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with it, and the flow of energy from that point (via S em ) must be accompanied by a
                        change in the stored energy at that point. This again gives a very useful and intuitively
                        satisfying point of view. Since we can associate the flow of energy with the propagation
                        of the electromagnetic fields, we can view the fields in any region of space as having the
                        potential to do work on charged particles in that region. If there are charged particles in
                        that region then work is done, accompanied by a transfer of energy to the particles and
                        a reduction in the amplitudes of the fields.
                          We must also remember that the association of stored electromagnetic energy density
                        W em with the mechanical energy density W k is only possible if the medium is nondisper-
                        sive. If we cannot make the assumptions that justify (2.295) and (2.296), then Poynting’s
                        theorem must take the form
                                                         ∂D       ∂B


                                    −    J · E dV =   E ·   + H ·     dV +   (E × H) · dS.    (2.301)
                                        V           V     ∂t      ∂t        S
                        For dispersive media, the volume term on the right-hand side describes not only the stored
                        electromagnetic energy, but also the energy dissipated within the material produced by
                        a time lag between the field applied to the medium and the resulting polarization or
                        magnetization of the atoms. This is clearly seen in (2.29), which shows that D(t) depends
                        on the value of E at time t and at all past times. The stored energy and dissipative terms
                        are hard to separate, but we can see that there must always be a stored energy term by
                        substituting D = 
 0 E + P and H = B/µ 0 − M into (2.301) to obtain


                                          −   [(J + J P ) · E + J H · H] dV =
                                             V
                                          1 ∂
                                                (
 0 E · E + µ 0 H · H) dV +  (E × H) · dS.   (2.302)
                                          2 ∂t  V                      S
                        Here J P is the equivalent polarization current (2.119) and J H is an analogous magnetic
                        polarization current given by

                                                                ∂M
                                                        J H = µ 0  .
                                                                ∂t
                        In this form we easily identify the quantity

                                                     1
                                                       (
 0 E · E + µ 0 H · H)
                                                     2
                        as the electromagnetic energy density for the fields E and H in free space. Any dissipa-
                        tion produced by polarization and magnetization lag is now handled by the interaction
                        between the fields and equivalent current, just as J · E describes the interaction of the
                        electric current (source and secondary) with the electric field. Unfortunately, the equiv-
                        alent current interaction terms also include the additional stored energy that results
                        from polarizing and magnetizing the material atoms, and again the effects are hard to
                        separate.
                          Finally, let us consider the case of static fields. Setting the time derivative to zero in
                        (2.299), we have


                                                 −   J · E dV =  (E × H) · dS.
                                                    V           S
                        This shows that energy flux is required to maintain steady current flow. For instance,
                        we need both an electromagnetic and a thermodynamic subsystem to account for energy
                        conservation in the case of steady current flow through a resistor. The Poynting flux


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