<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} \subsection*{ 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).

\subsection*{ 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.

\subsection*{ Collisional Excitation Cross-sections}

The Einstein analog: $$\overbrace{A}^{low\ E\ particle} + \overbrace{B_{fast}}^{high\ E\ e^-} \to \overbrace{A^*}^{excited} + B_{slow}$$ So the Rate of Excitations $R_{ex}$ is given by:

\def\rex{R_{ex}} \def\vrel{v_{rel}} $$\rex = n_An_B\sigma_{12}f(\vrel)\vrel$$ Suppose we have some distribution of relative velocities given by $f(v)dv$, where $f$ is the fraction of collisions occurring with relative velocities $[\vrel, \vrel +dv]$. Then: $$\rex = n_An_B\int_0^{\infty}f(v)dv\sigot(v)v$$ $$=n_An_B \mean{\sigot v}$$ \def\qot{q_{12}} \def\qto{q_{21}} $$=n_An_B\qot$$ where $\qot$ is the collisional rate coefficient $[cm^3s^{-1}]$. Then the Rate of de-excitation is given by: $$R_{deex} = n_An_B\int_0^\infty{f(v)v\,dv\sigma_{21}(v) = n_A^*n_B\qto}$$ We recognize now that $\sigot(v)n_An_Bf(v)dv\,v$ is the rate of excitations of A using B moving at relative velocity $v$. If we have detailed balance, then this has to be the same as the rate of de-excitation $n_A^*n_Bf(v^\prime)v^\prime\sigto(v^\prime)$. $$\overbrace{\frac12 m_rv^2}^{center\ of\ mass\ E} = hv_{12} + \frac12 m_rv^{\prime2}$$ Where $m_r$ is the reduced mass $m_Am_B\over m_A+m_B$. However many $B_{slow}$ are created by collisional excitation, the same number are used for the reverse de-excitation. This is {\bf detailed balance}.\par Second, under thermal equilibrium, particles have a {\bf Maxwellian} velocity distribution: $$\boxed{f(v)=4\pi\left({m_r\over 2\pi kT}\right)^{3\over 2}v^2 e^{-m_rv^2\over 2kT}}$$ \centerline{(Maxwellian velocity distribution)} In thermal equilibrium, $${n_A^*\over n_A} = {g_2\over g_1}e^{-h\nu_{21}\over kT}$$ Now, assuming detailed balance and thermal equilibrium, $$n_An_Bf(v)dv\,v\sigot=n_A^*n_Bf(v^\prime)dv^\prime v^\prime \sigto$$ \def\emvkt{e^{-m_rv^2\over2kT}} \def\emvpkt{e^{-m_rv^{\prime2}\over2kT}} \def\ehvkt{e^{-h\nu_{21}\over2kT}} $$\sigot v^2\emvkt = {g_2\over t_1}\ehvkt \nu^{\prime2}\emvpkt v^\prime dv^\prime\sigto$$ $$\sigot dv\,v^3 \emvkt = {g_2\over t_1}\ehvkt \ehvkt \emvkt v^{\prime3} dv^\prime \sigto$$ $$\boxed{g_1v^2\sigot(v) = g_2v^{\prime2}\sigto(v^\prime)}$$ This is the Einstein analog.\par For a specific case, $B=e^-$, $A=$ion with bound electron. \begin{itemize} \item Incident electron has kinetic energy $>h\nu_{21}$. $$\frac12 m_rv^2 \approx \frac12 m_ev^2 > h\nu_{21}$$ \item Coulomb focusing gives $\inv{v^2}$ cross-section. \end{itemize} We want to know how far away an electron with $v$ can be aimed and still hit the $a_0$ radius cloud around the ion. This is $b$, the {\bf impact parameter}. Our collision cross-section $=\pi b^2$. Our angular momentum is conserved, so $$m_ev\,b=m_ev_fa_0$$ We know that $v_f^2 = v^2 + v_\perp^2$, where $v_\perp$ is the velocity $\perp$ to the original electron velocity. This is a result of it falling toward the ion. Then: $$\frac12 m_ev_\perp^2 \sim {Ze^2\over a_0}$$ $$v_f^2 = v^2 + {Ze^2\over m_ea_0}$$ $$b={a_0v_f\over v}$$ $$\pi b^2 = {\pi a_0^2\over v^2}\left[v^2+{Ze^2\over m_ea_0}\right]$$ $$= \pi a_0^2\left[1+\underbrace{Ze^2\over m_ev^2a_0}_{Coulomb\ focusing\atop factor}\right]$$ Generally, the Coulomb focusing factor $>1$ because we want to excite, not ionize. $a_0={\hbar^2\over Ze^2m_e}$, so: $$\pi b^2 = {\pi\hbar^2\over m_ev^2}$$ $$\sigot = {\pi\hbar^2\over m_e^2v^2} \overbrace{\left({\Omega(1,2)\over g_1}\right)}^{quantum\ mechanical\atop correction\ factor}$$ $\Omega$ is the collisional strength, and generally is 0 below the $v$ threshold, goes to 1 at the threshold, and decreases for increasing $v$, with some occasional spikes. Generally, it is of order 1, with some slight temperature dependency. $$\qot=\mean{\sigot v} \propto \mean{\inv{v}} \propto \inv{\sqrt{T}}$$ \begin{itemize} \item 2000 K gas. $v_{term} \sim \sqrt{\gamma kT\over m}$, so $v\sim \sqrt{2000\over 100}42\cdot 1{km\over s} \approx 160{km\over s}$. Then $$\sigot \sim 10^{-14}cm^2\left({\Omega(1,2)\over g_1}\right)$$ $$\sigot\eval{osterbrock} \sim 10^{-15} cm^2$$ \end{itemize}

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