Scaling forms for Relaxation Times of the Fiber Bundle model

TL;DR

This study uses numerical analysis to explore the finite-size scaling of relaxation times in the Fiber Bundle Model, revealing specific scaling behaviors and critical exponents without any logarithmic dependence in the precritical state.

Contribution

The paper provides detailed finite-size scaling forms and critical exponents for relaxation times in the Fiber Bundle Model with equal load sharing, based on extensive numerical analysis.

Findings

No ln(N) dependence of average relaxation time in precritical state

Critical load approaches asymptotic value as N^{-1/nu} with nu=3/2

Relaxation time scales as N^{eta} at criticality with eta=1/3

Abstract

Using extensive numerical analysis of the Fiber Bundle Model with Equal Load Sharing dynamics we studied the finite-size scaling forms of the relaxation times against the deviations of applied load per fiber from the critical point. Our most crucial result is we have not found any $\ln (N)$ dependence of the average relaxation time $\langle T(\sigma,N) \rangle$ in the precritical state. The other results are: (i) The critical load $\sigma_c(N)$ for the bundle of size $N$ approaches its asymptotic value $\sigma_c(\infty)$ as $\sigma_c(N) = \sigma_c(\infty) + AN^{-1/\nu}$. (ii) Right at the critical point the average relaxation time $\langle T(\sigma_c(N),N) \rangle$ scales with the bundle size $N$ as: $\langle T(\sigma_c(N),N) \rangle \sim N^{\eta}$ and this behavior remains valid within a small window of size $|\Delta \sigma| \sim N^{-\zeta}$ around the critical point. (iii) When $1/N < |\Delta \sigma| < 100N^{-\zeta}$ the finite-size scaling takes the form: $\langle T(\sigma,N) \rangle / N^{\eta} \sim {\cal G}[\{\sigma_c(N)-\sigma\}N^{\zeta}]$ so that in the limit of $N \to \infty$ one has $\langle T(\sigma) \rangle \sim (\sigma - \sigma_c)^{-\tau}$. The high precision of our numerical estimates led us to verify that $\nu = 3/2$, conjecture that $\eta = 1/3$, $\zeta = 2/3$ and therefore $\tau = 1/2$.