Fourier, Gauss, Fraunhofer, Porod and the Shape from Moments Problem

We show how the Fourier transform of a shape in any number of dimensions can be simplified using Gauss’s law and evaluated explicitly for polygons in two dimensions, polyhedra three dimensions, etc. We also show how this combination of Fourier and Gauss can be related to numerous classical problems in physics and mathematics. Examples include Fraunhofer diffraction patterns, Porods law, Hopfs Umlaufsatz, the isoperimetric inequality and Didos problem. We also use this approach to provide an alternative derivation of Davis’s extension of the Motzkin-Schoenberg formula to polygons in the complex plane.

💡 Research Summary

The paper presents a unified framework that links the Fourier transform of an indicator function for a geometric shape to surface integrals via Gauss’s (divergence) theorem, and then exploits this connection to solve a variety of classical problems in physics and mathematics.
Starting from the definition θ_V(x)=1 inside a D‑dimensional region V and 0 outside, the normalized Fourier transform is φ̄(β)=v⁻¹∫_V e^{iβ·x}dx, where v=∫_V d^Dx is the volume (or area). By using the identity e^{iβ·x}= (i/β²)∇·(β e^{iβ·x}) the volume integral is rewritten as a surface integral:

∫V e^{iβ·x}dx = (i/2)β·∫{∂V} n̂(s) e^{iβ·R(s)} dS .

This is the central result (eq. 5). Expanding the exponential in powers of β shows that the coefficients are precisely the moments ⟨x₁^{p₁}…x_D^{p_D}⟩, establishing a direct algebraic bridge between the Fourier transform and all geometric moments (eq. 3).

In two dimensions the boundary is a closed curve R(s) parametrized by arclength s, with unit tangent t̂(s)=∂_sR(s) and outward normal n̂(s)=ε·t̂(s) (ε is the 2×2 antisymmetric Levi‑Civita matrix). Substituting into the surface integral and integrating by parts yields the familiar area formula

A = ½∮(R₁ dR₂ – R₂ dR₁) ,

and the “boundary of a boundary is zero” condition ∮ n̂ ds = 0 (eq. 10).

For a polygon with vertices v₁,…,v_N the surface integral can be evaluated exactly, giving a closed‑form expression (eq. 32):

φ(β)=−(1/β²)∑_{n=1}^N (β⊥·Δv_n)(β·Δv_n)