## 2 A Thermodynamic Model of Post Seismic Relaxation
Using Expression (4), \(d\tau/dE\)= \((\tau_{0}/k_{B}T)\)exp \((E/k_{B}T)\) = \(\tau/k_{B}T\), hence:  
\[f(\tau)=\frac{k_{B}TN(E)}{\tau} \tag{5}\]  
In the case of a uniform distribution of the activation energy, bounded by a minimum \(E_{min}\) and maximum \(E_{max}\) activation energy, \(N(E)\) has a constant value \(N_{o}\), which, according to Equation (5) for a temperature \(T\), leads to a distribution of relaxation times \(f(\tau)=(k_{B}TN_{0})/\tau\), which is proportional to \(1/\tau\). It is evident that the minimum \(\tau_{min}\) and maximum \(\tau_{max}\) relaxation times are related to the corresponding minimum and maximum activation energies by the expressions \(\tau_{min}=\tau_{0}e^{\frac{E_{min}}{k_{B}T}}\) and \(\tau_{max}=\tau_{0}e^{\frac{E_{max}}{k_{B}T}}\) and, thus, \(\Delta E=E_{max}-E_{min}=kT\,\ln(\frac{\tau_{max}}{\tau_{min}})\). The macroscopically observed relaxation is given by the superposition of relaxation processes, expressed as:  
\[R(t)=C\int_{\tau_{min}}^{\tau_{max}}\frac{1}{\tau}e^{-t/\tau}d\tau \tag{6}\]  
where \(C=T\,N_{o}\) is a constant for a given \(T\) and \(N_{o}\), which is omitted in the calculation merging it with the scaling parameter \(S\). We note that \(N_{o}\) is a constant value for a uniform distribution of energy \(N(E)\) between its minimum \(E_{min}\) and maximum \(E_{max}\) value. The weight factor \(1/\tau\) in Equation (6) is a result of the application of Arrhenius' law due to mesoscale relaxation processes operating on different timescales. In Figure 2, we present examples of the relaxation function with time on a logarithmic scale along the horizontal axis with a constant \(\tau_{min}\) and various values of \(\tau_{max}\). In a logarithmic scale, the curve in Figure 2 decreases almost linearly, up to a time \(t\approx\tau_{max}\), and at the end, it flattens out. Thus, the relaxation function \(R(t)\) is proportional to \(\ln(t)\) for \(t<<\tau_{max}\). Based on that, the maximum relaxation time \(\tau_{max}\) could be estimated as the time where the relaxation curve flattens out. The integral in Equation (6) is the exponential integral \(E_{1}(\mathbf{x})\)[38], which can not be solved analytically but behaves like a negative exponential for very large values of the argument and decays on a logarithmic mode in the intermediate range of the relaxation  
Figure 2: The relaxation function \(R(t)\) (see Equation (6) in the text) for \(\tau_{min}\) = 5 days and \(\tau_{max}\) = 200, 400, 600 days. curve \(R(t)\) while deviating from a logarithmic time dependence for \(t<\tau_{min}\). Furthermore, as results from Equation (6) when \(t=0\), \(R(t)\) can be calculated analytically:  
\[R_{0}=R(t=0)=\ln\left(\frac{\tau_{max}}{\tau_{min}}\right) \tag{7}\]  
supporting the physical condition to have a finite relaxation function for \(t=0\). Since \(\Delta E=E_{max}-E_{min}=kT\ln\left(\frac{\tau_{max}}{\tau_{min}}\right)=kTR_{0}\), the finite value of the relaxation function for \(t=0\) is required in order for the energy range \(\Delta E\) to be a finite quantity. Taking the derivative of \(R(t)\) with respect to \(t\), we form the following equation:  
\[\frac{dR}{dt}=\frac{1}{t}\bigg{(}e^{\frac{-t}{\tau_{min}}}-e^{\frac{-t}{\tau_{ max}}}\bigg{)} \tag{8}\]  
which, for \(\tau_{min}\approx\) t\(\ll\)\(\tau_{max}\), expands exponentially and leads to \(\left[\frac{dR(t)}{dt}\right]_{t=0}=-\left(\frac{1}{\tau_{min}}-\frac{1}{\tau_ {max}}\right)\) and  
\[R_{o}=1,\;u_{o}=u_{\infty}+S\;\text{and}\;u(t)=u_{\infty}+S\;e^{\frac{-t}{\tau}} \tag{2}\]0  
In the observed cases where \(\tau_{max}\gg\tau_{min}\), the slope of \(R(t)\) vs. \(t\) at \(t=0\) is well approximated as \(\left[\frac{dR(t)}{dt}\right]_{t=0}=-\frac{1}{\tau_{min}}\), while the ratio of the slope to \(R(0)\) follows the expression: \(\frac{1}{R(t=0)}\left[\frac{dR(t)}{dt}\right]_{t=0}=\frac{1}{\tau_{min}}\ \ln\left(\frac{\tau_{max}}{\tau_{min}}\right)=\frac{R_{o}}{\tau_{min}}\).