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Kramers-Kronig relations for the wavenumber

  1. Feb 10, 2014 #1
    Hi all,

    I am wandering if I can apply the Kramers-Kronig (KK) relations to the complex wavenumber k(ω) = k'(ω) + i k"(ω). I have a measurement that easily gives me k'(ω) for a certain range of frequencies, but where k"(ω) is unreliable. I would like to use KK to find k" from k'.

    According to Kristel Carolina Meza Fajardo's PhD thesis (located here: http://www.roseschool.it/files/get/id/4412 ), the damping coefficient α(ω) can be obtained from the phase velocity (section 3.4, page 34). Her equation (2.46) gives the relationship between the damping coefficient α(ω), the phase velocity V(ω) and the (complex) wavenumber k(ω) as:
    [itex]\tilde{k}(\omega) = \frac{\omega}{V(\omega)} - i \alpha(\omega) \equiv k' + i k"[/itex]​

    From another reference (Quantitative Seismology, Aki & Richards, 2nd edition, 2002), the complex wavenumber is given by (Box 5.8, equation (1), page 167):
    [itex]K = \frac{\omega}{c(\omega)} + i \alpha(\omega)[/itex]​

    Kramers-Kronig can be used to find the imaginary part from the real part of the wavenumber (Box 5.8, equation (10)):
    [itex]\alpha(\omega) = \frac{-1}{\pi} P \int_{-\infty}^{\infty} \xi \left( \frac{1}{c(\xi)} - \frac{1}{c_{\infty}} \right) \frac{d\xi}{\xi - \omega}[/itex]​
    where [itex]P[/itex] represents the principal value of the integral.

    So I have two references telling me I can use KK to find [itex]\alpha(\omega)[/itex] from [itex]\frac{\omega}{c(\omega)}[/itex].

    One problem is obviously the integration over infinite domain. I have discrete measurements for some (positive) frequencies. How should I proceed?

    But more importantly, can KK really apply to what I want? My ω axis is juste [itex]2 \pi f[/itex] where [itex]f[/itex] is the measurement's frequency, it's not the x-axis of a Fourier transform...

    Thanks for any hints!
     
  2. jcsd
  3. Feb 10, 2014 #2

    maajdl

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    Gold Member

    In principle this is possible.
    However, you need to be able to calculate this integral!
    This means a large-enough frequency domain of observation, and enough precision.
    This is more likely to be possible around some "resonance" where the "index of refraction" varies quickly.
    Then you will see the corresponding absorption peak.
    I suggest you to experiment with the idea.
     
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