# Difference between revisions of "Optical Depth"

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## 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.

## Examples

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.

A useful real world example of optical depth is fog. Within a fog cloud, nearby objects are clearly visible and distant objects are completely obscured. At intermediate distances, objects are difficult but not impossible to discern. In this image, the tree to the left corresponds to an optical depth of approximately 1, since it is mostly obscured, though still visible. The person in the foreground is at an optical depth of less than 1 (perhaps around 0.5), since they are easily discernable and there is very little fog blocking light between the person and the observer. Lastly, the end of the road is at an optical depth much greater than 1, since it is completely covered over by the fog and invisible to the observer.