Hankel Transform

Equivalent to a 2-D Fourier Transform with a radially symmetric Kernel, and also called the Fourier-Bessel Transform.

$$g(u,v)={\mathcal F}[f(r)] = \int_{-\infty}^\infty \int_{-\infty}^\infty f(r) e^{-2\pi i(ux+vy)}\,dx\,dy.$$ (1)

Let

$$x+iy = re^{i\theta}$$ (2)

$$u+iv = qe^{i\phi }$$ (3)

so that

$$x = r\cos\theta$$ (4)

$$y = r\sin\theta$$ (5)

$$r = \sqrt{x^2+y^2}$$ (6)

$$u = q\cos\phi$$ (7)

$$v = q\sin\phi$$ (8)

$$q = \sqrt{u^2+v^2}.$$ (9)

Then

$$g(q) = \int_0^\infty\int_0^{2\pi} f(r) e^{-2\pi i rq(\cos\phi\cos\theta+\sin\phi\sin\theta)} r\,dr\,d\theta$$

$$= \int_0^\infty\int_0^{2\pi} f(r)e^{-2\pi irq\cos (\theta-\phi)}r\,dr\,d\theta$$

$$= \int_0^\infty\int_{-\phi}^{2\pi-\phi} f(r)e^{-2\pi irq\cos \theta }r\,dr\,d\theta$$

$$= \int_0^\infty \int_0^{2\pi} f(r)e^{-2\pi irq\cos \theta }r\,dr\,d\theta$$

$$= \int_0^\infty f(r) \left[{\int_0^{2\pi} e^{-2\pi irq\cos \theta }\,d\theta }\right]r\,dr$$

$$= 2\pi\int_0^\infty f(r)J_0(2\pi qr)r\,dr,$$ (10)

where $J_0(z)$ is a zeroth order Bessel Function of the First Kind. Therefore, the Hankel transform pairs are

$$g(k) = \int_0^\infty f(x)J_0(kx)x\, dx$$ (11)

$$f(x) = \int_0^\infty g(k)J_0(kx)k\, dk.$$ (12)

See also Bessel Function of the First Kind, Fourier Transform, Laplace Transform

References

Arfken, G. Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, p. 795, 1985.

Bracewell, R. The Fourier Transform and Its Applications. New York: McGraw-Hill, pp. 244-250, 1965.