Gamma calorimeter for R3B: first simulation results

INDEX

● The calGamma Geant4 simulation ( a short introduction )

● Crystal and geometry selection:

– the full absorption explained

– the effect of the backwards aperture on the absorption

● Outlook

Héctor Alvarez Pol for the R3B collaboration

GENP - Univ. Santiago de Compostela 11 January 2005

● Introduction to the calGamma simulation

Starting the simulation: main features● Very simple geometry: G4Sphere (inner, outer radius, initial and delta and ).

● Still no cover, encapsulation or support in the code.

● A large set of materials is included both for the crystal and the environments; for the crystals: LaBr

3, CsI, NaI, PWO, BGO...; for the environment: air, vacuum...

● Full physics (em, hadronic) packages included, please test it!

● Primaries (gammas, protons, ..., coming from the origin) can be boosted according to the beam energy. Random emission (angle or energy) can be selected.

● Messenger commands for user control (next slides).

● ROOT libraries included for fully integrated analysis interface.

● Partially documented (code is documented in the C++ style, that is, readable ;-) some example macros and results directories available. More on request?

User interface / macros for batch

● Commands allow the control of the program from user interface or macros.

● Commands can be added easily in Messenger classes (requires code recompilation).

● OpenGL/(Dawn)/... viewers for graphical output.

● A ROOT file is created for the storage of TTree / TH1D / TH2D ...

● Online histograms are available while running an interactive session.

An example macro– Commands for output verbosity– Commands for geometry control– Commands for primaries control– ...

Note: about the kinematics...

Lorentz transformation (P||

is the component of P parallel to v):

E' = E + P|| =

E + Pcos

( remember, for gammas, E = P )For an isotropic angular emission the distribution in cos is flat!Then, for protons at T=700MeV ( =0.8197507, =1.74605 ):

E' = 1.74605 E + 1.43132571 E cos

In the limit: cos = 1 → E'=3.1774 E cos = -1 → E'=0.3147 Eand the energy distribution is also flat!

See, for instance, Simply Kinematics from G.I. Kopilov, p.124-128Gammas of 10 MeV

● Crystal and geometry selection

Crystal selection: full absorption results● Full sphere (4) with gamma emission from the center (no boost, 4-iso). Inner radius is always 0.● Results were obtained modifying the sphere outer radius and the energy of the emitted gamma.

We represent the percentage of gammas with full energy deposited on the crystal, as a function of the crystal

sphere radius and the gamma energy

Why the absorption drops at high energies?

25cm 25cm 25cm

25cm25cm

● Number of interactions (Photo, Compton or Pair Conversion) in a 500 mm thick crystal ball (250 < r < 750)● Conversion dominates for larger energies; conversion photons can escape from the crystal bulk.

Why the absorption drops at high energies? (2)

Sphere params:● inner radius: 0cm● outer radius: 20cm ● gamma energy: 30 MeV

Efficiency 65%

1.2% lost below 1% of gamma energy4.5% lost below 511 keV peak0.7% lost in 511 keV photons9.9% lost below 3% of gamma energy

Why the absorption drops at high energies? (3)

Sphere params:● inner radius: 0cm● outer radius: 20cm ● gamma energy: 10 MeV

Efficiency 86.5%

0.6% lost below 3% of gamma energy2.3% lost below 511 keV peak0.5% lost in 511 keV photons5.2% lost below 10% of gamma energy

Selecting the backwards opening angle

● Due to the boost, gammas are basically emitted in the forward direction. A large open zone in the backwards region reduces dramatically the crystal bulk needed.

● To check the effect on the efficiency, a set of simulations with different opening angles (“angle” in next slide) was made.

● Two different sets of gammas of 5 and 10 MeV are boosted as primary source.

Selecting the backwards opening angle

Outlook

● Simulation ready for user tests● First conclusions on geometry for LoI● Request for improvements and further details

Other topics already done...

● RPC GEANT4 simulation already working.● 3D drawings for RPC presentation (LoI).● 3D drawings for Gamma Calorimeter under development

Appendix: some calculations...

Proton mass: m = 938.27203 MeV and kinetic energy T = 700 MeV

Using T = E - m → T + m = m / √ (1-v2) Then: v2 = 1 – m2 / (T + m)2

KINETIC ENERGY (T) T+m (T+m)2 v gamma100 1038.27 1078008.81 0.183351 0.4281955 1.10658200 1138.27 1295663.21 0.320538 0.5661604 1.21316300 1238.27 1533317.62 0.425850 0.6525718 1.31974400 1338.27 1790972.03 0.508449 0.7130560 1.42632500 1438.27 2068626.43 0.574426 0.7579087 1.53289600 1538.27 2366280.84 0.627959 0.7924384 1.63947700 1638.27 2683935.24 0.671991 0.8197507 1.74605800 1738.27 3021589.65 0.708645 0.8418107 1.85263900 1838.27 3379244.06 0.739482 0.8599313 1.95921

1000 1938.27 3756898.46 0.765670 0.8750257 2.06579

Proton mass (m)938.27

Appendix: some formula

NotationRest mass: m, Energy(=mass): E = M = m

Momentum: P = M = m, kinetic energy: T = E-m = m – mAs usual: E2 = P2 + m2

Fotones -> m = 0 -> E = PLorentz transformation: E' = E + P

|| = E + Pcos

P||'= P

|| +E