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Fourier Transform of Polarization

  1. Feb 12, 2017 #1
    1. The problem statement, all variables and given/known data
    The problem is from an optics text, however I believe the problem to be a mathematical one.

    I'm trying to take the Fourier transform of
    P(t) = ε0∫ X(t-τ)E(τ) dτ which should equal
    P(ω) = ε0X(ω)E(ω)
    where ε0 is a constant
    X is the susceptibility
    E is the electric field

    2. Relevant equations
    P(t) = ε0∫ X(t-τ)E(τ) dτ <-starting point

    Fourier transforms:
    P(ω) = ∫ P(t)eiωt dt
    P(t) = 1/2π ∫ F(ω)e-iωt

    P(ω) = ε0X(ω)E(ω) <- end point (final solution)

    3. The attempt at a solution
    First off not sure if convolution is part of the answer where
    P(t) = ε0∫ X(t-τ)E(τ) dτ also equals
    P(t) = ε0∫ X(τ)E(t-τ) dτ
    but I will start my attempt with the first equation

    If I plug the P(t) time transforms into the left hand side I get

    1/2π ∫ F(ω)e-iωt dω = ε0∫ X(t-τ)E(τ) dτ

    now multiply by eiω'tdt and integrate over all time

    1/2π ∫ F(ω)e-iωt dω ∫ eiω'tdt= ε0∫ X(t-τ)E(τ) dτ ∫ eiω'tdt

    where the left hand side can be arranged to give

    1/2π ∫ F(ω) dω ∫ e-i(ω-ω')tdt= ε0∫ X(t-τ)E(τ) dτ ∫ eiω'tdt

    using ∫ e-i(ω-ω')tdt = 2πσ(ω-ω') <- Dirac Delta function

    ∫ F(ω)σ(ω-ω') dω = ε0∫ X(t-τ)E(τ) dτ ∫ eiω'tdt <-Dirac Delta function 'picks out' ω'

    F(ω') = ε0∫ X(t-τ)E(τ) dτ ∫ eiω'tdt

    and now I'm stuck, unless my approach is wrong to begin with.
    I also know the electric field is the real part of eiωt
    so the last equation could be rewritten

    F(ω') = ε0∫ X(t-τ)eiωτ dτ ∫ eiω'tdt

    Though I am not sure if this helps.
     
  2. jcsd
  3. Feb 12, 2017 #2

    blue_leaf77

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    It can be rearranged to give
    $$
    F(\omega') = \epsilon_0 \int E(\tau)\ d\tau \left( \int X(t-\tau)e^{i\omega' t} dt \right)
    $$
    The integral in the parentheses is the Fourier transform of ##X(t-\tau)##. Using the shifting property of Fourier transform you should be able to find its relation with the transform of ##X(t)##. If you can get the idea of this part, the rest should be pretty straight forward.
     
  4. Feb 13, 2017 #3

    vela

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    Just wanted to mention you could have gotten to this point without all that work. Just plug in
    $$P(t) = \varepsilon_0 \int X(\tau-t)E(t)\,d\tau$$ into
    $$F(\omega) = \int P(t)e^{i\omega t}\,dt.$$
     
  5. Feb 13, 2017 #4
    Thank you blue leaf for pointing me towards the shifting property of Fourier transforms, and vela you are certainly right I just didn't notice. This does save some time!

    As for the shifting property I still have a few questions about it,
    is the below true?

    ∫ X(t-τ) eiω't dt = ∫ X(u) eiω'(u+τ) du
    where u = t - τ and this then equals

    eiω'τ ∫ X(u) eiω'u = eiω'τ X(ω)

    This turns F(ω'), from my previous post, into

    F(ω') = ε0 X(ω)∫ E(τ) eiω'τ

    Does this mean

    ∫ E(τ) eiω'τ dτ = E(ω) ?
     
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