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Question Fictitious interaction, G4VEmProcess 

Keywords: Processes, G4VProcess, Materials
Forum: Fast Simulation, Transportation & Others
Date: 19 Oct, 2007
From: Niklas Rehfeld <Niklas.Rehfeld_REMOVE_THIS_@imed.jussieu.fr>

Hello,

I would like to introduce the VR reduction technique of fictitious interaction/delta tracking (see for example http://www.irs.inms.nrc.ca/EGSnrc/pirs701/node57.html) into a Geant4 toolkit (Gate). It is promising for voxel geometries where the voxel size is smaller than the smallest mean interaction path length that can occur for a given energy (or energy range). This is is very promising for PET photon tracking in patient geometries based on computed tomograph (CT) images (typically 512x512 x (20 to 100) voxels, up to 4096 densities/materials), where the smallest mean interaction path lengths of photons at 511 keV can be expected to be around a few centimeters (cortical bone: ca. 3.5 cm) and also for photons in the energy range of 100-1000 keV this is not much worse.

The ficititious process does not change the state of the particle, but accounts for the inhomogeneity of the geometry. For this reason, a process (the fictitious process) must be introduced that depends on the homogeneous replacement material and the actual material and the particle energy. It should be implemented to be quite efficient.

Where do I find further information how this can be done, or in other words:

- Should this process be better derived by G4VDiscreteProcess or G4VEmProcess?

- Or can this be done by only introducing a model?

- For the efficient implementation: It is important that the integral cross sections can be calculated fast. Is there a place where I can find information how this lambdaTable of G4VEmProcess can be used for that purpose (or how to tabulate/parameterize the process)?

- In order to reduce these tables, is there a way to store a table for a material type and calculate the cross section and mean free path length for the same material of other density on-the-fly (or is this done already?) ?

Thank you!

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