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Trodimensional Fourier transform

  1. Nov 5, 2013 #1
    [tex]\mathcal{F}\{f(r)\}=\int e^{i\vec{k}\cdot \vec{r}}f(r)d\vec{r}[/tex]
    in spherical polar coordinates
    [tex]\mathcal{F}\{f(r)\}=\int^{\infty}_0r^2dr\int^{\pi}_0\sin\theta d\theta\int^{\pi}_0d\varphi e^{ikr\cos \theta}f(r)[/tex]

    Why could I take ##e^{ikr\cos \theta}## and to take that ##\theta## is angle which goes from zero to ##\pi##. Thanks for the answer.
     
    Last edited: Nov 5, 2013
  2. jcsd
  3. Nov 5, 2013 #2

    mathman

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    You need to take φ from 0 to 2π in order to integrate over all space. As for your question (as best as I can understand it), the dot product between two vectors is the product of magnitude of the vectors and cosθ, where θ has the range [0,π].
     
  4. Nov 6, 2013 #3
    Could I use this also?
    [tex]\mathcal{F}\{f(\vec{r})\}=\int^{\infty}_0r^2dr\int^{\pi}_0\sin\theta d\theta\int^{\pi}_0d\varphi e^{ikr\cos \theta}f(\vec{r})[/tex]

    I thought that I can use ##e^{i\vec{k}\cdot \vec{r}}=e^{ikr\cos \theta}##, only when ##f(r)## is radial function.
     
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