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I am trying to build up intuition about what the fourier transform of a spherical shell will look like but I can't say I'm making much progress.

I've also tried to dumb down the problem in 2D and consider a circle (not a disc).

ie what is the fourier transform of:

$ f = \delta(x, y) \, \forall \, x^2 + y^2 = 1 \\ f = 0, \text{otherwise} $

And in 3D

$ f = \delta(x, y, z) \, \forall\, x^2 + y^2 + z^2 = 1 \\ f = 0, \text{otherwise} $

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    $\begingroup$ Since those functions are $0$ "almost everywhere" - that is, the sphere has measure zero - the Fourier transform is the same as the Fourier transform of the zero function. $\endgroup$
    – Thomas Andrews
    Feb 6, 2015 at 18:01
  • $\begingroup$ Thomas has the correct answer. If even means something other than this, he will have to explain. $\endgroup$
    – GEdgar
    Feb 6, 2015 at 18:12
  • $\begingroup$ @ThomasAndrews oops ok, sorry, formally what I meant is probably that it is a Delta function at those locations. I edited the question to hopefully make this clearer. $\endgroup$
    – evan54
    Feb 6, 2015 at 18:51
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    $\begingroup$ @GEdgar confused with your "even" comment $\endgroup$
    – evan54
    Feb 6, 2015 at 18:52
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    $\begingroup$ @evan54 What some of the comments mean is that what you wrote is not the same as $\delta(r - R)/(4\pi R^2)$ (in spherical coordinates) which is probably what you want. Unless you formalize the question it is going to be difficult to make progress. $\endgroup$
    – alfC
    Jul 5, 2017 at 15:06

2 Answers 2

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Let $\sigma$ be the normalized arc-length measure on the circle $\mathbb T$. The Fourier transform $\mathcal F(\sigma)$ of $\sigma$ makes sense.... $$ \mathcal F(\sigma)(s,t) = \int_{\mathbb T} e^{-2\pi i(sx+ty)} \;d\sigma(x,y) = \frac{1}{2\pi} \int_0^{2\pi}e^{-2\pi i (s\cos\theta+t\sin\theta)}\,d\theta =J_0(2\pi\sqrt{s^2+t^2 }\;) $$ Here $J_0$ is a Bessel function.

More on Fourier transform of a measure: LINK

Of course if you know nothing about the theory of measure and integration, this will not mean anything to you.

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  • $\begingroup$ indeed i know little about measure theory etc but I think this makes some sense. In 3D it would be $sx + ty + uz$ and then substitute $s, t, u$ with sin$\theta$ cos$\phi$, cos$\theta$ cos$\phi$, sin$\phi$ and do that integral over the full sphere? $\endgroup$
    – evan54
    Feb 6, 2015 at 22:57
  • $\begingroup$ meant substitute $x,y,z$ not $s,t,u$ $\endgroup$
    – evan54
    Feb 6, 2015 at 23:20
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In 3D: $\frac{\sin(kr)}{kr}$

Source:

Vembu, S. "Fourier transformation of the n-dimensional radial delta function." The Quarterly Journal of Mathematics 12.1 (1961): 165-168.

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