Geant4 simulation of SpaCal (update ) Jelena Ilic 18/11/2014 - - PowerPoint PPT Presentation

Jelena Ilic

Geant4 simulation of SpaCal (update )

18/11/2014

Analysis of scint.photons

Geometry

L ¡= ¡10 ¡cm ¡ 5 ¡cm ¡ 5 ¡cm ¡ x ¡ y ¡ z ¡ 1) Calo block: 5 x 5 x 10 cm 2) 100 µm air gap between W and quartz 3) 200 µm Al “mirror” at the front of calo block

Scintillation photons

NOTE

• In the earlier talks I was giving the number of scint. photons LEAVING the

calo block.

• But what we actually need is the number of scint. photons that arrive at the

back of the calo block.

• How many of them will leave the block depends on the material

surrounding the calo block, and in my simulations it is the AIR.

• Refractive index of the air is 1, refractive index of quartz (fused silica) is 1.47

and of our scintillator is 1.57. That means that the critical angle for the total internal reflection is around 42o

θc = arcsin(nair/n1)~42o

i.e. more than 50% of all scint. photons arriving at the back of the calo block will be totally reflected.

Scintillation photons

• The results on slide 5 show what is happening with scintill. photons for 3

different geometries. I used the following notation:

• 1. noAIR:

no air gap between quartz and W; no Al mirror

• 2. AIR:

100 µm gap between quartz tubes and W; no Al mirror

• 3. AIR & MIRROR:

100 µm gap between quartz tubes and W and 200 µm thick Al mirror at the front of the calo block

• eDep
• energy deposited in the scintillator
• Scint Produced - total number of scint. photons produced
• ScintOut
• the total number of scint. photons that arrive at the front and the back
• f the calo block
• ScintOutP
• the number of scint. photons that arrive at the back of the calo block
• Bulk Apsorp
• Bulk absorption of scintill. photons in material (i.e. in quartz and

scintillator itself)

• Absorbed @ Boundary
• number of scintill. photons absorbed at optical boundary

(i.e. boundary with W, since they cannot be absorbed at dielectric_dielectric boundaries)

ALSO:

Scintillation photons

• If we do not have an air gap between quartz tubes and tungsten around 6% of all

produced scint. photons will reach the end of calo block.

• leaving a small air gap we get 19% off all produced photons at the end of the

block

• Adding a thin Al mirror at the front that number rises to 35% of all scint.photons

produced.

• How many of them will arrive at pmt photodiode of course depends on the

refractive indices of glue/glass of the pmt we are going to use...

• In all 3 cases energy deposited in the scintillator is similar

For the result below I used 20 GeV electrons (1000 events) The impact point is the centre of the block (Tungsten) ¡

Analysis of X0

X0

• X0 is mean distance over which e- loses all but 1/e of its energy by

bremsstrahlung

• Depends on position of our calo-block in regard to the beam

Case 1)

electr electron sour

• n source:

ce: Point distance: distance: 3m dir direction: ection: Cone angular spr angular spread: ead: 3.2 mrad calo calo-block:

• block:

20x20x10 cm pitch: pitch: 1.2 mm tube tube φ: 0.8 mm AIR+ MIRROR

~1 ¡cm ¡ 3m ¡

1000 events

• 159 events (16%) electon didn’t produce shower; average energy deposited in

scintillator 10 MeV

• remaining 841 event-> X0=1.5 cm; average energy deposited in the scintillator

120 MeV ¡

with pitch=1.2 mm; φ holes =0.41 mm 33% holes 77% tungsten X0 mm

X0

Case 2)

electr electron sour

• n source:

ce: Plain sour source shape: ce shape: Circle sour source ce radious radious: 0.5 cm distance: distance: 20 cm e-

• dir

direction: ection: (0, 3deg, 0) calo calo-block:

• block:

10x10x10 cm pitch: pitch: 1.2 mm tube tube φ: 0.8 mm AIR+MIRROR

X0 ¡(cm) ¡ eDep ¡(MeV) ¡ Scint.Produced ¡ x103 ¡ 0.7 ¡ 141 ¡ 1622 ¡ Andrea ¡ Jelena ¡ ¡ pitch ¡=2mm, ¡X0 ¡= ¡0.38 ¡cm ¡ (Tungsten ¡X0 ¡=0.36 ¡cm) ¡ ¡