# Difference between revisions of "Optical Depth"

## Optical Depth

Solving for specific intensity ${\displaystyle I_{\nu }}$ in the Radiative Transport Equation

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

for zero emission (${\displaystyle j_{\nu }=0}$) gives a solution of the form:

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

where ${\displaystyle \tau _{\nu }}$ is the optical depth at ${\displaystyle \nu }$. Optical depth is often computed as:

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

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

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

where ${\displaystyle \Sigma }$ is the mass surface density.

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

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

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

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:

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

Therefore:

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

In fact, the radiative transport equation can be expressed in terms of optical depth. Dividing by ${\displaystyle \alpha _{\nu }}$ and recognizing ${\displaystyle d\tau _{\nu }=ds\alpha _{\nu }}$:

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

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

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

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

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

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

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

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

An example of the Mona Lisa at optical depth of ${\displaystyle \tau =0.1}$, for obscuring particles of various radii. To achieve the same optical depth, particles with a smaller cross-sectional area need to have a higher column density.

The Mona Lisa at various optical depths, illustrating how the transition from optically thin to optically thick erases the background picture.