# Radiation Lecture 06

<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}} \usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} \def\eb{e^{h\nu \over kT}}

\subsection*{ Flat Blackbody disk:}

$$\begin{aligned}\nu F_\nu&\propto\nu^4\int_{r_i}^{r_0}{2\pi r\,dr\over\eb-1}\\ &\propto \nu^3\int_{r_i}^{r_0}{r\,dr\over e^{h\nu\over kT}({r\over R_*})^{3\over 4}-1}\\ \end{aligned}$$ Define $x \equiv {h\nu\over kT_*}({r\over R_*})^{3/4}$. Then: \def\hnktTemplate:H\nu \over kT * \def\tqTemplate:3\over 4 $$\nu F_\nu \propto \nu^4\int{{\hnkt ({r_i\over R_*})^\tq}^{\hnkt ({r_o\over R_*})^\tq}{x^3dx\over e^x - 1}}$$ We are free to choose $\nu$, so we'll choose it such that: $$\nu_0\sim{kT_*\over h}({R_*\over r_0})^\tq\ll\nu\ll{kT_*\over h} ({R_*\over r_i})^\tq\sim\nu_i$$ Then the integral simplifies to: $$\nu F_\nu \propto \nu^{4\over 3}\int_{x\ll 1}^{x\gg 1}{x^3dx\over e^x-1}$$ This is, on the whole, insensitive to exact values on the bounds of the integral, so: $$\nu F_\nu \propto \nu^{4\over 3}$$ Thus, if we are plotting $\nu F_\nu$ vs. $\nu$ (like on the homework), we see a flattening of the spectrum for the blackbody disk because of the region where $\nu F_\nu \propto \nu^{4\over 3}$.

\subsection*{ 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

\subsection*{ 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*
\def\qabs{Q_{abs}}
\def\qscat{Q_{scat}}
\def\qemis{Q_{emis}}
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

\begin{itemize} \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 $P_{rad}\propto\qemis\propto\qabs$, $$\qabs\propto\left({a\over\lambda}\right)^2$$

\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. Recall that: $$\sigma_{abs} = {4\pi e^2\over m_e c}{w\gamma \over \left[(w_0^2-w^2)^2 + (w\gamma)^2\right]}$$ 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, $\qabs \propto \inv{\lambda^2}$. \end{itemize}

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

\def\qext{Q_{ext}}
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

\subsection*{ 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}

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