input: How does the book approach the topic of vector spaces?

answers: The book approaches the topic of vector spaces by introducing the postulates of vector spaces, discussing linear dependence, basis, and dimension. It also covers inner product, Euclidean vector space, and the Gram-Schmidt process.

context:
# VECTOR
ANALYSIS

LOUIS BRAND

DOVER PUBLICATIONS, INC.
Mineola, New York
_Bibliographical Note_

This Dover edition, first published in 2006, is an unabridged republication of the work originally published by John Wiley & Sons, Inc., New York, in 1957.

_Library of Congress Cataloging-in-Publication Data_

Brand, Louis, 1885–

Vector analysis / Louis Brand.

p. cm.

Originally published: New York : Wiley, 1957.

Includes index.

eISBN 13: 978-0-486-15484-8

1. Vector analysis. I. Title.

QC20.7.V4B73 2006

515'.63—dc22

2005049527

Manufactured in the United States of America
Dover Publications, Inc., 31 East 2nd Street, Mineola, N.Y. 11501
I dedicate this book to my former teacher and dean,
HERMAN SCHNEIDER

in tribute to the originator of cooperative education at its fiftieth anniversary, 1906–1956:

"JOINING THEORY AND PRACTICE
LINKING EDUCATION AND INDUSTRY
THROUGH KNOWLEDGE AND EXPERIENCE."

( _Inscription at west end of the Herman Schneider Quadrangle, University of Cincinnati._ )

##   **Preface**

A vectorial treatment of differential geometry, mechanics, hydrodynamics, and electrodynamics is now practically standard procedure. The use of vectors not only simplifies and condenses the exposition but also makes mathematical and physical concepts more tangible and easy to grasp. The day is not far distant when vector algebra will be introduced in analytic geometry, thus giving the student the double advantage of an early introduction to the subject and a welcome relief from a multiplicity of formulas to be memorized. The dot and cross product of Gibbs, so intimately involved in all questions of perpendicularity and parallelism, enable one to write the equations of lines and planes at will and to solve all distance problems in the most natural manner. Even the simple distributive law, _λ_ ( **u** \+ **v** ) = _λ_ **u** \+ _λ_ **v** , gives one the power to use the properties of similar triangles without figures and virtually in the dark. And then that tremendous step from two to three dimensions—which usually occurs so late in the course that the student misses it altogether—need never be taken at all. For we _start_ in three dimensions, the world we actually live in, and this on the fine axiom of J. Willard Gibbs: _The whole is simpler than its parts._

Vector analysis has also breached the walls of the calculus and now puts in an appearance in every modern textbook in that subject. The gradient, divergence, and rotation, in their capacity of invariants, form the natural language of spatial science. Avoidance of them is futile, and using long scalar notations every time they occur is wasteful of time and effort. I believe that the introduction of divergence and rotation from the beginning as invariants of a dyadic (the gradient of a vector) is not only the proper procedure but is actually the simplest in the long run. Although it requires a slight knowledge of dyadics, or tensors of valence two, the time is well spent; for the world is full of important tensors that the earnest student might just as well encounter head on. In this book the use of dyadics is held to a drastic minimum; and hence rigid dynamics (with its inertia dyadic) and the theory of elasticity (with its stress and strain tensors) have been regretfully omitted. Besides, this book was designed as a short course to give a beginning student the tools of vector algebra and calculus and a brief glimpse beyond into their manifold applications. Thus the equation for the angular rate of change of a radial unit vector is actually the key to the eccentricities of gyroscopic behavior. But in this book we have stopped short of gyroscopes as being beyond its proper scope. After all, a short book in vector analysis cannot pretend to be in addition a treatise on mathematical physics. But the applications have been developed to that extent so that the uses of the potential function, both scalar and vector, are fully illustrated. Moreover the basic postulates of the sciences dealt with have been brought to the foreground to put their logical structure in sharp relief.

Since the concept of a vector has been greatly generalized in geometry and mathematical physics, the final chapter gives a brief introduction to abstract vector spaces, together with the ideas of linear dependence, basis, and dimension. As this chapter is necessarily abstract, an especial effort was made to keep the exposition as simple and clear as possible.

In my more comprehensive book on _Vector and Tensor Analysis_ , I quoted the eloquent words of Bertrand Russell on mathematics as a science and as an art. They deserve to be quoted again:

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 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.

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 (8.3) means equation (3) of § 8. Figures are given the number of the article in which they appear followed by _a_ , _b_ , _c_ , ···; thus Fig. 2 _c_ is the third figure of § 2. In footnotes, references to my books, _Vectorial Mechanics, Vector and Tensor Analysis, Advanced Calculus_ ,* are simply made by title.

**Bold-face** type is used in the text to denote vectors or dyadics. The magnitude of a vector **v** is denoted by | **v** | or, on occasion, by _υ_. A displacement (or "arrow" vector) from _A_ to _B_ is written  . Superimposed arrows may also be used to denote vectors in manuscript.

Finally, I wish to express my grateful appreciation for the contribution of the Taft fund toward typing the manuscript, and to extend thanks to my colleagues Professors Jaeger and Restemeyer for their advice and suggestions. I am also indebted to Mrs. Anne Hagedorn for her efficient and painstaking help in the preparation of the manuscript.

LOUIS BRAND

_University of Cincinnati
January_ 1957

*Note: A reprint edition of Louis Brand's _Advanced Calculus_ [0-486-44548-8] is available from Dover Publications, Inc.

##   **Contents**

Chapter 1. VECTOR ALGEBRA

1. Vectors

2. Addition of Vectors

3. Subtraction of Vectors

4. Multiplication of Vectors by Numbers

5. Point of Division

6. Linear Relations Independent of the Origin

7. Centroid

8. Rectangular Components

9. Products of Two Vectors

10. Scalar Product

11. Vector Product

12. Vector Triple Product

13. Scalar Triple Product

14. Products of Four Vectors

15. Reciprocal Bases

16. General Bases

17. Equations of Line and Plane

18. Summary of Vector Algebra

Chapter 2. LINE VECTORS

19. Moment of a Line Vector

20. Addition of Line Vectors

21. Couples

22. Coplanar Line Vectors

23. Statics

Chapter 3. VECTOR FUNCTIONS OF ONE VARIABLE

24. Derivative of a Vector

25. Derivatives of Sums and Products

26. Curves

27. Tangent Vector

28. Unit Tangent Vector

29. Frenet's Formulas

30. Curvature and Torsion

31. Plane Curves

32. Kinematics of a Particle

33. Relative Velocity

34. Angular Velocity Vector

35. Kinematics of a Rigid Body

Chapter 4. DIFFERENTIAL INVARIANTS

36. Surfaces

37. Directional Derivative

38. Gradient of a Vector

39. Invariants of a Dyadic

40. Divergence and Rotation

41. Product Formulas

42. Curvilinear Coordinates

Chapter 5. INTEGRAL THEOREMS

43. Line Integrals

44. Line Integrals Independent of the Path

45. Irrotational Vectors

46. Simply Connected Regions

47. Green's Theorem in the Plane

48. Vector Forms of Green's Theorem

49. Surface Integrals

50. Stokes' Theorem

51. Generalized Stokes' Theorem

52. Divergence Theorem

53. Generalized Divergence Theorem

54. Solenoidal Vectors

55. Green's Identities

56. Harmonic Functions

57. Retrospect

Chapter 6. DYNAMICS

58. Dynamics of a Particle

59. Equations of Motion

60. Work and Energy

61. The Solar System

62. Systems of Particles

63. Body Revolving about a Fixed Axis

Chapter 7. FLUID MECHANICS

64. Perfect Fluids

65. Equation of Continuity

66. Euler's Equation

67. Fluid in Equilibrium

68. Vorticity

69. Circulation Theorem

70. Irrotational Motion

71. Steady Motion

72. Plane Motion

73. Fluid Mechanics

Chapter 8. ELECTRODYNAMICS

74. Maxwell's Equations

75. Integral Form of Maxwell's Equations

76. Ohm's Law

77. Units and Dimensions

78. Wave Equation

79. Coulomb's Law

80. Poynting's Vector

81. Potentials

82. Electrostatics

83. Conductors

84. Capacitance

85. Magnetostatics

86. Magnetostatic Potential

87. Forces on Charges and Currents

Chapter 9. VECTOR SPACES

88. Postulates

89. Linear Dependence

90. Subspaces

91. Bases and Dimension

92. Ordered Sets

93. Inner Product

94. Euclidean Vector Space

95. Gram–Schmidt Process

96. Reciprocal Bases

97. Hilbert Space

98. Linear Equations

APPENDIX

Quantities, Dimensions, and Units

Supplementary Problems

Answers to Problems

Index
Science and the applications of science are united together as the tree and its fruit.

PASTEUR
VECTOR
ANALYSIS

##   **Chapter 1**

## **Vector Algebra**

**1. Vectors.** _A vector is a directed line segment_. A vector from the point _A_ to the point _B_ is denoted by  . The vector   has the _length AB_ and the _direction_ from _A_ to _B_. Besides _proper vectors_ , which have a definite length and direction, we shall introduce the _zero vector_   which has zero length but no definite direction.

A _scalar_ is a number whose value is the same in all systems of reference. Thus the length of a vector is a scalar; but the direction cosines of a vector are not scalars as their values depend on the choice of coordinate axes.

In addition to the foregoing notation, in which we specify a vector by giving its end points, we shall also use single letters in boldface type (as **u** , **v** , **w** ) to denote vectors. The zero vector is then written **0** in bold type. The length of the vector **u** is denoted by  **u**  ; and  **0**  = 0. A vector of length 1 is called a _unit vector_. In manuscript, a vector may be denoted by a single letter with an arrow above it.

We shall now construct an algebra of vectors by defining equality, addition, and multiplication of vectors by real numbers or _scalars_.

_Two vectors are said to be equal when they have the same length and direction_. Thus a vector   is equal to all vectors obtained from   by a parallel displacement or _translation_. The initial point of **u** may be chosen at pleasure, say _P_ ; then   when the arrow   has the same length as   and is parallel to   in the same sense. In Fig. 2 _b_ for example, the vectors forming the opposite sides of a parallelogram are equal and may be represented by the same letter; thus

Vectors that conform to the above definition of equality are said to be _free_. But in some applications other definitions are advisable. Thus the forces acting on a rigid body may be regarded as vectors which are restricted to lie on a given line, their _line of action_. Such force vectors are not free; for they may only be shifted along a fixed line since other translations alter their dynamic effect. We shall call a vector confined to a definite line of action a _line vector_. Two line vectors are equal when and only when they have the same length and the same direction and lie on the same line.

Vectors that have a fixed position in space are called _bound vectors_. Thus the forces acting on a deformable body may be regarded as bound vectors, for any translation of such a force alters its effect on the body. Two bound vectors are equal when they have the same length, direction, and position in space.

In this chapter all vectors are regarded as _free_ ; and we begin our vector algebra by defining the addition of free vectors and their multiplication by numbers.

**2. Addition of Vectors.** A rectilinear displacement, or _translation_ , from _A_ to _B_ may be represented as a vector  . If a particle is given two displacements, 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 suggests the relation

We shall regard this equation as the _definition_ of vector addition; thus the sum of two vectors **u** , **v** , (Fig. 2 _a_) is defined by the

TRIANGLE CONSTRUCTION. _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**.

Fig. 2 _a_. Triangle construction.

Since any side of a triangle is less than the sum of the other two sides,

the equal sign holding only when **u** and **v** have the same direction.

_Vector addition is commutative:_

This is apparent from Fig. 2 _b_; for

_Vector addition is associative:_

Fig. 2 _b_. Commutative law.

Fig. 2 _c_. Associative law

This is apparent from Fig. 2 _c_; for

Since both sums in (4) are the same, they are simply written **u** \+ **v** \+ **w**.

From the commutative and associative laws we readily deduce the following

THEOREM. _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 directed segments represent those 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. If _A, B_ , _C_ , ···, _G, H_ are the successive vertices of the broken line (Fig. 2 _d_), then

Fig. 2 _d_. Vector sum.

If in constructing 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 _H_ coincides with _A_ in (5), we write

If  , we have

**3. Subtraction of Vectors.** The sum of two vectors is zero when and only when they have the same length and opposite directions:

If  , it is natural to write   in order that the characteristic equation for negatives

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** holds for vectors as well as for numbers.

The difference **u** − **v** of two vectors is defined by the equation

On adding − **v** to both sides, and using (1), we have

_subtracting a vector is the same as adding its negative_. The construction of **u** − **v** satisfying (2) is shown in Fig. 3 _a_.

Fig. 3 _a_. Vector subtraction.

Fig. 3 _b_. Difference of position vectors.

With respect to an origin _O_ any point _P_ in space is determined by its _position vector_  . Any vector   may be expressed as the difference of position vectors to its end points (Fig. 3 _b_.): thus

and, from (3),

**4. Multiplication of Vectors by Numbers.** The vector **u** \+ **u** is naturally denoted by 2 **u** ; similarly, we write − **u** \+ (− **u** ) = − 2 **u**. Thus 2 **u** and − 2 **u** denote vectors twice as long as **u** ; 2 **u** has the same direction as **u** , − 2 **u** the opposite direction. This notation is generalized as follows:

_The product of a vector_ **u** _by a number λ_ ( _written λ_ **u** or **u** _λ_ ) _is defined as a vector_  _λ_  _times as long as_ **u** _and having the same direction as_ **u** , _or the opposite, according as λ is positive or negative_.

In accordance with this definition

and

Moreover since the corresponding sides of similar triangles are proportional,

This is clear from Fig. 4 _a_, in which _α_ > 0. The distributive law (3) is a potent tool in obtaining geometric information by algebraic means.

The developments thus far show that:

_As far as addition, subtraction, and multiplication by numbers are concerned, vectors obey the laws of ordinary algebra_.

Fig. 4 _a_. Distributive law.

Fig. 4 _b_. Basis in a plane.

If **u** and **v** are parallel vectors, they have the same or opposite directions and **v** is a numerical multiple of **u** :

If **u** , **v** , **r** are coplanar vectors ( **u** and **v** not parallel), we can write

For, if we construct a parallelogram on **r** as diagonal with sides parallel to **u** and **v** , we have (Fig. 4 _b_)

If **u** , **v** , **w** , **r** are vectors in space ( **u** , **v** , **w** not coplanar), we can write

For, if we construct a parallelepiped on **r** as diagonal with sides parallel to **u** , **v** , **w** , we have (Fig. 4 _c_)

Fig. 4 _c_. Basis in space.

### **PROBLEMS**

**1.** Show that  , where _M_ is the mid-point of _BC_.

**2.** Show that  , where _M_ and _N_ are the mid-points of _AC_ and _BD_.

**3.**   and   are _line vectors_ (§1) acting along two intersecting lines. Show that   is the same for all positions of   and  .

**4.** If _A_ , _B_ , _C_ , _D_ are any four points, prove that

where _P_ and _Q_ are the mid-points of _AC_ and _BD_.

**5. Point of Division.** If _A, B, C_ are points of a straight line, C is said to divide the segment _AB_ in the ratio _λ_ if

Evidently _λ_ is positive or negative according as _C_ lies within or without the segment _AB_ ; to the left of _A_ , _λ_ varies between 0 and −1, to the right of _B_ , _λ_ varies between − ∞ and −1. As _C_ describes the entire line in the sense _AB_ , _λ_ varies as follows:

Thus _λ_ = 0, _λ_ = ± ∞, _λ_ = −1 when _C_ is at _A_ , at _B_ , or at the infinitely distant "point" of the line.

To find the position vector of _C_ relative to an origin _O_ (Fig. 5 _a_), write (1) in the form

then

Fig. 5 _a_. Point of division.

We shall denote the position vectors of _A_ , _B_ , _C_ by **a** , **b** , **c** ; then, if _λ_ = _β_ / _α_ , (2) becomes

This equation states that C divides _AB_ in the ratio _β_ / _α_. Thus the midpoint of _AB_ ( _α_ = _β_ = 1) has the position vector  .

From (3) we now deduce

THEOREM 1. _Three distinct points A, B, C will lie on a line when and only when there exist three nonzero numbers α, β, γ, such that_

_Proof_. If _A, B, C_ are collinear, _C_ divides _AB_ in some ratio _β_ / _α_ which is not 0, ∞, or −1; hence _α_ , _β_ and _α_ \+ _β_ are not zero. Thus (3) holds good; and on putting _γ_ = − ( _α_ \+ _β_ ) we obtain (4). Conversely, equations (4) give (3) on dividing

hence _C_ lies on the line _AB_.

The symmetrical relations (4) disclose how each point divides the segment formed by the other two. Thus (4) states that _A, B, C_ divide _BC, CA, AB_ , respectively, in the ratios _γ/β α/γ, β/α_ , whose product is 1.

THEOREM 2. _Four points A, B, C, D, no three of which are collinear, will lie in a plane when and only when there exist four nonzero numbers α, β, γ, δ, such that_

_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

where _λ, λ′_ are neither 0 nor −1. In both cases **a** , **b** , **c** , **d** are connected by a linear relation of the form (5).

Conversely let us assume that equations (5) hold good. If _α_ \+ _β_ = 0 (and hence _γ_ \+ _δ_ = 0), we have

and the lines _AB_ and _CD_ are parallel. If _α_ \+ _β_  0 (and hence _γ_ \+ _δ_  0),

where _P_ is a point common to the lines _AB, CD_ ; for (3) shows that _P_ divides _AB_ and _CD_ in the ratios _β_ / _α_ , _α_ / _γ_. In both cases _A, B, C, D,_ are coplanar.

Note that (5) is packed with information. If _β_ \+ _γ_  0,

and, if _γ_ \+ _α_  0,

The complete quadrangle _ABCD_ (Fig. 5 _b_) has four _vertices_ , and six _sides_ , the lines which join them; and the three pairs of opposite sides meet in the _diagonal points P, Q, R_ given by (6), (7), (8). These equations state how each diagonal point divides the sides on which it lies.

If _QR_ produced cuts _AB_ at _S_ , the points _S_ and _P_ divide _AB_ internally and externally in the numerically equal ratios ± _β_ / _α_. Such points are called _harmonic conjugates_ relative to _AB_ (see Ex. 1). To show this we eliminate **d** from

Fig. 5 _b_. Complete quadrangle.

by subtraction. We thus obtain

This states that _S_ divides _AB_ in the ratio − _β_ / _α_.

If _PR_ produced cuts _AD_ at _T, T_ and _Q_ divide _AD_ internally and externally in numerical equal ratios ± _δ_ / _α_. Thus _T_ and _Q_ are harmonic conjugates relative to _AD_.

**Example 1.** Two points _C, D_ divide the segment _AB_ internally and externally in numerically equal ratios ± _λ_ are said to be _harmonic conjugates_ relative to _A, B_. Then the points _A, B_ are also harmonic conjugates relative to _C, D_ ; for, if we solve

for **a** and **b** , we find that _A, B_ divide the segment _CD_ in the ratios ±(1 − _λ_ )/(1 + _λ_ ).

When _C_ is any point on the line _AB_ , the following construction will locate its harmonic conjugate _D_ relative to _AB_. Take a point _P_ not on the line _AB_ , and draw _PA, PB, PC_ (Fig. 5 _c_). Choose _Q_ at pleasure on _PC_ , draw the lines _AQ_ and _BQ_ , and let them cut _PB_ at _X, PA_ at _Y_. Then _XY_ produced will cut _AB_ at _D_. This follows from the properties of the complete quadrangle _ABXY_ deduced above.

Fig. 5 _c_. Construction of harmonic conjugates.

Fig. 5 _d_. Trisection of diagonal.

**Example 2.** In the parallelogram _ABCD, E_ and _F_ are the middle points of the sides _AB, BC_ (Fig. 5 _d_). How do the lines _DE_ and _DF_ divide the diagonal _AC?_

The hypotheses of our problem are expressed by the equations:

To find the point _X_ when _DE_ cuts _AC_ , eliminate **b** from the first and second equations. Thus

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 (3) now shows that _X_ divides _DE_ in the ratio 2/1, _AC_ in the ratio 1/2.

To find the point _Y_ where DF cuts _AC_ , eliminate **b** from the first and third equations. Thus

Hence _Y_ divides _DF_ in the ratio 2/1, _AC_ in the ratio 2/1.

**Example 3. The Trapezoid.** If _ABCD_ is a trapezoid with _AB_ parallel to _DC_ (Fig. 5 _e_), then

and hence

These equations may be written

where _P_ is the point where the diagonals _BD, AC_ meet, and _Q_ is the point where the sides _BC, AD_ meet. Evidently _P_ divides the diagonals in the same ratio _λ_ , and _Q_ divides the sides (externally) in the ratio − _λ_. Since

Fig. 5 _e_. Trapezoid.

_PQ_ produced will bisect _AB_ at _R_ ; for  .

In particular, if _λ_ = 1, the trapezoid becomes a parallelogram. The diagonals then bisect each other at _P_ , while _Q_ recedes to infinity.

### **PROBLEMS**

**1.** From (2.2) prove that  **a** − **b**      **a**  −  **b**  .

**2.** If _P_ lies on the line _ABC_ and divides _AB, BC, CA_ in the ratios _λ, μ, ν,_ show that _λμν_ = − 1.

**3.** If _ABC_ is any triangle and _L, M, N_ are the mid-points of its sides, show that, for any choice of _O_ ,

**4.** If  ,  , in what ratio does the point _P_ in which _AB_ and _A′B′_ intersect divide these segments?

**5.** If _ABCD_ is a quadrilateral (plane or skew), show that the mid-points of its sides are the vertices of a parallelogram whose center is ( **a** \+ **b** \+ **c** \+ **d** )/4.

**6.** In the triangle _ABC_ (Fig. 5 _f_) prove that the segments _AD, BE_ cannot bisect each other.

Fig. 5 _f_. Nonbisecting segments.

Fig. 5 _g_. Theorem of Ceva.

**7.** In a plane quadrilateral _ABCD_ the point _Q_ 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 _P_ , in which the sides _AB, CD_ intersect, divide these segments?

**8. Theorem of Ceva.** The point _D_ lies in the plane of the triangle _ABC_ (Fig. 5 _g_); then (Theorem 2)

If _AD, BD, CD_ meet the opposite sides in _A′, B′, C′,_ respectively, prove that _A′, B′, C′_ , divide _BC, CA, AB_ in the ratios _γ/β, α/γ, β/α_ whose product is 1.

**9.** In Prob. 8, prove that _A′, B′, C_ divide _DA, DB, DC_ in the ratios _α/δ, β/δ, γ/δ,_ whose sum is −1.

**10.** Justify _ab initio_ the construction for harmonic conjugates given in Ex. 1, Fig. 5 _c_.

**11.** The complete quadrangle _ABCD_ (Fig. 5 _b_) has the diagonals _AC, BD, PQ_. Prove that their mid-points  ,  ,   are collinear ( _Newton line_ ).

From [(6), (7), and (8),

hence show that

**12. Desargues' Theorem.** Let _ABC_ and _A′B′C′_ be two triangles, in the same or different planes; then, if the lines _AA′, BB′, CC′,_ meet at _S_ , the three pairs of corresponding sides ( _BC, B′C′_ ), ( _CA, C′A′_ ), ( _AB, A′B′_ ), intersect in collinear points _P, Q, R_ (Fig. 5 _h_).

Fig. 5 _h_. Desargues' theorem.

Fig. 5 _i_. Theorem of Menelaus.

[In view of (5.4) we may write

Find **p** , **q** , **r** , and show that

**13. Theorem of Menelaus.** A line cuts the sides _BC, CA, AB_ of the triangle _ABC_ in the points _P, Q, R_. If _P_ divides _BC_ in the ratio _γ_ / _β_ , and _Q_ divides _CA_ in the ratio − _α_ / _γ_ , prove that _R_ divides _AB_ in the ratio − _β_ / _α_. The product of the division ratios is thus −1 (Fig. 5 _i_).

**14.** Prove the converse of the theorem of Menelaus: If _P, Q, R_ divide _BC, CA, AB_ in the ratios _γ_ / _β_ , _α_ / _γ_ , − _β_ / _α_ whose product is −1, the points _P, Q, R_ are collinear.

**15. Pascal's Theorem.** The points _A, B, C_ and _A′, B′, C′,_ respectively, lie on two lines that intersect at _O_. If _BC′_ and _CB′, CA′_ and _AC′, AB′_ and _BA′_ cut in the points _P, Q, R_ , show that _P, Q, R_ are collinear.

[If **a** = _α_ **e** , **a** ′ = _α_ ′ **e** ′, etc., where **e** , **e** ′ are unit vectors along the lines, show that

**16.** _ABCD_ are the vertices of a parallelogram traversed counterclockwise. Take points _P_ and _Q_ on the sides _AB_ and _CD_ , respectively. If _PD_ meets _QA_ in _R, PC_ meets _QB_ in S, prove that the line _RS_ passes through _T_ , the intersection of the diagonals.

**6. Linear Relations Independent of the Origin.** We have seen that the position vectors of collinear or coplanar points satisfy a 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_

_connecting the position vectors of the points P_ 1, _P_ 2, ···, _P n, will be independent of the position of the origin O when and only when the sum of the scalar coefficients is zero:_

_Proof_. Change from _O_ to a new origin _O′_. Writing  ,  ,  , we have  ; hence (1) becomes

This equation will have the same form as (1) when and only when (2) is satisfied.

For two, three, and four points, such linear relations 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?

**7. Centroid.** If a point _P_ is associated with a real number _m_ , we speak of _P_ as a _weighted point_ and denote it by the symbol _mP_. 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 weighted points.

_The_ centroid _of n weighted points m iPi_ ( _i_ = 1, 2, ···, _n_ ) _for which_ Σ _m i_ 0 _is defined as the point P_ * _for which_

The case Σ _m i_ = 0 must be excluded; for then (1) would hold for any choice of _P_ * (§6).

We first show that, when Σ _m i_ 0, there is one and only one point _P_ * that satisfies (1). For, with an arbitrary origin _O_ , (1) can be written

and hence

This relation, independent of the origin _O_ , fixes the position of _P_ * relative to _O_. Thus _P_ * is _unique_ , and its position vector relative to any origin is

In particular, if _O_ = _P_ *, (3) reduces to (1). Thus relative to any origin, _the centroid of a set of weighted points has a position vector that equals the weighted average of the given set of position vectors_.

The centroid of the points _m iPi_ is not altered when the numbers _m i_ are replaced by another set _cm i_ proportional to them; for in (3) the constant _c_ may be canceled from numerator and denominator. In particular, if the numbers _m i_ are all equal, we may replace all _m i_ by 1. The centroid of _n_ equally weighted points is called their _mean center_ and is given by

If the set of points _m iPi_ has a subset whose weighted centroid is _m′P′_ , we may write (3) in the form

where Σ′, Σ″ denote summations over the subset and the remaining points, respectively. This equation shows that _P_ * is also the centroid of the point _m′P′_ , and the points _m iPi_ not included in the subset. Thus, in finding centroids, _any subset of points may be replaced by its weighted centroid_.

**Example 1.** The centroid of two points _αA_ , _βB_ is given by

hence _P_ * divides _AB_ in the inverse ratio _β_ / _α_. of their weights. In particular, if _α_ = _β_ , _P_ * is the mid-point of _AB_. _The mean center of two points lies midway between them_.

**Example 2.** The mean center _P_ * of three points _A_ , _B_ , _C_ is given by

Let _L_ , _M_ , _N_ denote the mid-points of _BC_ , _CA_ , _AB_ (Fig. 7 _a_); then

hence the segments _AL_ , _BM_ , _CN_ are all divided by _P_ * in the ratio 2/1. If _A_ , _B_ , _C_ are not collinear, they are the vertices of a triangle with _AL_ , _BM_ , _CN_ as medians. _The medians of a triangle meet at the mean center of its vertices and are all divided there in the ratio of_ 2/1.

Fig. 7 _a_. Mean center of three points.

Fig. 7 _b_. Mean center of four points.

**Example** 3. The mean center _P_ * of four points _A_ , _B_ , _C_ , _D_ is given by

Let _L_ 1, _L_ 2; _M_ 1 _M_ 2; _N_ 1, _N_ 2 be the mid-points of _AB_ , _CD_ ; _BC_ , _DA_ ; _AC_ , _BD_ , respectively (Fig. 7 _b_). Then

hence the segments _L_ 1 _L_ 2, _M_ 1 _M_ 2, _N_ 1 _N_ 2 are all divided in the ratio 1/1 (bisected) by _P_ *. _The bisectors of the opposite sides of the quadrangle ABCD_ ( _plane or skew_ ) _meet in the mean center of its vertices and are all bisected there_.

**Example** 4. If the points _A_ ′, _B_ ′, _C_ ′ divide the sides _BC_ , _CA_ , _AB_ of the triangle _ABC_ in the ratios _γ_ / _β_ , _α_ / _γ_ , _β_ / _α_ whose product is 1, the lines _AA_ ′, _BB_ ′, _CC_ ′ meet in a point, the centroid _P_ * of _αA_ , _βB_ , _γC_.

Since _A_ ′ is the centroid of _βB_ and _γC_ (Ex. 1),

Hence _P_ * lies on _AA_ ′ and, similarly, on _BB_ ′ and _CC_ ′. We have thus proved the converse of the theorem of Ceva (Prob. 5.8).

We can now answer the question raised at the end of §6. _A set of n weighted points m iP_ _i_ ( _m i_ 0) _which satisfy a linear relation_

_independent of the origin, has the intrinsic property that any point of the set is the centroid of the remaining weighted points_.

_Proof_. On dividing

by

we obtain

Thus _P_ 1 is the centroid of _m_ 2 _P_ 2, ···, _m nPn_.

A set of _n_ points having this property is readily constructed. Take any set of _n_ − 1 weighted points _m iPi_, and adjoin to the set their centroid _P n_ with the weight

Then, since

the set of _n_ weighted points _m iPi_ satisfies (5). We shall call such sets of points _self-centroidal_.

Theorem 5.1 shows that nonzero weights may be assigned to any three collinear points _A_ , _B_ , _C_ so that they form a self-centroidal set:

Theorem 5.2 shows that nonzero weights may be assigned to any four coplanar points _A_ , _B_ , _C_ , _D_ , no three of which are collinear, so that they form a self-centroidal set:

In both cases the _ratios_ of the weights are determined.

As a natural extension of these results we have the

THEOREM. _If A_ , _B_ , _C_ , _D_ , _E are five points, no four of which are coplanar, nonzero weights may be assigned to them so that they form a self-centroidal set_ :

_The_ ratios _of weights are uniquely determined by the relative positions of the points_.

_Proof_. We note first that, if any weight is zero, four points are coplanar contrary to hypothesis. Since

are not coplanar, we have a linear relation between these vectors (§4), say

With _ε_ = −( _α_ \+ _β_ \+ _γ_ \+ _δ_ ), this is a relation of the form (8).

Now let _α_ ′, _β_ ′, _γ_ ′, _δ_ ′, _ε_ ′ be a second set of weights, for which

Multiply (8) by _ε_ ′, (9) by _ε_ , and subtract. Then

thus _A_ , _B_ , _C_ , _D_ will be coplanar unless all the coefficients are zero. Since _A_ , _B_ , _C_ , _D_ are _not_ coplanar by hypothesis, we must have

### **PROBLEMS**

**1.** Prove that the sum of _n_ vectors

where _A_ * and _B_ * are the mean centers of the points _A i_ and _B i_, respectively.

**2.** Let _A_ , _B_ , _C_ , _D_ be the vertices of a tetrahedron and _P_ * their mean center. If _A_ ′, _B_ ′, _C_ ′, _D_ ′ are the mean centers of the triads _BCD_ , _CDA_ , _DAB_ , _ABC_ , show that _P_ * divides each of the segments _AA_ ′, _BB_ ′, _CC_ ′, _DD_ ′ in the ratio of 3/1.

**3.** Lines drawn through a point _E_ and the vertices _A_ , _B_ , _C_ , _D_ of a tetrahedron cut the planes of the opposite faces in _A_ ′, _B_ ′, _C_ ′, _D_ ′. Show that the sum of the ratios in which these points divide _EA_ , _EB_ , _EC_ , _ED_ is −1. Use [(8).]

**4. Carnot's Transversal Theorem.** A line cuts the sides of a plane polygon _P_ 1 _P_ 2 ··· _P n_ in _n_ distinct points _R_ 1 _R_ 2, ···, _R n_. Show that the product of the ratios in which _R_ 1 divides _P_ 1 _P_ 2, _R_ 2 divides _P_ 2 _P_ 3, ··· _R n_, divides _P nP_1, is (−1) _n_.

The sum of the weighted points on the left is ( _x_ 1 − _x_ _n_ \+ 1) **p** 1. If _x_ 1 − _x_ _n_ \+ 1 0, _P_ 1 will be their centroid; since this is impossible (why?), _x_ _n_ \+ 1 = _x_ 1.]

**5.** A plane cuts the sides of a skew polygon _P_ 1 _P_ 2 ··· _P n_ in _n_ distinct points _R_ 1, _R_ 2, ···, _R n_. Show that the product of the division ratios of Prob. 4 is (−1) _n_.

**8. Rectangular Components.** Let **i, j, k** denote a _dextral_ set of mutually perpendicular unit vectors; then the thumb, index, and middle fingers of the _right hand_ can be extended in the directions of **i, j, k**.† Let **i, j, k,** drawn from a common origin _O_ , give the positive directions along the system of rectangular axes _Oxyz_.

Fig. 8. Rectangular components of a vector.

Any vector **u** may now be expressed in the form

Draw   from the origin, and let  ,  ,   be the orthogonal projections of   on the axes. Then (Fig. 8)

or, since  ,  ,   are numerical multiples of **i, j, k** , we obtain   in the form (1). The numbers _u_ 1, _u_ 2, _u_ 3, are called the _rectangular components_ of **u**. Evidently _u_ 1 is the length _OP_ 1 taken positive or negative according as   has the direction of **i** of − **i** ; and, from the right triangle _OP_ 1 _P_ , _u_ 1 = | **u** | cos ( **i** , **u** ) where ( **i** , **u** ) is the angle between **i** and **u**. Thus the rectangular coordinates of a vector are the products of its length by its direction cosines:

From the Pythogorean theorem,

If we substitute from (2) into (3), we obtain the relation

between the direction cosines of any vector.

The _rectangular coordinates_ of any point _P_ are defined as the rectangular components of its position vector  . Thus, if

_P_ has the rectangular coordinates _x_ , _y_ , _z_. Since,  , this gives

_The components of_   are found by subtracting the coordinates of _P_ 1 _from the corresponding coordinates of P_ 2.

From (1) we have

We shall sometimes specify vectors by writing their components in brackets: **u** =  _u_ 1, _u_ 2, _u_ 3]. In this notation, [(7) and (8) become

and, in particular,

If **u** is a vector in the _xy_ -plane and _θ_ is the angle ( **i** , **u** ), then

hence the direction cosines of _u_ are

Thus **u** has the components:

and

**Example. Addition Theorems for the Sine and Cosine.** Let **a** and **b** be two unit vectors such that the angles ( **i** , **a** ) = _α_ , ( **i** , **b** ) = _α_ \+ _β_ ; then

If **b** is referred to rectangular axes along

we have also

On equating the components of **b** in (i) and (ii), we obtain

In particular, when  , we have

since cos _π_ = −1, sin _π_ = 0; therefore

These addition theorems, deduced without benefit of a figure, hold for all values of _α_ and _β_ , positive or negative.

**9. 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).† It is an offshoot of the _algebra of quaternions_ , adapted to the uses of geometry and physics. See Brand's _Vector and Tensor Analysis_ , Chapter X, for a brief account of quaternion algebra and its applications; in §184 it is shown how the scalar and vector products originated.

**10. 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:_

Here ( **u** , **v** ) denotes the smallest angle, taken positive, between **u** and **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,

When **u** and **v** are parallel,

according as the vectors have the same or opposite directions. The "square" of **u** is **u** · **u** = | **u** |2.

From (1) we see that

The last result is obvious when _α_ and _β_ are positive numbers; the other cases then follow from the equations preceding.

The definition (1) shows that _scalar multiplication is commutative_ :

As ( **u** · **v** ) · **w** is not defined (the dot is only used between _vectors_ ), the associative law need not be considered. We observe, however, that in general ( **u** · **v** ) **w**  **u** ( **v** · **w** ).

_Scalar multiplication is distributive with respect to addition_ :

_Proof_. If _l_ is a directed line whose sense is given by the _unit_ vector **e** , the orthogonal projection of a vector **u** upon _l_ (proj _l_ **u** ) is a numerical multiple of **e**. This number is called the _component_ of **u** upon _l_ (comp _l_ **u** ); its defining equation is therefore

We compute comp _l_ **u** as in §8: namely

Fig. 10 _a_. Projection is distributive

or, since | **e** | = 1,

Moreover, since (Fig. 10 _a_)

(6) shows that also

Now, if we write **w** = | **w** | **e** ,

By repeated applications of (5), we may expand the scalar product of two vector sums, just as in ordinary algebra. For example

If **c**  **0** in the equation,

we can conclude _either_ that **a** − **b** = **0** _or_ that **a** − **b** is perpendicular to **c**. Thus we cannot "cancel" **c** in (11) to obtain **a** = **b** unless we know that **a** − **b** and **c** are not perpendicular.

Since **i** , **j** , **k** are mutually perpendicular unit vectors,

Hence, if we expand the product,

we obtain

_The scalar product of two vectors is equal to the sum of the products of their corresponding rectangular components_.

Fig. 10 _b_. Right angle inscribed in semicircle.

Fig. 10 _c_. Plane triangle.

**Example 1.** The angle ( **u** , **v** ) between two vectors **u** , **v** may be found from (1):

In space, angles are regarded as positive; and ( **u** , **v** ) denotes the smallest angle _θ_ in the interval 0   _θ_   180°.

Thus, when **u** = [1, 2, −2], **v** = [2, 2, 1],

**Example 2.** If **u** \+ **v** \+ **w** = **0** and | **u** | = 3, | **v** | = 5, | **w** | = 7, find the angle ( **u** , **v** ).

Since **u** \+ **v** = − **w** ,

hence

**Example 3.** _An angle inscribed in a semicircle is a right angle_.

_Proof_. Let _P_ be any point on the semicircle with center at _O_ and _AOB_ as diameter (Fig. 10 _b_). Then

hence _AP_ and _BP_ are perpendicular.

**Example 4. The Cosine Law**. In the triangle _ABC_ , let _A_ , _B_ , _C_ denote the interior angles and _a_ , _b_ , _c_ the sides opposite them (Fig. 10 _c_). Then, from **a** \+ **b** \+ **c** = 0, we have

or

for the angle ( **b** , **c** ) = _π_ − _A_. This is the _cosine law_ for plane triangles.

**Example 5.** The identity

shows that the altitudes of a triangle _ABC_ meet in a point _H_ , the _orthocenter_ of the triangle (Fig. 10 _d_); for, if two terms of this equation are zero, the third is likewise.

Similarly, the identity

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,

Fig. 10 _d_. Euler line of a triangle.

On adding these, we obtain

or, on writing   for the position vector of the mean center _G_ of the triangle _ABC_ ,

Since this equation also holds when **a** − **b** is replaced by **b** − **c** and **c** − **a** , we conclude that

for, if **h** \+ 2 **k** − 3 **g**  0, the vector would be perpendicular to the three sides of _ABC_. 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 6. Theorems of Lagrange.** If _P_ * is the centroid of _n_ weighted points _m iPi_ for which Σ _m i_ = _M_ , we have the theorems:

where the left-hand sum is taken over all   combinations of _i_ , _j_ ( _i_  _j_ ).

_Proof of_ (15). Since

Multiply this equation by _m i_, and sum over the _n_ points. Then, on the right the first sum is _M_ ( _OP_ *)2 the second is Σ _m i_( _P_ * _P i_)2, and the third is zero since   by (7.1).

_Proof of_ (16). Since

Multiply this equation by _m imj_, and sum from _i_ , _j_ = 1 to _n_ ; then, on the left,

where the last sum if extended over all distinct combinations ( _ij_ ) in which _i_  _j_ (when _i_ = _j_ , _P iPj_ = 0). On the right, the first and second sums are equal and together contribute

while both factors in the third sum vanish by (7.1). On dividing by 2 we obtain (16).

In the special case when all weights _m_ _i_ = 1, _M_ = _n_. Hence, on dividing (15) by _n_ , we get

This equation holds as _n_ → ∞ and may be applied to continuous point aggregates if we compute the means by integration.

In (16) there are   terms in the left-hand sum, and _n_ terms in the right-hand sum; hence

On passing to the limit _n_ → ∞, 2 _n_ /( _n_ − 1) → 2 and we obtain the interesting formula

This may be applied to continuous point aggregates if we compute the means as integrals. For example, for the points forming a sphere of radius _a_ about _O_ , _P_ * _P_ = _r_ and

**11. Vector Product.** _The vector product of two vectors_ **u** _and_ **v** , _written_ **u** × **v** , _is defined as the vector_ ,

_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 right-handed screw revolved from **u** toward **v** will advance in its nut toward **u** × **v**.

When **u** and **v** are parallel, **e** is not defined; but in this case sin ( **u** , **v** ) = 0 and **u** × **v** = **0**. Moreover, if **u** and **v** are not zero, **u** × **v** = **0** only when sin ( **u** , **v** ) = 0. Hence, for proper vectors,

In particular, **u** × **u** = **0**.

From (1) we see that

The last result is obvious when _α_ and _β_ are positive numbers; the other cases then follow from the equations preceding.

If **u** and **v** are interchanged in (1), the scalar factors are not altered, but **e** is reversed; hence

_Vector multiplication is anticommutative_.

Fig. 11 _a_ Vector product.

Draw **u** and **v** from the point _A_ , and let _p_ be a plane perpendicular to **u** at _A_ (Fig, 11 _a_). Then **u** × **v** may be formed by a sequence of three operations:

(P) Project **v** on _p_ , and obtain **v** ′;

(M) Multiply **v** ′ by | **u** |, and obtain | **u** | **v** ′;

(R) Revolve | **u** | **v** ′ about **u** through +90°.

The resulting vector agrees with **u** × **v** in magnitude; for | **v** ′| = | **v** | sin ( **u** , **v** ), and also in direction (upward in the figure). We indicate this method of forming **u** × **v** by the notation

This means that **v** is _projected_ , and the projection _multiplied_ , and finally _revolved_ as previously described. Now each of these operators is _distributive_ : operating on the sum of two vectors is the same as operating on the vectors separately and adding the results; hence

Thus, from (5),

_Vector multiplication is distributive with respect to addition_.

By repeated applications of (6) we may expand the vector product of two vector sums just as in ordinary algebra, _provided that the order of the factors is not altered_. For example,

If **c**  **0** in the equation

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.

_Vector multiplication is not associative_ ; for we shall see in §12 that, in general, ( **u** × **v** ) × **w**  **u** × ( **v** × **w** ).

Since the unit vectors **i** , **j** , **k** form a _dextral_ orthogonal set, we have the cyclic relations

Hence, if we expand the product,

we obtain

The components of **u** × **v** are the determinants formed by columns 2 and 3, 3 and 1 (not 1 and 3), 1 and 2 of the array  ; hence we may write

For example, if **u** = [2, −3, 5], **v** = [−1, 4, 2], we compute the components of **u** × **v** from the array

Thus **u** × **v** = [−26, −9, 5]. As a check, we verify that **u** × **v** is perpendicular to both **u** and **v** : −52 + 27 + 25 = 0, 26 − 36 + 10 = 0.

A plane area _A_ , whose boundary is traced in a definite sense, corresponds to a _vector area A **e**_ normal to the plane; here **e** is the unit vector in the direction a right-handed screw would move if turned in the given sense. We may now interpret **u** × **v** as a vector area; for the area of the parallelogram whose sides are **u** , **v** is _A_ = | **u** | | **v** | sin ( **u** , **v** ), and its sense of circuit, shown by the arrows in Fig. 11 _b_, corresponds to the direction of **u** × **v** or **e**. Thus, from (1),

Fig, 11 _b_. Vector area.

**Example 1. Distance of a Point from a Plane.** If a plane is given by three of its points _A_ , _B_ , _C_ , we can find the shortest distance _d_ of a point _R_ to the plane directly. _The equation of the plane is not needed_.

First, find a vector normal to the plane, say  , and let **n** be a _unit_ vector in its direction. Then, if **u** is any vector from _R_ to the plane (as  ), _d_ is numerically equal to the component of _u_ on the normal:

The computation is checked by making another choice for **u** , say   or  .

Thus, if the plane is given by

we find that

The distance of the point _R_ (1, −2, 1) from plane is now

To check, compute

**Example 2. Distance of a Point from a Line.** If the line is given by two of its points _A_ , _B_ , we can find the shortest distance _d_ of a point _R_ to the line directly. _The equations of the line are not needed_.

First find from   a _unit_ vector **e** along the line. Then, if **u** is any vector from _R_ to the line (as  ),

The computation is checked by making another choice for **u** , say  .

Thus, if the line is given by _A_ (2, 3, 0), _B_ (−1, 2, 4),

To find the distance of the point _R_ (3, 1, −1) from the line, compute we

To check, compute  .

**Example 3. Distance between Two Lines.** If two nonparallel lines are given by the points _A_ , _B_ and _C_ , _D_ , we can find the shortest distance _d_ between them directly.

First find a vector   which is perpendicular to both lines, and let **n** be a _unit_ vector in its direction. Then, if **u** is any vector from one line to the other (as  ), _d_ is numerically equal to the component of **u** on the common perpendicular:

Thus let the lines _AB_ , _CD_ be given by the points

We have

The shortest distance _d_ between the lines is now

To check, compute  .

**Example 4.** In the triangle _ABC_ (Fig. 11 _c_) the position vectors of _A_ , _B_ , _C_ are **a** , **b** , **c** and the sides are _a_ = | **c** − **b** |, _b_ = | **a** − **c** |, _c_ = | **b** − **a** |; hence

where **n** is a unit normal to the plane _ABC_. But from the distributive law

From (11) we can make a series of deductions.

(i) The vector **a** × **b** \+ **b** × **c** \+ **c** × **a** is normal to the plane _ABC_.

(ii) The area of the triangle _ABC_ is

(iii) On dividing (11) by _abc_ we find that sin _A_ / _a_ equals an expression unaltered by cyclical permutation of its letters; hence

Fig. 11 _c_. Triangle defined by position vectors.

This is the _sine law_ for plane triangles.

**Example 5**. The equations

in which **a** , **b** , **c** , _α_ are given, have the unique solution

From **a** × ( **r** × **b** ) = **a** × **c** we have, from (12.1)

Hence the solution, if one exists, is given by (13). Direct substitution of (13) in (12) verifies the solution.

### **PROBLEMS**

**1.** Find the angle between the vectors:

( _a_ ) [1, 1, 0], [1, 0, 1];

( _b_ ) [2, −1, 3], [−1, 4, 2];

( _c_ ) [1, 1, 0], [−1, −1, 0];

( _d_ ) [1, 1, 1], [1, 1, −1];

( _e_ ) [ _a_ , _b_ , _c_ ], [ _b_ , _c_ , _a_ ]; [ _c_ , _a_ , _b_ ].

**2.** Show that **a** = [2, 3, −1], **b** = [1, 1, 5], **c** = [16, −11, −1] form a dextral set of mutually orthogonal vectors.

**3.** Find the perpendicular distance from _P_ (1,− 2, 1) to the plane _ABC_ , given the points

**4.** Find the shortest distance between the lines _AB_ , _CD_ , given the points

**5.** Rove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

**6.** If the vectors  ,  ,   are equal in length and coplanar, prove that

where _H_ is orthocenter of the triangle _ABC_ (the point in which its altitudes intersect).

[Dot-multiply **a** \+ **b** \+ **c** = **h** by **a** − **b** , **b** − **c** , **c** − **a** in turn.]

**7.** Prove _Schwarz's inequality_ :

the equal sigh holding only when _a_ 1/ _b_ 1 = _a_ 2/ _b_ 2 = _a_ 3/ _b_ 3.

**8.** If _A_ , _B_ , _C_ , _D_ are any four points in space, show that

is independent of _D_.

**9.** In the triangle _ABC_ (opposite sides _a_ , _b_ , _c_ ), the point _R_ divides _BC_ in the ratio _λ_ /1. Prove that

**10.** A face of a polyhedron of area _A_ is said to have the vector area **_An_** , where **n** is an outward unit normal to the face. Show that the four faces of a tetrahedron _ABCD_ have vector areas whose sum is zero.

[If the circuit _ABC_ is counterclockwise viewed from without, the face _ABC_ has the vector area

**11.** If _A_ , _B_ , _C_ , _D_ are the vertices of a square in circuital order, prove that

**12.** If _P_ and _P_ ′ are points on the surface of a sphere of radius _a_ , prove that

**13.** If _r_ is the radius of a sphere circumscribed about a regular tetrahedron of side _a_. prove that _r_ 2 = 3 _a_ 2/8. [Use (10.16′).]

**14.** If _P_ * and _Q_ * are the mean centers of _p_ points _P i_ and _q_ points _Q j_, respectively, prove that

**15.** The points _A_ ′, _B_ ′, _C_ ′ divide the sides _BC_ , _CA_ , _AB_ of the triangle _ABC_ in the ratio 1/2. The pairs of lines ( _AA_ ′ _BB_ ′), ( _BB_ ′, _CC_ ′), ( _CC_ ′, _AA_ ′) intersect at _P_ , _Q_ , _R_ , respectively. Show that the area of the triangle _PQR_ is 1/7 the area of _ABC_.

**16.** Prove that

**17.** Prove that

[The left member equals

in which each brace  0.]

**18.** If _ABC_ is a spherical triangle on unit sphere about _O_ , and  ,  ,   the vector **p** = _x_ **a** \+ _y_ **b** \+ _z_ **c** will pass through any given point _P_ of the sphere for a suitable choice of _x_ , _y_ , _z_. Show that the vectors

will pass through the points _U_ , _V_ , _W_ where the great circles _PA_ , _PB_ , _PC_ cut the sides _BC_ , _CA_ , _AB_ of the triangle; and that

( _Theorem of Ceva_ for spherical triangles).

[From **w** = _x_ **a** \+ _y_ **b** and | **a** | = | **b** | = 1, we have

on taking account of signs.]

**12. Vector Triple Product.** The vector ( **u** × **v** ) × **w** is perpendicular to **u** × **v** and therefore coplanar with **u** and **v** ; hence (§4),

But, since ( **u** × **v** ) × **w** is also perpendicular to **w** ,

All numbers _α_ , _β_ that satisfy this equation must be of the form _α_ = − _λ_ **v** · **w** , _β_ = _λ_ **u** · **w** , where _λ_ is arbitrary. Thus we have

In order to determine _λ_ , we use a special basis in which **i** is collinear with **u** , **j** coplanar with **u** , **v** ; then

On substituting these values, we obtain, after a simple calculation, _λ_ = 1. We therefore have the important expansion formulas,

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** × **v** ) × **w**  **u** × ( **v** × **w** ); for the former is coplanar with **u** and **v** , the latter with **v** and **w**. _Vector multiplication is not associative_.

From (1) we see that the sum of a vector triple product and its two cyclical permutations is zero:

If _l_ is a directed line carrying the unit vector **e** , we may express any vector **u** as the sum of its orthogonal projections on _l_ and on a plane _p_ perpendicular to _l_ :

**13. Scalar Triple Product.** Let the noncoplanar vectors

be drawn from a common origin _O_. Then **u** , **v** , **w** form a _dextral_ or _sinistral_ set according as the circuit around the triangle _UVW, as viewed from O_ , is clockwise or counterclockwise.* In Fig. 13 _a_, **u** , **v** , **w** , form a dextral set; for the circuit _UVW_ is counterclockwise as viewed by the reader and therefore clockwise as viewed from _O_ (which lies behind the plane _UVW_ ). The set **u** , **w** , **v** is sinistral.

Fig. 13 _a_. A dextral set: **u** , **v** , **w**.

Consider now the scalar product of **u** × **v** and **w** , written **u** × **v** · **w** or [ **uvw** ]. No ambiguity can arise from the omission of parentheses for **u** ×( **v** · **w** ) is not defined (the cross is only used between _vectors_ ). We shall also see that positions of the dot and cross are immaterial.

THEOREM. _The product_ **u** × **v** · **w** _is numerically equal to the volume V of a parallelepiped having_ **u** , **v** , **w** _as concurrent edges. Its sign is positive or negative according as_ **u** , **v** , **w** _form a dextral or a sinistral set_.

_Proof_. We may write the vector product **u** × **v** = _A_ **n** , where

is the area of the parallelogram having **u** and **v** as adjacent sides and **n** is a unit vector perpendicular to **u** and **v** pointing as a right-handed screw would advance when turned from **u** to **v**. When the set **u** , **v** , **w** is dextral (as in Fig. 13 _b_) the angle _θ_ = ( **n** , **w** ) is acute, and

where _h_ is the altitude of the box. When the set **u** , **v** , **w** is sinistral, _θ_ is obtuse, cos _θ_ is negative, and **u** × **v** · **w** = − _V_.

Fig. 13 _b_. Box product.

On account of the geometric meaning of the scalar triple product we shall call it the _box product._†

The dextral or sinistral character of a set **u** , **v** , **w** is not altered by a cyclical change in their order; hence

But a dextral set becomes sinistral, and vice versa, when the cyclical order is changed:

Thus, if the set **u** , **v** , **w** is dextral, the products in (1) all equal _V_ , while the products

all equal − _V_.

If **u** , **v** , **w** are proper vectors, _V_ = 0 when and only when the vectors are coplanar (parallel to the same plane). Therefore, for proper vectors,

In particular, a box product containing two parallel vectors is zero; for example, **u** × **v** · **u** = 0.

_The value of a box product is not altered by an interchange of the dot and cross_. For, from (1),

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 (10.3) and (11.3), we have

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** × **v** · **w** when the vectors are referred to the basis **i** , **j** , **k** , we obtain 27 terms of which all but 6 vanish as they contain box products with two or three equal vectors. The remaining 6 terms are those containing

and constitute the expansion of the determinant

This result also follows at once from (11.8).

### **PROBLEMS**

**1.** If **u** = [−1, 0, 2], **v** = [2, 1, −1], **w** = [1, 2, −2], compute

( _a_ ) **u** × v, **v** × **w** , **w** × **u** ;

( _b_ ) **u** × **v** · **w** , **v** × **w** **u** , **w** × **u** · **v** ;

( _c_ ) ( **u** × **v** ) × **w** , ( **v** × **w** ) × **u** , ( **w** × **u** ) × **v** ; and find their sum.

( _d_ ) Verify the results of ( _c_ ) by using (12.1).

**2.** Given the points _A_ (3, 2, 1), _B_ (2, 3, −1), _C_ (−1, 2, 3), find

( _a_ ) the area of the triangle _ABC_ ;

( _b_ ) the distance _p_ of _O_ from the plane _ABC_ ;

( _c_ ) the volume of the tetrahedron _OABC_ ;

( _d_ ) the volume of the box whose edges are _OA_ , _OB_ , _OC_.

**14. Products of Four Vectors.** We first consider the scalar product ( **a** × **b** ) · ( **c** × **d** ). If we interchange the first cross and dot, it becomes **a** · { **b** × ( **c** × **d** )}; and, since

This is _Lagrange's identity_ ; it may also be written

The special case

is also important.

We turn now to the vector product ( **a** × **b** ) × ( **c** × **d** ). If we expand it as the triple product of **a** , **b** , **c** × **d** by (12.1),

But, if we expand it as the triple product of **a** × **b** , **c** , **d** ,

On equating the right-hand members of (3) and (4), we obtain an equation connecting any four vectors:

When **a** , **b** , **c** are noncoplanar, [ **abc** ]  **0** and the formula

gives **d** explicitly in terms of **a** , **b** , **c**. This is an analytic solution of the problem solved geometrically in §4.

**Example. Spherical Trigonometry.** The unit vectors  ,  ,   define a spherical triangle _ABC_ on a unit sphere about _O_ ; and let [ **abc** ] > **0**. Let _α_ , _β γ_ denote the _sides_ (arcs of great circles) opposite the interior (dihedral) angles _A_ , _B_ , _C_ ; and let all sides and angles be < _π_. Now

Here **a** ′, **b** ′, **c** ′ are unit vectors normal to the planes _BOC_ , _COA_ , _AOB_ , respectively (Fig. 14), and include the angles

Thus _α_ ′, _β_ ′, _γ_ ′ are the _exterior_ dihedral angles at _A_ , _B_ , _C_. Moreover

a _positive_ multiple of **a**. Thus we have the equations

Thus there is complete reciprocity between the sets ( **a** , **b** , **c** ) and ( **a** ′, **b** ′, **c** ′), and also between the spherical triangles _ABC_ , _A_ ′ _B_ ′ _C_ ′ that they determine. Whereas _ABC_ has _α_ , _β_ , _γ_ for sides and _α_ ′, _β_ ′, _γ_ ′ for exterior angles, _A_ ′ _B_ ′ _C_ ′ has _α_ ′, _β_ ′, _γ_ ′ for sides and _α_ , _β_ , _γ_ for exterior angles. Since the vertices _A_ ′ _B_ ′ _C_ ′ are the poles of the sides _BC_ , _CA_ , _AB_ , and vice versa, the triangles _ABC_ , _A_ ′ _B_ ′ _C_ ′ are said to be _polar_.

Fig. 14. Spherical triangle.

From (8) and (10) we have

hence, on division, we obtain the _sine law_ for spherical triangles:

From (8) we have

hence, from (7) and (9), we obtain the _cosine law_ for the spherical triangle _ABC_ :

For the polar triangle this becomes

Of course _α_ , _β_ , _γ_ may be cyclically permuted in (12) and (13).

Since _α_ ′, _β_ ′, _γ_ ′ are exterior angles, _α_ ′ = _π_ − _A_ , _β_ ′ = _π_ − _B_ , _γ_ ′ = _π_ − _C;_ hence the sine and cosine laws may be written

In this version of the cosine laws the structural similarity of (12) and (13) is lost.

### **PROBLEMS**

Prove the formulas:

**1.** ( **a** × **b** ) × ( **b** × **c** ) · ( **c** × **a** ) = [ **abc** ]2.

**2.** ( **b** × **c** ) · ( **a** × **d** ) + ( **c** × **a** ) · ( **b** × **d** ) + ( **a** × **b** ) · ( **c** × **d** ) = 0.

**3.** ( **b** × **c** ) × ( **a** × **d** ) + ( **c** × **a** ) × ( **b** × **d** ) + ( **a** × **b** ) × ( **c** × **d** ) = — 2[ **abc** ] **d**.

**4**. ( **a** − **d** ) · ( **b** − **c** ) + ( **b** − **d** ) · ( **c** − **a** ) + ( **c** − **d** ) · ( **a** − **b** ) = 0.

**5.** ( **a** − **d** ) × ( **b** − **c** ) + ( **b** − **d** ) × ( **c** − **a** ) + ( **c** − **d** ) × ( **a** − **b** ) = 2( **a** × **b** ) + ( **b** × **c** \+ **c** × **a** ).

**6.** Prove that ( **a** × **b** ) × **c** = **a** × ( **b** × **c** ) when, and only when, ( **c** × **a** ) × **b** = **0**.

**7.** In the spherical triangle _ABC_ find the interior angles _A_ , _B_ , _C_ when

Check by the sine law.

**8.** If _C_ = _π_ /2 in the spherical triangle _ABC_ , prove that

in which _A_ , _B_ , _C_ are interior angles.

**9.** Prove that an equilateral spherical triangle is equiangular; and that cos _α_ \+ cos _α_ ′ + cos _α_ cos _α_ ′ = 0.

**10.** Prove _Hero's formula_ for the area of a plane triangle with sides _a_ , _b_ , _c_ :

[From **a** \+ **b** \+ **c** = **0** , _c_ 2 = _a_ 2 \+ _b_ 2 \+ 2 **a** · **b** ; now

**15. Reciprocal Bases.** We call a set of three noncoplanar vectors a _basis_ for space of three dimensions; for _all_ vectors may be expressed linearly in terms of them (§4).† Thus, if **a** , **b** , **c** form a basis, [ **abc** ] 0, and we can express any vector **d** as

where the numbers _α_ , _β_ , _γ_ are given explicitly in (14.6).

Now let **e** 1, **e** 2, **e** 3 form a basis in 3-space; then [ **e** 1 **e** 2 **e** 3] 0. We proceed to define a second basis **e** 1, **e** 2, **e** 3 closely related to the first; here the superscripts are not exponents, but mere identification tags.

DEFINITION. Two bases **e** 1, **e** 2, **e** 3 and **e** 1, **e** 2, **e** 3 are said to be _reciprocal_ when they satisfy the nine equations:

By use of the _Kronecker delta_   defined as

equations (1) condense to

Consider the three equations in the first column of (1). The second and third state that **e** 1 is perpendicular to both **e** 2 and **e** 3: that is, parallel to **e** 2 × **e** 3. Hence **e** 1 = _λ_ **e** 2 × **e** 3; and, from the first equation, we have 1 = _λ_ **e** 1 · **e** 2 × **e** 3. We thus obtain

**e** 2 and **e** 3 being derived from **e** 1 by cyclical permutation. From the symmetry of equations (1) in the two sets **e** _i_ , **e** _i_ we have also

Thus, either basis is expressed in terms of the other by precisely the same formulas.

From (4) and (5), we have

on making use of (14.1) and equations (1); hence

an equation that 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 dextral or both sinistral.

On dividing the identity

by [ **e** 1 **e** 2 **e** 3], we obtain the further relation

When a basis and its reciprocal are identical, **e** _i_ = **e** _i_ and equations (3) give

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 [ **e** 1 **e** 2 **e** 3] = 1, sinistral if [ **e** 1 **e** 2 **e** 3] = −1. Thus **i** , **j** , **k** and **i** , **j** , − **k** are typical dextral and sinistral self-reciprocal bases. _All_ such bases may be derived by revolving these types about axes through the origin.

We can now readily solve the problem of expressing a vector **u** in terms of a basis **e** 1, **e** 2, **e** 3. If we write

(the upper indices are not exponents), and dot-multiply this equation by **e** 1, **e** 2, **e** 3 in turn, we find

Similarly if we write

we find

When the basis is self-reciprocal

and we obtain the components of (8.2). For an arbitrary basis, we shall also call the three scalars

The base vectors now are not in general _unit_ vectors (as **i** , **j** , **k** ); but the _formulas_ for general components are analogous to those for rectangular components:

With a self-reciprocal basis the distinctions implied by upper and lower indices vanish and only subscripts need be used.

### **PROBLEMS**

**1.** Find the set of vectors reciprocal to

Find the components of **u** = [1, 2, 3] relative to the bases **e** 1, **e** 2, **e** 3 and **e** 1, **e** 2, **e** 3, and verify that | **u** |2 = _u_ 1 _u_ 1 \+ _u_ 2 _u_ 2 \+ _u_ 3 _u_ 3.

**2.** If **a** , **b** are vectors in the _xy_ plane, **a** × **b** = _λ_ **k**. Prove that the set reciprocal to

**3.** If **e** 1, **e** 2, **e** 3 is a set of mutually orthogonal vectors, show that the reciprocal set is **e** 1/| **e** 1|2, **e** 2/| **e** 2|2, **e** 3/| **e** 3|2.

**4.** If the sets **e** 1 **e** 2, **e** 3 and **e** 1, **e** 2, **e** 3 are reciprocal, show that the following sets are also:

(i) _α_ **e** 1, _β_ **e** 2, _γ_ **e** 3 and **e** 1/ _α_ , **e** 2/ _β_ , **e** 3/ _γ_ ;

(ii) **e** 2 × **e** 3, **e** 3 × **e** 1, **e** 1 × **e** 2 and **e** 2 × **e** 3, **e** 3 × **e** 1, **e** 1 × **e** 2.

**5.** Find the set reciprocal to **a** , **b** , **a** × **b**.

**6.** Solve the equations for **r** if [ **abc** ]  **0** :

**16. General Bases.** With a general basis **e** 1, **e** 2, **e** 3, it is convenient to express vectors in two forms:

To add such vectors or multiply them by numbers, we have

just as in §8.

In order to retain the simplicity of the formula (10.13) for **u** · **v** , we must refer the vectors to different bases that are reciprocal. Thus, if we expand

and use the relations   we find

In computing **u** × **v** , we refer both vectors to the same basis. Taking **e** 1, **e** 2, **e** 3 as basis, we have

Equations (15.5) now give

With these values **u** × **v** assumes the compact determinant form

analogous to (11.9).

If we dot-multiply (6) by **w** and note that **w** · **e** _i_ = _w i_, we obtain a determinant on the right with _w_ 1, _w_ 2, _w_ 3 in the first row. This can also be written

analogous to (13.6).

When **u** , **v** , **w** are referred to the basis **e** 1, **e** 2, **e** 3, formulas (6) and (7) have precisely the same form with _upper_ indices. With the self-reciprocal basis **i** , **j** , **k** , [ **ijk** ] = 1, and we obtain our previous results.

If we refer **u** , **v** , **w** to the basis **a** , **b** , **c** , we have

etc., and (7) yields the formula

If we refer **a** , **b** , **c** and **u** , **v** , **w** to the basis **i** , **j** , **k** , the left member of (8) becomes a product of determinants:

and (8) is equivalent to the "row-row" rule for multiplying determinants.

**17. Equations of Line and Plane.** A straight line through the point _R_ 1 and parallel to the vector **a** contains only the points _R_ for which   is parallel to **a**. Therefore its equation is ( **r** − **r** 1) × **a** = **0** , or, on writing **r** 1 × **a** = **a** _O_ ,

The vectors **a** , **a** _O_ must satisfy the relation

Since they determine the line, ( **a** , **a** _O_ ) are called coordinates of the line—its _Plücker coordinates_. When the origin is shifted from _O_ to _Q_ , **a** _O_ changes from

and, since  , we obtain the _shift formula_

The Plücker coordinates are _homogeneous_ ; that is, ( _λ_ **a** , _λ_ **a** _O_ ) determines the same line as ( **a** , **a** _O_ ), for equation (1) is unaltered. Thus a line is given by the ratios of six numbers (the components of **a** and **a** _O_ ) connected by a relation ( **a** · **a** _O_ = 0) and therefore depends on _four_ independent numbers. This is often expressed by saying: _There are_ ∞4 _lines in space_.

A plane through the point _R_ 1 and perpendicular to the vector **a** contains only the points _R_ for which   is perpendicular to **a**. Therefore its equation is ( **r** − **r** 1) · **a** = 0; or, on writing **r** · **a** = _α_ _O_ ,

Since **a** and _α_ _O_ determine the plane, ( **a** , _α_ _O_ ) are called coordinates of the plane. When the origin is shifted from _O_ to _Q_ , _α_ _O_ changes from

and, since  , we obtain the _shift formula_ ,

These coordinates are also homogeneous for ( _λ_ **a** , _λα_ O) determines the same plane as ( **a** , _α_ _O_ ), for equation (4) is unaltered. Thus, a plane is given by the ratios of four numbers (the components of **a** and _α_ _O_ ) and therefore depends on _three_ independent numbers. _There are_ ∞3 _planes in space_.

A point _R_ given by its position vector  . If we write **r** = **a** _O_ / _α_ , we may regard ( _α_ , **a** _O_ ) as the homogeneous coordinates of a point whose "equation" is

When the origin is shifted from _O_ to _Q_ , **a** _O_ changes from

and, since  , we obtain a shift formula

analogous to (3) and (5). The point coordinates ( _α_ , **a** _O_ ) are also homogeneous since ( _λα_ , _λ_ **a** _O_ ) represents the same point. Thus a point in space depends on _three_ independent numbers. _There are_ ∞3 _points in space_.

In the homogeneous coordinates for a

the first cannot vanish, and the second (with subscript _O_ ) depends upon the choice of origin. A change of origin _O_ → _Q_ entails the shift formulas

which have the same general form.

The point, line, and plane,

are at the distances

from the origin _O_. The first is obvious, and the others are easily proved. Thus, if **p** is the perpendicular vector from _O_ to the line,

hence

To find the distances from a point _Q_ , change _O_ to _Q_ in (10) and use the shift formulas (8). Thus, if  ,

**Example 1.** _Two lines_ ( **a** , **a** _O_ ), ( **b** , **b** _O_ ) _are coplanar when and only when_

_Proof_. If _R_ 1, _R_ 2 are points on ( **a** , **a** _O_ ), ( **b** , **b** _O_ ), the lines are coplanar when and only when the vectors **r** 1 − **r** 2, **a** , **b** are coplanar: that is, when

**Example 2.** _The shortest distance between the nonparallel lines_ ( **a** , **a** _O_ ), ( **b** , **b** _O_ ) _is_

_Proof_. If _R_ 1, _R_ 2 are points on the lines ( **a** , **a** 0), ( **b** , **b** 0), _d_ has the same numerical value as the component of   on their common normal **a** × **b** , namely

### **PROBLEMS**

1. Show that the line _AB_ has the coordinates ( **b** − **a** , **a** × **b** ).

2. Show that the plane _ABC_ has the coordinates ( **a** × **b** \+ **b** × **c** \+ **c** × **a** , [ **abc** ]).

3. If the lines ( **a** , **a** _O_ ) and ( **b** , **b** _O_ ) are coplanar, show that they meet in the point

**4.** Show that the lines _AA′_ and _BB′_ are coplanar, and find their point of intersection, given

**5.** Show that the nonparallel planes ( **a** , _α_ _O_ ), ( **b** , _β_ _O_ ) intersect in the line ( **a** × **b** , _β_ _O_ **a** − _α_ _O_ **b** ).

**6.** Show that the line ( **a** , **a** _O_ ) cuts the plane ( **b** , _β_ _O_ ) in the point

[Solve **r** × **a** = **a** _O_ , **r** · **b** = _β_ _O_ for **r**.]

**7.** If the lines ( **a** , **a** _O_ ) and ( **b** , **b** _O_ ) are parallel and **b** = **a** , show that their distance apart is | **b** _O_ − **a** _O_ |/| **a** |.

**8.** Find the point of intersection of the three planes

[Cf. Prob. 15.6.]

**9.** Show that the line ( **a** , **a** _O_ ) lies in the plane ( **b** , _β_ _O_ ) when and only when

[Eliminate **r** from **r** × **a** = **a** _O_ , **r** · **b** = _β_ _O_ when **a** · **b** = 0.]

**10.** If the three _points_ ( _α_ , **a** _O_ ), ( _β_ , **b** _O_ ), ( _γ_ , **c** _O_ ) lie on a line, Theorem 5.1 states that there are three numbers _x_ , _y_ , _z_ such that

If three _planes_ ( **a** , _α_ _O_ ), ( **b** , _β_ _O_ ), ( **c** , _γ_ _O_ ) pass through a line, is it true that there are three numbers _x_ , _y_ , _z_ such that

**18. Summary of Vector Algebra.**

_Sum:_ **a + b** by triangle construction.

_Difference:_ **a** − **b** = **a + (−b)**.

_Numerical Factor:_ If _λ_ > 0, ± _λ_ **a** has

_Dot Product:_ **a · b** = | **a** | | **b** | cos **(a, b).**

_Cross Product:_ **a** × **b** = | **a** | | **b** | sin ( **a, b** ) **e** ,

_Triple Product:_ **a** × ( **b** × **c** ) = ( **a · c** ) **b** − ( **a · b** ) **c**.

_Box Product:_

_Reciprocal Bases:_ **e** 1, **e** 2, **e** 3 and **e** 1, **e** 2, **e** 3;

_Components:_

_Equations in Homogeneous Coordinates:_

† A right-handed screw, revolved in a nut from **i** to **j** , will move in the direction **k**.

† 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_ , Yale University Press, 1901, based on Gibb's lectures, was written by Professor E. B. Wilson.

* Our previous definition of a dextral set in §8 applied to orthogonal sets such as i, j, k. This more general definition also shows that i, j, k form a dextral set.

† The name proposed by J. H. Taylor, _Vector Analysis_ , Prentice-Hall, 1939, p. 46.

† _Two_ nonparallel vectors **a** , **b** ( **a** × **b**  **0** ) form a basis for all vectors in their plane.

##   **Chapter 2**

## **Line Vectors**

**19. Moment of a Line Vector.** A vector which is restricted to lie in definite line is called a _line vector_ (§1). Two line vectors are equal when they have the same length, direction, and line of action.

If **u** is a line vector whose line of action passes through the point _R_ 1, the equation of the line is

where  .

DEFINITION 1. If _R_ 1 is any point of **u** 's line of action, the vector

is called the _moment of_ **u** _about the point O_.

The moment of **u** about the point _Q_ is therefore

and is given by the _shift formula_

We have also

_A line vector is always perpendicular to its moment about a point_.

We denote the line vector by the symbol ( **u** , **u** _O_ ) giving the coordinates of its line of action. Any pair of vectors ( **u** , **u** _O_ ) such that **u**  **0** , **u · u** _O_ = 0 determines uniquely a proper line vector.

In accordance with (1), the vector **u** = **0** has also a zero moment **u** _O_ = **0**. The symbol ( **0, 0** ) thus represents the zero line vector with an arbitrary line of action.

A _unit_ vector **e** acting in the line ( **e, e** _O_ ) is called an _axis_. Thus ( **i, 0** ), ( **j, 0** ), ( **k, 0** ) represent the rectangular axes of coordinates.

DEFINITION 2. If _l_ is an axis through _P_ with unit vector **e** , the rectangular component of **u** _P_ on _l_ is called the _moment of_ **u** _about the axis_.

This axial moment is therefore

or, since  ,

a number entirely independent of the choice of _P_ on the axis.

If _d_ is the perpendicular distance between ( **u** , **u** _O_ ) and the axis,

Hence the numerical value of the moment (5) is

This becomes | **u** | _d_ when **u** is perpendicular to **e**.

To summarize: If the line of a vector **u** passes through _Q_ , its moment about the point _P_ is

If _s_ is an axis through _P_ with the unit vector **e** , the moment of **u** about _s_ is the component of **u** _P_ on this axis:

_The vector moment_ **u** _P_ _represents a synthesis of the scalar moments of_ **u** _about all axes through P_. The various axial moments assume all values between ± | **u** _P_ |; and these extreme values correspond to axes having the direction of **u** _P_ and − **u** _P_.

**Example.** Consider the line vector **u** = [1, −1, 2] acting through the point _A_ (2, 4, −1). Find the moment of **u**

(a) about the point _P_ (3, −1, 2),

(b) about an axis through _P_ in the direction [2, −1, 2].

Since  , we have

A unit vector along the axis is  ; hence the moment of **u** about this axis is

Or we may compute

and use (5):

### **PROBLEMS**

**1.** The vector **f** = [2, 2, −3] acts through the point _A_ (1, −2, 1). Compute its moment

(i) about the origin ( **0, 0, 0** );

(ii) about the coordinate axes;

(iii) about an axis through _B_ (1, 3. −2) in the direction of the vector [2, 2, 1].

**2.** Find the moment of the vector   about the points _A_ and _B_ and about the axis _AB_ , given

**3.** If **f** 1 and **f** 2 are unit forces acting at _R_ 1 and _R_ 2, prove that the moment of either force about the axis of the other is ( **r** 1 − **r** 2) · **f** 1 × **f** 2.

**20. Addition of Line Vectors.** The sum of two line vectors ( **u** , **u** _O_ ) and ( **v** , **v** _O_ ) is defined by

If **u** \+ **v** = **0** , (1) is the zero line vector when **u** _O_ \+ **v** _O_ = **0**. If _R_ 1 and _R_ 2 are points on the lines of **u** and **v** ,

Since **u** _O_ \+ **v** _O_ = **0** only when _R_ 1 and _R_ 2 lie on the line of ( **u** , **u** _O_ ), we have

THEOREM 1. _The sum of two line vectors is zero when and only when the vectors have the same length and opposite directions and act along the same line_.

If **u** \+ **v**  **0** , (1) represents a line vector only when **u** \+ **v** is perpendicular to **u** _O_ \+ **v** _O_ (§19); then

and ( **u** , **u** _O_ ) and ( **v** , **v** _O_ ) are coplanar (Ex. 17.1). Hence we have

THEOREM 2. _The sum of two line vectors_ ( **u** , **u** _O_ ), ( **v** , **v** _O_ ) _is a_ proper _line vector when and only when_ **u** \+ **v**  **0** _and their lines of action are coplanar_.

Assume now that **u** \+ **v**  **0** and the lines of action intersect at _R_ 1. Then we may take **u** _O_ = **r** 1 × **u** , **v** _O_ = **r** 1 × **v**. Since the line of ( **u** \+ **v** , **u** _O_ \+ **v** _O_ ) has the equation

which is satisfied when **r** = **r** 1, the line passes through _R_ 1. Thus we have

THEOREM 3. _If two line vectors intersect, their sum is a line vector through their point of intersection_.

The construction, shown in Fig. 20 _a_, is aptly called the _parallelogram law_.

Fig. 20 _a_. Sum of intersecting line vectors

Next assume that **u** \+ **v**  **0** and the lines of action are parallel; then **v** = _λ_ **u** ( _λ_  −1). If _R_ 1, _R_ 2 are points on the lines of **u** and **v** ,

Since the line of ( **u** \+ **v** , **u** _O_ \+ **v** _O_ ) has the equation

or

which is satisfied when

the line passes through the point _R_ which divides _R_ 1 _R_ 2 in the ratio _λ_ /1 (5.2). We thus obtain

THEOREM 4. _If_ ( **u** , **u** _O_ ), ( **v** , **v** _O_ ) _are parallel line vectors and_ **v** = _λ_ **u** ( _λ_  −1), _the line of action of their sum is parallel to both and divides any segment from_ ( **u** , **u** _O_ ) to ( **v** , **v** _O_ ) _in the ratio λ_ /1.

The constructions when _λ_ = 2 and _λ_ = −2 are shown in Fig. 20 _b_.

Fig. 20 _b_. Sum of parallel line vectors.

**21. Couples.** The simplest case in which the sum

is not a line vector occurs when

for a zero vector must also have a zero moment. In this case the given vectors are _equal in magnitude, opposite in direction, and act along different parallel lines_. Their sum has the form ( **0, m**  **0** ); and the line vectors are said to form a _couple of moment_ **m**.

If the moments **u** _O_ , **v** _O_ are referred to another point _A_ , the moment of the couple is not changed; for, from (19.3),

_The moment of a couple is independent of the point about which moments are taken_.

If **u** \+ **v** = **0** , all pairs of line vectors ( **u** , **u** _O_ ), ( **v** , **v** _O_ ) which have the same sum ( **0** , **u** _O_ \+ **v** _O_ ) are regarded as equivalent couples of moment **m** = **u** _O_ \+ **v** _O_. Their moment may be computed about any point. Thus, if  ,   form a couple, _ABCD_ is a parallelogram (Fig. 21); the moment of the couple (about _D_ ) is   and equals the vector area of _ABCD_ (11.10). The couple  ,   has the same moment   (about _B_ ).

Thus  ,   and  ,   are equivalent couples.

The sum

Fig. 21. Equivalent couples.

is neither a line vector nor a couple when

In this case we may regard ( **w** , **w** _O_ ) as the sum of a line vector ( **w** , **0** ) through the origin and a couple ( **0**. **w** _O_ ) of moment **w** _O_ :

The symbol ( **w** , **w** _O_ ) now represents a new entity called a _motor_ which includes the line vector and couple as special cases.†

**Example.** Consider the line vectors  ,  ,   forming the triangle _ABC_. Their vector sum is zero; and their moment sum, computed about _A_ , is

Thus the three line vectors  ,  ,   are equivalent to a couple whose moment is twice the vector area of the triangle _ABC_.

### **PROBLEMS**

_Warning:_ Line vectors must be added in accordance with §20. To emphasize this point we use ≡ to denote equality of line vectors when their moments are omitted.

**1.** If  ,  ,   are line vectors and

prove that the points _A_ , _B_ , _C_ are collinear.

**2.** If _A_ , _B_ , _C_ , _D_ are the vertices of a quadrilateral in circuital order and _P_ , _Q_ are the mid-points of the diagonals _AC_ , _BD_ , prove that the line vectors

**3.** In a triangle _ABC_ , three forces of magnitude _AB_ , _BC_ , _CA_ are directed along the inward normals to these sides at their mid-points. Show that they form a system in equilibrium.

[The force **k** × ( **b** − **a** ) acts at the point  .]

**4.** Generalize the result of Prob. 3 to any plane polygon by dissecting it into triangles.

**5.** _ABCD_ are the vertices of a parallelogram taken in circuital order. Prove that the sum of the line vectors

**6.** If _A_ , _B_ , _C_ , _D_ are vertices of a plane quadrilateral taken in circuital order, prove that the line vectors  ,  ,  ,   are equivalent to a couple whose moment is twice the vector area of the quadrilateral.

Generalize this result.

**7.** If _A_ , _B_ , _C_ , _D_ are vertices of a skew quadrilateral and _PQRS_ are the mid-points of its sides taken in order, prove that the line vectors  ,  ,  ,   are equivalent to four times the couple formed by   and  .

**8.** The forces _X_ **i** , _Y_ **j** , _Z_ **k** act through the points (0, 0, _c_ ), ( _a_ , 0, 0), (0, _b_ , 0) respectively Show that they reduce to a single force if _a_ / _X_ \+ _b_ / _Y_ \+ _c_ / _Z_ = 0.

**9.** If the forces **i** , 2 **j** , − **k** acting at (0, 0, 2), (1, 0, 0), (0, 2, 0) respectively have a resultant, find its magnitude and axis.

**10.** The parallel forces _λ_ _i_ **e** act through the points _R_ _i_ ( _i_ = 1, 2, ···, _n_ ). Show that their resultant is as follows:

(i) If Σ _λ_ _i_  0, the system reduces to a single force acting through the centroid of the weighted points _λ_ _i_ _R_ _i_.

(ii) If Σ _λ_ _i_ = 0, the system reduces to zero or a couple according as the set of weighted points _λ_ _i_ _R_ _i_ is self-centroidal or not (§7).

**11.** The _plane_ system of forces **F** _i_ acting through the points _R_ _i_ has the resultant **F** = Σ **F** _i_  **0**. When all the forces **F** _i_ are revolved about these points through the same angle _θ_ , show that their resultant **F** also revolves through _θ_ about a point   (the _astatic center_ ) given by

When the forces are parallel ( **F** _i_ = _λ_ _i_ **e** ) show that the astatic center is the centroid of the weighted points _λ_ _i_ _R_ _i_.

**22. Coplanar Line Vectors.** If the set of line vectors   is _coplanar_ , we have, from (17.2) and (17.12),

Hence, if

give the sum or _resultant_ of the set, we obtain

on expanding  . Therefore:

_The resultant_ ( **f** , **f** _O_ ) _of a set of coplanar line vectors is_

This resultant may be obtained graphically by a succession of the constructions of §20. But, when the lines of action are parallel or nearly parallel, the following construction is much more convenient. Consider the case of three line vectors **f 1**, **f 2**, **f 3** acting along the lines shown in the space diagram Fig. 22 _a_. First construct the sum

by means of the vector polygon _ABCD_ (Fig. 22 _b_). Select any convenient point _O_ in the plane _ABCD_ as _pole_ and draw the vectors

Fig. 22 _a_. Link polygon.

Fig. 22 _b_. Vector polygon.

Now introduce the line vectors **a** and − **a** acting along a line _l_ || _OA_ into the space diagram. The lines of action of

are drawn in turn through _P_ 1, _P_ 2, _P_ 3, _P_ parallel to **b** , **c** , **d** , **f** in Fig. 22 _b_. The resultant line vector

acts along the line through _P_ where the lines of **d** and − **a** intersect.

The figure _PP_ 1 _P_ 2 _P_ 3, whose sides are parallel to the rays _OA_ , _OB_ , _OC_ , _OD_ drawn from the pole to the vertices of the vector polygon (Fig. 22 _b_), is called a _link_ (or _funicular_ ) polygon. Note that the sides of the link polygon that meet on the line of **f** _i_ are parallel to the rays that subtend **f** _i_ in the vector polygon.

The construction above corresponds to the case Σ **f** _i_  **0**. If, however, Σ **f** _i_ = **0** , the polygon formed by the vectors **f** _i_ will close ( _D_ = _A_ ), and the system will reduce to a couple or to zero. The construction for three vectors is shown in Fig. 22 _c_. As before we construct the lines of action of

Fig. 22 _c_. Reduction to a couple.

where **d** = **a** but acts along the line _l_ ′ parallel to _l_. The system now reduces to the _couple_ − **a** in _l_ , **d** = **a** in _l_ ′.

If the lines _l_ and _l_ ′ had coincided, the system would have reduced to the zero line vector ( **0** , **0** ).

The line vectors and couples of the _xy_ plane when written

suggest the one-to-one correspondence

with the space vectors **U** drawn from the origin; thus

Of the space vectors localized at the origin, those lying on the _z_ axis correspond to couples, all others to line vectors. Addition of coplanar line vectors now corresponds to the addition of free space vectors:

All line vectors parallel to ( **u** , **u** _O_ ) are given by ( _λ_ **u** , _λ_ ′ **u** _O_ ); hence the corresponding space vectors lie in plane parallel to **u** through the _z_ axis.

If the line vector ( **u** , **u** _O_ ) passes through the point ( _x_ , _y_ ), the corresponding

is a space vector perpendicular to [ _y_ , − _x_ , 1]; hence all line vectors through the point ( _x_ , _y_ ) correspond to space vectors **U** lying in the plane perpendicular to [ _y_ , − _x_ , 1].

We can now readily solve the following important problem:†

_Express a given line vector as the sum of three coplanar line vectors acting along specified lines, not concurrent and not all parallel_.

_Solution_. Write the given line vector ( **u** , **u** _O_ ) in the form (2), and let the three lines given by their Plücker coordinates

Since the lines are not concurrent and not all parallel, the corresponding space vectors **A** , **B** , **C** are not coplanar and admit a reciprocal set **A** ′, **B** ′, **C** ′. Now

where, from (15.8), _α_ = **U** · **A** ′, _β_ = **U** · **B** ′, _γ_ = **U** · **C** ′. This equality corresponds to

and the problem is solved. The solution, moreover, is unique.

**Example.** Resolve the force **f** = [1, 2] acting through the point (−3, 0) in the _xy_ -plane into forces acting along the lines _y_ = 0, _x_ = 0, _x_ \+ _y_ = 2.

Since **f** has the moment −6 about the _z_ -axis, **F** = [1, 2, −6]. The three lines correspond to space vectors

Since [ **ABC** ] = 2, their reciprocal set is

hence

Thus the given force **f** is equivalent to the force −2 **i** along the _x_ -axis, 5 **j** along the _y_ -axis, 3 **i** − 3 **j** along the line _x_ \+ _y_ = 2. Let the reader make a graphical check.

**23. Statics.** The forces acting on a rigid body may be treated as line vectors and replaced by their sum.† If the forces   are coplanar, there are _three_ possibilities, for their sum may represent a line vector, a couple, or zero (§22). In this case the link polygon gives a simple graphical method for obtaining the resultant.

With a general distribution of forces in space there are _four_ possibilities, for the sum

represents

The rigid body is in equilibrium only in the last case:

_A rigid body at rest will remain at rest when and only when the system of forces acting on it, treated as line vectors, have the zero line vector as their sum_.

### **PROBLEMS**

**1.** The line of action of a force **f** = [1, −1, 2] passes through the point _P_ (2, 4, −1). Find the moment of **f** about

(a) the point _A_ (3, −1, 2);

(b) an axis through _A_ in the direction [2, −1, 2];

(c) about the coordinate axes.

**2.** The forces **f** 1, **f** 2 are equal in magnitude and act at the points _P_ 1, _P_ 2. Prove that the moment of **f** 1 about the axis of **f** 2 is equal to the moment of **f** 2 about the axis of **f** 1.

**3.** Forces **f** 1 = 30 **j** , **f** 2 = −15 **j** , **f** 3 = −10 **j** , **f** 4 = 15 **j** at the points (0, 0), (2, 0), (3, 0), (4, 0) on the _x_ -axis. Find their resultant analytically and graphically.

**4.** Forces of 1, 2, and 3 lb, acting at the vertices _A_ , _B_ , _C_ , of an equilateral triangle having sides of 3 ft, are perpendicular to the sides _AB_ , _BC_ , _CA_. Each force tends to turn the triangle in the sense _ABC_ about its center. Show that their resultant is a force of   lb in the direction _BC_ and cutting _AB_ produced 8 ft from _A_.

Fig. 23. Rod supported by a cord.

**5.** The base plate of a crane is bolted to the floor by four bolts at the corners _A_ (1, 1), _B_ (−1, 1), _C_ (−1, −1), _D_ (1, −1) of a 2-ft square. If the crane load of 2000 lb acts vertically downward through the point _P_ (0, 10), find the reactions on the plate exerted by the bolts.

**6.** A uniform rod of weight _W_ rests on smooth planes at _A_ and _B_ and is supported by a cord at _C_ (Fig. 23). Show that the normal reactions at _A_ , _B_ and the tension of the cord at _C_ are given by

where _d_ = _p_ cos _α_ − _q_ sin _α_.

[Resolve the load   along the _unit_ line vectors

opposed to **a** , **b** , **c**. Then _a_ = **F** · **A** ′, etc. (§22).]

**7.** Solve Prob. 6 graphically when _W_ = 100 lb, _p_ = 8 ft, _q_ = 4 ft, _α_ = 30°, by means of a link polygon. Use the results of Prob. 6 to check the answers.

† See Louis Brand, _Vector and Tensor Analysis_ , Wiley, 1947, Chapter II, for a systematic account of the algebra of motors.

For the application of motors to line geometry consult

W. Blaschke, _Vorlesungen uber Differential Geometrie_ , Springer, Bd. 1, 1924, Kap. 7; and

E. A. Weiss, _Einführung in die Linien-geometrie und Kinematik_ , Teubner, 1935, p. 85.

A comprehensive account of the application of motor calculus to mechanics is given in the fundamental paper of

R. von Mises, "Motorrechnung, ein neues Hilfsmittel der Mechanik," _Z. angew. Math. u. Mech_., Bd. 4, 1924.

† In statics this is the _problem of three forces_. For a graphical solution see Brand, _Vectorial Mechanics_ , Wiley, 1930, §54.

† See _Vectorial Mechanics_ , Chapter II, for a development of statics from four fundamental principles.

##   **Chapter 3**

## **Vector Functions of One Variable**

**24. Derivative of a Vector.** Let **u** ( _t_ ) denote a vector function of a scalar variable _t_ over the interval  ; that is, when _t_ is given, **u** ( _t_ ) is uniquely determined.† Then **u** ( _t_ ) is said to be continuous for a value _t_ 0 if

The derivative of _u_ ( _t_ ) is defined as

and is also denoted by **u** ′( _t_ ).

If **u** ′( _t_ ) exists when _t_ = _t_ 0, **u** ( _t_ ) is said to be _differentiable_ at _t_ 0. Then we may write

where _ε_ → 0 as Δ _t_ → 0; hence

and, as Δ _t_ → 0, **u** ( _t_ 0 \+ Δ _t_ ) → **u** ( _t_ 0). Therefore:

_If the function_ **u** ( _t_ ) _is differ entiable at t_ 0, _it is also continuous there_.

Derivatives of higher order are defined as in the calculus:

**25. Derivatives of Sums and Products.** If **u** ( _t_ ), **v** ( _t_ ) are differentiate vector functions, and _f_ ( _t_ ) a differentiable scalar function, we can readily prove that

Formulas (3), (4), and (5) depend essentially on the distributive law for the respective products. Since the proofs follow the same lines as in the calculus of scalars, it will suffice to prove (5):

in which the _order_ of the factors must be preserved.

If **u** ( _t_ ) is referred to the constant basis **e** 1, **e** 2, **e** 3

we have from (1) and (3)

_The components of d_ **u** / _dt are the derivatives of the components of_ **u**.

THEOREM 1. _A necessary and sufficient condition that a proper vector_ **u** _have a constant length is that_

_Proof_. Since | **u** |2 = **u** · **u** , we have from (4)

Hence | **u** | = const implies (7), and conversely.

THEOREM 2. _A necessary and sufficient condition that a proper vector_ **u** _always remains parallel to fixed line is that_

_Proof_. Let **u** = _u_ ( _t_ ) **e** , where **e** is a _unit_ vector; then

If **e** is constant, _d_ **e** / _dt_ = **0** , and the condition follows. Conversely, since **u**  **0** , the condition implies that **e** × ( _d_ **e** / _dt_ ) = **0** ; but, since **e** · ( _d_ **e** / _dt_ ) = **0** from Theorem 1, these equations are contradictory unless _d_ **e** / _dt_ = **0** ; that is, **e** is constant.

When **u** is a function of the scalar variable _s_ , and _s_ in turn a function of _t_ , we have the familiar _chain rule_ of the calculus:

**26.** Curves. _A curve is an aggregate of points whose coordinates are functions of a single variable_. Thus the equations

represent a curve in space. The variable _t_ is called a _parameter_ , and each value of the parameter within a certain range _T_ :   corresponds to a definite point _P_ ( _x_ , _y_ , _z_ ) of the curve. If _x_ ( _t_ ), _y_ ( _t_ ), _z_ ( _t_ ) are continuous functions in _T_ , the curve is said to be _continuous_ in this interval.

To avoid having the curve degenerate into a point we exclude the case in which all three functions are constants. We also restrict the interval _T_ so that each _t_ in   corresponds to a different point 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  . When _t_ = _a_ and _t_ = _b_ correspond to the same point, the curve is said to be _closed_.

A parameter value _t_ corresponds to an _ordinary point_ of a curve when the three derivatives _x_ ′( _t_ ), _y_ ′( _t_ ), _z_ ′( _t_ ) exist and are continuous at _t_ and at least one is not zero. An arc of a curve that consists entirely of ordinary points is said to be _smooth_.

We can make the change of parameter

provided

(i) _φ_ ( _u_ ) > 0 in _U_ : _α_   _u_   _β_ ,

(ii) _φ_ ( _u_ ) is continuous in _U_.†

Then _φ_ ( _u_ ) is an increasing function in _U_ , and _t_ = _φ_ ( _u_ ) sets up a one-to-one correspondence between the _u_ -values in _U_ and the _t_ -values in _T_. On putting _t_ = _φ_ ( _u_ ) in (i), we get another set of parametric equations; and, since

the ordinary points are the same as before.

We may combine equations (1) into the single vector equation

Now each value to _t_ corresponds to a point _P_ of the curve whose position vector is  .

**Example.** The equations

are parametric equations of a unit circle with origin as center. Since _t_ = ± _π_ give the same point (−1, 0, 0) the curve is closed. It is obviously also continuous and smooth. If we make the change of parameter

and we obtain the new parametric equations for the circle:

**27. Tangent Vector.** If _P_ 0 and _P_ are neighboring points of a curve

and the line _P_ 0 _P_ tends to a limiting position as _P_ approaches _P_ 0 along the curve, the limiting line is called the _tangent_ to the curve at _P_ 0. Now, if

and, as _t_ → _t_ 0, we have

at all ordinary points of the curve; for the derivatives on the right exist, and not all are zero. Thus the limiting position of the line _P_ 0 _P_ as _t_ → _t_ 0

Fig. 27. Tangent vector.

is a line through _P_ 0 parallel to _d_ **r** / _dt_. Therefore _d_ **r** / _dt_ is a vector tangent to the curve at _P_ 0. If Δ _t_ > 0 in Fig. 27,   is a positive multiple of  ; but, if Δ _t_ < 0, Δ **r** /Δ _t_ is a negative multiple of  , and   in the figure must be reversed. In any case   approaches a limiting vector   which points in the direction of increasing _t_.

_The vector d_ **r** / _dt is a tangent vector to the curve_ **r** = **r** ( _t_ ) _at all ordinary points and points in the direction of increasing t_.

**Example.** Find the equations of the tangent to the curve

at the point _t_ = 1.

The point of tangency is _P_ 0(3, 3, 2), and the tangent vector at this point is

Every vector along the tangent line is a multiple of [1, 2, 2]; and, if _P_ ( _x_ , _y_ , _z_ ) is any point on the tangent,

Hence the equations of the tangent at (3, 3, 2) are

**28. Unit Tangent Vector.** It is proved in the calculus that an arc of a _smooth_ curve **r** = **r** ( _t_ ) is rectifiable, and its length from _t_ = _a_ to _t_ = _b_ is given by†

The arc _s_ between points corresponding to the parameter values _a_ and _t_ is therefore

and

Thus _s_ is a continuous increasing function of _t_. The same is true of the inverse function _t_ = _φ_ ( _s_ ), and consequently we can write the equation of the curve **r** = **r** ( _s_ ) with the arc _s_ as parameter.

We can now prove the important

THEOREM. _If_ **r** = **r** ( _s_ ) _is the vector equation of a smooth curve with the arc s_ = _P_ 0 _P as parameter, then_

_a unit vector tangent to the curve at P and pointing in the direction of increasing arcs_.

_Proof_. From §27 we know that _d_ **r** / _ds_ is a tangent vector to the curve in the direction of increasing _s_ ; and from (3)

COROLLARY. _The ratio of the chord of a smooth curve to the arc it subtends approaches_ 1 _as the arc approaches zero_.

_Proof:_ |Δ **r** |/|Δ _s_ | = |Δ **r** |/|Δ _s_ | → | _d_ **r** / _ds_ | = 1 as Δ _s_ → 0.

Fig. 28 _a_. Unit tangent vector: _d_ **r** / _ds_ = **T**.

If we _assume_ this corollary, the result (4) is evident from Fig. 28 _a_. For, if **e** denotes a unit vector along  ,

Now let _Q_ → _P_ ; then, as Δ _s_ → 0, **e** → **T** and chord/arc → 1 on the right; hence _d_ **r** / _ds_ = **T**.

If the position vector of the curve is

On dot-multiplying (5) by **i** , **j** , **k** in turn, we get

An important special case arises when **r** = **R** ( _θ_ ), a variable unit vector in a _plane_ making an angle of _θ_ radians with some fixed line as _x_ -axis. If **R** ( _θ_ ) is drawn from the origin, the locus of its end point is a circle of unit radius (Fig. 28 _b_), and _s_ = _θ_. Now (4) becomes

a unit vector perpendicular to **R** in the direction of increasing angles. Using this result, we have also

If we put

Fig. 28 _b_. Derivative of revolving unit vector: _d_ **R** / _dθ_ = **P**.

in (7), we find at one stroke the derivatives of both sine and cosine:

If _r_ = _f_ ( _θ_ ) is the equation of a plane curve in polar coordinates, its vector equation is

hence, on differentiating with respect to _s_ ,

On dot-multiplying by **R** and **P** , we find

**Example. Focal Properties of Conic Sections.** If _r_ 1 and **r** 2 are the focal distances of a point _P_ on the ellipse,

If we differentiate this with respect to _s_ , we have from (10)

where **R** 1 and **R** 2 are unit vectors along the focal radii. Thus **R** 1 \+ **R** 2 is normal to the ellipse at _P_ and bisects the interior angle between the focal radii at _P_ (draw the figure).

For the hyperbola

Thus **R** 1 − **R** 2 is normal to the hyperbola at _P_ and bisects the exterior angle between the focal radii at _P_ (draw the figure).

For a parabola whose axis is the _x_ -axis and directrix the _y_ -axis,

where **R** is a unit vector along the focal radius to the point _P_. Therefore **R** − **i** is normal to the parabola at _P_ and bisects the angle between **R** and − **i**. This is also an angle between focal radii if we regard the parabola as having a second focus on its axis at infinity.

### **PROBLEMS**

**1.** If the proper vector **u** ( _t_ ) is not parallel to a fixed line, prove that it will remain parallel to a fixed plane when and only when [ **u u** ′ **u** ″] = 0.

**2.** Find a tangent vector of the curve

What is the curve? Find its length from _t_ = 0 to _t_ = 1.

**3.** If   is the position vector to the

or

prove that  , where _OP_ and _OP_ ′ are conjugate semidiameters.

**4.** Show that the plane curve _f_ ( _x_ , _y_ ) = 0 has the normal vector

[Differentiate the equation with respect to _s_.]

**5.** Show that the plane curve _f_ ( _r_ , _θ_ ) = 0 has the normal vector

[Differentiate the equation with respect to _s_.]

**29. Frenet's Formulas.** For a rectifiable curve

the unit tangent at _P_. Since the length of **T** is constant, _d_ **T** / _ds_ , if not zero, must be perpendicular to **T** (Theorem 25.1). A directed line through _P_ in the direction of _d_ **T** / _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

where _κ_ is a non-negative scalar called the _curvature_ of the curve at _P_.

Now **B** = **T** × **N** is third unit vector such that **T** , **N** , **B** form a dextral set of orthogonal unit vectors at each point of the curve where _κ_  0. 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, _d_ **B** / _ds_ , if not zero, must be perpendicular to **B**. Differentiating **B** = **T** × **N** , we have

in view of (1). Hence _d_ **B** / _ds_ is perpendicular to **T** as well as **B** and therefore must be parallel to **N**. We therefore may write

Fig. 29. Moving trihedral.

where _τ_ is a scalar called the _torsion_ of the curve at _P_. The minus sign in (2) has this purpose: when _τ_ > 0, _d_ **B** / _ds_ has the direction of − **N** ; then, as _P_ moves along the curve in a positive direction, **B** revolves about **T** in the same sense as a right-handed screw advancing in the direction of **T** (Fig. 29).

We may now compute _d_ **N** / _ds_ by differentiating **N** = **B** × **T** and using (1) and (2):

Collecting (1), (2), and (3) we have the set of equations

known as _Frenet's formulas_.† They are fundamental in the theory of space curves.

From (4) we see that both _κ_ and _τ_ have the dimensions of the reciprocal of length; hence

have the dimensions of length and are called the _radius of curvature_ and _radius of torsion_ , respectively.

Since **N** , by definition, has the same direction as _d_ **T** / _ds_ , the curvature _κ_ is never negative. If _κ_ = 0 (identically), _d_ **T** / _ds_ = **0** , and **T** is a unit vector of constant direction; the curve is therefore a straight line. Conversely for a straight line **T** is constant, _d_ **T** / _ds_ = **0** and _κ_ = 0. _The only curves of zero curvature are straight lines_.

The torsion may be positive or negative. As _P_ traverses the curve in a positive direction, the trihedral **TNB** will resolve about **T** as a right-handed or left-handed screw according as _τ_ is positive or negative. The sign of _τ_ is independent of the choice of positive direction along the curve; for, if we reverse the positive direction, we must replace

and equations (4) maintain their form with unaltered _κ_ and _τ_.

If _τ_ = 0 (identically), _d_ **B** / _ds_ = **0** , and **B** is a constant vector; hence, from

and 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 normal to that plane; hence _d_ **B** / _ds_ = **0** at all points where **N** is defined ( _κ_  0) and _τ_ = 0. _The only curves of zero torsion are plane_.†

By introducing _Darboux's vector_

Frenet's formulas take on the memorable form

Darboux's vector proves to be the angular velocity of the moving trihedral **TNB** when the curve is traversed with constant speed (§34).

**30. Curvature and Torsion.** By use of Frenet's formulas we may compute the curvature and torsion of a curve from its parametric equations. Thus on differentiating **r** = **r** ( _t_ ) three times and denoting _t_ -derivatives by dots, we have

Hence

and, since   at an ordinary point,

If the positive direction on the curve is that of increasing _t_ ,  ; and the preceding equations show that

The planes through a point **r** ( _t_ ) of the curve and perpendicular to **T** , **N** , or **B** are called, respectively, the _normal, rectifying_ , and _osculating_ planes to the curve. Their equations are readily obtained from (3). Thus, if **q** is a variable position vector to the osculating plane, its equation is

If the parametric equations of a plane curve are _x_ = _x_ ( _t_ ), _y_ = _y_ ( _t_ ), we have

and from (1) and (2)

If the curve has the Cartesian equation _y_ = _f_ ( _x_ ), we can regard _x_ as the parameter: _x_ = _t_ , _y_ = _f_ ( _t_ ). Then _κ_ in (5) becomes

where the primes denote _x_ derivatives.

If the curve has the polar equation _r_ = _f_ ( _θ_ ), we take _t_ = _θ_ as parameter. Then

and since

we have from (1)

**Example 1.** At the point _t_ = 1 of the twisted cubic

find **T** , **N** , **B** , _κ_ , _τ_ and the equation of its osculating plane.

Since **r** ( _t_ ) = [2 _t_ , _t_ 2, _t_ 3/3],

hence, when _t_ = 1,

From (3)

the osculating plane is

and, from (1) and (2),  ,  .

Since  , the length of the curve is found from the equation

**Example 2. The Circular Helix.** A screw thread or helix, on a circular cylinder of radius _a_ , has the parametric equations

where _t_ is the angle measured counterclockwise from the _x_ -axis.

In order to find its curvature and torsion we compute

and, since

we have from (1) and (2)

Since _τ_ has the sign of _b_ , the helix is a right-handed if _b_ > 0, left-handed if _b_ < 0.

The length of the helix measured from _t_ = 0 is

### **PROBLEM**

**1.** Let **R** = **R** ( _t_ ) be a twisted curve on a unit sphere. If _t_ is its arc length, prove that the curve

and, if _t_ is an arbitrary parameter, the curve

**31. Plane Curves.** For plane curves Frenet's formulas reduce to

Since _d_ **T** / _ds_ is always directed to the concave side of a plane curve, the same is true of **N**. At points of inflection _d_ **T** / _ds_ = **0** , _κ_ = 0, and **N** is not defined. When _P_ passes through a point of inflection, **N** reverses its direction and **B** = **T** × **N** does the same. 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** as a _fixed_ unit vector normal to the plane of the curve, and define **N** = **B** × **T**. As before, **T** , **N** , **B** form a dextral set of orthogonal unit vectors. The curvature _κ_ , defined by (1), is now positive or negative according as _d_ **T** / _ds_ has the direction of **N** or the opposite. Equation (2) still holds good; for

Let _ψ_ be the angle from a fixed line in the plane to the tangent at _P_ , taken positive in the sense determined by **B**.† Then, from (28.7),

and hence, from (1),

The _intrinsic equation_ of a plane curve is the relation connecting _s_ and _ψ_ , say _s_ = _f_ ( _ψ_ ). When this relation is known, the radius of curvature _ρ_ = _f_ ′( _ψ_ ). For example, a circle of radius _a_ has the intrinsic equation _s_ = _aψ_ when _ψ_ is measured from the tangent at the point _s_ = 0; hence, for a circle, _ρ_ = _a_.

**Example 1. Evolute.** At any point _P_ of a plane curve where _κ_  0, the _center of curvature P_ 1 is given by

Fig. 31 _a_. Evolute of a plane curve.

The locus of _P_ 1 is called the _evolute_ of the curve. If _s_ = _AP_ and _s_ 1 = _A_ 1 _P_ 1 denote corresponding arcs on the curve Γ and its evolute Γ1 (Fig. 31 _a_), we have, on differentiating (5) with respect to _s_ ,

Choose the positive direction on Γ1 so that **T** 1 = **N** ; then

and, since Δ _s_ 1 = Δ _ρ_ , an arc of the evolute is equal to the difference in the values of _ρ_ at its end points. These properties show that a curve Γ may be traced by the end _P_ of a taut string unwound from its evolute Γ1; the string is always tangent to Γ1, and its free portion is equal to _ρ_. From this point of view, Γ is called the _involute_ of Γ1.

Since **T** 1 = **N** ,

hence the radius of curvature of the evolute is

**Example 2. Involute.** An involute Γ1 of a plane curve Γ is traced by the end _P_ 1 of a taut string _PP_ 1 unwound from Γ (Fig. 31 _b_). If Γ has the equation **r** = **r** ( _s_ ), an involute is given by

Fig. 31 _b_. Involute of a plane curve.

if the arcs _s_ and _s_ 1 are measured from the point _A_ where the curves meet. On differentiating (6) with respect to _s_ ,

If we choose the positive direction on Γ1 so that **T** 1 = − **N** ,

Thus, if Γ has the intrinsic equation _s_ = _f_ ( _ψ_ ) when _ψ_ is measured from a tangent at _A_ ( _s_ = 0), the arc of the involute

For example, from the intrinsic equation of a circle _s_ = _aψ_ we obtain

for the arc of its involute.

**Example 3. Envelopes.** Consider the plane vector function of two variables, **r** = **f** ( _u_ , _ν_ ). A one-parameter family of curves (with _u_ as parameter) is given by

If this family has a curve envelope given by _ν_ = _φ_ ( _u_ ), namely,

the vectors

are parallel at the points of contact; that is,

This condition must be fulfilled if the envelope exists. If (8) leads to a relation _υ_ = _φ_ ( _u_ ) (which does not make _∂_ **f** / _∂u_ or _∂_ **f** / _∂ν_ zero), this relation gives the envelope (7).

Consider, for example, the one-parameter family of normals to the plane curve **r** = **r** ( _s_ ), namely,

If they have an envelope,

Hence the envelope is given by _υ_ = 1/ _κ_ = _ρ_ , or

namely the locus of the centers of curvature of the curve.

### **PROBLEMS**

**1.** At the points _t_ = 0 and _t_ = 1 of the twisted cubic

find

( _a_ ) T, N, B;

( _b_ ) _κ_ and _τ_ ;

( _c_ ) the equations of the normal and osculating planes.

**2.** Find _κ_ and _τ_ at any point _t_ of the curve

( _a_ ) in Prob. 1;

( _b_ ) _x_ = _a_ (3 _t_ – _t_ 3), _y_ = 3 _at_ 3, _z_ = _a_ (3 _t_ \+ _t_ 3).

**3.** Show that the radius of curvature of the ellipse

is

**4.** If **r** = **r** ( _s_ ) is a plane curve Γ, show that **r** 1 = **r** \+ _c_ N ( _C_ const) is a _parallel curve_ ( **T** 1 = **T** at corresponding points); and

( _a_ ) _s_ 1 = _s_ − _cψ_ ) if _s_ 1 = _s_ = 0 when _ψ_ = 0;

( _b_ ) _ρ_ 1 = _ρ_ − _c_.

**5.** Prove that the intrinsic equation of the _catenary y_ = _c_ cosh ( _x_ / _c_ ) is _s_ = _c_ tan _ρ_ when _s_ is measured from the vertex ( _ρ_ = 0).

**6.** Prove that the intrinsic equation of the _cycloid_

formed by a circle of radius _a_ rolling along the _x_ -axis, is _s_ = 4 _a_ (1 − sin _ρ_ ) where _s_ and _ρ_ are measured from the _x_ -axis ( _ρ_ = _π_ /2).

**7.** Show that the curvature and torsion of the curve

are

**8.** Find the envelope of the family of straight lines in the _xy_ -plane

where **R, P** are the unit vectors of §28, _θ_ = angle ( **i** , **R** ), and _p_ **R** is the normal vector from _O_ to the line.

Show that the envelope is the curve

and that

**9.** 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_.

[Put _p_ = _c_ sin _θ_ cos _θ_ in Prob. 8.]

Show that the envelope has the parametric equations

and that the entire length of the curve is 6 _c_.

**10.** Prove that the Darboux vector **δ** is constant if _κ_ and _τ_ are constant, and that _κ_ has a fixed direction if _κ_ / _τ_ is constant.

**11.** A _helix_ is a twisted curve whose tangent makes a constant angle with a fixed direction ( **e** · **T** = cos _α_ where **e** is a unit vector in the fixed direction). Prove that:

( _a_ ) Helices are the only twisted curves whose Darboux vector has a constant direction.

( _b_ ) Helices are the only twisted curves for which _κ_ / _τ_ is constant.

( _c_ ) Circular helices are the only twisted curves for which _κ_ and _τ_ (and also the Darboux vector) are constant.

( _d_ ) The curve in Problem 1 is a helix with  , _α_ = _π_ /4.

**32. Kinematics of a Particle.** Let   be the position vector of a moving particle _P_ relative to some reference frame. If its path has the equation **r** = **r** ( _t_ ) in terms of the time _t_ as parameter, the _velocity_ **v** and _acceleration_ **a** of _P_ relative to this frame are defined as the time derivatives

By the chain rule

where the scalar _υ_ is a signed number whose absolute value, | _υ_ | = | **v** |, is called the _speed_ of the particle. _The velocity of P is a vector tangent to its path in the direction of motion and of length numerically equal to the speed_.

On differentiating (3) with respect to the time, we get

or, since **T** may be regarded as a function of the arc _s_ along the curve,

Therefore

_The acceleration of P is a vector in the plane of the tangent and principal normal to the path at P; its tangential and normal components are dυ/dt and υ 2/ρ._

The acceleration will be purely tangential when the motion is rectilinear ( _ρ_ = _∞_ ); it will be purely normal when the speed is constant ( _dυ_ / _dt_ = 0).

The velocity and acceleration vectors are regarded as bound to the moving particle.

With rectangular coordinates

and

Thus the rectangular components of **v** and **a** are the first and second time derivatives of the coordinates.

Motion in a plane may be referred to polar coordinates _r_ , _θ_ ; then

If we differentiate twice with respect to _t_ and use the equations

we find

The coefficients of **R** and **P** give the radial and transverse components of **v** and **a:**

If the acceleration is purely radial,

and

This means that the vector   sweeps out area at a constant rate; for the sectorial area measured from the initial line is

This result is known as the

LAW OF AREAS. _If the acceleration of a particle_ **_P_** _is always directed towards a fixed point O, the position vector_   _will sweep out area at a constant rate_.

**Example 1. Circular Motion.** A particle _P_ moves in a circle of radius _r_ , with variable angular speed _ω_ = _dθ_ / _dt_ (radians per second). Its position vector   is

and, from (28.7–8),

Its speed _υ_ = _rω;_ and its acceleration components

**Example 2. Uniformly Accelerated Motion.** If a particle has a constant acceleration **g** , and **r** = 0, **v** = **v** 0 when _t_ = 0, we have on integrating _dv_ / _dt_ = **g** twice:

The path is the result of superposing the displacement   due to the acceleration upon **v** 0 _t_ due to motion at constant velocity **v** 0. It is easily shown to be a parabola with axis parallel to **g**.

To find the path of a projectile under gravity (neglecting air resistance), we put

then the equations of the path are

Since _y_ = 0 at the time _t_ 1 = (2 _υ_ 0/ _g_ ) sin _α_ , the horizontal range _x_ 1 = ( _υ_ 02/ _g_ ) sin 2 _α_.

### **PROBLEMS**

**1.** A particle has the velocity **v** = _ω_ **k** × **r** where _ω_ is constant; prove that it travels in a circle with constant angular speed _ω_.

[Put **r** = _r_ **R**.]

**2.** A particle moves in an ellipse

so that the eccentric angle _φ_ increases at a constant rate _n_ ( _φ_ = _nt_ ). Prove that its acceleration **a** = — _n_ 2 **r** where **r** issues from the center of the ellipse.

**3.** A particle has the constant acceleration **a**. If its velocity is **v** 1 and **v** 2 at the times _t_ 1 and _t_ 2, show that **a** = ( **v** 2 − **v** 1)/( _t_ 2 − _t_ 1).

**4.** A particle moves along the curve

When _t_ = 1, find

( _a_ ) its velocity **v** and acceleration **a** ;

( _b_ ) **T, N, B** of the path;

( _c_ ) _κ_ and _τ_ of the path;

( _d_ ) the tangential and normal components of **a**.

**5.** A particle moves along the helix

When _t_ = _π_ /2, find the quantities required in Prob. 4.

**6.** A particle moves along the path

When _t_ = 0, find the quantities required in Prob. 4.

**33. Relative Velocity.** In §32 we have defined the velocity **v** of a particle _P_ relative to a given reference frame  . If   is a second reference frame, in motion with respect to  , how is the velocity **v** ′ of _P_ relative to   related to **v**?

At the instant _t_ let _P_ coincide with the point _Q_ fixed in  . Then on differentiating

with respect to _t_ , we have

in which the fixed point _Q_ is taken as origin; for, if _P_ is in motion with respect to.  ,   (originally zero) changes with time. We thus obtain

where **V** _Q_ , the velocity of the point _Q_ of   relative to  , 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  , 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  ; then **v** _Q_ is the _velocity of translation_ of  , and we may write

**Example 1. Wind Triangle.** An airplane _ρ_ 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. 33 _a_,

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 _θ_ and _θ_ ′ measured from the north around through the 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_.

In this example   is the earth,   the air in motion with respect to the earth.

Fig. 33 _a_. Wind triangle.

**Example 2. Interception.** A plane _p_ is flying over the track _PX_ with the ground speed _ep_ (Fig. 33 _b_). As plane _p_ passes the point _P_ , a plane _q_ departs from _Q_ to

Fig. 33 _b_. Interception.

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  .

Draw the vectors   and  , giving the velocities of plane _p_ and the wind _w_ relative to earth _e_. With _w_ 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   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  ; for   the velocity of plane _q_ relative to plane _p_ , has the direction  . 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. 33 _b_, plane _q_ can intercept _p_ on only one track. But, if the circle cuts the ray in two points, say _q_ 1 and _q_ 2, plane _q_ can intercept _p_ along two different tracks parallel to   and  , respectively.

### **PROBLEMS**

**1.** An airplane _p_ leaves a carrier _s_ at _O_ and patrols along the track _OY_ while the carrier follows the course _OX_ with constant speed of _υ_ miles per hour (Fig. 33 _c_). If the

Fig. 33 _c_. Plane returning to a carrier.

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 departure?

In Fig. 33 _c_,   and   give the velocities of carrier and wind, and the dotted circle, with center at _w_ , is the _air_ - _speed circle_. If a ray through _e_ in the direction _OY_ cuts this circle at _p_ 1, show that _p_ has the velocities relative to the earth and carrier:

If the carrier is at _A_ after _T_ hours, the turning point _B_ is the point where _AB_ || _ep_ 2 cuts _OY_. The track of the plane is _OBA_ ; and the carrier is at _C_ ( _BC_ || _p_ 1 _sp_ 2) when the plane is at _B_.

**2.** A plane _p_ , with _T_ hours of flying time, departs from _O_ along the track _OY_ (Fig. 33 _c_). Local bad weather makes landing at _Y_ dangerous, and the plane is directed to land at an alternative airport _A_. Show that the farthest turning point _B_ for the plane is located as in Fig. 33 _c_ if _A_ is regarded as a carrier traveling from _O_ to _A_ with the uniform velocity  .

**3.** In Prob. 1, _r_ = _CB_ is called the _radius of action_ of the plane. If the speeds of _p_ relative to the carrier _s_ are _u_ 1 and _u_ 2( _sp_ 1 and _sp_ 2), show that _r_ = _Tu_ 1 _u_ 2/( _u_ 1 \+ _u_ 2).

**34. Angular Velocity Vector.** 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 any instant may be specified by the angle 6 (radians) between an axial plane fixed in our frame of reference   and an axial plane fixed in the body. By choosing a positive direction on the axis (unit vector **e** ), we fix the positive sense of _θ_ by the right-handed screw convention. Then the _angular speed ω_ of the body at any instant is defined as

Thus _ω_ (radians per second) is positive or negative, according as _θ_ is increasing or decreasing at the instant in question.

The velocity distribution in the revolving body may be simply expressed if we define the _angular velocity_ as the vector,

Note that **ω** always is related to the instantaneous sense of rotation by the rule of the right-hand screw.

Choose an origin _O_ on the axis, and let   be the position vector of any particle of the body. Then (Fig. 34)

where **R** is a unit vector perpendicular to the axis and revolving with the body. Since _z_ **e** and _p_ are constant during the motion of _P_ , the velocity of _P_ is

Fig. 34. Body revolving about a fixed axis.

that is,

_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 _O_. Let **i** be a unit vector fixed in the body, **j** a unit vector in the direction of _d_ **i** / _dt_ (perpendicular to **i** ) and **k** = **i** × **j**. Then we may write

Hence _d_ **k** / _dt_ (perpendicular to **k** ) is also perpendicular to **i** and therefore parallel to **k** × **i** = **j**. Thus we have

or, on collecting results,

If we now write

these equations assume the same form:

Now the position vector **r** of any particle _P_ in the body may be referred to the orthogonal triple **i** , **j** , **k** fixed in the body; thus

where _x_ , _y_ , _z_ remain constant during the motion. Hence the velocity of _P_ is given by

Thus, at any instant, the velocity distribution in a rigid body with one point _O_ fixed is the same as if it were revolving about an axis through _O_ with angular velocity **ω**. The line through _O_ in the direction of to is called the _instantaneous axis of rotation_ , and **ω** is called, as before, the _angular velocity_. Now, however, **ω** may change in direction as well as in magnitude. With a fixed axis of rotation,

so that **ω** may be regarded as the time derivative of the vector angle _θ_ **e**. But with a variable axis of rotation **to** no longer can be expressed as a time derivative.

**35. Kinematics of a Rigid Body.** We next consider the 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 velocity distribution is given by  , the motion is said to be an _instantaneous rotation_ about an axis through _A_ in the direction of **ω**. 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.

Let _A_ be any point of the rigid body, and denote its velocity relative to a reference frame   by **v** _A_. Consider a second reference frame   having a translation of velocity **v** _A_ relative to  . Then the motion of the body relative to   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

relative to  , and, consequently,

relative to  . Moreover, for any other point _Q_ of the body,

and, on subtracting this from (1), we get

The content of these equations is stated in the following

THEOREM. _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 instantaneous translation_ **v** _A_ _and an instantaneous rotation_ **ω** _about an axis through A; and to is the same for any choice of A_.

From (2) we note that

_All velocities of the rigid body have the same projection on_ **ω** , _namely_ ( **e** · **v** _A_ ) **e**. This minimum velocity is actually attained at all points _R_ of the line for which

The vector equation of this line referred to _A_ as origin is therefore

or, since **e** = **ω** /| **ω** |,

The line, or _instantaneous axis_ , is parallel to **ω** and passes through the point _R_ , given by

and all of its particles have the same (minimum) velocity

in the direction of **ω**. Therefore:

_At any instant the motion of a rigid body is a rotation_ **ω** _about the instantaneous axis combined with a translation_ **V** _R_ _in the direction of_ **ω** ; _in other words, a screw motion_.

When to **ω** · **v** _A_ = 0, **v** _R_ = 0 and the motion reduces to a pure rotation to about the instantaneous axis

When **ω** = 0, the motion of the body is a pure translation **V** _A_.

**Example 1. Plane Motion.** When the motion is plane, to is perpendicular to the plane and v _R_ = 0. Thus at every instant plane motion is a pure rotation about an _instantaneous center R_ given by (5). The velocities of the body at that instant are given by

Therefore v _P_ is perpendicular to _RP_ and _υ_ _P_ = _ω_ ( _RP_.) A line perpendicular to **v** _P_ at _P_ passes through _R_. Similarly a line perpendicular to **v** _Q_ at _Q_ passes through _R_. We therefore have the following simple construction for the instantaneous center:

_If the points P, Q of a figure in plane motion have nonparallel velocities_ **v** _P_ , **V** _Q_ , _its instantaneous center is the point in which the normals to these velocities at P and Q meet_.

In Fig. 35 _a_, _AB_ represents the connecting rod of an engine; _A_ moves with the

Fig. 35 _a_. Connecting rod.

Fig. 35 _b_. Rolling wheel.

cross-head in a straight line, _B_ revolves with the crank in a circle. The instantaneous center of the connecting rod is at _R_.

**Example 2. Rolling Wheel.** In Fig. 35 _b_ a wheel of radius _r_ is rolling to the left with angular velocity **ω** = ω **k**. Its center has the velocity **v** _A_ = − _ωr_ **i** and the instantaneous center _R_ , given by

is at the point of contact with the track. The velocity of _B_ is therefore

### **PROBLEMS**

**1.** A vane on the masthead of a ship, going due east at 10 miles per hour, points NNE. When the ship stops the vane points NW. Find the speed of the wind.

**2.** At time zero ship _A_ is _d_ miles away from port and approaching it with the speed of _u_ miles per hour while ship _B_ is leaving port at the speed of _υ_ miles per hour. If their courses are straight and include an angle _α_ , show that the least distance _D_ between the ships and the time _T_ of reaching this position are given by

where   is the speed of _A_ relative to _B_.

**3.** A ladder _AB_ rests on a horizontal floor at _A_ and against a vertical wall at _B_. If it begins to slip down, find its instantaneous center _R_. If the ladder makes an acute angle _θ_ with the ground, show that the speeds of _A_ and _B_ are in the ratio _υ_ _A_ / _υ_ _B_ = tan _θ_.

**4.** The rigid frame   has the angular velocity **ω** referred to a frame  . If   and _P_ , _Q_ coincide with _A_ , _B_ of   at the instant _t_ , show that

where _d_ ′ **u** / _dt_ is the rate of **u** 's change referred to  .

**5.** If a point traverses a space curve with unit speed ( _ds_ / _dt_ = 1), show that the moving trihedral TNB has the angular velocity **ω** = **δ** , the Darboux vector. [Cf. (29.8).]

**6. Coriolis Acceleration.** A particle _P_ has the velocity and acceleration **v** , **a** and **v** ′, **a** ′ referred to the frames   and  , respectively. If,   has the angular velocity **ω** relative to  , and _P_ momentarily coincides with the point _Q_ in  , show that

Let _A_ be a fixed origin in  . Then, if  ,  ,   hence, from [(8),

When **v ′ = 0**, **a ′ = 0** these formulas give **v** _Q_ and **a** _Q_. The part **2ω** × **v** ′ of **a** is called the _Coriolis acceleration_ of _P_.]

**7.** The motion of a particle _P_ along a plane curve may be referred to a system of rectangular axes   revolving about _O_ so that the _x_ ′-axis always passes through _P_ ( **i** ′ = **R** , **j** ′ = **P** ). Let _P_ have the plane polar coordinates _r_ , _θ_ referred to the fixed frame  . Deduce (32.8) and (32.9) from (9) and (10) by showing that:

( _a_ ) The motion of _P_ relative of   is given by

( _b_ ) If _P_ coincides with the point _Q_ of  ,

( _c_ ) The Coriolis acceleration of _P_ is

† _Function_ is always used in the sense of _single-valued_ function.

† We may replace (i) by _φ_ ′( _u_ ) < **0** ; then _φ_ ( _u_ ) is a decreasing function, and the interval _U_ becomes _β_   _u_   _α_.

† See Louis Brand, _Advanced Calculus_ , Wiley, 1955, pp. 284–285.

† Or _Frenet-Serret formulas_.

† For a straight line ( _κ_ = **0** ), **N** is not determined. We then agree to give **N** any fixed direction normal to **T** and, as before, define **B** = **T** × **N**. Then **B** is constant, _d_ **B** / _ds_ = **0** and _τ_ = 0.

† A vector normal to a plane determines a positive sense of rotation by the right-handed screw convention. Thus, if **B** points _up_ from the paper, the positive sense is counterclockwise; then **N** is always 90° _ahead_ of **T**.

##   **Chapter 4**

## **Differential Invariants**

**36. Surfaces.** _A surface is an aggregate of points whose coordinates are functions of two variables_. Thus the equations

represent a surface. The variables _u_ , _υ_ are called _parameters_ or _surface coordinates;_ and each pair of values _u_ , _υ_ within a prescribed region corresponds to a definite surface point. If _x_ , _y_ , _z_ are functions of _t_ = _φ_ ( _u_ , _υ_ ), the equations (1) will represent a _curve_. In order to exclude this case, we shall require that the matrix

be of _rank two_ ; then at least one of its two-rowed determinants

is not identically zero.† However the three determinants may all vanish for certain surface points. Such points are called _singular_ in contrast to the _regular points_ where at least one determinant is not zero.

If **r** =  _x_ , _y_ , _z_ ] is the position vector to the surface, equations [(1) may be replaced by a single vector equation

and the matrix will be of rank two if

If we put _u_ = _f_ ( _t_ ), _υ_ = _g_ ( _t_ ) in (2), we obtain a curve on the surface. The curves _υ_ = _b_ , _u_ = _a_ are called the _parametric curves_ of the surface through the point _u_ = _a_ , _υ_ = _b_ ; their vector equations are

Their tangent vectors are **r** _u_ and **r** _υ_ and at any regular point **r** _u_ × **r** _υ_  0; that is, the tangents are not parallel and determine a normal vector **r** _u_ × **r** _υ_ to the surface. Hence the surface has a unique normal at every regular point.

**Example 1.** The equations

represent a sphere of radius _a_ about the origin. Referring to Fig. 42 _b_, we see that _u_ = _θ_ , the colatitude measured from the + _z_ -pole, while _υ_ = _φ_ is the longitude measured from the _xz_ -plane. The parametric curves _u_ = _a_ , _υ_ = _b_ are parallels of latitude and meridians, respectively. The matrix (2) is now

and hence

The vector in brackets is normal to the sphere and indeed a _unit normal:_

Since **r** _u_ × **r** _υ_ = 0 when sin _u_ = 0, the poles _u_ = 0 and _u_ = _π_ are singular points; these points however are not intrinsic singularities but are due to the parametric representation; in fact the unit normals at the poles are correctly given by **n.**

At a regular point _u_ 0, _υ_ 0 one of the Jacobians (3) is not zero. If _∂_ ( _x_ , _y_ )/ _∂_ ( _u_ , _υ_ ) 0, we may solve the first two equations of (1) for _u_ and _υ_ in terms of _x_ and _y_ ; † and, on substituting these values in the third equation, we obtain _z_ as a function of _x_ and _y_ :

This equation also represents the surface which now has the parametric form

Finally a surface may be given by the equation

This equation may be solved for _z_ in the neighborhood of any point for which _F_ _z_  0; we then again obtain an equation of the form (7).

Any three functions

that reduce (8) to an identity in _t_ correspond to a surface curve having the equations (9). If we substitute from (9) in (8) and differentiate the resulting equation with respect to _t_ , we obtain

Now  _x_ _t_ , _y_ _t_ , _z_ _t_ ] represents a tangent vector to curve [(9) at a point _P_ ( _t_ ); and, since (10) holds for _all_ surface curves through this point, the vector

is normal to all surface curves through _P_ and hence to the surface _F_ = 0 itself. This vector, denoted by ∇ _F_ , is called the _gradient_ of the function _F_ ( _x_ , _y_ , _z_ ).

If _Q_ ( _x_ , _y_ , _z_ ) is any point on the tangent plane to the surface (8) at the point _R_ 0( _x_ 0, _y_ 0, _z_ 0), the vector **q** − **r** 0 is perpendicular to ∇ _F_ at _R_ 0; hence the equation of the tangent plane to the surface at _R_ 0 is

**Example 2.** The plane

is normal to ∇ _F_ = [ _A_ , _B_ , _C_ ].

**Example 3.** The ellipsoid

has the gradient

The equation of a tangent plane at the point ( _x_ 0, _y_ 0, _z_ 0) is therefore

or, since ( _x_ 0, _y_ 0, _z_ 0) satisfies (13),

### **PROBLEMS**

**1.** If _u_ = _r_ , _υ_ = _φ_ are polar coordinates in the _xy_ -plane, show that the _surface of revolution_ obtained by revolving the curve _z_ = _f_ ( _x_ ), _y_ = 0, about the _z_ -axis has the parametric equations:

What are the parametric curves ? Find the unit normal vector to the surface.

**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_ = _r_ , _υ_ = _φ_ are polar coordinates in the _xy_ -plane, and the height of the line is given in terms of the angle turned, _z_ = _f_ ( _υ_ ), show that the conoid has the parametric equations

In particular, when _dz_ / _dv_ is constant, the conoid is a _right helicoid_ (a spiral ramp):

What are the parametric curves?

**3.** The parametric equations of a space curve Γ are

What surfaces are represented by the equations and what meaning has _υ_ in each case:

(i)  ;

(ii)  ?

What are the parametric curves _u_ = const?

**4.** Give parametric equations for the ellipsoid (13) patterned after equations (6) for the sphere.

**5.** Show that the equations

represent all cylinders whose rulings are parallel to the vector [ _a_ , _b_ , _c_ ]. What are the parametric curves _u_ = const?

**6.** Show that the equations

represent all cones whose vertices are at the point ( _a_ , _b_ , _c_ ). What are the parametric curves _u_ = const?

**7.** Show that the shortest distance _d_ from the point **_R_** 1( _x_ 1, _y_ 1, _z_ 1) to the plane _Ax_ \+ _By_ \+ _Cz_ \+ _D_ = 0 is

**8.** Show that the surface

is the parabolic cylinder _z_ = ( _x_ \+ _y_ )2. Find the direction of its rulings.

**37. Directional Derivative.** A scalar or vector that is uniquely defined at every point of a region is said to be a _point function_. If the point _P_ ( _x_ , _y_ , _z_ ) is given by the position vector  , we denote a scalar point function by _f_ ( **r** ) or _f_ ( _x_ , _y_ , _z_ ), a vector point function by **f** ( **r** ) or **f** ( _x_ , _y_ , _z_ ).

Consider now a scalar point function _f_ ( _x_ , _y_ , _z_ ) in neighborhood of a point _P_ 0 where it is continuous and differentiate. A ray (or half-line) through _P_ 0 in the direction of the unit vector

has the vector equation

Hence the parametric equations of the ray are

Along this ray _f_ ( _x_ , _y_ , _z_ ) is a function of _s_ alone whose right-hand derivative may be computed by the chain rule:

or, in view of equations (1),

If _f_ _x_ , _f_ _y_ , _f_ _z_ are computed at the point ( _x_ 0, _y_ 0, _z_ 0), formula (2) gives the _directional derivative_ of _f_ ( _x_ , _y_ , _z_ ) at _P_ 0 in the direction of **e**.

Since the direction cosines of **e** are equal to **e** · **i** , **e** · **j** , **e** · **k** , we may also write (2) in the form

in which the first factor indicates the direction of the derivative, while the second, the _gradient_

depends only upon the point where ∇ _f_ is computed. Thus the formula

sets up a one-to-one correspondence between directions **e** at a point and _df_ / _ds_ , the directional derivatives there. In effect, the _vector_ ∇ _f_ replaces the infinity of _scalars df_ / _ds_.

From (5) we see that at a given point _P_ 0 the maximum value of _df_ / _ds_ is assumed when **e** = **n** , a unit vector in the direction of ∇ _f_ , and this maximum is |∇ _f_ |. Moreover ∇ _f_ , at _P_ 0 is normal to the level surface

passing through _P_ 0 (§36). Thus the _direction_ of ∇ _f_ at _P_ 0 is normal to the level surface (6) and pointing in the direction of increasing _f_ ; and its magnitude,

is the derivative of _f_ in this direction. Hence in terms of this _normal derivative_ we have an expression for the gradient

_entirely independent of the coordinate system_. Since

the equations

hold in any system of rectangular coordinates.

If we replace _x_ , _y_ , _z_ by functions of other independent variables _u_ , _υ_ , _w_ , _f_ ( _x_ , _y_ , _z_ ) becomes _F_ ( _u_ , _υ_ , _w_ ). Along the ray (1)

or in view of (5)

But, since this holds for all vectors **e** ,

This expression for the gradient in _curvilinear coordinates u_ , _υ_ , _w_ obviously includes (10).

In particular, for a function _F_ ( _u_ ) of _u_ alone

For example, if _r_ denotes the distance from the origin,

**Example 1.** The polar distance

has the gradient

the unit radial vector in the direction  . This result also follows from the intrinsic definition (8); for the level surfaces of _r_ are spheres about the origin for which

**Example 2.** If the point _P_ has the plane polar coordinates _r_ , _φ_ we have

where **P** is the unit vector 90° ahead of **R** (§28). These results follow at once from (8). For example, the level curves of _φ_ are rays through the origin for which

If _f_ ( _r_ , _ω_ ) = _c_ is the equation of a curve in polar coordinates, a normal to the curve at any point is given by

### **PROBLEMS**

**1.** If _f_ ( _x_ , _y_ ) = _x_ 2 − 2 _xy_ \+ _y_ 2, find

( _a_ ) ∇ _f_ at (2,3);

( _b_ ) _df_ / _ds_ at (2, 3) along a ray inclined +45° to the _x_ -axis;

( _c_ ) the direction and magnitude of _df_ / _ds_ at (2, 3) when its value is greatest.

**2.** If _f_ ( _x_ , _y_ , _z_ ) = _xy_ \+ _yz_ \+ _zx_ , find

( _a_ ) ∇ _f_ at (1, 1, 3);

( _b_ ) _df_ / _ds_ at (1, 1, 3) in the direction of [1, 1, 1];

( _c_ ) the normal derivative _df_ / _dn_ at (1, 1, 3);

( _d_ ) the equations of the tangent plane and normal line to the surface _xy_ \+ _yz_ \+ _zx_ = 7 at (1, 1, 3).

**3.** If _f_ ( _r_ , _φ_ ) = _r_ − 2 _a_ sin _φ_ , find ∇ _f_ at the point ( _r_ , _φ_ ). What is the magnitude of the normal derivative at ( _r_ , _θ_ )?

**4.** If _u_ ( _x_ , _y_ ) and _υ_ ( _x_ , _y_ ) are functions that satisfy the equations _u_ _x_ = _υ_ _y_ , _u_ _υ_ = − _υ_ _x_ show that

( _a_ ) ∇ _u_ = ∇ _υ_ × **k** , ∇ _υ_ = **k** × ∇ _u_.

( _b_ ) At any point ( _x_ , _y_ ), _du_ / _ds_ = _dv_ / _ds_ ′ where _s_ ′ is a ray 90° in advance of _s_.

**5.** If  , show that

**6.** Show that the systems of equilateral hyperbolas _x_ 2 − _y_ 2 = _a_ , _xy_ = _b_ cut at right angles.

[The normal to the curve _f_ ( _x_ , _y_ ) = _c_ is parallel to ∇ _f_.]

**7.** Show that the systems of cardioids _r_ = _a_ (1 − cos _φ_ ), _r_ = _b_ (1 + cos _φ_ ) cut at right angles. Cf. [(16).]

**8.** Show that the curve _f_ ( _r_ 1, _r_ 2) = _c_ in bipolar coordinates has a normal parallel to _fr_ 1 **R** 1 \+ _fr_ 2 **R** 2.

Show that the systems of confocal ellipses and hyperbolas _r_ 1 \+ _r_ 2 = _a_ , _r_ 1 − _r_ 2 = _b_ cut at right angles.

**9.** Prove that the systems of coaxal circles _r_ 1/ _r_ 2 = _a_ , _φ_ 2 − _φ_ 1 = _b_ cut at right angles.

**38. Gradient of a Vector.** A vector point function **f** ( **r** ) is given by three scalar point functions _f_ 1, _f_ 2, _f_ 3 which form its rectangular components:

If these scalar functions are differentiable, we say that **f** ( **r** ) is differentiate.

If **f** ( **r** ) is differentiable, we may compute its directional derivative at _P_ 0 in the direction **e** just as in §37:

If we replace the direction cosines by **e** · **i** , **e** · **j** , **e** · **k** , we may write symbolically

Here the quantity in parenthesis has the same form as the gradient of a scalar. But since **f** is a _vector_ , the _gradient_

has as yet no meaning; for it consists of the sum of three ordered pairs of vectors. An expression of this sort is called a _dyadic_ , whereas the ordered vector pairs are called _dyads_. We could now develop an algebra of dyadics by defining equivalence, addition, and multiplication;† but for our purpose this is unnecessary. We shall merely regard ∇ **f** defined by (4) as an _operator_ that sets up a one-to-one correspondence between directions **e** at a point and _d_ **f** / _ds_ , the directional derivatives there:

In effect, the _dyadic_ ∇ **f** replaces an infinity of vectors _d_ **f** / _ds_.

We shall call any "sum" of dyads a _dyadic:_ thus the dyadic

represents a general dyadic in which the vectors **a** _i_ are _antecedents_ , **b** _i_. _consequents_. The dyadic

is called the _conjugate_ of **P** ; and, if (with equality defined as below)

Symmetric and skew dyadics are especially important since any dyadic **P** can be expressed as a sum of a symmetric and a skew dyadic in exactly one way, namely

The proof follows from **P** _cc_ = **P** and is left to the reader.

When vectors are cross-multiplied into a dyadic **P** = Σ **a** _i_ **b** _i_ new dyadics are formed; thus we define

DEFINITION. _Two dyadics are said to be equal when both transform an arbitrary vector in exactly the same way_ :

for any vector **u.**

From this definition we easily deduce the

THEOREM. _If_ **e** _i_ · **P** = **e** _i_ · **Q** _for three noncoplanar vectors_ **e** 1, **e** 2, **e** 3, _then_ **P = Q**.

_Proof_. As in §10 we can express any vector **u** in the form **u** = _u_ 1 **e** 1 \+ _u_ 2 **e** 2 \+ _u_ 3 **e** 3; hence

When **f** is the position vector **r** = _x_ **i** \+ _y_ **j** \+ _z_ **k** , **r** _x_ = **i** , **r** _υ_ = **j** , **r** _z_ = **k** , and, from (4),

The dyadic **I** is called the _idemfactor_ (Latin _idem_ = same) because it transforms any vector **u** into itself:

for from (15.10)

Moreover, since _u_ i = **u** · _e_ _i_ we have from (16.1)

We thus obtain the more general form of **I** :

Finally, to justify the use of the plus sign between dyads, we define the dyadic sum **P + Q** by

Thus, to add two dyadics, we merely add the dyads of one to those of the other.

The zero dyadic **O** has the property

and, from (13), we have also

**Example.** The dyadic **a** × **I** = **I** × **a** ; for

**39. Invariants of a Dyadic.** By definition the dyadic equation **P = Q** is equivalent to the vector equations

Hence it is also equivalent to the scalar equations

or to the vector equations

In these equations **u** is used as a _prefactor_ , **v** as a _postfactor_.

From (1) and (3) we have the dyadic distributive laws:

Thus (4) follows at once from

Using these laws, we may express a dyadic **P** = Σ **a** _i_ **b** _i_ various forms by substituting vector sums for **a** _i_ , **b** _i_ and expanding or collecting terms. In all these changes, however, there are certain quantities formed from the vectors of the dyadic which remain the same. Among these _invariants_ of **P** , the most important are the scalar and the vector

obtained by placing a dot or a cross between the vectors of each dyadic in **P**.

THEOREM 1. _The dyadic_ **P** = Σ **a** _i_ **b** _i_ _has the scalar invariant P_ _s_ = Σ **a** _i_ · **b** _i_ _and the vector invariant_ **p** = Σ **a** _i_ × **b** _i_.

_Proof_. Let **P = Q**. From (10.13)

hence

Since the corresponding terms in these sums are equal by (2), _P_ _s_ = _Q_ _s_.

Again, from (11.8)

hence the first components of **p** and **q** are

Since the corresponding terms on the right are equal by (2), _p_ 1 = _q_ 1; similarly _p_ 2 = _q_ 2, _p_ 3 = _q_ 3, and hence **p = q.**

THEOREM 2. _The scalar and vector invariants of the sum of two dyadics are the sums of their respective invariants_.

_Proof_ If **R = P + Q** , it is obvious that

It is this property that gives these invariants their special importance in geometry and physics.

**Example.** From (38.8) we see that the invariants of **I** are

Hence, from (38.12), we deduce that

### **PROBLEMS**

**1.** If **e** _i_ · **P** = **f** _i_ ( _i_ = l, 2, 3), prove that **P** = **e** 1 **f** 1 \+ **e** 2 **f** 2 \+ **e** 3 **f** 3.

**2.** If **P** · **e** _i_ = **g** _i_ ( _i_ = l, 2, 3), prove that **P** = **g** 1 **e** 1 \+ **g** 2 **e** 2 \+ **g** 3 **e** 3.

**3.** If **a** , **b** , **c** are arbitrary noncoplanar vectors, show that any dyadic may be expressed as the sum of three dyads having **a** , **b** , **c** either as antecedents or as consequents.

**4.** Prove that **a** × **I** is a skew dyadic.

**5.** Find the invariants of **a** × **I**.

**6.** Prove that  .

**7.** If the dyadic **P** is skew and has the vector invariant **p** , show that  .

Show that **P** transforms all vectors into vectors perpendicular to **p**.

**8.** Show that the vector invariant of any symmetric dyadic is zero.

**9.** Show that the scalar invariant of any skew dyadic is zero.

**10.** The dyadic product **P · Q** is defined by the equation

From this definition show that dyadic multiplication is

but not in general commutative: **P · Q**  **Q · P**.

[ _Vector and Tensor Analysis_ ,§65.]

**11.** From the definition (8) deduce the dyad product:

**12.** Use (9) to prove that

**13.** From (8) and (10) show that

**14.** The idemfactor **I** is often called the _unit dyadic_. Justify this term by proving that

**15.** Prove that, in _all_ cases,

**16.** If **P** _n_ = **P · P** ··· to _n_ factors, prove that

**40. Divergence and Rotation.** The gradient ∇ **f** (or grad **f** ) of a vector point function is a dyadic whose invariants are of the utmost importance " in geometry and mathematical physics.

The scalar invariant, formed by dot-multiplying the dyad vectors of ∇ **f** , is called the _divergence_ of **f** and is written ∇ · **f** or div **f**.

The vector invariant, formed by cross-multiplying the dyad vectors of ◯ **f** , is called the _rotation_ or _curl_ of **f** and is written ∇× **f** , rot **f** , or curl **f**.†

In terms of rectangular coordinates we therefore have the defining equations:

If **f** is resolved into its rectangular components

_f_ 1, _f_ 2, _f_ 3 are scalar functions of _x_ , _y_ , _z_ , and ∇ **f** assumes the nine-term form:

where _D x_ = _∂_ / _∂x_ , etc. This form is often abbreviated into the matrix,

in which the nine _unit dyads_ **ii** , **ij** , ···, **kk** are omitted.† By dotting and crossing between the vectors of (4), the invariants become

The last is easily remembered in the determinant form (11.9):

For the position vector **r** = _x_ **i** \+ _y_ **j** \+ _z_ **k** we have

the _idemfactor_ ; hence

When **f** = ∇ _φ_ , the gradient of a scalar _φ_ ( _x_ , _y_ , _z_ ), we have

With these components (5) and (6) give

In (9) the differential operator div ∇ or ∇ · ∇ is called the _Laplacian_ and often is written

The proof of (10) depends on the equality of the mixed second derivatives of _φ_ (as _φ xy_ = ( _φ yx_); this is always the case when the derivatives in question are continuous.‡

When **f** = rot **g** , the rotation of a vector **g** ( _x_ , _y_ , _z_ ),

In this case (5) and (6) give

The proofs are straightforward and depend upon the equality of mixed second derivatives. In (12) for example, the first component of rot rot **g** is

The four identities (9) through (12) are of the first importance. When written as

they may be remembered by treating ∇ formally as a vector. In (10), ∇ x ∇ = 0; in (11) ∇ enters twice in the box product; and in (12) the expansion theorem (12.1) gives the correct result. Thus in symbolic form, these basic formulas are easier to remember than to forget!

**Example 1.** The gradient of the vector function

is the dyadic

or, in matrix form (4)′,

Hence

or we may compute div **f** from (5), rot **f** from (6). Moreover the scalar function

has the gradient

**Example 2.** The velocity at any point _P_ of a rigid body is given by (35.1)

At any given instant _t_ , **v** is a vector point function; and from (3)

_At any instant the velocity distribution_ **v** _in a rigid body has a rotation equal to twice its instantaneous angular velocity_ ω.

**41. Product Formulas.** First note that

We next apply the defining equations (40.1–) to various products of functions.

If _λ_ is a scalar, **f** a vector point function, we have

From (40.1) we have

which is (3). This yields (4) and (5) by dotting and crossing between dyad vectors.

If **f** and **g** are vector point functions,

We give the proof of (7): from (40.3),

We next prove that

the left member equals

We can now show that

for, from (40.1),

which, in view of (8), gives (9).

**Example 1.** _The function_ 1/ _r is a solution of Laplace's equation:_

For, from (37.14),

and, from (4),

**Example 2.** If _u_ and _υ_ are scalar functions,

Since rot ∇ _u_ = rot ∇ _υ_ = 0, this follows at once from (6).

### **PROBLEMS**

**1**. Find the gradient of the vectors

in matrix form, and find their divergence and rotation.

**2**. If  , find

( _a_ ) ∇ _r_ 2 and ∇2 _r_ 2;

( _b_ ) ∇ _r_ −1 and ∇2 _r_ −1.

**3**. If **a** is a constant vector, prove that ∇( **r** × **a** ) = **I** × **a**.

**4**. If _u_ is a scalar point function, prove that rot ( _f_ ( _u_ ) ∇ _u_ ) = **0**. Hence show that rot ( _f_ ( _r_ ) **r** ) = **0**.

**5**. If **R** , **P** are the unit vectors of §28 in the _xy_ -plane and  , show that

**6**. If  , prove that

**7**. If ∇2 _f_ ( _r_ ) = 0, find _f_ ( _r_ ).

**8**. If **r** / _r_ is the unit radial vector in space, prove that

**9**. If **a** is a constant vector, prove that

**10**. If **e** is a unit vector, prove that

**42. Curvilinear Coordinates.** Let us change from rectangular coordinates ( _x_ , _y_ , _z_ ) to curvilinear coordinates ( _u_ , _υ_ , _w_ ) by means of the equations

where _F_ , _G_ , _H_ have continuous first partial derivatives in a certain region in which their Jacobian is not zero. The functions, therefore, are not connected by any functional relation. Since the Jacobian is continuous, its sign cannot change in the region; and, to be explicit, we shall suppose that the Jacobian

This involves no loss in generality; for, if the Jacobian were negative, an interchange of _υ_ and _w_ (for example) would make it positive.

Under these hypotheses the equations (1) have a unique inverse in some neighborhood _**I**_ 0 of any point ( _x_ 0, _y_ 0, _z_ 0) of the region:

and _f_ , _g_ , _h_ also have continuous first partial derivatives.† In the neighborhood _I_ 0 the correspondence ( _x_ , _y_ , _z_ ) ↔ ( _u_ , _υ_ , _w_ ) is one-to-one, so that a point _P_ ( _x_ , _y_ , _z_ ) may also be specified by the three numbers ( _u_ , _υ_ , _w_ ) given by (3).

When ( _u_ 0, _υ_ 0, _w_ 0) are given, _P_ 0 is the point of intersection of the three surfaces

These will intersect in three curves, the _coordinate curves_ , along which only one of the quantities _u_ , _υ_ , _w_ can vary. For this reason, _u_ , _υ_ , _w_ are called _curvilinear coordinates_ , in distinction to the rectangular coordinates _x_ , _y_ , _z_ , for which the coordinate curves are straight lines.

Consider now any vector point function **f** ( _u_ , _υ_ , _w_ ); its derivative in the direction **e** is

Since _d_ **f** / _ds_ = **e** · ∇ **f** , _du_ / _ds_ = **e** · ∇ _u_ , we have

for all vectors **e**. Hence (§38, Def.)

give the gradient, divergence, and rotation of a vector point function in general curvilinear coordinates.

When the position vector **r** is regarded as a function of _x_ , _y_ , _z_ , we have ∇ **r** = **I** (40.8); but, when **r** is regarded as a function of _u_ , _υ_ , _w_ , ∇ **r** is given by (4); hence

From this equation we can conclude that:

_The vector triads_ ∇ _u_ , ∇ _υ_ , ∇ _w and_ **r** _u_ , **r** _υ_ , **r** _w_ _form reciprocal sets_.†

_Proof_. Write **e** 1 = ∇ _u_ , **e** 2 = ∇ _υ_ , **e** 3 = ∇ _w_ , and let **e** 1, **e** 2, **e** 3 denote the reciprocal set. Then, on dot-multiplying (7) by **e** 1, **e** 2, **e** 3 as prefactors, we find that **e** 1 = **r** _u_ , **e** 2 = **r** _υ_ , **e** 3 = **r** _w_.

The vectors **r** _u_ are tangent to the _u_ -curves, the curves along which _υ_ and _w_ are constant. Thus at any point _**P**_ ( _u_ , _υ_ , _w_ ), **r** _u_ , **r** _υ_ , **r** _w_ are tangent to the three coordinate curves meeting there. Moreover from the properties of reciprocal sets:

In computing the invariants of ∇ _f_ it is usually more convenient to eliminate ∇ _u_ , ∇ _υ_ , ∇ _w_ from the above formulas by means of equations (8). Thus from (5) we obtain

or, in view of the identity,

In the important case when all coordinate curves cut at right angles, the coordinates are said to be _orthogonal_ ; then

We choose the notation so that

where **a** , **b** , **c** are a dextral set of (variable) unit vectors and _U_ , _V_ , _W_ are all positive; then _J_ = _UVW_. Now (11) becomes

When we replace **f** by the gradient of a scalar

we obtain the important formula for the Laplacian of a scalar function in orthogonal coordinates:

When the curvilinear coordinates are given by equations (1), we may compute _U_ , _V_ , _W_ from equations of the type

**Example 1. Cylindrical Coordinates.** The point _**P**_ ( _x_ , _y_ , _z_ ) projects into the point _**Q**_ ( _x_ , _y_ , 0) in the _xy_ -plane. If _ρ_ , _φ_ are polar coordinates of _**Q**_ in the _xy_ -plane, _u_ = _ρ_ , _υ_ = _φ_ , _w_ = _z_ are called the cylindrical coordinates of _**P**_ (Fig. 42 _a_). They are related to the rectangular coordinates by the equations

The level surfaces _ρ_ = _a_ , _φ_ = _b_ , _z_ = _c_ are cylinders about the _z_ -axis, planes through the _z_ -axis, and planes perpendicular to the _z_ -axis. The coordinate curves, for _ρ_ , are rays perpendicular to the _z_ -axis; for _φ_ , horizontal circles centered on the _z_ -axis; for _z_ , lines parallel to the _z_ -axis.

From **r** = [ _x_ , _y_ , _z_ ] we have

Since these vectors are mutually perpendicular and _J_ = _UVW_ = _ρ_ > 0, cylindrical coordinates form an orthogonal system which is dextral in the order _ρ_ , _φ_ , _z_.

From (16) the Laplacian

Fig. 42 _a_. Cylindrical coordinates: _ρ_ , _φ_ , _z_.

When _g_ = log _ρ_ , _ρg_ _ρ_ = 1; hence log _ρ_ is a particular solution of Laplace's equation ∇2 _g_ = 0.

**Example 2. Spherical Coordinates.** The spherical coordinates of a point _P_ ( _x_ , _y_ , _z_ ) are its distance _r_ = _OP_ from the origin, the angle _θ_ between _OP_ and the _z_ -axis, and the dihedral angle _φ_ between the _xz_ -plane and the plane _zOP_ (Fig. 42 _b_). They are related to the rectangular coordinates by the equations:

The level surfaces _r_ = _a_ , _θ_ = _b_ , _φ_ = _c_ are spheres about _O_ , cones about the _z_ -axis with vertex at _O_ , and planes through the _z_ -axis. The coordinate curves, for _r_ , are rays from the origin; for _θ_ , vertical circles centered at the origin; for _φ_ , horizontal circles centered on the _z_ -axis.

Fig. 42 _b_. Spherical coordinates: _r_ , _θ_ , _φ_.

From **r** = [ _x_ , _y_ , _z_ ] we have

Since these vectors are mutually perpendicular and _J_ = _UVW_ = _r_ 2 sin _θ_ > 0, spherical coordinates form an orthogonal system which is dextral in the order _r_ , _θ_ , _φ_.

From (16) the Laplacian

When _g_ = 1/ _r_ , _r_ 2 _g_ _r_ = − 1; hence 1/ _r_ satisfies Laplace's equation. The most general function of _r_ that satisfies Laplace's equation is the general solution

### **PROBLEMS**

**1.** With cylindrical coordinates _ρ_ , _φ_ , _z_ , show that

**2.** With spherical coordinates _r_ , _θ_ , _φ_ , show that

**3.** Deduce the result of Prob. 41.6 from (21).

**4.** Prove that the equations

represent a change from one set of rectangular coordinates _x_ , _y_ , _z_ into another _u_ , _υ_ , _w_ when the vectors **r** _u_ , **r** _υ_ , **r** _w_ form a self-reciprocal set.

[The identity _x_ 2 \+ _y_ 2 \+ _z_ 2 = _u_ 2 \+ _υ_ 2 \+ _w_ 2 implies the self-reciprocity of **r** _u_ , **r** _υ_ , **r** _w_ ; for it shows that _a_ _i_ _a_ _j_ \+ _b_ _i_ _b_ _j_ \+ _c_ _i_ _c_ _j_ = _δ_ _ij_ and hence **r** _u_ , **r** _υ_ , **r** _w_ are a set of orthogonal unit vectors.]

**5.** If _u_ , _υ_ , _w_ are curvilinear coordinates, prove the operational identities:

**6.** Deduce the formula for rot **f** in curvilinear coordinates

( _a_ ) analogous to (42.11):

( _b_ ) analogous to (40.7):

7. If **f** ( _u_ , _υ_ ) is a function defined on the surface **r** = **r** ( _u_ , _υ_ ) and **e** a unit vector in the tangent plane at the point ( _u_ , _υ_ ), prove that the directional derivative of **f** ( _u_ , _υ_ ) along all surface curves tangent to **e** is given by

where **a** , **b** , **n** is the set reciprocal to **r** _u_ , **r** _υ_ , **n**.

[The unit normal **n** has the direction of **r** _u_ × **r** _υ_ ; _Vector and Tensor Analysis_ , §95.]

**8**. On the surface **r** = **r** ( _u_ , _υ_ ) the _surface gradient_ , _divergence_ , and _rotation_ of a function **f** ( _u_ , _υ_ ) are defined as

Prove that

† In the excluded case _x_ _u_ = _x_ _t_ _t_ _u_ , _x_ _u_ = _x_ _t_ _t_ _υ_ , etc., and _A_ , _B_ , _C_ vanish identically.

† _Advanced Calculus_ ,§85.

† For a function _f_ ( _x_ , _y_ ) of two variables the direction **e** in the _xy_ -plane has the direction cosines

and

† See _Vector and Tensor Analysis_ , Chapter IV, for a systematic development of dyadic algebra.

† _Curl_ is favored by the British and American authors, _rotation_ by continental Europeans. In the interests of a uniform notation (for the language of mathematics is universal) we write rot **f** in this book although we may speak of "curl."

† When dyadics are written in matrix form, their algebra is the same as matric algebra. See _Vector and Tensor Analysis_ , §78. Note that the idemfactor I becomes the unit matrix, the zero dyadic the zero matrix.

‡ _Advanced Calculus_ , p. 163.

† For the theorems involved here see _Advanced Calculus_ , §§87–89.

† The nine equations ∇ _u_ · **r** _u_ = 1, ∇ _u_ · **r** _υ_ = 0, ··· also follow from the chain rule: for example

##   **Chapter 5**

## **Integral Theorems**

**43. Line Integrals.** Let the functions _P_ ( _x_ , _y_ , _z_ ), _Q_ ( _x_ , _y_ , _z_ ), _R_ ( _x_ , _y_ , _z_ ) be defined at all points of a smooth curve Γ given by the parametric equations

Then the line integral over Γ,

is computed as the Riemann integral between _t_ = _t_ 1 and _t_ = _t_ 2 by substituting from equations (1) in the integrand. If _P_ , _Q_ , _R_ become _P_ ( _t_ ), _Q_ ( _t_ ), _R_ ( _t_ ), the integral above becomes

If the functions _P_ ( _t_ ), _Q_ ( _t_ ), _R_ ( _t_ ) are integrable and the curve Γ is smooth (§26), the integral certainly exists. If Γ is merely _piecewise smooth_ , (2) holds over each smooth arc and hence, by addition, over the entire curve. The finite discontinuities of _x_ ′( _t_ ), _y_ ′( _t_ ), _z_ ′( _t_ ) do not prejudice the existence of the integral. Thus Γ may be any broken line, for example a _step line_ consisting of two adjacent sides of a rectangle. Moreover Γ may be a closed curve or polygon which does not cross itself; in this case

denotes the circuit integral about Γ in the positive sense. For a plane curve which encloses a region  , _the positive sense is one that leaves theregion to the left_. Thus the positive sense on a simple closed curve is counterclockwise.

If we regard _P_ , _Q_ , _R_ as the components of a vector point function

and write the equation of the curve in vector form

the integral (2) assumes the compact form

which applies equally well to two or three dimensions.

The value of a line integral is independent of the parametric representation of the curve. This follows from the general theorem on the change of variable in integrals.† With the arc _s_ along the curve as parameter, _d_ **r** / _ds_ = **T** (28.4), and the line integral becomes

the integral of the tangential component of **f** along the curve. If **f** is everywhere normal to Γ, the line integral is zero.

**Example 1.** Compute   from (0, 0) to (1, 1) over the

(i) straight line _y_ = _x_ ;

(ii) step path _y_ = 0,  ;

(iii) step path _x_ = 0,  ;

(iv) parabola _y_ = _x_ 2;

(v) circle _x_ = cos _t, y_ = 1 + sin _t_  .

(vi) Integrate also around the square (0, 0) − (1, 0) − (1, 1) − (0, 1) − (0, 0).

_Solutions_.

(i)  ;

(ii)  ;

(iii)  ;

(iv)  ;

(v)  .

(vi)  .

**Example 2.** The line integral over a curve Γ in Ex. 1 has a simple interpretation. If we change to polar coordinates,

we find that

where tan _φ_ 1 = _y_ 1/ _x_ 1, tan _φ_ 2 = _y_ 2/ _x_ 2, and _A_ is the area of the sector _OP_ 1 _P_ 2 subtended by Γ at the origin (Fig. 43). We may now express _A_ as a, _circuit integral_

taken over the complete boundary of the sector; for the radial lines, _φ_ = _φ_ 1 and _φ_ = _φ_ 2, contribute nothing. The area _A_ is positive or negative according as the circuit is counterclockwise or clockwise.

 .

Fig. 43. Area of a sector.

All the results of Ex. 1 are now readily verified.

**Example 3**. Compute   about the triangle whose vertices are (1, 0), (0, 1), (−1, 0).

Beginning the circuit at (1, 0), the sides of the triangle are segments of the lines

hence the circuit integral equals

**Example 4**. Compute   about the circle _x_ = cos _t_ , _y_ = 1 + sin _t_.

The integral taken counterclockwise equals

**Example 5.** Compute the line integral

( _a_ ) over the broken line (0, 0, 0) − (1, 0, 0) − (1, 1, 0) − (1, 1, 1);

( _b_ ) over the broken line (0, 0, 0) − (0, 1, 0) − (1, 1, 0) − (1, 1, 1);

( _c_ ) over the line _x_ = _y_ = _z_ = _t_ (0   _t_   1);

( _d_ ) over the twisted cubic _x_ = _t_ , _y_ = _t_ 2, _z_ = _t_ 3 (0   _t_   1).

_Solutions_.

( _a_ )  ;

( _b_ )  ;

( _c_ )  ;

( _d_ ) .

### **PROBLEMS**

**1.** Compute the line integral,

over the five paths given in Ex. 43.1.

**2.** Compute the circuit integral in Ex. 43.3 over

( _a_ ) the circle _x_ = _x_ 0 \+ _r_ cos _t_ , _y_ = _y_ 0 \+ _r_ sin _t_ ;

( _b_ ) the ellipse _x_ = _a_ cos _t_ , _y_ = _b_ sin _t_ ;

_c_ ) the square included by the lines _x_ = 0, _y_ = 0, _x_ = 1, _y_ = 1.

**3.** Compute   over the triangle included by the lines _y_ = 1, _x_ = 4, _y_ = _x_.

**4.** Compute the line integral

( _a_ ) over the step path (1, 0) − (2, 0) − (2, 1);

( _b_ ) over the step path (1, 0) − (1, 1) − (2, 1);

( _c_ ) over the line _y_ = _x_ − 1.

**5.** Compute the line integral

( _a_ ) over the step path (1, 2) − (3, 2) − (3, 4);

( _b_ ) over the step path (1, 2) − (1, 4) − (3, 4);

( _c_ ) over the line _x_ − _y_ \+ 1 = 0.

6. Compute   over one turn of the helix

**44. Line Integrals Independent of the Path.** We now inquire under what circumstances the line integral

taken between the points _P_ 1 and _P_ 2, is independent of the curve **r** = **r** ( _t_ ) joining the points. We consider only regular oriented curves, namely curves that are piecewise smooth, without double points,† and directed from _P_ 1 to _P_ 2.

We note that, if   is independent of the path in a region  , then   for all closed curves in the region. For, if _P_ 1 _AP_ 2 _BP_ 1 is a closed curve

hence, if   is independent of the path,  . Conversely, if   over all closed curves of  ,   is the same over any two paths from _P_ 1 to _P_ 2 that do not cross. But, even if the curves cross, the corresponding line integrals are equal; for both equal the integral over a third curve from _P_ 1 to _P_ 2 which cuts neither.‡

THEOREM 1. _If the continuous vector function_ **f** ( **r** ) _is the gradient_ ∇ _F of a single-valued scalar point function in a region_  , _the line integral_   _has the same value over all regular curves in_   _directed from P_ 1 _to P_ 2; _and_

_Proof_. Let **f** = [ _P_ , _Q_ , _R_ ] = ∇ _F_ ; then

by the chain rule; hence

**Example 1.** Since

We next prove the converse of Theorem 1.

THEOREM 2. _If_ **f** ( **r** ) = [ _P_ , _Q_ , _R_ ] _is a continuous vector function in a region_   _in which_   _has the same value over all regular curves directed between the same end points, then in_   **f** _is the gradient of the function_

_Proof_. With a fixed lower limit, the integral (2) is a function _F_ ( _x_ , _y_ , _z_ ) of the upper limit ( _x_ , _y_ , _z_ ); for the path of integration is immaterial. Now, if we compute _F_ ( _x_ \+ _h_ , _y_ , _z_ ) over a curve from ( _x_ 0, _y_ 0, _z_ 0) to ( _x_ , _y_ , _z_ ) and thence to ( _x_ \+ _h_ , _y_ , _z_ ) over a straight segment, we have

by the mean value theorem for integrals.† Therefore

since _P_ ( _x_ , _y_ , _z_ ) is continuous. In similar fashion we show that

and hence

In brief, these theorems state that the line integral   is independent of the path when and only when **f** = ∇ _F_ a gradient vector, or (what is the same thing) **f** · _d_ **r** = _dF_ , a perfect differential. In order to use this test we must learn how to _recognize_ a gradient vector. This will be our next task.

**Example 2.** If we compute the line integral

over two step paths from (1, 2) to (3, 4) and also over the line joining these points, we find in each case that _I_ = 48. This suggests that _I_ is independent of the path. To verify this we observe that

hence, if  ,

and

**Example 3.** If **f** = − **r** / _r_ 3,

Therefore, over any path from _P_ 1 to _P_ 2 that does not pass through the origin,

**45. Irrotational Vectors.** A vector function **f** ( **r** ) is said to be _irrotational_ in a region if rot **f** = **0** there. We have seen in §40 that, if the scalar function _φ_ ( **r** ) has continuous second partial derivatives in a region, its gradient ∇ _φ_ is irrotational:

This raises the question: Can every vector **f** that is irrotational in a region be expressed as a gradient vector? The answer is _yes_ provided the region is _simply connected_ in the sense of the following

DEFINITION. A closed curve is called _reducible_ in a region if it can be shrunk continuously to a point within the region. A region in which all regular closed curves are reducible is called _simply connected_.

Thus a plane region bounded by a single regular closed curve is simply connected; but the region between two concentric circles is not. In three dimensions, the region inside a sphere or rectangular prism is simply connected. The same is true of the region between two concentric spheres. But the region within a torus is not simply connected; for all closed curves that encircle its axis are irreducible.

If **f** = [ _P_ , _Q_ , _R_ ],

hence, rot **f = 0** implies

When _R_ = 0, **f** is a plane vector and rot **f** = **0** implies _Q_ _x_ = _P_ _y_.

We consider the plane case first and for a simple rectangular region.

THEOREM 1. _If the plane vector_ **f** = [ _P, Q_ , 0] _has continuous first partial derivatives in a rectangle and_

_then_ **f** = ∇ _F_ , _a gradient vector._

_Proof_. Define _F_ ( _x, y_ ) by the line integral

taken over the step path ( _x_ 0, _y_ 0) − ( _x, y_ 0) − ( _x, y_ ):

Now from the rules for differentiating an integral,† we have

By hypothesis _Q_ _x_ ( _x, t_ ) = _P_ _t_ ( _x, t_ ); hence

We thus have

Theorem 44.1 now shows that   is independent of the path; hence

over _any_ path in the rectangle from ( _x_ 0, _y_ 0) to ( _x, y_ ).

For example, the step path ( _x_ 0, _y_ 0) − ( _x_ 0, _y_ ) − ( _x, y_ ) yields

where the integrals are written in reverse order.

In three dimensions we have the corresponding

THEOREM 2. _If the vector_ **f** = [ _P_ , _Q_ , _R_ ] _has continuous first partial derivatives in a rectangular prism and_

_then_ **f** = ∇ _F_ , _a gradient vector._

_Proof_. Define _F_ ( _x_ , _y_ , _z_ ) by the line integral,

taken over the step path ( _x_ 0, _y_ 0, _z_ 0) − ( _x, y_ 0, _z_ 0) − ( _x_ , _y_ , _z_ 0) − ( _x_ , _y_ , _z_ ). Then

and we find _F_ _z_ = _R_ ( _x_ , _y_ , _z_ ), _F_ _y_ = _Q_ ( _x_ , _y_ , _z_ ) as before. Moreover

Therefore **f** = [ _P_ , _Q_ , _R_ ] = ∇ _F_ ;   is independent of the path and

over _any_ path in the rectangular prism from ( _x_ 0, _y_ 0, _z_ 0) to ( _x_ , _y_ , _z_ ).

For example, the step path ( _x_ 0, _y_ 0, _z_ 0) − ( _x_ 0, _y_ 0, _z_ ) − ( _x_ 0, _y, z_ ) − ( _x_ , _y_ , _z_ ) yields

where the integrals are written in reverse order.

Formulas (5) and (7) are easily remembered. Both give a function _F_ which vanishes at **r** 0 Changing the initial point **r** 0 merely alters _F_ by a constant. In practice **r** 0 is chosen to simplify the integrals as much as possible. If the region in which rot **f** = **0** includes the origin, this point is usually a good choice.

In mathematical physics it is customary to express an irrotational vector **f** as the _negative_ of a gradient. Thus, if rot **f** = **0** in a simply connected region,

and _φ_ is called the (scalar) _potential_ of **f**.

**Example 1.** When

Integrating over the step path (0, 0) − ( _x_ , 0) − ( _x, y_ ), we get

and, over (0, 0) − (0, _y_ ) − ( _x, y_ ),

where the integrals are written in reverse order.

We can now compute the line integral between any two points:

**Example 2.** When

Integrating over the step path (0, 0, 0) − ( _x_ , 0, 0) − ( _x, y_ , 0) − ( _x_ , _y_ , _z_ ), we get

and, over the step path (0, 0, 0) − (0, 0, _z_ ) − (0, _y, z_ ) − ( _x_ , _y_ , _z_ ),

where the integrals are written in reverse order.

Although step paths usually give the simplest calculation for _F_ ( _x_ , _y_ , _z_ ), a straight segment joining ( _x_ 0, _y_ 0, _z_ 0) to ( _x_ , _y_ , _z_ ) is sometimes preferable. If **f** ( **r** ) is continuous when **r** = **0** , we can integrate along the radial line from **0** to **r**. The points of this line have position vectors _t_ **r** (0   _t_   1), and hence

In particular when **f** =  _P_ , _Q_ , _R_ ] is homogeneous of degree _n_ , **f** ( _t_ **r** ) = _t_ _n_ **f** ( **r** ) and [(9) becomes

If _n_ < − 1, we may integrate from _t_ = ∞ to _t_ = 1 and again obtain (10).

**Example 3.** In Ex. 1, **f** is homogeneous of degree 2; hence from (10)

**Example 4.** In Ex. 2, **f** is not homogeneous, but (9) is applicable. Thus

**Example 5.** The vector **f** = _r_ _n_ −1 **r** is a gradient vector; for **r** / _r_ = ∇ _r_ (37.15), and

The function _F_ may also be found from (10) if _n_  −1.

**46. Simply Connected Regions.** When rot **f** = 0 in a rectangular region, we were able to construct a function   by integrating over a step path from _P_ 0 to an arbitrary point _P_ of the region. This construction cannot be used for simply connected regions in general; for, starting from _P_ 0, we may not be able to reach _P_ with a step path that lies in the region. To overcome this difficulty requires a rather detailed argument which we shall only sketch in outline.†

Consider a variable polygon Γ( _t_ ) in the region whose vertices are given by position vectors **r** 1( _t_ ), **r** 2( _t_ ), ···, **r** _k_ ( _t_ ) which are continuous functions of the time _t_ such that Γ( _t_ ) shrinks to a point as _t_ varies from 0 to 1. If we can show that the circuit integral

while Γ( _t_ ) shrinks to a point, the constant must be zero. Since the original polygon given by **r** 1(0), **r** 2(0), ···, **r** _k_ (0) is arbitrary, this means that   over any closed polygon in the region. Hence   is the same for all polygonal paths between **r** 0 and **r** and defines a function _F_ ( **r** ). We can now show as in §45 that **f** = ∇ _F_ and that the line integral has the same value over all regular curves between the same end points.

Thus the proof hinges on (1). To establish this result consider two polygons Γ( _t_ ) and Γ( _t_ ′) where 0   _t_ < _t_ ′   1. Choose _n_ points _A_ 1, _A_ 2, ···, _A_ _n_ on Γ( _t_ ), so that each pair _A_ _i_ _A_ _i_ +1 lies in a rectangle _R_ (or rectangular prism) inside our region. As _t_ → _t_ ′, let _A_ _i_ → _B_ _i_ on Γ( _t_ ′) (Fig. 46 _a_). Now choose _t_ ′ − _t_ so small that _B_ _i_ _B_ _i_ +1 also lies in _R_ _i_ ( _i_ = 1, 2, ···, _n_ ). When rot **f** = **0** , the circuit integral over the quadrangle _A_ _i_ _A_ _i_ +1 _B_ _i_ +1 _B_ _i_ _A_ _i_ is zero by Theorem 1 or  of §45; and, if we add all such circuit integrals, we have

Thus (1) is established.

Fig. 46 _a_. Polygon shrinking to a point.

Our final result is stated in the

THEOREM. _If the vector function_ **f** ( **r** ) _has continuous first partial derivatives in a simply connected region, the line integral_   _taken over any curve inside the region will be independent of the path when and only when_ rot **f** = **0**.

_Proof_. We have just shown that in a simply connected region rot **f** = **0** implies that   is independent of the path. Conversely, if   is independent of the path, **f** is a gradient vector (Theorem 44.2), and hence rot **f** = **0**.

If **f** satisfies the conditions of the theorem in a simply connected region _V_ except on a surface _S_ where **f** has finite discontinuities, we still have   over all closed curves in _V_ if _the tangential component of_ **f** _is continuous on S_.

Fig. 46 _b_. Surface of discontinuity.

_Proof_. Let _S_ divide _V_ into two parts, _V_ 1 and _V_ 2, and _PAQB_ be a closed curve cutting _S_ at _A_ and _B_ (Fig. 46 _b_). Then, if _AB_ is a regular curve on _S_ ,

and, on addition,

Hence   for all closed paths in _V_ When and only when the line integral

for _all_ paths on _S_. This requires that **f** 1 · **T** = **f** 2 · **T** at all points of _S_ ; that is, **f** · **T** is continuous on _S_.

**Example.** The vector function

is continuous everywhere except at (0, 0). In the region _R_ consisting of the entire plane outside of a small circle about the origin, **f** , **f** _x_ , **f** _y_ are all continuous; moreover

Thus   about any curve of _R_ that does not encircle the origin; for such a curve can be embedded in a simply connected subregion of _R_. But, for any curve about the origin,

on putting **_x_** = _r_ cos _φ_ , _y_ = _r_ sin _φ_ ; such a curve cannot be embedded in a simply connected subregion of _R_. Thus, in the doubly connected region _R_ ,   is _not_ independent of the path even though rot **f** = **0** everywhere in _R_.

### **PROBLEMS**

**1.** Express

as ∇ _F_ , and compute  .

**2.** If the integral

defines a function _F_ ( _x, y_ ) when _y_ > 0, find it.

**3.** Express

as ∇ _F_ , and compute  .

**4.** Compute

over any path that avoids the origin. When ( _x, y_ ) is ( _a, b_ ), find the value of the circuit integral. Is your answer correct if the circuit encloses the origin?

**5.** If the vector

prove that   over any closed curve.

Compute  .

**6.** Show that the vector

is continuous and irrotational throughout space. Find a _continuous_ function _F_ ( _r_ ) such that **f** = ∇ _F_.

Compute   over any path.

**47. Green's Theorem in the Plane.** Let   be a closed region that is bounded by one or more simple closed curves. Then the _positive sense_ in the boundary is the one that leaves   to the left. Thus, for the region within a closed curve _C_ (Fig. 47 _a_), the positive sense over _C_ is counterclockwise; but, for the region interior to _C_ 1, exterior to _C_ 2 and _C_ 3 (Fig. 47 _c_), the positive sense is counterclockwise for the outer boundary, clockwise for the inner boundaries.

GREEN'S THEOREM. _Let_   _be a closed region bounded by one or more closed regular_† _curves and capable of dissection into subregions whose boundaries are cut in at most two points by parallels to the axes. Then, if P_ ( _x, y_ ), _Q_ ( _x, y_ ), _P_ _y_ , _Q_ _x_ _are continuous in_  ,

Fig. 47 _a_. Region bounded by a simple closed curve.

_where the circuit integral is taken over the boundary of  in the positive sense_.

_Proof_. Suppose first that   is bounded by single closed curve _C_ which is cut in at most two points by any line parallel to the axes (Fig. 47 _a_).

We first compute the double integral   as a repeated integral,‡ integrating first with respect to _x_ between the curves _x_ = _x_ 1( _y_ ) and _x_ = _x_ 2( _y_ ), and then with respect to _y_ between the parallels _y_ = _c_ and _y_ = _d_ , which limit the curve below and above:

where the integrals are taken over the arcs _αβγ_ and _γδα_ , respectively; hence

We next compute   as a repeated integral, integrating first with respect to _y_ between the curves _y_ = _y_ 1( _x_ ) and _y_ = _y_ 2( _x_ , and then with respect to _x_ between the parallels _x_ = _a_ and _y_ = _b_ , which limit the curve to the left and right:

where the integrals are taken over the arcs _δαβ_. and _βγδ_ , respectively;

hence

On adding (2) and (3) we get (1).

We may now extend equation (3) to regions that may be dissected into a finite number of subregions having the property stated in the theorem. For each subregion equation (3) is valid; and, when these equations are added, the double integrals combine into an 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 circuit integral over the external boundary (Fig. 47 _b_).

Fig. 47 _b_. Dissected region.

A region   bounded by a single closed curve is _simply connected_ (§46); for any closed curve in   can be continuously contracted into a point within the region. But, if the boundary of   consists of two or more closed curves (Fig. 47 _c_), it is _multiply connected_ ; for a closed curve about _C_ 2 or _C_ 3 cannot be contracted to a point without crossing _C_ 2 or _C_ 3. But the region   of Fig. 47 _c_ can be made simply connected by two cross cuts, _aa_ ′ and _bb_ ′, joining _C_ 1 to _C_ 2 and _C_ 3; and, if we pass around the new boundary as shown by the arrows (each cross cut being traversed twice), the entire circuit is made in the positive sense.

We can now apply Green's theorem (1) to region of Fig. 47 _c_ made simply connected with cross cuts. In the left member the double integral covers the region interior to _C_ 1 and exterior to _C_ 2 and _C_ 3. In the right member the line integral comprises circuits in the positive sense over _C_ 1, _C_ 2 and _C_ 3; for the cross-cut integrals cancel as they are traversed twice and in opposite senses.

An internal boundary of a region may consist of a single point _P_ 0 which must be excluded from   to meet the continuity requirements of Green's theorem. Such a _punctured_ region is multiply connected; for a curve about _P_ 0 cannot be contracted to a point within the region without crossing _P_ 0. If the region consists of the entire plane punctured at _P_ 0, a cross cut from _P_ 0 to infinity will make it simply connected; for all closed curves not crossing the cut may be contracted to a point.

It can be shown by a rather intricate argument that _Green_ ' _s theorem holds for any region bounded by curves that are piecewise smooth_. This does not follow from the argument just given; for this is only valid for regions that may be dissected into subregions which are cut in at most two points by parallels to the axes. Thus a region, bounded in part by the curve _y_ = _x_ 2 sin (1/ _x_ ) (which is smooth when _x_ > 0), is cut by the _x_ -axis in an infinite number of points and could not be dissected as postulated above. Nevertheless Green's theorem holds good even with such eccentric regional boundaries. The proof depends on a limiting process.

Fig. 47 _c_. Multiply connected region made simply connected by cross cuts.

Green's theorem is also valid when _P_ ( _x, y_ ) and _Q_ ( _x, y_ ) are replaced by vector functions

whose components _P i_( _x, y_ ), _Q i_( _x_ , _y_ ) satisfy the stated requirements. To show this we need only apply the theorem to the three pairs of functions _P i_( _x, y_ ), _Q i_( _x_ , _y_ ), multiply the equations by **i** , **j** , **k** respectively, and add them.

We note, finally, that Green's theorem holds when _P_ and _Q_ are dyadics. The integrals of dyadics require no new definitions; for any dyadic **P** = ∑ **a** _i_ **b** _i_ can be reduced to the form

by expressing each antecedent **a** _i_ in the form (8.1), expanding into a sum of dyads, and then collecting the terms in **i** , **j** , **k** by means of the distributive laws (39.4–5). Thus, when **P** and **Q** are dyadics (such that **P** , **Q** , **P** _y_ , **Q** _x_ are continuous in _R_ )

and hence

**Example 1.** When _Q_ = _x_ /2 =, _P_ = − _y_ /2, _Q x_ − _P_ _y_ = 1; hence from Green's theorem

the area enclosed by the curve (43.5).

On putting _Q_ = _x_ , _P_ = 0 or _Q_ = 0, _P_ = − _y_ , we have also

**Example 2.** With _Q_ = _x_ 2, _P_ = − _y_ 2, _Q x_ − _P_ _y_ = 2( _x_ \+ _y_ ); hence from Green's theorem

where ( _x_ *, _y_ *) is the centroid of the area enclosed by the curve.

We may use (7) to check the results in Ex. 43.3–.

For the triangle of Ex. 43.3, _A_ = 1, _x_ * = 0,  . hence the circuit integral equals  .

For the circle of Ex. 43.4, _A_ = _π_ , _x_ * = 0, _y_ * = 1; hence the circuit integral equals −2 _π_.

**Example 3.** If the closed curve _C_ of Fig. 47 _a_ is revolved about the _x_ -axis, the volume generated is

If _C_ is revolved about the _y_ -axis, the volume generated is

If we apply Green's theorem to these circuit integrals we find

where ( _x_ *, _y_ *) is the centroid of the area _A_ enclosed by _C_.

These results are known as the theorem of Pappus.

### **PROBLEMS**

**1.** Solve Prob. 43.2 by Green's theorem.

**2.** Solve Prob. 43.3 by Green's theorem.

**3.** Find the area of a sector of the ellipse _x_ = _a_ cos _t_ , _y_ = _b_ sin _t_ from _t_ = 0 to _t_ = _φ_.

**4.** Find the area of a sector of the hyperbola _x_ = _a_ cosh _t_ , _y_ = _b_ sinh _t_ from _t_ = 0 to _t_ = _φ_.

**5.** Find the area of the loop of the _folium of Descartes_ :

**6.** The epicycloid formed by a circle of radius _b_ rolling over a circle of radius _a_ has the parametric equations

where _k_ = _b_ / _a_ and _θ_ is the angle through which the moving circle turns. Show that the area of one arch of the epicycloid exterior to the fixed circle is

**7.** Show that the vector area (§11) within a closed curve traversed counterclockwise is

Hence show that a polygon in the _xy_ -plane whose vertices have the position vectors **r** 1, **r** 2, ···, **r** _n_ has the vector area

**8.** Show that the area within a closed curve in the _xy_ -plane has the moments of inertia

about the axes of coordinates.

Find _I x_, _I y_, _I_ _z_ for the ellipse _x_ = _a_ cos _t_ , _y_ = _b_ sin _t_.

**9.** Prove that

the moment of inertia about the _z_ -axis of the enclosed area.

Find _I_ _z_ for the rectangle whose vertices are _x_ = ± _a_ , _y_ = ± _b_.

**10.** Compute

over the ellipse _x_ = 2 cos _t_ , _y_ = sin _t_ , and verify by Green's theorem.

**11.** Compute   over the area inside the square (0, 0) = (4, 0) = (4, 4) = (0, 4) − (0, 0) and outside of the rectangle (1, 1) − (2, 1) − (2, 3) − (1, 3) − (1, 1). Verify by transforming the integral into a line integral over the entire boundary.

**12.** Compute   over the boundary given in Prob. 11. Verify your answer.

**13.** Find the area between the parabolas 4 _y_ = _x_ 2, 4 _x_ = _y_ 2 by circuit integration.

**14.** Compute   over the triangle bounded by the lines _y_ = 0, _x_ = 1, _y_ = _x_ , and verify by Green's theorem.

**48. Vector Forms of Green's Theorem.** Green's theorem admits of two vector interpretations.

**1.** Let us put

Since the unit tangent vector to the boundary of   is

equation (47.1) becomes

where _dA_ = _dx dy_ is the element of area. We shall see that (1) is a special case of Stokes' theorem (§50).

The integral of the tangential component of a vector about a curve is called its _circulation_. Thus equation (1) states:

_The circulation of a vector about a plane curve is equal to the integral of the normal component of its rotation over the enclosed area. The sense of circulation and the normal form a right-handed screw_.

**2.** Let us put

Since the unit exterior normal vector to the boundary is

equation (47.1) becomes

We shall see that (2) is the plane version of the divergence theorem (§52).

The integral of the outward normal component of a vector about a curve is called its _normal flux_. Thus equation (2) states:

_The normal flux of a vector across a closed plane curve is equal to the integral of its divergence over the enclosed area_.

Thus, if we apply (1) and (2) to the position vector **r** = [ _x_ , _y_ , 0] in the _xy_ -plane, we obtain

Green's theorem in form (1) affords a simple proof of the

THEOREM. _If_ **f** _is a plane vector with continuous first partial derivatives in a region_  ,

_over all regular closed curves in the region; and_ , _conversely_ ,

_provided the region can be dissected into subregions whose boundaries are cut in at most two points by parallels to the axes._

_Proof_. When rot **f** = 0 throughout  , (1) shows that   over all regular closed curves in  .

Conversely, if   over all regular closed curves, then

For, if the _continuous_ function _Q x_ − _P_ _y_  0 at any interior point ( _x_ 0, _y_ 0) of the region, we can always find a circle about ( _x_ 0, _y_ 0) in which _Q x_ = _P_ _y_ keeps the same sign.† On integrating over its boundary we have   from (1) contrary to hypothesis. Thus _Q x_ − _P_ _y_ = 0 at all interior points, and, by continuity, also on the boundary. But _Q x_ − _P_ _y_ = 0 implies rot **f** = **0**.

**49. Surface Integrals.** The parametric equations of a surface _S_ ,

may be combined into a single vector equation

The _parametric curves_ on the surface,

have tangent vectors **r** _u_ and **r** _υ_ , respectively, and at a _regular point_ of the surface (§36) the normal vector **r** _u_ × **r** _υ_  0. The lengths of the vectors **r** _u_ _du_ , **r** _υ_ _dυ_ are the elements of arc on the parametric curves; and the vector area of the parallelogram they form,

is called the _vector element of area_ on the surface. Its magnitude

is the _scalar element of area_. If we regard the parameters _u_ , _υ_ as rectangular coordinates in the _uυ_ -plane, any region _S_ ′ of this plane corresponds to a portion of the surface whose area is _defined_ as the double integral

It can be shown that the surface area thus computed is independent of the choice of rectangular axes _x_ , _y_ , _z_ and also of the choice of parameters _u_ , _υ_.† Moreover _dS_ reduces to _dx dy_ on the _xy_ -plane: _x_ = _u_ , _y_ = _υ_ , _z_ = 0; for

The quantity | **r** _u_ × **r** _υ_ | may be expressed in terms of the _fundamental quantities_ for the surface, namely,

From (14.2),

and hence

On a _bilateral_ (two-sided) surface it is possible to distinguish one side from the other—by painting the sides with different colors for example.‡ On a bilateral surface **r** = **r** ( _u_ , _υ_ ), the positive side is the side toward which the normal **r** _u_ × **r** _υ_ points. If **f** ( **r** ) is any vector function defined over the surface, the integral

is called a _surface integral over the positive side of S_. On putting

where **n** is a _unit_ normal in the direction of **r** _u_ × **r** _υ_ , we may also write (8) in the form

where **f · n** is the normal component of **f** over the surface.

Compare now the line integral

with the surface integral

the former integrates _tangential_ components of a vector over a curve, the latter integrates _normal_ components of a vector over a surface.

**50. Stokes' Theorem.** Consider a bilateral surface that is _piecewise smooth_. Such a surface consists of portions _S_ having an equation **r** = **r** ( _u_ , _υ_ ) and over which the normal vector **r** _u_ × **r** _υ_ is continuous; furthermore, we assume that **r** _uυ_ = **r** _υu_.† The positive side of _S_ is the side toward which **r** _u_ × **r** _υ_ points; and a person, erect on the _positive_ side, will have _S_ to his left when he traverses its boundary in the _positive sense_.

The vector form of Green's theorem,

which converts the surface integral of rot **f** over a plane region into the circuit integral of **f** over its boundary, may be generalized to apply to curved surfaces; we need only replace the vector element of area,

where **n** is a unit normal in the direction of **r** _u_ × **r** _υ_

STOKES' THEOREM. _Let S be a surface consisting of portions that are piecewise smooth and bounded by regular closed curves. Then, if the vector function_ **f** ( **r** ) _has a continuous gradient_ ∇ **f** _over S_ , _the surface integral of_ rot **f** _over S is equal to the circuit integral of_ **f** _over its boundary C in the positive sense_ :

_Proof_. It will suffice to prove (1) for a smooth portion of the surface over which **r** _uυ_ = **r** _υu_ ; for, if such equations for all the smooth portions are added, we obtain (1). Note that in adding the circuit integrals all internal boundaries are traversed twice and in opposite directions so that the corresponding line integrals cancel.

Fig. 50 _a_. Surface _S_ mapped on a plane region _S_ ′.

We may therefore assume that _S_ is smooth and bounded by a regular closed curve _C_. Let _S_ ′ be the region of the _uυ_ -plane which contains all parameter values _u_ , _υ_ corresponding to points of _S_ (Fig. 50 _a_). Moreover, we suppose that _S_ is so restricted that the correspondence ( _u_ , _υ_ ) ↔ ( _x_ , _y_ , _z_ ) is one-to-one. If we put

the surface integral in (1) becomes

taken over the region _S_ ′ of the _uυ_ -plane. The curve _C_ forming the boundary of _S_ maps into a curve _C_ ′, the boundary of _S_ ′; and the circuit integral in (1) may be computed as a circuit integral over _C_ ′:

We now apply Green's theorem in the _uv_ -plane to show that the integrals (2) and (3) are equal.

On making use of formula (14.2),

we find that

and hence, from the chain rule,

since **r** _υu_ = **r** _uv_. If we substitute this value in (2), we may transform the integral by Green's theorem letting _u_ , _v_ , **r** _u_ · **f** , **r** _u_ · **f** correspond to _x_ , _y_ , _P_ , _Q_ in (47.1); thus we find

which is precisely the integral (3). Thus the proof is complete.

Fig. 50 _b_. Circuit integral computed by Stokes' theorem.

**Example**. Let us verify Stokes' theorem for the vector **f** =  _x_ \+ _y_ , 2 _x_ − _z_ , _y_ \+ _z_ ] taken over the triangle _ABC_ cut from the plane 3 _x_ \+ 2 _y_ \+ _z_ = 6 by the coordinate planes ([Fig. 50 _b_).

If the parametric equations of the plane are taken as

its positive normal **n** has the direction of

and the vector element of area is

Since rot **f** = [2,0, 1],

for the area of the triangle _AOB_ (the projection of _ABC_ on the _xy_ -plano) is 3.

Corresponding to the direction of **n** , the positive sense of circuit is _ABC_. The circuit integral,

is the sum of the line integrals over _AB_ , _BC_ , _CA_ which prove to be 1, 36, –16, respectively. Thus the theorem is verified (21 = 1 + 36 − 16).

### **PROBLEMS**

**1.** Verify Stokes' theorem for **f** = [2 _y_ \+ _z_ , _x_ − _z_ , _y_ − _x_ ] taken over the triangle _ABC_ cut from the plane _x_ \+ _y_ \+ _z_ = 1 by the coordinate planes.

[If _A_ (1, 0, 0), _B_ (0, 1, 0), _C_ (0, 0, 1), the line integrals over _AB_ , _BC_ , _CA_ , are  , 1, 1;

hence   over _ABC_.]

**2.** Verify Stokes' theorem for the vector **f** = [ _z_ , _x_ , _y_ ] taken over the half of the sphere _x_ 2 \+ _y_ 2 \+ _z_ 2 = _a_ 2 lying above the _xy_ -plane.

**3.** Compute   **f** · _d_ **r** around the circle ( _x_ − l)2 \+ _y_ 2 = 1, _z_ = 3 when **f** = [– _y_ , _x_ , 2].

**4.** If _u_ ( _x_ , _y_ , _z_ ), _v_ ( _x_ , _y_ , _z_ ) have continuous gradients, show that

where _S_ is a portion of a surface bounded by a closed curve _C_.

**5.** When _u_ = ρ2/2, _υ_ = _φ_ in Prob. 4, show that

[See Ex. 37.2; _ρ_ , _φ_ are the cylindrical coordinates of Ex. 42.1.]

**51. Generalized Stokes' Theorem.** With the same conditions on the function **f** ( **r** ) and the surface, we can prove the more general transformation

between the integrals of dyadics. In the surface integral, the vector _d_ **S** = ( **r** _u_ × **r** _υ_ ) _du dυ_ is cross-multiplied into the vectors of ∇ **f** on the left: thus

If we now make use of the formula

we find that

Hence, on applying Green's theorem (47.4) in which **P** and **Q** are dyadics, we have

as stated in (1). In this equation the _order_ of the vectors is essential.

The dyadic equation (1) gives two others on taking scalar and vector invariants (§39). Since dot and cross multiplication is distributive with respect to addition, either invariant of a dyadic integral is equal to integral of the invariant. Thus both scalar and vector invariants of the dyadic integrals in (1) may be taken _under the integral sign_.

On taking the scalar invariant, the integrand of the surface integral (apart from _du dυ_ ) becomes

and the circuit integral becomes   _d_ **r** · **f**. Thus, in view of (50.2), we again obtain Stokes' theorem.

On taking the vector invariant, the integrand of the surface integral becomes

and the circuit integral becomes   _d_ **r** × **f**. Thus, on writing **r** _u_ × **r** _υ_ _du dυ_ = **n** _dS_ , we obtain the new theorem:

For example, if **f** = **r** , the position vector to the surface, ∇ **r** = **I** , div **r** = 3 from (40.8), and (3) becomes

This expresses the vector area of a surface as a circuit integral about its boundary. Thus over the upper half _H_ of a hemisphere of radius _a_ , **r** = _a_ **R** , _d_ **r** = a **P** _dθ_ , and its vector area

### **PROBLEMS**

**1.** If the closed curve _C_ encloses a portion _S_ of a surface, show that

**2.** If **f** = **i** _u_ ( _x_ , _y_ ) + **j** _υ_ ( _x_ , _y_ ), prove that

where _C_ is a closed curve in the _xy_ -plane enclosing the region _A_.

**3.** Prove that the vector area of a closed surface is zero.

**52. Divergence Theorem.** A basic transformation of volume to surface integrals is given by the

DIVERGENCE THEOREM (GAUSS). _If the vector function_ **f** ( **r** ) _has continuous first partial derivatives in the region V bounded by a closed surface S over which the unit external normal_ **n** _is sectionally continuous, then the volume integral of_ div **f** _within S is equal to the external surface integral of_ **f** _over S:_

Fig. 52 _a_. Region bounded by a closed surface.

_Proof_. Since

the volume integral of div **f** is the sum of three integrals. In order to compute the integral of **k** · **f** _z_ , we first consider a surface _S_ which is cut in at most two points by a line parallel to the _z_ -axis (Fig. 52 _a_); denote them by ( _x_ , _y_ , _z_ 1) and ( _x_ , _y_ , _z_ 2) where _z_ 1 < _z_ 2. Then _S_ has a lower portion _S_ 1 consisting of the points ( _x_ , _y_ , _z_ 1) and an upper portion _S_ 2 consisting of the points ( _x_ , _y_ , _z_ 2). The points for which _z_ 1 = _z_ 2 form a closed curve separating _S_ 1 from _S_ 2. Now, if _S_ 1 and _S_ 2 project into an area _A_ of the _xy_ -plane,

If we regard _x_ , _y_ as the parameters _u_ , _v_ on the surfaces _S_ 1 and _S_ 2, the vector element of surface is **r** _x_ × **r** _y_ _dx dy_ (49.3). Now from

Over _S_ 2, _z_ = _z_ 2( _x_ , _y_ ), and the vector **r** _x_ × **r** _y_ has the direction of the external normal **n** ; but, over _S_ l, _z_ = _z_ 1( _x_ , _y_ ), and **r** _x_ × **r** _y_ has the direction of the internal normal – **n**. Hence, if **n** _dS_ denotes the _external_ vector element of area on _S_ ,

where the plus sign applies to _S_ 2, the minus to _S_ 1 The two integrals over _A_ may now be combined into a single integral over _S_ , and we may write

This formula is also valid when _S_ is bounded laterally by a part of a cylinder parallel to the _z_ -axis and separating _S_ 1 from _S_ 2. For (2) holds as before; and, in (3), **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 that do satisfy this condition and apply formula (3) to each point and add the results, the volume integrals will combine to the left member of (3); 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 of (3).

Finally we may extend (3) 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 (3) we make cyclic interchanges in _x_ , _y_ , _z_ , we obtain the corresponding formulas:

Now dot-multiply (3), (4), (5) by **k** , **i** , **j** , respectively, and add the resulting equations. Then the volume integral becomes

and the surface integral

This proves formula (1).

Volume and surface integrals are often denoted by a single integral sign when the differential _dV_ or _dS_ indicates their character. Moreover   ··· _dS_ is used to denote integration over a _closed_ surface. In this notation the divergence theorem becomes

With suitable conditions on **f** , the divergence theorem applies to an _infinite region V_ ∞ outside of a closed surface _S_.

THEOREM 2. _With the same conditions on_ **f** _and S as before, we have_

_provided_

_where **r** is the distance from any fixed point P, and the unit normal_ **n** , _external to the region V_ ∞, _points toward the interior of S_.

_Proof_. Let ∑ be a sphere about _P_ large enough to enclose _S_ completely. Then the divergence theorem applied to the region _V_ between _S_ and ∑ gives

**n** being an external normal with respect to the region _V_. The uniform approach of _r_ 2 **f** to zero means that

a suitably large value. If the radius _r_ of ∑ is now chosen > _R_ , we have

and hence the surface integral over ∑ approaches zero as _r_ → ∞. At the same time _V_ → _V_ ∞, and the theorem is proved.

**Example 1.** When **f** = **r** , div **r** = 3; then from (1)

where _p_ is the perpendicular from _O_ on the tangent plane to _S_ at the end point of **r**. Thus the volume within _S_ is expressed as a surface integral over _S_.

For a _cone_ of base _B_ , altitude _h_ , take _O_ at the vertex. Then _p_ = 0 over the lateral surface, _p_ = _h_ over the base, and  .

For a _sphere_ of radius _a_ , take _O_ at the center. Then _p_ = _a_ and  .

**Example 2. Solid Angle.** The rays from a point _O_ through the points of a closed curve generate a cone; and the surface of a unit sphere about _O_ intercepted by this cone is called the _solid_ angle Ω 1 of the cone.

Let us apply the divergence theorem to the vector

in the region interior to a cone of solid angle Ω, which is limited externally by a surface _S_ , internally by a portion σ of a small sphere of radius _a_ about _O_ (Fig. 52 _b_). Within this region **f** has continuous first partial derivatives, and from Ex. 41.1.

Over the small sphere the external normal **n** = **–r** / _a_ while over the conical surface **n** · **r** = 0; hence,

But, since _σ_ / _a_ 2 = Ω/12 we have

If _S_ is closed surface, we have

Fig. 52 _b_. Solid angle.

When _O_ is outside of _S_ , **f** has continuous partial derivatives throughout its interior, and (9) follows at once from (1). In this case the elements of solid angle _d_ Ω = **n** · **r** _dS_ / _r_ 3 corresponding to the same conical element cancel in pairs.

**Example 3.** If div **f** = 0 in a region _V_ , the normal flux of **f** is zero over any closed surface which satisfies the requirements of the divergence theorem:

Suppose now that div **f** = 0 in a simply connected region _V_ except on a surface _S_ dividing _S_ in two parts _V_ 1 and _V_ 2 and on which **f** has finite discontinuities. We can then show by an argument, which parallels that given at the end of §46, that   over all closed surfaces in _V_ if _the normal component of_ **f** _is continuous on S_.

### **PROBLEMS**

**1.** Verify the divergence theorem for the vector **f** = _x_ 2 **i** \+ _y_ 2 **j** \+ _z_ 2 **k** taken over the cube 0   _x_ , _y_ , _z_   1.

**2.** Find   for the vector **f** = _x_ **i** − _y_ **j** \+ 2 _z_ **k** over the sphere _x_ 2 \+ _y_ 2 \+ ( _z_ − l)2 = l.

**3.** Find a vector **f** = _g_ ( _r_ ) **r** such that div **f** = _r_ m( _m_ > − 3). Prove that

**4.** The vector **f** = _k_ ∇(1/ _r_ ), where _r_ is the distance from a point _O_ , represents an attraction which varies as 1/ _r_ 2 (inverse square law). If _S_ is a closed surface about _O_ , prove that

**5.** If **f** = ∇φ and div **f** = **0** , show that div ( _φ_ **f** ) = | **f** |2 and hence

where _V_ is the region within the closed surface _S_.

**6.** Prove that, when _m_ > − 1,

Apply ([53.3).]

**7.** If **f** = [ _x_ 2 − _z_ 2, 2 _xy_ , _y_ 2 \+ _z_ ], find   over

( _a_ ) the surface of a cube bounded by the six planes _x_ = _y_ = _z_ = 0, _x_ = _y_ = _z_ = 1.

( _b_ ) the surface bounded by the cylinder _y_ 2 \+ _z_ 2 = 4 and the planes _x_ = 0, _x_ = 2.

**8.** The vector

Show that   over any closed surface.

**9.** Apply the divergence theorem to **P** × rot **Q** to prove the identities

**10.** Compute   ( _ax_ 2 \+ _by_ 2 \+ _cz_ 2) _dS_ over the sphere _x_ 2 \+ _y_ 2 \+ _z_ 2 = 1. Use ([52.1); **n** = [ _x_ , _y_ , _z_ ]; find **f**.]

**11.** Compute   ( _a_ 2 _x_ 2 \+ _b_ 2 _y_ 2 \+ _c_ 2 _z_ 2)−1/2 _dS_ and   ( _a_ 2 _x_ 2 \+ _b_ 2 _y_ 2 \+ _c_ 2 _z_ 2)1/2 _ds_ over the ellipsoid _ax_ 2 \+ _by_ 2 \+ _cz_ 2 = 1. Use ([52.1); find **n** , then **f**.]

**53. Generalized Divergence Theorem.** We may use equations (3), (4), (5) of the preceding article to prove the more general

THEOREM. _If the vector_ **f** ( **r** ) _has continuous first partial derivatives in a region V bounded by a closed surface S_ ,

_Proof_. The volume integral in (1) equals

and, since

the surface integral equals

Since the corresponding integrals in (i) and (ii) are equal, (1) is established.

The dyadic equation (1) gives two others on taking scalar and vector invariants under the integral signs as in §52. Since the scalar invariant of ∇ **f** is div **f** , the first of these equations is again the divergence theorem. But the vector invariant of ∇ **f** is rot **f** , and we obtain the new theorem:

If in (1) we put **f** = _f_ **e** , where the scalar _f_ ( **r** ) has continuous first partial derivatives and **e** is a constant vector, we obtain

**Example 1.** Put _f_ = 1 in (3); then

Put **f** = **r** in (2); since rot **r** = **0** ,

From (4) and (5) we conclude that a closed surface is in equilibrium under any system of normal pressures − **n** _p_ of constant magnitude _p_ ; for their vector sum and vector moment about _O_ are zero.

**Example 2.** Put   in (3); then ∇ _f_ = **r** , and the left member

where **r** * locates the centroid of _V_ ; hence

**54. Solenoidal Vectors.** A vector function **f** ( **r** ) is said to be solenoidal in a region _V_ if div **f** = 0 in _V_.

THEOREM 1. _The vector rot_ **g** _is solenoidal in any region in which_ **g** ( **r** ) _has continuous second partial derivatives._

_Proof_. Let **g** = **i** _g_ 1 \+ **j** _g_ 2 \+ **k** _g_ 3; then (§40)

and

since the mixed derivatives cancel in pairs.

COROLLARY. _The vector_ ∇ _u_ × ∇u _is solenoidal_ ; for

We shall now prove the converse of Theorem 1. To avoid complications we consider the case when the region _V_ is a rectangular prism.

THEOREM 2. _If_ div **f** = 0 _in a rectangular prism V, a vector point function_ **g** _can be defined in V so that_

_Proof_. If **f** = **if** 1 \+ **j** _f_ 2 \+ **k** _f_ 3,

We shall first obtain a particular solution **G** of (2) for which _G_ s = 0:

In view of (1), **f** = rot **G** is equivalent to three scalar equations:

If ( _x_ 0, _y_ 0, _z_ 0) is any fixed point of _V_ , the first two equations of (4) are satisfied by

In these integrations _x_ and _y_ are regarded as constant parameters, and _α_ ( _x_ , _y_ ) is a function as yet undetermined. In order that _G_ 1, _G_ 2 satisfy the third equation of (4) we must have

or, in view of (3),

When we perform the integration this reduces to

an equation that is satisfied by taking

Hence the vector **G** whose components are

is a particular solution of (2)—as we readily verify by computing rot **G**.

If **g** is any other solution of (2),

Thus **g** − **G** is irrotational and may be expressed as a gradient vector ∇ _F_ (Theorem 45.2). Hence **g** − **G** = ∇ _F_ , and the general solution of (2) is

where _F_ ( **r** ) is an arbitrary twice-differentiable scalar function. The rectangular components of **g** are obtained by adding ∂ _F_ /∂ _x_ , ∂ _F_ /∂ _y_ , ∂ _F_ /∂ _z_ 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**.

If _C_ is a closed curve in the region _V_ , and _S_ is any surface spanning _C_ that satisfies the requirements of Stokes' theorem, the normal flux

Fig. 54. Closed conical region.

through _S_ of a solenoidal vector **f** = rot **g** depends only on _C_ and the vector potential **g** ; for from (50.1)

The vector potential of a solenoidal vector **f** may be found without resort to components.† From the divergence theorem

for any closed surface _S_. Let _S_ 1 denote the lateral surface of a cone with vertex at _O_ and having the closed curve _C_ as directrix and boundary (Fig. 54). We now apply (8) to the _closed_ surface formed by _S_ 1 and the "base" _S_ 2 of the cone (a portion of a surface bounded by _C_ ). By Stokes' theorem, _S_ 2 contributes

to the integral (8). If **r** = **r** ( _s_ ) is the equation of _C_ , _S_ 1 has the vector equation

and its vector element of area in the direction of the external normal **n** 1 is

Hence the contribution of _S_ 1 to (8) is

Since the sum of the integrals (9) and (10) is zero, we have

As this holds for an arbitrary closed curve _C_ , the vector in brackets is zero; hence,

whenever the integral exists.

When **f** ( **r** ) is homogeneous of degree _n_ , **f** ( _t_ **r** ) = _t_ _n_ **f** ( **r** ), and

When _n_ < −2, (12) is still valid; for we can integrate from _t_ = ∞ to _t_ = 1 over the portion of the cone beyond _S_ 2.

**Example 1.** When **f** =  _y_ , _z_ , _x_ ], div **f** = 0. Equation [(6) gives the vector potential

Since **f** is homogeneous of degree 1, (12) gives

The difference

is irrotational. We can readily express **G** − **g** as ∇ _F_ by means of (45.7):

**Example 2.** The vector **f** = **a** × **r** is solenoidal; for from (41.6)

Since **f** is homogeneous of degree 1, the vector potential given by (12) is

### **PROBLEMS**

**1.** Show that a constant vector **c** has the scalar potential _φ_ = − **c** · **r** and the vector potential  .

**2.** Find the vector potential of **f** = ( **r** − **a** ) × ( **r** − **b** ).

**3.** Find the vector potential of the given solenoidal vector **f** from (6) and also from (12), and verify that **g** − **G** is irrotational.

( _a_ ) **f** = [ _x_ , _x_ , − _z_ ];

( _b_ ) **f** = [ _z_ − _y_ , _x_ − _z_ , _y_ − _x_ ];

( _c_ ) **f** = [ _xy_ , − _y_ 2, _yz_ ];

( _d_ ) **f** = [ _z_ , _x_ , _y_ ].

**4.** Find the vector potential of **f** from (11) when

( _a_ ) **f** = _r_ 3 **c** × **r** ;

( _b_ ) **f** = [ _x_ , 1, − _z_ ].

**5.** If _ρ_ , _φ_ , _z_ are cylindrical coordinates (Ex. 42.1), show that ∇ log _ρ_ and ∇ _φ_ are solenoidal vectors in any region not containing the origin.

**6.** If _r_ , _θ_ , _φ_ are spherical coordinates (Ex. 42.2), show that ∇ log tan   is solenoidal in any region that does not include the _z_ -axis.

**7.** Show that a vector **f** is irrotational and solenoidal in a rectangular prism _R_ when and only when it is the gradient of a harmonic function in _R_.

**8.** Show that **f** = **r** / _r_ 3 irrotational and solenoidal at all points except the origin.

**55. Green's Identities.** We shall now use the divergence theorem to prove three important identities due to George Green.†

Let _φ_ and _ψ_ be scalar point functions having continuous derivatives of the first and second orders, respectively, in a region _V_ bounded by a closed surface _S_. The divergence theorem (52.1) applied to the vector **f** = _φ_ ∇ _ψ_ now gives

In case _ψ_ is not defined outside of _S_ , we replace _dw_ / _dn_ by the negative of the derivative along the internal normal − **n**. Now, from (41.4),

since div ∇ _ψ_ = ∇2 _ψ_ (40.9); therefore on writing **n** · ∇ _ψ_ = _dψ_ / _dn_ ,

This formula is known as _Green's first identity_.

If both _φ_ and _ψ_ have continuous derivatives of the first and second orders, we may interchange _φ_ and _ψ_ in (1). On subtracting this result from (1), we obtain _Green's second identity_ :

We now take _ψ_ = l/ _r_ , where _r_ is the distance from a point _P_. If _P_ is _interior_ to _S_ , we cannot apply (2) to the entire region enclosed for l/ _r_ becomes infinite at _P_. We therefore exclude _P_ by surrounding it by a small sphere _σ_ of radius _ε_ and apply (2) to the region _V_ ′ between _S_ and _σ_ ; then, since

we have

On the sphere _σ_ , _r_ = _ε_ , and the external normal is directed toward _P_ ; hence

and _dS_ = _ε_ 2 _d_ Ω, where Ω is the solid angle subtended by _dS_ at _P_. The integral over _σ_ is therefore

and approaches 4 _π φ_ ( _P_ ) as _ε_ → 0. Moreover, as _ε_ → 0, the volume integral

for the integrand remains finite since we may take _dV_ = _r_ 2 _d_ Ω _dr_ as the element of volume in the neighborhood of _P_. Thus, on passing to the limit ε →; 0, (2)′ yields _Green's third identity_.

When _P_ is _exterior_ to _S_ , we may put _ψ_ = 1/ _r_ directly in (2); we then obtain

In order to apply Green's identities to the infinite region _V_ ∞ consisting of a closed surface _S_ and all of space beyond it, we require that the functions _φ_ and _ψ_ vanish at infinity in accordance with the following

DEFINITION. A function _φ_ is said to be _regular at infinity_ when _r_ | _φ_ | and _r_ 2|∇ _φ_ | are uniformly bounded as _r_ → ∞:

a suitably large value; or, in the _O_ -notation†

This condition implies that the vectors _φ∇ψ_ and _ψ∇φ_ have the property (7) required in Theorem 52.2. For example, _r_ 3 | _φ_ ∇ _ψ_ | < _M_ 2 when _r_ > _R_ , and hence

Thus the divergence theorem may be applied to the vectors _φ ∇ ψ_ and _ψ_ ∇ ψ in the infinite region _V_ ∞ beyond _S_ ; then **n** is a unit normal external to _V_ ∞ and points toward the interior of _S_.

Noting that 1/ _r_ is regular at infinity, we conclude that Green's identities (1), (2), (3), (4) are all valid in _V_ ∞. It may be shown that (3) and (4) are even valid when ∇2φ is continuous in the _interior_ of the region, but not necessarily on its boundary _S_ , provided that the integral   converges.‡

Suppose now that _φ_ is regular at infinity and ∇2 _φ_ is continuous throughout space except at _P_. Then, if we apply (3) to the region _V_ ∞ beyond _S_ and let _S_ shrink down to the point _P_ , we obtain

_provided the integral converges_.

### **PROBLEMS**

**1.** Verify (3) by finding _φ_ (0) when _φ_ ( _r_ ) = _ar_ 2 \+ _br_ \+ _c_ and _S_ is the sphere _r_ = 1.

**2.** Verify (6) by finding _φ_ (0) when

Is ∇2 _φ_ continuous at _r_ = 0? at _r_ = _a_?

**3.** Verify (6) by finding _φ_ (0) when _φ_ ( _r_ ) = 1/( _a_ \+ _r_ ).

**4.** _φ_ ( _r_ ), ∇ _φ_ , and ∇2 _φ_ are everywhere continuous except that ∇2 _φ_ → ∞ as _r_ → 0. Show that the volume integral in (6) will converge to –4πφ(0), provided _φ_ is regular at infinity.

**56. Harmonic Functions.** A solution of Laplace's equation ∇ _φ_ 2 = 0 is called a _harmonic function_. The function _φ_ is said to be _harmonic at a point P_ if its second derivatives are continuous in some neighborhood of _P_ and satisfy Laplace's equation. If _φ_ is harmonic at all interior points of a region and continuous throughout the region, it is said to be _harmonic in the region_. In an _open_ region, all points are interior, and the continuity of _φ_ follows from the continuity of its derivatives. For a _closed_ region, the values of _φ_ in the interior must connect continuously with its boundary values.

In Green's identities we shall now suppose that _φ_ is harmonic in the region _V_ bounded by a closed surface _S_.

If we put _ψ_ = _φ_ in (55.1), we have

Consequently, if either

the volume integral is zero. Since |∇ _φ_ is _continuous_ and never negative, ∇ _φ_ = 0 throughout _V_ ; for, if |∇ _φ_ | were positive at a point _P_ , we could embed _P_ in a subregion _V_ ′ of _V_ in which |∇ _φ_ | > 0, and the integral of |∇ _φ_ |2 over _V_ ′, and hence also over _V_ , would be positive. Thus, in both cases ( _a_ ) and ( _b_ ),

In particular, _φ_ = 0 in case ( _a_ ).

Now let _φ_ 1 and _φ_ 2 be two harmonic functions in _V_ which have the same values on _S_ ; then _φ_ = _φ_ 1 − _φ_ 2 is harmonic in _V_ and zero on _S_ ; consequently _φ_ = 0, or _φ_ 1 = _φ_ 2, throughout _V_. We have thus proved

THEOREM 1. _A function that is harmonic in a closed region bounded by a surface S is completely determined by its values on the boundary_.

If _φ_ and _ψ_ are both harmonic in the region _V_ , the volume integral in (55.2) is zero, and we have

In particular, if _ψ_ = 1, (2) becomes

THEOREM 2. _If a function is harmonic in a closed region bounded by a surface S, the integral of its normal derivative over S is zero_.

This property is characteristic of harmonic functions; for it can be shown that, if (3) holds for all spheres in a region _V_ , _φ_ is harmonic in _V_.†

Finally, if we put _▽_ 2 _φ_ = 0 in (55.3–4) we obtain the important

THEOREM 3. _If φ is harmonic in a closed region V bounded by a surface S and **r** is the distance from a point P_,

Observe that (4) gives the value of a harmonic function at any interior point _P_ of a closed region in terms of the values of _φ_ and _dφ_ / _dn_ on its boundary. Thus the integrand of (4) is a function, say _F_ ( _x_ , _y_ , _z_ ), of the coordinates of _P_ (which enter in 1/ _r_ and its normal derivative). Now it may be shown that the integral can be differentiated under the integral sign, and, since _F_ ( _x_ , _y_ , _z_ ) admits derivatives of all orders, the same is true of the integral and _φ_ ( _P_ ). Thus _φ_ ( _P_ ) is a function _φ_ ( _x_ , _y_ , _z_ ) that admits derivatives of all orders; and, since the existence of all derivatives of order _n_ implies the continuity of all derivatives of order _n_ − 1, _φ_ ( _P_ ) has _continuous_ derivatives of all orders. This remarkable result flows from a mere condition ∇2 _φ_ = 0 on the _second_ derivatives of _φ_.

We have seen that (4) gives _φ_ ( _P_ ) in terms of the boundary values of _φ_ and _dφ_ / _dn_. But, since _φ_ is uniquely determined by its boundary values alone (Theorem 1), we might expect to find a formula giving _φ_ within _S_ in terms of its continuous boundary values on _S_. This more difficult problem was solved by Green by use of a function (known now as _Green's function_ ) which he defined as a potential due to certain electric charges. The purely mathematical treatment of this question is known as

DIRICHLET's PROBLEM. _Does a harmonic function φ exist in a closed region bounded by a surface S which assumes a given set of continuous values on the boundary ?_

The answer to this famous problem is _yes_ for surfaces of a very general type; but it has been shown that there are some surfaces for which there is no solution.

If we apply (4) to a sphere Σ of radius _r_ about _P_ ,

Since _r_ is constant over Σ the first integral in (4) vanishes by virtue of (3), and we obtain

Since the surface area of Σ is 4 _π_ 2, (6) expresses the _mean-value theorem of Gauss:_

THEOREM 4. _If a function is harmonic in a sphere, its value at the center is the arithmetic mean of its values on the surface_.

This property also proves to be characteristic of harmonic functions.† From this theorem it is easy to prove

THEOREM 5. _A function that is harmonic, but not constant, in a closed region attains its extreme values only on the boundary_.

_Proof_. If _φ_ is harmonic in a closed region _V_ , it must assume its extreme values in _V_ by virtue of its continuity.‡ Consider the set of points at which _φ_ = _M_ , its maximum value. Since _φ_ is not constant, this set cannot contain _all_ interior points. Moreover, the set cannot contain _any_ interior point; for, if it did, it would have a frontier point _P_ in the interior of _V_. Then there would be a sphere about _P_ lying entirely in _V_ and passing through some points where _φ_ < _M_. The mean of its surface values therefore could not be _φ_ ( _P_ ) = _M_ , as required by Gauss' theorem.

Formulas (1), (2), (4), (5) also hold for functions that are harmonic in the infinite region consisting of a closed surface and all of space beyond, provided _φ_ and _ψ_ are regular at infinity; for Green's identities from which they are derived hold for infinite regions (§55).

Formula (3), however, is _not_ in general true; for it was derived from (2) by taking _ψ_ = 1, a function not regular at infinity. Indeed (3) is false when _φ_ = l/ _r_ , where _r_ is the distance from a point _P_ within _S_ ; although l/ _r_ is harmonic beyond _S_ and regular at infinity, we have from (52.9)

We can, however, prove

THEOREM 6. _If φ is harmonic in the closed region between the closed surfaces S_ 1 _and S_ 2 ( _surrounding S_ 1), _then_

_where_ **n** 1 _and_ **n** 2 _both point from the inside to the outside of the surfaces_.

_Proof_. Apply the divergence theorem to the vector ∇ _φ_ in the closed region _V_ between _S_ 1 and _S_ 2. Then, since − **n** 1 and **n** 2 are external normals to the region _V_ (Fig. 56 _a_),

Fig. 56 _a_. Region between closed surfaces.

Thus (8) is proved.

**Example 1.** If we apply (4) and (5) to the harmonic function _φ_ = 1, we obtain

in agreement with (52.9).

**Example 2.** _If P is any point within a sphere_ Σ _of radius a and center O, and **r** is the_ distance from _P_ ,

If _R_ is the distance from _O_ (Fig. 56 _b_), the functions _φ_ = 1/ _r_ and _ψ_ = l/ _R_ are harmonic _outside_ of Σ and regular at infinity. Hence from (2)

If **n** denotes the external normal to Σ, this gives, since _R_ = _a_ on Σ,

Fig. 56 _b_. Region with a spherical boundary.

from (7). Thus (10) is proved; it also states that the mean of 1/ _r_ over Σ is 1/ _a_

**Example 3.** If _r_ 1 and _r_ 2 denote the distances from _P_ 1 and _P_ 2. the functions l/ _r_ 1 and l/ _r_ 2 are harmonic everywhere except at _P_ 1 and _P_ 2, respectively. Consider the integral over any closed surface _S_ :

If _P_ 1 and _P_ 2 lie outside of _S_ , 1/ _r_ 1 and 1/ _r_ 2 are harmonic inside _S_ and _I_ = 0 by (2). If _P_ 1 and _P_ 2 lie inside of _S_ , 1/ _r_ 1 and 1/ _r_ 2 are harmonic outside of _S_ and regular at infinity; again _I_ = 0. But, if _P_ 1 lies inside of _S_ , _P_ 2 outside of _S_ , 1/ _r_ 2 is harmonic within _S_ and, on putting _r_ = _r_ 1 _φ_ = 1/ _r_ 2 in (4),

### **PROBLEMS**

**1.** If _φ_ is harmonic in a closed region _V_ bounded by a surface _S_ , and _φ_ = _C_ in _S_ , prove that _φ_ = _C_ throughout _V_.

**2.** If _φ_ and _ψ_ are harmonic functions in a closed region _V_ bounded by a surface _S_ , and _dφ_ / _dn_ = _dψ_ / _dn_ on **S** , prove that _ψ_ = _φ_ \+ const throughout _V_.

**3.** If _φ_ and _f_ ( _φ_ ) are both harmonic in a region _V_ and _φ_ is not constant, show that

**4.** Prove that, if _φ_ is harmonic throughout space and regular at infinity, then _φ_ = 0.

Apply [(1) when _V_ is the region within a sphere Σ centered at _O_ and of radius _r_. Show the surface integral → 0 as _r_ → ∞.]

**57. Retrospect.** In computing the derivative of a scalar point function _f_ ( **r** ) in the direction **e** , we found that a single vector point function ∇ _f_ , the _gradient of f_ , combined all the directional derivatives _df_ / _ds_ at a point in a single package in accordance with the formula:

In this expression the unit vector **e** gives the _direction_ , whereas ∇ _f_ depends only on the _point_ where the derivatives are computed. Thus a single vector ∇ _f_ enables us to compute an infinite number of scalars. However the gradient ∇ _f_ is a more complex quantity than a single scalar; it is, in effect, defined by the three numbers, its components. Thus the three components of ∇ _f_ , enable us to combine an infinity of scalars _df_ / _ds_ in a single package. In order to simplify directional differentiation in space of _three_ dimensions, we have been led from scalars (with 30 = 1 component) to vectors (with 31 = 3 components). Technically speaking, we have advanced from _tensors of valence zero to tensors of valence one_.

Passing next to the directional differentiation of vector point functions **f** ( **r** ), we obtain a formula

of the same form as before. Here again the vector **e** gives the direction, whereas ∇ _f_ , the _gradient of_ **f** , depends only on the point where the derivatives are computed. The package that holds an infinite number of vector derivatives _d_ **f** / _ds_ is now a new kind of quantity which we call a _dyadic_. In the form of a 3 × 3 matrix it is fully represented by nine numbers. Through our efforts to simplify and understand directional differentiation of vectors in space of three dimensions, we have been led from vectors (with 31 = 3 components) to dyadics (with 32 = 9 components), We have again advanced, this time _from tensors of valence one to tensors of valence two_. The higher valence is the price we pay for lucidity, concision, and the better understanding that they bring about.

In order to differentiate a dyadic directionally we must introduce a _triadic_ , a tensor of valence three with 33 = 27 components. But tensors of higher valence do not concern us here although they are indispensable for formulating certain portions of geometry and mathematical physics—especially the theory of relativity, which is both geometry and physics.

The basic property of tensors is the law for changing their components when the coordinate system is changed. Thus a vector **f** = [ _f_ 1, _f_ 2, _f_ 3] has different components in each rectangular coordinate system; and the relation that connects the components in different coordinate systems is actually the essential character of a vector. The same is true for tensors of higher valence, such as dyadics and triadics. In all these changes, however, certain functions of the components do not change; these are the _invariants_ of the tensor. For a vector

is an invariant. For a dyadic

the two most important invariants are the scalar

and the vector

Since the invariants of tensors are independent of the system of coordinates employed, the theorems of geometry and the laws of physics are expressed in terms of invariants. This is the reason for studying them. This is the key to their importance.

The gradient of a vector **f** , namely the dyadic

has the scalar and vector invariants

They appear constantly in all branches of _field theory_ : fields of force, velocity fields in fluid flow, and the electromagnetic field. In the following chapters we shall study such fields.

But even in geometry the divergence and rotation are of paramount importance. The basic integral transformations all involve the gradients of tensors and are valid for tensors of any valence:†

Each transforms an integral over a region to an integral over its boundary. The prototype of all such transformations is the _fundamental theorem of the calculus_ given in ( _A_ ) in generalized form.

From these transformations others can be deduced by dotting and crossing within the tensor functions, for both of these operations are distributive with respect to addition and therefore may be carried out under the integral sign.

When **f** ( **r** ) is a vector, the integrals in ( _B_ ) and ( _C_ ) involve dyadics. From them, by dotting between the dyad vectors, we obtain _Stokes' theorem_

and the _divergence theorem_

From ( _B_ ·) we see that for all closed curves in a simply connected region _V_ ,

for, if _V_ is simply connected, a surface that lies entirely in _V_ may be spanned over the curve. _The circulation of an irrotational vector over any closed curve in a simply connected region is zero_.

From ( _C_ ·) we see that

_The normal flux of a solenoidal vector over any closed surface is zero_.

By crossing between the dyad vectors we get less famous, but still important, theorems:

We know that div **f** and rot **f** are independent of the frame of reference and express invariant properties of the field vector **f**. But what _are_ they? Can these abstractions be made more tangible? The answer is _yes;_ the divergence theorem and Stokes' theorem give divergence and curl the following meaning.

If _V_ is the volume of a small sphere about a point _P_ , we have from ( _C_ ·) and the mean value theorem for volume integrals,

where div   is computed at some point   within the sphere. Hence

thus, roughly speaking, div **f** is a _flux density_ :

through a small sphere about _P_.

As to rot **f** , consider its component **e** · rot **f** in the direction of the unit vector **e**. Draw a small circle of area _A_ about _P_ in a plane normal to **e** ; then, from ( _B_ ·) and the mean value theorem for surface integrals,

where rot   is computed at some point   within the circle. Hence

Again, speaking roughly, **e** • rot **f** is a _circulation density_ :

over a small circle about _P_ and in a plane normal to **e**. The _sense_ of circulation is that of a right-handed screw advancing in the direction **e**. The vector rot **f** represents a synthesis of all such circulation densities at a given point.

We have now a synopsis of motives, methods, and results in vector and tensor analysis. The student who keeps them in mind is in position to understand geometry and mathematical physics, not merely as a collection of formulas, but as a rich diversity of results all flowing from a few central principles.

† _Advanced Calculus_ , §123.

† Points at which a curve crosses itself or is tangent to itself are _double points_ of the curve.

‡ For _plane_ regions, this argument is fallacious, for a third curve which cuts neither may not exist. If the given curves cross but a _finite_ number of times the proof follows by considering each loop separately. But for an infinite number of crossings the proof is more difficult.

† _Advanced Calculus_ , §121.

† _Advanced Calculus_ , §119, §133.

† Cf. Courant, _Differential and Integral Calculus_ , Blackie, vol. 2, p. 357.

† A _regular curve_ is piecewise smooth and without double points.

‡ _Advanced Calculus_ , §159.

† _Advanced Calculus_ , §74.

† _Advanced Calculus_ , §164.

‡ Möbius strip, materialized by giving a strip of paper one twist and pasting its ends together, is a _unilateral surface_. If we attempt to paint one side we paint it all over; for it only _has_ one side.

† This is certainly the case when the functions in (49.1) have continuous second partial derivatives. _Advanced Calculus_ , §82.

† Cf. Louis Brand, " _The Vector Potential of a Solenoidal Vector," Am. Math. Monthly_ , vol. 57, 1950, pp. 161–167.

† George Green, _Essay on the Application of Mathematical Analysis to the Theory of Electricity and Magnetism_ , Nottingham, 1828.

† _Advanced Calculus_ , p. 129.

‡ Cf. O. D. Kellogg, _Foundations of Potential Theory_ , Berlin, 1929, pp. 119, 219.

† Cf. O. D. Kellogg, _Foundations of Potential Theory_ , Berlin, 1929, p. 227. A reprint edition of this book [0-486-60144-7] is available from Dover Publications, Inc.

† Cf. Kellogg, _Foundations of Potential Theory_ , p. 224–227

‡ Cf. _Advanced Calculus_ , p. 152, Theorem 4.

† Cf. _Vector and Tensor Analysis_ , Chapter 6.

##   **Chapter 6**

## **Dynamics**

**58. Dynamics of a Particle.** If in a certain dynamical problem the matter composing a body may be regarded as concentrated in a single point, the body is called a _particle_. Thus the motion of the earth around the sun may be determined quite accurately by treating earth and sun as particles. But a billiard ball, whose motion may involve rolling, sliding, and spinning ( _english_ ), certainly cannot be regarded as a particle. In brief, to know when a body can be treated as a particle depends upon the nature of the problem and the degree of accuracy to be attained. It calls for an exercise of sound judgment.

If however, the bodies involved may be treated as particles, their motion in response to forces acting upon them may be determined from three fundamental principles.

I. FORCE AND ACCELERATION. _A free particle acted upon by a force_ **f** _acquires an acceleration_ **a** _in the direction of the force; that is_ ,

_where m is a scalar constant whose value_ ( _for a given system of units_ ) _depends entirely on the nature of the body designated as a particle_.

II. VECTOR ADDITION OF FORCES. _A system, of forces acting simultaneously on the same particle may be replaced by a single force acting on this particle and equal to their vector sum_.

III. ACTION AND REACTION. _The interaction between two particles, whether in direct contact or at a distance from each other, may be represented_ _by two forces of equal magnitude and opposite direction acting along their joining line_.

Equation (1) is the _fundamental equation of dynamics_. It is Newton's _second law of motion_.

When there is no force acting on the particle, or when all the forces have a zero sum, the fundamental equation becomes (§32)

If the initial conditions are

integration of  yields **v** = **v** 0, or

Integrating again, we have

which is the equation of a _ray_ issuing from **r** 0 in the direction of **v** 0. If **v** 0 = **0** , **r** = **r** 0, and the body will remain at rest; otherwise it will move in a straight line with constant velocity. These results constitute the

LAW OF INERTIA. _A particle will continue in its state of rest or of uniform motion in a straight line unless acted upon by an unbalanced force_.

Galileo was the first to realize the truth of this fundamental law. It was stated in the above form by Newton as his first _law of motion_.

**59. Equations of Motion.** The central problem in particle dynamics is the solution of the second-order vector differential equation

subject to the initial conditions

In some problems, such as the motion of a planet around the sun, we can solve this vector equation directly (§61). More frequently, however, it is necessary to replace it by one, two, or three scalar equations according as the motion is in a straight line, a plane, or in 3-space. For this purpose we resolve both acceleration and force into components and equate like components in equation (1). Rectangular components of acceleration are given in (32.7), plane polar components in (32.9).

If a particle _P_ is moving in a straight line, we choose it as _x_ -axis, selecting an origin and positive direction ( **i** ) at pleasure. The position, velocity, and acceleration of _P_ , and the force acting upon it, are given by

and (1) is equivalent to the scalar equation  .† The _directions_ of **r** , **v** , **a** , and **f** are now given by the _signs_ of  , and _f_ ; and we call the _signed numbers_

the "velocity" and "acceleration." The positive number | _v_ | is called the speed.

In many problems the solution of   may be accomplished by solving two first-order equations.

When _f_ is a function of _t_ or _v_ , we solve

in turn to find _v_ and _x_ as functions of _t_. On eliminating _t_ from these equations, we obtain a relation between _v_ and _x_.

When _f_ is a function of _x_ or _v_ , we solve

in turn to find _v_ as a function of _x_ , _x_ as a function of _t_.

**Example 1. Falling Body.** For an unresisted free fall from rest under gravity we take the origin at the initial point, the _x_ -axis downward. The force of gravity is _mg_ i, and our differential equation is

On solving these with initial conditions _x_ = 0, _v_ = 0 when _t_ = 0, we find

For fall in which the air resistance is proportional to _v_ 2, we put _f_ = _m_ ( _g_ – _kv_ 2), in which _k_ > 0. The equation of motion is now

If the body falls from a point sufficiently high, _v_ will increase until _g_ – _kv_ 2 = 0; then _dv/dt_ = 0, and from then on the body will travel at a constant _terminal velocity V_. Thus _g_ – _kV_ 2 = 0 and _k_ = _g/V_ 2; and our equation of motion becomes

From these we find that

As _V_ → ∞, the right-hand members approach the values given in (6). Use l'Hospital's rule to show this.

**Example 2.** A particle slides from rest down a smooth concave curve (Fig. 59) under the influence of gravity. The equation of motion is

Fig. 59. Particle sliding down a curve.

where **R** is the normal reaction of the curve. On taking tangential and normal components, we have

Since

(8) admits the integral

where _x_ is the vertical distance fallen; and _R_ = _2mgx/p_ from (9).

### **PROBLEMS**

**1.** If a body falls from rest under gravity with air resistance proportional to _υ_ , show that the equation of motion is

and that

**2.** A body is projected vertically upward with the velocity _v_ 0. If the resistance of the air per unit mass is _kv_ 2, show that it will rise to the height

and will strike the earth again with the velocity

Check by dimensions.

**3.** A projectile has the equation of motion _d **v**_ / _dt_ = **g** when air resistance is neglected. If projected from the origin with **v** 0 = **i** _υ_ 0 cos α + **j** _υ_ 0 sin _α_ , show that it describes a parabola

and that its greatest height and horizontal range are

**4.** A body is attracted to the origin with a force − _k_ 2 ** _r_** per unit of mass. Show that its equation of motion

has the solution

If **r** 0 = _x_ 0 **i** , **v** 0 = _υ_ 0 **j** , show that the particle describes an ellipse

**5.** A body is repelled from the origin with a force _k_ 2 ** _r_** per unit of mass. Show that its equation of motion

has the solution

If **r** 0 = _x_ 0 **i** , **v** 0 = _υ_ 0 **j** , show that the orbit is a hyperbola

**6.** If the force per unit of mass is − _k_ **v** , show that the equation of motion

admits the integrating factor _e kt_. Hence deduce the solution

**7.** A falling body is subject to air resistance − _k_ **v** per unit of mass. Show that its equation of motion

admits the integrating factor _e kt_, and deduce the solution

Find the dimensions of _k_ and check.

**8.** If the force on a particle of mass _m_ is **c** × **v** ( **c** constant), show that

( _a_ ) the speed _υ_ is constant;

( _b_ ) **c** · **v** = _cυ_ cos α is constant;

( _c_ ) the acceleration is _kυ_ 2 **N** ;

( _d_ ) if **c** · **v**  0, the orbit is a circular helix whose curvature and torsion are _k_ = ( _c_ / _υ_ ) sin _α_ , _τ_ = ( _c_ / _mυ_ ) cos _α_ ;

( _e_ ) if **c** · **v** = 0, the orbit is a circle of radius _ρ_ = _mυ_ / _c_.

**9.** The equation of motion of a particle is

where _a_ and _b_ are real constants. Show that

and deduce therefrom that the orbit is a _plane_ curve. Show that, if

where **A** and **B** are constant vectors determined by the initial conditions: **r = r 0**, **v = v 0** when _t_ = 0. What are the dimensions of _a_ and _b_?

**10.** The equation of motion of a particle is _d_ **v** / _dt_ = **S** · **r** where **S** is a constant symmetric dyadic. If **r = 0** , **v = v 0** when _t_ = 0, show that

**60. Work and Energy.** If the forces acting on a particle have **f** as resultant, the equation of motion is

Dot-multiplying both members by **v** gives

and, on integrating this equation between _t_ 0 and _t_ , we get

The scalar   is called the _kinetic energy_ of the particle. If we regard **f** as a function of _t_ , the integral

is called the _work_ done by the force in the time interval. Equation (2) states the

PRINCIPLE OF WORK AND ENERGY. _The change in kinetic energy of a particle in any time interval is equal to the total work done by the forces acting on it during this interval_.

The equation

is called the _energy equation_.

In many cases the force **f** is known when the position of the particle is given. Then **f** is a vector point function **f** ( **r** ), and we speak of a _field of force_. Then the work may be computed as a line integral over the curve _C_ , **r = r** ( _t_ ), traversed by the particle. In general the work depends upon the path _C_ between **r** 0 and **r** ; but, if rot **f = 0.** in a simply connected region  , **f** is the gradient of a scalar point function − _φ_ ( **r** ),

and the work

is independent of the path. Hence, from (2),

that is,

at all points of the region  .

We say now that the force **f** is derived from the _potential φ_ , or that the particle has the _potential energy φ_ in the field of force **f**. Thus the left member in (6) is sum of two energies, kinetic and potential; the former is due to the _motion_ of the particle, the latter to its _position_. The minus sign in (4) was introduced so that the energies _add_ in the left member of (6). This equation is a special instance of the

LAW OF CONSERVATION OF ENERGY. _When a field of force has a potential function, the sum of the kinetic and potential energies of a particle moving in the field is constant_.

**A** field of force having a potential function is said to be _conservative_ because the total energy, kinetic plus potential, is conserved. If the function _φ_ ( **r** ) satisfies (4), all functions _φ_ ( **r** ) + const do likewise. The additive constant is usually determined, once for all, by prescribing the value of _φ_ ( **r** ) at some point **r** 0, say where _φ_ ( **r** 0) = 0.

**Example 1. The Localized Field of the Earth.** In the neighborhood of a given locality, the earth attracts a particle of mass _m_ with the force

where **g** is the local acceleration of gravity and **k** = ∇ _z_ is the unit vertical vector directed upward (37.9). For a particle at a distance _z_ above the earth we have

Thus the field of the earth is conservative and has the potential _φ_ = _mgz_. A particle moving in the earth's local field has constant total energy

Thus, if a particle is projected with an initial speed _υ_ 0 at the height _z_ = _h_ , it will strike the ground with a speed _V_ given by

**Example 2. The Field of the Sun.** The sun of mass _M_ attracts a particle of mass _m_ with the force

_r_ is the radial distance of _m_ from the sun's center, **r** / _r_ is a unit radial vector, and _γ_ the constant of gravitation. Since **r** / _r_ = ∇ _r_ (37.15),

Therefore, the field of the sun is conservative and has the potential

A particle moving in the sun's field has the constant total energy

Thus, if a meteor of mass _m_ is at rest when _r_ = ∞, _c_ = 0; and it will strike the sun's surface _r_ = **_R_** with speed _V_ given by

If _M_ and _R_ denote the mass and radius of the earth, (14) gives the speed _V_ at which a meteor enters the earth's atmosphere. Near the earth's surface _r_ = _R_ we have from (7) and (11)

Conversely, in order that a particle may leave the earth it must be projected with at least the speed _V_ (neglecting air resistance). Thus _V_ is often called the "speed of escape." Taking _R_ = 3960 miles, _g_ = 32.2/5280 miles per second per second, we find that _V_ is about 7 miles per second.

### **PROBLEMS**

**1.** Find the potential function _if one exists_ for the following fields of force:

( _a_ ) **f** = **c** (const);

( _b_ ) **f** = **c** × **r** ;

( _c_ ) **f** = ( **c · r** ) **r** ;

( _d_ ) **f** = _r_ _n_ −1 **r**.

**2.** When **f** = **c** × **r** show that the work done by **f** in making the circuit of a closed plane curve is **2 _Ac_ · n** where **_A_ n** is the vector area enclosed by the curve.

**3.** Show that the force

is conservative, and find its potential function _φ_ ( _x_ , _y_ , _z_ ) for which _φ_ (0, 0, 0) = 0.

Find the work done by **f** over a path from (1, 2, 3) to (3, 5, −1).

**4.** At a distance _r_ 0 from the center of the sun a body is launched into its field with a speed   Show that it cannot leave the solar system.

**5.** If the speed of a planet at perihelion ( _r_ = _r_ 1) is _υ_ 1 find its speed _υ_ 2 at aphelion ( _r_ = _r_ 2).

**6.** In elliptic harmonic motion a particle of mass _m_ is attracted to the origin with a force **f** = − _mk_ **r** ( _k_ > 0). Find the potential, and show that _υ_ 2 \+ _kr_ 2 = const.

**7.** A body of mass _m_ is projected vertically upward from the earth (radius **_R_** ) with the speed _υ_ 0. Find the height _h_ it attains if

( _a_ ) **f** = − _mg_ ∇ _z_ , where _z_ is the distance above the ground;

( _b_ )  , where _r_ is the distance from the earth's center.

**61. The Solar System.** Celestial mechanics is based upon Isaac Newton's

LAW OF UNIVERSAL GRAVITATION. _Any two bodies, whose dimensions are negligible in comparison with their distance apart, attract each other with forces directed along their joining line, and whose common magnitude is directly proportional to the product of the masses of the bodies and inversely proportional to the square of the distance between them_.

The consequences of this law have been verified by observation in innumerable cases. Einstein's law of gravitation, which dispenses altogether with forces acting at a distance, explains a slight discrepancy in the motion of the planet Mercury (the advance of its perihelion) and has also other theoretical advantages. But as a working tool astronomers still use Newton's law exclusively to deal with the motion of the planets and their satellites, the asteroids, the comets, and also to explain such terrestrial phenomena as the tides and the precession of the equinoxes. In brief, Newton's law is still a miracle of exactitude, and only in special problems needs to be replaced by Einstein's law.

We shall now turn to one of the most famous problems in dynamics: _How does a planet move in the sun's field of force?_ The solution of this problem was first obtained empirically by Johann Kepler early in the seventeenth century. About fifty years later Newton discovered that all of Kepler's laws were consequences of his law of gravitation and his second law of motion.† We now give this deduction in vectorial form, using Newton's laws and the incomparable tool he invented, the calculus.

We regard the sun, of mass _M_ , as fixed in space, and choose its center as the origin of position vectors. Let the planet of mass _m_ have the position vector **r** = **_r_ R**; since the motion turns out to be plane, we write **R** = **r/ _r_** for the unit radial vector. Treating sun and planet as particles, the equation of motion of the planet is

The right member represents the attraction of the sun on the planet according to Newton's law; and _γ_ is the gravitational constant. Our equation of motion is therefore

Cross-multiply (1) by **r** ; then, since

we have

If **h** = **0** , **v** is radial and the motion is rectilinear. If **h**  **0** , **r** and **v** are always perpendicular to **h** and the orbit lies in a plane through the sun. Moreover the vector area of the sector _OAP_ (Fig. 61 _a_) is

and hence

_The sectorial speed of a planet about the sun is constant_.

Since

(2) may be written

Fig. 61 _a_. Orbit of a planet.

On cross-multiplying (1) by this equation, we find

since **R** · ( _d_ **R** / _dt_ ) = 0. This may be integrated at once to give

where **∊** is vector constant of integration.

The equation of the orbit is now obtained by eliminating **v** from (2) and (4). Dot-multiply (2) by **h** , (4) by **r** ; then

Since the left members are equal (13.4), the right members are also; hence on writing **∊** · **r** = _εr_ cos _θ_ , where _θ_ is the angle ( **∊** , **r** ),

This is the polar equation of a conic section of eccentricity _ε_ referred to its focus as pole and axis as initial line. For from the focus-directrix definition of a conic, we have (Fig. 61 _a_)

where _p_ is the _parameter_ of the conic. Since _r_ = _p_ when _θ_ = _π_ /2, _p_ is the half-width of the conic at the focus. This is the same equation as (5) if we put _h_ 2/ _γM_ = _εd_. The orbit is therefore an ellipse, parabola, or hyperbola, according as the eccentricity _ε_ < 1, _ε_ = 1, or _ε_ > 1. Since the only _closed_ conies are ellipses, the _planets describe ellipses with the sun at one focus_. The orbits of comets may be ellipses, parabolas, or hyperbolas.

Since the sectorial speed   is constant, the period of revolution _T_ in an elliptic orbit is obtained by dividing the area of the ellipse by the sectorial speed. The major axis 2 _a_ of the orbit is _r_ 1 \+ _r_ 2 where _r_ 1 and _r_ 2 are the perihelion ( _θ_ = 0) and aphelion ( _θ_ = _π_ ) distances. Now, from (5),

and, on addition,

The area of an ellipse of semiaxes _a_ , _b_ is _πab_ or   hence the period of revolution is

and

_The ratio T_ 2/ _a_ 3 _is the same for all the planets of the solar system_. More generally this ratio is constant for all bodies revolving about the same center of attraction—the twelve moons of Jupiter for example ( _M_ in (7) is then the mass of Jupiter).

Thus the Newtonian law of gravitation has led to the proof of

KEPLER'S THREE LAWS OF PLANETARY MOTION.

I. _The planets describe ellipses with the sun at one focus_.

II. _The radius vector from the sun to a planet sweeps out equal areas in equal times_.

III. _The squares of the periods of the planets are proportional to the cubes of their mean distances† from the sun_.

These laws were stated by Kepler after a careful study of astronomical observations, the first two in 1609, the third in 1618. The discovery that these laws, so diverse in content, all flow from a single principle, the law of gravitation, is one of Newton's greatest achievements and an enduring monument to the mind of man.

The energy equation of a planet

gives its speed as a function of its distance from the sun. The _velocity_ of the planet may be found from (4). Equation (2) shows that the direction of **h** is related to the sense in which the planet describes its orbit by the rule of the right-handed screw. If we put **h** = _h_ **k** , where **k** is a unit vector perpendicular to the plane of the orbit, and cross-multiply (4) by **k** , we find

Since the polar angle _θ_ = ( **∊** , **R** ), the vector eccentricity _∊_ = _ε_ **i** , where **i** is a unit vector in the direction  .

If we draw   from a fixed point, the curve traced by _V_ is called a _hodograph_. Equation (9) now shows that

_The hodograph of a planet is a circle_.

_Proof_. Let us plot the velocity vectors   using _γ **M** /h_ as the unit of speed; and draw   Then (9) shows that

where _CV_ = 1 (Fig. 61 _b_). Thus the hodograph is a unit circle about _C_ as center.

To find the velocity at any point _P_ of the orbit, draw   parallel to the focal radius   of the planet, and   90° ahead of  ; then  . The velocities at perihelion and aphelion are   and  ; and obviously

Fig. 61 _b_. Hodograph of a planet.

where _r_ l, _υ_ 1 refer to perihelion, _r_ 2, _υ_ 2 to aphelion. These results agree with (2); for _υ_ 1 _r_ 2 = _υ_ 2 _r_ 2 = _h_.

Since   gives the direction of the tangent to the conic at **_P_** , Fig. 61 _b_ gives a simple method of drawing tangents to a conic when only _one_ focus is given. Note that, when _ε_ = 0, the orbit is a circle, when _ε_ = 1, a parabola.

### **PROBLEMS**

**1.** The orbit of a planet or comet is the conic (6) of eccentricity _ε_ and parameter _p_. Show that its sectorial speed is proportional to  .

**2.** If **h** = 0 in (2), show that **v** = _υ_ **R** ( **R** const) and

**3.** _Halley's Comet_ has a period of about 76 years. At perihelion its distance from the sun is 0.58 of an astronomical unit (the major semiaxis of the earth). Find its distance at aphelion and the eccentricity of its orbit.

**4.** Mercury's period is 88 days; what is its mean distance from the sun?

**62. Systems of Particles.** The forces acting on any system of particles may be divided into two classes: (1) the _external forces_ exerted by particles not in given system, and (2) the _internal forces_ consisting of the mutual actions between particles of the system. The internal forces may be grouped in pairs which represent the interaction of two particles of the system. Then, according to the principle of action and reaction, the forces of each pair are equal in magnitude, opposite in direction, and have a common line of action. Hence the vector sum of forces in each pair is zero; and the sum of their moments about any point is zero. Since the internal forces occur only in such pairs we conclude that

_In any system of particles the vector sum of the internal forces, and of their moments about any point, is zero_.

Consider now any particle _P_ of mass _m_ belonging to the system and acted on by the external forces **f** and internal forces **f** ′ then

Moreover, if _A_ is any point, moving or fixed, chosen as the center of moments and  ,

Here _m_ **a** is called the _mass acceleration_ of the particle and **r** × _m_ **a** the _moment of the mass acceleration_ about _A_. Now form these equations for each particle of the system, and add the equations of each set. Then, since both ∑∑ **f** ′ = **0** and ∑∑ **r** × **f** ′ = **0** , we obtain

These fundamental equations in the dynamics of systems may be stated as follows:

THEOREM 1. _For any system of particles, the vector sum of the mass accelerations is equal to the sum of the external forces_.

THEOREM 2. _For any system of particles, the vector sum of the moments of the mass accelerations about any point_ ( _fixed or moving_ ) _is equal to the sum of the moments of the external forces about the point_.

The double sums above indicate that the forces or moments are first added at each particle, and the sums then added for the entire system. In the future, however, we shall simply write ∑ **f** , ∑ **r** × **f** for these total sums.

The mass accelerations _m_ **a** are often regarded as fictitious forces acting on the particles—the so-called _inertia forces_. If we now write (1) and (2) in the form

we see from §23 that the external forces and the inertia forces form a system in equilibrium. This result is known as

D'ALEMBERT'S PRINCIPLE. _The external forces acting on a system of particles and the reversed inertia forces are equivalent to zero_.

The _momentum_ of a particle _P_ of mass _m_ is the vector _m_ **v** ; and its _moment of momentum_ about _O_ is  . For a system of particles, whether discrete or continuous, the momentum and moment of momentum are defined as the respective sums (or _integrals_ ) taken over all particles of the system:

where the integrals apply to continuous mass distributions.

We may now put the fundamental equations (1–2) in the form

for in computing _d_ **H** / _dt_ the term **v** × _m_ **v** = **0**

When ∑ **f** = **0** , we obtain from (5) the important theorem on

CONSERVATION OF MOMENTUM. _If the sum of the external forces acting on a system of particles is zero, the momentum of the system remains constant_.

When the particles are associated with numbers equal to their masses, their centroid (§7) is called the _center of mass_ of the system. From (7.3), the position vector **r*** of the center of mass is given by

On differentiating (3) twice with respect to the time, we obtain

These equations state that:

_The momentum of the system is the same as that of a particle having the total mass of the system and moving with its center of mass_.

_The center of mass of a system of particles moves like a free particle having the mass of the entire system and acted upon by the sum of all the external forces_.

In particular, if ∑ **f** = **0,** ,

_When the sum of the external forces on a material system is zero, its center of mass will remain at rest or move uniformly in a straight line_.

Finally the kinetic energy _E_ of a system of particles is defined as the sum of their kinetic energies; thus

**Example. Problem of Two Bodies:** To find the motions of two bodies subject only to their mutual Newtonian attractions.

In our treatment of the sun-planet problem in §61, we regarded the sun _as fixed_. If we regard all _external_ forces on sun and planet as negligible, their internal forces cancel by the principle of action and reaction (§58); hence the center of mass **_P_ *** of the sun and planet moves with constant velocity **v***. Hence, if we choose a frame of reference   moving with the velocity **v*** , we may take **P*** , which is at rest in this frame, as origin. If the position vectors of sun _S_ and planet _P_ are now **s** and **r** ,

The position vector of P relative to _S_ is **p** = **r − s**. The equations of motion of planet and sun in the frame   are now

where **R** is a unit vector in the direction  . If we divide (12) by _m_ , (13) by _M_ , and subtract, we get

This equation has the form of (61.1) with **p** in place of **r** , and _M_ \+ _m_ in place of _M_. The orbit of _P_ relative to the sun _S_ is therefore an ellipse in which _SP_ sweeps out equal areas in equal times. But in place of (61.7) we now have

Hence _T_ 2/ _a_ 3 is not strictly the same for all planets, as Kepler's third law asserts. But, since the mass of the sun is more than 1000 times that of Jupiter, the greatest planet, _M_ \+ _m_ in (15) may be replaced by _M_ without serious error. The third law is therefore very nearly true.

**63. Body Revolving about a Fixed Axis.** The particles of a rigid body revolving about the z-axis with the angular velocity **ω =** _ω_ **k** have the velocities **v = ω × r** (34.2). Its momentum is therefore

where **r*** locates its center of mass _P_ *. If _P_ * is on the axis of rotation, **M** = **0** , irrespective of its angular velocity **ω**. If forces **f** are applied to the body, the reactions **f** _b_ on the shaft bearings will be such that

In particular, if P* is on the axis, **M** = **0** , _d_ **M** / _dt_ = **0** , and ∑ **f** \+ ∑ **f** _b_ **= 0**.

The moment of momentum about _O_ is

therefore

where

is the _angular acceleration_ vector. The signed number _α_ = _dω_ / _dt_ is also called the angular acceleration (radians per sec.2).

If bearing friction is negligible, the bearing reactions **f** _b_ pass through the _z_ -axis, and we can get rid of them by taking moments about this axis. The acceleration of a particle _P_ at a distance _p_ from the axis (Fig. 63) has the radial and transverse components _ω_ 2 _p_ ,   (Ex. 32.1); hence the total moment of the inertia forces about the _z_ -axis is

where   is the _moment of inertia_ of the body about the axis of rotation. Moreover the moment (or _torque_ ) of the external forces about this axis is

since **k · r × f** _b_ = 0. The equation of motion is therefore

Fig. 63. Revolution about a fixed axis.

This equation may also be obtained by dot-multiplying (3) by **k** ; for the moment about the z-axis is the z-component of the moment about O(§19).†

Since the speed of a particle at a distance _p_ from the axis _ωp_ , the body has the kinetic energy

On multiplying (5) by _ω_ = _dθ_ / _dt_ , we have

integration between _t_ 1 and _t_ 2 now gives

The left-hand member is the change in kinetic energy from time _t_ 1 to _t_ 2; and, if the body turns through a total of _θ_ 2 − _θ_ 1 radians in this time,

is the work done by the torque _T_ during this interval (§60).

### **PROBLEMS**

**1.** If a constant torque acts on a body mounted on a smooth shaft, _α_ = _dω_ / _dt_ is constant. If _θ_ = 0, _ω_ = _ω_ 0 when _t_ = 0, show that

**2.** A thin uniform rod of mass _m_ and length _l_ is free to turn about a horizontal axis _O_ at one end. If it is released from a horizontal position, show that its angular velocity when vertical is   radians per second. [ _I_ = _ml_ 2/3.]

**3.** The rod in Prob. 2 is displaced slightly from its lowest position ( _θ_ = 0) and released. Show that its equation of motion is

and that, for _small_ vibrations its period is very nearly   seconds.

**4.** A flywheel and shaft slow down from 200 to 180 rpm in 42 seconds under axle friction. Under a braking torque of 18 foot-pounds, it slows down from 200 to 180 rpm in 18 seconds. Find its moment of inertia. Take the foot, radian, pound (force), slug (mass) as units.

**5.** If the rod in Prob. 2 is released from a vertical position above its axis, show that   after it has fallen through an angle _θ_.

† The dots indicate time derivatives (Newton's notation). In the Newton-Leibnitz controversy, this notation gave occasion for a celebrated pun in which the "pure deism" of Leibnitz was compared with the "dotage" of Newton.

† In Newton's own words: "All this was in the two plague years of 1665 and 1666, for in those days I was in the prime of my age for invention, and minded mathematics and philosophy more than at any time since." Newton was then 22 or 23 years old.

† Since the major semiaxis  , it is called the planet's _mean distance_ from the sun. For the earth, _a_  92,900,000 miles—an "astronomical unit."

†  .

##   **Chapter 7**

## **Fluid Mechanics**

**64. Perfect Fluids.** We assume as an experimental fact that the pressure exerted by a fluid _at rest_ on any surface element is normal to the surface. For fluids _in motion_ this is not true, since _viscous_ fluids in motion exert tangential (shearing) as well as normal forces. But in many problems the tangential forces are small and may be neglected. We shall develop fluid mechanics on the hypothesis of purely normal pressure, and imply this assumption by speaking _of perfect_ or _nonviscous_ fluids. In brief: _a perfect fluid at rest or in motion exerts only normal pressures_.

Consider now a small fluid tetrahedron with three faces parallel to the coordinates planes (Fig. 64). Let the inclined face, of area _A_ , have the unit external normal **n** ; then the faces normal to axes have areas

The volume of the tetrahedron is   where _h_ is its altitude above the face _A_. Let the fluid exert average pressures _p_ 1, _p_ 2, _p_ 3, _p_ on the faces _A_ 1, _A_ 2, _A_ 3, _A_ ; and let _ρ_ , **f** , **a** denote averages over the tetrahedron of the density, body force per unit mass, and acceleration. Then, from d'Alembert's principle applied to the fluid in the tetrahedron,

Fig. 64. Fluid tetrahedron.

Now replace _A_ 1, _A_ 2, _A_ 3 by their values, divide out _A_ , and let _h_ → 0; then the limiting pressures at _O_ satisfy the relation

If we dot-multiply this equation by **i** , **j** , **k** in turn, we obtain

_At any point within a perfect fluid the pressure is the same in all directions_.

**65. Equation of Continuity.** Let _ρ_ and **v** denote the density and velocity of a fluid at the point **r** = [ _x_ , _y_ , _z_ ] at the instant _t_. Both _ρ_ and _υ_ are functions of _x_ , _y_ , _z_ , and _t_. Consider now the mass of fluid within a fixed but arbitrary closed surface _S_. This mass is increasing at the rate

the time differentiation being local (fixed **r** ). This rate must equal the rate at which fluid is _entering S_ , namely

where **n** is the external unit normal to _S_. Hence

by the divergence theorem; and, since

for any choice of _S_ ,

if the left member is continuous.

This _equation of continuity_ may be put in another form by introducing the _substantial_ rate of change _dρ_ / _dt_ instead of the _local_ time rate _∂ρ_ / _∂t_. Along the actual path, or _line of motion_ , of a fluid particle,

the density _ρ_ ( _x_ , _y_ , _z_ , _t_ ) becomes a function of _t_ alone, and

Now

and hence

If we now put

equation (1) becomes

Since div **v** = −( _dρ_ / _dt_ )/ _ρ_ , we can interpret div **v** as the relative rate at which the density is decreasing. Thus a positive value of div **v** implies a negative _dρ_ / _dt_ and consequently an attenuation of the fluid at the point considered; hence the term _divergence_.

Liquids may be regarded as nearly incompressible. For an incompressible fluid the density of any fluid particle remains constant in time; hence _dρ_ / _dt_ = 0, and, from (3),

This equation has the integral equivalent

the flux of an incompressible fluid across the boundary of a fixed closed surface is zero. If an incompressible fluid is also homogeneous, _ρ_ is constant in time and space.

We note finally that a relation of the form (2) holds for any function _F_ ( **r** , _t_ ) of time and position associated with a fluid particle moving with the velocity **v** :

The three terms are called, respectively, _substantial_ , _local_ , and _stationary_ rates of change. Thus the actual acceleration of a moving fluid particle is

here _∂_ **v** / _∂t_ is the rate of change of fluid velocity at a fixed point.

The curves which at every instant are tangent to **v** are called the _streamlines_ of the fluid. They are not the actual paths, or _lines of motion_ , of the fluid particles except when the flow is _steady_ : that is, when **v** at any point does not vary with the time ( _∂_ **v** / _∂t_ = **0** ). Thus the stream-lines have the differential equation **v** × _d_ **r** = **0**

**66. Euler's Equation.** The _Eulerian_ or _statistical_ method of treating fluid motion aims at finding the velocity **v** , density _ρ_ , and pressure _p_ of the fluid as functions of _t_ and **r** , the time and position. Thus _x_ , _y_ , _z_ , _t_ are the independent variables.

Consider the fluid within a fixed closed surface _S_ at any instant _t_. By d'Alembert's principle (§62), the body and surface forces, together with the reversed inertia forces (− _m_ **a** ), may be treated as a system in statical equilibrium. In a perfect fluid the surface force is a normal pressure − _p_ **n**. If **f** denotes the body force per unit mass and **a** the acceleration of the fluid particles, then

The equations state that the sum of all forces (real and inertia) acting on the fluid inside _S_ , and also the sum of their vector moments about _O_ , is zero. We now transform the surface integrals into volume integrals:

and obtain

Since these volume integrals vanish for an arbitrary choice of _S_ , their integrands, if continuous, must be identically zero. We thus obtain the partial differential equations:

From (41.5),

and (4) may be written

thus the moment equation (4) is a mere consequence of (3). We thus obtain from (3) Euler's equation of motion for a perfect fluid:

When the density _ρ_ is a function of _p_ only, we introduce the function

Moreover, if the body forces are conservative,

where _Q_ ( **r** ) is a single-valued scalar potential (§45). Under these conditions Euler's equation (5) becomes

For a liquid with body forces due to gravity, we have

when the _z_ -axis is directed upward. Then (8) becomes

**Example.** If a liquid is revolving with constant angular velocity **ω** = _ω_ **k** about a vertical axis, its velocity distribution **v** = **ω** × **r** is that of a rigid body. Hence

Now

and

At a free surface _p_ is constant; a free surface is therefore a paraboloid of revolution.

**67. Fluid in Equilibrium.** If we put _d_ **v** / _dt_ = **0** in Euler's equation (66.5), we get the equilibrium condition

Thus a liquid will be in equilibrium under gravity when

with the _z_ -axis directed _downward_ ; hence, if _p_ = _p_ 0 when _z_ = 0,

If the origin is at a free surface, _p_ 0 is the atmospheric pressure and _ρgz_ is called the _hydrostatic pressure_.

**Example. Floating Body.** Let the surface _S_ of a floating body _V_ be divided by its plane section _A_ at the water line into two parts: _S_ 1 submerged, _S_ 2 in air (Fig. 67). Then, if _p_ 0 is the atmospheric pressure, _p_ 1 = _ρgz_ the hydrostatic pressure, _p_ 0 \+ _p_ 1 acts on _S_ 1, _p_ 0 on _S_ 2; or we may say that _p_ 0 acts over _S_ while _p_ 1 acts on the closed surface _S_ 1 \+ _A_ ( _p_ 1 = 0 on _A_ ) enclosing the volume _V_ 1 below the water line. Hence the equations of equilibrium of the body are

Fig. 67. Floating body.

We consider in turn the integrals in (3):

the weight of the displaced liquid. Therefore (3) states that _W_ = _W_ 1; this is the

PRINCIPLE OF ARCHIMEDES. _A floating body in equilibrium displaces its own weight of liquid_.

If we change signs throughout in (4), we have the following integrals to consider:

where **r*** , defined by

locates the center of mass _R*_ of the body;

where **r 1*** locates the center of mass **_R_ 1*** of the displaced liquid. Since _W_ = _W_ 1, (4) reduces to

so that **_R_ 1* _R_** is parallel to **W**. Therefore:

_The centers of mass of a floating body and of the displaced liquid lie on the same vertical_.

### **PROBLEMS**

**1.** A body (surface _S_ ) totally immersed in liquid is subject to the normal pressure _p_ = _p_ o \+ _gρz_ where _z_ is measured down from the free surface _z_ = 0. Prove that

the weight of the displaced liquid.

**2.** From the equation of fluid equilibrium ∇ _p_ = _ρ_ **f** , show that equilibrium is only possible when **f** · rot **f** = **0**

When _ρ_ is a function of _p_ alone show that **f** = ∇ _P_ and hence rot **f = 0.**

**3.** Assuming that the earth is a fluid sphere of radius _a_ , of constant density _ρ_ , and without rotation, show that the pressure at a distance _r_ from the center is

[The attraction on a unit mass at the distance _r_ is

where − _g_ **R** is the attraction when _r_ = _a_. Hence

Compute the pressure at the center if _ρ_ is taken as the mean density of the earth ( _ρg_ = 5.525 × 62.4 lb per ft3).

**4.** If a fluid is bounded by a fixed surface _F_ ( **r** ) = **0** , show that the boundary condition **v** · ∇ _F_ = **0** must be satisfied.

More generally, if the bounding surface _F_ ( **r** , _t_ ) = 0 varies with the time, show that the fluid satisfies the boundary condition

**5.** A sphere of radius _a_ is moving in a fluid with a constant velocity **u**. Show that at the surface of the sphere the velocity **v** of the fluid satisfies the condition

**6.** If the body forces are conservative ( **f** = −∇ _Q_ ), the integrals

represent the kinetic and potential energy of the fluid within the region of integration. Show that an incompressible fluid ( _ρ_ = const) has the energy equation

**68. Vorticity.** Starting with Euler's equation in the form

we transform **v** · ∇ **v** by means of (41.9):

Thus (1) may be written

and hence

Now, since **r** and _t_ are independent variables,

from (4.17); and, since

we have

A rigid body having the angular velocity **ω** has the velocities **v = ω × r** and rot **v = 2ω** (Ex. 40.2). For a liquid we may regard

as the molecular rotation or _vorticity_ of the fluid particles.

In (3) we now replace rot **v** by 2 **ω** ; and, from the equation of continuity (65.3),

Then (3), after division by 2/ _ρ_ , becomes

that is

This is known as _Helmholtz's equation_.

**69. Circulation Theorem.** The line integral of the velocity   over any path is called the _flow_ along that path. If the path is closed, the flow is called the _circulation_.

The flow over a path Γ0 at the instant _t_ = 0 is

where the arc _s_ on the path Γ0 is taken as parameter. At a later instant _t_ the fluid particles forming Γ0 move on to another curve Γ; and, since the particles that form Γ are in one-to-one correspondence with those of Γ0, we may still use the arc _s_ on Γ0 as parameter in computing the flow over Γ:

The velocity **v** is a function **v** ( _s_ , _t_ ) of _s_ and _t_ ; and **v = v** ( _s_ , 0) along Γ0, **v = v** ( _s_ , _t_ ) along Γ, _t_ having the given constant value. When Γ0 is given, _F_ ( _t_ ) is a function of _t_ alone; hence

where **v** = _∂_ **r** / _∂t_ and **a =** _∂_ **v** / _∂t_ are the velocity and acceleration of the fluid particle corresponding to a certain _s_ on Γ0. If the body forces are conservative and _ρ_ is a function of _p_ (or constant), we have **a = −∇( _Q_ \+ _P_ )** from (66.8) and

therefore, if Γ joins the points _A_ and _B_ ,

so that _dF_ / _dt_ depends only on the end points of Γ. In particular, if Γ is a closed curve, _F_ ( _t_ ) becomes the circulation over Γ, and

This is Lord Kelvin's

CIRCULATION THEOREM. _If the body forces are conservative and the density is a function of p alone or constant, the circulation over any closed curve moving with the fluid does not alter with the time_.

If, at any instant _t_ 0, rot **v = 0** in a simply connected region, Stokes' theorem shows that the circulation over any closed curve Γ0 in the region is zero:

and, as this curve moves with the fluid, the circulation around it remains zero. Thus at any later instant the circulation over any closed curve is zero, for its particles formed a closed curve Γ0 at the prior instant _t_ 0. This means that the flow   over any curve at any instant is independent of the path; hence **v** = −∇ _φ_ , a gradient vector, and rot **v = 0**.

Therefore:

_If the vorticity of a fluid vanishes at any instant, it will remain zero thereafter_.

**70. Irrotational Motion.** We have just seen that, if the body forces on a fluid are conservative and the density _ρ_ = _ρ_ ( _p_ ) or constant, the vanishing of the vorticity at any instant entails its vanishing thereafter. Then rot **v = 0** in space and time, and the motion is termed _irrotational_. In any simply connected portion of the fluid, **v** is a gradient vector (45.8):

is called the _velocity potential_.

The equation of continuity (65.3) now becomes

For an incompressible fluid, _dρ_ / _dt_ = 0, and hence ∇2 _φ_ = 0. Therefore:

_The velocity potential of an irrotational, incompressible fluid satisfies Laplace's equation_ ∇2 _φ_ = 0.

The equation of motion (68.2) now reduces to

since

for _local_ time differentiation ( **r** constant) and space differentiation ( _t_ constant) commute with each other. Hence

an arbitrary function of the time. If, in addition, the flow is steady, _∂φ_ / _∂t_ = 0 and

in time and space.

The solutions of Laplace's equation are called _harmonic functions_ (§56). Every harmonic function _φ_ represents some irrotational flow of a liquid whose velocity **v** = −∇ _φ_. The problem consists in finding a flow that conforms to the given conditions

**Example.** In a flow with central symmetry _φ_ must be a function of _r_ alone; hence, from (42.21),

Now

and the flux per second through any sphere of radius _r_ about _O_ is constant: 4 _πr_ 2( _a_ / _r_ 2) = 4 _πa_. When this flux is given, _a_ is determined; the value of _b_ is immaterial since it does not enter in **v.**

### **PROBLEMS**

**1.** For the radial flow **v** = ( _a_ / _r_ 3) **r** of Ex. 70, show that the normal flux through any closed surface about the origin is constant and equal to 4 _πa_.

**2.** A shearing flow of fluid between the parallel planes _y_ = **0** , _y_ = _b_ is given by **v** = _y_ **i**. Show that its vorticity   and that **v** has a vector potential.   for which div **A = 0**.

Prove that the circulation over any closed curve in the _xy_ -plane is numerically equal to the area within the curve.

**3.** A mass of liquid is revolving about the _z_ -axis with the angular velocity _f_ ( _r_ ) **k** where _r_ is the perpendicular distance from the axis. If **R** is a unit vector perpendicular to the axis, prove that the velocity of the liquid is

and

**4.** If the motion in Prob. 3 is irrotational, show that

and that the (multivalued) velocity potential

where _θ_ is the polar angle.

For a liquid of constant density _ρ_ under the action of gravity alone, show that the pressure is given by

**5.** Let _r_ , _θ_ , _z_ denote cylindrical coordinates. In _Rankine's combined vortex_ a body of liquid rotates as a whole when _r_   _a_ with the constant angular velocity _ω_ 0 **k** , but irrotationally with the angular velocity ( _ω_ 0 _a_ 2/ _r_ 2) **k** when _r_   _a_. Show that the velocity

has the potentials:

**6.** In Rankine's combined vortex (Prob. 5), let _z_ = _z a_ at the free upper surface. Show that the free surface is given by

Prove that the bottom of the vortex ( _r_ = 0) is at a distance _ω_ 02 _a_ 2/ _g_ below the general level ( _r_ = ∞) of the liquid.

**7.** If the vorticity is constant throughout an incompressible fluid, prove that ∇2 **v = 0**. Cf. [(40.12)].

**8.** If an incompressible liquid in irrotational motion occupies a simply connected region, show that

where _φ_ is the velocity potential and the normal derivative _dφ_ / _dn_ is taken externally to the bounding surface.

**9.** Under the conditions of Prob. 8, show that, if

( _a_ ) _φ_ = const over the boundary, or

( _b_ ) _dφ_ / _dn_ = 0 over the boundary,

_φ_ = const and **v = 0** throughout the region.

Hence prove that the irrotational motion of a liquid occupying a simply connected region is uniquely determined when the value of either _φ_ or _dφ_ / _dn_ is specified at each point of the boundary.

**10.** Prove _Kelvin's minimum_ \- _energy theorem:_ The irrotational motion ( **v** ) of an incompressible fluid occupying a simply connected region _V_ with finite boundaries has less kinetic energy

than any other motion ( **v** 1) satisfying the same boundary conditions.

[Put **v** 1 = **v** \+ **v** ′, then

Prove that _E_ 1 = _E_ \+ _E_ ′.]

**71. Steady Motion.** A flow is said to be _steady_ when it is invariable in time. Then all local time derivatives are zero. Thus _∂ρ_ / _∂t_ = 0, and the equation of continuity (65.1) is simply

Also _∂_ **v** / _∂t_ = **0** : _the stream_ \- _lines are also lines of motion_. Euler's equation (68.2) now becomes

If **T** is a unit tangent along a stream-line ( **T × v = 0** ) or a vortex-line ( **T** × rot **v = 0** ),

along a stream-line or vortex-line. We have thus proved

BERNOULLI'S THEOREM. _In a steady flow in which the body forces have a potential Q and the density ρ_ = _ρ_ ( _p_ ) _or constant_ ,

_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,

an absolute constant—the same throughout the fluid.

Now suppose that the fluid has a constant density _ρ_ ; this is approximately true for liquids. The equation of continuity is

Along any tube whose surface consists of stream-lines, let the normal cross section be denoted by _A_. If the tube is sufficiently thin, the normal flux through a portion between two sections _A_ 1 and _A_ 2 (Fig. 71 _a_ ) is nearly _υ_ 2 _A_ 2 − _υ_ 1 _A_ 1 = 0; thus

Fig. 11 _a_. Flow through a thin tube.

is the equation of continuity along a tube of flow.

If the liquid is subject only to gravitational body forces, their potential is _gz_ if the _z_ -axis points upward from a horizontal reference plane. Thus, with

we have

along a stream-line. If _z_ , _p_ , _υ_ have the values _z_ 0, _p_ 0, _υ_ 0 at the section _A_ 0, we have from (6)

For a thin tube we may take _υA_ = _υ_ 0 _A_ 0, and (7) becomes

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 area _A_ to the upper surface where its area is _A_ 0 (Fig. 71 _b_). At this surface _p_ 0 is the atmospheric pressure, and _υ_ 0 = _υA_ / _A_ 0; at the orifice _p_ = _p_ 0, and the outflow speed is _υ_. With these values we have, from (7),

where _h_ is the distance from the free surface to the orifice. When _A_ 0 is large compared with _A_ , we have approximately _υ_ 2 = 2 _gh_ , a result known as _Torricelli's law_.

Fig. 7 _1b_. Flow from a tank.

**72. Plane Motion.** When the flow is the same in all planes parallel to a fixed plane (the _xy_ -plane) and the velocity has no z-component, the motion is said to be _plane_. Velocity, density, and pressure are all functions of _x_ and _y_ alone; and

is normal to the _xy_ -plane.

The flux across any curve _C_ in the _xy_ -plane is defined as the volume of liquid per second crossing a right cylinder of unit height based on _C_.

For one who travels along the curve in the sense of increasing _s_ , the flux crossing _C_ from right to left is

here **k** , **T** , **k** × **T** from a dextral set of orthogonal unit vectors (Fig. 72 _a_).

In any simply connected portion of the fluid, the line integral of **v** × **k** will be independent of the path when rot ( **v** × **k** ) = **0**. From (41.7)

Fig. 72 _a_. Flow across a curve.

hence, for an incompressible fluid (div **v** = 0), the flux (1) will be independent or the path between end points. Thus

defines a scalar point function such that

Since ∇ _ψ_ is everywhere normal and **v** everywhere tangent to the curves _ψ_ = const, these curves are the stream-lines of the motion. Consequently the function _ψ_ ( **r** ) is called the _stream function_.

If we set **g** = − _ψ_ **k** , we have

thus the velocity has the vector potential − _ψ_ **k** (§54).

From (41.7) we have

Therefore: _The plane motion of an incompressible fluid will be irrotational when and only when the stream function is harmonic:_ ∇2 _ψ_ = 0. When rot **v** = **0** _j_ the velocity has a scalar potential _φ_ , **v** = −∇ _φ_ ; and, since

the velocity potential is also harmonic.

From

we have the relations

or, in terms of rectangular coordinates,

Thus (7) is equivalent to the equations

But these are precisely the _Cauchy_ – _Riemann equations_ which connect the real and imaginary parts of the analytic function

of the complex variable _x_ \+ _iy_.† We have thus proved the important

THEOREM. _In any plane, irrotational motion of an incompressible fluid, the velocity potential φ and the stream function ψ are two harmonic functions which combine into an analytic function φ_ \+ _iψ of a complex variable x_ \+ _iy_.

Since − _ψ_ \+ _iφ_ = _i_ ( _φ_ \+ _iψ_ ), we see that, if _φ_ \+ _iψ_ is analytic, − _ψ_ \+ _iφ_ is also; consequently, if _φ_ and _ψ_ are the velocity potential and stream function for an irrotational plane flow, − _ψ_ and _φ_ are the corresponding functions for another flow of this type.

From (7), we have ∇ _φ_ **·** ∇ _ψ_ = 0: the stream-lines ( _ψ_ = const) cut the equipotential lines ( _φ_ = const) at right angles.

Since the _complex potential w_ = _φ_ \+ _iψ_ is an analytic function of _z_ ,

or, since the velocity has the components,

where   denotes the conjugate of _w′_. Thus the velocity at any point is given by the complex vector  ; its magnitude is |w′|:

**Example 1.** Assume the complex potential _w_ = _az n_ (a real); then, if we write _z_ = _re θ_,

The stream-lines are the curves whose polar equations are _r n_ sin _nθ_ = const.

For the cases _n_ = 1, 2, −1, we put _z_ = _x_ \+ _iy_ , using rectangular coordinates.

The stream-lines are the lines _y_ = const, and the flow has the constant velocity   in the direction of − _x_.

The equipotential and stream-lines are the two families of equilateral hyperbolas,

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 regarded as a steady flow of liquid in the angle between two perpendicular walls. The velocity at any point is  ; its magnitude varies directly as the distance from the origin.

The equipotential and stream-lines are two families of circles:

Fig. 72 _b_. Flow about a cylinder.

tangent to _y_ -axis and _x_ -axis, respectively, at the origin. The velocity   becomes infinite at the origin.

**Example 2.** With the complex potential,

The stream-line _ψ_ = 0 includes the circle _r_ = _a_ and the _x_ -axis sin _θ_ = 0. Since _w′_ = _V_ ( _1_ − _a_ 2/ _z_ 2), the complex velocity

Therefore we may regard the motion as a flow to the left about an infinite cylindrical obstacle of radius _a_ (Fig. 72 _b_). At a great distance from the obstacle, the flow has a sensibly uniform velocity − _Vi_.

When body forces are neglected, the Bernoulli equation (71.6) gives  . 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.

### **PROBLEMS**

1. When the motion of an incompressible fluid is steady, deduce from Helmholtz's equation (68.5) that **ω · ∇v** = **v · ∇ω.**

2. A gas flows from a reservoir, in which the pressure and density are _p_ 0, _p_ 0, into a space where the pressure is _p_. If the expansion takes place adiabatically, _p_ / _p y_ = const, where _y_ is the ratio of specific heats, show that

Neglecting body forces and the velocity of the gas in the reservoir, show that the speed of efflux when the motion becomes steady is given by  ; whence

3. In Ex. 72.1 consider the motion when _n_ = _π/α_ (0 < _α_ < _π_ ). Since the lines _θ_ = 0, _θ_ = _α_ are parts of the same stream-line _ψ_ = 0, we have the steady irrotational motion of a liquid between two walls at an angle _α_. Find the radial and transverse components of **v** at any point ( _r_ , _θ_ ).

4. Discuss the plane irrotational motion when the complex potential _w_ = _φ_ \+ _iψ_ is given by _z_ = cosh _w_. Show that the equipotential lines and stream-lines are the families of confocal ellipses and hyperbolas:

with foci at (± _c_ , 0).

Show that the stream-lines _ψ_ = _nπ_ , where _n_ is any positive integer, correspond to the part of the _x_ -axis from _x_ = ± _c_ to _x_ = ± ∞. If we regard this as a wall, we have the case of a liquid streaming through a slit of breadth 2 _c_ in an infinite plane.

**73. Fluid Mechanics,** like ordinary mechanics, has its kinematics and its dynamics. The basic kinematic principle is the _equation of continuity_ , which is simply an expression of the conservation of mass. The dynamics of fluids may be studied in two ways. The one adopted here is the _Eulerian_ or _statistical_ method. This aims at finding the velocity, density, and pressure of the fluid as functions of place and time; the independent variables are **r** and _t_. In the other attack, the _Lagrangian_ or _historical_ method, the motion of the individual fluid particles are followed from their initial positions **r** 0 to their position **r** after a time _t_. Thus the history of each fluid particle is traced. In any function _f_ ( **r** ) associated with a fluid particle, **r** is a function of the independent variables **r** 0 and _t_. This plan leads to more subtle and difficult calculations and is not presented here; but the basic ideas and results are given in Brand's more comprehensive _Vector and Tensor Analysis_. The Lagrangian method is especially adapted to following the history of curves and surfaces composed of fluid particles. It leads moreover to a simple integral of the Helmholtz vortex equation.

Returning to the Eulerian method applied to perfect (or nonviscous) fluids, we first find by d'Alembert's principle, that at any point within the fluid, the pressure, normal to any surface element, is the same in all directions. If _F_ ( **r** , _t_ ) is any function (scalar or vector) associated with a fluid particle moving with the velocity **v** , its _substantial_ rate of change is

where _∂F/∂t_ is the _local_ rate of change. The Eulerian equation of continuity is

and, for an incompressible fluid ( _dp/dt_ = 0), becomes div **v** = 0. Euler's dynamical equation, also deduced from d'Alembert's principle, is

when the body forces have a potential _Q_ and the density is a function of _p_ alone. From this we can deduce Helmholtz's equation for the _vorticity_  **:**

From this equation it can be shown that, if the vorticity of the fluid vanishes at any instant, it remains zero thereafter. We however first proved Kelvin's circulation theorem:

_The circulation over a closed curve moving with the fluid is constant,_

to establish this result.

For _irrotational_ motion (rot **v** = **0** ), the velocity is derived from a potential,

and the circulation   over any reducible curve. Euler's equation ( _E_ ) now becomes

an arbitrary function of _t_.

For _steady motion_ ( _∂ **v** /∂t_ = **0** ), equation ( _E_ ) becomes

whence _Bernoulli's theorem:_

along a stream-line or vortex line, and this constant is absolute if the steady motion is also irrotational.

For _plane motion_ of an incompressible fluid, the integral independent of the path

defines the _stream function_. The curves _ψ_ = const are stream-lines ( **v** × _d_ **r** = **0** ).

For _irrotational plane motion_ , both stream function and velocity potential are harmonic: ∇2 _ψ_ = 0, ∇2φ = 0. Then ∇ _ψ_ = **k** × ∇φ; and this is the vector form of the Cauchy-Riemann equations which guarantee that _w_ = _φ_ \+ _iψ_ is an analytic function of _x_ \+ _iy_. The _complex potential w_ therefore has a unique derivative _dw/dz_ ; and the complex velocity vector

† _Advanced Calculus_ , §197.

##   **Chapter 8**

## **Electrodynamics**

**74. Maxwell's Equations.** A distribution of electric charges and currents in space gives rise to an _electromagnetic field_. The field may be characterized by means of four vectors:

**E,** the _electric intensity_ (electric field strength),

**H,** the _magnetic intensity_ (magnetic field strength),

**D** = _ε_ **E,** the _electric flux density_ (electric displacement),

**B** = _μ_ **H,** the _magnetic flux density_ (magnetic induction);

the terms in parenthesis are also widely used. The linear relations **D** = _ε_ **E,** **B** = _µ_ **H,** hold only on the case of isotropic media (whose properties are independent of direction). In nonisotropic media, such as crystals, _ε_ and _µ_ may be regarded as symmetric dyadics that transform **E** and **H** into **D** and **B.** Moreover for ferromagnetic substances, such as iron, cobalt and nickel, the relation **B** = _ψ_ **H** does not apply; and, as a further complication, **B** depends upon the past history of the body.

If the medium is homogeneous as well as isotropic, _ε_ and _µ_ are constants; then _ε_ is called the _permittivity_ (or _dielectric constant_ ) and _µ_ the _permeability_ of the medium. In free space these constants are denoted by _ε_ 0 and _µ_ 0.

In the neighborhood of an _ordinary point_ of space the physical properties of the medium are continuous. We assume also that the four field vectors are everywhere finite and, at all ordinary points, are continuous, differentiable functions of position and time. As the fundamental postulates of electrodynamics we now assume that the field vectors satisfy _Maxwell's equations:_

at all ordinary points, where the vector **J** is the _current density_.

If _p_ denotes the volume density of electric charge, the total charge _q_ ( _quantity_ of electricity) within a closed surface _S_ is the volume integral

over the region _V_ within _S_. To define the current density at a point _P_ , consider a surface through _P_ normal to the stream-lines of flow; then **J** has the direction of the current at _P_ and a magnitude equal to the net charge that crosses a unit area of the surface about _P_ in unit time. Both _p_ and **J** are the limits of averages; and, if _v_ is the average velocity of the charges near _P_ , we may put **J** = _p_ **v.**

If _S_ is a bilateral surface and **n** its unit normal pointing from side 1 to side 2, then the current flowing through _S_ from 1 to 2 is

If _S_ is a _closed_ surface with the _external_ normal **n** , the total current passing _outward_ through _S_ is

by the divergence theorem. Now the charge _q_ within _S_ is also diminishing at the rate

On equating these expressions we have

and, since this equation holds for any choice of _S_ ,

if the left member is continuous. This equation expresses the conservation of electric charge: If we put **J** = _p_ **v** , (5) has exactly the same form as the equation of continuity (65.1) in hydrodynamics. We shall therefore refer to (5) as the _equation of continuity_.

On taking the divergence of equations (1) and (2), we have

from (5). Since we may commute the spatial operator div with _∂/∂t_ , these equations become

Thus div **B** and div **D** − _p_ are constant in time; and, if in the past history of the field these quantities were ever zero, they must thenceforth remain zero. We thus add to the postulated equations (1) and (2) two others,

The entire set (1), (2), (6), (7) are usually called _Maxwell's equations_.

If _S_ is any closed surface enclosing a region in which **B** and **D** are continuous, we have from the divergence theorem,

where _q_ is the charge enclosed by _S_. If **B** and **D** change very rapidly but continuously at the bounding surfaces of bodies, these equations are still valid when _S_ encloses different media. Mathematically, however, it is more convenient to regard **B** and **D** as _discontinuous_ at such boundaries and _postulate_ the truth of (8) and (9) in general. Then, if we apply these equations to a "pillbox" (Fig. 74) of height _h_ , with one base in medium 1, the other in medium 2, and pass to the limit _h_ → 0, we get

Fig. 74. Surface separating two media.

where **n** is a unit normal to the bounding surface in the direction 1–2 and _w_ is the density of the surface charge (coulombs per square meter). Thus the normal component of **B** is continuous over a bounding surface; but the normal component of **D** has a finite jump equal to the local surface density of charge.

**75. Integral Form of Maxwell's Equations.** Let _S_ be a surface spanned over a closed curve _C_. Then, if _S_ and _C_ fulfill the requirements of Stokes' theorem (§50), we may express the field equations

in integral form by taking surface integrals over _S_ and transforming the left members by Stokes' theorem:

The second integral in (4),

the total current passing through _S_ in the direction of **n**. Moreover, if _C_ and _S_ are fixed, the operation _∂/∂t_ may be performed on the entire integral. We thus obtain

as integral equivalents of the field equations.

The integral

is called the _magnetic flux_ through _S_. Since div **B** = 0, the divergence theorem shows that Φ is the same for all surfaces spanned over _C_ ; and

is a statement of Faraday's

LAW OF INDUCTION. _The time rate of decrease of the magnetic flux through a closed curve is equal to the circulation of the electric intensity_ **E** _about the curve_.†

The partial derivative in (5) implies a flux whose density **B** varies with the time but which links a _fixed_ contour. But Faraday's experiments have shown that the law of induction is even valid when Φ is changed by varying the contour; and to express this fact (5) is usually written

For a stationary field _∂_ **D** /∂ _t_ = **0** , and from (4) or (6) we obtain

This equation states

AMPÈRES' LAW: _The conduction current linked by a circuit is equal to the circulation of the magnetic intensity_ **H** _about the curve_.

In general, when **D** is variable, (4) may be written

Thus the vector **D** contributes to the magnetic field a portion which would be produced by a conduction current of density _∂_ **D** / _∂t_. For this reason Maxwell called _∂_ **D** / _∂t_ the density of the "displacement current" and **D** the "electric displacement."

**76. Ohm's Law.** The current density **J** in a liquid or solid is related to the electric field strength **E** by _Ohm's law_ ,

where the number _σ_ is called the _conductivity_ of the medium. All bodies conduct electricity to some extent, but the conductivity _σ_ (mhos per meter) of various materials covers an enormous range of values; thus, for fused quartz _σ_ = 10−17 approximately, whereas for silver _σ_ = 6.1 × 107. Intermediate between such extreme values we have _σ_  10−4 for distilled water, _σψ_  4 for sea water.

For a stationary current

† As in Stokes' theorem, the circulation is right-handed relative to **n**.

The divergence theorem now shows that the surface integral of **J** over any thin tube formed by stream-lines of current between the normal sections _A_ 1 and _A_ 2 is zero:

thus the current _I_ = _AJ_ through any cross section is the same. Now, along any stream line of current of length _l_ ,

The factor

is called the _resistance_ of the conductor between the points 1 and 2. Thus the resistance of a homogeneous wire of length _l_ and uniform section _A_ is given by

and 1/ _σ_ is called the _resistivity_. The equation (2) may now be written

For _steady_ currents, _∂_ **B** /∂ _t_ = **0** and (74.1) gives rot **E** = **0**. Then **E** = −∇φ, where _φ_ is the scalar potential, and the last equation becomes

The potential difference _φ_ 1 − _φ_ 2 is written _V_ and called the _electromotive force_ (emf) between the points; and _V_ = _RI_ is the familiar form of Ohm's law.

In a homogeneous conductor _σ_ is constant; hence, from (1),

and the equation of continuity (74.5) becomes

But, since _ε_ div **E** = _p_ from (74.7), we obtain the differential equation

If _ρ_ = _ρ_ 0 when _t_ = 0, its solution

shows that within a conductor charge decays exponentially with the time. The time _τ_ = _ε_ / _σ_ required for the charge _ρ_ 0 to decay to _ρ_ 0/ _e_ is called the _relaxation time_ of the conductor. Even for poor conductors _τ_ is very small; thus for distilled water _τ_ is less than a millionth of a second. However for the best insulators _τ_ may assume large values.

If the conducting body lies in an insulating medium, the total charge on the conductor must be conserved. The decay of interior charge thus results in a rapid flow of current to the outer surface of the conductor, where it will be arrested by the negligible conductivity of the insulator. Thus charges on conductors appear almost immediately as surface charges.

**77. Units and Dimensions.** When the meter, the kilogram, and second are chosen as the fundamental units of length, mass, and time, the unit of force, the _newton_ , will accelerate one kilogram one meter per second per second; thus

The unit of work, or energy, the _joule_ , is the work done by 1 newton acting through one meter; hence

The unit of power, the _watt_ , is work done at the rate of 1 joule per second; thus

The dimensions of force ( **F** ), work ( _W_ ), and power (P) are

By adding a fourth electrical unit to the meter, kilogram, and second we obtain the mks system of Giorgi, which has found wide acceptance in electrotechnics as it leads naturally to the "practical" electrical units. We give here the mks system augmented by the coulomb,† the unit of charge; its dimensional symbol is _Q_. The dimensions of electromagnetic quantities may then be expressed in terms of _M_ , _L_ , _T_ , and _Q_.

† The international coulomb is the charge transported by a current of one ampere in one second. The international ampere deposits silver from a silver nitrate solution at the rate of 1.11800 × 10−6 kg per sec.

An electric field exerts a force **F** on a point charge _q_ proportional to _q_. The field strength **E** is defined by the equation

that is, **E** is force in newtons per coulomb. The dimensions of **E** are therefore

The dimensions of all other electromagnetic quantities now follow from Maxwell's equations. Noting that

we have from (74.2) and (74.7)

Now from (74.1)

Since [ _εμ_ ] = _L_ −2 _T_ −2,   has the dimensions of velocity:

Moreover   and   have the same dimensions; in fact

the dimensions of energy density: _ML_ 2 _T_ −2/ _L_ 3. In fact, we shall see that

may be interpreted as the energy per unit volume (joules per cubic meter) of the electromagnetic field.

For the conductivity we have from (76.1)

Thus _τ_ = _ε_ / _σ_ in (76.6) has the dimensions of time and is called the _relaxation time_ of the medium, for from (76.6) we see that a charge _ρ_ 0 will decay to _ρ_ 0/ _e_ in the time _τ_.

† Here _E_ = |E|; similarity _H_ , _D_ , _B_ , _J_ , denote magnitudes of **H** , **D** , **B** , **J**.

In the mks system the units of electrical quantities are precisely the practical units used in technology. Thus the unit of current _I_ , or coulomb per second, is the _ampere_. The unit of electromotive force _V_ (or difference of potential _φ_ 1 − _φ_ 2) is called the _volt_ ; and from **E** = −∇φ we have

The unit of resistance is the _ohm_ ; and from (76.5)

The unit of _magnetic flux_ Φ = ∫ **n** · **B** _dS_ is called the _weber_ ; and

Thus **B** , the magnetic flux density, is expressed in webers per square meter.

The dimensions of _capacitance C_ , defined in §84, are

its unit is the _farad_ (coulombs per volt). The _inductance_  † is the magnetic flux in webers per ampere through a surface spanning a closed current _I_ ; hence

and its unit is the _henry_ (webers per ampere). The dimensions of _ε_ and _μ_ given above show that

hence the units of _ε_ and _μ_ are _farads per meter_ and _henrys per meter_.

A systematic table of quantities, mechanical and electrical, their dimensions and units, is given at the end of the book. This table may be used to check any rational formula in physics; for

(i) all additive terms must have the same dimensions in _M, L, T_ and _Q;_ and

(ii) the arguments of all nonalgebraic functions (as the sine, exponential, logarithm) must be dimensionless.

### **PROBLEMS**

**1.** Derive the dimensions of **D, H** , and   by making use of Maxwell's [equations (74.1∇2)](0486450309_15_ch8.html#eq_74.1) only.

**2.** Find the dimensions of _γ_ , the constant of gravitation (§61). Check Kepler's third law (61.7) by dimensions.

**3.** Check Bernoulli's equation (71.4) by dimensions.

**4.** Check Coulomb's law (79.4) by dimensions.

**5.** In Prob. 59.9 find the dimensions of _a_ , _b_ , _y_ , **A** , **B** , **C**.

† We use   instead of _L_ , the usual symbol, to avoid confusion with length.

**78. Wave Equation.** In a medium in which _ε_ and _µ_ are constant and _ρ_ = 0, the electromagnetic field equations may be written

If we take the curl of (1) and eliminate rot **B** by means of (2), we find

In similar fashion the elimination of **E** gives

Now, from (40.12), we have

in view of (3) and (4). Hence, both **E** and **B** satisfy the same propagation equation:

The last term, which represents a dissipation of energy, is absent in an insulating medium for which _σ_ = 0; then (5) becomes the _wave equation_

for a speed of propagation

In §77 we saw that Maxwell's equations implied that   had the dimensions _LT_ −1 of velocity. Now experiment has shown that in free space

and hence the speed of propagation is

Within the limits of experimental error this is the speed of light in free space. Thus the field vectors **E** and **B** (and also **D** and **H** ) are propagated with the speed of light in free space.

**79. Coulomb's Law.** If we integrate the equation

over the volume within a closed surface _S_ and apply the divergence theorem, we have

where _q_ is the total charge enclosed by _S_. This result is known as

GAUSS' THEOREM. _The surface integral of the electric displacement_ **D** _over any closed surface equals the enclosed charge_.

In a homogeneous and isotropic medium the permittivity _ε_ is constant, **D** = _ε_ **E** , and (1) becomes

If the field **E** is due to a point charge _q_ 1 at _O_ , it must have radial symmetry about _O_ ; then **E** = _E_ ( _r_ ) **R** where **R** is a unit radial vector issuing from _O_. If the closed surface is a sphere of radius _r_ , **n** = **R** and (2) becomes

A charge of _q_ 2 coulombs at a distance of _r_ meters from _O_ will now experience a force **F** = _q_ 2 **E** (77.1) or

due to the field **E**. If _q_ 1 _q_ 2 > 0, the charge _q_ 2 will be repelled and **F** will have the direction of **R**. Equation (4) is _Coulomb's law_. Like the law of gravitation it is an inverse-square law; but the charges are repelled or attracted according as they have the same or opposite signs.

If we write (3) in the form

we see that the field of a point charge _q_ has the potential

We assume that **E** due to _n_ point charges _q_ 1, _q_ 2, ···, _q_ n is the sum ∑ **E** i of their separate intensities: then

where _r_ i is the distance from the charge _q_ i to the point _P_ where **E** is computed. Consequently the potential due to this system of charges is

The potential at _P_ due to a charge − _q_ at _Q_ and + _q_ at _Q′_ is

If _Q_ ′ → _Q_ and at the same time the product

remains constant, the limiting result is called a _dipole of moment_ **m**. The potential of this dipole at _P_ is

for the limit of the fraction in the second member is the directional derivative of 1/ _PQ_ at _Q_ in the direction  .

For a continuous charge distribution of density _p_ over a volume _V_ , (7) suggests the potential

at points outside of _V_. Similarly, for a charge distribution of surface density _co_ over a surface _S_ ,

A continuous distribution of dipoles over a surface with moments everywhere in the direction of the unit normal n is called a _double layer_.

If _τ_ **n** is the dipole moment per unit area over the surface _S_ , the potential at _P_ of this double layer, suggested by (8), is

where _r_ = _PQ_. From (52.8)

the solid angle subtended by _dS_ at _P_ ; hence (11) may be written

When _τ_ is constant over _S_ , this becomes

where Ω is the total solid angle subtended by _S_ at _P_.

**80. Poynting's Vector.** If we dot-multiply Maxwell's equations,

by **H** and **E** , respectively, and subtract, we obtain

Now

hence on writing

(3) becomes

If we integrate this expression over the volume included by a closed surface _S_ , we obtain, on using the divergence theorem,

Each term of this equation has the dimensions _ML_ 2 _T_ −3, a rate of doing work (joules per second). We therefore proceed to interpret (7) as an _energy equation_.

If we regard _w_ as the energy of the field per unit volume,

and this energy increases at the rate

Since − **n** is the _internal_ unit normal to the surface _S_ , we may regard the surface integral   as the energy in joules per second that flows into the region bounded by _S_. Therefore the vector **E** × **H** represents the intensity of energy flow in watts per square meter. Since this interpretation was first given by Poynting in 1884, the vector

is known as _Poynting's vector_.

On the other hand, the integral   represents the rate at which the energy of the electric field **E** is changing into heat. The energy of the magnetic field does not undergo a corresponding "relaxation."

We therefore interpret (7)′ as follows:

The rate of increase of electromagnetic energy within a closed surface equals the rate at which energy is flowing in through the boundary minus the rate at which the energy of the electric field within is dissipated into heat.

In a stationary field _∂w_ / _∂t_ = 0, and (6) reduces to

**81. Potentials.** Since div **B** = 0, the solenoidal vector **B** admits a vector potential **A:**

where **A** ( **r** , _t_ ), a function of position and time, is arbitrary to an additive gradient ∇ _ψ_ of a scalar function (§54). The first Maxwell equation now gives

hence the irrotational vector **E** \+ ∂ **A** /∂ _t_ admits a scalar potential _φ_ :

where _φ_ ( **r** , _t_ ) is arbitrary to an additive function of the time; and

If **A** 0, _φ_ 0 are particular values of the potentials, and _ψ_ ( **r** , _t_ ) is an arbitrary function, the potentials

will give the same field ( **B** and **E** ).

For a homogeneous isotropic medium _ε_ and _µ_ are constants, and

If we substitute these values in

we find that

Moreover, since div **D** = _p_ , we have from (5)

The degree to which **A** and _φ_ are arbitrary allows us to impose the further condition

on the potentials; for, from (3), we need only choose _ψ_ to satisfy the equation

When this is done, both **A** and _φ_ satisfy differential equations of the same form:

For _stationary fields_ the time derivatives are zero. Then **A** and _φ_ are functions of position alone and the field vectors are given by

The condition (6) becomes div **A** = 0. The potentials now satisfy the _Poisson equations_ :

Since div **J** = 0 for stationary currents (74.5), the condition div **A** = 0 is consistent with (8)′.

**82. Electrostatics.** In a stationary field the vectors **E** and **D** satisfy the Maxwell equations

Equation (1) shows that **E** has a scalar potential:

In a homogeneous isotropic medium, _ε_ is constant,

and (2) shows that _φ_ satisfies Poisson's equation

If we put ∇2 _φ_ = − _p_ / _ε_ in Green's identities (55.3–4), we have, for any closed surface _S_ enclosing a volume _V_ ,

In these formulas _r_ is the distance from _P_ , the point of observation, to a point _Q_ in the volume or surface element; hence the volume and surface integrals in (5) are not the same as those in (6).

If _φ_ is regular at infinity (55.5), (5) and (6) may be applied to the region _V_ ∞ outside of _S;_ the normal to _S_ exterior to _V_ ∞ is now − **n** , and we have

for when _P_ is outside _S_ it is inside _V_ ∞. If we now add (5) and (8), or (6) and (7), the surface integrals cancel, and we have for all positions of _P_

on integrating over the whole of space.

In any actual case the charges occupy only a finite region _V_ 1 beyond which _ρ_ = 0, say the region _V_ 1 bounded by a surface _S_ 1; then

We may now compute **E** = −∇ _φ_ at _P_ by taking the gradient of (10) under the integral sign; and, since ∇ _P_ _r_ −1 = − _r_ −2 **R** where **R** is a unit vector in the direction  ,

In differentiating functions of _r_ = _PQ_ we may vary either _P_ or _Q_ , holding the other point fixed. Thus

when _P_ and _Q_ , respectively, are varied; and in general

If _O_ is a fixed origin within _V_ 1 and _Q_ a variable point of _V_ 1, the ratio _QP_ / _OP_ → 1 as P → ∞; hence, at distances from _V_ 1 that are very great relative to the dimensions of _V_ 1, the potential

where _q_ = ∫ _ρ dV_ is the total charge. Similarly the electric intensity

where _R_ is a unit vector in the direction _OP_. If the charge is concentrated in a small volume, these expressions have a wide range of validity; and for a "point charge" at _O_ they may be regarded as exact. Thus we again obtain (79.3) and are led to _Coulomb's law_. We note also that (14) and (15) show that | _OP_ || _φ_ | and | _OP_ |2 | **E** | are uniformly bounded. Thus the potential due to a finite distribution of charge is _regular at infinity_ (55.5).

When there are charges inside and outside of _S_ , the potential at points inside of _S_ is given by (5). If we write

we have _φ_ = _φ_ 1 \+ _φ_ 2, where _φ_ 1 is a particular solution of ∇2 _φ_ = − _p_ / _ε_ valid within _S_ , and _φ_ 2 is a solution of ∇2 _φ_ = 0 corresponding to the boundary values of _φ_ and _dφ_ / _dn_ (Theorem 56.3). Thus _φ_ 1 and _φ_ 2 are the respective potentials due to the charges inside and outside of _S_. If we put

in (17), we have

Hence, in view of (79.10–), we may state

THEOREM 1. _The charges outside of a closed surface S produce the same potential at an interior point as a single layer of density _ε d _φ_ / _dn_ and a double layer of moment density_ − _εφ_ over S_.

In a similar fashion we may deduce, from (7) the

THEOREM 2. _The charges inside of a closed surface S produce the same potential at an exterior point as a single layer of density_ − _εφ_ / _dn and a double layer of moment density sφ over S_.

The surface _S_ divides space into two regions _V_ and _V_ ∞ (Fig. 82a). Theorem 2 follows from Theorem 1 if _V_ is regarded as "outside" of _V_ ∞; then _ω_ and _τ_ change sign because − **n** is now the external normal. In case _S_ is an equipotential surface, _τ_ is constant, and the integral for the

Fig. 82 _a_. Two regions formed by a closed surface.

Fig. 82 _b_. To illustrate continuity of _p_.

double layer vanishes by (56.3); for 1/ _r_ = l/ _PQ_ is harmonic within _S_ when _P_ is outside of _S_.

At the surface of separation _S_ of two media, **D** = _ε_ **E** , and **E** satisfy the _boundary conditions_ :

The argument for (19) was given in §74; and similar considerations lead to (20). For rot **E** = 0 implies that   over any reducible curve, provided we regard **E** and its space derivatives as changing rapidly but continuously in a transition layer about _S_. It is more convenient, however, to regard **E** as discontinuous at _S_ but in such a manner that   for all closed curves, even those cutting _S_. We then find as in §46 that the tangential component of **E** is continuous over _S_ , as stated in (20). Thus, if we write **E** = _E_ t **T** \+ _E_ _n_ **n** on _S_ ,

Nevertheless _the potential is continuous_. For, if _P_ l, _P_ 2 are points near _S_ but on opposite sides, the line integral over a curve Γ (Fig. 82 _b_) is

As _P_ 1 and _P_ 2 approach a point on _S_ ,

**Example.** Let us compute the field **E** due to a sphere of radius _a_ having a uniform charge density  . Since the field has radial symmetry, _E_ = _E_ ( _r_ ) **R** is a vector function of _r_ directed outward from the center. Instead of computing the integrals (10) and (11), it is simpler to use Gauss' theorem (79.1) to find **E**.

The normal flux over a sphere of radius _r_ is

where _q_ is the total charge on the sphere; hence

To find the potential _φ_ , we integrate

and determine the constants so that _φ_ (∞) = 0, and _φ_ ( _r_ ) is continuous when _r_ = _a_. We thus obtain

Equating these values when _r_ = _a_ , we find _C_ = 3 _q_ /8 _πεα_ ; hence

**83. Conductors.** In a _conductor_ interior charges move rapidly to its surface (§76). Consequently a conductor in electrostatic equilibrium has only surface charges. Thus, if a charge _q_ is placed inside an isolated conductor _K_ , the charge will distribute itself over the surface with a

density _ω_ so that  .

In the interior of a conductor **J** = **0** and

and, since **E** = −∇ _φ_ ,

Moreover _φ_ has this same constant value at the surface of the conductor by reason of its continuity (§82); hence

_The boundary of a conductor is an equipotential surface_.

If a conductor _K_ 1 is surrounded by a dielectric medium 2 of permittivity _ε_ , the boundary conditions (82.19–) take the form

where **n** is a unit normal external to _K_. In terms of potential, these become

hence at the surface of a conductor the potential satisfies the boundary conditions

A fundamental problem in electrostatics may be formulated as follows. A number of finite conductors bounded by surfaces _K i_ are placed in a dielectric of constant permittivity _ε_ which is free from charges. If each _K i_ is given a charge _q i_, together they produce an electrostatic field whose potential _φ_ satisfies Laplace's equation

in the uncharged ( _p_ = 0) dielectric. On the surface of each conductor _K i_, _φ_ satisfies the boundary conditions

where the constants _φ i_ are in general unknown. We then seek a solution of Laplace's equation which satisfies the boundary conditions and is regular at infinity. _If this problem has a solution it is unique_.

_Proof_ If _φ_ 1 and _φ_ 2 are solutions, _ψ_ = _φ_ 1 − _φ_ 2 satisfies ∇2 _ψ_ = 0 with the boundary conditions on each _K_ i:

For any function _ψ_ regular at infinity we may apply the divergence theorem to _ψ_ ∇ _ψ_ in the infinite region _V ∞_ outside of the conductors; since

we then obtain

from the boundary conditions (–); as to the minus sign, note that the normals external to _V_ ∞ are internal to the conductors. Since **E** = −∇ _φ_ is everywhere continuous in the dielectric, |∇ _ψ_ | is continuous and the volume integral cannot vanish unless ∇ _ψ_ = 0. Hence _ψ_ = const, and the constant is zero owing to the regularity of _ψ_ at infinity; thus _φ_ 1 = _φ_ 2.

When the conductors are given charges  , let the corresponding potential be _φ_ ′. It is then obvious that the charges   give rise to the potential _φ_ \+ _φ_ ′, and the corresponding field −∇( _φ_ \+ _φ_ ′) = **E** \+ **E** ′. This is the _law of superposition of electrical effects_.

If we apply (56.2) to the harmonic functions _φ_ and _φ_ ′ in the infinite region outside of the conductors, we have

or, in view of the conditions (8) and (9),

This result is known as _Green's reciprocation theorem_.

**Example 1.** If a charge _q_ is placed inside an isolated metal sphere of radius _α_ , it will spread to its surface with the density

If the sphere is surrounded by an unbounded medium of permittivity _ε_ , and we apply Gauss' theorem to a sphere of radius _r_ > _a_ , we have

hence

Thus _E_ jumps from 0 to _ω_ / _ε_ at the surface of the sphere. Since

we find

since _φ_ is continuous at the surface.

**Example 2. Circular Disk.** a thin circular disk of radius _α_ (Fig. 83 _a_) is given a uniform charge of surface density   on each side; the total charge is then _πα_ 2 _ω_. If

Fig. 83 _a_. Charged disk.

Fig. 83 _b_. Conducting plane.

_dV_ is the volume of a ring of area _dA_ = 2 _πy dy_ , _ρ dV_ = _ωdA_ , and at a point _P_ on the axis of the disk,

On the axis **E** = ± _E_ **k** according as _P_ lies to the right or left of the disk; and

Near the disk a   and **_E_**  _ω_ /2 _ε_ ; and, as we cross the disk from left to right, _φ_ is continuous, but **E** changes from

in conformity with (82.21).

As _a_ → ∞, the disk becomes an infinite plane and _E_ = _ω_ /2 _ε_ at _all_ points.

**Example 3. Induced Charge on a Conducting Plane.** Consider two point charges _q_ and − _q_ in a dielectric of permittivity _ε_ (Fig. 83 _b_). If _r_ and _r′_ denote the distance of a point _P_ from these charges, the potential of the field they produce at _P_ is

The median plane _K_ ( _r_ = _r_ ′) is the equipotential surface ( _φ_ = 0. Since _φ_ is regular at infinity, we may apply Green's theorem to the "closed" surface _K_. Then, by Theorem 82.2, the charge − _q_ to the left of _K_ produces the same potential to the right of _K_ as the surface density

spread over _K_. Now

and, as we approach the plane _r_ = _r_ ′

Therefore "Green's equivalent layer" is

Suppose now that the charge _q_ is placed to the right of a _conducting plane K_. Since _K_ extends to infinity, its potential will be zero. The field of _q_ must therefore be modified so that ( _φ_ = 0 on _K_. Transient currents will flow in _K_ and produce a charge density _ψ_ to attain this result. This surface charge induced by _q_ is most simply determined by replacing it by the equivalent point charge − _q_ , the _image of q_ in _K_. In passing through _K_ from left to right, **n** · **D** will jump from 0 to _ω_ by (3), provided _ω_ has the value given above.

To find the total charge   on _K_ we use annular elements of area

then

Since _ω_ = **n** · **D** , this is a consequence of Gauss' theorem (79.1):

**84. Capacitance.** _Case_ 1. Consider an isolated conductor _K_ with a charge _q_ in a dielectric of permittivity _ε_. The potential _φ_ due to this charge satisfies _∇_ 2 _φ_ = 0 outside of _K_ , the boundary conditions

and is regular at infinity. If the charge were _q_ ′ the potential _φ_ ′ would satisfy _∇_ 2 _φ_ = 0 outside of _K_ and the boundary conditions

If the _φ_ is a solution of the first problem,

is a solution of the second; and, as the solution is unique, it is the only solution. Hence the ratio of the charge on _K_ to its potential is a constant

The constant _C_ is called the _capacitance_ of _K_ , or, more precisely, of the _capacitor_ ( _K_ , _K_ ∞) where _K_ ∞ is a conductor at potential zero (the earth).

_Case_ 2. Consider next two isolated conductors _K_ 1 and _K_ 2. If _K_ 1 is given a charge _q_ , induced charges will appear on _K_ 2 of total amount zero, and the potential due to this system of charges satisfies ∇2 _φ_ = 0 and the boundary conditions

If _K_ 1 is given the charge _q_ ′, the potential _φ_ ′ is again given by (1); and, since _q_ ′, _φ_ ′1, _φ_ ′2 are _q_ ′/ _q_ times _q_ 1, _φ_ 1, _φ_ 2, the ratio

is the same in both cases. The constant _C_ is now called the capacitance of _K_ 1 and _K_ 2, or, more precisely, of the capacitor ( _K_ 1 _K_ 2, _K_ ∞).

_Case_ 3. With the same conductors _K_ 1, _K_ 2, let _K_ 2 be kept at a constant potential while _K_ 1 is being given the charge _q_. Thus _K_ 2 may be _grounded_ (connected to _K_ ∞) so that its potential is always zero. Then the charge _q_ on _K_ 1 will induce a negative charge _q_ 2 on _K_ 2. Then the potential _φ_ due to these charges satisfies ∇2 _φ_ = 0 and the boundary conditions

with _φ_ = 0 on _K_ 2. When _φ_ is known, the induced charge on _K_ 2 is given by

Evidently, if _q_ is replaced by _q_ ′ the potential _φ_ ′ is again given by (1); and the ratio

is the same in both cases. Now _C_ is called the _capacitance_ of the capacitor ( _K_ 1, _K_ 2 \+ _K_ ∞).

If _φ_ is the potential in case 2, _ψ_ = _φ_ − _φ_ 2 is evidently the potential in case 3. The capacitance in both cases are exactly the same; for _ψ_ 1 = _φ_ 1 − _φ_ 2.

**Example 1. Concentric Spheres**. A metal sphere _K_ 1 of radius _a_ has a charge _q_ and is surrounded by a concentric spherical shell _K_ 2 of radii _b_ , _c_ (Fig. 84 _a_). The field has radial symmetry, and _φ_ ( _r_ ) is a function of _r_ alone; and the most general function of _r_ that satisfies ∇2 _φ_ = 0 is

The constant _A_ is found from the boundary condition

Hence _A_ = _q_ /4 _πε_ and

the capacitance of ( _K_ 1, _K_ 2, _K_ ∞) is therefore

When _r_   _c_ , _φ_ ( _r_ ) = _A_ ′/ _r_ since _φ_ (∞) = 0; and the condition that the charge on _K_ 2 is zero, namely

Fig. 84 _a_. Spherical shell surrounding a sphere.

gives _A_ ′ = _A_ = _q_ /4πε. Finally _φ_ ( _b_ ) = _φ_ ( _c_ ) gives

The potential function is therefore

These results are more quickly computed by use of Gauss' theorem, but the derivation above illustrates the general procedure.

_When K_ 2 _is grounded_ and _K_ 1 has the charge _q_ , we apply Gauss' theorem to a sphere of radius _r_ between _K_ 1 and _K_ 2; then

and, since _φ_ ( _b_ ) = 0,

The capacitance of ( _K_ 1, _K_ 2 \+ _K_ ∞) is still given by (6); and, if we let _b_ → ∞, we obtain the capacitance 4 _πεa_ of the isolated sphere, or the capacitor ( _K_ 1, _K_ ∞).

_When K_ 1 _is grounded_ and _K_ 2 has the charge _q_ , we put _q_ = _q_ _b_ \+ _q_ _c_ , where _q_ _b_ is on the inner face _r_ = _b_ , _q_ _c_ on the outer face _r_ = _c_. If we apply Gauss' theorem to a sphere within _K_ 2, we find that _K_ 1 has the induced charge − _q_ _b_. Now from Gauss' theorem applied to a sphere of radius _r_ > _c_ , we find

and, since _φ_ (∞) = 0,

From Gauss' theorem applied to a sphere of radius _r_ ( _a_ < _r_ < _b_ ), we find

and, since _φ_ ( _a_ ) = 0,

But, since _φ_ ( _b_ ) = _φ_ (c),

The capacity of ( _K_ 1 \+ _K_ ∞, _K_ 2) is therefore

**Example 2. Parallel Plates**. Two parallel plates _K_ 1, _K_ 2 of equal area _A_ are separated by a distance _h_ very small relative to their dimensions. Neglecting fringe effects at the edges, we regard _K_ 1 as having a uniform surface charge _ω_. Except near the edges, the field between _K_ 1 and _K_ 2 is nearly perpendicular to the plates. If _K_ 1 is the plane _z_ = 0 and the _z_ -axis is drawn toward _K_ 2, we have from (83.4)

If _K_ 2 is grounded, _φ_ ( _h_ ) = 0 and the constant is determined; thus

and the capacitance of the plates is very nearly

### **PROBLEMS**

**1.** An uncharged conductor _K_ 2 is raised to the potential _φ_ by giving _K_ 1 a charge _q_. If _K_ 1 is uncharged, find its potential when _K_ 2 is given a charge _q_.Use [(83.10).]

**2.** A point charge _q_ at _P_ induces a charge _q_ 1 on the grounded conductor _K_ 1 ( _φ_ 1 = 0). When _K_ 1 is isolated and raised to the potential  , the potential of its field at _P_ is _φ′P_. Show that

**3.** In Prob. 2, _K_ 1 is a grounded sphere of radius _a_ and _P_ is at a distance _r_ from its center. Show that a charge _q_ at _P_ will induce a charge _q_ 1 = − _aq_ / _r_ on _K_ 1. [Ex. 83.1]

**4.** If _b_ = _a_ \+ _h_ in (6), where _a_ is large and _h_ relatively small, show that  , the capacitance of a parallel plate capacitor.

**5.** Three concentric spherical shells _A_ , _B_ , _C_ , of radii _a_ < _b_ < _c_ are separated by a dielectric of permittivity _ε_. If _A_ and _C_ are grounded and _B_ is given a charge _q_ , show that the inner and outer faces of _B_ will have the charges

and that the potential of _B_

Moreover _q_ / _φ_ ( _b_ ) is the sum of the capacitances of _AB_ and _BC_.

**6.** A straight thin wire of length 2 _a_ is uniformly charged with 2 _aλ_ coulombs. (Fig. 84 _b_). In a plane that bisects the wire show that the potential and field at a point _P_ at a distance _r_ from the wire are given by

Fig. 84 _b_. Uniform wire.

**7.** If the wire in Prob. 6 is infinite and charged with a linear density _λ_ , show that the potential and field at all points is

Find the constant _C_ if _φ_ = _φ_ 0 when _r_ = _r_ 0.

**8.** Let _O_ 1 and _O_ 2 represent two thin parallel wires (lines perpendicular to the paper) of infinite length and a distance 2 _b_ apart (Fig. 84 _c_). If their charge densities are _λ_ and − _λ_ per unit length, show that the potential in any plane normal to the wires is

Fig. 84 _c_. Two infinite parallel wires.

Show that the equipotential lines are a family of circles having _O_ 1 and _O_ 2 as inverse points, and that the field lines are the family of circles passing through _O_ 1 and _O_ 2.

[If _O_ 1, _O_ 2 are the origins of bipolar coordinates _r_ 1, _θ_ 1, and _r_ 2, _θ_ 2, the equipotential lines are the circles _r_ 1/ _r_ 2 = const; these are cut at right angles by the circles _φ_ 2 − _φ_ 1 = const, the field lines. Cf. Prob. 37.9.]

Fig. 84 _d_. Two parallel cylindrical conductors.

**9.** Two parallel cylindrical conductors _A_ , _B_ , of infinite length and radius _R_ , have their centers a distance 2 _h_ apart (Fig. 84 _d_). If their charge densities are _λ_ and − _λ_ per unit of length, show that their field may be computed by replacing them by similarly charged lines at a distance 2 _b_ apart (Prob. 8) where   Hence deduce from Prob. 3 that

[ _O_ 1 and _O_ 2 are inverse points to a circle of center _C_ and radius _R_ , provided CO1 · CO2 = ( _h_ − _b_ )( _h_ \+ _b_ ) = _R_ 2.]

**10.** In Fig. 84 _e_, _O_ represents a charged infinite wire parallel to the infinite conducting plane _p_. If _O_ has the linear charge density _λ_ , show that the potential in a plane normal to _O_ (the paper) and to the right of _p_ is

where _r_ , _r_ ′ are radial distances from _O_ and its image _O_ ′, and that

is the charge density induced on _p_ by the wire _O_.

Fig. 84 _e_. Infinite wire parallel to conducting plane.

**85. Magnetostatics.** In a stationary field the magnetic vectors **H** and **B** satisfy the equations

In addition the equation of continuity now becomes

**B** may now be derived from a vector potential

such that

If the medium is homogeneous, isotropic, and not ferromagnetic (§74), we have **B** = _μ_ **H**. Equation (1) now gives

or, since rot rot A = −∇2A in view of (5),

This vector equation is equivalent to three scalar Poisson equations,

where _A_ _i_ , _J_ _i_ denote the rectangular components of **A** and **J**.

If the steady currents occupy a finite volume _V_ 1 within a closed surface _S_ 1, we may deduce just as in §82 that

and, consequently,

This vector satisfies div A = 0. By varying _P_ we find that

Since div _Q_ **J** = 0, the integral equals

on applying the divergence theorem. But, if the flow of electricity is confined within the surface _S_ 1 **J** · **n** = 0 on _S_ 1 and hence

Thus the vector potential given by (7) is a solenoidal vector that satisfies (6). Now **B** = rot **A** and

since **J** in the integral does not depend on _P_. Putting **B** = _μ_ **H** and

where **R** is a unit vector directed from the current elements toward _P_ , we have

This formula is the analogue of (82.11).

We now apply (7) and (9) to a thin closed-current tube carrying a steady current _I_. Putting

where **T** is a unit tangent in the direction of the current,

Thus each current element _I ds_ contributes

to the total magnetic field.† In 1820 Biot and Savart deduced the essential facts of this formula from their experiments; it is therefore known as the _Biot-Savart law_.‡

The current flowing through a surface _S_ spanned over a closed curve _C_ is, from (1),

hence, from Stokes' theorem,

This important result, already proved in §75, is

AMPÈRE'S LAW. _The circulation of the magnetic intensity over any simple closed curve is equal to the total current linked by the circuit_.

**Example 1. Infinite Straight Current**. We shall compute **H** at the point _P_ due to an infinite straight current _I_ in the direction of the unit vector **e** (Fig. 85 _a_). If we write

 , where **N** is a unit normal to the wire, we have at _Q_

and from (12)

In the figure **e x N** points up from the page.

Fig. 85 _a_. Infinite straight current.

Fig. 85 _b_. Circular current.

Since _d_ **H** in (12) is regarded as an element due to a _closed_ current, one may doubt the validity of (14). But, if the current is closed through the dotted semicircle of radius   in the figure, we readily find that the field at _P_ is

As _b_ →, ∞ _α_ → _π_ /2, and we again obtain (14).

**H** has the same numerical value at all points of a cylinder coaxial with the current. Hence we may compute _H_ from Ampère's law; for, as _P_ describes a circle of radius _a_ about the current,

**Example 2. Axial Field of a Circular Current**. We shall find **H** at a point _P_ in the axis of a circular current _I_ (Fig. 85 _b_). If **T** , **N** are unit tangent and principal normal,

and _r_ = _a_ /sin _θ_ is constant; hence

Since

Fig. 85 _c_. Infinite wire forming angle 2 _α_.

**Example 3. Infinite Wire Making an Angle 2 _α_**. In Fig. 85 _c_ each leg of the angle contributes   at _P_. If **k** is a unit vector pointing up from the paper,

and, since _θ_ \+ _φ_ = _α_ , _dφ_ = − _dφ_ , we have

Hence

When  , **H = k** _I_ /2 _πa_ , in agreement with (14).

**86. Magnetostatic Potential**. A closed stationary current _I_ generates a magnetic field **H** , and in any region free from currents

over any reducible curve: that is, a curve that does not link the current. The entire region outside of the currents will be multiply connected due to their presence, but any simply connected portion will admit a single-valued scalar potential _ψ_ for **H** so that

To find this _magnetostatic potential_ at _P_ we rewrite the Biot–Savart law (85.11) in the form†

and then transform the integral by (51.3) into an integral over a surface _S_ spanning the current circuit. Since divQ∇Q(1/ _r_ ) = 0, we thus obtain

Now from (52.8)

the solid angle subtended by the current loop at _P_ ; for from Fig. 86 _a_ we see that **n** , the normal to _S_ , is directed toward the observer _P_ , whereas in (52.8) **n** is directed away from the observer. Therefore

and

is the desired scalar potential.

Note that, in Fig. 86 _a_, _P_ is on the positive of _S_ as determined by the sense of the circuit; in this case Ω is taken positive. If _P_ is on the negative side of _S_ , Ω must be taken negative.

Fig. 86 _a_. Current loop.

Fig. 86 _b_. Infinite straight current.

**Example 1. Infinite Straight Current.** In Fig. 86 _b_ the infinite straight current at _O_ is flowing up from the paper along the _z_ -axis. If **k** is a unit vector along this axis and **R** a unit radial vector toward the point _P_ in the _xy_ -plane, the field at _P_ , given by (85.14), is

Thus **H** is derived from the scalar potential

This result agrees with (5). For, if we regard the return current as situated to the _left_ of _O_ at an infinite distance, _P_ lies on the negative side of the circuit; hence the solid angle Ω subtended by the circuit at _P_ is the negative area of a lune of angle _φ_ ;: thus

where 4 _π_ is the area of the unit sphere.

If the return current were taken to the _right_ of _O_ , _P_ would have been on the positive side of the circuit. Now the solid angle is

but **H** = − ∇ _ψ_ is still given by (6).

**Example 2. Two Infinite Parallel Currents.** In Fig. 86 _c_ the paper represents the _xy_ -plane and the currents are parallel to the _z_ -axis (up at _O_ l down at _O_ 2). The point

Fig. 86 _c_. Two infinite parallel currents.

_P_ lies on the negative side of the circuit, and the solid angle subtended by the currents at _P_ is the negative area of a lune of angle _θ_ ; hence

Now from (5)

and **H** is given by − ∇ _ψ_ or

Since we can write

**B** = _μ_ **H** is derived from the vector potential

thus **B** = rot **A** , and div **A** = 0.

In the _xy_ -plane the equipotential lines _ψ_ = const are the family of circles _φ_ 1 − _φ_ 2 = const through _O_ 1 and _O_ 2. The field lines of **H** = − ∇ _ψ_ cut these circles at right angles, and, from (7), are the family of coaxial circles _r_ 2/ _r_ 1 = const having _O_ 1 and _O_ 2 as inverse points (cf. Prob. 37.9).

**Example 3.** The problem of Ex. 85.2 may also be solved by means of (4). Referring to Fig. 85 _b_, the solid angle Ω. subtended at _P_ by the current is equal to the area of a unit sphere about _P_ intercepted by a cone with the apical angle 2 _θ_ ; hence

Symmetry shows that, at points on the _z_ -axis, **H** has only a _z_ -component. This is given by

and, since _z_ = _a_ cot _θ_ ,

### **PROBLEMS**

**1.** Prove equation (85.15).

**2.** In Ex. 1 show that **B** = _μ_ **H** is derived from the vector potential

Hence deduce the vector potential (8) for two infinite parallel currents.

Fig. 86 _d_. Solenoid.

**3.** A _solenoid_ is a circle of wire wound into a circular helix. If the solenoid has _n_ closely spaced turns in a coil of length _l_ and radius _a_ , prove that the magnetic intensity at a point _P_ (Fig. 86 _d_) due to the current _I_ is of magnitude

Show that, at the center of a _very long_ solenoid  , _H_  _nI_ / _l_ ; and, at its ends, _H_  _nI_ /2 _l_.

Regard the current as a sheet of linear density _λ_ = _nI_ / _l_ amperes per meter, and use [(85.16):

**87. Forces on Charges and Currents.** In §77 we defined the electric field strength **E** so that a "point charge" _q_ would be subject to the force **F** = _q_ **E**. In terms of charges distributed with the volume density _ρ_ , we now define **E** so that the force exerted on the charges per unit volume is _ρ_ **E**. Then the total force on the charges within a volume _V_ is

If _V_ is so small that **E** may be regarded as constant within _V_ ,

in agreement with (77.1).

To relate the force exerted on a current to the magnetic field we observe from §77 that **J** x **B** (like _ρ_ **E** ) has the dimensions of force per unit volume:

Thus we define the vector **B** so that the force exerted on currents per unit volume is **J** x **B**. Then the total force on the currents within a volume _V_ is

If the current flows in a thin linear conductor, so that **B** may be regarded as constant over each cross section, we have

hence the force acting on a linear current element _I dr_ = _I_ **T** _ds_ is

In contrast to **F** e, which is exerted parallel to **E** , the force _d_ **F** _m_ is perpendicular to the plane of _d_ **r** and **B**.

**E** and **B** having been defined, the vectors **D** and **H** are given by _ε_ **E** and **B** / _μ_ respectively. The properties of the electromagnetic field defined in this manner are compatible with Maxwell's equations and are in complete accord with experiment.

Finally, if an element of charge _q_ is moving with the velocity **v** in an electromagnetic field, it constitutes a current element _q_ **v** the field therefore exerts the forces

upon it. The total force on the moving charge is therefore

**Example 1. Charged Particle in a Stationary Electric Field.** A particle of mass _m_ and charge _q_ has, from (5), the equation of motion

Dot-multiplying by **v** = _d_ **r** / _dt_ gives

or

Hence, on integrating over any path from **r** 0 to **r** , we have

for the energy acquired (joules) by a particle of _m_ kilograms and charged with _q_ coulombs in "falling" through a potential difference of _V_ = _φ_ 0 − _φ_ volts. If the initial velocity **v** 0 = **0** , the velocity acquired is

For an electron

and (7) gives

The energy acquired by an electron in falling through one volt is called one _electron_ - _volt_ ; from (6)

When **E** is constant, the equation of motion is readily integrated, giving

The orbit (as for any particle in a constant field of force) is a parabola if **v** 0 is not parallel to **E**.

In the discussion above, the mass _m_ is regarded as constant instead of having the relativistic value _m_ 0(1 − _υ_ 2/ _c_ 2)− 1/2, where _m_ 0 is the rest mass and _c_ the velocity of light. Hence, as _υ_ → _c_ , _m_ → ∞. But, if  , _m_  _m_ 0; thus, if _υ_ = _c_ /10,   1.0θ5 _m_ 0.

**Example 2. Charged Particle in a Constant Magnetic Field.** Write the constant field vector **H** = _He_ , where | **e** | = 1. Then, from (5), the charged particle has the equation of motion

Dot-multiplying by **v** and **e** now gives

hence _υ_ and **e** · **v** are constant. Since **v** = _υ_ **T** (32.3), we have

where _α_ is the constant angle between **e** and **v**. Since the speed _υ_ is constant, the acceleration is purely normal and equals _κυ_ 2N (32.5); hence the equation of motion is

On equating magnitudes and unit vectors, this gives two equations:

From (8) we see that _κ_ is constant and that the curvature of the path

On differentiating (9) with respect to the arc _s_ and using Frenet's formulas (29.4), we have

and, on dot-multiplying by **B** ,

Since both _κ_ and τ are constant, the path is a circular helix (Ex. 30.2). In particular, if **v** is perpendicular to **H** , _α_ = _π_ /2, and, from (10), the path is a circle of radius _ρ_ = _mυ_ / _μq **H**_. As a check, we note the _ρ_ has the dimensions of length (§77).

† If **f** is an arbitrary vector function, we can add **T** · ∇ **f** _ds_ to this expression for _d_ **H** without altering the value of the circuit integral. Although _d_ **H** given in (12) is by no means unique, it has at least the simplest form.

‡ Also called Ampere's law by some authors.

† In (85.11) **R** has the direction  ; since _Q_ is in the current element, ∇ _Q r_ = − **R**.

##   **Chapter 9**

## **Vector Spaces**

**88. Postulates.** In geometry and mathematical physics it is well to have a more general conception of a vector which is applicable to space whose dimensions are greater than three. We therefore formulate an abstract definition of a vector free from geometric concepts and applicable in many diverse cases.

A vector space   over a field†   of scalars is formed by a set of elements called _vectors_ which have the following properties.

A1. Two vectors **u** , **v** determine a unique sum **u** \+ **v**.

A2. Vector addition is associative:

A3. Vector addition is commutative:

A4. There is a null vector **0** such that

A5. Every vector **u** admits a negative – **u** such that

M1. A vector **u** from _V_ and a scalar _a_ from   determine a unique _scalar product_ **_au_** in _V_.‡

M2. Scalar multiplication is associative:

M3. Scalar multiplication is distributive both ways:

M4. The unit element _e_ of the field (denoted hereafter by 1) has the property

From these postulates we can deduce the cancelation laws

The null vector **0** and the zero scalar 0 are connected by the important identity

This follows from

on canceling _a_ **u**. Moreover, from

we have also

Note finally that the negative of **u** (postulated in A5) is given by

for

It should be observed that addition of vectors and their multiplication by scalars produce vectors, and that _V_ is _closed_ with respect to both operations.

**Example 1.** The directed line segments of Chapter 1 are vectors since they satisfy all the postulates. If these vectors **u** are represented by their components [ _u_ 1, _u_ 2, _u_ 3] written in order, such ordered triples obey the rules

and [0, 0, 0] = 0 is the null vector. The commutative, associative, and distributive laws A2, A3, M2, M3 are consequences of the corresponding laws in the field   of real numbers.

**Example 2.** In coplanar statics (§22), forces and couples may be fully represented by the symbol ( **f, f _O_** ). If **f**  **0** is a force, **f _O_** is the moment of **f** about _O_ ; if **f = 0** and **f _O_  0**, **f 0** is the moment of a couple. The laws of statics are comprised in the operational rules

and the fact that static equilibrium corresponds to

The elements ( **f, f _O_** ) thus constitute a vector space with ( **0, 0** ) as null vector. This vector space is isomorphic with the vector space of bound vectors in space issuing from the origin (§22).

**Example 3.** All real-valued functions of _x_ in the interval 0  _x_  1 constitute a vector space. For two such functions _f_ ( _x_ ), _g_ ( _x_ ) may be added to give a third function _f_ ( _x_ ) + _g_ ( _x_ ) of the set; and, if _λ_ is a number of the real field, _λf_ ( _x_ ) is also such a function. Although these functions cannot be represented by arrows, they nevertheless constitute a vector space; the value of _f_ ( _x_ ) at each point of the interval may be regarded as its "component" at that point.

**89. Linear Dependence.** The _n_ vectors **u 1, u2, ···, u _n_** of a vector space   are said to be _linearly dependent_ if there are _n_ scalars _λ_ 1, _λ_ 2, ···, _λ n_ in  , not all zero, such that

If the vectors are not linearly dependent, they are said to be _linearly independent_. Consequently, if the relation ((6) exists between _n_ linearly independent vectors, all _λ i_ = 0.

_If m vectors are linearly dependent, any greater number n of vectors including these are also linearly dependent_. For, if

we can give _λ_ 1, _λ_ 2, ···, _α m_ these values (at least one of which is not zero) and take _λ_ _m_ \+ 1 = ··· = _λ n_ = 0. Then (1) is satisfied, and the _n_ vectors **u** _i_ are linearly dependent.

If _λ_ **u = 0** and _λ_  0, then **u = 0** (cancel _λ_ in _λ_ **u** = _λ_ **0** hence _one vector is linearly dependent only when it is the null vector_. Thus the vectors of any set that includes the zero vector are linearly dependent.

THEOREM. _The nonzero vectors_ **u** 1, **u** 2, **u** 3, ···, **u** _n_ _are linearly dependent when and only when one vector is a linear combination of its predecessors in this sequence_.

_Proof_. If **u** _k_ \+ 1 = _λ_ 1 _u_ 1\+ ··· + _λ_ _k_ _u_ _k_ , the vectors **u** 1, ···, **u** _k_ \+ 1 are linearly dependent, and the same is true of any larger set containing them.

Conversely, if **u 1**,···, **u** _n_ are linearly dependent, let _k_ \+ 1 be the _highest_ index in (1) for which _λ_ _k_ \+ 1 0; we can then solve (1) for **u** _k_ \+ 1 in terms of **u** 1, ···, **u** _k_. Note that _k_ \+ 1 > 1; for, if _λ_ 1 0, _λ_ 2 = ··· = _λ n_ = 0, we would have _λ_ 1 **u** 1 = 0 and **u** 1 = 0, contrary to hypothesis.

**Example.** For the free arrow vectors of Chapter 1 we can readily show that:

1. _Two proper vectors are linearly dependent when they are parallel_.

2. _Three proper vectors are linearly dependent when they are parallel to a plane_.

_Proof of_ 2. If _λ_ 1 **u** 1 \+ _λ_ 2 **u** 2 \+ _λ_ 3 **u** 3 = 0 and _λ_ 3 0, division by _λ_ 3 gives **u** 3 = _α_ **u** 1 \+ _β_ **u** 2, and the parallelogram construction shows that **u** 3 lies in the plane of **u** 1 and **u** 2. Conversely, if **u** 1, **u** 2, **u** 3 are parallel to a plane, they are linearly dependent. For, if **u** 1 and **u** 2 are parallel, **u** 2 = _α_ **u** 1 ( _α_  0); and, if no two are parallel, we have **u** 3 = _α_ **u** 1 \+ _β_ **u** 2 ( _α_ , _β_  0) by the parallelogram construction.

**90. Subspaces.** A subset of vectors in   which itself forms a vector space is called a _vector subspace_ of

If **a** 1, **a** 2, ···, **a** _k_ are given vectors in  , the set of all linear combinations

is called a _linear system L_ _k_.

The linear system _L k_ is a vector subspace of  , which may possibly be   itself. For, if

the vectors

are members of _L k_. The subspace _L k_ is said to be _spanned_ by the vectors **a** 1, ···, **a** _k_.

Thus, for the arrow vectors of Chapter 1, the subspace _L_ 1 spanned by the vector **a  0** consists of all vectors _λ_ **a** 1 collinear with **a** 1. The subspace _L_ 2 spanned by two noncollinear vectors **a** 1 and **a** 2 consists of all vectors _λ_ 1 **a** 1 \+ _λ_ 2 **a** 2 coplanar with **a** 1 and **a** 2. Finally, if **a** 1, **a** 2, **a** 3 are not coplanar, the subspace _L_ 3 they span consists of all vectors _λ_ 1 **a** 1 \+ _λ_ 2 **a** 2 \+ _λ_ 3 **a** 3; and, since any arrow vector can be put in this form, **a** 1 **a** 2, **a** 3 span the entire vector space of three dimensions.

DEFINITION. _A set of linearly independent vectors which spans a vector space is called a_ basis _for that space_.

Thus, any three noncoplanar vectors **a** 1, **a** 2, **a** 3 form a _basis_ for the ordinary vector space of three dimensions.

We now have the fundamental

THEOREM. _If n vectors span a vector space containing k linearly independent vectors, n_   _k_.

_Proof_.† Let **a** 1, **a** 2, ···, **a** _n_ span the vector space _V_ which contains the _k_ linearly independent vectors **b** 1, **b** 2, ···, **b** _k_. Then **b** 1 0 and is given by a linear combination of **a** 1, **a** 2, ···, **a** _n_ ; hence the vectors of the sequence { **b** 1, **a** 1, **a** 2, ···, **a** _n_ } are linearly dependent. By Theorem 89 some **a** i is a linear combination of its predecessors in this sequence; and, if it is discarded, **b** 2, **b** 1 and the _n_ − 2 remaining **a** 's still span _V_.

If we call the discarded vector **a** 2 the _n_ vectors **b** 2, **b** 1, **a** 3, **a** 4 ···, **a** _n_ span the vector space _V_ which contains _k_ − 2 linearly independent vectors **b** 3, **b** 4, ···, **b** _k_. We now repeat the argument above. Now **b** 2 0 and is given by a linear combination of **b** 1, **a** 2, ···, **a** _n_ ; hence the vectors of the sequence { **b** 2, **b** 1 **a** 2, ···, **a** _n_ } are linearly dependent. Since **b** 1  _λ_ **b** 2 by hypothesis, Theorem 1 shows that some **a** i of this sequence is a linear combination of its predecessors; and, if it is discarded, **b** 2, **b** 1 and the _n_ − 2 remaining **a** 's still span _V_.

If we call the discarded vector **a** 2, the _n_ vectors **b** 2, **b** 1, **a** 3, **a** 4, ···, **a** _n_ span the vector space _V_ which contains _k_ – 2 linearly independent vectors **b** 3, **b** 4, ···, **b** _k_. We now repeat the above argument and find that we can discard another **a** i from the sequence { _b_ 3, **b** 2, **b** 1, **a** 3, ···, **a** _n_ }; for the **b** 's cannot be discarded as they are linearly independent.

This procedure is now repeated until

(a) all **a** 's are discarded, _or_

(b) all **b** 's are exhausted, but some **a** 's are left.

In case ( _a_ ), the **b** 's will be exhausted when all **a** 's are discarded; otherwise any remaining **b** would be a linear combination of the **b** 's already used, contradicting the hypothesis. Hence, in case ( _a_ ), _n_ = _k_. In case ( _b_ ), _n_ > _k_. Thus _n_   _k_ , and the theorem is proved.

**91. Bases and Dimension.** The following theorem enables us to define the dimension of a vector space which has a basis formed by a finite number of vectors.

THEOREM 1. _All bases of a vector space contain the same number of vectors provided this number is finite_.

_Proof_ Let **a** 1, **a** 2, ··· **a** _n_ and **b** 1, **b** 2, ···, **b** _k_ be two bases of a vector space _V_. Since the **a** 's span _V_ and the **b** 's are linearly independent, _n_   _k_ by Theorem 90. But, since the **b** 's span _V_ and the **a** 's are linearly independent, _k_   _n_. Hence, _n_ = _k_ , as stated in the theorem.

The _dimension_ of a vector space is now defined as the number of vectors (if finite) in any one of its bases. Any _n_ -dimensional vector space is a linear system _L n_.

THEOREM 2. _In an n_ - _dimensional vector space any n_ \+ 1 _vectors are linearly dependent_.

_Proof_. By Theorem 1, any basis of the vector space consists of exactly _n_ vectors. If any _n_ +1 vectors in _L n_ were linearly independent, Theorem 90 would require an impossibility: namely _n_   _n_ \+ 1. Hence any _n_ \+ 1 vectors of the space must be linearly dependent.

THEOREM 3. _Any set of n linearly independent vectors_ **e** 1, **e** 2, ···, **e** _n_ _of a vector space over a field_   _form a basis; and any vector_ **u** _of the space has a unique representation_

_where the scalars u i are in the field_  †.

_Proof_. The _n_ \+ 1 vectors **u** , **e** 1, **e** 2, ···, **e** _n_ are linearly dependent by Theorem 2; hence

where the lambdas are in   and at least one is not zero. If _λ_ = 0, the linear independence of **e** 1, **e** 2, ···, **e** _n_ would require _λ_ 1 = _λ_ 2 = ··· = _λ n_ = 0, and _all_ lambdas in (9)) would vanish. Hence, _λ_  0; and, on solving (2) for _u_ and writing _u i_ = − _λ i_/ _λ_ , we obtain (1). Since every vector of the space can be represented in this way, the vectors **e** 1, **e** 2, ···, **e** _n_ form a basis.

If there were a second expression for **u** of the form (1) with coefficients _u i_, we would have on subtraction

since **e** 1, **e** 2, ···, **e** _n_ are linearly independent. The expression (1) for **u** in terms of the basis is therefore unique.

The scalars _u_ 1, _u_ 2,···, _u n_ in (1) are called the _components_ of the vector **u** relative to the basis **e** 1, **e** 2, ···, **e** _n_. The null vector **0** has all of its components 0. The totality of vectors **u** form a linear system (§90); we therefore denote this vector space by. _L n_( ).

**92. Ordered Sets.** Consider now the ordered set of _n_ scalars _u i_ belonging to the field   :

We define addition of such sets, and their multiplication by scalars, by the equations

and denote the null set by

It is now easy to verify that all postulates for a vector space are satisfied by such _n_ -tuples. These _n_ -tuples are therefore vectors, and **0** is the null vector. In aggregate they constitute an _n_ -dimensional vector space which we denote by  ; for the vectors

are obviously linearly independent and form a basis; we have, in fact, from (1) and (2):

When a basis **e** 1, **e** 2, ···, **e** _n_ of a linear system   is given, any vector **u** of the system has the unique representation

We may now set up a one-to-one correspondence between the vectors **u** of   and the vectors **U** of   :

This correspondence is an _isomorphism_ between the vector spaces _L n_ and _V n_; for

Therefore: _Any n_ - _dimensional vector space is isomorphic with the vector space V n of ordered n_- _tuples over the same field_.

**93.Inner Product.** Since length and angle have not been defined in an abstract vector space, these concepts are not available in defining the analogue of the dot product **u** · **v** of two vectors. Since _λ_ **u** has been called the _scalar product_ , we henceforth term **u** · **v** the _inner product_. We now define the inner product **u** · **v** of two vectors of a vector space   over the field   of reals, as a scalar having the following properties:

From (1) and (3) we have also

and, since **u** · **v** = **u** · ( **v** \+ **0** ) = **u** · **v** \+ **u** · **0** ,

For the sake of compactness and brevity we shall in the following frequently use the

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_.

Consider now two vectors **u** , **v** of an _n_ -dimensional vector space,  :

Then, making use of properties (1), (2), (3) and (3)′, we have

The inner products **e** _i_ · **e** _j_ are now defined as

where the _n_ 2 real scalars _g ij_ are the coefficients of a _positive definite quadratic form:_

For such a form _g_ ( _x_ , _x_ )   0 for all sets ( _x_ 1, _x_ 2,···, _x n_); and _g_ ( _x_ , _x_ )= 0 only when all _x i_ = 0.† We now have

and property (4) is sustained.

Two nonzero vectors **u** , **v** are said to be _orthogonal_ when and only when **u** · **v** = 0. Thus orthogonal abstract vectors constitute a natural generalization of perpendicular arrow vectors.

A basis **a** 1, **a** 2, ···, **a** _n_ is said to be _orthogonal_ when each **a** _i_ is orthogonal to all the others; then

_The vectors of any nonzero orthogonal set_ **a** 1, **a** 2, ···, **a** _k_ _are linearly independent_. For, if the vectors are connected by a linear relation

dot-multiplying by **a** _i_ gives _λ_ _i_ **a** _i_ · **a** _i_ = 0; and, since **a** _i_ · **a** _i_ > 0, _λ i_ = 0.

_Any vector_ **v** _orthogonal to the set_ **a** 1, **a** 2, ···, **a** _k_ _is orthogonal to every vector of the subspace L k spanned by them:_ for

**94. Euclidean Vector Space.** In order to make _measurements_ in an abstract vector space, we next define the _length_ of a vector **u** to be

Moreover, in analogy with (10.14), we define the _angle_ ( **u** , **v** ) between **u** and **v** by

We now show that, when _g ij _x_ ixj_ is a positive definite quadratic form, length defined by (1) has the familiar Euclidean properties:

The _positiveness_ of length, expressed by (3), follows at once from (93.4). The multiplicative property (4) is an immediate consequence of (1):

To prove (5), consider

Since this may be written

we see that the left member will be   0 for all _λ_ when and only when the last fraction   0; that is,

Using this inequality, we now have

and, on taking the positive square root of both sides, we obtain (5).

As a consequence of the Schwarz inequality (6), we find from (2) that

hence the unique angle defined by (2) is always real.

We thus see that the positive definite form _g ijxixj_ enables us to make measurements of length and angle in our vector space. Therefore _g ijxixj_ is called the _metric quadratic form_ of the vector space; and the _n_ 2 numbers _g ij_† are said to define the _metric_ of the space.

If all vectors of an orthogonal basis **a** _i_ are of length 1, the basis **a** _i_ is called _orthonormal_. Thus the basis **a** 1, **a** 2, · ·, **a** _n_ is orthonormal when and only when

where _δ ij_ is the Kronecker delta   of §15. When vector components are referred to an orthonormal basis, the inner product assumes the simple form

_When the basis is orthonormal, the inner product of two vectors is the sum of the products of their corresponding components:_

This is the generalized form of (10.13).

In the vector space   of real _n_ -tuples (§92) the inner product is _defined_ as

With this definition the base vectors **E** _i_ , form an orthonormal set; for from (92.4) we obviously have **E** _i_ · **E** _j_ = _δ ij_.

In the vector space   of complex _n_ -tuples, the inner product **U** · **V** is defined as

where   is the complex conjugate of _υ i_. With this definition **U** · **U** > 0 if **U**  0; and | **U** |, the length of **U** , is given by the positive square root of **U** · **U** :

Unless **U** and **V** have real components, **U** · **V**  **V** · **U** ; instead we now have _Hermitian symmetry_ :

and for any scalar

The distributive laws,

are still valid. Moreover both the Schwarz and triangle inequalities hold good. To prove the former, consider

Since this may be written

we see that the left member will be 0 for _all λ_ when and only when the last fraction   0; we thus obtain the Schwarz inequality:

The triangle inequality

now follows from (14):

If **u** 1, **u** 2, ···, **u** _k_ are _k_ vectors of  , their _Gramian_ is the determinant

From (12) we see that   is obtained from _G k_ by an interchange of rows and columns. Thus  ; therefore _a Gramian is a_ real _scalar_.†

THEOREM. _A necessary and sufficient condition that_ **U** 1, **U** 2, . . ., **U** _k_ _be linearly dependent is that their Gramian vanish_.

_Proof_. The condition is necessary; for, if there are _k_ constants _c i_ not all zero, such that

we have the _k_ homogeneous equations

Since the _c_ 's are not all zero, the determinant _G k_ of the system (18) must vanish.

The condition _G k_ = 0 is also sufficient. For the system (18),

has the zero determinant _g k_ and therefore must admit a solution _c_ 1, _c_ 2, ···, _c k_ not consisting of zeros. If we multiply these equations by  , respectively, and add after using (13), the resulting equation,

implies (17).

**95. Gram–Schmidt Process.** If the vectors **e** 1, **e** 2 ···, **e** _n_ form a basis for an Euclidean vector space, we can derive from them by rational processes an orthogonal basis **a** 1, **a** 2, ···, **a** _n_ that spans the same space.

The construction, known as the _Gram_ – _Schmidt process_ , consists in forming the vector sequence

and, in general,

If **a** 1, **a** 2, ···, **a** _k_ +1 are not zero, these equations show that the subspace _L_ _k_ − 1 they span is the same as that spanned by **e** 1, **e** 2, ···, **e** _k_ − 1. Now, if **a** 1, **a** 2, ···, **a** _k_ − 1 form an orthogonal set, the same is true when **a** _k_ is added, for from (1)

Moreover **a** _k_  **0** ; for, if **a** _k_ = **0** , (1) implies that **e** _k_ is a vector of the subspace _L_ _k_ − 1 spanned by **a** 1, **a** 2, · ·*, **a** _k_... 1 (or by **e** 1, **e** 2, ···, **e** _k_ − 1) and **e** k could not be a base vector. Thus, if the construction succeeds for **a** 1, **a** 2, ···, **a** _k_ − 1 it will succeed for **a** 1, **a** 2, ···, **a** _k_ ; and, since **a** 2  **0** and **a** 2 · **a** 1 = 0, we have an induction proof for its general validity.

From the orthogonal basis **a** _i_ we may now form the _orthonormal_ basis **b** _i_ = **a** _i_ | **a** _i_ |; for the **b** _i_ are orthogonal and of length 1.

In view of (12.3) we may interpret the term

as the projection of **e** _k_ on **a** _i_ ; thus (1) states that

Thus the Gram-Schmidt construction simply amounts to this: Take **a** 1 = **e** 1; take **a** 2 as **e** 2 minus its projection on **a** 1; take **a** 3 as **e** 3 minus its projections on **a** 1 and **a** 2; and so on until the **e** 's are exhausted.

**96. Reciprocal Bases.** We can now generalize to abstract vector spaces the concept _of reciprocal bases_ (§15).

_Two bases_ **e** 1, **e** 2, ···, **e** _n_ and **e** 1, **e** 2, ···, **e** _n_ of the same vector space are said to be _reciprocal_ when they satisfy the _n_ 2 equations:

Since the **e** _i_ are vectors of the space spanned by **e** _i_ we have (using the summation convention)

Hence from (93.6)

or, from (1),

In order to solve these equations for the _n_ 2 lambdas, we introduce the reduced cofactor _g ij_ of _g ij_ in the determinant

The _cofactor G ij_ of _g i_ is obtained by striking out row _i_ and column _j_ in the determinant and giving the resulting minor the sign (− 1) _i_ \+ _j_ ; then the _reduced_ cofactor is

Note that _g ij_ = _g ij_ implies _g ij_ = _g ji_.

A fundamental theorem now states that, if the elements of a row (column) are multiplied by their corresponding cofactors we get _g_ ; but, if the elements of a row (column) are multiplied by the corresponding cofactors of _another_ row (column), we get zero. With the aid of the summation convention all this is expressed by

or, in terms of reduced cofactors,

In these equations _r_ is a summation index.

Return now to the _n_ 2 equations (3). Multiply (3) by _g jh_; then

Summing on _j_ and _k_ as required by convention, we have

hence _λ ih_ = _g ih_, and, from (2),

It is easy to verify (6); for

If we dot-multiply (6) by _**e** j_, we have

as a companion piece to **e** _i_ · **e** _j_ = _g ij_. If we regard **e** _i_ as the original base vectors with the metric _g ij_, the reciprocal basis is **e** _i_ and equations (6) become

This result is also easily verified by solving equations (6) for **e** _j_ by multiplying both members by _g ij_:

A vector **u** , referred to reciprocal bases, now has two forms:

These equations show that the components of **u** are

If we now put **e** _i_ = _g ik_ **e** _k_ , **e** _i_ = _g ik_ **e** _k_ in (10), we also obtain the equations

which enable us to pass from one type of component to the other without intervention of base vectors.

The simplicity of formula (94.9′) for the inner product **u** · **v** referred to an orthonormal basis may now be regained; for, if we refer **u** to one basis and **v** to the reciprocal basis, we find

or

hence

Finally we note that, for an orthonormal basis **a** _i_ ,

and hence **e** _i_ = **e** _i_ from (6). Conversely, if

Therefore: _A basis is self-reciprocal when and only when it is orthonormal_

### **PROBLEMS**

**1.** If the metric is given by

find the lengths and angles between the base vectors **e** _i_.

**2.** ( _a_ ) Find _g ij_ corresponding to _g ij_ in Prob. 1.

( _b_ ) Express **e** 1, **e** 2, **e** 3 linearly in terms of **e** 1, **e** 2, **e** 3.

**3.** If _u i_ = [0,1, 2], _υ i_ = [− 1, 0, 1] and _g ij_ has the values in Prob. 1, compute

( _a_ ) _u i_ and _υ i_;

( _b_ ) | **u** | and | **v** | in two ways;

( _c_ ) **u** · **v** in four ways;

( _d_ ) the angle ( **u** , **v** ).

**4.** If **e** 1 = **i** \+ **j** , **e** 2 = **i** − **j** , **e** 3 = **i** \+ **j** − **k** , compute

( _a_ ) _g ij_ and _g ii_;

( _b_ ) **e** 1, **e** 2, **e** 3;

( _c_ ) the components _u i_ and **u** _i_ of **u** = **i** \+ 2 **j** \+ **k**. Check by finding | **u** | in two ways.

**5.** If the basis **e** i is changed to   by the nonsingular linear transformation

prove that the components _u i_ of any vector **u** transform according to

owing to the invariance of the vector  .

Show also that

**6.** In Prob. 5 show that the vectors   actually form a new basis: i.e. that they are _linearly independent_.

**7.** If   is the reduced cofactor of   in det  , show that (14) and (16) may be written

**8.** If the basis **a** _i_ is orthonormal, and the transformation,

leads to a new orthonormal basis **ā** _i_ prove that  : i.e. _each element in the_   _determinant_ det   _is its own cofactor_.

**9.** A linear transformation between orthonormal bases is said to be _orthogonal_. If _C_ is the matrix ( ) of an orthogonal transformation and _C_ ′ is the transpose of _C_ (rows and columns interchanged), prove that

**10.** Prove that in three dimensions the determinant _c_ of an orthogonal transformation is 1 when the transformation _revolves_ the trihedral **a** 1, **a** 2, **a** 3 into **ā** 1, **ā** 2, **ā** 3.

**97. Hilbert Space.** A sequence _u_ 1, _u_ 2, ··· of real or complex scalars, for which the infinite series Σ | _u i_ | 2 converges, is called a _vector_

having _u_ 1, _u_ 2, ··· as components. The product of **U** by a scalar and the sum **U** \+ **V** are defined by

Since

_a_ **U** is a vector. Moreover the sum **U** \+ **V** of two vectors is a vector, for we shall prove that Σ | _u i_ \+ _υ i_ | 2 converges. The null vector is

and the negative of **U** ,

With these definitions all the postulates of §88 are satisfied; and the totality of such vectors constitutes the _Hilbert vector space_.

We now define the inner product **U** · **V** and the length | **U** | of **U** by a direct extension of the definitions of §94:

Since  , the existence of | **U** | is implied by the convergence of Σ | **U** |2. To deal with the series in (5), apply the Schwarz inequality (94.14) to the finite vectors:

then | **U** _n_ · **V** _n_ | | **U** _n_ | | **V** _n_ | becomes

Since this holds for every _n_ , it holds as _n_ → ∞ and since, we have

Thus the series in (5) converges absolutely and hence also converges in the usual sense† to a sum defined as **U** · **V**. Moreover

and hence, from (7),

this is the Schwarz inequality in Hilbert space.

The distributive law

obviously holds for three vectors; for we have just seen that the series for **U** · **V** and **U** · **W** converge. In particular

shows that the series for ( **U** \+ **V** ) · ( **U** \+ **V** ) converges; hence the sum of two vectors is a vector (as stated above). The triangle inequality

now follows from (8) as in §94.

The vectors **U** and **V** are said to be _orthogonal_ when **U** · **V** = 0. The set of vectors **A** 1, **A** 2, ···, **A** _n_ are said to be _orthonormal_ when

A vector **U** then is said to have the components **U** · **A** 1 ···, **U** · **A** _n_ relative to this set.

Now let _λ_ 1, _λ_ 2, ···, _λ n_ be _n_ arbitrary scalars, and consider the vector

We propose to find the values _λ i_ that make **T** _n_ the best approximation to **U** in the sense of the method of least squares; that is, those values that give the squared length

its minimum value. Making use of (10) and (94.13) we then find

Since  , this may be written

Thus _D n_ will assume its least value when _λ i_ = **U** · **A** _i_ ; then **T** _n_ becomes

and this least value is

We have thus proved

THEOREM 1. _For any given n, the sum_   _is the best approximation to_ **U** _when the scalars λ i_ = **U** · **A** _i_ _the components of_ **U** _on the orthonormal basis_ **A** _i_.

If the orthonormal basis **A** _i_ contains an infinite number of vectors, the terms for the best approximation for a given set **A** 1, ···, **A** _n_ are retained when this set is augmented by others. Since _D n_   0, and _n_ can be taken arbitrarily large, (13) also proves _Bessel_ ' _s inequality:_

and at the same time shows that the components **U** · **A** _i_ form a null sequence.

The orthonormal basis **A** 1, **A** 2, ··· is said to be _closed_ if

for all vectors **U**. From (13) we now have

THEOREM 2. _A necessary and sufficient condition that the orthonormal basis_ **A** _i_ _be closed is that_

This equation is called _Parseval_ ' _s equation_ ; it corresponds to the equal sign in Bessel's inequality and is a generalization of the Pythagorean theorem to space of a countable infinity of dimensions.

An orthonormal basis is said to be _complete_ when there is no nonzero vector which is orthogonal to all base vectors.

THEOREM 3. _Every closed basis is complete_.

_Proof_. If a vector **U** were orthogonal to each base vector **A** _i_ of an orthonormal set, **U** · **A** _i_ = 0 and **U = 0** from Parseval's equation (16).

Hence no nonzero vector is orthogonal to all base vectors.

DEFINITION. A sequence of vectors { **U** _n_ } is said to _converge strongly_ to a vector **U** when

and we write  .†

Thus the condition (15) for closure of an orthonormal basis **A** _i_ may be written

for all vectors **U**.

LEMMA 1. _A necessary and sufficient condition that_ Lim **U** _n_ _exist is that to every ε_ > 0 _there corresponds an integer N such that_

_Proof_. The condition is _necessary_ ; for, if Lim **U** _n_ = **U** ,

Hence when _m_ , _n_ > _N_ ,

To show that the condition is _sufficient_ , we first observe that, if **U** _n_ =  ** _u_** _n_ 1, **_u_** _n_ 2,···], [(18) shows that

hence

and   exists by Cauchy's criterion:† denote it by **u** _i_. Then, if we let _n_ → ∞ in the sum of _p_ terms

we obtain

As this holds for every _p_ , we also have

or upon writing **U** = ( _u_ 1, _u_ 2 ···),

Moreover **U** is a vector; for **U** − **U** _m_ ( _m_ > _N_ ) and **U** _m_ are vectors, and hence their sum is a vector.

LEMMA 2. If  ,  , then

_Proof of_ (19). Choose _N_ so that

Then by the triangle inequality (9),

_Proof of_ (20). Since

on making use of the Schwarz inequality (8). Now choose _N_ so that then

then

provided _ε_ < 2(| **U** | + | **V** |)2.

We are now in position to prove

THEOREM 4. _Every complete basis is closed_.

_Proof_ If **U** is any vector and **A** 1, **A** 2, ··· a complete orthonormal basis, consider the vector _S n_ in (12). As _n_ → ∞, _S n_ converges strongly to a limiting vector _S_. For, if _m_ < _n_ ,

and, since   converges by (14), we can choose an integer _N_ so that the last sum < _ε_ 2 when _m_ , _n_ > _N_. Hence,

and, by Lemma 1,  . Now from (19),

and, since

we have, on letting _n_ → ∞

from (20). Thus **U** − **S** is orthogonal to all base vectors **A** _i_ ; and, since the basis is complete, **U − S = 0**. Again from (20) we have

Since this implies (15), the basis is closed.

Relative to the orthonormal basis

the components of **U** = [ _u_ 1, _u_ 2,··· ] are obviously **U** · **E** , = _u i_; and, since

from (6), Parseval's equation (16) is satisfied. The basis (21) is therefore closed and complete.

**98. Linear Equations.** Consider the set of _n_ homogeneous equations

where **A** 1, **A** 2, ···, **A** _n_ are _n_ given vectors of Hilbert space and **X** is a vector to be determined.

THEOREM 1. _If **X** satisfies equations_ (1) _and is linearly dependent upon_ **A** 1, **A** 2, ···, **A** _n_ , _then_ **X = 0**.

_Proof_ If **X** = _c_ 1 **A** 1 \+ ··· + _c n_ **A** _n_ , is a solution of (1),

and hence X = 0.

THEOREM 2. _If the n vectors_ **A** 1 _are linearly independent, the general solution of equations_ (1) _is_

_where_ **U** _is an arbitrary vector and G_ ( **A** 1, **A** 2, ···, **A** _n_ ) _is the Gramian of_ **A** 1, ···, **A** _n_ ( _the cofactor of_ **U** _in the numerator_ ).

_Proof_ The proof of Theorem 94 applies without change to Hilbert space; hence _G_ ( **A** 1, **A** 2, ···, **A** _n_ ) 0. If we expand the numerator determinant of (2) by elements in the last row, we have

Since **U** is arbitrary, any vector, and consequently any solution of equations (1), can be put in this form. Now, if we form **X** · **A** , from (2), the numerator determinant will have the same elements in the _i_ th and last rows and **X** · **A** _i_ = 0, _i_ = 1, 2, ···, _n_.

Let us indicate the dependence of the solution **X** in (2) on **U** by writing **X(U)**. Now from (3) and Theorem 1 we see that **X(U) = 0** when and only when **U** is linearly dependent upon **A** 1, **A** 2, ···, **A** _n_. Moreover (3) shows that

hence **X(U 1)** = **X(U 2)** when and only when **U** 1 − **U** 2 is linearly dependent upon **A** 1, **A** 2, ···, **A** _n_.

From (3) we have

in view of equations (1). But, if we compute **X** · **U** from (2), we have

Now this equation applies to any _n_ \+ 1 vectors which are linearly independent, say **A** 1, **A** 2, ···, **A** _n_ , **A** _n_ \+ 1. Then both Gramians 0, and we have, with an obvious notation,

for **X  0** since **A** _n_ \+ 1, by hypothesis, is not linearly dependent on **A** 1, **A** 2, ···, **A** _n_. Consequently, if _G_ _n_ > 0, also _G_ _n_ \+ 1 > 0. But **A** 1 0, since the **A** 's are linearly independent; hence _G_ 1 = **A** 1 · **A** 1 > 0. With this starting point, we can now prove by induction

THEOREM 3. _The Gramian of any number of linearly independent vectors is positive_.

Finally let us consider the nonhomogeneous equations

THEOREM 4. _If_ **A** 1, **A** 2, ···, **A** _n_ _are linearly independent, the only solution of equations_ (7) _of the form_

_is given by_

_Proof_ If we compute **Y** 0 · **A** _i_ from (9), the last row of the upper determinant becomes

and, when the _i_ th row is subtracted from this, it becomes

hence

and equations (7) are satisfied. If there were a second solution **Y** 1 of the form (8), **Y** 0 − **Y** 1 would be a solution of equations (1), linearly dependent on **A** 1, **A** 2, ···, **A** _n_ , and **Y** 0 − **Y** 1 would vanish. Thus **Y** 0 is the only solution having the form (8).

The general solution **Y** of equations (7) is obtained by adding the particular solution **Y** 0 to the general solution **X** of the homogeneous equations (1):

We shall call **Y** 0 the _principal solution_ of equations (7). It is characterized by the following property.

THEOREM 5. _Of all the vectors which satisfy equations_ (7), _the principal solution has the smallest length._

_Proof_ From (10),

for, from (8),

and   Hence

the equal sign holding only when **X = 0** , in which case **Y** = **Y** 0.

The above results may be extended to a system of an infinite number of linear equations.†

### **PROBLEMS**

**1.** Show that, if equations (1) have a solution of the form (3), it must be given by (2). [Use Cramer's rule.]

**2.** Show that, if equations (7) have a solution of the form (8), it must be given by (9).

**3.** If the vectors **A** _i_ = [ _a_ _i_ 1, _a_ _i_ 2, ···, _a_ _in_ , 0, 0, ··· ], prove that

† See _Advanced Calculus_ ,§2, for the definition of a _field_ and the deduction of the cancelation laws. The rational, the real, and the complex numbers constitute three important fields. In this chapter _scalar_ means a number of  .

‡ Henceforth **u** · **v** will be called the _inner product_.

† We follow the argument in Birkhoff and MacLane, _A Survey of Modem Algebra_ , Macmillan, 1953. Since it makes no use of the commutative law _ab_ = _ba_ , the theorem is valid when the scalars form a _division ring_.

† Just as in §15, the upper indices are _not exponents_ but mere identification tags.

† Cf _Vector and Tensor Analysis_ ,§150. Note that _g ij_ > 0 ( _i_ = 1, 2, ···, _n_ ); for, if  .

† Owing to the symmetry _g ij_ = _g ji_ at most   of the coefficients _g ij_ have distinct values; for there are _n_ of the form _g ii_, and   of the form _g ij_ ( _i_  _j_ ) and

† _G 1_ = **U** 1 · **U** 1   0; and _G_ 2   0 from ((14). We shall see later (§98) that, in general, _G k_   0.

† _Advanced Calculus_ , p. 63.

† Note capital **L** in Lim.

† Advanced Calculus, §17.

† Throughout this article equations (94.13) must be kept in mind.

† Cf. Bôcher and Brand, "On Linear Equations with an Infinite Number of Variables", _Ann. Math_., vol. 13, 1912, pp. 167–186.

## **Appendix**

##  **Quantities, Dimensions, and Units.**

## **Supplementary Problems**

Verify the following statements:

**1.** If _P_ 1, _P_ 2, ···, _P_ 6; are six points in space and _Q_ 1, _Q_ 2, ···, _Q_ 6 are the mean centers of the triangles _P_ 1 _P_ 2 _P_ 3, _P_ 2 _P_ 3 _P_ 4, ···, _P_ 6 _P_ 1 _P_ 2, show that _Q_ 1 _Q_ 2 ··· _Q_ 6 is a hexagon whose opposite sides are equal and parallel.

**2.** The six planes passing through the mid-points of the edges of a tetrahedron _OABC_ meet in a point _P_ if they are perpendicular to these edges or to the opposite edges. If **a′, b′, c′** is the set reciprocal to the position vectors **a, b, c,**

or

**3.** If _P_ is a point in the plane of the triangle _ABC_ and

the numbers _α_ , _β_ , _γ_ are the _areal coordinates_ of _P_ , namely the ratios of the signed areas _PBC, PCA, PAB_ , to _ABC_. The signed areas are positive or negative according as the given circuit has the same sense as _ABC_ or the opposite.

**4.** The point _A_ ( _α_ , a _o_ ) and the line ( **b** , **b** _o_ ) determine the plane ( _α_ **b** _o_ − **a** _o_ × **b** , **a** _o_ · **b** _o_ ).

**5.** If ( **a** , **a** _o_ ) and ( **b** , **b** _o_ ) are nonintersecting line vectors, all tetrahedrons whose opposite edges are **a** , **b** have the same volume

irrespective of the position of **a** and **b** on their lines.

**6.** The line through the point _P_ and cutting two skew lines ( **a** , **a** _o_ ), ( **b** , **b** _o_ ) has the coordinates (§17)

**7.** If ( **a** , **a** _o_ ) and ( **b** , **b** _o_ ) are nonparallel lines and **n** is a unit vector along **a** × **b** , their common normal has the coordinates

**8.** The only curves of constant curvature that lie on a plane or sphere are circles.

**9.** In order that a twisted curve lie on a sphere it is necessary and sufficient that

[If ( **r** − **c** )·( **r** − **c** ) = _a_ 2 is the sphere,

**10.** The closed spherical curve

consists of two loops that cut at right angles at the double point _A_ ( _l_ , 0, 0); and _K_ = 1,   at _A_.

**11.** Two curves whose curvature and torsion are the same functions of the arc are _congruent_.

[If **T** 1 **N** 1 **B** 1 and **T** 2 **N** 2 **B** 2 coincide at the common origin of arcs and _f_ ( _s_ ) = **T** 1 · **T** 2 \+ **N** 1 · **N** 2 \+ **B** 1 · **B** 2, show that _f_ ′( _s_ ) = 0, _f_ ( _s_ ) = 3; **T** 1 = **T** 2, **N** 1 = **N** 2, **B** 1 = **B** 2 at corresponding points, and **r** 1( _S_ )= **r** 2( _s_ ).]

**12.** If **r** = **r** ( _s_ ) is a twisted curve of constant curvature _K_ , the locus **r** 1 = **r** \+ _ρ_ **N** of its centers of curvature has _k_ = _k_ , _τ_ 1 = _k_ 2/ _τ_.

**13.** The triangle _ABC_ is on the positive side of the lines which form its sides. If _x, y, z_ are the signed perpendicular distances from a point _P_ to the lines _BC, CA_ , _AB_ the locus of _αx_ \+ _βy_ \+ _γz_ = 0 is a straight line perpendicular to _α_ **N** 1 \+ _β_ **N** 2 \+ _γ_ **N** 3, where **N** 1, **N** 2, **N** 3 are unit vectors normal to _BC, CA, AB_ in the positive sense. Discuss the case when _α_ / _a_ = _β_ / _b_ = _γ_ / _c_ ( _a, b, c_ are sides of _ABC_ ).

As _P_ ranges over the triangle _ABC_ , show by a construction just where _x_ \+ _y_ \+ _z_ attains its least and greatest values.

**14.** A point _P_ in the plane of the triangle _ABC_ is at distances _r_ 1 _r_ 2, _r_ from _A, B, C_. Then _r_ 1 \+ _r_ 2 \+ _r_ 3 attains its minimum value at a point _Q_ for which the angles _BQC_ = _CQA_ = _AQB_ = 120°. Construct _Q_.

**15.** On the surface **r** = **r** ( _u_ , _v_ ) the unit normal **n** = ( **r** _u_ × **r** _υ_ )/ _H_ has the gradient

where **a** , **b** , **n** denotes the set reciprocal to **r** _u_ , **r** _υ_ , **n** (Prob. 42.7–8). The _mean curvature J_ and _total curvature K_ of the surface are defined as the invariants of Grad **n** :

Prove that _J_ and _K_ are determined by the equations

and are independent of the choice of parameters _u, v_.

Show that the total curvature of the surface of revolution of Prob. 36.1 is

and, for a sphere of radius _a, K_ = 1/ _a_ 2.

**16.** The points _P, Q, R_ divide the sides _BC, CA, AB_ of the triangle _ABC_ in the ratios _x_ /(1 − _x_ ), _y_ /(1 − _y_ ), _z_ /(1 − _z_ ). Then the signed areas _PQR_ and _ABC_ (sign determined by the sense of circuit) have the ratio

**17.** If _f_ = **i** _P_ ( _x_ , _y_ , _z_ ) + **j** _Q_ ( _x_ , _y_ , _z_ ) + **k** _R_ ( _x_ , _y_ , _z_ ), the differential equation

is said to be _exact_ when **f** = ∇ _φ_ , the gradient of a scalar.

The equation **f** · _d_ **r** = 0 is exact when and only when rot **f** = **0** ; and then its solution is

**18.** The function _λ_ ( _x_ , _y_ , _z_ ) is called a _multiplier_ of the differential equation **f** · _d_ **r** = 0 when _λf_ · _d_ **r** = 0 is an exact equation (Prob. 17).

If the equation **f** · _d_ **r** = 0 admits a multiplier, then **f** · rot **f** = 0.

When **f** = **i** _P_ ( _x_ , _y_ ) × **j** _Q_ ( _x_ , _y_ ) this condition is always fulfilled.

**19.** If an exact equation **f** · _d_ **r** = 0 admits a non-constant multiplier _λ_ ( _x_ , _y_ , _z_ ), its general solution is given by _λ_ ( _x_ , _y_ , _z_ ) = const.

**20.** If the differential equation _f_ · _d_ **r** = 0 admits two multipliers _λ_ , _μ_ whose ratio is not constant, its general solution is _λ_ / _μ_ = const.

[Use Prob. 19.]

## **Answers to Problems**

### **CHAPTER 1**

**5.4** − 4/3 ( _AB_ ); − 2/1 ( _A_ ′ _B_ ′).

**5.7** − 7/5 ( _AB_ ); − 7/10 ( _CD_ ).

**11.1** ( _a_ ) 60°; ( _b_ ) 90°; ( _c_ ) 180°; ( _d_ ) 70°33′; ( _e_ )  .

**11.2** **a** × **b** = **c**.

**11.3** 14/13.

**11.4** 3.

**13.1** ( _a_ ) [ − 2, 3, − 1], [0, 3, 3], [4, 0, 2];

( _b_ ) 6, 6, 6;

( _c_ ) [ − 4, − 5, − 7], [6, − 3, 3], [ − 2, 8, 4], [0, 0, 0];

**13.2** ( _a_ )  ; ( _b_ )  ; ( _c_ ) 5; ( _d_ ) 30.

**14.7** (i) _A_ = 60°, _B_ = 120°, _C_ = 60°;

(ii) _A_ = 45°, _B_ = 90°, _C_ = 60°.

**15.1** e1 = [1, − 1, 0], e2 = [0, 1, − 1], e3 = [0, 0, 1].

**15.5**  ,  ,   where _λ_ = | **a** × **b** |.

**15.6** **r** = _α_ **a** ′ + _β_ **b** ′ + _γ_ **c** ′, where **a** ′, **b** ′, **c** ′ is the set reciprocal to **a** , **b** , **c**.

**17.4** (8, − 11, − 11).

**17.8** **r** = _α_ 0 **a** ′ + _β_ 0 **b** ′ + _γ_ 0 **c** ′ (Prob. 15.6).

**17.10** Yes.

### **CHAPTER 2**

**19.1** (i) [4, 5, 6]; (ii) 4, 5, 6; (iii) 40/3.

**19.2** 23/3.

**21.9** **f** = [1, 2, − 1]; axis: _y_ \+ 2 _z_ = _z_ \+ _x_ = 2 _x_ − _y_ = 2.

**23.1** ( _a_ ) **f** A = [7, − 1, − 4j; ( _b_ )  ; ( _c_ ) **f** _o_ = [7, − 5, − 6].

**23.3** 20j at (0, 0).

**23.5** _Z_ _a_ = _Z b_ = 5500 lb, _Z c_ = _Z d_ = − 4500 lb.

### **CHAPTER 3**

**28.2** [a, b, c]; a straight line;  .

**31.1** ( _a_ ) T = i, N = j, B = k;
 ,  ,  .

( _b_ )  ;  .

( _c_ ) _x_ = 0, _z_ = 0; _x_ \+ 2 _y_ \+ 2 _z_ = 13, 2 _x_ − 2 _y_ \+ _z_ = 2.

**31.2** ( _a_ ) _κ_ = _τ_ = 2/3(1 + 2 _t_ 2)2;

( _b_ ) _κ_ = _τ_ = 1/3 _a_ (1 + _t_ 2)2.

**32.4** **v** = 3[2, 2, 1], **a** = 6[1,2,0];

 ,  ,  ;

_κ_ = _τ_ = 2/27; **a** · **T** = 12, **a** · **N** = 6.

**32.5** **v** = [0, − 1,2], a = [ − 1,0,0];

 , **N** = [ − 1,0,0],  ;

 ,  ; **a** · **T** = 0, **a** · **N** = 1.

**32.6**  , **a** = [1, 1, 0];

 ,  ,  ;

 ; **a** · **T** = 0,  .

**35.1** 10 mi/hr.

### **CHAPTER 4**

**36.1** Parallel circles _u_ = _a_ ; meridians _υ_ = _b_.

 .

**36.2** Circular helices _u_ = _a_ ; lines ⊥ _z_ -axis.

**36.3** (i) Ruled surface with rulings through _Γ_ and parallel to [cos _α_ , cos _β_ , cos _γ_.

(ii) Cone with vertex at O and rulings through _Γ_. Parametric curves _u_ = _a_ are rulings.

**36.4** _x_ = _a_ sin _u_ cos _υ_ , _y_ = _b_ sin _υ_ sin _υ_ , _z_ = _c_ cos _u_.

**36.5** Rulings.

**36.6** Rulings.

**36.8** [1, − 1,0].

**37.1** ( _a_ ) [− 2, 2]; ( _b_ ) 0; ( _c_ ) 135°,  .

**37.2** ( _a_ )[4, 4, 2]; ( _b_ )  ; ( _c_ ) 6;

( _d_ ) 2 _x_ \+ 2 _y_ \+ _z_ = 7, _x_ − 1 = _y_ − 1 = 2 _z_ − 6.

**37.3** R − 2 _ar_ − 1 cos _φ_ **p** ;  .

**39.5** Scalar 0; vector − 2a.

**41.1**  ; div **f** = 3, rot **f** = [1, 1, 1];

  div **g** = 2( _x_ \+ _y_ \+ _z_ ), rot **g** = **0**.

 ; div **h** = 0, rot **h** = [2 _xy_ − _x_ 2, 2 _yz_ − _y_ 2, 2 _zx_ − _z_ 2].

**41.2** ( _a_ ) 2 **r** , 6; ( _b_ ) − **r** / _r_ 3, 0.

### **CHAPTER 5**

**43.1** 7/3; 17/6; 11/6; 5/2; (22 + 3 _π_ )/12.

**43.2** ( _a_ ) − 2 _πr_ 2 ( _x_ 0 \+ _y_ 0); ( _b_ ) 0; ( _c_ ) − 2.

**43.3** 3.75 − 2 log 4.

**43.4** ( _a_ ) − 1/4; ( _b_ ) − 1/4; ( _c_ ) − 1/4.

**43.5** ( _a_ )  ; ( _b_ )  ; ( _c_ )  .

**43.6** − 2 _π_.

**46.1**  ;  .

**46.2** _F_ = ( _x_ 2/ _y_ ) − 3 _x_ \+ 4 _y_ − 4.

**46.3** _F_ = _x_ 2 _yz_ 3; 14.

**46.4**  ; 0; yes.

**46.5** 33/2.

**46.6** 3 _a_ 2/2.

**47.3**  .

**47.4**  .

**47.5** 3 _a_ 2/2.

**47.8**  ,  ,  .

**47.9**  .

**47.10** 2 _πab_.

**47.11**

**47.12** 14.

**47.13**  .

**47.14**

**50.3** 2 _π_.

**52.2** 8 _π_ /3.

**52.3** _g_ = _r m_ **r** /( _m_ \+ 3).

**52.7** ( _a_ ) 3; ( _b_ ) 40 _π_.

**52.10**  .

**52.11**  ;  .

**54.2**  .

**54.3** ( _a_ ) G = _xz_ [l, − 1, 0],  .

( _b_ )  ,  ,  .

**G** = [− _y_ 2 _z_ , − _xyz_ , 0],  .

( _d_ )  ,  .

**54.4** ( _a_ )  .

 .

**55.2**  , _r_ < _a_ ; ∇2 _φ_ = 0, _r_   _a_. No; yes.

### **CHAPTER 6**

**60.1** ( _a_ ) _φ_ = − **c** · **r** ( _b_ ) none; ( _c_ ) none;

( _d_ )  , _φ_ = log _r_ \+ _C_ , ( _n_ = − 1).

**60.3** _φ_ = _y_ \+ 2 _z_ − _xyz_ 2; _w_ = 2.

**60.5**  .

**60.6**  .

**60.7** ( _a_ )  .

**61.1**  .

**61.3** 35.30 ast. units; _ε_ = 0.97.

**61.4** 36,000,000 miles.

**63.4** 271 slug-ft2.

### **CHAPTER 7**

**72.3**  .

### **CHAPTER 8**

**77.2** [ _γ_ ] = _M_ − 1 _L_ 3 _T_ − 2.

**77.5** [ _a_ ] = [ _b_ ] = [ _γ_ ] = _T_ −1 [ _C_ ] = _L_ 2 _T_ −2 [ _A_ ] = _L_ , [ _B_ ] = _L_ ( _a_ ≡ _b_ [ _B_ ] = _LT_ − 1 ( _a_ = _b_ ).

**84.1** _φ_.

### **CHAPTER 9**

**96.1** | **e** 1 | = 1, | **e** 2 | = 2, | **e** 3 | = 3; all angles 60°.

**96.2**  ;

 , ,

**96.3** _u i_ = (4, 10, 21),  ,

  ; **U** · **V** =17; ( **U** , **V** ) = 27°.

**96.4**  ,

 ,   **e** 3 = − k.

 , _u i_ =(3, −1, 2).

## **Index**

Page numbers with asterisks locate definitions.

Acceleration ( **a** ), *,

angular ( **α** ),

components of,

of Coriolis, , *

of gravity ( **g** , _g_ ), ,

relative,

uniform, ,

Action and reaction,

Addition, of dyadics, *

of forces, *, *

of line vectors, *, , ,

of vectors, *, ,

theorems (sine, cosine),

Alternative airport,

Ampere, *

Ampére's law, ,

Angular acceleration ( **α** ),

speed, *

velocity ( **ω** ), *,

Antecedents of a dyadic, *

Arc, length of, ,

Archimedes, principle of,

Area, of a parallelogram,

of a polygon,

of a triangle, ,

vector, *,

within a closed curve,

Areas, law of, ,

Associative law, , , ,

Astatic center, *

Axes, rectangular,

Axis, *

instantaneous, *,

Bases, reciprocal, *, *,

Basis, *, , *, ,

closed, *

complete, *

orthogonal,

orthonormal, , *, , ,

self-reciprocal, ,

Bernoulli's theorem, ,

Bessel's inequality,

Bilateral surface,

Binormal ( **B** ), ,

Biot-Savart law, ,

Bôcher, Maxime,

Box product, *, ,

Capacitance, , *

Capacitor,

Carnot's theorem,

Catenary,

Cauchy–Riemann equations,

Cauchy's criterion,

Center, astatic, *

instantaneous, *,

mean, *

Center, of curvature, *

of mass, *,

Centroid, *

Ceva, theorem of,

Change, local, *

stationary, *

substantial, *

Charge, electric, , , , ,

Circular motion,

Circulation, *

density,

theorem,

Cofactor, *

reduced, *

Collinear points,

Commutative law, , ,

Complete quadrangle,

Complex potential, *,

velocity, *

Component, *

Components, general, , ,

of acceleration,

of velocity,

rectangular, , ,

Conductivity,

Conductor, , ,

Cones, equations of,

Conic sections,

Conjugate dyadic, *

Consequents of a dyadic, *

Conservation, of electric charge,

of energy,

of momentum,

Continuity, equation of, , ,

Convergence, strong, *

Coordinates, curvilinear, *

cylindrical, *

homogeneous, , ,

line, *

orthogonal, *

plane, *

Plücker, *

point, *, *

polar. ,

rectangular, *

spherical, *

Coplanar line vectors, , ,

Coplanar points,

Coplanar vectors,

Coriolis acceleration, , *

Cosine law, plane,

spherical,

Coulomb,

Coulomb's law, ,

Couple, ,

Courant, Richard,

Cross product ( **u** × **v** ), *, , ,

Cur  (curl **f** = rot **f** ). *

Current ( _I_ ), , , , , ,

Current density ( **J** ), *,

Curvature ( _κ_ ), *, ,

mean (surface), *,

total (surface), *,

Curve, *

closed, *

continuous, *

coordinate, *

of constant curvature, ,

of constant torsion, ,

parametric, *

plane,

reducible, *

regular, *,

smooth, *,

Cycloid,

Cylinders, equations of,

D'Alembert's principle, , ,

Darboux's vector, *, ,

Del (operator ∇), ,

Delta, Kronecker, *,

Derivative, directional, *, *

normal,

of **r** ( _d_ **r** / _ds_ = **T** ),

of **R** ( _d_ **R** / _dθ_ = **P** ),

of a vector, *,

Desargues' theorem,

Dextral set, *, *

Dielectric constant,

Dimensions, of a quantity, , ,  (table)

of a vector space, *

Dipole, *

Direction cosines, ,

Dirichlet's problem,

Distance, between lines, ,

between points,

from point to line, ,

from point to plane, , ,

Distributive laws, , , , ,

Divergence (div **f** , ∇ · **f** ), *, ,

Divergence, meaning of, *,

surface (Div **f** ), *

Divergence theorem,

generalized,

Dot product ( **U** · **V** ), ,

Double layer, *

Dyad,

Dyadic, *,

conjugate, *

invariants of, *

skew, *

symmetric, *

unit (idemfactor **I** ), *,

zero ( **O** ),

Dyadic equality, *

Dyadic product, *

Dyadic sum, *

Einstein, Albert,

Electric charge ( _q_ ), , , ,

Electric displacement ( **D** ),

Electric flux density ( **D** ),

at surface boundary, ,

Electric intensity ( **E** ), ,

at surface boundary, ,

Electromagnetic energy, , *

density of ( _w_ ), ,

potentials, , ,

Electromotive force (emf), *

Electron-volt, *

Electrostatics, ,

Energy, electromagnetic, , *

equation of, , ,

kinetic, *, , ,

potential, *,

Energy density ( _w_ ), ,

Envelopes,

Equation, Euler's, ,

Helmholtz's,

intrinsic, *

of continuity, , ,

Equations, Maxwell's, ,

of a curve,

of a surface, ,

of motion, ,

Equilibrium, of a fluid,

of a rigid body,

Equipotential surface,

Euclidean vector space,

Euler, Leonhard, , ,

Eulerian method, ,

Euler line,

Evolute,

Falling body, ,

Farad,

Field, of numbers,

of the earth,

of the sun,

Floating body,

Flow, *

Fluid, perfect, *

Flux, magnetic (Φ), ,

Flux density, electric ( **D** ), ,

magnetic ( **B** ), ,

Force,

body,

external, internal, ,

inertia, ,

on charges, , ,

on currents,

Form, metric quadratic,

positive definite,

Frenet's formulas,

Functional dependence,

Galileo Galilei,

Gauss, Carl Friedrich, ,

Gauss' theorem, *,

mean value theorem,

Gibbs, J. Willard, vii,

Gradient, of a scalar (∇ _F_ ), *, , ,

of a vector (∇ **f** ), *,

on a surface (Grad **f** ), *

Gramian, *,

Gram–Schmidt process,

Gravitation, law of,

Green, George, , ,

Green's identities, , ,

Green's reciprocation theorem,

Green's theorem, ,

dyadic form of, ,

vector form of, ,

Halley's comet,

Harmonic function, , , , *,

Helix,

circular, ,

Helmholtz's equation,

Henry,

Hermitian symmetry,

Hero's formula,

Hilbert space, *

Hodograph, *

Idemfactor ( **I** ), *, ,

Inductance,

Induction, law of,

Inertia, law of,

moment of,

Inertia forces, *

Inner product, *, , *

Integral, circuit, , ,

line, *, ,

surface, , *

volume, ,

Interception of a plane,

Intrinsic equation, *

Invariants of a dyadic, *, ,

Inverse square law, ,

Involute, of a circle,

of a plane curve,

Irrotational motion, *

Irrotational vector, *,

Jacobian ( _J_ ), , ,

Joule,

Kellogg, O. D.,

Kelvin's circulation theorem,

Kelvin's minimum-energy theorem,

Kepler, Johann, ,

Kepler's laws,

Kinematics, of a particle,

of a rigid body,

Kronecker delta, *,

Lagrange's identity,

Lagrange's theorems,

Lagrangian method,

Laplace's equation, , , , ,

Laplacian (∇2, div ∇), *, ,

Light, speed of,

Linear dependence, *,

Gramian test for, ,

Line integral, *, ,

Lines of motion, *,

Magnetic flux (Φ), *,

Magnetic flux density ( **B** ), , , ,

Magnetic flux density ( **B** ), at surface boundary,

Magnetic induction ( **B** ), , ,

Magnetic intensity ( **H** ), , ,

Magnetostatic potential ( _ψ_ ), *

Magnetostatics,

Matrix form of ∇ **f** , *,

Maxwell's equations, , , ,

integral form of,

Mean center, *

Mean curvature, *,

Menelaus, theorem of,

Metric for a vector space,

Mks (Giorgi) units, *

Möbius strip,

Moment, of a dipole, *

of a force, ,

of a line vector

about an axis, *

about a point, *

of inertia, *

of momentum ( **r** × _m_ **v** ), *

Momentum ( _m_ **v** ), *

Motion of a fluid,

irrotational,

line of, *,

plane,

steady, ,

Motion of a particle, , , ,

circular,

in earth's field,

in electromagnetic field, ,

in resisting medium,

in sun's field, ,

relative, ,

uniformly accelerated,

Motion of a rigid body,

about a fixed axis, ,

about a fixed point,

Motor,

Moving trihedral ( **TNB** ), *,

Newton (unit of force),

Newton, Isaac, , , ,

Newton line,

Normal, principal,

Normal derivative,

Normal plane, *

Normal vector to a surface, , ,

Null (zero) vector ( **0** ), , ,

Ohm,

Ohm's law,

Orbit of a planet,

Orthogonal coordinates, *

Orthogonal vectors, , *, *

Osculating plane, *

Parseval's equation, *

Particle, ,

Pascal's theorem,

Permeability ( _μ_ ), , ,

Permittivity ( _∊_ ), , ,

Plane, coordinates of,

equation of,

normal,

osculating,

rectifying,

tangent,

Plücker's coordinates, *

Point of division,

Poisson's equation,

Polygon, area of,

force,

link, ,

Position vector, *, , ,

gradient of, ,

Potential, complex, *,

electromagnetic ( _φ_ , **A** ), , ,

electrostatic ( _φ_ ), , , ,

magnetostatic ( _ψ_ ),

due to dipole,

due to double layer, ,

due to point charges, ,

due to space charges, , ,

due to surface charges, ,

scalar, *, , , , ,

vector, *, , , , ,

velocity, *,

Poynting's vector ( **S** = **E** × **H** ),

Product, box, *, ,

cross, *, , ,

dot, *, , , *,

dyadic, *

inner, *, , *

of determinants,

of four vectors,

scalar, *, , , *,

scalar triple, *, ,

vector, *, , ,

vector triple, *

Pythagorean theorem,

Quadrangle, complete,

Quadratic form, , ,

Quaternion,

Radius, of curvature ( _ρ_ ), *, ,

of torsion ( _σ_ ), *

Rate of change, local,

stationary,

substantial,

Reciprocal bases, *, , ,

Reduced cofactor, *

Regular point of a surface, *

Regular at infinity, *

Relative acceleration,

Relative velocity,

Relaxation time, ,

Resistance, *

Resistivity, *

Revolving body, ,

Revolving liquid, ,

Rolling wheel,

Rotation (rot **f** , ∇ × **f** ), *

meaning of, , *,

surface (Rot **f** ), *

Russell, Bertrand, viii

Scalar,

Scalar product ( **u** · **v** ), *, , , *,

Schwarz's inequality, , , ,

Self-centroidal set of points, *

Self-reciprocal basis, *,

Shift formulas, , ,

Simply connected region, *,

Sine law, plane,

spherical,

Singular point of a surface,

Sinistral set, *,

Skew dyadic, *

Solenoid,

Solenoidal vector, *, ,

Solid angle, *, , , ,

Speed, ,

Sphere, equations of,

Spherical coordinates, *

Spherical curve,

Spherical trigonometry, ,

Statics,

Step path, ,

Stokes' theorem,

generalized,

Stream function, *

Stream-line, *, , ,

Sum, of dyadics, *

of line vectors, *, , ,

of vectors, *, ,

Summation convention,

Surface, bilateral,

equations of, ,

Surface charges ( _ω_ ), , , ,

Surface divergence (Div **f** ), *,

Surface element,

Surface gradient (Grad **f** ), *,

Surface integral, , *

Surface normal, , , ,

Surface rotation (Rot **f** ), *

Symmetric dyadic, *

Tangent, plane to a surface,

to a curve,

Tangent vector, ,

Taylor, J. H.,

Tensors, viii,

Torricelli's law,

Torsion ( _τ_ ), *, ,

radius of ( _σ_ ), *

Total curvature, *,

Triangle construction,

inequality, , ,

Trigonometry, plane, , ,

spherical,

Uniform acceleration, ,

Unit dyadic ( **I** ), *,

Unit vector,

along binormal ( **B** ),

along principal normal ( **N** ),

along surface normal ( **n** ), , *,

along tangent ( **T** ), ,

radial, ( **R** , **r** / _r_ ),

transverse ( **P** ),

Units, mks, *

table of,

Valence of a tensor,

Vector, *, *, *

base, ,

bound,

free,

irrotational, *

line, , *

moment,

null, , ,

position, ,

proper,

solenoidal, *

unit,

zero ( **0** ), , , ,

Vector addition, ,

Vector product ( **u** × **v** ), *, , ,

Vector space, *,

Velocity ( **v** ), *.

absolute,

angular ( ** _ω_** ), *,

complex, *

components of,

relative,

transfer,

Volt,

von Mises, R.,

Vortex, Rankine's,

Vortex-line, *

Vorticity, *

Watt,

Wave equation,

Weber,

Wind triangle,

Zero dyadic ( **O** ),

Zero vector ( **0** ), , ,
www.doverpublications.com

label: mathematics

title: Vector Analysis

authors: Louis Brand