Page 96 - Electromagnetics Handbook
P. 96

they are the mechanism for exerting force on the charge distribution. Substituting for J
                        from (2.2) and for ρ from (2.3) we have

                                                                            ∂D
                                            f em = E(∇· D) − B × (∇× H) + B ×  .
                                                                             ∂t
                        Using

                                                    ∂D     ∂              ∂B
                                               B ×     =−    (D × B) + D ×
                                                    ∂t     ∂t             ∂t
                        and substituting from Faraday’s law for ∂B/∂t we have
                                                                            ∂
                          − [E(∇· D) − D × (∇× E) + H(∇· B) − B × (∇× H)] +  (D × B) =−f em .  (2.287)
                                                                           ∂t
                        Here we have also added the null term H(∇· B).
                          The forms of (2.287) and (2.279) would be identical if the bracketed term could be
                                                                  ¯
                        written as the divergence of a dyadic function T em . This is indeed possible for linear,
                                                                                        ¯
                        homogeneous, bianisotropic media, provided that the constitutive matrix [C EH ] in (2.21)
                        is symmetric [101]. In that case
                                                    1
                                               ¯                  ¯
                                              T em =  (D · E + B · H)I − DE − BH,             (2.288)
                                                    2
                        which is called the Maxwell stress tensor. Let us demonstrate this equivalence for a
                        linear, isotropic, homogeneous material. Putting D = 
E and H = B/µ into (2.287) we
                        obtain
                                                   1                            1
                             ∇· T em =−
E(∇· E) +   B × (∇× B) + 
E × (∇× E) −   B(∇· B).     (2.289)
                                                   µ                           µ
                        Now (B.46) gives
                                              ∇(A · A) = 2A × (∇× A) + 2(A ·∇)A

                        so that
                                                                                1    2
                                     E(∇· E) − E × (∇× E) = E(∇· E) + (E ·∇)E − ∇(E ).
                                                                                2
                        Finally, (B.55) and (B.63) give
                                                                          1

                                                                           ¯
                                          E(∇· E) − E × (∇× E) =∇ · EE − IE · E .
                                                                          2
                        Substituting this expression and a similar one for B into (2.289) we have
                                                       1

                                             ¯                       ¯
                                          ∇· T em =∇ ·  (D · E + B · H) I − DE − BH ,
                                                       2
                        which matches (2.288).
                                                                     ¯
                          Replacing the term in brackets in (2.287) by ∇· T em , we get
                                                        ¯    ∂g em
                                                    ∇· T em +     =−f em                      (2.290)
                                                              ∂t
                        where
                                                        g em = D × B.




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