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Reverse saturable absorption. Rate equations.

by ale_yoman
Tags: absorption, equations, rate, reverse, saturable
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ale_yoman
#1
Sep11-11, 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 long-lived 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] - 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 [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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