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\section{Radon transform}
\subsection*{9.A.1 Definition}
We decompose function on $\mathbb{R}^n$ into monochromatic plane waves: \begin{gather*} u(\boldsymbol{x})=\iiint \hat{u}(\xi) e^{i\boldsymbol{x}\cdot \xi}\,d\xi, \end{gather*} where \begin{gather*} \hat{u}(\xi,t)=(2\pi)^{-n}\iiint u(\boldsymbol{x},t) e^{-i\boldsymbol{x}\cdot \xi}\,dx. \end{gather*}
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(\boldsymbol{x})=\frac{1}{2}\int_{-\infty}^\infty \iint_{ \mathbb{S}^{n-1}} \hat{u}(\rho \omega)e^{i\rho \boldsymbol{x}\cdot \omega} \rho ^{n-1}\,d\omega , \label{eq-9.A.1} \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(\boldsymbol{x})=\iint_{\mathbb{S}^{n-1}} v(\boldsymbol{x}\cdot\omega,\omega)\,d\omega, \label{eq-9.A.2} \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 \boldsymbol{x}\cdot\omega+i\rho s} \,d\rho\Bigr)u(\boldsymbol{x})\,dx=\\ \frac{1}{2}(2\pi)^{1-n}(-i\partial_s)^{n-1} \iiint \delta(x\cdot\omega -s) u(\boldsymbol{x})\,dx=\\ \frac{1}{2}(2\pi)^{1-n}(-i\partial_s)^{n-1} \iint_{\boldsymbol{x}\cdot \omega =s } u(\boldsymbol{x})\,d\Sigma, \label{eq-9.A.3} \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 \begin{gather*} \int e^{-i(s'-s)\rho }\,d\rho=2\pi \delta (s-s'). \end{gather*}
Definition 9.A.1 \begin{gather} (Ru)(s ; \omega) :=\iint_{\boldsymbol{x}\cdot \omega =s } u(\boldsymbol{x})\,d\Sigma \label{eq-9.A.4} \end{gather} with $s\in \mathbb{R}$ and $\omega\in \mathbb{S}^{n-1}$ is a \emph{Radon transform} of $u(x)$.
Then \begin{gather} v(s,\omega)= \frac{1}{2}(2\pi)^{1-n}(-i\partial_s)^{n-1} (Ru)(s ; \omega). \label{eq-9.A.5} \end{gather}
Formulae (\ref{eq-9.A.2}) and (\ref{eq-9.A.4}) provide the inverse Radon transform.
Let $u(\boldsymbol{x},t)$ satisfy $n$-dimensional wave equation \begin{gather} u_{tt}-c^2\Delta u =0. \label{eq-9.A.6} \end{gather}
Then its Radon transform $(Ru)(s,\omega,t)$ satisfies $1$-dimensional wave equation \begin{gather} (Ru)_{tt}-c^2(Ru)_{ss} =0. \label{eq-9.A.7} \end{gather} Therefore, if $u(\boldsymbol{x},0)=0$ \begin{gather} (Ru)(s,\omega,t)= \frac{1}{2c}\int_{s-ct}^{s+ct} (R\phi)(s',\omega)\,ds',\qquad \phi (x)=u_t(x,0). \label{eq-9.A.8} \end{gather} Let $n=3$, define $v(s,\omega)$ by (\ref{eq-9.A.3}): \begin{multline*} v(s,\omega,t) = -\frac{1}{2}(2\pi)^{-2}(\partial_s)^{2} (Ru)(s, \omega,t)\\ =-\frac{1}{16c}\partial_s \Bigl( (R\phi)(s+ct,\omega) - (R\phi)(s-ct,\omega)\Bigr)\\ =-\frac{1}{16c^2\pi^2}\partial_t \Bigl( (R\phi)(s+ct,\omega) + (R\phi)(s-ct,\omega)\Bigr) \end{multline*} and according to (\ref{eq-9.A.2}) \begin{gather*} u(\boldsymbol{x},t)=-\frac{1}{16c^2\pi^2}\partial_t \iint_{\mathbb{S}^2} \Bigl( (R\phi)(\boldsymbol{x}\cdot\omega+ct,\omega) + (R\phi)(\boldsymbol{x}\cdot\omega-ct,\omega)\Bigr)\,d\omega. \end{gather*} From this one can derive (9.1.4).
The CT-scan is based on the following physical law: in homogeneous media intensity of the passed (refracted) electromagnetic ray decays exponentially \begin{gather*} I = I_0 e^{-\lambda s} \end{gather*} where $I_0$ is the intensity of the original ray and $\lambda$ is the linear attenuation coefficient often informally called optical density along the ray. In inhomogeneous media \begin{gather*} I = I_0 e^{-\int \lambda(s)\,ds} \end{gather*}
Remark
A larger attenuation (i.e., larger $\lambda$) produces lower transmitted intensity $I$, which results in a brighter region on a traditional X-ray film (radiographic film), because the film behaves as a \emph{negative} (reversing darkness). Modern systems use detector plates instead of film, but the mapping from intensity to displayed brightness still follows the same negative-like convention.
Thus, bones—which attenuate X-rays strongly—appear white or light gray, while soft tissues appear dark gray or black.
Remark
CT-scan (CT=Computer Tomography) a.k.a. CAT-scan (CAT=Computer Assisted Tomography) works by "illuminating” the body from many directions perpendicular to a chosen axis (say the $x$-axis) and recording, for each angle, the corresponding Radon transform in the $(y,z)$ plane. That is: we have $2$-dimensional Radon transform, $x$ is just a parameter.
By inverting this transform, one reconstructs $\lambda(x; y, z)$, the attenuation coefficient at each point, producing a 3D internal map of the body.
Read more: Radon transform and CT Scan.