# How do I find the Fourier transform of a function that is separable into a radial and an angular part?

how do I find the Fourier transform of a function that is separable into a radial and an angular part: $f(r, \theta, \phi)=R(r)A(\theta, \phi)$ ?

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Isn't the 3D Fourier transform with $e^{i\vec{k}\cdot\vec{r}}=e^{ik r\cos(k\angle r)}$? –  Raskolnikov Apr 20 '11 at 21:12
Thank you Raskolnikov! Introducing the angle between the k and r vector greatly simplifies the exponential, but on the other hand how do I then express $A(\theta, \phi)$ and $d\vec{r}$ in the Fourier transform? –  Andy Apr 21 '11 at 7:41
I think that if you try to work things out that way, that you'll eventually hit on spherical harmonics anyway. You see, the function $e^{ikr\cos(k\angle r)}$ is in fact just a generating function for the spherical harmonics. So, you'd need joriki's answer anyway for practical calculations. –  Raskolnikov Apr 21 '11 at 19:31
So, unless you can easily express $A(\theta,\phi)$ in terms of the angle between $\vec{k}$ and $\vec{r}$, you'll probably not gain much. –  Raskolnikov Apr 21 '11 at 19:47
You can use the expansion of a plane wave in spherical waves. If you integrate the product of your function with such a plane wave, you get integrals over $R$ times spherical Bessel functions and $A$ times spherical harmonics; you'll need to be able to solve those in order to get the Fourier coefficients.