As a physics student, I ran into the following problem. I left out a lot of context, if anything is unclear please ask me.

I quote:

The statistic that is observable is the angular correlation function $w(\theta)$, or its angular power spectrum $\Delta^{2}_{\theta}$. If the sky were flat, the relation between these would be the usual Hankel transform pair:

\begin{equation}w(\theta)=\int\limits^{\infty}_{0} \Delta^{2}_{\theta} J_{0}(K\theta) dK / K\end{equation}

My question is then twofold: (1) I have little intuition for Bessel functions, let alone this Hankel transform. In the text we are given this is the correct transform if we are looking at a flat sky but... (2)... in a flat sky, wouldn't a Fourier transform be the way to go? And if not, why not, and can a FT be related to this Hankel transform?

Forgive me the ignorance of a physicist regarding these concepts.


1 Answer 1


The relation between the Fourier transform on $\mathbb{R}^2$ and the Hankel transform is the following:

  • Given a function $f: \mathbb{R}^2 \mapsto \mathbb{C}$ which is radially symmetry that is it only depends on $|\mathbf{r}|$ with $$ g(r) = f(|\mathbf{r}|).$$

  • Then the Fourier transform $F$ of $f$ is related to the Hankel transform $G$ of $g$ as $$ F(\mathbf{k}) = \int \!d^2 r\,e^{i\mathbf{k}\cdot \mathbf{r}} f(\mathbf{r}) = \int_0^\infty \!dr\,r \,g(r) \underbrace{\int_0^{2\pi}\!d\varphi\, e^{i |\mathbf{k}| r \cos \varphi}}_{2\pi J_0(|\mathbf{k}| r)} = 2\pi G(|\mathbf{k}|).$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .