the scalar product of 0 with the function thus

V.F = divergence of F = div F..

The Curl of a vector function F is obtained by formingthe vector product of V with the function:

V -F = curl F = rotation of F (rot F).

The more important formulae of vector analysis are col-lected below for reference.

VECTOR ANALYSIS.

FORMULAS.*

Vectors.

a=aa,=aoa,= a,i + a2j + a3k.

r=xi+yj+zk.F = Xi + Yj +Zk.

9 = q1i +q2j +q3k=u4 +vj +wk.

w =wli +w2j + W3k

$i+1I j+Ck.Ea=iEa,+jEa2+kEa3.

1=a-' =a,a a

229

(1)

(4)

(7)

(2)

r=r,=icosa+jcos(3+kcosr,r

where a, ,9, r are the direction angles of the radius vector r.The equation of a straight line through the terminus of

b and parallel to a is, s being a scalar variable,r=b+sa. (11)

If it passes through the origin,r = s a. (10)

A line through the ends of a and br = s a + (1 - s) b

(12)or r=sb+(1 -s)a.

The condition that three vectors, a, b, and c, should endin the same straight line is

xa + y b + z c = 0.(13)x + y + z = 0.

* The numbering corresponds with that of the text.

280 VECTOR ANALYSIS.

The equation of a plane determined by the vectors aand b, and passing through the terminus of c, is

r=c+sa+tb. (15)

The plane through the origin and parallel to a and b isr=sa.+tb. (14)

The equation of a plane passing through the three pointsa, b, and c is

r=sa+tb+(1-s-t)c. (16)

The condition that four vectors, a, b, c, and d, should endin the same plane is

xa+yb+zc+wd=0.(17)x+y+z+w=0.

The vectorr_ma+nb

(18)m + n

divides the line joining the points a and b in the ratio of

mton.The vector to the center of gravity of the masses m, at

points a, isEmaEm

(20)

If the relationm,a,+m2a2+ . . .=0

is to be independent of the origin chosen, thenm,+m2+ . . .=0.

Vector and Scalar Products.

Products of Two Vectors.

(25)

ab cos (ab) = (26)

= alb, + a2b2 + a3b3. (30)a2 = a2=a,2+a22+a$2.

cos (ab).(a,)2 = 1.

VECTOR ANALYSIS. 231

If a 1 b. (27)

(a + d) = b.d. (28)

No attention need be paid to the order of the factors.a-b = e ab sin (ab) = - b- a (33)

= i(a2b, - a3b2) + j (a3b, - a,b3) + k(a,b2 + a2b,) (39)

i j ka, a2 a3 (40)b, b2 b3

e is a unit vector, perpendicular to the plane of a and band pointing in such a sense that as a is turned towards ba cork-screw would advance along e.

(a,Xb,)o= sin (ab).

a,a = 0. (34)

If axb = 0,

i2=j2=k2=1.then b is parallel to a.

(29)

ixi = jxj = kxk = 0. ixj = k, jxk = i, kxi = j. (35)

If a' is the component of a normal to b, thena,b = a'xb. (36)

(a + b)x(c + d) = axc + a-d + bxc + bxd. (38)

Great attention must be paid to the order of the factors.

Products of Three Vectors.The scalar

b, b2 b3

c, c2 c3

_ [abc] * (49)

a, a2 a3

is equal to the volume of the parallelopiped of which a, b,and c are the three determining edges.

* [abc] represents all arrangements of the triple scalar products, (48),having the same cyclical order of factors.

232 VECTOR ANALYSIS.

The vector a%(b%c) = b c a.b. (5.5)

Any vector r may be represented in terms of three othersby the formula

r [abc] _ [rbc] a + [rca] b + [rab] c. (61)

The plane normal to a and passing through the terminusof b is

a.(r - b) = 0.The perpendicular from the origin to this plane is

p =a-'

(64)

(68a)

The plane parallel to c and d through the end of b is

(cxd).(r - b) = 0. (70)

The plane through the three points, a, b, and c is

(r - b)-(b - c) = 0, (67)or 4.(r - a) = 0, where 4 - (axb + bxc + cxa).

The perpendicular from the origin to this plane is

p = +-' +.a. (68b)

The line through the end of b and parallel to a is

ax(r - b) = 0. (69)

The equation of the sphere (or circle) of radius a withcenter at the origin is

r2 = a2. (71)

If the origin is at the point c, it becomes

r2 - 2 a2 - c2 = const. (72)

If the origin lies on the circumference,

r2 - 2 0. (73)

VECTOR ANALYSIS. 233

DIFFERENTIATION OF VECTORS.

dna = i final + j dna2 + k d (77)dtn dtn dtn dtn

If p - d and pn do, this may be written

fit dtn

pna = i pna, + j pna2 + k p"a3, (78)p b. (79)

No attention need be paid to the order of the factors ina scalar product.

p(axb) = p axb + axp b. (80)

Great attention must be paid to the order of the factorsin a vector product.

p p bxc + c,

p [ax(bxc)] = p ax(bxc) + ax(p bxc) + ax(bxp c).(81)

The Operator V (del).

V=i a+ j a+ k a=de1, (102)ax ay Oz

dr = idx + j dy + kdz,

a fix + a dy + a dz = d ( ), (114)ax ay az

VV =iaV+jaV +k av =grad V (a vector) (106),ax ay az

(VV)0=do

yr=rV 1 = --r, , (109)

r r2

Vrn =nrn_ivr = nrn-'r, = nrn-zr, (108)

234 VECTOR ANALYSIS.

V(ab) = Va(ab) + Vb(ab) - V(ab)b+V(ab)a, (110)

(slv)V =s1(IV)=TS = s` ax +

s2 aV + s3 az' (112)

(s1V)F = s1(VF), (113)

(s1.V)F = (s1V) (i F1 + J F2 + k F3), (116)a(bVFF) = bVF(aF), (117)

l )vF = + + a = div F (119)(a sca ar .CIX y

The Divergence Theorem,

rfJn.qdS =ff f V.q dv, (121)

where q is any vector function and n is the externally-drawn unit normal to dS.

vxF =

i ; ka a a

ax ay az

F1 F2 F3

= curl F

(125)

I - a3 +k aF'-a'(Iay - aF2 + . aF F Fayaz) (az ax ) (ax ay )-

V-((o-r) = 2 co, if w = const. vector. (126)

General Differentiating Formulae for del.

V(u+v)=vu +Vv,V. (u + v) = vu + vv,vx(u + v) = Vxu + Vxv.

(127)

V (uv) = vvu + uvv,V.(uv) = Vuv + u vv, (128)

V. (uv) = Vuxv + u vXv.

VECTOR ANALYSIS. 235

V(u.v) = ux(Vxv) + vx(Vxu). (129)

V. (uxv) = U.Vxv, (130)

V.(uxv) = u(VUD.V) - v(V .u)= uV.v + vV.u - (131)

q(r + dr) = q(r) + +(Vxq)xdr. (135)

V(a.r) = a, a = const. vector.=w, (118)

3, (123)

2r

r 7j

1 = a-r =r r3 r2

r3 a b const. vectors.

74 r3

V2f(r) = f"(r) + 2 f'(r)r

Taylor's Theorem.

f(r + E) = e"°f(r) (141)

Stokes' Theorem.

. f f (136)

V.VV = V 2V = div grad V = del square V (a scalar)

= a2V + a2V +n- (147)c3x2 ay2 M

V2 is called the Laplwcian operator.V2F = V2(iF,+ jF2+kF3)= iV2F,+iV2F2+kV2F3

o2F o2F o'Faxe+8y2+az2.

236 VECTOR ANALYSIS.

VXVV = curl grad V = 0,div curl F = 0,

Vx(VxF) = curl2 F = V(V.F) -F - VZF,

V2rm = m(m + 1)rm-2, (148)

V2 1 = 0.r

Gauss' Integral.

ff!rds = 47r or = 0 (150)

according as the origin is taken inside or outside of theclosed surface S.

Laplace's Equation.V2V = 0. (157)

Poisson's Equation. According to system of units chosen,it is

V2V 4 7rp or V2V = - p. (156)

v

Its solution is

pdv

1

ep

dvres tV= (1674

7r Jff or , e .fff )

If V satisfies the equationsV2V=0

and nV,it is a spherical harmonic of degree n.

Green's Theorems.

JJn. UV VdS = fff UV Vdv +fff(vU .7 V)dv,

(158)

fffl.(uvv_vvu)ds=ffJ'1(uv2v_vv2u)dv.(159)

VECTOR ANALYSIS.

Green's Formula.Vo f ro

4 J' J Tidv

7r rRegion

+4- f

r

23.7

(161)

(168)

-41

irpot () is the inverse operator to V2 ( ). (169)

pot curl ( = curl pot ( ), for a vector function.V pot ( = pot V ( ), for a scalar function.

Theorem of Helmholtz.

W 4I7rvfffYdv+4

7rvx(ff!W dv(175)

_ -41-vpot (vW) + 417r vxpot(V'W) (178)

fqxdr=ff[(v.q)n - dS (201)

q = any vector function.

Maxwell's Electro-magnetic Equations for Media at Rest.

" = VxHat

F = eF,(192)

a3c = v-Fat

Linear Vector Function.If 4 is a linear vector function,

ae = pIi.

4(T f 17) _ 4T ±4a,4a 'r = a+T,

(223d(4T) _ 4 dT, )

T.40 _ Or+T.

is said to be the conjugate function to 4.

238 VECTOR ANALYSIS.

If

If

then + is said to be self-conjugate.

w = wli + w2j + w3kand +w = Awli + Bw2j + Cw3k, (232)

then +(4w) = 4)2w = A2wli + B2w2j + C2w3k,and +nw = Anwli + Bnw2j + Cnw3k.

If 4-1(4)w) == w, (233)then +-lo) = A-1wli + B-1&j2 j + C-1w3k.

Coriolis' Equation. A vector a in fixed space, whenreferred to a moving space which has an angular velocityof rotation w, satisfies the following relation :

(d )is(d9)ms o)-q. (239)

D'Alembert's Equation.

1 m

2dtd -F 8r= 0. (201)

Euler's Equation of Motion of a Rigid Body about aFixed Point.

dtI + wxH = M. (243)

Hamilton's Principle.

aft.

- W) dt = 0. (255)

Lagrange's Equations of Motion.

dtV, ,L - VnL = 0, (266)

or VL = Of (267)

where Q= (dt On -vn).

VECTOR ANALYSIS. 239

Hydrodynamics.

Equation of Continuity.

+ V.(pq)= 0,

or

at

+ pV.q = 0.D (271)t

Euler's Equation of Motion of a Fluid.

D(pg) = a(pq) pF -Vp.Dt at

Circulation along the path AB.JBq.dr.

OAB A

(275)

ADDITIONS TO APPENDIX.Note to § 4.

b\

P

_---------------------------

I B' P'-------------------------Note on Different Varieties of Vectors. - Consider a parti-

cle on the movable platform P. The particle is initiallyat A. If the particle remains at rest on the platform whilethe platform is displaced uniformly to a new position P',the particle will describe the path AA' relatively to the ground.This motion can be conveniently described by the vectorAA' or more concisely by the single symbol a.

If the platform remains at rest and the particle movesuniformly to B, the path thus discribed relatively to theground is AB, or more shortly b.

Evidently, if the particle moves uniformly from A to Bwhile the platform moves uniformly from P to P', theparticle will finally end up at B', and will have described thepath AB uniformly with respect to the ground, and in thesame time. This path is defined to be the sum of the dis-placements (or vectors) a and b and is written a + b.

(In this case also these displacements may take placeconsecutively in either order and the final position of theparticle will be the same.)

Now it is a fact that forces, velocities and many otherphysical quantities obey this same law. Hence they willobey the consequences of a calculus which follows from theabove, and other definitions consistent with the facts.

240

APPENDIX. 241

There is nothing else but convenience which obliges us todefine the sum of two vectors in the above fashion. Wecould have defined the sum in other ways, and by othernon-contradictory definitions obtained a consistent analysis.

We choose, of course, the definitions which seem to us mostnatural and best suited to our needs.

Free Vectors, Slide or Axial Vectors. - Quite often certainrestrictions must be placed upon our vectors. We mayrestrict them to a plane in which they may slide about,or to remain normal to some plane, or restrict them toslide back and forth in the lines of which they are segments,or even attach them to a fixed point allowing no motionwhatsoever.

For example : It is well known that couples may berepresented by vectors perpendicular to their plane and thatthe effect of any couple is the same as long as its represen-tative vector remains parallel to itself, however otherwisedisplaced. This is the freest kind of a vector and may becalled a Free Vector.

On the other hand, forces produce the same effect pro-vided only that they are not displaced out of their line ofaction. This kind of a vector with restricted freedom iscalled a Slide Vector, or Axial Vector.

Nevertheless, if we disregard for the time the known effectof displacing a force out of its line of action, i.e., changing itsmoment about any given axis, we may consider forces asfree vectors. For instance, in statics, one of the conditionsof equilibrium is that the resultant of all the forces, con-sidered acting at a common origin, shall vanish. Here weconsider the forces as free vectors for the time being. Therestriction to their line of action is intrinsically contained inthe remaining rule for the vanishing of the resultant moment.

Simultaneous angular velocities compound vectorially,while finite rotations do not. Thus a rotation of 0 aboutan axis followed by a rotation of 0 about another axis is

242 VECTOR ANALYSIS.

not, in general, equal to those rotations taken in reverseorder. And yet a finite rotation has an axis, which is adirection and an amount or magnitude and is, in that sense, avector, but it does not obey the laws of our vector analysis.

However if these rotations were to take place simultane-ously the resultant rotation would be correctly found byvector addition as defined above.

It is thus seen that we are endeavoring to deduce a calculuswhich coincides as nearly as possible with the fundamentalproperties of the majority of quantities to which we apply it.

Having thus constructed a consistent analysis coincidingas closely as may be with the facts, it can of course be taughtabstractly without reference to them. Later on when phy-sical quantities are shown to obey the same laws as vectorshave been defined to have, we may employ the results of thevector calculus to them without further ado.

Personally the writer does not approve of the teaching ofVector Analysis as an abstract science, nor even as a mathe-matical subject unless by a teacher who is thoroughlyfamiliar with the physical results to which it applies and forwhich it was designed.

The vector analysis as deduced in this book is that of freevectors.

Note to §38.

Normal, Normal Plane, Principal Normal, Binormal,Rectifying Plane.

Every line perpendicular to the tangent to a curve at thepoint of tangency, M, is called a Normal. These normalslie in a plane called the Normal Plane and from them two aresingled out for special mention: the normal which lies in theOsculating Plane called the Principal Normal, and the normalperpendicular to this plane called the Binormal.

The plane passing through the point of tangency perpen-dicular to the principal normal is called the Rectifying Plane.

APPENDIX. 243 12

Thus the tangent, the principal normal and the binormalform a rectangular system of vectors. Taking the pointof contact as origin, the directions of these vectors mayalways be taken so as to form a right-handed system. Theprincipal normal, however, is always chosen so as to pointtowards the center of curvature of the curve.

Let R be the magnitude of the radius of curvature and letT, V, P be unit vectors along the tangent, principal normaland binormal respectively. Then

drT - ds = r . (a)

The vector curvature c ( p-1 where p = vector radius ofcurvature) is, by definition, the rate of change of the unittangent per unit are or

dT ,T r=ds

_ p'

And as its direction is along the principal normal we maywrite it, where R is the magnitude of the radius of curvature,

T' = c =

R

= r(b)

Hencev=Rr".

As P is a unit vector 1 both to r and v, it isP=Txv=Rr'xr c)

The direction cosines of these lines are the coefficients ofi, j, k in the following equations.

T=r'=x'i+y'j+z'kv = Rr" = Rx"i + Ry"j + Rz"k,

i j k=Txv = R x' y' z'x y z

=RI i+ Rlz'x1 k.

244 VECTOR ANALYSIS.

Torsion or Tortuosity is defined on page 78 as

T=dp(d)(Here P is the n of page 78.)

The reciprocal of the torsion is a vector called the radius oftorsion S. Hence

T = S-'.

Formulae of Frenet for a Space Curve.

In the investigation of the properties of space curvescertain formulae due to Frenet are of fundamental importance.They enable us to express the derivatives of the unit vectorsalong the tangent, principal normal and binormal in termsof these vectors themselves.

Differentiate (a) givingT' = r

which by (b) becomesV.

T = R.

Differentiate (c) givingQQ

t' = T'XV +TXV',which by I becomes

I

VxvR -f- TXv' = TXV'.

From this equation we see that P' is L to T; and since P isa unit vector P' is L to P; it is therefore parallel to v and wemay write

S, II

where S is a scalar whose absolute value, as V is a unit vector,is the radius of torsion, by definition, equation (d).

Again, since v is at right angles to P and T, we may writeV=PxT,

APPENDIX. 245

which equation on differentiating becomesv' = P' x T + P x T'.

Using II and I we obtain_

vVxT + PxVS R

_- -T IIIS R

Equations I, II, III are Frenet's equations; they expressthe first derivatives of T, v and P in terms of themselves andthe scalars R and S.

Formula for the Torsion. -From the preceding we caneasily obtain a simple formula for the torsion of a curve.From II and (c) we have

Rrl rxr" + Rl.,xr",

The second term is identically zero. By multiplying byequation (b), v = Rr," we have

S = S = Rr'xr'"J.

Hence, as the first triple scalar product vanishes

=T=- R2[r'r"r",]. IV

Exercise: Express equations I, II, III, IV in Cartesiancoordinates. Show that for a plane curve the torsionvanishes, and conversely.

Path Described by an Electron in a Uniform Magnetic Field.

As an interesting application of the vector method con-sider the motion of an electron in a uniform magnetic fieldof strength H. It is well known by -the experiments ofRowland that a moving electrical charge is equivalent to acurrent. Let e be the value of the charge, m the mass of

246 VECTOR ANALYSIS.

the electron, and v its vector velocity. The force acting ona linear conductor of length dr is

F = i drx H.

Let this current i be due to the convection of electricitycarried by the n electrons contained in the element dr, mov-ing with the common velocity v, then

idr = ne v.Hence the force on a single electron is

F, _ ney-H = evxH.n

This force produces an acceleration v on the electron, sothat the equation of motion is

my = evxH.

Putting h = m H, the differential equation of the path is

v = vxh. (1)

From the equation we see that the acceleration is normal tothe path, hence the speed is constant. As the accelerationis also normal to h the velocity component parallel to h isconstant, and hence the whole acceleration is always parallelto the plane normal to h.

As h is a constant vector, we have from (1)

h.vxh= 0= dt (2)

so that the angle the path makes with the field is constant.From (1) again, since v and h and their included angle areconstant, the magnitude of the acceleration is constant.

The radius of curvature of any path is related to the speedand the normal acceleration by equation (92), page 81,

v2a=-F

APPENDIX. 247

Hencev2 y2P=v =vxh' (3)

And since the magnitudes of v and v x h are constant so isthe magnitude of p.

PO hsin (vh) (4)

The component of the speed normal to h is

vl = v sin (vh).

The radius of curvature of the path in the plane normal toh is, similarly to (3),

Pi -y12 l y2 sin2 (vh) _ v sin (vh) (5)

( vi -h /o - v sin (vh) h sin (vlh) h

as sin (v1h) = 1, and hence the path is a circle of radius pl.Thus the motion is completely determined. It is a curve

of constant curvature, described with constant speed whoseprojection on a plane normal to the magnetic lines is a circleof radius pl. The velocity parallel to the field is constant.It is therefore a right circular helix whose axis is parallel tothe lines of force.

Comparing equations (4) and (5) we see that the radiusof curvature of a curve and that of its projection on a planeare related by the formula,

pi = p sine (e), (6)where 0 is the angle between the curve and the normal tothe plane. This holds for every curve, because any smallportion of it may be considered to be a portion of some helix.This result is due to Euler.A circular helix is sometimes defined as a curve having (a)

constant curvature and (b) constant torsion. We can alsoprove that the above path is a helix by proving (b). Equa-tion (4) shows that the curvature is constant.

248 VECTOR ANALYSIS.

The torsion T is by IV of Frenet's formulae

T = p2 [r' r" (7)

where the primes denote differentiation with respect to thearc s.

Now r'=rds=v;and since v is constant

r = r (dt)2- vds v2

and

rift =v3

Substituting in (7)v2 2 v V V 1 [Vx(yxh)]_V_

T -(v/ [v v2 v3 v2 (vxh)2By differentiating (1)

v=vxh=(vxh)xh.Hence

T = 1v2 (vxh)2

which reduced by (58) becomes

T=tvhAs the two parts of the fraction are separately constant thetorsion is constant.

The result (4) obtained above, otherwise writtenme

v = Hp sin (vH),

is of great importance in obtaining a relation between (e ) ande

v for an electron, which in combination with other relationsenables us to determine their separate values.

APPENDIX. 249

Note to § 58.

Two Proofs of Stokes' Theorem.

1° The line integral around the bounding curve, Fig. 54, isequal to the sum of the line integrals around the elementaryparallelograms into which the surface may be considereddivided. For every side of these parallelograms is integratedtwice and in opposite directions, and the results cancel,except the sides which coincide with the bounding curve.

Consider an elementary parallelogram ABCD; let AB = dpiand AD = dp2. Let (dF)1 be the increment of F alongdpi and (dF)2 that along dp2. Then

(dF)1 = dp1.VF,(dF)2 = dp2.VF.

The vector point-function F has at the point A the value F;at B the value F + (dF)1; at D the value F + (dF) 2.

The line integral around the parallelogram is then bydefinition

[F + (dF)2].dp1 - [F + (dF)1].dP2= (dF)1dp2 - (dF)2dp1j= (dp1.VF)Fdp2 - (dP2-VF)Fdpl,= (dpi- dp2) (V x F), (by 58),= (nV x F) dS.

Because dplxdp2 = ndS where n is the unit normal to thiselementary area dS.

The theorem follows immediately.2° Again by means of the Divergence Theorem which is

the fundamental formula for the transformation of volumeinto surface integrals and vice versa, we can deduce Stokes'Theorem which is the fundamental formula for the trans-formation of surface into line integrals.

Apply the formula

fffv.F dv =f f Fn dS, (1)

250 VECTOR ANALYSIS.

(which is the Divergence Theorem and is to be taken through-out the volume of, and over the surface of, any closed space)to an infinitesimal right cylinder, of height h and base whosearea is S. Let ds be an element of the contour of the base,and c, a unit vector 1 to the base.

Suppose the cylinder so small that F may be considered constant throughout it, so that the two surface integrals overthe two plane ends cancel each other and their sum vanishes.

Replace F in (1) by c,-F, then

fffv.(cl%F)

A=

ffc1F.n ds.

By (130) 0.(c,xF) = F.V c, - c,.V.Fin which the first term of the right-hand number vanishes,as cl is a constant vector; hence

fffci.vxF dv = -ffciF.n dS

- f fF.nxc, dS.* (2)

But dv = h dS, and dS = h ds so that the volume and surfaceintegrals become surface and line integrals respectively, and

h ffci.vxF dS = hf ds.

Now suppose the base to be an element of area of a surfacebounded by a closed contour. Then c, becomes a unit vectornormal to the surface, which we call n, and what was formerlyn x c, in (2) becomes a unit vector in the direction of dr, sothat F.nxc, ds = F. dr.Summing up the elements

4 r rn.v x F dS = z fF.dr.

The summation of the surface integrals means simply thatthe integral ffn.vx F dS

is to be taken over the entire surface.In summing up the line integrals, the contour of every

elementary area is traced twice; but in opposite directionsexcept those forming part of the contour.

* The negative sign indicates relationship of direction around thecentour and the vector n.

APPENDIX 251

Hence the sum of the line integrals becomes the line inte-gral around the bounding curve and we have again Stokes'Theorem.

Note to § 65.

Another Proof of Gauss's Theorem.

By means of the Divergence Theorem an easy proof ofGauss's theorem may be given.

The problem is to evaluate the integral

ff'dsr02

over the closed surface S.This integral may be written

-f f ()ds, by (109).

There are three cases, according as the origin is without,within, or on the surface S.

CASE I. The origin is without. In this case r can neverbe zero, hence because by the Divergence Theorem

ffn.v dS =ffJV2rdv,

and because

we have

V2r = 0, by § 64,

ff'ds=o.re

CASE II. The origin is within. Surround the origin bya small sphere of radius e; then the origin is excluded fromthe region bounded by S and S'. Hence the required inte-gral taken over both surfaces is zero, by Case I, i.e.,

1 dS + Js, 1 d8 = 0.ff

252 VECTOR ANALYSIS.

But for the sphere, ro = e, 1,

ffds = 4 7re2;

n.rsfJ,'ds=_4.Hence

A n-r,dS = 47r.

rot

CASE III. The origin is on the surface. Exclude thepoint by an elementary hemisphere.

Then proceeding as in Case II, we find that the requiredintegral is equal to minus the integral over the hemisphere,i.e., to 2 7r.

Note to § 52.

Other Integration Theorems.

We can evaluate two volume integrals in a manner similarto that given by the Divergence Theorem. First, to find

fffvF dv,

where F is a scalar point-function. Let c be a constantvector,

fffc.VF dv = fJfv. (cF) dv.

But by the Divergence Theorem

fffv.(cF) dv = ffn.(CF) dS,

=cf fnFdS,

:. c. fffvF dv = C. ffnF dS,

APPENDIX. 253

or since c is perfectly arbitrary,

fffvFdv =f fnF dS.

Again

cf f fvxFdv= f f

ff fc (nx F) dS.

Hence finally

fffvxF dv =f fnxFdS.

By similar processes, starting with the Divergence Theorem,which is seen to be a formula of fundamental importance,many other relations between surface and volume integrals,and indeed also between line and surface integrals, usingthe device employed in the second proof of Stokes' Theoremabove, may be deduced.

INDEX.

Accelerated motion, 82.Acceleration, central, 84.

centripetal, 197.normal, 81.of moving space, 197.radial, 81.

Activity, equation of, 165Addition of vectors, 4.Ampere, 161, 165.Analytic solution of Euler's equa-

tions, 200.Angle, solid, 138, 166.Angular velocities, composition of,

41.Apparent inertia, 182.Appendix, 221.Applications to geometry, 73.

to mechanics, 39.Areal velocity, 86.Areas, description of, 85.Axes, moving, 194.

normal to surface for maximumand minimum, 189.

of a central quadric, 189.permanent, 193.principal, 186.

Axis, central, 65.instantaneous, 43, 183, 194.of a vector product, 34.

Body, rigid, 41, 181.system of forces on rigid, 63.

Book of Bucherer, 127.of Fehr, 80.of Foppl, 38.of Gans, 112, 127.of Gibbs-Wilson, 47, 56, 109, 127.of Joly, 123.

Book of Kelland and Tait, 63,of Lame, 104.of Routh, 41, 181.of Webster, 112.

Calculus of variations, 125.Cartesian expansion for scalar prod-

uct, 30.expansion for triple vector prod-

uct, 53.expansion for vector product, 38.expansion of divergence, 111.

Center of mass, 19.of mass, motion of, 179.

Central acceleration, 84.axis, 65.quadric, axes of, 189.

Centrifugal couple, 198.Centripetal acceleration, 197.Centroid, 18.Circle, equation of, 61.Circular motion, 82.Circulation, 214.

definition, 32.Collection of fprmulse, 229.Collinear vectors, 3.Combinations of three vectors, 48.

Commutativity of a and f , 125.

Comparison of various notations,223.

Complex variable, 15.Components of vector, 8.Composition of angular velocities, 41.Compound centripetal acceleration,

198.Condition for relation to be indepen.

dent of origin, 21.255

256 INDEX.

Condition of integrability, 130.of parallelism of vectors, 35of perpendicularity, 28.that four vectors terminate in

same plane, 18.that three vectors end in same

straight line, 13.that three vectors lie in a plane, 50.

Cone, polhode and herpolhode, 192.Conic sections, 63.

section orbit of planet, 87.Conjugate, linear vector-function,

186.self-, 183.

Conservation of circulation, 216.Conservation of motion of center of

mass, 179.of vortex motion, 213.

Conservative system of forces, 129,145, 203,

Continuity, equation of, 116, 207.of a scalar point-function, 7.of a vector point-function, 8.

Convergence, 112.Coordinates, curvilinear, 79.

isothermal, 80.of linear vector-function, 185.orthogonal, 80.polar, 61.

Coplanar vectors, 8.Coriolis, 198.

compound acceleration of, 198.theorem of, 194.

Cork-screw rule, 160, 183.Coulomb's law, 98, 143.Couple, minimum, 65.Cremieu and Pender, 162.Cross product, 34.

distributive law of, 35.Curl, example of, 118.

condition of vanishing of, 127.independent of axes, 117.of magnetic body, 120.the operator, 117.

Curvature, 76.Curve in space, 74.Curvilinear coordinates, 79.

D'Alembert, 178, 180, 202,Del, applications of, to scalars, 104,

applied to a vector, 109.formulae for use of, 121.rule for use of, 134.the operator, 94.

Derivative, directional, 106, 107.total, 107.

Description of areas, 85.Determinantal cubic, 190.Determinant form of vector product,

39.Determinant form of triple scalar

product., 50.Differential equation of geodetic, 78.

equation of harmonic motion, 86,operators, 94.perfect, 129.

Differentiation by V, 121.by V2, 135,of vectors, 70.

of vector and scalar products, 72.partial, 90.with respect to scalar variables,

70.Directional derivative, 106.Discontinuities, 95.Displacement current, 161.Distributive law for cross products.

35.law, physical proof of, 37.

Divergence, definition, 109.Cartesian expansion for, 111.interpretation of, 146.physical interpretation of, 109.theorem, 112.

Division of a line in a given ratio, 18,Dot or scalar product, 28.Dyad, reference to, 109.Dynamical equations of Euler, 200.Dynamics, 178.

included in a single formula, 207.

Electrical theory, 138.Electro-magnetic waves, 163.Electro-motive force definition, 32.Electro-dynamic potential, 173.

INDEX. 257

Electron theory, 162, 182.Elementary properties of linear vec-

tor-function, 183.Ellipse, equation of, 86.

equation relative to focus, 90.Ellipsoid, momental, 184.

of Poinsot, 184, 201.Energy of distribution in terms of

field intensity, 157.of system in terms of potential,

156.Equal vectors, 2.Equation of circle, 61.

of continuity, 208.of electro-magnetic field, 160.of hodograph, 83.of instantaneous axis, 43.of Lagrange, 205.of Laplace, 146.of plane, 17, 58, 59.of Poisson, 146.of sphere, 61.of straight line, 11, 60.

Equations, dynamical, of Euler, 200.of hydrodynamics, 207.of motion, 178.of polhode and herpolhode curves,

193.of surfaces, 79.

Equipotential surfaces, 98.Ether, 213.Euler's dynamical equations, 200.

equations of motion of a fluid,210.

theorem, on homogeneous func-tions, 131.

Expansion for moment of momen-tum, 185.

for triple vector product, 53.Exploding shell, motion of, 1.79.Extended vector, definitions, 204.Extension of vector to n-dimen-

sions, 204.

Faraday, 160.Fehr, book of, 80.Field due to a current, 165, 170.

Field, energy in terms of, 157.intensity, 157.

Fields, addition of, 6.Filament, vortex, 215.Fixed point, motion about, 198.

space, 195.Flow of heat, equation of, 114.Fluid, Euler's equations of motion

of a, 210.Flux of a vector, 33.

of heat, 114.Foppl, book by, 38.Force, central, 84.

centrifugal, 197.for Newtonian law, 143.

Formulae for use of V, 121.principal, of vector analysis, 229.

Foucault gyroscope, 199.Fourier's law, 104, 114.Frictionless fluid, 213, 217.Function, Green's, 148.

Lagrangian, 204.linear vector, 185.

Fundamental equations of hydro-dynamics, 207.

Gans, book of, 112, 127.Gauss's theorem, 138.

theorem for the plane, 140.theorem, second proof, 141.

General equations of motion ofLagrange, 205.

Generalized parameters of a system,205.

Geodetic lines on a surface, 78.Geometrical representation of mo-

tion of a rigid body, 191.Geometry, applications to, 73.Gibbs, Prof. Willard, 221, 224.Gibbs-Wilson, book of, 47, 56, 109,

127.Grad (gradient) of a scalar function,

102.independent of choice of axes,

103.

Graphical representation of diecontinuities, 96, 97.

258 . INDEX.

Grassmann, 50.Grassmann's notation, 221, 223.Gravity, center of, 20.Green's formula;, 148.

function, 148.theorem, 148, 158.

Gyration, radius of, 181.Gyroscope, 199.

Integration theorem, 174.Interpretation of products, 57.Invariable line, 192, 199.

plane, 184, 191.Inverse square law, 87, 105, 143.Irrotational motion, 119, 212.Isothermal surface, 104.

systenr of curves, 80.

Hamilton's Integral, 202, 204, 206.notation, 63, 221, 223.principle, 202.

Harmonic function, 147.motion, differential equation of, 86.

Heaviside, Oliver, 50, 155, 164.Heaviside's notation, 221, 223.Helmholtz, 125, 212, 213, 217.Helmholtz's theorem, 155, 173.Herpolhode curve, 192.Hodograph and orbit under New-

tonian forces, 87.Hodograph, definition, 81.

equation of, 83.of accelerated motion, 82.of a particle at rest, 82.of uniform circular motion, 82.of uniform motion, 82.

Homogeneous function, 131.Hydrodynamics, 207.

Incompressibility, condition of, 209.Independent of the origin, relations

of, 21.Inductance, 171.Induction, vector, 158.Inertia, apparent, 182.

moment of, 181, 185.products of, 185.

Instantaneous axis, 43, 183, 194.Integral due to Neumann, 171, 175.

line of a vector, 31.of Gauss, 138.surface, of a vector, 32.

Integrating factor, 129.operator pot, 152.

Integration, 83.as a vector sum, 5,

dolt', book of, 123.Joule, 164.

Kelland and Taft, book of, 63.Kelvin, Lord, 217.Kepler, laws of, 86.Kinematics of a particle, 80.Kinetic energy of rotation. 181.

of translation, 179.

Lagrange's equations of motion,205.

Lagrangian function, 204.Lame, book of, 104.

definition of, 6.Lamellar component of a vector

function, 154.vector, 128, 154.

Laplacian, the, operator, 134.Laplace's equation, 143.Law, distributive - for vector prod-

ucts, 35.of Coulomb, 98.of Fourier, 104.of Kepler, 86.of Lenz, 160.of Newton, 87, 105, 143.

Laws obeyed by I J k, 29, 35.obeyed by scalar products, 29.obeyed by triple scalar product,

49.Layers, equivalent, not equipotential,

151.Lenz, law of, 160.Level surfaces, 98.Linear vector-function, 182,183,238.Line integral of a vector, 31.

of normal component, 168.

INDEX.

Lines of force, 109.of vector-function, 32.

Liquid, fundamental equations, 207.Lord Kelvin, 217.

Magnetic field due to a current, 165.field due to element of current,

W.Magnitude of a vector, definition, 3.

of a vector, 29.Mathematical and physical discon-

tinuities, 95.Maxwell, 116, 128.Maxwell's equations, 160.McAulay, theorem of, 170.Mechanical force on element of cir-

cuit, 167.Mechanics, 178.

applications to, 39.Methods of solution of problems, 13.Minimum couple on central axis, 65.Moment about an axis, 180.

as vector, 40.definition, 39.of inertia, 181, 185.

Momental ellipsoid, 184.Moments of inertia, principal, 186.Momentum, moment of, 181.

of momentum, 181.Motion, circular, 82.

harmonic, 86.irrotational, 119.of a rigid body, 41.of center of mass, 179.Poinsot, 192.under no forces, 184.vortex, 212.

Moving axes, 190, 194.space, 195.

Multiple vector products, 55.Mutual energy of two circuits, 171,

175.Mutual inductance, 173, 176.

n-dimensions, extension of vectorto, 204.

Negative vector, 2.

259

Neumann's integral, 171, 175.Newtonian forces, 87.

law of force, 105, 143.Non-vortical motion, 120, 212.Normal acceleration, 81.

and tangent, 75.to tangent plane of quadric, 189.unit, 78.

Notation of Gans, 223.of Gibbs, 222, 223.of Grassmann, 221.of Hamilton, 221, 223.of Heaviside, 221, 223.of this book, 224.

Notations, comparison of, 223.various, 221.

Operator, integrating pot, 152.the, curl, 117.the, V, 94.the, " p ", 72, 73.involving V twice, 133.+(),187.

Orbit of a planet, 84.under Newtonian forces, 87.

Origin of operator V, 90.Orthogonal system of curves, 80.Osculating plane, 77.

Parabola, path of projectile, 179.Parabolic orbit, 84.Parallelism, condition of, 35.

condition of, of vectors, 35.Parallelogram law, 40.Parallelopiped principle, 50.Parentheses, 51.Partial differentiation of vectors,

90.differentiation with V, 105, 106.

Particle, kinematics of, 80.Perfect differential, 129.Permanent axes, 192.Perpendicular from origin to a plane,

59.Perpendicularity, condition of, 28.Physical discontinuities, 95.

proof of distributive law, 37.

260 INDEX.

Plane, equation of, 17, 58, 59.invariable, 184.osculating, 77.passing through end of a vector,

58.passing through ends of three vec-

tors, 59.perpendicular to a vector, 58.through ends of three given vec-

tors, 17.Planet, orbit of, 84.Poinsot ellipsoid, 184, 201.

motion, 192.Point-function, scalar, 6.

vector, 7.Poisson's equation, 143.

equation, solution of, 152.Polar coordinates, 61.Polarization, energy in terms of,159.Polhode and herpolhode curves, 192.Polygon of vectors, 4.Potential, 6.

definition, 98.derivatives of, 102, 149.forces having a, 129, 216.the, 6, 143.vector, 153, 173.velocity, 216.

Poynting's theorem, 165.Practical application of steady mo-

tion equation, 212.Principal axes, 186, 189.

axes, at right angles, 190.moments of inertia, 186.of D'Alembert, 178, 202.of Hamilton, 202.

Problems, method of solution, 13.to Chap. I, 22.to Chap. II, 43.to Chap. III, 66.to Chap. IV, 91.to Chap. V, 136.to Chap. VI, 176.to Chap. VII, 217.

Product, cross, 34.dot, 28.scalar, 28.

Product, vector, 34.Products of inertia, 185.

of two vectors, 28.of three vectors, 48.of more than three vectors, 55.

Projectile, path of, 179.Projections of vector, 29.Proof of del formula;, 121.

of normality of principal axes, 190.of expansion of triple vector prod-

uct, 51, 53, 54.Properties of frictionless fluid, 213,

217.

Quadric surface, 63, 86, 184.principal axes of, 189.tangent plane to, 188.

Quaternions, 63.

Radial acceleration, 81.Radiant-vector, 165.Radius of gyration, 181.Ratio, division of line in given, 18.Reciprocal operator, 187.

system of vectors, 57.

vector, 3.Rectangular coordinates, 62.Relation between any four vectors,

56.between force and potential, 100.

Relations independent of the origin,21.

Relative motion, 194.Remarkable formula, 207.Remarks on notation, 63, 221, 223.Representation of a vector-function,

74.of vector or cross product, 34.

Resolution of a system of forces, 63.of a vector-function into solenoidal

and lamellar components, 154.of velocity, 81.

Resumd of notation of this book,224.

Rigid body, 41, 181.body, motion of, 41.system, 180.

INDEX. 261

Rigidity, various kinds, 120.Rolling ellipsoid, 191, 193.Rotation as vector, 41.

equations for, 179.kinetic energy of, 181.

Routh, book of, 202.Rowland, 162.Rule, cork-screw, 160, 183.

Scalar and vector fields, 94.and vector functions of position,

95.definition of, 1.field, 6.product, 28.products of three vectors, 48.products, differentiation of, 72.variables, integration with respect

to, 83.Self-conjugate, 183.

inductance, 173, 176.Sink of heat, 112.Slope, 102.Solenoidal component of a vector

function, 154.distribution, 117.vector, 117.

Solid angle, 138, 166.Solution of differential equations,

86.Source of heat, 112.Sources and sinks, 114.Space curve, 74.Space of n-dimensions, 204.Special notation, 204.Speed, 81.Sphere, equation of, 61.Steady motion, 212.Step, 2.Stokes' theorem, 124, 160, 161, 168,

214.Straight line, equation of, 11, 60.

parallel to a given vector, 60.through end of a vector, 60.

Stroke, 2.Subtraction of vectors, 4.Surface, equations of, 79.

Surface integral of a vector, 32.of revolution, 191.

Surfaces, level or equipotential, 98.Symmetrical top, 199.System of forces on a rigid body,

63.Systems of units, 155.

Talt, 63, 170.Tangent and normal, 75.

plane to quadric, 188.unit, 74.

Tangential component in line inte-gral, 31.

Taylor's theorem, 124, 131, 213.Tensor of vector, 3.Termino-collinear vectors, 13.Termino-coplanar, 18.Theorem due to Helmholtz, 155,

173.of Coriolis, 194.of Euler, 131.of Gauss, 138, 140, 141.of Green, 148.of McAulay, 170.of Poynting, 164.of Stokes, 124, 160, 161, 168, 214.

Third integral of equations of rota-tion, 201.

Three axes in a rigid body, 193.Top, 199.Tortuosity, 78, 92.Total current, 162.

derivative, 107.kinetic energy of a system, 182.

Transformation of hydrodynamicequation, 211.

Translation, equation for, 178.Triple vector products, 27, 48.

vector product, expansion for, 53.Tube of vector function, 215.

Uniform motion, 82.Unit normal, 78.Units, other systems of, 155.

tangent, 74.vector, 3.

262 INDEX.

Variable, complex, 15.Variations, calculus of, 125.Vector and scalar fields, 94.Vector, components of, 8.

curvature, 76.definition of, 1.equations, 11, &8.field, 6.function, representation of, 74.graphical representation of, I.lamellar, 128.line integral of, 31.magnitude of, 29.negative, 2.perpendicular to a plane, 59.point-function, 7.-potential, 153, 173.product, 34.products, differentiation of, 72.products of three vectors, 48.radiant-, 165.reciprocal, 3.sum as an integration, 5.surface integral of, 32.-tortuosity, 78.unit, 3.velocity, 80.

Vectors, addition and subtrac-tion, 4.

collinear or parallel, 3.condition of parallelism of, 35.

Vectors, coplanar, S.decomposition of, S.differentiation of, 70.equality of, 2.products of three, 48.products of more than three, 55.reciprocal system of, 57.termino-collinear, 13.termino-coplanar, 18.the unit, I J k, 9, 29, 85.

Velocity along tangent and normalof a curve, 81.

angular, 41.areal, 86.of electric waves, 164.

Velocity-potential, 216.Volume of sphere by divergence

theorem, 114.Vortex-atom theory, 217.Vortex filament, 215.

motion, 212.Vorticity, 211.

indestructible, 213.uncreatable, 216.

Wave equation, 163.Ways in which a vector may vary;

70.Webster, book of, 112.Wilson-Gibbs, book of, 109.Work of a force, 31.