Radiation Lecture 26

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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\hf{\frac12} \def\^{\hat } \def\.{\dot } \usepackage{fullpage} \usepackage{amsmath} \usepackage{eufrak} \begin{document}

\subsection*{ The Saturated Mode }

Recall our masing equation: $${d\mathfrak{I}\over ds}={\beta(\mathfrak{I}+\hf)\over\beta+\mathfrak{J}}+S$$ In the unsaturated case, ${\beta\over\beta+\mathfrak{J}}\to1\iff\beta\gg\mathfrak{J}$. This is equivalent to saying: $$\hf\left(1+{\Gamma\over A}\right)\gg{BJ\over A}$$ And if $\Gamma\gg A$, then: $$A+\Gamma\gg BJ\iff\Gamma\gg BJ$$ In the saturated case, then $\beta\ll\mathfrak{J}$. In this case: $${d\mathfrak{I}\over ds}={\beta\mathfrak{I}\over\mathfrak{J}}+S={4\pi\beta\mathfrak{I}\over \int{\mathfrak{I}ds}}+S$$ where we used that $\mathfrak{J}\sim\inv{4\pi}\int{\mathfrak{I}d\Omega}$. If we consider a single beam of photons through a cloud, then $\int{\mathfrak{I}d\Omega}\approx \mathfrak{I}\Delta\Omega$, so: $$\boxed{ \begin{aligned}{d\mathfrak{I}\over ds}&={4\pi\beta\mathfrak{I}\over\mathfrak{I}\Delta\Omega}+S\\ &={4\pi\beta\over\Delta\Omega}+S\\ \end{aligned}}$$

\subsection*{ How We Get Population Inversions }

There is another dichotomy in masers: ones which are excited radiatively and those which are excited collisionally. We'll discuss a cloud of molecules which have 3 energy states which are populated by {\it simple collisional pumping}. To simplify our lives, we'll say that energy levels 3 and 2 can talk to each other via photon emission/absorption, as can 2 and 1, but we'll say that 1 and 3 cannot talk radiatively (say, because of parity violation, and note that they can still talk collisionally). 2 and 1 will be our masing levels. Then the rate of change of the population of energy state 1 is given by (sources - sinks): $${dn_1\over dt}=n_2\ato+n_2C_{21}+n_3C_{31}+n_2\bto J_{21}-n_1\bot J_{12} -n_1C_{12}-n_1C_{13}$$ Then $J_{12}=J_{21}$, and because of local collisional thermal equilibrium, $C_{12}=C_{21}{g_2\over g_1}e^{-{E_{21}\over kT}}$. Similarly, $C_{13}=C_{31}{g_3\over g_1}e^{-{E_{21}\over kT}}$. So dividing by $g$, and defining $N_1\equiv{n_1\over g_1}$, we have: $$\begin{aligned}{dN_1\over dt}=&\underbrace{n_2\over g_2}_{N_2}\underbrace{ {\ato\over g_1}g_2}_{\ato}+\underbrace{n_2\over g_2}_{N_2}\underbrace{ {C_{21}\over g_1}g_2}_{\equiv C}+\underbrace{n_2\over g_3}_{N_3}\underbrace{ {C_{31}\over g_1}g_3}_{\equiv C}+\underbrace{n_2\over g_2}_{N_2}\underbrace{ {\bto\over g_1}g_2}_{\bto}J_{21}\\ &-N_1\underbrace{\bto}_{\bto{g_2\over g_1}\equiv \bto}-N_1\underbrace{C_{21}{g_2\over g_1}}_{\equiv C}e^{-{E_{21}\over kt}}- N_1\underbrace{C_{31}{g_3\over g_1}}_{\equiv C}e^{-{E_{31}\over kT}}\\ \end{aligned}$$ Phew. Notice that we set $C_{21}=C_{31}$. This is just to make our lives easier. We can do the same for $dN_2\over dt$, but omitting $B_{23}$, because we're deciding not to have absorptions from $2\to3$ and no stimulated emission from $3\to2$. Then our total population is $N=N_1+N_2+N_3$. Without being careful, instinct tells us that in steady state, we'll have a population inversion ${N_2\over N_1}>1$ if $C\ll A_{32}$. This instinct is correct, but let's do this carefully. First we'll make some assumptions: \begin{itemize}\item $\ato\ll C$. For $H_2O$: $$\ato\sim10^8s^{-1}\left({1216\AA\over1.35cm}\right)^3 \sim10^8\e{-15}s^{-1} \sim10^{-7}s^{-1}\left({\mu\over ea_0}\right)^2$$ where $\mu$ is our way of accommodating the fact that the dipole moment for $H_2O$ might not be the same as for the fine structure of hydrogen. It turns out the answer is $\ato\sim2\e{-9}s^{-1}$.\par Estimating C: $$C\sim n_{H_2}\sigma v_{rel}\sim n_{H_2}\e{-15}\cdot(3{km\over s}) \sim n_{H_2}\cdot3\e{-10}$$ which is $\gg10^{-7}s^{-1}\left({\mu\over ea_0}\right)^2$ when $n_{H_2}\gg10^3cm^{-3}\left({\mu\over ea_0}\right)^2$. \item Next we'll assume $E_{21}\ll kT$. Define ${E_{21}\over kT}\equiv\delta \ll1$.\end{itemize} Now we have a 2-step: \begin{itemize}\item Step 1: Since $1\to3$ are not linked by radiation, $${N_3\over N_1}\approx e^{-{E_{31}\over kT}}\equiv\theta\le1$$ \item Step 2: Radiative decays from $3\to2$. To get an inversion, we'll argue that the sources into 2 are larger than the sinks out of 2 (this is a little weird because we're solving our steady-state equations, but whatever): $$N_3A_{32}+N_3C+N_1Ce^{-{E_{21}\over kT}}>N_2Ce^{-{E_{32}\over kT}}+N_2C$$ We can rewrite this as: $$\begin{aligned} \underbrace{N_2\over N_1}_{=\theta}{A_{32}\over C}&>\left[{N_2\over N_1} \underbrace{e^{-{E_{32}\over kT}}}_{E_{32}=E_{32}-E_{12}}-{N_3\over N_2}\right] +\left[{N_2\over N_1}-e^{-{E_{21}\over kT}}\right]\\ {A_{32}\theta\over C}&>\left[\underbrace{N_2\over N_1}_{1}\theta(1+\delta)- \theta\right]+\left[\underbrace{N_2\over N_1}_{1}-(1-\delta)\right]\\ &=\theta\delta+\delta=\delta(\theta+1)\\ \end{aligned}$$ Thus, the $\delta$ really helps get masing started. \end{itemize} Now one last thing: we'd chosen to ignore stimulated radiative transfer between energy states 3 and 2. In general, this process will tend to reduce the population inversion. However, for optically thick clouds $\tau\gg1$, photons only have a $P\sim\inv{\tau}$ probability of escaping, so we can describe this by ``diluting the $A_{32}$ term by $\inv{\tau}$.

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