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Strange Fourier transform into something else how?

  1. Feb 9, 2007 #1
    We have

    [tex]\int \frac{d^3 \texbf{q}}{(2 \pi)^3} \frac {e^{i \texbf{q} \dot \texbf{r}}} {q^2 + K^2} = \frac {e^{-Kr}} {4 \pi r}[/tex]

    How do we get from the left hand side to the right hand side?

    I've tried regular fourier transform of the function under the complex exponential I think that gives 1/(r^2 + K^2) which is nothing like the right hand side... also I thought about doing integration by parts but I only know the normal version of that so I wouldn't know how to tackle a triple integral like this.

    Any help would be much appreciated.
     
  2. jcsd
  3. Feb 9, 2007 #2

    Dick

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    Treat it as a contour integral. Complete the contour in the upper half plane and apply the residue theorem to the pole at q=iK.
     
  4. Feb 9, 2007 #3

    Dick

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    Oh, and first change it into spherical coordinates. Clearly.
     
  5. Feb 11, 2007 #4
    Thanks Dick. Ok so I changed the left hand side into spherical coords I think...

    [tex]\int \int \int \frac{ \sin \theta d \rho d \theta d \phi}{(2 \pi)^3} \frac {e^{i(r_x \rho \sin \theta \cos \phi + r_y \rho \sin \theta \sin \phi + r_z \rho \cos \theta)}}{\sin^2 \theta \cos^2 \phi + \sin^2 \theta \sin^2 \phi + \cos^2 \theta + K^2 / \rho^2} [/tex]

    Is that right? If so can I make that beast simpler?
     
    Last edited: Feb 11, 2007
  6. Feb 11, 2007 #5

    Dick

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    Well, the angular part in the denominator could be a LOT simpler. What is it? And the r*q factor in the exponential isn't a dot product. It's just rho*r in your notation. The integrand actually isn't angle dependent.
     
  7. Feb 11, 2007 #6

    Dr Transport

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    Use your trig identities to simplify the demoninator.....
     
  8. Feb 11, 2007 #7
    Dick: it is supposed to be a dot product of q and r in the exponential just didn't know how to write it correctly. I'll look it up and change it now.

    I'll have another look at trig identities and simplifying the denominator.

    cheers.

    EDIT: oh well can't edit the OP now.
     
  9. Feb 11, 2007 #8
    oh wow lol that was easy. ok so now i have:

    [tex]\int \int \int \frac{ \rho^2 \sin \theta d \rho d \theta d \phi}{(2 \pi)^3} \frac {e^{i(r_x \rho \sin \theta \cos \phi + r_y \rho \sin \theta \sin \phi + r_z \rho \cos \theta)}}{\rho^2 + K^2}[/tex]

    Next step is contour integration? I'll read some more on it..
     
    Last edited: Feb 11, 2007
  10. Feb 11, 2007 #9

    Dick

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    Oh, really? Well, notice that the direction of the vector r doesn't matter - only its magnitude (since you can always rotate the integration coordinates q). So put in in the z direction.
     
  11. Feb 11, 2007 #10
    Are we still free to rotate q given that f(q) and ~f(r) are fourier tranforms of each other?

    Assuming that we are we have...

    grr tex is playing up. but yes I now have an integrand with e^(i r rho) as the exponential :)
     
    Last edited: Feb 11, 2007
  12. Feb 11, 2007 #11

    Dick

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    You are free to rotate q. Now do the angular integrations.
     
  13. Feb 11, 2007 #12
    We arrive at

    [tex]\frac{1}{2 \pi^2} \int d \rho \frac{\rho^2 e^{i \rho r}}{\rho^2 + K^2}[/tex]
     
  14. Feb 11, 2007 #13

    Dick

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    Hmm. I'm not sure that's quite correct. I'll have to check it tomorrow. But in the meantime read up on contour integration.
     
  15. Feb 11, 2007 #14

    Dick

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    Ok. I don't like the rho^2 factor and I dislike the exponential. I think you missed some things in the angular integration. Check it again. Finally you should wind up with an integrand that is even in rho. This will let you change it into a contour.
     
  16. Feb 12, 2007 #15
    I'm pretty sure we have

    [tex]\int \frac{d \rho}{(2 \pi)^3} \frac{e^{i \rho r}}{1 + K^2 / \rho^2} \int \sin \theta d \theta \int d \phi[/tex].

    The exponential is fine because if we make q parallel to r we have

    [tex]exp(i |r| |q|) = exp(i |r| ( \rho^2 \sin^2 \theta \cos^2 \phi + \rho^2 \sin^2 \theta \sin^2 \phi + \rho^2 \cos^2 \theta)^{1/2})[/tex]

    and so using the trig identity cos^2 a + sin^2 a = 1 we find that the exponential has the form

    [tex]e^{ir \rho}[/tex]

    Carrying out the angular integrations between 0 < theta > pi and 0 < phi > 2 pi we arrive at

    [tex]\int \frac {d \rho}{2 \pi^2} \frac {e^{i \rho r}}{1 + K^2/ \rho^2}[/tex].

    I'm now comfortable with the concept of the residue theorem now...
     
    Last edited: Feb 12, 2007
  17. Feb 12, 2007 #16

    Dick

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    The exponential should have the form exp(i*rho*r*cos(theta)) - since you correctly informed me that it should be a dot product. And we rotated to set r_z=r, r_y=0, r_x=0. Right? It's not a completely trivial angular integral (just almost). You can't rotate all of the q's independently.
     
    Last edited: Feb 12, 2007
  18. Feb 12, 2007 #17
    OK... yes after thinking about the geometry I've convinced myself I can see that in some cases we would/could only hope to line up one of the components of r and q. Thank you for your patience.

    I'll get to work on that and produce the correct integral.

    One question about contour integration: the expression will have 2 singularities in the complex domain, one at q = iK and the other at q = - iK. Why is it sufficient to integrate over a closed contour that only encloses one of them?
     
    Last edited: Feb 12, 2007
  19. Feb 12, 2007 #18

    Dick

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    Well, contrary to what I was thinking before, you will need both of them. Once you get the integral, you'll see you have to split it into two parts and two different contours. One around each pole. You'll see...
     
  20. Feb 12, 2007 #19
    I arrive at

    [tex]\int \frac{\sin \theta d \rho d \theta}{4 \pi^2} \frac{e^{i \rho r \cos \theta}}{1 + k^2 / \rho^2}[/tex]

    I can't see how to go further to be honest :(
     
  21. Feb 12, 2007 #20

    Dick

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    Do the theta integration! Suggest substitution u=cos(theta)!!
     
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