THE EGS5 CODE SYSTEM

To find the limits of E, we first compute the limits for k ′ . Using Equation 2.167 with m 1 = m, E 1 =
Ĕ 0 , p 1 = ˘p 0 , m 3 = m 4 = 0, E 3 = p 3 = ˘k, which yields
cos θ = (Ĕ0 + m)˘k − Ĕ0m − m 2
˘p 0˘k
. (2.237)
Solving for ˘k and dividing by m we obtain
k ′ =
A
A − (˘p 0 /m) cos θ = 1
. (2.238)
1 −
√T 0 ′/A cos θ
Setting cos θ 3 = ±1 in Equation 2.238 we then see that
k ′ min =
k ′ max =
1
√ =
A
1 + T 0 ′/A A + p ′ 0
1
√ =
A
1 − T 0 ′/A A − p ′ 0
, (2.239)
. (2.240)
We also have
k min ′ + k max ′ 1
1
=
+
(2.241)
1 +
√T 0 ′/A 1 −
√T 0 ′/A √
= 1 − T 0
√T ′/A + 1 + 0 ′/A
2A
1 − T 0 ′/A =
γ + 1 − (γ − 1)
= A .
That is, when one photon is at the minimum, the other is at the maximum and the sum of their
energies is equal to the available energy (divided by m).
Because of the indistinguishability of the two photons, we have the restriction E < 1/2. Moreover,
we have E > E 0 , where
E 0 = k′ min
A = 1
√
A +
AT ′ 0
=
1
A + p ′ 0
. (2.242)
Given our change of variables, the cross section to be sampled is given as
d˘Σ Annih
dE
= mA [S 1 (AE) + S 1 (A(1 − E))] . (2.243)
Because of the symmetry in E, we can expand the range of E to be sampled from (E 0 , 1/2) to
(E 0 , 1 − E 0 ) and we can ignore the second S 1 , as we can use 1 − E if we sample a value of E greater
than 1/2. We can then express the distribution to be sampled as
d˘Σ Annih
dE
(
= mAC 1
[−1 + C 2 − 1 )/ ]
AE,
AE
71
, E ∈ (E 0 , 1 − E 0 ) . (2.244)