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Reverse saturable absorption. Rate equations. 
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#1
Sep1111, 02:53 PM

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 longlived strongly absorbing excited state (T0) which is rapidly populated from a ground state (S0), are given as: [itex]\frac{dS_{0}}{dt} =  \sigma_{01} S_{0} \phi + k_{30}T_{0} [/itex] [itex]\frac{dT_{0}}{dt} = \sigma_{01} S_{0} \phi  k_{30}T_{0} [/itex] [itex]\frac{d \phi}{dz} =  \sigma_{01} S_{0} \phi  \sigma_{34} T_{0} \phi [/itex] [itex]\phi[/itex] is the photon flux, [itex]\sigma_{ij} [/itex]  ijtransition crosssection. Thу last equation describes the light absorption in the sample depth. To solve these equations numerically with RKmethod it is necessary to reduce the last one to the form [itex]\frac{d \phi}{dt}[/itex]. In the literature i met the change of variable which led to [itex]\frac{1}{c}\frac{d \phi}{dt}[/itex]. Why is it correct? What is about refraction in the sample? 


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