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Part IA — Vector Calculus
Based on lectures by B
...
They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine
...

[1]
Integration in R2 and R3
Line integrals
...

[4]
Vector operators
Directional derivatives
...

Divergence, curl and ∇2 in Cartesian coordinates, examples; formulae for these operators (statement only) in cylindrical, spherical *and general orthogonal curvilinear*
coordinates
...

Vector derivative identities
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[5]
Laplace’s equation
Laplace’s equation in R2 and R3 : uniqueness theorem and maximum principle
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[4]
Cartesian tensors in R3
Tensor transformation laws, addition, multiplication, contraction, with emphasis on
tensors of second rank
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Symmetric and
antisymmetric tensors
...
Quotient
theorem
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[5]

1

Contents

IA Vector Calculus

Contents
0 Introduction

4

1 Derivatives and coordinates
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5
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2 Curves and Line
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2 Line integrals of vector fields
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3 Gradients and Differentials
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4 Work and potential energy
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3 Integration in R2 and R3
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3 Generalization to R3
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4 Further generalizations
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4 Surfaces and surface integrals
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1 Div, Grad, Curl and ∇
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2 Second-order derivatives
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1 Statement and examples
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7
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2 Stokes’ theorem
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8 Some applications of integral theorems
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49
9 Orthogonal curvilinear coordinates
51
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52

2

Contents

IA Vector Calculus

10 Gauss’ Law and Poisson’s equation
54
10
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54
10
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55
10
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57
11 Laplace’s and Poisson’s equations
11
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11
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11
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1 The mean value property
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2
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3 Integral solutions of Poisson’s equations
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3
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11
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2 Point sources and δ-functions*
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61
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12 Maxwell’s equations
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69
13 Tensors and tensor fields
13
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5 Tensor calculus
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70
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14 Tensors of rank 2
77
14
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14
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15 Invariant and isotropic tensors
81
15
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15
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82

3

0

0

Introduction

IA Vector Calculus

Introduction

In the differential equations class, we learnt how to do calculus in one dimension
...
We live in a
3 (or 4) dimensional world, and string theorists think that the world has more
than 10 dimensions
...

For example, the position of a particle in a three dimensional world can be
d
˙
given by a position vector x
...

This would require us to take the derivative of a vector
...
We can just differentiate the vector componentwise
...
We can
assign a number to each point in (3D) space, and ask how this number changes
as we move in space
...

In the most general case, we will assign a vector to each point in space
...

On the other side of the story, we also want to do integration in multiple
dimensions
...
For example, we can let v(x) be the velocity of some
fluid at each point in space
...

In this course, we are mostly going to learn about doing calculus in many
dimensions
...

Note that throughout the course (and lecture notes), summation convention
is implied unless otherwise stated
...
1

Derivative of functions

We used to define a derivative as the limit of a quotient and a function is differentiable if the derivative exists
...
So we want
an alternative definition of differentiation, which can be easily generalized to
vectors
...
It can be easily shown that the converse
is true — if f satisfies this relation, then f is differentiable
...
We say a function
F is differentiable if, when x is perturbed by δx, then the resulting change is
“something” times δx plus an o(δx) error term
...
Then that “something” will
be what we call the derivative
...

Definition (Vector function)
...

This takes in a number and returns a vector
...

Definition (Derivative of vector function)
...
F0 (x) is called the derivative of F(x)
...

Using differential notation, the differentiability condition can be written as
dF = F0 (x) dx
...
e
...

F0 (x) = Fi0 (x)ei
...

5

1

Derivatives and coordinates

IA Vector Calculus

Proposition
...

Example
...
It has position r(t), velocity r˙ (t)
and acceleration ¨r
...

Note that derivatives with respect to t are usually denoted by dots instead
of dashes
...

We can define the angular momentum about the origin to be
L = r × p = mr × r˙
...

which is the torque of F about the origin
...
A scalar function is a function f : Rn → R
...
g
...

Before we define the derivative of a scalar function, we have to first define
what it means to take a limit of a vector
...
The limit of vectors is defined using the norm
...
Similarly, f (r) = o(r) means |f|r|
→ 0 as r → 0
...
A scalar function f (r) is differentiable
at r if
def
δf = f (r + δr) − f (r) = (∇f ) · δr + o(δr)
for some vector ∇f , the gradient of f at r
...
But we will soon
give up on finding fancy names and just call everything the “derivative”!
Note also that here we genuinely need the new notion of derivative, since
“dividing by δr” makes no sense at all!
6

1

Derivatives and coordinates

IA Vector Calculus

The above definition considers the case where δr comes in all directions
...

Then taking δr = hn, with n a unit vector,
f (r + hn) − f (r) = ∇f · (hn) + o(h) = h(∇f · n) + o(h),
which gives
Definition (Directional derivative)
...

Using this expression, the directional derivative is maximized when n is in
the same direction as ∇f (then n · ∇f = |∇f |)
...

How do we evaluate ∇f ? Suppose we have an orthonormal basis ei
...

h→0 h
∂xi

ei · ∇f = lim
Hence
Theorem
...

∂xi

In differential notation, we write
df = ∇f · dr =

∂f
dxi ,
∂xi

which is the chain rule for partial derivatives
...
Take f (x, y, z) = x + exy sin z
...
So f increases/decreases
most rapidly for
√
n = ± √12 (1, 0, 1) with a rate of change of ± 2
...


7

1

Derivatives and coordinates

IA Vector Calculus

Now suppose we have a scalar function f (r) and we want to consider the rate
of change along a path r(u)
...

This shows that f is differentiable as a function of u and
Theorem (Chain rule)
...

du
du
∂xi du
Note that if we drop the du, we simply get
df = ∇f · dr =

∂f
dxi ,
∂xi

which is what we’ve previously had
...

Definition (Vector field)
...

Definition (Derivative of vector field)
...
M is the derivative of F
...

Given an arbitrary function F : Rn → Rm that maps x 7→ y and a choice
of basis, we can write F as a set of m functions yj = Fj (x) such that y =
(y1 , y2 , · · · , ym )
...

∂xi
and we can write the derivative as
Theorem
...

∂xi

Note that we could have used this as the definition of the derivative
...

Definition
...

This requires that all partial derivatives exist and are totally symmetric in i, j
and k (i
...
the differential operator is commutative)
...
g
...

8

1

Derivatives and coordinates

IA Vector Calculus

Theorem (Chain rule)
...
Suppose that
the coordinates of the vectors in Rp , Rn and Rm are ua , xi and yr respectively
...
Writing in matrix form,
M (f ◦ g)ra = M (f )ri M (g)ia
...

∂ua
∂ua ∂xi

1
...
e
...
Suppose
that f (x) = u and g(u) = x
...

Therefore we know that
M (g) = M (f )−1
...

∂ua
So by the chain rule,
∂ub ∂xi
= δab ,
∂xi ∂ua
i
...
M (f ◦ g) = I
...

Example
...
Then the function used to convert between the coordinate systems is
g(u1 , u2 ) = (u1 cos u2 , u1 sin u2 )
Then

 

∂x1 /∂ρ ∂x1 /∂ϕ
cos ϕ −ρ sin ϕ
M (g) =
=
∂x2 /∂ρ ∂x2 /∂ϕ
sin ϕ ρ cos ϕ
We can invert the relations between (x1 , x2 ) and (ρ, ϕ) to obtain
x2
ϕ = tan−1
x1
q
2
ρ = x1 + x22
We can calculate

M (f ) =

∂ρ/∂x1
∂ϕ/∂x1

∂ρ/∂x2
∂ϕ/∂x2



= M (g)−1
...

9

1

Derivatives and coordinates

IA Vector Calculus

Note that
det M (f ) det M (g) = 1
...
3

Coordinate systems

Now we can apply the results above the changes of coordinates on Euclidean
space
...
Then we can
define an arbitrary new coordinate system ua in which each coordinate ua is a
function of x
...


However, note that ρ and ϕ are not components of a position vector, i
...
they
are not the “coefficients” of basis vectors like r = x1 e1 + x2 e2 are
...


e2
eϕ

eρ

ρ
ϕ

e1

These are not “usual” basis vectors in the sense that these basis vectors vary
with position and are undefined at the origin
...

In three dimensions, we have cylindrical polars and spherical polars
...
1

Parametrised curves, lengths and arc length

There are many ways we can described a curve
...
For example, a circle can be
described by x2 + y 2 = 1
...
It is also often difficult to find a closed form like
this for a curve
...
So it is represented by a function x : R → Rn , and the curve itself is
the image of the function
...
In
addition to simplified notation, this also has the benefit of giving the curve an
orientation
...
Given a curve C in Rn , a parametrisation
of it is a continuous and invertible function r : D → Rn for some D ⊆ R whose
image is C
...
A parametrization is
regular if r0 (u) 6= 0 for all u
...

Example
...


ˆ
can be parametrised by 2 cos uˆi + sin uˆj + 3k
If we change u (and hence r) by a small amount, then the distance |δr| is
roughly equal to the change in arclength δs
...
Then we have
Proposition
...
Then
 
 dr 
ds
= ±   = ±|r0 (u)|
du
du
with the sign depending on whether it is in the direction of increasing or decreasing
arclength
...
Consider a helix described by r(u) = (3 cos u, 3 sin u, 4u)
...
i
...
the arclength from r(0) and r(u) is s = 5u
...
Then by the chain rule,
dr
dr
d˜
u
=
×
du
d˜
u du
dr
dr d˜
u
=
/
d˜
u
du du
11

2

Curves and Line

IA Vector Calculus

It is often convenient to use the arclength s as the parameter
...

ds

We call ds the scalar line element, which will be used when we consider integrals
...
The scalar line element of C is ds
...
ds = ±|r0 (u)|du

2
...
The line integral of a smooth vector field F(r) along
a path C parametrised by r(u) along the direction (orientation) r(α) → r(β) is
Z

Z

β

F(r(u)) · r0 (u) du
...
Note that the upper and lower
limits of the integral are the end point and start point respectively, and β is not
necessarily larger than α
...
Then we may divide the curve into many small segments
δr
...
Then the total work done across the curve is
Z
W =
F(r) · dr
...
Take F(r) = (xey , z 2 , xy) and we want to find the line integral from
a = (0, 0, 0) to b = (1, 1, 1)
...
Then r0 (u) =
2
(1, 2u, 3u2 ), and F(r(u)) = (ueu , u6 , u3 )
...
So r0 (t) =
(1, 1, 1)
...

3

We see that the line integral depends on the curve C in general, not just a, b
...
Since dr = t ds, with t
being the unit tangent vector, we have
Z
Z
F · dr =
F · t ds
...
We can
also integrate a scalar function as a function of s, C f (s) ds
...
In particular, we have
Z
1 ds = length of C
...
A closed curve is a curve with the same start and
H
end point
...

Sometimes we are not that lucky and our curve is not smooth
...
However, often we can
break it apart into many smaller segments, each of which is smooth
...
We call these piecewise smooth
curves
...
A piecewise smooth curve is a curve
C = C1 + C2 + · · · + Cn with all Ci smooth with regular parametrisations
...

C

C1

C2

Cn

Example
...
Then C = C1 + C3 is
piecewise smooth but not smooth
...

12 2

13

2

Curves and Line

IA Vector Calculus

b
C3

C1

a

2
...
However, for some nice functions, the integral does depend on the
end points only
...
If F = ∇f (r), then
Z
F · dr = f (b) − f (a),
C

where b and a are the end points of the curve
...
This is the vector counterpart of the fundamental
theorem of
H
calculus
...

Proof
...
Then
Z
Z
Z
dr
F · dr =
∇f · dr = ∇f ·
du
...

du

Definition (Conservative vector field)
...

The name conservative comes from mechanics, where conservative vector
fields represent conservative forces that conserve energy
...
e
...

It is convenient to treat differentials F · dr = Fi dxi as if they were objects
by themselves, which we can integrate along curves if we feel like doing so
...
A differential F · dr is exact if there is an f
such that F = ∇f
...

∂xi

To test if this holds, we can use the necessary condition

14

2

Curves and Line

IA Vector Calculus

Proposition
...

∂xj
∂xi
This is because both are equal to ∂ 2 f /∂xi ∂xj
...

C

C

Differentials can be manipulated using (for constant λ, µ):
Proposition
...

Example
...


C

We see that if we integrate the first term with respect to x, we obtain x3 y sin z
...
So this is
equal to
Z
d(x3 y sin z) = [x3 y sin z]b
a
...
4

Work and potential energy

R
Definition (Work and potential energy)
...
It is the limit of a sum of terms
F(r) · δr, i
...
the force along the direction of δr
...

Since the kinetic energy is defined as
T (t) =

1 2
m˙r ,
2

the rate of change of energy is
d
T (t) = m˙r · ¨r = F · r˙
...

C

So the work done on the particle is the change in kinetic energy
...
Given a conservative force F = −∇V , V (x) is
the potential energy
...

C

Therefore, for a conservative force, we have F = ∇V , where V (r) is the
potential energy
...
So
the total energy T + V is conserved, i
...
constant during motion
...
In fact, the converse
is true — the energy is conserved only for conservative forces
...
1

IA Vector Calculus

Integration in R2 and R3
Integrals over subsets of R2

Definition (Surface integral)
...
Let r = (x, y) be in Cartesian
coordinates
...
g
...
These shapes are labelled by I and have areas δAi
...
But we need a condition stronger than simply
δAi → 0
...
So we say that we find an ` such that each area can
be contained in a disc of diameter `
...
For a function f (r), we define the surface integral as
Z
X
f (r) dA = lim
f (ri )δAi
...
The integral exists if the limit
is well-defined (i
...
the same regardless of what Ai and ri we choose before we
take the limit) and exists
...

On the other hand, if we put z = f (x, y) and plot out the surface z = f (x, y),
then the area integral is the volume under the surface
...

However, the sensible thing is clearly to take Ai to be rectangles
...
We sum over subsets in a narrow horizontal strip of height δy
with
R y and δy held constant
...
We get a contribution
δy xy f (y, x) dx with range xy ∈ {x : (x, y) ∈ D}
...

Z

Z

!

Z

f (x, y) dx dy
...

Note that the range of the inner integral is given by a set xy
...
In this case,
Z
Z b1
Z b2
f (x) dx =
f (x) dx +
f (x) dx
...

y

x
We could also do it the other way round, integrating over y first, and come up
with the result

Z
Z Z
f (x, y) dA =
f (x, y) dy dx
...
If f is a continuous function and D is a compact
(i
...
closed and bounded) subset of R2 , then
ZZ
ZZ
f dx dy =
f dy dx
...

Definition (Area element)
...

Proposition
...

Example
...

We want to integrate the function f (x, y) = x2 y over the area
...

While this integral is tedious in general, there is a special case where it is
substantially easier
...
A function f (x, y) is separable if it can be
written as f (x, y) = h(y)g(x)
...
Take separable f (x, y) = h(y)g(x) and D be a rectangle {(x, y) :
a ≤ x ≤ b, c ≤ y ≤ d}
...
2

a

c

Change of variables for an integral in R2

Proposition
...
Then
Z
Z
f (x, y) dx dy =
f (x(u, v), y(u, v))|J| du dv,
D

D0

19

3

Integration in R2 and R3

IA Vector Calculus

where

 ∂x

∂(x, y)  ∂u
J=
=
∂(u, v)  ∂y

∂u
is the Jacobian
...

Proof
...

Suppose we start with an area δA0 = δuδv in the (u, v) plane
...

∂u
∂v

We have a similar expression for δy and we obtain
   ∂x ∂x   
δu
δx
∂v
≈ ∂u
∂y
∂y
δv
δy
∂u
∂v
Recall from Vectors and Matrices that the determinant of the matrix is how
much it scales up an area
...
Hence
dx dy = |J| du dv
...
We transform from (x, y) to (ρ, ϕ) with
x = ρ cos ϕ
y = ρ sin ϕ
We have previously calculated that |J| = ρ
...

Suppose we want to integrate a function over a quarter area D of radius R
...
Then
Z
Z
f dA = f ρ dρ dϕ
!
Z
Z
R

π/2

2

e−ρ

=
ρ=0

ϕ=0

20

/2

ρ dϕ δρ

3

Integration in R2 and R3

IA Vector Calculus

Note that in polar coordinates, we are integrating over a rectangle and the
function is separable
...

2
Note that the integral exists as R → ∞
...

Z
Z ∞ Z ∞
2
2
e−(x +y )/2 dx dy
f dA =
D
x=0 y=0
Z ∞
 Z ∞

−x2 /2
−y 2 /2
e
e
=
dx
dy
0

0

π
=
2
where the last line is from (*)
...

e
dx =
2
0

3
...
e
...
Consider a volume V ⊆ R3 with position vector
r = (x, y, z)
...
g
...

Assume that as ` → 0 and N → ∞, the union of the small subsets tend to
V
...

To evaluate this, we can take δVI = δxδyδz, and take δx → 0, δy → 0 and
δz in some order
...

V

D

Zxy

So we integrate f (x, y, z) over z at each point (x, y), then take the integral of
that over the area containing all required (x, y)
...

V

z

DZ

Again, if we take f = 1, then we obtain the volume of V
...
For
example, we might have mass density, charge density, or probability density
...
Then
R
ρ(r)
dV
is the total amount of quantity in V
...
The volume element is dV
...
dV = dx dy dz
...
Then
Proposition
...
In cylindrical coordinates,


∂x 
∂w 
∂y 
∂w 
∂z 

∂w

∂x
∂v
∂y
∂v
∂z
∂v

dV = ρ dρ dϕ dz
...

Proof
...

Example
...
Then
Z
Z a Z π Z 2π
f dV =
f (r)r2 sin θ dr dθ dϕ
V

r=0
a

θ=0 ϕ=0
Z π Z 2π

Z
=

dr
0

Z

dθ
0

a

dϕ r2 f (r) sin θ

0

h
iπ h i2π
r2 f (r)dr − cos θ
ϕ

=

0

0

Z
= 4π

0

a

f (r)r2 dr
...

Note that in the second line, we rewrote the integrals to write the differentials
next to the integral sign
...
in the limits of the integrals
...
We understand it as the sum of spherical
shells of thickness δr and volume 4πr2 δr
...

Example
...
The region is defined as
x2 + y 2 + z 2 ≤ a2
x 2 + y 2 ≥ b2
...
The second criteria gives
b ≤ ρ ≤ a
...

So the volume is
Z

Z

a

dV =
V

Z √a2 −ρ2

2π

Z
dρ

dϕ

b

√

−

0

dz ρ
a2 −ρ2

a

Z

p

a2 − ρ2 dρ

a
2
= 2π (a2 − ρ2 )3/2
3
b
4
2
2 3/2
= π(a − b )
...
Suppose the density of electric charge is ρ(r) = ρ0 az in a hemisphere
H of radius a, with z ≥ 0
...
So
r ≤ a,

0 ≤ ϕ ≤ 2π,

We have
ρ(r) =

0≤θ≤

π

...

a

The total charge Q in H is
Z

Z

Z

π/2

Z

2π

ρ0
r cos θr2 sin θ
a
0
0
0
Z
Z π/2
Z 2π
ρ0 a 3
=
r dr
sin θ cos θ dθ
dϕ
a 0
0
0
 a 
π/2
ρ0 r 4
1
2
=
sin θ
[ϕ]2π
0
a 4 0 2
0

ρ dV =
H

a

=

dr

dθ

ρ0 πa3

...
4

IA Vector Calculus

Further generalizations

Integration in Rn
R
Similar to the above, D f (x1 , x2 , · · · xn ) dx1 dx2 · · · dxn is simply the integration over an n-dimensional volume
...

Z
Z
f (x1 , x2 , · · · xn ) dx1 dx2 · · · dxn =

f ({xi (u)})|J| du1 du2 · · · dun
...
However, we use the following formula for
change of variables:
 
Z
Z
 dx 
f (x) dx =
f (x(u))   du
...
If D = [a, b] with a < b, we write a
...

To show that the modulus is the right thing to do, we check case by case: If
dx
a < b and α < β, then du
is positive, and we have, as expected
Z

b

Z

β

f (x) dx =

f (u)

a

If α > β, then

dx
du

Z

α

dx
du
...
So

b

Z

β

f (x) dx =
a

f (u)
α

dx
du = −
du

Z

α

f (u)
β

dx
du
...

This is not easily generalized to higher dimensions, so we don’t employ the
same trick in other cases
...
In practice, we integrate them componentwise
...


V

V

For example, if a mass has density ρ(r), then its mass is
Z
M=
ρ(r) dV
V

24

3

Integration in R2 and R3

IA Vector Calculus

and its center of mass is
R=

1
M

Z
rρ(r) dV
...
Consider a solid hemisphere H with r ≤ a, z ≥ 0 with uniform
density ρ
...

3
H
Now suppose that R = (X, Y, Z)
...
We can
find this formally by
Z
1
X=
xρ dV
M H
Z a Z π/2 Z 2π
ρ
xr2 sin θ dϕ dθ dr
=
M 0 0
0
Z a
Z π/2
Z 2π
ρ
2
3
=
r dr ×
sin θ dθ ×
cos ϕ dϕ
M 0
0
0
=0
as expected
...
Similarly, we obtain Y = 0
...

ρ
Z=
M

Z

a
3

Z

π/2

r dr
0

Z
sin θ cos θ dθ

0

dϕ
0

 
π/2
1
r a4
sin2 θ
2π
M 4
2
0
3a
=

...


25

2π

4

Surfaces and surface integrals

4

IA Vector Calculus

Surfaces and surface integrals

4
...

What we would like to do now is to study surfaces in R3
...
One way to specify a surface is to use an
equation
...
Then
f (r) = c defines a smooth surface (e
...
x2 + y 2 + z 2 = 1 denotes the unit sphere)
...
Then by the chain rule, if we differentiate
f (r) = c with respect to u, we obtain
d
dr
[f (r(u))] = ∇f ·
= 0
...
Since du
is the tangent to
the curve, ∇f is perpendicular to the tangent
...
Therefore

Proposition
...

Example
...
Then ∇f = 2(x, y, z) =
2r, which is clearly normal to the sphere
...
Then ∇f =
2(x, y, −z)
...
Here ∇f = 0, and we cannot find a meaningful direction of normal
...
A surface S can be defined to have a boundary ∂S
consisting of a piecewise smooth curve
...

A surface is bounded if it can be contained in a solid sphere, unbounded
otherwise
...
g
...

Example
...

Definition (Orientable surface)
...

If we can find a consistent choice of n that varies smoothly across S, then
we say S is orientable, and the choice of sign of n is called the orientation of the
surface
...
For example, for a sphere, we can
declare that the normal should always point outwards
...

For simple cases, we can describe the orientation as “inward” and “outward”
...
2

Parametrized surfaces and area

However, specifying a surface by an equation f (r) = c is often not too helpful
...
We write r(u, v) for the point labelled by (u, v)
...
Let S be part of a sphere of radius a with 0 ≤ θ ≤ α
...

We restrict the values of θ, ϕ by 0 ≤ θ ≤ α, 0 ≤ ϕ ≤ 2π, so that each point is
only covered once
...
This corresponds to a
region D of allowed values of u and v
...

When we have such a parametrization r, we would want to make sure this
indeed gives us a two-dimensional surface
...


The idea is that r has to depend on both u and v, and in “different ways”
...
By the chain rule, this is given by
δr =

∂r
∂r
δu +
δv + o(δu, δv)
...
What we want is for them to point in different directions
...
A parametrization is regular if for all
u, v,
∂r
∂r
×
6= 0,
∂u ∂v
i
...
there are always two independent tangent directions
...

Given a surface, how could we, say, find its area? We can use our parametrization
...
If we
want to find the area of D itself, we would simply integrate
Z
du dv
...
Instead,
suppose we produce a small rectangle in D by changing u and v by small δu, δv
...
In the surface S, these small
∂r
∂r
changes δu, δv correspond to changes ∂u
δu and ∂v
δv, and these span a vector
area of
∂r
∂r
δS =
×
δuδv = n δS
...

The actual area is then given by


 ∂r
∂r 
δS = 
×
δu δv
...
The vector area element is
dS =

∂r
∂r
×
du dv
...

∂u ∂v 
By summing and taking limits, the area of S is

Z
Z 
 ∂r
∂r 

dS =
×
du dv
...
Consider again the part of the sphere of radius a with 0 ≤ θ ≤ α
...

So we find

∂r
= aeθ
...

∂ϕ
Then

∂r
∂r
×
= a2 sin θ er
...

Our bounds are 0 ≤ θ ≤ α, 0 ≤ ϕ ≤ 2π
...

0

4
...
Suppose we have a surface
S parametrized by r(u, v), where (u, v) takes values in D
...

We might attempt to use the integral
Z
|F| dS
...
For example, if all the flow is tangential to the
surface, then nothing is really passing through the surface, but |F| is non-zero,
so we get a non-zero integral
...
e
...

Definition (Surface integral)
...

∂u ∂v
S
S
D
Intuitively, this is the total amount of F passing through S
...

R
For a given orientation, the integral F·dS is independent of the parametrization
...

Example
...
Then
∂r
= aeθ ,
∂θ

∂r
= a sin θeϕ
...

Suppose we want to calculate the fluid flux through the surface
...
Assume that
u depends smoothly on r (and t)
...

u δt
n
δS
So the amount of flow of u over at time δt through S is
Z
δt u · dS
...

S
For example, let u = (−x, 0, z) and S be the section of a sphere of radius a
with 0 ≤ ϕ ≤ 2π and 0 ≤ θ ≤ α
...

a

n·u=
Therefore
Z

Z

α

Z

u · dS =
S

r
1
= (x, y, z)
...

What happens when we change parametrization? Let r(u, v) and r(˜
u, v˜) be
two regular parametrizations for the surface
...

Since

d˜
u d˜
v=

IA Vector Calculus

∂(˜
u, v˜)
du dv,
∂(u, v)

We recover the formula




 ∂r
 ∂r
∂r 
∂r 


du dv = 
d˜
u d˜
v
...

∂u ∂v
∂u
˜ ∂˜
v

provided (u, v) and (˜
u, v˜) have the same orientation
...
4

Change of variables in R2 and R3 revisited

In this section, we derive our change of variable formulae in a slightly different
way
...

Consider a subset S of the plane R2 parametrized by r(x(u, v), y(u, v))
...
Then
∂r
∂r
×
= (0, 0, J),
∂u ∂v
with J being the Jacobian
...

Change of variable formula in R3
In R3 , suppose we have a volume parametrised by r(u, v, w)
...

∂u
∂v
∂w

Then the cuboid δu, δv, δw in u, v, w space is mapped to a parallelopiped of
volume



 ∂r

∂r
∂r

δV =  δu ·
δv ×
δw  = |J| δu δv δw
...


31

5

5

Geometry of curves and surfaces

IA Vector Calculus

Geometry of curves and surfaces

dr
Let r(s) be a curve parametrized by arclength s
...
Differentiating yields t · t = 0
...

We define the following:

Definition (Principal normal and curvature)
...
Then n(s) is called the principal normal and κ(s) is called
the curvature
...

We take a curve that can Taylor expanded around s = 0
...

2
We know that r0 = t and r00 = t0
...

2
How can we interpret κ as the curvature? Suppose we want to approximate the
curve near r(0) by a circle
...
So κ should be inversely proportional
to the radius of the circle
...

Consider the vector equation for a circle passing through r(0) with radius a
in the plane defined by t and n
...

We can expand this to obtain
1
r = r(0) + aθt + θ2 an + o(θ3 )
...

2a

As promised, κ = 1/a, for a the radius of the circle of best fit
...
The radius of curvature of a curve at a point
r(s) is 1/κ(s)
...

We can add a third normal to generate an orthonormal basis
...
The binormal of a curve is b = t × n
...
1
Definition (Torsion)
...
Then τ is the torsion
...
(b0 = t0 × n + t × n0 = t × n0 , so b0
is perpendicular to t; and b · b = 1 ⇒ b · b0 = 0
...

The geometry of the curve is encoded in how this basis (t, n, b) changes along
it
...

Surfaces and intrinsic geometry*
We can study the geometry of surfaces through curves which lie on them
...
The
intersection of the plane with the surface yields a curve on the surface through
P
...

If we choose different planes containing n, we end up with different curves of
different curvature
...
The principal curvatures of a surface at P are
the minimum and maximum possible curvature of a curve through P , denoted
κmin and κmax respectively
...
The Gaussian curvature of a surface at a
point P is K = κmin κmax
...
K is intrinsic to the surface S
...
which are measured entirely on the
surface
...

The is the start of intrinsic geometry: if we embed a surface in Euclidean
space, we can determine lengths, angles etc on it
...

For example, we can consider a geodesic triangle D on a surface S
...

Let θi be the interior angles of the triangle (defined by using scalar products
of tangent vectors)
...


33

5

Geometry of curves and surfaces

IA Vector Calculus

Theorem (Gauss-Bonnet theorem)
...


34

6

Div, Grad, Curl and ∇

IA Vector Calculus

Div, Grad, Curl and ∇

6
6
...


We can regard this as obtained from

∂
∂xi
for cartesian coordinates xi and orthonormal basis ei , where ei are orthonormal
and right-handed, i
...
ei × ej = εijk ek (it is left handed if ei × ej = −εijk ek )
...

∂x ∂y ∂z
∇ = ei

∇ (nabla or del ) is both an operator and a vector
...

Definition (Divergence)
...

∂xi
∂x1
∂x2
∂x3

Definition (Curl)
...
Let F = (xez , y 2 sin x, xyz)
...

∂x
∂y
∂z

and


∂
∂ 2
∇ × F = ˆi
(xyz) −
(y sin x)
∂y
∂z


∂
∂
+ ˆj
(xez ) +
(xyz)
∂z
∂x


ˆ ∂ (y 2 sin x) − ∂ (xez )
+k
∂x
∂y
= (xz, xez − yz, y 2 cos x)
...
For example,
F · ∇ = Fi

∂
∂xi

is a scalar differential operator, and
F × ∇ = ek εijk Fi
is a vector differential operator
...
Let f, g be scalar functions, F, G be vector functions, and µ, λ
be constants
...

Note that Grad and Div can be analogously defined in any dimension n, but
curl is specific to n = 3 because it uses the vector product
...
Consider rα with r = |r|
...
So r2 = xi xi
...

∂xi
r
So
∂r
∂ α
(r ) = ei αrα−1
= αrα−2 r
...

∇·r=
∂xi
and
∂xj
∇ × r = ek εijk
= 0
...
We have the following Leibnitz properties:
∇(f g) = (∇f )g + f (∇g)
∇ · (f F) = (∇f ) · F + f (∇ · F)
∇ × (f F) = (∇f ) × F + f (∇ × F)
∇(F · G) = F × (∇ × G) + G × (∇ × F) + (F · ∇)G + (G · ∇)F
∇ × (F × G) = F(∇ · G) − G(∇ · F) + (G · ∇)F − (F · ∇)G
∇ · (F × G) = (∇ × F) · G − F · (∇ × G)
which can be proven by brute-forcing with suffix notation and summation
convention
...

They can be derived when needed via suffix notation
...

∇ · (rα r) = (∇rα )r + rα ∇ · r
= (αrα−2 r) · r + rα (3)
= (α + 3)rα
∇ × (rα r) = (∇(rα )) × r + rα (∇ × r)
= αrα−2 r × r
=0
36

6

Div, Grad, Curl and ∇

6
...

∇ × (∇f ) = 0
∇ · (∇ × F) = 0
Proof
...

∂xi ∂xj

since if, say, k = 3, then
εijk

∂2f
∂2f
∂2f
=
−
= 0
...
If F is defined in all of R3 , then
∇ × F = 0 ⇒ F = ∇f
for some f
...
If F = ∇f ,
then f is the scalar potential
...

Similarly,
Proposition
...

Definition (Solenoidal field and vector potential)
...

Not that is is true only if F or H is defined on all of R3
...
The Laplacian operator is defined by
 2

∂2
∂
∂2
∂2
+
+

...


37

7

Integral theorems

7
7
...
There are all generalizations of the fundamental theorem of
calculus in some sense
...

We will first state all three theorems with some simple applications
...

7
...
1

Green’s theorem (in the plane)

Theorem (Green’s theorem)
...

∂x
∂y
C
A
Here C is assumed to be piecewise smooth, non-intersecting closed curve, traversed anti-clockwise
...
Let Q = xy 2 and P = x2 y
...

A

C

From example sheet 1, each side gives

104 4
105 a
...
Let A be a rectangle confined by 0 ≤ x ≤ a and 0 ≤ y ≤ b
...
We first consider the first term of Green’s theorem:
Z
Z aZ b
∂P
∂P
−
dA =
−
dy dx
∂y
∂y
0
0
Z a
=
[−P (x, b) + P (x, 0)] dx
Z0
=
P dx
C

38

7

Integral theorems

IA Vector Calculus

Note that we can convert the 1D integral in the second-to-last line to a line integral
around the curve C, since the P (x, 0) and P (x, b) terms give the horizontal part
of C, and the lack of dy term means that the integral is nil when integrating the
vertical parts
...

A ∂x
C
Combining them gives Green’s theorem
...
g
...

7
...
2

Stokes’ theorem

Theorem (Stokes’ theorem)
...

The direction of the line integral is as follows: If we walk along C with n
facing up, then the surface is on your left
...

39

7

Integral theorems

IA Vector Calculus

Example
...
In
spherical coordinates,
dS = a2 sin θer dθ dϕ
...
Then ∇ × F = (−x, 0, z)
...

S

Our boundary ∂C is
r(ϕ) = a(sin α cos ϕ, sin α sin ϕ, cos α)
...

So they agree
...
1
...
For a smooth vector field F(r),
Z
Z
∇ · F dV =
F · dS,
V

∂V

where V is a bounded volume with boundary ∂V , a piecewise smooth, closed
surface, with outward normal n
...
Consider a hemisphere
...

Take F = (0, 0, z + a) and ∇ · F = 1
...

40

7

Integral theorems

IA Vector Calculus

On S1 ,
dS = n dS =
Then
F · dS =

1
(x, y, z) dS
...

|
{z
}
a
dS

Then
Z

F · dS = a3

2π

Z

Z

π/2

sin θ(cos2 θ + cos θ) dθ

dϕ
0

0

S1

3



= 2πa

−1
1
cos3 θ − cos2 θ
3
2

π/2
0

5
= πa3
...
Then F · dS = −a dS
...

S2

So

Z
S1



Z
F · dS +

F · dS =
S2


5
2
− 1 πa3 = πa3 ,
3
3

in accordance with Gauss’ theorem
...
2

Relating and proving integral theorems

We will first show the following two equivalences:
– Stokes’ theorem ⇔ Green’s theorem
– 2D divergence theorem ⇔ Greens’ theorem
Then we prove the 2D version of divergence theorem directly to show that all of
the above hold
...

Proposition
...
Stokes’ theorem talks about 3D surfaces and Green’s theorem is about
2D regions
...

Let A be a region in the (x, y) plane with boundary C = ∂A, parametrised
by arc length, (x(s), y(s), 0)
...

ds ds
Given any P (x, y) and Q(x, y), we can consider the vector field
F = (P, Q, 0),
41

7

Integral theorems

IA Vector Calculus

So
∇×F=



∂Q ∂P
0, 0,
−

...

(∇ × F) · k dA =
∂x
∂y
A
A
Proposition
...

Proof
...
Hence to use Green’s to derive Stokes’ we need find some
2D thing to act on
...

Consider a parametrised surface S = r(u, v) corresponding to the region A in
the u, v plane
...
Then ∂S = r(u(t), v(t))
...


Doing some pattern-matching, we want
F · dr = Fu du + Fv dv
for some Fu and Fv
...

∂u
∂v

So we choose

∂r
∂r
, Fv = F ·

...

To match the right, recall that


∂r
∂r
(∇ × F) · dS = (∇ × F) ·
×
du dv
...

∂u
∂v
∂u ∂v

(∗)

Fortunately, this is true
...

∂u
∂u
∂v
∂u
∂v
∂xj ∂u
∂v
∂u∂v
42

7

Integral theorems

IA Vector Calculus

Similarly,
∂Fu
∂
=
∂u
∂u
So



∂r
F·
∂u



∂
=
∂u





∂xj
Fj
∂u



∂Fj ∂xi
∂xi ∂v



∂Fi
∂Fj
−
∂xj
∂xi



=

∂Fv
∂Fu
∂xj ∂xi
−
=
∂u
∂v
∂u ∂v



∂xj
∂xi
+ Fi

...


This is the left hand side of (∗)
...

∂xi
∂xj ∂u ∂v
So they match
...

Proposition
...

Proof
...

A

∂A

with an outward normal n
...
Then
∂Q ∂P
−

...
Then the normal,
being tangent to t, is n(s) = (y 0 (s), −x0 (s)) (check that it points outwards!)
...

ds
ds

Then we can expand out the integrals to obtain
Z
Z
G · n ds =
P dx + Q dy,
C

and

C

Z 

Z
(∇ · G) dA =
A

A

∂Q ∂P
−
∂x
∂y


dA
...
So the result follows
...
2D divergence theorem
...


7

Integral theorems

IA Vector Calculus

Proof
...

Furthermore, we assume that A is a simple, convex shape
...

Suppose G = G(x, y)ˆj
...

∂y

Z Z

Z
∇ · G dA =
A

X

Yx

∂G
dy
∂y


dx
...
Since I cannot draw, A will be pictured as a circle,
but the proof is valid for any simple convex shape
...

Hence by the Fundamental theorem of Calculus,
Z
Yx

∂G
dy =
∂y

Z

y+ (x)

y− (x)

∂G
dy = G(x, y+ (x)) − G(x, y− (x))
...
However, the
divergence theorem talks in terms of ds, not dx
...
If we move a distance δs, the change in x is δs cos θ, where
θ is the angle between the tangent and the horizontal
...
So cos θ = n · ˆj
...

In particular, G dx = G ˆj · n ds = G · n ds, since G = G ˆj
...
So,
actually, at C− , dx = −G · n ds
...

C

ˆ a purely vertical
To prove the 3D version, we again consider F = F (x, y, z)k,
vector field
...

V
D
Zxy ∂z
Again, split S = ∂V into the top and bottom parts S+ and S− (ie the parts
ˆ · n ≥ 0 and k
ˆ · n < 0), and parametrize by z+ (x, y) and z− (x, y)
...

V

D

S

45

8

Some applications of integral theorems

8

IA Vector Calculus

Some applications of integral theorems

8
...
The advantage of these alternative definitions is that they do not require a
choice of coordinate axes
...

Gauss’ theorem for F in a small volume V containing r0 gives
Z
Z
F · dS =
∇ · F dV ≈ (∇ · F)(r0 ) vol(V )
...

(∇ · F)(r0 ) =

1
diam(V )→1 vol(V )

Z
F · dS,

lim

∂V

where the limit is taken over volumes containing the point r0
...

∂A

A

So
Proposition
...

These are coordinate-independent definitions of div and curl
...
Suppose u is a velocity field of fluid flow
...
Taking V to be the volume occupied by a
fixed quantity of fluid material, we have
Z
V˙ =
u · dS
∂V

Then, at r0 ,

V˙
,
V →0 V

∇ · u = lim

the relative rate of change of volume
...


46

8

Some applications of integral theorems

IA Vector Calculus

Alternatively, take a planar area A to be a disc of radius a
...

∂A

∂A

(u · t is the component of u which is tangential to the boundary) We define the
quantity
1
ω = × (average of u · t)
...
As a → 0, a1 → ∞, but the
average of u · t will also decrease since a smooth field is less “twirly” if you look
closer
...
We have
Z
u · dr = 2πa2 ω
...
For example, if you have a washing machine
rotating at a rate of ω, Then the velocity u = ω × r
...


8
...
A vector field F is conservative if
(i) F = ∇f for some scalar field f ; or
R
(ii) C F · dr is independent of C, for fixed end points and orientation; or
(iii) ∇ × F = 0
...

We have previously shown (i) ⇒ (ii) since
Z
F · dr = f (b) − f (a)
...

So we want to show that (iii) ⇒ (ii) and (ii) ⇒ (i)
R
Proposition
...

Proof
...


47

8

Some applications of integral theorems

IA Vector Calculus

b

C
C˜
a

˜ By Stokes’ theorem,
If S is any surface with boundary ∂S = C − C,
Z
Z
Z
Z
∇ × F · dS =
F · dr =
F · dr −
F · dr
...
So

Z

Z
F · dr −

F · dr = 0,
˜
C

C

or

˜
C

C

Z

Z
F · dr =

F · dr
...
If (ii) C F · dr is independent of C for fixed end points and
orientation, then (i) F = ∇f for some scalar field f
...
We fix a and define f (r) = C F(r0 ) · dr0 for any curve from a to r
...
For small changes r to r + δr, there is a small
extension of C by δC
...

So
δf = f (r + δr) − f (r) = F(r) · δr + o(δr)
...

So we have F = ∇f
...
It also
works of F is defined on a simply connected domain D, ie a subspace of R3
without holes
...

˜ the process sweeps out a
If we have a smooth transformation from C to C,
˜ This is required by the proof that (iii) ⇒ (ii)
...

If D is not simply connected, then we obtain a multi-valued f (r) on D in
general (for the proof (ii) ⇒ (i))
...


48

8

Some applications of integral theorems

IA Vector Calculus

Example
...

x2 + y 2 x2 + y 2

This obeys ∇ × F = 0, and is defined on D = R3 \ {z-axis}, which is not
simply-connected
...

x
which is multi-valued
...
e
...
This shows that the simply-connected-domain criterion
is important!
However f can be single-valued if we restrict it to
f = tan−1

D0 = R3 − {half-plane x ≥ 0, y = 0},
which is simply-connected
...


8
...
Suppose we are interested in a quantity
Q
...

The conservation equation is
∂ρ
+ ∇ · j = 0
...
It says that Q cannot just disappear here and appear elsewhere
...

In particular, let V be a fixed time-independent volume with boundary
S = ∂V
...

dt
V ∂t
V
S
by divergence theorem
...
ie Q cannot just disappear
but must smoothly flow out
...
Then the surface
integral → 0, and the equation states that
dQ
= 0,
dt
49

8

Some applications of integral theorems

IA Vector Calculus

Example
...
e
...
j(r, t) is the electric
current density
...

Example
...
Then (ρu δt) · δS is equal
to the mass of fluid crossing δS in time δt
...
The conservation equation in this
case is
∂ρ
+ ∇ · (ρu) = 0
∂t
For the case where ρ is constant and uniform (i
...
independent of r and t), we
get that ∇ · u = 0
...


50

9

Orthogonal curvilinear coordinates

9

IA Vector Calculus

Orthogonal curvilinear coordinates

9
...
A coordinate system is,
roughly speaking, a way to specify a point in space by a set of (usually 3)
numbers
...

By the chain rule, we have
dr =

∂r
∂r
∂r
du +
dv +
dw
∂u
∂v
∂w

For a good parametrization,
∂r
·
∂u



∂r
∂r
×
∂v ∂w


6= 0,

∂r ∂r
∂r
, ∂v and ∂w
i
...
∂u
are linearly independent
...

Even better, they should be orthogonal:

Definition (Orthogonal curvilinear coordinates)
...

We can then set
∂r
= hu eu ,
∂u

∂r
= hv ev ,
∂v

∂r
= hw ew ,
∂w

with hu , hv , hw > 0 and eu , ev , ew form an orthonormal right-handed basis (i
...

eu × ev = ew )
...
Note that clearly by definition, we have
 
 ∂r 
hu =  
...

ˆ Then hx = hy = hz = 1,
(i) In cartesian coordinates, r(x, y, z) = xˆi + yˆj + z k
...

ˆ Then hρ = hz = 1,
(ii) In cylindrical polars, r(ρ, ϕ, z) = ρ[cos ϕˆi + sin ϕˆj] + z k
...

∂ϕ
The basis vectors eρ , eϕ , ez are as in section 1
...

Then hr = 1, hθ = r and hϕ = r sin θ
...
The vector
area element is
∂r
∂r
dS =
×
du dv = hu eu × hv ev du dv = hu hv ew du dv
...
The volume element is


∂r
∂r
∂r
dV =
·
×
du dv dw = hu hv hw du dv dw,
∂u
∂v ∂w
i
...
a small cuboid with sides hu δu , hv δv and hw δw respectively
...
2

Grad, Div and Curl

Consider f (r(u, v, w)) and compare
df =

∂f
∂f
∂f
du +
dv +
dw,
∂u
∂v
∂w

with df = (∇f ) · dr
...

1 ∂f
1 ∂f
1 ∂f
eu +
ev +
ew
...
Take f = r sin θ cos ϕ in spherical polars
...

Then we know that the differential operator is
Proposition
...

eu
ev
ew
hu ∂u hv ∂v hw ∂w
We can apply this to a vector field
∇=

F = Fu eu + Fv ev + Fw ew
using scalar or vector products to obtain
Proposition
...

∂u
52

9

Orthogonal curvilinear coordinates

IA Vector Calculus

There are several ways to obtain these formulae
...
(non-examinable)
(i) Apply ∇· or ∇× and differentiate the basis vectors explicitly
...
Then use ∇×∇f = 0 and ∇·(∇×f ) = 0
...

Recall that

Z
1
F · dr
...

n · ∇ × F = lim

Consider an area with W fixed and change u by δu and v by δv
...
Let C be its boundary
...
We split the curve C up into 4 parts
(corresponding to the four sides), and take linear approximations by assuming F and h are constant when moving through each horizontal/vertical
segment
...

∂u
∂v
Divide by the area and take the limit as area → 0, we obtain


Z
1
1
∂
∂
lim
F · dr =
hv Fv −
(hu Fu )
...

We can find the divergence similarly
...
Let A =

 er
∂
1

∇×A= 2
r sin θ  ∂r
0

1
r

tan θ2 eϕ in spherical polars
...

∂θ
∂ϕ
 r2 sin θ ∂θ
2
r
0 r sin θ · 1r tan θ2 
53

10

Gauss’ Law and Poisson’s equation

10
10
...
The total force is a sum of contributions from each part of the
mass distribution, and is proportional to m
...
g(r) is the gravitational field, acceleration due
to gravity, or force per unit mass
...

C

This means that if you walk around the place and return to the same position,
the total work done is 0 and you did not gain energy, i
...
gravitational potential
energy is conserved
...
Given any volume V bounded by closed
surface S,
Z
g · dS = −4πGM,
S

where G is Newton’s gravitational constant, and M is the total mass contained
in V
...

Example
...

Consider a total mass M distributed with a spherical symmetry about the
origin O, with all the mass contained within some radius r = a
...

ˆ = ˆr
...
Then n
So
Z
Z
Z
g · dS =
g(R)ˆr · ˆr dS = g(R)dS = 4πR2 g(R)
...

So
g(R) = −

GM
R2

for R > a
...

r2

If we take the limit as a → 0, we get a point mass M at the origin
...

54

10

Gauss’ Law and Poisson’s equation

The condition
theorem as

R
C

IA Vector Calculus

g · dr = 0 for any closed C can be re-written by Stoke’s
Z
∇ × g · dS = 0,
S

where S is bounded by the closed curve C
...
So
∇ × g = 0
...
But here we showed
that it is true for all cases
...
However, if the mass
distribution is not sufficiently symmetrical, Gauss’ law in integral form can be
difficult to use
...
Suppose
Z
M=
ρ(r) dV,
V

where ρ is the mass density
...

S

V

V

Since this is true for all V , we must have
Law (Gauss’ Law for gravitation in differential form)
...

Since ∇ × g = 0, we can introduce a gravitational potential ϕ(r) with
g = −∇ϕ
...

In the example with spherical symmetry, we can solve that
ϕ(r) = −

GM
r

for r > a
...
2

Laws of electrostatics

Consider a distribution of electric charge at rest
...

Definition (Electric field)
...

Again, this is conservative
...
It also obeys
55

10

Gauss’ Law and Poisson’s equation

IA Vector Calculus

Law (Gauss’ law for electrostatic forces)
...

Then we can write it in differential form, as in the gravitational case
...

∇·E=

ρ

...

So we can write E = −∇ϕ
...
If we write E = −∇ϕ, then ϕ is the
electrostatic potential, and
ρ
∇2 ϕ =
...
Take a spherically symmetric charge distribution about O with total
charge Q
...
Then similar
to the gravitational case, we have
E(r) =

Qˆr
,
4πε0 r2

and

−Q

...
From E, we can recover Coulomb’s law for the
force on another charge q at r:
ϕ(r) =

F = qE =

qQˆr

...
Consider an infinite line with uniform charge density
per unit length σ
...
e
...

Pick S to be a cylinder of length L and radius r
...

Also, the curved surface has area 2πrL
...

ε0
S
So
E(r) =

σ
ˆr
...
Intuitively, this is because we have
one more dimension of “stuff” compared to the point charge, so the field does
not drop as fast
...
3

Poisson’s Equation and Laplace’s equation

Definition (Poisson’s equation)
...

This is the form of the equations for gravity and electrostatics, with −4πGρ
and ρ/ε0 in place of ρ respectively
...
Laplace’s equation is
∇2 ϕ = 0
...
Since it is
incompressible, ∇ · u = 0 (cf
...
So ∇2 ϕ = 0
...

We’re concerned here mainly with cases exhibiting spherical or cylindrical
symmetry (use r for radial coordinate here)
...
e
...
Write ϕ = ϕ(r)
...

Then Laplace’s equation ∇2 ϕ = 0 becomes an ordinary differential equation
...

r
r
Then the general solution is
ϕ=
57

A
+ B
...

r
r
Then
ϕ = A ln r + B
...
e
...

For example, for a spherically symmetric solution of ∇2 ϕ = −ρ0 , with ρ0
constant, recall that ∇2 rα = α(α + 1)rα−2
...

r
6

To determine A, B, we must specify boundary conditions
...
If ϕ is defined on a bounded volume
V , then there are two kinds of common boundary conditions on ∂V :
– Specify ϕ on ∂V — a Dirichlet condition
– Specify n · ∇ϕ (sometimes written as
the outward normal on ∂V )
...
(n is

The type of boundary conditions we get depends on the physical content of
the problem
...

We can also specify different boundary conditions on different boundary
components
...
We might have a spherically symmetric distribution with constant
ρ0 , defined in a ≤ r ≤ b, with ϕ(a) = 0 and ∂ϕ
∂n (b) = 0
...

r
6

We apply the first boundary condition to obtain
A
1
+ B − ρ0 a2 = 0
...

2
b
3

These conditions give
1
A = − ρ 0 b3 ,
3

B=
58

1
1 b3
ρ0 a2 + ρ0
...
We might also be interested with spherically symmetric solution with
(
−ρ0 r ≤ a
2
∇ ϕ=
0
r>a
with ϕ non-singular at r = 0 and ϕ(r) → 0 as
r = a
...

r≤a
r > a
...
Since ϕ → 0 as r → ∞, D = 0
...

r
This is the gravitational potential inside and outside a planet of constant density
ρ0 and radius a
...
So we have
1
C
B + 4πρ0 Ga2 =
6
a
4
C
πGρ0 a = − 2
...
Substituting that into the first equation
to find B, we get
(


 r2
GM
−3
r≤a
2a
a
ϕ(r) =
GM
− r
r>a
Since g = −ϕ0 , we have
(
g(r) =

r
− GM
a3
− GM
r

r≤a
r>a

We can plot the potential energy:
ϕ(r)
r=a

r

We can also plot −g(r), the inward acceleration:
−g(r)

r=a

59

r

10

Gauss’ Law and Poisson’s equation

IA Vector Calculus

Alternatively, we can apply Gauss’ Law for a flux of g = g(r)er out of S, a
sphere of radius R
...

a3
For R ≥ a, we can simply apply Newton’s law of gravitation
...
This is called the Gauss Flux method
...
1

Uniqueness theorems

Theorem
...

Suppose ϕ satisfies either
(i) Dirichlet condition, ϕ(r) = f (r) on S
(ii) Neumann condition

∂ϕ(r)
∂n

= n · ∇ϕ = g(r) on S
...
Then
(i) ϕ(r) is unique
(ii) ϕ(r) is unique up to a constant
...

Since the proof of the cases of the two different boundary conditions are very
similar, they will be proved together
...

Proof
...
Then Ψ(r) = ϕ2 (r) − ϕ1 (r) satisfies ∇2 Ψ = 0 on V by
linearity, and
(i) Ψ = 0 on S; or
(ii)

∂Ψ
∂n

= 0 on S
...
So using
the divergence theorem,
Z
Z
∇ · (Ψ∇Ψ) dV = (Ψ∇Ψ) · dS = 0
...

=0

So

Z

|∇Ψ|2 dV = 0
...
So ∇Ψ = 0
...
So
(i) Ψ = 0 on S ⇒ c = 0
...

(ii) ϕ2 (r) = ϕ1 (r) + C, as claimed
...
How about existence? It turns out it isn’t difficult
to craft a boundary condition in which there are no solutions
...

V
∂S ∂n
Using Poisson’s equation and the boundary conditions, we have
Z
Z
ρ dV +
g dS = 0
V

∂V

So if ρ and g don’t satisfy this equation, then we can’t have any solutions
...
The result also
extends to unbounded domains
...
Then we just take the relevant limits to complete the proof
...
But we have to analyse
these case by case and see if the proof still applies
...

Z
Z
Z
(u∇v) · dS =
(∇u) · (∇v) dV +
u∇2 v dV,
S

V

V

By swapping u and v around and subtracting the equations, we have
Proposition (Green’s second identity)
...

S

V

These are sometimes useful, but can be easily deduced from the divergence
theorem when needed
...
2

Laplace’s equation and harmonic functions

Definition (Harmonic function)
...

These have some very special properties
...
2
...
Suppose ϕ(r) is harmonic on region V
containing a solid sphere defined by |r − a| ≤ R, with boundary SR = |r − a| = R,
for some R
...

4πR2 SR
Then ϕ(a) = ϕ(R)
...

Proof
...
We take spherical coordinates (u, θ, χ)
centered on r = a
...

So

dS
R2

is independent of R
...

R2

Differentiate this with respect to R, noting that dS/R2 is independent of R
...
So
d
1
ϕ(R)
¯
=
dR
4πR2

Z
∇ϕ · dS =
SR

1
4πR2

Z

∇2 ϕ dV = 0

VR

by divergence theorem
...

11
...
2

The maximum (or minimum) principle

In this section, we will talk about maxima of functions
...

Definition (Local maximum)
...

Proposition (Maximum principle)
...

Proof
...
Then there is
an ε such that for any r such that 0 < |r − a| < ε, we have ϕ(r) < ϕ(a)
...
Pick an ε
sufficiently small such that the region |r − a| < ε lies within V (possible since a
lies in the interior of V )
...

Z
1
ϕ(ε)
¯
=
ϕ(r) dS < ϕ(a),
4πR2 SR
which contradicts the mean value property
...
Suppose at r = a, we have ∇ϕ = 0
...
But since ϕ is harmonic, we have ∇2 ϕ = 0,
63

11

Laplace’s and Poisson’s equations

IA Vector Calculus

2

ϕ
i
...
∂x∂i ∂x
= Hii = 0
...
So
λi = 0
...
This clearly cannot happen if the sum is 0
...

(note we ignored the case where all λi = 0, where this analysis is inconclusive)

11
...
3
...
We start with a discrete case,
and try to generalize it to a continuous case
...

4π |r − a|

(we have λ = −4πGM for gravitation and Q/ε0 for electrostatics)
If we have many sources λα at positions rα , the potential is a sum of terms
ϕ(r) =

X 1
λα

...
It would be reasonable
to guess that the solution is what we obtain by replacing the sum with an integral:
Proposition
...

Example
...


Fix r and introduce polar coordinates r0 , θ, χ for r0
...

Then
Z
1
ρ0
ϕ(r) =
dV 0
...

We also have
|r − r0 | =

p

r2 + r02 − 2rr0 cos θ

64

11

Laplace’s and Poisson’s equations

IA Vector Calculus

by the cosine rule (c2 = a2 + b2 − 2ab cos C)
...
So
Z
ϕ(r) = ρ0
0

a

r02 0
r0 a3
dr =

...

r
6
2
r
0
11
...
2

Point sources and δ-functions*

Recall that
Ψ=

λ
4π|r − a|

is our potential for a point source
...


What about when r = a? Ψ is singular at this point, but can we say anything
about ∇2 Ψ?
For any sphere with center a, we have
Z
∇Ψ · dS = −λ
...

for V being a solid sphere with ∂V = S
...

65

11

Laplace’s and Poisson’s equations

IA Vector Calculus

In short, we have
∇2



1
|r − r0 |



= −4πδ(r − r0 )
...


66

12

Maxwell’s equations

12
12
...
Together with the Lorentz force law, these describe all
we know about (classical) electromagnetism
...
It is thus important to
understand the mathematical properties of these equations
...
These are denoted by E(r, t) and B(r, t)
respectively
...
The second is rather
straightforward, and is given by the Lorentz force law
...
A point charge q experiences a force of
F = q(E + r˙ × B)
...
To state
the equations, we need to introduce two more concepts
...
ρ(r, t) is the charge density, defined
as the charge per unit volume
...

Then Maxwell’s equations say
Law (Maxwell’s equations)
...

We can quickly derive some properties we know from these four equations
...

∂
∇ · (∇ × B) −µ0 ε0 (∇ · E) = µ0 ∇ · j
...

∂t
67

12

Maxwell’s equations

IA Vector Calculus

We can also take the volume integral of the first equation to obtain
Z
Z
1
Q
∇ · E dV =
ρ dV =
...

S

This roughly states that there are no “magnetic charges”
...
For example,
Z
Z
Z
d
B · dS,
E · dr =
∇ × E dS = −
dt S
C=∂S
S
where the first equality is from from Stoke’s theorem
...


12
...

We can solve the equations for electric fields:
∇ · E = ρ/ε0
∇×E=0
Second equation gives E = −∇ϕ
...

The equations for the magnetic field are
∇·B=0
∇ × B = µ0 j
First equation gives B = ∇ × A for some vector potential A
...
Making the transformation A 7→ A + ∇χ(x)
produces the same B, since ∇ × (∇χ) = 0
...

Then
∇2 A = ∇(∇
· A}) − ∇ × (∇ × A) = −µ0 j
...
g
...

µ0 sets the scale of magnetic effects,
e
...
force between two wires with currents
...
3

IA Vector Calculus

Electromagnetic waves

Consider Maxwell’s equations in empty space, i
...
ρ = 0, j = 0
...


∂B
∂
∂2E
= (∇ × B) = µ0 ε0 2
...

c ∂t

This is the wave equation describing propagation with speed c
...

c ∂t
So Maxwell’s equations predict that there exists electromagnetic waves in free
space, which move with speed c = √ε10 µ0 ≈ 3
...
1

IA Vector Calculus

Tensors and tensor fields
Definition

There are two ways we can think of a vector in R3
...
However,
the list of three numbers is simply a representation of the vector with respect to
some particular basis
...
In particular, when
we perform a rotation by Rip , the new components of the vector is given by
vi0 = Rip vp
...
Again, when we change basis, in order to represent the
same transformation, we will need a different array of numbers
...

We can think about this from another angle
...
Moreover,
we can write down a different set of numbers in a different basis
...
We
can do so because we have the power of the pen
...
e
...
By “sensibly”, we mean that it
has to follow the transformation rule A0ij = Rip Rjq Apq
...
While it is
something we can define and write down, it does not correspond to anything
meaningful
...
For example,
vectors and matrices (that transform according to the usual change-of-basis
rules) are tensors, but that Aij is not
...
In
order for a quantity Tij···k to be a tensor, we require it to transform according to
0
Tij···k
= Rip Rjq · · · Rkr Tpq···r ,

which is an obvious generalization of the rules for vectors and matrices
...
A tensor of rank n has components Tij···k (with n indices)
with respect to each basis {ei } or coordinate system {xi }, and satisfies the
following rule of change of basis:
0
Tij···k
= Rip Rjq · · · Rkr Tpq···r
...

– A tensor T of rank 0 doesn’t transform under change of basis, and is a
scalar
...
This is a vector
...
This is a matrix
...

(i) If u, v, · · · w are n vectors, then
Tij···k = ui vj · · · wk
defines a tensor of rank n
...
We do the case for n = 2 for simplicity of expression, and it should
be clear that this can be trivially extended to arbitrary n:
Tij0 = u0i vj0 = (Rip up )(Rjq vq )
= Rip Rjq (up vq )
= Rip Rjq Tpq
Then linear combinations of such expressions are also tensors, e
...
Tij =
ui vj + ai bj for any u, v, a, b
...
Also
ε0ijk = Rip Rjq Rkr εpqr = (det R)εijk = εijk ,
using results from Vectors and Matrices
...

Note that this relation entails that the resulting current need not be in the
same direction as the electric field
...

However, if the substance is isotropic, we have εij = σδij for some σ
...


13
...
Tensors T and S of the same rank can be added ;
T + S is also a tensor of the same rank, defined as
(T + S)ij···k = Tij···k + Sij···k
...

To check that this is a tensor, we check the transformation rule
...


71

13

Tensors and tensor fields

IA Vector Calculus

Definition (Scalar multiplication)
...
αT is a tensor of the same rank, defined by
(αT )ij = αTij
...

Definition (Tensor product)
...
The tensor product T ⊗ S is a tensor of rank n + m defined by
(T ⊗ S)x1 x2 ···xn y1 y2 ···ym = Tx1 x2 ···xn Sy1 y2 ···yn
...

We can similarly define tensor products for any (positive integer) number of
tensors, e
...
for n vectors u, v · · · , w, we can define
T = u ⊗ v ⊗ ··· ⊗ w
by
Tij···k = ui vj · · · wk ,
as defined in the example in the beginning of the chapter
...
For a tensor T of rank n with components
Tijp···q , we can contract on the indices i, j to obtain a new tensor of rank n − 2:
Sp···q = δij Tijp···q = Tiip···q
Note that we don’t have to always contract on the first two indices
...

To check that contraction produces a tensor, we take the ranks 2 Tij example
...
We have Tii0 = Rip Riq Tpq = δpq Tpq =
Tpp = Tii , since R is an orthogonal matrix
...
So our result above says that the trace is invariant under basis
transformations — as we already know in IA Vectors and Matrices
...


13
...
A tensor T of rank n is
symmetric in the indices i, j if it obeys
Tijp···q = Tjip···q
...

Again, a tensor can be symmetric or anti-symmetric in any pair of indices, not
just the first two
...
s
= Rki R`j Rrp · · · Rsq Tijp···q = ±Rki R`j Rrp · · · Rsq Tjip···q = ±T`kr···s

as required
...
A tensor is totally
(anti-)symmetric if it is (anti-)symmetric in every pair of indices
...
δij = δji is totally symmetric, while εijk = −εjik is totally antisymmetric
...
But in R3 ,
– Any totally antisymmetric tensor of rank 3 is λεijk for some scalar λ
...

Proof: exercise (hint: pigeonhole principle)

13
...
We will prove
an analogous fact for tensors
...
A map T that maps n vectors a, b, · · · , c to R
is multi-linear if it is linear in each of the vectors a, b, · · · , c individually
...

To show that tensors are equivalent to multi-linear maps, we have to show the
following:
(i) Defining a map with a tensor makes sense, i
...
the expression Tij···k ai bj · · · ck
is the same regardless of the basis chosen;
(ii) While it is always possible to write a multi-linear map as Tij···k ai bj · · · ck ,
we have to show that Tij···k is indeed a tensor, i
...
transform according to
the tensor transformation rules
...
So it is also a
tensor
...
e
...

To show the second property, assuming that T is a multi-linear map, it must
be independent of the basis, so
0
Tij···k ai bj · · · ck = Tij···k
a0i b0j · · · c0k
...
Substituting in gives
0
Tij···k (Rpi a0p )(Rqj b0q ) · · · (Rkr c0r ) = Tpq···r
a0p b0q · · · c0r
...

This shows that there is a one-to-one correspondence between tensors of rank
n and multi-linear maps
...

Note that the above is exactly what we did with linear maps and matrices
...

The converse is also true:
Proposition (Quotient rule)
...
Then Ti···jp···q is also a tensor
...

Proof
...
However, we can
reuse the result above to save some writing
...
By
assumption,
vi···j = Ti···jp···q cp · · · dq
is a tensor
...
Since Ti···jp···q ai · · · bj cp · · · dq is a
scalar and hence gives the same result in every coordinate system, Ti···jp···q is a
multi-linear map
...


13
...
A tensor field is a tensor at each point in space
Tij···k (x), which can also be written as Tij···k (x` )
...
g
...
We now
claim:
Proposition
...

Proof
...
e
...
Then
by the exact same argument we used to show that tensor products preserve
tensorness, we can show that the (∗) is a tensor
...
But the exact same
proof works!)
Since x0i = Riq xq , we have
∂x0i
= Rip
...
Similarly,
∂xq
= Riq
...

Hence by the chain rule,


∂
∂xq
∂
∂
=
= Riq

...
So done
...
Since we can sum tensors and take
limits, the definition
R of a tensor-valued integral is straightforward
...

For a physical example, recall our discussion of the flux of quantities for a
fluid with velocity u(x) through a surface element — assume a uniform density
ρ
...
So the flux of mass is ρuj nj δS
...
Then the flux through the surface S
is S Tij nj dS
...
We
can then use it to discuss conservation laws for tensor quantities
...
Then
Theorem (Divergence theorem for tensors)
...

The regular divergence theorem is the case where T has one index and is a
vector field
...
Apply the usual divergence theorem to the vector field v defined by
v` = ai bj · · · ck Tij···k` , where a, b, · · · , c are fixed constant vectors
...

Since a, b, · · · , c are arbitrary, therefore they can be eliminated, and the tensor
divergence theorem follows
...
1

IA Vector Calculus

Tensors of rank 2
Decomposition of a second-rank tensor

This decomposition might look arbitrary at first sight, but as time goes on, you
will find that it is actually very useful in your future career (at least, the lecturer
claims so)
...

2
2
Here Tij has 9 independent components, whereas Sij and Aij have 6 and 3
independent components, since they must be of the form




a d e
0
a b
(Sij ) = d b f  , (Aij ) = −a 0 c 
...
e
...
Then
Pij has 5 independent components while Q has 1
...
In fact, we can
write the antisymmetric part as
Aij = εijk Bk
for some vector B
...

Then


0
B3 −B2
0
B1 
(Aij ) = −B3
B2 −B1
0
To summarize,
1
Tij = Pij + εijk Bk + δij Q,
3
where Bk = 21 εpqj Tpq , Q = Tkk and Pij = Pji =

77

Tij +Tji
2

− 13 δij Q
...
The derivative of a vector field Fi (r) is a tensor Tij = ∂x
, a tensor
j
field
...

2
∂xj
2

and trace
Q=

∂Fk
= ∇ · F
...


14
...
So the velocities are vα = ω × rα
...


α

by vector identities
...
The inertia tensor is
X
Iij =
mα [|rα |2 δij − (rα )i (rα )j ]
...

V

By inspection, I is a symmetric tensor
...
Consider a rotating cylinder with uniform density ρ0
...


78

14

Tensors of rank 2

IA Vector Calculus

x3
a
2`
x1

x2
Use cylindrical polar coordinates:
x1 = r cos θ
x2 = r sin θ
x3 = x3
dV = r dr dθ dx3
We have
Z
I33 =

ρ0 (x21 + x22 ) dV

V

Z

a

Z

2π

Z

`

= ρ0
0

0


= ρ0 · 2π · 2`

r2 (r dr dθ dx2 )

−`

4 a

r
4

0

= ε0 π`a4
...


79

14

Tensors of rank 2

IA Vector Calculus

How about the off-diagonal elements?
Z
I13 = −
ρ0 x1 x3 dV
V

Z

a

`

Z

2π

Z

r2 cos θx3 dr dx3 dθ

= −ρ0
0

−`

0

=0
R 2π
Since 0 dθ cos θ = 0
...
So
the non-zero components are
1
M a2
2 

`2
a2
+
=M
4
3

I33 =
I11 = I22
In the particular case where ` =

√
a 3
2 ,

we have Iij = 21 ma2 δij
...


14
...

If T is symmetric, it can be diagonalized by such an orthogonal transformation
...
The directions defined by e1 , e2 , e3
are the principal axes for T , and the tensor is diagonal in Cartesian coordinates
along these axes
...
For the special case of the
inertia tensor, the eigenvalues are called the principal moments of inertia
...
With the axes we
chose, Iij was found to be diagonal by direct calculation
...
1

IA Vector Calculus

Invariant and isotropic tensors
Definitions and classification results

Definition (Invariant and isotropic tensor)
...
e
...

A tensor T which is invariant under every rotation is isotropic, i
...
the same
in every direction
...
The inertia tensor of a sphere is isotropic by symmetry
...
This ensures that the component
definitions of the scalar and vector products a·b = ai bj δij and (a×b)i = εijk aj bk
are independent of the Cartesian coordinate system
...

(i) There are no isotropic tensors of rank 1, except the zero tensor
...

(iii) The most general rank 3 isotropic tensor is Tijk = βεijk for some scalar β
...

We will provide a sketch of the proof:
Proof
...

(i) Suppose Ti is rank-1 isotropic
...

0
0 1
We want T1 = Rip Tp = R11 T1 = −T1
...
Similarly, T2 = 0
...

(ii) Suppose Tij is rank-2 isotropic
...
Then
T13 = R1p R3q Tpq = R12 R33 T23 = T23

81

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Invariant and isotropic tensors

IA Vector Calculus

and
T23 = R2p R3q Tpq = R21 R33 T13 = −T13
So T13 = T23 = 0
...

We also have
T11 = R1p R1q Tpq = R12 R12 T22 = T22
...

By picking a rotation about a different axis, we have T21 = T12 and
T22 = T33
...

(iii) Suppose that Tijk is rank-3 isotropic
...

So T133 = 0
...

So T111 = 0
...

Then we can show that Tijk = 0 unless all i, j, k are distinct
...
Then
T123 = R1p R2q R3r Tpqr = R12 R21 R33 T213 = −T213
...
Along with similar results for other indices and axes of
rotation, we find that Tijk is totally antisymmetric, and Tijk = βεijk for
some β
...
The most general isotropic tensor of rank 4 is
Tijk` = αδij δk` + βδik δj` + γδi` δjk
for some scalars α, β, γ
...
(we
might think we can write a rank-4 isotropic tensor in terms of εijk , like εijp εk`p ,
but this is just δik δj` − δi` δjk
...
2

Application to invariant integrals

We have the following very useful theorem
...
Let

IA Vector Calculus

Z
f (x)xi xj · · · xk dV
...

Given a rotation Rij , consider an active transformation: x = xi ei is mapped
to x0 = x0i ei with x0i = Rij xi , i
...
we map the components but not the basis, and
V is mapped to V 0
...
g
...

Then Tij···k is invariant under the rotation
...
First note that the Jacobian of the transformation R is 1, since it is
∂x0
simply the determinant of R (x0i = Rip xp ⇒ ∂xpi = Rip ), which is by definition
1
...

Then we have
Z
Rip Rjq · · · Rkr Tpq···r =
f (x)x0i x0j · · · x0k dV
V
Z
=
f (x0 )x0i x0j · · · x0k dV
using (i)
V
Z
=
f (x0 )x0i x0j · · · x0k dV 0 using (ii)
V0
Z
=
f (x)xi xj · · · xk dV
since xi and x0i are dummy
V

= Tij···k
...

Example
...
Our result applies with f = 1, which,
being a constant, is clearly invariant under rotations
...
So T must be isotropic
...
Hence we must have
Tij = αδij ,
and all we have to do is to determine the scalar α
...

5
V
0
So
Tij =

4
πa5 δij
...
But now we can do it (more) rigorously!
There is a closely related result for the inertia tensor of a solid sphere of
constant density ρ0 , or of mass M = 43 πa3 ρ0
...

V

R
We see that Iij is isotropic (since we have just shown that xi xj dV is isotropic,
and xk xk δij is also isotropic)
...
Then
Z
Iij =
ρ0 (xk xk δij − xi xj ) dV
V
 Z

Z
= ρ0 δij
xk xk dV −
xi xj dV
V

V

= ρ0 (δij Tkk − Tij )


4 5
4
5
= ρ0
πa δij − πa δij
5
15
8
ρ0 πa5 δij
=
15
2
= M a2 δij

---