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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
...
(By the way, flux here denotes mass
per unit time
...
5
...
(Divergence because it is an efflux not
an influx
...

(NB: The above does not constitute a rigorous proof of the assertion because we
have not proved that the quantity calculated is independent of the co-ordinate
system used, but it will suffice for our purposes
...
5

The Laplacian: div(gradU) of a scalar field

Recall that gradU of any scalar field U is a vector field
...
So we can certainly compute
div(gradU), even if we don’t know what it means yet
...

 
 

∂
∂
∂
∂
∂
∂
ˆı
∇U) = ˆı
+ ˆ
+ kˆ
·
+ ˆ
+ kˆ
U
∇ · (∇
∂x
∂y
∂z
∂x
∂y
∂z

 

∂
∂
∂
∂
∂
∂
ˆı
+ ˆ
+ kˆ
· ˆı
+ ˆ
+ kˆ
U
=
∂x
∂y
∂z
∂x
∂y
∂z
 2

∂
∂2
∂2
=
+
+
U
∂x 2 ∂y 2 ∂z 2
 2

∂ U ∂ 2U ∂ 2U
=
+
+
∂x 2
∂y 2
∂z 2
This last expression occurs frequently in engineering science (you will meet it next
in solving Laplace’s Equation in partial differential equations)
...
VECTOR OPERATORS: GRAD, DIV AND CURL

♣ Examples of ∇2 U evaluation
U

∇2 U

1) r 2 (= x 2 + y 2 + z 2 ) 6
2) x y 2z 3
2x z 3 + 6x y 2z
3) 1/r
0
Let’s prove example (3) (which is particularly significant – can you guess why?)
...
(x 2 + y 2 + z 2)−3/2
∂x ∂x
∂x
= −(x 2 + y 2 + z 2 )−3/2 + 3x
...
(x 2 + y 2 + z 2)−5/2
= (1/r 3)(−1 + 3x 2/r 2 )
Adding up similar terms for y and z


2
2
2
1
1
(x
+
y
+
x
)
∇2 = 3 −3 + 3
=0
r
r
r2

5
...

We are now overwhelmed by an irrestible temptation to
• cross it with a vector field ∇ × a
This gives the curl of a vector field
∇ × a ≡ curl(a)
We can follow the pseudo-determinant recipe for vector products, so that


 ˆı ˆ kˆ 
 ∂ ∂ ∂ 
 (remember it this way)
∇ × a =  ∂x
∂y ∂z 
a a a 
x
y
z






∂az ∂ay
∂ax ∂az
∂ay
∂ax ˆ
ˆı +
ˆ +
=
−
−
−
k
∂y
∂z
∂z
∂x
∂x
∂y

5
...
THE SIGNIFICANCE OF CURL

71

♣ Examples of curl evaluation
∇×a

a

1) −yˆı + xˆ
2kˆ
2) x 2y 2kˆ
2x 2yˆı − 2x y 2ˆ

5
...
The field a = −yˆı + xˆ is sketched in
Figure 5
...
(It is the field you would calculate as the velocity field of an object
rotating with ω = [0, 0, 1]
...
You can also see that a field
H like this must give a finite
value to the line integral around the complete loop C a · d r
...
3: (a) A rough sketch of the vector field −yˆı + xˆ
...


In fact curl is closely related to the line integral around a loop
...


72

LECTURE 5
...
3(b)
...

The fields in the x direction at the bottom and top are
ax (y )

and

ax (y + d y ) = ax (y ) +

∂ax
dy,
∂y

where ax (y ) denotes ax is a function of y , and the fields in the y direction at the
left and right are
ay (x )

and

ay (x + d x ) = ay (x ) +

∂ay
dx
∂x

Starting at the bottom and working round in the anticlockwise sense, the four
contributions to the circulation d C are therefore as follows, where the minus signs
take account of the path being opposed to the field:
d C = + [ax (y ) d x ] + [ay (x + d x ) d y ] − [ax (y + d y ) d x ] − [ay (x ) d y ]

  
 
∂ay
∂ax
= + [ax (y ) d x ] + ay (x ) +
d x d y − ax (y ) +
d y d x − [ay (x ) d y ]
∂x
∂y


∂ax
∂ay
−
dx dy
=
∂x
∂y
∇ × a) · d S
= (∇
where d S = d x d y kˆ
...


5
...

• A vector field with zero curl is said to be irrotational
...


Revised Oct 2013

Lecture 6
Vector Operator Identities
In this lecture we look at more complicated identities involving vector operators
...

There could be a cottage industry inventing vector identities
...
So why not leave it at that?
First, since grad, div and curl describe key aspects of vectors fields, they arise often
in practice, and so the identities can save you a lot of time and hacking of partial
derivatives, as we will see when we consider Maxwell’s equation as an example
later
...
So,
although this material is a bit dry, the relevance of the identities should become
clear later in other Engineering courses
...
1

Identity 1: curl grad U = 0


ˆ


ˆ
ˆ
ı

k


∇ × ∇U =  ∂/∂x ∂/∂y
∂/∂z 
 ∂U/∂x ∂U/∂y ∂U/∂z 
 2

∂ U
∂ 2U
= ˆı
−
+ ˆ () + kˆ ()
∂y ∂z ∂z ∂y
= 0 ,

as ∂ 2/∂y ∂z = ∂ 2/∂z ∂y
...

73

74

6
...
VECTOR OPERATOR IDENTITIES

Identity 2: div curl a = 0


 ∂/∂x ∂/∂y ∂/∂z 


∇ · ∇ × a =  ∂/∂x ∂/∂y ∂/∂z 
 ax
ay
az 
∂ 2 az
∂ 2 ay
∂ 2 az
∂ 2 ax
∂ 2 ay
∂ 2 ax
=
−
−
+
+
−
∂x ∂y ∂x ∂z ∂y ∂x ∂y ∂z ∂z ∂x ∂z ∂y
= 0

6
...
This is a vector field, so we can compute its divergence
and curl
...

The divergence (a scalar) of the product Ua is given by:
∇ · (Ua) = U(∇
∇ · a) + (∇
∇U) · a
= Udiva + (gradU) · a
In a similar way, we can take the curl of the vector field Ua, and the result should
be a vector field:
∇ × (Ua) = U∇
∇ × a + (∇
∇U) × a
...
4

Identity 4: div of a × b

Life quickly gets trickier when vector or scalar products ar

---