## Contents

### 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:

$F(\theta )={1-g^2\over 1+g^2-2g\cos \theta } \,\!$

where $g\equiv \left\langle \cos \theta \right\rangle ={\int {F\cos \theta d\Omega }\over \int {Fd\Omega }}$. Note that this formula fails for $\theta =100^\circ$. If g = 1, then we have isotropic scattering, and as $g\to 1$, F(θ) peaks increasingly in the θ = 0 direction (it is increasingly “forward throwing”).

### 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.

### 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 Rex is given by:

$R_{ex}= n_ An_ B\sigma _{12}f(v_{rel})v_{rel} \,\!$

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:

$R_{ex}= n_ An_ B\int _0^{\infty }f(v)dv\sigma _{12}(v)v \,\!$
$=n_ An_ B \left\langle \sigma _{12}v\right\rangle \,\!$
$=n_ An_ Bq_{12} \,\!$

where q12 is the “collisional rate coefficient” [cm3s − 1]. Then the Rate of de-excitation is given by:

$R_{deex} = n_ An_ B\int _0^\infty {f(v)v\, dv\sigma _{21}(v) = n_ A^*n_ Bq_{21}} \,\!$

We recognize now that $\sigma _{12}(v)n_ An_ Bf(v)dv\, v$ 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 $n_ A^*n_ Bf(v^\prime )v^\prime \sigma _{21}(v^\prime )$.

$\overbrace{\frac12 m_ rv^2}^{center\ of\ mass\ E} = hv_{12} + \frac12 m_ rv^{\prime 2} \,\!$

Where mr is the reduced mass $m_ Am_ B\over m_ A+m_ B$. However many Bslow are created by collisional excitation, the same number are used for the reverse de-excitation. This is detailed balance.

Second, under thermal equilibrium, particles have a Maxwellian velocity distribution:

${f(v)=4\pi \left({m_ r\over 2\pi kT}\right)^{3\over 2}v^2 e^{-m_ rv^2\over 2kT}} \,\!$

(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,

$n_ An_ Bf(v)dv\, v\sigma _{12}=n_ A^*n_ Bf(v^\prime )dv^\prime v^\prime \sigma _{21} \,\!$
$\sigma _{12}v^2e^{-m_ rv^2\over 2kT}= {g_2\over t_1}e^{-h\nu _{21}\over 2kT}\nu ^{\prime 2}e^{-m_ rv^{\prime 2}\over 2kT}v^\prime dv^\prime \sigma _{21} \,\!$
$\sigma _{12}dv\, v^3 e^{-m_ rv^2\over 2kT}= {g_2\over t_1}e^{-h\nu _{21}\over 2kT}e^{-h\nu _{21}\over 2kT}e^{-m_ rv^2\over 2kT}v^{\prime 3} dv^\prime \sigma _{21} \,\!$
${g_1v^2\sigma _{12}(v) = g_2v^{\prime 2}\sigma _{21}(v^\prime )} \,\!$

This is the “Einstein analog”.

For a specific case, B = e, A = ion with bound electron.

• Incident electron has kinetic energy > hν21.
$\frac12 m_ rv^2 \approx \frac12 m_ ev^2 > h\nu _{21} \,\!$
• Coulomb focusing gives ${1 \over v^2}$ cross-section.

We want to know how far away an electron with v can be aimed and still hit the a0 radius cloud around the ion. This is b, the impact parameter. Our collision cross-section = πb2. 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:

$\pi b^2 = {\pi \hbar ^2\over m_ ev^2} \,\!$
$\sigma _{12}= {\pi \hbar ^2\over m_ e^2v^2} \overbrace{\left({\Omega (1,2)\over g_1}\right)}^{quantum\ mechanical\atop correction\ factor} \,\!$

Ω 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.

$q_{12}=\left\langle \sigma _{12}v\right\rangle \propto \left\langle {1 \over v}\right\rangle \propto {1 \over \sqrt {T}} \,\!$
• 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
$\sigma _{12}\sim 10^{-14}cm^2\left({\Omega (1,2)\over g_1}\right) \,\!$
$\sigma _{12}\big |_{osterbrock} \sim 10^{-15} cm^2 \,\!$