Energy deposition by optical photons

Particles, Track, Event, Run and Biasing
1

Hi,
I am producing scintillation photons that propagate to the surface of a cristal where they are just killed because there is no refractive index there. The cristal is itself defined as a sensitive detector.
What I find is that in the step from the production to the surface, the total energy deposit is not zero (the non-ionising energy deposit is zero). Why do optical photons deposit ionising energy ?

Below a printout with the tracking verbose and the output of the following code in the SensitiveDetector::ProcessHits

if (aTrack->GetDefinition() == G4OpticalPhoton::OpticalPhoton()) {
G4cout << aStep->GetTotalEnergyDeposit()/MeV << " " << aStep->GetNonIonizingEnergyDeposit()/MeV << G4endl;
}:

  • G4Track Information: Particle = opticalphoton, Track ID = 2861, Parent ID = 1

Step# X(mm) Y(mm) Z(mm) KinE(MeV) dE(MeV) StepLeng TrackLeng NextVolume ProcName
0 0.521 1.17 -44 4.48e-06 0 0 0 Crystal initStep
1 4.03 -2.34 -29 4.48e-06 4.48e-06 15.8 15.8 Enclosure Transportation

4.477548327756716e-06 0

  • G4Track Information: Particle = opticalphoton, Track ID = 2860, Parent ID = 1

Step# X(mm) Y(mm) Z(mm) KinE(MeV) dE(MeV) StepLeng TrackLeng NextVolume ProcName
0 0.505 1.14 -44 3.23e-06 0 0 0 Crystal initStep
1 -1.1 20 -41.9 3.23e-06 3.23e-06 19 19 Shield Transportation

3.229576310666386e-06 0

2

I think the dE there is just the optical photon’s total energy when it is killed. Notice that in both cases you have

  • KinE at step 1 == KinE at step 0. That means the optical photon did not lose any energy during the step.
  • At step 1, dE == KineE.
    The latter, if I read your description correctly, is the stop-and-kill point.
3

I agree with your interpretation. But the point is that if I’am tallying the energy deposited in the crystal using aStep->GetTotalEnergyDeposit(), I will get this contribution summed… and this leads to total energy depositions greater than the initial particle’s energy…
Thanks