Radiation Lecture 08

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<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\qot{q_{12}} \def\qto{q_{21}} \def\ehvkt{e^{-h\nu_{21}\over2kT}} \def\hf{\frac12} \usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document} \subsection*{ More on Einstein Analog } Forgot something for the Einstein Analog. Recall: $$g_1v^2\sigma_{12}(v)=g_2v^{\prime2}\sigma_{21}(v^\prime)$$ $$\hf m_4v^2=h\nu_{21}+\hf m_r(v^\prime)^2$$ For the special case of $f(v)$ being Maxwellian, then: $${q_{12}\over q_{21}}\equiv{\mean{\sigma_{12}v}\over\mean{\sigma_{21}v}} ={g_2\over g_1}e^{-h\nu_{21}\over kT}$$ This has a Boltzmann factor, which makes you thing we're assuming LTE, but we're not. \par Last time, we were talking about electron-ion collisional excitation, given by: $$\qot\equiv\mean{\sigot v}\propto\inv{v}\propto T^{-\hf}$$ We can extend this for neutral-ion collisional excitation: $$\qot \propto T^\hf T^{-\hf} \propto T^0$$ $$\sigot\equiv{\pi\hbar^2\over m_ev^2}{\Omega(1,2)\over g}$$ where $\Omega(1,2)$ scales as $T^\hf$. Notice this means that for some neutral-ion collisional excitation, it is {\it temperature independent}. For neutral-neutral collisional excitation: $$\qot=\mean{\sigot v}\propto T^\hf$$ $$\sigot\propto T^0\sim \pi(Fa_0)^2$$ This is all we'll talk about bound-bound transitions.

\subsection*{ Bound-Free Transitions (Photoionization)}

\def\sigbf{\sigma_{bf}} We'll calculate the cross-section of a bound-free transition $\sigbf$: $$\sigbf\sim{\lambda^2\over 8\pi}{A_{21}\over \Delta\nu}$$ It turns out that $\Delta\nu$ is about $\nu$. $\lambda\approx 912\AA$, so scaling from Lyman-alpha: $$\sigbf\sim{(912\AA)^3\over c\cdot8\pi}A_{21,Ly\alpha} \left({1216\AA\over912\AA}\right)^3\sim 10^{-18}cm^2$$ It turns out that the real answer is $\sigbf\sim 6\cdot 10^{-18}cm^2$. In general: $$\sigbf=\sigma\eval{edge}\left({E_{photon\,in}\over E_{edge}}\right)^{-3}$$ That exponent (-3) is actually $-\frac83$ near the edge and goes to $-\frac72$ far from it. So you see $\sigbf$ spike up as the photon reaches the ionization energy, and then decrease exponentially as energy increases. However, you can see new spikes from ionizing electrons in inner shells.

\subsection*{ Radiative Recombination}

\def\sigfb{\sigma_{fb}} This is the inverse process of photoionization, so $\sigfb$ is the cross-section for an ion recapturing its electron and emitting a photon. We'll relate $\sigfb$ to $\sigbf$. This is called the Milne Relation. In this derivation, we'll start by assuming complete thermal equilibrium and derive a result which will end up being independent of thermal equilibrium. Let's start calculating the rate of radiative recombinations. Thermal equilibrium dictates that this must equal the rate of photoionization. For radiative recombination: $$rate\ of\ recombination =n_+n_e\sigfb(v)v[f(v)dv]={\#\ of\ recombinations\over volume\ time}$$ We'll set this equal to the rate of photoionization. This rate is: $$rate\ of\ photoionization ={B_\nu4\pi d\nu\over h\nu}n_0\sigbf \overbrace{\left(1-{g_0\over g_+}{n_+\over n_0}\right)}^{{correction\ for\atop stimulated}\atop recombination}$$ where $n_0$ is the \# density of neutrals. Note this has units of \# flux. Note also that $\sigbf$ depends on $\nu$, and $n_+\over n_0$ is evaluated at the relative velocity $v$ such that $h\nu=\hf m_ev^2+\chi$, where $\chi$ is the threshold ionization energy. In thermal equilibrium, we know that: $${n_+\over n_0}={g_+\over g_0}e^{-E\over kT}$$ where $E$ is the energy difference between state 1 (proton + unbound $e^-$) and state 2 (bound proton/electron pair). Thus $E$ is given by: $$E=\hf m_ev^2-(-\chi)=h\nu$$ So we can make our n's and g's go away. For $f(v)$, we'll use our Maxwellian: $$f(v)=4\pi\left({m_e\over 2\pi kT}\right)^{3\over 2}v^2e^{-m_ev^2\over 2kT}$$ Finally, Saha tells us in thermal equilibrium: $${n_+n_e\over n_0}=\left[{2\pi m_ekT\over h^2}\right]^{3\over 2}{2g_+\over g_n} e^{-\chi\over kT}$$ So now we're essentially done: $$1={n_+n_ef(v)v\sigfb(v)dv\over{4\pi B_\nu\over h\nu}d\nu(1-\ehvkt) n_0\sigbf(v)}$$ and plugging in all of our relations we get: $$\boxed{{\sigfb(v)\over\sigbf(\nu)}= {g_0\over g_+}\left({h\nu\over m_ecv}\right)^2}$$ \centerline{(Milne Relation)} Notice how all of the T's vanished. This result is independent of thermal equilibrium. However, you still have to pay attention to your statistical weights (g's).


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