Page 24 - Intro to Tensor Calculus
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               EXAMPLE 1.1-18.       Let Φ = Φ(r, θ)where r, θ are polar coordinates related to the Cartesian coordinates
                                                                                                          2
                                                                                               ∂Φ        ∂ Φ
               (x, y) by the transformation equations x = r cos θ  y = r sin θ. Find the partial derivatives  and
                                                                                               ∂x        ∂x 2
               Solution: The partial derivative of Φ with respect to x is found from the relation (1.1.9) and can be written
                                                    ∂Φ    ∂Φ ∂r   ∂Φ ∂θ
                                                        =       +       .                             (1.1.13)
                                                    ∂x    ∂r ∂x   ∂θ ∂x

               The second partial derivative is obtained by differentiating the first partial derivative. From the product
               rule for differentiation we can write
                                                                       2
                                      2
                                                2
                                     ∂ Φ    ∂Φ ∂ r   ∂r ∂    ∂Φ     ∂Φ ∂ θ  ∂θ ∂    ∂Φ
                                         =         +            +         +            .              (1.1.14)
                                     ∂x 2   ∂r ∂x 2  ∂x ∂x  ∂r     ∂θ ∂x 2  ∂x ∂x  ∂θ
               To further simplify (1.1.14) it must be remembered that the terms inside the brackets are to be treated as
               functions of the variables r and θ and that the derivative of these terms can be evaluated by reapplying the
               basic rule from equation (1.1.13) with Φ replaced by  ∂Φ  and then Φ replaced by  ∂Φ  . This gives
                                                              ∂r                       ∂θ
                                                      2
                                                                2
                                            2
                                                                          2
                                           ∂ Φ    ∂Φ ∂ r   ∂r    ∂ Φ ∂r  ∂ Φ ∂θ
                                               =         +            +
                                                                  2
                                           ∂x 2   ∂r ∂x 2  ∂x  ∂r ∂x    ∂r∂θ ∂x
                                                                                                      (1.1.15)


                                                      2
                                                                          2
                                                                 2
                                                  ∂Φ ∂ θ   ∂θ  ∂ Φ ∂r    ∂ Φ ∂θ
                                               +         +             +         .
                                                                           2
                                                  ∂θ ∂x 2  ∂x ∂θ∂r ∂x    ∂θ ∂x
                                                                                                   y
                                                                                2
                                                                       2
                                                                            2
                   From the transformation equations we obtain the relations r = x +y  and  tan θ =  and from
                                                                                                   x
               these relations we can calculate all the necessary derivatives needed for the simplification of the equations
               (1.1.13) and (1.1.15). These derivatives are:
                                             ∂r            ∂r   x
                                           2r   =2x or        =   =cos θ
                                             ∂x            ∂x   r
                                            ∂θ     y       ∂θ     y     sin θ
                                         2
                                      sec θ    = −     or     = −   = −
                                           ∂x     x 2      ∂x     r 2     r
                                                2
                                                            2
                              2
                            ∂ r         ∂θ    sin θ        ∂ θ   −r cos θ  ∂θ  +sin θ ∂r  2sin θ cos θ
                                                                        ∂x
                                                                                 ∂x
                                = − sin θ  =                   =                    =           .
                            ∂x 2        ∂x      r          ∂x 2          r 2              r 2
               Therefore, the derivatives from equations (1.1.13) and (1.1.15) can be expressed in the form
                            ∂Φ    ∂Φ        ∂Φ sin θ
                                =    cos θ −
                            ∂x    ∂r        ∂θ  r
                                                                                              2
                                                                                         2
                                                                          2
                                                             2
                                       2
                            2
                           ∂ Φ    ∂Φ sin θ    ∂Φ sin θ cos θ  ∂ Φ  2     ∂ Φ cos θ sin θ  ∂ Φ sin θ
                                =         +2             +      cos θ − 2             +          .
                           ∂x 2   ∂r   r      ∂θ    r 2     ∂r 2        ∂r∂θ     r      ∂θ 2  r 2
                                        1
                                               2
                          1
                                 2
               By letting x = r, x = θ, x = x, x = y and performing the indicated summations in the equations (1.1.9)
               and (1.1.12) there is produced the same results as above.
               Vector Identities in Cartesian Coordinates
                                                      2
                                                              3
                                              1
                   Employing the substitutions x = x, x = y, x = z, where superscript variables are employed and
               denoting the unit vectors in Cartesian coordinates by b e 1 , b e 2 , b e 3 , we illustrated how various vector operations
               are written by using the index notation.
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