# Basic Scattering

• TBD

### Reference Material

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\section{Scattering in the Radiative Transfer Equation}

Scattering is naturally handled in the Radiative Transfer Equation as an emissivity that is proportional to the background radiation field: $$j_\nu=\sigma_\nu J_\nu,$$ where $J_\nu\equiv\frac1{4\pi}\int{I_\nu d\Omega}$ is the isotropic radiation field, and $\sigma_\nu$ is a frequency-dependent cross-section for scattering. We'll focus on this cross-section, which can depend on the size and composition of the scattering material.

\section{ Plane waves through a lens onto a backdrop}

We are considering a case of sending an infinite plane wave through an infinite lens with focal length $f$ onto a backdrop which is distance $f$ away. Next, we consider what happens if we have: (a) an occulting object at the center of the lens, (b) the exact opposite of that object--an aperture at that point. We'll get diffraction patterns on our backdrop in both of these cases. What is interesting is that the sum of the apertures (in our case, an infinite aperture minus a spot plus that spot) should also sum the diffraction patterns. The diffraction pattern for an infinite, unblocked aperture is just a delta function (everything focused at the focal point). Thus the fringe patterns for each of the partially blocked apertures must be the same, but $180^\circ$ out of phase. Of course, the (infinite aperture minus a spot) contains a delta function in addition to its diffraction pattern.\par

\section{ Some special results}

\def\sabs{\sigma_{abs}} \begin{itemize} \item Crystalline dielectrics: $\qabs \propto {a\over \lambda^2}$ (for $\lambda \gg a$. Then: $$\alpha_\nu \eval{abs} = n_d\sabs = n_d\pi a^2\qabs \propto {a^2\over \lambda^2}$$ Now what we measure in the field is $\tau$, the optical thickness. However, $\tau_\nu = \sum{K_\nu}$, where $K_\nu = {\sabs \over m_{particle}} \propto a^0$. This is how the mass of particles can measured. Then $\qscat \sim ({a\over \lambda})^4$, like Rayleigh scattering. \item In general: $j_\nu\eval{emission} = \alpha_\nu \cdot B_\nu(T)$. Then: $$j_\nu\eval{scattering}(\hat n) = n_d(\qscat\pi a^2)\int{I_\nu(\hat n) F(\hat n-\hat n^\prime)d\Omega^\prime)}$$ The term $F(\hat n - \hat n^\prime)$ is called the scattering phase function, and is used to add up all of the incoming paths of light, taking into account their relative phases. $n_d$ is the \# density of scattering grains. For small enough grains ($a\ll \lambda$), F is independent of the properties of the material the grains are made of: $$F = {1+\cos^2\theta\over 2}$$ where $\theta$ is the angle from the propagation direction of the incident beam. This gives us a characteristic dog bone scattering pattern. Small grains are (to within a factor of 2) isotropic scatterers. The factor of 2 is the dog bone pattern. Note that the missing half at right angles to the propagation direction is the portion of incident light which had no polarization component which aligns with the characteristic polarization which must exist for light detected at a right angle from a scattering body. \end{itemize}

\section{ Increasing Grain Size}

We return to the model in which an infinite plane wave passes through an aperture, is focused by an infinite lens, and shines on a wall. In this model, as the slit aperture widens (a increases), then the diffraction pattern narrows. Thus, for larger grain sizes, there is more forward scattering than in other directions. The fitting formula for the power pattern for large grains is: $$F(\theta)={1-g^2\over1+g^2-2g\cos\theta}$$ where $g\equiv\mean{\cos\theta}={\int{F\cos\theta d\Omega}\over \int{Fd\Omega}}$. Note that this formula fails for $\theta=100^\circ$. If $g=1$, then we have isotropic scattering, and as $g\to1$, $F(\theta)$ peaks increasingly in the $\theta=0$ direction (it is increasingly forward throwing).

\section{ Forward Scattering}

Forward scattering can actually increase the intensity of light in some areas over if there were no scattering at all. If particles are smaller that the wavelength of light, there is isotropic scattering. For larger particles, the power is more concentrated in the perfectly forward and backward directions. This is why, where there is fog, you dim your headlights. The larger scattering particles in the fog actually increases the intensity of light reflected back at you and at oncoming cars.

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