2Vectors
IA Vectors and Matrices
2.8 Suffix notation
Here we are going to introduce a powerful notation that can help us simplify a
lot of things.
First of all, let
v ∈ R
3
. We can write
v
=
v
1
e
1
+
v
2
e
2
+
v
3
e
3
= (
v
1
, v
2
, v
3
).
So in general, the
i
th component of
v
is written as
v
i
. We can thus write
vector equations in component form. For example,
a
=
b → a
i
=
b
i
or
c
=
αa
+
βb → c
i
=
αa
i
+
βb
i
. A vector has one free suffix,
i
, while a scalar
has none.
Notation
(Einstein’s summation convention)
.
Consider a sum
x · y
=
P
x
i
y
i
.
The summation convention says that we can drop the
P
symbol and simply
write x · y = x
i
y
i
. If suffixes are repeated once, summation is understood.
Note that
i
is a dummy suffix and doesn’t matter what it’s called, i.e.
x
i
y
i
= x
j
y
j
= x
k
y
k
etc.
The rules of this convention are:
(i) Suffix appears once in a term: free suffix
(ii) Suffix appears twice in a term: dummy suffix and is summed over
(iii) Suffix appears three times or more: WRONG!
Example. [(a · b)c − (a · c)b]
i
= a
j
b
j
c
i
− a
j
c
j
b
i
summing over j understood.
It is possible for an item to have more than one index. These objects are
known as tensors, which will be studied in depth in the IA Vector Calculus
course.
Here we will define two important tensors:
Definition (Kronecker delta).
δ
ij
=
(
1 i = j
0 i 6= j
.
We have
δ
11
δ
12
δ
13
δ
21
δ
22
δ
23
δ
31
δ
32
δ
33
=
1 0 0
0 1 0
0 0 1
= I.
So the Kronecker delta represents an identity matrix.
Example.
(i) a
i
δ
i1
= a
1
. In general, a
i
δ
ij
= a
j
(i is dummy, j is free).
(ii) δ
ij
δ
jk
= δ
ik
(iii) δ
ii
= n if we are in R
n
.
(iv) a
p
δ
pq
b
q
= a
p
b
p
with p, q both dummy suffices and summed over.
Definition
(Alternating symbol
ε
ijk
)
.
Consider rearrangements of 1
,
2
,
3. We
can divide them into even and odd permutations. Even permutations include
(1
,
2
,
3), (2
,
3
,
1) and (3
,
1
,
2). These are permutations obtained by performing
two (or no) swaps of the elements of (1
,
2
,
3). (Alternatively, it is any “rotation”
of (1, 2, 3))
The odd permutations are (2
,
1
,
3), (1
,
3
,
2) and (3
,
2
,
1). They are the
permutations obtained by one swap only.
Define
ε
ijk
=
+1 ijk is even permutation
−1 ijk is odd permutation
0 otherwise (i.e. repeated suffices)
ε
ijk
has 3 free suffices.
We have
ε
123
=
ε
231
=
ε
312
= +1 and
ε
213
=
ε
132
=
ε
321
=
−
1.
ε
112
=
ε
111
= ··· = 0.
We have
(i) ε
ijk
δ
jk
= ε
ijj
= 0
(ii)
If
a
jk
=
a
kj
(i.e.
a
ij
is symmetric), then
ε
ijk
a
jk
=
ε
ijk
a
kj
=
−ε
ikj
a
kj
.
Since
ε
ijk
a
jk
=
ε
ikj
a
kj
(we simply renamed dummy suffices), we have
ε
ijk
a
jk
= 0.
Proposition. (a × b)
i
= ε
ijk
a
j
b
k
Proof. By expansion of formula
Theorem. ε
ijk
ε
ipq
= δ
jp
δ
kq
− δ
jq
δ
kp
Proof. Proof by exhaustion:
RHS =
+1 if j = p and k = q
−1 if j = q and k = p
0 otherwise
LHS: Summing over
i
, the only non-zero terms are when
j, k 6
=
i
and
p, q 6
=
i
.
If
j
=
p
and
k
=
q
, LHS is (
−
1)
2
or (+1)
2
= 1. If
j
=
q
and
k
=
p
, LHS is
(+1)(−1) or (−1)(+1) = −1. All other possibilities result in 0.
Equally, we have ε
ijk
ε
pqk
= δ
ip
δ
jq
− δ
jp
δ
iq
and ε
ijk
ε
pjq
= δ
ip
δ
kq
− δ
iq
δ
kp
.
Proposition.
a · (b × c) = b · (c × a)
Proof. In suffix notation, we have
a · (b × c) = a
i
(b × c)
i
= ε
ijk
b
j
c
k
a
i
= ε
jki
b
j
c
k
a
i
= b · (c × a).
Theorem (Vector triple product).
a × (b × c) = (a · c)b − (a · b)c.
Proof.
[a × (b × c)]
i
= ε
ijk
a
j
(b × c)
k
= ε
ijk
ε
kpq
a
j
b
p
c
q
= ε
ijk
ε
pqk
a
j
b
p
c
q
= (δ
ip
δ
jq
− δ
iq
δ
jp
)a
j
b
p
c
q
= a
j
b
i
c
j
− a
j
c
i
b
j
= (a · c)b
i
− (a · b)c
i
Similarly, (a × b) × c = (a · c)b − (b · c)a.
Spherical trigonometry
Proposition. (a × b) · (a × c) = (a · a)(b · c) − (a · b)(a · c).
Proof.
LHS = (a × b)
i
(a × c)
i
= ε
ijk
a
j
b
k
ε
ipq
a
p
c
q
= (δ
jp
δ
kq
− δ
jq
δ
kp
)a
j
b
k
a
p
c
q
= a
j
b
k
a
j
c
k
− a
j
b
k
a
k
c
j
= (a · a)(b · c) − (a · b)(a · c)
Consider the unit sphere, center O, with a, b, c on the surface.
A
B C
δ(A, B)
α
Suppose we are living on the surface of the sphere. So the distance from
A
to
B
is
the arc length on the sphere. We can imagine this to be along the circumference
of the circle through
A
and
B
with center
O
. So the distance is
∠AOB
, which we
shall denote by
δ
(
A, B
). So
a · b
=
cos ∠AOB
=
cos δ
(
A, B
). We obtain similar
expressions for other dot products. Similarly, we get |a × b| = sin δ(A, B).
cos α =
(a × b) · (a × c)
|a × b||a × c|
=
b · c − (a · b)(a · c)
|a × b||a × c|
Putting in our expressions for the dot and cross products, we obtain
cos α sin δ(A, B) sin δ(A, C) = cos δ(B, C) −cos δ(A, B) cos δ(A, C).
This is the spherical cosine rule that applies when we live on the surface of a
sphere. What does this spherical geometry look like?
Consider a spherical equilateral triangle. Using the spherical cosine rule,
cos α =
cos δ − cos
2
δ
sin
2
δ
= 1 −
1
1 + cos δ
.
Since
cos δ ≤
1, we have
cos α ≤
1
2
and
α ≥
60
◦
. Equality holds iff
δ
= 0, i.e. the
triangle is simply a point. So on a sphere, each angle of an equilateral triangle is
greater than 60
◦
, and the angle sum of a triangle is greater than 180
◦
.