Page 71 - Intro to Tensor Calculus
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               EXAMPLE 1.3-1.       Show the metric components g ij are covariant tensors of the second order.
                                               i
               Solution: In a coordinate system x ,i =1,... ,N the element of arc length squared is

                                                         2
                                                                 i
                                                       ds = g ij dx dx j                               (1.3.6)
                                          i
               while in a coordinate system x ,i =1,... ,N the element of arc length squared is represented in the form

                                                        2
                                                                 m
                                                                     n
                                                      ds = g mn dx dx .                                (1.3.7)
               The element of arc length squared is to be an invariant and so we require that
                                                                     i
                                                             n
                                                         m
                                                    g  dx dx = g ij dx dx j                            (1.3.8)
                                                     mn
               Here it is assumed that there exists a coordinate transformation of the form defined by equation (1.2.30)
               together with an inverse transformation, as in equation (1.2.32), which relates the barred and unbarred
                                        i
                                            i
               coordinates. In general, if x = x (x), then for i =1,... ,N we have
                                                   ∂x i  m          j   ∂x j  n
                                               i
                                             dx =    m  dx   and dx =     n  dx                        (1.3.9)
                                                   ∂x                   ∂x
               Substituting these differentials in equation (1.3.8) gives us the result
                                             i
                                                                                i
                                          ∂x ∂x  j   m  n                     ∂x ∂x j     m  n
                                m
                                    n
                          g  mn dx dx = g ij  m  n  dx dx    or      g mn  − g ij  m  n  dx dx =0
                                          ∂x  ∂x                             ∂x ∂x
                                                                           i
                                                                         ∂x ∂x j
               For arbitrary changes in dx m  this equation implies that g       and consequently g ij transforms
                                                                 mn  = g ij  m  n
                                                                         ∂x ∂x
               as a second order absolute covariant tensor.
               EXAMPLE 1.3-2. (Curvilinear coordinates) Consider a set of general transformation equations from
               rectangular coordinates (x, y, z) to curvilinear coordinates (u, v, w). These transformation equations and the
               corresponding inverse transformations are represented
                                                x = x(u, v, w)   u = u(x, y, z)

                                                y = y(u, v, w)   v = v(x, y, z)                       (1.3.10)
                                                z = z(u, v, w).  w = w(x, y, z)

                                   3
                                                               3
                                                 1
                                                        2
                     1
                            2
               Here y = x, y = y, y = z   and x = u, x = v, x = w are the Cartesian and generalized coordinates
               and N =3. The intersection of the coordinate surfaces u = c 1 ,v = c 2 and w = c 3 define coordinate curves
               of the curvilinear coordinate system. The substitution of the given transformation equations (1.3.10) into
                                r
               the position vector ~ = x b e 1 + y b e 2 + z b e 3 produces the position vector which is a function of the generalized
               coordinates and
                                     r ~ = ~(u, v, w)= x(u, v, w) b e 1 + y(u, v, w) b e 2 + z(u, v, w) b e 3
                                         r
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