# Fourier transform of radially symmetric functions + references

Let $$f:\mathbb{R}^2 \to \mathbb{C}$$ be a radially symmetric function, i.e., $$f(x,y) = g(r)$$, where $$r=\sqrt{x^2 + y^2}$$. How do I go about computing its Fourier transform? Specifically, Is there a way to express the two-dimensional Fourier transform of $$f$$ by means of a one-dimensional transform of $$g$$ in the radial variable $$r$$?

Both answers here and textbook references will be greatly appreciated.

• This is a special case of the Hankel transform. Apr 10, 2018 at 11:50

Just make a change of variables into polar coordinates. If $f(r\cos\theta,r\sin\theta)=g(r)$, then \begin{align} & \hat{f}(r\cos\theta,r\sin\theta)\\&=\iint_{\mathbb{R}^2} f(x,y)e^{-irx\cos\theta -iry\sin\theta}dxdy \\ & = \iint f(r'\cos\theta',r'\sin\theta')e^{-i rr'\cos\theta \cos\theta'-i rr'\sin\theta\sin\theta'}r'dr'd\theta' \\ & = \int_{0}^{\infty}g(r')\left( \int_{0}^{2\pi}e^{-i r'r\cos(\theta-\theta')}d\theta'\right) r' dr' \\ & = \int_{0}^{\infty}g(r')\int_{0}^{2\pi}e^{-irr'\cos(\theta')}d\theta'r'dr' \\ & = 2\pi\int_{0}^{\infty}g(r')J_0(rr')r'dr' \end{align} where $$J_0(r)=\frac{1}{2\pi}\int_{0}^{2\pi}e^{-ir\cos\theta'}d\theta'$$ is a known representation of the zeroeth order Bessel function. ( See Bessel Integrals on this page https://en.wikipedia.org/wiki/Bessel_function ) The resulting transform is a Hankel transform, and this transform is its own inverse. (See https://en.wikipedia.org/wiki/Hankel_transform)

• Thanks a lot! How do you prove the inner $d\theta$ integral equals $J(r)$? Apr 18, 2018 at 6:43
• @AmirSagiv : I was sloppy in the last post. I've written this out more carefully (and correctly) and have given refereces for the form of $J_0$. Apr 18, 2018 at 9:53
• For me this is a new definition of $J_0$, so double thanks for that! Apr 18, 2018 at 12:49

This answer uses Abel transform and is inspired by my recent investigation of the relationship between PSF and LSF, see https://physics.stackexchange.com/a/762565/68029

In the accepted answer is shown that the 2D Fourier transform of radially symmetric function $$f(x,y) = \bar{f}(\sqrt{x^2+y^2})$$ will be radially symmetric $$F^2(\xi_x, \xi_y) = \bar{F^2}(\sqrt{\xi_x^2 + \xi_y^2})$$.

Knowing this $$F^2(\xi_x,\xi_y) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y) e^{-2 \pi i (x \xi_x + y \xi_y)} dx dy$$ is equivalent to $$F^2(\xi_x,\xi_y) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(\sqrt{x^2+y^2}) e^{-2 \pi i (x \xi_x + y \xi_y)} dx dy$$ so from radial symmetry of $$F^2$$ $$\bar{F}^2(\xi_x)=F^2(\xi_x,0) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \bar{f}(\sqrt{x^2+y^2}) dy e^{-2 \pi i (x \xi_x)} dx$$

previous equation has the structure of 1D Fourier transform so $$g(x) : \mathcal{F}(g) = \bar{F}$$ is

$$g(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \bar{f}(\sqrt{x^2+y^2}) dy = \frac{2}{\sqrt{2\pi}} \int_x^\infty\frac{\bar{f}(r)r dr}{\sqrt{r^2-x^2}}$$

Note that it is Abel transform of $$\bar{f}$$. It might be easier to evaluate it than Hankel transform. The factor $$\sqrt{2\pi}$$ stems from the Fourier transform scaling, which I use.

• Thanks! I'm not sure I follow the last part, starting from So the function g(x)...", both in terms of the mathematical content and the English. Could you elaborate? May 8 at 2:00