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

◦

.