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Fourier transform

  1. Sep 16, 2008 #1
    A vector function can be decomposed to form a curl free and divergence
    free parts:



    [tex]\vec{f_{\parallel}}(\vec{r'}) = - \vec{\nabla} \left( \frac{1}{4
    \pi} \int d^3 r' \frac{\vec{\nabla'} \cdot
    \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} \right)[/tex]


    [tex]\vec{f_{\perp}}(\vec{r'}) = \vec{\nabla} \times \left( \frac{1}{4
    \pi} \int d^3 r' \frac{\vec{\nabla'} \times

    I am trying to take the Fourier transform of
    [itex]\vec{f_{\parallel}}(\vec{r'})[/itex] and

    I am starting at [itex]\vec{f_{\parallel}}(\vec{r'})[/itex]. We know
    that the fourier transform is given by:

    [tex] \vec{f}(\vec{k}) = \int_{-\infty}^{\infty} d^3r e^{- i \vec{k}
    \cdot \vec{r}} \vec{f}(\vec{r}) [/tex]

    [tex] \vec{f}(\vec{r}) = \frac{1}{(2 \pi)^3} \int_{-\infty}^{\infty}
    d^3k e^{- i \vec{k} \cdot \vec{r}} \vec{f}(\vec{k}) [/tex]

    I'm not exactly sure where to begin. If I just plug and chug , we'd have:

    [tex] \vec{f}(\vec{k}) = \int_{-\infty}^{\infty} d^3r e^{- i \vec{k}
    \cdot \vec{r}} \vec{f}(\vec{r}) [/tex]

    [tex] \vec{f}(\vec{k}) = \int_{-\infty}^{\infty} e^{- i \vec{k} \cdot
    \vec{r}} - \vec{\nabla} \left( \frac{1}{4 \pi} \int
    \frac{\vec{\nabla'} \cdot \vec{f}(\vec{r'})}{|\vec{r}-\vec{r'}|} d^3
    r' \right) d^3r [/tex]

    I just do not see a simple way of tacking this problem. Any thoughts
    would be appreciated.
  2. jcsd
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