Revisão | 0b7a88778adcf8906d0e1031b543f9f1e33b5924 (tree) |
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Hora | 2017-11-08 00:30:09 |
Autor | Lorenzo Isella <lorenzo.isella@gmai...> |
Commiter | Lorenzo Isella |
I added a file which contains a useful function to add comments to a tex file.
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45 | + | |
46 | +\pagestyle{plain} | |
47 | + | |
48 | +\begin{document} | |
49 | + | |
50 | +\title{Is the fragmentation kernel homogeneous?} | |
51 | + | |
52 | + | |
53 | +%\author{LI} | |
54 | +%\author{ADM} | |
55 | +\author{YD} | |
56 | +\email{yannis.drossinos@ec.europa.eu} | |
57 | +\affiliation{European Commission, Joint Research Centre, | |
58 | +I-21027 Ispra (VA), Italy} | |
59 | +%\author{Kostoglou?} | |
60 | +%\author{Konstandopoulos?} | |
61 | + | |
62 | +\date{\today, HomogeneousKernel\_yd2.tex} | |
63 | + | |
64 | +\maketitle | |
65 | + | |
66 | +I have been thinking about the fragmentation kernel and the way we split it into two parts. | |
67 | +In addition, or maybe triggered by my thoughts, I re-read (carefully this time) | |
68 | +Ref.~\cite{Odriozola2002}. A summary of my conclusions follows. | |
69 | + | |
70 | +\begin{enumerate} | |
71 | +\item We decided to express agglomerate sizes in terms of the largest size $y$ and one of | |
72 | +the fragment sizes $x$ (the other being, for binary fragmentation, $y-x$). This choice contrasts with | |
73 | +the more traditional, and more convenient for agglomeration, choice where the initial sizes | |
74 | +of the fragments are specified, ($i,j$), that upon | |
75 | +agglomeration give rise to a ($i+j$) agglomerate. The mapping is, of course, | |
76 | +$i=x$ (without loss of generality) and $j=y-x$. | |
77 | +\item We wrote the fragmentation kernel $K(x,y)$ in terms of the fragmentation rate $a(y)$ | |
78 | +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), | |
79 | +\beq | |
80 | +K(x,y) = a(y)\, b(x|y). | |
81 | +\label{eq:kernel} | |
82 | +\eeq | |
83 | +We, also, argued $b(x|y)$ should satisfy a number of constrains. I think it is important to | |
84 | +enumerate them here. | |
85 | +\begin{enumerate} | |
86 | +\item Mass (monomer number) conservation: the sum of the monomers in the fragments must equal | |
87 | +the number of monomers in the initial agglomerate. The algebraic manipulations arise | |
88 | +from the change of variable from $x$ with $0 \leq x \leq y$ to $z=x/y$ with | |
89 | +$0 \le z \le 1$. | |
90 | +\beq | |
91 | +\int_0^y dx \, x \, b(x|y) = y, \quad \Longrightarrow | |
92 | +\quad y^2\, \int_0^1 \, dz \, z \, b(z,y) = y \quad | |
93 | +\Longrightarrow \boxed{\int_0^1 dz \, z \, \tilde{b}(z) = 1 \quad \mbox{with} | |
94 | +\quad b(z,y) = y^{-1} \, \tilde{b}(z)}. | |
95 | +\label{eq:MassConservation} | |
96 | +\eeq | |
97 | +\item The number of particles arising from the fragmentation of a single size $y$ particle, | |
98 | +namely the expected number of particles, is (for binary fragmentation) | |
99 | +\beq | |
100 | +\int_0^y dx \, b(x|y) = 2, \quad \Longrightarrow \quad | |
101 | +\int_0^1 \, dz \, y \, b(z,y) = 2 \quad | |
102 | +\Longrightarrow \quad \boxed{\int_0^1 \, dz \, \tilde{b}(z) = 2 \quad | |
103 | +\mbox{as before,} | |
104 | +\quad b(z,y) = y^{-1} \, \tilde{b}(z)}. | |
105 | +\label{eq:NumberFragments} | |
106 | +\eeq | |
107 | +\end{enumerate} | |
108 | +\item We expressed the fragment size distribution as (I changed the notation slightly) | |
109 | +\begin{eqnarray} | |
110 | +\mathcolor{magenta}{b(x|y)} & = & \f{2}{B(\alpha,\alpha)} \, \f{1}{y} \, | |
111 | +\Big (\f{x}{y} \Big )^{\alpha -1} \, | |
112 | +\Big (1 - \f{x}{y} \Big )^{\alpha -1} | |
113 | += \frac{2}{B(\alpha,\alpha)} \, y^{-1} \, \Big[ z \, (1-z) \Big ]^{\alpha-1} \\ | |
114 | +& \equiv & 2 \, y^{-1} \, \pbeta(\frac{x}{y};\alpha,\alpha) | |
115 | +\equiv \mathcolor{magenta}{2 \, y^{-1} \, \pbeta(z;\alpha)} | |
116 | +\quad \mbox{with} \quad z = \frac{x}{y} \ \epsilon \ [0,1]. | |
117 | +\label{eq:FragBeta} | |
118 | +\end{eqnarray} | |
119 | +where $\pbeta(z; \alpha,\beta)$ is the beta distribution | |
120 | +\beq | |
121 | +\pbeta(z;\alpha,\beta) = \frac{z^{\alpha -1}(1-z)^{\beta-1}}{B(\alpha, \beta)} | |
122 | +\quad \mbox{with} \quad z \ \epsilon \ [0,1], | |
123 | +\label{eq:betadistr} | |
124 | +\eeq | |
125 | +and $B(\alpha, \beta)$ is the beta function. Note that I drop the third index in $\pbeta$ if it equals | |
126 | +the second, i.e. $\pbeta(z; \alpha) = \pbeta(z;\alpha,\alpha)$. | |
127 | +The constraints are trivially satisfied if one realizes that for | |
128 | +our fitting function | |
129 | +\beq | |
130 | +\tilde{b}(z) = 2 \pbeta(z; \alpha), | |
131 | +\eeq | |
132 | +and uses the following properties of the beta distribution | |
133 | +\begin{eqnarray} | |
134 | +\int_0^1 \, dz \, \pbeta (z;\alpha) & = & 1, \\ | |
135 | +\mu = E[X] = \int_0^1 \, dz \, z \, \pbeta (z;\alpha, \beta) & = & \frac{\alpha}{\alpha + \beta}. | |
136 | +\end{eqnarray} | |
137 | +\item Reference~\cite{Odriozola2002} makes two points: | |
138 | +\begin{enumerate} | |
139 | +\item They remark that the Brownian kernel for fractal-like agglomerates is (see, also, our | |
140 | +one-dimensional agglomeration VELA paper) | |
141 | +\beq | |
142 | +K_{ij}^{Br} = K^{Br}(x,y) = K^{Br}(z=x/y) = | |
143 | +\frac{2 k_B T}{3 \mu} \, \left [ z^{1/\df} + (1-z)^{1/\df} \right ] \, | |
144 | + \left [ z^{-1/\df} + (1-z)^{-1/\df} \right ] .\ | |
145 | +\label{eq:BrownianKernel} | |
146 | +\eeq | |
147 | +\item They argue that the fragmentation kernel $K_{ij}$, ``the mean rate constant at which | |
148 | +($i+j$)-size clusters break spontaneously into two $i$- and $j$-size clusters", may be expressed as | |
149 | +\beq | |
150 | +K_{ij} = \frac{1}{\tau} \, e_{ij} \, (1+ \delta_{ij}), | |
151 | +\label{eq:KernelBonds} | |
152 | +\eeq | |
153 | +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 | |
154 | +fragments". I think that this implies that $e_{ij}$ is nothing else than the fragment | |
155 | +size distribution-do you agree? | |
156 | + | |
157 | +\vspace*{2cm} | |
158 | + | |
159 | +\lorenzo{LI: I think this is the crucial part and this has | |
160 | + consequences on Eqs \eqref{eq:DefineKernel1} and | |
161 | + \eqref{eq:DefineKernel}. I have tried to write this several times, | |
162 | + and it never comes out very clear. I will focus only on the first | |
163 | + part of \eqref{eq:KernelBonds}, the one with $1/\tau$. | |
164 | + \begin{itemize} | |
165 | + \item I convinced myself we are looking at a Poisson process, or | |
166 | + waiting time or life and death etc... | |
167 | + We have to be super careful because when we think about decaying | |
168 | + atoms, we (or at least I) then transfer the reasoning to the | |
169 | + monomers, whereas the only decaying entities are the | |
170 | + bonds. True, there is (for our aggregates) a strict relation | |
171 | + between the aggregate size $y$ and the number of bonds $N_{b}$, | |
172 | + but conceptually they are two distinct entities. Also, | |
173 | + $N_{b}=N_{b}(t)$, seen that we are doing fragmentation, whereas | |
174 | + $y$ is conserved. | |
175 | + | |
176 | + On top of that, we are dealing with an inhomogeneous Poisson | |
177 | + process. Read what happens next. | |
178 | + \item For a single bond with | |
179 | + decay time $\tau$ if the bond is alive at $t=t'$, then | |
180 | + probability to find it alive at $t''=t'+s$ is $P(t''-t'=s)=\exp(-\lambda | |
181 | + s)$, where $\lambda=1/\tau$ and there is | |
182 | + nothing to say about $N_{b}$. We are dealing with a homogenous | |
183 | + Poisson process where $\lambda\neq \lambda(t)$. To make life | |
184 | + simple: if at $t=0$ all the $N_{b}$ bonds are intact, I can | |
185 | + expect to have $N_{b}(t)=N_{0}\exp(-\lambda t)$ bonds alive at | |
186 | + time $t$, where $N_{0}=N_{b}(t=0)$. | |
187 | +\item | |
188 | + Our aggregate breaks as soon as any of its $N_{b}$ bonds | |
189 | + breaks. The question is not when I can expect on average a bond | |
190 | + to break, but how long I can observe the aggregate without | |
191 | + observing any of its bonds break (a waiting/stopping time | |
192 | + problem, you are the expert here). I think that at the beginning we have a | |
193 | + superposition of $N_{b}$ independent Poisson processes each one | |
194 | + with its own mean $\lambda$. I say at the beginning because then | |
195 | + the number bonds will decrease in time and so will the number of | |
196 | + Poisson processes I can superimpose. The result at the beginning is a Poisson process | |
197 | + with mean $\lambda_{agg}=N_{b}\lambda$ and the aggregate expectation | |
198 | + lifetime is $\tau_{agg}=\tau/N_{b}$. As time goes by, $N_{b}$ | |
199 | + decreases as a result of the disintegration and so | |
200 | + $\lambda_{agg}=\lambda_{agg}(t)$. This is the definition of an | |
201 | + inhomogeneous Poisson process. So, if the aggregate is alive at | |
202 | + time $t=0$, it has a probability $P=\exp(-N_{b}\lambda s)$ to be | |
203 | + alive at time $t=s$. However, after each fragmentation, $N_{b}$ | |
204 | + decreases and the expected time to see the next fragmentation | |
205 | + increases and it tends to $\lambda$ when there is a single bond | |
206 | + left in the aggregate. | |
207 | + | |
208 | + \item The decreased life expectancy of an aggregate with respect | |
209 | + to its individual bonds makes sense because at time $t=s$ | |
210 | + each bond has a probability $P=\exp(-\lambda s)$ of not having | |
211 | + disintegrated so the probability that all the bonds are alive | |
212 | + (no disintegration, aggregate still whole) at time $s$ is $P=(\exp(-\lambda | |
213 | + s))^{N_{b}}$ which is the same result as the superposition of | |
214 | + Poisson processes. | |
215 | + \item Bottom line: neglecting the discrete nature of the problem, we | |
216 | + are dealing with an inhomogeneous Poisson process with | |
217 | + $\lambda_{agg}=\lambda N_{0}\exp(-\lambda t)$. How to deal with | |
218 | + this...I am not sure yet! From what I read online, the point should | |
219 | + be that the number of bonds that break in $[t, t+s]$ should be | |
220 | + given by $Poisson(\int_{t}^{(t+s)}\lambda(\alpha)d\alpha)$. | |
221 | +\item If the reasoning above is correct, then the more I enlarge the | |
222 | + aggregate, the shorter its lifetime meant as the waiting time before | |
223 | + any of its bonds breaks leading to a fragmentation event. I think in | |
224 | + my usual obscure way I mentioned it already in the past. | |
225 | + | |
226 | + \item Bottom line 2: $a=a(N_{b})$ and certainly it is too | |
227 | + simple to say that $a=1/\tau$, unless there is more in | |
228 | + $e_{ij}$, but I do not think that is the case. Back to your question: $e_{ij}$ appears to be linked if | |
229 | + not identical to the size distribution, seen that it counts how | |
230 | + many configurations of sizes $\{i,j\}$ I can generate from breaking | |
231 | + the aggregate. | |
232 | +\item Right now, I am not convinced about the $1/\tau$ business, but I | |
233 | + need some more thinking. However, this is ``just'' the kinetic of | |
234 | + the process, something we have no idea about (what is $\tau$ in the | |
235 | + first place), but I think they are not getting it right, unless... | |
236 | +\item they are in a sort of steady state. Think about the measurement | |
237 | + of radioactive decays on sample in a physics lab. You simply observe | |
238 | + on average e.g. 5 decays per second because you do not observe the | |
239 | + sample long enough to see it convert entirely into something non | |
240 | + radioactive. The number of radioactive atoms remains approximately | |
241 | + constant while you look at it. If we think about the fact that each | |
242 | + fragmentation even just decreases $N_{b}$ by 1, then perhaps it is | |
243 | + not meaningless to say that there is an approximately constant | |
244 | + $\tau$ for the aggregate, as long as I do not study it to its bitter | |
245 | + end of the disintegration into individual monomers. | |
246 | +\end{itemize} | |
247 | +} | |
248 | + | |
249 | +\vspace*{2cm} | |
250 | + | |
251 | +\item Based on the functional form of the Brownian agglomeration kernel | |
252 | +Eq.~(\ref{eq:BrownianKernel}) they used their synthetic agglomerates to propose | |
253 | +\begin{eqnarray} | |
254 | +\lefteqn{\left. e_{ij} \, (1+ \delta_{ij}) \right |_{\textrm{fit}} = | |
255 | +p_1 \Big ( i^{p_2} + j^{p_2} \Big ) \, \Big ( i^{p_3} + j^{p_3} \Big ) | |
256 | +\, \Big ( i j \Big )^{p_4},} \\ | |
257 | +& & p_1 = 0.439 \ ; \quad p_2 = 1.006 \ ; \quad p_3 = -1.007 \ ; \quad p_4 = -0.1363 | |
258 | +\label{eq:FitKernel} | |
259 | +\end{eqnarray} | |
260 | +where the numerical expressions for $p_i$ are the results of their fitting. | |
261 | + | |
262 | +\end{enumerate} | |
263 | + | |
264 | +\end{enumerate} | |
265 | + | |
266 | +What I suggest is to use this information to improve, possibly, our own estimate of the kernel. | |
267 | +Specifically, | |
268 | +\begin{enumerate} | |
269 | +\item Accept their kernel decomposition Eq.~(\ref{eq:KernelBonds}) so that | |
270 | +\begin{eqnarray}\label{eq:DefineKernel1} | |
271 | + a(y) & = & \f{1}{\tau}, \\ | |
272 | +\label{eq:DefineKernel} | |
273 | +b(x|y) & = & K_{ij} \, \tau = e_{ij} \, (1 + \delta_{ij} ) \equiv h_f(x|y), | |
274 | +\end{eqnarray} | |
275 | +namely, the fragmentation rate is independent of agglomerate size. Moreover, their proposed | |
276 | +fragment size distribution expressed in terms of our choice of variable (I will call it $h_f(x|y)$, | |
277 | +see Eq. ~(\ref{eq:DefineKernel})) | |
278 | +becomes | |
279 | +\beq | |
280 | +h_f (x|y) = \mathcolor{red}{p_1 \, B(1+p_4, 1+p_4)} \, | |
281 | +\mathcolor{blue}{y^{p_2 + p_3 + 2p_4}} | |
282 | +\, \mathcolor{magenta}{\pbeta(z; 1+p_4)} \, | |
283 | +\underbrace{\Big [z^{p_2} + (1-z)^{p_2} \Big ] \, \Big [z^{p_3} + (1-z)^{p_3} | |
284 | +\Big ]}_{\textrm{additional terms}}, | |
285 | +\label{eq:hf} | |
286 | +\eeq | |
287 | +with $z=x/y$. This form should be compared and contrasted to ours, reported again for easy comparison | |
288 | +\beq | |
289 | +b(x|y) = \mathcolor{red}{2} \, | |
290 | +\mathcolor{blue}{y^{-1}} \, \mathcolor{magenta}{\pbeta(z;\alpha)} | |
291 | +\quad \mbox{with} \quad z \ \epsilon \ [0,1]. | |
292 | +\label{eq:Ours} | |
293 | +\eeq | |
294 | +\item You might find interesting the values of the fitted numerical constants | |
295 | +\begin{eqnarray} | |
296 | +p_1 \ B(1+p_4, 1+p_4) & = & 0.581 \quad \quad \mbox{Our fit} \quad 2 \\ | |
297 | +p_2 + p_3 + 2p_4 & = & -0.2736 \quad \quad \mbox{Our fit} \quad -1 \\ | |
298 | +1+ p_4 & = & 0.8637 \quad \quad \mbox{Our fit} \quad \sim \frac{1}{2} \quad | |
299 | +\mbox{decreasing with increasing N} . | |
300 | +\end{eqnarray} | |
301 | +\end{enumerate} | |
302 | + | |
303 | +In short: What am I suggesting? | |
304 | +\begin{enumerate} | |
305 | +\item The similarity between Eq.~(\ref{eq:hf}) and (\ref{eq:Ours}) is I think apparent | |
306 | +(I chose colors to identify similar factors). | |
307 | +\item Would it be possible to use Eqs.~(\ref{eq:hf}) to fit our fragment size distribution? | |
308 | +I know that if more parameters are introduced it is easier to fit the empirical distribution. | |
309 | +However, we might argue, as they did, that there is a reason for our choice of the fitting function. | |
310 | +It might be worth a try if it does not involve a lot of work. | |
311 | + | |
312 | +\begin{enumerate} | |
313 | +\item I constructed a two-parameter function that emulates the Brownian agglomeration kernel | |
314 | +and satisfies the constraints mentioned earlier. My suggestion: | |
315 | +\begin{subequations} | |
316 | +\beq | |
317 | +\label{eq:Fit1} | |
318 | +\mathcolor{red}{ | |
319 | +\boxed{ | |
320 | +b(x|y) = \alpha_1 \, y^{-1} \, \Big [z^{\alpha_2} + (1-z)^{\alpha_2} \Big ] \, | |
321 | +\Big [z^{\alpha_3} + (1-z)^{\alpha_3} \Big ] \, | |
322 | +\Big [z \, (1-z) \Big ]^{\alpha_4} } } , | |
323 | +\eeq | |
324 | +\beq | |
325 | +\mathcolor{blue}{ | |
326 | +\boxed{ | |
327 | +\mbox{with} \quad \alpha_4 = - \frac{1}{2} \, \Big (1 + \alpha_2 + \alpha_3 \Big ) } } , | |
328 | +\label{eq:Fit1Constraint1} | |
329 | +\eeq | |
330 | +\beq | |
331 | +\mathcolor{blue}{ | |
332 | +\boxed{ | |
333 | +\mbox{and} \quad \alpha_1 = \Big [ \Gamma(1+\alpha_2+\alpha_4) \, \Gamma(1+\alpha_3 + \alpha_4) \, | |
334 | ++ \Gamma(1+\alpha_4) \, \Gamma(1+\alpha_2+\alpha_3+\alpha_4) \, \Big ]^{-1} } }. | |
335 | +\label{eq:Fit1Constraint2} | |
336 | +\eeq | |
337 | +\end{subequations} | |
338 | +I explicitly imposed the constraint $\alpha_2 + \alpha_3 + 2 \alpha_4 = -1$ (it arises from the | |
339 | +requirement that $b(z,y) = y^{-1} \tilde{b}(z)$, see | |
340 | +Eqs.~(\ref{eq:MassConservation} and \ref{eq:NumberFragments})), and I evaluated | |
341 | +(thanks to Mathematica) the normalization factor $\alpha_1$ such that the | |
342 | +two constraints are satisfied. | |
343 | + | |
344 | +Then, the proposed fitting function Eq.~(\ref{eq:Fit1}) has only two independent variables: $\alpha_2$ | |
345 | +and $\alpha_3$. If that seems difficult there is another option: see point 4. | |
346 | + | |
347 | +\item A parenthetical remark: if we take $\alpha_2 = \alpha_3 =0$ we should get something similar | |
348 | +to our current fitting function, | |
349 | +\beq | |
350 | +b(x|y) = y^{-1} \, \tilde{b}(z) = | |
351 | +\frac{2}{B(\alpha, \alpha)} \, y^{-1} \Big [ z \, (1-z) \Big ]^{\alpha -1}, | |
352 | +\eeq | |
353 | +with $\alpha$ the fitting parameter. For $\alpha_2 = \alpha_3 =0$ Eq.~(\ref{eq:Fit1}) becomes | |
354 | +\beq | |
355 | +b(x|y) = \frac{2}{B( 1/2, 1/2)} \, y^{-1} \, \Big [ z \, (1-z) \Big ]^{-1/2} = | |
356 | +2 y^{-1} \, \pbeta(z; \frac{1}{2}). | |
357 | +\eeq | |
358 | +Surprise, our parameter $\alpha$ is fixed at $\alpha = 1/2$ (and it is independent of $y$). | |
359 | + | |
360 | +\item Alternatively, one can use the function Reference~\cite{Odriozola2002} | |
361 | +proposes, | |
362 | +\beq | |
363 | +b (x|y) = \alpha_1 \, y^{\alpha_2 + \alpha_3 + 2 \alpha_4} | |
364 | +\Big [z^{\alpha_2} + (1-z)^{\alpha_2} \Big ] \, \Big [z^{\alpha_3} + (1-z)^{\alpha_3} | |
365 | +\Big ] \, \Big [z \, (1-z) \Big ]^{\alpha_4}. | |
366 | +\label{eq:Fit2} | |
367 | +\eeq | |
368 | +Here, the fitting parameters are three exponents and one normalization constant. The greater freedom | |
369 | +in fitting the fragment size distribution (four parameter) comes at the expense that there is | |
370 | +no guarantee that the numerical constants will be such that the constraints | |
371 | +hold (even approximately). As | |
372 | +I've been trying to argue, the constants of Ref.~\cite{Odriozola2002}) do not. | |
373 | +\end{enumerate} | |
374 | +\item There is one more reason to try to check their calculation. If we take the similarity of the | |
375 | +(random, binary) fragmentation kernel with the Brownian agglomeration kernel as valid then | |
376 | +we might expect | |
377 | +\beq | |
378 | +p_2 = - p_3 = 1/\df \quad \Longrightarrow p_2 = 1.006 \sim - p_3 \quad d_f \sim 0.99. | |
379 | +\eeq | |
380 | +In short, their numbers do not seem to be related to the fractal dimension of the fragments. | |
381 | +In addition, they only performed their calculations for DLCA agglomerates. | |
382 | +\item Latest addition: it just occurred to me that if we are really coragious we can just set | |
383 | +\beq | |
384 | +\alpha_2 = -\alpha_3 = 1/\df, \quad \Longrightarrow \quad | |
385 | +\alpha_4 = -\frac{1}{2}, \quad \alpha_1 = \Gamma(\frac{1}{2}) \, \Gamma(\frac{1}{2}) \, | |
386 | +\Big [ B(\frac{1}{2} + \frac{1}{\df}, \frac{1}{2} - \frac{1}{\df}) + 1 \Big ] | |
387 | +\eeq | |
388 | +where $\df$ is the fractal dimension of the fragments. There is nothing to fit. Will it work? | |
389 | +\item Lastly, it seems apparent that $h_f(z,y)$, defined in Eq.~(\ref{eq:hf}), does not | |
390 | +necessarily satisfy | |
391 | +the two constraints (ours, by construction and simplicity, does). | |
392 | +Specifically, the integral constraints become the algebraic constraints reported in | |
393 | +Eqs.~(\ref{eq:Fit1Constraint1} and \ref{eq:Fit1Constraint2}). The $p_i$ reported in | |
394 | +Ref.~\cite{Odriozola2002} give | |
395 | +\begin{subequations} | |
396 | +\beq | |
397 | +p_4 = - \frac{1}{2} \, (1 + p_2 + p_3) = -0.4995 , | |
398 | +\eeq | |
399 | +whereas the numerically determined value is $p_4 = -0.1363$ (first constraint), and | |
400 | +\beq | |
401 | +p_1 = \Big [ \Gamma(1+p_2+p_4) \, \Gamma(1+p_3 + p_4) \, | |
402 | ++ \Gamma(1+p_4) \, \Gamma(1+p_2+p_3+p_4) \, \Big ]^{-1} = -0.1630, | |
403 | +\eeq | |
404 | +whereas the numerically determined value is $p_1 = 0.439$ (second constraint). | |
405 | +\end{subequations} | |
406 | + | |
407 | +\end{enumerate} | |
408 | + | |
409 | +What triggered these investigations? Lorenzo's remark that the way we break up the | |
410 | +agglomerates, constant bond lifetime, might lead us to a process similar to radioactive | |
411 | +decay, and thence to a Poisson process. I have to admit, I've thought about it in the past, | |
412 | +but I did not reach any valuable conclusions. Maybe now I will be luckier. More in the future. | |
413 | + | |
414 | +\section{Addendum} | |
415 | + | |
416 | +I just realized that: | |
417 | +\begin{enumerate} | |
418 | +\item To say that $b(x|y)$ is a homogeneous function [which in our case implies | |
419 | +$b(x|y) = y^{-1} b(x/y)$] is not the same as stating that $b(x|y)$ is independent of | |
420 | +the initial particle size $y$. The functional form shows explicitly that the fragment | |
421 | +size distribution depends on the largest particle size as $y^{-1}$. | |
422 | +\item What I am implying is that our data show a dependence on $y$ but it is not the | |
423 | +trivial (expected?) $1/y$. Do you agree? | |
424 | +\item The trial (fitting) function may be modified to force it to remain a homogeneous | |
425 | +function and to obey a conservation law by dividing by $i+j = y$. In that case, the straight-chain | |
426 | +limit may be imposed, but no combination of exponents would give our beta distribution. | |
427 | +\item Our fitting function, the beta distribution, suggest that we've taken the | |
428 | +kernel to be proportional to $K_{ij} \sim (i j)^{\alpha -1}$. The proportionality constant | |
429 | +is important. | |
430 | +\item Remember this is thinking and writing, as if we were talking to each other in front | |
431 | +of a blackboard. Nothing conclusive, but I'm starting to appreciate the various subtleties. | |
432 | +\end{enumerate} | |
433 | + | |
434 | +\begin{thebibliography}{99} | |
435 | +\bibitem{Odriozola2002} G. Odriozola, A. Schmitt, A. Moncho-Jord\'{a}, J. Callejas-Fern\'{a}ndez, | |
436 | +R. Mart\'{i}nez-Garc\'{i}a, R. Leone, and R. Hidalgo-\'{A}lvarez, ``Constant bond breakup | |
437 | +probability model for reversible aggregation processes", Phys. Rev. E 65, 031405 (2002). | |
438 | + | |
439 | +%\bibitem{McGradyZiff1987} E.D. McGrady and R.M. Ziff, `` ``Shattering" transition in fragmentation", | |
440 | +%Phys. Rev. lett. 58, 892 (1987) | |
441 | +% | |
442 | +%\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). | |
443 | +% | |
444 | +%\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). | |
445 | +% | |
446 | +%\bibitem{Classify} P.G.J. van Dongen ad M.H. Ernst, ``Scaling solutions of Smoluchowski's coagulation | |
447 | +%equation", J. Stat. Phys. 50, 295 (1988). | |
448 | +% | |
449 | +%\bibitem{Ushape} S. Ito and S. Yukawa, ``Stochastic modeling on fragmentation process over | |
450 | +%lifetime and its dynamical scaling law of fragments distribution", J. Phys. Japan 83, 124005 (2014). | |
451 | +% | |
452 | +%\bibitem{FragTrees} Z. Kalay and E. Ben-Naim, ``Fragmentation of random tress", J. Phys. A: Math. Gen. \textbf{48}, 045001 (2015). | |
453 | +% | |
454 | +%%\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). | |
455 | + | |
456 | +\end{thebibliography} | |
457 | + | |
458 | +\end{document} | |
459 | +%%% Local Variables: | |
460 | +%%% mode: latex | |
461 | +%%% TeX-master: t | |
462 | +%%% End: |