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1

2A1 Vector Algebra and Calculus

8 Lectures MT 2013
Rev Oct 2013
Course Page

Stephen Roberts
sjrob@robots
...
ac
...
robots
...
ac
...
This course will remind you about that good stuff,
but goes on to introduce you to the subject of Vector Calculus which, like it says on the can,
combines vector algebra with calculus
...
Temperature T is a scalar, and will certainly be a function of a position vector x = (x , y , z )
and may also be a function of time t: T = T (x, t)
...

Suppose now that you kept y , z , t constant, and asked what is the change in temperature as you
move a small amount in x ? No doubt you’d be interested in calculating ∂T /∂x
...

But why restrict ourselves to movements up-down, left-right, etc? Suppose you wanted to know
what the change in temperature along an arbitrary direction
...
At each point x in the stream, at each time t, there will be a stream
velocity v(x, t)
...
The point now
is that v is a function that has the same four input variables as temperature did, but its output
result is a vector
...
That is, we will be interested in finding the
acceleration of the stream, the gradient of its velocity
...
Equally, we may be interested in the acceleration as a vector, so that
we can apply Newton’s law and figure out the force
...


2

Grey book
Vector algebra: scalar and vector products; scalar and vector triple products; geometric applications
...
Gradient, divergence and
curl - definitions and physical interpretations; product formulae; curvilinear coordinates
...
Derivation
of continuity equations and Laplace’s equation in Cartesian, cylindrical and spherical coordinate
systems
...

• Triple products, multiple products, applications to geometry
...

• Curvilinear coordinate systems
...

• Vector operators
...


• Gauss’ and Stokes’ Theorems
...


Learning Outcomes
You should be comfortable with expressing systems (especially those in 2 and 3 dimensions) using
vector quantities and manipulating these vectors without necessarily going back to some underlying
coordinates
...

You should have a good intuition of the physical meaning of the various vector calculus operators
and the important related theorems
...


References
Although these notes cover the material you need to know you should, wider reading is essential
...

• J Heading, ”Mathematical Methods in Science and Engineering”, 2nd ed
...
13, (Arnold)
...
, Ch
...

• E Kreyszig, ”Advanced Engineering Mathematics”, 6th ed
...
6, (Wiley)
...
P
...
J
...
6, 8 and 9, (CUP)
...
al
...
1, Ch
...

• H M Schey, “Div, Grad, Curl and all that”, Norton

3

Course WWW Pages
Pdf copies of these notes (including larger print versions), tutorial sheets, FAQs etc will be accessible
from

www
...
ox
...
uk/∼sjrob/Teaching/Vectors

4

Lecture 1
Vector Algebra
1
...
They are scalars
...
They are vectors
...

Obvious examples are velocity, acceleration, electric field, and force
...

There are three slightly different types of vectors:
• Free vectors: In many situtations only the magnitude and direction of a
vector are important, and we can translate them at will (with 3 degrees of
freedom for a vector in 3-dimensions)
...
The force vector can slide with 1 degree of freedom
...
The origins of position vectors all coincide at an overall
origin O
...
For example, if two free
vectors are equal we need only say that their magnitudes and directions are equal,
and that can be done with a drawing that is independent of any coordinate system
...
Try to spot things in the notes that are independent
5

6

LECTURE 1
...
1:

of coordinate system
...
1
...
These specified directions define a
coordinate frame
...
When we come to examine
vector fields later in the course you will use curvilinear coordinate frames, especially
3D spherical and cylindrical polars, and 2D plane polar, coordinate systems
...
2: Vector components
...
2
...
1
...
3: (a) Addition of two vectors is commutative, but subtraction isn’t
...
(b) Addition of three vectors is associative
...

Although we will be most often dealing with vectors in 3-space, you should not
think that general vectors are limited to three components
...
We shall use
the hat to denote a unit vector
...
1
...

If we use a coordinate frame, we might say that corresponding components of
the two vectors must be equal
...

1
...
3

Vector magnitude and unit vectors

Provided we use an orthogonal coordinate system, the magnitude of a 3-vector is
q
a = |a| = a12 + a22 + a32
To find the unit vector in the direction of a, simply divide by its magnitude
a

...
1
...
Thus
a + b = [a1 + b1,

a2 + b2 ,

a3 + b3 ]

8

LECTURE 1
...
3(a)
...
It is useful to remember that the vector
a − b goes from b to a
...
3(a))
(b) (a + b) + c = a + (b + c) = a + b + c (associativity) (Figure 1
...

(d) a + (-a) = 0
1
...
5

Multiplication of a vector by a scalar
...

It follows that:
p
|ca| = (ca1)2 + (ca2)2 + (ca3)2 = |c||a|
...

(By the way, a vector where the sign is uncertain is called a director
...
Coulomb’s law states that the electrostatic force on charged particle Q due
to another charged particle q1 is
Qq1
er
F=K 2 ˆ
r
where r is the vector from q1 to Q and ˆr is the unit vector in that same
direction
...
) The force between two particles is not modified by
the presence of other charged particles
...

A
...
The unit vector in that direction is
(R − ri )/|R − ri |, so the resultant force is
F(R) =

N
X
i =1

K

Qqi
(R − ri )
...


1
...
SCALAR, DOT, OR INNER PRODUCT

1
...

Note that
a · a = a12 + a22 + a32 = |a|2 = a2
...
2
...
4: (a) Cosine rule
...


Consider the square magnitude of the vector a − b
...
VECTOR ALGEBRA

But, by the cosine rule for the triangle OAB (Figure 1
...
It follows that
a · b = ab cos θ,
which is independent of the co-ordinate system used, and that |a · b| ≤ ab
...

1
...
2

Projection of one vector onto the other

Another way of describing the scalar product is as the product of the magnitude
of one vector and the component of the other in the direction of the first, since
b cos θ is the component of b in the direction of a and vice versa (Figure 1
...

Projection is particularly useful when the second vector is a unit vector — a · ˆı is
the component of a in the direction of ˆı
...

a2

In the particular case a · b = 0, the angle between the two vectors is a right angle
and the vectors are said to be mutually orthogonal or perpendicular — neither
vector has any component in the direction of the other
...

1
...
3

A scalar product is an “inner product”

So far we have been writing our vectors as row vectors a = [a1 , a2, a3 ]
...

a3
In matrix algebra

M11 M12
 M21 M22
M31 M32

vectors are more usually defined as column vectors, as in

 

v1
a1
M13
M23   a2  =  v2 
v3
a3
M33

1
...
SCALAR, DOT, OR INNER PRODUCT

11

and a row vector is written as a⊤
...

Why? Simply to point out at that the scalar product is also the inner product
more commonly used in linear algebra
...

b3
Here we treat a n-dimensional column vector as an n × 1 matrix
...
So for n-dimensional column vectors a and b, a⊤ is a 1 × n
matrix and b is n × 1 matrix, so the product a⊤ b is a 1 × 1 matrix, which is (at
last!) a scalar
...
A force F is applied to an object as it moves by a small amount δr
...
The work done is equal to the component of force in the direction of the displacement multiplied by the displacement itself
...

Q2
...
What is the angle
between an adjacent pair?
A2
...

The directions of the diagonals are
[±1, ±1, ±1]
...
The
angle is thus

[−1,1,1]

[1,1,1]

k

[1, 1, 1] · [−1, 1, 1]
−1 1
√
=
cos
θ = cos−1 √ 2
3
1 + 12 + 12 −12 + 12 + 12

j
i

12

LECTURE 1
...
A pinball moving in a plane with velocity s bounces (in a purely elastic impact)
from a baffle whose endpoints are p and q
...
Best to refer everything to a coordinate frame with principal directions
u
ˆ along and ˆ
v perpendicular to the
baffle:

q

q−p
|q − p|
ˆ
v = u⊥ = [−uy , ux ]

u
ˆ =

Thus the velocity before impact is
sbefore = (s
...
ˆ
v)ˆ
v
After the impact, the component of
velocity in the direction of the baffle is unchanged and the component
normal to the baffle is negated:

^u
p
^v
s

safter = (s
...
ˆ
v)ˆ
v
1
...
4

Direction cosines use projection

Direction cosines are commonly used in the field of crystallography
...
Since
a ·ˆı = a1 etc, it follows immediately that a = a(λˆı + µˆ + ν kˆ ) and λ2 + µ2 + ν 2 =
1 2
2
2
a2 [a1 + a2 + a3 ] = 1

1
...


1
...
VECTOR OR CROSS PRODUCT

13

k

j
i
Figure 1
...


It is MUCH more

 ˆı

a × b =  a1
 b1

easily remembered in terms of the (pseudo-)determinant

ˆ kˆ 
a2 a3 
b2 b3 

where the top row consists of the vectors ˆı , ˆ , kˆ rather than scalars
...

It is also easily verified that (a × b) · a = (a × b) · b = 0, so that a × b is orthogonal
(perpendicular) to both a and b, as shown in Figure 1
...

Note that ˆı × ˆ = kˆ , ˆ × kˆ = ˆı , and kˆ × ˆı = ˆ
...
This is left as an
exercise
...
Moreover the vector product
does not satisfy the associative law of multiplication either since, as we shall see
later a × (b × c) 6= (a × b) × c
...

Assuming that the co-ordinate vectors form a right-handed set in the order ˆı ,ˆ , kˆ
it can be seen that the sense of the the vector product is also right handed, i
...
VECTOR ALGEBRA

the vector product

 ˆı ˆ

ˆı × ˆ =  1 0
0 1

has the same sense as the co-ordinate system used
...

0

In practice, figure out the direction from a right-handed screw twisted from the
first to second vector as shown in Figure 1
...

ax b

axb

b

in right−hand screw sense

bsin θ

b
a

θ

Plane of vectors a and b

a

Figure 1
...
Twist from first to second and
move in the direction of a right-handed screw
...


1
...
1

Geometrical interpretation of vector product

The magnitude of the vector product (a × b) is equal to the area of the parallelogram whose sides are parallel to, and have lengths equal to the magnitudes of, the
vectors a and b (Figure 1
...
Its direction is perpendicular to the parallelogram
...
g is vector from A [1,2,3] to B [3,4,5]
...

Find m,
ˆ a UNIT vector along g × ˆℓ
Verify that m
ˆ is perpendicular to ˆℓ
...

A
...
3
...
VECTOR ALGEBRA

Lecture 2
Multiple Products
...
1

Triple and multiple products

Using mixtures of the pairwise scalar product and vector product, it is possible to
derive “triple products” between three vectors, and indeed n-products between n
vectors
...

2
...
1

Scalar triple product

This is the scalar product of a vector product and a third vector, i
...
a · (b × c)
...






 c1
 c1 c2 c3 
 a1 a2 a3 





 b1 b2 b3  1st SWAP  b1 b2 b3  2nd SWAP  a1





 b1
 a1 a2 a3 
 c1 c2 c3 
+
−

This says that

sign changes;

c2 c3 
a2 a3 
b2 b3 
+

(i) a · (b × c) = b · (c × a) = c · (a × b) (Called cyclic permutation
...
MULTIPLE PRODUCTS
...
(Called anti-cyclic permutation
...
This notation is not very helpful, and we will try to avoid
it below
...
1
...

We saw earlier that the vector product (a × b) has magnitude equal to the area
of the base, and direction perpendicular to the base
...
1(a)
...
1: (a) Scalar triple product equals volume of parallelopiped
...


2
...
3

Linearly dependent vectors

If the scalar triple product of three vectors is zero
a · (b × c) = 0

then the vectors are linearly dependent
...
For example,
a = λb + µc
where λ and µ are scalar coefficients
...
You’ll remember that by doing row operations, you could reach a row
of zeros, and so the determinant is zero
...
In this case all
three vectors lie in a plane, and so any one is a linear combination of the
other two
...
1b
...
1
...
1
...

Now, the vector triple product a × (b × c) must be perpendicular to (b × c), which
in turn is perpendicular to both b and c
...
It follows that
the vector triple product must be expressible as a linear combination of b and c:
a × (b × c) = λb + µc
...
Only the first component need be evaluated, the others then being obtained
by symmetry
...

bxc

a
In arbitrary direction
c
b

a x (bx c)
Figure 2
...


2
...
5

Projection using vector triple product

An example of the application of this formula is as follows
...
The component of v in the z
direction is v · kˆ , so the projection we seek is v − (v · kˆ )kˆ
...
MULTIPLE PRODUCTS
...


(Hot stuff! But the expression v − (v · kˆ )kˆ is much easier to understand, and
cheaper to compute!)
2
...
6

Other repeated products

Many combinations of vector and scalar products are possible, but we consider only
one more, namely the vector quadruple product (a × b) × (c × d)
...
On the other hand, regarding c × d as a single vector, we
see that it must also be a linear combination of a and b
...

[(a × b) · c]

This is not something to remember off by heart, but it is worth remembering that
the projection of a vector on any arbitrary basis set is unique
...

A1 Grinding away at the determinants, we find
[(a×b)·c] = −18; [(b×c)·d] = 6; [(c×a)·d] = −12; [(a×b)·d] = −12

So, d = (−a + 2b + 2c)/3
...
2
...
3: The projection of a (3-)vector onto a set of (3) basis vectors is unique
...


2
...
2
...
If you make b a unit vector, r = a + λb
represent metric length
...


a
r

Figure 2
...
With ˆ
b a unit vector, λ is in the length units established by the
definition of a
...
2
...
MULTIPLE PRODUCTS
...
5(a) the vector p from c to any point on the line is p =
ˆ − c = (a − c) + λb
ˆ which has length squared p 2 = (a − c)2 + λ2 + 2λ(a −
a + λb
ˆ
...
The
c) · b
minumum is found when d p 2/d λ = 0, ie when
ˆ
...

ˆ
p = (a − c) − ((a − c) · b)

You might spot that is the component of (a−c) perpendicular to ˆ
b (as expected!)
...
1
...

p=ˆ
b × [(a − c) × b]

ˆ is a unit vector, and is orthogonal to [(a − c) × b],
ˆ the modulus of the
Because b
vector can be written rather more simply as just
ˆ
...
5: (a) Shortest distance point to line
...


2
...
3

The shortest distance between two straight lines

If the shortest distance between a point and a line is along the perpendicular, then
ˆ and r = c + µd
ˆ
the shortest distance between the two straight lines r = a + λb
must be found as the length of the vector which is mutually perpendicular to the
lines
...

p
ˆ = (b
d)/|b
ˆ×ˆ
ˆ and d
ˆ are not orthogonal,
(Yes, don’t forget that b
d is NOT a unit vector
...

pmin = (a − c) · (b

2
...
GEOMETRY USING VECTORS: LINES, PLANES

23

♣ Example

Q Two long straight pipes are specified using Cartesian co-ordinates as follows:
Pipe A has diameter 0
...

Pipe B has diameter 1
...

Determine whether the pipes need to be realigned to avoid intersection
...
The vector equation of the axis
of pipe A is
√
r = [2, 5, 3] + λ′ [5, 5, 5] = [2, 5, 3] + λ[1, 1, 1]/ 3
The equation of the axis of pipe B is
√
r = [0, 6, 3] + µ′ [12, 6, 6] = [0, 6, 3] + µ[−2, −1, 1]/ 6
The perpendicular to the two axes has direction


ˆ
 ˆı
ˆ

k



[1, 1, 1] × [−2, −1, 1] =  1 1 1  = [2, −3, 1] = p
 −2 −1 1 

The length of the mutual perpendicular is
[2, −3, 1]
[2, −3, 1]
(a − c) · √
= [2, −1, 0] · √
= 1
...

14
14
But the sum of the radii of the two pipes is 0
...
5 = 0
...
Hence the pipes
do not intersect
...
2
...

1
...
Note that b and c are free vectors, so don’t
have to lie in the plane (Figure 2
...
)
2
...
6(b) shows the plane defined by three non-collinear points a, b and
c in the plane (note that the vectors b and c are position vectors, not free
vectors as in the previous case)
...
Figure 2
...


24

LECTURE 2
...
GEOMETRY USING VECTORS

r

^
n

r

c
b

a

c

O

b

a

NB that these are
parallel to the plane, not
necessarily in the plane

a

O

(a)

r

O

(b)

(c)

Figure 2
...
(b) Plane defined using three points
...
Vector r is the position vector of a general point in the
plane
...
2
...

Depending on how the plane is defined, this can be written as
D = |(d − a) · n
ˆ|

2
...

|b × c|

Solution of vector equations

It is sometimes required to obtain the most general vector which satisfies a given
vector relationship
...
The best
method of proceeding in such a case is as follows:
(i) Decide upon a system of three co-ordinate vectors using two non-parallel vectors
appearing in the vector relationship
...

(ii) Express the unknown vector x as a linear combination of these vectors
x = λa + µb + νa × b
where λ, µ, ν are scalars to be found
...

♣ Example

Q Solve the vector equation x = x × a + b
...

Step (ii): x = λa + µb + νa × b
...
4
...

⇒ λ = −ν(a · b)
µ = νa2 + 1
ν = −µ
so that
µ=

1
1 + a2

ν=−

1
1 + a2

λ=

a·b

...
4

1
((a · b)a + b − (a × b))
1 + a2

Rotation, angular velocity/acceleration and moments

A rotation can be represented by a vector whose direction is along the axis of
rotation in the sense of a r-h screw, and whose magnitude is proportional to the
size of the rotation (Fig
...
7)
...

Angular accelerations arise because of a moment (or torque) on a body
...

The vector equation for moment is
M=r×F
where r is the vector from Q to any point on the line of action L of force F
...

The instantaneous velocity of any point P on a rigid body undergoing pure rotation
can be defined by a vector product as follows
...
MULTIPLE PRODUCTS
...
7: The angular velocity vector ω is along the axis of rotation and has magnitude equal to
the rate of rotation
...
If r is the vector OP , where the origin O can
be taken to be any point on the axis of rotation, then the velocity v of P due to
the rotation is given, in both magnitude and direction, by the vector product
v = ω × r
...
8:

Revised Oct 2013

v

Lecture 3
Differentiating Vector Functions of a Single
Variable
Your experience of differentiation and integration has extended as far as scalar
functions of single and multiple variables — functions like f (x ) and f (x , y , t)
...
For example, suppose you were driving along a wiggly road with position
r(t) at time t
...
Let’s see how to do this
...
1

Differentiation of a vector

The derivative of a vector function a(p) of a single parameter p is
a(p + δp) − a(p)

...

dp
dp
dp
That is, in order to differentiate a vector function, one simply differentiates each
component separately
...

Thus, for example:
d
d
da
db
da
db
(a × b) =
×b+a×
(a · b) =
·b+a·

...
DIFFERENTIATING VECTOR FUNCTIONS OF A SINGLE VARIABLE

Note that d a/d p has a different direction and a different magnitude from a
...
If a = a(u) and u = u(t),
say:
da du
d
a=
dt
du dt

♣ Examples

Q A 3D vector a of constant magnitude is varying over time
...
We would guess that the
derivative a˙ is orthogonal to a
...

dt
dt
dt
dt
But (a · a) = a2 which we are told is constant
...

Q The position of a vehicle is r(u) where u is the amount of fuel consumed by
some time t
...

A The velocity is
v=

dr du
dr
=
dt
du dt

d dr
d 2r
a=
=
dt dt
d u2
3
...
1



du
dt

2

+

d r d 2u
d u d t2

Geometrical interpretation of vector derivatives

Let r(p) be a position vector tracing a space curve as some parameter p varies
...
(See
Fig
...
1
...
If the magnitude of this vector is 1 (i
...
a unit tangent), then

3
...
DIFFERENTIATION OF A VECTOR

29

|d r| = d p so the parameter p is arc-length (metric distance)
...
r
...

Of course if that parameter p is time, the magnitude |d r/d t| is the speed
...
Show that the tangent d r/d s
to the curve has a constant elevation angle w
...
t the x y -plane, and determine
its magnitude
...

We are expecting d r/d s = 1, and indeed
q
p
2
2
2
2
2
a sin () + a cos () + h / a2 + h2 = 1
...
1
...
Why not? The reason is that arc
length is special is that, whatever the parameter p,
Z p 
 dr 
  dp
...


30

LECTURE 3
...
1: Left: δr is a secant to the curve but, in the limit as δp → 0, becomes a tangent
...


So if a curve is parameterized in terms of p
s
ds
dx 2 dy 2 dz 2
=
+
+

...
So p/ a2 + h2 is arclength, not p
...
2

Integration of a vector function

The integration of a vector function of a single scalar variable can be regarded
simply as the reverse of differentiation
...

However, many other, more interesting and useful, types of integral are possible,
especially when the vector is a function of more than one variable
...
See later!

3
...
CURVES IN 3 DIMENSIONS

3
...
It
doesn’t have to be, but these are the main two of interest
...
r
...
time, but now let use look
more closely at 3D curves defined in terms of arc length, s
...
This is a
space curve
...

3
...
1

The Fr´
enet-Serret relationships

We are now going to introduce a local orthogonal coordinate frame for each point
s along the curve, ie one with its origin at r(s)
...
The ideas were first suggested by
two French mathematicians, F-J
...
A
...

1
...
We already know that
ˆt =

d r(s)
ds

2
...
Because
ˆt has constant magnitude but varies over s, dˆt/d s must be perpendicular to
ˆt
...

ds
κ is the curvature, and κ = 0 for a straight line
...

ˆ
3
...


32

LECTURE 3
...

ds
But this means that d ˆ
b/d s is along the direction of n
ˆ, or
b
dˆ
= −τ(s)ˆ
n(s)
ds
where τ is the torsion, and the negative sign is a matter of convention
...

ds
The Fr´
enet-Serret relationships:
dˆt/d s = κˆ
n
ˆ
dˆ
n/d s = −κ(s)ˆt(s) + τ(s)b(s)
dˆ
b/d s = −τ(s)ˆ
n(s)
♣ Example

Q Derive κ(s) and τ(s) for the helix
 
 
 
s
s ˆ
s
ˆı + a sin
ˆ + h
k;
r(s) = a cos
β
β
β
and comment on their values
...

ds
β
β
β
β
β
Differentiation gives

 
a
s
dˆt
= − 2 cos
,
κˆ
n=
ds
β
β

 

a
s
− 2 sin
, 0
β
β

β=

p

a2 + h 2

3
...
RADIAL AND TANGENTIAL COMPONENTS IN PLANE POLARS

Curvature is always positive, so

 
s
a
n
ˆ = − cos
,
κ= 2
β
β

33

 

s
− sin
, 0
...

Now use


ˆ  

 
 

ˆ
ˆ
ı

k


h
a
s
s
h
ˆ = ˆt׈
b
n =  (−a/β)S (a/β)C (h/β)  =
sin
, − cos
,
β
β
β
β
β

−C
−S
0 
ˆ to find an expression for the torsion
and differentiate b

 
 

h
s
s
h
−h
dˆ
b
=
cos
,
sin
,
0
=
n
ˆ
ds
β2
β
β2
β
β2

so the torsion is
τ=

h
β2

again a constant
...
4

Radial and tangential components in plane polars

In plane polar coordinates, the radius vector
of any point P is given by
r = r cos θˆı + r sin θˆ
= rˆ
er

ˆ
eθ
ˆ
er

where we have introduced the unit radial vector
ˆ
er = cos θˆı + sin θˆ
...


ˆı

34

LECTURE 3
...
Its velocity is
r˙ =

dr
dˆ
er
d
(rˆ
er ) =
ˆ
er + r
dt
dt
dt
dθ
dr
ˆ
er + r (− sin θˆı + cos θˆ )
=
dt
dt
dθ
dr
ˆ
er + r ˆ
eθ
=
dt
dt
= radial + tangential

The radial and tangential components of velocity of P are therefore d r /d t and
r d θ/d t, respectively
...
5
...
5

35

Rotating systems

Consider a body which is rotating with constant angular velocity ω about some
axis passing through the origin
...

If ρ is a vector of constant magnitude and constant direction in the rotating system,
then its representation r in the fixed system must be a function of t
...


ω
ρ

36

LECTURE 3
...
First, note that RR⊤ = I (the identity), so
Now let’s consider the term RR
differentiating both sides yields
˙ ⊤ + RR˙ ⊤ = 0
RR
˙ ⊤ = −RR˙ ⊤
RR
˙ ⊤ is anti-symmetric:
Thus RR


0 −z y
˙ ⊤= z
0 −x 
RR
−y x
0

Now you can verify for yourself that application of a matrix of this form to an
arbitrary vector has precisely the same effect as the cross product operator, ω ×,
where ω = [x y z ]⊤
...

3
...
1

Rotation: Part 2

Now suppose ρ is the position vector of a point P which moves in the rotating
frame
...

So, returning to the equations we derived earlier:
r(t) = R(t)ρρ(t)
and the instantaenous differential with respect to time:
dr
˙ ρ + Rρ˙ = RR
˙ ⊤r + Rρ˙
= Rρ
dt
Now ρ is not constant, so its differential is not zero; hence rewriting this last
equations we have that
The instantaneous velocity of P in the fixed frame is
dr
= Rρ˙ + ω × r
dt
The second term of course, is the contribution from the rotating frame which we
saw previously
...
5
...
5
...
Let us now consider the second differential:
¨r = ω˙ × r + ω × r˙ + R˙ ρ˙ + R¨
ρ
We shall assume that the angular acceleration is zero, which kills off the first term,
and so now, substituting for r˙ we have
¨r =
=
=
=

ω × r + Rρ˙ ) + R˙ ρ˙ + R¨
ω × (ω
ρ
ω × r) + ω × Rρ˙ + R˙ ρ˙ + R¨
ω × (ω
ρ
˙ ⊤ R)ρ˙ + R¨
ω × r) + ω × Rρ˙ + R(R
ω × (ω
ρ
ω × (ω
ω × r) + 2ω
ω × (Rρ˙ ) + R¨
ρ

The instantaneous acceleration is therefore
ω × (Rρ˙ ) + ω × (ω
ω × r)
¨r = R¨
ρ + 2ω
• The first term is the acceleration of the point P in the rotating frame measured in the rotating frame, but referred to the fixed frame by the rotation
R
• The last term is the centripetal acceleration to due to the rotation
...
Check it out
...
DIFFERENTIATING VECTOR FUNCTIONS OF A SINGLE VARIABLE

γt

r

ω = ωm
ˆ

m
ˆ
ˆℓ
n
ˆ

γ = γˆℓ

Figure 3
...


• The middle term is an extra term which arises because of the velocity of P
in the rotating frame
...

Because of the rotation of the earth, the Coriolis acceleration is of great importance in meteorology and accounts for the occurrence of high pressure anticyclones and low pressure cyclones in the northern hemisphere, in which the Coriolis
acceleration is produced by a pressure gradient
...

♣ Example

Q Find the instantaneous acceleration of a projectile fired along a line of longitude (with angular velocity of γ constant relative to the sphere) if the sphere
is rotating with angular velocity ω
...
3
...
We shall assume, without loss of generality,
that the fixed and rotating frames are instantaneously aligned at the moment
shown in the diagram, so that R = I, the identity, and hence r = ρ
...

¨r = γ × (γ
The first term is the centripetal acceleration due to the projectile moving
around the sphere — which it does because of the gravitational force
...
5
...
The middle term is the Coriolis acceleration
...
3
...

Likewise the last term can be obtained as
ω × r) = −ω 2 r sin(γt)ˆ
ω × (ω
n
Note that it is perpendicular to the axis of rotation m,
ˆ and because of the
minus sign, directed towards the axis)
The Coriolis term is derived as:
ω × ρ˙ = 2ω
ω × (γ
γ × ρ)
2ω
    

0
γ
0
= 2 ω  ×  0  × r cos γt 
0
0
r sin γt
= 2ωγr cos γtˆℓ
Instead of a projectile, now consider a rocket on rails which stretch north
from the equator
...
e
...
In the absence of the rails (or atmosphere)
the rocket’s tangetial speed (relative to the surface of the earth) is greater
than the speed of the surface of the earth underneath it (since the radius
of successive lines of latitude decreases) so it would (to an observer on the
earth) appear to deflect to the east
...

Revised Oct 2013

40

LECTURE 3
...
3: Rocket example

3
...
ROTATING SYSTEMS

Figure 3
...
DIFFERENTIATING VECTOR FUNCTIONS OF A SINGLE VARIABLE

Lecture 4
Line, Surface and Volume Integrals
...

We started off the course being concerned with individual vectors a, b, c, and so
on
...

In much of the rest of the course, we will be concerned with scalars and vectors
which are defined over regions in space — scalar and vector fields
In this lecture we introduce line, surface and volume integrals, and consider how
these are defined in non-Cartesian, curvilinear coordinates

4
...

Similarly, if a vector function v(r) is defined at each point, then v is a vector field
in that region
...

Familiar examples of each are shown in figure 4
...

In Lecture 1 we worked out the force F(r) on a charge Q arising from a number
of charges qi
...

|r − ri |3

(K =

1
)
4πǫr ǫ0

For example; you could work out the velocity field, in plane polars, at any point on
43

44LECTURE 4
...
CURVILINEAR COORDINATES
...
1: Examples of (a) a scalar field (pressure); (b) avector field (wind velocity)

a wheel spinning about its axis
v(r) = ω × r
or the fluid flow field around a wing
...
Of course, most
vector fields of practical interest in engineering science are not steady, and some
are unpredictable
...


4
...
Eg, given a map showing the
pollution density field in Oxford, you may wish to work out how much pollution
you breathe in when cycling from college to the Department via different routes
...
One assumes a set of n samples fi = f (xi ) spaced by δxi
...

f (x )d x =
lim
n → ∞ i =1
δxi → 0
For a smooth function, it is irrelevant how the function is subdivided
...
2
...
Each line segment is then

4
...
LINE INTEGRALS THROUGH FIELDS

45

F(r)
r

δr

Figure 4
...
In the diagram F(r) is a vector field, but it could be replaced with scalar
field U(r)
...

There are three types of integral we have to think about, depending on the nature
of the product:
1
...

!
Z
X
Ui δ r i
...
Integrand a(r) is a vector field dotted with d r hence the integral is a scalar:
!
Z
X
I = a(r) · d r
= lim
ai · δ r i
...
Integrand a(r) is a vector field crossed with d r hence vector result
...

I = a(r) × d r
= lim
L

δri →0

i

Note immediately that unlike an integral in a single scalar variable, there are many
paths L from start point rA to end point rB , and the integral will in general depend
on the path taken
...
If the force acts

46LECTURE 4
...
CURVILINEAR COORDINATES
...
d r, and so the total work done traversing
the path is
Z
WC =
F
...
d r = µ0 I
C

where I is the current enclosed by (closed) path C
...
So if a loop this wire C is placed in the field
then the total force will be and integral of type 3 above:
I
F = I dr × B
C

Note that the expressions above are beautifully compact in vector notation, and are
all independent of coordinate system
...
6
...
A force F = x 2yˆı + x y 2ˆ acts on a body as it
moves between (0, 0) and (1, 1)
...
along the line y = x
...
along the curve y = x n , n > 0
...
along the x axis to the point (1, 0) and then along the line x = 1
...
In planar Cartesians, d r =
ˆı d x + ˆ d y
...

L

L

1
...
So it is easiest to convert all
y references to x
...

(0,0)

x=0

x=0

4
...
LINE INTEGRALS THROUGH FIELDS

47

1,1
1
2 3

0,0

0,1

Figure 4
...
For the path y = x n we find that d y = nx n−1d x , so again it is easiest to
convert all y references to x
...
x
...
This path is not smooth, so break it into two
...


So in general the integral depends on the path taken
...


Q2 Repeat part (2) using the Force F = x y 2ˆı + x 2yˆ
...
LINE, SURFACE AND VOLUME INTEGRALS
...


A2 For the path y = x n we find that d y = nx n−1d x , so
Z

(1,1)
2

2

(y x d x + y x d y ) =
(0,0)

Z

x=1

(x 2n+1d x + nx n−1
...
x n d x )

x=0
x=1

=

Z

(x 2n+1d x + nx 2n+1d x )

x=0

n
1
+
2n + 2 2n + 2
1
independent of n
=
2

=

4
...
In fact it is wholly independent of path
...
Then using the definition of the perfect differential
dg =

∂g
∂g
dx +
dy
∂x
∂y

we find that
Z

B
2

2

(y x d x + y x d y ) =
A

=

Z

B

A
gB

dg
− gA

which depends solely on the value of g at the start and end points, and not at all
on the path used to get from A to B
...

One sort of line integral performs the integration around a complete loop and is
denoted with a ring
...

In electrostatics, if E is the electric field then the potential function is
Z
φ = − E · dr
...
4
...
3
...
They don’t appear
to involve vectors (but actually they are another form of type 2 defined earlier)
...
Convert the function to F (p),
writing
Z pend
ds
dp
I=
F (p)
dp
pstart
where
ds
=
dp

"

dx
dp

2

+



dy
dp

2

+



dz
dp

2#1/2


...
Then
"
 2  2 #1/2
Z xend
dz
dy
I=
dx
...
4

Surface integrals

These can be defined by analogy with line integrals
...

Again there are three possibilities:
R
• S Ud S — scalar field U; vector integral
...

R
• S a × d S — vector field a; vector integral
...


50LECTURE 4
...
CURVILINEAR COORDINATES
...
d S For example the mass of fluid
crossing a surface S in time d t is d M = ρv
...
The total mass flux can be expressed as
a surface integral:
Z
ΦM =
ρ(r )v(r)
...
Note, however, that in some problems, symmetry may lead us
to a different more natural coordinate system
...

d S is perpendicular to the surface
...
More often than not,
the surface will enclose a volume, and the
surface direction is taken as everywhere
emanating from the interior
...
5

Z Z

z
1
1

11111
00000
00000
11111
00000
11111
00000
11111
00000
11111
y 11111
00000
00000
11111
00000
11111
dS = dydz i
00000
11111
00000
11111
00000
11111
00000
11111
1 x

ydydz

1 21 1 1
y 0 z |0 =

...
One obvious difference though is
that the element of volume is a scalar (how could you define a direction with an
infinitesimal volume element?)
...
6
...

ad V — vector field; vector integral
...
The next section considers these again in the context of a change of
coordinates
...
6

Changing variables: curvilinear coordinates

Up to now we have been concerned with Cartesian coordinates x , y , z with coordinate axes ˆı ,ˆ , kˆ
...


The reason for using the basis ˆı ,ˆ , kˆ rather than any other orthonormal basis set is
that ˆı represents a direction in which x is increasing while the other two coordinates
remain constant (and likewise for ˆ and kˆ with y and z respectively), simplifying
the representation and resulting mathematics
...
We will see that the idea hinted at above – of defining a
basis set by considering directions in which only one coordinate is (instantaneously)
increasing – provides the approriate generalisation
...

4
...
1

Cylindrical polar coordinates

As shown in figure 4
...
LINE, SURFACE AND VOLUME INTEGRALS
...

z

Lines of
constant r

Lines of
constant z
^
k

y

^
j
r

^i

r
θ

Lines of
constant φ

x

(a)

(b)

Figure 4
...


Note that, by definition, ∂∂rr represents a direction in which (instantaneously) r is
changing while the other two coordinates stay constant
...
Likewise for ∂φ
and ∂z
, Thus the vectors:

∂r
= cos φˆı + sin φˆ
∂r
∂r
=
= −r sin φˆı + r cos φˆ
∂φ
∂r
= kˆ
=
∂z

er =
eφ
ez

Aside on notation: some texts
ˆ ,
...
Though I prefer the notation used here, where
the basis vectors are written as
eˆ with appropriate subscripts (as
used in Riley et al), you should be
aware of, and comfortable with,
either possibility
...
It is more usual, however, first to normalise the vectors to
obtain their corresponding unit vectors, ˆ
er , ˆ
eφ , ˆ
ez
...
In cartesian coordinates, a small change

4
...
CHANGING VARIABLES: CURVILINEAR COORDINATES

53

in (eg) x while keeping y and z constant would result in a displacement of
p
√
d s = |d r| = d r
...

Factors such as this r are known as scale factors or metric coefficients, and we
must be careful to take them into account when, eg, performing line, surface or
volume integrals, as you will below
...

Example: line integral in cylindrical coordinates

H
Q Evaluate C a · d l , where a = x 3ˆ − y 3ˆı + x 2y kˆ and C is the circle of radius r
in the z = 0 plane, centred on the origin
...
5
...

On the circle of interest
a = r 3 (− sin3 φˆı + cos3 φˆ + cos2 φ sin φkˆ )
and (since dz = dr = 0 on the path)
dr = r dφ ˆ
eφ
= r d φ(− sin φˆı + cos φˆ )
so that
I

C

since
Z

0

a · dr =

2π

4

Z

2π

r 4 (sin4 φ + cos4 φ)d φ =

0

sin φd φ =

Z

2π

cos4 φd φ =
0

3π
4

3π 4
r
2

54LECTURE 4
...
CURVILINEAR COORDINATES
...
5: Line integral example in cylindrical coordinates
Volume integrals in cylindrical polars

In Cartesian coordinates a volume element is given by (see figure 4
...
1
...
Thus the formula above can be
derived (even though it is “obvious”) as:
d V = d xˆı
...

In cylindrical polars a volume element is given by (see figure 4
...
(r d φˆ
eφ × d zˆ
ez ) = r d φd r d z
Note also that this volume, because it is a scalar triple product, can be written as
a determinant:
 

  ∂x ∂y ∂z 

 ˆ


∂r ∂r ∂r 
 er d r   er d r   ∂x
∂y ∂z 




eφ r d φ  =  eφ d φ  =  ∂φ ∂φ
dV =  ˆ
∂φ  d r d φd z

∂y
∂z 
 ˆ
ez d z   ez d z   ∂x
∂z ∂z ∂z

where the equality on the right-hand side follows from the definitions of ˆ
er = ∂∂rr =
∂x
ˆ ∂y ˆ ∂z ˆ
∂r ı + ∂r  + ∂r k , etc
...
6
...
6: Volume elements dV in (a) Cartesian coordinates; (b) Cylindrical polar coordinates
Surface integrals in cylindrical polars

Recall from section 4
...
In this case the plane is spanned by the vectors ˆ and kˆ and the area
of the element given by (see section 1
...
7) are given by:
d S = d rˆ
er × r d φˆ
eφ = r d r d φˆ
ez
d S = r d φˆ
eφ × d zˆ
ez = r d φd zˆ
er

(for surfaces of constant z )
(for surfaces of constant r )

Similarly we can find d S for surfaces of constant φ, though since these aren’t as
common this is left as a (relatively easy) exercise
...
LINE, SURFACE AND VOLUME INTEGRALS
...

dSz = r dr dφˆ
ez
z

er
r dφˆ
eφ drˆ
dzˆ
ez

dSr = r dφdzˆ
er
dzˆ
ez
r dφˆ
eφ

drˆ
er
dSφ = dr dzˆ
eφ

y

x
Figure 4
...
6
...

As shown in figure 4
...
2 a point in space P having cartesian coordinates x , y , z
can be expressed in terms of spherical polar coordinates, r, θ, φ as follows:
r = xˆı + yˆ + z kˆ
= r sin θ cos φˆı + r sin θ sin φˆ + r cos θkˆ
The basis set in spherical polars is obtained in an analogous fashion: we find unit
z

z

θ
P

kˆ

r

Lines of
constant φ
(longitude)

ˆ
er
ˆ
eφ
ˆ
eθ

Lines of
constant r

ˆ

y

y

ˆı

φ
x

x

Lines of
constant θ
(latitude)

4
...
CHANGING VARIABLES: CURVILINEAR COORDINATES

57

vectors which are in the direction of increase of each coordinate:
∂r
=
sin θ cos φˆı + sin θ sin φˆ + cos θkˆ
=ˆ
er
∂r
∂r
= r cos θ cos φˆı + r cos θ sin φˆ − r sin θkˆ = rˆ
eθ
eθ =
∂θ
∂r
eφ =
=
−r sin θ sin φˆı + r sin θ cos φˆ
= r sin θˆ
eφ
∂φ
er =

As with cylindrical polars, it is easily verified that the vectors ˆ
er , ˆ
eθ , ˆ
eφ form an
orthonormal basis
...

Volume integrals in spherical polars

In spherical polars a volume element is given by (see figure 4
...
(r d θˆ
eθ × r sin θd φˆ
eφ) = r 2 sin θd r d θd φ
Note again that this volume could be written as a determinant, but this is left as
an exercise
...
9)
...

Thus a surface element d S in spherical polars is given by
d S = r d θˆ
eθ × r sin θd φˆ
eφ = r 2 sin θˆ
er

58LECTURE 4
...
CURVILINEAR COORDINATES
...
8: Volume element dV in spherical polar coordinates

♣ Example: surface integral in spherical polars

R
Q Evaluate S a · d S, where a = z 3kˆ
and S is the sphere of radius A centred on the origin
...
6
...
9: Surface element dS in spherical polar coordinates

4
...
3

General curvilinear coordinates

Cylindrical and spherical polar coordinates are two (useful) examples of general
curvilinear coordinates
...
LINE, SURFACE AND VOLUME INTEGRALS
...

∂r
∂r
∂r
The metric coefficients are therefore hu = | ∂u
|, hv = | ∂v
| and hw = | ∂w
|
...
(hv d vˆ
ev × hw d wˆ
ew )

and simplifies if the coordinate system is orthonormal (since ˆ
eu
...
6
...
(ˆ
v × w)
ˆ
d S = hu hv d ud v ˆ
u׈
v (for surface element tangent to constant w )
u = ˆ
ˆ
eu =

Plane polar coordinates
x
r
hr
ˆ
er
dr
dS

=
=
=
=
=
=

r cos θ,
y = r sin θ
r cos θˆı + r sin θˆ
1,
hθ = r
cos θˆı + sin θˆ ,
ˆ
eθ = − sin θˆı + cos θˆ
d rˆ
er + r d θˆ
eθ
r d r d θkˆ

4
...
CHANGING VARIABLES: CURVILINEAR COORDINATES

61

Cylindrical polar coordinates
x
r
hr
ˆ
er
dr
dS
dS
dV

=
=
=
=
=
=
=
=

r cos φ,
y = r sin φ,
z =z
r cos φˆı + r sin φˆ + z kˆ
1,
hφ = r,
hz = 1
cos φˆı + sin φˆ ,
ˆ
eφ = − sin φˆı + cos φˆ ,
d rˆ
er + r d φˆ
eφ + d zˆ
ez
r d r d φkˆ (on the flat ends)
r d φd zˆ
er (on the curved sides)
r d r d φd z

ˆ
ez = kˆ

Spherical polar coordinates
x
r
hr
ˆ
er
ˆ
eθ
ˆ
eφ
dr
dS
dV

=
=
=
=
=
=
=
=
=

r sin θ cos φ,
y = r sin θ sin φ,
z = r cos θ
r sin θ cos φˆı + r sin θ sin φˆ + r cos θkˆ
1,
hθ = r,
hφ = r sin θ
sin θ cos φˆı + sin θ sin φˆ + cos θkˆ
cos θ cos φˆı + cos θ sin φˆ + sin θkˆ
− sin φˆı + cos φˆ
d rˆ
er + r d θˆ
eθ + r sin θd φˆ
eφ
r 2 sin θd r d θd φˆ
er (on a spherical surface)
2
r sin θd r d θd φ

IDR October 10, 2013

62LECTURE 4
...
CURVILINEAR COORDINATES
...
We introduce three field operators which reveal interesting
collective field properties, viz
...


There are two points to get over about each:
• The mechanics of taking the grad, div or curl, for which you will need to brush
up your multivariate calculus
...

In Lecture 6 we will look at combining these vector operators
...
1

The gradient of a scalar field

Recall the discussion of temperature distribution throughout a room in the overview,
where we wondered how a scalar would vary as we moved off in an arbitrary direction
...

If U(x , y , z ) is a scalar field, ie a scalar function of position r = [x , y , z ] in 3
dimensions, then its gradient at any point is defined in Cartesian co-ordinates by
∂U
∂U ˆ
∂U
ˆı +
ˆ +
k
...

∂x
∂y
∂z
63

64

LECTURE 5
...


Note immediately that ∇ U is a vector field!
Without thinking too carefully about it, we can see that the gradient of a scalar
field tends to point in the direction of greatest change of the field
...

♣ Worked examples of gradient evaluation
1
...


2
...

3
...



∂
∂
∂
⇒ ∇ U = ˆı
+ ˆ
+ kˆ
(c1x + c2y + c3z ) = c1ˆı +c2ˆ +c3 kˆ = c
...
U = f (r ), where r = (x 2 + y 2 + z 2 )
U is a function of r alone so df /d r exists
...

∂z
d r ∂z


∂r
∂r ˆ
df ∂r
ˆı + ˆ +
=
k
d r ∂x
∂y
∂z

∂f
∂f
∂f ˆ
⇒ ∇ U = ˆı + ˆ +
k
∂x
∂y
∂z
p
But r = x 2 + y 2 + z 2 , so ∂r /∂x = x /r and similarly for y , z
...

=
dr
r
dr r

5
...
THE SIGNIFICANCE OF GRAD

65

gradU

r
U(r)

dr
r + dr
U(r + dr)

Figure 5
...
2

The significance of grad

If our current position is r in some scalar field U (Fig
...
1), and we move an
infinitesimal distance d r, we know that the change in U is
dU =

∂U
∂U
∂U
dx +
dy +
dz
...

Now divide both sides by d s
dr
dU
= ∇U ·

...

This result can be paraphrased as:
• gradU has the property that the rate of change of U wrt distance in a
ˆ is the projection of gradU onto that direction
particular direction (d)
(or the component of gradU in that direction)
...

We could also say that

66

LECTURE 5
...

0
...
08

0
...
04

0
...

If we move a tiny amount within that iso-U surface, there is no change in U, so
d U/d s = 0
...

ds

But d r/d s is a tangent to the surface, so this result shows that
• gradU is everywhere NORMAL to a surface of constant U
...
3
...
3

67

The divergence of a vector field

The divergence computes a scalar quantity from a vector field by differentiation
...

♣ Examples of divergence evaluation
a
1) xˆı
2) r(= xˆı + yˆ + z kˆ )
3) r/r 3
4) r c, for c constant
We work through example 3)
...
(x 2 + y 2 + z 2)−3/2, and we need to find ∂/∂x of it
...
(x 2 + y 2 + z 2)−3/2 = 1
...
2x
∂x
2


−3
2 −2
= r
1 − 3x r

...
4

The significance of div

Consider a typical vector field, water flow, and denote it by a(r)
...

Now take an infinitesimal volume element d V and figure out the balance of the
flow of a in and out of d V
...
VECTOR OPERATORS: GRAD, DIV AND CURL

To be specific, consider the volume element d V = d x d y d z in Cartesian coordinates, and think first about the face of area d x d z perpendicular to the y axis
and facing outwards in the negative y direction
...
)

z

dz
dS = -dxdz j

dS = +dxdz j
y
dx
dy

x
Figure 5
...


The component of the vector a normal to this face is a · ˆ = ay , and is pointing
inwards, and so its contribution to the OUTWARD flux from this surface is
a · dS =

− ay (y )d z d x ,

where ay (y ) means that ay is a function of y
...
)
A similar contribution, but of opposite sign, will arise from the opposite face, but
we must remember that we have moved along y by an amount d y , so that this
OUTWARD amount is


∂ay
dy dxdz
ay (y + d y )d z d x = ay +
∂y
The total outward amount from these two faces is
∂ay
∂ay
dydxdz =
dV
∂y
∂y
Summing the other faces gives a total outward flux of


∂ax ∂ay ∂az
+
+
dV = ∇ · a dV
∂x
∂y
∂z
So we see that

5
...


THE LAPLACIAN: DIV(GRADU) OF A SCALAR FIELD

69

The divergence of a vector field represents the flux generation per unit
volume at each point of the field
...
)
Interestingly we also saw that the total efflux from the infinitesimal volume was
equal to the flux integrated over the surface of the volume
...
)

5
...
Recall also that we
can compute the divergence of any vector field
...

Here is where the ∇ operator starts to be really handy
...
For this reason, the
operator ∇2 is called the “Laplacian”

 2
2
2
∂
∂
∂
+
+
U
∇2 U =
∂x 2 ∂y 2 ∂z 2
Laplace’s equation itself is
∇2 U = 0

70

LECTURE 5
...

1/r = (x 2 + y 2 + z 2)−1/2
∂
∂ ∂ 2
(x + y 2 + z 2)−1/2 =
− x
...
x
...
6

The curl of a vector field

So far we have seen the operator ∇ applied to a scalar field ∇ U; and dotted with
a vector field ∇ · a
...
7
...
7

The significance of curl

Perhaps the first example gives a clue
...
3(a)
...
) This field has a curl of 2kˆ , which is in the r-h screw
sense out of the page
...

y

ax (y+dy)

y+dy

dy
ay (x)

x

y

ay (x+dx)

y

dx
x+dx

x
ax (y)

(a)

(b)

Figure 5
...
(b) An element in which to calculate
curl
...


The
circulation of a vector a round any closed curve C is defined to be
H
C a · dr
and the curl of the vector field a represents the vorticity, or circulation
per unit area, of the field
...
VECTOR OPERATORS: GRAD, DIV AND CURL

Our proof uses the small rectangular element d x by d y shown in Figure 5
...

Consider the circulation round the perimeter of a rectangular element
...

NB: Again, this is not a completely rigorous proof as we have not shown that the
result is independent of the co-ordinate system used
...
8

Some definitions involving div, curl and grad
• A vector field with zero divergence is said to be solenoidal
...

• A scalar field with zero gradient is said to be, er, constant
...

The main thing to appreciate it that the operators behave both as vectors and
as differential operators, so that the usual rules of taking the derivative of, say, a
product must be observed
...
HLT contains a lot
of them
...

Secondly, they help to identify other practically important vector operators
...


6
...

Note that the output is a null vector
...
2

LECTURE 6
...
3

Identity 3: div and curl of Ua

Suppose that U(r) is a scalar field and that a(r) is a vector field and we are interested in the product Ua
...
For example the density ρ(r) of a fluid is a scalar field, and the instantaneous velocity of the fluid v(r) is a vector field, and we are probably interested
in mass flow rates for which we will be interested in ρ(r)v(r)
...


6
...
bash out the products
...
5
...
5

75

Identity 5: curl(a × b)


ˆı
ˆ
kˆ

curl(a × b) = 
∂/∂x
∂/∂y
∂/∂z
 ay bz − az by az bx − ax bz ax by − ay bx

so the ˆı component is








∂
∂
(ax by − ay bx ) −
(az bx − ax bz )
∂y
∂z
which can be written as the sum of four terms:



 



∂by ∂bz
∂ay ∂az
∂
∂
∂
∂
ax
+
− bx
+
+ by
+ bz
ax − ay
+ az
bx
∂y
∂z
∂y
∂z
∂y
∂z
∂y
∂z
Adding ax (∂bx /∂x) to the first of these, and subtracting it from the last, and
doing the same with bx (∂ax /∂x) to the other two terms, we find that (you should
of course check this):
∇ × (a × b) = (∇
∇ · b)a − (∇
∇ · a)b + [b · ∇]a − [a · ∇]b
where [a · ∇ ] can be regarded as new, and very useful, scalar differential operator
...
6

This is a scalar operator, but it can obviously can be applied to a scalar field,
resulting in a scalar field, or to a vector field resulting in a vector field:


∂
∂
∂
[a · ∇ ] ≡ ax
+ ay
+ az

...
7

Identity 6: curl(curla) for you to derive

The following important identity is stated, and left as an exercise:
curl(curla) = graddiva − ∇2a
where
∇2a = ∇2 axˆı + ∇2ayˆ + ∇2 az kˆ

76

LECTURE 6
...
The entire
telecommunications industry is built on these
...

Now show that in a material with zero free charge density, ρ = 0, and with
zero conductivity, σ = 0, the electric field E must be a solution of the wave
equation
∇2 E = µr µ0 ǫr ǫ0 (∂ 2E/∂t 2 )
...
Imagine you are the first to do this — this is a tingle
moment
...

div(µr µ0 H) = µr µ0 divH = 0
⇒ divB = 0
−∂B/∂t = −µr µ0 (∂H/∂t)
J + ∂D/∂t = 0 + ǫr ǫ0 (∂E/∂t)

(a)
(b)
(c)
(d )

But we know (or rather you worked out in Identity 6) that curlcurl = graddiv−
∇2 , and using (c)
curlcurlE = graddivE − ∇2E = curl (−µr µ0 (∂H/∂t))
so interchanging the order of partial differentation, and using (a) divE = 0:
∂
(curlH)
∂t 

∂E
∂
ǫr ǫ0
= −µr µ0
∂t
∂t
∂ 2E
2
⇒ ∇ E = µr µ0ǫr ǫ0 2
∂t
−∇2E = −µr µ0

6
...
GRAD, DIV, CURL AND ∇2 IN CURVILINEAR CO-ORDINATE SYSTEMS

77

This equation is actually three equations, one for each component:
∂ 2 Ex
∇ Ex = µr µ0 ǫr ǫ0 2
∂t
2

and so on for Ey and Ez
...
8

Grad, div, curl and ∇2 in curvilinear co-ordinate systems

It is possible to obtain general expressions for grad, div and curl in any orthogonal
curvilinear co-ordinate system by making use of the h factors which were introduced
in Lecture 4
...
Similar expressions apply for the other co-ordinate directions
...


6
...
10

LECTURE 6
...

We consider an element of volume d V
...

However, it is not quite a cuboid: the area of two opposite faces will differ as the
scale parameters are functions of u, v and w in general
...
1: Elemental volume for calculating divergence in orthogonal curvilinear coordinates

So the net efflux from the two faces in the ˆ
v direction shown in Figure 6
...
e
...

By definition div is the net efflux per unit volume, so summing up the other faces:


∂(au hv hw )
∂(av hu hw )
∂(aw hu hv )
diva d V =
+
+
d ud v d w
∂u
∂v
∂w


∂(av hu hw )
∂(aw hu hv )
∂(au hv hw )
+
+
d ud v d w
⇒ diva hu hv hw d ud v d w =
∂u
∂v
∂w

6
...
CURL IN CURVILINEAR COORDINATES

79

So, finally,
1
diva =
hu hv hw

6
...
But the opposite
sides are no longer quite of the same length
...
2
is length hu (v )d u, but the upper is of length hu (v + d v )d u

y
au (v+dv)

v+dv

h u(v+dv) du
dv
hu (v) du
y

u+du

u
au (v)

Figure 6
...
VECTOR OPERATOR IDENTITIES

So the circulation per unit area is


1
∂(hv av ) ∂(hu au )
dC
=
−
hu hv d ud v
hu hv
∂u
∂v
and hence curl is



∂(hw aw ) ∂(hv av )
−
u
ˆ +
∂v
∂w


1
∂(hu au ) ∂(hw aw )
−
ˆ
v +
hw hu
∂w
∂u


∂(hv av ) ∂(hu au )
1
−
w
ˆ
hu hv
∂u
∂v

1
curla(u, v , w ) =
hv hw

You should check that this can be written as
Curl in curvilinear coords:

 h u
ˆ hv ˆ
v hw w
ˆ
1  u∂
∂
∂
curla(u, v , w ) =
∂v
∂w
hu hv hw  ∂u
hu au hv av hw aw

6
...

hu hv hw ∂u
hu ∂u
∂v
hv ∂v
∂w
hw ∂w

6
...
The position vector is r = r cos φˆı + r sin φˆ + z kˆ , and
hr = |∂r/∂r |, etc
...
14
...
14

Grad Div, Curl, ∇2 in spherical polars

Here (u, v , w ) → (r, θ, φ)
...

q
(sin2 θ(cos2 φ + sin2 φ) + cos2 θ) = 1
⇒ hr =
q
(r 2 cos2 θ(cos2 φ + sin2 φ) + r 2 sin2 θ) = r
hθ =
q
hφ =
(r 2 sin2 θ(sin2 φ + cos2 φ) = r sin θ
∂U
1 ∂U
1 ∂U
ˆ
er +
ˆ
eθ +
ˆ
eφ
∂r
r ∂θ
r sin θ ∂φ
1 ∂(r 2ar )
1 ∂(aθ sin θ)
1 ∂aφ
diva = 2
+
+
r
∂r
r sin θ
∂θ
 r sin θ ∂φ

∂
ˆ
eθ
∂
∂
∂
ˆ
er
(aφ sin θ) −
(aθ ) +
(ar ) − (aφr sin θ) +
curla =
r sin θ ∂θ
∂φ
r sin θ ∂φ
∂r


ˆ
eφ ∂
∂
(aθ r ) − (ar )
r ∂r
∂θ
2
∇ U = Tutorial Exercise

⇒ gradU =

♣ Examples

Q1 Find curla in (i) Cartesians and (ii) Spherical polars when a = x (xˆı + yˆ + z kˆ )
...

 x2
xy
xz 

82

LECTURE 6
...
So
a = r 2 sin θ cos φˆ
er
2
⇒ ar = r sin θ cos φ;

aθ = 0;

aφ = 0
...

Remember that we can work out the unit vectors ˆ
er and so on in terms of ˆı
etc using
ˆ
er =

1 ∂r
;
h1 d r

ˆ
eθ =

1 ∂r
;
h2 d θ

ˆ
eφ =

1 ∂r
h3 d φ

where r = xˆı +yˆ +z kˆ
...

So the result in spherical polars is

curla = (cos θ cos φˆı + cos θ sin φˆ − sin θkˆ )(−r sin φ) + (− sin φˆı + cos φˆ )(−r cos θ cos φ
= −r cos θˆ + r sin θ sin φkˆ
= −zˆ + y kˆ
which is exactly the result in Cartesians
...


6
...
GRAD DIV, CURL, ∇2 IN SPHERICAL POLARS

83

A2 (i) Using Cartesian coords:
∂ 2
(x + y 2 + z 2 )1/2cx +
...
(x 2 + y 2 + z 2 )−1/2cx +
...

r

diva =

(ii) Using Spherical polars
a = ar ˆ
er + aθ ˆ
eθ + aφˆ
eφ
and our first task is to find ar and so on
...
VECTOR OPERATOR IDENTITIES

For our particular problem, ax = r cx , etc, where cx is a constant, so now we
can write down
ar = r (sin θ cos φcx + sin θ sin φcy + cos θcz )
aθ = r (cos θ cos φcx + cos θ sin φcy − sin θcz )
aφ = r (− sin φcx + cos φcy )
Now all we need to do is to bash out
diva =

1 ∂(aθ sin θ)
1 ∂aφ
1 ∂(r 2 ar )
+
+
r 2 ∂r
r sin θ
∂θ
r sin θ ∂φ

In glorious detail this is
diva = 3 (sin θ cos φcx + sin θ sin φcy + cos θcz ) +


1
cos2 θ − sin2 θ)(cos φcx + sin φcy ) − 2 sin θ cos θcz +
sin θ
1
(− cos φcx − sin φcy )
sin θ
A bit more bashing and you’ll find
diva = sin θ cos φcx + sin θ sin φcy + cos θcz
= ˆ
er · c
This is EXACTLY what you worked out before of course
...

• Don’t forget about the vector geometry you did in the 1st year
...
Let the symmetry guide you
...
Two
theorems, both of them over two hundred years old, are explained:
• Gauss’ Theorem enables an integral taken over a volume to be replaced by
one taken over the surface bounding that volume, and vice versa
...


7
...
That is, we are interested in
calculating:
Z
a · dS
S

dS
dS
dS

dS

dS
Figure 7
...

85

86

LECTURE 7
...
1
...


7
...


If we sum over the volume elements, this results in a sum over the surface elements
...
2
...


Figure 7
...
One
can imagine building the entire volume up from the infinitesimal units
...
2
...

R
Q Derive S a · d S where a = z 3kˆ and S is the surface of a sphere of radius R
centred on the origin:
1
...
by applying Gauss’ Theorem

3
z k
2
R sin θ d θ d φ r
R

dz
z

R
Figure 7
...

Everywhere ˆr · kˆ = cos θ
...
R2 sin θd θd φˆ
er · kˆ
⇒
a · dS =
φ=0 θ=0
S
Z 2π Z π
=
R3 cos3 θ
...
cos θ
φ=0 θ=0
Z π
= 2πR5
cos4 θ sin θd θ
0

π 4πR5
2πR5 
5
=
− cos θ 0 =
5
5

(2) To apply Gauss’ Theorem, we need to figure out div a and decide how to
compute the volume integral
...
GAUSS’ AND STOKES’ THEOREMS

For the second, because diva involves just z , we can divide the sphere into
discs of constant z and thickness d z , as shown in Fig
...
3
...
3

div ad V = 3π
V

Z

R

z 2(R2 − z 2)d z

−R
2 3

z5
R z
−
= 3π
3
5
4πR5
=
5


R

−R

Surface versus volume integrals

At first sight, it might seem that with a computer performing surface integrals
might be better than a volume integral, perhaps because there are, somehow,
“fewer elements”
...
Imagine doing a surface integral
over a wrinkly surface, say that of the moon
...
With a volume integral,
most of the elements are not at the surface, and so the bulk of the integral is
done without accurate modelling
...


7
...
Then, as we showed in the previous lecture,
div a = gradU · c + Udiv c
= gradU · c
since divc = 0 because c is constant
...
4
...
This can be true only if the vector
equation
Z
Z
Ud S =
grad Ud V
S

V

is satisfied
...

Further “extensions” can be obtained of course
...

♣ Example of extension to Gauss’ Theorem
Q U = x 2 + y 2 + z 2 is a scalar field,
and volume V is the cylinder x 2 +
y 2 ≤ a2, 0 ≤ z ≤ h
...

A It is immediately clear from symmetry that there is no contribution
from the curved surface of the cylinder since for every vector surface element there exists an equal and opposite element with the same value
of U
...

Top face:

z=h
z=0

dS

U = x 2 + y 2 + z 2 = r 2 + h2 and d S = r d r d φkˆ
so
Z

a

1
1
Ud S =
(h2 + r 2 )2πr d r
d φkˆ = kˆ π h2 r 2 + r 4 = π[h2 a2 + a4]kˆ
2 0
2
r =0
φ=0
Z

a

Z

2π

90

LECTURE 7
...

On the other hand, using Gauss’ Theorem we have to compute
Z
grad Ud V

V

In this case, grad U = 2r,
Z
2 (xˆı + y kˆ + z kˆ )r d r d z d φ
V

The integrations over x and y are zero by symmetry, so that the only remaining
part is
Z h
Z a
Z 2π
2
zdz
r dr
d φkˆ = πa2 h2 kˆ
z=0

7
...


Stokes’ Theorem states:
I
Z
a · dl =
curl a · d S
C

S

where S is any surface capping the curve C
...
It just seems to be common usage
in line integrals!

7
...
INFORMAL PROOF

7
...

a · d l = d C = (∇
around elemental loop

Now if we add these little loops together, the internal line sections cancel out
because the d l’s are in opposite direction but the field a is not
...
7
...

y

ay (x)

dy

y

ay (x+dx)

ax (y+dy)

y+dy

dx
x+dx

x
ax (y)

Figure 7
...


For a given contour, the capping surface can be ANY surface bound by
the contour
...
See Fig
...
5
...
5: For a given contour, the bounding surface can be any shape
...


92

LECTURE 7
...

Q Vector field a = x 3ˆ − y 3ˆı and C is the circle of radius R centred on the origin
...

A(i) Directly
...


 ˆı
ˆ kˆ
 ∂
∂
∂
curl a =  ∂x ∂y
∂z
 −y 3 x 3 0

3π 4
R ,
2

3π
4




 = 3(x 2 + y 2 )kˆ = 3r 2 kˆ



We choose area elements to be circular strips of radius r thickness d r
...
7
...
7

93

An Extension to Stokes’ Theorem

Just as we considered one extension to Gauss’ theorem (not really an extension,
more of a re-expression), so we will try something similar with Stoke’s Theorem
...
Then
curl a = Ucurl c + grad U × c)
Again, curl c is zero
...

C

S

But c is arbitrary and so
I
Z
Ud l = − grad U × d S
C

7
...

Note that, for no special reason, we have used d r
here not d l
...

If the circle were centred at the origin, we would write d r = ad θˆ
eθ =
ad θ(− sin θˆı +cos θˆ )
...

However, in this example d r is not always in the direction of ˆ
eθ , and d r 6= 0
...
GAUSS’ AND STOKES’ THEOREMS

To avoid having to find an expression for r in terms of θ, we will perform a
coordinate transformation by writing r = [a, 0]⊤ + ρ
...

The line integral becomes
I
Z 2π
3
Ud r = 2a
(1 + cos α)(− sin αˆı + cos αˆ )d α = 2πa3ˆ
α=0

A(ii) Now using Stokes’
...

Be careful to note that x , y are specified for any point on the disc, not on its
circular boundary!
So


ˆ

ˆ
ˆ
ı

k


d S×gradU = 2ρ d ρ d α 
0
0
1  = 2ρ[−ρ sin αˆı +(a+ρ cos α)ˆ ] d ρ d α
 (a + ρ cos α) ρ sin α 0 
R 2π
R 2π
Both 0 sin αd α = 0 and 0 cos αd α = 0, so we are left with
Z a Z 2π
Z
d S × gradU =
2ρaˆ d ρ d α = 2πa3ˆ
S

Revised Oct 2013

ρ=0

α=0

Lecture 8
Engineering applications
In Lecture 6 we saw one classic example of the application of vector calculus to
Maxwell’s equation
...
As with Maxwell’s eqations, the examples show how vector calculus
provides a powerful way of representing underlying physics
...
Vector
calculus will, with practice, become a convenient shorthand for you
...
ENGINEERING APPLICATIONS

8
...

Now consider the H around a straight wire carrying current I
...
(You
might check this against Biot-Savart’s law
...


8
...
FLUID MECHANICS - THE CONTINUITY EQUATION

8
...
The continuity principle applied to any volume (called a control volume)
may be expressed in words as follows:
“The net rate of mass flow of fluid out of the control volume must equal
the rate of decrease of the mass of fluid within the control volume”

dS

q

Control Volume V
Figure 8
...
The element of rate-of-volume-loss through surface d S is d V˙ = q·d S,
so the rate of mass loss is
d M˙ = ρq · d S,
so that the total rate of mass loss from the volume is
Z
Z
∂
ρ(r)d V = ρq · d S
...

V ∂t
S

Now we use Gauss’ Theorem to transform the RHS into a volume integral
Z
Z
∂ρ
dV =
div (ρq)d V
...
This leads to the mathematical
formulation of

98

LECTURE 8
...

In this last case, we can say that the flow q is solenoidal
...
3

Thermodynamics - The Heat Conduction Equation

Flow of heat is very similar to flow of fluid, and heat flow satisfies a similar continuity equation
...

Assuming that there is no mass flow across the boundary of the control volume and
no source of heat inside it, the rate of flow of heat out of the control volume by
conduction must equal the rate of decrease of internal energy (constant volume)
or enthalpy (constant pressure) within it
...

In order to solve for the temperature field another equation is required, linking q
to the temperature gradient
...
Combining the two equations
gives the heat conduction equation:
−div q = κdiv grad T = κ∇2 T = ρc

∂T
∂t

where it has been assumed that κ is a constant
...


8
...
MECHANICS - CONSERVATIVE FIELDS OF FORCE

8
...
of path taken
...

But if true for any C containing A and B, it must be that
curl F = 0
Conservative fields are irrotational
All radial fields are irrotational
One way (actually the only way) of satisfying this condition is for
F=∇ U
The scalar field U(r) is the Potential Function

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LECTURE 8
...
5

The Inverse Square Law of force

Radial forces are found in electrostatics and gravitation — so they are certainly
irrotational and conservative
...
One reason why this
may be so is that it turns out to be the only central force field which is solenoidal,
i
...
has zero divergence
...

For div F = 0 we conclude
df
+ 3f = 0
r
dr
or
dr
df
+3
= 0
...

3
r
r

The condition of zero divergence of the inverse square force field applies everywhere
except at r = 0, where the divergence is infinite
...
This is
Z
Z
Z
F · dS = F
ˆr · d S = F
d = F 4πR2 = 4πA = Constant
...

But for all finite r , divF = 0, so divF must be infinite at the origin
...


8
...
GRAVITATIONAL FIELD DUE TO DISTRIBUTED MASS: POISSON’S EQUATION 101

8
...
4 for the gravitational field, A = Gm, where
m is the mass at the origin and G the universal gravitational constant, one would
run into the problem that there is no such thing as point mass
...

The mass contained in each small volume element d V is ρd V and this will make a
contribution −4πρGd V to the flux integral from the control volume
...

S

V

Transforming the left hand integral by Gauss’ Theorem gives
Z
Z
div Fd V = −4πG ρd V
V

V

which, since it is true for any V , implies that
−div F = 4πρG
...
e
...
It follows that the
gravitational potential U satisfies
Poisson’s Equation
∇2U = 4πρG
...
We
have
Z R
4πR2 F = 4πG
4πr 2 ρd r,
0

where F = |F|, or
Z R
G
MG
|F | = 2
4πr 2 ρd r = 2 ,
R 0
R
where M is the total mass inside radius R
...


102

8
...
ENGINEERING APPLICATIONS

Pressure forces in non-uniform flows

When a body is immersed in a flow it experiences a net pressure force
Z
Fp = − pd S,
S

where S is the surface of the body
...
The integral can be transformed using Gauss’ Theorem to give the
alternative expression
Z
Fp = − grad p d V,
V

where V is the volume of the body
...


z

V
Revised Oct 2013



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