0

I am not sure how find the Fourier Transform of:

$$f(r) = \frac{e^{-\alpha r}}{r}$$

where $r$ is the radial coordinate. And then I would like to find $\lim_{\alpha \to \infty}$.

Rio Alvarado
  • 979
Jackson Hart
  • 1,600

2 Answers2

1

I don't think your question is entirely clear. Are you working in regular 3-space so $r^2 = x^2 + y^2 + z^2$, and so are you doing a 3-dimensional Fourier transform from {x,y,z} to 3-dimensional k-space? So, is your Fourier transform supposed to be

$\tilde{f}(\vec{k}) = A \int d\vec{x} f(\vec{x}) \exp{(\vec{k}\cdot \vec{x})}$

where A is a prefactor which depends on your Fourier transform convention? If that's the case you can express $\vec{k} \cdot \vec{x}$ as $kr \cos{(\beta)}$, where $\beta$ is the angle between $k$ and $r$ and you can carry out the integral in polar coordinates.

gleedadswell
  • 325
  • Could you expand on this and show me how to do this? I assume your interpretation is correct. We often use the r^2 equation when converting to cartesian – Jackson Hart Aug 10 '15 at 23:54
1

OK, so since you are free to define your coordinates any way you want you could define the z-axis to point along the $\vec{k}$ vector. So now your Fourier transform is

$\tilde{f}(\vec{k}) = A \int d\vec{x} \frac{\exp{(-\alpha r + i kr \cos{(\theta)})}}{r}$

(note that I missed the "i" in my previous post) where I'm using $\theta$ now rather than $\beta$ because of the choice of axes, and where the integral is over all x,y,z. Also, the value of A depends on what your Fourier transform convention is. Look up the conventions and pick one, define it and stay consistent with it. You can transform $d\vec{x} = dx dy dz$ into polar coordinates so $d\vec{x} = r^2 \sin{\theta} dr d\theta d\phi$ so now the integral is

$\int_0^\infty dr \int_{0}^{\pi} d\theta \int_0^{2\pi} d\phi r \sin{(\theta)} \exp{(-\alpha r + i kr \cos{(\theta)})}$

That integral should not be too difficult (you will need to use a change of variables to do the theta integral).

A word of warning: not everyone uses the same convention for which angles $\theta$ and $\phi$ are in their definition of spherical coordinates. Check that you know which convention I've used and double check that I've been consistent.

gleedadswell
  • 325
  • How is it possible to take the integral of dr to INF? This value is Infinite correct? – Jackson Hart Aug 11 '15 at 03:17
  • I'm sorry, could you explain more about solving this integral? It seems to me the integral of dr to INF is just INF – Jackson Hart Aug 11 '15 at 15:55
  • Not really, though it looks so at first glance. There are other terms inside that depend on r, so the integral is not really infinite. It is a triple integration, not the product of 3 integrals. – Aritra Das Aug 11 '15 at 16:19
  • @AritraDas Could you expand the integral to show me? I am not clear on how it would not equal INF? First I solve the one on the right wrt phi ; however, phi is no where in the argument so it just adds 2pi. And then I do the middle integral. Maybe if you expanded, I could find my mistake? – Jackson Hart Aug 11 '15 at 16:34
  • @JacksonHart As you evaluate the middle integral, you would realise that the integral would contain a 'r' in the exponential part. Try it out once. – Aritra Das Aug 11 '15 at 18:45
  • 1
    @AritraDas As I am working it out, can you confirm the phi part would just introduce a factor of 2pi – Jackson Hart Aug 11 '15 at 18:47
  • 1
    An alternate way to write the integral would be $$ \int_0^\infty \int_{-\pi/2}^{\pi/2} \int_0^{2\pi} r \sin{(\theta)} \exp{(-\alpha r + i kr \cos{(\theta)})} \space d\phi d\theta dr $$ Is that of any help? – Aritra Das Aug 11 '15 at 18:48
  • @AritraDas I think the the dphi term is just 2pi. I am not sure how to integrate the theta term. We have sin(theta) and then a cos(theta) in exponential. should i let u = cos(theta) and du = -sin(theta)dtheta? – Jackson Hart Aug 11 '15 at 18:52
  • I mainly just hung up on the middle integral – Jackson Hart Aug 11 '15 at 18:54
  • 1
    Yes, the $d\phi$ therm is $2\pi$ and yes, your substitution is correct. – Aritra Das Aug 11 '15 at 18:54
  • @AritraDas Ok, but then when I do that it seems like the exponentials go away. I would get exp(-ar+ikrcos(theta)), but from pi/2 to -pi/2 this is 0 so I get exp(-ar) - exp(-ar) Right? – Jackson Hart Aug 11 '15 at 19:00
  • @AritraDas Do you see what I am saying? I have exp(-ar + ikrcos(pi/2))-exp(-ar+ikrcos(-pi/2)) – Jackson Hart Aug 11 '15 at 19:06
  • maybe i forgot to pull out a sin(Theta) in integral – Jackson Hart Aug 11 '15 at 19:10
  • Let us continue this discussion in chat. – Aritra Das Aug 11 '15 at 19:10
  • @gleedadswell I get 0 when I evaluate this integral. Can you show otherwise? – Jackson Hart Aug 11 '15 at 20:43
  • The $\phi$ part is $2\pi$ as you say. The r part isn't infinite because of the $\exp{(-\alpha r)}$. You can look up what an integral of the form $\int_0^\infty r \exp{(-ar)} dr$ is in any standard table of integrals (or Wolfram alpha). You'll find that it is related to a $\Gamma$-function. The $\theta$ part will require a change of variables like $u = \cos{\theta}$, but works out nicely once you do that. – gleedadswell Aug 12 '15 at 20:20
  • Oops! Overlooked something in my comment above, which is that the theta part still ends up depending on r after you carry out the theta integral. So the r-integral isn't quite what I said, but should still be not too bad. – gleedadswell Aug 12 '15 at 20:27
  • Ah, ha! ...and the limits of integration for theta should be 0 to pi, not -pi/2 to pi/2. That'll solve your difficulty with the theta integral coming out to zero. – gleedadswell Aug 12 '15 at 21:19
  • In the end you should come down to an r integral where the integrand is a sine times a negative exponential. This is a standard integral which you can find on tables, etc. – gleedadswell Aug 12 '15 at 21:26