Does the non-ionizing energy including the lost energy goes to binding energy

Dear expert,

I have a rather simple question about non-ionizing energy we can get in each step.
I have a large calorimeter and I shot a beam of pi- into it. In each step, if the “G4Track *theTrack = theStep->GetTrack(); theTrack->GetDefinition()->GetPDGEncoding()==2212”, i.e. protons, I will store the total energy and the non-ionizing energy (“theStep->GetNonIonizingEnergyDeposit()”). But I find most of them are 0.

We know when a proton hits a nucleus, some of its lost energy goes to binding energy, which is non-ionizing energy. But I didn’t see much non-ionizing energy for protons. So how should I understand the non-ionizing energy in GEANT? Is binding energy included?

Any comments would be helpful, Thanks!

Yihui

The non-ionizing energy (“theStep->GetNonIonizingEnergyDeposit()”) for protons are 0. So most of deposited energy of protons is ionizing energy.
I also tried to store the same information for neutrons. The result is that most of the deposited energy of neutrons is non-ionizing energy, which is opposite to protons. But shouldn’t it be similar for protons and neutrons?

Protons have charge, so they lose energy by interacting with atomic electrons (ionizing energy). Neutrons don’t have charge. If they lose energy without creating a visible secondary track, it’s energy that went into a nuclear recoil in the material, which is non-ionizing (at the microphysics scale, the displaced nucleus either stores energy in the lattice distortion, or in the form of phonons). If the nuclear recoil from a neutron is large enough, the nucleus may be tracked as a secondary (a G4GenericIon).

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Hello,

Non-ionizing energy loss (NIEL) is the energy which transfer by primary ion to atomic nucleus due to multiple scattering. In EMZ (option4) and EMY (option3) EM physics NIEL computation is enabled for protons and ions with kinetic energy below 1 MeV/nucleun. NIEL has nothing to do with nuclear binding.

VI

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Hi Vladimir, Thanks!

Previously I though the total Energy Deposit = non-ion energy + ion energy. So if Non-ionizing energy loss has nothing to do with nuclear binding, which is to say the total Energy Deposit = non-ion energy + ion energy + energy goes to binding energy ?

As an extreme example, very low-energy neutron breaks a nucleus. But the energy is so low that all the fragments stop after the nucleus break up. In this case, the kinetic energy of neutron purely goes into overcoming the binding energy of the nucleus. And this energy is not recorded as non-ion energy, right? This might just be a naive thought in my mind, Thanks for any corrections.

Regards,
Yihui

Hello,

it is not exact, more accurate: energy deposition at a step is the sum of ionisation energy loss, which includes energy transfer for atoms below production threshold of secondary particles: delta-electrons, bremsstrahlung gamma, low energy nuclear recoil after elastic scattering. Non-ionizing energy loss is an estimation of energy loss due to multiple scattering.

In very low energy neutron is captured by a nucleus, then extra energy (about 8 MeV) is delivered by production of gamma or electrons.

VI

In specific nuclei like Uranium also fission is possible.

VI

Hi Vladimir,

You say that the ionizing energy includes energies that would become secondaries that are below thresholds (as these secondaries would produce signal in the detector, but geant does not produce them due to thresholds). Does this mean that that there are no energy transfers by protons and charged p; ions that do not result in kinetic energy of secondaries, that would have produced signal if they were produced and tracked? That there are no energy transfers to the nuclei that are the “invisible energies” mentioned by Wigmans that just overcome the binding energy of the secondaries in the nuclei, and never appear as ionizing energy that would create a signal in a media llike scintillator or liquid argon, etc?

Regards,
Yihui

Hello,

secondaries are tracked by Geant4 and deposit energy during tracking.

VI

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