Strange Fourier transform into something else how?

  • Thread starter MadMax
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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.
 

Answers and Replies

  • #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.
 
  • #3
Dick
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Oh, and first change it into spherical coordinates. Clearly.
 
  • #4
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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?
 
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  • #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.
 
  • #6
Dr Transport
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Use your trig identities to simplify the demoninator.....
 
  • #7
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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.
 
  • #8
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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..
 
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  • #9
Dick
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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.

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.
 
  • #10
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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 :)
 
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  • #11
Dick
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You are free to rotate q. Now do the angular integrations.
 
  • #12
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We arrive at

[tex]\frac{1}{2 \pi^2} \int d \rho \frac{\rho^2 e^{i \rho r}}{\rho^2 + K^2}[/tex]
 
  • #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.
 
  • #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.
 
  • #15
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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...
 
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  • #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.
 
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  • #17
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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?
 
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  • #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...
 
  • #19
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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 :(
 
  • #20
Dick
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Do the theta integration! Suggest substitution u=cos(theta)!!
 
  • #21
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BTW dick can i suggest a good doctor for your high blood pressure???
 
  • #22
Dick
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Guess I Should Lay Off The Coffeee!!!!!!!
 
  • #23
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So:

[tex]- \int \frac{d \rho d u}{4 \pi^2} \frac{e^{i \rho r u}}{1 + K^2 / \rho^2}[/tex]

integrating and then substuting back for u = cos(theta) in the limits theta = 0 to theta = pi gives us:

[tex]\int \frac{i d \rho}{4 \pi^2} \frac{e^{-i \rho r} - e^{i \rho r}}{r/ \rho [\rho^2 + K^2]}[/tex]

? that doesn't look right.

sorry about being a bit slow at this.
 
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  • #24
Dick
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Looks about right to me. Why do you doubt it? Do you see that the integrand is symmetric in rho?
 
  • #25
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I didn't like the extra factors of i, r and rho popping up :P

but I can see that the function still shows the two same singularities..

So.. as far as I've understood contour integration I need to find the sum of the two residues of those poles, multiply that by 2 pi i and I have the value of the integral.. ?

So now I will read more about how to actually find the values of those residues..
 
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  • #26
Dick
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Stop! Wait! First you need to decide on the geometry of the contour(s)!
 
  • #27
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Oh.. doesn't Cauchy's theorem imply that whatever contour we choose that encloses both singularities the result of the integral will equal the integral of the function over a small circle centred around one singularity plus the integral of the function over a small circle centred around the other singularity?
 
  • #28
Dick
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Yes, but the contour had also better include the region of the integral you also want to evaluate or the answer will be useless. In your case the contour should include the positive real axis. I'm trying to get you to tell me it can be the whole real axis including the negative part (the function is symmetric in rho!).
 
  • #29
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so we wanted to integrate from -infinity to infinity in q.. so also from -infinity to infinity in rho... now if we pick an integration contour in the plane defined by the imaginary and real axes that is a circle of infinite radius centred around the origin we will include the whole positive real axis and also both singularities..

then cauchy's theorem tell's us this boils down to finding the integral of the function over a small circle centred around one singularity plus the integral of the function over a small circle centred around the other singularity..

and so it only remains to find the sum of the two residues of those poles, multiply that by (2 pi i) and I have the value of the integral.. ?

am I missing something? sorry if i am.
 
  • #30
Dick
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Yep, you're missing a little bit. The contours are from some large negative value of rho to the same large positive value of rho and then close with a semicircle in either the upper half plane or the lower half plane. The name of the game here is to be able to ignore the contribution from the semicircle to the integral when the radius of the semicircle goes to infinity. You'll want to break the integral up into two parts depending on the sign in the complex exponential. For one the upper half plane is the way to go, for the other one the lower half plane is good. Each one will enclose one pole. Hint: one exponential is small if rho has a positive imaginary part, the other is the opposite.
 

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