Page 416 - Electromagnetics
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Verify that the longitudinal component of the Laplacian of A is

                                                           2  	     2
                                                     ˆ u ˆ u ·∇ A = ˆ u∇ A u .
                         5.12  Verify the identities (B.82)–(B.93).
                         5.13  Verify the identities (B.94)–(B.98).

                         5.14  Derive the formula (5.112) for the transverse component of the electric field.

                         5.15  The longitudinal/transverse decomposition can be performed beginning with the
                        time-domain Maxwell’s equations. Show that for a homogeneous, lossless, isotropic region
                        described by permittivity   and permeability µ the longitudinal fields obey the wave
                        equations

                                  
  2       2
                                    ∂     1 ∂          ∂ H u        ∂E u    ∂J mt     ∂J t
                                       −        H t =∇ t    −   ˆ u ×∇ t  +     − ˆ u ×  ,
                                          2
                                    ∂u 2  v ∂t 2        ∂u           ∂t      ∂t       ∂u
                                  
  2       2
                                    ∂     1 ∂          ∂E u          ∂ H u     ∂J mt   ∂J t
                                       −         E t =∇ t   + µˆ u ×∇ t  + ˆ u ×   + µ    .
                                           2
                                    ∂u 2  v ∂t 2        ∂u            ∂t       ∂u      ∂t
                        Also showthat the transverse fields may be found from the longitudinal fields by solving

                                                 1 ∂        1 ∂ρ    ∂ J u
                                             2
                                           ∇ −         E u =     + µ    +∇ t × J mt ,
                                                 2
                                                v ∂t 2        ∂u     ∂t

                                             2  1 ∂         1 ∂ρ m   ∂ J mu
                                           ∇ −         H u =      +       −∇ t × J t .
                                                 2
                                                v ∂t 2      µ ∂u      ∂t
                                  √
                        Here v = 1/ µ .
                         5.16  Consider a homogeneous, lossless, isotropic region of space described by permittiv-
                        ity   and permeability µ. Beginning with the source-free time-domain Maxwell equa-
                        tions in rectangular coordinates, choose z as the longitudinal direction and showthat the
                        TE–TM decomposition is given by
                                             
  2       2         2       2
                                               ∂     1 ∂        ∂ E z    ∂ H z
                                                  −        E y =     + µ     ,                (5.194)
                                                     2
                                               ∂z 2  v ∂t 2     ∂z∂y     ∂x∂t
                                               ∂     1 ∂        ∂ E z    ∂ H z
                                             
  2       2         2       2
                                                  −        E x =     − µ     ,                (5.195)
                                                     2
                                               ∂z 2  v ∂t 2     ∂x∂z     ∂y∂t
                                             
  2       2           2      2
                                               ∂     1 ∂           ∂ E z  ∂ H z
                                                  −        H y =−       +     ,               (5.196)
                                                     2
                                              ∂z 2  v ∂t 2         ∂x∂t   ∂y∂z
                                             
  2       2         2      2
                                               ∂     1 ∂         ∂ E z  ∂ H z
                                                  −        H x =      +      ,                (5.197)
                                                     2
                                              ∂z 2  v ∂t 2       ∂y∂t   ∂x∂z
                        with
                                                    
         2
                                                           1 ∂
                                                       2
                                                     ∇ −         E z = 0,                     (5.198)
                                                           2
                                                          v ∂t 2
                                                           1 ∂
                                                    
         2
                                                       2
                                                     ∇ −         H z = 0.                     (5.199)
                                                           2
                                                          v ∂t 2
                                  √
                        Here v = 1/ µ .
                        © 2001 by CRC Press LLC
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