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Homework Help: 2D Fourier transform of Coulomb potenial

  1. Jun 14, 2010 #1
    The result is well known, but i need more details about the integral below
    [tex]
    \int \mathrm{d}^2x \frac{1}{|\mathbf{x}|} e^{- \mathrm{i} \mathbf{q} \cdot \mathbf{x}} = \frac{2 \pi}{q}[/tex]

    I've done the Fourier transform of the Coulomb potential in 3D. But failed to get the right answer in 2D.

    I did only a few practice about 2D integrals. Will anyone show me more details about it?

    Thanks in advance!
     
  2. jcsd
  3. Jun 14, 2010 #2

    gabbagabbahey

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    Well, what did you try? Show us your attempt.
     
  4. Jun 14, 2010 #3
    A little too long, and no latex in this computer.

    So, i upload a screenshot in the attachment.

    Thanks!
     

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  5. Jun 14, 2010 #4

    gabbagabbahey

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    :yuck:Yuck!:wink:

    I wouldn't write the complex exponential in terms of sines and cosines if I were you. Instead, just switch to polar coordinates right away (with your coordinate system chosen so that [itex]\textbf{q}[/itex] points in the positive x-direction) to get:

    [tex]\oint d^2 x \frac{e^{-i\textbf{q}\cdot\textbf{x}}}{|\textbf{x}|}=\int_0^\infty dr \left(\int_0^{2\pi} e^{-iqr\cos\theta}d\theta \right)[/tex]

    If you don't immediately recognize the angular integral, try defining [itex]\overline{r}=qr[/itex] and compute the first two derivatives of the integral w.r.t [itex]\overline{q}[/itex] to show that it satisfies a well known differential equation....:wink:
     
  6. Jun 15, 2010 #5

    gabbagabbahey

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    Also, unlike the 3D case, the integral converges without treating the coulomb potential as a limiting case of the Yukawa potential.
     
  7. Jun 15, 2010 #6
    Thanks for your reply.

    I'm not sure whether you want me to do the second derivatives of the angular integral w.r.t [itex]\overline{r}=qr[/itex] or something else. If i havent misunderstood, i'm sorry to tell you that i really dont know what the well known differential equation is.....

    Would you give me a little more details? Thanks a lot!
     
  8. Jun 15, 2010 #7

    gabbagabbahey

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    Well, I'll tell you that [tex]\int_0^{2\pi} e^{-i\overline{r}\cos\theta}d\theta=2\pi J_0(\overline{r})[/tex]....as for proving it, I'll leave that to you (Hint: What differential equation does the [itex]n=0[/itex] Bessel function of the 1st kind satisfy?)...
     
  9. Jun 16, 2010 #8
    Thank you very much!!!!
     
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