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ε /ε
                                                                  0
                                    60




                                    40
                                                       −ε /ε
                                                            0
                                    20



                                     0
                                      7       8       9       10      11      12
                                                           log    ( f )
                                                              10
                           Figure 4.5: Relaxation spectrum for water at 20 C found using Debye equation.
                                                                     ◦



                        Debye relaxation and the Cole–Cole equation.    In solids or liquids consisting of
                        polar molecules (those retaining a permanent dipole moment, e.g., water), the resonance
                        effect is replaced by relaxation. We can view the molecule as attempting to rotate in
                        response to an applied field within a background medium dominated by the frictional
                        term in (4.101). The rotating molecule experiences many weak collisions which continu-
                        ously drain off energy, preventing it from accelerating under the force of the applied field.
                        J.W.P. Debye proposed that such materials are described by an exponential damping of
                        their polarization and a complete absence of oscillations. If we neglect the acceleration
                        term in (4.101) we have the equation of motion

                                                 dl(r, t)  2         q e

                                              2        + ω l(r, t) =−  E (r, t),
                                                           r
                                                   dt                m e
                        which has homogeneous solution
                                                              ω 2 r
                                                                t
                                                 l(r, t) = l 0 (r)e  − 2  = l 0 (r)e −t/τ
                        where τ is Debye’s relaxation time.

                          By neglecting the acceleration term in (4.102) we obtain from (4.103) the dispersion
                        equation, or relaxation spectrum

                                                                   ω 2 p
                                                   ˜  (ω) =   0 +   0  2  .
                                                               ω + jω2
                                                                 0
                        Debye proposed a relaxation spectrum a bit more general than this, now called the Debye
                        equation:

                                                                  s −   ∞
                                                     ˜  (ω) =   ∞ +    .                      (4.106)
                                                                1 + jωτ



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