Fourier Slice Theorem
Theorem
For $f : \mathbb{R}^{2} \to \mathbb{R}$, the following equation holds:
$$ \begin{equation} \mathcal{F}_2 f(\xi \cos\theta,\ \xi \sin\theta)=\mathcal{F}(\mathcal{R}f)(\xi ,\ \theta) \label{thm1} \end{equation} $$
Here, $\mathcal{F}$ represents the 1-dimensional Fourier transform, $\mathcal{F}_2$ represents the 2-dimensional Fourier transform, and $\mathcal{R}$ is the Radon transform.
$$ \begin{align*} \mathcal{F}f (y) &= \int f(x) e^{-i xy } dx \\ \mathcal{F}_{2} f (y_{1}, y_{2}) &= \int \int f(x_{1}, x_{2}) e^{-i (x_{1}, x_{2}) \cdot (y_{1}, y_{2})} dx_{1} dx_{2} \\ \mathcal{R}f(s,\theta) &= \int_{t =-\infty}^{\infty} f\big( s\cos\theta-t\sin\theta,\ s\sin\theta + t\cos\theta \big) dt \end{align*} $$
Explanation
It is also called the projection slice theorem or the central slice theorem.
Proof
When calculating the left-hand side of $\eqref{thm1}$, it turns out like this:
$$ \begin{equation} \begin{aligned} \mathcal{F}_2 f(\xi \cos\theta,\ \xi \sin\theta) &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y)e^{-i(\xi \cos\theta \cdot x +\xi \sin\theta \cdot y) }dxdy \\ &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y)e^{-i \xi (x\cos\theta +y\sin\theta ) }dxdy \end{aligned} \label{eq1} \end{equation} $$
And to express a point on the plane in polar coordinates as shown below, let’s replace it as shown in the following figure.
$$ s=x\cos\theta + y\sin\theta \\ t=-x\sin\theta+y\cos\theta $$
Then
$$ x=s\cos\theta-t\sin\theta \\ y=s\sin\theta + t \cos \theta $$
And since $dxdy=dsdt$, when substituting for $\eqref{eq1}$
$$ \begin{align*} \mathcal{F}_2 f( \xi \cos\theta,\ \xi \sin\theta) &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(s\cos\theta-t\sin\theta,\ s\sin\theta + t \cos \theta)e^{-i\xi s}dtds \\ &= \int_{-\infty}^{\infty} \left( \int_{-\infty}^{\infty} f(s\cos\theta-t\sin\theta,\ s\sin\theta + t \cos \theta)dt \right)e^{-i \xi s} ds \\ &= \int_{-\infty} ^{\infty} \mathcal{R}f(s,\ \theta) e^{-i\xi s} ds \\ &= \mathcal{F} \mathcal{R}f(\xi,\ \theta) \end{align*} $$
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