PENELOPE 2003 - OECD Nuclear Energy Agency

138 Chapter 4. Electron/positron transport mechanics
and
∫
M k ′ ≡ N
[ ]
W k ∂σs (E; W )
dW =
∂E
E=E 0
The equations (4.70) with the boundary conditions 〈ω n 〉 s=0
sequentially to any order. For n = 1 we have
which yields
The equation for n = 2 reads,
and its solution is
Hence,
[ ] dMk
. (4.72)
dE
E=E 0
= 0 can now be solved
d
ds 〈ω〉 = S s(E 0 ) − S ′ s (E 0)〈ω〉, (4.73)
〈ω〉 = S s(E 0 ) {
1 − exp [−S
′
S s(E ′ 0 )
s (E 0 )s] } . (4.74)
d
ds 〈ω2 〉 = Ω 2 s(E 0 ) + [ 2S s (E 0 ) − Ω 2′ s (E 0 ) ] 〈ω〉 − 2S s(E ′ 0 )〈ω 2 〉, (4.75)
〈ω 2 〉 = Ω 2 s(E 0 ) 1 − exp[−2S′ s (E 0)s]
2S ′ s (E 0)
var(ω) = 〈ω 2 〉 − 〈ω〉 2
+ s [ 2S s (E 0 ) − Ω 2 s
= Ω 2 s(E 0 ) 1 − exp[−2S′ s (E 0)s]
2S ′ s(E 0 )
′ (E 0 ) ] [ ] 1 − exp[−S
′ 2
S s (E 0 ) s (E 0 )s]
. (4.76)
2S s(E ′ 0 )
− 2Ω 2 s
[ ] ′ 1 − exp[−S
′ 2
(E 0 )S s (E 0 ) s (E 0 )s]
. (4.77)
2S s(E ′ 0 )
Since these expressions are derived from the linear approximation, eq. (4.66), it is
consistent to evaluate 〈ω〉 and var(ω) from their Taylor expansions to second order,
[
〈ω〉 = S s (E 0 ) s 1 − 1 ]
2 S′ s(E 0 ) s + O(s 2 )
≃ S s (E 0 ) s
{
1 − 1 2
[ ] d ln Ss (E)
dE
E=E 0
S s (E 0 ) s
}
(4.78)
and
var(ω) = Ω 2 s (E 0) s −
≃ Ω 2 s(0) s
{
1 −
[ ]
1
2 Ω2 ′ s (E 0 ) S s (E 0 ) + Ω 2 s (E 0) S s ′ (E 0) s 2 + O(s 3 )
[ 1 d ln Ω 2 s(E)
+ d ln S ]
}
s(E)
S s (E 0 ) s , (4.79)
2 dE dE
E=E 0