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CERN Program Library Long Writeup W5013 - CERNLIB ...

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Geant 3.16 GEANT User’s Guide PHYS328<br />

Origin : G.Lynch Submitted: 23.08.93<br />

Revision : Revised: 16.12.93<br />

Documentation : F.Carminati<br />

1 Subroutines<br />

Plural scattering<br />

CALL GMCOUL<br />

(OMEGA,DIN*)<br />

OMEGA<br />

DIN<br />

(REAL) Ω 0 of the Molière theory;<br />

(REAL) array of dimension 3 containing the direction of the scattered particle with respect to<br />

the incoming direction;<br />

Generates single scatters in small angle approximation returning the new direction with respect to the old<br />

one. It is called by GMULTS when Ω 0 ≤ 20.<br />

2 Method<br />

In the Molière theory the average number of Coulomb scatters for a charged particle in a step is expressed<br />

by the parameter Ω 0 (see [PHYS325]). When Ω 0 ≤ 20, the Molière theory is not applicable any more, even<br />

if it has been noted [48] that it still gives reasonable results down to its mathematical limit Ω 0 = e. The<br />

range 1 < Ω 0 ≤ 20 is called the plural scattering regime. An interesting study of this regime can be found<br />

in [49].<br />

In GEANT, when Ω 0 ≤ 20, a direct simulation method is used for the scattering angle. The number of<br />

scatters is distributed according to a Poissonian law with average Ω=kΩ 0 = e 2γ−1 Ω 0 ≈ 1.167Ω 0 with<br />

γ =0.57721 ... Euler’s constant. Using the customary notations for the probability distribution function<br />

for small angle (sin θ ≈ θ) single scattering, we can write:<br />

f(θ)θdθ = dσ 1<br />

θdθ (1)<br />

θdθ σ<br />

where σ is the cross section for single elastic scattering. We use as cross section the one reported by Molière<br />

[43] [44]:<br />

(<br />

dσ<br />

θdθ =2π 2ZZinc e 2 ) 2<br />

1<br />

pv (θ 2 + χ 2 α )2<br />

This is the classical Rutherford cross section corrected by the screening angle χ α . This angle is described<br />

by Molière as a correction to the Born approximation used to derive the quantum mechanical form of the<br />

Rutherford cross section. We have then:<br />

∫ ( ) ∞<br />

dσ<br />

σ =<br />

0 θdθ θdθ =2π 2ZZinc e 2 2 ∫ ( ) ∞<br />

θdθ<br />

pv 0 (θ 2 + χ 2 α) 2 = π 2ZZinc e 2 2<br />

1<br />

pv χ 2 α<br />

and so equation (1) becomes:<br />

f(θ)θdθ = χ 2 2θdθ<br />

α<br />

(θ 2 + χ 2 α) 2 = 2ΘdΘ<br />

(1 + Θ 2 ) 2<br />

250 PHYS328 – 1

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