Urbán model

  The method for computing restricted energy losses with $\delta$ -ray production above given threshold energy in GEANT is a Monte Carlo method that can be used for thin layers. It is fast and it can be used for any thickness of a medium. Approaching the limit of the validity of Landau's theory, the loss distribution approaches smoothly the Landau form as shown in figure .

Figure: Energy loss distribution for a 3 GeV electron in Argon as given by standard GEANT. The width of the layers is given in centimeters.

It is assumed that the atoms have only two energy levels with binding energy $E$₁ and $E$₂ . The particle--atom interaction will then be an excitation with energy loss $E$₁ or $E$₂ , or an ionisation with an energy loss distributed according to a function $g(E)\; \sim 1/E2$ : $g(E)\; =\; \{\; (E$_max+ I)IE_max}{1E²}

The macroscopic cross-section for excitations (i=1,2) is $\Sigma$_i= C {f_iE_i}{ ln(2 m β²γ²/E_i) - β²ln(2 m β²γ²/ I) - β²}(1-r)

and the macroscopic cross-section for ionisation is $\Sigma$₃= C {E_maxI(E_max+I) ln({E_max+II})} r

$E$_max is the GEANT cut for $\delta$ -production, or the maximum energy transfer minus mean ionisation energy, if it is smaller than this cut-off value. The following notation is used:

$r,\; C$	parameters of the model
$E$i	atomic energy levels
$I$	mean ionisation energy
$f$i	oscillator strengths

The model has the parameters $f$_i , $E$_i , C and $r\; (0\le r\le 1)$ . The oscillator strengths $f$_i and the atomic level energies $E$_i should satisfy the constraints

$f$1+ f2	$=$	$1$
$f$1lnE1+ f2lnE2	$=$	$lnI$

The parameter C can be defined with the help of the mean energy loss $dE/dx$ in the following way: The numbers of collisions ($n$_i , i = 1,2 for the excitation and 3 for the ionisation) follow the Poissonian distribution with a mean number $⟨n$_i⟩ . In a step $\Delta x$ the mean number of collisions is $⟨n$_i⟩= Σ_iΔx

The mean energy loss $dE/dx$ in a step is the sum of the excitation and ionisation contributions $\{dEdx\}\Delta x\; =\; [\; \Sigma$₁E₁+ Σ₂E₂+ Σ₃∫_I^E_max+IE g(E) dE ]Δx

From this, using the equations (), (), () and (), one can define the parameter C $C\; =\; \{dEdx\}$

The following values have been chosen in GEANT for the other parameters:

With these values the atomic level $E$₂ corresponds approximately the K-shell energy of the atoms and $Zf$₂ the number of K-shell electrons. r is the only variable which can be tuned freely. It determines the relative contribution of ionisation and excitation to the energy loss.

The energy loss is computed with the assumption that the step length (or the relative energy loss) is small, and --- in consequence --- the cross-section can be considered constant along the path length. The energy loss due to the excitation is $\Delta E$_e= n₁E₁+ n₂E₂

where $n$₁ and $n$₂ are sampled from Poissonian distribution as discussed above. The loss due to the ionisation can be generated from the distribution $g(E)$ by the inverse transformation method:

$u\; =\; F(E)$	$=$	$\int$IEg(x) dx
$E\; =\; F-1(u)$	$=$	$\{I1\; -\; u\; \{E$maxEmax+I}}

where u is a uniform random number between $F(I)=0$ and $F(E$_max+I)=1 . The contribution from the ionisations will be $\Delta E$_i= ∑_j=1^n₃{I1 - u_j{E_maxE_max+ I}}

where $n$₃ is the number of ionisation (sampled from Poissonian distribution). The energy loss in a step will then be $\Delta E\; =\; \Delta E$_e+ ΔE_i .

Fast simulation for n3≥16

If the number of ionisation $n$₃ is bigger than 16, a faster sampling method can be used. The possible energy loss interval is divided in two parts: one in which the number of collisions is large and the sampling can be done from a Gaussian distribution and the other in which the energy loss is sampled for each collision. Let us call the former interval $[I,\; \alpha I]$ the interval A, and the latter $[\alpha I,E$_max] the interval B. $\alpha$ lies between 1 and $E$_max/I . A collision with a loss in the interval A happens with the probability $P(\alpha )\; =\; \int$_I^αIg(E)  dE = {( E_max+ I) (α- 1)E_maxα}

The mean energy loss and the standard deviation for this type of collision are $⟨\Delta E(\alpha )\; ⟩=\; \{1P(\alpha )\}\int$_I^αIE  g(E)  dE = {I αlnαα- 1}

and $\sigma 2(\alpha )\; =\; \{1P(\alpha )\}\int$_I^αIE² g(E)  dE = I²α( 1 - {αln²α(α- 1)²})

If the collision number is high , we assume that the number of the type A collisions can be calculated from a Gaussian distribution with the following mean value and standard deviation:

$⟨n$A⟩	$=$	$n$3P(α)
$\sigma$A2	$=$	$n$3P(α) ( 1 - P(α))

It is further assumed that the energy loss in these collisions has a Gaussian distribution with

$⟨\Delta E$A⟩	$=$	$n$A⟨ΔE(α) ⟩
$\sigma$E,A2	$=$	$n$Aσ2(α)

The energy loss of these collision can then be sampled from the Gaussian distribution.

The collisions where the energy loss is in the interval B are sampled directly from $\Delta E$_B= ∑_i=1^{n₃- n_A}{αI1 - u_i{E_max+ I - αIE_max+ I}}

The total energy loss is the sum of these two types of collisions: $\Delta E\; =\; \Delta E$_A+ ΔE_B

The approximation of equations ((), (), () and () can be used under the following conditions:

$⟨n$A⟩- c  σA	$\ge$	$0$
$⟨n$A⟩+ c  σA	$\le$	$n$3
$⟨\Delta E$A⟩- c  σE,A	$\ge$	$0$

where $c\; \ge 4$ . From the equations (), () and () and from the conditions () and () the following limits can be derived: $\alpha$_min= { (n₃+ c²)(E_max+I)n₃(E_max+ I) + c²I} ≤α ≤α_max= { (n₃+ c²)(E_max+I)c²(E_max+ I) + n₃I}

This conditions gives a lower limit to number of the ionisations $n$₃

for which the fast sampling can be done: $n$₃ ≥ c²

As in the conditions (), () and () the value of c is as minimum 4, one gets $n$₃ ≥16 . In order to speed the simulation, the maximum value is used for $\alpha$ .

The number of collisions with energy loss in the interval B (the number of interactions which has to be simulated directly) increases slowly with the total number of collisions $n$₃ . The maximum number of these collisions can be estimated as $n$_B,max= n₃- n_A,min≈n₃(⟨n_A⟩- σ_A)

From the previous expressions for $⟨n$_A⟩ and $\sigma$_A one can derive the condition $n$_B ≤ n_B,max= {2 n₃c²n₃+c²}

The following values are obtained with c=4:

$n$3	$n$B,max		$n$3	$n$B,max
16	16		200	29.63
20	17.78		500	31.01
50	24.24		1000	31.50
100	27.59			32.00

Special sampling for lower part of the spectrum

If the step length is very small ($\le 5$ mm in gases, $\le$ 2-3 $\mu$ m in solids) the model gives 0 energy loss for some events. To avoid this, the probability of 0 energy loss is computed $P(\; \Delta E=0)\; =\; e-(\; ⟨n$₁⟩+ ⟨n₂⟩+ ⟨n₃⟩)

If the probability is bigger than 0.01 a special sampling is done, taking into account the fact that in these cases the projectile interacts only with the outer electrons of the atom. An energy level $E$₀= 10 eV is chosen to correspond to the outer electrons. The mean number of collisions can be calculated from $⟨n\; ⟩=\; \{1E$₀}{ dEdx}Δx

The number of collisions n is sampled from Poissonian distribution. In the case of the thin layers, all the collisions are considered as ionisations and the energy loss is computed as $\Delta E\; =\; \sum$_i=1ⁿ{E₀ 1 - {E_maxE_max+ E₀}u_i}