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I need help with the fourier transform of Bessel function of first kind on 2 dimensions.

$$G(w_1,w_2) = F[J_0(asqrt{x^2+y^2})]$$

where $J_0$ is the bessel function of first kind of order 0, and $a$ is a possibly complex constant.

I think it will be easier in polar coordinates, but couldn't make much progress so far. Any kickstarter will be appreciated.

From Bracewell, the 2D cartesian Fourier Transform

$$F(u,v) = mathscr{F}left{f(x,y)right} = int_{-infty}^{infty} int_{-infty}^{infty} f(x,y) e^{-2pi i (xu+yv)} dx dy$$

for functions with circular symmetry, in polar coordinates, can be manipulated into a Hankel Transform of zero order

$$F(u,v) = F(q) = mathscr{H}_0left{f(r)right} = 2piint_{0}^{infty}f(r)J_0(2pi qr)rspace dr$$

with a strictly reciprocal inverse transform

$$f(x,y) = f(r) = mathscr{H}_0left{F(q)right} = 2piint_{0}^{infty}F(q)J_0(2pi qr)qspace dq$$

Note: $r = sqrt{x^2+y^2}$ and $q=sqrt{u^2+v^2}$

$ $

Looking at the inverse transform of $F(q) = dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)$, where $delta()$ represents a 1 dimensional delta function:

$$begin{align*}f(r) = mathscr{H}_0left{dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)right} &= 2piint_{0}^{infty}dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)J_0(2pi qr)qspace dq\
\
&= dfrac{2pi}{a}dfrac{a}{2pi}J_0left(2pi dfrac{a}{2pi}rright)\
\
&=J_0(ar)\
end{align*}$$

So that gives you your desired transform pair for real $a$:

$$mathscr{F}left{J_0(ar)right} = mathscr{H}_0left{J_0(ar)right} = dfrac{1}{a}deltaleft(q-dfrac{a}{2pi}right)$$

The transform is an impulse ring at a radius of $dfrac{a}{2pi}$ from the origin.