Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(lu) #1

VECTOR CALCULUS


For orthogonal coordinates this is given by


dV=|du 1 e 1 ·(du 2 e 2 ×du 3 e 3 )|

=|h 1 eˆ 1 ·(h 2 eˆ 2 ×h 3 ˆe 3 )|du 1 du 2 du 3

=h 1 h 2 h 3 du 1 du 2 du 3.

Now, in addition to the set{ˆei},i=1, 2 ,3, there exists another useful set of

three unit basis vectors atP.Since∇u 1 is a vector normal to the surfaceu 1 =c 1 ,


a unit vector in this direction isˆ 1 =∇u 1 /|∇u 1 |. Similarly,ˆ 2 =∇u 2 /|∇u 2 |and


ˆ 3 =∇u 3 /|∇u 3 |are unit vectors normal to the surfacesu 2 =c 2 andu 3 =c 3


respectively.


Therefore at each pointPin a curvilinear coordinate system, there exist, in

general, two sets of unit vectors:{ˆei}, tangent to the coordinate curves, and{ˆi},


normal to the coordinate surfaces. A vectoracan be written in terms of either


set of unit vectors:


a=a 1 ˆe 1 +a 2 ˆe 2 +a 3 eˆ 3 =A 1 ˆ 1 +A 2 ˆ 2 +A 3 ˆ 3 ,

wherea 1 ,a 2 ,a 3 andA 1 ,A 2 ,A 3 are the components ofain the two systems. It


may be shown that the two bases become identical if the coordinate system is


orthogonal.


Instead of theunitvectors discussed above, we could instead work directly with

the two sets of vectors{ei=∂r/∂ui}and{i=∇ui},whicharenot,ingeneral,of


unit length. We can then write a vectoraas


a=α 1 e 1 +α 2 e 2 +α 3 e 3 =β 1  1 +β 2  2 +β 3  3 ,

or more explicitly as


a=α 1

∂r
∂u 1

+α 2

∂r
∂u 2

+α 3

∂r
∂u 3

=β 1 ∇u 1 +β 2 ∇u 2 +β 3 ∇u 3 ,

whereα 1 ,α 2 ,α 3 andβ 1 ,β 2 ,β 3 are called thecontravariantandcovariantcom-


ponents ofarespectively. A more detailed discussion of these components, in


the context of tensor analysis, is given in chapter 26. The (in general) non-unit


bases{ei}and{i}are often the most natural bases in which to express vector


quantities.


Show that{ei}and{i}are reciprocal systems of vectors.

Let us consider the scalar productei·j; using the Cartesian expressions forrand∇,we
obtain


ei·j=

∂r
∂ui

·∇uj

=


(


∂x
∂ui

i+

∂y
∂ui

j+

∂z
∂ui

k

)


·


(


∂uj
∂x

i+

∂uj
∂y

j+

∂uj
∂z

k

)


=


∂x
∂ui

∂uj
∂x

+


∂y
∂ui

∂uj
∂y

+


∂z
∂ui

∂uj
∂z

=


∂uj
∂ui

.

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