Page 59 - Basic Structured Grid Generation

48 Basic Structured Grid Generation

Exercise 3. Show (by taking determinants of both sides of eqn (3.18)) that the property
of positive definiteness of a αβ is preserved under co-ordinate transformations provided
the Jacobian of the transformation is non-zero.
Vectors in the tangent plane to S at a point P of the surface are linear combinations
of the tangent vectors a 1 and a 2 , and may be called surface vectors. The vectors a α
thus serve as covariant base vectors for the tangent plane, and a surface vector A with
α
α
A = A a α has contravariant components A ,α = 1, 2. An equation analogous to
eqn (1.55) now gives the length of a surface vector A :

β
α
|A|= a αβ A A . (3.26)
α
We can also define surface contravariant base vectors a ,α = 1, 2, such that
α
α
a · a β = δ , (3.27)
β
and the surface contravariant metric tensor
α
β
a αβ = a · a . (3.28)
The symmetric 2 × 2 matrix array corresponding to (3.28) is the inverse of that
corresponding to eqn (3.17), that is,
γ
a αβ a αγ = δ . (3.29)
β
Thus we have, explicitly,
11 12 21 22
a = a 22 /a, a = a =−a 12 /a, a = a 11 /a. (3.30)
αβ
α
Exercise 4. Show that a = a a β with
1 1
1 2
a = (a 22 a 1 − a 12 a 2 ) and a = (−a 12 a 1 + a 11 a 2 ). (3.31)
a a
3
Exercise 5. A surface of revolution may be generated in E by rotating the curve in
the cartesian plane Oxz given in parametric form by x = f(u), z = g(u) about the
axis Oz. This gives the parametric form
x = f(u) cos v, y = f(u) sin v, z = g(u), (3.32)
where u and v may be used as surface co-ordinates. Show that the covariant surface
1
2
base vectors, with u = u and v = u ,are

a 1 = (f (u) cos v, f (u) sin v, g (u)), a 2 = (−f(u) sin v, f (u) cos v, 0)
in background cartesian co-ordinates and that the covariant metric tensor has
components
2 2
f + g 0
a αβ = 2 , (3.33)
0 f
which are functions of u but not v, while the contravariant metric tensor is
2 2 −1
αβ (f + g ) 0
a = −2 . (3.34)
0 f