next up previous contents
Next: Correcting the cross section Up: Common to All Charged Previous: Computing the Mean Energy   Contents

Subsections


Energy loss fluctuations

Fluctuations in thick absorbers

The total continuous energy loss of charged particles is a stochastic quantity with a distribution described in terms of a straggling function. The straggling is partially taken into account in the simulation of energy loss by the production of $\delta$-electrons with energy $T > T_c$. However, continuous energy loss also has fluctuations. Hence in the current GEANT4 implementation two different models of fluctuations are applied depending on the value of the parameter $\kappa$ which is the lower limit of the number of interactions of the particle in a step. The default value chosen is $\kappa = 10$. In the case of a high range cut (i.e. energy loss without delta ray production) for thick absorbers the following condition should be fulfilled:

\begin{displaymath}
\Delta E > \kappa \ T_{max}
\end{displaymath} (7.6)

where $\Delta E$ is the mean continuous energy loss in a track segment of length $s$, and $T_{max}$ is the maximum kinetic energy that can be transferred to the atomic electron. If this condition holds the fluctuation of the total (unrestricted) energy loss follows a Gaussian distribution. It is worth noting that this condition can be true only for heavy particles, because for electrons, $T_{max}=T/2$, and for positrons, $T_{max}=T$, where $T$ is the kinetic energy of the particle. In order to simulate the fluctuation of the continuous (restricted) energy loss, the condition should be modified. After a study, the following conditions have been chosen:
\begin{displaymath}
\Delta E > \kappa \ T_c
\end{displaymath} (7.7)

and


\begin{displaymath}
T_{max} <= 2 \ T_c
\end{displaymath} (7.8)

where $T_c$ is the cut kinetic energy of $\delta$-electrons. For thick absorbers the straggling function approaches the Gaussian distribution with Bohr's variance [4]:
\begin{displaymath}
\Omega^2 = 2\pi r^2_e m_e c^2 N_{el}\frac{Z_h^2}{\beta^2} T_c s
\left(1 - \frac{\beta^2}{2} \right),
\end{displaymath} (7.9)

where $r_e$ is the classical electron radius, $N_{el}$ is the electron density of the medium, $Z_{h}$ is the charge of the incident particle in units of positron charge, and $\beta$ is the relativistic velocity.

Fluctuations in thin absorbers

If the conditions 7.7 and 7.8 are not satisfied the model of energy fluctuations in thin absorbers is applied. The formulae used to compute the energy loss fluctuation (straggling) are based on a very simple physics model of the atom. It is assumed that the atoms have only two energy levels with binding energies $E_1$ and $E_2$. The particle-atom interaction can be an excitation with energy loss $E_1$ or $E_2$, or ionization with energy loss distributed according to a function $g(E) \sim 1/E^2$ :

\begin{displaymath}
\int_I^{T_{up}} g(E)\ dE = 1 \Longrightarrow
g(E) = \frac{I T_{up}}{T_{up}-I} \frac{1}{E^2} .
\end{displaymath} (7.10)

The macroscopic cross section for excitation $(i=1,2)$ is

\begin{displaymath}
\Sigma_i = C \frac{f_i}{E_i}
\frac{\ln[2mc^2 \ (\beta\gamm...
...E_i]-\beta^2}
{\ln[2mc^2 \ (\beta\gamma)^2/I]-\beta^2}\ (1-r)
\end{displaymath} (7.11)

and the ionization cross section is
\begin{displaymath}
\Sigma_3 = C \frac{T_{up}-I}
{I T_{up}\ln(\frac{T_{up}}{I})}\ r
\end{displaymath} (7.12)

where $I$ denotes the mean ionization energy of the atom, $T_{up}$ is the production threshold for delta ray production (or the maximum energy transfer if this value smaller than the production threshold), $E_i$ and $f_i$ are the energy levels and corresponding oscillator strengths of the atom, and $C$ and $r$ are model parameters. The oscillator strengths $f_i$ and energy levels $E_i$ should satisfy the constraints
\begin{displaymath}
f_1 + f_2 = 1
\end{displaymath} (7.13)


\begin{displaymath}
f_1 \cdotp lnE_1 + f_2 \cdotp lnE_2 = lnI .
\end{displaymath} (7.14)

The cross section formulae 7.11,7.12 and the sum rule equations 7.13,7.14 can be found e.g. in Ref. [1].

The model parameter $C$ can be defined in the following way. The numbers of collisions ($n_i$, $i=1,2$ for excitation and $3$ for ionization) follow the Poisson distribution with a mean value $\langle n_i \rangle$. In a step of length $\Delta x$ the mean number of collisions is given by

\begin{displaymath}
\langle n_i \rangle = \Delta x \ \Sigma_i
\end{displaymath} (7.15)

The mean energy loss in a step is the sum of the excitation and ionization contributions and can be written as
\begin{displaymath}
\frac{dE}{dx} \cdotp \Delta x =
\left \{ \Sigma_1 E_1 + \Sigma_2 E_2 +
\int_I^{T_{up}} E g(E) dE \right \} \Delta x .
\end{displaymath} (7.16)

From this, using eq. 7.11 - 7.14, one can see that
\begin{displaymath}
C = dE/dx .
\end{displaymath} (7.17)

The other parameters in the fluctuation model have been chosen in the following way. $Z \cdotp f_1$ and $Z \cdotp f_2$ represent in the model the number of loosely/tightly bound electrons

\begin{displaymath}
f_2 = 0 \hspace {15 pt} for \hspace {15 pt} Z = 1
\end{displaymath} (7.18)


\begin{displaymath}
f_2 = 2/Z \hspace {15 pt} for \hspace {15 pt} Z \geq 2
\end{displaymath} (7.19)


\begin{displaymath}
E_2 = 10 \mbox{ eV } Z^2 .
\end{displaymath} (7.20)

Using these parameter values, $E_2$ corresponds approximately to the K-shell energy of the atoms ( and $Z f_2 = 2 $ is the number of K-shell electrons). The parameters $f_1$ and $E_1$ can be obtained from Eqs.  7.13 and 7.14.

The parameter $r$ is the only variable in the model which can be tuned. This parameter determines the relative contribution of ionization and excitation to the energy loss. Based on comparisons of simulated energy loss distributions to experimental data, its value has been fixed at $r = 0.4$.

Sampling the energy loss.

The energy loss is computed in the model under the assumption that the step length (or relative energy loss) is small and, in consequence, the cross section can be considered constant along the step. The loss due to the excitation is
\begin{displaymath}
\Delta E_{exc} = n_1 E_1 + n_2 E_2
\end{displaymath} (7.21)

where $n_1$ and $n_2$ are sampled from a Poisson distribution. The energy loss due to ionization can be generated from the distribution $g(E)$ by the inverse transformation method :
\begin{displaymath}
u = F(E) = \int_I^E g(x) dx
\end{displaymath} (7.22)


\begin{displaymath}
E = F^{-1}(u) = \frac {I}{1-u \frac{T_{up}-I}{T_{up}}}
\end{displaymath} (7.23)

where $u$ is a uniformly distributed random number $\in [0,\ 1]$. The contribution coming from the ionization will then be
\begin{displaymath}
\Delta E_{ion} = \sum_{j = 1}^{n3} \frac {I}
{1-u_j \frac{T_{up}-I}{T_{up}}}
\end{displaymath} (7.24)

where $n_3$ is the number of ionizations sampled from the Poisson distribution. The total energy loss in a step will be $ \Delta E = \Delta E_{exc} + \Delta E_{ion}$ and the energy loss fluctuation comes from fluctuations in the number of collisions $n_i$.

Very thin layers

In the case of very small energy loss (small step lengths, $\sim 1$ mm in gases, $\sim 1$ micrometer in solids) this model calculation can give zero energy loss for some events. In order to avoid this nonphysical situation, the probability of zero energy loss is computed as
\begin{displaymath}
P(\Delta E = 0) = e^{-(\langle n_1 \rangle + \langle n_2 \rangle
+\langle n_3 \rangle)} .
\end{displaymath} (7.25)

If this probabillity is bigger than a limit $(0.005)$ a special sampling is done, taking into account the fact that in this case the projectile interacts only with the outer electrons of the atom. An energy level $E_0 = 10$ eV has been chosen to correspond to the outer electrons. The mean number of collisions can be computed as

\begin{displaymath}
\langle n \rangle = \frac{1}{E_0} \frac {dE}{dx} \Delta x .
\end{displaymath} (7.26)

All the collisions can be considered as ionizations for this case. The number of collisions is sampled according to a Poisson distribution and the energy loss is computed from the equation

\begin{displaymath}
\Delta E = \sum_{j = 1}^{n} \frac {E_0}
{1-u_j \frac{T_{up}-E_0}{T_{up}}} .
\end{displaymath} (7.27)

Thick layers

If the mean energy loss and step are in the range of validity of the Gaussian approximation of the fluctuation, the much faster Gaussian sampling is used to compute the actual energy loss.

Conclusions

This simple model of energy loss fluctuations is rather fast and can be used for any thickness of material. This has been verified by performing many simulations and comparing the results with experimental data, such as those in Ref.[2].
As the limit of validity of Landau's theory is approached, the loss distribution approaches the Landau form smoothly.

Status of this document

30.01.02 created by L. Urbán.
28.08.02 updated by V.Ivanchenko.
17.08.04 moved to common to all charged particles (mma)
04.12.04 spelling and grammar check by D.H. Wright
04.05.05 updated by L. Urbán.

Bibliography

  1. H.Bichsel Rev.Mod.Phys. 60 (1988) 663
  2. straggling.lassila K.Lassila-Perini, L.Urbán Nucl.Inst.Meth. A362(1995) 416
  3. GEANT3 manual Cern Program Library Long Writeup W5013 (1994).
  4. eloss.ICRU49ICRU (A. Allisy et al), Stopping Powers and Ranges for Protons and Alpha Particles, ICRU Report 49, 1993.


next up previous contents
Next: Correcting the cross section Up: Common to All Charged Previous: Computing the Mean Energy   Contents