# Re: Scattering cross sections

```You asked a few questions about photon scattering and CT contrast:

>For a start is it the mass attenuation or absorption coefficient?
>
>What exactly is a scattering cross section?
>
>How does all of this enable me to calculate the scatter factor?
>
>And finally (and prosaically), why does scattered radiation decrease
>contrast?

Let's start with the last question.  As I'm sure you understand, the reason
you can reconstruct an object with CT is because different structures
within the object attenuate the radiation beam differently. Let's say you
have a uniform water bath (CT no.= zero) with one highly attenuating object
in it (CT no. = 1000).  The scanner recognizes the different CT numbers for
these regions because less radiation reaches the detector when the beam
passes through the object compared with other projections.  Let's say the
detector sees 10 mR on a projection without the object and 7 mR on a
projection including the object.  The difference in transmission is 30
percent.  Now we add scatter to the picture.  Lets say there is an addition
of 10 mR due to scatter to all projectins.  Now we see 20 mR on projections
without the object and 17 mR on objects with it, a difference of only 15
percent.  The net result is a decreased contrast in the reconstructed image.

If you have a gamma source this can be dealt with electronically.  The
scattered photons have a lower energy, so you can count only those photons
that match the original gamma ray energy.  If you use a polychromatic
source, e.g. x-ray tube, you cannot use this trick.  Instead most medical
scanners solve this problem by another feature of scattered photons: after
scattering photons travel in a different direction.  The manufacturers
mount collimators on the detectors so that only photons that have travelled
along a straight path from the original point source will reach the
detector.  The simplest arrangement is a pencil ray beam and a linear
collimator.  (That's why the original RMI scanner used this geometry.)  It
is not much of a leap to guess the collimator design for a fan beam geometry.

1.  The attenuation and absorption coefficients are different things.  The
attenuation coefficient tells you how much energy fluency is removed from
the primary beam.  The absorption coefficient corrects this by adding back
the energy fluence that is not absorbed but scattered within the medium.
If you use a collimated system, scatter is mitigated and the important
quantity is the attenuation coeficient.

2.  The scattering cross section is the probability that a photon will be
scattered by an atom in the medium it passes through.  The numerical value
of the probability is given by the Kline-Nishina formula, which can be
found in almost any nuclear physics or radiation physics text book.
Another key word to look for is "Compton Scattering."

3.  Predicting the contrast degredation in an uncollimated CT system with a
broad spectrum, fan beam source is not a back-of-the-envelope calculation.
The scattered photons now have new energies and they can, in turn, be
absorbed or scattered in the medium.  You can't do it with simple
quantities like attenuation coefficient and absorption coefficient.  This
is a radiation transport problem (a three-dimensional integro-differential
equation). Many of the techniques developed for neutron transport can be
used with varying success in this case, but most people who perform these
calculations use the Monte Carlo technique.  It would be simpler to reject
the scatter with collimation rather than model it, especially since the
scatter does not give you any useful information (with the usual
reconstruction algorithms).

For more details on CT systems you can look in any number of books on
medical imaging or industrial CT.  There are books that range from the king
of pedestrian-level version I have given to high-speed productions.  Which
one is best depends on how much detail you need.

Hope this helps.  Contact me directly if you have more questions.

Dave Scherer

mailto:scherer@uiuc.edu
```