Page 39 - Intro to Tensor Calculus

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§1.2 TENSOR CONCEPTS AND TRANSFORMATIONS

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For b e 1 , b e 2 , b e 3 independent orthogonal unit vectors (base vectors), we may write any vector A as
~
A = A 1 b e 1 + A 2 b e 2 + A 3 b e 3

~
where (A 1 ,A 2 ,A 3 ) are the coordinates of A relative to the base vectors chosen. These components are the
~
projection of A onto the base vectors and
~
~
~
~
A =(A · b e 1 ) b e 1 +(A · b e 2 ) b e 2 +(A · b e 3 ) b e 3 .
~
~
~
Select any three independent orthogonal vectors, (E 1 , E 2 , E 3 ), not necessarily of unit length, we can then
write
~ ~ ~
E 1 E 2 E 3
b e 1 = , b e 2 = , b e 3 = ,
~
~
~
|E 1 | |E 2 | |E 3 |
~
and consequently, the vector A can be expressed as
! ! !
~ ~
~ ~
~ ~ A · E 2 A · E 3
A · E 1
~
~
~
~
A = E 1 + E 2 + E 3 .
~ ~ ~ ~ ~ ~
E 1 · E 1 E 2 · E 2 E 3 · E 3
Here we say that
~ ~
A · E (i)
, i =1, 2, 3
~ ~
E (i) · E (i)
~
~
~
~
are the components of A relative to the chosen base vectors E 1 , E 2 , E 3 . Recall that the parenthesis about
the subscript i denotes that there is no summation on this subscript. It is then treated as a free subscript
which can have any of the values 1, 2or 3.
Reciprocal Basis
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~
~
Consider a set of any three independent vectors (E 1 , E 2 , E 3 ) which are not necessarily orthogonal, nor of
~
1
3
2
unit length. In order to represent the vector A in terms of these vectors we must find components (A ,A ,A )
such that
1 ~
2 ~
3 ~
~
A = A E 1 + A E 2 + A E 3 .
This can be done by taking appropriate projections and obtaining three equations and three unknowns from
1
2
3
which the components are determined. A much easier way to find the components (A ,A ,A ) is to construct
~
~
~
~ 1 ~ 2 ~ 3
~ 1 ~ 2 ~ 3
a reciprocal basis (E , E , E ). Recall that two bases (E 1 , E 2 , E 3 )and (E , E , E ) are said to be reciprocal
if they satisfy the condition
1 if i = j
j
~
~ j
E i · E = δ = .
i
0 if i 6= j
~
~ 1
~
1
~ 1
~ 1
1
Note that E 2 · E = δ =0 and E 3 · E = δ = 0 so that the vector E is perpendicular to both the
2
3
~
~
vectors E 2 and E 3 . (i.e. A vector from one basis is orthogonal to two of the vectors from the other basis.)
~ 1
We can therefore write E = V −1 ~ ~
E 2 × E 3 where V is a constant to be determined. By taking the dot
~
~
~
~
product of both sides of this equation with the vector E 1 we find that V = E 1 · (E 2 × E 3 ) is the volume
~
~
~
of the parallelepiped formed by the three vectors E 1 , E 2 , E 3 when their origins are made to coincide. In a