Geant 4 Detector Simulation

Geant 4 Detector Simulation - Presentation Transcript

Uniform B-field 5 m Supposed that “Everything” is composed of “Vacuum” … 1.5 m detector # : 0 1 2 3 4 5 6 7

Sample: Signal Criteria

Generate 6’000 & 60’000 events

E total = 3 GeV – 7 GeV

 incident = 25 º - 35º

All e- (negatively charged, light particle)

No interactions w/ any matter, for example, the tracking chambers, medium and the target (point source).

Single hit per tracking chamber / detector

“ Purely” helix motion in an uniform B-field

Uniform B-field : Summary

X (t) = a*cos(  t+  )-a*cos  Y (t) = a*sin(  t+  )-a*sin  X (t) = C*cos  *t

X (t) = a*cos(  t)-a  Y (t) = a*sin(  t) X (t) = -0.5*a*  2 *t 2  Y (t) = a*  *t

 X = X7-X4 = -0.5*a*  2 *(t 7 2 – t 4 2 ) ;  X/  t = -0.5*a*  2 *(t 7 + t 4 ) ; X = (  X/  t)*t + 

 X -  (  X/  t)*t = 0.5*a*  2 *t 7 *t 4 = 0.5*(C*P ┴ /e*B)*(e*B/P) 2 *(Z 7 *Z 4 /C 2 *cos 2  )

 ~ sin  / P * cos 2   tan  rec ~ ( Y 7 - Y 4 )/( Z 7 - Z 4 ) = tan  ini

X’ (t) Y’ (t) = X (t) Y (t) cos  sin  -sin  cos  Rotation Transformation Physics Invariance  = 0 t << 1 /  S1 S2 Z  B = 1.5 T (  ) Y X

80.25 84.84 88.78 Z (per T) (m) 8.92 7.78 6.57 R (m) 3.27e-7 T (period) (sec) 7 GeV 57.32 60.60 63.42 Z (per T) (m) 6.37 5.56 4.70 R (m) 2.33e-7 T (period) (sec) 5 GeV 34.39 36.36 38.05 Z (per T) (m) 3.82 3.33 2.82 R (m) 1.40e-7 T (period) (sec) 35 º 30 º 25 º 3 GeV

   / (sin  / P*cos 2  Based on P.3    / (  / P) Based on P.7 Preliminary!   v.s. P   v.s. P   v.s.  ini   v.s.  ini    

Preliminary!    / (  / P) Based on P.7    / (sin  / P*cos 2  Based on P.3 K 2 ~ 14 – 15 P (  rec ,  ) = K 2 *sin  rec /  *cos 2  rec  ini (  rec ,  ) =  rec K 1 / P ini  = 0.017 K 1 ~ too small K 1 /  ini  = 0.045 K 1  abandoned! K 2 / P ini  = 14.1 K 2 is more reliable and stable! K 2 /  ini  = 15.3

Preliminary! S   S 1 / P ini  = 9.13 S 1 /  ini  = 15.3 S  v.s. P ini S  v.s.  ini

Preliminary! (S 1 /  ) / P ini  = 0.05 (S 1 /  ) is reliable! (S 1 /  ) /  ini 0.009 = 0.05    / (sin  / P*cos 2  Based on P.3 K 2 ~ 14 – 15 P (  rec ,  ) = K 2 *sin  rec /  *cos 2  rec  ini (  rec ,  ) =  rec (S 1 /  ) = 0.05 S 1 (P,  ini ) = 0.05*  = 0.05* K 2 *sin  rec / P*cos 2  rec S  /  v.s. P ini S  /  v.s.  ini S  / 

To-Do List :

Figure out the problem which caused  X and  Y not equal to zero. (in P.5 – P.6)

Apply a non-uniform B-field and as realistic materials as what the proposal described to our G4 simulation.

Compare the outcome with that derived from the uniform B-field case. See how much it may change with the impact of a non-uniform B-field as well as interactions between negatively charged electrons and materials on G4 simulation.