# Reverse saturable absorption. Rate equations.

by ale_yoman
Tags: absorption, equations, rate, reverse, saturable
 P: 2 Hello to all. Nice forum. Really like it. I hopee you'll can help me. The rate equaition in the case of reverse saturable absorber, with a long-lived strongly absorbing excited state (T0) which is rapidly populated from a ground state (S0), are given as: $\frac{dS_{0}}{dt} = - \sigma_{01} S_{0} \phi + k_{30}T_{0}$ $\frac{dT_{0}}{dt} = \sigma_{01} S_{0} \phi - k_{30}T_{0}$ $\frac{d \phi}{dz} = - \sigma_{01} S_{0} \phi - \sigma_{34} T_{0} \phi$ $\phi$ is the photon flux, $\sigma_{ij}$ - ij-transition cross-section. Thу last equation describes the light absorption in the sample depth. To solve these equations numerically with RK-method it is necessary to reduce the last one to the form $\frac{d \phi}{dt}$. In the literature i met the change of variable which led to $\frac{1}{c}\frac{d \phi}{dt}$. Why is it correct? What is about refraction in the sample?

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