Difference between revisions of "Basic Scattering"

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where $J_\nu\equiv\frac1{4\pi}\int{I_\nu d\Omega}$ is the isotropic radiation field, and $\sigma_\nu$ is a frequency-dependent
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.
cross-section for scattering.  We'll focus on this cross-section, which can depend on the size and composition of the scattering material.
\section{Geometric Interaction of Radiation with Grains: Babinet's Principle}
In general, we will be talking about a plane wave of wavelength $\lambda$
which is incident upon a particle (grain) of radius a.  The cross-section
for absorption, scattering, and emission will all be proportional to the
physical cross-section of the grain, with some {\it absorption efficiency}
$$\sigma_i=Q_i\pi a^2$$
Kirchkoff's Law requires that $Q_{emit}=Q_{abs}$.  If $a\gg\lambda$, then
we have the geometric optics limit, and $\qscat+\qabs\sim1$.  In fact,
Babinet's Principle says:
The proof of this goes as follows: suppose we have an infinite plane wave
focused by an infinite lens onto a point on a wall.  The power pattern
should be a delta function at that point on the wall.  Now suppose you
place a screen with an aperture of diameter $a$ between the plane wave
and the lens.  We should now see an interference pattern on the wall.  Call
the power incident on a point on the wall $P_1$.  Now put in a new aperture
which is the exact compliment of our previous one (it is a scattering
body of diameter $a$), we should see a new power at our point on the wall:
However, the sum of waves incident on the wall under these two apertures
should be the same as the wave from the sum of the apertures, which was
a delta function.  Thus:
The first aperture (the slit) represented the scattering power from
diffraction alone $(\qscat=1)$, and the second represented the absorbed
power.  The sum of these two powers, for everywhere but the true focus
point, is actually twice the incident power, and since $P_1=P_2$, we
Note that if $\qscat>1$, then the object is ``shiny'': more light gets
diffracted and less gets absorbed.  Thus, although $\qscat$ changes,
the sum remains the same.
\section{ Plane waves through a lens onto a backdrop}
\section{ Plane waves through a lens onto a backdrop}
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to its diffraction pattern.\par
to its diffraction pattern.\par
\section{ Small Grains}
Now let's consider the case that this spot (a small object obstructing a
plane wave) is small, so that $a \ll \lambda$.  This is called the ``Rayleigh
Limit''.  In this case, $Q_{scat} \ll 1$ and $Q_{em} = Q_{abs} \ll 1$.  If
we were to graph $\qabs$ vs. $\lambda$, we'd get something flat ($= 1$) out to
$2\pi a$, after which, $\qabs$ would decrease as $(\inv{\lambda})^\beta$, where
$\beta\eval{abs} = [1-3]$.  On the other hand, graphing $\qscat$ vs. $\lambda$
, we get that after $2\pi a$, $\qscat$ drops as $(\inv{\lambda})^4$.
Why does $Q$ decrease as $a\over \lambda$?\par
\item  Let's consider the case of emitting with a dipole antenna.  For a
region close to the antenna (the near zone, of order $\lambda$),
$P_{rad}\propto E^2$, and at the edge of the near zone,
$E\propto{a\over\lambda}$.  Since $P_{rad}\propto\qemis\propto\qabs$,
\item  Here's another crude way of looking at it:
if you're in a boat that's small compared
to the size of a wave coming at it, you aren't going to do anything to that
wave (you won't scatter, everything transmits).
\item  Another way of looking at it:
the damped simple harmonic oscillator.  In the context of dispersion relations, we wrote down the equation of motion of an electron in the
presence of a plane wave:
$$\ddot x+\gamma\dot x+\wz^2x={eE_0\eikrwt\over m_e}$$
where $\gamma$ is our dampening factor.  This equation had a steady-state solution:
$$x={-e\over m_e}{\left[(\wz^2-\omega^2)+i\omega\gamma\right]\over
The imaginary component of this solution is the absorptive (dissipative) part.  We can construct a
cross-section from this:
$$\sigma_{abs} = {4\pi e^2\over m_e c}{w\gamma \over \left[(w_0^2-w^2)^2 +
If $w \ll w_0$, then:
$$\sigma_{abs} \to {4\pi e^2\over m_e c}{w^2\gamma \over w_0^4} \propto w^2$$
Thus, $\qabs \propto \inv{\lambda^2}$.
Finally, there is Mie Theory, which states that if $\eta = n +ik$, where
$\eta$ is the complex index of refraction, then you can express $\qabs$ and
$\qscat$ in terms of the size parameter $x \equiv {2\pi a\over \lambda}$.
For this derivation that Mie did, $n, k$ are called ``optical constants''.  This
is a bad name, because both of these are actually functions of $\lambda$.\par
Now we look at the handout Eugene gave us (Chiang et al, 2001, ApJ, 547, 1077).
Notice on the $\qabs$ vs. $\lambda$ plots, in addition to the decay we
described, there are ``bumps and wiggles''.  These are characteristic of the
resonances of the systems (e.g. they leftrightarrowond to a rotational mode, etc.).
The varying lines on each graph are for different sized particles.  Notice the
characteristic Rayleigh Limit for small particles. \par
There's another graph about how extinction $\qext = \qabs + \qscat$ goes as
the size parameter is changed.  There is a large ``hump'' where $\qext$ goes to
about 4.  It can do this (we'd said it couldn't go above 2) because we are not
yet in the geometric optic limit.  Notice that on this big hump, there is a
region where increased wavelength causes increased extinction.  This is the
process which causes ``blue moons''.  If the size of particles in the atmosphere
is just right, it will pass blue wavelengths while extinguishing red.\par
\section{ Some special results}
\section{ Some special results}

Revision as of 11:36, 3 September 2015

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Reference Material

<latex> \documentclass[11pt]{article} \def\inv#1Template:1 \over \def\ddtTemplate:D \over dt \def\mean#1{\left\langle #1\right\rangle} \def\sigot{\sigma_{12}} \def\sigto{\sigma_{21}} \def\eval#1{\big|_{#1}} \def\tr{\nabla} \def\dce{\vec\tr\times\vec E} \def\dcb{\vec\tr\times\vec B} \def\wz{\omega_0} \def\ef{\vec E} \def\ato{{A_{21}}} \def\bto{{B_{21}}} \def\bot{{B_{12}}} \def\bfieldTemplate:\vec B \def\apTemplate:A^\prime \def\xp{{x^{\prime}}} \def\yp{{y^{\prime}}} \def\zp{{z^{\prime}}} \def\tp{{t^{\prime}}} \def\upxTemplate:U x^\prime \def\upyTemplate:U y^\prime \def\e#1{\cdot10^{#1}} \def\eikrwt{e^{i(\vec k\vec r-wt)}}

\usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} \def\lya{Ly\alpha}

\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: \begin{equation} j_\nu=\sigma_\nu J_\nu, \end{equation} 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.

\end{document} <\latex>