VECTOR and TENSOR ANALYSIS By
LOUIS BRAND, Ch.E., E.E., Ph.D. PROFESSOR OF MATHEMATICS UNIVERSITY OF CINCINNATI
New York JOHN WILEY & SONS, Inc. London CHAPMAN & HALL, Limited
VECTOR AND TENSOR ANALYSIS. By Louis Brand. 439 pages. 5Y2 by 8%. Cloth.
VECTORIAL MECHANICS. By Louis Brand. 544 pages. 5% by 8%. Cloth.
Published by John Wiley & Sons, Inc.
VECTOR and TENSOR ANALYSIS By
LOUIS BRAND, Ch.E., E.E., Ph.D. PROFESSOR OF MATHEMATICS UNIVERSITY OF CINCINNATI
New York JOHN WILEY & SONS, Inc. London CHAPMAN & HALL, Limited
COPYRIGHT, 1947 BY
Louis BRAND
All Rights Reserved This book or any part thereof must not
be reproduced in any form without the writ+en permission of the putlisher.
THIRD PRINTING, NOVEMBER, 1948
PRINTED IN THE UNITED STATES OF AMERICA
To My Wife
PREFACE
The vector analysis of Gibbs and Heaviside and the more general tensor analysis of Ricci are now recognized as standard tools in mechanics, hydrodynamics, and electrodynamics. These disciplines have also proved their worth in pure mathematics, especially in differential geometry. Their use not only materially simplifies and condenses. the exposition, but also makes mathematical concepts more tangible and easy to grasp. Moreover tensor analysis provides a simple automatic method for constructing invariants. Since a tensor equation has precisely the same form in all coordinate systems, the desirability of stating physical laws or geometrical properties in tensor form is manifest.. The perfect adaptability of the tensor calculus to the theory of relativity was responsible for its original renown. It has since won a firm place in mathematical physics and engineering technology Thus the British analyst E. T. Whittaker rates the discovery of the tensor calculus as one of the three principal mathematical advances in the last quarter of the 19th century. The first volume of this work not only comprises the standard vector analysis of Gibbs, including dyadics or tensors of valence two, but also supplies an introduction to the algebra of motors, which is apparently destined to play an important role in mechanics
as well as in line geometry. The entire theory is illustrated by many significant applications; and surface geometry and hydro-
dynamics * are treated at some length by vector methods in separate chapters. For the sake of concreteness, tensor analysis is first developed in 3-space, then extended to space of n dimensions. As in the case of vectors and dyadics, I have distinguished the invariant tensor from its components. This leads to a straightforward treatment of the affine connection and of covariant differentiation; and also to a simple introduction of the curvature tensor. Applications of tensor analysis to relativity, electrodynamics and rotating elec* For a systematic development of mechanics in vector notation see the author's Vectorial Mechanics, John Wiley & Sons, New York, 1930. vii
PREFACE
viii
tric machines are reserved for the second volume. The present volume concludes with a brief introduction to quaternions, the source of vector analysis, and their use in dealing with finite rotations. Nearly all of the important results are formulated as theorems,
in which the essential conditions are explicitly stated. In this connection the student should observe the distinction between necessary and sufficient conditions. If the assumption of a certain property P leads deductively to a condition C, the condition is necessary. But if the assumption of the condition C leads deductively to the property P, the condition C is sufficient. Thus wehave symbolically
P -* C (necessary),
C (sufficient) -> P.
When P C, the condition C is necessary and sufficient. The problems at the end of each chapter have been chosen not only to develop the student's technical skill, but also to introduce new and important applications. Some of the problems are mathematical projects which the student may carry through step by step and thus arrive at really important results. As very full cross references are given in this book, an article as well as a page number is given at the top of each page. Equations are numbered serially (1), (2), . . . in each article. A reference to an equation in another article is made by giving article
and number to the left and right of a point; thus (24.9) means article 2.4, equation 9. Figures are given the number of the article
in which they appear followed by a serial letter; Fig. 6d, for example, is the fourth figure in article 6. Bold-face type is used in the text to denote vectors or tensors of higher valence with their complement of base vectors. Scalar components of vectors and tensors are printed in italic type. The rich and diverse field amenable to vector and tensor methods is one of the most fascinating in applied mathematics. It is hoped
that the reasoning will not only appeal to the mind but also impinge on the reader's aesthetic sense. For mathematics, which Gauss esteemed as "the queen of the sciences" is also one of the great arts. For, in the eloquent words of Bertrand Russell: "The true spirit of delight, the exaltation, the sense of being more than man, which is the touchstone of the highest excellence,
is to be found in mathematics as surely as in poetry. What is
PREFACE
is
best in mathematics deserves not merely to be learned as a task, but also to be assimilated as a part of daily thought, and brought again and again before the mind with ever-renewed encouragement. Real life is, to most men, a long second-best, a perpetual compromise between the real and the possible; but the world of pure reason knows no compromise, no practical limitations, no barrier to the creative activity embodying in splendid edifices the passionate aspiration after the perfect from which all great work springs."
The material in this book may be adapted to several short Thus Chapters I, III, IV, V, and VI may serve as a
courses.
course in vector analysis; and Chapters I (in part), IV, V, and IX as one in tensor analysis. I But the prime purpose of the author was to cover the theory and simpler applications of vector and tensor analysis in ordinary space, and to weave into this fabric such concepts as dyadics, matrices, motors, and quaternions.
The author wishes, finally, to express his thanks to his colleagues, Professor J. W. Surbaugh and Mr. Louis Doty for their help with the figures. Mr. Doty also suggested the notation used in the problems dealing with air navigation and read the entire page proof. Louis BRAND
University of Cincinnati January 15, 1947
CONTENTS PREFACE .
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PAGE Vii
CHAPTER I VECTOR ALGEBRA ARTICLE
1. Scalars and Vectors . 2. Addition of Vectors .
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5 6 7 8
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3. Subtraction of Vectors
4. Multiplication of Vectors by Numbers . 5. Linear Dependence . . . . . . . . . . . . . . 6. Collinear Points . . . . 7. Coplanar Points .
8. Linear Relations Independent of the Origin. 9. Centroid . . . . . . . . . . . . . . . . 10. Barycentric Coordinates . . 11. Projection of a Vector . . . 12. Base Vectors . . . . . . . 13. Rectangular Components 14. Products of Two Vectors . . 15. Scalar Product . . . . . . 16. Vector Product . . . . . 17. Vector Areas . . . . . . . 18. Vector Triple Product . . . 19. Scalar Triple Product . . . 20. Products of Four Vectors . 21. Plane Trigonometry . . . 22. Spherical Trigonometry . . 23. Reciprocal Bases . . . . 24. Components of a Vector . . 25. Vector Equations . . . . 26. Homogeneous Coordinates . 27. Line Vectors and Moments 28. Summary: Vector Algebra . Problems . . . . . . . .
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24 24 26 29 29 34 37 40
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41
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43 44 44 46 48 50
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51
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55 57 59
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12 18 19 23
CHAPTER II MOTOR ALGEBRA
29. Dual Vectors . 30. Dual Numbers
31 Motors
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xi
63 64 65
CONTENTS
xii ARTICLE
PAGE
32. Motor Sum . . . . . . 33. Scalar Product . . . . . 34. Motor Product . . . . . 35. Dual Triple Product . . . 36. Motor Identities . . . 37. Reciprocal Sets of Motors 38. Statics . . . . . . . . 39. Null System . . . . . . 40. Summary: Motor Algebra Problems . . . . . . . .
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67 68 70 72 73 74 75 78 80 82
CHAPTER III VECTOR FUNCTIONS OF ONE VARIABLE
41. Derivative of a Vector . . . . . . 42. Derivatives of Sums and Products 43. Space Curves . . . . . . . . . 44. Unit Tangent Vector . . . . . . 45. Frenet's Formulas . . . . . . . 46. Curvature and Torsion . . . . . 47. Fundamental Theorem . . . . . 48. Osculating Plane . . . . . . . . 49. Center of Curvature . . . . . . 50. Plane Curves . . . . . . . . . . 51. Helices . . . 52. Kinematics of a Particle . . . . . 53. Relative Velocity . . . . . . . 54. Kinematics of a Rigid Body . . . 55. Composition of Velocities . . . . 56. Rate of Change of a Vector . . . 57. Theorem of Coriolis . . . . . . . 58. Derivative of a Motor . . . . . . 59. Summary: Vector Derivatives . . Problems . . . . . . . . . . .
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98 99 100 105 108 110 114 120
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121
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84 86 88 90 92 95 97
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123 126 128 130
CHAPTER IV LINEAR VECTOR FUNCTIONS
60. Vector Functions of a Vector . . . 61. Dyadics . . . . . . . . . . . . 62. Affine Point Transformation . . . 63. Complete and Singular Dyadics . . 64. Conjugate Dyadics . . . . . . . 65. Product of Dyadics . . . . . . . 66. Idemfactor and Reciprocal . . . . 67. The Dyadic 4) x v . . . . . . . . 68. First Scalar and Vector Invariant . 69. Further Invariants . . . . . . .
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135 136 138 139
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141
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142 144 146 147 148
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. .
CONTENTS
xiii
ARTICLE
PAGE
70. Second and Adjoint Dyadic . . . . 71. Invariant Directions . . . . . . . 72. Symmetric Dyadics . . . . . . . . 73. The- Hamilton-Cayley Equation . . 74. Normal Form of the General Dyadic . 75. Rotations and Reflections . 76. Basic Dyads . . . . . . . . . 77. Nonion Form . . . . . . . . . . 78. Matric Algebra . . . . . . . 79. Differentiation of Dyadics . . . . . . . . . 80. Triadics . . . . . . 81. Summary: Dyadic Algebra . . . . . .
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Problems
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153 156 160 162 164 166 167 169
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171
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172 172
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174
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151
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CHAPTER V DIFFERENTIAL INVARIANTS
82. Gradient of a Scalar . . . . . . . 83. Gradient of a Vector . . . . 84. Divergence and Rotation . . . . 85. Differentiation Formulas . . . . 86. Gradient of a Tensor . . . . . . 87. Functional Dependence . . . . . 88. Curvilinear Coordinates . . . . . . . 89. Orthogonal Coordinates . . 90. Total Differential . . . . . . . 91. Irrotational Vectors . . . . . . . 92. Solenoidal Vectors . . . . . . . 93. Surfaces . . . . . . . . . . . . 94. First Fundamental Form . . . 95. Surface Gradients . . . . . . . . 96. Surface Divergence and Rotation . 97. Spatial and Surface Invariants . 98. Summary: Differential Invariants . Problems . . . . . . . . . . . .
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178
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181
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183 186 187 190 191 194 197 198 201
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203 204 206 207 209
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211
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213
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CHAPTER VI INTEGRAL TRANSFORMATIONS
99. Green's Theorem in the Plane . . . . . 100. Reduction of Surface to Line Integrals . 101. Alternative Form of Transformation . . 102. Line Integrals . . . . 103. Line Integrals on a Surface . . . . . . 104. Field Lines of a Vector . . . . . . . . 105. Pfaff's Problem . . . . . . . 106. Reduction of Volume to Surface Integrals 107. Solid Angle . . . . . . . . . . . . .
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216 218
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221
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222 224 226 230 233 236
CONTENTS
xiv
PAGE
ARTICLE
108. Green's Identities . . . . . . . . . 109. Harmonic Functions . . . . . . . 110. Electric Point Charges . . . . . . 111. Surface Charges . . . . . . . . . 112. Doublets and Double Layers . . . . 113. Space Charges . . . . . . . . . . 114. Heat Conduction . . . . . 115. Summary: Integral Transformations . Problems . . . . . . . . . . . .
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237 239 241 242 243 245 247 248 250
CHAPTER VII HYDRODYNAMICS
116. Stress Dyadic . . . . . . . . . . . . 117. Equilibrium of a Deformable Body . . . 118. Equilibrium of a Fluid . . . . . . . . . 119. Floating Body . . . . . . . . . . . 120. Equation of Continuity . . . . . . . . 121. Eulerian Equation for a Fluid in Motion. 122. Vorticity . . . . . . . . . . . . . . 123. Lagrangian Equation of Motion . . . 124. Flow and Circulation . . . . . . . . . 125. Irrotational Motion . . . . . . . . . . 126. Steady Motion . . . . . . . . . . . 127. Plane Motion . . . . . . . . . . . . 128. Kutta-Joukowsky Formulas . . . . . . 129. Summary: Hydrodynamics . . . . . . . Problems . . . . . . . . . . . . . .
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253 255 257 257 258 260 262 263 266 267 268 270 273 278 280
CHAPTER VIII GEOMETRY ON A SURFACE
130. Curvature of Surface Curves . . . 131. The Dyadic On . . . . . . . . 132. Fundamental Forms . . . . . . 133. Field of Curves . . . . . . . . . 134. The Field Dyadic . . . . . . . . . 135. Geodesics . . . . . . . . . 136. Geodesic Field . . . . . . . . . 137. Equations of Codazzi and Gauss . 138. Lines of Curvature . . . . . . . 139. Total Curvature . . . . . . . . 140. Bonnet's Integral Formula . . . . 141. Normal Systems . . . . . . . . 142. Developable Surfaces . . . . . 143. Minimal Surfaces . . . . . . . . 144. Summary: Surface Geometry . . . . . . . Problems . . . . . . .
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283 285 289 293 293 297 299 300 302 306 308 313 314 316 319 321
CONTENTS
xv
CHAPTER IX TENSOR ANALYSIS ARTICLE
PAGE
145. The Summation Convention . . . . . . . . . . . . 146. Determinants . . . . . . . . . . . . . . . . . 147. Contragredient Transformations . . . . . . . . . 148. Covariance and Contravariance . . . . . . . . . . . 149. Orthogonal Transformations . . . . . . . . . . . 150. Quadratic Forms . . . . . . . . . . . . . . . . 151. The Metric . . . . . . . . . . . . . . . . . 152. Relations between Reciprocal Bases . . . . . . 153. The Affine Group . . . . . . . . . . . . . . 154. Dyadics . . . . . . . . . . . . . . . . . . 155. Absolute Tensors . . . . . . . . . . . . . . . . . 156. Relative Tensors . . . . . . . . . . . . . . . 157. General Transformations . . . . . . . . 158. Permutation Tensor . . . . . . . . . . . . . . 159. Operations with Tensors . . . . . . . . . . . . . 160. Symmetry and Antisymmetry . . . . . . . . . 161. Kronecker Deltas . . . . . . . . . . . 162. Vector Algebra in Index Notation . . . . . . . 163. The Affine Connection . . . . . . . . . . 164. Kinematics of a Particle . . . . . . . . . . . 165. Derivatives of e` and E . . . . . . . . . 166. Relation between Affine Connection and Metric Tensor 167. Covariant Derivative . . . . . . . . . . . 168. Rules of Covariant Differentiation . . . . . . 169. Riemannian Geometry . . . . . . . . . . 170. Dual of a Tensor . . . . . . . . . . . . . . . . . 171. Divergence . . . . . . . . . . . . . . . . . . . 172. Stokes Tensor . . . . . . . . . . . . . . . . . 173. Curl . . . . . . . . . . . . . . . . . . . . 174. Relation between Divergence and Curl . . . . . . 175. Parallel Displacement . . . . . . . . . . . . . 176. Curvature Tensor . . . . . . . . . . . . 177. Identities of Ricci and Bianchi . . . . . . . . . . . 178. Euclidean Geometry . . . . . . . . . . . . . 179. Surface Geometry in Tensor Notation . . . . . . . . 180. Summary: Tensor Analysis . . . . . . . . . . . . Problems . . . . . . . . . . . . . . . . . . .
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328 329 332 334 336 337 339 339 340 342 343 345 346 350 350 352 353 354 356 359 360
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361
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362 365 366 370
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371
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372 374 376 376 380 384 385 388 392 396
CHAPTER X QUATERNIONS
181. Quaternion Algebra . . . 182. Conjugate and Norm . 183. Division of Quaternions .
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403 406 409
CONTENTS
xvi
PAGE
ARTICLE
184. Product of Vectors . . . . . 185. Roots of a Quaternion . . . . . 186. Great Circle Arcs . . . . . . . 187. Rotations . . . . . . . . . . 188. Plane Vector Analysis . . . . . 189. Summary: Quaternion Algebra . Problems . . . . . . . . . .
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410 412 414 417
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421
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426
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427,
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431
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INDEX
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...
CHAPTER I VECTOR ALGEBRA 1. Scalars and Vectors. There are certain physical quantities, such as length, time, mass, temperature, electric charge, that may be specified by a single real number. A mass, for example, may be specified by the positive number equal to the ratio of the given mass to the unit mass. Similarly an electric charge may be specified by a number, positive or negative, according as the charge is "positive" or "negative." Quantities of this sort are called scalar quantities; and the numbers that represent them are often called scalars.
On the other hand, some physical quantities require a direction as well as magnitude for their specification. Thus a rectilinear displacement can only be completely specified by its length and direction. A displacement may be represented graphically by a segment of a straight line having a definite length and direction. We shall call such a directed segment a vector. Any physical quantity that involves both magnitude and direction, so that it may be represented by a line segment of definite length and direction, and which moreover conforms to the parallelogram law of addition (§ 2), is called a vector quantity. Velocity, acceleration, force, electric and magnetic field intensities are examples of vector quantities. It is customary, however, in applied mathematics, to speak of vector quantities as vectors. DEFINITIONS: A vector is a segment of a straight line regarded as having a definite length and direction. Thus we may represent a
vector by an arrow. The vector directed from the point A to the
point B is denoted by the symbol AB. With this notation AB
--
and BA denote different vectors; they have the same length but opposite directions. Besides the proper vectors just defined, we extend the term vector
to include the zero vector, an "arrow" of length zero but devoid of direction. 1
2
VECTOR ALGEBRA
§1
If the initial point of a vector may be chosen at pleasure, the vector is said to be free. If, however, its initial point is restricted to a certain set of points, the vector is said to be localized in this set. When the set consists of a single point (initial point fixed), the vector is said to be a bound vector. If the vector is restricted to the line of which it forms a part, it is called a line vector. For example, the forces acting upon rigid bodies must be regarded as line vectors; they may only be shifted along their line of action. Two vectors are said to be equal when they have the same length and direction. Vectors are said to be collinear when they are parallel to the same line. In this sense two parallel vectors are collinear. Vectors are said to be coplanar when they are parallel to the same plane. In this sense any two vectors are coplanar. A unit vector is a vector of unit length.
In addition to the foregoing notation, in which we denote a vector by giving its end points, we also shall employ single letters in heavy (bold-face) type to denote vectors. Thus, in Fig. 2b the vectors forming the opposite sides of the parallelogram are equal (they have the same length and direction) and may therefore be represented by the same symbol,
AB=DC=u,
AD=BC=v.
A vector symbol between vertical bars, as I AB I or I u 1, denotes the length of the vector. We shall also, on occasion, denote the lengths of vectors, u, v, F by the corresponding letters u, v, F in italic type. Bars about real numbers denote their positive magnitudes: thus I -3 1 = 3. The preceding definition of a vector is adequate for the elementary applications to Euclidean space of three dimensions. For
purposes of generalization, however, it is far better to define a vector as a new type of number-a hypernumber-which is given by a set of real numbers written in a definite order. Thus, in our ordinary space it will be -seen that, when a suitable system of reference has been adopted, a vector can be represented by a set of three real numbers, [a, b, c], called the components of the vector. With this definition the zero vector is denoted by [0, 0, 0]. In order to complete this definition of a vector, a rule must be given to enable us to compute the components when the system of reference is changed.
ADDITION OF VECTORS
§2
3
2. Addition of Vectors. To obtain a rule for adding vectors, let us regard them, for the moment, as representing rectilinear displacements in space. If a particle is given two rectilinear displace-
ments, one from A to B, and a second from B to C, the result is the same as if the particle were given a-single displacement from A to C. This equivalence may be represented by the notation,
AB + BC = AC.
(1)
We shall regard this equation as the definition of vector addition. The sum of two vectors, u, v, therefore is defined by the following triangle construction (Fig. 2a) : Draw v from the end of u; then the vector directed from the beginning of u to the end of v is the sum of u and v and is written u + v. C
Fia. 2b
FIG. 2a
Since any side of a triangle is less than the sum of the other two sides,
lu+vl _ lul +lvl, the equal sign holding only when u and v have the same direction. Vector addition obeys both the commutative and associative laws: (2)
u+v = v+u,
(3)
(u + v) + w = u + (v + w).
In the parallelogram formed with u and v as sides (Fig. 2b),
u+v=AB+BC=AC,
v+u=AD+DC=AC.
This proves (2). In view of this construction, the rule for vector addition is called the parallelogram law. To find u + v when u and v are line vectors whose lines intersect at A, shift the vectors along their lines so that both issue from A (Fig. 2b) and complete the parallelogram ABCD; then ---3
--3
---9
u+v=AB+AD= AC.
VECTOR ALGEBRA
4
§2
Since the diagonal of the parallelogram on u, v gives the line of action of u + v, the term "parallelogram law" is especially appropriate for the addition of intersecting line vectors. We shall speak of the addition of line vectors as statical addition. The associative law (3) is evident from Fig. 2c:
(u + v) + w = (AB -i- BC) -f- CD = AC -i- CD = AD,
u + (v -F w) = AB -f- (BC + CD) = AB + BD = AD. Since the grouping of the vectors is immaterial, the preceding swa
is simply written u + v + w.
B
Fia. 2c
FIG. 2d
From the commutative and associative laws we may deduce the following general result: The sum of any number of vectors is independent of the order in which they are added, and of their grouping to form partial sums. To construct the sum of any number of vectors, form a broken
line whose segments, in length and direction, are these vectors taken in any order whatever; then the vector directed from the beginning to the end of the broken line will be the required sum. The figure formed by the vectors and their sum is called a vector polygon. If A, B, C, , G, H are the successive vertices of a vector polygon (Fig. 2d), then (4)
When the vectors to be added are all parallel, the vector "polygon" becomes a portion of a straight line described twice. If, in the construction of a vector sum, the end point of the last vector coincides with the origin of the first, we say that the sum of the vectors is zero. Thus, if in (4) the point H coincides with A, we write (5)
-4 -- 3 -3 AB+BC++GA=O.
SUBTRACTION OF VECTORS
§3
5
This equation may be regarded as a special case of (4) if we agree --->
that AA = 0. The zero vector AA (or BB, etc.) is not a vector in the proper sense since it has no definite direction; it is an extension of our original vector concept. From --- f
----*
--3
AB+BB=AB,
-3 -3 -* AA+AB=AB
we have, on writing AB = u,
u+0=u,
0+u=u.
We shall refer to vectors which are not zero as proper vectors. 3. Subtraction of Vectors. The sum of two vectors is zero when, and only when, they have the same length and opposite directions:
AB + BA = 0. If AB = u, it is natural to write BA = -u in order that the characteristic equation for negatives,
u + (-u) = 0,
(1)
will hold for vectors as well as for numbers. Hence by definition: The negative of a vector is a vector of the same length but opposite direction.
Note also that - (-u) = u. The difference u - v of two vectors is defined by the equation,
(u - v) + v = U.
(2)
Adding -v to both sides of (2), we have
u-v=u+(-v);
(3)
that is, subtracting a vector is the same as adding its negative. The
construction of u - v is shown in Fig. 3a. A
0 Fia. 3a
Fia. 3b
If 0 is chosen as a point of reference, any point P in space may
be located by giving its position vector OP. Any vector AB may
VECTOR ALGEBRA
6
§4
be expressed in terms of the position vectors of its end points (Fig. 3b),
AB = AO + OB = OB + (-OA), and, from (3), (4)
-- -4 -4 AB = OB - OA.
4. Multiplication of Vectors by Numbers. The vector u + u is
naturally denoted by 2u; similarly, we write -u + (-u) = -2u. Thus, both 2u and -2u denote vectors twice as long as u; the former has the same direction as u, the latter the opposite direction. This definition is generalized as follows: The product au or ua of a vector u and a real number a is defined as a vector a times as long as u, and having the same direction as u, or the opposite, according as a is positive or negative. If a = 0,
au=0.
In accordance with this definition,
a(-u) = (-a)u = -au,
(-a)(-u) = au.
These relations have the same form as the rules for multiplying numbers. Moreover, the multiplication of a vector by numbers is commutative (by definition), associative, and distributive: (1)
au = ua,
(2)
(a/3)u = a(/3u),
(3)
(a + a)u = au + (3u.
The product of the sum of two vectors by a given number is also distributive:
a(u + v) = au + av.
(4)
The proof of (4) follows immediately from the theorem that the corresponding sides of similar triangles are proportional. Figure 4 applies to the case when a > 0. The quotient u/a of a vector by a number av u+V a (not zero) is defined as the product of u v
au by 1/a.
U
Fia. 4
The developments thus far show that: As far as addition, subtraction, and multi-
plication by numbers are concerned, vectors may be treated formally in accordance with the rules of ordinary algebra.
LINEAR, DEPENDENCE
§5
7
5. Linear Dependence. The n vectors u1, u2i , un are said to be linearly dependent if there exist n real numbers X1, X2, , An, not all zero, such that X1U1 + X2U2 + ... + XnUn = 0.
(1)
If the vectors are not linearly dependent, they are said to be linearly independent. Consequently, if a relation (1) exists between n linearly independent vectors, all the constants must be zero.
, um are linearly dependent, any greater If m vectors ul, U2, number n of vectors including these are also linearly dependent. For , um satisfy if u1i u2, X1U1 + A2U2 + ... + XrUm = 0,
we can give ?1, X2, , Xm the preceding values (at least one of these is not zero) and take X,n+1 = X m+2 = _ X n = 0. Then
(1) is satisfied, and the n vectors ui are linearly dependent. If Xu = 0 and X 0, u = 0; hence one vector is linearly dependent only when it is the zero vector. Hence the vectors of any set
that includes the zero vector are linearly dependent.
Conse-
quently, we need only consider sets of proper vectors in the theorems following.
If X1u1 + X2u2 = 0 and X1 0 0, we can write ul = au2 i hence ul and u2 are collinear. Conversely, if ul and u2 are collinear, U1 = au2 (a 0 0). Therefore: A necessary and sufficient condition that two proper vectors be linearly dependent is that they be collinear.
If X1u1 + A2u2 + X3u3 = 0 and X1 0, we can write ul = aU2 + ,13u3; the parallelogram construction (Fig. 5a) now shows that ul is parallel to the plane of u2 and u3. Conversely, if U1, U2, u3 are coplanar, they are linearly dependent. For (a) if two of the vectors are collinear, they are linearly dependent, and
the same is true of all three; and (b) if no two vectors are collinear, we can construct a parallelogram on ul as diagonal whose sides are parallel to u2 and u3 (Fig. 5a), so that
ul =AC=AB+BC=au2+$u3. Therefore: A necessary and sufficient condition that three proper vectors be linearly dependent is that they be coplanar.
VECTOR ALGEBRA
8
§6
In space of three dimensions, four vectors ul, u2i u3, u4 are always linearly dependent. For (a) if three of the vectors are co-
planar, they are linearly dependent, and the same is true of all four; and (b) if no three vectors are coplanar", we can construct a
FIG. 5a
FIG. 5b
parallelepiped on ut as a diagonal whose edges are parallel to U2i n3, U4 (Fig. 5b), so that
-3 ---3 --a - 3 ut = AD = AB + BC -}- CD = au2 + 9u3 -I- yu4.
Therefore: Any four vectors are linearly dependent.
6. Collinear Points. If A, B, P are points of a straight line, P is said to divide the segment AB in the ratio X when
-->
-3
AP=APB.
(1)
As P passes from A to B (Fig. 6a), X increases through all positive values from 0 to infinity. If P describes the line to the left of A,
X varies from 0 to -1; and, as P describes the line to the right of B, X varies from -oo to -1. Thus X = 0, X = =L00 , A = -1 correspond, respectively, to the points A, B, and the in0 FIG. 6a
finitely distant "point" of the line. The ratio A is positive or negative, according as P lies within or without the segment AB.
To find the position vector of P, relative to an origin 0, write (1) in the form, Then (2)
OP - OA = X(OB - OP).
-OA + A OB OP = 1 + A
COLLINEAR POINTS
§6
9
or, if we write X = (3/a,
-
-OPaOA +aOB =
(3)
a+0
In particular, if X = 1, P is the mid-point of AB.
In the following we shall denote the position vectors of the , p. Thus if C divides AB , P by a, b, c,
points A, B, C, in the ratio /3/a, (4)
c=ash-(3b
a+0
Thus the mid-point of AB has the position vector (a + b). 2 When the points C, D divide a segment AB internally and externally in the same numerical ratios ±X, we have
a+Ab c =1+x,
d=
a - Xb
1-x
If we solve these equations for a and b, we find that the points A, B also divide the segment CD in the same numerical ratios f(1 - X)/(l + X). Pairs of points A, B and C, D having this property are said to be harmonic conjugates; either pair is the harmonic conjugate of the other. A useful test for collinearity is given by the following: THEOREM.
Three distinct points A, B, C lie on a straight line
when, and only when, there exist three numbers a, /3, y, different from zero, such that (5)
aa+/3b-+-yc=0,
a+$+y=0.
Proof. If A, B, C are collinear, C divides AB in some ratio /3/a; hence on putting y = - (a + ,B) in (4) we obtain (5). Conversely,
from (5) we can deduce (4) since a + -y 0; hence C lies on the line AB. From (5) we conclude that C, A, B divide AB, BC, CA, respectively, in the ratios S/a, y/l3, a/y whose product is 1. If an equation of the form (5) subsists between three distinct non-
collinear points, we must conclude that a = (3 = y = 0. For at least one coefficient y = 0; and from
as+3b=0,
a+/3=0,
we have a = b (A coincides with B) unless a = 0 = 0.
§6
VECTOR ALGEBRA
10
Another criterion for collinearity may be based on the statical addition of line vectors (§ 2). THEOREM. The points A, B, C are collinear when the line vectors
AB, BC, CA are statically equal to zero:
AB +BC +CA =O.
(6)
Proof. If we use = to denote statical equivalence, AB + BC
BD, a vector through B; and, since BD + CA = 0, B, C, A are collinear. Example 1. In the parallelogram ABCD, E and F are the middle points of the sides AB, BC. Show that the lines DE, DF divide the diagonal AC into thirds and that AC cuts off a third of each line (Fig. 6b).
E
"
B Fia. 6c
Fia. 6b
The hypotheses of our problem are expressed by the equations:
d - a=c - b,
2e=a+b,
2f =b+c.
Let DE cut AC at X. To find x, eliminate b from the first and second equations. Thus d + 2e 2a + c = =a;
d-a+2e=a+c and
3
3
for the first member represents a point on DE, the second member a point on AC, and, since the points are the same, the point is at the intersection X of these lines. Comparison with (4) now shows that X divides DE in the ratio 2/1, AC in the ratio 1/2. Let DF cut AC at Y. To find y, eliminate b from the first and third equations. Thus
d-a+2f=2c and d+2f-a+2c_ 3 3
Hence Y divides DF in the ratio 2/1, AC in the ratio 2/1. Example 2. In a plane quadrilateral ABCD, the diagonals AC, BD intersect at P, the sides AB CD intersect at Q (Fig. 6c). If P divides AC and BD in the ratios 3/2 and 1/2, respectively, in what ratios does Q divide the segments A B, CD?
§6
COLLINEAR POINTS
11
By hypothesis 2a + 3c
2b + d
5
3
p hence
6a + 9c = 10b + 5d and
-
5d = lOb - 6a _ 4
for the first fraction represents a point
on CD, the second a point on AB, and both points are the same, that is, the point Q. Therefore Q divides CD
in the ratio -5/9, AB in the ratio -10/6. Example 3. Prove that the midpoints of the diagonals of a complete
quadrilateral are collinear.
In the complete quadrilateral ABCDLM (Fig. 6d) let P, Q, R be the mid-points of the diagonals AC, BD, LM. The sum of the line vectors,
--,
--->
AB + AD = 2 AQ,
--,
---
--i
CB + CD = 2 CQ,
--> ---> QA + QC = 2 QP;
hence we have the statical equivalence,
AB +AD +CB +CD =4PQ, for the quadrilateral ABCD with diagonals AC, BD. Similarly, for the rilateral BLDM with diagonals BD, LM,
ad-
--9 - 4 --+ BL + BM + DL + D31 = 4QR; -->
and, for the quadrilateral LA.11C with diagonals LM, AC,
-,
--4 --4 ----+ --+ LA + LC + MA + MC = 4 RP. On adding these three equations, we find that the entire left member is statically equal to zero; for --+ --- --> ---* ---
-9 DM+MA = 0,
=0,
-+ CB +BM +MC-0,
-+ CD +DL +LC -O, -4
are statical equations, since the vectors in each are collinear.
PQ +QR+RP =0, and I', Q, R are collinear.
Hence
VECTOR ALGEBRA
12
§7
7. Coplanar Points. THEOREM. If no three of the points A, B, C, D are collinear, they will lie in a plane when, and only when, there exist four numbers a, /3, y, S, different from zero, such that (1)
as+$b+yc+Sd = 0,
a+0+y+S = 0.
Proof. If A, B, C, D are coplanar, either AB is parallel to CD,
or AB cuts CD in a point P (not A, B, C, or D). In the respective cases, we have
b - a=K(d - c);
a+Ab
c+A'd
1+A = 1+A, -P,
where A, A' are neither 0 nor -1. In both cases, a, b, c, d are connected by a linear relation of the form (1). Conversely, let us assume that equations (1) hold good. If a + /3 = 0 (and hence
y + S = 0), we have
a(a - b) + y(c - d) = 0 and the lines AB and CD are parallel. If a + /3 5,16 0 (and hence
y+6
0),
as+/3b
(2)
a+/i
yc+Sd
y+6 -P
where P is a point common to the lines AB and CD. In both cases, A, B, C, D are coplanar.
Note that (2) states that the point P in which AB, CD intersect divides AB and CD in the ratios #/a, 6/y. Similarly, if
a+y; 0, (3)
as + yc
/3b + 6d
a -f-.y
#+ 6
- qi
thus Q, the point in which AC and BD intersect, divides AC and BD in the ratios y/a, S//i. What conclusion can be drawn if a + 3 F6 0? If an equation of the form (1) subsists between four distinct noncoplanar points, we must conclude that a = a = y = S = 0. For at least one coefficient 6 = 0; then, from
as+/3b+yc = 0,
a+0+y = 0,
and the fact that A, B, C are not collinear, we deduce (§ 6) that
a=0=y=0.
COPLANAR POINTS
§7
Example 1. The Trapezoid. DC (Fig. 7a), then
13
If ABCD is a trapezoid with AB parallel to
AB=XDC or b-a=A(c-d). Hence we have
b+Xd=a+Xc or b-Xc=a-Xd. These equations may be written
b+ad_a+Xc_-P 1+a +a
b - Ac_a - Ad
-q;
for the former expressions represent the point P where the diagonals BD, AC meet, and the latter the point Q where the sides BC, AD meet. Evidently P divides both BD and AC in the same ratio A; and Q divides both BC and AD in the same ratio -A. In what ratio does the line PQ divide AB? In particular, if A = 1, the trapezoid becomes a parallelogram. The diagonals then bisect each other at P, while Q recedes to infinity. Q
C
Fia. 7a Example 2. Theorem of Menelaus. If a line s cuts the sides BC, CA, AB of the triangle ABC in the points P, Q, R, respectively, the product of the ratios in which P, Q, R divide these sides equals - 1. Conversely, if P, Q, R divide the sides of the triangle in ratios whose product is - 1, the points are collinear. Proof. (Fig. 7b.) We lose no generality if we assume that P, Q divide
BC, CA in the ratios -y//3, -a/y: (i)
(/3-y)P=$b-yc,
(ii)
(y-a)q=yc-aa.
In order to locate R, which lies on the lines PQ, AB, we seek a linear relation
between p, q, a, b. Add (i) and (ii) to eliminate c and divide by S - a; then
(R - y)P + (y - a)q
Sb - as
Thus R divides AB in the ratio -S/a. The product of the division ratios
-y//3, -a/y, -/3/a is -1.
VECTOR ALGEBRA
14
§7
Conversely, let us assume that P, Q, R divide BC, CA, AB in the ratios
-y/S, -a/y, -a/a whose product is -1. Then we have equations (i), (ii) and also
(a - fl)r = as - 13b.
(iii)
From these we deduce the linear relation,
(0-y)p+(y-a)q+(a-$)r=0,
(iv)
in which the sum of the coefficients is zero. The points P, Q, R are therefore collinear. * Note. From (i), (ii),
(iii) it is easily proved that the three pairs of lines BQ, CR; CR, AP; AP, BQ meet in the points A', B', C' given by
(-a+0+y)a' _ -aa+ftb+yc, (a-S+y)b' =aa-$b+yc,
(a+9-y)c'=as+9b-yc. If we add 2aa, 2gb, 2yc, respectively, to these equations, we find that the point S given by s =
(v)
as +Ab+yc
a+#+y
is common to the lines AA', BB', CC'. Thus to every line s given by (iv) we have a corresponding point S given by (v)
-the pole of s relative to the triangle ABC.
C
Example 3. Theorem of Ceva. If S is a point in the plane of the triangle ABC, and the lines SA, SB, SC cut the sides opposite in the points A', B', C', then the product of the ratios
C,
Fte.7c
in which A', B', C' divide the sides BC, CA, AB equals 1. Conversely, if A', B', C' divide the sides BC, CA, AB in ratios whose product B is 1 , the lines AA', BB', CC meet in a p oint . Proof. (Fig. 7c.) Since A, B, C, S are coplanar,
aa+pb+yc+as =0,
a+0+y+S =0.
Hence
_
(i) a
(ii)
b' = c'
Sb + ye
as + SS
0+y
a+s
yc
as _ Rb + as
y+a
+b '
=as+Sb -yc+Ss. a+R y+S
*This conclusion is obvious; for the line PQ must meet AB in the point R for which the product of the division ratios is -1.
§7
COPLANAR POINTS
15
These equations state that A', B', C' divide BC, CA, AB in the ratios -y/fl, a/y, fl/a, whose product is 1. Incidentally, A', B', C' divide SA, SB, SC in the ratios a/a, fl/S, y/3 whose sum is -1. Conversely, let us assume that A', B', C' divide BC, CA, AB in the ratios y/$, a/-j, fl/a whose product is 1. Then we have equations (i), (ii), (iii).
From these we find that the vectors as + (0 + y)a', fib + (y + a)b', yc + (a + fl)c' are all equal to as + lb + yc; the point, s
(iv)
as+pb+yc a + fl + y
is therefore common to the lines AA', BY, CC'. Note. From (i), (ii), (iii) it is easily proved that the three pairs of lines BC, B'C'; CA, C'A'; AB, A'B' meet in the points P, Q, R given by
(Q-y)p=Rb-yc, (y - a)q = yc - aa,
(v)
(a-fl)r=as-/3b. From these equations we deduce the linear relation, (vi)
(fi-y)p+(y-a)q+(a-a)r=0,
in which the sum of the coefficients is zero. The points P, Q, R therefore lie
on a line s. Thus to every point S given by (iv) we have a corresponding line s whose points P, Q, R are given by (v)-the polar of S relative to the triangle ABC. Example 4. Let ABC and A'B'C' be two triangles, in the same or different
planes, so that the vertices A, B, C correspond to A', B', C', and the sides AB, BC, CA correspond to A'B', B'C', C'A'. We then have (Fig. 7d) P
FIG. 7d DESARGUES' THEOREM. If the lines joining the corresponding vertices of two
triangles are concurrent, the three pairs of corresponding sides intersect in collinear points, and conversely.
Let the lines AA', BY, CC' intersect at S; then as + a'a' = fib + fl'b' = yc + y'c' = s,
a+a =t3+fl' ='Y+y'=1.
VECTOR ALGEBRA
16
§7
From these equations we find in the usual manner the points P Q, R in which BC, B'C'; CA, C'A'; AB, A'B' intersect: (i)
Jr
=
(ii)
q
=
(iii)
'6',
fib
fl'b'
yC - as
y'C' - a'a'
- yc = fl_y
y_a
- y'c/
1
y,_«,
_«a-3b «a'-fl'b' at - of
Hence (iv) (v)
(l; - 'y)p + (-y - a)q + (a - 6)r = 0,
'y')p + ('y' - a')q + (a' - 6')r = 0;
either equation shows that P, Q, R are collinear. To prove the converse, we may start with the expressions (i), (ii), (iii) for p, q, r. These ensure that P, Q, R are collinear; but, in order that (iv) and (v) determine the same division ratios for P, Q, R, we must have
,-.y' 'y'-a' y
y - a
a -
= h,
or
(vii)
a'-ha =fl'-hfi=y'-hy=k,
h and k representing the values of the equal members of (vi) and (vii). In the usual way we now find from (i), (ii), (iii) that
aa'-haa=6'b'-hfib=y'c'-hyc=ks, where S is a point common to AA', BB', CC'.
FIG. 7e
Example 5. The Complete Quadrangle. A complete quadrangle consists of four coplanar points, its vertices, no three collinear, and the six lines, its sides,
which join them. The three pairs of sides which do not meet at a vertex
COPLANAR POINTS
37
17
arc said to be opposite; and the three points in which they meet are called rliagon
The properties of this configuration of points and lines must all be consequences of the fact that A, B, C, D are coplanar points, that is,
CA +(3b+yc+Sd =0,
(i)
a+$+y+a =0.
Since no three vertices are collinear, none of the scalars a, ,, y, S are zero. From equations (i) we locate at once the diagonal points at the intersections of the opposite sides:
Go +yc-aa+ad a+S
m
0+y _yc+aa
y+a
_aa+13b a+13
/3b+5d ,
13+S
yc+Sd 'Y+s
We here assume that no two of the scalars a, fl, y, S have a zero sum; the diagonal points L, M, N are then all in the finite plane. But if for example, a + 0 = y + S = 0, AB and CD are parallel, and N is the point at infinity in their common direction. To find the points X1, X2 where LM cuts AB and CD, we seek linear relations connecting 1, m, a, b and 1, m, c, d, respectively. Thus, from (ii) and (iii),
('Y+a)m- (S +y)1 (vi)
a-/3 (0+y)1-(S+S)m X2
y-S
as -/3b
a-0 '
yc - Sd
y-S
Equations (iv) and (v) show that N and X1 divide AB in the ratios '6/a and -/3/a; hence N, X1 are harmonic conjugates of A, B. Equations (iv) and (vi) show that N and X2 divide CD in the ratios S/y and -S/y; hence N, X2 are harmonic conjugates of C, D.
Equation (v) shows that X1 divides LM in the ratio -(y + a)/(3 + y);
and (vi) shows that X2 divides LM in the ratio -(S + 5)/(3 + y) _ (y + a)/(3 + y); hence X1, X2 are harmonic conjugates of L, M. We may now state the following harmonic properties of the complete quadrangle in the THEOREM. Two vertices of a complete quadrangle are separated harmonically
by the diagonal point on their side and by a point on the line joining the other two diagonal points. Two diagonal points are separated harmonically by points on the sides passing through the third diagonal point.
VECTOR ALGEBRA
18
§8
From (ii), (iii), (iv) we next find the points X1, Y1, Z1 in which AB, BC, CA
are cut by LM, MN, NL, respectively; thus $1 = Since
as - $b
a-/3
I
y1 =
yc - as
3b - yc
0-y
(a-/3)xl+(f -y)yl+(y-a)Z1 =0,
,
Z1 =
y-a
a-8+0-y+y-a =0,
the points Xl, Y1, Zl are collinear. This is also a consequence of Desargues'
Theorem applied to the triangles ABC, LMN, which are in perspective from D.
Each vertex of the quadrangle is the center of perspective of the triangles formed by the other three vertices and by the three diagonal points. Thus, with careful regard to exact correspondence, the triangle LMN is in perspec-
tive with DCB, CDA, BAD, ABC from A, B, C, D, respectively. Hence the coliesponding sides of these triangles intersect in four lines a, b, c, d (only d, the line X1Y1Z1, is shown in Fig. 7e). These lines are the polars of A, B, C, D with respect to the triangle LMN (ex. 3). Corresponding to the complete quadrangle given by four points A, B, C, D, we now have a complete quadrilateral given by four lines a, b, c, d. In these configurations the roles of points and lines are interchanged (Fig. 7f): The quadrangle has four vertices A, B, C, D and six sides consisting of three opposite pairs (BC, AD), (CA, BD), (AB, CD) which meet in the three diagonal points L, M, N. The quadrilateral has four sides a, b, c, d and six vertices consisting of three opposite pairs (bc, ad), (ca, bd), (ab, cd) which lie on the three diagonal lines 1, m, n.
The quadrangle and quadrilateral have the same diagonal triangle LMN or lmn; the sides 1, m, n are opposite the vertices L, M, N.
8. Linear Relations Independent of the Origin. We have seen that the position vectors of collinear or coplanar points satisfy a
CENTROID
§9
19
linear equation in which the sum of the scalar coefficients is zero. The significance of such relations is given by the THEOREM. A linear relation of the form,
XiP1 + X2P2 + ... + XnPn = 0,
(1)
connecting the position vectors of the points P1, P2, , Pn will be independent of the position of the origin 0 when, and only when, the sum of the scalar coefficients is zero:
X1 + X2 -{- ... + an = 0.
(2)
Change from 0 to a new origin 0'. Writing OPi = pi, O'Pi = pi, 00' = d, we have pi = d + pi; hence (1) becomes Proof.
(X1 + t2 +. +Xn)d+X1p1 +X2P2
XnPn' = 0.
This equation will have the same form as (1) when and only when (2) is satisfied.
For two, three, and four points relations independent of the origin have a simple geometric meaning; namely, the points are coincident, collinear, or coplanar, respectively. The question now arises: What geometric property relates five or more points whose position vectors satisfy a linear relation independent of the origin? 9. Centroid. We shall encounter problems in which each point
of a given set is associated with a certain number. The points, for example, may represent particles of matter and the numbers, their masses or electric charges. In the latter case, the numbers may be positive or negative. We shall now define a point P* called the centroid of a set of n points P1, P2, , Pn associated with the numbers ml, m2i , mn, respectively. Denote any one of these "weighted" points by the symbol m.iPi; then, if the sum of the numbers mi is not zero, the centroid of the entire set is defined as the point for which the sum of all the vectors m1P*Pi is zero.
The
defining equation for the centroid is thus (1)
2;miP*Pi = 0,
provided Mmi Fl- 0.
As we wish P* to be uniquely defined, the case Emi = 0 must be excluded; for then (1) is a relation independent of the position of P* (j 8).
i9
VECTOR ALGEBRA
20
When 2 mi 0 0 there is always a unique point P* which satisfies (1). written
For if we choose an origin 0 at pleasure, (1) can be ---4
-4
F.mi(OPi - OP*) = 0, or
(2mi)OP* = 1m1OPi.
(2)
This relation, independent of the origin, fixes the position of P* relative to 0. The point P* thus determined is the only point that satisfies (1) ; for if Q is a second point for which EmiQPi = 0, we have, on subtraction from (1), 2;mi(P*Pi
- QPi) = Timi(P*Pi + PQ) = (2;mi)P*Q = 0;
hence P*Q = 0, and Q coincides with P*.
If the position vectors of P*, Pi are written p* and pi, (2) becomes
(2;mi)P* = Emipi.
(3)
The centroid of the points miP1 is not altered when the numbers mi are replaced by any set of numbers cmi proportional to them;
for, in (2), the constant c may be canceled from numerator and denominator. In particular, if the numbers mi are all equal, we may replace them all by unity; the centroid of the n points then is called their mean center and is given by the equation: (4)
-3
1 --
1
OP* _ - ZOP1, or p* _ - Zpi. n n
In finding the centroid P* of any set of weighted points miP1, we may replace any subset of points for which the sum of the weights is a number m' 0 0 by their centroid P with the weight m'. For, if V and 1", respectively, denote summations extended over the points of the subset and over all the remaining points, we may write (3) in the form, (2;'mi + Z"mi)P* = 7,'mipi + E"miPi, or
(m' + I"mi)P* = m'p' + E"mipi.
CENTROID
§9
21
This equation shows that P* is also the centroid of the point m'P' and the points miPi not included in the subset.
Finally, let us consider the nature of a set of n points miPi If we attempt to find a point P*
(?Pi F4- 0) for which Ximi = 0. which satisfies
ZmiP*Pi = 0
(5)
when Xmi = 0,
we find that there are two possibilities. (a) For any arbitrary choice of the points miPi, subject only to
the condition Emi = 0, there will in general be no point P* that will satisfy (5). Thus in the cases n = 2, 3, 4, equations (5) imply,
respectively, that P1 and P2 coincide; P1, P2, P3 are collinear; P1, P2, P3, P4 are coplanar. Hence, if these conditions are not fulfilled, P* does not exist. (h) If, however, a point P* can be found which satisfies (5), any point whatever will serve (§ 8). In fact, we see, from (2), that the existence of P* implies that
1mi = 0,
,6dmiOPi = 0
(6)
for any choice of 0. In particular, we may take any one of the given points Pi as origin. Thus, if we take 0 at P1, (6) becomes V
i=2
miP1 Pi = 0,
2; mi = -m1
i=2
0.
Hence, from the defining equation (1), we see that P1 is the centroid of the points m2P2, m3P3, . . . , mnPn. Precisely the same , Pn. We now can answer conclusion may be drawn for P2, P3, the question raised at the end of § 8. Any set of weighted points miPi (mi F6 0) whose position vectors satisfy a linear relation (6) independent of the origin has the intrinsic property that any point of the set is the centroid of all the remaining weighted points.
A set of n points having this property may be readily constructed. Take any set of n - 1 weighted points miPi, such that 0, and adjoin to the set their centroid + mn_1 m1 + m2 + P* = Pn with the weight Mn = -(m1 + m2 + + mn-1). Then, since -mnPn = m1P1 + m2P2 + we have
+ mn-1Pn-1,
n
n
X mipi = 0,
2;mi = 0. t
1
VECTOR ALGEBRA
22
§9
Thus the relation,
as+Sb+yc=0,
(7)
a+$+-y=0,
between three collinear points shows that each point of the set aA, ,BB, yC is the centroid of the other two. Any point C on the line of A and B satisfies a relation of the type (7) and hence is the centroid of these points when suitably weighted. The relation, (8)
as+,9b+yc+8d = 0,
a+a+7+S = O,
between four coplanar points shows that each point of the set aA, ,BB, yC, 8D is the centroid of the other three. Any point D in the plane of A, B, C satisfies a relation of the type (8) and hence is the centroid of these points when suitably weighted. Example 1. Centroid of Two Points. The centroid of aA and pB is given by *
P
as+Rb a+13
Hence P* divides AB in the inverse ratio f/a of their weights. In particular, if a = S, P* divides AB in half. The mean center of two points lies midway between them.
Example 2. Mean Center of Three Points A, B, C. Let L, M, N be the midpoints of BC, CA, AB (Fig. 9a). Then the mean center P* of A, B, C is the centroid of A and 2L, B and 2M, C and 2N. Hence (ex. 1) the segments AL, BM, CN are all divided by P* in the ratio of 2/1. If A, B, C are not collinear, they are the vertices of a triangle. Its medians AL, BM, CN intersect at P* and are there divided in the ratio of 2/1. C
N Fia. 9a
a
M2 FIG. 9b
Example 3. Mean Center of Four Points A, B, C, D. Let L1, L2; M1, M2; N1, N2 be the mid-points of A B, CD; BC, DA; AC, BD, respectively (Fig. 9b). To find P* we may replace A, B, C, D by 2L1 and 2L2, or by 2M1 and 2M2,
or by 2N1 and 2N2. Therefore P* is the mid-point of the segments L1L2, M1M2, N1N2. The bisectors of the three pairs of opposite sides of the quad-
BARYCENTRIC COORDINATES
§ 10
23
rangle ABCD (which may be plane or skew) intersect at P* and are there divided in half.
If A, B, C, D are not coplanar, they determine a tetrahedron. Let A', B', C', D' be the mean centers of the triads BCD, CDA, DAB, ABC. Then P* is the centroid of A and 3A', B and 3B', C and 3C', D and 3D', and hence divides each of the segments AA', BB', CC', DD' in the ratio of 3/1. The preceding result also shows that the bisectors of the three pairs of opposite sides of the tetrahedron meet at P*. Example 4. The sum of n vectors AiB1 (i = 1, 2,
, n) is given by
Z(bi-a;) =nb*-na* = n(b* - a*) where A* and B* are the mean centers of the initial points Ai and the terminal points Bi, respectively; hence 2;A;Bi = n A*B*.
In particular,
-4 -A1B1 + A2B2 = 2 A*B*,
where A*, B* are the mid-points of A1A2 and B1B2, respectively.
10. Barycentric Coordinates. If P is any point in the plane of
-3-3 --3
the reference triangle ABC, the vectors AP, BP, CP are linearly dependent (§ 5) ; hence
«(p - a) + /3(p - b) + y(p - c) = 0, p=as+9b+-ic (1)
«+6 + y
The denominator is not zero; for, if
a+a+y=0, then aa+(3b+yc=0, and A, B, C would be collinear, contrary to hypothesis. Thus P is the centroid of the weighted points aA, $B, yC. The three numbers a, /3, y are called the barycentric coordinates of P; as they
all may be multiplied by the same number without altering p, their ratios determine P. The vertices of the reference triangle have the coordinates A(1, 0, 0), B(0, 1, 0), C(0, 0, 1); the unit point (1, 1, 1) is the mean center of A, B, C. If P is a point in space and ABCD a reference tetrahedron, the
vectors AP, BP, CP, DP are linearly dependent (§ 5); hence
«(p-a)+$(pp-b)+y(p-c)+a(p-d) =0, (2)
p
as+Ab+yc+Sd a+O+y+S
.
VECTOR ALGEBRA
24
§ 12
The denominator is not zero; for, if
a+8+y+S = 0, then as+,lib+yc+Sd =0 and the points A, B, C, D would be coplanar, contrary to hypothesis. Thus P is the centroid of the weighted points aA, $B, yC, W. The four numbers a, 0, y, S are called barycentric coordinates of P; as they all may be multiplied by the same number without altering p, their ratios determine P. The vertices of the reference tetrahedron have the barycentric coordinates A(1, 0, 0, 0), B(0, 1, 0,
0), C(0, 0, 1, 0), D(0, 0, 0, 1); the unit point (1, 1, 1, 1) is the mean center of A, B, C, D. 11. Projection of a Vector. To
find the projection of a vector Fla. 11
PQ upon a line x, a director plane 7r must be specified. Pass planes
through P and Q parallel to 7r and let them cut x in the points P1i Ql (Fig. 11). Then the vector P1Q1 is the 7r-projection of PQ upon x. If no director plane 7r is specified, we tacitly assume that IT is perpendicular to x; the projection is then orthogonal. .
Let the vectors PQ, QR and their sum PR have the projections P1Q1, Q1R1 and P1R1 on the line x; then, since P1Q1 + Q1R1 = P1R1,
the projection of the sum of two vectors on a line is equal to the sum of their projection on this line.
12. Base Vectors. Let e1, e2, e3 be three linearly independent vectors; they are then non-coplanar. If the vectors are drawn from a common origin 0 (Fig. 12a), we may pass a closed plane curve through their end points El, E2, E3. If this curve is viewed from the side of its plane opposite to that on which 0 lies, the order E1E2E3 defines a sense of circulation. If this sense is counterclockwise, the set e1, e2, e3 is said to be right-handed or dextral; for
it is then possible to extend the thumb, index, and middle fingers of the right hand so that they have the directions of e1, e2, e3,
BASE VECTORS
§ 12
2'r
respectively. If the sense defined by E1E2E3 is clockwise the set is said to be left-handed or sinistral.
Any vector u = PQ may be expressed as the sum of three vectors parallel to el, e2, e3, respectively. For, if we construct a parallelepiped on PQ as diagonal by passing planes through P and
R
Fia. 12a
FIG 12b
Q parallel to e2 and e3, e3 and el, el and e2 (Fig. 12b), its edges will be parallel to el, e2, e3, and
PQ = PQ1 + Q1R + RQ = PQ1 + PQ2 + PQ3 _ y
Since PQi is a scalar multiple of ei, say u`ei, (1)
u = ulel + u2e2 + u3e3,
where ul, u2, u3 are numbers called the components of u with respect to the basis el, e2, e3. Their indices are not exponents but mere identification tags; they are written as superscripts for reasons
given in Chapter IX. The vector u is often written [ul, u2, u3], with brackets to enclose its components.
If u is the position vector OP, the components ui are called the
Cartesian coordinates of the point P with respect to the basis e1, e2, e3-
A straight line upon which two directions are distinguished is called an axis. One direction is called positive, the other negative. In a figure the positive direction is marked with an arrowhead. With a given basis, lines drawn through the origin 0 parallel to
VECTOR ALGEBRA
26
§ 13
el, e2, e3 and with their directions positive are called the coordinate axes. The numbers u', u2, u3 often are called the components of u on these axes. The components of a zero vector are all zero; for, since el, e2, e3 are linearly independent, u = 0 implies u' = u2 = u3 = 0. If A is any scalar, Au = Aulel + Au2e2 + Au3e3.
(2)
If v = vie1 + v2e2 +.v3e3,
u + v = (u1 + v')el + (u 2 + v2)e2 + (u3 + v3)e3
(3)
With the bracket notation, (2) and (3) become (4)
A[ul, u2, u3] = [Au', Au2, Au3],
(5)
[ul, u2, u3] + [v', v2, v3] = [u' + v1, u2 + v2, u3 + v3].
To multiply a vector by a number, multiply its components by that number; to add vectors, add their corresponding components. In par-
ticular,
-[u', u2, u3]
(6) (7)
[ul, u2, u3]
= [-u', -u2, -u3],
- [v', v2, v3] = [u'
- v', u2 -
v2, u3
-
v3].
If u = v, then u - v = 0; hence (8) U=V implies u' = v', u2 = v2, u3 = v3. When referred to the same basis, equal vectors have their corresponding components equal.
13. Rectangular Components. Let i, j, k denote a dextral system of mutually perpendicular unit vectors. From an origin 0 draw the coordinate axes x, y, z with positive directions given by i, j, k (Fig. 13). Any vector u now is determined by giving its components u1, u2i u3 on these rectangular axes: (1)
u = uli + u2j + u3k.
Draw OP = u from the origin, and let OP1, OP2i OP3 be its orthogonal projections on the axes. Then ul is the length of OPI,
taken positive or negative according as OPI has the direction of
i or -i; hence u1 = I u I cos (i, u),
§ 13
RECTANGULAR COMPONENTS
27
where j u j denotes the length of u and (i, u) the angle between
i and u. The rectangular components of u thus are obtained by multiplying its length by the corresponding direction cosines: (2)
ul = I u I cos (i, u),
u2 = I U I cos (j, u),
u3 = I u I cos (k, u).
R
Fic. 13
The Pythagorean Theorem gives the length of u in terms of its components; from
0P2 = (OR)2 + (RP)2 = (OPl)2 + (OP2)2 + (OP3)2, (3)
I U 12 = (ul)2 + (u2)2 + (u3)2.
If we substitute from (2) in (3), we obtain the relation,
cost (i, u) + cost (j, u) + cost (k, u) = 1, satisfied by the direction cosines of any vector. The rectangular coordinates of any point P are defined as the
components of its position vector OP on the x-, y-, and z- axes. Thus, if (4)
OP=xi+yj+zk,
P has the rectangular coordinates (x, y, z). Since P1P2 = OP2
28
VECTOR ALGEBRA
§13
- OPI, this gives
PIP2 = (x2 - xl)i + (Y2 - yl)j + (z2 - zl)k
(5)
if (xi, y,, zi) are the coordinates of Pi.
The components of PIP2 are found by subtracting the coordinates of PI from the corresponding coordinates of P2.
With an orthogonal basis i, j, k, the components may be written with subscripts (as previously) or superscripts. Separately, they
are called the x-, y-, and z- components and often are written ul, U2, U3- Just as with a general basis, vectors are specified by giving their components: u = -[ul, u2, u3]. Thus the point (x, y, z) has the position vector [x, y, z]; and (5) may be written PIP2 = [x2 - xl, Y2 - yl, z2 - Z11The unit vectors i, j form a basis for all vectors u in their plane:
u = uli + u2j.
(6)
If e is a unit vector in the plane and 0 is the angle (i, e), reckoned positive in the sense from i to j, we define cos 0 and sin 0 as the components of e :
e=icos0+jsin0.
(7)
Any vector in tine plane may be written (8)
u= Iule= Jul {icos(i,u)+jsin(i,u)};
its rectangular components are (9)
ul = I U I cos (i, u),
u2 = I u I sin (i, u),
u3 = 0.
Example. Addition Theorems for the Sine and Cosine. Let a and b be two unit vectors such that the angles (i, a) = a, (a, b) = 0; then the angle (i, b)
= a + 14, and, from (7),
a =icosa+jsina, b = i cos (a + fl) + j sin (a + 0). If we refer b to the new basis i = a, j (a unit vector 90° ahead of a), we have
b=icos0+isin 0 = (i cos a + j sin a } cos l3 + I i cos (a a + 2) + j sin (a + 2)1 sin 6.
SCALAR PRODUCT
15
29
Comparing the components of b in the two preceding expressions, we find cos (a + 13) = cos a cos 8 + cos
(a +
sin (a + /3) = sin a cos l3 + sin
(a + 2) sin
sin [3,
2) With a =7r/2, these equations give cos (/3
+ 2)
sin $,
sin (p + 2) = cos 0;
hence
cos (a + 0) = cos a cos 0 - sin a sin 0, sin (a + )3) = sin a cos g + cos a sin g. These addition theorems hold for all values of a and fl, positive or negative.
14. Products of Two Vectors. Hitherto we have considered only the products of vectors by numbers. Next we shall define two operations between vectors, which are known as "products," because they have some properties in common with the products of numbers. These products of vectors, however, will also prove to have properties in striking disagreement with those of numbers.
Since one of these products is a scalar and the other a vector, they are called the scalar product and vector product, respectively. The definitions of these new products may seem rather arbitrary to one unfamiliar with the history of vector algebra. We present this algebra in the form and notation due to the American mathematical physicist, J. Willard Gibbs (1839-1903).t It is an offshoot of the algebra of quaternions, adapted to the uses of geometry and physics. In Chapter X quaternion algebra is developed briefly, and the origin of the foregoing products is revealed. 15. Scalar Product. The scalar product of two vectors u and v, written u v, is defined as the product of their lengths and the cosine of their included angle: (1)
Jul IvI cos(u,v).
The scalar product is therefore a number which for proper (nonzero) vectors is positive, zero, or negative, according as the angle (u, v) is acute, right, or obtuse. Hence, for proper vectors, (2)
means u1v.
t Professor of mathematical physics at Yale University. His pamphlet on the Elements of Vector Analysis was privately printed in 1881. A more complete treatise on Vector Analysis (New Haven, Yale University Press, 1901) based on Gibbs's lectures, was written by Professor E. B. Wilson.
30
VECTOR ALGEBRA
115
When u and v are parallel,
Jul lvl,
or
-Jul lvI,
according as the vectors have the same or opposite directions; thus
From (1) we see that (au) ($v) = ag u . V.
(3)
The last result is obvious when a and i3 are positive numbers; the other cases then follow from the equations preceding.
Besides the components of a vector on the coordinate axes (§§ 12, 13) we shall also use the orthogonal projection and compo-
nent of a vector on an arbitrary directed line 1. If the positive direction of 1 is given by the unit vector e, the projection of u upon 1 (proj 1 u) is a scalar multiple of e. This scalar is called the component of u upon 1 (comp1 u) ; its defining equation is therefore e comp j u = proj 1 u.
(4)
As in § 13, we compute comp j u as
comps u = l u i cos (e, u);
(5)
since I e
1, this may also be written
comps u = e u.
(6)
From the projection theorem of § 11, proj 1 (u + v) = proj 1 u + proj 1 v;
(7)
hence, from (3), (8)
comps (u + v) = comp] u + comps v.
The operations expressed by proj1 and comp] are distributive with respect to addition.
The definition (1) of u v now may be written (9)
U V= I u l compu V=IV l comp u.
Scalar or "dot" multiplication is commutative and distributive: (10) (11)
w(u+v) =wu+wv.
SCALAR, PRODUCT
§ 15
31
The last equation is nothing more than (8) multiplied by when 1 is taken along w. If c 0 in the equation,
or
we can conclude either that a - b = 0 or that a - b and c are perpendicular. We cannot "cancel" c to obtain a = b unless a - b and c are not perpendicular. It is obvious that, in general, (u v)w
u(v w).
Since i, j, k are mutually perpendicular unit vectors, i
(12)
Hence, if we expand the product,
u ' v = lull + u2j + u3k) (v1i + v2j + v3k) we obtain u v = u1v1 + u2v2 + u3v3
(13)
The scalar product of two vectors is equal to the sum of the products of their corresponding rectangular components. Example 1. Moreover, comp,, u =
If u = [2, -1, 3], v = [0, 2, 4], u v = 0 - 2 + 12 = 10.
urv7 v
cos (u, v) =
I
= uU
10
/20 v v
= 2 236,
=
10
/280 =
compu v = 0.5979,
uuv =
luI
10
= 2.673;
14
angle (u, v) = 530 18'.
Example 2. Identities involving scalar products may be given a geometric i.iterpretation. Thus (Fig. 15a)
gives cdcosw=a2-b2, and, if we write c = a + b, d = a - b, w = anL,Ie (c, d). If a = b, c d = 0; then PQRS is a rhombus and the angle PRT may be inscribed in a semicircle about Q. We thus have two geometric theorems:
1. The diagonals of a rhombus cut at right angles.
2. An angle inscribed in a semicircle is a right angle.
Moreover, from
(a-b) (a - b) = we have the cosine law: d2 = a2+0 - 2ab cos 0.
P
a
Q
Fro. 15a
VECTOR ALGEBRA
32
§15
Example 3. We also may interpret identities involving scalar products by
regarding the vectors a, b,
as position vectors OA, OB,
from an
arbitrary origin O. Then a - b = BA and a + b = 2 011 where M is the mid-point of AB. Consider, for example, the identity,
(b-a)2+(c-b)2+(d-c)2+(a-d)2 = (c-a)2+(d-b)2+(a+c-b-d)2, which can be verified on expansion; u2 means u u. If we regard a, b, c, d as the position vectors of the vertices of a space quadrilateral ABCD, p = (a + c), q = (b + d) locate the mid-points P, Q of the diagonals AC, BD; 2
hence
(AB)2 + (BC)2 + (CD)2 + (DA)2 = (AC)2 + (BD)2 + 4(QP)2. The sum of the squares of the sides of any space quadrilateral equals the sum of the squares of its diagonals plus four C times the square of the segment joining their middle points. Example 4.
The identity,
(a-b) (h-c) 0,
B
Fia. 15b
shows that the altitudes of a triangle
ABC meet in a point H, the ortho-
center of the triangle (Fig. 15b); for, if two terms of this equation are zero, the third is likewise. Similarly the identity,
(a-b) k- a +2 b \
b+c 2
c+a\ C
shows that the perpendicular bisectors of the triangle ABC meet in a point K, the circumcenter of the triangle. For the orthocenter H and circumcenter K the individual terms of the foregoing equations vanish; for example,
(a -
0,
(a -
0.
On adding these, we obtain
=0; or on writing g = 3 (a + b + c) for the position vector of the mean center G of the triangle ABC.
(a-b) (h+2k-3g) =0.
SCALAR PRODUCT
§ 15
33
Since this equation also holds when a - b is replaced by b - c and c - a, we conclude that
h + 2k - 3g = 0. Therefore the mean. center of a triangle lies on the line joining the orthocenter to the circumcenter and divides it in the ratio of 2/1. This line is called the Euler line of the triangle. Example 5. In order to interpret the identity,
(a + b - c - d)2- (a - b - c+ d)2 = 4(a - c) - (b-d), with reference to the plane quadrilateral ABCD, let P, Q, R, S denote the mid-points of AB, BC, CD, DA; then (Fig. 15e) we have
(PR)2 - (QS)2 = CA DB. Hence, if the diagonals of a quadrilateral cut at right angles, the lines joining the mid-points of opposite sides are equal. Since the lines PR, QS intersect D
P R' Fta. 15c
Fia.15d
in the mean center N of the points A, B, C, D and are bisected there (§ 9, ex. 3), a circle with N as center will pass through P, Q, R, S. If perpendiculars
from P, Q, R, S are dropped upon the sides opposite, their feet, P', Q', R', S also will lie on this circle. Thus we have proved the THEOREM. When the diagonals of a quadrilateral are perpendicular, the midpoints of its sides and the feet of the perpendiculars dropped from them on the opposite sides all lie on a circle described about the mean center of the vertices.
In the figure formed by a triangle ABC and its three altitudes meeting at the orthocenter H (Fig. 15d), three quadrilaterals, ABCH, BCAH, CABH, all have perpendicular diagonals. Their three eight-point circles are all the same. This circle, whose center is at the mean center N of A, B, C, H, is the famous nine-point circle of the triangle. THEOREM. For any triangle ABC whose orthocenter is H, a circle whose center is the mean center of A, B, C, H, passes through the nine points: the mid-points of the sides, the feet of its altitudes, and the mid-points of the segments joining H to the vertices.
VECTOR ALGEBRA
34
§ 16
The center N of the nine-point circle has the position vector n given by
4n =a+b-}-c+h=3g+h=2h+2k, in view of ex. 4. Therefore the center N of the nine-point circle is collinear with the mean center G of the triangle, its orthocenter H, and circumcenter K. Moreover N bisects the segment HK.
16. Vector Product. The vector product of two vectors u and v, written u x v, is defined as the vector,
uxv= Jul Ivlsin(u,v)e,
(1)
where e is a unit vector perpendicular to both u and v and forming with them a dextral set u, v, e. If u and v are not parallel, a righthanded screw revolved from u towards v will advance in its nut towards u x v.
When u and v are parallel, e is not defined; but in this case sin (u, v) = 0 and u x v = 0. Moreover, if u and v are not zero, u x v = 0 only when sin (u, v) = 0. Hence, for proper vectors,
uxv = 0 means u I I v.
(2)
In particular u x u = 0. From (1) we see that
(-u)xV = ux(-v) _ -uxv,
(-u). (-v) = uxv,
(au) x ((3v) = a# u x v.
(3)
The last result is obvious when a and /3 are positive numbers; the other cases then follow from the equations preceding.
If u and v are interchanged in (1), the scalar factor is not altered, but e is reversed; hence (4)
vxu = -uxv.
Vector multiplication is not commutative.
Draw u and v from the point A, and let p be a plane perpendicular to u at A (Fig. 16). Then u x v may be formed by a sequence of three operations: FIG. 16
(P)
Project v on p, and obtain v';
(M) Multiply v' by I u I, and obtain I u I v'; (R) Revolve I u I v' about u through +90°
VECTOR PRODUCT
316
The resulting vector agrees with u x v in magnitude, for I V, I v I sin (u, v), and also in direction (upward in the figure). We is dicate this method of forming u X v by the notation, u X v = RMPV.
(5)
This means that v is projected, and the projection multiplied, an finally revolved as previously described. Now each of these ope ators is distributive: operating on the sum of two vectors is the sarr.
as operating on the vectors separately and adding the result: hence
RMP(v + w) = RM(Pv + Pw) = R(MPv + MPw) = RMPV + RMPvc Thus, from (5), (6)
ux (v + w) = uxv + uxw,
(V + W) Xu = vxu + wxu
Vector or "cross" multiplication is distributive. By repeated appli
cations of (6) we may expand the vector product of two vecto sums just as in ordinary algebra, provided that the order of th, factors is not altered.
For example,
(a+b)x(c+d) =axc+axd+bxc+bxd. If c 7d 0 in the equation,
axc=bxc, or (a-b)xc=0, we can conclude either that a - b = 0 or that a - b and c are parallel. We cannot "cancel" c to obtain a = b unless a - b and c are not parallel. We shall see in § 18 that in general, (u x v) x w P6 u x (v x w).
Since the unit vectors i, j, k form a dextral orthogonal set, we have the cyclic relations. (7)
ixj=k, jxk=i, kxi=j;
ixi=jxj=kxk=0.
Hence, if we expand the product, u x v = (u1i + u2j + u3k) x (v1i + v2j + v3k), we obtain (8)
u x V = (u2v3 - u3v2)i + (u3v1 - u1v3)j + (ulv2 - u2v1)k.
VECTOR ALGEBRA
36
§ 16
The components of u x v are the determinants formed by columns ul U2 u3 °l and 3, 3 and 1 (not I and 3), 1 and l of the array (VI
hence we may write (9)
i
j
k
u1
U2
U3
V1
V2
V3
V2
V3
For example, if u = [2, -3, 5], v = [-1, 4, 2], we compute the components of u x v from the array,
2 -3 5 C
-1
42 '
2 5 -3 5 _ -9, = - 26, 2 -1 42
2 -3 -1
=5.
4
Thus u x v = [ -26, -9, 5 ]. As a check, we verify. that u x v is
perpendicular to both u and v: -52 + 27 + 25 = 0, 26 - 36 + 10 = 0. Example 1. To find the shortest distance d from a point A to the line BC. Method. Let e be the unit vector along BC. Then, if u is any vector from A to the line (as AB or AC),
d = I uI sin(u,e) =Iuxe1. Computation. If the points are A(3, 1, -1), B(2, 3, 0), C(-1, 2, 4),
BC=[-3,-1,4],
u=AB=[-1,2,1], [9, 1, 7] u x e = x/26
d = IuxeI
,
a=[-3, -1,4]
x/26 i s -_ -2.245
-4
Check. If we take u = AC = [-4, 1, 5], u x e again has the preceding value.
Example 2. To find the shortest distance d from a point A to the plane BCD. Method. Find a vector normal to the plane, such as BC x BD, and let n be the unit vector in its direction. Then, if u is any vector from A to the
plane (as AB or AC), d = n u Computation.
If the points are A(1, -2, 1), B(2, 4, 1), C(-1, 0, 1),
D(-1, 4, 2), -- 4
BC = [-3, -4, 0],
-i
u=AB=[1,6,0]; Check.
-, BD = [-3, 0, 1], n=
---->
---4
BC-BD = [-4,3, -12]; [-4,3, -121 _ 14 13
-4 If we take u = AC = [-2, 2, 0], n u =.
VECTOR AREAS
17
37
Example 3. To find the shortest distance d between two non-parallel lines
AB, CD; and to locate the shortest vector PQ from AB to CD. Method. Find the vector AB x CD which is perpendicular to both lines, and let n be a unit vector in its direction. Then, if u is any vector from AB
to CD (as AC orBD), d = In-
To find P and Q, write AP = a AB, CQ = y CD, and find the scalars a, y
from the condition that PQ = PA + AC + CQ is parallel to AB x CD. The length PQ = d. Computation. If the lines AB, CD are given by the points A(1, -2, -1),
B(4, 0, --3); C(1, 2, -1), D(2, -4, -5);
-4
AB = [3, 2, -2],
n=
--4 CD
= [1, -6, -4],
3[-2, 1, -21,
-->
--->
AB x CD = 10[-2, 1, -2];
d = n AC =
AC = [0, 4, 0],
To find P and Q, we have
--4 -+ -4 -PQ = -aAB +AC +yCD
= -a[3, 2, -2] + [0, 4, 0] + y[l, -6, -4]
= [-3a+y, -2a+4-6y,2a-4y]; and, since PQ is parallel to [-2, 1, -2],
-3a + y
-2
-2a + 4 - 6y
2a - 4y
1
-2
These equations give a = y = $; hence
OP=OA+AP=[1,-2,-1]+$[3,2,-2]=$[21,-10,-17], OQ = OC + CQ = [1, 2, -1] + $[1, -6, -4] _ $[13, -6, -25]. Check.
The distance PQ =
.
17. Vector Areas. Consider a plane area A whose boundary is traced in a definite sense-shown by arrows in Fig. 17a. We shall
V
U
FIG. 17a
Fia. 17b
associate such an area with a vector of magnitude A, normal to its plane, and pointing in the direction a right-handed screw would move if turned in the given sense. Thus the parallelogram whose sides
VECTOR ALGEBRA
38
$ 17
are the vectors u, v (Fig. 17b) and whose sense agrees with the rotation that carries u into v is associated with the vector u x v; for its area is
A = Iul Ivlsin(u,v) = Iuxvl.
All plane areas associated with the same vector will be regarded as equal. The sum of two plane areas associated with the vectors,
u, v, is defined as the plane area associated with u + v.
A directed plane is a plane associated with a definite normal vector, say the unit normal n. A circuit in a directed plane is positive when it is counterclockwise relative to n; a clockwise circuit is negative. Consider now a plane area A
associated with the vector u. The projection of A on a directed plane p is A' = A cos 8
Fm. 17c
The circuital sense of A is projected on A'. We shall give A' the sign corresponding to its circuit on p, and call this signed area the component of A on p. Since I u = A and (n, u) = 8 or ir - 0, (Fig. 17c).
n- u = u I cos (n, u) = A cos 0 or -A cos 8 according as (n, u) is acute or obtuse. In both cases n u gives the component of A on p:
compp A = n u.
(1)
THEOREM. When the vector areas of the faces of a closed poly-
hedron are all drawn in the direction of the
C
outward normals, their sum is zero. Proof.
Any polyhedron may be subdi-
vided by planes into a finite number of
A
tetrahedrons. Let ABCD (Fig. 17d) be one FIG. 17d such tetrahedron. If we imagine ABC to lie in the plane of the paper while D is above the paper, the outward vector areas of the triangular faces, DAB,
DBC,
DCA,
ABC,
VECTOR AREAS
§ 17
39
are given by one half of the respective vectors, DA X DB,
DB X DC,
-AB-AC.
DC X DA,
If we choose D as origin, the sum of these vectors is given by
axb+bxc+cxa - (b - a)X(c - a) = 0. Thus the theorem is true for a tetrahedron. Now write such an equation for all the tetrahedrons that make up the polyhedron, and add the results. The vector areas over all inner faces cancel, for each appears twice but with opposed directions. The net result on the left is double the sum of the outward vector areas of the polyhedron's faces. Since this sum is zero, the theorem follows. Consider now a polygon P1P2
. Ph of area A lying in a directed plane of unit normal n (Fig. 17e), and suppose that the D circuit P1P2 . Ph is positive. In the figure n points up from the plane of the paper, so that a positive circuit is .
counterclockwise. Choose an origin 0 at pleasure above the plane (towards P
the reader), and let r1,
r2,
,
rh
''
denote the position vectors of the vertices. These vectors are the edges of P, a pyramid having 0 as vertex and the Fla. 17e given polygon as base. The triangular faces of this pyramid have as their outward vector areas 2r1 x r2i ire X r3, , irh x r1, while the outward vector area of the base is -An. From the preceding theorem the sum of these vector areas is zero; hence
An = (r1 x r2 + r2 x r3 + ... + rh x r1). i As the vertex 0 approaches the plane of the base, the terms on the right of (2) vary continuously, but their sum is always An. Hence when 0 is in the plane of the polygon, (2) remains valid. If 0 is any origin of rectangular coordinates in the plane, let the (2)
vertices of the polygon be (x1, yl), (x2, y2), counterclockwise order. Now n = k, and
,
(xh, yh) taken in
x1
x2
Y1
Y2
r1 X r2 = (x1i + yll) X (x2i + y2i) =
k,
VECTOR ALGEBRA
40
§ 18
so that (2) gives
2A = I xl
(3)
Y1
+
x2
I
Y2
I
f
x2
x3
y2
Y3
-+ -. ... + I
I
Xh
xl
Yh
Y1
J
Example 1. The sum of the outward vector areas of the triangular prism (Fig. 17f) is
uxw +vXw - (u +v) xw = 0. The vector areas of the triangular bases cancel, for they are equal in magnitude but opposite in direction.
Example 2. To find the area A of a triangle
u
whose vertices are (a, 0, 0), (0, b, 0), (0, 0, c), put r1 = ai, r2 = bj, r3 = ck in (2); then
FIG. 17f
An = (abk + bci + caj), A = 11/a2b2 + b2c2 + OA i Example 3. To find the area of the polygon whose vertices in counterclockwise order are (1, 2), (5, 4), (-3, 7), (-5, 5), (-1, -3), we form the array, (x)
1
5
(y)
2
4
repeating the first column.
-3 -5 -1 7
5
-3
1
2,
Then, from (3),
2A = -6+47+20+20+1 =82,
A
=41.
18. Vector Triple Product. The vector (u x v) x w is perpendicular to u x v and therefore coplanar with u and v; hence (§ 5), (u x v) x w = au + /3v.
But, since (u x v) x w is also perpendicular to w, aU
All numbers a,
a = -A v w,
0.
that satisfy this equation must be of the form = X u w, where X is arbitrary. Thus we have
(uxv)xw = In order to determine X, we use a special basis in which i is collinear with u, j coplanar with u, v; then V = v1i + v2j, w = wli + w2j + w3k. u = u1i, Substituting these values gives, after a simple calculation, A = 1 We therefore have the important expansion formulas, (1)
(uxv)xw = u-WV
wx(uxv) = w-vu
-wuv.
SCALAR TRIPLE PRODUCT
§ 19
41
In the left-hand members of (1), one of the vectors in parenthesis is adjacent to the vector outside, the other remote from it. The right-hand members may be remembered as (Outer dot Remote) Adjacent - (Outer dot Adjacent) Remote. In general (u x v) x w -/- u x (v x w); for the former is coplanar with u and v, the latter with v and w. Cross multiplication of vectors is not associative.
From (1) we see that the sum of a vector triple product and its two cyclical permutations is zero: (axb)xc -{- (bxc) x a + (cxa)xb = O. (2) If 1 is a directed line carrying the unit vector e, we may express any vector u as the sum of its orthogonal projections on 1 and on a plane p perpendicular to 1: (15.6), projl u = e comps u = (e u) e (3) projp u = u - (e u) e = e x (u x e). (4)
19. Scalar Triple Product. The scalar product of u x v and w is written u x v - w or [uvw]. No ambiguity can arise from the parentheses being omitted, since u x (v w) uXv
is meaningless. THEOREM. The product u x v - w is numerically equal to the volume V of a parallelepiped having u, v, w as concur-
rent edges. Its sign is positive or negative according as u, v, w form a right-handed or left-handed set.
Fic. 19
Proof. The volume V (Fig. 19) may be computed by multiplying the area of a face parallel to u and v,
A = J u l H v l sin (u, v) = l u x e 1, by the corresponding altitude h = I w I cos B:
V=
IuxvIIwIcos6.
Now the angle between u x v and w is B or 7r - B according as u, v, w form a dextral (as in the figure) or a sinistral set (§ 12). The definition of a scalar product now shows that (1)
uxvw=
V
V}
when the set u, v, w is
dextral. sinistral.
VECTOR ALGEBRA
42
§ 19
The dextral or sinistral character of a set u, v, w is not altered
by a cyclical change in their order. Hence, from (1), (2)
But a dextral set becomes sinistral and vice versa, when the cyclical order is changed : (3)
Thus, if the set u, v, w is dextral, the products in (2) all equal V, while the products, U"W V = all equal - V. On account of the geometric meaning of the scalar triple product, we shall call it the "box product." $ If u, v, w are proper vectors, V = 0 only when the vectors are coplanar. Therefore: three proper vectors are coplanar (parallel to the same plane) when and only when their box product is zero. In particular, a box product containing two parallel vectors is zero; for example u x v u = 0. The value of a box product is not altered by an interchange of the dot and cross. For (4)
as dot multiplication is commutative. The notation [uvw] often is used for the box product, as the omission of dot and cross causes no ambiguity. From (15.3) and (16.3), we have (5)
(au) x (av) . (tiw) = a(3y u x v w.
Finally, the distributive law for scalar and vector products shows that a box product of vector sums may be expanded just as in ordinary algebra, provided that the order of the vector factors is not altered. Thus, if we expand the product u x v w when the vectors are referred to the basis i, j, k, we obtain 27 terms of which all but six vanish as they contain box products with two or three equal vectors. The remaining six terms are those containing
[ijk] = [jki] = [kij] = 1,
[ikj] = [kji] = [jik] = -1,
$ The name proposed by J. H. Taylor, Vector Analysis, New York, 1939, p. 46.
PRODUCTS OF FOUR VECTORS
3 20
43
and constitute the expansion of the determinant, [uvw] =
(6)
261
U2
U3
v1
v2
v3
W1
W2
W3
.
This result also follows at once from (16.8). Example. To find the point P where the line AB pierces the plane CDE.
Method. AP = it AB; the scalar it is then determined by the equation, CP CD X CE = (CA + it AB) . CD x CE = 0
(i)
which expresses that P is coplanar with C, D, E. Computation. With the points, A(1, 2, 0),
B(2, 3, 1);
C(2, 0, 3),
D(0, 4, 2),
E(-1, 2, -2);
CP =CA+XAB =[-1,2, -3]+1`[1,1,1] =[7,-1,it+2,A-3];
CD xCE = [-2,4, -1]x[-3,2, -5] = [-18, -7,8]; hence, from (i),
-18(X - 1) -7(X+2)+8(x-3) =17x-20=0,
7;
OP = OA + X AB = [1, 2,0] - -4[1, 1, 1] = .[-3, 14, -20].
-> --4
---,
Check. DP CD X CE = 0; for
DP = --[1,18,18], 20. Products of Four Vectors. Since (a x b)
(c x d) = a b x (c x d) = a
(b d c - b c d),
(1)
x d) may be regarded as a triple product of a x b, c, d or of a, b, c x d. We thus are led to the two expan(a x b) x
sions, (2)
(a x b) x (c x d) = [acd]b - [bcd]a = [abd]c - [abc]d,
which give, in turn, the following equation connecting any four vectors: (3)
a[bcd] - b[cda] + c[dab] - d[abc] = 0.
§ 22
VECTOR ALGEBRA
44
When [abc]
0,
[abc]d = [dbc]a + [adc]b + [abd]c
(4)
gives d explicitly in terms of the basis a, b, c.
21. Plane Trigonometry. If the vectors a, b, c form a closed triangle when placed end to end (Fig. 21),
a+b+c=0.
(1)
We shall denote the lengths of the sides by a, b, c and the interior angles opposite them by A, B, C; then the angles (b, c), (c, a), (a, b) are equal, respec-
tively,to7r- A,7r-B,,r-C. From the identity,
a
(a+b) (a+b) =
FIG. 21
we obtain the cosine law for plane triangles:
c2 = a2 + b2 - tab cos C.
(2)
Cyclical interchanges of the letters in (2) give two other forms of this law. On multiplying (1) first by a x, then by b x, we find
bxc = cxa = axb.
(3)
Division by abc gives the sine law for plane triangles: sin A
sin B
sin C
a
b
c
(4)
Since each product in (3) is double the vector area of the triangle, its (5)
Area = be sin A 2
= I ca sin B = 2 ab sin C.
22. Spherical Trigonometry. Consider a spherical triangle ABC on a sphere of unit radius, and let a, b, c, denote the position
FIG. 22
vectors of the vertices referred to its center (Fig. 22). The notation is so chosen that a, b, c, form a dextral set: then [abc] > 0.
SPHERICAL TRIGONOMETRY
22
45
Let a, /3, y denote the sides (arcs of great circles) opposite the vertices, A, By C. The interior dihedral angles at these vertices also are denoted by A, B, C. We shall consider only spherical triangles in which the sides and angles are each less than 180°. Now (1) (2)
axb =
cxa = sin a b',
bxc = sin a a',
where a', b', c' are unit vectors. The vectors b' and c' are perpendicular to the planes of c, a and a, by respectively, and include an
angle a' = it - A, the exterior dihedral angle at A. Moreover, from (2),
b'xc' _
(c x a) x (axb)
[abc]
sin /3sin y
sin f3sin y
a,
a positive multiple of a. Hence if a', 0', y' denote the exterior dihedral angles at A, By C, we have (3)
b' c' = cos all
(4)
b'xc' = sin a' a,
a' b' = cos y'
Ca= cos
c'xa' = sin#'b,
a' b' = sin y'c.
Thus there is complete reciprocity between the vector sets a, by c, and a', b', c', and also between the spherical triangles they determine. While ABC has a, S, y for sides and a', (3', y' for exterior angles, A'B'C' has a', 0', y' for sides and a, /3, y for exterior angles. Since the vertices of one triangle are poles of the corresponding sides of the other, the triangles are said to be polar. From (2) and (4), we have
[abc] = sin a a a' = sin/3b - b' = sin
a'
sin /3'b - by = sin y' c. c';
hence, on division, [abc] (5)
[a'b'c']
_ sin a _ sin / _ sin y sin a'
sin S'
sin y'
This is the sine law for spherical triangles. Again, from (2), we have sin /3 sin y by c' = (c x a)
(axb) = (c a) (a b) - b c;
VECTOR ALGEBRA
46
§ 23
hence, from (1) and (3), (6)
cos a = cos 0 cos y - sin 0 sin y cos a'.
Similarly, from (4), we deduce (7)
cos a' = cos 0' cos y' - sin f3' sin y' cos a.
Equations (6) and (7) and the four others derived from them by cyclical permutation constitute the cosine laws for spherical triangles.
By replacing a', 0', y' in (5), (6), and (7) by 7r - A, 7r - B, it - C we may express the sine and cosine laws in terms of the sides and interior angles. Thus, we find, for the sine law, sin a
sin j
sin y
sin A
sin B
sin C '
and, for the cosine laws,
cosa = cos,3cosy + sin $ sin -y cos A, cos A = - cos B cos C + sin B sin C cos a.
In this version of the cosine laws, the structural similarity exhibited by (6) and (7) is lost. 23. Reciprocal Bases. Two bases, el, e2, e3 and e', e2, e3, are said to be reciprocal when they satisfy the nine equations:
(1)
el e'=1,
el e2=0,
e2
e' = 0,
e2
e2 = 1,
e2
e3 = 0,
e3
e' = 0,
e3
e2 = 0,
e3
e3 = 1.
el e3=0,
The superscripts applied to the base vectors are not exponents, but mere identification tags. By use of the Kronecker delta 6j, defined as when (2)
'
0
when
i=j i 5 j,
equations (1) condense to (3)
ei e' = Si
(i, j = 1, 2, 3).
Consider the three equations in the first column of (1). The second and third state that e' is perpendicular to both e2 and e3,
RECIPROCAL BASES
§ 23
47
that is, parallel to e2 x e3. Hence el = X e2 x e3; and, from the first equation, 1 = X el e 2 x e3. We thus obtain (4)
e2 = e3 x el
el = e2 x e3
e3 = el x e2
[ele2e3]
[ele2e3]
[ele2e3]
e2 and e3 being derived from el by cyclical permutation. From the symmetry of equations (1) in the two sets e1, ei, we have also elxe2 e2xe3 e3xel (5)
el = [ele2e3] ,
e3 = [ele2e3] .
e2 = [ele2e3] ,
Thus, either basis is expressed in terms of the other by precisely the same formulas. From (4) and (5), we have e
1
(e2 x e3)
el =
(e2 x e3)
[ele2e3][ele2el]
_ [eIe2e3][ele2e3]
on making use of (20.1) and equations (1); hence (6)
[ele2e3][ele2e3] = 1,
an equation which gives further justification for the name reciprocal applied to the sets. Since the box-products in (6) must have the same sign, a basis and its reciprocal are both righthanded or both left-handed. As to the orientation of reciprocal sets, we see from (1) that e' is perpendicular to the
plane of e2 and e3 in the direction which makes an acute angle with el. Similar statements apply to e2 and e3. If the vectors et, et are all drawn from the same point 0 and cut
by a sphere s about 0 in the
Fla. 23
points Ei, Et, respectively, the three planes OEiE; cut a spherical triangle E1E2E3 from s; and the three planes OEE' cut out a second spherical triangle E'E2E3 (Fig. 23). Either triangle is the polar of the other; for E', E2, E3 are poles of the great-circle arcs, E2E3, E3E1, E1E2i and similarly for El, E2, E3.
VECTOR ALGEBRA
48
§ 24
On dividing the identity, ei x (e2 x e3) + e2 x (e3 x el) + e3 x (el x e2) = 0 (18.2), by [ele2e3], we obtain the relation,
elxel +e2xe2 +e3xe3 = 0. This has an interesting geometric interpretation. The great circle (7)
EIEI is perpendicular to both great circles E2E3 and E2E3; it thus contains the altitudes of both triangles through the vertex labeled 1. Similarly, the great circles E2E2, E3E3 contain the altitudes of both triangles through the vertices labeled 2 and 3. The planes of the great circles EIEI, E2E2, E3E3 are perpendicular to el x el, e2 x e2, e3 x e3, respectively; and these vectors, in view of (7), lie in a plane p. The poles of this plane (the ends of the diameter of s perpendicular to p) are points common to all of the great circles EiEi. Therefore, the three altitudes EiEi of the polar triangles EIE2E3, E1E2E 3 meet in a pole P of the plane p. When a basis and its reciprocal are identical, the basis is called self-reciprocal. The equations (3) then become (8)
ei - e; = bi;. j'
These equations characterize an orthogonal triple of unit vectors. Hence a basis is self-reciprocal when and only when it consists of a mutually orthogonal triple of unit vectors. The triple will be dextral if [ele2e3] = 1 (§ 19), sinistral if [ele2e3] = -1. Thus i, j, k and i, j, -k are typical dextral and sinistral orthogonal bases. 24. Components of a Vector. We now supplement the notation of § 12 by writing the components of a vector u ui or ui according as the basis is ei or et. The components ui are called contravariant, the components ui covariant, for reasons given in § 148. Any vector now may be written in two forms:
u = ulel + u2e2 + u3e3. u = ulel + u2e2 + u3e3i From (1) and (2), we obtain equations of the type u - e' = ul, (1) (2)
u - el = ul ; all six are included in ui = u - ei ui = u ei, (3) (4)
(i = 1, 2, 3). When a basis is given, a vector u is completely specified by giving its components written in order. With the notation of t With a self-reciprocal basis there is no need for superscripts; we therefore write S as Sii.
COMPONENTS OF A VECTOR
§ 24
49
§ 12, we write a vector u as a number triple, [ul, u2, u31 or [ul, U2, u31 according as it is referred to the basis ei or et. We denote the (non-zero) box products of the base vectors by 1/E = [ele2e31.
E = [ele2e3l,
(5)
Addition and multiplication by scalars follow the formulas of § 12; thus, with covariant components, (6)
X [u1, u2, u31 = [Xu1, Xu2, Xu3]
(7)
[u1, u2, u31 + [v1, v2, u31 _ [u1 + v1, u2 + v2, u3 + vJ
A simple expression for the scalar product u v is obtained when u and v are referred to different, but reciprocal, bases. Thus, if we compute J (u'e1 + u2e2 + u3e3) . (vie1 + v2e2 + v3e3) u . V = (ulel + u2e2 + u3e3) - (vlel + v2e2 + v3e3) l
we obtain, by virtue of equations (23.1), (8)
u V = u1V1 + u2V2 + u3e3 = u1v1 + u2e2 + u3V3.
We next compute the components of u x v, relative to bases e2 and ei. Using covariant components gives u x v = (uie' + u2e2 + u3e3) x (v1e1 + v2e2 + v3e3) U2
U3
V2
V3
U2
U3
V2
V3
e2 x e3 + e1
u3
u1
V3
V1
e3xel +
u1
U2
V1
V2
U3
u1
e2
e3
V3
V1
E
E
e1xe2
or, more compactly, (9)
u x v = E-1
Using contravariant components, we find, in similar fashion, (10)
uxv=E
VECTOR ALGEBRA
50
§ 25
From (9) and (10), we see that the components of q = u x v are (11)
q' = E-1
Uj
uk
I
qt
Vi
u'
11k
v'
ak
1
= E
vk
where the indices ijk form a cyclical permutation of 123. From (9) and (10), we next obtain two expressions for the box product u x v w: u2
U1
(12)
[uvw] = E-1
u1
U3
V1
V2
V3
W1
w2
w3
=E
V1
W1
u2
u3
v2
v3
w2
w3
The first formula of (12) may be written (13)
[uvw][e1e2e3]
v el
=
v . e2
v - e3
f
whereas the second gives the analogous equation obtained by replacing ei by ei. Since e1i e2, e3 may be any linearly independent set of vectors a, b, c, we also have
ua [uvw][abc] =
(14)
uc
va vb vc
wa wb wc
When the basis ei is self-reciprocal (an orthogonal triple of unit
vectors) e i = ei and E = +1. The two sets of components of u then coalesce into a single set which we arbitrarily write U. If the self-reciprocal basis is dextral, we can write el = i, e2 = j, e3 = k, E = 1; then U V = U1V1 + U2V2 + u03
U X S7 =
,
[uvw] =
in agreement with our previous results. 25. Vector Equations. A vector is uniquely determined when its scalar and vector products with two known non-perpendicular vectors are given. Thus let (1)
ua=a,
uxb=c,
(ab-/- 0);
HOMOGENEOUS COORDINATES
§ 26
51
the scalar a and the vectors a, b, c are regarded as known, and
b c = 0.
We have
ax(uxb) =a -c, (2)
or
ab
By direct substitution, (2) is seen to be a solution of equations (1). A vector is also determined when its scalar products with three known non-coplanar vectors are given. For example, if u - e1 = u1,
u - e2 = u2, u - e3 = u3, and the sets et, e i are reciprocal, we have, from (24.2), (3)
(4)
u = ulel + 112e2 + u3e3.
By direct substitution, (4) is seen to be a solution of equations (3). 26. Homogeneous Coordinates. Coordinates of a point, line or plane are called homogeneous if the entity they determine is not altered when the coordinates are multiplied by the same scalar. A coordinate that depends upon the choice of origin 0 bears the subscript 0; such coordinates are written after those independent of the origin. In equations r denotes any position vector from 0 to the entity in question, whereas r1i r2 denote position vectors to points R1, R2 given in advance. In this article we use Latin letters for vectors, Greek for scalars. Point Coordinates. If r = OA, we write r = ao/a. The "equation" of the point A is then (1)
ra = ao,
and its homogeneous coordinates are (a, ao). Note that (Xa, Xao)
determine the same point; hence the coordinates (a, ao) depend on three independent scalars: "there are o03 points in space." Plane Coordinates. The equation of a plane through R1 and perpendicular to the vector a is (r - r1) a = 0; or, on writing
ao=r1-a, r a = ao. The homogeneous coordinates of the plane are (a, ao). Since (2)
(Xa, Xao) determine the same plane, the coordinates (a, ao) depend on three independent scalars: "there are oo 3 planes in space."
VECTOR ALGEBRA
52
§ 26
Line Coordinates. The equation of a line through Rl and parallel to the vector a is (r - r1) x a = 0; or, on writing ao = rl x a, (3) r x a = ao. The homogeneous (or Plucker) coordinates of the line are (a, ao) and are connected by the relation,
a ao = 0.
(4)
In view of. (4) and the fact that (Xa, Xao) determine the same line, vile coordinates (a, ao) depend on four independent scalars: "there are ooa lines in space." We thus have the homogeneous coordinates: Plane (a, ao),
Point (a, ao),
Line (a, ao).
The first coordinate cannot vanish. When the second coordinate (subscript 0) is zero, the origin 0 is on the point, plane, or line, respectively.
The distances of the point, plane, or line from the origin are, respectively,
ao!
Iai'
(5)
laol Jai
aol '
Ia!
The first result is obvious. If p is the vector from 0 perpendicular
to the plane r a = ao,
p= p
ao
p is the vector from
a
0 perpendicular to the line r x a = ao, p x a = ao,
p a= 0;
hence
p=
axa0
from (25.2), and I p I = ( ao I/I a I
If the origin is shifted from 0 to P, we must replace r in (1),
(2), (3) by PO + r to obtain the second coordinate referred to P. We thus obtain the shift formulas: (6)
ap = ao + P0 a,
(7)
ap =
(8)
ap=ao+POxa,
in the respective cases.
-3
HOMOGENEOUS COORDINATES
§ 26
53
Two Elements Determine a Third. We have the following cases: Two distinct points determine a line: (a, ao),
(9)
(a, bo)
(abo - $ao, ao x bo).
Two non-parallel planes determine a line: (b, $o) --> (a x b, 00a - aob).
(a, ao),
(10)
A point and line determine a plane: (a, ao),
(11)
(b, bo) -> (abo - aoxb, ao bo),
unless abo - ao x b = 0; then also ao bo = 0, and the point lies on the line. A line and plane determine a point: (a, ao),
(12)
(b, $o) -* (a b, 0oa - aoxb),
unless a b = 0. If 0oa - ao x b = 0 (and hence a b = 0), the line lies in the plane. Proofs in outline. For (9) : The line AB has the equation, as Cr
/
as
x( a
J
= 0, or r x (abo - gao) = ao x b0.
For (10) : The line is parallel to a x b; and
rx(axb) =
$oa -aob.
For (11) : Refer the line (b, bo) to the point A(a, ao); then, from (8),
bA=bo - OAxb=bo -
aoxb a
The plane through A normal to bA has the equation
(r -aoa-
(abo - ao x b) = 0,
or r (abo - ao x b) = ao bo.
For (12) : The equations of line and plane, r x a = ao and r b 13o, have the solution (§ 25) :
r=-Qoaa- aob x b
VECTOR ALGEBRA
54
§26
Two Lines. If Rl and R2 are points on the lines (a, ao), (b, bo),
the lines are coplanar when and only when the vectors r1 - r2, a, b are coplanar. (13)
Since
(r1
a necessary and sufficient condition that the lines be coplanar is (14)
If the lines are not coplanar, let e be a unit vector in the direction a x b of their common normal; then, from (13),
a bo + b ao
(15)
IaxbI
This is the component of R2R1 in the direction of e; its numerical value gives the shortest distance between the lines. Equations in Point and Plane Coordinates. If the point (a, so) lies on the plane (t, To) we have (16)
on eliminating r from ra = so and r t = To. This may be regarded as the equation of the point (a, so) in plane coordinates; or as the equation of the plane (t, To) in point coordinates. If the line (p, po) contains the point (a, so) we have (17)
soxP -apo = 0,
on eliminating r from ra = so, r x p = po. This is the equation of the line (p, po) in point coordinates. If the line (p, po) lies on the plane (t, TO) we have p t = 0 and hence (18)
Top - Po x t = 0,
on eliminating r from r x p = po and r t = To. This is the equation of the line (p, po) in plane coordinates. Example. If three points (a, ao), (S, bo), (y, co) lie on the line (p, po), we have, from (17), ao co bo
-xP =
XP = _XP = po;
16
Y
hence (i)
y)+µ(0
y) =0,
LINE VECTORS AND MOMENTS
§ 27
55
since both vectors in parenthesis are parallel to p. This is a linear relation between ao/a, bo/)3, co/y in which the sum of the coefficients is zero (§ 6). If the three planes (a, ao), (b, ,o), (c, yo) pass through the line (p, po), we have, from (18), a b c
POX-=POX-=Pox-=P, ao lio yo
provided ao, 00, yo are not zero; hence Op
\ao
yoi + µ \ljo
-Yo
-
since both vectors in parenthesis are parallel to po. This is a linear relation connecting a/ao, b/3o, C/-Yo in which the sum of the coefficients is zero. A shift in origin does not alter the coefficients in (i). In (ii), however, the
coefficients are changed but their sum still remains zero; if we write v = -A - IA and shift the origin to P,
v-= X o+ AA Po -+ -Yo
0
becomes X'
p +lz'-+v op
'YP
-=0,
where X' = Xap/ao, u' = -flP/0o, d = vyp/yo Let the student prove that
x'+A '+v'=0. 27. Line Vectors and Moments. A vector which is restricted to lie in a definite line is called a line vector. If f is the vector, and (1)
r x f = fo,
(f fo = 0)
the equation of its line of action, the line vector is completely specified by the Pliicker coordinates f, fo. The vector fo is called the moment of f about the point 0. If
f=AB, (2)
fo = OA x AB
is twice the vector area of the triangle OAB (§ 17); fo remains constant as f is shifted along its line of action.
If the origin is shifted from 0 to P, the moment of f about P is PA x AB = (PO + OA) x AB; hence (3)
fp = fo + PO x f.
If s is any axis through P with the unit vector e, the component of fp on s, namely, e fp (15.6), is called the moment of f about the axis s. We speak of moment about an axis because e fp is inde-
VECTOR ALGEBRA
56
§ 27
pendent of the position of P on this axis; for, if the axis is given by the unit line vector (e, eo),
(4)
e fp as the component of twice the vector
area PAB on a directed plane of unit normal e (§ 17). In Fig. 27, A1B1 is the projection of AB on the plane;
the moment of f about s is numerically equal to twice the area PA1B1, that is, to the product of the length A1B1 and the perpendicular distance h of A1B1 from P; its sign is plus or minus according as a turn in
the sense PA1B1 would advance a righthanded screw in the direction of s or -s. We repeat these important definitions.
The moment of the line vector f = AB about a point P is A
Fro. 27
fp=PAxAB=rxf,
(5)
where r is a vector from P to any point on f's line of action. The moment of f about an axis is the component of r x f on the axis, where r is a vector from any point on the axis to any point on the line of action. The moment about a point is a vector; about an axis, a scalar. Example. The line of action of the force f = [1, -1, 2] passes through the point A (2, 4, -1). Find its moment about an axis through the point P(3, -1, 2) and having the direction of the vector [2, - 1, 2].
0
Since PA = [-1, 5, -3] and e = 3[2, -1, 21 is a unit vector along the axis, fp = PA x f =
i
i
-1
5
1
is the moment of f about P; and
is the moment of f about the axis.
-1
k
-3 = [7, -1, -4] 2
SUMMARY: VECTOR ALGEBRA
§ 28
57
28. Summary: Vector Algebra. Equal vectors have the same length and direction. Free vectors are added by the triangle construction. The negative of a vector is the vector reversed. To subtract a vector, add its negative. The product Au is a vector A I times as long as u; its direction is the same as u if X > 0, the reverse if X < 0. Vectors may be added, subtracted, and multiplied by real numbers in conformity with the laws of ordinary algebra. The n vectors ui are linearly dependent if there exist n real num-
bers Ai, not all zero, such that EAiui = 0. When n = 2, 3, linear dependence implies that the vectors are collinear or coplanar, respectively; and conversely. In space of three dimensions any four vectors are linearly dependent.
A point P divides a segment AB in the ratio X = /3/a when
-3
----3
AP = APB; then
-3
-->
-3
(a + $)OP = aOA + SOB. ----3
A linear relation2;A1OP1 = 0 connecting n position vectors will
hold for any origin when and only when :Ai = 0. When n = 2, 3, 4, the points Pi are coincident, collinear, or coplanar, respectively.
A set of weighted points miPi has a unique centroid P if 2;mi $ 0; its defining equation is
--3 -* XmiOPi and OP* =
lmiP*Pi = 0;
Emi
If all mi = 1, P* is called the mean center. The scalar product u v is defined as
Jul Ivlcos(u,v). Laws: V. Up
uv
U. (v+w) =
uv uv=lu
is a e
a dextral set.
I vl
u
v
VECTOR ALGEBRA
58
§ 28
Laws :
UXV = `VxU,
Ux(V+w) = uxv+uxw.
If u, v 0, u x v = 0 implies that u l v and conversely. Expansion rule: l
ux(vxw) = Cross multiplication is not associative. The box product, u x v w or [uvw], is numerically equal to the volume of a "box" having u, v, w as concurrent edges; its sign is + or - according as u, v, w form a dextral or sinistral set. The value of u x v w is not affected by a change in cyclical order of the vectors, or by an interchange of dot and cross. If u, v, w 5-4- 0, [uvw] = 0 implies that u, v, w are coplanar, and conversely. A set of three vectors ej forms a basis if E = [ele2e3] 0. To
every basis there corresponds a unique basis ei such that ej e' = S1; such bases are called reciprocal. If the indices i, j, k form a cyclical permutation of 12 3, ei x ek
e,xek
ei = [ele2e3] ;
ei =
[ele2e3][e'e2e3] = 1.
[e1e2e3]
Both bases are dextral if E > 0; sinistral, if E < 0. Given a basis e1, a vector u may be written as Euiei or ruiei; the numbers ui are contravariant components of u, ui covariant components.
u+v = E(ui'+vi')ei = E(ui+vt)ei; Xu = 2:xuiei = 2;xuie`;
uV=
uIv1
+ u2v2 + u3v3 = u1v1 + u2v2 + u3v3; el
U x v = E-1
e2
e3
I
I
e3
v2
v3
E ul
U2 V3
[uvw] = E-1 I
e2 el
vl U1
u2
U3
V1
V2
V3
W1
W2
W3
Ut
E
u2
1vi
u3 I
w2
w3 When the basis ej is self-reciprocal (ei = ei), its vectors form a mutually orthogonal set of unit vectors; E = 1, if the basis is dextral, E = -1, if sinistral. A dextral self-reciprocal basis is written
PROBLEMS
59
i, j, k. For such a basis ui = ui; these rectangular components are given by ui = I u I cos (ei, u),
and
= (ul)2 + (u2)2 + (u3)2. The two formulas previously given for u + v, Au, u - v, u x v, u x v w in each case become identical. I U I2
The equations of a point, line, and plane, in terms of their homogeneous coordinates (a, ao), (b, bo), (c, -yo), are
ra=ao, respectively.
rxb=bo,
r - c=yo,
When the origin is shifted to P,
ap = ao+POa,
by = bo+POxb,
yP =
The moment of the line vector (f, fo) about the point P is fp = r x f, where r is any vector from P to its line of action. The moment of (f, fo) about an axis through P is the component of fp on this axis; if the axis is given by the unit line vector (e, eo), this axial moment
PROBLEMS
1. If ABC is any triangle and L, M, N are the mid-points of its sides, show that, for any choice of 0, a -- 3
-3
-3
-4
--3
-3
OA + OB + OC = OL + OM + ON.
---
---, -->
-9
2. If OA' = 3 OA, OB' = 2 OB, in what ratio does the point P in which AB and A'B' intersect divide these segments? 3. Show that the mid-points of the four sides of any quadrilateral (plane or skew) form the vertices of a parallelogram.
4. P and Q divide the sides CA, CB of the triangle ABC in the ratios --4 -3 x/(1 - x), y/(1 - y). If PQ = X AB, show that x = y = X. 5. E, F are the mid-points of the sides A B, BC of the parallelogram A BCD.
Show that the lines DE, DF divide the diagonal AC into thirds and that AC cuts off a third of each line. 6. OAA', OBB', OCC', ODD' are four rays of a pencil of lines through 0 cut by two straight lines ABCD and A'B'C'D'. If C and D divide AB in
the ratios r and s, C' and D' divide A'B' in the ratios r' and s', prove that
r/s = r'/s'. 7. The points A, B, C and A', B', C' lie, respectively, on two intersecting Show that BC', CB'; CA', AC'; AB', BA' intersect in the collinear points P, Q, R (Pascal's Theorem). lines.
VECTOR ALGEBRA
60
8. Lines drawn through a point P and the vertices A, B, C, D of a tetrahedron cut the planes of the opposite faces at A', B', C', D'. Show that the sum of the ratios in which these points divide the segments PA, PB, PC, PD
is -1. [Equation (10.2), with e = - (a + 0 + y + S) may be written
a+i9+y+s+e =0.
as+8b+yc+ad+ep =0,
From this we conclude, as in § 7, ex. 3, that a' _ (aa + ep)/(a +,E), etc.] 9. The line DE is drawn parallel to the base AB of the triangle ABC and is included between its sides. If the lines AE, BD meet at P, show that the line CP bisects AB. 10. The points P, Q, R divide the sides BC, CA, AB of the triangle ABC in the ratios a/(1 - a), l3/(1 - 0), y/(1 - y). If P, Q, R are collinear,
x+y+z=0;
xp+yq+zr=0,
putting p = (1 - a)b + ac, etc. in this equation, we obtain a linear relation in a, b, a whose coefficients have a zero sum. Show that this implies that the separate coefficients vanish; hence deduce the Theorem of Menelaus (§ 7, ex. 2):
afy/(1 - a) (1 - p)(1 - y) = -1. 11. Using an argument patterned after that in Problem 10, prove Carnot's Theorem:
If a plane cuts the sides AB, BC, CD, DA of a skew quadrilateral ABCD
in the points P, Q, R, S, respectively, the product of the ratios in which P, Q, R, S divide these sides equals 1.
12. Id a plane quadrilateral ABCD, the point P in which the diagonals AC, BD intersect divides these segments in the ratios 4/3 and 2/3, respectively. In what ratio does the point Q, in which the sides AB, CD meet, divide these segments?
13. If el = O 1, e2 = OE2i e3 = 0
0
3E
form a basis, the reciprocal set
el, e2, e3 are vectors perpendicular to the planes OE2E3, OE3E1i OE1E2 and
having lengths equal to the reciprocals of the distances of El, E2, E3 from these planes, respectively. Prove this theorem.
14. Prove that a necessary and sufficient condition that four points A, B, C, D be coplanar is that [dbc] + [add] + [abd] - [abc) = 0. 15. Show that the shortest distance from the point A to the line BC is
Iaxb +bxc +cxal/Ib - cI.
16. If the vectors el, e2, e3; el, e2, e3 form reciprocal sets, show that
the vectors e2 x e3, e3 x el, el x e2; e2 x e3, e3 x e1, el x e2 do likewise.
17. Prove the formulas: (a)
[a x b, bxc, c x a] = [abc]2,
(b)
(bxc) (a x d) +(c x a) (b x d) +(a x b)
(c)
(bxc) x (a x d) +(c x a) x (b x d) +(a x b) x (c x d) = -2[abc]d,
(c x d) =0,
(d)
(d-a) (b-c)+(d-b) (c-a)+(d-c) (a-b)=0,
(e)
(a-d) x (b-c)+(b-d) x (c-a)+(c-d) x (a-b) =2(a x b+b x c+c x a).
PROBLEMS
61
18. Find the shortest distance between the straight lines AB and CD when (a)
A(-2, 4, 3), B(2, -8, 0);
(b)
A(2, 3, 1),
B(0, -1, 2);
C(1, -3, 5),
D(4, 1, -7),
C(1, 2, 5),
D(-3, 1, 0).
19. Show that the lines AB, CD are coplanar, and find the point P in which they meet:
A(-2, -3, 4),
B(2, 3, 0);
C(-2, 3, 2),
D(2, 0, 1).
--4
[CD is parallel to CP = AP - AC = XAB - AC; find X and then P from
-4
--->
OP = OA + XAB.] 20. If the points P, Q, R divide the sides BC, CA, AB of the triangle ABC in the same ratio, show that A, B, C and P, Q, R have the same mean center. 21. If the vectors OA, OB, OC lie in a plane and are equal in length, prove
that OA + OB + OC = OH, where H is the orthocenter of the triangle ABC. [§ 15, ex. 4.] 22. The power of a point P with respect to a sphere of center C and radius r is defined as (CF)2 - r2. Prove that the power of P with respect to a sphere
having AB as diameter is PA I'B. , P,,, 23. Prove that the sum of the n2 powers of n given points, P1, P21 with respect to the n spheres having for diameters the n segments joining the given points to a variable point P in space, is constant. [Am. Math. Monthly, vol. 51, p. 96.1
24. The lines (a, ao), (b, bo)'are coplanar; then a bo + b ao = 0 (26.14). Show that they determine the plane (a x b, ao b) if a x b 0 0; and the point
00.
aoxbo) if
Solve Problem 19 using these results.
25. Show that the three planes (a, ao), (b, $o), (c, yo) meet in the point, ([abc],
aobxc +Rocxa +yoaxb),
provided [abc] 0 0. 26. Show that the three points (a, ao), (0, bo), (y, co) determine the plane, (a bo x co + 0 co x ao + y ao x bo,
[aobocol),
provided aboxco +3coxao +yaoxbo 0 0. 27. The points P, Q, R divide the sides BC, CA, AB of the triangle ABC in the ratio of 1/2. The pairs of lines (AP, BQ), (BQ, CR), (CR, AP) intersect at X, Y, Z, respectively. Show that the area of the triangle XYZ is 1 /7 of the area of ABC. [Twice the vector area of XYZ is y x z + z x a +
xxyl
When P, Q, R divide the sides in the ratio t/1, the area of XYZ is
(1 - t)2/(1 + t + t2) times the area of ABC.
VECTOR ALGEBRA
62
28. Using (26.16), prove that (a) If four points (a, ao), (0, bo), (y, co), (S, do) lie on a plane, the vectors ao/a, , do/S are connected by a linear relation in which the sum of the coefficients is zero;
(b) If four planes (a, ao), (b, 3o), (c, yo), (d, So) pass through a point, the vectors a/ao, , d/So are connected by a linear relation in which the sum of the coefficients is zero, provided ao, Qo, 'yo, So 54 0.
29. If a sphere S with a fixed center cuts two concentric spheres S1, S2i prove that the distance between the planes of intersection SS, and SS2 is independent of the radius of S. Extend this theorem to cover the case when S fails to cut one or both of the concentric spheres. [Consider the radical planes SS1, SS2.1
30. Prove that the radical planes of the three spheres (r - c1)2 = P1 (1 = 1, 2, 3) meet in the line J
`C1 X C2 + C2 X C3 + C3 X c1,
2c1(e2 - e3
- P2 + A3) + cycl.;,
their radical axis. Consider the case when e1 X C2 + C2 X C3 + C3 X cl = 0. 31. A plane system of forces Fi acting through the points ri is equivalent to a single force F = 2;Fi. When all the forces Fi are revolved about these points through an angle B, show that their resultant F also revolves through B about the point f (the astatic center) given by
t=
F X M - (11ri Fi)F 2 F
where M = 'ri x Fi.
When the forces are parallel (Fi = Xie), show that the astatic center is the centroid of the weighted points Xiri. 32. P* is the centroid of a set of n weighted points miPi for which Tmi = in. Prove the Theorems of Lagrange: (a) (b)
1mi(OPi)2 = m(OP*)2 + 2:mi(P*Pi)2.
Vmim1(PPi)2 = in mi(P*Pj)2,
where ij ranges over 2n(n - 1) combinations.
[OPi = OP* + P*Pi;
PiP7 = P*Pi - P*Pi.]
33. Deduce the following results from Prob. 32. (a) If ABCD are the vertices of a square in circuital order, (OA)2 + (OC)2 _ (OB)2 + (OD)2 for any 0. Generalize. (b) If r is the radius of a sphere circumscribed about a regular tetrahedron of side a, r2 = 3x2/8.
(c) The mean square of the mutual distances of all the points within a sphere of radius r is 6r2/5.
34. If P* and Q* are the mean centers of p points Pi and q points Q respectively, prove that Mean (PA)2 = Mean (P*Pi)2 + Mean (Q*Q1)2 + (P*Q*)2.
CHAPTER II MOTOR ALGEBRA
29. Dual Vectors. The vector f, bound to the line whose equation, referred to the origin 0, is
rxf = fo,
(1)
is completely determined by the two vectors f, fo, its Plucker coordinates. These obviously satisfy the relation, f fo = 0.
(2)
The vector f does not depend upon 0; but fo, the moment of f about 0, becomes (3)
fp=fo+POxf,
when the origin is shifted from 0 to P. We now amalgamate f and fo into a dual vector.
F=f+efo
(4)
where a is an algebraic unit having the property e2 = 0. If f = 1, F is called a unit line vector. The unit line vectors,
A=a+eao, stand in one-to-one correspondence with the c0 4 lines of space. A line vector F of length X always may be written F = XA, where A is a unit line vector. Unit line vectors depend upon four independent scalars, general line vectors upon five. Finally, for applications in mechanics, we consider dual vectors F without the restriction (2). The vectors f, fo then involve six independent scalars. The dual vector (or the entity it represents) then is called a motor, provided its resultant vector f is independent of the choice of 0, while its moment vector fo changes in accordance with (3) when the origin is shifted to P. Line vectors are thus special motors for which f fo = 0; for unit line vectors also f f = 1. 63
MOTOR ALGEBRA
64
§ 30
30. Dual Numbers. In analogy with the complex numbers x + ix', W. K. Clifford introduced dual numbers x + ex', in which x, x' are real and e is a unit with the property e2 = 0.
In x + ex', x is called the real part and x' the dual part. We write
x-{-ex'=y+ey' when x=y,x'=y'; x+ex'=0 when x = 0, x' = 0.
Addition and multiplication of dual numbers are defined by the equations : (1) (2)
(x+ex')+(y+ey') =x+y+e(x'+y'), (x + ex') (y + ey') = xy + e(xy' + x'y)
Observe that (2) may be obtained by distributing the product on the left and putting e2 = 0. From these definitions we see that addition and multiplication are commutative and associative and that multiplication is distributive with respect to addition. In fact, the formal operations are precisely those of ordinary algebra followed by setting e2 = e3 = . . . = 0. The negative of x + ex' is defined as -x - ex'.
If A = a + ea', B = b + eb', the difference X = A - B and quotient Y = A/B satisfy, by definition, the equations B + X = A, BY = A. We find that (3)
A - B = a - b + e(a' - b');
and that BY = A has the unique solution, a'b - ab' A a
B=b+ e
(4)
b2
when and only when b F6 0. Division by a pure dual number eb' is not defined. The quotient (4) may be remembered by means of
the device (a + ea') (b - eb')/ (b + eb') (b - eb') used in complex algebra.
A dual number a + ea' in which a 5=1 0 is said to be proper; the product and quotient of proper dual numbers are also proper. If the dual product (2) is zero, there are three alternatives:
x=x'=0;
y=y'=0;
x=y=0.
The last shows that a dual product can vanish when neither factor is zero; for any two pure dual numbers have zero as their product.
MOTORS
§ 31
65
If the function f(x) has the derivative f'(x), we define its value for the dual argument X = x + Ex' by writing down its formal = 0; thus Taylor expansion and setting E2 = E3 = Ax + Ex') = AX) + Ex'f' (x)
(5)
In particular, (6)
sin (x + Ex') = sin x + Ex' cos x,
(7)
cos (x + ex') = cos x - ex' sin x.
When x = 0, we have sin Ex' = Ex', cos ex' = 1; consequently (6) and (7) have the form of the usual addition theorems of the sine and cosine. Note also that sine X + cost X = 1. 31. Motors. We now can characterize a motor as a dual multiple of a unit line vector. Thus, on multiplying the unit line vector
A = a + Eao (a a = 1, a ao = 0) by the dual number X + EX', we obtain the dual vector, M = m + Emo = (X + EX') (a + Ea.o).
(1)
On equating the real and dual parts of both members, we obtain m = Xa,
(2)
mo = Xao + A'a.
To show that M is a motor we need only verify that mo transforms in accordance with (29.3):
MP =mo+POxm.
(3)
In view of (2), this result follows from
mp = Nap + A'a = X (ao + PO -a) + X'a _ (Xao + X'a) + PO x (Xa).
From (3), m mp = m mo; hence the scalars m m,and m mo are invariants in the sense that they are not altered by a shift of origin.
From (2),
(4)
m
0, we call the invariant, X'
(5)
mmo
mm
MOTOR ALGEBRA
66
the pitch of the motor.
§31
Choosing A > 0, we have the unique
solution,
A=ImI, Aa=m,
A'=,zim1, Aao=mo-µm.
When m 0 0, M has the dual length, A + EX, = I m { (1 + eµ) (6) and its axis is along the line vector, (7)
ImIA=m+e(mo-µm).
The equation of the axis is therefore
rxm =mo -gym=
(8)
(mxmo)Xm
;
this shows that the axis passes through the point Q given by --f mxmo
r=OQ=
(9)
;
and, from (3),
mQ=mo -
(10)
(mxmo)Xm
=µm.
At all points on its axis M has the same moment µm, a fact also apparent from (7). Only for points P on the axis is the moment mp parallel to m.
Motors for which m
0 (A 0) are called proper. Proper motors are screws if m mo 0 0 (A, A' 0 0) ; line vectors if m mo = 0 (A 0 0, A' = 0). Only proper motors have a definite axis.
If m = 0, mo 0 0 (A = 0, A' 0), M = emo is pure dual; then mo is not altered by a change of origin. In this case M is called a couple of moment mo. A couple may be regarded as a screw- of infinite pitch with an axis of given direction but arbitrary position in space.
Finally if m = 0, mo = 0, the motor M = 0, then A = A' = 0 from (2). Hence M = 0 only when its dual length A + eA' = 0. Our classification of motors is therefore as follows: Screw
0;1
} Proper
Line vector Couple Zero
(m=0, (m=0,
m0 =0):A=0,A' =0.
MOTOR SUM
§ 32
67
If two proper motors M, N are connected by a linear relation with proper coefficients,
(a + ia')M + ($ + ia')N = 0,
as 34 0,
either may be expressed as a dual multiple of the other; hence both M and N are multiples of the same unit line vector and are therefore coaxial. Conversely, if M and N are coaxial, they satisfy a linear relation with proper coefficients.
32. Motor Sum. The sum of the motors M = m + em0, N = n + eno is defined as the motor,
M + N = m + n + e(mo + no).
(1)
That M + N is a motor follows from (29.3) :
mp+nr = mo+no+POx(m+n) THEOREM 1.
The sum of two line vectors is a line vector only when
their axes are coplanar and their vectors have anon-zero sum. Proof.
If M and N are line vectors, M + N is a line vector
when and only when
m+n
0,
and (m + n) (mo + no) = 0.
Since m mo = 0, n no = 0, the latter condition reduces to (2)
which is precisely the condition (26.14) that the axes be coplanar. THEOREM 2. If two line vectors intersect, their sum is a line vector
through the point of intersection. Proof. If r1 is the position vector of the point Pl in which the
line vectors M and N intersect, we may take mo = rl x m, no = r1 x n. Since the axis of M + N has the equation,
r- (m + n) = mo + no = r1 x (m + n), it passes through the point P1. THEOREM 3.
If the line vectors M and N are parallel and n =
Xm (X F6 -1), the axis of their sum divides any segment from M to N in the ratio X/1.
68
MOTOR ALGEBRA
§ 33
Proof. If Pl, P2 are points on the axes of M and N, we may
take mo = rl x m, no = r2 x n. The line vector,
M + N = (1+A)m+E(r,+Ar2)xm; and the equation of its axis, r x (1 + A)m = (r1 + Xr2) x m,
is satisfied by the point r = (r1 + Ar2)/(1 + X) which divides PiP2 in the ratio A/1. The division is internal when m and n have the same direction (X > 0), external when they have opposite directions (A < 0). THEOREM 4. couple.
The sum of two couples, if net zero, is another
Proof. If M and N are pure dual, M + N, if not zero, is also pure dual.
If M and N are line vectors such that n = -m and P1, P2 are points on their respective axes, M + N is a couple of THEOREM 5.
moment (r1 - r2) x m. Proof.
Since M = m + er1 x m, N = -m + ere x (-m), M + N = e(rl - r2) x m.
33. Scalar Product. The scalar product of the motors M = m + emo, N = n + eno is defined as the result of distributing the product,. (m + emo) + (n + eno), namely, (1)
This dual number is independent of the choice of origin; for on computing mp, np from (29.3), we find that (2)
The definition (1) shows that (3)
The definition (1) for M M. N differs from that given by R. von Mises ("Motorrechnung, ein neues Hilfsmittel der Mechanik," Z. angew. Math. Mech., vol. 4, 1924, p. 163) who defines M N as the invariant real scalar (2). The definition (2) is suggested by applications in mechanics; its consequences are (a) the familiar rules of vector algebra do not all carry over into motor algebra, and (b) the elegant generalization of the scalar product of unit vectors, given in (5), is lost.
SCALAR PRODUCT
§ 33
69
In order to compute the scalar product of two unit line vectors A = a + ea0i B = b + - ebo, we shall suppose first that they are not parallel and write axb = esin
exb,
A = a,
E = e.
Then we have
cos cp - ep sin cp.
If we now define the dual angle t between A and B as 'P + Ecp
(4)
a reference to (30.7) shows that (5)
in complete analogy with a b = cos
=cos(p-ecp'sin(p=0, when and only when
a
cos,p=0, rp'sin
=0;
that is, when the line vectors cut at right angles. t The concept of dual angle is due to Study, Geometrie der Dynamen, Leipzig, 1903, p. 205.
MOTOR ALGEBRA
70
§ 34
THEOREM. In order that the axes of two proper motors cut at right angles, it is necessary and sufficient that their scalar product vanish.
Proof. We express the motors in the form,
M = (a + ea )A,
N = (0 + E#')B,
where A, B are unit line vectors along their axes and a0 - 0. Then
M - N = [a$+ and, since the first factor is neither zero nor pure dual, M N = 0 implies A B = 0, and conversely. We denote the dual part of M N by (6)
As previously noted, this is von Mises' definition of the scalar product. In common with M - N, the product M c N is commutative and distributive with respect to addition. The preceding calculation shows that
AoB = -
hence A o B = 0 implies p' = 0 or sin p = 0, that is, the line vectors intersect or are parallel, and conversely. The same criterion applies to line vectors aA, OB of arbitrary length. Consequently a necessary and sufficient condition that two line vectors 1V ,JI be coplanar is that M e N = 0. This is the condition (26.14). Writing the motor M = (a + ea')A gives
(a+ea')2=a2+2eaa' as the square of its dual length; and M o M = 2aa'. Evidently M o M = 0 implies that M is a line vector (a' = 0) or a couple (a = 0). 34. Motor Product. The motor product of the motors M = m + ano, N = n + eno is defined as the dual vector obtained by distributing the product (m + emo) x (n + eno), namely, (1)
MxN = mxn + e(mxno +moxn).
When the origin is shifted to P, the dual part of M x N becomes
mxnp + mp x n = m x (no + POxn) + (mo + POxm) xn
= mxno + moxn + POx (mxn),
MOTOR PRODUCT
§ 34
71
in view of the identity, PO x (m x n) + m x (n x PO) + n x (PO x m) = 0.
This transformation conforms with (29.3) and shows that M x N is a motor. The definition (1) shows that (2)
Lx(M+N) =LxM+LxN.
MxN = -NxM,
In order to compute the motor product of two unit line vectors A = a + eao, B = b + ebo, we shall suppose, first, that they are not parallel. With the same notation and choice of origin as in § 33, we have
AxB =a xb + &p ax (exb) = e(sin (P + ecp' cos (p)
or, in terms of the dual angle 4 = c + eqp ,
AxB = Esin4
(3)
in complete analogy with a x b = e sin cp. When A and B are parallel, let e be any unit vector perpendicular to both A and B. If the origin is chosen on A, the foregoing com-
putation still applies, and
AxB =
e
e + Eeo, an arbitrary line vector in the direction of e. This conforms with the fact that A x B, being pure dual a couple), has its axis determined in directionlbut not as tolposition in space. When A and B are collinear, we choose an origin on their common
line;thenA=a,B bandAxB=axb=0andEisentirely arbitrary. Formula (3) covers these cases in which sin 4 = f Ev' and 0, respectively. We note that A x B = sin (D E = (sin p + erp cos cp) E = 0,
(or sin 4) = 0) when and only when
sin(p=0,
gyp'=0,
that is, when the line vectors are collinear. THEOREM. In order that two proper motors be coaxial, it is necessary and sufficient that their motor product vanish.
MOTOR ALGEBRA
72
Proof.
§ 35
With the notation of § 33, M x N = [a/3 + e(af' + a'f )]A x B,
(a/3 X 0).
Since the first factor is neither zero nor pure dual,,M x N = 0 implies sin 4) = 0, and conversely. Finally let us consider the motor product, (4)
M x N = [a$ + E(a3' + a'/3) ] [sin gp +
cos go] E,
for any proper motors M, N. If their axes coincide, M x N = 0. If their axes are parallel, sin
motors L = I + do, M = m + em0i N = n + eno is defined as the dual number obtained by distributing the product L x M N; thus (1)
This definition shows that (2)
LxM N =
LxM N =
0,
in exact analogy with vector products. With von Mises' definition of a scalar product, the preceding triple product becomes
LxMoN = (lxm) - no + (lxmo +loxm) - n; this is the dual part of L x M N. Equations (2), (3), (4) also (5)
apply to L x M o N.
We now propose to find under what conditions L x M N = C when L, M, N are proper motors. If the axes of L, M, N are all parallel, l x m= m x n= n x 1= 0 and L xM N= 0.
If their
axes are not all parallel, choose the notation so that 1 x m
0.
Then L x M is a proper motor whose axis cuts the axes of L and M
at right angles. The condition L x M N = 0 then holds when and only when the axis of N cuts the axis of L x M at right angles.
MOTOR IDENTITIES
§ 36
73
In this case the axes of L, M, N have a common normal (the axis of L x M). We therefore may state the The dual triple product of three proper motors will vanish only when (a) their axes are all parallel, or (b) their axes have THEOREM.
a common normal.
If the proper motors L, M, N satisfy a linear relation with proper coefficients, (6)
(a+ecx')L+($+E$')M+(y+ey')N=0,
aIiy
0,
L x M N = 0, and we can apply the preceding theorem. More precise information, however, is given by the equations,
MxN
NxL
a+ea'=a+e0' =
LxM +e1''
which follow from (6). These show that the three motors M x N, N x L, L x M are all proper or all couples. When proper these
motors are all coaxial; when couples, their moments are all parallel. Hence from (6) we conclude that either (a) the axes of L, M, N, no two of which are parallel, have a common normal;
or (b) their axes are all parallel and coplanar. Thus the linear relation (6) implies that the axes of L, M, N have a common normal. Conversely, if the axes of L, M, N have a common normal, we can infer a linear relation (6), provided the case in which two axes are
parallel to each other but not to the third is excluded. We omit the proof.
36. Motor Identities. All the identities of vector algebra are still valid when the vectors are replaced by motors. For the motor products are defined as the results of applying the distributive law to dual vectors and subsequently equating e2, e3, . . . to zero. Now the identities apply to the ordinary vectors forming the dual to zero are vectors, so that the results before equating e2, e3,
true for any value of the scalar e. In particular, the identities are replaced by zero. Thus we have the hold when e2, e3, motor identities: (1)
Lx(MxN)
(2)
Lx (M N) + Mx (NHL) +Nx (L M) = 0,
(3)
(L M) (NxP)
=LNMP -LPMN.
MOTOR ALGEBRA
74
§ 37
The identity (2) has an interesting geometric interpretation when all nine motors,
L,M,N;MxN,NxL,LXM;Lx(MxN),Mx(NxL),Nx(LxM), are proper. In this case, the axes of the three last motors have a common normal (§ 35). This common normal and the axes of the preceding nine motors form a configuration of ten lines, each one of which is normal to three others. This is essentially the theorem of Peterson and Morley. $ When L, M, N are line vectors along the sides of a plane triangle, this theorem implies the concurrence of its altitudes. The theorem of Peterson and Morley therefore may be regarded as a generalization of this result. 37. Reciprocal Sets of Motors. Consider two sets of proper motors M1i M2, M3; M1, M2, M3 for which ml m2 x m3 0, m1 m2 x m3 /- 0. The sets are said to be reciprocal when the nine equations, (1) Mi Mi = S (i, j = 1, 2, 3), are satisfied. The indices are merely labels, and S is the Kronecker delta. Let the set Mi be given; we then may compute the reciprocal
set M' uniquely. Since M2 M1 = 0, M3 M1 = 0, the axis of M1 is the common normal to the axes of M2 and M3; hence M1 = (X + eX')M2 x M3i and, since MI-M' = 1, A + eX' = 1/M1 M2 X M3; the condition ml m2 x m3 0 0 ensures that M1 M2 X M3 is not pure dual. Thus we find (2)
MX Mk M2 = M1 M2 x M3 ,
Mi =
M'xMk M1
M2 x M3
where ijk represent any cyclical permutation of 123. That the motors (2) satisfy equations (1) is shown by direct substitution. Equations (2) have the same form as the corresponding equations in vector algebra. Moreover, in (M2xM3) . (M2xM3) (M1 M2 x M3) (M1 M2 x M3) the left member is 1, and, from (30.3) the numerator on the right is 1; hence
M1,M1 =
(3)
(M1 M2 x M3)
(Ml M2 x M3) = 1.
$ E. A. Weiss, Einfuhrung in die Linien-geometrie and Kinematik, Leipzig 1935, p. 85.
STATICS
§ 38
75
The axes of two reciprocal sets taken in the order M1M3M2M'M3M2 form a skew hexagon with six right angles. Since (4)
M1 x M1 + M2 x M2
+ M3 x M3 = 0,
the common normals of its opposite sides themselves admit a common normal. The six sides, the three common normals of the opposite sides, and their common normal, thus form a PetersonMorley configuration (§ 36).
If we put Mi = Mi in (1), we find that the only self-reciprocal motor sets are formed by three mutually orthogonal unit line vectors through a point. Any motor M may be expressed linearly with dual coefficients in terms of M1, M2, M3, provided m1 m2 x m3 X 0. To show that the representation is possible, we note that (5)
M = (a + ea')M1 + ($ + ea')M2 + (y + ey')M3
is equivalent to the two vector equations, (6)
m = amt + /3m2 + 'Ym3,
(7)
mo = amlo + /3m20 + 'Ym30 + a'm1 + /3'm2 + 7'm3.
Since m1 m2 x m3 ; 0, the vectors ml, m2, m3 have a reciprocal set m1, m2, m3. From (6),
a = m- m1,
/3 = m-m2,
y = m-m3;
with a, 0, y known, we may determine a', 0', y' in the same way from (7). The actual computation, however, can be effected most simply from (5) by using the reciprocal set m1, m2, m3. Thus we find (8)
M = M M1M1 +M M2M2 +M M3M3
38. Statics. The statics of rigid bodies may be developed independently of dynamics from four fundamental principles: t Principle I (Vector Addition of Forces). Two forces acting on the same particle may be replaced by a single force, acting on the particle, equal to their vector sum. Principle II (Transmissibility of a Force). A force acting on a rigid body may be shifted along its line of action so as to act on any particle of that line. f Sec Brand, L., Vectorial Mechanics, New York, 1930, Chapter II.
MOTOR ALGEBRA
76
§ 38
Principle III (Static Equilibrium). If the forces acting on a rigid body, initially at rest, can be reduced to zero by means of principles I and II, the body will remain at rest. Principle IV (Action and Reaction). This need not concern us here.
Principle II states in effect that a force acting on a rigid body is Such a force is determined by the vector f giving its magnitude and direction and by its moment fo about 0 (§ 27). Thus the dual vector, a line vector.
(1)
F=f+efo
represents a force f acting along the line whose equation is r x f = fo.
If two forces P, Q act on a particle with position vector r,
P=p+Erxp,
Q=q+E1xq;
principle I asserts that both forces may be replaced by a single
resultant, whose vector and moment are p + q, r x (p + q) ; that is, the intersecting line vectors P and Q are force, their
equivalent to the line vector P + Q, their motor sum (§ 32). Two parallel forces P, Q also have a resultant P + Q provided p, + q 0. To see this, we need only introduce a pair of opposed forces F, -F (equivalent to zero) acting along any line cutting the lines of P and Q. We may now apply principle I to find the resultants of P + F and Q - F. These coplanar forces intersect, since
(p+f)x(q-f) =fx(p+q) 96 0, and therefore have the resultant,
(P + F) + (Q - F) = P + Q. If Q = XP (X 96 -1), the line of P + Q divides all segments from P to Q in the ratio X/1 (§ 32, theorem 3). If p + q = 0, po + qo 0, the parallel forces P, Q are equal
in magnitude and opposite in direction and are said to form a couple of moment po + qo. In any case, when a system of forces Fa acting on a rigid body
is changed into an equivalent system G; by the application of principles I and II, the force-sum and moment-sum remain unaltered : 21,; = 2;g1,
lfto = 2;gio.
STATICS
§ 38
77
For each application of principle I leaves these sums unaltered; and shifting a force in accordance with principle II does not alter its vector or moment. Now it can be shown that any system of forces Fi acting on a rigid body can be reduced by means of principles I and II to two forces, * say P= p + epo, Q = q + eqo. Hence, if EF1 = m + emo, we have
p+q=m,
po+qo=mo.
0, m mo
If m
0, the system Fi is equivalent to a screw M = m + emo (§ 31) which may be expressed in many ways as two non-coplanar forces. A screw cannot be reduced to a single force. The screw M may be regarded as a force m acting through the origin and a couple of moment mo. If m 5,4- 0, m mo = 0, the forces P and Q are coplanar; for m mo = 0 implies p qo + q po- = 0 (26.14). The forces P, Q
now may be reduced to the single force M = m + ano, their resultant.
If m = 0, mo
0, the system F1 is equivalent to the couple
P, Q of moment mo. Finally, if m = 0, mo = 0, the system F, reduces to zero, and, in accordance with principle III, the rigid body is in equilibrium. In brief, the system of forces Ft acting on a rigid body is equivalent to the motor M = EFz j this may be a screw, a single force, a couple, or zero; only in the last case is the rigid body in equilibrium.
The moment of a force F = f + do about an axis along the unit
line vector A = a + eao was defined (27.4) as a fp, P being a point on the axis of A. The moment of a system of forces, equivalent to the motor M = m + emo, about this axis is therefore
Now a
and, since OP * a = ao, (2)
We call this the moment of the motor M about the axis A. This moment is independent of the choice of P on the axis. * Brand, L., Vectorial Mechanics, p. 143.
MOTOR ALGEBRA
78
§ 39
39. Null System. A line is called a null line with respect to a motor M if the moment of M about this line is zero. In order that
a line, given by the unit line vector A = a + eao, be a null line with respect to M, it is necessary and sufficient that (1)
If L = XA, L o M = XA o M; hence the axis of any line vector L will be a null line when and only when L o M = 0. The totality of such lines constitutes the null system of the motor M. A motor M can be replaced in many ways by two line vectors F, G. When M is a screw (m mo F6 0), we shall prove that the line of one may be chosen at pleasure, provided it is not a null
line or parallel to the axis of M; after this choice F and G are uniquely determined.
Write F = aA, G = oB as multiples of unit line vectors, and choose A as any line not in the null system. Now M = F + G is equivalent to
m = as + ob,
mo = aao + obo
Since a ao = b bo = 0,
a =
(2)
thus F = aA is known, and G = M - F. Since A is not a null
line, a mo + m ao 7 0; and, since A is not parallel to m, o 7 0, and G is actually a line vector. All null lines through a point P are perpendicular to mp and therefore lie in a plane p, the null plane of P. Moreover all the null lines on any plane p pass through a point P, the null point of p, where mp is perpendicular to p. Analytic Proof. Consider a null line through the points (a, ao), (o, bo); the coordinates are (abo - aoo, ao x bo) from (26.9). Since it is a null line, (3)
m ao X bo + mo (abo - aoo) = 0.
This may be written (4)
(amo - ao x m) bo = ao moo,
NULL SYSTEM
§ 39
79
hence, if the point (a, ao) is fixed and ($, bo) varies over all null lines through (a, ao), (,l, bo) always will lie in the plane whose coordinates are (amo - ao x m, ao mo). Next, let the null line be common to the planes (a, ao), (b, Rio); its coordinates are (a x b, a00 - aob) from (26.10). Since it is a null line, (5)
m (a(3o - aob) + mo a x b = 0.
This may be written (mao - mo x a) b = m a $o;
(6)
hence, if the plane (a, ao) is fixed and (b,,30) turns about all null lines in (a, ao), (b, Qo) always will pass through the point whose coordinates are (m a, mao - mo x a). Equations (3) and (5) are not altered by an interchange of the two points or two planes; hence the same is true of (4) and (6). Therefore: If the null plane of point A passes through B, the null plane of B passes through A.
If the null point of plane a lies on b, the null point of b lies on a. From the preceding results: (7)
Point (a, ao) - null plane
(amo - ao x m, ao mo) ;
(8)
Plane (b, Ro) - null point
(m b, mao - mo x b).
Indeed, if we solve
b=amo - aoxm, for (a, ao), we obtain the null point in (8). Note also that ao b = a0o; let the student interpret this relation in the light of (26.16). When M is a line vector, the null lines are lines that intersect its axis. Then (7) gives the plane through the point and M (26.11); and (8) 'the point on the plane and M (26.12). If M is a screw (m mo F!5 0), not only are points associated with planes, but also lines with lines in general. For if A is any unit line vector not in the null system of M or parallel to its axis, we can determine uniquely a unit line vector B, its conjugate, such
that (9)
M = aA + (3B,
MOTOR ALGEBRA
80
§ 40
where a = m mo/A o M from (2). Any line vector C that cuts two conjugate lines is a null line; for
MoC = aAoC+$B°C = 0. In view of this property we may characterize conjugate lines as follows: The conjugate of a line A is the line B: (i) which is common to the null planes of all points on A, or (ii), which contains the null
points of all planes through A. With this point of view, null lines are self-conjugate.
Along a line parallel to the axis of M, the
moment of M is constant; since the null planes of its points are all parallel, we may say that its conjugate is at infinity. 40. Summary: Motor Algebra. The number X = x + ex' is called dual if x and x' are real and a is a unit with the property E2 = 0. If x 0, the dual number is proper. Operations with dual numbers are carried out as in ordinary algebra and then . set equal to zero. Division, however, is only defined for proper dual numbers. The function f(x + ex') is defined by means of. its formal Taylor expansion; only the first two terms, f (x) + Ex' f'(x), appear since E2 = e3 = . . 0. A line vector with the Pliicker coordinates a, ao (a ao = 0) is e2, E3 ,
.
written as a dual vector, A = a + eao; for a unit line vector I a I = 1.
When the origin is shifted to P, ao becomes
ap = ao + PO- a. A motor M = m + emo is a dual multiple of a unit line vector, M = (X + EA')A = (X + EA') (a + eao)
now m mo = 0 only when A' = 0. A shift of origin to P changes the moment vector as before:
mp=mo+P0xm, but m mp = m mo is invariant. Motors for which X F 0 are said to be proper. Proper motors are screwf when X' 0 0, line vectors when A' = 0. Proper motors have a definite axis given by the unit line vector A. When X = 0, A' 0 0 we have a pure dual motor emo; it is called a couple of moment mo. The moment of a couple is a free vector, determined in direction but not in position. When X = A' = 0 (X + EX' = 0), the motor reduces to zero.
SUMMARY: MOTOR ALGEBRA
§ 40
S1
The sum of the motors M = m + em0, N = n + en0 is defined as
M + N = m + n + e(mo + no). If M and N are line vectors, M + N is a line vector only when
m no+n mo=0.
The scalar and motor products, M N and M x N are obtained by distributing the products as in ordinary vector analysis and then setting e2 = 0:
M N = m - n+ MxN = mxn+ e(mxno+moxn). If A and B are line vectors making an angle p and at a normal distance apart (reckoned positive if a x b points from A to B), the dual angle between A and B is defined as F + &p'. For unit line vectors,
A B= cos
A x B= sin -tE
where E is a unit line vector in the direction a x b and cutting A, B at right angles. These results are perfect analogues of If M and N are non-parallel proper motors, the common normal
to their axes is the axis of M x N. When M and N are parallel, M x N is a couple whose moment is perpendicular to their axes.
If M and N are proper motors, M N = 0 implies that their axes cut at right angles; M x N = 0 that their axes coincide; and conversely.
For three motors, the dual number L x M N is obtained by distributing the product (1 + el0) x (m + emo) (n + eno) :
If L, M, N are proper, L x M N = 0 implies that their axes are all parallel or have a common normal, and conversely.
Two sets of three motors Mti, M' whose triple products are proper are said to be reciprocal when M; M' = S (i, j = 1, 2, 3). All formulas for reciprocal vectors have exact analogues for reciprocal motors. The forces acting on a rigid body are line vectors. A system F;
of such forces is equivalent to their motor sum M = SFj, which
MOTOR ALGEBRA
82
may be a screw, a single-force, a couple, or zero; only when M = 0 is the body in equilibrium. All the rules of calculation in "real" vector algebra have exact analogues in motor algebra. PROBLEMS
1. If A is a unit line vector, show that the screw M = (A + ea')A may be expressed as the sum of a line vector and a couple whose moment is parallel to the line vector, namely, M = aA + eµm, where A= X'/X is the pitch of the screw.
2. Prove that the axes of the motors M, N and M + N either are parallel or have a common normal.
3. Express the screw M = 2i + e(i + j + k) (a) As the dual multiple of a unit line vector. (b) As the sum of a line vector and a couple whose moment is parallel to the line vector. (c) Find the equation of the axis of M.
--
---,
4. In the tetrahedron OABC, el = OA, e2 = OB, e3 = OC. If the forces Pet, Qe2, Re3;
P'(e3 - e2), Q'(e1 - e3), R'(e2 - el)
acting along its six edges are equivalent to the motor m + emo, show that they are uniquely determined by the six equations: P1 = mo e1
...
P + Q' - R' = m , e1,
[eie2e31
5. Show that the n forces P1Q1, P2Q2,
, P. Q are equivalent to the
motor,
n(q* -p*) +expi gi, where p*, q* give the mean centers of the points pi and q;, respectively. [Cf. (9.4)].
6. In order that the axes of the motors, M = m + emo, N = a + eao, be_ coplanar, it is necessary and sufficient that
where µ and v denote the pitch of M and N. 7. If three line vectors, a + eao, b + ebo, c + eco, are parallel to a plane, as +,6b + -yc = 0 (§ 5). Prove that any line vector d + edo meeting them is parallel to a fixed plane normal to aao + Rbo + yco. 8. The forces Xi, Yj, -Zk act through the points (0, 0, 0), (a, b, c), (0, b, 0), respectively. Show that they reduce to a single force if
X/a + Y/b + Z/c = 0.
PROBLEMS
83
9. If ABCD are the vertices of a skew quadrilateral and PQRS are the midpoints of its sides taken in order, show that the motor equivalent to the forces
AB, BC, CD, DA is also equivalent to four times the couple formed by PQ and RS.
10. If the moment of the motor M about each of the six edges of a tetrahedron is zero, show that M = 0. [Cf. (38.2)]. 11. If the origin 0 is taken on the axis of a screw M of pitch µ show that the equation of the null plane of the point Ti is
If the axis of M is chosen as z-axis, this equation becomes xyl - yxl = A(z - zi).
CHAPTER III VECTOR FUNCTIONS OF ONE VARIABLE
41. Derivative of a Vector. Let u(t) denote a vector function of a scalar variable t over the interval a < t < b; that is, when t is given, u(t) is uniquely determined.* The function u(t) is said to be continuous for the value t if u(t + At) -+ u(t) as At -* 0. To obtain a clear idea of the way in which u varies with t, we may regard u(t) = OP as a position vector issuing from the origin.
Then as. t varies, the point P traces a certain curve r in space.
For the values t and t + At, let u(t) = OP, u(t + At) = OQ; the change in u for the increment At is
--4
-*
--4
Du=u(t+At) -u(t) =OQ-OP=PQ. The average change per unit of t is Au/At. When At > 0, Au/At
is a vector (PA' in Fig. 41) in 'R
R
*
the direction of Au and 1/At times
as long; but if At < 0, Au/At has
the direction of -Au (then PA' in Fig. 41 must be reversed). If PA' approaches a limiting vector
PA as At -+ 0, we call PA the derivative of u(t) with respect to t
and denote it by du/dt or u'(t). The equation defining the de-
Fia. 41
(1)
rivative is therefore
du = lim u(t + At) - u(t) Du = lim dt
w -o
At
'U - o At
* The word function implies a single-valued function. 84
DERIVATIVE OF A VECTOR
§ 41
85
If the limit du/dt exists for the value t, u(t) is said to be differentiable at t. Then we may write Au/.t = du/dt + h where h --> 0 as At -* 0; hence
0u =
At
and u(t + At) ---> u(t). continuous there.
(du
-dt + h) --). 0, as
At --> 0,
Thus if u(t) is differentiable at t, it is also
If u(t) is differentiable at P, Q describes the arc QP of the curve r as At -* 0, and the limiting direction of the chord PQ, and hence of the vector PA', is along the tangent at P. The limiting vector
PA = du/dt is therefore tangent to I' at P. Since Du/At has the same direction as Au when At > 0, the opposite when At < 0, du/dt is a vector tangent t o r in the direction of increasing t. We restate this important result as follows:
0
If the vector u(t) = OP varies with t, so that P describes the curve r when 0 is held fast, the derivative du/dt, for any value of t, is a vector tangent to r at P in the direction of increasing t. Example 1. If u = OP is a variable vector of constant direction, P will move on a straight line when 0 is held fast; hence du/dt, being tangent to this line, will be parallel to u. Example 2. If u = OP is a variable vector of constant length, P will describe
a curve r on the surface of a sphere when 0 is held fast; hence du/dt, being tangent to r at P, will be perpendicular to the radius OP of the sphere. In brief : If j u I is constant, du/dt is perpendicular to u.
If u is a constant vector, that is, constant in both length and direction,
Du = 0, Ot = 0, and dt = 0. The derivative of a constant vector is zero.
When u is a function of a scalar variable s, and s in turn a function of t, a change of At in t will produce a change As in s and therefore a change Au in u. On passing to the limit At --* 0 in the identity, Du At
du du ds -, we get = --As At dt ds dt Du As
the familiar "chain" rule.
VECTOR FUNCTIONS OF ONE VARIABLE
86
§ 42
The higher derivatives of u(t) are defined as in the calculus: U"'(t) = du" , etc.
u"(t) = du' ,
dt
dt
42. Derivatives of Sums and Products. Let u(t) and v(t) be two differentiable vector functions of a scalar t. When t changes by an amount At, let Au, Av, and O(u + v) denote the vectorial changes in u, v, and u + v. Then
u + v + O(u+v) = u+ Du+v+ Ov, O(u + v) = Du + AV, A(u + v)
Du
AV
At
At
At
and passing to the limit At - 0 gives du
d v) (1)
dt
(u +
dt
dv
+
dt
Consequently, the derivative of the sum of two vectors is equal to the sum of their, derivatives. This result may be generalized to the sum
of any number of vectors. Consider next the product f(t)u(t) of a differentiable scalar and a vector function. When t changes by an amount At, let Af, Au,
and A(fu) denote the increments of f, u, and fu, respectively. Then, since multiplication of vectors by scalars is distributive (§ 4),
fu + o(fu) = (f + Af)(u + Au) = fu + f Du + Af U + Of Au,
O(fu)=fAu+Afu+OfAu, A(otu)
f =
AU-u+Of
-;
passing to the limit At - 0, and noting that Af -* 0, we have (2)
dt (f u) = f dt
+
dt u.
This is formally the same as the rule for differentiating a product of scalar functions.
DERIVATIVES OF SUMS AND PRODUCTS
§ 42
87
Important special cases of (2) arise when either f or u is constant: du
d
df
d
=cdt,
-(cu)
-(fc) =atc.
If the components of a vector u(t) are ui(t) when referred to a constant basis ei, u(t) = 2;ui(t) e,,
(3)
du
dug
dt
dt
ei.
The components of the derivative of a vector are the derivatives of its components.
If we pass now to the products u v and u x v, where u and v are vector functions of t, the same type of argument used in proving (2) shows that d
d (5)
dt
dv
du
dv
du
-(u.v)=u-at+at-V,
(4)
(u xv - U x
- - x v.
dt
+
dt
The proofs depend essentially upon the distributive laws for the dots and cross product. In (5) the order of the factors must be preserved. If 1 is an axis with unit vector e, the orthogonal component of u
on 1 is e u (15.6) ; hence d dt
comp1 u = e
du -
du
comps dt = dt
THEOREM 1. A necessary and sufficient condition that a proper
vector u be of constant length is that
u
(6)
Proof.
du dt
=0.
Since u 12 = u u, we have, from (4), I
d
-1
du u12=2u'dt
If I u I is constant, the condition follows: conversely, the condition implies that I u I is constant.
VECTOR FUNCTIONS OF ONE VARIABLE
88
§ 43
THEOREM 2. A necessary and sufficient condition that a proper vector always remain parallel to a fixed line is that du
(7)
Proof.
uX
dt
= 0.
Let u = u(t)e where e is a unit vector; then uX
du
= ue x
-- +e u -de\J = u2e X de
Cdu dt
dt
dt
dt
If e is constant, de/dt = 0, and the condition follows. Conversely, since u 0 0, the condition implies that de
ex-=0. dt
Also
de dt
from theorem 1. These equations are contradictory unless de/dt = 0; that is, e is constant. 43. Space Curves. Consider a space curve whose parametric equations are (1)
x = x(t),
y = y(t),
z = z(t),
where x(t), y(t), z(t) are analytic functions of the real variable t defined in a certain interval T of t values. To avoid having the curve degenerate into a point, we explicitly exclude the case in which all three functions are constants. We also restrict the interval T so that there is just one value of t corresponding to each point P of the curve. Then equations (1) set up a one-to-one correspondence between the points of the curve and the values of t in the interval T. The requirement that the functions x(t), y(t), z(t) be analytic in T ensures the continuity of the functions and their derivatives of all orders and also guarantees a Taylor expansion for each function about any point of T. A curve which admits a representation (1) with functions thus restricted is called an analytic space curve. Moreover, if all three derivatives dx/dt, dy/dt, dz/dt do not vanish simultaneously for any value of t in the interval T, the curve is said to be regular, and the parameter t is said to be a regular parameter. bet us make a change of parameter, t = P(u), where jp(u) is an analytic function of u in a certain interval U; we
SPACE CURVES
§ 43
89
assume also that t just covers T as u ranges over U. Then the equations,
x = x[,P(u)l,
z = z[co(u)] y = y[O(u)l, constitute a new parametric representation of the curve. Writing dt/du =
dx
dy
dy lp
dz
dz
du = at
du = at (u)' du = (u)' a regular parameter, u is a regular parameter when and
only when (u) 0 in U. When this condition is fulfilled, the implicit-function theorem ensures the existence of the inverse function u = 4'(t) which is also single valued and analytic in T. Thus
the points of the curve not only correspond one to one to the t values in T, but also to the u values in U: P -* t -> fi(t) = U,
u --> (P(u) = t --+ P.
Let equations (1) define a regular analytic space curve. The position vector of the point P(t) is (2)
r = i x(t) + j y(t) + k z(t) = r(t);
and, if we write t, z for dr/dt, dx/dt, (3)
t = it(t) + j y(t) + k i(t) F6 0.
If Po(t = to) is a fixed point on the curve, the length of the arc s from Po to P is defined as (4)
s=
fV
dt,
to
an analytic function of t which is positive or negative, according as t > to or t < to; moreover (5)
ds/dt = VT-t > 0,
The inverse function t =
0. Putting t =
VECTOR FUNCTIONS OF ONE VARIABLE
90
§ 44
44. Unit Tangent Vector. Let r = r(t) be a regular analytic space curve on which s = arc POP is reckoned positive in the direction of increasing t. Then (Fig. 44a) dr
Ar
-- = lim - = lim
PQ
A -o As Q -. P arc PQ is a unit vector (§ 43), and hence ds
Em
chord PQ
Q -. P arc PQ
The tangent line to the curve at P is defined as the limiting position of the secant PQ as Q approaches P. Hence from its definition
FIG. 44b
FIG. 44a
we conclude that dr/ds is a unit vector T tangent to the curve at P and pointing in the direction of increasing arcs; dr
= T.
(1)
ds
An important special case arises when r is a unit vector revolving
in a plane. If we imagine this vector R always drawn from the same initial point 0, its end point will describe a circle of unit radius (Pig. 44b) and s = 0, where 0 denotes the angle in radians between a fixed line OA and R. If P is a unit vector perpendicular to R in the direction of increasing angles, (1) now becomes dR/dO = P. If k is a unit vector normal to the plane and pointing in the direction a right-handed screw advances when revolved in the positive sense of 0, P = k x R, and (2)
dR dB
= k x R = P.
UNIT TANGENT VECTOR
§ 44
91
In Fig. 44b, k points upward from the paper. Moreover, since dP/d6 = k x (dR/d6), we have dP
-=kxP= -R. d6
(3)
We state these results in the THEOREM.
The derivative of a unit vector revolving in a plane,
with respect to the angle that it makes with a fixed direction, is another
unit vector perpendicular to the first in the direction of increasing angles.
If P(x, y) is a variable point on a plane curve, r = xi + yj, T=
dr
dx
dy
as = as'
+ ds ,
and i T = dx/ds, j T = dy/ds; hence ds = cos (i, T),
(4)
ds = sin 01 T).
If (r, 0) are the polar coordinates of P, we write r = rR, where the polar distance r and the unit vector R are functions of 0. Now dr
dR d9
dr
d6
ds
d6 ds
ds
ds
T = -- R+r --- = - R+r--P, and R T = dr/ds, P T = r d6/ds; hence dr
-=
(5)
cos (R, T),
r
do
--
= sin (R, T).
Example 1. If 0 is measured counterclockwise from the x axis,
R=icos0+jsin0,
P=icos(0+ 117r)
From (2), the corresponding components of dR/d9 and P must be equal; hence dB
cos 0 = cos (0 + 4r)
sin B,
d sin B = sin (0 + fir) = cos 0.
Example 2. If rl, r2 are the distances of a point P on an ellipse from the foci,
rl + r2 = const. On differentiating this equation with respect to a, we have, from (5), drl ds
+
dr2
ds
= (RI + R2) T = 0.
92
VECTOR FUNCTIONS OF ONE VARIABLE
§ 45
Since RI + R2 is perpendicular to T, the normal to the ellipse at P has the direction of RI + R2. The normal therefore bisects the angle between the focal radii.
45. Frenet's Formulas. Let r = r(s) be a space curve, s = are POP, and T = dr/ds the unit tangent vector at P. Since T is of constant length, dT/ds, if not zero, must be perpendicular to T. A directed line through P in the direction of dT/ds is called the principal normal of the curve at P. Let N denote a unit vector in the direction of the principal normal; then we may write (1)
dT
= KN,
ds
where K is a non-negative scalar called the curvature of the curve
at P. Now B = T x N is a third unit vector perpendicular to both T and N. Thus at each point of the curve we have a right-handed set of orthogonal unit vectors, T, N, B, such that
TxN = B,
NxB = T,
BxT = N.
As P traverses the curve, we speak of the moving trihedral TNB. A directed line through P in the direction of B is called the binormal to the curve at P. Since B is a unit vector, dB/ds, if not zero, must be perpendicular to B. Differentiating B
T x N, we have
dB
dT
dN
dN
ds
dsxN+Txds = Txds
in view of (1). Hence dB/ds is perpendicular to T as well as B and therefore must be parallel to -B x T = N. We therefore may write dB
7-7I/
__
(2)
= -TN,
ds
where T is a scalar called the torsion
of the curve at P. The minus sign Torsion positive
Fla. 45
is introduced in (2) so that, when 7is positive, dB/ds has the direction
of - N; then, as P moves along the curve in the positive direction, B revolves about T in the same sense as a right-handed screw advancing in the direction of T (Fig. 45).
FRENET'S FORMULAS
§ 45
93
We may now compute dN/ds from N =
by use of (1) and
(2) ; thus dN
dB
dT
ds
ds
ds
(3)
-xT+Bx-= -rNxT+KBxN.
Collecting (1), (2), and (3), we have the set of equations: dT ds
dN (4)
ds
dB ds
known as Frenet's Formulas, which are fundamental in the theory of space curves. If we write (3) in the form, dN
ds
= (rT + KB) x N,
and introduce the Darboux vector,
S = rT + KB,
(5)
we have SxT = KN,
SxB = -TN.
Hence Frenet's Formulas may be put in the symmetric form: dT
(6)
ds
= S x T,
dN
dB
ds
ds
- =SxN
SxB.
Since S N = 0, we may write the Darboux vector in the form: (7)
S = Nx(SxN) =
dN .
Nxd8
From (5), we see that the curvature K and torsion r are the components of the Darboux vector on the binormal and tangent, respectively.
From (4), it is clear that both K and r have the
dimensions of the reciprocal of length; hence
p=1/K,
a=1/r
VECTOR FUNCTIONS OF ONE VARIABLE
94
§ 45
have the dimensions of length and are called the radius of curvature and the radius of torsion, respectively.
Since N, by definition, has the same direction as dT/ds, the curvature K is never negative. If K vanishes identically, dr
dT
ds
r=as+b,
T=ds=a,
=0,
where a and b are vector constants. The curve is then a straight line. Conversely, for a straight line, T is constant and K = 0. The only curves of zero curvature are straight lines. For a straight line the preceding definition fails to determine N. We therefore agree to give N any fixed direction normal to T, and as before define B = T x N. Since B is also constant, dB/ds = 0 and T = 0. For a straight line the Darboux vector is zero. The torsion may be positive or negative. As P traverses the curve in the positive direction, the trihedral TNB will revolve about
T as a right-handed or left-handed screw, according as r is positive or negative. The sign of r is independent of the choice of positive direction along the curve; for, if we reverse the positive direction, we must replace dT s,
T, ds ,
' dB
dN N,
By
ds ,
-8) -T,
by
ds
dT
dN N,
ds ,
ds
,
- B,
dB
ds '
and equations (4) maintain their form with unaltered K and r. If r vanishes identically, dB/ds = 0, and B is a constant vector; hence, from BT=B
dr ds
0,
B
(r - ro) = 0,
that is, the curve lies in a plane normal to B. Conversely, for a plane curve, T and N always lie in a fixed plane, while B is a unit vector normal to that plane; hence dB/ds = 0 at all points where N and B are defined (K* 0) and -r = 0. The only curves of zero torsion are plane. Example. Parallel Curves. Two curves r and rl are called parallel if a plane normal to one at any point is also normal to the other.
The common
normal plane cuts r and rl in corresponding points P and PI; then PP, = rl - r lies in the plane of N and B and may be written XN + µB, where X, µ are scalars. Thus we have (i)
rl=r+XN+pB;
CURVATURE AND TORSION
§ 46
95
on differentiating (i) with respect to s and using Frenet's Formulas, we have (ii)
ds1 _ Tids
dX =T+dN +X(-KT
+rB)+dsduB -uTN
_ (1 - AK)T + (A' - µr)N + (A' + Xr)B. Since T1 and T are normal to the same plane, we may choose the positive sense on t1 so that T1 = T; hence, from (ii), we conclude that
^ = 1 - AK,
ds = Jlr,
ds
= -Ar.
Moreover dX
du
1 d
2ds()2+u2) =0,
and A2 + µ2 is constant. Consequently, the distance PP1 between corresponding points of parallel curves is always the same. On differentiating T1 = T with respect to s, we have also (iv)
KjN1
1 = KN;
d3
hence
ds =
Kl ,
and N1 = N, B1 = B. Comparison with (iii) gives K/KJ = 1 - AK
or pl = p - A.
Finally, on differentiating B1 = B with respect to s, we have (v)
-r1N1
dsi
= -rN;
hence, if r 0 0, - =
r 7-1
From (iv) and (v), K/r = K1/7-1 at corresponding points.
46. Curvature and Torsion. By use of Frenet's Formulas, the curvature and torsion are readily computed from the parametric equations of the curve. Thus, on differentiating r = r(t) three times and denoting t derivatives by dots, we have
_drds ds dt
=
$T,
f=ST+ dTSZ
ds
= ST + S2KN, dT
ds
+(2saK+a2K)N+S3K
dN CAS
= ST + SSKN + (2SSK + S2K)N + S3K(-KT + TB)
= (S - S3K2)T + (3MK +
S2K)N + S3KTB.
VECTOR FUNCTIONS OF ONE VARIABLE
96
§ 46
Hence t x It y= S6K2T,
t x y= S3KB,
t = IsI
and, since
K=
(1), (2)
0, xr
T=
I
Itt x
It t I 3
I2
If the positive direction on the curve is that of increasing t, ds/dt > 0; and the preceding equations show that T has the direction of t, B has the direction of t x i', and, since N = B x T, N has the direction of (f x Y) x t.
If the parametric equations of a plane curve are x = x(t), y = y(t), we have
t=
r = xi + yj,
r = xi + yl,
+ yJ,
and, from (1), its curvature is
-
I xy + yx I K = (x2 y2)
(3)
If the curve has the Cartesian equation y = y(x), we can regard
x as parameter: x = t, y = y(t). Then (3) becomes K =
(4)
(1 + y,2)I'
where the primes indicate differentiation with respect to x.
If the curve has the polar equation r = r(9), we may write r = rR, where R is a unit radial vector. Then regarding 0 as the parameter t, we have, from (44.2) and (44.3),
t = rR + rP, hence, from (1), K_
(5)
f = (r - r)R + 2rP;
Ire+2r2-rrI (r2+t2)§
Example. Find the vectors T, N, B and the curvature and torsion of the twisted cubic
x=2t,
at the point where t = 1.
y=t2,
z=t3/3
FUNDAMENTAL THEOREM
1 47
97
Since r(t) _ [2t, t2, 0/31, we have f = [2, 2t, t2],
7 = [0, 0, 2];
1' = [0, 2, 2t],
hence, when t = 1, f = [2, 2, 1],
i = [0, 2, 2],
f = [0, 0, 2],
fxi' f = 8.
t x 1' = [2, -4,41,
Since T, B are unit vectors in the directions of t, t x t, and N = B x T,
N=1t2,-1,-2], 3
T = 13 [2,2,11,
K = $,
Moreover, from (1) and (2),
B=1[1,-2,21. 3
r=$
47. Fundamental Theorem. Two curves for which the curvature and torsion are the same functions of the arc are congruent. Proof. Let the curves have the equations r = r1(s), r = r2(s). Bring the origins of are and the trihedrals T1N1B1, T2N2B2 at these
points into coincidence. At the points P1i P2 of the curves, corresponding to the same value of s, consider the function,
f(s) =
(1)
By Frenet's Formulas, d = ds
K1N1
T2
+
K2N2
T1
(-K1T1 + 71B1) N2 + (-K2T2 -- T2B2) . N1 T1N1 - B2 - T2N2 B1;
and, since K1 = K2, T1 = T2 for the same value of s,
-ds
= 0,
f (s) = C,
a const.
But at the point s = 0, the trihedrals coincide, and
C=f(0)=1+1+1=3. Hence for any value of s, f(s) = 3, an equation that implies that each scalar product in (1) equals one; consequently, N1 = N2,
T1 = T2,
B1 = B2.
From the first of these equations, dr1
dr2
ds
ds
'
rl = r2 + a;
and a, the vector constant of integration, is zero, since r1 = r2 when s = 0. Therefore r1(s) = r2(s), and the curves coincide.
VECTOR FUNCTIONS OF ONE VARIABLE
98
§ 48
48. Osculating Plane. The osculating plane of a curve at a point PI is defined as the limiting plane through three of its points P1, P2, P3 as P2 and P3 approach P1. Let the points P1, P2, P3 of the curve r = r(s) correspond to the are values 81, s2i 83 (sI < s2 < s3). Then, if n is the unit normal to the plane P1P2P3, the function,
f(s) = [r(s) - rl] n vanishes when s = s1i 82, s3. Hence from Rolle's Theorem,
f(s) = r'(s) n vanishes twice, and
f"(s) = r"(s) n vanishes once in the interval s1 < s < 83. Consequently, as P2 and P3 approach P1, n approaches a limiting vector n1i such that
r'(sl) n1 = T1 n1 = 0, r"(sl) n1 = K1N1 nl = 0. If ICI 76 0, n1 is perpendicular to both T1 and N1i that is, parallel to B1. The osculating plane at a point of a curve is the plane of the
tangent and principal normal at this point. For a plane curve T and 'N lie in the plane of the curve-the osculating plane at all of i
points.
F r the curve r = r(t), B is parallel to t x 'r (§ 46); hence the equation of the osculating plane at any point r is (1)
her e R is the position vector to any point of the osculating plane. The osculating planes of a curve form a one-parameter family. The osculating planes at the points P1 and P2 intersect in a straight line; and, as P2 approaches P1, the limiting position of this line is called the characteristic of the osculating plane at P1. To find the characteristics of the osculating planes to the curve r = r(s), we must adjoin to their equation (R - r) B = 0, the equation obtained by differentiating it with respect to s, namely, or
if r 0 0. The two equations,
(R-r)B=0,
(R-r)N=0,
CENTER OF CURVATURE
§ 49
99
represent planes through the point r of the curve perpendicular to B and 11, respectively; together, they represent a line through the point r parallel to N x B = T. The characteristics of the osculating planes of a skew curve are tangents to the curve. Example. For the twisted cubic
x=2t,
y=t
z=t3/3
of the example in § 46, t x 'r = 2[1, -2, 2] at the point (2, 1, for which t = 1. Hence, from (1), the equation of the osculating plane to3)the curve at this point is
(x-2)-2(y-1) }2(z-3)=0.
49. Center of Curvature. The center of curvature of a curve at the point P1 is defined as the center of the limiting circle through
three of its points P1, P2, P3, as P2 and P3 approach P1. Since the limiting plane of P1, P2, P3 is the osculating plane through P1, the limiting circle lies in this plane. If c and a denote the position vector of the center and the radius of the circle through
the points P1, P2, P3 of the curve r = r(s), the function,
f(s) = [r(s) - c] [r(s) - c] - a2 vanishes for s = 81, 82i 83 (s1 < 82 < s3).
Hence
f(s) vanishes twice, and
f"(s) vanishes once in the interval s1 < s < s3. As P2 and P3 approach P1, c and a approach limits c1, a1, such
that (r1 - c1) (rl - c1) - ai = 0, (r1 - c1) Tl = 0,
(r1 - c1) Nl + pi = 0.
Since r1 - cl lies in the osculating plane, its components on T1, N1, B1 are 0, -pl, 0, respectively, and r1 - cl = -p1R1. Thus the center and radius of the limiting circle are cl = r1 + p1 N1 and a1 = p1. The center of curvature at any point P of the curve has the position vector, (1)
c=r+pN.
100
VECTOR FUNCTIONS OF ONE VARIABLE
§ 50
This is a point on the principal normal at a distance p from P in its positive direction. Example. Curves of Constant Curvature. Let r = r(s) be a twisted curve t of constant curvature K. The locus t1 of its centers of curvature has the equation,
rl=r+pN.
(2)
Differentiating with respect to s, we have ds1 T1
ds
= T + P(rB - KT) =
1
B. K
Choose the positive direction on rl so that Ti = B; then dal (3)
(8
rK
Differentiating T1 = B with respect to s now gives K1N1ds1_ ds
-7,N,
or
K1N1 = - KN;
hence N1 = -N and K1 = K. Thus r1 is also a curve of constant curvature. Since (2) may be written r = r1 + p1N1,
(4)
r is also the locus of the centers of curvature of r1. Consequently from (3) we also have d8/ds1 = r,/K1; hence rr1 = KK1. The two curves thus have the following relations: (5)
T1 = B,
Ni = -N, B1 = T;
K1 = K,
771 = KK1.
Twisted curves of constant curvature may be associated in pairs, each curve being the locus of the centers of curvature of the other.
50. Plane Curves. For plane curves the torsion is zero and Frenet's Formulas reduce to (1)
dT
ds
= KN
dN
ds
= -KT.
Since dT/ds is always directed towards the concave side of a plane curve, the same is true of N. The osculating plane at any point P is the plane of the curve; and, if K 0 0, the center of curvature P1 is given by (2)
rl = r + pN.
At points of inflection dT/ds = 0, K = 0, and N ceases to be defined. As we pass through a point of inflection, N abruptly reverses its direction, and hence B = T x N does the same.
PLANE CURVES
150
101
To remedy this discontinuous behavior of N and B at points of inflection, the following convention often is adopted in the differential geometry of plane curves. Take B once and for all as a
fixed unit vector normal to the plane of the curve, and define N = B x T. As before T, N, B form a right-handed set of orthog-
onal unit vectors. The curvature K is defined by (1) and is posi-
Fia. 50a
FIG. 50b
tive or negative, according as N has the direction of dT/ds or the opposite. Equations (1) still hold good; for dN
-
dT
=BxdB =BXKN=-KT.
Let 4, be the angle from a fixed line in the plane to the tangent
at P, taken positive in the sense determined by B (Fig. 50a). Then, from (44.2), dT
dT do
ds = dt' ds
d4, d# = BxT= N-
ds
ds
and hence, from (1), d4, K=a,
(3)
p=dds .
The locus of the centers of curvature of a plane curve is called Let P and P1 be corresponding points of a curve r and its evolute r1 (Fig. 50b); and s = AP and sl = A1P1 denote corresponding arcs. On differentiating the equation (2) of r1 with respect to s, we have its evolute.
ds1
dN
dp
dp =T+pds+dsN=-N.
T1
102
VECTOR FUNCTIONS OF ONE VARIABLE
§50
Choose the positive direction on F1 so that T1 = N; then ds1
dp
ds
ds'
and s1 = p + const.
Hence the tangent to F1 is normal to r; and, since As1 = Op, an arc of F1 is equal to the difference in the radii of curvature of r to its end points. These properties show that a curve may be traced by a taut string unwound from its evolute; the string is always tangent to F1 and its free portion equal to p. From this point of view, r is called the involute of Fl. From T1 = N, we have '1 = 4, + 7r/2; hence, from (3), ds1
p1=d4,1 =
dp
d
d2s d . 4,2
Example 1. The only plane curves of constant non-zero curvature are circles. For from dN dN dr = -KT, or = -p , ds ds ds
we have, on integration,
r - c = -pN,
or Ir-cl2 =p2
This is the equation of a circle of radius p and center c.
FIG. 50c
Example 2. An involute r1 of a plane curve r is generated by the point P1 of a taut string unwound from r (Fig. 50c). If t is the curve r - r(e), the involute is given by r1 =P r - sT.
PLANE CURVES
§ 50
103
Hence, on differentiation with respect to 8, dsl
Tlds =T - T - sKN = -8rcN. If we choose the positive direction on r, so that Ti = -N, dsl
(i)
dsl
d4,
= sx = s - ,
or
= s.
dp
d8
The intrinsic equation of a plane curve is the relation connecting s and ', say s = s(4,). For the are sl along its involute we have, from (i),
sl = f
provided sl = 0,
when ¢ = VO.
0
The intrinsic equation of a circle of radius r is s = ry& when >G is measured
from the tangent at the origin of axes. For its involute we have sl = rp2/2, provided sl = 0 when' = 0. The intrinsic equation of a catenary of parameter c is s = c tan when a is measured from the vertex (V, - 0). t The are sl, measured along the involute from the vertex of the catenary, is
sl = c
f tan k dy, = clog see ¢. 1k
0
Example 3. Envelopes. Consider the plane vector function of two variables, r = f(u, v). The one-parameter family of curves u = c (constant) is given by
rl = f(u, v),
u = const.
If this family has a curve envelope given by v = p(u), namely,
R = f[u,,p(u)],
(ii) the vectors,
dR u_ of au
of
+ av `P (u)
and
drl _ of dv
av
are parallel at the points of contact; that is, of
of
- x -- = 0. au av
This condition must be fulfilled if the envelope exists. If (iii) leads to a relation v = p(u) (which does not make of/au oraf/av zero), this relation gives the envelope (ii).
t See Brand, Vectorial Mechanics, equation (97.3). In Fig. 97b (p. 202), the string PA will unwrap into the position PL; as P moves along the catenary, the locus of L is the involute of the catenary. The line LQ is tangent to the
involute (Tj = -N); and, since the x-axis intercepts a constant segment LQ = c on this tangent, the involute of the catenary is a tractrix, a curve characterized by this property.
VECTOR FUNCTIONS OF ONE VARIABLE
104
§ 50
Consider, for example, the one-parameter family of normals to the plane curve r = r(s), namely,
rl = r(s) + v N(s)
(8 = const.)
If they have an envelope,
or,xari = (T - v,cT)xN = (1 -vK)B = 0. as
av
Hence the envelope is given by v = 1/K = p, or
R = r(s) +'N(s), namely the locus of the centers of curvature of the curve. Example 4. Plane Caustic by Reflection. Light issuing from a point 0 is
reflected from a mirror r (Fig. 50d). The curve enveloped by the reflected rays is called a caustic. If R and e are
TR _
the unit vectors along the incident and reflected rays, the reflected rays are the one-parameter family of lines,
ri = r(s) + ve(s).
To find their envelope, we form the equation : (iv)
Fia. 50d
arl arl _ (T as x 49V
de)
+ v dsJ x e =
0.
Denote the angles (i, R) = 0, (R, T) - y; then (T, e) = y by the. law of reflection, and
4,=(i,T) -0+y, Now
.a
de
from (44.5).
de
p=(i,e)=0+2y-24, -0.
sin yl d0\ =kxe(2'--)=kxe(2Kr / dy&
Substitution in (iv) now gives
ksin y-kv(2K- sinr y =0; whence 1
1
2
v
r
p sin y
(v)
This equation determines v and the caustic curve. Letting r --, co, we obtain a beam of parallel rays; then v - 1p sin y. When r is a circular are of radius p with center on Ox, the caustic has a cusp on Ox at a distance vi from r given by 1/vl + 1/r - 2/p.
HELICES
§ 51
105
Example 5. The tractrix is a plane curve for which the segment of any tangent between the point of contact and a fixed line is constant. In Fig. 50e
the fixed line is the x-axis, the constant length c, and, if r = OP, rl = OP1,
r1=r+cT. Hence if we differentiate with respect to s, i
ds,
= T + cKN;
F iu. 50e
on multiplying by j we have
sink, or
ds
d = c tan V,. Ifs =Owhen# -0, this gives s = c log sec
for the intrinsic equation of the tractrix (cf. ex. 2).
51. Helices. A helix is a twisted curve whose tangent makes a constant angle with a fixed direction.
If e is a unit vector in the
fixed direction, and a is the constant angle, the defining equation of a helix is (1)
(0
106
VECTOR FUNCTIONS OF ONE VARIABLE
§ 51
Differentiating (1) twice with respect to s gives
ea
dN 0,
=0;
hence the Darboux vector 6 = N x (dR/ds) is parallel to e. Conversely, if S for a curve is always parallel to a fixed vector e, dT a--=0, ds
e-N=0,
and e - T=cosa,
where cos a is a constant of integration; the curve is therefore a helix. The only twisted curves whose Darboux vector has a fixed direction are helices. Thus helices are characterized by the prop-
erty, dS SX-=0
(2)
(§ 42, theorem 2).
ds
The fixed direction of
S = TT+KB is along the axis of the helix. Now dS
dr
ds
ds
dK
B
(3)
T +
ds
for T dT/ds + K dB/ds = 0 (45.4); hence dS
d-
d-\
d (K-
sx- = KT - T -- a = -'r2 N-ds C ds A ds
TJ
Thus 6 x (dS/ds) = 0 implies that the ratio K/r is constant, and conversely. Helices are the only twisted curves for which the ratio of curvature to torsion is constant.
If we put T = dr/ds in (1) and integrate, we obtain
e - r=scosa+c for the component of r in the direction of e. Hence, if r1 denotes the projection of r on a plane perpendicular to e, (4)
r = r1 + (s cos a + c)e.
Every helix therefore lies on a cylinder with generators parallel to e. The plane curve r1 traced by r1 is a normal section of the cylinder.
§ 51
HELICES
107
If s = POP is the arc along the helix, and s1 = POP1 the arc along the normal section r1 through P0, we have, on differentiating (4) twice with respect to s, (5)
(6)
ds1 T=T1-+cosae ds
KN = K1N1
dsl
2
C ds
From (5), ds1 2 = ds
(T - e cos a) - (T - e cos a)
= 1 -2cos2a+C082a
sing a;
and, if we choose the positive direction on r1 so that s1 with s,
increases
ds1 (7)
- = sin a. ds
Remembering that K and K1 are essentially positive, we now have, from (6), N = N1i and (8)
K = K1 sin2 a.
On differentiating e = T cos a+B sin a we find K/r =tan a; hence (9)
r = K1 sin a cos a.
Finally, on integrating (7), we obtain (10)
s1 = s sin a,
the constant of integration being zero, since both s and s1 are measured from the-same point Po. The only twisted curve of constant curvature and torsion is the circular helix. For, since K/r is constant, such curves are helices;
and, from (8), K1 is constant; that is, the normal section is a circle.
The circular helix is the only twisted curve for which the Darboux vector is constant. For from (3) we see that 6 is constant when and only when K and r are constant.
108
VECTOR FUNCTIONS OF ONE VARIABLE
§ 52
Example. The curve, x = a cos t,
y = a sin t,
z = bt,
is a circular helix; for the curve lies on the cylinder x2 + y2 = a2, and, since
t=[-asint,acost,b], cos (k, T) = k t/I t I = b/1/a2.+ b2 (const.).
Moreover
[-a cos t, -a sin t, 0],
T = [a sin t, -a cost, 0], and, from (46.1) and (46.2), s
_
ItXrI II I3
_
a
a2 +b2'
r
b
ItxiI2
a2+ b2
We now find the Darboux vector S = k/1/a2 + b2; it is constant in magnitude and direction.
52. Kinematics of a Particle. To define the position P of a particle moving along a curve r, choose a point PO of r from which to measure arcs, and take a definite direction along r as positive. Then, if the arc s = POP is given as a function of the time, s = f (t), the motion of the particle is determined; for its position is given
at every instant. The speed v of the particle at the instant t then is defined as (1)
v=
ds dt
Thus the speed will be positive or negative, according as s is increasing or decreasing at the instant in question. The speed measures the instantaneous rate at which the particle is moving along its path, but gives no information about its instantaneous direction of motion. We therefore introduce a vector quantity, the velocity, which gives the rate at which the particle is changing its position in both magnitude and direction. Let 0 be a definite point of a reference frame a, say the origin of a system of rectangular axes fixed in a rigid body. Then if r = OP denotes the position vector of P, the velocity v of P, relative to , is defined as the time derivative of its position vector: (2)
v=
KINEMATICS OF A PARTICLE
§ 52
109
As P moves along r, r may be regarded as a function of the arc s; hence dr
dr ds
dt = ds
ds
dt =
Tt
where T is the unit tangent vector to r at P in the direction of increasing arcs. Thus the velocity and speed are connected by the relation, V = VT.
(3)
The velocity of P is represented by a vector tangent to the path at P in the direction of instantaneous motion and of length numerically equal to the speed.
Finally we define the acceleration a of the particle as the time derivative of its velocity :
a
(4)
dv
d2r
- dt
dt2
From (3), we find dv dT a=-T+V t;
or, since T may be regarded as a function of s, dT
dT ds
dt
ds dt
a=
(5)
dv
dt
v
= (KN)V=-N p
T+
v2
N.
p
The components of the acceleration in the positive direction the tangent, principal normal, and binormal, are therefore dv
(6)
ae=- t,
of
v2
an- ,
as = 0.
P
The acceleration of a particle P is a vector lying in the plane to the tangent and principal normal to the path at P. The tangential component is the time derivative of the speed, the normal component the square of the speed divided by the radius of curvature at P.
110
VECTOR FUNCTIONS OF ONE VARIABLE
§ 53
The acceleration will be purely tangential when the motion is rectilinear (p = x) ; it will be purely normal when speed is constant (dv/dt = 0). The velocity and acceleration vectors are regarded as localized at the moving particle. With rectangular axes, r = xi + yj + zk, and (7 )
v
(8)
a
dr
dx
dt
dt 1 +
dv
d2x
dt
dt2
dy
l+ dt d2y
dz dt
k, d2z
,
1 + dt2 J +
dt2
k.
The rectangular components of v and a are the first and second time derivatives of x, y, z. Example 1. Circular Motion. In the case of motion in a circle of radius r.
s=rO,
de
dv
v=rdt,
_
d20
dt=rdt2,
the angle 0 being expressed in radians. On writing w = dB/dt for the angular speed, we have
v = rw,
at = r
dw ,
an = rw2.
These results may also be deduced directly from the equation r = rR of the circle if we make use of (44.2) and (44.3).
Example 2. Uniformly Accelerated Motion. If a particle has a constant acceleration a, and r = TO, v - vo when t - 0, we have, on integrating dv/dt = a twice; v = at + vo, r = jat2 + vot + TO. The path is the result of superposing the displacement "'ate, due to the accelera-
tion, upon TO + vot due to rectilinear motion at constant velocity vo. It is easily shown to be a parabola having its axis parallel to a. At the vertex of
the parabola, v is perpendicular to a; the condition v a = 0 gives t = -a vo/a a for the time of passing the vertex.
53. Relative Velocity. In § 52, we have seen how to find the velocity v of a particle P relative to any given reference frame .
If ' is a second reference frame, in motion with respect to , how is the velocity v' of P relative to a' related to v? At any instant t let P coincide with the point Q of a'. At a later.instant t1 = t + At, let P and Q have the positions P1, Q1;
RELATIVE VELOCITY
§ 53
111
here Q is a fixed point of a', and its motion is due to the motion of ' relative to a. Then PP1 is the displacement of P relative to
,
QQ1 is the displacement of Q relative to a, Q1P1 is the displacement of P relative to
',
and
PP1 = PQ1 + Q1P1 = QQ1 + Q1P1 If we divide this equation by At and pass to the limit At -* 0, we obtain VP = VQ + VP,
(1)
-,-
where VQ, the velocity of the point Q of R' relative to a, is called the transfer velocity of P. If we regard the frame as "fixed" and velocities referred to it
as "absolute," while velocities referred to a' are "relative," we may state (1) as follows: The absolute velocity of a particle is equal to the sum of its transfer and relative velocities.
In many applications all points of the frame ' have the same velocity relative to a; then vQ is the velocity of translation of ', and we may write Vp = Vtr + VP-
(2)
Example 1. Wind Triangle. An airplane p has the velocity v relative to the ground (the earth e), v' relative to the air; and the air (the wind w) has
the velocity V relative to the ground; then v = V + v', from (2). In Fig. 53a, v = ep,
V = ew,
v' = up;
thus vectors from e and w represent velocities
relative to the earth and wind, respectively. The magnitudes of v and v' give the ground speed (ep) and air speed (wp) of the plane. The directions of v and v', given as angles a and B'
w
measured from the north around through the Fla. 53a east (clockwise), determine the track and heading of the plane. The plane is pointed along its heading but travels over the ground along its track. The angle (v', v) from heading to track is the drift angle.
112
VECTOR FUNCTIONS OF ONE VARIABLE
§ 53
Example 2. Interception. A plane p is flying over the track PX with the ground speed ep (Fig. 53b). As plane p passes the point P, a plane q departs from Q to intercept plane p. If the air speed of plane q is given, over what track shall q fly in order to intercept p? Solution. In order that plane q may intercept plane p, the velocity of q
relative to p must have the direction QP. Draw the vectors ep and ew, giving the velocities of plane p and the wind w relative to earth e. With to as center describe a circle having the known air speed of plane q as radius. If a ray drawn through point p in the direction of
,,q
Fia. 53b
-4
QP cuts the circle at point q, plane q will intercept plane p on flying with the ground speed eq over the track QY parallel to eq; for pq, the velocity of plane
q relative to plane p, has the direction QP. Interception occurs at I after a flying time of QI/eq (or PI/ep) hours. Interception is impossible if the air-speed circle of plane q fails to cut the ray. If the circle cuts the ray in just one point, as in Fig. 53b, plane q can intercept p on only one track. But if the circle cuts the ray in two points, say ql and q2, plane q can intercept p along two different tracks, parallel to eql and eq2, respectively.
Example 3. Plane Returning to a Carrier. An airplane p leaves a carrier a at 0 and patrols along the track OY while the carrier follows the course OX with constant speed of v miles per hour (Fig. 53c). If the fuel in the tank allows the plane T hours of flying time at a given air speed, at what point B
must the plane turn in order to rejoin the carrier at A, T hours after its departure? Find also the time ti, the heading, and the ground speed vl on the leg out; and the time t2, the heading, and the ground speed v2 on the leg back. Solution. Let the vectors es and ew give the velocities of carrier and wind. With to as center describe a circle having the known air speed of p as radiusthe air-speed circle. If this circle cuts the ray through e in the direction OY
at pl, epl is the velocity of plane p on the leg out (ground speed vl = epl).
RELATIVE VELOCITY
§ 53
113
The course of plane p relative to the carrier s is out and back along the same
straight line. Now sp1 is the velocity of p relative to s on the leg out; hence the velocity of p relative to s on the leg back is a vector sp2 whose direction is
opposed to sp1. Thus the point p2 is at the intersection of the air-speed circle with the line pis prolonged, and ep2 is the velocity of p on the leg back (ground speed V2 = ep2). p2
Fro. 53c
After T hours the carrier will travel the course OA = T es. The return track
of p is along a line BA parallel to ep2. Thus p travels t1 = OB/ep1 hours on
the track OB, t2 = BA/ep2 on the track BA, rejoining the carrier after t1 + t2 = T hours. On the two legs the plane has the speeds ul = spli u2 = sp2i relative to the carrier. At any time carrier and plane lie on a line parallel to p18p2. When the plane is at B, the farthest point out, the carrier is at C(BC 11 pisp2). The distance r = CB is called the radius of action of the plane. Relative to the carrier the plane travels the course CB out, BC back; hence
-+_=T, r= T r
r
UIU2
u1
U2
U1 + u2
t1 = r =T u1 U2+ u2 U1
These times agree with those previously
t1=-=-, spl CB
t2=r=T u2
u1
.
u1 + u2
given; for
- = BA _-
OB
t2 = BC
epl
8P2
ep2
Example 4. Alternative Airport. Let a plane p depart from the airport 0 along the track OY (Fig. 53c). If a landing at an airport Y is rendered danger-
114
VECTOR FUNCTIONS OF ONE VARIABLE
§54
ous by local bad weather, the plane may be directed to land at an alternative airport A. If the fuel supply allows T hours flying time to the plane, how far may the plane fly on the track OY in order still to reach the port A by changing its course? This problem is reduced to the one preceding if the airport A is regarded
as a carrier traveling from 0 to A with the uniform velocity OA/T. Lay off
es = OA/T and proceed precisely as in the carrier problem. The line CB, drawn parallel to pisp2, fixes the farthest point B at which the plane can change its course and still reach the airport A. The time t1 when the turning point is reached is given in ex. 3.
54. Kinematics of a Rigid Body. We shall now investigate the velocity distribution of the particles of a rigid body moving in any manner. Consider first a rigid body having a fixed line or axis; its motion
is then a rotation about this axis. The position of the body at
I
Fia. 54b
Fia. 54a
any instant may be specified by the angle 0 between an axial plane o fixed in our frame of reference and an axial plane p fixed in the body (Fig. 54a). By choosing a positive direction on the axis (unit vector e), we fix the positive sense of 0 by the right-handed
screw convention. Then the angular speed w of the body at any instant is defined as de
(1)
dt
Thus w is positive or negative, according as 0 is increasing or decreasing at the instant in question.
KINEMATICS OF A RIGID BODY
§ 54
115
The velocity distribution in the revolving body may be simply expressed if we define the angular velocity as the vector, do w = -- e. dt
Note that w always is related to the instantaneous sense of rotation by the rule of the right-hand screw.
--
Choose an origin 0 on the axis and let r = OP be the position vector of any particle of the body. Then (Fig. 54b)
r=OQ+QP=ze+pR, where R is a unit vector perpendicular to the axis and revolving with the body. Since ze and p are constant during the motion of P, the velocity of P is dr V
(do \ e J x (ze + pR) do dt = p dt e x R = --
dR do
dt =
p
do
that is,
v = wxr.
(2)
The velocity of any particle of a body revolving about a fixed axis is equal to the vector product of the angular velocity and the position vector of the particle referred to any origin on the axis. Let us next consider a rigid body having one fixed point 0. Let i be a unit vector fixed in the body, j a unit vector in the direction of di/dt (perpendicular to i) and k = i x j. Then we may write di dt
dk
dj di dj -=-xj+ix-=ix-
=aj,
dt
dt
dt
dt
Hence dk/dt (perpendicular to k) is also perpendicular to i and therefore parallel to k x i = j. Thus we have dk dt =
j3j,
dj
d
dt
dt
(k x i) = /3j x i + k x aj,
or, on collecting results di
-=
aj,
dj dt
= (ak - #i) x j,
dk dt
= Rj.
116
VECTOR FUNCTIONS OF ONE VARIABLE
If we now write
54
w=ak - ii,
these equations assume the same form: di
dt = wxl'
dj
dk
wxjf dt =
dt
=
Now the position vector r of any particle P in the body may be Now referred to the orthogonal triple i, j, k fixed in the body; thus
r=OP=xi+yj+zk, where x, y, z remain constant during the motion. Hence the velocity of P is given by dr
di
dj
dt
dk
=xd6+ydt+zdt = wx(xi+yj+zk),
V = co x OP.
(3)
Thus, at any instant, the velocity distribution in a rigid body with one point 0 fixed is the same as if it were revolving about an axis through 0 with angular velocity co. The line through 0 in the direction of co is called the instantaneous axis of rotation, and co is called, as before, the angular velocity. Now, however, co may change
in direction as well as in magnitude. With a fixed axis of rotation,
w=
do
dt
e=
d
dt
(oe),
so that co may be regarded as the time derivative of the vector But with a variable axis of rotation co no longer can be expressed as a time derivative. Finally let us consider the general motion of a free rigid body. If, at any instant, all points of the body have the same velocity v, the motion is said to be an instantaneous translation. When the angle Be.
velocity distribution is given by v = co -A P, the motion is said to be an instantaneous rotation about an axis through A in the direction of co. We now shall show that, in the most general motion of a rigid body, the velocities may be regarded as compounded of an instantaneous translation and rotation.
§ 54
KINEMATICS OF A RIGID BODY
117
Let A be any point of the rigid body, and denote its velocity relative to by VA. Consider a second reference frame ' having a translation of velocity VA relative to . Then the motion of the
body relative to a' is an instantaneous rotation about an axis through A, since A has zero velocity relative to '. The velocity of any particle P of the body is therefore
vy = wxAP, relative to a', and, consequently, VP = V.4 + wxAP, relative to R. Moreover, for any other point Q of the body, (4)
(4)'
VQ = vA + w x AQ;
and, on subtracting this from (4), we get VP = VQ + 0) x QP.
The content of these equations is stated in the following If A is any point of a free rigid body, the velocities of its points are the same as if they were compounded of an instanTHEOREM 1.
taneous translation VA and an instantaneous rotation w about an axis through A; and w is the same for any choice of A.
Thus the instantaneous velocity distribution of a rigid body is determined by two vectors, its angular velocity w and the velocity VA of any point A of the body. These may be combined into the velocity motor,
V = Co + EVA;
(5)
for, from (4), (6)
VP= VA +PA xw,
in accordance with (31.3).
If w VA 0 0, the motor Visa screw. Since w VP = CO
VA,
the velocities of all particles of the body have the same projection on Co. The axis of V is called the instantaneous axis of velocity; its equation, from (31.8), is (7)
rxw =
(wxv4)xw
ww
(origin A),
118
VECTOR FUNCTIONS OF ONE VARIABLE
§ 54
a line parallel to co and passing through the point Q given by AQ = Co x VA/0) - uo; and, from (4)',
_ w.VA
(8)
VQ = VA + w x AQ = -co = projw vA. w-w
All points of the instantaneous axis have the velocity VQ; for, if R is another of its points, VR = VA -I- w x (AQ -{- QR) = vQ
since
w x QR = 0.
Moreover vQ is characterized by being parallel to co; and, since all particle velocities have the same projection on Co, their least nu-
merical value at any instant is j vQ I. Referred to a point Q on the instantaneous axis, the velocity motor becomes (9)
V = to -f- evQ.
This form, in which co and vQ are parallel, symbolizes THEOREM 2.
At any instant, the particle velocities of a rigid body
may be represented by a screw motion-a rotation about an instantaneous axis combined with a velocity of translation along this axis. 0, VA = 0, we have vQ = 0. The motion is then a pure If Co rotation of angular velocity w about the instantaneous axis. V becomes a line vector along this axis.
If w = 0, Co VA 0, we see from (4) that all points of the body have the same velocity. The instantaneous motion is then a pure translation, and V = EVA is a pure dual motor. Finally let us consider the case of plane motion of a rigid body. 0, the plane of the motion is perpendicular to w and vQ = 0; If co the motion is then an instantaneous rotation about an axis. The point Q where the instantaneous axis cuts the (reference) plane of motion is called the instantaneous center; and, as previously, (10)
-- 9
AQ =
wxVA w - w
If the velocities of the points A, B of the body are known and VA, VB are not parallel, Q is at the intersection of the lines AQ, BQ drawn perpendicular to VA and VB, respectively.
KINEMATICS OF A RIGID BODY
§ 54
Example.
119
Let the curve r roll without slipping over a
Rolling Curve.
fixed curve r1 (Fig. 54c). The points A and Al were originally in contact;
and, if I is the instantaneous point of
contact, let s = arc Al, sl = arc All. The conditions for pure rolling are S = Si,
(1)
at I.
T = Tj
Vt=V
If I is regarded as a moving particle, its velocity relative to fixed and moving frames
Al
attached to r2 and r is the same; for
Fia. 54c ds
dsl
Ti =
VI =
T = Vj. dt
dt
Hence, if I coincides with the fixed point Q of r, we have, from (53.1),
vl=VQ+Vj,
VQ=0.
The instantaneous center of the rolling curve is at its point of contact with the fixed curve.
The speed of rolling, v
_ ds _
ds1
dt
dt
The angular speed is co = de/dt, where 0 is the angle between the tangents at A and A1. We now differentiate the equation Ti = T with respect to t. Since T is a function of s and B, we have dT1 dsl
8T de
OT ds
dsl dt = as dt + ae dt ' or, since Nl = N,
KgN1v = KNv + Nw,
(iii)
1
1
w
P1
P
v
Here N is ir/2 in advance of T and p and P1 are positive when N points to the respective centers of curvature (in Fig. 54c, P1 < 0, p > 0 and w < 0).
r
(0>0;p1>0,p<0
6)>O;p1<0,p<0
Fla. 54d
VECTOR FUNCTIONS OF ONE VARIABLE
120
§ 55
If r and ri are circles of radius r and rl (Fig. 54d), w V
V
_
1) =
1
` r)
ri ri
\
I
+
r
r/
I
ri
r
for external contact,
ri
for internal contact.
When ri -+ co, ri becomes a straight line and v = wr.
55. Composition of Velocities. Let a rigid body have the velocity motor, V' = w' + EVA,
relative to a frame a'; and let the frame a' have the motor, V = CO + EVB,
relative to a "fixed" frame , where B is the point of a' which coincides for the instant with the point A of the body. What is the motion of the body relative to a? The velocity of any point P of the body relative to a' is (1)
Vp = VA + W' X AP.
At the instant in question let P coincide with the point Q of a'. Then the velocity of Q relative to is (2)
vQ=vB+WXBQ.
Now the velocities of P and A relative to Vp = VQ + Vp,
are (53.1)
VA = VB + VA.
On adding (1) and (2) and observing that AP = BQ at the instant considered, we get
VP = VA + (w+W')XAP.
Thus the motion of the body relative to a is compounded of the velocity of translation VA and the angular velocity w + w' about an axis through A ; that is, the motion is given by the motor,
V+V'=w+w'+evp. THEOREM. If the motion of a rigid body relative to a' is given by the motor V' = w' + evA, and the motion of a' relative to by V = w + eVB, the motion of the body relative to a- is given by the
RATE OF CHANGE OF A VECTOR
§ 56
121
motor sum V + V' provided A and B coincide at the instant in question.
Two translations, ev' and evB thus compound into a translation e(vA + VB).
If V' = co' + evA and V = w + evB represent pure rotations, these motors are line vectors; hence the motion of the body relative to a will be a pure rotation when and only when w + w' 0 and the axes of rotation are coplanar (§ 32, theorem 1). If the axes of rotation intersect at A, the motors referred to A are
V=w;
V'=w',
and V+V'=w+w'.
Angular velocities about intersecting axes may be compounded by vector addition.
If V and V' represent pure rotations about parallel axes and w + w' = 0, V + V' = e(vB + vi); the motion of the body relative to is then a pure translation of velocity vB + V. 56. Rate of Change of a Vector. Referred to a fixed origin 0
in the frame a, the vector u = PQ = OQ - OP; hence du
(1)
dt
= VQ - VP1
where vp, vQ are velocities relative to
.
Similarly, if a' is a second frame in motion with respect to the rate of change of u relative to a' is d'u (2)
dt
,
= ve - vp,
where v p', vQ are velocities relative to
Let the motion of a' relative to
'.
be given by the motor
co + EVA, A being a fixed point of a'; and, at the instant in question, let P and Q coincide with the points R and S of a'. Then, from (53.1) and (54.4),
Vp = oP+OR =VP+oA + WXAR,
VQ = VQ+Vs =VQ+vA+wxAS; and, on subtraction, du
d'u
dt = dt
-> -
+ w X (AS - AR).
- --
VECTOR FUNCTIONS OF ONE VARIABLE
122
-->
--->
du
du
§ 56
But, since AS - AR = RS = PQ = u, d =
(3)
+ 0).U; dt
in particular du
(4)
dt
=wxu
if u is fixed in a'.
When u = w, the angular velocity of the frame a', we have, from (3), dw
d'w
dt
dt
(5)
The vector dw/dt is denoted by a and called the angular acceleration vector of a. Example. Kinematic Interpretation of Frenet's Formulas. At any point P of a space curve, the trihedral TNB may be used as a frame of reference.
If P moves along the curve with unit speed, ds/dt = 1 and s = t if t = 0 at the origin of arcs. Then the arc s may be interpreted as the time, and dr/ds = T is the velocity of P. If w is the angular velocity of TNB referred to a "fixed" frame a, we have, from (4), dT
ds =
dN
, x T,
ds = p1 x
N,
dB
ds = "
x B.
A comparison with Frenet's Formulas (44.6) shows that 0) = S, the Darboux vector. The Darboux vector of a space curve is the angular-velocity vector of its moving trihedral TNB. Since the vertex P of the trihedral has the velocity T, its motion is represented completely by the velocity motor,
V =S+eT=rT+KB +Er. From (54.7) and (54.8), we see that the axis of this motor passes through the point Q given by --->
PQ =
8xT S
8
K
ST
= ,,2+,2N and vQ = S-S S =
r K2 +
'28.
In general, the trihedral TNB has, at every instant, a screw motion: a combina-
tion of the angular velocity S about an axis through Q and the velocity vQ along this axis. The velocity of translation vanishes when and only when the curve is plane (r = 0); then S = KB, PQ = pN. For plane curves TNB has, at
every instant, a pure rotation of angular velocity KB about the center of curvature.
THEOREM OF CORIOLIS
§ 57
123
57. Theorem of Coriolis. Let v and a denote the velocity and
acceleration of a particle P, relative to a frame , while v' and a' denote these vectors relative to a frame ' in motion with respect to . Then if 0 and A are origins fixed in and a', we have
r= OP,
dr
a= dv-;
v= -- ,
dt
dt
r' = AP
v'=
d'r' dt
d'v'
a
dt
,
If w is the angular velocity of ' with respect to , we have, on differentiating,
-- 3
r = OA + r', twice with respect to the time, and, on making use of (56.3), dr'
V =VA+
dt
= VA +wxr'+v', dr' dv' a=aA+axr'+wx+ dt dt
/
(a
dw
dt
J
a A+ a x r' + w x ((o x r' + V) + w x v'+ a'
a4+axr'+wx(wxr')+2wxv'+a'. The velocity and acceleration of the point Q of the frame j' with which P momentarily coincides are called the transfer velocity and transfer acceleration of P. To find them, put v' = 0, a' = 0 in the foregoing equations; thus VQ = VA + wxr',
aQ = aA + ax r' +wx Our equations now read (1)
V=vQ+V',
(2)
a=aQ+2wxv'+a'.
Equation (1) restates the theorem on the composition of velocities already proved in § 55. Equation (2) shows that an analogous theorem for the composition of accelerations is not in general true;
VECTOR FUNCTIONS OF ONE VARIABLE
124
§ 57
we have, in fact, the additional term 2w x v', known as the Coriolis acceleration.
If we regard the frame
as "fixed" and rates of
change referred to it as "absolute," while the corresponding rates
referred to ' are "relative," we may state (2) as the The absolute acceleration of a particle is equal to the sum of its transfer acceleration, Coriolis acceleration, and relative acceleration. THEOREM OF CORIOLIS.
The Coriolis acceleration, 2w x v', vanishes in three cases only: (a) co = 0; the motion of ' relative to is a translation.
(b) v' = 0; the particle is at rest relative to '. (c)
v' is parallel to co.
A particle of mass m, acted on by forces of vector sum F, has the equation of motion F = ma. When the motion is referred to a rotating frame the equation of motion becomes N U
P(Q)
(3)
ma' = F - maQ - 2mw x v'.
If we regard -maQ and -2mw x v' as fictitious forces, ma' equals a sum of forces just as in the case of a fixed frame. The term
V O(A)
Fio. 57a
-2mw x v' is called the Coriolis force. When ' has the constant angular velocity w about
a fixed axis through A, a = 0, aA = 0, and aQ = w x (w x r'). Taking 0 at A and the z-axis along w (Fig. 57a), we have
-maQ = mw2 (r' - k r' k) = mw2 (OP - ON) = mw2 NP; this vector perpendicular to the axis and directed outward is called the centrifugal force on the particle. Example. Particle Falling from Rest. Refer the particle, originally at 0, to the revolving frame Oxyz attached to the earth: Oz points to the zenith (along a plumb line), Ox to the south, Oy to the east (Fig. 57b). The earth
revolves from west to east about the axis SN and its angular velocity at north latitude a is _ 2r (k sin X - i cos a) radians/see. 24 X 602
When the particle is at rest relative to the earth, the force acting upon it is its local weight mg; hence in (3) F - maQ = mg. Therefore the equation of motion becomes (i)
a'=g-2o,xv',
§ 57
THEOREM OF CORIOLIS
125
where a' and v' are the acceleration and velocity relative to the earth. We shall integrate (i) by successive approximations under the initial conditions = 0, v' = 0 when t = 0 and with g regarded as constant. 1. If we neglect the term in w,
a' = g,
v' = gt,
=zgt2.
The displacement r' is along the plumb line.
Fla. 57b
2. With v' - gt, (i) gives a' = g - 2tw x g,
v'=gt-12wxg, ' = zr
gt2 - 3 taw x g.
Since g = -gk, the second term gives an eastward deflection, - 3 taw x g = 3 t3wg cos Xj.
3. With the last value of v', a' = g - 2tw x g + 2t2w x (w x g), v' = gt - t2w x g + 3 taw x (w x g), = 2gt2 - 3t3w x g + Gt4w x (w x
g).
VECTOR FUNCTIONS OF ONE VARIABLE
126
The last term is a deflection in the meridian or xz-plane.
§ 58
Since w x g = -wg
cos X j, wng
w x (w X g) =
r=
1 s tow" 9 sin X cos X i +
cos x (i sin X + k cos X),
1 taw 3 9
cos X J' 1 t2 - 114w2 9 cos'- X) k. (29 6
The first and second terms give deflections to the south and east; the third shows that the particle in t seconds falls through a distance -z = 2gt2 - I t4w2g cost X.
58. Derivative of a Motor. Let the motor M= m + EmA be referred to the frame a' in which A is a fixed point; and let ' itself have the motion V= o + EV A with respect to the frame . If m and mA are functions of the real variable t, the derivative of
M relative to a' is
dm
d'M (1)
dt
dt
d'mA
+
e
dt
In order to compute dM/dt relative to a we must remember that mA depends not only upon t but also upon the point A, which is in motion relative to a. If A moves to B in the interval At, Om.4
mB(t + At) - m4(t)
At
'At
MA (t + At) + AB X m(t + At) - m.4 (t) At
mA (t + At) - mA (t) At
+ OB -- At- OA X m(t + At),
and hence, if rA ° OA, dmA
AMA
c l-.o
dt -
At
drA
+
xm.
dt
Relative to the frame a, we therefore define dM
dm
dmA
dt
dt
dt
(2)
drA
+
dt
m)
To verify that dM/dt is a motor, we must show that dmP dt
drp
+
dt x m =
dmA
+ dt -
dr,4
dt
-) dm
x m + PA X
dt
(31.3)
DERIVATIVE OF A MOTOR
§ 58
127
This equation, in fact, follows from
mp = mA + (rA - rp) x m on differentiation. If t denotes time, drA
VA; dt =
and
dm
wxm+
dt =
d'm
d'mA
dmA
dt =
dt '
dt
'
Substituting these results in (2) gives
from (56.3).
dM (3)
dt
d'M
= (0-M+ E((axmA + VA xm)
+
dt
or, in view of (34.1),
d'M
dM (4)
V X M +
dt
dt
This is the motor analogue of (56.3). Making use of the definition (2), we may verify that the following rules of differentiation are valid : dM
d ( 5)
dt ( M
+ N)
dt
d ( 7)
dt d
( 8)
dt
+
dt
dt '
dM
d (6)
dN
(XM) = X
(M -N) = M (M
x
N)
=
dt
dA
+
dt
M, dM
dN '
dt
+N
dt
-N dM
dN Mx
dt '
+
.
dt
Example 1. If the motion of a rigid body is given by the velocity motor V = w + EVA (A a fixed point of the body), its acceleration motor is dV
= a+e(aA +VAXW).
The axis of this motor is called the instantaneous axis of acceleration; its equation may be written from (31.8).
VECTOR FUNCTIONS OF ONE VARIABLE
128
§ 59
Example 2. Let F = f + EfA be a force acting on a rigid body whose velocity motor is V = w + EvA. Then dF
(i)
dt + E ( dt + VA x
f)
.
If F acts at the point P of the body, fp = 0 and dF
(ii)
= df + Evp x f, referred to P.
Hence, referred to A,
d
(iii)
do+E(vPxf+APxdt/
let the reader show that (i) and (iii) are consistent. Example 3. The moving trihedral TNB at the point P of a space curve r = r(s) may be regarded as a rigid body having the velocity motor,
V=S+Evp=8+ET, as the point P traverses the curve with unit speed. The three line vectors T = T, N = N, B = B through P are fixed in the trihedral; hence, from (4),
ds =VxT,
dN =VxN,
B
=VxB.
These are Frenet's Formulas in motor form. Written out in full they become
T = KN,
dN = -7-T + KB + EB,
d- _ -rN - EN;
for example dB
- =(S+ET)xB=-dB- EN=-TN-EN. 59. Summary: Vector Derivatives. The derivative du/dt of a vector function u(t) is defined as the limit of Au/At as At approaches zero. If u = OP is drawn from a fixed origin and P describes the
curve C as t varies, du/dt is a vector tangent to C at P in the direction of increasing t. If I u I is constant, C is a circle and du/dt is perpendicular to u. The derivative of a constant vector is zero. The derivatives of
the sum u + v and the products fu, u v, u x v are found by the familiar rules of calculus; but for u x v the order of the factors must be preserved.
§ 59
SUMMARY: VECTOR DERIVATIVES
129
If s is the are along a curve r = r(s), dr/ds = T, a unit vector tangent to the curve the direction of increasing s. The unit prin-
cipal normal ft to the curve has the direction of dT/ds; and the unit binormal B = T X N; then [TNB] = 1. The vectors of the moving trihedral TNB change conformably to Frenet's Formulas: dT
= SXT,
Kft
ds dN
-KT
+ TB = 6 X ft,
ds
dB
-rft
= SXB;
ds
K is the curvature, r the torsion of the curve; and the Darboux vector
o = TT + KB is the angular velocity of the moving trihedral as its vertex traverses the curve with unit speed (ds/dt = 1). For plane curves r = 0. If t denotes time, a particle P traveling along the curve r = r(t) has the velocity and acceleration, dr
v
dt '
a
-
dv
d2r
dt
dt2
If v = ds/dt is the speed, and p = 1/K the radius of curvature of the path, V = VT,
dv
v2
dt
p
a = -- T+ - ft.
If A is any point of a free rigid body, the velocities of its points are given by the velocity motor, V = w + EvA ;
here co, the angular velocity vector, is the same for any choice of A.
Since V is a motor, for any point P of the body, Vp = VA + PA Xw = vA +(0 XAP.
If, at any instant, co 0, the body has a screw motion about the axis of V, the instantaneous axis of velocity; this reduces to a pure rotation if V is a line vector (co VA = 0). If co = 0 the motion is an instantaneous translation of velocity VA.
130
VECTOR FUNCTIONS OF ONE VARIABLE
If the frame a' has the angular velocity relative to , the rates of change of a vector u relative to these frames are connected by du
d'u
dt
dt
+
w .U.
A particle P has the velocity and acceleration v', a' relative to frame a-', v, a relative to the frame a; then, if the motion of a' relative to a is given by the motor V = co + EvQ, where Q is the point of ' coinciding at the instant with P,
a = aQ +2wxv' + a'.
v = VQ +v',
The term 2w X v' is the acceleration of Coriolis. When a' is in trans-
lation relative to , the velocity equation may be written v = vl, + V'. The derivative of the motor M = m(t) + emA(t) is defined as dM
dm
dmA
at = at +
at-_
drA
+
Xm
at
If t denotes time, and d'M/dt refers to a frame a' having the motion V = co + evA, relative to a frame , then d'M
dM dt
V X M +
dt
PROBLEMS
1. If r and x are the distances of a point on a parabola from the focus and
directrix, r - x = 0. Show that (R - i) - T = 0, and interpret the equation. 2. Prove that the tangent to a hyperbola bisects the angle between the
focal radii to the point of tangency. [Cf. § 44, ex. 2.1 3. An equiangular spiral cuts all vectors from its pole 0 at the same angle a (R T = cos a). If (r, 0) are the polar coordinates of a variable point P on the spiral, show that (a) (b)
ds/dr = sec a, 1 dr
rd0
= cot a,
(c)
s - so = (r - ro) sec a; log r = (9 - Bo) cot a; ro
p = ds/dB = r/sin a;
and that the center of curvature is the point where the perpendicular to OP
at 0 cuts the normal at P.
PROBLEMS
13 1
4. If r = r(s) is a plane curve. r, show that (a) r1 = r + eN (c cont.) is a parallel curve r1 (§ 45, ex.l at it tance c from r; (b) Si = s - c, provided Si = s = 0, when ,y = 0;
ii: l dis-
(C) P1 = P - C.
5. A curve r = r(s) has the property that the locus r1 = r + rT (e const.) is a straight line. Prove that the curve is plane, in fact, a tractrix. [Cf. § 50, ex. 5.)
6. An involute of a curve r, r = r(s), is a curve r1 which cuts the tangents of r at right angles. Prove that r has the one-parameter family of involutes,
r1 = r + (c - s)T
(c = const.);
and, if we take T1 = N, ds1/ds = (c - s)K. 7. Show that the involute of the circle,
y = a sin t,
x = a cos t,
obtained by unwrapping a string from the point t = 0 is
y1 = a(sint - tonst);
x1 = a(cost +tsint),
and that s1 = Zat2 gives the arc along the involute. 8. The cylinders x2 + y2 = a2, y2 + z2 = a2 intersect in two ellipses, one of which is x = a cos t, y = a sin t, z = a cos t.
Show that its radius of curvature is
P = a(l + sine t)I/''2. 9. The equation of a cycloid is
y = all - cos t).
x = a(t - sin t), Prove that
t
1
2
2J
(a)
T = [sin , cos
(b)
>y = (i, T) = 2 - 2 ,
,
ds/dt = 2a sin
2
P = -4a sin 2
(44.3);
(c) The equation of its evolute is x1 = a(t + sin t),
y1 = -a(1 - cost).
10. Find the vectors T, N, B and the curvature and torsion of the twisted
cubic,
x = 3t, y = 312,
z = 20,
at the points where t = 0 and t = 1. Write the equations for the normal and osculating planes to the curve at the point t = 1. 11. Show that curvature and torsion of the curve, x = a(3t - t3),
y = 3at2,
z = a(3t + t3),
are
K = r = 1/3a(1 +
t2)2.
132
VECTOR FUNCTIONS OF ONE VARIABLE
12. Find the envelope of the family of straight lines in the xy-plane, r = p(O)R + XP,
where R and P = k x R are the unit vectors of § 44, 0 = angle (i, R), and p is the perpendicular distance from the origin to the line. Show that the curve,
r, = pR + p'P is the envelope, and that
dsl/de = p + p".
Ti = T,
13. Find the envelope of the family of lines for which the segment included between the x-axis and y-axis is of constant length c. (In Problem 11 put p = c sins cos 0.1 Show that the envelope has the parametric equations, x = c sin3 0,
y = C C083 0;
and that the entire length of the curve is 6c. 14. Show that the curvature and torsion of the curve, are
/2 t,
x = et, y = e-t,
z=
Sc = -T
+ a-t)2.
15. Verify that the curve, z = a cos t,
x = a sin2 t, y = a sin t cos t,
lies on a sphere. Show that the curve has a double point at (a, 0, 0) for the parameter values t = ±r/2, and that the tangents to the curve at this point are perpendicular.
16. If a curve r = r(s) lies on a sphere (r - c) (r - c) = a2, show that dp r - C = -pN - TI--B. ds
Hence, prove that
d (ldP)
pr+ ds
r ds
0
is a necessary and sufficient condition that a twisted curve (r d 0) lie on a sphere. 17. The points on two curves r and r, are in one-to-one correspondence.
If T = T, at corresponding points, prove that N = N1,
B = B1,
dsl/ds = K/Kt = r/T,.
When both curves cut the rays from 0 at the same angle, show that r, = cr; and that s, = as, if both arcs are measured from the same ray. 18. A curve r is called a Bertrand curve if its principal normals are principal normals of another curve r1; then r1 is also a Bertrand curve, and
rl = r + XN, Ni = eN, where e = 1 or -1.
PROBLEMS
133
Show, in turn, that
Ti d = (1 - cK)T + cTB;
X = c (const.),
(a)
11
(b) If
= angle (B, Bl), taken positive relative to N, B1 = B cos (P + T sin p,
ET1 = T cos rp - B sin rp;
= const., and
(c)
-ET1
ds1
ds =
K sin p - r COS gyp, ds1
Eds- =
(d)
COS cp + T sin gyp;
ds
1 - CK -Cr _ cos p sin
;
K1 = -E(Ti cot (p + 1/c);
C2TT1 = Sin2 (P,
(e)
ds1
Kl
since K1 = 0, the last equation determines E. '(f) As to the Darboux vectors, Si ds1/ds = S. 19. The characteristics of a one-parameter family of planes whose homogeneous coordinates (§ 26) are (a(t), ao(t)), are the limiting lines of inter-
section of the planes corresponding to the parameter values t and t + h as h -- 0. The line of intersection of the two planes has the homogeneous coordinates (26.10),
ao(t + h)a(t) - ao(t)a(t + h) [;
{a(t) x a(t + h),
and, if we divide both by h and pass to the limit h -+ 0, show that we obtain
(a x a', apa - aoa') as the coordinates of the characteristics. 20. From the result of Problem 19 show that the three families of planes
associated with the twisted (r 0 0) curve r = r(s), namely, (T, r T),
normal planes,
(N, r N),
rectifying planes,
(B, r B),
osculating planes,
have as characteristics the respective families of lines: [B, (r + pN) x B],
parallel to B through the centers of curvature;
(8, r x 8),
parallel to S through points of the curve;
(T, r x T),
tangent to the curve (§ 48).
21. If a particle P of mass m is subject to a central force F = mf (r)R, where
r = OP and R is the unit vector along OP, its equation of motion is z
m dt2 = F or
dv
= f (r)R.
Prove that r x v = h, a vector constant; hence, show that the motion is plane
and that r = OP sweeps out area at a constant rate (Law of Areas).
134
VECTOR FUNCTIONS OF ONE VARIABLE
22. When mf(r) = --ymM/r2 in Problem 21 the particle P is attracted towards a mass M at 0 according to the law of inverse squares: dv dt
(a)
Show, in turn, that
= - -k R
(k
rxv = h
(b)
h = r2 R x dR
(c)
,
=kdR,
(d)
dt xh
(e)
v x h = k(R + e),
where a is a constant vector.
From (b) and (e),
r x v h = h2,
r vxh = kr(1 +acose)
where 0 = angle (e, r); thus obtain the equation of the orbit, hz/k
(f)
1 + e Cos 0
a conic section of eccentricity a referred to a focus as pole. When e < 1, prove that the orbital ellipse is described in the periodic time,
7, - area of ellipse _ 2a
k a'
h/2 and, hence, (g)
T2/a3 = 4,r2/-yM
(Kepler's Third Law).
(See Brand's Vectorial Mechanics, § 177.)
23. If the plane Ax + By + Cz = 1, fixed in space, is referred to rectangular axes rotating with the angular velocity w = [wl, w2, w3], show that
rdA dB dC] l dt
'
dt '
dt
=
[A, B, C] x [WI, (02Y w3].
24. A particle P has the cylindrical coordinates p, p, z (cf. § 89, ex. 1). If P moves in the pz-plane so that dp/dt and dz/dt are constant while the plane itself revolves about the z-axis with the constant angular speed dip/dt = w, find the acceleration of P, (a) By direct calculation. (b) By use of the Theorem of Coriolis. 25. A real, everywhere convex, closed plane curve with a unique tangent at each point is called an oval. It can be shown that an oval has just two tangents parallel to every direction in the plane. The distance between these tangents is the width of the curve in the direction of the perpendicular. Prove that the perimeter of an oval of constant width b is 7rb (Barbier's Theorem).
CHAPTER IV LINEAR VECTOR FUNCTIONS
60. Vector Functions of a Vector. A vector v is said to be a function of a vector r if v is determined when r is given; and we write v = f(r). Since r is determined by its components, f(r) is a function of two or three scalar variables according as r varies in a plane or in space. A vector function may be given by a formula, as f(r) _ (a X r) x r; or it may be defined geometrically. Thus if
r = OP varies over the points P of a given surface, and v = OQ is the vector perpendicular on the tangent plane to the surface at
P, v = f(r). A vector function f(r) is said to be continuous for r = ro if (1)
lim f(r) = f(ro). r - ro
This means that, when the components of r approach those of ro in any manner, the components of f (r) approach those of f (ro). A vector function is said to be linear when (2)
(3)
f(r + s) = f(r) + f(s),
f (Xr) = Af (r),
for arbitrary r, s, X. For example, linear vector functions are defined by the formulas kr, a X r, a b r, in which k, a, b are constant. It can be shown that, when a continuous vector function satisfies the relation (2), it also satisfies (3) and is therefore linear.
Since we assume that (2) holds when r = s = 0, f(0) = 2f(0); hence, for any linear vector function, (4)
f(0) = 0.
A linear vector function is completely determined when f(a,), f(a2), f(a3) are given for any three non-coplanar vectors al, a2, a3. For, if
we express r in terms of al, a2, a3 as a basis,
r = xlal + x2a2 + x3a3, 135
136
LINEAR VECTOR FUNCTIONS
§ 61
we have, from (2) and (3), (5)
f(r) = x'f(a,) + x2f(a2) + x3f(a3)
Let a', a2, a3 denote the set reciprocal to a,, a2, a3 (§ 23); then, if we write f(ai) = b1, xt = r at,
f(r) may be written in either of the forms: (6)
f(r) = r
(7)
f(r) = (b,al + b2a2 + b3a3) r.
(albs + a2b2 + a3b3),
These formulas represent the most general linear vector function.
61. Dyadics. In linear vector functions of the form,
f(r) = alb, r+a2b2.r+...+anbn.r, we now regard f (r) as the scalar product of r and the operator,
= alb, + a2b2 + ... + anbn.
(1)
Assuming the distributive law for such products, we now write
f (r) = ' r.
(2)
Following Willard Gibbs, we call the operator 4) a dyadic and each of its terms aib1 a dyad. The vectors ai are called antecedents, the vectors bi consequents. While a b is a scalar and a x b a vector, the dyad ab represents a new mathematical entity. Gibbs regarded ab as a new species of product, the "indeterminate product." We shall find indeed that this product conforms to the distributive and associative laws, but
is not commutative (in general ab 0 ba). Since r follows. in (2), we call r a postfactor. If we use r as a prefactor, we get, in general, a different linear vector function:
g(r) = r
4).
We proceed to develop an algebra for dyadics, laying down for this purpose definitions for equality, addition, and multiplication of dyadics. Definition of Equality. We write 4) = NY when (3)
4>
r = 'Y r for every vector r.
DYADICS
§ 61
137
If s is an arbitrary vector, we have from (3) s - (4) - r) = s -
or
r
Since s 1 and s ' are two vectors that yield the same scalar product with every vector r, these vectors must be equal; thus s iD = s - T for every vector s.
(4)
Conversely, from (4) we may deduce (3). Therefore 4) = T when either (3) or (4) is fulfilled. The Zero Dyadic. We write = 0 when (5)
4)
r = 0 for every r.
The preceding argument shows that 4) = 0 also when s S. 4) = 0
(6)
for every s.
Definition of Addition. The sum 4) +' of two dyadics is defined by the property,
(4) +'Y) r=
(7)
If
for every r.
is given by (1), and ' = c1d1 + C2d2 + ... + Cmd,n,
4) +' = a1b1 + ... + anbn + c1d1 + ... + c..d,,,..
In the sense of this definition, 1 (or ') is the sum of its dyad terms, thus justifying our notation. The order of these dyads is immaterial; but the order of the vectors in each dyad must not be altered, for, in general, ab r ba r and hence ab 54 ba. The Distributive Laws,
(8), (9)
a(b + c) = ab + ac,
(a + b)c = ac + bc,
are valid. The proof follows at once from the definition of equality; thus from a(b + c) r = (ab + ac) r,
we deduce (8). We may now perform expansions as in ordinary algebra, if the order of the vectors is not altered; for example:
(a+b)(c+d) =ac+ad+bc+bd.
LINEAR VECTOR FUNCTIONS
138
§ 62
If X is any scalar, (Xa)b = a(Xb),
(10)
and we shall write simply Xab for either member.
From (60.6) or (60.7) we conclude that any dyadic D can be reduced to the sum of three dyads. To effect this reduction on 4) as given by (1), express each antecedent ai in terms of the basis ei, e2, e3, ai = ale, + aie2 + a;e3,
expand by the distributive law, and collect the terms which have the same antecedent: thus 4) _
(a1 ei
+ a2e2 + a3e3)bi = elf, + e2f2 + e3f3,
where f;
We also may reduce 4) to the sum of three dyads by expressing each consequent bi in terms of the basis el, e2, e3, and then expanding and collecting terms which have the same consequent; then 4) assumes the form, 4) = giei + 92e2 + 93e3-
Thus it is always possible to express any dyadic so that its antecedents or its consequents are any three non-coplanar vectors chosen at pleasure.
If a,, a2, a3 are non-coplanar, a dyadic 4) is completely determined by giving their transforms (§ 60). If
4) - a, =b,,
4> - a2 = b2,
(D - a3= b3,
and the set a', a2, a3 is reciprocal to al, a2, a3, we have explicitly (11)
4) = bla' + b2a2 + b3a3.
For a physical example of a dyadic the reader may turn to § 116, where the stress dyadic is introduced. The name tensor, now used in a much more general sense, originally was ajTplied to this dyadic. 62. Affine Point Transformation. If we draw the position vectors,
r=OP,
r'=4 r=OQ,
from a common origin 0, the dyadic 4) defines a certain transformation of the points of space: to each point P corresponds a defi-
COMPLETE AND SINGULAR DYADICS
§_63
139
nite point Q. If, when P ranges over all space, Q does likewise, this transformation is called affine; the dyadic 4) then is called complete.
Important properties of an affine point transformation follow at once from the equations,
4) (a+b) = which characterize a linear vector function. Since - 0 = 0, the transformation leaves the origin invariant. Lines and planes are transformed into lines and planes. Thus, for variable x, the Line
r = a + xb - Line r' = a' + xb';
and, for variable x, y, the Plane r = a + xb + yc - Plane r' = a' + xb' + yc'. The transformed equations always represent lines and planes when 4) is complete; for we shall show in § 70 that b 5-4 0 implies b' F6 0, and b x c 5x-I 0 implies b' x c' 54 0.
63. Complete and Singular Dyadics. Given an arbitrary basis, e1, e2, e3, we can express any dyadic in the form, (1)
4) = glee + 92e2 + g3e3
If we express r in terms of the reciprocal basis,
r = xle' + x2e2 + x3e3, then, by virtue of the equations, et e' = St, 4)
r = x1g1 + x282 + x383
When g1, g2, g3 are non-coplanar, r' = 4) r assumes all possible
vector values as r ranges over the whole of space. If we put ---3 --a r = OP, r' = OQ, the dyadic defines an affine transformation r' _ ( r of space into itself; 3-dimensional P-space goes into 3-dimensional Q-space. A dyadic having this property is said to be complete. A complete dyadic cannot be reduced to a sum of less than three dyads; if, for example, we could reduce 4) to the sum of two dyads, ab + cd, all vectors r would transform into .vectors r' = a b r + c d r parallel to the plane of a and c.
LINEAR VECTOR FUNCTIONS
140
§ 63
If, however, g1, g2, g3, are coplanar, but not collinear, we can express each gi in terms of two non-parallel vectors f1, f2, and reduce
c to the sum of two dyads: 4, = f1h1 + f2h2.
(2)
This dyadic transforms all vectors r into vectors r' = 1 - r in the plane of f1 and f2; 3-dimensional P-space goes into 2-dimensional
Q-space. A dyadic having this property is said to be planar. A planar dyadic cannot be reduced to a single dyad ab; for then all vectors r would transform into vectors a b . r parallel to a. If 91, g2, g3 are collinear, we can replace each gi by a multiple of a single vector f and reduce 4' to a single dyad:
4 = fh.
(3)
This dyadic transforms all vectors r into vectors r' = fi - r parallel to f; 3-dimensional P-space goes into 1-dimensional Q-space. Such a dyadic is called linear. Finally, if g1, g2, g3 are all zero, 4' = 0. Planar, linear, and zero dyadics collectively are called singular. The point transformation, OQ = (D -OP,
corresponding to a singular dyadic 4' reduces 3-dimensional P-space to a 2-, 1-, or 0-dimensional Q-space.
This discussion shows that, when - is reduced to the form (1) in which the consequents are non-coplanar, then 4, is complete,
planar, linear, or zero, according as the antecedents are noncoplanar, coplanar but not collinear, collinear, or zero. In particular, we have the
THEOREM. A necessary and sufficient condition that a dyadic al + bm + cn be complete is that the antecedents a, b, c and consequents 1, m, n be two sets of non-coplanar vectors.
As a corollary, (4)
4) - r = 0 implies r = 0 when - is complete.
For, if - = al + bm + cn,
al-r+bm-r+cn-r=0, and hence r = 0.
1 - r = m - r = n - r = 0,
CONJUGATE DYADICS
§ 64
64. Conjugate Dyadics.
141
The dyadics,
4) = alb, + a2b2 + ... + anon, + bn.an,
,D, = b1a1 + b2a2 +
are said to be conjugates of each other. In general 4) and (D, are different dyadics; but evidently
4) r=r
(1)
define the same linear vector function. If two dyadics are equal,
their conjugates are equal; for 4, =' implies that 4)
by definition.
r, and hence 4), A dyadic is called : Symmetric
if
Antisymmetric if
4), = 4),
4 = - 4).
The importance of these special types of dyadics is due to the THEOREM. Every dyadic can be expressed in just one way as the sum of a symmetric and an antisymmetric dyadic. Proof.
For any dyadic 4), Ave have the identity, =
(1)
4)+4), 2
+
4)
2
=
+ 52;
since
= *I
4rc =
= -Q,
St, =
2
2
41 and Sl are, respectively, symmetric and antisymmetric. Moreover 4) can be so expressed in only one way. For, if 41
+Q_N,,+9,
gave two such decompositions, we have, on taking conjugates, q/ - SZ =
and hence
_ V, Sl = S2'.
LINEAR VECTOR FUNCTIONS
142
§ 65
65. Product of Dyadics. If the transformations corresponding to two linear vector functions,
v=4) - r, are applied in succession, their resultant,
is a third linear vector function; for w1 + w2 corresponds to r1 + r2, and Xw to Xr.
This function is written
w= and 'I' ( is called the product of the dyadics ' and 4), taken in 4) is therethis order. The defining equation for the product fore (1)
(*
for every r.
From (1) we find that the distributive and associative laws hold for the products of dyadics: (2)
4) ('I' + S2)
-{- 4
(3)
((F +')
+
S2,
(4)
Proofs.
For every vector r,
(I' + l)] r = 4 [( + 0)
r]
r+l r] r)+ (St
r+ [( +'F)
St]
r = ((F +'I')
r; r)
(St
r)+
r=
[(`I'
r)
' ) ' r] r)
r)
PRODUCT OF DYADICS
§ 65
143
Equations t2), (3), (4) follow from these results by the definition of equality. In the proofs of (2) and (3), definition (1) justifies the first and last steps; in the proof of (3), (1) is applied in every step.
explicitly, we first find the In order to compute a product product of two dyads. By definition,
for every r, and hence (ab) (cd) = (b c) ad.
(5)
The product of ab and cd, in this order, is the scalar b c times the dyad ad.
Similarly,
(ab) = (d a) cb,
(cd)
(6)
which in general differs from (5). Making use of the distributive law, we now may form the product of any two dyadics, m
n
' = E c;d;,
4) = E aibi, i=1
i=1
by expanding into nm dyads. n
(7)
m
=E i=1 j=1
(bi c,)aidr
The conjugate of 4) ' is m
n
71 (c; (I' ) = Ej=1 i=1
(8)
bi)d;ai
The conjugate of the product of two dyadics is the product of their conjugates taken in reverse order.
Making use of (8), we now find that
r
(4)
- *) = (4)
'P ) ,
r = ( * ,-( 1 ) , )-r
hence
r ((D 'Y) = (r (D) 4, for every r, an associative law analogous to (1) but with r as prefactor. If el, e2, e3 form a basis and e', e2, e3 the reciprocal basis, we can express any two dyadics in the form, (9)
4) = fie, + f2e2 + f3e3,
' = elg1 + e2g2 + e3g3
LINEAR VECTOR FUNCTIONS
144
§ 66
In the product 4 - ', six dyads vanish, and we find (10)
4)
`I' = figs + f2g2 + f393-
If 4) and' are complete, fl, f2, f3 and g1, g2, g3 are non-coplanar sets; then (- 4, is also complete. But if (F or 4, is singular, one of these sets must be coplanar, and 4) 'Y is likewise singular. Therefore the product of two dyadics is complete when and only when both dyadic factors are complete.
If 4) is complete, it may be "canceled" from equations such as
4.4,=O,4, (F=0, to giveT =O. Thus, if are non-coplanar in (10), and hence g1 = g2 = g3 = 0. We also = 4) St; for this is equivalent to (4, - 9) = 0, and hence 4, - SZ = 0. 66. Idemfactor and Reciprocal. The unit dyadic or idemfactor I is defined by the equation,
may "cancel" - in (F
I
(1)
r=r
for every r.
The idemfactor is unique (§ 60). For any basis e1, e2, e3, we have I e; = e I e' = e'; hence, from (60.7), (2)
I = e1e' + e2e2 + e3e3 = e'e1 + e2e2 + e3e3;
in particular the self-reciprocal basis i, j, k gives (3)
I = ii + jj + kk.
Evidently I is symmetric and complete. THEOREM. In order that a, b, c and u, v, w form reciprocal sets, it is necessary and sufficient that (4)
au+bv+cw=I.
When the sets are reciprocal, we have just proved (4). Conversely, if (4) holds, a, b, c are non-coplanar, since I is complete; let a', b', c' denote the reciprocal set. Using these vectors as prefactors on (4) gives u = a', v = b', w = c'. Multiplying a dyadic by I leaves it unaltered; for, from
r (I.4)) _ we conclude that (5)
4I=I
(F = 4.
IDEMFACTOR AND RECIPROCAL
§ 66
145
If 4 I = I, 4) and ' are both complete, since I is complete From
(§ 65).
T
or
(4,
4) = I. Two dyadics 4), 4, are said to be reciprocals of each other when
4) 41 _I (D =I,
(6)
and we write 4, = 4)-1, 4) = T-1 A complete dyadic 4) has a unique reciprocal (D-1. If (7)
(D-1 = flee + f2e2 + f3e3,
4) = elf, + e2f2 + e3f3,
for their products in either order give I: 4) .
4>-I
= e,e' + e2e2 + e3e3,
4)-1
4) = f If, + f'f2 + f3f3.
Since dyadic multiplication is associative 4,-I) _ 4) ('F -1) 4,-1 = 4) (4) F) (T-1
(D-1 = I;
(4) . *)-I = 'F-I .,I>-,.
(8)
Making use of (8), we have (4) . * . Q) -1 = 2-1 , ggeneral,
and, in
(4)
q') -1 =
the reciprocal of the product of n dyadics is the
product of their reciprocals taken in reverse order. For positive integral n we define 4,n
= (P . 4)-n = (4)-i)n = (4)n)-I;
4)°=I. With these definitions, (12)
4)m . 4,n = Pm+n,
(4)m) n.
_
4)mn,
for all integral exponents. But owing to the non-commutativity of the factors in a dyadic product, (4>. ')n is not in general equal ,0 cpn . *n.
LINEAR VECTOR FUNCTIONS
146
§ 67
By means of reciprocal dyadics, we readily may solve certain vector and dyadic equations. Thus, if (F is complete, the equations, (F
r=v,
r 4 ) =v,
4 )-4 1=2,
2,
(F
have the respective unique solutions:
r=4)-1 v, If (F = Maibi, we define the dyadics:
67. The Dyadic (1) x v.
- x v = Ea1bi x v,
(1)
v x 4) = Tv x aibi.
From these definitions we have the relations: (2)
(4) x v)
r = 4) (vxr),
r (v x (F) = (rxv)
(F;
(r-(P)-v,
(3)
When 4, = I, these give (4)
r = vxr.
(Ixv) = Since these equations hold for any r, (6)
Ixv = vxI,
(7)
(I x v), _ -Ixv.
Thus I x v is antisymmetric and planar; it transforms all vectors f into vectors perpendicular to v. Since
aibi (I x v) = aibi x v,
(I x v)
aibi = v x aibi,
the definitions (1) give (8)
(P
(Ixv) _
xv,
(Ixv) - 4) = vx(F.
Consequently, the operations v x and x v on vectors or dyadics may be replaced by dyadic products (I x v) and (I x v). For any vectors u, v, r we have [I x (u x v)]
r = (u x v) x r = (vu - uv) - r,
and hence (9)
1 x (u x v) = vu - uv.
FIRST SCALAR AND VECTOR IN\'AItIANT
§ 68
147
From this identity we arrive at the general form of any antisymmetric dyadic. THEOREM.
If the dyadic f = 2;aibi is antisymmetric, and the
vector co = tai x bi,
l = -21xw.
(10)
Proof.
The conjugate of f is -St = 2;biai; hence
-2SZ = 2; (biai - aibi) = EI x (ai x bi) = I x w. Every antisymmetric dyadic is planar. 68. First Scalar and Vector Invariant. We may express a dyadic (P = 2;aibi in various forms by substituting vector sums for ai, bi, and expanding and collecting terms by applying the distributive law. For all these forms there are certain functions of the vectors, in terms of which is expressed, which remain the same. These functions, which may be scalar, vector, or dyadic, are called invariants of the dyadic. Each step of the process in changing = 2;aibi from one form to another may be paralleled by the same step in transforming the scalar and the vector, Corollary.
(1)
Cpl = tai bi,
(2)
4) = 1dai x bi,
obtained by placing a dot or cross between the vectors of each dyad of (D. Each application of the distributive law in the transformation of 4) is also valid in the corresponding transformations of 01
and 4; and, just as 1 is not altered by these changes, the same is true of the scalar cpj and vector 4. These quantities are therefore invariants of 4 with respect to the transformations in question. They are called, respectively, the scalar (or first scalar invariant) and vector of the dyadic.
For example, if P = ij + jk + kk, c1
1,
4)=ixj+jxk+kxk=k+i.
For the idemfactor I = ii + fl + kk, the scalar is 3 and the vector 0; and we obtain these same values if I is expressed in terms of an arbitrary basis: I = ele' + e2e2 + e3e3. Again, for the dyadic,
S2 = IxV = iixv+jjxv+kkxv,
LINEAR VECTOR FUNCTIONS
148
§ 69
we have co, = 0, and
w=ix(ixv)+Jx(Jxv)+kx(kxv)
-2v;
thus S2 = -ZI x w, as in (67.10). The scalar or vector of the sum of two dyadics is the sum of their scalars or vectors: thus, if (3)
'Pi=4i+Wi, 4=4+w.
4) =`I'+0;
It is principally to this property that these invariants owe their importance. In (64.1), we have expressed any dyadic 4) as the sum of a symmetric and antisymmetric dyadic: (4)
The antisymmetric part S2 =
z
(4) - 4) has the vector,
From the theorem of § 67 we now have
SZ = -2Ix4.
(5)
Hence, from (4), (6)
24=4) +F,,-Ix4, +Ix
From (6), we have the THEOREM. A necessary and sufficient condition that a dyadic be symmetric is that its vector invariant vanish.
69. Further Invariants. We may obtain further invariants of the dyadic c = 1aibi by processes that are distributive with respect to addition. The most important of these are the dyadic, (1)
4)2 = 2;aixajbixbj, i.l
called by Gibbs the second of I ; its scalar invariant, (2)
(P2 = 2 (ai x a3) (bi x bj); i,)
and the scalar, (3)
(P3 = * i.i.k
(aixaj .ak)(bixbj .bk).
FURTHER INVARIANTS
§ 69
149
In (1) the summation is taken over all permutations i, j. When
i = j, the dyad vanishes; when j - i, the permutations i, j and j, i give the same dyad, so that each dyad occurs twice in the final sum; this doubling is avoided by the factor 2.
In (3) the summation is taken over all permutations i, j, k. When two subscripts are the same, the term vanishes; when i, j, k all differ, the 3! = 6 permutations of these subscripts give the same term, so that each term occurs six times in the final sum; i this multiplication of terms is avoided by the factor s Let CF be reduced to the three-term form:
CF = al+bm+cn.
(4)
The invariants considered thus far are now (5)
(7)
4 = axl+bxm+cxn, 42 = bxcmxn+cxanxl+axblxm,
(8)
IP2 = (b x c)
(9)
(p3 = (a x b
(6)
(m x n) + (c x a)
(n x 1) + (a x b)
(l x m),
c) (1 x m n).
The numbers 01, P2, 03 often are called the first, second, and third scalars of 4).
If CF is singular, the antecedents a, b, c or the consequents = 0, 1, m, n in (4) will be coplanar, and p3 = 0. Conversely, if a, b, c or 1, m, n are coplanar sets and 4) is singular. Therefore a dyadic is singular when and only when its third scalar is zero. If V3 = 0, 4) must be planar, linear, or zero. * When CF is linear it can be reduced to the form al and cF2 = 0. Conversely, if 4'2 = 0, 4) will be linear or zero; for, if we choose a non-coplanar set a, b, c as antecedents in (4), b x c, c x a, a x b are also noncoplanar, and `1'2 = 0 implies that
mxn = nxl = lxm = 0; then 1, m, n are parallel or zero. Therefore we may state the THEOREM. Necessary and sufficient conditions that a dyadic CF be Complete
'P3 3-' 0,
Planar
are that
Linear
CF2 = 0,
CF2
(1)
0, 0.
LINEAR, VECTOR FUNCTIONS
150
§ 69
We next compute the invariants of the dyadic
_ 'D2. From
(7),
*2 = (c x a) x (a x b) (n x 1) x (1 x m) + cyclical terms
= [abc] [lmn] (al + bm + cn) ; hence, from (9), *2 = t03-1-
(10)
We now may compute the three scalars of 4,: (11)
V1 = P2,
412 = '3'P1,
4'3 =
2;
the first follows from (8), the second from (10), and the third from (b x c)
(c x a) x (a x b) = [abc]2,
(m x n) (n x 1) x (1 x m) = [lmn]2.
Finally, the vector invariant of ' is
4 = (bxc) x (mxn) + (cxa) x (nx1) + (axb) x (lxm). If we express 4r in terms of a, b, c, the term in a is
ab lxm = al - (bxm+cxn) = al
4
from (6) ; hence
4r = (al + bm + cn)
(12)
It can be shown that all scalar invariants of 4) may be expressed in terms of the six scalars, 'PI,'02,'3,'
(13)
This property is expressed by saying that these six scalars form a complete system of invariants. When 4) is symmetric, 4 _ 4, = 0, and the last three scalars vanish. The third scalar of the product of two dyadics is equal to the product of their third scalars. For any two dyadics 4), ' can be put in the form, = e'g1 + e2g2 + e3g3, 4) = f1e1 + f2e2 + f3e3,
where e1, e2, e3 and e1, e2, e3 are reciprocal sets (§ 65); hence [f1f2f3][e1e2e3][e'e2e3][919293]
'P43 =
= [f1f2f3][g1g2g3],
which is the third scalar of
4' ' = f1g1 + f2g2 + f3g3
SECOND AND ADJOINT DYADIC
§ 70
1.51
Example. For the idemfactor I = ii + jj + kk, the second 12 = I, the vector is zero, and the three scalars are 3, 3, and 1. From
I = eiel + e2e2 + e3e3,
(i)
we have, for 12,
I = e2xe3e2xe3 +e3xe1e3xel +elxe2e1xe2,
(ii)
and the vector, elxel + e2xe 2 + e3 x e3 = 0.
The third scalar gives [ele2e3][ele2e3] = 1.
From (ii) we have, for example, I I. e3 = el x e2[e'e2e3],
e3 = el x e2/[ele2e3]
Thus (i) and (ii) give a handy compendium of the properties of reciprocal vector sets.
70. Second and Adjoint Dyadic.
The second of the dyadic
4) = Eaibi, namely, (1)
(D2 =
ialai i.;
x a; bi x b;,
has the property,
(4 u) x (4) v)
(2)
= -4)2
(u x v).
Proof. We have v) i
;
= Mai x a; (bi u) (b; v) i;
= Ma; x ai (b, . u) (bi v), i;
on interchanging i and j. The left member also equals half the sum of the two last expressions :
-u)
ij or, with regard to (20.1),
"M(aixa;)(bixb;) (uxv) = 4)2 (uxv). i;
LINEAR VECTOR FUNCTIONS
152
§ 70
The conjugate of 4'2 is called the adjoint of 4) and written 4'a. The adjoint satisfies the important relation, 4)=gC3I.
(3)
Proof. (4)
Write (D = al + bm + cn; then
4)a = mxnbxc+nx1cxa+lxmaxb.
Choose for the antecedents a, b, c of 4) a non-coplanar set. (§ 61).
Then, if a', b', c' denote the reciprocal set, we have, by direct multiplication,
D 4)a = (1 mxn)(abxc +b cxa + caxb) _ (1 m x n) (a b x c) (aa' + bb' + cc') =(P3I.
If we choose the consequents 1, m, n of 4) as a non-coplanar set (then a, b, c may or may not be coplanar), let 1', m', n' denote the reciprocal set. Then
(a b x c)
x n) (1'l + m'm + n'n)
= (P31-
0; then (3) shows that
If 4) is complete, P3
4'-1 = 4'a/1P3
(5)
From (3) we also may show that, if u, v, w are any three vectors, (6)
[4)
u, 4) v, 4) w] = ,P3[uvw].
Proof. Since 4'a is the conjugate of 4>2i we have, from (2),
4)a
(U
V)
(U
V)
4)a,
4) by P3I, we obtain (6).
We now can deduce important properties of the affine trans. formation (§ 62) : (7)
r'
r
((P3 $ 0).
The vector area u x v (§ 17) is transformed into (8)
u' x v' = 4'2 (u x v).
INVARIANT DIRECTIONS
§ 71
153
From (4), the third scalars of ad and P2 equal [abc]2[lmn]2 = Since 4) and c2 are complete, r 7-4 0 and u X v 0 imply 0, u' x v' 0 (thus filling a gap in § 62). An affine trans0.
,p3
r'
formation invariably changes lines into lines, planes into planes. Moreover (7) transforms any parallelepiped [uvw] into another [u'v'w'] whose volume is p3[uvw] according to (6). As any volume
can be regarded as the limit of a sum of parallelepiped elements, the affine transformation alters all volumes in the constant ratio of 93/1.
71. Invariant Directions. We next seek those vectors r which are transformed by 4) into scalar multiples of r, say - r = Ar.
(1)
If we write this equation,
It - r = 0 where '1 = 4) - Al, it is clear that the multiplier A must make the dyadic T singular; its third scalar 43 is then zero (§ 69). Conversely, if A is a root of the equation 03 = 0, A is a multiplier of 4); for, when 4, is planar or linear, 4, - r = 0 for all vectors r normal to the plane of the consequents of T. Let us write
I = aa' + bb' + cc',
4) = al + bm + cn,
where a, b, c are non-coplanar and a', b', c' the reciprocal set; then
4, = a(l - Xa') + b(m - Xb') + c(n - Ac'), 4'3 = [abc] [(1 - Xa')(m - Xb')(n - Xc')]. The second box product in 43 gives, on expansion,
[lmn] - A { [a'mn] + [b'nl] + [c'lm] }
+ A2 { [b'c'l] + [c'a'm] + [a'b'n] l - A3[a'b'c']. Substituting for the primed vectors gives bxc ... ' .I a= [abc] '
b
x
a .. . c = [abc] '
LINEAR VECTOR FUNCTIONS
154
§71
and then multiplying the entire expansion by [abc] yields L'3 = [abc][1mn]
- A{(bxc) (mxn) + (cxa) (nx1) + (axb) (lxm)} -A3, or, on making use of (69.5), (69.8), (69.9), 43 =
(2)
Denote the right member of (2) by f (A) ; then the cubic equation,
'P3 = f(A) = 0,
(3)
is called the characteristic equation of 4,. Since 4) - AI is singular when f (X) = 0, the coefficients of f (X) depend only upon the nature
of the dyadic 4) and not upon the particular form in which it is expressed. We thus have an independent proof of the invariance of
The three roots A1, A2i A3 of (3) are called the multipliers or characteristic numbers of -4). From the relations between the roots
and coefficients of an algebraic equation, we have (4)
IP1 = Al + A2 + A3,
P3 = A1A2A3.
The cubic (3) may have three real roots (not necessarily distinct) or one real root and two conjugate complex roots. To find the invariant direction corresponding to a root A1i we consider in turn the three cases in which 4) - X1I is singular. 1. If 4 - X11 is planar, let 4' - A1I = ch + dk; then the postfactor r1 = h x k reduces the right member to zero and gives the invariant direction for A1. If 4) - X1I has more than two dyads, the cross product of any two non-parallel consequents will be normal to their plane and give a vector parallel to r1. If Al is real, r1 is a real vector. But if Al is complex, say Al = a + i(3, then r1 = a + ib is also complex. Then, on equating the real and imaginary parts of (5)
4)
(a + Zb) _ (a + ia) (a + Zb),
we have (6)
(D
a = as - $b,
4)
b = #a + ab.
INVARIANT DIRECTIONS
§ 71
155
In this case we know a second multiplier,
X2= a-i$, with I2=a-2b, For, from (6),
as invariant direction.
(a - ib) = (a - i(3) (a - ib).
4)
From (5), we conclude that a and b are not parallel; for, if b = ka, 4) a = (a + i(3)a, which is impossible unless # = 0. 2. If 4) - X1I is linear, it may be reduced to a single dyad ch.
Then, if r1 is any vector in the plane perpendicular to h, 4,
r1
= X1r1. In this case there is a whole plane of invariant directions
corresponding to X1.
3. If (D - X1I is zero, 4 r1 = X1r1 for any vector r1. All directions are invariant with the multiplier X1. If we write r' _ 4) r, we have, from (70.2), (Uxv). U" V' = (D2 Now 4) transforms all vectors in the plane of u, v into vectors of the plane u', V. These planes will be the same if u' x v' = x u x v, (7)
that is, if u x v is an invariant direction of 42. Hence the invariant planes of 4) are normal to the invariant directions of (P2. Example 1.
= ii + j(i + 2j) + k(j + 2k). Then
fit = i(4i - 2j + k) + j(2j - k) + 2kk,
pp1= 1+2+2=5,
I'2=4+2+2=8,
S3=4.
The characteristic equation,
f(T)=4-8a+5X2-X3=(1-X)(2-A)2=0, has the roots, X1 = 1, For the root, X1 = 1,
X2 = X3 = 2.
- XiI = ji + jj + kj + kk = j(i + j) + k(j + k), and the corresponding invariant direction is
r1= (i+j)x(j+k) =i-j+k. For the double root, X2 = X3 = 2,
'P -1\2I= -ii +ji+kj = (-i+j)i+kj, and r2 = i x j = k. Example 2.
4) = ij + jk + Id. Then 'Pi=0,
92=0,
IP3=1.
LINEAR VECTOR FUNCTIONS
156
§ 72
The characteristic equation, 1 - X3 = 0, has the roots, X1 = 1,
X2 =
For X1 = 1,
+
=
2
X3 =
w,
-1 -2
= w2.
4, -all=i(j-i)+j(kj)+k(i-k)
is planar- its consequents are all normal to
r1=(j-i)x(k-j)=i+j+k, which is the corresponding invariant direction. For X2 = w, the consequents of
4 -X21=i(j-wi)+j(k-wj)+k(i-wk) are all normal to
r2 = (j-wi)x(k-wj) =i+wj+w2k. For X3 = w2, the consequents of
4' -X3I=i(j-w2i)+j(k-w2j)+k(i-w2k) are all normal to
r3 = (j - w2i) x (k - w2j) = i + w2j + wk.
72. Symmetric Dyadics. The characteristic numbers of a symmetric dyadic are all real. Proof. If X1 = a + i(3 is a complex multiplier and rl = a + 2b the corresponding invariant direction of the symmetric dyadic 4), we have (§ 71)
(a+ib) = (a+ii)(a+ib); c a = as -(3b, 4 ) b = 0a+ab. ,D
Since cD is symmetric, b
(D
a=a
4)
b; hence
=0.
or
But a and b are real vectors, and a a + b b is positive; hence (3 = 0, and X1 is real. The invariant directions corresponding to two distinct characteristic numbers of a symmetric dyadic are perpendicular. Proof. From the equations,
4 rl = X1rl,
`k
r2 = X2r2
(X1 0 X2),
and
r2, we deduce (X1 - X2)rl r2 = 0,
rl r2 = 0.
SYMMETRIC DYADICS
§ 72
1,57
Consider now the following cases.
1. If f (X) = 0 has three distinct roots, the corresponding invariant directions are mutually perpendicular and may be denoted by i, j, k. Then 4, - i = X1i,
4 'j = X2j,
4) - k = X3k,
and, from (61.11), 4) = X1ii + X2JJ + X3kk.
(1)
2. If two roots of f (X) = 0 are equal, let X1 5-' X2 = X3. Then, if i, j are the perpendicular invariant directions corresponding to Xl and X2, we may write 4, - i = X1i,
4, - k = v,
(D 'j = X2j,
v being the (unknown) transform of k; then, from (61.11), 4) = X1ii + X2jj + vk.
Now the symmetry of 4) requires that vk = kv, and hence v must be a multiple of k, say yk. Thus
4' = X1ii + X2jj +ykk,
but, since Cpl = X1 + 2X2 by hypothesis, -y = 1\2, and 'D = X1ii + X2jj + X2kk.
(2)
Corresponding to the double root X2 = X3, we have a whole plane of invariant directions.
3. If all three roots of f(X) = 0 are equal, X1 = X2 = X3, let i denote an invariant direction corresponding to X1, and write 4)
i = Xli,
4>
j = u,
4 ) k = v;
then, from (61.11),
= X1ii + uj + vk.
Now the symmetry of 4) requires that uj + vk = ju + kv, and this in turn shows that u and v have the form,
u = aj + yk,
v = yj + $k.
Thus
4) = X1ii + (ai + yk)j + (yj + $k)k, (Pi = X1 + a + 0,
o3 = Xl (a$ - y2);
LINEAR VECTOR FUNCTIONS
158
§ 72
but, since
a(3 -
a + /3 = 2X1,
y2 =
ai
Elimination of X1 gives the relation,
(a + 0)2 - 4(a/3 - y2) = (a - (3)2 + 4y2 = 0; and, since neither (a - 0)2 nor 4y2 can be negative, their sum can
vanish only if a-i3=0,y=0. Hence a = 0 = X1, and 4> reduces to 4, = X1ii + A1jj + X1kk = XII.
(3)
In the case of a triple root all directions are invariant. Evidently all cases are included in (1); by making two or three
of the roots equal, we obtain (2) and (3). THEOREM. Every symmetric dyadic may be reduced to the form,
4, = aii + $jj + 7kk,
(4)
in which a, 0, y are the real multipliers and i, j, k corresponding invariant directions. The reciprocal of 4) is
=-ii+-jj+-kk;
(15)
for .4 4)-1 = I.
1
1
1
a
a
y
Moreover,
4'2 = f7ii + yajj + a$kk
(6)
and, by direct multiplication,
4)" = a"n + a"jj + y"kk.
(7)
If
4)
is symmetric and 4)" = 0, then 4) = 0.
Moreover, if
r 4 r = 0 for every r, - = 0; for, if we choose r = i, j, k in
turn, a=/3=y=0. Example 1.
For the symmetric dyadic,
(D = i(j +k) +j(k +i) +k(i +j), w1 = 0, v2 = -3, 93 = 2, and the characteristic equation,
2+3X-X3=(2-a)(1+x)2=0, has the roots, X1 = 2, X2 = X3 = -1.
SYMMETRIC DYADICS
§ 72
159
For X1 = 2,
-XiI=i(-2i+j+k)+j(-2j+k+i)+k(-2k +i+j), The consequents are all normal to
(-2i+j+k)x(-2j+k+i) =3(i+j+k); hence r1 = i + j + k.
For a2=X3= -1,
-X2I= (i + j + k) (i + j + k) is linear. Hence any vector in the plane perpendicular to i + j + k is an invariant direction. If we choose the unit vectors,
i' = (i + j + k)/V'3,
j' = (i - j)/V'2,
as invariant directions for the roots 2 and -1, respectively, then the unit vector,
k' =i'xj' = (i+j -2k)/V,
gives a second invariant direction for -1.
Therefore
4) =2i'i'-j'j'-kk', as we may readily verify. Example 2. Inertia Dyadic. Let 0 be any point of a rigid body and s an axis through 0 in the direction of the unit vector e. Then, if dm is an element of mass at P, at a distance p from the axis s, the moment of inertia of the body
about s is defined as the integral f p2 dm over the body. If r = OP,
p2 =r2-
=e (r2I
hence, if we introduce the dyadic,
K = I fr2dm - frr dm,
(8)
known as the inertia dyadic of the body for the point 0,
e-
(9)
fp2dm.
For example, i.
f(r2-x2)dm= f(y2 +z2)dm
is the moment of inertia about the x-axis. Thus K effects a synthesis of the moments of inertia of a body about all axes through 0. This dyadic also has the property that, for any pair of perpendicular unit vectors el, e2,
- el . K . e2 = f
(el
r) (r . e2) dm
LINEAR VECTOR FUNCTIONS
160
§ 73
is the product of inertia for the corresponding axes; thus
j=f
j) dm = fxydm.
The inertia dyadic K is evidently symmetric. Hence we always can find three mutually perpendicular axes x, y, z through 0 such that
K=Ali+Bjj+Ckk.
(10)
These axes are called the principal axes of inertia at 0, and A, B, C are the moments of inertia of the body about these principal axes. The principal axes are characterized by the property that the product of inertia for any pair is zero. The ellipsoid,
(11)
r K . r = 1, or Axe + Bye + Cz2 = 1,
is called the ellipsoid of inertia at 0; its principal axes are the principal axes of inertia at O. It has the property that the moment of inertia about any axis s
through 0 and cutting the ellipsoid at P is 1/(OP)2. For, if OP = r = re,
73. The Hamilton-Cayley Equation. The identity (70.3) applied to the dyadic ' = - AI gives (1)
`y
'Ya = f(X)I.
The form of ' given in § 71 shows that we may write (2)
'a = A + BA + CA2,
where A, B, C are dyadics independent of A; hence, from (1), (3)
(4 - XI) (A + BA + CA2) = (-P3 - V2A +
V1A2
- A3)I.
Since (3) is an identity in A, the dyadic coefficients of like powers of A in the two members must be equal, hence 4) A = (P3I, -V217
(4)
-C= -I. If we multiply these equations in order by I, 4), 42, (3 and add, the first members camel, and we get (5)
te3I - (P24) + (P1 cy - (1,3 = 0.
THE HAMILTON-CAYLEY EQUATION
§ 73
161
Every dyadic 4) satisfies this cubic equation, the Hamilton-C,ayley Equation; it evidently is formed by replacing X by 4) in the characteristic equation, f(X) = V3 - cP2X + P1
(6)
X2
-
X3
= 0,
and inserting I in the constant term. If the X1, X2, X3 are the characteristic numbers of 4), AX) = (X1 - X) N - X) (X3 - X) hence (5) also may be written
(4) - X1I) . (1 - X2I)
(7)
(4 - X31) = 0,
in which the dyadic factors are commutative. From equations (4), we find
C=I,
B=4) -q'1I,
hence, from (2), q,. = (8)
4)2
A=4)2-914) +92I;
(,P1 - X)4) + (V2 - (PI X + X2)I.
When X = X1, a characteristic number, 91 - X1 = X2 + X3,
'P2 - 'P1X1 + X1 = X2X3,
and (8) becomes (9)
(4) - a1I)a = (4) - X21) (4) - X3I).
Although every dyadic 4) satisfies the cubic (5), 4) will satisfy an equation of lower degree when `"a = (4) - XI)a vanishes for a characteristic number X j. The equation of lowest degree satisfied by 4) is called its minimum equation. *
If 4, = 4) - XI = 0 for a characteristic number X1 (then also Ta = 0), the minimum equation is linear, namely, (10)
4) -X1I=0.
When 4) = X11, 4, = (A1 - X)I, and 03 = f(A) = (X1 - X)3; the Hamilton-Cayley Equation is therefore (11)
(4) - X11)3 = 0.
* For its formation and properties see Maeduflee, C. C., An Introduction to Abstract Algebra, New York, 1940, pp. 224-6. This treatment for n X n matrices also applies to dyadics, regarded as 3 X 3 matrices.
LINEAR VECTOR FUNCTIONS
162
§ 74
If 4, does not vanish for any characteristic number, but *a, = 0 when X = X1, we see from (9) that the minimum equation is the quadratic, (,b - X31) = 0.
(4, - X21)
(12)
When Fd = 0, 4'2 = 0, and, consequently, 4, is linear (theorem, § 69) ; hence we may write 4) = AiI + uv. If we take u = i, v = ai + /3j ± 'Yk, I = ii + jj + kk, we have
-A)I+aIi+flij+yik, and the determinant of 'F's matrix is ¢3 = f(A) = (A1 - X)2(X1 + a - A). Thus the characteristic numbers of b are X1, A2 = A1, A3 = Al + a,
and its Hamilton-Cayley Equation is ((D - A1I)2 . (4, - A3I) = 0.
(13)
From 'I = A1I + uv, we see that all vectors perpendicular to v have the multiplier A1i whereas vectors parallel to u have the multiplier Xi + u v = Al + a. When u v = a = 0, (13) reduces to (11).
We note that, in every case, the minimum equation and the Hamilton-Cayley Equation have the same linear factors and differ only in their degree of multiplicity. 74. Normal Form of the General Dyadic. Every complete dyadic transforms at least one set of mutually orthogonal directions (its principal directions) into another set of the same kind. To find the principal directions of the complete dyadic -1,, conThe latter is complete (§ 65) and symsider the dyadic (D. metric; for, from (65.8),
'
(c. 4)). = We therefore may write (§ 72) (1)
(DC
= A1u + A2jj + A3kk,
(Ai
0).
Consequently,
i.cc.4).j = j.4)a.(D .k = k.4)a.4) .i = 0, or
(-t
i) =0.
§ 74
NORMAL FORM OF THE GENERAL DYADIC
163
The vectors 4 i, 4) j, 4 k are thus mutually orthogonal, and hence i, j, k give a set of principal directions of 4). If we write
4 i = ai',
(2)
4,
j = #j',
4, . k = yk',
where i', j', k' is a second dextral set of unit vectors, we have, from (63.11),
4) = ai'i + (3j'j + yk'k.
(3)
Moreover, we always can arrange so that a, fl, y have the same sign. If, for example, a and 0 have one sign, -y the opposite, we
can replace i', j' by -i', -j', and the set -i', -j', k' still will be dextral. From (3) and
,Pc = aii' + 3jj' + ykk', we have, by direct multiplication, (4)
4rD'
4) = a2ii + a2jj + y2kk,
= a2i'i' + 132j'j' + y2k'k'.
(5)
These symmetric dyadics, which in general are different, have the same multipliers, evidently all positive. We have therefore proved the THEOREM. ,I)c
If (P is complete, any three invariant directions of
4) that are mutually orthogonal are principal directions of (D, and
The principal directions of 4) transform into invariant directions of 4' 4,. Any complete dyadic 4) can be reduced to the normal form (3) in which the scalars a, (3, y are square roots of the multipliers of (D, 4) (or 4) (D,) having the same sign. conversely.
Example. Homogeneous Strain. In distinction to the ideal rigid body, the particles of a deformable body are capable of displacements relative to one another. The totality of such relative displacements is said to constitute its state of strain.
Suppose that a particle at P moves to P' under the strain; then r' = OP' is a continuous function of r = OP. The simplest type of strain occurs when r' is a constant linear vector function of r, (6)
r' = 4' r (4 complete);
the strain is then said to be homogeneous. We have seen in § 62 and § 70 that a homogeneous strain transforms lines into lines and planes into planes; and
LINEAR VECTOR FUNCTIONS
164
§ 75
evidently parallelism is preserved. Moreover all volumes are altered in the constant ratio of (P3/1. Since 4' is complete, r = 4-1 r'. The particles originally on a sphere about 0 are displaced so as to lie upon an ellipsoid; for r r = a2 transforms into
When 4' is reduced to the form (3),
+jj'+ kk', 1
4'-1
12
k'k' a i'i' + j'j' + 2 ti
,
and the foregoing strain ellipsoid (with a = 1) has the equation, x'2 cr2
z,2
,2
+ -2 +
= 1. r2
The principal directions i, j, k of 4) are called the principal axes of strain; they transform into ad', /3j' 7k', the principal semiaxes of the strain ellipsoid.
75. Rotations and Reflections. In order that a dyadic 4 transform all vectors so that their lengths are unchanged, it is necessary and sufficient that its inverse be equal to its conjugate:
-1 = b'.
(1)
If, for any vector r,
Proof. (2)
r)
(4)
then
r
4)c
and, since I
cb
r=r
I
r
r,
-I) r= 0,
4) - I is symmetric, it must be zero: 4),
4' = I, or 4), _
-1.
Conversely, if (1) is fulfilled, (1
A dyadic that preserves the lengths of vectors also preserves the angles between them: for, by virtue of (1),
(4) r)(4) - s) =r I
I
and, since lengths are unaltered, cos (c r, 4) s) = cos (r, s).
§ 75
ROTATIONS AND REFLECTIONS
165
Condition (1), although sufficient to ensure preservation of angles, is by no means necessary. Thus the dyadic XI preserves angles but multiplies all lengths by X. Since fi preserves lengths and angles, any orthogonal set of unit vectors is transformed into another such set. Thus there are two
possible cases: the dextral set i, j, k is transformed into another
dextral set i.', j', k', or into a sinistral set i', j', -k'. Hence we have two types of length-preserving dyadics:
fi = i'i + j'j ± k'k.
(3)
Their third scalar is V3 = =L1. Moreover,
fit = ±i'i ± j'j + k'k = -fi. This shows that the vector invariant of fit is f+; but, from (69.12), this vector is fi
Equating these values, we obtain
fi + _+;
(4)
the vector invariant of fi gives an invariant direction of multiplier, 'P3 = f 1. If we choose k in this direction, (3) becomes IP3 = f 1. fi = Vi + j'j + p3kk, When V3 = 1, fi r transforms i, j, k into i', j', k. But a rotation 0 about k as axis, through an angle X such that i, j revolve (5)
into i', j', transforms i, j, k in the same way; and, since a is linear vector function, fi = e (§ 60). Thus
0=i'i+j'j+kk
(6)
is a rotation about the axis k through an angle X determined by its scalar and vector invariants: (7)
01 = 1 + 2 cos X,
0 = -2 sin X k.
When P3 = -1,
fi = i'i + j'j - kk = (i'i + j'j + kk) (ii + jj - kk). The dyadic in the first factor is the rotation e. The dyadic ii + jj - kk in the second factor transforms i, j, k into i, j, -k. But a reflection E in the plane of i, j transforms i, j, k in the same way; and, since E is a linear vector function, (8)
E = ii + jj - kk = I - 2kk.
Thus, when P3 = -1, fi = 0 E.
166
LINEAR VECTOR FUNCTIONS
76
THEOREM. A dyadic 4) that preserves lengths is a rotation e when (p3 = 1, and a rotation O followed by a reflection in the plane per-
pendicular to its axis when c03 = -1. In the latter case n E will reduce to Z, a pure reflection, when O = I.
76. Basic Dyads. If we express both antecedents and consequents of a dyadic 4) in terms of a given basis e1, e2, e3, we obtain upon expansion 3 X 3 = 9 types of basic dyads eie; (i, j = 1, 2, 3). On collecting terms we may write 4: (1)
4) _ (p11e1e, + c'21e2e,
(p12e1e2
+ cp22e2e2
w e3e, +
032
+ +
e3e2 +
P13ele3 P23e2e3
+ +
(p33e3e3.
The nine coefficients 9 are called the contravariant components of
4) relative to the basis ei (cf. § 23). If we drop the nine basic dyads e;e; in (1), we can represent 4) by the 3 X 3 matrix, =
13
11
1p12
p21
(p22
(p23
031
032
(p33
(P
V
a skeleton of numbers arranged in a definite order, which stands for the full expression (1). This is analogous to the use of a number triple (u', u2, u3), or 1 X 3 matrix, to represent the vector a = ule, + u2e2 + u3e3. If we express the vectors of 4) in terms of the reciprocal basis
e we write (2)
4)
The nine numbers pij are called the covariant components of (b relative to the basis ei. Just as before, we may represent 4) by a matrix of the components cpjj. This corresponds to the use of the number triple (ul, u2, u3) to represent the vector u = u,e1 + u2e2 + u3e3.
But with the same basis e1, we can represent 4) in two other ways. First, we may express the antecedents of 4) in terms of ei, the consequents in terms of e'; the basic dyads are then eie', and (3)
c _ Z2;oz j eie'.
3 77
NONION FORM
167
Or we may express the antecedents in terms of ei and consequents in terms of e;, so that the basic dyads are eie;; then = XMv 5eie;.
(4)
If we represent 1 by matrices of mixed components, we must indicate the order of the subscripts as shown in the preceding notation; for, in general, e. p, i However, if we use the full notations (3) or (4), the The components gyp';, cpti' are called mixed.
components can be written gyp;; for the order of the indices then is shown by the base vectors.
77. Nonion Form. When the self-reciprocal orthogonal set i, j, k is used as a basis, all four representations of 4) given in § 76 become the same. Since upper and lower indices no longer are needed, we write the orthogonal components of D arbitrarily as Vij.
When no basis is indicated, the components of the matrix
(
+ 'P33kk =
_ 'P11ii + 'P121j +
'P11
'P12
'P13
rP21
'P22
923
'P31
'P32
'P33
The conjugate of I corresponds to the transpose of this matrix:
(2)
'11
V21
'P31
'P12
'P22
'P32
'P13
'P23
'P33
Hence c is symmetric when 'Pij = 'Pji, antisymmetric when 'Pij = ('i1 = 0). c° The first scalar and vector invariant of P are readily computed: (3) (4)
'P1 = 911 + V22 + 'P33, = ('P23 - 'P32)1 + ('P31 - VP13)i + ('P12 - (p21)k
In order to compute 'P2 and
= i(
LINEAR VECTOR FUNCTIONS
168
§ 77
Then, since [ijk] = 1,
namely the determinant of matrix ():
(6)
Thus
'P11
'P12
'P13
'P21
'P22
'P23
'P31
'P32
'P33
is singular when the determinant of its matrix vanishes.
As to 4)2i the terms with antecedent i = j X k have as consequents, (11i + (p1) + (D13k,
('P211 + 'P223 + 'P23k) x (cP311 +
where 'ij denotes the cofactor t of 9i j in the determinant (6) ; hence (7)
,11
,12
,13
4)21
,22
X23
X31
,32
,33
2
'P2 =
(8)
(D11
+
X22
and (% is the transpose of (7).
+ X33,
Moreover, since (D-1 = 4)a/cp3
(70.5), 11
'P21
4031
P22
032
'23
033
(P
-1 =
(9)
'P13
('P 12
where (P''j = (I)ij/(p3 is the reduced cofactor of 'ij.
Making use of
the well-known relations in determinant theory, 'Pi1'Pj1 +
we may verify that (10)
'P1i'P1j +
+
ati
4-' or 4)-1 4) give the idemfactor: 1
0
0
I= 0
1
0
0
0
1
The characteristic equation of 4) is obtained by equating the third scalar of 4' - XI to zero; hence, with 4) in nonion form, it t The cofactor of vii in the determinant I pif I is defined as the coefficient of Pii in the expansion of the determinant; it equals the minor obtained by striking out the ith row and jth column with the sign (-1)'+i affixed.
MATRIC ALGEBRA
§ 78
169
becomes
f(X) =
(11)
(P11 - X
'P12
'P13
'P21
'P22 - X
'P23
(P31
IP33 - X
= 0.
Example. If the dyadic 4) in (1) is symmetric, and r = xi + yj + k, then (i)
r$
r = Pllx2 + 2co12xy + P22y2 + 2w13x + 2-P23Y + V33 = 0
represents a conic section-an ellipse, parabola, or hyperbola, according as V12 - IPiisv2z < 1, = 1, or >1. Let this conic cut the sides BC, CA, AB of a triangle ABC in the points R1, Ri; R2, R2; R3, R3, respectively, and let the corresponding division ratios be Plr p l; P2r P2i Par P3'- If we put r = (b + Pc)/(1 + p) in (i) we find that P1, P1 are the roots of the quadratic equation b 4) b +2Pb 4) - C +P2c (D c = 0; hence
a'4
c'4) 'c
and
a b
follow in the same way. On multiplying these equations, we have P1P1P2P2P3P3 = 1
(ii)
Now let R2'R3i R3R1, R'R2 meet BC, CA, AB in the points S1, S2, S3 which divide the respective sides in the ratios al, 02, 03. Then by the Theorem of Menelaus (§ 7, ex. 2) P2P3°1 = - 1r
P3P1Q2 = - 1r
P1P20'3 = - 1
On multiplying these equations together and making use of (ii), we have 0.10.20.3 = -1; hence S1, S2,83 are collinear. We thus have proved Pascal's Theorem: $ The opposite sides of a hexagon (R1R'1R2RZR3R3) inscribed in a conic meet in three collinear points (Si, S2, S3)-
78. Matric Algebra. The sum of two dyadics A + B in nonion form obviously is obtained by adding corresponding elements of their matrices. As to the product C = A B, let us consider the formation of a certain dyad of C, say c12ij. This evidently results from the product of terms in A with antecedent i and terms in B with consequent j: thus C12 = a11b12 + a12b22 + a13b32 = 2;alb,2r
The general result is therefore Cij = 2;airbrj ; r
$ This proof is due to Wedderburn, Am. Math. Monthly, vol. 52, 1945, p. 383.
170
LINEAR VECTOR FUNCTIONS
§ 78
the element in the ith row and jth column of A B is the sum of the products of the elements in the ith row of A by the corresponding elements in the jth column of B-the "row-column rule." The foregoing rules for the sum and product of dyadics in nonion form are precisely the classic definitions for the sum and product of square matrices. These definitions may be extended to rectangular matrices with m rows and n columns (m X n matrices).
1. The sum A + B of two m X n matrices is them X n matrix obtained by adding their corresponding elements. 2. The product of an m X p matrix A and a p X n matrix B is the m X n matrix C = AB, whose element in the ith row and jth column is P cjj = E airbrj.
(1)
r=1
Note that only similar matrices can be added; whereas in a product the second matrix must have the same number of rows as the first has columns. This extension enables us to interpret scalar products of dyadics and vectors in terms of matric algebra. We regard a vector u with rectangular components ui either as 1 X 3 matrix (row vector) or a 3 X 1 matrix (column vector). Since a dyadic A in nonion
form is a 3 X 3 matrix, u must be a row vector in u A, a column vector in A u. With this proviso, the rules of matric algebra give values of vector components in full agreement with vector algebra: (2)
(3)
v = u A, w = A u,
vi = u1a1i + u2a2i + u3a3i;
wi = ai1u1 + ai2u2 + ai3u3
The matric product of a row vector u into a column vector v is a 1 X 1 matrix consisting of a single element, the scalar product: (4)
u v = u1v1 + u2V2 + u3v3
But the matric product vu of a column vector into a row vector is a 3 X 3 matrix, namely the dyad, (5)
Vu =
v1u1
v1u2
v1u3
v2u1
v2u2
v2u3
v3u1
v3u2
V3u3
DIFFERENTIATION OF DYADICS
§ 79
171
in nonion form. Matric multiplication, although associative and distributive with respect to addition, is in general not commutative. The
proofs follow readily from (1). 79. Differentiation of Dyadics. If a dyadic (F is a function of a scalar variable t, we define
d(=
lim
(F(t
+ At) -
it -o
dt
(F(t)
At
For a single dyad ab, let a and b become a + Aa, b + Ab when t becomes t + At; then, since
(a + Aa) (b + Ab) - ab
Ab Aa Aa = a-+-b+-Ab, At At At
At
we have, on passing to the limit At - 0, d
-
(1)
(ab) =a
db dt
da
+
dt
b.
This formula suggests the usual product rule with the order of the factors preserved.
The derivative of the dyadic (D = 2;aibi is evidently the sum of the derivatives of its dyads. If 4, is given in the nonion form (77.1)) 'P12
d( (2)
'P22
dt 'P32
where the primes denote derivatives with respect to t. Note that
the derivative of <03, the determinant of the matrix ('i;), is not the determinant of the matrix The derivatives of products such as - r, fi x r, s (- r, which conform to the distributive law, are computed just as in the calculus when the order of the factors is preserved. d
dr
d(F r+(D
(3) (4)
dt
For example,
dt;
d
ds
d(F
dr
dt
dt
dt
dt
LINEAR VECTOR FUNCTIONS
172
§ 81
80. Triadics. A triadic is defined as a sum of triads, Eaib,ci. A triad abc consists of three vectors written in a definite order. We may regard a triadic as an operator which converts vectors r into dyadics; thus 4)
r = Eaibici r,
r 4) = 2;r aibici.
If 4) and' are two triadics, we write - = T, when (1)
4)
r = 4, r for every vector r.
Then, for any vector s, s
(D
r=s
4,
r, and, from the defini-
tion of equality for dyadics (61.3), (2)
s 4) = s - ' for every vector s.
Conversely, from (2) we may deduce (1). Thus from either (1) or (2) we may conclude that 4) = T. Using a given basis ei and its reciprocal ei to form basic dyads, we have seen in § 76 that the 32 = 9 components of a dyadic are of 22 = 4 types. For a triadic 4), the 33 = 27 components are of
23 = 8 types; for, for each vector in the basic, triads may be chosen from the set ei or e' giving 23 types; and, for a given type, each index on the base vectors may be chosen in three ways, giving 33 components. Similarly, we define a tetradic as the sum of tetrads Maibicidi.
Two tetradics 4), I are equal when either (1) or (2) holds good. A tetradic has 34 = 81 components of 24 = 16 types. collecWe shall speak of scalars, vectors, dyadics, triadics, The equality of two tively as tensors of valence 0, 1, 2, 3, tensors of valence n, say 4) = 'Y, depends upon the equality of two tensors of valence n - 1, as required by equations (1) or (2). 81. Summary: Dyadic Algebra. A linear vector function f(r) is characterized by the properties, f(a + b) = f(a) + f(b),
f(Xa) = Xf(a).
A dyadic 4) = X.aibi is the sum of dyads aibi; the vectors ai are antecedents, bi are consequents. The conjugate of 4) is 4 = 2;biai.
Any linear vector function may be expressed as 4 r (or r 0 and f (ai) = bi (i = 1, 2, 3), and, if [ala2a3]
f(r) _ 4 r,
where
4) = blal + b2a2 + b3a3.
SUMMARY: DYADIC ALGEBRA
81
173
Basic definitions (r an arbitrary vector) : 4)
NP :
(D r0;
(+'1')r= r =
r = r;
I
I (idemfactor) :
c
-i (reciprocal) :
(I' r);
(I)-1
4) xv:
Taibixv
V x (D:
My x aibi.
= 4'-1 4 = I; if
4) = Eaibi;
Every dyadic may be reduced to the sum of three dyads,
4) =al+bm+cn, in which either antecedents or consequents may be an arbitrary non-coplanar set. In this form, the principal invariants of 4) are:
4, = bxcmxn+cxanxl+axblxm = = axl+bxm+cxn,
Dyadic: Vector:
(D2;
*= (b x c) x (m x n) + cycl Scalar:
(P1 =
'P2 = `I'1 = (b x c) (m x n) + cycl, (P3 = [abc][lmn], 4PO
4) is complete if 'P3 0 0; only complete dyadics have reciprocals. 4) is planar if
p), = *' 4)" ((D ,F)-1 = j-1 4)-1; Ixv = vxl, (Ixv)c = -Ixv, I x(uxv) = vu -uv; (4,
(Da = (Da
1431 - 'P24) + tP1 (D2
-
`I' = (P3I 4)3 = 0
(the adjoint 4)a = (h,); (Hamilton-Cayley Equation).
LINEAR VECTOR FUNCTIONS
174
4) is symmetric if 4), = 4), antisymmetric if 4), = - 4). Every dyadic 4) can be expressed uniquely as the sum of a symmetric and antisymmetric dyadic; the latter is - 2I x The vector r1 is an invariant direction of - with multiplier Al if 4) r1 = A1r1. The multipliers of 4) satisfy the cubic,
.
503 - <02X +
P1A2
-
A3
= 0 (characteristic equation).
The invariant direction rl makes (4) - X11) r1 = 0. If 4) is symmetric its multipliers Ai are all real; and, if Al rl 1 r2. A symmetric 4) always may be reduced to the form,
A2i
`1) = A1u + A2JJ + X3kk;
the multipliers Ai need not be distinct. Any dyadic 4) can be reduced to the normal form,
(P = ai'i + $j'j + yk'k, in which a, 0, y all have the same sign; i, j, k and i', j', k' are two dextral sets of orthogonal unit vectors. When - is complete, the point transformation, r' = 4) r, changes lines into lines and planes into planes, preserves parallelism, and alters all volumes in the ratio 'P3/1. It preserves
lengths when and only when 4-1 = 4,; it is then a rotation if 03 = 1, a rotation followed by a reflection if p3 = -1. PROBLEMS
1. Prove that
abxc +bcxa +caxb = [abc)I.
2. For any dyadic 4> show that
U.4> v - v 4> u =4. uxv. 3. If 4> has the characteristic numbers X1, X2, X3 and the corresponding invariant directions r1i r2, r3, prove that 4)" (n an integer) has the characteristic numbers X', 02, 03 corresponding to the salve invariant directions. 4. If 4> has the characteristic numbers X1, X2, 1\3 for the directions r1r r2, r3, prove that 4>2 has the characteristic numbers X2X3, X3X1, X1X2 corresponding to the directions r2 x r3i r3 x r1, r1 x r2. [Cf. (70.2).]
5. If 4, = 4,, prove that *2 = 4>2,,, + = -4 and h = cir >k2 = v2, 4,3 = v3 6. Prove that the scalar invariants of 4' = 4>2 are 01 = 'c1 - 2'P2, 2 = 'P2 - 2'vl'P3, '3 = 'P3 [Use Prob. 3 and (71.4).]
PROBLEMS
7. Given the invariants c2, +, iants for (a) k4),
17 5
, v3 of 4), find the corresponding invar-
,
(c) 4-1.
(b) P x u,
8. Compute (P2, +, and the six scalar invariants of (69.13), namely,
+ ' +, + ' 4) ' +, + ' p2 - +,
'Pl, V2, 4'3;
for the dyadic 1
3
-2
2
0
4
-1
2
3
=
9. If D = I x e and e is a unit vector, prove that 4)2
= - (I - ee), 4'3 = -4), D4 = I - ee, 4)5 = (P.
10. Show that the symmetric part of (P, namely, 4' = 1(4) +4,), has the scalar invariants
h = c1, 42 = V2 - 4+'
11
4
413 = V3
11. Prove that ('t - ')2 = `1'2 *212. Prove that the first three scalar invariants of 4
1
and 4 4) are the
same. m
13. Ifs
n.
a;b,, 4' 1=1
c;d1,
J=1
we define the double-dot product 'F:4' as the scalar m
n
E(a`.c1)(b`-d,). 1-I j_1 Prove that
++4':
(a) (b)
(uv) :4' = u
4)
v,
4 ) :1
(c) If 4) is given in the nonion form (77.1), 'P11 + V12 +'P13 +'P2I + P22 + V23 +
'P21
+'P32
+P2
Hence 4, = 0 when and only when 4):4' = 0. 14. Any central quadric surface with its center at the origin has an equation of the form, (1)
r- 'F r = 1,
where 4, is a symmetric dyadic; for, if 4) is reduced to the standard form (72.4), we have
axe + $y2 + yz2 = 1.
This represents an ellipsoid, an hyperboloid of one sheet, or an hyperboloid of two sheets, according as a, j3. y include no, one, or two negative constants.
176
LINEAR VECTOR FUNCTIONS
If p 5*1 0, the equation (1) associates with every point (1, p) the plane p, 1), its polar plane. Prove that: (a) If the point (1, p) lies on the quadric surface, its polar plane (4' p, 1)
(4'
is tangent to the quadric at (1, p). [Find the points where the line r = p + Xe cuts the quadric when e 4' p = 0.]
(b) If the polar plane of (1, p) passes through (1, q), the polar plane of (1, q) passes through (1, p).
15. The diametral plane of any point (1, p) on the quadric surface (1) is the locus of the mid-points of all chords parallel to p. Show that the diametral plane of (1, p) is (4). p, 0). 16. If three points (1, u), (1, v), (1, w) on the ellipsoid r 4) r = 1, satisfy the equations, U. the position vectors u, v, w are said to form a conjugate set. Show that: (a) The vectors u, v, w and 4' - u, 4) v, 4) w form reciprocal sets.
(b) 4'-1 = uu + vv + ww. (c) For any conjugate set u, v, w, of the ellipsoid r (D r = 1, the sum u u + v- v + w- w and the product u x v- w are constant. 17. Verify by direct computation that the dyadic 4' in Problem 8 satisfies its Hamilton-Cayley Equation. 18. A rigid body with one point 0 fixed has the inertia dyadic K relative to 0 (§ 72, ex. 2). If the angular velocity of the body at any instant is w, show that its moment of momentum H (defined as f r x v dm) and kinetic energy T (defined as
if v
v dm) are given by
19. If the forces acting on the body of Problem 18 have the moment sum M about 0, the equation of motion is dH/dt = M. Show, from (56.3), that dH dt
=K dw dt
+wxK
w.
Let K = Aii + Bjj + Ckk (72.10) when referred to the principal axes of inertia (fixed in the body). Then if w = [WI, w2i w3], M = [Ml, M2, M3] referred to these axes, deduce Euler's Equations of Motion: Awl - (B - C)w2w3 = M1, Bm2 - (C - A )w3w1 = M2, (A - B)w1w2 = M3-
20. In Problem 19 show that d7'/dt = M w (the energy equation). 21. In Problem 18, suppose that the only forces acting on the body are its weight W and the reaction R at the support 0; then if the center of mass is at 0,
W passes through 0 and M = 0. (a) H is a constant vector. (b) T is a constant scalar.
Prove in turn that
PROBLEMS
17 7
(C) If w = OP, the locus of P in space is the invariable plane w H = 2T; :uul the locus of P in the body is the energy ellipsoid w K w = 27'. (d) The energy ellipsoid is always tangent to the invariable plane at P. (c) The body moves so that its energy ellipsoid rolls without slipping on
the invariable plane (Poinsot's Theorem). [See Brand's Vectorial Mechanics, § 219.]
22. The vectors of a dyadic 'F are fixed in a rigid body having the instantaneous angular velocity w, relative to a "fixed" frame. Show that, relative to this frame, d'F/dt=ca x4) -4) xw. 23. A rigid body revolving about its fixed point 0 has the angular velocity w = [wl, W2, w3] referred to fixed rectangular axes through O. If I1, 12, I3 are
the moments of inertia about the fixed axes x, y, z and 123, I31, 112 are the products of inertia for yz, zx, xy, show that
dIl/dt = 2(I12w3 - 113-2), ..
,
dI23/dt = (13 - 12)wl + 113W3 - 112-2, ... ,
[If K is the inertia dyadic of the body relative to 0, Il = i K i, 123 = -j K K. k, etc. See § 72, ex. 2.] 24. The axis of a homogeneous solid of revolution has the direction of the unit vector e. If its principal moments of inertia at the mass center G are A, A, C, show that the inertia dyadic at G is
KG =AI+(C-A)ee. Hence find the moments and products of inertia with respect to fixed rectangular axes x, y, z through G if e = [1, in, n].
25. If G is the mass center of a rigid body of mass in and r* = OG, show that the inertia dyadic at 0 is Ko = m(r* r* I - r*r*) + KG. Hence compare moments and products of inertia for parallel axes at 0 and G. 26. If X is arbitrary parameter < a2 but b2 or c2, the dyadic,
x211 + b2 J1
c2kk {
(a > b > c),
defines a one-parameter family of confocal quadric surfaces r % r = 1. These are ellipsoids if X < c2, hyperboloids of one sheet if c2 < \ < b2, hyperbolas of two sheets if b2 < X < a2. Prove that
(a) The central quadrics r T r = 1 and r O r = 1 are confocal when and only when I-' - 0-1 = kI. (b) Two confocal central quadrics of different species intersect at right angles.
[O - 4, = ke *.]
CHAPTER V DIFFERENTIAL INVARIANTS
82. Gradient of a Scalar. The points P of a certain region may be specified by giving their position vectors r = OP; and we shall on occasion refer to P as the "point r." A scalar, vector, or dyadic which is uniquely defined at every point P of a certain region is called a point function in this region and will be denoted by f (r). For example, the temperature and velocity of a fluid at the points of a three-dimensional region are scalar and vector point functions respectively.
A scalar point function f (r) is said to be continuous at a point P1 if to each positive number e, arbitrarily small, there corresponds a positive number S such that
If(r) - f(r1)
then, as r approaches r1 in any manner, limf(r) = f(r1). From a point P1 draw a ray in the direction of the unit vector,
e =icosa+jcos/3+kcosy. Along this ray r = r1 + se, where s denotes the distance P1P, and f (r) is a function of s. We now define of = lim
(1)
ds
f (r, + se) - f(r1)
s -.o
s
as the directional derivative of f (r) at P, in the direction e. If this limit exists on all rays issuing from P1i f (r) is said to be diferentiable at P1.
The rectangular coordinates of any point P on the ray r = r1 -}- se (s > 0) are
x = x1 + s cos a,
y = y1 -}- s cos 178
z = z1 -{- s cos y.
GRADIENT OF A SCALAR
82
179
If f(r) is given as a function f(x, y, z), df
of dx
of dy
of dz
ds
ax ds
ay ds
az ds
-- _ - - --F - - + -- --
(2)
=
af
ax
cos a +
of ay
cos i3 +
of az
cos y;
or, since
e i = cos a,
e
j = cos a,
\1
ax+iafy +kaz/
e - k = cosy,
= (3)
A
e
The vector in parenthesis is called the gradient of f (r) and is written grad f or (4)
Vf
49
19
ax+jay+kc3z
Vf is a vector point function; thus, when (3) is written as df (5)
ds
the direction enters only through the factor e.
The directional
derivative of a scalar function at a point P is the component of its gradient at P in the given direction. The gradient Vf at P effects a synthesis of all the directional derivatives of f at P. In effect, the vector of replaces the infinity of scalars df/ds. When P varies along a curve tangent to e at P1, f(r) is a func-
tion of the are s = P1P along the curve, and df/ds is still given by (2) ; for, at P1, dr/ds = e (44.1), and hence dx/ds = cos a,
dy/ds = cos a,
dz/ds = cos y.
Since df/ds, as defined by (1), does not depend upon any specific choice of coordinates, (5) shows that the gradient Vf has the same property. In fact, we determine Vf by giving the directional derivatives df /dsi in three non-coplanar directions e1. For, since ei Of = df/dsi are the covariant components of Vf, we have, from (24.2), (6)
of = el
df ds1
+ e2 df + e3 ds:2
df ds3
.
DIFFERENTIAL INVARIANTS
180
§ 82
We proceed to specify the length and direction of Vf independently of the coordinate system. The points for which f has a constant value lie on a level surface of f. In any direction e tangent to the level surface at P, df/ds = e - Vf = 0; hence Vf is normal to the level surface at P. If n is a unit vector normal to the level
surface and directed towards increasing values of f, n Vf > 0. Hence Vf has the direction of n, and its magnitude is the value of df,/ds in this direction. Writing this normal derivative df/dn, we have (7)
For example, if f = r, the distance OP from the origin, the level surfaces are spheres about 0 as center, and n = R, the unit radial vector; the normal derivative dr/dn = 1, and
Vr=R.
(8)
When f is a function f (x, y, z) of rectangular coordinates, Vf is given by (4). In particular, (9)
Vx=i,
Vy = j,
Vz = k.
If f is a function f(u, v, w) of variables which themselves are functions of x, y, z, we have df
of du
of dv
of dw
ds
au ds
av ds
aw ds
or, in view of (5),
\
Vu Of
au
+Vvaf +Vwa-f1 aw
av
As this equation holds for every e, (10)
Vf = Vu
of
+ Vv
Ou
af
av
+ Vw
Of .
aw
When f = f (u, v) or f = f (u), Of
Vf = Vu of + Vv , av au
Vf = Du
Of
au
,
§ 83
C1t.ADIENT OF A VECTOR
181
respectively; for example,
Vu' = nu"-1 Vu,
V(uv) = v Vu + it Vv,
1
V log u = - Vu. It
When f is constant, Vf = 0; conversely, Vf = 0 implies of/ax = of/ay = of/az = 0, and f is constant. If f (r) is differentiable in a region R, Vf is defined at all points of R. If moreover Vf is continuous in R, we say that f (r) is continuously differentiable in R. Example. Gradients in a Plane. The gradient of a point function f(x, y) in the xy-plane is
vf=afi+Of j. ay ax At any point P, Vf is normal to the level curve f(x, y) = c through P. For a function f(r, 0) of plane polar coordinates, Cf
of = or R
of-
+00r'
in the notation of § 44; for Vr = R and V0 = P/r (from (44.5) dB/ds = 1/r in the direction perpendicular to R). For a function f(rl, r2) of bipolar coordinates,
Vf =OrlR1+_R2, where R1, R2 are unit radial vectors from 01, 02 directed to the point in question.
The families of curves u = c, v = c cut at right angles when Cu Cv = 0. For example, the ellipses rl + r2 = c and hyperbolas r, - r2 = c cut orthogonally since (R1 + R2) (R1 - R2) = 0-
83. Gradient of a Vector. Let f(r) denote a vector point function in a certain region. It is said to be continuous at a point P1
if to each e > 0 there corresponds a S > 0 such that I f (r) - f(r1) I < e when
Ir
- r1 I< S.
If f(r) is given by its rectangular components,
f(r) = f1i + f2j + f3k,
f(r) is continuous at P1 when the three scalar point functions fi, f2, f3 are continuous there.
DIFFERENTIAL INVARIANTS
182
§ 83
Let P1P be a ray drawn from P1 in the direction of the unit vector,
e = icosa+jcos$+kcosy. Along the ray r = r1 + so (s > 0) and f(r) is a function of s. We now define df
-- = lim ds
f(rl + se) - f(r1)
s-.o
s
as the directional derivative of f (r) at P1 in the direction e. If this limit exists on all rays issuing from P1, f(r) is said to be differentiable at Pl. Evidently f (r) is differentiable if its rectangular components f, are differentiable. If f(r) is given as a function f(x, y, z) of rectangular coordinates, we have. just as in § 82, (2)
of dz
of dy
df
of dx
ds
axds+ayds+azds of
=-
cosa +
of ay
of
-
cosa+
cosy.
On replacing the cosines by e i, e j, e k, df (3)
of
ds =
of
of
e1-+ja +kaz y
,
a formula entirely analogous to (82.3). The dyadic in parenthesis is called the gradient of f(r) and is written grad f or (4)
of
of Vf = ofi-+ j - +k --
ax
ay
az
Vf is a dyadic point function; thus, when (3) is written (5)
df
the direction enters only through the prefactor e. The gradient Vf at P effects a synthesis of all the directional derivatives of f at P. In effect, the dyadic Vf replaces the infinity of vectors df/ds. Since df/ds, as defined by (1), does not depend upon any specific
choice of coordinates, (5) shows that Vf has the same property.
DIVERGENCE AND ROTATION
§ 84
In fact Vf is determined by giving the directional derivatives df/dsi in three non-coplanar directions ei: df
+ e2 df + e3 df
of _ e'
(6)
ds3
ds2
ds1
For, if we put df/dsi = ei Vf, the right member becomes (e'ei + e2e2 + e3e3) Vf = I I. Vf = Vf. When f is function of rectangular coordinates f (x, y, z), Vf is given by (4). More generally, for f(u, v, w) we have df ds
of du au ds
of dv
of dw
av ds
aw d s
of
C
au
+ VV
of
of
av
aw
for any e; and, from the definition of dyadic equality, of of of Vf = Vu-+Vv--+Vw-
(7)
au
aw
av
84. Divergence and Rotation. The first scalar invariant of the gradient Vf of a vector f is called the divergence of f and is written
The vector invariant of Vf is called the rotation or curl of f and is written V x f, rot f, or curl f.
In terms of rectangular coordinates, we therefore have the defining equations: (1)
Vf=gradf=iaxof - +j ayof- +k -azof
,
= i ax -of + j
,
(2)
V f = divf
3)
xf = rotf
of
of
ay
az
of of of = ix-+ jx-+kx__. ax ay az
If f is resolved into rectangular components,
f=ifl+jf2+kf3,
DIFFERENTIAL INVARIANTS
184
§ 84
we obtain from (1) its gradient of in nonion form with the matrix:
afl/ax af2/ax af3/ax of = afl/ay af2/ay af31ay
(4)
af2/az
8fl/az
af3/az
The first scalar and vector invariants of of now become
v
(5) (6)
axl+ a19h
+
f=
az
,
vxf=i --- -f-j(az---ax -}-k(ax-- ay)/ Gay f3
afl
af2
af3
af2
afl
az
The last expression is easily remembered when written in determinant form: j k (7)
Vxf =
a
a
a
ax
ay
az
fl
f2
f3
These expressions for divergence and rotation may be written down
at once, if we regard V - f and V x f as products of the vector operator del (or nabla), (8)
a a V =axai-+j-+k--, ay az
with the vector f = if, + jf2 + kf3. For the position vector r = xi + yj + zk, we have Or yr = i-+ j -Or+k --Or= ii+ jj +kk,
ax
ay
az
the idemfactor; hence (9)
When f =
Vr = I,
rot r = 0.
div r = 3,
the gradient of a scalar, (9 (P
fl=ax,
f2=ay
a4'
,
f3=az-
DIVERGENCE AND ROTATION
§ 84
185
Using these components, (5) and (6) gives a2
( 10 )
+
ax2
.
(11)
a2V
(P
v v'P
a2,p
+
ay 2
az2
V X VV = 0.
The differential operator V V or div grad is called the Laplacian and often is written a2
a2
__
v2
(12)
ax2
a2
+ ay2 +az2
When f = V x g, the rotation of a vector g, A
C193
092
ay
az
_a91
a93
ax '
az
f2
f3
a92
C191
ax
ay
In this case (5) and (6) give (13)
0,
(14)
Vx(Vxg) _ V(V.g) - O29.
Proof of (14) : From (6),
-
r a292
ay ax
x291
-
a291
aye
az2
= i j a tag, + a92 + lax
=i
ax
ay
a
ax
(O. g) - V291
a293
+
a931
az J
az ax Jt
+ .. .
- v291 I + .. .
+...
= V(V . g) - V2g.
Note that, if we regard V as an actual vector and expand V x (V x g), but keep the V's to the left of g, we obtain V(V g) (V V)g, the correct result. The proofs of (11), (13), and (14) depend upon changing the order of differentiations in mixed second derivatives; this is always valid when the derivatives in question are continuous.
DIFFERENTIAL INVARIANTS
186
§ 85
The foregoing differential relations of the second order also may be written (10)
div grad p = V2
(11)
rot grad cp = 0,
(13)
div rot g = 0,
(14)
rot rot g = grad div g - V2g.
Example. The velocity distribution in a rigid body is given by (54.4): vP = vo -}- w x OP = vo + w x r. Hence, from (3),
rotvp = rot(wxr) = ix(wxi) + =3w- w =2w; the rotation of the velocity of the particles of a rigid body at any instant is equal to twice its instantaneous angular velocity.
85. Differentiation Formulas. Starting from the defining equations (84.1), (84.2), (84.3), we now can deduce some useful identities. If X is a scalar point function: V(Xf) = (OX)f + XOf,
(1)
(2)
div (hf) = (VX) f + X div f,
(3)
rot (Xf) = (VX) x f + X rot f.
For the vector point function f x g: (4)
V (f x g) = (of) x g - (Vg) x f,
(5)
div (f x g) = g rot f - f rot g,
(6)
rot (f x g) = g Of - f Vg + f div g - g div f.
We give the proof of (6) :
rot(fxg) =1x
Caf
ag/
axxg+ fxax
+...
+. .. axg =g iax-- (i f -f.i g \ -g+(i ax! \ ax
= g - Vf - (div f)g + (div g)f - f . Vg.
GRADIENT OF A TENSOR
§ 86
187
For the scalar point function f g:
0(f - g) _ (of) g+ (7g) . f
(7)
+Ix
From (68.6), D, =
hence
(8)
(of), = Vf + I x rot f,
(9)
(Vf) - a = a - Vf + a x rot f.
Making use of (9), we also can write (7) as (10)
If '1 is a constant dyadic,
v(r
(11)
=I.4) =4);
-1))
(12)
div (r D) = cpl,
(13)
rot (r
4) _+.
Finally, if 4) is a constant symmetric dyadic, r.
(14)
From this result we can prove the THEOREM. A symmetric dyadic I transforms any vector r = OP
into a vector r' = 4) r normal to the real central quadric surface r 4) r = f1 * at the point P; and r' = 1/p where p is the distance from 0 to the tangent plane to the quadric at P. Proof.
From (72.4), we see that
ax2+(3y2+yz2
= f1
represents a central quadric (an ellipsoid if a, y have the same sign). From (14), r' has the direction of n, a unit normal to the quadric at P; hence
r'n= r' =
1/p. 86. Gradient of a Tensor. If f(r) is a tensor point function of 1/I
valence v (cf. § 80), the derivative of f(r) at Pl in the direction of the unit vector e is defined by (1)
df
-- = lim ds
s -o
f(rl + se) - f(rl) s
* The sign is chosen so that the quadric is real.
DIFFERENTIAL INVARIANTS
188
§ 86
If this limit exists for all rays drawn from P1, f(r) is said to be differentiable at P1. When referred to a constant basis, f(r) is differentiable when all its scalar components are differentiable. If f(r) is given as a function f(x, y, z) of rectangular coordinates, df
of dx
ds
Ox
of dy
of dz
ds+ay ds+azds
Since dx/ds = e - i, etc., we have, just as in § 83, df
(2)
ds
= e vf,
of of of of Vf=i-+j-+k-=i,.t ay ax az axr
(3)
Here Vf, the gradient of f (grad f) is tensor point function of valence v + 1; of effects a synthesis of all the directional derivatives (valence v) of the tensor f. Thus, if f is dyadic, the triadic of replaces the infinity of dyadics df/ds. of is independent of the choice of coordinates. At any point, of
is completely determined when df/dsi is given for three noncoplanar directions ei: df
of = e1df- + e2 - + e3 df
(4)
dsi '
ds2
ds1
for, if we put df/dsi = ej Vf, the right member becomes I of = Vf. When f(r) is a vector, we obtain from the dyadic of the invariants div f and rot f by putting dots and crosses, respectively, between the vectors of each term. Similarly, if f(r) is any tensor of valence v > 0, we obtain from of the further invariants,
=i -of +j -of +k ay
(5)
ax
(6 )
Vx
f = ix
of + j x of ax
ay
of
of
az
axr
+ k x of _ Oz
it
%
Of , axr
of valence v - 1 and v, respectively. In forming these invariants from Vf, the dots and crosses are inserted between the first and second vectors to the left-we dot and cross in the first position. t Here i, j, k; x, y, z are written il, i2, i3; xl, x2, x3, and we employ the sumnmation convention: a repeated index (as r) denotes summation over the index range 1, 2, 3. See § 145.
86
GRADIENT OF A TENSOR
1S9
For example consider the dyadic
f(r) = fiiii + fi2ij + ... = fstisit where s and t are summation indices. Then if D, denotes a/8x Vf = i,D,f = D,fst i,isit, a triadic;
V f = i, D,f = Dfst i,
is it
= (Difit + D2f2t + D3f3t)it, a vector; here i, is = 6,s, the Kronecker delta (23.2) ;
Vxf = i,>D,f = Drfsti,xisit = (D2f3t - D3f2t)iiit + (D3fit - Dif3t)i2it + (Dlf2t - D2f3t)i3it, a dyadic. Again, if X(r) is a variable scalar and 4) a constant dyadic, V(X4)) = (VX)4), a triadic; (7) V V. (a4)) = (VX) 4), a vector; (8) (9)
Ox (A4)) = (VX) x 4),
a dyadic.
For any tensor f(r) with continuous second derivatives we have identities which are generalizations of (10), (11), (13), (14) of § 84: (10)
V Vf = V2f,
(11)
V X Vf = 0;
and, if f(r) is not a scalar, (12) Vx
x f) _ V(V f) - V2f.
The proofs are straightforward applications of (3), (5) and (6): V Vf = i, D,(isDsf) = i, is DrDsf = DrDrf = V2f; V x Vf = it x D,(isDsf) = i, x is DrDsf = 0; V (V x f) = it Dr(ls x Dsf) = i, x is DrDsf = 0; V x (V x f) = i, x Dr(is x Dsf) = i, x (is x DrDsf) = is i, DrDsf - i, is DrDsf
= i8D5(i, Df) - D,D,f = V(V f) - V2f. Since i, x is = -is x i,, DrDsf = DsD,f, all non-zero terms in
V x of and V (V x f) cancel in pairs.
DIFFERENTIAL INVARIANTS
190
§ 87
87. Functional Dependence. THEOREM 1. A necessary and sufficient condition that two continuously differentiable functions u(x, y), v(x, y) satisfy identically a
functional relation f(u, v) = 0 is that their Jacobian vanish: (1)
a(u, v)
a(x, y)
=
au/ax au/ay I
av/a. av/ay
I
= 0 or Vu x Vv = 0.
Proof. The condition is necessary ; for, from f(u, v) = 0, we have
of
of
of
au
av
av
-Vu+--Vv = 0,
Vu. Vv = 0.
If it is constant, Vu = 0, and Vu x Vv = 0. If u is not constant, f (u, v) must contain v, of/ft is not identically zero, and again VuxVV=0. Conversely, suppose that Vu x Vv = 0. This relation is satisfied
if either u or v is constant; thus, if u = c, we may take f (u, v) = it - c. If u and v are not constant, Vv is parallel to Vu; hence Vv is normal to the curves u = c. Along these curves, dv,/ds = 0, and v is constant. In other words, a level curve it = a is also a level curve v = b; when u is given, v is determined, and v is a function of it.
A necessary and sufficient condition that two continuously differentiable functions u(x, y, z), v(x, y, z) satisfy identically a functional relation f (u, v) = 0 is that THEOREM 2.
(2)
Vu x Vv =
k
j
i
au/ax au/ay au./az av/ax
av/ay
= a.
av/az
The proof is essentially the same as in theorem 1.
Instead of
level curves of u and v, we now have level surfaces. THEOREM 3. A necessary and sufficient condition that three continuously differentiable functions u(x, y, z), v(x, y, z), w(x, y, z) sat-
isfy identically a functional relation f (u, v, w) = 0, is that their Jacobian vanish: (3)
a( u, v, w) a(x, y, z)
au/ax
= av/ax Ow/ax
au/ay
au/az
av/ay av/az aw/ay aw/az
= Vu x Vv Vw = 0.
CURVILINEAR COORDINATES
88
191
Proof. The condition is necessary for, from f (u, v, w) = 0, we have of of -of Vv+ -of Vw = O, 0. aw au av aw
-Vu+
If u and v alone satisfy a relation f (u, v) = 0, Vu x Vv = 0, and also Vu x Vv Vw = 0. If this is not the case, f (u, v, w) must contain w, of/aw is not identically zero, and again Vu x Vv Vw = 0. Conversely, suppose that Vu x Vv Vw = 0. If Vu x Vv = 0, f(u, v) = 0 from theorem 2. If Vu x Vv F- 0, consider the curve of intersection of the level surfaces u = a, v = b. Vu and Ov are normal to this curve; and, since Vu, Vv, Vw are coplanar, Vw is also normal to the curve. Therefore dw/ds = 0 and w = c along
the curve u = a, v = b; in other words, when u and v are given, w is determined: w is a function of u and v. 88. Curvilinear Coordinates. In a given region let (1)
u = u(x, y, z),
v = v(x, y, z),
w = w(x, y, z)
be three continuously differentiable functions whose Jacobian Vu x Vv Vw F4- 0 at all points. The functions, therefore, are not connected by a relation f (u, v, w) = 0. Since the Jacobian is continuous, its sign cannot change in the region; and, to be explicit, we shall suppose that (2)
This involves no loss in generality; for, if the Jacobian were nega-
tive, an interchange of v and w (for example) would make it positive.
Under the foregoing hypotheses a well-known theorem $ states that in the neighborhood of any point (x0, yo, zo) the equations (1) have a unique inverse (3)
x = x(u, v, w),
y = y(u, v, w),
z = z(u, v, w);
and that these functions are also continuously differentiable. At least in a suitably restricted region of uvw space, each set of values it, v, w yields, through equations (3), a unique set x, y, z. In this region the correspondence (x, y, z) - (u, v, w) effected by (1) and $ Cf. Sokolnikoff, Advanced Calculus, New York, 1939, p. 434.
DIFFERENTIAL INVARIANTS
192
488
(3) is one-to-one. Instead of specifying a point Po by the coordinates (xo, yo, zo), we may use instead the three numbers, uo = u(xo, yo, zo),
v0 = v(xo, yo, zo),
wo = w(xo, yo, zo)
When (uo, vo, wo) are given, Po is located at the point of intersection of the three coordinate surfaces, w(x, y, z) = wo. v(x, y, z) = vo, u(x, y, z) = uo, These, in general, will intersect in three curves, the coordinate curves, along which only one of the quantities u, v, w can vary. For this reason u, v, w are called curvilinear coordinates, in distinction to the rectangular coordinates x, y, z, for which the coordinate curves are straight lines.
If the position vector r is regarded as a function of x, y, z, Vr = I (84.9). But, if r is regarded as a function of u, v, w, ar au
ar
car
av
aw,
Vr = Vu-+ Vv-+ Vw-, from (83.7). We therefore have the fundamental relation, Vur,,
(4)
on writing ru = ar/au, etc. From the theorem of § 66, we now conclude that: The vector triples Vu, Vv, Vw and ru, r, rw form reciprocal sets. The vectors ru are tangent to the u curves, the coordinate curves along which v and w are constant. Thus at any point P(u, v, w), r, r, rw are tangent to the three coordinate curves meeting there; and, from (23.6), Vv Vw] = 7
(5)
Here [rurrw] is the Jacobian a(x, y, z)/a(u, v, w) of the inverse transformation (3); and, from (2),
J=
(6)
0.
Moreover, from the properties of reciprocal sets, (7)
Vu = rv
J
x Jrw,
Vv =
rJ
x
ru,
r = J Vw
Vw =
ru x
J
x Vv.
CURVILINEAR COORDINATES
§ 88
193
The volume of the parallelepiped whose edges are ru du, r dv, rw dw is called the element of volume dV; thus dV = [rurvrw] du dv dw = J du dv dw.
(9)
To find the gradient of a tensor f (u, v, w), we first compute (10)
df
of du
ds
au ds
of dw
of dv
+
av ds
+
aw ds
= e (Vufu + Cvfv+ Vwfw). But, from (86.2), df/ds = e Vf for every e; hence of = Vu fu + Vv fv + Vw fw.
(11)
Thus, in curvilinear coordinates the operator del becomes a
a
a
au
ov
aw
V = Vu-+ Vv -- + Vw-
(12)
From (11), we obtain V f and V x f by dotting and crossing: (13)
Vxf = Vuxfu+ Cvxfv+ VwXfw.
(14)
For purposes of computation, it is usually more convenient to eliminate Vu, Vv, Vw from these formulas by use of equations (7). We thus obtain 1
J
{rvxrwfu + rwxrufv + ruxrvfw}
,
and corresponding equations for V f and V x f. In view of the identity, (rv x rw)u + (rw x ru) v + (ru x rv) w = 0, these also may be written (16)
Of =
1
J
{ (rv x r f) u
+ (rw x ru f) v + (ru x rv f)w },
1
(17)
(18)
=J Vxf=
J
1 [(rv x rw) X flu + [(rw x ru) x f]v + [(ru x rv) x f]w }.
194
DIFFERENTIAL INVARIANTS
§ 89
When the triple products in (18) are expanded, the brace becomes (rw r71
f).u - (r rw f)u. + (r.u rw f) - (rw ru f)v
+ (rv ru f) w - (ru r f) w
= rw(rv f)u - rv(rw f), + ru(rw f) - rw(ru f)
+
1,
f)w - ru(rv f)w.
With this value for the brace, (18) may be written compactly in determinant form: (19)
Vxf
ru
r
rw
1
a
a
a
J
au
av
aw
This equation reduces to (84.7) when ru, r,,, rw are replaced by rx = i, rv = j, rz = k. 89. Orthogonal Coordinates. The curvilinear coordinates u, v, w are said to be orthogonal if the coordinate curves (along which one coordinate only can vary) cut at right angles. Since ru, r, r.v, are tangent to the coordinate curves, the coordinates are orthogonal when and only when (1)
We choose the notation so that (2)
ru = Ua,
r = Vb,
rw = Wc,
where a, b, c are a dextral set of orthogonal unit vectors and U, V, W are all positive; then [abc] = 1, and (3)
J = [rurvrv,] = UVW.
The set of vectors reciprocal to ru, rv, rw are the gradients of the coordinates: b c a . Vw = (4) Vv = , Vu = , U
V
W
From the general formulas (11), (17), (19) of the preceding article we obtain Vf, V f and V x f in orthogonal curvilinear coordinates:
ORTHOGONAL COORDINATES
§ 89
(5)
Vf =
1
U
1
(6)
1
afu +
V f =UVW
V
bf +
1
W
195
cf a
i18
(VW a f) -F a (WU b f) Lou
+_ (7)
Vxf =
We
Ua
Vb
a
a
8
av
aw
1
UVW au
If we put f = Vg in (6), we obtain the Laplacian V2g = V. Vg in orthogonal coordinates,
la
1
(8)
o2g
a WU a UV \I I/V\) au \ U gu + av \ V gv/ + aw \ W gw/ 1 \1
1/
1/
UVTV
When the curvilinear coordinates are given by the equations, x = x(u, v, w),
y = y(zt, v, w),
z = z(u, v, w),
we may compute the (positive) functions U, V, TV from equations of the type, (9)
z2 U = I ru _ xui + yuJ + zuk = 1/x2z u + yu + u
The element of volume (88.9) is now (10)
dV = UVW du dv dw.
Example 1. Cylindrical Coordinates. The point P(x, y, z) projects into the point Q(x, y, 0) in the xy-plane. If p,p are polar coordinates of Q in the xyplane, u = p, v = gyp, uw = z are called the cylindrical coordinates of P (Fig. 89a). (10)
They are related to rectangular coordinates by the equations: x = p cos (p,
y = p sin gyp,
z = Z.
From r = ix + j y + kz, we have rp = [cos v, sin gyp, 0],
r, = [ -p sin rp, p cos gyp, 0],
r. = [0, 0, 11.
Since these vectors are mutually perpendicular and [rrr=] > 0, cylindrical
DIFFERENTIAL INVARIANTS
196
§ 89
coordinates form an orthogonal system which is dextral in the order P, c, Z. Moreover, (11)
U=
rn
= 1,
V = rw I = p, J = UVW = P.
(12)
W
The level surfaces p = a, v = b, z = c are cylinders about the z-axis, planes through the z-axis, and planes perpendicular to the z-axis. The coordinate
FIG. 89a
curves for p are rays perpendicular to the z-axis; for p, horizontal circles centered on the z-axis; for z, lines parallel to the z-axis. The element of volume
dV =Pdpdpdz. From (8), the Laplacian is (13)
V2g =
- -8 (Pgp) + 1
1
P aP
Pa
= log P, Pgv = 1; hence log p satisfies Laplace's Equation V2g = 0. Such a function is called harmonic. g
Example 2. Spherical Coordinates. The spherical coordinates of a point P(x, y, z) are its distance r = OP from the origin, the angle 0 between OP and the z-axis, and the dihedral angle ' between the xz-plane and the plane
z OP (Fig. 89b). They are related to rectangular coordinates by the equations: (14) y = r sin 0 sin gyp, x = r sin 0 cos p, z = r cos 0.
From r = ix + jy + its, we have rr = [sin 8 cos p, sin B sin gyp, cos 0],
rs = [r cosBcosgyp,rcos©singyp, -rsin0],
r, =
rsin0cos p, 0].
TOTAL DIFFERENTIAL
90
197
since these vectors are mutually perpendicular and [rrror,] > 0, spherical coordinates form an orthogonal system which is dextral in the order r, 0, Moreover, (15)
U=jrr1,
Iiir,rsin o;
V=Irol=r,
J = UVIV = r2 sin 0.
(16)
The level surfaces r = a, 0 = b, = c are spheres about 0, cones about the z-axis with vertex at 0, and planes through the z-axis. The coordinate curves
Fia. 89b
for r are rays from the origin; for B, vertical circles centered at the origin; for gyp, horizontal circles centered on the z-axis. The element of volume dV = 7-2sin0drdodrp. From (8), we now have (17)
V2g
r2 sin 0
isin
0
or
(r2gr) +
aB
(sin 0 go) +
1
sin 0
9
0`°
If g = 1 /r, r2gr = -1; hence 1 /r satisfies Laplace's Equation V2g = 0.
O. Total Differential. In passing from the point P to P, the ---4
--->
position vector r = OP changes by the increment Ar = PP. Then, if f(r) is any differentiable tensor point function, the total differential of f (r) is defined by the equation : (1)
In particular, when f = r, Vf = I, and dr = Or. (2) The differential of the position vector is the same as the increment The defining equation (1) therefore may be written (3)
df=drVf.
DIFFERENTIAL INVARIANTS
198
§91
If f is a function f (u, v, w) of curvilinear coordinates,
Vf = Vufu+VvfV+Vwfw, and, from (3), since dr Vu = du, etc., df = fu du + f dv + fw dw. (4)
91. Irrotational Vectors. A vector function f (r) is said to be irrotational in a region R when rot f = 0 in R. If V(r) is a scalar function with continuous second derivatives, its gradient VV is irrotational; for, from (84.10),
rot grad ,p = 0.
(1)
Conversely, if rot f = 0, and f is continuously differentiable in R, we shall show that f may be expressed as the gradient of a scalar V(r). Using rectangular coordinates, let f = f1i + f2j + f3k; then, if rot f = 0, the three determinants of the matrix, aax a/ay
(2)
f2
(fl,
a/az\ f3
all vanish (84.7). Under this condition we shall determine a scalar function ,p(r), so that Vv = f, that is, (3)
= f1(x, y, z),
ax
ay
= f2(x, y, z),
az
= f3 (x, y, z)
Let (xo, yo, zo) be an arbitrary point of R. On integrating app/ax = f1 with respect to x and regarding y and z as constant parameters, we have x
P=
(4)
J
f1(x, y, z) dx + «(y, z),
o
_ -dx+-=
where «(x, y) is a function of x and y as yet undetermined. Hence x afl a« /' af2 a« app ay
,,
-_J 49Z
1o
ay
ay
-dx+-, or ay
Ixo ax
f
a
f2(x, y, z) = f2(x, y, z) - f2(xo, y, z) + xafl
a«
- dx + - _ az az
xaf3
-
- dx + ax az
,
ay
or
a«
f3 (x, y, z) = f3 (x, y, z) - f3 (xo, y, z) +
-
IRROTATIONAL VECTORS
§ 91
199
Instead of three equations (3) for gp(x, y, z), we now have two equations, a«
a« (5)
ay
= f2(xo, y, z),
az
= f3(xo, y, z)
for «(x, y), with the condition aft/az = af3/ay. The problem has been reduced from three to two dimensions. On integrating a«/ay = f2 with respect to y, we have
f f v
a=
(6)
f2(xo, y, z) dy + #(z),
bo
where $(z) is a function of z to be determined.
a«fY af2 dy az
vo
az
d(3
+
dz
Y a af3
-
y
dy +
-
,
dz
Hence
or
f3(xo, y, z) = f3(xo, y, z) - f3(xo, yo, z) +
d
da (7)
dz
= f3 (xo, y0, z)
We now have one equation (7) to determine $(z); and J f3 (xo, yo, z) dz.
(8)
o
f
On collecting the results (4), (6), and (8), we have finally (9)
'=
ffi
(x, y, z) dx +
z
Y
f2 (xo, y, z) dy + f f3 (xo, yo, z) dz.
Yo
D irect substitution shows that Vip = f; and, if
is a second func-
tion for which V4, = f, V(¢ - cp) = 0, and 4 - (p is a constant. Thus (9) gives the solution of equations (3), determined uniquely except for an additive constant. There are evidently five other forms for ,p which may be obtained from (9) by permuting 1, 2, 3 and making the corresponding permutation on x, y, z; for example, (10)
c = ff2(x
z
y , z) dy + ffs(x, yo, z) dz + ffi(x, yo, zo) dx.
Y
Moreover, xo, yo, zo may be given any values that do not make the integrands infinite.
DIFFERENTIAL INVARIANTS
200
§91
In mathematical physics, it is customary to express an irrotational vector f as the negative of a gradient. Thus, if 4, we have
f = -V ';
(11)
¢ is then called the scalar potential of f. Example 1. When
f = 2xzi + 2yz2j + (x2 + 2y2z - 1)k,
we find that rot f = 0; hence f = V. With xo = yo = zo = 0, we have, from (9),
=f X2xz dx +f 2yz2 dy 0
0
- f Zdz = x2z + y2z2 - z. 0
The differential equation,
Example 2. Exact Equation. (i)
is said to be exact when f dr = dp, the differential of a scalar. If (i) is exact,
we have, from (90.3), dr f = dr vv; and, since dr is arbitrary, f = When f is continuously differentiable, f = V implies rot f = 0, and conversely.
Therefore we may state the
THEOREM. If f is a continuously differentiable vector, in order that f dr = 0 be exact it is necessary and sufficient that rot f - 0. Thus, in view of ex. 1, the equation,
2xz dx + 2yz2 dy + (x2 + 2y2z - 1) dz = 0,
is exact and may be put in the form dp = 0. Its general solution is that is,
= c,
x2z + y2z2 - z = C. If (i) is not exact, a scalar X which makes Xf dr = 0 exact is called an integrating factor of (i). The preceding theorem shows that X must satisfy the equation, (ii)
rotaf = 0 or VX x f + X rot f = 0.
On multiplying (ii) by f , we have
rot f = 0. Hence, when f dr = 0 admits an integrating factor, f rot f = 0. (iii)
f
Con-
versely, when f is continuously differentiable, the condition (iii) ensures the existence of an integrating factor. We shall prove this in § 105. For this reason f rot f = 0 is called the integrability condition for f dr = 0. When f dr = 0 is integrable, ?f = V (p. Then the vector field f(r), being parallel to Dip, is everywhere normal to the level surfaces rp = c. The condition (iii) therefore implies the existence of a one-parameter family of surfaces everywhere normal to f. Thus the geometrical content of the condition f rot f = 0 is that the vector field f (r) is surface-normal.
SOLENOIDAL VECTORS
§ 92
201
92. Solenoidal Vectors. A vector function f(r) is said to be solenoidal in a region R when div f = 0 in R. If g(r) is a vector function whose components have continuous second derivatives, rot g is solenoidal, for, from (84.13), div rot g = 0.
(1)
Conversely, if div f = 0, we shall show that f may be expressed as the rotation of a vector g. Using rectangular coordinates, let f = f1i + f2j + f3k, and (84.5) af3 aft div f aftax= -+-+= 0. az ay
(2)
We now shall determine a vector g = g1i + g2j + g3k, so that i (3)
f = rot g = a/ax 91
j
k
c1/ay
a/az
92
93
We first find a particular solution of (3) for which g3 = 0; then (3) is equivalent to the scalar equations: (4)
a91
4992
fl = -
az
f2 =
,
az
f3 =
,
4992
ax
_
1991
ay
The first two equations of (4) are satisfied when z
(5)
92 = - f f i (x, y, z) dz,
91 = f f2 (x; y, z) dz + 49(x, y) ;
w z
Zo
in these integrations x and y are regarded as constant parameters, and a(x, y) is a function as yet undetermined. In order that these functions satisfy the third equation of (4),
- Zaf + J dz - as = f3, ay (ax ay/ .
aJ2
Zp
or, in view of (2),
I.
Zafs az
as
dz - - = f3 (x, y, z). ay
When we perform the integration, this reduces to -f3 (x, y, z0) -
as ay
= 0,
DIFFERENTIAL IN VARIANTS
202
§ 92
an equation which is satisfied by taking v
a (X, y) = -
f3 (x, y, zo) dy. Yo
Hence the vector g, whose components are z
(6)
v
gi = ff2(x, y, z) dz - ff3(x, y, zo) dy, a
z
92 = - f fi (x, y, z) dz,
93 = 0,
zo
is a particular solution of our problem. If G is any other solution, rot G = rot g = f, and hence
rot (G - g) = 0.
(7)
But any irrotational vector may be expressed as the gradient of a scalar p (§ 91); hence the general solution of (7) is G - g = o,p, where ,p(r) is an arbitrary twice-differentiable scalar. When div f = 0, the general solution of rot g = f is therefore g + Vip; its rectangular components are obtained by adding c /cx, 8,p/cy, 8,p/8z to the components of g given in (6). In mathematical physics, the solenoidal vector f = rot g is said to'be derived from the vector potential g. Example 1. When
f = x(z - y)i + y(x - z)j + z(y - x)k, we find that div f = 0. Therefore f = rot g; if we take zo - yo = 0, the particular solution (6) is 91 = f
Zy(x - z) dz =xyz -
Zyz2,
0
z
g2
f x(z - y) dz =
- Zxz2 + xyz,
ga = 0.
0
Example 2. If u and v are continuously differentiable scalars, the vector Vu x Vv is solenoidal, for, from (85.3), Vu x Vv = rot (uVv).
Conversely, we shall show in § 104 that a continuously differentiable solenoidal vector always can be expressed in the form Vu x Vv.
SURFACES
§ 93
203
93. Surfaces. A surface is represented in parametric form by the equations: x = x(u, v),
(1)
y = y(u, v),
z = z(u, v).
We assume that the three functions of the surface coordinates u, v
are continuous and have continuous first partial derivatives, a requirement briefly expressed by saying that the functions are continuously differentiable. In order that equations (1) represent a proper surface, we must exclude the two cases: (i) the functions x.(u, v), y(u, v), z(u, v) are constants: equations
(1) then represent a point; (ii) these functions are expressible as functions of a single variable t = t(u, v); equations (1) then represent a curve. In case (i) all the elements of the Jacobian matrix,
(2)
vanish. In case (ii), the rows of the matrix are dx/dt, dy/dt, dz/dt
multiplied by at/au and at/av, respectively, and all of its tworowed determinants vanish identically. We exclude these cases by requiring that the matrix (2) be, in general, of rank two; then, at least one of its two-rowed determinants, (3)
A =
yu
zu
yv
zv
B =
zu
xu
Zv
xv
C
_
I xu
Yu
xv
,
yv
is not identically zero.
Even when the matrix in general is of rank two, the three determinants A, B, C, all may vanish for certain points u, v. Such points are called singular, in contrast to the regular points, where at least one determinant is not zero. If we introduce the position vector to the surface, (4)
r = xi + yj + zk = r(u, v),
(5)
and the condition for a regular point may be written (6)
ru x r, /- O.
DIFFERENTIAL INVARIANTS
204
§94
If we introduce new parameters u, v by means of the equations,
u = u(u, v),
(7)
v = v(u, v),
we shall require that the Jacobian of this transformation, a (u, v)
J =
(8)
FK 0.
a(u, i)
Then equations (7) may be solved for u and v, yielding
u = u(u, v),
(9)
v = 17(u, v),
and the correspondence between u, v and u, v will be one to one. Since
(lp)
_
X
rU
au au
av) x (ru au -av au
ru - - rU -
av\
r -- J = J ru
X
ry,
aU
the requirement J 0 0 makes ru x rv 0 a consequence of (6). Thus a point which is regular with respect to the parameters u, v is also regular with respect to u, D.
94. First Fundamental Form. A curve on the surface r(u, v) may be obtained by setting u and v equal to functions of a single variable t:
u = u(t),
(1)
v = v(t).
A tangent vector along the curve (1) is given by dr dt
and (2)
The are s along the curve is defined as in (43.4) :
fro'
t t dt;
and
ds
dt
=
Since du = ft dt, dv = v dt by definition, on multiplying (2) by dt2, we have (3)
ds2 = ru ru du2 + 2ru r du dv + r r, dv2.
This first fundamental quadratic form usually is written (4)
ds2 = E du2 + 2F du dv + Gdv2,
FIRST FUNDAMENTAL FORM
§ 94
where
E = ru
(5)
F = ru
ru,
G = r r,,.
r,,,
Moreover, from (20.1), (ru x rv)
(ru x rv) _
ru - ru
ru rv
hence
ru x rv 12 = EG - F2
is positive at every regular point. The curves v = const (u-curves) and u = const (v-curves) are called the parametric curves on the surface. For these curves dv = 0 and du = 0, respectively, and the corresponding elements of arc are (7)
dsl = 1/E du,
ds2 = N/G dv.
Since the vectors ru, rv are tangent to the u-curves and v-curves, respectively, the parametric curves will cut at right angles when and only when .(8)
The vector ru x rv is normal to the surface. The parallelogram formed by the vectors ru du, rv dv, tangent to the parametric curves and of length dsl, ds2i has the vector area (§ 16), (9)
dS = ru x rv du dv.
We shall call this the vector element of area; the scalar element of area is (10)
dS = I ru x rv I du dv =
E
du dv.
The unit normal n to the surface will be chosen as
n=
ruxrt E
=
ruxrv
H
then dS = n dS. At every regular point the vectors ru, r, n form a dextral set; for (12)
[rurvn] = H = VEG
> 0.
DIFFERENTIAL INVAItIANTS
206
§ 95
95. Surface Gradients. Let f(u, v) be a differentiable function,
scalar, vector, or dyadic, which is defined at the points of the surface r = r(u, v). We shall compute the derivative of f(u, v) with respect to the are s along a surface curve u = u(t), v = v(t). Along this curve, df
du
ds
fu ds
(1)
dv
+ fv ds '
where du/ds = is/s, dv/ds = v/s, and, from (93.4),
-
ds
ds = V L ic2 + 2Ficv + Gv2. dt
(2)
If we apply (1) to the position vector r(u, v), we obtain the unit tangent vector e to the curve: du dv e=r,,,+rv ds ds
(3)
Let a, b, c denote the set reciprocal to ru, r,,, n; then, since [rur ,n] = H,
_ rvxn H
b
_ nxru II
,
c =
ruxrv
= n.
H Now from (3) a e = du/ds, b e = dv/ds; hence (1) may be (4)
a
,
written df
ds
We now define afu + bf as the surface gradient of f and denote it by V8f t or Grad f: (5)
V8f = Grad f = afu + b f,,.
Grad f has the characteristic properties: (6)
e- Grad f =
(7)
ru Grad f =
df
ds
,
of -au ,
n Grad f = 0;
r Grad f =
-
of av
t In surface geometry we shall write of for the surface gradient.
96
SURFACE DIVERGENCE AND ROTATION
207
If f (r) is a tensor of valence v, Grad f is of valence v + 1. At any point (u, v) of the surface, Grad f is in effect a synthesis of all the values of df/ds for surface curves through this point. Since Grad f depends only on the point.(u, v), df/ds is the same for all
surface curves having the unit tangent vector e at this point. At any point (u, v) of the surface Grad f is completely determined when df/dsi is given for two directions ei in the tangent plane at (u, v). If the set reciprocal to el, e2, n is e', e2, n,*
Grad f = el
(8)
df + e2 df dsi
ds2
for, by virtue of equations (6), the right member of (8) may be written
(elel + e2e2 + nn) Grad f = I Grad f where I is the idemfactor (§ 66). In particular if el and e2 = n x el
are perpendicular unit vectors, el = el, e2 = e2. In view of (8), Grad f is independent of the coordinates x, y, z and of the surface parameters u, v.
96. Surface Divergence and Rotation. If f(r) is a tensor point function defined over the surface r = r(u, v), its surface gradient,
V8f=afu+bf,,,
(1)
has the invariants, (2)
V,-f = axfu +bxf,,.
(3)
We recall that the set a, b, n is reciprocal to ru, rv, n. When f (r) is a vector, V8f is a planar dyadic; then the scalar and vector invariants of oaf are called the surface divergence and surface rotation and are written (4)
V8
f = Div f,
V8 x f = Rot f.
The second of V8f is the linear dyadic, 1
(V3f)2 = axbfuxfv = Hnfuxfv; the second scalar of V8f is therefore n f,, x f v/H. * Since el x e2 = An, e3 = n, we have e3 = Xn/X = n.
DIFFERENTIAL INVAItIANTS
208
196
For the position vector r to the surface, we have V8r = ar.u + br v =
(6)
Div r = 2,
(7)
I - nn,
Rot r = 0.
For the unit normal n,
Rot n = 0.
(8)
To prove this, put f = n, a = r x n/H, b = n x ru/H in (3); then H Rot n = (rz, x n) x nu + (n x ru) x n,,.
= Now, from n n = 1, we have, on differentiation with respect to u and v,
n nu = n nv = 0; and, from ru n = r n = 0, ru nv = rv nu = -n ruv; (10) (9)
hence H Rot n = 0. From the defining equations (1), (2), (3), we may derive various expansion formulas. Thus, if X is a scalar, f a tensor, V3(af) = (V3X)f + XV3f,
(11)
(Af) _ (V8A) f + XV8 f,
(12)
V.
(13)
V8 x (Xf) _ (V8X) x f + XV3 x f.
If g and f are vectors, (14)
Div (g x f) _ (Rot g)
f - g Rot f;
and, in particular,
Div (n x f) = -n Rot f.
(15)
Proof of (14) :
Div(gxf) (Rot g)
f - g Rot f.
SPATIAL AND SURFACE INVARIANTS
§ 97
209
97. Spatial and Surface Invariants. If f(r) is a tensor function of valence v defined over a 3-dimensional region including the
surface r = r(u, v), its spatial gradient at a point (u, v) of the surface may be computed from (86.4). If el, e2 are vectors in the
tangent plane at (u, v) and e3 = n, then, at all points of the surface,
df df df df Vf = e'-+e2-+n= V8f +n ds2 do do dsl
(1)
;
here the set e', e2, n is reciprocal to el, e2, n and df/dn denotes the derivative of f in the direction of n. Moreover, if v > 0, df
df
df
df
ds.l
ds2
do
do
(2)
V
(3)
df df df df Vxf = elx-+e2x-+nx-= V8xf +nx-. ds2 do dsl do
From (1), we see that the tensors of valence v + 1,
nx Vf = nx V j,
(4)
are the same over the surface; therefore both may be written Thus n x Vf may be computed solely from the values of f on the surface. The same is true of the invariants obtained from n x Vf by dotting and crossing in the first position. Since
n x Vf.
nxVf = nxafu+nxbft,,
(5)
from (96.1), this process yields (n x a) fu
+
(nxb) f = n
(a x fu
+ b x f,) = n V. x f,
n208f -nV8 f. Here n V8f means that n is dotted into the second vector from the left in each term of V8f. These invariants remain the same when computed from the corresponding spatial quantities. Thus we verify at once, from (3), (6)
and, from (1) and (2), (7)
nVf-nV.f=n2V8f-nV8f.
DIFFERENTIAL INVARIANTS
210
§ 97
When f is a vector, these invariants become (6)'
n rot f = n Rot f,
(7)'
(grad f) n - n div f = (Grad f) n - n Div f.
We now express n x Vf and its invariants (6) and (7) in terms of the surface coordinates u, v. Since a, b, n and ru, r,,, n are reciprocal sets, we have, from (5), 1
-
nx Vf
(8)
1
(r,fu - ruf
II
{(rvf)u - (ruf),, },
since ruv = r,,,, if these derivatives are continuous. Dotting and crossing now yields
n Vxf
(9)
1
II II{(rvxf)u-(ruxf)v}.
(10)
Since ru, rv are tangent to the surface, (9) shows that: If a vector f is everywhere normal to the surface, n rot f 0. With f = V3g in (9) we obtain the important identity,
=0;
(11) for then
ru f = gu f = gv, When f = pq, a dyad, we have, from (9),
(95.7).
rv
1
n V x (pq) =
II
Pq) u - (ru pq)
(r,, {
1
1
=
H
P)u - (ru
{ (r,,
P) v } q +
H
p (rvqu - rugv),
that is,
n V x (pq) = (n rot p)q + p n x Vq. For future use we compute the invariant (10) when f = ng and
(12)
g is an arbitrary tensor: n2 Vf
-nVf=
H
1(rvxng)u - (ruxng)v} 1
= agu + bg,, +
H
{(rvxn)u - (ruxn)v} g.
§ 98
SUMMARY: DIFFERENTIAL INVARIANTS
211
Now
Grad g = agu + bg n Grad n = (an. +
(95.5),
2
n=0
(96.9) ;
hence, on putting f = n in (10), we have
-n Div n =
1
H
{ (r x n)u - (ru x n) v
}
.
Thus, with f = ng, (13)
n? Vf - n V f = Grad g - (Div n) ng.
98. Summary : Differential Invariants. Let f (r) be a tensor point function of valence v; its derivative at the point P and in the direction of the unit vector e is defined as df
-- = lim ds
f (r + se) - f (r) (s > 0).
s
s -- o
If df/dsi denote the directional derivatives corresponding to three non-coplanar unit vectors ei, the gradient of f, namely,
grad f = Vf = e'
ds 1
+ e2
df
+ e3
2
has the property,
,
63
df
-- =e V.
ds
Vf, a tensor of valence v + 1, thus gives a synthesis of all the directional derivatives of f at point r. If the vectors ei are i, j, k,
Vf = if., + jfv + kfz
(fz = of/ax, etc.).
From Vf (valence v + 1) we derive the invariants V f (valence v - 1) and V x f (valence v) by dotting and crossing in the first position. When f is a vector (v = 1),
V f = div f, the divergence of f, V x f = rot f,
the rotation of f.
When r is a function r(u, v, w) of curvilinear coordinates Vf = Vu fu + Vv fv + Vw f.W,
Vr = Vu ru + Vv r + Vw rw = I.
DIFFERENTIAL INVARIANTS
212
The sets Vu, Vv, Vw and ru,
r',,,
ru, are reciprocal.
§ 98
With J =
we have
J Vf = (r x ru, f)u + cycl,
J V f = (r, x ru, f)u + cycl,
ru
J V x f = ((r x ru,) x f)u + cycl =
r
ru,
a/au a/av a/aw
ru, ru, are mutually orthogonal, the coordinates u, v, w are called orthogonal. Then if a, b, c denote a dextral set of or-
thogonal unit vectors, ru = Ua, r = Vb, ru, = Wc, and J = UVW in the preceding formulas. For any tensor f (r),
V Vf = V2f,
V (V f) = 0,
V. Vf = 0, Vx (Vxf) = V(V _ f) -
V2f.
The operator V V. V is called the Laplacian.
A vector f is called irrotational if rot f = 0, solenoidal if div f = 0. An irrotational vector f can be expressed as the gradient of a scalar (f = Dip) ; a solenoidal vector f can be expressed as the rotation of a vector (f = rot g). For the surface r = r(u, v), the fundamental quadratic form is ds2 = E due + 2F du dv + G dv2, where
At a regular point,
E>0,
and the unit surface normal is defined as n = ru x The derivative of a tensor f (r) along any surface curve tangent to the unit vector e at the point (u, v) is df
ds
the set a, b, n is reciprocal to ru, r, n. The surface gradient,
V8f = Grad f = a fu + b f,,,
PROBLEMS
213
has the properties,
e. Grad f =
df
n Grad f = 0;
,
ds
df
Grad f = - ,
r.u
r, Grad f =
au
df av
From V3f we derive the invariants,
Oaxf = axfu+bxfv,
O3 f =
by dotting and crossing in the first position. When f is a vector, V8
f = Div f, the surface divergence of f,
Og x f = Rot f,
the surface rotation of f.
The tensor n x of and its dot and cross invariants,
n?of are not altered when V is replaced by V8. They may be computed
solely from the values f(r) assumes on the surface.
Thus, if
H= n x of =
1
H
{
(ru.f) v } ;
and the invariants follow by dotting and crossing between r,, r and f. PROBLEMS
1. If R = r/r is the unit radial vector, prove that div R = 2/r, rot R = 0. 2. For any scalar function f (r) of r alone prove that
V2f(r) = frr + 2fr/r. If v2f (r) = 0, show that f = A/r + B. S. If a is a constant vector, prove that
grad (a. r) = a, div (a x r) = 0, rot (a x r) = 2a. 4. If a is a constant vector, prove that grad (a - f)
div (a x f) = -a rot f, rot(axf) Specialize these results when f = r.
DIFFERENTIAL INVARIANTS
214
5. For any vector point function f prove that 6. If Sp and ¢ are scalar point functions, prove that
div (Vp x v) = 0 7. If e is a unit vector, prove that div (e r)e = 1,
rot (e r)e = 0;
div (e x r) x e = 2,
rot (exr) xe = 0.
8. If a is a constant vector, prove that V(r x a) = I x a. 9. Show that Laplace's Equation V2f = 0 in cylindrical and spherical coordinates is and
1 a2f
1 of
a2f
a2f
+ p (9p + p2
a22
a/ af\ r{ -- sin0-J+
sine 0 5;i
apt
a2(rf) are
1
1
sin 0 a0
a0
a2f
= 0,
=0.
10. Prove that 1
1
- (3RR - I)
vC T =
where R is a unit radial vector. 11. If f is a vector point function, prove that V
(Vf), = grad div f;
V x (Of), _ (V rot f),.
12. If Vf is antisymmetric, prove that rot f is constant and that the dyadic Vf itself is constant. 13. If X is a scalar point function, prove that
V(,I) = VXI,
V V. (XI) = VX,
V x (XI) = VX x I.
14. If f is a vector point function, prove that V
v x (I x f) = (Vf), - I div f.
(I x f) = rot f,
15. For the dyad fg prove that V V. (fg) = (div f)g + f
V x (fg) _ (rot f)g - f x Vg.
Vg,
In particular, if r is the position vector, V V. (rr) = 4r,
16. If
V x (rr) _ -I x r.
is a constant dyadic, prove that div 4 . r = 91;
rot e r =
-+ (§ 69).
17. If the scalar function f(p, gyp) in plane polar coordinates is harmonic, prove that f(p-', tip) is also harmonic. 18. If the scalar function f(r, 0, Sp) in spherical coordinates is harmonic, prove that r 'f (r ', 0, gyp) is harmonic.
PROBLEMS
215
19. If f, g, h are scalar point functions, prove that V2(fg) = g02f + 2Vf Vg + fV2g;
V2(fgh) = gh V2f + hf V2g + fg v2h + 2fvg vh + 2gvh of + 2hvf vg. 20. If u, v, w are orthogona' coordinates and f(u), g(v), h(w) are scalar functions of a single variable, show that
V2(fgh) = fgh { f + egg + hh 21. If a is a constant vector and f = ar", prove that
of = nrs-2ra,
e'-f = n(n + 1)ri-2a.
22. If f = r"r, prove that
of = r"I + nrn-err,
V2f = n(n + 3)rn-2r.
23. If f = r"r, find a scalar function p such that f = vcp. 24. Prove that rot f = 0, and find V so that f = vp: (a)
f = yzi + zxj + xyk;
(b)
f = 2xyi + (x2 + log z)j + y/z k;
f = c r, cD a constant symmetric dyadic.
(c)
25. If a is a constant vector and f = r" a x r, prove that
div f = 0,
rot f = (n + 2)rna - nrn-2(a r)r.
26. Find a vector g such that rot g = a x r (cf. § 92). 27. Prove that
a r
= - rot --
.
r'
28. The position vectors from 0 to the fixed points P1, P2 are r1, r2.
(a) If f = (r - r1) x (r - r2), show that
divf =0, (b) Prove that
rotf =2(r2 -r1).
V(r-rl) (r-r2) =2r-r1-r2.
29. If f(u, v) is a vector function defined over the surface r = r(u, 1'), prove
that
H Div f =n (ruxf -r, xfu).
30. If the vector f(u, v) is always normal to the surface r = r(u, v), prove that Rot f is tangent to the surface or zero. 31. If u, v, w are curvilinear coordinates, prove the operational identities: a a a 'le = ruxrw-+ru,xru-+ruxrv-; av aw au [rurvru7
a
a (ruxr,,)xV =rv -- ruav-; au
a ru.V =_.
au
CHAPTER VI INTEGRAL TRANSFORMATIONS
99. Green's Theorem in the Plane. Let R be a region of the xj-plane bounded by a simple closed curve C which consists of a finite number of smooth arcs. Then, if P(x, y), Q(x, y) are continuous functions with continuous first partial derivatives, (1)
J
I1
CIQ
x
- y) dx dy =
f
(P dx F Q dy),
where the circuit integral on the right is taken in the positive sense; then a person making the circuit C will always have the region R to his left.
Fra. 99a
Consider first a region R in which boundary C is cut by every line parallel to the x- or y-axis in two points at most (Fig. 99a). Then, if a horizontal line cuts C in the points (x1, y), (x2, y),
f f aQ dx dy =
f
a
{Q(x2, Y) - Q(xi, y)) dy
a fQ(X2,
= =
y) dy + fQ(xiy) dy
fQ(x, y) dy. 216
GREEN'S THEOREM IN THE PLANE
§ 99
217
And, if a vertical line cuts C in the points (x, yl), (x, y2), b
If
ay
dy dx = f I P(x, Y2) - P(x, Y1)) dx
fb(
yl) dx -
J
P(x, y2) dx
6
= - fP(x, y) dx. On subtracting this equation from the preceding, we get (1). We now can extend this formula to more general regions that may be divided into a finite number of subregions which have the property that a line parallel to the x- or y-axis cuts their boundary
Frc. 99b
Fia. 99c
in at most two points. For in each subregion formula (1) is valid, and, when these equations are added for all the subregions, the surface integrals add up to the integral over the entire region; but the line integrals over the internal boundaries cancel, since each is traversed twice, but in opposite directions, .leaving only the line integral over the external bounding traversed in a positive direction (Fig. 99b). The boundary of R may even consist of two or more closed curves: thus in Fig. 99c the region R is interior to Cl but exterior to C2; we may now make a cut between Cl and C2 and traverse the entire boundary of R in the positive sense as shown by the arrows.
The line integrals over the cut cancel, for it is traversed twice and in opposite directions; and the resultant line integral over C consists of a counterclockwise circuit of Cl and a clockwise circuit of Cam.
INTEGRAL TRANSFORMATIONS
218
§ 100
Although Green's Theorem is commonly stated for scalar func-
tions P, Q, it is evidently true when P and Q are tensors with continuous first partial derivatives; for, when P and Q are referred to a constant basis, the theorem holds for each scalar component. 100. Reduction of Surface to Line Integrals. A surface is said to be bilateral if it is possible to distinguish one of its sides from
the other. Not all surfaces are bilateral. The simplest unilateral surface is the Mobius strip; this b may be materialized by taking a a,
a
b'
rectangular strip of paper, giving it one twist, and pasting the ends ab and a'b' together (Fig. 100a). This surface has but one side; if we move a point P along its median line and make a complete FIG. 100a
circuit of the strip it will arrive at the point P directly underneath. Since we can travel on a continuous path from one side of the Mobius Strip to the opposite, these sides cannot be distinguished. This is not the case with a spherical surface, which has an inside and an outside, or any portion S of the surface botinded by a simple closed curve C; for we cannot pass from one side of S to the other without crossing C. Let S be a portion of a bilateral surface r = r(u, v) bounded by' a simple closed curve C which consists of a finite number of smooth arcs. The surface itself is assumed to consist of a finite number
of parts over which the unit normal n is continuous. The positive sense on C is such that a person, erect in the direction of n, will have S to the left on making the circuit C. If such an oriented surface is continuously deformed into a portion of the xy-plane bounded by a curve C', n becomes k (the unit vector in the direction +z) and the positive circuit on C' forms with n = k a righthanded screw.
Let f (r) be a tensor point function (scalar, vector, dyadic, etc.) whose scalar components have continuous derivatives over S. Such a tensor is said to be continuously differentiable. When the foregoing conditions on S and C are fulfilled, we then have the BASic THEOREM I. If f(r) is a continuously differentiable tensor point function over S, the surface integral of n x Of over S is equal to the integral of Tf taken about its boundary C in the positive sense: (1)
fn
Vf dS =
JTf ds.
§ 100
REDUCTION OF SURFACE TO LINE INTEGRALS
Proof.
219
Since
n x Vf =
1
II
dS = H du dv,
(ref),},
{
fn x Vf dS =
{ (r ,f).u -
du dv,
IS, s'
where S' is the region of the uv-plane which contains all parameter
values u, v corresponding to points of S.
We now may apply
U
FIG. 100b
Green's Theorem in the plane to the last integral, letting u and v play the roles of x and y; thus
fs, { (rtf)u -
du dv =
f(rtf du +
dv),
where the circuit integral is taken about the curve C', which forms the boundary of S', in the positive sense, that is, in the direction of a rotation of the positive u-axis into the positive v-axis. If we
regard u and v as the functions of the are s on the curve C, it becomes
I
/
du dv ruf-+rf-)ds =ic Tfds, ds
ds
\
ru
du ds
+
ry
dv
dr
ds
A
= T,
the positive unit tangent along C. Formula (1) thus is established.
INTEGRAL TRANSFORMATIONS
220
§100
If we introduce the vector elements of surface and of arc,
dS = n dS,
dr = T ds,
the basic theorem for transforming surface to line integrals becomes
fdS X Vf =fdr f,
(2)
in which dS and dr must be written as prefactors.
If f is a vector, dyadic, etc., we may obtain the other integral transformations from (1) by placing a dot or a cross between the first two vectors in each term (dyad, triad, etc.) of the integrands; for both of these operations are distributive with respect to addition and therefore may be carried out under the integral sign. This process applied to the tensor n x Vf gives n V X f and n z Vf - nV f; we thus obtain the important formulas: (3)
ffl.VxfdsfT.fds,
(4)
f (n z Vf - n V f) dS = if T X f ds.
When f is a vector function, (3) becomes Stokes' Theorem:
fn.rotfds=fT.fds
(3)'
,
a result of first importance in differential geometry, hydrodynamics and electrodynamics.
If S is a closed surface which is divided into two parts S1, S2 by the curve C, we may choose n as the unit external normal; then, from (1),
fnxVfds = fTlf ds, c
fn,< Vf dS = fT2f ds,
where T2 = - T1, since the positive sense on C regarded as bounding S1 is reversed when regarded as bounding S2. On adding these integrals, we find that the integral of n X Vf over the entire surface is zero. We indicate this by the notation, (5)
fnx Of dS = 0,
§ 101
ALTERNATIVE FORM OF TRANSFORMATION
321
which denotes an integral over a closed surface. On taking dot and cross invariants in (5), we have also (6)
fn.VxfdS=0,
(7)
f(n'4Vf_nV.f)dS=0.
Example 1. Put f = r in (4) and (7); then, since grad r = I, div r = 3,
J'ndS = 2irxdr,
fnds = 0.
The last result, which states that the vector area of a closed surface is zero, generalizes the polyhedron theorem of § 17.
Example 2. When div f = 0, we have seen that f = rot g (§ 92); hence
f Thus, if f = Vu x Vv, div f = 0, and we may take g as uVv or -vVu; hence
Jn.
Vu x Vv dS =
it dv = -
Jv du = 2 f(u dv - v du).
In particular, when u = x, v = y, n = k, this formula expresses an area in the
xy-plane as a circuit integral 2
J(x dy - y dx) over its boundary.
101. Alternative Form of Transformation. If we replace f by ng in (100.4), where g is any continuously differentiable tensor, we have, from (97.13), (1)
f { Grad g - (Div n) ng } dS =
fT x n g is.
This integral transformation is quite as general as that given by basic theorem I. For, if we replace g by of and then cross in the first position, the integrand on the left becomes a x (nuf + n f.u) + b X
n fv) _ -n x Grad f,
since Rot n = 0; whereas, on the right,
(Txn) Xnf = -Tf. We thus retrieve the basic transformation (100.1).
INTEGRAL TRANSFORMATIONS
222
§ 102
We now shall express (1) in slightly different notation. On the right the vector T x n = m is the unit normal to C tangent to S and pointing outward. If we write f instead of g and J = - Div n (J is not the Jacobian of § 88), (1) now becomes
f
(2)
(vsf + J nf) dS =
fm f ds.
On dotting and crossing in the first position of the tensor integrands, we obtain the additional formulas:
f(vs.f+Jn.f)ds =
(3)
f
f
f f= Div n is called the mean curvature of the surface (§ 131).
Surfaces for which J = 0 are called minimal surfaces; for such surfaces the integrands on the left reduce to the first term. On the plane, n = k, a constant vector, and Div n = 0; then J = 0 in (2), (3), and (4). When f is a vector, these equations become (if we write dA for dS) : (5)
fGrad f dA =
(6)
fDivfdA
(7)
JRotfdA = fm f ds, -
i(m f ds,
where m is the external unit normal to the closed plane curve forming the boundary. We may deduce (6) from (7) by replacing
fbykxf;for
rf
m x (k x f) = km f. 102. Line Integrals. When f(r) is a vector, consider the line Rot (k x f) = k Div f,
integral, (1)
J'f.dr= o
to
dr dt
dt,
ta ken over a curve C: r = r(t) from the point P0 to P. The value of this integral depends on the curve C and the end points Po and
LINE INTEGRALS
§ 102
223
P, but not on the parameter t. For, if we make the change of parameter t = t(T), t
Jcu
f -dr dt= dt
pT TO
f---drf - dr. dt dr ar r
dr dt
dr
,,
Let us consider under what conditions the line integral (1) is independent of the path C-that is, when its value is the same for all paths from Po to P. We shall call a closed curve reducible in a region R if it can be
shrunk continuously to a point without passing outside of R. Thus in the region between two concentric spheres all closed curves are reducible. However in the region composed of the points within a torus, all curves that encircle the axis of the torus are irreducible. A region in which all closed curves are reducible is called simply connected. Thus the region between two concentric spheres is simply connected, whereas the interior of a torus is not. If f is continuously differentiable, and rot f = 0 in a region R, the line integral of f around any reducible closed curve in R is zero;
for we can span a surface S over R that lies entirely within R (shrinking the curve to a point generates such a surface), and then, by Stokes' Theorem,
ff . dr = fn. rotfdS=0. We may now prove the THEOREM.
If f is a continuously differentiable vector, and rot f
= 0 in a simply connected region, the line integral
J'f
dr between
any two points of the region is the same for all paths in the region joining these points. Proof. If C1 and C2 are any two curves POAP, POBP joining PO to P, the line integral of f around the circuit POAPBPO is zero; for all closed curves in a simply connected region are reducible. Hence
J
f dr=0.
fPBP, oA P
P
Po
INTEGRAL TRANSFORMATIONS
224
§ 103
Under the conditions of this theorem, the line integral (1) from a fixed point Po to a variable point P of the region depends only upon P; in other words the integral defines a scalar point function,
p(r) = Jf dr.
(2)
o
Let us compute d
-
p(r) =
f
+8e
f
8
0
since dr = e ds along the ray from r to r + se. By the law of the mean for integrals,
ff.eds = sef(r+e),
(0 < 9 < s),
and hence d(P
= lim p(r ds s -o
+ se) - p(r) s
= e f(r).
Since e - Vp = e f for any e, (3)
Thus an irrotational vector f may be expressed as the gradient of scalar gyp, given by (2). The determination of p given in § 91 is the line integral of f taken over a step path from (xo, yo, zo) to (x, y, z). Thus in (91.9) the integrals, in reverse order, are taken over the straight segments from P0(xo, Yo, zo) - (x0, Yo, z) - (x0, y, z) - P(x, y, z).
There are six such step paths; for the first segment can be chosen in three ways and the second in two. In mathematical physics it is customary to express an irrotational vector as the negative of a gradient: (4)
f = - V4,;
then ¢(r) =
jTO
dr
is called the scalar potential of f.
103. Line Integrals on a Surface. We next consider line and circuit integrals over curves restricted to lie on a given surface, or within a certain region S of the surface. A region S of a surface
LINE INTEGRALS ON A SURFACE
§ 103
225
is called simply connected if all closed surface curves in S are reducible-that is, can be shrunk continuously to a point without passing outside of S. In the plane, for example, the region within a circle is simply connected, but the region between two concentric circles is not. If the vector f is continuously differentiable, the circuit integral
if
dr about any reducible curve of S is zero when rot f is every-
where tangential to the surface; for
f
When S is a portion of the level surface u(x, y, z) = c of a scalar function, Vu on the surface is parallel to the surface normal and
the condition n rot f = 0 may be written (1)
0.
Just as in § 101, we may now prove the
If f is a continuously differentiable vector, and n rot f = 0 in a simply connected portion S of a surface, the line THEOREM 1.
integral
ff
dr between any two points of S is the same for all
surface curves joining these points.
Under the conditions of this theorem the line integral, r
p(r) =
(2)
f dr, ra
is the same over all surface curves from Po to P and therefore defines a scalar point function V(r) in S. Along any definite curve
r = r(s) issuing from P0 (s = 0) c is a function of the are s, dr
s
p(s)
8
=J'ff Tds,
where T is the unit tangent vector; hence at any point P d(p/ds = f T. Since the curve may be varied so that T assumes any direction e at P, we have the relation, (3)
d s
INTEGRAL TRANSFORMATIONS
226
§ 104
for all unit vectors e in the tangent plane at P. Now if e1, e2 are two such vectors, and el, e2, n the set reciprocal to e1, e2, n, we have, from (95.8), Grad cp = el
d + e2 d
ds2
ds1
= ele1 f + e2e2 f.
Since f = (ele1 + e2e2 + nn) f, (4) Grad 'P = ft, the projection of f on the tangent plane at P. We have thus proved the
THEOREM 2.
If f is a continuously differentiable vector, and
n rot f = 0 in a simply connected region of a surface, the tangential projection of f on the surface is
ft = Grad 9 where 9 =
f dr. ro
104. Field Lines of a Vector. When f(r) is a continuously differentiable vector function, the curves tangent to f(r) at all their points are called the field lines of f. If r = r(t) is a field line, dr/dt is tangent to the curve (§ 41) and consequently a multiple of f. The field lines have therefore the differential equation: fxdr = 0. (1)
If f = f1i + f2j + f3k, (1) is equivalent to the system, (2)
dxdydz fl
f2
f3
Any integral u(x, y, z) = a of this system represents a surface locus of field lines. For, from du = dr Vu = 0 and (1), we have (3)
and, since Vu is normal to the surface u = a, the vector f at any point of the surface is in the tangent plane. If v(x, y, z) = b is a second integral of (2), we have also (4)
If v is independent of u, Vu x Vv 0 0 (§ 87, theorem 1). From (3) and (4) we conclude that f is parallel to Vu x Vv, and hence (5)
Vu x Vv = Xf.
FIELD LINES OF A VECTOR,
104
227
But the curve in which the surfaces u = a, v = b intersect is everywhere tangent to Vu X Vv. Hence we have the THEOREM 1.
If u = a and v = b are independent integrals of the
system f x dr = 0, the surfaces they represent intersect in the field lines of f.
From (3) and (4), we see that u and v are independent solutions of the partial-differential equation: aw
(6)
aw
caw
f- Vw =flax + f2 a + f3 az = 0ax
In view of (5), this equation is equivalent to (7)
Vu X Vv Vw = 0.
From § 87, theorem 3, (7) is satisfied when and only when u, v and w are connected by a functional relation. Hence we have THEOREM 2. The general solution of the partial differential equation f Vw = 0 is w = cp(u, v), where cp is an arbitrary function and u = a, v = b are independent integrals of f X dr = 0.
When the vector field f(r) is solenoidal, div f = 0. In § 92 we found that we could express f as rot g. We now deduce another form for f which gives at once its field lines. THEOREM 3. A solenoidal vector f which is continuously differentiable can be expressed in the form: f = Vu X Vv.
(8)
Proof.
If we assume the relation (8), we have f Vu = 0,
Vv = 0; thus both u and v are solutions of f Vw = 0. If we choose for u some integral of the system (2), f Vu = 0. Now,
f
if dr is a differential on the level surface u = a., we have, from (8), (9)
f X dr = Vv du - Vu dv = - Vu dv.
In the function u(x, y, z), at least one variable is actually present. If z is present, au/az = uz is not identically zero; hence, on multiplying (9) by k , we have k f X dr = -uz dv and (10)
v=
-
kxf dr ,
uz
INTEGRAL TRANSFORMATIONS
228
§ 104
the integral being taken over a curve on the surface u = a. This integral is independent of the path; for, from (85.3) and (85.6), rot
kxf uz
1\
1
= O - J X (kXf) + -rot (kXf) uz
U,
Of
2(Vuz)X(kxf) -k
uZ
uZ
-(f V i )k + (k
Vu,)f
f2
2
ui
Vu -rot
kXf uZ
uz
fZ - cu
=
-1 a _-- (f - Vu) = 0. uZ az
uZ
If u(x, y, z) contains x or y, we may replace k and uz in (10) by i and ux or j and uv. With the values of it and v thus obtained, we now have, from theorem 2 of § 103, Vu X Vv = Vu x Grad v = - V it X
(k x f)c
-1
= - Vu x (k X f) = f,
uZ
uZ
where Grad v refers to the surface u = a, and (k X f)t is a tangential projection upon it. Moreover f - Vv = 0. The field lines of the vector Vu X Vv are the curves in which the surfaces u = a, v = b intersect. Example 1. The field lines of the vector,
f = xzi + yzj + xyk, have the differential equations: dx
(i)
dy
dz
From these, we obtain the equations,
ydx-xdy=0,
y dx + x dy - 2z dz = 0,
which have the integrals:
y/x = a,
xy - z2 = b.
These one-parameter families of surfaces intersect in the two-parameter family of field lines.
FIELD LINES OF A VECTOR
§ 104
229
When one integral, as y/x = a, is known, we may find a second by obtaining the field lines on the surface y/x = a. These must satisfy the equation obtained from (i) by putting y = ax, namely, ax dx = z dz. Its integral ax2 - z2 = b gives the field lines on the surface y/x = a. On replacing a by y/x, we obtain a second family of surfaces xy - z2 = b which intersects the first family y/x = a in the field lines. Example 2. The field lines of the vector,
f = xi + 2x 2j + (y + z)k, have the differential equations: dx x
(ii)
dy
dz
2x2
y+z
From dy = 2x dx, we obtain the family of surfaces,
y-x2=a. As no other integrable combination is evident, we put y = a + x2 in (ii); then dx x
_
dz dz z a a+x2+z' or dx-x=x-I-x.
This linear equation, with the integrating factor 11x, has the solution,
z = -a+x2+bx. A second family of integral surfaces is therefore
z +Y x
Example 3.
- 2x = b.
The vector,
f = x(y - z)i + y(z - x)j + z(x - y)k, is solenoidal; for div f = 0. In order to express f as Vu x Vv we choose for u an integral of the system, dx
x(y - z)
_
dy
y(z - x)
_
dz
z(x - y)
Since the sum of the denominators is zero,
dx+dy+dz=0,
x+y+z=a;
hence we may take u = x + y + Z.
We now use (10) to compute v. On the surface u = a, z = a - x - y, and
-kxf dr = -x(y -z)dy+y(z - x)dx = y(a-2x-y)dx+x(a-x-2y)dy.
INTEGRAL TRANSFORMATIONS
230
§ 105
Since this is an exact differential, the method of § 91 gives v=
f(aY_2xY_Ydx=axY_?J_x?J
= xy(a - x - y) = xyz. With u = x + y + z, v = xyz, we may readily verify that f = Vu x vv.
105. Pfaff 's Problem. Since rot f is solenoidal, we have from theorem 3 of § 104, (1)
rot f = Vu x Vv = rot (u Vv),
rot (f - uVv) = 0; hence (§ 91) f - uVv is the gradient Vw of a scalar, and (2)
f = Vw + uvv.
When f is a continuously differentiable vector, we may find three scalars u, v, w so that (2) holds good. The determination of these scalar functions is known as Pfaff's Problem. We proceed to give a simple and direct solution. Assuming the truth of (2), we at once deduce (1); hence (3)
Vu rot f = 0,
Vv
rot f = 0,
so that both It and v satisfy the same partial differential equation : V. rot f = 0. Let v = a be some integral of the system dr x rot f
= 0; then Vv rot f = 0. Now on the surface v = a we have, from (2),
this integral, taken over a curve of the surface v = a, is independent of the path since Vv rot f = 0. On substituting the functions v and w thus obtained in (2), this equation uniquely determines u.
We now must show that, with It, v, w thus determined, Vw + uVv = f. Let Grad w and n denote the gradient and unit normal on a surface of the family v = const; then (103.4) dw
Vw=Grad w+ndn
dw
ft+ndn
PFAFF'S PROBLEM
§ 105
231
and the projection of Vw + uVv tangential to the surface is ft. Moreover u was chosen to make the normal projections of Vw + uVv and f the same, that is, dw n-+uVv=f-ft. do
From (1) and (2), a (u, v, w)
f- rot f =
C )(X' y, z)
hence f rot f = 0 implies that u., v, w are functionally dependent (§ 87, theorem 3). In this case, (4)
f dr = dw -i-- u(v, w) dv = 0
is an ordinary differential equation which admits solutions under general conditions-as when u(v, w) and au/caw are continuous in a region R. * Hence the condition
f-rot f=0
(5)
shown in § 91, ex. 2, to be necessary for the integrability of f dr = 0, is also sufficient.
When (5) is fulfilled, let (p(w, v) = C be the general solution of Then
(4).
d(p =
amp
aw
dw + - dv = 0 av
must yield (4) upon division by app/8w. In other words, X = a,p/8w is an integrating factor of (4) :
Then Xf = VV, and f is everywhere normal to the surfaces p = const. The field lines of f are then the orthogonal trajectories of these surfaces.
A family of curves is said to form a field in a region R if just one curve of the family passes through every point of R. The unit vectors T, N) B along the curves are then vector point functions in R; and the first Frenet Formula dT/ds = KN can be written (6)
T- VT = KN.
* See Agnew, Differential Equations, New York, 1942, p. 310, et seq.
INTEGRAL TRANSFORMATIONS
232
§ 105
Now, from (85.7) and (85.9),
TT= T = -KN. For a surface-normal field
T rot T = 0,
(T x rot T) X T = rot T,
and, from (7), we have rot T I = K.
rot T = KB,
(8)
The Darboux vector of a surface-normal field of curves is therefore 8 = rT + rot T. Example. Find u, v, w so that
f = [2yz, zx, 3xy] = Vw + u Vv. Solution.
rot f = [2x, -y, -z]; since the system, dx
_ dy _
dz
2x
-y
-z
has the solution z/y = a, we take v = z/y.
On the surface v = a, z = ay;
hence w
=ff
dr = f (2yz dx + zx dy + 3xy dz) = f (2ay2 dx + 4axy dy) 2axy2 = 2xyz.
Now u is given by [2yz, zx, 3xy] = [2yz, 2zx, 2xy] + u[0, -z/y2, 1/y], whence u = xy2. Thus our solution is
u = xy2,
v = z/y,
w = 2xyz.
Since f U. rot f = 0, u, v, and w must be functionally related; in fact w = 2uv. The total differential equation, (i)
2yzdx+zxdy+3xydz =0,
is thus equivalent to
dw + udv = 3udv + 2vdu = 0, and has the integral u9 = C, or x2yz3 = C.
106 REDUCTION OF VOLUME TO SURFACE INTEGRALS
233
106. Reduction of Volume to Surface Integrals. Let S be a closed surface and V the volume it encloses. Then S has two sides, an inside and an outside, and at all regular points of S we have a definite unit normal n directed towards the outside. We shall suppose that S consists of a finite number of parts over which n is continuous.
Consider, first, a surface S which is cut in at most two points by a line parallel to the z-axis; denote them by (x, y, z1) and (x, y, z2), where zl < z2. Then S has a lower portion Sl conz
*_y
FIG. 106
silting of all the points (x, y, z1), and an upper portion S2 consisting of the points (x, y, z2). We suppose also that the points for which zl = z2 form a closed curve separating Sl from S2 (Fig. 106). The equations of Sl and S2 may be taken as z = zl(x, y), z = z2(x, y). Now let f (x, y, z) be a tensor function of valence v whose scalar
components have continuous partial derivatives throughout V. Then, if the volume integral of of/8z over V is written as a triple integral with the element of volume dV = dx dy dz, we may effect a first integration with respect to z,
(i) ff
az dx dy dz
=
fff
( x, y, z2) dx dy
-fff
(x, y, z1) dx dy,
whe re the double integrals are taken over the common projection A of Sl and S2 on the xy-plane.
234
INTEGRAL TRANSFORMATIONS
§106
If we regard x, y as the parameters u, v on the surfaces S1 and S2, the vector element of surface is ry x ry dx dy (94.9). Now from
r = ix + jy + kz(x, y),
rz = i + kzz,
ry = j -}- kzy,
Over S2 (z = z2) this vector has the direction of the external nor-
mal n; but over Sl (z = zl) it has the direction of the internal normal -n. Hence, if we denote the vector element of area in the direction of the external normal by dS = n dS,
n dS = ± (k - izz - jzy) dx dy,
k . n dS = fdx dy,
where the plus sign applies to S2 and the minus to S1. The two integrals over A now may be combined into a single integral over 8, so that we may write (1)
j/,f
dx dy dz
=f
I
This formula is also valid when S is bounded laterally by a part of a cylinder parallel to the z-axis and separating Sl from S2. For (i) holds as before; and, in (1), k n = 0 over the cylinder, so that it contributes nothing to the integral over S.
We now may remove the condition that S is cut in only two points by a line parallel to the z-axis. For, if we divide V into parts bounded by surfaces which do satisfy this condition and apply formula (1) to each point and add the results, the volume integrals will combine to the left member of (1); the surface integrals over the boundaries between the parts cancel (for each appears twice but with opposed values of n), whereas the remaining surface integrals combine to the right member df (1). Finally we may extend (1) to regions bounded by two or more closed surfaces, that is, regions with cavities in them, by this same process of subdivision. Additional surfaces must be introduced so that the parts of V are all bounded by a single closed surface, and the surface integrals over these will cancel in pairs as before. When x, y, z form a dextral system of axes, the same is true of y, z, x and z, x, y. Hence, if in (1) we make cyclic interchanges in x, y, z, we obtain the corresponding formulas: (2)
fff a dx dy dz =
y, z) i n dS,
§ 106 REDUCTION OF VOLUME TO SURFACE INTEGRALS (3)
fff ay dx dy dz =
ff
235
f(x, y, z) j n dS.
If we insert the prefactors k, i, j in the integrands of (1), (2), and (3), respectively, add the resulting equations, and note that
iof +jof +kof ax
az ay we obtain finally
= Vf,
f of d V
(4)
=in f dS,
using J( to denote integration over a closed surface. The integrands are tensors of valence v + 1. We have thus proved BASic THEOREM II. If f(r) is a continuously differentiable tensor
point function over the region V bounded by a closed surface S, whose
unit external normal n is sectionally continuous, then the integral of of over the volume V is equal to the integral of of over the surface S.
If f(r) is a vector or tensor of higher valence, we may obtain from (5) other integral transformations by placing a dot or a cross between the first two vectors in each term of the integrands; for both of these operations are distributive with respect to addition and therefore may be carried out under the integral sign. We thus obtain the important formulas: (5)
fv.fdv=Jn.fds,
(6)
fv
f
When f is a vector, (5) is known as the divergence theorem; the integral
n f dS then is called the normal flux of f through the
surface. Example. If rot f = 0 in a simply connected origin, f =
ffdV =f
102), and
= fncodS.
Thus the volume integral of an irrotational vector can be expressed a surface integral over the boundary.
236
INTEGRAL TRANSFORMATIONS
§ 107
For example, the center of mass of a homogeneous body of mass m is fixed by the position vector,
r*=mfrdm=yfrdV; for m = pV, where p denotes the constant density. Since rot r = 0, we have, from (102.2), r
r=V,
e
where P= U
Hence
Vr* = i 0 nr2 dS.
107. Solid Angle. The rays from a point 0 through the points of a closed curve generate a cone; and the surface of a unit sphere about 0 intercepted by this cone is called the solid angle & of the cone.
The reciprocal of the distance r = OP is a harmonic function (§ 89, ex. 2) ; hence
div grad
1
= 0, and grad
r
R = r r2 1
is a solenoidal vector. Let us now apply the divergence theorem to the vector f = R/r2 in the region interior to a cone of solid angle 0 and limited exter-
nally by a surface S, internally by a small sphere a about 0 of radius a. Within this region f is continuously differentiable, div f = 0 and
fn
f dS = 0. The external normal n = - R over
0', and over the conical surface n R = 0; hence a R (1)
r2
J
1
dS
Sa
=,a2 dS = a2 = S2,
where Sa is the area cut from or by the cone. Note that the ratio Sa/a2 is independent of the radius and may be computed with
a=1.
If S is a closed surface, we have
nR (2)
r
2
dS =
14r,
l0,
0 inside of S 0 outside of S.
When 0 is outside of S, f is continuously differentiable throughout its interior, and the foregoing result is immediate. In this case
GREEN'S IDENTITIES
§ 108
237
the elements of solid angle, d1t =
(3)
2
dS,
r
corresponding to the same ray cancel in pairs. 108. Green's Identities. We now apply the divergence theorem (106.5) to the vector f = p0j,, where
div (,p0') _ V p V4, + p div V ' n
(85.2),
d¢/dn,
where the normal derivative d¢/dn is in the direction of the external normal to the bounding surface. Hence, on writing the operator div grad as V2, we have V p V. V4, dV +
(1)
f
g' V24,dV
do dS.
In case ' is not defined outside of S, we replace d ,/dn by the negative of the derivative along the internal normal -n. Formula (1) is known as Green's first identity.
If both cp and 4, have continuous derivatives of the first and second orders, we may interchange (p and 4, in (1). On subtracting this result from (1), we obtain Green's second identity:
f(v2G -
(2)
4'V2
dd4,
v) dV =
\ do - 41 dnl dS.
We now take ,' = 1/r, where r is the distance OP. If 0 is interior to S, we cannot apply (2) to the entire region enclosed since 4, becomes infinite at 0. We therefore exclude 0 by surrounding it by a small sphere v of radius e and apply (2) to the region R' between S and v; then 1
V24pdV
=
dl 1dipdS. dl ldc dS- cP----co----J, dn r r dn JS \ dn r r dn
On the sphere r = e, d
(1l
do \ r/
d
(1l
dr \ r)
1
1
dp
r2
e2 '
do
ar '
dS = eQ, 2d
INTEGRAL TRANSFORMATIONS
238
§ 108
dil denoting the solid angle subtended by dS. The integral over v is therefore
J( o
E2
+
dSt +
1 E ate) ar dS
E
f
a
ar
Ef
M=
a
or
dg,
where p is a value of
d1 do r
1
r do
dS -
-v2(p dV-- -f-v2(P dV Rr R1r
j
for the integrand remains finite if we use the spherical element of volume dV = r2 sin 0 dr dB d4p. We thus obtain Grreen's third identity: (3)
Jr v2r dV +
i dip
d it
\r do
do r/
dS.
When 0 is exterior to S we may put ¢ = 1/r directly in (2); in this case the right member of (3) equals zero. We may deduce three analogous identities from (101.6),
fDivfdA the divergence theorem in the plane. With f = (p Grad ¢, we find (4)
fv.vdA+fv2dA = f (P do ds,
where we now interpret V and V2 as Grad and Div Grad and di/dn as a derivative in the direction of the external normal m. By an interchange of cp and , we find as before (5)
f(cv24, - ,'tv2 p) dA
d
=
/C do -
dip
# d.n) ds.
To obtain the third identity from (5), we take 4, = log p, where p is the distance OP in the plane. Since log p is a solution of Laplace's Equation in the plane,
Div Grad # = 0,
HARMONIC FUNCTIONS
§ 109
239
v24, = 0 in (5). As before, we must exclude the origin from the region by surrounding it by a small circle y of radius E. Applying (5) to the region remaining, we have
- flog p
\
v2cp dA =
R
c
d
do
log p - log p
ds +
do
d (P do log p - log p
,.
J
do
ds.
On the circle p = E, d
d
log p
do
dp
1
1
p
E
dip
log p
do
'
=
aP
- ap '
ds = E d6;
the integral over y is therefore
f\ where
log E - - ap
acv
- d8 - 27r gyp, ap
E dB = E log c-
is a value of p at some point of y. When e -> 0, E log t
--* 0, and we obtain (0)
27r p(0) =
d
log p V2,p dA + C
R
(`p
do
duo
log p - log p -) ds.
/
109. Harmonic Functions. A solution of Laplace's Equation, div grad cp =
(1)
0,
is called a harmonic function. A function
in a region R if it has continuous derivatives of the first and second orders and V2p = 0 in R. If a vector f is both irrotational and solenoidal, its scalar potential 4, is harmonic; for, from (102.4),
f = -Vi,
divf = -V2' = 0.
If in Green's second identity (108.2) cp and 4, are harmonic throughout R, we have
f\odn -
(2)
Thus, if (3)
'=
dn/dS=0.
1, we have, for any harmonic function cp, do dS = 0.
240
INTEGRAL TRANSFORMATIONS
§ 109
If cp is harmonic in the region bounded by a closed surface S, Green's third identity (108.3) gives the value v(P) at any interior point P, v (4)
l dS,
is \r do -
do r/
where r = PQ, the distance from P to points Q on the surface. Thus a function gyp, harmonic in the region enclosed by S, is deter-
mined completely at any interior point P by the values of
0
- .Js (r do
do r) dS'
on putting i = 11r in (2). If S is a sphere of radius r about P as center and lying entirely within the region R, we have, from (4),
4irv(P) = or, in view of (3), (6)
r
f
do dS
c'(P) = 4r2 r
+ r2 JCp dS;
--
f p dS.
Since the surface of the sphere is 47rr2, we have the MEAN VALUE THEOREM. The value of a function, harmonic in a
region R, at any point P is equal to the mean of its values on any sphere about P as center and lying entirely within R.
This theorem shows that a function which is harmonic in a closed bounded region R, but not constant, attains its extreme values only on the boundary. For let P be a frontier point of the set for which cp attains its minimum value m. If P were an interior
point of R, there would be a sphere about P, lying within R, on which V > m at some points; hence the mean value of c over the sphere would be greater than (p(P) = m. If in Green's first identity (108.1) we take cp = 4' and assume that (p is harmonic in R, we have (7)
fi
o1 2 dV
=
- is
dS.
do
§ 110
ELECTRIC POINT CHARGES
241
Hence, if either cp = 0 or dcp/dn = 0 on S, the volume integral in (7) vanishes; and, since I nip 12 is continuous and never negative, we must conclude that V p = 0, and cp has a constant value throughout R. If 'p = 0 on S, 'p = 0 in R; and, if d'p/dn = 0 on S, p = C in R. Now let 'i and 'P2 be two harmonic functions in R; then their difference
F = el2 R f
(1)
r2
upon a charge e2 at P; r = OP, and R is a unit vector in the direction OP. Thus the charges repel when elel > 0 (charges of same sign) and attract when ele2 < 0. The force exerted by a charge e at 0 upon a unit charge at P, namely, e
E =
(2)
R, a r
is called the electric intensity at P due to the charge e.
Since
R = Vr, e
E _ -V-, r
(3)
so that E has the scalar potential (102.4), (4)
The vector E is both irrotational and solenoidal; for (5)
rot E = 0,
1
div E _ -eV2 - = 0. r
t In a vacuum, if the charges are measured in statcoulombs and the distance in centimeters, the force is given in dynes.
242
INTEGRAL TRANSFORMATIONS
§ III
If S is a closed surface, we have, from (107.2), (6)
n E dS = e
J
nR r2
dS =
0 inside of S, 0 outside of S.
147re
0
We assume that the intensity due to a system of point charges , e,, is the sum of their separate intensities: E = EEi. Since Ei = - Vei/ri, where ri is the distance from the charge ei to P, e1, e2,
E _ - -+-+ e1
e2
r1
r2
r, en/
and the potential of the system is el
e2
e
r1
r2
rn
(7)
If this system of charges is within the closed surface S, we have, from (6),
Jn.EdS = 47r1ei.
(8)
The normal flux of the electric intensity through a closed surface is equal to 4ir times the sum of the enclosed charges.
111. Surface Charges. If a surface S carries a distributed charge a per unit of area, the potential at a point P due to the charged element dS at Q (regarded as a point charge a dS) is a dS/r, where r is the distance QP. If we assume that the surface density a is continuous or piecewise continuous over S, the total potential at P is dS
The electric intensity at a point P outside of S, due to the charge element a dS at Q, is -a dS Vp 1/r; the total electric intensity at P is therefore (2)
E
fov-r dS = -Vpp.
The notation Op means that P must be varied in computing the gradient. Since P is outside of S, 1/r and its first partial deriva-
3 112
DOUBLETS AND DOUBLE LAYERS
243
tives are continuous for all positions of Q on S; hence in computing
E = -Vpcp, we may differentiate (1) under the integral sign. In differentiating functions of r = PQ, we may vary either P or Q, holding the other point fixed. Thus, if R is a unit vector in
-4
the direction PQ,
Vpr = - R,
VQr = R,
when Q and P, respectively, are varied; and, in general, (3)
Vpf(r) = f'(r)Vpr = -f'(r) VQr = -VQf(r)
Consequently, we also may write (2) as
E_
(2)'
1 dS. r
In integrals, such as this, the subscript on V may be omitted on the understanding that the variation is at dS. From (1) and (2), we see that cp and E are continuous at all points P not on the surface. At such points cp is harmonic; for
/'
vp"p = I crop dS = 0. s
r
At a point P on the surface, the integrals for
E+ - E_ = 47rv n,
(4)
where o- and n are the surface density and unit normal at Q. Thus the normal component of E experiences a jump of 4irv as P passes through the surface.
112. Doublets and Double Layers. The potential at P due to a charge -e at Q and a charge +e at Q' is (110.7) -e
e
//1
QQ'PG
1
r
t See O. D Kellogg, Foundations of Potential Theory, Berlin, 1929, Chapter 3, §5.
244
INTEGRAL TRANSFORMATIONS
§ 112
If Q' approaches Q and at the same time the charges increase, so that the product, e QQ' = m,
remains constant, the limiting result is called a doublet of moment m.
The potential of this doublet is 1
1
1
(1)
P=QimQje(QQ')PQQQ,PQ} =m - OQ - ;
r
frallction
in the second member is the directional for the limit of the derivative of 1/r in the direction of m. A continuous distribution of doublets over a surface with mo-
ments everywhere in the direction of the normal n is called a double layer. If /in dS is the moment of the doublet at the surface
element dS at Q, the potential of the double layer at an outside point P is (2)
_
1
fs
r
where r = PQ. Since 1
dS=dSl
r
r2
is the solid angle subtended by dS (107.3), we also may write (3)
1P _
-f
Ada
s
When µ is constant, this reduces to -µf2, where Sl is the total solid angle subtended by S at P. When µ, the moment density, t is piecewise continuous, p is continuous at all points P not on the surface. At such points cp is harmonic; for
r At a point Q where µ is continuous and the surface has contins
uous curvature, it can be shown * that cp has definite limits cp+, (p_ t In the case of a magnetic shell, u is called the density of magnetization. * See 0. D. Kellogg, op. cit., Chapter 6, § 6.
SPACE CHARGES
113
245
according as P approaches Q from the positive or negative side of
the surface, and that p+ - Cpl = 4rµ.
(4)
At a point P outside of S, the electric intensity E = -Vpca is continuously differentiable. Since pn is a function of Q (not P), we have, from (2), (5)
cep = -
E
But, from (85.6), GP X
n x G'Q (-1
1
fs
\
=
r
µn CQ VP
dS.
r
1
-n VP6Q - ,
so that we also may write E = Vp x 1 µn x VQ
(6)
-1
dS.
r Consequently the intensity due to a double layer has, besides the s
scalar potential gyp, also a vector potential A (§ 92) : (7)
E = rot pA,
(8)
A = fLn
VQ
r
dS.
When ja is constant, A may be transformed into a circuit integral about the boundary of S; for, from the basic theorem (100.1), (9)
A=A fnxV s
r
cr
Example. Let (P be a function harmonic in a region bounded by a closed surface S; then, at any interior point P (109.4), w(P)
47rr
dS
d (1 do
r/
dti.
Comparison with (111.1) and (112.2) shows that p may be regarded as the potential due to a surface change of density a = (d'/dn)/4a and a double layer of moment density IA = -c/4,r.
113. Space Charges. If a region V carries a distributed charge p per unit of volume, the potential at a point P due to the charged element dV at Q (regarded as a point charge p dV) is p dV/r, where
246
INTEGRAL TRANSFORMATIONS
§ 113
r is the distance QP. If we assume that volume density p is piecewise continuous over V, the total potential at a point P outside of V is (1)
(P
= I aT
The electric intensity at P due to the charge element p dV at Q is -p dV Vp 1/r; the total electric intensity at P is therefore (2)
E
fp X ©p dV = - Vpp.
Since P lies outside of V, 1/r and its first partial derivatives are continuous for all positions of Q within V; hence, in computing Vpcp, we may differentiate (1) under the integral sign. When P lies within the charged region V, the integrals for and E are improper since r passes through zero. But it can be shown ¶ that, when p is piecewise continuous, (p and E exist at the points of V and are continuous throughout space. Moreover the potential p is everywhere differentiable, and E = - Vptp. The equation (110.8) also may be proved for space charges; namely, (3)
fn.EdS = 47r
f
p dV,
the integral on the right covering all space charges within S. The closed surface S may either completely enclose the charged region V or cut through it. If S encloses no charges, the integral (3) is zero.
When p is continuously differentiable, it can be shown that E has the same property. Now if Sl is any closed surface enclosing a subregion Vl of V, the divergence theorem shows that
fn.EdS = fdiv E dV. Yi
But, from (3),
fdiv Ed V = 47r J pdV; Yi
¶ See O. D. Kellogg, op. cit., Chapter 6, § 3.
HEAT CONDUCTION
§ 114
247
and, since this holds for any subregion V1 of V, div E = 47rp,
(4)
or, if we put E -4Trp.
(5)
This partial-differential equation is called Poisson's Equation. At points outside of the charged region V, p = 0, and the potential P satisfies Laplace's Equation: (6)
v2 P = 0.
114. Heat Conduction. In mathematical physics the integral theorems often are used in setting up differential equations. As an illustration, consider the flow of heat at a point P of a body. According to Fourier's Law, the direction of flow is normal to the isothermal surface through P; and the flow F (calories per second) per unit of surface is proportional to the temperature gradient at P. Thus F may be represented by the vector, (1)
F = -k VT cal./sec./cm.2,
where T is the temperature and k the thermal conductivity of the body at P. Since k is positive, the minus sign in (1) expresses the fact that heat flows in the direction of decreasing temperature. Let p and c denote the density and specific heat of the body at P. Then the rate at which heat is being absorbed in a region R bounded by a closed surface S is a
at
cpT dV =
dV. fcp aT at
If n is the outward unit normal to the surface S, the rate at which heat flows into R through S is
- in
FdS =
f
div (kVT) dV.
Hence, if heat is being generated in R at the rate of h calories per unit of volume,
a=f
fcp!dv
{div (kVT) + h} dV.
INTEGRAL TRANSFORMATIONS
248
§ 115
Since this holds for an arbitrary region R within the body, OT
cp - = div (kVT) + h.*
(2)
at
This is the differential equation of heat conduction. If the body is homogeneous, k is constant, and (1) becomes (3)
cp
aT
= k V2T + h;
at
and, if there is no internal generation of heat, we have
aT
(4)
= K V2 T,
at
where K = k/cp is called by Kelvin the "diffusivity." When the heat flow becomes steady the temperature distribution is constant in time. Hence in the steady state (4) reduces to Laplace's Equation V2T = 0. Example. Find the temperature distribution in a homogeneous hollow sphere whose inner and outer surfaces are held at constant temperatures. When the flow is steady, T is a function of r alone; hence, from (89.17), V 2T
r dr
(r2T r),
dr
(r2)__O,
T=
r + B.
A and B may be determined from the conditions T = Tl when r = rl, T = T2 when r = r2.
115. Summary : Integral Transformations. If f (r) is a continuously differentiable tensor point function, the basic integral transformations are: (A)
fvfdv=fnfds
(B)
fri x Vf dS = fTf ds
(
(
f f
dV Vf
=ids f);
dS x Vf =
fdr f).
* If f(r) is a continuous scalar function in a region V and ff(r) dV = 0 for an arbitrary subregion of V, then f(r) = 0 throughout V. For, if f(r) 96 0 at a point P, we can surround P with a sphere a so small that f does not change
sign in a, and hence f f (r) dV 0
0, contrary to hypothesis.
§ 115
SUMMARY: INTEGRAL TRANSFORMATIONS
249
In addition,
J"T2ofds= (C)
J
rl
l
may be regarded as a third basic type. From these, other integral transformations may be deduced by dotting and crossing within tensor functions. If we dot and cross in the first position, we get, from (A) :
fv.fdv=fn.fds, fvxfdv =inxfdS.
(Ax) Similarly, from (B),
fn.vxfds=fT.f, (Bx)
f(n? Vf - nV -f) dS =
fT
f ds,
in which V may be replaced by V8. In the B transformations the surface integrals vanish when taken over a closed surface.
When f is a vector function, (A - ) is known as the divergence theorem and (B - ) as Stokes' Theorem. If we replace f by of in (B x) we obtain (B')
f
(osf + J nf) dS =
fm f ds,
a transformation of the same scope as (B); J = - Div n is the mean curvature of the surface, and m = T x n is the unit external normal to the bounding circuit. From (B') we derive
(B'.)
f(vs.f-i-Jn.f)ds=fm.fds,
(BI X)
f(Vsxf+Jllxf)dS
=
f
mxfds.
On a plane, n is constant, and J = 0; if we write dA for dS, the B' transformations become (P)
fV8f dA = fm f ds,
INTEGRAL TRANSFORMATIONS
250
fv3.fdA fvsxfdA=fmxf(1s.
(Px)
When f is a vector, (P - ) is the divergence theorem in the plane. If f (r) is a continuously differentiable vector and rot f = 0 in a simply connected region of 3-space, the line integral of f - dr is independent of the path, and r
f=V
p(r) =
,
f dr. ro
If f (r) is a continuously differentiable vector, and a rot f = 0 in a simply connected portion of a surface, the line integral of f - dr over surface curves is independent of the path, and rf
ft = Grad (p,
=frf, .
.
(r) =
dr
dr,
ra
where ft = f - (n f)n is the tangential projection of f on the surface. PROBLEMS
1. Show that a closed curve lying in a plane with unit normal a encloses an area A given by
nA =
frxdr
2. Compute the integral / n rot f dS over that portion of the sphere r - a lying above the xy-plane when f = p(r)c.
3. Prove that
J'r x n dS = 0 over any closed surface S. If a body bounded
by S is subjected to a uniform pressure -pn per unit of area, show that these forces have zero moment about any point in space. 4. If the closed curve C encloses a portion of a surface S, show that
I
n x r dS = I JTr2ds.
5. If f = ia(x, y) + jv(x, y), prove that
ff x dr = kf f
div f dx dy
where C is closed curve in the xy-plane enclosing the region A.
PROBLEMS
251
6. If c is a constant vector, prove that
nx(cxr)dS=2Vc is
where V is the volume enclosed by the surface S. 7. The closed surface S encloses a volume V. If the vector f is everywhere
rot f dV = 0.
normal to S, prove that V
8. Show that the centroid of a volume V bounded by a closed surface S is given by
Vr* = z _fr2n dS. Apply this formula to find the centroid of a hemisphere of radius a, center 0, lying above the xy-plane.
9. If p denotes the distance from the center of the ellipsoid r 4) r = 1 to the tangent plane at the point r, show that the integrals of p and 1/p over its surface have the values
fPds = 3V,
IdS/p = cv1V,
where V is the enclosed volume [p = r n, 1/p = r
4)
n].
Express these
results in terms of the semiaxes a, b, c of the ellipsoid.
10. Find a vector v = f(r)r such that div v = rm (in 96 -3).
frmdv
Prove that
rmr n dS.
11. Show that
(a) rrmR dV = M
(b)
R
r
dV =
1
(m76 -1);
rm+in dS
flog r n dS.
12. If p and q are vector functions, prove that (a)
13. If r = PQ, the solid angle subtended at P by the surface S (over which Q ranges) is
Slp = -
i
is
n VQ - dS r
(107.1),
252
INTEGRAL TRANSFORMATIONS
Prove that
where V0l /r is computed at Q. Vp12p
r 1TXVQdS
ra
where r = PQ and the curve C is the boundary of S.
14. If C and C' are two unlinked closed curves, show that if r = PP' the double-circuit integral,
r TxT' r
a
ds (is' = 0
.
If C and C' are two simple loops, linked as in a chain, show that the preceding
double integral is ±4a, the sign depending on the sense of C'. 15. Show that tangential forces of constant magnitude o- acting along a closed plane curve C are equivalent to a couple of moment 2QA, where A is the vector
area enclosed by C. What is the moment of the couple when C is a twisted curve? [Put f = r in (100.4).] 16. Let f(r) be a solenoidal vector function. Prove that f = rot g (§ 92) where
"OD
t
g = -r X
fxi(,r) dX o
or
r x j Xf(ar) dX t
provided the integrals exist. (The integrand of the former may become infinite at the lower limit.) In particular if f(r) is homogeneous of degree n 0 -2, f(,\r) = X"f(r) and g is given by g = (f x r) /(n + 2). Prove this formula directly by making use of r Vf = of and (85.6).
CHAPTER VII HYDRODYNAMICS
116. Stress Dyadic. Let S be any closed surface inside a deformable body. It divides the body into two portions, A within S, and B without S. The forces acting on A are of two kinds: (i) body or mass forces which act on the interior particles, and (ii) surface forces, which act on the bounding surface S. Let k denote the average body force on the element of mass Am; then, if Am shrinks to zero while always enclosing a point P, the limit of P/Om is defined as the body force R, per unit of mass,
at P. Let F,, denote the average surface force acting on A over the vector element of surface nOS of the boundary S; then, if AS shrinks to zero while always enclosing a point P, the limit of Fn/AS is defined as the surface force F, per unit of area, acting on A at P. Here n denotes a directed line normal to S at P and directed from A to B. The notation Fn thus associates a surface force with a surface element of unit normal n. The force F dS is exerted on the element n dS by the matter toward which n points. The surface force acting on B on the same element n dS is, from the law of action and reaction, (1)
F_n = -Fn.
Consider now a small tetrahedron (Fig. 116) with three faces parallel to the coordinate planes; and let outwardly directed lines
normal to the faces be denoted by -x, -y, -z, n. If the area of the inclined face is A, the faces normal to x, y, z have areas A cos (n, x), A cos (n, y), A cos (n, z), and the volume of the tetrahedron is 3Ah where h is the altitude of P above the base A. If body and surface forces are replaced by averages and p denotes the average density, the equilibrium of the tetrahedron requires
that F,,A + F_IA cos (n, x) + F_NA cos (n, y) + F_ZA cos (n, z) + 3Ahpr2 = 0. 253
HYDRODYNAMICS
254
§I II
If we divide out A and let h ---. 0 so that the tetrahedron shrink, to the vertex P, we have in the limit
F,, + F_x cos (n, x) + F_ cos (n, y) + F_Z cos (n, z) = 0.
If n is a unit vector along the line n, cos (n, x) = n i, etc.; and, since F_x = -Fr, we have Fn = n (i Fx + i Fy + k Fe). (2) All surface forces in (2) now refer to the point P. Thus a plan( through P normal to n divides the body into two parts; and thE
-x Fla. 116
part toward which n points acts upon the other part with the fort( F,, per unit of surface at P. The dyadic.,
c= i Fx + j Fy + k F=,
(3)
is called the stress dyadic at P; it effects a synthesis of all surfac( forces at P by means of the relation, (4)
If the deformable body is a fluid, we assume as an experimenta fact that the stress across any surface element of a fluid in equilibrium is a pressure normal to the element; hence
Fn = -pn;
F. =
- 711,
Fv = - p2J,
Fz = - p3k.
With these values, (2) becomes
pa=nipli-f-nj P2j+nkp3k;
117
EQUILIBRIUM OF A DEFORMABLE BODY
and, on multiplying by i
,
j
,
k
,
255
in turn, we have
p=P1=P2=P3
(5)
At any point within a fluid the pressure is the same in all directions. At any point where the pressure is p, the stress dyadic is
`, = -p(ii + ii + kk) = -pI.
(6)
For fluids in motion it is no longer true that only normal stresses exist. Viscous fluids in motion do exert tangential stresses; but in
many problems these tangential stresses are small and may be neglected. We shall develop the mechanics of fluids on the hypothesis of purely normal stress. We imply this assumption by speak-
ing of perfect or non-viscous fluids; for a perfect fluid the stress
dyadic is -pI. 117. Equilibrium of a Deformable Body. If a body in a strained state is in equilibrium, any portion of it bounded by a closed surface S is in equilibrium under its body forces and the surface forces acting on S. Let the body forces be R per unit of mass, and let the surface forces Fn per unit of area be given by the stress dyadic 4. Then, from (116.4), F,, = n 4, and the equations of equilibrium are
fRdv+fn.ds=o
(1)
J'r x Rp dV + jr x (n
(2)
,
(b) dS = 0,
where (2) is the moment equation about the origin. If we transform the surface integrals by means of (106.5),
in
- -1) dS =
frx(n.4)ds=
f
v 4dV,
xrdS= -
(1) and (2) become
f(R + V
)dV=0,
f[PRxr+ V.(xr)]dV = 0.
HYDRODYNAMICS
256
§ 117
Since these integrals vanish for an arbitrary choice of S, their integrands must be identically zero. The equations of equilibrium are, therefore (3)
=0,
(4)
pRxr+V.(cxr) =0.
On eliminating pR from these equations, we obtain
V.(4)xr) - (V -4)) -r = 0.
(5)
From (5) we conclude that 4), the vector invariant of c, is zero; for, if we write 4) = i F,z + j Fy + k F2, a a (aFx a yy aFZ\
(Fxxr) +a (F,-r) +az ax y
(FZxr)
-
+Fyx-+F2xar=
Fx'<
ay
ax
az
+
ax
a
+
az
xr-0,
Fxxi+Fyxj+FLxk=0,
and hence 4) = 0. In view of the theorem of § 68, we see that the moment equation (2) requires that the stress dyadic (D be symmetric. If we write Fx = i Xx -f - j Yx + k Zx, . . .
the stress dyadic may be written in the nonion form, Xx
c = Xy
(6)
XZ
Yx
Zx
Y, Zy YZ
,
ZZ
where, by virtue of its symmetry, (7)
Xy = Yx,
YZ = Zy,
Zx = XZ.
The normal components of stress Xx, Yy, ZZ occur in the principal
diagonal, whereas the tangential components, or components of shear, are equal in pairs. More specifically, in perpendicular planes, the components of shear perpendicular to their line of intersection are equal.
Since 4) is symmetric, we always can find three mutually orthogonal vectors i, j, k, so that
4) = a ii + j3 jj + y kk Then i, j, k give the principal directions of stress, and a, (8)
the principal stresses at the point in question.
are
FLOATING BODY
§ 119
257
118. Equilibrium of a Fluid. For a fluid the stress dyadic is
4_ -pI and
-Vp,
from (86.8); the equation of equilibrium (117.3) is therefore Vp = pR. (1)
Consider a liquid of constant density p in equilibrium under gravity. The body force per unit mass is then g, the acceleration of gravity, and Op = pg = pgk = pg Oz = V (pgz)
(2)
when the z-axis is directed downward along a plumb line
Hence
p = p.9z + po, where p = po when z = 0. If the origin is at a free surface, po is the atmospheric pressure, and pgz is called the hydrostatic pressure. 119. Floating Body. Let the sur(3)
face S of a floating body V be divided by its plane section A at the water line into two parts: S1 submerged, S2 in air (Fig. 119). If po is
the atmospheric pressure, pi = pgz
the hydrostatic pressure at depth z, po + pi acts on S, po on S2; or we may say that po acts over S, while p1 acts on the closed surface Fia. 119 S1 + A (pl = 0 on A) enclosing the volume V1 below the water line. Hence the equations of equi. librium of the body are: (1)
(2)
fgdm -ispondS -i+ p1ndS = 0, frxgdrn
-
f
is rxnpodS - f rnpl dS = 0. A
S,+A
S
We consider in turn the integrals in (1) :
fgdm
= W, the weight of the body;
fpondS = po
in dS = 0;
HYDRODYNAMICS
258
n pl dS =
Vpi
§ 120
dV = fpg dV = W1,
the weight of the displaced liquid. W = W1i this is the
Therefore (1) states that
PRINCIPLE OF ARCHIMEDES: A floating body in equilibrium displaces its own weight of liquid.
In order to interpret (2), we note that the center of mass of a discrete set of particles Pi of mass mis defined as a centroid (9.3); its position vector r* is given by
mr* =
(3)
where m is the total mass. Similarly, the center of mass of a continuous body is defined by mr* =
-(4)
f
r dm.
If we change signs throughout in (2), we have the following integrals to consider:
f
gxrdm = gx
f
rdm = gxmr* = Wxr*,
v
v
ojnxrdS = po Ifrot r dV = 0, s
s
n x pir dS = Trot (p1r) dV = vt s+A
j(Opl) x r dV v
pr dV = g x m1r= W1 x r.
=gx vt
Thus (2) reduces to W x (r* - r) = 0; this states that the center of mass of the body and of the displaced liquid are on the same vertical.
Let p and v denote the density and velocity of a fluid at the point P and at the instant t. Con120. Equation of Continuity.
sider the mass of fluid
dV within a fixed but arbitrary closed
surface S. This mass is increasing at the rate, a
at
dV =
ap
J at dV'
§ 120
EQUATION OF CONTINUITY
259
the time differentiation being local. This rate must equal the rate
at which fluid is entering S, namely, -
in
pv dS, where n de-
notes the outward unit normal. Hence
Iapat d V = - fn
pv dS = - fdiv (pv) dV,
when the divergence theorem (106.5) is applied. Since the integral of ap/at + div (pv) over any closed region within the fluid is zero, we conclude that ap (1)
at
+ div
(pv) = 0.
This equation of continuity may be put in another form by intro, ducing the substantial rate of change dp/dt instead of the local time rate Op/at. Along the actual path, or line of motion, of a fluid particle, z = z(t), y = y(t), x = x(t),
the density p(t, x, y, z) becomes a function of t alone, and dp
OP
dt
at
+
OP dx
ap dy
ap dz
Ox dt
+ ay dt +
az dt
OP
OP
at
+° \l
+
ap OP
+ k OZ
gives the time rate at which the density of a moving fluid particle is changing. We thus obtain the important relation, dp
OP
dt
at
(2)
+ v - op'
connecting the substantial and local changes of p. For any tensor function f(t, r) of time and position, associated with a fluid particle moving with the velocity v, we have, in the same way, (3)
df
of
dt
at
+v
Of.
260
HYDRODYNAMICS
§ 121
From (85.2), div (pv) = v - Vp + p div v;
hence the equation of continuity (1) also may be written dp dt+pdivv=0.
(4)
Since div v = - (dp/dt)/p we can interpret div v as the relative time rate of decrease of density of a fluid particle having the velocity v. Thus a positive value of div v implies a negative dp/dt and, consequently, an attenuation of the fluid at the point considered; hence the term divergence.
For an incompressible fluid, dp/dt = 0, and (5)
div v = 0.
This equation has the integral equivalent, (6)
in - v dS =
fdiV v dV = 0;
the "flux" of an incompressible fluid across the boundary of a fixed closed surface within the fluid is zero. If an incompressible fluid is also homogeneous, p is constant. 121. Eulerian Equation for a Fluid in Motion. In the Eulerian or statistical method of treatment we aim at finding the velocity, density, and pressure (v, p, p) of the fluid as functions of the time at all points of space (r) occupied by the fluid. Consider the fluid within a fixed closed surface S at any instant t. By D'Alembert's Principle, the body and surface forces, together with the reversed inertia forces (-ma), may be treated as a system in statical equilibrium. In a perfect fluid the surface force is a normal pressure: Fn = -pn (§ 116). Hence, if the body force per unit mass is denoted by R, D'Alembert's Principle applied to the fluid enclosed by S gives the equation, (1)
(2)
f(R_a)Pdv_fnpds=o, frx(R - a)pdV -
f
rXn pdS = 0,
5 121
EULERIAN EQUATION FOR A FLUID IN MOTION
261
where a denotes the acceleration of the fluid particles. On transforming the surface integrals in (1) and (2), we have
f {(R-a)p-Vp}dV=0, f {r x (R - a)p + rot (pr) } dV = 0. Since these integrals vanish for any choice of S, the integrands are identically zero. From (85.3), rot (pr) = (Vp) x r; the equations of motion are therefore (3)
(R - a)p - Vp = 0,
(4)
rx(R-a)p-rxVp=0.
When (3) holds good, (4) is identically satisfied and may be omitted. The Eulerian Equation of Motion is therefore 1
a=R--Vp.
(5)
P
Here a = dv/dt is the acceleration of a moving fluid particle and must be distinguished from av/at, the rate of change of fluid velocity at a fixed point. These substantial and local rates of change are connected by the relation (120.3) : a
(6)
dv
av
dt
at
+v - Vv.
When the density p is a function of p only, we introduce the function: P dp
(7)
P = fe - ;
dP
1
dp
P
then VP = - Vp = - Vp.
P
Moreover, if the body force R has a single-valued scalar potential Q (§ 102),
R = - VQ;
(8)
such forces are said to be conservative.
Under these conditions
(5) becomes dv (9)
dt
= - v (Q + P).
HYDRODYNAMICS
262
§ 122
In the Eulerian method r and t are independent variables, so
that ar/at = 0. Equation (120.3) applied to r, dr
or
I=v,
dt
is simply an identity. The.lines of the fluid which at any instant are everywhere tangent to v are called its stream-lines. They are not the actual paths of the fluid particles except when the flow is steady, that is, when v is constant in time (av/at = 0). The stream-lines have the differential equation v x dr = 0. Example. Revolving Fluid. If the fluid is revolving with constant angular velocity w = wk about a vertical axis, its velocity distribution is that of a rigid body; hence, with the origin on the axis of revolution, we have, from
-4
(54.3), vp = w x OP, or simply v = w x r. Then a = w x v, and (9) becomes
wx(wxr) = -V(Q+P).
(i)
Now co x (w x r) = r - (ww - w2I)
- 2v{r (w2I - ww) r}
-
1 w2
v(r2
(85.14)
- z2)
- 2w2 0(x2 + y2); hence, from (i),
V[Q + P -
2W2(x2 + y2)1= 0,
Q + P - 21w2(x2 + y2) = cont. For gravitational body forces, R = g, and Q = gz, if the z-axis is directed upward; and, if the density is constant, P = p/p. In this case, (ii) becomes
gz + p/p - iw2(x2 + y2) = cont. At a free surface p is constant; a free surface is therefore a paraboloid of revolution.
122. Vorticity. Starting with the Eulerian Equation in the form. av (1)
vv = - V (Q + P), at + v
we transform v Vv by means of (85.9) and (85.7) :
v Vv = (Vv) v - v x rot v = 2 V(v v) - v x rot v;
LAGRANGIAN EQUATION OF MOTION
§ 123
263
hence (1) may be written av
- - vxroty = -o(Q + P + 2v2). at
(2)
Now, from (84.11),
rot av -- vxroty = 0, at
arot V = rot (v x rot v) at = (rot v) Vv - v - V rot v - (rot v) (div v), when we make use of (85.6) and (84.13) ; and, since
arot y
drot y
at
dt
drot y dt
(3)
'
+ (rot v) (div v) = (rot v)
Vv.
When v gives the velocities of a rigid body having the instantaneous angular velocity w, rot v = 2w (§ 84, ex.). For a liquid we may regard
w=yrot V
(4)
as the molecular rotation or vorticity of the fluid particles. In (3) we now replace rot v by 2w; and, from the equation of continuity (120.4), (5)
dp
dt
+
p diV v =0,
diV v = p
d
/1\
dt p
Then (3), after division by p, becomes
ldw p dt
d(1\ +wdt\p/
w
=
P
d (w _
co
vv,
This is known as Helmholtz's Equation. 123. Lagrangian Equation of Motion. In the Lagrangian or historical method of dealing with a moving fluid, the motion of the individual fluid particles is followed from their initial positions ro
HYDRODYNAMICS
264
§ 123
to their position r after a time t. Thus the history of each fluid particle is traced. If the function f(r) is associated with a fluid particle, r is a function of the independent variables ro and t. In space differentiation we may form gradients relative to r or ro; these are denoted by the symbols V and vo. Thus we may compute df (§ 90) as either
df = dro vo f or df = dr Vf; and, since dr = dro vor,
(1)
and
vof = dro Dor of for arbitrary displacements dro, we have dro
vol = vor vf.
(2)
In particular, when f = ro, we have
I = vor
(3)
vro,
so that the dyadics vor and vro are reciprocal. Consequently, if we multiply (2) by vro as prefactor, we have
of = vro vof
(4)
An element of fluid volume dVo is altered by the transformation
(1) in the ratio J/1, where J denotes the third scalar invariant of vro (§ 70) : thus (5)
dV = J dVo.
If we use rectangular coordinates,
(6)
axo/ax
ayo/ax
azo/ax
vro = axo/ay
ayo/ay
azo/ay
axo/az
ayo/az
azo/az
and J is the determinant of this matrix, namely, the Jacobian a(xo, yo, zo)/a(x, y, z). The element of mass Po dVo becomes p dV = pJ dVo; and, since mass is conserved, d
(7)
PJ = Po
or
dt
(pJ) = 0.
This is the equation of continuity in the Lagrangian method. Since
LAGRANGIAN EQUATION OF MOTION
§ 123
265
dp/dt = -p div v from (120.4), d
(dJ
dt (PJ) = P
J div v = 0,
dt
and (7) is equivalent to dJ - = J div v.
(8)
dt
The Dynamical Equation of Lagrange corresponding to the Eulerian Equation (121.9) is obtained by multiplying the latter by Vor as prefactor; thus (9)
Vor
dv
dt
= -Vor V(Q + P) _ -oo(Q + P),
from (2). Since ro and t are the independent variables, the symbols Vo and d/dt commute; hence the left member of (9) may be written d
-{
d
v} - (Vov) v =
(Vor)
dt
{ (cor)
v} - vozv2,
and (9) becomes (10)
d
( (oor) v}
= - Vo(Q +
P-
zv2).
With the aid of the dyadic Vor we may integrate Helmholtz's Equation (122.6). If we make use of (4), this becomes
d/ w\ dt
p
w
w
d
w
d
p
p
dt
p
dt
) Vor,
since, from (3), the time derivative of Vro Vor is zero. Multiplying this equation by Oro as postfactor now gives
dt \p10 /
'
Oro +
d p
\dt orb/
dt \p
w
(00
P
PO
fro/ - 0,
- Vro = -, the constant wo/Po being the value of (w/p) Vro when t = 0. Multiplication by Vor gives finally P
PO
This is Cauchy's integral of Helmholtz's Equation.
HYDRODYNAMICS
266
124
To verify this integral, we need only differentiate (11) with respect to t: thus
d(w
=
P
wo
pr =
0
Ov=
G'r Po
Po
P0
coo
P
The lines of a fluid which are everywhere tangent to w are Their differential equation is cox dr = 0. Consider a vortex line at the instant t = 0, and let its differential equation be wo x dro = 0. After a time t, coo and dro become called its vortex-lines.
P
w=
dr = dro
Vor,
coo
Vor.
Po
Hence, if coo = X dro, then co = (p/po) X dr; that is, wo X dro = 0 implies w X dr = 0. Thus we have proved the THEOREM.
If the body forces have a potential and p = f (p) or
constant, the vortex-lines move with the fluid. consist of the same fluid particles.
Vortex-lines always
124. Flow and Circulation. The tangential line integral of the velocity of a fluid along any path is called the flow along that path. If the path is closed, the flow is called the circulation.
If the path AB at time t was AoBo when t = 0, the flow over AB is
f
B
Bo
f dro
v dr
(Vor)
v;
Ao
and, since ro and t are independent variables,
f v6 dr = jOdro
{ (Vor)
v
a
When the body forces have a potential Q and p = f(p), we have, from (123.10),
J.B
Bp
dr= -
vo(Q-- I+ Ao
' -v)
Bo
fo
LBdr V( Q+P-2v2);
IRROTATIONAL MOTION
§ 125
267
or, since the last integrand is a perfect differential, d
--
(1)
fv.dr=
[-Q - P + v2]and,
in particular, (2)
d
dt
fv.dr=0.
The last result is Kelvin's CIRCULATION THEOREM. If the body forces have a potential and
p = f(p) or constant, the circulation over any closed curve moving with the fluid does not alter with the time.
125. Irrotational Motion. When the body forces have a potential and p = f(p) or constant, Cauchy's Equation (123.11) shows
that, if the vorticity of a fluid vanishes at any instant, it will remain zero thereafter. Then rot v = 0 in space and time, and the motion is termed irrotational. In any simply connected por-
tion of the fluid, we may express v as the gradient of a scalar (§ 102) ; thus (1)
v = -Vs ,
fr
P=-
v dr, ro
and (p is called the velocity potential. Then the velocity is everywhere normal to the equipotential surfaces (p = const and is directed toward decreasing potentials. Hence, along any line of motion, cp continually decreases; in a simply connected region the lines of motion cannot form closed curves.
By Stokes' Theorem "the circulation
iv
dr = 0 over any re-
ducible curve (§ 102) ; and, as this curve moves with the fluid, the circulation around it remains zero (circulation theorem, § 124). From this fact we again may deduce that, if the motion is irrota-
tional at any instant, it remains irrotational thereafter. The equation of continuity (120.4) now becomes (2)
dt -
p02(p
= 0.
For an incompressible fluid, dp/dt = 0, and harmonic function.
V2(P = 0; then
HYDRODYNAMICS
268
§ 126
The equation of motion (122.2) reduces to
v (Q
(3)
+P+
2v2
- ail = 0,
since av/at = - V (a(p/at) ; for a local time differentiation, a/at (r constant) and a space differentiation V (t constant) commute with each other. Hence
Q+P+1
(4)
2
- at = f (t)
an arbitrary function of the time; and, if the flow is steady,
Q + P + 2 v2 is an absolute constant.
Every harmonic function cp represents some irrotational flow of an incompressible liquid whose velocity v = -VV. The problem consists in finding a solution of V2V = 0 which conforms to the given conditions. For example, in a flow with central symmetry,
1d
2
d`pl
r2dr Cr drJl
0,
=a, r dr=b-,r 2
d`p
a
Now v = aR/r2 where R is a unit radial vector; and the flux per second through any sphere of radius r about the origin is constant: 47rr2 (a/r2) = 47ra. When this flux is given, a is determined. The value of b is immaterial, since it disappears in v = - vcp; as it is customary to have
(1)
Also av/at = 0 and the stream-lines are also lines of motion. The Eulerian Equation (122.2) now becomes
vxroty = v(Q+P+1v2).
(2)
If T is a unit tangent along a stream-line (T x v = 0) or a vortex-
line (TX rot v = 0), T v X rot v = 0; hence T V(Q + P + 2v2) _ 0, or (3)
d
ds
(Q + P + Zv2) = 0 along a stream- or vortex-line.
STEADY MOTION
3126
269
We thus have proved BERNOULLI'S THEOREM. Let the body forces have a potential and
P = f(p) or constant; then, for a steady flow,
Q + P + v2 = const
(3)
z
along any 'stream-line or vortex-line.
The constant will vary, in general, from one line to another. However, if the motion is irrotational as well as steady,
V(Q+P+zv2) = 0,
Q+P+zv2=C,
(4)
where C is an absolute constant-the same throughout the fluid. Now suppose that the fluid has a constant density p; this is nearly fulfilled in the case of a liquid. The equation of continuity
is div v = 0 or
fn - v dS = 0.
Along any tube whose surface
consists of stream-lines, let the normal cross section be denoted
by A. If the tube is sufficiently thin, and we apply -f n - v dS = 0 to the portion between A 1 and A2, we have approximately v1A1 - v2A2 = 0,
or vA = const
as the equation of continuity along a tube of flow. If the liquid is subject only to gravitational body forces, their
potential is gz, where z is measured upward from a horizontal reference plane. Thus with
P=p/P,
Q=gz,
we have
gz + p
(5)
P
+
2
v2 = const
If z, p, v are known at the section A0 (zo, po, vo), we have, from (5),
along a stream-line. (6)
Pg(z-Z)+(p-po)+zP(v2 -vo)=0.
For a thin tube we may take vA = v0A0, and (6) becomes 2
(7)
p - po = Pg(zo - z) - Pvo C2 - 1) 2
.
HYDRODYNAMICS
270
§ 127
Thus the pressure is least at the narrowest part of the tube. Example. Torricelli's Law. When liquid escapes from an orifice near the bottom of a vessel which is kept filled to a constant level, the flow may be regarded as steady. Consider a stream tube extending from the orifice of the area A to the upper surface where its area is Ao. At this surface PO is the atmospheric pressure, and vo = vA/Ao; at the orifice p = po, and the outflow speed is v. With these values we have, from (6),
v2(1 - A2/A0) = 2g(zo - z) = 2gh, where h is the distance from the free surface to the orifice. When Ao is large compared with A, we have approximately v2 = 2gh, a result known as Torricelli's Law.
127. Plane Motion. When the flow is the same in all planes parallel to a fixed plane it and the velocity has no component normal to Tr, the motion is said to be plane. Such motion is completely determined by the motion in 7r, which may be taken as the xy-plane. Velocity, density, and pressure are all functions of x and y alone; and, for any tensor function f(x, y), we have, from (97.1), Of
grad f = Grad f + k - = Grad f; 49Z
and, similarly,
div f = Div f,
rot f = Rot f.
The flux across any curve C in the xy-plane is defined as the volume of liquid per second crossing a right cylindrical surface of unit height based on C. For an observer who travels the curve in the positive sense of s, the flux crossing C from (his) right to left is (1) x
FIG. 127
fkxT.vds=fvxk.dr.
here k, T, k x T are unit vectors along the z-axis, the tangent and normal to the curve (Fig. 127).
In any simply connected portion of the fluid the flux across a plane curve joining two points will be independent of the path, provided
k rot (v x k) = 0
(§ 103, theorem 1).
PLANE MOTION
§ 127
271
From (85.6),
rot (v x k) = k - Vv - k div v = -k div v. For an incompressible fluid, div v = 0 (120.5) and F will be independent of the path. Hence ¢(r)
(2)
ro
defines a scalar point function; and, from § 103, theorem 2, Grad 4, = v x k, since v x k lies in the xy-plane. Writing V for Grad, we have (3)
V,k=vxk,
v=kxVik.
V,' is everywhere normal and v everywhere tangent to the curves t = const; these curves are therefore stream-lines (§ 121). Consequently, the function is called the stream function.
If we set g = -4k, we have (4)
rot g=kxV,,=v,
divg=0;
thus the velocity has the vector potential -,pk (§ 92). From (85.6), we have (5)
rot v = rot (k x
k V2VI.
Therefore the plane motion of an incompressible fluid will be irrotational when and only when the stream function is harmonic: V2W = 0.
(6)
In this case v = - Vq: the velocity has a scalar potential (p; and, since
V2(p= -dive=0,
(7)
the velocity potential is also harmonic. From
v= -V
Vp = (VP) x k,
VL. = k x Vsp;
or, in terms of rectangular coordinates, i
--=i--japp
a,k
491P
ay
ay
ax
3
ax
HYDRODYNAMICS
272
5 127
Thus (8) is equivalent to the equations: (9)
ax
at ay '
49 (P
alp
ay
ax
But these are precisely the Cauchy-Riemann Equations which connect the real and imaginary parts of the analytic function w = y of the complex variable z = x + iy. We thus have proved the important THEOREM. In any plane, irrotational motion of an incompressible fluid, the velocity potential p and the stream function ' are two harmonic functions which combine into an analytic function V + i¢ of a complex variable x + iy.
Since -,y + icp =
i1G), we see that, if cp + iy is analytic, -4, + i
and stream function for an irrotational plane flow, -4, and V are the corresponding functions for another flow of this type. From (8), we have Ocp 0¢ = 0: the stream-lines (4i = const) cut the equipotential lines (,p = const) at right angles. Since the complex potential w = cp + i>G is an analytic function of z,
a w'=-=-+i-=--i-; dw
app
dz
ax
ax
.9,P
ag'
ax
ay
or, since the velocity has the components, vx
(10)
-1 Vc
ag
ax
,
w' _ -vz+ivy,
vY
-j Vc
app
ay
,
- w' = vx+ivy)
where w' denotes the conjugate of w'. Thus the velocity at any point is given by the complex vector -w'; its magnitude is I w' 1. Example 1. Assume the complex potential w = azn (a real); then, if we write z = rete, w = arner'ne = arn(COS n9 + i sin no);
,p = amcosn9,
4, = amsinn9.
The stream-lines are the curves whose polar equations are rn sin n9 = const.
For the cases n = 1, 2, -1, we put z = x + iy, using rectangular coordinates. (a)
n = 1:
w = az = ax + iay.
§ 128
KUTTA-JOUKOWSKY FORMULAS
273
The stream-lines are the lines y = const, and the flow has the constant ve-
locity -t3' = -a in the direction of -x. (b) n = 2: w = az2 = a(x2 - y2) + i2axy. The equipotential and stream-lines are the two families of equilateral hyperbolas, x2 - y2 = const,
xy = const.
Since the stream-line xy = 0 may be taken as the positive halves of the x-axis and y-axis, these may be considered as fixed boundaries and the motion re-
garded as a steady flow of liquid in the angle between two perpendicular walls. The velocity at any point is -zb' _ -2a2; its magnitude varies directly as the distance from the origin. (c)
n = -1:
w = a/z = a(x - iy)/(x2 + y2).
The equipotential and stream-lines are two families of circles:
x/(x2 + y2) = const,
y/(x2 + y2) = const,
tangent to y-axis and x-axis, respectively, at the origin. The velocity -w' = a/22 becomes infinite at the origin. Example 2. With the complex potential,
w = V(z +a2/Z) = V(reie + a2r 'e-'°), (V real), a2
= V r + a2 r
cos B,
>G=
V r-a2a2r sin 0.
The stream-line 4, = 0 includes the circle r = a and the x-axis sin 0 = 0. Since w' = V(1 - a2/z2), the complex velocity,
/
-2b'=-V(1-a22/-.-V
as
z --+ oo.
Therefore we may regard the motion as a flow to the left about an infinite cylindrical obstacle of radius a. At a great distance from the obstacle, the flow has a sensibly uniform velocity - Vi.t When body forces are neglected, the Bernoulli Equation (126.4) gives p/p + 1v2 = const. From the symmetry of the flow about the cylinder, it is clear that the total pressure exerted by the fluid on the cylinder is zero. We shall consider this matter for cylinders of arbitrary section in the next article.
128. Kutta-Joukowsky Formulas. Consider an incompressible fluid flowing past an infinite cylindrical obstacle of arbitrary cross section. The flow, in a plane perpendicular to the generators of the cylinder, is assumed to have the sensibly uniform velocity v
at a great distance from the obstacle. Then, if the motion is t See Lamb, Hydrodynamics, Cambridge, 1916, p. 75, for the stream-lines.
HYDRODYNAMICS
274
§ 128
steady and irrotational, and the body forces on the fluid are neglected, we have, from (126.4),
p=K --v - v,
(1)
where K is an absolute constant. If n is a unit internal normal to the boundary C of the obstacle (Fig. 128), F
(2)
=jnpds,
M
c
=frxnpds c
give the resultant force and moment about the origin exerted by Ay
0
x
FIG. 128
fluid pressure on a unit length of cylinder. Substituting p from (1) in these integrals and noting that
in ds = 0, Jr x n ds = J rot r dA = 0, from (101.5), (101.7), we have (3)
F=
2
J
When the flow is given by the complex potential w = + i4,, we shall compute these integrals after converting them into circuit integrals in the complex z-plane.
The vectors r = xi + yj and r = xi - yj correspond to z and its conjugate 2:
r
z = x + iy = r(cos d + i sin 0) = re'6,
7
2 = x - iy = r(cos 0 - i sin 0) =
re-t0.
KUTTA-JOUKOWSKY FORMULAS
§ 128
275
By definition,
r2 = r1r2 cos (02 - 01),
r1
' it
hence
r Thus
r1
1
r2
1
r1 x r2 = rlr2 sin (02 - 01)k;
x rk = 2
12
rre,(B2--B,)
-= 202-
r2 and r1 x r2 k correspond to the real and imaginary
parts of 21z2: (4)
r1
(5)
r2 - (R(21x2) = 2 (21z2 + x122),
r1 x r2 . k -9(202) = 2 (21z2 -
Note also that k x r ti iz; for the multiplications by k x and i (= eia/2) both revolve a vector in the xy-plane through 7r/2. When r1 and r2 are perpendicular or parallel we have, respectively, r2 = 0 - 21z2 + z122 = 0,
(6)
r1
(7)
rl x r2 = 0 - 21z2 - z122 = 0.
Turning now to the integrals (3), we have the internal normal n = k x T for a counterclockwise circuit of C; hence
nds = kxTds = kxdr-idz. Moreover, if v'' v, then v v ,..' v v, and c
The moment M, normal to the plane, is completely specified by the scalar,
-2
P
c
and, since
k r x (k x T) ds = (k x r) (k x T) ds = r dr '-' 01(2 dz),
M
v v 0? (2 dz)
P
2
c
2 P
G1
c
vi 2 dz.
t The real and imaginary parts of any complex number z are (R(z) _ 2(z + 2) and 4(z) = 2(z - 2). Note also that the conjugate of a product is the product of the conjugates: z
= 2122.
HYDRODYNAMICS
276
§ 128
But v and dz are parallel along C, which forms part of a stream, line; hence v dz = v d2, from (7), and we have P
F = -
i
JV2 dz,
M
2
dz
c
2
for the corresponding complex force and moment. Finally we form F and replace v2z dz in M by its conjugate v2z dz; thus P
(8)
j J02 dz=
2JC
v2
M = - 2 6l
(9)
P -z
2
dw 2dz, c
dz
z dz = -
C
P
2
(dw
a? C
\
2
dz) z dz,
since v = -dw/dz (127.10). These integrals exist if dw/dz remains finite over C. Now, outside of C, v is an analytic function which approaches v at infinity. The Laurent series for v therefore has the form:
2' = P. + al/z + a2/z2 + a3/z3 + .. .
(10)
If C. is a large circle of radius R about the origin and enclosing Cwe have, by Cauchy's Integral Theorem,
ff(z) dz = ff(z) dz,
if f (z) is analytic between C and C
.$
Thus the counterclockwise circulation about C is
y=
fcv
dr ti i J(P dz - v dz)
dz.
dz
c
cW
Moreover the static moment of the circulation is
µ=
fry dr-fzvdz = fz3dz. C
If we replace v by the series in (10), all integrals vanish except those involving dz
Reie i dB
JC. z -
Reie
21r
- i f dB = 2ri.
t See, for example, Franklin, A Treatise on Advanced Calculus, New York, 1940, §§ 267, 268.
KUTTA-JOUKOWSKY FORMULAS
§ 128
277
We thus find
y = 27ri al,
(11)
µ = 27ri a2.
We now can compute the integrals in (8) and (9) in terms of From (10), we have
v,,, y, and /2. v2
= v2 + 2a1v /z + (2a25 + a12 ) /z2 + .. .
When this series is substituted in (8) and (9), all integrals vanish except
fdz/z = 2rri; hence
(12)
F=2i
ipyv,,,
F = -ipyv,,;
(13) M = - 2 6?(2a2v + a2l)2-7ri = -pR?()u0) since a2 = -y2/4r is real, and ff?(2iri a2) = 0. In vector notation these give (14)
(15)
F = -pykxv,,,
M= -pL
xF, y
for the resultant force and moment on a unit length of cylinder. From (15) we see that F acts through the point, (16)
µ/y =
Jr
v
dr/ Jv - dr, c
which may be called the centroid of the circulation.
The equations (14) and (15) are called the Kutta-Joukowsky Formulas. From (14) we see that, for a counterclockwise circulation (y > 0), the force F is upward if the flow is horizontal and to
the left (v = - Vi); we then have a lift of pyVj per unit length of cylinder. The counterclock circulation diminishes the flow velocity below the cylinder and increases it above; by Bernoulli's Theorem this results in an excess of pressure below with a consequent upward lift. In the absence of circulation about the cylinder (y = 0), F = 0, and there can be no lift. This is the case in § 127, ex. 2, where
w = V(z + a2/z),
v = -dw/dz = - V(1 - a2/z2),
and a1 = 0 in the Laurent series.
HYDRODYNAMICS
278
§ 129
129. Summary: Hydrodynamics. For a perfect (non-viscous)
fluid the stress dyadic is -pI; at any point within a fluid the pressure is the same in all directions and exerted normal to a surface element. If f (t, r) is any tensor function associated with a fluid particle moving with the velocity v, its substantial rate of change is df dt
of
Vf, at + V
where of/8t is the local rate of change. In the Eulerian method of dealing with fluid motion, the aim is to compute v, p, and p as functions of the independent variables r, t. The (kinematic) equation of continuity is 8p
at
+div(pv) =
dp
dt
+pdivv=0;
and, for an incompressible fluid (dp/dt = 0), becomes div v = 0. The Eulerian Equation of Motion is
P=f dp,
dt=-V(Q+P),
(E)
when the body forces have a potential Q and the density a function of p only. From this we can deduce the Differential Equation of Helmholtz for the vorticity co = 1 rot v:
dt \p/ =
P
Vv.
In the Lagrangian method the aim is to follow the motion of the fluid particles from their initial positions ro (t = 0) to their positions r after a time t. The independent variables are now ro and
t; the position r of a particle at time t is a function of ro and t. We may take gradients relative to ro(V0) or r(V); and these conform to the relations, Vof = Vor Of,
Vf = Vro Vof;
Vor
Vro = I.
If J is the third scalar of the dyadic Vro, the equation of continuity becomes
pJ = po,
or dJ/dt = J div v.
SUMMARY: HYDRODYNAMICS
§ 129
279
The Lagrangian Equation of Motion corresponding to the Eulerian Equation (E) is (L)
Vor
dv
dt
= - Vo(Q + P).
From this we deduce Cauchy's integral of Helmholtz's Equation, 0)0 _=-.Vor;
(0
P
Po
f
and this in turn shows that vortex-lines (co x dr = 0) move with
the fluid. We find, moreover, that the circulation
v dr over
any closed curve moving with the fluid does not alter with the time.
When the motion is irrotational (rot v = 0), v = - VV, where p is the velocity potential, and
iv
dr = 0 over any reducible
curve. When the body forces are conservative and p is a function of p alone,
Q+P+
Zv2
- a = f (t), an arbitrary function of t.
For a steady-motion equation (E) gives
vxroty=V(Q+P+Zv2); whence Bernoulli's Theorem: Q + P + Zv2 is constant along a
stream-line or vortex-line, and this constant is absolute if the steady motion is also irrotational. For an incompressible fluid in plane motion, the integral independent of the path,
¢(r) =
x k dr (k I. plane of motion),
defines the stream function. The curves 4, = const are the streamlines (v x dr = 0). If the motion is also irrotational, both stream function and velocity potential are harmonic (V2# = 0, V2V = 0), and V4, = k x VV. This is the vector equivalent of the Cauchy-
Riemann Equations which guarantee that w = p + iii is an analytic function of the complex variable z = x + iy in the plane of motion. The complex potential w therefore has a unique derivative dw/dz; and the negative of its conjugate gives the complex velocity
vector: v = -w'.
HYDRODYNAMICS
280
PROBLEMS 1. A mass of liquid is revolving about a vertical axis with the angular speed
f(r), where r is the perpendicular distance from the axis. With cylindrical coordinates r, 0, z (we replace p, p in § 89, ex. 1 by r, 0 to avoid conflict with
the notation of Chapter VII), let a, b, c be unit vectors in the directions of r,., rg, rZ (Fig. 89a). If the angular velocity w = f(r)c, prove that
v = rf(r)b,
rot v = r dr [ref (r)]c.
[Cf. (89.7).]
2. If the motion in Problem 1 is irrotational, show that
a
PC,
and that the velocity potential v =
a v=rb, - ao is not single valued (a, $ are
constants). For a liquid of constant density p under the action of gravity alone, show that the pressure is given by
gz+p+-const. 3. If a fluid is bounded by a fixed surface F(r) = 0, show that the fluid must
satisfy the boundary condition v VF = 0. More generally, if the bounding surface F(r, t) = 0 varies with the time, show that the fluid satisfies the boundary condition
F + v 7F = 0
[Cf. (120.3).]
4. A sphere of radius a is moving in a fluid with the constant velocity u. Show that at the surface of the sphere the velocity of the fluid satisfies the condition
!v-u) (r - ut) = 0.
5. A fluid flows through a thin tube of variable cross section A. Show that for a tube PoP of length s a
J8PA ds + pAv
0;
hence deduce the equation of continuity at (pA) + as (pAv) = 0.
6. From the equation of fluid equilibrium (118.1) show that when p is a function of p alone, rot R = 0 and that the potential of R is -P. If p is not a function of p alone, show that equilibrium is only possible when R R. rot R = 0. [Cf. (121.7) for P.]
PROBLEMS
281
7. A gas flows from a reservoir in which the pressure and density are po, po into a space where the pressure is p. If the expansion takes place adiabatically,
p/py = const. (y is the ratio of specific heats), show that
P=
y
p
y-1p
Neglecting body forces and the velocity of the gas in the reservoir, show that the velocity v of efflux when the motion becomes steady is given by v2 =
2y
y=1
1 - (p/po) 7 } .
7?0
y - 1 PO
The velocity of sound in a gas is c = 1'yp/p; hence show that 2
v2 _
- 1
(c0 - c2).
y
8. If a body of liquid rotates as a whole from r = 0 to r = a with the constant angular velocity coo, and rotates irrotationally with the angular velocity to = woa2/r2 when r = a, we have the so-called "combined vortex" of Rankine. If z = za at the free surface when r = a, show that the free surface is given by
Z = za.+-(r2-a2) a2`
Z = Z. +w2a2 - 1! 1 - 2J , 2 \ r
r 5 a, r
a.
Prove that the bottom of the vortex (r = 0) is a distance w2a2/g below the general level (r = oc) of the liquid. 9. If the vorticity is constant throughout an incompressible fluid, prove that V2v = 0. [Cf. (84.14).] 10. When the motion of an incompressible fluid is steady, deduce from Helmholtz's Equation (122.6) that w Vv = v Vw. 11. If an incompressible liquid in irrotational motion occupies a simply connected region, show that
fv2dv
=
f
dodS'
where p is the velocity potential and the normal derivative dp/dn is in the direction of the external normal to the bounding surface. [Cf. (108.1).] 12. Prove Kelvin's theorem: The irrotational motion (v) of an incompressible fluid occupying a simply connected region S with finite boundaries has
less kinetic energy (T =
p
f
z the same boundary conditions.
v v dV) than any other motion (vl) satisfying
[Put vi = v + v'; then div v' = 0 and n v' = 0, where n is the unit normal vector to the boundary S. Show that Ti = T + T'.]
HYDRODYNAMICS
282
13. Under the conditions of Problem 11, show that if (a) v = const. over the boundary, or (b) dip/dn = 0 over the boundary, then V is constant throughout the region and v = 0. Hence show that the irrotational motion of a liquid occupying a simply connected region is uniquely determined when the value of either w or d'p/dn is specified at each point of the boundary. 14. If the body forces have a potential Q, the integrals, T
f a pv2 dV,
U= f pQ dV,
represent the kinetic and potential energy of the fluid within the region of
integration.
Show that for an incompressible fluid
d(T-[-U)_15. Assuming that the earth is a sphere of incompressible fluid of constant density p and without rotation, show that the pressure at a distance r from its center is p = 2gpa(1 - r2/a2), where a is the radius of the earth.
[If -y is the constant of gravitation, the attraction on a unit mass at the distance r is 47rtipr3/r2
= gr/a
where g is the attraction when r = a; hence the body force -grR/a has the potential 'gr2/a.] Compute the pressure at the center if p is taken equal to the mean density of the earth, (pg = 5.525 X 62.4 lb./ft.3). 16. In example 1 of § 127, consider the motion when n = r/a(0 < a < r). Since the lines 0 = 0, 0 = a are parts of the same stream-line = 0, we have the steady irrotational motion of a liquid within two walls at an angle a. Find the radial and transverse components of v at any point (r, 0).
17. Discuss the plane, irrotational motion when the complex potential w = rp + i¢ (§ 127) is given by z = cosh w. Show that the equipotential lines and stream-lines are the families of confocal ellipses and hyperbolas: y2
xz
c2 cosh' v
+
x2 C2 COS2 '
c2 sinh2 ,p -
1,
2
c2 sin2
- 1'
with foci at (fc, 0). Show that the stream-lines 4, = nor, where n is any positive integer, correspond to the part of the x-axis from x = fc to x = f co. If we regard this as a wall; we have the case of a liquid streaming through a slit of breadth 2c in an infinite plane.
CHAPTER VIII GEOMETRY ON A SURFACE
130. Curvature of Surface Curves. Let C be any curve on the surface r = r(u, v) with unit normal n (94.11). As the point P traverses C with unit speed, the Darboux vector, 6 = TT+KB,
(1)
gives the angular velocity of the moving trihedral TNB. Consider now the motion of the dextral trihedral Tnp (p = T x n) associated with the curve (Fig. 130). Since the trihedrals Tnp and
T
T points upward from paper
FIG. 130
TNB have T in common, the motion of Tnp relative to TNB is a rotation about T. If p = angle (N, n), taken positive in the sense of T, the angular velocity of Tnp relative to TRB is (d
w=S
ds
\T + ds /
-{- KB.
Now N X n = T sin p from the definition of p and n x (N x n) - p sin gyp; hence
N=ncosV - psingyp,
B =TXN=pcosp+nsincp.
Substituting this value of B in (2) gives (3)
(T -f dco/ds) T + K sin p n + K cos (p p. 283
GEOMETRY ON A SURFACE
284
§ 130
The three scalar coefficients in (3) are written t, y, k and are named as follows: (4)
t = r -}- d
(5)
y = K sin (p,
(6)
k= K cos rp, the normal curvature.
the geodesic torsion,
the geodesic curvature,
With this notation,
w = tT+-yn+kp.
(3)'
If we reverse the positive sense on the curve, we must replace T, s,
do
-=wxn= -kT-tnxT. ds
Now n is uniquely defined by (94.11) at all regular points of the
surface and do/ds = T Grad n is the same for all surface curves through a point having a common unit tangent T. From (7), (8)
k = --do T, ds
t = --do nxT, ds
we may therefore state At a given point of a surface, the normal curvature and geodesic torsion are the same for all surface curves having a THEOREM 1.
common tangent there.
Since cos p = n N, we see from (6) that the curvature K = k/cos rp is completely determined by T and N at the point considered, provided p 5-6 a/2. But, since T and N determine the osculating plane at the point (§ 48), we have All curves of a surface passing through a point and having a common osculating plane, not tangent to the surface, have the same curvature there, namely, the curvature of the plane curve cut from the surface by the osculating plane. THEOREM 2.
THL DYADIC vn
§ 131
285
In view of this theorem, we now confine our attention to the curvature of plane sections of the surface. Consider now the normal section Cn of the surface cut by a plane through n and T at P. For this curve Nn = fn, cos cp = =E-1, and Kn = ±k, according as N,, and n have the same or opposite directions. If k 0, the center of curvature of C at P, namely,
Cn = r + N. 1K. = r + n/k, is called the center of normal curvature for the direction T.
But
any surface curve C tangent to T at P has c = r + N/K for its center of curvature. Hence c - cn = N/K - n/k, and (C - Cn.)
T = 0,
(c - C,,)
1
cost
K
k
N=--
= 0,
from (6), so that c - cn is perpendicular to the osculating plane of C. THEOREM 3 (Meusnier). The center of curvature of any surface curve is the projection of the corresponding center of normal curvature upon its osculating plane if the latter is not tangent to the surface.
Curves on a surface along which t, y, or k vanish are named as follows:
t = 0: lines of curvature, -y = 0: geodesic lines or geodesics,
k = 0: asymptotic lines. When t = 0,. do/ds = -k T, from (7); hence (9)
dr do - X - = 0, along a line of curvature. ds ds
When k = 0, do/ds = -t n x T, from (7); hence (10)
dr do
-ds - ds= 0,
along an asymptotic line.
The parametric line v = const is a line of curvature when ru x nu = 0, and an asymptotic line when r,, n,, = 0. 131. The Dyadic Vn. Let e and e' = n X e be two perpendicular
unit tangent vectors to the surface r = r(u, v) at the point Pl.
GEOMETRY ON A SURFACE
286
If k, t and k', t' are the normal curvature and geodesic torsion associated with these directions, we have, from (130.7), do
ds do
ds'
(1)
Grad n = e
_ -ke - tnxe = -ke - te', _ - k'e' - t' n X e' do
ds
+ e'
k'e' + t'e;
do
-- = e(-ke - te') + e'(-k'e' + t'e). ds'
This dyadic is symmetric, since its vector invariant, Rot n = 0 (96.8). (2)
(3)
We have, in fact,
Rot n = - (t + t')n = 0, t' = - t; Vn = -k ee - k'e'e' - t(ee' + e'e).
For brevity, here and elsewhere in this chapter, we use V instead of Grad since no misunderstanding is possible. We state the result (2) as THEOREM 1 (Bonnet). The geodesic torsions associated with any two perpendicular tangents at a point of a surface are equal in magnitude but opposite in sign.
From (1) we find that the first and second scalar invariants of
On are - (k + k') and kk' + tt'. We write (4)
J = k + k' = - Div n,
(5)
K = kk' - t2;
J is called the mean curvature and K the total curvature of the surface at P. Since On and its invariants are point functions over the surface, the values of J and K are independent of the choice of e; we therefore have THEOREM 2. For any pair of perpendicular tangents at P, k + k' and kk' + tt' have the same value.
From (3), we have (6)
k=
t= -eVne';
THE DYADIC C-n
§ 131
287
hence k and t are not altered when e is replaced by -e. If we assume that k remains finite at P and is not constant, there are certain directions for which k attains its extreme values. To find these we examine the variation of k as e revolves about P in the tangent plane. If the angle 0 between e and a fixed line in the tangent plane is taken positive in the sense determined by n, we have (§ 44) de
do
=nXe = e',
de'
-=nXe'= -e; do
and, from (6), (7)
dk de
The extremes of k therefore occur when t = 0. If the direction el gives an extreme value k1, tl = 0. Then the perpendicular direction e2 = n x e1, for which t2 = 0 from (2), gives another extreme value k2. But since k + k' is constant, if k = k1 is a maximum, k' = k2 is a minimum, and vice versa. If we take e = e1, e' = e2 in (3), we have Vn = -k1e1e1 - k2e2e2, (8) and the symmetric dyadic Vn appears in the standard form (72.4). Evidently el and e2 are the invariant directions of Vn with multi-
pliers -k1, -k2. Moreover, if k is not constant, ±e1 and fee are the only invariant directions at P. We state these results in THEOREM 3. If the normal curvature is finite and not constant at a point of the surface, it attains its maximum and minimum values for just two normal sections, at right angles to each other, and char-
acterized by the vanishing of the geodesic torsion.
The orthogonal directions e1, e2 are called the principal directions at P, and the corresponding normal curvatures k1, k2, the principal curvatures. From (4) and (5), (9)
J = kl + k2,
K = k1k2;
consequently k1 and k2 are roots of the quadratic, (10)
k2-Jk+K=0,
the characteristic equation (§ 71) for the symmetric planar dyadic On.
GEOMETRY ON A SURFACE
288
§ 131
Consider finally the case when k is constant at P; from (7), t = 0 for all directions, and (3) becomes
On = -k(ee + e'e') = -k(I - nn).
(11)
Any direction in the tangent plane at P is an invariant direction with multiplier - k, and P is called an umbilical point or simply an umbilic.
If k is constant over the entire surface, we have
Vn = -k or,
V (n + kr) = 0,
and n + kr is constant over the surface. If k = 0, n is a constant, and the surface is a plane. If k 0,
kr=c, -n
Ir-c12 = 1/k2;
and the surface is a sphere (center c, radius 1/k). Therefore the only surfaces whose points are all umbilical are the plane and sphere.
All curves on the plane or sphere are lines of curvature (t = 0). Example 1. Formulas (6) give k and t for any direction e. If the angle (ele) = 0 is reckoned positive in the sense of n (el x e = n sin 0),
e =elcos0+e2sin0,
e' = -elsin0+e2cosO.
Taking Vn in the form (8), we have (12)
k = e (klelel + k2e2e2) e = ki eos2 0 + k2 sin2 0,
(13)
t=e
(k1elel + k2e2e2) e' _ (k2 - k1) sin 0 cos 0.
These equations are due, respectively, to Euler and Bonnet. Since 2t = (k2 - ki) sin 20, it is clear that t attains its extreme values ±(k2 - kl)/2 for the directions 0 = fir/4. At a surface point the directions for which k = 0 are called From (12), the asymptotic directions are given by tan20 = -ki/k2 and are real and distinct only when K = klk2 < 0. When kl = 0, Example 2.
asymptotic.
k2 0 0, both asymptotic directions coalesce with the principal direction el. In an asymptotic direction, t2 = -K, from (5). Along an asymptotic line, k = K cos p = 0; and, if K , 0, cos v = 0, and ,p = =1= 7r/2. Along a curved asymptotic line, the osculating plane remains tangent to the surface; moreover -y = ±K, and t = T. Referred to the principal direction el the asymptotic directions have the and are perpendicular when and only when these slopes slopes f
are ±1; then -kl/k2 = 1, and J = kl + k2 = 0.
FUNDAMENTAL FORMS
§ 132
289
Example 3. If e is the unit tangent vector of a surface curve, the trihedral ee'n has the angular velocity,
w = to - ke' + yn
(14)
(130.3)',
along the curve, and de
ds
de' ds
= cu x e = kn + ye',
= w x e' = to - ye.
On differentiating equations (6), we now have
UdVn / e,
ds ds
d-s
Since dVn/ds is the same along all surface curves having the common tangent e at a point, the same is true of the expressions,
ds - 27t
(15)
and
d-s
+ y(k - k'),
respectively discovered by Laguerre and Darboux.
132. Fundamental Forms. On the surface r = r(u, v), the gradient of a tensor function f(u, v) is given by (95.5),
Vf = afu + bf,
(1)
where a, b, n is the set reciprocal to ru, r, n:
aru + br + nn = I.
(2)
From (1), we have, in particular,
Vu=a, Vv=b; Vr = aru + br = I - nn.
(3)
(4)
The dyadic Vr is symmetric and may also be written rua + rub; it transforms any vector f(u, v) into its projection ft on the tangent plane at (u, v) : (5) Vr acts as an idemfactor on vectors tangent to the surface at the point (u, v) and on dyadics whose vectors lie in the
tangent plane. If we form the product, (Vr)
(Vr) = (aru +
(rua + r,;b),
GEOMETRY ON A SURFACE
290
§ 132
we obtain (6)
yr = E as + Flab + ba) + G bb,
where (7)
E = ru ru,
F = ru
rv,
G = rv r,
are the coefficients of the first fundamental form, (8)
dr dr = E due + 2F du dv + G dv2,
In fact (8) follows from (6) when we form dr yr dr and note that a . dr = du, b dr = dv, from (3). defined in § 94.
If we compute
(vn) - (vr) = (anu + bn,) (rua + we obtain, in similar fashion, (9)
- Vn = L as + M(ab + ba) + N bb,
where
L = - nu ru = n ruu, (10)
M = -nu r _ -nv ru = n ruv,
The relations between scalar products in (10) are obtained by differentiating the equations,
=0, with respect to u and v :
nu ru + n ruu = 0,
nv ru + n ruv = 0,
nu rv+n rvv=0,
n,,
From (9) we obtain the second fundamental form, (11)
-dr do = L du2 + 2M du dv + N dv2,
by computing the product dr Vn dr. Remembering that Vn is symmetric, we next compute (Vn)
(Vn) = (anu + bnv) (nua + nab)
to obtain (12)
(Vn)2 = e as + flab + ba) + g bb,
FUNDAMENTAL FORMS
§ 132
291
where
e = nu nu,
(13)
f = nu
9=nn
n,,,
From (12), we obtain the third fundamental form,
do do = e due + 2f du dv + g dv2,
(14)
by computing the product dr - (Vn)2 dr. The quantities (7), (10), and (13) are known, respectively, as the fundamental quantities of the first, second, and third orders. If e1, e2 are unit vectors in the principal directions at P, Vr = e1e1 + e2e2,
Vn = -k1 e1e1 - k2 e2e2, (on)2 = ki e1e1 -1- k2 e2e2,
and, on multiplying these equations in turn by K = k1k2, J. _ k1 + k2, 1, and adding, we get
(on)2+JVn+KVr=0.
(15)
This is the Hamilton-Cayley Equation for the planar dyadic Vn; for its first and second scalar invariants are -J, K, and Vr is the idemfactor in the tangent plane. Substituting from (6), (9), and (12) for the dyadics in (15), we obtain the following relations between the fundamental quantities: (16)
e-JL+KE=0, f -JM+KF=0, g-JN+KG=0.
Example 1. (17)
From the reciprocity of a, b, n and ru, r,., n, r,, x ,
ru x r,,
bnxru,
H
n; hence
(18)
We now may compute the invariants -J and K of Vn: Vn = anu +bn,,,
(On)2 = axbnuxn,,;
ruxrv n K = (a x b) . (nu x ne,) = nu x nv a
ruxr,- n
292
GEOMETRY ON A SURFACE
Since the cross products in these equations are all parallel to n, these equations may be written also in the vector form: -J ru x r,. = nu x r,, + ru x n,.,
(19)
K ru x rn = nu x nv.
(20)
If we multiply these equations by (ru x r,;) , apply (20.1), and introduce the fundamental quantities from (7) and (10), we have (21)
(22)
J(EG - F2) = GL - 2F31 + EN, K(EG - F2) = LN - 312.
These values of J and K also may be found from (9)
-Vn = a(La + Mb) + b(Ma + Nb),
by making use of (18). Example 2. The surface z = z(x, y), with the position vector,
r=xi+yl+z(x,y)k, can be written in the parametric form
x=u, y=v, z=z(u,v). If we denote the partial derivatives z.,, zy, zxx, zxy = zyx, zyy by p, q, r, s, t, respectively, we have
ru=rx=i+pk,
=ry=l+qk;
r,,
hence, from (7),
E = 1 + p2, F = pq, G = 1 + q2;
H=EG-F2=1+ p2+g2; Hn=ruxru= -pi - gl+k. Furthermore,
ruu = rk, ru = sk, hence, from (10),
r,.y. = tk;
L=r/H, M=s/H, N=t/H.
We may now compute the mean curvature J from (21), the total curvature K from (22): (23)
T- (1+q2)r-2pgs+(1+p2)t (1 + p2 + q2)
(24)
rt - s2
K = (1+p2+q2)2
The principal curvatures k1, k2 are the roots of the quadratic (131.10): (25)
k2 - Jk + K = 0.
§ 134
TI-HE' FIELD DYADIC
293
133. Field of Curves. A one-parameter family of curves on a surface r = r(u, v) is said to form a field over a portion S of the surface if one and only one curve of the family passes through every point of S. After the positive direction on one of the curves
has been chosen, the positive direction on the others is taken so that the unit tangent vector to the curves is a continuous vector point function over S.
If the curves of the field have the equation p(u, v) = const, their differential equation is ccudu
From this we may compute Al = dv/du at all points of S. Then the vector, ar av dr ar ru -I Alrv, du a u + av au is tangent to the field curve through the point (u, v). Similarly if A2 = dv/du for a second one-parameter family of
curves, the vector ru + A2r is a tangent to the curve through (u, v). When the second family cuts the first everywhere at right angles,
(ru + Air,.) (ru + A2r,.) = 0,
or, in terms of fundamental quantities, (1)
E + (A1 + A2)F + A1A2 G = 0.
Since Al = -,pu/p is known, (1) is the differential equation of the orthogonal trajectories of the field curves. Standard existence theorems for differential equations of the first order guarantee, under very general conditions, a solution of (1). Since ru.,r are tangent vectors to the curves v = const, u =
const, the parametric curves cut at right angles when (2)
134. The Field Dyadic. Consider a field of curves C1 and their orthogonal trajectories C2 on a portion S of the surface r = r(u, v). At every point (u, v) their unit tangents e1, e2 = n x e1, and the surface normal n form a dextral trihedral of unit vectors e1e2n.
As a point traverses CI with unit speed, e1e2n has the angular velocity (130.3), (1)
wl = t1e1 - k1e2 + yin;
GEOMETRY ON A SURFACE
294
§ 134
since p1 = el x n = -e2. Similarly, along C2, (02 = t2e2 + k2e1 + y2n,
(2)
since p2 = e2 x n = e1. Therefore de1
= w1 x e1 = kin + yle2,
ds1
de1
ds2
= w2 x e1 = -t2n + y2e2
We now have, from (95.8), de1 de1 Grad e1 = e1 ds1 + e2 ds2
= e1(k1n + y1e2) + e2(t1n + 72e2) = (k1e1 + t1e2)n + (y1e1 + y2e2)e2,
since t2 = -t1 by Bonnet's Theorem (§ 131). Now, from (130.7), do
-k1e1 - t1 n x e1;
ds1
and, if we write R = y1e1 + 72e2,
(3)
the field dyadic Grad e1 for the curves Cl becomes do vet =
(4)
-
n + R e2ds1
If we replace e1 and e2 in (3) by n x e1 = e2, n x e2 = -e1, R is unchanged, since y2e2 + (-y1)(-e1) = R. Therefore the field dyadic for the curves C2 is do vet =- ds2 -n - Re,.
(5)
We have moreover (6)
do do do do Vn=e1-+e2-=-e1+-e2. ds2 ds2 ds1 ds1
With the notation, (7)
P =
do ds1
do ds2
THE FIELD DYADIC
§ 1.34
295
the preceding equations become (4)'
Vet = Re2 - Pn,
(5)'
De2 = - Re 1 - Qn,
(6)'
Vn = Pel + Qe2.
From (4), we have
or, in view of (3),
y1 = n rot el. The geodesic curvature is therefore a surface invariant. THEOREM. A field of curves having the unit tangent vector e(u, v) has at every point the geodesic curvature, (8)
y = n rot e =
1
H
{ (r e),, - (ru
The last expression follows from (97.9).
Consider now a, curve C which cuts the curves Cl at an angle 6 = (eli e), reckoned positive in the sense of n. Along C the trihedral ee'n has the angular velocity
w = to - ke' + yn
(9)
(131.14);
and, since the trihedral e1e2n has the angular velocity - (de/ds) n relative to ee'n, the angular velocity of e1e2n along C is (10)
dB\
dB
w -asn=to - ke'-F
hence
de1 ds
y
-ds n'
\ / w -n- Xe1, ds
dO \
dO
The left member of (11) shows that y - dB/ds is the same for all surface having e as common tangent vector.
GEOMETRY ON A SURFACE
296
§ 134
Since Vel e2 = R, from (4), we may write (11) in the form: de
(12)
e R = ds
If C is a member of a field of curves, the angle 0 = (el, e) is a point function over this field, and
=ye+y'e'-OB.
R=
(13)
If the field curves C, and C cut everywhere at the same angle,
VO=0,and R=ye+y'e'. Example. For a field of surface curves p(u, v) = c, the unit tangent vec-
tor is
where puu.+cvv =0.
e =dr/ds Therefore y
-U _ - 'Pu _ ` ;
u _ cv- ,
1
X
'Pv
(Pa
v A
and, if we substitute these values in
e e =Eci2+2Fuv+Gi,2 = 1, we have
= E Vn - 2F vupt, + G wo to find X. Thus
e=
where the sign is chosen to give the positive sense desired. The geodesic curvature now is given by (8) :
(FP,, (14)
H lau
a
(EPv - Fvu)1
Gpu) _ av
This formula is due to Bonnet. Let us apply it to find the geodesic curvature of the parametric curves in the sense of increasing u and v. For the curves v = const, vu = 0, pv = 1, a = 1/E, el = ru/N/-r,,; for the curves u = const, vu = 1, 'RU = 0, a
e3 = r/1'G; hence in the respective cases, (15)
Y1 =
H
S
av
E}
ya =
-G 1 } l iau
G
av
\1
These formulas also follow at once from (8) when the foregoing values of el and e3 are used.
The subscript 3 refers to the curves u = const when they cut the curves v = const at an arbitrary angle; when this angle is it/2, we use the subscript 2.
GEODESICS
§ 135
297
135. Geodesics. The geodesic curvature of a curve C has been defined (§ 130) as (1)
K sin gyp,
where p, the angle (N, n), is taken positive in the sense of T. Since dT/ds = KN, dT
-xn = KNXn = KSin VT = yT;
(2)
ds
and, as dT/ds is unaltered by a change in positive direction, y must change sign with T (§ 130). A surface curve for which y = 0 is called a geodesic. Thus, if a straight line can be drawn on a surface, it is necessarily a geodesic since K = 0. If a geodesic is curved (K 0), y = 0 implies sin
Along a curved geodesic the principal normal is
always normal to the surface.
Along a curved geodesic, (3)
dip
t = T + - = T;
k = K COS p = ±K,
ds
and, if the geodesic is straight (K = r = 0) and the same equations
apply (k = t = 0). Along a geodesic the normal curvature is numerically the same as the curvature and the geodesic torsion equals the THEOREM 2.
torsion.
On putting y = 0 in (2), we have (4)
d2r
- x n = 0, ds2
or
or
d2r ds2
= 0,
since the tangential projection of d2r/ds2 is zero (132.5). If we replace Vr by aru + br,,, we obtain the scalar differential equations of a geodesic: (5)
ru
d2 r d82
= 0,
d2r
r - = 0. dS2
These differential equations of the second order show that the geodesics on a surface form a two-parameter family. In general, a geodesic may be determined by two conditions, by specifying (a)
298
GEOMETRY ON A SURFACE
a 1:35
that it shall pass through a given point in a given direction, or (b), that it shall pass through two given points. That a geodesic on a surface in general can be found to fullill conditions of the type (a) or (b) is plausible in view of the follow. ing theorems from mechanics. THEOREM 3. A particle constrained to move on a surface and fief from the action of any tangential forces will describe a geodesic writi constant speed.
Proof. In view of (52.5), the equation of motion ma = F may
be written (ddt
m
T + Kv2 N)
= pn,
where v is the speed and pn the normal force. Multiplying by T
gives dv/dt = 0, v = const. If K = 0, the particle describes a straight line. If K F4- 0, N = ±n, and sin cp = 0. Since either K or sin (p must vanish, y = 0 in both cases, and the particle describes a geodesic.
The pressure I p I = mKV2.
THEOREM 4. If a weightless flexible cord is stretched over a smooth
surface between two of its points, its tension is constant, and the line of contact is a geodesic. Proof. Since the cord is in equilibrium, the vector sum of all the forces acting upon any portion of it vanishes. Let F denote
FT
Fia. 135
the magnitude of the tension at P, distant s (arc POP) from the fixed end of the cord, and pn the normal reaction of the surface per unit length. Then from the equilibrium of the length s of the cord (Fig. 135),
FT - FOTO +
o s pn 0
ds = 0.
GEODESIC FIELD
§ 136
299
On differentiating this equation with respect to s, we have dF
- T+FKN+pn = 0; ds
hence dF/ds = 0, and F is constant. If K = 0, the line of contact is straight; if K n= ±n, and I p I = KF. In either case, -y = 0, and the line of contact is a geodesic. As a cord can be stretched between any two points of a convex surface, the line of contact is a geodesic fulfilling condition (b). On concave surfaces we must imagine the cord replaced by a thin strip of spring steel laid flatwise. 136. Geodesic Field. Consider now a one-parameter family of geodesics that form a field of tangent vector el over a portion S of the surface.
Then, since
(1)
0,
y1
we have, from Stokes' Theorem,
fT
e1 ds =
fn rote,dS=0,
for any sectionally smooth closed curve lying entirely within S. Writing 0 = angle (e1, T) gives
fcos 0 ds = 0
(2)
as the integral equivalent of (1). This formula leads to THEOREM 1. An arc of a geodesic that is one of the curves of a geodesic field is shorter than any other surface curve joining its end points and lying entirely within, the field.
Proof.
Let AI'B be a geodesic arc of the field and AQB any
other curve of surface covered by the field (Fig. 136a). Applying (2) to the circuit APBQB formed by these arcs, we have
f ds + PB
BQA
f PB(ls = f QBcos 0 ds < /QBI cos 0 1 ds < fQBds;
that is, are APB < are AQB.
GEOMETRY ON A SURFACE
300
§ 137
We next consider the orthogonal trajectories of a geodesic field. Their basic property is given by THEOREM 2 (Gauss). The orthogonal trajectories of a geodesic field intercept equal arcs on the geodesics. Q
B!
_A'
B-
-A
FIG. 136b
FIG. 136a
Proof. Let AB, A'B' be geodesic arcs intercepted between the
curves AA', BY cutting them at right angles (Fig. 136b). Then applying (2) to the circuit ABB'A'A, and, noting that
cos0 = 1,0, -1,0 over AB, BY, B'A', A'A, we have are AB = arc A'B'. 137. Equations of Codazzi and Gauss. These celebrated equations in the theory of surfaces simply state that (1)
n - Qx(Vn) = 0,
(2)
n - V x (Ve1) = 0.
If we remember that V means V8, both identities follow from (97.11).
In order to express these identities in terms of the vectors, (3)
P = do/dsl,
Q = dn/ds2,
R = y1e1 + '2e2,
we make use of (97.12) :
n - Vx(pq) _ On substituting
(4)
Vn = Pe1 + Qe2
(5)
in (1), we have, from (4),
(n - rotP)e1 +
0;
and, if we put (§ 134) (6)
Vet = Re2 - Pn,
Ve2 = -Re1 - Qn,
EQUATIONS OF CODAZZI AND GAUSS
§ 137
301
we have
rot P - Q n x R)e, + (n rot Q + P n x R)e2 = 0.
(n
This equivalent to the two equations: (7)
n- rot P = -n - QxR,
(8)
n rot Q = -n R x P.
These are the Equations of Codazzi.
We next put Ve, = Re2 - Pn in (2) and obtain (n - rot R)e2
- (n -
Replacing vet and Vn by the foregoing values, we have
This is also equivalent to two equations. One of these is the same as (7) ; the other, (9)
is the Equation of Gauss.
In the right-hand member, K,
the second scalar invariant of Vn or the total curvature Gauss's Equation thus becomes
(§ 131).
n rot R = -K.
(10)
This is perhaps the most important result in the theory of surfaces. The equations of Codazzi and Gauss can be expressed in terms of the quantities ki, tt;_,yi along the field curves.t From (130.7), do
-
i
= -kie, - tin x e,,
and, since t.2 _ (11)
n,,
-P = kie, + tie2i
do
as = -k2e2 - t2n x e2, 2
-ei, -Q = tie, + k2e2
(134.7).
In order to compute the left members of (7), (8), (10), we apply the identity, (12) t Now k1 and k2 denote the normal curvatures of the orthogonal field curves; only when t1 = t2 = 0 are k1 and k2 principal curvatures (§ 131).
GEOMETRY ON A SURFACE
302
§ 13S
which follows at once from (85.3), to both terms of -P, -Q and R. Remembering that n rot ei = yt, we thus obtain dk1
dt1
-n rot P = y1k1 + y2t1ds2- - +dsl-n rot Q = 'Y1t1 + -12k2 -
yi+y2
dt1
ds2
+
n Q x R = ylk2 - y2tl,
dsl
dye - dyl ds + ds 2
and, since
dk2
1
n R x P = -yltl + y2k1,
the Equations of Codazzi and Gauss become (13)
y1(k1 - k2) + 2724 +
(14)
72(k2 - k1) + 2y1t1 +
dt1
dk1
dk2
dt1
-ds1- -ds2= 0,
- ds = 0, dsl yi+yi+---+h'=0. ds2 2
(15)
dye
2
d-11
ds1
Equation (14) states the same property for the e2-field that (13) states for the el-field. To show this, change the subscripts 1, 2 in (13) into 2, -1 (n x e2 = -e1) ; then, since k-1 = k1i ds_1 = -ds1, t2 = -t1, y-1 = -71, we obtain (14). 138. Lines of Curvature. The curves on the surface which are everywhere tangent to the principal directions (§ 131) are called lines of curvature. Along a line of curvature the geodesic torsion
is zero (t = 0), and the torsion z = - dcp/ds. From (130.7), we have do/ds = -kT along lines of curvature. Their differential equation is therefore dr do
= 0,
or
dr do
= 0;
n'dsxds for (dr/ds) x (dn/ds) is always parallel to n. In particular, the (1)
dsxds
parametric curves are lines of curvature, if
ruxnu = 0,
0;
LINES OF CURVATURE
§ 138
303
then, if k1, k2 are the principal curvatures,
nu = -klru,
(2)
n = -k2r,.
On the plane and sphere, all curves are lines of curvature (§ 131).
On other surfaces there are in general two orthogonal principal directions at each point; and the lines of curvature form two fields cutting each other at right angles. If el and e2 = n X el are the corresponding unit field vectors, the Equations of Codazzi become dk1
dk2
//
= y2(kl -
= 'Y1 \ki - k2),
(3)
k2).
dsl
These lead to a simple proof of THEOREM 1.
If K = 0, J 0 0, the lines of curvature along which
the normal curvature vanishes are straight lines. Proof. Since K = klk2 = 0, J = k1 + k2 0 0, we may suppose
that k1 = 0, k2 0 0. Hence yl = 0, from (2); and, since kl = K1 COS cp = 0,
yj = Kj sin c = 0,
we conclude that K1 = 0. Along the rulings of the surface (which are asymptotic lines as
well as lines of curvature) t1 = 0, kl = 0, and do/dsl = 0 (130.7); hence n has a fixed direction along a ruling. This also follows from (134.1); for co = 0, and the trihedral ele2n has a fixed orientation along a ruling. In general a surface has a different tangent plane at each point and is therefore the envelope of a two-parameter family of planes. However the ruled surface under considera-
tion has the same tangent plane along an entire ruling and is therefore the envelope of a one-parameter family of planes. Such a surface is called developable.
Consider now the lines of curvature cutting the rulings orthogonally. Their tangent vector e2 = n x el is constant along a ruling. From (137.14) and (137.15), dye
dk2 ds1
122)
dsl
= -y2i2.
_ d
_
dsl
k2
'Y2
=
0.
But y2/k2 = tan P2; hence X02 = (N2, n) is constant along a ruling, and N2 as well. This property lends plausibility to the fact that
the developable surface may be bent into a plane (after certain cuts are made) without stretching or tearing.
GEOMETRY ON A SURFACE
304
§138
We now may amplify theorem 1: If K = 0, J 54 0, the surface is developable. We shall see in § 142 that developable surfaces are of three types; cylinders (including planes), cones, and tangent surfaces (generated by the tangents to a curve which is not a straight line). The normals to a surface along one of its curves have
THEOREM 2.
an envelope when and only when the curve is a line of curvature; in this case the envelope is the locus of the centers of normal curvature along the curve.
Proof. Let s denote the are measured along the curve C of the surface S, and r(s) and n(s) the position vector and unit surface normal at a point of C. The points on the family of normals have position vectors r(s) + X n(s), where X is a variable scalar. If we take X as a function of s, say X = X(s), the curve C,
f = r(s) + X(s)n(s) will be the envelope of the normals if df
-
=T+X
do
dX
+
dsn
or
do
T+X --
is parallel to n. But since both T and do/ds are perpendicular to n, the envelope will exist only when T+X
do
= (1 - Xk)T - Xt n x T = 0
(130.7),
that is, when t = 0 and X = 1/k. Therefore the normals have an envelope only along lines of curvature; and then the envelope is the curve f = r + n/k, the locus of the centers of normal curvature along C. We turn now to some theorems relative to the lines of intersection of surfaces. THEOREM 3.
If two surfaces S1, S2 cut under a constant angle 0,
the curve of intersection has the same geodesic torsion whether regarded
as a curve of Sl or of S2. Proof.
If N is the unit principal normal along the curve of
intersection,
P2 - Pl = (N, n2) - (N, nl) = (n1, N) + (N, n2) = (n1, n2) = 0,
LINES OF CURVATURE
§ 138
305
and
t2-tl =T -}---- T -(lg'ds-_-do =O. ds d.s
From this result, we have at once THEOREM 4 (Joachimsthal). If two surfaces cut under a constant angle, their curve of intersection is a line of curvature of both or of neither; and, conversely, if the line of intersection is a line of curvature of both, the surfaces cut under a constant angle.
We are now in position to prove the celebrated THEOREM 5 (Dupin). Two surfaces belonging to different families of a triple orthogonal system cut one another in lines of curvature of each.
Proof. Denote the geodesic torsion at a point P on the curve of intersection of the surfaces Si and Sj by tij or tji, according as the curve is regarded as belonging to Si or Sj. Now at the point P we have
tij = -tik (§ 131, theorem 1),
tij = tji (theorem 3),
where i, j, k represents any permutation of the indices 1, 2, 3. Then tij = - tik = - tki = tkj = tik = - tji = - tij,
so that tij = 0. Example. A surface of revolution about the z-axis has the parametric equations,
z = u cos v,
z = z(u),
y = u sin v,
where u, v are the plane polar coordinates (p, gyp). equation, r = u R(v) + k z(u),
where R is a unit radial vector (R k = 0).
ru=R+kz',
These give the vector
Now
r =uP;
ruXr k - z'R n = ru x r _ (t + z,2)
nu = -
z" (1 + z,2)2
ru,
n
z'
u(1
+
, r,,.
z'2)=
GEOMETRY ON A SURFACE
306
§ l39
The last equations show that the parametric lines (the parallels and meridians of our surface) are the lines of curvature, and that the principal curvaturv are Z/1 z' k2 = kl = (1 + z) 1' u(1 + z,2)z Consequently, z' z" K = k l k2 =
(4)
u(l + z,2)2
J=k1+k2= uz" + z'(l + z'2)
(5)
u(l + z'2)2
Let us apply these results to find K and J for the torus generated by revolving the circle,
(u-a)2+z2=b2, about the z-axis.
On differentiation, we find
zt =
a-u
b2
1 +z'2
z
b2
z" _ -
z2
2
3
hence, from (4) and (5), _u - a
K
J
a-2u bu
b2u
If we introduce the latitude 0 on the generating circle, u = a + b cos 0, and
K_
(6)
139. Total Curvature. (1)
cos0 -J ba+bcos0 +
Cos0
1
b(a+bcose)'
1Y1
f- yz
The Gauss Equation (137.15),
+ dyl
ds1
+ K = 0, ds2
shows that, if orthogonal geodesics can be chosen as fielt ^urves,
71 = 72 = 0, and K = 0; hence THEOREM 1. Orthogonal geodesic fields can exist only on of zero total curvature.
We now compute K from the Gauss Equation (137.10); using (97.9), we have (2)
- K = n rot R =
1
- { (r R)u - (ru H
TOTAL CURVATURE
§ 139
307
Let the parametric curves cut at an angle 0 = (el, e3) where el = Then, if e2 = n x el, e4 = n x e3i we have,
ru/-/E, e3 = r, from (134.13),
R = ylel + y2e2 = y3e3 + y4e4 - V0;
and
yj
-/ ae y3VG-av
-Vi, /
(95.7).
Substitution in (2) now gives the elegant Formula of Liouville for thi' total curvature, (3)
K
1 ja - H 1av (yl
1/E)
-a
au
020
(-I"-\/G' ) +
au av.
in which II = 1/EG - F2, yl and 73 are given by (134.15), and F
(4)
-\/EG
EG
When the parametric curves are orthogonal, 0 = it/2 and F = 0, EG. Equations (134.15) now give
II = (5)
1 aE/av
yl = - 21i NIT '
73
1 aG/au
- 211 \/G
and Liouville's Formula becomes (6)
h'
1
211
aau (il_ I aG a / 1 aE)l au + av \II a v f
From (3), we obtain an immediate proof of The total curvature K of a surface depends only upon ' e coefficients E, F, G of the first fundamental form and their fir f .,and second partial derivatives with respect to u and v. THEOREM 2 (Gauss).
This iu.idamental theorem was first proven by Gauss after long and t' iious calculations. Gauss's name of "Theorema egregium" (La' n egregius means literally "out of the herd") for this result shows that he was fully aware of its importance. Example. To find a surface of revolution of constant negative total curva-
ture, we set K = -1/a2 in (138.4). The resulting differential equation, 2z'z" (1 + z'2)2
2u
a2 '
GEOMETRY ON A SURFACE
308
§ 140
has the first integral, 1
u2
1 + z'2
a2
A or cost i = u2 a2
+ A,
where >G = tan -l z' is the inclineftion of the tangent to the u-axis (Fig. 139).
If we impose the condition u = -a when 4, = 0, A = 0 and u = -a Now dz d,k
=
dz du du dik
= tank a sin 0 = a(sec ¢ - cos 0,
z = a log (sec V, + tan p) - a sin ¢ + B;
FIG. 139
and B = 0 if z = 0 when ¢ = 0. Therefore the meridian of our surface of constant negative K has the parametric equations:
u = -a cos ¢,
z = a log (sec V, + tan ik) - a sin ik.
This curve, asymptotic to the z-axis as 4, --> a/2, is called a tractrix. The equation u = -a cos yG shows that the segment of any tangent between the curve and z-axis has the constant length a. This property is characteristic of the tractrix (ef. § 50, ex. 5).
140. Bonnet's Integral Formula. The differential formula for the total curvature, (1)
K = -n rot R,
has an important integral equivalent. If we integrate K over a portion of a surface S bounded by a simple closed curve which consists of a finite number of smooth arcs, we have, by Stokes'
BONNET'S INTEGRAL FORMULA
§ 140
309
Theorem,
fKdS= from (134.12); hence (2)
fKds =
f f do -
f(. -
ds,
y ds.
If the bounding curve has a continuously turning tangent,
fdo
= 2ir and (2) becomes (3)
fKdS
= 2ir
fds.
f
This very important formula was discovered by the French geom-
eter, Bonnet, in 1848. The integral
K dS over S is called the
integral curvature of S. Let us first apply (2) to the figure bounded by two geodesic arcs
APB and AQB meeting at A and B. Then denoting the interior angles at the corners by A and B (Fig. 140a), we have (4)
fK dS = fdo = 2ir - (ir - A) - (ir - B) =A+ B.
Since A + B > 0 this equation is impossible when K = 0; hence, on a surface of negative or zero total curvature, two geodesic arcs
Fu}. 140a
FIG. 140b
GEOMETRY ON A SURFACE
310
3 14t)
cannot meet in two points so as to enclose a simply connected area. We may state (4) as follows: THEOREM 1. The integral curvature of a geodesic lune is equal to the sum of its interior angles.
We next apply (2) to a geodesic triangle ABC (Fig. 140b) : that is, the figure enclosed by three geodesic arcs. Again denoting the interior angles at the corners by A, B, C, we have
fKdS = fdO=27r- (7r-A) - (ir-B) - (rr-C), (5)
fKds=A+B+C_.
THEOREM 2 (Gauss).
The integral curvature of a geodesic triangle
is equal to its "angular excess," that is, the excess of the sum of its angles over 7r.
When K is constant, the integral curvature
f
K dS is the prod-
uct of K by the area S. If K is identically zero, as on a plane, the sum of the angles of a geodesic triangle is 7r. If K 34 0, we have THEOREM 3 (Gauss). On a surface of constant non-zero total curva-
ture, the area of a geodesic triangle is equal to the quotient of its angular excess by the total curvature. Example 1. A sphere of radius r has the constant total curvature K = I /r2. For all normal sections at a point are great circles of curvature 1/r; all points are umbilical points and k1k2 = 1/r2. Formula (5) now states that: The area of a spherical triangle is r2 times its angular excess in radians. If e denotes the angular excess in right angles, we therefore have the formula,
Area = r2 2 e =
4,rr2
8
e,
that is, the area of a spherical triangle equals the area of a spherical octant times the angular excess in right angles. Example 2. Let us decompose a closed bilateral surface S, consisting entirely of regular points (§ 93), into a number f of curvilinear "polygons" Si whose sides are analytic arcs. These sides meet in vertices at which at least three arcs come together. Since S is regular and bilateral, the surface has a continuous unit external normal n which defines a positive sense of circuit on each polygon by the rule of the right-handed screw. When positive circuits
BONNET'S INTEGRAL FORMULA
§ 140
311
are made about two adjoining polygons, their common side is traversed in opposite directions. Now for each Si we have, by Bonnet's Theorem (2),
J(i KdS=27r-
(7r-ai;)-Js
where ail are the interior angles at the vertices of Si. If these equations for all of the f polygons Si are added, we have
fKdS
i
j
(a -
2irf -
for the integrals f y ds over adjoining sides cancel in pairs since y changes sign with change of direction. In each polygon the number of interior angles equals the number of sides; hence 2;2:7r = 2eir, where e is the total number of sides or edges. Moreover the sum of the interior angles about any vertex is
27r; hence TdIai; = 2v7r, where v is the total number of vertices. We thus obtain
f-e+v=1
(6)
KdS.
is,
From the right member of (6) we see that the number on the left is independent of the manner in which S is subdivided into polygons. From the left member
we conclude that cb K dS is not altered when S is deformed into another completely regular surface. If S can be continuously deformed into a sphere, we have
fK dS =
(7)
47rr2 = 4a
for a sphere, and (8)
f - e + v = 2.
Evidently f - e + v is not altered by any continuous deformation of S, even though the resulting surface is not completely regular. Thus, if S is transformed into a polyhedron (a closed solid with plane polygons for faces), the relation (8) still holds good and constitutes the famous Polyhedron Formula that Euler discovered in 1752: Faces + vertices - edges = 2. Next suppose that S can be continuously deformed into a torus. From (138.6), for a torus, (9)
f"KdS=ffKb(a+bcoso)dedv
2
= 0
f
2"
cos e do dv, = 0;
0
hence f - e + v = 0 for any surface continuously deformable into a torus. This relation holds, for example, for any ring surface with polyhedral faces.
GEOMETRY ON A SURFACE
312
§140
Example 3. Parallel Displacement. A vector f, that always remains tangent to a surface S, is said to undergo a parallel displacement along a surface curve C when the component of df/ds tangential to the surface is zero; then df
df nx--=0 or
(10)
ds=>`n.
Since
the length of f must remain constant during a parallel displacement. Now the trihedral Tnp associated with the curve C has the angular velocity,
w = tT + yn + kp
(130.3').
If the angle 0 = (T, f) is reckoned positive in the sense of n, the trihedral fng (g = f x n) has the angular velocity n dB/ds relative to Tnp. Hence, as f moves along C with unit speed, the trihedral fng revolves with the angular velocity,
SZ = to + n - = tT + (y + dB/ds)n + kp, hence df/ds = St x f, and
dfxn = (12 xf) xn = (n - fl)f =
+do> f. C'Y
Hence f will undergo a parallel displacement along the curve C when and only when de
y+ - =0.
(11)
If f is tangent to C (or, more generally, when 0 remains constant), f will undergo a parallel displacement along C only when C is a geodesic. THEOREM. If a tangential surface vector is given a parallel displacement about a smooth closed curve C, it will revolve through an angle, (12)
27r - if y ds = f K dS.
Proof. From (11) we see that f revolves through the angle,
if
ds
-
fy ds
relative to T, the unit tangent vector of C. Since T itself revolves through 2r
in passing around C, the total rotation of f is 27r Bonnet's Integral Formula (3).
fy
ds,
orJK dS by JJJJJ
NORMAL SYSTEMS
§ 141
313
141. Normal Systems. At every point of the surface S, r = r(u, v), a straight line is defined by the unit vector m(u, v). The points of this two-parameter family of lines are given by
rl = r + Am.
(1)
Under what circumstances do these lines admit an orthogonal surface? That is, when do they form the system of normals to a surface?
When A is a definite continuous function A(u, v), r1 is the position vector of a surface S1. If V denotes a surface gradient relative to S, we have, from (1), Vr1 = Or + (VX)m + AOm,
+VA,
since m m = 1. In Vr1 = arlu + brl,,, the postfactors are tangent to S1; hence, in order that m be normal to S1, it is necessary and sufficient that (Or1) m = 0, that is,
- VA = (Or) m = m - (m n)n
(2)
(132.5).
But (2) implies that
n rot m = 0;
(3)
conversely, from § 103, theorem 2, (3) implies that (4)
mg = (Or) m = -VA where X= -
f m dr, ro
the integral being taken over any path from ro to r(u, v) on S. We have thus proved THEOREM 1.
In order that a two-parameter family of lines
r(u, v) + Xm(u, v) form a normal system, it is necessary and suffi-
cient that n rot m = 0, where n is the unit normal to the surface r = r(u, v). When m fulfils condition (3), we can find a one-parameter family
of surfaces normal to the lines r + Am. We need only determine A(u, v) from (4) with some definite choice of ro. Since ro may be chosen at pleasure, all possible values of A are then given by
314
GEOMETRY ON A SURFACE
§ 142
X + C, where C is an arbitrary constant. We thus obtain a oneparameter family of surfaces, r1 = r + (X + C)m, normal to the lines. Consider now a geodesic field over S with the unit tangent vector e. Since -y = n rot e = 0 at every point of S, we have The tangents to the geodesics of a field form a normal system of lines. THEOREM 2 (Bertrand).
We conclude with an important theorem in geometrical optics. THEOREM 3 (Malus and Dupin). If a normal system of lines is reflected or refracted at any surface, it still remains a normal system. n
FIG. 141
Let el denote the unit vectors along the incident rays and e2 the unit vectors along the refracted (or reflected) ray (Fig. 141). If µi denotes an absolute refractive index, µ1 sin 01 = Proof.
µ2 sin 02 (Snell's Law) and e1, e2, n are coplanar; hence,
(µ2e2 - µ1e1) x n = 0 and n-rot (µ2e2 - µ1e1) = 0, by the theorem following (97.10). For the reflected ray (§ 75),
e2 = (I - 2nn)
e1
and n - rot (e2 - e1) = 0.
In either case: n rot e1 = 0 implies n - rot e2 = 0142. Developable Surfaces. A developable surface (§ 138) is the envelope of one-parameter family of planes. The limiting line of intersection of two neighboring tangent planes is a line, or ruling, on the surface. A developable is therefore a ruled surface. The equation of a ruled surface may be written (1)
r = p(u) + ve(u),
DEVELOPABLE SURFACES
§ 142
315
where p = p(u) is a curve C crossing the rulings and e(u) is a unit vector along the rulings. The surface normal is parallel to ru x rv = (Pu + veu) x e.
When the normal maintains the same direction along a ruling, the surface is developable. For the same plane is tangent to the sur-
face along an entire ruling; and, as each value of u corresponds to a ruling and hence to a tangent plane, the surface is enveloped by a one-parameter family of planes. Consequently the surface (1) is developable when and only when pu x e and eu x e are parallel; that is, when pu, e and eu are coplanar:
pu e x eu = 0.
(2)
Case 1: e x eu = 0. Since e e = 1, we have e . eu = 0 and hence eu = 0. Therefore e is constant, and (1) represents a general cylinder. Case 2: e x eu 34 0. We then may write (§ 5)
Pu = a(u)e + 5(u)eu.
(3)
With a function A(u), as yet undetermined, let us write (1) as (4)
r= (P+Xe)+(v-X)e=q+(v-X)e,
where (5)
q(u) = p(u) + X(u)e(u).
Then
qu = pu + Xue + Xeu = (a + Xu)e + (S + x)eu; and, if we choose X = -,6, (6)
qu = (a - {3u)e.
If a = l3u along C, qu = 0 and q is constant. Then (4) represents a cone of vertex q if the vectors e(u) are not coplanar, a plane if they are coplanar.
In general, however, a 0 $u; then (6) shows that q(u) traces a curve having e as tangent vector. The surface (4) then is generated by the tangents of the curve q = q(u); it is the tangent surface of this curve. Developable surfaces are planes, cylinders, cones or tangent surfaces.
GEOMETRY ON A SURFACE
316
§143
143. Minimal Surfaces. In § 131 the mean curvature of a sur-
face was defined as J = - Div n = kl + k2. With f = 1, the integral theorem (101.2) becomes
fJnds=Jmde.
(1)
Next put f = r in (101.4) ; then, since Or = I - nn, Rot r = 0, and we have (2)
fr
x Jn dS
=Jrxmds.
These equations admit of a simple mechanical interpretation: THEOREM 1. The normal pressures Jn (per unit of area) over any simply connected portion of surface S bounded by a closed curve C are statically equivalent to a system of unit forces (per unit of length) along the external normals to C and tangent to S.
Consider, now, a soap film with a constant surface tension q per
unit of length and subjected to an unbalanced normal force pn per unit of area. Since any portion of the film bounded by a closed curve C is in equilibrium, we must have
fpn dS + 2q Jmds= f (p + 2Jq)n dS = 0; the surface tension q is doubled, because it is exerted on both sides
of the film. We therefore conclude that (3)
p = -2Jq.
Thus, for a soap bubble of radius r, the normal curvature is everywhere
k = «x - n = rc cos Tr =- 11r
and J = -2/r;
thus the pressure inside of a soap bubble exceeds the pressure out-
side by an amount p = 4q/r. In particular, if the film is exposed to the same pressure on both sides, p = 0; then J = 0 at all points of the film. A surface whose mean curvature J is everywhere zero is called a minimal surface. Thus a soap film spanned over a wire loop of any shape and with atmospheric pressure on both sides materializes a minimal surface. Let So be a minimal surface bounded by a closed curve C. Its existence is guaranteed by its physical counterpart, the soap film
MINIMAL SURFACES
§ 143
317
spanned over C. Imagine now that So is embedded in a field of minimal surfaces, that is, a one-parameter family of surfaces in a certain region R enclosing So, such that through every point of R there passes one and only one surface of the family. Such a field may be generated, for example, by the parallel translation of So.
At each point of R, the unit normal no to the minimal surfaces may be so chosen that no is a continuous vector-point function; and, from (97.2),
div no = Div no + no
dno
do
= -J + 0 = 0.
Now let S be any other surface in R spanning the curve C and enclosing with So a volume V of unit external normal n. From the divergence theorem,
n=
f
n no
over
n no
cos
dS <
fdS.
We have thus proved THEOREM 2. A minimal surface spanning a closed curve and belonging to a field of minimal surfaces has a smaller area than any other surface spanning the same curve and lying entirely in the field. Example 1. In order to find a surface of revolution which is also minimal, we set J = 0 in (138.5). The resulting differential equation,
uz" + z'(1 + z'2) = 0 may be written z'z" z72
z'z"
1
+z'2+u=0
and has the first integral uz'/1/1 + z'2 = C. Solving for z', we have z' _ C/1/u2 - C2; hence
z + k =Ccosh-r C ,
u
=Ccoshz Ck.
This represents a catenary in the uz-plane; when revolved about the z-axis it generates a surface known as the catenoid. Catenoids are the only minimal surfaces of revolution.
GEOMETRY ON A SURFACE
318
§ 143
Example 2. The right helicoid is a surface generated by a line which always cuts a fixed axis at right angles while revolving about and sliding along the axis at uniform rates. In other words, the right helicoid is a spiral ramp. A right helicoid about the z-axis has the parametric equations,
z = as,
y = u sin v,
x = u cos v,
(4)
where u, v are plane polar coordinates p, p in the xy-plane. The lines v = const are the horizontal rulings on the surface; the lines u = const are circular helices on the cylinder x2 + y2 = u2. Equations (4) give the vector equation,
r = u R(v) + av k,
(5)
where R is a unit radial vector (R . k = 0).
Now
r = uP + ak; ruxr uk - aP
ru = R,
n-
_ I ru x rv I
=
a2
a
nu = (u2 + a2)2 rv,
a nv = (u2 + a2)2 ru.
Since ru r, = 0, the parametric lines are orthogonal; and, since ru . nu = 0,
r n = 0, they are also asymptotic lines. From nu x n = K ru x r (132.20), we have
K=
- a2 (u2 + a2)2
and, since nu x r + ru x n = 0, J = 0 (132.19). The right helicoid is a minimal surface. Its negative total curvature is constant along any helix u - const.
Along the lines of curvature,
(rudu +rvdv) x (nudu
ruxnudu2 +rvxnndv2 = 0,
or du2 - (u2 + a2) dv2 = 0. The two families of these lines therefore satisfy the differential equations:
du/
u2 + a2 = fdv.
On integration, these give u sinh-1- = c f v or u = a sinh (c f v) a
as the finite equations of the lines of curvature. Example 3. On a minimal surface, the asymptotic lines form an orthogonal system (§ 131, ex. 2). If we choose them as our field curves, k1 = k2 = 0, and the Codazzi Equations (137.13), (137.14) become 27211 +
i = 0,
2y1t1 -
-
ds2
= 0.
SUMMARY: SURFACE GEOMETRY
144
319
Since K = -ti, dK dsl
dt1 = -2t1_ -472K,
dK ds2
ds1
dt1 = -2t1= 4y1K; ds2
dK dK VK = el - + e2 - = 4K(y1e2 - 72e1) = 4K n x R dsl ds2
(134.3).
144. Summary: Surface Geometry. On a surface r = r(u, v) of
unit normal n, the angular velocity of the trihedral Tnp (p = T x n) along a curve is w = tT + yn + kp. If (p = angle (N, n), positive in the sense of T,
t = r+ dv/ds,
y= K sin cp,
k = K cos V
are the geodesic torsion, geodesic curvature, and normal curvature, respectively. From do (1)
ds
=wxn= -kT - tnxT,
we conclude that k and t are the same for all surface curves having
a common tangent at a point. Important surface curves and their differential equations: Lines of curvature (t = 0) : Asymptotic lines (k = 0) : Geodesics (y = 0) :
d2r nx
ds2
dr do -x= 0; ds ds dr do -ds-ds= 0; = 0 or Vr
d2r ds2
= 0.
The equations for geodesics follow from dT
-xn=KNxn=yT. ds
If e and e' = n x e are perpendicular directions in the tangent plane,
-Vn = k ee + k' e'e' + t ee' + t' e'e. Since Rot n = 0, Vn is symmetric, and t + t' = 0. The first and second scalar invariants of - Vn,
J = - Div n = k + k',
the mean curvature;
K = kk' + tt' = kk' - t2, the total curvature,
GEOMETRY ON A SURFACE
320
§ 144
have the same value for any pair of perpendicular directions. If the point is not an umbilic (k constant for all directions), k attains its extreme values in the orthogonal principal directions e1, e2 for which t1 = 0, t2 = 0. In terms of the principal curvatures k1, k2,
J = k1+'k2iK=k1k2. If a, b, n are reciprocal to ru, r,,, n, the first and second fundamental forms are
Or = E as + F(ab + ba) + G bb,
-Vn = Las+M(ab+ba) -}-Nbb; and the fundamental quantities are given by
L = -ru nu,
M = -ru nv = -rv nu,
N = -rv nv.
If e1i e2 = n x e1 are the unit tangent vectors along a field of surface curves and their orthogonal trajectories, we have
Vet = Re2 - Pn,
Ve2 =-Re1 - Qn,
Vn = Pet + Qe2,
where
P = do/ds1,
Q = dn/ds2,
R = y1e1 + y2e2.
Codazzi Equations: n V x (Vn) = 0; or
Gauss Equation: n V x (De1) = 0; or
n rot R = -K. For any field of curves of tangent vector e,
y = n- rot e. The last two equations show that y and K may be expressed in terms of E, F, G and their partial derivatives. When embedded in a geodesic field, the arc of a geodesic is shorter than any other curve lying within the field and having the same end points.
If a field of curves of tangent vector e cuts the field e1 at the angle 0 = (eli e), positive in sense of n, R = y1e1 + y2e2 = ye + y'e' - V0.
PROBLEMS
321
Bonnet's Integral Formula,
fK d S= fdo -
ds,
is the integral ("Stokian") equivalent of the Gauss Equation, n - Rot R = -K. If the simple closed curve C has a continuously turning tangent
fdo = 27r, but if the path has angles (in-
terior angles ai),
fdo = 2ir If C is a geodesic triangle,
- as).
fK dS = al + a2 + a3 - ir; and, on
a surface of constant K 3-1 0, the area of a geodesic triangle is
(al + a2 + a3 - ir)/K. A sphere of radius a has the constant positive curvature 1/a2; a tractrix of revolution for which the tangential distance to the asymptote is a has the constant negative
curvature -1/a2. A minimal surface (J = 0) embedded in a field of minimal surfaces and spanning a closed curve C has a smaller area than any other surface lying within the field and spanning C. Soap films spanned over wire framework materialize minimal surfaces. The right helicoid is a ruled minimal surface. The catenoid is the only minimal surface of revolution. PROBLEMS
1. If u, v, are plane polar coordinates r, 0, prove that the surface obtained by revolving the curve z = f(x) about the z-axis has the parametric equations:
x = u cos v,
y = u sin v, z = z(u).
2. A straight line, which always cuts the z-axis at right angles, is revolved about and moved along this axis. The surface thus generated is called a conoid.
If u and v are plane polar coordinates in the xy-plane, show that the conoid has the parametric equations: x = u cos v,
z = z(v).
y = u sin v,
In particular when dz/dv is constant the conoid is a right helicoid: x = U cos v, y = u sin v, z = av. 3. The central quadric surface, x2
y2
a2 ± b2 ±
z2
_
1,
GEOMETRY ON A SURFACE
322
is an ellipsoid, a hyperboloid of one sheet, or a hyperboloid of two sheets according
as the terms on the left have the signs (+, +, +), (+, +, -), (+, -, -). Show that the corresponding parametric equations are x = a sin u cos v, y = b sin u sin v,
z = c cos u;
x = a cosh u cos v, y = b cosh u sin v, z = c sinh u; y = b cosh u sinh v,
x = a cosh u cosh v, 4. The paraboloid,
x2 a2
z = c sinh u.
t -=-, y2
2z
b2
c
is elliptic or hyperbolic, according as the terms on the left have the signs (+, +) or (+, -). Show that the corresponding parametric equations are
x = au cos v, y = bu sin v, z = Zcu2; x = au cosh v, y = bu sinh v, z =.
4cu2.
5. The position vector,
r = f(u) + g(v), traces a surface of translation. Show that any curve u = a (const.) may be obtained by giving the curve rl = g(v) a translation f(a); and that any curve v = b may be obtained by giving the curve r2 = f(u) a translation g(b). What are the surfaces,
r=f(u)+bv,
r=au+bv+c?
6. Show that a right circular cylinder of radius a has the constant mean curvature J = 1/a. 7. For the surface xyz = a3 show that 3a6
Jr = 2a3(x2 + y2 + z2)
(X2y2 + y2z2 + 22x2)2 '
(x2y2 + y2z2 + 22x2)7
[Cf. § 132, ex. 2.]
8. For the elliptic paraboloid, x2/a2 + y2/b2 = 2z/c
prove that the total curvature K 5 c2/a2b2. 9. For the surface of revolution about the z-axis,
r = iu cos v + ju sin v + kz(u), show that
L=
E=1+z'2 F=O, G=u2; 2'2)1, M = 0, N = uz'/(1 + 2'2)1.
[Cf. § 132.1
From (132.21) and (132.22) deduce the values of K and J and from (132.25) find the principal curvatures. Check with (138.4) and (138.5).
PROBLEMS
323
10. For the conoid,
r = iu cos v + ju sin v + kz(v), show that
E=1, F=0, G=u2+z'2=H2; L = 0, M = -z'/H, N = uz"/H; K = -z'2/H4, J = uz"/H3.
[Cf. § 132.1
When z = av, show that the resulting right helicoid is minimal (§ 143); and that its principal curvatures are fa/(u2 + a2). 11. Show that the curvatures J and K have the dimensions of (length) -1 and (length) -22. Verify this in Problems 6, 7, 8, 9, 10.
12. The position vector of a twisted curve r is given as a function ro(s) of the The surface generated by its tangents is
are.
r=rc(v)+uT(v) where v = s and u is the distance along a tangent measured from r. Show that
this tangent surface has the curvatures K = 0, J = z/i u Jx. What are the principal curvatures?
13. If a parabola is revolved about its directrix, show that the principal curvatures of the surface of revolution satisfy the relation 2k1 + k2 = 0.
[With the notation of § 138, ex., the parabola has the equation z2 _ 4a(u - a) when the directrix is the z-axis and u = 2a at the focus.] 14. The vectors a, b, n and ru, r,,, n form reciprocal sets of vectors over the surface r = r(u, v) [Cf. § 95]. Hence show that any vector f(u, v), defined over the surface, may be written f 15. Prove that
H2a = Gr - Fr,,,
H2b = -Fru + Er,,.
[Use Prob. 14 and (132.18).]
16. Prove that
FM - GL H2
II,
a
'b=
'
H2
FM - EN
FN - GM H2
FL - EM
I
H2
[Use Prob. 15.]
17. Prove the derivative formulas of Weingarten:
HZnu = (FM - GL)ru + (FL - EM)r,,,
H2n _ (FN - GM)ru + (FM - EN)r,,. [Use Prob. 14 and Prob. 16.]
;
GEOMETRY ON A SURFACE
324
18. Show that the asymptotic lines of the surface r = r(u, v) have the differential equation,
(ru du + r dv) (nu du + n dv) = 0 [Cf. (130.10)]; or, in terms of fundamental quantities,
Ldu2+2Mdudv+Ndv2 = 0. 19. Prove that the asymptotic lines on a right helicoid (Prob. 10) are the parametric curves and form an orthogonal net. 20. For the ruled surface,
r = p(u) + ve(u), prove that M = pu e x eu/H, N = 0.
(142.1).
21. Prove that a ruled surface is developable when, and only when, M = 0. [Cf. (142.2).] 22. Prove that the asymptotic lines of a developable surface, not a plane, are the generating lines (counted twice). [Differential equation: du 2 = 0.]
23. Prove that, along a curved asymptotic line, (a) The osculating plane is tangent to the surface. (b) The geodesic torsion equals the torsion: t = r. (c) The geodesic curvature is numerically equal to the curvature: y = ±K. (d) The square of the torsion is equal to the negative of the total curvature:
2 = -K. 24. Prove that
(a) If an asymptotic line is a plane curve other than a straight line, it is also a line of curvature. (b) An asymptotic line of curvature is plane. 25. Prove that the following conditions are necessary and sufficient in order that the surface be
(i) A plane: L = M = N = 0. (ii) A sphere: E/L = F/M = GIN. 26. Show that the lines of curvature of the surface r = r(u, v) have the differential equation, n (r, du + r dv) x (nu du + n,, dv) = 0
[Cf. (138.1)]; or, in terms of fundamental quantities, I
du2
E
-du dv F
L
M
N
dv2
I
G
= 0.
27. Show that the parametric curves are lines of curvature when, and only
when, F = 0, M = 0. 28. In the notation of § 132, ex. 2, prove that the lines of curvature on the surface z = z(x, y) have the differential equation, dy2
-dx dy
dx2
1 + p2
pq
1 + q2
= 0.
PROBLEMS
325
29. Show that the hyperbolic paraboloid of Prob. 4 has the parametric equations:
x = 2a(u + v), y = 2b(u - v), z = cuv. Prove that the differential equations of its asymptotic lines and lines of curvature are, respectively, dvl2
du dv = 0,
\du /
a2 + b2 + c2y2 a2 + b2 + c2u2
and find the uv-equations of the asymptotic lines and lines of curvature. 30. If the parametric curves on a surface are its lines of curvature, show that
nu = -kiru,
nv = -k2rv.
31. Two surfaces that have the same normal lines are called parallel. Surface points on the same normal are said to correspond; thus
r(u, v) = r(u, v) + An(u, v) determines corresponding points on the parallel surfaces S and S. Prove that the distance X between the parallel surfaces is constant. [If we take the lines of curvature of S as parametric curves,
ru = (1 - Aki)ru + Xun,
rv = (1 - Xk2)rv + Xn
(Prob. 30); and, on cross multiplication,
ffn- = (1 - Xki)(1 - Ak2)Hn + Xu(1 - Xk2)n x rv + X ,(l - Xkl)ru x n.
Now ii = En where e = ±1; and as n, ru, r are mutually orthogonal,
H = E(1 - Xki)(1 - Xk2)H,
Xu = Xv = 0.1
32. Prove that the lines of curvature on parallel surfaces correspond and that the corresponding principal curvatures satisfy the equations, kl
_
Eke
Ekl
1 - Xkl '
T2 =
1 - xk2
Hence the mean and total curvature of S are given by E(J - 2XK) J _ 1rJ+X2K' -
K
_
K
1-xJ+x2K
33. Show that surfaces parallel to a surface of revolution are also surfaces of revolution.
34. Prove the "Theorema egregium" of Gauss by establishing Baltzer's formula for the total curvature K:
Fuv - UGuu -Evv H4K=I Fv--2Gu
2Eu
Fu - 2Ev
E
F
F
G
0 ZEv
-'Gu
U Ei,
Gu
E F
F G
GEOMETRY ON A SURFACE
326
From (132.10) and (132.22) show that
Method.
H4K = [ruurx][rv,;r r,;] - [ruvrurv]2. Using (24.14), the right member becomes ruu
ruu ru
rvv
ruu
rv
ruv
ruv
ru
ruv
ruv
ru rvv
E
F
ruv
ru
E
F
rv
F
G
ruv
rv
F
G
rvv
rv
In both determinants the upper left element has the cofactor EG - F2; hence we may replace these elements by ruu rvv - ruv ruv and 0, respectively. Now, by differentiating we may show that rut,,
ru
rvv
rv =
i
2F!u, zGv,
ruu rvv - ruv
ruv
ruv
ru = 2iEv,
rv =
ruv - Fuv
2Gu>
ru = F,,
rvv
ruu
r,, - Fu -
i2Gu,
2Ev;
- 2!Evv. vv
1Guu
35. On the surface r = r(u, v) let us write u1 = u, u2 = v; el = ru,
e2 = rv;
D1 =
el = a,
aau
,
D2 = ava ;
e2 = b;
where the Greek indices have the range 1, 2.
then e"
Moreover let
gap = ea ep, 9ap = ea e$, g = det gap =
hap = n DaDpr = hpa;
ep = SQ,
911
912
921
922
h = det ha p.
Prove that
(a) 9u= E,g12=F,922=G;g=EG-F2. (b) 911 =
922
g12
-
9
-912 922 = 9
922
9
that is, g"p is the reduced cofactor of gap in det gap.
(c) h11=L,h12=M,h22=N;h=LN-M2;
(d) Daep - Dpea;
(e) Dagpy + Dog-y. - D.rgap = 2ey Daep.
(f) Daep = ey(Dagp , + Dpgya - Dygap) + nhap (summed over y = 1, 2). (g) e" = g"pep (summed over (3 = 1, 2). [Cf. § 145.] 36. The Christoffel symbols Papa, g, y = 1, 2) for a surface r = r(ul, u2) having gap du" dup as first fundamental form (§ 132) are computed from the equations: Pap = ey . Daep.
PROBLEMS
327
With reference to Prob. 35 prove that
(a) ra0 = r'a (b) rP# =
Diggya - Dyga#) (summed over p = 1, 2).
(c) 29r1i = 922 D1911 - 2g12 D1912 + 912 D29uu, 29ri2 = 922 D2911 - 912 D1922,
29r22 = -912 D2922 + 2g22 D2912 - 922 D1922
[Use the formula of part b; for example, 21'1ii
= 911(D1911 + D1g11 - D1911) + 91Y(D1912 + D1921 - D2911),
1
29t1 = 922 Dig,, - 912(2D1912 -
(d) From the formulas for r,# obtain the formulas for rap. (e) From Prob. 35 (f) prove the derivative formulas of Gauss: X Daep = raryea + heron (summed over a = 1, 2).
37. From n efl = 0 show that eo Dan = -han. Hence prove the derivative formulas of Weingarten;
Dan = -hafieO (summed over S = 1, 2). Check these results with Prob. 17. 38. With the notations of Prob. 35 show that equations (132.21) and (132.22) may be written J = g0h,,#, K = h/g (a, 0 summed over 1, 2). 39. Show that the tangential projection of the vector curvature dT/ds = KNi of a surface curve is d2r
dua duDr
= ey
ds2 + raA ds ds (--
summed over a, 0, y = 1, 2. Hence prove that the equations of a geodesic on a surface are y
d3
B
dua
+ rad
ds
ds
=0 a
1 T = ep ds ,
= ep ds2 + Daeg d
(Y = 1, 2). 8
,
Vr = eyes. Cf. Prob. 36. ]
CHAPTER IX TENSOR ANALYSIS
145. The Summation Convention. Expressions which consist of a sum of similar terms may be condensed greatly, without any essential loss of clarity, by indicating summation by means of repeated indices. This usage, originally due to Einstein, is stated precisely in the following SUMMATION CONVENTION. Any term, in which the same index
(subscript or superscript) appears twice, shall stand for the sum of all such terms obtained by giving this index its complete range of values. This range of values, if not understood, must be specified in advance. By way of illustration, we repeat some of the formulas of § 24, using the summation convention. The index range is 1, 2, 3. The two forms of a vector (24.1) and (24.2) become (1)
u = u;e'.
U = uiei,
In these equations i and j are summation (or dummy) indices. But any other letter will do as well; thus u = carer = ule1+ u2e2 + u3e3.
In order to compute u - v, we first recall the defining equation for reciprocal sets, (2)
ei e' = Si
(23.3).
Now (3)
u . v = (uiei) . (v,e') = uiv; 'Vi' = uiei;
we here use different indices in expressing u and v in order to get the 32 = 9 terms in the expanded product (six are zero). Similarly, (4)
u v = (uiei) . (vie;) = uiv' S = u1vi. 328
DETERMINANTS
§ 146
329
Note that the summation indices in the preceding examples appear once as subscript and once as superscript. The significance of this arrangement (not required by the summation convention) soon will be apparent. 146. Determinants. A permutation of the first n natural numbers is said to be even or odd, according as it can be formed from 123 n by an even or odd number of interchanges of adjacent numbers. The total number of permutations of n different numbers is n!; one half of these are even and one half odd. For example, when n = 3, 123, 231, 312 are even permutations, 213, 132, 321 are odd. Consider now a permutation ijk r of the numbers 123 n.
We then define the permutation symbols ijk...r and as equal to 1 or -1, according as ijlc . r is an even or an odd pereijk""r
mutation of 123 . . . n; but, if any index is repeated, the epsilon is zero. For example, when n = 3, E123 = 231 = 312 = 1
E213 = 132 = 321 = - 1
112 = 122 = 222 =
0.
The epsilons just defined are useful in dealing with determinants. To be specific, we shall take n = 3; but all the formulas apply without change of form for any value of n. From the definition of a determinant : 1
(1)
a = det a _
2
3
a1
a1
a1
2
as
a2
3
1
2
a3
a3
a2 3
a3
ei jka 12a3
e'k ai1 aj2ak3
The implied summations on i, j, k produce 33 = 27 terms; of these,
21 involve repeated indices and therefore give a zero epsilon, whereas 6 involve permutations of 123 and give precisely the 3! = 6 terms of the determinant. (Write them out!) The theorems relative to an interchange of rows and columns
i
are given by (2)
Proof.
eijkarasat
aer8t =
'
ijkarL atk a8
When r, s, t are 1, 2, 3, equations (2) reduce to (1).
Since rst changes sign when two adjacent indices are interchanged,
TENSOR ANALYSIS
330
§ 146
(2) will beestablished when the right members are shown to have
this same property. Now Eijkarasat = eijkasarat = - ejikasa r aki
and, if we interchange the summation indices i, j, the last expression becomes - eijkasajat . If we multiply the first equation of (2) by e" and sum on rst, we obtain 3! a =
(3)
eijkerstarasae.
On the left er8tergt, summed over the 3! permutations of 123, equals 3!; on the right, the summation extends over all six indices and produces 36 = 729 terms. Many with zero epsilons vanish; and each non-vanishing term such as a1la22a3 3 appears six times, corresponding to the 3! permutations of its factors. Equation (3) thus is not useful for computation; its importance rests on the information it gives about the determinant when its elements are transformed.
The cofactor of an element az in the determinant (1) is defined as its coefficient in the expansion of the determinant. The cofactor of a; is denoted by A'. To find A',* strike out the row and
column in which ai stands; then A equals the resulting minor taken with the sign (-1) i+j. If the elements of any row or column are multiplied by their respective cofactors and added, we obtain the determinant-its Laplace expansion; but, if the elements of any row (column) are multiplied by the corresponding cofactors of another row (column) and added, the sum is zero. These important properties both are included in aiA; = a'Ai = a S .
(4)
Here r is the summation index. If in (4) we put j = i, both r and i are summation indices, and we get a7AT = a; A, = a Si,
(5)
where S = Si + S2 + S3 = 3 (n = 3).
But, if we put j = i in
(4) and suspend summation on i, we obtain the Laplace expansions, (6)
a7,Ar = arAi = a (i fixed),
for S = 1 for a fixed i.
DETERMINANTS
§ 146
331
A cofactor divided by the value of the determinant is called a reduced cofactor. The reduced cofactor of a2 is therefore
a; = A,/a
(7)
(a F6 0).
In terms of reduced cofactors, equations (4) become aTc = 8a.
(8)
When the elements of a determinant are written aid, its definition becomes a = E'.'kalia2Ja3k = E1Jkailaj2ak3-
(9)
The reduced cofactor of ai3 then is written ai
and equations (8)
become aira'r = aria'' = SL.
(10)
If the elements ai; are functions of a variable x, we have, from (9), da
_ ilk (dali dal; da3k\ a2ja3k + a1i a3k + aiia2i = x x dx ) E
dx
=
dal-Ali+da2jA21+da3kA3k dx
dx
dx
'
where A 'j denotes the cofactor of aij. Summing on two indices, we may write (11)
-da= daii Ai' dx
(i, j = 1, 2, 3).
d.r,
The derivative of a determinant is the sum of the products formed by differentiating each clement and multiplying by the cofactor of the element.
The properties (8) of reduced cofactors enable us to solve a system of linear equations when the determinant of the coefficients is not zero. Consider, for example, the equations: (12)
ax' = yt
(i, j = 1, 2, 3; a
0).
To solve these for x', multiply (12) by the reduced cofactor ak and sum on i : the left member becomes S; x' = xk, and we obtain xk = a;yi, or, on replacing k by j, (13)
x' = aiyi
(i, j = 1, 2, 3).
TENSOR ANALYSIS
332
5 147
To find the product of the determinants a = det a, b = det b;, we have, from (1) and (2), ab = EiJkaia2a3b
(14)
s t ia2 ja3 kEr8tbsrbjbk
= a1
= Erst(albi)(aabj)(a3kbk)
r
=
8
ErstC1C2Ct3,
where c% = aabj = aib1 -F a2b2 -f- a2b33.
(15)
From (15), we see that the element in the ith row and jth column of the product ab is given by the sum of the products of the corresponding terms in the ith row of a and the jth column of b-the so-called "row-column" rule. Since the value of a determinant is not altered by an interchange
of rows and columns, we also can compute ab by "row-row," "column-row," and "column-column" rules. For example, the product 1
a1
(16)
2
al
3
al
1
a1
2
1
1
a2
a3
1
0
0
0
1
0
0
0
1
a2
a2
a2
ai
a2
a3
1
z
3
3
3
3
a3
a3
a3
a1
a2
a3
= 1.
by use of the row-row rule. THEOREM. If a determinant a 0, the determinant formed by replacing each element by its reduced cofactor is 1/a.
If we solve equations (13) for yi by using the reduced cofactors in det a, we will obtain (12); in other words, the reduced cofactor
of a in its determinant is a. The two determinants in (16) thus are reciprocally related: each is formed from the reduced cofactors of the other. .'. 147. Contragredient Transformations. Let us now introduce a new basis e1i e2, e3 by means of the linear transformation, 61 = cie1 + ciez + Cie3, (1)
e2 = c2e1 + c2e2 + 4e3,
e3 = c3e1 + c3e2 + c3e3,
CONTRAGREDIENT TRANSFORMATIONS
§ 147
333
where the coefficients ci are real constants whose determinant 0.
c
In brief, ei = c'ej,
(1)
c = det ca /- 0.
The condition c 0 ensures that e1, e2, e3 are linearly independent. For, if there were three constants At such that Aiei = 0,
Aicej = 0;
then
the linear independence of the vectors ej requires that c'Ai = 0, and, since c P 0, these equations only admit the solution Ai = 0. This argument applies without change to space of n dimensions, in which a basis consists of n linearly independent vectors.
In the present case (n = 3), we also may argue as follows. From (1), we have 161
. e2 x e3
=
cic2c3 ei ej x ek = C'iC2iC3Eijk e1 ' e2 x e3;
hence, on writing E = el e2 x e3, we have
E _ (det c')E = cE.
(2)
Since the vectors ei form a basis E 96 0, hence c E
0 implies
0 and the linear independence of the vectors e1. For any vector u, u = 'uiei = ujej;
(3)
hence, on substitution from (1), ujej,
or, since the vectors ej are linearly independent, u.l = Ci,u
(4)
If y is the reduced cofactor of c in the determinant c, we have on solving equations (4),
ut =
(5)
or, written out in full, ul = riu1 + 'Y2u2 + Y3u3, (5)
u2
= 'Ylui + Y2u2 + Y3 u3,
u3 = 'iu1 + Y32 u2 +
'Y3u3.
3
TENSOR ANALYSIS
334
§ 148
The linear transformations (1) and (5) are said to be contragredient.
Their matrices,
C=
Cl
C1
c1
1
C2
2 C2
3 C2
C31
2 C3
3 C3
,
71
72
r = 71
2
72
3
3
72
71
73 2
73 3
73
are so related that any element of r is the reduced cofactor of the corresponding element of C in its determinant c; and any element of C is the reduced cofactor of the corresponding element of r in its determinant y.
In order to state this definition analytically, we remind the reader of the following definitions from matrix algebra. The product AB of two square matrices A and B is the matrix whose element
in the ith row and jth column is the sum of the corresponding products of the elements in the ith row of A and jth column of (row-column rule). The transpose of a matrix A is a matrix A' obtained from A by interchanging rows and columns. The unit matrix I has the elements Si (ones in the principal diagonal, zeros elsewhere).
If we compute Cr' or rC' and make use of the equations, (6)
cz7r = S,
7rc1 = bj,
we find that
Cr'=rC'=I.
(7)
These equations characterize contragredient transformations.
Two matrices whose product is I are said to be reciprocal. Thus C and r' or C' and r are reciprocal matrices. From (7), we see that contragredient matrices are so related that the transpose of either is the reciprocal of the other.
148. Covariance and Contravariance. When new base vectors are introduced by means of the transformation, (1)
ei = c e5,
det c
0.
the components of any vector u are subjected to the contragredient transformation, (2)
ui = 7;u'.
In view of (24.3), this may be written
u-ei=7jue';
§ 148
COVARIANCE AND CONTRAVARIANCE
335
and, as this holds for every vector u, (3)
The basis ei, reciprocal to ei, thus is transformed in the same way as ui. To find the transformation for the components ui, multiply (1) by u ; then, from (24.4),
ui = cu
(4)
a transformation cogredient with (1). Thus quantities written with subscripts transform in the same way (cogrediently); and quantities written with superscripts also transform in the same way-but the latter transformations are contragredient to the former.
Thus the position of the index indicates the character of the transformation.
A vector u is said to have the contravariant components ui, the covariant components ui. These terms suggest variation unlike and like that of the base vectors ei; in other words, the transformation
(1) is regarded as a standard. Thus [u', u2, u3] and [U1, u2i u3] are two ways of representing the same vector u; in the first it is referred to the basis ei, in the second to the reciprocal basis e' [u1, u2, u3J often is called a contravariant vector, [u1, u2, u3] a co-'
Actually, the vector u is neither contravariant or covariant, but invariant. variant vector.
By using the properties of the reduced cofactors, eir'Yr i
r
i
= Yi = ai, we may solve equations (1) to (4) for the original base vectors and components (§ 146). Thus the equations, (5)
(6)
ei = c,ei,
ui = ciuj,
et = 'Yie',
ui = Yiui;
e' = c ,%J,
uz
- yiu',
have the solutions, (7)
ei = -?j-6j,
ut = c ui.
Note that the matrices c, y in (7) are the matrices cz, -y' of (6) transposed. To be quite clear on this point the reader should write out equations ei = (.'e; and et = c;ei in full. From (1) and (3), we find that ei ' e' = cier yie8 = ciYaSr = c 'Yr = 31j,
which shows that the new bases have the fundamental property of reciprocal sets.
TENSOR ANALYSIS
336
§ 149
An expression, such as uiv i, uie i, ei e i, summed over the same upper and lower indices, maintains its form under the transformation. Take, for example, the scalar product uivi: making use of (2) and (4), we have uiv2 = Ciur78vs = bryurv8 = u'1T.
The index notation thus automatically indicates quantities that are invariant to affine transformations of the base vectors. 149. Orthogonal Transformations. When the basis ei and the new basis ei both consist of mutually orthogonal triples of unit vectors, the transformation, ei = ciei,
(1)
is called orthogonal. Since both bases are self-reciprocal (§ 23), the corresponding transformation (148.3) between the reciprocal bases, ei = 7e', now becomes ei = 7e;.
Since this transformation must be the same as (1), C11
(2)
C=
c2
ci c2
c3
C3
3 C3
1
C3
ci
7i 72
=
2
71
72
3
3
71
72
73
7sq
=r
3
73
The matrix of an orthogonal transformation is called an orthogonal matrix. From (2), we have the THEOREM. Every element of an orthogonal matrix is its own reduced cofactor in the determinant of the matrix.
In view of the properties of reduced cofactors, the coefficients of an orthogonal transformation satisfy the equations: (3)
crc; = Cic; = 8L.
The matric equation (147.7) characterizing contragredient transformations becomes (4)
CC' = I
for orthogonal transformations. Thus a real matrix is orthogonal when its transpose equals its reciprocal.
If c denotes the determinant of C, we have, from (4), (5)
c2 = 1,
c = f 1.
QUADRATIC FORMS
§ 150
337
If the bases ei and ei are both dextral or both sinistral, we can bring the trihedral e1e2e3 into coincidence with e1e2e3 by continu-
ous motion-in fact, by a rotation about an axis through the origin. At any stage of the motion, the base vectors are related
by equations of the form (1); and, as the motion progresses, the coefficients c change continuously from their initial to their final values S , and c = det ci becomes det S = 1. But, since this determinant equals =E1 at all stages of the motion and must change continuously if at all, c = 1 when the bases have the same orientation. If one basis is dextral, the other sinistral, e1e2e3 and ele2(-e3) have the same orientation. Hence the determinant of the transformation, el = c1er,
-e3 = -caer, namely, -c, must equal 1; thus c = -1 when the bases have dife2 = C3er,
ferent orientations. 150. Quadratic Forms. A real quadratic polynomial, (1)
A(x, x) = aijxixj
(i, j = 1, 2,
,
n),
for which aij = aji is called a real quadratic form in the variables The symmetry requirement aij = aji entails no x1, x2, , x"`. loss in generality; for, if aij 0 aji, the form is not altered if we replace aij and aji by (aij + aji). The determinant of2 the coefficients, a = det aij, is called the discriminant of the form. The form is said to be singular if det aij = 0, non-singular if det aij 0 0. Associated with the quadratic form A (x, x) is the bilinear form, (2)
A(x, y) = aijx'y' = A(y, x),
known as its polar form.
The expansion of aij(xi + Xyi) (x' + Xyj) shows that a quadratic form and its polar are related by the identity, (3)
A (x + X y, x + Xy) = A (x, x) + 2XA (x, y) + X2A (y, y).
If we make the linear transformation, (4)
xi=c yr,
c=detc;F-- 0,
the form (1) becomes (5)
B(y, y) = br4yry8 where
bra = aijcTc'.
TENSOR ANALYSIS
338
§150
From the rule for multiplying determinants, det bra = det (ai;cT) det c8 = det ai; det cr det c8;
the discriminant of B(y, y) is therefore (6)
b = det bra = c2a.
Linear transformations of non-zero determinant do not alter the singe lar or non-singular character of a quadratic form.
If the form A (x, x) is singular, it can be expressed in terms of fewer than n variables which are linear functions of x. For since det ai, = 0, the system of n linear equations ai;xi = 0 has a solution (xa, xo, , xo) which does not consist entirely of zeros; thus we may assume that xo 5-4- 0. Now A(xo, y) = ai;xoy' = 0, and, from (3), A (x + Axo, x + Axo) = A (x, x),
irrespective of the value of A.
If we now choose A = -xl/xo
and write y i = x, i + Axp = x i - x4x1 /x0,
we have A (y, y) = A (x, x). Since y' = 0, A (y, y) is expressed in
, y". terms of the n - 1 variables y2, y3, Henceforth we shall consider only non-singular forms. A nonsingular form is said to be definite when it vanishes only for
xl = x2 =
= x" = 0. For all other sets of values, the sign of
a definite quadratic form is always the same. To prove this, sup, x" and A(y, y) < 0 pose that A(x, x) > 0, for the set x1, x2, Then, from (3), for the set y', y2, , y".
(7)
A (x + Ay, x + Ay) = 0
is a quadratic equation in A having two distinct real roots A1i A2; for
{A(x, y) 12 - A(x, x)A(y, y) > 0.
Consequently, the form vanishes for two different sets of values, xi + Alyi and xi + A2yi and therefore cannot be definite. For a definite quadratic form, the quadratic equation (7) in A must have a pair of either complex roots or equal roots; hence, for a definite form, (8)
JA (x, y) 12 - A(x, x)A(y, y) < 0,
the equal sign corresponding to the case xi + Ayi = 0, (i = 1. , n), in which the sets xi and yi are proportional.
§ 152
RELATIONS BETWEEN RECIPROCAL BASES
339
A definite quadratic form is called positive definite or negative definite according as its sign (for non-zero sets) is positive or negative. A non-singular form which is not definite is called indefinite: such a form may vanish for values x` other than zero. When A (x, x) is positive definite, it is well known that we can find a real linear transformation (4) which will reduce A (x, x) to a sum of squares: (9)
B(x, x) = Siyiyj = ylyl + y2y2 + ... + ynyn
The discriminant of a positive definite form is therefore positive;
for, from (6), b = 1 = c2a, and a = 1/c2 > 0. 151. The Metric.
Using the notation,
9i; = ei ej = gji,
(1)
for the nine scalar products of the base vectors, we have (2)
u . v = (uiei) . (viej) = giptivj,
u u = gi;uiu'.
Thus u u is a real quadratic form, giiU'u1 + g22u2u2 + 933u3u3 + 2g12u'u2 +
2923u2u2
+ 2931u3u1r
in the variables u1, u2, u3; and, since u u > 0 when u 0, this form is positive definite. Moreover, u v is given by the associated bilinear (polar) form in ui, vj. From (2), we have (3) (4)
I
gijl(itljr
UI=
Cos (u, V) =
gijuw' UV _ VgijuiujVVgijvivi lUl VI
Thus, when the six constants gij are known, we can find the lengths of vectors and the angles between them when a fixed unit of length is adopted. The quantities gjj, since they make measurements possible, are said to determine the metric of our 3-space; and gijuiuj is called the metric (or fundamental) quadratic form. 152. Relations between Reciprocal Bases. The discriminant c' the metric form is (1)
9=
911
912
913
921
922
923
931
932
933
:" Bother, Introduction to Higher Algebra, New York, 1907, p. 150.
TENSOR ANALYSIS
340
§ 153
The reduced cofactor of gij in this determinant is denoted by gij; and we have, from (146.10), the relations:
= 9rigr' = Since any vector may be written girg.r
(2)
u=
Si.
ei,
ei = gije' We may solve these equations for ej by multiplying by gik and (3)
summing on i; thus
k j = e k, 9ikei = 9ikgijej = 3je or, if we interchange i and k, ei = 9kiek-
(4)
From (4), we have also ei ej = gkiek ej = gkiak = gji (5)
gij =
ez
e' = 9'i
Making use of (3) and (4), we now have, for any vector u, (6)
(7)
ui = ei
u = gijuj
The equations enable us to convert contravariant components of a vector to covariant and vice versa. Finally, from (24.14), we have (8)
(9)
g = det gij = det (e1 ej) = [eie2e312 = E2,
det gij = det (ei ej) = [ele2e312 = E-2 = 1 9
153. The Affine Group. When an origin 0 is given, the constant basis el, e2, e3 defines a Cartesian coordinate system x1, x2, x3; for any point P is determined by the components of its position vector :
OP = x'el + x2e2 + x3e3. The components xi of OP are called the Cartesian coordinates of P relative to the basis ei.
THE AFFINE GROUP
§ 153
341
When the base vectors ei are subjected to the transformation,
et=ce;,
(1)
the invariance of OP, namely, i''es = xZei,
(2)
induces a contragredient transformation on the coordinates. For,
if we substitute from (1) in (2), we find that x' = ci"xi; that is, xi = c;x'.
(3)
The matrix c? in (1) is transposed in (3); and, while (1) expresses
the new base vectors in terms of the old, (3) expresses the old coordinates in terms of the new. We now can solve (3) for the new coordinates by multiplying by the reduced cofactors ti; and summing; thus we find xiy, = xk, or (4)
x' = y2xi,
-y = det yj = 1/c.
The transformation (3) is called affine, or, more specifically, centered affine, since the origin 0 has not been altered. The centered-affine transformations with non-zero determinant form a group, that is,
(a) the set includes the identity transformation: ri = S ; a ; ' = x`,
det 6 = 1;
(b) each transformation of the set has an inverse; (c) the succession of two transformations of the set is equivalent to a single transformation of the set. Thus the transformations, a x',
x' = bj;xk, give xa = Ckx1,
where ck = a bk and
det ck = (det a) (det bk) F4- 0.
Our present point of view is that a transformation of the base vectors induces a definite transformation of coordinates. But the reverse point of view is adopted when transformation groups more general than the affine are under consideration : the coordinates are transformed, and we then inquire as to corresponding transformation of the base vectors.
TENSOR ANALYSIS
342
a 151
In the case of the affine group the transformation (3) of coordi-
nates obviously entails the transformation (1) of base vector.,. For, if we put x' = x'cy in (2), we have
xi(czej - ei) = 0 for all xi, that is, (1) must hold. 154. Dyadics. Under the affine transformation, (1)
ei = c ei,
xi = y;xi
ei = y;ei,
ti = e xi
(c
0),
we have also (2)
(148.6).
Contravariant and covariant components of a vector, ut and ui, transform like x i and xi, respectively. The sets of components u i and ui often are called contravariant and covariant vectors. Actually they are different representations of the same invariant vector,
U = uiei = uiei.
(3)
Nine (3 X 3) numbers Tii which transform like the nine products of vector components uivi, namely,
Tii =
(4)
yryBT'8
are called contravariant components of a dyadic. Similarly, nine numbers Ti; which transform like uiei, namely, T,i = ('icjTra,
(5)
are called covariant components of a dyadic. Finally, nine num-
bers Ti; or T;i which transform like u ivi or uivi, respectively, namely, (6)
7,'i =
yrc;TB,
T:' =
are called mixed components of a dyadic. Note that in Tti the upper index comes first, in Tii the lower index. Just as ui and ui are two representations of a single invariant vector u, the components Tii, Tii, Ti, T;i are four representations of one and the same dyadic, (7)
T = Tiieiei = Tiieiei = Ttieiei = Ti eiei.
Note that the indices on components and base vectors always are placed in the same order from left to right. All forms of T given
ABSOLUTE TENSORS
§ 155
343
in (7) are invariant; for example, on using (1) and (4), we have T'16x.e
i ckek = 3,h6kTr8e1,ek = Thkehe b, = YrYsjTrsc'eh i r
from the properties of reduced cofactors (146.8). The coefficients gij of the metric quadratic form are tensor components; for, from (151.1), (8)
9ij = ei ' ej = (tier) ' (cjes) = c'- jug,.
in agreement with (5). The same is true of gij; for, from (152.5), (9)
9i7
= e ' e' = (Yrer) - (Yses) = Y:Y39r8, b, S transforms
in agreement with (4). Again, if we define as a mixed tensor,
ei ej = (Yrer) - (c;es) = Yrc;a8.
(10)
Indeed, gij, gij and b are all components of the same tensor, (11)
gijeiej = gijeiej = beiej = 8jeiej = eiei = eiei,
namely, the idemfactor (§ 66). In fact, if we use the formulas,
ei =
(12)
girer,
ei = girer
(§ 152),
and remember the properties of gij, g" as reduced cofactors, any two members of (11) can be shown to be equal; for example, gijeie' = 9ij9irere' = b;ere-i = eje',
gije ej = gc,e gjrer = Sie er = e ei. i
i
r
ti
i
Since eiei and eiei represent the same tensor, the order of the indices in its mixed components, the Kronecker deltas, is immaterial. Any component of T may be expressed in terms of components of another type by means of the metric form. For example, we have, from (7) and (12),
Tit = ei,T.ej = 9trer'T'ej = 9irT,J, = 9'r ei'T -er = girgjs er - T - e8 = girgj8Trs
155. Absolute Tensors. We next define absolute tensors with respect to the centered-affine group of transformations: xi = cxj
TENSOR ANALYSIS
344
Scalars p(x', x2, x3) which have one component in each coor(linate system given by
gxl, x2, x3) = P(x', x2, x3)
(1)
are said to be absolute tensors of valence zero.
Vectors have three components, and these may be of two types u,, ui. The laws of transformation, 2li = ciuj,
(2)
ui = Yu
characterize absolute tensors of valence one. The vector itself,
u = uiei = uiei, is invariant to the transformation: u = u. Dyadics have 32 = 9 components, and these may be of 22 = 4 types: Ti;, Tij, T'j, Tij. If these components transform, respectively, like the products ujv;, uivj, uiv;, uivj of components of two absolute vectors, they are said to form absolute tensors of valence two. The dyadic itself,
T = Tijeie' = Ti'eiej = TZjeie' = T'jetej, is invariant to the transformation: T = T. In general, an absolute tensor of valence m is a set of 3' components that transform like the product of m absolute-vector components; and, since each of these may be covariant or contravariant, the components are of 2' types. One of these types is purely contravariant (T1 .k), another purely covariant (Tij...k); the remaining types are mixed and have both upper and lower indices.
Consider, for example, a tensor T of valence three (a triadic); its 33 = 27 components may be of 23 = 8 types. Its covariant components Ti;k will transform like uiv;wk, namely, Tijk = c2cjckT,.t,
(3)
and the mixed components Tijk like uivjwk, namely,
Ti
(4)
jk
rjk
= ea'Ya t Tr
at
The tensor itself,
T = T.;keiejek = Tijkeiejek =
,
RELATIVE TENSORS
§ 156
345
is invariant to the transformation; for example,
T = Tijkeiejek (CiciCkTrst) (7ea) (' beb) (} cec) SaSbS,'Trsteaebec
= Trstereset = T. Note that the indices on the base vectors in T have the same order as the indices on the component but occupy opposed positions. We may solve the 27 equations (3) for the original components Trst by multiplying by yayby, and summing; we thus find
j k (5)-
r s t
yaybycTijk = SaSbScTrst = Tabc
In similar fashion we find, from (4), (6)
yaCbCkijk
= brbbb -'r = Tab`.
156. Relative Tensors. When the coordinates are subjected to an affine transformation, (1)
xi = Cr,
;xti
(c = det c s 0),
= yrxr
the transformation equations, for all tensor components thus far considered contain only the coefficients c, y. More generally, we may have equations of transformation, such as (2)
T.jk = CNCr j i
t
r ,
which involve the Nth power of the determinant c. In this case the quantities in question are said to be components of a relative tensor of weight N. When N = 0, equations (2) reduce to (155.4) for an absolute tensor. As a consequence of (2), the relative tensor T = Trseere8et becomes (3)
T = cNT
when the transformation (1) is effected. The law of transformation of a tensor component is determined by (i) its valence: the number of its indices; (ii) its type: the position of its indices from left to right; (iii) its weight: the power of c that enters into the transformation equations.
TENSOR ANALYSIS
346
§ 157
If the weight of a tensor is not specified, it is assumed to be zero; the tensor is then absolute. A relative scalar so of weight N (valence zero) is transformed according to gxl, x2, x3) = CNc,(x1) x2, x3).
(4)
The box product of the base vectors, E = el e2 X e3, is a relative scalar of weight 1; for, from (147.2),
E = cE.
(5)
The box product of the reciprocal base vectors, el . e2. e3 = E-1, is a relative scalar of weight - 1; for E-1 = c-1E-1, from (5). The determinant,
g = det gij = E2
(6)
(152.8),
is a relative scalar of weight 2; for E2 = c2E2. THEOREM. The permutation symbols Eijk and eijk, regarded as the same set of numbers in all coordinate systems, are components of relative tensors of weight - 1 and 1, respectively.
Proof. If the formulas (146.2) are applied to the determinants,
y = det -?j- = 1/c ,
c = det c2j
we have CEijk =
iCjCkerat'
YEzlk = yr"Ys7
Brat.
Since Eijk = Eijk, Eijk = Eijk by definition, these equations assume the form, e
= C8Ctk jarat, z
Eijk
j kerat _ Y iryayt
required by relative tensors of weight -1 and 1. 157. General Transformations. Three equations, (1)
x' = f z(xls x2
x3) f
in which the functions f i are single-valued for all points of a region
R and which can be solved reciprocally to give the three equations, (2)
xt = gi(xl 22,
3)
in which the functions gi also are single valued, determine a oneto-one correspondence between the sets of numbers xi and xi. If we regard both sets of numbers as coordinates of the same point,
GENERAL TRANSFORMATIONS
§ 157
347
(1) defines a transformation of coordinates. The affine transforma-
tion is the particular case of (1) in which the functions f i are linear and homogeneous in x', x2, x3. Consider now the totality of such transformations in which the functions f i(x', x2, x3) are analytic functions having a non-vanishing Jacobian in R: ax
azi
=det- 3,6 0.
(3)
ax' The implicit-function theorem ensures the existence of the solutions (2) of equations (1) in a sufficiently restricted neighborhood of any point. In order to have a transformation of coordinates OX
in R, we must assume that the solution (2) exists and is single valued throughout R. The coordinate transformations (1) thus defined form a group; for
(a) they include the identity transformation xi = xi whose Jacobian is 1; (b) each transformation (1) has an inverse (2) whose Jacobian ax/ax I is the reciprocal of (3); (c) the succession of two transformations of the set is equivalent to another transformation of the set. As to (c), the two transformations, xL = fi(xl x2 x3)
xz = h'(x' x2 x3)
are equivalent to
xz = h'[f`(x),f2(x),f3(x)] = ji(xl, x2, x3), for which the Jacobian, /0xi axr axi
detax'- = det\axr(-ax'-/
-
/
axi
\
axr
= (det
/ \
( det -)
L
0.
ax'/ ax Moreover, since f i and hi are analytic functions, the functions ji are likewise analytic. In the affine transformation, (4)
xi
= yrxt, i
xi
i-r , = crx
we have axi (5)
axr
yr
=r axr axi
z;
TENSOR ANALYSIS
348
§ 157
the Jacobians, axi
(6)
det - = y, axr
axi
det - = c,
and yc = 1.
axr
A tensor of weight N, with respect to the affine group, transforms according to the pattern, (7)
TU'.k =
CNyTy3CtkTrst.
In view of (5) and (6), this equation may be written (8)
axi ax' axt
ax
Ti'k- at
IN
axr axs axk
Now, by definition, a set of 33 functions Tr. $t is said to form a tensor
of valence 3 and weight N with respect to the general transforma-
tions (1) provided the components transform according to the pattern (8). This equation becomes (7) when the transformation is affine and constitutes a natural generalization of (7). The corresponding equation for a tensor of any valence or type is now obvious. In particular, for contravariant and covariant components of an absolute vector, axi
(9), (10)
uy =axr- ur,
_
axr
ii = ax- ur. i
Since the new tensor components are linear and homogeneous in terms of the old components, we have the important result: If the components of a tensor vanish in one coordinate system, they vanish in all coordinate systems.
More generally, tensor equations maintain their form in all coIf any geometrical or physical property is expressed by means of an equation between tensors, this equation in an arbitrary coordinate system expresses the same property. Although the coordinates themselves are not vector components in the general group (1), their differentials dxi are the components of a contravariant vector; for, from (1), axi axi axi axi dxI + (11) dxi = dx2 + - dx3 = - dxr. ax' axi axr ax2 We regard the differentials dxr as the prototype of contravariant vectors; and the rule of total differentiation gives the correct pattern for their transformation. ordinate systems.
- -
GENERAL TRANSFORMATIONS
§ 157
349
The partial derivatives app/axi of an absolute scalar p(x', x2, x3) are the components of a covariant vector; for, since
0(.'
x2, x3) =
P(xl, x2,
x3),
when we replace xi on the right by the values (2), a P ax2
(12)
ax
ax
1
ax i
_ axr aV ax ax r
a,p ax3 ax3 at i
+ ax2 ax i
We regard the derivatives ap/axr as the prototype of covariant vectors; and the chain rule for partial differentiation gives the correct pattern for their transformation. In the affine transformation, the position vector r = xiei, and
ei = ar/axi.
(13)
We now adopt (13) as the definition of ei in any coordinate system
x'; then
ar ar axr ei=-=--=-er. axr
(14)
axraxt
axi
axz
The base vectors ei thus transform after the pattern in (10) and therefore merit a subscript.
The base vectors ei must transform after the pattern in (9), namely, a. t
ey = - er;
(15)
axr
for, since er e8 = S8, ei
axr axj
axr axj
axj
49V ax8
a.2'i ax'
axi
and the sets ei, e' are also reciprocal. We now can show that (16)
gij =
gij =
S
transform as tensor components; we need only make the replacements indicated by (5) in the proofs of § 154. In fact gij, gi', S; are all components of the idemfactor ere' = erer; thus S; sometimes is written gg. (17)
Moreover,
ei = ei ere' = gire',
ei = ei. erer = girer.
TENSOR ANALYSIS
350
158. Permutation Tensor. The box product E = el
e2 x e3 is
a relative scalar of weight 1: ax
E
ax
for the proof of (147.2) applies when we replace c; by axe/a.V. Moreover, E-' = e' e2 x e3 is a relative scalar of weight -1. The permutation tensor is defined by (2)
e1 - ej x ek eiejek = ei ej x ek eiejek.
These triadics are equal absolute tensors.
ej =
ei = gireT)
For, if we put ek = gkter,
9ise3,
the left member becomes er es x e` ereset. Moreover, (157.14) and (157.15) show that (3)
= ei ej x ek eiejek.
C-
From (2), we see that the covariant and contravariant components of the permutation tensor are (4)
e1 - ej x ek = EijkE,
ei ej x ek =
EijkE-1
Since E and E-' are relative scalars of weight 1 and -1, Eijk and Eijk are components of relative tensors of weight - 1 and 1. This is the theorem of § 156; the former proof still holds when obvious changes are made.
159. Operations with Tensors. The three basic operations on tensors are addition, multiplication, and contraction. 1. Addition of tensor components of the same valence, weight, and
type generates a tensor component of precisely the same characters. Example.
If Pij, Qij are both of weight N,
Fij + Qij _
ax N axr ax" ax
-(Pre + Qr.)a.ri azj
Hence Tit = Pij + Qij transforms in exactly the same manner as Pij and Qij.
2. Multiplication of tensor components of valence m1i m2, of weight N1, N2, and of arbitrary type generates a tensor component of valence m1 + m2i of weight N1 + N2, and of a type which is defined by the position of the indices in the factors.
OPERATIONS WITH TF''\SORS
§ 159
Example.
351
Let Pi' and Uk have the weights 2 and 1; then
ax ' ati a-' P'uk = I ai ly axr axe Pr8
at'.
ax
Ox
axt axkut
az
=
as
3
az' axt
ax'' ax8 D.
PT8 ut.
Hence T'jk = P''uk transforms as a tensor of valence 3, of weight 3, and of the type indicated by its indices. Even though P''uk = ukPi', the product tensors,
uP = ukPz'ekeie
Pu = Pi'ukeie?ek, are not the same.
3. Contraction. In any mixed tensor component an upper and lower index may be set equal and summed over the index range this generates a tensor component of the same weight and of valence two less. Its type is determined by the remaining indices, not involved in the summation. Example 1.
In the absolute component T'!.k set j = k; then
u=_VV jT1.11'+'V.22+T133 is a contravariant vector, for Tt =
a ax ax T3 = az ax T ..t .. - ax"" ax8 all Of i
i
i
t
t
ax8
all =-38T'st = ax,*
axr
T
Trt al'
"` = ax"
ur.
A tensor component may also be contracted with respect to two indices on the same level; we need only raise or lower one index and then contract as previously. Example 2. To contract the absolute component V% on the indices i, j we first lower the index j 4 !r T.jk = T..k9i*;
setting i = j now gives the covariant vector
! Vk = T ::..kgir = T ..kgij. We may also obtain this vector by lowering the index i and setting i = j: Tick =
Tt"k = V" k9ri
Contraction on a pair of indices is equivalent to the scalar multiplication of the corresponding base vectors in the complete tensor.
In the preceding examples, T = T!.ke,e ek,
u = Ti'.keiej ek = r'.ks;ei = T .,ei, v = T:'.kei ejek = Tt'.kgi,ek
352
TENSOR ANALYSIS
§ 160
The proof that contraction produces tensors follows from the equation : T'.'.keie;ek = TY?.keie,ek.
When the new base vectors on the left are expressed in terms of the old, this becomes an identity; and, since scalar multiplication is distributive with respect to addition, it remains an identity when base vectors in the same position are multiplied on both sides. Thus, in ex. 2, v. V = T'.kgiiek = As long as a contracted tensor has two or more indices, the foregoing process may be repeated, each contraction reducing the valence by two but leaving the weight unaltered. Thus the tensor T!.kh may be contracted twice to yield two different scalars 7".'.i; and T'!.;i, each consisting of nine terms. These are obtained from
the invariant tensor TT'.kheiejekeh by the formation of ei - ek, e; eh and ei - eh, e; ek, respectively.
Contraction often is combined with multiplication. For example, if we multiply the vectors u, v.; and then contract, we obtain their scalar product u'vi. Again the product A B of two dyadics defined in § 65 is equivalent to tensor multiplication followed by contraction on the two inner indices. Thus, if A = Ati,eV , B = Bkhekeh, A - B = Ai;Bkhe'e' . ekeh = Aj;B'he'eh. The product of the idemfactor I = ere? and a tensor T of valence In general IT differs from TI; but the contracted products reproduce T:
m is a tensor IT of valence m + 2.
160. Symmetry and Antisymmetry. A tensor is said to be symmetric in two indices of the same type (both covariant or both contravariant) if the value of any component is not changed by permuting them. It is antisymmetric or alternating in two indices
of the same type if permuting them in any component merely changes its sign. Thus, if Tabc = Tbac,
Tabc = -Tcba,
the tensor T is symmetric in a, b, alternating in a, c.
KRONECKER DELTAS
§ 161
353
Symmetry or antisymmetry are properties that subsist after a general transformation of coordinates. Thus, for the preceding example, we have Tijk =
axa axb axe a
ax ax axk
axb axa a.T.e
Tabc =
ax ax £ axk
Tbac = Tjik
axe axb axa
axk axj axi
(-Tcba)
A tensor cannot be symmetric or alternating in two indices of different types; for a property such as Ti j = Vi i does not subsist after a transformation of coordinates. A tensor is said to be symmetric in any set of upper or lower indices if its components are not altered in value by any permutation of the set. The subsistence of this property in one coordinate system ensures it in all. A tensor is said to be alternating (or antisymmetric) in any set of upper or lower indices if its components are not altered in value by any even permutation of the indices and are merely changed in sign by an odd permutation of these indices. A tensor Tij or Tijk (T ij, Tijk) which is alternating in all indices is called a bivector or trivector, respectively. In such alternating tensors, all components having two equal indices are zero. In a trivector Tijk, the non-zero components can have but two values, ±T123; moreover, the contracted product, EijkTijk = 3!T123
In this connection we remind the reader that a given permutation of n indices from some standard order can be accomplished by a succession of transpositions of two adjacent indices and that the total number of transpositions required to bring about a definite permutation is always even or always odd. The permutation in question is said to be even or odd in the respective cases. 161. Kronecker Deltas. We now generalize the simple Kronecker delta S by introducing two others, defined as follows: aiajbkc
(1)
- Eabr E
(2)
aab = EabrE
ij
ijk ijr ijr = aabr
Since the epsilons have weights of 1 and -1, ba"bk, is an absolute tensor of valence six; hence b b, formed by contracting S, is an absolute tensor of valence four.
TENSOR ANALYSIS
354
§ 162
Those definitions show that S b and bay can only assume the values 0, 1, -1; they are evidently alternating in both upper and lower indices. Their precise values in any case are as follows: If both upper and lower indices of a generalized delta consist of the same distinct numbers chosen from 1, 2, 3, the delta is 1 or -1 according as the upper indices form an even or odd permutation of the lower; in all other cases the delta is zero. This rule is a direct consequence of the properties of the epsilons. We have, for example, 32
012
12 =
023 - - 1,
1,
123
6213
231
x123 = 0123 - 1,
23
13
011 = 021
o. ;
v321
8123 = 0
= 0123
162. Vector Algebra in Index Notation. The three operations on tensors enable us to give a succinct account of vector algebra. Vectors are to be regarded as absolute unless stated to be relative; and we shall often speak of components as vectors.
If w = u + v, we have
vi =ui+vi, or wi=ui+vi; obviously vector addition is commutative and associative. The tensor product of two vectors u, v is the dyad uv. On contraction, uv yields the scalar product, (1)
u . v = uivi = uivi = gijuivj = giju1vj.
From (1), we have
U. (v+w) = The antisymmetric dyadic (bivector) P = uv - vu is called the outer product of u and v; its covariant components are Pjk = UjVk - UkVj = 8jkuavb.
(2)
The dual of Pjk (cf. § 170) is the contravariant vector of weight 1: p =
(3)
2
e ijk Pik = E''ikujvk;
Z
its components, U2V3 - u02,
U3V1 - u1V3,
u1V2 - u2v1Y
are the same as the non-zero components Pik. Since E-1 is a scalar of weight -1, the components E-'p' are absolute. The corresponding vector,
§ 162
VECTOR ALGEBRA IN INDEX NOTATION
(4) E-lpiei = E-1EijkeiujVk = E-1
355
= uXv (24.9). v1
V2
V3
Similarly, from the outer product,
P'k = bab26avb = u'vk - ukvJ, we obtain as dual the covariant vector of weight - 1: (2)'
(3)'
i pi =_ZEijkPJk = eijklljvk.
The absolute components Epi again give u X v : (4)'
el
e2
e3
Epiei = Eeijkeiujvk = E ul
u2
u3
vl
v2
v3
= uxv
(24.10).
From (4) or (4)', we have uXv = - V X u,
U- (v + w) . u x v + U X W.
The components of the triple product u x (v X w) are (u X (v X w))' = E-1EijkUj(V X w)k
(4)
(4)'
= E-hE'fkujEEkabvawb
=
EijkEabkujvawb
(161.2)
= SabUjva'Uib
= uj(viwj - vjwi) = (u w)vi - (u - V)wi; hence
u X (v X w) = (u w)v - (u v)w. The box product u x v w is given by either of the absolute scalars, (5)
(5)'
(U X V)'wi = E-'Cijk,ujvkwi = E-1
(u X v)z-w = Eeijku'vkw'
=E
ul
u2
U3
Vi
V2
V3
W1
W2
W3
U1
u2
u3
V1
V
w1
in agreement with (24.12).
2
w2
V3
w3
TENSOR ANALYSIS
356
§ m3
163. The Afne Connection. Any vector can be expressed as a linear combination of the base vectors. When the 32 derivatives of the base vectors ae;/axi = Die; are thus expressed, (1)
Die; = ref + r e2 + r 'e3 = ri,er,
the 33 coefficients r, are called components of the affine connection. If we multiply (1) by ek , the right-hand member becomes rzJj r = r ; hence
r = ek
(2)
Die;.
The law of transformation for r is given by (3)
r = ek
axk ec
Die; =
ax c
\ / -- DQ I I axb eb ), /
l ax i
/ \ax
'
where (4)
a axb Di =axia - _axa _ Da; axi axi axa
The differential operator Di transforms like a covariant vector (hence
the subscript). In (3), Da acts on both scalar and vector factors following; hence, from (1), axk
ax`
e
axa axb ( a2xb eb+--raber
axiax'
axi axi
Since e° eb = bb, ec er = 6'r, this gives a2xb axk
axa axb axk + - - - rab axiaxj axb axi ax' ax`
(5)
for the desired law of transformation. Therefore r is a tensor component when and only when a2xb
49-Pi axi
= 0,
axb
- = CJ (const), ax'
and hence xb = c;x', if the coordinates xi and xi have a common origin.
The components r of the affine connection (between the deriva-
tives of the base vectors and the vectors themselves) are tensor components only with respect to affine transformations.
THE AFFINE CONNECTION
§ 163
357
Although the Irk are not tensor components for general transformations, we shall see that the index notation still serves a useful purpose. j Since e1 = D;r (157.13),
P = ek DiD;r = ek D;Dir = Pk;
(6)
and, from (5), we see that the symmetry of P in the subscripts persists after a transformation of coordinates. If we transform coordinates from x to IF, we have, from (3), (7)
P4 - \axk k/
\ax1'
\4924
or, on making the replacements, axk
k
Di = - Da,
_r _ PPQ
-
e;
axi
ax,
(8)
axb
(9x,
e = - es,
axa
axr
ax'
eb,
(axb
l
l
ax, ec/ \a2p Da/ \ai7 eb/
a relation of the same form as (3). Consequently, the succession of transformations r --p I` -> f produces a transformation r -- r of the same form as r -+ 1. We express this property by saying that the transformation (3), or its equivalent (5), is transitive. If, in particular, x i = x', we may delete all the tildas (-) in (7) ; this equation then gives, on expansion, (9)
a2xJ P,,
air
axP axq ax'
a.
a. axr
k
I'2J.
axP a:rq axk
These equations constitute the transformation inverse to (5) and also may be derived by solving these equations for P'ab. This solution may be effected by the multiplication of (5) by axi ax' axr axp ax4 axk
t The coordinates xi present a similar situation. For afflne transformations they are components of a vector; for general transformations this is not the case, but the indices still serve to indicate that their differentials dxi are vector components.
T1E;NSOR ANALYSIS
358
§163
On writing y = x, the equation of transformation (5) becomes:
ax, a2-' + - - rab lay, ayj ayi ay' i
(10)
axb
ayk
1
axc
If we now make the change of coordinates,
xr = xo + yT -
(11)
2(rPq)oypy4,
where the gammas are computed for x = xo, the point x" = xo corresponds to yr = 0. Since ay'r/ayi = ax,
ayi =
Si
-
a2xr
ayi a y.i
- - (r)o;
hence, at the point x' = xo (yr = 0), the brace in (10) becomes
-(r )0 +aj(rab)o = 0. Consequently, all the gammas I'- = 0 vanish at the origin y'' = 0 of the new coordinates. Such a system of coordinates is termed geodesic. Since the gammas are not tensors, the equations riI; = 0
(y = 0) do not imply r = 0 (x = x0). Example 1.
For cylindrical coordinates p, p, z, we have (§ 89, ex. 1) r = [x, y, z] = [P cos ip, P sin ,p, z].
If we put x1 = p, x2 ='P,x3 =z, el = Dlr = [cos p, sin p, 0], e2 = D2r = p[- sin p, cos {p, 0],
e3=D3r=[0,0,1]; Die, = 0,
D2ei = [- sin gyp, cos (a, 0] = 1 e2,
D3ei = 0,
P
D2e2 = -p[cos rp, sin p, 0] = -pei,
D3e2 = 0,
D3e3 = 0.
Hence all gammas are zero except
rig = r2, = 1/P,
r22 = -P-
Example 2. For spherical coordinates r, sp, B, we have (§ 89, ex. 2)
r = [x, y, z] = r[sin 0 cos p, sin 0 sin p, cos 0].
KINEMATICS OF A PARTICLE
§ 164
359
If we putxi =r,x2= V,x3=0,* ei = Djr = [sin 0 cos gyp, sin 0 sin gyp, cos 01,
e2 = D2r = r sin 01 - sin gyp, cos p, 0], e3 = D3r = r[cos 0 cos gyp, cos 6 sin gyp, - sin 0]; D2e1
D1e1 = 0,
=
1 1
D3e1
e2,
r D2e2 = -r sin2 6 e1 - sin 0 cos 0 e3,
=
1
I e3,
r
D2e3 = cot 0 e2,
D3e3 = -rel. The non-zero gammas are therefore
r12=r21=1/r,
r13=r31= 1/r,
r22 = - sin 0 cos 0,
r23
=
1132
= Cot 0,
r22= -r sin2 B, r33 = -r.
164. Kinematics of a Particle. We now may find velocity and acceleration of a particle in general coordinates:
a = dv/dt = akek. In rectangular coordinates x1 = x, x2 = y, x3 = z we have vk = V = vkek,
(1)
dxk/dt (52.7). Hence, in general coordinates 2i, these components become 8xk dxi
dxk
vk = - - _ axi dt dt The time derivatives of the coordinates, which we write ±k, thus transform as contravariant vector components. For the acceleration we have, from (1), (2)
a=
dvk
dt
ek + vk
aek dxi
ax i
dt =
vkek +
vtvkr:kei;
and, on interchanging the summation indices j, k in the last term, (3)
a = (vk + viv'r ,)ek
Therefore, in any coordinate system the velocity and acceleration components are vk = xk ak = xk + x°'x'I'). (4) Since the base vectors ei are not, in general, unit vectors, we must distinguish between the components vk and ak and the numerical values of the terms in vkek, akek. Thus the terms v'e1, a'e1 have the numerical values vllell, a'lell. * The order r, p, 0 is sinistral and [ele2e3] < 0; cf. p. 197.
TENSOR ANALYSIS
360
Example 1.
Cylindrical Coordinates p, ,p, z.
§ 1&
From § 163, ex.
1, el, e2, e3
have the lengths 1, p, 1; hence the values of the velocity terms in (1) are a, pp, z
From the non-zero gammas r22 = -p, rig = 1/p, we have
a1 = a - Psp2, a2 = sp + 2Pcp/P, a3 = z; and the numerical values of the acceleration terms are
p' + 2p',
P - Pcp2,
z.
Example 2. Spherical Coordinates r, p, 0. From § 163, ex. 2, e1, e2, e3 have
the lengths 1, r sin 0, r; hence the velocity terms in (1) have the values, r, rcc sin B, r9.
With the non-zero gammas r22 = -r sin 2 0, r33 = -r we have, from (4), al = r - p2r22 + O2r33 = r - rrp2 sin2 B - r92;
with r12 = 1/r, r23 = cot 6,
a2 = + 2r,pr12 + 2,pOr23 = + 2r(p/r + 24 cot 0; and, with r13 = 1/r, r22 = - sin 6 cos 0, a3 = 9 + 2rOri3 + ,,2r22 = 9 + 2r9/r - ,p2 sin B cos 0. Hence the values of the acceleration terms are r,p sin 0 + 2i ,p sin 0 + 2r09 cos 0,
1' - r,p2 sin2 0 - r92,
r9 + 2r9 - r02 sin B cos 0.
165. Derivatives of e i and E. From the relation e' er = S*, we have, on differentiation, (Die')
er = - e' Deer = - rsr,
and hence
Die'
(1)
rrer
From Die; = r jer and (1), we have
Di(e,e') = rz,ere' - rreer = 0,
(2)
on interchanging the summation indices r, j in the second term. Since the product E = el e2 x e3 is distributive with respect to addition, its partial derivatives may be found by the familiar rule for differentiating a product, DiE = (Diet) . e2 x e3 + el (Die2) x e3 + el e2 x (D:e3) = r,r'1er e 2 x e3 + ri2e1 er x e3 + ri3e1 1
2
3
_ (ri1 + ri2 + ri3)e1 . e2 x e3,
e2- er
§ 166
AFFINE CONNECTION AND METRIC TENSOR
361
or, if we apply the summation convention,
DiE = rirE. The derivative of E-i = el e2 e3 is therefore (3)
(4)
DiE-i =
-E-2DiE
rirE-i
From (3), we have
DjDiE = (Djrir + rarr .,)E; and, since DjDiE = DiDjE, (5)
Dj rir = Dir;r
166. Relation between Affine Connection and Metric Tensor. On differentiating gij = ei ej, we have
Dk9ij = rkier - ej + rkjei - er = rkjJrj + rkjgir Since Dkgij and F are both symmetric in the indices ij, there are 3 X 6 = 18 quantities in each set. Equations (1), 18 in number and linear in the 18 gammas, may be solved for the latter. We first introduce the notation, (1)
(2)
F9rk = rij,k,
just as if r k were a tensor whose upper index was lowered. Then we have also (3)
rij,rgrk = r ,
for the left member equals
ri,gar9rk = I't S = r We note that Fij, k is also symmetric in the indices i, j. Moreover (4)
Diej = F jer = rij,se9,
if we put e,. = g,.ses. We may now write (1) in the form: (5)
rki,j + rkj,i = Dkgij.
If we permute i, j, k cyclically in (5), we obtain the equations,
rij,k + rik,j = Di9jk,
rjk,i + rji,k = Dj9ki;
TENSOR ANALYSIS
362
§ 167
and, upon subtracting (5) from their sum, we obtain ri.i.k =
(6)
Z
(Digjk + D;gki - Dkgi1)
We may now compute r from (3). In the older literature, the components of the affine connection are denoted by [ } = r., [ii,k] = rij,k, .V
Z)
These Christoffel symbols of the first and second kind, respectively, therefore are defined by the equations: J
l
7
} = gkr[ij, r]. [ [ij, k] = (Digik. + D7gki - Dkgii), z 167. Covariant Derivative. The gradient of an absolute tensor T is defined as aT (7)
vT=eh
(1) Since T = T,
eh
aT axh
axh
aT aT ax8 8 r-=axh_ ar__ =e = Sre axr
ax8 axh
8
aT axr ;
hence VT is a tensor of valence one greater than T. If the components of T are T ;.'.'.k (the order of the indices is not specified), the components of VT are written (2)
Vh
Tab
i;...k,
the index h on V corresponding to the differentiation a/axh. This is a covariant index, for the operator Dk = a/axh transforms like a covariant vector: (3)
axr axr a Dh=----=-Dr. axh axr a
axh
axh
For this reason the components (2) are called covariant derivatives of Tab
If T is a relative tensor of weight N and valence m, E-NT is an absolute tensor, whose gradient, ehDh(E-NT),
is an absolute tensor of valence m + 1. If this is multiplied by EN, we again obtain a relative tensor of weight N; this tensor is defined to be the gradient of T and written (4)
VT = ENehDh(E-NT).
COVARIANT DERIVATIVE
§ 167
363
When N = 0, (4) reduces to (1). The components of VT, denoted by prefixing Vh to the components of T as in (2), are called covariant derivatives.
From (165.3), we have DhE_N
and hence
= -NE-'V-1DhE = -NrhrE-N,
Dh(E-'YT) = E-v(DhT - NrhrT),
VT = eh(Dh - Nrhr)T. Thus the operator, (5)
V
= eh(Dh - Nrhr), r
applied to any invariant tensor T (with its complement of base vectors) generates another tensor VT of the same weight and valence one greater. We next compute the components of VT explicitly, where
T = Tq '...ceaeb ... e,eiej ... ek. If T is a relative tensor of weight N, VT contains the term, ' e k. k eheaab e,e iej N rrhr It remains to compute the part of VT due to the operator ehDh. (6)
-
Tab'
ij...
Now Dh acts on the "product" of T ;.'.'.k and a series of base vectors. Since this product is distributive with respect to addition, DhT may be computed by the usual rule for a product; hence c ab DhT = (DhTi;...k)eaeb . . . eye iaj
+
. .
. ek
eceiej
ab
In the second line put Dhea = rhaer, ...
... ek + .. .
i j ec(Dhe )e ... ek +
,
Dhe, =
and in the successive terms interchange the summation indices In the third line put r and a, r and b, , r and c. i r , ... , k r Dhe kDhei - -rhre - - rhre , and in the successive terms interchange the summation indices r and i, r and j, , r and k. When this is done, each term of VT contains the same complex of base vectors, eheaeb ... eceiej ... ek,
TENSOR ANALYSIS
364
§ 167
and the component VhTt6.'.k of VT equals the sum of their scalar coefficients :
v Tab e = D
17
k-
Tab . C q- k
+ Tij.k rchr
a Tij...krhr + Trj...k rhi -
- Tij...r rhk
r
-
r
Nrr rTab c hr
This is the general formula for the covariant derivative of any tensor component, absolute or relative. The last term is absent when T is absolute (N = 0). For every upper index
hik
v Ta*:
contains a term
T°::.r:: t k r* hr
y
and, for every lower index *, OhTa' *:: °k
contains a term
- T"
'
a rr.
We consider now some important special cases. If p is a relative scalar of weight N, vhcp = Dh-p - Nrhr(c
(8)
Since E and E-1 are scalars of weight 1 and -1, we have, from (165.3) and (165.4), (9) (10)
vhE = DhE - rhrE = 0,
vhE-1 = DhE-' + rhrE-1 = 0. Again, since g = det gij = E2 is a relative scalar of weight 2, (11) Vhg = DhE2 - 2r;,rE2 = 0. If v = vie1 = vie' is an absolute vector, (12)
VhV = Dhvti + vrrhr,
r vhvi = Dhvi - vrrhi. These expressions are the mixed and covariant components of one and the same dyadic Vv. For an absolute dyadic T, the components of VT may have the (13)
forms :
(14)
VhT2 = DhT'' + Tr1rhr + Tirrhr,
(15)
VhT`j = DhVj + T'jrhr - Tirrhj, VhTij = DhTij - Trjrhi - T,rrhj.
(16)
§ 168
RULES OF COVARIANT DIFFERENTIATION
365
For the metric tensor, (17)
G = gijeiej = gijeiej = eiei = I,
we have, from (165.2), (18)
VG = e'DhG = ehDh(eiei) = 0.
The components of VG therefore vanish: (19)
Chgij = 0,
Vhgii = 0,
Vhbi = 0.
Note that (166.1) is equivalent to Vkgij = 0. Example. We can write
VT = eh(Dh - Nrhr)T = ehvhT,
i20)
if we regard Vh as an operator that acts only on scalars. Covariant differentiation then is defined by this operational equation. With this convention, we have also
VVT = ei(Di - Nrir)ej(Dj - Nrn)T = eiejVjVjT. The second member may be written eiej(Di
- Nrit) (Dj - Nrjr)T - eirise'(Dj - Nrr )T = eiej ( (Di - Nrit) (Dj - Nrir) - rij(D.
Nr )1T
when indices j and s are interchanged; hence
viv jT = (Di - N rir) (Dj - Nrrr,)T - rijV,T, ViViT = (D, - Nrjr) (Di - Nrsr)T - r-! v.T, and, on subtraction, (21)
(CiV1 - Vjvi)T = (D1Dj - DjDi)T,
in view of (165.5). On the left the operators Vr act only upon the scalar components of T. We note that (21) applies to relative as well as absolute tensors.
168. Rules of Covariant Differentiation. 1. The covariant derivative of the sum or product of two tensors may be computed by the rules for ordinary differentiation.
If we introduce a system of geodesic coordinates yi 1163), the corresponding gammas 1 , will all vanish at the point yi = 0; hence, from (167.7), OTab...r=DTab...r (1) h ij...k h ij...k,
at the origin of geodesic coordinates, which moreover can be chosen at pleasure. For example, we have the tensor equation, Vh(Tt'uk) = (VhTz3)uk + T''Vhuk,
TENSOR ANALYSIS
366
§ 169
in geodesic coordinates and therefore in any coordinates (§ 157). Consequently, the sum and product rules of ordinary differentiation also apply in covariant differentiation. 2. The covariant derivatives of the epsilons and Kronecker deltas are zero.
Since these tensors have constant components, their covariant derivatives vanish at the origin of a system of geodesic coordinates; hence they vanish in all coordinate systems. 3. The operation of contraction is commutative with covariant differentiation. For example, if we contract T=jk on the indices i, j to form T?ik = SiT -jk,
we have OhVik = BiZVO
-jk.
4. The components of the metric tensor (gij, gij, S) may be treated as constants in covariant differentiation. This follows at once from (167.19). For example, Civj = Vi(g7rvr) = gjiViv'.
Thus we may find Vivj by lowering the index j in Divj; in other words, Divj and Vivj are components of one and the same tensor: Vv = V vieiej = V1vjeiei. Evidently the operation of raising or lowering an index is commutative with covariant differentiation. 5. The relative scalars E, E-1 and g may be treated as constants in covariant differentiation.
Since E, E-1 and g = E2 are relative scalars of weight 1, -1, 2, respectively, we have, from (167.4), V Vg = 0; VE = 0, for in each case Dh(E-NT) = DO = 0169. Riemannian Geometry. Any set of objects which can be placed in one-to-one correspondence with the totality of ordered . sets of real numbers (.r.1, x2, , x") satisfying certain inequalities, I xi - ai I < ki (a1 and ki > 0 are constants), is said to form a region of space of n dimensions.$ We speak of
-'=0,
(x1) x2,
. ,
x") as a point; but the actual objects may be very
$ Veblen, Invariants of Quadratic Differential Forms, Cambridge, 1933, p. In some applications the numbers xi may be complex.
13.
RIEMANNIAN GEOMETRY
§ 169
diverse.
367
Thus an event in the space-time of relativity may be
pictured as a point in four-space; and the configurations of a dynamical system, determined by n generalized coordinates, often are regarded as points in n-space. If we associate the space (x', x2, , x") with an arbitrary non-singular quadratic form, 9ii dxi dx',
g = det gij 0), (9ij = 9ii, we have a Riemannian space with a definite system of measure(1)
ment prescribed by this form (§ 151) ; and the geometry of this metric space is called Riemannian geometry. We assume that the coefficients gij are continuous twice-differentiable functions of x. The base vectors ei are not specified; but their lengths and the angles between them may be found from the relations: 9ij = ei e,. The reciprocal base vectors now are given by
(2)
ei =9irer,
(3)
where gii is the reduced cofactor of gij in det gij; for (4) ei e; = girgir = d
(152.2).
Moreover (5)
ei e' = girS* = gij
If we now transform coordinates from xi to xi, we assume that the new base vectors are given by (6)
axr
ei = - er,
axi
ei = - es;
axb
ax8
then ei and e' still form reciprocal sets, for axr a'V
8r = o.
a:C i ax8
In view of (2) and (5), equations (6) show that gjj and gi' transform like absolute dyadics. These tensors often are called the fundamental covariant and contravariant tensors of Riemannian geometry. By use of the equations, (7) e = g,rer, ei = they permit us to raise and lower indices (§ 154) and thus represent any tensor of valence m by 2' types of components. girer,
TENSOR ANALYSIS
368
§ lti'9
At a given point, gij gives the orientation of the base vectors ei relative to each other. In order to determine the relative orientation of sets of base vectors at different points, we must know the components r of the affine connection, defined by
Diej = rer.
(8)
Then, just as in § 165, we have also
Die'
(9)
Tier
(165.1).
We now assume that the affine connection is symmetric (r = r ). The gammas then are determined by the metric tensor (§ 166) : (10)
rkV = 129kr(Digjr + Djgri - Drgij) I
The epsilons in n-space, defined in § 146, have n subscripts or superscripts. The equations, ax .. (9.t xi axj axk Eb... (11)
-
. .
ax ax
axa axb IEij ...k =
axa axb
.ax`-
'
ax°
...axk Eab...ej,
ax ax` 49V t We shall call the base vectors constant if Die, = 0 (i, j = 1, 2,. , n); then raf = 0 and the functions gij = ei e; are also constant. Conversely, (12)
when gii are constants, (10) shows that r!J = 0, and hence Die, = 0. In Riemannian space it is not, in general, possible to introduce coordinates xi for which the base vectors ei are constant. Only flat space (§ 175) is compatible with such Cartesian coordinates; then each point has the position vector
r = xiei and ei = ar/axi. Moreover, for any coordinates 2i in flat space, we have (cf. § 163, ex. 1, 2) (i)
ei =
ax' 021 e'
=
or ax'ar ax' &V
=
.
at i
The geometry on a surface with the metric tensor gi; is Riemannian (n = 2). Unless the surface is flat (a plane, for example), constant base vectors cannot be introduced. If we regard the surface as immersed in Euclidean 3-space each surface point has the position vector r = xi + yj + zk, where
x = x(u, v), y = y(u, v), z = z(u, v) are the Cartesian equations of the surface. If we write xl = u, x2 = v, the
base vectors on the surface may be taken as ei = ar/axi; for equation (i) shows that these vectors transform in the manner prescribed in (6). Here r is a position vector in Euclidean 3-space; but, in general, the surface points have no position vector in the Riemannian 2-space they define. Any Riemannian space of n dimensions may be regarded as immersed in a Euclidean space (§ 178) of n(n + 1)/2 dimensions. This theorem has not as yet been rigorously proved; see Veblen, op. cit., p. 69.
RIEMANNIAN GEOMETRY
§ 169
369
generalized from (146.2), show that eij"'k and eij...k are relative tensors of weight 1 and -1; for these are the powers of the Jacobian I ax/ax I when it is transferred to the right-hand side. There are n types of Kronecker deltas in n-space: S,, S b, S bc, up to Sa'bc'...f with 2n indices. As in § 161, they are defined in terms of the epsilons. In the case n = 4, for example, i 1 ibcd Sa = - eabcde
1
Sab - - eabcde
ijk
1
abc
ijcd ,
2!
3!
eabcde
ijkd
,
ijkh
aabcd = eabcde
ijkh
The rule given in § 161 for the value (0, 1, or -1) of a generalized delta still applies. Moreover the preceding definitions show that all the deltas are absolute tensors, alternating in both upper and lower indices. See Prob. 42. The n-rowed determinant, (13)
g = 1t etii...ke,s...tgir9js ... gkt n!
[cf. (146.3)],
generalized from (152.1), is the contracted product of two epsilons of weight 1 and n absolute dyadics. Hence the discriminant g of the fundamental quadratic form is a relative scalar of weight 2. The cofactor of gij in g is ggij; hence, from (146.11) and (166.5), we have
Dhg = gg"Dhgij = ggi (rhi, j + rhj, or, in view of (166.3), (14)
Dhg = grhi + gr,. = 2grhr-
From (14), we have also (15)
DhV_ -\'g-r,.,,,
a result of the same form as (165.3) with E replaced by -/g. In defining the gradient of a tensor, we replace E by -\/g in (167.4); thus, in Riemannian n-space, N
N
VT = g2ehDh(g 2T). The components of VT again are given by (167.7). Hence this formula for the covariant derivative is still valid in Riemannian (16)
geometry.
TENSOR ANALYSIS
370
§ 170
Since the metric tensor G = gjkejek = ekek still has the property DhG = 0, VG and its components vanish as before: Vhb = 0. Moreover, from (16), vq = gehDh1 = 0, and Ohg J = 0,
Ohgij = 0,
(17)
Ohg = 0.
(18)
170. Dual of a Tensor. An m-vector is a tensor of valence m which is alternating in all indices (cf. § 160). For convenience in
wording, we also regard scalars (m = 0) and vectors (m = 1) as m-vectors. In n-space we can associate with any m-vector (0 < m < n) an (n - m)-vector, its dual, defined as follows: If and Qij...k are m-vectors of weight N, their duals are the (n - m)-vectors, 1
(1)
ab...c ij...k
m.
li
gab...c -
(2)
Pj...k,
Qij...kEij...k
m.
ab...c,
of weights N + 1 and N - 1, respectively. Note that the epsilons have n indices in n-space; and, in the contracted products, the contravariant tensor is written first, and the summation indices are adjacent.
A scalar Sp has two duals, Eab .htp, Eab . hV, according as we use (1) or (2); they have the same numerical components but different weights. The dual of an n-vector Tijk...h is the scalar: ijk...h
1 E
n!
Tijk...h = T123...n.
m < n), dual dual T = T. (3) Proof. Write T = P, dual T = p. Using (2) and (1), we have THEOREM.
If T is an m-vector (0
1
(dual
pab
c
(n -- m) !
=
1
Eab...c
(n - m) !m! 1
i;...k
_ -m!brs tPi;...k = Pra...t)
rs.
ra...t
[Cf. Prob. 42.]
DIVERGENCE
§ 171
371
since Pip -.k is alternating in all indices. Hence dual p = P; and, similarly, dual q = Q. When T is an n-vector, say Tijk...h, (1) gives the dual T123 . . n now (2) gives dual dual T = Eijk...hT123...n =
Thus the theorem holds in this case also, provided both dualizing equations (1), (2) are used. 171. Divergence. The gradient of v = viei is Vv = vhviehei.
The first scalar of this dyadic is V11) div v = %ib = If vi is an absolute vector, the divergence is the absolute scalar Viv'; this definition applies in n-space and for any coordinate (1)
system. When vi is a relative vector of weight 1, VhV'
= Dhvi + 17rrhr - vtrhr
(167.7).
On contracting with h = i, the second and third terms cancel; for vrrsr = virir, since r and i are summation indices. Hence (2) div v = Divi (wt. v = 1). vi has the weight 1 imparted by the scalar
If vi is absolute,
/; hence
viva = vi(9-i91vi) = 9-lvd(91v=) = 9-1Di(91vi), in view of (2). Thus (3)
div v = 9
Di( / vi)
(wt. v = 0).
The Laplacian V2p of the scalar p is defined as div Vp. If rp is absolute, v = V p is an absolute vector; then Vr = Drop,
Vi
= 9irDrcp,
and, from (3), (4)
v2(p
Di(V -J 9irDrcp)
The divergence of any tensor T is defined as the gradient VT contracted on the first and last indices. Thus if T has the com-
TENSOR ANALYSIS
372
1 172
porients T 'j- kh of valence m and weight N, (divT)ij...k
(5)
VhTij...kh
=
is a tensor of valence m - 1 and weight N. THEOREM. When T i' ., kh is an m-vector of weight 1, div T is the (m - 1)-vector, (div T) ij...k = DhTij...kh, (6) where V in the defining equation is replaced by D. Moreover,
div div T = 0
(7)
From (167.7), VhTij...kh = DhTij...kh +
(m > 1).
Proof.
+ rhrT
... + rk
1,hrTrj...kh +
Tij...rh
ij" . kr - rhrT ij... kh.
The two final terms cancel, as we see on" interchanging the summation indices h, r in the last term. The remaining terms containing 1'hr vanish separately on summing over h and r; for rh,. is
symmetric and T antisymmetric in these indices. Thus (6) is proved.
From (6) and the alternating character of T,
(div div T) 'j... =
DkDhT'j... kh = 0.
172. Stokes Tensor. The gradient of the covariant vector vk is the dyadic Vhvk. From this we form the antisymmetric dyadic, Pij = Sf V hvk,
(1)
known as the Stokes tensor. t Ohvk =
and, since az4r (2)
When Vk is albsolute, hVk -
rrhkvr, .
= 0 owing to the symmetry of rr , we have
Pij = SfDhvk = Divj - Djvi
(wt. v = 0).
In 2-space we can form from Pij the relative scalar of weight 1, (3)
2eijPij = P12 = Dlv2 - D2v1
t Veblen, op. cit., p. 64.
(wt. V = 0),
STOKES TENSOR
§ 172
373
and from this the absolute scalar, 1
(4)
-Vg
(Dlv2 - D2v1)
(wt. V = 0).
This is the absolute invariant on the surface with the fundamental form gig dxi dx', written n rot v or n V X v $ in § 97. In 3-space we can form from Pii the relative vector of weight 1,
-2
wi = 1EijkP7k = Ei'kD7'v k,
(5)
having the components, D2v3 - D3v2i
D3v1 - D1v3,
D1v2 - D2v2.
The absolute vector,
rot v =
(6)
1
wiei =
1
1/9
Ei'keiDjvk,
may be written as a symbolic determinant (cf. § 146): el
rot v =
(7)
1
V9
D1 V1
e2
e3
D2 D3 V2
V3
Comparing this with (88.19) now shows that rot v is in fact the rotation of v previously defined.*
In § 91 we proved that a vector v is the gradient of a scalar in 3-space when and only when rot v = 0. In n-space we have the corresponding R.
THEOREM. Let v be a continuously differentiable vector in a region Then, in order that v be a gradient vector,
(8)
V = OAP,
vi = Dip,
it is necessary and sufficient that the Stokes tensor vanish in R:
Pi,=Diva-D;vi=0.
(9)
Proof. The condition is necessary; for, if vi = Dip, Pi; vanishes identically, owing to the continuity of the second derivatives of cp.
In (97.9), H=-v19, u=x1, v=x2, ru=el, rv=e2,
r,.f
= f2 in our present notation.
* In (88.19), J notation.
u = x1, ru = el, ru f = fl, etc., in our present
TENSOR ANALYSIS
374
§ 173
The condition is sufficient. For the method of § 91, extended to case of n coordinates, leads to the function,
f
z1
(10)
(P =
vj (xl,
al
I.
x2
+
x2)
... , xn) dxl
v2 (a', x2, ... xn) dx2
a2
" f"
f
3
v3 (ale a21 x3,
xn) dx3 ...
(ai a2 ..
an-1 xn) dxn
zn
n
v
where all a2, , an are the coordinates of a fixed but arbitrary point. In the ith integral xi is the variable of integration while xi+1' ... xn are regarded as constant parameters. We now can , show from (9) and (10) that Dig = vi. Let us compute, for example, D3V. Only the first three integrals in (10) contain x3; their derivatives with respect to x3 are, respectively, v3(xl, x2) ... , xn) _ v3(al) x2, ... , x"), v3(a1 ,
x2, ... , xn)
- v3(al, a2, ... 'X n),
v3(al, a2, ... , xn),
when we make use of D3vi = Dlv3, D3v2 = D2v3 in the first and second; hence D34P = v3(xl, x2,
,
xn).
173. Curl. We define the curl of a covariant tensor Tb...d of valence m (m < n) as the (m + 1)-vector: (1)
(curl T)hij...k =
1
m.-
bhij
.k VaTbc. d
When m = 0, T = P (a scalar), 0! = 1, and curl cp = ahV aco = V MP
is the gradient of gyp.
When m = 1, T = v (a vector), curl v is the Stokes tensor (172.1).
In general, we have the THEOREM. When Tab.. d is an absolute tensor of valence m < n, (2)
(curl T)hij ... k =
Shaijc b ......kD..Tbc...d,
M.
CURL
§ 173
375
where V in the defining equation is replaced by D. Moreover,
(m < n - 1).
curl curl T = 0
(3)
Proof.
From (167.7), we have
VaTbc...d = DaTbc...d - rabTrc...d
1'adTbc...r.
Hence in the right member of (1) there are m terms of the type, Sabc...dl,r 7,
1
ab
rc...d,
m!
these all vanish separately when we sum over the subscripts of I'a*. Thus (2) is proved. From (2) we have Sahi kD (-1(,'_ Sab d D T 1 = (curl curl T) 1
1)'
(m + 1
m!
a
g
.
.
Sfet...a DgDaTb...d
which vanishes when we sum over g and a. When Tbc...d is an m-vector, the summation in (1) or (2) may
be carried out in m + 1 stages by setting a = h, i, j,
k in turn and summing over the other m indices. Thus, from (1), we ,
have (4)
(curl T)hij...k = VhTij...k + (-1)mViTj...kh + ... ,
taking the m + 1 cyclical permutations of the subscripts hij .. k in order and placing (- 1)' before the second, fourth, terms. In the cases m = 1, 2, 3, we thus obtain
(curl T)ij = VjTj - ojTi, (curl T)ijk = ViTjk + VJTki + VkTij, (curl T)hijk = VhTijk - ViTjkh + VVTkhi - VkThij
When Tbc...d is an absolute m-vector, we obtain, from (2), an equation of the form (4) with V replaced by D.
When S is an absolute tensor of valence m - 1, T = curl S is an absolute m-vector and curl T = 0. Then (4) becomes (8)
0=
(-1)'DiTjk...h + .
TENSOR ANALYSIS
376
§ 175
Thus if vi is absolute, T i j = S Davb is an absolute bivector, and
DiTjk + DjTki + DkTij = 0.
(9)
174. Relation between Divergence and Curl. For alternating tensors, we have the THEOREM.
If T be d is an n2-vector (n2 < n),
dual curl T = div dual T,
(1)
provided dual T is taken contravariant when T is a scalar. Proof.
From § 170, 1
(dual curl T)pq"'r
ii ... k(CUrlT) ij...k Epq...r ij...kaab ..kdOaTb...d
(m + 1)!m! pq... r ab... dVaT
b...d
M! Va
(1. E pq "r
Tb ...d
= Va(dual T)P . ra
= (div dual T)pq*'*.
On taking duals of both members of (1), we have
curl T = dual div dual T.
(2)
Moreover, if we replace T in (1) by dual T, we have also
div T = dual curl dual T.
(3)
175. Parallel Displacement. A vector p is said to undergo a parallel displacement along a curve xi = pi(t) when dp/dt = 0 along this curve. dp
dt
If p = pkek, we have
dxi dxi = dpk -dpkek + pk aek - ek + pk rikej, axi dt dt dt
dt
or, on interchanging summation indices j, k in the last term, dp dt
(+pr)ek. dt
PARALLEL DISPLACEMENT
§ 175
377
Hence, if dp/dt = 0, the components pk satisfy the differential equations: dpk
(1)
dt
pj
+
dx'
dt
(k = 1, 2, ... , ii).
0
r". =
A solution pk(t) of this system, satisfying the arbitrary initial conditions pk(0) = ak, defines a vector at each point t of the curve. The vector ak at the point PO (t = 0) is said to undergo a parallel displacement along the curve into the vector pk(t) at the point P. In (1) dxi/dt depends upon the functions pi(t) defining the curve; hence, in general, the solutions pk(t) will change when the curve is altered. In other words, the vector pk obtained by a parallel displacement of ak from PO to P depends, in general, upon the path connecting these points. The length of a vector p and the angle between two vectors p, q remain constant during a parallel displacement; for, if dp/dt and dql dt vanish along a curve, we have also d
d
dt(P - q) = 0.
dt(P.P) =0,
We shall say that a vector remains constant during a parallel displacement.
If s is the are along the curve, ds2
= gi; dxi dx1 = g;xix' dt2.
If we choose the are as parameter (t = s), we have gjjziz' = 1, an equation which states that the tangent vector dxi/ds to the curve is of unit length. If a curve has the property that its unit tangent vectors dxi/ds are parallel with respect to the curve, it is said to be a path curve for the parallel displacement. With t = s, pk = dxk/ds, (1) gives the differential equations of the path curves, d2xk
dxi dx'
ds2
t' ds ds -
0.
(2)
The path curves are the "straightest" curves in our Riemannian space-the analogues of straight lines in Euclidean geometry. When equations (1) can be satisfied by functions pk(x', , x") of the coordinates alone, the parallel displacement is independent of the path, and the space is said to be flat. Then dpk
apk dxi
dt
axe dt
TENSOR ANALYSIS
378
§ 175
and the ordinary differential equations (1) are replaced by the partial-differential equations: ap''
axi + p'1' = V pk = 0.
(3)
Since Vipk are the components of Vp = e'Dip (§ 167), we see that a flat space contains vectors p(x', , X'), such that Vp = 0, or
(i = 1, 2, ... , n). Dip = 0 Since pk = p ek, we see that (4) is equivalent to (4)
D,Pk = p Dzek. For any fixed value of k, the n functions p D,ek are components of a gradient vector, and for this it is necessary and sufficient that (5)
Di(p D,ek) - D,(p Diek) = 0
(§ 172, theorem) ;
or, since Dip = 0,
p (DiD, - D,Di)ek = 0. These equations must hold independently of the initial conditions imposed upon p and are therefore equivalent to
(DiD; - D,Di)ek = 0.
(6)
Since Di transforms like a covariant vector (167.3), the operator
Di, = DiD; - D,Di = a bDaDb,
(7)
transforms like a covariant dyadic; for
DiD' -
(at.,
a2xb
axb
ax,
Da) \ax1 Dbl
Di, _ DiD, - D,Di =
axa axb
axiaxI Db + azs a
D0Db,
ax, axb
-. Dabat, ax'
Moreover, axa axb
Di;ek
\1
axa axb axc
(axc
Dabec,
Caxi at-, Dab/ \axk ec/
at - 4921 axk
since D,,Dkxc = 0; hence eh
axa axb axc axh
DiJek = - -axk at t at,
axd
ed Dabec
§ 175
PARALLEL DISPLACEMENT
379
This equation shows that e' Di;ek is an absolute mixed tensor of valence four, say (8)
Rijk-h = eh Di;ek.
The components of this curvature tensor R are thus the coefficients in the equation, (9)
Rijkheh;
Dijek =
the condition (6), necessary for flat space, now assumes the tensor form, (10) h Rtijk = 0.
If it holds in one coordinate system, it holds in all. When the space is flat, we can choose n linearly independent vectors ai at a point P and, by giving them parallel displacements to neighboring points obtain a set of constant base vectors ei = ai in a region about P. For these base vectors, we have
gi; = ai a; = const,
I'k = 0
(166.6),
and the corresponding coordinate system x is called Cartesian. To determine the Cartesian coordinates y = x corresponding to the base vectors ai, we have ayr
ek =axk a,., aek
c12yr
OX,
ax' axk
ar,
and, on dot-multiplying (11), member for member, by ay8
- er = a8 , axr
ays (12)
r
ax, rA
a2y3
t
axi axk
From (11), we have the necessary conditions for the integrability of equations (12): D;ek = Dke;,
Di,jek = 0;
t This equation also follows from (163.9) with Tj = 0.
TENSOR, ANALYSIS
380
that is,
r r rjk = rkj,
Rijk
h
§ 176
= 0.
These conditions are also sufficient for the complete integrability of equations (12).$ When these conditions are fulfilled, (12) ad, x") which with ay/axi take on arbimits solutions y(x', x2, trary values at a given point. If we place the origin of the Car-
tesian coordinate system at the point x0, we have y = 0 when x = x0; and, if we choose n linearly independent sets of initial values,
axi - pi,
a
= p2,
... ,
ax"
(i = 1, 2, ... , n),
= pn
when x = x0, we obtain n corresponding solutions y'(x) whose Jacobian I ay/ax I = det p 0 when x = x0. In the region about x = x0 for which I ay/ax 19 0, the n functions yj(x) thus obtained define a Cartesian coordinate system. In brief, we have the important THEOREM. A necessary and sufficient condition that a Riemann space, with symmetric afne connection, be flat is that its curvature tensor vanish identically.
We may readily compute the components Rijkh from (8):
*.k'; eh
(DiDjek - D;Diek)
= eh {Di(r;ker) - Dj(riker)I eh
{(Dirjk)er
- (Djrik)er} +
and, on putting eh e, (13)
eh
{rjkrires - rikrjres},
eh e8 = 887 we have
Ri;kh = D2-r' - Djr k + r; r r -
176. Curvature Tensor. From the defining equation for the curvature tensor, Dijek = ji rer, we obtain the covariant components, (1)
eh Dijek = Rijk r9rh = Rijkh
$ Cf. Veblen, op. cit., p. 70-1.
CURVATURE TENSOR
§ 176
:381
We now can express the covariant curvature tensor Rijkh in terms of the gammas :
Rijkh = eh (DIDjek - D;Diek) (166.4) = eh' {Di(rjk,re') - Dj(rik,rer)} Sh{Dirjk,r - Djrik,rI - 0h{ rjk,rria - rik,rrja};
(2)
Rijkh = Dirjk,h - Djrik,h - rjk,rrik + rik,rrjh
Since
rjk,rrih = grerikrih = rjkrih,s, we also may write (2) in the form: (3)
Rijkh = Dirjk,h - Djrik,h - rikrih,r + rr,krjh,r.
Rijkh has the following types of symmetry:
Rijkh + Rijkh = 0; Rijkh + Rijhk = 0; Rijkh + Rjkkh + Rkiih = 0;
Rijkh - Rkhij = 0. Proofs.
(I) follows from Dij = -Dji. Since the scalars ek
eh
= gkh have continuous second derivatives (§ 169),
gives (II). Thus Rijkh is alternating in its first two and last two indices.
Since the affine connection is symmetric (§ 169), we have (4)
Die; = r er = r;ier = Djei.
Hence, on adding the identities,
Dk(D1ej - Djei) = 0,
Di(Djek - Dkej) = 0,
Dj(Dkei - Diek) = 0, we obtain
Dijek + Djkei + Dkjej = 0, which, on multiplication by e h- , gives (III).
TENSOR ANALYSIS
382
§ 176
Now (IV) is a consequence of (I), (II), and (III). From (III), we have
Rijkh + Rikih + Rkiih = 0, Rjkhi + Rkhji + Rhjki = 0,
-Rkhij - Rhhkj - Rikhj = 0, - Rhijk - Rijkk - Rjhik = 0. When we add these equations and make use of (I) and (II), only the underlined terms survive, namely, 2Rijkh - 2Rkhij, and we obtain (IV). The symmetry relations (I) to (IV) reduce the number of independent components of Rijkh to i22n2(n2 - 1). Proof. Rijkh = 0 when i = j or k = h (I, II); hence, in general, the number of non-zero components is (nC2)2 = n2(n - 1)2/4. But, when ij ; kh, these are paired, because Rijkh = Rkhij (III); hence, if we add the number nC2 of unpaired components Rjjij to the preceding total, we obtain double the number of components with distinct values. The number of components with distinct values thus is reduced to 1n2(n
4
n(n - 1)
1)2
+
2
1=*n(n-1)(n2-n+2).
These are still further reduced by the ,,C4 relations (III); for i, j, k, h must all be different in order to get a new relation. If, for example, i = j, Riikh + Rikih + Rkiih = Rikih + Rkiih = 0
is already included in (I). The number of linearly independent components is therefore
n(n - 1)(n2 - n + 2) -
n(n - 1)(n - 2)(n - 3) 24
=
n2(n2 - 1).
For n = 2, 3, 4 this gives 1, 6, 20 linearly independent components Rijkh, respectively. Example 1. When n = 2, the contracted product, eijEkhR,,kh
= 4R1212,
CURVATURE TENSOR
§ 176
383
is a relative scalar of weight 2 (§ 169); hence R1212/g is an absolute scalar. Now, from (2), R1212 = D1r21,2 - D2r11,2 - r21,.ri2 + rll,rr22
(5)
We shall compute this expression when the base vectors are orthogonal: g12 = 0.
From (166.6),
rll,1 = !Digll, r12,1
=
r11,2 = -iD2g11, r12,2 = 2D1g22,
ZD2911,
r22,, = -2D1922,
r22,2 = 3Dzg22.
Moreover, since g = 911922, 911 = 1/911, g2 = 1/g22; hence
r21,rr12 = r21,,ri2 + r21,2r12 = r21,,r,2,1 911 + r21,2r,2,2 g22
-
(D2g11)2
+
(D1922 )2
4911
4922
r11,rr22 = rll,lr22 + r11,2r22 = r11.1r22,1 911 + rll,2r22,2 g22 (Dig,,) (D1922) 491,
(D2911) (D2g22)
4922
Substituting these results in (5) gives 81212 - DiD1922 + 2 D2D2911
Dlgll + D1922 D19zz + Dw11 + D2922 D2g11 l j
St l\ 911
922
922
911
J
2
\
The absolute scalar, (6)
K
_81212 9
(
1
219
Dl
1
(.\/g-
D19zz +
J
\
D2
D2 911 J
,
is precisely the total curvature of the surface whose fundamental differential form is ds2
= gll dx1 dx1 + 922 dx2 dx2;
for, if we put x1 = u, x2 = v, 911 = E, g22 = G, H =, (6) agrees with (139.6).
Example 2. We may contract the tensor, (7)
R41k
in essentially two different ways.
h
= ghrRijkr,
TENSOR ANALYSIS
384
§ 177
With h = k, we have (8)
Rt1x
k
= gkrRiikr = 0,
since gkr and Rijkr are, respectively, symmetric and antisymmetric in k, r. From (175.13), we see that (8) is equivalent to the identity:
Dirjr = Djr;,.
(9)
With h = i, we obtain the Ricci tensor, (10)
Rik = R0 ' = 9ihRijkhi this is a symmetric dyadic; for, from (IV), (11)
Rkj = 9''`Rikih = 9'Rihik = ghiRhiki = Rik.
The first scalar of this dyadic, R = gikRik = gikgihRijkh,
(12)
is an absolute invariant. In the case n = 2, g12 = 0 considered in ex. 1, we have (13)
R = 2g11g22R2112
= -2R1212/9 = 2K.
177. Identities of Ricci and Bianchi. Ricci Identity. In analogy with
Dij = D,Dj - DjDj, we also write Vij = ViVj - V, Vi. With this notation, (167.21) becomes
VijT = DijT
(1)
for any tensor, absolute or relative, with its base vectors. On the left Vii acts only on scalars (cf. § 167, ex.); on the right Dij acts only on base vectors, for Dij
Dijek = Rij'heh Dijek = - Rijh
(175.9), keh.
Equation (3) is proved as follows: Dijek = Dijgkrer = gkrR,;r.aea = ha _ -g gkrRijhrea = R;jhkeh .
gkryh8Rijrhee
For the vector vkek, we have Dijvkek = vkDijek = vkRijk aea = vaRija kek, (4)
Vijvk
=
v8RtijB'k.
EUCLIDEAN GEOMETRY
§ 178
385
Similarly, for vkek, Vijvkek
= vkDijek = -vkRijs
ke3
= -v3Rijk'ek,
Vijvk = -v8Rijk8.
(5)
The general Ricci identity now is readily established. Thus, if the components of T are Thk:::;,a, for every upper index * in T, DijTh::...m contains a term,
Th..' ..mRij,;
and, for every lower index * in T, vijTh::;:: ,,
contains a term,
,
Bianchi Identity. At the origin of geodesic coordinates (§ 163), we have, from (174.13),
DiRjkhm = DiRjkhm = DiDjrkh - DiDkrjh.
By permuting ijk cyclically in this equation, we obtain two others.
On adding the three equations, we find that the right members cancel; we thus obtain the Bianchi identity: (6)
DiRjkhm + OjRkihm + OkRijh-m - 0.
Since this tensor equation holds at any point, it is also true for general coordinates.
178. Euclidean Geometry. When the space is Rat, we can determine a Cartesian coordinate system xi (§ 175). The corresponding metric tensor gij has then constant components.. If in addition the metric form gijxixj is positive definite, the space and its geometry are termed Euclidean. We can then always make a real linear transformation to coordinates yi for which gij = b, the Kronecker delta, and the metric form becomes a sum of squares (§ 150) :
b yiyj = y'y' + y2y2 +.... + yny+l.
The corresponding base vectors ai then form an orthogonal set (ai - aj = b), and the coordinates yi are said to form an orthogonal system. Let xi denote a Cartesian coordinate system with the base vectors ai. If we transform to another Cartesian system xi with the base vectors Ai, we have
-
8xt
aj =
ai
TENSOR ANALYSIS
386
§ 178
and, since both a, and ai are constant, aa;
a2xi
a2k
aXk a2
ai = 0,
exi
82k a2'
= 0.
On integrating this equation twice, we have
xi=c;2'+Ci,
(1)
where c and Ci are sets of constants. The transformation between any two Cartesian coordinate systems is therefore linear with constant coefficients. As in the general transformations of § 157, we require that the Jacobian,
ax' det - = det c 54 0.
(2)
a2'
The transformations (1) with non-vanishing determinant form a group-the affine group. When the Cartesian coordinate systems y and y are both orthogonal, the law of transformation, aya
Ji- aya = gi; = ayi ayb - gab, becomesaya
S;.
09)
If we multiply this equation by ayi/ayk and sum with respect to i, we obtain (3)
ayk
ay' = ayk
ay;
Since orthogonal coordinate systems are also Cartesian, the transformation between two orthogonal systems has the form: (4)
yk = c;4j' + Ck.
The inverse transformation is (5)
yj = Yk(yk - Ck),
where Yk is the reduced cofactor of c; in det c;. Equation (3) thus becomes (6)
This is precisely the condition that the matrix c; be orthogonal (§ 149, theorem); thus a transformation between orthogonal coordi-
EUCLIDEAN GEOMETRY
§ 178
387
nate systems is orthogonal (has an orthogonal matrix). Condition (6), which characterizes an orthogonal matrix, also implies that c'iCj = cic = Sj,
(7)
in view of the relations (146.8) between reduced cofactors. Conditions (7), in turn, imply (6); either (6) or (7) is a necessary and sufficient condition that the matrix c; be orthogonal. Two orthogonal transformations,
yz=ay'+Ay,
yj =bkyk+Bj,
have an orthogonal resultant, yi
i
i
s
i, j;
we have, for example, s a r s a r a r r cic; = (aebi)(atbj) = Stbib; = bib; = S;. t
t
i
Moreover (5), the inverse of (4), and the identity transformation yi = S;yj are orthogonal. Consequently, the orthogonal transformations form a group.
In view of (3), the equations, ayj U' =ayk tlk
ayk
Vi = ayj vk,
show that covariant and contravariant vectors transform alike under orthogonal transformations. Within the orthogonal group, the distinction between covariance and contravariance vanishes, and
tensor components may be written indifferently with upper or lower indices. For example, we may write Sij or Sij for the Kronecker delta.
The orthogonal group of transformations admits as a subgroup those transformations for which (8)
det c =
ayy a
= 1.
If we regard (4) as a transformation between the points y and y in the same Cartesian coordinate system, the transformation is called a displacement or rigid motion. In fact in 3-space the trans-
formation (5) may be written (9)
s=r-a,
TENSOR ANALYSIS
388
§ 179
where the dyadic CF in non ion form is given by the matrix (ck). Since this matrix is orthogonal and its determinant is 1, the transformation (9) is a translation followed by a rotation (§ 75, theorem), in brief, a displacement. The subgroup characterized by det c = 1 is therefore called the displacement group. In view of (8), we see that, within the displacement group, the distinction between absolute and relative tensors also vanishes.
A displacement which leaves the origin invariant is called a Thus the transformation y' = c;y' is a rotation if the matrix (c) is orthogonal and has the determinant 1. Rotations rotation.
form a subgroup of the displacement group. 179. Surface Geometry in Tensor Notation. The equations,
x' = xi(ul) u2)
(i = 1, 2, 3),
define a surface embedded in Euclidean 3-space. The space coordinates xi are rectangular Cartesian and are designated by italic indices (range 1, 2, 3); the surface coordinates u" are curvilinear and are designated by Greek indices (range 1, 2). If we write
al = i, a2 = j, a3 = k, the position vector to the surface is r = xiai. The metric tensor in space is then Si; = ai - a;.
(1)
If we limit the coordinate transformations x - x to the displacement group (§ 178) the distinction between covariance and contravariance as well as the distinction between absolute and relative tensors does not exist in 3-space. First fundamental form.
The base vectors on the surface are
ar axi ar a"=-=--=x.ai, axi au" au"
(2)
where x" = axi/au". Since
e"-e#=
=x'xp'si; =xaxs,
the metric tensor for the surface is (3)
gap = e., ep = x;.xP.
This tensor defines the first fundamental form on the surface: ds2 = g"g du"du#.
SURFACE GEOMETRY IN TENSOR NOTATION
§ 179
389
Note that x, is a covariant surface vector; for if we make the transformation u -+ u, we have axe
axi auo
aic"
aufl au"
If a vector v has the "surface components" v" (a = 1, 2),
v = vae« = and vi = v"xa (i = 1, 2, 3) are the "space components" of v. Unit surface normal. The space vector,
N = el x e2 = Eijkaixix2, has the components Ni = Eijk xix2; moreover N2 = EijkEist xix2 x1x2 = sit xix2 XIX2 = xix2 xix2 - xix2 xix2 = 911922 - g12
Thus N2 = det ga# = g t; hence the unit normal n to the surface
has the components Ni/\:
ni = ni =
(4)
eijk xix2.
Second fundamental form. On the surface with metric tensor gang, the Christoffel symbols are given by (cf. § 166)
is = ZgX7(D«goti + DRgy« - D7g«#)
(5)
Covariant derivatives are then computed from the formulas of § 167.
In particular we have, from (167.13),
(6)
Vaxkp
= DaDftxk -
VPxQ.
Since the covariant derivative of the metric tensor gij is zero (167.19), (7)
Vagoti =
V. xksxy
= 0.
If the product rule (§ 168, 1) is used, this equation and its cyclical permutations give (7a)
xyV 4 + x0Vaxy = 0,
(7b)
xgVVxy + xyVflxx = 0,
(7c)
xkovyxa + xxVyx1
= 0.
t This also follows from the expansion of (el x e2)
(e1 x e2) given in (20.1).
390
TENSOR, ANALYSIS
§ 179
Subtract the third equation from the sum of the first two; then, in view of (6), we obtain xyDax3 = 0.
This equation states that the space vector Vax is perpendicular to both el and e2 (whose space components are 4, x2); that is Vax, is a multiple of the unit normal nk, say oax3 = ha#nk.
(8)
The symmetric covariant tensor, (9)
k
ha0 = nkVaXQ = hha,
defines the second fundamental form on the surface: ha#duadus.
Derivative formulas. From (6) and (8) we have (10)
Daxkg
=F
xk\ + haonk;
these equations are the derivative formulas of Gauss. If we adjoin
the (constant) base vector ak to each term they become (10)' Dae# = raaea + hafln. f On multiplying (10) by xry and summing on k, we have also (11)
'xk Da
X = raggay = rap,7.
Since nk is a space vector with no components along el or e2, Vank = Dank. On differentiating nknk = 1, we obtain (12)
nkVank
= nkDank = 0;
hence Dank (1 nk) is a tangential surface vector. Similarly, from nkx' = 0, we obtain nkvaxp
i
xpvank
= 0;
or, in view of (9) and the symmetry of hao, (13)
hp = -xxVank = -xaV nk.
Hence hapxa 90XhaflxX
= -9aavpnk, o Vpnk = - Yank,
-
Dank = Dank -ham xa. (14) t Note that the term hapn in this equation is in apparent disagreement with
(163.1); this is due to the fact that our 2-dimensional geometry is not intrinsic but that of a 2-space embedded in a 3-space.
SURFACE GEOMETRY IN TENSOR NOTATION
§ 179
391
These equations are the derivative formulas of Weingarten. If we adjoin the (constant) base vector ak to each term, they become
Dan =
(14)'
Equations of Codazzi and Gauss. Osxy VaV#xky
=
From (8) we have
hsynk,
= (Vahs.y)nk + hsyVank,
= (Vahsy)nk - hsyhaIxk
in view of (14). Now form V$Vaxk, and subtract it from the last we thus equation; writing Vas for the operator VaVs obtain VasXY = (Vahsy - V$hay)nk + (hayha"' - hsyha")xx. If we replace Vasx1' by the value, (15)
Vasxak
= -Raay
X
xkx,
given by Ricci's Identity (177.5) and adjoin ak to each term, (15) becomes (16)
-Rasyx ex = (Vahsy - oshay)n +
(hayhft"
- hsyhal')ex.
This vector equation is equivalent to the scalar equations : (17) (18)
0 = Vahsy - vshay, ROT" = hsyha)` -
Equations (17) are the Equations of Codazzi. When a = /3 the right member vanishes identically; and an interchange of a and 0 repeats the same equation. Hence there are but two independent Codazzi equations; these may be written with a = 1, S = 2, y = 1, 2; (19)
V1h21 - V2h11 = 0,
V1h22 - V2h12 = 0.
On multiplying (18) by gas and summing on X, we obtain the covariant curvature tensor: (20)
Rasys = hsyhas - hayhss.
From the symmetry has = hsa we may verify at once the four symmetry relations of Rasys : Rasys = - Rsays,
Rasys = - Rassy,
Rasys + Rsyas + Ryass = 0,
Rasys = R,ysas.
TENSOR ANALYSIS
392
§180
As Greek indices range over 1, 2, these relations show that there is but one independent equation (20). This Equation of Gauss may be taken as (21)
- 81212 = h11h22 -
22
= h,
where h = det han.
The contracted product,
Total and mean curvature.
and g = 811922 - 912
E"S Erya R,,,#,3 = 4 R1212
are both relative scalars of weight 2; hence,
K = -R1212/9 = h/9 is an absolute scalar, namely the total curvature of the surface (22)
(§ 176, Ex. 1).
The mean curvature of the surface is defined as the absolute scalar J = g"0hg. Since g"R is the cofactor of gag in det ga#, (23)
912 = 921
911 = 922/9, 922 = 911/9,
= -912/9;
hence (24)
J =
9a,6haQ
= (922h11 - 2g12h12 + 911h22)/9
180. Summary: Tensor Analysis. Under general transformations,
.ti ={i(x1, J
x2, . . ,
xn)
ax
001
,
ax
the component of a relative tensor of weight N transforms accord-
ing to the pattern: Ti'.k =
ax
N axi ax' axc
ax
axb axk
axa
b
When N = 0, the tensor is absolute. For brevity, components of tensors often are called tensors.
The number of indices on a tensor component is called its In n-space a tensor of valence m has nm components. A tensor of valence zero is a scalar. A scalar p(x) has one component in each coordinate system given by valence.
ax N ,7p (.t)
=
I
ax
'P(x) I
SUMMARY: TENSOR ANALYSIS
§ 180
393
A tensor of valence one is a vector. The differentials of the coordinates and the gradients of an absolute scalar are the prototypes of absolute contravariant and covariant vectors : axz
dx' _ - dzr, axr
-- _ a.-axraxi
axraxi
Measurement is introduced into Riemannian geometry by the non-singular quadratic form: ds2
= gi; dxi dx.'
(gi; = gji,
g = det gi; F6 0).
The character of the geometry depends upon the choice of the n(n + 1)/2 functions gij(x) of the coordinates. The relations ei e; = gij determine the lengths of the base vectors ei and the angles between them. Any vector in n-space at the point x is linearly dependent upon the n base vectors ei at this point. The reciprocal base vectors e' are defined by e' . e; = S . If g = det gij and g'' is the reduced cofactor of gij in g, (1}
(2)
ei = girer, ei e; = gi;,
ei = girer;
ei e' = S,
e' e = gi'.
In passing from coordinates x to x, the transformation of the base vectors is prescribed by (3)
ei=-axr er, axi
axi ei=-er. axr
These equations show that gjj, M, g2' are components of an absolute dyadic, the metric tensor G = gi,eiel = e;e'. Use of equations (1) permits indices to be raised or lowered on tensor components:
T.i. = girT*r. 9irT.r., If T is a tensor of weight N (say T = T?'.ke,ejek), T =
ax
IN
- T. at l
Addition of tensor components of the same valence, weight, and type produces a component of this same character.
TENSOR ANALYSIS
394
§180
Multiplication of tensor components of valence ml, m2i of weight
N1, N2, and of arbitrary type produces a component of valence m1 + m2 and of weight N1 + N2. Contraction of a tensor of valence m > 1 results on forming the dot product of any two of its base vectors. If the vectors in question are ei e' = s, the components of the contracted tensor are obtained from the original components by putting i = j and performing the implied summation. The components of the affine connection rk are functions of the coordinates defined by
Die; = I'er
(4)
(rk = ek Diei).
Then also (5)
Die' = - rrer;
(6)
DiG = 0
(G = eye');
Dig = 2grir
(7)
By definition,
then
ri,,k = gkrr,
rk. = gkrri;,r.
When r k is symmetric in ij, 13
ri7,k = 2(Digik + Digki - Dkgi.l) The gradient of a tensor of weight N and valence m is defined as the tensor, N
(9)
N
VT = g2ehDh(g 2T)
of weight N and valence m + 1. When T is absolute, VT = ehDhT.
The components of VT, denoted by prefixing Vh to the components of T, are called covariant derivatives. For any tensor we have (10)
VhT : = DhT; - NrhrT:
+ (T -rhr + ...) - (TK:rr + the first parenthesis contains one term for every upper index, the second contains one term for every lower index. The metric tensor G is absolute and VG = 0, from (6); hence Vhgii = 0,
Vhba = 0,
Vhgii = 0.
SUMMARY: TENSOR ANALYSIS
§ 180
395
Since g is a relative scalar of weight 2, Vg = 0 from (9), and
Ohg=0. The covariant derivatives of the epsilons and Kronecker deltas are zero. The divergence of a tensor T is defined as the gradient VT con-
tracted on the first and last indices: thus VhTij...kh; = (divT)ij...k
when T is an m-vector of weight 1, we may replace Oh by Dh.
The curl of a covariant tensor fined as the (m + 1)-vector:
of valence m < n is de-
1 abc...d (curl T)hii...k = ' Shij...kVaTbc ...d. m.
When T is absolute, we may replace Va by Da.
A Riemannian n-space xi with the metric tensor gij and base vectors ei has the associated curvature tensor, RiJk h
-eh (DiDj - DjDi)ek.
Its covariant components, Rijkh = eh (DZDj - D;Di)ek,
have four types of symmetry: Rijkh + Rjikh = 0,
Rijkh + Riihi = 0,
- Rkhij = 0. These relations reduce the number of linearly independent components Rijkh to n2(n2 - 1)/12. When n = 2, there is but one independent component, say R1212; and the absolute scalar -R1212/g = K, the total curvature of a surface whose fundaRijkh + Rikkh + Rki h = 0,
Rijkh
mental form is gij dxi dxj. A Cartesian coordinate system yi is one in which the components
gij of the metric tensor are constants; then all t = 0, and the base vectors ai remain invariable in space (aaj/aye = 0). The Riemannian space xi with metric tensor gjj is said to be flat if it is possible to transform to a Cartesian coordinate system. When the affine connection is symmetric, a necessary and sufficient condition for a flat space is that the curvature tensor vanish. If the space is flat and the metric form gijxixj is positive definite, the space and its geometry are termed Euclidean. We then can
TENSOR ANALYSIS
396
make a real linear transformation to orthogonal coordinates yi for which gij = S; the metric form then becomes a sum of squares. A transformation y i = c)y' + C' between orthogonal coordinate systems is characterized by the relations: ayi ay' or c; = y= (orthogonal matrix); ay' ay, then
and det c; _ ± 1.
crx; = cTcT = S,
Within this orthogonal group of transformations, the distinction between covariance and contravariance vanishes. Orthogonal transformations for which det c) = 1 form the displacement subgroup in
which the distinction between absolute and relative tensors vanishes.
When yi and y' are regarded as points in the same coordinate system, the transformation yi = c)y' is a rotation when its matrix is orthogonal and its determinant +1. PROBLEMS Summation Convention. Index range is 1, 2, 3 unless otherwise stated. 1. Prove the following: ETjk = 2! Si;
(a)
Etjk
(b)
Eijk Srst = 3! Erst;
Si = 3,
(c)
S
=3
Stjk = 3!
2,
[§ 161.1
2. Show that u1
u2
143
V1
v2
v3
W1
1U2
= Eijk UiVjWk.
,W3
3. Show that the two-rowed determinant formed by columns i, j of the matrix,
xi x2 ... xn Y
is
y2 ... yn)
ST8 xrys.
4. For the dyadic vi, in (77.1) show that the scalar invariants are 'P1 = 'Pit,
'P2 = fit,
'P3 = 1 Eijk Erat 'Pzr'Pjs'Pkt; 3!
and that the vector invariant has the components z fi k'Pik. 5. Show that the general solution of the equations: aix' = 0,
bix' = 0
is
x' = X
ajbk.
PROBLEMS
397
6. Show that the cofactor Arl of the element a= in a = det a
is
Atr = 1 Eiik Erat
2
[
' At ali -=2l E rat Eiik aPaa : f k
= 'll =
E
rat
E pat
a = b r a.]
7. If yl, y2, y3 are functions of x1, x2, x3, det (ay'/axi), is called their Java bian and written a(yl, y2, y3)
,
or, more briefly,
a(xl, x2, x3)
ay
x
If z1, z2, z3 are functions of yl, y2, y3, prove that Oz
az
ax
ay
ay ax
If, in particular, the functions z` (yl,
y2, y3) = x`,
show that ay
ax
ax
ay
= 1.
8. Prove that
a(y', y', yk) = Stik a(yl, Y2, y3) x3) . rat a(xl' x,' xt) a( xr' x2,
9. Prove that the t
ay
Cofactor of ay. in
isax' -. I ay ay' ax
I
Ox
axi
ay i = = a yk ax' a yk
ay' axi
r
t ak
10. If the elements of a = det aii are functions of x1, x2, x3, prove that as _ aa,t Aat axr
axr
11. Prove that a
ay
axr
ax
axi axr axi ayi 02y'
ay ax
[Apply Probs. 9 and 10.] 12. Prove that the bordered determinant: vl
V2
V3
0
all
a12
an
ul
a21
a22
a23
U2
a31
a32
a33
u3
= uiviAii.
This determinant is formed by bordering det aii with the vectors vi and ui. If aii is symmetric it also equals viuiA `i.
TENSOR ANALYSIS
398
13. When the index range is from 1 to n show that the bordered determinant (written compactly), V,
0
aii ui
_
= 1, 2,
(i,
1)n+lusy7iAi7
14. If Aij is the cofactor of aij in a = det aii, and A = det Aii, show that aA = a3 (and hence A = a2 when a 96 0). State the corresponding theorem when the index range is from 1 to n. 15. Show that the n-rowed determinant,
a = det apq = ij"'k aii a2i ... ank =
j8 ...
E
n!
ii...k
ari aaj ... aek,
and that the cofactor of ari is
Ari =
1 (n - 1)!
Erd... Eij...k a8.
. .
. atk
'
16. If det aii is symmetric (aii = aii), show that det Aii is also (A ii = Aii). 17. If the n-rowed determinant a = det aii is antisymmetric (aii = -aii), show that a = 0 when n is odd. 18. Show that the linear equations,
aij xi = 0,
(i, j = 1, 2,
, n),
for which det aij = 0 and not all the cofactors Aij vanish has a non-zero solution of the form xi = Aki for some value of k. 19. If aij and gii are symmetric dyadics and gijx`xi is a positive definite quadratic form (§ 150), prove that the roots of the cubic equation, (1)
(i, j = 1, 2, ...
det (aii - X gii) = 0,
,
n),
are all real. [The system of n linear equations,
(aij - X gii)zj = 0, must have a solution zi = xi + iyi other than (0, 0, is a root of (1). Hence
[aii - (a +
, 0) when X = a + 1$
l(x1 + iy') = 0; i$)gii11
and, on equating real and imaginary parts to zero, we have (2) (3)
aii xi - a gii x' +,6 gii y' = 0, aij yi - a gii yj - 0 gii xi = 0.
On multiplying (2) by yi and (3) by xi, and subtracting, we find that
0(gii xx' + gii yy') = 0, and hence 0 = 0; for if gii xx' = gij y'y' = 0, xi = yi = 0 and consequently zi = 0, contrary to hypothesis.]
PROBLEMS
399
Tensor Character. 20. Prove the theorem: If aii, bii, cj are absolute dyadics,
a = det aii, b = det bii, c = det cj are scalars of weight 2, -2, 0; the cofactors Aii, Bii, Ct in these determinants are dyadics of weight 2, -2, 0; and the reduced cofactors A"/a, Bii/b, Cs/c are all absolute dyadics. [Cf. Prob. 15.1 21. Show that, if u, v, w are absolute vectors, ui
u2
u3
Vi
V2
V3
W1
W2
W3
and
u1
U2
u3
vl
V2
V3
W1
W2
w3
are relative scalars of weight 1 and -1. 22. If aii and b" are absolute antisymmetric dyadics in 3-space, show that (a) a23, a31, a12 are components of a contravariant vector ui of weight 1; b23, b31, b12 are components of a covariant vector vi of weight -1:
.(b)
[ux. = 1 Eijk aik; Z
Vi = 2 Eiik bik].
23. If T1. ..,,, Sz1"'h are absolute n-vectors (§ 170) in n-space, prove that T12...n and S12 ' are relative scalars of weight 1 and -1. ti ...h
[n! T12... , = e
T"...n.]
24. If vi, ui, aii are absolute tensors, prove that the bordered determinant in Prob. 13 is a relative scalar of weight 2.
25. Prove the "quotient law": If the set of functions viTi:: a are tensor components of the type indicated by the indices for all absolute vectors vi, then Tz:: c is a tensor of the same weight. 26. If uk is a covariant vector, prove that the total differential equation uk dxk = 0 has the same form in all coordinate systems. This equation is said to be integrable when there exists a function X such that Auk = aw/axk; for the equation is then equivalent to dp = 0 and _ const is an integral. Show that eilk uiDjuk = 0
is a necessary condition for the integrability of ilk dxk = 0 and that the form of this condition is the same in all coordinate systems. 27. Let u, v be quantities (scalars, base vectors, tensors) whose "product" uv is distributive with respect to addition but not necessarily commutative.
Show that the differential operator Dii = DiDi - DiDi has the property, Dii(uv) = (Disu)v + u(Di v).
28. Show by direct calculation that the operator Dii transforms like a covariant dyadic:
Dii =
ax, axb
. - Dab.
a2i a2i
TENSOR ANALYSIS
400
29. Deduce (177.3) from (177.2) by applying the operator Dii to eh = Bt-
ek
[Cf. Prob. 27.1
30. If T is a tensor (complete with base vectors) of weight N and valence in, prove that eiei DiiT is a tensor of weight N and valence m + 2. [Since E-NT is an absolute tensor, R5'-NT = E-NT; hence, from Prob. 28, eie_i
N_
Di (R-,T) =
.
e
,.-axa
-axb
eaeb
N
axi ax'
N
Thus (Prob. 26), Dii(E-NT)
eie1
= E-N eiei DiiT
is an absolute tensor, which multiplied by EN yields the relative tensor eiei DiiT of weight N.] 31. In Riemannian n-space with the metric tensor gii show that the gradient of an absolute scalar p(xl, . , xn) is given by
Vv = eiDi,p = eigii Div = 1 (Div)e7Gi1, 9
or, in view of Prob. 13, by the bordered n-rowed determinant vw =
(1)
(-1)n+i ei 19ii
9
0 Disc
Show that the scalar product of (1) by VV = elDkv gives (-1)n+1 Disc 0
(2)
9
Div
19ii
I
32. In Riemannian n-space with the metric tensor gii show that the divergence of an absolute vector v is given by div v
Di(9
-
D' (-!'L G t') V9-
or by the bordered n-rowed determinant, (1)
dlv v =
(-1)n+l Di 1/9
0
9 1
9ii
Vi
where the determinant is to be expanded according to the elements of the first row and the operators Di applied to differentiate their cofactors. In particular, if v = Vv, vi = Dip, we obtain the Laplacian, (_1)n+1
Di 0 1
V9-
9ii
9
Di'v
33. The equations of a surface in 3-space are xi = xi(ul, u2). Show that
za =
au t au
(i = 1, 2, 3; « = 1, 2)
PROBLEMS
401
is a contravariant vector in 3-space (1 = 1, 2, 3; a fixed) and a covariant vector in the 2-space (a = 1, 2; i fixed) formed by the surface. Prove that
i x9k pi = 21eao ziik xa
is an absolute covariant space vector normal to the surface. Write out its three components in full. 34. In special relativity it is customary to use the independent variables,
xl=x, x2=y, x3=Z, x4=ct where c is the speed of light. The interval ds between two events in space time is defined by ds2 = CO dt2 - dx2 - dye - dz2 = gii dxi dxi.
(1)
Hence the metric tensor is given by 9i1 = g22 = 933 = -1,
(2)
941 = 1,
gii = 0 (i
j).
If v is the speed of a particle relative to a frame j5, we have (ds/dt)2 = c2 - v2
from (1), and hence ds
(3)
/1--
c
v' \\
where y=(
dt=y
)
In a rest frame ao attached to the particle a clock registers the proper time r. Putting v = 0, t = r in (3) we have ds (4)
_
dr =
c
and
dt
dr = y.
Corresponding to the position vector (x, y, z) in space we now have the event vector xi = (x, y, z, ct) in space time. The velocity and acceleration fourvectors are now defined as (5) U =dxi- , W, =duidr
dr
Denoting t derivatives with dots, prove that Eli _ _y (1, y, Z, c),
w'
ui = 'Y(-x, -y, -Z, c);
'Y(7x + -Yx, 'Yy + 'Yy, 'YZ + ?'z, -ic);
utui = c2; wtui = 0.
Thus the velocity four-vector has the constant magnitude c and is always perpendicular to acceleration four-vector. Show also that in the rest frame ao (t = r), uo = (0, 0, 0, c),
uo = (x, y, z, 0).
Covariant Differentiation. 35. Prove the product rule (§ 168, 1) for covariant differentiation without resort to geodesic coordinates.
TENSOR ANALYSIS
402
36. If Ti iis an absolute bivector, show that div
1
T = vi 7'ij = - Dj(V'g Tij).
37. If u and v are absolute vectors, prove that Dh(gij UV) = ui. Vhv2 + vi Vhuh.
If I u I is the length of u, prove that
DI I-
Ui Vh U
38. Prove that Vijvk + vjkvi + Vkivj = 0.
[Cf. (177.5).
39. Prove that Rijkh + RJkih T RkjJh --0-
By contracting this identity with h = j, show that the Ricci tensor Sik (§ 176, ex. 2) is symmetric.
40. In the Bianchi identity (177.6) raise the index h and contract with j = h, k = m to obtain vi Rjk'k + Vi Rki + vk Ri1'k = 0. On introducing the mixed Ricci tensor,
Rij =
R = Rii = Rr4i
(its first scalar), show that this equation implies that the divergence of the tensor R*j - 1SjR is zero. 41. Check the results of § 164, ex. 1, by differentiating r = pR(,p) + zk twice with respect to the time. [Apply (44.2) and (44.3).] 42. The generalized Kronecker delta 6a'.".' d has k subscripts and k superscripts (1 5 k 5 n), each ranging from 1 to n. Its value is defined as follows: If both upper and lower indices consist of the same set of distinct numbers, chosen from 1, 2, , n, the delta is 1 or -1 according to the upper indices from an even or an odd permutation of the lower; in all other cases the delta is zero. Prove that if Som.': d has 2k indices, (n - k)! Sab...d - E
where the epsilons necessarily have n indices. Hence show that the delta is an absolute tensor.
CHAPTER X QUATERNIONS
181. Quaternion Algebra. The problem of extending 3-dimensional vector algebra to include multiplication and division was first solved by Sir William Rolvan Hamilton in 1843. He found that it was necessary to invent an algebra for quadruples of numbers, or quaternions, before a serviceable algebra for number triples, or vectors, was possible. Without attempting to motivate Ham-
ilton's invention, we proceed to a brief account of quaternion algebra. A real quaternion is a quadruple of real numbers written in a definite order. We shall designate quaternions by single letters,
p, q, r; thus q = (d, a, b, c), q' = d', a', b', c'). The fundamental definitions are the following.
Equality: q = q' when and only when d = d', a = a', b = b',
c=c'.
Addition: (1)
q+q' = (d+d',a+a',b+b') c+c').
Multiplication by a Scalar X: Xq = (ad, Xa, Xb, Xc).
(2)
Negative: -q = (-1)q. Subtraction: q - q' = q + (- q'). Hence
q - q' = (d-d',a-a',b-b',c-c'). The zero quaternion (0, 0, 0, 0) is denoted simply by 0. From these definitions it is obvious that, as far as addition, subtraction, and multiplication by scalars are concerned, quaternions obey the rules of ordinary algebra: (3)
p + q = q+p,
(p+q) +r = p+ (q+r);
(4)
Xq = qX,
(Xu)q = X(µq);
(5)
(X + µ)q = Xq +µq,
X(p + q) = Xp + Xq 403
404
QUATERNIONS
§ isi
In order to define the product qq' of two quaternions in a convenient manner, we shall denote the four quaternion units as follows:
1 = (1, 0, 0, 0), i = (0, 1, 0, 0), j = (0, 0, 1, 0),
k = (0, 0, 0, 1).
Then, in view of the preceding definitions, we can write any quaternion in the form:
q = (d, a, b, c) = dl + ai + bj + ck.
(6)
Definition of Multiplication: The quaternion product,
qq' = (dl + ai + bj + ck) (d'l +a'i+b'j+c'k), is obtained by distributing the terms on the right as in ordinary algebra, except that the order of the units must be preserved, and then replacing each product of units by the quantity given in the following multiplication table:
(7)
First factor
1
i
j
k
1
i
j
k
i
-1
k
j k
-k -1 j -i
-j i
-1
Note that i2 = j2 = k2 = -1, and the cyclic symmetry of the equations :
ji = - k, kj = - i, ik = -j.
ij = k, jk = i, ki = j;
With this definition we find that (8)
qq'=dd'-aa'-bb'-cc' +d(a'i+b'j+c'k)+d'(ai+bj+ck) +
Ii
j
k
a
b
c
a'
b'
c'
If we form q'q by interchanging primed and unprimed letters, the first two lines above remain unchanged, but the interchange of rows in the determinant is equivalent to changing its sign; hence
QUATERNION ALGEBRA
§ 181
4U3
q'q = qq' only when the determinant is zero. That q'q and qq' differ
in general was to be expected; for, from the table, ij = k, ji = - k. The table shows that multiplying a unit by 1 leaves it unchanged ;
hence (dl)q = q(dl) = dq, (dl)(d'l) = dd'l; and, from (1), dl + d'1 = (d + d')1. Quaternions of the form dl therefore behave exactly like real scalars and may be identified with them. Henceforth we shall write dl or (d, 0, 0, 0) simply as d; in particular, 1 or (1, 0, 0, 0) is regarded as the real unit 1. We also may identify i, j, k with a dextral set of orthogonal unit vectors. For, if we make the orthogonal transformation, i = c11i + e12j + c13k, c21i + c22j + c23k, I6
= c3ii + c32i +
c33k,
we have the relations cirri, = S,j (149.3) and i2 = - Clr Clr = - 1,
j
k
C21
C22
C23
C31
C32
C33
li
3k =
C2. C3r +
= c11i + c12j + C13k = i,
since the elements of det cii (= 1) are their own cofactors (§ 149).
Thus every quaternion q = d + ai + bj + ck is the sum of a scalar d and vector v = ai + bj + ck. With Hamilton we use the symbols Sq and Vq to denote the scalar and vector parts of q; thus (9)
Sq = d,
Vq = ai + bj + ck,
q = Sq + V q.
The operations S and V are evidently distributive with respect to addition. We are now in position to prove the fundamental THEOREM. Quaternion multiplication is associative and distributive with respect to addition; but the commutative law pq = qp holds only when one factor is a scalar, or the vector parts of both factors are proportional. In symbols:
(pq)r = p(qr);
(p + q)r = pr + qr; p(q + r) = pq + pr, (12) pq = qp only when Vp = 0, or Vq = 0, or Vp = XVq.
QUATERNIONS
406
§ 182
Proof of (10). It will suffice to verify (10) for all possible combinations of the units i, j, k. Since the multiplication table is unchanged under a cyclical permution of i, j, k, we need examine (10) only for those products whose left factor is i; thus we find
Proof of (11). If we form p(q + r) and pq + pr by formal expansion and without use of the multiplication table, the two expressions will agree term for term; hence they will agree also after the table is used. Proof of (12). We have seen that q'q = qq' only when the de. terminant in (8) vanishes. This occurs only in the three cases: a=b=c=0; a'=b'=c'=0; a'a b'b c'c that is, either q or q' must be scalar, or Vq and Vq' must be pro-
-_-_-;
portional. From (8), we have (13)
S(qq') = S(q'q),
for both scalars equal dd' - aa' - bb' - cc'. From (13), we may prove that, the value of the scalar part of a quaternion product is not changed by a cyclical permutation of its factors. We have, for example, S(p . qr) = S(qr . p) = S(q . rp) = S(rp - q), and hence (14)
S(pqr) = S(qrp) = S(ppq).
182. Conjugate and Norm. written Kq, is defined as (1)
The conjugate of a quaternion q,
Kq = Sq - Vq.
CONJUGATE AND NORM
§ 182
407
The conjugate of a sum of quaternions is evidently the sum of their conjugates:
K(q + q') = Kq + Kq'.
(2)
Since the vector parts of q and Kq differ only in sign, q(Kq) _ (Kq) q. This product is known as the norm of q and is written Nq.
If
q=d+ai+bj+ck,
Kq=d-ai-bj-ck,
we have, from (181.8), (3)
Nq = q(Kq) = (Kq)q = d2 + a2 + b2 + c2.
Therefore Nq is a scalar; and Nq = 0 implies that a = b = c = d = 0, that is, q = 0. If Nq = 1, q is called a unit quaternion. If v = ai + bj + ck, we have, from (181.8), i j (4) vv' (aa' + bb' + cc') + a b a'
b'
k c c'
Since changing the sign of the determinant is equivalent to interchanging the second and third rows, K(vv') = v'v. (5) On taking the conjugate of every term in
qq' _ (d+v)(d'+v') = dd' + dv' + d'v + vv', we see that (6)
K(qq') = dd' - dv' - d'v + vv = (d' - v') (d - v) = (Kq') (Kq)
Therefore the conjugate of the product of two quaternions is equal to the product of their conjugates taken in reverse order. Since Kv =
- v, (5) is a special case of (6). We now use this property to compute the norm of a product. From (3), we have
N(pq) = pq - K(pq) = pq . Kq . Kp = p . Nq . Kp = pKp - Nq, since Nq is a scalar and therefore commutes with Kp; hence (7)
N(pq) = Np Nq.
The norm of the product of two quaternions is equal to the product of their norms.
QUATERNIONS
408
§ 182
By mathematical induction we immediately may extend (6) and (7) to products of n quaternion factors: (8) (9)
K(gig2 ... TO = Kgn . Kg.-1 ... Kq1, N(g1g2 ... qn) = Nq1 . Nq2 ... Nqn.'
From (6), we conclude that the product of two quaternions is zero only when one factor is zero. Thus, if pq = 0, Np - Nq = 0; and,
since the norms are scalars, Np = 0 or Nq = 0, whence p = 0 or q = 0. We now can appreciate Hamilton's exceptionally happy choice of a multiplication table for the quaternion units. t For quaternion algebra has a unique place among the algebras of hypernumbers,
x = x1e1 + x2e2 + ... + xnen, linear in the units ei, with coefficients xi in the field of real numbers, and for which the associative law of multiplication holds good. For it can be shown $ that the most general linear associative algebra over the field of reals, in which a product is zero only when one factor is zero, is the algebra of real quaternions.
Quaternions include the real numbers (x, 0, 0, 0) with a single unit 1, and the complex numbers (x, y, 0, 0) with two units 1, i. Both real and complex numbers form a field, that is, a set of numbers in which the sum, difference, product, and quotient (the divisor not being zero) of two numbers of the set must be definite numbers belonging to the set. Moreover, quaternions include vectors (0, x, y, z) in space of three dimensions. But (4) shows that the product of two vectors is not in general a vector, but a quater* This theorem applied to qq' gives Euler's famous identity: (d2 + a2 + b2 + c2)
(d,2 + a,2 + b12 + c,2)
(dd'-as'-bb'-cc')2+(ad'+da'+bc'-cb')2+ (bd' + db' + ca' - ac')2 + (cd' + dc' + ab' - ba')2.
t As to the genesis of quaternions, Hamilton himself has written: "They started into life, or light, full grown, on the 16th of October, 1843, as I was walking with Lady Hamilton to Dublin, and came up to Brougham Bridge, which my boys have since called the Quaternion Bridge. That is to say, I then and there felt the galvanic circuit of thought close; and the sparks which fell from it were the fundamental equations in i, j, k; exactly such as I have used them ever since." t Dickson, Linear Algebras, London, 1914, p. 10.
DIVISION OF QUATERNIONS
§ 183
409
nion. Unlike addition-the sum of two vectors is always a vector
-the multiplication of vectors leads outside of their domain. Vector multiplication is not closed, and, consequently, a pure vector
algebra having all the desirable properties of quaternion algebra is not possible. 183. Division of Quaternions. If q is not zero, Nq is a non-zero scalar; and we may write the defining equation for the norm (182.3) as Kq q = 1, or qq = 1.
We therefore define Kq/Nq as the reciprocal of q and write (1)
q-1
=
Kq Nq
then
;
qq-1
= q-'q = 1.
The last equations show that Nq - Nq-1 = 1, or (2)
Nq-1 =
In order to divide p by q ( (3) (4)
-
Nq
0), we must solve the equation
rq = p or qr = p
for r. This is easily done by multiplying by q-1 on the right in (3), on the left in (4). We thus obtain the two solutions, r2 = q-1p, r1 = pq-1, (5)(6) which are in general different. For this reason the symmetrical notation p/q will not be used. The notations (5), (6) are unambiguous: r1 satisfies (3) and may be called the left-hand quotient of p by q; and r2, the right-hand quotient, satisfies (4). These solutions are unique; if, for example, rq = r1q,
(r - r1)q = 0, and, since q 0 0, r - r1 = 0, or r = r1. On taking norms in (5) and (6), we have Np (7) Nrl = Nr2 = Nq
The norm of either quotient of two quaternions is equal to the quotient of their norms.
QUATERNIONS
410
§ 184
From (182.8) and (182.9), (8)
. qn 11 ... qi 1. (gig2 ... qn.)-i = N(gig2 ... q) nn = qn i
The reciprocal of the product of n quaternions is equal to the product of their reciprocals taken in reverse order.
The definition (1) shows that the reciprocal of a unit quaternion is its conjugate; the reciprocal of a unit vector is its negative. Example.
Solve the equations (3) and (4) when
q=2-i-2k.
p=1+3i-j+k,
From (1), we have q-' = (2 + i + 2k)/9; hence
ri=pq 1=1(1+3i-j+k)(2+i+2k) = 1(-3 + 5i - 7j + 5k); r2 =q 1p = 1(2+i+2k)(1 +3i - j+k) = y(-3+9i+3j+3k). Note that Sri = Sr2, Nri = Nr2 in conformity with (181.13) and (182.7).
.84. Product of Vectors. The product of two vectors vv' is the quaternion (182.4) whose scalar and vector parts are ,
(1)
S(vv') = - (aa' + bb' + cc'), i j k
(2)
V (vv') =
a
b
c
a'
b'
c'
In order to find their geometric meaning we adopt a special basis i, j, k. Choose i as the unit vector along
v, j as the unit vector perpendicular to i in the plane of v, v' and so directed that the angle (j, v') is not greater than 900 (Fi g. 184 ). Th en k i s the un it vec t or
v' v °
k
Fia. 184
which completes the right-handed orthogo-
nal basis i, j, k. If the angle (v, v') = 0, we have vli, v' = Iv'I(icos0+jsin0), v vv' =
vIv'I(-cos0+ksin0);
(3)
S(vv') = - I v I v' I cos 0,
(4)
V(vv')
I
= IvI
v'
sin 8 k.
Noting that k is the unit vector perpendicular to the plane of v and v' and directed so that v, v', k form a right-handed set, we see
PRODUCT OF VECTORS
§ 184
411
that (3) and (4) are geometric expressions for the scalar and vector parts of vv', entirely independent of the basis i, j, k. They form the cornerstones of the vector algebra of J. Willard Gibbs, who defined the scalar and vector product of two vectors u, v as
u- v=-S(uv),
(5)(6)
u X v = V(uv),
using the dot and cross to distinguish between these two types of "multiplication." Henceforth we shall use Suv, Vuv, Kuv to denote S(uv), V(uv), K(uv); similarly Suvw = S(uvw), etc. A change in the order of the vectors in (3) has no effect, but in (4) reverses the direction of k; hence Svu = Suv,
(7) (8)
Vvu = - Vuv.
From u(v + w) = uv + uw and the distributive character of S and V,
Su(v + w) = Suv + Suw,
(9) (10)
Vu(v + w) = Vuv + Vuw.
Since Kuv = (- v) (- u) = vu, uv = Suv + Vuv, (11)(12)
Suv =
2(uv + vu),
vu = Suv - V UV;
Vuv = 1(uv - vu).
Turning now to products of three vectors, we have (13)
Suvw = Svwu = Swuv
(181.14).
Since Kuvw = (-w)(-v)(--u) = -wvu (§ 182), (14) (15)
Suvw = SKuvw = -Swvu Vuvw = - VKuvw = Vwvu.
Also from uvw = u(Svw + Vvw), (16)
Suvw = SuVvw,
(17)
Vuvw = uSvw + VKVvw.
We now compute Vuvw in another way:
2Vuvw = uvw - Kuvw = uvw + wvu
= uvw + vuw - vuw - vwu + vwu + wvu = (uv + vu)w - v(uw + wu) + (vw + wv)u;
QUATERNIONS
412
§ 185
hence, from (11),
V uvw = uSvw + wSuv - vSuw.
(18)
Comparison with (17) now gives the important formula: VuVvw = wSuv - vSuw.
(19)
Finally, we express these results in the dot and cross notation of Gibbs:
Jul Ivl cos(u,v), uXv = Jul lvlsin(u,v)k;
(5)(6) (7)(8)
vxu= -u-V;
v,
(9)
u.(v+w) =uXv+uxw;
(10)
(13) (14)
uX(vXw) =
(19)
From (5) and (6) we have also
uv =
(20)
as the connecting link between quaternion and vector algebra. 185. Roots of a Quaternion.t Every quaternion
q = d + ai + bj + ck with real coefficients may be written as a real multiple of a unit quaternion : (1)
q = h(cos 0 + e sin
B),
0 < 0 < 27r.
Here
h=
sin 9 = f \/a2 + b2 + c2/h
cos B = d/h,
and, when a2 + b2 + c2 (2)
e
d2+a2+b2+c2, 0, e is the unit vector:
= f ai+bj±ck
1/a2+b2+c2
t In this and following articles we again denote vectors with bold-face letters; but ab denotes the quaternion (not dyadic) product.
ROOTS OF A
§ I35
413
When q is a real number, sin 0 = 0, and e may be chosen at pleasure. Since e2 = - 1, we have, by De'Xloivre's Theorem,
q' = h"(cos no + e sin no).
(3)
We now may find the nth roots of a real quaternion (4)
Q=H
0 < (P < ,r;
p -I- a sin gyp),
the angle always may be taken in the interval from 0 to it by choosing the appropriate e in (2). In solving q' = Q, we consider two cases: 1. sin 0; we choose the e in q the same as in Q. Then
hn = H,
cos no = cos cp,
sin no = sin gyp,
and n nth roots of Q are given by (1), provided (5) (6)
h = H' I', the positive root, 0 = (gyp + 2irm)/n
(m = 0, 1,
,n-
1).
These n values of 0 comprise all values in the interval 0 < 0 < 2ir which satisfy the preceding equations. 2. sin p = 0: the e in q is then an arbitrary unit vector.
IfQ>0: o =0,
0 =2mir/n
(m=0,1, ,n-1).
When n = 2, the values 0 = 0, Tr give just two roots =LA/Q-, both real. When n > 2, some values of 0 (0 0 or ir) give non-real roots q with which any e may be associated.
(n=0,1, ,n-1).
IfQ<0: p =ir, 0=(2m+1)ir/n
In every case some values of 0 (0 ir) give non-real roots q with which any e may be associated. We summarize these results in the THEOREM. A quaternion with real coefficients, but not a real number, has exactly n nth roots. If Q is a positive real number, it has
just two square roots ± /; in all other cases a real number has infinitely many quaternion roots with real coefficients.
In all cases the roots may be computed from (5) and (6). For
example, if Q = 1 + i + j + k, we write Q = 2(cos 60° + e sin
60°),
e = (i + j + k)/1'3.
QUATERNIONS
414
§186
The cube roots of Q are then
q=
0 = 20',140',260'.
(cos 0 + e sin 0),
186. Great Circle Arcs. Every unit quaternion
q=d+ai+bj+ck
(Nq=1)
can be expressed in the form q = cos 0 + e sin 0.
(1)
Here e is given by (185.2); and 0 satisfies
sine =
cos 0 = d,
(2)
a2 + b2 + c2.
If we choose the plus sign in these formulas, 0 < 0 < 7r. In particular, if q = 1, -1, e, the angle 0 = 0, zr, 2 7r, respectively. THEOREM. The unit quaternion cos 0 + e sin 0 may be expressed as the quotient ba-1 of any two vectors which satisfy the conditions:
(i)lal=lbl, (ii) Angle (a, b) = 0, (iii) Plane a, b is perpendicular to e, (iv) a, b, e form a dextral set. Proof.
In view of (i), we may write
cos9+esin9=
Ial bIcos0+Ial bIsin0e
IaI2 hence, if we choose the vectors a and b so that conditions (ii), (iii) and (iv) are fulfilled,
cosh+esin0=
-Sab + Vab Na
_
-Sba - Vba Na
ba Ka = --=bNa Na
or, in view of (183.1), (3)
cos 0 + e sin 0 = ba-1.
From Fig. 186 we see that, when a, b, e form a right-handed set, the angle (a, b), when less than 7r, is counterclockwise, viewed from the tip of e. We shall describe the sense of (a, b) as positive relative to e. When the angle (a, b) = 0 or 7r, q is 1 or -1, re-
spectively, and e is entirely arbitrary.
§.,%
G'ItEAT
CIRCLE ARCS
415
To every unit quaternion, q = cos 0 + e sin 0 corresponds to a great circle are AB of a sphere centered at 0, provided OA = a
and OB = b satisfy the preceding conditions (i) through (iv). Thus q corresponds to an are of a great circle whose plane is normal to e and whose central angle 0 has the positive sense relative to e. All such arcs of this great circle are equally valid representations. If the are AB represents q, AB is free to move about in its great circle, provided its length and sense remain unaltered. The cases q = 1 (B = 0) and
q = -1
(0 = -7r),
in which e
q
FIG. 186
is
arbitrary, are exceptional. Any point of the sphere represents q = 1; and any great semicircle represents q = - 1. A unit vector q = e (0 = 2 r) corresponds to a quadrantal are in the plane through 0 normal to e. If q = cos 0 + e sin 0 corresponds to the are AB,
qI = Kq = cos 6 - e sin B
(4)
corresponds to an are having the same plane and angle, but whose
sense is positive relative to -e. Hence q-' corresponds to the are BA. (5)
Moreover,
- q = - cos 0 - e sin 0 = cos (ir - 0) - e sin (7r - 0) ;
hence, if q corresponds to the are AB, and AOA' is a diameter, -q corresponds to are A'B, the supplementary are reversed in Sense.
Using the sign '' to denote correspondence, we sum up our findings as follows:
1 ti point,
-1 - semicircle, e
quart.ercircle;
q - are AB, q-I - are BA, -q '' are A'B. The utility of this representation is due to a simple analytical method of adding great circle arcs "vectorially." To add two arcs, shift them along their great circles until the terminal point of the first
(AB) coincides with the initial point of the second (BC); then the
great circle arc AC is defined as their vector sum.
If a = OA,
QUATERNIONS
416
§186
b = OB, c = OC, the preceding theorem shows that the arcs AB, BC, AC are represented by the quaternions ba-1, cb-1, ca-1, respectively. Write
p = ba-1, q = cb-1;
then
ca-1
=
cb-lba-1
= qp;
and the equation,
are AB + are BC = arc AC, may be written are p + arc q = are qp.
(6)
For three arcs, (7)
arc p + are q + are r = arc qp + are r = arc rqp;
and, in general, the vector sum of any number of great circle arcs is given by the arc corresponding to the product of their representative quaternions taken in reverse order.
In interpreting such arc-quaternion equations, remember that are q = 0 means that q = 1; and, if are q is any great semicircle, q = -1. For example, if arc p, are q, are r form the sides of a spherical triangle taken in circuital order,
arc p + arc q + arc r = 0,
arc rpq = 0,
rpq = 1.
In general, the arcs representing the quaternions q1, q2, , qn, taken in this circuital order, will form a closed spherical polygon when and only when qnqn-1 '
(8)
g2g1 = 1.
Example. Spherical Trigonometry. Consider again the spherical triangle ABC of § 22. With the notation of this article
are BC ' cb-1 = cos a + a' sin a, are CA ' ac-1 = cos $ + b' sin Q,
are AB - ba 1 = cos y + c' sin y. The "vector" equation
are CA + arc A B = arc CB
corresponds to the quaternion equation (bat) (ac 1) = be 1, or (i)
(cosy + c' sin y) (cos i3 + b' sin 3) = cos a - a' sin a.
ROTATIONS
§ 187
417
If we expand the left member, put
c'b' = - b' c' - b' x c'
cos a' - a sin a';
and equate scalar parts in both members, we have cos 0 Cos y - sin 3 sin y cos a' = cos a.
(ii)
This is the cosine law (22.6) of spherical trigonometry. On equating the vector parts in both members of (i) we have, (iii)
a' sin a + b' sing cos y + c' cos l; sin y = a sin a' sin 0 sin y
and hence, on multiplication by a
,
sin a' sin /i sin y = a a' sin a = a bxc, or
sin a'
a- bxc
sina
sin a sin#sin y
Since the right member is unchanged by a cyclical permutation, we have (iv)
sin a'
sin /3'
sina
sing
sin y' sin y
the sine law (22.5) of spherical trigonometry.
187. Rotations. With the aid of quaternion algebra, finite rotations in space may be dealt with in a simple and elegant manner. This application depends upon the fundamental THEOREM.
If q and r are any non-scalar quaternions, then
r' = qrq-1
(1)
is a quaternion whose norm and scalar are the same as for r. The vector Vr' is obtained by revolving Vr conically about Vq through twice the angle of q. Thus if q = Nq(cos 0 + i sin 6), Vr' is obtained by revolving Vr conically about i through an angle 26. Proof. (2) (3)
The norm and scalar of r' are Nq-1 N(qrq-1) = Nq Nr = Nr S(qrq-1) = S(q 'qr) = Sr.
Moreover on writing r = Sr + Vr, qrq-1
= Sr + q(Vr)q-1;
(182.9), (181.14).
QUATERNIONS
418
§ 187
Now, from (3), fore a vector; hence
q(Vr)q-1 has
(4)
V (qrq-1) = q(Vr)q-1
the same scalar as Vr and is there-
Let us now write
r = Nr(cos c + e sin (p); then from (4),
Vr' = Nr sin
e' where
e' = qeq-1.
If we choose j in the plane of e and i (Fig. 187a) and k to complete the dextral
set i, j, k, then
e = icosA+ Jsin A e' = (qiq-1) cos A + (qjq 1) sin A. Since Vq is parallel to i, qi = iq and
Flo. 187a
qiq--1
=
iqq-1
= i.
Moreover, qjq-1
= (cos 0 + i sin 0)j(cos 0 - i sin 0) = (j cos 0 + k sin 0) (cos 0 - i sin 0) = j (cost 0 - sine 0) + k(2 sin 0 cos 0) ;
hence j goes into
j' = j cos 20 + k sin 20,
a vector obtained by revolving j about i through an angle 20 in the positive sense.
Consequently
e = icosA+ jsinX -*e' = icosA + j'sinA and also Vr -. Vr' by a conical revolution about i of the same amount. This completes the proof. We note that vectors transform into vectors.
In particular, if
q = a, a unit vector,. 0 = 90°, and
e' = aea-1 = -aea is obtained by revolving e conically through 180° about a.
The
transformation -a( )a thus gives vectors a half-turn about a.
ROTATIONS
§ 187
419
The transformation a( )a may be regarded as a half-turn followed by a reversal; a vector thus transformed is simply reflected in a plane normal to a (Fig. 187b). Thus aea is the reflection of e
in the plane normal to a.
a
A rotation through an angle a about a implies that the angle turned has the positive sense relative to a. If
p = Cos. a+asin1a, q= cos 2 S -{- b sin 3 z are unit quaternions, the operators p( )p-1
Fia. 187b
and q( )q-1 effect rotations of a about a and ,3 about b; for brevity we call these the rotations p and q. The succession of rotations p, q corresponds to the operator, qp(
)p-'q-1
= qp( ) (qp)-1;
since qp is also a unit quaternion, say
qp = cos zy + c sin
21
'Y,
the resultant is equivalent to the single rotation qp, that is, a rotation through the angle y about c. Similarly, the resultant of the rotations q and p corresponds to the operator, pq(
)q-'p-1 = pq( )(pq)-1
Since pq = qp only when Vp and Vq are parallel (the values p, q = -4-1 are excluded), the composition of rotations is non-commutative except when they have the same axis. The rotation p followed by the rotation q is equivalent to the single rotation qp. More generally, the succession of rotations q1, q2, qn is equivalent to the single rotation gnq,,-1 . g2g1. (-q)-1 Since = -q-1, the rotations q( )q-1 and (-q)( )(-q) are the same. If q = cos B + a sin B,
-q = cos(Tr-9)+(-e)sin(7r-8); thus the rotation -q is a rotation through 27r - 20 about -e; this produces the same result as the rotation q, namely 20 about e. Since q-lq( )q-lq = 1( )1, the rotation q-1( )q is the reverse of q( )q-1; this is also evident from q-1 = cos 0 - e sin 0.
QUATERNIONS
420
§ 187
Example 1. The rotation of 90° about j followed by a rotation of 90° about i is represented by the quaternion product
(cos 45° + i sin 45°) (cos 45° + j sin 45°) = !(I + i + j + k); that is, by 1
+
i+j+kV' V3
- = cos 60° +
i + j + k
-
sin
60°.
-\/3 The resultant rotation is therefore a rotation of 120° about an axis equally inclined to the (positive) axes of x, y and z. 2
2
Example 2. The resultant of two reflections in planes normal to the unit vectors a, b corresponds to the operator
ba( )ab = ba( ) (ba)-1.
But if a b = cos 0, a x b = e sin 0, we have
ba = -a b - axb = -(cosB+esin0)
(184.20).
Now the rotation q( )q-' = (-q)( )(-q)-1; hence successive reflections in two plane mirrors is equivalent to a rotation about their line of intersection of double the angle between them. Example 3. From (181.13) we know that S(qp) = S(pq); moreover from (4) V (qp) = V (gpgq-1) = qV (pq)q-1
Hence V(qp) is obtained by revolving V(pq) about Vq through double the angle of q. Thus, if u is a vector, V(up) is obtained by revolving V(pu) 180° about u; that is, the vector u bisects the angle between V(pu) and V(up).
Now let a, b, c be three radial vectors from the center of a sphere to its Then a, b, c bisect the angles between Vabc, Vbca; Vbca, Vcab; Vcab, Vabc respectively. In other words, if we form a spherical triangle
surface.
whose vertices are Vabc, Vbca, Vcab, the middle points of the sides opposite lie on the vectors b, c, a respectively. Example 4. If the sides of a spherical polygon are represented by the quaternions qi, q2, , qn taken in this circuital order, gnqn-1 g2gt = 1 (186.8). Hence the succession of rotations,
gnqn-1 ... q2ql( )gig2 ... q.-1qn = 1( )1, about axes through a point 0 will restore a body to its original position. We state this result for the case of a triangle as follows: THEOREM (Hamilton and 1)onkin). If ABC is any spherical triangle, three successive rotations represented by the directed arcs 2 BC, 2 CA, 2 AB (about their
polar axes) will restore a body to its original position.
This same theorem applies to the polar triangle A'B'C'. Since the side B'C' = a' = 7r - A (§ 22) and has OA as polar axis, successive rotations of 2ir - 2A, 27r - 2B, 2ir - 2C about OA, OB, OC will restore a body to its original position. Since the rotations 2ir - 2A and -2A about OA give the same displacement, we may state the
PLANE VECTOR ANALYSIS
§ 188
421
THEOREM (Hamilton). If ABC is any spherical triangle on a sphere centered at 0, three successive rotations about OA, OB, OC through the angles 2A, 2B, 2C in the sense of CBA will restore a body to its original position.
188. Plane Vector Analysis. The three-term quaternion c + ai + bj has given rise to two types of vector analysis in the plane. The one interprets c + ai as a vector w in the complex plane; then the product wlw2 is always a complex vector. The other interprets ai + bj as a "real" vector w and decomposes the quaternion product w1w2 into its scalar and vector parts, which are used separately as "products." To indicate the interpretation used, we denote complex and real vectors by italic and bold-face letters, respectively. Thus the plane vector whose components are u, v may be written as
w = u + iv,
or w=ui+vj.
In the first case,
w1w2 = (ulna - vlv2) + (ulv2 + u2v0i;
(1)
in the second,
w1w2 = - (ulu2 + vlv2) + (u1v2 - u2v1)k.
(2)
In Gibbs's notation, ulu2 + V1V2 = w1 w2i
u1v2 - u2v1 = k w1 x W2.
Let w = u - vi denote the conjugate of w; then, from (1) and (2), W2.+ik.wlxw2, w1W2 = W1 W1W2 = W1 W2 - i k W1 x w2, and hence
(3)
W1 W2 = 2 (wlw2 + W1w2),
(4)
21i(w1w2
- wlw2)
The conditions (128.6) and (128.7) for perpendicular and parallel vectors may be read from these equations. We next consider corresponding differential invariants. The operator, a a a a
V =i-+j--+iax
ay
ax
ay
QUATERNIONS
422
§188
(read - as corresponds to). If we introduce the conjugate variables,
2=x-iy,
z=x+iy,
-+i- _ -+i-ay az-+(ax-+i-ay a2a
a
az
ax
ay
ax
az
a
(1 + i2)
a
az
+ (1 - i2)
az
az
a
a
a2,
or
a (5)
_
a
a-
ax + Z y
a
0
2 az '
ax
8
- Z ay
_
a
2 az
follows in the same way. Corresponding to the gradient V of a real function p(x, y), we have 2 a(p/az in the complex plane, in which the variables x, y in are replaced by the values,
y = 2i(2 - z). For example, x2 + y2 = z2, and hence 0(x2 + y2) = 2z in the x = 2 (2 + z),
complex plane.
The unit vector, (6)
e = icos0+jsin8NeiO = cos0+isin9.
In view of (3), the operator for differentiation in the direction e, namely (7)
Therefore
-
dw
-, - = e Vw -e t8 aw + e-i0aw az
(8)
ds
az
and hence dz/ds ti eie. If w is a complex function, dw/dz in the direction 0 corresponds to the ratio of dw/ds to dz/ds; hence dw
(9)
dz
aw
az
+
e
_2. aw az
in the direction 0.
When dw/dz is independent of 0, w is said to be an analytic func, tion of z; for this, it is necessary and sufficient that aw
az
=0,
or
-+i(u+iv)=0, ay ax a
a
PLANE VECTOR ANALYSIS
§ 188
423
in view of (5). The last condition is equivalent to the familiar Cauchy-Riemann Equations : (10)
au
av
ax
ay
au -+-=0. av
=0,
ay
ax
We next find the correspondents for div w = V w and k rot w = k V x w by making use of (3), (4) and (5) : 3z' divwti+aw-, az
a2
aw k rot w /aw tii(--az \az
(12)
Moreover, for the Laplacian V2 = V V, we have
V2=2
(13)
a2
a2
a2
azaz+2azaz - 4azaz
Thus a real function cp(z, 2) is a harmonic when a2,p/3z ai = 0; for example, log I z I = 2 log zz = (log z + log z) 2 is harmonic. If the plane vector w - w(z, 2), the condition, Ow aw rotw=ON---=0.
(14)
az
az
When rot w = 0,
w dr;
a=
w = VX, where
:o
hence, when w(z, 2) is irrotational, (15)
w=
z
,
where
p
=
f
(w dz + w dz)
the real function 2X(z, 2). The field lines cut the curves const at right angles. If the vector w is plane, the condition, is
Ow (16)
aw
div w = 0 - - + - = 0. az
a2
QUATERNIONS
424
§ 188
From (85.6), rot (k x w) = k div w; hence, when w is solenoidal, k X w is irrotational. Since k X w is w revolved through +7r/2, k x w - iw. Thus when w(z, 2) is solenoidal, (15) applies when w
is replaced by iw (and iv by -iw). From this result, we conclude
that (17)
w=i
az
=
where
ifZ(w dz - w d2) zo
is a real function. The field lines are the curves From these results we have
= const.
THEOREM 1. If
The preceding results give a simple method for decomposing a plane vector function into an irrotational and a solenoidal part. For, if w = W1 + w2i
rot W1 = 0,
div W2 = 0,
there exist real functions gyp,' such that
a, w=-+i-=-( o +ii'); a
app
a2
a2
az
' + i¢ = Jw(z, 2) d2
(z const)
is determined to an arbitrary additive f (z) and w1 = app/az, w2 = i ay/az. If the vector field w(z, 2) is both irrotational and solenoidal, rot w = 0 implies that
w=-; az
az p then =w,asp divw=2 =0. az dz dz
Hence p is a real harmonic function, and w is an analytic function of z. Conversely, if w is an analytic function of z, (9w/a2 = 0; hence aw/az = 0, and (11) and (12) show that w is solenoidal and irrotational. THEOREM 2. In order that the complex vector w(z, 2) be irrotational
and solenoidal, it is necessary and sufficient that its conjugate be an analytic function of z.
PLANE VECTOR. ANALYSIS
§ 188
425
The vector
Example 1.
w = 22 =
(x2
- y2) - 2xy i,
is both irrotational and solenoidal; for its conjugate z2 is an analytic function of Z.
Example 2. To decompose the vector,
w = z2 = (x2 - y2) + 2xy i,
into its irrotational and solenoidal parts, we may take
v + ik = fz2 d2 = z22. Since
,P - i= 22Z, = 12z2(2 + z), wl
_
av a2
w2 = i
¢ = Zi z2(2 - z); (3x2 + y2) + xy it
= z2 + 2z2 =
ak a2
2
= -z2 + 2z2 = - (x2 + 3y2) + xy i.
i
Stokes' Theorem in the plane,
-
fk . rot w dA =
fw . dr,
corresponds to
i
aw 492
- -aw dA = a az
(w d2 + w dz),
by virtue of (12) and (3). If we replace w by -iw (and iv by i4 b), we obtain, after canceling i,
-+ i f(afv az
-aw dA = az
2 J(-w dz + w dz).
On adding these equations and then replacing w by w, we have (18)
dA. fw dz = 2if -aw az
When w is analytic in the region within the circuit, aw/a2 = 0, and (18) reduces to Cauchy's Integral Theorem:
Jw dz = 0.
QUATERNIONS
426
Example 3.
2iA =
When w = z in (18),
12 dz = J(x
- iy) (dx + i dy) = i
§ 189
f
(x dy - y dx).
This gives the well-known circuit integral for a plane area. With w = z2, we obtain the static moments of a plane area about the axes, expressed as circuit integrals.
189. Summary : Quaternion Algebra is a linear, four-unit (1, i, j, k) associative algebra over the field of reals. The unit 1 has the properties of the real one; and
i2=j2=k2= -1,
ij=k,
ji= -k,
the last equations admitting cyclical permutations. Quaternions
q = d + ai + bj + ck include real and complex numbers (d, d + ai). Since i, j, k may be interpreted as dextral set of orthogonal unit vectors, quaternions also include vectors v = ai + bj + ck. Thus q = d + v, a scalar plus a vector. Quaternion multiplication is associative and distributive, but not in general commutative; in fact pq = qp holds only when p or q is a scalar or when the vector parts of p and q are proportional. The product vv' of two vectors is the quaternion :
vv' = - (aa' + bb' + cc') +
= -v - v' + v x v' in Gibbs' notation. The quaternion q = d + v has the conjugate Kq = d - v. The conjugate of the product qq' is K(qq') = (Kq') (Kq). The norm of a non-zero quaternion q is the positive real number,
Nq = q(Kq) = d2 + a2 + b2 + c2; and N(qq') = (Nq) (Nq'). The equation q = 0 implies Nq = 0, and conversely; hence, if qq' = 0, either q = 0 or q' = 0. A unit quaternion q (Nq = 1) may be put in the form
q = cos0+esin9
(IeI = 1),
and associated with the great circle are of angle 0 and pole e on the unit sphere. On a fixed great circle, all arcs of the same length
PROBLEMS
427
and sense correspond to the same q and are denoted by arc q. The scalars 1 and - 1 correspond, respectively, to any point and to any great semicircle of the sphere. Great circle arcs may be added vectorially; and
are p + are q = arc qp. If three arcs form a spherical triangle, say-
arcp+arcq+arcr = 0, then rpq= 1. If q = Nq(cos 0 + e sin 0), the operator q(
)q-1
effects a conical
revolution of 20 about e on the vector of the operand; thus if r' = qrq-1, then Nr' = Nr, Sr' = Sr, T Y = Vr revolved about e through 20.
When q = e, a unit vector, q-1 = -e; the operator -e( )e gives vectors a half-turn about the axis e. The operator e( )e reflects vectors in the plane normal to e. PROBLEMS
1. Solve the quaternion equations, rq = p, qs = p, for r and s when
q=2-i-2k,
p= 1+3i-j+k.
Verify that Vr = Ns. 2. Every quaternion q satisfies the quadratic,
q2 - 2qSq + Nq = 0, known as its principal equation. The conjugate Kq satisfies the same equation.
3. Show that 2 + 5i and 2 + 3j + 4k have the same principal equation; hence factor its left member in two different ways. 4. Show that if we identify the quaternion units with the 2 X 2 matrices
j 1 - (0
1
),
(
0
0)'
L
\-1 0),
-
\-c0
0/'
where L (iota) is the complex unit, L2 = -1, these matrices satisfy the multiplication table (181.7). 5. If u, v, w are vectors prove the following identities: (a)
u22 _ -Nu;
(u - v)(u + v) = u'- + 2Vuv - v2;
(b)
2 Suvw = uvw - wvu;
(c)
2 Vuvw = uvw + wvu;
(d)
S(u + v) (v + w) (w + u) = 2 Suvw.
428
QUATERNIONS
6. Show that the multiplication table of the units i, j, k is completely given by
i2=j2=k2=ijk= -1. 7. Interpret the equations, ij = k, jk = i, ki = j and kji = 1, in terms of "arc vectors."
8. Solve the equation aq + qb = c for the quaternion q if a, b, c are known quaternions and Na 5x, Nb. [q = (ac - cb)/(Nb - Na).) 9. Solve the equation qa + bq = q2. [Reduce to aq-1 + q -1b = 1.] 10. If a, b, c are vectors for which Vabc = 0, prove that a, b, c are mutually orthogonal.
[Vabc = aSbc + VaVbc and VaVbc 1 a; hence Sbc = 0, VaVbc = 0.1 11. If a, b, c, d are vectors for which Vabcd = 0, prove that (a) bed is a vector parallel to a; (b) Vbcda = Vcdab = Vdabc = 0; (c) the vectors a, b, c, d are coplanar; (d) cda is a vector parallel to b; and thus cyclically. 12. A body is revolved through 90 degrees about two axes e1, e2 which intersect at an angle 0 (cos 0 = e1 e2). Show that the equivalent single rotation is through an angle 2 cos' z (1 - cos 0) and about an axis parallel to e1 + e2 - e1 x e2. 13. If a and b are unit vectors along intersecting axes, show that the half-
turns -a( )a, -b( )b in this order are equivalent to a rotation of twice the angle from a to b about their common perpendicular. 14. The quaternion q = a1a2 a,,_1 an is the product of it unit vectors.
Show that if Sq = 0, q is the vector q = ±anan-1 ... a2a1 (+ when it is odd, - when n is even).
Hence show that the successive reflections ai( )ai in the order i = 1, 2, 3, , n reduce to the single reflection q( )q when n is odd, and to a half-turn -q( )q when n is even. 15. If the successive reflections ai( )ai in the order i = 1, 2, 3,. , it reduce to a single reflection or half-turn, show that this is true for any cyclical permutation of this order. 16. Show that the succession of reflections in three coaxial planes reduces to a single reflection. 17. Prove the famous theorem of Euler (1776): Any displacement of a rigid body which leaves the point 0 fixed is equivalent to a rotation about an axis through
0. [Let the displacement move the trihedral ijk in the body to the new position ijj1k1. This may be accomplished by two reflections: the first takes i -+ i1, j -s j'; the second takes j' - ji, leaving i1 undisturbed.] 18. If q = cos 0 + e sin 0 (not scalar), show that the rotation q( )q-1 followed by the reflection a( )a reduces to a single reflection b( )b when and only when a e = 0; moreover, b = aq = a cos 0 + a x e sin 0. 19. If we define qn for arbitrary real n by (185.3), show that any quaternion may be expressed as the power of a vector. When q = cos 0 + e sin 0, prove
PROBLEMS
429
that q = e2e/r; and that the r( ti tion of angle about the axis e is represented by the operator e ' n ( )e -" Oi*. 20. Show that successive h ili-turns about three mutually orthogonal and intersecting axes will restore a body to its original position. 21. ABC is a spherical triangle and P, Q are the middle points of the sides AB, BC, respectively. Prove .hat two successive rotations represented by the arcs AB, BC are equivalent tc the rotation represented by twice the great circle are PQ. 22. Prove the theorem: Th,ee successive rotations represented by the arcs AB, BC, CA of a spherical triangle ABC are equivalent to a rotation about OA through an angle equal to the spherical excess (A + B + C - 7r) of ABC. 23. Prove Rodriques' Cons'ruction for the composition of two rotations through the angles a, f3 about he axes OA, OB: Draw great circle arcs AC m.nd BC such that
angle (AB, AC) = -!a, angle (BA, BC) = 2 determining the spherical triangle ABC; then rotation a about OA followed by 0 about OB is equivalent to the rotation -y = 2(CA, CB) about OC. 117se Hamilton's Theorem given in §187, Ex. 4.]
Give the construction when rotation 0 about OB is followed by rotation a about OA. Draw a figure to illustrate both constructions. 24. Prove that successive rotations through angles of gyp, ,r/2, o about the axes of x, y and z respectively are equivalent to a rotation of 7r/2 about the y-axis.
25. If r = xi + yj + zk and r' = x'i + y'j + z'k, are position vectors and q is an arbitrary quaternion, show that the transformation r' = qr(Kq) represents the most general rotation and expansion of 3-dimensional space. What is the ratio of expansion? If q = d + ai + bj + ck, express x', y', z' in terms of x, y, Z.
INDEX The numbers refer to pages. A starred number locates the definition of the term in question. The letter f after a number means "and following pages." Terms under a noun (key word) are to be read in before this word, unless preceded by a preposition such as "of" or "for." Terms under an adjective (key word) are to be read in after this word; the repeated adjective is indicated by dash. Absolute tensors, 343f Acceleration, 109* angular, 122* four-vector, 401 general components of, 359 normal component of, 109 of Coriolis, 124* of gravity, 124 rectangular components of, 110 relative, 123 tangential component of, 109 transfer, 123* uniform, 110 Addition, of dyadics, 137* of forces, 75 of great circle arcs, 415f of motors, 67* of quaternions, 403* of tensors, 350* of vectors, 3* statical, 4* Addition theorems (sine, cosine), 28 Adjoint dyadic, 152* Af ine connection, 356*f, 361, 368
Antecedents of a dyadic, 136* Antisymmetric dyadic, 141 *
-tensor, 352* Archimedes, Principle of, 258 Arc vector, 415f Area, of a polygon, 39 of a spherical triangle, 310 of a triangle, 44 vector, 37*f Areas, Law of, 133 Astatic center, 62 Asymptotic lines, 285*, 288, 303 Axes, rectangular, 26 principal, of a dyadic, 162 of inertia, 160 of strain, 164 of stress, 256 Axis, 25* instantaneous, of velocity, 117* of acceleration, 127* of a motor, 66* radical, 62
Barbier's Theorem, 134 Base vectors, 24, 147, 349, 367, 368 dextral and sinistral, 24, 25, 41 Basis, 25 Bernoulli's Theorem, 279 Bertrand, J., 132, 314 Bianchi identity, 385 Bilateral surface, 218 Binormal, 92* Bivector, 353* B6cher, M., 339
-group, 340f, 386 -transformation, 138 Alternative airport, 113 Analytic function, of (real) coordinates, 347 of complex variable, 272, 422* Angular acceleration, 122*
-excess, 310 -speed, 114* -velocity, 115* 431
INDEX
432
Bonnet, 0., 286, 288, 296, 309 Bonnet's Integral Formula, 309 Box product, 42* Bracket notation for vectors, 25
Carnot's Theorem, 60 Cartesian coordinate system, 379 *f, 385f
Catenary, 103, 317 Catenoid, 317 Cauchy-Riemann Equations, 272, 423
Cauchy's integral, 265 -Integral Theorem, 276, 425 Caustic, 104 Center, astatic, 62 instantaneous, 118* mean, 20* of curvature, 99* of mass, 258*, 236 of normal curvature, 285* Central forces, 133, 134 Centrifugal force, 124 Centroid, 9* Ceva, Theorem of, 14 Characteristic equation, 154*, 168f -numbers, 154*, 156 Characteristics, 98*, 99, 133 Christoffel symbols, 326, 362*, 389 Circle, nine-point, 33 of curvature, 99 rolling, 120 Circular motion, 110 Circulation, 266*f Clifford, W. K., 64 Codazzi, Equations of, 300f, 391 Cofactor, 168*, 330*f reduced, 331 *
Cogredient transformations, 335 Collinear points, 8f -vectors, 2* Complete quadrangle, 16
-quadrilateral, 18 Components, contravariant, 335, 342, 344
covariant, 335, 342, 344 mixed, 342, 344 of acceleration, 164, 359 of affine connection, 356
Components of a dyadic, 166, 342 of stress, 256 of a tensor, 344 of a vector, 26, 48 of velocity, 164, 359 Composition of velocities, 120f Conditions, necessary and sufficient, viii* Cone, 304, 315
Conjugate, of a complex number 274, 422
of a dyadic, 141 of a quaternion, 406* Cc njugate lines, 79, 80 Conoid, 321 *, 323
Consequents of a dyadic, 136 Continuity, Equation of, 258f, 264 Contraction, 351 *f Covariant derivative, 362*f, 365f, 401 of dyadics, 364 of epsilons, 366 of Kronecker deltas, 366 of metric tensor, 366 of scalars, 364, 366 of tensor product, 365 of tensor sum, 365 of vectors, 364 Covariant vector, 335* Cross product, 34 Curl, of a tensor, 374*, 376 of a vector, 183* Curvature (K), 92*, 96, 97 center of, 99*, 285 circle of, 99 geodesic (y), 284* integral, 309*f lines of, 285*, 302f mean (J), 222, 286*, 287, 292, 392 normal (k), 284*, 287 of plane curves, 101 of surface curves, 283f principal, 287* radius of (p), 93* total (K), 286*, 287, 292, 306f, 392 Curvature tensor, 379, 380 *f Contragredient transformations, 334* Contravariant components, of a dyadic, 342 of a tensor, 344
I \ DEX Contravariant component;;, c a ve:-tor, 48, 335 Contravariant vector, 334i Coordinates, barycentric-, '23* Cartesian, 25*, 340*, 335 curvilinear, 191 cylindrical, 195 geodesic, 358* homogeneous, 51f line, 52
orthogonal, 89f plane, 51 Pliicker, 52, 55, 63 point, 51 spherical, 196 Coplanar line vectors,
-points, 12f -vectors, 2 Coulomb's Law, 241 Coriolis, acceleration of, 124" Theorem of, 124 Cosine law, for plane triangles, 44 for spherical triangles, 46 417 Covariant components, of a ,ayadie, 342
of a tensor, 344 of a vector, 48, 335 Curves, Bertrand, 132* congruent, 97 field of, 231*, 293* of constant curvature, '100 parallel, 94*, 131 parametric, 205* plane, 100f reducible, 223* space, 88f Cycloid, 131 Cylinder, 304, 315
D'Alembert's Principle, 260 Darboux, G., 289 Darboux vector, 93*, 105, 1:!2, 232 Del (V), 184* Deltas, Kronecker, 46*, 253:',366, 402
Derivative, covariant, 36:2*f 365f directional, 178, 182, :187 normal, 180 of a determinant, 331
433
Derivative, of a dyadic, 171 of a motor, 126 of a vector, 84f
Derivative formulas, of Gauss, 327, 390
of Weingarten, 323, 327, 391 Desargues' Theorem, 15 Determinants, 329*f bordered, 397, 400 Developable surface, 303, 304, 314*f Dickson, L. E., 408 Differential, total, 90* Direction cosines, 27 Discriminant of a quadratic form, 337 Displacement group, 388* Distributive Laws, 30, 35, 137 Divergence, of a vector (div f, V f), 183*f, 260*, 371 of a tensor, 371 *f, 376 surface (Div f, V f), 207* Divergence Theorem, 235* Division of quaternions, 409*
Dot product, 29 Double-dot product, 175 Double layer, 244 Doublet, 244 Dual angle, 69* -number, 64 *
-vector, 63* Dual of a tensor, 354, 370*, 376 Dupin, Theorem of, 305 Dyad, 136*, 166 Dyadic, 136*, 342 adjoint, 152* antisymmetric, 141 complete, 139* conjugate, 141* field, 293f inertia, 159* invariants of, 147, 148f, 150* linear, 140* planar, 140* second of a, 151* singular, 140* stress, 253f symmetric, 140*, 156 unit, 144* zero, 137*, 140 Dyadic equality, 136*
INDEX
434
E = [eie2ea], 49* E, F, G, 205*, 290*
Form, polar, 337* quadratic, 337f second fundamental, 290*, 389
e,f,g,291* Einstein, A., 328 Electric intensity, 241* Ellipsoid, of inertia, 160* Energy, of a fluid, 282 (kinetic) of a top, 176 Energy ellipsoid, 177 Envelopes, 103, 132 Epsilons 366,369
singular, 337 *
third fundamental, 291 * Franklin, P., 276 Frenet's Formulas, 93*, 100, 122 Functional dependence, 190f Fundamental quantities, 291
-quadratic form, 339 329*, 346,
Equation, exact, 200 integrable, 200, 231 intrinsic, 103* Equiangular spiral, 130 Equilibrium, of a cord, 298 of a deformable body, 255f of a fluid, 257 of a rigid body, 76 Euclidean geometry, 385f -space, 368, 385* Euler, L., 176, 260, 288, 311, 408, 428 Eulerian Equations, for fluid motion, 260f
for a top, 176 Evolute, of a plane curve, 101 * Falling body, 124 Field dyadic, 293f Field, of curves, 231*, 293*, 299 of geodesics, 299 of numbers, 408* of surfaces, 317 Field lines, 226f Flat space, 377*f, 285 Floating body, 257 Flow, 266* Force, body, 253, 255 centrifugal, 124* conservative, 261 Coriolis, 124* on rigid body, 75f surface, 253, 255 transmissibility of, 75 Form, definite, 338* first fundamental, 204*, 290*, 388 indefinite, 339* metric, 339*
g = det gij, 339*, 369* y (geodesic curvature), 284*, 295, 297
Gammas (rte), 326, 356*, 368 Gauss, C. F., 300, 301, 307, 310 derivative formulas of, 327, 390 Equation of, 301, 302, 306, 320, 392 Theorem of, 307 Geodesic, 285*, 297f Geodesic coordinates, 358*
-curvature (y), 284*, 295, 297 -field, 299 -line, 285*
-lune, 310 -torsion (t), 284*, 286, 305 -triangle, 310 Gibbs, J. W., 29, 136, 421, 426 Gradient, of a scalar, 179* of a tensor, 187* of a vector, 181 * surface, 206 *
Green's identities, 237, 238
-Theorem, 216f Group (definition), 341* displacement, 387f of affine transformations, 340f of general transformations, 347 of orthogonal transformations, 387 of rigid motions, 387f
H = [rrn] = E(, 205* 291 Half-turn, 427*, 429 Hamilton, W. R., 403, 405, 408, 420, 421
Hamilton-Cayley Equation,
160f,
176, 291
Harmonic function, 196, 197, 214, 239*, 244f, 247, 423f, 424
INDEX Heat conduction, 247 Helix, 105*f circular, 107
Helmholtz's Equation, 263 Homogeneous coordinates, 51
-strain, 163
435
Kinematics, of it, particle, 108f, 359 of a rigid body, 114E Kinetic energy, of a fluid, 281, 282 of a top, 176 Kronecker delta (Sji), 46*, 343, 366 generalized, 353*, 366, 369, 402 Kutta-Joukowsky Formulas, 273f
i, j, k, 26* i, j, k, 404 * Idemfactor (I), 144*, 151, 34i3
Inertia, dyadic, 159* moment of, 159* product of, 160* Integral, circuit, 216f line, 222f, 224f surface, 218f, 221f, 233f volume, 233f Integrability condition, 20), 231 Interception (of a plane), 11': Intrinsic equation, 103* Invariable plane, 177 Invariant, differential, 181, :.83, 207,
L, M, N, 290* Lagrange's Theorems, 62 Lagrangian Equations for fluid motion, 265 Laguerre, E., 289 Lamb, H., 273 Laplace expansion of a determinant, 330
Laplace's Equation, 196, 197, 214, 238, 247 *, 248
Laplacian (V2), 185*, 400, 423 Laurent series, 276 Lift, 277 Linear dependence of vectors, 7*
-relation between vectors, 9, 12, 19,
209
first scalar, 147*, 149 1'13 of a dyadic, 147*f, 149 second scalar, 149*, 173 third scalar, 149*, 17.3 vector, 147*, 149, 173 Invariant directions, 153f, 156.'
-planes, 155
21, 22
-relation between motors, 73 -vector function, 135* Line integral, 216f, 218f, 222f, 224f Liouville, Formula of, 307 Macduffee, C. C., 161 Magnetic shell, 244 Males and Dupin, Theorem of, 314 Matric algebra, 169f Matrix, 167, 169f, 203, 334, 336 Mean center, 20*
Inverse square law, 134 Involute of a catenary, 103 of a circle, 103, 131 of a plane curve, 102* of a space curve, 131 * Irrotational vector, 198*f
-curvature (J), 222, 286*, 287, 292,
J=
-value theorem (harmonic
392 192*
J (mean curvature), 286* Jacobian, 190*f, 264, 347 Joachimsthal, Theorem of, ;305
K (total curvature), 286* k (normal curvature), 284* K (curvature), 92*, 96 Kelvin's Theorem, on circu' ation, 267
on minimum energy, 281 Kepler's Laws, 134
func-
tions), 240 Menelaus, Theorem of, 13 Metric, 339* Metric quadratic form, 339*
-tensor, 349, 366 Meusnier's Theorem, 285 Minimal surface, 316*f Minimum equation (of a dyadic), 161*
Mises, von, R., 68, 70 Mobius strip, 218*
436
INDEX
Moment, of inertia, 159* of momentum, 176* of a motor, about an axis, 77*f about a point, 63*f of a vector, about an axis, 55*f about a point, 55 *f Motion, of a fluid, 260f, 263f irrotational, 267 line of, 267, 268 plane, 270f steady, 268f of a particle, 108f circular, 110 relative, 110f, 117, 123f under gravity, 124 uniformly accelerated, 110 of a rigid body, 114f, 127 Motor, 63, 65* acceleration, 127* axis of, 66 *
derivative of, 126* force, 77* pitch of, 66* proper, 66* velocity, 117*, 120, 128 Motor identities, 73f
-product, 70*
-sum, 67 Moving trihedral, 92*, 122, 128, 283, 312 m-Vector, 370*
Nabla (V), 184*, 363, 421 Nonion form of a dyadic, 167*f Norm, of a quaternion, 407* Normal, principal, 92* Normal derivative, 180*
-form of a dyadic, 162f -plane, 133 -system of lines, 232, 313*f -vector to a surface, 205*, 208, 222, 286, 316, 389 Null line, 78*
-plane, 78* -system, 78*f Operators:
ta()a,418,419 Di = a/axi, 356, 362
Operators:
Dij = DID, - D,Di, 378, 384 V (grad), 184, 215, 363 V2, 185
Vh, 362, 365
Dij = vivj
Vjvi, 384 V. (Grad), 206 q( )q-', 419 Orbit, of a planet, 134 Orthogonal coordinates, 194*f
-group, 387, 396 -surfaces (triple system), 305 -trajectories, 231, 300 -transformations, 336*, 386, 387 -triple of unit vectors, 26*, 48, 194, 336
Outer product, of vectors, 354* P, Q, R, 294* Parallel displacement, 312, 376f -curves, 94 * Parallelogram law, 3 Parametric equations: of a circular helix, 108 of a conoid, 321, 323 of an ellipsoid, 322 of a general curve, 88 of a general surface, 203 of a hyperboloid, 322 of a paraboloid, 322 of a right helicoid, 318 of a surface, of revolution, 305, 322 of translation, 322 Particle, falling, 124 Pascal's Theorem, 169 Path curves, 377* Permutation symbols, 329, 346 -tensor, 350 * Peterson and Morley, Theorem of, 74 Pfaff's Problem, 230f Pitch, of a motor, 66* Plane, diametral, 176 equation of, 51 normal, 133 osculating, 98*, 133 polar, 176 radical, 62 rectifying, 133 Planet, motion of, 134
INDEX PI(icker coordinates, , Poinsot's Theorem, 1; Point of division, 8*i
'.
Poisson's Equation, 2.t"' Polar form, 337 -triangles, 45, 420 Polygon, plane, 39 spherical, 416, 420 Polyhedron, 38 Polyhedron Formula )f l:, ler, 311 * Position vector, 5*, 8 Postfactor, 136* Potential, complex, 27'4' * scalar, 200*, 241f vector, 202*, 245 velocity, 267* Prefactor, 136* Pressure, fluid, 255, '& 5i'
Principal axes, of ine 1t.a, .60* of strain, 164* of stress, 256* Principal curvatures, 28(1 Principal directions, of a 1 yadic, 162* on a surface, 287*
Principal normal, 92' -stresses, 256 * Principles, fundamental c f statics, 75f
Product, box, 42 cross, 34 dot, 29 double-dot, 175* indeterminate, 13(. motor, 70* of dual numbers, 64'' of dyadics, 142* of four vectors, 43 of matrices, 170* of quaternions, 404" of tensor componcnl., 1150* of vectors, 410* by numbers, 6* outer, 354* scalar, of motors, 6,s" of vectors, 29*f scalar triple, 41 * vector, 29, 34*f vector triple, 40* Pythagorean Theoroii-., 2 7
437
Quadrangle, complete, 16f Quadratic forms, 337*f, 367 Quadric surfaces, 175, 321, 322 Quadrilateral, complete, 18 Quaternion, conjugate of (Kq), 406 *
norm of (Nq), 407* real, 403 reciprocal of, 409* roots of, 412f scalar part of (Sq), 405 unit, 407*, 412, 414 vector part of (Vq), 405 zero, 403 Quaternion algebra, 403f
-division, 409* -multiplication, 404*f -units (i, j, k), 404*, 408 Quotient law, 399 R, 294 * R, 90*
Radical axis, 62
-plane, 62 Radius, of curvature (p), 94* of torsion (v), 94* Rank of a matrix, 203 Rate of change, local, 259*, 261 of a vector, 121 substantial, 259*, 261 Reciprocal bases, 46*f, 335, 367
-dyadics, 145* -quaternions, 409* -sets of motors, 74*f -sets of vectors, 46*f, 144, 192, 207, 209
Reduced cofactor, 331 * Reflections, 164f, 419
Regular point of a surface, 203* Relative acceleration, 123*
-motion, 117 -scalar, 346* -tensor, 345*f -velocity, 110*f, 123* Relativity, special, 401 Revolving fluid, 262
-unit vector, 90, 91 Ricci identity, 384
-tensor, 384
INDEX
438
Riemannian geometry, 366f -space, 366*f, 377 Rodrigues' construction, 429 Rolling curve, 119 Roots of a quaternion, 412 Rotation (rot f, V x f), 183*f, 194, 195, 263 *
surface (Rot f, V. x f), 207* Rotations in space, 164f, 396, 417f, 420, 421
Ruled surface, 303, 314 Ruling, 303, 314 Scalar, 1 Scalar product: of two motors, 68*f of two vectors, 29*f of three vectors, 41f Shear, components of, 256 Shift formulas, 52 Singular dyadic, 140*
-point of a surface, 203* -quadratic form, 337 Solenoidal vector, 201 *f, 227 Solid angle, 236* Space charges, 245f
Spatial invariants, 209 Speed, 108* Spherical trigonometry, 44f, 416
Static Equilibrium, Principle of, 76 Statics, 75f Step path, 224 Stokes tensor, 372* -Theorem, 220*, 309 Strain, homogeneous, 163 Stream-lines, 262*, 268, 269 Stress dyadic, 253f for a fluid, 255 Sum, of dual numbers, 64* of dyadics, 137* of matrices, 170* of motors, 67* of quaternions, 403* of tensors, 350* of vectors, 3*f Summation convention, 188, 328*f Surface, developable, 303, 304, 314*f minimal, 316*f
Surface, of constant total curvature, 307f, 310 of revolution, 305, 307, 322 of translation, 322 parallel, 325 parametric equations of, 203* quadric, 175, 321, 322 ruled, 303, 314* tangent, 315*, 323 Surface charges, 242f
-divergence (Div f), 207* -gradient (Grad f), 206*f -integral, 218f, 233f -invariants, 209 -normal, 205 *
-rotation (Rot f), Symmetric dyadics, 141*, 156f -tensors, 352 * Symmetry, of affine connection, 357, 361
of curvature tensor, 381 of inertia dyadic, 160 of stress dyadic, 256 T, N, B, 92* t (geodesic torsion), 284 z (torsion), 92*, 96 Tangent vector, unit (T), 90*, 92 Taylor, J. 11., 42 Tensor, 172, 344, 348* absolute, 343f algebra, 350f antisymmetric, 352, 353 contraction of, 351 contravariant, 344 covariant, 344 covariant derivative of, 362*, 365*f curl of, 374 *
curvature, 379, 380*f differential invariants of a, 209f
V f,Vxf,188* f, V. x f, 207* divergence of, 372* dual of, 370* gradient of, 362* Of, grad f, 188* V8f, Grad f, 206* metric, 339, 349, 361, 366 mixed, 344 V.
INDEX Tensor, relative, 345f Iticci, 384* Stokes, 372*f
transformation equations of a, in afline group, 344, 345 in general, 348* Tensor equations, 348
-operations, 350 Tetradic, 172 Theorems egregium (Gauss), 307 Torricclli's Law, 270 Torsion (r), 92* geodesic (t), 284*, 286 radius of (v), 93* Total curvature (K), 286*, 287, 292, 306f, 392
139
Vector, bound, 2* contravariant, 48*, 335 covariant, 48*, 335 Darboux, 93*, 106, 122, 133, 232 dual, 63 *
free, 2* irrotational, 198*f line, 2*, 55, 63, 66 moment, 63* position, 5*, 8 proper, 1 rate of change of, 121
resultant, 63* solenoidal, 201 *f, 227
unit, 2* zero, 1 *, 4
-differential, 197*
Vector addition, 3*
Tractrix, 105*, 308 Transformation of integrals, surface to line, 216, 218f, 221f volume to surface, 233f Transpose, of a matrix, 167 Triadic, 172* Trigonometry, plane, 21 spherical, 21f, 416 Trivector, 353 *
-algebra, 3f, 29f, 354 -equations, 50 -quantity, 1 *
U, V, IV, 194* Umbilical point, 288* Unit dyadic (I), 144*
-quaternion, 407*, 412, 414 Unit vector, 2* along binormal (B), 92* along principal normal (N), 92* along surface normal (n), 205* along tangent (T), 90*, 92 derivative of, 91* radial (R), 90* revolving, 90 transverse (P), 90* Uniformly accelerated motion, 110
Velocity, 108* absolute, 111 angular, 115*, 116*f complex, 272* general components of, 359 of sound, 281 rectangular components of, 110 relative, 110f transfer, 111 * Velocity four-vector, 401
-motor, 117*, 120 Vortex, combined, 281 Vortex lines, 2fi6*, 269 Vorticity, 263*
Wedderburn, J. 11. M., 169 Weingarten, Derivative formulas of, 323, 327, 391
Weight of a tensor, 345* Wind triangle, 111 Zero dual number, 64*
Valence of a tensor, 172*, 344, 345 Veblen, 0., 366, 368, 380 Vector, 1 *
are, 415 base, 24f
-dyadic, 137* -motor, 66* -quaternion, 403* -tensor, 348 -vector, 1*, 2*, 4
