The Geometry of the Slice Theorems

When studying the Radon transform, we saw that we could reconstruct the Fourier transform of a function from its Radon Transform:

  • The Fourier Transform integrates the product of a function with “waves” that are constant on hyperplanes.
  • The Radon Transform computes the integral of a function over these hyperplanes.
  • So the Fourier Transform of a function is a sort of phase-weighted sum of integrals over hyperplanes, i.e. a phase-weighted sum of the Radon transform.

When evaluating the Fourier Transform of a function f at a point \rho\omega, the hyperplane x\cdot\omega = s appears with “weight” e^{-i\rho x\cdot\omega} = e^{-i\rho s}. So we can write the Fourier transform of f in terms of the Radon Transform as follows:

\hat{f}(\rho\omega) = \int_{-\infty}^{\infty}e^{-i\rho s}Rf(s, \omega) ds = \mathfrak{F}_sRf(\rho, \omega)

The same arguments apply to the X-Ray transform.  In fact, we can compute the Radon transform from the X-Ray transform at a hyperplane by integrating Xf over a set of parallel lines that covers the hyperplane (there is more than one way to do this!).

To compute the Fourier Transform directly from the X-Ray transform, consider the following.  Say we want to compute the Fourier Transform of f at a frequency vector \eta.  We can pick any orthogonal \theta, and have \eta \in \theta^\perp.  Now the “wave” with frequency \eta will be constant in direction \theta, so to compute the Fourier Transform of f at \eta we just need to add up the integrals of f along the lines in direction \thetaXf(y, \theta) — weighted it by the value of the wave on these lines.  In other words

\hat{f}(\eta) = \int_{\theta^\perp}e^{-iy\cdot\eta}Xf(y,\theta)dH_{\theta^\perp}(y) = \mathfrak{F}_{\theta^\perp}Xf(\eta, \theta)

for \eta \in \theta^\perp.  Notice that our choice of \theta was arbitrary, so this really gives us a continuum of formulas, one for each \theta \in \eta^\perp.