Page 158 - Electromagnetics
P. 158

Figure 3.13: A quadrupole distribution.


                        Carrying through the details, we find that the first two moments of ρ vanish, while the
                        third is given by
                                                                            ¯
                                              ¯
                                              Q = q[−3(d 1 d 2 + d 2 d 1 ) + 2(d 1 · d 2 )I].
                        As expected, it is independent of r 0 .
                          It is tedious to carry (3.92)beyond the quadrupole term using the Taylor expansion.
                        Another approach is to expand 1/R in spherical harmonics. Referring to Appendix E.3
                        we find that
                                                        n
                                                    ∞
                                          1        
 
       1   r  n
                                                                      ∗


                                              = 4π                  Y (θ ,φ )Y nm (θ, φ)
                                                                     nm

                                       |r − r |            2n + 1 r n+1
                                                   n=0 m=−n
                        (see Jackson [91] or Arfken [5] for a detailed derivation). This expansion converges for
                        |r| > |r m |. Substitution into (3.61)gives
                                                                   n
                                                   ∞
                                                 1  
  1      1
                                           (r) =       n+1            q nm Y nm (θ, φ)         (3.94)
                                                 	    r     2n + 1
                                                   n=0            m=−n
                        where

                                                               n


                                                                ∗


                                                q nm =   ρ(r )r Y (θ ,φ ) dV .
                                                                nm
                                                       V
                        We can now identify any inverse power of r in the multipole expansion, but at the price
                        of dealing with a double summation. For a charge distribution with axial symmetry (no
                        φ-variation), only the coefficient q n0 is nonzero. The relation

                                                             2n + 1
                                                 Y n0 (θ, φ) =     P n (cos θ)
                                                               4π
                        allows us to simplify (3.94)and obtain
                                                            ∞
                                                        1  
    1
                                                 (r) =            q n P n (cos θ)              (3.95)
                                                       4π	    r n+1
                                                           n=0
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