Irreversible Clustering

Abstract

The merging of two clusters to produce a new cluster that conserves mass is one of the most basic mechanisms of size change in dispersed populations. The process is known by many different names: aggregation, agglomeration, coagulation, polymerization, flocculation, and others that refer to the physical aspects of the particular problem at hand. We will use aggregation.

Appendix: Derivations

8.1.1 Equation (8.20)

Here we derive Eq. (8.20), which gives the evolution of the partition function during aggregation. We begin with

$$\displaystyle \begin{aligned} P(\mathbf{n}|\mathbf{n'}) &= C_{\mathbf{n'}} \frac{n^{\prime}_{i-j} (n^{\prime}_j - \delta_{i-j,j}) } {1+\delta_{i-j,j}} \, {K}_{i-j,j}, \end{aligned} $$

([8.16])

$$\displaystyle \begin{aligned} P(\mathbf{n}) &= \mathbf{n!}\frac{W(\mathbf{n})}{\,\Omega_{M,N}} \end{aligned} $$

([8.18])

We insert these into the master equation (8.15) and rearrange the result in the form,

$$\displaystyle \begin{aligned} \frac{ W(\mathbf{n})}{\Omega_{M,N}}= \frac{2}{N(N+1)} \sum_{\mathbf{n'}} \frac{\mathbf{n'!} }{\mathbf{n!}}\cdot \frac{{K}_{i-j,j}}{{\bar K}(\mathbf{n'})}\cdot \frac{n^{\prime}_{i-j} (n^{\prime}_j - \delta_{i-j,j}) } {1+\delta_{i-j,j}}\cdot \frac{W(\mathbf{n'})}{\Omega_{M,N+1}} \end{aligned}$$

The ratio of the factorials involves the offspring distribution in the numerator, and the parent distribution in the denominator. These differ only in the number of cluster masses involved in the aggregation event:

$$\displaystyle \begin{aligned} \frac{\mathbf{n'!}}{\mathbf{n!}} = \frac{(N+1)!}{N!}\cdot\frac{n_i!n_j!n_{i-j}!}{n^{\prime}_i!n^{\prime}_j!n^{\prime}_{i-j}!} = \frac{(N+1)n_i}{n^{\prime}_{i-j}(n^{\prime}_j-\delta_{i-j,j})} \end{aligned}$$

with the last result obtained by application of the parent–offspring relationships in Eq. (8.12). Combining these results we obtain

$$\displaystyle \begin{aligned} \frac{W(\mathbf{n})}{\Omega_{M,N}}= \frac{2}{N} \sum_{i=2}^\infty\sum_{j=1}^{i/2} \frac{{K}_{i-j,j}}{{\bar K}(\mathbf{n'})}\cdot \frac{n_i} {1+\delta_{i-j,j}}\cdot \frac{W(\mathbf{n'})}{\Omega_{M,N+1}} . \end{aligned}$$

Here, the summation over parents of n is organized as follows: for every cluster mass in n we first sum over the i∕2 parents of that cluster mass (inner summation), then run the outer summation over all members of n. Cluster mass i = 1 cannot be formed by aggregation and thus the outer summation begins at i = 2. Given the symmetry of the kernel, K _ij = K _ji, the inner summation is extended over the range j = 1 to j = i − 1; this amounts to double-counting all terms of this summation and so the result is

$$\displaystyle \begin{aligned} \frac{W(\mathbf{n})}{\Omega_{M,N}}= \frac{1}{N} \sum_{i=2}^\infty\sum_{j=1}^{i-1} n_i \frac{{K}_{i-j,j}}{{\bar K}(\mathbf{n'})}\cdot \frac{W(\mathbf{n'})}{\Omega_{M,N+1}} . \end{aligned}$$

This leads directly into Eq. (8.20) of the text. In this form, the summation over all (i − j, j)-parents of n is converted into a double summation in which the outer sum goes over all clusters i > 1 that can be formed by aggregation, while the inner sum goes over all possible binary events that can form the cluster mass i. The construction of this summation is illustrated with a numerical example below.

Note 8.5 (Construction of the Double Summation)

We consider the distribution

$$\displaystyle \begin{aligned} \mathbf{n} = (1,1,1,0,0,0) \end{aligned}$$

in the ensemble M = 6, N = 3 (see Fig. 8.6). The distribution consists of one monomer, one dimer, and one trimer. It has two parents, one that forms the trimer by the reaction 2 + 1, and one that forms the dimer by the reaction 1 + 1. The (2, 1)-parent is counted twice, therefore the number of ordered parent–offspring events is 3, also equal to M − N. The double summation in Eq. (8.20) contains the following terms:

$$\displaystyle \begin{aligned} n_2\frac{{K}_{1,1}}{{\bar K}({\mathbf{n}}_2)}\frac{W({\mathbf{n}}^{\prime}_2)}{W(\mathbf{n})} + n_3\frac{{K}_{2,1}}{{\bar K}({\mathbf{n}}_1)}\frac{W({\mathbf{n}}^{\prime}_1)}{W(\mathbf{n})} + n_3\frac{{K}_{1,2}}{{\bar K}({\mathbf{n}}_1)}\frac{W({\mathbf{n}}^{\prime}_1)}{W(\mathbf{n})} . \end{aligned}$$

In this example, n ₃ = n ₂ = 1, and the summation finally expands to M − N = 6 − 3 = 3 terms. In general, if the multiplications by n _i are expanded to a sum of n _i identical terms, the total number of terms in the double summation will always be M − N. Each (i − j, j)-parent contributes the quantity

$$\displaystyle \begin{aligned} n_i\frac{{K}_{i-j,j}}{{\bar K}(\mathbf{n'})} \frac{W(\mathbf{n'})}{W(\mathbf{n})}(2-\delta_{i-j,j}) \end{aligned}$$

to the summation. Then an alternative expression for the double sum in Eq. (8.20) is

$$\displaystyle \begin{aligned} \sum_{i=2}^\infty\sum_{j=1}^{i-1} n_i \frac{{K}_{i-j,j}}{{\bar K}(\mathbf{n'})}\,\frac{W(\mathbf{n'})}{W(\mathbf{n})} = \sum_{\mathbf{n'}} n_i\frac{{K}_{i-j,j}}{{\bar K}(\mathbf{n'})} \frac{W(\mathbf{n'})}{W(\mathbf{n})}(2-\delta_{i-j,j}) \end{aligned} $$

(8.119)

in which the summation is now over all (i − j, j) parents of distribution n.

Fig. 8.6

Illustration of the summations in the recursion for the partition function in Eq. (8.20) for M = 6. Distribution n belongs in the ensemble M = 6, N = 3, and has two parents in the ensemble M = 6, N = 2

8.1.2 Smoluchowski Eq. (8.87)

We begin with Eq. (8.85)

$$\displaystyle \begin{aligned} \left\langle n_k \right\rangle_{M,N} = \frac{2}{N(N+1)} \left\langle \sum_{i=2}^\infty\sum_{j=1}^{i/2} \left(n_k + \delta_{k,i} - \delta_{k,i-j} - \delta_{k,j}\right) T_{i-j,j} \right\rangle_{M,N+1}, \end{aligned} $$

([8.85])

and work the summation through the Kronecker deltas to express the result in the form

$$\displaystyle \begin{aligned} \begin{array}{rcl} \left\langle n_k \right\rangle_{M,N} &\displaystyle &\displaystyle = \frac{2}{N(N+1)}\times \\ &\displaystyle &\displaystyle \left\langle n_k\sum_{i=2}^\infty \sum_{j=1}^{i/2} T_{i-j,j} + \sum_{j=1}^{k/2} T_{k-j, j} - \sum_{j=1}^\infty (1 + \delta_{kj} ) T_{k,j} \right\rangle_{M,N+1} \end{array} \end{aligned} $$

(8.120)

All terms on the right-hand side refer to the parent ensemble, $\mu \mathcal C(M,N+1)$, as indicated by the subscript on the ensemble average, therefore we drop the primes as unnecessary. The first term on the right-hand side reduces to the mean number of cluster mass k in the parent ensemble:

$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \frac{2}{N(N+1)} \left\langle n_k \sum_{i=2}^\infty\sum_{j=1}^{i/2} T_{i-j,j} \right\rangle_{M,N+1}= \\ &\displaystyle &\displaystyle \qquad \frac{2}{N(N+1)} \left\langle n_k\sum_{i=2}^\infty \sum_{j=1}^{i/2} \frac{n_{i-j} (n_j - \delta_{i-j,j})}{1+\delta_{i-j,j}} \, \frac{{K}_{i-j,j}}{{\bar K}(\mathbf{n})} \right\rangle_{M,N+1} = \left\langle n_k \right\rangle_{M,N+1} \end{array} \end{aligned} $$

The second term is

$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{2}{N(N+1)}\left\langle \sum_{j=1}^{k/2} T_{k-j,j} \right\rangle &\displaystyle =&\displaystyle \frac{2}{N(N+1)}\left\langle \sum_{j=1}^{k/2} \frac{n_{k-j} (n_j - \delta_{k-j,j}) } {1+\delta_{k-j,j}} \, \frac{{K}_{k-j,j}}{{\bar K}(\mathbf{n})} \right\rangle \\ &\displaystyle =&\displaystyle \frac{1}{N(N+1)} \left\langle \sum_{j=1}^{k-1} n_{k-j} (n_{j} - \delta_{k-j,j}) \frac{{K}_{k-j,j}}{{\bar K}(\mathbf{n})} \right\rangle\qquad \; \end{array} \end{aligned} $$

The third term is

$$\displaystyle \begin{aligned} \frac{2}{N(N+1)} \left\langle \sum_{j=1}^\infty (1 + \delta_{kj} ) T_{k,j} \right\rangle = \frac{2}{N(N+1)} \left\langle \sum_{j=1}^\infty n_k (n_j - \delta_{k,j}) \frac{{K}_{k,j}}{{\bar K}(\mathbf{n})} , \right\rangle \end{aligned}$$

Combining these results we obtain Eq. (8.87) of the main text.

8.1.3 Aggregation in the Continuous Domain

8.1.3.1 Derivation 1

The partition function is the product of the volume of the ensemble and a term that represents the effect of the kernel,

$$\displaystyle \begin{aligned} \Omega_{M,N} = {\mathcal V}_{M,N} \, \mathcal {K}_{M,N} \end{aligned} $$

(8.121)

with

$$\displaystyle \begin{aligned} {\mathcal V}_{M,N} = \binom{M-1}{N-1} \end{aligned} $$

(8.122)

and

$$\displaystyle \begin{aligned} \mathcal {K}_{M,N} = \prod_{N'=N+1}^{M}\left\langle {K}_{M,N'} \right\rangle. \end{aligned} $$

(8.123)

In this form the partition function is expressed as a product of two terms, one that represents the unbiased ensemble (${\mathcal V}$), and one that represents the effect of the aggregation kernel ($\mathcal K$). This factorization carries over to the parameters β and $\log q$, which may also be expressed as the product of the unbiased case times a factor that includes the effect of the kernel.

The $\log$ of the first term in the ThL is obtained using the Stirling approximation $\log x!\to x\log x-x$

$$\displaystyle \begin{aligned} \frac{\log {\mathcal V}_{M,N}}{N} = {\bar x}\log {\bar x}-({\bar x}-1)\log({\bar x}-1). \end{aligned} $$

(8.124)

For large ${\bar x}$ this further reduces to the derivative of ${\bar x}\log {\bar x}$:

$$\displaystyle \begin{aligned} \frac{\log {\mathcal V}_{M,N}}{N} = 1+ \log {\bar x}. \end{aligned} $$

(8.125)

For $\log \mathcal K$ we have

$$\displaystyle \begin{aligned} \log \mathcal {K}_{M,N} &= \sum_{N'=N+1}^{M}\log\left\langle {K}_{M,N'} \right\rangle.\end{aligned} $$

(8.126)

The ensemble average kernel in the ThL is an intensive function of ${\bar x}=M/N$:

$$\displaystyle \begin{aligned} \left\langle K \right\rangle_{M,N} = \int\int f(x)f(y)k(x,y)\,d x\, d y \equiv \bar K({\bar x}). \end{aligned} $$

(8.127)

The summation in (8.126) becomes an integral in dN′

$$\displaystyle \begin{aligned} \log \mathcal {K}_{M,N} = \int\limits_{N+1,M=\mathrm{const.}}^M \log \bar K({\bar x})\, d N', \end{aligned} $$

(8.128)

which is to be evaluated at constant M. Using $N'=M/{\bar x}$, we have

$$\displaystyle \begin{aligned} d N'\Big|{}_M = -M \frac{d{\bar x}}{{\bar x}^2} . \end{aligned} $$

(8.129)

We insert into Eq. (8.128) and switch the integration variable to $y={\bar x}$. Noting that the integration limits are now from $y={\bar x}$ (current state) to y = 1 (initial state), we have

$$\displaystyle \begin{aligned} \log \mathcal {K}_{M,N} = -M\int_{\bar x}^1 \log \bar K(y)\frac{d y}{y^2} \end{aligned} $$

(8.130)

and finally

$$\displaystyle \begin{aligned} \frac{\log \mathcal {K}_{M,N}}{N} = {\bar x} \int_1^{\bar x} \log \bar K(y)\frac{d y}{y^2}. \end{aligned} $$

(8.131)

Combining Eqs. (8.124) and (8.131) we obtain the full partition function:

$$\displaystyle \begin{aligned} \log\omega\equiv \frac{\log\Omega_{M,N}}{N} = {\bar x}\log {\bar x}-({\bar x}-1)\log({\bar x}-1) + {\bar x} \int_1^{\bar x} \log \bar K(y)\frac{d y}{y^2}. \end{aligned} $$

(8.132)

We obtain β and $\log q$ from their relationship to the partition function:

$$\displaystyle \begin{aligned} \beta = \frac{d\log\omega}{d{\bar x}} =\log\frac{{\bar x}}{{\bar x}-1} + \int_1^{\bar x} \frac{d\log \bar K(y)}{y}, \end{aligned} $$

(8.133)

$$\displaystyle \begin{aligned} \log q=\log\omega - {\bar x} \frac{d\log\omega}{d{\bar x}} = \log ({\bar x}-1)-\log \bar K({\bar x}). \end{aligned} $$

(8.134)

Identity

The following identity was used:

$$\displaystyle \begin{aligned} \int_1^{\bar x} \frac{d\log \bar K(y)}{y} = \frac{\log \bar K({\bar x})}{{\bar x}} + \int_1^{\bar x} \frac{\log \bar K(y)}{y^2}d y. \end{aligned} $$

(8.135)

This is obtained by integration by parts noting that $\log \bar K(1)=0$.

8.1.3.2 Alternative Derivation

As an exercise we try an alternative derivation. We start with the discrete quantities

$$\displaystyle \begin{aligned} \beta &= \log\frac{\Omega_{M+1,N}}{\Omega_{M,N}} \\ \log q &= \log\frac{\Omega_{M,N+1}}{\Omega_{M,N}} \end{aligned} $$

to obtain their form in the thermodynamic limit. We may obtain each of these terms for each partition function, ${\mathcal V}_{M,N}$ and $\mathcal {K}_{M,N}$, separately.

By direct application of the multinomial term, the M-derivative is

$$\displaystyle \begin{aligned} \beta_V &= \binom{M+1}{N}\Big/\binom{M}{N} = \log\frac{M}{M-N+1} \to \log\frac{{\bar x}}{{\bar x}-1}. \end{aligned} $$

(8.136)

Similarly, the derivative with respect to N is

$$\displaystyle \begin{aligned} \log q_V &= \binom{M}{N+1}\Big/\binom{M}{N} = \log\frac{M-N}{N} \to \log({\bar x}-1). \end{aligned} $$

(8.137)

The finite difference of $\log \mathcal {K}_{M,N}$ with respect to M is

$$\displaystyle \begin{aligned} \beta_{\mathcal K} &= \log\frac{\mathcal {K}_{M+1,N}}{\mathcal {K}_{M,N}} = \log\left\langle K \right\rangle_{M+1,M+1} + \sum_{N'=N+1}^M \log \frac{\left\langle K \right\rangle_{M+1,N'}}{\left\langle {K}_{M,N'} \right\rangle}, \end{aligned} $$

which is obtained by application of Eq. (8.123) and careful arrangement of the terms in the summation. For the first term we have

$$\displaystyle \begin{aligned} \log\left\langle K \right\rangle_{M+1,M+1} = \log {K}_{11} = 0 \end{aligned}$$

because by normalization of the kernel, K ₁₁ = 1. The ratios in the summation turn into the partial derivative of $\log \left \langle K \right \rangle _{M,N}$ with respect to M:

$$\displaystyle \begin{aligned} \log \frac{\left\langle K \right\rangle_{M+1,N'}}{\left\langle {K}_{M,N'} \right\rangle} \to \left(\displaystyle\frac{\partial \log\left\langle {K}_{M,N'} \right\rangle}{\partial M} \right)_{N'} \end{aligned}$$

and the summation becomes an integral in dN′ at constant M ¹⁷:

$$\displaystyle \begin{aligned} \sum_{N'=N+1}^M \log \frac{\left\langle K \right\rangle_{M+1,N'}}{\left\langle {K}_{M,N'} \right\rangle} \to \int_{N+1,M=\mathrm{const.}}^{M} \left(\displaystyle\frac{\partial \log\left\langle {K}_{M,N'} \right\rangle}{\partial M} \right)_{N'}\, d N'. \end{aligned}$$

Setting

$$\displaystyle \begin{aligned} y = M/N' \end{aligned}$$

we obtain

$$\displaystyle \begin{aligned} \partial M = N' d y, \quad \partial N'= -M \frac{d y}{y^2} \end{aligned}$$

and with these results, the integral in the ThL becomes

$$\displaystyle \begin{aligned} \int\limits_{N+1,M=\mathrm{const.}}^{M} \left(\displaystyle\frac{\partial \log\left\langle {K}_{M,N'} \right\rangle}{\partial M} \right)_{N'}\, d N' = \int_{\bar x}^1 \frac{1}{N'} \frac{d\log \bar K(y)}{d y}\left(-M \frac{d y}{y^2}\right). \end{aligned}$$

Finally, the result for $\beta _{\mathcal K}$ is

$$\displaystyle \begin{aligned} \beta_{\mathcal K} = \int_1^{\bar x}\frac{d\log \bar K(y)}{y} = \int_1^{\bar x} \frac{\log \bar K(y)}{y^2} \, d y+\frac{\log \bar K({\bar x})}{{\bar x}}. \end{aligned} $$

(8.138)

This can be expressed in the equivalent form by performing integration by parts:

$$\displaystyle \begin{aligned} \beta_{\mathcal K} &= \frac{\log \bar K({\bar x})}{x}\Big|{}_1^x - \int_1^x \log \bar K({\bar x})d\left(\frac{1}{y}\right), \end{aligned} $$

which gives

$$\displaystyle \begin{aligned} \beta_{\mathcal K} = \frac{\log \bar K({\bar x})}{x} + \int_1^x \log \bar K({\bar x}) \frac{d y}{y^2}. \end{aligned} $$

(8.139)

The N-derivative of $\log \mathcal {K}_{M,N}$ is much easier and follows from Eq. (8.123):

$$\displaystyle \begin{aligned} \log q_{\mathcal K} &= \log\frac{\mathcal {K}_{M,N+1}}{\mathcal K_{N}} = \log\frac{1}{\left\langle {K}_{M,N+1} \right\rangle} \to -\log \bar K({\bar x}). \end{aligned} $$

The final results for β and $\log q$ are obtained by combining the above results:

$$\displaystyle \begin{aligned} \beta = \beta_V+\beta_{\mathcal K} = \int_1^{\bar x} \frac{\log \bar K(y)}{y^2} \,dy + \frac{\log \bar K({\bar x})}{{\bar x}} + \log\frac{{\bar x}}{{\bar x}-1}, \end{aligned}$$

$$\displaystyle \begin{aligned} \log q=\log q_V+\log q_{\mathcal K} = \log({\bar x}-1) -\log \bar K({\bar x}). \end{aligned}$$

These results agree with those in Eqs. (8.132), (8.133), and (8.134).