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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 $\lim_{\alpha \to \infty}$.

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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.

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  • $\begingroup$ How is it possible to take the integral of dr to INF? This value is Infinite correct? $\endgroup$ – Jackson Hart Aug 11 '15 at 3:17
  • $\begingroup$ I'm sorry, could you explain more about solving this integral? It seems to me the integral of dr to INF is just INF $\endgroup$ – Jackson Hart Aug 11 '15 at 15:55
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    $\begingroup$ @AritraDas As I am working it out, can you confirm the phi part would just introduce a factor of 2pi $\endgroup$ – Jackson Hart Aug 11 '15 at 18:47
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    $\begingroup$ 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? $\endgroup$ – Aritra Das Aug 11 '15 at 18:48
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    $\begingroup$ Yes, the $d\phi$ therm is $2\pi$ and yes, your substitution is correct. $\endgroup$ – Aritra Das Aug 11 '15 at 18:54
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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.

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  • $\begingroup$ 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 $\endgroup$ – Jackson Hart Aug 10 '15 at 23:54

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