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CERN Program Library Long Writeup W5013 - CERNLIB ...

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where<br />

α 1 = (0.5 − ɛ 0) 2<br />

F 10 α 2 = 1 3<br />

2 F 20<br />

3<br />

f 1 (ɛ) =<br />

(0.5 − ɛ 0 ) 3 (ɛ − 1<br />

0.5)2 f 2 (ɛ) =<br />

0.5 − ɛ 0<br />

g 1 (ɛ) = F 1 /F 10 g 2 (ɛ) = F 2 /F 20<br />

and<br />

F 1 = F 1 (δ) =3Φ 1 (δ) − Φ 2 (δ) − F (Z) F 10 = F 1 (δ min )<br />

F 2 = F 2 (δ) = 3 2 Φ 1(δ)+ 1 2 Φ 2(δ) − F (Z) F 20 = F 2 (δ min )<br />

δ min = 4 136m<br />

Z 1 3 3E<br />

δ min is the minimal value of the screening variable δ. It can be seen that the functions f i (ɛ) are normalised<br />

and that the functions g i (ɛ) are “valid” rejection functions.<br />

Therefore, if r i are uniformly distributed random numbers (0 ≤ r i ≤ 1), we can sample the ɛ (x in the<br />

program) in the following way:<br />

1. select i to be 1 or 2 according to the following ratio:<br />

BPAR = α 1<br />

α 1 + α 2<br />

If r 0 < BPAR then i =1, otherwise if r 0 ≥ BPAR i =2;<br />

2. sample ɛ from f 1 (ɛ). This can be performed by the following expressions:<br />

⎧<br />

1<br />

⎨ 0.5 − (0.5 − ɛ 0 )r 1 3<br />

( )<br />

when i =1<br />

ɛ = 1<br />

⎩ ɛ 0 +<br />

2 − ɛ 0 r 1 when i =2<br />

3. calculate the rejection function g i (ɛ). Ifr 2 ≤ g i (ɛ), accept ɛ, otherwise return to step 1.<br />

It should be mentioned that we need a step just after sampling ɛ in the step 2, because the cross-section<br />

formula becomes negative at large δ and this imposes an upper limit for δ;<br />

[ ]<br />

42.24 − F (Z)<br />

δ max =exp<br />

− 0.952<br />

8.368<br />

If we get a δ value using the sampled ɛ such that δ>δ max , we have to start again from the step 1. After the<br />

successful sampling of ɛ, GPAIRG generates the polar angles of the electron with respect to an axis defined<br />

along the direction of the parent photon. The electron and the positron are assumed to have a symmetric<br />

angular distribution. The energy-angle distribution is given by Tsai [13, 14] as following:<br />

dσ<br />

dpdΩ = 2α2 e 2 {[ 2x(1 − x)<br />

πkm 4<br />

+<br />

]<br />

12lx(1 − x)<br />

(1 + l) 4 (Z 2 + Z)<br />

(1 + l) 2 −<br />

[ ]<br />

2x 2 − 2x +1 4lx(1 − x) [<br />

(1 + l) 2 +<br />

(1 + l) 4 X − 2Z 2 f((αZ) 2 )<br />

where k is the photon energy, p and E are the momentum and the energy of the electron of the e + e − -pair,<br />

x = E/k and l = E 2 θ 2 /m 2 . This distribution is quite complicated to sample and, furthermore, considered<br />

214 PHYS211 – 3<br />

] }

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