Topics: Purdue ME606: Equation of Radiative Transfer (RTE): Lecture 11

Purdue ME606: Equation of Radiative Transfer (RTE): Lecture 11

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Equation of Radiation Transfer (RTE)

Sum the contribution from self-emission absorption, at scattering and in scattering for $d I η$ .

First, we can write:

$dI_\eta = \frac{{\partial I_\eta }}{{\partial t}}dt + \frac{{\partial I_\eta }}{{\partial s}}ds$

Devide all former terms by $d s$

$\frac{{dt}}{{ds}}\frac{{\partial I_\eta }}{{\partial t}} + \frac{{\partial I_\eta }}{{\partial s}} = \kappa _\eta I_{b\eta } - \kappa _\eta I_\eta - \sigma _\eta I_\eta + \frac{{\sigma _\eta }}{{4\pi }}\int\limits_{4\pi } {I_\eta \phi _\eta dw_i }$

$\frac{{dI_\eta }}{{ds}} = \kappa _\eta I_{b\eta } - \beta _\eta I_\eta + \frac{{\sigma _\eta }}{{4\pi }}\int\limits_{4\pi } {I_\eta \phi _\eta dw_i }$

$= \hat s \cdot \nabla I_\eta$

This is an integro-differential equation -tough to solve

Another form: introduce scattering albedo

$\equiv w_\eta = \frac{{\sigma _{_\eta } }}{{\chi _{_\eta } + \sigma _{_\eta } }} = \frac{{\sigma _{_\eta } }}{{\beta _{_\eta } }}$

$\frac{1}{{\beta _\eta }}\frac{{dI_\eta }}{{ds}} = - I_\eta + (1 - w_\eta )I_{b\eta } + \frac{{w_\eta }}{{4\pi }}\int\limits_{4\pi } {I_\eta \phi _\eta dw_i }$

$= \frac{{dI_\eta }}{{d\tau _\eta }}$

$τ η$ = optical distance and

$d I η = β η d s$

Last modified on 12 Oct, 2008

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Purdue ME606: Equation of Radiative Transfer (RTE): Lecture 11

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