For a function $f(r, \vartheta, \varphi)$ given in spherical coordinates, how can the Fourier transform be calculated best? Possible ideas:

  • express $(r,\vartheta,\varphi)$ in cartesian coordinates, yielding a nonlinear argument of $f$
  • express $\vec k,\vec r$ in the $e^{i\vec k\vec r}$ term in spherical coordinates, yielding a nonlinear exponent in $\vartheta$ and $\varphi$
  • decompose $f$ into Spherical Harmonics and then change base to Fourier space, requiring the Fourier transform of the Spherical Harmonics (it is obviously not possible to calculate them using this very method..., can that be be found somewhere?)
  • 4
    $\begingroup$ Why do you want to do this? The standard Fourier transform does not really apply to functions defined on a sphere. In this context, the analogues of Fourier modes are in fact precisely the spherical harmonics. $\endgroup$
    – user856
    Jan 14, 2011 at 11:22
  • $\begingroup$ @Rahul Narain: I know what you mean, but Fourier space provides for example the big advantage that you can describe translations by mere phase factors while in spherical Harmonics you have to use rather messy expressions involving Clebsch–Gordan coefficients... $\endgroup$ Jan 14, 2011 at 11:40

1 Answer 1


Tobias, your notation makes it look like your function $f:\mathbb{R}^3\to\mathbb{R}$, in other words, it takes in points in three-dimensional space and spits out real numbers. In that case, as you note, it can be written as a function of $(x,y,z)$. So there doesn't seem to be any reason not to go with your first option. Maybe you could write down the function so I can see the difficulty. If on the other hand your function takes in points on the sphere $\{(x,y,z):\, x^2+y^2+z^2=1\}$, then it makes sense to use spherical harmonics. Your second option doesn't seem reasonable--if you want translation to correspond to phase shifts, then you need to integrate along lines in $\mathbb{R}^3$, and then after a change of coordinates you would be back in your first situation.

  • $\begingroup$ yes, I implicitly meant $\mathbb R^3\to \mathbb R$. The problem with the first solution is, take e.g. a spherical harmonic $Y_1^0=\cos\vartheta=\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2+z^2}}$ multiplied by some arbitrary radial function $R(r)=R(\sqrt{x^2+y^2+z^2})$. I don't think there is a easy solution in most cases, maybe it's not even analytic... $\endgroup$ Jan 31, 2011 at 8:14
  • $\begingroup$ Yeah, I guess I seldom hope for a solution in closed form. I'm usually interested in properties of the function. I'd say that the technique you use really should depend on the function you're starting with. If it's difficult (it seems to be), then it's usually some trick...you could try looking through books to get an idea. Also, I realize my answer was more of a comment... $\endgroup$ Jan 31, 2011 at 20:08

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