Page 219 - Electromagnetics
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4.5.3   The energy theorem
                          A convenient expression for the time-average stored energies (4.60) and (4.61) is found
                        by manipulating the frequency-domain Maxwell equations. Beginning with the complex
                        conjugates of the two frequency-domain curl equations for anisotropic media,

                                                                ∗
                                                        ˜ ∗
                                                   ∇× E = jω ˜ ¯µ · H ,
                                                                   ˜ ∗
                                                                    ∗
                                                   ∇× H = J − jω ˜ ¯  · E ,
                                                        ˜ ∗
                                                                       ˜ ∗
                                                            ˜ ∗
                        we differentiate with respect to frequency:
                                                           ∗
                                               ∂E     ∂[ω ˜ ¯µ ]       ∗ ∂H
                                                                           ˜ ∗
                                                ˜ ∗
                                                              ˜ ∗
                                           ∇×      = j       · H + jω ˜ ¯µ ·  ,                (4.66)
                                               ∂ω       ∂ω                ∂ω
                                                                ∗
                                               ∂H    ∂J ˜ ∗  ∂[ω ˜ ¯  ]    ∗ ∂E
                                                                               ˜ ∗
                                                ˜ ∗
                                                                    ˜ ∗
                                          ∇×       =     − j      · E − jω ˜ ¯  ·  .           (4.67)
                                               ∂ω    ∂ω       ∂ω              ∂ω
                        These terms also appear as a part of the expansion
                                            ∂H     ∂E

                                                    ˜ ∗
                                              ˜ ∗
                                         ˜               ˜
                                    ∇· E ×       +     × H =
                                             ∂ω    ∂ω
                                     ∂H                   ∂H           ∂E    ∂E
                                                                               ˜ ∗
                                      ˜ ∗
                                                           ˜ ∗
                                                                         ˜ ∗
                                                                                       ˜
                                               ˜
                                                   ˜
                                                                ˜
                                         · [∇× E] − E ·∇ ×    + H ·∇ ×     −     · [∇× H]
                                     ∂ω                   ∂ω            ∂ω    ∂ω
                                                                                               ˜
                        where we have used (B.44). Substituting from (4.66)–(4.67) and eliminating ∇× E and
                            ˜
                        ∇× H by Maxwell’s equations we have
                                1        ∂H     ∂E
                                                 ˜ ∗
                                           ˜ ∗
                                                      ˜
                                      ˜
                                 ∇· E ×      +     × H =
                                4         ∂ω    ∂ω
                                 1       ∗ ∂E    ∂E            1        ∗ ∂H     ∂H
                                                   ˜ ∗
                                             ˜ ∗
                                                                                  ˜ ∗
                                                                            ˜ ∗
                                     ˜
                                                                                         ˜
                                                         ˜
                                                                    ˜
                                j ω E · ˜ ¯  ·  −    · ˜ ¯  · E + j ω H · ˜ ¯µ ·  −  · ˜ ¯µ · H +
                                 4          ∂ω    ∂ω           4           ∂ω    ∂ω
                                            ∗
                                                            ∗
                                  1     ∂[ω ˜ ¯  ]     ∂[ω ˜ ¯µ ]     1     ∂J ˜ ∗  ∂E
                                                                                     ˜ ∗
                                                    ˜
                                                                                 ˜
                                                                         ˜
                                      ˜
                               + j   E ·      · E + H ·       · H  −    E ·    + J ·     .
                                                               ˜ ∗
                                                ˜ ∗
                                  4       ∂ω             ∂ω          4     ∂ω       ∂ω
                        Let us assume that the sources and fields are narrowband, centered on ω 0 , and that ω 0
                        lies within a transparency range so that within the band the material may be considered
                                                                             †
                                                                  †
                        lossless. Invoking from (4.49) the facts that ˜ ¯  = ˜ ¯  and ˜ ¯µ = ˜ ¯µ , we find that the first two
                        terms on the right are zero. Integrating over a volume and taking the complex conjugate
                        of both sides we obtain
                                         ˜
                                              ˜
                             1         ∂H    ∂E
                                                  ˜ ∗
                                  ˜ ∗
                                  E ×     +     × H   · dS =
                             4  S      ∂ω    ∂ω
                                                                                          ˜
                                1        ∂[ω ˜ ¯ ]     ∂[ω ˜ ¯µ]       1        ∂J ˜     ∂E
                                                              ˜
                                                ˜
                                     ˜ ∗
                             − j     E ·      · E + H ·      · H  dV −      E ·    + J ·     dV.
                                                                             ˜ ∗
                                                                                     ˜ ∗
                                                   ˜ ∗
                                4  V      ∂ω            ∂ω             4  V     ∂ω       ∂ω
                        Evaluating each of the terms at ω = ω 0 and using (4.60)–(4.61) we have
                                                        ˜
                                                   ˜
                                       1         ∂H    ∂E
                                             ˜ ∗
                                            E ×      +    × H        · dS =
                                                            ˜ ∗
                                       4  S      ∂ω    ∂ω         ω=ω 0
                                                                            ˜

                                                         1        ∂J ˜     ∂E
                                                               ˜ ∗
                                                                       ˜ ∗
                                       − j [ W e  + W m  ] −  E ·    + J ·          dV         (4.68)
                                                         4  V     ∂ω       ∂ω    ω=ω 0
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