Page 16 - Intro to Tensor Calculus
P. 16

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               We next sum on the index j which repeats itself in each term of equation (1.1.3). This gives


                                             C i = e i11 A 1 B 1 + e i21 A 2 B 1 + e i31 A 3 B 1
                                                + e i12 A 1 B 2 + e i22 A 2 B 2 + e i32 A 3 B 2        (1.1.4)
                                                + e i13 A 1 B 3 + e i23 A 2 B 3 + e i33 A 3 B 3 .

               Now we are left with i being a free index which can have any of the values of 1, 2or3. Letting i =1, then
               letting i =2, and finally letting i = 3 produces the cross product components

                                                     C 1 = A 2 B 3 − A 3 B 2

                                                     C 2 = A 3 B 1 − A 1 B 3
                                                     C 3 = A 1 B 2 − A 2 B 1 .
                                                                   ~
                                                               ~
               The cross product can also be expressed in the form A × B = e ijk A j B k b e i . This result can be verified by
               summing over the indices i,j and k.



               EXAMPLE 1.1-13.       Show


                                                               for  i, j, k =1, 2, 3
                                             e ijk = −e ikj = e jki
               Solution: The array ik j represents an odd number of transpositions of the indices ij k and to each
               transposition there is a sign change of the e-permutation symbol. Similarly, jk i is an even transposition
               of ij k and so there is no sign change of the e-permutation symbol. The above holds regardless of the
               numerical values assigned to the indices i, j, k.



               The e-δ Identity
                   An identity relating the e-permutation symbol and the Kronecker delta, which is useful in the simpli-
               fication of tensor expressions, is the e-δ identity. This identity can be expressed in different forms. The
               subscript form for this identity is


                                      e ijk e imn = δ jm δ kn − δ jn δ km ,  i,j,k,m,n =1, 2, 3

               where i is the summation index and j, k, m, n are free indices. A device used to remember the positions of
               the subscripts is given in the figure 1.1-3.
                   The subscripts on the four Kronecker delta’s on the right-hand side of the e-δ identity then are read


                                                 (first)(second)-(outer)(inner).

               This refers to the positions following the summation index. Thus, j, m are the first indices after the sum-
               mation index and k, n are the second indices after the summation index. The indices j, n are outer indices
               when compared to the inner indices k, m as the indices are viewed as written on the left-hand side of the
               identity.
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