Page 153 - Semiconductor For Micro- and Nanotechnology An Introduction For Engineers
P. 153

The Electromagnetic System
                                               2
                                      ∇ 2 A µε ∂ A  –  ∇  ∇• A +  µε ∂Φ   =  – µJ  (4.15a)
                                          –
                                               t ∂  2         t ∂ 
                                               ∂  ∇•  ∇ 2     ρ
                                                              ε
                                                t ∂  A +  Φ =  – ---             (4.15b)
                                                                                    ⁄
                             We  now consider the case for a vacuum; recall that  µ ε =  1 c 2  .
                                                                             0 0
                             Because of the relation B =  ∇× A B
                                                         ,   is unchanged through the addition
                             of the gradient of an arbitrary scalar function  Λ  , because
                             ∇× (  A +  ∇ Λ) =  ∇× A  . However, through equation (4.14), which under
                                                 A
                             the same substitution for   results in
                                                   ∂A     ∂Λ   
                                              E =  –  –  ∇   +  Φ                 (4.16)
                                                    t ∂    t ∂   

                             we see that the scalar potential must be transformed according to
                                        ⁄
                             Φ ⇒  Φ ∂Λ ∂t   to achieve invariance of the electric field. Now choos-
                                   –
                                Λ
                             ing   such that
                                                         2
                                                        ∂Λ
                                                ∇ 2 Λ +  µε  =  0                 (4.17)
                                                         t ∂  2

                             implies that

                                                      1 ∂Φ
                                                ∇• A +  ----  =  0                (4.18)
                                                      c 2  t ∂

                             which completely de-couples the remaining two vacuum Maxwell equa-
                             tions to give

                                                        2
                                                     1 ∂Φ     ρ
                                               ∇ 2 Φ –  ----  =  – ---           (4.19a)
                                                     2
                                                     c ∂ t 2  ε
                                                       2
                                                    1 ∂ A
                                              ∇ 2 A –  ----  =  – µJ             (4.19b)
                                                     2
                                                    c ∂ t  2




                150          Semiconductors for Micro and Nanosystem Technology
   148   149   150   151   152   153   154   155   156   157   158