Hello! i am simulating a detector with liquid argon as active medium. The active medium is crossed by muons. I want to know the fraction of energy deposited by ionisation and the fraction of energy deposited by radiation (scintillation and cerenkov separately). Is geant4 able to do this?, if yes, how can i ask for these (three) quantities?. I have been doing some tests, and i saw that if i throw a muon fully contained in liquid argon, with scintillation and cerenkov processes switched off, if i add total energy deposited step by step (theStep->GetTotalEnergyDeposit()) i recover the muon vertex kinetic energy, but if i switch on scintillation or cerenkov or both processes, this is not so, because of energy loss by radiation processes, i suppose. But i have been looking at all these and i don`t understand how this exactly occur. I have done the following test: I switch off cerenkov and i switch on scintillation,i calculate the total energy deposited step by step (theStep->GetTotalEnergyDeposit()) (less than vertex energy) by a totally contained muon, and i count the number of scintillation photons emited by the muon. I know that more or less the energy needed to emit a scintillation photon in liquid argon is about 19.5 eV. I add to total energy deposited step by step the quantity 19.5*NumberofPhotons eV and i do not recover the vertex energy. In this way i obtain that the energy for a photon emission must be 32.8 eV with a scintillation yield of 1000/MeV, but if i change the yield this energy value (for the photon emission) change. So i don`t understand. Am i doing something wrong? Thanks!!
Cerenkov photons are generated in Geant4 without conserving energy; i.e. the track doesn't lose any additional energy over and above ionization, on account of the sum of Cerenkov photons generated and the comparatively little energy they carry off. The number of scintillation photons are generated depending on the ionization energy lost during the step. If it takes more or less 19.5eV to emit a scintillation photon then the scintillation yield is roughly 51000/MeV. The energy of each individual photon is sampled from the scintillation spectrum. You mustn't expect that the sum of photons times their individual energy add up to the ionization energy lost by a contained muon. What is true is that the number of photons will be proportional to the ionization energy, the proportionality is the scintillation yield, with the actual number statistically distributed.
