Page 21 - Basic Structured Grid Generation

10 Basic Structured Grid Generation

Exercise 4. Show that if we define the matrix B as that whose i-j element is equal to
i
j
∂x /∂x ,then
T
AB = I (1.60)
and
det B = J −1 . (1.61)

We obtain new covariant base vectors, which transform according to the rule
∂r ∂r ∂x j ∂x j
g = i = j i = i g j , (1.62)
i
∂x ∂x ∂x ∂x
with the inverse relationship
∂x j
g i = g . (1.63)
j
∂x i
In background cartesian components, eqn (1.62) may be written in matrix form as
L = BL, (1.64)
i
where L is the matrix with i-j component given by ∂y j /∂x , and from eqn (1.23) and
eqn (1.60) we deduce that
M = AM, (1.65)
i
where M is the matrix with i-j component ∂x /∂y j .
The new system of co-ordinates has associated metric tensors given, in comparison
with eqn (1.26), by
T ij T
(g ) = L L , (g ) = M M , (1.66)
ij
so that the corresponding determinant g = det(g ) is given by
ij
2
g = (det L) .
√ √
Hence det L = g,det L = g, and det L = det B det L = J −1 det L from
eqn (1.64). Thus we have
g

J = . (1.67)
g
Equation (1.65) yields the relation between corresponding contravariant base vectors:

∂x i
i j
g = g . (1.68)
∂x j
Expressing u as a linear combination of the base vectors in the new system gives
i
i
u = u g = u i g . (1.69)
i
We now easily obtain, using eqn (1.62), the transformation rule for the covariant
components of a vector:
j j j
∂x ∂x ∂x
u i = u·g = u· i g j = i u · g = i u j , (1.70)
j
i
∂x ∂x ∂x