Fragmentation and Shattering

Abstract

Binary fragmentation is the reverse process of binary aggregation: a cluster with i monomers splits into two clusters with masses j and i − j such that 1 ≤ j ≤ i − 1. This is represented schematically by the irreversible mass-conserving reaction

$$\displaystyle {} (i) \xrightarrow {B_{i-j,j}} (i-j) + (j) . $$

Appendix: Derivations

10.1.1 Equation (10.24)

Consider parent and offspring distributions for the fragmentation reaction

$$\displaystyle \begin{aligned} (i+j) \to (i) + (j) . \end{aligned}$$

Only the cluster masses involved in this reaction have different number of clusters in the parent and offspring distributions. Then we have

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

where we have used the parent–offspring relationships with i ≠ j. Repeating with i = j, n _i > 1:

$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{\mathbf{n^\prime!}}{\mathbf{n!}} =\frac{1}{N} \frac{(n_i!) (n_{2i})!}{(n^{\prime}_i!) (n^{\prime}_{2i})!} =\frac{1}{N} \frac{(n_i!) (n^{\prime}_{2i}-1)!}{(n_i-2)! (n^{\prime}_{2i})!} =\frac{1}{N} \frac{n_i (n_i-1)}{n^{\prime}_{2i}} \end{array} \end{aligned} $$

and we notice that we no longer need to specify n _i > 1 since n _i = 1 makes this ratio 0. The results can be summarized into a single expression as follows:

$$\displaystyle \begin{aligned} \frac{\mathbf{n^\prime!}}{\mathbf{n!}} = \frac{1}{N} \frac{n_i (n_j-\delta_{ij})}{n^{\prime}_{i+j}} \end{aligned} $$

(10.74)

which is Eq. (10.24) in the text.

10.1.2 Equation (10.44)

We start with Eq. (10.42)

$$\displaystyle \begin{aligned} \left\langle n_k \right\rangle &= \left\langle \sum_{i=2}^\infty\sum_{j=1}^{i-1} \mathcal P^{\prime}_{i-j,j} (n^{\prime}_k - \delta_{k,i} + \delta_{k,i-j} + \delta_{k,j}) \right\rangle \\ &=\left\langle \sum_{i=2}^\infty\sum_{j=1}^{i-1} P^{\prime}_{i-j,j} n^{\prime}_k \right\rangle -\left\langle \sum_{i=2}^\infty\sum_{j=1}^{i-1} P^{\prime}_{i-j,j}\delta_{i,k} \right\rangle \\ &\quad +\left\langle \sum_{i=2}^\infty\sum_{j=1}^{i-1} P^{\prime}_{i-j,j}\delta_{k,i-j} \right\rangle +\left\langle \sum_{i=2}^\infty\sum_{j=1}^{i-1} P^{\prime}_{i-j,j} \delta_{k,j} \right\rangle \end{aligned} $$

([10.42])

and calculate each term on the right-hand side separately. For the first term we have

$$\displaystyle \begin{aligned} \left\langle \sum_{i=2}^\infty\sum_{j=1}^{i-1} P^{\prime}_{i-j,j} n^{\prime}_k \right\rangle = \left\langle n^{\prime}_k\sum_{i=2}^\infty\sum_{j=1}^{i-1} P^{\prime}_{i-j,j} \right\rangle = \left\langle n^{\prime}_k \right\rangle \end{aligned}$$

with the last result obtained by the normalization condition for $\mathcal P^{\prime }_{i-j,j}$. The second term gives

$$\displaystyle \begin{aligned} -\left\langle \sum_{i=2}^\infty\sum_{j=1}^{i-1} P^{\prime}_{i-j,j}\delta_{i,k} \right\rangle =-\left\langle \sum_{j=1}^{k-1} P^{\prime}_{k-j,j} \right\rangle =-\left\langle \sum_{j=1}^{k-1} \frac{n^{\prime}_k B_{k-j,j}}{(M-N+1){\bar B}{(\mathbf{n'})}} \right\rangle . \end{aligned}$$

The third and fourth terms are identical due to symmetry between all ordered pairs of fragments. The two terms combined are

$$\displaystyle \begin{aligned}\displaystyle \left\langle \sum_{i=2}^\infty\sum_{j=1}^{i-1} P^{\prime}_{i-j,j} (\delta_{k,i-j}+\delta_{k,j}) \right\rangle = 2\left\langle \sum_{i=2}^\infty\sum_{j=1}^{i-1} P^{\prime}_{i-j,j}\delta_{k,j} \right\rangle = 2\left\langle \sum_{i=k+1}^\infty P^{\prime}_{i-k,k} \right\rangle \\\displaystyle = 2\left\langle \sum_{i=k+1}^\infty \frac{n_i B_{i-k,k}}{(M-N+1){\bar B}(\mathbf{n'})} \right\rangle \end{aligned} $$

Incorporating these results into Eq. (10.42) we obtain

$$\displaystyle \begin{aligned} \left\langle n_k \right\rangle = \left\langle n^{\prime}_k \right\rangle - \left\langle \sum_{j=1}^{k-1} \frac{n^{\prime}_k B_{k-j,j}}{(M-N+1){\bar B}{\mathbf{n'}}} \right\rangle + 2\left\langle \sum_{i=k+1}^\infty \frac{n_i B_{i-k,k}}{(M-N+1){\bar B}(\mathbf{n'})} \right\rangle . \end{aligned}$$

which leads directly to Eq. (10.44) in the main text.