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Introduction ( Introduction to: Integral geometry))

Basics of Integral geometry - primer

The underlying problem of tomography can be stated as:
$  g\left ( L \right )=\int_{L}f\left (\textbf{x}  \right )d\textbf{x}Radon-transform (1)
here L stands for a line, f(x) for some object, of which x spatial coordinates are meant in two or three dimensions. On the left-hand-side g(L) is a line integral along L of the f(x) spatial distribution. The task of image reconstruction is reconstructing f(x) if g(L) is given.

We can start mulling over many questions about this formula, like how should we understand that a function's variable is a line, or what objects are in this context in general, or how should we carry out a line integral in two or more dimensions; all in all how are these symbols meant? But we may ask a more fundamental one: why is this formula the basic problem of tomographic imaging?

Line integral and the CT

As a first example we can take Computed Tomography (CT). When a narrow X-ray or gamma beam is transmitted through homogenous medium the original $ I_{0} intensity decreases exponentially according to the Beer-Lambert law with distance d :

$  I\left ( d \right )=I_{0}e^{-\mu d}Beer-Lambert law for homogenous medium

where $ \mu is the linear attenuation coefficients, a constant depending on the material in question. These constants are determined by the isotope constituents, the values of these coefficients can be found here(external link).

If the medium is not homogenous, the attenuation coefficient will also depend on
x (a perhaps multi-D) spatial coordinate, $ \mu = \mu\left ( \textbf{x} \right ), but also on other variables like energy and time. Instead of the symbol $  \mu in transport theory the more general $ \Sigma\left ( \textbf{x} \right ) cross section is applied for describing the same quantity.

In an inhomogeneous medium, when a particle proceeds along a certain L line , the attenuation law will take a form of:

 I=I_{0}e^{-\int_{L} { \mu\left ( \mathbf{x} \right ) dx}}

Now we can take the logarithm of both sides:

 -\ln \frac{ I}{I_{0}}  =\int_{L} { \mu\left ( \mathbf{x} \right ) dx}(2)

and we immediately see that the attenuation can be given as a line integral.

Line integral and nuclear medicine

Our second example is nuclear medicine, when a compound labelled with a radioactive isotope is injected into an organism and accumulates differently in the organs following the specific ways of metabolism. We detect the emitted radiation with a tool capable of recording the position of the incoming photon, moreover it also records the incoming direction.A position and a direction together determines a line uniquely, and each point of this line is a possible emission point and with that a possible location for the radioactive isotope.

Now let us choose in each x point the symbol C(x) for the isotope concentration, then if a detector can detect photons emitted along an L line, the number of counts D(L) is a sum of the concentrations in each point of the line, more precisely their integral:

$  D\left ( L \right )=\int_{L}C\left (\textbf{x}  \right )d\textbf{x}Detector response as a line integral (3)

The Radon transform

When we take Eq. (1), (2) and (3), and perform the integral for every possible L line, we obtain the Radon transform, named after Johann Radon (1887-1956) a bohemian mathematics professor of Vienna.

A possible L line of integration
Intensity distribution along the L line: this is the line we integrate

In the coming chapter we introduce the concepts above more precisely, show how the L line may be represented, the properties of the Radon transform, and with some integral transform, that occur in imaging problems.

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