Method

The double differential cross-section for the process can be written []:

$\{d2\sigma d\nu d\; \rho \}$

$=$

$\alpha 4\{23\; \pi \}(Z\; \lambda )2\{1-\nu \nu \}[\phi$e+ (m/M)2φμ]

All the quantities in the expression above are defined in [PHYS450]. By computing this cross-section for different ($\nu ,\rho$ ) points, it can be seen that:

1. the shape of the functions $\{d2\sigma d\nu d\rho \}$

   and $\{d\; \sigma d\nu \}\int d\; \rho \{d2\sigma d\nu d\rho \}$

   practically does not depend on Z

2. the dominant contribution comes from the low $\nu$ region: $\nu$_min= (4m/E) ≤ν≤100*ν_min

3. in this low region ($d2\sigma /d\nu d\rho$ ) is flat as a function of $\rho$

1. In the low $\nu$ region the differential cross-section

   $\{d\; \sigma d\nu \}=\int d\; \rho \{d2\sigma Kd\nu d\sigma \}$

   can be approximated as:

   

   We can write: $\{d\; \sigma d\nu \}\approx f(\nu )\; g(\nu )$

   where, $f(\nu )\; =\; \{(a-1)\{1\nu$_c^{a- 1}}- ({1ν_max})^a-1}{1ν^a}

   is the normalised distribution in the interval $[\nu$_c,ν_max] and $g(\nu )\; =\; [1-\{\nu$_minν}]^1/2

   is the rejection function.

2. $r$₁ and $r$₂ being two uniformly distributed random numbers in the interval $[0,1]$ :

   

3. Then compute $\rho$_max(ν) = [1 -{6M²E²(1-ν)}][1-{4mνE}]^1/2

   and generate $\rho$ uniformly in the range $[-\rho$_max,+ρ_max] .

$\phi +$ is generated isotropically and $\phi -=\; \phi ++\; \pi$

F. Carminati PHYS460