Page 372 - Electromagnetics
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The Hertzian potentials.  With a little manipulation and the introduction of a new
                        notation, we can maintain the wave nature of the potential functions and still provide a
                        decomposition into purely lamellar and solenoidal components. In this analysis we shall
                        assume lossless media only.
                          When we chose the Lorentz gauge to remove the arbitrariness of the divergence of the
                        vector potential, we established a relationship between A e and φ e . Thus we should be
                        able to write both the electric and magnetic fields in terms of a single potential function.
                        From the Lorentz gauge we can write φ e as
                                                           1     t
                                               φ e (r, t) =−     ∇· A e (r, t) dt.
                                                          µ   −∞
                        By (5.17) and (5.18) we can thus write the EM fields as
                                                      1     t          ∂A e
                                                 E =   ∇     ∇· A e dt −   ,                   (5.33)
                                                     µ                  ∂t
                                                          −∞
                                                 B =∇ × A e .                                  (5.34)
                          The integro-differential representation of E in (5.33) is somewhat clumsy in appear-
                        ance. We can make it easier to manipulate by defining the Hertzian potential
                                                            1     t
                                                     Π e =        A e dt.
                                                           µ   −∞
                        In differential form
                                                                ∂Π e
                                                        A e = µ    .                           (5.35)
                                                                dt
                        With this, (5.33) and (5.34) become

                                                                    ∂ 2
                                                  E =∇(∇· Π e ) − µ    Π e ,                   (5.36)
                                                                    ∂t 2
                                                             ∂Π e
                                                  B = µ ∇×       .                             (5.37)
                                                              ∂t
                          An equation for Π e in terms of the source current can be found by substituting (5.35)
                        into (5.31):
                                                    
           2
                                                  ∂    2       ∂            i
                                               µ     ∇ Π e − µ    Π e  =−µJ .
                                                 ∂t            ∂t 2
                        Let us define
                                                              ∂P i
                                                           i
                                                          J =     .                            (5.38)
                                                               ∂t
                        For general impressed current sources (5.38) is just a convenient notation. However, we
                        can conceive of an impressed polarization current that is independent of E and defined
                                                          i
                        through the relation D =   0 E + P + P . Then (5.38) has a physical interpretation as
                        described in (2.119). We nowhave
                                                             ∂ 2      1  i
                                                    2
                                                  ∇ Π e − µ    Π e =− P ,                      (5.39)
                                                            ∂t 2
                        which is a wave equation for Π e . Thus the Hertzian potential has the same wave behavior
                        as the vector potential under the Lorentz gauge.




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