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diff --git a/Master/texmf-dist/doc/latex/hyperref/test/test3.tex b/Master/texmf-dist/doc/latex/hyperref/test/test3.tex deleted file mode 100644 index cbfe6d3a3c7..00000000000 --- a/Master/texmf-dist/doc/latex/hyperref/test/test3.tex +++ /dev/null @@ -1,556 +0,0 @@ -\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} |