Mathematical Methods for Physics and Engineering - Matematica.NET

VECTOR CALCULUSFor orthogonal coordinates this is given bydV = |du 1 e 1 · (du 2 e 2 × du 3 e 3 )|= |h 1 ê 1 · (h 2 ê 2 × h 3 ê 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 {ê i }, i =1, 2, 3, there exists another useful set ofthree unit basis vectors at P .Since∇u 1 is a vector normal to the surface u 1 = c 1 ,a unit vector in this direction is ˆɛ 1 = ∇u 1 /|∇u 1 |. Similarly, ˆɛ 2 = ∇u 2 /|∇u 2 | andˆɛ 3 = ∇u 3 /|∇u 3 | are unit vectors normal to the surfaces u 2 = c 2 and u 3 = c 3respectively.Therefore at each point P in a curvilinear coordinate system, there exist, ingeneral, two sets of unit vectors: {ê i }, tangent to the coordinate curves, and {ˆɛ i },normal to the coordinate surfaces. A vector a can be written in terms of eitherset of unit vectors:a = a 1 ê 1 + a 2 ê 2 + a 3 ê 3 = A 1ˆɛ 1 + A 2ˆɛ 2 + A 3ˆɛ 3 ,where a 1 , a 2 , a 3 and A 1 , A 2 , A 3 are the components of a in the two systems. Itmay be shown that the two bases become identical if the coordinate system isorthogonal.Instead of the unit vectors discussed above, we could instead work directly withthe two sets of vectors {e i = ∂r/∂u i } and {ɛ i = ∇u i },whicharenot,ingeneral,ofunit length. We can then write a vector a asa = α 1 e 1 + α 2 e 2 + α 3 e 3 = β 1 ɛ 1 + β 2 ɛ 2 + β 3 ɛ 3 ,or more explicitly as∂r ∂r ∂ra = α 1 + α 2 + α 3 = β 1 ∇u 1 + β 2 ∇u 2 + β 3 ∇u 3 ,∂u 1 ∂u 2 ∂u 3where α 1 ,α 2 ,α 3 and β 1 ,β 2 ,β 3 are called the contravariant and covariant componentsof a respectively. A more detailed discussion of these components, inthe context of tensor analysis, is given in chapter 26. The (in general) non-unitbases {e i } and {ɛ i } are often the most natural bases in which to express vectorquantities.◮Show that {e i } and {ɛ i } are reciprocal systems of vectors.Let us consider the scalar product e i · ɛ j ; using the Cartesian expressions for r and ∇, weobtaine i · ɛ j = ∂r · ∇u j∂u i)=( ∂x∂u ii + ∂y∂u ij + ∂z∂u ik= ∂x∂u i∂u j∂x + ∂y∂u i∂u j∂y + ∂z∂u i∂u j∂z = ∂u j∂u i.366( ∂uj·∂x i + ∂u j∂y j + ∂u )j∂z k