Commit: 0b7a88778adcf8906d0e1031b543f9f1e33b5924 - myprojects-hg-reborn (hg)

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• Revision: 0b7a88778adcf8906d0e1031b543f9f1e33b5924 (tree)
• Zeit: 2017-11-08 00:30:09
• Autor: Lorenzo Isella <lorenzo.isella@gmai...>
• Commiter: Lorenzo Isella

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I added a file which contains a useful function to add comments to a tex file.

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• add: latex-documents/HomogeneousKernel_yd2.tex

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diff -r fa117086f5ea -r 0b7a88778adc latex-documents/HomogeneousKernel_yd2.tex
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+++ b/latex-documents/HomogeneousKernel_yd2.tex	Tue Nov 07 16:30:09 2017 +0100
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+%\documentclass[titlepage,preprint,showpacs,superscriptaddress,amsmath,amssymb,aps,pre]{revtex4-1}
+\documentclass[notitlepage,a4paper,superscriptaddress,amsmath,amssymb,aps,pre,nofootinbib,10pt]{revtex4-1}
+\usepackage{graphicx}
+\usepackage[usenames]{color}
+\usepackage{url}
+\usepackage{longtable}
+\usepackage{natbib}
+
+\usepackage{xcolor}
+
+\makeatletter
+\def\mathcolor#1#{\@mathcolor{#1}}
+\def\@mathcolor#1#2#3{%
+  \protect\leavevmode
+  \begingroup
+    \color#1{#2}#3%
+  \endgroup
+}
+\makeatother
+
+\newcommand{\lorenzo}[1]{{\leavevmode\color{blue}[#1]}}
+
+\newcommand{\s}{\sum}
+\newcommand{\f}{\frac}
+\newcommand{\rro}{\right)}
+\newcommand{\lro}{\left( }
+\newcommand{\lsq}{\left[}
+\newcommand{\rsq}{\right]}
+
+\newcommand{\beq}{\begin{equation}}
+\newcommand{\eeq}{\end{equation}}
+
+\newcommand{\Pe}{\textrm{Pe}}
+\newcommand{\St}{\textrm{St}}
+\newcommand{\DaC}{\textrm{Da}_1}
+\newcommand{\DaD}{\textrm{Da}_2}
+
+\newcommand{\vth}{v_{th}}
+\newcommand{\taup}{\tau_p}
+\newcommand{\la}{\langle}
+\newcommand{\ra}{\rangle}
+
+\newcommand{\pbeta}{p_{\beta}}
+\newcommand{\df}{d_f}
+
+\pagestyle{plain}
+
+\begin{document}
+
+\title{Is the fragmentation kernel homogeneous?}
+
+
+%\author{LI}
+%\author{ADM}
+\author{YD}
+\email{[e-mail]}
+\affiliation{European Commission, Joint Research Centre,
+I-21027 Ispra (VA), Italy}
+%\author{Kostoglou?}
+%\author{Konstandopoulos?}
+
+\date{\today, HomogeneousKernel\_yd2.tex}
+
+\maketitle
+
+I have been thinking about the fragmentation kernel and the way we split it into two parts.
+In addition, or maybe triggered by my thoughts, I re-read (carefully this time)
+Ref.~\cite{Odriozola2002}. A summary of my conclusions follows.
+
+\begin{enumerate}
+\item We decided to express agglomerate sizes in terms of the largest size $y$ and one of
+the fragment sizes $x$ (the other being, for binary fragmentation, $y-x$). This choice contrasts with
+the more traditional, and more convenient for agglomeration, choice where the initial sizes
+of the fragments are specified, ($i,j$), that upon
+agglomeration give rise to a ($i+j$) agglomerate. The mapping is, of course,
+$i=x$ (without loss of generality) and $j=y-x$.
+\item We wrote the fragmentation kernel $K(x,y)$ in terms of the fragmentation rate $a(y)$
+of a particle of size $y$ and $b(x|y)$ the distribution of particles of size $x$ resulting from the breakup of a particle of size $y$ (i.e., the fragment size distribution), 
+\beq
+K(x,y) = a(y)\, b(x|y).
+\label{eq:kernel}
+\eeq
+We, also, argued $b(x|y)$ should satisfy a number of constrains. I think it is important to 
+enumerate them here.
+\begin{enumerate}
+\item Mass (monomer number) conservation: the sum of the monomers in the fragments must equal
+the number of monomers in the initial agglomerate. The algebraic manipulations arise
+from the change of variable from $x$ with $0 \leq x \leq y$ to $z=x/y$ with
+$0 \le z \le 1$.
+\beq
+\int_0^y dx \, x \, b(x|y) = y, \quad \Longrightarrow
+\quad y^2\, \int_0^1 \, dz \, z \, b(z,y) = y \quad
+\Longrightarrow \boxed{\int_0^1 dz \, z \, \tilde{b}(z) = 1 \quad \mbox{with}
+\quad  b(z,y) = y^{-1} \, \tilde{b}(z)}.
+\label{eq:MassConservation}
+\eeq
+\item The number of particles arising from the fragmentation of a single size $y$ particle,
+namely the expected number of particles, is (for binary fragmentation)
+\beq
+\int_0^y dx \, b(x|y) = 2, \quad \Longrightarrow \quad
+\int_0^1 \, dz \, y \, b(z,y) = 2 \quad
+\Longrightarrow \quad \boxed{\int_0^1 \, dz \, \tilde{b}(z) = 2 \quad
+\mbox{as before,}
+\quad  b(z,y) = y^{-1} \, \tilde{b}(z)}.
+\label{eq:NumberFragments}
+\eeq
+\end{enumerate} 
+\item We expressed the fragment size distribution as (I changed the notation slightly)
+\begin{eqnarray}
+\mathcolor{magenta}{b(x|y)} & = & \f{2}{B(\alpha,\alpha)} \, \f{1}{y} \,
+\Big (\f{x}{y} \Big )^{\alpha -1} \,
+\Big (1 - \f{x}{y} \Big )^{\alpha -1} 
+= \frac{2}{B(\alpha,\alpha)} \, y^{-1} \, \Big[ z \, (1-z) \Big ]^{\alpha-1} \\
+& \equiv & 2 \, y^{-1} \, \pbeta(\frac{x}{y};\alpha,\alpha)
+\equiv \mathcolor{magenta}{2 \, y^{-1} \, \pbeta(z;\alpha)}
+\quad \mbox{with} \quad z = \frac{x}{y} \ \epsilon \ [0,1].
+\label{eq:FragBeta}
+\end{eqnarray}
+where $\pbeta(z; \alpha,\beta)$ is the beta distribution
+\beq
+\pbeta(z;\alpha,\beta) = \frac{z^{\alpha -1}(1-z)^{\beta-1}}{B(\alpha, \beta)}
+\quad \mbox{with} \quad z \ \epsilon \ [0,1],
+\label{eq:betadistr}
+\eeq
+and $B(\alpha, \beta)$ is the beta function. Note that I drop the third index in $\pbeta$ if it equals
+the second, i.e. $\pbeta(z; \alpha) = \pbeta(z;\alpha,\alpha)$.
+The constraints are trivially satisfied if one realizes that for
+our fitting function
+\beq
+\tilde{b}(z) = 2 \pbeta(z; \alpha),
+\eeq
+and uses the following properties of the beta distribution
+\begin{eqnarray}
+\int_0^1 \, dz \, \pbeta (z;\alpha) & = & 1, \\
+\mu = E[X] = \int_0^1 \, dz \, z \, \pbeta (z;\alpha, \beta) & = & \frac{\alpha}{\alpha + \beta}.
+\end{eqnarray}
+\item Reference~\cite{Odriozola2002} makes two points:
+\begin{enumerate}
+\item They remark that the Brownian kernel for fractal-like agglomerates is (see, also, our
+one-dimensional agglomeration VELA paper)
+\beq
+K_{ij}^{Br} = K^{Br}(x,y) = K^{Br}(z=x/y) =
+\frac{2 k_B T}{3 \mu} \, \left [ z^{1/\df} + (1-z)^{1/\df} \right ] \,
+ \left [ z^{-1/\df} + (1-z)^{-1/\df} \right ] .\
+\label{eq:BrownianKernel}
+\eeq
+\item They argue that the fragmentation kernel $K_{ij}$, ``the mean rate constant at which
+($i+j$)-size clusters break spontaneously into two $i$- and $j$-size clusters", may be expressed as
+\beq
+K_{ij} = \frac{1}{\tau} \, e_{ij} \, (1+ \delta_{ij}),
+\label{eq:KernelBonds}
+\eeq
+where $\tau$ is the average bond lifetime, and ``$e_{ij}$ the number of bonds contained in $n$-size clusters which, on breakup, lead to $i$- and $j$-size
+fragments". I think that this implies that $e_{ij}$ is nothing else than the fragment
+size distribution-do you agree?
+
+\vspace*{2cm}
+
+\lorenzo{LI: I think this is the crucial part and this has
+  consequences on Eqs \eqref{eq:DefineKernel1} and
+  \eqref{eq:DefineKernel}. I have tried to write this several times,
+  and it never comes out very clear. I will focus only on the first
+  part of \eqref{eq:KernelBonds}, the one with $1/\tau$.
+  \begin{itemize}
+    \item I convinced myself we are looking at a Poisson process, or
+      waiting time or life and death etc...
+      We have to be super careful because when we think about decaying
+      atoms, we (or at least I) then transfer the reasoning to the
+      monomers, whereas the only decaying entities are the
+      bonds. True, there is (for our aggregates) a strict relation
+      between the aggregate size $y$ and the number of bonds $N_{b}$,
+      but conceptually they are two distinct entities. Also,
+      $N_{b}=N_{b}(t)$, seen that we are doing fragmentation, whereas
+      $y$ is conserved.
+
+      On top of that, we are dealing with an inhomogeneous Poisson
+      process. Read what happens next.
+    \item  For a single bond with
+      decay time $\tau$ if the bond is alive at $t=t'$, then
+      probability to find it alive at $t''=t'+s$ is $P(t''-t'=s)=\exp(-\lambda
+      s)$, where $\lambda=1/\tau$ and there is
+      nothing to say about $N_{b}$. We are dealing with a homogenous
+      Poisson process where $\lambda\neq \lambda(t)$. To make life
+      simple: if at $t=0$ all the $N_{b}$ bonds are intact, I can
+      expect to have $N_{b}(t)=N_{0}\exp(-\lambda t)$ bonds alive at
+      time $t$, where $N_{0}=N_{b}(t=0)$. 
+\item 
+      Our aggregate breaks as soon as any of its $N_{b}$ bonds
+      breaks. The question is not when I can expect on average a bond
+      to break, but how long I can observe the aggregate without
+      observing any of its bonds  break (a waiting/stopping time
+      problem, you are the expert here).  I think that at the beginning we have a
+      superposition of $N_{b}$ independent Poisson processes each one
+      with its own mean $\lambda$. I say at the beginning because then
+      the number bonds will decrease in time and so will the number of
+      Poisson processes I can superimpose. The result at the beginning is a Poisson process
+      with mean $\lambda_{agg}=N_{b}\lambda$ and the aggregate expectation
+      lifetime is $\tau_{agg}=\tau/N_{b}$. As time goes by, $N_{b}$
+      decreases as a result of the disintegration and so
+      $\lambda_{agg}=\lambda_{agg}(t)$. This is the definition of an
+      inhomogeneous Poisson process. So, if the aggregate is alive at
+      time $t=0$, it has a probability $P=\exp(-N_{b}\lambda s)$ to be
+      alive at time $t=s$. However, after each fragmentation, $N_{b}$
+      decreases and the expected time to see the next fragmentation
+      increases and it tends to $\lambda$ when there is a single bond
+      left in the aggregate.
+
+      \item The decreased life expectancy of an aggregate with respect
+        to its individual bonds makes sense because at time $t=s$
+      each bond has a probability $P=\exp(-\lambda s)$ of not having
+      disintegrated so the probability that all the bonds are alive
+      (no disintegration, aggregate still whole) at time $s$ is $P=(\exp(-\lambda
+      s))^{N_{b}}$ which is the same result as the superposition of
+      Poisson processes.
+ \item Bottom line: neglecting the discrete nature of the problem, we
+   are dealing with an inhomogeneous Poisson process with
+   $\lambda_{agg}=\lambda N_{0}\exp(-\lambda t)$. How to deal with
+   this...I am not sure yet! From what I read online, the point should
+   be that the number of bonds that break in $[t, t+s]$ should be
+   given by $Poisson(\int_{t}^{(t+s)}\lambda(\alpha)d\alpha)$.   
+\item If the reasoning above is correct, then the more I enlarge the
+  aggregate, the shorter its lifetime meant as the waiting time before
+  any of its bonds breaks leading to a fragmentation event. I think in
+  my usual obscure way I mentioned it already in the past.
+   
+  \item Bottom line 2: $a=a(N_{b})$ and certainly it is too
+    simple to say that $a=1/\tau$, unless there is more in
+    $e_{ij}$, but I do not think that is the case. Back to your question: $e_{ij}$ appears to be linked if
+    not identical to the size distribution, seen that it counts how
+    many configurations of sizes $\{i,j\}$ I can generate from breaking
+    the aggregate.
+\item Right now, I am not convinced about the $1/\tau$ business, but I
+  need some more thinking. However, this is ``just'' the kinetic of
+  the process, something we have no idea about (what is $\tau$ in the
+  first place), but I think they are not getting it right, unless...
+\item they are in a sort of steady state. Think about the measurement
+  of radioactive decays on sample in a physics lab. You simply observe
+  on average e.g. 5 decays per second because you do not observe the
+  sample long enough to see it convert entirely into something non
+  radioactive. The number of radioactive atoms remains approximately
+  constant while you look at it. If we think about the fact that each
+  fragmentation even just decreases $N_{b}$ by 1, then perhaps it is
+  not meaningless to say that there is an approximately constant
+  $\tau$ for the aggregate, as long as I do not study it to its bitter
+  end of the disintegration into individual monomers.
+\end{itemize}
+}
+
+\vspace*{2cm}
+
+\item Based on the functional form of the Brownian agglomeration kernel
+Eq.~(\ref{eq:BrownianKernel}) they used their synthetic agglomerates to propose
+\begin{eqnarray}
+\lefteqn{\left.  e_{ij} \, (1+ \delta_{ij}) \right |_{\textrm{fit}} =
+p_1 \Big ( i^{p_2} + j^{p_2} \Big ) \,  \Big ( i^{p_3} + j^{p_3} \Big )
+\, \Big ( i j \Big )^{p_4},} \\
+& & p_1 = 0.439 \  ; \quad p_2 = 1.006 \ ; \quad p_3 = -1.007 \ ; \quad p_4 = -0.1363
+\label{eq:FitKernel}
+\end{eqnarray}
+where the numerical expressions for $p_i$ are the results of their fitting.
+
+\end{enumerate}
+
+\end{enumerate}
+
+What I suggest is to use this information to improve, possibly, our own estimate of the kernel.
+Specifically,
+\begin{enumerate}
+\item Accept their kernel decomposition Eq.~(\ref{eq:KernelBonds}) so that
+\begin{eqnarray}\label{eq:DefineKernel1}
+  a(y) & = & \f{1}{\tau}, \\
+\label{eq:DefineKernel}   
+b(x|y) & = & K_{ij} \, \tau = e_{ij} \, (1 + \delta_{ij} ) \equiv h_f(x|y),
+\end{eqnarray}
+namely, the fragmentation rate is independent of agglomerate size. Moreover, their proposed
+fragment size distribution expressed in terms of our choice of variable (I will call it $h_f(x|y)$,
+see Eq. ~(\ref{eq:DefineKernel}))
+becomes
+\beq
+h_f (x|y) = \mathcolor{red}{p_1 \, B(1+p_4, 1+p_4)} \, 
+\mathcolor{blue}{y^{p_2 + p_3 + 2p_4}} 
+\, \mathcolor{magenta}{\pbeta(z; 1+p_4)} \,
+\underbrace{\Big [z^{p_2} + (1-z)^{p_2} \Big ] \, \Big [z^{p_3} + (1-z)^{p_3}
+\Big ]}_{\textrm{additional terms}},
+\label{eq:hf}
+\eeq
+with $z=x/y$. This form should be compared and contrasted to ours, reported again for easy comparison
+\beq
+b(x|y) = \mathcolor{red}{2} \,
+\mathcolor{blue}{y^{-1}} \, \mathcolor{magenta}{\pbeta(z;\alpha)}
+\quad \mbox{with} \quad z \ \epsilon \ [0,1].
+\label{eq:Ours}
+\eeq
+\item You might find interesting the values of the fitted numerical constants
+\begin{eqnarray}
+p_1 \ B(1+p_4, 1+p_4) & = & 0.581  \quad \quad \mbox{Our fit} \quad 2 \\
+p_2 + p_3 + 2p_4 & = & -0.2736 \quad \quad \mbox{Our fit} \quad -1 \\
+1+ p_4 & = & 0.8637 \quad \quad \mbox{Our fit} \quad  \sim \frac{1}{2} \quad
+\mbox{decreasing with increasing N} .
+\end{eqnarray}
+\end{enumerate}
+
+In short: What am I suggesting?
+\begin{enumerate}
+\item The similarity between Eq.~(\ref{eq:hf}) and (\ref{eq:Ours}) is I think apparent
+(I chose colors to identify similar factors).
+\item Would it be possible to use Eqs.~(\ref{eq:hf}) to fit our fragment size distribution?
+I know that if more parameters are introduced it is easier to fit the empirical distribution.
+However, we might argue, as they did, that there is a reason for our choice of the fitting function.
+It might be worth a try if it does not involve a lot of work.
+
+\begin{enumerate}
+\item I constructed a two-parameter function that emulates the Brownian agglomeration kernel
+and satisfies the constraints mentioned earlier. My suggestion:
+\begin{subequations}
+\beq
+\label{eq:Fit1}
+\mathcolor{red}{
+\boxed{
+b(x|y) = \alpha_1 \, y^{-1} \, \Big [z^{\alpha_2} + (1-z)^{\alpha_2} \Big ] \,
+\Big [z^{\alpha_3} + (1-z)^{\alpha_3} \Big ] \,
+\Big [z \, (1-z) \Big ]^{\alpha_4} } } ,
+\eeq
+\beq 
+\mathcolor{blue}{
+\boxed{
+\mbox{with} \quad \alpha_4 = - \frac{1}{2} \, \Big (1 + \alpha_2 + \alpha_3 \Big ) } } ,
+\label{eq:Fit1Constraint1}
+\eeq
+\beq
+\mathcolor{blue}{
+\boxed{
+\mbox{and} \quad \alpha_1 = \Big [ \Gamma(1+\alpha_2+\alpha_4) \, \Gamma(1+\alpha_3 + \alpha_4) \,
++ \Gamma(1+\alpha_4) \, \Gamma(1+\alpha_2+\alpha_3+\alpha_4) \, \Big ]^{-1} } }.
+\label{eq:Fit1Constraint2}
+\eeq
+\end{subequations}
+I explicitly imposed the constraint $\alpha_2 + \alpha_3 + 2 \alpha_4 = -1$ (it arises from the
+requirement that $b(z,y) = y^{-1} \tilde{b}(z)$, see
+Eqs.~(\ref{eq:MassConservation} and \ref{eq:NumberFragments})), and I evaluated
+(thanks to Mathematica) the normalization factor $\alpha_1$ such that the
+two constraints are satisfied.
+
+Then, the proposed fitting function Eq.~(\ref{eq:Fit1}) has only two independent variables: $\alpha_2$
+and $\alpha_3$. If that seems difficult there is another option: see point 4.
+
+\item A parenthetical remark: if we take $\alpha_2 = \alpha_3 =0$ we should get something similar
+to our current fitting function,
+\beq
+b(x|y) = y^{-1} \, \tilde{b}(z) =
+\frac{2}{B(\alpha, \alpha)} \, y^{-1} \Big [ z \, (1-z) \Big ]^{\alpha -1}, 
+\eeq
+with $\alpha$ the fitting parameter. For $\alpha_2 = \alpha_3 =0$ Eq.~(\ref{eq:Fit1}) becomes
+\beq
+b(x|y) = \frac{2}{B( 1/2, 1/2)} \, y^{-1} \, \Big [ z \, (1-z) \Big ]^{-1/2} =
+2 y^{-1} \, \pbeta(z; \frac{1}{2}).
+\eeq
+Surprise, our parameter $\alpha$ is fixed at $\alpha = 1/2$ (and it is independent of $y$).
+
+\item Alternatively, one can use the function Reference~\cite{Odriozola2002}
+proposes,
+\beq
+b (x|y) = \alpha_1 \, y^{\alpha_2 + \alpha_3 + 2 \alpha_4}
+\Big [z^{\alpha_2} + (1-z)^{\alpha_2} \Big ] \, \Big [z^{\alpha_3} + (1-z)^{\alpha_3}
+\Big ] \, \Big [z \, (1-z) \Big ]^{\alpha_4}.
+\label{eq:Fit2}
+\eeq
+Here, the fitting parameters are three exponents and one normalization constant. The greater freedom
+in fitting the fragment size distribution (four parameter) comes at the expense that there is
+no guarantee that the numerical constants will be such that the constraints
+hold (even approximately). As
+I've been trying to argue, the constants of Ref.~\cite{Odriozola2002}) do not.
+\end{enumerate}
+\item There is one more reason to try to check their calculation. If we take the similarity of the
+(random, binary) fragmentation kernel with the Brownian agglomeration kernel as valid then
+we might expect
+\beq
+p_2 = - p_3 = 1/\df \quad \Longrightarrow p_2 = 1.006 \sim - p_3 \quad d_f \sim 0.99.
+\eeq
+In short, their numbers do not seem to be related to the fractal dimension of the fragments.
+In addition, they only performed their calculations for DLCA agglomerates.
+\item Latest addition: it just occurred to me that if we are really coragious we can just set
+\beq
+\alpha_2 = -\alpha_3 = 1/\df, \quad \Longrightarrow \quad
+\alpha_4 = -\frac{1}{2}, \quad \alpha_1 = \Gamma(\frac{1}{2}) \, \Gamma(\frac{1}{2}) \,
+\Big [ B(\frac{1}{2} + \frac{1}{\df}, \frac{1}{2} - \frac{1}{\df}) + 1 \Big ]
+\eeq
+where $\df$ is the fractal dimension of the fragments. There is nothing to fit. Will it work?
+\item Lastly, it seems apparent that $h_f(z,y)$, defined in Eq.~(\ref{eq:hf}), does not
+necessarily satisfy
+the two constraints (ours, by construction and simplicity, does). 
+Specifically, the integral constraints become the algebraic constraints reported in
+Eqs.~(\ref{eq:Fit1Constraint1} and \ref{eq:Fit1Constraint2}). The $p_i$ reported in
+Ref.~\cite{Odriozola2002} give
+\begin{subequations}
+\beq
+p_4 = - \frac{1}{2} \, (1 + p_2 + p_3) = -0.4995 ,
+\eeq
+whereas the numerically determined value is $p_4 = -0.1363$ (first constraint), and
+\beq
+p_1 = \Big [ \Gamma(1+p_2+p_4) \, \Gamma(1+p_3 + p_4) \,
++ \Gamma(1+p_4) \, \Gamma(1+p_2+p_3+p_4) \, \Big ]^{-1} = -0.1630,
+\eeq
+whereas the numerically determined value is $p_1 = 0.439$ (second constraint).
+\end{subequations}
+ 
+\end{enumerate}
+
+What triggered these investigations? Lorenzo's remark that the way we break up the
+agglomerates, constant bond lifetime, might lead us to a process similar to radioactive
+decay, and thence to a Poisson process. I have to admit, I've thought about it in the past,
+but I did not reach any valuable conclusions. Maybe now I will be luckier. More in the future.
+
+\section{Addendum}
+
+I just realized that:
+\begin{enumerate}
+\item To say that $b(x|y)$ is a homogeneous function [which in our case implies
+$b(x|y) = y^{-1} b(x/y)$] is not the same as stating that $b(x|y)$ is independent of
+the initial particle size $y$. The functional form shows explicitly that the fragment
+size distribution depends on the largest particle size as $y^{-1}$.
+\item What I am implying is that our data show a dependence on $y$ but it is not the
+trivial (expected?) $1/y$. Do you agree?
+\item The trial (fitting) function may be modified to force it to remain a homogeneous
+function and to obey a conservation law by dividing by $i+j = y$. In that case, the straight-chain
+limit may be imposed, but no combination of exponents would give our beta distribution.
+\item Our fitting function, the beta distribution, suggest that we've taken the
+kernel to be proportional to $K_{ij} \sim (i j)^{\alpha -1}$. The proportionality constant
+is important.
+\item Remember this is thinking and writing, as if we were talking to each other in front
+of a blackboard. Nothing conclusive, but I'm starting to appreciate the various subtleties.
+\end{enumerate}
+
+\begin{thebibliography}{99}
+\bibitem{Odriozola2002} G. Odriozola, A. Schmitt, A. Moncho-Jord\'{a}, J. Callejas-Fern\'{a}ndez,
+R. Mart\'{i}nez-Garc\'{i}a, R. Leone, and R. Hidalgo-\'{A}lvarez, ``Constant bond breakup
+probability model for reversible aggregation processes", Phys. Rev. E 65, 031405 (2002).
+
+%\bibitem{McGradyZiff1987} E.D. McGrady and R.M. Ziff, `` ``Shattering" transition in fragmentation",
+%Phys. Rev. lett. 58, 892 (1987)
+%
+%\bibitem{Kostoglou1997} M. Kostoglou, S. Dovas, and A.J. Karabelas, ``On the steady-state size distribution of dispersion in breakage processes", Chem. Engng. Sci. \textbf{52}, 1285 (1997). 
+%
+%\bibitem{Redner} S. Redner, ``Statistical theory of fragmentation", Chapter 3 in \textit{``Disorder and fracture"}, eds. J.C. Charment, S. Roux, and E. Guyon, Plenum Press, New York (1990).
+%
+%\bibitem{Classify} P.G.J. van Dongen ad M.H. Ernst, ``Scaling solutions of Smoluchowski's coagulation
+%equation", J. Stat. Phys. 50, 295 (1988).
+%
+%\bibitem{Ushape} S. Ito and S. Yukawa, ``Stochastic modeling on fragmentation process over
+%lifetime and its dynamical scaling law of fragments distribution", J. Phys. Japan 83, 124005 (2014).
+%
+%\bibitem{FragTrees} Z. Kalay and E. Ben-Naim, ``Fragmentation of random tress", J. Phys. A: Math. Gen. \textbf{48}, 045001 (2015).
+%
+%%\bibitem{FragRandom} E. Metin El\c{c}i, M. Weigel, and N.G. Fytas, ``Fragmentation of fractal random %structures", Phys. Rev. lett. \textbf{114}, 115701 (2015).
+
+\end{thebibliography}
+
+\end{document}
+%%% Local Variables:
+%%% mode: latex
+%%% TeX-master: t
+%%% End: