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Lecture 4: Suffix Notation
Introductory Fields and Waves (PHYS08053)
The University of Edinburgh
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IFW: Fields and Vector Calculus
Section 4: Using Suffix Notation
- More onδij
Last lecture we defined the Kronecker delta symbol such that an orthornormal basis sete i
i= 1, 2 ,3 obeys
e i
·e j
= δ ij where
δ ij = 1 wheni=j
= 0 wheni 6 =j
Notes
- Since there are two free indicesiandj, e i
·e j
=δij is equivalent to 9 equations.
- δ ij =δ ji . We sayδ ij issymmetricin its indices.
3.
3 ∑
i=
δ ii =δ 11 +δ 22 +δ 33
= 3
4.
3 ∑
j=
a j δ jk =a 1 δ 1 k +a 2 δ 2 k +a 3 δ 3 k
To go further, first note thatkis a free index.
Ifk= 1, then only the first term on the RHS contributes and the RHS =a 1. Similarly,
ifk= 2 then the RHS =a 2 , and ifk= 3 the RHS =a 3 . Hence
3 ∑
j=
a j δ jk =a k
In other words, Kronecker deltaδ jk picks out thek
th term in the sum overj.
Generalising the reasoning in 4 the so-calledsifting property of Kronecker delta
3 ∑
j=
(anything)jδjk= (anything )k
where (anything) k denotes any expression that has a free indexk(appearing once).
Examples of the use of Kronecker delta
- a·e j
=
( 3 ∑
i=
a i e i
)
·e j
=
3 ∑
i=
a i (e i
·e j
)
=
3 ∑
i=
a i δ ij = a j because terms withi 6 =jvanish.
- a·b =
( 3 ∑
i=
a i e i
)
·
3 ∑
j=
b j e j
=
3 ∑
i=
3 ∑
j=
a i b j (e i
·e j
) =
3 ∑
i=
3 ∑
j=
a i b j δ ij
=
3 ∑
i=
aibi
or
3 ∑
j=
ajbj
- Free Indices and Summation Indices
Consider, for example, the vector equation
a−(b·c)d+ 3n= 0 (1)
The basis vectors are linearly independent, so this equation must hold for each component
separately
a i −(b·c)d i + 3n i = 0 for i= 1, 2 , 3 (2)
The free indexioccursonceandonly oncein each term of the equation (2). In general
every term in the equation must be of the same kind,i.e the same free indices.
Now suppose that we want to write the scalar product that appears in the second term of
equation (2) in suffix notation. As we have seen, summation indices are ‘dummy’ indices
and can be relabelled. For example
b·c=
3 ∑
i=
b i c i
=
3 ∑
k=
b k c k
This freedom shouldalways be used to avoid confusion with other indices in the equation.
In this case, we avoid usingias a summation index, as we have already used it as a free
index, and rewrite equation (2) as
a i
−
( 3 ∑
k=
b k c k
)
d i + 3n i = 0 for i= 1, 2 , 3
rather than
ai−
( 3 ∑
i=
bici
)
di+ 3ni= 0 WRONG
which would lead to great confusion and inevitably lead to mistakes when the brackets are
removed – as they will be very soon.
Then thedeterminantof the matrixAis defined as
detA ≡
∣ ∣ ∣ ∣ ∣ ∣ ∣
a 11 a 12 a 13
a 21 a 22 a 23
a 31 a 32 a 33
∣ ∣ ∣ ∣ ∣ ∣ ∣
=a 11
∣ ∣ ∣ ∣ ∣
a 22 a 23
a 32 a 33
∣ ∣ ∣ ∣ ∣
−a 12
∣ ∣ ∣ ∣ ∣
a 21 a 23
a 31 a 33
∣ ∣ ∣ ∣ ∣
+a 13
∣ ∣ ∣ ∣ ∣
a 21 a 22
a 31 a 32
∣ ∣ ∣ ∣ ∣
= a 11 (a 22 a 33 −a 23 a 32 )−a 12 (a 21 a 33 −a 23 a 31 ) +a 13 (a 21 a 32 −a 22 a 31 )
Now the expression for the scalar triple product is
a·(b×c) =
3 ∑
i=
a i (b×c) i
= a 1 (b 2 c 3 −c 2 b 3 )−a 2 (b 1 c 3 −c 1 b 3 ) +a 3 (b 1 c 2 −c 1 b 2
)
and we deduce the convenient formula to be memorised
a·(b×c) =
∣ ∣ ∣ ∣ ∣ ∣ ∣
a 1 a 2 a 3
b 1 b 2 b 3
c 1 c 2 c 3
∣ ∣ ∣ ∣ ∣ ∣ ∣
- Summary of the algebraic approach to vectors
We are now able to define vectors and the various products of vectors in an algebraic way
(as opposed to the geometrical approach of lectures 1 and 2).
Avectorisrepresented(in some orthonormal basise 1
, e 2
, e 3
) by an ordered set of 3 numbers
with certain laws of addition. For example
a is represented by
a 1
a 2
a 3
a+b is represented by
a 1 +b 1
a 2 +b 2
a 3 +b 3
It is important to remember that the representationdepends on the basis. We usually use a
column vector, as we do here. Sometimes we use a row vector which we denotea
T .
The various ‘products’ of vectors are now defined as follows:
The Scalar Productis denoted bya·banddefined as
a·b ≡
∑
i
a i b i
a·a = a
2 defines the magnitudeaof the vector.
The Vector Productis denoted bya×banddefinedin a right-handed basis as
a×b≡
∣ ∣ ∣ ∣ ∣ ∣ ∣
e 1
e 2
e 3
a 1 a 2 a 3
b 1 b 2 b 3
∣ ∣ ∣ ∣ ∣ ∣ ∣
The Scalar Triple Product
(a, b, c) ≡
∑
i
a i (b×c) i
=
∣ ∣ ∣ ∣ ∣ ∣ ∣
a 1 a 2 a 3
b 1 b 2 b 3
c 1 c 2 c 3
∣ ∣ ∣ ∣ ∣ ∣ ∣
In all the above formula the summations imply sums over dummyindexitaking values
1 , 2 ,3.
- The Einstein Summation Convention
The novelty of writing out summations soon wears thin. The standard way to avoid this
tedium is to adopt the Einstein summation convention. By adheringstrictlyto the following
conventions or “rules” the summation signs are suppressed completely.
Rules of the summation convention
Omitallsummation signs.
If a suffix appearstwice, a summation is implied,e.g=a 1 b 1 +a 2 b 2 +a 3 b 3.
Hereiis adummyorrepeatedindex.
- If a suffix appears onlyonceit can take any valuee.g i =b i holds fori= 1, 2 ,3.
Hereiis afreeindex. Note that there may be more than one free index.
Alwayscheck that the free indices match on both sides of an equation.
For example,a j =b i is WRONG.
- A given suffix mustnotappear more thantwicein any term in an expression.
Alwayscheck that there aren’t more than two identical indicese.g is simply
WRONG.
Examples
a=aie i
(iis a dummy index)
a·e j
=a i e i
·e j
=a i δ ij =a j (iis a dummy index, butjis a free index)
a·b= (a i e i
)·(b j e j
) =a i b j δ ij =a j b j (i, jare both dummy indices)
(a·b)(a·c) =a i b i a j c j (againi, jare dummy indices)
Armed with the summation convention one can rewrite many of the equations from the
previous sections without summation signs,e.g sifting property ofδ ij now becomes
[.. .]jδjk= [.. .]k
The repeated indexjis implicitly summed over, so that, for example,δ ij δ jk =δ ik
.
From now on, except where indicated, the summation convention will be assumed.
You should make sure that you are completely at ease with it.
Lecture 4: Suffix Notation
Module: Introductory Fields and Waves (PHYS08053)
University: The University of Edinburgh
