I need help finding the Fourier transform of the function

$$ \rho(\vec{r}) = \alpha \delta_{\vec{r},0} \left(\lambda\lambda' J_1 (\beta |\vec{r}|)Y_1(\beta |\vec{r}|) - \pi^2 J_0 (\beta |\vec{r}|)Y_0(\beta |\vec{r}|) \right), $$

where $\alpha,\beta \in \mathcal{R}$ and $\lambda,\lambda'=\pm1$. The $J_i(x)$ are the $i$-th Bessel functions of the first, the $Y_i(x)$ are the $i$-th Bessel functions of the second kind. $\delta_{\vec{r},0}$ denotes the Kronecker delta.

The vector $\vec{r} = (x,y)^T$ is discrete because I'm working on a discrete set of points. I'm a physicist, so please don't hesitate to ask for more specific information.

I tried something like

$$ \rho(\vec{k}) = \sum_\vec{r} e^{-i\vec{k}\cdot\vec{r}}\rho(\vec{r}), $$

which should in fact not be too difficult because the only vector that contributes to the sum is $\vec{r} = \vec{0}$ due to the Kronecker delta in $\rho(\vec{r})$. The problem is that $\rho(\vec{r})$ diverges at this point. I need some mathematical advice here. Is there anybody who can show me how to compute the Fourier transform of $\rho(\vec{r})$?

  • Please confirm that this problem has strict radial symmetry. The reason I ask is that $\delta^2(x,y)$ and $\delta(r)$ are slightly different (for continuous axes), and a 2-dimensional Fourier Transform with radial symmetry is different than a one dimensional Fourier Transform. – Andy Walls 2 days ago
  • @AndyWalls: I think we can assume that the problem is radially symmetric. But I wouldn't say no to seeing both Fourier transforms :) – MeMeansMe yesterday

Your Answer

 

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.