Kinetics and Final Transformations in Self-Similar Processes

We propose a theory for the kinetics and final transformations in two types of self-similar processes, when the operators describing the process at each instant of time are inversely or directly proportional to the difference between the parameters corresponding to the final state and the state achieved. In the first case, the parameter governing the stability of the system reaches a critical value, and the final stage occurs in avalanche fashion. In the second case, there is no critical state, the final stage occurs asymptotically, and the final state is an equilibrium state. As the model processes, we take the fracture process in the composites and their strengthening process. The critical parameter in the fracture process is a measure of the stress tensor. For this process, we obtain an equation relating the initial and final degree of damage in the composite and the operator generating the self-similar process; we construct the theoretical time-to-fracture distribution functions vs. the applied stress, including zero fracture probability. The distribution functions can be constructed without long-term testing. We have obtained an equation describing the kinetics of the strengthening process and an expression defining the equilibrium state. The theory has been confirmed by experiment. The maximum testing time was ~20000 h.

Appendix

Let us consider some cases of the application of Eqs. (6) and (11) to solution of specific problems.

• 1) Limiting case $\overline{m}$ → ∞. A suitable model is the solar system, the existence of which is determined by the Sun. The radius of the Sun is R _s = 696·10⁶ m. The size of the clusters from which the Sun is formed is unknown. The basic parameter characterizing the relative scale of the Sun is the relative area of the spherical surface of the Sun, expressed in units of the area of a spherical element with radius conveniently taken to be equal to ΔR ₀ = $\sqrt{4\uppi}$m, from which we obtain $\overline{m}$ = R _c /ΔR ₀ = 696·10⁶ = 696·10⁶. Let us assume a time t ₀ = 1 and an extent of undamaged regions in the system at the instant it appears as 〈q ₀〉 = 1.

  Comparing expressions (10) and (11) for 〈q ₀〉 = 1 and taking into account the equality $1/\overline{m}\cong 1/\left(\overline{m}+1\right)$ for $\overline{m}$ → ∞, we obtain

  $$ \begin{array}{c}\hfill \left\langle q*\right\rangle =\frac{1}{{\left(\overline{t}*\right)}^{1/\overline{m}}}={\left(\frac{1}{\overline{t}*}\right)}^{1/\overline{m}}={\left(\frac{1}{{\overline{m}}^2}\right)}^{1/\overline{m}},\hfill \\ {}\hfill \overline{t}*\approx {\overline{m}}^2={\left({R}_s/\varDelta {R}_0\right)}^2=4\uppi {(6.96)}^2\cdot {10}^{16}/{\left(\sqrt{4\uppi}\right)}^2.\hfill \end{array} $$

  Thus the time of existence for the solar system t* on the average is equal to about 14.2 billion years, which is consistent with the age of our Universe and justifies the assumption made concerning the cluster size ΔR ₀.

• 2) Limiting case $\overline{m}$ → 0. In this case, Eq. (6) takes on the form

  $$ \left\langle q*\right\rangle {\left[1- \exp \left(-\frac{2}{\left\langle q*\right\rangle +1}\right)\right]}^{-1}=\left\langle {q}_0\right\rangle \left[\frac{1+\left\langle {q}_0\right\rangle }{2}\right] $$

  (A1)

  A model for such a process can be transition of an electron in an atom from one energy level to another. According to quantum mechanics, in the case of the hydrogen atom the relative width of two adjacent energy levels is equal to

  $$ \frac{\Delta E}{E_n}=\frac{E_{n+1}-{E}_n}{E_n}=-\left[1-\frac{n^2}{{\left(n+1\right)}^2}\right] $$

  Here E _n is the energy level corresponding to the number n (n = 1, 2, 3,…). Using this formula, let us establish the correspondence between the degree of electronic order and the number of the energy levels as follows. We will assume a degree of electronic order in the ground state level 〈q ₀〉 = 1. During a transition to an excited level, the electron absorbs energy, its entropy increases, and the degree of order decreases. Let us rewrite the preceding formula as follows:

  $$ \frac{E_{n+1}-{E}_n}{E_n}=-\left(1-\frac{E_{n+1}}{E_n}\right)=-\left[1-\frac{n^2}{{\left(n+1\right)}^2}\right] $$

  We will consider the expression 1 – E _n+1/E _n as the ratio of the degree of order for the level under consideration to the degree of order for the preceding level. Thus we have

  $$ 1-\frac{E_{n+1}}{E_n}=\frac{\left\langle {q}_{0,n+1}\right\rangle }{\left\langle {q}_{0,n}\right\rangle }<1 $$

  The ratio obtained is always less than unity for any n. Hence we obtain the relationship between the ratio of the degree of order for the level under consideration 〈q _0,n + 1〉 to the degree of order for the preceding level and the ratio of the corresponding numbers for the energy levels in the following form:

  $$ \frac{\left\langle {q}_{0,n+1}\right\rangle }{\left\langle {q}_{0,n}\right\rangle }=1-{\left(\frac{n}{n+1}\right)}^2,\kern0.24em \left\langle {q}_{0,1}\right\rangle =1,\kern0.24em n=1,2,3,\dots $$

  (A2)

  Here for n = 1 (closest to the core of the level), we assume of degree of order 〈q _0,1〉 = 1. The results of estimates of the degree of order using formula (A1) in the MatLab environment and formula (A2) within 0.1 accuracy are given in Table 1.

  Table 1 Degree of Electronic Order According to Energy Level n

• 3) Economic analysis of a hypothetical macrosystem when destructive processes are present.

  Let us assume that at the instant of time $\overline{t}$ = 1− 〈q ₀〉 = 1, we have 〈q ₀〉 = 1, and at the instant of time ${\overline{t}}_{\mathrm{D}}$ = 25 − 〈q _D〉 = 0.6, we have, t ₀ = 1 year. Considering the state of the system as close to destruction, let us estimate the scale of the system under the specified conditions, using formula (11)

  $$ {\overline{t}}_{\mathrm{D}}=1/{\left\langle {q}_{\mathrm{D}}\right\rangle}^{\overline{m}} $$

  from which we have

  $$ \overline{m}=- \log {\overline{t}}_{\mathrm{D}}/ \log \left\langle {q}_{\mathrm{D}}\right\rangle =- \log 25/ \log 0.6\cong 1.398/0.222\cong 6.3. $$

  Thus at the instant of time ${\overline{t}}_{\mathrm{D}}$ = 25, for 〈q _D〉 = 0.6 we have $\overline{m}$ ≅ 6.3 . These data allow us to estimate the time of existence for the system under consideration, taking the instant of time ${\overline{t}}_{\mathrm{D}}$ = 25 as the starting point. From Eq. (6), considering that the expression with the exponential in this case is practically equal to zero, we obtain

  $$ \frac{\left\langle q*\right\rangle }{\left\langle {q}_{\mathrm{D}}\right\rangle }={\left(\frac{1+\left\langle {q}_{\mathrm{D}}\right\rangle }{2{\left(\overline{m}+1\right)}^2}\right)}^{1/\left(\overline{m}+1\right)}={\left(\frac{1+\left\langle 0.6\right\rangle }{2{\left(6.3+1\right)}^2}\right)}^{1/\left(6.3+1\right)}\cong 0.57, $$

  from which according to formula (11), we have

  $$ \left\langle \overline{t}*\right\rangle ={\left(\frac{\left\langle {q}_{\mathrm{D}}\right\rangle }{\left\langle q*\right\rangle}\right)}^{\overline{m}}={\left(\frac{1}{0.57}\right)}^{6.3}\approx 34.5\kern0.24em \mathrm{years} $$

  Considering that $\overline{t}*$ ≥ 0, and that the value obtained is an average, as a rough estimate we find that 0 ≤ $\overline{t}*$ ≤ 70 years. For a more exact estimate, let us construct the Weibull distribution function of the quantity $\overline{t}*$. Note that near the mathematical expectation value, the Weibull distribution can be approximated by a normal distribution, and we can make use of the “three sigma” rule to estimate the mean-square deviation of the Weibull distribution. Let us use these assumptions and set $\overline{t}*=\left\langle \overline{t}*\right\rangle$, $s\left(\overline{t}*\right)\approx \overline{t}*/3$ and ${\left(\overline{t}*\right)}_{\min }=0$. Then from the system of equations

  $$ \begin{array}{l}\left\langle \overline{t}*\right\rangle ={\upsigma}_{\overline{m}}^{*}\varGamma \left(1+1/{\uplambda}_{\overline{m}}\right),\hfill \\ {}s\left(\overline{t}*\right)={\upsigma}_{\overline{m}}^{*}\sqrt{\varGamma \left(1+2/{\uplambda}_{\overline{m}}\right)-{\varGamma}^2\left(1+1/{\uplambda}_{\overline{m}}\right)},\hfill \end{array} $$

  (A3)

  where Γ(x) is a gamma function:

  $$ \begin{array}{cc}\hfill \mathrm{and}\kern9em \hfill & \hfill \left[\varGamma (x)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-t}{t}^{x-1}dt;s\left(\overline{t}*\right)}\right]\hfill \end{array} $$

  ais the mean-square deviation, we can determine the parameters of the Weibull distribution function ${\upsigma}_{\overline{m}}$*, ${\uplambda}_{\overline{m}}$. The parameter ${\uplambda}_{\overline{m}}$ is determined from solution of the nonlinear equation

  $$ \left\langle \overline{t}*\right\rangle /\varGamma \left(1+1/{\uplambda}_{\overline{m}}\right)-s\left(\overline{t}*\right)/\sqrt{\varGamma \left(1+2/{\uplambda}_{\overline{m}}\right)-{\varGamma}^2\left(1+1/{\uplambda}_{\overline{m}}\right)}=0 $$

  (A4)

  Then

  $$ {\upsigma}_{\overline{m}}^{*}=\left\langle \overline{t}*\right\rangle /\varGamma \left(1+1/{\uplambda}_{\overline{m}}\right) $$

  (A5)

  Having constructed the Weibull distribution function $\mathrm{P}\left(\overline{t}*\left|\left\langle {q}_0\right\rangle \right.\right)$, we find the corresponding confidence intervals ${\overline{t}}_1^{*}$ ≤ $\overline{t}*$ ≤ ${\overline{t}}_2^{*}$ for the specified confidence level α from the equations

  $$ {\overline{t}}_1^{*}={\upsigma}_{\overline{m}}^{*}{\left[ \ln \frac{1}{1-\mathrm{P}\left(\overline{t}*\le {\overline{t}}_1^{*}\left|\left\langle {q}_0\right\rangle \right.\right)}\right]}^{1/{\uplambda}_{\overline{m}}} $$

  (A6)

  where $\mathrm{P}\left(\overline{t}*\le {\overline{t}}_1^{*}\left|\left\langle {q}_0\right\rangle \right.\right)=\left(1-\upalpha \right)/2$, and

  $$ \Delta {\overline{t}}_2^{*}={\upsigma}_{\overline{m}}^{*}{\left[ \ln \frac{1}{1-\mathrm{P}\left(\Delta \overline{t}*\le {\overline{t}}_2^{*}\left|\left\langle {q}_0\right\rangle \right.\right)}\right]}^{1/{\uplambda}_{\overline{m}}} $$

  (A7)

  where $\mathrm{P}\left(\overline{t}*\le {\overline{t}}_2^{*}\left|\left\langle {q}_0\right\rangle \right.\right)=\left(1+\upalpha \right)/2$.

  Solving the nonlinear equation (A4) in the MatLab environment to determine ${\uplambda}_{\overline{m}}$ and Eq. (A5), we obtain ${\uplambda}_{\overline{m}}$ ≅ 3.0, ${\upsigma}_{\overline{m}}^{*}$ = 38.6 years. Then from Eqs. (P6) and (P7) for confidence level α = 0.90, we have the estimates: $\Delta {\overline{t}}_1^{*}$ = 14 2 years and $\Delta {\overline{t}}_2^{*}$ = 55 6 years. Thus the time of existence for the destructive processes with specified probability are found within the range 14.2 ≤ $\overline{t}*$ ≤ 55.6 years.

• 4) Calculations similar to the preceding show that if a person has lived to 80 years old and the overall degree of age-related changes (general loss of fitness in the entire body) has reached approximately 90%, the remaining life expectancy can be calculated within the range 2.5 ≤ $\overline{t}*$ ≤ 9.7 years with confidence level α = 0.90.

• 5) Let us consider processes of the second type for solving the problem of predicting inflation. With this goal, in Eq. (17) instead of the degree of orientation s we will use the concept of order/disorder 〈q〉. Then we will represent Eq. (17) as

$$ \left\langle {q}_t\right\rangle =\left\langle {q}^{\infty}\right\rangle \times \frac{1-\left(\left\langle {q}^{\infty}\right\rangle -\left\langle {q}_0\right\rangle \right){\left[\left(1-\left\langle {q}_0\right\rangle \right)\left\langle {q}^{\infty}\right\rangle \right]}^{-1}{e}^{-\left(1-\left\langle {q}^{\infty}\right\rangle \right)\overline{t}}}{1-\left(\left\langle {q}^{\infty}\right\rangle -\left\langle {q}_0\right\rangle \right){\left(1-\left\langle {q}_0\right\rangle \right)}^{-1}{e}^{-\left(1-\left\langle {q}^{\infty}\right\rangle \right)\overline{t}}} $$

(A8)

for initial conditions 〈q _t 〉 = 〈q ₀〉, ${\overline{t}}_0$ = 0. Here 〈q _t 〉 is the instantaneous degree of order/disorder, 〈q ₀〉 is the initial degree of order/disorder, and ${\overline{t}}_0$ − is the initial time. The 〈q _t 〉 approaches the equilibrium value 〈q ^∞〉, corresponding to minimum business activity.

Let us use Eq. (A8) for analysis of the inflation process in a hypothetical macrosystem. At the initial instant of time $\overline{t}$ = 0, let inflation be 3%, i.e., 〈q ₀〉 = 0.03, while at the instant of time $\overline{t}$= 0.5 (t ₀ = 1 year) it is 6.5%, $\left\langle {q}_{\overline{t}=0,5}\right\rangle$ = 0 065. This condition makes it possible, from solving Eq. (A8), to find the value of 〈q ^∞〉, corresponding to minimum business activity in this time interval. Solution of nonlinear equation (A8) in the MatLab environment gave the estimate 〈q ^∞〉 ≈ 0.13. This result allows us to predict inflation at the end of the year ($\overline{t}$ = 1.0), using Eq. (A8). As a result, we obtain the estimate $\left\langle {q}_{\overline{t}=1,0}\right\rangle$ ≡ 0 09. Considering that this value is an average, let us use expressions (9) and (10) to determine the confidence interval for this estimate 0 03 ≤ ${q}_{\overline{t}=1.0}$ ≤ 0 13 with confidence level α = 0.90. The corresponding interval for the equilibrium value is equal to 0.05 ≤ q ^∞ ≤ 0.19 .