Part IA Vector Calculus
Based on lectures by B. Allanach
Notes taken by Dexter Chua
Lent 2015
These notes are not endorsed by the lecturers, and I have modified them (often
significantly) after lectures. They are nowhere near accurate representations of what
was actually lectured, and in particular, all errors are almost surely mine.
Curves in R
3
Parameterised curves and arc length, tangents and normals to curves in
R
3
, the radius
of curvature. [1]
Integration in R
2
and R
3
Line integrals. Surface and volume integrals: definitions, examples using Cartesian,
cylindrical and spherical coordinates; change of variables. [4]
Vector operators
Directional derivatives. The gradient of a real-valued function: definition; interpretation
as normal to level surfaces; examples including the use of cylindrical, spherical *and
general orthogonal curvilinear* coordinates.
Divergence, curl and
2
in Cartesian coordinates, examples; formulae for these oper-
ators (statement only) in cylindrical, spherical *and general orthogonal curvilinear*
co ordinates. Solenoidal fields, irrotational fields and conservative fields; scalar poten-
tials. Vector derivative identities. [5]
Integration theorems
Divergence theorem, Green’s theorem, Stokes’s theorem, Green’s second theorem:
statements; informal proofs; examples; application to fluid dynamics, and to electro-
magnetism including statement of Maxwell’s equations. [5]
Laplace’s equation
Laplace’s equation in
R
2
and
R
3
: uniqueness theorem and maximum principle. Solution
of Poisson’s equation by Gauss’s method (for spherical and cylindrical symmetry) and
as an integral. [4]
Cartesian tensors in R
3
Tensor transformation laws, addition, multiplication, contraction, with emphasis on
tensors of second rank. Isotropic second and third rank tensors. Symmetric and
antisymmetric tensors. Revision of principal axes and diagonalization. Quotient
theorem. Examples including inertia and conductivity. [5]
Contents
0 Introduction
1 Derivatives and coordinates
1.1 Derivative of functions
1.2 Inverse functions
1.3 Coordinate systems
2 Curves and Line
2.1 Parametrised curves, lengths and arc length
2.2 Line integrals of vector fields
2.3 Gradients and Differentials
2.4 Work and potential energy
3 Integration in R
2
and R
3
3.1 Integrals over subsets of R
2
3.2 Change of variables for an integral in R
2
3.3 Generalization to R
3
3.4 Further generalizations
4 Surfaces and surface integrals
4.1 Surfaces and Normal
4.2 Parametrized surfaces and area
4.3 Surface integral of vector fields
4.4 Change of variables in R
2
and R
3
revisited
5 Geometry of curves and surfaces
6 Div, Grad, Curl and
6.1 Div, Grad, Curl and
6.2 Second-order derivatives
7 Integral theorems
7.1 Statement and examples
7.1.1 Green’s theorem (in the plane)
7.1.2 Stokes’ theorem
7.1.3 Divergence/Gauss theorem
7.2 Relating and proving integral theorems
8 Some applications of integral theorems
8.1 Integral expressions for div and curl
8.2 Conservative fields and scalar products
8.3 Conservation laws
9 Orthogonal curvilinear coordinates
9.1 Line, area and volume elements
9.2 Grad, Div and Curl
10 Gauss’ Law and Poisson’s equation
10.1 Laws of gravitation
10.2 Laws of electrostatics
10.3 Poisson’s Equation and Laplace’s equation
11 Laplace’s and Poisson’s equations
11.1 Uniqueness theorems
11.2 Laplace’s equation and harmonic functions
11.2.1 The mean value property
11.2.2 The maximum (or minimum) principle
11.3 Integral solutions of Poisson’s equations
11.3.1 Statement and informal derivation
11.3.2 Point sources and δ-functions*
12 Maxwell’s equations
12.1 Laws of electromagnetism
12.2 Static charges and steady currents
12.3 Electromagnetic waves
13 Tensors and tensor fields
13.1 Definition
13.2 Tensor algebra
13.3 Symmetric and antisymmetric tensors
13.4 Tensors, multi-linear maps and the quotient rule
13.5 Tensor calculus
14 Tensors of rank 2
14.1 Decomposition of a second-rank tensor
14.2 The inertia tensor
14.3 Diagonalization of a symmetric second rank tensor
15 Invariant and isotropic tensors
15.1 Definitions and classification results
15.2 Application to invariant integrals
0 Introduction
In the differential equations class, we learnt how to do calculus in one dimension.
However, (apparently) the world has more than one dimension. We live in a
3 (or 4) dimensional world, and string theorists think that the world has more
than 10 dimensions. It is thus important to know how to do calculus in many
dimensions.
For example, the position of a particle in a three dimensional world can be
given by a position vector
x
. Then by definition, the velocity is given by
d
dt
x
=
˙
x
.
This would require us to take the derivative of a vector.
This is not too difficult. We can just differentiate the vector componentwise.
However, we can reverse the problem and get a more complicated one. We can
assign a number to each point in (3D) space, and ask how this number changes
as we move in space. For example, the function might tell us the temperature at
each point in space, and we want to know how the temperature changes with
position.
In the most general case, we will assign a vector to each point in space. For
example, the electric field vector E(x) tells us the direction of the electric field
at each point in space.
On the other side of the story, we also want to do integration in multiple
dimensions. Apart from the obvious “integrating a vector”, we might want to
integrate over surfaces. For example, we can let
v
(
x
) be the velocity of some
fluid at each point in space. Then to find the total fluid flow through a surface,
we integrate v over the surface.
In this course, we are mostly going to learn about doing calculus in many
dimensions. In the last few lectures, we are going to learn about Cartesian
tensors, which is a generalization of vectors.
Note that throughout the course (and lecture notes), summation convention
is implied unless otherwise stated.
1 Derivatives and coordinates
1.1 Derivative of functions
We used to define a derivative as the limit of a quotient and a function is differ-
entiable if the derivative exists. However, this obviously cannot be generalized
to vector-valued functions, since you cannot divide by vectors. So we want
an alternative definition of differentiation, which can be easily generalized to
vectors.
Recall, that if a function
f
is differentiable at
x
, then for a small perturbation
δx, we have
δf
def
= f (x + δx) f(x) = f
0
(x)δx + o(δx),
which says that the resulting change in
f
is approximately proportional to
δx
(as opposed to 1
x
or something else). It can be easily shown that the converse
is true if f satisfies this relation, then f is differentiable.
This definition is more easily extended to vector functions. 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. In the most general case,
δx
will
be a vector and that “something” will be a matrix. Then that “something” will
be what we call the derivative.
Vector functions R R
n
We start with the simple case of vector functions.
Definition (Vector function). A vector function is a function F : R R
n
.
This takes in a number and returns a vector. For example, it can map a time
to the velocity of a particle at that time.
Definition
(Derivative of vector function)
.
A vector function
F
(
x
) is differen-
tiable if
δF
def
= F(x + δx) F(x) = F
0
(x)δx + o(δx)
for some F
0
(x). F
0
(x) is called the derivative of F(x).
We don’t have anything new and special here, since we might as well have
defined F
0
(x) as
F
0
=
dF
dx
= lim
δx0
1
δx
[F(x + δx) F(x)],
which is easily shown to be equivalent to the above definition.
Using differential notation, the differentiability condition can be written as
dF = F
0
(x) dx.
Given a basis
e
i
that is independent of
x
, vector differentiation is performed
componentwise, i.e.
Proposition.
F
0
(x) = F
0
i
(x)e
i
.
Leibnitz identities hold for the products of scalar and vector functions.
Proposition.
d
dt
(fg ) =
df
dt
g + f
dg
dt
d
dt
(g · h) =
dg
dt
· h + g ·
dh
dt
d
dt
(g × h) =
dg
dt
× h + g ×
dh
dt
Note that the order of multiplication must be retained in the case of the cross
product.
Example.
Consider a particle with mass
m
. It has position
r
(
t
), velocity
˙
r
(
t
)
and acceleration
¨
r. Its momentum is p = m
˙
r(t).
Note that derivatives with respect to
t
are usually denoted by dots instead
of dashes.
If F(r) is the force on a particle, then Newton’s second law states that
˙
p = m
¨
r = F.
We can define the angular momentum about the origin to be
L = r × p = mr ×
˙
r.
If we want to know how the angular momentum changes over time, then
˙
L = m
˙
r ×
˙
r + mr ×
¨
r = mr ×
¨
r = r × F.
which is the torque of F about the origin.
Scalar functions R
n
R
We can also define derivatives for a different kind of function:
Definition. A scalar function is a function f : R
n
R.
A scalar function takes in a position and gives you a number, e.g. the potential
energy of a particle at different positions.
Before we define the derivative of a scalar function, we have to first define
what it means to take a limit of a vector.
Definition
(Limit of vector)
.
The limit of vectors is defined using the norm.
So v c iff |v c| 0. Similarly, f (r) = o(r) means
|f(r)|
|r|
0 as r 0.
Definition
(Gradient of scalar function)
.
A scalar function
f
(
r
) is differentiable
at r if
δf
def
= f (r + δr) f(r) = (f ) · δr + o(δr)
for some vector f , the gradient of f at r.
Here we have a fancy name “gradient” for the derivative. 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!
The above definition considers the case where
δr
comes in all directions.
What if we only care about the case where
δr
is in some particular direction
n
?
For example, maybe
f
is the potential of a particle that is confined to move in
one straight line only.
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). The directional derivative of f along n is
n · f = lim
h0
1
h
[f(r + hn) f(r)],
It refers to how fast f changes when we move in the direction of n.
Using this expression, the directional derivative is maximized when
n
is in
the same direction as
f
(then
n · f
=
|∇f|
). So
f
points in the direction
of greatest slope.
How do we evaluate
f
? Suppose we have an orthonormal basis
e
i
. Setting
n = e
i
in the above equation, we obtain
e
i
· f = lim
h0
1
h
[f(r + he
i
) f(r)] =
f
x
i
.
Hence
Theorem. The gradient is
f =
f
x
i
e
i
Hence we can write the condition of differentiability as
δf =
f
x
i
δx
i
+ o(δx).
In differential notation, we write
df = f · dr =
f
x
i
dx
i
,
which is the chain rule for partial derivatives.
Example. Take f(x, y, z) = x + e
xy
sin z. Then
f =
f
x
,
f
y
,
f
z
= (1 + ye
xy
sin z, xe
xy
sin z, e
xy
cos z)
At (x, y, z) = (0, 1, 0), f = (1, 0, 1). So f increases/decreases most rapidly for
n
=
±
1
2
(1
,
0
,
1) with a rate of change of
±
2
. There is no change in
f
if
n
is
perpendicular to ±
1
2
(1, 0, 1).
Now suppose we have a scalar function
f
(
r
) and we want to consider the rate
of change along a path
r
(
u
). A change
δu
produces a change
δr
=
r
0
δu
+
o
(
δu
),
and
δf = f · δr + o(|δr|) = f · r
0
(u)δu + o(δu).
This shows that f is differentiable as a function of u and
Theorem (Chain rule). Given a function f(r(u)),
df
du
= f ·
dr
du
=
f
x
i
dx
i
du
.
Note that if we drop the du, we simply get
df = f · dr =
f
x
i
dx
i
,
which is what we’ve previously had.
Vector fields R
n
R
m
We are now ready to tackle the general case, which are given the fancy name of
vector fields.
Definition (Vector field). A vector field is a function F : R
n
R
m
.
Definition
(Derivative of vector field)
.
A vector field
F
:
R
n
R
m
is differen-
tiable if
δF
def
= F(x + δx) F(x) = Mδx + o(δx)
for some m × n matrix M . M is the derivative of F.
As promised, M does not have a fancy name.
Given an arbitrary function
F
:
R
n
R
m
that maps
x 7→ y
and a choice
of basis, we can write
F
as a set of
m
functions
y
j
=
F
j
(
x
) such that
y
=
(y
1
, y
2
, ··· , y
m
). Then
dy
j
=
F
j
x
i
dx
i
.
and we can write the derivative as
Theorem. The derivative of F is given by
M
ji
=
y
j
x
i
.
Note that we could have used this as the definition of the derivative. However,
the original definition is superior because it does not require a selection of
coordinate system.
Definition.
A function is smooth if it can be differentiated any number of times.
This requires that all partial derivatives exist and are totally symmetric in
i, j
and k (i.e. the differential operator is commutative).
The functions we will consider will be smooth except where things obviously
go wrong (e.g. f (x) = 1/x at x = 0).
Theorem
(Chain rule)
.
Suppose
g
:
R
p
R
n
and
f
:
R
n
R
m
. Suppose that
the coordinates of the vectors in
R
p
, R
n
and
R
m
are
u
a
, x
i
and
y
r
respectively.
By the chain rule,
y
r
u
a
=
y
r
x
i
x
i
u
a
,
with summation implied. Writing in matrix form,
M(f g)
ra
= M(f)
ri
M(g)
ia
.
Alternatively, in operator form,
u
a
=
x
i
u
a
x
i
.
1.2 Inverse functions
Suppose
g, f
:
R
n
R
n
are inverse functions, i.e.
g f
=
f g
=
id
. Suppose
that f (x) = u and g(u) = x.
Since the derivative of the identity function is the identity matrix (if you
differentiate x wrt to x, you get 1), we must have
M(f g) = I.
Therefore we know that
M(g) = M(f)
1
.
We derive this result more formally by noting
u
b
u
a
= δ
ab
.
So by the chain rule,
u
b
x
i
x
i
u
a
= δ
ab
,
i.e. M (f g) = I.
In the n = 1 case, it is the familiar result that du/dx = 1/(dx/du).
Example.
For
n
= 2, write
u
1
=
ρ
,
u
2
=
ϕ
and let
x
1
=
ρ cos ϕ
and
x
2
=
ρ sin ϕ
. Then the function used to convert between the coordinate systems is
g(u
1
, u
2
) = (u
1
cos u
2
, u
1
sin u
2
)
Then
M(g) =
x
1
/∂ρ x
1
/∂ϕ
x
2
/∂ρ x
2
/∂ϕ
=
cos ϕ ρ sin ϕ
sin ϕ ρ cos ϕ
We can invert the relations between (x
1
, x
2
) and (ρ, ϕ) to obtain
ϕ = tan
1
x
2
x
1
ρ =
q
x
2
1
+ x
2
2
We can calculate
M(f) =
ρ/∂x
1
ρ/∂x
2
ϕ/∂x
1
ϕ/∂x
2
= M(g)
1
.
These matrices are known as Jacobian matrices, and their determinants are
known as the Jacobians.
Note that
det M(f) det M (g) = 1.
1.3 Coordinate systems
Now we can apply the results above the changes of coordinates on Euclidean
space. Suppose
x
i
are the coordinates are Cartesian coordinates. Then we can
define an arbitrary new coordinate system
u
a
in which each coordinate
u
a
is a
function of x. For example, we can define the plane polar coordinates ρ, ϕ by
x
1
= ρ cos ϕ, x
2
= ρ sin ϕ.
However, note that
ρ
and
ϕ
are not components of a position vector, i.e. they
are not the “coefficients” of basis vectors like
r
=
x
1
e
1
+
x
2
e
2
are. But we can
associate related basis vectors that point to directions of increasing
ρ
and
ϕ
,
obtained by differentiating
r
with respect to the variables and then normalizing:
e
ρ
= cos ϕ e
1
+ sin ϕ e
2
, e
ϕ
= sin ϕ e
1
+ cos ϕ e
2
.
e
1
e
2
ρ
e
ρ
e
ϕ
ϕ
These are not “usual” basis vectors in the sense that these basis vectors vary
with position and are undefined at the origin. However, they are still very useful
when dealing with systems with rotational symmetry.
In three dimensions, we have cylindrical polars and spherical polars.
Cylindrical polars Spherical polars
Conversion formulae
x
1
= ρ cos ϕ x
1
= r sin θ cos ϕ
x
2
= ρ sin ϕ x
2
= r sin θ sin ϕ
x
3
= z x
3
= r cos θ
Basis vectors
e
ρ
= (cos ϕ, sin ϕ, 0) e
r
= (sin θ cos ϕ, sin θ sin ϕ, cos θ)
e
ϕ
= (sin ϕ, cos ϕ, 0) e
ϕ
= (sin ϕ, cos ϕ, 0)
e
z
= (0, 0, 1) e
θ
= (cos θ cos ϕ, cos θ sin ϕ, sin θ)
2 Curves and Line
2.1 Parametrised curves, lengths and arc length
There are many ways we can described a curve. We can, say, describe it by
a equation that the points on the curve satisfy. For example, a circle can be
described by
x
2
+
y
2
= 1. However, this is not a good way to do so, as it is
rather difficult to work with. It is also often difficult to find a closed form like
this for a curve.
Instead, we can imagine the curve to be specified by a particle moving along
the path. So it is represented by a function
x
:
R R
n
, and the curve itself is
the image of the function. This is known as a parametrisation of a curve. In
addition to simplified notation, this also has the benefit of giving the curve an
orientation.
Definition
(Parametrisation of curve)
.
Given a curve
C
in
R
n
, a parametrisation
of it is a continuous and invertible function
r
:
D R
n
for some
D R
whose
image is C.
r
0
(
u
) is a vector tangent to the curve at each point. A parametrization is
regular if r
0
(u) 6= 0 for all u.
Clearly, a curve can have many different parametrizations.
Example. The curve
1
4
x
2
+ y
2
= 1, y 0, z = 3.
can be parametrised by 2 cos u
ˆ
i + sin u
ˆ
j + 3
ˆ
k
If we change
u
(and hence
r
) by a small amount, then the distance
|δr|
is
roughly equal to the change in arclength
δs
. So
δs
=
|δr|
+
o
(
δr
). Then we have
Proposition. Let s denote the arclength of a curve r(u). Then
ds
du
= ±
dr
du
= ±|r
0
(u)|
with the sign depending on whether it is in the direction of increasing or decreasing
arclength.
Example. Consider a helix described by r(u) = (3 cos u, 3 sin u, 4u). Then
r
0
(u) = (3 sin u, 3 cos u, 4)
ds
du
= |r
0
(u)| =
p
3
2
+ 4
2
= 5
So s = 5u. i.e. the arclength from r(0) and r(u) is s = 5u.
We can change parametrisation of
r
by taking an invertible smooth function
u 7→ ˜u
, and have a new parametrization
r
(
˜u
) =
r
(
˜u
(
u
)). Then by the chain rule,
dr
du
=
dr
d˜u
×
d˜u
du
dr
d˜u
=
dr
du
/
d˜u
du
It is often convenient to use the arclength
s
as the parameter. Then the tangent
vector will always have unit length since the proposition above yields
|r
0
(s)| =
ds
ds
= 1.
We call d
s
the scalar line element, which will be used when we consider integrals.
Definition (Scalar line element). The scalar line element of C is ds.
Proposition. ds = ±|r
0
(u)|du
2.2 Line integrals of vector fields
Definition
(Line integral)
.
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
C
F(r) · dr =
Z
β
α
F(r(u)) · r
0
(u) du.
We say d
r
=
r
0
(
u
)d
u
is the line element on
C
. 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.
For example, we may be moving a particle from
a
to
b
along a curve
C
under a force field
F
. Then we may divide the curve into many small segments
δr
. Then for each segment, the force experienced is
F
(
r
) and the work done is
F(r) · δr. Then the total work done across the curve is
W =
Z
C
F(r) · dr.
Example.
Take
F
(
r
) = (
xe
y
, z
2
, xy
) and we want to find the line integral from
a = (0, 0, 0) to b = (1, 1, 1).
a
b
C
1
C
2
We first integrate along the curve
C
1
:
r
(
u
) = (
u, u
2
, u
3
). Then
r
0
(
u
) =
(1, 2u, 3u
2
), and F(r(u)) = (ue
u
2
, u
6
, u
3
). So
Z
C
1
F · dr =
Z
1
0
F · r
0
(u) du
=
Z
1
0
ue
u
2
+ 2u
7
+ 3u
5
du
=
e
2
1
2
+
1
4
+
1
2
=
e
2
+
1
4
Now we try to integrate along another curve
C
2
:
r
(
t
) = (
t, t, t
). So
r
0
(
t
) =
(1, 1, 1).
Z
C
2
F · dr =
Z
F · r
0
(t)dt
=
Z
1
0
te
t
+ 2t
2
dt
=
5
3
.
We see that the line integral depends on the curve C in general, not just a, b.
We can also use the arclength
s
as the parameter. Since d
r
=
t
d
s
, with
t
being the unit tangent vector, we have
Z
C
F · dr =
Z
C
F · t ds.
Note that we do not necessarily have to integrate
F ·t
with respect to
s
. We can
also integrate a scalar function as a function of
s
,
R
C
f
(
s
) d
s
. By convention,
this is calculated in the direction of increasing s. In particular, we have
Z
C
1 ds = length of C.
Definition
(Closed curve)
.
A closed curve is a curve with the same start and
end point. The line integral along a closed curve is (sometimes) written as
H
and is (sometimes) called the circulation of F around C.
Sometimes we are not that lucky and our curve is not smooth. For example,
the graph of an absolute value function is not smooth. However, often we can
break it apart into many smaller segments, each of which is smooth. Alternatively,
we can write the curve as a sum of smooth curves. We call these piecewise smooth
curves.
Definition
(Piecewise smooth curve)
.
A piecewise smooth curve is a curve
C
=
C
1
+
C
2
+
···
+
C
n
with all
C
i
smooth with regular parametrisations. The
line integral over a piecewise smooth C is
Z
C
F · dr =
Z
C
1
F · dr +
Z
C
2
F · dr + ··· +
Z
C
n
F · dr.
Example.
Take the example above, and let
C
3
=
C
2
. Then
C
=
C
1
+
C
3
is
piecewise smooth but not smooth. Then
I
C
F · dr =
Z
C
1
F · dr +
Z
C
3
F · dr
=
e
2
+
1
4
5
3
=
17
12
+
e
2
.
a
b
C
1
C
3
2.3 Gradients and Differentials
Recall that the line integral depends on the actual curve taken, and not just the
end points. However, for some nice functions, the integral does depend on the
end points only.
Theorem. If F = f (r), then
Z
C
F · dr = f(b) f(a),
where b and a are the end points of the curve.
In particular, the line integral does not depend on the curve, but the end
points only. This is the vector counterpart of the fundamental theorem of
calculus. A special case is when C is a closed curve, then
H
C
F · dr = 0.
Proof.
Let
r
(
u
) be any parametrization of the curve, and suppose
a
=
r
(
α
),
b = r(β). Then
Z
C
F · dr =
Z
C
f · dr =
Z
f ·
dr
du
du.
So by the chain rule, this is equal to
Z
β
α
d
du
(f(r(u))) du = [f(r(u))]
β
α
= f(b) f (a).
Definition
(Conservative vector field)
.
If
F
=
f
for some
f
, the
F
is called a
conservative vector field.
The name conservative comes from mechanics, where conservative vector
fields represent conservative forces that conserve energy. This is since if the
force is conservative, then the integral (i.e. work done) about a closed curve is 0,
which means that we cannot gain energy after travelling around the loop.
It is convenient to treat differentials
F ·
d
r
=
F
i
d
x
i
as if they were objects
by themselves, which we can integrate along curves if we feel like doing so.
Then we can define
Definition
(Exact differential)
.
A differential
F ·
d
r
is exact if there is an
f
such that F = f. Then
df = f · dr =
f
x
i
dx
i
.
To test if this holds, we can use the necessary condition
Proposition. If F = f for some f , then
F
i
x
j
=
F
j
x
i
.
This is because both are equal to
2
f/∂x
i
x
j
.
For an exact differential, the result from the previous section reads
Z
C
F · dr =
Z
C
df = f(b) f(a).
Differentials can be manipulated using (for constant λ, µ):
Proposition.
d(λf + µg) = λdf + µdg
d(fg) = (df)g + f(dg)
Using these, it may be possible to find f by inspection.
Example. Consider
Z
C
3x
2
y sin z dx + x
3
sin z dy + x
3
y cos z dz.
We see that if we integrate the first term with respect to
x
, we obtain
x
3
y sin z
.
We obtain the same thing if we integrate the second and third term. So this is
equal to
Z
C
d(x
3
y sin z) = [x
3
y sin z]
b
a
.
2.4 Work and potential energy
Definition
(Work and potential energy)
.
If
F
(
r
) is a force, then
R
C
F ·
d
r
is
the work done by the force along the curve
C
. It is the limit of a sum of terms
F(r) · δr, i.e. the force along the direction of δr.
Consider a point particle moving under
F
(
r
) according to Newton’s second
law: F(r) = m
¨
r.
Since the kinetic energy is defined as
T (t) =
1
2
m
˙
r
2
,
the rate of change of energy is
d
dt
T (t) = m
˙
r ·
¨
r = F ·
˙
r.
Suppose the path of particle is a curve C from a = r(α) to b = r(β), Then
T (β) T (α) =
Z
β
α
dT
dt
dt =
Z
β
α
F ·
˙
r dt =
Z
C
F · dr.
So the work done on the particle is the change in kinetic energy.
Definition
(Potential energy)
.
Given a conservative force
F
=
−∇V
,
V
(
x
) is
the potential energy. Then
Z
C
F · dr = V (a) V (b).
Therefore, for a conservative force, we have
F
=
V
, where
V
(
r
) is the
potential energy.
So the work done (gain in kinetic energy) is the loss in potential energy. So
the total energy T + V is conserved, i.e. constant during motion.
We see that energy is conserved for conservative forces. In fact, the converse
is true the energy is conserved only for conservative forces.
3 Integration in R
2
and R
3
3.1 Integrals over subsets of R
2
Definition
(Surface integral)
.
Let
D R
2
. Let
r
= (
x, y
) be in Cartesian
coordinates. We can approximate
D
by
N
disjoint subsets of simple shapes, e.g.
triangles, parallelograms. These shapes are labelled by I and have areas δA
i
.
x
y
D
To integrate a function
f
over
D
, we would like to take the sum
P
f
(
r
i
)
δA
i
,
and take the limit as
δA
i
0. But we need a condition stronger than simply
δA
i
0. We won’t want the areas to grow into arbitrarily long yet thin strips
whose area decreases to 0. So we say that we find an
`
such that each area can
be contained in a disc of diameter `.
Then we take the limit as
`
0,
N
, and the union of the pieces tends
to D. For a function f(r), we define the surface integral as
Z
D
f(r) dA = lim
`0
X
I
f(r
i
)δA
i
.
where
r
i
is some point within each subset
A
i
. The integral exists if the limit
is well-defined (i.e. the same regardless of what
A
i
and
r
i
we choose before we
take the limit) and exists.
If we take f = 1, then the surface integral is the area of D.
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.
The definition allows us to take the
δA
i
to be any weird shape we want.
However, the sensible thing is clearly to take A
i
to be rectangles.
We choose the small sets in the definition to be rectangles, each of size
δA
I
=
δy
. We sum over subsets in a narrow horizontal strip of height
δy
with
y
and
δy
held constant. Take the limit as
δx
0. We get a contribution
δy
R
x
y
f(y, x) dx with range x
y
{x : (x, y) D}.
x
y
δy
y
x
y
Y
D
We sum over all such strips and take δy 0, giving
Proposition.
Z
D
f(x, y) dA =
Z
Y
Z
x
y
f(x, y) dx
!
dy.
with x
y
ranging over {x : (x, y) D}.
Note that the range of the inner integral is given by a set
x
y
. This can be an
interval, or many disconnected intervals, x
y
= [a
1
, b
1
] [a
2
, b
2
]. In this case,
Z
x
y
f(x) dx =
Z
b
1
a
1
f(x) dx +
Z
b
2
a
2
f(x) dx.
This is useful if we want to integrate over a concave area and we have disconnected
vertical strips.
x
y
We could also do it the other way round, integrating over
y
first, and come up
with the result
Z
D
f(x, y) dA =
Z
X
Z
y
x
f(x, y) dy
dx.
Theorem
(Fubini’s theorem)
.
If
f
is a continuous function and
D
is a compact
(i.e. closed and bounded) subset of R
2
, then
ZZ
f dx dy =
ZZ
f dy dx.
While we have rather strict conditions for this theorem, it actually holds in many
more cases, but those situations have to be checked manually.
Definition (Area element). The area element is dA.
Proposition. dA = dx dy in Cartesian coordinates.
Example.
We integrate over the triangle bounded by (0
,
0)
,
(2
,
0) and (0
,
1).
We want to integrate the function f(x, y) = x
2
y over the area. So
Z
D
f(xy) dA =
Z
1
0
Z
22y
0
x
2
y dx
dy
=
Z
1
0
y
x
3
3
22y
0
dy
=
8
3
Z
1
0
y(1 y)
3
dy
=
2
15
We can integrate it the other way round:
Z
D
x
2
y dA =
Z
2
0
Z
1x/2
0
x
2
y dy dx
=
Z
2
0
x
2
1
2
y
2
1x/2
0
dx
=
Z
2
0
x
2
2
1
x
2
2
dx
=
2
15
Since it doesn’t matter whether we integrate
x
first or
y
first, if we find it
difficult to integrate one way, we can try doing it the other way and see if it is
easier.
While this integral is tedious in general, there is a special case where it is
substantially easier.
Definition
(Separable function)
.
A function
f
(
x, y
) is separable if it can be
written as f (x, y) = h(y)g(x).
Proposition.
Take separable
f
(
x, y
) =
h
(
y
)
g
(
x
) and
D
be a rectangle
{
(
x, y
) :
a x b, c y d}. Then
Z
D
f(x, y) dx dy =
Z
b
a
g(x) dx
!
Z
d
c
h(y) dy
!
3.2 Change of variables for an integral in R
2
Proposition.
Suppose we have a change of variables (
x, y
)
(
u, v
) that is
smooth and invertible, with regions D, D
0
in one-to-one correspondence. Then
Z
D
f(x, y) dx dy =
Z
D
0
f(x(u, v), y(u, v))|J| du dv,
where
J =
(x, y)
(u, v)
=
x
u
x
v
y
u
y
v
is the Jacobian. In other words,
dx dy = |J| du dv.
Proof.
Since we are writing (
x
(
u, v
)
, y
(
u, v
)), we are actually transforming from
(u, v) to (x, y) and not the other way round.
Suppose we start with an area
δA
0
=
δv
in the (
u, v
) plane. Then by
Taylors’ theorem, we have
δx = x(u + δu, v + δv) x(u, v)
x
u
δu +
x
v
δv.
We have a similar expression for δy and we obtain
δx
δy
x
u
x
v
y
u
y
v
δu
δv
Recall from Vectors and Matrices that the determinant of the matrix is how
much it scales up an area. So the area formed by
δx
and
δy
is
|J|
times the area
formed by δu and δv. Hence
dx dy = |J| du dv.
Example. We transform from (x, y) to (ρ, ϕ) with
x = ρ cos ϕ
y = ρ sin ϕ
We have previously calculated that |J| = ρ. So
dA = ρ dρ dϕ.
Suppose we want to integrate a function over a quarter area D of radius R.
x
y
D
Let the function to be integrated be
f
=
exp
(
(
x
2
+
y
2
)
/
2) =
exp
(
ρ
2
/
2). Then
Z
f dA =
Z
fρ dρ dϕ
=
Z
R
ρ=0
Z
π/2
ϕ=0
e
ρ
2
/2
ρ dϕ
!
δρ
Note that in polar coordinates, we are integrating over a rectangle and the
function is separable. So this is equal to
=
h
e
ρ
2
/2
i
R
0
[ϕ]
π/2
0
=
π
2
1 e
R
2
/2
. ()
Note that the integral exists as R .
Now we take the case of x, y and consider the original integral.
Z
D
f dA =
Z
x=0
Z
y=0
e
(x
2
+y
2
)/2
dx dy
=
Z
0
e
x
2
/2
dx
Z
0
e
y
2
/2
dy
=
π
2
where the last line is from (*). So each of the two integrals must be
p
π/2, i.e.
Z
0
e
x
2
/2
dx =
r
π
2
.
3.3 Generalization to R
3
We will do exactly the same thing as we just did, but with one more dimension:
Definition
(Volume integral)
.
Consider a volume
V R
3
with position vector
r
= (
x, y, z
). We approximate
V
by
N
small disjoint subsets of some simple
shape (e.g. cuboids) labelled by
I
, volume
δV
I
, contained within a solid sphere
of diameter `.
Assume that as
`
0 and
N
, the union of the small subsets tend to
V . Then
Z
V
f(r) dV = lim
`0
X
I
f(r
I
)δV
I
,
where r
I
is any chosen point in each small subset.
To evaluate this, we can take
δV
I
=
δz
, and take
δx
0,
δy
0 and
δz in some order. For example,
Z
V
f(r) dv =
Z
D
Z
Z
xy
f(x, y, z) dz
!
dx dy.
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).
Alternatively, we can take the area integral first, and have
Z
V
f(r) dV =
Z
z
Z
D
Z
f(x, y, z) dx dy
dz.
Again, if we take f = 1, then we obtain the volume of V .
Often,
f
(
r
) is the density of some quantity, and is usually denoted by
ρ
. For
example, we might have mass density, charge density, or probability density.
ρ
(
r
)
δV
is then the amount of quantity in a small volume
δV
at
r
. Then
R
V
ρ(r) dV is the total amount of quantity in V .
Definition (Volume element). The volume element is dV .
Proposition. dV = dx dy dz.
We can change variables by some smooth, invertible transformation (
x, y, z
)
7→
(u, v, w). Then
Proposition.
Z
V
f dx dy dz =
Z
V
f|J| du dv dw,
with
J =
(x, y, z)
(u, v, w)
=
x
u
x
v
x
w
y
u
y
v
y
w
z
u
z
v
z
w
Proposition. In cylindrical coordinates,
dV = ρ dρ dϕ dz.
In spherical coordinates
dV = r
2
sin θ dr dθ dϕ.
Proof. Loads of algebra.
Example.
Suppose
f
(
r
) is spherically symmetric and
V
is a sphere of radius
a
centered on the origin. Then
Z
V
f dV =
Z
a
r=0
Z
π
θ=0
Z
2π
ϕ=0
f(r)r
2
sin θ dr dθ dϕ
=
Z
a
0
dr
Z
π
0
dθ
Z
2π
0
dϕ r
2
f(r) sin θ
=
Z
a
0
r
2
f(r)dr
h
cos θ
i
π
0
h
ϕ
i
2π
0
= 4π
Z
a
0
f(r)r
2
dr.
where we separated the integral into three parts as in the area integrals.
Note that in the second line, we rewrote the integrals to write the differentials
next to the integral sign. This is simply a different notation that saves us from
writing r = 0 etc. in the limits of the integrals.
This is a useful general result. We understand it as the sum of spherical
shells of thickness δr and volume 4πr
2
δr.
If we take
f
= 1, then we have the familiar result that the volume of a sphere
is
4
3
πa
3
.
Example.
Consider a volume within a sphere of radius
a
with a cylinder of
radius b (b < a) removed. The region is defined as
x
2
+ y
2
+ z
2
a
2
x
2
+ y
2
b
2
.
a
b
We use cylindrical coordinates. The second criteria gives
b ρ a.
For the x
2
+ y
2
+ z
2
a
2
criterion, we have
p
a
2
ρ
2
z
p
a
2
ρ
2
.
So the volume is
Z
V
dV =
Z
a
b
dρ
Z
2π
0
dϕ
Z
a
2
ρ
2
a
2
ρ
2
dz ρ
= 2π
Z
a
b
2ρ
p
a
2
ρ
2
dρ
= 2π
2
3
(a
2
ρ
2
)
3/2
a
b
=
4
3
π(a
2
b
2
)
3/2
.
Example.
Suppose the density of electric charge is
ρ
(
r
) =
ρ
0
z
a
in a hemisphere
H of radius a, with z 0. What is the total charge of H?
We use spherical polars. So
r a, 0 ϕ 2π, 0 θ
π
2
.
We have
ρ(r) =
ρ
0
a
r cos θ.
The total charge Q in H is
Z
H
ρ dV =
Z
a
0
dr
Z
π/2
0
dθ
Z
2π
0
dϕ
ρ
0
a
r cos θr
2
sin θ
=
ρ
0
a
Z
a
0
r
3
dr
Z
π/2
0
sin θ cos θ dθ
Z
2π
0
dϕ
=
ρ
0
a
r
4
4
a
0
1
2
sin
2
θ
π/2
0
[ϕ]
2π
0
=
ρ
0
πa
3
4
.
3.4 Further generalizations
Integration in R
n
Similar to the above,
R
D
f
(
x
1
, x
2
, ···x
n
) d
x
1
d
x
2
···
d
x
n
is simply the integra-
tion over an n-dimensional volume. The change of variable formula is
Proposition.
Z
D
f(x
1
, x
2
, ···x
n
) dx
1
dx
2
··· dx
n
=
Z
D
0
f({x
i
(u)})|J| du
1
du
2
··· du
n
.
Change of variables for n = 1
In the
n
= 1 case, the Jacobian is
dx
du
. However, we use the following formula for
change of variables:
Z
D
f(x) dx =
Z
D
0
f(x(u))
dx
du
du.
We introduce the modulus because of our natural convention about integrating
over
D
and
D
0
. If
D
= [
a, b
] with
a < b
, we write
R
b
a
. But if
a 7→ α
and
b 7→ β
,
but
α > β
, we would like to write
R
α
β
instead, so we introduce the modulus in
the 1D case.
To show that the modulus is the right thing to do, we check case by case: If
a < b and α < β, then
dx
du
is positive, and we have, as expected
Z
b
a
f(x) dx =
Z
β
α
f(u)
dx
du
du.
If α > β, then
dx
du
is negative. So
Z
b
a
f(x) dx =
Z
β
α
f(u)
dx
du
du =
Z
α
β
f(u)
dx
du
du.
By taking the absolute value of
dx
du
, we ensure that we always have the numerically
smaller bound as the lower bound.
This is not easily generalized to higher dimensions, so we don’t employ the
same trick in other cases.
Vector-valued integrals
We can define
R
V
F
(
r
) d
V
in a similar way to
R
V
f
(
r
) d
V
as the limit of a sum over
small contributions of volume. In practice, we integrate them componentwise. If
F(r) = F
i
(r)e
i
,
then
Z
V
F(r) dV =
Z
V
(F
i
(r) dV )e
i
.
For example, if a mass has density ρ(r), then its mass is
M =
Z
V
ρ(r) dV
and its center of mass is
R =
1
M
Z
V
rρ(r) dV.
Example.
Consider a solid hemisphere
H
with
r a
,
z
0 with uniform
density ρ. The mass is
M =
Z
H
ρ dV =
2
3
πa
3
ρ.
Now suppose that
R
= (
X, Y, Z
). By symmetry, we expect
X
=
Y
= 0. We can
find this formally by
X =
1
M
Z
H
dV
=
ρ
M
Z
a
0
Z
π/2
0
Z
2π
0
xr
2
sin θ dϕ dθ dr
=
ρ
M
Z
a
0
r
3
dr ×
Z
π/2
0
sin
2
θ dθ ×
Z
2π
0
cos ϕ dϕ
= 0
as expected. Note that it evaluates to 0 because the integral of
cos
from 0 to 2
π
is 0. Similarly, we obtain Y = 0.
Finally, we find Z.
Z =
ρ
M
Z
a
0
r
3
dr
Z
π/2
0
sin θ cos θ dθ
Z
2π
0
dϕ
=
r
M
a
4
4
1
2
sin
2
θ
π/2
0
2π
=
3a
8
.
So R = (0, 0, 3a/8).
4 Surfaces and surface integrals
4.1 Surfaces and Normal
So far, we have learnt how to do calculus with regions of the plane or space.
What we would like to do now is to study surfaces in
R
3
. The first thing to
figure out is how to specify surfaces. One way to specify a surface is to use an
equation. We let
f
be a smooth function on
R
3
, and
c
be a constant. Then
f
(
r
) =
c
defines a smooth surface (e.g.
x
2
+
y
2
+
z
2
= 1 denotes the unit sphere).
Now consider any curve
r
(
u
) on
S
. Then by the chain rule, if we differentiate
f(r) = c with respect to u, we obtain
d
du
[f(r(u))] = f ·
dr
du
= 0.
This means that
f
is always perpendicular to
dr
du
. Since
dr
du
is the tangent to
the curve,
f
is perpendicular to the tangent. Since this is true for any curve
r(u), f is perpendicular to any tangent of the surface. Therefore
Proposition. f is the normal to the surface f(r) = c.
Example.
(i)
Take the sphere
f
(
r
) =
x
2
+
y
2
+
z
2
=
c
for
c >
0. Then
f
= 2(
x, y, z
) =
2r, which is clearly normal to the sphere.
(ii)
Take
f
(
r
) =
x
2
+
y
2
z
2
=
c
, which is a hyperboloid. Then
f
=
2(x, y, z).
In the special case where
c
= 0, we have a double cone, with a singular apex
0. Here f = 0, and we cannot find a meaningful direction of normal.
Definition
(Boundary)
.
A surface
S
can be defined to have a boundary
S
consisting of a piecewise smooth curve. If we define
S
as in the above examples
but with the additional restriction
z
0, then
S
is the circle
x
2
+
y
2
=
c
,
z
= 0.
A surface is bounded if it can be contained in a solid sphere, unbounded
otherwise. A bounded surface with no boundary is called closed (e.g. sphere).
Example.
The boundary of a hemisphere is a circle (drawn in red).
Definition
(Orientable surface)
.
At each point, there is a unit normal
n
that’s
unique up to a sign.
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.
Most surfaces we encounter are orientable. For example, for a sphere, we can
declare that the normal should always point outwards. A notable example of a
non-orientable surface is the obius strip (or Klein bottle).
For simple cases, we can describe the orientation as “inward” and “outward”.
4.2 Parametrized surfaces and area
However, specifying a surface by an equation
f
(
r
) =
c
is often not too helpful.
What we would like is to put some coordinate system onto the surface, so that
we can label each point by a pair of numbers (
u, v
), just like how we label points
in the x, y-plane by (x, y). We write r(u, v) for the point labelled by (u, v).
Example. Let S be part of a sphere of radius a with 0 θ α.
α
We can then label the points on the spheres by the angles θ, ϕ, with
r(θ, ϕ) = (a cos ϕ sin θ, a sin θ sin ϕ, a cos θ) = ae
r
.
We restrict the values of
θ, ϕ
by 0
θ α
, 0
ϕ
2
π
, so that each point is
only covered once.
Note that to specify a surface, in addition to the function
r
, we also have
to specify what values of (
u, v
) we are allowed to take. This corresponds to a
region
D
of allowed values of
u
and
v
. When we do integrals with these surfaces,
these will become the bounds of integration.
When we have such a parametrization
r
, we would want to make sure this
indeed gives us a two-dimensional surface. For example, the following two
parametrizations would both be bad:
r(u, v) = u, r(u, v) = u + v.
The idea is that
r
has to depend on both
u
and
v
, and in “different ways”.
More precisely, when we vary the coordinates (
u, v
), the point
r
will change
accordingly. By the chain rule, this is given by
δr =
r
u
δu +
r
v
δv + o(δu, δv).
Then
r
δu
and
r
v
are tangent vectors to curves on
S
with
v
and
u
constant
respectively. What we want is for them to point in different directions.
Definition
(Regular parametrization)
.
A parametrization is regular if for all
u, v,
r
u
×
r
v
6= 0,
i.e. there are always two independent tangent directions.
The parametrizations we use will all be regular.
Given a surface, how could we, say, find its area? We can use our parametriza-
tion. Suppose points on the surface are given by
r
(
u, v
) for (
u, v
)
D
. If we
want to find the area of D itself, we would simply integrate
Z
D
du dv.
However, we are just using
u
and
v
as arbitrary labels for points in the surface,
and one unit of area in
D
does not correspond to one unit of area in
S
. Instead,
suppose we produce a small rectangle in
D
by changing
u
and
v
by small
δu, δv
.
In
D
, this corresponds to a rectangle with vertices (
u, v
)
,
(
u
+
δu, v
)
,
(
u, v
+
δv
)
,
(
u
+
δu, v
+
δv
), and spans an area
δv
. In the surface
S
, these small
changes
δu, δv
correspond to changes
r
u
δu
and
r
v
δv
, and these span a vector
area of
δS =
r
u
×
r
v
δv = n δS.
Note that the order of u, v gives the choice of the sign of the unit normal.
The actual area is then given by
δS =
r
u
×
r
v
δu δv.
Making these into differentials instead of deltas, we have
Proposition. The vector area element is
dS =
r
u
×
r
v
du dv.
The scalar area element is
dS =
r
u
×
r
v
du dv.
By summing and taking limits, the area of S is
Z
S
dS =
Z
D
r
u
×
r
v
du dv.
Example. Consider again the part of the sphere of radius a with 0 θ α.
α
Then we have
r(θ, ϕ) = (a cos ϕ sin θ, a sin θ sin ϕ, a cos θ) = ae
r
.
So we find
r
θ
= ae
θ
.
Similarly, we have
r
ϕ
= a sin θe
ϕ
.
Then
r
θ
×
r
ϕ
= a
2
sin θ e
r
.
So
dS = a
2
sin θ dθ dϕ.
Our bounds are 0 θ α, 0 ϕ 2π.
Then the area is
Z
2π
0
Z
α
0
a
2
sin θ dθ dϕ = 2πa
2
(1 cos α).
4.3 Surface integral of vector fields
Just computing the area of a surface would be boring. Suppose we have a surface
S
parametrized by
r
(
u, v
), where (
u, v
) takes values in
D
. We would like to ask
how much “stuff” is passing through
S
, where the flow of stuff is given by a
vector field F(r).
We might attempt to use the integral
Z
D
|F| dS.
However, this doesn’t work. 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. Instead, what we should do is to consider the
component of F that is normal to the surface S, i.e. parallel to its normal.
Definition
(Surface integral)
.
The surface integral or
flux
of a vector field
F
(
r
)
over S is defined by
Z
S
F(r) · dS =
Z
S
F(r) · n dS =
Z
D
F(r(u, v)) ·
r
u
×
r
v
du dv.
Intuitively, this is the total amount of
F
passing through
S
. For example, if
F
is the electric field, the flux is the amount of electric field passing through a
surface.
For a given orientation, the integral
R
F·
d
S
is independent of the parametriza-
tion. Changing orientation is equivalent to changing the sign of
n
, which is in
turn equivalent to changing the order of
u
and
v
in the definition of
S
, which is
also equivalent to changing the sign of the flux integral.
Example. Consider a sphere of radius a, r(θ, ϕ). Then
r
θ
= ae
θ
,
r
ϕ
= a sin θe
ϕ
.
The vector area element is
dS = a
2
sin θe
r
dθ dϕ,
taking the outward normal n = e
r
= r/a.
Suppose we want to calculate the fluid flux through the surface. The velocity
field
u
(
r
) of a fluid gives the motion of a small volume of fluid
r
. Assume that
u
depends smoothly on
r
(and
t
). For any small area
δS
, on a surface
S
, the
volume of fluid crossing it in time δt is u · δS δt.
δS
u δt
n
So the amount of flow of u over at time δt through S is
δt
Z
S
u · dS.
So
R
S
u · dS is the rate of volume crossing 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. Then
dS = a
2
sin θn dϕ dθ,
with
n =
r
a
=
1
a
(x, y, z).
So
n · u =
1
a
(x
2
+ z
2
) = a(sin
2
θ cos
2
ϕ + cos
2
θ).
Therefore
Z
S
u · dS =
Z
α
0
Z
2π
0
a
3
sin θ[(cos
2
θ 1) cos
2
ϕ + cos
2
θ] dϕ dθ
=
Z
α
0
a
3
sin θ[π(cos
2
θ 1) + 2π cos
2
θ] dθ
=
Z
α
0
a
3
π(3 cos
3
θ 1) sin θ dθ
= πa
3
[cosθ cos
3
θ]
α
0
= πa
3
cos α sin
2
α.
What happens when we change parametrization? Let r(u, v) and r(˜u, ˜v) be
two regular parametrizations for the surface. By the chain rule,
r
u
=
r
˜u
˜u
u
+
r
˜v
˜v
u
r
v
=
r
˜u
˜u
v
+
r
˜v
˜v
v
So
r
u
×
r
v
=
(˜u, ˜v)
(u, v)
r
˜u
×
r
˜v
where
(˜u,˜v)
(u,v)
is the Jacobian.
Since
d˜u d˜v =
(˜u, ˜v)
(u, v)
du dv,
We recover the formula
dS =
r
u
×
r
v
du dv =
r
˜u
×
r
˜v
d˜u d˜v.
Similarly, we have
dS =
r
u
×
r
v
du dv =
r
˜u
×
r
˜v
d˜u d˜v.
provided (u, v) and (˜u, ˜v) have the same orientation.
4.4 Change of variables in R
2
and R
3
revisited
In this section, we derive our change of variable formulae in a slightly different
way.
Change of variable formula in R
2
We first derive the 2D change of variable formula from the 3D surface integral
formula.
Consider a subset
S
of the plane
R
2
parametrized by
r
(
x
(
u, v
)
, y
(
u, v
)). We
can embed it to R
3
as r(x(u, v), y(u, v), 0). Then
r
u
×
r
v
= (0, 0, J),
with J being the Jacobian. Therefore
Z
S
f(r) dS =
Z
D
f(r(u, v))
r
u
×
r
v
du dv =
Z
D
f(r(u, v))|J| du dv,
and we recover the formula for changing variables in R
2
.
Change of variable formula in R
3
In R
3
, suppose we have a volume parametrised by r(u, v, w). Then
δr =
r
u
δu +
r
v
δv +
r
w
δw + o(δu, δv, δw).
Then the cuboid
δu, δv, δw
in
u, v, w
space is mapped to a parallelopiped of
volume
δV =
r
u
δu ·
r
v
δv ×
r
w
δw
= |J| δu δv δw.
So dV = |J| du dv dw.
5 Geometry of curves and surfaces
Let
r
(
s
) be a curve parametrized by arclength
s
. Since
t
(
s
) =
dr
ds
is a unit vector,
t · t
= 1. Differentiating yields
t · t
0
= 0. So
t
0
is a normal to the curve if
t
0
6
= 0.
We define the following:
Definition
(Principal normal and curvature)
.
Write
t
0
=
κn
, where
n
is a unit
vector and
κ >
0. Then
n
(
s
) is called the principal normal and
κ
(
s
) is called
the curvature.
Note that we must be differentiating against
s
, not any other parametrization!
If the curve is given in another parametrization, we can either change the
parametrization or use the chain rule.
We take a curve that can Taylor expanded around s = 0. Then
r(s) = r(0) + sr
0
(0) +
1
2
s
2
r
00
(0) + O(s
3
).
We know that r
0
= t and r
00
= t
0
. So we have
r(s) = r(0) + st(0) +
1
2
κ(0)s
2
n + O(s
3
).
How can we interpret
κ
as the curvature? Suppose we want to approximate the
curve near
r
(0) by a circle. We would expect a more “curved” curve would be
approximated by a circle of smaller radius. So
κ
should be inversely proportional
to the radius of the circle. In fact, we will show that
κ
= 1
/a
, where
a
is the
radius of the best-fit circle.
Consider the vector equation for a circle passing through
r
(0) with radius
a
in the plane defined by t and n.
a
r(0)
t
n
θ
Then the equation of the circle is
r = r(0) + a(1 cos θ)n + a sin θt.
We can expand this to obtain
r = r(0) + t +
1
2
θ
2
an + o(θ
3
).
Since the arclength s = , we obtain
r = r(0) + st +
1
2
1
a
s
2
n + O(s
3
).
As promised, κ = 1/a, for a the radius of the circle of best fit.
Definition
(Radius of curvature)
.
The radius of curvature of a curve at a point
r(s) is 1(s).
Since we are in 3D, given
t
(
s
) and
n
(
s
), there is another normal to the curve.
We can add a third normal to generate an orthonormal basis.
Definition (Binormal). The binormal of a curve is b = t × n.
We can define the torsion similar to the curvature, but with the binormal
instead of the tangent.
a
Definition (Torsion). Let b
0
= τn. Then τ is the torsion.
Note that this makes sense, since
b
0
is both perpendicular to
t
and
b
, and
hence must be in the same direction as n. (b
0
= t
0
× n + t × n
0
= t × n
0
, so b
0
is perpendicular to t; and b ·b = 1 b · b
0
= 0. So b
0
is perpendicular to b).
The geometry of the curve is encoded in how this basis (
t, n, b
) changes along
it. This can be specified by two scalar functions of arc length the curvature
κ
(
s
) and the torsion
τ
(
s
) (which determines what the curve looks like to third
order in its Taylor expansions and how the curve lifts out of the t, r plane).
Surfaces and intrinsic geometry*
We can study the geometry of surfaces through curves which lie on them. At a
given point
P
at a surface
S
with normal
n
, consider a plane containing
n
. The
intersection of the plane with the surface yields a curve on the surface through
P . This curve has a curvature κ at P .
If we choose different planes containing
n
, we end up with different curves of
different curvature. Then we define the following:
Definition
(Principal curvature)
.
The principal curvatures of a surface at
P
are
the minimum and maximum possible curvature of a curve through
P
, denoted
κ
min
and κ
max
respectively.
Definition
(Gaussian curvature)
.
The Gaussian curvature of a surface at a
point P is K = κ
min
κ
max
.
Theorem
(Theorema Egregium)
. K
is intrinsic to the surface
S
. It can be
expressed in terms of lengths, angles etc. which are measured entirely on the
surface. So
K
can be defined on an arbitrary surface without embedding it on a
higher dimension surface.
The is the start of intrinsic geometry: if we embed a surface in Euclidean
space, we can determine lengths, angles etc on it. But we don’t have to do so
we can “live in the surface and do geometry in it without an embedding.
For example, we can consider a geodesic triangle
D
on a surface
S
. It consists
of three geodesics: shortest curves between two points.
Let
θ
i
be the interior angles of the triangle (defined by using scalar products
of tangent vectors). Then
Theorem (Gauss-Bonnet theorem).
θ
1
+ θ
2
+ θ
3
= π +
Z
D
K dA,
integrating over the area of the triangle.
a
This was not taught in lectures, but there is a question on the example sheet about the
torsion, so I might as well include it here.
6 Div, Grad, Curl and
6.1 Div, Grad, Curl and
Recalled that
f
is given by (
f
)
i
=
f
x
i
. We can regard this as obtained from
the scalar field f by applying
= e
i
x
i
for cartesian coordinates
x
i
and orthonormal basis
e
i
, where
e
i
are orthonormal
and right-handed, i.e. e
i
× e
j
= ε
ijk
e
k
(it is left handed if e
i
× e
j
= ε
ijk
e
k
).
We can alternatively write this as
=
x
,
y
,
z
.
(nabla or del) is both an operator and a vector. We can apply it to a vector
field F(r) = F
i
(r)e
i
using the scalar or vector product.
Definition (Divergence). The divergence or div of F is
· F =
F
i
x
i
=
F
1
x
1
+
F
2
x
2
+
F
3
x
3
.
Definition (Curl). The curl of F is
× F = ε
ijk
F
k
x
j
e
i
=
e
1
e
2
e
3
x
y
z
F
x
F
y
F
z
Example. Let F = (xe
z
, y
2
sin x, xyz). Then
· F =
x
xe
z
+
y
y
2
sin x +
z
xyz = e
z
+ 2y sin x + xy.
and
× F =
ˆ
i
y
(xyz)
z
(y
2
sin x)
+
ˆ
j
z
(xe
z
) +
x
(xyz)
+
ˆ
k
x
(y
2
sin x)
y
(xe
z
)
= (xz, xe
z
yz, y
2
cos x).
Note that is an operator, so ordering is important. For example,
F · = F
i
x
i
is a scalar differential operator, and
F × = e
k
ε
ijk
F
i
x
j
is a vector differential operator.
Proposition.
Let
f, g
be scalar functions,
F, G
be vector functions, and
µ, λ
be constants. Then
(λf + µg) = λf + µg
· (λF + µG) = λ · F + µ · G
× (λF + µG) = λ × F + µ × G.
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.
Example.
Consider
r
α
with
r
=
|r|
. We know that
r
=
x
i
e
i
. So
r
2
=
x
i
x
i
.
Therefore
2r
r
x
j
= 2x
j
,
or
r
x
i
=
x
i
r
.
So
r
α
= e
i
x
i
(r
α
) = e
i
αr
α1
r
x
i
= αr
α2
r.
Also,
· r =
x
i
x
i
= 3.
and
× r = e
k
ε
ijk
x
j
x
i
= 0.
Proposition. We have the following Leibnitz properties:
(fg) = (f)g + f(g)
· (fF) = (f ) · F + f ( · F)
× (fF) = (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.
There is absolutely no point in memorizing these (at least the last three).
They can be derived when needed via suffix notation.
Example.
· (r
α
r) = (r
α
)r + r
α
· r
= (αr
α2
r) · r + r
α
(3)
= (α + 3)r
α
× (r
α
r) = ((r
α
)) × r + r
α
( × r)
= αr
α2
r × r
= 0
6.2 Second-order derivatives
We have
Proposition.
× (f) = 0
· ( × F) = 0
Proof. Expand out using suffix notation, noting that
ε
ijk
2
f
x
i
x
j
= 0.
since if, say, k = 3, then
ε
ijk
2
f
x
i
x
j
=
2
f
x
1
x
2
2
f
x
2
x
1
= 0.
The converse of each result holds for fields defined in all of R
3
:
Proposition. If F is defined in all of R
3
, then
× F = 0 F = f
for some f .
Definition
(Conservative/irrotational field and scalar potential)
.
If
F
=
f
,
then f is the scalar potential. We say F is conservative or irrotational.
Similarly,
Proposition.
If
H
is defined over all of
R
3
and
· H
= 0, then
H
=
× A
for some A.
Definition
(Solenoidal field and vector potential)
.
If
H
=
× A
,
A
is the
vector potential and H is said to be solenoidal.
Not that is is true only if F or H is defined on all of R
3
.
Definition (Laplacian operator). The Laplacian operator is defined by
2
= · =
2
x
i
x
i
=
2
x
2
1
+
2
x
2
2
+
2
x
3
3
.
This operation is defined on both scalar and vector fields on a scalar field,
2
f = · (f),
whereas on a vector field,
2
A = ( · A) × ( × A).
7 Integral theorems
7.1 Statement and examples
There are three big integral theorems, known as Green’s theorem, Stoke’s theorem
and Gauss’ theorem. There are all generalizations of the fundamental theorem of
calculus in some sense. In particular, they all say that an
n
dimensional integral
of a derivative is equivalent to an
n
1 dimensional integral of the original
function.
We will first state all three theorems with some simple applications. In the
next section, we will see that the three integral theorems are so closely related
that it’s easiest to show their equivalence first, and then prove just one of them.
7.1.1 Green’s theorem (in the plane)
Theorem
(Green’s theorem)
.
For smooth functions
P
(
x, y
),
Q
(
x, y
) and
A
a
bounded region in the (x, y) plane with boundary A = C,
Z
A
Q
x
P
y
dA =
Z
C
(P dx + Q dy).
Here
C
is assumed to be piecewise smooth, non-intersecting closed curve, tra-
versed anti-clockwise.
Example.
Let
Q
=
xy
2
and
P
=
x
2
y
. If
C
is the parabola
y
2
= 4
ax
and the
line x = a, both with 2a y 2a, then Green’s theorem says
Z
A
(y
2
x
2
) dA =
Z
C
x
2
dx + xy
2
dy.
From example sheet 1, each side gives
104
105
a
4
.
Example. Let A be a rectangle confined by 0 x a and 0 y b.
x
y
a
b
A
Then Green’s theorem follows directly from the fundamental theorem of calculus
in 1D. We first consider the first term of Green’s theorem:
Z
P
y
dA =
Z
a
0
Z
b
0
P
y
dy dx
=
Z
a
0
[P (x, b) + P (x, 0)] dx
=
Z
C
P dx
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 d
y
term means that the integral is nil when integrating the
vertical parts.
Similarly,
Z
A
Q
x
dA =
Z
C
Q dy.
Combining them gives Green’s theorem.
Green’s theorem also holds for a bounded region
A
, where the boundary
A
consists of disconnected components (each piecewise smooth, non-intersecting
and closed) with anti-clockwise orientation on the exterior, and clockwise on the
interior boundary, e.g.
The orientation of the curve comes from imagining the surface as:
and take the limit as the gap shrinks to 0.
7.1.2 Stokes’ theorem
Theorem (Stokes’ theorem). For a smooth vector field F(r),
Z
S
× F · dS =
Z
S
F · dr,
where
S
is a smooth, bounded surface and
S
is a piecewise smooth boundary
of S.
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.
It also holds if
S
is a collection of disconnected piecewise smooth closed
curves, with the orientation determined in the same way as Green’s theorem.
Example.
Let
S
be the section of a sphere of radius
a
with 0
θ α
. In
spherical coordinates,
dS = a
2
sin θe
r
dθ dϕ.
Let F = (0, xz, 0). Then × F = (x, 0, z). We have previously shown that
Z
S
× F · dS = πa
3
cos α sin
2
α.
Our boundary C is
r(ϕ) = a(sin α cos ϕ, sin α sin ϕ, cos α).
The right hand side of Stokes’ is
Z
C
F · dr =
Z
2π
0
a sin α cos ϕ
| {z }
x
a cos α
| {z }
z
a sin α cos ϕ dϕ
| {z }
dy
= a
3
sin
2
α cos α
Z
2π
0
cos
2
ϕ dϕ
= πa
3
sin
2
α cos α.
So they agree.
7.1.3 Divergence/Gauss theorem
Theorem (Divergence/Gauss theorem). For a smooth vector field F(r),
Z
V
· F dV =
Z
V
F · dS,
where
V
is a bounded volume with boundary
V
, a piecewise smooth, closed
surface, with outward normal n.
Example. Consider a hemisphere.
S
2
S
1
V is a solid hemisphere
x
2
+ y
2
+ z
2
a
2
, z 0,
and V = S
1
+ S
2
, the hemisphere and the disc at the bottom.
Take F = (0, 0, z + a) and ·F = 1. Then
Z
V
· F dV =
2
3
πa
3
,
the volume of the hemisphere.
On S
1
,
dS = n dS =
1
a
(x, y, z) dS.
Then
F · dS =
1
a
z(z + a) dS = cos θa(cos θ + 1) a
2
sin θ dθ dϕ
| {z }
dS
.
Then
Z
S
1
F · dS = a
3
Z
2π
0
dϕ
Z
π/2
0
sin θ(cos
2
θ + cos θ) dθ
= 2πa
3
1
3
cos
3
θ
1
2
cos
2
θ
π/2
0
=
5
3
πa
3
.
On S
2
, dS = n dS = (0, 0, 1) dS. Then F · dS = a dS. So
Z
S
2
F · dS = πa
3
.
So
Z
S
1
F · dS +
Z
S
2
F · dS =
5
3
1
πa
3
=
2
3
πa
3
,
in accordance with Gauss’ theorem.
7.2 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. A sketch of the proof of the 3D version of divergence theorem
will be provided, because it is simply a generalization of the 2D version, except
that the extra dimension makes the notation tedious and difficult to follow.
Proposition. Stokes’ theorem Green’s theorem
Proof.
Stokes’ theorem talks about 3D surfaces and Green’s theorem is about
2D regions. So given a region
A
on the (
x, y
) plane, we pretend that there is a
third dimension and apply Stokes’ theorem to derive Green’s theorem.
Let
A
be a region in the (
x, y
) plane with boundary
C
=
A
, parametrised
by arc length, (x(s), y(s), 0). Then the tangent to C is
t =
dx
ds
,
dy
ds
, 0
.
Given any P (x, y) and Q(x, y), we can consider the vector field
F = (P, Q, 0),
So
× F =
0, 0,
Q
x
P
y
.
Then the left hand side of Stokes is
Z
C
F · dr =
Z
C
F · t ds =
Z
C
P dx + Q dy,
and the right hand side is
Z
A
( × F) ·
ˆ
k dA =
Z
A
Q
x
P
y
dA.
Proposition. Green’s theorem Stokes’ theorem.
Proof.
Green’s theorem describes a 2D region, while Stokes’ theorem describes
a 3D surface
r
(
u, v
). Hence to use Green’s to derive Stokes’ we need find some
2D thing to act on. The natural choice is the parameter space, u, v.
Consider a parametrised surface
S
=
r
(
u, v
) corresponding to the region
A
in
the
u, v
plane. Write the boundary as
A
= (
u
(
t
)
, v
(
t
)). Then
S
=
r
(
u
(
t
)
, v
(
t
)).
We want to prove
Z
S
F · dr =
Z
S
( × F) · dS
given
Z
A
F
u
du + F
v
dv =
Z
A
F
v
u
F
u
v
dA.
Doing some pattern-matching, we want
F · dr = F
u
du + F
v
dv
for some F
u
and F
v
.
By the chain rule, we know that
dr =
r
u
du +
r
v
dv.
So we choose
F
u
= F ·
r
u
, F
v
= F ·
r
v
.
This choice matches the left hand sides of the two equations.
To match the right, recall that
( × F) · dS = ( × F) ·
r
u
×
r
v
du dv.
Therefore, for the right hand sides to match, we want
F
v
u
F
u
v
= ( × F) ·
r
u
×
r
v
. ()
Fortunately, this is true. Unfortunately, the proof involves complicated suffix
notation and summation convention:
F
v
u
=
u
F ·
r
v
=
u
F
i
x
i
v
=
F
i
x
j
x
j
u
x
i
v
+ F
i
x
i
u∂v
.
Similarly,
F
u
u
=
u
F ·
r
u
=
u
F
j
x
j
u
=
F
j
x
i
x
i
v
x
j
u
+ F
i
x
i
u∂v
.
So
F
v
u
F
u
v
=
x
j
u
x
i
v
F
i
x
j
F
j
x
i
.
This is the left hand side of ().
The right hand side of () is
( × F) ·
r
u
×
r
v
= ε
ijk
F
j
x
i
ε
kpq
x
p
u
x
q
v
= (δ
ip
δ
jq
δ
iq
δ
jp
)
F
j
x
i
x
p
u
x
q
v
=
F
j
x
i
F
i
x
j
x
i
u
x
j
v
.
So they match. Therefore, given our choice of
F
u
and
F
v
, Green’s theorem
translates to Stokes’ theorem.
Proposition. Greens theorem 2D divergence theorem.
Proof. The 2D divergence theorem states that
Z
A
( · G) dA =
Z
A
G · n ds.
with an outward normal n.
Write G as (Q, P ). Then
· G =
Q
x
P
y
.
Around the curve
r
(
s
) = (
x
(
s
)
, y
(
s
)),
t
(
s
) = (
x
0
(
s
)
, y
0
(
s
)). Then the normal,
being tangent to
t
, is
n
(
s
) = (
y
0
(
s
)
, x
0
(
s
)) (check that it points outwards!). So
G · n = P
dx
ds
+ Q
dy
ds
.
Then we can expand out the integrals to obtain
Z
C
G · n ds =
Z
C
P dx + Q dy,
and
Z
A
( · G) dA =
Z
A
Q
x
P
y
dA.
Now 2D version of Gauss’ theorem says the two LHS are the equal, and Green’s
theorem says the two RHS are equal. So the result follows.
Proposition. 2D divergence theorem.
Z
A
( · G) dA =
Z
C=A
G · n ds.
Proof.
For the sake of simplicity, we assume that
G
only has a vertical component,
noting that the same proof works for purely horizontal
G
, and an arbitrary
G
is
just a linear combination of the two.
Furthermore, we assume that
A
is a simple, convex shape. A more complicated
shape can be cut into smaller simple regions, and we can apply the simple case
to each of the small regions.
Suppose G = G(x, y)
ˆ
j. Then
· G =
G
y
.
Then
Z
A
· G dA =
Z
X
Z
Y
x
G
y
dy
dx.
Now we divide
A
into an upper and lower part, with boundaries
C
+
=
y
+
(
x
)
and
C
=
y
(
x
) respectively. Since I cannot draw,
A
will be pictured as a circle,
but the proof is valid for any simple convex shape.
C
+
C
dy
Y
x
x
y
We see that the boundary of
Y
x
at any specific
x
is given by
y
(
x
) and
y
+
(
x
).
Hence by the Fundamental theorem of Calculus,
Z
Y
x
G
y
dy =
Z
y
+
(x)
y
(x)
G
y
dy = G(x, y
+
(x)) G(x, y
(x)).
To compute the full area integral, we want to integrate over all
x
. However, the
divergence theorem talks in terms of d
s
, not d
x
. So we need to find some way
to relate d
s
and d
x
. 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. But
θ
is also the angle
between the normal and the vertical. So cos θ = n ·
ˆ
j. Therefore dx =
ˆ
j · n ds.
In particular, G dx = G
ˆ
j · n ds = G · n ds, since G = G
ˆ
j.
However, at
C
,
n
points downwards, so
n ·
ˆ
j
happens to be negative. So,
actually, at C
, dx = G · n ds.
Therefore, our full integral is
Z
A
· G dA =
Z
X
Z
y
x
G
y
dY
dx
=
Z
X
G(x, y
+
(x)) G(x, y
(x)) dx
=
Z
C
+
G · n ds +
Z
C
G · n ds
=
Z
C
G · n ds.
To prove the 3D version, we again consider
F
=
F
(
x, y, z
)
ˆ
k
, a purely vertical
vector field. Then
Z
V
· F dV =
Z
D
Z
Z
xy
F
z
dz
!
dA.
Again, split
S
=
V
into the top and bottom parts
S
+
and
S
(ie the parts
with
ˆ
k · n
0 and
ˆ
k · n <
0), and parametrize by
z
+
(
x, y
) and
z
(
x, y
). Then
the integral becomes
Z
V
· F dV =
Z
D
(F (x, y, z
+
) F (x, y, z
)) dA =
Z
S
F · n dS.
8 Some applications of integral theorems
8.1 Integral expressions for div and curl
We can use these theorems to come up with alternative definitions of the div and
curl. The advantage of these alternative definitions is that they do not require a
choice of coordinate axes. They also better describe how we should interpret div
and curl.
Gauss’ theorem for F in a small volume V containing r
0
gives
Z
V
F · dS =
Z
V
· F dV (· F)(r
0
) vol(V ).
We take the limit as V 0 to obtain
Proposition.
( · F)(r
0
) = lim
diam(V )1
1
vol(V )
Z
V
F · dS,
where the limit is taken over volumes containing the point r
0
.
Similarly, Stokes’ theorem gives, for A a surface containing the point r
0
,
Z
A
F · dr =
Z
A
( × F) · n dA n · ( × F)(r
0
) area(A).
So
Proposition.
n · ( × F)(r
0
) = lim
diam(A)0
1
area(A)
Z
A
F · dr,
where the limit is taken over all surfaces A containing r
0
with normal n.
These are coordinate-independent definitions of div and curl.
Example. Suppose u is a velocity field of fluid flow. Then
Z
S
u · dS
is the rate of which fluid crosses
S
. Taking
V
to be the volume occupied by a
fixed quantity of fluid material, we have
˙
V =
Z
V
u · dS
Then, at r
0
,
· u = lim
V 0
˙
V
V
,
the relative rate of change of volume. For example, if
u
(
r
) =
αr
(ie fluid flowing
out of origin), then · u = 3α, which increases at a constant rate everywhere.
Alternatively, take a planar area A to be a disc of radius a. Then
Z
A
u · dr =
Z
A
u · t ds = 2πa × average of u · t around the circumference.
(
u · t
is the component of
u
which is tangential to the boundary) We define the
quantity
ω =
1
a
× (average of u · t).
This is the local angular velocity of the current. As
a
0,
1
a
, but the
average of
u · t
will also decrease since a smooth field is less “twirly” if you look
closer. So ω tends to some finite value as a 0. We have
Z
A
u · dr = 2πa
2
ω.
Recall that
n · × u = lim
A0
1
πa
2
Z
A
u · dr = 2ω,
ie twice the local angular velocity. For example, if you have a washing machine
rotating at a rate of ω , Then the velocity u = ω × r. Then the curl is
× (ω ×r) = 2ω,
which is twice the angular velocity.
8.2 Conservative fields and scalar products
Definition (Conservative field). A vector field F is conservative if
(i) F = f for some scalar field f; or
(ii)
R
C
F · dr is independent of C, for fixed end points and orientation; or
(iii) × F = 0.
In R
3
, all three formulations are equivalent.
We have previously shown (i) (ii) since
Z
C
F · dr = f(b) f(a).
We have also shown that (i) (iii) since
× (f) = 0.
So we want to show that (iii) (ii) and (ii) (i)
Proposition. If (iii) × F = 0, then (ii)
R
C
F · dr is independent of C.
Proof.
Given
F
(
r
) satisfying
× F
= 0, let
C
and
˜
C
be any two curves from
a
to b.
a
b
˜
C
C
If S is any surface with boundary S = C
˜
C, By Stokes’ theorem,
Z
S
× F · dS =
Z
S
F · dr =
Z
C
F · dr
Z
˜
C
F · dr.
But × F = 0. So
Z
C
F · dr
Z
˜
C
F · dr = 0,
or
Z
C
F · dr =
Z
˜
C
F · dr.
Proposition.
If (ii)
R
C
F ·
d
r
is independent of
C
for fixed end points and
orientation, then (i) F = f for some scalar field f.
Proof.
We fix
a
and define
f
(
r
) =
R
C
F
(
r
0
)
·
d
r
0
for any curve from
a
to
r
.
Assuming (ii),
f
is well-defined. For small changes
r
to
r
+
δr
, there is a small
extension of C by δC. Then
f(r + δr) =
Z
C+δC
F(r
0
) · dr
0
=
Z
C
F · dr
0
+
Z
δC
F · dr
0
= f(r) + F(r) · δr + o(δr).
So
δf = f(r + δr) f(r) = F(r) · δr + o(δr).
But the definition of grad is exactly
δf = f · δr + o(δr).
So we have F = f.
Note that these results assume
F
is defined on the whole of
R
3
. It also
works of
F
is defined on a simply connected domain
D
, ie a subspace of
R
3
without holes. By definition, this means that any two curves
C,
˜
C
with fixed
end points can be smoothly deformed into one another (alternatively, any loop
can be shrunk into a point).
If we have a smooth transformation from
C
to
˜
C
, the process sweeps out a
surface bounded by C and
˜
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)). However, we can choose to restrict to a subset
D
0
D such that f (r) is single-valued on D
0
.
Example. Take
F =
y
x
2
+ y
2
,
x
x
2
+ y
2
, 0
.
This obeys
× F
= 0, and is defined on
D
=
R
3
\ {z-axis}
, which is not
simply-connected. We can also write
F = f,
where
f = tan
1
y
x
.
which is multi-valued. If we integrate it about the closed loop
x
2
+
y
2
= 1
, z
= 0,
i.e. a circle about the
z
axis, the integral gives 2
π
, as opposed to the expected 0
for a conservative force. This shows that the simply-connected-domain criterion
is important!
However f can be single-valued if we restrict it to
D
0
= R
3
{half-plane x 0, y = 0},
which is simply-connected. (Draw and check!) Any closed curve we can draw in
this area will have an integral of 0 (the circle mentioned above will no longer be
closed!).
8.3 Conservation laws
Definition
(Conservation equation)
.
Suppose we are interested in a quantity
Q
. Let
ρ
(
r, t
) be the amount of stuff per unit volume and
j
(
r, t
) be the flow rate
of the quantity (eg if Q is charge, j is the current density).
The conservation equation is
ρ
t
+ · j = 0.
This is stronger than the claim that the total amount of
Q
in the universe is
fixed. It says that
Q
cannot just disappear here and appear elsewhere. It must
continuously flow out.
In particular, let
V
be a fixed time-independent volume with boundary
S = V . Then
Q(t) =
Z
V
ρ(r, t) dV
Then the rate of change of amount of Q in V is
dQ
dt
=
Z
V
ρ
t
dV =
Z
V
· j dV =
Z
S
j · ds.
by divergence theorem. So this states that the rate of change of the quantity
Q
in
V
is the flux of the stuff flowing out of the surface. ie
Q
cannot just disappear
but must smoothly flow out.
In particular, if
V
is the whole universe (ie
R
3
), and
j
0 sufficiently rapidly
as
|r|
, then we calculate the total amount of
Q
in the universe by taking
V
to be a solid sphere of radius
R
, and take the limit as
R
. Then the surface
integral 0, and the equation states that
dQ
dt
= 0,
Example.
If
ρ
(
r, t
) is the charge density (i.e.
ρδV
is the amount of charge in
a small volume
δV
), then
Q
(
t
) is the total charge in
V
.
j
(
r, t
) is the electric
current density. So j · dS is the charge flowing through δS per unit time.
Example.
Let
j
=
ρu
with
u
being the velocity field. Then (
ρu δt
)
·δS
is equal
to the mass of fluid crossing δS in time δt. So
dQ
dt
=
Z
S
j · dS
does indeed imply the conservation of mass. The conservation equation in this
case is
ρ
t
+ · (ρu) = 0
For the case where
ρ
is constant and uniform (i.e. independent of
r
and
t
), we
get that · u = 0. We say that the fluid is incompressible.
9 Orthogonal curvilinear coordinates
9.1 Line, area and volume elements
In this chapter, we study funny coordinate systems. A coordinate system is,
roughly speaking, a way to specify a point in space by a set of (usually 3)
numbers. We can think of this as a function r(u, v, w).
By the chain rule, we have
dr =
r
u
du +
r
v
dv +
r
w
dw
For a good parametrization,
r
u
·
r
v
×
r
w
6= 0,
i.e.
r
u
,
r
v
and
r
w
are linearly independent. These vectors are tangent to the
curves parametrized by u, v, w respectively when the other two are being fixed.
Even better, they should be orthogonal:
Definition
(Orthogonal curvilinear coordinates)
. u, v, w
are orthogonal curvi-
linear if the tangent vectors are orthogonal.
We can then set
r
u
= h
u
e
u
,
r
v
= h
v
e
v
,
r
w
= h
w
e
w
,
with
h
u
, h
v
, h
w
>
0 and
e
u
, e
v
, e
w
form an orthonormal right-handed basis (i.e.
e
u
× e
v
= e
w
). Then
dr = h
u
e
u
du + h
v
e
v
dv + h
w
e
w
dw,
and
h
u
, h
v
, h
w
determine the changes in length along each orthogonal direction
resulting from changes in u, v, w. Note that clearly by definition, we have
h
u
=
r
u
.
Example.
(i)
In cartesian coordinates,
r
(
x, y, z
) =
x
ˆ
i
+
y
ˆ
j
+
z
ˆ
k
. Then
h
x
=
h
y
=
h
z
= 1,
and e
x
=
ˆ
i, e
y
=
ˆ
j and e
z
=
ˆ
k.
(ii)
In cylindrical polars,
r
(
ρ, ϕ, z
) =
ρ
[
cos ϕ
ˆ
i
+
sin ϕ
ˆ
j
] +
z
ˆ
k
. Then
h
ρ
=
h
z
= 1,
and
h
ϕ
=
r
ϕ
= |(ρ sin ϕ, ρ sin ϕ, 0)| = ρ.
The basis vectors e
ρ
, e
ϕ
, e
z
are as in section 1.
(iii) In spherical polars,
r(r, θ, ϕ) = r(cos ϕ sin θ
ˆ
i + sin θ sin ϕ
ˆ
j + cos θ
ˆ
k).
Then h
r
= 1, h
θ
= r and h
ϕ
= r sin θ.
Consider a surface with
w
constant and parametrised by
u
and
v
. The vector
area element is
dS =
r
u
×
r
v
du dv = h
u
e
u
× h
v
e
v
du dv = h
u
h
v
e
w
du dv.
We interpret this as
δS
having a small rectangle with sides approximately
h
u
δu
and h
v
δv. The volume element is
dV =
r
u
·
r
v
×
r
w
du dv dw = h
u
h
v
h
w
du dv dw,
i.e. a small cuboid with sides h
u
δ
u
, h
v
δ
v
and h
w
δ
w
respectively.
9.2 Grad, Div and Curl
Consider f (r(u, v, w)) and compare
df =
f
u
du +
f
v
dv +
f
w
dw,
with df = (f ) · dr. Since we know that
dr =
r
u
du +
r
v
dv +
r
w
dw = h
u
e
u
du + h
v
e
v
dv + h
w
e
w
dv,
we can compare the terms to know that
Proposition.
f =
1
h
u
f
u
e
u
+
1
h
v
f
v
e
v
+
1
h
w
f
w
e
w
.
Example. Take f = r sin θ cos ϕ in spherical polars. Then
f = sin θ cos ϕ e
r
+
1
r
(r cos θ cos ϕ) e
θ
+
1
r sin θ
(r sin θ sin ϕ) e
ϕ
= cos ϕ(sin θ e
r
+ cos θ e
θ
) sin ϕ e
ϕ
.
Then we know that the differential operator is
Proposition.
=
1
h
u
e
u
u
+
1
h
v
e
v
v
+
1
h
w
e
w
w
.
We can apply this to a vector field
F = F
u
e
u
+ F
v
e
v
+ F
w
e
w
using scalar or vector products to obtain
Proposition.
× F =
1
h
v
h
w
v
(h
w
F
w
)
w
(h
v
F
v
)
e
u
+ two similar terms
=
1
h
u
h
v
h
w
h
u
e
u
h
v
e
v
h
w
e
w
u
v
w
h
u
F
u
h
v
F
v
h
w
F
w
and
· F =
1
h
u
h
v
h
w
u
(h
v
h
w
F
u
) + two similar terms
.
There are several ways to obtain these formulae. We can
Proof. (non-examinable)
(i) Apply ∇· or ∇× and differentiate the basis vectors explicitly.
(ii)
First, apply
∇·
or
∇×
, but calculate the result by writing
F
in terms of
u, v
and
w
in a suitable way. Then use
×f
= 0 and
·
(
×f
) = 0.
(iii) Use the integral expressions for div and curl.
Recall that
n · × F = lim
A0
1
A
Z
A
F · dr.
So to calculate the curl, we first find the e
w
component.
Consider an area with
W
fixed and change
u
by
δu
and
v
by
δv
. Then
this has an area of h
u
h
v
δv with normal e
w
. Let C be its boundary.
u
δu
v
δv
C
We then integrate around the curve
C
. We split the curve
C
up into 4 parts
(corresponding to the four sides), and take linear approximations by assum-
ing
F
and
h
are constant when moving through each horizontal/vertical
segment.
Z
C
F · dr F
u
(u, v)h
u
(u, v) δu + F
v
(u + δu, v)h
v
(u + δu, v) δu
F
u
(u, v + δv)h
u
(u, v + δv) δu F
v
(u, v)h
v
(u, v) δv
u
h
v
F
v
v
(h
u
F
u
)
δv.
Divide by the area and take the limit as area 0, we obtain
lim
A0
1
A
Z
C
F · dr =
1
h
u
h
v
u
h
v
F
v
v
(h
u
F
u
)
.
So, by the integral definition of divergence,
e
w
· × F =
1
h
u
h
v
u
(h
v
F
v
)
v
(h
u
F
u
)
,
and similarly for other components.
We can find the divergence similarly.
Example. Let A =
1
r
tan
θ
2
e
ϕ
in spherical polars. Then
× A =
1
r
2
sin θ
e
r
re
θ
r sin θe
ϕ
r
θ
ϕ
0 0 r sin θ ·
1
r
tan
θ
2
=
e
r
r
2
sin θ
θ
sin θ tan
θ
2
=
1
r
2
e
r
.
10 Gauss’ Law and Poisson’s equation
10.1 Laws of gravitation
Consider a distribution of mass producing a gravitational force
F
on a point
mass
m
at
r
. The total force is a sum of contributions from each part of the
mass distribution, and is proportional to m. Write
F = mg(r),
Definition
(Gravitational field)
. g
(
r
) is the gravitational field, acceleration due
to gravity, or force per unit mass.
The gravitational field is conservative, ie
I
C
g · dr = 0.
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.e. gravitational potential
energy is conserved.
Gauss’ law tells us what this gravitational field looks like:
Law
(Gauss’ law for gravitation)
.
Given any volume
V
bounded by closed
surface S,
Z
S
g · dS = 4πGM,
where
G
is Newton’s gravitational constant, and
M
is the total mass contained
in V .
These equations determine g(r) from a mass distribution.
Example.
We can obtain Newton’s law of gravitation from Gauss’ law together
with an assumption about symmetry.
Consider a total mass
M
distributed with a spherical symmetry about the
origin
O
, with all the mass contained within some radius
r
=
a
. By spherical
symmetry, we have g(r) = g(r)
ˆ
r.
Consider Gauss’ law with
S
being a sphere of radius
r
=
R > a
. Then
ˆ
n
=
ˆ
r
.
So
Z
S
g · dS =
Z
S
g(R)
ˆ
r ·
ˆ
r dS =
Z
g(R)dS = 4πR
2
g(R).
By Gauss’ law, we obtain
4πR
2
g(R) = 4πGM.
So
g(R) =
GM
R
2
for R > a.
Therefore the gravitational force on a mass m at r is
F(r) =
GMm
r
2
ˆ
r.
If we take the limit as
a
0, we get a point mass
M
at the origin. Then we
recover Newton’s law of gravitation for point masses.
The condition
R
C
g ·
d
r
= 0 for any closed
C
can be re-written by Stoke’s
theorem as
Z
S
× g · dS = 0,
where S is bounded by the closed curve C. This is true for arbitrary S. So
× g = 0.
In our example above,
×g
= 0 due to spherical symmetry. But here we showed
that it is true for all cases.
Note that we exploited symmetry to solve Gauss’ law. However, if the mass
distribution is not sufficiently symmetrical, Gauss’ law in integral form can be
difficult to use. But we can rewrite it in differential form. Suppose
M =
Z
V
ρ(r) dV,
where ρ is the mass density. Then by Gauss’ theorem
Z
S
g · dS = 4πGM
Z
V
· g dV =
Z
V
4π dV.
Since this is true for all V , we must have
Law (Gauss’ Law for gravitation in differential form).
· g = 4πGρ.
Since
× g
= 0, we can introduce a gravitational potential
ϕ
(
r
) with
g = −∇ϕ. Then Gauss’ Law becomes
2
ϕ = 4πGρ.
In the example with spherical symmetry, we can solve that
ϕ(r) =
GM
r
for r > a.
10.2 Laws of electrostatics
Consider a distribution of electric charge at rest. They produce a force on a
charge q, at rest at r , which is proportional to q.
Definition
(Electric field)
.
The force produced by electric charges on another
charge q is F = qE(r), where E(r) is the electric field, or force per unit charge.
Again, this is conservative. So
I
C
E · dr = 0
for any closed curve C. It also obeys
Law (Gauss’ law for electrostatic forces).
Z
S
E · dS =
Q
ε
0
,
where ε
0
is the permittivity of free space, or electric constant.
Then we can write it in differential form, as in the gravitational case.
Law (Gauss’ law for electrostatic forces in differential form).
· E =
ρ
ε
0
.
Assuming constant (or no) magnetic field, we have
× E = 0.
So we can write E = −∇ϕ.
Definition
(Electrostatic potential)
.
If we write
E
=
−∇ϕ
, then
ϕ
is the
electrostatic potential, and
2
ϕ =
ρ
ε
0
.
Example.
Take a spherically symmetric charge distribution about
O
with total
charge
Q
. Suppose all charge is contained within a radius
r
=
a
. Then similar
to the gravitational case, we have
E(r) =
Q
ˆ
r
4πε
0
r
2
,
and
ϕ(r) =
Q
4πε
0
r
.
As
a
0, we get point charges. From
E
, we can recover Coulomb’s law for the
force on another charge q at r:
F = qE =
qQ
ˆ
r
4πε
0
r
2
.
Example (Line charge). Consider an infinite line with uniform charge density
per unit length σ.
We use cylindrical polar coordinates:
z
r =
p
x
2
+ y
2
E
By symmetry, the field is radial, i.e.
E(r) = E(r)
ˆ
r.
Pick
S
to be a cylinder of length
L
and radius
r
. We know that the end caps do
not contribute to the flux since the field lines are perpendicular to the normal.
Also, the curved surface has area 2πrL. Then by Gauss’ law in integral form,
Z
S
E · dS = E(r)2πrL =
σL
ε
0
.
So
E(r) =
σ
2πε
0
r
ˆ
r.
Note that the field varies as 1
/r
, not 1
/r
2
. Intuitively, this is because we have
one more dimension of “stuff” compared to the point charge, so the field does
not drop as fast.
10.3 Poisson’s Equation and Laplace’s equation
Definition (Poisson’s equation). The Poisson’s equation is
2
ϕ = ρ,
where ρ is given and ϕ(r) is to be solved.
This is the form of the equations for gravity and electrostatics, with
4
π
and ρ/ε
0
in place of ρ respectively.
When ρ = 0, we get
Definition (Laplace’s equation). Laplace’s equation is
2
ϕ = 0.
One example is irrotational and incompressible fluid flow: if the velocity is
u
(
r
), then irrotationality gives
u
=
ϕ
for some velocity potential
ϕ
. Since it is
incompressible, · u = 0 (cf. previous chapters). So
2
ϕ = 0.
The expressions for
2
can be found in non-Cartesian coordinates, but are a
bit complicated.
We’re concerned here mainly with cases exhibiting spherical or cylindrical
symmetry (use
r
for radial coordinate here). i.e. when
ϕ
(
r
) has spherical or
cylindrical symmetry. Write ϕ = ϕ(r). Then
ϕ = ϕ
0
(r)
ˆ
r.
Then Laplace’s equation
2
ϕ = 0 becomes an ordinary differential equation.
For spherical symmetry, using the chain rule, we have
2
ϕ = ϕ
00
+
2
r
ϕ
0
=
1
r
2
(r
2
ϕ
0
)
0
= 0.
Then the general solution is
ϕ =
A
r
+ B.
For cylindrical symmetry, with r
2
= x
2
1
+ x
2
2
, we have
2
ϕ = ϕ
00
+
1
r
ϕ
0
=
1
r
(rϕ
0
)
0
= 0.
Then
ϕ = A ln r + B.
Then solutions to Poisson’s equations can be obtained in a similar way, i.e. by
integrating the differential equations directly, or by adding particular integrals
to the solutions above.
For example, for a spherically symmetric solution of
2
ϕ
=
ρ
0
, with
ρ
0
constant, recall that
2
r
α
=
α
(
α
+ 1)
r
α2
. Taking
α
= 2, we find the particular
integral
ϕ =
ρ
0
6
r
2
,
So the general solution with spherical symmetry and constant ρ
0
is
ϕ(r) =
A
r
+ B
1
6
ρ
0
r
2
.
To determine
A, B
, we must specify boundary conditions. If
ϕ
is defined on all
of
R
3
, we often require
ϕ
0 as
|r|
. 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
ϕ
n
): a Neumann condition. (
n
is
the outward normal on V ).
The type of boundary conditions we get depends on the physical content of
the problem. For example, specifying
ϕ
n
corresponds to specifying the normal
component of g or E.
We can also specify different boundary conditions on different boundary
components.
Example.
We might have a spherically symmetric distribution with constant
ρ
0
, defined in a r b, with ϕ(a) = 0 and
ϕ
n
(b) = 0.
Then the general solution is
ϕ(r) =
A
r
+ B
1
6
ρ
0
r
2
.
We apply the first boundary condition to obtain
A
a
+ B
1
6
ρ
0
a
2
= 0.
The second boundary condition gives
n · ϕ =
A
b
2
1
3
ρ
0
b = 0.
These conditions give
A =
1
3
ρ
0
b
3
, B =
1
5
ρ
0
a
2
+
1
3
ρ
0
b
3
a
.
Example.
We might also be interested with spherically symmetric solution with
2
ϕ =
(
ρ
0
r a
0 r > a
with
ϕ
non-singular at
r
= 0 and
ϕ
(
r
)
0 as
r
, and
ϕ, ϕ
0
continuous at
r = a. This models the gravitational potential on a uniform planet.
Then the general solution from above is
ϕ =
(
A
r
+ B
1
6
ρ
0
r
2
r a
C
r
+ D r > a.
Since
ϕ
is non-singular at
r
= 0, we have
A
= 0. Since
ϕ
0 as
r
,
D
= 0.
So
ϕ =
(
B
1
6
ρ
0
r
2
r a
C
r
r > a.
This is the gravitational potential inside and outside a planet of constant density
ρ
0
and radius a. We want ϕ and ϕ
0
to be continuous at r = a. So we have
B +
1
6
4πρ
0
Ga
2
=
C
a
4
3
π
0
a =
C
a
2
.
The second equation gives
C
=
GM
. Substituting that into the first equation
to find B, we get
ϕ(r) =
(
GM
2a

r
a
2
3

r a
GM
r
r > a
Since g = ϕ
0
, we have
g(r) =
(
GMr
a
3
r a
GM
r
r > a
We can plot the potential energy:
r
ϕ(r)
r = a
We can also plot g(r), the inward acceleration:
r
g(r)
r = a
Alternatively, we can apply Gauss’ Law for a flux of
g
=
g
(
r
)
e
r
out of
S
, a
sphere of radius R. For R a,
Z
S
g · dS = 4πR
2
g(R) = 4πGM
R
a
3
So
g(R) =
GMR
a
3
.
For R a, we can simply apply Newton’s law of gravitation.
In general, even if the problem has nothing to do with gravitation or electro-
statics, if we want to solve
2
ϕ
=
ρ
with
ρ
and
ϕ
sufficiently symmetric, we
can consider the flux of ϕ out of a surface S = V :
Z
S
ϕ · dS =
Z
V
ρ dV,
by divergence theorem. This is called the Gauss Flux method.
11 Laplace’s and Poisson’s equations
11.1 Uniqueness theorems
Theorem.
Consider
2
ϕ
=
ρ
for some
ρ
(
r
) on a bounded volume
V
with
S = V being a closed surface, with an outward normal n.
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.
where f, g are given. Then
(i) ϕ(r) is unique
(ii) ϕ(r) is unique up to a constant.
This theorem is practically important - if you find a solution by any magical
means, you know it is the only solution (up to a constant).
Since the proof of the cases of the two different boundary conditions are very
similar, they will be proved together. When the proof is broken down into (i)
and (ii), it refers to the specific cases of each boundary condition.
Proof.
Let
ϕ
1
(
r
) and
ϕ
2
(
r
) satisfy Poisson’s equation, each obeying the boundary
conditions (N) or (D). Then Ψ(
r
) =
ϕ
2
(
r
)
ϕ
1
(
r
) satisfies
2
Ψ = 0 on
V
by
linearity, and
(i) Ψ = 0 on S; or
(ii)
Ψ
n
= 0 on S.
Combining these two together, we know that Ψ
Ψ
n
= 0 on the surface. So using
the divergence theorem,
Z
V
· Ψ) dV =
Z
S
Ψ) · dS = 0.
But
· Ψ) = (Ψ) · (Ψ) + Ψ
2
Ψ
|{z}
=0
= |(Ψ)|
2
.
So
Z
V
|∇Ψ|
2
dV = 0.
Since
|∇
Ψ
|
2
0, the integral can only vanish if
|∇
Ψ
|
= 0. So
Ψ = 0. So Ψ =
c
,
a constant on V . So
(i) Ψ = 0 on S c = 0. So ϕ
1
= ϕ
2
on V .
(ii) ϕ
2
(r) = ϕ
1
(r) + C, as claimed.
We’ve proven uniqueness. How about existence? It turns out it isn’t difficult
to craft a boundary condition in which there are no solutions.
For example, if we have
2
ϕ
=
ρ
on
V
with the condition
ϕ
n
=
g
, then by
the divergence theorem,
Z
V
2
ϕ dV =
Z
S
ϕ
n
dS.
Using Poisson’s equation and the boundary conditions, we have
Z
V
ρ dV +
Z
V
g dS = 0
So if ρ and g don’t satisfy this equation, then we can’t have any solutions.
The theorem can be similarly proved and stated for regions in
R
2
, R
3
, ···
, by
using the definitions of grad, div and the divergence theorem. The result also
extends to unbounded domains. To prove it, we can take a sphere of radius
R
and impose the boundary conditions
|
Ψ(
r
)
|
=
O
(1
/R
) or
|
Ψ
n
(
r
)
|
=
O
(1
/R
2
) as
R . Then we just take the relevant limits to complete the proof.
Similar results also apply to related equations and different kinds of boundary
conditions, eg
D
or
N
on different parts of the boundary. But we have to analyse
these case by case and see if the proof still applies.
The proof uses a special case of the result
Proposition (Green’s first identity).
Z
S
(uv) · dS =
Z
V
(u) · (v) dV +
Z
V
u
2
v dV,
By swapping u and v around and subtracting the equations, we have
Proposition (Green’s second identity).
Z
S
(uv vu) · dS =
Z
V
(u
2
v v
2
u) dV.
These are sometimes useful, but can be easily deduced from the divergence
theorem when needed.
11.2 Laplace’s equation and harmonic functions
Definition
(Harmonic function)
.
A harmonic function is a solution to Laplace’s
equation
2
ϕ = 0.
These have some very special properties.
11.2.1 The mean value property
Proposition
(Mean value property)
.
Suppose
ϕ
(
r
) is harmonic on region
V
containing a solid sphere defined by
|ra| R
, with boundary
S
R
=
|ra|
=
R
,
for some R. Define
¯ϕ(R) =
1
4πR
2
Z
S
R
ϕ(r) dS.
Then ϕ(a) = ¯ϕ(R).
In words, this says that the value at the center of a sphere is the average of
the values on the surface on the sphere.
Proof.
Note that
¯ϕ
(
R
)
ϕ
(
a
) as
R
0. We take spherical coordinates (
u, θ, χ
)
centered on r = a. The scalar element (when u = R) on S
R
is
dS = R
2
sin θ dθ dχ.
So
dS
R
2
is independent of R. Write
¯ϕ(R) =
1
4π
Z
ϕ
dS
R
2
.
Differentiate this with respect to
R
, noting that d
S/R
2
is independent of
R
.
Then we obtain
d
dR
¯ϕ(R) =
1
4πR
2
Z
ϕ
u
u=R
dS
But
ϕ
u
= e
u
· ϕ = n · ϕ =
ϕ
n
on S
R
. So
d
dR
¯ϕ(R) =
1
4πR
2
Z
S
R
ϕ · dS =
1
4πR
2
Z
V
R
2
ϕ dV = 0
by divergence theorem. So
¯ϕ
(
R
) does not depend on
R
, and the result follows.
11.2.2 The maximum (or minimum) principle
In this section, we will talk about maxima of functions. It should be clear that
the results also hold for minima.
Definition
(Local maximum)
.
We say that
ϕ
(
r
) has a local maximum at
a
if
for some ε > 0, ϕ(r) < ϕ(a) when 0 < |r a| < ε.
Proposition
(Maximum principle)
.
If a function
ϕ
is harmonic on a region
V
,
then ϕ cannot have a maximum at an interior point of a of V .
Proof.
Suppose that
ϕ
had a local maximum at
a
in the interior. Then there is
an ε such that for any r such that 0 < |r a| < ε, we have ϕ(r) < ϕ(a).
Note that if there is an
ε
that works, then any smaller
ε
will work. Pick an
ε
sufficiently small such that the region
|r a| < ε
lies within
V
(possible since
a
lies in the interior of V ).
Then for any r such that |r a| = ε, we have ϕ(r) < ϕ(a).
¯ϕ(ε) =
1
4πR
2
Z
S
R
ϕ(r) dS < ϕ(a),
which contradicts the mean value property.
We can understand this by performing a local analysis of stationary points
by differentiation. Suppose at
r
=
a
, we have
ϕ
= 0. Let the eigenvalues of the
Hessian matrix
H
ij
=
2
x
i
x
j
be
λ
i
. But since
ϕ
is harmonic, we have
2
ϕ
= 0,
i.e.
2
ϕ
x
i
x
i
=
H
ii
= 0. But
H
ii
is the trace of the Hessian matrix, which is the
sum of eigenvalues. So
P
λ
i
= 0.
Recall that a maximum or minimum occurs when all eigenvalues have the
same sign. This clearly cannot happen if the sum is 0. Therefore we can only
have saddle points.
(note we ignored the case where all
λ
i
= 0, where this analysis is inconclusive)
11.3 Integral solutions of Poisson’s equations
11.3.1 Statement and informal derivation
We want to find a solution to Poisson’s equations. We start with a discrete case,
and try to generalize it to a continuous case.
If there is a single point source of strength λ at a, the potential ϕ is
ϕ =
λ
4π
1
|r a|
.
(we have λ = 4πGM for gravitation and Q/ε
0
for electrostatics)
If we have many sources
λ
α
at positions
r
α
, the potential is a sum of terms
ϕ(r) =
X
α
1
4π
λ
α
|r r
α
|
.
If we have infinitely many of them, having a distribution of
ρ
(
r
) with
ρ
(
r
0
) d
V
0
being the contribution from a small volume at position
r
0
. It would be reasonable
to guess that the solution is what we obtain by replacing the sum with an integral:
Proposition.
The solution to Poisson’s equation
2
ϕ
=
ρ
, with boundary
conditions |ϕ(r)| = O(1/|r|) and |∇ϕ(r)| = O(1/|r|
2
), is
ϕ(r) =
1
4π
Z
V
0
ρ(r
0
)
|r r
0
|
dV
0
For
ρ
(
r
0
) non-zero everywhere, but suitably well-behaved as
|r
0
|
, we can
also take V
0
= R
3
.
Example. Suppose
2
ϕ =
(
ρ
0
|r| a
0 |r| > a.
Fix
r
and introduce polar coordinates
r
0
, θ, χ
for
r
0
. We take the
θ
= 0 direction
to be the direction along the line from r
0
to r.
Then
ϕ(r) =
1
4π
Z
V
0
ρ
0
|r r
0
|
dV
0
.
We have
dV
0
= r
02
sin θ dr
0
dθ dχ.
We also have
|r r
0
| =
p
r
2
+ r
02
2rr
0
cos θ
by the cosine rule (c
2
= a
2
+ b
2
2ab cos C). So
ϕ(r) =
1
4π
Z
a
0
dr
0
Z
π
0
dθ
Z
2π
0
dχ
ρ
0
r
02
sin θ
r
2
+ r
02
2rr
0
cos θ
=
ρ
0
2
Z
a
0
dr
0
r
02
rr
0
h
p
r
2
+ r
02
rr
0
cos θ
i
θ=π
θ=0
=
ρ
0
2
Z
a
0
dr
0
r
0
r
(|r + r
0
| + |r r
0
|)
=
ρ
0
2
Z
a
0
"
dr
0
r
0
r
(
2r
0
r > r
0
2r r < r
0
!#
If r > a, then r > r
0
always. So
ϕ(r) = ρ
0
Z
a
0
r
02
r
dr
0
=
r
0
a
3
3r
.
If r < a, then the integral splits into two parts:
ϕ(r) = ρ
0
Z
r
0
dr
0
r
02
r
+
Z
a
r
dr
0
r
0
= ρ
0
1
6
r
2
+
a
2
2
.
11.3.2 Point sources and δ-functions*
Recall that
Ψ =
λ
4π|r a|
is our potential for a point source. When r 6= a, we have
Ψ =
λ
4π
r a
|r a|
3
,
2
Ψ = 0.
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
S
Ψ · dS = λ.
By the divergence theorem, we have
Z
2
Ψ dV = λ.
for
V
being a solid sphere with
V
=
S
. Since
2
Ψ is zero at any point
r 6
=
a
,
we must have
2
Ψ = λδ(r a),
where δ is the 3d delta function, which satisfies
Z
V
f(r)δ(r a) dV = f(a)
for any volume containing a.
In short, we have
2
1
|r r
0
|
= 4πδ(r r
0
).
Using these, we can verify that the integral solution of Poisson’s equation we
obtained previously is correct:
2
Ψ(r) =
2
1
4π
Z
V
0
ρ(r
0
)
|r r
0
|
dV
0
=
1
4π
Z
V
0
ρ(r
0
)
2
1
|r r
0
|
dV
0
=
Z
V
0
ρ(r
0
)δ(r r
0
) dV
0
= ρ(r),
as required.
12 Maxwell’s equations
12.1 Laws of electromagnetism
Maxwell’s equations are a set of four equations that describe the behaviours
of electromagnetism. Together with the Lorentz force law, these describe all
we know about (classical) electromagnetism. All other results we know are
simply mathematical consequences of these equations. It is thus important to
understand the mathematical properties of these equations.
To begin with, there are two fields that govern electromagnetism, known
as the electric and magnetic field. These are denoted by
E
(
r, t
) and
B
(
r, t
)
respectively.
To understand electromagnetism, we need to understand how these fields
are formed, and how these fields affect charged particles. The second is rather
straightforward, and is given by the Lorentz force law.
Law (Lorentz force law). A point charge q experiences a force of
F = q(E +
˙
r × B).
The dynamics of the field itself is governed by Maxwell’s equations. To state
the equations, we need to introduce two more concepts.
Definition
(Charge and current density)
. ρ
(
r, t
) is the charge density, defined
as the charge per unit volume.
j
(
r, t
) is the current density, defined as the electric current per unit area of
cross section.
Then Maxwell’s equations say
Law (Maxwell’s equations).
· E =
ρ
ε
0
· B = 0
× E +
B
t
= 0
× B µ
0
ε
0
E
t
= µ
0
j,
where
ε
0
is the electric constant (permittivity of free space) and
µ
0
is the
magnetic constant (permeability of free space), which are constants determined
experimentally.
We can quickly derive some properties we know from these four equations.
The conservation of electric charge comes from taking the divergence of the last
equation.
· ( × B)
| {z }
=0
µ
0
ε
0
t
( · E)
| {z }
=ρ/ε
0
= µ
0
· j.
So
ρ
t
+ · j = 0.
We can also take the volume integral of the first equation to obtain
Z
V
· E dV =
1
ε
0
Z
V
ρ dV =
Q
ε
0
.
By the divergence theorem, we have
Z
S
E · dS =
Q
ε
0
,
which is Gauss’ law for electric fields
We can integrate the second equation to obtain
Z
S
B · dS = 0.
This roughly states that there are no “magnetic charges”.
The remaining Maxwell’s equations also have integral forms. For example,
Z
C=S
E · dr =
Z
S
× E dS =
d
dt
Z
S
B · dS,
where the first equality is from from Stoke’s theorem. This says that a changing
magnetic field produces a current.
12.2 Static charges and steady currents
If ρ, j, E, B are all independent of time, E and B are no longer linked.
We can solve the equations for electric fields:
· E = ρ/ε
0
× E = 0
Second equation gives E = −∇ϕ. Substituting into first gives
2
ϕ = ρ/ε
0
.
The equations for the magnetic field are
· B = 0
× B = µ
0
j
First equation gives
B
=
× A
for some vector potential
A
. But the vector
potential is not well-defined. Making the transformation
A 7→ A
+
χ
(
x
)
produces the same
B
, since
×
(
χ
) = 0. So choose
χ
such that
· A
= 0.
Then
2
A = ( · A
|{z}
=0
) × ( × A
| {z }
B
) = µ
0
j.
In summary, we have
Electrostatics Magnetostatics
· E = ρ/ε
0
· B = 0
× E = 0 × B = µ
0
j
2
ϕ = ρ/ε
0
2
A = µ
0
j.
ε
0
sets the scale of electrostatic effects,
e.g. the Coulomb force
µ
0
sets the scale of magnetic effects,
e.g. force between two wires with cur-
rents.
12.3 Electromagnetic waves
Consider Maxwell’s equations in empty space, i.e.
ρ
= 0,
j
=
0
. Then Maxwell’s
equations give
2
E = ( · E) × ( × E) = ×
B
t
=
t
( × B) = µ
0
ε
0
2
E
2
t
.
Define c =
1
µ
0
ε
0
. Then the equation gives
2
1
c
2
2
t
2
E = 0.
This is the wave equation describing propagation with speed
c
. Similarly, we
can obtain
2
1
c
2
2
t
2
B = 0.
So Maxwell’s equations predict that there exists electromagnetic waves in free
space, which move with speed
c
=
1
ε
0
µ
0
3.00 × 10
8
m s
1
, which is the speed
of light! Maxwell then concluded that light is electromagnetic waves!
13 Tensors and tensor fields
13.1 Definition
There are two ways we can think of a vector in
R
3
. We can either interpret it as
a “point” in space, or we can view it simply as a list of three numbers. However,
the list of three numbers is simply a representation of the vector with respect to
some particular basis. When we change basis, in order to represent the same
point, we will need to use a different list of three numbers. In particular, when
we perform a rotation by R
ip
, the new components of the vector is given by
v
0
i
= R
ip
v
p
.
Similarly, we can imagine a matrix as either a linear transformation or an
array of 9 numbers. Again, when we change basis, in order to represent the
same transformation, we will need a different array of numbers. This time, the
transformation is given by
A
0
ij
= R
ip
R
jq
A
pq
.
We can think about this from another angle. To define an arbitrary quantity
A
ij
, we can always just write down 9 numbers and be done with it. Moreover,
we can write down a different set of numbers in a different basis. For example,
we can define
A
ij
=
δ
ij
in our favorite basis, but
A
ij
= 0 in all other bases. We
can do so because we have the power of the pen.
However, for this
A
ij
to represent something physically meaningful, i.e. an
actual linear transformation, we have to make sure that the components of
A
ij
transform sensibly under a basis transformation. By “sensibly”, we mean that it
has to follow the transformation rule
A
0
ij
=
R
ip
R
jq
A
pq
. For example, the
A
ij
we defined in the previous paragraph does not transform sensibly. While it is
something we can define and write down, it does not correspond to anything
meaningful.
The things that transform sensibly are known as tensors. For example,
vectors and matrices (that transform according to the usual change-of-basis
rules) are tensors, but that A
ij
is not.
In general, tensors are allowed to have an arbitrary number of indices. In
order for a quantity
T
ij···k
to be a tensor, we require it to transform according to
T
0
ij···k
= R
ip
R
jq
···R
kr
T
pq···r
,
which is an obvious generalization of the rules for vectors and matrices.
Definition
(Tensor)
.
A tensor of rank
n
has components
T
ij···k
(with
n
indices)
with respect to each basis
{e
i
}
or coordinate system
{x
i
}
, and satisfies the
following rule of change of basis:
T
0
ij···k
= R
ip
R
jq
···R
kr
T
pq···r
.
Example.
A tensor
T
of rank 0 doesn’t transform under change of basis, and is a
scalar.
A tensor T of rank 1 transforms under T
0
i
= R
ip
T
p
. This is a vector.
A tensor
T
of rank 2 transforms under
T
0
ij
=
R
ip
R
jq
T
pq
. This is a matrix.
Example.
(i) If u, v, ···w are n vectors, then
T
ij···k
= u
i
v
j
···w
k
defines a tensor of rank
n
. To check this, we check the tensor transformation
rule. 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:
T
0
ij
= u
0
i
v
0
j
= (R
ip
u
p
)(R
jq
v
q
)
= R
ip
R
jq
(u
p
v
q
)
= R
ip
R
jq
T
pq
Then linear combinations of such expressions are also tensors, e.g.
T
ij
=
u
i
v
j
+ a
i
b
j
for any u, v, a, b.
(ii) δ
ij
and
ε
ijk
are tensors of rank 2 and 3 respectively with the special
property that their components are unchanged with respect to the basis
coordinate:
δ
0
ij
= R
ip
R
jq
δ
pq
= R
ip
R
jp
= δ
ij
,
since R
ip
R
jp
= (RR
T
)
ij
= I
ij
. Also
ε
0
ijk
= R
ip
R
jq
R
kr
ε
pqr
= (det R)ε
ijk
= ε
ijk
,
using results from Vectors and Matrices.
(iii)
(Physical example) In some substances, an applied electric field
E
gives rise
to a current density
j
, according to the linear relation
j
i
=
ε
ij
E
j
, where
ε
ij
is the conductivity tensor.
Note that this relation entails that the resulting current need not be in the
same direction as the electric field. This might happen if the substance
has special crystallographic directions that favours electric currents.
However, if the substance is isotropic, we have
ε
ij
=
σδ
ij
for some
σ
. In
this case, the current is parallel to the field.
13.2 Tensor algebra
Definition
(Tensor addition)
.
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
= T
ij···k
+ S
ij···k
.
in any coordinate system.
To check that this is a tensor, we check the transformation rule. Again, we
only show for n = 2:
(T + S)
0
ij
= T
0
ij
+ S
0
ij
= R
ip
R
jq
T
pq
+ R
ip
R
jq
S
pq
= (R
ip
R
jq
)(T
pq
+ S
pq
).
Definition
(Scalar multiplication)
.
A tensor
T
of rank
n
can be multiplied by
a scalar α. αT is a tensor of the same rank, defined by
(αT )
ij
= αT
ij
.
It is trivial to check that the resulting object is indeed a tensor.
Definition
(Tensor product)
.
Let
T
be a tensor of rank
n
and
S
be a tensor of
rank m. The tensor product T S is a tensor of rank n + m defined by
(T S)
x
1
x
2
···x
n
y
1
y
2
···y
m
= T
x
1
x
2
···x
n
S
y
1
y
2
···y
n
.
It is trivial to show that this is a tensor.
We can similarly define tensor products for any (positive integer) number of
tensors, e.g. for n vectors u, v ··· , w, we can define
T = u v ··· w
by
T
ij···k
= u
i
v
j
···w
k
,
as defined in the example in the beginning of the chapter.
Definition
(Tensor contraction)
.
For a tensor
T
of rank
n
with components
T
ijp···q
, we can contract on the indices
i, j
to obtain a new tensor of rank
n
2:
S
p···q
= δ
ij
T
ijp···q
= T
iip···q
Note that we don’t have to always contract on the first two indices. We can
contract any pair we like.
To check that contraction produces a tensor, we take the ranks 2
T
ij
example.
Contracting, we get
T
ii
,a rank-0 scalar. We have
T
0
ii
=
R
ip
R
iq
T
pq
=
δ
pq
T
pq
=
T
pp
= T
ii
, since R is an orthogonal matrix.
If we view
T
ij
as a matrix, then the contraction is simply the trace of
the matrix. So our result above says that the trace is invariant under basis
transformations as we already know in IA Vectors and Matrices.
Note that our usual matrix product can be formed by first applying a tensor
product to obtain M
ij
N
pq
, then contract with δ
jp
to obtain M
ij
N
jq
.
13.3 Symmetric and antisymmetric tensors
Definition
(Symmetric and anti-symmetric tensors)
.
A tensor
T
of rank
n
is
symmetric in the indices i, j if it obeys
T
ijp···q
= T
jip···q
.
It is anti-symmetric if
T
ijp···q
= T
jip···q
.
Again, a tensor can be symmetric or anti-symmetric in any pair of indices, not
just the first two.
This is a property that holds in any coordinate systems, if it holds in one,
since
T
0
k`r...s
= R
ki
R
`j
R
rp
···R
sq
T
ijp···q
= ±R
ki
R
`j
R
rp
···R
sq
T
jip···q
= ±T
0
`kr···s
as required.
Definition
(Totally symmetric and anti-symmetric tensors)
.
A tensor is totally
(anti-)symmetric if it is (anti-)symmetric in every pair of indices.
Example. δ
ij
=
δ
ji
is totally symmetric, while
ε
ijk
=
ε
jik
is totally antisym-
metric.
There are totally symmetric tensors of arbitrary rank n. But in R
3
,
Any totally antisymmetric tensor of rank 3 is λε
ijk
for some scalar λ.
There are no totally antisymmetric tensors of rank greater than 3, except
for the trivial tensor with all components 0.
Proof: exercise (hint: pigeonhole principle)
13.4 Tensors, multi-linear maps and the quotient rule
Tensors as multi-linear maps
In Vectors and Matrices, we know that matrices are linear maps. We will prove
an analogous fact for tensors.
Definition
(Multilinear map)
.
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.
We will show that a tensor
T
of rank
n
is a equivalent to a multi-linear map
from n vectors a, b, ··· , c to R defined by
T (a, b, ··· , c) = T
ij···k
a
i
b
j
···c
k
.
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.e. the expression
T
ij···k
a
i
b
j
···c
k
is the same regardless of the basis chosen;
(ii)
While it is always possible to write a multi-linear map as
T
ij···k
a
i
b
j
···c
k
,
we have to show that
T
ij···k
is indeed a tensor, i.e. transform according to
the tensor transformation rules.
To show the first property, just note that the
T
ij···k
a
i
b
j
···c
k
is a tensor
product (followed by contraction), which retains tensor-ness. So it is also a
tensor. In particular, it is a rank 0 tensor, i.e. a scalar, which is independent of
the basis.
To show the second property, assuming that
T
is a multi-linear map, it must
be independent of the basis, so
T
ij···k
a
i
b
j
···c
k
= T
0
ij···k
a
0
i
b
0
j
···c
0
k
.
Since
v
0
p
=
R
pi
v
i
by tensor transformation rules, multiplying both sides by
R
pi
gives v
i
= R
pi
v
0
p
. Substituting in gives
T
ij···k
(R
pi
a
0
p
)(R
qj
b
0
q
) ···(R
kr
c
0
r
) = T
0
pq···r
a
0
p
b
0
q
···c
0
r
.
Since this is true for all a, b, ···c, we must have
T
ij···k
R
pi
R
qj
···R
rk
= T
0
pq···r
Hence T
ij···k
obeys the tensor transformation rule, and is a tensor.
This shows that there is a one-to-one correspondence between tensors of rank
n and multi-linear maps.
This gives a way of thinking about tensors independent of any coordinate
system or choice of basis, and the tensor transformation rule emerges naturally.
Note that the above is exactly what we did with linear maps and matrices.
The quotient rule
If T
i ···j
|{z}
n
p ···q
|{z}
m
is a tensor of rank n + m, and u
p···q
is a tensor of rank m then
v
i,···j
= T
i···jp···q
u
p···q
is a tensor of rank
n
, since it is a tensor product of
T
and
u
, followed by
contraction.
The converse is also true:
Proposition
(Quotient rule)
.
Suppose that
T
i···jp···q
is an array defined in each
coordinate system, and that
v
i···j
=
T
i···jp···q
u
p···q
is also a tensor for any tensor
u
p···q
. Then T
i···jp···q
is also a tensor.
Note that we have previously seen the special case of
n
=
m
= 1, which says
that linear maps are tensors.
Proof.
We can check the tensor transformation rule directly. However, we can
reuse the result above to save some writing.
Consider the special form
u
p···q
=
c
p
···d
q
for any vectors
c, ···d
. By
assumption,
v
i···j
= T
i···jp···q
c
p
···d
q
is a tensor. Then
v
i···j
a
i
···b
j
= T
i···jp···q
a
i
···b
j
c
p
···d
q
is a scalar for any vectors
a, ··· , b, c, ··· , d
. Since
T
i···jp···q
a
i
···b
j
c
p
···d
q
is a
scalar and hence gives the same result in every coordinate system,
T
i···jp···q
is a
multi-linear map. So T
i···jp···q
is a tensor.
13.5 Tensor calculus
Tensor fields and derivatives
Just as with scalars or vectors, we can define tensor fields:
Definition
(Tensor field)
.
A tensor field is a tensor at each point in space
T
ij···k
(x), which can also be written as T
ij···k
(x
`
).
We assume that the fields are smooth so they can be differentiated any
number of times
x
p
···
x
q
T
ij···k
,
except for where things obviously fail, e.g. for where
T
is not defined. We now
claim:
Proposition.
x
p
···
x
q
| {z }
m
T
ij ···k
| {z }
n
, ()
is a tensor of rank n + m.
Proof.
To show this, it suffices to show that
x
p
satisfies the tensor transfor-
mation rules for rank 1 tensors (i.e. it is something like a rank 1 tensor). Then
by the exact same argument we used to show that tensor products preserve
tensorness, we can show that the (
) is a tensor. (we cannot use the result of
tensor products directly, since this is not exactly a product. But the exact same
proof works!)
Since x
0
i
= R
iq
x
q
, we have
x
0
i
x
p
= R
ip
.
(noting that
x
p
x
q
= δ
pq
). Similarly,
x
q
x
0
i
= R
iq
.
Note that R
ip
, R
iq
are constant matrices.
Hence by the chain rule,
x
0
i
=
x
q
x
0
i
x
q
= R
iq
x
q
.
So
x
p
obeys the vector transformation rule. So done.
Integrals and the tensor divergence theorem
It is also straightforward to do integrals. Since we can sum tensors and take
limits, the definition of a tensor-valued integral is straightforward.
For example,
R
V
T
ij···k
(
x
) d
V
is a tensor of the same rank as
T
ij···k
(think of
the integral as the limit of a sum).
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
. The flux of volume is
u · nδs
=
u
j
n
j
δS
. So the flux of mass is
ρu
j
n
j
δS
.
Then the flux of the
i
th component of momentum is
ρu
i
u
j
n
j
δS
=
T
ij
n
j
kδS
(mass times velocity), where
T
ij
=
ρu
i
u
j
. Then the flux through the surface
S
is
R
S
T
ij
n
j
dS.
It is easy to generalize the divergence theorem from vectors to tensors. We
can then use it to discuss conservation laws for tensor quantities.
Let
V
be a volume bounded by a surface
S
=
V
and
T
ij···k`
be a smooth
tensor field. Then
Theorem (Divergence theorem for tensors).
Z
S
T
ij···k`
n
`
dS =
Z
V
x
`
(T
ij···k`
) dV,
with n being an outward pointing normal.
The regular divergence theorem is the case where
T
has one index and is a
vector field.
Proof.
Apply the usual divergence theorem to the vector field
v
defined by
v
`
= a
i
b
j
···c
k
T
ij···k`
, where a, b, ··· , c are fixed constant vectors.
Then
· v =
v
`
x
`
= a
i
b
j
···c
k
x
`
T
ij···k`
,
and
n · v = n
`
v
`
= a
i
b
j
···c
k
T
ij···k`
n
`
.
Since
a, b, ··· , c
are arbitrary, therefore they can be eliminated, and the tensor
divergence theorem follows.
14 Tensors of rank 2
14.1 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).
Any second rank tensor can be written as a sum of its symmetric and
anti-symmetric parts
T
ij
= S
ij
+ A
ij
,
where
S
ij
=
1
2
(T
ij
+ T
ji
), A
ij
=
1
2
(T
ij
T
ji
).
Here
T
ij
has 9 independent components, whereas
S
ij
and
A
ij
have 6 and 3
independent components, since they must be of the form
(S
ij
) =
a d e
d b f
e f c
, (A
ij
) =
0 a b
a 0 c
b c 0
.
The symmetric part can be be further reduced to a traceless part plus an isotropic
(i.e. multiple of δ
ij
) part:
S
ij
= P
ij
+
1
3
δ
ij
Q,
where
Q
=
S
ii
is the trace of
S
ij
and
P
ij
=
P
ji
=
S
ij
1
3
δ
ij
Q
is traceless. Then
P
ij
has 5 independent components while Q has 1.
Since the antisymmetric part has 3 independent components, just like a usual
vector, we should be able to write
A
i
in terms of a single vector. In fact, we can
write the antisymmetric part as
A
ij
= ε
ijk
B
k
for some vector
B
. To figure out what this
B
is, we multiply by
ε
ij`
on both
sides and use some magic algebra to obtain
B
k
=
1
2
ε
ijk
A
ij
=
1
2
ε
ijk
T
ij
,
where the last equality is from the fact that only antisymmetric parts contribute
to the sum.
Then
(A
ij
) =
0 B
3
B
2
B
3
0 B
1
B
2
B
1
0
To summarize,
T
ij
= P
ij
+ ε
ijk
B
k
+
1
3
δ
ij
Q,
where B
k
=
1
2
ε
pqj
T
pq
, Q = T
kk
and P
ij
= P
ji
=
T
ij
+T
ji
2
1
3
δ
ij
Q.
Example.
The derivative of a vector field
F
i
(
r
) is a tensor
T
ij
=
F
i
x
j
, a tensor
field. Our decomposition given above has the symmetric traceless piece
P
ij
=
1
2
F
i
x
j
+
F
j
x
i
1
3
δ
ij
F
k
x
k
=
1
2
F
i
x
j
+
F
j
x
i
1
3
δ
ij
· F,
an antisymmetric piece A
ij
= ε
ijk
B
k
, where
B
k
=
1
2
ε
ijk
F
i
x
j
=
1
2
( × F)
k
.
and trace
Q =
F
k
x
k
= · F.
Hence a complete description involves a scalar
· F
, a vector
× F
, and a
symmetric traceless tensor P
ij
.
14.2 The inertia tensor
Consider masses
m
α
with positions
r
α
, all rotating with angular velocity
ω
about
0. So the velocities are v
α
= ω ×r
α
. The total angular momentum is
L =
X
α
r
α
× m
α
v
α
=
X
α
m
α
r
α
× (ω × r
α
)
=
X
α
m
α
(|r
α
|
2
ω (r
α
· ω )r
α
).
by vector identities. In components, we have
L
i
= I
ij
ω
j
,
where
Definition (Inertia tensor). The inertia tensor is
I
ij
=
X
α
m
α
[|r
α
|
2
δ
ij
(r
α
)
i
(r
α
)
j
].
For a rigid body occupying volume
V
with mass density
ρ
(
r
), we replace the
sum with an integral to obtain
I
ij
=
Z
V
ρ(r)(x
k
x
k
δ
ij
x
i
x
j
) dV.
By inspection, I is a symmetric tensor.
Example.
Consider a rotating cylinder with uniform density
ρ
0
. The total
mass is 2a
2
ρ
0
.
x
1
x
3
x
2
2`
a
Use cylindrical polar coordinates:
x
1
= r cos θ
x
2
= r sin θ
x
3
= x
3
dV = r dr dθ dx
3
We have
I
33
=
Z
V
ρ
0
(x
2
1
+ x
2
2
) dV
= ρ
0
Z
a
0
Z
2π
0
Z
`
`
r
2
(r dr dθ dx
2
)
= ρ
0
· 2π · 2`
r
4
4
a
0
= ε
0
π`a
4
.
Similarly, we have
I
11
=
Z
V
ρ
0
(x
2
2
+ x
2
3
) dV
= ρ
0
Z
a
0
Z
2π
0
Z
`
`
(r
2
sin
2
θ + x
2
3
)r dr dθ dx
3
= ρ
0
Z
a
0
Z
2π
0
r
r
2
sin
2
θ [x
3
]
`
`
+
x
3
3
3
`
`
!
dθ dr
= ρ
0
Z
a
0
Z
2π
0
r
r
2
sin
2
θ2` +
2
3
`
3
dθ dr
= ρ
0
2πa ·
2
3
`
3
+ 2`
Z
a
0
r
2
dr
Z
2π
0
sin
2
θ
= ρ
0
πa
2
`
a
2
2
+
2
3
`
2
By symmetry, the result for I
22
is the same.
How about the off-diagonal elements?
I
13
=
Z
V
ρ
0
x
1
x
3
dV
= ρ
0
Z
a
0
Z
`
`
Z
2π
0
r
2
cos θx
3
dr dx
3
dθ
= 0
Since
R
2π
0
d
θ cos θ
= 0. Similarly, the other off-diagonal elements are all 0. So
the non-zero components are
I
33
=
1
2
Ma
2
I
11
= I
22
= M
a
2
4
+
`
2
3
In the particular case where ` =
a
3
2
, we have I
ij
=
1
2
ma
2
δ
ij
. So in this case,
L =
1
2
Ma
2
ω
for rotation about any axis.
14.3 Diagonalization of a symmetric second rank tensor
Recall that using matrix notation,
T = (T
ij
), T
0
= (T
0
ij
), R = (R
ij
),
and the tensor transformation rule T
0
ij
= R
ip
R
jq
T
pq
becomes
T
0
= RT R
T
= RT R
1
.
If
T
is symmetric, it can be diagonalized by such an orthogonal transformation.
This means that there exists a basis of orthonormal eigenvectors
e
1
, e
2
, e
3
for
T
with real eigenvalues
λ
1
, λ
2
, λ
3
respectively. The directions defined by
e
1
, e
2
, e
3
are the principal axes for
T
, and the tensor is diagonal in Cartesian coordinates
along these axes.
This applies to any symmetric rank-2 tensor. For the special case of the
inertia tensor, the eigenvalues are called the principal moments of inertia.
As exemplified in the previous example, we can often guess the correct
principal axes for
I
ij
based on the symmetries of the body. With the axes we
chose, I
ij
was found to be diagonal by direct calculation.
15 Invariant and isotropic tensors
15.1 Definitions and classification results
Definition
(Invariant and isotropic tensor)
.
A tensor
T
is invariant under a
particular rotation R if
T
0
ij···k
= R
ip
R
jq
···R
kr
T
pq···r
= T
ij···k
,
i.e. every component is unchanged under the rotation.
A tensor
T
which is invariant under every rotation is isotropic, i.e. the same
in every direction.
Example. The inertia tensor of a sphere is isotropic by symmetry.
δ
ij
and
ε
ijk
are also isotropic tensors. This ensures that the component
definitions of the scalar and vector products
a·b
=
a
i
b
j
δ
ij
and (
a×b
)
i
=
ε
ijk
a
j
b
k
are independent of the Cartesian coordinate system.
Isotropic tensors in R
3
can be classified:
Theorem.
(i) There are no isotropic tensors of rank 1, except the zero tensor.
(ii) The most general rank 2 isotropic tensor is T
ij
= αδ
ij
for some scalar α.
(iii)
The most general rank 3 isotropic tensor is
T
ijk
=
βε
ijk
for some scalar
β
.
(iv)
All isotropic tensors of higher rank are obtained by combining
δ
ij
and
ε
ijk
using tensor products, contractions, and linear combinations.
We will provide a sketch of the proof:
Proof.
We analyze conditions for invariance under specific rotations through
π
or π/2 about coordinate axes.
(i) Suppose T
i
is rank-1 isotropic. Consider a rotation about x
3
through π:
(R
ij
) =
1 0 0
0 1 0
0 0 1
.
We want
T
1
=
R
ip
T
p
=
R
11
T
1
=
T
1
. So
T
1
= 0. Similarly,
T
2
= 0. By
consider a rotation about, say x
1
, we have T
3
= 0.
(ii) Suppose T
ij
is rank-2 isotropic. Consider
(R
ij
) =
0 1 0
1 0 0
0 0 1
,
which is a rotation through π/2 about the x
3
axis. Then
T
13
= R
1p
R
3q
T
pq
= R
12
R
33
T
23
= T
23
and
T
23
= R
2p
R
3q
T
pq
= R
21
R
33
T
13
= T
13
So T
13
= T
23
= 0. Similarly, we have T
31
= T
32
= 0.
We also have
T
11
= R
1p
R
1q
T
pq
= R
12
R
12
T
22
= T
22
.
So T
11
= T
22
.
By picking a rotation about a different axis, we have
T
21
=
T
12
and
T
22
= T
33
.
Hence T
ij
= αδ
ij
.
(iii)
Suppose that
T
ijk
is rank-3 isotropic. Using the rotation by
π
about the
x
3
axis, we have
T
133
= R
1p
R
3q
R
3r
T
pqr
= T
133
.
So T
133
= 0. We also have
T
111
= R
1p
R
1q
R
1r
T
pqr
= T
111
.
So
T
111
= 0. We have similar results for
π
rotations about other axes and
other choices of indices.
Then we can show that T
ijk
= 0 unless all i, j, k are distinct.
Now consider
(R
ij
) =
0 1 0
1 0 0
0 0 1
,
a rotation about x
3
through π/2. Then
T
123
= R
1p
R
2q
R
3r
T
pqr
= R
12
R
21
R
33
T
213
= T
213
.
So
T
123
=
T
213
. Along with similar results for other indices and axes of
rotation, we find that
T
ijk
is totally antisymmetric, and
T
ijk
=
βε
ijk
for
some β.
Example. The most general isotropic tensor of rank 4 is
T
ijk`
= αδ
ij
δ
k`
+ βδ
ik
δ
j`
+ γδ
i`
δ
jk
for some scalars
α, β, γ
. There are no other independent combinations. (we
might think we can write a rank-4 isotropic tensor in terms of
ε
ijk
, like
ε
ijp
ε
k`p
,
but this is just
δ
ik
δ
j`
δ
i`
δ
jk
. It turns out that anything you write with
ε
ijk
can be written in terms of δ
ij
instead)
15.2 Application to invariant integrals
We have the following very useful theorem. It might seem a bit odd and arbitrary
at first sight if so, read the example below first (after reading the statement
of the theorem), and things will make sense!
Theorem. Let
T
ij···k
=
Z
V
f(x)x
i
x
j
···x
k
dV.
where f (x) is a scalar function and V is some volume.
Given a rotation
R
ij
, consider an active transformation:
x
=
x
i
e
i
is mapped
to
x
0
=
x
0
i
e
i
with
x
0
i
=
R
ij
x
i
, i.e. we map the components but not the basis, and
V is mapped to V
0
.
Suppose that under this active transformation,
(i) f (x) = f(x
0
),
(ii) V
0
= V (e.g. if V is all of space or a sphere).
Then T
ij···k
is invariant under the rotation.
Proof.
First note that the Jacobian of the transformation
R
is 1, since it is
simply the determinant of
R
(
x
0
i
=
R
ip
x
p
x
0
i
x
p
=
R
ip
), which is by definition
1. So dV = dV
0
.
Then we have
R
ip
R
jq
···R
kr
T
pq···r
=
Z
V
f(x)x
0
i
x
0
j
···x
0
k
dV
=
Z
V
f(x
0
)x
0
i
x
0
j
···x
0
k
dV using (i)
=
Z
V
0
f(x
0
)x
0
i
x
0
j
···x
0
k
dV
0
using (ii)
=
Z
V
f(x)x
i
x
j
···x
k
dV since x
i
and x
0
i
are dummy
= T
ij···k
.
The result is particularly useful if (i) and (ii) hold for any rotation
R
, in
which case T
ij···k
is isotropic.
Example. Let
T
ij
=
Z
V
x
i
x
j
dV,
with
V
being a solid sphere of
|r| < a
. Our result applies with
f
= 1, which,
being a constant, is clearly invariant under rotations. Also the solid sphere is
invariant under any rotation. So
T
must be isotropic. But the only rank 2
isotropic tensor is αδ
ij
. Hence we must have
T
ij
= αδ
ij
,
and all we have to do is to determine the scalar α.
Taking the trace, we have
T
ii
= 3α =
Z
V
x
i
x
i
dV = 4π
Z
a
0
r
2
· r
2
dr =
4
5
πa
5
.
So
T
ij
=
4
15
πa
5
δ
ij
.
Normally if we are only interested in the
i 6
=
j
case, we just claim that
T
ij
= 0
by saying “by symmetry, it is 0”. 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 ρ
0
, or of mass M =
4
3
πa
3
ρ
0
.
Recall that
I
ij
=
Z
V
ρ
0
(x
k
x
k
δ
ij
x
i
x
j
) dV.
We see that
I
ij
is isotropic (since we have just shown that
R
x
i
x
j
d
V
is isotropic,
and x
k
x
k
δ
ij
is also isotropic). Let I
ij
= βδ
ij
. Then
I
ij
=
Z
V
ρ
0
(x
k
x
k
δ
ij
x
i
x
j
) dV
= ρ
0
δ
ij
Z
V
x
k
x
k
dV
Z
V
x
i
x
j
dV
= ρ
0
(δ
ij
T
kk
T
ij
)
= ρ
0
4
5
πa
5
δ
ij
4
15
πa
5
δ
ij
=
8
15
ρ
0
πa
5
δ
ij
=
2
5
Ma
2
δ
ij
.