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merely a mapping from the original vector fields of Minkowski’s form, we still have four
                        vector fields with which to contend. And with these must also be a mapping of the
                        constitutive relationships, which now link the fields E, B, P, and M. Rather than argue
                        the actual physical existence of the equivalent sources, we note that a real benefit of
                        the new view is that under certain circumstances the equivalent source quantities can be
                        determined through physical reasoning, hence we can create physical models of P and M
                        and deduce their links to E and B. We may then find it easier to understand and deduce
                        the constitutive relationships. However we do not in general consider E and B to be in
                        any way more “fundamental” than D and H.


                        Covariance of the Boffi form.   Because of the linear relationships (2.117) and (2.118),
                        covariance of the Maxwell–Minkowski equations carries over to the Maxwell–Boffi equa-
                        tions. However, the conversion between fields in different moving reference frames will
                        now involve P and M. Since Faraday’s law is unchanged in the Boffi form, we still have

                                                    E = E   ,                                 (2.124)


                                                    B = B   ,                                 (2.125)


                                                    E = γ(E ⊥ + β × cB ⊥ ),                   (2.126)
                                                     ⊥

                                                   cB = γ(cB ⊥ − β × E ⊥ ).                   (2.127)
                                                     ⊥
                        To see how P and M convert, we note that in the laboratory frame D = 
 0 E + P and
                        H = B/µ 0 − M, while in the moving frame D = 
 0 E + P and H = B /µ 0 − M .Thus







                                                P = D − 
 0 E = D   − 
 0 E   = P

                        and

                                             M = B /µ 0 − H = B   /µ 0 − H   = M   .



                        For the perpendicular components

                                D = γ(D ⊥ + β × H ⊥ /c) = 
 0 E + P = 
 0 [γ(E ⊥ + β × cB ⊥ )] + P ;
                                  ⊥                        ⊥    ⊥                         ⊥
                        substitution of H ⊥ = B ⊥ /µ 0 − M ⊥ then gives
                                 P = γ(D ⊥ − 
 0 E ⊥ ) − γ
 0 β × cB ⊥ + γ β × B ⊥ /(cµ 0 ) − γ β × M ⊥ /c

                                  ⊥
                        or

                                                   cP = γ(cP ⊥ − β × M ⊥ ).
                                                     ⊥
                        Similarly,

                                                   M = γ(M ⊥ + β × cP ⊥ ).
                                                     ⊥
                        Hence


                                      E = E   ,  B = B   ,  P = P   ,  M = M   ,  J = J ⊥ ,   (2.128)



                                                                               ⊥
                        and

                                                    E = γ(E ⊥ + β × cB ⊥ ),                   (2.129)
                                                     ⊥

                                                   cB = γ(cB ⊥ − β × E ⊥ ),                   (2.130)
                                                     ⊥
                                                   cP = γ(cP ⊥ − β × M ⊥ ),                   (2.131)

                                                     ⊥
                                                   M = γ(M ⊥ + β × cP ⊥ ),                    (2.132)

                                                     ⊥

                                                    J = γ(J   − ρv).                          (2.133)

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