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

Thanks in advance for any answers!

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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
up vote 5 down vote accepted

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.

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Thank you for your answer joriki! – Andy Apr 21 '11 at 7:09
You're welcome. – joriki Apr 21 '11 at 7:10
BTW do you know how the separable property "appears" in the Fourier space? As an analogy; in 2D rectangular coordinates: f(x,y)=X(x)Y(y)=>F(kx, ky)=F{X(x)}F{Y(y)}, which means that the separable property is a convolution? The separable property in spherical coordinates may also be thought of as a "radial convolution" I guess? If so do you know if there exists a convolution theorem for this kind of convolution? – Andy Apr 21 '11 at 8:11
Please explain what you mean by "the separable property". (BTW, you can delete your earlier truncated comment to clear things up, by clicking on the little cross that appears at the end of it when you hover over it.) – joriki Apr 21 '11 at 8:28
In cartesian coordinates it is "worth gold" to know that a function is separable; can be written as a product of a x dependent, y dependent and z dependent function: f(x,y,z)=X(x)Y(y)Z(z). E.g. rect(x, y, z) = rect(x)*rect(y)*rect(z) => F{ rect(x, y, z) } = sinc(x)*sinc(y)*sinc(z), where rect is the boxcar function. I am just a bit disappointed that the "separable property" (don't know what to call this properly) is not worth so much in spherical coordinates. The thing is that I map my problem to cartesian coordinates and use FFT, but I would like to utilize the "separable property" somehow. – Andy Apr 21 '11 at 8:48

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