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Simulation of Energy Loss StragglingMaria Physicist
January 17, 1999
Introduction${(a+b)}^{2}$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}{{E}_{\mathrm{max}}}$${E}_{\mathrm{max}}$ is the
maximum transferable energy in a single collision with an atomic electron.
${E}_{\mathrm{max}}=\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:
$\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\phantom{\rule{12pt}{0ex}}\mathrm{keV},\text{}$
where
$z$charge of the incident particle ${N}_{Av}$Avogadro's number $Z$atomic number of the material$A$atomic weight of the material $\rho $density $\delta x$thickness of the material $\kappa $
measures the contribution of the collisions with energy transfer close to
${E}_{\mathrm{max}}$. 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
:
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).
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.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 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. Figure 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.
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.
Landau theoryFor 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 \left(\lambda \right)$
as:
$f(\epsilon ,\delta x)=\frac{1}{\xi}\phi \left(\lambda \right)\text{}$
where
$\phi \left(\lambda \right)=\frac{1}{2\pi i}{\int}_{c+i\infty}^{c-i\infty}exp\left(u,ln,u,+,\lambda ,u\right)du\phantom{\rule{2cm}{0ex}}c\ge 0\text{}$$$$\lambda =\frac{\epsilon -\underset{}{\overset{\u2305}{\epsilon}}}{\xi}-\gamma \prime -{\beta}^{2}-ln\frac{\xi}{{E}_{\mathrm{max}}}\text{}$$$$\gamma \prime =0.422784...=1-\gamma \text{}$$$$\gamma =0.577215...\text{(Euler's constant)}\text{}$$$$\underset{}{\overset{\u2305}{\epsilon}}=\text{average energy loss}\text{}$$$$\epsilon =\text{actual energy loss}\text{}$
RestrictionsThe Landau formalism makes two restrictive assumptions :
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
section
).
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
is
necessary
to
accurately
simulate
data
distributions.
In
GEANT,
a
parameterised
model
by
L.
Urbán
is
used
(see
section
).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 GLANDO routine is:
$\delta dE/dx=\epsilon -\underset{}{\overset{\u2305}{\epsilon}}=\xi (\lambda -\gamma \prime +{\beta}^{2}+ln\frac{\xi}{{E}_{\mathrm{max}}})$
In order for this to have average 0, we must impose that:
$\underset{}{\overset{\u2305}{\lambda}}=-\gamma \prime -{\beta}^{2}-ln\frac{\xi}{{E}_{\mathrm{max}}}$This is realised introducing a ${\lambda}_{\mathrm{max}}\left(\underset{}{\overset{\u2305}{\lambda}}\right)$
such that if only values of $\lambda \le {\lambda}_{\mathrm{max}}$
are accepted, the average value of the distribution is
$\underset{}{\overset{\u2305}{\lambda}}$.
A parametric fit to the universal Landau distribution has been performed, with following result:
${\lambda}_{\mathrm{max}}=0.60715+1.1934\underset{}{\overset{\u2305}{\lambda}}+(0.67794+0.052382\underset{}{\overset{\u2305}{\lambda}})exp(0.94753+0.74442\underset{}{\overset{\u2305}{\lambda}})$ only values
smaller than ${\lambda}_{\mathrm{max}}$
are accepted, otherwise the distribution is resampled.
Vavilov theoryVavilov derived a more accurate straggling distribution by introducing the kinematic
limit on the maximum transferable energy in a single collision, rather than using
${E}_{\mathrm{max}}=\infty $. Now
we can write:
$f\left(\epsilon ,,,\delta ,s\right)=\frac{1}{\xi}{\phi}_{v}\left({\lambda}_{v},,,\kappa ,,,{\beta}^{2}\right)\text{}$
where
${\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\phantom{\rule{2cm}{0ex}}c\ge 0\text{}$$$$\phi \left(s\right)=exp\left[\kappa ,(1+{\beta}^{2}\gamma )\right]exp\left[\psi ,\left(s\right)\right],\text{}$$$$\psi \left(s\right)=sln\kappa +(s+{\beta}^{2}\kappa )\left[ln,(s/\kappa ),+,{E}_{1},(s/\kappa )\right]-\kappa {e}^{-s/\kappa},\text{}$
and
${E}_{1}\left(z\right)={\int}_{\infty}^{z}{t}^{-1}{e}^{-t}dt\phantom{\rule{1cm}{0ex}}\text{(the exponential integral)}\text{}$$$${\lambda}_{v}=\kappa \left[\frac{\epsilon -\underset{}{\overset{\u2305}{\epsilon}}}{\xi},-,\gamma ,\prime ,-,{\beta}^{2}\right]\text{}$The Vavilov parameters are simply related to the Landau parameter by
${\lambda}_{L}={\lambda}_{v}/\kappa -ln\kappa $. It can be shown that
as $\kappa \to 0$, the distribution of
the variable ${\lambda}_{L}$ approaches
that of Landau. For $\kappa \le 0.01$
the two distributions are already practically identical. Contrary to what many textbooks
report, the Vavilov distribution does not approximate the Landau distribution for small
$\kappa $, but rather the
distribution of ${\lambda}_{L}$
defined above tends to the distribution of the true
$\lambda $ from
the Landau density function. Thus the routine GVAVIV samples the variable
${\lambda}_{L}$ rather
than ${\lambda}_{v}$.
For $\kappa \ge 10$
the Vavilov distribution tends to a Gaussian distribution (see next section).
Gaussian TheoryVarious conflicting forms have been proposed for Gaussian straggling functions, but most
of these appear to have little theoretical or experimental basis. However, it has been shown
that for $\kappa \ge 10$
the Vavilov distribution can be replaced by a Gaussian of the form:
$f(\epsilon ,\delta s)\approx \frac{1}{\xi \sqrt{\frac{2\pi}{\kappa}\left(1,-,{\beta}^{2},/,2\right)}}exp\left[\frac{{(\epsilon -\underset{}{\overset{\u2305}{\epsilon}})}^{2}}{2},\frac{\kappa}{{\xi}^{2}(1-{\beta}^{2}/2)}\right]\text{}$
thus implying
$\mathrm{mean}=\underset{}{\overset{\u2305}{\epsilon}}\text{}$$$${\sigma}^{2}=\frac{{\xi}^{2}}{\kappa}(1-{\beta}^{2}/2)=\xi {E}_{\mathrm{max}}(1-{\beta}^{2}/2)\text{}$
Urbán modelThe method for computing restricted energy losses with
$\delta $-ray
production above given threshold energy in GEANT is a Monte Carlo method that
can be used for thin layers. It is fast and it can be used for any thickness of a
medium. Approaching the limit of the validity of Landau's theory, the loss
distribution approaches smoothly the Landau form as shown in Figure .
It is assumed that the atoms have only two energy levels with binding energy
${E}_{1}$ and
${E}_{2}$.
The particle--atom interaction will then be an excitation with energy loss
${E}_{1}$ or
${E}_{2}$, or
an ionisation with an energy loss distributed according to a function
$g\left(E\right)\sim 1/{E}^{2}$:
$g\left(E\right)=\frac{({E}_{\mathrm{max}}+I)I}{{E}_{\mathrm{max}}}\frac{1}{{E}^{2}}(1)$The macroscopic cross-section for excitations
($i=1,2$) is
${\Sigma}_{i}=C\frac{{f}_{i}}{{E}_{i}}\frac{ln(2m{\beta}^{2}{\gamma}^{2}/{E}_{i})-{\beta}^{2}}{ln(2m{\beta}^{2}{\gamma}^{2}/I)-{\beta}^{2}}(1-r)(2)$and
the macroscopic cross-section for ionisation is
${\Sigma}_{3}=C\frac{{E}_{\mathrm{max}}}{I({E}_{\mathrm{max}}+I)ln\left(\frac{{E}_{\mathrm{max}}+I}{I}\right)}r(3)$${E}_{\mathrm{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}+{f}_{2}=1\text{(4)}$$$${f}_{1}ln{E}_{1}+{f}_{2}ln{E}_{2}=lnI\text{(5)}$
The parameter $C$
can be defined with the help of the mean energy loss
$dE/dx$ in the following way: The
numbers of collisions (${n}_{i}$,
i = 1,2 for the excitation and 3 for the ionisation) follow the Poisson distribution with a mean
number $<{n}_{i}>;$. In a step
$\Delta x$ the mean number
of collisions is $<{n}_{i}>;={\Sigma}_{i}\Delta x(6)$The
mean energy loss $dE/dx$
in a step is the sum of the excitation and ionisation contributions
$\frac{dE}{dx}\Delta x=\left[{\Sigma}_{1},{E}_{1},+,{\Sigma}_{2},{E}_{2},+,{\Sigma}_{3},{\int}_{I}^{{E}_{\mathrm{max}}+I},E,g,\left(E\right),d,E\right]\Delta x(7)$From
this, using the equations (), (), () and (), one can define the parameter
$C$$C=\frac{dE}{dx}(8)$The following values have been chosen in GEANT for the other parameters:
$\begin{array}{ccc}{f}_{2}=\left\{\begin{array}{cc}0& \mathrm{if}Z\le 2\\ 2/Z& \mathrm{if}Z>2\\ \end{array}\right)& \twoheadrightarrow & {f}_{1}=1-{f}_{2}\\ {E}_{2}=10{Z}^{2}\mathrm{eV}& \twoheadrightarrow & {E}_{1}={\left(\frac{I}{{E}_{2}^{{f}_{2}}}\right)}^{\frac{1}{{f}_{1}}}\\ r=0.4& & \\ \end{array}$ With these values
the atomic level ${E}_{2}$
corresponds approximately the K-shell energy of the atoms and
$Z{f}_{2}$ 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
$\Delta {E}_{e}={n}_{1}{E}_{1}+{n}_{2}{E}_{2}(9)$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
$g\left(E\right)$ by
the inverse transformation method:
$u=F\left(E\right)={\int}_{I}^{E}g\left(x\right)dx\text{}$$$$E={F}^{-1}\left(u\right)=\frac{I}{1-u\frac{{E}_{\mathrm{max}}}{{E}_{\mathrm{max}}+I}}\text{(10)}$$$$\text{(11)}$
where $u$ is a uniform random
number between $F\left(I\right)=0$ and
$F({E}_{\mathrm{max}}+I)=1$. The contribution from the
ionisations will be $\Delta {E}_{i}={\sum}_{j=1}^{{n}_{3}}\frac{I}{1-{u}_{j}\frac{{E}_{\mathrm{max}}}{{E}_{\mathrm{max}}+I}}(12)$where
${n}_{3}$ is the
number of ionisation (sampled from Poisson distribution). The energy loss in a step will
then be $\Delta E=\Delta {E}_{e}+\Delta {E}_{i}$.
Fast simulation for ${n}_{3}\ge 16$If the number of ionisation ${n}_{3}$
is bigger than 16, a faster sampling method can be used. The possible energy loss
interval is divided in two parts: one in which the number of collisions is large and the
sampling can be done from a Gaussian distribution and the other in which
the energy loss is sampled for each collision. Let us call the former interval
$[I,\alpha I]$ the interval A,
and the latter $[\alpha I,{E}_{\mathrm{max}}]$ the
interval B. $\alpha $ lies
between 1 and ${E}_{\mathrm{max}}/I$.
A collision with a loss in the interval A happens with the probability
$P\left(\alpha \right)={\int}_{I}^{\alpha I}g\left(E\right)dE=\frac{({E}_{\mathrm{max}}+I)(\alpha -1)}{{E}_{\mathrm{max}}\alpha}(13)$The
mean energy loss and the standard deviation for this type of collision are
$<\Delta E\left(\alpha \right)>;=\frac{1}{P\left(\alpha \right)}{\int}_{I}^{\alpha I}Eg\left(E\right)dE=\frac{I\alpha ln\alpha}{\alpha -1}(14)$and
${\sigma}^{2}\left(\alpha \right)=\frac{1}{P\left(\alpha \right)}{\int}_{I}^{\alpha I}{E}^{2}g\left(E\right)dE={I}^{2}\alpha \left(1,-,\frac{\alpha {ln}^{2}\alpha}{{(\alpha -1)}^{2}}\right)(15)$If the
collision number is high, we assume that the number of the type A collisions can be
calculated from a Gaussian distribution with the following mean value and standard
deviation:
$<{n}_{A}>;={n}_{3}P\left(\alpha \right)\text{(16)}$$$${\sigma}_{A}^{2}={n}_{3}P\left(\alpha \right)(1-P\left(\alpha \right))\text{(17)}$
It is further assumed that the energy loss in these collisions has a Gaussian
distribution with
$<\Delta {E}_{A}>;={n}_{A}<\Delta E\left(\alpha \right)>;\text{(18)}$$$${\sigma}_{E,A}^{2}={n}_{A}{\sigma}^{2}\left(\alpha \right)\text{(19)}$
The energy loss of these collision can then be sampled from the Gaussian
distribution.
The collisions where the energy loss is in the interval B are sampled directly from
$\Delta {E}_{B}={\sum}_{i=1}^{{n}_{3}-{n}_{A}}\frac{\alpha I}{1-{u}_{i}\frac{{E}_{\mathrm{max}}+I-\alpha I}{{E}_{\mathrm{max}}+I}}(20)$The
total energy loss is the sum of these two types of collisions:
$\Delta E=\Delta {E}_{A}+\Delta {E}_{B}(21)$The approximation of equations (), (), () and () can be used under the following
conditions:
$<{n}_{A}>;-c{\sigma}_{A}\ge 0\text{(22)}$$$$<{n}_{A}>;+c{\sigma}_{A}\le {n}_{3}\text{(23)}$$$$<\Delta {E}_{A}>;-c{\sigma}_{E,A}\ge 0\text{(24)}$
where $c\ge 4$. From
the equations (), () and () and from the conditions () and () the following limits can be
derived: ${\alpha}_{\mathrm{min}}=\frac{({n}_{3}+{c}^{2})({E}_{\mathrm{max}}+I)}{{n}_{3}({E}_{\mathrm{max}}+I)+{c}^{2}I}\le \alpha \le {\alpha}_{\mathrm{max}}=\frac{({n}_{3}+{c}^{2})({E}_{\mathrm{max}}+I)}{{c}^{2}({E}_{\mathrm{max}}+I)+{n}_{3}I}(25)$This
conditions gives a lower limit to number of the ionisations
${n}_{3}$ for which the fast
sampling can be done: ${n}_{3}\ge {c}^{2}(26)$As
in the conditions (), () and () the value of
$c$ is as minimum
4, one gets ${n}_{3}\ge 16$.
In order to speed the simulation, the maximum value is used for
$\alpha $.
The number of collisions with energy loss in the interval B (the number of interactions
which has to be simulated directly) increases slowly with the total number of collisions
${n}_{3}$.
The maximum number of these collisions can be estimated as
${n}_{B,max}={n}_{3}-{n}_{A,min}\approx {n}_{3}(<{n}_{A}>;-{\sigma}_{A})(27)$From the previous
expressions for $<{n}_{A}>;$ and
${\sigma}_{A}$ one can derive the
condition ${n}_{B}\le {n}_{B,max}=\frac{2{n}_{3}{c}^{2}}{{n}_{3}+{c}^{2}}(28)$The following
values are obtained with $c=4$:
${n}_{3}$${n}_{B,max}$${n}_{3}$${n}_{B,max}$16 16 200 29.6320 17.78 500 31.0150 24.24 1000 31.50100 27.59 $\infty $ 32.00
Special sampling for lower part of the spectrumIf 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 $P(\Delta E=0)={e}^{-(<{n}_{1}>;+<{n}_{2}>;+<{n}_{3}>;)}(29)$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
$<n>;=\frac{1}{{E}_{0}}\frac{dE}{dx}\Delta x(30)$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
$\Delta E={\sum}_{i=1}^{n}\frac{{E}_{0}}{1-\frac{{E}_{\mathrm{max}}}{{E}_{\mathrm{max}}+{E}_{0}}{u}_{i}}(31)$
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Landau
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