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  Page 131

  • • ...+ 2iabdv0 = [1 + i(1 + 2a)]bdv0.
    • ...+ bd) + i(1 + 2a)bd . (A21) The above expressions...= [1 + i(1 + 2a)]bd (1 + bd) + i(1 + 2a)bd ηdu0.
    • ...2Mω0x 1 + i(1 + 2a) (1 + bd) + i(1 + 2a)bd = iηdbd 2M 1 + i(1 + 2a) (1 + bd) + i(1 + 2a)bd = 1 2M √ ρlηdω 2 −(1 + 2a) + i (1 + bd) + i(1 + 2a)bd = 1 2M √ ρlηdω 2 −(1 + 2a) + i{1 + [1 + (1 + 2a)2]bd} 1 + 2bd + [1 + (1 + 2a)2]b2d . (A23) Recalling that Z...√ ρlηdω 2 1 + 2a 1 + 2bd + [(1 + 2a)2 + 1]b2d , (A24) ∆D...2 1 + [(1 + 2a)2 + 1]bd 1 + 2bd + [(1 + 2a)2 + 1]b2d .
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  Page 132

  • • ...∆f f0 = 1 + 2a 1 + 2bd + [(1 + 2a)2 + 1]b2d ∆f f0 ∣∣∣...= 1 + [(1 + 2a)2 + 1]bd 1 + 2bd + [(1 + 2a)2 + 1]b2d ∆D 2π ∣∣∣...
    • ...√ ρlηdω 2 1 + 2a 1 + 2bd + [(1 + 2a)2 + 1]b2d , (A28) ∆D...2 1 + [(1 + 2a)2 + 1]bd 1 + 2bd + [(1 + 2a)2 + 1]b2d .
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  Page 57

  • • ...= 1 + [(1 + 2a)2 + 1]bd 1 + 2a , (3.30) ∆D 2π −...√ ρlηdω 2 [(1 + 2a)2 + 1]bd − 2a 1 + 2bd + [(1 + 2a)2 + 1]b2d . (3.31) In...
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  Page 55

  • • ...√ ρlηdω 2 1 + 2a 1 + 2bd + [(1 + 2a)2 + 1]b2d , (3.28) ∆D...2 1 + [(1 + 2a)2 + 1]bd 1 + 2bd + [(1 + 2a)2 + 1]b2d . (3.29) To...
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  Page 58

  • • ...follows ∆D/2π |∆f/f0| ≈ 2bd + 1− 2a. (3.32) The above expression can be furthermore...
    • ...∣∣∣∣∆ff0 ∣∣∣∣] = sign { [(1 + 2a)2 + 1]bd − 2a } ≈ sign(bd − a), (3.33) which...