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PENELOPE 2003 - OECD Nuclear Energy Agency

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4.1. Elastic scattering 131<br />

The simple distribution (4.30) is flexible enough to reproduce the combinations of first<br />

and second moments encountered in the simulations [notice that 〈µ〉 (s) , eq. (4.28), is<br />

always less than 1/2] and allows fast random sampling of µ.<br />

4.1.3 Simulating with the MW model<br />

penelope simulates elastic scattering by using the MW model (see section 3.1), which<br />

allows the formulation of the mixed simulation algorithm in closed analytical form.<br />

The mean free path λ (h)<br />

el between hard elastic events and the cutoff deflection µ c =<br />

(1 − cos θ c )/2 are related through [see eqs. (3.18) and (4.9)]<br />

1<br />

λ (h)<br />

el<br />

This equation can be easily inverted to give<br />

where<br />

= 1<br />

λ el<br />

∫ 1<br />

µ c<br />

p MW (µ) dµ. (4.33)<br />

µ c = P −1<br />

MW (ξ c ) , (4.34)<br />

ξ c ≡ 1 − λ el<br />

λ (h)<br />

el<br />

(4.35)<br />

and P −1<br />

MW is the inverse of the single scattering cumulative distribution function given<br />

by eqs. (3.31) and (3.36).<br />

In the following, we assume that the MW distribution is that of case I, eq. (3.24);<br />

the formulae for case II can be derived in a similar way. The random sampling of the<br />

angular deflection µ in hard collisions is performed by the inverse transform method<br />

(section 1.2.2); random values of µ are obtained from the sampling equation<br />

∫ µ<br />

µ c<br />

p MW (µ ′ ) dµ ′ = ξ<br />

∫ 1<br />

µ c<br />

p MW (µ ′ ) dµ ′ . (4.36)<br />

With the MW distribution, eq. (3.24), this equation can be solved analytically to give<br />

µ = P −1<br />

MW<br />

(<br />

1 − λ )<br />

el<br />

(1 − ξ) . (4.37)<br />

To determine the angular distribution of soft events F a (s; µ), eq. (4.30), we need the<br />

first and second transport mean free paths for soft collisions, which are given by<br />

λ (h)<br />

el<br />

with<br />

( )<br />

(s) −1 2<br />

λ el,1 = T 1 (µ c )<br />

λ el<br />

and<br />

( )<br />

(s) −1 6<br />

λ el,2 = [T 1 (µ c ) − T 2 (µ c )] (4.38)<br />

λ el<br />

T 1 (µ c ) =<br />

∫ µc<br />

0<br />

µp MW (µ) dµ and T 2 (µ c ) =<br />

∫ µc<br />

0<br />

µ 2 p MW (µ) dµ. (4.39)

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