deletions | additions       diff --git a/Second-order decay.tex b/Second-order decay.tex  index 5b638ce..2dc5f74 100644  --- a/Second-order decay.tex  +++ b/Second-order decay.tex 
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\section{Second-order decay, $A+A\to P$} $A+A\to\textrm{products}$}  These forms of $k(t)$ satisfy the bound $\mathcal{J}-\mathcal{L}^2 = 0$ in the absence of disorder, when $k_i(t)\to\omega$. This is straightforward to show for the case of an $i^{th}$-order reaction, with the traditional integrated rate law  \begin{equation} 
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%\begin{equation}  % \mathcal{L}(\Delta{t})^2=\left[\int_{t_i}^{t_f}k(t)dt\right]^2  %\end{equation}  $\mathcal{L}_2(\Delta{t})\frac{1}{S(t)}\big|_{S_(t_f)}^{S_(t_i)}$, $\mathcal{L}_2(\Delta{t}) = \frac{1}{S(t)}\big|_{S_(t_f)}^{S_(t_i)}$,  which measures the cumulative rate coefficient, same as in first order irreversible decay. As seen in first order, the statistical length is also dependent on the time interval, with the statistical length being infinite in an infinite time interval. In first order irreversible decay, the inequality between the statistical length squared and Fisher divergence determines when a rate coefficient is constant, which is only when the inequality turns into an equality.[cite] A similar inequality is found in second order kinetics.   Putting in the time dependent rate coefficient for a second order irreversible decay, the inequality becomes  \begin{equation}