Page updated: October 26, 2020
Author: Curtis Mobley
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# The Volume Scattering Function

We now take into account the angular distribution of the scattered power, with two assumptions. We ﬁrst assume that the medium is isotropic, i.e., its inﬂuence on light is the same in all directions at a given point. This is a reasonable assumption for natural waters in which the particles are randomly oriented by turbulence. We also assume that the light is unpolarized. If these two assumptions are true, then the scattering process is azimuthally symmetric. This means that the scattering depends only on the scattering angle $\psi$, which is measured from the direction of the unscattered beam as shown in Fig. 1. Clearly 0 $\le \psi \le \pi$.

### Conceptual Deﬁnition of the Volume Scattering Function

With these two assumptions, let $\Delta B\left(\psi ,\lambda \right)$ be the fraction of incident power scattered out of the beam through an angle $\psi$ into a solid angle $\Delta \Omega$ centered on $\psi$, as $\Delta r\to 0$ and $\Delta \Omega \to 0$, as shown in the ﬁgure. The solid angle $\Delta \Omega$ now includes all directions within the two red rings shown in the ﬁgure, corresponding to all directions between scattering angles $\psi$ and $\psi +\Delta \psi$. The angular scatterance per unit distance and unit solid angle, $\beta \left(\psi ,\lambda \right)$, is then

$\begin{array}{llll}\hfill \beta \left(\psi ,\lambda \right)\equiv & \underset{\Delta r\to 0}{lim}\underset{\Delta \Omega \to 0}{lim}\phantom{\rule{2.6108pt}{0ex}}\frac{\Delta B\left(\psi ,\lambda \right)}{\Delta r\phantom{\rule{0.3em}{0ex}}\Delta \Omega }\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill =& \underset{\Delta r\to 0}{lim}\underset{\Delta \Omega \to 0}{lim}\phantom{\rule{2.6108pt}{0ex}}\frac{\Delta {\Phi }_{\text{s}}\left(\psi ,\lambda \right)}{{\Phi }_{\text{i}}\left(\lambda \right)\phantom{\rule{0.3em}{0ex}}\Delta r\phantom{\rule{0.3em}{0ex}}\Delta \Omega }\phantom{\rule{2em}{0ex}}\left({m}^{-1}\phantom{\rule{2.6108pt}{0ex}}s{r}^{-1}\right)\phantom{\rule{2.6108pt}{0ex}}.\phantom{\rule{2em}{0ex}}& \hfill \text{(1)}\end{array}$

Recalling the deﬁnition of spectral radiant intensity, the spectral power scattered into the given solid angle $\Delta \Omega$ is just the intensity scattered into direction $\psi$ times the solid angle: $\Delta {\Phi }_{\text{s}}\left(\psi ,\lambda \right)$ = ${I}_{\text{s}}\left(\psi ,\lambda \right)\Delta \Omega$. Moreover, if the incident power ${\Phi }_{\text{i}}\left(\lambda \right)$ falls on an area $\Delta A$, then the corresponding incident irradiance is ${E}_{\text{i}}\left(\lambda \right)$ = ${\Phi }_{\text{i}}\left(\lambda \right)∕\Delta A$. Noting that $\Delta V=\Delta r\Delta A$ is the volume of water that is illuminated by the incident beam gives

 $\beta \left(\psi ,\lambda \right)=\underset{\Delta V\to 0}{lim}\phantom{\rule{2.6108pt}{0ex}}\frac{{I}_{\text{s}}\left(\psi ,\lambda \right)}{{E}_{\text{i}}\left(\lambda \right)\phantom{\rule{0.3em}{0ex}}\Delta V}\phantom{\rule{2.6108pt}{0ex}}.$ (2)

This form of $\beta \left(\psi ,\lambda \right)$ suggests the name volume scattering function (commonly abbreviated as VSF) and the physical interpretation of scattered intensity per unit incident irradiance per unit volume of water. In the language of a physicist, the VSF also can be interpreted as the diﬀerential scattering cross section per unit volume.

### Other Measures of Scattering

Integrating $\beta \left(\psi ,\lambda \right)$ over all directions (solid angles) gives the total scattered power per unit incident irradiance and unit volume of water, in other words the scattering coeﬃcient:

 $b\left(\lambda \right)={\int }_{\Xi }\phantom{\rule{0.3em}{0ex}}\beta \left(\psi ,\lambda \right)\phantom{\rule{2.6108pt}{0ex}}d\Omega =2\pi {\int }_{0}^{\pi }\phantom{\rule{0.3em}{0ex}}\beta \left(\psi ,\lambda \right)\phantom{\rule{0.3em}{0ex}}sin\psi \phantom{\rule{0.3em}{0ex}}d\psi \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}.$ (3)

The $2\pi$ in the last equation follows from our assumption that the scattering is azimuthally symmetric about the incident direction. This integration is often divided into forward scattering, 0 $\le \psi <\pi ∕2$, and backward scattering, $\pi ∕2\le \psi \le \pi$, parts. The corresponding forward and backward scattering coeﬃcients are, respectively,

 ${b}_{\text{f}}\left(\lambda \right)\equiv 2\pi {\int }_{0}^{\pi ∕2}\phantom{\rule{0.3em}{0ex}}\beta \left(\psi ,\lambda \right)\phantom{\rule{0.3em}{0ex}}sin\psi \phantom{\rule{0.3em}{0ex}}d\psi$ (4)
 ${b}_{\text{b}}\left(\lambda \right)\equiv 2\pi {\int }_{\pi ∕2}^{\pi }\phantom{\rule{0.3em}{0ex}}\beta \left(\psi ,\lambda \right)\phantom{\rule{0.3em}{0ex}}sin\psi \phantom{\rule{0.3em}{0ex}}d\psi \phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}.$ (5)

The backscatter fraction, deﬁned by

 ${B}_{b}\left(\lambda \right)\equiv \frac{{b}_{\text{b}}\left(\lambda \right)}{b\left(\lambda \right)}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}},$ (6)

gives the fraction of scattered light that is deﬂected through scattering angles greater than 90 deg. This quantity is fundamental to remote sensing because most of the light leaving the ocean in an upward direction comes from sunlight that was originally heading downward, but which was backscattered into upward directions.

The volume scattering phase function, $\stackrel{̃}{\beta }\left(\psi ,\lambda \right)$, is deﬁned by

 $\stackrel{̃}{\beta }\left(\psi ,\lambda \right)\equiv \frac{\beta \left(\psi ,\lambda \right)}{b\left(\lambda \right)}\phantom{\rule{2em}{0ex}}\left(s{r}^{-1}\right)\phantom{\rule{2.6108pt}{0ex}}.$ (7)

Writing the VSF $\beta \left(\psi ,\lambda \right)$ as the product of the scattering coeﬃcient $b\left(\lambda \right)$ times the phase function $\stackrel{̃}{\beta }\left(\psi ,\lambda \right)$ partitions $\beta \left(\psi ,\lambda \right)$ into a factor giving the magnitude of the total scattering, $b\left(\lambda \right)$ with units of ${m}^{-1}$, and a factor giving the angular distribution of the scattered light, $\stackrel{̃}{\beta }\left(\psi ,\lambda \right)$, with units of ${sr}^{-1}$. Combining Eqs. (3) and (7) gives the normalization condition for the phase function:

 $2\pi {\int }_{0}^{\pi }\phantom{\rule{0.3em}{0ex}}\stackrel{̃}{\beta }\left(\psi ,\lambda \right)\phantom{\rule{0.3em}{0ex}}sin\psi \phantom{\rule{0.3em}{0ex}}d\psi =1\phantom{\rule{2.6108pt}{0ex}}.$ (8)

This normalization implies that the backscatter fraction can be computed from

 ${B}_{b}\left(\lambda \right)\equiv 2\pi {\int }_{\pi ∕2}^{\pi }\phantom{\rule{0.3em}{0ex}}\stackrel{̃}{\beta }\left(\psi ,\lambda \right)\phantom{\rule{0.3em}{0ex}}sin\psi \phantom{\rule{0.3em}{0ex}}d\psi \phantom{\rule{2.6108pt}{0ex}}.$ (9)

Comment: The term “phase function” has its historical origins in astronomy, where the phase angle is the angle between the directions of the light incident onto and reﬂected from an object. In the case of sunlight being reﬂected by the moon and seen on earth, this angle depends on the phase of the moon—hence the name phase function for the function that gives the phase angle as a function of time. In our terminology, the phase angle is $\pi$ minus the scattering angle. The phase function $\stackrel{̃}{\beta }\left(\psi ,\lambda \right)$ deﬁned here has nothing whatsoever to do with the phase of an electromagnetic wave.

The asymmetry parameter g or mean cosine of the phase function is average over all scattering directions of the cosine of the scattering angle $\psi$, namely

 $g=⟨cos\psi ⟩\equiv 2\pi {\int }_{0}^{\pi }\phantom{\rule{0.3em}{0ex}}\stackrel{̃}{\beta }\left(\psi \right)\phantom{\rule{0.3em}{0ex}}cos\psi \phantom{\rule{0.3em}{0ex}}sin\psi \phantom{\rule{0.3em}{0ex}}d\psi \phantom{\rule{2.6108pt}{0ex}}.$ (10)

The asymmetry parameter is a convenient measure of the “shape” of the phase function. For example, if $\stackrel{̃}{\beta }\left(\psi \right)$ is very large for small $\psi$, then $g$ is near one. If $\stackrel{̃}{\beta }\left(\psi \right)$ is symmetric about $\psi$ = 90, then $g=0$. Typical ocean waters have $g$ values in the range of 0.8 to 0.95.

Comment: We have assumed above that the scattering is azimuthally symmetric, so that the directional pattern or angular shape of the VSF depends only on the scattering angle $\psi$. This is not the case for polarized light, even if the medium is isotropic. Thus a linearly polarized laser beam will scatter diﬀerently for diﬀerent azimuthal directions (measured relative to the plane of polarization), even if the medium is isotropic. If the medium contains non-spherical particles that are not randomly oriented, even unpolarized light will scatter diﬀerently for diﬀerent azimuthal directions. This is sometimes the case, for example, in cirrus clouds and ice fogs, which can have non-spherical ice crystals that become oriented in a particular way as the crystals fall through a calm atmosphere. The atmosphere is then an optically anisotropic medium, and scattering is not azimuthally symmetric. Scattering of sunlight in such an atmosphere gives rise to phenomena such as “sun dogs” or parhelia, which are bright spots to either side of the sun.