9.A. Radon transform

$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$

9.A. Radon transform

We decompose function on $\mathbb{R}^n$ into monochromatic plane waves: $$ u(x)=\iiint \hat{u}(\xi) e^{ix\cdot \xi}\,d\xi, $$ where $$\hat{u}(\xi,t)=(2\pi)^{-n}\iiint u(x,t) e^{-ix\cdot \xi}\,dx. $$

Introducing spherical coordinates in $\mathbb{R}^n\ni \xi$, $\xi=\rho \omega$ with $\rho=|\xi|$ and $\omega\in \mathbb{S}^{n-1}$, which is $(n-1)$-dimensional sphere of radious $1$ in $\mathbb{R}^n$, we see that for odd $n$ \begin{equation} u(x)=\frac{1}{2}\int_{-\infty}^\infty \iint_{ \mathbb{S}^{n-1}} \hat{u}(\rho \omega)e^{i\rho x\cdot \omega} \rho ^{n-1}\,d\omega , \end{equation} where we replaced $\int_0^\infty \,d\rho$ by $\frac{1}{2}\int_{-\infty}^\infty \,d\rho$, because $(\rho,\omega)\mapsto (-\rho,-\omega)$ does not change the integrand.

Then \begin{equation} u(x)=\iint_{\mathbb{S}^{n-1}} v(x\cdot\omega,\omega)\,d\omega, \end{equation} with \begin{multline} v(s,\omega)=\frac{1}{2}\int_{-\infty}^{\infty} \hat{u}(\rho\omega)e^{i\rho s}\rho^{n-1}\,d\rho=\\ \frac{1}{2}(2\pi)^{-n}\iiint \Bigl(\int \rho^{n-1} e^{-i\rho x\cdot\omega+i\rho s}\,d\rho\Bigr)u(x)\,dx=\\ \frac{1}{2}(2\pi)^{1-n}(-i\partial_s)^{n-1}\iiint \delta(x\cdot\omega -s) u(x)\,dx=\\ \frac{1}{2}(2\pi)^{1-n}(-i\partial_s)^{n-1} \iint_{x\cdot \omega =s } u(x)\,d\Sigma, \end{multline} where $d\Sigma$ is an area element on the $(n-1)$-dimensional plane $\{x\colon x\cdot\omega=s\}$. Here we used that $$ \int e^{-i(s'-s)\rho }\,d\rho=2\pi \delta (s-s'). $$

Definition

Radon transform


$\Leftarrow$  $\Uparrow$  $\Rightarrow$