14.6 Vector Identities And Notation
To begin with consider u×
and it is desired to simplify this expression. It turns
out this expression comes up in many different contexts. Let
be defined similarly.
Next consider u×
which is given by
When you multiply this out, you get
and if you are clever, you see right away that
A related formula is
This derivation is simply wretched and it does nothing for other identities which may
arise in applications. Actually, the above two formulas, 14.24
are sufficient for
most applications if you are creative in using them, but there is another way. This other
way allows you to discover such vector identities as the above without any creativity
or any cleverness. Therefore, it is far superior to the above nasty and tedious
computation. It is a vector identity discovering machine and it is this which is
the main topic in what follows. I cannot understand why it is not routinely
presented in calculus texts. The engineers I have known seem to know all about it.
There are two special symbols, δij and εijk which are very useful in dealing with
vector identities. To begin with, here is the definition of these symbols.
Definition 14.6.1 The symbol δij, called the Kroneker delta symbol is defined
With the Kroneker symbol i and j can equal any integer in
n ∈ ℕ.
Definition 14.6.2 For i,j, and k integers in the set,
, εijk is defined as
The subscripts ijk and ij in the above are called indices. A single one is called an index.
This symbol εijk is also called the permutation symbol.
The way to think of εijk is that ε123 = 1 and if you switch any two of the numbers in
the list i,j,k, it changes the sign. Thus εijk = −εjik and εijk = −εkji etc. You should
check that this rule reduces to the above definition. For example, it immediately implies
that if there is a repeated index, the answer is zero. This follows because εiij = −εiij and
so εiij = 0.
It is useful to use the Einstein summation convention when dealing with
these symbols. Simply stated, the convention is that you sum over the repeated
index. Thus aibi means ∑
iaibi. Also, δijxj means ∑
jδijxj = xi. When you
use this convention, there is one very important thing to never forget. It is
this: Never have an index be repeated more than once. Thus aibi is all right but aiibi
is not. The reason for this is that you end up getting confused about what is
meant. If you want to write ∑
iaibici it is best to simply use the summation
notation. There is a very important reduction identity connecting these two
Lemma 14.6.3 The following holds.
then every term in the sum on the left must have either
contains a repeated index. Therefore, the left side equals zero. The right side also
equals zero in this case. To see this, note that if the two sets of indices are not equal, then
there is one of the indices in one of the sets which is not in the other set. For
example, it could be that j
is not equal to either r
Then the right side equals
Therefore, it can be assumed
then there is
exactly one term in the sum on the left and it equals 1. The right also reduces
to 1 in this case. If i
there is exactly one term in the sum on
the left which is nonzero and it must equal −
1. The right side also reduces to
1 in this case. If there is a repeated index in
then every term in the
sum on the left equals zero. The right also reduces to zero in this case because
and so the right side becomes
Proposition 14.6.4 Let u,v be vectors in ℝn where the Cartesian coordinates of u
and the Cartesian coordinates of v are
. Then u ⋅ v
= uivi. If
u,v are vectors in ℝ3, then
Also, δikak = ai.
Proof: The first claim is obvious from the definition of the dot product. The second
is verified by simply checking it works. For example,
From the above formula in the proposition,
the same thing. The cases for
are verified similarly. The last
claim follows directly from the definition. ■
With this notation, you can easily discover vector identities and simplify expressions
which involve the cross product.
Example 14.6.5 Discover a formula which simplifies
From the above reduction formula,
Since this holds for all i,
it follows that