Topics: Purdue ME606: Emission and In-scattering (Lecture-11)

Purdue ME606: Emission and In-scattering (Lecture-11)

$(d I η) e m = χ η I b η d s$

Ex.

Consider a non-scattering layer that absorbs and emits

Find

$d I η = κ η (I b η - I η) d s$

Integrate:

$I_\eta (s) = I_\eta (o)e^{ - \chi _\eta s} + I_\eta (1 - e^{ - \kappa _\eta s} )$

Radiation impinging on dA from dW

$I_\eta (\hat s)dA(\hat s_i \cdot s)dw_i = E_{\eta i}$

This energy travel through a distance

$\frac{{ds}}{{\hat s_i \cdot s}}$

with in a volume $d\forall = dsdA$

Therefore, the total energy scattered away from $\hat s_i$

$\sigma _\eta E_{\eta i} \frac{{ds}}{{\hat s_i \cdot s}} = \sigma _\eta I_\eta (\hat s_i )dAdw_i ds$

Of this, how much is scattered into direction $\hat s_i$ ?

$\sigma _\eta I_\eta (\hat s_i )dAdw_i ds = \frac{{\phi _n (s_i ,\hat s)}}{{4\pi }}$

where $\phi _\eta \left( {\hat s_i ,\hat s} \right)$ is the scattering phase function

We then integrate overall incoming direction to find

$dI_\eta = ds\frac{{\sigma _\eta }}{{4\pi }}\int\limits_{4\pi } {I_\eta (\hat s_i )\phi _\eta \left( {\hat s_i ,\hat s} \right)dw_i }$

Note that energy conservation requires

$frac{1}{{4\pi }}\int\limits_{4\pi } {\phi _\eta \left( {\hat s_i ,\hat s} \right)dw_i }$

Last modified on 12 Oct, 2008