Radiative Transfer Equation

Short Topical Videos

• Radiative Transport Equation (Aaron Parsons)

Reference Material

• Radiative Transfer (Condon & Ransom)

1 The Fundamental Equation of Radiative Transfer

The fundamental equation of radiative transfer is governed by emission and extinction. Extinction is brought about by absorption (which changes photon energy) or by scattering (which does not). Examples of scattering are Thomson scattering of light off of cold electrons, Rayleigh scattering in the atmosphere, and Line scattering (reemission in a different direction). An example of absorption is photoionization (where a photon ionizes an atom, say by knocking off an electron).

1.1 Absorption

Let’s say radiation $I_{\nu }$ passes through a region $ds$ of absorption/scattering on its way to us. Then:

$dI_{\nu }=-\alpha _{\nu }I_{\nu }ds\,\!$

where $\alpha _{\nu }$ is the extinction coefficient (units of $cm^{-1}$). We may compute $\alpha _{\nu }$ a couple different ways:

$\alpha _{\nu }=\overbrace {n} ^{\#\ density}\overbrace {\sigma _{\nu }} ^{cross\ section}=\overbrace {\rho } ^{mass\ density}\overbrace {\kappa _{\nu }} ^{opacity}\,\!$

Solving for intensity:

${\begin{aligned}I_{\nu }(s)&=I_{\nu }(0)e^{-n\sigma _{\nu }s}\\&=I_{\nu }(0)e^{-\tau _{\nu }}\\\end{aligned}}\,\!$

where $\tau _{\nu }$ is the optical depth at $\nu$.

Optical depth is often computed as:

$\tau _{\nu }=n\sigma _{\nu }s=N\sigma _{\nu }\,\!$

where $N$, the column density, is in $cm^{-2}$ and is the # of extinguishers per unit area. Similarly,

$\tau _{\nu }=\rho \kappa _{\nu }s=\Sigma \kappa _{\nu }\,\!$

where $\Sigma$ is the mass surface density and $\kappa _{\nu }$ is the density-weighted extinction coefficient.

$\tau _{\nu }{\begin{cases}\ll 1&optically\ thin\\\gg 1&optically\ thick\end{cases}}\,\!$

The Mean Free Path is given by: $\lambda _{mfp,\nu }=\alpha _{\nu }^{-1}={\frac {1}{n\sigma _{\nu }}}={\frac {1}{\rho \kappa _{\nu }}}$. Thus:

$\tau _{\nu }={s \over \lambda _{mfp,\nu }}\,\!$

That is, the optical depth is the number of mean-free-paths deep a medium is. For Poisson processes, the probability of absorption is given by:

$P(n)={e^{-{s \over \lambda _{mfp,\nu }}}\left({s \over \lambda _{mfp,\nu }}\right)^{n} \over n!}\,\!$

Therefore:

$I_{\nu }(s)=I_{\nu }(0)e^{-\alpha _{\nu }s}\,\!$

1.2 Emission

If $j_{\nu }$ is the emissivity, then the contribution of the emissivity of a medium to the flux is:

$dI_{\nu }=j_{\nu }ds\,\!$

Emission and Extinction together:

${{dI_{\nu } \over ds}=j_{\nu }-\alpha _{\nu }I_{\nu }}\,\!$

(Fundamental Equation of Transfer)

It is often convenient to express this in terms of optical depth. Dividing by $\alpha _{\nu }$ and recognizing $d\tau _{\nu }=ds\alpha _{\nu }$:

${\begin{aligned}{dI\nu \over d\tau _{\nu }}&={j_{\nu } \over \alpha _{\nu }}-I_{\nu }\\&=S_{\nu }-I_{\nu }\\\end{aligned}}\,\!$

where $S_{\nu }$ is a “source function”. In general,

$S_{\nu }{\big |}_{scattering}\propto \int {I_{\nu }d\Omega }\,\!$

There is a formal solution for $I_{\nu }$. Let’s define ${\tilde {I}}\equiv Ie^{\tau \nu }$ and ${\tilde {S}}\equiv Se^{\tau \nu }$. Then:

${d{\tilde {I}} \over d\tau _{\nu }}={\tilde {S}}\,\!$ ${\tilde {I}}(\tau _{\nu })={\tilde {I}}(0)+\int _{0}^{\tau _{\nu }}{{\tilde {S}}d{\tilde {\tau }}_{\nu }}\,\!$ ${I_{\nu }(\tau _{\nu })=\overbrace {I_{\nu }(0)e^{-\tau _{\nu }}} ^{atten\ bg\ light}+\overbrace {\int _{0}^{\tau _{\nu }}{S_{\nu }(\tau _{\nu }^{\prime })\underbrace {e^{-(\tau _{\nu }-\tau _{\nu }^{\prime })}} _{self-absorption}d\tau _{\nu }^{\prime }}} ^{glowing\ medium}}\,\!$

If $S_{\nu }$ is constant with $\tau _{\nu }$, then:

$I_{\nu }(\tau _{\nu })=I_{\nu }(0)e^{-\tau _{\nu }}+S_{\nu }(1-e^{-\tau _{\nu }})\,\!$

That second term on the righthand side can be approximated as $S_{\nu }\tau _{\nu }$ for $\tau _{\nu }\ll 1$, since self-absorption is negligible. Similarly, for $\tau _{\nu }\gg 1$, it may be approximated as $S_{\nu }$. The source function $S_{\nu }$ is everything. It has both the absorption and emission coefficients embedded in it.