[Gate-users] Questions about LET and average LET calculation in Gate

李广儒 liguangru at impcas.ac.cn
Tue Mar 29 02:34:05 CEST 2022

Dear Gate users,

I’m trying to calculate LET and average LET(including dose-average and track-average) by Gate. But I met some problems.

First, I found that in the user guide, it says that

“DoseAveraged (default) or TrackAveraged LET both calculated by using the Geant4 EMCalculator method -GetElectronicStoppingPowerDEDX”. But I checked the manual of Geant4(Application Developers Guide) and found that there is a “ComputeElectronicDEDX method” in G4EmCalculator class as shown below:

So I think maybe it use ComputeElectronicDEDX method ,not GetElectronicStoppingPowerDEDX. It maybe a little mistake in the manual of Gate in my opinion.

Second, if Gate use ComputeElectronicDEDX method, the LET  is calculated by kinetic energy of particles as  below:

“Instead of computing the LET as the quotient between the actual electronic energy loss and step length

computed during the simulation, we calculated the LET of the step, Lsn, as the corresponding

mean energy loss per unit path length in the material according to the proton kinetic energy

at the step. In our code, Lsn was obtained with the method ComputeElectronicDEDX() of

G4EmCalculator class.”

These are indicated in the Cortés-Giraldo M A and Carabe A 2015 Phys. Med. Biol. 60 2645–69.But I wonder what method was used to calculate DoseAveraged and TrackAveraged LET in Gate.

And it’s very confusing to me. Because it also said in the Gate manual that if I use parallel computation, there will be “2 output images for each LET actor and run, a file labeled as ‘-numerator’ and one labeled as ‘-denominator’. Building the quotient of these two images results in the averaged LET image. Note that the numerator and denominator images have to be summed up before the division.”So why there are 2 outputs? And what are physical meanings of  denominator and  numerator ?

Hope to get the interpretation of these confusing methods. Any suggestions and answers are welcomed and helpful to me. Thank you in advance!


Guangru Li

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