Page 19 - Basic Structured Grid Generation
P. 19

8  Basic Structured Grid Generation

                           1.4 Line, area, and volume elements


                        Lengths of general infinitesimal line-elements are given by eqn (1.21). An element of
                             1                           2      3                            2
                        the x co-ordinate curve on which dx = dx = 0 is therefore given by (ds) =
                              1 2
                                                         i
                        g 11 (dx ) . Thus arc-length along the x -curve is
                                                           √      i
                                                      ds =   g ii dx                       (1.42)
                        (with no summation over i).
                                                  1
                                                                                      1
                          A line-element along the x -curve may be written  ∂r  dx 1  = g 1 dx ,and simi-
                                                                        ∂x  1
                                                                2
                                                   2
                        larly a line-element along the x -curve is g 2 dx . The infinitesimal vector area of the
                        parallelogram of which these two line-elements form the sides is the vector product
                                     2
                              1
                        (g 1 dx × g 2 dx ), which has magnitude
                                                                     2
                                                                  1
                                                  dA 3 =|g 1 × g 2 | dx dx .               (1.43)
                        Again by the Lagrange vector identity we have
                                      2
                              |g 1 × g 2 | = (g 1 × g 2 ) · (g 1 × g 2 ) = (g 1 · g 1 )(g 2 · g 2 ) − (g 1 · g 2 )(g 1 · g 2 )
                                                     2
                                       = g 11 g 22 − (g 12 ) .
                                    
                 1  2
                                                  2
                        Hence dA 3 =  g 11 g 22 − (g 12 ) dx dx , giving the general expression

                                                                               k
                                                                   k
                                                                           j
                                                               j
                                                            2
                                        dA i =  g jj g kk − (g jk ) dx dx = G i dx dx ,    (1.44)
                        using eqn (1.34), where i, j, k must be taken in cyclic order 1, 2, 3, and again there is
                        no summation over j and k.
                                                                      1
                                                                                   3
                                                                             2
                          The parallelepiped generated by line-elements g 1 dx , g 2 dx , g 3 dx , along the co-
                        ordinate curves has infinitesimal volume
                                                                                     3
                                            1
                                                                                  2
                                                   2
                                                           3
                                                                              1
                                  dV = g 1 dx · (g 2 dx × g 3 dx ) ={g 1 · (g 2 × g 3 )}dx dx dx .
                        By eqn (1.31) we have
                                                        √     1  2  3
                                                   dV =   g dx dx dx .                     (1.45)
                           1.5 Generalized vectors and tensors
                        A vector field u (a function of position r) may be expressed at a point P in terms of
                                                                                               1
                        the covariant base vectors g 1 , g 2 , g 3 , or in terms of the contravariant base vectors g ,
                         2
                             3
                        g , g . Thus we have
                                                    1     2     3      i
                                               u = u g 1 + u g 2 + u g 3 = u g i           (1.46)
                                                            2
                                                     1
                                                                  3
                                                                         i
                                                = u 1 g + u 2 g + u 3 g = u i g ,          (1.47)
                               i
                        where u and u i are called the contravariant and covariant components of u, respect-
                                                                                j
                        ively. Taking the scalar product of both sides of eqn (1.46) with g gives
                                                   j    i    j   i j    j
                                               u · g = u g i · g = u δ = u .
                                                                   i
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