DOVER BOOKS ON MATHEMATICS

HANDBOOK OF MATHEMATICAL FUNCTIONS: WITH FORMULAS, GRAPHS, AND MATHEMATICAL TABLES, Edited by Milton Abramowitz and Irene A. Stegun. (0-486-61272-4)

ABSTRACT AND CONCRETE CATEGORIES : THE JOY OF CATS, Jiri Adamek, Horst Herrlich, George E. Strecker. (0-486-46934-4)

NONSTANDARD METHODS IN STOCHASTIC ANALYSIS AND MATHEMATICAL PHYSICS, Sergio Albeverio, Jens Erik Fenstad, Raphael Høegh-Krohn and Tom Lindstrøm. (0-486-46899-2)

MATHEMATICS: ITS CONTENT, METHODS AND MEANING, A. D. Aleksandrov, A. N. Kolmogorov, and M. A. Lavrent’ev. (0-486-40916-3)

COLLEGE GEOMETRY: AN INTRODUCTION TO THE MODERN GEOMETRY OF THE TRIANGLE AND THE CIRCLE. Nathan Altshiller-Court. (0-486-45805-9)

REAL VARIABLES WITH BASIC METRIC SPACE TOPOLOGY, Robert B. Ash. (0-486-47220-5)

INTRODUCTION TO DIFFERENTIABLE MANIFOLDS, Louis Auslander and Robert E. MacKenzie. (0-486-47172-1)

PROBLEM SOLVING THROUGH RECREATIONAL MATHEMATICS, Bonnie Averbach and Orin Chein. (0-486-40917-1)

THEORY OF LINEAR OPERATIONS, Stefan Banach. Translated by F. Jellett. (0-486-46983-2)

VECTOR CALCULUS, Peter Baxandall and Hans Liebeck. (0-486-06620-5)

INTRODUCTION TO VECTORS AND TENSORS: SECOND EDITION-TWO VOLUMES BOUND AS ONE, Ray M. Bowen and C.-C. Wang. (0-486-46914-X)

ADVANCED TRIGONOMETRY, C. V. Durell and A. Robson. (0-486-43229-7)

FOURIER ANALYSIS IN SEVERAL COMPLEX VARIABLES, Leon Ehrenpreis. (0-486-44975-0)

THE THIRTEEN BOOKS OF THE ELEMENTS, VOL. 1, Euclid. Edited by Thomas L. Heath. (0-486-60088-2)

THE THIRTEEN BOOKS OF THE ELEMENTS, VOL. 2, Euclid. (0-486-60089-0)

AN INTRODUCTION TO DIFFERENTIAL EQUATIONS AND THEIR APPLICATIONS, Stanley J. Farlow. (0-486-44595-X)

POINT SET TOPOLOGY, Steven A. Gaal. (0-486-47222-1)

DISCOVERING MATHEMATICS: THE ART OF INVESTIGATION, A. Gardiner. (0-486-45299-9)

LATTICE THEORY: FIRST CONCEPTS AND DISTRIBUTIVE LATTICES, George Grätzer. (0-486-47173-X)

ORDINARY DIFFERENTIAL EQUATIONS, Jack K. Hale. (0-486-47211-6)

METHODS OF APPLIED MATHEMATICS, Francis B. Hildebrand. (0-486-67002-3)

BASIC ALGEBRA I: SECOND EDITION, Nathan Jacobson. (0-486-47189-6)

BASIC ALGEBRA II: SECOND EDITION, Nathan Jacobson. (0-486-47187-X)

GEOMETRY AND CONVEXITY: A STUDY IN MATHEMATICAL METHODS, Paul J. Kelly and Max L. Weiss. (0-486-46980-8)

TRIGONOMETRY REFRESHER, A. Albert Klaf. (0-486-44227-6)

THE PHILOSOPHY OF MATHEMATICS: AN INTRODUCTORY ESSAY, Stephan Körner. (0-486-47185-3)

See every Dover book in print at www.doverpublications.com

As far as the laws of mathematics.
refer to reality, they are not certain;
and as far as they are certain,
they do not refer to reality.

ALBERT EINSTEIN

Books by Banesh Hoffmann

The Strange Story of the Quantum
The Tyranny of Testing
About Vectors
Albert Einstein, Creator and Rebel
(with Helen Dukas)

Copyright © 1966 by Banesh Hoffmann.

All rights reserved.

This Dover edition, first published in 1975, is an unabridged, corrected republication of the work originally published by Prentice-Hall, Inc., New Jersey, in 1966.

Library of Congress Catalog Card Number: 75-11081

9780486151694

Manufactured in the United States by Courier Corporation
60489610
www.doverpublications.com

PREFACE

This book is written as much to disturb and annoy as to instruct. Indeed, it seeks to instruct primarily by being disturbing and annoying, and it is often deliberately provocative. If it should cause heated discussion and a re-examination of fundamentals in classroom and mathematics club it will have achieved one of its main purposes.

It is intended as a supplement and corrective to textbooks, and as collateral reading in all courses that deal with vectors. Because the exercises call for no great manipulative skill, and the book avoids using the calculus, it may at first sight seem to be elementary. But it is not. It has something for the beginner, to be sure. But it also has something for quite advanced students—and something, too, for their instructors.

I have tried to face awkward questions rather than achieve a spurious simplicity by sweeping them under the rug. To counteract the impression that axioms and definitions are easily come by and that mathematics is a thing of frozen beauty rather than something imperfect and growing, I have mixed pure and applied mathematics and have made the problem of defining vectors a developing, unresolved leitmotif. The book is unconventional, and to describe it further here would be to blunt its intended effect by giving away too much of the plot. A brief word of warning will not be amiss, however. There are no pat answers in this book. I often present ideas in conventional form only to show later that they need modification because of unexpected difficulties, my aim being to induce a healthy skepticism. But too much healthy skepticism can be decidedly unhealthy. The reader should therefore realize that the ideas could have been presented far more hearteningly as a sequence of ever-deepening insights and, thus, of successive mathematical triumphs rather than defeats. If he reads between the lines he will see that, in a significant sense, they are indeed so presented.

To my friends Professors Arthur B. Brown and Václav Hlavatý, who read the manuscript, go my warmest thanks. It is impossible to express the depth of my indebtedness to them for their penetrating comments, which have led to major improvements in the text. They should not be held accountable for the views expressed in the book: on some issues I resisted the urgent advice of one or the other of them. A ground-breaking book of this sort is unlikely to be free of debatable views and outright errors, and for all of these I bear the sole responsibility.

BANESH HOFFMANN

1

INTRODUCING VECTORS

1. DEFINING A VECTOR

Making good definitions is not easy. The story goes that when the philosopher Plato defined Man as “a two-legged animal without feathers,” Diogenes produced a plucked cock and said “Here is Plato’s man.” Because of this, the definition was patched up by adding the phrase “and having broad nails”; and there, unfortunately, the story ends. But what if Diogenes had countered by presenting Plato with the feathers he had plucked?

Exercise 1.1 What? [Note that Plato would now have feathers.]

Exercise 1.2 Under what circumstances could an elephant qualify as a man according to the above definition?

A vector is often defined as an entity having both magnitude and direction. But that is not a good definition. For example, an arrow-headed line segment like this

has both magnitude (its length) and direction, and it is often used as a drawing of a vector; yet it is not a vector. Nor is an archer’s arrow a vector, though it, too, has both magnitude and direction.

To define a vector we have to add to the above definition something analogous to “and having broad nails,” and even then we shall find ourselves not wholly satisfied with the definition. But it will let us start, and we can try patching up the definition further as we proceed—and we may even find ourselves replacing it by a quite different sort of definition later on. If, in the end, we have the uneasy feeling that we have still not found a completely satisfactory definition of a vector, we need not be dismayed, for it is the nature of definitions not to be completely satisfactory, and we shall have learned pretty well what a vector is anyway, just as we know, without being able to give a satisfactory definition, what a man is—well enough to be able to criticize Plato’s definition.

Exercise 1.3 Define a door.

Exercise 1.4 Pick holes in your definition of a door.

Exercise 1.5 According to your definition, is a movable partition between two rooms a door?

2. THE PARALLELOGRAM LAW

The main thing we have to add to the magnitude-and-direction definition of a vector is the following:

Figure 2.1

Let us think of vectors as having definite locations. And let the arrow-headed line segments and in Figure 2.1 represent the magnitudes, directions, and locations of two vectors starting at a common point O. Complete the parallelogram formed by and , and draw the diagonal OR. Then, when taken together, the two vectors represented by and are equivalent to a single vector represented by the arrow-headed line segment . This vector is called the resultant of the vectors represented by OP and , and the above crucial property of vectors is called the parallelogram law of combination of vectors.

Exercise 2.1 Find (a) by drawing and measurement, and (b) by calculation using Pythagoras’ theorem, the magnitude and direction of the resultant of two vectors and if each has magnitude 3, and points thus – → while points perpendicularly, thus ↑ . [Ans. The magnitude is 3√2, or approximately 4.2, and the direction bisects the right angle between and .]

Exercise 2.2 Show that the resultant of two vectors and that point in the same direction is a vector pointing in the same direction and having a magnitude equal to the sum of the magnitudes of and . [Imagine the parallelogram in Figure 2.1 squashed flat into a line.]

Exercise 2.3 Taking a hint from Exercise 2.2, describe the resultant of two vectors and that point in opposite directions.

Exercise 2.4 In Exercise 2.3, what would be the resultant if and had equal magnitudes? [Do you notice anything queer when you compare this resultant vector with the definition of a vector?]

Exercise 2.5 Observe that the resultant of and is the same as the resultant of and . [This is trivially obvious, but keep it in mind nevertheless. We shall return to it later.]

In practice, all we need to draw is half the parallelogram in Figure 2.1—either triangle OPR or triangle OQR. When we do this it looks as if we had combined two vectors and (or and ) end-to-end like this, even though they do not have the same starting point. Actually, though, we have merely combined and by the parallelogram law.¹ But suppose we were dealing with what are called free vectors—vectors having the freedom to move from one location to another, so that and in Figure 2.2, for example, which have the same magnitude and the same direction, are officially counted not as distinct vectors but as the same free vector. Then we could indeed combine free vectors that were quite far apart by bringing them end-to-end, like and in Figure 2.2. But since we could also combine them according to the parallelogram law by moving them so that they have a common starting point, like and in Figure 2.1, the parallelogram law is the basic one. Note that when we speak of the same direction we mean just that, and not opposite directions—north and south are not the same direction.

Figure 2.2 (For clarity, the arrow heads meeting at R have been slightly displaced. We shall occasionally displace other arrow heads under similar circumstances.)

Exercise 2.6 Find the resultant of the three vectors , and in the diagram.

Solution We first form the resultant, of and like this:

and then we form the resultant, , of and like this:

This figure looks complicated. We can simplify it by drawing only half of each parallelogram, and then even omitting the line OR, like this:

From this we see that the resultant can be found quickly by thinking of the vectors as free vectors and combining them by placing them end-to-end: which has the same magnitude and direction as , starts where ends; and then , which has the same magnitude and direction as starts where ends.

Exercise 2.7 Find, by both methods, the resultant of the vectors in Exercise 2.6, but by combining and first, and then combining their resultant with . Prove geometrically that the resultant is the same as before.

Exercise 2.8

The above diagram looks like a drawing of a box. Show that if we drew only the lines OA, AR, RS, and OS we would have essentially the last figure in Exercise 2.6; that if we drew only the lines OB, BT, TS, and OS we would have a corresponding figure for Exercise 2.7; and that if we drew only OA, AU, US, and OS we would have a figure corresponding to our having first combined with and then their resultant with .

Exercise 2.9 In Exercises 2.6, 2.7, and 2.8, is it essential that the three vectors , and lie in a plane? Give a rule for finding the resultant of three noncoplanar vectors , and that is analogous to the parallelogram law, and that might well be called the parallelepiped law. Prove that their resultant is the same regardless of the order in which one combines them.

Exercise 2.10 Find the resultant of the three vectors , and below by combining them in three different orders, given that vectors and have equal magnitudes and opposite directions. Draw both the end-to-end diagrams and the full parallelogram diagrams for each case.

3. JOURNEYS ARE NOT VECTORS

It is all very well to start with a definition. But it is not very enlightening. Why should scientists and mathematicians be interested in objects that have magnitude and direction and combine according to the parallelogram law? Why did they even think of such objects? Indeed, do such objects exist at all —outside of the imaginations of mathematicians?

There are, of course, many objects that have both magnitude and direction. And there are, unfortunately, many books about vectors that give the reader the impression that such objects obviously and inevitably obey the parallelogram law. It is therefore worthwhile to explain carefully why most such objects do not obey this law, and then, by a process of abstraction, to find objects that do.

Suppose that I live at A and my friend lives 10 miles away at B. I start from A and walk steadily at 4 m.p.h. for hours. Obviously, I walk 10 miles. But do I reach B?

You may say that this depends on the direction I take. But what reason is there to suppose that I keep to a fixed direction? The chances are overwhelming that I do not—unless I am preceded by a bulldozer or a heavy tank. Most likely I walk in all sorts of directions; and almost certainly, I do not arrive at B. I may even end up at home.

Exercise 3.1 Where are all the possible places at which I can end, under the circumstances?

Now suppose that I start again from A and this time end up at B. I may take four or five hours, or I may go by bus or train and get there quickly. Never mind how I travel or how long I take. Never mind how many times I change my direction, or how tired I get, or how dirty my shoes get, or whether it rained. Ignore all such items, important though they be, and consider the abstraction that results when one concentrates solely on the fact that I start at A and end at B. Let us give this abstraction a name. What shall we call it? Not a “journey.” That word reminds us too much of everyday life—of rain, and umbrellas, and vexations, and lovers meeting, and all other such items that we are ignoring here; besides, we want to preserve the word “journey” for just such an everyday concept. For our abstraction we need a neutral, colorless word. Let us call it a shift.

Here are routes of four journeys from A to B :

Figure 3.1

All four journeys are different—with the possible but highly improbable exception of (b) and (c).

Exercise 3.2 Why “highly improbable”?

But though the four journeys are not all the same, they yield the same shift. We can represent this shift by the arrow-headed line segment AB. It has both magnitude and direction. Indeed, it seems to have little else. Is it a vector? Let us see.

Consider three places A, B, and C as in Figure 3.2. If I walk in a straight line from A to B and then in a straight line from B to C, I make a journey from A to C, but it is not the same as if I walked directly in a straight line from A to C: the scenery is different, and so is the amount of shoe leather consumed, most likely, and we can easily think of several other differences.

Figure 3.2

Exercise 3.3 Why “most likely”?

Thus, though we could say that the walks from A to B and from B to C combine to give a “resultant” journey from A to C, it is not a journey in a straight line from A to C: the walks do not combine in a way reminiscent of the way in which vectors combine; they combine more in the tautological sense that 2 + 1 = 2 + 1 than 2 + 1 = 3.

Journeys, then, are not vectors. But when we deal with shifts we ignore such things as the scenery and the amount of shoe leather consumed. A shift from A to B followed by a shift from B to C is indeed equivalent to a shift from A to C. And this reminds us so strongly of the vectorial situation in Figure 2.2 that we are tempted to conclude that shifts are vectors. But there is a crucial difference between the two situations. We cannot combine the above shifts in the reverse order (compare Exercise 2.5). There is no single equivalent to the shift from B to C followed by the shift from A to B. We can combine two shifts only when the second begins where the first ends. Indeed, in Figure 2.1, just as with journeys, we cannot combine a shift from O to P with one from O to Q in either order. Thus shifts are not vectors.

4. DISPLACEMENTS ARE VECTORS

Now that we have discovered why shifts are not vectors, we can easily see what further abstraction to make to obtain entities that are. From the already abstract idea of a shift, we remove the actual starting point and end point and retain only the relation between them: that B lies such and such a distance from A and in such and such a direction.² Shifts were things we invented in order to bring out certain distinctions. But this new abstraction is an accepted mathematical concept with a technical name: it is called a displacement. And it is a vector, as we shall now show.

In Figure 4.1, the arrow-headed line segments AB and LM are parallel and

Figure 4.1

of equal length. Any journey from A to B is bound to be different from a journey from L to M. Also, the shift from A to B is different from that from L to M because the starting points are different, as are the end points. But the two shifts, and thus also the various journeys, yield the same displacement: if, for example, B is 5 miles north-northeast of A, so too is M 5 miles north-northeast of L, and the displacement is one of 5 miles in the direction north-northeast.

Exercise 4.1 Starting from a point A, a man bicycles 10 miles due east to point B, stops for lunch, sells his bicycle, and then walks 10 miles due north to point C. Another man starts from B, walks 4 miles due north and 12 miles due east and then, feeling tired, and having brought along a surplus of travellers’ checks, buys a car and drives 6 miles due north and 2 miles due west, ending at point D in the pouring rain. What displacement does each man undergo? [Ans. 10√2 miles to the northeast.]

Now look at Figure 2.1. The shift from O to P followed by the shift from P to R is equivalent to the shift from O to R. The shift from P to R gives a displacement that is the same as the displacement . Therefore the displacement followed by the displacement is equivalent to the displacement .

Exercise 4.2 Prove, similarly, that the displacement followed by the displacement is also equivalent to the displacement .

Thus, displacements have magnitude and direction and combine according to the parallelogram law. According to our definition, they are therefore vectors. Since displacements such as and in Figure 4.1 are counted as identical, displacements are free vectors, and thus are somewhat special. In general, vectors such as and are not counted as identical.

5. WHY VECTORS ARE IMPORTANT

From the idea of a journey we have at last come, by a process of successive abstraction, to a specimen of a vector. The question now is whether we have come to anything worthwhile. At first sight it would seem that we have come to so pale a ghost of a journey that it could have little mathematical significance. But we must not underestimate the potency of the mathematical process of abstraction. Vectors happen to be extremely important in science and mathematics. A surprising variety of things happen to have both magnitude and direction and to combine according to the parallelogram law; and many of them are not at all reminiscent of journeys.

This should not surprise us. The process of abstraction is a powerful one. It is, indeed, a basic tool of the mathematician. Take whole numbers, for instance. Like vectors, they are abstractions. We could say that whole numbers are what is left of the idea of apples when we ignore not only the apple trees, the wind and the rain, the profits of cider makers, and other such items that would appear in an encyclopedia article, but also ignore even the apples themselves, and concentrate solely on how many there are. After we have extracted from the idea of apples the idea of whole numbers, we find that whole numbers apply to all sorts of situations that have nothing to do with apples. Much the same is true of vectors. They are more complicated than whole numbers—so are fractions, for example—but they happen to embody an important type of mathematical behavior that is widely encountered in the world around us.

To give a single example here: forces behave like vectors. This is not something obvious. A force has both magnitude and direction, of course. But this does not mean that forces necessarily combine according to the parallelogram law. That they do combine in this way is inferred from experiment.

It is worthwhile to explain what is meant when we say that forces combine according to the parallelogram law. Forces are not something visible, though their effects may be visible. They are certainly not arrow-headed line segments, though after one has worked with them mathematically for a while, one almost comes to think they are. A force can be represented by an arrow-headed line segment that starts at the point of application O of the force, points in the direction of the force, and has a length proportional to the magnitude of the force—for example, a length of x inches might represent a magnitude of x pounds. When a force is represented in this way, we usually avoid wordiness by talking of “the force .” But let us be more meticulous in our wording just here. To verify experimentally that forces combine according to the parallelogram law, we can make the following experiment. We arrange stationary weights and strings, and pulleys A and B, as shown, the weight W being the sum of the weights W₁ and W₂. Then along OA we mark off a length OP of W₁ inches, where W, is the number of pounds in the weight on the left and, thus, a measure of the force with which the string attached to it pulls on the point O where the three pieces of string meet. Similarly, we mark off on OB a length OQ of W₂ inches. We then bring a vertical piece of paper up to the point O, and on it complete the parallelogram defined by OP and OQ. We find that the diagonal OR is vertical and that its length in inches is W, the number of pounds in the weight in the middle. We conclude that the resultant of the forces W₁ and W₂ in the strings would just balance the weight W. Since the forces W, and W₂ also just balance the weight W, we say that the resultant is equivalent to the two forces. We then do the experiment over again, with different weights, and reach a similar conclusion. After that, we do it yet again; and we keep at it till our lack of patience overcomes our skepticism, upon which we say that we have proved experimentally that forces combine according to the parallelogram law. And we bolster our assertion by pointing to other experiments, of the same and different types, that indicate the same thing.

Figure 5.1

We all know that it is much easier to get through a revolving door by pushing near the outer edge than by pushing near the central axis. The effect of a force depends on its location. Home runs are scarce when the bat fails to make contact with the ball. Thus forces do not behave like free vectors. Unlike displacements, vectors representing forces such as and in Figure 4.1, though they have the same magnitude and the same direction, are not counted as equivalent. Such vectors are called bound vectors.

Perhaps it worries us a little that there are different kinds of vectors. Yet we have all, in our time, survived similar complications. Take numbers, for example. There are whole numbers and there are fractions. Perhaps you feel that there is not much difference between the two. Yet if we listed the properties of whole numbers and the properties of fractions we would find considerable differences. For instance, if we divide fractions by fractions the results are always fractions, but this statement does not remain true if we replace the word “fractions” by “whole numbers.” Worse, every whole number has a next higher one, but no fraction has a next higher fraction, for between any two fractions we can always slip infinitely many others. Even so, when trying to define number we might be inclined to insist that, given any two different numbers, one of them will always be the smaller and the other the larger. Yet when we become more sophisticated and expand our horizons to include complex numbers like , we have to give up even this property of being greater or smaller, which at first seemed an absolutely essential part of the idea of number. With vectors too, not only are there various types, but we shall learn that not every one of their attributes that seems at this stage to be essential is in fact so. One of the things that gives mathematics its power is the shedding of attributes that turn out not to be essential, for this, after all, is just the process of abstraction.

Exercise 5.1 Find the resultants of the following displacements:

3 ft. due east and 3 ft. due north. [Ans. 3√2 ft. to the northeast.]

5 ft. due north and 5 ft. due east.

9 cm. to the right and 9√3 cm. vertically upwards. [Ans. 18 cm. in an upward direction making 60° with the horizontal towards the right.]

9 cm. to the left and 9√3 cm. vertically downward.

the resultants in parts (c) and (d).

x units positively along the x-axis and y units positively along the y-axis. [Ans. units in the direction making an angle tan – 1 y/x with the positive x-axis.]

Exercise 5.2 Like Exercise 5.1 for the following:

8 km. to the left and 3 km. to the left.

5 fathoms vertically downward and 2 fathoms vertically upward.

a units to the right and β units to the left. [There are three different cases. What are they? Show how they can be summed up in one statement.]

h miles 60° north of east and h miles 60° south of east.

Exercise 5.3 What single force is equivalent to the following three horizontal forces acting on a particle at a point O? (1) magnitude 1 lb. pulling to the north; (2) magnitude 1 lb. pulling to the east; (3) magnitude √2 lb. pulling to the northwest. [Ans. 2 lbs. acting at point O and pulling to the north.]

Exercise 5.4 What force combined with a force at a point 0 of 1 lb. pulling to the east will yield a resultant force of 2 lbs. pulling in a direction 60° north of east?

Exercise 5.5 Vector has magnitude 2a and points to the right in a direction 60° above the horizontal. What vector combined with it will yield a vertical resultant, , of magnitude 2√3 a?

Exercise 5.6 Find two forces at a point O, one vertical and one horizontal, that have a resultant of magnitude h, making 45° with the horizontal force. [Ans. The forces have magnitude h/√2 .]

Exercise 5.7 Find two forces at a point O, one vertical and one horizontal, that have a resultant of magnitude h that makes an angle of 30° with the horizontal force.

Exercise 5.8 Find two displacements, one parallel to the x-axis and the other to the y-axis, that yield a resultant displacement of magnitude h ft. making a positive acute angle α with the positive x-direction.

Exercise 5.9 What is the resultant of n vectors, each starting at the point O, each having magnitude h, and each pointing to the pole star? [We could have shortened this by asking for the resultant of n equal vectors. But we have not yet defined “equal” vectors—even though we have already spoken of the equality of free vectors! You may find it instructive to try to do so here; but be warned that it is not as easy as it seems, and that there is something lacking in the wording of the question.]

Exercise 5.10 A particle is acted on by two forces, one of them to the west and of magnitude 1 dyne, and the other in the direction 60° north of east and of magnitude 2 dynes. What third force acting on the particle would keep it in equilibrium (i. e., what third force would make the resultant of all three forces have zero magnitude)? [Ans. Magnitude √3 dynes pointing due south.]

6. THE CURIOUS INCIDENT OF THE VECTORIAL TRIBE

It is rumored that there was once a tribe of Indians who believed that arrows are vectors. To shoot a deer due northeast, they did not aim an arrow in the northeasterly direction; they sent two arrows simultaneously, one due north and the other due east, relying on the powerful resultant of the two arrows to kill the deer.

Skeptical scientists have doubted the truth of this rumor, pointing out that not the slightest trace of the tribe has ever been found. But the complete disappearance of the tribe through starvation is precisely what one would expect under the circumstances; and since the theory that the tribe existed confirms two such diverse things as the NONVECTORIAL BEHAVIOR OF ARROWS and the DARWINIAN PRINCIPLE OF NATURAL SELECTION, it is surely not a theory to be dismissed lightly.

Exercise 6.1 Arrow-headed line segments have magnitude and direction and are used to represent vectors. Why are they nevertheless not vectors?

Exercise 6.2 Given the three vectors represented by , , and in Figure 2.1, form three new entities having the same respective directions, but having magnitudes equal to five times the magnitudes of the respective vectors. Prove geometrically that these new entities are so related that the third is a diagonal of the parallelogram having the other two as adjacent sides.

Exercise 6.3 If in Exercise 6.2 the new entities had the same respective directions as the vectors represented by , and , but had magnitudes that were one unit greater than the magnitudes of the corresponding vectors, show that the new entities would not be such that the third was a diagonal of the parallelogram having the other two as adjacent sides.

Exercise 6.4 Suppose we represented vectors by arrow-headed line segments that had the same starting points and directions as the vectors, but had lengths proportional to the squares of the magnitudes of the vectors, so that, for example, if a force of 1 lb. were represented by a segment of length 1 inch, then a force of 2 lbs. would be represented by one of 4 inches. Show that, in general, these representations of vectors would not obey the parallelogram law. Note that the statement of the parallelogram law in Section 2 therefore needs amending, and amend it accordingly. [If you think carefully, you will realize that this is a topsy-turvy question since, in proving the required result, you will assume that the vectors, when “properly” represented, obey the parallelogram law; and thus, in a sense, you will assume the very amendment you are seeking. But since you have probably been assuming the amendment all this while, you will be able to think your way through. The purpose of this exercise is to draw your attention to this rarely mentioned, usually assumed amendment.]

7. SOME AWKWARD QUESTIONS

When are two vectors equal? The answer depends on what we choose to mean by the word “equal”—we are the masters, not the word. But we do not want to use the word in an outrageous sense: for example, we would not want to say that two vectors are equal if they are mentioned in the same sentence.

Choosing a meaning for the word “equal” here is not as easy as one might imagine. For example, we could reasonably say that two vectors having the same magnitudes, identical directions, and a common starting point are equal vectors. And if one of the vectors were somehow pink and the other green, we would probably be inclined to ignore the colors and say that the vectors were still equal. But what if one of the vectors represented a force and the other a displacement? There would then be two difficulties.

The first difficulty is that the vector representing a displacement would be a free vector, but the one representing the force would not. If, in Figure 4.1, we counted free vectors represented by and as equal, we might find ourselves implying that forces represented by and were also equal, though actually they have different effects. [Even so, it is extremely convenient to say such things as “a force acts at A and an equal force acts at L.” We shall not do so in this book. But one can get by with saying such things once one has explained what is awkward about them, just as, in trigonometry, one gets by with writing sin² θ after one has explained that this does not stand for sin(sin θ) but (sin θ)².]

As for the second difficulty about the idea of the equality of vectors, it takes us back to the definition of a vector. For if, in Figure 2.1, represents a force and a displacement, the two vectors will not combine by the parallelogram law at all. We know this from experiments with forces. But we can appreciate the awkwardness of the situation by merely asking ourselves what the resultant would be if they did combine in this way. A “disforcement”?³ [Compare Exercise 5.9.]

If two vectors are to be called equal, it seems reasonable to require that they be able to combine with each other. The situation is not the same as it is with numbers. Although 3 apples and 3 colors are different things, we can say that the numbers 3 are equal in the sense that, if we assign a pebble to each of the apples, these pebbles will exactly suffice for doing the same with the colors. And in this sense we can indeed combine 3 apples and 3 colors—not to yield 6 apples, or 6 colors, or 6 colored apples [it would surely be only 3 colored apples], but 6 items. There does not seem to be a corresponding sense in which we could reasonably combine a vector representing a force with one representing a displacement, quite apart from the question of bound versus free vectors: there does not seem to be a vectorial analogue of the numerical concept of a countable item.⁴

Though and do not combine according to the parallelogram law if, for example, represents a force and a displacement, they nevertheless represent vectors. Evidently our definition of a vector needs even further amendment. We might seek to avoid trouble by retreating to the definition of a vector as “an entity having both magnitude and direction,” without mentioning the parallelogram law. But once we start retreating, where do we stop? Why not be content to define a vector as “an entity having direction,” or as “an entity having magnitude,” or, with Olympian simplicity, as just “an entity”? Alternatively, we could make the important distinction between the abstract mathematical concept of a vector and entities, such as forces, that behave like these abstract vectors and are called vector quantities. This helps, but it does not solve the present problem so much as sweep it under the rug. We might amend our definition of a vector by saying that vectors combine according to the parallelogram law only with vectors of the same kind: forces with forces, displacements with displacements, accelerations (which are vectors) with accelerations, and so on. But even that is tricky since, for example, in dynamics we learn that force equals mass times acceleration. So we would have to allow for the fact that though a force does not combine with an acceleration, it does combine with a vector of the type mass-times-acceleration in dynamics.

We shall return to this matter. (See Section 8 of Chapter 2.) But enough of such questions here. If we continue to fuss with the definition we shall never get started. Even if we succeeded in patching up the definition to meet this particular emegency, other emergencies would arise later. The best thing to do is to keep an open mind and learn to live with a flexible situation, and even to relish it as something akin to the true habitat of the best research.

2

ALGEBRAIC NOTATION AND BASIC IDEAS

1. EQUALITY AND ADDITION

Instead of denoting vectors by symbols like and , it is often convenient to denote them by single letters printed in bold-face type, like this: U, V. (In written work, some people place semi-arrows above ordinary capital letters, as Others write double lines on the letters, like this: . Yet others place lines underneath the letters: .)

Let us agree to say that two vectors, U and V, are equal if they have equal magnitudes, identical directions, and a common starting point, and represent types of entities that can combine with each other according to the parallelogram law. If they are free vectors we need not insist that they have a common starting point.

We denote this equality by the familiar symbol =, writing, for example,

(1.1)

We shall also use this symbol = in the broader sense of “equivalent to” as the phrase was used in the statement of the parallelogram law in Chapter I. Also, of course, we shall use the symbol in its usual arithmetical and algebraic sense, as in 2 + 2 = 4.

Exercise 1.1 In the definition of equal vectors, what do you think we mean by “equal magnitudes”? Consider, for example, magnitudes of 6 ft., 6 lbs., and 2 yds.

The magnitude of a vector is usually denoted by |OA| or just plain OA, and that of a vector V by V.

Exercise 1.2 If U and V have the same starting point and the same direction, show that the magnitude of their resultant is U + V.

Note that we did not say in Exercise 1.2 that U and V were capable of having a resultant: U might have been a force and V a displacement, for example. Let us agree to assume that vectors in a given problem or discussion can combine unless the contrary is either obvious or else explicitly stated.

When vectors are related as in Exercise 1.2, their resultant is akin to a sum. For this reason, and for others that will appear later, it is appropriate to denote the combination of vectors according to the parallelogram law by the familiar symbol +. Thus, if is the resultant of and , as in Figure 2.1 of Chapter 1, we write:

(1.2)

and we even use arithmetical language, saying that we add the vectors and and that their sum is .

From Exercise 2.5 of Chapter 1, we see that

(1.3)

Technically, we refer to this fact by saying that vectorial addition is commutative.

Exercise 1.3 Show geometrically that vectorial addition is associative, that is,

(1.4)

Also write out what this relation says using the language of Chapter 1, e. g., resultant, combination, etc. [Note that we have not defined the symbol “(”and the symbol“)”: their significance is easily guessed here.]

2. MULTIPLICATION BY NUMBERS

The resultant of V and V is a vector identical with V except that its magnitude is twice as great. We denote it by 2V, and this allows us to write the satisfactory equation:

V + V = 2V.

Similarly, we denote the resultant of V, V, and V by 3V, and so on.

Exercise 2.1 Show that 2V + V = 3V.

Exercise 2.2 Describe 5V in terms of V.

If

3U = 2V

we write

more generally, if a and b are positive whole numbers, and

(2.1)

we write

(2.2)

Exercise 2.3 Show that 3(V ) = 2V. [Hint. Start by expressing the left-hand side in terms of U as in the preceding paragraph.]

Exercise 2.4 Show that if a and b are positive whole numbers,

Exercise 2.5 Describe V in terms of V.

Exercise 2.6 Show that if a and b are positive whole numbers, then (b/a)V is a vector identical with V except that its magnitude is (b/a) V.

If h is a positive number, we write hV for the vector that is identical with V except that its magnitude is h times that of V. It may seem that this is merely a repetition of the result of Exercise 2.6, and contains no more than is contained in Equations (2.1) and (2.2); for we have only to express h as a fraction b/a, where a and b are whole numbers, in order to link up with those equations. But not every number can be expressed as such a fraction. Those that can are called rational numbers, and the amount of rational numbers is negligible compared with the amount of numbers that cannot be so expressed. The latter are called irrational numbers, examples being √2 and π. A book about vectors is not the place for a discussion of irrational numbers. One can do arithmetic with them much as one does with rational numbers, and the general case hV can be regarded as a valid extension of the ideas embodied in Equations (2.1) and (2.2). If we feel so inclined, we can amend the definition of a vector to include the idea of hV where h is irrational. We say that hV is the result of multiplying the vector V by the number h.

Exercise 2.7 Let the vector V point along the x-axis. Form two new vectors, the first by rotating V about its starting point through 45° towards the positive y-direction, and the second by rotating V similarly through 45° towards the negative y-direction. Show that the resultant of these two vectors is

Exercise 2.8 What would be the resultant if we did Exercise 2.7 starting not with V but with ?

Exercise 2.9 Show by means of a diagram that multiplication of vectors by numbers obeys the distributive law:

(2.3)

[This can not be proved just by noting its similarity to a familiar algebraic equation. For a hint, compare Exercise 6.2 of Chapter 1, which is a special case of this theorem.]

Exercise 2.10 If U and V both start at the origin and have equal magnitudes, U pointing along the positive x-axis and V along the positive y-axis, and if a and b are positive numbers, what is the tangent of the angle that the vector aU + bV makes with the x-axis?

3. SUBTRACTION

Having managed to relate the symbol + to the parallelogram law without doing violence to its usual arithmetical meaning, we wonder whether we can be as successful with the symbol – . It turns out that we can. If V is a vector, we define its negative, which we write – V, as a vector identical with it except that it points in the opposite direction. Thus in Figure 3.1, if U is V turned

Figure 3.1

through 180°, then U = – V. Note that V and – V have the same magnitude: the magnitude is always taken to be positive (when it is not zero, of course). To consolidate the link between the arithmetical and vectorial uses of the symbol –, we define ( – h)V for positive h as h( – V).

Exercise 3.1 If U = – V, show that V = – U. What is - (- V)?

Exercise 3.2 What is the resultant of 2V and – V? [Ans. V.]

Exercise 3.3 Prove that if h is positive, h( – V) = – (hV).

Exercise 3.4 What is the resultant of aV and – bV?

We denote a vector of zero magnitude by the symbol 0, without bothering to use bold-face type. A zero vector, as it is called, does not quite fit our definition of a vector since it has no direction or, thought of in another way, it has all directions. We count it as a vector nevertheless, and we shall not pause to patch up the definition of a vector to accomodate it.

Exercise 3.5 What is the resultant of V and – V? What is the resultant in Exercise 3.4 if a = b?

Exercise 3.6 Show that V + O = O + V = V.

We now define U – V as U + ( – V), and say that to subtract V we add – V. Since this subtraction is thus essentially a type of vectorial addition, it obeys the same rules as vectorial addition. For example,

(U + V) – W = U + (V – W)

and

h(U – V) = hU – hV.

Exercise 3.7 Consider the discussion of the equality of vectors in Section 7 of Chapter 1 in the light of the reasonable requirement that if U = V then U – V = 0.

Exercise 3.8 If U + V = W, show that U = W – V, first by means of a diagram involving two overlapping parallelograms, and second by the purely algebraic manipulation of adding – V to both sides of the equation and using the results of Exercises 3.5 and 3.6.

For displacements it is easy to see that

(3.1)

though if were not a free vector this would, in general, not be true, since starts at B while starts at A.

Again, the resultant of the three displacements , and is the displacement

. (Note that this is true even if A, B, C, and D do not lie in a plane.) Let D coincide with A. Then the resultant is zero. Therefore for any displacements ,

(3.2)

Suppose we replace the displacement by the algebraic expression c – b, by a – c, and by b – a. Then from Equations (3.1) and (3.2) we obtain the formally valid equations

(3.3)

and

(3.4)

Exercise 3.9 For displacements, show that , write down the analogous algebraic equation, and verify that it is satisfied.

Later we shall see that this sort of relationship between displacement equations and algebraic equations is not accidental. We can begin to see why by considering vectors in a one-dimensional space.

Exercise 3.10 If A and B are points on the x-axis with x-coordinates a and b respectively, and b > a, show that the magnitude of the displacement is b – a.

In one dimension, there are only two possible directions. Vectors confined to the x-axis, for example, can point to the right or to the left, but in no other direction. We could give these directions labels such as “right” and “left,” or “first” and “second.” But it is much more pleasing to label them “+” and “ – .” For example, in Exercise 3.10, the quantity b – a, being positive, can now be taken to describe not just the magnitude of the displacement but also its direction; while if we similarly relate the displacement to a – b, this quantity, being negative, gives both the magnitude and direction of .

Exercise 3.11 Show that b – a in Exercise 3.10 represents both the magnitude and direction of the displacement whether b > a or not.

Exercise 3.12 In Exercise 3.11, if O is the origin, show that = , no matter what the relative positions of O, A, and B on the x-axis. [Do this first by deducing it from Equation (3.2) with O replacing C, and second, by considering the corresponding algebraic relation.]

No longer confining ourselves to one dimension, we have, from Equations (3.1) and (3.2),

so that, by Equation (3.1),

(3.5)

This gives us a convenient way to draw the difference of two displacements, . We merely draw and and join their tips from C to B (not B to C), as in Figure 3.2.

Figure 3.2

Figure 3.3

For vectors that are not free vectors but are tied to a particular starting point, the preceding still gives the magnitude and direction of their difference though not its location. Thus in Figure 3.3, if vectors U, V are represented by and we wish to find U – V, we should first construct – V by turning about A through 180° to the position AC′. Then we should complete the parallelogram AC′ DB to obtain U – V as the vector represented by the diagonal . But since triangles AC’D and CAB are congruent, and CAC′ is a straight line, it is easy to prove that CB and AD are parallel and of equal length. So we can find the magnitude and direction of U – V precisely as in Figure 3.2, and this is all we need since we already know it starts at A. [Compare Exercise 3.8.]

Exercise 3.13 Prove that triangles AC’D and CAB are congruent, and that CB and AD are parallel and of equal length.

Exercise 3.14 Find the magnitudes and directions of U – V in the following cases. [Use the method of Figure 3.2.]

U has magnitude 5 and points northeast; V has magnitude 5 and points northwest. [Ans. magnitude 5√2; pointing east.]

U has magnitude 8√3 and points east; V has magnitude 16 and points 30° north of east.

U has magnitude 7 and points 20° north of east; V has magnitude 7√2 and points 65° north of east.

Exercise 3.15 A force of 8 1bs. acts along the positive direction of the x-axis. If it acts on a particle at the origin, what force combined with it would produce on the particle a resultant force of 8 1bs. pulling along the positive y-axis? [If the resultant force is R, the given force X, and the required force Y, then X + Y = R; so Y = R – X, whose magnitude and direction can be found by the method of Figure 3.2.]

Exercise 3.16 Like Exercise 3.15 but with the resultant making 60° with the positive x-direction.

4. SPEED AND VELOCITY

In ordinary conversation we usually make no distinction between speed and velocity. In fact, most dictionaries treat the two words as synonymous. But in the physical sciences, speed and velocity have different meanings. Speed is the rate at which distance is covered per unit time, but velocity involves the direction also. Thus 5 m. p. h. is a speed, but 5 m. p. h. vertically upward is a velocity, and it is a different velocity from 5 m. p. h. in any other direction. A point moving with constant speed in a circle does not have an unchanging velocity: the velocity changes because its direction changes.

Figure 4.1 shows a platform with a line AB marked on it. Suppose a point

Figure 4.1

moves uniformly along this line from A to B at 2 ft./sec., and that the platform moves uniformly to the east at 5 ft./sec. relative to the ground. Then the point is being subjected to two different velocities at the same time. But it does not, therefore, move relative to the ground in two different directions at once. To say that it did would be to speak nonsense (except, perhaps, if we were talking of quantum phenomena, which we are not). It would remind us of the humorist Stephen Leacock’s famous account of the angry man who “flung himself from the room, flung himself upon his horse and rode madly off in all directions.”

How does the point actually move? We can find out by considering what happens in one second. For convenience, let us assume that AB is of length 2 ft., so that in one second the point goes from A to B. During this second, the platform moves 5 ft. to the east, and the ends of the line AB drawn on it move to new positions, A′, B′, that are 5 ft. to the east of their original positions, as in Figure 4.2. The moving point travels from A to B′ relative to the ground.

Figure 4.2

The distance AB′ depends on the angle BAA’; for simplicity, let us suppose that the distance comes to 6 ft. It is easy to see that the point travels from A to B′ in a straight line, with uniform speed relative to the ground.

Exercise 4.1 Prove this. [To get the idea, consider first what happens in, say, sec., using similar triangles. Then consider what happens in t secs.]

Figure 4.2 tells what happens in I sec. Let us, so to speak, divide it by 1 sec.—that is, let us replace ft. by ft./sec. wherever it occurs there. Then will represent the velocity of the point along AB relative to the platform, the velocity of the platform relative to the ground, and that of the point relative to the ground. And we see that the first two combine by the parallelogram law to yield the third. In this sense, then, we may regard velocities as vector quantities.

Exercise 4.2 Show that had Figure 4.2 been drawn to illustrate what happens in t sec., the distances would be 2t ft., 5t ft., and 6t ft., but the ultimate relationship between the velocities would be the same as before.

Exercise 4.3 A train is going at 50 m. p. h. due north, and a fly in one of the compartments flies at 20 m. p. h. northeast relative to the compartment. Find, by means of an accurate diagram, the actual velocity of the fly relative to the rails.

Exercise 4.4 In Exercise 4.3, if the fly had been flying at 20 m. p. h. northeast relative to the rails, find, graphically, its velocity relative to the compartment.

Exercise 4.5 A cyclist goes at 8 m. p. h. due east, and the wind is blowing at 10 m. p. h. due east. What is the velocity of the wind relative to the cyclist? What would it be if he turned around and travelled due west at 8 m. p. h.?

Exercise 4.6 Like Exercise 4.5, but with the wind blowing due south at 8 m. p. h.

Exercise 4.7 A cyclist goes at 6 m. p. h. due north and feels the wind coming against him with a relative velocity of 4 m. p. h. due west. What is the actual velocity of the wind?

Exercise 4.8 The wind is blowing due south with a speed of 9 ft./sec. How fast must a car travel due east in order that the wind shall have, relative to the car, a speed of 15 ft./sec.?

Exercise 4.9 A gun mounted on a train is pointed vertically upward. The train moves horizontally due east at 80 m. p. h. and the gun fires a bullet with a muzzle velocity of 80 m. p. h. What is the velocity of the bullet relative to the ground? [Ans. Since the bullet has the velocity of the gun, and thus of the train, as well as its vertical muzzle velocity relative to the gun, its velocity relative to the ground is 80 √2 m. p. h. toward the east at 45° to the horizontal.]

Exercise 4.10 Rain is falling vertically with a speed of u ft./sec. At what angle does it appear to be falling as viewed from a train moving horizontally at v ft./sec.? [Ans. At tan⁻¹ (u/v) to the horizontal.]

Exercise 4.11 The wind is blowing steadily from the east at 5m.p.h. A boat in still water is travelling in a circle, with a constant speed of 5 m. p. h. A pennant is flying at the top of the mast. Describe in words the way the direction of the pennant changes (a) relative to the deck of the boat, and (b) relative to the water. [It was a situation like this that led the astronomer Bradley to an understanding of the phenomenon of aberration, a topic you may find well worth looking into on your own. Note that the velocities in parts (a) and (b) both change.]

Exercise 4.12 In Exercise 4.10, if the rain were falling at an angle of θ to the horizontal, show that its apparent direction would make with its actual direction an angle sin⁻¹ {(u/v) sin θ}. [Compare this with the astronomical phenomenon of aberration which may be briefly described as follows: To catch the light from a star, a telescope must point in the apparent direction of the incoming light. Consequently, as the earth moves in its orbit the stars seem to move once a year in tiny ovals (actually ellipses) whose shapes depend on θ, and thus on the positions of the stars.]

It is customary to define velocity as the rate of change of displacement with respect to time. Let us consider it from this point of view. If a point moves uniformly along a straight line from P to Q, a distance of s ft., in t sec., it has a speed of s ft./t sec. or ft./sec. Its velocity is a quantity having the direction of the displacement but a magnitude of ft./sec., which is 1/t times the magnitude of and is, in fact, the speed. The displacement combines with other displacements of the point according to the parallelogram law; and changing the magnitudes of these displacements by a factor 1/t will clearly yield entities that also combine according to the parallelogram law. So velocities, as here defined, are vector quantities.

The velocity vector above can be denoted by (1/t) , and this is reminiscent of the multiplication of a vector by a number. There is a difference, though. Vectors 5V and V can combine with one another, but and (1/t) can not: the former is a displacement, and we have been measuring its magnitude here in ft.; but the latter is a velocity and its magnitude is not a length but a speed, here measured in ft./sec.

Suppose the motion is not uniform. Then when we divide the distance gone by the time taken we get the average speed. Correspondingly, we can define the average velocity as (1/t) . But this average velocity has an unexpected property. If I walk 10 miles in 2 hours my average speed is 5 m. p. h. But unless I walk in a straight line, my average velocity does not have a magnitude of 5 m. p. h. This is because my displacement is less than 10 miles if I change direction. If I walk a circuitous path and, having gone my 10 miles, end up where I started, I may end up exhausted from averaging 5 m. p. h. up hill and down dale for 2 solid hours, but my average velocity comes to zero.

I think we simply have to accept the fact that the vectorial concept of average velocity has unpleasant aspects when applied to motion not in a straight line. When motion is uniform, the speed is the same at every instant. When the motion is not uniform, the concept of instantaneous speed is a rather sophisticated one. It involves ideas belonging to the calculus into which we can not go here; but roughly speaking, the instantaneous speed is the value that the average speed approaches as the time interval gets smaller and smaller. When these ideas are applied to the average velocity, they yield an instantaneous velocity that, except in highly artificial situations, behaves much as one would expect it to behave, and this despite the peculiar characteristics of the average velocity from which it is obtained. The instantaneous velocity is a vector quantity, and its magnitude is the instantaneous speed. At every instant, it has the direction of the line tangent to the path of the moving point at the instantaneous position of the point.

There is a subtle difference between velocity considered in terms of the moving platform and velocity defined as the rate of change of displacement with respect to time. Look back at the discussion of the former, and at the exercises based on it, and you will see that in each case the three velocities that are involved in the parallelogram law are not all measured with respect to the same standard of reference. In the platform situation, for example, the 5ft./sec. and the 6 ft./sec. are velocities relative to the ground, but the 2 ft./sec. is a velocity relative to the moving platform. With the definition in terms of displacements, the three velocities in a parallelogram-law relationship are really velocities with respect to the same standard of reference.⁵ This may well seem like an unimportant distinction, especially since the two approaches to the concept of velocity both lead to the theorem that velocity is a vector quantity, and both yield the same resultant for two given velocities. But actually the distinction is a basic one physically. The resultants are the same only because of the nature of Newtonian space and time. In Einstein’s special theory of relativity, the resultants are different.

When defined in terms of displacements, velocities, like displacements, are free vectors. Yet in physical situations it seems essential to give velocity a specific location. For example, the velocity of a particle should surely be located at the particle and thus be represented by an arrow-headed line segment starting at the particle. In a storm, the velocity of the air is different at different places. To regard velocity as a free vector under the circumstances would be inappropriate. If a thrown object spins as it flies through the air, different points of it have different velocities; one can not free these velocities and transfer them from point to point and still describe the motion correctly.

Exercise 4.13 What is the resultant of a velocity of 2 ft./sec. due east and a velocity of 2√2 ft./sec. northwest? [Note that the wording of this exercise lets it pertain to the definition of velocity as the time rate of change of displacement.]

Exercise 4.14 A certain boat of negligible length has a maximum speed of 4 yds./sec. in still water. The water in a straight canal, 100 yds. wide, flows at 3 yds./sec. If the boat goes at maximum speed while pointing perpendicularly to the bank, what is its speed relative to the bank? How long does it take to cross the canal? [The answer is neither secs. nor 20 secs.] In what direction should the boat be pointed in order to cross the canal perpendicular to the banks, and how long would it take to cross? [It would take 100/√7 secs.]

5. ACCELERATION

Acceleration is the rate of change of velocity with respect to time. The awkwardness associated with a changing velocity is compounded when we consider acceleration; but, just as in the case of velocity, when ideas belonging to the calculus are brought in, these awkward aspects no longer obtrude. Suffice it to say here that acceleration is a vector quantity. The average acceleration of a particle may be said to be the change in velocity divided by the time taken, the velocity being regarded as a free vector. Thus if a point has a velocity of 15√3 ft./sec. due east at noon and a velocity of 30 ft./sec. 30° north of east at 5 secs. past noon, the change in its velocity is a velocity of 15 ft./sec. due north, as is easily seen from a figure. Since this change of velocity took place in 5 secs., the average acceleration is 3 ft./sec.² due north; we obtain this by dividing the 15 ft./sec. by 5 sees.

Exercise 5.1 A point changes velocity from 144 ft./sec. due east to 144 ft./sec. due north in 8 secs. What is the average acceleration? [Ans. 18√2 ft./sec.² to the northwest.]

Exercise 5.2 A man has a velocity of 5 ft./sec. due east and an acceleration of 5 ft./sec.² due north. Describe the resultant. [One word suffices.]

Exercise 5.3 A point moves in a circle of radius r with constant speed v. Find its average acceleration for (a) one revolution, (b) half of a revolution, and (c) a quarter of a revolution. [Remember that the instantaneous velocity has the direction of the tangent to the circle at the position of the particle. Remember also that though the speed is constant, the velocity changes because its direction changes. In case (c) the average acceleration comes out to have a magnitude of 2√2 v²/πr.]

Exercise 5.4 A point moves in a circle with constant speed. Prove geometrically that the average acceleration between two positions, A and B, of the point has a direction parallel to the bisector of the angle between the radii to A and B.

From Exercise 5.4 we see that, in finding the average acceleration of a point moving in a circle with constant speed, the smaller we take the interval of time, the closer together will be the two points, A and B, and the closer will the direction of the average acceleration be to the directions of the radii to A and B. When the time interval is so small that these radii are close to being coincident, the direction of the average acceleration will also be close to being coincident with them. It will come as no great surprise, then, that when one applies ideas belonging to the calculus one finds that the instantaneous acceleration at the instant when the point is at A lies along the radius to A and points towards the center of the circle. If the speed were not constant, the instantaneous acceleration would, in general, not lie along the radius.

6. ELEMENTARY STATICS IN TWO DIMENSIONS

If a body placed at rest⁶ remains at rest, it is said to be in equilibrium, and the forces acting on it are also said to be in equilibrium. In statics we study the conditions under which bodies and forces are in equilibrium.

If a body is so small that we can neglect its size and obtain satisfactory results by treating it mathematically as a point having mass, we call it a particle. (In some investigations quite large bodies—stars, for example—are treated as particles.) According to Newtonian dynamics, a particle of mass m acted on by forces having a resultant F will undergo an acceleration a where

(6.1)

For a particle to be in equilibrium, it must be unaccelerated, and therefore, by Equation (6.1), the resultant of the forces acting on it must be zero.

If a particle is in equilibrium under the influence of three forces, the forces must be coplanar. This is easily proved. Denote the forces by F₁, F₂, F₃, and let R be the resultant of F₂ and F₃. Then the three forces are equivalent to F₁, and R. For equilibrium, these two forces must have zero resultant. So R = – F₁. But the parallelogram construction ensures that R, F₂, and F₃ are coplanar. Therefore F₁, F₂, and F₃ must be too.

Exercise 6.1 A particle is acted on by three forces. Prove that if arrow-headed line segments representing, in the usual way, their magnitudes and directions (but not necessarily their locations) can be appropriately arranged to form a triangle, the particle is in equilibrium. Also prove the converse.

Consider now a rigid body that is too large to be treated as a particle. If

Figure 6.1

it is acted on at points A and B by two parallel forces F₁, and F₂ pointing in opposite directions as shown on the left in Figure 6.1, experiments show that it will be apt to turn and, if friction is present (but not in excessive amounts), it will take up the final position shown on the right,⁷ with the two forces acting along a common line; and if F₁ and F₂ have equal magnitudes the forces will then be in equilibrium.

From such experiments we infer two things. The first is that a rigid body can transmit the effect of a force unchanged from its point of application to any other point in the body that lies on the line along which the force acts. Because force is a vector quantity that has a definite point of application, it is often referred to as a bound vector (a term we have already been using) to distinguish it from a free vector. But when a force acts on a rigid body in equilibrium, its point of application can, in a certain sense, be thought of as free to move along the line of action of the force to another position within the body. Therefore, it is sometimes referred to as a sliding vector. Note, though, that if on the left in Figure 6.1 we moved the point of application of the force F₁, from A to another point A′ in the body lying on the line of action of A, the body would end up on the right in a position having A’B instead of AB lined up with F₁, and F₂; this position would not be the same as the one illustrated. There is danger, therefore, in thinking of forces as other than bound vectors, though the presence of danger need not prevent us from thinking of them as sliding vectors in appropriate circumstances, such as when a rigid body is held immovable.

The second thing we infer from these experiments is that the conditions for the equilibrium of a rigid body involve the locations of the forces in a significant way, since even if F₁ and F₂ had equal magnitudes, the rigid body would not be in equilibrium when it was in the position on the left, though it would be when in the position on the right.

Two conditions must be fulfilled by the forces acting on a rigid body to produce equilibrium. They are consequences of Equation (6.1), but we shall not give their derivation here. One condition of equilibrium is that if the forces acting on the body are treated as free vectors their resultant must be zero—just as if they were all acting on a particle. This condition ensures that the body has no overall translational acceleration. But it must be supplemented by another condition since the forces may still tend to twist the body.

Consider the simple case of a seesaw of negligible mass, as shown, with a boy of weight 75 1 bs. at B, x ft. from O, and a man of weight 150 1bs. at M

Figure 6.2

y ft. from O. The turning effect, or moment, of the boy’s weight is counterclockwise, and it is measured by the product of his weight and the perpendicular distance OC from O to the line of action of the weight. It is therefore 75x cos θ, where θ is as shown. Similarly the turning effect of the man’s weight is 150y cos θ, the effect being clockwise. For equilibrium, the clockwise and counterclockwise turning effects must balance. So

(6.2)

from which we see that x = 2y. Thus, the boy balances the man if he sits twice as far as the man does from the point of support; and the balance then holds for all values of θ.

By regarding counterclockwise moments as positive and clockwise moments as negative, we can rewrite Equation (6.2) in the form:

(6.3)

It then exemplifies the second condition for the equilibrium of a rigid body: the total moment of the forces acting on the body must be zero. [All moments must be about the same point, but this point can be anywhere.] For simplicity we are confining ourselves here to situations in which all the forces lie in a plane. The two conditions of equilibrium, asserted here without proof, hold for the general case; we shall discuss this case when we have developed the necessary vectorial mathematics.

Note that the weights of the boy and the man add up to 225 1bs. so that, at first sight, it seems as if the first condition of equilibrium is not satisfied. But there is a third force acting on the seesaw, namely the reaction on it of the fulcrum at O. (We are assuming that the mass of the seesaw is negligible.) Denote the reaction by R, and its magnitude by R. Then the first condition tells us that R is vertical, and

R – 75 – 150 = 0

so that R = 225 1bs.

Exercise 6.2 Show that if we took moments about B instead of O the total moment would still be zero. [Note that R will now have a nonzero moment. Its moment about O is zero, so it would not have made any difference to Equation (6.2) or (6.3), even if we had thought of it at the time.]

Exercise 6.3 Like Exercise 6.2, but taking moments about M.

Exercise 6.4 Prove that if a rigid body is in equilibrium under the influence of three nonparallel coplanar forces, the lines of action of these three forces must meet in a point. [Take moments about the point of intersection of two of these lines of action.]

Exercise 6.5 ABCD is a rigid square of side a. Force F₁ acts at B in the direction from A to B, and force F₂, of equal magnitude, acts at C in the direction from B to C. Find the magnitude, direction, and location of the single additional force that would maintain the square in equilibrium. [Use Exercise 6.4.] Verify that the total moment of the three forces about D is zero.

Exercise 6.6 If n parallel coplanar forces acting on a rigid body have zero total moment about a point O in their plane, prove that they do not have zero total moment about every other point in their plane unless they are in equilibrium. [Let the forces have magnitudes F₁, F₂, . . . , Fn and be at distances x₁, x₂, . . . , xn from O. Then x₁ F₁ + x₂ F₂ + . . . + xn Fn = 0. But (x₁ + a)F₁ + (x₂ + a)F₂ + . . . + (xn+ a)Fn will not, in general, be zero.]

7. COUPLES

Consider two parallel forces F₁ and F₂ acting on a rigid body at points A and B as in Figure 7.1, the forces having equal magnitudes, F, and opposite directions. Let their distances apart be a. Because the magnitudes of the forces are equal, the first condition of equilibrium is satisfied since, when treated as free vectors, F₁ and F₂ have zero resultant. The second condition is not satisfied, though. The forces do not have zero total moment. They tend to turn the body. If we take moments about A, we obtain the values 0 and aF for a total of aF. If we take moments about B we obtain aF and 0, also yielding a total of aF. If we take moments about a point E, in the plane containing the forces, that is distant x from the line of action of F₁ and thus a – x from that of F₂, we obtain xF + (a – x) F which gives the same total, aF, as before.

Figure 7.1

A pair of parallel forces of equal magnitudes and opposite directions constitutes a couple. The magnitude of the moment of a couple is measured by the product of the magnitude of either force and the perpendicular distance between their lines of action. Thus the two forces in Figure 7.1 constitute a couple of moment aF acting counterclockwise. Usually the total moment of a system of forces depends on the position of the point about which the moments are taken. With a couple, though, as we have seen, the moment does not change when the point is changed. Therefore there is no need to specify a point when talking about the moment of a couple.

Suppose a force F₁ of magnitude F acts on a rigid body at point A, as in Figure 7.2. At point B, let us introduce two self-cancelling forces F₂ and F₃ parallel to F₁ and having the same magnitude F. Now let us regroup these three forces in our minds: let us think of F₁ and F₃ as forming a couple, and F₂ as standing apart. Then we see that the single force F₁ at A is equivalent to the combination of the single parallel force F₂ of equal magnitude acting at B and the couple formed by F₁ and F₃.

Figure 7.2

In effect, then, when we move a force to a different line of action (parallel to the original one), we automatically introduce a couple.

Exersise 7.1 In Figure 7.2, if the forces make an angle θ with the line AB, and the distance AB is a, what is the moment of the couple ? [Ans. aF sin θ.]

Exercise 7.2 A, B, C, and D are the vertices of a square of side a. A force of magnitude 101bs. acts at A and points in the direction . What couple would be introduced if the force were moved parallel to itself so as to act (a) at B, (b) at C, and (c) at D? [Ans. 0, – 10a, – 10a.]

Exercise 7.3 A force of magnitude F 1bs. pointing in the positive x-direction acts at the origin. A couple in the xy-plane has a moment of G ft. lbs. Where in the xy-plane could the point of application of the force be moved to in order to yield a total couple of H ft. lbs.?

8. THE PROBLEM OF LOCATION. VECTOR FIELDS

The problem of the location of vectors is a vexing one for which there seems to be no wholly satisfactory elementary solution. The easy way out is to ignore the matter, or to say that such an entity as a force is more than a vector, being a vector equipped with the additional, nonvectorial attribute of location. The purely vectorial aspect of such an entity—its vector, so to speak —is then regarded as a free vector, all vectors being taken to be free vectors.

There is something highly attractive about this approach. It makes for a neat mathematical theory of vectors. But we really do not understand vectors until we grapple with the problem of location, and ultimately realize that we do not understand it. In this book, therefore, we shall continue to look into the awkward aspects of this problem. The reader who feels that he is uninterested in the problem of location is invited, at any stage in this book, to make the ultimate (or seemingly ultimate) abstraction of denying vectors location and then to ignore all further mentions of the problem. If he does this he will be happier than those who try to face the issue. But he will understand vectors less intimately in the end.

Let us recall the discussion of velocity as the rate of change of displacement with respect to time. Since displacement is a free vector, we at first thought that velocity must be too. But then we thought of the velocity of a particle and felt that this should be located at the particle. And after that, we thought of the velocity of the wind in a storm and realized that there was a different velocity vector at each point.

There are many situations analogous to that of the wind in a storm. The force exerted by a given magnet on another magnet, for instance, is different at different locations of the other magnet. The force exerted by an electric charge on another charge stationary with respect to it also depends on the location. When we have such a situation—an enormous number of vectors, one at each place—we say we have a vector field. Mathematically, we would say that there were infinitely many vectors, one at each point. Evidently, in a nonuniform field, we can not switch vectors around to different locations without altering the field. In this sense, the vectors of a vector field have to be considered bound vectors.

By a process of abstraction similar to that that led us from shifts to displacements, we can imagine free vectors associated with the vectors of a field. But then we wonder how much it will profit us to do so. For in a field of force, for example, the force at P does not combine physically by the parallelogram law with the force at a different point Q.

Nevertheless we can ask what is the difference of these two forces in magnitude and direction. And in answering we could go off into a margin and work the problem out, treating the two forces as free vectors and moving them to a common starting point. In working in the margin in this way, we would be working with what the mathematicians call a linear vector space. When dealing with forces, we can work in one linear vector space. When dealing with, say, displacements, we can work in another. By judiciously using different linear vector spaces for different situations, we can make the structures of all the linear vector spaces the same, and say that in a given linear vector space not only do all vectors have a common starting point, but every vector combines with every other vector therein according to the parallelogram law.

Pure mathematicians often prefer to start by defining vectors as abstract elements in a linear vector space that obey certain rules laid down as axioms. These axioms, indeed, simultaneously define the vectors and the vector space that they inhabit. Often pure mathematicians think of vectors in a vector space as “ordered sets of numbers,” by which they mean unpictorial entities like (1, 9, 1, ) that are counted as different from, say, (9, 1, 1, ). the amount of numbers in each ordered set depending on the dimensionality of the vector space. By taking this abstract stance, they gain enormously in precision and elegance and power—as generally happens when good mathematicians make abstractions. But the person who studies vectors only as a branch of abstract algebra, regarding them solely as elements in a vector space, obtains the precision and elegance and power at a price. For they come to him too patly for him to realize their worth. He gains no inkling of the motives that led to the choice of just these axioms and abstractions rather than others, and he is apt not to realize how ugly are the seams that join such neat algebraic abstractions to their applications—even to applications to other branches of mathematics such as geometry.

Note, for example, that in asking about the difference of the two forces at P and Q we were not being as reasonable as we thought at the time. We were assuming not only that such a comparison could be made, but also that it could be made in a unique way. That sounds innocent enough. But in effect we were assuming the validity of Euclid’s axiom of parallels (among other things) and these days we all know that this axiom can be dispensed with. To pursue this particular aspect of the matter further would be outside the scope of this book. We have pushed it this far in order to show that the problem is not as simple as it sometimes seems.

3

VECTOR ALGEBRA

1. COMPONENTS

Let V be a vector starting at a point O. Take any plane containing V, and in it draw any two noncoincident lines, A′A and B′B, passing through O. Then we can always find two vectors, X and Y, lying respectively along these

Figure 1.1

two lines, such that X and Y are together equivalent to V——that is, such that X + Y = V. Moreover, the vectors X and Y are uniquely determined by V and the two lines. For, through the end R of V we can draw lines RT and RS respectively parallel to A′A and B′B, intersecting those lines at S and T respectively. Then OSRT is a parallelogram. Therefore, the vectors represented by and are uniquely the X and Y we seek.

Exercise 1.1 Prove the uniqueness of X and Y.

Let ex and ey be nonzero vectors starting at O and lying respectively along the lines of action of X and Y. Denote the ratio of the magnitude of X to that of ex by Vx, taking Vx to be positive if X and ex have the same direction, and negative if they have opposite directions. Similarly, denote the ratio of the magnitude of Y to that of ey by Vy, these quantities, Vx and Vy, being, of course, not vectors. Then we can write:

(1.1)

Vx and Vy are called the components of the vector V relative to (or with respect to, or in) the frame of reference defined by ex and ey. The point O is called the origin of the frame of reference, and ex, ey its base vectors. Some books use the word components for the vectors X and Y, and then fail to make a sharp verbal distinction between X, Y and Vx, Vy. To avoid confusion, we shall refer to X and Y as component vectors of V.

Exercise 1.2 A vector of magnitude 144 makes 45° with each of two perpendicular lines. What are its component vectors along these lines? [There are four possible answers to the question as worded.]

Exercise 1.3 In Exercise 1.2, if ex and ey lie along the perpendicular lines, and ex has a magnitude of 4 and eya magnitude of 6, what are the components of the vector relative to the reference frame defined by the base vectors ex and ey? [Ans. ± 18√2, ± 12√2.]

Exercise 1.4 If ex and ey have unit magnitude and are perpendicular, what are the components of a vector of magnitude h pointing along the positive direction of ex? Would the components be different if ey were not perpendicular to ex? [In situations of this sort we shall assume that the vectors have the same starting point unless they are free vectors, in which case the assumption is unnecessary.]

Exercise 1.5 If ex has a fixed direction but ey a varying one, is Vx constant or not?

Exercise 1.6 Find the component vectors of a vector of magnitude 10 pointing 30° north of east relative to (a) the east and north directions, (b) the east and northwest directions, and (c) the east direction and the direction 30° north of east.

Exercise 1.7 In Exercise 1.6, find the component of the given vector relative to the reference frame ex, ey, if ex has magnitude 2 and ey magnitude 1, and if, in each part, ex points to the east while ey points in the second direction mentioned. [Ans. (b) 5(√3 + 1)/2, 5√2.]

Figure 1.2

If, in Figure 1.1, the vector V did not lie in the plane of the lines A′A and B′B, we could not have constructed the parallelogram. But if we take three lines OA, OB, and OC that do not lie in a plane, we can always⁸ find, along the respective lines, vectors X, Y, and Z that are together equivalent to V; these three vectors are uniquely determined by V and the lines. The method is to construct a parallelepiped with edges along and parallel to the three chosen lines OA, OB, and OC, as in Figure 1.2.

Exercise 1.8 Explain how to construct the parallelepiped in Figure 1.2 by passing appropriate planes through R.

Exercise 1.9 Prove that X + Y + Z = V in Figure 1.2 (a) by first combining X and Y and combining their resultant with Z; (b) by first combining Y and Z; (c) by first combining Z and X; and (d) by using the end-to-end method of finding the resultant of several vectors.

If we introduce three nonzero base vectors ex, ey, ez along OA, OB, and OC, respectively, we can write:

(1.2)

the quantities Vx, Vy, and Vz being the components of V with respect to the reference frame defined by ex, ey, and ez. Then, since V = X + Y + Z, we have:

(1.3)

Exercise 1.10 In two dimensions, if U = Uxex + Uyey and V = Vxex + Vyey, show by means of a diagram that U + V = (Ux + Vx)ex + (Uy + Vy)ey and U – V = (Ux – Vx)ex + (Uy – Vy)ey. [Note how neatly the symbols, + and –, applied to vectors turn out to be related to the algebraic symbols, + and –, applied to their components.]

Exercise 1.11 Prove the relations in Exercise 1.10 algebraically by noting that U + V is really the sum of four vectors Uxex, Uyey, Vxex, and Vyey, adding the first and third, then the second and fourth, and combining the results.

Exercise 1.12 Relative to ex and ey, let U have components Ux and Uy, let V have components Vx and Vy, and let W have components Wx and Wy. What are the components of U + V – W?

Exercise 1.13 In Exercise 1.12, what must be true of the components of U, V, and W if U + V – W (a) points in the ex direction, (b) points in the negative ey direction (i. e., in the direction opposite to that in which ey points), (c) lies along the direction of the resultant of ex and ey?

Exercise 1.14 In the situation epitomized in Equation (1.3), what will happen to the components of V if the magnitudes of the base vectors ex, ey, and ez, are increased to 5 times their present values? [Note that the components become smaller, not larger.]

Exercise 1.15 In Equation (1.3) what new magnitudes for ex, ey, and e, would make the components of V all equal to 1? Are there exceptional cases?

Exercise 1.16 What are the components of ex, ey, and e, respectively in the reference frame defined by ex, ey, and ez?

Exercise 1.17 If U = Uxex + Uyey + Uz,ez, and V = Vxex + Vyey + Vzez, show that the components of aU ± bV are aUx ± bVx, etc., a and b being numbers.

Exercise 1.18 If, relative to ex, ev, and ez, U has components (6, 2, – 4), what can you say about the components of V (a) if U + 2V lies along the ex direction? (b) if 3U – V lies in the plane defined by ev and ez? (c) if U + V is the same as 6ex + 2ev – 4ez? (d) if ex + U = 3ey + V? [Ans. (a) Since ex has components (1, 0, 0), we must have 6 + 2Vx = some constant, 2 + 2Vv = 0, – 4 + 2Vz = 0. So we can say that Vy = – 1, VZ = 2.]

Exercise 1.19 Relative to ex, ey, and ez, the velocity of an airplane has components (100, 150, 20) and the wind velocity has components (30, – 20, 0). What are the components of the velocity with which the wind passes the plane? [Ans. (30 – 100, – 20 – 150, 0 20), or ( – 70, – 170, – 20), just like that! Note how much easier it is here to work with components than with arrow-headed line segments and three-dimensional diagrams.]

Exercise 1.20 A particle in equilibrium is acted on by three forces, two of which have components (6, 5, – 2) and (3, – 10, 8) respectively, relative to a particular reference frame. What are the components of the third force? [Ans. (- 9, 5, – 6).]

2. UNIT ORTHOGONAL TRIADS

The formidable title of this section has to do with something that is simpler than what was studied in the preceding section, being a special case of it.

If we know the components of a vector with respect to a reference frame, we know the vector, for it is given by Equation (1.3). Theoretically, then, we ought to be able to calculate its magnitude once we know its components in a given reference frame. And indeed we can. But when the reference frame is a general one of the type considered in the preceding section, the formula is rather complicated. Consider, for example, the two-dimensional case, with the reference vectors ex and ey making an angle θ with each other. If the components of V are Vx and Vv, and the magnitudes of ex and ey are ex and ey, then the lines OS and OT in Figure 2.1 are respectively of length Vzex and Vvey. In triangle OSR, SR has the length of OT, namely Vvey, and angle OSR is 180° – θ. So, by the law of cosines, since cos (180° – θ) = – cos θ, we have:

(2.1)

a none-too-pleasant formula even without our taking the square roots of both sides. And this is just for the two-dimensional case.

Figure 2.1

In many situations, of which the preceding is one, the work is significantly simplified if we use a special type of reference frame, though the simplification is obtained at a price: loss of generality and of depth of understanding.

The idea is to use reference frames in which ex, ey, and ez are mutually perpendicular and of unit magnitude; hence the title of this section. It is customary to denote the vectors ex, ey, and ez of such a unit orthogonal reference triad by the special symbols i, j, and k. Note that i does not stand for here. For one thing, it is printed in boldface type.

The parallelepiped in Figure 1.2 now becomes rectangular, as in Figure 2.2. [Note the change in letters.] And instead of Equation (1.3) we have:

Figure 2.2

(2.2)

What is the magnitude of V in terms of Vx, Vy, and Vz? If V is represented by in Figure 2.2, Equation (2.2) tells us that

(2.3)

since the magnitudes of i, j, and k are all unity. [But we shall have to return to this in the next chapter.] Applying-Pythagoras’ theorem to the right triangle OAD, we have OD² = OA² + AD². And applying it to the right triangle ODR, we have OR² = OD² + DR². So OR² = OA² + AD² + DR² = OA² + OB² + OC². But OR is the magnitude of V. Therefore, by Equation (2.3),

(2.4)

from which we obtain

(2.5)

Since the magnitude is always counted as non-negative, we do not use a ± sign in Equation (2.5).

Exercise 2.1 A vector has components (1, 1, 1) relative to i, j, and k. What is its magnitude? Describe as best you can its direction in relation to i, j, and k.

Exercise 2.2 Like Exercise 2.1 for a vector having components (1, 1, – 1).

Exercise 2.3 What are the magnitudes of the vectors having the following components relative to i, j, and k? (a) (1, 2, 2); (b) ( – 4, – 4, 7); (c) (√3, √7, – √6); (d) (10, – 20, – 30); (e) (Vx, Vy, 0).

Exercise 2.4 Relative to a unit orthogonal triad, a vector U has components (5, 12, 5) and a vector V has components (4, – 3, z). What can you say about z if U and V have equal magnitudes?

Exercise 2.5 If U and V have components (Ux, Uy, Uz) and (Vx, Vy, Vz,) relative to i, j, and k, what are the magnitudes of U + V and U – V?

Exercise 2.6 A particle is acted on by two forces F and F′ having components (Fx, Fy, Fz) and () relative to i, j, and k. What are the components and magnitude of the force that will keep the particle in equilibrium?

Exercise 2.7 What is the locus of the tip of a vector of fixed magnitude r and fixed initial point O? If the components of the vector relative to i, j, and k are (x, y, z) what equation must these quantities satisfy?

For given i-and j, we can take k in either of the two opposite directions perpendicular to the plane of i and j, as shown in Figure 2.3. The two configurations (a) and (b) are distinct. We can not maneuver one of them as a rigid body so as to make it coincide with the other. We therefore give the configurations different names; we refer to (a) as a right-handed system and to (b) as a left-handed system.

Figure 2.3

Imagine a right-handed corkscrew pointing along the k-direction in the right-handed system in (a). If we twist the handle as if we were trying to make i take up the position now occupied by j (via the 90° route, not the 270° one!) the corkscrew will move in the direction in which k points. In the left-handed system (b) it will move in the direction opposite to that in which k points.

Exercise 2.8 Verify that the preceding paragraph will remain valid if we replace i, j, and k by j, k, and i respectively, or by k, i, and j respectively.

Exercise 2.9 In Figure 2.2, do the lines OA, OB, and OC form a right-handed or a left-handed system? What if we took them in the order OB, OC, and OA? What if we took them in the order OB, OA, and OC? [Ans. right-handed; right-handed; left-handed.]

Exercise 2.10 Starting with a right-handed system, what does it become if we reverse the direction of one axis? Of two? Of three?

Exercise 2.11 Give a definition of right-handedness that would apply to a general reference frame defined by ex, ey, and ez in that order [Note the importance of the words in that order. We usually take them for granted without mentioning them. The definition is a simple generalization of that for the orthogonal case; but one must be careful to specify that, for example, the rotation of ex towards ey is via the shorter route. What is the maximum angle one could permit in this connection?]

3. POSITION VECTORS

Let the period at the end of this sentence represent a point P.

Where is the point P? We could say it is right there, at the end of the sentence. But where is the sentence? On page 40 of a book? Then where is the book? On a table in a certain room on the earth? Then where is the earth?

After pushing this silent dialogue with ourselves a little further, we begin to realize that the concept of “where” is fundamentally a puzzling one.

But suppose we erect a frame of reference by means of three base vectors ex, ey, and ez. Then, without knowing where this frame of reference is in any deep sense of the word, we can nevertheless ask where a point is in relation to the reference frame; and this relative sort of where is a far more tractable concept.

To specify the location of a point relative to a frame of reference, it suffices to give the components of the vector that starts at O and ends at the point. For example, in Figure 1.2 the location of the point R relative to the reference frame is given by the components of the vector V relative to the frame.

Exercise 3.1 Point R is the midpoint of ex. Denote the vector by V. What are the components of V relative to ex, ey, and ez?

Exercise 3.2 What would the components of the vector V in Exercise 3.1 become if, keeping the same origin, we rotated to a new reference frame with base vectors , and such that ? [Ans. (0, 0, ).]

We could dispense with a reference frame, retaining only its origin, O, and specifying the point R by means of the vector V itself rather than by means of its components relative to some reference frame. Or so it would seem. Indeed, the idea becomes quite attractive when we recall that if we change from one reference frame to another having the same origin, the components of V change but the vector itself does not.

But what do we mean when we say that the vector V does not change? Change relative to what? How do we know it isn’t whirling around O at an alarming rate? You say you know because you are holding your book steady? Then hurl it into the air, and now try answering the question.

It is clear that when we think we are dispensing with a reference frame we really are not. In doing Exercise 3.2 above, did we so much as pause to ask ourselves what was meant by “the vector V in Exercise 3.1”? Did we not assume that the meaning was obvious? The meaning is indeed obvious. But let us face what it implies. In thinking of V as being the same in Exercises 3.1 and 3.2, we have in mind some master reference frame relative to which V is indeed unchanging. It could be the frame ex, ey, and ez, but it need not be; and the chances are that we do not think of it as being that reference frame. However, we do here think of ex, ey, and ez as being unchanging vectors relative to this master frame, and we think of , and as being so too. And if we should have occasion to speak of a moving reference frame, we would really mean that it was moving relative to some other frame, whether we mentioned this other frame or not.

Having looked into the situation, we may, for the most part, go on speaking about fixed vectors, fixed points, and the like as we have been doing all along, realizing now that we are really doing so in relation to some mentioned or unmentioned master reference frame.

Given a fixed reference point O, then, we can specify the position of a point P by means of a vector having magnitude and direction represented by the arrow-headed line segment . This vector is called the position vector of P relative to O.

In what sense can we locate a point P relative to a point O by means of a vector? In thinking of not just as an arrow-headed line segment but as a vector, we commit ourselves to seemingly nonsensical things about the positions of points; for example, that if OPRQ is a parallelogram, the position of P relative to O and the position of Q relative to O are together somehow equivalent to the position of R relative to O. Stated this way, it is indeed nonsense. It makes sense if we regard these positions as displacements, even though they were all conceived as positions relative to the same point O; but it does not make the sort of sense we had in mind when we thought we were specifying the locations of points relative to O. Even so, the idea of representing these positions by means of vectors that we then regard as displacements is a powerful one, as the following exercises will indicate.

Exercise 3.3 If points A and B have position vectors a and a and b relative to O, what is the position vector of the mid-point M of AB?

Solution Denote the position vector of M by r. Then, regarding a and b as displacements, which are free vectors, we have:

Figure 3.1

and

But the displacements and are equal. Therefore,

r – a = b – r,

from which we find that

(3.1)

This relation is called the MID-POINT FORMULA.

Exercise 3.4 Points A, and A₂ have position vectors a₁ and a₂ relative to O. A point P on A₁ A₂ is such that A₁P/PA₂ = m₁/m₂, where m₁ and m₂ are numbers. Show that the position vector, r, of the point P is given by:

(3.2)

[This relation is called the POINT-OF-DIVISION FORMULA.]

Exercise 3.5 Using vectors, prove that the medians of a triangle meet at a point that divides each median in the ratio 2:1 starting from its vertex.

Figure 3.2

Solution Denote the position vectors of the vertices A, B, and C of the triangle by a, b, and c respectively. Then, by the mid-point formula, the position vector of the midpoint A′ of side BC is Let G be the point of AA’ such that AG = 2GA’, so that AG: GA′ = 2 : 1. Then, by the point-of-division formula, the position vector of G is:

which is just . This quantity is symmetrical in a, b, and c: though we found it by going two-thirds of the way down the median through A, it contains the vector a on an equal footing with b and c. Consequently, if we did the corresponding calculations for the other medians they would have to give essentially the same result——if we made no mistakes, or all our mistakes cancelled. Therefore, the point G is common to all three medians.

Exercise 3.6 Do Exercise 3.5 in terms of the median through B.

Exercise 3.7 Using vectors, prove that the line segments joining the mid-points of opposite sides of any quadrilateral, whether plane or skew, bisect each other.

A position vector is a curious thing. It is akin to a displacement, but instead of being free it is tied to a particular origin O. If the position vector of a point with respect to an origin O has the same magnitude and direction as that of P′ relative to O′, the points P and P′ will not coincide unless O and O′ do. With O and O′ distinct, the position vectors and do not define the same position, though they would be counted as equal vectors if they were displacements. Yet if we think of a position vector as a bound vector starting at O we have to wonder in what sense, for example, the position vectors and in Figure 3.1 can combine to give a vector that not only does not start at O but does not even pass through O. And we may conclude that does not yield but a vector starting at O having the same magnitude and direction as . This is awkward, but Exercise 3.3 and others like it can, in fact, be worked out in terms of vectors all of which start at O.

Exercise 3.8 Work out Exercise 3.3 in terms of such vectors. (Note that a crucial step is to require that the vectors defined by and be identical. Draw the two parallelograms involved.)

Perhaps we find it distasteful to work in this way here in terms of bound vectors all starting at O. But before we decide to reject them, let us consider Figure 3.3. What does it represent? Three guesses: A design with match sticks? A lopsided equation telling that X = 1? A preliminary study for a modern work of art? All wrong. It is Figure 3.1 done in terms of free vectors (with letters and arrowheads omitted).

Figure 3.3

When one is dealing with position vectors, there seems to be no way of avoiding awkwardness, except by ignoring it. But that is not a bad method.

Exercise 3.9 It is instructive to try to do Exercise 3.3 solely in terms of shifts, using only relations of the type OA + AM = OM. Try this, and note why the attempt does not succeed [For example, AM and MB are not identical shifts, and Equation (3.1) is not applicable to shifts, though it is to displacements corresponding to shifts. One sees from this how powerful is the concept of a vector as compared with that of a shift, even though the shift seems to contain more data of importance.]

Suppose, in Exercise 3.3, we changed from the origin O to a new origin O′. Then all the position vectors would be changed. Yet (relative to our master reference frame) the points A, M, and B would not move, and M would remain the mid-point of the line segment AB. Clearly, the mid-point formula (3.1 ) will have to hold for the new position vectors as well as for the old. If we are dubious (though the foregoing is really a sufficient proof), we can check very easily. Let the position vector of O′ relative to O be h. Then, from Figure 3.4, we see that the old position vector a of point A is related to its new position vector a′ by

(3.3)

Figure 3.4

Similar relations hold for all other position vectors. Let us introduce new position vectors for old ones in Equation (3.1), namely r = (a + b). Then we have

which reduces to

as we foretold that it must.

Exercise 3.10 Show by the two methods above that the point-of-division formula (3.2) also holds for new position vectors when the origin is changed.

Exercise 3.11 Like Exercise 3.10 for the formulas for the position vectors of the ultimate points of intersection in Exercises 3.5 and 3.7.

There is a simple theorem that tells us what sort of linear relation between position vectors preserves its form when the origin is changed. Let the relation be:

(3.4)

where the a’s are position vectors relative to O, and the α’s are numbers. Make the change to a new origin O′ by means of equations of which Equation (3.3) is a specimen. Then Equation (3.4) becomes:

or

and this will reduce to Equation (3.4), with a“s replacing a’s, if and only if

(3.5)

Exercise 3.12 Verify that the condition (3.5) is satisfied by Equation (3.1). [Write Equation (3.1) as and note that the coefficients 1, and add up to zero.]

Exercise 3.13 Like Exercise 3.12 for the results of Exercises 3.4, 3.5, and 3.7. [Note that in Figure 3.2 we did not even bother to draw the origin.]

If, without drawing the origin, we can draw a diagram that nevertheless contains all the essential data, the expression of these data in terms of position vectors will retain the same form when the origin is changed. And this will be true even if the relations between the vectors are nonlinear. The reason is so simple as to be almost primitive. The position vectors depend on the origin. If they are used in order to represent a configuration that does not even involve an origin, then they must do so in a way that does not give one particular origin preference over any other. Therefore, the relation between the position vectors must retain the same form when the origin is changed, since otherwise all origins would not enjoy equal status—or perhaps we should say here, lack of status.

Examples of such situations are easy to find. For instance, we can draw a diagram of the fact that M is the mid-point of the line segment AB without needing to draw in an origin. Other such situations have been encountered in Exercises 3.4, 3.5, and 3.7. Yet others are that four given points lie in a plane, or that they lie at the vertices of a regular tetrahedron, or that one of the points is the intersection of the angle bisectors of the triangle formed by the other three. [The vectorial formula for this last would probably be encountered as a formula for the position vector of the point of intersection of the angle bisectors in terms of the position vectors of the vertices of the triangle.]

Exercise 3.14 Show by means of Equation (3.3) that , and therefore, that any vector relationship built up solely of combinations of the form will retain its form when the origin is changed. [Note how this is related to the preceding discussion of diagrams in view of the fact that .]

Exercise 3.15 Show that if a₁, a₂, and a₃ are position vectors and β₂₃, β₃₁, and β₁₂ are numbers, the linear relation among differences of a’s,

β₂₃(a₂ – a₃) + β₃₁(a₃ – a,) + β₁₂(a₁ – a₂) = 0,

satisfies the condition (3.5) for being independent of the origin.

Exercise 3.16 Extend Exercise 3.15 to the case of n position vectors a₁, a₂, . . . , an.

Exercise 3.17 Show that if a₁, a₂, and a₃ are position vectors and the linear relation

α₁a₁ + α₂a₂ + α₃a₃ = 0

is independent of the origin so that α₁ + α₂ + α₃ = 0, then the linear relation can be written as a linear relation among differences of position vectors. [Thus we can rearrange the left-hand side by writing it as α₁(a₁ – a₂) + (α₁ + α₂)a₂ + α₃a₃, which is equal to α₁(a₁ – a₂) + (α₁ + α₂) (a₂ – a₃) + (α₁ + α₂ + α₃)a₃, and the last term vanishes because of the conditions on the α’s. Now work the problem starting with a₃ instead of a,, and note that the resulting relation has a different form.]

Exercise 3.18 At first sight, it may seem surprising that there should be various ways of expressing a linear relation among position vectors in terms of differences of such vectors, in view of the fact that these differences directly represent line segments in a diagram that does not contain O. To understand what is involved, show that the mid-point formula in Exercise 3.3 can be rewritten in the following three forms, and state their geometric interpretation in terms of line segments involving the points A, M, and

Exercise 3.19 Denoting the position vector of the point G in Exercise 3.5 by g, express the relation in a form involving only differences of position vectors, and interpret the result geometrically. [One such form is (g – a) + (g – b) + (g – c) = 0. This tells us that the vectors represented by , and have zero resultant. So if, for example, we complete the parallelogram CGBK, we see that GK is an extension of AG and of the same length as AG. Thus, by merely rearranging the original relation between position vectors in terms of differences of these vectors (a rearrangement that is only possible if the original relation is independent of the origin), we come out with a geometrical theorem.]

Exercise 3.20 In Exercise 3.19, one could rewrite the relation in the form 3(g – a) + 2(a – b) + (b – c) = 0. Interpret this geometrically. [It is easier to see what is involved if one rewrites this as

Exercise 3.21 A, B, C, and D are four points not necessarily in the same plane. Using position vectors, prove that if the line segments AC and BD bisect each other, ABCD is a parallelogram. [Hint: rearrange (a + c)/2 = (b + d)/2 to obtain a – b = c – d.]

Exercise 3.22 Extend Exercise 3.17 to the case of n position vectors a₁ a₂, . . . , an.

Let A be a fixed point and c a given vector. How can we represent the line through A parallel to c? If P is any point on this line, and the position vectors of A and P are a and r, respectively, then, since AP is parallel to c,

(3.6)

where t is a parameter. We may rewrite this as

(3.7)

and this represents the line in the sense that it gives the position vector of every point on the line in terms of the parameter t.

Consider now the analogous problem of representing the straight line through two points A and B with position vectors a and b. Here we have merely to replace c in Equation (3.6) or (3.7) by b – a, since the line is certainly parallel to the vector b – a inasmuch as it actually lies along AB. So we have

(3.8)

or

(3.9)

Now a seeming puzzle arises. If we write Equation (3.7) in the form:

r – a – tc = 0

and Equation (3.9) in the form:

r + (t – 1)a – tb = 0,

we see that the sum of the coefficients in the latter is 1 + (t – 1) – t, which is zero, whereas the corresponding sum in the former is 1 – 1 – t, which is not zero. It appears, then, that Equation (3.9) is independent of the origin but (3.7) is not. This is not true, though, or rather, it is true that this appears to be the case, but it is actually not the case. Both equations are independent of the origin. This is clear from the fact that (3.6) involves only (r – a) and c, each of which is independent of the origin. Where is the fallacy? [Hint: Why is c independent of the origin? Is it a position vector? What is the nature of the vectors a₁, ... , an in Equation (3.5)? Does Equation (3.7) fall within the category of (3.5), or is it more general?]

All this preoccupation with vectorial relations that are independent of the origin can be misleading. It is a standard preoccupation, and it is forced on us because we are using those hybrid things, position vectors. But it obscures a crucial point. With what we may refer to as nonposition vectors, i. e., vectors of the usual type, relations such as

are automatically independent of any choice of origin. They are, in fact, objective. And the objectivity of such vectorial relations is one of the most valuable attributes of vectors. Bear it in mind. It will play a crucial role later.

4. COORDINATES

Let r be the position vector of a point P relative to a reference frame defined by vectors ex ey, and ez starting at an origin O. The components of r relative to this frame are called the coordinates of P relative to the frame, and they are usually denoted by (x, y, z). [We are here considering the three-dimensional case, of course.] Thus,

(4.1)

Figure 4.1

Note that while all vectors have components, only the components of position vectors are called coordinates. The coordinates here defined are of a restricted sort called Cartesian coordinates. Polar coordinates, and other curvilinear or nonuniform types of coordinates are not of this sort.

If we use Cartesian coordinates in two dimensions we are, in effect, using graph paper of the type shown in Figure 4.1, the point P having the coordinates (5, 2). Note that the scales along the two axes, though uniform, are not necessarily the same.

Exercise 4.1 In Figure 4.1, which vector has the greater magnitude, ex or ey?

It is not so easy to draw a corresponding diagram of a three-dimensional Cartesian coordinate mesh: the diagram tends to become confusing.

Exercise 4.2 Draw such a diagram and see for yourself.

Exercise 4.3 Two points A and B have coordinates (x₁, y₁, z₁), (x₂, y₂, z₂) relative to a given reference frame. Prove that the coordinates of the mid-point, M, of AB are: . [Hint: Merely consider the coordinates of the position vectors in the vectorial form of the mid-point formula in Equation (3.1).]

Exercise 4.4 In Exercise 4.3, if P is a point on AB such that AP/PB = m₁/m₂, show that the coordinates of P are:

Exercise 4.5 Work out Exercises 3.5 and 3.7 using coordinates throughout. Note that, in effect, one does the same thing three times, once for the x-coordinates, once for the y, and once for the z.

Exercise 4.6 Prove, first in terms of vectors and then in terms of coordinates, that the lines joining the vertices of any tetrahedron to the points of intersection of the medians of the opposite faces intersect at a point that divides each of the lines in the ratio 3:1.

Exercise 4.7 Show that, as t changes, the point with coordinates (ax + tcx, ay + tcy, az + tcz) moves along a straight line, the a’s and c’s being constants. [Use Equation (3.7).]

When the reference frame is defined by a unit orthogonal triad of vectors, the corresponding coordinates are called rectangular Cartesian coodinates. [I suppose that, strictly speaking, this term should apply also to the case in which ex, ey, ez are mutually perpendicular but do not have unit magnitudes or even equal magnitudes. But we shall use it only for the above coordinates.] In two dimensions, the corresponding graph paper is of the familiar sort.

Exercise 4.8 Show that, in rectangular Cartesian coordinates, the distance from the origin of a point with coordinates (x, y, z) is .

Exercise 4.9 Points A and B have rectangular Cartesian coordinates (x₁, y₁, z₁), (x₂, y₂, z₂). Show that the vector represented by has components (x₂ – x₁, y₂ – y₁, z₂ – z₁) and thus that the distance AB is .

5 DIRECTION COSINES

In Figure 5.1 the lines OX, OY, and OZ are mutually perpendicular and lie along the positive directions of the vectors (not shown) i, j, and k of a unit orthogonal triad. These three lines are called coordinate axes. [They extend indefinitely on both sides of O, of course. The diagram shows only a

Figure 5.1

portion of each coordinate axis.] The line segment OP is a diagonal of the rectangular parallelepiped shown. There are many right angles in the diagram that are not readily recognized as such.

Exercise 5.1 Which pairs of lines meeting at the point D are at right angles? [There are four such pairs.] Also, which pairs of lines meeting at the point A are at right angles? [There are four—and that is without counting AX and AO as distinct lines; otherwise the count would be seven.]

We wish to specify in a convenient way the direction of OP relative to i, j, and k. There are various possible ways. For example, we could state the angle through which the plane OPC has swung about OC from the plane AOC, and the angle in the plane OPC that the line OP makes with the line OC; that is, we could state the two angles AOD and COP. Alternatively, we could use other appropriate pairs of angles. In certain cases describing the direction in terms of two angles is both convenient and useful. Indeed, the two angles AOD and COP are those used in what are called spherical polar coordinates.

There is a different method of specifying the direction of OP. It has the advantage of being symmetrical. But it uses three angles, and these angles, when first encountered, seem about as awkward a trio as one could imagine. They are the three angles AOP, BOP, and COP, that OP makes with the three axes OX, OY, and OZ. These angles, denoted respectively by α, β, and γ, do not lie in one plane. Moreover, since any two of them suffice to determine the direction of OP, the three can not be independent of one another and, therefore, there must be a relation between them.

Exercise 5.2 Is it strictly correct to say that any two of these angles suffice to determine the direction of OP? [For a given α, OP will be on a certain cone. For given β, it will be on another cone. Consider how these cones intersect.]

We may by now be wondering what merits the angles α, β, and y can have that will outweigh the above disadvantages. But this is because we have, in a sense, told the story backwards. Mathematicians did not arbitrarily pick three queer angles to do the work of two seemingly more sensible ones. Let us look at the situation from a different point of view.

Consider OP as a vector V with components (Vx, Vv, Vz) relative to i, j, and k. Then if we know the components, we know the vector, and therefore, in particular, its direction. But if h is any number (greater than zero) the vector hV has the same direction as V. Let us specify the direction, then, by means of the components of a vector of unit magnitude lying along V. This unit vector is (1/V) V, where V is the magnitude of V:

(5.1)

The components of the unit vector, denoted by /, m, and n, are given by

(5.2)

and we easily see that

(5.3)

so that any two of the quantities l, m, and n automatically determine the third, to within a sign.

Having thus come upon these quantities l, m, and n, we now ask what they look like on a diagram. Let us look at triangle POA in Figure 5.1. Since OA is perpendicular to the plane PDA, it is perpendicular to every line in that plane, and therefore to the line AP. So despite appearances to the contrary, angle OAP is a right angle. In the right triangle OAP, the length of OA is Vx and the length of the hypotenuse OP is V. Therefore Vx/V, which is just l, is the cosine of the angle AOP that we have called α.

The coordinate planes OYZ, OZX, and OXY (called respectively the yz-plane, the zx-plane, and the xy-plane) separate the whole three-dimensional space into eight regions called octants. We have here discussed only the case in which OP points into what is called the first octant. When OP points into other octants, some or all of the angles AOP, BOP, and COP are obtuse and the corresponding cosines are negative.

Exercise 5.3 Prove that angle OBP is a right angle, and that m, which is Vy/V, equals cos β. Prove also that n = cos y.

We see, then, that

(5.4)

and from Equation (5.3) we see that

(5.5)

The angles α, β, and γ that the line OP makes with the coordinate axes are called the direction angles of OP with respect to the reference frame; and the cosines of these angles are called its direction cosines. In practice, one works much more with the direction cosines than directly with the direction angles.

Though the line segments OP and PO are the same, the directions OP and PO are opposite, and if the direction cosines of the direction OP are l, m, and n, those of the direction PO are – l, – m, and – n. When we wish to stress that we are thinking of a given line (or line segment) as pointing in one rather than the other of the two directions associated with it, we call it a directed line (or line segment). Thus the x-axis, in its role as a coordinate axis, is a directed line rather than just a line, though we often think of it as just a line —as when we say that a point is 5 units away from it.

If a directed line or line segment does not pass through O, its direction can still be given in terms of direction cosines, since the direction is the same as that of a parallel directed line that does pass through O. Let points A and B have position vectors ra and rb relative to i, j, and k, and let the coordinates of these points be (xa, ya, za), (xb, yb, zb). Then the displacement has components (xb – xa, yb – ya, zb – za), and therefore its direction cosines are

(5.6)

where

(5.7)

Exercise 5.4 What are the direction cosines of i? Work this out (a) by means of a diagram, and (b) purely algebraically, by means of Equation (5.2).

Exercise 5.5 If OP has direction cosines such that l and m are positive but n is negative, into which octant does it point? What if l and m were negative and n positive?

Exercise 5.6 A line lies in the plane of i and j, and makes equal angles with the lines containing i and j. What are its three direction cosines? Express them in terms of α only [There are four possible cases —two pairs of opposite directions. Had we spoken of vectors instead of lines, there would have been only two cases. Why?]

Exercise 5.7 A line makes equal angles with i, j, and k. What are its direction cosines if all are positive?

Exercise 5.8 A vector of magnitude V has direction cosines l, m, and n. What are its components? [Ans. (lV, mV, nV).]

Exercise 5.9 If two of the direction cosines of a vector are cos θ cos ϕ and cos θ sin ϕ, what is the third direction cosine? [Ans. ± sin θ.]

Exercise 5.10 If the direction cosines of a line are positive and in the ratio 2: 3: 4, what are their actual values? [By Equation (5.3), the sum of the squares of the three direction cosines must be unity. But 2² + 3² + 4² = 29. So we divide 2, 3, and 4 by √29 to obtain the direction cosines 2/√29, 3/√29, and 4/√29. Three numbers that are proportional to direction cosines l, m, and n are called direction numbers.]

Exercise 5.11 A vector has direction numbers 1, 2, and 4. What are its direction cosines? [Two possibilities.]

6. ORTHOGONAL PROJECTIONS

We can get an idea of the usefulness and convenience of direction cosines by deriving some fundamental formulas of analytical geometry. We shall later derive these formulas by more compact vectorial methods, but the latter tend to obscure what is going on behind the scenes.

In deriving these formulas by means of direction cosines, we must make use of the idea of an orthogonal projection. Given a point P and a plane π, drop a perpendicular from P to π to meet π at the point P′. Then P′ is called the orthogonal projection of P on the plane π. Given point P and a line λ, the foot, P′, of the perpendicular from P to λ is called the orthogonal projection of P on the line λ.

Figure 6.1

There are other types of projection, but it will be safe in this book for us to drop the word “orthogonal” when speaking of orthogonal projections. The idea of such projections seems so simple that one wonders how it could possibly be worth considering. Yet it is actually a powerful mathematical concept, as we shall see.

If P traces out a curve, its projection, P′, on a plane π traces out a curve called the projection on π of the original curve. If the curve is a plane curve and closed, the area enclosed by its projection on π is called the projection on π of the area enclosed by the original curve.

Exercise 6.1 What is the shape of the projection on π of a straight line? What of a triangle? What of a circle? [Ans. A straight line or a point; a triangle or a line segment; a circle, or an ellipse, or a line segment.]

Exercise 6.2 A rectangle ABCD has AB lying in the plane π and BC making an angle θ with the plane. If K is the area of ABCD, show that the area of its projection on π is K cos θ.

No matter how a point P may move, its projection, P′, on a line λ. cannot move off the line. Two simple theorems form the basis of the applications of projections on a line.

Theorem 1. If a directed line segment AB of length d makes an angle θ with a directed line λ, the length and sign of its projection on λ are given by d cos θ; the sign is positive if the projection points in the same direction as λ and negative if it points in the opposite direction.

To prove this, pass planes through A and B that are perpendicular to λ and let them cut λ at A′ and B′. Then A′ and B′ are the projections of A and B whether the line AB is coplanar with λ or not. If the segment AB is moved parallel to itself, with A and B remaining on the respective planes, the projections of A and B will be unaltered. So move AB to the position A′B₁, A′ being the above-mentioned projection of A. Then from the right triangle A′B′B, we see that since cos θ its negative when θ is obtuse, the length and sign of the projection, A′B′, are given by d cos θ.

Figure 6.2

Theorem 2. If the projections of A and B on line λ are A′ and B′, and a point P starting at A moves on a zigzag line ending at B, the algebraic sum of the projections on λ of the line segments forming the zigzags is just A′B′.

This powerful theorem may seem, at first glance, difficult to prove, especially when one realizes that the zigzag line need not lie in a plane but could even jut into four dimensions, or more. Yet the proof is almost a triviality, a fact that by no means lessens the theorem’s importance. As P traces out its zigzag path from A to B, its projection, P′, moves to and fro on the line λ starting at A′ and ending at B′; and when, for example, P′ retraces to the right ground previously traced out to the left, it cancels it. The algebraic sum of the projections is therefore A′B′. And that is that.

Figure 6.3

Exercise 6.3 Let O be the origin and A the point (x, 0, 0). Let ON be a directed line segment having direction cosines l, m, and n. Show that the length and sign of the projection on ON of the directed line segment OA are given by lx whatever the signs of l and x. Also, if K is the point (x, y, 0), show that the length and sign of the projection on ON of the directed line segment AK are similarly given by my. [Use Theorem 1. Note how conveniently direction cosines enter.]

To apply these two theorems, we first consider a plane such that ON, the perpendicular to it from the origin, has length p and direction cosines /, m, and n. Let any point P on the plane have rectangular Cartesian coordinates (x, y, z). We wish to find an equation that x, y, and z must satisfy—an equation of the plane, as it is called. Drop a perpendicular from P to the xy-plane meeting it at K. Draw KA parallel to the y-axis to meet the x-axis at A. Then OA = x, AK = y, and KP = z. If and only if P lies in the plane, PN will be perpendicular to ON. Therefore:

projection of OP on ON = ON.

Figure 6.4

Using Theorem 2, we replace OP by the zigzag OAKP. Then:

(6.1)

Applying Exercise 6.3, and remembering that the length of ON is p, we have immediately

(6.2)

which is the equation we sought.

Exercise 6.4 How far from the origin is the plane whose equation is 3x + 2y – 6z = 63?

Solution The first step is to resist the temptation to say, by comparison with Equation (6.2), that l = 3, m = 2, n = 6 and p = 63. Obviously something is wrong with such a method, since we could have rewritten the given equation as, say, 300x + 200y – 600z = 6300, and thus have come to the conclusion that l is not 3 but 300, and similarly for m, n, and p. We have to remember that l² + m² + n² = 1. The numbers 3, 2, and – 6 are not direction cosines but direction numbers (see Exercise 5.10). Since = √49 = 7, we divide the given equation by 7 to obtain 9. We may now make the identifications , and therefore, also, p = 9.

Exercise 6.5 Like Exercise 6.4 for the planes x – 2y + 2z = 108, x + y + z = 9, and 3y + 4z = 100.

Consider next the problem of finding a formula for the angle between two directed lines having direction cosines l, m, n and l′, m′, n′ respectively. [If two lines do not intersect, the angle between them is defined as the angle between lines parallel to them that do intersect. Actually there are infinitely many angles, both positive and negative, between two given lines. When we talk of “the angle” between them, we presumably have some specific one in mind. Here, as on previous occasions, we mean the smallest positive angle between the positive directions of the lines regarded as directed lines.]

Denote the angle between the two lines by θ, and for convenience (though it is not really necessary), imagine the lines emanating from the origin. Take points P and P′ on the lines such that OP and OP’ are of unit length. Then, by Theorem 1, the length and sign of the projection of OP on OP′ will be given by just cos θ. But we can also compute the length and sign of the projection by using the zigzag path OAKP instead of OP. Since OP is of unit length, the coordinates of p are just (l, m, n), so that OA = l, AK = m, and KP = n. These line segments, being parallel to the coordinate axes, make with OP′ angles whose cosines are respectively l′, m′, and n′. Therefore the algebraic sum of the lengths of their projections on OP′ is ll′ + mm′ + nn′. So we must have:

(6.3)

Figure 6.5

Exercise 6.6 In the above, did we use the fact that OP′ was of unit length? Derive Equation (6.3) by projecting OP′ onto OP.

Exercise 6.7 Derive Equation (6.3) by projecting OP onto OP′, but taking OP to be of arbitrary length C.

Exercise 6.8 Find the cosine of the angle between two lines with direction numbers 1, 3, 5 and 2, – 4, 3.

Exercise 6.9 Prove that two lines are perpendicular if their direction cosines satisfy

(6.4)

Exercise 6.10 Find the cosine of the angle between the planes x + 3y + 5z = 10 and 2x – 4y + 3z = 15. [The angle is the same as that between the normals to the planes, and these normals have direction numbers 1, 3, 5 and 2, – 4, 3 respectively.]

Exercise 6.11 The xy-plane has an equation z = 0, and the yz-plane x = 0. Using Equation (6.4), verify that these planes are perpendicular.

Exercise 6.12 What is the distance between the two parallel planes x + 2y + z = 2 and x + 2y + z = 8?

7. PROJECTIONS OF AREAS

In Exercise 6.2 we found that if rectangle ABCD makes an angle θ with a plane π through AB, the area of its projection is cos θ times the area of ABCD. This is strongly reminiscent of the way projections of line segments behave, as given in Theorem

Figure 7.1

1. And the result is not confined to rectangles having one side in the plane of projection. It is not even confined to rectangles.

If we have a plane area of quite irregular shape, as in Figure 7.2, we can crisscross it with lines to break it up into a lot of rectangles—with some left-overs of irregular shape next to the boundary. Ignore the left-overs for the moment and consider the resulting serrated-edged region. Since the area of the projection of its constituent rectangles is cos θ times the area of the original, the total area of the projection is cos θ times that of the original serrated-edged region. So, for this region, the area of the projection is cos θ times that of the original. The finer the crisscross mesh, the more nearly the serrated-edged region fills the over-all region, and it will hardly come as a surpise that, by methods belonging to the calculus, one can prove that the result actually holds for the over-all region too. Thus we have the important theorem that if a plane region of any shape has an area K, the area of its projection on a plane making an angle θ with it is K cos θ. We shall use this result in a later chapter.

Figure 7.2

Exercise 7.1 In Figure 7.1, take two points M and L on the line DC such that ML = DC. Show that the area of the parallelogram ABLM is equal to that of the rectangle ABCD. Hence, show that the area of the projection of this parallelogram is cos θ times the area of the parallelogram.

Exercise 7.2 A circular disc of radius 5 inches touches the ground, which is horizontal, and lies in a plane that is inclined at 45° to the ground. If the sun is vertically overhead, what is the area of the shadow of the disc, what is the time of day, and where on earth can the disc be?

4

SCALARS. SCALAR PRODUCTS

1. UNITS AND SCALARS

The components of a vector V depend on the frame of reference. They change when we go from one frame to another. For example, if a displacement V lies in the plane of ex and ev as indicated on the left of Figure 1.1, its components (Vx, Vy, Vz) are such that Vx, Vy ≠ 0, Vz = 0. If we go over to a new reference frame in which and are the same as ex and ez but lies along V, as on the right of Figure 1.1, then the new components () are such that , and . Even when we merely change the magnitudes of the base vectors ex, ey, and ex without altering their directions, the components of V are altered—unless V is the zero vector, of course. Nevertheless, no matter how we change the reference system, the vector itself remains the same [remember, here, the discussion we had earlier of the master reference frame], and in particular the magnitude of V remains the same.

Figure 1.1

Exercises 1.1 A displacement V points along the ex direction and has magnitude V. Its components are therefore (V/ex, 0, 0), where ex is the magnitude of ex. What will its components be if we go over to a new reference frame in which has twice the magnitude of ex? [Ans. (V/2ex, 0, 0).]

We must now make a subtle distinction. Suppose that the displacement V has a magnitude of 108 inches, that the base vectors ex, ey, and ez have each a magnitude of 1 inch, and that the components of V are (36, 24, 72). If we go over to a new reference frame in which the base vectors , and differ from ex, ey, and ez only in that their magnitudes are 12 inches instead of 1 inch, the components of V change to one-twelfth of their former values, i. e., to (3, 2, 6). Nevertheless the magnitude of V is still 108 inches. Its numerical value, 108, has not changed. Suppose, though, that we go back to the original reference frame and without altering the base vectors or the vector V, we use a different scale of measurement for all lengths, going from inches to feet. Then the components of V will still be (36, 24, 72) since they are the ratios of the magnitudes of the component vectors of V to the magnitudes of the e’s and thus are independent of the scale of measurement. But, though the magnitude of V is the same as before, its numerical value has changed from 108 to 9, because we are now measuring the magnitude in feet instead of inches.

Exercise 1.2 What will the components of V be relative to the foregoing and , when we go over from measuring in inches to measuring in feet? [Ans. (3, 2, 6), just as they were when we used inches in this reference frame.]

Exercise 1.3 A force F has a magnitude of 6 1bs., and is represented by an arrow-headed line segment of length 6 inches. Its components relative to ex, ey, and e, are (4, 1, 3). If the reference frame is changed to one with new base vectors , and where and , what will be the new components of F? [Ans. ( ).]

If the components of the displacement V are pure numbers, how can Equation (2.5) of chapter 3, namely,

give us a length? How can it give us anything but a pure number? It is a standard equation, found in one guise or another in practically all textbooks on vectors. Let us consider it more carefully.

The fact is that we glossed over the question of units in deriving it. The legerdemain occurs in Equation (2.3) of Chapter 3. To keep the wording simple, let us discuss here the case in which all quantities involved are positive. If OA means the length of the line segment OA, then it is not a pure number and so can not be equal to Vx. It is really equal to Vxi, where i is the magnitude of i and thus a length. Since i has the numerical value 1, it can be omitted—provided we realize that it is really there. Equation (2.5) of Chapter 3 can thus be regarded as yielding a length after all, though this would have been clearer had we written it as

which corresponds (except for the taking of the square root) to Equation (2.1) of Chapter 3 for the case of a unit orthogonal triad. It was this sort of thing, among others, that we had in mind when we said there that the price of using such triads is loss of generality and loss of depth of understanding.

For a displacement V, which has a magnitude that is a length, the problem of units is relatively straightforward. What of a different sort of vector, such as a force F? The remark in Exercise 1.3 that the 6 lbs. is represented by 6 inches is less innocent than it seems. Forces combine by the parallelogram law with forces, not with displacements, whether the latter are thought of as localized or not. Therefore, a force cannot be equivalent to the vectorial sum of three such displacements. Strictly speaking, instead of writing an equation like

(1.1)

and thinking of the e’s as geometrical vectors having magnitudes that are lengths, we should write something like

(1.2)

where F and the f’s are all forces, and the components Fx, Fy, and Fz are pure numbers. If we did this consistently with all types of vectors, the components would always be dimensionless numbers. But when working with vectors we like to draw arrow-headed line segments and parallelograms, and it is difficult to resist the temptation to think of the arrow-headed line segments as lengths. [We can hardly formulate the parallelogram law without thinking geometrically in this way. Compare page 9, where we explained the sense in which forces are vectors.] So we usually express F as in Equation (1.1) of this chapter. But, to clarify matters, let us here make a distinction between F, the force, and its geometrical counterpart, which we shall denote by . This has a magnitude that is a length; for example, each pound of force could correspond to an inch of . Then if we write

(1.3)

the components x, y, and z are pure numbers and are, in fact, the same as the Fx, Fy, and Fz in Equation (1.2)—but not the same as those in Equation (1.1). For we can go from Equation (1.2) to Equation (1.1) by changing all lbs. into inches. But when we write Equation (1.1), we are apt to think of F as in 1bs., not inches. In this case we have to say that the components Fx, Fy, and Fz in Equation (1.1) are not pure numbers, but have the dimensions of a force divided by a length. When we then calculate the magnitude of F in terms of its components, we have to remember to bring in the magnitudes of the base vectors, the quantities xex, etc. being thus forces. Similar considerations apply to other vector quantities such as velocities, accelerations, and others we shall come upon later.

Exercise 1.4 Force F has a magnitude of 6 1bs. and is represented by an arrow-headed line segment of length 6 inches. Its components Fx, Fy, and Fz relative to ex, ey, and ez are (4, 1, 3). (a) What are its components relative to corresponding vectors fx, fy, and fz with magnitudes that are in lbs? (b) What are the components, x, y, and z, of relative to ex, ey, and ez? [Ans. (4, 1, 3) in each case.]

Exercise 1.5 In Exercise 1.4, let the unit of force be changed from the 1b. to the oz. (but with each 1b. represented by 1 inch as before). What are the answers to (a) and (b) in this case? [Ans. The same as before. Which components do change?]

Exercise 1.6 In Exercise 1.4, if we changed from one inch per lb. to one inch per oz., what would happen to the three sets of components mentioned therein? [Ans. Those in (a) and (b) would be unaltered. Those of F relative to ex, ey, and ez would become (64, 16, 48).]

If a force has a magnitude of 6 lbs. and we change the unit of force from lbs. to oz., the numerical value of its magnitude changes to 96.

A magnitude that is objective in the sense that it does not change simply because we change the reference frame, though it may change its numerical measure when we change the scale of measurement, is called a scalar. Thus the magnitude of a vector is a scalar. But a scalar need not be associated with a vector. For example, the number of fingers on one’s hand is a scalar. So are the mass of a body and the time interval between two events, according to Newtonian physics. A scalar need not be constant. For example, the height of a growing child is a scalar.

Exercise 1.7 Name various other scalars. [Do not be content with “the number of fingers on two hands,” “the number of fingers on three hands if one finger has been chopped off,” and the like. Give various types of scalars including variable ones.]

Exercise 1.8 Could the number of fingers on one’s hand change because of a change of units? [This would make an excellent true-false question, because the answer is both yes and no. It depends on what we mean by number. If we mean that we can put the fingers in one-to-one correspondence with five pebbles, say, then the number will not change. But we can change the unit of counting: for example, we can count in dozens, in which case the number of fingers becomes 5/12. Thus, even a dimensionless number can be changed by a change of unit.]

Exercise 1.9 Is the angle between two vectors having the same starting point a scalar? Is the area of the parallelogram defined by the two vectors a scalar?

Not everything that has magnitude is a scalar. For instance, the components of a vector are not scalars since they change when the reference system is changed. [But the “components” relative to a particular ex, ey, ez are scalars since these “components” are really related not to a reference frame regarded as such but to a particular set of fixed vectors ex, ey, ez.] The number of vertices of a given triangle, namely 3, is a scalar, but the number of vertices lying in coordinate planes is not.

Exercise 1.10 Why not?

Note how long it has taken us to get to a place where we could try to define a scalar satisfactorily. Even so we have cheated. In Chapter 1 we pointed out the difficulty of giving satisfactory definitions that are brief. Have you noticed that though we have used the word magnitude freely, we have never defined it? We have no intention of doing so. But you might like to try for yourself.

Exercise 1.11 Define magnitude. [You may come up with a definition along these lines: magnitude is something that can be represented by a single number. This is not satisfactory, though, for two opposite reasons. One is that we may need different numbers for the same magnitude in different units, so that the word “single” might need qualification despite the word “can.” The other is that, by means of a code, we can represent pretty well anything by a single number. Anyway, what about such a thing as your telephone number? Perhaps we have to insist that numbers representing magnitudes of the same sort shall combine according to the usual arithmetical laws of addition. This sounds promising. It would exclude catalog numbers, telephone numbers, and other code numbers. But suppose we chose to measure weights by the logarithms of the numbers of pounds in them. Then, since for example log 2 + log 5 = log 2 × 5 = log 10, which is not log 7, would this fact mean that we could not regard the weight of a piece of matter as a magnitude? One would prefer that it did not. Evidently some rewording is necessary. Keep at it —if you are so inclined. And when you have overcome this particular difficulty, ask yourself whether temperature is a magnitude.]

Exercise 1.12 Is the magnitude of a position vector a scalar? [This is not a question to be answered with a simple yes or no. It has facets. Think it through.]

Given a vector V of magnitude V, we can form a new vector by multiplying the magnitude by a scalar, say h, and keeping the same direction and location. This new vector is denoted by hV, and we say that it is the result of multiplying the vector V by the scalar h. We have already discussed this for the case in which h is a dimensionless number, and we have seen instances in which h was not. In the latter case, hV cannot combine with V according to the parallelogram law since it is not of the same sort as V. For example, if a is the acceleration vector of a particle of mass m, then ma is not an acceleration vector; by Newton’s second law of motion, we find that it actually combines with forces (see Equation (6.1) of Chapter 2).

2. SCALAR PRODUCTS

We have seen how the arithmetical symbols + and – can be used to represent the combination of vectors according to the parallelogram law. Having seen in this way how we can talk of adding and subtracting vectors, we naturally ask ourselves how we can multiply and divide vectors. But actually there is no reason—except perhaps misguided optimism—why we should expect to be able to do so usefully. For example, though we can “add” colors (as on a color television screen) or “subtract” them (as in amateur color photography), it does not occur to us to ask how we could multiply or divide colors by colors. Again, we can add and subtract money; and if we are astute and lucky, we can multiply money—but by numbers, not by money. If we start with $ 100 and triple our money, we end up with $ 300 which is $ 100 multiplied by 3, not multiplied by $ 3. We could contemplate the idea of multiplying $ 100 by $ 3 and getting 300 “square dollars,” but even apart from the name we would not feel that the idea was likely to be particularly useful. What right have we, then, to expect that there is even any meaning to “U times V” or “U divided by V”?

Let us see what we can think of that might be somehow worthy of the phrase “U times V.” Given two vectors U and V with magnitudes U and V, we could form the scalar UV which would be a sort of product of the vectors U and V. But it is really just the product of the two scalars U and V, and it does not involve the directions of the vectors at all.

We can bring in the directions to some extent by considering the angle between U and V. Let us call it θ. The angle θ does not tell us everything about the directions of U and V. It tells us only one quarter of the total amount of information about the directions, since two angles are needed to specify each direction; thus, θ is one angle out of a needed four. It has the advantage, though, of being a scalar.

Using θ as well as the magnitudes U and V, we can form all kinds of scalars that involve the product of the magnitudes. For example, UVθ, UVθ⁵, UV sin θ, UV cos θ, UV tan θ, UV sec θ, UV sin³ θ cos θ, UV sin θ/θ², and so on. Out of all these possibilities, there is one that has, shall we say, a pleasing geometrical significance: the quantity UV sin θ gives the area of the parallelogram defined by U and V, as is easily seen. We might think, therefore, that this particular scalar quantity formed by multiplying various scalars connected with the vectors U and V would so merit our attention as to be worthy of some such title as the “scalar product” of the vectors U and V, a title that graces the name of this chapter. Why do we shrink from so regarding it? One reason, and it is a good one, is that we happen to know (as perhaps you do too) that the term “scalar product” is used for a different combination of U, V, and θ. But that is hardly an enlightening reason. Let us see what makes us reluctant to give UV sin θ the particular honor of the title “scalar product.”

The trouble with UV sin θ comes to light when we consider the parallelogram law. Suppose vectors A, B, and C are such that C is the resultant of A and B. We write this fact in the form

(2.1)

and we say that C is equivalent to A and B combined. Now if we really mean the word “equivalent,” we will want the “scalar product” of a vector V with C to be the same as the scalar product of V with A + B. We can always ensure this by saying that A + B is just another way of saying C. But we would prefer to be able to say the more stringent, and more significant thing that the “scalar product” of V with the resultant (A + B) is the same as that of V with A plus that of V with B; that is, we would like the “scalar product” to obey the distributive law.

It is easy to see that UV sin θ does not obey this law. Take, for simplicity, a vector V that is perpendicular to A, B, and C, as in Figure 2.1. Then, unless A and B have the same direction, the area of the rectangle formed by V and C is clearly smaller than the sum of the areas of the rectangle formed by V and A and the rectangle formed by V and B. This is because the magnitude of C is smaller than the sum of the magnitudes of A and B. Naturally, we are disappointed. The quantity UV sin θ has such a pleasant geometrical significance that we are reluctant to see it fail our test. But it is not lost to us. We shall find a way to make it pass the test—at a price. That, however, comes later.

Figure 2.1

Meanwhile the foregoing gives us an important hint. Though the magnitudes of A and B do not add up to the magnitude of C, we recall Theorem 2 about projections, and we realize that it tells us, in effect, that the sum of the projections of A and B—on any line—is equal to the projection of C on that line. Let us see how we can profit from this fact. Take a vector V and project the vectors A, B, and C onto it. Then:

projection of C = projection of A + projection of B.

If θAV, θBV, and θCV are the angles that A, B, and C make with V, this can be written, by Theorem 1, as

(2.2)

Therefore, if we define the scalar product of V with U as U cos θUV, and denote it by V·U, we see that Equation (2.2) yields:

V·C = V · (A + B) = V·A + V·B.

This is good as far as it goes, but such a quantity V·U would not be the same as U·V: the former would be U cos θ, the latter V cos θ, where θ is the angle between U and V. There is no reason why we should insist that V·U = U·V, but it happens that there is a simple way of making this relation true, so we might as well make the most of our opportunity. We modify the above by regarding U·V not as U cos θ but as UV cos θ. Let us make this official:

“Let U and V be two vectors with magnitudes U and V, the angle between the vectors being θ. The scalar UV cos θ its denoted by the symbol U·V and is called the scalar product, or the dot product, of U and V.”

Thus,

(2.3)

In particular, we see that

(2.4)

Exercise 2.1 Is (A · B) ·C = A · (B·C)? [Ans. In a nonsensical sense, yes, since neither has any meaning. Why not?]

Exercise 2.2 Prove that

(2.5)

Exercise 2.3 Prove that if C = A + B, then V·C = V·A + V·B, i. e., that

(2.6)

Exercise 2.4 What is the value of U·V if U and V are perpendicular? What if they have the same direction?

Exercise 2.5 Prove that if two opposite edges of a tetrahedron are perpendicular, and two other opposite edges are perpendicular, then the remaining two are also perpendicular. [Denote the vertices by P, Q, R, and S. Then we are given that , and , say, and we wish to prove that We express everything in terms of vectors emanating from P. Thus , etc. So, given and , we wish to prove that = 0, but this is an immediate consequence, as you can see by adding the two preceding relations.]

As happens often, though not always, that which is mathematically elegant turns out to be that which is important in applications. In the field of physics, consider the work done by a constant force F with magnitude F lbs. If it acts on a particle that moves through a distance D ft. in its own direction, it does an amount of work FD ft. lbs. Suppose that the displacement of the particle, D, is not along the line of action of the force. Then we can resolve the force F into two forces F// and F⊥, the former lying along D and the latter perpendicular to it. The particle does not move in the direction of the force F⊥ at all, and thus F⊥ does no work during the displacement. The force F// lies along the displacement of the particle and since the magnitude of F// is F cos θ, it does an amount of work FD cos θ ft. lbs., which is thus the amount of work done by F. We see, then, that the work done is none other than the scalar product of F and D:

(2.7)

Figure 2.2

The above probably sounds like a convincing and enlightening argument. It may be enlightening, but it is not really convincing. It assumes, for example, that the work done by F can be found by adding the work done by F⊥ and the work done by F//, merely because F = F⊥ + F//; it also assumes that the work done by F⊥ is zero because the particle does not move in the direction of F⊥ at all.

Actually, the above argument is a valid consequence of the definition in Equation (2.7) of the work done by a constant force. It is not a derivation of that equation. From Equation (2.7) we can deduce the various assumptions made in the above argument. Without using calculus, it is difficult to show why physicists should be sufficiently interested in the quantity F·D to give it a name. The fact is that, by making certain manipulations on the Newtonian equation of motion of a particle, namely F = ma, we can derive from it, for the case of a constant force, the scalar equation,

where v and v₀ are the magnitudes of the velocities of the particle at the end and the beginning of the displacement. [A more general equation, involving integration, results when F is not constant.] Because this equation is a mathematically simple consequence of the equation of motion and has various pleasant properties, it takes on important physical significance. Accordingly, physicists give names to the quantities therein. They call the kinetic energy of the particle, and, as we have already said, F·D the work done by F in the displacement D. This enables them to state the equation in words: the work done by the force is equal to the change it produces in the kinetic energy of the particle.

Exercise 2.6 A constant force F of magnitude F makes 45° with the x-axis. If it acts on a particle moving along the x-axis through a distance D, what is the work done?

Exercise 2.7 Prove that if, in Exercise 2.6, F had been perpendicular to the x-axis, the work done would have been zero.

Exercise 2.8 Show that if two constant forces F₁ and F₂ act on a particle, the work done by their resultant is the sum of the work done by F₁ and the work done by F₂.

Exercise 2.9 A constant force having components (Fx, 0, 0) acts on a particle that undergoes a displacement with components (Dx, 0, 0), all components being relative to i, j, and k. What is the work done by the force?

Exercise 2.10 The same as Exercise 2.9, but with the force having components (Fx, Fy, Fz). [Ans. FxDx.]

Exercise 2.11 Show that the work done by a constant force F acting on a particle that moves through a displacement D can be regarded as the product of the magnitude of F and the length of the projection of D on the line of action of F.

3. SCALAR PRODUCTS AND UNIT ORTHOGONAL TRIADS

Using Equations (2.6) and (2.5), we have

and by repeated application of the same maneuvers, we see that a dot product of the type (A + B + . . .)·(U + V + ...) is equal to the sum of all the individual dot products of each member of the first parenthesis with each member of the second.

Now consider two vectors U and V expressed in terms of their components relative to a unit orthogonal triad i, j, and k:

(3.1)

(3.2)

How do we find the value of U·V in terms of the components? We could draw a diagram and work out the projection of V on U using a zigzag path. But there is an algebraic method that gives the result without any appeal to zigzag paths or even to a diagram, though, in fact, it is basically the same method.

We begin by noting that, from the definition of the scalar product, since the vectors i, j, and k have unit magnitude,

(3.3)

We now write,

U·V = (Uxi + Uyj + Uzk)·(Vxi + Vyj + Vzk),

and use the distributive law just discussed. There will be nine individual terms in the distributed product on the right, since each of the first three must multiply each of the second three. Of these nine terms, six will vanish because of the relations (3.3): for example, the term (Uxi)·(Vyj) equals Ux Vyi·j and this vanishes because i·j = 0. Typical of the remaining three terms is (Uxi)· (Vxi) which equals Ux Vxi·i, or just Ux Vx since i·i = 1. So we find in the end the neat-looking and powerful formula:

(3.4)

Exercise 3.1 In addition to the distributive law, we used the fact that (Uxi)·(Vxi) = Ux Vxi·i, etc. Note that this is obviously valid.

In particular, using Equation (2.4), we see from this that

(3.5)

a result we already knew. Also, if U and V are perpendicular, cos θ will be zero and therefore U·V will be zero so that we have this condition for perpendicularity:

(3.6)

Exercise 3.2 Find the value of U·V if U = 2i + 3j – k and V = 3i + 4j + 2k.

Exercise 3.3 Find the value of U·V if U = 4i + 2j – k and V = 3i – 5j + 2k, and so show that U and V are perpendicular. [Imagine trying to prove this without the aid of Equation (3.4).]

Exercise 3.4 Do Exercises (2.9) and (2.10) using Equation (3.4).

Exercise 3.5 For what value of x will 3i – 2j + 3k and xi – 2j + 2k be perpendicular?

Exercise 3.6 In Exercise 3.2, find the values of U and V, and thus, using Equation (2.3), show that for these vectors, .

Exercise 3.7 Show that when U and V are vectors of unit magnitude, Equation (3.4) is just cos θ = ll′ + mm′ + nn′ in disguise. Then compare this method of derivation with the previous zigzag method and see if you can trace out the underlying identity of the two methods.

From Equations (2.3) and (3.4) we see that

so that

(3.7)

Compare this with Exercise (3.7).

Exercise 3.8 Using Equations (3.5) and (3.7), find cos θ for the vectors 2i + j + k and 3i + 2j + 2k. [Ans. .]

Exercise 3.9 Find the angle (not just its cosine) between j + k and i + k, and draw a diagram showing the two vectors in relation to i, j, and k. [Note the bearing of this exercise on the following one.]

Exercise 3.10 From a vertex, O, of a cube, diagonals are drawn of the three square faces that meet at O. Prove that these diagonals make angles of 60° with each other. [While this may seem like an excellent demonstration of the power of the formula (3.5), the angles are easily found by ordinary geometry. Find them in that way too.]

Exercise 3.11 Prove that the vectors i, j + k, and i + j + k are coplanar, and find the cosines of the angles between them. Also draw an appropriate diagram involving a cube, and from it check the angle between i and j + k and the fact that the three vectors are coplanar. [To see without the diagram that the vectors are coplanar, note that one is the resultant of the other two.]

Exercise 3.12 Would Equation (3.4) be valid if (Ux, Uv, Uz) and (Vx, Vv, Vz) had been the components of U and V relative to a reference frame defined by general base vectors ex, ey, and e,? [This is probably the nastiest question in the book. According to all that has been said here, the answer is no, so give yourself full credit if that is your answer. However, there is more to the concept of a scalar product than we are in a position to explain here, and in a significant sense it could actually be correct to say that Equation (3.4) is valid in all reference frame.]

Let us derive Equation (6.2) of Chapter 3 by using scalar products. Here is the relevant part of the diagram:

Figure 3.1

Denote by r the position vector of a general point P on the plane through N that is perpendicular to ON. Denote the unit vector along ON by n. Then, since the length of the projection of OP on ON is p, we have

(3.8)

which is the required equation of the plane. To put it in the less compact, nonvectorial form, we note that r has components (x, y, z) and the components of the unit vector n are just the direction cosines (l, m, n) of the normal to the plane.

It follows at once that

lx + my + nz = p.

Exercise 3.13 Derive this equation by expressing the fact that PN is perpendicular to ON. [Hint. The displacement is given by therefore, the perpendicularity can be expressed as = 0. This gives . Now use the fact that and have components (lp, mp, np) and (x, y, z).]

Equation (3.7) for cos θ can be used to obtain proofs of familiar formulas in trigonometry and two-dimensional analytical geometry. By way of orientation do this exercise first.

Exercise 3.14 Prove that the vectors i + 3j and 4i + 2j make 45° with each other. [Note that these vectors are in the xy-plane and can thus be regarded as belonging to two-dimensional analytical geometry in that plane. Note, too, how powerful the cos θ formula is by contemplating the problem of trying to prove that the angle is 45° using only the methods of ordinary geometry. Draw the diagram and see for yourself.]

We shall now derive the well-known condition in two-dimensional analytical geometry that two lines be perpedicular. We recall that the slope of a line is defined as the tangent of the angle it makes with the x-axis, and that it is usually denoted by m (which, of course, is not to be confused with the m denoting a direction cosine). If a line in the xy-plane makes an angle α with the x-axis, its slope is given by m = tan α, and its direction cosines are easily seen to be cos α, sin α, and 0. Consider two lines making angles α and α′ with the x-axis, and having slopes m and m′. We have, for perpendicularity, by Equation (3.6)

cos α cos α′ + sin α sin α′ = 0.

Divide this by cos α cos α′ and we find that 1 + mm′ = 0, or

mm′ = – 1,

showing that the slopes are negative reciprocals.

Exercise 3.15 Using the scalar product, derive the well-known trigonometric formula cos (A – B) = cos A cos B + sin A sin B. [Consider the direction cosines of two lines in the xy-plane making angles A and B with the x-axis, and note that the angle between them is A – B.]

It is possible to derive the so-called law of cosines by means of the scalar product. In triangle ABC let the sides be of length a, b, and c as shown. Using displacements, we have . So,

Therefore,

5

VECTOR PRODUCTS. QUOTIENTS OF VECTORS

1. AREAS OF PARALLELOGRAMS

In the preceding chapter we noted that the quantity UV sin θ gives the area of the parallelogram defined by the vectors U and V. We were therefore tempted to regard it as a “scalar product” of these vectors. We were discouraged from doing so when we found that this product would not obey the distributive law, but that UV cos θ would. One of the purposes of this chapter is to rescue the quantity UV sin θ from the danger of vectorial oblivion.

Figure 1.1

At this stage we do not know how to do this. But let us cross our fingers and agree to denote whatever it is we are seeking by the symbol U × V (to distinguish it from the U·V discussed in the preceding chapter) and to call it the cross product of U and V.

In Figure 1.1, the vectors A, B, and C are such that

(1.1)

To make our exploratory task easier, we consider first the special case in which vector V is perpendicular to the vectors A, B, and C.

By the symbol V × A we mean somehow to refer to the area of the parallelogram defined by V and A. [In the present special case, this and other parallelograms are actually rectangles, but let us use the word parallelogram nevertheless.] Here V × A, the area of the parallelogram defined by V and A, is just VA. Similarly, the areas corresponding to the cross products V × B and V × C are VB and VC. We recall from the previous chapter that if we take V × A to mean just the area of the parallelogram defined by V and A, then V × C = V × (A + B) ≠ V × A + V × B. This is because the magnitude of A + B is rarely equal to the sum of the magnitudes of A and B; that is, despite Equation (1.1) we have, in general,

(1.2)

Exercise 1.1 What is wrong with the following argument? Since V, A, and B are scalars, we have, by ordinary algebra,

V(A + B) = VA + VB.

So if V × A stands for just VA, etc. we can write this as

V × (A + B) = V × A + V × B.

[Hint. Does the + on the left in the second equation mean the same thing as the other three +’s?]

One of the most important things in life is the ability to extract success from failure. This usually involves a seemingly contradictory combination of stubbornness and flexibility: one must be stubborn in one’s determination to reach a solution, but highly flexible in seeking it.

Our failure above gives us an idea. We have, in effect, been multiplying Equation (1.2) by V. Why not work on Equation (1.1) instead? We shall then get

(1.3)

So if we agree to look on V × A, for example, not as a scalar representing the area of the parallelogram defined by V and A but as a vector lying along A and having this area as magnitude, we shall be able to rewrite Equation (1.3) in the form

(1.4)

and apparently our troubles will be over—at least for the special case we are considering here.

But there is something unsatisfying and inelegant about representing the area of a parallelogram by a vector pointing along one of its sides. Why one side rather than the other? Why should V × A, for instance, point along A rather than along V? Is there some way by which we can avoid playing favorites? We had better try choosing a neutral direction—a direction perpendicular to both A and V. How does Equation (1.4) fare if we think of V × A as having the same magnitude as before, but as pointing in a direction perpendicular to the plane of the parallelogram? Surprising as it may seem, Equation (1.4) survives. You may think this will be hard to prove. But actually it is obvious. For a glance at Figure 1.1 shows that the new V × A, V × B, and V × C are merely the old ones turned as a rigid body through 90° about V as an axis. Since Equation (1.3) holds for the A, B, and C in Figure 1.1, it obviously has to hold for the rotated vectors. Therefore Equation (1.4) still holds.

Figure 1.2

But we have been making things easy for ourselves by considering the special case in which V is perpendicular to A, B, and C. Will our luck hold when V is not perpendicular to them? It is easy to see that it will. Consider Figure 1.2, in which, though the plane of the vectors A, B, and C is not perpendicular to V, the plane of the vectors A′, B′, and C′ is; and the lines that look parallel to V are, in fact, parallel to V. The area of the (undrawn) parallelogram defined by V and

A is equal to the area of the coplanar rectangle defined by V and A′. Therefore V × A′ = V × A. Similarly V × B′ = V × B, and V × C′ = V × C, and that is about all we need. For the upper part of Figure 1.2 apes Figure 1.1, with A′, B′, and C′ replacing the A, B, and C in Figure 1.1. So relation (1.4) holds with A′, B′, and C′ replacing A, B, and C, and therefore, by the above equalities, it holds without the primes too, even though the present A, B, and C are not perpendicular to V.

Exercise 1.2 Go through the above proof for the special case in which V, A, B, and C are all coplanar, making the necessary adjustments in the diagram. [Since V × A, V × B, V × C have the same direction, Equation (4.1) reduces to a relation between the areas of three parallelograms in a plane. These areas are given by the products of a common base V with three altitudes the sum of the lengths of two of which is easily shown to be equal to the length of the third.]

What specially pleases us here is that when we look at the way Equation (1.4) was obtained, we see that it plays a dual role. Not only does it show that the distributive law is valid, but it also shows in what way we can regard the new V × A and V × B as combining according to the parallelogram law: if C is the vectorial resultant of A and B, then V × C will be the vectorial resultant of V × A and V × B. We begin to sense that despite the change to the perpendicular direction, cross products can still be regarded as vectors. But what about the combination not of V × A and V × B but of, say, L × M and R × S, in which there is no common letter, the vectors L, M, R, and S, all starting at O? Our luck still holds. we can reduce this to the simpler case quite easily as follows. Let the plane defined by L and M cut the plane defined by R and S in the line λ. Take any nonzero vector V along λ. Then we can always find a vector A in the plane of L and M such that the area of the parallelogram defined by V and A is the same as that of the parallelogram defined by L and M. Since V and A lie in the same plane as L and M, the quantity V × A will be the same as L × M. Similarly we can replace R × S by V × B where V is the same V as before. Consequently L × M + R × S reduces to the case V × A + V × B that we have already considered.

Exercise 1.3 In the above, if L and M were i and j respectively, and V was i + j, what would A be? [Ans. It could be i. It also could be j. Or . And there are infinitely many other possibilities. The fact is that the A and B above were by no means unique.]

Exercise 1.4 Deduce from the above discussion that L × M + R × S = R × S + L × M.

By now we may be fairly well convinced that we can regard these cross products as vectors. But a minor trouble comes to mind: which way should V × A point? There are two opposite directions perpendicular to a given plane. If we are not consistent in our choice of direction, we shall lose Equation (1.4).

Exercise 1.5 Why?

We shall therefore have to adopt a special convention. Let us agree to say that U, V, and U × V, in that order, form a right-handed system. This then takes care of the problem. But it does so at a price:

(1.5)

Exercise 1.6 Prove Equation (1.5). [Note that, by the convention, V, U, and V × U, in that order, must also form a right-handed system; but the third member of a right-handed triad starting with V, U points in the opposite direction to that of a triad starting with U, V.]

We are now ready for the official definition: the cross product U × V denotes a vector having magnitude UV sin θ and pointing in a direction perpendicular to U and V in such a way as to make U, V, and U × V, in that order, a right-handed system. The cross product is often also called the vector product. [We shall have more to say about the right-handed system later.]

Perhaps you think that this standard definition is long overdue. But let us go into this matter of the vector product with our eyes open. We have indeed shown how cross products can be regarded as vectors. But, as you have doubtless uneasily realized, in so doing we have had to agree to some curious things about the way areas are to be regarded as combining. We shall have to look into this in more detail later. Meanwhile, let us learn how to work with vector products.

Exercise 1.7 Show that A × A = 0, and that A × (αA) = 0, where α is a scalar.

Exercise 1.8 If U is of magnitude 3 and points horizontally due east, and V is of magnitude 5 and points vertically upward, describe the vector U × V. [Ans. Magnitude 15, and pointing—well, let’s see—due south—I hope!]

Exercise 1.9 If U is of magnitude 8 and points horizontally due west, and V is of magnitude 7 and points horizontally 45° north of west, describe the vector U × V.

Exercise 1.10 By applying Equation (1.5) to Equation (1.4) show that

(1.6)

Exercise 1.11 Using Equations (1.4) and (1.6) show that

(A + B) × (U + V) = A × U + A × V + B × U + B × V.

and generalize this to the case in which each parenthesis contains the sum of any number of vectors.

Exercise 1.12 Triangle ABC has sides of length a, b, and c. Using vector products, prove that a/sin A = b/sin B = c/sin C. [Hint. Using displacements, we have . Take the cross product of this with , and use Equation (1.5). Then do the same with .]

Exercise 1.13 Show that (A – B)·(A + B) = A² – B² and that (A – B) × (A + B) = 2A × B, and interpret these results geometrically by means of appropriate diagrams.

2. CROSS PRODUCTS OF i, j, AND k

From the definition of a vector product, since sin 0° = 0 and sin 90° = 1, and i, j, and k are unit vectors, we have for a right-handed unit orthogonal triad the important relations

(2.1)

These relations are worth comparing with the corresponding relations for dot products given in Equation (3.3) of the preceding chapter.

Exercise 2.1 If α and β are scalars, show from the definition of the cross product that (αi) × (βi) = 0, (αj) × (βk) = αβi, (αk) × (βj) = – αβi, etc., and in general that (αU) × (βV) = αβ U × V.

Exercise 2.2 Using the relations (2.1), show that

2i × (3i + 4j + 5k) = – 10j + 8k.

Exercise 2.3 As in Exercise 2.2, find the components relative to i, j, and k of the vector (Uxi) × (Vxi + Vyj + Vzk).

Exercise 2.4 Like Exercise 2.3 for (Uxi + Uyj) × (Vxi + Vyj). Note that the result is perpendicular to i and j, and explain why this was to have been expected. [Note that, for example, Uxi + Uyj lies in the same plane as i and j.]

Exercise 2.5 Using the scalar product, find the cosine of the angle between i and i + j + k. Then, using the vector product, find the sine of this angle. Check that the sum of their squares comes to unity. [To find the sine, calculate the magnitudes of i and i + j + k and multiply by the unknown sin θ. The result must be equal to the magnitude of the cross product i × (i + j + k). The sine comes to and the cosine to 1/√3.]

Exercise 2.6 Like Exercise 2.5 for i + j and i + j + k.

Exercise 2.7 By drawing a cube and noting two congruent triangles, explain why the cosine in Exercise 2.5 equals the sine in Exercise 2.6.

3. COMPONENTS OF CROSS PRODUCTS RELATIVE TO i, j, AND k

If

(3.1)

(3.2)

we can find the components of U × V by much the same method we used for finding U·V under similar circumstances. We have

(3.3)

Of the nine terms that arise from the product on the right, three are zero, namely UxVxi × i, UyVyj × j, and UzVzk × k, because of Equations (2.1). The other six terms fall naturally into three pairs. For example, Uy Vzj × k and UzVyk × j belong together because j × k = – k × j = i. In fact, these two terms may be combined as (UyVz – UzVy)i. Two other pairs may be similarly treated, and we ultimately find that

(3.4)

Exercise 3.1 Check that the other two pairs do yield the two terms written in Equation (3.4).

Comparing this formula with that for U·V, we see that the products UxVx, UyVy, UzVz that survive in U·V are absent from U × V, while the products Ux Vy, etc. that are absent from U·V are the ones that survive in U × V. The formula for U × V is not as pleasant as the one for U·V; but its structure is not at all as complicated as it seems, and writing it down from memory is a fairly simple matter. Note first that the coefficient of i does not contain Ux and Vx, the components along i of the original vectors; and similarly for j and k. Next note that if we take x, y, and z, in cyclic order (that is, x → y → z → x, as would be the case if x, y, and z were written in a circle), then the positive terms have the suffixes in this order (when we write the components of the first vector first, as we would naturally do). The positive terms may thus be written down easily, and if gaps are left for the negative terms, we can fill them in later by simply reversing the order of the suffixes that appeared in the corresponding positive terms.

Exercise 3.2 Study the above. Then, without looking at Equation (3.4), fill in the missing terms in the following:

Exercise 3.3 Write the positive terms in the formula for L × Q, leaving gaps for the negative terms. Then fill in the latter.

Exercise 3.4 Write the formula for V × U and verify that it is the negative of that for U × V.

Exercise 3.5 Without looking at the full formula, write down the coefficient of j in U × V and that of k in B × A.

There is another way of remembering how to write the formula for a vector product in terms of components. First write

(3.5)

[The arrows may be omitted once you are used to the method.] For the component of U × V along i, consult the j and k columns, and form the two products indicated by the arrows, counting the downward products positive and the upward ones negative. For the coefficient of j, do the same with the k and i columns (the third and fourth), and for the coefficient of k use the i and j columns (the first and second).

If you have studied determinants, you will already have realized that, formally,

(3.6)

and there in no reason why you should not find cross products by means of this formula.

Choose whichever method you find most congenial. When dealing with symbols having appropriate suffixes, I prefer the first since it lets one write down the formula or any part of it immediately without bothering with Equation (3.4), or schema (3.5), or Equation (3.6). In numerical problems involving i, j, and k, we can use the distributive law and the relations (2.1 For example, to work out (2i + 3j + k) × (3i + 2j + 5k), we first note that the i × i, j × j, and k × k terms will vanish. Then we group the 15j × k and the 2k × j to give 15i – 2i or 13i, and similarly for the other terms. The result is 13i – 7j – 5k.

Exercise 3.6 Show that (i + 2j + 3k) × (4i + 5j + 6k) = 3( – i + 2j – k).

Exercise 3.7 Show that (i – 2j + 3k) × ( – 4i + 5j + 6k) = – 3 (9i + 6j + k).

Exercise 3.8 Find the sine of the angle between i + j + k and i + 2j + 3k. Check the answer by finding the cosine from the scalar product and verifying that the sum of the squares of the sine and cosine is unity. [Ans. 1/√7, √6/√7, the signs being correct for 0° ≤ θ ≤ 180°.]

Exercise 3.9 Like Exercise 3.8 for 3i + 4k and 2i – j – 2k. [Ans. ]

Exercise 3.10 Show that i cos α + j sin α is a vector of unit magnitude in the ij-plane, making an angle α with i. Then, by considering the cross product of this vector with a similar one making an angle β with i, derive the well-known trigonometrical formula sin (β – α)= sin β cos α – cos β sin α.

Exercise 3.11 1 If W = U × V, then W is perpendicular to both U and V. Consequently W·U = 0 and W·V = 0. Express these two equations in terms of the components of U, V, and W, and by solving them, show that the components of W must be in the ratio

(Uy Vz – Uz Vy):(Uz Vx – UxVz):(UxVy – Uy Vx).

4. TRIPLE PRODUCTS

In ordinary algebraic multiplication of scalars, the expression uvw has a definite meaning. The situation is different when we are dealing with scalar products or vector products of vectors. For example, U·V·W has no meaning at all: if we form U·V first we obtain a scalar, and we can not form a dot product of this scalar with W; nor are we any better off if we form, say, V·W first. With cross products the situation is different. Consider, as a simple case, the expression i × j × j. If we first form the product i × j we obtain k, and since this is a vector, we can certainly form its cross product with the remaining j: the result is – i. But if we had first formed the product j × j we would have obtained the result zero. Let us write this in symbols:

(i × j) × j = k × j = – i but i × (j × j) = i × 0 = 0.

We see from this that an expression like U × V × W is ambiguous, its value depending on whether we first form U × V or V × W. In general,

(4.1)

a fact that we express by saying that vector products do not obey the associative law. Therefore, whenever we have dealings with a quantity like U × V × W, we have to include parentheses, as (U × V) × W or U × (V × W), to show what we mean. Notice that Equation (4.1) preserved the order of the vectors U, V, and W. We are not concerned here with the changes of sign that could arise if we altered the order.

Exercise 4.1 Find the values of (j × k) × k and j × (k × k).

Exercise 4.2 Find the values of (i × j) × i and i × (j × i) and show that they happen to be equal.

Exercise 4.3 Find the values of all of the following that have meaning: (i + j + k) × k, (i·i) × j, i·(i × j), (i·i)j, (i.i).j, (i – j)·(i + j), [(i – j) × (i + j)] × i, [(i – j) × (i + j)] × k, (i·i)(i·i), (i·i) × (i·i), (2i)·(3i). [Three are meaningless.]

Exercise 4.4 Find the values of [(i + k) × j] × (i + k) and (i + k) × [j × (i + k)] and show that they happen to be equal.

Exercise 4.5 From Exercises, 4.2 and 4.4 we begin to suspect that

(U × V) × U = U × (V × U). Using Equation (1.5) twice, prove that this relation is indeed true. [Use Equation (1.5) once on (U × V) and then on the pair (U × V) × U.]

Exercise 4.6 Show that i·(j × k) = (i·j) × k = i × (j.k) = (i × j).k = 1. [This is not a coincidence. It is a special case of what we are about to prove.]

Consider the parallelepiped defined by three vectors U, V, and W having a common starting point O.

The cross product U × V is perpendicular to the parallelogram defined by U and V, and its magnitude is equal to the area of the parallelogram. To find the volume of the parallelepiped, we multiply the area of the base by the altitude. But the altitude is W cos θ, where θ is the angle that W makes with the normal to the base. The vector U × V happens conveniently to be normal to the base. Therefore (U × V)·W will give us the volume of the parallelepiped, provided U, V, and W form a right-handed set. If not, (U × V)·W will come out negative and will give minus the volume. Since the dot product is commutative, (U × V)·W = W·(U × V). Also, since the expression U × (V·W) would be nonsense, we can omit the parentheses in (U × V)·W and write it without ambiguity as U × V·W.

It is clear that we could also find the volume using different base parallelograms, namely, those defined by V and W, and by W and U (note the order of this last). We thus have, for a right-handed set of vectors U, V, and W, the following seemingly formidable array of equalities which in fact tells us something simple:

(4.2)

What these relations tell us is that when we form a triple product of vectors using one dot and one cross, it does not matter where we put the dot and cross (within reason, of course—we would not want them on another page, nor even like this: UVW ×). Moreover, it does not matter whether we start with U, or V, or W, just so long as we keep the letters in cyclic order. And if we should break the cyclic order, the only penalty would be a minus sign. Because of this flexibility, the six positive combinations in Equation (4.2) are often denoted by the single symbol [UVW].

Exercise 4.7 Rewrite a sampling of the relations (4.2) for the case of a left-handed set of vectors U, V, and W. [U × V W will now have a minus sign in front of it, and V × U W a plus sign, for example. In practice, it is simplest to avoid left-handed sets U, V, and W by rearranging their order so as to form a right-handed set. If U × V W comes out negative, we know that the set was left-handed. Let us assume in this section that our sets of vectors are right-handed. We shall have to look into the matter of handedness in more detail later.]

Exercise 4.8 Show by means of Equation (4.2) that the volume of the cube defined by i, j, and k is unity. Work it out several ways.

Exercise 4.9 Find the volume of the parallelepiped defined by j + k, k + i, and i + j. [Note how complicated this looks when one tries to prove the simple result geometrically by means of a diagram.]

Exercise 4.10 Show that i × (i + j) (i + j + k) = 1. Interpret the result geometrically in connection with a cube of unit edge, and then prove the theorem geometrically. [Once you see it, the geometrical proof is easy. But it may take you some time to see it.]

Exercise 4.11 Find the volume of the parallelepiped defined by i + 2j + 3k, 2i + 3j + 4k, and i + 3j + k. [Ans. 4.]

Exercise 4.12 If you have studied determinants show that

Exercise 4.13 What must be true geometrically of the three nonzero vectors A, B, and C, all starting at O, if [ABC] = 0?

Exercise 4.14 What must be true geometrically of the four nonzero vectors A, B, C, and D, all starting at O, if (A × B).(C × D) = 0? What if (A × B) × (C × D) = 0?

Exercise 4.15 Show that [UVU] = 0. Using this and similar results, show that [(V + W)(W + U)(U + V)] = 2[UVW], and interpret this relation geometrically. [Compare Exercise 4.9.]

As an example of the use of Equation (3.4), consider the not altogether pleasant problem of working out a formula for U × (V × W). The result is clearly a vector. Let us therefore concentrate on the i component. This will come from Uyj multiplying the k term in V × W, and Uzk multiplying the j term in V × W. So it is:

Vy(Vx Wy – Vy Wx) – Uz (Vz Wx – Vx Wz).

Two of the terms have Vx as a factor and two others have Wx. So we regroup this as:

Vx (Uy Wy + Uz Wz – Wx (Uy Vy + Uz Vz).

Now we add Ux Vx Wx – Ux Vx Wx, which is zero, and obtain

Vx (Vx Wx + Uy Wy + Uz Wz) – Wx (Ux Vx + Uy Vy + Uz Vz),

which is the i component of V(U-W) - W(U V) or (U · W)V – (U V)W. By symmetry, the other components will fit the same scheme. So

(4.3)

Exercise 4.16 Work out the j component by the same method.

5. MOMENTS

In Chapter 2, in connection with a seesaw, we became acquainted with the idea of the moment of a force. We now extend the discussion to the three-dimensional case.

Consider a rigid body acted on by a force F at a point A in it as shown in Figure 5.1. Let O be a pivotal point in the body. [The point O need not be a pivotal point, nor need there be a rigid body. One can take moments about any point, with or without a rigid body connecting the point and the force. For example, one can take the moment about the center of the sun of a force acting on a planet. However, we shall more easily get the feel of what is going on if we think here of O as an actual pivot in a rigid body acted on by the force F.] The turning effect of the force F about O is measured by the product of its magnitude F and the perpendicular distance, p, between O and its line of action. Thus the magnitude of the moment of F about O is pF.

Figure 5.1

Now suppose we take any point P on the line of action of F. Let the position vector, r, of P relative to O make an angle θ with F, as shown. Then p = r sin θ. Therefore, the magnitude of the vector product r × F, which is rF sin θ, is just pF, the magnitude of the moment of F about O.

The relation between r × F and the moment of F about O is even closer than this. It extends to direction as well as magnitude. For the effect of F is to turn the rigid body in such a way as not to tilt the plane containing O and F; thus the effect is not just to turn the body “about O” but rather to turn it about an axis through O that is perpendicular to the plane of r and F. So we see that the moment, or turning effect, of F about O is, in this sense, given both as to its magnitude and the direction of its axis by r × F. Moreover, if we reversed the direction of F, thus producing a moment of opposite sign, the vector product r × F would automatically mirror this change of sign since, by the right-hand convention, r × F would now point in the direction diametrically opposite to its former direction.

Apparently, then, we may take r × F as representative of the moment of F about O. But before we do this, one thing remains to be verified. The quantity r × F is a vector. Do moments combine according to the parallelogram law? The answer is yes. But it is not an automatic yes. Simply agreeing to represent moments by vector products of the type r × F gives us no guarantee that the physical turning effect of an r₁ × F₁ and an r₂ × F₂ together is the same as the turning effect of their vectorial resultant. That it is the same can be deduced from equations F = ma which are based on experiment. It can also be inferred directly from experiment. We omit the details.

This parallelogram-law behavior of moments of forces is a far-reaching thing. Consider, for example, the rigid cube shown in Figure 5.2, which we shall assume has edges of unit length. Let the cube be pivoted at O and be acted on by two forces F₁ and F₂ as shown in Figure 5.2. What is the combined turning effect of the two forces? The forces act somewhat, but not entirely, against each other; and it is not easy to visualize offhand what sort of a twist they would tend to give to the cube when acting together. We can find their combined turning effect by forming the resultant of two cross products; a convenient way to do this is to use components and then stop visualizing while we perform some routine calculations.

Figure 5.2

Let us introduce i, j, and k along OA, OB, and OC respectively. Then the moment vector of F₁ is easily seen to be – F₁ k. [Why the minus?] To find the moment vector of F₂ about O, we note that the position vector, r, of the point D has components (1, 0, 1), while the components of F₂ are (0, F₂, 0). So, by a standard procedure, we find that r₂ x F₂ = – F₂i + F₂k. [The easiest procedure is to use the distributive law on (i + k) × (F₂ j).] Thus the resultant moment is given by the vector – F₂i + (F₂ – F₁)k.

In particular, if the two forces have the same magnitudes, so that F₁ = F₂, we see that the resultant is just – F₂i, a fact that may at first sight seem surprising. But if we look at the problem in a different way, this result becomes rather obvious. We simply ask ourselves what the combined turning effects of the forces are about the three axes OA, OB, and OC individually. (The turning effect of a force F about OA, for example, is measured by the moment about OA of the component of F perpendicular to OA.) About OA, the force F₁ has zero moment but F₂ has moment – F₂ because the perpendicular distance AD from OA to the line of action of F₂ is unity. So the i component of the resultant moment is – F₂. About OB, the moments are both zero, because the line of action of F₁ intersects OB, and F₂ is parallel to OB. About OC, the moments are – F₁ and F₂, and when F, = F₂, the sum is zero.

Exercise 5.1 Explain how the distributive law of vector products justfies the second method above.

Exercise 5.2 Find the resultant moment about O in Figure 5.2 if F₁ acts along CD and F₂ along the diagonal BC. [Ans. (F₂/√2)i + Flj.]

Exercise 5.3 Like Exercise 5.2, but with F₂ acting along the diagonal BD. [Ans. – (F₂/√3)i + F₁ j – (F₂/√3)k.]

Exercise 5.4 Show that if two nonzero forces act on a rigid body they will have zero resultant moment about a point O only if they lie in a plane through O.

Exercise 5.5 If the sum of the moments of n forces about a point O is zero, show that it will be zero about every other point, provided the resultant of the forces moved (as though they were free vectors) to act at a common point is zero. [Compare Exercise 6.6 of Chapter 2.]

We already know from Chapter 2 that a couple consists of two parallel forces having equal magnitudes and opposite directions but, in general, different lines of action. It is clear from the present discussion of the moment of a force about a point, that if P₁ and P₂ are any points on the respective lines of action of the forces F₁ and F₂ of a couple (these forces having equal magnitudes), the moment of the couple is given by the vector product (which is the same as ); and that if F is the common magnitude of the forces and d the perpendicular distance between them, the magnitude of the moment of the couple is Fd. Its direction, of course, is perpendicular to the plane of the couple.

Figure 5.3

Exercise 5.6 Given a couple and any point O not necessarily in the plane of the forces comprising the couple, show that the resultant of the moments of the forces about O is equal to the moment vector of the couple. [Use the fact tht .]

Exercise 5.7 Find the moment of the resultant couple of the two couples shown in the diagram, if all forces have magnitude 5 and the square has sides 2 units long. [Ans. The individual couples have moments – 10i – 10k, – 10i + 10k, so their resultant is – 20i.]

Exercise 5.8 Like Exercise 5.7 but with the couples as shown, the forces and having magnitudes of 2 1bs, and and 2√2 lbs. [Why do we get the wrong answer if we argue that the three vectors , and are together equivalent to the vector so that we have, in effect, just a single couple with forces along and

Exercise 5.9 A force F acts at a point with position vector r relative to O. By introducing at O a pair of cancelling forces parallel to F and having the same magnitude as F, show that the original force is equivalent to a parallel force of equal magnitude through O and a couple having a moment r × F.

By applying the technique outlined in Exercise 5.9 to each force in succession, we can reduce any system of forces to a single resultant force at any point O and a single resultant couple. Let us denote the single force at O by F and the couple by G. In general F and G will not be parallel. If we go to a new point P, the new force will be parallel to the old and, indeed, identical with it except for location. But the couple will in general be altered; for if the position vector of P relative to O is r, there will be an additional couple – r × F arising from the change from O to P, so the resultant couple will be G – r × F. We therefore have a chance of finding a point P for which the resultant force and resultant couple of the system are parallel—in which case we say we have a wrench, this being the sort of influence one customarily exerts on a screwdriver. Let us see if we can find such a point P. We want G – r × F to be parallel to F. Therefore, we want F × (G – r × F) = 0, i.e.,

(5.1)

As it stands, this is not an easy equation to solve for r. We therefore seek help from equation (4.3) which tells us that

F × (r × F) = (F F)r – (F r)F.

This lets us write equation (5.1) in the form

(5.2)

We are still not out of the woods. The next step is to decompose r into the sum of two vectors n and p, n being parallel to F × G and thus normal to F, and p being parallel to F:

(5.3)

Then, since F n = 0, we have, from equation (5.2),

(5.4)

The component vectors of the vector 0 are zero in all directions. The component vectors of the resultant of the left hand side must therefore also be zero. Since F × G and n are perpendicular to p and F, we see that we can extract two distinct equations from equation (5.4), one for the component vectors in the direction of n, the other for the component vectors in the direction of p. They are

(5.5)

So

(5.6)

At first we are inclined to think that these equations (5.6), (5.3) fix the vector r. But the second equation in (5.6) does not give p; it gives only its direction —which we already knew was parallel to F.

Exercise 5.10 Why does it give only the direction? [Hint: Try to find the magnitude of p from the equation and see what happens.]

Since Equation (5.1) will hold for all vectors p satisfying Equation (5.6), we see that instead of finding a single point P relative to which the forces of the system form a wrench, we have found a whole line of such points. This line is parallel to F, and is called Poinsot’s central axis.

Exercise 5.11 1 The above mathematics breaks down if F = 0. Does it matter? What could one say about the whole idea of searching for a central axis when F = 0?

Exercise 5.12 Given one point relative to which a system of forces becomes a wrench, show that it is obvious that any point on the line through this point parallel to F will also be one relative to which the system is a wrench.

Exercise 5.13 Show that the magnitude of the resultant moment of the forces of a system is a minimum when the moment is taken relative to points on the central axis. [Hint: F does not change its magnitude or direction as one goes from point to point. G changes by – r × F which is perpendicular to F. So G F does not change value. For given F and G F, what angle should G make with F if G is to be a minimum?]

Exercise 5.14 Find the central axis of the forces in Figure 5.2. [We have F = F₁i – F₂j, and, taking moments about O, G = – F₂i + (F₂ – F₁)k. So , + F × G = – F₂(F₂ – F₁)i - F₁(F₂ – F₁)j – . Then calculate n using equation (5.6). Note how complicated the result is for even so simple a case as this.]

Exercise 5.15 Like Exercise 5.14 if the forces on the cube are those described in Exercise 5.2.

Exercise 5.16 Relative to a point O, a wrench pointing along i has force and couple of magnitude 1 and 1. Another wrench, pointing along j, has force and couple of magnitude 1, 2 respectively. Find the resultant central axis. [Ans. A horizontal line through the point (0,0, ) making 45° with the i and j directions.]

6. ANGULAR DISPLACEMENTS

If we rotate a rigid body about a fixed axis through an angle α radians, it undergoes an angular displacement. We can represent this angular displacement by means of an arrow-headed line segment of length α units (say inches) lying along the axis of rotation and pointing in the direction a right-handed corkscrew would move if similarly rotated.⁹ Since the angular displacement has magnitude and direction and can be represented by an arrow-headed line segment, we begin to suspect that it behaves like a vector. Indeed, if we combine a rotation about a particular axis through an angle α radians with one about the same axis through an angle β radians, the resulting angular displacement is one of α + β radians about the axis, and this is just what we would obtain by combining the corresponding arrow-headed line segments according to the parallelogram law—the parallelogram, in this special case being, of course, squashed into a straight line. Again, if we combine a rotation about an axis through an angle α radians with a rotation about the same axis through an angle – α radians (which is through an angle α radians in the opposite direction and would thus pertain to an arrow-headed line segment pointing in a direction opposite to that of the line segment representing the original angular displacement), we end up with a zero angular displacement, and this result, too, we would get from the parallelogram law.

Figure 6.1

Add to all this our vectorial experiences not only with moments but even with such unlikely things as areas of parallelograms, and we begin to feel confident that angular displacements can be treated as vectors. It seems hardly necessary to check further whether the arrow-headed line segments representing angular displacements obey the parallelogram law.

But suppose we rotated the rigid body through α radians about an axis I₁, and then through – α radians about a different, but parallel axis I₂. Then the two angular displacements together would not leave the body in its original position. They would displace it, without rotation, to a new position.

Exercise 6.1 Check this assertion by considering the simple case of a rod AB in a plane, first rotating it about A and then rotating it through an equal angle in the opposite direction about the new position of B.

Evidently we have to tread carefully if we wish to regard angular displacements as vectors. Two ways of saving the situation seem open to us. One is to say that the resulting rotation above was zero, as predicted by the parallelogram law, and that the bodily displacement is not our present concern. The other is to say that then we combine angular displacements vectorially they must be angular displacements about the same axis. Clearly, the second is too drastic. If we combine only rotations having a common axis we shall never use the full parallelogram law but only the case in which the parallelogram is squashed into a line. While we would then be able to say correctly that rotations about a given axis behave like vectors, this would hardly be a worthwhile assertion.

We therefore try to combine the first possibility with a less stringent version of the second. We say that we want to look upon angular displacements not as free vectors but as bound (or at least sliding) vectors, and that we will limit ourselves to combining rotations about axes that have a common point. This immediately rules out the case of parallel axes that caused the trouble discussed earlier. Presumably, with this restriction, our luck will hold. Let us see.

Consider two rotations, each through π/2 radians,¹⁰ about fixed perpendicular axes that, for convenience, we shall take to be the x-and y-axes of a three-dimensional rectangular Cartesian coordinate system. To simplify the discussion, take the rigid body to be a playing card, say the ace of spades. Place it face down in the position shown in Figure 6.2a. On rotating it about the x-axis through π/2 radians, we bring it into the position shown in Figure 6.2b. And on now rotating it about the y-axis through π/2 radians, we bring it finally to the position shown in Figure 6.2c.

Figure 6.2

The two rotations, if treated as vectors, would be (π/2)i and (π/2)j, and their resultant would be (π/2)(i + j), whose magnitude is π/√2 and whose direction bisects the angle between the positive x- and y-axes. But it is easy to see that if we had rotated the card in one move through the unpleasant angle π/√2 radians about the bisector of the foregoing angle we would not bring it into anything like the final position above. Therefore the arrow-headed line segments representing angular displacements have failed the parallelogram test, and they have done so even though we have limited ourselves to axes of rotation having a common point O. There is no avoiding the fact that we just cannot treat angular displacements about intersecting but nonidentical axes as vector quantities.

This fact becomes strikingly clear when we consider the effect of performing the above rotations in the reverse order. The sequence of positions of the card is shown in Figure 6.3, and the final position is quite different from

Figure 6.3

that in Figure 6.2. Thus, if we denote the rotations about the x- and y-axes respectively by X and Y, we see that the addition of angular displacements does not even obey the commutative law:

X + Y ≠ Y + X.

Exercise 6.2 Starting with the card in the same position as above, rotate it first through π/2 radians about the y-axis and then through π/2 radians about the z-axis, and compare the resulting position with that obtained when these two rotations are perfomed in the reverse order. [The final positions are not the same.]

7. ANGULAR VELOCITY

Fresh from the chastening experience of the preceding section, let us consider a rigid body turning about a fixed axis at a constant rate of ω) radians per second. Associated with the motion are a magnitude and a direction, the magnitude beings ω and the direction that of the axis in, say, the right-hand corkscrew sense. Can it be properly regarded as a vector quantity?

Exercise 7.1 Place your bet: yes or no? [Whichever you choose, you will be surprised—either in this section or later.]

There are two ways of thinking of the motion. One is as a single over-all rotation; the other is as the set of motions of all the individual particles making up the rigid body. Let us think of it in the latter sense.

Take a fixed point O on the axis of rotation, and denote by r the position vector relative to O of a general point P of the rigid body. Draw PN perpendicular to the axis of rotation. The rotation of the body causes P to move in a circle perpendicular to the axis of rotation, the center of the circle being N. If r makes an angle θ with the axis, the radius of the circle, PN, is r sin θ. The speed of P is easily calculated. In any time t secs. the body turns through an angle of ωt radians. Therefore P traverses an arc length of (ωt)(r sin θ). Dividing this arc length by the time taken, namely t, we find that the speed of P is ωr sin θ. [Note the advantage of using radians here. Had ω been given in deg./sec. the angle would have been ωtπ/180 radians, and the speed of P would have been given by the ugly expression (πωr sin θ)/180.]

Figure 7.1

Though the speed of P is constant, its velocity is not. At each instant it lies along the tangent at P to the circle traversed by P, and its direction changes from instant to instant. The fact that the instantaneous velocity of P has a continually changing direction is a considerable complication. To avoid trouble with the changing direction in what follows, we would like to talk about what happens at a particular instant. But a genuine instant affords no time for anything at all to happen. It was precisely this sort of dilemma that the calculus was designed to overcome; but the dilemma is of such subtlety that even when Newton and Leibnitz succeeded in constructing a calculus that gave formalized mathematical procedures for dealing with it, a century was to pass before a significant beginning was made to put these procedures on an acceptable mathematical basis. Nevertheless, during that century enormous progress was made by means of the calculus.

Taking a hint from the pioneers, then, let us cross our fingers and try to get the feel of what is going on without striving for mathematical rigor. We ask ourselves whether we can regard the above motion of a rigid body as a vector. Let us see what happens if we think of it as a vector starting at O, call it angular velocity, and denote it by ω.

The first thing we observe is that we may now represent the instantaneous velocity, v, of the point P by ω × r:

(7.1)

Exercise 7.2 Verify Equation (7.1) considering both the magnitude and direction of v, and noting that r × ω would have given the wrong sign for v.

Exercise 7.3 A rigid body is spinning about the z-axis with a constant angular speed of 10 radians/sec. At a particular instant, a point P in the body has coordinates (2, 2, 5). What is its velocity at that instant? [Ans. 10k × (2i + 2j + 5k), i. e., – 20i + 20j.]

Exercise 7.4 Like Exercise 7.3 if the body was rotating about a line through the origin whose positive direction makes equal acute angles with the positive directions of the axes. [Ans. 10√3i – 10√3j. Note that the instantaneous velocity has a horizontal direction. Try to visualize the situation without thinking of vector products to see why the velocity of P at this instant has to be horizontal. You will then begin to realize the power of Equation (7.1).]

Now suppose that the rigid body was somehow subjected to two angular velocities ω₁ and ω₂ simultaneously, these having, in general, different axes but axes with a common point O. Since we are thinking of ω₁ and ω₂ as vectors starting at O, they will have a resultant, ω, also starting at O, that is given by:

(7.2)

This equation is, of course, an expression of the fact that ω₁ and ω₂ yield ω according to the parallelogram law.

Now ω₁ and ω₂ individually impart to P the respective velocities v₁ and v₂, these being given by:

(7.3)

The resultant angular velocity ω would impart to P the velocity v given by Equation (7.1). From Equation (7.2) we obtain

(7.4)

which is also an expression of the parallelogram law. But by Equations (7.1) and (7.2), this equation is just

(7.5)

which tells us that if we represent angular velocities by vectors as above, the instantaneous velocities imparted to P by ω₁ and ω₂ have a vectorial resultant that is just the velocity that would be imparted to P by the vectorial resultant of ω₁ and ω₂. Since this holds for every point P of the rigid body, we are justified in treating angular velocities as vectors.

Note in the above how important it is that the position vector r of P should be the same throughout. We could not have obtained Equation (7.4) from Equation (7.2) otherwise. Therefore, we must here regard the angular velocities ω, ω₁, and ω₂ as having lines of action all passing through a common point O. As here regarded, they are, thus, not free vectors but sliding vectors.

That they are not free vectors is clear from the fact that if we spin a rigid body about an axis we do not cause the same motion as when we spin it with the same angular speed about a parallel axis that does not coincide with the first. Despite this, if we are willing to be content with the changes of orientation produced by angular velocities and to ignore the actual motions of individual points, we can think of angular velocities as free vectors.

Exercise 7.5 Verify the above by considering the motion of a rigid rod AB in a plane if the rod is spun about an axis through A perpendicular to the plane, and if it is spun about a parallel axis through B.

Exercise 7.6 You are probably familiar with Foucault’s pendulum. It is merely a pendulum suspended in such a way that the support does not influence the direction in which the pendulum swings. What makes this simple piece of apparatus fascinating is that it reveals the rotation of the earth without our needing to look at the stars or other external objects to take our bearings. For example, if the pendulum were mounted at a pole, the vertical plane in which it swings would appear to rotate once every 24 hours. What really happens is that the pendulum swings in a plane that may be described as fixed in direction relative to the so-called fixed stars, and the earth rotates once every 24 hours beneath it. At the equator, the pendulum would not exhibit any such rotation relative to the earth. Why not?

Exercise 7.7 At what rate relative to the earth would a Foucault pendulum seem to rotate if it were at a place of latitude λ°? [Hint: Let O be the center of the earth and P the point of support of the pendulum. Then we can resolve the angular velocity vector of the earth, say ω, into two components through O, one of magnitude ω sin λ having its axis along OP, the other of magnitude ω cos λ perpendicular to OP. Only the former will be revealed by the Foucault pendulum. Note how difficult it would be to visualize the situation without the concept of the angular velocity vector.]

We must now return to the deliberately vague phrase “suppose that the rigid body was somehow subjected to two angular velocities ω₁ and ω₂ simultaneously” [see page 89]. What does it mean? There are two interpretations. Imagine the rigid body mounted on another body that is also rotating. Then if the first has angular velocity ω₁ relative to the second, and the second has angular velocity ω₂ relative to some master reference frame, the first is being subjected simultaneously to the two angular velocities ω₁ and ω₂, and its angular velocity relative to the master frame is the vectorial resultant of ω₁ and ω₂. We are using two reference frames here, just as we did for ordinary velocities when we considered a point moving on a moving platform.

But we can also take the mathematical instead of the physical point of view, just as we did when we looked upon ordinary velocity as the rate of change of displacement with respect to time and could thus combine velocities that were all measured relative to the same reference frame. With angular velocities the situation is as follows. Suppose a rigid body has angular velocity ω about an axis through O. If we introduce a reference frame ex, ey, and ez and write:

(7.6)

where the components ωx, ωy, and ωz are found in the usual vectorial way, then the instantaneous velocity of each point of the rigid body, as given by Equation (7.1), will also be correctly given by applying Equation (7.1) to the vectors ωxex, ωyey, and ωzez to obtain three velocity vectors, and then forming the resultant of these component velocity vectors.

Exercise 7.8 Relative to i, j, and k a rigid body has angular velocity ω having components (ω/√2, ω/√2, 0). A point P on the x-axis has a position vector with components (r, 0, 0). Draw a picture to see what is going on. Then calculate the instantaneous velocity of P (a) by working directly with ω and forming ω × r and (b) by finding the component velocities (ωxi) × r and (ωy j) × r and finding their resultant. In which direction does the instantaneous velocity of P point?

Exercise 7.9 Like Exercise 7.8, but with the components of the position vector of P given by (0, 0, r).

Exercise 7.10 Like Exercises 7.8 and 7.9, but with the components of the position vector given by (r/√2, 0, r/√2). [Note the sudden increase in complexity here so far as visualization of the motions is concerned. The mathematical manipulations, on the contrary, are almost as easy as before.]

All this time we must have been wondering how it happens that angular velocities behave like vectors despite the fact that angular displacements do not. Here is the essential reason: If we make an angular displacement of a rigid body about a fixed axis, the points of the body move on arcs of circles. Suppose the point P moves on the circle with center N to the position P‘, as shown in Figure 7.2. Then the displacement PP’ is not, in general, perpendicular to the position vector , nor does it make a fixed angle with for all positions of P‘—or for all positions of P.

Figure 7.2

Look back at the discussion of the vectorial character of angular velocity and you will see that it depended crucially on the fact that the instantaneous velocity of P was always perpendicular to the position vector of P. We might, if pressed, have managed even if the angle had not been a right angle, provided it had been a fixed angle. But with a varying angle we are beaten.

Even so, how does it happen that the instantaneous velocity of P, which is, after all, just a time rate of change of a displacement, is always perpendicular to P? The reason is that we are dealing with the instantaneous velocity. This is, roughly speaking, the time rate of change of an extremely small displacement. If we take PP’ extremely small in Figure 7.2 we see that it does indeed make approximately 90° with OP, being approximately tangent to the circle at P. And the fact is that extremely small angular displacements do indeed behave approximately like vectors, so that one often hears it said that infinitesimal angular displacements are vectors even though finite angular displacements are not.

8. MOMENTUM AND ANGULAR MOMENTUM

If a particle of mass m has velocity v, the vector mv is called its momentum. The combination mv is a rather natural one. Newton regarded it as measuring the quantity of motion. For example, if two particles of mass m are moving with velocity v, there is twice as much motion as with one; and the total momentum adds up to 2mv, accurately reflecting the fact.

In collisions momentum is conserved; that is, the total momentum vector of the system before the collision is equal to the total momentum vector after the collision.

Exercise 8.1 A particle of mass m moves in the positive direction along the x-axis with speed v. Another particle, of mass 2m moves in the positive direction along the y-axis with speed v. The two collide at the origin and coalesce into a single particle. What is the velocity of this particle?

Solution The momenta of the original particles are mvi and 2mvj respectively. The resulting particle has mass 3m. So if its velocity is V, its momentum is 3mV. But the resultant of the original momenta is mvi + 2mvj. So, since the momentum is conserved in the collision, we must have 3mV = mvi+2mvj; and from this we see that V= (v/3)i + (2 v/3)j. Thus, the resulting particle moves with speed √5 v/3 in a direction making an angle tan – 1 2 with the positive x-axis.

Exercise 8.2 Like Exercise 8.1 but with a third particle, of mass m, moving with speed 2v in the positive direction along the z-axis, all three particles meeting simultaneously at O and coalescing into a single particle. [Ans. Speed 3v/4, and moving in a direction having direction cosines ( ). Note that had the resulting particle in Exercise 8.1 later been struck by a particle of mass m moving parallel to the z-axis in the positive direction with speed 2v, and had this particle coalesced with the one it struck, the resulting speed and direction would be the same as before, though the velocity vector would have a different location.]

Exercise 8.3 A particle of mass m moving with speed v in the i direction suddenly splits apart into two pieces. One piece, of mass m₁, moves with speed v, in the j direction. How does the other piece move?

[Ans. Its velocity is .]

Exercise 8.4 Like Exercise 8.3 if the piece having mass m moves with speed v in a direction having direction cosines cos α, cos β, and cos γ.

The momentum vector is often denoted by the symbol p. If a particle has momentum p and, relative to a point O, position vector r, the vector product r × p is called the moment of momentum of the particle about O. It is often also called the angular momentum of the particle about O.

From the basic equation of motion F = ma, which is valid in an unaccelerated reference frame, we have:

(8.1)

It happens that r × (ma) is the rate of change of r × (mv) with respect to time. This is not obvious, despite the fact that a is the rate of change of v with respect to time, since not only v but also r changes with time. However it does happen to be true (it can be easily proved using the calculus), so let us accept it as such here. Then Equation (8.1) tells us that the moment about an unaccelerated point O of the force acting on a particle is equal to the rate of change of the angular momentum of the particle about O. In particular, if the force has zero moment about O the angular momentum will be constant, since its rate of change is zero.

Exercise 8.5 A particle acted on by zero force moves in a straight line with uniform speed. Verify that its moment of momentum about a fixed point O is constant.

Exercise 8.6 If a planet is acted on only by the force of gravity directed towards the center of the sun, show that its angular momentum about the center of the sun must be constant, and therefore, that it travels fastest in its orbit at the place where the tangent line to the orbit is closest to the sun.

A gyroscope is essentially just a top, though usually one of special shape and mounted in a special way. Consider the gyroscope shown, consisting of a massive squat cylinder spinning on its axis and mounted in such a way that its axis can turn freely in any direction about the center of symmetry O, which is kept in a fixed position. In practice, this requires a moderately elaborate mounting mechanism, but we can here think of the gyroscope as freely pivoted at O. Each particle of the gyroscope has a moment of momentum about O, and the sum of all these moments of momentum is the moment of momentum of the gyroscope about O. Denote it by vector M lying along the axis.

Equation (8.1) holds for each particle of the gyroscope, and by adding up the equations for all the particles (the mathematicians can do this rigorously) we find that the total moment of the forces acting on the gyroscope is equal to the rate of change of its total angular momentum.

Since the gyroscope is mounted at its center of symmetry, the gravitational forces on it have zero resultant moment about O. Suppose we point the axis of the gyroscope horizontally towards the right as shown in Figure 8.1. Since the resultant moment of the forces acting on it is zero, its angular momentum vector M will not change. Therefore, it will continue to point in the same direction. Now suppose we exert a constant pressure downward on the tip, P, of the axis, say by hanging a weight on it. How will the axis move? One’s first guess—provided one has never played with a gyroscope—is that the tip of the axis will drop vertically. But it will not.

The weight on the tip of the axis exerts on the gyroscope a force whose moment vector about O is in the direction of the horizontal line OQ perpendicular to the axis OP, as shown. It will, therefore, cause the angular momentum vector of the gyroscope to change vectorially at a rate equal to this moment. We can think of the force as continually pumping into the gyroscope angular momentum that always points horizontally at right angles to the position of OP as indicated in the figure by the arrow-headed line segment OQ. By forming the vectorial resultant of M and a small horizontal additional angular momentum perpendicular to it, we see that the effect of this pumping is to turn the axis of the gyroscope horizontally counterclockwise as viewed from above.

Figure 8.1

Exercise 8.7 In what direction would the turning effect be if, instead of tying a weight to the tip of the axis, we exerted an upward pressure on it?

Exercise 8.8 Like Exercise 8.7 if we pushed the tip of the axis horizontally into the page.

Exercise 8.9 Assume that the propellers of an airplane are all turning in a clockwise sense as viewed from behind. If, in flight, the pilot turns the airplane towards the right, what is the gyroscopic effect of the propellers? [Ans. It tends to turn the nose of the airplane downward.]

We have seen that, when a weight is tied to the tip of the axis of the gyroscope in Figure 8.1, the axis will tend to rotate horizontally. When the weight is applied abruptly, the tip of the axis has a tendency to dip as well as to move horizontally and the resulting motion is somewhat complex, the tip of the axis bobbing up and down as the vertical plane containing it sweeps around like a revolving door. However, if the weight is applied gradually and the axis is allowed to find its appropriate motion one can manage, in principle, to end up with the axis turning steadily in a horizontal plane. Since the manifest cause of this horizontal motion of the axis is the vertical force exerted by the weight, this is likely to strike us as curious, to say the least. Let us therefore consider what is happening without making use of cross products. For simplicity, regard the gyroscope as a circular disc and consider the motion of a point A on the rim of the gyroscope that is vertically above O, as in Figure 8.2. Because the gyroscope is spinning about its axis, the point A has an instantaneous velocity tangent to the disc that we can represent by the arrow-headed line segment . When we apply the weight at P we tend to pull OP downward; but this will tend to move the point A horizontally, giving it an additional velocity that we can represent by . The resultant velocity of A, given by , thus lies outside the plane of the disc and shows that the disc tends to turn about a vertical axis in the manner already discussed. Had we taken a point diametrically opposite to A, both component velocities would be reversed in direction as, therefore, the resultant would be too, thus confirming the turning effect. Had we taken intermediate points the turning effect would be in the same general direction, except for points on the line OQ which would seem to be unaffected. But since the gyroscope is reasonably rigid, even these points must join in the general turning. Indeed, the whole turning is a compromise among discordant local turning effects, and we see that when we tie a weight to the tip of the axis of a spinning gyroscope and thereby cause its axis to rotate in a horizontal plane we introduce considerable internal stresses of a quite complicated sort. But then, even if we did not tie a weight to the tip, the spinning of the gyroscope on its axis would still cause internal stresses, these being sometimes so great as to cause the gyroscope to explode into pieces.

Figure 8.2

9. AREAS AND VECTORIAL ADDITION

When we consider vector products like r × F that yield moments, we have no qualms about accepting the physical fact that they combine according to the parallelogram law. But what about vector products that we think of as representing areas? Having decided that U × V represents the area of a parallelogram both in magnitude and, via the normal to the area, in direction, and having irrevocably agreed that cross products combine vectorially, we must face the fact that actual areas do not seem to fit into the vectorial scheme. For example, in Figure 9.1, if the cross product U × V represents the area of the parallelogram OLMN, then V × U represents the area of the parallelogram ONML, and we have to count this as the negative of the former area, even though the amount of paint we would need to paint the one is equal to and not the negative of the amount we would need to paint the other. But this is not too far removed from the idea of treating distances along a line as positive in one direction and negative in the opposite direction; and we have learned to live with that and even to approve of it as a valuable mathematical concept. So for the sake of mathematical harmony, and possible benefit, let us accept the official mathematical convention of counting an area as positive if, when “walking” around its rim (we may be somewhat upside down when doing so), we keep the area on our left, and negative if we keep it on our right.

Figure 9.1

Immediately, we find ourselves rewarded by something pretty. For let us place two rectangles LMNP and MQRN side by side as in Figure 9.2 so that the sum of their areas is the area of the rectangle LQRP. Then consider the “sum” of the rims of the original rectangles. The part of the first can be regarded as cancelling the part of the second so that, in a reasonable sense, we can say that the rim of the sum of the areas is equal to the sum of their rims.

Figure 9.2

Exercise 9.1 Show that it is not necessary for LMNP and MQRN to be rectangles, or even to have straight sides, so long as the areas are contiguous. [Actually they need not be contiguous, but then the prettiness is gone.]

Exercise 9.2 Show that if an irregularly shaped region is crisscrossed by lines, as in Figure 9.3, and the areas of the small regions into which it is divided are all taken positively, then the sum of the rims of these small regions is just the rim of the whole region. Would this be true if all the areas were taken negatively?

Figure 9.3

Exercise 9.3 Observe that the sum of the areas of OLMN and ONML in Figure 9.1 is zero, and that the sum of their rims is too. Is this true of any two areas, one of which is counted as positive and the other, having the same number of square inches, is counted as negative? [Consider U × V and × (2U). Note here, incidentally, that two areas of different shapes can nevertheless cancel each other vectorially.]

Exercise 9.4 In Figure 9.1, if the parallelogram were rotated about O while remaining in the plane, what would be the effect on U × V?

Well, we have weathered the first crisis, and even benefited from it. But Exercises 9.3 and 9.4 have reminded us of something else that needs to be explored: the same vector can represent the areas of all sorts of parallelograms —a fact that we have already used, for example, in discussing A × B + C × D.

figure 9.4

Suppose we have three parallelograms in a plane, as in Figure 9.4. Each of them can be represented by a vector through O perpendicular to the plane. The resultant of the three vectors represents the area of the shaded jagged region—which is not a parallelogram.

Indeed, if we are willing to treat the vector representing a plane area as a free vector, we can represent a plane area of quite general shape by means of a vector—simply by breaking the area up into a sufficient number of parallelograms and taking the sum of the corresponding vectors, all of which are parallel to one another. Infinitely many tiny parallelograms may be needed and a proper investigation would involve ideas belonging to the calculus. But the fact is that we can represent by means of a vector a plane area of any shape—even a curved one.

Exercise 9.5 Must the plane area be all in one piece?

Exercise 9.6 A circle of radius 10 ft. lies on the horizontal ground. What are the magnitude and direction of the vector by which a man would represent the circular area if he were walking around it with the area on his right? [Ans. Magnitude 100π ft²., and pointing vertically downward.]

Let us recall that all this began in an innocent attempt to use UV sin θ because it had geometrical interest: the area of the parallelogram defined by U and V. Soon we found ourselves looking on it as a vector, and now we are finding all sorts of unexpected consequences. We are no longer confined to parallelograms, nor even to cross products U × V. Willy-nilly, we have had to give vectorial character in their own right to plane areas of all shapes, and this was forced on us by a special case only: the combination of cross products pointing in the same direction. We have still to consider what is forced on us by our combining areas vectorially in the general case.

We start with the relatively simple case of the rectangle CDEB shown in Figure 9.5. Let the area of the rectangle be K. If vector represents this area (we have drawn it as a free vector) its magnitude is K and its direction cosines are 0, sin θ, and cos θ.

Figure 9.5

Exercise 9.7 Prove that the direction cosines are as stated. [Note that is parallel to the jk-plane.]

Consequently the components of are (0, K sin θ, K cos θ). But since the rectangle OAEB is the projection of CDEB on the ij-plane its area is K cos θ; so we see that the k component of is the area of the projection of CDEB on the ij-plane. More, we see that the component vector of in the k direction, namely (K cos θ)k, represents both in magnitude and direction the projection of CDEB on the ij-plane. Similarly, the component vector of in the j direction represents the projection of the original area CDEB on the ki-plane.

Exercise 9.8 Prove the last assertion.

Exercise 9.9 Prove that a corresponding assertion is valid for the component vector in the i direction and the projection of the area on the jk-plane.

We can easily generalize this result to the case of any plane region oriented in any way relative to i, j, and k. The crucial fact that we have to use is that the angle between two planes is the same as the angle between their normals. Let the region have area K and let a line normal to it have direction cosines cos α, cos β, and cos γ. Then the vector V representing the area of the plane region is given by:

(9.1)

Since the normal makes an angle γ with k (which is normal to the ij-plane), the plane region makes the same angle γ with the ij-plane. So its projection on the ij-plane has area K cos γ, and we see that THE COMPONENT VECTOR IN THE k DIRECTION OF THE VECTOR V, NAMELY (K cos γ)k, REPRESENTS IN MAGNITUDE AND DIRECTION THE PROJECTION OF THE REGION ON THE ij-PLANE. SIMILAR RESULTS HOLD FOR THE OTHER COMPONENT VECTORS.

While this is an important theorem, and a useful one, it is stated in terms of i, j, and k, and it applies, at most, to reference frames in which the base vectors ex, ey, and ez are mutually orthogonal. (For more general reference frames it can be restated in terms of oblique instead of orthogonal projections, but then it takes on a somewhat different character.) Meanwhile, Equation (9.1) confronts us with a stark fact that we have been trying to avoid: when we treat areas vectorially we commit ourselves to agreeing, for example, that three areas lying respectively in the jk-plane, the ki-plane, and the ij-plane are somehow to be regarded as together equivalent to a single area that lies in none of these planes, and that contains fewer square inches than the three do together.

Our first reaction is that this is preposterous. But we recall that when we said of ordinary vectors that , the corresponding parallelogram diagram seemed equally preposterous. By a process of abstraction we came, via the idea of shifts, to the idea of displacements, and then at last we could accept the vectorial law of combination geometrically. We have to try to do something of the same sort with areas.

Figure 9.6

Instead of returning to Figure 9.5, let us avoid page-turning by considering a similar but even simpler situation that lends itself better to what we wish to discuss. The cross products j × k and k × i are respectively i and j and they can represent the areas of the squares OBDC and OCEA in Figure 9.6, though they could also represent differently shaped areas. The resultant of i and j is i + j, which is shown as This latter vector can represent the area of the rectangle ABDE if we think of it as a free vector.

Exercise 9.10 Show that the magnitude of i + j does indeed give the area of ABDE, that i + j is normal to ABDE, and that it points in the direction appropriate to ABDE and not to AEDB.

When we say that j × k + k × i = i + j, we say that, vectorially speaking, the areas of the squares OBDC and OCEA are together somehow equivalent to the area of the rectangle ABDE (though not to that of AEDB). We notice that triangles DCE and BOA have the same area. Since they are parallel, we can represent them by equal free vectors. Then if we reverse the sense of one of them, their sum will be zero. So let us add to the two squares OBDC and OCEA the pair of self-cancelling triangles DEC and BOA. [Note the order: DEC, not DCE.] Then, from the vectorial point of view, we have to regard these four patches taken together as equivalent to the area ABDE. And what we now notice with the sudden delight that attends discovery, is that the four patches taken together have the same rim as the area ABDE.

Exercise 9.11 1 Prove that the rims are as stated. [Note the various cancellations: for example, the of the square OBDC cancels the of the triangle DEC.]

Figure 9.7

Now the question arises: is this just a coincidence or an instance of a general theorem? The matter is worth exploring. Consider a tetrahedron with vertices P, A, B, and C, as shown in Figure 9.7. Working with displacements, we see that and . So, using the fact that , we have . [Note the change of sign when we reversed the order of the factors in the term .] This relation has an immediate geometrical interpretation. The cross product represents vectorially the area of the triangle ABC, the vector pointing upward if we think of ABC as horizontal.

Exercise 9.12 Why upward? [Note that .]

Also represents vectorially the area of the triangle PBC, the vector pointing in an upwardish direction—out of, not into the tetrahedron. Similarly for and . Therefore the above relation, based on the distributive law, here tells us that we must regard the area ABC as vectorially equivalent to the three areas PBC, PCA, and PAB taken together. And we note that the rim of the former is the same as the rim of the latter three taken together.

Exercise 9.13 Prove that the rims are as stated.

Exercise 9.14 Prove the corresponding theorem for a pyramid PABCD with a four-sided base. [Do it in two ways: (a) without cross products, by regarding the pyramid as two contiguous tetrahedra and applying the above theorem, and (b) by means of cross products as above. In method (b) you will, in effect, be considering two tetrahedra, since you will have to regard the area ABCD as the sum of two triangles, such as ABC and BCD.]

Exercise 9.15 By method (a), extend the result in Exercise 9.14 to the case of a pyramid with an n-sided base.

Figure 9.8

Exercise 9.16 In Figure 9.8, the plane A’B’C’ is not necessarily parallel to the plane ABC. By “subtracting” the tetrahedron PA’B’C’ from the tetrahedron PABC, show, without resorting to cross products, that the area ABC is vectorially equivalent to the sum of the areas of ABB’ A’, BCC’ B’, CAA’C’, and A’B’C’, and that the rim of ABC is the rim of the four latter areas taken together. [Hint: We have, using a self-explanatory notation, ABC = PBC + PCA + PAB, and A’B’C’ = PB’C’ + PC’A’ + PA‘B’. Subtract, using the vectorial fact that, for example, PBC – PB‘C’ = BCC‘B’ because these areas are coplanar.]

Now consider any plane area bounded by a simple closed curve r, as in Figure 9.9, and think of it as made of a ductile metal. By means of an appropriately shaped tool and a hammer, beat a tetrahedral dent PABC into the surface. Then, vectorially, the area is equivalent to what it was before, because area ABC is vectorially equivalent to the three areas PBC, PCA, and PAB taken together.

Figure 9.9

What about the rim of the distorted surface? In a sense it is obviously the same as before since the dent, not being a perforation, hardly qualifies as part of the rim. But let us convince ourselves of this. Had we punched out a triangular hole ABC, the rim would have been Γ plus the rim of the triangle taken in the order ACB—remember, we have to walk the rim with the area on our left. When we add the cap PABC we, in effect, add the rim ABC, and this just cancels the rim of the punched hole that we had to take in the opposite order.

So we have the important result that by punching the tetrahedral dent we changed neither the vectorial area nor the rim. We now become “punch drunk,” making tetrahedral dents all over the place, and we end up with the realization that we can grossly distort the surface without altering its vectorial area, provided we leave its rim undistorted, which means, incidentally, that we may not punch holes in the surface. And we are ready to accept without proof the fact that the result holds even if the original surface is distorted into the shape of a bulging balloon, or a convoluted, semi-inflated one, or worse. The proof involves ideas belonging to the calculus.

At last we seem to understand the situation: areas are to be counted as vectorially equivalent if they have the same boundary. This is not the whole story, however.

Exercise 9.17 Show that the boundary curve need not be a plane curve. [Consider, for simplicity, a boundary lying in a plane and in the form of a closed polygon. Now replace one side AB by two lines AP and PB not in the plane. Consider two surfaces having this distorted rim. Add to each the triangle APB and they will now have the same flat rim and, so, will be equivalent. So they must have been equivalent before.]

Exercise 9.18 In Figure 9.6, what is the vector that would represent the area of a surface whose boundary was OBDCEAO?[Ans. i + j.]

This is the crux of the matter, but not the whole of it. In our discussion we had to treat areas as free vectors. So we must be ready to agree that two areas with congruent, similarly oriented rims are equivalent even if the rims do not coincide. Also we have to remember that, for example, a triangle and a parallelogram of the same area and in parallel planes are represented by the same vector and are thus regarded as equivalent; and that each could be equivalent to an aggregate of several isolated patches of area of suitable size and orientation. Since the exploration of all these possibilities frightens me, I shall pretend I have said enough about them.

There is a remarkable parallelism between the vectorial concept of area and the idea of a displacement. A displacement is equivalent to any sequence of zigzag displacements starting at A and ending at B (or, indeed, starting at C and ending at D if ; but for simplicity let us stick to the points A and B here). The points A and B can be regarded as the rim or boundary of the displacement . So when we wish to treat line segments vectorially we have to agree that all zigzag line segments—or indeed any curved segments—having the same boundary are to be counted as equivalent. Analogously, when we wish to treat surface areas vectorially, we have to agree that, no matter how greatly convoluted they may be, all surfaces having the same boundary are to be counted as equivalent. [But see below.] When we stay with the same end points A and B rather than allowing also equivalent pairs of end points, like C and D, we are dealing with shifts rather than vectors. When we correspondingly keep to areas having the same rim we are dealing with an areal analogue of shifts. In a restricted case, this “areal shift” has a simple pictorial significance. Consider a closed curve Γ that is not necessarily plane. Imagine it capped by a suitably bent iron plate, and consider how we would go about protecting one side of the plate from rusting because of exposure to rain. We could attach a tarpaulin to the rim Γ in a watertight way, and no matter how large we made the tarpaulin or how we arranged its billows and folds, it would protect the specified side of the iron from the rain. If we wished to protect the other side of the plate instead, we would attach the tarpaulin differently, though to the same rim. How can we specify the side we wish to protect? In conformity with what has gone before, we can associate with each side a sense of walking around the rim. If we walk with the immediate area of the iron cap on our left, we will agree to protect the side on which our feet tread; if on our right, the other side.

Exercise 9.19 Show that the above is consistent in that, if we walk around Γ in the opposite sense, the criterion will yield the opposite side in either case.

From all tarpaulin surfaces that protect a given side of the iron plate, we abstract everything except their common rim Γ and the sense in which we go around Γ. The resulting abstraction is the areal shift, and all these surfaces yield the same areal shift.

The difficulty with this picture is that the surfaces we have been likening to tarpaulins can cross the cap. To save the analogy we could make the cap bulge so that none of the protecting surfaces crosses it. But since we have to contemplate all surfaces having the given rim, the bulge would have to be infinitely great. Now that the idea of protection from rusting has served its pictorial purpose for us, we might as well abandon the idea of an iron cap and say that the areal shift is basically the common rim of all the surfaces, but with a particular sense of going around it. Thus it is A RIM AND A SENSE; but with it goes a faint reminiscence of the idea of a covering area, just as an ordinary shift is a pair of points and a sense, but with a faint reminiscence of journeying.

When we go from areal shifts to areal vectors we encounter the frightening pictorial complications already mentioned. But these are analogous to those that arise in the transition from ordinary shifts to displacements, which require us to agree that

are together equivalent to

and if the situation here is less complex than it is for areas, it is nonetheless analogous to the areal situation.

It is interesting to speculate on the likelihood of our coming to the idea of a vector product had we started out with this concept of areal shifts.

Exercise 9.20 In Figure 9.7, show that if one adopts the convention of representing the areas by normals that point outward from the tetrahedron, the vectorial sum of the four faces is zero. Show also that with this convention the total rim of the four faces together is zero.

Exercise 9.21 Show that VECTORIALLY THE AREA OF A CLOSED POLYHEDRAL SURFACE IS ZERO IF THE VECTORS ALL POINT OUTWARD FROM THE REGION ENCLOSED BY THE FACES. [A simple method is to slice the region into two parts by means of a plane. Then the surface area of each half is equivalent vectorially to the plane area of the polygon of intersection, but the two halves are equivalent to this plane area with opposite signs.]

10. VECTOR PRODUCTS IN RIGHT- AND LEFT-HANDED REFERENCE FRAMES

We have been hinting for some time that there was something yet to be told about the direction of the vector product of two vectors. Suppose we say, as we have been saying all along, that the vector product A × B points in the direction such that A, B, and A × B, in that order, form a right-handed system. Let us take a right-handed reference frame—for simpticity, a unit orthogonal triad, i, j, k—and consider the components relative to it of A, B, and C where

C = A × B. (10.1)

Figure 10.1

If A and B have components (Ax, Ay, Az) and (Bx, By, Bz) respectively, we have seen that C has the components (AyBz – AzBy, AzBx – AxBz, AxBy – AyBx).

Now suppose that we reflect Figure 10.1 in a mirror, so that it looks like Figure 10.2—except for the fact that Figure 10.2 has, for example, B instead of . Clearly, the components of the A and B in Figure 10.2 with respect to i, j, and k in Figure 10.2 will still be (Ax, Ay, Az) and (Bx, By, Bz) respectively. We would therefore like the components of their cross product C relative to this i, j, and k frame to be the same as before. But they obviously will not be, because the vector we have labelled C in Figure 10.2 is not really A × B if we adhere to the requirement that A, B, and A × B, in that order, shall form a right-handed system. The reflection in the mirror has turned A, B, and C into a left-handed system, and if we adhered to our right-handed convention for the direction of the cross product, we should say that the true A × B in Figure 10.2 points in the direction opposite to that of the vector labelled C in that figure.

Figure 10.2

We can confirm the existence of the dilemma by forming the cross product (Axi + Ayj + Azk) × (Bxi + Byj + Bzk). At first we imagine we get the same components as before. But that is because we are using the relations i × j = k, etc. Those relations do not hold here, though, if, in calculating i × j etc. we use the convention that A, B, and A × B form a right-handed system. Instead of these relations between the i, j, and k in Figure 10.2, we should, by this convention, use i × j = – k, etc; and when we do this we find that the components come out to be the negatives of what they were before.

Now this is very strange behavior for a vector: that when both it and its reference frame are reflected in a mirror its components change sign. The vectors A and B did not behave in this unseemly way.

To preserve some semblance of order we appear to be driven to say that A × B is a vector perpendicular to A and B and pointing in a direction that makes A, B, and A × B a right-handed system when we are working in terms of a right-handed reference frame, but that it points in a direction that makes A, B, and A × B a left-handed system when we are working in terms of a left-handed reference frame. At least, under this convention, if A has the same components relative to both frames and B has the same components relative to both reference frames, then A × B will also. But is it wholly satisfactory? It means that if we draw A and B flat on the page as in Figure 10.3, we cannot say whether A × B, which is perpendicular to A and B, points upward toward us or downward away from us until we decide whether, on future occasions, we shall use right-handed reference frames or left-handed ones. And if we simultaneously consult two bystanders, one of whom prefers right-handed frames, and the other left-handed frames we shall be in a grave quandary as to the direction of A × B in Figure 10.3.

Figure 10.3

Equation (10.1), if true in a right-handed frame, would be false in a left-handed one, unless we said that though C was in a right-handed frame it had to switch to in a left-handed one, and this would hardly be the sort of vectorial behavior we have hitherto regarded as proper.

11. LOCATION AND CROSS PRODUCTS

Now that our suspicions have been aroused, we begin to think of other curious things about cross products. There is the problem of location, for instance, and while this problem is not confined to cross products, we might as well bring it in here while we are on the warpath.

When we began by thinking of A × B as representing the area of a parallelogram we had A and B starting at a common point O. But in the end, we found ourselves regarding the vectors representing areas as free vectors despite the frightening pictorial consequences. Contrast this with the situation in which we deal with moments. Physically, the moment r × F of a force F about a point O would seem to have to be a vector passing through O. Yet if cross products are free vectors, the moment should presumably be a free vector. In contemplating A × B as an area, we could at least think of A and B as displacements, and thus as free vectors, thereby consoling ourselves for having to take A × B as a free vector. But with moments, the situation is different, because r and F are not free vectors. Indeed, they do not start from a common point—except in the case of zero moment. If we overlook this and still form r × F, which is what we have been doing up to now, we can not assign a reasonable location to r × F because the point P, which is the only reasonable candidate as a localizing point, is any point on the line of action of F. If we wish to use O, we may be tempted to add two cancelling forces at O, thus replacing F by a parallel force F₁, of equal magnitude passing through O, and a couple of moment r × F. Then we can form r × F, and reasonably insist that its line of action pass through O. But what shall we then do with the extra, and seemingly superfluous couple? There does not seem to be a satisfactory way out of the dilemma. We would be much better off not talking about it.

Exercise 11.1 Consider the problem of location for the relation v = ω × r.

12. DOUBLE CROSS

Are cross products really vectors? By now we may be beginning to wonder. They have magnitude and direction, they combine according to the parallelogram law, and they obey all the other addenda to the definition that we have thought of so far. Yet they behave queerly in relation to right-handed and left-handed reference frames. Perhaps we need to add another amendment to the definition. Let us not try to do so, though. Making good definitions is far too difficult. Instead, let us use our general feeling for what is appropriate and what is not, without seeking to formulate sharp criteria.

OUR SUSPICIONS SHOULD HAVE BEEN AROUSED LONG AGO BY THE SEEMINGLY INNOCENT RELATIONS i × j = k, ETC. There is something obviously wrong with them. If the magnitudes of i, j, and k are lengths measured in, say, feet, then i × j has a magnitude of one square foot and so cannot be the same sort of thing as k, whose magnitude is one foot. Perhaps we counter by arguing that since i, j, and k are of unit length, the magnitudes of i × j and k have the same numerical value, namely 1. But this will not do. If we change to inches, the former becomes 144 square inches and the latter 12 inches. Clearly, i × j cannot really be k, even though we have been able to get extremely important equations by assuming it is. [Look back at Exercise 9.6 in this connection.]

Again, suppose we consider cross products in two dimensions instead of in three. Then, given two vectors U and V, we say that their cross product has magnitude UV sin θ and that it points—where? Why, nowhere. There is no third dimension into which it can point when we have only two dimensions in all. Clearly the cross product is not a vector in two dimensions.

What about four dimensions? Here again we start by forming the parallelogram defined by two vectors U and V starting at O. But when we seek a line through O perpendicular to both U and V, instead of finding no possible directions as in the two-dimensional case, we now find far too many—infinitely many, in fact, and all of them on an equal footing. For in four dimensions, the lines through O perpendicular to both U and V in general fan out into a plane. So here too the cross product is not a vector. And our conclusion is that in the three-dimensional case too, the cross product is not really a vector, even though by some sort of good fortune it has many of the characteristics of a vector and can be handled much as regular vectors are handled, provided it is handled with care.

Since the cross product in three dimensions has so many of the characteristics of a vector, a good name for an entity of this sort would be pseudovector, and this is a particularly apt choice since it happens to be the official technical term for it. [Such entities are sometimes referred to as axial vectors.]

13. DIVISION OF VECTORS

The discussion in this section is confined to the three-dimensional case.

Having explored the multiplication of vectors, we naturally wonder how to divide one vector by another. But here too we have to realize that there is no reason why such division should be possible. We have to see if we can find a useful meaning for the quotient of two vectors, and if we cannot, we shall probably wish to keep mum about the matter. There is, of course, no guarantee that the quotient, if it exists, will be a vector: let us recall that we found two reasonable types of “products” of vectors, one yielding scalars and the other pseudovectors. We must therefore be prepared for almost anything here.

Let U and V be two vectors starting at a common point O. Denote their quotient, if it exists and whatever it may be, by Q, so that we may write:

(13.1)

Then, since we want to keep as close as possible to our usual ideas of what constitutes the operation of division, we ask that Equation (13.1) mean that

(13.2)

So the quantity Q is to be something that acts on the vector U so as to convert it into the vector V.

If U and V have the same line of action, Q is obviously just the ratio of their magnitudes. But we have to consider the case in which they have different directions. To convert U to V we have to change its magnitude by a factor V/U and also to rotate it so that its direction coincides with that of V. We find, therefore, that Q involves a ratio of magnitudes and an angular displacement about an axis through O perpendicular to U and V. Since the specification of the latter requires three numbers, we see that Q requires four. It was therefore called a quaternion by W. R. Hamilton, who originated the concept.

This book is not the place to develop the quaternion calculus in any detail, but an extremely sketchy indication of the general idea as it was developed by Hamilton will not be without interest. In what follows, we make various reasonable assertions that, in a fuller treatment, would be listed as axioms.

If triangles OAB and OLM are coplanar, and similar in such a way that OB/OA equals OM/OL rather than OL/OM, then the operation that converts into is the same as that that converts into . So the quaternions and are equal. Thus, given a quaternion, we can express it in a form that has as denominator any vector we wish in the plane of the two vectors involved. This being so, we can add two quaternions by first putting them on a common denominator, this latter being a vector lying in the line of intersection of their planes. We then have something of the form ( ) + (); and since and are vectors, these quaternions combine by the parallelogram law to yield , where is the resultant of and

Figure 13.1

We now reverse the idea. Given a quaternion , we split up into two components: lying along , and perpendicular to it, as in Figure 13.1. Then,

(13.3)

The quantity was called by Hamilton the scalar of the quaternion . Note that, in general, it is not the same as the ratio of the magnitudes of the vectors and . It is, in fact, their scalar product divided by the square of the magnitude of . [Hamilton called the ratio of the magnitudes the tensor, a word that is now used in a quite different sense, as will be seen in the next chapter.]

Exercise 13.1 Prove the above assertion about the scalar product.

Exercise 13.2 Prove that the scalar of plus the scalar of is equal to the scalar of , where OR is the resultant of and .

The quaternion in Equation (13.3) is of a special type since it involves a rotation through a right angle. Such quaternions are called right quaternions. They turn out to be of special importance.

Hamilton introduced three basic right quaternions which he denoted by i, j, and k. These are not the same as the vectors i, j, and k, though they have distant kinship with them. They are defined in terms of three mutually perpendicular vectors X, Y, and Z, starting at O, having unit magnitudes, and forming a right-handed system. The right quaternion Z/Y converts Y into Z. It is denoted by i, and it involves a rotation through 90° about X. From a diagram, one easily sees that i also converts Z into – Y. So we may write:

Similarly, j and k are defined by:

Exercise 13.3 Show that the scalars of i, j, and k are zero.

By i² we mean the result of applying i twice. Since i converts Y into Z, and the second i converts Z into – Y, i² converts Y into – Y and we therefore write i² = – 1.

Exercise 13.4 Prove, similarly, that j² = – 1 and k² = – 1.

By ij we mean the result of applying j and then i. Since j converts X into – Z, and i converts Z into – Y (and thus, as is easily seen, – Z into Y), ij converts X into Y, which is just what k does. So we write ij = k. However, since i converts Y into Z, and j converts Z into X, ji converts Y into X. But k converts Y into – X. Therefore we write ji = – k.

Exercise 13.5 Prove, similarly, that jk = i, kj = – i, ki = j, and ik = – j.

Exercise 13.6 Prove that ijk = – 1; also, that i(ij) = (ii)j = – j, i(ji) = (ij)i = j, and i(jj) = (ij)j = – i. [Results such as these show that i, j, and k obey the associative law of multiplication, even though they do not obey the commutative law.]

Consider now a right quaternion, such as in Equation (13.3). We can associate with it an entity called its index. This consists of an arrow-headed line segment that lies, in the right-handed sense, along the axis of the rotation performed by the quaternion , and it has a length representing the number OP/OA. Its direction is thus perpendicular to the plane of of and . Since all right quaternions involve rotations through the same angle, 90° , their indices combine according to the parallelogram law.

Exercise 13.7 Prove this. [If , then . We obtain the indices of these three quaternions by multiplying , , and by the same quantity, 1/OA, and rotating them in their plane through 90° about an axis through O perpendicular to their plane. Compare the analogous discussion in connection with cross products of perpendicular vectors.]

Exercise 13.8 Show that the index of i is the vector X.

Thus, provided we stay with right-handed systems, the indices behave like vectors. (They are actually pseudovectors.) So we can express the index of the quaternion in the form,

(13.4)

where (x, y, z) are the coordinates of P’ relative to the base vectors X, Y, and Z. Since the indices X, Y, and Z belong respectively to the quaternions i, j, and k, we have, corresponding to Equation (13.4), the quaternion equation:

(13.5)

and, therefore, such quaternion quantities obey the same algebraic rules as their vectorial indices. If we denote the scalar of the quaternion in Equation (13.3), namely , by w, we see from Equation (13.5) that this quaternion, which is any quaternion Q, can be expressed in the form:

(13.6)

The significant thing about this is that the quantities on the right, and thus quaternions in general, obey all the usual rules of algebra except the commutative law of multiplication. Quaternions are thus of considerable theoretical importance quite apart from their relationship to vectors. They constitute what algebraists call a noncommutative field (not to be confused with what physicists call a field—when they are talking physics or agriculture rather than algebra). Thus quaternions, like complex numbers, are a natural extension of the idea of number, and are often called hypercomplex numbers.

It is worthwhile contemplating Hamilton’s achievement. You and I, if the idea had ever occurred to us of discussing the quotient of two vectors, would have noted, perhaps, that it involved the ratio of their magnitudes and also something akin to an angular displacement. The former we would have thought of as a scalar of no particular charm, and the latter would have scared us away from pursuing the idea further. Hamilton, by defining a new scalar, and focusing on right quaternions and relating them to their indices, not only created the essentials of scalar and vector products, but also constructed an algebra of extreme elegance.

Scalar and vector products were conceived by Hamilton and H. Grassmann independently in the 1840’s. Since Hamilton’s quaternion calculus proved unwieldy for applications to the physics of the nineteenth century, J. W. Gibbs and, independently, O. Heaviside, extracted from it the much more convenient vector calculus that we have been studying in this book. They did so in a creative way, and in the face of bitter opposition from the supporters of quaternionic methods.

When we tried to regard angular displacements as vectors, we were shocked and thwarted by their properties of noncommutativity and worse. We may well wonder, then, that these properties do not destroy Hamilton’s structure. One reason is that, for example, the quaternions i, j, and k are not angular displacements but merely relations between pairs of vectors; and another is that Hamilton made the noncommutativity apply to multiplication rather than addition. Even so he was, for a long time, frustrated by the lack of commutativity of angular displacements, and it came to him as a staggering discovery that the key to progress lay in the heretical renunciation of the commutative law of multiplication. When, finally, the idea did come to him he realized immediately that it was a momentous one. But he could not know that almost a century later noncommutative multiplication would prove to be of crucial importance in the quantum theory, and that the spin of atomic particles would actually be represented by symbols satisfying equations corresponding to:

ij + ji = 0,

as would other fundamental quantities of atomic physics.

Exercise 13.9 Imagine unit vectors X, Y, and Z along the respective coordinate axes in Figures 6.2 and 6.3. Show that the vector X in Figure 6.3c occupies the position of Y in Figure 6.3a. Denote this by X → Y. Show similarly that Y → Z, and Z → X. Show that the corresponding results for Figure 6.2 are X → – Z, Y → X, and Z → – Y. Note that X → Y is the effect of the quaternion k, while Y → X is that of – k, these results being in conformity with ij = – ji = k; but note that X = – Z is not the effect of k but of j. Consider the other effects similarly. [This exercise indicates the subtlety and boldness of Hamilton’s concept: though i, j, and k are related to angular displacements, ij does not yield an angular displacement about the z-axis. Despite this, Hamilton, with profound mathematical instinct, did not shy away from quaternion relations ij = k and the like, as you and I would have. He later found a way to represent general angular displacements by means of quaternions.]

6

TENSORS

1. HOW COMPONENTS OF VECTORS TRANSFORM

The time has come for a new approach. To make further progress with vectors, we concentrate on their components. This may seem an unpromising tactic, since the components change whenever we change the reference frame, whereas relations among vectors are independent of the choice of the frame. But by studying how such changeable things as components can embody this objectivity of vectors, we come to an important generalization: the concept of a tensor.

We do not propose to discuss location. For this and other reasons we use a fixed origin O and assume that all vectors and the like start at O. To get the feel of how components of vectors behave, we start with a simple numerical example in two dimensions. Suppose we are given

(1.1)

and we change to a new reference frame with base vectors x and y where

(1.2)

To find how the components change, we first solve Equations (1.2) for ex and ey in terms of x and y; the result is:

(1.3)

Then from Equations (1.1) and (1.3) we have:

(1.4)

But the components and in the new reference frame must satisfy an equation corresponding to Equation (1.1), namely,

(1.5)

So Equation (1.4) shows that the new components of V are given by

(1.6)

Compare the coefficients in Equations (1.6) and (1.3).

Exercise 1.1 If V = Vxex + Vyey, find its components relative to x and y, where x. = 4ex + ey and y = 2ex + ey. [Ans. From and ey = – x + 2y, we obtain and

. Explain why the coefficients of the V’s automatically come out to be the same as those of the e’s, though in different places.]

Exercise 1.2 Like Exercise 1.1 for x = x – 2ey and , and compare with the example worked out in the text.

2. THE INDEX NOTATION

In three dimensions, we have been denoting the base vectors by ex, ey, and ez, and the components of a vector by Vx, Vy, and V,. In four dimensions we already find ourselves in a squeeze when we try to think of what letter to add to x, y, and z, and in n dimensions, especially when n > 26, the inconvenience of using different letters becomes considerable. So we use numbers instead, denoting the base vectors, for example, by e₁, e₂, ..., en. Then we denote all these e’s collectively by the single symbol eα, where a takes on the values 1, 2, ..., n, in turn. Thus, in three dimensions, a takes on the values 1, 2, and 3, and eα stands for e₁, e₂, and e₃, which are our new symbols for ex, ey, and ez. Note that eb, with b running over the same range as a, stands for the same set of e’s as ea did.

For a reason that will become apparent later, we use an upper index when denoting the components of a vector V [but more about this later]. Thus we denote the components collectively by Va, which stands for V¹, V², . . . , Vn. Note that, for example, V² does not mean “V squared.” It means the second component of V, and is what we have been calling Vy in three dimensions.

Using the index notation, we can write:

(2.1)

where the symbols means that we must let a go from 1 to n and must then add the terms that result. For n = 2, Equation (2.1) yields essentially Equation (1.1).

The index notation plays an important role in the theory of relativity. Einstein noticed that, when an index letter occurred in an upper position and the same index letter occurred in a lower position in the same term, then almost always a Σ occurred too. So he introduced the summation convention according to which an index repeated in this way automatically implies a summation. This removes the unsightly Σ’s and makes the work look neater and simpler.

Using the summation convention, we write Equation (2.1) in the form:

(2.2)

which stands for

Exercise 2.1 If Va = (3, 2) and Wa = (1, 5), show that Va Wa = 13.

Exercise 2.2 If Vb = (1, 3, 6) and Wa = (2, 7, – 3), show that VaWa = Vb Wb = 5.

Exercise 2.3 If is such that and = 4, show that = 9. What is the value of ?

Exercise 2.4 How many quantities are there in if a and b go from 1 to 3? [Ans. 9.] How many if they go from 1 to n?

A linear change from one set of base vectors ea to another set a with the same origin can be written in the form:

(2.3)

For example, if a, b have the range 1, 2, and

(2.4)

then equation (2.3) is essentially the two equations in (1.2); for when a = 1 it yields

and similarly for a = 2.

Exercise 2.5 Find the values of for Exercises 1.1 and 1.2.[Ans. In Exercise 1.1,

Exercise 2.6 Write out Equation (2.3) for the three-dimensional case. [Ans. One of the equations is

When we solve Equation (2.3) for the e’s in terms of the ’s, we get an equation of the form,

(2.5)

Thus if a, b have the range 1, 2, and if

(2.6)

Equation (2.5) becomes essentially the pair of Equations (1.3).

Exercise 2.7 Write down the values of for Exercises 1.1 and 1.2 and save the results for later.

Substituting from Equation (2:5) into Equation (2.2), we have

(2.7)

But if the components of V relative to a are we must have (compare Equation (1.5))

(2.8)

When we compare this with Equation (2.7) we are puzzled by the fact that appears there but α here. But then we remember that we can equally well write Equation (2.8) as

(2.9)

And comparing this with Equation (2.7) we see that

(2.10)

This equation means exactly the same thing as do , etc. For the indices that occur on both sides of the equality signs merely label n equations one after the other, while the summed indices indicate the same summation in each case. Indices that are summed according to the summation convention are called dummy indices.

Exercise 2.8 In deriving Equation (2.7), we moved parentheses and changed the order of terms in a product. Our experiences with cross products make us wary of such maneuvers. Show, for the two-dimensional case, that and similarly that . Then extend the results to the n-dimensional case.

From Equations (2.10) and (2.5), we see that is given in terms of by means of the same that gives eα in terms of α (not α in terms of ea). Accordingly V is called a contravariant vector. Note, though, that in Equation (2.10) the summation is over the lower index of while in Equation (2.5) it is over the upper index. It is this that causes the displacement of the numbers noted when we compared Equations (1.3) and (1.6).

How are, , and related? To find out, we introduce an index c having the same range as a and b, and we write Equation (2.5) in the form:

Applying this to the right-hand side of Equation (2.3) we see that

(2.11)

For convenience, let us write,

(2.12)

Then Equation (2.11) takes the form:

(2.13)

Since this is summed over c, we must have

(2.14)

Because of its importance, this quantity is called the Kronecker delta, in honor of a famous mathematician who did not invent it. Equation (2.12), with defined by Equation (2.14), gives us one set of the relations we seek between and . It can be written, of course, in the equivalent form:

(2.15)

Exercise 2.9 Why are Equations (2.15) and (2.12) equivalent?

To obtain another relation, we repeat the process in reverse order, applying Equation (2.3) to Equation (2.5). Thus,

from which it follows that

(2.16)

While this can be written in the equivalent form

(2.17)

it is not the same as Equations (2.12) and (2.15). Note that the summation is over differently situated indices.

Exercise 2.10 Show that in two dimensions . What is it in n dimensions? .

Exercise 2.11 Show that , and only the first term survives. Similarly, in only the second term survives. And so on.]

Exercise 2.12 Show that , and that

Exercise 2.13 Show that

Exercise 2.14 Show that and note that the effect of a Kronecker δ is merely to change one index letter on a T to a new letter.

Exercise 2.15 Using the remark in Exercise 2.14, show that What would be the value of ?

Working with the base vectors eα, we found that A cancels the effect of A, and A cancels the effect of A. Because of this, each is said to be the inverse of the other. This cancelling is not confined to the e’s. For example, we can easily see that if we solve Equation (2.10) for va in terms of the result must be:

(2.18)

For, using Equation (2.17), we have:

Exercise 2.16 Starting with Equation (2.10) and applying Equation (2.18) to the Va there, show that the A cancels the A, and note that this follows from Equation (2.15), with appropriate indices, rather than from Equation (2.17).

3. THE NEW CONCEPT OF A VECTOR

[In this book we do not consider coordinate systems that are curvilinear or marked off nonuniformly. What follows would need significant modification before it could apply to such systems.]

Take a reference frame with origin O and base vectors eα. The components of the position vector, r, of a point P are the coordinates of P relative to this frame. In the index notation, we denote the coordinates by xa, and this lets us write the compact equation:

(3.1)

Equation (2.10) holds for the components of any contravariant vector. Applying it to r, we see that if we go over to a new reference frame having the same origin but different base vectors α given, as in Equation (2.3), by

the coordinates of P change to where

(3.2)

This suggests a new way of thinking about vectors. We do not harp on arrow-headed line segments. Instead, we say first that a contravariant vector is an entity that has a different set of n components in each coordinate system —one set per system. And then we add that, if the components are Vα when the coordinates are xa, and when the coordinates are , and if is related to xa by Equation (3.2), then

(3.3)

this last being essentially none other than Equation (2.10).

As thus expressed, this is disconcertingly unpictorial, but therein lies its strength, as we shall see. Inasmuch as it deals with components, it embodies an important aspect of the parallelogram law. But for full embodiment, we must add that if, in a particular coordinate system, the components of two vectors are Ua and Va, then the components of their resultant are (Ua + Vα).

We have now to settle an important question: how do we know that we can get away with our statement that a vector, as here regarded, has a unique set of components in a given coordinate system? Not simply because we asserted as much: the law of transformation (3.3) might betray us. For example, if we went from coordinates xa to coordinates and from to , the components would go from Va to and then to . But if we went directly from xa to , would the corresponding be the same as the previous ? The answer is obviously yes, because Va transforms exactly as xa transforms. So the uniqueness of the components in a given coordinate system is assured.

Now comes the crucial theorem: If Ua and Va are the components of two contravariant vectors in a given coordinate system, and

(3.4)

then a corresponding relation holds in all coordinate systems.

This is easily proved. For since

(3.5)

we have, by subtraction,

(3.6)

But the right-hand side of Equation (3.6) vanishes, by Equation (3.4), since Ub – Vb is zero for each value of b. So the left-hand side also vanishes and, therefore,

(3.7)

We now understand how the changeable components can embody objectivity. For if Ua = Va, this equation tells us something that holds for all coordinate systems and, thus, something that is actually independent of the coordinate system used.

Exercise 3.1 By cUa we mean (cU¹, cU²,..., cUn), c being a number. Prove that if Ua are the components of a contravariant vector, cUa are components of one also. [Simply check that the components of cUa in any given coordinate system are unique, that they transform according to the law of transformation (3.3), and that they combine correctly with other contravariant vector components. The proofs are extremely easy.]

Exercise 3.2 Let l, m, and n be numbers, and Ua, Va, and Wa components of contravariant vectors in a given coordinate system. Show that if lUa + mVa – nWa = 0, a corresponding relation holds in every coordinate system.

Our new way of looking at vectors suggests new possibilities, for we can think of other ways of attaining uniqueness of components and objectivity. Thus, suppose we have an entity that has components Va when the coordinates are xa, and suppose that when we go to new coordinates by means of Equation (3.2), the components transform according to the law:

(3.8)

Then the components in a given coordinate system are obviously unique, because Va transforms as the base vectors ea do. Also, if Ua and Va are components of two such entities in the xa coordinate system, and Ua = Va, then a corresponding relation holds in every other coordinate system. The proof so closely parallels that for Ua = Va that we shall not bother to write it out.

Clearly, this objective entity with components Va is entitled to the name vector. It is called a covariant vector.

We have only begun, and already we have learned something surprising: there are two types of vectors. Do not make the mistake of thinking that covariant vectors are vectors of a sort that we knew about before, say, pseudovectors. They are not. If we confine ourselves to ijk reference frames, including both right-handed and left-handed kinds, it turns out that contravariant and covariant vectors transform in the same way and are thus indistinguishable. But consider, for example, how Va and Va behave under the simple change of reference frame α = 12ea. From the laws of transformation, we easily see that but a. [Note that ea are contravariant vectors, not covariant vectors. They are printed in boldface type and the index a does not label their components.]

Exercise 3.3 In the xa coordinate system in two dimensions, a contravariant vector and a covariant vector have the same components: Va = (3, 5), Va= (3,5). The coordinates are changed to where 2x² and . Find and . [Ans. , . Note that because Equation (2.18) applies to the x’s one can obtain by solving for the x’s in terms of the . One finds . Note also that = 34. This is no accident.]

Exercise 3.4 Like Exercise 3.3 if Va = (2, 3), Va = (2, 3), and . [Ans. . If you mix up the values of and you will get the wrong answer . Note that VaVa = .]

4. TENSORS

Later we shall tell how to visualize a covariant vector. Meanwhile we find other ideas crowding in on us. For example, if Uα and Vα are components of two contravariant vectors, we can form their various products two at a time, namely U¹ V¹, U¹ V², U¹ V³,..., U¹ Vn, U² V¹, U² V²,..., up to Un Vn, there being n² of them. The index notation lets us represent all n² of them by the single symbol Ua Vb. Since Ua and Va are unique in each coordinate system, the same must be true of Ua Vb, and since Ua and Va have the objective property, there is a chance that Ua Vb will too. The crucial thing is the way Ua Vb transforms, and we easily see that

(4.1)

From this, much as before, we find that if Ra and Sa are also components of contravariant vectors, and Ua Vb = Ra Sb, then a corresponding relation holds in every coordinate system.

Exercise 4.1 Prove this.

But now the thought strikes us that we do not need products of vector components in order to have uniqueness and objectivity. We could imagine a a more general entity having n² components Tab in the xa coordinate system, these components transforming as Ua Vb did, namely, according to the law:

(4.2)

Such an entity will have only one set of components per coordinate system. Also, if two such entities have components Sab and Tab in the xa coordinate system, and Sab = Tab, then a corresponding relation will hold in every coordinate system. For from Equation (4.2) and a similar one for Sab we obtain:

But the n² quantities Scd – Tcd are all zero. Hence .

Obviously, an analogous entity with components Tab that transform as the quantities UaVb do will have similar objectivity. But why stop here? We can have analogous objective entities with components . All such entities are called tensors.

A tensor is an entity having a unique set of components in a given coordinate system xa and such that these components transform according to the law:

(4.3)

when the coordinates are changed as in Equation (3.2).¹¹

Unfortunately we shall have scant opportunity in this book to indicate the crucial role of tensors. Though the above definition is abstract to the point of aridness, tensors are of major importance since they are objective entities. If we wish to represent objective phenomena mathematically, we can hardly escape them.

The tensor law of transformation is easy to remember. For each contravariant index write an A, and for each covariant index an A. Then first put in indices as shown:

and after that, fill in the dummy indices in pairs; note that if, for example, a dummy is placed on A c, the corresponding dummy goes on T in the position that c occupied on T.

Exercise 4.2 Cover up Equation (4.3) and write down the laws of transformation for , and Tabc.

Since it becomes tedious to say “the tensor having components in a given coordinate system xa, ” mathematicians say simply “the tensor and use similar contractions on other occasions, as in the following exercise.

Exercise 4.3 If and are tensors [note the language], and [again, note the language], prove that a corresponding relation holds in every coordinate system. Also prove that if and are tensors, l, m, and n are numbers, and , a corresponding relation holds in every coordinate system. [The meaning of is reasonably clear.]

Exercise 4.4 Show that the results in Exercise 4.3 would not hold for , and .

Exercise 4.5 Generalize Exercises 4.3 and 4.4 to the case of tensors having any number of indices.

Exercise 4.6 If and Tcd are tensors, show that is a tensor. [Hint: From the laws of transformation of and Tcd we see at once that , = and that is practically all there is to it.]

Exercise 4.7 Generalize Exercise 4.6 to the general case of and .

5. SCALARS. CONTRACTION

The number of indices of a tensor is called the order of the tensor, and sometimes its valence. Thus is of order 3. Often we say that such a tensor has contravariant order 2 and covariant order 1, or that it is of the second contravariant order and the first covariant order. Vectors are tensors of order 1. What about tensors of order zero? Since they have no indices we write them as T (or S, or Q, or some other index-free symbol). The corresponding law of transformation, by Equation (4.3), is just

Thus, tensors of zero order are scalars, a fact that probably heightens our respect for the concept of a tensor.

If Ua and Va are vectors, what can we say about Ua Va? By Exercise (4.7) we know that Ua Vb is a second-order tensor; but UaVa has a pair of dummy indices and thus, in a sense, no indices: Ua Vb has n² components, but Ua Va has only one. The notation suggests, as did Exercises (3.3) and (3.4), that Ua Va is a scalar, and we can easily prove it is one. For we may write the laws of transformation of Ua and Va as and Therefore, using Equation (2.17), we have:

Exercise 5.1 If is a tensor, prove that is a scalar.

Exercise 5.2 If is a tensor, prove that is a contravariant vector.

Exercises 5.1 and 5.2 are special cases of the following general theorem:

If is a tensor of order r, then is a tensor of order r – 2.

We omit the proof. It is the same as that in Exercise 5.1, except for the presence of dots all over the place. The process of summing over a repeated index, one contravariant and the other covariant, to obtain a tensor of lower order is called contraction.

The proof that contraction yields a tensor depends crucially on the presence of an A and an A that combine to form a δ. So, even if an expression like Taa or Taa were summed over a, the result would not be a scalar. We can easily verify this by referring to Exercise (3.3). There = 9 + 25 = 34, but which is too unpleasant to be worth evaluating but is obviously not 34. Also, was equal to 34 there too, but comes to 85. There is something disquieting about this, as we shall see later.

6. VISUALIZING TENSORS

The title of this section almost constitutes fraudulent labelling. The fact is that, with a few exceptions, tensors cannot be readily visualized.

Scalars we are familiar with, and therefore, perhaps we feel that we can visualize them. Contravariant vectors we picture as arrow-headed line segments, albeit with a curious mode of combination. But what of covariant vectors?

If Va is a covariant vector, the quantity Vaxa is a scalar, since xa is a contravariant vector. So if we write the equation,

(6.1)

it will mean the same thing in all coordinate systems. In two dimensions it is, in the x, y notation, just

which represents a straight line. So we can visualize Va in two dimensions as this straight line. But do not jump to conclusions. It is not a line segment—with or without an arrowhead. It is a whole line. Moreover, in three dimensions Equation (6.1) becomes

which represents a plane, not a line. So we can there picture Va as this plane. For the record we mention, without elaboration, that in n dimensions we can picture Va as an (n – 1)-dimensional hyperplane.

A second-order covariant tensor Tab can be pictured with the aid of the equation:

(6.2)

In two dimensions this represents an ellipse, a hyperbola, or a pair of lines.¹² In three dimesions it represents an ellipsoid, a hyperboloid, a cylinder, or a pair of planes.¹² As we see, the game of visualization becomes more complicated the greater the number of indices, and we shall pursue it no further except to remark that the above process does not apply to contravariant indices.

Exercise 6.1 In two dimensions if T₁₁ = T₂₂ = 1, T₁₂ = T₂₁ = 0, in rectangular Cartesian coordinates, show that Equation (6.2) represents a circle.

7. SYMMETRY AND ANTISYMMETRY. CROSS PRODUCTS

If a tensor Tab is such that

(7.1)

in one coordinate system, a similar relation holds in every coordinate system. For we have

(7.2)

(7.3)

But in Equation (7.3) we can replace Tcd by Tdc, because of Equation (7.1). Therefore,

Since the c’s and d’s are dummy indices, we can change them to other letters.

So we change both of the present c’s to d’s and, at the same time, both of the present d’s to c’s. The result is:

which is identical with

Comparing this with Equation (7.2) we see that .

Thus the relation Tab = Tba is an objective one. We say that this Tab is symmetric in the indices a and b. In a similar way, one finds that the relationship

(7.4)

is also an objective one. Such a tensor is said to be antisymmetric, or skew symmetric.

Exercise 7.1 If T , then Tab = Tbα. Calculate if , and verify that

Exercise 7.2 If , then Tab = - Tba. Calculate for the coordinate transformation in Exercise 7.1 and verify that .

Exercise 7.3 Prove that the relationships Tab = Tba and Tab = – Tba are objective.

Given two vectors Ua and Va, we can form the second-order antisymmetric tensor:

(7.5)

Let us write out its components in the three-dimensional case. We easily see that T¹¹, T²², and T³³ are zero. As for T²³, T³¹, and T¹², they are respectively U² V³ – U³ V², V³ V¹ – U¹ V³, U¹ V² – U² V¹; and if we look closely we recognize that THEY ARE OF THE SAME FORM AS THE COMPONENTS OF THE CROSS PRODUCT U x V RELATIVE TO A RIGHT-HAND UNIT ORTHOGONAL TRIAD i, j, k. This rouses our interest, and we naturally wonder what the remaining components of Tab, namely T³², T¹³, and T²¹, will be. But, since Tab is antisymmetric, they are obviously the negatives of the above—a fact that reminds us of the trouble we had with the changing sign of U x V when we went from a right-handed to a left-handed reference frame.

It can be shown that if we confine ourselves to right-handed unit orthogonal reference frames the components T²³, T³¹, and T¹² transform like the components of a vector (as also do the components T³², T¹³, and T²¹). It is this fact that caused our luck to hold when we dealt with cross products as though they were vectors. If you look back you will see that we worked with their components only with respect to ijk reference frames. Had we used more general frames we would have encountered difficulties. Indeed, even if we had used base vectors that were not of unit magnitude, though still mutually perpendicular, we would have encountered trouble—as we realize from the awkwardness connected with i × j = k.

Exercise 7.4 Write out the four components of UaVb – Ub Va for the two-dimensional case, and note that two of them are necessarily zero, and the other two, if they are not zero, are equal except for sign. Show that this holds for any second order antisymmetric tensor Tab or Tab in two dimensions.

Exercise 7.5 Show that in four dimensions a second-order antisymmetric tensor Tab or Tab has four components that are necessarily zero, and twelve components which, if they are not zero, are equal in pairs except for sign.

Exercise 7.6 In two dimensions, if Tab is an antisymmetric tensor it involves, essentially, the single numerical quantity T¹²(= – T²¹). Show that T¹² does not transform like a scalar but that

[Use the facts that T¹¹ = T²² = 0; T²¹ = – T¹².]

In Exercise 7.6, if we restrict ourselves to transformations between right-handed unit orthogonal base vectors, it can be shown that . For such reference frames, then, Tab does behave like a scalar. But if we go from a right-handed frame of this sort to a left-handed one, or vice versa, it changes sign. It is called a pseudoscalar.

Usually the components of a tensor change when the coordinates are changed. But consider a tensor that has components in the xa coordinate system. In a new system its components are:

by Equation (2.15). So this particular tensor has the unusual property that its components are the same in every coordinate system. We therefore speak of “the tensor .”

Consider a Kronecker delta with two covariant indices:

(7.6)

If a tensor Tab has components δab in one coordinate system, does it too have the same components in all systems? We have

but we cannot continue as we did before. The right side does not, in general, reduce to δab. So the answer to our question is no. Putting it another way, we can say that “δab is not a tensor.”

Exercise 7.7 Show that “δab is not a tensor.”

8. MAGNITUDES. THE METRICAL TENSOR

If (Vx, Vy, Vz) are the components of a contravariant vector V relative to i, j, k, the square of the magnitude of V is given by:

(8.1)

This fact gave us no qualms when we first encountered it. But now it puzzles us. For we regard (V)² as a scalar, but the right-hand side of Equation (8.1) is not a scalar. We could express it as Σ Va Va, but that does not fit the summation convention. So we try δab Va Vb. This does fit our notation, and if δab were a tensor, this expression would indeed be a scalar, being the double contraction of a tensor. But as we have just seen δab is not a tensor.

If we insist on (V)² being a scalar, we have to say that, though δab is not a tensor (in the sense that it does not retain the components δab in all coordinate systems) there nevertheless must exist a tensor, say gab, that has the components δab in an ijk reference frame:

(8.2)

Then we can write Equation (8.1) in the form

(8.3)

with the right-hand side now manifestly a scalar. Since this tensor equation holds in one coordinate system, it holds in all coordinate systems. The tensor gab is called the metrical tensor. As defined, it is obviously symmetric, since δab = δba and symmetry in one reference frame implies symmetry in all. And we note that we can make Equations (8.2) and (8.3) apply to the n-dimensional case by letting a, b have the range 1, 2, . . . , n, the coordinates being the analogue of those in an ijk frame.

Let us get an idea of the role of the metrical tensor by discussing a simple case. Consider a vector V with components Va relative to ijk. Suppose we stretch the x coordinate by the transformation

(8.4)

Then

(8.5)

Also, since , we have , and from the law of transformation,

we find after a short calculation that

(8.6)

Had we taken the sum of the squares of the components they would not have given the same result as the sum of the squares of the components Va. But the gab acts as a sort of counterpoise to annul the effect of the stretching of the x¹ coordinate, and we have:

Exercise 8.1 Apply the transformation γx³ to the components of gab in Equation (8.2) and find Also find in terms of Va. Then verify that comes to the same thing as δabVa Vb.

Exercise 8.2 In one dimension, if V¹ = 2 and g₁₁ = 9, what is V? [Ans. 6.]

Exercise 8.3 If Va = (3, -1) and , show that V = √3.

Exercise 8.4 If Va = (1, 2, 2, 4) and gab = ,

show that V = 5.

Exercise 8.5 Given the coordinate transformation = x², and , solve for the x’s in terms of the ’s and thus obtain . Using this, show that the gab in Equation (8.2) transforms into ab where ₂₂ = 2 and ₁₂ = ₂₁ = -1, while the other components are unchanged. Then show that . [The coefficient – 2 in the last term arises from the addition of two terms, one involving ₁₂ and the other, ₂₁.]

9. SCALAR PRODUCTS

Given two contravariant vectors Ua and Va, we can form the scalar gab Ua Vb. What does it represent?

From the analogue of Equation (8.2) in the n-dimensional case, we see that in uniform unit rectangular coordinates,

which, in the three-dimensional case, is just the scalar product U.V. Since gab UaVb is a scalar, it has the same value in all coordinate systems. Therefore even when the coordinates are such that terms like 5U¹ V¹ and 3U¹ V² appear on the right, it still represents the scalar product. Indeed, we call it the scalar product of U and V even in the n-dimensional case.

If we think of U and V as arrow-headed line segments starting at O, we may write, therefore,

(9.1)

where θ is the angle between them. Therefore,

(9.2)

This argument is all very well for three dimensions. But isn’t it a little reckless for n dimensions?

Not really. One of the nice things about a tensor equation is that we can check its validity by using any convenient coordinate system. So let us take coordinates such that U lies along the x¹-axis and has components Ua = (U¹, 0, 0, . . . , 0), and V lies along the x²-axis and has components Va =

(0, V², 0, . . . , 0). Then Equations (9.1) and (9.2), when expanded, yield the same equations we would get if we worked two-dimensionally in the plane containing U and V.

Exercise 9.1 Show that gab Ua Vb = gab Va Ub. [Use the symmetry of gab.]

Exercise 9.2 If Ua = (2, 1) and Va, gab are as in Exercise 8.3, find gab Ua Vb and cos θ. [Ans. 2, √2/3√3.]

Exercise 9.3 If Ua = (3, 1), Va = (2, 5), and , find U,

V, and cos θ. [Ans. .]

Exercise 9.4 If Ua = (5, 0, 0, 0), and Va and gab are as in Exercise 8.4, find gab Ua Vb and cos θ. [Ans. 5, 1/5√5.]

Exercise 9.5 Show that the angle θ between the x¹- and x²-axes is given by cos . [Consider Ua = (U¹, 0, 0, . . . , 0) and Va =

(0, V², 0, . . . , 0).]

Exercise 9.6 Show that if gab = 0 when a ≠ b, the coordinate axes are mutually perpendicular. [Use the preceding exercise.]

Exercise 9.7 If Ua = (m, 2), and Va and gab are as in Exercise 8.3, what must be the value of m if Ua is perpendicular to Va? [Ans. 0.]

Exercise 9.8 Using Exercise 9.5, show that the angle between the and in Exercise 8.5 is 135°. Then see if you can obtain the same result by considering the diagrammatical significance of x². [The second part is easy once you see it. But there is an enticing false trail that leads to the incorrect answer 45°.]

Exercise 9.9 Show that in Exercise 8.5 the scale on the is √2 times that along the and . [Hint: Consider a vector with components (0, 1, 0) which thus lies along the and stretches from the origin to the “1” mark on that axis. Find its magnitude, using. the values of , and compare it with the magnitudes of vectors with components (1, 0, 0) and (0, 0, 1).]

Exercise 9.10 If xa are coordinates in three dimensions associated with an ijk frame, and we define new coordinates by find the cosine of the angle between the and axes, and the relationship of the scales along these axes. [Ans. We find that and , and the ratio of the scales is √5.]

Exercise 9.11 What are the components of the metrical tensor in two dimensions when the scale along the x¹-axis is such that the “1” mark on it is 2 inches from the origin, the scale along the x²-axis is such that the ”1” mark on it is √3 inches from the origin, and the cosine of the angle between the axes is ? [Ans. g₁₁ = 4, g₂₂ = 3, g₁₂ = √3.]

When we work with ijk frames, the metrical tensor has components δab. This makes it practically invisible, because we rarely write coefficients that are 1 or 0. When, for example, in the two-dimensional case we write

(V)² = (V¹)² + (V²)²,

we are not likely to realize that this is really

(V)² = 1V¹ V¹ + 0V¹ V² + 0V² V¹ + 1V² V² = δab Va Vb = gab Va Vb.

Since the components of a displacement are pure numbers but its magnitude is a length, we see that the components of the metrical tensor must have the dimensions of the square of a length. And this brings us to something that may have been worrying you for quite a while. When we complained about i × j = k, should we not also have complained about ii = 1 ? That equation, too, is unsatisfactory, since the left-hand side is the square of a length, but the right hand side looks like a pure number.

With the aid of the metrical tensor, we can clarify the situation. The vector i has components (1, 0, 0) relative to the ijk frame; the 1 is a pure number, not a length. Denote these components of i by ia. Then,

ii = gabiaib = g₁₁i¹i¹ = g₁₁,

and though g₁₁ here has the value 1, the right-hand side is now explicitly the square of a length, as it has to be.

Exercise 9.12 A vector Va lies along the x¹-axis and has components (V¹, 0, 0, . . . , 0). Show that its magnitude is not V¹ but V¹.

10. WHAT THEN IS A VECTOR?

This being a book about vectors, we have presented only the sketchiest account of tensors—barely enough to illustrate the advantages of thinking of vectors in terms of the way their components transform.

We have one final point to make. Notice that we defined contravariant vectors and covariant vectors—indeed, tensors of all ranks—before we introduced the metrical tensor. Suppose there were no metrical tensor. What could we then say about the magnitudes of vectors? Or about the cosines of the angles between them?

You may be tempted to argue that such questions prove that there has to be a metrical tensor. But actually there does not. Mathematicians often work with spaces that do not possess one; they call them nonmetrical spaces.

Thus vectors do not have to have magnitudes. And this is as good a place as any to stop.

INDEX

Aberration of stars

Abstract vectors

Acceleration

average

definition

force and

instantaneous

Ace of spades

Addition of vectors

areas and

Algebra, vector (see Vector algebra)

Algebraic’ notation

for addition of vectors

for equality of vectors

for multiplication of vectors by numbers

for subtraction of vectors

for vectors of zero magnitude

Analytical geometry, formulas of

Angles, direction

Angular displacements:

noncommutativity

parallelogram law and

Angular momentum

Angular velocity

as vector

Antisymmetric tensors

Apples

Areal shift

Areas:

addition of vectors and

projections

vectorially equivalent

vector products and

Arithmetical symbols

Arrows

Associative law:

in addition of vectors

in products of vectors

Astronomical applications of vectors

Atomic theory, noncommutative, multiplication in

Average acceleration

Average speed, definition

Average velocity, definition

Axial vectors

Bound vectors

definition

forces as

subtraction

Boundary curves

Calculus

vector

Cartesian coordinates

Commutative law:

in addition of vectors

in products of vectors

Components:

of cross products

definition of

reference frames and

scalars and

of tensors

of vectors

Component vectors

Conservation of momentum

Contraction

Contravariant vectors:

components

covariant vectors compared with

definition

scalar products and

visualizing

Coordinates:

Cartesian

curvilinear

definition

nonuniform

polar

spherical polar

Cosines:

direction

law, derivation

Countable items, concept(see also Numbers)

Couples

Covariant vectors, definition of

Cross products (see also Vector products)

components

as free vectors

called vector products

of i, j, and k

as tensors

Curvilinear coordinates

Diogenes

Directed lines, definition

Direction angles, definition

Direction cosines

definition

uses

“Disforcement,”

Displacement equations

Displacements

definition

subtraction

as vectors

velocity and

Distributive law

in products of vectors

Division of vectors

Dummy indices

Einstein

Equality of vectors

notation for

Equilibrium

Euclid’s axiom of parallels

Forces

acceleration and

combination

as couples

magnitude and

moments

nature

parallelogram law and

rigid bodies and

as vectors

Foucault’s pendulum

Four dimensions

index notation and

Fractions

Free vectors

cross products as

definition

displacements as

equality

starting point for

velocities as

Friction

Geometrical vectors

Geometry, analytical (see Analytical Geometry)

Gibbs, J. W.

Grassmann, H.

Gyroscopes

Hamilton, W. R.

Heaviside, O.

Hypercomplex numbers

i × j = k obviously wrong(see also 104, 124)

Importance of vectors

Index notation

Instantaneous acceleration

Instantaneous speed

Instantaneous velocity

Irrational numbers

Items, countable, concept(see also Numbers)

Journeys and vectors

Kronecker delta

Leacock, Stephen

Left-handed reference frames vector products in

Leibnitz, Gottfried Wilhelm

Linear vector space

Location(see also Position vectors)

vector products and

of vectors

Magnitude:

definition

forces and

metrical tensor and

vectors and

Master reference frame

Mass of particles

Metrical tensor

definition

Mid-point formula

Moments of force

parallelogram law and

Moments of momentum:

definition

of gyroscope

Momentum

angular

conservation of

Momentum vector

Motion, Newton’s laws

Moving reference frames

Multiplication of vectors (see also Scalar products, and Vector products):

distributive law in

by numbers

Newton, Isaac

Newtonian physics: laws of motion in

space and time in

Noncommutativity:

of angular displacements

of quaternions

in science

Nonmetrical spaces

Nonuniform coordinates

Notation:

for addition of vectors

arithmetical

for equality

index

for multiplication of vectors by numbers

for subtraction of vectors

for vector of zero magnitude

Numbers:

complications with.

properties

hypercomplex

irrational

multiplication by

ordered sets

quaternions and

rational

types

vectors compared with

Octants

One-dimensional space, vectors in

Order of tensors

Ordered sets of numbers

Origin:

definition

independence

Orthogonal projections

Orthogonal triads, unit

right-handed

scalar products and

vector products and

Parallelepiped

volume

Parallelogram law

angular displacements and

in combination of forces

in combination of vectors

in equality of vectors

moments of force and

quaternions and

vector products and

velocities and

Parallels, Euclid’s axiom

Particle:

acceleration

definition

in equilibrium

momentum

Newtonian equation of motion

velocity

Plane curves

Plato

Poinsot’s central axis

Point-of-division formula

Polar coordinates

spherical

Position vectors

components

Projections:

of areas

orthogonal

Pseudoscalar

Pseudovector

Quantum physics

noncommutivity in

Quaternions:

definition

importance

noncommutativity

number and

parallelogram law and

Rational numbers, defintion

Reference frames

components and

left-handed

master

moving

right-handed

Relativity, theory of

index notation in

Resultant

Right quaternions

Right-handed reference frames vector products in

Right-handed unit orthogonal triads

Rigid body, forces acting on

Rusting

Scalar products

contravariant vectors and

tensors and

trigonometric formula derived by

unit orthogonal triads and

work and

Scalars:

components and

definition

tensors and

Second-order tensors

covariant

Shift

Space:

Newtonian

one-dimensional

four-dimensional

Special theory of relativity

Speed:

average

instantaneous

velocity compared with

Spherical polar coordinates

Statics, two-dimensional

Subtraction of vectors

Summation convention

Symmetric tensors

Tarpaulin

Tensors

antisymmetric

components

contraction

cross products as

definition

index notation for

law of transformation

metrical

new concept of vectors and

order

scalar products and

scalars and

second-order

symmetric

transformation of vector components

valence

vector products as

visualizing

Tetrahedron

Time, Newtonian

Transformation, tensor law

Trigonometric functions and formulas

direction cosines

Triple products

Two-dimensional statics

Unit orthogonal triads(see also 128)

right-handed

scalar products and

in vector algebra

vector products and

Units

Valence of tensors

Vector calculus

Vector field, definition

Vector products (see also Cross products)

addition of vectors and

angular displacements and

angular momentum and

angular velocity and

areas and

associative law and

called cross products

commutative law and

distributive law and

left-handed reference frames and

location and

moments and

parallelogram law and

right-handed reference frames and

trigonometric formula and

Vector quantities

Vectorial tribe

Vector space, linear

Velocity

of air in storm

angular

average

definition

displacements and

as free vector

instantaneous

parallelogram law and

of particles

speed compared with

Whole numbers

Work

Wrench

Zero magnitude, vector of

1

Have you noticed that we have been careless in sometimes speaking of “the vector represented by ,” at other times calling it simply “the vector ,” and now calling it just “”? This is deliberate—and standard practice among mathematicians. Using meticulous wording is sometimes too much of an effort once the crucial point has been made.

2

We retain, too, the recollection that we are still linked, however tenuously, with journeying, for we want to retain the idea that a movement has occurred, even though we do not care at all how or under what circumstances it occurred.

3

·Actually, of course, lack of a name proves no more than that if the resultant exists, it has not hitherto been deemed important enough to warrant a name.

4

Even with numbers there are complications. For example, 3 ft. and 3 inches can be said to yield 6 items ; yet in another sense they yield 39 inches, ft., and so on—and each of these can also be regarded as a number of items, though the involves a further subtlety. Consider also 3 ft. and 3 lbs., and then 2.38477 ft. and 2.38477 lbs.

5

The unpleasant aspects of average velocity are not exorcized by the moving platform concept.

6

Let us not ask awkward questions here about the word “rest.”

7

You may think friction is unnecessary. But the situation is not the same as it would be if we were pulling on strings attached to A and B. Can you see why not? The point is a subtle one.

8

Beware of the word “always.” It is almost “always” wrong. What if we were discussing three lines in four dimensions, for example?

9

Except, of course, that we were not contemplating any motion by the rigid body along the axis of rotation.

10

This is, of course, 90°. But we use radians here with an eye on the next section.

11

Strictly speaking. it is incorrect to say that the components in a given coordinate system are unique. If the components are not pure numbers, they will change when the units of length, time, and mass are changed.

12

Which may coincide.