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1
Vector algebra revisited
1.1 Prerequisites
Before starting on the book proper in section 1.2, the reader may welcome an outline of the formal vector mathematics which he should already have met, together
with some revision exercises. The topics all concern vector algebra or vector functions of one variable. We shall write vectors either in bold type (e.g. a) or in the
--+
form PQ, a directed line segment from P to Q. The corresponding italic type (e.g. a)
denotes the magnitude of a vector, as does the 'modulus' notation (e.g. 1a I).
Basic vector algebra
If I, m are scalars and a, b, c vectors, then
a +b
= b + a,
a + (b + c)
=
(a + b)
+ c,
vector addition being governed by the triangle or parallelogram laws (see figure 1.1),
I a = el is a vector parallel to a and of I times its magnitude, - a is a vector antiparallel to a and of the same magnitude,
I(ma)
=
(1m) a,
(I + m)a = la
+ rna,
I(a + b) = la
+ lb.
A unit vector is a vector of unit magnitude, for expressing direction only.
a
a
a
(a)
(b)
(c)
Fig. 1.1. (a) and (b): Additions by the triangle law, with vectors head-to-tail, (c):
addition by the parallelogram law, with both vectors rooted at 0 (i.e, both tails at 0).
The right-hand screwconvention
When a vector is used to represent a physical quantity that has a rotational quality
(e.g. angular velocity) the vector is directed along the axis of rotation in the direction of advance of a right-hand screw under the rotation implied.
Cartesian axes
To avoid ambiguity it is essential to use right-handed axes as shown in figure 1.2.
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Vector algebra revisited
a
I
I
..,., [1
x
Fig. 1.2. Right-handed axes and the components (direction cosines) of a unit vector.
These are characterised by the fact that a rotation about Oz of Ox towards Oy
would cause a right-hand screw to advance in the z-direction. The three axes are
mutually perpendicular.
Unit base vectors: i, j, k
These are of unit magnitude, oriented respectively in the x,y and z directions. The
components (ax, ay , az ) of a vector a are such that
if we use
~
to denote a sum over three cartesian terms.
Direction cosines
If unit vector a makes angles a, {3, 'Y with Ox, Oy, Oz, respectively, then ax = cos a,
a y = cos {3, az = cos 'Y. (See figure 1.2.) The quantities (cos a, cos (3, cos 'Y) are the
'direction cosines' of the vector and of its direction. Note that ~ cos? a = 1, which
expresses the fact that only two of a, (3, 'Yare independent.
Vector multiplication
(i) The scalar product of two vectors a and b is written a •b and equals ab cos () ,
where () is the angle between the two vectors when rooted at the same point as in
figure 1.1(c). The product is negative when () is obtuse. Ifb is a unit vector, a·b is
the component of a in the direction of b. If a • b = 0, then a and b are perpendicular, if both are non-zero.
Other results: a· b = b •a, a· (b + c) = a •b + a· c, (la)· b = l(a •b) = a· (lb),
i · i = j •j = k •k = 1, but i· j = j •k = k· i = 0, a· b = ~ axb x and a· a = ~ a~ = a2 •
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1.1 Prerequisites
The conventional shorthand a 2 is normally used for a-a. The point P, (x, y, z), lies
on the plane through Q, (xo,Yo, zo), perpendicular to the vector v = Ii + mj + nk
if the scalar product
~
v-QP
=
l(x-xo)+m(y-yo)+n(z-zo)
Hence the plane lx
nk.
=
O.
+ my + nz = const. is perpendicular to the
vector v = li + mj +
(ii) The vector product of two vectors a and b is written a x b and is a vector perpendicular to both a and b having the direction in which a right-hand screw would
advance if, with a and b rooted at the same point, a were rotated towards b. The
magnitude of ax b is ab sin (), where () is the angle (less than 180°) between a and
b, if rooted at the same point. If () is zero because a and b are parallel, a x b
vanishes and the ambiguity of its direction is immaterial. Ifax b = 0, then a and b
are parallel, if both are non-zero.
Other results: a x b = - b x a, a significant departure from normal algebra.
= axb+axc,
= jxj = kxk = 0
ax(b+c)
ixi
= (la)xb = a x (lb),
= k,jxk = i,kxi = j,
l(axb)
but
ixj
i;
j
k
by
bz
if the idea of a determinant is generalised for mnemonic purposes.
(iii) Triple products of vectors. There are three kinds:
A. (a e b)c, a vector. This is not equal to a(b -c) in general.
B. (a x b) x c, a vector. TWs is not equal to a x (b x c) in general. However,
(a x b) x c == (a - c) b - (b -c) a is a general identity.
C. a-fb x c), a scalar. This equals c-(a x b) and -(a x c) -b, etc. (the sign changes
if the cyclic order abcabc ... changes), the very important transformation property of the scalar triple product. Note that a- b x c = a x b -c (Le., the dot and
cross may be interchanged) and that
a-b xc
(The brackets round b x c can be safely omitted.)
Vector functions ofone variable such as time t. If a
general parallel to a, equal to ~ i(dax/dt),
d
dt(a+b)
=
da db
dt + dt'
= a(t), da/d t is a vector, not in
d
dl
da
dt(la) = dt a+l dt'
3
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d
da
db
-(a-b) = --b+a-dt
dt
dz '
,
d
da
db
-(axb) = -xb+axdt
dt
dr '
---+
Ifr is the position vector OPof a point P, then dr/dt = v, its velocity, and d 2r/dt2 =
dv/dt = a, its acceleration, which has components dv/dt parallel to v and v 2/R perpendicular to v (along the principal normal) where R is the radius of curvature of
P's path.
Problems lA
1.1
Identify the following as either vectors or scalars: momentum, kinetic
energy, pressure, angular velocity, acceleration, mass density, weight,
electric charge, frequency, voltage, light beam intensity, equilibrium
radiation in a black-body enclosure.
1.2
Figure 1.3 shows a plan of two. roads X, Yand two points A, B which must
be connected by the shortest sewer, subject to the constraint that the
sewer must cross each road at right angles (because of the cost of breaking
concrete). Find how to construct the route.
y
Fig. 1.3
1.3
The normals to two planes have direction cosines' (cos Q1, cos (31, cos "11)
and (cos Q2, cos (32, cos "12). Find an expression for cos 0 where 0 is the'
angle between the planes. What happens if "11 = "12 = 1(/2?
1.4
What is the condition for the planes ax + by '+ cz = d and lex + ly + mz
n to be perpendicular?
1.5
Find the unit vector that is perpendicular to both of the vectors i + 2j +
3k and 4i + 5j + 6k.
1.6
Find the shortest distance between the line joining the points (1, 1, 1) and
(2,3,4) and the line joining the points (3,1, 1) and (1,3,2).
1.7
For a parallelogram defined by two adjacent sides a and b, show that the
sum of the squares of the diagonals equals the sum of the squares of the
sides, and that the diagonals are orthogonal if the sides are equal in length.
Find a simplified expression for the vector product of the diagonals. Is it
in the right direction?
=
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Problems lA
~
~
1.8
If 0 1 and O2 are fixed points, P is a movable point, 01P = fl) 02P = f2, a
is a fixed vector and k is a fixed scalar, find the geometrical constraints on
P implied by (i) fl + f2 = a, (ii) fl X f2 = a or (iii) fl • f2 = k.
1.9
(a) Show that the volume of a parallelepiped is a •b )( c if a, band care
three adjacen t edges.
(b) What is the result if a, b and c are coplanar? What is then the angle
between a and b x c?
(c) Use (b) above to find the cartesian equation of the plane defined by
the points PI P2 P3 with position vectors fl, f2, f3 (components Xl, X2, X3,
etc.), by noting that the position vector r of a point P on the plane is such
~~~
that PtP, P2P t ,P3 P2 are coplanar. Express the result as a determinant.
What happens if P, P2 P3 are collinear?
1.10
From the result (u x v) x w == (u ·w)v - (v·w)u, show that
(a x b) • (c x d) == (a •c)(b • d) - (a •d)(b •c).
Interpret this result simply in the case where a and c are identical unit
vectors and band d are identical unit vectors, inclined at () to a.
1.11
A bead of mass m slides on a curved wire lying in a plane. The coefficient
of friction is J.l and gravity is negligible. Show that the speed of the bead is
proportional to e-IJ,'IJ, where 1/1 = angle between tangent and some
reference direction. What if there is a point of inflexion?
1.12
(a) On a curve or trajectory, the position vector r can be regarded as a
function of s, the distance along the curve from some datum. Show that
dr/ds is a unit vector, T say, in the tangent direction and that dT/ds is
perpendicular to the tangent, but is zero if the curve is locally straight.
(The direction of dT/ds is called the principal normal.)
(b) Find the unit principal normal vector to the curve y = x 2 , Z = x 3 at
the point (1, 1, 1). (Note that dT/dx is also in the principal normal direction.)
(c) If the radius of curvature R is defmed such that l/R = 1dT/ds I, show
that, for a particle traversing a curve, the acceleration d/dt(vT) has components vdv/ds and v 2/R along the tangent and principal normal respectively, v being the magnitude of the velocity. (Note that v = ds/dt.)
(d) A point P moves arbitrarily on a sphere of radius p, centred at the
origin. By successively differentiating the equation r •r = p2 show that the
velocity is perpendicular to the radius vector r and that the radial component of acceleration is - v 2/ p. Note that p will not be the radius of curvature of the locus in general. Examine the case where the locus is a circle of
radius smaller than p.
1.13
A particle of unit mass, initially moving with unit velocity in the positive
z -direction at the origin, suffers a force
5
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i sin 2 t
+j
cos 2 t - k sin t,
expressed in terms of time t. Show that its acceleration and velocity are
always perpendicular. Verify that its speed is constant and find the radius
of curvature of its path as a function of time. (See problem 1.12(c).)
1.2 Vectors and scalars
Before we move on to our main business of vector calculus, it is necessary to spend
two chapters extending physical appreciation of vectors and vector algebra.
A major step in the systematisation of knowledge of the three-dimensional
world was the realisation that most physical quantities may be classified as scalars
or vectors. Scalars have magnitude only (e.g. mass or temperature) whereas vectors
have magnitude and direction (e.g. force or displacement). To specify a vector we
must state three independent quantities, which might be its three cartesian components or its magnitude and direction cosines (only two of which are independent).
One consequence is that a vector equation is really three pieces of information
written concisely to look like one. A primitive example is a = b, which implies the
three independent facts that each of the three cartesian components of a equals the
corresponding one for b.
There are physical quantities of practical importance which are neither scalars
nor vectors. These are tensors, to specify which 6, 9 or even more magnitudes are
necessary. A good example is stress in a solid or viscous fluid. This book avoids any
discussion of tensors.
1.3 Addition of vectors
Not all physical quantities that have magnitude and direction are vectors, however.
Vectors must also obey the familiar triangle or parallelogram addition laws for
finding the combined physical effect (whatever that may mean in each case) of two
vectors acting together. These addition laws function by representing vectors as
directed line segments, whose lengths represent their magnitudes to some arbitrary
scale. Only in the case of vectors that are actual displacements is this representation
literal rather than conventional. It does however always lead to the correct addition
of vectors. Figure 1.1 shows the addition laws applied to two vectors a and b, represented by directed line segments, the direction being indicated by an arrowhead.
The statement that velocity, for instance, is a vector therefore includes the fact
that the combined physical effect of two velocities acting simultaneously, a familiar
notion which is easily interpreted, can be calculated from the triangle or parallelogram laws. One obvious consequence of these laws is that
a + b = b + a.
Through the failure of this relation, problem 1.14 reveals a counter-example which
shows that there are physical quantities which have magnitude and direction but
which are not vectors.
The notion of 'combined effect' must be used with discretion. Like can only be
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1.4 The essential nature of vectors; 'cosine-quality'
added to like, and the combined effect of a current J with a magnetic field B at the
same point is certainly not J + B. Later we shall find that a kind of multiplication is
appropriate here, instead.
1.4 The essential nature of vectors; 'cosine-quality'
As well as their addition properties, physical quantities that are vectors have another
characteristic quality. This emerges when we need to consider the 'contribution',
'effect' or 'influence' that a vector quantity makes or exerts in some direction of
interest, different from the vector's own direction. If we are crossing the road
diagonally, our speed directly across the road is of great relevance to our chance
of survival. This leads to the formal idea of resolution.
a
a
~~~~~
direction of interest
Fig. 1.4. Resolution along and normal to a direction of interest.
T
p
tangent plane
at 0
p
Fig. 1.5. A vector's 'sphere of influence'.
Figure 1.4 shows how the triangle addition law enables us to write a vector a as
the combined effect of all (the contribution of a in the direction of interest) and al,
a vector with no contribution in that direction. We call this resolving a into components all and al'
;
The key pointto notice is that the magnitude of all is a cos e, i.e. a vector is a
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physical entity whose influence in a particular direction varies like cos (J, where (J is
the angle between that direction and the unique direction of maximum influence.
Weshall frequently exploit this characteristic cosine-quality as a means of recognising vectors.
If we plot in three dimensions how all varies with (J, for a given vector a, we get
figure 1.5, which shows how the tip T of all traverses a sphere based on a as diameter. The influence pattern is axisymmetric about a. It is characteristic of a vector
that it exerts no influence in all directions OP that are perpendicular to it, and
makes a maximum contribution in just one direction (its own).
1.5 Localisation of vectors
Can vectors be moved about with impunity? In figure 1.1 we appear to be assuming
that the vector a is still the same entity whether we put it at the bottom or the top
of the picture. The normal situation is that the vector is a physical effect which
prevails ('is localised at') some particular point in space, whereas for visual-aid and
addition purposes we habitually represent vectors as arrowed segments of straight
lines. Where do we then put the line segments in relation to the location of the
effects? The usual convention is to put the tails of the arrows there, as at 0 in
figure 1.1(c) - but it is pure convention - and then the parallelogram law seems
the more natural one for addition, as for instance when we seek the combined
effect of two superposed forces, localised at the same point.
On the other hand, when there is physical meaning in adding two vectors not
localised at the same point, as for example with two successive displacements, the
triangle law seems more appropriate, indeed obvious. But either law may be used
for the addition process if we are prepared to move the vectors about, preserving
their magnitude and direction.
There is one case particularly likely to cause confusion, namely, that of the position vector r which is the displacement vector OP, used to specify the position of
a point P relative to some origin O. The vector is rooted at 0 (the arrow-tail) but
can be regarded as localised at P (the arrow-head) in the sense that it describes P,
and is liable to be combined with other vectors or operations localised at P. The
same is true of variants of r such as the inverse-square law field rr-3 • So be warned!
There are circumstances where a vector need not be regarded as localised at a
point. In rigid dynamics, to determine the instantaneous state of acceleration of a
body we need specify only the lines of action along which the forces acting are
localised, while any couples acting can be represented as vectors that are not
localised at all, as regards their instantaneous effect on the dynamics. But if we are
interested in the subsequent motion of the body or its state of stress, or if it is not
rigid, then it matters greatly where the forces or couples act and we are back to the
normal situation of localisation at a point. Problem 1.15 takes the discussion further.
~
1.6 Scalar and vector fields
In this book we shall be mostly concerned with phenomena which are distributed in
space. The distributions in space of the physical quantities that are relevant we shall
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1.7 Vectorarea
call fields, and these are usually either vector or scalar fields. Consider the air in a
room: it moves about in a manner described bya velocity field (a vector field)
under the action of variations in the temperature field (a scalar field) which affects
the density (a scalar field) and consequently also the weight distribution (a vector
field) in the fluid. In problems of this kind there is no ambiguity about localisation,
as all vectors are localised at the points for which they are specifying the conditions
that prevail, as are the scalars, also.
/
/
area element a
projection
a cosO
\
Fig. 1.6
1.7 Vector area
The reader will be accustomed to force and velocity vectors and able to interpret
the meaning of 'contribution' or 'influence' in a particular direction in such contexts.
But many other kinds of directed quantities will concern us in this book. For
example a plane area of finite extent and arbitrary shape has a magnitude and an
orientation. Can these aspects (if not the details of th~ shape) be represented by a
vector? Some clue is given by the fact that the vector product a x b of two directed
line segments a and b, localised at a point 0, equals in magnitude the area of the
parallelogram defined by a and b, as in figure 1.1(a), and its direction is perpendicular to the area in question. This suggests that for representing a plane area of any
shape by a vector, the natural choice is to make the vector a normal to the plane
and its magnitude a that of the area element, as in figure 1.6. At this stage the
direction is arbitrary.
To confirm true vector behaviour, the idea of adding areas to check the addition
rule is not obviously meaningful except in the degenerate case of coplanar elements,
so instead we examine whether area has the crucial cosine-quality.
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If we view the area element in an arbitrary direction PS, inclined at () to the
normal as in figure 1.6, we see an effective, or projected, area equal to a cos (), for
() is also the angle between the plane PRT containing the element and the plane
QRTperpendicular to PS. Thus cosine-quality is indeed exhibited; the component
or 'effect' of the vector a in an arbitrary direction finds a ready interpretation as
the projected area in that direction. (See problem 1.16(b).)
This new concept of vector area is particularly convenient in connection with
fluid pressure p. The pressure-force on a plane surface element defined by vector a
can now be succinctly written as ± pa, where the sign depends on which side the
fluid lies.
Example 1.1
Show that fluid pressure is isotropic in the absence of shear stresses.
Let figure 1.6 now represent a small element of fluid, assumed to have a typical
linear dimension e. The pressures on the curved face have no component in the
direction PS. If we denote the pressures on the faces containing P and Q as Pp and
PQ, then the balance of forces in the direction PS requires that
pQ(a cos ()) - (ppa) cos ()
the body or inertia forces on the element
= O(€3),
for the terms on the right-hand side must be of the same order as the volume or
mass of the element, if accelerations, etc. are to be finite. But a is of order €2 and so
dividing by a cos () gives the result:
PQ <p»
=
O(€)
and tends to zero in the limit e ~ 0, i.e. the pressure is isotropic.
Just as direction cosines are used to describe the direction of a vector or a straight
line, so the direction of a plane or plane area element can be specified by the direction cosines of a vector normal to it. One then speaks of 'the direction cosines of
the area'.
1.8 Vector flow intensity
Another important new kind of vector concerns physical phenomena involving the
flow, transport or diffusion of some entity such as heat, electric charge, seeping
fluid or a foreign species (such as dopant in a semiconductor) through a region of
space or a material medium. In the case of heat, we can specify that at a given point
it passes at an intensity Q per unit area of a plane perpendicular to the direction of
flow, for it is obvious that heat flow does have a direction; heat going from left to
right is not going from right to left, or upwards! But is heat flow intensity a true
vector Q (or, more strictly, a vector field, since it is distributed in space)?
It is not meaningful to discuss whether the heat flow intensity can be added by
the triangle or parallelogram law so as to qualify for being deemed a vector. Adding
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