Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm sorry because I'm not a mathematician so that my question may look a little bit messy.

I have tabulated values [1] of a 3 dimensional radial function $f(r)$: $f(x_1,x_2,x_3)=f(\sqrt{x_1^2+x_2^2+x_3^2})=f(r)$

I'd like to Fourier transform it in order to get $f(k)$ (or when having $f(k)$ transforming it to $f(r)$. I have looked around and understood this is a Hankel transform of order? (2 or 3, did not get it). And I did not find a practical way of doing this transformation. I have access to web, fortran and mathematica 7.

Could you please, please, please help me?

[1] To be more practical: I have a picture where is plotted the static structure factor $S(k)$ of a inhomogeneous fluid as obtained by neutron diffraction. I can extract points $(k:S(k))$ from this picture. $S(k)$ is related to the direct correlation function $C(k)$ as defined by Ornstein-Zernike through $S(k)=(1-nC(k))^{-1}$ where $n$ is a constant. I'd like to plot $C(r)$.

share|improve this question
Thank you very much for your effort. I found the result here: casa.colorado.edu/~ajsh/FFTLog/app.ps.gz and it has been confirmed by someone else. Your formulae is not exactly true while I am not able to demonstrate it :(. The formulae for a 3 dimensional sphericaly symetric function $f$ are: $ f(k)=\int_0^{\infty} f(r)\dfrac{\sin(kr)}{kr}4\pi r^2dr$ and $ f(r)=\int_0^{\infty} f(k)\dfrac{sin(kr)}{kr} \dfrac{4\pi}{(2\pi)^3} k^2 dk$ –  max Jul 19 '11 at 9:26
Two things: (a) you should probably post comments as comments, and not as answers. (b) What Alice wrote and what you wrote are exactly the same. Perhaps you are not familiar with the unnormalised sinc function? –  Willie Wong Jul 19 '11 at 19:24
add comment

1 Answer

up vote 1 down vote accepted

Radial symmetric Fourier transforms in 2 D are Hankel transforms. Spherically symmetric ones in 3D I think give you something like

$$ f(k) = 4 \pi \int f(r) {\rm sinc}(kr) ~ r^2 dr $$

To show this you need to do the integral of $$f(k) = 2\pi \int \int f(r) e^{i kr\cos \theta} \sin \theta d \theta r^2 dr $$

Which follows from $$ f(k,\theta', \phi') = \int \int \int e^{i {\bf k \cdot r}} f(r,\theta,\phi) r^2 dr \sin \theta d\theta d\phi $$ and ${\bf k \cdot r} = k r (\cos \theta \cos \theta' + \sin \theta \sin \theta' \cos(\phi -\phi')) $.

share|improve this answer
At first sight I misunderstood your answer because I'm not familiar with the unnormalized sinc function (thanks @Willie-Wong for the link). But your answer is exactly what I was looking for. Thank you very much @Alice! –  max Jul 20 '11 at 9:53
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.