Page 50 - Electromagnetics
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∂      ∂       ∂      ∂
                                               =    − v x   − v y  − v z
                                                 ∂t  	  ∂x  	   ∂y 	   ∂z
                                                  ∂

                                               =    − (v ·∇ ).                                 (2.44)
                                                 ∂t
                        Similarly
                                             ∂     ∂       ∂    ∂       ∂    ∂
                                                =    ,       =    ,       =    ,
                                             ∂x   ∂x  	   ∂y   ∂y  	   ∂z   ∂z
                        from which



                                      ∇× A(r, t) =∇ × A(r, t),   ∇· A(r, t) =∇ · A(r, t),      (2.45)
                        for each vector field A.
                          Newton was aware that the laws of mechanics are invariant with respect to Galilean
                        transformations. Do Maxwell’s equations also behave in this way? Let us use the Galilean
                        transformation to determine which relationship between the primed and unprimed fields
                        results in form invariance of Maxwell’s equations. We first examine ∇ ×E, the spatial rate

                        of change of the laboratory field with respect to the inertial frame spatial coordinates:
                                                             ∂B      ∂B

                                           ∇ × E =∇ × E =−       =−     + (v ·∇ )B
                                                              ∂t     ∂t
                        by (2.45) and (2.44). Rewriting the last term by (B.45) we have


                                                   (v ·∇ )B =−∇ × (v × B)
                        since v is constant and ∇ · B =∇ · B = 0, hence

                                                                      ∂B

                                                   ∇ × (E + v × B) =−    .                     (2.46)
                                                                      ∂t
                        Similarly
                                                       ∂D       ∂D

                                  ∇ × H =∇ × H = J +      = J +    +∇ × (v × D) − v(∇ · D)
                                                       ∂t       ∂t

                        where ∇ · D =∇ · D = ρ so that
                                                                 ∂D

                                                ∇ × (H − v × D) =   − ρv + J.                  (2.47)
                                                                 ∂t
                        Also
                                                            ∂ρ     ∂ρ

                                             ∇ · J =∇ · J =−   =−     + (v ·∇ )ρ
                                                             ∂t    ∂t
                        and we may use (B.42) to write




                                                (v ·∇ )ρ = v · (∇ ρ) =∇ · (ρv),
                        obtaining
                                                                    ∂ρ

                                                     ∇ · (J − ρv) =−   .                       (2.48)
                                                                    ∂t


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