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-\documentclass{article}
-\usepackage{graphicx,ae}
-\usepackage[T1]{fontenc}
-\usepackage[latin1]{inputenc}
-%\def\rmdefault{mbv}
-\usepackage{url}
-%\textwidth3in
-\let\SetRowColor\relax
-%\usepackage[times,symbolmenu,spaced=false,zebra,paperwidth=6in,paperheight=4in]{screenpdf}
-\usepackage[]{hyperref}
-\title{Simulation of Energy Loss Straggling}
-\author{Maria Physicist}
-\newcommand{\Emax}{\ensuremath{E_{\mathrm{max}}}}
-\newcommand{\GEANT}{\texttt{GEANT}}
-\begin{document}
-\maketitle
-
-\section{Introduction}
-
-Due to the statistical nature of ionisation energy loss, large
-fluctuations can occur in the amount of energy deposited by a particle
-traversing an absorber element. Continuous processes such as multiple
-scattering and energy loss play a relevant role in the longitudinal
-and lateral development of electromagnetic and hadronic
-showers, and in the case of sampling calorimeters the
-measured resolution can be significantly affected by such fluctuations
-in their active layers. The description of ionisation fluctuations is
-characterised by the significance parameter $\kappa$, which is
-proportional to the ratio of mean energy loss to the maximum allowed
-energy transfer in a single collision with an atomic electron
-\[
-\kappa =\frac{\xi}{\Emax}
-\]
-\Emax{}
-is the maximum transferable energy in a single collision with
-an atomic electron.
-\[
-\Emax =\frac{2 m_e \beta^2\gamma^2 }
-{1 + 2\gamma m_e/m_x + \left ( m_e/m_x\right)^2},
-\]
-where $\gamma = E/m_x$, $E$ is energy and
-$m_x$ the mass of the incident particle,
-$\beta^2 = 1 - 1/\gamma^2$ and $m_e$ is the electron mass.
-$\xi$ comes from the Rutherford scattering cross section
-and is defined as:
-\begin{eqnarray*} \xi = \frac{2\pi z^2 e^4 N_{Av} Z \rho \delta x}
- {m_e \beta^2 c^2 A} = 153.4 \frac{z^2} {\beta^2} \frac{Z}{A}
- \rho \delta x \quad\mathrm{keV},
-\end{eqnarray*}
-where
-
-\begin{tabular}{ll}
-\SetRowColor $z$ & charge of the incident particle \\
-\SetRowColor $N_{Av}$ & Avogadro's number \\
-\SetRowColor $Z$ & atomic number of the material \\
-\SetRowColor $A$ & atomic weight of the material \\
-\SetRowColor $\rho$ & density \\
-\SetRowColor $ \delta x$ & thickness of the material \\
-\end{tabular}
-
-$\kappa$ measures the contribution of the collisions with energy
-transfer close to \Emax. For a given absorber, $\kappa$ tends
-towards large values if $\delta x$ is large and/or if $\beta$ is
-small. Likewise, $\kappa$ tends towards zero if $\delta x $ is small
-and/or if $\beta$ approaches 1.
-
-The value of $\kappa$ distinguishes two regimes which occur in the
-description of ionisation fluctuations :
-
-\begin{enumerate}
-\item A large number of collisions involving the loss of all or most
- of the incident particle energy during the traversal of an absorber.
-
- As the total energy transfer is composed of a multitude of small
- energy losses, we can apply the central limit theorem and describe
- the fluctuations by a Gaussian distribution. This case is
- applicable to non-relativistic particles and is described by the
- inequality $\kappa > 10 $ (i.e. when the mean energy loss in the
- absorber is greater than the maximum energy transfer in a single
- collision).
-
-\item Particles traversing thin counters and incident electrons under
- any conditions.
-
- The relevant inequalities and distributions are $ 0.01 < \kappa < 10
- $, Vavilov distribution, and $\kappa < 0.01 $, Landau distribution.
-\end{enumerate}
-
-An additional regime is defined by the contribution of the collisions
-with low energy transfer which can be estimated with the relation
-$\xi/I_0$, where $I_0$ is the mean ionisation potential of the atom.
-Landau theory assumes that the number of these collisions is high, and
-consequently, it has a restriction $\xi/I_0 \gg 1$. In \GEANT{}
-(see URL \url{http://wwwinfo.cern.ch/asdoc/geant/geantall.html}), the
-limit of Landau theory has been set at $\xi/I_0 = 50$. Below this
-limit special models taking into account the atomic structure of the
-material are used. This is important in thin layers and gaseous
-materials. \autoref{fg:phys332-1} shows the behaviour of $\xi/I_0$
-as a function of the layer thickness for an electron of 100 keV and 1
-GeV of kinetic energy in Argon, Silicon and Uranium.
-
-\begin{figure}
- \centering
- \includegraphics[width=.6\linewidth]{phys1}
- \caption{The variable $\xi/I_0$ can be used to measure the
- validity range of the Landau theory. It depends
- on the type and energy of the particle, $Z$, $A$
- and the ionisation potential of the material and
- the layer thickness.
- }
- \label{fg:phys332-1}
-\end{figure}
-
-In the following sections, the different theories and models for the
-energy loss fluctuation are described. First, the Landau theory and
-its limitations are discussed, and then, the Vavilov and Gaussian
-straggling functions and the methods in the thin layers and gaseous
-materials are presented.
-
-\section{Landau theory}
-\label{sec:phys332-1}
-
-For a particle of mass $m_x$ traversing a thickness of material
-$\delta x $, the Landau probability distribution may be written in
-terms of the universal Landau function $\phi(\lambda)$
-as\cite{bib-LAND}:
-\begin{eqnarray*}
-f( \epsilon , \delta x ) & = &\frac{1}{\xi} \phi ( \lambda )
-\end{eqnarray*}
-where
-\begin{eqnarray*}
-\phi(\lambda )& = & \frac{1} {2 \pi i}\int^{c+i\infty}_{c-i\infty}
-\exp \left ( u \ln u + \lambda u \right ) du \hspace{2cm} c \geq 0 \\
-\lambda & = & \frac{\epsilon -\bar{\epsilon} }{\xi}
- - \gamma' - \beta^2 - \ln \frac{\xi}{\Emax} \\
-\gamma' & = & 0.422784\dots = 1 - \gamma \\
-\gamma & = & 0.577215\dots \mbox{(Euler's constant)} \\
-\bar{\epsilon} & = & \mbox{average energy loss} \\
-\epsilon & = & \mbox{actual energy loss}
-\end{eqnarray*}
-
-\subsection{Restrictions}
-
-The Landau formalism makes two restrictive assumptions :
-\begin{enumerate}
-\item The typical energy loss is small compared to the maximum energy
- loss in a single collision. This restriction is removed in the
- Vavilov theory (see \autoref{vavref}).
-\item The typical energy loss in the absorber should be large compared
- to the binding energy of the most tightly bound electron. For
- gaseous detectors, typical energy losses are a few keV which is
- comparable to the binding energies of the inner electrons. In such
- cases a more sophisticated approach which accounts for atomic energy
- levels\cite{bib-TALM} is necessary to accurately simulate data
- distributions. In \GEANT, a parameterised model by L. Urb\'{a}n is
- used (see section \ref{urban}).
-\end{enumerate}
-
-In addition, the average value of the Landau distribution is infinite.
-Summing the Landau fluctuation obtained to the average energy from the
-$dE/dx$ tables, we obtain a value which is larger than the one coming
-from the table. The probability to sample a large value is small, so
-it takes a large number of steps (extractions) for the average
-fluctuation to be significantly larger than zero. This introduces a
-dependence of the energy loss on the step size which can affect
-calculations.
-
-A solution to this has been to introduce a limit on the value of the
-variable sampled by the Landau distribution in order to keep the
-average fluctuation to 0. The value obtained from the \texttt{GLANDO}
-routine is:
-\[
-\delta dE/dx = \epsilon - \bar{\epsilon} = \xi ( \lambda - \gamma'
-+\beta^2 +\ln \frac{\xi}{\Emax})
-\]
-In order for this to have average 0, we must impose that:
-\[
-\bar{\lambda} = -\gamma' - \beta^2 -\ln \frac{\xi}{\Emax}
-\]
-
-This is realised introducing a $\lambda_{\mathrm{max}}(\bar{\lambda})$
-such that if only values of $\lambda \leq \lambda_{\mathrm{max}}$ are
-accepted, the average value of the distribution is $\bar{\lambda}$.
-
-A parametric fit to the universal Landau distribution has been
-performed, with following result:
-\[
-\lambda_{\mathrm{max}} = 0.60715 +
- 1.1934\bar{\lambda}+(0.67794+0.052382\bar{\lambda})
- \exp(0.94753+0.74442\bar{\lambda})
-\]
-only values smaller than $\lambda_{\mathrm{max}}$ are accepted, otherwise the
-distribution is resampled.
-
-
-
-\section{Vavilov theory}
-\label{vavref}
-
-Vavilov\cite{bib-VAVI} derived a more accurate straggling distribution
-by introducing the kinematic limit on the maximum transferable energy
-in a single collision, rather than using $ \Emax = \infty $.
-Now we can write\cite{bib-SCH1}:
-\begin{eqnarray*}
-f \left ( \epsilon, \delta s \right ) & = & \frac{1}{\xi} \phi_{v}
-\left ( \lambda_{v}, \kappa, \beta^{2} \right )
-\end{eqnarray*}
-where
-\begin{eqnarray*}
-\phi_{v} \left ( \lambda_{v}, \kappa, \beta^{2} \right ) & = &
-\frac{1}{2 \pi i} \int^{c+i\infty}_{c-i\infty}\phi \left( s \right )
-e^{\lambda s} ds \hspace{2cm} c \geq 0 \\
-\phi \left ( s \right ) & = &
-\exp \left [ \kappa ( 1 + \beta^{2}\gamma ) \right ]
-~ \exp \left [ \psi \left ( s \right ) \right ], \\
-\psi \left ( s \right ) & = & s \ln \kappa + ( s + \beta^{2} \kappa )
-\left [ \ln (s/\kappa) + E_{1} (s/\kappa) \right ] - \kappa e^{-s/\kappa},
-\end{eqnarray*}
-and
-\begin{eqnarray*}
-E_{1}(z) & = & \int^{\infty}_{z} t^{-1} e^{-t} dt
-\mbox{\hspace{1cm} (the exponential integral)} \\
-\lambda_v & = & \kappa \left [ \frac{\epsilon - \bar{\epsilon}}{\xi}
-- \gamma' - \beta^2 \right]
-\end{eqnarray*}
-
-The Vavilov parameters are simply related to the Landau parameter by
-$\lambda_L = \lambda_v/\kappa - \ln\kappa $. It can be shown that as
-$\kappa \rightarrow 0$, the distribution of the variable $\lambda_L$
-approaches that of Landau. For $\kappa \leq 0.01$ the two
-distributions are already practically identical. Contrary to what many
-textbooks report, the Vavilov distribution \emph{does not} approximate
-the Landau distribution for small $\kappa$, but rather the
-distribution of $\lambda_L$ defined above tends to the distribution of
-the true $\lambda$ from the Landau density function. Thus the routine
-\texttt{GVAVIV} samples the variable $\lambda_L$ rather than
-$\lambda_v$. For $\kappa \geq 10$ the Vavilov distribution tends to a
-Gaussian distribution (see next section).
-
-\section{Gaussian Theory}
-
-Various conflicting forms have been proposed for Gaussian straggling
-functions, but most of these appear to have little theoretical or
-experimental basis. However, it has been shown\cite{bib-SELT} that
-for $\kappa \geq 10 $ the Vavilov distribution can be replaced by a
-Gaussian of the form :
-\begin{eqnarray*}
-f( \epsilon , \delta s) \approx \frac{1}
-{\xi \sqrt{\frac{2 \pi}{\kappa} \left( 1 - \beta^2/2 \right)}}
- \exp \left [ \frac{( \epsilon - \bar{\epsilon} )^2}{2} \frac{\kappa}
- {\xi^2 (1- \beta^2/2)}\right ]
-\end{eqnarray*}
-thus implying
-\begin{eqnarray*}
-\mathrm{mean} & = & \bar{\epsilon} \\
-\sigma^2 & = & \frac{\xi^2}{\kappa} (1-\beta^2/2) = \xi
- \Emax (1-\beta^2/2)
-\end{eqnarray*}
-
-\section{Urb\'an model}
-\label{urban}
-
-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 \autoref{fg:phys332-2}.
-\begin{figure}
- \centering
- \includegraphics[width=.6\linewidth]{phys2}
- \caption{Energy loss distribution for a 3 GeV electron in
- Argon as given by standard \GEANT. The width of the layers is
- given in centimeters.}
- \label{fg:phys332-2}
-\end{figure}
-
-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 $g(E) \sim 1/E^2$:
-\begin{equation}
-g(E) = \frac{(\Emax + I)I}{\Emax} \frac{1}{E^2}
-\end{equation}
-
-The macroscopic cross-section for excitations ($i=1,2$) is
-\begin{equation}
-\label{eq:sigex}
-\Sigma_i = C \frac{f_i}{E_i} \frac{\ln (2 m \beta^2 \gamma^2/E_i) - \beta^2}
- {\ln (2 m \beta^2 \gamma^2/ I) - \beta^2}(1-r)
-\end{equation}
-and the macroscopic cross-section for ionisation is
-\begin{equation}
-\label{eq:sigion}
-\Sigma_3 = C \frac{\Emax}{I(\Emax+I) \ln(\frac{\Emax+I}{I})}
- ~ r
-\end{equation}
-\Emax{} 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:
-
-\begin{tabular}{ll}
-\SetRowColor $r, C$ & parameters of the model \\
-\SetRowColor $E_i$ & atomic energy levels \\
-\SetRowColor $I$ & mean ionisation energy \\
-\SetRowColor ${f_i}$ & oscillator strengths
-\end{tabular}
-
-The model has the parameters $f_i$, $E_i$, $C$ and $r ~(0\leq r\leq
-1)$. The oscillator strengths $f_i$ and the atomic level energies
-$E_i$ should satisfy the constraints
-\begin{eqnarray}
-f_1 + f_2 & = & 1 \label{eq:fisum}\\
-f_1 \ln E_1 + f_2 \ln E_2 & = & \ln I \label{eq:flnsum}
-\end{eqnarray}
-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 Poisson
-distribution with a mean number $ \langle n_i \rangle $. In a step
-$\Delta x$ the mean number of collisions is
-\begin{equation}
-\langle n_i \rangle = \Sigma_i \Delta x
-\end{equation}
-The mean energy loss $dE/dx$ in a step is the sum of the excitation
-and ionisation contributions
-\begin{equation}
-\frac{dE}{dx} \Delta x = \left[ \Sigma_1 E_1 + \Sigma_2 E_2 +
- \Sigma_3 \int_{I}^{\Emax+I} E~g(E)~dE \right]
- \Delta x
-\end{equation}
-From this, using the equations (\ref{eq:sigex}), (\ref{eq:sigion}),
-(\ref{eq:fisum}) and (\ref{eq:flnsum}), one can define the parameter
-$C$
-\begin{equation}
-C = \frac{dE}{dx}
-\end{equation}
-
-The following values have been chosen in \GEANT{} for the other
-parameters:
-\[
-\begin{array}{lcl}
-f_2 = \left\{ \begin{array}{ll}
- 0 & \mathrm{if} Z \leq 2 \\
- 2/Z & \mathrm{if} Z > 2 \\
- \end{array} \right. & \Rightarrow & f_1 = 1 - f_2 \\
-E_2 = 10 Z^2 \mathrm{eV} & \Rightarrow & E_1 = \left(\frac{I}{E_{2}^{f_2}}
- \right)^{\frac{1}{f_1}} \\
-r = 0.4 & & \\
-\end{array}
-\]
-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
-\begin{equation}
-\Delta E_e = n_1 E_1 + n_2 E_2
-\end{equation}
-where $n_1$ and $n_2$ are sampled from Poisson distribution as
-discussed above. The loss due to the ionisation can be generated from
-the distribution $g(E)$ by the inverse transformation method:
-\begin{eqnarray}
-u = F(E) & = & \int_{I}^E g(x) dx \nonumber \\
-E = F^{-1}(u) & = & \frac{I}{1 - u \frac {\Emax}{\Emax+I}} \\
-\end{eqnarray}
-where $u$ is a uniform random number between $F(I)=0$ and
-$F(\Emax+I)=1$. The contribution from the ionisations will be
-\begin{equation}
-\Delta E_i = \sum_{j=1}^{n_3} \frac{I}
- {1 - u_j \frac {\Emax}{\Emax + I}}
-\end{equation}
-where $n_3$ is the number of ionisation (sampled from Poisson
-distribution). The energy loss in a step will then be $\Delta E =
-\Delta E_e + \Delta E_i$.
-
-
-\subsection{Fast simulation for $n_3 \geq 16$}
-
-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
-$[\alpha I,\Emax]$ the interval B. $\alpha$ lies between 1 and
-$\Emax/I$. A collision with a loss in the interval A happens with
-the probability
-\begin{equation}
-\label{eq:phys332-5}
-P(\alpha) = \int_I^{\alpha I} g(\!E\!) \, dE =
- \frac {( \Emax + I) (\alpha - 1)}{\Emax \alpha}
-\end{equation}
-The mean energy loss and the standard deviation for this type
-of collision are
-\begin{equation}
-\langle \Delta E(\alpha) \rangle = \frac{1}{P(\alpha)}
- \int_I^{\alpha I} E \, g(\!E\!) \, dE =
- \frac{I \alpha \ln \alpha}{\alpha - 1}
-\end{equation}
-and
-\begin{equation}
-\sigma^2(\alpha) = \frac{1}{P(\alpha)}
- \int_I^{\alpha I} E^2 \, g(\!E\!) \, dE =
- I^2 \alpha \left( 1 - \frac{\alpha \ln \! ^2 \alpha}{(\alpha - 1)^2} \right)
-\end{equation}
-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:
-\begin{eqnarray}
-\label{eq:phys332-1}
-\langle n_A \rangle & = & n_3 P(\alpha) \\
-\label{eq:phys332-2}
-\sigma_A^2 & = & n_3 P(\alpha) ( 1 - P(\alpha))
-\end{eqnarray}
-It is further assumed that the energy loss in these collisions
-has a Gaussian distribution with
-\begin{eqnarray}
-\label{eq:phys332-3}
-\langle \Delta E_A \rangle & = & n_A \langle \Delta E(\alpha) \rangle \\
-\label{eq:phys332-4}
-\sigma_{E,A}^2 & = & n_A \sigma^2(\alpha)
-\end{eqnarray}
-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
-\begin{equation}
-\Delta E_B = \sum_{i=1}^{n_3 - n_A} \frac{\alpha I}
- {1 - u_i \frac{\Emax + I - \alpha I}{\Emax + I}}
-\end{equation}
-The total energy loss is the sum of these two types of collisions:
-\begin{equation}
-\Delta E = \Delta E_A + \Delta E_B
-\end{equation}
-
-The approximation of equations (\ref{eq:phys332-1}),
-(\ref{eq:phys332-2}), (\ref{eq:phys332-3}) and (\ref{eq:phys332-4})
-can be used under the following conditions:
-\begin{eqnarray}
-\label{eq:phys332-6}
-\langle n_A \rangle - c \, \sigma_A & \geq & 0 \\
-\label{eq:phys332-7}
-\langle n_A \rangle + c \, \sigma_A & \leq & n_3 \\
-\label{eq:phys332-8}
-\langle \Delta E_A \rangle - c \, \sigma_{E,A} & \geq & 0
-\end{eqnarray}
-where $c \geq 4$. From the equations (\ref{eq:phys332-5}),
-(\ref{eq:phys332-1}) and (\ref{eq:phys332-3}) and from the conditions
-(\ref{eq:phys332-6}) and (\ref{eq:phys332-7}) the following limits can
-be derived:
-\begin{equation}
-\alpha_{\mathrm{min}} = \frac{(n_3 + c^2)(\Emax+I)}
- {n_3 (\Emax + I) + c^2 I} \; \leq \alpha \; \leq
-\alpha_{\mathrm{max}} = \frac{(n_3 + c^2)(\Emax+I)}
- {c^2 (\Emax + I) + n_3 I}
-\end{equation}
-This conditions gives a lower limit to number of the ionisations $n_3$
-for which the fast sampling can be done:
-\begin{equation}
-n_3 \; \geq \; c^2
-\end{equation}
-As in the conditions (\ref{eq:phys332-6}), (\ref{eq:phys332-7}) and
-(\ref{eq:phys332-8}) the value of $c$ is as minimum 4, one gets $n_3
-\; \geq 16$. 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
-\begin{equation}
-n_{B,max} = n_3 - n_{A,min} \approx n_3 (\langle n_A \rangle
- - \sigma_A)
-\end{equation}
-From the previous expressions for $\langle n_A \rangle$ and $\sigma_A$
-one can derive the condition
-\begin{equation}
-n_B \; \leq \; n_{B,max} = \frac{2 n_3 c^2}{n_3+c^2}
-\end{equation}
-The following values are obtained with $c=4$:
-
-\begin{tabular}{llcrr}
-$n_3$ & $n_{B,max}$ & & $n_3$ & $n_{B,max}$\\
-\hline
-\SetRowColor 16 & 16 & & 200 & 29.63\\
-\SetRowColor 20 & 17.78 & & 500 & 31.01 \\
-\SetRowColor 50 & 24.24 & & 1000 & 31.50 \\
-\SetRowColor 100 & 27.59 & & $\infty$ & 32.00
-\end{tabular}
-
-\subsection{Special sampling for lower part of the spectrum}
-
-If the step length is very small ($\leq 5$ mm in gases, $\leq$ 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
-\begin{equation}
-P( \Delta E=0) = e^{-( \langle n_1 \rangle + \langle n_2 \rangle
- + \langle n_3 \rangle )}
-\end{equation}
-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
-\begin{equation}
-\langle n \rangle = \frac{1}{E_0} \frac{dE}{dx} \Delta x
-\end{equation}
-The number of collisions $n$ is sampled from Poisson distribution.
-In the case of the thin layers, all the collisions are considered as
-ionisations and the energy loss is computed as
-\begin{equation}
-\Delta E = \sum_{i=1}^n \frac{E_0}
- {1 - \frac {\Emax}{\Emax + E_0} u_i}
-\end{equation}
-
-
-\begin{thebibliography}{10}
-\bibitem{bib-LAND}
-L.Landau.
-\newblock On the Energy Loss of Fast Particles by Ionisation.
-\newblock Originally published in \emph{J. Phys.}, 8:201, 1944.
-\newblock Rerpinted in D.ter Haar, Editor, \emph{L.D.Landau, Collected
- papers}, page 417. Pergamon Press, Oxford, 1965.
-
-\bibitem{bib-SCH1}
-B.Schorr.
-\newblock Programs for the Landau and the Vavilov distributions and the
- corresponding random numbers.
-\newblock \emph{Comp. Phys. Comm.}, 7:216, 1974.
-
-\bibitem{bib-SELT}
-S.M.Seltzer and M.J.Berger.
-\newblock Energy loss straggling of protons and mesons.
-\newblock In \emph{Studies in Penetration of Charged Particles in
- Matter}, Nuclear Science Series~39, Nat. Academy of Sciences,
- Washington DC, 1964.
-
-\bibitem{bib-TALM}
-R.Talman.
-\newblock On the statistics of particle identification using ionization.
-\newblock \emph{Nucl. Inst. Meth.}, 159:189, 1979.
-
-\bibitem{bib-VAVI}
-P.V.Vavilov.
-\newblock Ionisation losses of high energy heavy particles.
-\newblock \emph{Soviet Physics JETP}, 5:749, 1957.
-
-\end{thebibliography}
-
-\end{document}